NEET Physics

Centre of Mass and Rotation

Question 1. Consider a system of two particles of masses m₁ and m₂. If is pushed towards the centre of mass of the system through a distance d, by what distance should the second particle be moved so as to keep the position of the centre of mass unchanged?

1. \(\left(\frac{m_1}{m_1+m_2}\right) d\)
2. \(\left(\frac{m_2}{m_1}\right) d\)
3. \(\left(\frac{m_1}{m_2}\right) d\)
4. \(\left(\frac{m_2}{m_1+m_2}\right) d\)

Answer: 3. \(\left(\frac{m_1}{m_2}\right) d\)

The position of the centre of mass is given by

⇒ \(x_{\mathrm{CM}}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}\)

For \(\Delta x_{\mathrm{CM}}=0\), we have

⇒ \(\frac{m_1 \Delta x_1+m_2 \Delta x_2}{m_1+m_2}=0\)

or, m₁d+ m₂ΔX₂ = 0

or, \(\Delta x_2=-\left(\frac{m_1}{m_2}\right) d\)

∴ \(\left|\Delta x_2\right|=\left(\frac{m_1}{m_2}\right) d\)

Question 2. A cubical block of ice of mass m and edge L is placed in a big plate of mass M. If the whole ice melts, and the centre of mass of the ice-tray system will come down by

1. \(\frac{M L}{m+M}\)
2. \(\frac{m L}{2(m+M)}\)
3. \(\frac{M L}{2(M+m)}\)
4. \(\frac{m L}{m+M}\)

Answer: 2. \(\frac{m L}{2(m+M)}\)

If the block of ice melts fully, water will spread over the plate. So, the CM will go down by \(\frac{L}{2}\), while the position of the CM of the plate will remain unchanged.

Since \(x_{\mathrm{CM}}=\frac{m x_1+M x_2}{m+M}\) the shiftin the CM of the system will be

⇒ \(\Delta x_{\mathrm{CM}}=\frac{1}{m+M}\left(m \Delta x_1+M \Delta x_2\right)\)

⇒ \(\frac{1}{m+M}\left(m \cdot \frac{L}{2}+M \cdot 0\right)\)

= \(\frac{m L}{2(m+M)}\)

Question 3. A block of mass m is placed on the top of a bigger wedge of mass 5m, as shown in the figure. All the surfaces are frictionless. The system is released from rest. At the instant when the smaller block reaches the ground, the bigger block will move through

1. \(\frac{d}{6}\) towards left
2. \(\frac{d}{6}\) towards right
3. \(\frac{d}{5}\) towards left
4. \(\frac{d}{5}\) towards right

Answer: 2. \(\frac{d}{5}\) towards left

Since there is no external force along the horizontal, the net displacement of the CM of the system will be zero.

Thus,

⇒ \(x_{\mathrm{CM}}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}=0\)

∴ \(\Delta x_{\mathrm{CM}}=0=\frac{1}{m_1+m_2}\left(m_1 \Delta x_1+m_2 \Delta x_2\right)\)

Here, m₁ = m, m₂ = 5m, Δx₁=d-x and Δx₂ = x, where x = displacement of the wedge to the right so as to keep the position of the CM unchanged.

Thus,

m(d-x) = 5mx

x =\(\frac{d}{6}\) (towards right).

Question 4. On a truck of mass M, moving with uniform velocity on a horizontal road, a rock of mass m drops from above and strikes the floor of the truck vertically. The final velocity of the truck will be

1. \(\frac{M v}{m}\)
2. \(\frac{(M+m) v}{m}\)
3. \(\frac{M v}{M+m}\)
4. \(\frac{m v}{M+m}\)

Answer: 3. \(\frac{M v}{M+m}\)

Conserving the linear momentum along the horizontal,

we have Mv =(M+ m)v’

⇒ \(v^{\prime}=\frac{M v}{M+m}\)

Question 5. A circular plate of diameter a is kept in contact with a square plate of edge in the same plane, as shown in the figure. The surface density of the material and the thickness are the same everywhere. The centre of mass of the combined system will be

1. Inside the circular plate
2. Inside the square plate
3. At the point of contact
4. At the midpoint of the radius of the circular plate

Answer: 2. Inside the square plate

Let p be the surface mass density. So, the mass of the circular plate,

m₁ = \(\pi\left(\frac{a^2}{4}\right) \rho\) and mass of the square plate = m₂ = a²p.

Taking the centre of the circular plate as the origin,

⇒ \(x_{\mathrm{CM}}=\frac{m_1 \cdot 0+m_2 x_2}{m_1+m_2}\)

=\(\frac{a^2 \rho a}{\pi\left(\frac{a^2}{4}\right) \rho+a^2 \rho}\)

=\(\frac{a}{1+\frac{\pi}{4}}=\frac{a}{1.785}\)

=\(0.56 a>\frac{a}{2}\)

Thus, the CM lies inside the square plate

Question 6. Two blocks of masses 10 kg and 30 kg are placed along a vertical line. The first block is raised vertically through 7 cm. By what distance should the second block be moved so as to raise the centre of mass of the system by 1 cm?

1. 1 cm downwards
2. 1 cm upwards
3. 2 cm downwards
4. 2 cm upwards

Answer: 1. 1 cm downwards

⇒ \(x_{\mathrm{CM}}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}\)

⇒ \(\Delta x_{\mathrm{CM}}=\frac{m_1 \Delta x_1+m_2 \Delta x_2}{m_1+m_2}\)

Hence, mx- 10 kg, m2 = 30 kg, AxCM =1 cm (upward) and Ax1-7 cm.

∴ \(1 \mathrm{~cm}=\frac{(10 \mathrm{~kg})(7 \mathrm{~cm})+(30 \mathrm{~kg}) \Delta x_2}{40 \mathrm{~kg}}\)

30 ΔX₂ = 40 cm- 70 cm

= -30 cm

Δx₂ =-1 cm.

Hence, the second block must be moved by 1 cm downwards.

Question 7. The balloon (of mass M), the light string (of length L) and the monkey (of mass m) shown in the figure are at rest in the air. If the monkey reaches the top of the rope, the balloon will descend vertically through the distance

1. \(\frac{m L}{M}\)
2. \(\frac{M L}{m}\)
3. \(\frac{M L}{M+m}\)
4. \(\frac{m L}{M+m}\)

Answer: 4. \(\frac{m L}{M+m}\)

Since the system is initially at rest, the position of the CM of the system must be unchanged. Let the balloon move down by x. Then,

Mx = m(L- x)

=> (M + m)x= mL

⇒ \(x=\frac{m L}{M+m}\)

Question 8. A boy of mass m is standing on one end of a long boat of mass M and length L. Neglecting friction, how far does the boat move if the boy slowly moves to the other end?

1. \(\frac{m L}{M}\)
2. \(\frac{M L}{m}\)
3. \(\frac{m L}{M+m}\)
4. \(\frac{M L}{m+M}\)

Answer: 3. \(\frac{m L}{M+m}\)

As the boy moves towards the right, let the boat move towards the left by x so as to keep the position of the CM of the system unchanged.

Thus,

m(L-x)=Mx

=> mL = (M+ m)x

⇒ \(x=\frac{m L}{M+m}\)

Question 9. The linear mass density of a rod of length 3m is directly proportional to the distance x from one end. The distance of the centre of mass of the rod from that end will be

1. 1.5m
2. 2 m
3. 2.5 m
4. 3.0m

Answer: 2. 2 m

⇒ \(x_{\mathrm{CM}}=\frac{1}{M} \int_0^L x d m\), where the total mass is

⇒ \(M=\int d m=\int_0^L \lambda x d x=\frac{\lambda L^2}{2}\)

Now, \(\int_0^L x d m=\int_0^L x(\lambda x d x)=\lambda \int_0^L x^2 d x=\frac{\lambda L^3}{3}\)

⇒ \(x_{\mathrm{CM}}=\frac{\frac{\lambda L^3}{3}}{\frac{\lambda L^2}{2}}=\frac{2}{3} L=\frac{2}{3} \cdot(3 \mathrm{~m})\)

= 2m.

Question 10. Three particles of the same mass lie in the xy-plane. The x and y-coordinates of their positions are given by (1, 1), (2, 2) and (3, 3) respectively. The x and y-coordinates of the centre of mass of the system are given by

1. (1,2)
2. (2,2)
3. (2, 3)
4. (4,5)

Answer: 2. (2,2)

The x- and y-coordinates of the CM are respectively,

⇒ \(x_{\mathrm{CM}}=\frac{1}{3 m}(m \cdot 0+m \cdot a+m \cdot 0)=\frac{a}{3}\)

and \(y_{\mathrm{CM}}=\frac{1}{3 m}(m \cdot 0+m \cdot 0+m \cdot b)=\frac{b}{3}\)

= \(\frac{1}{3}\) (1+2+3)

= 2.

Thus, the x- and y-coordinates of the CM are given by (2, 2)

Question 11. Three particles, each of mass m, are placed at the vertices of a right-angled triangle OAB, as shown in the figure. If OA = a and OB = b then the position vector of the centre of mass of the given system is

1. \(\frac{1}{3}(a \hat{i}+b \hat{j})\)
2. \(\frac{1}{3}(a \hat{i}-b \hat{j})\)
3. \(\frac{2}{3}(a \hat{i}+b \hat{j})\)
4. \(\frac{2}{3}(a \hat{i}-b \hat{j})\)

Answer: 1. \(\frac{1}{3}(a \hat{i}+b \hat{j})\)

With reference to the origin O,

⇒ \(x_{\mathrm{CM}}=\frac{1}{3 m}(m \cdot 0+m \cdot a+m \cdot 0)=\frac{a}{3}\)

and \(y_{\mathrm{CM}}=\frac{1}{3 m}(m \cdot 0+m \cdot 0+m \cdot b)=\frac{b}{3}\)

∴ the position vector of the centre of mass is

⇒ \(\vec{r}_{C M}=\left(\frac{a}{3}\right) \hat{i}+\left(\frac{b}{3}\right) \hat{j}=\frac{1}{3}(a \hat{i}+b \hat{j})\)

Question 12. Three particles, each of mass m, are placed at the vertices of an equilateral triangle of side a, as shown in the figure. The position vector of the centre of mass of the system is

1. \(\frac{a}{2}\left(\hat{i}+\frac{1}{\sqrt{3}} \hat{j}\right)\)
2. \(\frac{a}{2}(3 \hat{i}+\sqrt{3} \hat{j})\)
3. \(\frac{a}{2}\left(3 \hat{i}+\frac{1}{\sqrt{3}} \hat{j}\right)\)
4. \(\frac{a}{2}(\hat{i}+\hat{j})\)

Answer: 1. \(\frac{a}{2}\left(\hat{i}+\frac{1}{\sqrt{3}} \hat{j}\right)\)

With reference to the origin O,

⇒ \(x_{\mathrm{CM}}=\frac{1}{3 m}\left(m \cdot 0+m \cdot a+m \cdot \frac{a}{2}\right)\)

= \(\frac{a}{2}\)

and \(y_{\mathrm{CM}}=\frac{1}{3 m}\left(m \cdot 0+m \cdot 0+m \cdot \frac{\sqrt{3}}{2} a\right)\)

= \(\frac{\sqrt{3} a}{2}\)

∴ \(\vec{r}_{\mathrm{CM}}=\left(\frac{a}{2}\right) \hat{i}+\left(\frac{\sqrt{3} a}{2}\right) \hat{j}\)

= \(\frac{a}{2}\left(\hat{i}+\frac{\sqrt{3}}{3} \hat{j}\right)\)

= \(\frac{a}{2}\left(\hat{i}+\frac{1}{\sqrt{3}} \hat{j}\right)\)

Question 13. A one-metre-long uniform-disbent at a right angle at its midpoint. The distance of the centre of mass from the centre of the rod is

1. 17.7 cm
2. 25.2 cm
3. 34.1cm
4. Zero

Answer: 1. 17.7 cm

⇒ \(x_{\mathrm{CM}}=\frac{\frac{m}{2}\left(\frac{L}{4}\right)}{m}=\frac{L}{8} \text { and } y_{\mathrm{CM}}=\frac{L}{8}\)

∴ \(\vec{r}_{\mathrm{CM}}=\frac{L}{8}(\hat{i}+\hat{j})\)

∴ the distance of the CM from O is

⇒ \(\left|\vec{r}_{\mathrm{CM}}\right|=\frac{\sqrt{2}(1 \mathrm{~m})}{8}\)

= 0.177m

= 17.7 cm.

Question 14. Two particles of mass m₁ and m₂, initially at rest, move towards each other under a mutual force of attraction. At any instant when their respective speeds are v₁ and v₂ the speed of the centre of mass of the system is

1. \(\frac{m_1 v_1}{m_1+m_2}\)
2. \(\frac{m_2 v_2}{m_1+m_2}\)
3. \(\frac{v_1+v_2}{2}\)
4. Zero

Answer: 4. Zero

Since there is no external force, the linear momentum of the system will be conserved.

The initial momentum = 0,

so, \(m_1 \vec{v}_1+m_2 \vec{v}_2=0\)

Now, the velocity of the CM is

⇒ \(\vec{v}_{\mathrm{CM}}=\frac{m_1 \vec{v}_1+m_2 \vec{v}_2}{m_1+m_2}=0\)

Question 15. Two blocks of masses 10kg and 4kg are connected by a light spring and placed on a frictionless horizontal surface. An impulse to the heavier block imparts a velocity of 14 m s^-1 in the direction of the lighter block. The velocity of the centre of mass of the system is

1. 5 ms^-1
2. 10 m s^-1
3. 15m s^-1
4. 20 m s^-1

Answer: 2. 10 m s^-1

Given that m₁ = 10 kg and m₂ = 4 kg.

The initial velocity imparted to m1 is w₁ = 14m^-1 and u₂ = 0.

∴ the velocity of the CM of the system is

⇒ \(\vec{v}_{\mathrm{CM}}=\frac{1}{m_1+m_2}\left(m_1 \vec{u}_1+m_2 \vec{u}_2\right)\)

⇒ \(\frac{1}{14 \mathrm{~kg}}\left[(10 \mathrm{~kg})\left(14 \mathrm{~m} \mathrm{~s}^{-1}\right)+0\right]\)

= 10 ms^-1.

Question 16. A wheel rotates with a constant acceleration of 2.0 rad s^-2 starting from rest. The number of revolutions made in the first 10 seconds is

1. 5
2. 10
3. 16
4. 20

Answer: 3. 16

Given that angular acceleration = a = 2.0 rad s^-2, time = t = 10 s and angular frequency = ω₀ = 0.

For angular motion,

⇒ \(\theta=2 \pi N=\omega_0 t+\frac{1}{2} \alpha t^2=\frac{1}{2}\left(2 \mathrm{rad} \mathrm{s}^{-2}\right)\left(100 \mathrm{~s}^2\right)\)

⇒ \(N=\frac{100}{2 \pi}=15.9 \approx 16\)

Question 17. A motor wheel, accelerated uniformly from rest, rotates through 2.5 rad during the first second. The angle rotated through the next second is

1. 5 rad
2. 10 rad
3. 7.5 rad
4. 12.5 rad

Answer: 3. 7.5 rad

ω₀ = θ and θ₁ = 2.5 rad in the first second.

For the first second,

⇒ \(\theta_1=\omega_0 t+\frac{1}{2} \alpha t^2\)

or, \(2.5 \mathrm{rad}=\frac{1}{2} \alpha(1 \mathrm{~s})^2\)

∴ angular acceleration= α = 5.0 rad s^-2.

For the first two seconds,

⇒ \(\theta_2=\frac{1}{2} \alpha(2 \mathrm{~s})^2=\frac{1}{2}(5.0)(4) \mathrm{rad}\)

= 10 rad

and \(\theta_1=\frac{1}{2} \alpha(1 s)^2=\frac{1}{2}(5)(1)\)

= 2.5 rad.

θ₂ – θ₁ = (10- 2.5)rad

= 7.5 rad.

Question 18. A wheel revolves about its axis with uniform angular acceleration. Starting from rest, it attains an angular velocity of 100 rev s^-2 in 4 seconds. The angular acceleration of the wheel is

1. 10 revs^-2
2. 15 revs^-2
3. 20 revs^-2
4. 25 rev s^-2

Answer: 4. 25 rev s^-2

⇒ \(\omega=\omega_0+\alpha t\)

⇒ \(100 \mathrm{rps}=0+\alpha(4 \mathrm{~s})\)

∴ angular acceleration= a = 25 rev s^-2

Question 19. In the preceding question, the angle rotated by the wheel during the first four seconds is 6

1. 200ft rad
2. 400ft rad
3. 100ft rad
4. 300ft rad

Answer: 2. 400ft rad

⇒ \(\theta=\omega_0 t+\frac{1}{2} \alpha t^2\)

= \(0+\frac{1}{2}\left(25 \times 2 \pi \mathrm{rad} \mathrm{s}^{-2}\right)(4 \mathrm{~s})^2\)

= 400n rad.

Question 20. A disc of radius 10 cm is rotating uniformly about its axis at an angular speed of 20rad s^-1. The linear speed of a point on the rim of the disc is

1. 0.5 ms^-1
2. 1.0ms^-1
3. 1.5 ms^-1
4. 2.0 ms^-1

Answer: 4. 2.0 ms^-1

Angular speed = to = 20 rad s^-1 and radius = r

= 10 cm

= 0.1 m.

∴ linear speed, v = ωr

= (20rad s^-1)(0.1 m)

= 2m s^-1

Question 21. In the preceding question, what is the ratio of the linear speed at the rim to that at the midpoint of the radius of the disc?

1. 1:1
2. 1:2
3. 2:1
4. None of these

Answer: 3. 2:1

⇒ \(\frac{v_r}{v_{r / 2}}=\frac{\omega r}{\omega(r / 2)}=2: 1\)

Question 22. Three-point masses m₁ m₂ and m₃ are located at the vertices of an equilateral triangle of side a. What is the moment of inertia of the system about an axis along the altitude of the triangle passing through m₁?

1. \(\left(m_1+m_2\right) \frac{a^2}{4}\)
2. \(\left(m_2+m_3\right) \frac{a^2}{4}\)
3. \(\left(m_1+m_3\right) \frac{a^2}{4}\)
4. \(\left(m_1+m_2+m_3\right) \frac{a^2}{4}\)

Answer: 2. \(\left(m_2+m_3\right) \frac{a^2}{4}\)

The mass lies on the axis of rotation, so it does not contribute to the moment of inertia.

∴ \(I=m_2 r_2^2+m_3 r_3^2\)

⇒ \(\left(m_2+m_3\right)\left(\frac{a}{2}\right)^2\)

⇒ \(\left(m_2+m_3\right) \frac{a^2}{4}\)

Question 23. Four particles of masses m, 2m, 3m and 4m are connected by a rod of negligible mass, as shown in the figure. The radius of gyration of the system about the axis AB is

1. 2a
2. √5a
3. √3a
4. √2a

Answer: 2. √5a

The moment of inertia of the given system about the axis AB is

I = m(0) + 2m(a²) + 3m(2a)² + 4m(3a)²

= 50ma²

= (m + 2m + 3m + 4m)k²

= 10mk².

∴ the radius of gyration = k

⇒ \(\sqrt{\frac{50}{10} a^2}=\sqrt{5} a\)

Question 24. A light rod of length l has two masses m₁ and m₂ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the length of the rod and passing through the centre of mass is

1. \(\left(m_1+m_2\right) l^2\)
2. \(\sqrt{m_1 m_2} l^2\)
3. \(\left(\frac{m_1 m_2}{m_1+m_2}\right) l^2\)
4. \(\left(\frac{m_1+m_2}{m_1 m_2}\right) l^2\)

Answer: 3. \(\left(\frac{m_1 m_2}{m_1+m_2}\right) l^2\)

Let x be the distance of the CM from the A; m1 end. So,

⇒ \(x=\frac{m_1 \cdot 0+m_2 l}{m_1+m_2}=\left(\frac{m_2}{m_1+m_2}\right) l\)

∴ the distance between the CM and m₂,

⇒ \(l-x=\left(\frac{m_1}{m_1+m_2}\right) l\)

Hence, the moment of inertia of the given system about AB will be

⇒ \(I=m_1 x^2+m_2(l-x)^2\)

⇒ \(m_1 \cdot \frac{m_2{ }^2 l^2}{\left(m_1+m_2\right)^2}+m_2 \cdot \frac{m_1{ }^2 l^2}{\left(m_1+m_2\right)^2}=\left(\frac{m_1 m_2}{m_1+m_2}\right) l^2\)

Question 25. Three thin metal rods, each of mass M and length l are welded to form an equilateral triangle. The moment of inertia of the composite system about an axis passing through the centre of mass of the structure and perpendicular to its plane is

1. \(\frac{M l^2}{2}\)
2. \(\frac{M l^2}{4}\)
3. \(\frac{M l^2}{8}\)
4. \(\frac{M l^2}{12}\)

Answer: 1. \(\frac{M l^2}{2}\)

From the given figure,

⇒ \(\frac{O N}{B N}=\tan 30^{\circ}\)

⇒ \(O N=B N \tan 30^{\circ}=\frac{l}{2 \sqrt{3}}\)

The moment of inertia of BC about an axis through N and perpendicular to the plane of the rods is

⇒ \(I_{\mathrm{N}}=\frac{1}{12} M l^2\)

From the parallel-axes theorem,

⇒ \(I_{\mathrm{O}}=I_{\mathrm{N}}+M(O N)^2=\frac{M l^2}{12}+M\left(\frac{l}{2 \sqrt{3}}\right)^2=\frac{M l^3}{6}\)

For the total system comprising the triangle,

⇒ \(I=3 I_{\mathrm{O}}=3\left(\frac{M l^2}{6}\right)=\frac{M l^2}{2}\)

Question 26. Four metal rods, each of mass M and length L, are welded to form a square-shaped structure ABCD, as shown in the figure. The moment of inertia of the composite structure about the file axis bisecting the rods AB and CD and coplanar with the structure is

1. \(\frac{M L^2}{6}\)
2. \(\frac{M L^2}{3}\)
3. \(\frac{M L^2}{12}\)
4. \(\frac{2}{3} M L^2\)

Answer: 4. \(\frac{2}{3} M L^2\)

About the given axis, the moment of inertia of AB is

⇒ \(I_{\mathrm{AB}}=\frac{M L^2}{12}\),

that for \(\mathrm{BC} \text { is } I_{\mathrm{BC}}=M\left(\frac{L}{2}\right)^2=\frac{M L^2}{4}\)

that for \(\mathrm{CD} \text { is } I_{\mathrm{CD}}=\frac{M L^2}{12}\),

and that for \(\mathrm{AD} \text { is } I_{\mathrm{AD}}=\frac{M L^2}{4}\)

∴ the moment of inertia of the system is

⇒ \(I=2\left(\frac{M L^2}{4}\right)+2\left(\frac{M L^2}{12}\right)=\frac{2}{3} M L^2\)

Question 27. The moment of inertia of a thin rod of mass M and length L about an axis passing perpendicularly at a distance \(\frac{L}{4}\) from one end of the rod is

1. \(\frac{7}{48} M L^2\)
2. \(\frac{1}{3} M L^2\)
3. \(\frac{1}{12} M L^2\)
4. \(\frac{1}{8} M L^2\)

Answer: 1. \(\frac{7}{48} M L^2\)

The moment of inertia of the rod about the axis through the centre of mass (O) is

⇒ \(I_{\mathrm{CM}}=\frac{M L^2}{12}\)

∴ the moment of inertia about the parallel axis through Ais.

⇒ \(I_{\mathrm{A}}=I_{\mathrm{CM}}+M d^2=\frac{M L^2}{12}+M\left(\frac{L}{4}\right)^2\)

=\(\frac{7}{48} M L^2\)

Question 28. A uniform rod OA of mass M and length L is inclined at an angle θ to the vertical. What is its moment of inertia about the vertical axis passing through its end O, as shown in the figure?

1. \(\frac{1}{3} M L^2\)
2. \(\frac{1}{3} M L^2 \sin ^2 \theta\)
3. \(\frac{1}{3} M L^2 \cos ^2 \theta\)
4. \(\frac{1}{12} M L^2\)

Answer: 2. \(\frac{1}{3} M L^2 \cos ^2 \theta\)

The mass of an element dx at a distance x from one end O is dm = \(\frac{M}{L}\)dz, and its perpendicular distance from the axis of

AN= OA sin θ = xsin θ.

∴ the moment of inertia is

⇒ \(I=\int d^2 d m=\int_0^L(x \sin \theta)^2\left(\frac{M}{L} d x\right)\)

⇒ \(\frac{M}{L}\left(\sin ^2 \theta\right) \frac{L^3}{3}=\frac{1}{3} M L^2 \sin ^2 \theta\)

Question 29. A uniform rod AB of mass M and length L is inclined at an angle θ to the vertical. What is its moment of inertia about the vertical axis passing through its midpoint O, as shown in the figure?

1. \(\frac{1}{3} M L^2\)
2. \(\frac{1}{12} M L^2\)
3. \(\frac{1}{12} M L^2 \sin ^2 \theta\)
4. \(\frac{1}{12} M L^2 \cos ^2 \theta\)

Answer: 3. \(\frac{1}{12} M L^2 \sin ^2 \theta\)

As solved in the preceding question,

⇒ \(d I=\left(\frac{M}{L} d x\right)(x \sin \theta)^2\)

For the total,

⇒ \(I=\left(\frac{M}{L} \sin ^2 \theta\right)_{-l / 2}^{l / 2} x^2 d x\)

⇒ \(2\left(\frac{M}{L} \sin ^2 \theta\right)\left[\frac{x^3}{3}\right]_0^{l / 2}\)

⇒ \(\frac{1}{12} M L^2 \sin ^2 \theta\)

Question 30. Two identical uniform rods, each of mass M and length L, are welded together at their centre and placed in a horizontal plane. The rods are inclined at an angle of θ, as shown in the figure. The moment of inertia of the combination about a vertical axis (perpendicular to the plane of the rods) passing through their midpoint O is

1. \(\frac{1}{3} M L^2\)
2. \(\frac{1}{6} M L^2\)
3. \(\frac{1}{12} M L^2 \sin ^2 \theta\)
4. \(\frac{1}{6} M L^2 \cos ^2 \theta\)

Answer: 2. \(\frac{1}{6} M L^2\)

The moment of inertia of each rod about the given axis is

⇒ \(\frac{M L^2}{12}\)

The moment of inertia is a scalar quantity and does not depend on their inclination θ. So,

⇒ \(I_{\text {net }}=2 I=2\left(\frac{M L^2}{12}\right)\)

=\(\frac{M L^2}{6}\)

Question 31. A thin rod AB of mass M and length L is hinged at one end A to a horizontal floor. It is gently allowed to fall from its initial vertical position. The angular velocity of the rod when its free end strikes the floor is

1. \(2 \sqrt{\frac{g}{L}}\)
2. \(\sqrt{\frac{2 g}{L}}\)
3. \(\sqrt{\frac{g}{L}}\)
4. \(\sqrt{\frac{3 g}{L}}\)

Answer: 4. \(\sqrt{\frac{3 g}{L}}\)

As the rod falls, its PE is converted to rotational KE. Thus,

⇒ \(M g \cdot \frac{L}{2}=\frac{1}{2} I \omega^2\)

⇒ \(\frac{1}{2}\left(\frac{1}{3} M L^2\right) \omega^2\)

⇒ \(\omega=\sqrt{\frac{3 g}{L}}\)

Question 32. A uniform thin wire of length L and linear mass density X is bent into a circular loop. The moment of inertia of the loop about its tangent is

1. \(\frac{\lambda L^3}{8 \pi^2}\)
2. \(\frac{3 \lambda L^3}{8 \pi^2}\)
3. \(\frac{\lambda L^3}{16 \pi^2}\)
4. \(\frac{\lambda L^3}{2 \pi^2}\)

Answer: 2. \(\frac{3 \lambda L^3}{8 \pi^2}\)

The moment of inertia of a circular loop about its diameter is given by

⇒ \(I_{\mathrm{dia}}=\frac{1}{2} M R^2\),

where \(M=\lambda L \text { and } R=\frac{L}{2 \pi}\) =radius.

∴ the moment of inertia about the tangents given by

⇒ \(I_{\mathrm{tan}}=I_{\mathrm{dia}}+M R^2\)

=\(\frac{3}{2} M R^2=\frac{3}{2}(\lambda L)\left(\frac{L}{2 \pi}\right)^2\)

=\(\frac{3 \lambda L^3}{8 \pi^2}\)

Question 33. A tube of uniform cross-section and length L is completely filled with an incompressible liquid of mass M and closed at both ends. The tube is then revolved in a horizontal plane about an axis passing perpendicularly through its one end with uniform angular velocity ω. The force exerted by the liquid at the other end is

1. \(\frac{1}{4} M \omega^2 L\)
2. \(\frac{1}{2} M \omega^2 L\)
3. \(\omega^2 L\)
4. \(\frac{M \omega^2 L^2}{2}\)

Answer: 2. \(\frac{1}{2} M \omega^2 L\)

In circular motion, the centripetal force exerted by an element of length dx at a distance x is \(d F=\left(\frac{M}{L} d x\right) \omega^2 x\)

∴ total force \(F=\int d F=\frac{M \omega^2}{L} \int_0^L x d x\)

=\(\frac{1}{2} M \omega^2 L\)

Question 34. A uniform circular disc of mass M and radius R is rotating about its axis of symmetry at a uniform rate. A lump of wax of mass m is gently stuck on the disc at a distance r = R/2 from its centre. The new angular velocity of the system is

1. \(\frac{2 \omega M}{2 M+m}\)
2. \(\frac{2 \omega M}{M+2 m}\)
3. \(\frac{\omega m}{M}\)
4. \(\frac{\omega M}{m}\)

Answer: 1. \(\frac{2 \omega M}{2 M+m}\)

Conserving the angular momentum, we have

⇒ \(I \omega=I^{\prime} \omega^{\prime} \quad\)

⇒ \(\quad \frac{1}{2} M R^2 \omega=\left[\frac{1}{2}\left(M R^2\right)+\frac{m R^2}{4}\right] \omega^{\prime}\)

Simplifying, we get

⇒ \(\omega^{\prime}=\frac{2 M \omega}{2 M+m}\)

Question 35. From a disc of radius R and mass M, a circular hole of diameter is cutItsrimpassesthrough the centre of the disc. What is the moment of inertia of the remaining part of the disc about a perpendicular axis passing through the centre?

1. \(\frac{15}{32} M R^2\)
2. \(\frac{11}{32} M R^2\)
3. \(\frac{13}{32} M R^2\)
4. \(\frac{9}{32} M R^2\)

Answer: 3. \(\frac{13}{32} M R^2\)

The total mass of the disc is M. Then, the mass of the cutout portion is

⇒ \(m=\left(\frac{M}{\pi R^2}\right) \pi\left(\frac{R}{2}\right)^2=\frac{M}{4}\)

The moment of inertia of the total disc about the given axis through O is

⇒ \(I=\frac{1}{2} M R^2\)

The moment of inertia of the cutout portion about the perpendicular axis through its centre

⇒ \(I_{\mathrm{O}^{\prime}}=\frac{1}{2}\left(\frac{M}{4}\right)\left(\frac{R}{2}\right)^2=\frac{M R^2}{32}\)

From the theorem of parallel axes, its moment of inertia about the perpendicular axis through O is

⇒ \(I_{\mathrm{O}}=I_{\mathrm{O}^{\prime}}+\frac{M}{4}\left(O O^{\prime}\right)^2=\frac{M R^2}{32}+\frac{M R^2}{16}=\frac{3}{32} M R^2\)

Now, the net moment of inertia is

⇒ \(I_{\text {net }}=I-I_0=\frac{M R^2}{2}-\frac{3}{32} M R^2=\frac{13}{32} M R^2\)

Question 36. Three identical spherical shells, each of mass m and radius r, are placed touching each other, as shown in the figure. Consider an axis XX’, which is touching the two shells and passing through the file diameter of the fluid shell. The moment of inertia of the system consisting of these three shells about XX’is

1. \(\frac{16}{5} m r^2\)
2. \(\frac{11}{5} m R^2\)
3. 4mr²
4. 3mr²

Answer: 3. 4mr²

The moment of inertia of the shell A about XX’ is

⇒ \(I_{\mathrm{A}}=\frac{2}{3} m r^2\)

Similarly, \(I_{\mathrm{B}}=\frac{2}{3} m r^2+m r^2=\frac{5}{3} m r^2\)

∴ the net moment of inertia of the system is

⇒ \(I=I_{\mathrm{A}}+I_{\mathrm{B}}+I_{\mathrm{C}}=4 m r^2\)

Question 37. The ratio of the radii of gyration of a circular ring and a circular disc of the same mass and radius about an axis passing through their centres and perpendicular to their planes is

1. 1:√2
2. √2:1
3. 2:1
4. 3:2

Answer: 2. √2:1

Moment of inertia of the ring = \(I_{\text {ring }}=M R^2=M k_{\text {ring }}^2\)

Moment of inertia of the disc = \(I_{\text {disc }}=\frac{1}{2} M R^2=M k_{\text {disc }}^2\)

∴ \(\frac{I_{\text {ring }}}{I_{\text {disc }}}=\frac{M R^2}{\frac{1}{2} M R^2}=\frac{M k_{\text {ring }}^2}{M k_{\text {disc }}^2}=\frac{k_{\text {ring }}^2}{k_{\text {disc }}^2}\)

⇒ \(k_{\text {ring }}: k_{\text {disc }}\)

= \(\sqrt{2}: 1\)

Question 38. The moment of inertia of the uniform circular disc shown in the figure is maximum about an axis perpendicular to the disc and passing through

1. A
2. D
3. B
4. C

Answer: 3. B

According to the theorem of parallel axes, \(I=I_{\mathrm{CM}}+M a^2\)

The distance from the centre is maximum for the point B, so I is a maximum of about B.

Question 39. The instantaneous angular position of a point on a rotating wheel is given by the equation θ(t) = 2t- 6t². The torque on the wheel becomes zero at

1. t = 1s
2. t = 0.5s
3. t = 2s
4. t = 1.5s

Answer: 1. t = 1s

Given that θ = 2t³- 6t². Hence, angular velocity = \(\omega=\frac{d \theta}{d t}=6 t^2-12 t\) and

angular acceleration \(\alpha=\frac{d \omega}{d t}=12 t-12\)

For torque \(\tau=I \alpha\) to be zero,

α = 12t – 12 = 0

=> t =1 s.

Question 40. From a circular disc of radius R and mass 9M, a small disc of mass M and radius R/3 is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

1. 8MR²
2. 4MR²
3. \(\frac{40}{9} M R^2\)
4. \(\frac{4}{9} M R^2\)

Answer: 3. \(\frac{40}{9} M R^2\)

The moment of inertia of the remaining disc is

⇒ \(I=\frac{1}{2}(9 M) R^2-\frac{1}{2}(M)\left(\frac{R}{3}\right)^2\)

= \(\frac{1}{2} M R^2\left(9-\frac{1}{9}\right)\)

= \(\frac{40}{9} M R^2\)

Question 41. The moment of inertia of a uniform disc of mass M and radius R about an axis touching the disc at its edge and normal to the plane of the disc is

1. \(\frac{1}{2} M R^2\)
2. MR²
3. \(\frac{3}{2} M R^2\)
4. 2MR²

Answer: 3. \(\frac{3}{2} M R^2\)

From the theorem of parallel axes,

⇒ \(I=I_{\mathrm{CM}}+M R^2=\frac{1}{2} M R^2+M R^2\)

= \(\frac{3}{2} M R^2\)

Question 42. Two bodies A and B have their moments of inertia I and 21 about their axes of rotation. If their kinetic energies of rotation are equal, their angular velocities will be in the ratio

1. 2: 1
2. 1: 2
3. √2: 1
4. 1: √2

Answer: 3. √2: 1

The kinetic energies of the bodies A and B are respectively

⇒ \(E_{\mathrm{A}}=\frac{1}{2} I \omega_{\mathrm{A}}^2 \text { and } E_{\mathrm{B}}=\frac{1}{2}(2 I) \omega_{\mathrm{B}}^2 \text {. }\)

Given that EA = EB. So,

⇒ \(\omega_A{ }^2=2 \omega_B{ }^2\)

∴ \(\frac{\omega_A}{\omega_B}=\frac{\sqrt{2}}{1}\)

⇒ \(\omega_A: \omega_B=\sqrt{2}: 1\)

Question 43. A circular disc of the moment of inertia I₂ about its axis perpendicular to its plane and passing through its centre is gently placed over another disc of the moment of inertia I₁ rotating with an angular velocity GO about the same axis. The final angular velocity of the combination of discs is

1. \(\frac{I_2 \omega}{I_1+I_2}\)
2. \(\frac{I_1 \omega}{I_1+I_2}\)
3. \(\frac{\left(I_1+I_2\right) \omega}{I_1}\)
4. \(\frac{I_1 \omega}{I_2}\)

Answer: 2. \(\frac{I_1 \omega}{I_1+I_2}\)

Initial angular momentum = I₁ω

and final angularmomentum=(I₁ + I₂)ω.

Conserving the angularmomentum \(\omega^{\prime}=\frac{I_1 \omega}{I_1+I_2}\)

Question 44. The ratio of the radius of gyration of a disc about a coplanar tangential axis to that of a ring of the same radius about a coplanar tangential axis is

1. 2:3
2. 2:1
3. √5: √6
4. √1: √3

Answer: 3. \(\frac{I_1 \omega}{I_1+I_2}\)

The moment of inertia of a uniform disc about its tangent is given by

⇒ \(I_{\text {disc }}=\frac{1}{4} M R^2+M R^2=\frac{5}{4} M R^2\)

where \(\frac{1}{4} M R^2\) is the moment ofinertia of the disc aboutits diameter.

Similarly, for the ring,

⇒ \(I_{\text {ring }}=\frac{M R^2}{2}+M R^2=\frac{3}{2} M R^2\)

∴ \(\frac{I_{\text {disc }}}{I_{\text {ring }}}=\frac{\frac{5}{4} M R^2}{\frac{3}{2} M R^2}=\frac{5}{6}=\frac{M k_{\text {disc }}^2}{M k_{\text {ring }}^2}\)

⇒ \(\frac{k_{\text {disc }}}{k_{\text {ring }}}=\sqrt{\frac{5}{6}} \Rightarrow k_{\text {disc }}: k_{\text {ring }}=\sqrt{5}: \sqrt{6}\)

Question 45. A heavy disc is rotating with an angular speed co. If a child gently sits on it, what is conserved?

1. Linear momentum
2. Angular momentum
3. Kinetic energy
4. Potential energy

Answer: 2. Angular momentum

Since there is no external torque, the angular momentum will be conserved.

Question 46. ABC is a triangular plate of uniform thickness. The sides are in the ratio 3: 4: 5, as shown in the figure. I_AB, I_BC and I_CA are the moments of inertia of the plate about AB, BC and CA respectively. Which of the following is correct?

1. \(I_{\mathrm{AB}}+I_{\mathrm{BC}}=I_{\mathrm{CA}}\)
2. ICA is maximum
3. \(I_{\mathrm{AB}}>I_{\mathrm{BC}}\)
4. \(I_{\mathrm{BC}}>I_{\mathrm{AB}}\)

Answer: 4. \(I_{\mathrm{BC}}>I_{\mathrm{AB}}\)

Adding the shaded part APC, we get a rectangular plate for which the moment of inertia is

⇒ \(I_{\mathrm{AB}}=\frac{1}{3}(2 M)(3)^2=6 M \text { units, }\)

⇒ \(I_{\mathrm{BC}}=\frac{1}{3}(2 M)(4)^2=\frac{32}{3} M=10.67 M \text { units, }\)

and I_CA=minimum.

Thus, I_BC > I_AB.

Question 47. In a rectangular plate ABCD A (where BC = 2AB), the moment of inertia is minimum along the axis passing through the points E

1. B and C
2. B and D
3. H and F
4. E and G

Answer: 2. B and D

The moment of inertia depends on the mass distribution of the rotational axis.

For the axis BD or AC, the maximum distance of the points A and C is less than the smaller side AB.

Hence, I_BD is minimal.

Question 48. The moment of inertia of a body about a given axis is 1.2 kg m². Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of 25 rad s^-2 must be applied about that axis for a duration of

1. 4 s
2. 2 s
3. 8 s
4. 10 s

Answer: 2. 2 s

⇒ \(\mathrm{KE}=\frac{1}{2} I \omega^2=\frac{1}{2} I\left(\omega_0+\alpha t\right)^2=\frac{1}{2} I \alpha^2 t^2\)

⇒ \(\text { time }=t=\sqrt{\frac{2(\mathrm{KE})}{I \alpha^2}}=\frac{1}{\alpha} \sqrt{\frac{2(\mathrm{KE})}{I}}\)

Substituting the given values,

⇒ \(t=\frac{1}{25} \sqrt{\frac{2(1500)}{1.2}} \mathrm{~s}=\frac{1}{25} \times 50 \mathrm{~s}\)

= 2s.

Question 49. The moment of inertia of a uniform circular disc about its diameter is I. Its moment of inertia about an axis perpendicular to its plane and touching a point on its rim will be

1. 5I
2. 3I
3. 4I
4. 6I

Answer: 4. 6I

The moment of inertia of the disc about its diameter is \(I=\frac{M R^2}{4}\) and

that about the axis of symmetry is

⇒ \(I_{\mathrm{O}}=\frac{M R^2}{2}=2 I\)

From the theorem of parallel axes,

I = I_o+MR²

= 2I+ 4I

= 6I

Question 50. A ring of mass m and radius r rotates about an axis passing through its centre and perpendicular to its plane with an angular velocity co. Its kinetic energy is

1. mr²ω²
2. \(\frac{1}{2} m r^2 \omega^2\)
3. mrω²
4. \(\frac{1}{2} m r \omega^2\)

Answer: 2. \(\frac{1}{2} m r^2 \omega^2\)

The kinetic energy of the ring about the given axis is

⇒ \(\frac{1}{2} I \omega^2=\frac{1}{2} m r^2 \omega^2\)

Question 51. Two discs of moments of inertia I₁ and I₂ rotating about their respective axes, with angular frequencies co₁ and co₂ respectively, are gently brought into contact face to face, with their axes of rotation coincident. The angular frequency of the composite disc will be

1. \(\frac{I_1 \omega_1-I_2 \omega_2}{I_1-I_2}\)
2. \(\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}\)
3. \(\frac{I_1 \omega_2+I_2 \omega_1}{I_1+I_2}\)
4. \(\frac{I_1 \omega_2-I_2 \omega_1}{I_1-I_2}\)

Answer: 2. \(\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}\)

Conserving the angular momentum, we have

⇒ \(I_1 \omega_1+I_2 \omega_2=\left(I_1+I_2\right) \omega^{\prime}\)

⇒ \(\omega^{\prime}=\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}\)

Question 52. If the earth were to suddenly contract to half its present size without any change in its mass, the duration of the day would have been

1. 12 hours
2. 6 hours
3. 18 hours
4. 30 hours

Answer: 2. 6 hours

The earth revolves in its orbit due to the gravitational attraction of the sun, and the torque is zero. So, the angular momentum is conserved.

∴ \(I \omega=I^{\prime} \omega^{\prime} \Rightarrow \frac{2}{5} M R^2 \cdot \frac{2 \pi}{24 \mathrm{~h}}=\frac{2}{5} M\left(\frac{R}{2}\right)^2 \cdot \frac{2 \pi}{T}\)

∴ 24h= 4T

T= 6h

Question 53. A solid cylinder of mass M and radius R rotates about its axis with a constant angular speed a. Its kinetic energy is

1. 2MR²ω²
2. \(\frac{1}{2} M R^2 \omega^2\)
3. \(\frac{1}{4} M R^2 \omega^2\)
4. MR²ω²

Answer: 3. \(\frac{1}{4} M R^2 \omega^2\)

The KE of the solid cylinder is

⇒ \(\frac{1}{2} I \omega^2=\frac{1}{2}\left(\frac{1}{2} M R^2\right) \omega^2=\frac{1}{4} M R^2 \omega^2\)

Question 54. A circular disc of mass m and radius rerolling on a rough horizontal surface with a constant speed v. Its kinetic energy is

1. \(\frac{1}{4} m v^2\)
2. \(\frac{1}{2} m v^2\)
3. \(\frac{3}{4} m v^2\)
4. mv²

Answer: 3. \(\frac{3}{4} m v^2\)

KE of a rolling disc \(\mathrm{KE}_{\text {trans }}+\mathrm{KE}_{\text {rot }}=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2\)

The condition for rolling is v = ωr. Hence

⇒ \(E_{\mathrm{tot}}=\frac{1}{2} m v^2+\frac{1}{2}\left(\frac{1}{2} m R^2\right) \frac{v^2}{R^2}=\frac{3}{4} m v^2\)

Question 55. A solid sphere is rotating about its diameter at a constant angular velocity ω. If its size is slowly reduced to 1/n of its original value without any change in its mass, its angular velocity becomes

1. \(\frac{\omega}{n}\)
2. nω
3. \(\frac{\omega}{n^2}\)
4. n²ω

Answer: 4. n²ω

Conserving the angular momentum,

⇒ \(I \omega=I^{\prime} \omega^{\prime} \Rightarrow\left(\frac{2}{5} M R^2\right) \omega=\frac{2}{5} M\left(\frac{R}{n}\right)^2 \omega^{\prime}\)

ω = n²ω

Question 56. A circular disc of radius R is free to oscillate about an axis passing through a point on its rim and perpendicular to its plane. The disc is turned through an angle of 60° and released. Its angular velocity when it reaches the equilibrium position will be

1. \(\sqrt{\frac{g}{3 R}}\)
2. \(\sqrt{\frac{2g}{3 R}}\)
3. \(\sqrt{\frac{2g}{R}}\)
4. \(\sqrt{\frac{3g}{2 R}}\)

Answer: 2. \(\sqrt{\frac{2g}{3 R}}\)

Loss in PE = mgR(1- cos 60°) = \(\frac{m g R}{2}\)

Gain in KE = \(\frac{1}{2} I \omega^2=\frac{1}{2}\left(\frac{1}{2} m R^2+m R^2\right) \omega^2\)

Conserving the energy,

⇒ \(\frac{1}{2}\left(\frac{3}{2} m R^2\right) \omega^2=\frac{m g R}{2}\)

⇒ \(\omega=\sqrt{\frac{2 g}{3 R}}\)

Question 57. The moment of inertia of a thick shell of mass and internal and external radii R and 2R respectively, about its diameter is

1. \(\frac{13}{35} M R^2\)
2. \(\frac{3}{2} M R^2\)
3. \(\frac{62}{35} M R^2\)
4. \(\frac{31}{35} M R^2\)

Answer: 3. \(\frac{62}{35} M R^2\)

We consider the given hollow sphere as if a solid sphere of radius R and mass M₂ be removed from a solid sphere of radius 2R and mass M₁.

Thus,

⇒ \(M=M_1-M_2=\frac{4}{3} \pi(2 R)^3 \rho-\frac{4}{3} \pi R^3 \rho=\frac{28}{3} \pi R^3 \rho\)

∴ the moment of inertia of the given hollow sphere is

⇒\(\frac{2}{5} M_1(2 R)^2-\frac{2}{5} M_2 R^2\)

⇒ \(\frac{2}{5} \times \frac{32}{3} \pi R^3 \cdot \rho \cdot 4 R^2-\frac{2}{5} \cdot \frac{4}{3} \pi R^3 \rho \cdot R^2\)

⇒ \(\frac{2}{5}(32-1) \frac{4}{3} \pi R^5 \rho=\frac{62}{5}\left(\pi R^3\right)\left(\frac{4}{3} R^2 \rho\right)\)

⇒ \(\frac{62}{5}\left(\frac{3 M}{28 \rho}\right)\left(\frac{4}{3} R^2 \rho\right)=\frac{62}{35} M R^2\)

Question 58. A disc of mass M and radius 2R has a concentric hole of radius R. Its moment of inertia about an axis through its centre is

1. \(\frac{3}{2} M R^2\)
2. \(\frac{5}{3} M R^2\)
3. \(\frac{5}{2} M R^2\)
4. \(\frac{2}{5} M R^2\)

Answer: 3. \(\frac{5}{2} M R^2\)

The disc with a concentric hole may be considered as a solid disc of radius 2R and mass M₁ with a concentric cutout disc of radius R and mass M₂.

The mass of the given discs

⇒ \(M=M_1-M_2=\pi(2 R)^2 \sigma-\pi R^2 \sigma=3 \pi R^2 \sigma,\)

where is the surface mass density?

The moment of inertia of the given disc is

⇒ \(I=\frac{1}{2} M_1(2 R)^2-\frac{1}{2} M_2 R^2\)

⇒ \(\frac{1}{2}\left(4 \pi R^2 \sigma\right)\left(4 R^2\right)-\frac{1}{2}\left(\pi R^2 \sigma\right) R^2\)

⇒ \(\frac{1}{2} \pi R^4 \sigma(16-1)=\frac{15}{2}\left(\pi R^4\right) \frac{M}{3 \pi R^2}\)

⇒ \(\frac{5}{2} M R^2\)

Question 59. A mass m is moving at a constant velocity along a straight line parallel to the x-axis away from the origin. It’s angular with respect to the origin

1. Goes on increasing
2. Goes on decreasing
3. Remains constant
4. Is zero

Answer: 3. Remains constant

The angular momentum of the mass m about the origin is

⇒ \(\vec{L}=\vec{r} \times \vec{p}=\vec{r} \times(m \vec{v})=m(\vec{r} \times \vec{v})\)

Position vector = \(\overrightarrow{O P}=(x \hat{i}+y \hat{j})\), where (x, y) gives the coordinates of P, and

velocity = \(\vec{v}=v \hat{i}\)

∴ \(\vec{L}=m(x \hat{i}+y \hat{j}) \times v \hat{i}=m y v(-\hat{k})\)

Here m and v are constant, and y is also constant as the particle moves parallel to the x-axis.

Thus, \(|\vec{L}|\) = myv

= constant.

Question 60. A disc of mass and radius R is rolling with an angular speed ω on a horizontal plane, as shown in the figure. The magnitude of the angular momentum of the disc about the origin O is

1. \(\frac{1}{2} M R^2 \omega\)
2. MR²ω
3. \(\frac{3}{2} M R^2 \omega\)
4. 2MR²ω

Answer: 3. \(\frac{3}{2} M R^2 \omega\)

Let \(\overrightarrow{O C}=\vec{r} \text { and } \vec{v}_{\mathrm{CM}}\) = velocity of the centre of mass of the disc.

Hence, the linear momentum of the disc is

⇒ \(\vec{p}_{\mathrm{C}}=m \vec{v}_{\mathrm{CM}}\)

If Lc = angular momentum about C, the angular momentum about the origin is

⇒ \(\vec{L}_{\mathrm{O}}=\vec{L}_{\mathrm{C}}+\vec{r} \times \vec{p}_{\mathrm{C}}\)

⇒ \(I_{C^\omega} \omega+r \times M v_{C M} \sin \theta\)

⇒ \(\frac{1}{2} M R^2 \omega+M R v_{\mathrm{CM}}\) [∵ r sinθ = R]

⇒ \(\frac{1}{2} M R^2 \omega+M R^2 \omega\) [∵ v_CM = Rω]

⇒ \(\frac{3}{2} M R^2 \omega\)

Question 61. A cubical block of mass m and side L rests on a rough horizontal surface having the coefficient of friction μ, A horizontal force F is applied on the block, as shown in the figure. The coefficient of friction is sufficiently large so that the block does not slide before toppling. The minimum force required to topple the block is

1. mg
2. \(\frac{m g}{2}\)
3. \(\frac{m g}{4}\)
4. mg (1-μ)

Answer: 2. \(\frac{m g}{2}\)

By the applied force, the cubical block will turn about the edge at A.

Hence, the torque of the force F about A is

⇒ \(\tau_1=F L\) (clockwise).

The torque due to the weight mg about A is

⇒ \(\tau_2=m g\left(\frac{L}{2}\right)=\frac{m g L}{2}\) (anticlockwise).

For turning, must slightly exceed \(\tau_2\).

So, in the limit

⇒ \(\tau_1=\tau_2 \Rightarrow F L=m g\left(\frac{L}{2}\right) \Rightarrow F=\frac{m g}{2}\)

Question 62. A solid metallic sphere of radius R has a moment of inertia I about its diameter. The sphere is melted and recast into a solid disc of radius r and uniform thickness. The moment of inertia of the disc about an axis passing through its edge and perpendicular to its plane is also equal to I. The ratio r/R is

1. \(\frac{2}{\sqrt{10}}\)
2. \(\frac{2}{\sqrt{5}}\)
3. \(\frac{2}{\sqrt{15}}\)
4. \(\frac{1}{\sqrt{2}}\)

Answer: 3. \(\frac{2}{\sqrt{15}}\)

Themomentofinertia of the solid sphere is \(I_1=\frac{2}{5} M R^2\) , and the moment of inertia of the disc is

Given that I₁ = I₂, we have

⇒ \(\frac{2}{5} M R^2=\frac{3}{2} M r^2 \Rightarrow \frac{r}{R}=\frac{2}{\sqrt{15}}\)

Question 63. A circular disc rolls down an inclined plane without slipping. What fraction of its total energy is translational?

1. \(\frac{1}{\sqrt{2}}\)
2. \(\frac{1}{2}\)
3. \(\frac{1}{3}\)
4. \(\frac{2}{3}\)

Answer: 4. \(\frac{2}{3}\)

At any instant during rolling,

v = ωr,

KE of translation = \(\frac{1}{2}\)mv²

and KE of rotation = \(\frac{1}{2} I \omega^2=\frac{1}{2}\left(\frac{1}{2} m r^2\right)\left(\frac{v^2}{r^2}\right)=\frac{1}{4} m v^2\)

∴ total energy = \(\frac{1}{2} m v^2+\frac{1}{4} m v^2=\frac{3}{4} m v^2\)

Hence, the ratio is

⇒ \(\frac{\mathrm{KE} \text { of translation }}{\text { total KE }}=\frac{\frac{1}{2} m v^2}{\frac{3}{4} m v^2}=\frac{2}{3}\)

Question 64. A solid sphere rolls down an inclined plane without slipping. What fraction of its total energy is rotational?

1. \(\frac{2}{7}\)
2. \(\frac{3}{7}\)
3. \(\frac{2}{3}\)
4. \(\frac{3}{5}\)

Answer: 1. \(\frac{2}{7}\)

KE of translation = \(\frac{1}{2}\) mv²

and KE of rotation = \(\frac{1}{2} I \omega^2=\frac{1}{2}\left(\frac{2}{5} m R^2\right)\left(\frac{v}{R}\right)^2\)

⇒ \(\frac{1}{5}\) [∵ v = Rω for rolling].

∴ total energy = \(\frac{1}{2} m v^2+\frac{1}{5} m v^2=\frac{7}{10} m v^2\)

Hence, \(\frac{\text { rotational KE }}{\text { total } \mathrm{KE}}=\frac{\frac{1}{5} m v^2}{\frac{7}{10} m v^2}=\frac{2}{7}\)

Question 65. A solid cylinder of mass M and radius R rolls down without slipping on a rough inclined plane from a height of h. The rotational kinetic energy of the cylinder, when it reaches the bottom of the plane, will be

1. Mgh
2. \(\frac{M g h}{2}\)
3. \(\frac{M g h}{3}\)
4. \(\frac{2M g h}{3}\)

Answer: 3. \(\frac{M g h}{3}\)

The initial PE (= mgh) will change into the total KE.

Now,KE of rotation = \(=\frac{1}{2} I \omega^2\)

=\(\frac{1}{2}\left(\frac{1}{2} m R^2\right) \omega^2\)

=\(\frac{1}{4} m R^2\left(\frac{v^2}{R^2}\right)\)

⇒ \(\frac{1}{4} m v^2\)

and KE of translation = \(\frac{1}{2}\)mv².

∴ total KE = \(\mathrm{KE}=\frac{1}{2} m v^2+\frac{1}{4} m v^2\)

=\(\frac{3}{4} m v^2\)

Now, \(\frac{3}{4} m v^2=m g h \Rightarrow v^2\)

=\(\frac{4}{3} g h\)

∴ the KE of rotation at the bottom is

⇒ \(\frac{1}{4} m v^2=\frac{1}{4} m\left(\frac{4}{3} g h\right)\)

=\(\frac{1}{3} m g h\)

Question 66. A solid sphere rolls down from the top of a rough inclined plane. Its velocity on reaching the bottom of the incline is v. When the same sphere slides down from the top of a smooth but similar inclined plane, its velocity is v’. The ratio v’/v is equal to

1. \(\sqrt{\frac{3}{5}}\)
2. \(\sqrt{\frac{7}{5}}\)
3. 1
4. \(\frac{3}{\sqrt{5}}\)

Answer: 2. \(\sqrt{\frac{7}{5}}\)

While rolling down the inclined plane,

total KE= mgh

⇒ \(\frac{1}{2} m v^2+\frac{1}{2} I \omega^2=m g h\)

⇒ \(\frac{1}{2} m v^2+\frac{1}{2}\left(\frac{2}{5} m R^2\right) \frac{v^2}{R^2}=m g h\)

⇒ \(\frac{7}{10} m v^2=m g h\)

While slipping,

⇒ \(\frac{1}{2} m v^{\prime 2}=m g h\)

⇒ \(\frac{1}{2} m v^{\prime 2}=\frac{7}{10} m v^2\)

⇒ \(\frac{v^{\prime}}{v}=\sqrt{\frac{7}{5}}\)

Question 67. A solid sphere rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout the motion). The directions of the static frictional force acting on the cylinder are

1. Up the incline while ascending and down the incline while descending
2. Up the incline while ascending as well as descending
3. Down the incline while ascending and up the incline while descending
4. Down the incline while ascending as well as descending

Answer: 2. Up the incline while ascending as well as descending

When the sphere rolls up the inclined plane, friction opposes rolling and acts up along the plane so as to produce an anticlockwise torque.

When the sphere rolls down the inclined plane, friction supports rolling and acts up along the plane, thus producing an anticlockwise torque, which increases the angular speed.

Question 68. A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a constant force of 30 N?

1. 0.25 rad s^-2
2. 25 rad s^-2
3. 5 ms^-2
4. 25 ms^-2

Answer: 2. 25 ms^-2

Torque = FR =Ia = MR²a.

∴ angular acceleration

⇒ \(\alpha=\frac{F R}{M R^2}\)

= \(\frac{F}{M R}=\frac{30 \mathrm{~N}}{(3 \mathrm{~kg})\left(\frac{40}{100} \mathrm{~m}\right)}\)

= \(25 \mathrm{rad} \mathrm{s}^{-2}\)

Question 69. Two discs of the same moment of inertia rotate about their regular axes passing through the centre and perpendicular to their planes with angular velocities ω₁ and ω₂. They are brought into contact face to face coinciding their axes of rotation. The expression for the loss of energy during this process is

1. \(\frac{1}{4} I\left(\omega_1-\omega_2\right)^2\)
2. \(I\left(\omega_1-\omega_2\right)^2\)
3. \(\frac{1}{8} I\left(\omega_1-\omega_2\right)^2\)
4. \(\frac{1}{2} I\left(\omega_1-\omega_2\right)^2\)

Answer: 1. \(\frac{1}{4} I\left(\omega_1-\omega_2\right)^2\)

Conserving the angular momentum,

⇒ \(I \omega_1+I \omega_2=2 I \omega \Rightarrow \omega=\frac{\omega_1+\omega_2}{2}\)

Initial KE = \(\frac{1}{2} I \omega_1{ }^2+\frac{1}{2} I \omega_2{ }^2\)

and final KE = \(\frac{1}{2}(2 I) \omega^2=I\left(\frac{\omega_1+\omega_2}{2}\right)^2\)

∴ loss of energy = KE_i-KE_f

= \(\frac{I}{2}\left[\left(\omega_1^2+\omega_2^2\right)-\frac{\left(\omega_1+\omega_2\right)^2}{2}\right]\)

=\(\frac{1}{4} I\left(\omega_1-\omega_2\right)^2\)

Question 70. Two rotating bodies A and B of respective masses m and 2m with moments of inertia I_A and I_B (where I_A > I_A) have the same kinetic energy of rotation. If L_A and L_B are their angular momenta respectively then

1. \(L_{\mathrm{A}}=\frac{L_{\mathrm{B}}}{2}\)
2. L_A = 2L_B
3. L_B>L_A
4. L_A>L_B

Answer: 3. L_B>L_A

KEin terms of angular momentum (L) is \(\frac{L^2}{2 I}\)

∴ \(\frac{L_{\mathrm{A}}^2}{2 I_{\mathrm{A}}}=\frac{L_{\mathrm{B}}^2}{2 I_{\mathrm{B}}}\)

⇒ \(\frac{L_{\mathrm{A}}^2}{L_{\mathrm{B}}^2}=\frac{I_{\mathrm{A}}}{I_{\mathrm{B}}}\)

Since I_B > I_A, we have L_B > L_A

Question 71. A solid sphere of mass m and radius R is rotating about its diameter. A solid cylinder of the same mass and the same radius is also rotating about a geometrical axis with an angular speed of twice that of the sphere. The ratio of their kinetic energies of rotation E_Asph:E_Acyl will be

1. 2:3
2. 1:5
3. 1:4
4. 3:1

Answer: 2. 1:5

The kinetic energy of the solid sphere is \(E_1=\frac{1}{2} I \omega^2=\frac{1}{2}\left(\frac{2}{5} M R^2\right) \omega^2\)

The kinetic energy of the solid cylinder is \(E_2=\frac{1}{2} I(2 \omega)^2=\frac{1}{2}\left(\frac{M R^2}{2}\right) 4 \omega^2\)

∴ \(\frac{E_1}{E_2}=\frac{\frac{1}{5} M R^2 \omega^2}{M R^2 \omega^2}\)

= 1: 5.

Question 72. A disc and a sphere of the same radius roll down on two inclined planes from the same altitude and length. Which of the two objects gets to the bottom of the plane first?

1. Both read at the same time.
2. The disc
3. It depends on their masses.
4. The sphere

Answer: 4. The sphere

The acceleration of a body rolling down an inclined plane is given

⇒ \(a=\frac{g \sin \theta}{1+\frac{k^2}{R^2}}\)

For the disc, \(\frac{k^2}{R^2}=\frac{1}{2} \Rightarrow a_{\mathrm{disc}}=\frac{2}{3} g \sin \theta\)

For the sphere, \(\frac{k^2}{R^2}=\frac{2}{5} \Rightarrow a_{\mathrm{sph}}=\frac{5}{7} g \sin \theta\)

Since \(a_{\mathrm{sph}}>a_{\text {disc }}\), the sphere will reach the bottom before the disc.

Question 73. A uniform circular di9c of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s^-2. Its net acceleration at the end of 2.0 s is approximately

1. 6.0 ms^-2
2. 3.0 m s^-2
3. 7.0 ms^-2
4. 8.0 ms^-2

Answer: 4. 8.0 ms^-2

Given that radius = R = 50 cm = 0.5 m and angular acceleration

α = 2.0 rad s^-2.

The angular velocity (ω) at the end of 2.0 s will be

ω = ω₀ + at = 0 + (2.0rad s^-2)(2.0 s)= 4rad s^-1.

Tangential acceleration a_tan = Rα = (0.5m)(2.0rad s^-2) = 1.0m s^-2

and radial acceleration= a_rad = ω2R = (4rad s^-1)2(0.5m) = 8.0m s^-2.

∴ net acceleration = \(\sqrt{a_{\mathrm{tan}}^2+a_{\mathrm{rad}}^2}=\sqrt{\left(1.0 \mathrm{~m} \mathrm{~s}^{-2}\right)^2+\left(8.0 \mathrm{~m} \mathrm{~s}^{-2}\right)^2}\)

⇒ \(\sqrt{65 \mathrm{~m}^2 \mathrm{~s}^{-4}} \approx 8.0 \mathrm{~m} \mathrm{~s}^{-2}\)

Question 74. A force \(F=\alpha \hat{i}+3 \hat{j}+6 \hat{k}\) is acting at a point \(\vec{r}=2 \hat{i}-6 \hat{j}-12 \hat{k}\). The value of α for which the angular momentum about the origin is conserved is

1. 1
2. -1
3. 2
4. Zero

Answer: 2. -1

The angular momentum is conserved when the torque \((\vec{\tau}=\vec{r} \times \vec{p})\) is zero.

Given that \(\vec{r}=2 \hat{i}-6 \hat{j}-12 \hat{k} \text { and } \vec{F}=\alpha \hat{i}+3 \hat{j}+6 \hat{k}\)

∴ \(\vec{\tau}=\vec{r} \times \vec{p}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -6 & -12 \\
\alpha & 3 & 6
\end{array}\right|\)

⇒ \((-36+36) \hat{i}+(-12 \alpha-12) \hat{j}+(6+6 \alpha) \hat{k}\)

⇒ \(-\hat{j}(12+12 \alpha)+\hat{k}(6+6 \alpha)\)

Since x = 0, we have,

12 + 12α = 0

or, α =-1 and 6 + 6α = 0

or, α= -1.

Question 75. A solid cylinder of mass 50kg and radius 0.5m is free to rotate about the horizontal axis. A light string is wound around the cylinder with one end attached to it and the other hanging freely. The tension in the string required to produce an angular acceleration of 2 rev s^-2 is

1. 25N
2. 50N
3. 78.5N
4. 157N

Answer: 4. 157N

Given that M = 50 kg, R = 0.5 m

and angular acceleration = α = 2 rev s^-2

= 4π rad s^-2.

Let T be the required tension.

∴ torque = \(\tau=T R=I \alpha\)

tension = \(T=\frac{I \alpha}{R}=\frac{\left(\frac{1}{2} M R^2\right) \alpha}{R}=\frac{1}{2} M R \alpha\)

⇒ \(\frac{1}{2}(50 \mathrm{~kg})\left(\frac{1}{2} \mathrm{~m}\right)\left(4 \pi \mathrm{rad} \mathrm{s}^{-2}\right)\)

= 50πN

= 157N.

Question 76. A rod PQ of mass m and length L is hinged at the end P. The rod is kept horizontal by a massless string tied to the point Q, as shown in the figure. When the string is cut, the initial angular acceleration of the rod is

1. \(\frac{3 g}{2 L}\)
2. \(\frac{2 g}{3 L}\)
3. \(\frac{g}{L}\)
4. \(\frac{2 g}{L}\)

Answer: 1. \(\frac{3 g}{2 L}\)

When the string is cut, the weight mg acting at the centre of the rod will produce a torque of \(\tau=m g\left(\frac {L}{2}\right)\)

But \(\tau=I \alpha=\frac{1}{3} m L^2 \alpha\)

∴ \(\frac{1}{3} m L^2 \alpha=m g\left(\frac{L}{2}\right)\)

Question 77. A small object of uniform density rolls up a curved surface with an initial velocity of v. It reaches up to a maximum height of \(\frac{3 v^2}{4 g}\) with respect to the initial position. The object is a

1. Ring
2. Shell
3. Disc
4. Solid sphere

Answer: 3. Disc

The kinetic energy at the base gets converted into the PE at the maximum height.

Thus, at the base,

⇒ \(\frac{1}{2} m v^2\left(1+\frac{k^2}{R^2}\right)=m g h=m g\left(\frac{3 v^2}{4 g}\right)\)

or, \(1+\frac{k^2}{R^2}=\frac{3}{2} \quad\)

or, \(\quad \frac{k^2}{R^2}=\frac{1}{2}\), which is true for a disc

Question 78. A solid cylinder and a hollow cylinder, both of the same mass and the same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which will reach the bottom first?

1. The hollow cylinder
2. The solid cylinder
3. Both together
4. Both together only when the angle of inclination of the plane is 45°

Answer: 2. The solid cylinder

The acceleration of a body rolling down an inclined plane is

⇒ \(a=\frac{g \sin \theta}{1+\frac{k^2}{R^2}}\)

For a solid cylinder, \(\frac{k^2}{R^2}=\frac{1}{2}\)

For a hollow cylinder, \(\frac{k^2}{R^2}=1\)

⇒ \(a_{\text {solid }}=\frac{g \sin \theta}{1+\frac{1}{2}}=\frac{2}{3} g \sin \theta=0.66 g \sin \theta\)

and \(a_{\text {hollow }}=\frac{g \sin \theta}{2}=0.5 g \sin \theta\)

Since \(a_{\text {solid }}>a_{\text {hollow }}\), the solid cylinder will reach the bottom before the hollow one.

Question 79. A cylindrical shell of mass M and radius R rolls down without slipping along an inclined plane of inclination θ. The frictional force

1. Dissipates energy as heat
2. Decreases the rotational motion only
3. Decreases both the rotational and translational motion
4. Converts translational energy to rotational energy

Answer: 4. Converts translational energy to rotational energy

In the case of pure rolling, there is no slipping or relative motion between the rolling body and the surface at the point of contact. Only static friction exists. It supports rolling downwards but does not dissipate energy. It simply converts translational energy to rotational energy due to the torque produced by static friction.

Question 80. Consider a point P in contact with the wheel on the ground. The wheel rolls on the ground without slipping. The magnitude of displacement of the point P when the wheel completes half the revolution is (where the radius of the wheel is 1 m)

1. 2 m
2. πm
3. \(\sqrt{\pi^2+4} \mathrm{~m}\)
4. \(\sqrt{\pi^2+2} \mathrm{~m}\)

Answer: 3. \(\sqrt{\pi^2+4} \mathrm{~m}\)

During the rotation, the contact point moves horizontally and through d= 2r vertically as shown.

Hence, the magnitude of the displacement is

⇒ \(P P^{\prime}=\sqrt{(P O)^2+\left(O P^{\prime}\right)^2}=\sqrt{(\pi r)^2+(2 r)^2}\)

⇒ \(\sqrt{\pi^2(1 m)^2+(2 m)^2}=\sqrt{\pi^2+4} m\)

Question 81. The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height h from rest without slipping is

1. \(\sqrt{\frac{4}{3} g h}\)
2. √gh
3. \(\sqrt{\frac{6}{5} g h}\)
4. \(\sqrt{\frac{10}{7} g h}\)

Answer: 4. \(\sqrt{\frac{10}{7} g h}\)

⇒ \(\frac{1}{2} m v^2\left(1+\frac{k^2}{R^2}\right)=m g h\)

⇒ \(\frac{1}{2} m v^2\left(1+\frac{2}{5}\right)=m g h\)

⇒ \(v=\sqrt{\frac{10}{7} g h}\)

Question 82. A solid sphere rolls up along a rough inclined plane of inclination θ = 30° with an initial velocity of 2.8 m s^-1. The maximum distance covered by the sphere is approximately

1. 5.45 m
2. 2.74 m
3. 1.10m
4. 3.25 m

Answer: 3. 1.10m

From the work-energy theorem, work done by gravity = change in kinetic energy

⇒ \(-m g s \sin 30^{\circ}=-\frac{1}{2} m v^2\left(1+\frac{k^2}{R^2}\right)\)

⇒ \(g s\left(\frac{1}{2}\right)=\frac{1}{2} v^2\left(1+\frac{2}{5}\right)=\frac{7}{10} v^2\)

⇒ \(s=\frac{7}{10}\left(2.8 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\left(\frac{2}{10 \mathrm{~m} \mathrm{~s}^{-2}}\right)=1.097 \mathrm{~m} \approx 1.10 \mathrm{~m}\)

Question 83. Consider a system of two blocks A and B connected by a light, inextensible string passing over a pulley of mass M = 2 kg and radius R = 10 cm. Find the acceleration of the block A if m_A =1 kg and mB = 0.5 kg.

1. 3.5m s^-2
2. 2ms^-2
3. 1.5 ms^-2
4. 2.5 ms^-2

Answer: 2. 2ms^-2

The magnitudes of the accelerations of both the blocks are equal (a), but the tensions will be different (T₁ > T₂).

For thepulley, \(\left(T_1-T_2\right) R=I \alpha=\left(\frac{M R^2}{2}\right)\left(\frac{a}{R}\right)\)

⇒ \(T_1-T_2=\left(\frac{M}{2}\right) a\)…..(1)

For theblock A, m_Ag – T₁ = m_Aa…..(2)

For theblock B, T₁-m_Bg = m_Ba…..(3)

Adding (1), (2) and (3),

⇒ \(\left(m_{\mathrm{A}}-m_{\mathrm{B}}\right) g=\left(m_{\mathrm{A}}+m_{\mathrm{B}}+\frac{M}{2}\right) a\)

Substituting the given values,

(1.0kg- 0.5 kg)(10m s^-2) = (1.0kg + 0.5kg +1 kg)a

⇒ \(\left(\frac{1}{2} \mathrm{~kg}\right)\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)\)

= (2.5 kg)a

a = 2ms^-2

Question 84. A disc of mass 0.5 kg and radius 20 cm rolls down an inclined plane. Find the force of friction required for pure rolling.

1. \(\frac{5}{3 \sqrt{2}} \mathrm{~N}\)
2. \(\frac{5 \sqrt{2}}{3} \mathrm{~N}\)
3. \(\frac{5}{2 \sqrt{3}} \mathrm{~N}\)
4. \(\frac{5}{\sqrt{2}} \mathrm{~N}\)

Answer: 1. \(\frac{5}{3 \sqrt{2}} \mathrm{~N}\)

The force of friction for pure rolling down a rough inclined plane is given by

⇒ \(f=\frac{m g \sin \theta}{1+\frac{R^2}{k^2}}\)

⇒ \(\frac{(0.5 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right) \sin 45^{\circ}}{1+\frac{2}{1}}\)

= \(\frac{5}{3} \times \frac{1}{\sqrt{2}} \mathrm{~N}\)

= \(\frac{5}{3 \sqrt{2}} \mathrm{~N}\)

Question 85. A disc of mass 2 kg and radius 10 cm undergoes pure rolling on a rough horizontal surface, as shown. Find the ratio of the linear and angular accelerations (a/α).

1. \(\frac{1}{15} m\)
2. \(\frac{1}{10} m\)
3. \(\frac{1}{20} m\)
4. \(\frac{1}{5} m\)

Answer: 2. \(\frac{1}{10} m\)

Due to the force F at the centre of mass, the point of contact A has a tendency to slip towards the right. Hence, static friction/acts towards the left. This friction produces a torque for pure rolling.

For translational motion,

F-f= ma and forrotation,

⇒ \(\tau=f R=I \alpha\)

or, \(f R=\frac{1}{2} m R^2\left(\frac{a}{R}\right) \quad \text { or } \quad f=\frac{1}{2} m a\)

or, \(f=\frac{1}{2} m a\)

∴ \(F=\frac{3}{2} m a\)

⇒ \(f=\left(\frac{1}{2} m\right)\left(\frac{2 F}{3 m}\right)=\frac{F}{3}\)

Now, \(\frac{F-f}{f R}=\frac{m a}{I \alpha}\)

⇒ \(\frac{F-\frac{F}{3}}{\left(\frac{F}{3}\right) R}=\left(\frac{a}{\alpha}\right)\left(\frac{m}{\frac{1}{2} m R^2}\right)\)

⇒ \(\left(\frac{a}{\alpha}\right)\left(\frac{2}{R^2}\right)\)

⇒ \(\frac{2}{R}=\left(\frac{a}{\alpha}\right)\left(\frac{2}{R^2}\right)\)

⇒ \(\frac{a}{\alpha}=R=10 \mathrm{~cm}\)

= \(\frac{1}{10} \mathrm{~m}\)

Question 86. A disc of radius 2 m and mass 100 kg rolls on a horizontal floor. Its centre of mass has a speed of 20 cm s^-1. How much work is needed to stop it?

1. 1J
2. 2J
3. 3J
4. 20 J

Answer: 3. 3J

The kinetic energy of a rolling body is \(\frac{1}{2} M v_{\mathrm{CM}}^2\left(1+\frac{k^2}{R^2}\right)\)

Given that M = 100kg and \(v_{\mathrm{CM}}=20 \mathrm{~cm} \mathrm{~s}^{-1}=0.2 \mathrm{~m} \mathrm{~s}^{-1}\)

and \(\frac{k^2}{R^2}=\frac{1}{2}\)

Hence, \(\mathrm{KE}=\frac{1}{2}(100 \mathrm{~kg})\left(0.2 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\left(\frac{3}{2}\right)\)

= 3J.

According to the work-energy theorem,

| work done | = | KE_f-KE_i |

= 3J.

Question 87. A solid cylinder of mass 2kg and radius 4 cm rotating about its axis at a rate of 3 rpm. The torque required to stop after 2π rev is

1. 2 x 10^-6Nm
2. 2 x 10^-3N m
3. 12 x 10^-4Nm
4. 2 x 10⁴N m

Answer: 1. 2 x 10^-6Nm

The moment of inertia of the solid cylinder is

⇒ \(I=\frac{1}{2} M R^2=\frac{1}{2}(2 \mathrm{~kg})\left(4 \times 10^{-2} \mathrm{~m}\right)^2=16 \times 10^{-4} \mathrm{~kg} \mathrm{~m}^2\)

Initial angular speed = \(=\omega_0=\frac{3 \times 2 \pi}{60} \mathrm{rad} \mathrm{s}^{-1}\)

=\(\frac{\pi}{10} \mathrm{rad} \mathrm{s}^{-1}\)

Angular displacement θ = 2θ rev = 2π(2π) rad = 4π² rad

Now, \(\omega^2=\omega_0^2-2 \alpha \theta\)

∴ angular acceleration \(\alpha=\frac{\omega_0^2}{2 \theta}\)

=\(\frac{\left(\frac{\pi}{10}\right)^2}{2 \times 4 \pi^2} \mathrm{rad} \mathrm{s}^{-2}\)

=\(\frac{1}{800} \mathrm{rad} \mathrm{s}^{-2}\)

Hence, the required torque is

⇒ \(\tau=I \alpha=\left(16 \times 10^{-4} \mathrm{~kg} \mathrm{~m}^2\right)\left(\frac{1}{800} \mathrm{rad} \mathrm{s}^{-2}\right)\)

=\(2 \times 10^{-6} \mathrm{~N} \mathrm{~m}\)

Question 88. A disc of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass, as shown in the figure. The system is initially at rest. A constant torque required to rotate the disc about AB at 25 rps in 5 s is close to

1. 4.0 x 10^-66N m
2. 1.6 x 10^-5N m
3. 2.0 x 10^-5N m
4. 6.5 x 10^-4N m

Answer: 3. 2.0 x 10^-5N m

The moment of inertia of the disc about the axis AB (a tangent) is

⇒ \(I=\frac{M R^2}{4}+M R^2=\frac{5}{4} M R^2\)

Angular acceleration

⇒ \(\alpha=\frac{d \omega}{d t}=\frac{25 \mathrm{rps}}{5 \mathrm{~s}}=10 \pi \mathrm{rad} \mathrm{s}^{-2}\)

Hence, the required torque is

⇒ \(\tau=I \alpha=\left(\frac{5}{4} M R^2\right)(10 \pi)\)

∴ \(\tau=\frac{5}{4}\left(5 \times 10^{-3} \mathrm{~kg}\right)\left(10^{-2} \mathrm{~m}\right)^2\left(10 \pi \mathrm{rad} \mathrm{s}^{-2}\right)\)

= 19.6 x 10^-6 Nm ≈ 2 x 10^-5N m.

Question 89. The magnitude of the torque on a particle of mass 1 kg is 2.5 N m about the origin. If the force acting on it is 1 N and the distance of the particle from the origin is 5 m. Then, the angle between the force and the position vector is

1. \(\frac{\pi}{3}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{\pi}{8}\)
4. \(\frac{\pi}{6}\)

Answer: 4. \(\frac{\pi}{6}\)

Torque = \(|\vec{\tau}|=|\vec{r} \times \vec{F}|\) = rF sin0

=> (2.5Nm) = (5m)(lN) sinθ

⇒ \(\sin \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{6}\)

Question 90. A rod of length 0.5 m is pivoted at one end. It is raised such that it makes an angle of 30° from the horizontal, as shown, and then released from rest. Its angular speed when it passes through the horizontal will be

1. \(\frac{\sqrt{30}}{2} \mathrm{rad} \mathrm{s}^{-1}\)
2. 30 rad s^-1
3. \(\frac{\sqrt{20}}{2} \mathrm{rad} \mathrm{s}^{-1}\)
4. √15 rad s^-1

Answer: 2. √30 rad s^-1

Work done by gravity = \(W_{\text {grav }}=M g\left(\frac{L}{2}\right) \sin 30^{\circ}=\frac{M g L}{4}\)

Change in KE = \(\frac{1}{2} I \omega^2=\frac{1}{2}\left(\frac{M L^2}{3}\right) \omega^2\)

From the work-energy theorem,

⇒ \(\frac{1}{2}\left(\frac{M L^2}{3}\right) \omega^2=\frac{M g L}{4} \Rightarrow \omega=\sqrt{\frac{3 g}{2 L}}\)

Substituting the values,

⇒ \(\omega=\sqrt{\frac{3\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)}{2(0.5 \mathrm{~m})}}=\sqrt{30} \mathrm{rad} \mathrm{s}^{-1}\)

Question 91. Two identical solid spheres of mass Mandradius R each are stuck on the two ends of a uniform rod of length 2R and mass M, as shown in the figure. The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is

1. \(\frac{137}{15} M R^2\)
2. \(\frac{209}{15} M R^2\)
3. \(\frac{17}{15} M R^2\)
4. \(\frac{152}{15} M R^2\)

Answer: 1. \(\frac{137}{15} M R^2\)

The moment of inertia of a solid cp sphere about its own diameter is given by

I_CM = \(\frac{2}{5}\)MR².

From the theorem of parallel axes, the moment of inertia of each sphere about the parallel axis through O is given

⇒ \(I=I_{\mathrm{CM}}+M d^2\)

= \(\frac{2}{5} M R^2+M(2 R)^2\)

= \(\frac{22}{5} M R^2\)

So, the moment of inertia of the whole system about the axis through O is

⇒ \(2\left(\frac{22}{5} M R^2\right)+\frac{M(2 R)^2}{12}=\frac{137}{15} M R^2\)

Question 92. A solid sphere and a solid cylinder of equal radii approach an incline with the same linear velocity. Both roll without slipping throughout their motion. The two respectively reach the maximum heights of h_sph and h_cyl on the incline. The ratio \(\frac{h_{\mathrm{sph}}}{h_{\mathrm{cyl}}}\) is equal to

1. \(\frac{4}{5}\)
2. \(\frac{14}{15}\)
3. \(\frac{2}{\sqrt{5}}\)
4. 1

Answer: 2. \(\frac{14}{15}\)

For a solid sphere, the KE of rolling motion is

⇒ \(E_{\mathrm{sph}}=\frac{1}{2} M v^2\left(1+\frac{k^2}{R^2}\right)\)

=\(\frac{1}{2} M v^2\left(1+\frac{2}{5}\right)\)

If the height reached on the incline is h_sph then

⇒ \(M g h_{\mathrm{sph}}=\frac{1}{2} M v^2\left(\frac{7}{5}\right)\)……..(1)

Similarly, for a solid cylinder, we have

⇒ \(M g h_{\mathrm{cyl}}=\frac{1}{2} M v^2\left(1+\frac{k^2}{R^2}\right)=\frac{1}{2} M v^2\left(1+\frac{1}{2}\right)\)

⇒ \(M g h_{\mathrm{cyl}}=\frac{1}{2} M v^2\left(\frac{3}{2}\right)\)……(2)

Now, by dividing (1) by (2), we get the ratio

⇒ \(\frac{h_{\text {sph }}}{h_{\text {cyl }}}=\frac{\frac{7}{5}}{\frac{3}{2}}=\frac{14}{15}\)

Question 93. A circular disc of radius a has a hole of radius b around its centre. If surface density (mass per unit area) varies as \(\sigma=\frac{\sigma_0}{r}\) (where r is the radius of a concentric ring), the radius of gyration (k) of the disc about the axis passing through its centre and perpendicular to its plane is equal to

1. \(\frac{a+b}{2}\)
2. \(\frac{a+b}{3}\)
3. \(\sqrt{\frac{a^2+b^2+a b}{2}}\)
4. \(\sqrt{\frac{a^2+b^2+a b}{3}}\)

Answer: 4. \(\sqrt{\frac{a^2+b^2+a b}{3}}\)

Let us consider a concentric ring of radius r and width dr. Its mass is i

⇒ \(d M=2 \pi r d r\left(\frac{\sigma_0}{r}\right)\)

Then, the total mass of the disc is

⇒ \(M=2 \pi \sigma_0 \int_b^a d r=2 \pi \sigma_0(a-b)\)……(1)

The moment of inertia ofthisringis

⇒ \(d I=(\text { mass }) r^2=(2 \pi r d r)\left(\frac{\sigma_0}{r}\right) r^2=2 \pi \sigma_0 r^2 d r\)

∴ the moment of inertia of the disc is

⇒ \(I=\int d I=2 \pi \sigma_0 \int_b^a r^2 d r=\frac{2}{3} \pi \sigma_0\left(a^3-b^3\right)\)…..(2)

But I = Mk². Hence, the radius of gyration is

⇒ \(k=\sqrt{\frac{I}{M}}=\sqrt{\frac{\frac{2}{3} \pi \sigma_0\left(a^3-b^3\right)}{2 \pi \sigma_0(a-b)}}=\sqrt{\frac{a^2+b^2+a b}{3}}\)

Question 94. Two coaxial discs of moments of inertia I and I/2 are rotating with angular velocities ω and ω/2 about their common axis. They are brought in contact with each other and then they rotate with a common angular velocity. If Ef and be the final and initial total energies respectively, E_f– E_i is equal to

1. \(-\frac{I \omega^2}{12}\)
2. \(\frac{I \omega^2}{6}\)
3. \(-\frac{I \omega^2}{24}\)
4. \(\frac{3}{8} I \omega^2\)

Answer: 3. \(-\frac{I \omega^2}{24}\)

Conserving the angular momentum,

⇒ \(I \omega+\left(\frac{I}{2}\right)\left(\frac{\omega}{2}\right)=\left(I+\frac{I}{2}\right) \omega_0 \Rightarrow \omega_0=\frac{5}{6} \omega\)……(1)

Initial KE = \(E_{\mathrm{i}}=\frac{1}{2} I \omega^2+\frac{1}{2}\left(\frac{I}{2}\right)\left(\frac{\omega}{2}\right)^2=\frac{9}{16} I \omega^2\)

Final KE = E_f = \(E_{\mathrm{f}}=\frac{1}{2}\left(I+\frac{I}{2}\right) \omega_0^2=\frac{1}{2}\left(\frac{3}{2} I\right)\left(\frac{5 \omega}{6}\right)^2\) [from (1)]

= \(\frac{25}{48} I \omega^2\)

∴ \(E_{\mathrm{f}}-E_{\mathrm{i}}=\left(\frac{25}{48}-\frac{9}{16}\right) I \omega^2=-\frac{I \omega^2}{24}\)

Question 95. Four particles A, B, C and D of masses m, 2m, 3m and 4m respectively are at the comers of a square. They have accelerations of the same magnitude \(|\vec{a}|\) and the directions as shown in the figure. The acceleration of the centre of mass of the system of particles is

1. \(\frac{a}{5}(\hat{i}-\hat{j})\)
2. \(\frac{a}{5}(\hat{i}+\hat{j})\)
3. Zero
4. \(a(\hat{i}+\hat{j})\)

Answer: 1. \(\frac{a}{5}(\hat{i}-\hat{j})\)

The acceleration of the centre of mass of the system is

⇒ \(\vec{a}_{\mathrm{CM}}=\frac{1}{M}\left(m_1 \vec{a}_1+m_2 \vec{a}_2+m_3 \vec{a}_3+m_4 \vec{a}_4\right)\)

⇒ \(\frac{[m a(-\hat{i})+2 m a(\hat{j})+3 m a(\hat{i})+4 m a(-\hat{j})]}{m+2 m+3 m+4 m}\)

⇒ \(\frac{a}{10}(2 \hat{i}-2 \hat{j})\)

=\(\frac{a}{5}(\hat{i}-\hat{j})\)

Question 96. A thin smooth rod of length L and mass M is rotating freely with an angular speed ω₀ about an axis perpendicular to the length of the rod and passing through its centre of mass. Two beads of mass m each and of negligible sizes are at the centre of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod, will be

1. \(\frac{M \omega_0}{M+2 m}\)
2. \(\frac{M \omega_0}{M+3 m}\)
3. \(\frac{M \omega_0}{M+6 m}\)
4. \(\frac{M \omega_0}{M+m}\)

Answer: 3. \(\frac{M \omega_0}{M+6 m}\)

Conserving the angular momentum,

⇒ \(I \omega_0=I^{\prime} \omega \Rightarrow \frac{M L^2}{12} \omega_0=\left[\frac{M L^2}{12}+2 m\left(\frac{L}{2}\right)^2\right] \omega\)

⇒ \(\frac{M \omega_0}{12}=\left(\frac{M}{12}+\frac{m}{2}\right) \omega=\left(\frac{M+6 m}{12}\right) \omega\)

⇒ \(\omega=\left(\frac{M}{M+6 m}\right) \omega_C\)

Question 97. A man of mass 80 kg is standing at the edge of a circular disc of mass 200 kg. The disc rotates about the vertical axis through its centre at 5 revolutions per second. If the moves to the centre of the disc, the angular speed of the disc becomes

1. 9rps
2. 3rps
3. 6rps
4. 12rps

Answer: 1. 9rps

The initial moment of inertia of the man-disc system is

⇒ \(I_1=\frac{1}{2} M R^2+m R^2=\left(\frac{M}{2}+m\right) R^2\)

where M and m are the masses of the disc and them respectively.

Finally, when the man is at the centre,

⇒ \(I_2=\frac{1}{2} M R^2\)

Conserving the angular momentum,

⇒ \(\left(\frac{M}{2}+m\right) R^2 \omega_1=\frac{1}{2} M R^2 \omega^{\prime}\)

final angular speed

⇒ \(\omega^{\prime}=\frac{(180 \mathrm{~kg})(5 \mathrm{rps})}{100 \mathrm{~kg}}\)

= 9rps

Question 98. A one-metre-long uniform rod AB of mass 2mpivoted atAishithorizontally at the free end B by a ball of mass m moving at 6 m s^-1. If the ball sticks to the rod after a collision, find the maximum angular displacement (θ) of the rod. (Given that cos 63° = 0.46.)

1. 53°
2. 63°
3. 59°
4. 69°

Answer: 2. 63°

Conserving the angular momentum,

⇒ \(\frac{1}{2} I \omega^2=2 m g\left(\frac{l}{2}\right)(1-\cos \theta)+m g l(1-\cos \theta)\)

the initial angular speed of the rod is

Conserving the mechanical energy,

⇒ \(\frac{1}{2}\left(\frac{5}{3} m l^2\right)\left(\frac{3 v}{5 l}\right)^2=2 m g l(1-\cos \theta)\)

⇒ \(\frac{1}{2}\left(\frac{5}{3} m l^2\right)\left(\frac{3 v}{5 l}\right)^2=2 m g l(1-\cos \theta)\)

Substituting l=1 m and v = 6ms^-1, we have

⇒ \(\cos \theta=\frac{23}{50}=0.46=\cos 63^{\circ}\)

⇒ \(\theta \approx 63\)

Question 99. A block of mass 3m is suspended by a metre scale of mass m, as shown. If the tension in the string A is kmg A in equilibrium, then the value of k will be

1. \(\frac{5}{4}\)
2. \(\frac{4}{5}\)
3. \(\frac{1}{5}\)
4. \(\frac{3}{5}\)

Answer: 1. \(\frac{5}{4}\)

The forces acting on the system are:

1. Weight of the scale through its centre of
2. Tension TA and T acting vertically upward.
3. Weight of the suspended block = 3mg.

Since the system is in equilibrium, the torque of all forces about any point must be zero.

Taking the torque about O, kmg (100cm)

= mg(50cm) + 3mg(25cm)

⇒ \(k(100)=125\)

⇒ \(k=\frac{5}{4}\)

Question 100. Two discs having the moments of inertia I₁ = 0.10 kg m² and I₂ = 0.20 kg m² and rotating with the angular velocities ω₁ = 10 rad s^-1 and ω₂ = 5 rad s^-1 respectively are brought into contact coaxially and thereafter they rotate together. What is the kinetic energy of the system when they have a common angular velocity?

1. Zero
2. 10J
3. 5 J
4. \(\frac{20}{3}\)J

Answer: 4. \(\frac{20}{3}\)J

Conserving the angular momentum,

⇒ \(L=I_1 \omega_1+I_2 \omega_2=\left(I_1+I_2\right) \omega\)

The kinetic energy of the system when they rotate together will be

⇒ \(E=\frac{L^2}{2 I}\)

=\(\frac{\left(I_1 \omega_1+I_2 \omega_2\right)^2}{2\left(I_1+I_2\right)}\)

=\(\frac{(0.1 \times 10+0.2 \times 5)^2}{2(0.1+0.2)} \mathrm{J}=\frac{20}{3} \mathrm{~J}\)

Question 101. A square of side a/2 is drilled in a uniform disc of radius a, as shown in the adjoining figure. The position of the centre of mass of the remaining part is at

1. \(x=-\frac{2 a}{\pi}\)
2. \(x=-\frac{4 a}{3 \pi}\)
3. \(x=-\frac{a}{8 \pi-2}\)
4. \(x=\frac{3 a}{4 \pi}\)

Answer: 3. \(x=-\frac{a}{8 \pi-2}\)

Mass per unit area of the disc = \(\frac{M}{\pi R^2}\)

Mass of the square area \(\left(\frac{M}{\pi R^2}\right)\left(\frac{a^2}{4}\right)=\frac{M}{4 \pi}\) [∵ a = R]

The distance of the centre of mass of the remaining part of the disc from the centre O is

⇒ \(x=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}=\frac{M \cdot 0+\left(-\frac{M}{4 \pi}\right)\left(\frac{a}{2}\right)}{M+\left(-\frac{M}{4 \pi}\right)}=\frac{-a}{8 \pi-2}\)

Question 102. Two discs A and B of the same mass have the radii R and R/2. A rotates with an angular speed of co; while B is at rest. When A is placed gently on B, both rotate with the same angular velocity. In the process, the percentage loss of kinetic energy is

1. 20
2. 40
3. 10
4. 30

Answer: 1. 20

For disc A, \(I_1=\frac{1}{2} m R^2 \text { and } \omega_1=\omega\)

For disc B, \(I_2=\frac{1}{2} m\left(\frac{R^2}{4}\right)=\frac{m R^2}{8}=\frac{I_1}{4}\)

and \(\omega_2=0\)

The angular momentum remains constant.

Initial KE = \(E_1=\frac{L^2}{2 I_1}\)

and final,

⇒ \(\mathrm{KE}=E_2=\frac{L^2}{2\left(I_1+I_2\right)}=\frac{L^2}{2\left(\frac{5 I_1}{4}\right)}\)

∴ fractional change in KE, \(\frac{\Delta E}{E_1}=1-\frac{E_2}{E_1}=1-\frac{4}{5}=\frac{1}{5}\)

Hence, the percentage decrease

⇒ \(\frac{\Delta E}{E_1} \times 100 \%=\frac{1}{5} \times 100 \%=20 \%\)

Question 103. Two discs of radii R and aR are of the same thickness and made of the same material. If their moments of inertia about their own axis are in the ratio 1: 16, the value of α is

1. \(\frac{1}{2}\)
2. 2
3. \(\frac{1}{4}\)
4. 1

Answer: 2. 2

Let a denote the mass per unit area.

Hence,\(m_1=\pi R^2 \sigma \text { and } m_2=\pi \alpha^2 R^2 \sigma\)

∴ \(\frac{I_1}{I_2}=\frac{\frac{1}{2} m_1 R^2}{\frac{1}{2} m_2 \alpha^2 R^2}\)

=\(\frac{\sigma \pi R^4}{\sigma \pi \alpha^4 R^4}\)

=\(\frac{1}{\alpha^4}\)

=\(\frac{1}{16} \text { (given) }\)

∴ a = 2

Question 104. The ratio of the moments of inertia of the rectangular plate of uniform thickness about the axes through O and O’ perpendicular to the plane of the plate is

1. \(\frac{1}{2}\)
2. \(\frac{1}{3}\)
3. \(\frac{1}{4}\)
4. \(\frac{1}{5}\)

Answer: 3. \(\frac{1}{4}\)

About the axis through O,

⇒ \(I_0=M\left(\frac{a^2+b^2}{12}\right)\)

About the axis through O’

⇒ \(I^{\prime}=I_0+M r^2=M\left(\frac{a^2+b^2}{12}\right)+M\left(\frac{a^2+b^2}{4}\right)\)

⇒ \(M\left(\frac{a^2+b^2}{3}\right)\)

∴ \(\frac{I_0}{I^{\prime}}=\frac{1}{4}\)

Question 105. Find the torque about the origin when a force \(\vec{F}=(3 \mathrm{~N}) \hat{j}\); acts on a particle whose position vector is \((2 \mathrm{~m}) \hat{k}\).

1. \(6 \hat{j} \mathrm{Nm}\)
2. \(-6 \hat{i} \mathrm{Nm}\)
3. \(6 \hat{k} \mathrm{Nm}\)
4. \(6 \hat{i} \mathrm{Nm}\)

Answer: 2. \(-6 \hat{i} \mathrm{Nm}\)

Given that \(\vec{F}=(3 \mathrm{~N}) \hat{j} \text { and } \vec{r}=(2 \mathrm{~m}) \hat{k}\)

∴torque = \(\vec{\tau}=\vec{r} \times \vec{F}=(3 \times 2 \mathrm{~N} \mathrm{~m})(\hat{k} \times \hat{j})\)

⇒ \((6 \mathrm{Nm})(-\hat{i})\)

⇒\(-6 \hat{i} \mathrm{Nm}\)

Current Electricity

Question 1. A wire of resistance 4 is stretched to twice its original length. The resistance of the stretched wire would be

1. 4
2. 6
3. 8
4. 16

Answer: 4. 16

Resistance = R = \(\rho \frac{l}{A}\).

When l is doubled, A is reduced to \(\frac{A}{2}\), so that volume remains constant.

∴ \(R^{\prime}=\rho \frac{l^{\prime}}{A^{\prime}}=\rho \frac{2 l}{A / 2}=4\left(\rho \frac{l}{A}\right)=4 R=4(4 \Omega)=16 \Omega\)

Question 2. A wire of resistance 12 SI m”1 is bent to form a complete circle of radius 10 cm. The resistance between its two diametrically opposite points A and B as shown in the figure is

1. 0.6
2. 3
3. 6
4. 6

Answer: 1. 0.6

The total resistance of the wire,

⇒ \(R=\left(12 \Omega \mathrm{m}^{-1}\right) \text { (total length) }=\left(12 \Omega \mathrm{m}^{-1}\right)(2 \pi r)\)

⇒ \(\left(12 \Omega \mathrm{m}^{-1}\right)\left(2 \pi \times 10 \times 10^{-2} \mathrm{~m}\right)=2.4 \pi \Omega\)

Resistance of each half across AB is \(\frac{R}{2}\) = 1.2π Ω

∴ equivalent resistance across AB will be 0.6π Ω.

Question 3. A wire of a certain material is stretched slowly by 10 percent. What will be its resistance and resistivity respectively?

1. 1.2 times and 1.1 times
2. 1.2 times and remains unchanged
3. 1.1 times and 1.2 times
4. None of the above

Answer: 2. 1.2 times and remains unchanged

\(R=\rho \frac{l}{A}=\rho \frac{l}{V n}=\frac{\rho}{V} l^2\) (where volume V is constant).

Fractional change in the resistance,

\(\frac{d R}{R}=2 \frac{d l}{l}\) = 2(10%) = 0.2.

Increase in resistance, dR = 0.2R.

Hence, resistance becomes R + 0.2R = 1.2R, but resistivity remains unchanged.

Question 4. When a uniform wire of resistance R is stretched so that its radius r becomes r/2 then its resistance becomes

1. R
2. 4R
3. 12R
4. 16R

Answer: 4. 16R

Since volume V = Al remains constant,

⇒ \(A_1 l_1=A_2 l_2 \Rightarrow \frac{l_1}{l_2}=\frac{A_2}{A_1}=\frac{\pi(r / 2)^2}{\pi r^2}=\frac{1}{4}\)

⇒ Now, \(\frac{R_1}{R_2}=\frac{\rho l_1 / A_1}{\rho l_2 / A_2}=\left(\frac{l_1}{l_2}\right)\left(\frac{A_2}{A_1}\right)=\left(\frac{1}{4}\right)\left(\frac{1}{4}\right)=\frac{1}{16}\)

∴ \(R_2=16 R_1=16 R\)

Question 5. Three conductors have their conductances 2 S, 4 S, and 6 S. When they are joined in parallel, their equivalent conductance Will be

1. 12 S
2. \(\frac{1}{12} \mathrm{~S}\)
3. \(\frac{12}{11} \mathrm{~s}\)
4. \(\frac{11}{12} \mathrm{~S}\)

Answer: 1. 12 S

Conductance is reciprocal of the resistance and measured in Siemens ormhos. When connected in parallel,

\(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=(2+4+6) \mathrm{S}\).

∴ Equivalent conductance = \(G=\frac{1}{R}=12 \mathrm{~S}\).

Question 6. A and B are two points on a uniform ring of resistance R. The ∠ACB = θ, where C is the center of the ring. The equivalent resistance between A and B is

1. \(\frac{R \theta(2 \pi-\theta)}{4 \pi^2}\)
2. \(R\left(1-\frac{\theta}{2 \pi}\right)\)
3. \(\frac{R \theta}{2 \pi}\)
4. \(\frac{R(2 \pi-\theta)}{4 \pi}\)

Answer: 1. \(\frac{R \theta(2 \pi-\theta)}{4 \pi^2}\)

Let σ be the resistance per unit length of the wire. Hence, the resistance of the upper part of APB is

R₁ = σ (length APB)

= σ (θr), where r is the radius of the circle.

Similarly, the resistance of the lower part of AQB is

R₂ = σr(2π-θ).

These segments are joined in parallel, so their equivalent resistance is

⇒ \(R_{\text {eq }}=\frac{R_1 R_2}{R_1+R_2}=\frac{\sigma(\theta r) \sigma r(2 \pi-\theta)}{\sigma \theta r+\sigma r(2 \pi-\theta)}=\frac{\sigma r \theta(2 \pi-\theta)}{2 \pi} .\)

Substituting σ \(\sigma=\frac{R}{2 \pi r}\), we get

∴ \(R_{\text {eq }}=\left(\frac{R}{2 \pi r}\right)\left(\frac{r \theta}{2 \pi}\right)(2 \pi-\theta)=\frac{R \theta(2 \pi-\theta)}{4 \pi^2} .\)

Question 7. The masses of three wires of copper are in the ratio of 1 : 3: 5 and their lengths are in the ratio of 5 : 3: 1. The ratio of their resistances is

1. 1:3:5
2. 5:3:1
3. 1:25:125
4. 125:15:1

Answer: 4. 125:15:1

The Ratio of the masses = m₁: m₂: m₃ = 1:3:5

m₁=m, m₂ : 3m, m₃ = 5m.

Ratio of the lengths = l₁ : l₂ : l₃ = 5:3:1

l₁=5l, l₂=3l, l₃=l.

Now, resistance \(R=\rho \frac{l}{A}=\rho \frac{l^2}{V}=\rho D \frac{l^2}{m}\)

where V= volume and D = density.

Hence,

⇒ \(\rho D=\text { constant, so } R \propto \frac{l^2}{m} \text {. }\)

⇒ \(R_1: R_2: R_3=\frac{l_1^2}{m_1}: \frac{l_2^2}{m_2}: \frac{l_3^2}{m_3}=\frac{25 l^2}{m}: \frac{9 l^2}{3 m}: \frac{l^2}{5 m}\)

∴ \(25: 3: \frac{1}{5}=125: 15: 1\)

Question 8. N equal resistors are first connected in series and then connected in parallel. What is the ratio of the maximum to the minimum resistance?

1. N
2. \(\frac{1}{N^2}\)
3. \(N^2\)
4. \(\frac{1}{N}\)

Answer: 3. \(N^2\)

Equivalent resistance is maximum when connected in series, so

⇒  \(R_{\max }=(R+R+\ldots)_{N \text { times }}=N R\).

It will be minimal when connected in parallel, so

⇒ \(\frac{1}{R_{\min }}=\left(\frac{1}{R}+\frac{1}{R}+\ldots\right)_{N \text { times }}=\frac{N}{R} \Rightarrow R_{\min }=\frac{R}{N}\)

∴ Hence, the ratio \(\frac{R_{\max }}{R_{\min }}=\frac{N R}{R / N}=N^2\).

Question 9. Two wires of the same metal have the same length, but their cross sections are in the ratio 3:1. They are joined in series. The resistance of the thicker wire is 10 Ω. The total resistance of the combination will be

1. 10
2. 20
3. 40
4. 100

Answer: 3. 40

Given that \(l_1=l_2 \text { and } A_1: A_2=3: 1\)

Ratio of resistances, \(\frac{R_1}{R_2}=\frac{l_1}{l_2} \frac{A_2}{A_1}=\frac{1}{3}\)

Resistance of the thicker wire, \(R_1=10 \Omega\)

Hence, \(R_2=3 R_1=30 \Omega\)

Equivalent resistance when connected in series is \(R=R_1+R_2=10 \Omega+30 \Omega=40 \Omega\).

Question 10. A wire of resistance R is melted and recast to half of its length. The resistance of the new wire becomes

1. \(\frac{R}{4}\)
2. \(\frac{R}{2}\)
3. R
4. 2R

Answer: 1. \(\frac{R}{4}\)

When the length is halved, the area of the cross-section will be doubled so that volume remains constant

⇒ \(\frac{R^{\prime}}{R}=\frac{\rho l^{\prime} / A^{\prime}}{\rho l / A}=\left(\frac{l^{\prime}}{l}\right)\left(\frac{A}{A^{\prime}}\right)=\left(\frac{l / 2}{l}\right)\left(\frac{A}{2 A}\right)=\frac{1}{4}\)

∴ \(\text { Hence, } R^{\prime}=\frac{R}{4}\)

Question 11. Three resistances, each of 4 Q, are connected to form a triangle. The resistance between any two terminals will be

1. 12
2. 2
3. 6
4. \(\frac{8}{3} \Omega\)

Answer: 4. \(\frac{8}{3} \Omega\)

The combination across any two comers (say B and C) consists of 8 Ω and 4 Ω in parallel, for which the equivalent resistance is

∴ \(R=\frac{(8 \Omega)(4 \Omega)}{(8 \Omega)+(4 \Omega)}=\frac{8}{3} \Omega\)

Question 12. When a wire of uniform cross-section a, length L, and resistance R is bent into a circle, the resistance between two of its diametrically opposite points will be

1. 4R
2. 2R
3. \(\frac{R}{4}\)
4. \(\frac{R}{2}\)

Answer: 3. \(\frac{R}{4}\)

Total resistance = R. This is equally divided into two parts across the diameter, each equal to \(\frac{R}{2}\).

Hence, for parallel combination, \(\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R / 2}+\frac{1}{R / 2}=\frac{4}{R} \Rightarrow R_{\mathrm{eq}}=\frac{R}{4}\)

Question 13. N resistors, each of rΩ, when connected in parallel give an equivalent resistance of R Ω. If these resistors were connected in series, the combination would have a resistance in ohms equal to

1. \(N^2 R\)
2. \(\frac{R}{N^2}\)
3. \(\frac{R}{N}\)
4. NR

Answer: 1. \(N^2 R\)

Equivalent resistance of N resistances, each of resistance r, connected in parallel is \(R=\frac{r}{N} \Omega\).

When connected in series, the equivalent resistance is \(R^{\prime}=N r \Omega=N(N R) \Omega\)

\(N^2 R \Omega\).  [∵\(R=\frac{r}{N}\)]

Question 14. A circuit consisting of five resistors, each of resistance R, forms a Wheatstone bridge. What is the equivalent resistance of the circuit?

1. 2R
2. R
3. \(\frac{R}{2}\)
4. \(\frac{2 R}{3}\)

Answer: 2. R

Equivalent resistance means resistance across the terminals of the cell which is a combination of two resistance of 2R in parallel. This is equal to R.

Question 15. Two metal wires of identical dimensions are connected in series. If \(\sigma_1 \text { and } \sigma_2\) are the conductivities of the metal wires respectively, the effective conductivity of the combination is

1. \(\frac{\sigma_1+\sigma_2}{\sigma_1 \sigma_2}\)
2. \(\frac{\sigma_1 \sigma_2}{\sigma_1+\sigma_2}\)
3. \(\frac{\sigma_1+\sigma_2}{2 \sigma_1 \sigma_2}\)
4. \(\frac{2 \sigma_1 \sigma_2}{\sigma_1+\sigma_2}\)

Answer: 4. \(\frac{2 \sigma_1 \sigma_2}{\sigma_1+\sigma_2}\)

Resistances of the component resistors are \(R_1=\rho_1 \frac{l}{A}\)

and \(R_2=\rho_2 \frac{l}{A}\)

∴ \(R_{\mathrm{eq}}=R_1+R_2=\left(\rho_1+\rho_2\right) \frac{l}{A}\)

If this is replaced by a single wire of equivalent resistivity p then \(\rho \frac{2 l}{A}=\left(\rho_1+\rho_2\right) \frac{l}{A}\)

\(2 p=\rho_1+\rho_2\).

Since conductivity \(\sigma=\frac{1}{\rho}\), we have

⇒ \(\frac{2}{\sigma}=\frac{1}{\sigma_1}+\frac{1}{\sigma_2}=\frac{\sigma_1+\sigma_2}{\sigma_1 \sigma_2}\)

∴ \(\sigma=\frac{2 \sigma_1 \sigma_2}{\sigma_1+\sigma_2}\)

Question 16. A 12-cm-long wire is given the shape of a right-angled triangle ABC having sides 3 cm, 4 cm, and 5 cm as shown in the figure. The resistance between two ends (AB, BC, CA) of the respective sides are measured one by one by a multimeter, the resistances will be in file ratio

1. 9:16:25
2. 3:4:5
3. 27:32:35
4. 21:24:25

Answer: 3. 27:32:35

Let r be the resistance per unit length of the wire, so the three sides have the resistances 3r, 4r, and 5r respectively.

Resistances across AB, \(R_1=\frac{9 \times 3}{12} r\);

across BC, \(R_2=\frac{8 \times 4}{12} r \text { and across AC, } R_3=\frac{7 \times 5}{12} r\).

Hence, the ratio \(R_1: R_2: R_3=27: 32: 35\).

Question 17. A ring is made of a wire having a resistance of 12 Q. Find the points A and B, as shown in the figure at which a current-carrying conductor should be connected so that the resistance R of the sub-circuit between these two points is equal to (8/3)

1. \(\frac{l_1}{l_2}=\frac{5}{8}\)
2. \(\frac{l_1}{l_2}=\frac{1}{2}\)
3. \(\frac{l_1}{l_2}=\frac{1}{3}\)
4. \(\frac{l_1}{l_2}=\frac{3}{8}\)

Answer: 2. \(\frac{l_1}{l_2}=\frac{1}{2}\)

Let r be the resistance per unit length.

∴ Resistance of length\(l_1 \text { is } R_1=r l_1 \text { and } R_2=r l_2\).

Total resistance = \(\left(R_1+R_2\right)=r\left(l_1+l_2\right)=12 S\).

Equivalent resistance across AB is \(\frac{R_1 R_2}{R_1+R_2}=\frac{\left(r l_1\right)\left(r l_2\right)}{r\left(l_1+l_2\right)}=\frac{8}{3} \Omega\)

⇒ \(R_1 R_2=\frac{8}{3}\left(R_1+R_2\right)=\frac{8}{3} \times 12 \Omega=32 \Omega\)

Hence, \(R_1=8 \Omega, R_2=4 \Omega \text { and } \frac{R_1}{R_2}=\frac{l_1}{l_2}=\frac{8}{4}=\frac{2}{1}\).

The ratio is 2: l or 1: 2.

Question 18. Three resistors P, Q, and R each of 2Ω, and an unknown resistance S form the four arms of a Wheatstone bridge circuit When a resistor of 6 Ω is connected parallel to S, the bridge gets balanced. What is the value of S?

1. 1
2. 3
3. 6
4. 2

Answer: 2. 3

For the bridge to be balanced, the fourth arm must have equivalent resistance = \(2 \Omega\)

⇒ \(\frac{(6 \Omega) S}{6 \Omega+S}=2 \Omega\)

⇒ \(12 \Omega+2 S=6 S\)

⇒ \(4 S=12 \Omega \Rightarrow S=3 \Omega\)

Question 19. In a Wheatstone bridge, all four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is

1. \(\frac{R}{4}\)
2. \(\frac{R}{2}\)
3. R
4. 2R

Answer: 3. R

The equivalent resistance as seen by the battery is equivalent to the effective resistance across the battery, which is \(\frac{(2 R)(2 R)}{(2 R)+(2 R)}=R\). The galvanometer resistance R will be ineffective as current through the galvanometer, \(I_g=0\).

Question 20. A fuse wire is a wire of

1. High resistance and high melting point
2. High resistance and low melting point
3. Low resistance and low melting point
4. Low resistance and high melting point

Answer: 2. High resistance and low melting point

A fuse wire is used to reduce the damage of electrical appliances when a high current passes through the wire. A fuse wire should have high resistance and a low melting point so that it may melt easily if the current suddenly becomes high.

Question 21. The resistance of each arm of a Wheatstone bridge is 10 Ω. If a resistance of Ω is connected in series with the galvanometer, the equivalent resistance across the battery will be

1. 10
2. 15
3. 20
4. 40

Answer: 1. 10

When the die Wheatstone bridge is balanced, no current flows through the galvanometer, so galvanometer resistance remains ineffective. Any addition of resistance \((10 \Omega)\) connected in series or across the galvanometer will not change the effective resistance across the battery. In this case,

⇒ \(R_{\mathrm{eq}}=\frac{(20 \Omega)(20 \Omega)}{40 \Omega}=10 \Omega\)

Question 22. What is the equivalent resistance between A and B in the given figure?

1. 40
2. 10
3. 20
4. 50

Answer: 3. 20

The given network of resistors represents a balanced Wheatstone bridge whose each branch has the same resistance of \(20 \Omega\). Hence, the equivalent resistance is \(R=\frac{(40 \Omega)(40 \Omega)}{40 \Omega+40 \Omega}=\frac{1600}{80} \Omega=20 \Omega\).

Question 23. The resistance across the terminal points A and B of the given infinitely long circuit will be

1. \((\sqrt{3}-1) \Omega\)
2. \((2-\sqrt{3}) \Omega\)
3. \((\sqrt{3}+1) \Omega\)
4. \((2+\sqrt{3}) \Omega\)

Answer: 3. \((\sqrt{3}+1) \Omega\)

Let the equivalent resistance of the given infinitely long circuit be x. Hence, the given circuit can be replaced by the circuit shown in the adjoining figure. The equivalent resistance across AB will also be x.

Hence, \(x=1 \Omega+\frac{(1 \Omega) x}{1 \Omega+x}+1 \Omega\)

⇒ \(x^2-2 x-2=0\)

⇒ \(x=(\sqrt{3}+1) \Omega\).

Question 24. The equivalent resistance between points X and Y in the given figure will be

1. 10
2. 7
3. 5
4. 3

Answer: 3. 5

The given network of resistances can be redrawn in a simple form as shown in the adjoining diagram. 3 Ω and 7 Ω in series is equal to 10 Ω which in parallel with 10 Ω is equivalent to 5 Ω. Next, this 5 Ω x and 5 Ω in series is 10 Ω which finally combines in parallel with 10 Ω on XY and gives R = 5 Ω.

Question 25. Five identical resistors, each of resistance r, are connected as shown in the figure. A battery of V volt is connected between A and B. The current flowing through the branch AFCEB will be

1. \(\frac{3 V}{r}\)
2. \(\frac{V}{r}\)
3. \(\frac{V}{2 r}\)
4. \(\frac{2 V}{r}\)

Answer: 3. \(\frac{V}{2 r}\)

The given network of resistors can be redrawn as shown in this figure. This constitutes a balanced Wheatstone bridge for which the equivalent resistance across AB is \(\frac{(2 r)(2 r)}{4 r}=r\).

Current delivered by the cell is I = \(\frac{V}{r}\)

This current is divided equally into the upper and lower branches. Hence, current through the branch AFCEB is \(\frac{I}{2}=\frac{V}{2 r}\).

Question 26. The net resistance of a circuit between A and B is

1. \(\frac{8}{3} \Omega\)
2. \(\frac{14}{3} S\)
3. \(\frac{16}{3} \Omega\)
4. \(\frac{22}{3} \Omega\)

Answer: 2. \(\frac{14}{3} S\)

The given circuit represents a balanced Wheatstone bridge, since \(\frac{P}{Q}=\frac{3}{4}=\frac{R}{S}=\frac{6}{8}\).

Hence, the cross-connected resistance of 7 Ω is ineffective.

Resistance of upperbranch = 7 Ω

and that oflowerbranch =14 Ω.

∴ equivalent resistance = \(R=\frac{(7 \Omega)(14 \Omega)}{(7+14) \Omega}=\frac{14}{3} \Omega\)

Question 27. In the given figure, each resistor has its resistance r =10. What will be the equivalent resistance between the terminals A and D?

1. 10
2. 20
3. 30
4. 40

Answer: 3. 30

In order to find the equivalent resistance across AD, we imagine a cell whose terminals are connected to A and D respectively. The equivalent circuit is shown in the figure. The equivalent resistance across AD is \(R=r+\frac{(2 r)(2 r)}{2 r+2 r}+r=3 r=3(10 \Omega)=30 \Omega\).

Question 28. Six resistors, each of 3 Q., are connected along the sides of a hexagon and three resistors of 6Q each are connected along AC, AD, and AE as shown in the figure. The equivalent resistance F between A and B is equal to

1. 6
2. 2
3. 3
4. 9

Answer: 2. 2

Resistors AF and FE, each of 3 Ω in series, give 6 £2. This 6 Ω with AE (= 6 Ω) in parallel gives 3 Ω. This, 3 £2 with ED (=3 Ω) gives 6 Ω which in parallel with AD (=6 Ω) gives 3 Ω. Proceeding in the same way, the equivalent resistance across AC is found to be Ω. Finally, we are left with 6 Ω and 3 Ω across AB in parallel which gives \(R_{\mathrm{AB}}=\frac{(6 \Omega)(3 \Omega)}{6 \Omega+3 \Omega}=2 \Omega\)

Question 29. In the given circuit, if the current through the 3 Q resistor is 0.8 A then the potential drop across the 4Q resistor is

1. 9.6 V
2. 1.2V
3. 4.8 V
4. 2.6 V

Answer: 3. 4.8 V

Potential difference (p.d.) across the 3-Ω resistor,

V-IR = (0.8 A)(3 ft) = 2.4 V.

p.d. is the same across the 6-Ω resistor, J'(6 Q) = 2.4 V.

Hence, current through the 6-Ω resistor,\(I^{\prime}=\frac{2.4 \mathrm{~V}}{6 \Omega}=0.4 \mathrm{~A}\).

Total current through the circuit = \(\left(I+I^{\prime}\right)=(0.8 \mathrm{~A})+(0.4 \mathrm{~A})=1.2 \mathrm{~A}\)

∴ p.d. across 4 Q resistor = (1.2 A)(4 £2) = 4.8 V

Question 30. In the given network, each resistance r =1 £1 The effective resistance between A and B is

1. \(\frac{4}{3} \Omega\)
2. \(\frac{8}{7} \Omega\)
3. \(7 \Omega\)
4. \(\frac{3}{2} \Omega\)

Answer: 2. \(\frac{8}{7} \Omega\)

By symmetry, the current (I₁ + I₂), which enters at A must leave at B. So, does not get divided at O. Hence, point O can be detached from the arm. This circuit can be redrawn equivalently as shown in figure (2), for which the upper and lower arms have the resistances \(r+\frac{2 r}{3}+r=\frac{8}{3} r \text { and } 2 r\)

equivalent resistance across AB \(R_{\mathrm{AB}}=\frac{\left(\frac{8}{3} r\right)(2 r)}{\frac{8}{2} r+2 r}=\frac{8}{7} r=\frac{8}{7}(1 \Omega)=\frac{8}{7} \Omega\)

Question 31. In the network shown in the figure, each resistance is equal to r. The equivalent resistance between terminals A and b is

1. 3r
2. 6r
3. \(\frac{3 r}{2}\)
4. \(\frac{2 \pi}{3}\)

Answer: 4. \(\frac{2 \pi}{3}\)

When the current enters the network through a, it gets equally divided through branches. Points c, d, and e will be at the same potential. Hence, no current flows through cd and de, and these two resistors are ineffective. The circuit is thus equivalent to three equal resistors in parallel, each equal to 2r.

∴ \(\frac{1}{R}=\frac{1}{2 r}+\frac{1}{2 r}+\frac{1}{2 r}=\frac{3}{2 r} \Rightarrow R=\frac{2}{3} r\)

Question 32. In the given figure, let the equivalent resistance between terminals A and B be (when A°- switch S is open) and Rc (when switch S is closed). The ratio R₀:R_c will be

1. 9: 8
2. 8:9
3. 4:3
4. 3:4

Answer: 1. 9:8

The two adjoining figures have been drawn with switch S open and with switch S closed.

So, \(R_{\mathrm{o}}=\frac{(18 \Omega)(18 \Omega)}{(18 \Omega)+(18 \Omega)}=9 \Omega\)

⇒ and \(R_c=\frac{(12 \Omega)(6 \Omega)}{(12 \Omega)+(6 \Omega)}+\frac{(6 \Omega)(12 \Omega)}{(6 \Omega)+(12 \Omega)}=8 \Omega\)

∴ \(\frac{R_0}{R_c}=\frac{9}{8}\)

Question 33. The equivalent resistance between terminals A and B in the given figure is

1. 2
2. 5
3. 4
4. 6

Answer: 2. 5

Resistance of 3 Ω each in DE and EF in series gives 6 Ω which joined in

parallel with 3 Ω in branch DG gives \(\frac{6 \times 3}{9} \Omega=2 \Omega\).

This 2 Ω with 4 Ω in series with branch CD gives 6 ft which joined in parallel with 3 Ω in branch CH gives 2 ft.

⇒ Finally, \(R_{\mathrm{AB}}=3 \Omega+2 \Omega=5 \Omega\).

Question 34. In the given circuit, what should be the value of R so that no current flows through it?

1. 9
2. 12
3. 18
4. Any value

Answer: 4. Any value

The given network of resistors constitutes a balanced Wheatstone bridge since the ratio \(\frac{3 \Omega}{6 \Omega}=\frac{4 \Omega}{8 \Omega}=\frac{1}{2}\).

Hence, no current will flow through R for any value of R.

Alternative method

The voltage of 5 V of the source is divided in the upper and lower branches in the same proportion; so potential difference, \(V_c-V_d=0\). Hence, current through R is zero for any value of R.

Question 35. The equivalent resistance for the given network of resistors between terminals a and b are R1 and R2 respectively. The ratio R1/R2 is

1. \(\frac{4}{3}\)
2. \(\frac{3}{16}\)
3. \(\frac{2}{3}\)
4. \(\frac{1}{2}\)

Answer: 2. \(\frac{3}{16}\)

In the first case, one end of all resistors is connected at a common point a, and the other ends to a common point b, so they are in parallel,

Hence, \(R_1=\frac{r}{4}\).

In the second case, the network can be redrawn as shown in the given figure.

Here,\(R_2=r+\frac{r}{3}=\frac{4}{3} r\).

Hence, \(\frac{R_1}{R_2}=\frac{r / 4}{4 r / 3}=\frac{3}{16}\)

Question 36. In the given circuit, the potential difference between A and B is

1. zero
2. 5 V
3. 10 V
4. 15 V

Answer: 3. 10 V

The p-n junction is forward biased, hence its resistance is negligible and the circuit can be redrawn as shown.

Equivalent resistance, \(R=10 \mathrm{k} \Omega+5 \mathrm{k} \Omega=15 \mathrm{k} \Omega\).

Maincurrent, \(I=\frac{30 \mathrm{~V}}{15 \mathrm{k} \Omega}\).

This current is equally divided atA, so p.d. across AB is

∴ \(V_{\mathrm{AB}}=\left(\frac{I}{2}\right)(10 \mathrm{k} \Omega)=\frac{(30 \mathrm{~V})(10 \mathrm{k} \Omega)}{2(15 \mathrm{k} \Omega)}=10 \mathrm{~V}\).

Question 37. For the given circuit, the value of the current I is

1. 10 A
2. 5 A
3. 2.5 A
4. 20 A

Answer: 1. 10 A

The four branches of the resistors7 network have equal resistance (=5 Ω), so they satisfy the balance condition of the Wheatstone bridge. The cross-connected 5-Ω resistor is ineffective, so the equivalent circuit can be redrawn as shown.

For a parallel combination of resistors,

⇒ \(\frac{1}{R}=\frac{1}{10}+\frac{1}{10}+\frac{1}{5}=\frac{2}{5}\)

⇒ \(R=\frac{5}{2} \Omega\)

Hence, \(I=\frac{25 \mathrm{~V}}{5 / 2 \Omega}=10 \mathrm{~A}\).

Question 38. Across a metallic conductor of a nonuniform cross-section, a constant potential difference is applied. The quantity which remains constant along the conductor is

1. Current density
2. Electric field
3. Travelocity
4. Current

Answer: 4. Current density

In a current-carrying metallic conductor, there is no accumulation of charge anywhere. So, the rate of flow of charge (ΔQ/Δt), or the current, remains constant Drift speed, current density, and electric field vary with change in cross-sectional area.

Question 39. A, B, and C are voltmeters of resistances R, 1.5R, and . 3R respectively as shown in the figure. x When some potential difference is maintained between X and Y, the voltmeter readings are VA, VB, and Vc respectively. Then,

1. \(V_{\mathrm{A}}=V_{\mathrm{B}}=V_{\mathrm{C}}\)
2. \(V_{\mathrm{A}} \neq V_{\mathrm{B}}=V_{\mathrm{C}}\)
3. \(V_{\mathrm{A}}=V_{\mathrm{B}} \neq V_{\mathrm{C}}\)
4. \(V_{\mathrm{A}} \neq V_{\mathrm{B}} \neq V_{\mathrm{C}}\)

Answer: 1. \(V_{\mathrm{A}}=V_{\mathrm{B}}=V_{\mathrm{C}}\)

The equivalent resistance of B and C is

⇒ \(R_{e q}=\frac{(1.5 R)(3 R)}{(1.5 R)+(3 R)}=\frac{(1.5)(3 R)}{(4.5)}=R \Omega\)

If l is the current through A then

⇒ \(V_{\mathrm{X}}-V_{\mathrm{a}}=V_{\mathrm{A}}=I R, V_{\mathrm{a}}-V_{\mathrm{b}}=V_{\mathrm{B}}=V_{\mathrm{C}}=I R_{\mathrm{eq}}=I R\)

∴ Thus, \(V_A=V_B=V_C\)

Question 40. The reading of the voltmeter in the given circuit is

1. 2.25 V
2. 4.25 V
3. 2.75 V
4. 6.25 V

Answer: 1. 2.25 V

Equivalent resistance in the given circuit is

⇒ \(R+R_{\text {eq }}=40 \Omega+\frac{(60 \Omega)(40 \Omega)}{60 \Omega+40 \Omega}=40 \Omega+24 \Omega=64 \Omega\)

Current through the cell is

⇒ \(I=\frac{6 \mathrm{~V}}{64 \Omega}=\frac{3}{32} \mathrm{~A}\)

Hence, the reading of the voltmeter is

⇒ \(V=I\left(R_{\text {eq }}\right)=\left(\frac{3}{32} \mathrm{~A}\right)(24 \Omega)=2.25 \mathrm{~V}\)

Question 41. The internal resistance of a cell of emf 2 V is 0.1 Ω. It is connected to a resistance of 3.9 Ω. The voltage across the cell (terminal voltage) will be

1. 1.90 V
2. 1.95 V
3. 0.5 V
4. 2 V

Answer: 2. 1.95 V

Given that \(\varepsilon=2 \mathrm{~V}, r=0.1 \Omega, R=3.9 \Omega\)

Current through the circuit is \(I=\frac{\varepsilon}{R+r}=\frac{2 \mathrm{~V}}{4 \Omega}=0.5 \mathrm{~A}\).

∴ The terminal voltage across the cell is

⇒ \(V=\varepsilon-I r=2 \mathrm{~V}-(0.5 \mathrm{~A})(0.1 \Omega)\)

= 2 V-0.05 V =1.95 V.

Question 42. In the given circuit, the current flowing through the 25 V cell is

1. 7.2 A
2. 14.2 A
3. 10 A
4. 12 A

Answer: 4. 12 A

The current through each branch is shown in the figure. Applying Kirchhoff’s voltage loop rule, we have for each closed loop formed with a 25 V cell,

-30+11 I_i -25 = 0;

20+5 I₂ -25 = 0;

-5 +10 I₃ -25 = 0;

10 + 5 I₄ -25 = 0.

⇒ \(I_1=\frac{55}{11} \mathrm{~A}=5 \mathrm{~A}, I_2=1 \mathrm{~A}, I_3=3 \mathrm{~A}, I_4=3 \mathrm{~A}\)

Hence, current through the 25-V cell is

⇒ \(I=I_1+I_2+I_3+I_4=12 \mathrm{~A}\)

Question 43. A meter bridge is set up to determine resistance X using a standard 10 Ω resistor. The galvanometer shows a null point when the tapping key is at the 52 cm mark. The end-corrections are 1 cm and A 2 cm respectively for the ends A and B. The determined value of X is

1. 10.2
2. 10.8
3. 10.6
4. 11.1

Answer: 3. 10.6

From the balance condition,

⇒ \(\frac{X}{10 \Omega}=\frac{(52+1) \mathrm{cm}}{(48+2) \mathrm{cm}}=\frac{53}{50}\)

Unknown resistance is

⇒ \(X=\frac{53}{50} \times 10 \Omega=10.6 \Omega\).

Question 44. When a current is passed through a conductor, a 5°C rise in temperature is observed. If the strength of the current is made three times, the rise in temperature will be

1. 5°C
2. 15 °C
3. 45 °C
4. 20 °C

Answer: 3. 45 °C

Amount of heat produced, \(H=I^2 R t\)

⇒ \(\frac{H_1}{H_2}=\frac{I_1^2}{I_2^2}\)

⇒ \(\frac{H_1}{H_2}=\frac{m c \theta_1}{m c \theta_2}=\frac{I_1^2}{9 I_1^2}=\frac{1}{9} \quad\left(\text { given } I_2=3 I_1\right)\)

Hence, \(\theta_2=9 \theta_1=9\left(5^{\circ} \mathrm{C}\right)=45^{\circ} \mathrm{C}\)

Question 45. The amount of heat developed by a 100-W bulb in 1 minute will be

1. 100 J
2. 1000 J
3. 600 J
4. 6000 J

Answer: 4. 6000 J

Heat energy developed is

H = power x time = (100 W)(l m)

⇒ \(\left(100 \mathrm{~J} \mathrm{~s}^{-1}\right)(60 \mathrm{~s})=6000 \mathrm{~J}\)

Question 46. A bulb of 25 W, 200 V’ and another bulb of 100 W, 200 V’ are connected in series with a supply line of 220 V. Then

1. Both the bulbs will glow with the same brightness
2. Both the bulbs will get fused
3. The 25-W bulb will glow more brightly
4. The 100-W bulb will glow more brightly

Answer: 3. The 25-W bulb will glow more brightly

Since watt = volt x ampere, so

⇒ \(\text { watt }=V(I)=V\left(\frac{V}{R}\right)=\frac{V^2}{R}\)

⇒ \(R=\frac{V^2}{\text { watt }}\)

Hence, the resistance of the 25-watt bulb will be more than that of the 100-watt bulb. Since the bulbs are in series, the same current flows through them.

Hence, heat \(\left(H=I^2 R t\right)\) will be more in the 25-W bulb and it will glow more brightly.

Question 47. A uniform wire consumes power P when connected across a given potential difference V. If it is cut into two equal halves and joined in parallel with the same potential difference V, it will consume power

1. P
2. 2P
3. 4P
4. \(\frac{P}{4}\)

Answer: 3. 4P

Power (in watt) \(P=V I=\frac{V^2}{R}\)

When the wire is cut and joined the equivalent resistance is \(R^{\prime}=\frac{R}{4}\)

∴ \(P^{\prime}=\frac{V^2}{R^{\prime}}=\frac{V^2}{R / 4}=4\left(\frac{V^2}{R}\right)=4 P\).

Question 48. An electric bulb marked ’40 W, 200 V’ is connected to a circuit of supply voltage 100 V. The power consumed will be

1. 100 W
2. 20 W
3. 10 W
4. 40 W

Answer: 3. 10 W

Resistance of the bulb, \(R=\frac{V^2}{P}\)

Here, V = 200 V, P = 40 W, hence \(R=\frac{(200 \mathrm{~V})^2}{(40 \mathrm{~W})}=1000 \Omega\)

When connected across a 100-V supply, the power consumed is

⇒  \(P=\frac{V^2}{R}=\frac{(100 \mathrm{~V})^2}{(1000 \Omega)}=10 \mathrm{~W}\)

Question 49. The potential difference \(\left(V_{\mathrm{A}}-V_{\mathrm{B}}\right)\) between the points A and B as shown in the figure is

image

1. -3 V
2. +3 V
3. +6 V
4. +9 V

Answer: 4. +9 V

Consider two points a and b across the cell.

Hence, \(V_{\mathrm{A}}-V_{\mathrm{B}}=\left(V_{\mathrm{A}}-V_{\mathrm{a}}\right)+\left(V_{\mathrm{a}}-V_{\mathrm{b}}\right)+\left(V_{\mathrm{b}}-V_{\mathrm{B}}\right)\)

⇒ \((2 A)(2 \Omega)+(3 V)+(2 A)(1 \Omega)=4 V+3 V+2 V=9 V\)

Question 50. Two cells, one of emf 18 V and internal resistance 2 and the other of emf 12 V and internal resistance 1 ft, are connected as shown. The reading of the voltmeter will be

1. 15 V
2. 30 V
3. 14 V
4. 18 V

Answer: 3. 14 V

The given circuit is redrawn as shown in the figure. Let I be the current in the closed loop consisting of two cells only.

Applying the loop rule,

12 V +1(1 Ω)+(2 Ω)1-18 V =0

⇒ 1 = 2 A.

Voltmeter reading will be the p.d. across a and b, which is

⇒ \(V_{\mathrm{a}}-V_{\mathrm{b}}=(18 \mathrm{~V})-(2 \Omega)(2 \mathrm{~A})=18 \mathrm{~V}-4 \mathrm{~V}=14 \mathrm{~V}\)

Alternative method

The equivalent emf of the cells is

⇒ \(\varepsilon=\frac{e_1 r_2+e_2 r_1}{r_1+r_2}=\frac{(18 \mathrm{~V})(1 \Omega)+(12 \mathrm{~V})(2 \Omega)}{2 \Omega+1 \Omega}\)

⇒ \(\frac{(18+24) V \Omega}{3 \Omega}=\frac{42}{3} V=14 \mathrm{~V}\).

Question 51. The current through the cell in the circuit is

1. 1 A
2. \(\frac{2}{3} \mathrm{~A}\)
3. \(\frac{2}{9} \mathrm{~A}\)
4. \(\frac{1}{8} \mathrm{~A}\)

Answer: 1. 1 A

The circuit is redrawn as shown. Equivalent resistance across the cell is R, where \(\frac{1}{R}=\frac{1}{3}+\frac{1}{6}=\frac{2+1}{6}=\frac{1}{2} \Rightarrow R=2\)

Current through the cell is

⇒ \(I=\frac{\varepsilon}{R}=\frac{2 \mathrm{~V}}{2 \Omega}=1 \mathrm{~A}\).

Question 52. From the graph of current I and voltage V as shown here, identify the region corresponding to negative resistance.

1. DE
2. CD
3. BC
4. AB

Answer: 2. CD

Resistance is. measured by \(\frac{\Delta V}{\Delta I}\) which is the slope of the V-I graph.

When the slope is negative, resistance is negative, hence the region where R is negative is CD.

Question 53. In a Wheatstone bridge, the resistance of each of the four arms is 10 Q. If the resistance of the galvanometer is also 10 £2 then the effective resistance of the bridge will be

1. 10
2. 5
3. 20
4. 40

Answer: 1. 10

When each of the four arms has the same resistance of 10 Ω, the bridge is balanced, and no current flows through the galvanometer. Hence, the effective resistance is

⇒ \(R=\frac{(20 \Omega) \times(20 \Omega)}{20 \Omega+20 \Omega}=10 \Omega\)

Question 54. A cell has an emf of 1.5 V. When connected across an external resistance of 2 Ω, the terminal voltage falls to 1.0 V. The internal resistance of the cell is

1. 2
2. 1.5
3. 1.0
4. 0.5

Answer: 3. 1.0

Given that ε =1.5 V, R = 2 Ω, V =1.0 V.

Terminal voltage, V = ε -Ir. →(1)

But V = IR => \(I=\frac{\dot{V}}{R}=\frac{1 \mathrm{~V}}{2 \Omega}=0.5 \mathrm{~A}\).

SubstitutingI = 0.5 A in (1),

⇒ \(1.0 \mathrm{~V}=1.5 \mathrm{~V}-0.5 r \Rightarrow r=\frac{0.5 \mathrm{~V}}{0.5 \mathrm{~A}}=1.0 \Omega\).

Question 55. Two batteries of emf 4 V and 8 V with internal resistances 1 and 2 are connected in a circuit with an external resistance R = 9 as shown in the figure. The current and potential differences between points P and Q are

1. \(\frac{1}{3} \mathrm{~A} \text { and } 3 \mathrm{~V}\)
2. \(\frac{1}{12} \mathrm{~A} \text { and } 12 \mathrm{~V}\)
3. \(\frac{1}{6} \mathrm{~A} \text { and } 4 \mathrm{~V}\)
4. \(\frac{1}{9} \mathrm{~A} \text { and } 9 \mathrm{~V}\)

Answer: 1. \(\frac{1}{3} \mathrm{~A} \text { and } 3 \mathrm{~V}\)

Net resistance, R = l Ω + 2 Ω + 9 Ω = 12 Ω.

Net voltage, V =8 V- 4 V = 4 V.

Current in the direction Q to P is

⇒ \(I=\frac{V}{R}=\frac{4 \mathrm{~V}}{12 \Omega}=\frac{1}{3} \mathrm{~A}\)

Now, the potential difference between P and Q is

⇒ \(V_{\mathrm{P}}-V_{\mathrm{Q}}=\left(V_{\mathrm{P}}-V_{\mathrm{a}}\right)+\left(V_{\mathrm{a}}-V_{\mathrm{b}}\right)+\left(V_{\mathrm{b}}-V_{\mathrm{c}}\right)+\left(V_{\mathrm{c}}-V_{\mathrm{Q}}\right)\)

⇒ \(-\left(\frac{1}{3} A\right)(1 \Omega)+(-4 V)+\left(-\frac{1}{3} A\right)(2 \Omega)+8 V\)

⇒ \(-\frac{1}{3} V-4 V-\frac{2}{3} V+8 V=3 V\).

Question 56. The resistivity of the material of a potentiometer wire is 10^-7m and its area of cross section is 10^-6 m². When a current I = 0.1 A flows through the wire, its potential gradient is

1. \(10^{-2} \mathrm{Vm}^{-1}\)
2. \(10^{-4} \mathrm{~V} \mathrm{~m}^{-1}\)
3. \(10 \mathrm{~V} \mathrm{~m}^{-1}\)
4. \(0.1 \mathrm{~V} \mathrm{~m}^{-1}\)

Answer: 1. \(10^{-2} \mathrm{Vm}^{-1}\)

Resistance, \(R=\rho \frac{l}{A}\)

Hence, the potential difference across the ends of the wire is

⇒ \(V=I R=I\left(\rho \frac{l}{A}\right)\)

Potential gradient is

⇒ \(\frac{V}{l}=\frac{I \rho}{A}=\frac{(0.1 \mathrm{~A})\left(10^{-7} \Omega \mathrm{m}\right)}{10^{-6} \mathrm{~m}^2}=10^{-2} \mathrm{~V} \mathrm{~m}^{-1}\)

Question 57. The resistances of the four arms P, Q, R, and S of a Wheatstone bridge are 10 Ω, 30 Ω, 30, and 90 respectively. The emf and internal resistance of the cell are 7 V and 5respectively. If the galvanometer resistance is 50 Ω, the current drÿwn from the cell will be

1. 1.0 A
2. 0.2 A
3. 0.1 A
4. 2.0 A

Answer: 2. 0.2 A

With the given values of the four arms of the bridge, the balance condition \(\frac{P}{Q}=\frac{R}{S}\) is. satisfied.

So, the current through the galvanometer is zero for all values of G.

The equivalent resistance of the bridge is

⇒ \(R=\frac{(40 \Omega)(120 \Omega)}{40 \Omega+120 \Omega}=30 \Omega\)

Net resistance in the circuit is

= R + r = 30 Ω + 5 Ω = 35 Ω.

Hence, the current drawn from the cell is

⇒ \(I=\frac{\varepsilon}{R+r}=\frac{7 \mathrm{~V}}{35 \Omega}=0.2 \mathrm{~A}\).

Question 58. The resistance for balance in the two arms of a meter bridge is 5 and R respectively. When the resistance R is shunted with an equal resistance, the balance point is at 1.6 l₁. The resistance R is

1. 10
2. 15
3. 25
4. 20

Answer: 2. 15

In the first case of balance,

⇒ \(\frac{5 \Omega}{R}=\frac{l_1}{100 \mathrm{~cm}-l_1}\) → (1)

When R is shunted by an equal resistance R, its equivalent resistance will be

⇒  \(\frac{R}{2}\).

For this new condition of balance,

⇒ \(\frac{5 \Omega}{R / 2}=\frac{1.6 l_1}{100 \mathrm{~cm}-1.6 l_1}\)  →(2)

Taking the ratio \(\frac{l}{2}\), we get,

⇒ \(\frac{1}{2}=\frac{l_1}{100 \mathrm{~cm}-l_1} \times \frac{100 \mathrm{~cm}-1.6 l_1}{1.6 l_1}\)

⇒ \(\frac{1.6}{2}=\frac{100 \mathrm{~cm}-1.6 l_1}{100 \mathrm{~cm}-l_1} \Rightarrow l_1=25 \mathrm{~cm}\)

∴ From (1) \(R=5 \Omega \frac{(100 \mathrm{~cm}-25 \mathrm{~cm})}{25 \mathrm{~cm}}=15 \Omega\)

Question 59. Three resistors P, Q, and R, each of 2 Ω, and an unknown resistance S form the four arms of a Wheatstone bridge circuit. When a resistance of 6 is connected in parallel to S, the bridge gets balanced. The value of S is

1. 2
2. 3
3. 1
4. 6

Answer: 2. 3

For balance, the fourth arm must be of 2.

Hence, \(\frac{1}{S}+\frac{1}{6 \Omega}=\frac{1}{2 \Omega}\)

⇒ S = 3Ω.

Question 60. In the circuit shown, if a conducting wire is connected between the points A and B, the current in this wire will

1. Flow from A to B
2. Flow from B to A
3. Be zero
4. Flow in the direction which will be decided by the value of V

Answer: 2. Flow from B to A

Let the current through the upper and lower branches be I₁ and I2 respectively.

Hence,

VP – VQ = I₁ (4+4) =8 I₁ ,

VP – VQ = I₂ (1 + 3) = 4 I₂.

So, 4I2 = 8 I₁ ⇒ I₂  = 2 I_{1  →} (1)

Now, VP- VA = 4 I₁ ⇒ VA = Vp- 4 I₁. → (2)

And Vp-VB = 1 ( I₂ ) = 2 I₁ ⇒ VB = VP – 2 I₁. → (3)

From (2) and (3), VB>VA.

Hence, on joining A and B by a conducting wire, current will flow from B to A

Question 61. A cell can be balanced against 110 cm and 100 cm of a potentiometer wire respectively, with and without being short-circuited through a resistance of 10 Ω. Its internal resistance is

1. 1.0
2. 0.5
3. 2.0
4. 0.2

Answer: 1. 1.0

With the switch open, balance l =110 cm is obtained across emf £ of
the cell. Thus, ε=kl.

When S is closed, the cell is shunted by a 10-Ω resistance so that the terminal voltage V is balanced at l’=100 cm.

Thus, V = kl’.

So, \(\frac{\varepsilon}{V}=\frac{l}{l^{\prime}}=\frac{110 \mathrm{~cm}}{100 \mathrm{~cm}}\)  → (1)

But \(\varepsilon=I(R+r) \text { and } V=I R\)

∴ \(\frac{\varepsilon}{V}=\frac{R+r}{R}=\frac{10 \Omega+r}{10 \Omega}\)  → (2)

Equating (1) and (2),

⇒ \(\frac{10 \Omega+r}{10 \Omega}=\frac{110}{100} \Rightarrow r=1 \Omega\)

Question 62. A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery used across the potentiometer wire has an emf of 2.0 V and negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R, connected across the given cell, has values of

• infinity and
• 9.5 ft, the balancing lengths on the potentiometer wire are found to be 3 m and 2.85 m respectively. The value of the internal resistance of the cell is

1. 0.25
2. 0.95
3. 0.5
4. 0.75

Answer: 3. 0.5

When the cell is without any parallel resistance, the balance length is at l =3 m and when shunted with R = 9.5 Ω, the balance length l’ = 2.85 m.

We know that internal resistance is

⇒ \(r=\left(\frac{l}{l^{\prime}}-1\right) R=\left(\frac{3 \mathrm{~m}}{2.85 \mathrm{~m}}-1\right) 9.5 \Omega=\frac{0.15 \mathrm{~m}}{2.85 \mathrm{~m}} \times 9.5 \Omega=0.5 \Omega\)

Question 63. A potentiometer wire of length l and resistance r is connected in series with a battery of emf80 and a resistance rx. A cell of unknown emf8 is balanced against the length of the potentiometer wire. The emf if will be given by

1. \(\frac{L \varepsilon_0 r}{l r_1}\)
2. \(\frac{\varepsilon_0 r}{\left(r+r_1\right)} \frac{l}{L}\)
3. \(\frac{\varepsilon_0 l}{L}\)
4. \(\frac{L \mathcal{E}_0 r}{\left(r+r_1\right) l}\)

Answer: 2. \(\frac{\varepsilon_0 r}{\left(r+r_1\right)} \frac{l}{L}\)

Current through the potentiometer wire by the driving cell ε₀ is

⇒ \(I=\frac{\varepsilon_0}{r+r_1}\)

The potential drop across the ends of the potentiometer wire is

⇒ \(V=I r=\frac{\varepsilon_0 r}{r+r_1}\)

Hence, potential gradient = \(\frac{V}{L}=\frac{\mathcal{E}_0 r}{\left(r+r_1\right) L}\)

The balance point is obtained at distance Z, so the emf of the cell is

⇒ \(\varepsilon=\frac{V l}{L}=\frac{\varepsilon_0 r l}{\left(r+r_1\right) L}\)

Question 64. A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite directions. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf is

1. 5:4
2. 3:2
3. 3:4
4. 5 :1

Answer: 2. 3:2

When the cells are in supporting order,

⇒ \(\left(\varepsilon_1+\varepsilon_2\right)=k l_1=k(50 \mathrm{~cm})\)

When the cells are in reverse order

⇒ \(\begin{aligned}
\left(\varepsilon_1-\varepsilon_2\right)=k l_2=k(10 \mathrm{~cm})\end{aligned} \)

⇒ \(\begin{aligned}\frac{\varepsilon_1+\varepsilon_2}{\varepsilon_1-\varepsilon_2}=5, \text { hence } \frac{\varepsilon_1}{\varepsilon_2}=\frac{3}{2}
\end{aligned}\).

Question 65. A circuit contains an ammeter, a battery of 30 V, and a resistance of 40.8 all connected in series. If the ammeter has a coil of resistance 480 and a shunt of 20 then the reading in the ammeter will be

1. 0.5 A
2. 1 A
3. 2 A
4. 0.25 A

Answer: 1. 0.5 A

Equivalent resistance across ab is

⇒ \(R_{\mathrm{eq}}=\frac{G S}{G+S}=\frac{480 \times 20}{500} \Omega=19.2 \Omega\)

Total resistance in the circuit,

R = 40.8 Ω +19.2 Ω = 60 Ω.

Main currents

∴ \(I=\frac{30 \mathrm{~V}}{60 \Omega}=0.5 \mathrm{~A}\)

The current measured by the ammeter is the current flowing through the circuit, so the ammeter reading will be 0.5 A.

Question 66. A millivoltmeter of 25 mV range is to be converted into an ammeter of 25 A range. The value (in ohm) of the required shunt will be

1. 0.001
2. 0.01
3. 1
4. 0.05

Answer: 1. 0.001

For full-scale deflection, the current through the meter,

⇒ \(I_g=\frac{V}{G}=\left(\frac{25 \mathrm{mV}}{G}\right) A\) where G = resistance of the meter.

In order to increase the range to 25 A when used as an ammeter, the value of the required shunt is

⇒ \(S=\frac{I_{\mathrm{g}} \mathrm{G}}{I-I_{\mathrm{g}}} \approx \frac{I_{\mathrm{g}} \mathrm{G}}{I}\) [∵ Ig<<I]

⇒ \(\frac{25 \times 10^{-3} \mathrm{~V}}{25 \mathrm{~A}}=0.001 \Omega\).

Question 67. A galvanometer which has a coil of resistance 100 Ω, gives a full-scale deflection for 30 mA current. If it is to work as a voltmeter in the 30-V range, the resistance required to be added will be

1. 900
2. 1800
3. 500
4. 1000

Answer: 1. 900

Given that G =100 Ω, Ig = 30 mA.

The voltage across the galvanometer for full-scale deflection is

⇒ \(V=I_{\mathrm{g}} \cdot G=(30 \mathrm{~mA})(100 \Omega)=3 \mathrm{~V}\).

The factor n by which its measuring range as a voltmeter is increased is

⇒ \(n=\frac{30 \mathrm{~V}}{3 \mathrm{~V}}=10\)

The resistance required to be added in series with the meter will be

⇒ \(R=(n-1) G=(10-1)(100 \Omega)=900 \Omega\).

Question 68. The resistance of an ammeter is 13 Ω and its scale is graduated for a current up to 100 A. After an additional shunt has been connected to this ammeter, it becomes possible to measure currents up to 750 A by this meter. The value of the resistance of this shunt is

1. 20
2. 2
3. 0.2
4. 2k

Answer: 2. 2

Given that G =13 Ω, Ig =100 A.

p.d. across the ammeter for full-scale deflection is

V = Ig x G = (100 A)(13 Ω) =1300 V

When the measuring range is increased (I = 750 A), let the shunt resistance be S. Equivalent resistance of the parallel combination of G and S is \(\frac{G S}{G+S}\) and potential difference across this = \(I\left(\frac{G S}{G+S}\right)\)

For full-scale deflection, the p.d. across this combination must be V =1300 V.

⇒ \(1300 \mathrm{~V}=(750 \mathrm{~A}) \frac{13 \Omega \times S}{13 \Omega+S}\).

Hence, shunt=S = 2Ω.

Question 69. A battery is charged at a potential of 15 V for 8 h when the current flowing is 10 A. The battery during discharge supplies a current of 5 A for 15 h. The mean terminal voltage during discharge is 14 V. The watt-hour efficiency of the battery is

1. 82.5%
2. 87.5%
3. 80%
4. 90%

Answer: 2. 87.5%

During the charging process, the input energy is

E₁ = coulomb x volt = (It)V J

= (10 A)(8 h)(15 V) =1200 W h.

During discharge, the energy output is

E₀ = (5 A)(15 h)(14 V) =1050 W h

⇒ Efficiency = \(\eta=\frac{\text { output }}{\text { input }} \times 100 \%=\frac{1050}{1200} \times 100 \%=87.5 \%\)

Question 70. Two cells having the same emf are connected in series through an external resistance R. The cells have internal resistances r₁ and r₂ (r₁ > r₂) respectively. When the circuit is closed the potential difference across the first cell is zero. The value of R is

1. \(r_1-r_2\)
2. \(\frac{r_1-r_2}{2}\)
3. \(\frac{r_1+r_2}{2}\)
4. \(r_1+r_2\)

Answer: 1. \(r_1-r_2\)

Current through the circuit is

⇒ \(I=\frac{2 e}{R+r_1+r_2}\)

The potential difference across the first cell is

⇒ \(V_{\mathrm{a}}-V_{\mathrm{b}}=e-I r_1=0 \text { (given) } \Rightarrow e=I r_1\)

⇒ \(e=\frac{2 e r_1}{R+r_1+r_2}\)

⇒ \(R+r_1+r_2=2 r_1 \Rightarrow R=r_1-r_2\)

Question 71. Refer to the electrical circuit shown in the figure. Which of the following equations is the correct representation for the given circuit?

1. \(\varepsilon_1-\left(I_1+I_2\right) R-I_1 r_1=0\)
2. \(\varepsilon_2-I_2 r_2-\varepsilon_1-I_1 r_1=0\)
3. \(-\mathcal{E}_2-\left(I_1+I_2\right) R+I_2 r_2=0\)
4. \(\varepsilon_1-\left(I_1+I_2\right) R+I_1 r_1=0\)

Answer: 1. \(\varepsilon_1-\left(I_1+I_2\right) R-I_1 r_1=0\)

According to Kirchhoff’s junction rule, the current through the resistance R is ( I₁ + I₂ ). Applying Kirchhoff’s loop rule for the upper closed loop,

⇒ \(\begin{aligned}
R\left(I_1+I_2\right)+I_1 r_1-\varepsilon_1=0\end{aligned}\)

⇒ \(\begin{aligned}\varepsilon_1-\left(I_1+I_2\right) R-I_1 r_1=0
\end{aligned}\)

Question 72. In the given circuit, the cells A and B have negligible resistances. For VA =12 V, Rx = 500 Q, and R =100 Q, the galvanometer G shows null deflection. The value of VB is

1. 4 V
2. 2 V
3. 12 V
4. 6 V

Answer: 2. 2 V

For null deflection in the galvanometer, a and b must be at the same potential, and

⇒ \(\begin{aligned}
V_{\mathrm{B}}  =V_{\mathrm{a}}-V_{\mathrm{c}}=I(100 \Omega)\end{aligned} \)

⇒ \(\begin{aligned}\frac{12 \mathrm{~V} \times 100 \Omega}{600 \Omega}=2 \mathrm{~V}
\end{aligned}\).

Question 73. Six identical bulbs are connected as shown in the figure with a DC source of emf E and zero internal resistance. The ratio of power consumed by the bulbs when

1. All are glowing and
2. Two from section A and one from section B are glowing and will be

1. 4:9
2. 9:4
3. 1:2
4. 2:1

Answer: 2. 9:4

Equivalent watt in series is given by \(\frac{1}{W_S}=\frac{1}{W_1}+\frac{1}{W_2}+\frac{1}{W_3}+\ldots\),and in parallel \(W_P=W_1+W_2+W_3+\ldots\)

If the voltage of each bulb is W then in case (1)

⇒ \(W_1=\frac{W}{2}+\frac{W}{2}+\frac{W}{2}=\frac{3}{2} W\)

and in case (2),

⇒ \(\frac{1}{W_2}=\frac{1}{2 W}+\frac{1}{W}=\frac{3}{2 W} \Rightarrow W_2=\frac{2 W}{3}\)

∴ Required ratio, \(\frac{W_1}{W_2}=\frac{3 W / 2}{2 W / 3}=\frac{9}{4}\).

Question 74. Which of the following acts as a circuit-protecting device?

1. Conductor
2. Inductor
3. Switch
4. Fuse

Answer: 4. Fuse

A fuse is an electrical safety device that operates to provide over-current protection to an electrical circuit. It has high resistance and low melting point so that when a high current flows through the circuit, the fuse wire melts and the current stops flowing.

Question 75. In the circuit shown below, the readings of voltmeters and ammeters will be

1. \(V_2>V_1 \text { and } I_1=I_2\)
2. \(V_1=V_2 \text { and } I_1>I_2\)
3. \(V_1=V_2 \text { and } I_1=I_2\)
4. \(V_2>V_1 \text { and } I_1>I_2\)

Answer: 3. \(V_1=V_2 \text { and } I_1=I_2\)

An ideal voltmeter has infinite resistance whereas an ideal ammeter has effectively zero resistance. Under this ideal condition,

⇒ \(I_1=\frac{10 \mathrm{~V}}{10 \Omega}=1 \mathrm{~A}, V_1=(1 \mathrm{~A})(10 \Omega)=10 \mathrm{~V}\),

⇒ and \(I_2=\frac{10 \mathrm{~V}}{10 \Omega}=1 \mathrm{~A}, V_2=(1 \mathrm{~A})(10 \Omega)=10 \mathrm{~V}\).

Thus, V₁ = V₂ and I₁ = I₂

Question 76. The resistive network shown is connected to a DC source of 16 V. The power consumed by the network is 4 W. The value of R is

1. l
2. 16
3. 8
4. 4

Answer: 3. 8

The equivalent resistance of the given combination of resistors is

R_net = 2R+ 2R + 4R = SR.

Power consumed = \(4 \mathrm{~W}=\frac{V^2}{R_{\text {net }}}\)

∴ \(4=\frac{(16)^2}{8 R}, \text { hence } R=\frac{16 \times 16}{4 \times 8} \Omega=8 \Omega\)

Question 77. In the given circuit, an ideal voltmeter connected across the 10-Q resistor reads 2 V. The internal resistance r of each cell is

1. 0.5
2. 0
3. 1.5
4. 1

Answer: 1. 0.5

The potential difference across the 10-Ω and 15-Ω resistors is 2 V, so the current is

⇒ \(i_1=\frac{V}{R_1}=\frac{2}{10} \mathrm{~A}\)

Similarly, \(i_2=\frac{V}{R_2}=\frac{2}{15} \mathrm{~A}\).

∴ total current \(I=i_1+i_2=\left(\frac{2}{10}+\frac{2}{15}\right) \mathrm{A}=\frac{1}{3} \mathrm{~A}\)

Applying Kirchhoff’s loop rule,

(10 Ω)i₁ + (2 Ω)I + (2r)I = 3 V

⇒ \(\left(10 \times \frac{2}{10}\right) V+\left(2 \times \frac{1}{3}\right) V+(2 r)\left(\frac{1}{3} A\right)=3 V\)

⇒ \(\frac{2}{3} r=\left(3-2-\frac{2}{3}\right) \Omega\)

∴ internal resistance = r = 0.5 Ω.

Question 78. An ideal battery of 4 V and a resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5 Ω. The value of R to give a potential difference of 5 mV across 10 cm of the potentiometer wire is

1. 480
2. 490
3. 495
4. 395

Answer: 4. 395

The total resistance in the main circuit is (R +5 Ω) and current through the l primary circuit is

⇒ \(I=\frac{4 V}{R+5 \Omega}\)

Potential gradient is

⇒ \(\frac{V}{L}=\frac{I \times 5 \Omega}{100} \mathrm{~V} \mathrm{~cm}^{-1}\)

Hence, p.d. across the 10-cm wire will be

⇒ \(\left(\frac{I}{20} \times 10\right) \mathrm{V}=5 \mathrm{mV}=5 \times 10^{-3} \mathrm{~V}\)

From (1),

⇒ \(\frac{1}{2}\left(\frac{4 V}{R+5 \Omega}\right)=5 \times 10^{-3} V\)

⇒ \(\frac{2}{R+5 \Omega}=\frac{5}{1000} \Rightarrow R=395 \Omega\)

Question 79. For the circuit shown with R₁=1 Ω, R₂ =2 Ω, ε₁=2 V, and ε₂ = 4 V, the potential difference between points a and b is

1. 2.7 V
2. 3.3 V
3. 2.3 V
4. 3.7 V

Answer: 1. 2.7 V

From the Kirchhoff loop rule for the closed path ABCDA,

1(x)+2 +(1)x-y-2-y = 0

⇒ 2x-2y = 0 ⇒ x = y  → (1)

For the closed-loop AbaDA,

x-4+2(x + y)+x+2 = 0

⇒ 4x + 2y = 2

⇒ 2x+y =1 → . (2)

From (1) and (2),

⇒ \(x=\frac{1}{3}=y\)

⇒ \(V_{\mathrm{a}}-V_{\mathrm{b}}=-2(x+y)+4=4 \mathrm{~V}-2\left(\frac{2}{3}\right) \mathrm{V}=\frac{8}{3} \mathrm{~V}=2.67 \mathrm{~V} \approx 2.7 \mathrm{~V}\)

Question 80. A current of 2 mA was passed through an unknown resistor which dissipated 4.4 W of power. When a supply of 11 V is connected across this resistor, the dissipated power will be

1. \(11 \times 10^{-4} \mathrm{~W}\)
2. \(11 \times 10^{-5} \mathrm{~W}\)
3. \(11 \times 10^{-6} \mathrm{~W}\)
4. \(11 \times 10^{-3} \mathrm{~W}\)

Answer: 2. \(11 \times 10^{-5} \mathrm{~W}\)

Power dissipation = 4.4 W = I2R

⇒ \(R=\frac{4.4 \mathrm{~W}}{\left(2 \times 10^{-3} \mathrm{~A}\right)^2} \Omega=11 \times 10^5 \Omega\)

With a supply voltage of 11 V, the power dissipated is

⇒ \(\frac{V^2}{R}=\frac{(11 \mathrm{~V})^2}{11 \times 10^5 \Omega}=11 \times 10^{-5} \mathrm{~W}\).

Question 81. The drift speed of electrons, when A of electric current flows through a copper wire of cross section 5 mm², is v. If the number density of electrons in copper is 9 x l0²⁸ m^-3, the value of v in mm s^-1 is close to

1. 0.02
2. 0.2
3. 2
4. 3

Answer: 1. 0.02

Current 1 through a conductor is given by I = Avdne, where I =1.5 A, rt = number density = 9 x1028 m-3, and A- cross-sectional area =5 mm2.

Drift speed is

⇒ \(v_{\mathrm{d}}=\frac{I}{A n e}\)

⇒ \(\frac{1.5 \mathrm{~A}}{\left(5 \times 10^{-6} \mathrm{~m}^2\right)\left(9 \times 10^{28} \mathrm{~m}^{-3}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)}\)

⇒ \(\frac{1.5 \mathrm{~A}}{72 \times 10^3 \mathrm{C} \mathrm{m}^{-1}}=0.02 \times 10^{-3} \mathrm{~m} \mathrm{~s}^{-1}=0.02 \mathrm{~mm} \mathrm{~s}^{-1}\)

Question 82. A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance is

1. 2.0%
2. 1.0%
3. 0.5%
4. 2.5%

Answer: 2. 1.0%

Resistance = \(R=\rho \frac{l}{A}=\rho \frac{l^2}{V}\), where Z = length and V = volume.

⇒ \(\frac{\Delta R}{R}=2 \frac{\Delta l}{l}=2(0.5 \%)=1.0 \%\).

Question 83. When the switch S in the circuit shown is closed, the value of the current I will be

1. 4 A
2. 3 A
3. 2 A
4. 5 A

Answer: 4. 5 A

Let V be the potential at junction C.

We have, \(i_1=\frac{20-V}{2}, i_2=\frac{10-V}{4}\)

∴ i₁ + i₂ = i = \(\frac{V}{2}\)

⇒ \(\frac{20-V}{2}+\frac{10-V}{4}=\frac{V}{2}\) ⇒ V = 10 volts.

∴ Current = i \(\frac{V}{2 \Omega}=\frac{10 \mathrm{~V}}{2 \Omega}\) = 5 A.

Question 84. Two equal resistances when connected in series to a battery consume 60 W of electric power. If these resistances are connected in parallel combination to the same battery, the electric power consumed will be

1. 240 W
2. 120 W
3. 30 W
4. 60 W

Answer: 1. 240 W

In a series combination, equivalent wattage is

⇒ \(W_S=\frac{W_1 W_2}{W_1+W_2} \Rightarrow 60 \mathrm{~W}=\frac{W_1 \times W_2}{2 W_1}\)

The wattage of each resistor is

W₁ = 120 W

Hence, the consumption of power when connected in parallel is

WP = W₁ + W₂ = 2 W₁ = 2(120 W) = 240 W.

Question 85. The resistance of a galvanometer is 50and the maximum current of 0.002 A gives full-scale deflection. What resistance must be connected to it so as to convert it into an ammeter of range 0-0.5 A?

1. 0.5
2. 0.02
3. 0.2
4. 0.002

Answer: 3. 0.2

The initial measuring range is

I = 0.002 A and the final range is nl = 0.5 A.

∴ \(n=\frac{0.5 \mathrm{~A}}{0.002 \mathrm{~A}}=250\)

Required shunt is

⇒ \(S=\frac{G}{n-1}=\frac{50 \Omega}{249} \approx 0.2 \Omega\).

Question 86. A metal wire of resistance 3 is elongated to make it a uniform wire of double its previous length. This new wire is now bent and its ends are joined to make a circle. If two points P and Q on this circle make an angle of 60° at the center, the equivalent resistance across these two points will be

1. \(\frac{5}{2} \Omega\)
2. \(\frac{5}{3} \Omega\)
3. \(\frac{12}{5} \Omega\)
4. \(\frac{7}{2} \Omega\)

Answer: 2. \(\frac{5}{3} \Omega\)

Initial resistance, R =3 Ω.

When the length is doubled, its new resistance will become 4R =12 Ω.

Resistance of part \(P Q=\frac{1}{6}(12 \Omega)=2 \Omega\)

and that oflower part=12Ω- 2 Ω =10 Ω.

Equivalent resistance across PQ

∴ \(R_{\mathrm{eq}}=\frac{r_1 r_2}{r_1+r_2}=\frac{(2 \Omega)(10 \Omega)}{12 \Omega}=\frac{5}{3} \Omega\).

Question 87. A potentiometer wire AB having length L and resistance 12r is joined to a cell D of emf £ and internal resistance r. A cell C of emf \(\) and internal resistance 3r is connected to A as shown. The length AJ at which the galvanometer shows null deflection is

1. \(\frac{5 L}{12}\)
2. \(\frac{11 L}{12}\)
3. \(\frac{13 L}{24}\)
4. \(\frac{11 L}{24}\)

Answer: 3. \(\frac{13 L}{24}\)

The total resistance in the main circuit = r +12r =13r.

∴ current through the potentiometer wire = \(I=\frac{\varepsilon}{13 r} .\)

The potential drop across length L of the wire AB is

⇒ \(V=I(12 r)=\frac{\varepsilon}{13 r}(12 r)=\frac{12}{13} \varepsilon\)

Potential gradient = \(\frac{V}{L}=\frac{12}{13} \frac{\varepsilon}{L} \mathrm{~V} \mathrm{~m}^{-1}\).

When the galvanometer shows null deflection at length AJ = x, we have

⇒  \(\frac{\varepsilon}{2}=\frac{12}{13} \frac{\varepsilon}{L} x \Rightarrow x=\frac{13 L}{24}\)

Question 88. In the. given circuit, the cells have. zero internal resistance. The current (in ampere) passing through resistances JRj and R2 respectively are

1. 0.5,0
2. 1.0,2
3. 2, 2
4. 0,1.5

Answer: 1. 0.5,0

In the closed loop A,

-10 +(x + y) 20 = 0

⇒ \(x+y=\frac{1}{2}\)  → (1)

In the closed loop B,

(x + y)20 + 20y-10 = 0

⇒ 2(x+y)+2y =l

⇒ 2x+4y =l

⇒ \(x+2 y=\frac{1}{2}\) → (2)

Solving (1) and (2),

⇒ \(x=\frac{1}{2}\) A and y = 0.

∴ Current in R₁ is \((x+y)=\frac{1}{2}\) A and in R₂ is y = 0.

Question 89. The given circuit contains two ideal diodes, each with a forward resistance of 50 ft. If the battery voltage is 6 V, the current through the 100-12 resistance will be

1. 0.027 A
2. 0.036 A
3. 0.020 A
4. 0.030 A

Answer: 3. 0.020 A

In the given circuit, diode D2 is reverse-biased. Hence, nonconducting. D1 will be conducted with forward resistance of 50.

Thus, the total resistance is

R = 50 Ω+150 Ω +100 Ω = 300 Ω.

⇒ So, \(I=\frac{6 \mathrm{~V}}{300 \Omega}=0.02 \mathrm{~A}\)

Question 90. In the given circuit, the currents I₁=-03 A, I₄ = 0.8 A, and I₅ = 0.4 A are flowing as shown. The currents I₂, I₃ and I₆ are respectively

1. 1.1 A, 0.4 A, 0.4 A
2. 1.1 A, -0.4 A, 0.4 A
3. 0.4 A, 1.1 A, 0.4 A
4. 0.4 A, 0.4 A, 1.1 A.

Answer: 1. 1.1 A, 0.4 A, 0.4 A

Applying junction Me:

At R, I₂ +(-0.3 A) = 0.8 A

⇒ I₂ = 1.1 A.

At S, I₃ + 0.4 A = 0.8 A

⇒ I₃ = 0.4 A.

At P, I₆ = 0.4 A.

Thus, the values are 1.1 A, 0.4 A, and 0.4 A.

Question 91. In the figure shown, what is the current (in ampere) drawn by the battery?

1. \(\frac{7}{38}\)
2. \(\frac{20}{3}\)
3. \(\frac{9}{32}\)
4. \(\frac{13}{24}\)

Answer: 3. \(\frac{9}{32}\)

The given circuit can be redrawn in a simplified form as shown. The equivalent resistance of the parallel combination is

⇒ \(R_1=\frac{50 \times 10}{60} \Omega=\frac{25}{3} \Omega .\)

Net resistance in the circuit is

⇒ \(R=\frac{25}{3} \Omega+15 \Omega+30 \Omega=\frac{160}{3} \Omega\)

current delivered by the battery is

⇒ \(I=\frac{\varepsilon}{R}=\frac{15 \mathrm{~V}}{\frac{160}{3} \Omega}=\frac{9}{32} \mathrm{~A}\)

Question 92. The potential difference across A and B in the given circuit is

1. 1 V
2. 3 V
3. 2 V
4. 6 V

Answer: 3. 2 V

The current distribution in the different branches is shown in the diagram.

No current in 5 Ω and 10 Ω. Thus, V_A-V_B = V_a-V_b.

For the upper loop, x+y+1-2+y = 0 => x + 2y = 1A.

For the lower loop, -y+2-3+x=0 => x-y =1 A.

Solving, we get x=1 A, y =0.

Now for any branch across ab,

V_a-V_b = -(1 Ω)x +3V =-1V +3V = 2V.

Question 93. The Wheatstone bridge shown in the figure gets balanced when the carbon resistor has the color code orange, red, and brown. The resistors R₂ and R₄ are 80 Ω and 40 Ω respectively. Assuming that the color code for the carbon resistor gives accurate values, the color code for R₃ would be

1. Brown, blue, black
2. Brown, blue, brown
3. Grey, black, brown
4. Red, green, brown

Answer: 2. Brown, blue, brown

R₁ has the color code

Orange – 3

Red – 2

Brown – 1

⇒ \(R_1=32 \times 10^1=320 \Omega\) .

⇒ For balance \(\frac{R_1}{R_2}=\frac{R_3}{R_4}\)

⇒ \(\quad R_3=\frac{R_4}{R_2} R_1=\frac{40 \Omega}{80 \Omega} \times 320 \Omega=160 \Omega \text {. }\)

The corresponding color code is

Brown – 1

Blue – 6

Brown -10¹

⇒ 16 x 10¹ Ω = 160 Ω

Question 94. A carbon resistor has the color code resistor indicated in the figure. What is the value of the resistance?

1. 5.3 M ± 5%
2. 64 k±10%
3. 6.4 M ±5%
4. 64 k±10%

Answer: 4. 64 k±10%

From the given color code

Green – 5

Orange – 6

Yellow – 4

Golden – 5%

⇒ R = 53 x 10⁴ ± 5% = 530 kΩ ± 5% .

Question 95. A resistor is shown in the figure. Its value and tolerance are given respectively as

1. 270Ω and 5%
2. 27kΩ and 20%
3. 27kΩ and 10%
4. 270kΩ and 10%

Answer: 3. 27kΩ and 10%

The color rings are

Red – 2

Violet – 7

Orange – 3

Silver – 10%

∴ Resistance = R = 27 x 10³ Ω,10% = 27 kΩ,10%.

Question 96. A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when

1. R = O.Olr
2. R = 2r
3. R = r
4. R =10r

Answer: 3. R = r

According to the maximum power output theorem/ the external resistance R must be equal to the internal resistance r of the voltage source.

Question 97. A moving-coil galvanometer has a resistance of 50 Ω and it indicates full deflection at 4-mA current. A voltmeter is made using this galvanometer and a 5-kΩ resistance. The maximum voltage that can be measured using this voltmeter will be close to

1. 10 V
2. 15 V
3. 40 V
4. 20 V

Answer: 4. 20 V

Given that G = 50 Ω, I_g = 4 mA.

While using this galvanometer ‘ as a voltmeter (with measuring range V), a resistance of 5 kΩ is connected in series so as to get the same current I_g (=4 x10^-3A) for full-scale deflection

∴ Hence, \(I_{\mathrm{g}}=\frac{V}{G+R}\)

⇒ V = (G + R)I_g = (50 Ω + 5000 Ω)(4 x10-3 A) = 20.2 V ≈ 20 V.

Question 98. A 200 resistor has a certain color code. If one replaces the red color with green in the code then the new resistance will be

1. 100
2. 200
3. 400
4. 500

Answer: 4. 500

For the 200-Ω resistor, the color code is red, black, and black. Red corresponds to 2 which is replaced by green (≡ 5) so the new resistance will be 500 Ω.

Question 99. A 2 W-carbon resistor is color-coded with green, black, red, and brown respectively. The maximum current that can be passed through the resistor is

1. 0.4 mA
2. 63 mA
3. 20 mA
4. 100 mA

Answer: 3. 20 mA

The color codes are

Green – 5

Black – 0

Red – 2

Brown – 1

R_min = 50 x 10²

⇒ \(I_{\max }=\sqrt{\frac{P}{R_{\min }}}=\sqrt{\frac{2 \mathrm{~W}}{50 \times 10^2 \Omega}}=2 \times 10^{-2} \mathrm{~A}=20 \mathrm{~mA}\)

Question 100. The given figure shows a cell of emf S =s 3 V and internal resistance r connected across an external resistance R. If power dissipation in R is 0.5 W and the terminal voltage across the cell is 2.5 V then the power dissipation in the internal resistance r is

1. 0.5 W
2. 0.1W
3. 0.3 W
4. 1.0 W

Answer: 2. 0.1W

Power dissipation in external resistance is

W = V x I ⇒ 0.5 W = (2.5V) I.

∴ \(\text { current }=I=\frac{0.5 \mathrm{~W}}{2.5 \mathrm{~V}}=\frac{1}{5} \mathrm{~A}\)

The potential drop across internal resistance is

V_r = 3 V- 2.5 V = 0.5 V.

Power dissipation in the internal resistance r is

⇒ \(W_r=V_r \cdot I=(3 \mathrm{~V}-2.5 \mathrm{~V})\left(\frac{1}{5} \mathrm{~A}\right)=0.5 \times \frac{1}{5} \mathrm{~W}=0.1 \mathrm{~W}\)

101. The resistance of the voltmeter in the given circuit is 10 k£l. Find the voltmeter reading.

1. 4 V
2. 3.25 V
3. 1.25 V
4. 1.95 V

Answer: 4. 1.95 V

Let V_A-V_B = V = voltmeter reading.

The main current is

⇒ \(I=I_1+I_2=\frac{V}{10000}+\frac{V}{400}\)

V + 7(800 Ω) = 6 V

⇒ \(V+V\left(\frac{1}{10000}+\frac{1}{400}\right)(800 \Omega)=6 \mathrm{~V}\)

⇒ \(V+\frac{8 V}{100}+2 V=6 \mathrm{~V} \Rightarrow V=1.95 \mathrm{~V}\).

Question 102. In the given circuit, 1.02 V is balanced at 51 cm from end A of a 1.0-m-long potentiometer wire. Find the potential gradient along AB.

1. \(0.1 \mathrm{~V} \mathrm{~cm}^{-1}\)
2. \(0.02 \mathrm{~V} \mathrm{~cm}^{-1}\)
3. \(0.3 \mathrm{~V} \mathrm{~cm}^{-1}\)
4. \(0.4 \mathrm{~V} \mathrm{~cm}^{-1}\)

Answer: 2. \(0.02 \mathrm{~V} \mathrm{~cm}^{-1}\)

For null deflection, potential drop = ΔV =1.02 V for length l = 51 cm of the potentiometer wire.

Hence, the potential gradient is

⇒ \(\frac{\Delta V}{\Delta l}=\frac{1.02 \mathrm{~V}}{51 \mathrm{~cm}}=0.02 \mathrm{~V} \mathrm{~cm}^{-1}\)

Question 103. A cylindrical shell of length l with inner and outer radii Rx and R2 is made of a material of resistivity p. Find its resistance if current flows radially outward.

1. \(\frac{\rho}{2 \pi l} \ln \frac{R_2}{R_1}\)
2. \(\frac{\rho}{\pi l} \ln \frac{R_2}{R_1}\)
3. \(\frac{\rho}{4 \pi l} \ln \frac{R_2}{R_1}\)
4. \(\frac{\rho}{3 \pi l} \ln \frac{R_2}{R_1}\)

Answer: 1. \(\frac{\rho}{2 \pi l} \ln \frac{R_2}{R_1}\)

Resistance of a coaxial cylindrical shell of radius r and thickness dr is

⇒ \(d R=\rho \frac{d r}{A}=\rho \frac{d r}{2 \pi r l}\)

total resistance is

⇒ \(R=\int d R=\frac{\rho}{2 \pi l} \int_{R_1}^{R_2} \frac{d r}{r}=\frac{\rho}{2 \pi l} \ln \frac{R_2}{R_1}\)

Question 104. Which of the following graphs represents the variation of resistivity (p) with temperature (T) for copper?

Answer: 2.

With the increase in temperature of a conductor, thermal agitation increases and the collisions become more frequent. Thus, copper being a good conductor, its conductivity decreases and resistivity(ρ)increases nonlinearly with temperature. Hence, graph (b).

105. The color code of a resistor is given in the figure. The value of its resistance with tolerance is

1. 47 kΩ, 10%
2. 4.7 kΩ, 5%
3. 470 Ω, 5%
4. 470 kΩ, 5%

Answer: 3. 470Ω, 5%

The sequence of colors is

Yellow – 4

Violet – 7

Brown – 10

Gold – 5%

Hence, R =(47×10) Ω ± 5% = 470 Ω, 5%.

Question 106. The resistance wire connected in the left gap of a meter bridge balances a 10-Ω resistance in the right gap at a point that divides the bridge wire in the ratio 3: 2. If the length of the resistance wire is 1.5 m then the length of 1 Ω of the resistance wire is

1. \(1.0 \times 10^{-1} \mathrm{~m}\)
2. \(1.5 \times 10^{-1} \mathrm{~m}\)
3. \(1.5 \times 10^{-2} \mathrm{~m}\)
4. \(1.0 \times 10^{-2} \mathrm{~m}\)

Answer: 1. \(1.0 \times 10^{-1} \mathrm{~m}\)

For balance: \(\frac{R}{10 \Omega}=\frac{3}{2}\) R =15Ω.

⇒ Since R<*l, hence \(\frac{R^{\prime}}{R}=\frac{l^{\prime}}{l} \Rightarrow \frac{R^{\prime}}{15 \Omega}=\frac{l^{\prime}}{1.5 \mathrm{~m}}\)

⇒ \(l^{\prime}=\frac{(1.5 \mathrm{~m})}{(15 \Omega)} \times(1 \Omega)=\frac{1}{10} \mathrm{~m}=1.0 \times 10^{-1} \mathrm{~m}\).

Question 107. A charged particle having a drift velocity of \(7.5 \times 10^{-4} \mathrm{~m} \mathrm{~s}^{-1}\) in an electric field of3 x10-10 V m-1 has mobility (in m² V^-1 s^-1) of

1. \(2.5 \times 10^6\)
2. \(2.5 \times 10^{-6}\)
3. \(2.25 \times 10^{-15}\)
4. \(2.25 \times 10^{15}\)

Answer: 1. \(2.5 \times 10^6\)

Mobility is defined as drift velocity per unit electric field \(\left(=\frac{v}{E}\right)\)

Given that

⇒ \(v=7.5 \times 10^{-4} \mathrm{~m} \mathrm{~s}^{-1}, E=3 \times 10^{-10} \mathrm{Vm}^{-1}\).

∴ Hence, mobility = \(\frac{7.5 \times 10^{-4}}{3 \times 10^{-10}} \text { units }=2.5 \times 10^6 \text { units. }\)

Question 108. In the given circuit, the cell has negligible internal resistance. The potential difference across BD is

1. 4 V
2. 3 V
3. 2 V
4. 1 V

Answer: 3. 2 V

Current in the upper branch ABC is

⇒ \(I_1=\frac{40 \mathrm{~V}}{(40+60) \Omega}=\frac{40}{100} \mathrm{~A}=0.4 \mathrm{~A}\)

Similarly, for the lower branch,

⇒ \(I_2=\frac{40 \mathrm{~V}}{90 \Omega+110 \Omega}=\frac{40}{200} \mathrm{~A}=0.2 \mathrm{~A}\)

Now, V_A-V_B = l₁ (40 Ω) = 0.4 x 40 V =16 V;

V_A-V_D = l₂ (90Ω) = 0.2 x 90 V =18 V.

∴ (V_A-V_D)-(V_A-V_B) = V_B-V_D = 2V.

Question 109. For the given circuit, find the potential at B with respect to A.

1. +2V
2. -2 V.
3. +1 V
4. -1 V

Answer: 3. +1 V

From the junction rule at D, the current in branch DC is 1 A.

Now,

V_B-V_A=(V_B-V_D)+(V_D-V_c)-(V_c-V_A)

= (-2V)+(2 V) + (1V) = +1 V.

Question 110. The value of the current l₁ flowing from A to C in the circuit diagram is

1. 1 AB
2. 5 A
3. 4 A
4. 2 A

Answer: 1. 1 AB

Potential difference = V_A-V_c=8 V

∴ \(I_1=\frac{V}{R}=\frac{8 \mathrm{~V}}{8 \Omega}=1 \mathrm{~A}\).

Work Energy Power And Collisions

Question 1. A particle moves from a point \((-2 \hat{i}+5 \hat{j}) \text { to }(4 \hat{j}+3 \hat{k})\) when a force of \((4 \hat{i}+3 \hat{j})\)N is applied. How much work has been done by this force?

1. 8J
2. 11J
3. 5J
4. 2J

Answer: 3. 5J

The position vector of the initial position of the particle is,

⇒ \(\vec{r}_1=-2 \hat{i}+5 \hat{j}\)

and that of the final position is,

⇒ \(\overrightarrow{r_2}=4 \hat{j}+3 \hat{k}\).

∴ the displacement vector is \(\vec{s}=\vec{r}_2-\vec{r}_1=(2 \hat{i}-\hat{j}+3 \hat{k}) \mathrm{m}\).

Force \(\vec{F}=(4 \hat{i}+3 \hat{j}) \mathrm{N}\)

∴ work done is \(W=\vec{F} \cdot \vec{s}=(4 \hat{i}+3 \hat{j}) \mathrm{N} \cdot(2 \hat{i}-\hat{j}+3 \hat{k}) \mathrm{m}\)

= (8 – 3) J

= 5 J

Question 2. A body of mass 1 kg is thrown upwards with a velocity of 20ms^-1. It momentarily comes to rest after attaining a height of 18m. How much energy is lost due to air friction? (Take g = 10ms^-2).

1. 30J
2. 40J
3. 10J
4. 20J

Answer: 4. 20J

Work done by gravity = mgh = (1 kg)(-10 m s^-2)(18 m)

= -180 J.

Initial \(\mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2}(1 \mathrm{~kg})\left(20 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\)

= 200 J.

By the work-energy theorem,

⇒ \(W_{\text {grav }}+W_{\text {air }}=\mathrm{KE}_{\mathrm{f}}-\mathrm{KE}_{\mathrm{i}}\)

=> -180 J + W_air = 0J -200J.

∴ work done by air friction = 180 J- 200 J

= 20 J.

Hence, the energy lost due to air friction is 20 J.

Question 3. A child is swaying on a swing. Its minimum and maximum heights from the ground are 0.75 are 2.0m respectively. Its maximum speed will be

1. 10 m s^-1
2. 5 m s ^-1
3. 8 m s^-1
4. 15 m s^-1

Answer: 2. 5 m s ^-1

At the maximum height (h₂= 2.0m), the KE is zero, and at the minimum height (h₁ = 0.75 m), the KE is the maximum

⇒ \(\left(=\frac{1}{2} m v_{\max }^2\right)\)

Hence, by the principle of conservation of energy

⇒ \(\frac{1}{2} m v_{\max }^2=m g\left(h_2-h_1\right)\)

⇒ \(v_{\max }=\sqrt{2 g\left(h_2-h_1\right)}=\sqrt{2\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(1.25 \mathrm{~m})}\)

= 5ms^-1

Question 4. A particle having a kinetic energy K is projected making an angle of 60 with the horizontal. The kinetic energy at the highest point will be

1. K
2. \(\frac{K}{2}\)
3. \(\frac{K}{4}\)
4. \(\frac{K}{8}\)

Answer: 3. \(\frac{K}{4}\)

The horizontal component of the velocity of projection remains unchanged.

At the highest point, the velocity is along the horizontal and equals u cos 60°= \(\frac{u}{2}\)

Initial kinetic energy

⇒ \(K=\frac{1}{2} m u^2\)

At the highest point,

⇒ \(K^{\prime}=\frac{1}{2} m\left(\frac{u}{2}\right)^2=\frac{1}{2} m u^2 \times \frac{1}{4}=\frac{K}{4}\)

Question 5. A body moves through a distance of 10 m long in a straight line under the action of a 5-N force. If the work done is 25J, the angle between the force direction of the body is

1. 30°
2. 45°
3. 60°
4. 75°

Answer: 3. 60°

Work done = W = Fs cos θ.

∴ 25 J= (5N)(10 m)cos θ

⇒ \(\cos \theta=\frac{25 \mathrm{~J}}{50 \mathrm{~J}}=\frac{1}{2}\)

θ = 60°.

Question 6. A force on a 3-g particle in such a way that the position of the particle as a function of time is given by \(x = 3t- At² + t³\), where x is in meters and t is in seconds. The work done during the first four seconds is

1. 490 3J
2. 450 mJ
3. 528 mJ
4. 530 mJ

Answer: 3. 528 mJ

Given that position x = 3t- At² + t³.

So, velocity \(0=\frac{d x}{d t}=3-8 t+3 t^2\)

Initial velocity \(u=\left.\frac{d x}{d t}\right|_{t=0}=3 \mathrm{~m} \mathrm{~s}^{-1}\)

Final velocity

⇒ \(v=\left.\frac{d x}{d t}\right|_{t=4 \mathrm{~s}}=(3-32+48) \mathrm{m} \mathrm{s}^{-1}=19 \mathrm{~m} \mathrm{~s}^{-1}\)

∴ the change in kinetic energy is

⇒ \(\Delta \mathrm{KE}=\frac{1}{2} m\left(v^2-u^2\right)=\frac{1}{2}(3 \mathrm{~g})\left[\left(19 \mathrm{~m} \mathrm{~s}^{-1}\right)^2-\left(3 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\right]\)

= 528mJ

Question 7. Two bodies with their kinetic energies in the ratio 4:1 are moving with equal linear momenta. The ratio of their masses is

1. 4: 1
2. 1: 2
3. 2: 1
4. 1: 4

Answer: 4. 1: 4

Let m₁ and m₂ be the masses of the bodies and p be their equal momentum.

Now, \(\mathrm{KE}=\frac{1}{2} m v^2=\frac{m^2 v^2}{2 m}=\frac{p^2}{2 m}\)

Given that,

⇒ \(\frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{4}{1}=\frac{\frac{p^2}{2 m_1}}{\frac{p^2}{2 m_2}}=\frac{m_2}{m_1}\)

⇒\(\frac{m_1}{m_2}=\frac{1}{4} \Rightarrow m_1: m_2\)

= 1: 4

Question 8. Two bodies of masses m and 4m are moving with equal kinetic energies. The ratio of their linear momenta is

1. 1: 2
2. 1: 4
3. 4: 1
4. 1: 1

Answer: 1. 1: 2

Given that m₁ = m, m₂ = 4m and K₁ = K₂

⇒ \(\frac{p_1^2}{2 m_1}=\frac{p_2^2}{2 m_2} \Rightarrow \frac{p_1}{p_2}=\sqrt{\frac{m_1}{m_2}}=\sqrt{\frac{m}{4 m}}=\frac{1}{2}\)

= 1: 2

Question 9. A body constrained to move along the y-axis is subjected to a force given by \(\vec{F}=(-2 \hat{i}+15 \hat{j}+6 \hat{k})\)N. The work done by this force in moving the body through a distance of \(10 \hat{j}\) m along the y-axis is

1. 160J
2. 20J
3. 150J
4. 190J

Answer: 3. 150J

Work done \(W=\vec{F} \cdot \vec{s}=(-2 \hat{i}+15 \hat{j}+6 \hat{k}) \mathrm{N} \cdot(10 \hat{j}) \mathrm{m}\)

= 150J

Question 10. A position-dependent force F=(7- 2x + 3X²)N acts on a small body of mass 2kg and displaces it from x= 0 to x= 5m. The work done is

1. 35 J
2. 70 J
3. 135 J
4. 270 J

Answer: 3. 135 J

The work done by the given variable force is

⇒ \(W=\int d W=\int F d x=\int_0^{5 \mathrm{~m}}\left[\left(7-2 x+3 x^2\right) \mathrm{N}\right] d x\)

⇒ \(\left[\left(7 x-x^2+x^3\right) \mathrm{N}\right]_0^{5 \mathrm{~m}}=(35-25+125) \mathrm{J}\)

= 135J

Question 11. A bullet of mass 10 g leaves a rifle with an initial velocity of 1000m s^-1 and strikes the ground at the same horizontal level with a velocity of 500m s^-1. The work done in overcoming the air resistance will be

1. 375 J
2. 500 J
3. -3750 J
4. 5000 J

Answer: 3. -3750 J

By the work-energy theorem, the work done by the bullet in overcoming the air resistance is

W = change in \(\mathrm{KE}=\frac{1}{2} m v^2-\frac{1}{2} m u^2\)

⇒ \(\frac{1}{2}(10 \mathrm{~g})\left[\left(500 \mathrm{~m} \mathrm{~s}^{-1}\right)^2-\left(10^3 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\right]\)

Question 12. The kinetic energy acquired by a mass m in traveling through a distance d, starting from rest, under the action of a constant force is directly proportional to

1. m
2. m°
3. √m
4. \(\frac{1}{\sqrt{m}}\)

Answer: 2. m°

v² = u² + 2as

= 2as = 2ad [here, u = θ and s = d].

∴ \(\mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2} m(2 a s)\)

= mas

= F.s

Hence, the KE acquired is independent of m and is thus proportional to m°

Question 13. Water falls from a height of 60 m at the rate of 15 kg s^-1 to operate a turbine. The losses due to frictional forces are 10% of the energy. How much power is generated by the turbine? (Takeg = 10m s^-2.)

1. 12.3 kW
2. 7.0 kW
3. 8.1 kW
4. 10.2 kW

Answer: 3. 8.1 kW

The mass of water falling per second is 15 kg, the height is 60 m, and g = 10 ms^-2.

Now, the loss of energy is 10%.

∴ available energy = 90%.

∴ power generated = \(\left(15 \mathrm{~kg} \mathrm{~s}^{-1}\right)\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(60 \mathrm{~m})\left(\frac{90}{100}\right)\)

= 8100W

= 8.1 W.

Question 14. An engine pumps water continuously through a hose. Water leaves the hose with a velocity of v, and m is the mass per unit length of the water jet. What is the rate at which the kinetic energy is imparted to the water?

1. mv³
2. \(\frac{1}{2} m^2 v^2\)
3. \(\frac{1}{2} m v^2\)
4. \(\frac{1}{2} m v^3\)

Answer: 4. \(\frac{1}{2} m v^3\)

Given that velocity of water = v and mass flowing per unit length = m. In one second, the length of the water jet is v, so the mass contained is mv.

∴ the KE imparted to water in one second is

⇒ \(\frac{1}{2}(m v) v^2=\frac{1}{2} m v^3\)

Question 15. A block of mass M is attached to the lower end of a vertical spring, which is hung from the ceiling and has a force constant of k. The block is released from rest with the initial stretch. The maximum extension produced in the length of the spring will be

1. \(\frac{M g}{2 k}\)
2. \(\frac{2 M g}{k}\)
3. \(\frac{4 M g}{k}\)
4. \(\frac{M g}{k}\)

Answer: 2. \(\frac{2 M g}{k}\)

Let x be the extension in the spring.

So, PE lost by the block = PE gained by the spring

⇒ \(M g x=\frac{1}{2} k x^2 \Rightarrow x=\frac{2 M g}{k}\)

Question 16. A block2kof mass m, starting from rest, undergoes a uniform acceleration. If the speed acquired in a time t be v, the power delivered to the block is

1. \(\frac{m v^2}{t}\)
2. \(\frac{m v^2}{2 t^2}\)
3. \(\frac{m v^2}{t^2}\)
4. \(\frac{m v^2}{2 t}\)

Answer: 4. \(\frac{m v^2}{2 t}\)

Power delivered = \(\frac{\text { work done }}{\text { time }}=\frac{\text { change in the } \mathrm{KE}}{t}\)

⇒ \(\frac{\frac{1}{2} m v^2}{t}=\frac{m v^2}{2 t}\)

Question 17. A block of mass 10 kg, moving along the x-axis with a constant speed of 10 m s^-1, is subjected to a retarding force F = -0.1x J m^-1 during its travel from x = 20 m to x = 30 m. Its final kinetic energy will be

1. 275 J
2. 450 J
3. 475J
4. 250 J

Answer: 3. 475J

By the work-energy theorem, work done by the retarding force = change in the KE.

∴ \(W=\int F d x=-(0.1) \int x d x=\frac{1}{2} m u^2-\frac{1}{2} m v^2\)

∴ final \(\mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2}(10 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right)^2-0.1\left[\frac{x^2}{2}\right]_{20 \mathrm{~m}}^{30 \mathrm{~m}}\)

⇒ \(=500 \mathrm{~J}-\frac{0.1}{2}[900-400] \mathrm{J}\)

= 50Q J- 25 J

= 475 J.

Question 18. The force F acting on an object varies with the distance x as shown in the adjoining figure. The forceFisinnewtons and x are in metres. The work done by the force in moving the object from x = θ to x = 6m is

1. 9.0 J
2. 4.5 J
3. 13.5 J
4. 18.0 J

Answer: 3. 13.5 J

The work done by a variable force is given by

W = \(\int d W=\int F d x\) = area under the F-x graph.

From the given figure, the total work done is

W= area of the rectangle + area of the triangle

⇒ \((3 m)(3 N)+\frac{1}{2}(3 m)(3 N)\)

= 13.5J.

Question 19. The force F on a particle moving along a straight line varies with the distance d as shown in the following figure. The work done on the particle during its displacement of 12m is

1. 13 J
2. 18 J
3. 21 J
4. 30 J

Answer: 1. 13 J

From the given figure, the force from d = 0 to d = 3m is zero.

So, the work done is

⇒ \(W=(7-3) m \times(2 N)+\frac{1}{2}(12 m-7 m)(2 N)\)

= 8J + 5J

= 13J.

Question 20. A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude P₀. The instantaneous velocity of this car is proportional to

1. \(t^2 P_0\)
2. t^1/2
3. t ^-1/2
4. \(\frac{t}{\sqrt{m}}\)

Answer: 2. t^1/2

Instantaneous power is given by

⇒ \(P_0=F v=\left(m \frac{d v}{d t}\right) v\)

∴ P₀dt = mv dv.

Integrating, \(P_0 t=\frac{m v^2}{2}\)

∴ velocity = \(v=\sqrt{\frac{2 P_0 t}{m}}\)

∴ \(v \propto t^{1 / 2}\).

Question 21. A particle of mass m is driven by a machine that delivers a constant power P. If the particle starts from rest, the force on the particle at a time t is

1. \(\sqrt{\frac{2 m P}{t}}\)
2. \(\frac{1}{2} \sqrt{\frac{m P}{t}}\)
3. \(\sqrt{\frac{m P}{2 t}}\)
4. \(\sqrt{\frac{m P}{t}}\)

Answer: 3. \(\sqrt{\frac{m P}{2 t}}\)

Constant power = P = \(\frac{d W}{d t}\)

So, W = \(\int d W=\int P d t=P t\)

Now, work done = change in the KE = \(\frac{1}{2} m v^2-0=\frac{1}{2} m v^2\)

∴ \(P t=\frac{1}{2} m v^2 \text { or } v=\sqrt{\frac{2 P t}{m}}\)

Now, acceleration = a = \(\frac{d v}{d t}=\sqrt{\frac{2 P}{m}}\left(\frac{1}{2 \sqrt{t}}\right)\)

The force on the particle is

⇒ \(F=m a=m \cdot \sqrt{\frac{2 P}{m}}\left(\frac{1}{2 \sqrt{t}}\right)=\sqrt{\frac{m P}{2 t}}\)

Question 22. A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 x 10^-4J at the end of the second revolution after the beginning of the motion?

1. 0.2 ms^-2
2. 0.15 m s^-2
3. 0.1 m s^-2
4. 0.18 m s^-2

Answer: 3. 0.1 m s^-2

Given: mass = m = 10 g = 10 x 10^-3 kg,

radius = R = 6.4 cm = 6.4 x 10^-2 m,

final KE = 8x 10^-4J

and initial KE = 0.

By the work-energy theorem,

work done = change in the KE

=> maT. 2(2πR) = KE_f.

Hence, tangential acceleration = aT = \(\frac{\mathrm{KE}_{\mathrm{f}}}{4 \pi R m}\)

⇒ \(\frac{8 \times 10^{-4} \mathrm{~J}}{4 \times 3.14 \times 6.4 \times 10^{-2} \times 10^{-2} \mathrm{~kg} \mathrm{~m}}\)

= 0.099m s^-2

= 0.1m s^-2

Question 23. Consider a drop of rainwater of mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m s^-1 Take g equal to 10 m s^-2 The work done by the gravitational force and that by the resistive force of air are respectively

1. 1.25 J and -8.25 J
2. 100 J and 8.75 J
3. 10 J and -8.75 J
4. -10 J and -8.25 J

Answer: 3. 10 J and -8.75 J

The work done by the gravitational force is

⇒ \(W_{\text {grav }}=m g h=\left(1 \times 10^{-3} \mathrm{~kg}\right)\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(1000 \mathrm{~m})\)

= 10 J.

If W_fric = work done by the resistive force then

⇒ \(W_{\text {grav }}+W_{\text {fric }}=\text { change in the } \mathrm{KE}=\frac{1}{2} m v^2-0\)

⇒ \(10 \mathrm{~J}+W_{\text {fric }}=\frac{1}{2}\left(10^{-3} \mathrm{~kg}\right)\left(50 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\)

= 1.25 J.

Hence, \(W_{\text {fric }}=(1.25-10) \mathrm{J}\)

= -8.75 J.

Question 24. A body of mass 1 kg begins to move under the action of a time-dependent force \(\vec{F}=\left(2 t \hat{i}+3 t^2 \hat{j}\right) \mathrm{N}, \text { where } \hat{i} \text { and } \hat{j}\) are the unit vectors along the x- and y-axes. The power developed by the force at a time t will be

1. (2t³ + 3t⁴)W
2. (2t³ + 3t⁵)W
3. (2t² + 3t³)W
4. (2t² + 4t⁴)W

Answer: 2. (2t³ + 3t⁵)W

Given: force = \(F=\left(2 t \hat{i}+3 t^2 \hat{j}\right)\) N and mass = m =1 kg.

∴ acceleration = \(\vec{a}=\frac{\vec{F}}{m}=\frac{\left(2 t \hat{i}+3 t^2 \hat{j}\right) \mathrm{N}}{1 \mathrm{~kg}}\)

Hence, the velocity at the time t is

⇒ \(\vec{v}=\int_0^t \vec{a} d t=\int_0^t\left(2 t \hat{i}+3 t^2 \hat{j}\right) d t=\left(t^2 \hat{i}+t^3 \hat{j}\right) \mathrm{m} \mathrm{s}^{-1}\)

∴ the instantaneous power delivered at the time t will be

⇒ \(P=\vec{F} \cdot \vec{v}=\left(2 t \hat{i}+3 t^2 \hat{j}\right) \mathrm{N} \cdot\left(t^2 \hat{i}+t^3 \hat{j}\right) \mathrm{m} \mathrm{s}^{-1}=\left(2 t^3+3 t^5\right) \mathrm{W}\)

Question 25. What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

1. √gR
2. √2gR
3. √5gR
4. √3gR

Answer: 3. √5gR

⇒ At the highest point, the centripetal force (mv2/R) is provided by the weight (mg).

So,

⇒ \(\frac{m v^2}{R}=m g \text { and } \mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2} m g R\)

By the work-energy principle,

⇒ \(W_{\text {grav }}=\mathrm{KE}_{\mathrm{f}}-\mathrm{KE}_{\mathrm{i}}\)

or \(m g \cdot 2 R=\frac{1}{2} m v_{\mathrm{f}}^2-\frac{1}{2} m g R\)

∴ the required speed is \(v_{\mathrm{f}}=\sqrt{5 g R}\)

Question 26. The heart of a man pumps 5 liters of blood through the arteries per minute at a pressure of 150 mmHg. If the density of mercury be 13.6 x 10³ kg m^-3 and g = 10 m s^-2 then the power is

1. 1.50 W
2. 2.35 W
3. 3.0 W
4. 1.70 W

Answer: 3. 3.0 W

The flow rate of blood is \(5 \mathrm{~L} \mathrm{~m}^{-1}=\frac{5 \times 10^{-3} \mathrm{~m}^3}{60 \mathrm{~s}}\)

Blood pressure =p = 150mmHg = hpg

= (150 x 10^-3m)(13.6 x 10³kgm^-3)(10m s^-2)

= 20.4 x 10³Pa.

∴ the power of the heart is

⇒ \(P=\frac{W}{t}=\frac{p V}{t}=\frac{\left(20.4 \times 10^3 \mathrm{~Pa}\right)\left(5 \times 10^{-3} \mathrm{~m}^3\right)}{60 \mathrm{~s}}\)

= 1.70 W.

Question 27. A bullet is fired from a rifle which recoils after firing. The ratio of the kinetic energy of the rifle to that of the bullet is

1. One
2. Zero
3. Less Than One
4. More Than One

Answer: 3. Less Than One

Let M = mass of the gun, m = mass of die bullet, V = recoil speed of the gun, and v = speed of the bullet.

By the principle of conservation of linear momentum, MV = mv = p.

⇒ \(\mathrm{KE}_{\text {gun }}=\frac{p^2}{2 M} \text { and } \mathrm{KE}_{\text {bullet }}=\frac{p^2}{2 m}\)

Now,

∴ \(\frac{\mathrm{KE}_{\text {gun }}}{\mathrm{KE}_{\text {bullet }}}=\frac{m}{M}<1\)

Question 28. A body of mass 0.5 kg, moving at a speed of 1.5 m s^-1 on a smooth horizontal surface, collides with a light spring of force constant k = 5ON m^-1. The maximum compression of the spring will be

1. 1.5 m
2. 0.15 m
3. 0.5 m
4. 0.12 m

Answer: 2. 0.15 m

By the principle of conservation of energy,

KE of the body = PE in the spring

⇒ \(\frac{1}{2} m v^2=\frac{1}{2} k x^2\)

Here, the maximum compression of the spring is

⇒ \(x=\sqrt{\frac{m v^2}{k}}=\sqrt{\frac{(0.5)(1.5)^2}{50}} \mathrm{~m}\)

= 0.15m.

Question 29. A block of mass m moves up an inclined plane of inclination 0. If p is the coefficient of friction between the block and the inclined plane, the work done in moving the block through a distance s up along the plane is

1. mgs (sin θ + μ cos θ)
2. mgs sin θ
3. mgs(sm θ- μ cos θ)
4. μmgs cos θ

Answer: 1. mgs(sin 0 + μ cos 0)

Since the block is pulled up the incline, the applied force will overcome the components of the weight (mg sin 9) and the kinetic friction (mmg cos 0).

Hence, the work done by the pulling force is

W = Fs = (mg sin θ + μmg cos θ)s

= mgs(sin θ + μ cos θ).

Question 30. Amovingbullethits a solid target resting on a frictionless horizontal surface and gets embedded in the target. What is conserved in this process?

1. The momentum and the kinetic energy
2. The momentum alone
3. The kinetic energy alone
4. Neither the momentum nor the kinetic energy

Answer: 2. The momentum alone

In all types of collisions, it is only the momentum that is conserved.

Question 31. The potential energy of a particle in a force field is \(U=\frac{A}{r^2}-\frac{B}{r}\), where A and B are positive constants and r is the distance of the particle from the center of the field. For stable equilibrium, the distance of the particle is

1. \(\frac{B}{2 A}\)
2. \(\frac{B}{A}\)
3. \(\frac{2 A}{B}\)
4. \(\frac{A}{B}\)

Answer: 3. \(\frac{2 A}{B}\)

Given that PE = \(U(r)=\frac{A}{r^2}-\frac{B}{r}\)

For equilibrium,

⇒ \(\frac{d U}{d r}=0=-\frac{2 A}{r^3}+\frac{B}{r^2} \Rightarrow r=\frac{2 A}{B}\)

For stable equilibrium,

⇒ \(\frac{d^2 U}{d r^2}>0 \Rightarrow \frac{d^2 U}{d r^2}=\frac{6 A}{r^4}-\frac{2 B}{r^3}\)

Substituting r = \(\frac{2A}{B}\), we have

⇒ \(\frac{d^2 U}{d r^2}=6 A\left(\frac{B}{2 A}\right)^4-2 B\left(\frac{B}{2 A}\right)^3=\frac{B^4}{8 A^3}>0\)

Thus, for stable equilibrium,

⇒ \(r=\frac{2 A}{B}\)

Question 32. A body projected vertically from the earth reaches a height equal to the earth’s radius before returning to the earth’s surface. When is the power delivered by the gravitational force of the earth the greatest?

1. At the highest position of the body
2. At the instant before the body hits the ground
3. Constant all through
4. At the instant after the body is projected

Answer: 2. At the instant before the body hits the ground

The instantaneous power delivered is P= Fv.

Just before hitting the ground, the velocity v is maximum, so the power delivered is also maximum.

Question 33. A body of mass 5 kg rests on a rough horizontal surface having a coefficient of friction of 0.2. The body is pulled through a distance of 10 m by a horizontal force of 25 N. The kinetic energy acquired by it is (assuming g = 10 m s^-2)

1. 200 J
2. 150 J
3. 100 J
4. 50 J

Answer: 2. 150 J

The work done by the applied force is

⇒ \(W_1=\vec{F} \cdot \vec{s}=F s \cos \theta\)

= (25 N)(10 m) cos θ°

= 250 J,

and the work done by friction is

⇒ \(W_2=-(\mu m g) s\)

= -(0.2)(5)(10)(10) J

= -100 j.

Net work done = change in the KE.

∴ KE acquired = W₁ + W₂

= 250 J- 100 J

= 150 J.

Question 34. A body is moving up an inclined plane of angle 0 with an initial kinetic energy E. The coefficient of friction between the plane and the body is p. The work done against the friction before the body comes to rest is

1. \(\frac{\mu \cos \theta}{E \sin \theta+\cos \theta}\)
2. \(\mu E \cos \theta\)
3. \(\frac{\mu E \cos \theta}{\mu \cos \theta+\sin \theta}\)
4. \(\frac{\mu E \cos \theta}{\mu \cos \theta-\sin \theta}\)

Answer: 3. \(\frac{\mu E \cos \theta}{\mu \cos \theta+\sin \theta}\)

By the work-energy theorem,

gain in KE = E = total work done

⇒ \(W_{\text {grav }}+W_{\text {fric }}=m g s \sin \theta+W_{\text {fric }}\)

∴ \(W_{\text {fric }}=E-m g s \sin \theta\)……(1)

Now, the upward acceleration along the rough plane is

a = g(sin0 + mcos 0).

Hence, u² = 2as= 3g (sinθ+ m cos θ)s

or, \(s=\frac{u^2}{2 g(\sin \theta+\mu \cos \theta)}\)

Substituting s in(1),

⇒ \(W_{\text {fric }}=E-\frac{m g \sin \theta \cdot u^2}{2 g(\sin \theta+\mu \cos \theta)}\)

⇒ \(E-\frac{E \sin \theta}{\sin \theta+\mu \cos \theta}=\frac{\mu E \cos \theta}{\mu \cos \theta+\sin \theta}\)

Question 35. A bullet is fired normally on an immovable wooden plank. It loses 25% of its momentum penetrating a thickness of 3.5 cm. The total thickness penetrated by the bullet is

1. 7 cm
2. 8 cm
3. 10 cm
4. 12 cm

Answer: 2. 8 cm

KE = \(\frac{p^2}{2 m}\)

Loss in momentum = \((25 \% \text { of } p)=\frac{p}{4}\)

∴ remaining momentum = \(\frac{3 p}{4}\)

∴ final \(\mathrm{KE}=\frac{\left(\frac{3 p}{4}\right)^2}{2 m}=\frac{9 p^2}{32 m}\)

According to the work-energy theorem,

⇒ \(F(3.5 \mathrm{~cm})=\Delta \mathrm{KE}=\frac{p^2}{2 m}-\frac{9 p^2}{32 m}=\frac{7 p^2}{32 m}\)

At the moment when the bullet finally comes to rest,

⇒ \(F(s)=\Delta K E=\frac{p^2}{2 m}-0=\frac{p^2}{2 m}\)

⇒ \(\frac{F(s)}{F(3.5 \mathrm{~cm})}=\frac{\frac{p^2}{2 m}}{\frac{7 p^2}{32 m}}=\frac{16}{7}\)

⇒ \(s=(3.5 \mathrm{~cm}) \frac{16}{7}\)

= 8 cm

Question 36. A uniform rod of mass m and length is held inclined at an angle of 60° with the vertical. What will be its potential energy in this position?

1. \(\frac{m g l}{4}\)
2. \(\frac{m g l}{3}\)
3. \(\frac{m g l}{2}\)
4. mgl

Answer: 1. \(\frac{m g l}{4}\)

PE of the uniform rod

= weight x height of the CM from the ground

⇒ \(m g\left(\frac{l}{2} \cos 60^{\circ}\right)\)

⇒ \(\frac{1}{4} m g l\)

Question 37. An engine pulls a car of mass 1500 kg on a level road at a constant speed of 18 km h^-1. If the frictional force is 1500N, what power does the engine generate?

1. 5.0kW
2. 7.5 kW
3. 10 kW
4. 12.5 kW

Answer: 2. 7.5 kW

Since the car moves at a constant velocity (c = 5ms ^-2), acceleration = 0.

Hence, F_net = 0.

The force F exerted by the engine must be equal and opposite to the frictional force (F = 1500 N).

∴ the power delivered by the engine is

P = Fv = (1500N)(5m s^-1)

= 7500 W

= 7.5kW

Question 38. In the preceding question, what extra power must the engine develop to maintain the same speed up along an inclined plane having a gradient of 1 in 10? (Take g = 10 m s^-2.)

1. 2.5 kW
2. 5.0 kW
3. 7.5 kW
4. 10 kW

Answer: 3. 7.5 kW

Extra power required = (mg sin θ)v

⇒ \((1500 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)\left(\frac{1}{10}\right)\left(5 \mathrm{~m} \mathrm{~s}^{-1}\right)\)

= 7500 W

= 7.5kW.

Question 39. Each of two identical cylindrical vessels, with their bases at the same level, contains a liquid of density p. The height of the liquid in one vessel is hx and that in the other is h₂. The area of either base is A. What is the work done by gravity in equalizing the levels when the vessels are interconnected?

1. \(A \rho g\left(h_1-h_2\right)^2\)
2. \(A \rho g\left(h_1+h_2\right)^2\)
3. \(A \rho g\left(\frac{h_1-h_2}{2}\right)^2\)
4. \(A \rho g\left(\frac{h_1+h_2}{2}\right)^2\)

Answer: 3. \(A \rho g\left(\frac{h_1-h_2}{2}\right)^2\)

The work done by gravity is equal to the change in the PE of the system.

The total initial PE of the system is

⇒ \(U_i=m_1 g\left(\frac{h_1}{2}\right)+m_2 g\left(\frac{h_2}{2}\right)\)

⇒ \(A h_1 \rho g\left(\frac{h_1}{2}\right)+A h_2 \rho g\left(\frac{h_2}{2}\right)=\frac{A \rho g}{2}\left(h_1^2+h_2^2\right)\)

When the vessels are interconnected, the height of the liquid in each

⇒ \(\frac{h_1+h_2}{2}\).

Hence, the final PE of the system is

⇒ \(U_{\mathrm{f}}=2\left[A\left(\frac{h_1+h_2}{2}\right) \rho g\left\{\frac{\left(h_1+h_2\right) / 2}{2}\right\}\right]=\frac{A \rho g}{4}\left(h_1+h_2\right)^2\)

∴ the change in the PE is

⇒ \(\Delta U=U_{\mathrm{f}}-U_{\mathrm{i}}=\frac{A \rho g}{4}\left(h_1+h_2\right)^2-\frac{A \rho g}{2}\left(h_1^2+h_2{ }^2\right)\)

⇒ \(\frac{A \rho g}{4}\left[\left(h_1+h_2\right)^2-2\left(h_1^2+h_2^2\right)\right]=-\frac{A \rho g}{4}\left(h_1-h_2\right)^2\)

∴ the work done by gravity is

⇒ \(W=-\Delta U=A \rho g\left(\frac{h_1-h_2}{2}\right)^2\)

Question 40. An electric pump on the ground floor of a building takes 10 minutes to fill a tank of volume 30 m3 with water. If the tank is 60 m above the ground and the efficiency of the engine is 30%, how much electric power is consumed by the pump in filling the tank? (Takeg = 10 m s^-2.)

1. 100 kW
2. 150 kW
3. 200 kW
4. 250 kW

Answer: 1. 100 kW

Output power = \(\frac{\text { work }}{\text { time }}=\frac{m g h}{t}=\frac{V \rho g h}{t}\)

⇒ \(\frac{\left(30 \mathrm{~m}^3\right)\left(10^3 \mathrm{~kg} \mathrm{~m}^{-3}\right)\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(60 \mathrm{~m})}{10 \times 60 \mathrm{~s}}\)

= 30 x 10³ W.

⇒ \(\text { Efficiency }=\frac{\text { output power }}{\text { input power }}\)

⇒ \(30 \%=\frac{30}{100}=\frac{30 \times 10^3 \mathrm{~W}}{\text { input power }}\)

⇒ input power = power consumed by the engine

⇒ \(\left(30 \times 10^3 \mathrm{~W}\right) \times \frac{100}{30}\)

= 100kW.

Question 41. A ball of mass m is thrown vertically upwards with a velocity of v. The height at which the kinetic energy of the ball will reduce to half its initial value is given by

1. \(\frac{v^2}{g}\)
2. \(\frac{v^2}{2 g}\)
3. \(\frac{v^2}{3 g}\)
4. \(\frac{v^2}{4 g}\)

Answer: 4. \(\frac{v^2}{4 g}\)

Initial KE = E = \(\frac{1}{2} m v^2\)

At the height h,KE = \(\frac{E}{2}=\frac{1}{4} m v^2\)

By the work-energy theorem,

⇒ \(W_{\text {grav }}=\Delta \mathrm{KEE}=\frac{E}{2}-E=-\frac{E}{2}\)

⇒ \(-m g h=-\frac{1}{4} m v^2 \Rightarrow h=\frac{v^2}{4 g}\)

Question 42. In a hydroelectric power station, the height of the dam is 10 m. How many kilograms of water must fall per second on the blades of a turbine so as to generate 1 MW of electrical power? (Take g = 10 m s^-2.)

1. 10³
2. 10⁴
3. 10⁵
4. 10⁶

Answer: 2. 10⁴

Power generated =1 MW = 10⁶ W = \(\frac{m g h}{t}\)

Given that h = 10m,g = 10m s^-2 and t = 1 s.

∴ \(m=\frac{\left(10^6 \mathrm{~W}\right)(1 \mathrm{~s})}{\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(10 \mathrm{~m})}=10^4 \mathrm{~kg} \mathrm{~s}^{-1}\)

Question 43. A metal ball of mass 2 kg, moving at a speed of 36 km h, undergoes a perfectly inelastic head-on collision with a stationary ball of mass 3 kg. The loss in its kinetic energy during the collision is

1. 40 J
2. 100 J
3. 140 J
4. 60 J

Answer: 4. 60 J

By the conservation of linear momentum,

⇒ \(m_1 u=\left(m_1+m_2\right) v \Rightarrow v=\frac{m_1 u}{m_1+m_2}\)

Initial KE = \(\frac{1}{2}(2 \mathrm{~kg})\left(36 \times \frac{5}{18} \mathrm{~m} \mathrm{~s}^{-1}\right)^2=100 \mathrm{~J}\)

and final KE = \(\frac{1}{2}\left(m_1+m_2\right) \frac{m_1^2 u^2}{\left(m_1+m_2\right)^2}=\frac{1}{2}\left(\frac{m_1^2 u^2}{m_1+m_2}\right)\)

⇒ \(\frac{\frac{1}{2}(2 \mathrm{~kg})^2\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right)^2}{5 \mathrm{~kg}}=40 \mathrm{~J}\)

∴ loss in KE = 100j – 40J

= 60J.

Question 44. A ball is thrown vertically downwards from a height of 20 m with an initial velocity v₀. It collides with the ground, loses 50% of its energy during the collision, and rebounds to the same height. The initial velocity v₀ is equal to

1. 10 ms^-1
2. 14 ms^-1
3. 28 ms^-1
4. 20 ms^-1

Answer: 4. 20 ms^-1

Let the velocity of the ball just before it strikes the ground be v.

⇒ \(v^2=v_0^2+2 g h\)

⇒ \(\frac{1}{2} m v^2=\frac{1}{2} m v_0^2+m g h\)

⇒ \(E=E_0+m g h\)

The remaining energy E’ after the 50% loss will be

⇒ \(E^{\prime}=\frac{E}{2}=\frac{1}{2}\left(E_0+m g h\right)\)

∴ \(\frac{1}{2}\left(\frac{1}{2} m v_0^2+m g h\right)=m g h\) [ … at h, velocity = 0]

⇒ \(v_0=\sqrt{2 g h}=\sqrt{2\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(20 \mathrm{~m})}\)

= 20m s^-1.

Question 45. A bullet of mass 10 g, moving horizontally at a velocity of 400m s^-1, strikes a wooden block of mass 2 kg, which is suspended by a light, inextensible string of length 5 m. As a result, the center of gravity of the block is found to rise vertically through 10 cm. The speed of the bullet after it emerges horizontally from the block will be

1. 80 ms^-1
2. 100 ms^-1
3. 120 ms^-1
4. 160 ms^-1

Answer: 3. 120 ms^-1

Given that m = 10 g, u = 400m s^-1, and M = 2 kg.

If V₁ = velocity of the block just after the collision,

⇒ \(\frac{1}{2} M v_1^2=M g h\)

⇒ \(\dot{v}_1=\sqrt{2 g h}=\sqrt{2(10)\left(10 \times 10^{-2}\right)} \mathrm{ms}^{-1}\)

⇒ \(\sqrt{2} \mathrm{~ms}^{-1}\)

By the conservation of linear momentum,

mu =Mv₁ + mv

⇒ \(\left(10 \times 10^{-3} \mathrm{~kg}\right)\left(400 \mathrm{~m} \mathrm{~s}^{-1}\right)=(2 \mathrm{~kg})\left(\sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\right)+\left(10 \times 10^{-3} \mathrm{~kg}\right) v\)

⇒ \(v=\frac{4-2 \sqrt{2}}{0.01} \mathrm{~m} \mathrm{~s}^{-1}=117.15 \mathrm{~m} \mathrm{~s}^{-1} \approx 120 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 46. A block of mass m, moving at a speed v on a frictionless horizontal surface, collides elastically with another block of the same mass m, initially at rest. After the collision, the first block moves at an angle 0 to the initial direction and has a speed of  The speed of the second block after the collision is 3

1. \(\frac{\sqrt{3}}{2} v\)
2. \(\frac{3}{4} v\)
3. \(\frac{3}{\sqrt{2}} v\)
4. \(\frac{2 \sqrt{2}}{3} v\)

Answer: 4. \(\frac{2 \sqrt{2}}{3} v\)

Let vx be the speed of the second block. In an elastic collision, the kinetic energy is conserved.

So,

⇒ \(\frac{1}{2} m v^2+0=\frac{1}{2} m\left(\frac{v}{3}\right)^2+\frac{1}{2} m v_1^2\)

⇒ \(v_1^2=v^2-\frac{v^2}{9}=\frac{8}{9} v\)

⇒ \(v_1=\frac{2 \sqrt{2}}{3} v\)

Question 47. A body of mass 4m is lying in the xy-plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass m, fly off perpendicular to each other with equal speeds of v. The total kinetic energy generated due to the explosion is

1. mv²
2. \(\frac{3}{2} m v^2\)
3. 2mv²
4. 4 muv²

Answer: 2. \(\frac{3}{2} m v^2\)

The resultant of the momentum vectors of A and B is √2mv.

Before the explosion, the momentum is zero.

So, for the final momentum to be zero, we have from the adjoining figure,

⇒ \(2 m v^{\prime}=\sqrt{2} m v \Rightarrow v^{\prime}=\frac{v}{\sqrt{2}}\)

∴ the KE generated due to the explosion is

⇒ \(\frac{1}{2} m v^2+\frac{1}{2} m v^2+\frac{1}{2}(2 m)\left(\frac{v}{\sqrt{2}}\right)^2\)

⇒ \(\frac{3}{2} m v^2\)

Question 48. A mass m, moving horizontally (along the *-axis) at a velocity v, collides with sand sticks to a mass of 3m, moving vertically upwards (along die y-axis) at a velocity 2v. The final velocity of the combination is

1. \(\frac{1}{3} v \hat{i}+\frac{2}{3} v \hat{j}\)
2. \(\frac{2}{3} v \hat{i}+\frac{1}{3} v \hat{j}\)
3. \(\frac{3}{2} v \hat{i}+\frac{1}{4} v \hat{j}\)
4. \(\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}\)

Answer: 4. \(\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}\)

The initial momentum of the mass m is \(\vec{p}_1=m v \hat{i}\), and that of the mass

⇒ \(3 m \text { is } \vec{p}_2=3 m(2 v) \hat{j}\)

∴ the total initial momentum is

⇒ \(\vec{p}_1=\vec{p}_1+\vec{p}_2=m v \hat{i}+6 m v \hat{j}\)

Finally, when the masses stick and move together, let the final momentum be \(\vec{p}_{\mathrm{f}}=4 m \vec{v}_0\)

Conserving the momentum, we have

⇒ \(4 m \vec{v}_0=m v \hat{i}+6 m v \hat{j} \Rightarrow \vec{v}_0=\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}\)

Question 49. A body of mass 2 kg has an initial velocity \(\vec{v}_i=(\hat{i}+\hat{j}) \mathrm{m} \mathrm{s}^{-1}\). After a collision with another body, its velocity becomes \(\vec{v}_{\mathrm{f}}=(5 \hat{i}+6 \hat{j}+\hat{k}) \mathrm{m} \mathrm{s}^{-1}\). If the time of impact is 0.02 s, the average force of impact on the body is

1. \(100(4 \hat{i}+5 \hat{j}-\hat{k}) \mathrm{N}\)
2. \(100(4 \hat{i}+5 \hat{j}+\hat{k}) \mathrm{N}\)
3. \(50(4 \hat{i}-5 \hat{j}-\hat{k}) \mathrm{N}\)
4. \(50(4 \hat{i}+5 \hat{j}+\hat{k}) \mathrm{N}\)

Answer: 2. \(100(4 \hat{i}+5 \hat{j}+\hat{k}) \mathrm{N}\)

Initial momentum = \(\vec{p}_{\mathrm{i}}=(2 \mathrm{~kg})(\hat{i}+\hat{j}) \mathrm{m} \mathrm{s}^{-1}\)

and final momentum = \(\vec{p}_{\mathrm{f}}=(2 \mathrm{~kg})(5 \hat{i}+6 \hat{j}+\hat{k}) \mathrm{m} \mathrm{s}^{-1}\)

∴ changeinmomentum = \(\Delta \vec{p}=\vec{p}_{\mathrm{f}}-\vec{p}_{\mathrm{i}}\)

⇒ \((2 \mathrm{~kg})(4 \hat{i}+5 \hat{j}+\hat{k}) \mathrm{m} \mathrm{s}^{-1}\)

∴ average force = \(\vec{F}_{\mathrm{av}}=\frac{\Delta \vec{p}}{\Delta t}=\frac{(2 \mathrm{~kg})(4 \hat{i}+5 \hat{j}+\hat{k}) \mathrm{m} \mathrm{s}^{-1}}{0.02 \mathrm{~s}}\)

⇒ \(100(4 \hat{i}+5 \hat{j}+\hat{k}) \mathrm{N}\)

Question 50. A ball of mass m moving with a velocity v undergoes an oblique elastic collision with another ball of the same mass at rest. After the collision, if the two balls move at equal speeds, the angle between their directions of motion will be

1. 60°
2. 30°
3. 90°
4. 120°

Answer: 3. 90°

Let \(\vec{p}_1 \text { and } \vec{p}_2\) be the momenta of the two balls after the elastic collision.

Hence, conserving the momentum, we have

⇒ \(\vec{p}=\overrightarrow{p_1}+\overrightarrow{p_2} \text {, where } \vec{p}\) = initial momentum.

Taking the self-dot-product,

⇒ \(\vec{p} \cdot \vec{p}=\left(\overrightarrow{p_1}+\overrightarrow{p_2}\right) \cdot\left(\overrightarrow{p_1}+\overrightarrow{p_2}\right)\)

or \(p^2=p_1^2+p_2^2+2 \vec{p}_1 \cdot \vec{p}_2\)

or \(\frac{p^2}{2 m}=\frac{p_1^2}{2 m}+\frac{p_2^2}{2 m}+\frac{2 \vec{p}_1 \cdot \overrightarrow{p_2}}{2 m}\) ……(1)

⇒ In an elastic collision, the KE (=p²/2m) is conserved.

∴ \(E=E_1+E_2 \text { or } \frac{p^2}{2 m}=\frac{p_1^2}{2 m}+\frac{p_2^2}{2 m}\) …(2)

From (1) and(2),

⇒ \(\overrightarrow{p_1} \cdot \overrightarrow{p_2}=0 \text {, i.e., } \overrightarrow{p_1} \text { is perpendicular to } \overrightarrow{p_2}\)

So, the angle between the velocities of the two balls is 90°.

Question 51. A body of mass 2 kg moving at a velocity \((\hat{i}+2 \hat{j}-3 \hat{k}) \mathrm{m} \mathrm{s}^{-1}\) collides with another body of mass 3 kg moving at a velocity \((2 \hat{i}+\hat{j}+\hat{k}) \mathrm{m} \mathrm{s}^{-1}\). If they stick together, the velocity of the composite body will be

1. \(\frac{1}{5}(8 \hat{i}+7 \hat{j}-3 \hat{k}) \mathrm{m} \mathrm{s}^{-1}\)
2. \(\frac{1}{5}(-4 \hat{i}+\hat{j}-3 \hat{k}) \mathrm{m} \mathrm{s}^{-1}\)
3. \(\frac{1}{5}(8 \hat{i}+\hat{j}-\hat{k}) \mathrm{m} \mathrm{s}^{-1}\)
4. \(\frac{1}{5}(-4 \hat{i}+7 \hat{j}-3 \hat{k}) \mathrm{m} \mathrm{s}^{-1}\)

Answer: 1. \(\frac{1}{5}(8 \hat{i}+7 \hat{j}-3 \hat{k}) \mathrm{m} \mathrm{s}^{-1}\)

Conserving the momentum,

⇒ \((2 \mathrm{~kg})(\hat{i}+2 \hat{j}-3 \hat{k}) \mathrm{m} \mathrm{s}^{-1}+(3 \mathrm{~kg})(2 \hat{i}+\hat{j}+\hat{k}) \mathrm{m} \mathrm{s}^{-1}\)

⇒ \((2 \mathrm{~kg}+3 \mathrm{~kg}) \vec{v}\)

or, \((5 \mathrm{~kg}) \vec{v}=(8 \hat{i}+7 \hat{j}-3 \hat{k}) \mathrm{kg} \mathrm{m} \mathrm{s}^{-1}\)

Hence, the velocity of the composite body is

⇒ \(\vec{v}=\frac{1}{5}(8 \hat{i}+7 \hat{j}-3 \hat{k}) \mathrm{m} \mathrm{s}^{-1}\)

Question 52. A steel ball falls from a height on a floor for which the coefficient of restitution is e. The height attained by the ball after two rebounds is

1. eH
2. e²H
3. e³H
4. e⁴H

Answer: 4. e⁴H

We know that the coefficient of restitution (e) is expressed by the relation velocity of separation

= e(velodty of approach).

The initial speed (u) when the ball hits the ground is given by

⇒ \(u^2=2 g H \quad \Rightarrow \quad u=\sqrt{2 g H}\)

The first recoil speed is u₁ = eu and the second recoil speed is

⇒ \(u_2=e u_1=e(e u)=e^2 u=e^2 \sqrt{2} g H .\)

If H² is the maximum height,

⇒ \(u_2^2=2 g H_2 \Rightarrow 2 g H_2=e^4(2 g H)\)

=> H² = e⁴H.

Question 53. A body of mass m, moving at a constant speed, undergoes an elastic collision with a body of mass m₂, initially at rest. The ratio of the kinetic energy of the first body after the collision to that before the collision is

1. \(\left(\frac{m_1-m_2}{m_1+m_2}\right)^2\)
2. \(\left(\frac{m_1+m_2}{m_1-m_2}\right)^2\)
3. \(\left(\frac{2 m_1}{m_1+m_2}\right)^2\)
4. \(\left(\frac{2 m_1}{m_1-m_2}\right)^2\)

Answer: 1. \(\left(\frac{m_1-m_2}{m_1+m_2}\right)^2\)

Conserving the momentum,

⇒ \(m_1 u=m_1 v_1+m_2 v_2\)….(1)

where u is the initial speed of the ball before the collision, and v₁ and v₂ are the speeds of the balls after the collision.

For an elastic collision, the coefficient of restitution = e = 1.

∴ \(\left(v_2-v_1\right)=1 \times u \Rightarrow v_2=u+v_1\)

Substituting for v₂ in (1), we get

⇒ \(m_1 u=m_1 v_1+m_2\left(u+v_1\right)\)

=> (m₁– m₂)u = (m₁ + m₂)v₁

⇒ \(v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right) u\)

⇒ \(\frac{\mathrm{KE}_{\text {after }}}{\mathrm{KE}_{\text {before }}}=\frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_1 u^2}=\left(\frac{v_1}{u}\right)^2=\left(\frac{m_1-m_2}{m_1+m_2}\right)^2\)

Question 54. A radioactive nucleus of mass number A, initially at rest, emits an a-particle with a speed v. What will be the recoil speed of the daughter nucleus?

1. \(\frac{2 v}{A-4}\)
2. \(\frac{2 v}{A+4}\)
3. ⇒ \(\frac{4 v}{A-4}\)
4. \(\frac{4 v}{A+4}\)

Answer: 3. \(\frac{4 v}{A-4}\)

The scheme of decay is shown in the adjoining figure. After the decay,

themomentaofthe fragments are \(4 m v \hat{i} \text { and }(A-4) m v^{\prime}(-\hat{i})\) respectively.

Conserving the momentum,

⇒ \(4 m v \hat{i}-(A-4) m v^{\prime} \hat{i}=0 \Rightarrow 4 v=(A-4) v^{\prime}\)

⇒ \(v^{\prime}=\frac{4 v}{A-4}\)

Question 55. Consider a system comprising a light string, a light spring, and a block of mass M suspended from a rigid support, as shown in the adjacent figure. The spring constant is k and the block is released when the spring

1. \(\frac{M g}{2}\)
2. Mg
3. 3Mg
4. 2Mg

Answer: 4. 2Mg

The initial of the block is zero, and finally, at the maximum extension (x) of the file spring, it comes to momentary rest.

Thus, by the work-energy theorem,

⇒ \(W_{\mathrm{grav}}+W_{\mathrm{sp}}=\Delta \mathrm{KE}=0\)

⇒ \(M g x+\left(-\frac{1}{2} k x^2\right)=0 \Rightarrow x=\frac{2 M g}{k}\)

∴ the maximum tension in the spring is

⇒ \(F_{\max }=k x_{\max }=k\left(\frac{2 M g}{k}\right)\)

= 2Mg.

Question 56. In the given figure, the spring has its natural length, the blocks are at rest, and there is no friction anywhere. If the spring constant is k= 20Nm^-1, the maximum extension in the spring will be

1. \(\frac{10}{3} \mathrm{~cm}\)
2. \(\frac{20}{3} \mathrm{~cm}\)
3. \(\frac{40}{3} \mathrm{~cm}\)
4. \(\frac{19}{3} \mathrm{~cm}\)

Answer: 1. \(\frac{10}{3} \mathrm{~cm}\)

When the extension in the spring is maximum (= x0), the blocks have the same velocity and the same acceleration (= a).

The system can be considered as connected bodies, and we apply Newton’s second law from the given free-body diagram.

For A, F-kx₀ = m₁a…(1)

For B, kx₀ = m₂a….(2)

Adding (1) and (2),

⇒ \(a=\frac{F}{m_1+m_2}\)….(3)

Substituting a from (3) in (2), the maximum extension becomes

⇒ \(x_0=\frac{m_2 a}{k}=\frac{m_2 F}{k\left(m_1+m_2\right)}\)

Substituting the given values

⇒ \(x_0=\frac{(1 \mathrm{~kg})(1 \mathrm{~N})}{\left(20 \mathrm{~N} \mathrm{~m}^{-1}\right)\left(\frac{3}{2} \mathrm{~kg}\right)}=\frac{10}{3} \mathrm{~cm}\)

Question 57. Find the maximum force if the potential energy is \(U=\frac{a}{r^2}-\frac{b}{r}\) (Given that a = 2 and b = 4.)

1. \(-\frac{16}{27} \mathrm{~N}\)
2. \(-\frac{32}{27} \mathrm{~N}\)
3. \(+\frac{32}{27} \mathrm{~N}\)
4. \(+\frac{16}{27} \mathrm{~N}\)

Answer: 1. \(-\frac{16}{27} \mathrm{~N}\)

By definition, \(F=-\frac{d U}{d r} \Rightarrow F=\frac{2 a}{r^3}-\frac{b}{r^2}\)

For F to be maximum,

⇒ \(\frac{d F}{d r}=0 \Rightarrow-\frac{6 a}{r^4}+\frac{2 b}{r^3}=0\)

⇒ \(r=\frac{3 a}{b}=3\left(\frac{2}{4}\right)=\frac{3}{2}\)

∴ \(F_{\max }=\frac{2 a}{r^3}-\frac{b}{r^2}\)

⇒ \(\frac{2(2)}{\left(\frac{3}{2}\right)^3} \mathrm{~N}-\frac{4}{\left(\frac{3}{2}\right)^2} \mathrm{~N}\)

⇒ \(\frac{32}{27} \mathrm{~N}-\frac{16}{9} \mathrm{~N}\)

⇒ \(-\frac{16}{27} \mathrm{~N}\)

Question 58. Initially when the spring is relaxed, having its natural length, block A of mass 0.25 kg is released. Find the maximum force exerted by the system on the floor.

1. 20N
2. 15N
3. 30N
4. 25N

Answer: 4. 25N

By the work-energy theorem,

⇒ \(W_{\text {grav }}+W_{\text {sp }}=\Delta \mathrm{KE}=0\)

⇒ \(m_{\mathrm{A}} g x_0-\frac{1}{2} k x_0^2=0\), where x0= maximum compression

⇒ \(x_0=\frac{2 m_A g}{k}\)

∴ the maximum force on the floor is

⇒ \(F_{\max }=m_{\mathrm{B}} g+k x_0=m_{\mathrm{B}} g+2 m_{\mathrm{A}} g\)

⇒ \(\left(m_{\mathrm{B}}+2 m_{\mathrm{A}}\right) \mathrm{g}=(2 \mathrm{~kg}+0.5 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)\)

= 25N.

Question 59. A force F = 20x + 10y acts on a particle in the y-direction, where F is in newtons and y is in meters. The work done by this force to move the particle from y = 0 to y = 1 m is

1. 30 J
2. 5 J
3. 25 J
4. 20 J

Answer: 3. 25 J

The total work done is

⇒ \(W=\int d W=\int \vec{F} \cdot d \vec{y}=\int_0^1m(20+10 y) d y\)

⇒ \(\left[20 y+5 y^2\right]_0^{1 m}\)

= 25J

Question 60. A constant force acts on a 2-kg object so that its position is given as a function of time by x = 3t²-5. What is the work done by this force in the first five seconds?

1. 850 J
2. 875 J
3. 900 J
4. 950 J

Answer: 3. 900 J

Since x = 3t2 – 5, the instantaneous velocity is

⇒ \(v=\frac{d x}{d t}=6 t\)

and the acceleration is

⇒ \(a=\frac{d v}{d t}=6 \mathrm{~m} \mathrm{~s}^{-2}=\text { constant }\)

∴ force = F = ma = (2kg)(6m s^-2)

= 12N.

The distance covered in 5 s is

⇒ \(x=u t+\frac{1}{2} a t^2=0 \cdot t+\frac{1}{2}\left(6 \mathrm{~m} \mathrm{~s}^{-2}\right)\left(25 \mathrm{~s}^2\right)\)

= 75m.

∴ work done =Fx = (12N)(75m) = 900 J

Question 61. A particle undergoes a displacement under the action of a \(\vec{r}=4 \hat{i}\) constant force \(\vec{F}=3 \hat{i}-12 \hat{j}\);. If the initial kinetic energy of the particle is 3 J, the kinetic energy at the end of the displacement will be

1. 9 J
2. 15 J
3. 12 J
4. 10 J

Answer: 2. 15 J

Work done = \(\vec{F} \cdot \vec{r}=(3 \hat{i}-12 \hat{j}) \mathrm{N} \cdot 4 \hat{i} \mathrm{~m}\)

=12 J.

∴ change in the KE = KE_f – KE_i = work done.

∴ \(\mathrm{KE}_{\mathrm{f}}=3 \mathrm{~J}+12 \mathrm{~J}=15 \mathrm{~J}\)

Question 62. Ablockofmassm, lying on a smooth horizontal surface, is attached to a light spring of spring constant k. The other end of the spring is fixed, as shown in the adjoining figure. The block is initially at rest in its equilibrium position. If the block is now pulled with a constant force F, the maximum speed of the block will be

1. \(\frac{\pi F}{\sqrt{m k}}\)
2. \(\frac{F}{\sqrt{m k}}\)
3. \(\frac{2 F}{\sqrt{m k}}\)
4. \(\frac{F}{\pi \sqrt{m k}}\)

Answer: 2. \(\frac{F}{\sqrt{m k}}\)

As the force produces elongation (x) in the spring, a retarding force (-kx) develops and we have

F_net= F-kx = ma.

When the velocity is maximum, the acceleration \(\left(\frac{d v}{d t}\right)\) is zero.

Thus,

⇒ \(F-k x=0 \Rightarrow x=\frac{F}{k}\)

According to the work-energy theorem,

⇒ \(W_{\text {fric }}+W_{\text {sp }}=\mathrm{KE}_{\max }=\frac{1}{2} m v_{\max }^2\)

⇒ \(F x-\frac{1}{2} k x^2=\frac{1}{2} m v_{\max }^2\)……(1)

Substituting x = \(\frac{F}{k}\) in (1) the condition for the maximum velocity,

⇒ \(F\left(\frac{F}{k}\right)-\frac{1}{2} k\left(\frac{F}{k}\right)^2=\frac{1}{2} m v_{\max }^2\)

⇒ \(\frac{F^2}{2 k}=\frac{1}{2} m v_{\max }^2 \Rightarrow v_{\max }=\frac{F}{\sqrt{m k}}\)

Question 63. A block of mass m is kept on a platform that starts from rest with a constant acceleration \(\frac{g}{2}\) upwards, as shown in the adjoining figure. The work done by the normal reaction on the block in a time t will be

1. \(\frac{m g^2 t^2}{8}\)
2. \(\frac{3 m g^2 t^2}{8}\)
3. \(-\frac{m g^2 t^2}{8}\)
4. Zero

Answer: 2. \(\frac{3 m g^2 t^2}{8}\)

The height reached by the block in the time t is

⇒ \(h=\frac{1}{2}\left(\frac{g}{2}\right) t^2=\frac{g t^2}{4}\)

and the velocity at that time is

⇒ \(v=\left(\frac{g}{2}\right) t=\frac{g t}{2}\)

The work done by gravity is

⇒ \(W_{\mathrm{grav}}=-(m g) h=-m g\left(\frac{g t^2}{4}\right)\)

Change in the

⇒ \(\mathrm{KE}=\frac{1}{2} m\left(\frac{g t}{2}\right)^2=\frac{1}{2}\left(\frac{m g^2 t^2}{4}\right)\)

From the work-energy theorem,

⇒ \(\mathrm{W}_{\mathrm{grav}}+W_{O V}=\frac{1}{2} m v^2\)

⇒ \(-\frac{m g^2 t^2}{4}+W_{\delta V}=\frac{1}{2} m\left(\frac{g^2 t^2}{4}\right)\)

∴ the work done by the normal reaction is

⇒ \(W N =\frac{m g^2 t^2}{4}+\frac{m g^2 t^2}{8}=\frac{3}{8} m g^2 t^2\)

Question 64. Throe blocks A, B, and C are placed on a smooth horizontal surface, as shown in the given figure. A and B have equal masses of m, while C has a mass of M. Block A is given an initial speed of v towards B, due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically. If in the whole process, \(\frac{5}{6}\) of its initial kinetic energy is lost, the value of the ratio \(\frac{M}{m}\) is

1. 4
2. 2
3. 3
4. 5

Answer: 1. 4

Since momentum is conserved during each impact, p = mu = (M + 2m) v₀, where v₀ is the final velocity of the combined system.

Initial kinetic energy of A is \(\frac{p^2}{2 m}\) and final kinetic energy of the combined system is

⇒ \(\frac{1}{6}\left(\frac{p^2}{2 m}\right), \text { since } \frac{5}{6} \text { part in loss }\)

⇒ \(\frac{1}{6}\left(\frac{p^2}{2 m}\right)=\frac{p^2}{2(M+2 m)}\)

or, 6m = (M+2m)

Hence \(\frac{M}{m}=4 .\)

Question 65. A gun exerts a time-dependent force on a bullet given by

F = [100- (0.5 x 10⁵)t] N.

If the bullet emerges with a speed of 400 m s^-1, find the impulse delivered to the bullet till the force F is reduced to zero.

1. 0.1 Ns
2. 0.3Ns
3. 0.5Ns
4. 0.4Ns

Answer: 1. 0.1 Ns

When F is reduced to zero at a time t, we have

⇒ \(t=\frac{100}{(0.5) 10^5} \mathrm{~s}=\frac{2}{1000} \mathrm{~s}\)

The impulse delivered to the bullet is

⇒ \(I=\int_0^t F d t=\left[100 t-\left(0.5 \times 10^5\right) \frac{t^2}{2}\right]_0^{(2 / 1000)}\)

⇒ \(100\left(\frac{2}{1000}\right)-\frac{1}{2} \times \frac{10^5}{2}\left(\frac{2}{10^3}\right)^2\)

=(0.2- 0.1)N s

= 0.1N s.

Question 66. A block A of mass 4m, moving towards block B of mass 2m at rest, undergoes an elastic head-on collision. The fraction of energy lost by the colliding body A is

1. \(\frac{4}{9}\)
2. \(\frac{5}{9}\)
3. \(\frac{8}{9}\)
4. \(\frac{2}{9}\)

Answer: 3. \(\frac{8}{9}\)

Let v₁ and v₂ be the velocities of the blocks A and B after the collision.

Conserving the momentum,

4mu₁ + 2m(0) = 4mv₁ +2mv₂

=> 2u₁= 2v₁+v₂…..(1)

For an elastic collision, e = 1.

Thus,

v₂-v₁ = e(u₁-u₂)

=> v₂-v₁ = u₁…..(2)

Subtracting(2) from(1),

⇒ \(3 v_1=u_1 \Rightarrow v_1=\frac{u_1}{3}\)

The fractional loss of energy the block A is

⇒ \(\frac{\Delta \mathrm{KE}_{\mathrm{A}}}{\mathrm{KE}_{\mathrm{A}}}=\frac{\frac{1}{2} m_{\mathrm{A}} u_1^2-\frac{1}{2} m_{\mathrm{A}} v_1^2}{\frac{1}{2} m_{\mathrm{A}} u_1^2}\)

⇒ \(1-\left(\frac{v_1}{u_1}\right)^2=1-\left(\frac{1}{3}\right)^2\)

⇒ \(\frac{8}{9}\)

Question 67. A block rests on a rough horizontal surface. A horizontal force that increases linearly with time (t) starts acting on the block at t = 0. Which of the following velocity-time graphs will be correct?

Answer: 2.

Given that \(F=m \frac{d v}{d t}=\alpha t \Rightarrow d v=\frac{\alpha}{m}, t d t\)

Integrating, \(v=\frac{\alpha}{2 m} \cdot t^2\)

Due to its roughness, the block starts moving after some time when

F≥f_k.

Question 68. A particle of mass m is moving at a speed 2v and collides with a particle of mass 2m, moving at a speed v in the same direction. After the collision, the first particle is stopped, while the second particle splits up into two smaller particles, each of mass m, which move at 45° with respect to the original direction. The speed of each moving particle will be

1. \(\frac{v}{2 \sqrt{2}}\)
2. 2 √2v
3. \(\frac{v}{\sqrt{2}}\)
4. √2v

Answer: 2. 2 √2v

The situations before and after the collision are shown in the figure below.

Conserving the momentum along the direction of motion,

m.2v + 2m.v = mv’cos45°+ mv’cos45°

⇒ \(4 m v=2 m v^{\prime}\left(\frac{1}{\sqrt{2}}\right)\)

⇒ \(v^{\prime}=\frac{4}{\sqrt{2}} v=2 \sqrt{2} v\)

Question 69. A body of mass m, moving at a velocity \(\hat{u i}\), collides elastically with a stationary body of mass 3m. If the velocity of the lighter body after the collision is \(\hat{v j}\) then this velocity has the magnitude After the collision

1. \(v=\frac{u}{2}\)
2. \(v=\frac{u}{\sqrt{3}}\)
3. \(v=\frac{u}{\sqrt{2}}\)
4. \(v=\frac{4}{3}\)

Answer: 3. \(v=\frac{u}{\sqrt{2}}\)

Conserving the KE, we have

⇒ \(\frac{1}{2} m u^2=\frac{1}{2} m v^2+\frac{1}{2}(3 m) v^{\prime 2}\)

or, \(u^2-v^2=3 v^2\)

Conserving the momentum, we have along the

x-axis: mu = 3mv’cosθ

and along the y-axis: 0 = mv-3mv’sin θ.

∴ \(u^2+v^2=9 v^{\prime 2}\)

From (1) and (2)

⇒ \(u^2+v^2=3\left(u^2-v^2\right) \Rightarrow v=\frac{u}{\sqrt{2}}\)

Question 70. Five blocks are placed on a smooth, horizontal, plane surface along the same line, as shown. The first block (of mass m) is given a velocity v so that the blocks successively undergo a perfectly inelastic collision. The percentage loss in the kinetic energy after the last collision is

1. 88.5%
2. 93.75%
3. 90.2%
4. 85.5%

Answer: 3. 90.2%

The momentum (p) is conserved during the collision.

However, due to the perfectly inelastic collision, the mass has changed from m to 16m.

Initial KE = \(\frac{p^2}{2 m} \text { and final } \mathrm{KE}=\frac{p^2}{2(16 m)}\)

∴ AKE = changee in the KE = \(\)

∴ % loss in KE = \(\frac{p^2}{2 m}-\frac{p^2}{2(16 m)}\)

Question 71. The force-displacement graph of a particle moving along the x-axis is shown in the given figure. The work done by the force from x = 0 to x = 30m is

1. 4550 J
2. 5250 J
3. 4250 J
4. 7500 J

Answer: 2. 5250 J

The area under the force-displacement graph gives the work done.

Hence,

W = area of the rectangle + area of the trapezium

= (200N)(15 m) + \(\frac{1}{2}\) (200N + 100 N)(15m)

= 3000 J + 2250 J

= 5250 J

Question 72. A block of mass m starts slipping down an inclined plane from the topmost point B and finally comes to rest at the lowermost point A. If BC = 2AC and the friction coefficient of the rough part ACis A m = k tan 6 then the value of k is

1. 0.5
2. 0.35
3. 0.75
4. 3.0

Answer: 4. 3.0

Let AC = x and BC = 2x.

∴ AB = 3x.

Work done by gravity = \(W_{\mathrm{grav}}=m g h=m g(3 x \sin \theta)\)

Work done by friction = \(W_{\text {fric }}=-f x=-(\mu m g \cos \theta) x\)

From the work-energy theorem,

⇒ \(W_{\text {grav }}+W_{\text {fric }}=\Delta K E=K_f-K_1=0\)

=> (mg)(3x) sin θ = (μmg cosθ)x

=> 3 tan θ = m = k tan 0 [ given ]

=> k=3.

Question 73. A 100-g bullet moving horizontally at 20m s^-1 strikes a stationary block of mass 1.9 kg and comes to rest inside. The kinetic energy of the block just before hitting the ground is

1. 11 J
2. 21 J
3. 24 J
4. 32 J

Answer: 2. 21 J

Conserving the linear momentum,

(100 g)(20m s^-1) = (100 g + 1.9 kg)v

=> v = 1m s^-1.

From the work-energy principle,

Wgrav = changein the KE

⇒ \((2 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(1 \mathrm{~m})=\mathrm{KE}_{\mathrm{f}}-\mathrm{KE}_{\mathrm{i}}\)

=\(\mathrm{KE}_{\mathrm{f}}-\frac{1}{2}(2 \mathrm{~kg})\left(1 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\)

Hence, KE_f = 20J +1J

= 21J.

Question 74. A block is pushed up along a rough inclined plane with a speed V₀, as shown in the given figure. After some time, it is again back to the starting point with a reduced Find the coefficient of speed \(\frac{v_0}{2}\) fiction (μ).

1. 0.15
2. 0.45
3. 0.35
4. 0.75

Answer: 3. 0.35

Applying the work-energy theorem,

⇒ \(W_{\text {grav }}+W_{\text {fric }}=\Delta K E=K_f-K_i\)

The work done by gravity is zero, and the work done by friction is

⇒ \(W_{\text {fric }}=-\mu \delta N(2 s)=-\left(\mu m g \cos 30^{\circ}\right)(2 s)\)

The change in KE is

⇒ \(\frac{1}{2} m\left(\frac{v_0}{2}\right)^2-\frac{1}{2} m v_0^2=-\frac{1}{2}\left(\frac{3}{4} m v_0^2\right)\)

∴ \(\frac{1}{2} \cdot \frac{3}{4} m v_0^2=(\mu m g \sqrt{3}) s\)

⇒ \(v_0^2=\frac{8}{\sqrt{3}} \mu g s\)…..(1)

For demotion down the plane,

⇒ \(\left(\frac{v_0}{2}\right)^2=0+2 g(\sin \theta-\mu \cos \theta) s\)

⇒ \(v_0^2=8 g\left(\frac{1}{2}-\frac{\sqrt{3}}{2} \mu\right) s\) …..(2)

Equating (1) and(2),

⇒ \(4 g s(1-\sqrt{3} \mu)=\frac{8}{\sqrt{3}} \mu g s\)

⇒ \(\mu\left(\sqrt{3}+\frac{2}{\sqrt{3}}\right)=1\)

⇒ \(\mu=\frac{\sqrt{3}}{5}\)

= 0.35

Question 75. A block of mass 2 kg at rest is driven by a constant power of 1.0 W. The distance covered by the block in 6 s is

1. 2√6 m
2. 4√6 m
3. 3√3 m
4. 4√3 m

Answer: 2. 4√6 m

Instantaneous power P = Fv = \(\left(m \frac{d v}{d t}\right) v\)

∴ \(\frac{P}{m} d t=v d v\)

Integrating, \(\left(\frac{P}{m}\right) t=\frac{y^2}{2}\)

Hence, velocity = \(v=\frac{d s}{d t}=\sqrt{\frac{2 P}{m}} t^{1 / 2}\)

Integrating again,

⇒ \(s=\sqrt{\frac{2 P}{m}}\left(\int t^{1 / 2} d t\right)=\sqrt{\frac{2 P}{m}}\left(\frac{2}{3} t^{3 / 2}\right)=\frac{2}{3} \sqrt{\frac{2 \times 1}{2}}\left(6^{3 / 2}\right)=4 \sqrt{6} \mathrm{~m}\)

Friction and Circular Motion

Question 1. A block released on a rough inclined plane of inclination 0 = 30° Or slides down the plane with a constant acceleration of \(\frac{g}{4}\), where g is the acceleration due to gravity. What is the coefficient of friction between the block and the inclined plane?

1. \(\frac{1}{\sqrt{3}}\)
2. \(\frac{1}{2}\)
3. \(\frac{1}{2 \sqrt{2}}\)
4. \(\frac{1}{2 \sqrt{3}}\)

Answer: 4. \(\frac{1}{2 \sqrt{3}}\)

The acceleration of a block sliding down an inclined plane is given by a =g(sin θ- cos θ).

Given that a = \(\frac{g}{4}\) and θ = 30°.

∴ \(\frac{g}{4}=g\left(\sin 30^{\circ}-\mu \cos 30^{\circ}\right)=\frac{g}{2}(1-\sqrt{3} \mu) .\)

∴ Coefficient offriction = \(\mu=\frac{1}{2 \sqrt{3}}\)

Question 2. A block takes twice as much time to slide down a rough 45°-inclined plane as it takes to slide down an identical but smooth 45°-inclined plane. The coefficient of kinetic friction between the block and the rough inclined plane is

1. 0.25
2. 0.50
3. 0.75
4. 0.90

Answer: 3. 0.75

The acceleration down a rough incline is

⇒ \(a_{\text {rough }}=g(\sin \theta-\mu \cos \theta)=\frac{g}{\sqrt{2}}(1-\mu)\) [∵ θ = 45°]

and that down a smooth incline is

⇒ \(a_{\text {smooth }}=g \sin \theta=\frac{8}{\sqrt{2}} \text {. }\)

Time to slide, \(t=\sqrt{\frac{2 s}{a}}\)

∴ \(t \propto \frac{1}{\sqrt{a}} .\)

Now, \(\frac{t_{\text {rough }}}{t_{\text {smooth }}}=\sqrt{\frac{a_{\text {smooth }}}{a_{\text {rough }}}}=\sqrt{\frac{\frac{g}{\sqrt{2}}}{\frac{g}{\sqrt{2}}(1-\mu)}}=\sqrt{\frac{1}{1-\mu}}\)

⇒ \(2=\frac{1}{\sqrt{1-\mu}}\)

⇒ \(\mu=\frac{3}{4}=0.75\)

Question 3. The upper half of an inclined plane of inclination θ is perfectly smooth, while the lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom of the coefficient of friction between the block and the lower half of the plane is given by

1. p = 2 tan θ
2. p = tan θ
3. \(\mu=\frac{2}{\tan \theta}\)
4. p = cot θ

Answer: 1. p = 2 tan θ

For the motion along the upper (smooth) half,

⇒ \(v_0^2=2 g \sin \theta \cdot \frac{s}{2}=g \sin \theta\)

For the motion on the lower (rough) half,

⇒ \(v^2=v_0^2-2 a\left(\frac{s}{2}\right)=g s \sin \theta-2(\sin \theta-\mu \cos \theta) g \cdot \frac{s}{2}\)

⇒ \(2(\sin \theta-\mu \cos \theta)\left(\frac{g s}{2}\right)=g^s \sin \theta\) [∵ v = 0]

Simplifying, we gel \(\mu\) = 2tanθ.

Question 4. A 20-kg block is pulled by a force of 100 N acting at an angle of 30° above the horizontal with a uniform speed on a horizontal surface. The coefficient of friction between the surface and the block is

1. 0.75
2. 0.85.
3. 0.43
4. 0.58

Answer: 4. 0.58

For vertical equilibrium, N + F sin 30° = mg.

Hence, the normal reaction is

N = mg-Fsin 30°

= 200 N-50 N

= 150 N.

Since the block moves with a uniform velocity,

⇒ \(F_{\text {net }}=0 \Rightarrow F \cos 30^{\circ}=f=\mu \nu\)

∴ \(\mu=\frac{100\left(\frac{\sqrt{3}}{2}\right)}{150}=\frac{\sqrt{3}}{3}=0.58\)

Question 5. A uniform chain of length L lies on the horizontal surface of a table. The coefficient of static friction between the chain and the table is p. The maximum length of the chain that can hang over the edge of the table so that the entire chain can remain at rest will be

1. \(\frac{\mu L}{1+\mu}\)
2. \(\frac{\mu L}{1-\mu}\)
3. \(\frac{L}{1-\mu}\)
4. \(\frac{L}{1+\mu}\)

Answer: 1. \(\frac{\mu L}{1+\mu}\)

Let X be the mass per unit length of the chain and xL be its maximum length that can hang.

∴ The pulling force on the chain is \(\lambda\)xLg, while the weight of the chain lying on the table is \(\lambda(1-x) L g\), which is equal to the normal reaction N.

For the limiting static equilibrium,

⇒ \(\lambda x L g=\mu \delta V=\mu \lambda(1-x) L g\)

Simplifying, we get \(x=\frac{\mu}{1+\mu}\)

∴ The maximum length of the Overhanging part is

⇒ \(x L=\frac{\mu L}{1+\mu}\)

Question 6. A boy weighing 40 kg is climbing a vertical pole at a constant speed. If the coefficient of friction between his palms and the pole is 0.8 and g = 10 m s^-2, the force exerted by him on the pole is

1. 500 N
2. 300 N
3. 400 N
4. 600 N

Answer: 1. 500 N

Given that the weight of the boy is mg = 40 kg x g = 400 N.

The force exerted by the boy on the pole is F = N.

Hence, the force of friction = \(f=\mu \nu =\mu F.\)

For uniform motion,

⇒ \(F_{\text {net }}=m g-\mu \nu=0 \Rightarrow \mu F=m g\) [∵ F = N]

⇒ \(F=\frac{m g}{\mu}=\frac{400 \mathrm{~N}}{0.8}=500 \mathrm{~N}\)

Question 7. A boy of mass m is sliding down a vertical pole by pressing it with a horizontal force f. If \(\mu\) is the coefficient of friction between his palms and the pole, the acceleration with which he slides down will be

1. g
2. \(\frac{\mu f}{m}\)
3. \(g-\frac{\mu f}{m}\)
4. \(g+\frac{\mu f}{m}\)

Answer: 3. \(g-\frac{\mu f}{m}\)

Normal reaction = force exerted horizontally.

The upward force of friction = μf.

∴ \(F_{\text {net }}=m g-\mu f=m a\)

∴ downward acceleration

⇒ \(a=\frac{F_{\text {net }}}{m}=g-\frac{\mu f}{m} .\)

Question 8. A block B is pushed momentarily along a horizontal surface with an initial velocity of v. If (A is the coefficient of sliding friction between B and the surface, B will come to rest after a time

1. \(\frac{\mu g}{v}\)
2. \(\frac{g}{v}\)
3. \(\frac{v}{g}\)
4. \(\frac{v}{\mu g}\)

Answer: 4. \(\frac{v}{\mu g}\)

Initial velocity = v.

Frictional force = \(f=\mu \nu =\mu m g\)

∴ acceleration = \(a=\frac{L}{m}=\mu g\)

Now, applying v = u-at, we have

⇒ \(0=v-\mu g t \Rightarrow t=\frac{v}{\mu g}\)

Question 9. The coefficient of static friction between μ_s the block A (of mass 2 kg) and the table is shown in the. figure is 0.2. What should be the maximum mass of the block B so that the two blocks do not move? (The string and the pulley are assumed to be smooth and light, and g = 10 m s^-2.)

1. 2.0 kg
2. 4.0 kg
3. 0.2 kg
4. 0.4 kg

Answer: 4. 0.4 kg

The limiting static friction for block A is

⇒ \(\mu m_A g=0.2(2 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)=4 \mathrm{~N}\)

The tension in the string is also T = 4 N.

For the block B,

⇒ \(m_{\mathrm{B}} g=T=4 \mathrm{~N} \Rightarrow m_{\mathrm{B}}=\frac{4 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-2}}{10 \mathrm{~m} \mathrm{~s}^{-2}}=0.4 \mathrm{~kg}\)

Question 10. The system consists of three masses m_1, m₂, and m₃, connected by a string passing over a pulley P. The mass m₁ hangs freely, and m₂ and m₃ are on a rough horizontal table (the coefficient of friction being p). The pulley is frictionless and of negligible mass. The downward acceleration of mass m₁ is (assuming m₁= m₂ = m₃ = m)

1. \(\left(\frac{1-\mu}{9}\right) g\)
2. \(\frac{2}{3} \mu g\)
3. \(\left(\frac{1-2 \mu}{3}\right) 8\)
4. \(\left(\frac{1-2 \mu}{2}\right) g\)

Answer: 3. \(\left(\frac{1-2 \mu}{3}\right) 8\)

Let the common acceleration of the system be a, the tension connecting the hanging mass ml be T₁, and that connecting m₂ and m₃ be T₂. From Newton’s law, we obtain the following.

For the mass m₁, m₁g-T₁ = m₁a.

For the mass \(m_2, T_1-T_2-\mu m_2 g=m_2 a\)

For the mass \(m_3, T_2-\mu m_3 g=m_3 a\)

Adding the three equations and putting m₁ = m₂ = m₃ = m, we get

⇒ \(m g(1-2 \mu)=3 m a\) = 3ma.

Hence, acceleration = \(a=\left(\frac{1-2 \mu}{3}\right) g\)

Question 11. A block released from rest from the top of a smooth inclined plane of inclination 0 has a speed of v when it reaches the bottom. The same block, released from the top of a rough inclined plane of the same inclination 9, acquires a speed \(\frac{v}{n}\) on reaching the bottom, where n > 1. The coefficient of friction is given by

1. \(\mu=\sqrt{1-\frac{1}{n^2}} \cdot \tan \theta\)
2. \(\mu=\sqrt{1-\frac{1}{n^2}} \cdot \cot \theta\)
3. \(\mu=\sqrt{1-\frac{1}{n^2}} \cdot \sin \theta\)
4. \(\mu=\left(1-\frac{1}{n^2}\right) \tan \theta\)

Answer: 4. \(\mu=\left(1-\frac{1}{n^2}\right) \tan \theta\)

For the motion down the smooth inclined plane,

v² = 2(gsinθ)s……(1)

For the rough inclined plane, the acceleration down the plane is

⇒ \(a=g(\sin \theta-\mu \cos \theta)\).

Hence,

⇒ \(\left(\frac{v}{n}\right)^2=2 g(\sin \theta-\mu \cos \theta) s\)…….(2)

Dividing (1) by (2),

⇒ \(n^2=\frac{\sin \theta}{\sin \theta-\mu \cos \theta}\)

⇒ \(\mu=\left(1-\frac{1}{n^2}\right) \tan \theta\)

Question 12. Two blocks of masses m₁-m and m₂-m are connected by a light inextensible string that passes over a smooth pulley fixed on the top of an inclined plane, as shown in the figure. When θ = 30°, the block of mass ml just begins to move up along the inclined plane. The coefficient of kinetic friction between the block of mass m₁ and the inclined plane is

1. \(\frac{1}{\sqrt{3}}\)
2. \(\frac{1}{\sqrt{2}}\)
3. \(\frac{1}{2 \sqrt{2}}\)
4. \(\frac{1}{3 \sqrt{2}}\)

Answer: 1. \(\frac{1}{\sqrt{3}}\)

Since the block of mass m₁ just begins to move, it is friction, and the net force on each block is zero.

For m₂, m₂g – T = 0.

For \(m_1, T-m_1 g \sin \theta-\mu m_1 g \cos \theta=0\)

Taking m₁ = m₂ = m and adding the two equations,

⇒ \(m g-m g \sin \theta-\mu m g \cos \theta=0\)

⇒ \(\mu=\frac{1-\sin \theta}{\cos \theta}=\frac{1-\sin 30^{\circ}}{\cos 30^{\circ}}=\frac{1}{\sqrt{3}}\)

Question 13. A block rests on an inclined plane. If the angle of inclination is gradually increased, the block just begins to slide down the plane when the angle of inclination is 30°. The coefficient of friction between the block and the inclined plane is

1. \(\frac{1}{3}\)
2. \(\frac{1}{2 \sqrt{3}}\)
3. \(\frac{1}{2 \sqrt{3}}\)
4. \(\frac{1}{2 \sqrt{2}}\)

Answer: 2. \(\frac{1}{2 \sqrt{3}}\)

According to the given question, θ is the angle of repose, for which

⇒ \(\mu=\tan \theta=\tan 30^{\circ}=\frac{1}{\sqrt{3}}\)

Alternative method:

⇒ \(m g \sin \theta=f=\mu \sigma V=\mu m g \cos \theta\)

⇒ \(\mu=\tan \theta=\tan 30^{\circ}=\frac{1}{\sqrt{3}}\)

Question 14. A plank with a block on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30°, the block just starts to slip down the plank and slides 4.0 m in 4.0 s. The coefficients of static and kinetic frictions between the block and the plank will respectively be

1. 0.5 and 0.6
2. 0.4 and 0.3
3. 0.6 and 0.6
4. 0.6 and 0.5

Answer: 4. 0.6 and 0.5

When the die block just tends to slip down, it is a case of static limiting friction, for which

⇒ \(\mu_{\mathrm{s}}=\tan \theta=\tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}=0.58 \approx 0.6\)

When the block slides down the plane, the acceleration is

⇒ \(a=g\left(\sin 30^{\circ}-\mu \cos 30^{\circ}\right)=\frac{g}{2}(1-\sqrt{3} \mu) .\)

∵ \(s=u t+\frac{1}{2} a t^2\)

⇒ \(4.0 \mathrm{~m}=0+\frac{1}{2} \cdot 5\left(1-\sqrt{3} \mu_{\mathrm{k}}\right)(4 \mathrm{~s})^2\)

⇒ \(\sqrt{3} \mu_{\mathrm{k}}=1-\frac{1}{10}=\frac{9}{10}\)

Hence, \(\mu_k=\frac{0.9}{\sqrt{3}}=0.5\)

Question 15. A car is negotiating a curved road of radius R. The road is banked at an angle θ. The coefficient of friction between the tire of the car and the road is μ_s. The maximum safe velocity on this road is

1. \(\sqrt{\frac{g}{R} \cdot \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}\)
2. \(\sqrt{\frac{g}{R^2} \cdot \frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}}\)
3. \(\sqrt{g R^2 \cdot \frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}}\)
4. \(\sqrt{g R \cdot \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}\)

Answer: 4. \(\sqrt{g R \cdot \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}\)

The forces acting on the car moving on a banked road are the weight (mg), the normal reaction (N), and the friction (f).

Resolving f and N along the vertical and horizontal directions (as shown in the adjoining figure), we obtain the following.

For vertical equilibrium,

⇒ \(\mathcal{N} \cos \theta=m g+f \sin \theta\)

or, \(m g=\alpha N \cos \theta-f \sin \theta\) ….. (1)

The centripetal force is provided by

⇒ \(f \cos \theta+\propto N \sin \theta=\frac{m v^2}{R}\)….(2)

Dividing (2) by (1),

⇒ \(\frac{f \cos \theta+\delta N \sin \theta}{\delta N \cos \theta-f \sin \theta}=\frac{v^2}{R g}\)

In the limiting case, when \(f=f_{\max }=\mu_{\mathrm{s}} \nu\)

⇒ \(\frac{v_{\max }^2}{R g}=\frac{\mu_{\mathrm{s}} N \cos \theta+N \sin \theta}{\alpha N \cos \theta-\mu_{\mathrm{s}} \nu \sin \theta}\)

⇒ \(v_{\max }=\sqrt{R g\left(\frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}\right)}\)

Question 16. A block of mass 10 kg is placed on a rough horizontal surface having a coefficient of friction μ = 0.5. If a horizontal force of 100 N acts on it, the acceleration of the block will be

1. 10ms^-2
2. 5ms^-2
3. 15 ms^-2
4. 0.5ms^-2

Answer: 2. 5ms^-2

Maximum kinetic friction = \(f_{\max }=\mu \subset N=(0.5)(10 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)\)

= 50 N.

Applied force = F = 100 N.

∴ \(F_{\text {net }}=F-f_{\max }=m a\)

100 N – 50 N = (10kg)4.

∴ acceleration = a = 5ms^-2.

Question 17. A car is negotiating a curved level road of radius r. If the coefficient of friction between the tyre and the road is μ_s, the car will skid if its speed exceeds

1. \(\sqrt{2 \mu_s r g}\)
2. \(\sqrt{3 \mu_s r g}\)
3. \(\sqrt{\mu_s r g}\)
4. \(2 \sqrt{\mu_s r g}\)

Answer: 3. \(\sqrt{\mu_s r g}\)

Since \(\mu_{\mathrm{s}}\) is the coefficient of static friction between the tyre and the road, its maximum value is \(f_{\max }=\mu_s m g\); and for a safe turn (without skidding),

⇒ \(\frac{m v^2}{r} \leq \mu_s m g \Rightarrow v \leq \sqrt{\mu_s r g}\)

∴ \(v_{\max }=\sqrt{\mu_s r g}\)

Question 18. A 20-kg box is placed gently on a horizontal conveyor belt moving at a speed of 4 m s^-1. If the coefficient of friction between the box and the belt is 0.8, through what distance will the block slide on the belt?

1. 0.6 m
2. 0.8 m
3. 1.0 m
4. 1.2 m

Answer: 3. 1.0 m

Theforce offrictionon theboxis \(f=\mu m g\), so the accelerationisa \(a=\frac{f}{m}=\mu g\)

The box will slide on the belt until it attains the speed of the belt (v).

The distance moved by the box is (applying v² = u²- 2as)

⇒ \(s=\frac{v^2}{2 a}\) [∵ u=0]

⇒ \(\frac{\left(4 \mathrm{~m} \mathrm{~s}^{-1}\right)^2}{2(0.8)\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)}=1 \mathrm{~m}\)

Question 19. A car is negotiating a circular horizontal track of radius 10 m with a constant speed of 10 m s^-1. A bob is suspended from the roof of the car by a light thread of length 1.0 m. The angle made by the thread with the vertical is

1. \(\frac{\pi}{3} \mathrm{rad}\)
2. \(\frac{\pi}{6} \mathrm{rad}\)
3. \(\frac{\pi}{4} \mathrm{rad}\)
4. 0°

Answer: 3. \(\frac{\pi}{4} \mathrm{rad}\)

Let the thread make an angle 0 with the vertical when T is the tension.

Resolving T, we have

⇒ \(T \sin \theta=\frac{m v^2}{R} \text { and } T \cos \theta=m g\)

Dividing, we get

⇒ \(\tan \theta=\frac{v^2}{R g}=\frac{\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right)^2}{(10 \mathrm{~m})\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)}=1\)

⇒ \(\theta=45^{\circ}=\frac{\pi}{4} \mathrm{rad}\)

Question 20. A Block A of mass m₁ rests on a horizontal table. A light string connected to it passes over a frictionless pulley fixed at the edge of the table, and a block B of mass m₂ is suspended from its other end. The coefficient of kinetic friction between block A and the table is μ_k. When the block A is sliding on the table, the tension in the string is

1. \(\frac{m_1 m_2\left(1+\mu_{\mathrm{k}}\right) g}{m_1+m_2}\)
2. \(\frac{m_1 m_2\left(1-\mu_k\right) g}{m_1+m_2}\)
3. \(\frac{m_2+\mu_{\mathrm{k}} m_1 g}{m_1+m_2}\)
4. \(\frac{m_2-\mu_{\mathrm{k}} m_1 g}{m_1+m_2}\)

Answer: 1. \(\frac{m_1 m_2\left(1+\mu_{\mathrm{k}}\right) g}{m_1+m_2}\)

The given figure shows the situation. Let a be the acceleration of the system and T be the tension in the connecting string.

For the block of mass m₁, T – μ_km₁g= m₁a,

and for the block of mass m₂, m₂g – T = m₂a.

Dividing the first equation by the second,

⇒ \(\frac{T-\mu_{\mathrm{k}} m_1 g}{m_2 g-T}=\frac{m_1}{m_2}\)

Simplifying, we get

⇒ \(T=\frac{m_1 m_2\left(1+\mu_{\mathrm{k}}\right) g}{m_1+m_2}\)

Question 21. A block of mass m is in contact with a cart C, as shown in the figure. The coefficient of static friction between the block and the cart is μ. The acceleration α of the cart that will prevent the block from falling down satisfies the relation

1. \(\alpha>\frac{m g}{\mu}\)
2. \(\alpha>\frac{g}{\mu m}\)
3. \(\alpha \geq \frac{g}{\mu}\)
4. \(\alpha<\frac{g}{\mu}\)

Answer: 3. \(\alpha \geq \frac{g}{\mu}\)

The pseudo-force acting on the block is

Fps = -(mass of the block)(acceleration of the frame)

= -mα towards right

= mα towards the left,

This pressing force produces the normal reaction N, and in the limiting case, c \(f_{\max }=\mu \propto N=\mu m \alpha,\), which will act against the weight mg. The block will not fall as long as \(f \geq m g \text { or } \mu m \alpha \geq m g \text { or } \alpha \geq g / \mu\)

Question 22. On a horizontal surface (p = 0.6) of a truck, a block of mass1 kg is placed, and the truck is moving with an acceleration of 5 m s^-2. The frictional force exerted on the block is

1. 5 N
2. 6 N
3. 5.88 N
4. 8 N

Answer: 1. 5 N

Here the truck is the accelerated frame of reference with a = 5 m s^-2.

The pseudo-force acting on the block is

F_ps = ma = ( 1 kg)(5ms^-2)

= 5 N (towards left).

The friction/acts to the right and has the maximum value

⇒ \(f_{\max }=\mu N=(0.6)(1 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)=6 \mathrm{~N}\)

Since friction is a self-adjusting force, the frictional force

f = 5 N

Question 23. A bridge over a river is in the form of an arc of radius of curvature R. If m is the total mass of the bike and the rider, and the rider is crossing the bridge at a speed v, the thrust on the bridge at the highest point will be

1. \(\frac{m v^2}{R}\)
2. mg
3. \(\frac{m v^2}{R}-m g\)
4. \(m g-\frac{m v^2}{R}\)

Answer: 4. \(m g-\frac{m v^2}{R}\)

The thrust on the topmost point of the river bridge will be the normal reaction N by the bike.

The centripetal force \(\left(\frac{m v^2}{R}\right)\) is provided by mg – N

Thus, \(\frac{m v^2}{R}=m g-N\)

⇒ \(\mathcal{N}=m g-\frac{m v^2}{R}\) = thrust on the bridge.

Question 24. A block is placed on a smooth hemispherical surface of radius R. The block is given a small horizontal push so that it slides down along the curved spherical surface. The block loses its contact with the hemisphere at a point whose height from file ground is

1. \(\frac{3}{4} R\)
2. \(\frac{2}{3} R\)
3. \(\frac{R}{2}\)
4. \(\frac{2}{\sqrt{3}} R\)

Answer: 2. \(\frac{2}{3} R\)

Let the block start from point A and leave its contact at point P, where ΔAOP = 9.

Now, gain in KE = \(\text { loss in } \mathrm{PE} \Rightarrow \frac{1}{2} m v^2=m g \cdot A B=m g R(1-\cos \theta)\)

⇒ \(v^2=2 g R(1-\cos \theta)\)……(1)

At P, centripetal force = \(\frac{m v^2}{R}=m g \cos \theta-N\)

Here, N = θ (for no contact). So,

v² = gR cosθ…….(3)

Equating (1) and (2),

⇒ \(2 g R(1-\cos \theta)=g R \cos \theta \Rightarrow 3 \cos \theta=2\)

⇒ \(\cos \theta=\frac{2}{3} .\)

∴ height = \(O B=R \cos \theta=\frac{2}{3} R\)

Question 25. A block is placed on a smooth hemispherical surface of radius R. What minimum horizontal velocity must be imparted to the block so that it leaves the hemisphere without sliding over it?

1. \(\frac{\sqrt{8 R}}{2}\)
2. \(2 \sqrt{8 R}\)
3. \(\sqrt{\frac{g R}{2}}\)
4. \(\sqrt{g R}\)

Answer: 3. \(\sqrt{\frac{g R}{2}}\)

Let u be the velocity imparted to the block so that it strikes the ground without touching the sphere. The motion of the block will be like that of a projectile, for which

⇒ \(R=\frac{1}{2} g t^2 \Rightarrow t=\sqrt{\frac{2 R}{g}}\)

Further, \(R=u t=u \sqrt{\frac{2 R}{g}}\)

⇒ \(u=R \sqrt{\frac{g}{2 R}}=\sqrt{\frac{g R}{2}} .\)

Question 26. A 1-m-long string is fixed to a rigid support and carries a mass of 100 g at its free end. The string makes √5/π revolutions per second about a vertical axis passing through the fixed end of the string. What is the angle of inclination of the string with the vertical? (Take g = 10 m s^-2.)

1. 30°
2. 45°
3. 60°
4. 75°

Answer: 3. 60°

Given that m = 100 g = 0.1 kg,

⇒ \(\omega=2 \pi n=\frac{2 \pi \sqrt{5}}{\pi}=2 \sqrt{5} \mathrm{rad} \mathrm{s}^{-1}\) and l =1 m.

The horizontal component of the tension T is

⇒ \(T \sin \theta=m \omega^2 r\)…..(1)

and the vertical component is

⇒ \(T \cos \theta=m g\)…..(2)

Dividing (1) by (2)

⇒ \(\tan \theta=\frac{\omega^2 r}{g}=\frac{(2 \sqrt{5})^2(l \sin \theta)}{10}\)

⇒ \(\frac{\sin \theta}{\cos \theta}=\frac{(4 \times 5)(1 \times \sin \theta)}{10}\)

⇒ \(\cos \theta=\frac{1}{2} \Rightarrow \theta=60^{\circ} \text {. }\)

Question 27. One end of a string of length l is connected to a particle of mass m, and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in a circle at a uniform speed v, the net force on the particle (directed towards the center) will be (T being the tension in the string) 

1. T
2. \(T-\frac{m v^2}{l}\)
3. T + mg
4. \(T+\frac{m v^2}{l}\)

Answer: 1. T

The forces acting on the particle of mass m are the weight (mg), the normal reaction (N), and the tension (T) in the string. Here mg and N balance. So, the net force on the particle will be T, which will provide the required centripetal force.

Question 28. A string can break due to a tension just exceeding 25N. The great. the speed with which a body of mass1 kg can be whirled in a horizontal circle using such a string of length m (without breaking it) is

1. 10 m s^-1
2. 2.5 m s^-1
3. 5 ms^-1
4. 7.5 ms^-1

Answer: 3. 5 ms^-1

Gravity acts vertically downwards. Its component along the horizontal is zero, so only the tension T along the string provides the required centripetal force

⇒ \(\left(\frac{m v^2}{R}\right)\)

∴ \(25 \mathrm{~N}=\frac{(1 \mathrm{~kg}) v^2}{1 \mathrm{~m}} \Rightarrow v=5 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 29. A simple pendulum with a bob of mass m swings with an angular amplitude \(\phi\). When the angular displacement is θ (where θ < \(\phi\)), the tension in the string is

1. mg cos θ
2. mg sin θ
3. Greater than mg cos θ
4. Greater than mg sin θ

Answer: 3. Greater than mg cos θ

When the bob is at P with the angular displacement θ (where θ < Φ), the forces acting are the weight (mg) and the tension (T). The component of mg along the string is mg cosθ.

Then, the net force along the string directed toward the center is

T- mg cos θ, which provides the required

centripetal force \(\left(\frac{m v^2}{r}\right)\)

Thus,

T- mgcosθ = \(\frac{m v^2}{r}\)

⇒ \(T=m g \cos \theta+\frac{m v^2}{r}\)

∴ \(T>m g \cos \theta\)

Question 30. A bridge over a river is in the form of an arc of radius of curvature 10 m. The highest speed with which a motorcyclist can cross the bridge without his bike losing contact with the ground is

1. 10 ms^-1
2. 10 2 ms^-1
3. 10 √3 m s^-1
4. 20 ms^-1

Answer: 1. 10 ms^-1

At the highest point of the river bridge, the forces acting are the weight mg (downward) and the normal reaction N (upward).

The net force (mg- N) towards the center provides the required centripetal force \(\left(\frac{m v^2}{R}\right)\)

Thus, \(m g – N=\frac{m v^2}{R}\)

⇒ \(\mathcal{N}=m g-\frac{m v^2}{R}\)

An increase in the speed (v) of the bike will reduce N and for N = 0 (when the contact will just go),

⇒ \(\frac{m v_{\max }^2}{R}=m g \Rightarrow v_{\max }=\sqrt{g R}=\sqrt{\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(10 \mathrm{~m})}=10 \mathrm{~m} \mathrm{~s}^{-1} .\)

Question 31. A mass m attached to a thin wire is whirled in a vertical circle. The wire is most likely to break when the

1. Mass is at the highest point
2. Wire is horizontal
3. Mass is at the lowest point
4. The wire is inclined at an angle of 60° to the vertical

Answer: 3. Mass is at the lowest point

At the lowest point of the vertical circle, the centripetal \(\left(\frac{m v^2}{l}\right)\) force is provided by T- mg.

Thus, \(T-m g=\frac{m v^2}{l}\)

⇒ \(T=m g+\frac{m v^2}{l}\)

Hence, the tension is maximum at the lowest point, so the chance of breaking is maximum at this point.

Question 32. A particle moving at a velocity v is acted upon by three forces P, Q, and R, shown by the vector triangle in the p figure. The velocity of the particle will

1. Increase
2. Decrease
3. Remain constant
4. Change with time periodically

Answer: 3. Remain constant

According to the triangle rule for vector addition, \(\vec{P}+\vec{Q}+\vec{R}=\overrightarrow{0}\). So, for net force = θ, acceleration = θ; hence the velocity will remain constant.

Question 33. Two particles A and B are undergoing uniform circular motion in concentric circles of radii r_A and r_B with speeds v_A and v_B respectively. Their time periods of revolution are equal. The ratio of the angular speed of A to that of B will be

1. r_A:r_B
2. v_A: v_B
3. r_B:r_A
4. 1:1

Answer: 4. 1:1

The particles A and B are synchronous since they have the same period of revolution (T_A = T_B). Hence, the ratio of their angular speeds is

⇒ \(\frac{\omega_{\mathrm{A}}}{\omega_{\mathrm{B}}}=\frac{\frac{2 \pi}{T_{\mathrm{A}}}}{\frac{2 \pi}{T_{\mathrm{B}}}}=\frac{T_{\mathrm{B}}}{T_{\mathrm{A}}}=1=1: 1 .\)

Question 34. A block of mass 10 kg is in contact with the inner surface of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the surface is 0.1. The minimum angular speed ω required for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis will be

1. √10 rad s^-1
2. \(\frac{10}{2 \pi} \mathrm{rad} \mathrm{s}^{-1}\)
3. 10 rad s^-1
4. 10π rad s^-1

Answer: 3. 10 rad s^-1

The forces acting on the block are its weight mg (vertically downward), the normal reaction \(\mathcal{N}=m \omega^2 r\) (towards the center), and the force of friction

⇒ \(f_{\max }=\mu \propto N\) vertically upward.

For the block to remain stationary,

⇒ \(f_{\max } \geq m g \Rightarrow \mu\left(m \omega^2 r\right) \geq m g\)

⇒ \(\omega_{\min }=\sqrt{\frac{g}{\mu r}}\)

⇒ \(\sqrt{\frac{10 \mathrm{~m} \mathrm{~s}^{-2}}{(0.1)(1 \mathrm{~m})}}\)

= 10 rad s^-1

Question 35, A smooth parabolic wire rotates about the vertical y-axis (expressed as y= 4cx²), as shown below. If a bead of mass m does not slip at (a,b), the value of to is

1. \(2 \sqrt{\frac{2 g c}{b}}\)
2. \(2 \sqrt{2 g c}\)
3. \(2 \sqrt{\frac{2 g c}{a b}}\)
4. \(\sqrt{\frac{g c}{2 a b}}\)

Answer: 4. \(\sqrt{\frac{g c}{2 a b}}\)

Let the bead stay in equilibrium at P. The forces acting on the bead are the weight (mg) and the normal reaction (N). The vertical component N cos θ balances the weight (mg), while the horizontal component N sin 0 provides the centripetal force \(\left(\frac{m v^2}{r}\right)\) for the circular motion.

Hence,

⇒ \(\propto N \sin \theta=\frac{m v^2}{r} \text { and } \alpha \cos \theta=m g\)

∴ \(\tan \theta=\frac{v^2}{r g}\)

From the figure, the slope at P is

⇒ \(\tan \theta=\frac{d y}{d x}=\frac{d}{d x}\left(4 c x^2\right)=8 c x=8 c r\)

Hence,

⇒ \(8 c r=\frac{\omega^2 r}{g} \Rightarrow \omega=\sqrt{8 c g}=2 \sqrt{2 c g}\)

Question 36. An insect is at the bottom of a hemispherical ditch of radius 1 m. It crawls up the ditch but starts slipping after it reaches a height h above the bottom. If the coefficient of friction between the ground and the insect is 0.75 then h is (assuming g = 10 m s^-2)

1. 0.60 m
2. 0.20 m
3. 0.80 m
4. 0.45 m

Answer: 4. 0.45 m

The insect starts slipping when

⇒ \(m g \sin \theta \geq f\)

or, \(m g \sin \theta \geq \mu m g \cos \theta\)

or, \(\tan \theta=\mu=\frac{3}{4}\)

or, \(\frac{P A}{O A}=\mu\)

or, \(\mu=\frac{R \sin \theta}{R-h} .\)

Substituting the values,

⇒ \(\frac{3}{4}=\frac{\frac{3}{5}}{1-h}\)

or, \(1-h=\frac{3}{5} \times \frac{4}{3}=\frac{4}{5}\)

or, \(h=\left(1-\frac{4}{5}\right)=\frac{1}{5} \mathrm{~m}\)

= 20cm

= 0.20m

Motion in a Straight Line

Question 1. If the velocity of a particle is v = At+Bt², where A and B are constants, the distance traveled by it between 1 s and 2 s is

1. \(\frac{3}{2} A+4 B\)
2. \(\frac{A}{2}+\frac{B}{3}\)
3. 3A + 7B
4. \(\frac{3}{2} A+\frac{7}{3} B\)

Answer: 4. \(\frac{3}{2} A+\frac{7}{3} B\)

Given that the velocity at the time t is

⇒ \(v=A t+B t^2 \text { or } \quad \frac{d x}{d t}=A t+B t^2\)

Integrating,

⇒ \(\int d x=\int\left(A t+B t^2\right) d t\)

or, \(x=\frac{A t^2}{2}+\frac{B t^3}{3}+c\)

At t=1s, \(x_1=\frac{A}{2}+\frac{B}{3}+c\)

and at t = 2s, x₂ =2A + \(\frac{8}{3}\)B + c.

∴ The distance covered in the given time interval is

⇒ \(x_2-x_1=\left(2 A+\frac{8}{3} B+c\right)-\left(\frac{A}{2}+\frac{B}{3}+c\right)=\frac{3}{2} A+\frac{7}{3} B\)

Question 2. The motion of a particle along a straight line is described by the equation x =8 +12t-t², where x is in meters and t is in seconds. The retardation of the particle when its velocity becomes zero is

1. 6ms^-2
2. zero
3. 12 m s^-2
4. 24 m s^-2

Answer: 3. 12 m s^-2

The equation representing the position (x) as a function of time (t) is

x = 8 +12t-t³.

Hence, the instantaneous velocity is

⇒ \(v=\frac{d x}{d t}=12-3 t^2\)

and the instantaneous acceleration

⇒ \(a=\frac{d v}{d t}=-6 t\)

When = 0, 12- 3t² = 0 and t = 2 s.

Hence, the magnitude of the acceleration at t = 2 s is

⇒ \(|a|_{t=2 s}=6(2) \mathrm{ms}^{-2}=12 \mathrm{~m} \mathrm{~s}^{-2} \text {. }\)

∴ retardation =12 ms^-2

Question 3. Preeti reached a metro station and found that the escalator was not working. She walked up the stationary escalator in a time tv On other days, if she remains stationary on the moving escalator, the escalator takes her up in a time t₂. The time taken by her to walk up the moving escalator is

1. t₁ – t₂
2. \(\frac{t_1 t_2}{t_1+t_2}\)
3. \(\frac{t_1 t_2}{t_2-t_1}\)
4. \(\frac{t_1+t_2}{2}\)

Answer: 2. \(\frac{t_1 t_2}{t_1+t_2}\)

Let the velocity of Preeti on the stationary escalator be u₀ and the length of the escalator be s. Hence,

⇒ \(u_0=\frac{s}{t_1}\) ….(1)

When the escalator is moving with a velocity u_e,

⇒ \(u_{\mathrm{e}}=\frac{s}{t_2}\) …(2)

Finally, Preeti’s resultant velocity relative to the ground when both are moving will be u₀ + u_e. So,

⇒ \(u_0+u_{\mathrm{e}}=\frac{s}{T}\) …(3)

Adding (1) and (2),

⇒ \(u_0+u_{\mathrm{e}}=s\left(\frac{1}{t_1}+\frac{1}{t_2}\right)\)

Equating (3) and (4),

⇒ \(\frac{s}{T}=s\left(\frac{t_1+t_2}{t_1 t_2}\right)\)

Hence, the time T needed for Preeti to walk up the moving escalator is

⇒ \(T=\frac{t_1 t_2}{t_1+t_2}\)

Question 4. The displacement x (in meters) of a particle of mass m (in kilograms) moving in one dimension under the action of a force is related to the time t (in seconds) by t = √x + 3. The displacement of the particle when its velocity is zero will be

1. 2 m
2. Zero
3. 4 m
4. 6 m

Answer: 2. Zero

Time (t) as a function of position (x) is t = √x +3 or √x =t- 3.

Squaring, x = t²- 6t + 9 = (t- 3)².

∴ instantaneous velocity, \(v=\frac{d x}{d t}=2 t-6\)

For 0 = 0, 2t = 6 or t = 3.

Hence, the position at t = 3 will be x =(t- 3)² = 0.

Question 5. Two cars P and Q start from a point at the same time in a straight line, and their positions are represented by \(x_{\mathrm{P}}(t)=a t+b t^2\) and \(x_{\mathrm{Q}}(t)=f t-t^2\) respectively. At what time do the cars have the velocity?

1. \(\frac{a+f}{2(1+b)}\)
2. \(\frac{a+f}{2(b-1)}\)
3. \(\frac{a-f}{1+b}\)
4. \(\frac{f-a}{2(1+b)}\)

Answer: 4. \(\frac{f-a}{2(1+b)}\)

For the car P, x_p(t) = at + bt², and for the car Q, x_Q(t) =ft- f².

∴ The velocity of P is

⇒ \(v_{\mathrm{P}}=\left(\frac{d x}{d t}\right)_{\mathrm{P}}=a+2 b t\)

and the velocity of Q is

⇒ \(v_{\mathrm{Q}}=\left(\frac{d x}{d t}\right)_{\mathrm{Q}}=f-2 t\)

At the time t, let P and Q have the same velocity. So,

V_p = V_Q

or, a + 2bt = t – 2t or 2t(b + 1) = f-a.

∴ \(t=\frac{f-a}{2(1+b)}\)

Question 6. A particle covers half its journey with a speed vl and the other half with a speed v₂. Its average speed during the complete journey is

1. \(\frac{v_1 v_2}{v_1+v_2}\)
2. \(\frac{2 v_1 v_2}{v_1+v_2}\)
3. \(\frac{v_1+v_2}{2}\)
4. \(\frac{v_1^2 v_2^2}{v_1^2+v_2^2}\)

Answer: 3. \(\frac{v_1+v_2}{2}\)

Let s be the total distance covered.

For the first half,

⇒ \(\frac{s}{2}=v_1 t_1 \Rightarrow t_1=\frac{s}{2 v_1}\)

Similarly, in the second half,

⇒ \(\frac{s}{2}=v_2 t_2 \Rightarrow t_2=\frac{s}{2 v_2}\)

Hence, the average velocity is

⇒ \(v_{\mathrm{av}}=\frac{\text { total displacement }}{\text { total time taken }}=\frac{s}{t_1+t_2}=\frac{s}{\frac{s}{2}\left(\frac{1}{v_1}+\frac{1}{v_2}\right)}=\frac{2 v_1 v_2}{v_1+v_2} \text {. }\)

Question 7. A particle starts moving from rest with a uniform acceleration of (4/3) m s^-2. The distance traveled by the particle during the third second of its motion is

1. 4 m
2. 6 m
3. \(\frac{10}{3} \mathrm{~m}\)
4. \(\frac{19}{3} m\)

Answer: 3. \(\frac{10}{3} \mathrm{~m}\)

The distance covered in the nth second is

⇒ \(s_n=u+\left(\frac{2 n-1}{2}\right) a\)

Given that u =0,n = 3, a = \(\frac{10}{3}\) ms^-2.

∴ The distance covered in the third second is

⇒ \(s_3=0+\left(\frac{2 \times 3-1}{2}\right) \frac{4}{3} \mathrm{~m}=\frac{10}{3} \mathrm{~m}\)

Question 8. A car moves from X to Y with a uniform speed v₁ and returns to X with a uniform speed v₂. The average speed during this round trip is

1. \(\sqrt{v_1 v_2}\)
2. \(\frac{v_1 v_2}{v_1+v_2}\)
3. \(\frac{v_1+v_2}{2}\)
4. \(\frac{2 v_1 v_2}{v_1+v_2}\)

Answer: 4. \(\frac{2 v_1 v_2}{v_1+v_2}\)

Average speed = \(\frac{\text { total distance covered }}{\text { total time taken }}\)

⇒ \(\frac{s+s}{\frac{s}{v_1}+\frac{s}{v_2}}=\frac{2 s}{s\left(\frac{1}{v_1}+\frac{1}{v_2}\right)}=\frac{2 v_1 v_2}{v_1+v_2}\)

Question 9. A particle moves along a straight line OX. At a time t (in seconds), the distance x (in meters) of the particle from O is given by \(x=40+12 t-t^3\). How long would the particle travel before coming to rest?

1. 16 m
2. 24 m
3. 40 m
4. 50 m

Answer: 1. 16 m

Given that x = 40+12t-t3.

∴ instantaneous velocity, v = \(\frac{d x}{d t}\) =12-3t².

For v = 0,

12-3t2 = 0 => t = 2s.

The distance covered by the particle before coining to rest will be

⇒ \(s=\int d s=\int_{t=0}^{t=2 s} v d t=\int_0^{2 s}\left(12-3 t^2\right) d t\)

⇒ \(\left[12 t-t^3\right]_0^2=\left(12 \times 2-2^3\right) \mathrm{m}\)

= 16 m.

Question 10. The position s (in meters) of a particle as a function of time t (in seconds) is expressed as \(s=3 t^3+7 t^2+14 t+8\) The value of the acceleration of the particle at time t =1 s is

1. 16 ms^-2
2. 32 ms^-2
3. 10 ms^-2
4. 30 ms^-2

Answer: 2. 32 ms^-2

Given that s = 3³ +7t² +14t +8.

Hence, velocity = \(v=\frac{d s}{d t}=9 t^2+14 t+14\)

and acceleration = \(a=\frac{d v}{d t}=18 t+14\)

The value of the acceleration t =1 s is

⇒ \(|a|_{t=1 \mathrm{~s}}=|18 \mathrm{t}+14|_{t=1 \mathrm{~s}}\)

=32 ms^-2.

Question 11. The position x of a particle varies with time t as \(x=a t^2-b t^3\). The time at which acceleration will be zero is equal to

1. \(\frac{a}{3 b}\)
2. \(\frac{3 a}{2 b}\)
3. \(\frac{a}{b}\)
4. Zero

Answer: 1. \(\frac{a}{3 b}\)

The position (x) of the given particle as a function of time (t) is

⇒ \(x=a t^2-b t^3\)

∴ instantaneous velocity = \(v=\frac{d x}{d t}=2 a t-3 b t^2\)

and instantaneous acceleration = \(a=\frac{d v}{d t}=2 a-6 b t\)

When A = 0, the required time is t = \(\frac{2 a}{6 b}=\frac{a}{3 b}\)

Question 12. The displacement * of a particle varies with the time t as \(x=a e^{-\alpha t}+b e^{\beta t}\), where a, b, a and p are positive constants. The velocity of the particle will

1. Drop to zero when a = p
2. Go on decreasing with time
3. Be independent of p
4. Go on increasing with time

Answer: 4. Go on increasing with time

Given that \(x=a e^{-\alpha t}+b e^{\beta t}\)

Hence, velocity = \(v=\frac{d x}{d t}=-a \alpha e^{-\alpha t}+b \beta e^{\beta t}\)

The velocity of the particle will increase with time, being an exponential function

Question 13. A car is moving along a straight road with a uniform acceleration. It passes two points P and Q separated by some distance with a velocity of 30 km h^-1 and 40 km h^-1 respectively. The velocity of the car while crossing the midpoint between P and Q will be

1. 33.5 km h^-1
2. \(20 \sqrt{2} \mathrm{~km} \mathrm{~h}^{-1}\)
3. \(25 \sqrt{2} \mathrm{~km} \mathrm{~h}^{-1}\)
4. 38 km h^-1

Answer: 3. \(25 \sqrt{2} \mathrm{~km} \mathrm{~h}^{-1}\)

Given that \(v_{\mathrm{P}}=30 \mathrm{~km} \mathrm{~h}^{-1} \text { and } v_{\mathrm{Q}}=40 \mathrm{~km} \mathrm{~h}^{-1}\)

Let the velocity at the midpoint O be v0 and the uniform acceleration
be a.

For the motion from P to Q,

⇒ \(v_{\mathrm{Q}}^2-v_{\mathrm{P}}^2=2 a L\)

or, \(40^2-30^2=2 a L\) ….(1)

For the motion from to Q,

⇒ \(v_{\mathrm{Q}}{ }^2-v_{\mathrm{O}}{ }^2=2 a\left(\frac{L}{2}\right)\)

or, \(40^2-v_{\mathrm{O}}^2=a L .\) ….(2)

Dividing (1) by (2),

⇒ \(\frac{40^2-30^2}{40^2-v_O^2}=2\)

Solving, we get \(v_0=25 \sqrt{2} \mathrm{kmh}^{-1}\).

Question 14. The acceleration of a particle is increasing linearly with time as bt. The particle starts from the origin with an initial velocity v₀. The distance traveled by the particle in time t will be

1. \(v_0 t+\frac{1}{3} b t^2\)
2. \(v_0 t+\frac{1}{6} b t^3\)
3. \(v_0 t+\frac{1}{2} b t^2\)
4. \(v_0 t+\frac{1}{3} b t^3\)

Answer: 2. \(v_0 t+\frac{1}{6} b t^3\)

Acceleration = a = bt = \(\frac{dv}{dt}\).

So,dv = bt dt.

Integrating,

⇒ \(\int^v d v=b \int^t t d t \Rightarrow v-v_0=\frac{b}{2} \cdot t^2 \Rightarrow v=\frac{d x}{d t}=v_0+\frac{b t^2}{2}\)

Integrating again,

⇒ \(x=\int v d t=\int_0^t\left(v_0+\frac{b t^2}{2}\right) d t=\left[v_0 t+\frac{b t^3}{6}\right]_0^t\)

∴ the distance traveled in the time this

⇒ \(x=v_0 t+\frac{b t^3}{6}\)

Question 15. A particle starts sliding down a smooth inclined plane. If sn is the distance traversed by it from a time t = (n-1) s to a time t = n s then the ratio \(s_n / s_{n+1}\) is

1. \(\frac{2 n-1}{2 n+1}\)
2. \(\frac{2 n+1}{2 n}\)
3. \(\frac{2 n}{2 n+1}\)
4. \(\frac{2 n+1}{2 n-1}\)

Answer: 1. \(\frac{2 n-1}{2 n+1}\)

Given that u = 0. So, the distance covered during the nth second is

⇒ \(s_n=u+(2 n-1) \frac{a}{2}=(2 n-1) \frac{a}{2}\)

and that during the (n + l)th second is

⇒ \(s_{n+1}=[2(n+1)-1] \frac{a}{2}\)

∴ \(\frac{S_n}{S_{n+1}}=\frac{2 n-1}{2 n+1}\)

Question 16. A particle is moving with a velocity \(\vec{v}=k(y \hat{i}+x \hat{j})\), where k is constant. The general equation for its path is

1. y² – x² + constant
2. y = x² + constant
3. y² = x + constant
4. xy = constant

Answer: 1. y² – x² + constant

Given that \(\vec{v}=k(y \hat{i}+x \hat{j})=v_x \hat{i}+v_y \hat{j}\)

∴ \(v_x=\frac{d x}{d t}=k y \text { and } v_y=\frac{d y}{d t}=k x\)

∴ \(\frac{v_y}{v_x}=\frac{x}{y}=\frac{d y / d t}{d x / d t}=\frac{d y}{d x}\)

∴ ydy = xdx.

Integrating,

⇒ \(\frac{y^2}{2}=\frac{x^2}{2}\) + constant

y² = x² + constant.

Question 17. A particle, located at x = 0 at a time t = 0, starts moving along the positive x-direction with a velocity v, which varies as v = a√x. The displacement of the particle varies with time as

1. t³
2. t²
3. t
4. t½

Answer: 2. t²

Instantaneous velocity = \(v=\frac{d x}{d t}=\alpha \sqrt{x}\)

∴ \(\frac{d x}{\sqrt{x}}=\alpha d t\)

Integrating,

⇒ \(2 \sqrt{x}=\alpha t \Rightarrow 4 x=\alpha^2 t^2 \Rightarrow x \propto t^2\).

Question 18. The velocity of a particle is \(v=v_0+g t+f t^2\). If its position at t= 0 gt is x = 0 then its displacement after a unit time (f = 1 s) is

1. \(v_0+2 g+3 f\)
2. \(v_0+\frac{g}{2}+\frac{f}{3}\)
3. \(v_0+g+f\)
4. \(v_0+\frac{g}{2}+f\)

Answer: 2. \(v_0+\frac{g}{2}+\frac{f}{3}\)

Given that \(v=\frac{d x}{d t}=v_0+g t+f^2\).

So, the displacement at a time t will be

⇒ \(x=\int d x=\int v d \dot{t}=\int\left(v_0+g t+f t^2\right) d t\)

⇒ ⇒ \(\left[v_0 t+g \frac{t^2}{2}+f \frac{t^3}{3}\right]_{t=0}^{t=1 \mathrm{~s}}\)

⇒ \(v_0+\frac{g}{2}+\frac{f}{3}\)

Question 19. The relation between the time t and the displacement x is t = ax² + bx, where a and b are constants. The acceleration is

1. -2abv²
2. 2bv³
3. -2av³
4. 2av²

Answer: 3. -2av³

Given that t = ax² + bx.

Differentiating with respect to time t,

⇒ \((2 a x+b) \frac{d x}{d t}=1\)

or v(2ax + b) =1

or \(v=\frac{1}{2 a x+b}\)

Differentiating again with respect to time t,

acceleration = \(\frac{d v}{d t}=\frac{-2 a v}{(2 a x+b)^2}=-(2 a v) v^2=-2 a v^3\)

Question 20, The speeds of two identical cars are u and 4u at a given instant. The ratio of their respective distances at which the two cars are stopped after that instant is

1. 1:1
2. 1:4
3. 1:8
4. 1:16

Answer: 4. 1:16

If d = stopping distance then

⇒ \(v^2=0=u^2-2 a d \text { or } d=\frac{u^2}{2 a}\)

∴ \(\frac{s_1}{s_2}=\frac{u_1^2}{u_2^2}\) [∵ a is the same for the two identical cars]

\(\frac{u^2}{(4 u)^2}\)

= 1:16.

Question 21. An object moving with a speed of 6.25 m s^-1 is decelerated at a rate given \(\frac{d v}{d t}=-2.5 \sqrt{v}\), where v is the instantaneous speed. The time taken by the object to come to rest would be

1. 2 s
2. 4 s
3. 8 s
4. 1 s

Answer: 1. 2 s

Given that \(\frac{d v}{d t}=-2.5 \sqrt{v}\)

∴ \(\frac{d v}{\sqrt{v}}=-2.5 d t\)

Integrating, 2√v = -2.5f + c.

At t = 0, v = 6.25ms^-1, soc = 5.

For the object to come to rest,

2V0 = -2.5t + 5

=> t = 2s

Question 22. A particle is moving eastwards with a velocity of 5 m s^-1. In 10 s the velocity changes to 5 m s^-1 northwards. The average acceleration during tins time is

1. \(\frac{1}{\sqrt{2}} \mathrm{~m} \mathrm{~s}^{-2} \text { towards } \mathrm{NE}\)
2. \(\frac{1}{2} \mathrm{~m} \mathrm{~s}^{-2} \text { towards north }\)
3. Zero
4. \(\frac{1}{\sqrt{2}} \mathrm{~m} \mathrm{~s}^{-2} \text { towards NW }\)

Answer: 4. \(\frac{1}{\sqrt{2}} \mathrm{~m} \mathrm{~s}^{-2} \text { towards NW }\)

The average acceleration is

⇒ \(\vec{a}_{\mathrm{av}}=\frac{\text { change in velocity }}{\text { time }}=\frac{\vec{v}_{\mathrm{f}}-\vec{v}_{\mathrm{i}}}{t}\)

⇒ \(\frac{\left(5 \mathrm{~ms}^{-1}\right) \hat{j}-\left(5 \mathrm{~ms}^{-1}\right) \hat{i}}{10 \mathrm{~s}}=\frac{1}{2}(\hat{j}-\hat{i}) \mathrm{ms}^{-2}\)

∴ \(\left|\vec{a}_{\mathrm{av}}\right|=\frac{\sqrt{2}}{2} \mathrm{~ms}^{-2}=\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2} \text { (towards NW)}\)

Question 23. The adjoining figure shows the variation of the acceleration of a particle with time. The particle is moving along a straight line and has a velocity v = 2 m s^-1 at f = 0. The velocity after 2 s will be

1. 2 ms^-1
2. 4 ms^-1
3. 6 ms^-1
4. 8 m s^-1

Answer: 3. 6 ms^-1

The area enclosed under the given-t graph is

⇒ \(A=\frac{1}{2}(2 \mathrm{~s})\left(4 \mathrm{~ms}^{-2}\right)=4 \mathrm{~ms}^{-1}\)

But this area = \(\int a d t=\int \frac{d v}{d t} d t=[v]_{v_{\mathrm{i}}}^{v_{\mathrm{i}}}\)

∴ 4 m s^-1 = v_f – v_i = v_f-2 m s^-1

∴ The velocity after 2 s is v_f = 6 m s^-1.

Question 24. The given figure shows the velocity-time graph of two particles A and B moving along the same straight line in the same direction. Which of the following statements is true?

1. The relative velocity is zero.
2. The relative velocity is nonzero but constant.
3. The relative velocity increases continuously.
4. The relative velocity decreases continuously.

Answer: 3. The relative velocity increases continuously.

The relative velocity (ur) and relative acceleration (ar) of two particles

are related by vr = ur + art.

Here, ur = 0 and at = (aA – aB) = constant.

∴ relative velocity = vT = (aA- aB)t.

∴ \(v_{\mathrm{r}} \propto t .\)

Hence, the relative velocity increases continuously with time.

Question 25. The position of a particle moving along a straight line is expressed as a function of time (t) as x = 6 + 12t- 2t², where x is in meters and t is in seconds. The distance covered by the particle in the first 5 s is

1. 32 m
2. 24 m
3. 20 m
4. 26 m

Answer: 4. 26 m

Given that x = 6 + 12t- 2t².

Hence, velocity = u = \(v=\frac{d x}{d t}=12-4 t\)

and acceleration = \(a=\frac{d v}{d t}=-4 \mathrm{~m} \mathrm{~s}^{-2}\)

Initial velocity = \(u=[12-4 t]_{t=0}=12 \mathrm{~m} \mathrm{~s}^{-1}\)

Due to the retardation of -4 ms^-2, the particle comes to rest when
u = 0 =12-4t or t =3s.

The distance covered in3s is

⇒ \(s_1=u t+\frac{1}{2} a t^2=(12 \times 3) \mathrm{m}+\frac{1}{2}(-4)(9) \mathrm{m}=18 \mathrm{~m}\)

The distance covered backward in 2s is

⇒ \(s_2=\frac{1}{2}(4)(2)^2 \mathrm{~m}=8 \mathrm{~m}\)

Hence, total distance = s = s1 + s2

= 18 m + 8 m

= 26 m.

Question 26. A particle starts moving along a straightlinesuchthattheacceleration varies with the displacement (s) as shown in the given figure. Which of the following represents the velocity-displacement graph?

Answer: 2.

The graph showing the variation of acceleration (a) with displacement (s) is linear. Hence, a = ks, where k- constant,

∴ \(a=\frac{d v}{d t}=\frac{d v}{d s} \frac{d s}{d t}=k s\)

or, \(v \frac{d v}{d s}=k s\) [∵ \(\frac{d s}{d t}=v\)]

or, vdv = ks ds.

Integrating,

⇒ \(\frac{v^2}{2}=\frac{k s^2}{2} \text { or } v \propto s\).

⇒ Question 27. Two particles moving along the same straight line and leaving the same point O at the same time have their initial velocities u and 2u and uniform accelerations 2a and a respectively. The distance of the two particles from O when one particle overtakes the other is

1. \(\frac{u^2}{a}\)
2. \(\frac{4 u^2}{a}\)
3. \(\frac{6 u^2}{a}\)
4. \(\frac{8 u^2}{a}\)

Answer: 3. \(\frac{6 u^2}{a}\)

When one particle overtakes the other at a time t, both travel the same distance s.

Hence, for the first particle,

⇒ \(s=u t+\frac{1}{2}(2 a) t^2\)

and for the second particle,

⇒ \(s=2 u t+\frac{1}{2} a t^2\)

Equating,

⇒ \(u t+a t^2=2 u t+\frac{a}{2} t^2 \Rightarrow \frac{a}{2} t=u \Rightarrow t=\frac{2 u}{a} .\)

Hence,

⇒ \(s=u t+\frac{1}{2}(2 a) t^2=u\left(\frac{2 u}{a}\right)+\frac{1}{2}(2 a)\left(\frac{2 u}{a}\right)^2=\frac{2 u^2}{a}+\frac{4 u^2}{a}=\frac{6 u^2}{a}\)

Question 28. A particle moves a distance x in a time t according to the equation x = {t + 5)^-1. The acceleration of the particle is proportional to

1. (velocity)^2/3
2. (distance)²
3. (distance)^-2
4. (velocity)^3/2

Answer: 1. (distance)²

The displacement (x) of the given particle at a time t is x = (t + 5)^-1.

∴ velocity = \(v=\frac{d x}{d t}=-(t+5)^{-2}\)

and acceleration = \(a=\frac{d v}{d t}=2(t+5)^{-3}\)

∴ \(a=2\left[(t+5)^{-1}\right]^3=2 x^3 \Rightarrow a \propto x^3\)

But

⇒ \(v^{3 / 2}=-(t+5)^{-3} \text {. Hence, } a \propto v^{3 / 2} \text {. }\)

Question 29. A particle starts moving from rest under the action of a constant force. If it covers a distance of sx in the first s and a distance s² in the first 20 s then

1. s₂ = s₁
2. s₂ = 4s₁
3. s₂ = 3s₁
4. s₂ = 2s₁

Answer: 2. s₂ = 4s₁

Initial velocity = u = 0.

The distance covered in the first s is

⇒ \(s_1=\frac{1}{2} a\left(10^2\right)=50 a,\)

and the distance covered in the first 20 s is

⇒ \(s_2=\frac{1}{2} a\left(20^2\right)=200 a=4(50 a)=4 s_1\)

∴ \(s_2=4 s_1\)

Question 30. A particle moves in a straight line with a constant acceleration. It changes its velocity from 10ms^-1 to 20ms^-1 while passing through a distance of 135 m in a time t. The value of t is

1. 10 s
2. 9 s
3. 12 s
4. 1.8 s

Answer: 2. 9 s

Given that, u = 10 m s^-1, v = 20m s^-1, and s = 135 m.

∴ acceleration = \(a=\frac{v^2-u^2}{2 s}=\frac{400-100}{2 \times 135} \mathrm{~ms}^{-2}=\frac{10}{9} \mathrm{~ms}^{-2} .\)

Now, v = u + at, so the required time is

⇒ \(t=\frac{v-u}{a}=\frac{20 \mathrm{~ms}^{-1}-10 \mathrm{~ms}^{-1}}{\frac{10}{9} \mathrm{~ms}^{-2}}=9 \mathrm{~s}\)

Question 31. The position x of a particle with respect to the time t along the x-axis is given by \(x=9 t^2-t^3\), where x is in meters and t is in seconds. What will be the position of the particle when it achieves the maximum speed in the positive

1. 54 m
2. 32m
3. 24 m
4. 81m

Answer: 1. 54 m

The position as a function of time (t) is x = 9t²-t³ and thus the velocity

is \(v=\frac{d x}{d t}=18 t-3 t^2\)

For v to be the maximum,

\(\frac{d v}{d t}\) = 0 or 18 — 6f = 0 or t = 3s.

The position at f = 3 s willbe

⇒ \(|x|_{t=3 \mathrm{~s}}=\left|9 t^2-t^3\right|_{t=3 \mathrm{~s}}\)

= 81m-27m

= 54m.

Question 32. A bus is moving at a speed of 10 ms^-1 on a straight road. A scooterist wishes to overtake the bus in 100s. If the bus is at a distance of 1 km from the scooterist, with what speed should the scooterist chase the bus?

1. 20 ms^-1
2. 10 ms^-1
3. 40 ms^-1
4. 25 ms^-1

Answer: 1. 20 ms^-1

Let the velocity of the scooter be v sc.

Given that the velocity of the bus = 10 m s^-1 and its initial (relative) separation between the two vehicles =1 km = 1000 m.

The acceleration values of both are zero. So, for the relative separation to reduce to zero,

⇒ \(s_{\mathrm{r}}=u_{\mathrm{r}} t \Rightarrow t=\frac{s_{\mathrm{r}}}{u_{\mathrm{r}}}=\frac{1000 \mathrm{~m}}{v_{\mathrm{sc}}-10 \mathrm{~m} \mathrm{~s}^{-1}}=100 \mathrm{~s}\)

⇒ \(v_{\mathrm{sc}}-10 \mathrm{~m} \mathrm{~s}^{-1}=10 \mathrm{~m} \mathrm{~s}^{-1} \Rightarrow v_{\mathrm{sc}}=20 \mathrm{~m} \mathrm{~s}^{-1}\)

Hence, the velocity of the scooter is v sc = 20 m s^-1.

Question 33. If a car at rest accelerates uniformly to a speed of 144 km h^-1 in 20 s, it covers a distance of

1. 2980 m
2. 1440 m
3. 400 m
4. 20 m

Answer: 3. 400 m

Given that u = 0, v = 144 km h^-1  = 144 x \(\frac{5}{18}\) m s^-1  = 40 m s^-1  and the time taken = 20 s.

∵ v = u + at,

∴ \(a=\frac{v}{t}=\frac{40 \mathrm{~m} \mathrm{~s}^{-1}}{20 \mathrm{~s}}=2 \mathrm{~m} \mathrm{~s}^{-2}\)

∴ The distance covered is

⇒ \(s=u t+\frac{1}{2} a t^2=0+\frac{1}{2}\left(2 \mathrm{~m} \mathrm{~s}^{-2}\right)(20 \mathrm{~s})^2\)

= 400m.

Question 34. A car accelerates from rest at a constant rate for some time, after which it decelerates at a constant rate of P and comes to rest. If the total time elapsed is t, the maximum velocity acquired by the car will be

1. \(\left(\frac{\alpha^2+\beta^2}{\alpha \beta}\right) t\)
2. \(\left(\frac{\alpha \beta}{\alpha+\beta}\right) t\)
3. \(\left(\frac{\alpha^2-\beta^2}{\alpha \beta}\right) t\)
4. \(\left(\frac{\alpha+\beta}{\alpha \beta}\right) t\)

Answer: 2. \(\left(\frac{\alpha \beta}{\alpha+\beta}\right) t\)

Let v max be the maximum velocity. While accelerating, sa

⇒ \(v_{\max }=\alpha t_1 \Rightarrow t_1=\frac{v_{\max }}{\alpha} .\)

Similarly, while decelerating to stop finally,

⇒ \(\frac{0-v_{\max }}{t_2}=-\beta \text { or } t_2=\frac{v_{\max }}{\beta}\)

Total time = \(t=t_1+t_2=v_{\max }\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\)

∴ \(v_{\max }=\left(\frac{\alpha \beta}{\alpha+\beta}\right)\)

Question 35. 11M? displacement 9 (in meters) of a particle moving along a straight line is expressed as a function of time t (in seconds) by the equation \(s=\left(t^3-6 t^2+3 t+4\right)\). When the acceleration is zero, the velocity will be

1. 42 m s^-1
2. -9 m s^-1
3. 5ms^-1
4. -12 m s^-1

Answer: 2. -9 m s^-1

Given that s = t³-6t² + 3t + 4.

Hence, velocity = \(y=\frac{d s}{d t}=3 t^2-12 t+3\)

and acceleration = a = \(\frac{d v}{d t}\)

= 6t-12 = 0.

So, t = 2 s.

The required velocity is

⇒ \(v=\left|3 t^2-12 t+3\right|_{t=2 \mathrm{~s}}\)

= (12-24+3) ms^-1

= -9 ms^-1.

Question 36. A bus travels the first one-third distance at a speed of 10 km h^-1 the next one-third distance at 20 km h^-1 and the last one-third distance at 60 km h^-1. The average speed of the bus is

1. 10 km h^-1
2. 16 km h^-1
3. 18 km h^-1
4. 40 km hr^-1

Answer: 3. 18 km h^-1

Let the total distance covered by s.

∴ \(T=t_1+t_2+t_3=\frac{s}{3 u_1}+\frac{s}{3 u_2}+\frac{s}{3 u_3}=\frac{s}{3}\left(\frac{1}{u_1}+\frac{1}{u_2}+\frac{1}{u_3}\right)\)

Hence, the total time of travel is

⇒ \(\frac{s}{3}\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{60}\right)=\frac{s \times 10}{3 \times 60}\)

∴ average speed \(=v_{\mathrm{av}}=\frac{\mathrm{s}}{T}=18 \mathrm{~km} \mathrm{~h}^{-1}\)

Question 37. A car covers a distance of 200 m. It covers the first half of the distance at a speed of 40 km h^-1 and the second half at a speed v. If the average speed of the car is 48 km h^-1, the value of vis

1. 48 km h^-1
2. 45 km h^-1
3. 60 km h^-1
4. 50 km h^-1

Answer: 3. 60 km h^-1

Since the motion of the car is uniform, its acceleration is zero. During the first half,

⇒ \(\frac{s}{2}=u_1 t_1 \quad \text { or } \quad t_1=\frac{s}{2 u_1}\)

and during the second half,

⇒ \(t_2=\frac{s}{2 u_2}\)

the total time of travel is

⇒ \(T=t_1+t_2=\frac{s}{2}\left(\frac{1}{u_1}+\frac{1}{u_2}\right)\)

the average speed is

⇒ \(v_{\mathrm{av}}=\frac{s}{T}=\frac{2}{\frac{1}{u_1}+\frac{1}{u_2}}\)

Substituting the values,

⇒ \(48=\frac{2}{\frac{1}{40}+\frac{1}{v}}\)

or, \(\frac{1}{40}+\frac{1}{v}=\frac{1}{24}\)

or, v = 60 km^-1

Question 38. A car starting from rest and moving with a constant acceleration covers a distance of s₁ in the fourth second and a distance of s₂ in the sixth second The ratio s₁/s₂ is

1. \(\frac{4}{9}\)
2. \(\frac{6}{11}\)
3. \(\frac{7}{11}\)
4. \(\frac{2}{3}\)

Answer: 3. \(\frac{7}{11}\)

Given that u = 0. So, in the fourth second,

⇒ \(s_4=u+\left(\frac{2 n-1}{2}\right) a=\frac{7}{2} a\)

and in the sixth second

⇒ \(s_6=\frac{11}{2} a .\)

∴ \(\frac{s_4}{s_6}=\frac{\frac{7}{2} a}{\frac{11}{2} a}=\frac{7}{11}\)

Question 39. The acceleration a of a body starting from rest varies with the time t according to the relation a = 3t + 4. The velocity of the body at the time t = 2s will be

1. 10 ms^-1
2. 12 ms^-1
3. 14 m s^-1
4. 16 m s^-1

Answer: 3. 14 m s^-1

Given that

⇒ \(a=\frac{d v}{d t}=3 t+4\)

∴ \(v=\int_0^t(3 t+4) d t=\frac{3}{2} t^2+4 t\)

Aat t=2s, \(v=\left[\frac{3}{2}\left(2^2\right)+4(2)\right] \mathrm{ms}^{-1}\)

= 143s ^-1.

Question 40. The displacement y meters) of a body varies with the time t Cm seconds) as \(y=-\frac{2}{3} t^2+16 t+2\) How long does the body take to come to rest? 

1. 8 s
2. 10 s
3. 12s
4. 16 s

Answer: 3. 12s

The position is given by

⇒ \(y=-\frac{2}{3} t^2+16 t+2\)

Comparing the equation with the standard equation of a uniformly accelerated motion,

⇒ \(s=u t+\frac{1}{2} a t^2\) we have

u = 16ms^-1

and \(\frac{a}{2}=-\frac{2}{3} \text { or } a=-\frac{4}{3} \mathrm{~ms}^{-2}\)

For the body to come to rest, v = 0.

∴ \(v=u+a t \Rightarrow 0=16 \mathrm{~m} \mathrm{~s}^{-1}+\left(-\frac{4}{3} \mathrm{~m} \mathrm{~s}^{-2}\right) t\)

t = 12s.

Question 41. The given figure represents the speed-time graph of a body moving along a straight line. How much distance does it cover during the last 10 seconds of its motion?

1. 40m
2. 100 m
3. 60 m
4. 120 m

Answer: 3. 60 m

The area under the v-t graph gives the displacement. So, the distance covered in the last 10 s will be the area of ΔABC, i.e.,

⇒ \(\frac{1}{2}(B C)(A B)\)

⇒ \(\frac{1}{2}(20 \mathrm{~s}-10 \mathrm{~s})\left(20 \mathrm{~m} \mathrm{~s}^{-1}\right)\)

= 100m.

Question 42. The velocity-time graph of a stone thrown vertically upwards with an initial velocity of 30 m s^-1 is shown in the given figure. The velocity in the upward direction is taken as positive. What is the maximum height to which the stone rises?

1. 30m
2. 45m
3. 80m
4. 90m

Answer: 2. 45m

Given that u = 30m s^-1.

Acceleration = slope of the line AB, so a = \(\frac{0-30 \mathrm{~ms}^{-1}}{3 \mathrm{~s}}=-10 \mathrm{~m} \mathrm{~s}^{-2}\)

At the maximum height, v = 0.

Hence, by \(v^2=u^2-2 g h\)

⇒ \(h=\frac{u^2}{2 g}=\frac{900}{20} \mathrm{~m}\)

= 45m

Question 43. The driver of train A moving at a speed of 30 m s^-1 sights another train B moving on the same track at a speed of 10 m s^-1 in the same direction. He immediately applies the brakes and achieves a uniform retardation of 2 m s^-2. To avoid a collision, what must be the minimum distance between the two trains?

1. 100 m
2. 120 m
3. 60m
4. 160 m

Answer: 1. 100 m

The relative speed of the train A with respect to the train B,

vAB = vA -vB

= (30- 10) m s^-1

= 20 m s^-1.

To avoid a collision, let the minimum separation between them be s.

This means that the relative speed must reduce to zero when the distance covered is s.

Thus,

⇒ \(v_{\mathrm{r}}^2-u_{\mathrm{r}}^2=2 a_{\mathrm{r}} s\)

⇒ \(0-\left(20 \mathrm{~m} \mathrm{~s}^{-1}\right)^2=2\left(-2 \mathrm{~m} \mathrm{~s}^{-2}\right) \mathrm{s}\)

s = 100m.

Question 44. Three particles P, Q, and R are projected from the top of a tower with the same speed u. P is thrown straight up, Q is thrown straight down and R is thrown horizontally. They hit the ground with speeds of v_p, v_Q, and v_R respectively. Then,

1. \(v_{\mathrm{P}}>v_{\mathrm{Q}}=v_{\mathrm{R}}\)
2. \(v_{\mathrm{P}}=v_{\mathrm{Q}}>v_{\mathrm{R}}\)
3. \(v_{\mathrm{P}}=v_{\mathrm{Q}}=v_{\mathrm{R}}\)
4. \(v_{\mathrm{P}}>v_{\mathrm{Q}}>v_{\mathrm{R}}\)

Answer: 3. \(v_{\mathrm{P}}=v_{\mathrm{Q}}>v_{\mathrm{R}}\)

The net vertical displacement for each of the three particles is the same (= h).

For P, work done by gravity is mph.

Hence, change in KE = \(\frac{1}{2} m_{\mathrm{P}}\left(v_{\mathrm{P}}^2-u^2\right) \Rightarrow \frac{1}{2} m_{\mathrm{P}}\left(v_{\mathrm{P}}^2-u^2\right)=m_{\mathrm{P}} g h\)

vp² =2gh+u².

Similarly, \(v_{\mathrm{O}}^2=2 g h+u^2 \text { and } v_{\mathrm{R}}^2=2 g h+u^2\)

∴ \(v_{\mathrm{P}}=v_{\mathrm{Q}}=v_{\mathrm{R}}\)

Question 45. A body dropped from a tower of height h covers a distance h/2inthe last second of its motion. The height of the tower is approximately (taking g = 10 ms^-1)

1. 50 m
2. 55 m
3. 58 m
4. 60 m

Answer: 3. 58 m

Let the total time of fall be t. So,

⇒ \(h=\frac{1}{2} g t^2\) ……(1)

The distance covered in the final second is h/2. So, for the upper half (h/2), time = t^-1.

∴ \(\frac{h}{2}=\frac{1}{2} g(t-1)^2\) …..(2)

Now, dividing (1) by (2),

⇒ \(2=\left(\frac{t}{t-1}\right)^2\)

Solving, we get

⇒ \(t=(2+\sqrt{2}) \mathrm{s}\)

the height of the tower is

⇒ \(h=\frac{1}{2}(10)(2+\sqrt{2})^2 m=58.3 m \approx 58 m\)

Question 46. A body is projected vertically upwards with a velocity u. It crosses three points A, B, and C in succession in its upward journey with velocities u/2, u/3, and u/4. The ratio AB/BC is

1. \(\frac{10}{7}\)
2. \(\frac{20}{7}\)
3. \(\frac{3}{2}\)
4. \(\frac{2}{1}\)

Answer: 2. \(\frac{20}{7}\)

For the motion along AB,

⇒ \(\left(\frac{u}{3}\right)^2-\left(\frac{u}{2}\right)^2=-2 g(A B) \Rightarrow A B=\frac{5 u^2}{72 g}\)

Similarly, for the motion along BC,

⇒ \(\left(\frac{u}{4}\right)^2-\left(\frac{u}{3}\right)^2=-2 g(B C)^{\prime} \Rightarrow B C=\frac{7 u^2}{288 g}\)

Hence,

⇒ \(\frac{A B}{B C}=\frac{5}{72} \times \frac{288}{7}=\frac{20}{7}\)

Question 47. From the top of a 40-m-high tower, a stone is projected vertically upwards with an initial velocity of 10 m s^-1. After how much time will the stone hit the ground? (Take g = 10 m s^-2.)

1. 4 s
2. 1 s
3. 2 s
4. 3 s

Answer: 1. 4 s

Let us take the point of rejection as the origin. Applying the coordinate sign convention, velocity of projection = u = +10 ms^-1 net displacement

= h = -40 m and g =-10 m s^-2.

Thus, \(-h=+u t-\frac{1}{2} g t^2\)

or, -40 = 10t -5t²

or t²-2t-8=0 or (t-4)(t + 2)=0.

The acceptable solution is = 4s.

Question 48. In the preceding question, what will be the speed of the stone when it hits the ground?

1. 20 ms^-1
2. 30 m s^-1
3. 35m sr^-1
4. 40 m s^-1

Answer: 3. 35m srl

Applying the standard equation v = u+gt, we have

-v =10 m s^-1-10 m s^-2 x 4s

= -30 m s^-1

v = 30ms^-1.

Question 49. A stone dropped from the top of a tower hits the ground after 4 s. How much time does it take to travel the first half of the distance from the top of the tower?

1. 1 s
2. 2√2 s
3. 2 s
4. √3 s

Answer: 2. 2√2 s

For the total flight of the stone,

⇒ \(h=\frac{1}{2} g t^2\) [∵ u=0]

⇒ \(h=\frac{1}{2} \times 10 \times 4^2=80\)

⇒ \(\frac{h}{2}=40=\frac{1}{2} \cdot(10) t^2=5 t^2\)

So, the required time

⇒ \(t=2 \sqrt{2} \mathrm{~s}\)

Question 50. A stone is projected vertically downwards with a velocity u from the top of a tower. It strikes the ground with a velocity of 3M. The time taken by the stone to reach the ground is

1. \(\frac{u}{g}\)
2. \(\frac{3 u}{g}\)
3. \(\frac{2 u}{g}\)
4. \(\frac{4 u}{g}\)

Answer: 3. \(\frac{2 u}{g}\)

For the downward motion,

v = u+gt

⇒ 3u = u +gt

⇒ \(t=\frac{2 u}{g}\)

Question 51. In the preceding question, what is the height of the tower?

1. \(\frac{u^2}{g}\)
2. \(\frac{2 u^2}{g}\)
3. \(\frac{3 u^2}{8}\)
4. \(\frac{4 u^2}{g}\)

Answer: 4. \(\frac{4 u^2}{g}\)

Let h be the height of the tower.

Hence,

⇒ \(h=u t+\frac{1}{2} g t^2=u\left(\frac{2 u}{g}\right)=\frac{1}{2} g\left(\frac{2 u}{g}\right)^2=\frac{4 u^2}{g}\)

Question 52. A stone falls freely under gravity. It covers the distances hv h₂ and h₃ in the first 5 s, the second 5 s, and the third 5 s respectively. The relation between h₁, h₂ and h₃ is given by

1. \(h_2=3 h_1 \text { and } h_3=2 h_2\)
2. \(h_1=h_2=h_3\)
3. \(h_1=2 h_2=3 h_3\)
4. \(h_1=\frac{h_2}{3}=\frac{h_3}{5}\)

Answer: 4. \(h_1=\frac{h_2}{3}=\frac{h_3}{5}\)

The stone is dropped from O.

So, \(O A=h_1=\frac{1}{2} g\left(5^2\right)=\frac{25 g}{2}\)

⇒ \(O B=h_1+h_2=\frac{1}{2} g\left(10^2\right)=\frac{100 g}{2}\)

and \(O C=h_1+h_2+h_3=\frac{1}{2} g\left(15^2\right)=\frac{225 g}{2}\)

∴ \(h_2 \doteq O B-O A=\frac{75 g}{2} \text { and }\)

⇒ \(h_3=O C-O B=\frac{125 g}{2}\)

Hence,

⇒ \(h_1: h_2: h_3=25: 75: 125=1: 3: 5\)

∴ \(h_1=\frac{h_2}{3}=\frac{h_3}{5}\)

Question 53. A ball is dropped from a high-rise platform at t = 0. After 6 s, another ball is thrown downwards from the same platform with a speed v. The two balls meet at t – 18 s. What is the value of 7 (Take g = 10 m s^-2.)

1. 40 ms^-1
2. 75ms^-1
3. 60 ms^-1
4. 55 ms^-1

Answer: 2. 75ms^-1

The two balls meet at time t=18s. Hence, for the first ball,

⇒ \(h=\frac{1}{2} g t^2=\frac{1}{2}\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(18 \mathrm{~s})^2\)

= 5 (18)² m.

For the second ball, time = (t-6) s when the two balls meet.

Hence,

h = v(t-6)+ \(\frac{1}{2}\)g(t- 6)²

= v(t8- 6) + 5(18- 6)²

=12v + 5 x 12².

Equating the two expressions for h, we have

12v + 5 x 144 = 5 x 18 x 18

⇒ \(v=\frac{5 \times 18^2}{12} \mathrm{~ms}^{-1}-\frac{5 \times 144}{12} \mathrm{~ms}^{-1}=75 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 54. Two bodies A (of mass 1 kg) and B (of mass 3 kg) are dropped from the heights of 16 m and 25 m respectively. The ratio of the time taken by them to reach the ground is

1. \(\frac{12}{5}\)
2. \(\frac{5}{4}\)
3. \(\frac{5}{12}\)
4. \(\frac{4}{5}\)

Answer: 3. \(\frac{5}{12}\)

Theoretically, the time of fall is independent of mass.

So, applying \(h=\frac{1}{2} g t^2\)

or, \(t=\sqrt{\frac{2 h}{g}}\), we obtain

⇒ \(\frac{t_{\mathrm{A}}}{t_{\mathrm{B}}}=\sqrt{\frac{h_{\mathrm{A}}}{h_{\mathrm{B}}}}=\sqrt{\frac{16 \mathrm{~m}}{25 \mathrm{~m}}}=\frac{4}{5}\)

Question 55. A ball is projected vertically upwards. It attains a speed of 10 m s^-1 when it reaches half its maximum height. How high does the ball rise? (Take g = 10 m s^-2.)

1. 5 m
2. 20 m
3. 10 m
4. 15 m

Answer: 3. 10 m

v² = u² – 2gh, where o = 0 at the maximum height (h) and u = 10 m s^-1.

∴ \(0=10^2-2 g\left(\frac{h}{2}\right)=100-10 h\)

Hence, h = 10 m.

Question 56. If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is

1. \(\frac{1}{2} g t^2\)
2. ut
3. \(u t-\frac{1}{2} g t^2\)
4. (u+gt)t

Answer: 1. \(\frac{1}{2} g t^2\)

Let A be the point at the maximum height, where the speed is reduced to zero. Let AB be the distance during the last f seconds. Hence, time of ascent = time of descent, where BA = AB = h

∴ \(h=u_{\mathrm{A}} t+\frac{1}{2} g t^2=0+\frac{1}{2} g t^2=\frac{1}{2} g t^2\)

Note This value of h is independent of u.

Question 57. A particle is thrown vertically upwards. Its velocity at half the maximum height is 10 m s^-1. The maximum height attained is (taking g = 10 m s^-2)

1. 8 m
2. 10 m
3. 16 m
4. 20 m

Answer: 2. 10 m

During the descent of the particle from the maximum height, the speed will be the same (= 10 m s^-1) at half the height. Hence,

⇒ \(v^2=u^2+2 g\left(\frac{h}{2}\right)(\text { from the top) }\)

⇒ \(\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right)^2=0^2+2\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right) \frac{h}{2}\)

⇒ \(h=\frac{100}{10} \mathrm{~m}=10 \mathrm{~m}\)

Question 58. A man throws balls with the same speed vertically upwards, one after another, at an interval of 2 s. What should be the speed of projection so that more than two balls are in the sky at any time? (Given that g = 9.8m s^-2.)

1. More than 19.6 m s^-1
2. At least 9.8 m s^-1
3. Any speed less than 19.6 m s^-1
4. 19.6 m s^-1 only

Answer: 1. More than 19.6 m s^-1

The time interval between the two balls thrown is 2 s. If more than two balls (or at least three balls) remain in the air, the time of flight of the first ball must be greater than 2 x 2 s = 4 s.

∴ \(T>4 \mathrm{~s} \Rightarrow \frac{2 u}{g}>4 \mathrm{~s} \Rightarrow u>2 g=19.6 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 59. A rubber ball is dropped from a height of 5 m on the ground. On bouncing, it rises to 1.8 m. The fractional loss in the velocity bouncing is (taking g – 10 m s^-2)

1. \(\frac{9}{25}\)
2. \(\frac{16}{25}\)
3. \(\frac{2}{5}\)
4. \(\frac{3}{5}\)

Answer: 3. \(\frac{2}{5}\)

The velocity of the ball when it hits the ground for the first time is

⇒ \(v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 5} \mathrm{~m} \mathrm{~s}^{-1}=10 \mathrm{~m} \mathrm{~s}^{-1}\)

The velocity of rebmmd is \(v^{\prime}=\sqrt{2 g \times 1.8} \mathrm{~m} \mathrm{~s}^{-1}=6 \mathrm{~m} \mathrm{~s}^{-1}\)

∴ the fractional loss in velocity is

⇒ \(\frac{\Delta v}{v}=\frac{v-v^{\prime}}{v}=\frac{10-6}{10}=\frac{2}{5}\)

Question 60. A body dropped from a height h strikes the ground with a velocity of 3 m s^-1. Another body of the same mass is projected from the same height h with an initial velocity of 4 m s^-1. The final velocity of the second body with which it strikes the ground is (assuming g = 10 m s^-2)

1. 3 ms^-1
2. 5ms^-1
3. 4 ms^-1
4. 12 m s^-1

Answer: 2. 5ms^-1

For the first body,

⇒ \(v^2=u^2+2 g h\)

or (3 ms^-1)² = 0² + 2gh = 2gh.

For the second body,

⇒ \(v^2=\left(4 \mathrm{~m} \mathrm{~s}^{-1}\right)^2+2 g h\)

or, \(2 g h=v^2-\left(4 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\)

Equating (1) and (2),

v²-(4ms^-1)² = (3ms^-1)²

v = 5m s^-1.

Question 61. Water drops fall at regular intervals from a tap 5 m above the ground. The third drop leaves the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant? (Take g = 10 m s^-2.)

1. 1.25 m
2. 4.00 m
3. 3.75 m
4. 2.50 m

Answer: 3. 3.75 m

Given that h = 5 m. For the time of flight t of the first water drop,

⇒ \(5 \mathrm{~m}=\frac{1}{2} g t^2 \Rightarrow t^2=1 \mathrm{~s}^2\) t = 1 s. Hence, the third drop leaves one second after the first drop.

So, the time gap between two successive drops is (1/2) s.

Now, the distance covered by the second drop after (1/2) s will be

⇒ \(\frac{1}{2} g t^2=5\left(\frac{1}{2}\right)^2 \mathrm{~s}=\frac{5}{4} \mathrm{~s}\)

Thus, the distance of the second drop above the ground will be

⇒ \(5 \mathrm{~m}-\frac{5}{4} \mathrm{~m}=\frac{15}{4} \mathrm{~m}\)

= 3.75 m.

Question 62. A body dropped from the top of a tower fell through 40 m during the last two seconds of its fall. The height of the tower is (assuming g = 10 m s^-2)

1. 45 m
2. 50 m
3. 80 m
4. 60 m

Answer: 1. 45 m

Let h = height of the tower and t = time of flight.

Now, \(h=\frac{1}{2} g t^2 \text { and } h-40=\frac{1}{2} g(t-4)^2\)

Subtracting,

⇒ \(40=\frac{g}{2}\left[t^2-(t-4)^2\right]\)

= 5(2t-4)(4)

=> 2t = 6s

=> t = 3s.

∴ \(h=\frac{1}{2} g t^2=5\left(3^2\right) \mathrm{m}\)

= 45 m

Question 63. What will be the ratio of the distances covered by a freely falling body from rest in the fourth and fifth seconds of its fall?

1. 1:1
2. 7:9
3. 16:25
4. 4:5

Answer: 2. 7:9

The distance covered in the nth second is given by

⇒ \(s_n=u+\left(\frac{2 n-1}{2}\right) a\)

Hence, for the fourth second,

⇒ \(s_4=\frac{7 g}{2}\)

and for the fifth second,

⇒ \(s_5=\frac{9 g}{2}\) [∵ u = 0]

∴ \(\frac{s_4}{s_5}=\frac{7}{9} \quad \text { or } \quad s_4: s_5=7: 9\)

Question 64. A balloon is rising vertically upwards with a velocity of 10 m s^-1, When it is at a height of 45 m above the ground, a parachutist bails out from it. After 3 s, he opens the parachute and decelerates at a constant rate of Sms^-2. What was the height of the parachute above the ground when he opened the parachute? (Take g = 10 m s^-2.)

1. 30 m
2. 15 m
3. 60 m
4. 45 m

Answer: 1. 30 m

When the parachutist bails out, he has the velocity of the balloon, which is 10 m s^-1 upwards.

∴ the net displacement in 3 s is

⇒ \(s=u t+\frac{1}{2}\left(-g t^2\right)=10 \times 3-\frac{10}{2}(3)^2\)

= -15 m.

Here the negative sign indicates a downward displacement.

Hence, the height above the ground when he opens the parachute is

= 45m-15m

= 30m.

Question 65. In the previous problem, how far was the parachutist from the balloon at t^-3 s?

1. 15 m
2. 30 m
3. 45 m
4. 60 m

Answer: 3. 45 m

The balloon rises up in 3 s by a height of (10 m s^-1)(3 s) = 30 m. Hence, the parachutist is now at a distance of 30 m + 15 m = 45 m from the balloon.

Question 66. In the preceding problem, how much time does the parachutist take to hit the ground after his exit from the balloon?

1. 4 s
2. 5 s
3. 6 s
4. 7 s

Answer: 2. 5 s

The total time the parachutist takes after his exit from the balloon to hit the ground is 3 s + 2 s = 5 s.

Question 67. From the top of a tower, a stone is thrown up and it reaches the ground in a time tx. A second stone is thrown down with the same speed and it reaches the ground in a time t². A third stone is released from rest and it reaches the ground in a time t³. The relationship of t, t², and t³ is given by

1. \(\frac{1}{t_3}=\frac{1}{t_1}+\frac{1}{t_2}\)
2. \(t_3=\sqrt{t_1 t_2}\)
3. \(t_3=\frac{1}{2}\left(t_1+t_2\right)\)
4. \(t_3{ }^2=t_1{ }^2-t_2{ }^2\)

Answer: 2. \(t_3=\sqrt{t_1 t_2}\)

Taking the coordinate sign convention into consideration, we have

for the firststone, \(-h=u t_1-\frac{1}{2} g t_1^2\) …(1)

for the second stone, \(h=u t_2+\frac{1}{2} g t_2^2\) ….(2)

and for the third stone, \(h=\frac{1}{2} g t_3{ }^2\)…(3)

In order to eliminate u, multiplying (1) by t2 and (2) by itself and then subtracting, we obtain

⇒ \(-h\left(t_2+t_1\right)=-\frac{1}{2} g\left(t_1+t_2\right) t_1 t_2 \Rightarrow h=\frac{g}{2}\left(t_1 t_2\right)\)

But from (3) \(h=\frac{1}{2} g t_3{ }^2\)

∴ \(t_3{ }^2=t_1 t_2 \Rightarrow t_3=\sqrt{t_1 t_2}\)

Question 68. Water drops fall at regular intervals from a tap in the roof. At an instant when a drop is about to leave the tap, the separation between three successive drops is in the ratio

1. 1:2:3
2. 1:4:9
3. 1:3:5
4. 1:5:13

Answer: 3. 1:3:5

Let the time interval between two successive drops be f.

For the first drop (being detached), h₁=0.

For the second drop, \(h_2=\frac{1}{2} g t^2\)

For the third drop, \(h_3=\frac{1}{2} g(2 t)^2=\frac{g}{2}\left(4 t^2\right)\)

For the fourth drop, \(h_4=\frac{1}{2} g(3 t)^2=\frac{g}{2}\left(9 t^2\right)\)

Hence, the separations between two successive drops are

⇒ \(h_2-h_1=\frac{1}{2} g t^2, h_3-h_2=\frac{1}{2} g\left(3 t^2\right) \text { and } h_4-h_3=\frac{1}{2} g\left(5 t^2\right)\)

∴ The required ratio is \(\frac{1}{2} g t^2: \frac{1}{2} g\left(3 t^2\right): \frac{1}{2} g\left(5 t^2\right)\)

= 1:3:5.

Question 69. A balloon starts rising up from the ground with an acceleration of 1.25 m s^-2. After 8 s, a stone is released from the balloon. The stone will

1. Cover a distance of 40 m
2. Have a displacement of 50 m
3. Reach the ground in 4 s
4. Begin to move down after being released

Answer: 3. Reach the ground in 4 s

The balloon starts moving upwards with an acceleration of 1.25 m s^-2.
Hence, its height at t = 8s will be

⇒ \(h=\frac{1}{2} a t^2=\frac{1}{2}\left(1.25 \mathrm{~m} \mathrm{~s}^{-2}\right)(8 \mathrm{~s})^2=40 \mathrm{~m}\)

and its velocity at that instant will be

⇒ \(n=a t=\left(1.25 \mathrm{~m} \mathrm{~s}^{-2}\right)(8 \mathrm{~s})=10 \mathrm{~m} \mathrm{~s}^{-1}\)

At this instant, the stone shares the motion of the balloon and falls

freely under gravity for which

⇒ \(u=10 \mathrm{~m} \mathrm{~s}^{-1}, h=-40 \mathrm{~m} \text { and } g=-10 \mathrm{~m} \mathrm{~s}^{-2}\)

∵ \(h=u t+\frac{1}{2} g t^2 \Rightarrow-40=10 t-\frac{1}{2}(10) t^2\)

⇒ \(5 t^2-10 t-40=0 \Rightarrow t^2-2 t-8=0\)

(t- 4)(t + 2) = 0

=> t = 4 s.

Thus, the stone strikes the ground in 4 s.

Question 70. A body is thrown vertically upwards- Which of the following graphs correctly represents the variation of its velocity against time?

Answer: 4.

The slope of the v-t graph is the acceleration (g). At t = 0, the velocity is maximum and positive.

It goes on decreasing (remaining positive), and at the highest point (A), v 0.

Further, it becomes negative and increases with the same acceleration.

Hence, the correct graph is given in (d).

Question 71. A parachutist after bailing out falls 50 m without any friction. When the parachute opens, it decelerates at 2 m s^-2. He reaches the ground at a speed of 3 m s^-1. At what height has he bailed out?

1. 91m
2. 182 m
3. 293 m
4. 111 m

Answer: 3. 293 m

When the parachutist falls freely through a height of 50 m, let his
speed be u.

So, u² = 2gh

= 2(9.8 m s^-2)(50 m)

= 980 (m s^-1)².

When he opens the parachute, he falls through a height of h, where

32- 980 + 2(-2)h.

∴ \(h=\frac{980-9}{4} \mathrm{~m}=242.75 \mathrm{~m} \approx 243 \mathrm{~m}\)

Hence, the height at bailing out is H = 50 m + 243 m

= 293 m.

Question 72. Two identical balls are thrown upwards with the same initial velocity of 40 m s^-1 in the same vertical direction at an interval of 2 s. The balls collide at a height of (taking g = 10 m s^-2)

1. 210 m
2. 125 m
3. 75 m
4. 40 m

Answer: 3. 75 m

Given that u = 40 m s^-1.

Let the height at which the balls collide be h.

For the first ball, \(h=u t-\frac{1}{2} g t^2\)

and for the second ball, \(h=u(t-2)-\frac{1}{2} g(t-2)^2\)

Equating the expressions for h, we obtain

⇒ \(u(t-2)-\frac{1}{2} g(t-2)^2=u t-\frac{1}{2} g t^2\)

Simplifying, we get

2u = 2g(t^-1).

Substituting the values,

2(40 m s^-1) = (20 ms^-1)(t – Is).

∴ The time of collision is t- 5 s.

Hence, the balls collide at a height of

⇒\(h=u t-\frac{1}{2} g t^2=(40 \times 5-5 \times 25) \mathrm{m}\)

= 75m.

Question 73. A ball is projected vertically upwards from the ground. It experiences a constant air resistance of 2 m s^-2 directed opposite to the direction of its motion. The ratio of the time of ascent to the time of descent equals (assuming g = 10 m s^-2)

1. 1
2. \(\frac{2}{3}\)
3. \(\sqrt{\frac{4}{3}}\)
4. \(\sqrt{\frac{2}{3}}\)

Answer: 4. \(\sqrt{\frac{2}{3}}\)

During the ascent,

⇒ \(v=0=u-(g+a) t_1 \Rightarrow t_1=\frac{u}{g+a}\)… (1)

During the descent,

⇒ \(h=0\left(t_2\right)+\frac{1}{2}(g-a) t_2{ }^2=\frac{1}{2}(g-a) t_2{ }^2 .\)

For the upward motion,

⇒ \(h=u t_1-\frac{1}{2}(g+a) t_1^2\)

Substituting for t₁ from (1),

⇒ \(h=\frac{u^2}{g+a}-\frac{1}{2}(g+a) \cdot \frac{u^2}{(g+a)^2}=\frac{u^2}{2(g+a)}\)

Equating the expressions for h,

⇒ \(\frac{1}{2}(g-a) t_2^2=\frac{u^2}{2(g+a)} \Rightarrow t_2=\frac{u}{\sqrt{g^2-a^2}}\)

Hence,

⇒ \(\frac{t_1}{t_2}=\frac{u}{g+a} \cdot \frac{\sqrt{g^2-a^2}}{u}=\sqrt{\frac{g-a}{g+a}}=\sqrt{\frac{10-2}{10+2}}=\sqrt{\frac{2}{3}}\)

Question 74. A ball is projected vertically upwards from the foot of a tower. It crosses the top of the tower twice after a time interval of 8s and strikes the ground after 16 seconds. The height of the tower is (taking g = 10 m s^-2)

1. 140 m
2. 240 m
3. 100 m
4. 200 m

Answer: 2. 240 m

The velocity of the ball while crossing the top of the tower during its
descent is

⇒ \(v=0+g t=10 \mathrm{~m} \mathrm{~s}^{-2} \times \frac{8}{2} \mathrm{~s}=40 \mathrm{~m} \mathrm{~s}^{-1}\)

The time taken to cover the height of the tower is \(\frac{1}{2}(16 s-8 s)\) = 4s.

∴ height of the tower is

⇒ \(h=u t+\frac{1}{2} g t^2=\left(40 \mathrm{~m} \mathrm{~s}^{-1}\right)(4 \mathrm{~s})+\frac{1}{2}\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(4 \mathrm{~s})^2\)

= 160 m + 80 m

= 240 m.

Question 75. When an object is shot from the bottom of a long and smooth inclined plane kept at an angle of 60° with the horizontal, it can travel a distance xx along the plane. But when the inclination is decreased to 30° and the same object is shot with the same velocity, it can travel a distance of x₂. The ratio of x₁ to x₂ will be

1. 1:√2
2. √2:1
3. 1:√3
4. 1:2√3

Answer: 3. 1:√3

In Case I, work done by gravity

⇒ \(-m g h_1=-m g x_1 \sin 60^{\circ}\)

and change in KE = \(-\frac{1}{2} m v^2\)

Hence, \(m g x_1\left(\frac{\sqrt{3}}{2}\right)=\frac{1}{2} m v^2\)

Similarly, in Case 2

⇒ \(m g x_2 \sin 30^{\circ}=\frac{1}{2} m v^2 .\)

∴ \(m g x_1\left(\frac{\sqrt{3}}{2}\right)=m g x_2\left(\frac{1}{2}\right)\)

⇒ \(\frac{x_1}{x_2}=\frac{1}{\sqrt{3}} \Rightarrow x_1: x_2=1: \sqrt{3}\)

Question 76. The speed of a swimmer in still water is 20 m s^-1. The speed of the water of a river, flowing due east, is 10 m s^-1. If the swimmer is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes relative to the north is given by

1. 30° west
2. 0°
3. 60° west
4. 45° west

Answer: 1. 30° west

Given that the velocity of the swimmer relative to the river is v st = 20 m s^-1 and the velocity of the river relative to the ground is urg = 10 m s^-1. These are shown vectorially in the adjoining diagram.

According to the vector diagram,

⇒ \(\vec{v}_{\mathrm{sg}}=\vec{v}_{\mathrm{sr}}+\vec{v}_{\mathrm{rg}}\)

and \(\sin \theta=\frac{v_{\mathrm{rg}}}{v_{\mathrm{sr}}}=\frac{10 \mathrm{~ms}^{-1}}{20 \mathrm{~ms}^{-1}}=\frac{1}{2} \Rightarrow \theta=30^{\circ}\)

Hence, the direction of the strokes is 30° due west.

Question 77. A particle is moving along a circular path with a constant speed of 10 m s^-2. What is the magnitude of the change in velocity of the particle when it moves through an angle of 60° around the center of the circular path?

1. 10√2 m s^-2
2. 1.m s^-1
3. 10√3 m s^-1
4. Zero

Answer: 2. 1.m s^-1

The initial velocity at A is

⇒ \(\vec{u}=\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right) \hat{i}\)

and the final velocity at B is

⇒ \(\vec{v}=\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right)\left(\cos 60^{\circ}\right) \hat{i}+\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right)\left(\sin 60^{\circ}\right) \hat{j}\)

⇒ \(\left(5 \mathrm{~m} \mathrm{~s}^{-1}\right) \hat{i}+\left(5 \sqrt{3} \mathrm{~m} \mathrm{~s}^{-1}\right) \hat{j}\)

∴ the change in velocity is

⇒ \(\Delta \vec{v}=\vec{v}-\vec{u}=(5-10) \mathrm{m} \mathrm{s}^{-1} \cdot \hat{i}+5 \sqrt{3} \mathrm{~m} \mathrm{~s}^{-1} \cdot \hat{j}\)

Hence, the magnitude of change in velocity is

⇒ \(|\Delta \vec{v}|=|-5 \hat{i}+5 \sqrt{3} \hat{j}| \mathrm{m} \mathrm{s}^{-1}\)

⇒ \(\sqrt{25+75} \mathrm{~m} \mathrm{~s}^{-1}=10 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 78. The position (x) of a particle as a function of time (t) is given by x(t) = at + bt² -ct³, where a, b, and c are constants. When the particle attains zero acceleration, its velocity will be

1. \(a+\frac{b^2}{2 c}\)
2. \(a+\frac{b^2}{c}\)
3. \(a+\frac{b^2}{3 c}\)
4. \(a+\frac{b^2}{4 c}\)

Answer: 3. \(a+\frac{b^2}{3 c}\)

Given that x = at + bt²- ct³.

∴ velocity = \(v=\frac{d x}{d t}=a+2 b t-3 c t^2\) …..(1)

and acceleration = \(a=\frac{d v}{d t}=2 b-6 c t\) …..(2)

For a = 0, 2b-6cf = 0

⇒ \(t=\frac{2 b}{6 c}=\frac{b}{3 c}\)

From (1), the velocity at this time t is

⇒ \(v=a+2 b\left(\frac{b}{3 c}\right)-3 c\left(\frac{b}{3 c}\right)^2=a+\frac{2 b^2}{3 c}-\frac{b^2}{3 c}=a+\frac{b^2}{3 c}\)

Question 79. A helicopter is rising vertically upwards from the ground with an acceleration g starting from rest. When it reaches a height of h, a packet is dropped at t = 0. Find the time t when the packet strikes the ground.

1. \(\sqrt[4]{\frac{2 g}{h}}\)
2. \(\sqrt{\frac{2 h}{8}}\)
3. \(2 \sqrt{\frac{2 h}{g}}\)
4. \((1+\sqrt{2}) \sqrt{\frac{2 h}{g}}\)

Answer: 4. \((1+\sqrt{2}) \sqrt{\frac{2 h}{g}}\)

Let the velocity of the helicopter at tire height h be v.

∴ \(v^2=2 g h \Rightarrow v=\sqrt{2 g h} \text { (upward). }\)

For free fall of the packet from the height h,

⇒ \(-h=u t-\frac{1}{2} g t^2=\sqrt{2 g h} t-\frac{1}{2} g t^2\)

⇒ \(\left(\frac{g}{2}\right) t^2-\sqrt{2 g h} t-h=0\)

Solving for t, we get

⇒ \(t=\frac{1}{g}[\sqrt{2 g h}+\sqrt{2 g h+2 g h}]=\frac{(2+\sqrt{2}) \sqrt{g h}}{g}\)

⇒ \((1+\sqrt{2}) \sqrt{\frac{2 h}{g}}\)

Units Dimensions and Errors

Question 1. The physical quantity having the same dimension as that of length that can be formed out of c, G and e²/4πε₀ (where c is the speed of light, G is the universal constant of gravitation and e is the electric charge) is:

1. \(e^2\left(G \cdot \frac{e^2}{4 \pi \varepsilon_0}\right)^{1 / 2}\)
2. \(\frac{1}{c^2}\left(\frac{e^2}{G \cdot 4 \pi \varepsilon_0}\right)^{1 / 2}\)
3. \(\frac{1}{c} \cdot G \cdot \frac{e^2}{4 \pi \varepsilon_0}\)
4. \(\frac{1}{c^2}\left(G \cdot \frac{e^2}{4 \pi \varepsilon_0}\right)^{1 / 2}\)

Answer: 4. \(\frac{1}{c^2}\left(G \cdot \frac{e^2}{4 \pi \varepsilon_0}\right)^{1 / 2}\)

F = \(F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{e^2}{d^2}\). Hence, the dimension of \(\frac{e^2}{4 \pi \varepsilon_0}\) is

⇒ \(\left[F d^2\right]=\mathrm{ML}^3 \mathrm{~T}^{-2},[G]=\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\)

and [c] = LT^-1.

Let length l = \(=\left(\frac{e^2}{4 \pi \varepsilon_0}\right)^p G^q c^r\)

Writing the dimensions of both sides,

⇒ \(\mathrm{M}^0 \mathrm{LT}^0=\left(\mathrm{ML}^3 \mathrm{~T}^{-2}\right)^p\left(\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right)^q\left(\mathrm{LT}^{-1}\right)^r\)

Comparing both sides and solving,

⇒ \(p=\frac{1}{2}, q=\frac{1}{2} \text { and } r=-2\)

∴ \(l=\left(\frac{e^2}{4 \pi \varepsilon_0}\right)^{1 / 2} G^{1 / 2} c^{-2}=\frac{1}{c^2}\left(\frac{G e^2}{4 \pi \varepsilon_0}\right)^{1 / 2}\)

Question 2. If force (F), velocity (v) and time (t) be chosen as the fundamental quantities, the dimensional formula for mass is:

1. \(\left[F v t^{-1}\right]\)
2. \(\left[F v t^{-2}\right]\)
3. \(\left[F v^{-1} t^{-1}\right]\)
4. \(\left[F v^{-1} t\right]\)

Answer: 4. \(\left[F v^{-1} t\right]\)

Let \(m \propto F^a v^b t^c\).

∴ mass \(m=k P^a v^b t^c\), where k is a dimensionless constant and a, b and c are the exponents of powers.

∴ \(\mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0=\left(\mathrm{MLT}^{-2}\right)^a\left(\mathrm{LT}^{-1}\right)^b \mathrm{~T}^c\)

⇒ \(\mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0=\mathrm{M}^a \mathrm{~L}^{a+b} \mathrm{~T}^{-2 a-b+c}\)

Equating the exponents, a = l,a + b = 0 and-2a-b +c = 0.

Solving, we get a =1, b=-1 and c =1.

∴ [m] =[Fv^-1t].

Question 3. The Planck constant (h), the speed of light in vacuum and Newton’s gravitational constant (G) are three fundamental constants. Which of the following combinations of these has the same dimension as that of length?:

1. \(\frac{\sqrt{h G}}{c^{3 / 2}}\)
2. \(\frac{\sqrt{h G}}{c^{5 / 2}}\)
3. \(\sqrt{\frac{h c}{G}}\)
4. \(\sqrt{\frac{G c}{h^{3 / 2}}}\)

Answer: 1. \(\frac{\sqrt{h G}}{c^{3 / 2}}\)

Given that \(l \propto h^a c^b G^c\), Hence,

⇒ \(\mathrm{M}^0 \mathrm{LT}^0=\left(\mathrm{ML}^2 \mathrm{~T}^{-1}\right)^a\left(\mathrm{LT}^{-1}\right)^b\left(\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right)^c\)

⇒ \(\mathrm{M}^0 \mathrm{LT}^0=\left(\mathrm{M}^{a-c}\right)\left(\mathrm{L}^{2 a+b+3 c}\right)\left(\mathrm{T}^{-a-b-2 c}\right)\).

Equating the exponents, a-c = 0, 2a + b + 3c =1 and-a-b-2c = 0.

Solving, we have \(a=\frac{1}{2}, b=-\frac{3}{2} \text { and } c=\frac{1}{2}\).

Thus, \(l=\frac{\sqrt{h G}}{c^{3 / 2}}\)

Question 4. If energy (E), velocity (v)and time (t) are chosen as the fundamental quantities, the dimensional formula for surface tension will be:

1. \(\left[E v^{-2} t^{-2}\right]\)
2. \(\left[E^{-2} v^{-1} t^{-3}\right]\)
3. \(\left[E v^{-2} t^{-1}\right]\)
4. \(\left[E v^{-1} t^{-2}\right]\)

Answer: 1. \(\left[E v^{-2} t^{-2}\right]\)

Let the surface tension be S = \(k E^a v^b t^c\)

∴ \(\mathrm{ML}^0 \mathrm{~T}^{-2}=\left(\mathrm{ML}^2 \mathrm{~T}^{-2}\right)^a\left(\mathrm{LT}^{-1}\right)^b(\mathrm{~T})^c=\mathrm{M}^a \mathrm{~L}^{2 a+b} \mathrm{~T}^{-2 a-b+c}\)

Equating the exponents, we get a =1, 2a + b = 0 and -2a-b + c =-2.

Solving, a =1, b =-2 and c =-2.

Thus, \([S]=\left[E v^{-2} t^{-2}\right]\).

Question 5. Among the following, the pair of physical quantities having the same dimension is:

1. Impulse and surface tension
2. Angular momentum and work
3. Work and torque
4. The Young modulus and energy

Answer: 3. Work and torque

Torque = \(\vec{\tau}=\vec{r} \times \vec{F} \text { and work }=\vec{F} \cdot \vec{s}\).

Both have the dimension of force x length.

Let us check by finding the dimension of each individual term below:

⇒ \(\left.\begin{array}{l}
\text { [impulse }]=[\text { force }][\text { time }]=\mathrm{MLT}^{-1}, \\
\text { [surface tension }]=\frac{[\text { force }]}{[\text { length }]}=\mathrm{MT}^{-2},
\end{array}\right\} \text { different }\)

[angular momentum] = \([\vec{r} \times \vec{p}]=\mathrm{ML}^2 \mathrm{~T}^{-1}\) , different

⇒ \(\left.\begin{array}{l}
{[\text { work }]=[\vec{F}] \cdot[\vec{s}]=\mathrm{ML}^2 \mathrm{~T}^{-2},} \\
\text { [torque] }]=[\vec{r} \times \vec{F}]=\mathrm{ML}^2 \mathrm{~T}^{-2},
\end{array}\right\} \text { identical }\)

⇒ \(\left.\begin{array}{l}
\text { [energy] }=[\text { work }]=\mathrm{ML}^2 \mathrm{~T}^{-2}, \\
{[\text { Young modulus }]=\frac{\text { [stress] }}{[\text { strain] }}=\left[\frac{F}{A}\right]=\mathrm{ML}^{-1} \mathrm{~T}^{-2},}
\end{array}\right\} \text { different }\)

Question 6. The density of a material in the c.g.s. system of units is 4 g cm-3. In a system of units in which the unit of length is 10 cm and the unit of mass is 100 g, the numerical value of the density of the material will be:

1. 0.04
2. 0.4
3. 40
4. 400

Answer: 3. 40

∴ \(n_1 \mathrm{u}_1=n_2 \mathrm{u}_2\)

∴ \(4 \frac{\mathrm{g}}{\mathrm{cm}^3}=n_2 \frac{100 \mathrm{~g}}{(10 \mathrm{~cm})^3} \Rightarrow n_2=40\)

Question 7. If the dimension of the critical velocity vc of a liquid flowing through a horizontal tube is expressed as \(\left[\eta^x \rho^y r^z\right]\), where n = viscosity coefficient of the liquid, p = density of the liquid and r= radius of the tube then the values of x, y and z are given respectively by:

1. 1,1 and 1
2. 1,-1 and -1
3. -1, -1 and -1
4. -1,1 and -1

Answer: 3. -1, -1 and -1

Given that \(\left[v_{\mathrm{c}}\right]=\left[\eta^x \rho^y r^z\right]\)

Expressing in dimensions,

⇒ \(\mathrm{M}^0 \mathrm{LT}^{-1}=\left(\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right)^x\left(\mathrm{ML}^{-3}\right)^y \mathrm{~L}^z=\mathrm{M}^{x+y} \mathrm{~L}^{-x-3 y+z} \mathrm{~T}^{-x}\)

Equating the exponents from both sides, x + y = 0, -x- 3y + 2 = 1
and -x =-1.

Solving, we get x = 1, y =-1 and z = -1.

Question 8. The dimension of \(\frac{1}{2} \varepsilon_0 E^2\) where e0 = permittivity of free space and E = electric field, is:

1. \(\mathrm{ML}^2 \mathrm{~T}^{-2}\)
2. \(\mathrm{ML}^2 \mathrm{~T}^{-1}\)
3. \(\mathrm{ML}^{-1} \mathrm{~T}^{-2}\)
4. \(\mathrm{ML}^{-1} \mathrm{~T}^{-1}\)

Answer: 3. \(\mathrm{ML}^{-1} \mathrm{~T}^{-2}\)

The energy density (= energy/volume) in an electric field is given by

⇒ \(U_E=\frac{1}{2} \varepsilon_0 E^2\)

∴ \(\left[U_E\right]=\left[\frac{1}{2} \varepsilon_0 E^2\right]=\frac{\text { [energy] }}{\text { [volume] }}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^3}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}\)

Question 9. The damping force on a harmonic oscillator is directly proportional to its velocity. The SI unit of the constant of proportionality is:

1. kg ms^-1
2. kg ms^-2
3. kgs^-1
4. kg s

Answer: 3. kgs^-1

Given that F α v or F = kv, where k is the constant of proportionality.

∴ \(k=\frac{F}{v} \equiv \frac{\mathrm{N}}{\mathrm{ms}^{-1}}=\frac{\mathrm{kg} \mathrm{m} \mathrm{s}^{-2}}{\mathrm{~m} \mathrm{~s}^{-1}}=\mathrm{kg} \mathrm{s}^{-1}\)

Question 10. If the dimensional formula of a physical quantity is \(M^a L^b T^c\) then the physical quantity will be:

1. Velocity if a =1, b- 0, c = -1
2. Acceleration if a = 1, h = 1, c = -2
3. Force if a = 0, b = -1, c = -2
4. Pressure if a = 1, b = -1, c = -2

Answer: 4. Pressure if a = 1, b = -1, c = -2

Velocity and acceleration do not have mass in their dimensions (a ≠ 1). But force has a mass in its dimension (a =1). Thus, the only option left is pressure, for which

⇒ \([p]=\frac{[\text { force }]}{\text { [area] }}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}=\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}=\mathrm{M}^a \mathrm{~L}^b \mathrm{~T}^c\)

∴ a = 1, b = -1 and c = -2.

Question 11. The dimensional formula of \(\left(\mu_0 \varepsilon_0\right)^{-1 / 2}\) is:

1. L^1/2T^1/2
2. L^-1T
3. LT^-1
4. L^1/2T^1/2

Answer: 4. L^1/2T^1/2

The speed of light in free space is given by

⇒ \(c_0=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=\left(\mu_0 \varepsilon_0\right)^{-1 / 2} \Rightarrow\left[\left(\mu_0 \varepsilon_0\right)^{-1 / 2}\right]=\left[c_0\right]=\mathrm{LT}^{-1}\)

Question 12. What is the dimensional formula of electrical resistance in terms of mass (M), length (L), time (T) and electric current (I)?:

1. \(\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{I}^{-2}\)
2. \(\mathrm{ML}^2 \mathrm{~T}^{-1} \mathrm{I}^{-1}\)
3. \(M L^2 T^{-2} I^0\)
4. \(\mathrm{ML}^2 \mathrm{~T}^{-1} \mathrm{I}\)

Answer: 1. \(\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{I}^{-2}\)

According to Ohm’s law

⇒ \(I=\frac{V}{R} \Rightarrow[R]=\frac{[V]}{[I]}\)

The dimension of potential difference is

⇒ \([V]=\frac{[W]}{[q]}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{IT}}\)

∴ The dimension of resistance is

⇒ \([R]=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2} / \mathrm{IT}}{\mathrm{I}}=\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{I}^{-2}\)

Question 13. If R = resistance and C = capacitance then the dimension of RC is the:

1. Square of time
2. Square of the inverse of time
3. Same as time
4. Inverse of time

Answer: 4. Inverse of time

During the discharge of a charged capacitor of capacitance C through a resistor of resistance R, we have

⇒ \(Q=Q_0 e^{-t / R C}\)

The exponent of an exponential (or any number) is dimensionless. Hence, [RC] = [time] = T.

Question 14. Which of the following physical quantities have the same dimension?:

1. \(\frac{L}{R}\)
2. \(\frac{C}{L}\)
3. LC
4. \(\frac{R}{L}\)

Answer: 1. \(\frac{L}{R}\)

During the decay of current in an LR circuit, the instantaneous current

⇒ \(I=I_0 e^{-t /(L / R)}\)

The exponent of e is dimensionless and has no unit.

Hence,

⇒ \(\left[\frac{L}{R}\right]=[t]=\mathrm{T}\)

Question 15. Newton’sformulafor viscous force acting between two liquid layers of area A and velocity gradient Av/Az is given by \(F=\eta A \Delta v / \Delta z\), where \(\eta\) is the coefficient of viscosity of the liquid. The dimensional formula for t) is:

1. \(\mathrm{ML}^2 \mathrm{~T}^{-2}\)
2. \(\mathrm{ML}^{-1} \mathrm{~T}^{-1}\)
3. \(\mathrm{ML}^{-2} \mathrm{~T}^{-2}\)
4. \(\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\)

Answer: 2. \(\mathrm{ML}^{-1} \mathrm{~T}^{-1}\)

Given that F = \(F=-\eta A \frac{\Delta v}{\Delta z}\)

∴ \([\eta]=\frac{[F]}{[A]\left[\frac{\Delta v}{\Delta z}\right]}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2 \cdot \frac{\mathrm{LT}^{-1}}{\mathrm{~L}}}=\mathrm{ML}^{-1} \mathrm{~T}^{-1}\)

Hence, the dimensional formula for the viscosity coefficient n is ML^-1T^-1.

Question 16. The time-dependence of a physical quantity Q is given by \(Q=Q_0 \mathrm{e}^{-a t^2}\), where a is a constant and t is the time. The constant a:

1. Is dimensionless
2. Has the dimension T^-2
3. Has the dimension T²
4. Has the same dimension as that of Q

Answer: 2. Has the dimension T^-2

Given that \(Q=Q_0 e^{-a t^2}\).

The exponent at 2 is dimensionless.

∴ \([a]=\left[\frac{1}{t^2}\right]=\frac{1}{\mathrm{~T}^2}=\mathrm{T}^{-2}\)

Question 17. Among the following physical quantities, the dimensional formula of which is different from that of the remaining three?:

1. Energy per unit volume
2. Force per unit area
3. Product of voltage and charge per unit volume
4. Angular momentum

Answer: 4. Angular momentum

Let us find the dimensions below:

⇒ \(\frac{\text { [energy] }}{\text { [volume] }}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^3}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}\),

⇒ \(\frac{[\text { force] }}{\text { [area] }}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}\),

⇒ \(\frac{\text { [voltage } \times \text { charge] }}{\text { [volume] }}=\left[\frac{\frac{\text { work }}{\text { charge }} \times \text { charge }}{\text { volume }}\right]=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^3}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}\)

and [angular momentum] = ML²T^-1.

Hence, the dimension of angular momentum is different from the dimensions of the remaining three quantities given.

Question 18. Let P represent the radiation pressure, c represent the speed of light and S represents the radiant energy striking per unit area per unit time. The values of the nonzero integers x, y, and z such that \(P^x S^y c^z\) is dimensionless are given by:

1. x = 1, y=1, z = 1
2. x = 1, y = -1, 2 =1
3. x =-1, y = 1, z = 1
4. x = 1, y = 1, z = -1

Answer: 3. x = 1, y = 1, z = -1

Given that \(P^x S^y c^z\) is dimensionless.

Hence, \(\left[P^x S^y c^z\right]=\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\)

Now, \([P]=\frac{[\text { force] }}{\text { [area] }}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}\)

⇒ \([S]=\frac{\text { [energy] }}{\text { [area][time] }}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^2 \mathrm{~T}}=\mathrm{MT}^{-3}\)

and (3) = LT^-1

Substituting these dimensions in the given expression, we obtain

⇒ \(M^0 L^0 T^0=\left(M^{-1} T^{-2}\right)^x\left(M^{-3}\right)^y\left(L^{-1}\right)^z=M^{x+y} L^{-x+z} T^{-2 x-3 y-z}\)

Equating the exponents, x + y =0,-x+z=0and-2x-3y- z = 0.

Solving, we get x =1, y=-1 and z=1

Question 19. The dimensional formula of impulse is equal to that of:

1. Force
2. Linear Momentum
3. Pressure
4. Angular Momentum

Answer: 2. Linear Momentum

From the impulse-momentum theorem,

impulse = change in linear momentum.

Hence, impulse and linear momentum have the same dimensional formula.

Alternative method:

[Impulse] = [force][time] = (MLT^-2)(T) = MLT^-1

and [linear momentum] = [mass][velocity] = MLT^-1.

∴ [impulse] = [linear momentum].

So, both have the same dimension.

Question 20. In the equation \(p+\frac{a}{V^2}=\frac{b \Theta}{V}\) = pressure, V = volume and \(\Theta\)= absolute temperature. The dimensional formula of the constant:

1. ML⁵T²
2. ML⁵T
3. ML^-5T^-1
4. M^-1L⁵T²

Answer: 1. ML⁵T²

Given \(p+\frac{a}{V^2}=\frac{b \Theta}{V}\)

Dimensionally \(\frac{a}{V^2}\) and pare equal.

Since both are added, we have

⇒ \(\left[\frac{a}{V^2}\right]=[p] \Rightarrow[a]=[p]\left[V^2\right]=\left(\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right)\left(\mathrm{L}^6\right)=\mathrm{ML}^5 \mathrm{~T}^{-2}\)

Question 21. The velocity of a particle at a time t is given by v = at +, where a, b and c are constants. The dimensions of, b and c are respectively:

1. L², T and LT^-2
2. L, LT and LT^-2
3. LT^-2, LT and L
4. LT^-2, L and T

Answer: 4. LT^-2,L and T

Given that velocity = \(v=a t+\frac{b}{t+c}\)

Since is added to the time t, [c] =T.

∴ \([v]=\frac{[b]}{[t+c]} \Rightarrow \mathrm{LT}^{-1}=\frac{[b]}{\mathrm{T}} \Rightarrow[b]=\mathrm{L}\)

Similarly, \([v]=[a][t] \Rightarrow[a]=\frac{[v]}{[t]}=\mathrm{LT}^{-2}\)

Hence, the dimensions of, band care respectively LT-2, L and T

Question 22. The ratio of the dimension of the Planck constant to that of the moment of inertia is equal to the dimension of:

1. Angular momentum
2. Velocity
3. Time
4. Frequency

Answer: 4. Frequency

⇒ \(\frac{[\text { Planck constant }]}{[\text { moment of inertia] }}=\frac{[h]}{[I]}=\frac{\text { [angular momentum] }}{[I]}\)

⇒ \(\frac{[I \omega]}{[I]}=[\omega]=\mathrm{T}^{-1}=[\text { frequency }]\)

Question 23. The dimension of the Planck constant is equal to that of:

1. Energy
2. Linear momentum
3. Power
4. Angular momentum

Answer: 4. Angular momentum

⇒ \(E=h v \Rightarrow[h]=\frac{[E]}{[v]}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~T}^{-1}}=\mathrm{ML}^2 \mathrm{~T}^{-1}\)

Now, [angular momentum] = \(=[I][\omega]=\left(\mathrm{ML}^2\right)\left(\mathrm{T}^{-1}\right)=\mathrm{ML}^2 \mathrm{~T}^{-1}\)

∴ [Planck constant] = [angular momentum]

Question 24. Which of the following pairs of physical quantities do not have the same dimension?:

1. Force and impulse
2. Energy and torque
3. Angular momentum and the Planck constant
4. The Young modulus and pressure

Answer: 1. Force and impulse

[Force] = MLT^-2 and [impulse] = [F][f] = MLT^-1.

Hence, force and impulse do not have the same dimension.

Question 25. Which two of the following five physical parameters have the same dimension?:

1. Energy density

2. Refractive index

3. Dielectric constant

4. Young modulus

5. Magnetic field

1. (1)and(4)
2. (1) and (5)
3. (2)and(4)
4. (3) and (5)

Answer: 1. (1)and(4)

Let us find the dimensions of the given physical quantities below:

[energy density] = \(\frac{\text { [energy] }}{\text { [volume] }}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^3}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}\)

and [Young modulus] = \(\frac{[\text { stress }]}{[\text { strain }]}=\frac{[F / A]}{[\Delta L / L]}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}\)

The refractive index and dielectric constant are dimensionless
constants.

Also, \([B]=\frac{[F]}{[I][L]}=\mathrm{MT}^{-2} \mathrm{I}^{-1}\)

Hence, only energy density and the Yoimg modulus are dimensionally equivalent.

Question 26. The dimensions of the universal gravitational constant (G) are:

1. \(\mathrm{ML}^2 \mathrm{~T}^{-1}\)
2. \(M^{-1} L^3 T^{-2}\)
3. \(\mathrm{M}^{-2} \mathrm{~L}^3 \mathrm{~T}^{-1}\)
4. \(M^{-1} L^2 T^{-3}\)

Answer: 2. \(M^{-1} L^3 T^{-2}\)

Gravitational force \(F=\frac{G m_1 m_2}{r^2}\)

∴ \([G]=\frac{[F]\left[r^2\right]}{\left[m_1 m_2\right]}=\frac{\left(\mathrm{MLT}^{-2}\right)\left(\mathrm{L}^2\right)}{\mathrm{M}^2}=\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\)

Question 27. The dimensional formula of magnetic flux is:

1. \(\mathrm{ML}^0 \mathrm{~T}^{-2} \mathrm{I}^{-2}\)
2. \(\mathrm{M}^0 \mathrm{~L}^{-2} \mathrm{~T}^{-2} \mathrm{I}^{-2}\)
3. \(M L^2 T^{-1} I^3\)
4. \(M L^2 T^{-2} I^{-1}\)

Answer: 4. \(M L^2 T^{-2} I^{-1}\)

[Magnetic flux] = \([\Phi]=[B][A]=\frac{[F]}{[I][l]} \cdot[A]=\frac{\mathrm{MLT}^{-2}}{\mathrm{IL}} \cdot \mathrm{L}^2=\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{I}^{-1}\)

Question 28. The dimensional formula of the permeability of free space (PO) is:

1. \(\mathrm{MLT}^{-2} \mathrm{I}^{-2}\)
2. \(\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-1} \mathrm{I}^2\)
3. \(\mathrm{M}^0 \mathrm{LT}^{-1} \mathrm{I}^{-1}\)
4. \(\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{I}^{-2}\)

Answer: 1. \(\mathrm{MLT}^{-2} \mathrm{I}^{-2}\)

The force between two straight currents is

⇒ \(F=\frac{\mu_0 I_1 I_2 l}{2 \pi d}\)

∴ \(\mu_0=\frac{F(2 \pi d)}{I_1 I_2 l} \Rightarrow\left[\mu_0\right]=\frac{[F][d]}{\left[I_1 I_2\right][l]}=\frac{\mathrm{MLT}^{-2} \mathrm{~L}}{\mathrm{I}^2 \mathrm{~L}}=\mathrm{MLT}^{-2} \mathrm{I}^{-2}\)

Question 29. The dimensional formula of self-inductance is:

1. \(\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{I}^{-2}\)
2. \(\mathrm{MLT}^{-2} \mathrm{I}^{-2}\)
3. \(\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{I}^{-1}\)
4. \(M L^2 T^{-1} I^{-2}\)

Answer: 2. \(\mathrm{MLT}^{-2} \mathrm{I}^{-2}\)

The magnetic energy stored in an inductor is given by

⇒ \(U=\frac{1}{2} L I^2\)

∴ \([L]=\frac{\text { [energy] }}{\left[I^2\right]}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{I}^2}=\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{I}^{-2}\)

Question 30. If h and e respectively represent the Planck constant and electronic charge then the die dimension of h/e is the same as that of:

1. Magnetic field
2. Magnetic flux
3. Electric field
4. Electric flux

Answer: 3. Electric field

⇒ \(\left[\frac{h}{e}\right]=\frac{\mathrm{ML}^2 \mathrm{~T}^{-1}}{\mathrm{IT}}=\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{I}^{-1}\)

Now, the dimension of the magnetic field is [B] = MT^-2I^-1.

[magnetic flux] = [B][area] = ML²T^-2I^-1

Question 31. If E and B respectively represent the electric field and magnetic field then the ratio E/B has the dimension of:

1. Displacement
2. Velocity
3. Acceleration
4. Angular momentum

Answer: 2. Velocity

The force \(\vec{F}\) on a charged particle having a charge q moving with a velocity \(\vec{v}\) through a region containing both \(\vec{E}\) and \(\vec{B}\)is given by

⇒ \(\vec{F}=q(\vec{E}+\vec{v} \times \vec{B})\)

Hence, the dimension of \(\vec{E}\) will be the same as that of the product

∴ \(\left[\frac{E}{B}\right]=[v]\)

Question 32. If L, R, C and V respectively represent inductance, resistance, capacitance and potential difference, the dimension of L/RCV is the same as that of:

1. Current
2. \(\frac{1}{\text { current }}\)
3. Charge
4. \(\frac{1}{\text { charge }}\)

Answer: 2. \(\frac{1}{\text { current }}\)

RC has the dimension of time. The potential difference V has the dimension of e.m.f. linked with L as

\([V]=\left[L \frac{d I}{d t}\right] \Rightarrow \frac{[L]}{[V]}=\frac{1}{\left[\frac{d I}{d t}\right]}=\left[\frac{d t}{d I}\right]=\frac{\mathrm{T}}{\mathrm{I}}\)

Hence, \(\frac{[L]}{[R C V]}=\frac{1}{[R C]} \cdot\left[\frac{L}{V}\right]=\frac{1}{\mathrm{~T}} \cdot \frac{\mathrm{T}}{\mathrm{I}}=\mathrm{I}^{-1}=\frac{1}{\text { [current] }}\)

Question 33. If e, e0, h and c respectively represent the electronic charge, permittivity of free space, Planck constant and speed of light, \(e^2 / \varepsilon_0 h c\) has the dimension of:

1. Current
2. Pressure
3. Angular momentum
4. Angle

Answer: 4. Angle

⇒ \(F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2} \Rightarrow[F]=\frac{(I T)^2}{\left[\varepsilon_0\right]\left[r^2\right]}\)

∴ \(\frac{1}{\left[\varepsilon_0\right]}=\frac{\left(\mathrm{MLT}^{-2}\right) \mathrm{L}^2}{\mathrm{I}^2 \mathrm{~T}^2}=\mathrm{ML}^3 \mathrm{~T}^{-4} \mathrm{~T}^{-2}\)

⇒ \(\left[\varepsilon_0\right]=\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{I}^2\)

Also, \([h]=\mathrm{ML}^2 \mathrm{~T}^{-1} \text { and }[c]=\mathrm{LT}^{-1}\)

∴ \(\frac{\left[e^2\right]}{\left[\varepsilon_0 h c\right]}=\frac{\mathrm{I}^2 \mathrm{~T}^2}{\left(\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{I}^2\right)\left(\mathrm{ML}^2 \mathrm{~T}^{-1}\right)\left(\mathrm{LT}^{-1}\right)}\)

⇒ \(\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{I}^0\) (dimensionless) = [angle].

Question 34. When a wave propagates in a medium, the displacement y of a particle located at x at a time t is given by y = a sin(bt- cx), where &,b,c are the constants of the wave. The dimensions of a, b and c are respectively:

1. L, T and T^-1
2. L^-1, T and T^-1
3. L, T^-1 and L^-1
4. L², T and L^-1

Answer: 3. L, T^-1 and L^-1

Given that displacement = y = a sin(bt- cx).

Trigonometrical functions are dimensionless, so [y] = [a] = L.

Here, bt and cx are angles, so [b] = T-l and [c] = L^-1.

Hence, [a] = L, [b] =T-1 and [c] = L-1

Question 35. What is the dimensional formula of specific latent heat?:

1. \(\mathrm{ML}^2 \mathrm{~T}^{-2}\)
2. \(\mathrm{ML}^{-2} \mathrm{~T}^{-2}\)
3. \(\mathbf{M}^0 \mathrm{LT}^{-2}\)
4. \(\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}\)

Answer: 4. \(\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}\)

According to calorimetry,

⇒ \(H=m L \Rightarrow[L]=\frac{[H]}{[m]}=\frac{[\text { heat energy] }}{[\text { mass }]}=\frac{\mathbf{M L}^2 \mathrm{~T}^{-2}}{\mathbf{M}}=\mathbf{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}\)

Question 36. The dimensional formula of specific heat capacity is:

1. \(\mathrm{MLT}^{-2} \Theta^{-1}\)
2. \(\mathrm{ML}^2 \mathrm{~T}^{-2} \Theta^{-1}\)
3. \(M^0 L^2 T^{-2} \Theta^{-1}\)
4. \(\mathrm{M}^0 \mathrm{LT}^{-2} \Theta^{-1}\)

Answer: 3. \(M^0 L^2 T^{-2} \Theta^{-1}\)

According to the principles of calorimetry,

heat = (mass)(specific heat capacity)(change in temperature)

⇒ \(H=m c(\Delta \theta) \Rightarrow c=\frac{H}{m(\Delta \theta)}\)

Thus, \([c]=\frac{[H]}{[m][\Delta \theta]}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{M} \Theta}=\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2} \Theta^{-1}\)

Question 37. The van der Waals equation for n moles of a real gas is \(\left(p+\frac{a}{V^2}\right)(V-b)=n R T\), where p = pressure, V = volume, T = absolute temperature, R = gas constant, and a, b are van der Waals constants. The dimensional formula of ab is:

1. ML²T^-2
2. ML⁶T^-2
3. ML⁴T^-2
4. ML⁴T^-2

Answer: 4. ML⁸T^-2

Dimension of p = dimension of \(\frac{a}{V^2}\)

∴ \([a]=[p]\left[V^2\right]=\left(\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right)\left(\mathrm{L}^6\right)=\mathrm{ML}^5 \mathrm{~T}^{-2}\)

Also, [b] = [V] = L³.

∴ the dimensional formula for the product is

⇒ \([a b]=[a][b]=\left(\mathrm{ML}^5 \mathrm{~T}^{-2}\right)\left(\mathrm{L}^3\right)=\mathrm{ML}^8 \mathrm{~T}^{-2}\)

Question 38. The frequency f of vibrations of a uniformly stretched string of length f under a tension of F is given by \(f=\frac{p}{2 l} \sqrt{\frac{F}{\mu}}\), where p is I the number of loops in the vibrating string and p is a constant of the string. The dimension of p is:

1. ML^-1T^-1
2. ML^-1T⁰
3. ML^-1T⁰
4. ML²T^-1

Answer: 2. ML^-1T⁰

Squaring the given equation, we get

⇒ \(\mu=\frac{p^2 F}{4 l^2 f^2}\)

Here, p is a number, so it is dimensionless; F = tension, so [F] = MLT-2;

l = length, so [l²] = L²; f = frequency, so [f] = T^-1.

Substituting the dimensions, we have

⇒ \([\mu]=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2 \mathrm{~T}^{-2}}=\mathrm{ML}^{-1} \mathrm{~T}^0\)

Question 39. A student measured the diameter of a small steel ball using a screw gauge of least count 0,001 cm. The main-scale reading Is 5 mm, and the zero mark of the circular-scale division coincides with 25 divisions above the reference line. If the screw gauge has a zero error of -0.004 cm, the correct diameter of the ball Is:

1. 0.521mm
2. 0.529 mm
3. 0.525 cm
4. 0,053 cm

Answer: 2. 0.529 mm

The diameter of the ball is

D = main-scale reading + (circular-scale reading x LC)- zero error

= 0.5 cm + (25 x 0.001 cm)- (-0.004 cm)

= (0.5 + 0.025 + 0.004) cm

= 0.529 cm.

Question 40. The period of oscillations of a simple pendulum is \(T=2 \pi \sqrt{\frac{L}{g}}\). The measured value of L is 20.0 cm (known to 1 mm accuracy) and the time for 100 oscillations of the pendulum is found to be 90 s using a wristwatch of 1 s resolution. The accuracy in the determination of g is nearly:

1. 3%
2. 1%
3. 5%
4. 2%

Answer: 1. 3%

⇒ \(T=2 \pi \sqrt{\frac{l}{g}} \text { and } g=4 \pi^2\left(\frac{l}{T^2}\right)\)

∴ \(\frac{\Delta g}{g}=\frac{\Delta l}{l}+2\left(\frac{\Delta T}{T}\right)=\frac{0.1}{20}+2\left(\frac{1}{90}\right)=0.027\)

∴ accuracy = \(\frac{\Delta g}{g} \times 100 \%=2.7 \% \approx 3 \%\)

Question 41. The errors in the measurement of mass and speed are 2% and 3% respectively. The error in the estimation of the kinetic energy obtained by measuring the mass and speed will be:

1. 2%
2. 8%
3. 10%
4. 12%

Answer: 2. 8%

Kinetic energy Ek = \(\frac{1}{2} m v^2\)

the error in the measurement of KE is

⇒ \(\frac{\Delta E_{\mathrm{k}}}{E_{\mathrm{k}}}=\frac{\Delta m}{m}+2 \frac{\Delta v}{v}=2 \%+2 \times 3 \%=8 \%\)

Question 42. A wire of length l = (6 ± 0.06) cm and radius r = (0.5 ±0.005) cm has a mass of m (0.3 ± 0.003) g. The maximum error in the measurement of density is:

1. 4%
2. 2%
3. 1%
4. 6.8%

Answer: 1. 4%

Density \(\rho=\frac{\text { mass }}{\text { volume }}=\frac{m}{\pi r^2 l}\)

∴ the maximum error in p is

⇒ \(\frac{d \rho}{\rho}=\frac{d m}{m}+2\left(\frac{d r}{r}\right)+\frac{d l}{l}\)

⇒ \(\frac{0.003}{0.3}+2\left(\frac{0.005}{0.5}\right)+\frac{0.06}{6}=\frac{4}{100}=4 \%\)

Question 43. The resistance of a given wire is obtained by measuring the current flowing through it and the potential difference across it. If the error in the measurement of both is 3% each, the error in the value of the resistance of the wire will be:

1. 6%
2. 1%
3. 3%
4. zero

Answer: 1. 6%

Resistance R = \(\frac{V}{I}\)

The error in the measurement of resistance is

⇒ \(\frac{\Delta R}{R} \times 100 \%=\frac{\Delta V}{V} \times 100 \%+\frac{\Delta I}{I} \times 100 \%=3 \%+3 \%=6 \%\)

Question 44. In an experiment, four quantities a, b, c and d are measured with the percentage errors 1%, 2%, 3% and 4% respectively. If a quantity P is calculated as P = aV/cd, the error in P will be:

1. 4%
2. 7%
3. 14%
4. 10%

Answer: 3. 14%

Given that P = \(P=\frac{a^3 b^2}{c d}\)

∴ The error in P is

⇒ \(\frac{\Delta P}{P} \times 100 \%=\left[3\left(\frac{\Delta a}{a}\right)+2\left(\frac{\Delta b}{b}\right)+\frac{\Delta c}{c}+\frac{\Delta d}{d}\right] \times 100 \%\)

= 3 x 1% + 2 x 2% + 3% + 4%

= 14

Question 45. A student measures the distance covered in the free fall of a body, initially at rest, in a given time. He uses the data to estimate g, the acceleration due to gravity. If the maximum percentage errors in the measurement of distance and time are ex and e2 respectively, the percentage error in die estimation of g is:

1. e₂-ex
2. ex + 2e₂
3. ex + e₂
4. ex-2e₂

Answer: 2. ex + 2e₂

We know that

⇒ \(h=u t+\frac{1}{2} g t^2=\frac{1}{2} g t^2\) [.∴ u=0]

⇒ \(g=\frac{2 h}{t^2}\)

∴ \(\frac{d g}{g}=\frac{d h}{h}+2\left(\frac{d t}{t}\right)\)

∴ The error ing will be

⇒ \(\frac{d h}{h} \times 100 \%+2\left(\frac{d t}{t}\right) \times 100 \%=e_1+2 e_2\)

Question 46. If a wire is stretched to make it 0.1% longer, its resistance will:

1. Increase by 0.2%
2. Decreaseby0.05%
3. Decrease by 0.2%
4. Increase by 0.05%

Answer: 1. Increase by 0.2%

Volume = \(V=l A \Rightarrow \text { area }=A=\frac{V}{l}\)

∴ resistance =\(R=\rho \frac{l}{A}=\frac{\rho l^2}{V}=k l^2, \text { where } k=\frac{\rho}{V}\)

∴ \(\frac{\Delta R}{R}=2\left(\frac{\Delta l}{l}\right)\)

= 2(0.1%) = 0.2% (increase).

Question 47. A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is: 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is1 s then the reported mean time should be:

1. (92 ± 5.0) s
2. (92±1.8)s
3. (92±3)s
4. (92±2)s

Answer: 4. (92±2)s

Mean value = \(\bar{x}=\frac{1}{N} \sum x_i=\frac{1}{4}(90+91+95+92)\)

= 92 s.

Mean deviation = \(\frac{1}{N} \Sigma\left|\bar{x}-x_i\right|=\frac{2+1+3+0}{4} \mathrm{~s}\)

= 1.5s.

∴ least count =1 s,

∴ required mean time = (92 ± 2) s.

Question 48. The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively15% and1%, the maximum error in determining the density is:

1. 2.5%
2. 35%
3. 45%
4. 6%

Answer: 3. 45%

Density = d = \(\frac{m}{V}=\frac{m}{l^3}\).

∴ \(\frac{\Delta d}{d}=\frac{\Delta m}{m}+3\left(\frac{\Delta l}{l}\right)\)

Hence, the maximum error in the measurement of density is 1.5% + 3(1%) = 4.5%.

Question 49. A wire when heated shows a 2% increase in length. The increase in the cross-sectional area is:

1. 1%
2. 2%
3. 4%
4. 3%

Answer: 3. 4%

Volume = V = Al.

∴ \(\frac{\Delta V}{V}=\frac{\Delta A}{A}+\frac{\Delta l}{l}\)

Dividing throughout by AT (rise in temperature), we have

\(\frac{1}{T} \frac{\Delta V}{V}=\frac{1}{T} \frac{\Delta A}{A}+\frac{1}{T} \frac{\Delta l}{l}\)

or, \(\gamma=\frac{1}{T} \frac{\Delta A}{A}+\alpha\)

or, \(3 \alpha-\alpha=\frac{1}{T} \frac{\Delta A}{A}\)

or, \(2 \alpha=\frac{1}{T} \frac{\Delta A}{A}\)

∴ The increase in the cross-sectional area is

⇒ \(\frac{\Delta A}{A}=2 \alpha \cdot T=\frac{2}{T}\left(\frac{\Delta l}{l}\right) \cdot T=2(2 \%)=4 \% .\)

Question 50. A force is applied on a square-shaped plate of side L. If the error in die determination of L is 2% and that in F is 4%, what is the maximum permissible error in the pressure?:

1. 2%
2. 4%
3. 6%
4. 8%

Answer: 4. 8%

Pressure = p = \(\frac{F}{A}=\frac{F}{l^2}\)

∴ \(\frac{\Delta p}{p}=\frac{\Delta F}{F}+2\left(\frac{\Delta l}{l}\right)=4 \%+2(2 \%)=8 \%\)

Question 51. The relative density of the material of a body is found by weighing it first in air and then in water. If the weight in air is (5.00 ± 0.05) g and that in water is (4.00 ± 0.05) g then the relative density along with the maximum permissible error is:

1. 5.0 ±11%
2. 5.0 ± 1%
3. 5.00 ±6%
4. 1.25 ±5%

Answer: 1. 5.0 ±11%

Relative density \(=\frac{\text { weight in air }}{\text { loss of weight in water }}\)

⇒ \(d=\frac{w_{\mathrm{A}}}{w_{\mathrm{A}}-w_{\mathrm{w}}}=\frac{5}{5-4}=5\)

∴ \(\frac{\Delta d}{d}=\frac{\Delta w_{\mathrm{A}}}{w_{\mathrm{A}}}+\frac{\Delta\left(w_{\mathrm{A}}-w_{\mathrm{w}}\right)}{w_{\mathrm{A}}-w_{\mathrm{w}}}\)

⇒ \(\frac{\Delta w_{\mathrm{A}}}{w_{\mathrm{A}}}+\frac{\Delta w_{\mathrm{A}}}{w_{\mathrm{A}}-w_{\mathrm{w}}}+\frac{\Delta w_{\mathrm{w}}}{w_{\mathrm{A}}-w_{\mathrm{w}}}\)

⇒ \(\frac{0.05}{5} \times 100 \%+\frac{0.05}{5-4} \times 100 \%+\frac{0.05}{5-4} \times 100 \%\)

= 1% + 5% + 5%

= 11%.

Hence, d = 5.0 ±11%.

Question 52. How many nanometres are there in one kilometre?:

1. 10¹⁰
2. 10¹²
3. 10
4. 109

Answer: 2. 10¹²

1 km = 10³ m

= 10³ x 10⁹ nm

= 10¹² ran.

Question 53. One nanometre is equal to:

1. 10⁹m
2. 10^-9 m
3. 10^-2cm
4. 10⁶cm

Answer: 2. 10^-9 m

1 nm = 10^-9 m.

Question 54. How many wavelengths of Kr-86 are there in one metre?:

1. 652189.63
2. 1553164.13
3. 1650763.73
4. 2348123.73

Answer: 3. 1650763.73

The wavelength of krypton-86 is known to be nearly 605.78021 nm in the orange region.

Hence, the number of wavelengths in one metre is

⇒ \(\frac{1}{605.78021 \times 10^{-9}} \approx\) 0.0016507637 x10⁹

=1650763.73.

Question 55. How many significant figures are there in 80.00?:

1. 3
2. 1
3. 4
4. 2

Answer: 3. 4

In a given number with a decimal point, all the zeros to the right of the last nonzero digit are significant.

Hence, 80.00 has four significant digits – 8, 0, 0 and 0.

Question 56. How many significant figures are there in 0.00125?:

1. 5
2. 3
3. 4
4. None of these

Answer: 2. 3

The given number 0.00125 =12.5 x 10^-4 has only three significant digits, 1,2 and 5.

Question 57. Length cannot be measured by the unit:

1. Light-year
2. Micron
3. Debye
4. Fermi

Answer: 4. Fermi

The debye is the unit of electric dipole moment (1 debye = 3.336 x 10^-30 Cm). All the remaining three units are used to express length.

Question 58. The parsec is a unit of:

1. Frequency
2. Time
3. Distance
4. Angular acceleration

Answer: 3. Distance

The parsec (symbol: pc) is a unit of length used to measure large distances to astronomical objects away from the solar system. One parsec is equal to about 3.2616 light-years.

Question 59. In SI units, the dimensional formula for \(\sqrt{\varepsilon_0 / \mu_0}\) is:

1. ML³TI^-1
2. M^-1L^-1T2I
3. ML^3/2T-3I
4. M^-1L^-2T³I²

Answer: 4. M^-1L^-2T³I²

The force between two charges is given by

⇒ \(F_e=\frac{1}{4 \pi \varepsilon_0} \frac{Q_1 Q_2}{r^2}\)

Hence, \(\left[\varepsilon_0\right]=\left[\frac{Q^2}{F_e r^2}\right]=\frac{\mathrm{I}^2 \mathrm{~T}^2}{\mathrm{MLT}^{-2} \mathrm{~L}^2}=\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{I}^2\)

The force between two parallel currents is given by

⇒ \(F_{\mathrm{m}}=\frac{\mu_0 I_1 I_2 l}{2 \pi d}\)

∴ \(\left[\mu_0\right]=\left[\frac{F_{\mathrm{m}} d}{I^2 l}\right]=\mathrm{MLT}^{-2} \mathrm{I}^{-2}\)

Therefore, \(\sqrt{\frac{\varepsilon_0}{\mu_0}}=\sqrt{\frac{\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{I}^2}{\mathrm{MLT}^{-2} \mathrm{I}^{-2}}}=\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^3 \mathrm{I}^2\)

Question 60. The density of a material in SI units is 120 kg m^-3. In a certainty of units in which the unit of length is 5 cm and the unit of mass is 50 g, the numerical value of the density of the material is:

1. 640
2. 40
3. 410
4. 16

Answer: 2. 40

[density] = ML^-3.

∴ \(n_1 \mathrm{M}_1 \mathrm{~L}_1^{-3}=n_2 \mathrm{M}_2 \mathrm{~L}_2^{-3}\)

or 128 kg m^-3 = n2(50 g)(25 cm)^-3

⇒ \(n_2=128\left(\frac{1 \mathrm{~kg}}{50 \mathrm{~g}}\right)\left(\frac{25 \mathrm{~cm}}{1 \mathrm{~m}}\right)^3\)

⇒ \(128\left(\frac{1000 \mathrm{~g}}{50 \mathrm{~g}}\right)\left(\frac{25 \mathrm{~cm}}{100 \mathrm{~cm}}\right)^3\)

= 40

Question 61. Taking into account the significant figures, what is the value of 9.99 m- 0.0099 m?:

1. 9.980 m
2. 9.9 m
3. 9.98 m
4. 9.9801 m

Answer: 3. 9.98 m

9.99 m- 0.0099 m = 9.9801 m.

The answer is to be expressed in die least number of significant digits, which is three.

Hence, the answer is 9.98.

Question 62. If force (F), velocity (v) and area are considered as the fundamental physical quantities then find the dimensional formula of the Young modulus:

1. \([Y]=\left[F^{-1}\right][v]\left[A^{-1 / 2}\right]\)
2. \([Y]=[F]\left[v^{-1}\right]\left[A^{1 / 2}\right]\)
3. \([Y]=[F]\left[v^0\right]\left[A^{-1}\right]\)
4. \([Y]=[F][v]\left[A^{1 / 2}\right]\)

Answer: 3. \([Y]=[F]\left[v^0\right]\left[A^{-1}\right]\)

Young modulus \(Y=\frac{\text { stress }}{\text { strain }}\).

Now, stress = \(\frac{\text { force }}{\text { area }}\) and strain = \(\frac{\Delta L}{L}\) (dimensionless).

Hence, [stress] = \([F]\left[A^{-1}\right]=[F]\left[v^0\right]\left[A^{-1}\right]\)

Question 63. If area {A), time (f) and momentum (p) are chosen as fundamental quantities, the dimensional formula for energy will be:

1. \(\left[A t^{-2} p\right]\)
2. \(\left[A^{1 / 2} t^{-1} p^2\right]\)
3. \(\left[A^{1 / 2} t^{-1} p\right]\)
4. \(\left[A t^{-1 / 2} p^2\right]\)

Answer: 3. \(\left[A^{1 / 2} t^{-1} p\right]\)

Energy and work have the same dimension.

Hence, [energy] = [work] = [force][distance]

⇒ \(\left[\frac{\text { momentum }}{\text { time }}\right][\sqrt{\text { area }}]=\frac{p}{T} \sqrt{A}\)

Thus, [energy] = \(=\left[A^{1 / 2} T^{-1} p\right]\)

Question 64. Given that \(x=\sqrt{\frac{1}{\varepsilon_0 \mu_0}}, y=\frac{E}{B} \text { and } z=\frac{1}{R C}\) Which of the following statements is correct?:

1. x and z have the same dimension.
2. x and y have the same dimension.
3. y and z have the same dimension.
4. x, y and z have different dimensions.

Answer: 2. x and y have the same dimension.

Both x and y represent the speed of light.

Hence, they have the same dimension.

But z is the time constant of an RC circuit.