Friction and Circular Motion Multiple Choice Question And Answers

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

“circular motion questions “

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

⇒ \(x L=\frac{\mu L}{1+\mu}\)

vertical circular motion class 11

 

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

⇒ \(0=v-\mu g t \Rightarrow t=\frac{v}{\mu g}\)

vertical circular motion class 11

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),

⇒ \(n^2=\frac{\sin \theta}{\sin \theta-\mu \cos \theta}\)

⇒ \(\mu=\left(1-\frac{1}{n^2}\right) \tan \theta\)

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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:

⇒ \(m g \sin \theta=f=\mu \sigma V=\mu m g \cos \theta\)

⇒ \(\mu=\tan \theta=\tan 30^{\circ}=\frac{1}{\sqrt{3}}\)

vertical circular motion class 11

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]

⇒ \(\frac{\left(4 \mathrm{~m} \mathrm{~s}^{-1}\right)^2}{2(0.8)\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)}=1 \mathrm{~m}\)

vertical circular motion class 11

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 \(f \geq m g \text { or } \mu m \alpha \geq m g \text { or } \alpha \geq g / \mu\)

vertical circular motion class 11

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, \(R=u t=u \sqrt{\frac{2 R}{g}}\)

⇒ \(u=R \sqrt{\frac{g}{2 R}}=\sqrt{\frac{g R}{2}} .\)

vertical circular motion class 11

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 = \(\frac{m v^2}{r}\)

⇒ \(T=m g \cos \theta+\frac{m v^2}{r}\)

∴ \(T>m g \cos \theta\)

vertical circular motion class 11

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