Units Dimensions and Errors Multiple Choice Question And Answers

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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. L1/2T1/2
  2. L-1T
  3. LT-1
  4. L1/2T1/2

Answer: 4. L1/2T1/2

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. ML3/2T-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.

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