Question

The physical quantities not having same dimensions are:
(A) torque and work
(B) Momentum and Planck’s constant
(C) Stress and Young’s modulus
(D) Speed and ${\left( {{\mu _0}{ \in _0}} \right)^{ - 1/2}}$

Answer 1

Hint:Derive the dimensions of the parameters given in the option, when their unit is known and if the unit is not known, write the formula of the parameters and derive the unit and the dimensions from that. The physical dimension of all the units consists of only $MLT$ .

Useful formula:
(1) The formula of the work is given by

$w = F \times d$

Where $w$ is the work done, $F$ is the force required for the work to be done and $d$ is the distance.

(2) The formula of the force is given as

$P = \dfrac{F}{A}$

Where $P$ is the pressure and $A$ is the area.

Complete step by step solution:
(A) The unit of the torque is $Kg{m^2}{s^{ - 2}}$ and hence its dimension is written as $M{L^2}{T^{ - 1}}$ . The dimension if the work is calculated as follows.

$w = F \times d$

The dimension of the force is $ML{T^{ - 1}}$ and the dimension if the distance is $L$ , hence the dimension of the work is $M{L^2}{T^{ - 1}}$ . Hence torque and work have the same dimension.

(B) The unit of momentum is $Kgm{s^{ - 1}}$ and thus its dimension is $ML{T^{ - 1}}$ . The unit of the Planck’s constant is obtained by

$ = ML{T^{ - 1}} \times L = M{L^2}{T^{ - 1}}$

Thus the dimensions of the momentum and the Planck’s constant are not the same.

(C) By using the formula of the stress,

$P = \dfrac{F}{A}$

The dimension of the pressure is calculated. $ML{T^{ - 2}} \times {L^{ - 2}} = M{L^{ - 1}}{T^{ - 2}}$ . The dimension of the Young’s modulus is also $M{L^{ - 1}}{T^{ - 2}}$ . Hence the dimensions of the pressure and the young’s modulus is also the same.

(D) $\dfrac{1}{{\sqrt {{\mu _0}{ \in _0}} }} = \sqrt {{L^2}{T^{ - 2}}} $

$ = L{T^{ - 1}}$

The dimension of the speed is $L{T^{ - 1}}$ . Hence the physical dimensions of both the parameters are the same.
From the above explanation, it is clear that the dimension of the momentum and the Planck’s constant are not the same.

Thus the option (B) is correct.

Note:The physical dimensions of the Planck’s constant is obtained by multiplying the dimensions of the momentum and the distance or the energy and the time. The unit of the speed is obtained from the known formula $\dfrac{{{\text{distance}}}}{{{\text{time}}}}$ and its unit is $m{s^{ - 1}}$ .