A pair of physical quantities having same dimensional formula is

Options

A. force and torque
B. work and energy
C. force and impulse
D. linear momentum and angular momentum

Answer

B

Explanation

Force = Mass `xx` acceleration
or `F = ma`
`= [M] [LT^(-2)] = [MLT^(-2)]`
Torque = Moment of inertia `xx` angular acceleration
or `tau = I xx alpha = [ML^(2)] [T^(-2)] = [ML^(2)T^(-2)]`
(b) Work = Force `xx` displacement
or `W = F xx d = [MLT^(-2)] [L] = [ML^(2)T^(-2)]`
Energy `=(1)/(2) xx "mass" xx ("velcotiy")^(2)`
or `K = (1)/(2) mv^(2) = [M] [LT^(-1)]^(2) = [ML^(2)T^(-2)]`
(c) Force as discussed above
`[F] = [MLT^(-2)]`
Impulse = Force `xx` time-interval
`:. [I] = [MLT^(-2)] [T] = [MLT^(-1)]`
(d) Linear momentum = Mass `xx` velocity
or `p = mv`
`:. [p] = [M] [LT^(-1)] = [MLT^(-1)]`
Angular momentum = Momentum of inertia `xx` angular velocity
or `[L] = [I] xx [omega]`
`:. [L] = [ML^(2)] [T^(-1)] = [ML^(2)T^(-1)]`
Hence, we observe that choice (b) is correct.
NOTE
In this problem, the momentum of inertia and impulse are denoted by same symbol I.