The pair of quantities having same dimensions is

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Impulse and Surface Tension

Angular momentum and Work

Work and Torque

Young’s modulus and Energy

answer is C.

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Detailed Solution

Impulse = Force × time $= [M L T^{- 2}] [T] = [M L T^{- 1}]$

Surface tension = $\frac{Force}{length} = \frac{[{MLT}^{- 2}]}{[L]} = [{ML}^{0} T^{- 2}]$

Angular momentum
= Moment of inertia × angular velocity

$= [M L^{2}] [T^{- 1}] = [M L^{2} T^{- 1}]$

$W o r k = F o r c e \times d i s \tan c e = [M L T^{- 2}] [L] = [M L^{2} T^{- 2}]$

$E n e r g y = [M L^{2} T^{- 2}] T o r q u e = F o r c e \times d i s \tan c e = [M L T^{- 2}] [L] = [M L^{2} T^{- 2}] Y o u n g ’ s m o d u l u s$

$= \frac{Force / Area}{Change in length / original length}$ $= \frac{[{MLT}^{- 2}] / [L^{2}]}{[L] / [L]} = [{ML}^{- 1} T^{- 2}]$

Hence, among the given pair of physical quantities work and torque have the same dimensions $[M L^{2} T^{- 2}]$ .

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