Correct Option (c) Work and Torque
Explanation:
Impulse = Force \(x\) time
$$
=\left[\mathrm{MLT}^{-2}\right][\mathrm{T}]=\left[\mathrm{MLT}^{-1}\right]
$$
Surface tension \(=\) Force/length \(=\left[\mathrm{MLT}^{-2}\right] /[\mathrm{L}]=\left[\mathrm{ML}^{0} \mathrm{~T}^{-2}\right]\)
Angular momentum \(=\) Moment of inertia \(x\) angular velocity
$$
=\left[\mathrm{ML}^{2}\right]\left[\mathrm{T}^{-1}\right]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]
$$
Work Force \(x\) distance \(=\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\)
Energy \(=\left[M L^{2} T^{2}\right]\)
Torque \(=\) Force \(x\) distance \(=\left[M L T^{-2}\right][L]=\left[M L^{2} T^{-2}\right]\)
Young's modulus
$$
\begin{aligned}
&=\frac{\text { Force / Area }}{\text { Change in length / original length }} \\
&=\frac{\left[\mathrm{MLT}^{-2}\right] /\left[\mathrm{L}^{2}\right]}{[\mathrm{L}] /[\mathrm{L}]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]
\end{aligned}
$$
Hence, among the given pair of physical quantities work and torque have the same dimensions \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\).
