The pair of quantities having same dimensions is

Impulse = Force × time $= \left[MLT{ }^{-2}\right]\left[T\right] = \left[ML{T}^{-1}\right]$

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

Angular momentum
= Moment of inertia × angular velocity

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

$Work = Force \times dis\tan ce = \left[ML{T}^{-2}\right]\left[L\right] = \left[M{L}^2{T}^{-2}\right]$

$$\begin{gathered} Energy = \left[M{L}^2{T}^{-2}\right] Torque = Force \times dis\tan ce = \left[ML{T}^{-2}\right]\left[L\right] = \left[M{L}^2{T}^{-2}\right] \\ Young’s modulus \end{gathered}$$

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

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