Which two of the following quantities are dimensionally equivalent?

We know that, Force $=$ Mass $\times$ Acceleration $\Rightarrow \quad F=m \times a$ $\Rightarrow \quad[F]=[\mathrm{m}][\mathrm{a}]=[\mathrm{M}]\left[\mathrm{LT}^{-2}\right]=\left[\mathrm{MLT}^{-2}\right]$ Pressure $(p)=\frac{\text { Force }(F)}{\operatorname{Area}(A)}$ $\Rightarrow \quad[p]=\frac{[\mathrm{F}]}{[\mathrm{A}]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^{2}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$ Young's modulus, $Y=\frac{\text { Stress }}{\text { Strain }}$ $\Rightarrow \quad[Y]=\frac{[\text { Stress }]}{[\text { Strain }]}=\frac{[F / A]}{[\text { Strain }]}$ $=\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]}{\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]}$ $=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$ Energy $=$ Work done $=$ Force $\times$ Displacement $\Rightarrow[$ Energy $]=[$ Force $] \times[$ Displacement $]$ $=\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]$ Hence, we see that dimensions of pressure and Young's modulus is same.