Which of the following units denotes the dimensions ML²Q⁻², where Q denotes the electric charge?

To determine which unit corresponds to the dimensions ${\mathrm{ML}}^2{Q}^{-2}$, we can derive the dimensional formula for the self-inductance ($L$), whose SI unit is the henry (H).

The magnetic potential energy ($U$) stored in an inductor of self-inductance $L$ carrying a current $I$ is given by the formula:
$U=\frac{1}{2}L{I}^2$

We can rearrange this formula to solve for the self-inductance $L$:
$L=\frac{2U}{I^2}$

Now, let's write down the dimensions of each quantity in terms of mass ($M$), length ($L$), and charge ($Q$):

1. Energy ($U$) has the same dimensions as work done (force × distance):
$\left[U\right]=\left[\text{Force}\right]\times \left[\text{Distance}\right]=\left({\mathrm{MLT}}^{-2}\right)\times \left(L\right)={\mathrm{ML}}^2{T}^{-2}$

2. Electric current ($I$) is defined as the rate of flow of charge ($Q$) per unit time ($T$):
$I=\frac{Q}{T}$
Therefore, the dimensional formula for current in terms of charge and time is:
$\left[I\right]={\mathrm{QT}}^{-1}$

Now, substitute the dimensions of $U$ and $I$ back into the expression for self-inductance $L$:
$\left[L\right]=\frac{\left[U\right]}{{\left[I\right]}^2}$
$\left[L\right]=\frac{{\mathrm{ML}}^2{T}^{-2}}{{\left({\mathrm{QT}}^{-1}\right)}^2}$
$\left[L\right]=\frac{{\mathrm{ML}}^2{T}^{-2}}{Q^2{T}^{-2}}$

Simplifying the expression by cancelling out the time terms ($T^{-2}$):
$\left[L\right]={\mathrm{ML}}^2{Q}^{-2}$

This matches the dimensions given in the question. Since the SI unit of self-inductance is the henry (H), the correct unit is henry (H).