A quantity $X$ is given by $\varepsilon_0 L \frac{\Delta V}{\Delta t}$ where $\varepsilon_0$ is the permittivity of the free space, $L$ is a length, $\Delta V$ is a potential difference and $\Delta t$ is a time interval. The dimensional formula for $X$ is the same as that of

Step 1: State the given formula and variables.

The quantity $X$ is defined by the relation:

$$X = \varepsilon_0 L \frac{\Delta V}{\Delta t}$$

where $\varepsilon_0$ is the permittivity of free space, $L$ is length, $\Delta V$ is potential difference, and $\Delta t$ is a time interval.

Step 2: Determine the dimensions of each component.

Using the base dimensions of mass ($M$), length ($L$), time ($T$), and current ($I$), the dimensional formulas for each quantity are:

$$[\varepsilon_0] = M^{-1} L^{-3} T^{4} I^{2}$$ $$[L] = L^{1}$$ $$[\Delta V] = M L^{2} T^{-3} I^{-1}$$ $$[\Delta t] = T^{1}$$

Step 3: Calculate the dimensional formula of $X$.

Substitute the individual dimensions into the expression for $X$.

$$[X] = [\varepsilon_0] [L] [\Delta V] \left[\frac{1}{\Delta t}\right]$$ $$[X] = \left(M^{-1} L^{-3} T^{4} I^{2}\right) \left(L^{1}\right) \left(M L^{2} T^{-3} I^{-1}\right) \left(T^{-1}\right)$$ $$[X] = M^{-1+1} L^{-3+1+2} T^{4-3-1} I^{2-1}$$ $$[X] = M^{0} L^{0} T^{0} I^{1}$$ $$[X] = I$$

Step 4: Compare with the dimensions of the given options.

Analyze the dimensional formulas of the options:

$$[\text{resistance}] = M L^{2} T^{-3} I^{-2}$$ $$[\text{charge}] = I T$$ $$[\text{voltage}] = M L^{2} T^{-3} I^{-1}$$ $$[\text{current}] = I$$

The dimensional formula of $X$ matches that of current.

Answer: Current