The quantity `X = (epsilon_(0)LV)/(t)` where `epsilon_(0)` is the permittivity of free space, L is length, V is the potential difference and t is time. The dimensions of X are the same as that of

Options

A. Resistance
B. Charge
C. Voltage
D. Current

Answer

To solve the problem, we need to find the dimensions of the quantity \( X = \frac{\epsilon_0 L V}{t} \), where \( \epsilon_0 \) is the permittivity of free space, \( L \) is length, \( V \) is the potential difference, and \( t \) is time.

### Step-by-Step Solution:

1. **Identify the Dimensions of Each Quantity**:
- The dimension of \( \epsilon_0 \) (permittivity of free space) is given by:
\[ [\epsilon_0] = \frac{[Q]^2}{[M][L]^3[T]^2} = \frac{A^2}{M L^3 T^2} \]
- The dimension of \( L \) (length) is:
\[ [L] = L \]
- The dimension of \( V \) (potential difference) can be derived from the relationship \( V = \frac{W}{Q} \), where \( W \) is work done (energy). The dimension of energy is \( [E] = [M][L]^2[T]^{-2} \), thus:
\[ [V] = \frac{[E]}{[Q]} = \frac{ML^2T^{-2}}{Q} = \frac{ML^2T^{-2}}{A} = \frac{ML^2T^{-2}}{A} = \frac{M L^2 T^{-2}}{A} \]
- The dimension of \( t \) (time) is:
\[ [t] = T \]

2. **Combine the Dimensions**:
- Now, we can substitute these dimensions into the expression for \( X \):
\[ [X] = \frac{[\epsilon_0][L][V]}{[t]} = \frac{\left(\frac{A^2}{M L^3 T^2}\right)(L)\left(\frac{M L^2 T^{-2}}{A}\right)}{T} \]

3. **Simplify the Expression**:
- Combine the dimensions:
\[ [X] = \frac{A^2 L^2 M T^{-2} L}{M L^3 T^2 A T} = \frac{A^2 L^2}{A L^3 T^2} = \frac{A L^2}{L^3 T^2} \]
- This simplifies to:
\[ [X] = \frac{A}{L T^2} \]

4. **Recognize the Result**:
- The dimension \( \frac{A}{L T^2} \) can be interpreted as the dimension of current (\( A \)) per unit length and time squared. However, we can also express it in terms of current:
\[ [X] = A \cdot T^{-1} = \text{Current} \]

### Conclusion:
The dimensions of \( X \) are the same as that of current, which is represented as \( [I] \).