The dimensions of the quantity `L//C` is identical to

Options

A. `("resistance")^(-1)`
B. `("timet")^(-2)`
C. `("resistance")^(2)`
D. None of these

Answer

To find the dimensions of the quantity \( \frac{L}{C} \), where \( L \) is inductance and \( C \) is capacitance, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Dimensions of Inductance (L)**:
The dimensions of inductance \( L \) are given by:
\[ [L] = M^1 L^2 T^{-2} I^{-2} \]

2. **Identify the Dimensions of Capacitance (C)**:
The dimensions of capacitance \( C \) are given by:
\[ [C] = M^{-1} L^{-2} T^4 I^2 \]

3. **Set Up the Division**:
We need to find the dimensions of \( \frac{L}{C} \):
\[ \frac{L}{C} = \frac{M^1 L^2 T^{-2} I^{-2}}{M^{-1} L^{-2} T^4 I^2} \]

4. **Perform the Division**:
When dividing the dimensions, we subtract the exponents of the same base:
- For mass \( M \):
\[ 1 - (-1) = 1 + 1 = 2 \quad \Rightarrow \quad M^2 \]
- For length \( L \):
\[ 2 - (-2) = 2 + 2 = 4 \quad \Rightarrow \quad L^4 \]
- For time \( T \):
\[ -2 - 4 = -2 - 4 = -6 \quad \Rightarrow \quad T^{-6} \]
- For current \( I \):
\[ -2 - 2 = -2 - 2 = -4 \quad \Rightarrow \quad I^{-4} \]

5. **Combine the Results**:
Thus, the dimensions of \( \frac{L}{C} \) are:
\[ \frac{L}{C} = M^2 L^4 T^{-6} I^{-4} \]

6. **Identify the Equivalent Quantity**:
We can express the dimensions in a more compact form, if needed, but we have identified that the dimensions correspond to the square of resistance:
\[ \frac{L}{C} \sim [R^2] \]

### Final Answer:
The dimensions of the quantity \( \frac{L}{C} \) are identical to the dimensions of resistance squared, \( R^2 \).

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