Let l,r,c and v represent inductance, resistance, capacitance and voltage, respectively. The dimension of `(l)/(rcv)` in SI units will be :

Options

A. `[LA^(-2)]`
B. `[A^(-1)]`
C. `[LTA]`
D. `[LT^(2)]`

Explanation

To find the dimension of the expression \( \frac{L}{R \cdot C \cdot V} \), where \( L \) is inductance, \( R \) is resistance, \( C \) is capacitance, and \( V \) is voltage, we will follow these steps:

### Step 1: Write down the dimensions of each quantity.
- Inductance \( L \): The dimension of inductance is given by
\[ [L] = M L^2 T^{-2} I^{-2} \]
- Resistance \( R \): The dimension of resistance is given by
\[ [R] = M L^2 T^{-3} I^{-2} \]
- Capacitance \( C \): The dimension of capacitance is given by
\[ [C] = M^{-1} L^{-2} T^4 I^2 \]
- Voltage \( V \): The dimension of voltage is given by
\[ [V] = M L^2 T^{-3} I^{-1} \]

### Step 2: Substitute the dimensions into the expression.
We need to find the dimensions of \( \frac{L}{R \cdot C \cdot V} \):
\[ \frac{[L]}{[R] \cdot [C] \cdot [V]} = \frac{M L^2 T^{-2} I^{-2}}{(M L^2 T^{-3} I^{-2}) \cdot (M^{-1} L^{-2} T^4 I^2) \cdot (M L^2 T^{-3} I^{-1})} \]

### Step 3: Calculate the denominator.
First, we will multiply the dimensions in the denominator:
\[ [R] \cdot [C] \cdot [V] = (M L^2 T^{-3} I^{-2}) \cdot (M^{-1} L^{-2} T^4 I^2) \cdot (M L^2 T^{-3} I^{-1}) \]

Calculating this step by step:
1. Multiply \( [R] \) and \( [C] \):
\[ (M L^2 T^{-3} I^{-2}) \cdot (M^{-1} L^{-2} T^4 I^2) = (M^{1-1}) (L^{2-2}) (T^{-3+4}) (I^{-2+2}) = M^0 L^0 T^1 I^0 = T \]
2. Now multiply the result with \( [V] \):
\[ T \cdot (M L^2 T^{-3} I^{-1}) = M^1 L^2 T^{1-3} I^{-1} = M L^2 T^{-2} I^{-1} \]

### Step 4: Substitute back into the expression.
Now substituting back into the expression:
\[ \frac{[L]}{[R \cdot C \cdot V]} = \frac{M L^2 T^{-2} I^{-2}}{M L^2 T^{-2} I^{-1}} = \frac{M L^2 T^{-2} I^{-2}}{M L^2 T^{-2} I^{-1}} = I^{-1} \]

### Final Result:
The dimension of \( \frac{L}{R \cdot C \cdot V} \) is:
\[ \boxed{I^{-1}} \]