If `L and R` denote inductance and resistance , respectively , then the dimensions of `L//R` are

Options

A. ` M^(1) L^(0) T^(0) Q^(-1)`
B. ` M^(0) L^(0) T Q^(0)`
C. ` M^(0) L^(1) T^(-1) Q^(0)`
D. ` M^(-1) L T^(0) Q^(-1)`

Explanation

To find the dimensions of \( \frac{L}{R} \), where \( L \) is inductance and \( R \) is resistance, we will first determine the dimensions of both \( L \) and \( R \) separately.

### Step 1: Find the dimension of voltage
Voltage \( V \) is defined as work done per unit charge. The dimension of work done can be expressed as:

\[ \text{Dimension of Work} = \text{Force} \times \text{Displacement} = (M L T^{-2}) \times L = M L^2 T^{-2} \]

The dimension of charge \( Q \) is represented as \( Q \). Therefore, the dimension of voltage is:

\[ [V] = \frac{\text{Dimension of Work}}{\text{Dimension of Charge}} = \frac{M L^2 T^{-2}}{Q} = M L^2 T^{-2} Q^{-1} \]

### Step 2: Find the dimension of inductance \( L \)
The voltage across an inductor is given by:

\[ V = L \frac{di}{dt} \]

Where \( \frac{di}{dt} \) is the rate of change of current. The dimension of current \( I \) is represented as \( A \). Therefore, the dimension of \( \frac{di}{dt} \) is:

\[ \left[\frac{di}{dt}\right] = \frac{A}{T} \]

Now, substituting the dimensions into the equation for voltage:

\[ M L^2 T^{-2} Q^{-1} = L \left(\frac{A}{T}\right) \]

Rearranging to find the dimension of inductance \( L \):

\[ [L] = \frac{M L^2 T^{-2} Q^{-1} \cdot T}{A} = M L^2 T^{-1} Q^{-1} A^{-1} \]

### Step 3: Find the dimension of resistance \( R \)
According to Ohm's law, resistance is defined as:

\[ R = \frac{V}{I} \]

Substituting the dimensions of voltage and current:

\[ [R] = \frac{M L^2 T^{-2} Q^{-1}}{A} = M L^2 T^{-2} Q^{-1} A^{-1} \]

### Step 4: Find the dimensions of \( \frac{L}{R} \)
Now we can find the dimensions of \( \frac{L}{R} \):

\[ \frac{L}{R} = \frac{M L^2 T^{-1} Q^{-1} A^{-1}}{M L^2 T^{-2} Q^{-1} A^{-1}} \]

Cancelling out the common terms:

- The \( M \) cancels out.
- The \( L^2 \) cancels out.
- The \( Q^{-1} \) cancels out.
- The \( A^{-1} \) cancels out.

This simplifies to:

\[ \frac{L}{R} = T^{1} \]

Thus, the dimensions of \( \frac{L}{R} \) are:

\[ \boxed{T} \]