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 solve the problem of finding the dimensions of \( \frac{L}{R} \) where \( L \) is inductance and \( R \) is resistance, we will follow these steps:

### Step 1: Identify the dimensions of inductance \( L \)
Inductance \( L \) can be expressed in terms of its fundamental dimensions. The dimension of inductance is given by:
\[ [L] = M^1 L^2 T^{-2} I^{-2} \]
where:
- \( M \) is mass,
- \( L \) is length,
- \( T \) is time,
- \( I \) is electric current.

### Step 2: Identify the dimensions of resistance \( R \)
Resistance \( R \) also has its own dimensional formula. The dimension of resistance is:
\[ [R] = M^1 L^2 T^{-3} I^{-2} \]

### Step 3: Calculate the dimensions of \( \frac{L}{R} \)
Now, we can find the dimensions of \( \frac{L}{R} \) by dividing the dimensions of inductance by the dimensions of resistance:
\[ \frac{L}{R} = \frac{M^1 L^2 T^{-2} I^{-2}}{M^1 L^2 T^{-3} I^{-2}} \]

### Step 4: Simplify the expression
When we divide the dimensions, we can simplify:
\[ \frac{L}{R} = M^{1-1} L^{2-2} T^{-2 - (-3)} I^{-2 - (-2)} = M^0 L^0 T^{1} I^{0} \]
This simplifies to:
\[ \frac{L}{R} = T^1 \]

### Step 5: Conclusion
Thus, the dimensions of \( \frac{L}{R} \) are:
\[ [T] \]
This indicates that \( \frac{L}{R} \) has the dimension of time.

### Final Answer
The dimensions of \( \frac{L}{R} \) are \( M^0 L^0 T^1 Q^0 \) or simply \( [T] \).

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