The dimension of combination `(L)/(CVR)` are same as dimensions of :-

Options

A. Change
B. Current
C. `"Charge" ^(-1)`
D. `"Current "^(-1)`

Answer

To solve the problem of finding the dimensions of the combination \( \frac{L}{CVR} \) and determining what it is equivalent to, we will follow these steps:

### Step 1: Identify the dimensions of each quantity
- **Inductance (L)**: The dimension of inductance is given by \([L^2 M^{-1} T^{-2} A^{-2}]\).
- **Capacitance (C)**: The dimension of capacitance is given by \([M^{-1} L^{-2} T^4 A^2]\).
- **Resistance (R)**: The dimension of resistance is given by \([M L^2 T^{-3} A^{-2}]\).

### Step 2: Write the expression for \( CVR \)
The expression for \( CVR \) can be combined as follows:
\[ CVR = C \cdot R = \left[M^{-1} L^{-2} T^4 A^2\right] \cdot \left[M L^2 T^{-3} A^{-2}\right] \]

### Step 3: Multiply the dimensions
Now, we will multiply the dimensions of capacitance and resistance:
\[ CVR = \left[M^{-1} L^{-2} T^4 A^2\right] \cdot \left[M L^2 T^{-3} A^{-2}\right] = M^{-1+1} L^{-2+2} T^{4-3} A^{2-2} = [L^0 M^0 T^1 A^0] = [T] \]

### Step 4: Substitute into the original expression
Now substitute \( CVR \) back into the expression \( \frac{L}{CVR} \):
\[ \frac{L}{CVR} = \frac{L}{[T]} = \frac{[L^2 M^{-1} T^{-2} A^{-2}]}{[T]} = [L^2 M^{-1} T^{-3} A^{-2}] \]

### Step 5: Identify the equivalent dimension
The dimensions \( [L^2 M^{-1} T^{-3} A^{-2}] \) represent the dimensions of a quantity that is equivalent to the inverse of current (since current has dimensions of \([A]\)). This means that:
\[ \frac{L}{CVR} \text{ has dimensions equivalent to } [I^{-1}] \]

### Final Answer
Thus, the dimensions of the combination \( \frac{L}{CVR} \) are the same as the dimensions of \( \frac{1}{\text{Current}} \).

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