The dimension of `1//CR` is, where `C` is capacitance and `R` is electrical resistance :

Options

A. `M^(0)L^(0)T^(2)`
B. `M^(1)L^(2)T^(-1)A^(-1)`
C. `M^(0)L^(0)T^(-1)`
D. `M^(0)L^(1)T^(-1)`

Answer

To find the dimension of \( \frac{1}{RC} \), where \( C \) is capacitance and \( R \) is electrical resistance, we can follow these steps:

### Step 1: Understand the dimensions of capacitance (C) and resistance (R)

1. **Capacitance (C)**: The dimension of capacitance can be derived from its formula:
\[ C = \frac{Q}{V} \]
where \( Q \) is charge and \( V \) is voltage. The dimension of charge \( Q \) is \( [I][T] \) (current times time), and the dimension of voltage \( V \) is \( [M][L^2][T^{-3}][I^{-1}] \) (derived from \( V = \frac{W}{Q} \), where \( W \) is work done). Therefore, the dimension of capacitance is:
\[ [C] = \frac{[I][T]}{[M][L^2][T^{-3}][I^{-1}]} = [M^{-1}][L^{-2}][T^4][I^2] \]

2. **Resistance (R)**: The dimension of resistance can be derived from Ohm's law:
\[ R = \frac{V}{I} \]
Using the dimensions of voltage and current, we have:
\[ [R] = \frac{[M][L^2][T^{-3}][I^{-1}]}{[I]} = [M][L^2][T^{-3}][I^{-2}] \]

### Step 2: Find the dimension of the product \( RC \)

Now, we can find the dimension of the product \( RC \):
\[ [RC] = [R] \cdot [C] = ([M][L^2][T^{-3}][I^{-2}]) \cdot ([M^{-1}][L^{-2}][T^4][I^2]) \]

### Step 3: Simplify the dimensions

When we multiply these dimensions together:
\[ [RC] = [M][L^2][T^{-3}][I^{-2}] \cdot [M^{-1}][L^{-2}][T^4][I^2] \]
This simplifies to:
\[ [RC] = [M^{1-1}][L^{2-2}][T^{-3+4}][I^{-2+2}] = [M^0][L^0][T^1][I^0] = [T] \]

### Step 4: Find the dimension of \( \frac{1}{RC} \)

Now, we need to find the dimension of \( \frac{1}{RC} \):
\[ \left[\frac{1}{RC}\right] = \frac{1}{[RC]} = \frac{1}{[T]} = [T^{-1}] \]

### Conclusion

Thus, the dimension of \( \frac{1}{RC} \) is:
\[ [M^0][L^0][T^{-1}] \]

### Final Answer
The dimension of \( \frac{1}{RC} \) is \( [M^0][L^0][T^{-1}] \).

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