If `g` is the acceleration due to gravity and `R` is the radius of earth, then the dimensional formula for g `R` is

Options

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

Answer

To find the dimensional formula for the product of `g` (acceleration due to gravity) and `R` (radius of the Earth), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Dimensional Formula for `g`:**
- Acceleration is defined as the change in velocity per unit time. The formula for acceleration is:
\[ g = \frac{dv}{dt} \]
- Velocity (`v`) is defined as displacement per unit time:
\[ v = \frac{ds}{dt} \]
- Displacement (`s`) has the dimension of length, which is denoted as `[L]`.
- Therefore, the dimensional formula for velocity is:
\[ [v] = \frac{[L]}{[T]} = [L][T^{-1}] \]
- Substituting this back into the formula for acceleration, we get:
\[ [g] = \frac{[L][T^{-1}]}{[T]} = [L][T^{-2}] \]
- Thus, the dimensional formula for `g` is:
\[ [g] = [M^0 L^1 T^{-2}] \]

2. **Identify the Dimensional Formula for `R`:**
- The radius `R` is a measure of length, so its dimensional formula is:
\[ [R] = [L^1] \]

3. **Calculate the Dimensional Formula for `gR`:**
- Now, we need to find the dimensional formula for the product `gR`:
\[ [gR] = [g] \times [R] \]
- Substituting the dimensional formulas we found:
\[ [gR] = [M^0 L^1 T^{-2}] \times [L^1] \]
- This simplifies to:
\[ [gR] = [M^0 L^{1+1} T^{-2}] = [M^0 L^2 T^{-2}] \]

4. **Final Result:**
- The dimensional formula for `gR` is:
\[ [gR] = [M^0 L^2 T^{-2}] \]

### Conclusion:
The dimensional formula for the product of acceleration due to gravity `g` and the radius of the Earth `R` is \([M^0 L^2 T^{-2}]\).