If unit of length and time is double, dthe numerical valuure of `'g'` (acceleration due to gravity) will be

Options

A. doubled
B. halved
C. four times
D. remain same

Answer

To solve the problem of how the numerical value of acceleration due to gravity \( g \) changes when the units of length and time are doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Dimensional Formula of \( g \)**:
The acceleration due to gravity \( g \) has the dimensional formula of:
\[ g = [L][T^{-2}] \]
where \( L \) represents length and \( T \) represents time.

2. **Define New Units**:
If we double the units of length and time, we can denote the new units as:
\[ L' = 2L \quad \text{and} \quad T' = 2T \]

3. **Express the New Value of \( g \)**:
The new acceleration due to gravity \( g' \) in terms of the new units becomes:
\[ g' = \frac{L'}{(T')^2} \]
Substituting the new units:
\[ g' = \frac{2L}{(2T)^2} \]

4. **Simplify the Expression**:
Simplifying the denominator:
\[ g' = \frac{2L}{4T^2} = \frac{2L}{4} \cdot \frac{1}{T^2} = \frac{1}{2} \cdot \frac{L}{T^2} \]
Recognizing that \( \frac{L}{T^2} \) is the original value of \( g \):
\[ g' = \frac{1}{2} g \]

5. **Conclusion**:
Therefore, when the units of length and time are doubled, the numerical value of \( g \) becomes:
\[ g' = \frac{1}{2} g \]

### Final Answer:
The numerical value of \( g \) will be half of its original value, or \( \frac{1}{2} g \).

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