If the unit of length be doubled then the numerical value of the universal gravitation constant G will become (with respect to present value)

Options

A. Double
B. Half
C. `8` times
D. `1//8` times

Explanation

To solve the problem, we need to analyze how the universal gravitational constant \( G \) changes when the unit of length is doubled.

### Step-by-Step Solution:

1. **Understand the formula for \( G \)**:
The universal gravitational constant \( G \) has the dimensional formula:
\[ G = \frac{F \cdot r^2}{m_1 \cdot m_2} \]
where \( F \) is the gravitational force, \( r \) is the distance between the masses \( m_1 \) and \( m_2 \).

2. **Identify the dimensions of \( G \)**:
The dimensional formula for \( G \) is:
\[ [G] = M^{-1} L^3 T^{-2} \]
where \( M \) is mass, \( L \) is length, and \( T \) is time.

3. **Change in units**:
If the unit of length is doubled, we denote the new unit of length as \( L_2 = 2L_1 \).

4. **Substituting the new length into the formula**:
The new value of \( G \) can be expressed as:
\[ G_2 = \frac{F \cdot (L_2)^2}{m_1 \cdot m_2} \]
Since \( L_2 = 2L_1 \), we have:
\[ G_2 = \frac{F \cdot (2L_1)^2}{m_1 \cdot m_2} = \frac{F \cdot 4L_1^2}{m_1 \cdot m_2} \]

5. **Relate the new \( G \) to the old \( G \)**:
Since the force \( F \) remains the same, we can express \( G_2 \) in terms of \( G_1 \):
\[ G_2 = 4 \cdot \frac{F \cdot L_1^2}{m_1 \cdot m_2} = 4G_1 \]

6. **Final expression**:
Therefore, when the unit of length is doubled, the new value of \( G \) becomes:
\[ G_2 = \frac{G_1}{4} \]

### Conclusion:
The numerical value of the universal gravitational constant \( G \) will become \( \frac{1}{4} \) of its present value when the unit of length is doubled.