Two spherical balls of 1 kg and 4kg are seprated by a distance of 12 cm. Find the distance of a point from the 1 kg mass at which the gravitational force on any mass becomes zero.

Options

A. 4 cm
B. 12 cm
C. 6 cm
D. 8 cm

Answer

To solve the problem, we need to find the distance \( x \) from the 1 kg mass where the gravitational force on any mass becomes zero.

### Step-by-Step Solution:

1. **Identify the masses and distance**:
- Let \( m_1 = 1 \, \text{kg} \) (the mass of the first ball).
- Let \( m_2 = 4 \, \text{kg} \) (the mass of the second ball).
- The distance between the two masses is \( d = 12 \, \text{cm} \).

2. **Set up the problem**:
- We need to find a point \( P \) such that the gravitational forces due to both masses on a test mass placed at point \( P \) are equal and opposite.
- Let \( x \) be the distance from the 1 kg mass to point \( P \). Therefore, the distance from the 4 kg mass to point \( P \) will be \( d - x = 12 - x \).

3. **Write the gravitational force equations**:
- The gravitational force \( F_1 \) exerted by the 1 kg mass on the test mass is given by:
\[ F_1 = \frac{G \cdot m \cdot m_1}{x^2} \]
- The gravitational force \( F_2 \) exerted by the 4 kg mass on the test mass is given by:
\[ F_2 = \frac{G \cdot m \cdot m_2}{(12 - x)^2} \]
- Here, \( G \) is the gravitational constant and \( m \) is the mass of the test mass (which will cancel out).

4. **Set the forces equal**:
- For the forces to balance (i.e., \( F_1 = F_2 \)):
\[ \frac{G \cdot m \cdot 1}{x^2} = \frac{G \cdot m \cdot 4}{(12 - x)^2} \]
- Cancel \( G \) and \( m \) from both sides:
\[ \frac{1}{x^2} = \frac{4}{(12 - x)^2} \]

5. **Cross-multiply to solve for \( x \)**:
- Cross-multiplying gives:
\[ (12 - x)^2 = 4x^2 \]
- Expanding the left side:
\[ 144 - 24x + x^2 = 4x^2 \]
- Rearranging the equation:
\[ 144 - 24x + x^2 - 4x^2 = 0 \]
\[ -3x^2 - 24x + 144 = 0 \]
\[ 3x^2 + 24x - 144 = 0 \]

6. **Divide the entire equation by 3**:
\[ x^2 + 8x - 48 = 0 \]

7. **Use the quadratic formula**:
- The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 1, b = 8, c = -48 \):
\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-48)}}{2 \cdot 1} \]
\[ x = \frac{-8 \pm \sqrt{64 + 192}}{2} \]
\[ x = \frac{-8 \pm \sqrt{256}}{2} \]
\[ x = \frac{-8 \pm 16}{2} \]

8. **Calculate the two possible values for \( x \)**:
- \( x = \frac{8}{2} = 4 \, \text{cm} \) (valid since it's positive)
- \( x = \frac{-24}{2} = -12 \, \text{cm} \) (not valid since distance cannot be negative)

9. **Conclusion**:
- The distance from the 1 kg mass at which the gravitational force on any mass becomes zero is \( \boxed{4 \, \text{cm}} \).