Two masses 'M' and '4M' are at a distance 'r' apart on the line joining them. 'P' is point where the resultant gravitational force is zero (such a point is called as null point). The distance of 'P' from the mass '4M' is

Options

A. `r/5`
B. `r/3`
C. `(2r)/3`
D. `(4r)/5`

Explanation

To find the distance of point 'P' from the mass '4M' where the resultant gravitational force is zero, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Setup**:
- We have two masses: \( M \) and \( 4M \) separated by a distance \( r \).
- Let point \( P \) be at a distance \( x \) from the mass \( 4M \). Therefore, the distance from mass \( M \) to point \( P \) will be \( r - x \).

2. **Gravitational Forces**:
- The gravitational force exerted by mass \( M \) on a mass \( m \) at point \( P \) is given by:
\[ F_1 = \frac{G \cdot M \cdot m}{(r - x)^2} \]
- The gravitational force exerted by mass \( 4M \) on the mass \( m \) at point \( P \) is given by:
\[ F_2 = \frac{G \cdot 4M \cdot m}{x^2} \]

3. **Setting Forces Equal**:
- For the gravitational force to be zero at point \( P \), the magnitudes of \( F_1 \) and \( F_2 \) must be equal:
\[ F_1 = F_2 \]
- This leads to the equation:
\[ \frac{G \cdot M \cdot m}{(r - x)^2} = \frac{G \cdot 4M \cdot m}{x^2} \]
- We can cancel \( G \) and \( m \) from both sides since they are common factors:
\[ \frac{M}{(r - x)^2} = \frac{4M}{x^2} \]

4. **Simplifying the Equation**:
- Cancel \( M \) from both sides:
\[ \frac{1}{(r - x)^2} = \frac{4}{x^2} \]
- Cross-multiplying gives:
\[ x^2 = 4(r - x)^2 \]

5. **Expanding and Rearranging**:
- Expanding the right side:
\[ x^2 = 4(r^2 - 2rx + x^2) \]
- This simplifies to:
\[ x^2 = 4r^2 - 8rx + 4x^2 \]
- Rearranging gives:
\[ 0 = 3x^2 - 8rx + 4r^2 \]

6. **Solving the Quadratic Equation**:
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 3 \), \( b = -8r \), and \( c = 4r^2 \).
\[ x = \frac{8r \pm \sqrt{(-8r)^2 - 4 \cdot 3 \cdot 4r^2}}{2 \cdot 3} \]
\[ x = \frac{8r \pm \sqrt{64r^2 - 48r^2}}{6} \]
\[ x = \frac{8r \pm \sqrt{16r^2}}{6} \]
\[ x = \frac{8r \pm 4r}{6} \]
- This gives two possible solutions:
\[ x = \frac{12r}{6} = 2r \quad \text{or} \quad x = \frac{4r}{6} = \frac{2r}{3} \]

7. **Choosing the Valid Solution**:
- Since \( P \) must be between the two masses, the valid solution is:
\[ x = \frac{2r}{3} \]

8. **Finding the Distance from Mass \( 4M \)**:
- The distance of point \( P \) from mass \( 4M \) is:
\[ x = \frac{2r}{3} \]

### Final Answer
The distance of point \( P \) from the mass \( 4M \) is \( \frac{2r}{3} \).