A large solid sphere of diameter d attracts a small particle with a force F. If the central portion of the sphere of diameter `(d)/(2)` be removed leaving behind a cavity, then the new force of attraction becomes

Options

A. `(9F)/(8)`
B. `(7F)/(8)`
C. `(3F)/(2)`
D. `(3F)/(4)`

Answer

To solve the problem step by step, we will analyze the situation before and after the removal of the central portion of the sphere.

### Step 1: Understand the initial conditions
- We have a solid sphere with a diameter \( d \). Therefore, the radius \( R \) of the sphere is:
\[ R = \frac{d}{2} \]
- A small particle with mass \( m \) is attracted to the sphere with a force \( F \).

### Step 2: Determine the initial force of attraction
- The gravitational force \( F \) between the sphere and the particle is given by Newton's law of gravitation:
\[ F = \frac{G M m}{(R + x)^2} \]
where \( M \) is the mass of the sphere and \( x \) is the distance from the surface of the sphere to the particle.

### Step 3: Remove the central portion of the sphere
- A central portion of the sphere with a diameter \( \frac{d}{2} \) is removed. This means the radius of the removed cavity is:
\[ r = \frac{d}{4} \]

### Step 4: Calculate the mass of the removed portion
- The volume of the original sphere is:
\[ V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi \left(\frac{d}{2}\right)^3 = \frac{4}{3} \pi \frac{d^3}{8} = \frac{\pi d^3}{6} \]
- The volume of the removed cavity is:
\[ V_{removed} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left(\frac{d}{4}\right)^3 = \frac{4}{3} \pi \frac{d^3}{64} = \frac{\pi d^3}{48} \]
- The mass of the removed portion \( M_{removed} \) can be calculated using the density of the sphere:
\[ M_{removed} = \frac{M}{V} \times V_{removed} = \frac{M}{\frac{\pi d^3}{6}} \times \frac{\pi d^3}{48} = M \times \frac{6}{48} = \frac{M}{8} \]

### Step 5: Calculate the remaining mass of the sphere
- The mass of the remaining portion \( M' \) of the sphere after removing the cavity is:
\[ M' = M - M_{removed} = M - \frac{M}{8} = \frac{7M}{8} \]

### Step 6: Calculate the new force of attraction
- The new force of attraction \( F' \) between the small particle and the remaining mass of the sphere is given by:
\[ F' = \frac{G M' m}{(R + x)^2} = \frac{G \left(\frac{7M}{8}\right) m}{(R + x)^2} \]
- Since \( F = \frac{G M m}{(R + x)^2} \), we can express \( F' \) in terms of \( F \):
\[ F' = \frac{7}{8} F \]

### Conclusion
- The new force of attraction after removing the central portion of the sphere is:
\[ F' = \frac{7}{8} F \]