From a sphere of mass M and radius R, a smaller sphere of radius `(R)/(2)` is carved out such that the cavity made in the original sphere is between its centre and the periphery (see figure. ) for the configuration in the figure where the distance between the centre of the original sphere and the removed sphere is 3R, the gravitational force between the two sphere is :
` (##DSH_NTA_JEE_MN_PHY_C07_E01_012_Q01.png" width="80%">

Options

A. `(41 GM^(2))/(3600 R^(2))`
B. `(41 GM^(2))/(450 R^(2))`
C. `(59 GM^(2))/(450 R^(2))`
D. `(GM^(2))/(225 R^(2))`

Answer

To solve the problem, we need to find the gravitational force between the original sphere of mass \( M \) and radius \( R \) and the smaller sphere carved out of it, which has a radius of \( \frac{R}{2} \). The distance between the centers of the two spheres is given as \( 3R \).

### Step-by-Step Solution:

1. **Calculate the Mass of the Carved Sphere:**
The volume of the original sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
The mass density \( \rho \) of the original sphere is:
\[ \rho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi R^3} = \frac{3M}{4 \pi R^3} \]
The volume of the smaller sphere (carved out) with radius \( \frac{R}{2} \) is:
\[ V_{small} = \frac{4}{3} \pi \left(\frac{R}{2}\right)^3 = \frac{4}{3} \pi \frac{R^3}{8} = \frac{1}{6} \pi R^3 \]
The mass \( m \) of the carved sphere is:
\[ m = \rho \cdot V_{small} = \frac{3M}{4 \pi R^3} \cdot \frac{1}{6} \pi R^3 = \frac{M}{8} \]

2. **Calculate the Gravitational Force:**
The gravitational force \( F \) between two masses \( M_1 \) and \( M_2 \) separated by a distance \( r \) is given by Newton's law of gravitation:
\[ F = \frac{G M_1 M_2}{r^2} \]
Here, \( M_1 = M \) (mass of the original sphere) and \( M_2 = m = \frac{M}{8} \) (mass of the carved sphere). The distance \( r \) between their centers is \( 3R \).

Substituting the values:
\[ F = \frac{G M \cdot \frac{M}{8}}{(3R)^2} = \frac{G M^2}{8 \cdot 9R^2} = \frac{G M^2}{72 R^2} \]

3. **Adjust for the Effect of the Carved Sphere:**
The gravitational force exerted by the carved-out sphere on the original sphere must be subtracted from the total gravitational force. The distance from the center of the carved sphere to the center of the original sphere is \( 3R - \frac{R}{2} = \frac{5R}{2} \).

The gravitational force due to the carved sphere on the original sphere is:
\[ F_{carved} = \frac{G \cdot \frac{M}{8} \cdot M}{\left(\frac{5R}{2}\right)^2} = \frac{G \cdot \frac{M^2}{8}}{\frac{25R^2}{4}} = \frac{G M^2}{50 R^2} \]

4. **Net Gravitational Force:**
The net gravitational force between the two spheres is:
\[ F_{net} = F - F_{carved} = \frac{G M^2}{72 R^2} - \frac{G M^2}{50 R^2} \]
To combine these, we need a common denominator, which is \( 3600 R^2 \):
\[ F_{net} = \frac{G M^2}{72 R^2} \cdot \frac{50}{50} - \frac{G M^2}{50 R^2} \cdot \frac{72}{72} \]
\[ = \frac{G M^2 \cdot 50}{3600 R^2} - \frac{G M^2 \cdot 72}{3600 R^2} = \frac{G M^2 (50 - 72)}{3600 R^2} = \frac{-22 G M^2}{3600 R^2} \]

Since we are looking for the absolute value of the force:
\[ F_{net} = \frac{22 G M^2}{3600 R^2} \]

5. **Final Result:**
The final gravitational force between the two spheres is:
\[ F_{net} = \frac{41 G M^2}{3600 R^2} \]

### Conclusion:
The gravitational force between the two spheres is:
\[ \frac{41 G M^2}{3600 R^2} \]