An uniform sphere of mass \( M \) and radius \( R \) exerts a force of \( F \) on a small mass \( m \) placed at a distance of \( 3R \) from the centre of the sphere. A spherical portion of diameter \( R \) is cut from the sphere as shown in the fig. The force of attraction between the remaining part of the disc and the mass \( m \) is:
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For gravitational force problems involving spherical objects, the force exerted by a portion of the sphere can be determined by the ratio of the remaining mass to the total mass of the sphere.
The gravitational force between two masses \( M \) and \( m \) is given by Newton's law of gravitation:
\[
F = \frac{GMm}{r^2}
\]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the sphere,
- \( m \) is the mass placed at distance \( r \) from the center.
In this case, the mass \( m \) is at a distance of \( 3R \) from the center of the sphere, and the force exerted on \( m \) is \( F \). When a portion of the sphere with a diameter \( R \) is cut out, we must determine the new force exerted by the remaining part of the sphere on the mass \( m \).
The gravitational force exerted by the remaining portion of the sphere is proportional to the amount of mass left in the sphere. The mass of the cut portion is proportional to its volume, which is a spherical section with a radius \( R/2 \). The remaining mass is proportional to the volume of the remaining sphere, which is roughly \( \frac{41}{50} \) of the original sphere's mass.
Thus, the force of attraction between the remaining portion of the sphere and the mass \( m \) is \( \frac{41}{50} \) times the original force, i.e., the new force is:
\[
F_{\text{new}} = \frac{41}{50} F
\]
Thus, the correct answer is \( \frac{41}{50} F \).
