Question

A gravity of radius $R/2$ is made inside a solid sphere of radius R. The center of the gravity is located at a distance $R/2$ from the center of the sphere. The gravitational force on a particle of mass m at a distance $R/2$ from the center of the sphere on the line joining both the centers of the sphere and the gravity is (opposite to the center of the gravity)[Here $g=\left(GM\right)/R^2$, where M is the mass of the sphere]

• A. $\frac{mg}{2}$
• B. $\frac{3mg}{8}$
• C. $\frac{mg}{16}$
• D. None of these

Answer 1

The correct option is B $\frac{3mg}{8}$

Consider the figure

Gravitational force inside a solid sphere $=\frac{GMm}{R^3}\times r.....\left(1\right)$

Where $r$ is distance from centre.

Gravitational force due to solid sphere outside$=\frac{GMm}{r^2}....\left(2\right)$ where $r$ is distance from centre.

Gravitational force due to remaining body $=$ gravitational force due to $9$ solid sphere due to cavity$)$

Mass of material inside cavity

$=M^1=M\times \frac{\frac{4}{3}\pi (\frac{R}{2}{)}^3}{\frac{4}{3}\pi R^3}=\frac{M}{8}$

$\therefore F_{body}=\frac{GMm}{R^3}\times \left(\frac{R}{2}\right)-\frac{GMm}{8{\left(\frac{R}{1}\right)}^2}$

$\frac{GMm}{2R^2}-\frac{GMm}{8R^2}\frac{GM}{R^2}.m(\frac{1}{2}-\frac{1}{8})mg.\left(\frac{4-1}{8}\right)=\frac{3mg}{8}=F_{body}$