Question

A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F₁ on a particle placed at P, distance 2R from the centre O of the sphere. A spherical cavity of radius R/2 is now made in the sphere as shown in figure. The sphere with cavity now applies a gravitational force F₂ on same particle placed at P. The ratio $\frac{F_2}{F_1}$ will be

• A. 1/2
• B. 7/9
• C. 3
• D. 7

Answer 1

The correct option is B

7/9

$F_1$ = $\frac{G\frac{4}{3}\pi r^3\rho m}{4r^2}$ = $\frac{1}{3}$ $\pi$ $\rho$ Gm R

$F_2$ = $\frac{G\frac{4}{3}\pi r^3\rho m}{4r^2}$ - $\frac{G\frac{4}{3}\pi \frac{r^3}{8}\rho m}{\frac{9}{4}{R}^2}$

= $\frac{1}{3}$ $\pi$ $\rho$ GmR - $\frac{2}{27}$ $\pi$ $\rho$ Gm R

$\therefore$ $\frac{F_1}{F_2}$ = $\frac{\frac{1}{3}}{\frac{1}{3}-\frac{2}{27}}$ = $\frac{1}{1-\frac{2}{9}}$ = $\frac{9}{7}$

$\to$ $\frac{F_1}{F_2}$ = $\frac{7}{9}$