A small point mass m is placed at a distance | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The gravitational force on mass $m$ due to the complete sphere is given by:$F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2} \quad ...(1)$When a spheri

Vedclass · Accepted Answer

$12: 11$

Vedclass · Answer

(C) The gravitational force on mass $m$ due to the complete sphere is given by:$F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2} \quad ...(1)$When a spherical part of radius $r = R/3$ is removed,its mass $M'$ is calculated by considering the density $ho$ of the sphere:$M' = ho \cdot V' = \left( \frac{M}{\frac{4}{3}\pi R^3} ight) \cdot \left( \frac{4}{3}\pi (R/3)^3 ight) = \frac{M}{27}$The center of the removed sphere is at a distance $d = R - R/3 = 2R/3$ from the center $O$. The distance of this center from the point mass $m$ is $2R - 2R/3 = 4R/3$.The force $F_2$ exerted by the remaining part is the original force minus the force that would have been exerted by the removed part:$F_2 = F_1 - F_{	ext{removed}} = \frac{GMm}{4R^2} - \frac{G(M/27)m}{(4R/3)^2}$$F_2 = \frac{GMm}{4R^2} - \frac{GMm}{27 \cdot (16R^2/9)} = \frac{GMm}{4R^2} - \frac{GMm}{48R^2}$$F_2 = \frac{GMm}{R^2} \left( \frac{1}{4} - \frac{1}{48} ight) = \frac{GMm}{R^2} \left( \frac{12-1}{48} ight) = \frac{11}{48} \frac{GMm}{R^2}$Now,the ratio $F_1 : F_2$ is:$\frac{F_1}{F_2} = \frac{GMm / 4R^2}{11GMm / 48R^2} = \frac{1}{4} \cdot \frac{48}{11} = \frac{12}{11}$Thus,$F_1 : F_2 = 12 : 11$.

Similar Questions

$A$ study of binary stars is most helpful in

The solar constant on the surface of the earth is $S$. What will be its value on the surface of another planet which is about $5.3 \, AU$ away from the sun?

$A$ tunnel is dug in the Earth across one of its diameters. Two masses $m$ and $2m$ are dropped from the ends of the tunnel. The masses collide, stick to each other, and perform $S.H.M.$ The amplitude of the $S.H.M.$ will be:

$A$ planet has a core of density $3\rho$ and an outer crust of density $\rho$. There is a small tunnel through the core. $A$ small particle of mass $m$ is released from end $A$. What is the time required to reach end $B$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module