A small point of mass m is placed at a distance 2R from the centre 'O' of a big uniform solid sphere of mass M and radius R. The gravitational force on 'm' due to M is . A spherical part of radius R/3 is removed from the big sphere as shown in the figure

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A small point of mass m is placed at a distance 2R from the centre 'O' of a big uniform solid sphere of mass M and radius R. The gravitational force on 'm' due to M is $F_{1}$ . A spherical part of radius R/3 is removed from the big sphere as shown in the figure and the gravitational force on m due to remaining part of M is found to be $F_{2}$ . The value of ratio $F_{1} : F_{2}$ is [2025]

1 16 : 9

2 11 : 10

3 12 : 11

4 12 : 9

Ans.
(3)

$F_{1} = \frac{G M m}{4 R^{2}}$

$F_{2} = \frac{G M m}{4 R^{2}} - G (\frac{M}{27}) \frac{m}{{(\frac{4 R}{3})}^{2}}$

$\Rightarrow \frac{G M m}{R^{2}} (\frac{1}{4} - \frac{1}{48}) = \frac{11}{48} \frac{G M m}{R^{2}}$

$\Rightarrow \frac{F_{1}}{F_{2}} = \frac{12}{11}$

Want Us to Email you About Special Offers & Updates?