JEE MAIN - Physics (2025 - 22nd January Morning Shift - No. 1)
A small point of mass $m$ is placed at a distance $2 R$ 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 $\mathrm{F}_1$. A spherical part of radius $\mathrm{R} / 3$ 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
_22nd_January_Morning_Shift_en_1_1.png)
Explanation
_22nd_January_Morning_Shift_en_1_2.png)
Given mass of whole sphere = $M$
$M_2$ = mass of removed sphere
$$ = {M \over {{4 \over 3}\pi {R^3}}}{4 \over 3}\pi {\left( {{R \over 3}} \right)^3} = {M \over {27}}$$
Due to whole sphere,
$${F_1} = {{GMm} \over {{{(2R)}^2}}}$$ (using Newton's gravitational law)
$${F_2}$$ =Force due to whole sphere $-$ Force due to removed sphere
$$ = {{GMm} \over {{{(2R)}^2}}} - {{G{M_2}m} \over {{{\left( {{{4R} \over 3}} \right)}^2}}}$$
$$ = {{GMm} \over {4{R^2}}} - {{G\left( {{M \over {27}}} \right)m} \over {{{16} \over 9}{R^2}}}$$
$$ = {{GMm} \over {4{R^2}}}\left[ {1 - {1 \over {12}}} \right] = {{4Mm} \over {4{R^2}}} \times {{11} \over {12}}$$
So, $$ = {{{F_1}} \over {{F_2}}} = {{GMm} \over {4{R^2}}}/{{GMm} \over {4{R^2}}} \times {{11} \over {12}} = {{12} \over {11}}$$
$$ \Rightarrow {F_1}:{F_2} = 12:11$$
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