JEE Main 2025 Jan 22 Shift 1, Physics Q13: Universal Law Of Gravitation And Field Intensity

JEE Main2025Jan 22, Shift 1Physics

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

A
16 : 9
B
11 : 10
C
12 : 11
D
12 : 9

Solution

$F_{1}$ comes from treating the full sphere as a point mass at O at distance $2 R$ :

$F_{1} = \frac{GM m}{( 2 R ) ^{2}} = \frac{GM m}{4 R ^{2}}$ .

Mass of removed sphere: $M^{'} = M \cdot \frac{( R /3 ) ^{3}}{R ^{3}} = \frac{M}{27}$ . Its centre sits at distance $2 R /3$ from O toward P, so distance from m is $2 R - 2 R /3 = 4 R /3$ .

Force the removed piece would have exerted on m: $F^{'} = \frac{G ( M /27 ) m}{( 4 R /3 ) ^{2}} = \frac{GM m}{27} \cdot \frac{9}{16 R ^{2}} = \frac{GM m}{48 R ^{2}}$ .

$F_{2} = F_{1} - F^{'} = \frac{GM m}{4 R ^{2}} - \frac{GM m}{48 R ^{2}} = \frac{12 - 1}{48} \cdot \frac{GM m}{R ^{2}} = \frac{11 GM m}{48 R ^{2}}$ .

$\frac{F _{1}}{F _{2}} = \frac{GM m /4 R ^{2}}{11 GM m /48 R ^{2}} = \frac{12}{11}$ .

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JEE Main 2025 Jan 22 Shift 1, Physics Q13: Universal Law Of Gravitation And Field Intensity

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