JEE MAIN - Physics (2021 - 25th February Morning Shift - No. 2)
A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force F1. Now a spherical cavity of radius $$\left( {{R \over 2}} \right)$$ is made in the sphere (as shown in figure) and the force becomes F2. The value of F1 : F2 is
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36 : 25
41 : 50
50 : 41
25 : 36
Explanation
$${g_1} = {{GM} \over {{{(3R)}^2}}} = {{GM} \over {9{R^2}}}$$
$${g^2} = {{GM} \over {9{R^2}}} - {{G\left( {{M \over 8}} \right)} \over {{{\left( {3R - {R \over 2}} \right)}^2}}}$$
$$ = {{GM} \over {9{R^2}}} - {{GM} \over {{R^2}50}} = {{41} \over {9 \times 50}}{{GM} \over {{R^2}}}$$
$${{{g_1}} \over {{g_2}}} = {{41} \over {50}}$$
$$ \therefore $$ $$ {{{F_1}} \over {{F_2}}} = {{m{g_1}} \over {m{g_2}}} = {{41} \over {50}}$$
$${g^2} = {{GM} \over {9{R^2}}} - {{G\left( {{M \over 8}} \right)} \over {{{\left( {3R - {R \over 2}} \right)}^2}}}$$
$$ = {{GM} \over {9{R^2}}} - {{GM} \over {{R^2}50}} = {{41} \over {9 \times 50}}{{GM} \over {{R^2}}}$$
$${{{g_1}} \over {{g_2}}} = {{41} \over {50}}$$
$$ \therefore $$ $$ {{{F_1}} \over {{F_2}}} = {{m{g_1}} \over {m{g_2}}} = {{41} \over {50}}$$
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