Physics · 2021
JEE Main · 25 February 2021, Shift 1 · Q20
A solid sphere of radius R gravitationally attracts a particle placed at 3 R from its centre with a force F_1. Now a spherical cavity of radius (R/2)…
A solid sphere of radius $\displaystyle R$ gravitationally attracts a particle placed at $\displaystyle 3 R$ from its centre with a force $\displaystyle \mathrm{F}_1$. Now a spherical cavity of radius $\displaystyle \left(\frac{\mathrm{R}}{2}\right)$ is made in the sphere (as shown in figure) and the force becomes $\displaystyle \mathrm{F}_2$. The value of $\displaystyle \mathrm{F}_1: \mathrm{F}_2$ is :
Official answer
(2)
$\displaystyle 50$ : $\displaystyle 41$
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JEE Main 2021 Physics question, with the answer from NTA’s final answer key. Where our answers come from.