Q60 / 75 · JEE Main 2025 (22 Jan Shift 1) · Physics

A small point of mass $m$ is placed at a distance $2 R$ from the centre ' $O^{\prime}$ 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 $\mathrm{F}_2$. The value of ratio $\mathrm{F}_1: \mathrm{F}_2$ is Figure

  • Correct answer
    A
    $12: 11$
  • B
    $11: 10$
  • C
    $12: 9$
  • D
    $16: 9$

Solution

$\begin{aligned} & \mathrm{F}_1=\frac{\mathrm{GMm}}{(2 \mathrm{R})^2} .....(1)\\ & \mathrm{~F}_2=\frac{\mathrm{GMm}}{(2 \mathrm{R})^2}-\left(\frac{\mathrm{G}\left(\frac{\mathrm{M}}{27}\right) \mathrm{m}}{\left(\frac{4 \mathrm{R}}{3}\right)^2}\right) \\ & \mathrm{F}_2=\frac{11}{48} \frac{\mathrm{GMm}}{\mathrm{R}^2} ....(2)\\ & \mathrm{~F}_1: \mathrm{F}_2=12: 11 \end{aligned}$

Hence, the correct answer is option: 'A'.

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