JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2025
- Hard
- MCQ
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 _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(12: 11\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{F}_1=\frac{\mathrm{GMm}}{(2 \mathrm{R})^2} .....(1)\\
& \mathrm{~F}_2=\frac{\mathrm{GMm}}{(2 \mathrm{R})^2}-\left(\frac{\mathrm{G}\left(\frac{\mathrm{M}}{27}\right) \mathrm{m}}{\left(\frac{4 \mathrm{R}}{3}\right)^2}\right) \\
& \mathrm{F}_2=\frac{11}{48} \frac{\mathrm{GMm}}{\mathrm{R}^2} ....(2)\\
& \mathrm{~F}_1: \mathrm{F}_2=12: 11
\end{aligned}\)
& \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}\)
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