Verified Solution • 2025–26 Session moderate

QJEEM25Phe888: JEE Main 2025 Physics - Class 11 JEEM25Phe888 Solution: A small point of mass is placed at a distance from the centre ' …

A small point of mass mm is placed at a distance 2R2 R from the centre ' OO ' of a big uniform solid sphere of mass M and radius R . The gravitational force on ' m ' due to M is F1\mathrm{F}_1 . A spherical part of radius R/3\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 MM is found to be F2F_2 . The value of ratio F1:F2F_1: F_2 is

JEE Main 2025 (Online) 22nd January Morning Shift Physics - Gravitation Question 6 English

Multiple Choice Options

  • A 11:10
  • B 12:11
  • C 16:09
  • D 12:09

Step-by-Step Verified Solution

amp;F1=GMm(2R)2.... (1)amp; F2=GMm(2R)2−(G(M27)m(4R3)2)amp;F2=1148GMmR2.... (2)amp; F1:F2=12:11\begin{aligned}& \mathrm{F}_1=\frac{\mathrm{GMm}}{(2 \mathrm{R})^2} \quad\text{.... (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} \quad\text{.... (2)}\\& \mathrm{~F}_1: \mathrm{F}_2=12: 11\end{aligned}

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