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JEE MAIN 2021
25-02-21 S1
Question
A solid sphere of radius $R$ gravitationally attracts a particle placed at $3 R$ form its centre with a force $F_1$. Now a spherical cavity of radius $\left(\frac{R}{2}\right)$ is made in the sphere (as shown in figure) and the force becomes $F_2$. The value of $F_1: F_2$ is:
Question Image
Select the correct option:
A
25 : 36
B
36 : 25
C
50 : 41
D
41 : 50
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Main
Physics
Medium
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