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Gravitation question

2019 · 9 Apr · Shift 1 · Q63
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  5. /2019 · 9 Apr · Shift 1 · Q63

Gravitation question

2019 · 9 Apr · Shift 1 · Q63

JEE MainPhysicsGravitationMCQ+4 / −1
A solid sphere of mass 'M' and radius 'a' is surrounded by a uniform concentric spherical shell of thickness 2a and mass 2M. The gravitational field at distance '3a' from the centre will be :
  1. A
    GM3a2{{GM} \over {3{a^2}}}3a2GM​
  2. B
    2GM9a2{{2GM} \over {9{a^2}}}9a22GM​
  3. C
    GM9a2{{GM} \over {9{a^2}}}9a2GM​
  4. D
    2GM3a2{{2GM} \over {3{a^2}}}3a22GM​
View written solutionFree

Correct answer: A

  1. Identify the mass distribution
  • A solid sphere has mass MMM and radius aaa.
  • It is surrounded by a concentric spherical shell of thickness 2a2a2a and mass 2M2M2M.

So the shell extends from radius aaa to radius 3a3a3a.

We need the gravitational field at distance r=3ar = 3ar=3a from the centre.


  1. Use the shell theorem / spherical symmetry

For any spherically symmetric mass distribution, the gravitational field at radius rrr depends only on the total mass enclosed within radius rrr:

g(r)=GMenclosedr2g(r)=\frac{G M_{\text{enclosed}}}{r^2}g(r)=r2GMenclosed​​

directed toward the centre.

At r=3ar=3ar=3a, the entire solid sphere and the entire shell are enclosed.

Thus,

Menclosed=M+2M=3MM_{\text{enclosed}} = M + 2M = 3MMenclosed​=M+2M=3M
  1. Compute the field at r=3ar=3ar=3a
g=G(3M)(3a)2g = \frac{G(3M)}{(3a)^2}g=(3a)2G(3M)​ g=3GM9a2=GM3a2g = \frac{3GM}{9a^2} = \frac{GM}{3a^2}g=9a23GM​=3a2GM​
  1. Match with the options

The value is

GM3a2\boxed{\frac{GM}{3a^2}}3a2GM​​

This corresponds to Option A.


  1. Verification with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.

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