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Gravitation question

2022 · 27 Jun · Shift 2 · Q50
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  5. /2022 · 27 Jun · Shift 2 · Q50

Gravitation question

2022 · 27 Jun · Shift 2 · Q50

JEE MainPhysicsGravitationMCQ+4 / −1
Four spheres each of mass m from a square of side d (as shown in figure). A fifth sphere of mass M is situated at the centre of square. The total gravitational potential energy of the system is : JEE Main 2022 (Online) 27th June Evening Shift Physics - Gravitation Question 96 English
  1. A
    −Gmd[(4+2)m+42M]- {{Gm} \over d}\left[ {(4 + \sqrt 2 )m + 4\sqrt 2 M} \right]−dGm​[(4+2​)m+42​M]
  2. B
    −Gmd[(4+2)M+42m]- {{Gm} \over d}\left[ {(4 + \sqrt 2 )M + 4\sqrt 2 m} \right]−dGm​[(4+2​)M+42​m]
  3. C
    −Gmd[3m2+42M]- {{Gm} \over d}\left[ {3{m^2} + 4\sqrt 2 M} \right]−dGm​[3m2+42​M]
  4. D
    −Gmd[6m2+42M]- {{Gm} \over d}\left[ {6{m^2} + 4\sqrt 2 M} \right]−dGm​[6m2+42​M]
View written solutionFree

Correct answer: A

  1. Use the formula for gravitational potential energy

For two point masses m1m_1m1​ and m2m_2m2​ separated by distance rrr,

U=−Gm1m2rU=-\frac{Gm_1m_2}{r}U=−rGm1​m2​​

The total potential energy of the system is the sum over all distinct pairs.

  1. Identify all interacting pairs

There are 5 masses total:

  • 4 corner spheres, each of mass mmm
  • 1 central sphere of mass MMM

So total energy =

  • energy among the 4 corner masses
  • energy between the centre mass and each corner mass
  1. Potential energy among the 4 corner masses

For a square of side ddd:

  • 4 sides of length ddd
  • 2 diagonals of length d2d\sqrt{2}d2​

(a) Side pairs

There are 4 such pairs, each contributing

−Gm2d-\frac{Gm^2}{d}−dGm2​

So total from sides:

U1=4(−Gm2d)=−4Gm2dU_1 = 4\left(-\frac{Gm^2}{d}\right)=-\frac{4Gm^2}{d}U1​=4(−dGm2​)=−d4Gm2​

(b) Diagonal pairs

There are 2 such pairs, each contributing

−Gm2d2-\frac{Gm^2}{d\sqrt{2}}−d2​Gm2​

So total from diagonals:

U2=2(−Gm2d2)=−2Gm2dU_2 = 2\left(-\frac{Gm^2}{d\sqrt{2}}\right) = -\frac{\sqrt{2}Gm^2}{d}U2​=2(−d2​Gm2​)=−d2​Gm2​

Thus, total energy among corner masses is

Ucorners=U1+U2=−Gm2d(4+2)U_{\text{corners}}=U_1+U_2 = -\frac{Gm^2}{d}(4+\sqrt{2})Ucorners​=U1​+U2​=−dGm2​(4+2​)
  1. Potential energy between the central mass and the 4 corner masses

Distance from centre of square to any corner is half the diagonal:

r=d22=d2r=\frac{d\sqrt{2}}{2}=\frac{d}{\sqrt{2}}r=2d2​​=2​d​

Energy for one corner-centre pair:

−GmMd/2=−2GmMd-\frac{GmM}{d/\sqrt{2}}=-\frac{\sqrt{2}GmM}{d}−d/2​GmM​=−d2​GmM​

There are 4 such pairs, so

Ucentre-corners=4(−2GmMd)=−42GmMdU_{\text{centre-corners}}=4\left(-\frac{\sqrt{2}GmM}{d}\right) = -\frac{4\sqrt{2}GmM}{d}Ucentre-corners​=4(−d2​GmM​)=−d42​GmM​
  1. Total potential energy

Add both contributions:

U=−Gm2d(4+2)−42GmMdU= -\frac{Gm^2}{d}(4+\sqrt{2}) - \frac{4\sqrt{2}GmM}{d}U=−dGm2​(4+2​)−d42​GmM​

Factor out −Gmd-\frac{Gm}{d}−dGm​:

U=−Gmd[(4+2)m+42M]U=-\frac{Gm}{d}\left[(4+\sqrt{2})m+4\sqrt{2}M\right]U=−dGm​[(4+2​)m+42​M]
  1. Match with options

This matches Option A:

−Gmd[(4+2)m+42M]- \frac{Gm}{d}\left[(4+\sqrt 2)m + 4\sqrt 2 M\right]−dGm​[(4+2​)m+42​M]

So the correct answer is A.

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