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Gravitation question

2005 · Shift 0 · Q167
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Gravitation question

2005 · Shift 0 · Q167

JEE MainPhysicsGravitationMCQ+4 / −1
A particle of mass 10g10g10g is kept on the surface of a uniform sphere of mass 100kg100kg100kg and radius 10cm.10cm.10cm. Find the work to be done against the gravitational force between them to take the particle far away from the sphere (you may take G=6.67×10−11  Nm2/kg2G = 6.67 \times {10^{ - 11}}\,\,N{m^2}/k{g^2}G=6.67×10−11Nm2/kg2)
  1. A
    3.33×10−10 J3.33 \times {10^{ - 10}}\,J3.33×10−10J
  2. B
    13.34×10−10 J13.34 \times {10^{ - 10}}\,J13.34×10−10J
  3. C
    6.67×10−10 J6.67 \times {10^{ - 10}}\,J6.67×10−10J
  4. D
    6.67×10−9 J6.67 \times {10^{ - 9}}\,J6.67×10−9J
View written solutionFree

Correct answer: C

  1. Given data
  • Mass of particle: m=10 g=0.01 kgm = 10\,g = 0.01\,kgm=10g=0.01kg
  • Mass of sphere: M=100 kgM = 100\,kgM=100kg
  • Radius of sphere: R=10 cm=0.1 mR = 10\,cm = 0.1\,mR=10cm=0.1m
  • Gravitational constant: G=6.67×10−11 N m2/kg2G = 6.67 \times 10^{-11}\,\text{N m}^2/\text{kg}^2G=6.67×10−11N m2/kg2
  1. Concept used

For a particle kept on the surface of a uniform sphere, the gravitational potential energy is

U=−GMmRU = -\frac{GMm}{R}U=−RGMm​

To take the particle from the surface to infinity, the final potential energy is

U∞=0U_\infty = 0U∞​=0

Hence, the work done against gravity is

W=U∞−U=0−(−GMmR)=GMmRW = U_\infty - U = 0 - \left(-\frac{GMm}{R}\right) = \frac{GMm}{R}W=U∞​−U=0−(−RGMm​)=RGMm​

  1. Substitute the values

W=(6.67×10−11)(100)(0.01)0.1W = \frac{(6.67\times 10^{-11})(100)(0.01)}{0.1}W=0.1(6.67×10−11)(100)(0.01)​

Now simplify step by step:

100×0.01=1100 \times 0.01 = 1100×0.01=1

So,

W=6.67×10−110.1W = \frac{6.67\times 10^{-11}}{0.1}W=0.16.67×10−11​

Since dividing by 0.10.10.1 is multiplying by 101010,

W=6.67×10−10 JW = 6.67\times 10^{-10}\,JW=6.67×10−10J

  1. Match with options

This corresponds to:

Option C: 6.67×10−10 J6.67 \times 10^{-10}\,J6.67×10−10J

  1. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So, the derived answer agrees with the stored answer.

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