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Gravitation question

2022 · 28 Jun · Shift 2 · Q55
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  5. /2022 · 28 Jun · Shift 2 · Q55

Gravitation question

2022 · 28 Jun · Shift 2 · Q55

JEE MainPhysicsGravitationMCQ+4 / −1
Two objects of equal masses placed at certain distance from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other object, then the new force will be :
  1. A
    29{2 \over 9}92​ F
  2. B
    169{16 \over 9}916​ F
  3. C
    89{8 \over 9}98​ F
  4. D
    F
View written solutionFree

Correct answer: C

  1. Initial masses and force

Let each object initially have mass mmm, and the distance between them be rrr.

By Newton’s law of gravitation,

F=Gm⋅mr2=Gm2r2F = \frac{Gm\cdot m}{r^2} = \frac{Gm^2}{r^2}F=r2Gm⋅m​=r2Gm2​
  1. Transfer of one-third mass

One-third of the mass of one object is transferred to the other.

So the new masses become:

  • First object: m−m3=2m3m - \frac{m}{3} = \frac{2m}{3}m−3m​=32m​
  • Second object: m+m3=4m3m + \frac{m}{3} = \frac{4m}{3}m+3m​=34m​
  1. New gravitational force

The distance remains the same, so the new force is

F′=G(2m3)(4m3)r2F' = \frac{G\left(\frac{2m}{3}\right)\left(\frac{4m}{3}\right)}{r^2}F′=r2G(32m​)(34m​)​

Simplify:

F′=Gr2⋅8m29F' = \frac{G}{r^2}\cdot \frac{8m^2}{9}F′=r2G​⋅98m2​

But since

F=Gm2r2,F = \frac{Gm^2}{r^2},F=r2Gm2​,

we get

F′=89FF' = \frac{8}{9}FF′=98​F
  1. Option check
  • A: 29F\frac{2}{9}F92​F ❌
  • B: 169F\frac{16}{9}F916​F ❌
  • C: 89F\frac{8}{9}F98​F ✅
  • D: FFF ❌

Therefore, the correct answer is

89F\boxed{\frac{8}{9}F}98​F​
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