Question:

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

Updated On: Mar 19, 2025
  • \(\frac{2}{9}F\)

  • \(\frac{16}{9}F\)

  • \(\frac{8}{9}F\)

  • \(F\)

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The Correct Option is C

Solution and Explanation

The correct answer is (C) : \(\frac{8}{9}F\)

Let the masses are m and distance between them is l, then
\(F=\frac{Gm^2}{I^2}\)
When 1/3rd mass is transferred to the other then masses will be 4m/3 and 2m/3. So new force will be
\(F^′=\frac{G\frac{4m}{3}×\frac{2m}{3}}{I^2}\)
\(=\frac{8}{9}\frac{Gm^2}{I^2}=\frac{8}{9}F\)

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Questions Asked in JEE Main exam

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Concepts Used:

Gravitation

In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.

Newton’s Law of Gravitation

According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,

  • F ∝ (M1M2) . . . . (1)
  • (F ∝ 1/r2) . . . . (2)

On combining equations (1) and (2) we get,

F ∝ M1M2/r2

F = G × [M1M2]/r2 . . . . (7)

Or, f(r) = GM1M2/r2

The dimension formula of G is [M-1L3T-2].