A uniform ring of mass m and radius $r$ is placed directly above a uniform sphere of mass $M$ and of equal radius. The centre of the ring is directly above the centre of the sphere at a distance $r\sqrt{3}$ as shown in the figure. The gravitational force exerted by the sphere on the ring will be
- $\frac{GMm}{8r^{2}}$
- $\frac{GMm}{4r^{2}}$
- $\sqrt{3}\frac{GMm}{8r^{2}}$
- $\frac{GMm}{8r^{3}\sqrt{3}}$
The Correct Option is C
Solution and Explanation
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(R) The mass of the pendulum remains unchanged at Earth and the other planet.
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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].