Question:

A point $P(\sqrt3R,0,0)$ lies on the axis of a ring of a mass 'M' and radius 'R'. The ring is located in y-z plane with its centre at origin 'O'. A small particle of mass 'm' starts from 'P' and reaches 'O' under gravitational attraction only. Its speed at 'O' will be

Updated On: Jul 5, 2022
  • $\sqrt\frac{GM}{R}$
  • $\sqrt\frac{Gm}{R}$
  • $\sqrt\frac{GM}{\sqrt2R}$
  • $\sqrt\frac{GM}{\sqrt {R}}$
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The Correct Option is A

Solution and Explanation

According to law of conservation of energy $\frac{1}{2} m V^2 = m(\Delta V)$ $\Rightarrow \frac{1}{2} mV^2 = m \left[ \frac{-GM}{2R} - \left( - \frac{GM}{R}\right) \right]$ $\Rightarrow \frac{1}{2} m V^2 = m \left(\frac{GM}{2R}\right) \, \Rightarrow V = \sqrt{\frac{GM}{R}}$
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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].