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JEE Main 20188 April
Exam
JEE Main
Year
2018
Date
8 April
Subject
Physics
Chapter
Gravitation

JEE Main 2018 · 8 April · Physics · Gravitation

A particle is moving with a uniform speed in a circular orbit of...

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth{n}^{th} power of R. If the period of rotation of the particle is T, then:

Options

  1. AT∝Rn/2T ∝ {R}^{n / 2}
  2. BT∝R3/2T ∝ {R}^{3 / 2} for any n
  3. CT∝Rn2+1T ∝ {R}^{\dfrac{n}{2} + 1}
  4. DT∝Rn+12T ∝ {R}^{\dfrac{n + 1}{2}}

Correct answer

D. T∝Rn+12T ∝ {R}^{\dfrac{n + 1}{2}}

Step-by-step solution

F=KRn⇒mV2r=KRn⇒V2∝KRn−1F = \dfrac{K}{R^{n}}⇒ \dfrac{mV^{2}}{r} = \dfrac{K}{Rn}⇒ V^{2} ∝ \dfrac{K}{Rn - 1}
⇒V∝1Rn−12⇒T=2\piRV∝RV⇒ V ∝ \dfrac{1}{R\dfrac{n - 1}{2}}⇒ T = \dfrac{2\piR}{V} ∝ \dfrac{R}{V}
⇒T∝R1Rn−12⇒T∝Rn+12⇒ T ∝ \dfrac{R}{1R\dfrac{n - 1}{2}}⇒ T ∝ R\dfrac{n + 1}{2}
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