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JEE Main 202118 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
18 March
Shift
Evening Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2021 · 18 March · Evening Shift · Physics · Work, Power and Energy

A particle of mass m moves in a circular orbit under the central...

A particle of mass mm moves in a circular orbit under the central potential field, U(r)=−CrU\left(r\right) = \dfrac{- C}{r}, where CC is a positive constant. The correct radius - velocity graph of the particle's motion is :A particle of mass m moves in a circular orbit under the central potential field, U(r) = (- C)/(r) , where C is aA particle of mass m moves in a circular orbit under the central potential field, U(r) = (- C)/(r) , where C is aA particle of mass m moves in a circular orbit under the central potential field, U(r) = (- C)/(r) , where C is aA particle of mass m moves in a circular orbit under the central potential field, U(r) = (- C)/(r) , where C is a

Options

  1. AThe given circuit shows a uniform straight wire A B of 40 cm length fixed at both ends. In order to get zero reading in
  2. BThe given circuit shows a uniform straight wire A B of 40 cm length fixed at both ends. In order to get zero reading in
  3. CThe given circuit shows a uniform straight wire A B of 40 cm length fixed at both ends. In order to get zero reading in
  4. DThe given circuit shows a uniform straight wire A B of 40 cm length fixed at both ends. In order to get zero reading in

Correct answer

A. The given circuit shows a uniform straight wire A B of 40 cm length fixed at both ends. In order to get zero reading in

Step-by-step solution

U = - \dfrac{C}{r}</p><p></p><p>F = - \dfrac{dU}{dr} = - \dfrac{C}{{r}^{2}}</p><p>∣F∣=mv2r</p><p>\left|F\right| = \dfrac{m{v}^{2}}{r}

Cr2=mv2r\dfrac{C}{{r}^{2}} = \dfrac{m{v}^{2}}{r}

v2∝1r{v}^{2} ∝ \dfrac{1}{r}

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