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JEE Main 202529 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Morning Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2025 · 29 January · Morning Shift · Physics · Work, Power and Energy

A body of mass ' m ' connected to a massless and unstretchable...

A body of mass ' mm ' connected to a massless and unstretchable string goes in verticle circle of radius ' RR 'under gravity gg. The other end of the string is fixed at the center of circle. If velocity at top of circular path is (n \sqrt{g R}), where, (n \geqslant 1), then ratio of kinetic energy of the body at bottom to that at top of the circle is

Options

  1. A(\frac{n^2}{n^2+4})
  2. B(\frac{n^2+4}{n^2})
  3. C(\frac{n+4}{n})
  4. D(\frac{n}{n+4})

Correct answer

B. (\frac{n^2+4}{n^2})

Step-by-step solution

v0=v2+2g(2R)v0=n2gR+4gR∴kbottom ktop =v02v2=n2+4n2\begin{aligned} & v_0=\sqrt{v^2+2 g(2 R)} \\ & v_0=\sqrt{n^2 g R+4 g R} \\ & \therefore \quad \frac{k_{\text{bottom }}}{k_{\text{top }}}=\frac{v_0^2}{v^2}=\frac{n^2+4}{n^2}\end{aligned}
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