Verified Solution • 2025–26 Session moderate

QJEEM25Phe33a: JEE Main 2025 Physics - Class 11 JEEM25Phe33a Solution: A body of mass ‘m’ connected to a massless and unstretchable…

A body of mass ‘m’ connected to a massless and unstretchable string goes in vertical circle of radius ‘R’ under gravity g. The other end of the string is fixed at the center of circle. If velocity at top of circular path is ngRn\sqrt{ g R} , where, n ≥ 1, then ratio of kinetic energy of the body at bottom to that at top of the circle is :

Multiple Choice Options

  • A n2+4n2\frac{n^2 + 4}{n^2}
  • B n+4n\frac{n + 4}{n}
  • C n2n2+4\frac{n^2}{n^2 + 4}
  • D nn+4\frac{n}{n + 4}

Step-by-Step Verified Solution

VTop =n2gR VBotom =n2gR+4gR Ratio =n2+4n2

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring