Verified Solution • 2025–26 Session moderate

QJEEM24Ph095b: JEE Main 2024 Physics - Class 11 JEEM24Ph095b Solution: A bob of mass ' ' is suspended by a light string of length ' '. …

A bob of mass ' mm ' is suspended by a light string of length ' LL '. It is imparted a minimum horizontal velocity at the lowest point AA such that it just completes half circle reaching the top most position B. The ratio of kinetic energies (K.E)A(K.E)B\frac{(K . E)_A}{(K . E)_B} is :

JEE Main 2024 (Online) 29th January Evening Shift Physics - Work Power & Energy Question 23 English

Multiple Choice Options

  • A 5:01
  • B 3:02
  • C 1:05
  • D 2:05

Step-by-Step Verified Solution

Apply energy conservation between A & B

amp;12mVL2=12mVH2+mg(2 L)amp;∵VL=5gL\begin{aligned}& \frac{1}{2} \mathrm{mV}_{\mathrm{L}}^2=\frac{1}{2} \mathrm{mV}_{\mathrm{H}}^2+\mathrm{mg}(2 \mathrm{~L}) \\& \because \mathrm{V}_{\mathrm{L}}=\sqrt{5 \mathrm{gL}}\end{aligned}

So, VH=gL\mathrm{V}_{\mathrm{H}}=\sqrt{\mathrm{gL}}

(K.E)A(K.E)B=12 m(5gL)212 m(gL)2=51\frac{(\mathrm{K} . \mathrm{E})_{\mathrm{A}}}{(\mathrm{K} . \mathrm{E})_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{~m}(\sqrt{5 \mathrm{gL}})^2}{\frac{1}{2} \mathrm{~m}(\sqrt{\mathrm{gL}})^2}=\frac{5}{1}

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