Verified Solution • 2025–26 Session moderate

QJEEM21Ph8d2b: JEE Main 2021 Physics - Class 11 JEEM21Ph8d2b Solution: A cord is wound round the circumference of wheel of radius r.…

A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance 'h', the square of angular velocity of wheel will be :

Multiple Choice Options

  • A 2mghI+2mr2{{2mgh} \over {I + 2m{r^2}}}
  • B 2mghI+mr2{{2mgh} \over {I + m{r^2}}}
  • C 2gh
  • D 2ghI+mr2{{2gh} \over {I + m{r^2}}}

Step-by-Step Verified Solution

JEE Main 2021 (Online) 26th February Evening Shift Physics - Rotational Motion Question 112 English Explanation

Using energy conservation between A and B point

mgh=12m(wR)2+12Iω2mgh = {1 \over 2}m{(wR)^2} + {1 \over 2}I{\omega ^2}

2mgh=(MR2+I)ω22mgh = (M{R^2} + I){\omega ^2}

ω2=2mghI+MR2{\omega ^2} = {{2mgh} \over {I + M{R^2}}}

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