Verified Solution • 2025–26 Session moderate

QJEEM17Phf4e9: JEE Main 2017 Physics - Class 11 JEEM17Phf4e9 Solution: A uniform disc of radius R and mass M is free to rotate only…

A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is :

JEE Main 2017 (Online) 8th April Morning Slot Physics - Rotational Motion Question 183 English

Multiple Choice Options

  • A 2  mg2 m+M{{2\,\,mg} \over {2\,m + M}}
  • B 2  Mg2 m+M{{2\,\,Mg} \over {2\,m + M}}
  • C 2  mg2 M+m{{2\,\,mg} \over {2\,M + m}}
  • D 2  Mg2 M+M{{2\,\,Mg} \over {2\,M + M}}

Step-by-Step Verified Solution

JEE Main 2017 (Online) 8th April Morning Slot Physics - Rotational Motion Question 183 English Explanation
From figure, we can say

ma = mg −- T . . . . . (1)

Moment of inertia of a uniform disc,

I = 12{1 \over 2} MR2

and the acceleration of the disc, a = ∝\propto R

We know,

Torque = I ∝\propto = RT

∴   \therefore\,\,\, RT = 12{1 \over 2} M R2 ×\times aR{a \over R} = MaR2{{MaR} \over 2}

⇒\Rightarrow    \,\,\, T = Ma2{{Ma} \over 2}

Putting the value of T in eq (1).

ma = mg −- Ma2{{Ma} \over 2}

⇒\Rightarrow    \,\,\, a ( m + Ma2{{Ma} \over 2} ) = mg

⇒\Rightarrow    \,\,\, a = 2mg2m+M{{2mg} \over {2m + M}}

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