Verified Solution • 2025–26 Session moderate

QJEEM19Ph405a: JEE Main 2019 Physics - Class 11 JEEM19Ph405a Solution: Moment of inertia of a body about a given axis is 1.5 kg m 2 .…

Moment of inertia of a body about a given axis is 1.5 kg m2. Initially the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular accleration of 20 rad/s2 must be applied about the axis for a duration of :-

Multiple Choice Options

  • A 2.5 s
  • B 3 s
  • C 5s
  • D 2 s

Step-by-Step Verified Solution

KE = 12Iω2=1200{1 \over 2}I{\omega ^2} = 1200 (given)

⇒ω=40 rad/s\Rightarrow \omega = 40\,rad/s

⇒ω=ω0+αt\Rightarrow \omega = {\omega _0} + \alpha t

⇒40=0+(20)t\Rightarrow 40 = 0 + (20)t

⇒t=2 sec⁡\Rightarrow t = 2\,\sec

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