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Rotational Motion question

2019 · 9 Apr · Shift 2 · Q60
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Rotational Motion question

2019 · 9 Apr · Shift 2 · Q60

JEE MainPhysicsRotational MotionMCQ+4 / −1
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 :-
  1. A
    2.5 s
  2. B
    3 s
  3. C
    5s
  4. D
    2 s
View written solutionFree

Correct answer: D

  1. Given data
  • Moment of inertia: I=1.5 kg m2I = 1.5\ \text{kg m}^2I=1.5 kg m2
  • Initial angular velocity: ω0=0\omega_0 = 0ω0​=0
  • Required rotational kinetic energy: K=1200 JK = 1200\ \text{J}K=1200 J
  • Angular acceleration: α=20 rad/s2\alpha = 20\ \text{rad/s}^2α=20 rad/s2
  1. Use rotational kinetic energy formula

The rotational kinetic energy is

K=12Iω2K = \frac{1}{2} I \omega^2K=21​Iω2

Substitute the given values:

1200=12(1.5)ω21200 = \frac{1}{2}(1.5)\omega^21200=21​(1.5)ω2

1200=0.75ω21200 = 0.75\omega^21200=0.75ω2

ω2=12000.75=1600\omega^2 = \frac{1200}{0.75} = 1600ω2=0.751200​=1600

ω=40 rad/s\omega = 40\ \text{rad/s}ω=40 rad/s

  1. Use angular acceleration relation

Since the body starts from rest,

ω=ω0+αt\omega = \omega_0 + \alpha tω=ω0​+αt

40=0+20t40 = 0 + 20t40=0+20t

t=4020=2 st = \frac{40}{20} = 2\ \text{s}t=2040​=2 s

  1. Match with options

The required duration is

2 s\boxed{2\ \text{s}}2 s​

So the correct option is D.

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