Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Rotational Kinetic Energy, Work & Power›Question 55

NEET 1990 (AIPMT) · Rotational Kinetic Energy, Work & Power

Rotational Kinetic Energy, Work & Power question from NEET 1990 (AIPMT)

Q55NEET 1990 · AIPMTEasy

Rotational Kinetic Energy, Work & Power

The moment of inertia of a body about a given axis is 1.2 kg m21.2\text{ kg m}^{2}. Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 joule, an angular acceleration of 25 radian/sec2 must be applied about that axis for a duration of

Show answer and solution

Answer: (b)

  1. 12Iω2=1500⇒ω2=30001.2=2500\frac{1}{2}I\omega^2 = 1500 \Rightarrow \omega^2 = \frac{3000}{1.2} = 2500, ω=50\omega = 50 rad/s
  2. From rest: ω=αt⇒t=5025=2\omega = \alpha t \Rightarrow t = \frac{50}{25} = 2 s
  3. So the answer is (b) 2 s2\text{ s}.
Report a mistake

More questions on Rotational Kinetic Energy, Work & Power