A disc of moment of inertia (9.8/π²) kg m² is rotating at 600 rpm. If the frequency

⇦

⇨

A disc of moment of inertia (9.8/π²) kg m² is rotating at 600 rpm. If the frequency of rotation changes from 600 rpm to 300 rpm then what is the work done?

Options

(a) 1470 J
(b) 1452 J
(c) 1567 J
(d) 1632 J

Correct Answer:

1470 J

Explanation:

Given: Moment of inertia I = (9.8/π²) kgm²

ʋ₁ = 600 rpm = 10 rps; ʋ₂ = 300 rpm = 5 rps

.·. ω₁ = 2π ʋ₁ = 20π rad s⁻¹

.·. ω₂ = 2π ʋ₂ = 10π rad s⁻¹

Kinetic energy of rotation= (1/2) Iω²

Work done W = change in rotational kinetic energy

.·. work done W = (1/2).I [ω₂² – ω₂²]

W = (1/2) x (9.8/π²).[(10π)² – (20π)²]

= (1/2) x (9.8/π²).[-300 π²] = -1470 J

-ve sign show that rotational kinetic energy decreases.

Related Questions:

1. If dimensions of critical velocity of a liquid flowing through a tube are expressed
2. The half-life period of a radio-active element X is same as the mean life time
3. A body of mass m is orbiting the earth at a radius r from the center of earth
4. If two soap bubbles of different radii are connected by a tube
5. A spring has time period T. It is cut into n equal parts. The time period of each part

Topics: Motion of system of Particles and Rigid Body (73)
Subject: Physics (2479)

---

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

---

Share this page with your friends