Skip to main content
Back to SYSTEM OF PARTICLES

JEE PHYSICS MCQ: SYSTEM OF PARTICLES

1 Medium
1 of 1
Application +4 | -1
A body is rotating with angular velocity 20 rad/s. If
its kinetic energy is 400 J, then its moment of
inertia is
View answer and explanation
Correct option: A
2 kg m22\,kg\,{m^2}
Option A (2 kg m22\,kg\,{m^2}) is correct because the moment of inertia is calculated using the formula K.E=12Iω2K.E = \frac{1}{2}I\omega^2.

More Questions from SYSTEM OF PARTICLES

Practice All Chapter MCQs →
It is easier for a swimmer jumping into water from a height to describe a loop in the air by A uniform thin rod AB\mathrm{AB} of length L\mathrm{L} has linear mass density μ(x)=a+bxL\mu(x)=a+\frac{b x}{L}, where xx is measured from A\mathrm{A}. If the CM\mathrm{CM} of the rod lies at a distance of (712)L\left(\frac{7}{12}\right) \mathrm{L} from A\mathrm{A}, then a and b\mathrm{b} are related as : A solid metal ring and a disc of same radius and mass are rotating about their diameters with same angular frequency. The ratio of their respective rotational kinetic energy values is When a torque acting upon a system is zero, then which of the following will be constant A point particle of mass 1.0 g1.0 \mathrm{~g} is rotating in a circle of diameter 8 cm8 \mathrm{~cm}. Its moment of inertia will be The angular velocity of a particle is given by ω=t4+2t3\omega = {t^4} + 2{t^3} then its angular acceleration of the particle at t=1 sect = 1\,sec is (rad s−2)\left( {rad\,{s^{ - 2}}} \right)
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →