JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2023
- Medium
- Numerical
A solid sphere of mass \(1\,kg\) rolls without slipping on a plane surface. Its kinetic energy is \(7 \times 10^{-3}\,J\). The speed of the centre of mass of the sphere is \(.........cm s ^{-1}\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(10\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{2} mv ^2+\frac{1}{2} I \omega^2=7 \times 10^{-3}\)
\(\frac{1}{2} mv ^2+\frac{1}{2}\left(\frac{2}{5} MR ^2\right)\left(\frac{ V }{ R }\right)^2=7 \times 10^{-3}\)
\(\frac{1}{2} MV ^2\left[1+\frac{2}{5}\right]=7 \times 10^{-3}\)
\(\frac{1}{2}(1)\left( V ^2\right)\left(\frac{7}{5}\right)=7 \times 10^{-3}\)
\(V ^2=10^{-2}\)
\(V =10^{-1}=0.1\,m / s =10\,cm / s\)
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