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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 \(.........cm s ^{-1}\).

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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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