A solid sphere of mass 1,kg rolls without slipping on a plane surface. Its kinetic energy is 7 × 10

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6.System of Particles and Rotational Motion

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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 $.........cm s ^{-1}$.

A

$10$

B

$9$

C

$8$

D

$7$

(JEE MAIN-2023)

Solution

$\frac{1}{2} mv ^2+\frac{1}{2} I \omega^2=7 \times 10^{-3}$

$\frac{1}{2} mv ^2+\frac{1}{2}\left(\frac{2}{5} MR ^2\right)\left(\frac{ V }{ R }\right)^2=7 \times 10^{-3}$

$\frac{1}{2} MV ^2\left[1+\frac{2}{5}\right]=7 \times 10^{-3}$

$\frac{1}{2}(1)\left( V ^2\right)\left(\frac{7}{5}\right)=7 \times 10^{-3}$

$V ^2=10^{-2}$

$V =10^{-1}=0.1\,m / s =10\,cm / s$

Standard 11

Physics

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