JEE MAIN - Physics (2023 - 31st January Morning Shift - No. 22)
A solid sphere of mass $$1 \mathrm{~kg}$$ rolls without slipping on a plane surface. Its kinetic energy is $$7 \times 10^{-3} \mathrm{~J}$$. The speed of the centre of mass of the sphere is __________ $$\operatorname{cm~s}^{-1}$$
Answer
10
Explanation
$\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=7 \times 10^{-3}$
$$ \Rightarrow $$ $\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{MR}^{2}\right)\left(\frac{\mathrm{V}}{\mathrm{R}}\right)^{2}=7 \times 10^{-3}$
$$ \Rightarrow $$ $\frac{1}{2} \mathrm{MV}^{2}\left[1+\frac{2}{5}\right]=7 \times 10^{-3}$
$$ \Rightarrow $$ $\frac{1}{2}(1)\left(\mathrm{V}^{2}\right)\left(\frac{7}{5}\right)=7 \times 10^{-3}$
$$ \Rightarrow $$ $\mathrm{V}^{2}=10^{-2}$
$$ \Rightarrow $$ $\mathrm{V}=10^{-1}=0.1 \mathrm{~m} / \mathrm{s}=10 \mathrm{~cm} / \mathrm{s}$
$$ \Rightarrow $$ $\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{MR}^{2}\right)\left(\frac{\mathrm{V}}{\mathrm{R}}\right)^{2}=7 \times 10^{-3}$
$$ \Rightarrow $$ $\frac{1}{2} \mathrm{MV}^{2}\left[1+\frac{2}{5}\right]=7 \times 10^{-3}$
$$ \Rightarrow $$ $\frac{1}{2}(1)\left(\mathrm{V}^{2}\right)\left(\frac{7}{5}\right)=7 \times 10^{-3}$
$$ \Rightarrow $$ $\mathrm{V}^{2}=10^{-2}$
$$ \Rightarrow $$ $\mathrm{V}=10^{-1}=0.1 \mathrm{~m} / \mathrm{s}=10 \mathrm{~cm} / \mathrm{s}$
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