A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving…

1. $8.75\times {10}^{-4}\mathrm{J}$
2. $8.75\times {10}^{-3}\mathrm{J}$
3. $6.25\times {10}^{-4}\mathrm{J}$
4. $1.13\times {10}^{-3}\mathrm{J}$

Solution

$K.E$ of the sphere $=$ Translational $K.E+$ Rotational $K.E$
$=\frac{1}{2}{mv}^2\left(1+\frac{K^2}{R^2}\right) ,K=$ Radius of gyration
$= \frac{1}{2}\times \frac{1}{2}\times {\left(\frac{2}{100}\right)}^2\left(1+\frac{2}{5}\right)$
$= \frac{35}{4}\times {10}^{-4} \mathrm{J}$
$\mathrm{=\; 8.75}\times {10}^{-4}\mathrm{J}$

Asked in: JEE Main 2020 (08 Jan Shift 2)

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