Question 172 of 280

A uniform sphere of mass $500g$ rolls without slipping on a plane horizontal surface with its centre moving at a speed of $5.00cm{s}^{-1}$. Its kinetic energy is:

A$8.75\times {10}^{-4} J$B$8.75\times {10}^{-3} J$C$6.25\times {10}^{-4} J$D$1.13\times {10}^{-3} J$

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Answer: (A) $8.75\times {10}^{-4} J$

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

JEE Main Physics practice question on Rotational Motion (Rotational kinematics), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.