question_answer

A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm/s. Its kinetic energy is                                                  [JEE MAIN Held on 08-01-2020 Evening]

A)   \[6.25\times {{10}^{4}}J\]     

B)        \[1.13\times {{10}^{3}}J\]

C)   \[8.75\times {{10}^{\,4}}\text{ }J\]

D)        \[8.75\times {{10}^{3}}\text{ }J\]

Correct Answer: C

Solution :

[c]

\[KE\,\,=\,\,\frac{1}{2}\,\,\left( \frac{2}{5}\,\,m{{R}^{2}}\,\,+\,\,m{{R}^{2}} \right)\,\,{{\left( \frac{v}{R} \right)}^{2}}\]

\[\Rightarrow \,\,KE\,\,=\,\,\frac{1}{2}\,m{{R}^{2}}\,\,\times \,\,\frac{7}{5}\,\,\times \,\,\frac{{{v}^{2}}}{{{R}^{2}}}\,\,=\,\,\frac{7}{10}\,\,\times \,\,\frac{1}{2}\,\,\times \,\,\frac{25}{{{10}^{4}}}\]

\[KE\,\,=\,\,\frac{35}{4}\,\,\times \,\,{{10}^{-\,4}}\,\,Joule\]

\[KE\,\,=\,\,8.75\,\,\times \,\,{{10}^{\,4}}\,\,Joule\]