A sphere of mass 3 kg and radius 0.2 m rolls down from the slope of height 7 m, its rotational kinetic energy will be ………….

A sphere of mass 3 kg and radius 0.2 m rolls down from the slope of he

1.	A sphere of mass 3 kg and radius 0.2 m rolls down from the slope of height 7 m, its rotational kinetic energy will be ………….
Answer» <html><body><p>`42J`<br/>`60J`<br/>`36J`<br/>`70J`</p>Solution :Velocity of body at the <a href="https://interviewquestions.tuteehub.com/tag/bottom-401202" style="font-weight:bold;" target="_blank" title="Click to know more about BOTTOM">BOTTOM</a> slope of height h when the rolls is <br/> `v=sqrt((2gh)/(1+(k^(2))/(R^(2))))=sqrt((2gh)/(1+(2)/(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)))[because k=sqrt((2)/(5))Rrarr" for sphere"]` <br/> `therefore v=sqrt((10gh)/(<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>))=sqrt((10xx10xx7)/(7))=10m//s` <br/> Now rotational <a href="https://interviewquestions.tuteehub.com/tag/kinetic-533291" style="font-weight:bold;" target="_blank" title="Click to know more about KINETIC">KINETIC</a> energy <br/> `therefore k=(1)/(2)Iomega^(2)` <br/> `therefore k=(1)/(2)xx(2)/(5)<a href="https://interviewquestions.tuteehub.com/tag/mr-549185" style="font-weight:bold;" target="_blank" title="Click to know more about MR">MR</a>^(2)xx(v^(2))/(R^(2))[because" For sphere "I=(2)/(5)MR^(2),omega=(v)/(R)]` <br/> `therefore k=(1)/(5)Mv^(2)=(1)/(5)xx3xx(10)^(2)=(300)/(5)` <br/> `therefore k=60J`</body></html>