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» `42J` `60J` `36J` `70J` Solution :Velocity of body at the BOTTOM slope of height h when the rolls is `v=sqrt((2gh)/(1+(k^(2))/(R^(2))))=sqrt((2gh)/(1+(2)/(5)))[because k=sqrt((2)/(5))Rrarr" for sphere"]` `therefore v=sqrt((10gh)/(7))=sqrt((10xx10xx7)/(7))=10m//s` Now rotational KINETIC energy `therefore k=(1)/(2)Iomega^(2)` `therefore k=(1)/(2)xx(2)/(5)MR^(2)xx(v^(2))/(R^(2))[because" For sphere "I=(2)/(5)MR^(2),omega=(v)/(R)]` `therefore k=(1)/(5)Mv^(2)=(1)/(5)xx3xx(10)^(2)=(300)/(5)` `therefore k=60J`