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 ………….

Options

A. `42J`
B. `60J`
C. `36J`
D. `70J`

Answer

B

Explanation

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`