If the mass of the earth is kept constant and its radius is made half, then the duration of the day will be
where T is the present duration of the day.

Options

A. `(T)/(3)`
B. `(4T)/(3)`
C. `(T)/(4)`
D. `(3T)/(4)`

Answer

C

Explanation

By the principle of conservation of angular momentum,
`I_(1)omega_(1)=I_(2)omega_(2)`
`therefore" "(2)/(5)MR^(2)xx(2pi)/(T)=(2)/(5)M((R)/(2))^(2)xx(2pi)/(T')`
`therefore" "T=4T'" "therefore T'=(T)/(4)`
Note: If the density is kept constant and the radius is made `R//2`, then the mass will change
`M=(4)/(3)piR^(3)rho and M'=(4)/(3)pi((R^(3))/(8))rho" "therefore M'=(M)/(8)`
`therefore" Total K.E. (E) "=(1)/(2)m omega^(2)(R^(2)+K^(2))`
`therefore" "(K_(R))/(E)=((1)/(2)mK^(2).omega^(2))/((1)/(2)momega^(2)(K^(2)+R^(2)))=(K^(2))/(K^(2)+R^(2))`