Q: If the angular momentum of any rotating body increases by 200 %, then the increase in its kinetic energy

(a) 400 %

(b) 800 %

(c) 200 %

(d) 100 %

Ans: (b)

Sol: Angular Momentum , L’ = L + 2 L = 3 L

Kinetic Energy $K = \frac{L^2}{2 I}$

$ K \propto L^2 $

Percentage increase in K.E will be

$ \frac{K’ – K}{K} \times 100 = \frac{L’^2 – L^2}{L^2} \times 100$

$ = (\frac{L’^2}{L^2} – 1 ) \times 100 $

$ = (\frac{(3L)^2}{L^2} – 1 ) \times 100 $

= 800 %