Q : If the angular momentum of a rotating body about a fixed axis is increased by 10 %. Its kinetic energy will be increased by

(a) 10 %

(b) 20 %

(c) 21 %

(d) 5 %

Ans: (c)

Sol: L’ = L + L/10 = 11L/10

Rotational K.E $ = \frac{L^2}{2I}$

$K.E \propto L^2 $ (Since I = constant)

Percentage increase in K. E = $\frac{K’ – K}{K} \times 100 = (\frac{K’}{K} – 1)\times 100$

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

$ = [(\frac{11}{10})^2 – 1 ]\times 100 $

$= \frac{21}{100} \times 100$

= 21 %