Question

If the angular momentum of any rotating ball increases by $200\%$, then the increase in kinetic energy is?

• A. $400\%$
• B. $800\%$
• C. $200\%$
• D. $100\%$

Answer 1

The correct option is B $800\%$

$E=\frac{L^2}{2l}$
$\therefore E\propto L^2$
$\frac{E_2}{E_1}={\left(\frac{L_2}{L_1}\right)}^2$
$\frac{E_2}{E_1}=\left[\frac{L_1+200\%of{L}_1}{L_1}\right]$
$={\left[\frac{L_1+2L_1}{L_1}\right]}^2=(3{)}^2$
$E_2=9E_1$
Increment in kinetic energy
$\Delta E=E_2-E_1$
$=9E_1-E_1$
$\Delta E=8E_1$
$\frac{\Delta E}{E_1}=8$
or percentage increase$=800\%$.