Question

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

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

Answer 1

Answer: (B)

$E=\frac{L^{2}}{2 l}$
$\therefore E \propto L^{2}$
$\frac{E_{2}}{E_{1}} =\left(\frac{L_{2}}{L_{1}}\right)^{2}$
$\frac{E_{2}}{E_{1}} =\left[\frac{L_{1}+200 \% \text { of } L_{1}}{L_{1}}\right]$
$=\left[\frac{L_{1}+2 L_{1}}{L_{1}}\right]^{2}=(3)^{2}$
$E_{2} =9 E_{1}$
Increment in kinetic energy
$\Delta E=E_{2} -E_{1}$
$=9 E_{1}-E_{1}$
$\Delta E =8 E_{1}$
$\frac{\Delta E}{E_{1}} =8$
or percentage increase $=800 \%$