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6.System of Particles and Rotational Motion
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If the rotational kinetic energy of a body is increased by $300\ \%$ then the percentage increase in its angular momentum will be .......... $\%$
A
$600$
B
$150$
C
$100$
D
$1500$
Solution
Percentage increase in momentum $=\frac{L_{2}-L_{1}}{L_{1}} \times 100$
$L \propto \sqrt{E} \Rightarrow E_{1}=E$
and $E_{2}=E+\frac{300}{100} E$
$\Rightarrow E_{2}=4 E$
Increase in momentum $=\frac{\sqrt{E_{2}}-\sqrt{E_{1}}}{\sqrt{E_{1}}} \times 100$
$=\frac{\sqrt{4 E}-\sqrt{E}}{\sqrt{E}} \times 100=100 \%$
Standard 11
Physics