KVPY SX SB Exam 20141 Question Paper

J = mv .... (1)
where v is the velocity of centre of mass.
After impulse rod get angular velocity ω
Angular impulse = Iω
$\mathit{J}\times \mathit{L}=\frac{\mathit{M}{\left(2\mathit{L}\right)}^2}{12}\times \mathit{\omega }$....(2)
$\mathit{J}=\frac{\mathit{M}{\mathit{L}}^2\mathit{\omega }}{3}$
$\mathit{\omega }=\frac{3\mathit{J}}{\mathit{M}\mathit{L}}$
from equation (1); $\mathit{v}=\frac{1}{\mathit{m}}$
Kinetic energy = $\mathit{K}\mathit{E}=\frac{1}{2}\mathit{m}{\mathit{v}}^2+\frac{1}{2}\mathit{I}{\mathit{\omega }}^2$
⇒ $\frac{1}{2}\mathit{m}\frac{{\mathit{J}}^2}{{\mathit{m}}^2}+\frac{1}{2}\times \frac{\mathit{m}\times 4{\mathit{L}}^2}{12}\times \frac{9{\mathit{J}}^2}{{\mathit{m}}^2{\mathit{L}}^2}$
⇒ $\frac{{\mathit{J}}^2}{2\mathit{M}}+\frac{36{\mathit{J}}^2}{24\mathit{M}}$
$\frac{48{\mathit{J}}^2}{24\mathit{M}}\Rightarrow \frac{2{\mathit{J}}^2}{\mathit{m}}$