One fourth length of a uniform rod of length 2l and mass m is placed on a horizontal table and the rod is held horizontal. The rod is released from rest. Find the normal reaction on the rod as soon as the rod is released

The correct option is C

The torque applied by gravity about the edge $O=T=mg\left(\frac{31}{4}\right)$

$\Rightarrow I_o\propto =\frac{3}{4}mgl$ ...(i) where $I_g+m{\left(\frac{3l}{4}\right)}^2=\frac{m{l}^2}{12}+\frac{9}{16}m{l}^2=\frac{m{l}^2}{4}(\frac{1}{3}+\frac{9}{4})I_o=\frac{31m{l}^2}{48}$ ...(ii)

(i) And (ii) yield, $\propto =\frac{\frac{3mgl}{4}}{\frac{31m{l}^2}{48}}$ $\Rightarrow \propto =\frac{144g}{124l}=\frac{36g}{31l}$ ....(iii)

Newton's second law of motion : mg - N = ma

Kinematics: $\alpha =\left(\frac{3l}{4}\right)\propto =\frac{36g}{31}$
Using (iii) & (iv) we obtain, $N=mg-m\left(\frac{27g}{31}\right)\Rightarrow N=\frac{4mg}{27}$.