Question

A uniform rod of mass $m$ and length $l$ is in equilibrium under the action of constraint forces, gravity and tension in the string.

	Column-I		Column-II
i.	Frictional force acting on the rod	p.	$\frac{mg}{3}$
ii.	Tension in the string	q.	$\frac{mg}{6}$
iii.	Normal reaction on the rod	r.	$\frac{2mg}{7}$
iv.	Normal reaction on the rod just after the string is cut	s.	$\frac{\sqrt{3}mg}{2}$

• A. i-p ; ii-q ; iii-r ; iv-s
• B. i-s ; ii-q ; iii-p ; iv-r
• C. i-s ; ii-p ; iii-q ; iv-r
• D. i-p ; ii-r ; iii-q ; iv-s

Answer 1

The correct option is B i-s ; ii-q ; iii-p ; iv-r


Resolving the forces parallel and perpendicular to the rod. Along the length of the rod,
$f=mg\sin {60}^{\circ }=\frac{\sqrt{3}}{2}mg$
Hence, friction force action on rod at support $P$ will be $\frac{\sqrt{3}mg}{2}$ .
For finding tension, we can take torque about support $P$

$T\left(\frac{3}{4}l\right)=mgcos{60}^0\left(\frac{l}{4}\right)$

$\Rightarrow T=\frac{mg}{6}$

Just before cutting the string, the rod is in equilibrium. Considering the forces perpendicular to rod length.
$$N+T=mg\cos {60}^{\circ }\Rightarrow N=\frac{mg}{2}-\frac{mg}{6}=\frac{mg}{3}$$Just after cutting the string, the equilibrium of the rod will be
disturbed. The rod will have angular acceleration.
As the rod does not slip, point $P$ will be at rest at the time just after cutting the string. We can apply torque equation about $P$

${\tau }_P=I_P\alpha$
$\left(mg\cos {60}^{\circ }\right)\cdot \frac{l}{4}=(\frac{m{l}^2}{12}+m{\left(\frac{l}{4}\right)}^2)\alpha$ $\Rightarrow \alpha =\frac{6g}{7l}$

If we apply torque equation about centre of mass,

$N\cdot \frac{l}{4}=\left(\frac{m{l}^2}{12}\right)\alpha$ $=\left(\frac{m{l}^2}{12}\right)\left(\frac{6g}{7l}\right)$

$\Rightarrow N=\frac{2}{7}mg$