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=mgsin60∘=√32mg
Hence, friction force action on rod at support
P will be
√3mg2 .
For finding tension, we can take torque about support
P
T(34l)=mg cos600(l4)
⇒T=mg6
Just before cutting the string, the rod is in equilibrium. Considering the forces perpendicular to rod length.
N+T=mgcos60∘⇒N=mg2−mg6=mg3Just 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
τP=IPα
(mgcos60∘)⋅l4=(ml212+m(l4)2)α ⇒α=6g7l
If we apply torque equation about centre of mass,
N⋅l4=(ml212)α =(ml212)(6g7l)
⇒N=27mg