Question

In the column -I there are some physical situations. Match them with statement in column - II.

• A. A - P, Q,R ; B - P,Q,S,T; C-P,Q,R; D-P,Q,S
• B. A-P,Q,S;B-P,Q,R,S,T ; C-Q,R,T;D-P,Q,S,T
• C. A- P, Q,R; B-P,Q,R,T; C-Q,R; D-P,Q
• D. A-P, Q,S; B-P,Q,R,S,T; C-P, Q,R; D-P,Q,S

Answer 1

The correct option is D A-P, Q,S; B-P,Q,R,S,T; C-P, Q,R; D-P,Q,S

For (A):
${\tau }_{net}$ about $O{O}^{\prime }=0\Rightarrow L\text{is conserved}$ (P)
${\tau }_{gravitational}$ about O on A is non zero. (Q)
Since KE of system increase
$\Rightarrow$ work done by internal force is not zero (S).
The centre of mass is moving in circles constantly, which also means the linear momentum of the system is not conserved.

For (B):
${\tau }_{net}\text{about}O{O}^1=0\Rightarrow L$ is conserved (P)
${\tau }_{mg}$ about O is non-zero on A. (Q)
$F_{net}$ is horizontal direction is zero. (R) Since KE of system changes $\Rightarrow$ work is done by internal forces (S)
And the centre of mass of the system remains at rest as there is no external force on the system to accelerate it.

For (C):
The external forces acting on the system are either along the axis or radially away from the axis. Therefore, the torque due to them along the axis is zero. So, the liner momentum is conserved. (P)
${\tau }_{mg}$ is clearly non zero about point $O$. (Q)
The external force is zero on the system in the horizontal direction. (R)
Internal forces are just normal reactions between rigid bodies, so the work done by internal forces is zero. The centre of mass is moving downwards because the ball is rolling downwards.


For (D):
$F_{ext}$ (Friction) is present between boy & ground in horizontal direction (Not R & T).