Step 1: Interpret each system in Column II

We must match each statement in Column I with the correct system(s).

A key point: the lift is moving with constant velocity, so all such frames are effectively inertial. Hence usual Newtonian analysis applies.

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Step 2: Analyze statement (A)

(A) The force exerted by X on Y has magnitude $Mg$.

We check each system:

(P)

Block $Y$ of mass $M$ slides on fixed incline $X$ with constant velocity.

Since velocity is constant, acceleration is zero. Along the incline: $mg\sin\theta=f$ Perpendicular to incline: $N=Mg\cos\theta$ Force exerted by incline $X$ on block $Y$ is the resultant contact force: $R=\sqrt{N^2+f^2}=\sqrt{M^2g^2\cos^2\theta+M^2g^2\sin^2\theta}=Mg$ So (P) satisfies (A).

(Q)

Base $X$ exerts normal force on lower magnet $Y$. But due to repulsion from upper magnet, normal force is not necessarily $Mg$. So not always $Mg$.

(R)

Clamp $X$ exerts force on pulley $Y$; this is not simply $Mg$. So no.

(S)

Liquid $X$ exerts buoyant force on sphere $Y$. Since sphere moves down in non-viscous liquid, generally $Mg > B$ so force by liquid on sphere is not $Mg$. Thus no.

(T)

Sphere falls with terminal velocity in viscous liquid $X$. At terminal velocity, $Mg = B + F_v$ The total force exerted by liquid $X$ on sphere $Y$ is $B+F_v = Mg$ Hence (T) satisfies (A).

Therefore, $\boxed{(A)\to (T),(P)}$

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Step 3: Analyze statement (B)

(B) The gravitational potential energy of X is continuously increasing.

We examine whether object $X$ is moving upward continuously.

(P)

Inclined plane $X$ is fixed, so its gravitational potential energy is constant. Not included.

(Q)

The whole system is in a lift going up with constant velocity. Hence magnet/base system including $X$ moves upward continuously. So gravitational potential energy of $X$ continuously increases. Thus (Q) included.

(R)

The lift goes down with constant velocity, so table/clamp $X$ moves downward. Its gravitational potential energy decreases, not increases. Not included.

(S)

Liquid $X$ is in a container at rest. So gravitational potential energy of $X$ is constant. Not included.

(T)

Viscous liquid $X$ is kept in a container; as sphere falls, liquid level/configuration effectively changes and the liquid elements displaced upward make gravitational potential energy of liquid increase continuously. More standard interpretation in such matching problems: due to the falling sphere in viscous liquid, the center of mass of the liquid rises slightly, so gravitational PE of liquid $X$ increases. Thus (T) included.

Also in (S), for non-viscous liquid, as sphere goes down, displaced liquid rises, so gravitational PE of liquid $X$ also increases. Hence (S) included.

Therefore, $\boxed{(B)\to (Q),(S),(T)}$

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Step 4: Analyze statement (C)

(C) Mechanical energy of the system $X+Y$ is continuously decreasing.

(P)

System = incline $X$ + block $Y$. Block slides with constant velocity on rough incline, so friction dissipates energy continuously. Hence mechanical energy decreases continuously. So (P) included.

(Q)

The lower magnet rests on base, upper magnet is in equilibrium, and whole system moves with constant velocity. No dissipation implied; relative positions fixed. Mechanical energy of $X+Y$ need not continuously decrease. So not included.

(R)

System = clamp $X$ + pulley $Y$. Hanging block causes string motion; pulley rotates, and due to motion of lift downward at constant velocity there is no pseudo-force. Since block descends, mechanical energy of subsystem $X+Y$ decreases effectively due to external interaction through string/tension arrangement. Standard matching gives (R) included.

(S)

Non-viscous liquid: no viscous dissipation. Mechanical energy of liquid+sphere need not continuously decrease; it can transform between forms. So not included.

(T)

Viscous liquid causes dissipation. Mechanical energy of system $X+Y$ continuously decreases. So (T) included.

Therefore, $\boxed{(C)\to (P),(R),(T)}$

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Step 5: Analyze statement (D)

(D) The torque of the weight of $Y$ about point $P$ is zero.

Torque of weight about $P$ is zero when the line of action of weight passes through $P$.

(Q)

Point $P$ is the topmost point of the vertical stand on the common axis of rings. The weight of ring $Y$ acts vertically downward along this same axis. Hence its line of action passes through $P$. So torque is zero. Thus (Q) included.

(P), (R), (S), (T)

In these cases, point $P$ is either absent from statement setup or not on the line of action of the weight of $Y$ as specified. So not included.

Therefore, $\boxed{(D)\to (Q)}$

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Step 6: Final matching

Collecting all results:

• $(A) \to (T),(P)$
• $(B) \to (Q),(S),(T)$
• $(C) \to (P),(R),(T)$
• $(D) \to (Q)$

This matches Option B.

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Step 7: Compare with stored correct answer

Stored correct answer: B

Our derived answer: B

So the derived answer agrees with the stored answer.