Column-II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown. Column-I gives some statements about X and/or Y. Match these statements to the appropriate system(s) from Column II.

Question

Column-II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown. Column-I gives some statements about X and/or Y. Match these statements to the appropriate system(s) from Column II. [2009]

	Column-I		Column-II
(A)	The force exerted by X on Y has a magnitude	(p)	Block Y of mass M left on a fixed inclined plane X, slides on it with a constant velocity.
(B)	The gravitational potential energy of X is continuously increasing.	(q)	Two ring magnets Y and Z, each of mass MMM, are kept in a frictionless vertical plastic stand so that they repel each other. Y rests on the base X and Z hangs in air in equilibrium. P is the topmost point of the stand on the common axis of the two rings. The whole system is in a lift that is going up with a constant velocity.
(C)	Mechanical energy of the system X + Y is continuously decreasing.	(r)	A pulley Y of mass m0m_0m0? is fixed to a table through a clamp X. A block of mass MMM hangs from a string that goes over the pulley and is fixed at point P of the table. The whole system is kept in a lift that is going down with a constant velocity.
(D)	The torque of the weight of Y about point P is zero.	(s)	A sphere Y of mass M is put in a non-viscous liquid X kept in a container at rest. The sphere is released and moves down in the liquid.
		(t)	A sphere Y of mass M is falling with its terminal velocity in a viscous liquid X kept in a container.

Smart Achievers · Accepted Answer

(2)

(p) As the velocity is constant

$∴$ $f = m g \sin θ$

$But f = μ N = μ m g \cos θ$

$∴$ $μ m g \cos θ = m g \sin θ \Rightarrow μ = \tan θ$

$The force by X on Y is the resultant of f and N .$

$\sqrt{f^{2} + N^{2}} = \sqrt{μ^{2} N^{2} + N^{2}} = \sqrt{μ^{2} + 1} N$

$= (\sqrt{\tan^{2} θ + 1}) m g \cos θ = \sec θ m g \cos θ = m g$

$= weight of Y .$

Again, due to the presence of frictional force between Y and X, the mechanical energy of the system (X + Y) decreases continuously as Y slides down.

(q) Lift moves up, X also moves up and therefore the gravitational energy of X is continuously increasing.

T of weight of Y about P as the perpendicular distance of the line of action of force from the point P is zero. Force exerted by X on Y $= M g + M g = 2 M g$

where $M g$ is wt. of Y and $M g$ is the force on Y due to Z.

(r)

In this case the force exerted by X on Y = force exerted by Y on X. The force on X due to Y is

$R = \sqrt{{(M g)}^{2} + [{(m_{0} + M) g]}^{2}} \neq M g$

The mechanical energy of the system (X + Y) is continuously decreasing as the system is coming down and its potential energy is decreasing, the kinetic energy remaining the same.

The torque of the weight of Y about $P \neq 0$

(s) Force on Y by X is = wt. of liquid displaced which cannot be equal to $M g$ as the density of Y > density of X (Y is sinking)

The gravitational potential energy of X increases continuously because as Y moves down, the centre of mass of X moves up.

(t) Sphere Y is moving with terminal velocity $V_{T}$

$∴$ Net force on Y is zero i.e. $M g = B + F_{v}$

The $B + F_{v}$ are exerted by X on Y.

The gravitational potential energy of X is continuously increasing because as Y moves down, the centre of mass of X moves up.

The mechanical energy of the system (X + Y) is continuously decreasing to overcome the viscous forces.

Solution

1 $A \to (q, p); B \to (q, s, t); C \to (p, r, t); D \to (p, t)$

2 $A \to (p, t); B \to (q, s, t); C \to (p, r, t); D \to (q, p)$

3 $A \to (q, p); B \to (p, r, t); C \to (q, s, t); D \to (p, t)$

4 $A \to (p, r, t); B \to (q, p); C \to (q, s, t); D \to (p, t)$

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Solution

1 A→(q,p); B→(q,s,t); C→(p,r,t); D→(p,t)

2 A→(p,t); B→(q,s,t); C→(p,r,t); D→(q,p)

3 A→(q,p); B→(p,r,t); C→(q,s,t); D→(p,t)

4 A→(p,r,t); B→(q,p); C→(q,s,t); D→(p,t)

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1 $A \to (q, p); B \to (q, s, t); C \to (p, r, t); D \to (p, t)$

2 $A \to (p, t); B \to (q, s, t); C \to (p, r, t); D \to (q, p)$

3 $A \to (q, p); B \to (p, r, t); C \to (q, s, t); D \to (p, t)$

4 $A \to (p, r, t); B \to (q, p); C \to (q, s, t); D \to (p, t)$