General Physics Question 10

Question 10

  1. 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

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

Column II

(p) Block Y of mass M left on a fixed inclined plane X, slides on it with a constant velocity.

(q) Two ring magnets Y and Z, each of mass M, are kept in 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 m0 is fixed to a table through a clamp X. A block of mass M 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 it 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.

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Answer:

Correct Answer: 11. Ap,t, B-q, s, t, Cp,r,s,t,Dq

Solution:

  1. (p) Constant velocity means, net acceleration or net force =0.

Net force exerted by X and Y=Mg in upward direction (opposite to its weight Mg ). Since Y is moving with constant velocity, some friction is there between X and Y.

Therefore, some work is done against friction and mechanical energy of (X+Y) is continuously decreasing.

(s) Potential energy of Y is decreasing but same volume of X rises up. Hence, potential energy of X is increasing. Some part of mechanical energy of X+Y is lost in the form of heat in doing work against viscous forces. Net force on Y in this case is downwards before Y attains its terminal velocity.

(t) After attaining the terminal velocity, net force on Y becomes zero. Hence, force on Y from X is Mg, upwards. Rest of the logics are same as discussed in part (s).



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