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PhysicsMatrix match+8Question 60 of 60

IIT-JEE 2009 · Paper 1, Physics question 60

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.

Column I

  • A
    The force exerted by X on Y has a magnitude MgMg.
  • B
    The gravitational potential energy of X is continuously increasing.
  • C
    Mechanical energy of the system X + Y is continuously decreasing.
  • D
    The torque of the weight of Y about point P is zero.

Column II

  • P
    Block Y of mass MM left on a fixed inclined plane X, slides on it with a constant velocity.
    A fixed inclined plane X (right-angled wedge rising to the right) with a block Y sliding down it; P is at the lower left end of the incline.
  • Q
    Two ring magnets Y and Z, each of mass MM, 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.
    A vertical plastic stand on a base X: ring magnet Y rests on the base and ring magnet Z floats above it on the central rod; P is at the top of the rod; an arrow shows the lift moving up.
  • R
    A pulley Y of mass m0m_0 is fixed to a table through a clamp X. A block of mass MM 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.
    A table with a clamp X holding a pulley Y at its edge; a string fixed at P on the table runs over the pulley and holds a hanging block; an arrow shows the lift moving down.
  • S
    A sphere Y of mass MM is put in a nonviscous liquid X kept in a container at rest. The sphere is released and it moves down in the liquid.
    A container of liquid X with a sphere Y moving down in it; P is at the bottom left of the container.
  • T
    A sphere Y of mass MM is falling with its terminal velocity in a viscous liquid X kept in a container.
    A container of viscous liquid X with a sphere Y falling at terminal velocity; P is at the bottom left of the container.
Mark every match for each entry
PQRST
A
B
C
D
Answer and worked solution

Official answer: A → P, T; B → Q, S, T; C → P, R, T; D → Q

  • A, force of X on Y equals Mg: when Y moves with constant velocity the net force on it is zero, so the total force from X balances its weight: P (incline: normal plus friction) and T (buoyancy plus viscous drag). In Q the base pushes Y with 2Mg; in S the sphere accelerates.
  • B, potential energy of X increasing: in Q the lift carries the base upward; in S and T the sinking sphere displaces liquid upward. → Q, S, T.
  • C, mechanical energy of X + Y decreasing: friction in P and viscosity in T dissipate energy; in R the system descends at constant speed, losing potential energy with no gain in kinetic energy. → P, R, T.
  • D, torque of Y's weight about P is zero: only in Q does P lie on the vertical line through Y (the common axis of the rings). → Q.

Answer: A → P, T; B → Q, S, T; C → P, R, T; D → Q

Topics:Conservation of mechanical energyNewton's lawsViscosity and Stokes' lawTorque and rotational dynamics
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