Skip to content
dronacharya.ai
← IIT-JEE 2008 · Paper 2
PhysicsMatrix match+6Question 42 of 66

IIT-JEE 2008 · Paper 2, Physics question 42

Column I gives a list of possible set of parameters measured in some experiments. The variations of the parameters in the form of graphs are shown in Column II. Match the set of parameters given in Column I with the graphs given in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.

Column I

  • A
    Potential energy of a simple pendulum (yy axis) as a function of displacement (xx axis)
  • B
    Displacement (yy axis) as a function of time (xx axis) for a one dimensional motion at zero or constant acceleration when the body is moving along the positive xx-direction
  • C
    Range of a projectile (yy axis) as a function of its velocity (xx axis) when projected at a fixed angle
  • D
    The square of the time period (yy axis) of a simple pendulum as a function of its length (xx axis)

Column II

  • P
    (graph p)
    y against x: a U-shaped curve (parabola-like) with its minimum above the x-axis, to the right of the y-axis.
  • Q
    (graph q)
    y against x: a straight line through the origin with positive slope.
  • R
    (graph r)
    y against x: a straight line with positive slope and a positive y-intercept.
  • S
    (graph s)
    y against x: a curve from the origin rising slowly and then steeply (concave up).
Mark every match for each entry
PQRS
A
B
C
D
Answer and worked solution

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

  • A. A pendulum's potential energy is quadratic in displacement, U∝x2U\propto x^2 about the mean position: a parabola. Graph S (from the origin, concave up) and graph P (a U-shaped curve, i.e. the same parabola with the reference shifted). → P, S.
  • B. x=x0+ut+12at2x=x_0+ut+\frac12at^2 with u>0u>0: with a=0a=0 a straight line through the origin (Q) or with an intercept (R); with constant a>0a>0 a curve rising ever more steeply (S). → Q, R, S.
  • C. R=v2sin⁡2θg∝v2R=\dfrac{v^2\sin2\theta}g\propto v^2: a parabola from the origin. → S.
  • D. T2=4π2gL∝LT^2=\dfrac{4\pi^2}gL\propto L: a straight line through the origin. → Q.

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