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Saturday, 31 October

JEE Main 2020 · PhysicsMultiple choiceOtherMediumData or graph

JEE Main 2 September 2020, Shift 2, Physics Q2

Question 2 of 75 in this shift, Physics question 2 of 25.

The displacement time graph of a particle executing S.H.M. is given in figure : (sketch is schematic and not to scale) Which of the following statements is/are true for this motion ? (A) The force is zero at t=3T4t = \frac{3T}{4} (B) The acceleration is maximum at t=Tt = T (C) The speed is maximum at t=T4t = \frac{T}{4} (D) The P.E. is equal to K.E. of the oscillation at t=T2t = \frac{T}{2}

The figure, in words

Displacement (vertical axis) versus time (s) (horizontal axis) graph. Curve starts above the axis at t = 0 (point O marked on the axis), falls to cross the time axis at T/4, dips below the axis to a trough above which the label 2T/4 is printed (dot on the axis at 2T/4), returns to the axis at 3T/4, then rises into a hump above the axis that ends on the axis at 5T/4. Labels printed under the axis: T/4, 3T/4, T (dot on axis), 5T/4.
  1. (1)(A), (B) and (D)
  2. (2)(A) and (D)
  3. (3)(A), (B) and (C)Official answer
  4. (4)(B), (C) and (D)

Official answer

Option 3

NTA final key (Sep 2020).