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← IIT-JEE 2012 · Paper 1
PhysicsMultiple correct+4Question 13 of 60

IIT-JEE 2012 · Paper 1, Physics question 13

A small block of mass of 0.1 kg lies on a fixed inclined plane PQPQ which makes an angle θ\theta with the horizontal. A horizontal force of 1 N acts on the block through its center of mass as shown in the figure. The block remains stationary if (take g=10 m/s2g=10\ \mathrm{m/s^2})

A right-triangular wedge OPQ with the incline PQ rising from P (bottom right) to Q (top left) at angle θ at P; a block on the incline is pushed by a horizontal 1 N force pointing towards the vertical side OQ (into the incline).
Choose one or more options
Answer and worked solution

Official answer: A, C

Forces along the incline. The weight is mg=1mg=1 N; its component down the slope (towards P) is sin⁡θ\sin\theta N. The horizontal 1 N force points into the incline, and its component up the slope (towards Q) is cos⁡θ\cos\theta N.

  • (A) At θ=45°\theta=45° these are equal, so the block stays at rest even without friction. True.
  • (B), (C) For θ>45°\theta>45°, gravity wins, so friction must act up the slope, towards Q. (C) true, (B) false.
  • (D) For θ<45°\theta<45°, the push wins, so friction acts towards P, not Q. False.

(A), (C)

Topics:FrictionFree body diagrams
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