Physics Circle home Student Login

Home›Chapters›Laws of Motion PYQs›Friction on an Incline›Question 17

JEE Main 2022 (27 Jul, Shift 2) · Friction on an Incline

Friction on an Incline question from JEE Main 2022 (27 Jul, Shift 2)

Q17JEE Main 2022 · 27 JUL · SHIFT 2Difficult

Friction on an Incline

Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10 m. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point C from point A is t(2+1)t(\sqrt{2}+1) s. The value of t is ___________.

(use g=10 m/s2g = 10\text{ m/s}^{2})

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

  1. AB (smooth, 45∘45^\circ): length =102= 10\sqrt{2} m, retardation gsin⁡45∘=52g\sin 45^\circ = 5\sqrt{2} m/s2^2, and the block just stops at B.
  2. Reverse motion: t1=2La=2×10252=2t_1 = \sqrt{\frac{2L}{a}} = \sqrt{\frac{2 \times 10\sqrt{2}}{5\sqrt{2}}} = 2 s
  3. BC (30∘30^\circ): length =10sin⁡30∘=20= \frac{10}{\sin 30^\circ} = 20 m, a=gsin⁡30∘=5a = g\sin 30^\circ = 5 m/s2^2, starts from rest.
  4. t2=2×205=22t_2 = \sqrt{\frac{2 \times 20}{5}} = 2\sqrt{2} s
  5. t1+t2=2+22=2(2+1)t_1 + t_2 = 2 + 2\sqrt{2} = 2(\sqrt{2}+1) s, so t=2t = 2.
Report a mistake

More questions on Friction on an Incline