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JEE Main 2024 (31 Jan, Shift 1) · Friction on an Incline

Friction on an Incline question from JEE Main 2024 (31 Jan, Shift 1)

Q11JEE Main 2024 · 31 JAN · SHIFT 1Difficult

Friction on an Incline

In the given arrangement of a doubly inclined plane two blocks of masses MM and mm are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25. The value of mm, for which M=10 kgM = 10\text{ kg} will move down with an acceleration of 2 m/s22\text{ m/s}^{2}, is: (take g=10 m/s2g = 10\text{ m/s}^{2} and tan⁡37∘=3/4)\tan 37^\circ = 3/4)

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Answer: (a)

  1. M on the 53∘53^\circ side moving down: Mgsin⁡53∘−T−μMgcos⁡53∘=MaMg\sin 53^\circ - T - \mu Mg\cos 53^\circ = Ma
  2. 80−T−0.25(60)=10(2)⇒T=4580 - T - 0.25(60) = 10(2) \Rightarrow T = 45 N
  3. m on the 37∘37^\circ side moving up: T−mgsin⁡37∘−μmgcos⁡37∘=maT - mg\sin 37^\circ - \mu mg\cos 37^\circ = ma
  4. 45−6m−0.25(8m)=2m⇒45=10m⇒m=4.545 - 6m - 0.25(8m) = 2m \Rightarrow 45 = 10m \Rightarrow m = 4.5 kg
  5. So the answer is (a) 4.5 kg.
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