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JEE Main 2019 (12 Jan, Shift 2) · Friction on an Incline

Friction on an Incline question from JEE Main 2019 (12 Jan, Shift 2)

Q21JEE Main 2019 · 12 JAN · SHIFT 2Medium

Friction on an Incline

A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force 2 N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10 N. The coefficient of static friction between the block and the plane is: [Take g = 10 m/s2]

Figure

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Answer: (c)

  1. Max force down (friction up): 2+mgsin⁡30∘=μmgcos⁡30∘2 + mg\sin 30^\circ = \mu mg\cos 30^\circ.
  2. Max force up (friction down): 10=mgsin⁡30∘+μmgcos⁡30∘10 = mg\sin 30^\circ + \mu mg\cos 30^\circ.
  3. Subtracting: 8=2mgsin⁡30∘⇒mg=8 N8 = 2mg\sin 30^\circ \Rightarrow mg = 8\text{ N}; then μmgcos⁡30∘=6\mu mg\cos 30^\circ = 6.
  4. μ=68×3/2=32\mu = \dfrac{6}{8 \times \sqrt{3}/2} = \dfrac{\sqrt{3}}{2}
  5. So the answer is (c) 32\dfrac{\sqrt{3}}{2}.
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