JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Laws of Motion

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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 / s^{2}

]

(A) $\frac{3}{2}$ ✓ Correct
(B) $\frac{3}{4}$
(C) $\frac{1}{2}$
(D) $\frac{2}{3}$

Answer: (A)

Solution

Take the incline at

3 0^{\circ}

. When the block is on the verge of sliding down under an extra

2

N down the plane, friction acts up the plane:

m g sin 3 0^{\circ} + 2 = μ m g cos 3 0^{\circ}

. When it is on the verge of moving up under

10

N up the plane, friction acts down:

10 = m g sin 3 0^{\circ} + μ m g cos 3 0^{\circ}

. Adding,

12 = 2 μ m g cos 3 0^{\circ} = μ m g 3

; subtracting,

8 = 2 m g sin 3 0^{\circ} = m g

, so

m g = 8

N. Then

μ = \frac{12}{8 3} = \frac{3}{2 3} = \frac{3}{2}

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Laws of Motion

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