JEE Main 12 Jan 2019 Evening Question 1
Question: 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=10m/s^{2} $ ] [JEE Main Online Paper Held On 12Jan2019 Evening]
Options:
A) $ \frac{1}{2} $
B) $ \frac{\sqrt{3}}{2} $
C) $ \frac{\sqrt{3}}{4} $
D) $ \frac{2}{3} $
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Answer:
Correct Answer: B
Solution:

Case I: When 2 N force is acting on the block resolve the forces along and perpendicular the plane. $ 2=mg\sin \theta \mu mg,\cos \theta $ …(i)
Case II. When the force 10 N is acting on the block, free body diagram of the block is given as follow: $ mg\sin \theta +\mu mg,\cos \theta =10 $ …(ii)
Dividing eq. (i) by eq. (ii), $ \frac{g\sin \theta \mu g,\cos \theta }{g\sin \theta +\mu g,\cos \theta }=\frac{2}{10};\frac{1\mu \sqrt{3}}{1+\mu \sqrt{3}}=\frac{1}{5}$
$\Rightarrow \mu =\frac{\sqrt{3}}{2} $ c