A) 1.6
B) 4.0
C) 2.0
D) 2.5
Correct Answer: C
Solution :
[c] The angle of repose, \[\alpha ={{\tan }^{-1}}(\mu )=ta{{n}^{-1}}(0.8)=37{}^\circ \] Here, the angle of inclined plane is \[\theta =37{}^\circ \]. It means that block is at rest and therefore the friction should be static in nature. It will balance the component of weight parallel to inclined plane. Static friction=component of weight in downward direction\[=mg\sin \theta =10N\] \[\therefore m=\frac{10}{9\times \sin 30{}^\circ }=2kg\]
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