A) 9.8 N
B) \[0.7\times 9.8\times \sqrt{3}\,N\]
C) \[9.8\times \sqrt{3}\,N\]
D) \[0.8\times 9.8\,N\]
Correct Answer: A
Solution :
Limiting friction \[{{F}_{l}}=\mu \ mg\cos \theta \] \[{{F}_{l}}=0.7\times 2\times 10\times \cos 30{}^\circ =12\ N\](approximately) But when the block is lying on the inclined plane then component of weight down the plane \[=mg\sin \theta \] \[=2\times 9.8\times \sin 30{}^\circ =9.8\ N\] It means the body is stationary, so static friction will work on it \[\therefore \] Static friction = Applied force = 9.8 N
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