A) \[1.7\times \sqrt{2}\times \frac{1}{\sqrt{3}}N\]
B) \[1.7\times \sqrt{3}\times \frac{1}{2}N\]
C) \[1.7\times \sqrt{3}N\]
D) \[1.7\times \sqrt{2}\times \frac{\sqrt{2}}{\sqrt{3}}N\]
Correct Answer: B
Solution :
Since the body is sliding, force of friction\[F=\mu mg\cos \theta \] \[=1.7\times 0.1\times \frac{\sqrt{3}}{2}\] Note: If the body on inclined plane is in equilibrium, the force of friction is\[mg\sin \theta \].
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