The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. The coefficient of friction, when the angle of inclination of the plane is $60^{\circ}$ is

• A

  $\frac{1}{3}$
• B

  $\frac{1}{\sqrt{2}}$
• Correct answer
  C

  $\frac{1}{\sqrt{3}}$
• D

  $\frac{1}{2}$

Solution

For upward motion, $F_{\mathrm{up}}=m g(\sin \theta+\mu \cos \theta)$ For downward motion, $F_{\text {down }}=m g(\sin \theta-\mu \cos \theta)$ ...(ii)
According to the question, $$ \begin{aligned} & F_{\text {up }}=2 F_{\text {down }} \\ & m g(\sin \theta+\mu \cos \theta)=2 m g(\sin \theta-\mu \cos \theta) \\ & \sin \theta+\mu \cos \theta=2 \sin \theta-2 \mu \cos \theta \\ & 3 \mu \cos \theta=\sin \theta \\ & \mu=\frac{1}{3} \tan \theta \\ & \mu=\frac{1}{3} \times \tan 60^{\circ} \\ & \Rightarrow \quad \mu=\frac{1}{3} \sqrt{3}=\frac{1}{\sqrt{3}} \\ & \end{aligned} $$

Hence, the correct answer is option: 'C'.