Question

A block kept on a rough inclined plane, as shown in the figure, remains at rest after maximum force $2\text{N}$ down the inclined plane. The maximum external force up the inclined plane that does not move the block is $10\text{N}$. The coefficient of static friction between the block and the plane is:
$[\text{Take}g=10{\text{ms}}^{-2}]$

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{\sqrt{3}}{4}$
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$

Answer 1

The correct option is A $\frac{\sqrt{3}}{2}$

When the block is pulled down the inclined plane, the friction acts up the inclined plane,

$\Rightarrow 2+mg\sin {30}^{\circ }=\mu mg\cos {30}^{\circ }$

$\Rightarrow 2=mg(\mu \times \frac{\sqrt{3}}{2}-\frac{1}{2})--\left(1\right)$

And, when the block is pulled up, then the friction acts down the inclined plane,

$\Rightarrow 10=mg\sin {30}^{\circ }+\mu mg\cos {30}^{\circ }$

$\Rightarrow 10=mg(\mu \times \frac{\sqrt{3}}{2}+\frac{1}{2})--\left(2\right)$

Dividing $\left(2\right)$ by $\left(1\right)$,

$5=\frac{\sqrt{3}\mu +1}{\sqrt{3}\mu -1}$

$\Rightarrow 4\sqrt{3}\mu =6$

$\Rightarrow \mu =\frac{\sqrt{3}}{2}$

Hence, option $\left(A\right)$ is correct.