Question

In figure shown, minimum mass of block B (at a particular angle between horizontal and string AP) to just slide the block A on rough horizontal surface is $\frac{m}{2}$ as shown in figure. If $\mu$ is the coefficient of friction between block A and ground then $\frac{1}{{\mu }^2}$ will be

• A. $3$
• B. $9$
• C. $\frac{1}{3}$
• D. $\frac{1}{9}$

Answer 1

The correct option is A $3$

If the block A is pulled at an angle $\theta$ with the horizontal then the minimum force required is
$F=\frac{\mu mg}{\cos \theta +\mu \sin \theta }$

For sliding on rough surface, minimum force $F$ should be equal to $mg/2$.

Thus, $\frac{\mu mg}{\cos \theta +\mu \sin \theta }=\frac{mg}{2}...\left(1\right)$

We know that, $\theta ={\tan }^{-1}\left(\mu \right)\Rightarrow \tan \theta =\mu$

It will give $AC=\mu ,AB=1$

So, $BC=\sqrt{1+{\mu }^2}$

Now, $\cos \theta =\frac{AB}{BC}=\frac{1}{\sqrt{1+{\mu }^2}}$ and $\sin \theta =\frac{AC}{BC}=\frac{\mu }{\sqrt{1+{\mu }^2}}$

Eqn (1) becomes, $\frac{\mu mg}{\frac{1}{\sqrt{1+{\mu }^2}}+\mu .\frac{\mu }{\sqrt{1+{\mu }^2}}}=\frac{mg}{2}$

$\frac{\mu mg}{\sqrt{1+{\mu }^2}}=\frac{mg}{2}$

$1+{\mu }^2=4{\mu }^2$

${\mu }^2=\frac{1}{3}$

$\frac{1}{{\mu }^2}=3$