In diagram, BAC is a rigid fixed rough wire and angle BAC is 60^o . P and Q are two identi

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4-2.Friction

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In the diagram, $BAC$ is a rigid fixed rough wire and angle $BAC$ is $60^o$. $P$ and $Q$ are two identical rings of mass $m$ connected by a light elastic string of natural length $2a$ and elastic constant $\frac{mg}{a}$. If $P$ and $Q$ are in equilibrium when $PA = AQ = 3a$ then the least coefficient of friction between the ring and the wire is $\mu$. Then value of $\mu + \sqrt 3 $ is :-

A

$2$

B

$3$

C

$4$

D

$7$

Solution

$\mathrm{mg} \sin \theta-\mathrm{f}-\mathrm{kx} \cos \theta=0$

$\mathrm{N}-\mathrm{kx} \sin \theta-\mathrm{mg} \cos \theta=0$

$\mathrm{f}=\mu \mathrm{N}$

Standard 11

Physics

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