question_answer

The string just touch the block A in the figure as shown, such that A and B of equal mass are in equilibrium. All the surfaces have same coefficient of friction. Find it:             

A) \[\frac{\sqrt{2}}{\sqrt{3}}\]          

B)                                    \[\frac{1}{\sqrt{3}}\]

C) \[\frac{2}{\sqrt{3}}\]                                     

D) \[\frac{\sqrt{3}}{2}\]

Correct Answer: B

Solution :

Bod A, \[mg=2\mu N\]                ?(1) Bod B, \[N=T\cos {{60}^{o}}=\frac{T}{2}\]           ?(2) \[mg+\mu N=T\sin {{60}^{o}}\] \[=T\frac{\sqrt{3}}{2}\] Putting value of N \[T=\frac{2mg}{\left( \sqrt{3}-\mu  \right)}\] From (1) and (2), \[T=\frac{mg}{\mu }\] Comparing \[\frac{2mg}{(\sqrt{3}-\mu )}=\frac{mg}{\mu }\]or\[\mu =\frac{1}{\sqrt{3}}\]