question_answer

A block of mass 15 kg is resting on a rough inclined plane as shown in the figure. The block is tied by a horizontal sting, which has a tension of 50 N, The coefficient of friction between the surfaces of contact is \[(g=10\,m/{{s}^{2}})\]

A) \[\frac{1}{2}\]                        

B) \[\frac{3}{4}\]            

C) \[\frac{2}{3}\]            

D)      \[\frac{1}{4}\]   

Correct Answer: A

Solution :

The figure shows free body diagram of the block. For equilibrium, along the plane \[\mu R+T\cos {{45}^{o}}=mg\sin {{45}^{o}}\] \[\mu R+T\frac{1}{\sqrt{2}}=\frac{mg}{\sqrt{2}}\]                                     (i) For equilibrium, in direction \[\bot \]to inclined plane,\[R=T\sin {{45}^{o}}+mg\cos {{45}^{o}}\] \[=\frac{T}{\sqrt{2}}+\frac{mg}{\sqrt{2}}\] Put this value of R in (i), \[\frac{\mu }{2}(T+mg)=\frac{1}{\sqrt{2}}(mg-T)\] \[\mu (50+15\times 10)=(15\times 10-50)\] \[\mu =\frac{100}{200}=\frac{1}{2}\] Hence, the correction option is (a).