A) \[\frac{1}{\sqrt{3}}\]
B) \[\frac{1}{2\sqrt{3}}\]
C) \[\frac{1}{3\sqrt{3}}\]
D) \[\frac{1}{4\sqrt{3}}\]
Correct Answer: B
Solution :
From the relation, force required to move on the inclined plane is given by \[{{F}_{1}}=mg(sin\theta +\mu cos\theta )\] ?(i) But, when the body slides down then force required to slide down is given by \[{{F}_{2}}=mg(sin\theta -\mu cos\theta )\] ?(ii) (because force of friction acts opposite) The condition given \[{{F}_{1}}=3{{F}_{2}}\] ?(iii) Now, putting the values from (i) and (ii) Eqs. (iii), we obtain \[mg(sin\theta +\mu cos\theta )=3mg(sin\theta -\mu cos\theta )\] \[\sin \theta +\mu \cos \theta =3\sin \theta -3\mu \cos \theta \] \[2\sin \theta =4\mu \cos \theta \] \[\mu =\frac{1}{2}\tan \theta \] (given\[\theta ={{30}^{o}}\]) \[\mu =\frac{1}{2}\tan {{30}^{o}}\] \[\mu =\frac{1}{2\sqrt{3}}\]
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