A) \[1/(1-{{n}^{2}})\]
B) \[1/(1-{{n}^{2}})\]
C) (c)Ö\[\{1-(1/{{n}^{2}})\}\]
D) Ö\[\{1/(1-{{n}^{2}})\}\]
Correct Answer: B
Solution :
[b] Acceleration when there is no friction is a = \[g\sin \theta \] and acceleration when friction is there is a' = \[(g\sin \theta -\mu g\cos \theta )\] We know that \[s=ut+\frac{1}{2}\]\[a{{t}^{2}},\]under condition of u = 0, gives, t=Ö(2s/a). Therefore, t'/t =Ö (a/a'). For our case we get n = Ö \[[g\sin \theta /(g\sin \theta -\mu g\cos \theta )]\]\[\Rightarrow {{n}^{2}}=g\sin \theta /(g\sin \theta -\mu g\cos \theta )=1\]\[(1-\mu g\cot \theta )\]But \[\theta ={{45}^{o}}\]\[\Rightarrow \]\[\cot \theta =1\therefore {{n}^{2}}=1/(1-\mu )\]or\[\mu =1-(1/{{n}^{2}})\]
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