A) \[(1-1/{{n}^{2}})\]
B) \[1/(1-{{n}^{2}})\]
C) \[\sqrt{(1-1/{{n}^{2}})}\]
D) \[1/\sqrt{(1-{{n}^{2}})}\]
Correct Answer: A
Solution :
For smooth surface, \[s=\frac{1}{2}g\sin \theta t_{1}^{2}\] | ..(i) |
For rough surface,\[a=g(\sin \theta -\mu \cos \theta )_{t_{2}^{2}}^{{}}\] | |
\[\therefore s=\frac{1}{2}g{{(sin\theta -\mu cos\theta )}_{t_{2}^{2}}}\] ..(ii) |
From (i) and (ii) |
\[\therefore s=\frac{1}{2}g\sin {{\theta }_{t_{2}^{2}}}=\frac{1}{2}g{{(\sin \theta -\mu \cos \theta )}_{t_{2}^{2}}}\] |
Given, \[\theta =45{}^\circ \therefore t_{1}^{2}=(1-\mu )t_{2}^{2}\] |
Also, given that, \[{{t}^{2}}=n{{t}_{1}}\therefore t_{1}^{2}=\left( 1-\mu \right){{n}^{2}}t_{1}^{2}\] |
\[\frac{1}{{{n}^{2}}}=1-\mu \therefore \mu =\left( 1-\frac{1}{{{n}^{2}}} \right)\] |
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