A) \[\left( 1-1/{{n}^{2}} \right)\]
B) \[1/\left( 1-{{n}^{2}} \right)\]
C) \[\sqrt{\left( 1-1/{{n}^{2}} \right)}\]
D) \[1/\sqrt{\left( 1-{{n}^{2}} \right)}\]
Correct Answer: A
Solution :
[a] \[Smooth\text{ }surface,\,\,S=\frac{1}{2}g\,\,\sin \,\,\theta \,\,{{t}_{1}}^{2}....(i)\] \[For\text{ rough }surface,\text{ a=g(}sin\,\theta -\mu \,\,\cos \theta )\text{ }{{\text{t}}_{\text{2}}}^{\text{2}}\] \[\therefore \,\,s=\frac{1}{2}g\left( sin\theta -\mu cos\theta \right){{t}_{2}}^{2}\] ...(ii) From (i) and (ii), \[\therefore \,\,s=\frac{1}{2}gsin\theta {{t}_{1}}^{2}=\frac{1}{2}\left( sin\theta -\mu cos\theta \right){{t}_{2}}^{2}\] Given \[\theta ={{45}^{o}}\therefore {{t}_{1}}^{2}=\left( 1-\mu \right){{t}_{2}}^{2}\] Also, given that. \[{{t}_{2}}=n{{t}_{1}}\therefore {{t}_{1}}^{2}=(1-\mu ){{n}^{2}}{{t}_{1}}^{2}\] \[\frac{1}{{{n}^{2}}}=1-\mu \therefore \mu =\left( 1-\frac{1}{{{n}^{2}}} \right)\]You need to login to perform this action.
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