A) \[\frac{1}{(1-v_{1}^{2}/{{v}^{2}})}\]
B) \[\frac{1}{(1+v_{1}^{2}/{{v}^{2}})}\]
C) \[(1-v_{1}^{2}/{{v}^{2}})\]
D) \[\left( 1+\frac{v_{1}^{2}}{{{v}^{2}}} \right)\]
Correct Answer: A
Solution :
Let d be the width of lake to be crossed then \[{{T}_{0}}=\frac{2d}{v}\] and \[T=\frac{d}{(v+{{v}_{1}})}-\frac{d}{(v-{{v}_{1}})}\] \[=\frac{2d\,v}{{{v}^{2}}-v_{1}^{2}}\] \[=\frac{2dv}{{{v}^{2}}(1-v_{1}^{2}/{{v}^{2}})}\] \[=\frac{2d}{v(1-v_{1}^{2}/{{v}^{2}})}\] So, \[\frac{T}{{{T}_{0}}}=\frac{1}{(1-v_{1}^{2}/{{v}^{2}})}\]You need to login to perform this action.
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