A) Case [a]
B) Case [b]
C) Same in both
D) Nothing can be predicted
Correct Answer: B
Solution :
If the breadth of the lake is l and velocity of boat is\[{{v}_{b}}\]. Time in going and coming back on a quiet day \[tQ=\frac{l}{{{v}_{b}}}+\frac{l}{{{v}_{b}}}=\frac{2l}{{{v}_{b}}}\] ?. (i) Now if \[{{v}_{a}}\] is the velocity of air-current then time taken in going across the lake, \[{{t}_{1}}=\frac{l}{{{v}_{b}}+{{v}_{a}}}\] [As current helps the motion] and time taken in coming back \[{{t}_{2}}=\frac{l}{{{v}_{b}}-{{v}_{a}}}\] [As current opposes the motion] So \[{{t}_{R}}={{t}_{1}}+{{t}_{2}}=\frac{2l}{\nu b[1-{{({{\nu }_{a}}/{{\nu }_{b}})}^{2}}]}\] ? (ii) From equations (i) and (ii) \[\frac{{{t}_{R}}}{{{t}_{Q}}}=\frac{1}{[1-{{({{\nu }_{a}}/{{\nu }_{b}})}^{2}}]}>1\,\,\left[ as\,\,1\,-\frac{\nu _{a}^{2}}{\nu _{b}^{2}}<1 \right]\] i.e., \[{{\operatorname{t}}_{R}}\, >\, {{t}_{Q}}\] i.e., time taken to complete the journey on quiet day is lesser than that on rough day.
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