Answer:
Let
l be the width of lake and v be the velocity of steam boat. On a quiet day,
time taken by steam boat in going and coming back is
\[{{\text{t}}_{\text{Q}}}=\frac{l}{\upsilon }+\frac{l}{\upsilon
}=\frac{2l}{\upsilon }\] ....(i)
On a rough day, let \[\upsilon '\] be the velocity of air current.
As in going across the lake, the air current helps the motion, so time taken is
\[{{\text{t}}_{\text{1}}}=\frac{l}{\upsilon +\upsilon '}\]
In coming back, as the air current opposes the motion, so time
taken is \[{{\text{t}}_{2}}=\frac{l}{\upsilon -\upsilon '}\] Total time in
going and coming back on a rough day
\[{{\text{t}}_{\text{r}}}={{\text{t}}_{\text{1}}}+\text{
}{{\text{t}}_{\text{2}}}=\frac{2l\upsilon }{\left( {{\upsilon }^{2}}-\upsilon
{{'}^{2}} \right)}=\frac{2l}{\upsilon [1-{{\left( \upsilon '/\upsilon \right)}^{2}}]}\]
....(ii)
From (i) and (ii), we have
\[\frac{{{t}_{R}}}{{{t}_{Q}}}=\frac{1}{[1-{{(\upsilon
'/\upsilon )}^{2}}]}>1\] or \[{{\text{t}}_{\text{R}}}>\text{
}{{\text{t}}_{\text{Q}}}\]
Therefore the time taken to complete the journey on quiet day is
less than on a rough day.
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