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# A steam boat goes across a lake and comes back (a) on a quiet day, when the water is still and(b) on a rough day when there is uniform air current so as to help onward and to impede the journey back. If the speed of the launch on both days was the same, in which case it will complete the journey in less time.A. Case (a)B. Case (b)C. same in bothD. nothing can be predicted

Last updated date: 04th Aug 2024
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Hint: We need to calculate the time taken for the journey on a quiet day then for the rough day. For the quiet day, the velocity of the boat remains the same but for the rough the velocity of the boat will increase or decrease depending on whether the air current is aiding or resisting the motion of the boat.

We are given a steam boat travelling across a river. Let the width of this river be l. Consider the first case when the steam boat goes across a lake and comes back on a quiet day, when the water is still. This means that for both the onward and backward journey, the time taken is the same as the velocity of the boat will be the same. Let this velocity be v. Then the total time taken for the onward and backward journey can be written as
$t = \dfrac{l}{v} + \dfrac{l}{v} = \dfrac{{2l}}{v}$
Now consider the second case of a rough day when there is uniform air current so as to help onward and to impede the journey back. Let v’ be the velocity of this air current. For the onward journey, the air current aids the motion of the boat by increasing its velocity. SO, we can write the time taken for onward journey as follows:
${t_1} = \dfrac{l}{{v + v'}}$
Now for the backward journey, the air current opposes the motion of the boat reducing its velocity. In this case the time taken can be written as
${t_2} = \dfrac{l}{{v - v'}}$
Now the total time taken for the onward and backward journey can be written as follows:
\begin{align} & t' = {t_1} + {t_2} \\ & = \dfrac{l}{{v + v'}} + \dfrac{l}{{v - v'}} \\ & = l\left( {\dfrac{{v - v' + v + v'}}{{\left( {v + v'} \right)\left( {v - v'} \right)}}} \right) \\ & = \dfrac{{2lv}}{{{v^2} - v{'^2}}} \\ & \Rightarrow \dfrac{{2l}}{{v\left[ {1 - {{\left( {\dfrac{{v'}}{v}} \right)}^2}} \right]}} \\ \end{align}
Now we can compare t and t’ by dividing t’ with t. Doing so, we get
$\dfrac{{t'}}{t} = \dfrac{1}{{\left[ {1 - {{\left( {\dfrac{{v'}}{v}} \right)}^2}} \right]}}$
This value will be greater than one. This means that $t' > t$.

So, the correct answer is “Option A”.

Note:
It should be noted that in case the motion of the boat had been upstream against the flow of the river then the velocity of the boat would have reduced. While if the motion of the boat had been downstream along the direction of flow of the river then the velocity of the boat would have increased.