Question

A steamboat goes across a lake and comes back

$\left(a\right)$ On a quiet day when the water is still and

$\left(b\right)$ On a rough day when there is a uniform air current so as to help the journey 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 lesser time

• A. Case $\left(a\right)$
• B. Case $\left(b\right)$
• C. Same in both
• D. Nothing can be predicted

Answer 1

The correct option is B

Case $\left(b\right)$

Step 1: Find the time taken by the boat to go and come back in quiet water, ${\mathit{t}}_{\mathbf{Q}}$

Let the width of the lake$=l$ and the speed of the stream boat$=v$

On a quiet day, the time taken by the boat to go and come back is $t_Q=\frac{l}{v}+\frac{l}{v}=2l/v$ $($As, time$t=\frac{D}{S}$,where, $D$ is distance, and $S$ is speed$)$

Step 2: Find the time taken by the boat to go and come back on a rough day, $\mathbf{}{\mathit{t}}_{\mathbf{R}}$

On a rough day, the velocity of the air current is $v,$

The time taken by the boat to go is$t_1=\frac{1}{(v+v’)}$, since the air current helps in increasing the speed of motion.

The time taken by the boat to return is$t_2=\frac{1}{(v–v’)}$, since the air current helps in decreasing the speed of motion.

Total time to go and come on a rough day is

$$\begin{gathered} t_R=t_1+t_2 \\ =\frac{2vl}{(v^2–v{\text{'}}^2)} \\ =\frac{2l}{v}\left[1–{\left(\frac{v’}{v}\right)}^2\right] \end{gathered}$$

Step 3: compare ${\mathit{t}}_{\mathbf{Q}}$and $\mathbf{}{\mathit{t}}_{\mathbf{R}}$

We get, $\frac{t_R}{t_Q}=\frac{1}{\left[1–{\left(\frac{v’}{v}\right)}^2\right]}<1$

$\Rightarrow t_Q>t_R$

Hence, option $\text{'}B\text{'}$ is correct.