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Q.

River stream velocity grows in proportion to the distance from the bank and reaches its maximum velocity to $2 m s^{- 1}$ in the middle. Near the bank velocity is zero. The velocity of a swimmer in still water is $5 m s^{- 1}$ and is directed perpendicular to river stream. The width of river is 100 m. The drifting in swimming is 5n metre. Find the value of n.

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answer is 4.

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Detailed Solution

$v_{x} = k y a n d v_{y} = 5$
y = 5t, at middle point, 50 = 5t; t = 10 s; $v_{x} = k y$
$v_{x} = \frac{2}{50} = y = \frac{y}{25} = \frac{5 t}{25} = \frac{t}{5}$
$\frac{d x}{d t} = \frac{t}{5} \Rightarrow \int_{0}^{X_{0}} d x = \int_{0}^{10} \frac{t}{5} d t$
$x_{0} = 10 m$ upto middle
Total drifting $= 2 \times 10 = 20 m$
$5 n = 20, n = 4$

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