A man can swim with speed 5ms- 1 in still river while the river is also flowing with speed 10ms- 1 . If the width of the river is 100m then minimum possible drift is

Question

A man can swim with speed 5ms- 1 in still river while the river is also flowing with speed 10ms- 1 . If the width of the river is 100m then minimum possible drift is (A) 100m (B) 0 (C) 100√3 m (D) 100√2m

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-ry3froamwpsl.png" /> Let the man swim with an angle θ with upstream , By using the concept of relative velocity with respect to river and substituting given values, Drift=(10 - 5 cos θ )(100/5 sin â¡ θ ) Differentiate this find θ at extremum then put θ value in the formula Drift =100√3 m .

Q. A man can swim with speed $5 m s^{- 1}$ in still river while the river is also flowing with speed $10 m s^{- 1}$ . If the width of the river is $100 m$ then minimum possible drift is

$100 m$

0

$1003 m$

$1002 m$

Let the man swim with an angle $θ$ with upstream , By using the concept of relative velocity with respect to river and substituting given values,
$Dr i f t = (10 - 5 cos θ) \frac{100}{5 s in θ}$
Differentiate this find $θ$ at extremum & then put $θ$ value in the formula
Drift $= 1003 m$ .

Q. A man can swim with speed 5ms−1 in still river while the river is also flowing with speed 10ms−1 . If the width of the river is 100m then minimum possible drift is

100m

0

1003​m

1002​m

Let the man swim with an angle θ with upstream , By using the concept of relative velocity with respect to river and substituting given values, Drift=(10−5cosθ)5sin⁡θ100​ Differentiate this find θ at extremum & then put θ value in the formula Drift =1003​m .

Q. A man can swim with speed $5 m s^{- 1}$ in still river while the river is also flowing with speed $10 m s^{- 1}$ . If the width of the river is $100 m$ then minimum possible drift is

$100 m$

$1003 m$

$1002 m$

Let the man swim with an angle $θ$ with upstream , By using the concept of relative velocity with respect to river and substituting given values,
$Dr i f t = (10 - 5 cos θ) \frac{100}{5 s in θ}$
Differentiate this find $θ$ at extremum & then put $θ$ value in the formula
Drift $= 1003 m$ .