A swimmer crosses a river with minimum possible time of 10 s and when he reaches the other end, he starts swimming in the direction towards the point from where he started swimming. Keeping the direction fixed, the swimmer crosses the river in 15 s. The ratio of speed of swimmer with respect to water to the speed of river flow is √(x/y). Find (x+y). (Assume constant speed of river and swimmer)

Question

A swimmer crosses a river with minimum possible time of 10 s and when he reaches the other end, he starts swimming in the direction towards the point from where he started swimming. Keeping the direction fixed, the swimmer crosses the river in 15 s. The ratio of speed of swimmer with respect to water to the speed of river flow is √(x/y). Find (x+y). (Assume constant speed of river and swimmer) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/b7d3872b3791918ddd4e451f80922122-.png" /> 15=(10 VS R/VS R cos θ)=(10 VS R/VS R((10 VS R)10 √VR2+VS R2)) ⇒ 15=10 √((VR/VS R))2+(1) ⇒ (VS R/VR)=(2/√5)

15=VSR​cosθ10VSR​​=VSR​(10VR2​+VSR2​​10VSR​​)10VSR​​ ⇒15=10(VSR​VR​​)2+(1)​ ⇒VR​VSR​​=5​2​

$15 = \frac{10 V _{SR}}{V _{SR} c o s θ} = \frac{10 V _{SR}}{V _{SR} ( \frac{10 V _{SR}}{10 V _{R}^{2} + V _{SR}^{2}} )}$
$\Rightarrow 15 = 10 (\frac{V _{R}}{V _{SR}})^{2} + (1)$
$\Rightarrow \frac{V _{SR}}{V _{R}} = \frac{2}{5}$