Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A man crosses the river perpendicular to river flow in time t seconds and travels an equal distance down the stream in T seconds. The  ratio  of  man's speed in still water to th

A man crosses the river perpendicular to river flow in time t seconds and travels an equal distance down the stream in T seconds. The  ratio  of  man's speed in still water to the speed of river water will be :

Option: 1

\frac{t^{2}-T^{2}}{t^{2}+T^{2}}


Option: 2

\frac{T^{2}-t^{2}}{T^{2}+t^{2}}


Option: 3

\frac{t^{2}+T^{2}}{t^{2}-T^{2}}


Option: 4

\frac{T^{2}+t^{2}}{T^{2}-t^{2}}


Answers (1)

best_answer

 

 

Let the velocity of man in still water be v and that of water with respect to the ground be u.

       So The velocity of man perpendicular to river flow with respect to ground =  \sqrt{v^{2}-u^{2}}

Velocity of man downstream = v + u

As given, \sqrt{v^{2}-u^{2}}\; t=\left ( v+u \right )T

\Rightarrow \left ( v^{2}-u^{2} \right )t^{2}=\left ( v+u \right )^{2}T^{2}

\Rightarrow \left ( {v-u} \right )t^{2}=\left ( v+u \right )T^{2}

\therefore \frac{v}{u}=\frac{t^{2}+T^{2}}{t^{2}-T^{2}}

Posted by

rishi.raj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 session 2 answer key out for BArch, BPlanning; response sheet on jeemain.nta.nic.in
anam-khan

JEE Mains 2025 Paper 2 Answer Key LIVE: BArch final answer key PDF expected soon at jeemain.nta.nic.in
anam-khan

JEE Main 2025 response sheet errors: Delhi HC allows student to apply for JEE Adv, seeks NTA explanation

Latest Question