
A man can swim in still water at $4m/s$. River is flowing at $2m/s$. The angle downstream at which he should swim to cross the river with minimum drift.
A) $120^\circ $
B) $150^\circ $
C) $30^\circ $
D) $60^\circ $
Answer
214.2k+ views
Hint: If a man wants to cross the river with minimum drift then he must be reached at the point which is in front of his starting point. In this condition the drift is zero which is minimum drift.
Its means he should not have any velocity in the direction of river or opposite direction he must have velocity perpendicular to river flow
Complete step by step solution:
In our question the given values are ${v_{mr}} = 4m/s$ and ${v_{rg}} = 2m/s$
Where ${v_{mr}} \Rightarrow $ velocity of man with respect to river
And ${v_{rg}} \Rightarrow $ velocity of river with respect to ground
If we see man from ground frame then see the diagram

If man wants to reach at point B from point A then he should swim with an angle $\theta $ with vertical direction or with y direction as mention in diagram
River flow in x direction with velocity $2m/s$
Component of velocity of man with respect to river in x and y direction is $4\sin \theta $ and $4\cos \theta $ as and velocity of river is in x direction as shown in diagram.

If man reached exactly at point B then the x component of velocity of man must equal to the velocity of the river.
$ \Rightarrow 4\sin \theta = 2$ Means net velocity in x direction must be zero.
$ \Rightarrow \sin \theta = \dfrac{2}{4}$
$ \Rightarrow \sin \theta = \dfrac{1}{2}$
So $\sin \theta = \sin 30^\circ $
$\therefore \theta = 30^\circ $
$\theta \Rightarrow$ Angle made from y direction
So angle made with downstream is $ = 30^\circ + 90^\circ $
$ \Rightarrow 120^\circ $
Hence the angle with downstream is $120^\circ $
Therefore, Option (A) is correct.
Note: Man has zero velocity in x direction means that the component of velocity of man in x direction must cancel out the velocity of river in x direction. Only in this condition man do not travel any distance along the river and can reach exactly the opposite point to the initial point. Means zero drift or minimum drift.
Its means he should not have any velocity in the direction of river or opposite direction he must have velocity perpendicular to river flow
Complete step by step solution:
In our question the given values are ${v_{mr}} = 4m/s$ and ${v_{rg}} = 2m/s$
Where ${v_{mr}} \Rightarrow $ velocity of man with respect to river
And ${v_{rg}} \Rightarrow $ velocity of river with respect to ground
If we see man from ground frame then see the diagram

If man wants to reach at point B from point A then he should swim with an angle $\theta $ with vertical direction or with y direction as mention in diagram
River flow in x direction with velocity $2m/s$
Component of velocity of man with respect to river in x and y direction is $4\sin \theta $ and $4\cos \theta $ as and velocity of river is in x direction as shown in diagram.

If man reached exactly at point B then the x component of velocity of man must equal to the velocity of the river.
$ \Rightarrow 4\sin \theta = 2$ Means net velocity in x direction must be zero.
$ \Rightarrow \sin \theta = \dfrac{2}{4}$
$ \Rightarrow \sin \theta = \dfrac{1}{2}$
So $\sin \theta = \sin 30^\circ $
$\therefore \theta = 30^\circ $
$\theta \Rightarrow$ Angle made from y direction
So angle made with downstream is $ = 30^\circ + 90^\circ $
$ \Rightarrow 120^\circ $
Hence the angle with downstream is $120^\circ $
Therefore, Option (A) is correct.
Note: Man has zero velocity in x direction means that the component of velocity of man in x direction must cancel out the velocity of river in x direction. Only in this condition man do not travel any distance along the river and can reach exactly the opposite point to the initial point. Means zero drift or minimum drift.
Recently Updated Pages
Chemical Equation - Important Concepts and Tips for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

Conduction, Transfer of Energy Important Concepts and Tips for JEE

JEE Analytical Method of Vector Addition Important Concepts and Tips

Atomic Size - Important Concepts and Tips for JEE

JEE Main 2022 (June 29th Shift 1) Maths Question Paper with Answer Key

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

JEE Main Correction Window 2026 Session 1 Dates Announced - Edit Form Details, Dates and Link

Equation of Trajectory in Projectile Motion: Derivation & Proof

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Hybridisation in Chemistry – Concept, Types & Applications

Angle of Deviation in a Prism – Formula, Diagram & Applications

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

Collision: Meaning, Types & Examples in Physics

