A river is flowing from east to west at a speed of `5(m)/(min)`. A man on south bank of river, capable of swimming `(10m)/(min)` ini still water, wants to swim across the river in shortest time. He should swim

Options

A. Due north
B. due north-east
C. due north-east with double the speed of river
D. none of these

Explanation

To solve the problem of the man swimming across the river in the shortest time, we need to analyze the situation step by step.

### Step-by-Step Solution:

1. **Understand the Problem**:
The river flows from east to west at a speed of 5 m/min. The man can swim at a speed of 10 m/min in still water. He is on the south bank and wants to swim directly across the river to the north bank.

2. **Identify the Directions**:
- The river's flow is horizontal (east to west).
- The man's desired direction to swim is vertical (north).

3. **Determine the Required Angle**:
To minimize the time taken to cross the river, the man must swim at an angle upstream (towards the east) to counteract the downstream flow of the river. This means he will not swim directly north but at an angle.

4. **Use Vector Components**:
Let’s denote:
- \( V_r = 5 \, \text{m/min} \) (speed of the river)
- \( V_m = 10 \, \text{m/min} \) (speed of the man in still water)

The man's swimming velocity can be broken into two components:
- \( V_{mx} \) (east-west component)
- \( V_{my} \) (north-south component)

Since he wants to swim straight across, the north-south component \( V_{my} \) must equal the speed of the river \( V_r \) to ensure he reaches the north bank directly.

5. **Set Up the Equation**:
The relationship between the components can be expressed as:
\[ V_{my} = V_m \sin(\theta) \]
\[ V_{mx} = V_m \cos(\theta) \]
To ensure he reaches directly across, we set:
\[ V_{mx} = V_r \]
Thus,
\[ V_m \cos(\theta) = 5 \]

6. **Calculate the Angle**:
From the above equations, we have:
\[ 10 \cos(\theta) = 5 \implies \cos(\theta) = \frac{1}{2} \]
This gives us:
\[ \theta = 60^\circ \]

7. **Calculate the Time Taken**:
The width of the river (let's denote it as \( d \)) is not given, but the time taken to cross can be calculated using the north-south component:
\[ V_{my} = V_m \sin(60^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/min} \]
The time taken to cross the river is:
\[ t = \frac{d}{V_{my}} = \frac{d}{5\sqrt{3}} \]

### Final Answer:
The man should swim at an angle of \( 60^\circ \) upstream relative to the north direction to cross the river in the shortest time.