A boat which has a speed of 6km/h in still water crosses a river of width 1 km along the shortest possible path in 20 min. the velocity of the river water in km/h is

Options

A. 1
B. 3
C. 4
D. `3sqrt(3)`

Explanation

To solve the problem, we need to find the velocity of the river water given the speed of the boat in still water and the time taken to cross the river.

### Step-by-Step Solution:

1. **Identify Given Information:**
- Speed of the boat in still water, \( V_b = 6 \) km/h
- Width of the river, \( d = 1 \) km
- Time taken to cross the river, \( t = 20 \) minutes = \( \frac{20}{60} = \frac{1}{3} \) hours

2. **Determine the Effective Speed of the Boat:**
The boat crosses the river along the shortest path, which means it must be angled upstream to counteract the downstream flow of the river.
- Let \( V_r \) be the velocity of the river.
- The boat's velocity can be broken down into two components:
- \( V_b \cos(\theta) \) (across the river)
- \( V_b \sin(\theta) \) (against the river's current)

3. **Use the Time to Cross the River:**
The distance across the river is given by:
\[ d = V_b \cos(\theta) \cdot t \]
Substituting the known values:
\[ 1 = 6 \cos(\theta) \cdot \frac{1}{3} \]
Simplifying gives:
\[ 1 = 2 \cos(\theta) \implies \cos(\theta) = \frac{1}{2} \]
This implies:
\[ \theta = 60^\circ \]

4. **Determine the Velocity of the River:**
From the earlier breakdown of velocities, we have:
\[ V_r = V_b \sin(\theta) \]
Substituting \( V_b = 6 \) km/h and \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[ V_r = 6 \cdot \sin(60^\circ) = 6 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3} \text{ km/h} \]

5. **Final Answer:**
The velocity of the river water is \( 3\sqrt{3} \) km/h.