A swimmer crosses a river with minimum possible time 10 Second. And when he reaches the other end starts swimming in the direction towards the point from where he started swimming. Keeping the direction fixed the swimmer crosses the river in 15 sec. The ratio of speed of swimmer with respect to water and the speed of river flow is (Assume contant speed of river & swimmer)

Options

A. `(3)/(2)`
B. `(9)/(4)`
C. `(2)/(sqrt5)`
D. `(sqrt5)/(2)`

Explanation

To solve the problem, we need to analyze the swimmer's motion in the river and derive the relationship between the speed of the swimmer with respect to water (v) and the speed of the river flow (u).

### Step-by-Step Solution:

1. **Understanding the scenario**:
- The swimmer crosses the river in the minimum possible time of 10 seconds. This means he swims directly perpendicular to the flow of the river.
- When he swims back towards the starting point, it takes him 15 seconds, but he is swimming at an angle against the current.

2. **Setting up the equations**:
- Let the width of the river be \( d \).
- The speed of the swimmer with respect to water is \( v \).
- The speed of the river flow is \( u \).

3. **Equation for the first crossing**:
- For the first crossing (minimum time), the swimmer's time \( t_1 = 10 \) seconds.
- The distance \( d \) is covered in this time, so we have:
\[ d = v \cdot t_1 = v \cdot 10 \quad \text{(1)} \]

4. **Equation for the return journey**:
- For the return journey, the swimmer swims at an angle \( \theta \) against the current.
- The time taken for this journey is \( t_2 = 15 \) seconds.
- The effective velocity in the direction of the river's width is \( v \cos \theta \), so:
\[ d = v \cos \theta \cdot t_2 = v \cos \theta \cdot 15 \quad \text{(2)} \]

5. **Equating the two equations**:
- From equations (1) and (2), we can set them equal since both equal \( d \):
\[ v \cdot 10 = v \cos \theta \cdot 15 \]
- Dividing both sides by \( v \) (assuming \( v \neq 0 \)):
\[ 10 = 15 \cos \theta \]
- Rearranging gives:
\[ \cos \theta = \frac{10}{15} = \frac{2}{3} \quad \text{(3)} \]

6. **Using the cosine relation**:
- From the definition of cosine in a right triangle:
\[ \cos \theta = \frac{v}{\sqrt{v^2 + u^2}} \]
- Substituting from (3):
\[ \frac{v}{\sqrt{v^2 + u^2}} = \frac{2}{3} \]
- Cross-multiplying gives:
\[ 3v = 2\sqrt{v^2 + u^2} \]

7. **Squaring both sides**:
- Squaring both sides results in:
\[ 9v^2 = 4(v^2 + u^2) \]
- Expanding and rearranging gives:
\[ 9v^2 = 4v^2 + 4u^2 \implies 5v^2 = 4u^2 \]

8. **Finding the ratio**:
- Rearranging gives:
\[ \frac{v^2}{u^2} = \frac{4}{5} \]
- Taking the square root gives:
\[ \frac{v}{u} = \frac{2}{\sqrt{5}} \]

### Final Answer:
The ratio of the speed of the swimmer with respect to water (v) to the speed of the river flow (u) is:
\[ \frac{v}{u} = \frac{2}{\sqrt{5}} \]