To cross the river in shortest distance, a swimmer should swimming an angle `theta with the upsteram. What is the ratio of the time taken to swim across in the shortest time to that in swimming across over shortest distance. [Asume that the speed of swimmer in still water is greater than the speed of river flow]

Options

A. `cos theta`
B. `sin theta`
C. `tan theta`
D. `cot theta`

Answer

To solve the problem, we need to analyze the swimmer's motion across the river in two different scenarios: one where the swimmer aims to cross in the shortest distance and another where the swimmer aims to cross in the shortest time.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let \( V_s \) be the speed of the swimmer in still water.
- Let \( V_r \) be the speed of the river current.
- Let \( x \) be the width of the river.
- The swimmer makes an angle \( \theta \) with the upstream direction.

2. **Case 1: Shortest Distance**:
- To cross the river in the shortest distance, the swimmer must swim at an angle \( \theta \) such that the horizontal component of the swimmer's velocity cancels out the river's current.
- The horizontal component of the swimmer's velocity is \( V_s \cos \theta \).
- For the swimmer to not drift downstream, we set:
\[ V_s \cos \theta = V_r \]
- The vertical component of the swimmer's velocity, which helps him cross the river, is:
\[ V_{s_y} = V_s \sin \theta \]
- The time taken to cross the river (shortest distance) is:
\[ T_1 = \frac{x}{V_{s_y}} = \frac{x}{V_s \sin \theta} \]

3. **Case 2: Shortest Time**:
- To cross the river in the shortest time, the swimmer should swim directly across the river, perpendicular to the banks.
- In this case, the time taken to cross the river is:
\[ T_2 = \frac{x}{V_s} \]

4. **Finding the Ratio**:
- We need to find the ratio of the time taken to swim across in the shortest time \( T_2 \) to that in swimming across the shortest distance \( T_1 \):
\[ \text{Ratio} = \frac{T_2}{T_1} = \frac{\frac{x}{V_s}}{\frac{x}{V_s \sin \theta}} = \frac{V_s \sin \theta}{V_s} = \sin \theta \]

5. **Conclusion**:
- The ratio of the time taken to swim across in the shortest time to that in swimming across the shortest distance is:
\[ \frac{T_2}{T_1} = \sin \theta \]

### Final Answer:
The ratio of the time taken to swim across in the shortest time to that in swimming across the shortest distance is \( \sin \theta \).

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