A swimmer can swim in still water with speed `v` and the river is flowing with velociyt `v//2`. To cross the river in shortest distance, he should swim making angle `theta` with the upstream. What is the ratio of the time taken to swim across the shortest time to that is swimming across over shortest distance

Options

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

Answer

To solve the problem, we need to find the ratio of the time taken to swim across the river in the shortest time to the time taken to swim across the river in the shortest distance.

### Step-by-Step Solution:

1. **Identify Variables:**
- Let \( v \) be the speed of the swimmer in still water.
- Let \( v_r = \frac{v}{2} \) be the speed of the river.
- Let \( d \) be the width of the river.
- Let \( \theta \) be the angle the swimmer makes with the upstream.

2. **Time Taken to Swim Across the Shortest Time (T1):**
- To cross the river in the shortest time, the swimmer should swim directly across the river. The effective speed across the river is given by the component of the swimmer's speed in the direction perpendicular to the river flow.
- The speed of the swimmer in the direction across the river is \( v \sin \theta \).
- The time taken to cross the river is given by:
\[ T_1 = \frac{d}{v \sin \theta} \]

3. **Time Taken to Swim Across the Shortest Distance (T2):**
- To swim across the shortest distance, the swimmer should swim at an angle \( \theta \) upstream to counteract the river's flow. The swimmer's effective speed across the river is still \( v \sin \theta \).
- However, the swimmer's effective speed in the direction of the river flow is \( v \cos \theta \).
- The time taken to swim across the river in this case is:
\[ T_2 = \frac{d}{v \cos(90^\circ - \theta)} = \frac{d}{v \cos \theta} \]

4. **Finding the Ratio of Times:**
- Now, we can find the ratio of the time taken to swim across the shortest time to the time taken to swim across the shortest distance:
\[ \text{Ratio} = \frac{T_1}{T_2} = \frac{\frac{d}{v \sin \theta}}{\frac{d}{v \cos \theta}} = \frac{\cos \theta}{\sin \theta} = \cot \theta \]

### Final Answer:
The ratio of the time taken to swim across the shortest time to the time taken to swim across the shortest distance is:
\[ \text{Ratio} = \cot \theta \]