A swimmer can swim in still water with speed v and the river is flowing with speed `v//2`. What is the ratio of the time taken to swimming across the river in shortest time to that of swimming across the river over the shortest distance ?

Options

A. (a)`sqrt(3)/2`
B. (b)1
C. (c)2
D. (d)`sqrt(3)`

Explanation

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 over the shortest distance.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let the speed of the swimmer in still water be \( v \).
- Let the speed of the river be \( \frac{v}{2} \).
- Let the width of the river be \( d \).

2. **Determine the Shortest Time**:
- To swim across the river in the shortest time, the swimmer must swim directly downstream (in the direction of the river flow).
- The effective speed of the swimmer in the direction across the river is \( v \) (since he swims perpendicular to the flow).
- The time taken to cross the river in the shortest time \( T_2 \) is given by:
\[ T_2 = \frac{d}{v} \]

3. **Determine the Shortest Distance**:
- To swim across the river over the shortest distance, the swimmer must swim at an angle \( \theta \) such that the resultant velocity across the river is maximized.
- The swimmer's velocity can be broken down into two components: one across the river and one downstream.
- The component of the swimmer's velocity across the river is \( v \cos \theta \) and the downstream component is \( v \sin \theta \).
- The river's current adds to the swimmer's downstream velocity, so the effective downstream velocity is \( v \sin \theta + \frac{v}{2} \).
- The time taken to swim across the river over the shortest distance \( T_1 \) is given by:
\[ T_1 = \frac{d}{v \cos \theta} \]

4. **Finding the Ratio of Times**:
- Now, we need to find the ratio \( \frac{T_2}{T_1} \):
\[ \frac{T_2}{T_1} = \frac{\frac{d}{v}}{\frac{d}{v \cos \theta}} = \frac{v \cos \theta}{v} = \cos \theta \]

5. **Using the Given Speeds**:
- Given that the speed of the river is \( \frac{v}{2} \), we can use the relationship of the angles:
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\frac{v}{2}}{v} = \frac{1}{2} \]
- This means \( \theta = 30^\circ \).
- Therefore, \( \cos 30^\circ = \frac{\sqrt{3}}{2} \).

6. **Final Calculation**:
- The ratio of the time taken to swim across the river in the shortest time to the time taken to swim across the river over the shortest distance is:
\[ \frac{T_2}{T_1} = \cos 30^\circ = \frac{\sqrt{3}}{2} \]

### Conclusion:
The ratio of the time taken to swimming across the river in shortest time to that of swimming across the river over the shortest distance is \( \frac{\sqrt{3}}{2} \).