A swimmer swins in still water at a speed `=5 km//hr`. He enters a `200m` wide river, having river flow speed `=4km//hr` at point `A` and proceeds to swim at an angle of `127^(@) (sin 37^(@) = 0.6)` with the river flow direction. Another point `B` is located directly across `A` on the other side. the swimmer lands on the other bank at a point `C`, form which he walks the distance `CB` with a speed `= 3km//hr`. The total time in which he reaces from `A` to `B` is

Options

A. 5 minutes
B. 4 minutes
C. 3 minutes
D. None

Answer

To solve the problem, we will break it down into steps and calculate the total time taken by the swimmer to reach from point A to point B via point C.

### Step 1: Understand the scenario
- The swimmer swims at a speed of \( V_s = 5 \, \text{km/hr} \).
- The river flows with a speed of \( V_r = 4 \, \text{km/hr} \).
- The width of the river is \( d = 200 \, \text{m} = 0.2 \, \text{km} \).
- The swimmer swims at an angle of \( 127^\circ \) with respect to the river flow direction. This means he swims at an angle of \( 37^\circ \) with respect to the direction directly across the river.

### Step 2: Calculate the vertical and horizontal components of the swimmer's velocity
1. The vertical component of the swimmer's velocity (across the river) is:
\[ V_{s_y} = V_s \cdot \cos(37^\circ) = 5 \cdot \frac{4}{5} = 4 \, \text{km/hr} \]
2. The horizontal component of the swimmer's velocity (downstream) is:
\[ V_{s_x} = V_s \cdot \sin(37^\circ) = 5 \cdot 0.6 = 3 \, \text{km/hr} \]

### Step 3: Calculate the time taken to swim from A to C (across the river)
- The time \( T_1 \) taken to swim directly across the river can be calculated using:
\[ T_1 = \frac{d}{V_{s_y}} = \frac{0.2 \, \text{km}}{4 \, \text{km/hr}} = 0.05 \, \text{hr} \]

### Step 4: Calculate the drift (distance from B to C)
- The effective horizontal velocity (downstream) of the swimmer is the sum of the swimmer's horizontal component and the river's flow:
\[ V_{effective} = V_{s_x} + V_r = 3 + 4 = 7 \, \text{km/hr} \]
- The drift distance \( X \) can be calculated as:
\[ X = V_{effective} \cdot T_1 = 7 \cdot 0.05 = 0.35 \, \text{km} \]

### Step 5: Calculate the time taken to walk from C to B
- The distance \( CB \) is equal to the drift distance \( X \), which is \( 0.35 \, \text{km} \).
- The time \( T_2 \) taken to walk this distance at a speed of \( 3 \, \text{km/hr} \) is:
\[ T_2 = \frac{CB}{\text{walking speed}} = \frac{0.35 \, \text{km}}{3 \, \text{km/hr}} = 0.1167 \, \text{hr} \]

### Step 6: Calculate the total time from A to B
- The total time \( T_{total} \) is the sum of \( T_1 \) and \( T_2 \):
\[ T_{total} = T_1 + T_2 = 0.05 + 0.1167 = 0.1667 \, \text{hr} \]

### Step 7: Convert total time into minutes
- To convert hours into minutes:
\[ T_{total} \text{ (in minutes)} = 0.1667 \times 60 \approx 10 \, \text{minutes} \]

### Final Answer
The total time taken for the swimmer to reach from point A to point B is approximately **10 minutes**.