A swimmer wishes to reach directly opposite bank of a river, flowing with velocity 8 m/s. The swimmer can swim 10 m/s still water. The width of the river is 480 m. Time taken by him to do so:

Options

A. 60 sec
B. 48 sec
C. 80 sec
D. None of these.

Answer

To solve the problem of the swimmer trying to reach the opposite bank of a river flowing at 8 m/s while swimming at 10 m/s in still water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Velocity of the river (\(V_r\)) = 8 m/s
- Velocity of the swimmer in still water (\(V_s\)) = 10 m/s
- Width of the river (\(d\)) = 480 m

2. **Determine the Components of the Swimmer's Velocity:**
- The swimmer must swim at an angle (\(\theta\)) such that the component of his swimming velocity that counters the river's current allows him to swim directly across.
- The component of the swimmer's velocity in the direction of the river's flow is given by:
\[ V_s \cos(\theta) = V_r \]
- The component of the swimmer's velocity perpendicular to the river flow (which helps him cross) is:
\[ V_s \sin(\theta) \]

3. **Set Up the Equation:**
- From the first equation, we can express \(\cos(\theta)\):
\[ \cos(\theta) = \frac{V_r}{V_s} = \frac{8}{10} = 0.8 \]
- Using the Pythagorean identity, we can find \(\sin(\theta)\):
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
\[ \sin^2(\theta) + (0.8)^2 = 1 \]
\[ \sin^2(\theta) + 0.64 = 1 \]
\[ \sin^2(\theta) = 0.36 \]
\[ \sin(\theta) = 0.6 \]

4. **Calculate the Effective Velocity Across the River:**
- The effective velocity of the swimmer in the direction across the river is:
\[ V_{\text{cross}} = V_s \sin(\theta) = 10 \times 0.6 = 6 \text{ m/s} \]

5. **Calculate the Time Taken to Cross the River:**
- The time taken (\(t\)) to cross the river can be calculated using the formula:
\[ t = \frac{d}{V_{\text{cross}}} \]
- Substituting the values:
\[ t = \frac{480 \text{ m}}{6 \text{ m/s}} = 80 \text{ seconds} \]

### Final Answer:
The time taken by the swimmer to reach the opposite bank is **80 seconds**.

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