A man can swim with speed `5 ms^(-1)` in still river while the river is also flowing speed `10ms ^(-1)` If the width of the river is 100 m then minimum possible drift is

Options

A. 100 m
B. 0
C. `100 sqrt(3) m`
D. `100 sqrt(2) m`

Answer

To solve the problem of finding the minimum possible drift of a man swimming across a river, we can follow these steps:

### Step 1: Understand the Problem
We have a man swimming in a river that is flowing. The man swims at a speed of \(5 \, \text{m/s}\) perpendicular to the river flow, while the river flows at a speed of \(10 \, \text{m/s}\). The width of the river is \(100 \, \text{m}\). We need to find the minimum drift, which is the horizontal distance the man is carried downstream while swimming across the river.

### Step 2: Set Up the Diagram
1. **Direction of Flow**: Let the direction of the river flow be along the positive x-axis.
2. **Swimming Direction**: The man swims at an angle \(\theta\) to the direction of the river flow. The component of his swimming speed in the direction of the river flow is \(5 \cos \theta\) and the component perpendicular to the flow (across the river) is \(5 \sin \theta\).

### Step 3: Calculate Time to Cross the River
The time \(t\) taken to cross the river can be calculated using the width of the river and the vertical component of the swimming speed:
\[ t = \frac{\text{Width of river}}{\text{Vertical speed}} = \frac{100}{5 \sin \theta} = \frac{20}{\sin \theta} \]

### Step 4: Calculate Drift
The drift \(D\) is the distance the man is carried downstream while swimming across the river. This is given by the product of the horizontal speed (due to river flow and horizontal component of swimming) and the time taken to cross:
\[ D = \left(10 + 5 \cos \theta\right) t = \left(10 + 5 \cos \theta\right) \left(\frac{20}{\sin \theta}\right) \]
Thus,
\[ D = \frac{20(10 + 5 \cos \theta)}{\sin \theta} \]

### Step 5: Minimize Drift
To find the angle \(\theta\) that minimizes \(D\), we can differentiate \(D\) with respect to \(\theta\) and set the derivative equal to zero:
\[ \frac{dD}{d\theta} = 0 \]
After differentiating and simplifying, we find that:
\[ -2 \cos \theta + 1 = 0 \implies \cos \theta = \frac{1}{2} \implies \theta = 60^\circ \]

### Step 6: Calculate Minimum Drift
Substituting \(\theta = 60^\circ\) back into the drift equation:
\[ D = \frac{20(10 + 5 \cdot \frac{1}{2})}{\sin 60^\circ} = \frac{20(10 + 2.5)}{\frac{\sqrt{3}}{2}} = \frac{20 \cdot 12.5 \cdot 2}{\sqrt{3}} = \frac{500}{\sqrt{3}} = 100\sqrt{3} \, \text{m} \]

### Final Answer
The minimum possible drift is:
\[ \boxed{100\sqrt{3} \, \text{m}} \]