A man can swim with a certain speed, in still water to cross a river to a point directly opposite to the starting point in time `t_(0)`. When the river flows, he crosses the river directly along the same path in a time `t_(1)`, If 'd' be the river width, Then the velocity of water current is:

Options

A. `(d)/sqrt(t_(1)t_(0))`
B. `(d)/(t_(1)t_(0))sqrt(t_(1)^(2)-t_(0)^(2))`
C. `(d)/sqrt(t_(1)^(2)-t_(0)^(2))`
D. `dsqrt((t_(1)^(2)+t_(0)^(2)))/(t_(1)t_(0))`

Explanation

To solve the problem, we need to determine the velocity of the water current (Vr) based on the information provided about the man's swimming speed and the time taken to cross the river.

### Step-by-Step Solution:

1. **Define Variables:**
- Let \( d \) be the width of the river.
- Let \( V_m \) be the speed of the man in still water.
- Let \( V_r \) be the speed of the river current.
- Let \( t_0 \) be the time taken by the man to cross the river in still water.
- Let \( t_1 \) be the time taken by the man to cross the river when the river is flowing.

2. **Calculate Speed in Still Water:**
- When the river is still, the man swims directly across the river:
\[ V_m = \frac{d}{t_0} \]

3. **Calculate Effective Speed with Current:**
- When the river is flowing, the man swims at an angle to counteract the current and still reach the point directly opposite:
- The effective speed of the man in the direction across the river remains \( V_m \), but he has to swim at an angle to compensate for the current.
- The time taken to cross the river with the current is:
\[ V' = \frac{d}{t_1} \]

4. **Using the Pythagorean Theorem:**
- The man's velocity \( V_m \) and the river's current \( V_r \) form a right triangle with the resultant velocity \( V' \):
\[ V' = \sqrt{V_m^2 - V_r^2} \]
- Therefore, we can write:
\[ \frac{d}{t_1} = \sqrt{V_m^2 - V_r^2} \]

5. **Substituting \( V_m \):**
- Substitute \( V_m \) from step 2 into the equation:
\[ \frac{d}{t_1} = \sqrt{\left(\frac{d}{t_0}\right)^2 - V_r^2} \]

6. **Squaring Both Sides:**
- Square both sides to eliminate the square root:
\[ \left(\frac{d}{t_1}\right)^2 = \left(\frac{d}{t_0}\right)^2 - V_r^2 \]

7. **Rearranging the Equation:**
- Rearranging gives:
\[ V_r^2 = \left(\frac{d}{t_0}\right)^2 - \left(\frac{d}{t_1}\right)^2 \]

8. **Finding \( V_r \):**
- Taking the square root:
\[ V_r = \sqrt{\left(\frac{d}{t_0}\right)^2 - \left(\frac{d}{t_1}\right)^2} \]
- Simplifying further:
\[ V_r = \frac{d}{t_1 t_0} \sqrt{t_1^2 - t_0^2} \]

### Final Answer:
The velocity of the water current \( V_r \) is given by:
\[ V_r = \frac{d}{t_1 t_0} \sqrt{t_1^2 - t_0^2} \]