Water is flowing in a river at `2 ms^(-1)`. The river is `50 m` wide and has an average depth of `5 m`. The power available from the current in the river is (Density of water `= 1000 kg m^(3)`

Options

A. 0.5 MW
B. 1 MW
C. 1.5 MW
D. 2 MW

Explanation

To solve the problem of calculating the power available from the current in the river, we can follow these steps:

### Step 1: Calculate the Cross-Sectional Area of the River
The cross-sectional area (A) of the river can be calculated using the formula:
\[ A = \text{width} \times \text{depth} \]

Given:
- Width of the river = 50 m
- Depth of the river = 5 m

Calculating the area:
\[ A = 50 \, \text{m} \times 5 \, \text{m} = 250 \, \text{m}^2 \]

### Step 2: Calculate the Volume of Water Flowing per Second
The volume of water flowing through the cross-sectional area in one second can be calculated using the formula:
\[ \text{Volume} = A \times \text{velocity} \]

Given:
- Velocity of the river = 2 m/s

Calculating the volume:
\[ \text{Volume} = 250 \, \text{m}^2 \times 2 \, \text{m/s} = 500 \, \text{m}^3/\text{s} \]

### Step 3: Calculate the Mass of Water Flowing per Second
Using the density of water, we can calculate the mass of water flowing per second:
\[ \text{Mass} = \text{density} \times \text{volume} \]

Given:
- Density of water = 1000 kg/m³

Calculating the mass:
\[ \text{Mass} = 1000 \, \text{kg/m}^3 \times 500 \, \text{m}^3/\text{s} = 500000 \, \text{kg/s} = 5 \times 10^5 \, \text{kg/s} \]

### Step 4: Calculate the Kinetic Energy of the Water
The kinetic energy (KE) of the flowing water can be calculated using the formula:
\[ \text{KE} = \frac{1}{2} m v^2 \]

Substituting the values:
\[ \text{KE} = \frac{1}{2} \times (5 \times 10^5 \, \text{kg}) \times (2 \, \text{m/s})^2 \]

Calculating:
\[ \text{KE} = \frac{1}{2} \times 5 \times 10^5 \times 4 = 10 \times 10^5 = 10^6 \, \text{J} \]

### Step 5: Calculate the Power Available from the Current
Power (P) is the rate of doing work or the rate of energy transfer. Since we are looking for the power available from the current, we can use the kinetic energy calculated above. Since this energy is transferred in one second, the power is:
\[ P = \text{KE} = 10^6 \, \text{W} = 1 \, \text{MW} \]

### Final Answer
The power available from the current in the river is **1 Megawatt (MW)**.

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