Water falls from a 40 m high dam at the rate of `9xx10^(4)kg` per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydro electric energy number of 100 W lamps, that can be lit, is :
(Take `g=10ms^(-2)`)

Options

A. 25
B. 50
C. 100
D. 18

Explanation

To solve the problem, we need to calculate the number of 100 W lamps that can be lit using the hydroelectric energy generated from water falling from a dam. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the mass of water falling in one second
Given that water falls at a rate of \( 9 \times 10^4 \) kg per hour, we need to convert this to kg per second.

\[ \text{Mass flow rate} = \frac{9 \times 10^4 \text{ kg}}{3600 \text{ s}} = \frac{9 \times 10^4}{3600} \approx 25 \text{ kg/s} \]

### Step 2: Calculate the gravitational potential energy (GPE)
The gravitational potential energy (GPE) of the falling water can be calculated using the formula:

\[ \text{GPE} = mgh \]

Where:
- \( m \) is the mass of water falling per second (25 kg/s),
- \( g \) is the acceleration due to gravity (10 m/s²),
- \( h \) is the height of the dam (40 m).

Substituting the values:

\[ \text{GPE} = 25 \text{ kg/s} \times 10 \text{ m/s}^2 \times 40 \text{ m} = 10000 \text{ J/s} = 10000 \text{ W} \]

### Step 3: Calculate the electrical energy available
Since only 50% of the gravitational potential energy can be converted into electrical energy, we calculate:

\[ \text{Electrical energy} = \frac{1}{2} \times \text{GPE} = \frac{1}{2} \times 10000 \text{ W} = 5000 \text{ W} \]

### Step 4: Calculate the number of 100 W lamps that can be lit
To find the number of 100 W lamps that can be powered by the available electrical energy, we divide the total electrical energy by the power of each lamp:

\[ N = \frac{\text{Electrical energy}}{\text{Power of each lamp}} = \frac{5000 \text{ W}}{100 \text{ W}} = 50 \]

### Final Answer
The number of 100 W lamps that can be lit is **50**.

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