Question

Water falls from a \( 40 \) m high dam at the rate of \( 9 \times 10^4 \) kg per hour. Fifty percent of gravitational potential energy can be converted into electrical energy. The number of \( 100W \) lamps that can be lit is:

• A. \( 25 \)
• B. \( 50 \)
• C. \( 100 \)
• D. \( 18 \)

Hint:

For hydroelectric power:
- Use \( P = \frac{mgh}{t} \) to find power output.
- Only a fraction of energy is converted into electricity.
- Power is measured in watts (W) or kilowatts (kW).

Answer 1

The Correct Option is B

To determine the number of \( 100W \) lamps that can be lit, we first calculate the available power from the falling water.
Step 1: Given Data
- Height of the dam: \( h = 40 \) m
- Mass flow rate of water: \( m = 9 \times 10^4 \) kg per hour
- Efficiency of conversion: \( \eta = 50\% = 0.5 \)
- Acceleration due to gravity: \( g = 9.8 \) m/s²
- Power of each lamp: \( P_{{lamp}} = 100 \) W
Step 2: Calculate Gravitational Potential Energy Per Second
The gravitational potential energy released per second is:
\[ P_{{input}} = \frac{mgh}{t} \] Since mass flow rate is given per hour, we convert it to per second:
\[ \dot{m} = \frac{9 \times 10^4}{3600} = 25 { kg/s} \] Thus,
\[ P_{{input}} = 25 \times 9.8 \times 40 \] \[ = 9800 { W} = 9.8 { kW} \] Step 3: Calculate Electrical Power Output
Only \( 50\% \) of this energy is converted into electrical energy:
\[ P_{{output}} = 0.5 \times 9800 = 4900 { W} = 4.9 { kW} \] Step 4: Calculate Number of Lamps
Each lamp requires \( 100 \) W. The number of lamps that can be powered is:
\[ N = \frac{P_{{output}}}{P_{{lamp}}} = \frac{4900}{100} = 50 \] Thus, the correct answer is:
\[ {50} \]