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Step 1: Calculate the potential energy of the water in the bathtub.
The formula for potential energy (PE) is given by PE = mgh, where m is the mass of the water, g is the acceleration due to gravity, and h is the height.
The volume of water is given as 200 liters, which can be converted to mass (assuming the density of water is 1000 kg/m³) as follows:
Calculate mass:
$$ m = ext{Density} \times ext{Volume} = 1000 \text{ kg/m}^3 \times 0.2 \text{ m}^3 = 200 \text{ kg} $$
Now, applying the values to the potential energy formula:
$$ PE = mgh = 200 \text{ kg} \times 10 \text{ m/s}^2 \times 10 \text{ m} = 20000 \text{ J} $$
Step 2: Calculate the usable energy considering the generator's efficiency.
The efficiency of the generator is 90%, thus:
$$ ext{Usable energy} = PE \times \frac{\text{Efficiency}}{100} = 20000 \text{ J} \times 0.9 = 18000 \text{ J} $$
Step 3: Determine the power consumption of the bulb.
We know that the bulb consumes 40 watts, which is 40 joules per second.
Step 4: Calculate the duration the bulb can be kept on using the usable energy.
$$ \text{Time} (t) = \frac{\text{Usable energy}}{\text{Power}} = \frac{18000 \text{ J}}{40 \text{ W}} = 450 \text{ seconds} \approx 7.5 \text{ minutes} $$
Step 5: Calculate the rate at which the water should drain.
For the total energy to be used in 450 seconds:
$$ \text{Rate of draining} = \frac{\text{Total Volume}}{\text{Time}} = \frac{0.2 \text{ m}^3}{450 \text{ s}} = \frac{200 \text{ liters}}{450 \text{ s}} \approx 0.444 \text{ liters/s} $$
Therefore, to achieve the desired output of lighting the bulb for 450 seconds, the water should drain at a rate of approximately 0.444 liters per second.