A perosn decides to use his bath tub water to generate electric power to run a 40W bulb. The bath tub is located at a height of 10m from the ground and it holds 200 litres of water. He instals a water driven wheel generator on the ground. At what rate should the water drain from the bath tub to light the bulb? How long can he keep the bulb on , if bath tub was full initially ? Efficiency of generator is `90%`. Take `g=9.8m//s^(2)`

Answer

To solve the problem step by step, we need to calculate the rate at which water should drain from the bathtub to generate enough power for a 40W bulb, and then determine how long the bulb can be kept on with the initial amount of water.

### Step 1: Calculate the output power required from the generator
The power required to run the bulb is given as 40W. Since power is measured in joules per second, the generator must output 40 joules of energy every second.

**Hint:** Remember that power is defined as energy per unit time.

### Step 2: Account for the efficiency of the generator
The efficiency of the generator is given as 90%. This means that only 90% of the input energy is converted to output energy. Therefore, the input energy required can be calculated using the formula:

\[ \text{Input Power} = \frac{\text{Output Power}}{\text{Efficiency}} = \frac{40W}{0.9} \approx 44.44W \]

**Hint:** Efficiency is the ratio of useful output energy to input energy, expressed as a percentage.

### Step 3: Calculate the energy required per second
Since the input power is approximately 44.44W, this means that the generator needs to convert this amount of energy from the falling water every second.

**Hint:** Remember that 1 Watt = 1 Joule/second.

### Step 4: Calculate the gravitational potential energy per unit mass
The gravitational potential energy (PE) for a mass \(m\) falling from a height \(h\) is given by:

\[ PE = m \cdot g \cdot h \]

Where:
- \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity)
- \(h = 10 \, \text{m}\) (height of the bathtub)

Thus, the energy per unit mass (for 1 kg of water) falling from 10m is:

\[ PE = 1 \cdot 9.8 \cdot 10 = 98 \, \text{J/kg} \]

**Hint:** Gravitational potential energy depends on the height from which the mass falls and the acceleration due to gravity.

### Step 5: Calculate the mass flow rate required
To find the mass flow rate (\( \dot{m} \)) required to generate the necessary power, we can use the relationship:

\[ \text{Power} = \text{mass flow rate} \cdot \text{energy per unit mass} \]

Rearranging gives:

\[ \dot{m} = \frac{\text{Power}}{\text{Energy per unit mass}} = \frac{44.44 \, \text{W}}{98 \, \text{J/kg}} \approx 0.453 \, \text{kg/s} \]

**Hint:** The mass flow rate tells us how much mass needs to fall per second to generate the required power.

### Step 6: Calculate the total mass of water in the bathtub
The bathtub holds 200 liters of water. Since the density of water is 1 kg/L, the total mass of water is:

\[ m_i = 200 \, \text{kg} \]

**Hint:** The density of water is a key factor in converting volume to mass.

### Step 7: Calculate the time the bulb can be kept on
Now we can find out how long the bulb can be kept on by dividing the total mass of water by the mass flow rate:

\[ t = \frac{m_i}{\dot{m}} = \frac{200 \, \text{kg}}{0.453 \, \text{kg/s}} \approx 441 \, \text{s} \]

**Hint:** The time the bulb can be kept on is determined by the total mass of water and the rate at which it is used.

### Final Answers
1. The rate at which the water should drain from the bathtub is approximately **0.453 kg/s**.
2. The bulb can be kept on for approximately **441 seconds**.