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Solved: A water pump rated 400 W, has an efficiency of 75%. If it is employed to raise water throu [Physics]

Answer

the final result for the volume of water drawn in 10 minutes is: 0.45 m³.

Explanation

To determine the volume of water drawn by the pump in 10 minutes, we can follow these steps:

1. Calculate the output power of the pump using its efficiency. The formula for output power ($$P_{\text{out}}$$P out​) is given by:

$$P_{\text{out}} = \eta \times P_{\text{input}}$$P out​=η×P input​

where $$\eta$$η is the efficiency (0.75) and $$P_{\text{input}}$$P input​ is the input power (400 W). Thus,

$$P_{\text{out}} = 0.75 \times 400 = 300 \, \text{W}$$P out​=0.75×400=300 W

2. Relate the power to the work done in raising water. The power can also be expressed as:

$$P = \frac{dW}{dt} = \frac{d(mgh)}{dt}$$P=d t d W​=d t d(m g h)​

where $$m$$m is the mass of water, $$g$$g is the acceleration due to gravity (approximately $$9.81 \, \text{m/s}^2$$9.81 m/s 2), and $$h$$h is the height (40 m). Rearranging gives:

$$P = \frac{dm}{dt} \cdot g \cdot h$$P=d t d m​⋅g⋅h

3. Rearranging for the mass flow rate ($$\frac{dm}{dt}$$d t d m​):

$$\frac{dm}{dt} = \frac{P}{gh}$$d t d m​=g h P​

Substituting the known values:

$$\frac{dm}{dt} = \frac{300}{9.81 \times 40} \approx 0.765 \, \text{kg/s}$$d t d m​=9.81×40 300​≈0.765 kg/s

4. Calculate the total mass of water drawn in 10 minutes (600 seconds):

$$m = \frac{dm}{dt} \times t = 0.765 \times 600 \approx 459 \, \text{kg}$$m=d t d m​×t=0.765×600≈459 kg

5. Convert mass to volume using the density of water ($$\rho = 1000 \, \text{kg/m}^3$$ρ=1000 kg/m 3):

$$V = \frac{m}{\rho} = \frac{459}{1000} \approx 0.459 \, \text{m}^3$$V=ρ m​=1000 459​≈0.459 m 3

Since the closest option is 0.45 m³, we conclude that: