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An engine pumps water through a hosepipe. Water passes through the pipe and leaves it with a velocity of 2 m/s2\text{ m/s}. The mass per unit length of water in the pipe is 100 kg m−1100\text{ kg m}^{-1}. What is the power of the engine?

A

400 W

B

200 W

C

100 W

D

800 W

Step-by-Step Solution

  1. Determine Mass Flow Rate: The rate at which mass flows (dmdt\frac{dm}{dt}) is the product of mass per unit length (μ\mu) and the velocity (vv). dmdt=μ×v=100 kg m−1×2 m s−1=200 kg s−1\frac{dm}{dt} = \mu \times v = 100\text{ kg m}^{-1} \times 2\text{ m s}^{-1} = 200\text{ kg s}^{-1}
  2. Calculate Power: Power (PP) is the rate at which kinetic energy is imparted to the water. P=dKdt=12dmdtv2P = \frac{dK}{dt} = \frac{1}{2} \frac{dm}{dt} v^2 [Class 11 Physics, Ch 6, Sec 6.10] Alternatively, substituting dmdt=μv\frac{dm}{dt} = \mu v: P=12(μv)v2=12μv3P = \frac{1}{2} (\mu v) v^2 = \frac{1}{2} \mu v^3
  3. Substitute Values: P=12(100)(2)3=50×8=400 WP = \frac{1}{2} (100) (2)^3 = 50 \times 8 = 400\text{ W}
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