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NEET PHYSICSWORK, ENERGY AND POWERMedium

Question

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 m1100\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 m1×2 m s1=200 kg s1\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}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWORK, ENERGY AND POWERenginethroughhosepipepassesthrough

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