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ExamsJEE MainPhysics

Water of density rho flows through a fire hydrant at a volume flow rate L. The water rises vertically inside the hydrant and then turns through 90 degrees to leave horizontally with speed V. The pipe has the same uniform cross-section throughout. Find the magnitude of the force exerted by the water on the bend (corner) of the hydrant.

  1. rho*V*L
  2. zero
  3. 2*rho*V*L
  4. sqrt(2)*rho*V*L

Correct answer: sqrt(2)*rho*V*L

Solution

Mass flow rate = rho*L. Incoming momentum is vertical (rho*L*V upward) and outgoing is horizontal (rho*L*V). The force on the water has components rho*L*V in each direction, magnitude sqrt(2)*rho*L*V. By Newton's third law the water exerts the same magnitude on the corner.

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