Exams › JEE Main › Physics
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.
- rho*V*L
- zero
- 2*rho*V*L
- 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.
Related JEE Main Physics questions
⚔️ Practice JEE Main Physics free + battle 1v1 →