Question

On opposite sides of a wide vertical vessel filled with water (density \(ρ\)), two identical holes are drilled, each having cross-section area \(A\). The height difference between holes is \(x\). The resultant force of reaction of water flowing out of vessel is (\(g\) = acceleration due to gravity)

• A. \(ρAgx\)
• B. \(2ρAgx\)
• C. \(3ρAgx\)
• D. \(4ρAgx\)

Hint:

Each jet pushes the vessel back with force rho A v squared; the jets on opposite sides partly cancel.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Water leaving a hole carries momentum. By Newton's third law, the water pushes the vessel in the opposite direction with a force equal to the rate of change of momentum.

Step 2: Force from one hole
Speed of outflow (Torricelli): \(v=\sqrt{2gh}\). Mass leaving per second: \(\rho Av\). So
\[ F=\rho Av\cdot v=\rho Av^2=2\rho Agh \]

Step 3: Two holes on opposite sides
The two reaction forces point in opposite directions, so they subtract. The holes are at depths \(h_1\) and \(h_2\) with \(h_2-h_1=x\):
\[ F_{net}=2\rho Ag(h_2-h_1)=2\rho Agx \]

Step 4: Conclusion
The resultant force is \(2\rho Agx\), option (B).

Final Answer:
Each jet gives 2 rho A g h and the opposite jets subtract, so the net is 2 rho A g x, option (B). \[ \boxed{2\rho Agx} \]