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

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)

TestHub · Accepted Answer

Let the depth of the upper hole from the free surface of water be $h_{1}$ and the depth of the lower hole be $h_{2}$ .

The velocity of efflux from the upper hole is $v_{1} = 2 g h_{1}$ and from the lower hole is $v_{2} = 2 g h_{2}$ .

The force of reaction due to water flowing out from a hole is given by the rate of change of momentum, $F = v \frac{d m}{d t} = ρ A v^{2}$ .

The reaction force from the upper hole is $F_{1} = ρ A (2 g h_{1}) = 2 ρ A g h_{1}$ .

The reaction force from the lower hole is $F_{2} = ρ A (2 g h_{2}) = 2 ρ A g h_{2}$ .

Since the holes are on opposite sides, the water jets emerge in opposite directions, meaning the reaction forces also act in opposite directions.

The resultant force of reaction is $F_{n e t} = F_{2} - F_{1} = 2 ρ A g (h_{2} - h_{1})$ .

Given that the height difference between the holes is $x$ , we have $h_{2} - h_{1} = x$ .

Therefore, $F_{n e t} = 2 ρ A gx$ .

Answer: $2 ρ A gx$

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