The Question is from neet-physics mechanical-properties-of-fluids torrecellis-throrem, and it was created by the author VENKATESWARARAO

Question

A vertical tube of cross-sectional area $A_1$ is filled with an ideal fluid of density $\rho$.  A small orifice of area $A_2$ ($A_2 << A_1$) is located at a depth $h$ below the surface of the fluid.  A horizontal tube of cross-sectional area $A_3$ is connected to the orifice.  If $A_3 = 2A_2$, what is the horizontal force exerted by the fluid jet on the horizontal tube? (Assume atmospheric pressure $P_0$)

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION ✓ **Answer**: B ## CONCEPT FRAMEWORK **Primary Concept**: Fluid dynamics, Bernoulli's principle, momentum conservation **Related Topics**: Efflux velocity, impact force **Prerequisites**: Understanding of fluid pressure, velocity, and area relationships in fluid flow. ## SOLUTION PATHWAY **Given Information**: $A_1$: Cross-sectional area of vertical tube $ ho$: Density of the fluid $A_2$: Area of the orifice ($A_2 << A_1$) $h$: Depth of the orifice $A_3$: Area of the horizontal tube ($A_3 = 2A_2$) $P_0$: Atmospheric pressure [Important conditions]: Ideal fluid, negligible viscosity, steady flow. **Method & Steps**: 1. Find the efflux velocity ($v$) of the fluid jet using Bernoulli's equation. Reasoning: Applying Bernoulli's equation between the fluid surface and the orifice, we can find the velocity of the fluid as it exits the orifice. Formula/Rule: $P_1 + \frac{1}{2} ho v_1^2 + ho g h_1 = P_2 + \frac{1}{2} ho v_2^2 + ho g h_2$. Since $A_1 >> A_2$, $v_1 \approx 0$. Also, $P_1 = P_2 = P_0$ (atmospheric pressure). Thus, $ ho g h = \frac{1}{2} ho v^2$, where $v$ is the efflux velocity. 2. Calculate the mass flow rate ($\dot{m}$) of the fluid. Working: $v = \sqrt{2gh}$. Mass flow rate is given by $\dot{m} = ho A_2 v = ho A_2 \sqrt{2gh}$ 3. Determine the horizontal force ($F$) exerted by the fluid jet on the horizontal tube using the momentum equation. Reasoning: The change in momentum of the fluid per unit time is equal to the force exerted on the horizontal tube. Since the fluid initially has no horizontal momentum and then gains momentum in the horizontal direction, the force is simply the rate of change of momentum. Formula/Rule: $F = \dot{m}v = ( ho A_2 \sqrt{2gh})(\sqrt{2gh}) = 2 ho g h A_2$ Result: $F = 2 ho g h A_2$ Verification: Units are consistent (N), and the result is physically reasonable. ## OPTION ANALYSIS [A] :x: Incorrect. This only accounts for the pressure force, not the momentum change. [B] ✓: Correct. This correctly calculates the force based on the momentum change of the fluid jet. [C] :x: Incorrect. Uses $A_3$ instead of $A_2$ for the area of the jet. [D] :x: Incorrect. Uses $A_3$ instead of $A_2$ and doubles the result incorrectly. ## LEARNING AIDS **Common Mistakes**: 1.Ignoring the change in momentum of the fluid: This leads to underestimating the force. 2.Using the wrong area: The area of the jet ($A_2$) is crucial, not the area of the horizontal tube ($A_3$). **Quick Tips**: 1.Remember that the force on the tube is due to the change in momentum of the fluid. 2.Bernoulli's equation is key to finding the efflux velocity.

VENKATESWARARAO · Answer

$\rho g h A_2$

VENKATESWARARAO · Answer

The correct Option is "$2\rho g h A_2$"

VENKATESWARARAO · Answer

$\rho g h A_3$

VENKATESWARARAO · Answer

$2\rho g h A_3$

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