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The area of two holes \(A\) and \(B\) are \(2 a\) and \(a\), respectively. The holes are at height \((H / 3)\) and \((2 H / 3)\) from the surface of water. Find the correct option(s): (1) The velocity of efflux at hole \(B\) is 2 times the velocity of efflux at hole \(A\). (2) The velocity of efflux at hole \(B\) is \(\sqrt{2}\) time the velocity of efflux at hole \(A\). (3) The discharge is same through both the holes. (4) The discharge through hole \(A\) is \(\sqrt{2}\) time the discharge through hole \(B\)

Short Answer

Expert verified
Option (4) is correct: The discharge through hole A is \( \sqrt{8} \) or 2.83 times the discharge through hole B.

Step by step solution

01

Identify Key Variables

The problem involves two holes, A and B, with areas \( 2a \) and \( a \) respectively, positioned at heights \( \frac{H}{3} \) and \( \frac{2H}{3} \) from the water surface. We are required to determine the relationship between the velocity of efflux and discharge through these holes.
02

Apply Torricelli's Theorem for Efflux Velocity

Torricelli's theorem states that the velocity \( v \) of efflux for a fluid under gravity from a hole at depth \( h \) is given by: \( v = \sqrt{2gh} \). Here, for hole \( A \), \( h_A = H - \frac{H}{3} = \frac{2H}{3} \). For hole \( B \), \( h_B = H - \frac{2H}{3} = \frac{H}{3} \). Applying the theorem: - Velocity of efflux at hole \( A \), \( v_A = \sqrt{2g \times \frac{2H}{3}} = \sqrt{\frac{4gH}{3}} \).- Velocity of efflux at hole \( B \), \( v_B = \sqrt{2g \times \frac{H}{3}} = \sqrt{\frac{2gH}{3}} \).
03

Compare Velocities of Efflux

We need to find \( \frac{v_B}{v_A} \):\[ \frac{v_B}{v_A} = \frac{\sqrt{\frac{2gH}{3}}}{\sqrt{\frac{4gH}{3}}} = \frac{\sqrt{2}}{\sqrt{4}} = \frac{1}{\sqrt{2}}\approx 0.71 \]. Therefore, \( v_B \) is not twice nor \( \sqrt{2} \) times \( v_A \).
04

Calculate the Discharge Through Each Hole

Discharge (or flow rate) \( Q \) through a hole is given by \( Q = A \cdot v \), where \( A \) is the area of the hole.- Discharge through hole \( A \), \( Q_A = 2a \cdot v_A = 2a \cdot \sqrt{\frac{4gH}{3}} \).- Discharge through hole \( B \), \( Q_B = a \cdot v_B = a \cdot \sqrt{\frac{2gH}{3}} \).
05

Compare Discharges

We calculate \( \frac{Q_A}{Q_B} = \frac{2a \cdot \sqrt{\frac{4gH}{3}}}{a \cdot \sqrt{\frac{2gH}{3}}} = 2 \times \sqrt{2} = \sqrt{8} \). Therefore, the discharge through hole \( A \) is \( \sqrt{8} \approx 2.83 \) times the discharge through hole \( B \), not \( \sqrt{2} \). Thus, option (4) is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torricelli's Theorem
Torricelli's theorem is a fundamental principle in fluid mechanics that describes the rate at which fluid exits an opening beneath a body of fluid under the influence of gravity. The theorem is named after Italian scientist Evangelista Torricelli. It provides a way to calculate the velocity of a fluid flowing out of a hole in a tank filled with liquid. According to the theorem, the velocity of efflux from the hole is equal to the velocity that a drop of the same fluid would acquire if it were allowed to free-fall under gravity from the same height as the fluid above the hole.

Mathematically, this is expressed as:\[ v = \sqrt{2gh} \]where:
  • \(v\) is the velocity of efflux,
  • \(g\) is the acceleration due to gravity, and
  • \(h\) is the vertical distance between the fluid surface and the hole where the fluid exits.
This theorem is crucial for understanding how fluids behave when passing through openings, allowing us to predict the speed at which they exit. Overall, it's a powerful application in various engineering and fluid dynamics scenarios.
Velocity of Efflux
The velocity of efflux refers to the speed at which a fluid exits through an orifice. By using Torricelli's theorem, it's possible to determine this velocity based on the height of the fluid column above the orifice. The height dictates the potential energy available to convert into kinetic energy, which is why the formula for efflux velocity involves the square root of twice the gravitational acceleration multiplied by this height.

For instance, in the exercise, we have two holes, A and B, located at different heights. By applying the theorem:
  • Hole \(A\) has a height of \(\frac{2H}{3}\) from the bottom, resulting in a velocity \(v_A = \sqrt{\frac{4gH}{3}}\).
  • Hole \(B\)'s height is \(\frac{H}{3}\), giving it a velocity \(v_B = \sqrt{\frac{2gH}{3}}\).
Comparing these, it’s evident that different heights produce different efflux velocities, explaining why fluids exit with different speeds from orifices at varying depths in a container.
Flow Rate Calculation
Flow rate, also known as discharge, refers to the volume of fluid passing through a hole in a given time. It's calculated using the formula:\[ Q = A \cdot v \]where:
  • \(Q\) is the flow rate,
  • \(A\) is the area of the hole, and
  • \(v\) is the velocity of the fluid flowing out.
In cases with multiple holes or varying areas, like holes A and B in the exercise, the area significantly affects the flow rate. Since hole A has an area twice as large as hole B \( (2a) \) and a greater efflux velocity \( v_A \), it also exhibits a much larger discharge compared to hole B.
To understand this relationship, we compare flow rates:
  • The discharge through hole \(A\) is \(Q_A = 2a \cdot v_A = 2a \cdot \sqrt{\frac{4gH}{3}}\),
  • while that through hole \(B\) is \(Q_B = a \cdot v_B = a \cdot \sqrt{\frac{2gH}{3}} \).
Upon calculating, it becomes clear that the discharge through hole \(A\) is markedly higher than through hole \(B\), precisely at \(\sqrt{8}\) times, reinforcing the connection between area, velocity, and resulting flow rate.

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Most popular questions from this chapter

Length of horizontal arm of a uniform cross section \(U\)-tube is \(l=21 \mathrm{~cm}\) and ends of both of the vertical arms are open to surrounding of pressure \(10 \times 500 \mathrm{~N} / \mathrm{m}^{2}\). A liquid of density \(\rho=10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) is poured into the tube such that liquid just fillg is sealed and the tube is then rotated about a vertical axiy arm with angular val \(\omega_{0}=10 \mathrm{rad} / \mathrm{s} .\) If length of each vertical arm is \(a=6 \mathrm{~cm}\) calculate the length of air column in the sealed arm. (in cm).

For Problems 31-32 A U-tube of uniform cross section contains a liquid of density d. Diameter of the tube is \(D\). Height of liquid in the left arm and also in the right arm is \(h\) and length of horizontal portion of the tube and hence the horizontal column of liquid is \(L\). Consider the following situations. (i) The tube is given a uniform acceleration \(a\) towards left. (ii) The tube is mounted on a horizontal table that is made to rotate with a uniform angular speed \(\omega\) with one of the arms (left or right) on the axis of rotation. Take \(M\) as the total mass of the liquid and \(g\) as the acceleration due to gravity. Difference in height between the liquid columns in the vertical arms (left and right) in situation (i) will be (1) \(\sqrt{\frac{L a}{g}} \times h\) (2) \(\frac{L^{2} h^{2} d a}{M g}\) (3) \(\frac{L a}{g}\) (4) \(\frac{L a D}{g h}\)

1\. Three identical vessels \(A, B\) and \(C\) contain same quantity of liquid. In each vessel balls of different densities but same masses are placed. In vessel \(A\), the ball is partly submerged; in vessel \(B\), the ball is completely submerged but floating and in vessel \(C\), the ball has sunk to the base. If \(F_{A}, F_{B}\) and \(F_{c}\) are the total forces acting on the base of vessels \(A, B\) and \(C\), respectively, then (1) \(F_{A}=F_{B}=F_{C}\) (2) \(F_{A}

For Problems 33-34 A piston of mass \(M=3 \mathrm{~kg}\) and radius \(R=\) \(4 \mathrm{~cm}\) has a hole into which a thin pipe of radius \(r=1 \mathrm{~cm}\) is inserted. The piston can enter a cylinder tightly and without friction, and initially it is at the bottom of the cylinder. \(750 \mathrm{~g}\) of water is now poured into the pipe so that the piston and pipe are lifted up as shown. The height \(h\) of water in the pipe is (1) \(\frac{3}{2 \pi} \mathrm{m}\) (2) \(\frac{3}{\pi} \mathrm{m}\) (3) \(\frac{1}{\pi} \mathrm{m}\) (4) \(\frac{2}{\pi} \mathrm{m}\)

For Problems 1-5 If the container filled with liquid gets accelerated horizontally or vertically, pressure in liquid gets changed. In liquid for calculation of pressure, effective \(g\) is used. A closed box with horizontal base \(6 \mathrm{~m}\) by \(6 \mathrm{~m}\) and a height \(2 \mathrm{~m}\) is half filled with liquid. It is given a constant horizontal acceleration \(g / 2\) and vertical downward acceleration \(g / 2\). The angle of the free surface with the horizontal is equal to (1) \(30^{\circ}\) (2) \(\tan ^{-1} \frac{2}{3}\) (3) \(\tan ^{-1} \frac{1}{3}\) (4) \(45^{\circ}\)

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