/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 A large open tank has two holes ... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 44

A large open tank has two holes in the walls. One is a square hole of side \(L\) at a depth \(y\) from the top and other is a circular hole of radius \(R\) at a depth \(4 y\) from the top. When the tank is completely filled with water the quantities of water flowing out per second from both the holes are the same. Then \(R\) is equal to (a) \(2 \pi L\) (b) \(\frac{L}{\sqrt{2 \pi}}\) (c) \(L\) (d) \(\frac{L}{2 \pi}\)

Short Answer

Expert verified
Option (b): \(R = \frac{L}{\sqrt{2 \pi}}\).

Step by step solution

01

Understand the problem

We have a tank with two holes: one square (side \(L\)) and one circular (radius \(R\)). The depths from the top to these holes are \(y\) for the square hole and \(4y\) for the circular hole. The flow rates from both holes are equal when the tank is filled.
02

Express the flow rates

The flow of water through each hole can be calculated using Torricelli's Law, which states that \( v = \sqrt{2gh} \), where \( h \) is the depth of the hole. Therefore, the flow rate for the square hole is \(Q_1 = A_1 \cdot v_1 = L^2 \sqrt{2gy}\) and for the circular hole, it is \(Q_2 = A_2 \cdot v_2 = \pi R^2 \sqrt{2g(4y)}\).
03

Equate the flow rates

Since the flow rates are equal, set \(Q_1 = Q_2\). This gives: \[ L^2 \sqrt{2gy} = \pi R^2 \sqrt{2g(4y)} \]
04

Simplify the equation

Cancel \(\sqrt{2g}\) from both sides and simplify: \[ L^2 \sqrt{y} = \pi R^2 \sqrt{4y} \] This further simplifies to:\[ L^2 = 2\pi R^2 \]
05

Solve for R

Now, solve \( L^2 = 2\pi R^2 \) for \( R \):\[ R^2 = \frac{L^2}{2\pi} \]Take the square root of both sides:\[ R = \frac{L}{\sqrt{2\pi}} \]
06

Choose the correct answer

Compare the expression for \( R \) to the given options: (a) \(2 \pi L\), (b) \(\frac{L}{\sqrt{2 \pi}}\), (c) \(L\), (d) \(\frac{L}{2 \pi}\). The correct option is (b): \(R = \frac{L}{\sqrt{2 \pi}}\).

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Key Concepts

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

Torricelli's Law
Named after the Italian scientist Evangelista Torricelli, Torricelli's Law is a principle in fluid dynamics that predicts the speed of fluid flowing out of an orifice. According to this law, the speed (\( v \)) of fluid flowing under the force of gravity through a small hole is given by the equation:\[ v = \sqrt{2gh} \] Here,
  • \( g \), gravity's acceleration (approximately 9.81 m/s虏)
  • \( h \), the height of the liquid above the hole.
This equation implies that the velocity of efflux is directly proportional to the square root of the depth of the liquid. Torricelli's Law is essential for calculating the flow of fluids in various engineering problems.
Flow Rate
Flow rate refers to the volume of fluid that passes through a cross-section per unit time. In fluid dynamics, it's typically expressed in cubic meters per second \((m^3/s)\). Understanding flow rate is crucial as it determines how quickly a fluid can be transported through systems such as pipes or open channels.
The mathematical expression for flow rate \(Q\) is given by:\[ Q = A \cdot v \]where:
  • \( A \) = the cross-sectional area of the orifice
  • \( v \) = the velocity of the fluid as given by Torricelli's Law
This relationship helps in situations like tank drainage, where determining the rate at which a fluid exits is crucial for system design.
Circular Orifice
A circular orifice is simply a round hole through which fluid flows. The flow through a circular orifice can be described using both the area of the circle and Torricelli's Law. To calculate the area of a circular orifice:\[ A = \pi R^2 \]Here:
  • \( R \) is the radius of the orifice
When the fluid flows through this orifice, the velocity can be calculated using Torricelli鈥檚 formula. Circular orifices are common in many applications, such as nozzles and drains, where engineers need to determine how fluid exits a system.
Square Orifice
A square orifice is characterized by having four equal sides through which fluid flows. Just like other shapes, the flow through a square orifice is determined by its area and the velocity of fluid flow. The area \(A\) of a square orifice is given by:\[ A = L^2 \]where:
  • \( L \) is the length of the side of the square
Understanding this helps in calculating the flow rate from square orifices when combined with the velocity determined by Torricelli's Law. They are often used in various industrial applications where accurate fluid discharge is necessary.

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Most popular questions from this chapter

The profile of advancing liquid in a tube is a (a) straight line (b) circle (c) parabola (d) hyperbola

A large tank is filled with water to a height \(H .\) A small hole is made at the base of the tank. It takes \(T_{1}\) time to decrease the height of water to \(\frac{H}{\eta}(\eta>1)\); and it takes \(T_{2}\) time to take out the rest of water. If \(T_{1}=T_{2}\), then the value of \(\eta\) is (a) 2 (b) 3 (c) 4 (d) \(2 \sqrt{2}\)

A liquid of density \(\rho\) is completely filled in a rectangular box. The box is accelerating horizontally with acceleration \(a\). What should be the gauge pressure at four points \(P, Q, R, S ?\) (a) \(P_{P}=0, P_{Q}=0, P_{R}=\rho g h, P_{S}=0\) (b) \(P_{P}=\rho g h, P_{o}=0, P_{k}=\rho g a, P_{s}=\rho a L\) (c) \(P_{P}=0, P_{Q}=\rho g h, P_{k}=\rho g h+\rho a L, P_{S}=\rho a L\) (d) \(P_{P}=\rho g h, P_{\varrho}=0, P_{R}=\rho g h-\rho a L, P_{s}=\rho g L\)

An astronaut is standing at the north pole of a newly discovered planet of radius \(R .\) He holds a container full of liquid having mass \(m\) and volume \(V .\) At the surface of the liquid pressure is \(P_{e}\) and at a depth \(d\) below the surface, pressure is \(P\). From this information determine the mass of the planet (a) \(\frac{\left(P+P_{0}\right) V R^{2}}{G d m}\) (b) \(\frac{\left(P+P_{0}\right) R^{2} V}{2 G d m}\) (c) \(\frac{\left(P-P_{0}\right) R^{2} V}{2 G d m}\) (d) \(\frac{\left(P-P_{0}\right) R^{2} V}{G d m}\)

A cube of mass \(m\) and density \(D\) is suspended from the point \(P\) by a spring of stiffness \(K .\) The system is kept inside a beaker filled with a liquid of density \(d\). The elongation in the spring, assuming \(D>d\), is (a) \(\frac{m g}{K}\left(1-\frac{d}{D}\right)\) (b) \(\frac{m g}{K}\left(1-\frac{D}{d}\right)\) (c) \(\frac{m g}{K}\left(1+\frac{d}{D}\right)\) (d) None of these
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