
A large tank filled with water has two holes in the bottom, one with twice the radius of the other. In steady flow the speed of water leaving the larger hole is _______________the speed of the water leaving the smaller.
(A) twice
(B) four times
(C) half
(D) the same as
Answer
235.8k+ views
Hint: We know that Bernoulli's theorem, in fluid dynamics, relation among the pressure, velocity, and elevation in a moving fluid (liquid or gas), the compressibility and viscosity (internal friction) of which are negligible and the flow of which is steady, or laminar. The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one point along its flow. Because the Bernoulli equation is equal to a constant at all points along a streamline, we can equate two points on a streamline.
Complete step by step answer
We know that Bernoulli's equation can be viewed as a conservation of energy law for a flowing fluid. We saw that Bernoulli's equation was the result of using the fact that any extra kinetic or potential energy gained by a system of fluid is caused by external work done on the system by another non-viscous fluid.
Velocity of water surface $_{1}=0$
From Bernoulli equation $\mathrm{P}+\dfrac{\rho \mathrm{v}^{2}}{2}+\rho \mathrm{gh}=\mathrm{constant}$
$\mathrm{P}_{\mathrm{atm}}+\dfrac{\rho \mathrm{v}_{1}^{2}}{2}+\rho \mathrm{gh}_{1}=\mathrm{P}_{\mathrm{atm}}+\dfrac{\rho \mathrm{v}_{2}^{2}}{2}+\rho \mathrm{gh}_{2}$
$\dfrac{\rho \mathrm{v}_{2}^{2}}{2}=\rho \mathrm{g}\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right)$
$\mathrm{v}_{2}=\sqrt{2 \mathrm{g}\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right)}$
Velocity of water is $\sqrt{2 \mathrm{g}\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right)}$
Speed of water from hole is independent of the area. Hence, in steady flow the speed of water leaving the larger hole is the same as the speed of the water leaving the smaller.
So, the correct answer is option D.
Note: We can conclude that Bernoulli's principle is then cited to conclude that since the air moves slower along the bottom of the wing, the air pressure must be higher, pushing the wing up. However, there is no physical principle that requires equal transit time and experimental results show that this assumption is false. According to Bernoulli's theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, non-viscous fluid in a streamlined flow remains constant. He proposed a theorem for the streamline flow of a liquid based on the law of conservation of energy.
Complete step by step answer
We know that Bernoulli's equation can be viewed as a conservation of energy law for a flowing fluid. We saw that Bernoulli's equation was the result of using the fact that any extra kinetic or potential energy gained by a system of fluid is caused by external work done on the system by another non-viscous fluid.
Velocity of water surface $_{1}=0$
From Bernoulli equation $\mathrm{P}+\dfrac{\rho \mathrm{v}^{2}}{2}+\rho \mathrm{gh}=\mathrm{constant}$
$\mathrm{P}_{\mathrm{atm}}+\dfrac{\rho \mathrm{v}_{1}^{2}}{2}+\rho \mathrm{gh}_{1}=\mathrm{P}_{\mathrm{atm}}+\dfrac{\rho \mathrm{v}_{2}^{2}}{2}+\rho \mathrm{gh}_{2}$
$\dfrac{\rho \mathrm{v}_{2}^{2}}{2}=\rho \mathrm{g}\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right)$
$\mathrm{v}_{2}=\sqrt{2 \mathrm{g}\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right)}$
Velocity of water is $\sqrt{2 \mathrm{g}\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right)}$
Speed of water from hole is independent of the area. Hence, in steady flow the speed of water leaving the larger hole is the same as the speed of the water leaving the smaller.
So, the correct answer is option D.
Note: We can conclude that Bernoulli's principle is then cited to conclude that since the air moves slower along the bottom of the wing, the air pressure must be higher, pushing the wing up. However, there is no physical principle that requires equal transit time and experimental results show that this assumption is false. According to Bernoulli's theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, non-viscous fluid in a streamlined flow remains constant. He proposed a theorem for the streamline flow of a liquid based on the law of conservation of energy.
Recently Updated Pages
A uniform ring of mass M is lying at a distance sqrt class 11 physics JEE_Main

Two wires A and b are of the same materials Their lengths class 11 physics JEE_Main

Which type of lens has negative power A Convex lens class 11 physics JEE_Main

The centre of mass of a system of particles is at the class 11 physics JEE_Main

A faulty thermometer reads 5 circ C in melting ice class 11 physics JEE_Main

External forces are A Always balanced B Never balanced class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Session 1 Results Out and Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Electromagnetic Waves and Their Importance

Understanding Uniform Acceleration in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

CBSE Notes Class 11 Physics Chapter 14 - Waves - 2025-26

CBSE Notes Class 11 Physics Chapter 9 - Mechanical Properties of Fluids - 2025-26

CBSE Notes Class 11 Physics Chapter 11 - Thermodynamics - 2025-26

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

