Two communicating vessels contain mercury. The diameter of one vessel is n times larger than the diameter of the other. A column of water of height h is poured into the left vessel. The mercury level will rise in the right hand vessel (s=relative density of mercury and ρ =density of water) by:
$\eqalign{
& {\text{A) }}\dfrac{{{{\text{n}}^{\text{2}}}{\text{h}}}}{{{{{\text{(n + 1)}}}^{\text{2}}}{\text{s}}}} \cr
& {\text{B)}}\dfrac{{\text{h}}}{{{\text{(}}{{\text{n}}^{\text{2}}}{\text{ + 1)s}}}} \cr
& {\text{C)}}\dfrac{{\text{h}}}{{{{{\text{(n + 1)}}}^{\text{2}}}{\text{s}}}} \cr
& {\text{D)}}\dfrac{{\text{h}}}{{{{\text{n}}^{\text{2}}}{\text{s}}}} \cr} $
Answer
Verified
116.4k+ views
Hint: We will follow Pascal's principle, also called Pascal's law, any force applied to a confined fluid is transmitted uniformly in all directions throughout the fluid regardless of the shape of the container.
Complete step by step solution:
If the level in narrow tubes goes down by $h_1$, then in broader tube goes up to $h_2$
$\eqalign{
& \Rightarrow \pi {r^2}h_1 = \pi {(nr)^2}h_2 \cr
& \Rightarrow h_1 = {n^2}h_2 \cr
& {\text{Now pressure at point A = pressure at point B}} \cr
& {\text{h}}\rho {{g = (h_1 + h_2)}}\rho {\text{'g}} \cr
& \Rightarrow {\text{h = (}}{{\text{n}}^2}h_2 + h_2)sg \cr
& {\text{also s = }}\dfrac{{\rho {\text{'}}}}{\rho } \cr
& h_2 = \dfrac{h}{{({n^2} + 1)s}} \cr} $
Option B is correct.
Additional Information: A solid surface can exert pressure, but liquids (ie liquid or gas) can also exert pressure. It may sound strange if you think about it because it is hard to imagine hitting a hammer in a nail with liquid. To understand this, imagine sinking to some depth in water. Due to the force of gravity, the water above you will be pulled down and this will increase the pressure on you. If you go deeper, you will have more water, so the weight and pressure will also increase with the water.
The pressure in a fluid depends on the height of the fluid column, but not on the shape of the vessel in which it is contained, when we assume that wherever the fluid is in contact with the vessel wall, the force between it and the wall Perpendicular to the surface.
Note: Pressure exerted by liquid changes with the depth in liquid. Value of pressure is small just below the surface of the liquid, but as we go deeper, the pressure increases with the depth. All liquids have weight and exerts pressure at the bottom of the container. In SI units, the unit of pressure is Pascal (Pa), which is equal to $N / {m^2}$.The primary mathematical expression for pressure is given by: ${\text{pressure = }}\dfrac{{force}}{{area}}$.
Complete step by step solution:
If the level in narrow tubes goes down by $h_1$, then in broader tube goes up to $h_2$
$\eqalign{
& \Rightarrow \pi {r^2}h_1 = \pi {(nr)^2}h_2 \cr
& \Rightarrow h_1 = {n^2}h_2 \cr
& {\text{Now pressure at point A = pressure at point B}} \cr
& {\text{h}}\rho {{g = (h_1 + h_2)}}\rho {\text{'g}} \cr
& \Rightarrow {\text{h = (}}{{\text{n}}^2}h_2 + h_2)sg \cr
& {\text{also s = }}\dfrac{{\rho {\text{'}}}}{\rho } \cr
& h_2 = \dfrac{h}{{({n^2} + 1)s}} \cr} $
Option B is correct.
Additional Information: A solid surface can exert pressure, but liquids (ie liquid or gas) can also exert pressure. It may sound strange if you think about it because it is hard to imagine hitting a hammer in a nail with liquid. To understand this, imagine sinking to some depth in water. Due to the force of gravity, the water above you will be pulled down and this will increase the pressure on you. If you go deeper, you will have more water, so the weight and pressure will also increase with the water.
The pressure in a fluid depends on the height of the fluid column, but not on the shape of the vessel in which it is contained, when we assume that wherever the fluid is in contact with the vessel wall, the force between it and the wall Perpendicular to the surface.
Note: Pressure exerted by liquid changes with the depth in liquid. Value of pressure is small just below the surface of the liquid, but as we go deeper, the pressure increases with the depth. All liquids have weight and exerts pressure at the bottom of the container. In SI units, the unit of pressure is Pascal (Pa), which is equal to $N / {m^2}$.The primary mathematical expression for pressure is given by: ${\text{pressure = }}\dfrac{{force}}{{area}}$.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs
How to find Oxidation Number - Important Concepts for JEE
How Electromagnetic Waves are Formed - Important Concepts for JEE
Electrical Resistance - Important Concepts and Tips for JEE
Average Atomic Mass - Important Concepts and Tips for JEE
Chemical Equation - Important Concepts and Tips for JEE
Trending doubts
Charging and Discharging of Capacitor
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
Inductive Effect and Acidic Strength - Types, Relation and Applications for JEE
Which of the following is the smallest unit of length class 11 physics JEE_Main
JEE Main 2025 Maths Online - FREE Mock Test Series
JEE Main 2024 Physics Question Paper with Solutions 27 January Shift 1
Other Pages
NCERT Solutions for Class 11 Physics Chapter 7 Gravitation
NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids
Thermodynamics Class 11 Notes CBSE Physics Chapter 11 (Free PDF Download)
NCERT Solutions for Class 11 Physics Chapter 3 Motion In A Plane
Ideal and Non-Ideal Solutions Raoult's Law - JEE