A boy can reduce the pressure in his lungs to 750 mm of mercury. Using a straw, he can drink water from glass up to the maximum depth of (atmospheric pressure = 760 mm of mercury; density of mercury = $13.6gc{m^{ - 3}}$)
(A) 13.6 cm
(B) 9.8 cm
(C) 10 cm
(D) 76 cm
(E) 1.36 cm
Answer
271.8k+ views
Hint: Pressure drop in the lungs is used to pull the fluid in a straw. We will first calculate the pressure difference of the lungs of the student and atmosphere. Then we will use $P = \rho \,h\,g$ to calculate he can drink water from glass up to the maximum depth.
Complete step by step answer:
Static fluid pressure:
It is pressure exerted by fluids on the walls of the container when it is at rest. It is independent of the shape, mass and surface of the container. It only depends on density of fluid, depth of container, and acceleration due to gravity.
In general pressure is force per unit area.
Pressure = $\frac{{Force}}{{Area}}$
$ \Rightarrow P = \frac{{mass \times acceleration}}{{Area}}$
$\left[ {density\,\,(\rho ) = \frac{{mass\,(m)}}{{volume\,\,(V)}}} \right]$
$ \Rightarrow m = \rho V$
Using mass formula in pressure formula, we get
Also,
$volume = area\, \times \,height$
Height=h
Area=A
Acceleration due to gravity= g
$P = \,\frac{{m \times \,g}}{A}$
$ \Rightarrow P = \frac{{\rho V \times g}}{A}$
$ \Rightarrow P = \frac{{\rho \times A \times h \times g}}{A}$
$ \Rightarrow P = \rho \,h\,g$
Pressure of lungs of student = 750 mm of Hg
Atmospheric pressure = 760 mm of Hg
Difference in pressure of lungs of student and atmosphere = (760 mm - 750 mm) of Hg
= 10 mm of Hg
(1 mm = 0.1 cm)
= 1 cm of Hg
This pressure difference helps the student to take water inside his body.
Pressure created due to water column = 1 cm of Hg
Now we can compare this pressure in terms of Hg column with the pressure in terms of water column:
Density of mercury $ = 13.6gc{m^{ - 3}}$
Putting the values in equation $P = \rho \,h\,g$, we get
$1 = 13.6 \times g$ … (1)
Similarly, pressure exerted by water,
$1 = 1 \times h \times g$ … (2)
Equating (1) and (2), we get
h=13.6 cm
Therefore, the correct solution is option A.
Note:
If the units are used in mm only and not converted to cm then this might create trouble and solution will get altered.
Secondly considering pressure difference (10 mm) directly as height is again a wrong concept. Option C is also wrong.
Also, the value of g is 9.8 and we know that acceleration due to gravity is not equal to the height of the container, so option B is also not correct.
Therefore, the correct option is A.
Complete step by step answer:
Static fluid pressure:
It is pressure exerted by fluids on the walls of the container when it is at rest. It is independent of the shape, mass and surface of the container. It only depends on density of fluid, depth of container, and acceleration due to gravity.
In general pressure is force per unit area.
Pressure = $\frac{{Force}}{{Area}}$
$ \Rightarrow P = \frac{{mass \times acceleration}}{{Area}}$
$\left[ {density\,\,(\rho ) = \frac{{mass\,(m)}}{{volume\,\,(V)}}} \right]$
$ \Rightarrow m = \rho V$
Using mass formula in pressure formula, we get
Also,
$volume = area\, \times \,height$
Height=h
Area=A
Acceleration due to gravity= g
$P = \,\frac{{m \times \,g}}{A}$
$ \Rightarrow P = \frac{{\rho V \times g}}{A}$
$ \Rightarrow P = \frac{{\rho \times A \times h \times g}}{A}$
$ \Rightarrow P = \rho \,h\,g$
Pressure of lungs of student = 750 mm of Hg
Atmospheric pressure = 760 mm of Hg
Difference in pressure of lungs of student and atmosphere = (760 mm - 750 mm) of Hg
= 10 mm of Hg
(1 mm = 0.1 cm)
= 1 cm of Hg
This pressure difference helps the student to take water inside his body.
Pressure created due to water column = 1 cm of Hg
Now we can compare this pressure in terms of Hg column with the pressure in terms of water column:
Density of mercury $ = 13.6gc{m^{ - 3}}$
Putting the values in equation $P = \rho \,h\,g$, we get
$1 = 13.6 \times g$ … (1)
Similarly, pressure exerted by water,
$1 = 1 \times h \times g$ … (2)
Equating (1) and (2), we get
h=13.6 cm
Therefore, the correct solution is option A.
Note:
If the units are used in mm only and not converted to cm then this might create trouble and solution will get altered.
Secondly considering pressure difference (10 mm) directly as height is again a wrong concept. Option C is also wrong.
Also, the value of g is 9.8 and we know that acceleration due to gravity is not equal to the height of the container, so option B is also not correct.
Therefore, the correct option is A.
Recently Updated Pages
JoSAA Counselling 2026: JoSAA 2026 Mock Seat Allotment LIVE: Round 2 Result Released, Registration, Choice Filling and Ranks

Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

