
A novice scuba diver practicing in a swimming pool takes enough air from his tank to fully expand his lungs before abandoning the tank at depth l and swimming to the surface, failing to exhale during his ascent. At the surface, the difference $\Delta $ p between the external pressure on him and the air pressure in his lungs is $9.3kPa$ . From what depth does he start? What potentially lethal danger does he face?
Answer
162.9k+ views
Hint: Here in this question, it is said that The diver is diving at a distance I in the swimming pool. As he intake oxygen from the tank and left his tank beyond the pool. After which he comes upwards in the pool so a pressure difference takes place. Here our external pressure decreased because we were moving in the upwards direction in the pool but the diver would not be able to exhale at that time because the pressure on his lungs got constant. Now, here we have to find the depth by which the diver is diving upwards.
Formula used:
$p = {p_ \circ } + \rho gL$
Complete step by step solution:
As we know that from the above conclusion that,
Here, when the diver fills his lungs at depth L, the air pressure inside his lungs is higher than usual because of the external pressure on him.
As we know the equation as,
$p = {p_ \circ } + \rho gL$
Where $\rho $ represents the density of water. The external pressure on him drops as he ascends, reaching atmospheric pressure ${p_ \circ }$ at the surface. His blood pressure also falls till it returns to the surface. The air pressure in his lungs, however, stays at the level it was at depth L since he does not exhale. The pressure differential $\Delta p$ at the surface is
$\Delta p = p - {p_ \circ } = \rho gL$
Here in this question, we need the value of $L$ so rearranging the formula as we need,
$L = \dfrac{{\Delta p}}{{\rho g}}$
As by putting all values from the question we get,
$L = \dfrac{{9.3 \times 1000}}{{1000 \times 9.81}}$
By doing further solution we get,
$L = 0.95m$
Hence, we get the depth where he starts from is $L = 0.95m$ .
This is not profound! However, the diver's lungs are enough to burst from the $9.3kPa$ pressure difference (or around $9\% $ of atmospheric pressure) and release air into the depressurized blood, which subsequently transports the air to the heart, killing the diver. There is no risk if the diver follows directions and exhales gradually as he ascends, allowing the pressure in his lungs to equalize with the external pressure.
Note: While solving such questions, always convert all the needed physical quantities into the same standard units in order to avoid any calculation mistakes and kilopascal is the bigger unit of pressure which is thousand times larger than a pascal.
Formula used:
$p = {p_ \circ } + \rho gL$
Complete step by step solution:
As we know that from the above conclusion that,
Here, when the diver fills his lungs at depth L, the air pressure inside his lungs is higher than usual because of the external pressure on him.
As we know the equation as,
$p = {p_ \circ } + \rho gL$
Where $\rho $ represents the density of water. The external pressure on him drops as he ascends, reaching atmospheric pressure ${p_ \circ }$ at the surface. His blood pressure also falls till it returns to the surface. The air pressure in his lungs, however, stays at the level it was at depth L since he does not exhale. The pressure differential $\Delta p$ at the surface is
$\Delta p = p - {p_ \circ } = \rho gL$
Here in this question, we need the value of $L$ so rearranging the formula as we need,
$L = \dfrac{{\Delta p}}{{\rho g}}$
As by putting all values from the question we get,
$L = \dfrac{{9.3 \times 1000}}{{1000 \times 9.81}}$
By doing further solution we get,
$L = 0.95m$
Hence, we get the depth where he starts from is $L = 0.95m$ .
This is not profound! However, the diver's lungs are enough to burst from the $9.3kPa$ pressure difference (or around $9\% $ of atmospheric pressure) and release air into the depressurized blood, which subsequently transports the air to the heart, killing the diver. There is no risk if the diver follows directions and exhales gradually as he ascends, allowing the pressure in his lungs to equalize with the external pressure.
Note: While solving such questions, always convert all the needed physical quantities into the same standard units in order to avoid any calculation mistakes and kilopascal is the bigger unit of pressure which is thousand times larger than a pascal.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Charging and Discharging of Capacitor

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
