1800-150-456-789
FlagBACK
Flag Thermal Physics> thermo...
question mark

THE VOLUME OF AIR BUBBLE IS DOUBLED WHEN IT RISES FROM BOTTOM OF LAKE TO ITS SURFACE. THE ATMOSPHERIC PRESSURE IS 75CM OF MERCURY AND RATIO OF DENSITY OF MERCURY TO THAT OF LAKE IS 40/3.FIND THE DEPTH OF THE WATER.

akhil samir killawala , 14 Years ago
Grade 11
anser 2 Answers
Askiitians Expert Soumyajit IIT-Kharagpur

Last Activity: 14 Years ago

Dear Akhil Killawala,

Ans:- The pressure at the bottom of the lake is = The pressure due to hte air + The pressure due to the water column=P1

The pressure at the top is only the atmosphereic pressure=P2 Initial Vol=V1 and final Vol is =V2

NOW YOU HAVEN'T MENTIONED ABOUT THE TEMP SO I AM CONSIDERING IT TO BE CONSTANT

So from gas eq we get P1V1=P2V2

V2=2V1

Putting the values in the eq 1 we get,

H=10 meter

Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We
are all IITians and here to help you in your IIT JEE preparation.


All the best Akhil Killawala !!!

 


Regards,

Askiitians Experts
Soumyajit Das IIT Kharagpur

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,
Please find the attached answer to your question.
 
Let us assume the depth of water to be H
The pressure at the bottom of the lake, P1 = atmospheric pressure + The pressure due to the water column
                                                              = Patm + ρwgH
                                                              = ρmg(0.75) + ρmgH x 3/40   [ since, ρm/ρw = 40/3]
The pressure at the top of lake,P2 = atmospheric pressure
                                                  = Patm
                                                  = ρmg(0.75)
Initial Vol, V1 = V
final Vol, V2 = 2V
 
Considering the temperature of water to be constant, we have from gas eq
P1V1 = P2V2
Since, V2 = 2V1
Hence, P1 = 2P2
Hence, ρmg(0.75) + ρmgH x 3/40 = 2 x ρmg(0.75)
or, 0.75 + 3H/40 = 1.5
or, 3H/40 = 0.75
or, H = 10 m
 
Hope it helps.
Thanks and regards,
Kushagra

 

 

 

Provide a better Answer & Earn Cool Goodies

Enter text here...
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments


Ask a Doubt

Get your questions answered by the expert for free

Enter text here...

Other Related Questions on Thermal Physics

question mark
A container filled with a sample of an Idea gas at the pressure of 1.5atm. the gas is compressed isothermal ly to one fourth of its original volume. What is the new pressure of the gas?
Thermal Physics
0 Answer Available

Last Activity: 1 Year ago(s)

question mark
how do we know the electric field due to an electric dipole of dipole moment P at distance r from the centre of dipole is fiven by .. where theta is angle between p amd r vector..
Thermal Physics
0 Answer Available

Last Activity: 2 Year ago(s)

question mark
A A mole of helium gas initially at stp undergoes an isovolumetric process in which its pressure falls to half its initial value. The work done by gas is
Thermal Physics
0 Answer Available

Last Activity: 2 Year ago(s)

question mark
5000 J of heat are added to two moles of an ideal monatomic gas, initially at a temperature of 500 K, while the gas performs 7500 J of work. What is the final temperature of the gas? step by step please
Thermal Physics
1 Answer Available

Last Activity: 2 Year ago(s)

question mark
Please help me with theanswer of this questionrelated to radiation graph
Thermal Physics
0 Answer Available

Last Activity: 2 Year ago(s)