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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.

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.

Grade:11

2 Answers

Askiitians Expert Soumyajit IIT-Kharagpur
28 Points
11 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

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Regards,

Askiitians Experts
Soumyajit Das IIT Kharagpur

Kushagra Madhukar
askIITians Faculty 629 Points
11 months 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, ρmw = 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

 

 

 

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