wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A soap bubble, having radius 1 mm, is blown from a detergent solution having a surface tension of 2.5×102 N/m. The pressure inside the bubble equals at a point Zo below the free surface of water in a container. Taking g=10 m/s2, density of water =103 kg/m3, the value of Z0 is
[Take g=10 m/s2]

A
5 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
8 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1 cm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
3 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 1 cm
Given,
Radius of soap bubble, R=1 mm=103 m
Surface Tension, T=2.5×102 N/m
Value of g=10 m/s2
Density of water, ρ=1000 kg/m3
Depth of water at which pressure is equal to pressure inside bubble =Zo= ?
Let us suppose, atmospheric pressure is po.
We know pressure inside soap bubble is given by
p=po+4TR
=po+4×2.5×102103
=po+100 ...........(1)
Pressure at depth Zo below free surface of water
p=p0+ρgZ0
=p0+1000×10×Zo
=p0+104Zo ...........(2)
Given, pressure at depth Zo is equal to pressure inside soap bubble.
From (1) and (2),
po+100=p0+104Zo
or Zo=102 m=1 cm

flag
Suggest Corrections
thumbs-up
103
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon