What is the excess pressure inside a liquid drop of radius 2mm at room temperature ? Surface tension of the liquid at that temperature is 0.596 N/m
Answers
Answered by
0
Answer: = [2×4.65×10
−1
/(3×10
−3
)]=310 Pa
Explanation:
Radius of the mercury drop, r=3.00mm=3×10
−3
m
Surface tension of mercury, S=4.65×10
−1
N/m
Atmospheric pressure, P
o
=1.01×10
5
Pa
Total pressure inside the mercury drop
= Excess pressure inside mercury + Atmospheric pressure
= 2S/r+P
o
= [2×4.65×10
−1
/(3×10
−3
)]+1.01×10
5
=1.0131×10
5
Excess pressure = 2S / r
= [2×4.65×10
−1
/(3×10
−3
)]=
Similar questions