Respuesta :
Answer:
Gauge pressure
R(1.95) =4.25
R(3.85) =2.15
R(4.00 merged bubble radius)=2.07
Explanation:
As we know that pressure increment after crossing a curved surface is [tex]\frac{gamma}{radius of curvature at that point}[/tex]
here we cross two surface thus two times of above
thus in soap bubble gauge pressure = [tex]\frac{2*gamma}{radius}[/tex]
- So in first bubble of radius 1.95
plug in the values into above formula[tex]\frac{2*gamma}{radius}[/tex]
[tex]\frac{2*0.0415}{0.0385}[/tex] =4.25 newton/[tex]meter^{2}[/tex]
- In second bubble of radius 3.85
Similarly as above[tex]\frac{2*0.0415}{0.0385}[/tex] =2.15 newton/[tex]meter^{2}[/tex]
Finally we merge them to form one bubble then volume of both two combined is equal to final one.
[tex]\frac{4}{3}[/tex]*π[tex]1.95^{3}[/tex] + [tex]\frac{4}{3}[/tex]*π[tex]3.85^{3}[/tex] = [tex]\frac{4}{3}[/tex]*π[tex]radius^{3}[/tex]
(By volume conservation )
Final radius = 4.00 cm
Putting this value in [tex]\frac{2*gamma}{4.00}[/tex]
Gauge pressure =2.07
