Practice free →
HomeAP Physics 2PhysicsFluid Statics › A soap bubble of radius 2.00 x 10^-4 m is blown …

A soap bubble of radius 2.00 x 10^-4 m is blown from soapy water whose surface tension is 0.0370 N/m. The gauge pressure inside a spherical bubble is four times the surface tension divided by the radius. What is that pressure?

A185 Pa
B370 Pa
C1480 Pa
D740 Pa
Answer & Solution
Correct answer: D. 740 Pa
1. Gauge pressure inside a spherical bubble = 4 x surface tension / radius. 2. Numerator = 4 x 0.0370 N/m = 0.148 N/m. 3. Divide by the radius: (0.148 N/m) / (2.00 x 10^-4 m). 4. This gives 740 N/m^2, that is 740 Pa above the outside air. 5. Note that this is a gauge pressure, so the absolute pressure inside is 740 Pa above atmospheric. 6. The value 370 Pa uses a factor of two instead of four, and 185 Pa drops the factor entirely. 7. The value 1480 Pa doubles the correct result, which is what a factor of eight would give. _Source: OpenStax College Physics (CC BY 4.0), Ch 11 "Fluid Statics", section 11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action_
Solve this in the app — AP Physics 2 practice & 24k+ MCQs →
Related questions