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Mechanical Properties Of Fluids

Question
CBSEENPH11018364

A soap bubble of radius 10mm is blown from soap solution of surface tension 0.06 N/m. Find the work done in blowing the bubble. What addition work will be done in further blowing to double the radius?

Solution
The soap bubble is formed from the soap solution.
Therefore, increase in the surface area of soap bubble is equal to total surface area of soap bubble.
Since the soap bubble has two free surfaces, therefore increase in area of free surface of bubble is,
ΔA subscript 1 equals 2 cross times 4 πr squared
We have,
Radius of the soap bubble, r = 1mm=10-3 
therefore

I n c r e a s e space i n space a r e a comma space increment straight A subscript 1 equals 2 cross times 4 straight pi left parenthesis 10 to the power of negative 3 end exponent right parenthesis squared

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 2.51 cross times 10 to the power of negative 6 end exponent straight m squared 
Now the work done in blowing the bubble
straight W subscript 1 equals capital delta A subscript 1 cross times straight T

space space space space space equals 2.51 cross times 10 to the power of negative 6 end exponent cross times 0.06

space space space space space equals 1.5 cross times 10 to the power of negative 7 end exponent straight J
space space space space space equals 1.5 space e r g s
Additional work done in doubling the radius of bubble is,
straight W space equals space ΔA cross times straight T

space space space space space equals 7.53 cross times 10 to the power of negative 6 end exponent cross times 0.06

space space space space space equals 4.5 cross times 10 to the power of negative 7 end exponent straight J equals 4.5 space ergs