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Calculate the work down against surface tension in blowing a soap bubble from a radius of `10 cm` to `20 cm`, if the surface tension of soap solution in `25xx10^(-3)N//m`. |
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Answer» Original total surface area `=2xx4pir_(1)^(2)=2xx4pixx(0.1)^(2)m^(2)` (as bubble has two surfaces). Final total surface area`=2xx4pir_(2)^(2)=2xx4pixx(0.2)^(2)m^(2)` Therefore, extension in area `=2xx4pi[(0.2)^(2)-(0.1)^(2)]=0.24pim^(2)` Now, work done `W = `surface tension `xx` extension in area `=25xx10^(-3)xx0.14pi=6pixx10^(-3)J` |
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