Question

A soap bubble was slowly enlarged from a radius of 4cm to 6cm. The amount of work necessary for enlargement was 1.5 x 10^-4 joules. Calculate the surface tension of the soap bubble.​

Answer 1

Answer:

$T=3*10^-3 N/m$

Explanation:

From the question we are told that:

Radius :

$R_1=4=>0.04\\\\R_2=6=>0.06$

Work $W=1.5 * 10^{-4}$

Generally the equation for Work done  is mathematically given by

$W=TdA$

Where

$dA=A_2-A_1\\\\dA=(2 \pi r_2^2)(2 \pi r_1^2)$

$dA=8 \pi*(r_2^2-r_1^2)\\\\dA=8*3.142*(0.06^2-0.04^2)$

$dA=0.050m^2$

Therefore

$W=TdA$

$T=\frac{1.5 * 10^{-4}}{0.05m^2}$

$T=3*10^-3 N/m$