Find out the work done (in times 10^{-3} ; J) | vedclass

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(B) soap bubble has two surfaces (inner and outer),so the change in surface area is $2 	imes (4 \pi R^2)$.The work done $W$ is given by the product o

Vedclass · Accepted Answer

$6.28$

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(B) soap bubble has two surfaces (inner and outer),so the change in surface area is $2 	imes (4 \pi R^2)$.The work done $W$ is given by the product of the change in surface area and the surface tension $T$:$W = 2 	imes (4 \pi R^2) 	imes T$Given:$R = 5 \; cm = 5 	imes 10^{-2} \; m$$T = 0.1 \; N/m$Substituting the values:$W = 2 	imes 4 	imes \pi 	imes (5 	imes 10^{-2})^2 	imes 0.1$$W = 8 	imes \pi 	imes 25 	imes 10^{-4} 	imes 0.1$$W = 200 	imes \pi 	imes 10^{-5}$$W = 2 	imes \pi 	imes 10^{-3} \; J$Using $\pi \approx 3.14$:$W = 2 	imes 3.14 	imes 10^{-3} \; J = 6.28 	imes 10^{-3} \; J$.

Find out the work done (in $\times 10^{-3} \; J$) to expand the soap bubble to radius $R = 5 \; cm$ (Surface tension of water $= 0.1 \; N/m$).

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