Surface tension of a soap solution is 1.9 tim | vedclass

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(B) The surface tension $T = 1.9 	imes 10^{-2} 	ext{ N/m}$.The diameter of the bubble $d = 2.0 	ext{ cm}$,so the radius $R = 1.0 	ext{ cm} = 1.0 \

Vedclass · Accepted Answer

$15.2 	imes 10^{-6} \pi 	ext{ J}$

Vedclass · Answer

(B) The surface tension $T = 1.9 	imes 10^{-2} 	ext{ N/m}$.The diameter of the bubble $d = 2.0 	ext{ cm}$,so the radius $R = 1.0 	ext{ cm} = 1.0 	imes 10^{-2} 	ext{ m}$.$A$ soap bubble has two surfaces (inner and outer),so the work done $W$ is given by $W = 2 	imes (4 \pi R^2 T) = 8 \pi R^2 T$.Substituting the values: $W = 8 	imes \pi 	imes (1.0 	imes 10^{-2})^2 	imes 1.9 	imes 10^{-2}$.$W = 8 	imes \pi 	imes 10^{-4} 	imes 1.9 	imes 10^{-2} = 15.2 	imes 10^{-6} \pi 	ext{ J}$.

Surface tension of a soap solution is $1.9 \times 10^{-2} \text{ N/m}$. The work done in blowing a soap bubble of $2.0 \text{ cm}$ diameter will be:

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