The surface tension of a soap solution is 3.5 | vedclass

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Question

(A) The surface tension $T = 3.5 	imes 10^{-2} \, N m^{-1}$.Initial radius $r_1 = 10 \, cm = 0.1 \, m$.Final radius $r_2 = 20 \, cm = 0.2 \, m$.$A$ s

Vedclass · Accepted Answer

$264$

Vedclass · Answer

(A) The surface tension $T = 3.5 	imes 10^{-2} \, N m^{-1}$.Initial radius $r_1 = 10 \, cm = 0.1 \, m$.Final radius $r_2 = 20 \, cm = 0.2 \, m$.$A$ soap bubble has two surfaces (inner and outer),so the change in surface area is $\Delta A = 2 	imes (4 \pi r_2^2 - 4 \pi r_1^2) = 8 \pi (r_2^2 - r_1^2)$.The work done $W = T 	imes \Delta A = T 	imes 8 \pi (r_2^2 - r_1^2)$.Substituting the values: $W = 3.5 	imes 10^{-2} 	imes 8 	imes 3.14159 	imes (0.2^2 - 0.1^2)$.$W = 3.5 	imes 10^{-2} 	imes 8 	imes 3.14159 	imes (0.04 - 0.01) = 3.5 	imes 10^{-2} 	imes 8 	imes 3.14159 	imes 0.03$.$W = 0.28 	imes 3.14159 	imes 0.03 = 0.026389 \, J$.Converting to the required format: $W \approx 264 	imes 10^{-4} \, J$.

The surface tension of a soap solution is $3.5 \times 10^{-2} \, N m^{-1}$. The amount of work done required to increase the radius of a soap bubble from $10 \, cm$ to $20 \, cm$ is $..... \times 10^{-4} \, J$.

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