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  • The surface tension of soap solution is <img alt="3.5 \times 10^{-2} \mathrm{Nm}^{-1}" src="https://entrancecorner.oncodecogs.com/gif.latex?3.5%20%5Ctimes%2010%5E%7B-2%7D%20%5Cmathrm%7BNm%7D%5E%7B-

The surface tension of soap solution is 3.5 \times 10^{-2} \mathrm{Nm}^{-1}. The amount of work done required to increase the radius of soap bubble from 10 \mathrm{~cm}$ to $20 \mathrm{~cm} is_________ \times 10^{-4} \mathrm{~J} .(take \; \pi=22 / 7)

Option: 1

264


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{W}=\Delta \mathrm{U}

\mathrm{W}=2\times S \times\left(\mathrm{A}_{\mathrm{f}}-\mathrm{A}_{\mathrm{i}}\right)
      =2 \times S \times 4 \pi\left(\mathrm{r}_{\mathrm{f}}^{2}-\mathrm{r}_{\mathrm{i}} 2\right)
      =2 \times 3.5 \times 10^{-2} \times 4 \times \frac{22}{7} \times 10^{-4}(300)
      =264 \times 10^{-4} \mathrm{~J}

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Anam Khan

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