Question

Objective Type QuestionsThe amount of work done in forming a soap film ofsize 5 cm x 15 cm if surface tension is 10-2 N/m(1) 3 x 104(3)/1.5 x 10-4 J2x 4.5 x 10-50(4) 10-3 J​

Answer 1

Answer:

$7.5\times 10^{-5}$ Joule

Explanation:

Given

Surface Tension $T=10^{-2}$ N/m

Increase in area of the soap film

$\Delta A=5\times 15\times 10^{-4}\text{ m}^2$

$\Delta A=75\times 10^{-4}\text{ m}^2$

Work done in forming the soap bubble

$=T\times \Delta A$

$=10^{-2}\times 75\times 10^{-4}$

$=75\times 10^{-6}$ N-m

$=7.5\times 10^{-5}$ Joule

Note: this is not matching any of the options given because some data might be missing. Th approach will be the same.

Hope this helps.