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The amount of work done in blowing a soap bubble such that its diameter increases from d to D is (T=surface tension of the solution)A. `2 pi (D^(2)-d^(2))S`B. `pi(D^(2)-d^(2))S`C. `4pi (D^(2)-d^(2))S`D. `8pi(D^(2)-d^(2))S` |
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Answer» Correct Answer - A Change in surface area `=2 xx 4pi(D/2)^(2) -(d/2)^(2)=2pi(D^(2)-d^(2))` `therefore` Work done = surface x change in area `=2piS(D^(2)-d^(2))` |
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