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Work done in increasing the size of a soap bubble from a radius of `3 cm` to `5 cm` is nearly. (surface tension of soap solution `= 0.3 Nm^(-1))`A. `4pimJ`B. `0.2pimJ`C. `2pimJ`D. `0.4 pi mJ` |
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Answer» Correct Answer - D `W = TDeltaA` `= 0.03 (2 xx 4pi xx (5^(2) - 3^(2))10^(-4)` `= 24pi(16)xx 10^(-6)` `= 0.384 pi xx 10^(-3)` Joule `-= 0.4 pimJ` |
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