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The ratio of radii of two soap bubbles is 4 : 9. Find the ratio of (i) excess pressure inside these bubbles (ii) work done to blow these bubbles. |
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Answer» Solution :`R_(1)/R_(2)=4/9` T=Surface tension of solution Excess pressure inside bubble of radius `R_(1)` `P_(1)=(4T)/(R_(1))` Similarly, `P_(2)=(4T)/(R_(2))` `P_(1)/P_(2)=(4T)/(R_(1)) R_(2)/(4T)=R_(2)/R_(1)=9/4` Work done in blowing up soap bubble of radius R, `=W_(1)=2 XX 4 pi R_(1)^(2)T` Similarly, `W_(2)=2 xx 4PI R_(2)^(2) T` `W_(1)/W_(2)=R_(1)^(2)/R_(2)^(2)=(4/9)^(2)=(16)/(81)` |
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