1.

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.

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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