What is the excess pressure inside (i) a drop (ii) a bubble of water of diameter 2mm. If the atmospheric pressure is `1.01 xx 10^(5) `Pa, calculate the total pressure inside them `= sigma = 72 xx 10^(-3) N//m`.

Explanation

`r =1 m m = 10^(-3) m`
Excess pressure inside the drop ` = ( 2 sigma )/ (r ) = (2xx 72 xx 10^(-3))/( 10^(-3))`
= 144Pa
Excess pressure inside a bubble `=( 4 sigma )/( r )`
`= ( 4 xx 72 xx 10^(-3))/( 10^(-3)) = 288 Pa`
Total pressure inside the drop = Atmospheric pressure `+` excess pressure
`= 10100 + 144 = 101144 pa.`
Total pressure inside the bubble `= 10100 + 288 = 1101288Pa`