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Calculate the excess pressure withing a bubble of air of radius 0.1 mm in water. If te bubble had been formed 10 cm below the water surface on a day when the atmospheric pressure was `1.013xx10^(3)Pa` then what would have been the total pressure inside the bubble? surface tension of water `=73xx10^(-3)N//m` |
Answer» Excess pressure `P_("excess")=(2T)/(r)=(2xx73xx10^(-3))/(0.1xx10^(-3))=1460Pa` The pressure at a depth d, in liquid is `P=hdg`. Therefore, the total pressure inside the air bubble is `P_("in")=P_(atm)+hdg+(2T)/(r)=(1.013xx10^(5))+(10xx10^(-2)xx10^(3)xx9.8)+1460` `=101300+980+1460=103740=1.037xx10^(5)Pa` |
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