An air bubble of radius 2.0mm is formed at the bottom of a 3.3 deep river. Calculate the radius of the bubble as it comes to the surface. Atmospheric pressure `=1.0xx10^(5)Pa` and desnity of water`=1000kg m^(-3).

Explanation

Given,
P_1 = (10^5) + rho g h = (10^5)+ 1000`
` = 1.33 xx (10^5)Pa`
` (P_2) = (10^5)Pa`
` T_1 = (T_2) = T `
` V_1 = (4/ (3 pi))((2 xx (10^-3))^3), `
` V_2 (4/(3pi(r^3)))`
` We know, ((P_1)(V_1)/T_1) = ((P_2)(V_2)/T_2)`
` ( 1.33 xx (10^5) xx (4/ 3pi) xx ((2 xx (10^-3))^3))/T_1`
` = ((10^5) xx (4/3 pi (r^3))/ T_2)`
` rArr = (1.33 xx 8 xx (10^5) xx (10^-9))`
` = (10^5) xx (r^3)`
` rArr r = (3 sqrt (10.64 xx (10^-9)))`
` = (2.19 xx (10^-3)) = 2.2 mn.`