A small spherical monoatomic ideal gas bubble `(gamma=5//3)` is trapped inside a liquid of density `rho` (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is `T_0`, the height of the liquid is H and the atmospheric pressure `P_0` (Neglect surface tension).

As the bubble moves upwards, besides the buoyancy force the following forces are acting on it

Options

A. `rho_l n Rg T_0 (P_0 + rho_l gH_^(2//5))/((P_0 + rho_l gy)^(7//5))`
B. `(rho_l n RgT_0)/((P_0 + rho_l gH)^(2//5)[(P_0 + rho_l (H-Y))]^(3//5))`
C. `rho_l nRgT_0 ((P_0 + rho_l gH)^(3//5))/((P_0 + rho_l gy)^(8//5))`
D. `(rho_l nRgT_0)/((P_0 + rho_l gH)^(3//5)[P_0 + rho_l g(H - Y)]^(2//5))`

Answer

B

Explanation

`F_B = rho_1 g` (Volume of bubble)
Volume can be calculated by
`PV^(gamma)` = Const
`((P_0 + rho_1 g(H-Y))/(P_0 + rho_1 gH))(V/V_0)^(5//3)=1 " "rArr" "V=V_0((P_0+rho_1 gH)/(P_0 + rho_1 g (H-y)))^(3//5)`
So, `F_B = (nRT_0 rho_1 g)/((P_(0)+rho_1 gH)^(2//5)(P_0 + rho_1g(H-Y))^(3//5))`