here we know that volume of the drop is given as
[tex]V = \frac{4}{3}\pi R^3[/tex]
[tex]V = \frac{4}{3}\pi (1.35\times 10^{-3})^3[/tex]
[tex]V = 1.03 \times 10^{-8} m^3[/tex]
now the mass of the drop is given as
[tex]m = \rho V[/tex]
[tex]m = 1.03 \times 10^{-5} kg[/tex]
now the net buoyancy force on the drop is given as
[tex]F_b = \rho_{air} V g[/tex]
[tex]F_b = 1.2 (1.03\times 10^{-8})9.8[/tex]
[tex]F_b = 1.21\times 10^{-7} N[/tex]
now the net force on the drop is
[tex]F = F_g - F_b[/tex]
[tex]F = 1.03\times 10^{-5}(9.8) - 1.21\times 10^{-7}[/tex]
now acceleration of the drop is given as
[tex]a = \frac{F}{m}[/tex]
[tex]a = 9.79 m/s^2[/tex]
now the speed of the drop is given as
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]v_f^2 - 0 = 2(9.79)(3950)[/tex]
[tex]v_f = 278 m/s[/tex]
