The terminal speed of a sphere of gold (density = 19.5 kg `m^(-3)`) is 0.2 `ms^(-1)` in a viscous liquid (density = 1.5 kg `m^(-3)`). Then, the terminal speed of a sphere of silver (density = 10.5 kg `m^(-3)`) of the same size in the same liquid is

Options

A. `0.1m//s^(-1)`
B. `0.2m//s^(-1)`
C. `0.4m//s^(-1)`
D. `0.133m//s^(-1)`

Answer

C

Explanation

The terminal speed of a sphere of radius r, and density `sigma`, moving down jin a liquid of density `rho` and viscosity `eta` is given by
`V_t=2/9(r^2g(sigma-rho))/(eta)`
In this problem, `2/9(r^2g)/(eta)` is constant
`:. " "V_t=K(sigma-rho)`
For a gold sphere,
`V_g=K[19.5-1.5]xx10^3=18xx10^3` K
For a silver sphere,
`V_s=K[10.5-1.5]xx10^3=9xx10^3K`
`V_s=(V_g)/2=(0.2)/2=0.1m//s`