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. ` 0.4 m//s `
B. `0.133 m//s `
C. `0.1 m//s `
D. `0.2 m//s `

Answer

C

Explanation

The terminal speed of the spherical body in a viscous liquid is given by `v_(r)=(2r^(2)(rh0-sigma)g)/(9eta)`
Where `rho` the density of the substance of the body is `sigma` is the density of the liquid.
From the given data, we have `(v_(T)(Ag))/(v_(T)(Gold))=(rho_(Ag)-sigma_(1))/(rho_(Gold)-sigma_(1))`
`Rightarrowv_(T)(Ag)=(10.5-1.5)/(19.5-1.5)xx0.2=(9)/(18)xx0.2=0.1m//s`