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. `0.1 ms^(-1)`
B. `0.4 ms^(-1)`
C. `0.2 ms^(-1)`
D. `0.3 ms^(-1)`

Answer

A

Explanation

The terminal speed of a spherical body of radius r, density `rho` falling through a medium of density `sigma` is given by `v= (2)/(9) (r^(2) (rho - sigma)g)/(eta)`
where `eta` is the coefficient of viscosity of the medium.
`:. V_(g) = (2)/(9) (r_(g)^(2) (rho_(g) - sigma_("liquid"))g)/(eta_("liquid")) and v_(S) = (2)/(9) (r_(S)^(2) (rho_(S) - sigma_("liquid"))g)/(eta_("liquid"))`
where the subscripts g and s represent gold and silver spheres respectively. Since both the spheres are of same size and falling in the same liquid,
`:. (v_(g))/(v_(s)) = (rho_(g) - sigma_("liquid"))/(rho_(s) - sigma_("liquid")) = ((19.5 - 1.5) kg m^(-3))/((10.5 - 1.5) kg m^(-3)) = (18)/(9) = 2`
or `v_(s) = (v_(g))/(2) = (0.2)/(2) ms^(-1) = 0.1 ms^(-1)`