Two metal spheres are falling through a liquid of density 2.5 × 1 0 3 kg / m 3 with the same uniform speed. The density of material of first sphere and second sphere is 11.5 × 1 0 3 kg / m 3 and 8.5 ×

Question

Two metal spheres are falling through a liquid of density 2.5 × 1 0 3 kg / m 3 with the same uniform speed. The density of material of first sphere and second sphere is 11.5 × 1 0 3 kg / m 3 and 8.5 × 1 0 3 kg / m 3 respectively. The ratio of the radius of first sphere to that of second sphere is

TestHub · Accepted Answer

The terminal velocity of a sphere falling through a viscous liquid is given by $v = \frac{2}{9} \frac{r ^{2} ( σ - ρ ) g}{η}$

Since both spheres fall with the same uniform speed, their terminal velocities are equal.

$v_{1} = v_{2}$

$\frac{2}{9} \frac{r _{1}^{2} ( ρ _{1} - ρ ) g}{η} = \frac{2}{9} \frac{r _{2}^{2} ( ρ _{2} - ρ ) g}{η}$

$r_{1}^{2} (ρ_{1} - ρ) = r_{2}^{2} (ρ_{2} - ρ)$

$\frac{r _{1}^{2}}{r _{2}^{2}} = \frac{ρ _{2} - ρ}{ρ _{1} - ρ}$

Substituting the given values:

$\frac{r _{1}^{2}}{r _{2}^{2}} = \frac{8.5 \times 1 0 ^{3} - 2.5 \times 1 0 ^{3}}{11.5 \times 1 0 ^{3} - 2.5 \times 1 0 ^{3}}$

$\frac{r _{1}^{2}}{r _{2}^{2}} = \frac{6.0}{9.0} = \frac{2}{3}$

$\frac{r _{1}}{r _{2}} = \frac{2}{3}$

Answer: $\frac{2}{3}$

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