Two metal spheres are falling through a liquid of density $2 \times 10^{3} k g / m^{3}$ with the same uniform speed. The material density of sphere 1 and sphere 2 are $8 \times 10^{3} k g / m^{3}$ and $11 \times 10^{3} k g / m^{3}$ respectively. The ratio of their radii is

\frac{11}{8}

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\sqrt{\frac{11}{8}}

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\frac{3}{2}

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\sqrt{\frac{3}{2}}

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Solution

The correct option is D $\sqrt{\frac{3}{2}}$
$\begin{matrix} W e k n o w t h a t v_{r} = \frac{2}{9} \frac{R^{2} (ρ - σ) g}{η} w h e r e η i s t h e c o e f f i c i e n t o f v i s o c i t y o f t h e l i q u i d ∴ v_{r_{1}} = \frac{2 R_{1}^{2} (ρ - σ) g}{9 η} a n d v_{r_{2}} = \frac{2 R_{2}^{2} (ρ - σ) g}{9 η} A c c o r d i n g t o t h e g i v e n p r o b l e m, v_{r_{1}} = v_{r_{2}} \frac{2 R_{1}^{2} (ρ - σ) g}{9 η} = \frac{2 R_{2}^{2} (ρ - σ) g}{9 η} 2 R_{1}^{2} (ρ - σ) g = 2 R_{2}^{2} (ρ - σ) g o r, \frac{R_{1}^{2}}{R_{2}^{2}} = \frac{(11 \times 10^{3} - 2 \times 10^{3})}{(8 \times 10^{3} - 2 \times 10^{3})} = \frac{9}{6} = \frac{3}{2} ∴ \frac{R_{1}^{2}}{R_{2}^{2}} = \frac{3}{2} \frac{R_{1}}{R_{2}} = \sqrt{\frac{3}{2}} . \end{matrix}$
Hence, the correct answer is option $D$ .

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Q. A metallic sphere floats in immiscible mixture of water (density

10^{3} k g / m^{3}

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8 \times 10^{3} k g / m^{3}

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part in the liquid. The density of the metal is

A metallic sphere floats in an immiscible mixture of water (density 10 $^{3}$ kg/m $^{3}$ ) and a liquid (density $8 \times 10^{3} k g / m^{3}$ ) such that its ( $\frac{2}{3}$ ) part is in water and ( $\frac{1}{3}$ ) part in the liquid. The density of the metal is