Two metal spheres are falling through a liquid of density 2.5 × 10³ kg/m³ with the same uniform speed. The density of material of first sphere and second sphere is 11.5 × 10³ kg/m³ and 8.5 × 10³ kg/m³ respectively. The ratio of the radius of first sphere to that of second sphere is (a) 2/3 (b) √(2/3) (c) 3/2 (d) √(3/2)

Answer: (b). v ∝ r²(ρ - σ); equal speeds give r₁²(ρ₁ - σ) = r₂²(ρ₂ - σ). (r₁²)/(r₂²) = (8.5 - 2.5)/(11.5 - 2.5) = 6/9 = 2/3 (r₁)/(r₂) = √(2/3) So the answer is (b) √(2/3).

Two metal spheres are falling through a liquid of density 2.5 × 10³… | MHT CET 2024 (10 May, Shift 2)

Q26MHT CET 2024 · 10 MAY · SHIFT 2Medium

Terminal Velocity

Two metal spheres are falling through a liquid of density $2.5 \times 10^{3}\text{ kg/m}^{3}$ with the same uniform speed. The density of material of first sphere and second sphere is $11.5 \times 10^{3}\text{ kg/m}^{3}$ and $8.5 \times 10^{3}\text{ kg/m}^{3}$ respectively. The ratio of the radius of first sphere to that of second sphere is

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Answer: (b)

$v \propto r^2(\rho - \sigma)$ ; equal speeds give $r_1^2(\rho_1 - \sigma) = r_2^2(\rho_2 - \sigma)$ .
$\dfrac{r_1^2}{r_2^2} = \dfrac{8.5 - 2.5}{11.5 - 2.5} = \dfrac{6}{9} = \dfrac{2}{3}$
$\dfrac{r_1}{r_2} = \sqrt{\dfrac{2}{3}}$
So the answer is (b) $\sqrt{\frac{2}{3}}$ .

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Terminal Velocity question from MHT CET 2024 (10 May, Shift 2)

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