A solid steel ball of diameter 3.6 mm acquired terminal velocity 2.45 × 10⁻² m/s while falling under gravity through an oil of density 925 kg m⁻³. Take density of steel as 7825 kg m⁻³ and g as 9.8 m/s². The viscosity of the oil in SI unit is (a) 2.18 (b) 1.68 (c) 2.38 (d) 1.99

Answer: (d). vt = (2r²(ρ - σ)g)/(9η) ⇒ η = (2r²(ρ - σ)g)/(9vt) r = 1.8 × 10⁻³ m, ρ - σ = 7825 - 925 = 6900 kg/m³ η = 2 × 3.24 × 10⁻⁶ × 6900 × 9.89 × 2.45 × 10⁻² = (0.438)/(0.2205) ≈ 1.99 Pa s So the answer is (d) 1.99.

A solid steel ball of diameter 3.6 mm acquired terminal velocity 2.45… | JEE Main 2025 (3 Apr, Shift 2)

Q18JEE Main 2025 · 3 APR · SHIFT 2Medium

Terminal Velocity

A solid steel ball of diameter 3.6 mm acquired terminal velocity $2.45 \times 10^{-2}\text{ m/s}$ while falling under gravity through an oil of density $925\text{ kg m}^{-3}$ . Take density of steel as $7825\text{ kg m}^{-3}$ and g as $9.8\text{ m/s}^{2}$ . The viscosity of the oil in SI unit is

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Answer: (d)

$v_t = \dfrac{2r^2(\rho - \sigma)g}{9\eta} \Rightarrow \eta = \dfrac{2r^2(\rho - \sigma)g}{9v_t}$
$r = 1.8 \times 10^{-3}$ m, $\rho - \sigma = 7825 - 925 = 6900\ \text{kg/m}^3$
$\eta = \dfrac{2 \times 3.24 \times 10^{-6} \times 6900 \times 9.8}{9 \times 2.45 \times 10^{-2}} = \dfrac{0.438}{0.2205} \approx 1.99$ Pa s
So the answer is (d) 1.99.

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