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

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Fluid Mechanics

A solid steel ball of diameter 3.6 mm acquired terminal velocity

2.45 \times 1 0^{- 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 / s^{2}

. The viscosity of the oil in SI unit is

(A)2.18
(B)2.38
(C)1.68
(D)1.99✓ Correct

Answer: (D)

Solution

At terminal velocity the viscous drag balances the net downward force, giving

v_{T} = \frac{2}{9} \frac{( ρ _{s} - ρ _{f} ) r ^{2} g}{η}

. With radius

r = 1.8 \times 1 0^{- 3} m

ρ_{s} - ρ_{f} = 6900 kg/m^{3}

g = 9.8 m/s^{2}

and

v_{T} = 2.45 \times 1 0^{- 2} m/s

, rearranging gives

η = \frac{2}{9} \cdot \frac{6900 ( 1.8 \times 1 0 ^{- 3} ) ^{2} ( 9.8 )}{2.45 \times 1 0 ^{- 2}}

. Evaluating, the viscosity of the oil is about

1.99 Pa\cdots

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Single correct option

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