JEE Main 2025 Apr 3 Shift 2, Physics Q12: Hydrodynamic

JEE Main2025Apr 3, Shift 2Physics

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
$1.99$

Solution

For a sphere in Stokes flow at terminal velocity:

$v_{T} = \frac{2 r ^{2} ( ρ _{s} - ρ _{ℓ} ) g}{9 η} \Rightarrow η = \frac{2 r ^{2} ( ρ _{s} - ρ _{ℓ} ) g}{9 v _{T}}$ .

Use $r = 1.8 \times 1 0^{- 3}$ m, $ρ_{s} - ρ_{ℓ} = 7825 - 925 = 6900$ kg/m $^{3}$ :

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

Numerator: $2 \cdot 3.24 \times 1 0^{- 6} \cdot 6900 \cdot 9.8 \approx 0.4382$ .

Denominator: $9 \cdot 2.45 \times 1 0^{- 2} = 0.2205$ .

$η \approx 1.99$ Pa $\cdot$ s.

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JEE Main 2025 Apr 3 Shift 2, Physics Q12: Hydrodynamic

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