JEE Main 2026 — Apr 2, Shift 2 — Physics solution

JEE Main2026Apr 2, Shift 2Physics

If an air bubble of diameter $2$ mm rises steadily through a liquid of density $2000$ kg/m $^{3}$ at a rate of $0.5$ cm/s, then the coefficient of viscosity of liquid is_____Poise. (Take $g = 10$ m/s $^{2}$ )

A
$0.88$
B
$8.8$
C
$88.8$
D
$0.088$

Solution

At terminal (steady) velocity, the buoyant force balances the viscous drag (the bubble's own weight is negligible). Equating $\frac{4}{3} π R^{3} ρ g = 6 π η R v$ yields $η = \frac{2 ρ R ^{2} g}{9 v}$ .

With $R = 1 0^{- 3}$ m, $ρ = 2000$ kg/m $^{3}$ , $v = 5 \times 1 0^{- 3}$ m/s, $g = 10$ m/s $^{2}$ :

$η = \frac{2 \cdot 2000 \cdot 1 0 ^{- 6} \cdot 10}{9 \cdot 5 \times 1 0 ^{- 3}} = \frac{0.04}{0.045} = \frac{8}{9}$ Pa $\cdot$ s.

Converting to Poise ( $1$ Pa $\cdot$ s $= 10$ Poise): $η = \frac{80}{9} \approx 8.88$ Poise, i.e. $\approx 8.8$ Poise.

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