An air bubble of radius 1 cm rises from the bottom portion through a liquid of density 1.5 g/cc at a constant speed of 0.25 cm s-1. If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (In Pa)

Question

An air bubble of radius 1 cm rises from the bottom portion through a liquid of density 1.5 g/cc at a constant speed of 0.25 cm s-1. If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (In Pa) (A) 13000 (B) 1300 (C) 130 (D) 13

Tardigrade Admin · Accepted Answer

v =(2/9) (r2 ρ g/η) ⇒ η =(2/9) ⋅ (r2 ρ g/v) =(2/9) ((1 × 10-2)2 ×(1.5 × 103) × 9.8/0.25 × 10-2) =130 Pa - s

Q. An air bubble of radius $1 c m$ rises from the bottom portion through a liquid of density $1.5 g / cc$ at a constant speed of $0.25 c m s^{- 1}$ . If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (In Pa)

13000

1300

130

13

$v = \frac{2}{9} \frac{r ^{2} ρ g}{η}$
$\Rightarrow η = \frac{2}{9} \cdot \frac{r ^{2} ρ g}{v}$
$= \frac{2}{9} \frac{( 1 \times 1 0 ^{- 2} ) ^{2} \times ( 1.5 \times 1 0 ^{3} ) \times 9.8}{0.25 \times 1 0 ^{- 2}}$
$= 130 P a - s$

Q. An air bubble of radius 1cm rises from the bottom portion through a liquid of density 1.5g/cc at a constant speed of 0.25cms−1. If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (In Pa)

13000

1300

130

13

v=92​ηr2ρg​ ⇒η=92​⋅vr2ρg​ =92​0.25×10−2(1×10−2)2×(1.5×103)×9.8​ =130Pa−s

Q. An air bubble of radius $1 c m$ rises from the bottom portion through a liquid of density $1.5 g / cc$ at a constant speed of $0.25 c m s^{- 1}$ . If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (In Pa)

$v = \frac{2}{9} \frac{r ^{2} ρ g}{η}$
$\Rightarrow η = \frac{2}{9} \cdot \frac{r ^{2} ρ g}{v}$
$= \frac{2}{9} \frac{( 1 \times 1 0 ^{- 2} ) ^{2} \times ( 1.5 \times 1 0 ^{3} ) \times 9.8}{0.25 \times 1 0 ^{- 2}}$
$= 130 P a - s$