An air bubble of radius 5mm rises through a vat of syrup at a steady speed of 4mm/s. If the syrup has a density of 1.4 × 10 3 k g / m 3 , what is its viscosity? Options: A . &nbsp 1.15 kg/m-s B . &nbsp 15kg/m-s C . &nbsp 2.25 kg/m-s D . &nbsp 19.05 kg/m-s

Answer: Option D : D In this case the terminal velocity is upward. Since σ > ρ , the density of the air inside the bubble may be neglected completely. From equation v = 2 a 2 ( σ − ρ ) g g η = 2 a 2 ρ g 9 η ⇒ η = 2 a 2 ρ g 9 v where, a = 5mm, ρ = 1.4 × 10 3 k g / m 3 , v = 4 m m / s ∴ η = 2 × 25 × 10 − 6 × 1.4 × 10 3 × 9.8 9 × 4 × 10 − 3 k g / m − s = 19.05 k g / m − s Hence, (D) is correct.

An air bubble of radius 5mm rises through a vat of syrup at a steady speed of

Discussion Forum : Mechanical Properties Of Fluids

Question -

An air bubble of radius 5mm rises through a vat of syrup at a steady speed of 4mm/s. If the syrup has a density of $1.4 \times 10^{3} k g / m^{3}$ , what is its viscosity?

Options:

A . 1.15 kg/m-s

B . 15kg/m-s

C . 2.25 kg/m-s

D . 19.05 kg/m-s

Answer: Option D
:
D

In this case the terminal velocity is upward. Since $σ > ρ$ , the density of the air

inside the bubble may be neglected completely.

From equation

$v = \frac{2 a^{2} (σ - ρ) g}{g η} = \frac{2 a^{2} ρ g}{9 η} \Rightarrow η = \frac{2 a^{2} ρ g}{9 v}$

where, a = 5mm, $ρ = 1.4 \times 10^{3} k g / m^{3}, v = 4 m m / s$

$∴ η = \frac{2 \times 25 \times 10^{- 6} \times 1.4 \times 10^{3} \times 9.8}{9 \times 4 \times 10^{- 3}} k g / m - s = 19.05 k g / m - s$

Hence, (D) is correct.

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