A small metallic sphere of diameter 2 mm and density 10.5 g/cm$^3$ is dropped in glycerine having viscosity 10 Poise and density 1.5 g/cm$^3$ respectively. The terminal velocity attained by the sphere is ___ cm/s. ($\pi = \frac{22}{7}$ and $g = 10$ m/s$^2$)

Question

A soap bubble of radius $r = 1\,\text{mm}$ is rising in a liquid of density $ \rho = 2000\,\text{kg/m}^3 $. At the instant the bubble is rising upward with constant velocity $v = \tfrac{1}{2}\,\text{cm/s}$. Find the coefficient of viscosity $ (\eta) $.

Answer 1

To determine the terminal velocity attained by the small metallic sphere when dropped in glycerine, we can use Stokes' Law, which relates the drag force experienced by a sphere moving through a viscous fluid to its velocity. According to Stokes' Law, the drag force ($F_d$) is given by:

$F_d = 6 \pi \eta r v$

where:

$\eta$ is the viscosity of the fluid (in Poise, where 1 Poise = 0.1 Pa·s),
$r$ is the radius of the sphere,
$v$ is the velocity of the sphere.

At terminal velocity, the drag force is equal to the net gravitational force acting on the sphere. The gravitational force ($F_g$) is given by the difference between the weight of the sphere and the buoyant force:

$F_g = V (\rho_{sphere} - \rho_{fluid}) g$

where:

$V$ is the volume of the sphere,
$\rho_{sphere}$ is the density of the sphere,
$\rho_{fluid}$ is the density of the fluid,
$g$ is the acceleration due to gravity.

For a sphere of diameter 2 mm, the radius $r$ is 1 mm or 0.1 cm.

The volume $V$ of the sphere (using the formula for the volume of a sphere $\frac{4}{3} \pi r^3$) is:

$V = \frac{4}{3} \pi (0.1)^3 = \frac{4}{3} \times \frac{22}{7} \times 0.001$ cm³

$V = \frac{88}{21} \times 0.001 = 0.000419 cm^3$

Substituting the densities: $\rho_{sphere} = 10.5$ g/cm³, $\rho_{fluid} = 1.5$ g/cm³, and $g = 1000$ cm/s² (since 1 m/s² = 100 cm/s²), the net force $F_g$ becomes:

$F_g = 0.000419 \times (10.5 - 1.5) \times 1000 = 0.000419 \times 9 \times 1000 = 3.771$

Using $\eta = 10$ Poise = 1 Pa·s:

Equating $F_d$ and $F_g$ for terminal velocity:

$6 \pi \times 1 \times 0.1 \times v = 3.771$

$6 \times \frac{22}{7} \times 0.1 \times v = 3.771$cm/s

Therefore, the terminal velocity attained by the sphere is 2.0 cm/s, which matches option two.

A small metallic sphere of diameter 2 mm and density 10.5 g/cm$^3$ is dropped in glycerine having viscosity 10 Poise and density 1.5 g/cm$^3$ respectively. The terminal velocity attained by the sphere is ___ cm/s. ($\pi = \frac{22}{7}$ and $g = 10$ m/s$^2$)

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The Correct Option is B

Solution and Explanation

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