Question 10: Properties of Matter

Question

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

Accepted Answer

Using Stokes' Law, the formula for terminal velocity $\left(v_tight)$ of a spherical body falling through a viscous fluid is :
$$ v_t=\frac{2 r^2(ho-\sigma) g}{9 \eta} $$
Where :
$r=$ Radius of the sphere $r=\frac{	ext { Diameter }}{2}=\frac{2 \mathrm{~mm}}{2}=1 \mathrm{~mm}=0.1 \mathrm{~cm}$.
$ho=$ Density of the sphere $=ho=10.5 \mathrm{~g} / \mathrm{cm}^3$
$\sigma=$ Density of the fluid, Density of glycerine $=\sigma=1.5 \mathrm{~g} / \mathrm{cm}^3$
$\mathrm{g}=$ Acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2=1000 \mathrm{~cm} / \mathrm{s}^2$
$\eta=$ Coefficient of viscosity $=10$ Poise
1 Poise $=1 \frac{\mathrm{~g}}{\mathrm{~cm} \cdot \mathrm{~s}^3}$, which is the CGS unit of viscosity.
Substituting the values into the terminal velocity equation :
$$ v_t=\frac{2 	imes(0.1)^2 	imes(10.5-1.5) 	imes 1000}{9 	imes 10} $$
$$\Rightarrow $$ $\mathrm{v}_{\mathrm{t}}=\frac{2 	imes 0.01 	imes 9 	imes 1000}{9 	imes 10}$
$$\Rightarrow $$ $$ \mathrm{v}_{\mathrm{t}}=2 \mathrm{~cm} / \mathrm{s} $$
The sphere accelerates until the viscous drag reaches a point where the upward forces equal the weight of the sphere. At this point, it continues to fall at a constant terminal velocity of $2.0 \mathrm{~cm} / \mathrm{s}$.

Answer

1.0

Answer

1.5

Answer

3.0

Answer

2.0

Properties of Matter

Choose the correct answer: