Question

3. A steel shot of diameter 2 mm is dropped in a viscous liquid filled in a drum. Find the terminal
speed of the shot. Density of the material of the shot = 8.0 x 100 kg/m”, density of liquid = 1.0 x 103
kg/mº. Coefficient of viscosity of liquid = 1.0 kg/(m-s), g = 10 m/s?.
​

Answer 1

Given that,

Diameter = 2 mm

Density of the material $\rho_{m}=8.0\times10^{3}\ kg/m^3$

Density of liquid $\rho_{l}=1.0\times10^{3}\ kg/m^3$

Coefficient of viscosity of liquid =10 kg/ms

We need to calculate the terminal speed of the shot

Using formula of terminal speed

$v=\dfrac{2}{9}\dfrac{r^2}{\eta}(\rho_{m}-\rho_{l})g$

Where, r = radius

$\eta$= viscosity of liquid

$\rho_{m}$ = density of the material

$\rho_{l}$ = density of liquid

Put the value into the formula

$v=\dfrac{2}{9}\times\dfrac{(1\times10^{-3})^2}{10}\times(8.0\times10^{3}-1.0\times10^{3})\times10$

$v=0.0015\ m/s$

$v=0.15\ cm/s$

Hence, The terminal speed of the shot is 0.15 cm/s.

Answer 2

Explanation:

Given that,

Diameter = 2 mm

Density of the material \rho_{m}=8.0\times10^{3}\ kg/m^3ρ

m

=8.0×10

3

kg/m

3

Density of liquid \rho_{l}=1.0\times10^{3}\ kg/m^3ρ

l

=1.0×10

3

kg/m

3

Coefficient of viscosity of liquid =10 kg/ms

We need to calculate the terminal speed of the shot

Using formula of terminal speed

v=\dfrac{2}{9}\dfrac{r^2}{\eta}(\rho_{m}-\rho_{l})gv=

9

2

η

r

2

(ρ

m

−ρ

l

)g

Where, r = radius

\etaη = viscosity of liquid

\rho_{m}ρ

m

= density of the material

\rho_{l}ρ

l

= density of liquid

Put the value into the formula

v=\dfrac{2}{9}\times\dfrac{(1\times10^{-3})^2}{10}\times(8.0\times10^{3}-1.0\times10^{3})\times10v=

9

2

×

10

(1×10

−3

)

2

×(8.0×10

3

−1.0×10

3

)×10

v=0.0015\ m/sv=0.0015 m/s

v=0.15\ cm/sv=0.15 cm/s

Hence, The terminal speed of the shot is 0.15 cm/s.