You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Terminal Velocity

A spherical b...

Question

A spherical ball of iron of radius $2 mm$ is falling through a column of glycerine. If densities of glycerine and iron are respectively $1.3 \times 10^{3} {kg/m}^{3}$ and $8 \times 10^{3} {kg/m}^{3}$ . $η f o r g l y c e r i n e = 0.83 {Nm}^{- 2} sec,$ then the terminal velocity is:

0.7 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.07 m/s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.007 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.0007 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $0.07 m/s$
Terminal velocity $, v_{0} = \frac{2 r^{2} (ρ - ρ_{0}) g}{9 η}$
$v_{0} = \frac{2 \times (2 \times 10^{- 3})^{2} \times (8 - 1.3) \times 10^{3} \times 9.8}{9 \times 0.83} = 0.07 {ms}^{- 1}$

Suggest Corrections

Similar questions

Q. A smooth spherical ball of radius 1 cm and density

4 \times 10^{3} \frac{k g}{m^{3}}

is dropped gently in a large container containing viscous liquid of density

2 \times 10^{3} \frac{k g}{m^{3}}

and

η = 0.1 \frac{N - s}{m^{2}}

. The distance travelled by the ball in t = 0.2s after it attains terminal velocity is

Q. A smooth spherical ball of radius

1

cm and density

4 \times 10^{3} k g / m^{3}

is dropped gently in a large container containing viscous liquid of density

2 \times 10^{3} k g / m^{3}, η = 0.1 N - s / m^{2}

. The distance moved by the ball in

t = 0.1 s e c

after it attains terminal velocity is :

Q. If the rate of flow of glycerine of density

1.25 \times 10^{3} k g / m^{3}

through the conical section of a pipe is

x \times 10^{- 5} m^{3} / s

and the radii of its ends are

0.1 m

and

0.04 m

and the pressure drop across its length is

10 N / m^{2}

. Find

x :

The terminal velocity of a steel ball of 4 mm in diameter falling through a glycerin of viscosity 3.3 poise, given relative density of steel is 8, relative density of glycerin is 1.3 is

Q. The terminal velocity of a steel ball 2 mm in diameter falling through glycerin is

44 \times 10^{- 2} c m / s

(Given that specific gravity of steel = 8, specific gravity of steel =8, specific gravity of glycerin a 1.3, viscosity of glycerine 8.3 poise.)

Join BYJU'S Learning Program

A spherical ball of iron of radius 2 mm is falling through a column of glycerine. If densities of glycerine and iron are respectively 1.3×103 kg/m3 and 8×103 kg/m3. η for glycerine=0.83 Nm−2 sec, then the terminal velocity is:

The correct option is A 0.07 m/sTerminal velocity,v0=2r2(ρ−ρ0)g9ηv0=2×(2×10−3)2×(8−1.3)×103×9.89×0.83=0.07 ms−1

A spherical ball of iron of radius $2 mm$ is falling through a column of glycerine. If densities of glycerine and iron are respectively $1.3 \times 10^{3} {kg/m}^{3}$ and $8 \times 10^{3} {kg/m}^{3}$ . $η f o r g l y c e r i n e = 0.83 {Nm}^{- 2} sec,$ then the terminal velocity is:

The correct option is A $0.07 m/s$
Terminal velocity $, v_{0} = \frac{2 r^{2} (ρ - ρ_{0}) g}{9 η}$
$v_{0} = \frac{2 \times (2 \times 10^{- 3})^{2} \times (8 - 1.3) \times 10^{3} \times 9.8}{9 \times 0.83} = 0.07 {ms}^{- 1}$