A smooth spherical ball of radius 1cm and density 4 - 103 kg-m3 is dropped gently in a large container containing viscous liquid of density 2 - 103 kg-m3- - - 0-1 N-s-m2 - The distance moved by the ball in t - 0-1 sec after it attains terminal velocity is -

The correct option is C 49m down V _ T = 2 gr ^ 2 ( d l ) #160; 9η #160; V _ T = 2× 10× 10 ^ 4 ( 2× 10 ^ 3 ) #160; 9× 0.1 V _ T = ...

You visited us 6 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Reynold's Number

A smooth sphe...

Question

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 :

\frac{4}{5} m

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

\frac{4}{9} m

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

\frac{2}{3} m

down

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

\frac{4}{9} m

down

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

Open in App

Solution

The correct option is C $\frac{4}{9} m$ down
$V_{T} = \frac{2 {g r}^{2} (d - l)}{9 η}$
$V_{T} = \frac{2 \times 10 \times 10^{- 4} (2 \times 10^{3})}{9 \times 0.1}$
$V_{T} = \frac{40}{9}$ m/s
Distance travelled $= \frac{40}{9} \times 0.1 = \frac{4}{9} m$
As velocity is down ward so $D = \frac{4}{9} m$ downward.

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 spherical ball of diameter

1

cm and density

5 \times 10^{3}

m^{- 3}

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

3 \times 10^{3}

m^{- 3}

and coefficient of viscosity

0.1

m^{- 2}

. The distance, the ball moves in

1

s after attaining terminal velocity is

(g = 10 m s^{- 2})

Q. A spherical ball of density

ρ

and radius 3 mm is dropped into a tube containing a viscous liquid filled a upto 0 mark as shown in the figure. Viscosity of the liquid is 1.260

N - s / m^{2}

and its density is

ρ_{L} = ρ / 2 = 1260 k g / m^{3}

. Assume the ball reaches a terminal speed by the 10 cm mark. The time taken by the ball to travel the distance between the 10 cm and 20 cm mark is
(take g= 10

m / s^{2}

)

Q. A spherical ball of density

ρ

and radius

0.003 m

is dropped into a tube containing a viscous fluid up to the

0 c m

mark as shown in the figure. Viscosity of the fluid

= 1.26 N - s / m^{2}

and its density

ρ_{L} = \frac{ρ}{2} = 1260 k g / m^{3}

Assume the ball reaches a terminal speed at

10 c m

mark. The time taken by the ball to travel the distance between the

10 c m

and

20 c m

mark is

(g = 10 m / s^{2})

Q. Find out the terminal velocity of rain water drop of radius 1 mm. Viscous coefficient for air is

1.8 \times 10^{- 5} N s / m^{2}

and density

1.2 k g / m^{3}

, density of water is

10^{3} k g / m^{3}

, (

g = 10 m / s^{2}

Join BYJU'S Learning Program

A smooth spherical ball of radius 1cm and density 4×103kg/m3 is dropped gently in a large container containing viscous liquid of density 2×103kg/m3,η=0.1N−s/m2 . The distance moved by the ball in t=0.1sec after it attains terminal velocity is :

The correct option is C 49m downVT=2gr2(d−l)9ηVT=2×10×10−4(2×103)9×0.1VT=409 m/sDistance travelled =409×0.1=49mAs velocity is down ward so D=49m downward.