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 = ...

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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

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\frac{4}{9} m

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\frac{2}{3} m

down

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\frac{4}{9} m

down

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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.

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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.