A smooth spherical ball of radius 1 cm and density 4 - 103 kg - m 3 is dropped gently in a large container containing viscous liquid of density 2 - 103 kg - m 3 and -0-1 N - s - m 2 - The distance travelled by the ball in t -0-2 s after it attains terminal velocity isA- 8-0 m upB- 8-5 m upC- 8-9 m downD- 8-5 m down

The correct option is D 8/5 m downV_t = 2/9r^2 (ρ σ )g/η = 2/9(10^ 2 )^2 ×(4 2 ) × 10^3 × 10/0.1 = 40/9m/s ∴ distance travelled = v × t = 40/9× 0.2 = ...

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Byju's Answer

Standard XII

Physics

Terminal Velocity

A smooth sphe...

Question

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

\frac{8}{9}

m up

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\frac{8}{5}

m up

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\frac{8}{9}

m down

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\frac{8}{5}

m down

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Solution

The correct option is D $\frac{8}{5}$ m down
$V_{t} = \frac{2}{9} \frac{r^{2} (ρ - σ) g}{η} = \frac{2}{9} \frac{{(10^{- 2})}^{2} \times (4 - 2) \times 10^{3} \times 10}{0.1} = \frac{40}{9} \frac{m}{s} ∴ d i s t a n c e t r a v e l l e d = v \times t = \frac{40}{9} \times 0.2 = \frac{8}{9} m d o w n$

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A smooth spherical ball of radius 1 cm and density 4×103kgm3 is dropped gently in a large container containing viscous liquid of density 2×103kgm3 and η=0.1N−sm2. The distance travelled by the ball in t = 0.2s after it attains terminal velocity is

The correct option is D 85 m downVt=29r2(ρ−σ)gη=29(10−2)2×(4−2)×103×100.1=409ms∴distancetravelled=v×t=409×0.2=89m down