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Standard XII

Physics

Floating

A solid ball ...

Question

A solid ball of density $ρ_{1}$ and radius r falls vertically through a liquid of density $ρ_{2}$ . Assume that the viscous force acting on the ball is $F = - k r v$ , where $k$ is a constant and $v$ its velocity. What is the terminal velocity of the ball?

\frac{4 π r^{2} (ρ_{1} - ρ_{2})}{3 k}

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\frac{2 π r (ρ_{1} - ρ_{2})}{3 g k}

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\frac{2 π g (ρ_{1} + ρ_{2})}{3 g r^{2} k}

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None of these

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Solution

The correct option is A $\frac{4 π r^{2} (ρ_{1} - ρ_{2})}{3 k}$

Balnacing the forces, we get
$F_{u} + F_{v} = m g$
$v . ρ_{2} g + K r v_{T} = m g; V_{T}$ = terminal velocity
$V_{T} = \frac{(m - v ρ_{2}) g}{K r}$
$V_{T} = \frac{v g (ρ_{1} - ρ_{2})}{K r} {m = ρ \times volume}$
$V_{T} = \frac{4 π r^{2} (ρ_{1} - ρ_{2})}{3 K} (∵ v = \frac{4}{3} π r^{3})$

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A solid ball of density ρ1 and radius r falls vertically through a liquid of density ρ2. Assume that the viscous force acting on the ball is F=−krv, where k is a constant and v its velocity. What is the terminal velocity of the ball?

The correct option is A 4πr2(ρ1−ρ2)3k Balnacing the forces, we get Fu+Fv=mg v.ρ2g+KrvT=mg;VT= terminal velocity VT=(m−vρ2)gKr VT=vg(ρ1−ρ2)Kr{m=ρ×volume} VT=4πr2(ρ1−ρ2)3K(∵v=43πr3)

A solid ball of density $ρ_{1}$ and radius r falls vertically through a liquid of density $ρ_{2}$ . Assume that the viscous force acting on the ball is $F = - k r v$ , where $k$ is a constant and $v$ its velocity. What is the terminal velocity of the ball?