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A spherical solid ball of volume V is made of a material of density $\rho _{1}$ . It is falling through a liquid of density $\rho _{2}\left (\rho _{2}< \rho _{1} \right )$ Assume that the liquid applie a viscous force on the ball that is proportional to the square of its speed $\nu ,i.e.,F_{viscous}= -kv^{2}\left ( k> 0 \right )$ . The terminal speed of the ball is

Option 1)
$\frac{V g\left ( \rho _{1}- \rho _{2}\right ) }{k}$

Option 2)
$\sqrt{\frac{V g\left ( \rho _{1}- \rho _{2}\right ) }{k}}$

Option 3)
${\frac{Vg\rho _{1}}{k}}$

Option 4)
$\sqrt{\frac{Vg\rho _{1}}{k}}$

Answers (1)

best_answer

As we learnt in

Net force on the body -

$F_{B}+F_{v}= W$

$\rightarrow 6\pi \eta rv+\frac{4}{3}\pi r^{3}\sigma g= \frac{4}{3}\pi r^{3}\rho g$

$\rightarrow 6\pi \eta rv=\frac{4}{3}\pi r^{3} g\left ( \rho -\sigma \right )$

$\rightarrow v_{t}=\frac{2}{9}\frac{ r^{2} \left ( \rho -\sigma \right )}{\eta }g$

- wherein

$F_{B}-Buoyant \: force$

$F_{v}-viscous \: force$

$w-weight$

$\rho \rightarrow density \: of \: ball$

$\sigma \rightarrow density \: of \: water$

$V_{T} =terminal \: velocity$

The ball acquire terminal speed in its equilibrium state.

$v\rho_{2}g+kv^{2}=v \rho_{1}g$

$kv^{2}=vg\left(\rho_{1}-\rho_{2} \right )$

$\Rightarrow\ \;v=\sqrt{\frac{vg\left(\rho_{1}-\rho_{2} \right )}{k}}$

Correct option is 2.

Option 1)

$\frac{V g\left ( \rho _{1}- \rho _{2}\right ) }{k}$

This is an incorrect option.

Option 2)

$\sqrt{\frac{V g\left ( \rho _{1}- \rho _{2}\right ) }{k}}$

This is the correct option.

Option 3)

${\frac{Vg\rho _{1}}{k}}$

This is an incorrect option.

Option 4)

$\sqrt{\frac{Vg\rho _{1}}{k}}$

This is an incorrect option.

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