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If the terminal speed of a sphere of gold $\left ( density = 19.5 kg/m^{3} \right )$ is 0.2 m/s in a viscous liquid

$\left ( density = 1.5 kg/m^{3} \right )$ find the terminal speed of a sphere of silver $\left ( density = 10.5 kg/m^{3} \right )$ of the same size in the same liquid

Option 1)
$0.2 m/s$

Option 2)
$0.4 m/s$

Option 3)
$0.133 m/s$

Option 4)
$0.1 m/s$

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

$F_{B}-Buoyant \: force$

$F_{v}-viscous \: force$

$w-weight$

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

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

$V_{T} =terminal \: velocity$

Terminal velcity $=v_{1}=\frac{2r^{2} \left(d_{1}-d_{2} \right )g}{9\eta}$

$\frac{v_t_{2}}{0.2}=\frac{(10.5-1.5)}{(19.5-1.5)}$

$\Rightarrow\ \;v_{t_{2}}=0.5\times\frac{9}{18}$

$\therefore \;v_{t_{2}}=0.1\ ms^{-1}$

Corrrect option is 4.

Option 1)

$0.2 m/s$

This is an incorrect option.

Option 2)

$0.4 m/s$

This is an incorrect option.

Option 3)

$0.133 m/s$

This is an incorrect option.

Option 4)

$0.1 m/s$

This is the correct option.

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