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)

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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