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)

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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