Question

A metal ball of radius $2mm$ and density $10.5{ g/c}{{{m}}^{3}}$ is dropped in a glycine of coefficient of velocity. $9.8$ paise and density $1.5{ g/c}{{{m}}^{3}}$ The terminal velocity of the ball in ${cm/s}$ is:
A) $2$
B) $4$
C) $6$
D) $8$

Answer 1

Hint: Terminal velocity is the steady speed achieved by an object freely falling through a gas or liquid. After some time the velocity of the ball will become constant and will not change.

Formula used:
Terminal velocity
$V=\dfrac{2g{{r}^{2}}\left( T-\sigma \right)}{9\eta }$
Where $V$ is velocity, $g$ is specific gravity, $r$ is radius, $T$ is density of metal, $\sigma $is density of glycerine, $\eta $ is coefficient of velocity.

Complete solution:
When a metal ball is falling freely the initial velocity is zero there is final velocity as the energy is always conserved. Thus initially there is only potential energy which by the end of the motion gets converted to kinetic energy. No other force except self weight of the ball is acting in this case so there is a constant acceleration due to gravity.
However, when the body is falling freely in a liquid or gas medium there is a buoyant force that will act on the body and as result there will be an acceleration due this force.So the velocity will keep on changing until there comes a point when the accelerations will be balanced. That is the point when the ball will acquire a steady velocity known as the terminal velocity.
We are given that;
$\Rightarrow r=2mm=0.2m$
$\Rightarrow T=10.5{ g/c}{{{m}}^{3}}$
$\Rightarrow \sigma =1.5{ g/c}{{{m}}^{3}}$
$\Rightarrow \eta =9.8{ paise}$
So applying the formula and substituting the values we get;
$\Rightarrow V=2g{{r}^{2}}\dfrac{\left( T-\sigma \right)}{9\eta }=\dfrac{2\times {{0.2}^{2}}\times \left( 10.5-15 \right){ cm/s}}{9\times 9.8}$
$\Rightarrow V=8{ cm/s}$

Hence option (D) is the correct answer.

Note: When the temperature is increased its density decreases, thus the fluid becomes less viscous. When terminal velocity is reached the downward force of gravity is equal to the sum of object buoyancy and drug force.