
An air bubble of radius $0.4mm$rise up in water. Determine the terminal speed of bubble density of air is negligible.
Answer
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Hint: Terminal velocity is the maximum velocity attained by an object as it flows through a fluid. Terminal velocity can be achieved when an object's acceleration is zero which means when the speed of the object is constant.
To solve this question, we must know about the coefficient of viscosity.
Viscosity is defined as the extent up to which fluid can resist the flow under the influence of an applied force. The viscosity is measured in terms of coefficient of viscosity. Coefficient of viscosity of water is ${10^{ - 3}}kgm{s^{ - 1}}$
Complete step by step solution:
We are given that
$r = 0.4mm$ $\eta = {10^{ - 3}}kgm{s^{ - 1}}$ $\rho = {10^3}kg{m^{ - 3}}$ $\sigma = 0$
Where r is the radius of the bubble
$\eta {\text{ }}$is the coefficient of viscosity.
$\rho $is the density of water.
$\sigma {\text{ }}$is the density of air
Terminal velocity $v = \dfrac{2}{9}g{r^2}\left( {\dfrac{{\rho - \sigma }}{\eta }} \right)$
$ \Rightarrow v = \dfrac{2}{9} \times 9.8 \times {\left( {4 \times {{10}^{ - 4}}} \right)^2} \times \left( {\dfrac{{{{10}^3} - 0}}{{{{10}^3}}}} \right)$
$ \Rightarrow v = \dfrac{{2 \times 9.8 \times 16 \times {{10}^{ - 2}}}}{9}$
$ \Rightarrow v = 0.348m{s^{ - 1}}$
Additional Information:
An important thing to be noted is that coefficient of viscosity of fluids decreases as the temperature increases while coefficient of viscosity of gases increases with increase in temperature.
Terminal velocity occurs when the sum of drag force and the buoyancy becomes equal to downward gravitational force that is acting on the body. When the speed of an object increases drag force also increases. This force acts in the direction opposite to the motion of the body in the fluid. According to Stoke’s law the intensity of drag force increases with increase in velocity.
Note:
Terminal velocity is inversely proportional to square root coefficient of viscosity of the given medium and it is directly proportional to square root of the radius of the body. It also depends on the densities of the given mediums.
To solve this question, we must know about the coefficient of viscosity.
Viscosity is defined as the extent up to which fluid can resist the flow under the influence of an applied force. The viscosity is measured in terms of coefficient of viscosity. Coefficient of viscosity of water is ${10^{ - 3}}kgm{s^{ - 1}}$
Complete step by step solution:
We are given that
$r = 0.4mm$ $\eta = {10^{ - 3}}kgm{s^{ - 1}}$ $\rho = {10^3}kg{m^{ - 3}}$ $\sigma = 0$
Where r is the radius of the bubble
$\eta {\text{ }}$is the coefficient of viscosity.
$\rho $is the density of water.
$\sigma {\text{ }}$is the density of air
Terminal velocity $v = \dfrac{2}{9}g{r^2}\left( {\dfrac{{\rho - \sigma }}{\eta }} \right)$
$ \Rightarrow v = \dfrac{2}{9} \times 9.8 \times {\left( {4 \times {{10}^{ - 4}}} \right)^2} \times \left( {\dfrac{{{{10}^3} - 0}}{{{{10}^3}}}} \right)$
$ \Rightarrow v = \dfrac{{2 \times 9.8 \times 16 \times {{10}^{ - 2}}}}{9}$
$ \Rightarrow v = 0.348m{s^{ - 1}}$
Additional Information:
An important thing to be noted is that coefficient of viscosity of fluids decreases as the temperature increases while coefficient of viscosity of gases increases with increase in temperature.
Terminal velocity occurs when the sum of drag force and the buoyancy becomes equal to downward gravitational force that is acting on the body. When the speed of an object increases drag force also increases. This force acts in the direction opposite to the motion of the body in the fluid. According to Stoke’s law the intensity of drag force increases with increase in velocity.
Note:
Terminal velocity is inversely proportional to square root coefficient of viscosity of the given medium and it is directly proportional to square root of the radius of the body. It also depends on the densities of the given mediums.
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