Explanation
A small sphere of radius r falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to:
A small sphere of radius r falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to
Let r is the radius of sphere and $${{v}_{t}}$$ is its terminal speed. Then the weight of sphere is balanced by the buoyant force and viscous force such that:
Weight,
$$ w=mg $$
$$ \because \rho =\dfrac{m}{V} $$
$$ m=\dfrac{4}{3}\pi {{r}^{3}}\rho g........(1) $$
$$ so, $$
$$ w=\dfrac{4}{3}\pi {{r}^{3}}\rho $$
Buoyant force,
$${{F}_{B}}=\dfrac{4}{3}\pi {{r}^{3}}\sigma g..........(2)$$
Where, $$\sigma $$ is density of water.
Viscous force, $$F=6\pi \eta rvt..............(3)$$
Where, $$\eta $$is viscosity.
From equation (1) (2) and (3)
$$ w={{F}_{B}}+{{F}_{v}} $$
$$ \dfrac{4}{3}\pi {{r}^{3}}\rho g=\dfrac{4}{3}\pi {{r}^{3}}\sigma g+6\pi \eta rvt $$
$$ {{v}_{t}}=\dfrac{2}{9}\dfrac{{{r}^{2}}\left( \rho -\sigma \right)g}{\eta }.......(4) $$
The rate of production of heat when the sphere attains its terminal velocity is
equal to work done by the viscous forces.
$$ W=\dfrac{dQ}{dt}={{F}_{v}}\times {{v}_{t}} $$
$$ W=6\pi \eta rv_{t}^{2} $$
$$ W=6\pi \eta r{{\left( \dfrac{2}{9}\dfrac{{{r}^{2}}\left( \rho -\sigma \right)g}{\eta } \right)}^{2}} $$
$$ \dfrac{dQ}{dt}\propto {{r}^{5}} $$
As an object falls through the atmosphere, there is air resistance acting on it. Air resistance is dependent on the velocity of the falling object, so as it speeds up, the force acting on it gets larger and larger.
Eventually, the force from air resistance will equal the force due to gravity. At that point the forces, acting in opposite directions, will sum to zero meaning that the object will travel at a constant velocity: terminal velocity.
The terminal velocity varies based on drag coefficients and masses.
$$ \dfrac{dv}{dt}=6-3v $$
$$ \int{\dfrac{dv}{6-3v}}=\int{dt} $$
$$ -\dfrac{1}{3}\log \left| 2-v \right|=dt $$
$$ -\log \left| 2-v \right|=3t $$
$$ \left| 2-v \right|={{e}^{-3t}} $$
$$ 2-{{e}^{-3t}}=v $$
Initial acceleration is $$0$$.
At $$t=0$$, terminal velocity is $$v=2m/s$$
Velocity of water surface,$${{v}_{1}}=0$$
From Bernoulli equation
$$ P+\dfrac{\rho {{v}^{2}}}{2}+\rho gh=\text{constant} $$
$$ {{P}_{atm}}+\dfrac{\rho v_{1}^{2}}{2}+\rho g{{h}_{1}}={{P}_{atm}}+\dfrac{\rho v_{2}^{2}}{2}+\rho g{{h}_{2}} $$
$$ \dfrac{\rho v_{2}^{2}}{2}=\rho g({{h}_{1}}-{{h}_{2}}) $$
$$ {{v}_{2}}=\sqrt{2g({{h}_{1}}-{{h}_{2}})} $$
Velocity of water is $$\sqrt{2g({{h}_{1}}-{{h}_{2}})}$$
Speed of water from hole is independent on area.
Hence, In steady flow the speed of water leaving the larger hole is the same as the speed of the water leaving the smaller.
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