Question

derive an expression for the terminal velocity attained by a spherical body falling through a viscous medium

Answer 1

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Answer 2

TERMINAL VELOCITY DERIVATION:

• Let us consider a spherical body of mass m and density $\rho$. Let density of liquid be $\sigma$.

Now, mass of body is :

$m = volume \times \rho$

$\implies \: m = \dfrac{4}{3} \pi {r}^{3} \times \rho$

Now, mass of fluid liquid displaced :

$\implies \: m = \dfrac{4}{3} \pi {r}^{3} \times \sigma$

At terminal velocity, the body is in equilibrium, so we can say that :

$F_{b} + stroke \: force = W$

$\implies \: \dfrac{4}{3} \pi {r}^{3} \sigma g + 6\pi \eta rv = \dfrac{4}{3} \pi {r}^{3} \rho g$

$\implies \: 6\pi \eta rv = \dfrac{4}{3} \pi {r}^{3} ( \rho - \sigma)g$

$\boxed{\implies \: v = \dfrac{2}{9} \dfrac{{r}^{2} g( \rho - \sigma)}{ \eta}}$

[Hence derived].