A small spherical ball of steel falls through a viscous medium with terminal velocity$v$. If a ball of twice the radius of the first one but of the same mass is dropped through the same method, it will fall with a terminal velocity (neglect buoyancy)

Question

Dr. Abubacker · Accepted Answer

(a) Given, $v = \frac{2r^{2}\rho g}{9\eta}$ …(i)
Mass $= \frac{4}{3}\pi r^{3}\rho = \frac{4}{3}\pi(2r)^{3}\rho_{1}$
Or $\rho_{1} = \rho/8$
Terminal velocity of second ball is
$v_{1} = \frac{2(2r)^{2}(\rho/8)g}{8\ \eta} = \frac{v}{2}$

Q. A small spherical ball of steel falls through a viscous medium with terminal velocity$v$. If a ball of twice the radius of the first one but of the same mass is dropped through the same method, it will fall with a terminal velocity (neglect buoyancy)

Explanation

Reviewed by Dr. Abubacker

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