Question

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)

• A. $\frac{v}{2}$
• B. $\frac{v}{\sqrt{2}}$
• C. v
• D. 2 v

Answer 1

Answer: (A)

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