Question

A very small ball of radius r and mass m is released in a large tank having a liquid of density p and moving with an acceleration a as shown in the figure. If the coefficient of viscosity the liquid is $\eta$, then terminal speed of the ball will be (neglect buoyant force)

• A. mg/$6\pi \eta r$
• B. ma/$6\pi \eta r$
• C. m/$6\pi \eta r$ $\sqrt{g^2+a^2}$
• D. m/$6\pi \eta r$ $\sqrt{g+a}$

Answer 1

The correct option is C m/$6\pi \eta r$ $\sqrt{g^2+a^2}$

Terminal speed of ball will be $=\frac{mass}{6\pi nr}$ (total acceleration)

$=\frac{m}{6\pi nr}\sqrt{g^2+a^2}$

Total acting on ball in is $\sqrt{g^2+a^2}$