Q. At a given place where, acceleration due to gravity is $g\ ms^{- 2}$, a sphere of lead of density $d\ kgm^{- 2}$ is gently released in a column of liquid of density $\rho$ $kgm^{- 3}$. If $d > \rho$, the sphere will
A) Fall vertically with an acceleration of $g\ ms^{- 2}$
B) Fall vertically with no acceleration
C) Fall vertically with an acceleration g$\left( \frac{d - \rho}{d} \right)$
D) Fall vertically with an acceleration $\rho/d$
Explanation
(c) When a sphere of mass $m$ is released in a liquid, it falls vertically down with acceleration $= \frac{mg - F_{B}}{m}$
$\frac{\frac{4}{3}\pi r^{3}dg - \frac{4}{3}\pi r^{3}\rho g}{\frac{4}{3}\pi r^{2}d} = \frac{(d - \rho)g}{d}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.