Question

The acceleration due to gravity at a place is $gm/s^2$. A lead sphere of density $dkg/m^3$ is dropped into a liquid column of density $\rho kg/m^3$. If $d>\rho$ , then the sphere will fall down with

• A. an acceleration '$g$'
• B. without acceleration
• C. acceleration $a=g(d-\rho )/d$
• D. with an acceleration $\frac{g\rho }{d}$

Answer 1

Answer: (C)

acceleration $a=g(d-\rho )/d$

Given density of sphere and liquid are $dkg{m}^{-3}$ and $\rho radius=rkg{m}^{-3}$.
Acceleration due to gravity is $gm{s}^{-2}$
Let, volume of sphere be $v{m}^3\to massofsphere=dvkg$
$Netforce=F_g-F_b$
$dva=dvg-\rho vg$
$\Rightarrow a=g(d-\rho )/d$
Given that $d>\rho$, $a$ is in downward direction as shown. Thus C is correct.