Question

A spherical ball of diameter $1$cm and density $5\times {10}^3$kg $m^{-3}$ is dropped gently in a large tank containing viscous liquid of density $3\times {10}^3$kg $m^{-3}$ and coefficient of viscosity $0.1$Ns $m^{-2}$. The distance, the ball moves in $1$s after attaining terminal velocity is $(g=10m{s}^{-2})$.

• A. $\frac{10}{9}$m
• B. $\frac{2}{3}$m
• C. $\frac{4}{9}$m
• D. $\frac{4}{5}$m
• E. $\frac{9}{10}$m

Answer 1

Answer: (A)

$\frac{10}{9}$m

Given, diameter of spherical ball$\left(d\right)=1cm=0.01$m
Radius of spherical ball(r)$=\frac{0.01}{2}$m
Density of spherical ball$\left(\rho \right)=5\times {10}^3$kg $m^{-3}$
Density of liquid$\left(\sigma \right)=3\times {10}^{3}kg{m}^{-3}$
Coefficient of viscosity $\left(\eta \right)=0.1$N$s{m}^{-2}$

We know that,
$v=\frac{2}{9}\frac{g{r}^2(\varphi -\sigma )}{\eta }$
$v=\frac{2\times 10\times (0.01{)}^2\times (5\times {10}^3-3\times {10}^3)}{9\times (2{)}^2\times 0.1}$
$v=\frac{10}{9}m{s}^{-1}$
Now, S$=$vt
S$=\frac{10}{9}\times 1$
S$=\frac{10}{9}$m.