Question

A spherical ball of density $\text{ρ}$ and radius $0.003 \, m$ is dropped into a tube containing a viscous fluid, filled up to the $0 \, cm$ mark as shown in the figure. Viscosity of the fluid = $1.260 \, N \, m^{- 2} \, s^{- 1}$ and its density $\text{ρ}_{\text{L}}$ = $\text{ρ}/2=1260 \, kg \, m^{- 3}$ . Assume the ball reaches a terminal speed by the $10 \, cm$ mark. Find the time taken by the ball to traverse the distance between the $10 \, cm$ and $20 \, cm$ mark. [ $g$ = acceleration due to gravity = $10 \, m \, s^{- 2}$ ^​​ ]

• A. $2 \, s$
• B. $3s$
• C. $5 \, s$
• D. $1.5s$

Answer 1

Answer: (C)

Terminal Velocity:
$V_{T} = \frac{2}{9} \times \frac{\left(\text{r}\right)^{2} \left(\text{ρ} - \left(\text{ρ}\right)_{\text{L}}\right) \text{g}}{\text{η}}$
$\Rightarrow V_{\text{T}}=\frac{2}{9}\times \frac{\left(3 \times 1 0^{- 3}\right)^{2} \times 1 2 6 0 \times 10}{1 \text{.} 2 6 0}$
$\text{V}_{\text{T}}=2\times 10^{- 2} \, m \, s^{- 1}$
$\Rightarrow \text{V}_{\text{T}}=2 \, cm \, s^{- 1}$
$\text{t}=\frac{10 \, \text{cm}}{\text{V}_{\text{T}}}$
$\Rightarrow \text{t}=5 \, \text{s}$