Answer:
Option C
Explanation:
Given,
Diameter of pipe (d)= 5mm= $5 \times 10^{-3}$ m
Density of gasoline ($\rho$)=720 kg/m3
Viscosity of gasoline ($\eta$)= $6 \times 10^{-3}$ Poise
We know that,
$V_{c}=\frac{\eta}{\rho d}=\frac{6 \times 10^{-3}}{720 \times 5 \times 10^{-3}}$
$=\frac{6 \times 10^{-3}}{3600\times10^{-3}}=\frac{1}{600}=\frac{1}{6}\times10^{-2}$
$=1.66 \times 10^{-3} m/s$