Question: 16: The terminal velocity of a copper ball of radius 5 mm falling through a tank of oil at room temperature is 10 \mathrm{~cm} \mathrm{~s}^{-1}. If the viscosity of oil at room temperature is 0.9 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}, the viscous drag force is :
(1) 8.48 \times 10^{-3} \mathrm{~N}
(2) 8.48 \times 10^{-5} \mathrm{~N}
(3) 4.23 \times 10^{-3} \mathrm{~N}
(4) 4.23 \times 10^{-6} \mathrm{~N}
Answer: Option (1)
Explanation:
When a small spherical body moves through a viscous fluid with low speed, the viscous drag force acting on it is given by Stokes’ law.
The viscous drag force is
F = 6 \pi \eta r vwhere \eta is the coefficient of viscosity, r is the radius of the sphere,
and v is the velocity of the sphere.
Given radius of the copper ball,
r = 5\ \text{mm} = 5 \times 10^{-3}\ \text{m}Terminal velocity,
v = 10\ \text{cm s}^{-1} = 0.1\ \text{m s}^{-1}Viscosity of oil,
\eta = 0.9\ \text{kg m}^{-1}\text{s}^{-1}Substituting these values in Stokes’ formula,
F = 6 \pi \times 0.9 \times 5 \times 10^{-3} \times 0.1 F = 6 \pi \times 4.5 \times 10^{-4} F = 27 \pi \times 10^{-4}Using \pi \approx 3.14,
F = 8.48 \times 10^{-3}\ \text{N}Therefore, the viscous drag force acting on the copper ball is
8.48 \times 10^{-3}\ \text{N}.
Hence, the correct option is (1).