A steel ball of diameter \(3.2 \mathrm{~mm}\) falls under gravity through an oil of density \(920 \mathrm{~kg} \mathrm{~m}^{-3}\) and viscosity $1.64 \mathrm{Ns} \mathrm{m}^{-2}\(. The density of steel. may be taken as \)7820 \mathrm{~kg} \mathrm{~m}^{-3}$. What is the terminal velocity of the ball ? (A) \(3.45 \mathrm{cms}^{-1}\) (B) \(4.25 \mathrm{cms}^{-1}\) (C) \(1.85 \mathrm{~cm}^{-1}\) (D) \(2.39 \mathrm{cms}^{-1}\)

Since we are given the diameter of the steel ball, we can find the radius, and consequently, the volume of the steel ball. The formula for the volume of a sphere is: \[V = \frac{4}{3}\pi r^3\] where \(r\) is the radius of the sphere. Then, using the density of steel, we can calculate the mass of the steel ball. The formula for mass is: \[m = ρV\] where \(ρ\) is the density of the material. First, we calculate the radius (in meters): \(r = \frac{3.2}{2}\times10^{-3} \mathrm{~m} = 1.6\times10^{-3} \mathrm{~m}\) Now, let's find the volume of the ball: \[V = \frac{4}{3}\pi(1.6\times10^{-3})^3 = 1.717\times10^{-8} \mathrm{~m}^3\] And finally, we compute the mass of the steel ball: \[m = (7820 \mathrm{~kg} \mathrm{~m}^{-3})(1.717\times10^{-8} \mathrm{~m}^3) = 0.1342 \mathrm{~kg}\]

The buoyant force (\(F_b\)) is equal to the weight of the fluid (oil) displaced by the steel ball. We will use the formula for buoyant force: \[F_b = ρ_o V g\] where \(ρ_o\) is the density of the oil, \(V\) is the volume of the steel ball, and \(g\) is the gravitational acceleration (\(9.81 \mathrm{ms}^{-2}\)). Now, let's compute the buoyant force: \[F_b = (920 \mathrm{~kg} \mathrm{~m}^{-3})(1.717\times10^{-8} \mathrm{~m}^3)(9.81 \mathrm{~m} \mathrm{~s}^{-2}) = 0.0156 \mathrm{N}\]

At terminal velocity, the net force acting on the steel ball is equal to the drag force (\(F_d\)) exerted by the oil due to its viscosity. The net force (\(F_n\)) is given by: \[F_n = F_g - F_b\] where \(F_g\) is the gravitational force acting on the steel ball and can be calculated as: \[F_g = mg\] Now, let's compute the gravitational force: \[F_g = (0.1342 \mathrm{~kg})(9.81 \mathrm{~m} \mathrm{~s}^{-2}) = 1.316 \mathrm{N}\] Compute the net force: \[F_n = 1.316 \mathrm{N} - 0.0156 \mathrm{N} = 1.3004 \mathrm{N}\]

At terminal velocity, the net force acting on the ball is equal to the drag force exerted by the viscous oil according to the equation: \[F_n = 6\pi rηv\] where \(η\) is the viscosity of the oil and \(v\) is the terminal velocity. Rearrange equation to solve for \(v\): \[ v = \frac{F_n}{6\pi r η} \] Now, let's calculate the terminal velocity (\(v\)) of the steel ball: \[ v = \frac{1.3004 \mathrm{N}}{6 \pi (1.6\times10^{-3} \mathrm{~m})(1.64 \mathrm{~Ns} \mathrm{~m}^{-2})} = 0.0247 \mathrm{~m} \mathrm{~s}^{-1}\] To express the result in cm/s, multiply by 100: \[ v = 0.0247 \mathrm{~m} \mathrm{~s}^{-1} \times 100 = 2.47 \mathrm{~cm} \mathrm{~s}^{-1}\] The terminal velocity of the steel ball is approximately \(2.47 \mathrm{~cm} \mathrm{~s}^{-1}\). The closest answer in the given options is (D) \(2.39 \mathrm{cms}^{-1}\).