MHTCET 10 May 2024 Shift 2 Paper

Concept:Terminal velocity of a sphere in a viscous liquid depends on the square of its radius and the difference between the density of the sphere and the liquid.For equal terminal velocities, the quantity $r^2(\rho - \sigma)$ must be the same for both spheres.Explanation:Terminal velocity is given by:$$v = \frac{2r^2(\rho - \sigma)g}{9\eta}$$Here, $\rho$ is the density of the sphere, $\sigma$ is the density of the liquid, and $\eta$ is the viscosity of the liquid.Since both spheres have the same uniform speed, their terminal velocities are equal.Thus, $r_1^2(\rho_1 - \sigma) = r_2^2(\rho_2 - \sigma)$.Rearranging gives:$$\frac{r_1}{r_2} = \sqrt{\frac{\rho_2 - \sigma}{\rho_1 - \sigma}}$$Substitute the given densities:$$\frac{r_1}{r_2} = \sqrt{\frac{8.5 \times 10^3 - 2.5 \times 10^3}{11.5 \times 10^3 - 2.5 \times 10^3}}$$$$= \sqrt{\frac{6}{9}} = \sqrt{\frac{2}{3}}$$Answer:The ratio of the radius of the first sphere to that of the second sphere is $\sqrt{\frac{2}{3}}$.Therefore, the correct option is B.