Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Two small spherical metal balls, having equal masses, are made from materials of densities $\rho_1$ and $\rho_2$ such that $\rho_1 = 8\rho_2$ and having radii of $1 \text{ mm}$ and $2 \text{ mm}$ , respectively. They are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals $\eta$ and whose density is $0.1\rho_2$ . The ratio of their terminal velocities would be:

\frac{79}{72}

\frac{19}{36}

\frac{39}{72}

\frac{79}{36}

Step-by-Step Solution

Terminal Velocity Formula: The terminal velocity ( $v_t$ ) of a spherical body falling through a viscous fluid is given by Stokes' Law: $v_t = \frac{2r^2 (\rho - \sigma)g}{9\eta}$ where $r$ is the radius of the sphere, $\rho$ is the density of the sphere, $\sigma$ is the density of the fluid, and $\eta$ is the coefficient of viscosity .
Parameters for Ball 1:

Radius $r_1 = 1 \text{ mm}$
Density $\rho_1 = 8\rho_2$
Fluid density $\sigma = 0.1\rho_2$ $v_1 \propto r_1^2 (\rho_1 - \sigma) = (1)^2 (8\rho_2 - 0.1\rho_2) = 7.9\rho_2$

Parameters for Ball 2:

Radius $r_2 = 2 \text{ mm}$
Density $\rho_2 = \rho_2$
Fluid density $\sigma = 0.1\rho_2$ $v_2 \propto r_2^2 (\rho_2 - \sigma) = (2)^2 (\rho_2 - 0.1\rho_2) = 4(0.9\rho_2) = 3.6\rho_2$

Calculate Ratio: $\frac{v_1}{v_2} = \frac{7.9\rho_2}{3.6\rho_2} = \frac{79}{36}$

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