NEET UG PHYSICS MCQ: MECHANICAL PROPERTIES OF FLUIDS

Solution approach

1. Correct Answer + Core Reason: $27:8$. The ratio of volumes is the cube of the ratio of radii, which is related to the terminal velocities.
   
   Step-by-Step Solution:
   
   Step 1: The terminal velocity $v_t$ of a raindrop is given by the formula $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$, where $r$ is the radius of the raindrop, $\rho$ is the density of the raindrop, $\sigma$ is the density of air, $g$ is the acceleration due to gravity, and $\eta$ is the coefficient of viscosity of air.
   
   Step 2: Since the raindrops are made of the same material and falling through the same medium, $\rho$, $\sigma$, $g$, and $\eta$ are constant. Therefore, $v_t \propto r^2$.
   
   Step 3: Let $v_{t1}$ and $v_{t2}$ be the terminal velocities of the two raindrops, and $r_1$ and $r_2$ be their radii. We are given that $\frac{v_{t1}}{v_{t2}} = \frac{9}{4}$.
   
   Step 4: Since $v_t \propto r^2$, we have $\frac{v_{t1}}{v_{t2}} = \frac{r_1^2}{r_2^2}$. Therefore, $\frac{r_1^2}{r_2^2} = \frac{9}{4}$, which means $\frac{r_1}{r_2} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.
   
   Step 5: The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$. Let $V_1$ and $V_2$ be the volumes of the two raindrops. Then $\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \frac{r_1^3}{r_2^3}$.
   
   Step 6: Substituting $\frac{r_1}{r_2} = \frac{3}{2}$, we get $\frac{V_1}{V_2} = (\frac{3}{2})^3 = \frac{27}{8}$.
   
   Why other options are wrong:
   - Option A: $3:2$ is the ratio of the radii, not the volumes.
   - Option B: $4:9$ is the inverse of the ratio of the squares of the radii (or the inverse of the ratio of the terminal velocities).
   - Option C: $9:4$ is the ratio of the terminal velocities, not the volumes.
   
   Key Concepts:
   - Terminal Velocity
   - Stokes' Law
   - Volume of a Sphere
   
   Memory Aids:
   - $v_t \propto r^2$ and $V \propto r^3$
   
   Quick Tips:
   - Remember the relationship between terminal velocity and radius.