The Question is from neet-physics mechanical-properties-of-fluids coefficient-of-viscosity-and-viscous-force, and it was created by the author SUBHANI

Question

Two spherical balls of radii $r_1$ and $r_2$ ($r_1 < r_2$) are falling through a viscous fluid with terminal velocities $v_1$ and $v_2$ respectively. If the densities of the balls are the same and Stokes' Law applies, which of the following relationships is correct?

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
 **Primary Concept**: Stokes' Law and Terminal Velocity
 **Related Topics**: Viscous drag, buoyancy, fluid dynamics
 **Prerequisites**: Understanding of forces, viscosity, and terminal velocity.
## SOLUTION PATHWAY
**Given Information**:
  * Two spherical balls with radii $r_1$ and $r_2$ ($r_1 < r_2$)
  * Terminal velocities $v_1$ and $v_2$ respectively
  * Same density of balls
  * Stokes' Law applies
**Method & Steps**:
 1. Understand Stokes' Law and Terminal Velocity:
  Reasoning: When a sphere falls through a viscous fluid, it experiences three forces: gravity ($F_g$), buoyancy ($F_b$), and viscous drag ($F_d$). At terminal velocity, the net force is zero.  Stokes' law gives the viscous drag force: $F_d = 6πηrv$, where η is the dynamic viscosity of the fluid, r is the radius of the sphere, and v is its velocity.
  Formula/Rule:  At terminal velocity, $F_g - F_b - F_d = 0$.
 2. Apply Stokes' Law to both balls:
  Working: For ball 1: $F_{g1} - F_{b1} - 6πηr_1v_1 = 0$. For ball 2: $F_{g2} - F_{b2} - 6πηr_2v_2 = 0$. Since the densities are the same, the gravitational and buoyant forces are proportional to the volume, which is proportional to $r^3$. Thus, $F_{g1} - F_{b1} = k r_1^3$ and $F_{g2} - F_{b2} = k r_2^3$ for some constant k.
  Key Point: The difference between gravitational and buoyant forces is proportional to the cube of the radius.
 3. Determine the relationship between $v_1$ and $v_2$:
  Result: Substituting into the equations from step 2, we get $kr_1^3 = 6πηr_1v_1$ and $kr_2^3 = 6πηr_2v_2$.  Solving for $v_1$ and $v_2$, we get $v_1 = \frac{kr_1^2}{6πη}$ and $v_2 = \frac{kr_2^2}{6πη}$. Therefore, $\frac{v_2}{v_1} = \frac{r_2^2}{r_1^2} = (\frac{r_2}{r_1})^2$.
  Verification: The relationship shows that the terminal velocity is proportional to the square of the radius, which aligns with our understanding of Stokes' Law and the forces involved.
 ## OPTION ANALYSIS
A :x: Incorrect. This suggests a linear relationship between velocity and radius.
B ✓: Correct. This correctly reflects the quadratic relationship between terminal velocity and radius derived from Stokes' Law.
C :x: Incorrect. This is the inverse of the correct relationship.
D :x: Incorrect. This suggests a square root relationship.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Confusing the relationship between drag force and radius:  Remember that the drag force is proportional to the radius, but the net force (and thus terminal velocity) involves the cubic relationship through the volume term.
  2.Ignoring buoyancy: Buoyancy plays a crucial role in determining the net force acting on the sphere.
 **Quick Tips**:
  1.Always start by writing down the relevant equations and forces involved.
  2.Remember the relationship between volume and radius for a sphere ($V = \frac{4}{3}\pi r^3$) and how it affects gravity and buoyancy.

SUBHANI · Answer

$v_2/v_1 = r_2/r_1$

SUBHANI · Answer

The correct Option is "$v_2/v_1 = (r_2/r_1)^2$"

SUBHANI · Answer

$v_2/v_1 = (r_1/r_2)^2$

SUBHANI · Answer

$v_2/v_1 = (r_2/r_1)^{1/2}$

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