The Question is from neet-physics , and it was created by the author P.NARASIMHA

Question

A small air bubble of radius $r$ rises steadily through a liquid of density $\rho$ and viscosity $\eta$ at a constant speed $v$. If $g$ is the acceleration due to gravity and the surface tension of the liquid is $T$, which of the following expressions is dimensionally consistent for the viscous force acting on the bubble? (Assume the bubble is small enough that its shape remains spherical.)

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
 **Primary Concept**: Dimensional analysis
 **Related Topics**: Stokes' law, viscous force, surface tension
 **Prerequisites**: Understanding of dimensions of physical quantities (length, mass, time)
## SOLUTION PATHWAY
**Given Information**:
  Radius of bubble = $r$ [L]
  Density of liquid = $ho$ [$ML^{-3}$]
  Viscosity of liquid = $\eta$ [$ML^{-1}T^{-1}$]
  Speed of bubble = $v$ [$LT^{-1}$]
  Acceleration due to gravity = $g$ [$LT^{-2}$]
  Surface tension = $T$ [$MT^{-2}$]
**Method & Steps**:
 1. Determine the dimensions of each option.
  Reasoning: We need to find which option has the dimensions of force, which is $[MLT^{-2}]$.
 2. Analyze the dimensions of each option:
  Option A: $\frac{4\pi r^2 T}{3}$ has dimensions $[L^2][MT^{-2}] = [ML^2T^{-2}]$, which is incorrect.
  Option B: $6\pi \eta rv$ has dimensions $[ML^{-1}T^{-1}][L][LT^{-1}] = [MLT^{-2}]$, which is the dimension of force.
  Option C: $\frac{4}{3} \pi r^3 ho g$ has dimensions $[L^3][ML^{-3}][LT^{-2}] = [ML^2T^{-2}]$, which is incorrect.
  Option D: $4\pi r^2 T$ has dimensions $[L^2][MT^{-2}] = [ML^2T^{-2}]$, which is incorrect.
 3. Identify the dimensionally consistent expression.
  Result: Only option B has the dimensions of force ($[MLT^{-2}]$).
  Verification: The viscous force on a sphere moving through a fluid is given by Stokes' law: $F = 6\pi \eta rv$, which has the correct dimensions.
## OPTION ANALYSIS
A :x: Incorrect dimensions. This represents a surface tension force.
B ✓: Correct dimensions. This is consistent with Stokes' law for viscous drag.
C :x: Incorrect dimensions. This represents a buoyant force.
D :x: Incorrect dimensions. This represents a surface tension force.
## LEARNING AIDS
 **Common Mistakes**:
  1.Confusing the dimensions of viscosity and surface tension: Remember that viscosity has dimensions of $[ML^{-1}T^{-1}]$ while surface tension has dimensions of $[MT^{-2}]$.
  2.Forgetting to check the dimensions of all the constants: Pi ($\pi$) is dimensionless, and numerical constants don't affect the overall dimension.
 **Quick Tips**:
  1.Always write down the dimensions of each physical quantity before starting the calculation.
  2.Use a consistent system of units (e.g., SI units) to avoid confusion.

P.NARASIMHA · Answer

$\frac{4\pi r^2 T}{3}$

P.NARASIMHA · Answer

The correct Option is "$6\pi \eta rv$"

P.NARASIMHA · Answer

$\frac{4}{3} \pi r^3 \rho g$

P.NARASIMHA · Answer

$4\pi r^2 T$

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