The Question is from neet-physics , and it was created by the author SUBHANI

Question

A soap bubble with radius $r$ has an internal pressure $P$.  If the radius doubles to $2r$, what happens to the internal pressure?

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D. Decreases
## CONCEPT FRAMEWORK
 **Primary Concept**: Pressure inside a soap bubble and its relation to radius.
 **Related Topics**: Laplace's Law, Surface Tension
 **Prerequisites**: Understanding of inverse proportionality and Laplace's Law.
## SOLUTION PATHWAY
**Given Information**:
  Initial radius: $r$
  Initial internal pressure: $P$
  Final radius: $2r$
**Method & Steps**:
 1. According to Laplace's law, the pressure difference across a soap bubble's surface is inversely proportional to its radius: $ΔP = \frac{4γ}{r}$, where $γ$ is the surface tension.
  Reasoning:  Since the external pressure remains constant, the change in $ΔP$ reflects the change in internal pressure.
  Formula/Rule: $ΔP = \frac{4γ}{r}$
 2. If the radius doubles ($2r$), the pressure difference becomes $ΔP' = \frac{4γ}{2r} = \frac{1}{2} \frac{4γ}{r} = \frac{1}{2}ΔP$.
  Working:  The pressure difference is halved.
  Key Point:  Assuming constant external pressure, the internal pressure also decreases.
 3. Therefore, when the radius doubles, the internal pressure decreases (though not necessarily halved, as we're considering the absolute pressure, not just the pressure difference).
  Result: The internal pressure decreases.
  Verification:  The inverse relationship in Laplace's law confirms the decrease in pressure.
 ## OPTION ANALYSIS
A :x: Doubling the radius would halve the pressure difference, not double the internal pressure.
B :x: The internal pressure more than halves; it decreases by a factor related to the pressure difference.
C :x: The pressure difference is inversely proportional to the radius; therefore, it cannot remain the same.
D ✓: Correctly identifies that the internal pressure decreases as the radius increases.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Direct proportionality: Mistaking the relationship between radius and pressure as direct instead of inverse.
  2.Ignoring external pressure: Failing to consider that the external pressure remains constant.
 **Quick Tips**:
  1.Focus on the inverse relationship between radius and pressure difference.
  2.Remember that the internal pressure is always slightly greater than the external pressure.

SUBHANI · Answer

Doubles

SUBHANI · Answer

Halves

SUBHANI · Answer

Remains the same

SUBHANI · Answer

The correct Option is "Decreases"

NEET Physics MCQs

Select Chapter

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

NEET Subjects

Our Bots

Company