The Question is from neet-physics mechanical-properties-of-fluids work-and-surface-energy, and it was created by the author Mr. ANIL KUMAR.G

Question

A liquid drop of radius 'r' breaks into 'n' identical smaller drops. If the surface tension of the liquid is 'T', the change in surface energy is:

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [B]
## CONCEPT FRAMEWORK
	**Primary Concept**: Surface Energy and Surface Tension
	**Related Topics**:  Surface Area, Volume Conservation
	**Prerequisites**: Basic understanding of surface tension and geometry of spheres.
## SOLUTION PATHWAY
**Given Information**:
	Initial radius of the drop: r
	Number of smaller drops: n
	Surface tension of the liquid: T
	Smaller drops are identical
**Method & Steps**:
	1. Calculate the initial surface area of the large drop.
		Reasoning: The surface area of a sphere is given by $4\pi r^2$.
		Formula/Rule:  $A_{initial} = 4\pi r^2$
	2. Calculate the volume of the large drop.
		Reasoning: The volume of a sphere is given by $\frac{4}{3}\pi r^3$.
		Formula/Rule: $V_{initial} = \frac{4}{3}\pi r^3$
	3. Calculate the volume of a single small drop.
		Reasoning: The total volume remains constant.  The volume of the large drop is equal to n times the volume of a small drop.
		Formula/Rule: $V_{small} = \frac{V_{initial}}{n} = \frac{4}{3}\pi \frac{r^3}{n}$
	4. Calculate the radius of each small drop.
		Reasoning: Let the radius of each small drop be $r'$. Volume of small drop is $ \frac{4}{3}\pi r'^3 = \frac{4}{3}\pi \frac{r^3}{n}$ 
		Formula/Rule: $r' = \frac{r}{n^{1/3}}$
	5. Calculate the surface area of each small drop.
		Reasoning: Surface area of a sphere is $4\pi r^2$. The radius of the small drop is $r'$.
		Formula/Rule: $A_{small} = 4\pi r'^2 = 4\pi (\frac{r}{n^{1/3}})^2 = 4\pi \frac{r^2}{n^{2/3}}$
	6. Calculate the total surface area of the n small drops.
		Reasoning: The total surface area is the surface area of one small drop multiplied by the number of drops.
		Formula/Rule: $A_{final} = n A_{small} = n(4\pi \frac{r^2}{n^{2/3}}) = 4\pi r^2 n^{1/3}$
	7. Calculate the change in surface area.
		Reasoning: Change in surface area is the final surface area minus the initial surface area.
		Formula/Rule: $\Delta A = A_{final} - A_{initial} = 4\pi r^2 n^{1/3} - 4\pi r^2 = 4\pi r^2(n^{1/3} - 1)$
	8. Calculate the change in surface energy.
		Reasoning: Change in surface energy is surface tension multiplied by the change in surface area.
		Formula/Rule: $\Delta E = T \Delta A = 4\pi r^2 T(n^{1/3} - 1)$
 ## OPTION ANALYSIS
[A] :x: Incorrect, the change in surface energy should be multiplied by T and the power of n should be 1/3, not 1.
[B] ✓: Correct, the change in surface energy is given by 4\pi r^2 T(n^{1/3} - 1)
[C] :x: Incorrect, the power of n is incorrect.
[D] :x: Incorrect, the power of n is incorrect and also the sign of 1 is incorrect.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Incorrectly calculating the radius of the smaller drops]: [Make sure to use volume conservation to find the radius of the smaller drops and not assume it to be r/n]
		2.[Forgetting to multiply by 'n' when calculating the total surface area of smaller drops]: [Remember that there are 'n' small drops, so you need to multiply the surface area of one small drop by 'n'.]
	**Quick Tips**:
		1.[Remember that the volume is conserved when a drop breaks into smaller drops]: [Use this to find the radius of the smaller drops]
		2.[Surface energy is directly proportional to the surface area]: [The change in surface energy is equal to the surface tension multiplied by the change in surface area.]

Mr. ANIL KUMAR.G · Answer

The correct Option is "4\pi r^2 T(n^{1/3} - 1)"

Mr. ANIL KUMAR.G · Answer

4\pi r^2 T(n^{2/3} - 1)

Mr. ANIL KUMAR.G · Answer

4\pi r^2 T(n - 1)

Mr. ANIL KUMAR.G · Answer

4\pi r^2 T(n^{-1/3} - 1)

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