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

Question

If the length of a wire is doubled and its radius is halved, keeping the force constant, how does the elongation change?

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [Option C]
## CONCEPT FRAMEWORK
	**Primary Concept**: Hooke's Law and the relationship between elongation, length, radius, and force in a wire
	**Related Topics**: Young's modulus, material properties of solids
	**Prerequisites**: Understanding of stress, strain, and elasticity
## SOLUTION PATHWAY
**Given Information**:
		- Initial length of wire: $L$
		- Initial radius of wire: $r$
		- Final length of wire: $2L$ (doubled)
		- Final radius of wire: $\frac{r}{2}$ (halved)
		- Force applied: $F$ (constant)
**Method & Steps**:
	1.	Determine the original area of cross-section:
		Reasoning: The area of a circle is given by $A = \pi r^2$.
		Formula/Rule: $A_1 = \pi r^2$
	2.	Determine the new area of cross-section after modification:
		Working: New radius is $\frac{r}{2}$, so the new area is:
		$$A_2 = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} = \frac{1}{4} A_1$$
		Key Point: The area has decreased to a quarter of the original area.
	3.	Using Hooke's Law, relate elongation ($\Delta L$) to the parameters:
		$$\Delta L = \frac{F L}{A Y}$$
		Where $Y$ is Young's modulus. Substituting for both the original and modified conditions:
		$$\Delta L_{original} = \frac{F L}{A_1 Y}$$
		$$\Delta L_{new} = \frac{F (2L)}{A_2 Y}$$
		Substituting $A_2$:
		$$\Delta L_{new} = \frac{F (2L)}{(\frac{1}{4} A_1) Y} = \frac{8 F L}{A_1 Y} = 8 \Delta L_{original}$$
		Result: The elongation increases by 8 times.
		Verification: Since both force and Young's modulus remain constant, the elongation is directly proportional to the length and inversely proportional to the cross-sectional area.
## OPTION ANALYSIS
[A] :x: Increases by 2 times: This does not consider the effects of both the length and area changes.
[B] :x: Increases by 4 times: This only considers the area change, not the length change.
[C] ✓: Increases by 8 times: Correct, as derived from the calculations above.
[D] :x: Remains the same: Incorrect, as elongation changes with length and area adjustments.
## LEARNING AIDS
	**Common Mistakes**:
		1. Confusing the effects of changes in length and radius: Always analyze how both parameters interact.
		2. Neglecting the relationship between force, area, and elongation: Remember that they are interconnected through Young's modulus.
	**Quick Tips**:
		1. Always write down the original and modified parameters before solving.
		2. Use a dimensional analysis to check if the units make sense in the context of the problem.

P.NARASIMHA · Answer

Increases by 2 times

P.NARASIMHA · Answer

Increases by 4 times

P.NARASIMHA · Answer

The correct Option is "Increases by 8 times"

P.NARASIMHA · Answer

Remains the same

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