Question

Young’s modules of material of a wire of length ‘L’ and cross-sectional area A is Y. If the length of the wire is doubled and cross-sectional area is halved then Young’s modules will be :

• A. \( \frac{Y}{4} \)
• B. \( 4Y \)
• C. \( Y \)
• D. \( 2Y \)

Answer 1

The Correct Option is C

To address this problem, we will first define Young's modulus and examine its relationship with wire dimensions.

Definition: Young's modulus quantifies a material's resistance to length changes under axial stress. It is calculated using the formula:

\(Y = \frac{FL}{A \Delta L}\)

where:

• \(F\) represents the applied force,
• \(L\) is the initial length,
• \(A\) is the initial cross-sectional area,
• \(\Delta L\) is the resulting change in length.

Analysis:

The problem states that a wire of original length \(L\) and cross-sectional area \(A\) has a Young's modulus of \(Y\). We must determine the effect on Young's modulus if the wire's length is doubled (\(2L\)) and its cross-sectional area is halved (\(\frac{A}{2}\)).

Young's modulus is an intrinsic material property, independent of dimensions, and is constant for a given material under minor deformations. Therefore, altering the wire's dimensions will not impact its Young's modulus.

Conclusion:

Consequently, the Young's modulus remains \(Y\) even when the wire's length is doubled and its cross-sectional area is halved. The correct answer is: \(Y\)