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 :
- \( \frac{Y}{4} \)
- \( 4Y \)
- \( Y \)
- \( 2Y \)
The Correct Option is C
Solution and Explanation
To solve this question, let's first understand the basic concept of Young's modulus and how it is affected by changes in the dimensions of a wire.
Concept: Young's modulus is a measure of the ability of a material to withstand changes in length when under lengthwise tension or compression. It is defined by the formula:
\(Y = \frac{FL}{A \Delta L}\)
where:
- \(F\) is the force applied,
- \(L\) is the original length,
- \(A\) is the original cross-sectional area,
- \(\Delta L\) is the change in length.
Analysis:
The question provides that Young’s modulus of a material is \(Y\) for a wire of original length \(L\) and cross-sectional area \(A\). If the length of the wire is doubled (\(2L\)) and the cross-sectional area is halved (\(\frac{A}{2}\)), we need to determine how Young's modulus will change.
Since Young's modulus is a property of the material itself and is only dependent on the material properties (not on its dimensions), changing the dimensions of the wire will not affect Young's modulus. Young’s modulus remains constant for a given material under small deformations.
Conclusion:
Thus, even after doubling the length and halving the cross-sectional area of the wire, Young’s modulus remains \(Y\). Therefore, the correct answer is: \(Y\)
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