AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

The expression for the extension of the wire is given by,
∆L=‌

FL

YA

∆L=‌

4F

Yπ

×‌

L

D2

∆L∝‌

L

D2

Substitute the values given in the option one by one.
For option (1), the extensionis calculated as
∆L‌‌=‌

4F

Yπ

×‌

0.5

(0.5×10−3)2

‌‌=‌

4F

Yπ

×2×106
Similarly, for option (2), the extension is,
∆L=‌

4F

Yπ

×1.0×106
For option (3), the extension is,
∆L=‌

4F

Yπ

×0.5×106
For option (4), the extension is,
∆L=‌

4F

Yπ

×0.33×106
The highest value of extension is associated with the option (1)