AP EAMCET Engineering 21 Sep 2020 Shift 2 Paper

(c) For copper wire, Y=12×1010Nm−2
Length, l=5m,
F=mg=40×10=400N
We know that, Y=‌

Fl

A∆l

⇒‌‌∆l‌‌=‌

Fl

AY

=‌

400×5

πr2×12×1010

∆l‌‌=‌

2000

12πr2×1010

‌‌‌.......(i)
For aluminium wire, Y=7×1010Nm−2 Diameter, d=3mm=3×10−3m
∴‌‌r=‌

d

2

=‌

3×10−3

2

m=1.5×10−3m Similarly, ‌‌Y=‌

Fl

A∆l

⇒‌‌∆l=‌

Fl

AY

‌‌=‌

400×5

π(1.5×10−3)2×7×1010

∆l‌‌=‌

2000

15.75π×104

‌‌‌.....(ii)
According to given condition, ∆l is same in both cases. From Eqs. (i) and (ii), we get
∴‌‌‌

2000

12πr2×1010

‌‌=‌

2000

15.75π×104

12r2‌‌=15.75×10−6
r‌‌=1.15×10−3m=1.15mm
⇒‌‌‌ Diameter, ‌d=2r‌‌=2×1.15=2.3mm