AP EAPCET 04-Jul-2022 Shift 1 Paper

We know that, young's modulus,
Y=‌

Fl

A∆l

Where,
A= Area of cross-section,
l= length of wire and
∆l= elongation in the longest wire. Here, A=π˙r2
Here, A=π˙r2
=π(‌

d

2

)2=π‌

d2

4

‌‌(∵d=2r)
‌∴‌‌Y=‌

Fl

πd2 4 ∆l

‌⇒∆l=‌

4Fl

πd2Y

=(4‌

F

πY

)‌

l

d2

⇒∆l=k‌

l

d2

where, k=‌

4F

πy

= constant
For wire given in option (a)
(∆l)1=k⋅‌

50

(0.05)2

=20000k
similarly,
For wire given in option (b)
(∆l)2=k×‌

200

(0.)2

=5000k
For wire given in option (c)
∆l3=k⋅‌

300

(0.3)2

=3333.33k
For wire given in option (d)
∆l4=‌

100

(0.1)2

=10000k
Hence, we see that wire given in option (a) has maximum elongation.