Refer figure, let x be the depression at the mid point i.e. CD=x.
In figure, AC=CB=l=0.5m;
m=100g=0.100 kg,
AD=BD=(l2+x2 )1∕2
Increase in length,
Δl=AD+DB–AB=2AD–AB
=2(l2+x2)1∕2−2l
=2l(1+

x2

l2

)1∕2−2l
=2l[1+

x2

2l2

]−2l
=

x2

l

∴ Strain =

Δl

2l

=

x2

2l2

If T is the tension in the wire, then 2T‌cos‌θ=mg or T=

mg

2‌cos‌θ

Here, cos‌θ=

x

(l2+x2)1∕2

=

x

l(1+ x2 l2 )1∕2

=

x

l (1+ 1 2 x2 l2 )

As, x<<l, so 1>>

1

2

x2

l2

and 1+

1

2

x2

l2

≈1∴cos‌θ=

x

l

Hence, T=

mg

2( x l )

=

mgl

2x

,
Stress =

T

A

=

mgl

2Ax

Y=

‌stress

‌strain

=

mgl

2Ax

×

2l2

x2

=

mgl3

Ax3

∴x=l[

mg

YA

]1∕3
=0.5[

0.1×10

20×1011×0.5×10−6

]1∕3
=1.074×10−2
=1.074cm