Given, The change in the length of the wire A,∆LA=2mm=0.002m The change in the length of the wire B,∆LB=4mm=0.004m The force subjected to the wire, F=2N The radius of the wire B is 4 times the radius of the wire A, i.e.,
rB
rA
=
4
1
Since, the wire is made of the same material, so the Young's modulus of the elasticity of the wire is same. ⇒YA=YB Using Hooke's law, Stress =Y (Strain)
F
A
=Y(
∆L
L
) ⇒L=
Y∆LA
F
⇒
LA
LB
=
YA
YB
×
∆LA
∆LB
×
AA
AB
×
FB
FA
⇒
LA
LB
=
YA
YB
×
0.002
0.004
×
πrA2
πrB2
×
2
2
⇒
LA
LB
=
0.002
0.004
×
rA2
16rA2
⇒
LA
LB
=
a
b
=
1
32
Comparing this equation with 1∕x, we get the value of the x is 32 .