Let I0 be the original length, A be the area of cross-section, α be the coefficient of linear expansion, ∆l be the change in length and Y be the Young's modulus of elasticity. As, ‌‌I1=I0(1+α∆T) ⇒‌‌I1−I0=I0α∆T⇒∆I=I0α∆T Initially, Y=‌
‌ Stress ‌
‌ Strain ‌
=‌
T∕A
∆l∕I0
⇒‌‌Y=‌
T1∕A
(I1−I0)∕I0
. . . (i) Finally, Y=‌
T2∕A
(I2−I0)∕I0
. . . (ii) Now, from Eqs. (i) and (ii), we get ‌‌