JEE Main 26 Feb 2021 Shift 2 Solved Paper

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
‌‌

T1∕A

(I1−I0)∕I0

‌‌=‌

T2∕A

(I2−I0)∕I0

⇒‌‌

T1

I1−I0

=‌‌‌

T2

I2−I0

⇒‌T1I2−T1I0‌‌=T2I1−T2I0
⇒‌T1I2−T2I1‌‌=(T1−T2)I0
⇒‌I0=‌

T1I2−T2I1

T1−T2

‌ or ‌‌I0=‌

T2I1−T1I2

T2−T1