TS EAMCET 4-Aug-2021 Shift 2 Question Paper

Given, initial length of wire, L1=L
Inital tension, T1=T
Increased tension, T2=T+∆T
Final length, L2=L+∆L
As we know that, Y=‌

Tl0

A∆l

where Y is Young's modulus of elasticity, T∕A is stress, ∆l is change in length and l0 is the natural length.
∴‌‌

T1l0

A(L1−l0)

=‌

T2l0

A(L2−l0)

⇒‌T1(L2−l0)=T2(L1−l0)
⇒‌T1L2−T1l0=T2L1−T2l0
⇒‌T1L2−T2L1=(T1−T2)l0
⇒‌‌l0‌‌=‌

T1L2−T2L1

T1−T2

⇒‌‌l0‌‌=‌

T(L+∆L)−(T+∆T)L

−∆T

‌‌=‌

TL+T∆L−TL−∆TL

−∆T

‌‌=‌

T∆L−∆TL

−∆T

=‌

L∆T−T∆L

∆T