TS EAMCET 18-July-2022 Shift 2 Solved Paper

A wire is stretched such that its resistance increases by 6% .
If we assume initial resistance to be 100Ω Then the new resistance will be given by
R‌new ‌=100Ω+‌

6

100

×100Ω=106Ω
We know that, the resistance is given by
R=ρ‌

I

A

‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
Where, aréa (A) can be expressed in terms of volume (V) is
A=‌

V

l

Equation (i) then gives us, R=ρ‌

l2

V

Since, the resistivity and volume are going to remain the same, only change in length will be observed.
So, ‌‌l2=‌

RV

ρ

‌‌........ (ii) (Original length)
Let new length be l‌new ‌ , so
l‌new ‌2=‌

R‌new ‌V

ρ

‌‌‌⋅⋅⋅⋅⋅⋅⋅(iii)
Dividing Eq. (iii) by Eq. (ii), we get
‌

l‌new ‌2

l2

=‌

R‌new ‌

R

=‌

106

100

‌. ‌
⇒‌‌l‌new ‌=1.029l
% change in length
‌=(‌

l‌new ‌−l

l

)×100%=(‌

1.029l−l

l

)×100
‌=0.029×100=29%
‌‌≈3%