TS EAMCET 5-Aug-2021 Shift 1 Question Paper

Given,
Initial resistance of wire be R,
Initial length of wire be l,
Final resistance of wire be R′
and final length of wire be l′.
∵‌‌R=‌

ρl

A

where, R is resistance,
l is length,
ρ is resistivity
and A is area of cross-section.
As we know that, on stretching wire, volume remains the same.
∴ Change in volume, ∆V= constant
⇒Area(A)× Length (l)= constant
By using percentage error calculation,
‌

∆A

A

+‌

∆l

l

‌‌=0
‌

∆A

A

‌‌=−‌

∆l

l

. . . (i)
From Eq. (i), we get
⇒‌‌‌

∆R

R

×100=(‌

∆ρ

ρ

+‌

∆l

l

−‌

∆A

A

)×100
⇒‌‌‌

∆R

R

×100=(0+‌

∆l

l

+‌

∆l

l

)×100
4=2‌

∆l

l

⇒‌

∆l

l

=2%