BITSAT 2021 Paper

Let we have a wire of length l, cross-sectional area A, and Resistivity ρ
Then Resistance of wire is
R=ρ‌

l

A

−(1)
The volume of cylindrical wire is the product of its length and cross-sectional area.
V=I.A
If length is increased by 10 percent, the new length l' is
I′=∣+10%‌ of ‌I=1.1∣−(2)
After stretching the cross-sectional area will also change, as volume remains constant. Let the new cross-sectional area be A'
V=I.A=I′A′...‌ (3) ‌
Putting (2) in (3) we have
I⋅A=1.1I⋅A′
A′=‌

A

1.1

....(4)
So, the Resistance of this stretched wire is
R′=ρ‌

l′

A′

⋯‌ (5) ‌
Putting (2) and (4) in (5)
R′=ρ(‌

1.1l

A∕1.1

)
⇒R′=ρ(‌

1.1×1.1l

A

)
⇒R′=ρ(‌

1.21l

A

)
⇒R′=1.21(ρ‌

l

A

)−(6)
Comparing Eq(6) and Eq(1) We get R′=1.21R
So, the resistance increases 1.21 times.