JEE Main 26 Feb 2021 Shift 2 Solved Paper

Given, resistance of wire, R=1Ω
Increase in length, ∆l=25%
Let I1,I2A1,A2,V1 and V2 be the length, area and volume, respectively.
As, volume remains same and V=Al
∴‌‌A1l1=A2l2
⇒‌‌A2=‌

A1I1

I2

a
nd resistance, R=‌

ρl

A

∴‌‌‌

R2

R1

‌‌=‌

ρI2

A2

×‌

A1

ρI1

⇒‌‌‌

R2

R1

‌‌=‌

I2A1

A2l1

=‌

1.25I1⋅A1×I2

A1l1×I1

(∵I2=1.25I1)
⇒‌‌‌

R2

R1

‌‌=‌

(1.25I1)2A1

I12A1

⇒R2=(1.25)2R1
‌ Now, ‌‌

∆R

R

×100‌‌=‌

R2−R1

R1

×100
‌‌=‌

(1.25)2R1−R1

R1

×100
∴‌‌‌

∆R

R

×100‌‌=‌

(1.25)2−1

1

×100
‌‌=56.25%∼eq56%