JEE Main 18 March 2021 Shift 2 Solved Paper

Given, specific resistance for wire 1 ,
ρ1=6Ω−cm
Specific resistance for wire 2,
ρ2=3Ω−cm
Resistance,
R=‌

ρl

A

For parallel connections,
R‌net ‌=‌

R1R2

R1+R2

⇒‌

ρI

2A

=‌

‌ ρ1I A × ρ2I A

ρ1I A +‌ ρ2I A

⇒‌‌‌

ρ

2

=‌

ρ1ρ2

ρ1+ρ2

⇒‌‌‌

ρ

2

=‌

6×3

6+3

ρ=4Ω−cm
Hence, the value of ρ to the nearest integer is 4 .