JEE Main 18 March 2021 Shift 1 Solved Paper

Given, the potential difference applied across the ends of the conductor, V=5V
The length of the conductor, L=10cm
The measured value of the current in the conductor, I=2A
The diameter of the conductor, d=5mm
As we know that, R=‌

ρl

A

Using Ohm's law,
V=IR⇒R=‌

V

I

‌

ρl

A

=‌

V

l

⇒ρ=‌

V

ll

(‌

πd2

4

)
In error form,
‌

∆ρ

ρ

=‌

∆V

V

+‌

∆l

l

+‌

∆l

l

+2‌

∆d

d

‌

∆ρ

ρ

×100=3.9%
The maximum permissible percentage error in the resistivity of the conductor is 3.9%.