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
=
0.1
5
+
0.01
2
+
0.1
10
+2
(0.01)
(5)
⇒
∆ρ
ρ
=0.039
∆ρ
ρ
×100=3.9% The maximum permissible percentage error in the resistivity of the conductor is 3.9%.