Question

The temperature coefficient of resistivity of a material is 0.0004 °CWhen the temperature of
material is increased by 50°C, its resistivity increases by 2 x 10-60 m. The initial resistivity
of the material in ohm-metre is​

Answer 1

Hii Dear,

◆ Answer -

ρ0 = 10^-58 ohm-m

● Explaination -

# Given -

α = 0.0004 /℃

∆ρ = 2×10^-60 ohm-m

∆T = 50 ℃

# Solution -

Increase in resistivity of material is given as -

∆ρ = α.ρ0.∆T

Rearranging we get -

ρ0 = ∆ρ/(α.∆T)

ρ0 = 2×10^-60 / (0.0004 × 50)

ρ0 = 10^-58 ohm-m

Therefore, initial resistivity of the material is 10^-58 ohm-m.

Hope this is helpful...

Answer 2

Answer:

100×10^-8

Explanation:given

α=0.0004/°C

ρ0=2×10^-60 ohm-m

∆T=50°C

∆ρ=α×ρ0×∆T

ρ0= ∆ρ/α×∆T

=2×10^-60/0.0004×50

=2×10^-60/0.02

=2×10^-60/2×10^-2

=10^-60×10^2

= 10^-58