Question

The resistance of a wire at 20 degree celsius and 100 degree celsius is 3 ohm-m and 4 ohm-m respectively. The resitivity of the wire at 0 degree celsius is?

Answer 1

answer : option (1) 11/4 ohm.m

explanation : relation between resistivity and temperature is given by,

$\rho=\rho_0[1+\alpha\Delta(T-T_0)]$

where $\rho$ is resistivity at temperature T, $\rho_0$ is resistivity at temperature $T_0$ and $\alpha$ is coefficient of temperature.

Let $\rho_0$ is initial resistivity at 0°C then,

at 20°C , $\rho=3\Omega.m$

$3=\rho_0[1+\alpha(20-0)]$....(1)

at 100°C, $\rho=4\Omega.m$

$4=\rho_0[1+\alpha(100-0)]$....(2)

from equations (1) and (2),

4 - 3 = 1 = $80\rho_0\alpha$...(3)

and 3/4 = [1 + 20α]/[1 + 100α]

or, 3 + 300α = 4 + 80α

or, 220α = 1

or, α = 1/220 , putting it in equation (3),

1 = 80 × $\rho_0$ × 1/220

220/80 = 11/4 = $\rho_0$

hence, option (1) is correct choice.

Answer 2

Answer:

answer : option (1) 11/4 ohm.m

explanation : relation between resistivity and temperature is given by,

\rho=\rho_0[1+\alpha\Delta(T-T_0)]ρ=ρ

0

[1+αΔ(T−T

0

)]

where \rhoρ is resistivity at temperature T, \rho_0ρ

0

is resistivity at temperature T_0T

0

and \alphaα is coefficient of temperature.

Let \rho_0ρ

0

is initial resistivity at 0°C then,

at 20°C , \rho=3\text{\O}mega.mρ=3Ømega.m

3=\rho_0[1+\alpha(20-0)]3=ρ

0

[1+α(20−0)] ....(1)

at 100°C, \rho=4\text{\O}mega.mρ=4Ømega.m

4=\rho_0[1+\alpha(100-0)]4=ρ

0

[1+α(100−0)] ....(2)

from equations (1) and (2),

4 - 3 = 1 = 80\rho_0\alpha80ρ

0

α ...(3)

and 3/4 = [1 + 20α]/[1 + 100α]

or, 3 + 300α = 4 + 80α

or, 220α = 1

or, α = 1/220 , putting it in equation (3),

1 = 80 × \rho_0ρ

0

× 1/220

220/80 = 11/4 = \rho_0ρ

0

hence, option (1) is correct choice