__Question 11.3__:*R*=

*R*

_{o}[1 + α (

*T*–

*T*

_{o})]

The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?

__Solution__:*R*=

*R*

_{0}[1 + α (

*T*–

*T*

_{0})] … (

*i*)

Where,

*R*

_{0}and

*T*

_{0}are the initial resistance and temperature respectively

*R*and

*T*are the final resistance and temperature respectively

α is a constant

At the triple point of water,

*T*

_{0}= 273.15 K

Resistance of lead,

*R*

_{0}= 101.6 Ω

At normal melting point of lead,

*T*= 600.5 K

Resistance of lead,

*R*= 165.5 Ω

Substituting these values in equation (

*i*), we get:

*R*=

*R*

_{o}[1 + α (

*T*–

*T*

_{o})]

165.5 = 101.6 [ 1 + α(600.5 - 273.15) ]

1.629 = 1 + α (327.35)

∴ α = 0.629 / 327.35 = 1.92 X 10

^{-3}K

^{-1}

For resistance,

*R*

_{1}= 123.4 Ω

*=*

*R*_{1}*R*

_{0}[1 + α (

*T*–

*T*

_{0})]

Where T is the temperature when the resistance of lead is 123.4 Ω

123.4 = 101.6 [ 1 + 1.92 X 10

^{-3}( T - 273.15) ]

Solving for T, we get

T = 384.61 K

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