### Q.3:- The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:

*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?

**Answer:-**It is given that:

*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 × 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 × 10

^{-3}( T – 273.15) ]

Solving for

*T*, we get

T = 384.61 K.