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### Thermal Properties of Matter NCERT Solutions Class 11 Physics - Solved Exercise Question 11.3

Question 11.3:
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
R = Ro [1 + α (T To)]
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:
It is given that:
R = R0 [1 + α (T T0)] … (i)
Where,
R0 and T0 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, T0 = 273.15 K
Resistance of lead, R0 = 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 = Ro [1 + α (T To)]
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, R1 = 123.4 Ω
R1 = R0 [1 + α (T T0)]
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