NCERT solution Class 11 thermal properties of matter

NCERT Solutions

Class 11 Physics

Thermal Properties of Matter

NCERT Questions

Q. 10.1
Q. 10.2
Q. 10.3
Q. 10.4
Q. 10.5
Q. 10.6
Q. 10.7
Q. 10.8
Q. 10.9
Q. 10.10
Q. 10.11
Q. 10.12
Q. 10.13
Q. 10.14
Q. 10.15
Q. 10.16
Q. 10.17
Q. 10.18
Q. 10.19
Q. 10.20

Q. 10.3

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?

Given:

R = R_o [1 + α (T – T_o)] … (i)

Where R_o = initial resistance, R = final resistance, T_o = initial temperature and

T = final temperature; α = 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 = R_o [1 + α (T – T_o)]

165.5 = 101.6[1 + α (600.5 - 273.15)]

1.629 = (1 + α (327.35))

Therefore, α = (0.629)/ (327.35)

=1.92 x 10^-3 K^-1

For resistance, R₁ = 123.4 Ω

R₁ = R_o [1 + α (T – T_o)] where T=temperature when the resistance of lead is 123.4 Ω

123.4 = 101.6[1 + (1.92 x 10^-3) (T – 273.15)]

1.214 = (1 + (1.92 x 10^-3) (T – 273.15))

(0.214)/ (1.92 x 10^-3) = (T – 273.15)

Therefore T = 384.61K

Video solution:

Thermal Properties of Matter Class 11 Physics NCERT Chapter 10 NCERT Solutions

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