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. 10.8

A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10^–5 K^–1.

Initial temperature, T₁ = 27.0°C

Diameter of the hole at T₁, d₁ = 4.24 cm

Final temperature, T₂ = 227°C

Diameter of the hole at T₂ = d₂

Co-efficient of linear expansion of copper, α^Cu= 1.70 × 10^–5 K^–1

For co-efficient of superficial expansion β, and change in temperature ΔT, we have the relation:

(Change in area (ΔA))/ (Original Area (A) = β ΔT

(π (d²₂/4) – π (d₁²/4))/ (π (d₁²/4)) = (ΔA)/ (A)

(ΔA)/ (A) = (d²₂ - d²₁)/ (d²₁)

β = 2α

(d²₂ - d²₁)/ (d²₁) = 2 α ΔT

((d²₂ / d²₁) – 1) = 2 α (T₂ – T₁)

((d²₂) / (4.24)²) = 2 x 1.7 x 10^-5(227-21) +1

(d²₂) = 17.98 + 1.0068 =18.1

Therefore,

d₂ = 4.2544cm

Change in diameter = (d₂ – d₁) = (4.2544 – 4.24) = 0.0144 cm

Hence, the diameter increases by 1.44 × 10^–2 cm.

Video solution:

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

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