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. 10.10

A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter, what is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10^–5 K^–1, steel = 1.2 × 10^–5 K^–1).

Initial temperature, T₁ = 40°C

Final temperature, T₂ = 250°C

Change in temperature, ΔT = (T₂ – T₁) = 210°C

Length of the brass rod at T₁, l₁ = 50 cm

Diameter of the brass rod at T₁, d₁ = 3.0 mm

Length of the steel rod at T₂, l₂ = 50 cm

Diameter of the steel rod at T₂, d₂ = 3.0 mm

Coefficient of linear expansion of brass, α₁ = 2.0 × 10^–5 K^–1

Coefficient of linear expansion of steel, α₂ = 1.2 × 10^–5 K^–1

For the expansion in the brass rod, we have:

(Change in length (Δ l₁)) / (Original length (l₁)) = α₁ ΔT

Therefore,

Δl₁ = (50 x (2.1 x 10^-5) x 210)

=0.2205cm

For the expansion in the steel rod, we have:

(Change in length (Δ l₂)) / (Original length (l₂)) = α₂ ΔT

Therefore,

Δl₂ = (50 x (1.2 x 10^-⁵) x 210)

=0.126cm

Total change in the lengths of brass and steel,

Δl = Δl₁ + Δl₂

= 0.2205 + 0.126

= 0.346 cm

Total change in the length of the combined rod = 0.346 cm

No thermal stress is developed so the ends of rod are free to expand.

Video solution:

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

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