# Thermal Expansion - Long Question Answer | Physics Grade 11 Solution

### Long Question Solution # Thermal Expansion - Long Question Answer 2

Question: Define linear and cubical expansivity of solids. Establish a relation between the coefficient of linear and cubical expansions.

Ans: Linear expansivity (α): The coefficient of linear expansion is defined as the ratio of the increase in length per unit original length per degree rise in temperature. It is denoted bα and is given by
α = (l2-l1) / l2(θ21)
= ∆l / l1∆θ
Where, ∆l is the change in length, lis original length and ∆θ is the change in temperature
Its unit is K-1 or oC-1

Cubical expansion (g): The coefficient of cubical expansion is defined as the fractional increase in volume per unit change in temperature. It is denoted by g and is given by
γ = (V2-V1)/V121)
= ∆V/(V1∆θ)
where, ∆V is the change in volume, V1 is original volume and ∆θ is the change in temperature
Its unit is K-1 or oC-1

Relation between a and g Let us consider a metal cube of length l1 and its volume V1 at the initial temperature θ1. When it is heated up to higher temperature θ2 its length becomes l2 and volume becomes V2 after expansion. Hence expansion for length and volume at higher temperature can be written as
l2 = l1 (1 + α ∆θ)  …....(1)
V2 = V1 (1 + γ ∆θ)  …....(2)
Also, volume of metal cube at θand θ2 can be written as
V1 = l13  …....(3)
V2 = l23   ….....(4)

Putting the value of l2 from eqn. 1 in eqn.4 we get,
V= {l1 (1 + α ∆θ)}3
= l1(1 + α ∆θ)3
= l1(1 + 3α2∆θ + 3α(∆θ)2 + α3(∆θ)3)

Since the value of α is small that is of the order of 10-5, hence of α2 and α3 are very small and that can be neglected so the above equation can be written as:
V= l1(1 + 3α∆θ) .....…(5)
Putting the value of V2 from eqn.2 to eqn.5 we get,
V1 (1 + γ ∆θ) = l1(1 + 3α∆θ)
or  l1(1 + γ ∆θ) = l1(1 + 3α∆θ)
or 1 + γ ∆θ = 1 + 3α∆θ
or γ ∆θ = 3α∆θ
γ
Thus, the coefficient of cubical expansion of a solid is three times its coefficient of linear expansion.

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