Question

derive a relation between coefficient of linear expansion and coefficient of cubical expansion

Answer 1

original length=l
change in length=Δl
original volume=v
change in volume=Δv
initial volume=l³
final volume(v+Δv)=(l+Δl)³
change in volume=final volume-initial voluma
                          Δv=l³+Δl³+3lΔl(l+Δl)-l³
                             Δv=l³+Δl³+3lΔl²+3l²Δl-l³
Δl³is neglected bcoz it is too small
Δv=3l²Δl+3lΔl²
we will also ignore 3lΔl²
Δv=3l²Δl
we know that
coefficient of volumetric expansion
y=Δv/v×Δt
y=3l²Δl/l³×Δt
y=3Δl/l×Δt                                             {Δl/l×Δt=α}
y=3α

Answer 2

Answer:

Explanation:

Consider a metallic sheet of area A₁ and Length L₁ at temperature θ₁.

∴A₁=L₁²

On heating to θ₂, its area becomes A₂ and length becomes L₂.

∴A₂=L₂²

We have,

L₂=L₁(1+αΔθ)

Here, α = Linear Expansivity

L₂²=L₁²(1+αΔθ)²

L₂²=L₁²(1+2αΔθ+α²Δθ²)

Here, α is a small quantity and higher power of it is very small which can be neglected.

∴L₂²=L₁²(1+2αΔθ)

A₂=A₁(1+2αΔθ)___(1)

Also,

A₂=A₁(1+βΔθ)___(2)

Here, β = Superficial Expansivity

​Comapring (1) and (2), we get,

β=2α

∴α=β2

Relation between Linear And Cubical Expansivities

Consider a metallic cube of volume V₁ and Length L₁ at temperature θ₁.

∴V₁=L₁³

On heating to θ₂, its volume becomes V₂ and length becomes L₂.

∴V₂=L₂³

We have,

L₂=L₁(1+αΔθ)

L₂³=L₁³(1+αΔθ)³

L₂³=L₁³(1+3αΔθ+3α²Δθ²+α³Δθ³)

Here, α is a small quantity and higher power of it is very small which can be neglected.

∴L₂³=L₁³(1+3αΔθ)

V₂=V₁(1+3αΔθ)___(3)

Also,

V₂=V₁(1+γΔθ)___(4)

Here, γ = Cubical Expansivity

​Comapring (3) and (4), we get,

γ=3α

∴α=γ3

​Thus, we have the relation,

α=β2=γ/3