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

### Long Question Solution

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#### Viva Voice Questions # Thermal Expansion - Long Question Answer

Question: Define the coefficients of real expansion and apparent expansion and hence derive the relation them.

Ans: Real and Apparent Expansion of Liquid: let us consider a round bottom flask with a long narrow neck containing liquid in it up to mark A. when the liquid is heated at first the vessel expands and level of liquid fall to mark B. So volume AB represents the expansion of vessel. Sooner the liquid gains heat and it expands so the level of liquid raises up to C higher than A. We observe the level of liquid raises from A to C hence volume AC is called apparent expansion of liquid. Infact liquid expands from B to C hence volume BC represents a real expansion of the liquid. From figure,
BC = AC + AB
Real expansion of liquid = apparent expansion of liquid + expansion of the vessel.

Coefficient of Real expansion of liquid
It may be defined as a real increase in the volume of liquid per unit original volume per degree rise in temperature.
i.e. γ= real increase in volume of liquid/(original volume * rise in temperature) = ∆Vr/(V*∆θ)

Coefficient of Apparent expansion of liquid
It may be defined as an apparent increase in the volume of liquid per unit original volume per degree rise in temperature.
i.e. γ= apparent increase in volume of liquid/(original volume * rise in temperature) = ∆Va/(V*∆θ)

Relation between Coefficient of Real and Apparent Expansion of Liquid
From the figure, it is clear that,
Real expansion of liquid = apparent expansion of liquid + expansion of vessel.
∆Vr = ∆V+ ∆V...........(i)

Let us consider a glass vessel of volume V filled with some liquid at temperature θ1. When the liquid is heated to θ2, it expands and its real increase in volume is
We know, γ = ∆Vr/(V*∆θ)
∆Vγ V ∆θ = γ V (θ2 - θ1) ….……..(ii)
Similarly apparent increase in the volume of the liquid,
∆Vγ V ∆θ = γ V (θ- θ1) ...........(iii)
and expansion of vessel,
∆Vγ V ∆θ = γ V (θ- θ1) ...........(iv)

Now putting the value of ∆Vr∆Va and ∆Vg from eqn. (ii), (iii) and (iv) respectively in eqn. (i) we get,
γ V ∆θ = γ V ∆θ + γ V ∆θ
or, γ V ∆θ = V ∆θ (γ+ γg)
or, γ = γ+ γg

This equation gives the relation between real and apparent coefficients of a liquid. If α is the coefficient of linear expansion of the glass vessel, then
γg = 3α
γ = γ+ 3α

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