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Question: A sinker of weight \[{W_0}\] was an apparent weight of \({W_1}\) when placed in a liquid at a temper...

A sinker of weight W0{W_0} was an apparent weight of W1{W_1} when placed in a liquid at a temperature T1{T_1} and weight W2{W_2} when weighed in the same liquid at a temperature T2{T_2} . The coefficient of cubical expansion of the material of the sinker is γ\gamma . What is the coefficient of volume expansion of the liquid ?

Explanation

Solution

When an object is heated to some temperature its dimensions of length is increased due to thermal heating and agitation of molecules takes place which result in increase in dimension of lengths and its increased area and volume is directly proportional to temperature and hence these proportionality constant is known as coefficient of volume expansion.

Complete step by step answer:
Let us assume that θ=T2−T1\theta = {T_2} - {T_1} be the change in temperature and γ\gamma be the coefficient of volume expansion of the liquid. Let the density of liquid at temperature at T1{T_1} is ρ1{\rho _1} and at temperature T2{T_2} be ρ2{\rho _2} respectively.So we get,
ρ1=ρ2(1+γθ)→(i){\rho _1} = {\rho _2}(1 + \gamma \theta ) \to (i)

Let us assume that the volume of sinker at T1{T_1} is V1{V_1} and at T2{T_2} is V2{V_2} . So,
V2=V1(1+γsθ)→(ii){V_2} = {V_1}(1 + {\gamma _s}\theta ) \to (ii)
Now, the loss in weight can be written as W0−W1=V1ρ1g→(iii){W_0} - {W_1} = {V_1}{\rho _1}g \to (iii)
And in another case, it can be written as W0−W2=V2ρ2g→(iv){W_0} - {W_2} = {V_2}{\rho _2}g \to (iv)
Divide the equation (iii) by (iv)(iii)\,by\,(iv) we get,
W0−W1W0−W2=V1ρ1V2ρ2\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}} = \dfrac{{{V_1}{\rho _1}}}{{{V_2}{\rho _2}}}
⇒1+γθ=W0−W1W0−W2+γs(W0−W1W0−W2)θ\Rightarrow 1 + \gamma \theta = \dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}} + {\gamma _s}(\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}})\theta
∴γ=(W0−W1W0−W2)1T2−T1+(W0−W1W0−W2)γs\therefore \gamma = (\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}})\dfrac{1}{{{T_2} - {T_1}}} + (\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}}){\gamma _s}

Hence, the coefficient of volume expansion is γ=(W0−W1W0−W2)1T2−T1+(W0−W1W0−W2)γs\gamma = (\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}})\dfrac{1}{{{T_2} - {T_1}}} + (\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}}){\gamma _s}.

Note: It should be remembered that, the basic relation between weight, density and volume of a body is related as M=ρVgM = \rho Vg and the coefficient of volume expansion is the proportionality factor which shows rate of increasing of volume of a body when the temperature is increased. In the case of the cube all three dimensions increase with an equal amount in their respective directions.