Answer
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Hint: 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 $\theta = {T_2} - {T_1}$ be the change in temperature and $\gamma $ be the coefficient of volume expansion of the liquid. Let the density of liquid at temperature at ${T_1}$ is ${\rho _1}$ and at temperature ${T_2}$ be ${\rho _2}$ respectively.So we get,
${\rho _1} = {\rho _2}(1 + \gamma \theta ) \to (i)$
Let us assume that the volume of sinker at ${T_1}$ is ${V_1}$ and at ${T_2}$ is ${V_2}$ . So,
${V_2} = {V_1}(1 + {\gamma _s}\theta ) \to (ii)$
Now, the loss in weight can be written as ${W_0} - {W_1} = {V_1}{\rho _1}g \to (iii)$
And in another case, it can be written as ${W_0} - {W_2} = {V_2}{\rho _2}g \to (iv)$
Divide the equation $(iii)\,by\,(iv)$ we get,
$\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}} = \dfrac{{{V_1}{\rho _1}}}{{{V_2}{\rho _2}}}$
$\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 $
$\therefore \gamma = (\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}})\dfrac{1}{{{T_2} - {T_1}}} + (\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}}){\gamma _s}$
Hence, the coefficient of volume expansion is $\gamma = (\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}})\dfrac{1}{{{T_2} - {T_1}}} + (\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}}){\gamma _s}$.
Note:It should be remembered that, the basic relation between weight, density and volume of a body is related as $M = \rho Vg$ 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.
Complete step by step answer:
Let us assume that $\theta = {T_2} - {T_1}$ be the change in temperature and $\gamma $ be the coefficient of volume expansion of the liquid. Let the density of liquid at temperature at ${T_1}$ is ${\rho _1}$ and at temperature ${T_2}$ be ${\rho _2}$ respectively.So we get,
${\rho _1} = {\rho _2}(1 + \gamma \theta ) \to (i)$
Let us assume that the volume of sinker at ${T_1}$ is ${V_1}$ and at ${T_2}$ is ${V_2}$ . So,
${V_2} = {V_1}(1 + {\gamma _s}\theta ) \to (ii)$
Now, the loss in weight can be written as ${W_0} - {W_1} = {V_1}{\rho _1}g \to (iii)$
And in another case, it can be written as ${W_0} - {W_2} = {V_2}{\rho _2}g \to (iv)$
Divide the equation $(iii)\,by\,(iv)$ we get,
$\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}} = \dfrac{{{V_1}{\rho _1}}}{{{V_2}{\rho _2}}}$
$\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 $
$\therefore \gamma = (\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}})\dfrac{1}{{{T_2} - {T_1}}} + (\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}}){\gamma _s}$
Hence, the coefficient of volume expansion is $\gamma = (\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}})\dfrac{1}{{{T_2} - {T_1}}} + (\dfrac{{{W_0} - {W_1}}}{{{W_0} - {W_2}}}){\gamma _s}$.
Note:It should be remembered that, the basic relation between weight, density and volume of a body is related as $M = \rho Vg$ 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.
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