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# The coefficient of volume expansion $\gamma$ is _________ that of the coefficient of linear expansion.A) doubleB) one-thirdC) halfD) equal to

Last updated date: 16th Jul 2024
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Hint: To understand this question, we have to consider a sample body and then study the expansion in it by application of heat. The expansion in the body happens in three ways: linear, superficial and volumetric. Their definitions must be understood to solve this problem.

There are three kinds of expansion of a solid when heat is applied to it:
i) Linear: Change in the length
ii) Superficial or Areal: Change in the area
iii) Volumetric: Change in the volume.
To understand the expansion, let us take an example of a solid rectangle of dimensions a, b, and c heated from ${0^ \circ }C$ to a temperature ${t^ \circ }C$ where the new dimensions are A, B and C.
The original volume of the solid at ${0^ \circ }C$ = $abc$
The volume of the solid at temperature ${t^ \circ }C$ = ABC
The coefficient of linear expansion is represented by $\alpha$. If A is the length at temperature ${t^ \circ }C$ and a is the length at ${0^ \circ }C$, the coefficient of linear expansion is given by the relation –
$A = a\left( {1 + \alpha t} \right)$
Similarly, for the other dimensions we have –
$B = b\left( {1 + \alpha t} \right)$
$C = c\left( {1 + \alpha t} \right)$
Final volume at temperature ${t^ \circ }C$ –
$ABC = a\left( {1 + \alpha t} \right) \times b\left( {1 + \alpha t} \right) \times c\left( {1 + \alpha t} \right)$
$\Rightarrow ABC = abc{\left( {1 + \alpha t} \right)^3}$
$\Rightarrow ABC = abc\left( {1 + 3\alpha t + 3{\alpha ^2}{t^2} + {\alpha ^3}{t^3}} \right)$
The value of coefficient of linear expansion is very low. Hence, the higher powers of $\alpha$ are negligible.

Therefore, we have –
$ABC = abc\left( {1 + 3\alpha t} \right)$
The coefficient of volumetric expansion is represented by $\gamma$. If V is the length at temperature ${t^ \circ }C$ and v is the length at ${0^ \circ }C$, the coefficient of linear expansion is given by the relation –
$V = v\left( {1 + \gamma t} \right)$
Comparing this equation with the above, we get the following relationship between the coefficients as –
$\gamma = 3\alpha$
$\therefore \alpha = \dfrac{\gamma }{3}$
Hence, the coefficient of linear expansion is one-third the coefficient of volumetric expansion.

Hence, the correct option is Option B.

Note: There is another coefficient of expansion known as the coefficient of superficial expansion represented by $\beta$. The one equation relating all the three coefficients of expansion is:
$\dfrac{\alpha }{1} = \dfrac{\beta }{2} = \dfrac{\gamma }{3}$