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**Hint:**When a matter is subjected to heat or a temperature change, its atoms gain energy and move very fast, leading to thermal expansion. When we heat things, not only their length changes, but there is also a change in their area and volume. This gives rise to the need to define three different coefficients of thermal expansion, namely, linear, areal, and volumetric coefficients of thermal expansion.

**Complete step by step solution:**

For a small value of $\Delta V$:

$\Delta V \propto \Delta T$ and $\Delta V \propto V$

$\Rightarrow \Delta V \propto V\Delta T$

$\therefore \Delta V = \gamma V\Delta T$……………….. Equation (1)

where,

$\gamma =$ coefficient of volumetric expansion

$V =$ volume of body

$\Delta T =$ change in temperature

$\Delta V =$ a small change in volume [as $\Delta V \ll V$]

Now, $\Delta V = {V_T} - {V_0}$ as

${V_0} =$ the initial volume of the body and

${V_T} =$ the volume of the body after increasing temperature, i.e., at the temperature $T$

$\Rightarrow {V_T} - {V_0} = \gamma {V_0}\Delta T$

$\Rightarrow {V_T} = {V_0} + \gamma {V_0}\Delta T$

Therefore, the change in the volume of a body is given by:

$\Rightarrow {V_T} = {V_0}(1 + \gamma \Delta T)$

Here, we have $\dfrac{{\Delta V}}{V} = 0.24\% = \dfrac{{0.24}}{{100}}$ and $\Delta T = 40{}^ \circ C$

Using equation (1) and rearranging it, we get

$\Rightarrow \gamma = \dfrac{{\Delta V}}{{V\Delta T}}$

Substituting the values to find the coefficient of volumetric expansion:

$\Rightarrow \gamma = \dfrac{{0.24}}{{100}} \times \dfrac{1}{{40}}$

$\Rightarrow \gamma = 0.06 \times {10^{ - 3}}$

$\Rightarrow \gamma = 6 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$

Just like we derived the coefficient of volumetric expansion, similarly according to equation (1), linear expansion of a body is expressed as:

$\Delta L = \alpha L\Delta T$……………………….Equation (2)

where $\alpha =$ coefficient of linear expansion

Let us assume a cube of the side $L$ whose volume is ${L^3}$.

Now, $\Delta V = {(L + \Delta L)^3} - {L^3}$

Using the mathematical formula, ${(a + b)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}$

We get $\Delta V = [{L^3} - 3L{(\Delta L)^2} + 3{L^2}.\Delta L - {(\Delta L)^3}] - {L^3}$

Since, $\Delta L \ll L$,${(\Delta L)^3} \approx 0$ and ${(\Delta L)^2} \approx 0$

$\Rightarrow \Delta V = {L^3} + 3{L^2}.\Delta L - {L^3}$

$\Rightarrow \Delta V = 3{L^2}.\Delta L$

Substituting the value of $\Delta L$ from equation (2);

$\Rightarrow \Delta V = 3{L^2}(\alpha L\Delta T)$

$\Rightarrow \Delta V = 3\alpha {L^3}\Delta T$……….Equation (3)

Comparing equations (1) and (3), we get

$\Rightarrow \gamma = 3\alpha$

$\Rightarrow \alpha = \dfrac{\gamma }{3}$

$\Rightarrow \alpha = \dfrac{{6 \times {{10}^{ - 5}}{}^ \circ {C^{ - 1}}}}{3}$

$\Rightarrow \alpha = 2 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$

**The correct answer is [A], $2 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$.**

**Note:**Every material has its unique coefficient of volumetric expansion, but all three different coefficients of thermal expansion are interdependent, further indicating that these coefficients are also a characteristic property of every material (mostly metal).

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