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# The volume of a metal sphere increases by $0.24\%$ when its temperature is raised by ${40^0}C$. What is the coefficient of linear expansion of the metal?A) $2 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$ B) $6 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$C) $18 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$ D) $1.2 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$

Last updated date: 13th Jul 2024
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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).