Question

An Aluminium sphere of $20cm$ diameter is heated from ${{0}^{\circ }}C$ to ${{100}^{\circ }}C$. Its volume changes by (given that coefficient of linear expansion of Aluminium is ${{\alpha }_{Al}}=23\times {{10}^{-6}}{{/}^{\circ }}C$).
A. $49.8\text{ cc}$
B. $\text{28}\text{.9 cc}$
C. $2.89\text{ cc}$
D. $9.28\text{ cc}$

Answer 1

Hint: When a body is heated, expansion takes in the case of length, area and volume of the body. The coefficient of expansion is the ratio of change in dimension to the original dimension of a body during a unit temperature change. So if a sphere is heated there will be an increase in the volume of the sphere and as the volume increases the length of radius also increases.

Complete step by step answer:

So as the sphere is heated its volume increases. How much of the volume is increased depends on the coefficient of volume expansion, the initial volume of the sphere and the temperature change the sphere had to undergo.

In the problem we are given the diameter of the sphere, from that we can calculate the volume of the sphere given by the formula, $V=\dfrac{4}{3}\pi {{R}^{3}}$.
The linear expansion coefficient $\left( \alpha \right)$ and the volume expansion coefficient $\left( \gamma \right)$ are related by the equation, $\gamma =3\alpha$.
So we can write the change in volume of the sphere due to thermal expansion as,
$\Delta V=\gamma V\Delta T$
Where,
$\gamma$ is the thermal coefficient of volume expansion.
V is the volume of the sphere.
$\Delta T$ is the change in temperature.
So we know the volume of the sphere and relation between the linear coefficient and the volume coefficient. So, we rewrite the above equation as,
$\Delta V=\left( 3\alpha \right)\times \left( \dfrac{4}{3}\pi {{\left( \dfrac{d}{2} \right)}^{3}} \right)\Delta T$

Substituting the values given in the problem into the equation we get,

$\Delta V=\left( 3\times 23\times {{10}^{-6}}{{/}^{\circ }}C \right)\times \left( \dfrac{4}{3}\pi {{\left( \dfrac{20cm}{2} \right)}^{3}} \right)\times {{\left( 100-0 \right)}^{\circ }}C$
$\Rightarrow \Delta V=\left( 69\times {{10}^{-6}}{{/}^{\circ }}C \right)\times \left( \dfrac{4}{3}\pi {{\left( 10cm \right)}^{3}} \right)\times {{100}^{\circ }}C$
$\therefore \Delta V=2.89\text{ cc}$
So, the volume change of the sphere is $\Delta V=2.89\text{ cc}$.
So the answer to the question is option (C)- 2.89 cc.

Note: The coefficient of thermal expansion shows how the shape of an object changes with temperature. Which of the changes is evident depends on the object. For a long rod, whose surface area and volume are negligible, the thermal expansion of length is prominent. For a surface having surface area A, the thermal expansion is happening on the surface area. In other cases, it is the volume that is being increased or decreased.
For an isotropic material, the expansion taking place is uniform in all the directions. Since it is uniform, we can express the coefficient of thermal expansion of area or volume in terms of linear coefficient of thermal expansion. We can write it as,

$\beta =2\alpha$
Where,
$\beta$ is the coefficient of thermal expansion in surface area.
$\alpha$ is the coefficient of thermal expansion in length.
$\gamma =3\alpha$
$\gamma$ is the coefficient of thermal expansion in volume.