Question

The radius of a metal sphere at room temperature \[T\] is \[\;R\], and the coefficient of linear expansion of the metal is \[\alpha \]. The sphere is heated a little by a temperature \[\Delta T\] so that its new temperature is \[\left( {T + \Delta T} \right).\;\] the increase in the volume of the sphere is approximate:

A. \[2\pi R\alpha T\]
B. \[{\pi ^2}R\alpha T\]
C. \[4\pi {R^3}\alpha \Delta T/3\]
D. \[4\pi {R^3}\alpha \Delta T\]

Answer 1

Hint: Thermal expansion is defined as the tendency of matter to change its shape, area, or volume in response to a change in temperature. This expansion occurs when the substance is heated up. When we heat the substance the particles inside it vibrate faster which creates more space in between the particles. This results in the expansion of the substance.

Complete step by step solution:
Given a three-dimensional body. Let the body be the metal sphere. The body has room temperature \[T\] and volume \[V\]. Its radius at the room temperature \[T\] is \[R\].
When we heat a three-dimensional body its volume increases. This is caused by thermal expansion as we already said.
Given that we have increased the temperature by \[T + \Delta T\]. Also given that the coefficient of linear expansion of the metal is \[\alpha \].
The volumetric thermal expansion is given as,
\[{\alpha _{\text{V}}} = \dfrac{1}{{\text{V}}}\dfrac{{{\text{dV}}}}{{{\text{dT}}}}\]
Here, \[\alpha \] is the coefficient of thermal expansion with respect to volume.
\[{\text{V}}\] is the volume of the material
\[\dfrac{{{\text{dV}}}}{{{\text{dT}}}}\] is the rate of change of volume with respect to temperature.
Similar to volume expansion we also have a linear expansion that applies to one dimension bodies.
Coefficient of Linear expansion= \[\alpha \]
Coefficient of Volume expansion= \[3\alpha \]
We can substitute the formula for the coefficient of thermal expansion with respect to volume from above. So we get,
\[\dfrac{1}{{\text{V}}}\dfrac{{{\text{dV}}}}{{{\text{dT}}}} = 3\alpha \]
We need to find only the volume change\[{\text{dV}}\]. Therefore rearranging the above equation,
\[{\text{dV = }}3{\text{V}}\alpha {\text{dT}}\]
We know the volume of the sphere, \[{\text{V}}\]=\[\dfrac{4}{3}\pi {R^3}\alpha {\text{dT}}\]
Also change in the temperature is given as \[{\text{dT}}\]=\[\Delta T\]
Therefore,
\[{\text{dV}} = 4\pi {R^3}\alpha \Delta T\]
Hence the correct option is D.

Note:
Generally, the coefficient of volume expansion is givens as,
\[{\alpha _{\text{V}}} = \dfrac{1}{{\text{V}}}{(\dfrac{{\partial {\text{V}}}}{{\partial {\text{T}}}})_{\text{p}}}\]
The subscript \[p\] indicates that the pressure is held constant during the expansion. In the case of a gas, keeping pressure constant is important, as the volume of a gas will vary appreciably with pressure as well as with temperature. But in the above case, we didn’t consider pressure. Because in solids we can ignore the effects of the pressure of the material.