91Ó°ÊÓ

Initial increase in diameter is by $\frac{∆\mathrm{D}}{\mathrm{D}}=0.18\%$atdata-custom-editor="chemistry" $∆\mathrm{T}={100}^{\mathrm{o}}\mathrm{C}$

When an object's temperature changes, it expands and grows larger, a process known as thermal expansion. Thermal expansion can occur due to an increase in temperature. For the given problem, we have to use the formula for linear expansion to calculate the thermal expansion coefficient. Mass does not depend on thermal expansion.

Formula:

The linear expansion of a body,$∆\mathrm{L}=\mathrm{³¢Î\pm }∆\mathrm{T}$ …(¾±)

Where,${\mathrm{Î\pm }}_{\mathrm{Pb}}$ is the coefficient of linear expansion of body, L is length and$∆\mathrm{T}$ is change in temperature

The area of a coin of or radius, R or diameter, D,

$$\begin{gathered} \mathrm{A}={\mathrm{Ï€¸é}}^2 \\ \mathrm{A}=\mathrm{Ï€}{\left(\frac{\mathrm{D}}{2}\right)}^2 \end{gathered}$$ …(¾±¾±)

As we differentiate the equation (ii) by D, we can get

$$\begin{gathered} \frac{d\mathrm{A}}{d\mathrm{t}}=\frac{2\mathrm{Ï€\P Ù}}{4} \\ \mathrm{dA}=\frac{\mathrm{Ï€\P ÙdD}}{2} \end{gathered}$$

Using the equation (ii) for area, we can get, the above equation as:

$\frac{\mathrm{dA}}{\mathrm{A}}=\frac{2\mathrm{dD}}{\mathrm{D}}$

In terms of$∆,$we can write,

$\frac{∆\mathrm{A}}{\mathrm{A}}=\frac{2∆\mathrm{D}}{\mathrm{D}}$

Since area is a two dimensional quantity, so from the above equation, we can conclude that the area increases by the factor 2 i.e. the area increases by$2×0.18\%=0.36\%$

Hence, the percent increase in area of the face is 0.36%

Since thickness is a one dimensional quantity, so, if 0.18% is the increase in diameter, then 0.18% is the increase in its thickness.

If the linear increase in diameter is 0.18%, then the volume increase is three times of that result, so, Volume increase is given by: $3×0.18\%=0.54\%$

Hence, the percent increase in volume is 0.54%

From the formula of thermal expansion of equation (i), we can conclude that the mass is independent of thermal expansion i.e. total mass always remains constant in expansion.

Fromthelinear expansion formula of equation (i), we can calculatethethermal expansion coefficient as:

$\mathrm{Î\pm }=\frac{∆\mathrm{D}}{\mathrm{D}∆\mathrm{T}}$

where,

role="math" localid="1661876703196" $\begin{array}{rcl}\frac{∆\mathrm{D}}{\mathrm{D}}& =& 0.18\\ & =& 0.18×{10}^{-2}\end{array}$

is the percentage increase in diameter.

And fordata-custom-editor="chemistry" $∆\mathrm{T}={100}^{\mathrm{o}}\mathrm{C}$

role="math" localid="1661876748017" $\begin{array}{rcl}\mathrm{Î\pm }& =& \frac{0.18×{10}^{-2}}{{100}^{\mathrm{o}}\mathrm{C}}\\ & =& 1.8×{10}^{-5}/{}^{\mathrm{o}}\mathrm{C}\end{array}$

Hence, the coefficient of linear expansion isrole="math" localid="1661876763362" $1.8×{10}^{-5}/{}^{\mathrm{o}}\mathrm{C}$