Question

A non-isotropic solid metal cube has coefficient of linear expansion as $5\times {10}^{-5}/{}^{0}C$ along the x-axis and$5\times {10}^{-6}/{}^{0}C$ along y-axis and z-axis. If the coefficient of volumetric expansion of the solid is $C\times {10}^{-6}/{}^{0}C$then the value of $C$ is.

Answer 1

Step 1 : Given data

It is a non-isotropic solid metal cube

Coefficient of linear expansion

Along $X$ axis $=5\times {10}^{-5}/{}^{0}C$

Along $YandZ$ axes $=5\times {10}^{-6}/{}^{0}C$

Coefficient of volumetric expansion $=C\times {10}^{-6}/{}^{0}C$

Step 2 : To find the volume expansion and value of $C$

Let the initial cube has sides $l_1,l_2,l_3$ in $x,y,z$ direction respectively.

Initial volume $V=l_1{l}_2{l}_3$ ………$\left(1\right)$

Let the change in volume be $∆V$ , then the volume expansion given by $\frac{∆V}{V∆t}$

Taking $\log$ of $\left(1\right)$ on both sides

$\ln \left(V\right)=\ln \left(l_1\right)+\ln \left(l_2\right)+\ln \left(l_3\right)$

Multiplying the whole equation by $\frac{1}{∆t}$

$\frac{1}{V}\frac{∆V}{∆t}=\frac{1}{l_1}\frac{∆l_1}{∆t}+\frac{1}{l_2}\frac{∆l_2}{∆t}+\frac{1}{l_3}\frac{∆l_3}{∆t}$

We know that,

$$\begin{gathered} \frac{1}{l_1}\frac{∆l_1}{∆t}=5\times {10}^{-5}/{}^{0}C \\ \frac{1}{l_2}\frac{∆l_2}{∆t}=\frac{1}{l_3}\frac{∆l_3}{∆t} \\ =5\times {10}^{-6}/{}^{0}C \end{gathered}$$

Therefore, after putting the values, we get

$$\begin{gathered} \frac{1}{V}\frac{∆V}{∆t}=5\times {10}^{-5}+10\times {10}^{-6} \\ =6\times {10}^{-5} \\ 6\times {10}^{-5}=C\times {10}^{-6} \\ C=60 \end{gathered}$$

Hence in the case of volume expansion the value of $C$ is $60$