Question

When the temperature of a metal wire is increased from $0°C$ to $10°C$, its length increases by $0.02\%$. The percentage change in its mass density will be closest to:

• A. $0.06$
• B. $0.008$
• C. $2.3$
• D. $0.8$

Answer 1

The correct option is A

$0.06$

Explanation for correct answer:

Step 1: Formula used:

We know that the coefficient of linear expansion is ${\alpha }_l=\frac{\delta l}{l.\delta T}$ where $\delta l$ is change in the length, l is the original length and $\delta T$ is change in the temperature.

Step 2: Solution:

We know that co-efficient of volume expansion is three times the co-efficient of linear expansion i. e,$\gamma =3\alpha$=$3(2\times {10}^{-5})$=$6\times {10}^{-5}$. The coefficient of linear expansion equation becomes, $\frac{0.02}{100\times 10}=2\times {10}^{-5}$

If $\rho$ is the density of the material, $m$ is its mass and V the volume, We have,

$$\begin{gathered} \rho =\frac{m}{v} \\ \delta \rho =\frac{\delta m}{\delta v} \end{gathered}$$

$\frac{\delta \rho }{\rho }=\frac{\delta m}{m}-\frac{\delta v}{v}$

here,$\frac{\delta m}{m}=0$

$\frac{\delta \rho }{\rho }=-\frac{\delta v}{v}$

now

$$\begin{gathered} \frac{\delta v}{v}\times 100=\gamma \delta T\times 100 \\ =6\times {10}^{-5}\times 10\times 100 \\ =.06 \end{gathered}$$

Hence, option (A) is the correct answer.