Question

A rod is heated from $0^{\circ }\text{C}$ to ${10}^{\circ }\text{C}$ and its length is changed by $0.02\%$. By what percent does the mass density change?

• A. $0.02\%$
• B. $0.04\%$
• C. $0.06\%$
• D. $0.08\%$

Answer 1

The correct option is C $0.06\%$

Given,
Change in temperaure, $\Delta T={10}^{\circ }\text{C}-0^{\circ }\text{C}={10}^{\circ }\text{C}$

Percentage change in length, $\frac{\Delta L}{L}\times 100=0.02\times {10}^{-2}$ [given]

We know that change in length,

$\Delta L=\alpha L\Delta T$

$\Rightarrow \alpha =\frac{\Delta L}{L\Delta T}$

$\Rightarrow \alpha =\frac{0.02\times {10}^{-2}}{10}$

$\Rightarrow \alpha =2\times {10}^{-5}{/}^{\circ }\text{C}$

Now, for constant mass, we can write,

$\text{Change in mass density = - Change in volume}$

[since $m=\rho V$]

$\frac{d\rho }{\rho }=-\frac{dV}{V}....\left(1\right)$

For volume change $\Delta V$, we have

$\Delta V=\gamma V\Delta T....\left(2\right)$

where, $\gamma =3\alpha =6\times {10}^{-5}{/}^{\circ }\text{C}$

Substituting the values in $\left(2\right)$,

$\Rightarrow \frac{\Delta V}{V}\times 100=\gamma \Delta T\times 100$

$\Rightarrow \frac{\Delta V}{V}\times 100=6\times {10}^{-5}\times 10\times 100=6\times {10}^{-2}$

$\Rightarrow \frac{\Delta V}{V}\times 100=0.06\%....\left(3\right)$

From $\left(1\right)$ and $\left(3\right)$,

$\frac{d\rho }{\rho }\times 100=-0.06\%$

Thus, mass density changes by $0.06$ percent.