The change in density of mercury, when it is heated from $10^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$…

The change in density of mercury, when it is heated from $10^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ is (The coefficient of volume expansion of mercury is $18.2 \times 10^{-5} \mathrm{~K}^{-1}$ )
  1. $1.82 \%$
  2. $0.91 \%$
  3. $9.1 \%$
  4. $0.45 \%$

Solution

$\rho=\frac{\mathrm{m}}{\mathrm{V}}$ $\frac{\Delta \rho}{\rho}=\frac{\Delta V}{V}=\frac{V_0 Y \Delta T}{V_0}=Y \Delta T$ So, $\frac{\Delta \rho}{\rho} \times 100=\mathrm{Y} \Delta \mathrm{T} \times 100$ $\begin{aligned} & =18.2 \times 10^{-5} \times 50 \times 100 \\ & =0.91 \%\end{aligned}$

Asked in: AP EAMCET 2022 (08 Jul Shift 1)

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