Question

The change in density of mercury when it is heated from 10 °C to 60 °C. (The coefficient of volume expansion of mercury is \(18.2 \times 10^{-5} \, \text{K}^{-1}\))

• A. 1.82 %
• B. 0.91 %
• C. 9.1 %
• D. 0.45 %

Hint:

For liquids, \(\frac{\Delta \rho}{\rho} = - \beta \Delta T\), where \(\beta\) is volume expansion coefficient.

Answer 1

The Correct Option is B

Step 1: Start from the volume expansion instead of quoting the density-change shortcut.
Heating a fixed mass $m$ of mercury lets its volume grow as $V_2 = V_1(1+\beta\Delta T)$, since $\beta$ is defined as the fractional change in volume per unit rise in temperature.

Step 2: Write the density before and after in terms of this.
\[ \rho_1 = \frac{m}{V_1}, \qquad \rho_2 = \frac{m}{V_1(1+\beta\Delta T)} = \frac{\rho_1}{1+\beta\Delta T} \]

Step 3: Since $\beta\Delta T$ is tiny, use $\dfrac{1}{1+\beta\Delta T}\approx 1-\beta\Delta T$, giving the fractional drop in density directly.
\[ \frac{\rho_1-\rho_2}{\rho_1} \approx \beta\Delta T \]

Step 4: Substitute $\beta = 18.2\times10^{-5}\ \text{K}^{-1}$ and $\Delta T = 60-10 = 50$ K.
\[ \frac{\Delta\rho}{\rho} \approx 18.2\times10^{-5}\times50 = 9.1\times10^{-3} \]

Step 5: Convert to a percentage.
\[ 9.1\times10^{-3}\times100\% = 0.91\% \]

Step 6: Conclusion.
\[ \boxed{0.91\%} \]