Question

The coefficient of volume expansion of a material is \(5\times10^{-4}\left(^{\circ}\mathrm{C}\right)^{-1}\). The fractional change in its density for a \(40^\circ\mathrm{C}\) rise in temperature is nearly:

• A. \(0.01\)
• B. \(0.02\)
• C. \(0.03\)
• D. \(0.04\)

Hint:

Since \(\rho=\dfrac{m}{V}\), density varies inversely with volume. For small thermal expansions, \[ \frac{\Delta \rho}{\rho}\approx-\beta \Delta T \] where the negative sign indicates a decrease in density.

Answer 1

The Correct Option is B

Step 1: Find fractional change in volume.
\[ \frac{\Delta V}{V} = \beta\,\Delta T = 5\times10^{-4}\times40 = 0.02 \]

Step 2: Relate to fractional change in density.
Since \( \rho = m/V \) and mass is constant, \( \frac{\Delta\rho}{\rho} \approx -\frac{\Delta V}{V} = -0.02 \). The magnitude of the fractional change in density is \[ \boxed{0.02} \]