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:
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:
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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.
Updated On: Jun 26, 2026
- \(0.01\)
- \(0.02\)
- \(0.03\)
- \(0.04\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the relation for volume expansion.
The coefficient of volume expansion is given by \[ \beta=\frac{\Delta V}{V\,\Delta T} \] Hence, \[ \frac{\Delta V}{V}=\beta \Delta T \] Given, \[ \beta=5\times 10^{-4}\,^\circ\text{C}^{-1} \] and \[ \Delta T=40^\circ\text{C} \] Therefore, \[ \frac{\Delta V}{V} =(5\times 10^{-4})(40) \] \[ \frac{\Delta V}{V} =2\times 10^{-2} \] \[ \frac{\Delta V}{V} =0.02 \]
Step 2: Relate density and volume.
Density is given by \[ \rho=\frac{m}{V} \] Since mass remains constant, \[ \rho \propto \frac{1}{V} \] Therefore, for small changes, \[ \frac{\Delta \rho}{\rho} = -\frac{\Delta V}{V} \] Substituting the value obtained above, \[ \frac{\Delta \rho}{\rho} = -0.02 \] The negative sign indicates that density decreases as temperature increases.
Step 3: Find the fractional change in density.
The magnitude of the fractional change in density is \[ \left|\frac{\Delta \rho}{\rho}\right| =0.02 \]
Step 4: Final conclusion.
Therefore, the fractional change in density is nearly \[ \boxed{0.02} \]
The coefficient of volume expansion is given by \[ \beta=\frac{\Delta V}{V\,\Delta T} \] Hence, \[ \frac{\Delta V}{V}=\beta \Delta T \] Given, \[ \beta=5\times 10^{-4}\,^\circ\text{C}^{-1} \] and \[ \Delta T=40^\circ\text{C} \] Therefore, \[ \frac{\Delta V}{V} =(5\times 10^{-4})(40) \] \[ \frac{\Delta V}{V} =2\times 10^{-2} \] \[ \frac{\Delta V}{V} =0.02 \]
Step 2: Relate density and volume.
Density is given by \[ \rho=\frac{m}{V} \] Since mass remains constant, \[ \rho \propto \frac{1}{V} \] Therefore, for small changes, \[ \frac{\Delta \rho}{\rho} = -\frac{\Delta V}{V} \] Substituting the value obtained above, \[ \frac{\Delta \rho}{\rho} = -0.02 \] The negative sign indicates that density decreases as temperature increases.
Step 3: Find the fractional change in density.
The magnitude of the fractional change in density is \[ \left|\frac{\Delta \rho}{\rho}\right| =0.02 \]
Step 4: Final conclusion.
Therefore, the fractional change in density is nearly \[ \boxed{0.02} \]
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