Question

If the coefficient of volume expansion of honey is $150 \times 10^{-5} K^{-1}$, then the percentage change in its density for $10^{\circ}C$ rise in its temperature is

• A. 0.5
• B. 1.5
• C. 0.75
• D. 0.15

Hint:

For solids/liquids, $\Delta \rho / \rho \approx -\gamma \Delta T$. The magnitude of percentage change is simply $\gamma \Delta T \times 100$.

Answer 1

The Correct Option is B

Step 1: Formula for Density Variation with Temperature:
Density decreases as temperature increases. \[ \rho' = \frac{\rho}{1 + \gamma \Delta T} \approx \rho (1 - \gamma \Delta T) \] Change in density $\Delta \rho = \rho' - \rho = -\rho \gamma \Delta T$. Fractional change (magnitude) $= \frac{\Delta \rho}{\rho} = \gamma \Delta T$.
Step 2: Calculate Percentage Change:
\[ % \text{ Change} = (\gamma \Delta T) \times 100 \] Given: $\gamma = 150 \times 10^{-5} K^{-1}$ $\Delta T = 10 K$ (or $10^{\circ}C$) \[ % \text{ Change} = (150 \times 10^{-5}) \times 10 \times 100 \] \[ = 150 \times 10^{-5} \times 1000 \] \[ = 150 \times 10^{-2} \] \[ = 1.5 \]
Step 4: Final Answer:
The percentage change is 1.5%.