The coefficient of volume expansion of glycerine is $$49×10^{-5}\text{K}^{-1}$$​. Find the fractional change in its density for a 20° C rise in temperature?

Correct option is D

​$$\text{The formula for the increase in volume due to temperature change is:} \\V_f = V_0 \left( 1 + \gamma \Delta T \right) \\\text{Where:} \\V_f = \text{Final volume,} \\V_0 = \text{Initial volume,} \\\gamma = \text{Coefficient of volume expansion,} \\\Delta T = \text{Change in temperature.} \\\text{For the fractional change in density, we use:} \\\dfrac{\Delta \rho}{\rho_0} = \dfrac{1}{1 + \gamma \Delta T} - 1 \\\text{Where:} \\\rho_0 = \text{Initial density,} \\\rho_f = \text{Final density.}$$

$$\text{Given:} \\\gamma = 49 \times 10^{-5} \, \text{K}^{-1}, \\\Delta T = 20^\circ \text{C}. \\\text{ the final volume using the given formula:} \\V_f = V_0 \left( 1 + 49 \times 10^{-5} \times 20 \right) \\V_f = V_0 \left( 1 + 0.0196 \right) = V_0 \times 1.0098 \\\text{ the fractional change in density:} \\\dfrac{\Delta \rho}{\rho_0} = \dfrac{1}{1 + 49 \times 10^{-5} \times 20} - 1 \\\dfrac{\Delta \rho}{\rho_0} = \dfrac{1}{1.0098} - 1 = -0.0098 \\\text{Therefore:} \\\dfrac{\Delta \rho}{\rho_0} = -9.8 \times 10^{-3} \approx -0.98 \times 10^{-2}$$​​​