Let the initial volume be $V$ and initial density be $\rho$. Since mass $M$ remains constant, $\rho = \frac{M}{V}$. When temperature increases by $\Delta T$, the new volume $V'$ is given by: $V' = V(1 + \gamma \Delta T)$ The new density $\rho'$ is: $\rho' = \frac{M}{V'} = \frac{M}{V(1 + \gamma \Delta T)} = \frac{\rho}{1 + \gamma \Delta T}$ The fractional change in density is given by: $\frac{\Delta \rho}{\rho} = \frac{\rho - \rho'}{\rho} = 1 - \frac{\rho'}{\rho} = 1 - \frac{1}{1 + \gamma \Delta T} = \frac{\gamma \Delta T}{1 + \gamma \Delta T}$ Since $\gamma \Delta T \ll 1$, we can approximate this as: $\frac{\Delta \rho}{\rho} \approx \gamma \Delta T$ Given: $\gamma = 5 \times 10^{-4} \text{ K}^{-1}$ $\Delta T = 40^\circ\text{C} = 40 \text{ K}$ Fractional change $\approx 5 \times 10^{-4} \times 40 = 200 \times 10^{-4} = 0.020$