(a) We know that the rate of decrease of density $\rho$ of air is directly proportional to density $\rho$ i.e. $-\frac{d\rho}{dy} \propto \rho$ or $\frac{d\rho}{dy} = -K \rho$ where K is a constant of proportionality. Here –ve sign shows that $\rho$ decreases as $y$ increases.$\therefore \frac{d\rho}{\rho} = -K \, dy$Integrating it within the conditions, as $y$ changes from 0 to $y,$ density changes from $\rho_0 \to \rho$, we have $\int\limits_{\rho_0}^{\rho} \frac{d\rho}{\rho} = -\int\limits_{0}^{y} K \, dy$$[\log_e \rho]_{\rho_0}^{\rho} = -K y$ or $\log_e \rho - \log_e \rho_0 = -K y$or $\log \frac{\rho}{\rho_0} = -K y$ or $\frac{\rho}{\rho_0} = e^{-K y}$ or $\rho = \rho_0 e^{-K y}$Here $K$ is a constant. Suppose $y_0$ is a constant such that $K = \frac{1}{y_0}$ , then$\rho = \rho_0 e^{-\frac{y}{y_0}}$(b) The balloon will rise to a height, where its density becomes equal to the air at that height.Density of balloon,$\rho = \frac{\text{ mass }}{\text{ volume }} = \frac{\text{ pay load }+\text{ mass of He }}{\text{ volume}}$$= \frac{(400+1425 \times 0.18) \,\mathrm{kg}}{1425 \,\mathrm{m}^3}$$= \frac{656.5}{1425} \,\mathrm{kg}/\mathrm{m}^3$$\text{As}, \quad \rho = \rho_0 e^{-\frac{y}{y_0}}$$\therefore \frac{656.5}{1425} = 1.25 e^{-\frac{y}{8000}}$or $e^{-\frac{y}{8000}} = \frac{656.5}{1425 \times 1.25}$ or $e^{\frac{y}{8000}} = \frac{1425 \times 1.25}{656.5}$$= 2.7132$Taking log on both the sides$\frac{y}{8000} = \log_e 2.7132$$= 2.3026 \log_{10} 2.7132$$= 2.3026 \times 0.4335 \approx 1$$y = 8000 \times 1 = 8000 \,\mathrm{m}$$= 8 \,\mathrm{km}.$