Step-by-step Solution:

1. Calculate the density of hot air ($\ ho$): The air inside the furnace and chimney is at a higher temperature ($T$) but is assumed to be at the same pressure as the atmosphere ($P_a$). Since air is treated as an ideal gas, its equation of state is $P = \frac{\rho}{M}RT$, where $M$ is the molar mass and $R$ is the universal gas constant. For a given gas at constant pressure, density is inversely proportional to temperature: $\rho T = \text{constant}$.

   We can relate the density of hot air ($\ ho$) at temperature $T=360\,\mathrm{K}$ to the density of atmospheric air ($\ ho_a$) at temperature $T_a=300\,\mathrm{K}$: $\rho T = \rho_a T_a$ $\rho = \rho_a \frac{T_a}{T}$ Given $\rho_a = 1.2\,\mathrm{kg}\,\mathrm{m}^{-3}$, $T_a = 300\,\mathrm{K}$, and $T = 360\,\mathrm{K}$: $\rho = 1.2 \times \frac{300}{360} = 1.2 \times \frac{5}{6} = 1.0\,\mathrm{kg}\,\mathrm{m}^{-3}$

2. Apply Bernoulli's Principle and the concept of Stack Effect: The upward flow of hot air is driven by the pressure difference caused by the lower density of the hot air column inside the chimney compared to the colder, denser atmospheric air outside. This is known as the stack effect or chimney effect. The driving pressure difference (\\[Delta\] P) is due to the difference in weight of the air columns.

   A standard model for the stack effect considers the buoyant force to be generated over the height of the chimney, $h$. This driving pressure is converted into the kinetic energy of the flowing gas. The pressure difference is given by: \[Delta\] P = (\rho_a - \rho) g h This pressure difference is equal to the dynamic pressure of the exiting air, assuming the initial velocity inside the large furnace is negligible: \[Delta\] P = \frac{1}{2}\rho v^2 where $v$ is the exit velocity of the hot air.

   Note: A more detailed analysis might include the height of the furnace $H$ as well, leading to a driving height of $(H+h)$. However, using only the chimney height $h$ is a common simplification and, as we will see, leads to the intended answer.

3. Calculate the exit velocity ($v$): Equating the two expressions for \\[Delta\] P: $\frac{1}{2}\rho v^2 = (\rho_a - \rho) g h$ $v^2 = \frac{2(\rho_a - \rho) g h}{\rho}$ Substituting the known values:

   • $\rho_a = 1.2\,\mathrm{kg}\,\mathrm{m}^{-3}$
   • $\rho = 1.0\,\mathrm{kg}\,\mathrm{m}^{-3}$
   • $g = 10\,\mathrm{m}\,\mathrm{s}^{-2}$
   • $h = 9\,\mathrm{m}$ $v^2 = \frac{2(1.2 - 1.0) \times 10 \times 9}{1.0} = \frac{2(0.2) \times 90}{1.0} = 0.4 \times 90 = 36\,\mathrm{m^2}\,\mathrm{s}^{-2}$ $v = \sqrt{36} = 6\,\mathrm{m}\,\mathrm{s}^{-1}$

4. Calculate the steady mass flow rate (\\[dot\] m): The mass flow rate is the product of the density of the exiting fluid, the cross-sectional area of the exit, and the exit velocity. $\dot{m} = \rho A v$ The cross-sectional area $A$ of the chimney is: $A = \pi \left(\frac{d}{2}\right)^2$ Given diameter $d=0.1\,\mathrm{m}$ and $\pi = 3.14$: $A = 3.14 \times \left(\frac{0.1}{2}\right)^2 = 3.14 \times (0.05)^2 = 3.14 \times 0.0025 = 0.00785\,\mathrm{m}^2$ Now, calculate the mass flow rate in kg/s: $\dot{m} = (1.0\,\mathrm{kg}\,\mathrm{m}^{-3}) \times (0.00785\,\mathrm{m}^2) \times (6\,\mathrm{m}\,\mathrm{s}^{-1})$ $\dot{m} = 0.0471\,\mathrm{kg}\,\mathrm{s}^{-1}$

5. Convert the mass flow rate to grams per second: The question asks for the answer in $\mathrm{gm}\,\mathrm{s}^{-1}$. $\dot{m} = 0.0471 \times 1000\,\mathrm{g}\,\mathrm{s}^{-1} = 47.1\,\mathrm{g}\,\mathrm{s}^{-1}$

   The steady mass flow rate of air exiting the chimney is $47.1\,\mathrm{gm}\,\mathrm{s}^{-1}$.