Step-by-step Solution

1. Analyze the initial state:

   • Let the total height of the cylindrical vessel be L = 500 mm = 0.5 m.
   • Water is initially filled to a height H.
   • The top of the vessel is then sealed. This traps a column of air above the water.
   • The height of this air column is (L - H).
   • Since the vessel was sealed at atmospheric conditions, the initial pressure of the trapped air is the atmospheric pressure, $P_1 = P_{atm} = 1.0 \times 10^5$ N/m$^2$.
   • Let A be the cross-sectional area of the cylinder. The initial volume of the trapped air is $V_1 = A(L - H)$.

2. Analyze the final state:

   • The orifice at the bottom is opened, and water flows out until the water level becomes steady at a final height $h_f = 200$ mm = 0.2 m.
   • The water flow stops when the total pressure at the orifice level inside the vessel equals the atmospheric pressure outside.
   • In the final state, the height of the air column is $(L - h_f)$.
   • The final volume of the trapped air is $V_2 = A(L - h_f)$.
   • Let the final pressure of the trapped air be $P_2$.
   • The pressure balance equation at the orifice is: $P_2 + \rho g h_f = P_{atm}$ where $\rho$ is the density of water and g is the acceleration due to gravity.
   • From this, we can express the final air pressure as: $P_2 = P_{atm} - \rho g h_f$

3. Apply Boyle's Law:

   • As the water level drops, the trapped air expands from volume $V_1$ to $V_2$. Assuming the temperature of the air remains constant during this process, we can apply Boyle's Law: $P_1 V_1 = P_2 V_2$
   • Substituting the expressions from the previous steps: $P_{atm} \times A(L - H) = (P_{atm} - \rho g h_f) \times A(L - h_f)$
   • The cross-sectional area A cancels out: $P_{atm}(L - H) = (P_{atm} - \rho g h_f)(L - h_f)$

4. Calculate the initial height H:

   • We are given the following values:
     • $P_{atm} = 1.0 \times 10^5$ N/m$^2$
     • $\rho = 1000$ kg/m$^3$
     • g = 10 m/s$^2$
     • L = 0.5 m
     • $h_f = 0.2$ m
   • First, let's calculate the hydrostatic pressure term $\rho g h_f$: $\rho g h_f = 1000 \times 10 \times 0.2 = 2000 \text{ N/m}^2$
   • Now, substitute all the values into the equation from Step 3: $1.0 \times 10^5 \times (0.5 - H) = (1.0 \times 10^5 - 2000) \times (0.5 - 0.2)$ $10^5 (0.5 - H) = (98000) \times (0.3)$ $10^5 (0.5 - H) = 29400$
   • Solve for (0.5 - H): $0.5 - H = \frac{29400}{10^5} = 0.294$
   • Solve for H: $H = 0.5 - 0.294 = 0.206 \text{ m}$
   • The initial height of the water was H = 206 mm.

5. Calculate the fall in height:

   • The fall in height is the difference between the initial height and the final height. $\text{Fall in height} = H - h_f$ $\text{Fall in height} = 206 \text{ mm} - 200 \text{ mm} = 6 \text{ mm}$

Thus, the fall in height of the water level is 6 mm.