A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it up to height H. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 mm. Find the fall in height (in mm) of water level due to opening of the orifice. [Take atmospheric pressure $= 1.0 x 10^{5} N/m^{2},$ density of water $= 1000$ kg$/m^{3}$ and $g = 10 m/s^{2}.$ Neglect any effect of surface tension.]

Step 1: Identify the given parameters.

The physical parameters of the cylindrical vessel and the fluid are:

$$L = 500\,\text{mm}$$ $$h_f = 200\,\text{mm} = 0.2\,\text{m}$$ $$P_{\text{atm}} = 1.0 \times 10^{5}\,\text{N/m}^{2}$$ $$\rho = 1000\,\text{kg/m}^{3}$$ $$g = 10\,\text{m/s}^{2}$$

Step 2: Analyze the steady-state condition.

When the water level becomes steady, the pressure at the bottom of the vessel just inside the orifice must equal the atmospheric pressure outside.

$$P_{1} + \rho g h_f = P_{\text{atm}}$$

Here, $P_{1}$ is the final pressure of the trapped air.

Step 3: Calculate the final pressure of the trapped air.

Rearrange the steady-state equation to solve for $P_{1}$.

$$P_{1} = P_{\text{atm}} - \rho g h_f$$

Substitute the given values into the equation.

$$P_{1} = 10^{5} - (1000)(10)(0.2)$$ $$P_{1} = 9.8 \times 10^{4}\,\text{Pa}$$

Step 4: Apply Boyle's Law to the trapped air.

Assuming the temperature of the trapped air remains constant during the expansion, apply the isothermal gas equation.

$$P_{0} V_{0} = P_{1} V_{1}$$

Since the vessel has a uniform cross-sectional area $A$, the volume of the air column is proportional to its height. The initial air pressure is $P_{0} = P_{\text{atm}}$.

$$P_{\text{atm}} A (L - H) = P_{1} A (L - h_f)$$ $$P_{\text{atm}} (L - H) = P_{1} (L - h_f)$$

Step 5: Calculate the initial air column height.

Find the final air column height.

$$L - h_f = 500\,\text{mm} - 200\,\text{mm} = 300\,\text{mm}$$

Substitute the known values to find the initial air column height $L - H$.

$$L - H = \frac{P_{1} (L - h_f)}{P_{\text{atm}}}$$ $$L - H = \frac{9.8 \times 10^{4}}{10^{5}} \times 300\,\text{mm}$$ $$L - H = 294\,\text{mm}$$

Step 6: Determine the initial water height and the fall in water level.

Calculate the initial water height $H$.

$$H = 500\,\text{mm} - 294\,\text{mm} = 206\,\text{mm}$$

Calculate the fall in the water level.

$$\text{Fall} = H - h_f$$ $$\text{Fall} = 206\,\text{mm} - 200\,\text{mm} = 6\,\text{mm}$$

Answer: The fall in height of the water level is $6\,\text{mm}$.