To solve this problem, we will use the principles of fluid mechanics, specifically the equation of continuity and Bernoulli's equation.

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Given Data:

1. Pressure at the storage tank: $P_1 = 3.0 \times 10^5 \, \mathrm{Pa}$
2. Speed at the storage tank: $v_1 = 1.6 \, \mathrm{m/s}$
3. Radius of the pipe at the storage tank: $r_1 = 1.0 \, \mathrm{cm} = 0.01 \, \mathrm{m}$
4. Radius of the pipe on the ground floor: $r_2 = 0.50 \, \mathrm{cm} = 0.005 \, \mathrm{m}$
5. Height difference: $h = 9.0 \, \mathrm{m}$
6. Density of water: $\rho = 1000 \, \mathrm{kg/m^3}$ (standard assumption)

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(a) Speed of Flow on the Ground Floor ($v_2$):

The equation of continuity relates the speed and cross-sectional areas of the fluid at two points:

$A_1 v_1 = A_2 v_2$

The cross-sectional area of a pipe is:

$A = \pi r^2$

Thus, the areas are:

$A_1 = \pi (r_1)^2 = \pi (0.01)^2 = 3.14 \times 10^{-4} \, \mathrm{m^2}$

$A_2 = \pi (r_2)^2 = \pi (0.005)^2 = 7.85 \times 10^{-5} \, \mathrm{m^2}$

From the continuity equation:

$A_1 v_1 = A_2 v_2$

$v_2 = \frac{A_1 v_1}{A_2} = \frac{(3.14 \times 10^{-4})(1.6)}{7.85 \times 10^{-5}}$

$v_2 = 6.4 \, \mathrm{m/s}$

So, the speed of flow on the ground floor is:

$\boxed{v_2 = 6.4 \, \mathrm{m/s}}$

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(b) Pressure in the Pipe on the Ground Floor ($P_2$):

We apply Bernoulli's equation:

$P_1 + \frac{1}{2} \rho v_1^2 + \rho g h = P_2 + \frac{1}{2} \rho v_2^2$

Rearranging for $P_2$:

$P_2 = P_1 + \frac{1}{2} \rho v_1^2 - \frac{1}{2} \rho v_2^2 - \rho g h$

Substituting the known values:

$P_1 = 3.0 \times 10^5 \, \mathrm{Pa}, \quad \rho = 1000 \, \mathrm{kg/m^3}, \quad g = 9.8 \, \mathrm{m/s^2}, \quad h = 9.0 \, \mathrm{m}$

$P_2 = (3.0 \times 10^5) + \frac{1}{2} (1000)(1.6)^2 - \frac{1}{2} (1000)(6.4)^2 - (1000)(9.8)(9.0)$

First, calculate each term:

1. Dynamic pressure at the top:

$\frac{1}{2} \rho v_1^2 = \frac{1}{2} (1000)(1.6)^2 = 1280 \, \mathrm{Pa}$

2. Dynamic pressure at the bottom:

$\frac{1}{2} \rho v_2^2 = \frac{1}{2} (1000)(6.4)^2 = 20480 \, \mathrm{Pa}$

3. Gravitational potential energy term:

$\rho g h = (1000)(9.8)(9.0) = 88200 \, \mathrm{Pa}$

Now substitute:

$P_2 = (3.0 \times 10^5) + 1280 - 20480 - 88200$

$P_2 = 300000 + 1280 - 20480 - 88200 = 231600 \, \mathrm{Pa}$

Thus, the pressure on the ground floor is:

$\boxed{P_2 = 2.32 \times 10^5 \, \mathrm{Pa}}$

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Final Answers:

1. $v_2 = 6.4 \, \mathrm{m/s}$
2. $P_2 = 2.32 \times 10^5 \, \mathrm{Pa}$

Would you like further clarification or help with related problems? Here are some possible follow-ups:

1. How does the height difference affect the pressure and speed?
2. Why does the pressure decrease when the speed increases?
3. What assumptions are made in Bernoulli's equation?
4. How would using a fluid other than water (e.g., oil) change the result?
5. What happens if the pipe radius on the ground floor increases?

Tip: The conservation of energy is key in fluid