# Step-by-Step Solution Let's break down the problem: **Given:** - Hole diameter (\( d_h \)) = 5.4 mm - Tube internal diameter (\( D \)) = 72 mm - Height of water above the hole (\( h \)) = 25 cm = .25 m We are to **estimate the time for the bottle to drain**. --- ## 1. **Calculate the Areas** - **Area of hole (\( A_h \)):** \[ d_h = 5.4\, \text{mm} = .0054\, \text{m} \] \[ A_h = \frac{\pi}{4} d_h^2 = \frac{\pi}{4} (.0054)^2 \approx 2.29 \times 10^{-5}\, \text{m}^2 \] - **Area of tube (\( A_t \)):** \[ D = 72\, \text{mm} = .072\, \text{m} \] \[ A_t = \frac{\pi}{4} D^2 = \frac{\pi}{4} (.072)^2 \approx 4.07 \times 10^{-3}\, \text{m}^2 \] --- ## 2. **Torricelli’s Law for Flow Speed** The speed of water exiting the hole (neglecting viscosity, assuming atmospheric pressure at both ends): \[ v = \sqrt{2gh} \] Where \( g = 9.81\, \text{m/s}^2 \) and \( h \) is the height above the hole. --- ## 3. **Volume and Draining Rate** - **Initial volume of water above the hole:** \[ V = A_t h = 4.07 \times 10^{-3} \times .25 = 1.02 \times 10^{-3}\, \text{m}^3 \] - **Rate of change of water level:** The rate of outflow through the hole is: \[ Q = A_h v = A_h \sqrt{2gh} \] But as the water drains, \( h \) decreases. --- ## 4. **Set Up the Differential Equation** Let \( H \) be the varying height of the water above the hole. \[ A_t \frac{dH}{dt} = -A_h \sqrt{2gH} \] Negative sign: \( H \) decreases as \( t \) increases. Rearrange: \[ \frac{dH}{\sqrt{H}} = -\frac{A_h}{A_t} \sqrt{2g} dt \] --- ## 5. **Integrate Both Sides** Integrate from \( H = h \) to \( H = \) (from \( t = \) to \( t = T \)): \[ \int_h^ \frac{dH}{\sqrt{H}} = -\frac{A_h}{A_t} \sqrt{2g} \int_^T dt \] \[ \left[ 2\sqrt{H} \right]_h^ = -\frac{A_h}{A_t} \sqrt{2g} T \] \[ 2( - \sqrt{h}) = -\frac{A_h}{A_t} \sqrt{2g} T \] \[ -2\sqrt{h} = -\frac{A_h}{A_t} \sqrt{2g} T \] \[ T = \frac{2 A_t \sqrt{h}}{A_h \sqrt{2g}} \] --- ## 6. **Plug in the Numbers** \[ T = \frac{2 \times 4.07 \times 10^{-3} \times \sqrt{.25}}{2.29 \times 10^{-5} \times \sqrt{2 \times 9.81}} \] First, calculate \(\sqrt{.25} = .5\). Calculate \(\sqrt{2 \times 9.81} \approx \sqrt{19.62} \approx 4.43\). Now, plug in: \[ T = \frac{2 \times 4.07 \times 10^{-3} \times .5}{2.29 \times 10^{-5} \times 4.43} \] \[ T = \frac{2 \times 4.07 \times .5 \times 10^{-3}}{2.29 \times 4.43 \times 10^{-5}} \] \[ T = \frac{4.07 \times 10^{-3}}{10.15 \times 10^{-5}} \] \[ T = \frac{4.07}{.1015} \approx 40.1 \text{ seconds} \] --- ## **Final Answer** \[ \boxed{T \approx 40 \text{ seconds}} \] **Summary:** It will take approximately **40 seconds** for the bottle to drain.