Water flows out from a large tank of cross-sectional area \( A_t = 1 \ \text{m}^2 \) through a small rounded orifice of cross-sectional area \( A_o = 1 \ \text{cm}^2 \), located at \( y = 0 \). Initially the water level, measured from \( y = 0 \), is \( H = 1 \ \text{m} \). The acceleration due to gravity is \( 9.8 \ \text{m/s}^2 \).
Neglecting any losses, the time taken by water in the tank to reach a level of \( y = H/4 \) is \(\underline{\hspace{2cm}}\) seconds (round off to one decimal place).
Show Hint
Torricelli's law gives the discharge velocity of water flowing from an orifice and can be used to calculate the time it takes for the water to reach a particular level.
We can use Torricelli's law to model the flow rate of the water from the tank. The equation for the discharge from an orifice is: \[ Q = A_o \sqrt{2gH} \] Where: - \( Q \) is the flow rate, - \( A_o = 1 \ \text{cm}^2 = 1 \times 10^{-4} \ \text{m}^2 \), - \( g = 9.8 \ \text{m/s}^2 \) is the acceleration due to gravity, - \( H \) is the initial height of the water. The volume of water leaving the tank is \( V = A_t H \), and the time \( t \) to reach a certain height can be calculated by integrating the flow rate over the height. The differential equation governing the process is: \[ \frac{dH}{dt} = -\frac{A_o}{A_t} \sqrt{2gH} \] Rearranging the equation: \[ \frac{dH}{\sqrt{H}} = -\frac{A_o}{A_t} \sqrt{2g} dt \] Integrating from \( H = 1 \ \text{m} \) to \( H = \frac{1}{4} \ \text{m} \): \[ \int_1^{1/4} \frac{dH}{\sqrt{H}} = -\frac{A_o}{A_t} \sqrt{2g} \int_0^t dt \] The solution is: \[ 2\sqrt{H} \Bigg|_1^{1/4} = -\frac{A_o}{A_t} \sqrt{2g} t \] Substituting the values: \[ 2(\sqrt{1} - \sqrt{1/4}) = -\frac{1 \times 10^{-4}}{1} \times \sqrt{2 \times 9.8} \times t \] Simplifying: \[ 2(1 - 0.5) = -10^{-4} \times 4.43 \times t \] \[ 1 = -4.43 \times 10^{-4} \times t \] Solving for \( t \): \[ t = \frac{1}{4.43 \times 10^{-4}} = 2257.3 \ \text{seconds} \] Thus, the time is: \[ \boxed{2257.3 \ \text{seconds}} \]
