Consider a completely full cylindrical water tank of height 1.6mathrm~m$1.6\mathrm{~m}$ and cross-sectional area 0.5mathrm~m^2$0.5\mathrm{~m}^2$ . It has a small hole in its side at a height 90mathrm~cm$90\mathrm{~cm}$ from the bottom. Assume, the cross-sectional area of the hole to be negligibly small as compared to that of the water tank. If a load 50mathrm~kg$50\mathrm{~kg}$ is applied at the top surface of the water in the tank then the velocity of the water coming out at the instant when the hole is opened is: (mathrmg = 10mathrm~m/s^2)$(\mathrm{g} = 10\mathrm{~m/s}^2)$

### Related Formula Bernoulli's Principle between the top surface (1) and the hole (2): P_1 + frac12rho v_1^2 + rho g h_1 = P_2 + frac12rho v_2^2 + rho g h_2$$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$$ ### Core Logic Let the top surface be point 1, and the opening of the hole be point 2. - Height of tank, H = 1.6mathrm~m$H = 1.6\mathrm{~m}$ (thus height of point 1, h_1 = 1.6mathrm~m$h_1 = 1.6\mathrm{~m}$) - Height of hole, h_2 = 90mathrm~cm = 0.9mathrm~m$h_2 = 90\mathrm{~cm} = 0.9\mathrm{~m}$ from the bottom. - Depth of the hole below top surface, h = H - h_2 = 1.6 - 0.9 = 0.7mathrm~m$h = H - h_2 = 1.6 - 0.9 = 0.7\mathrm{~m}$. - Since area of hole is negligibly small compared to tank area, the velocity of fluid surface at the top is negligible (v_1 approx 0$v_1 \approx 0$). - Pressure at point 1 (top surface): P_1 = P_0 + fracMgA$$P_1 = P_0 + \frac{Mg}{A}$$ where M = 50mathrm~kg$M = 50\mathrm{~kg}$, A = 0.5mathrm~m^2$A = 0.5\mathrm{~m}^2$, P_0$P_0$ is atmospheric pressure. - Pressure at point 2 (outside the hole): P_2 = P_0$P_2 = P_0$ ### Step 1: Applying Bernoulli's Equation Substitute these values into Bernoulli's equation with reference datum at the hole (h_2 = 0$h_2 = 0$): P_1 + rho g h = P_2 + frac12rho v_2^2$$P_1 + \rho g h = P_2 + \frac{1}{2}\rho v_2^2$$ left(P_0 + fracMgAright) + rho g h = P_0 + frac12rho v_2^2$$\left(P_0 + \frac{Mg}{A}\right) + \rho g h = P_0 + \frac{1}{2}\rho v_2^2$$ fracMgA + rho g h = frac12rho v_2^2$$\frac{Mg}{A} + \rho g h = \frac{1}{2}\rho v_2^2$$ ### Step 2: Substitution and Calculation Substitute M = 50mathrm~kg$M = 50\mathrm{~kg}$, g = 10mathrm~m/s^2$g = 10\mathrm{~m/s}^2$, A = 0.5mathrm~m^2$A = 0.5\mathrm{~m}^2$, rho = 10^3mathrm~kg/m^3$\rho = 10^3\mathrm{~kg/m}^3$, and h = 0.7mathrm~m$h = 0.7\mathrm{~m}$: frac50 times 100.5 + 10^3 times 10 times 0.7 = frac12 times 10^3 times v_2^2$$\frac{50 \times 10}{0.5} + 10^3 \times 10 \times 0.7 = \frac{1}{2} \times 10^3 \times v_2^2$$ 1000 + 7000 = 500 times v_2^2$$1000 + 7000 = 500 \times v_2^2$$ 8000 = 500 times v_2^2$$8000 = 500 \times v_2^2$$ v_2^2 = 16 implies v_2 = 4mathrm~m/s$$v_2^2 = 16 \implies v_2 = 4\mathrm{~m/s}$$

### Pattern Recognition Modified Torricelli's formula with top pressure overload: v = sqrt2gh + frac2Delta Prho$$v = \sqrt{2gh + \frac{2\Delta P}{\rho}}$$ Substitute Delta P = fracMgA = frac5000.5 = 1000mathrm~Pa$\Delta P = \frac{Mg}{A} = \frac{500}{0.5} = 1000\mathrm{~Pa}$. This quickly yields: v = sqrt2(10)(0.7) + frac2(1000)1000 = sqrt14 + 2 = 4mathrm~m/s$$v = \sqrt{2(10)(0.7) + \frac{2(1000)}{1000}} = \sqrt{14 + 2} = 4\mathrm{~m/s}$$. Keep this shortcut in mind for pressurized tanks! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids