Two water drops each of radius r$r$ coalesce to form a bigger drop. If T$T$ is the surface tension, the surface energy released in this process is:

### Related Formula Bernoulli's pressure balance equation for aerofoils: Delta P = P_1 - P_2 = frac12rho(v_2^2 - v_1^2)$$\Delta P = P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$$ Dynamic Lift force balancing plane weight: F_textlift = Delta P cdot A_texttotal = mg$$F_{\text{lift}} = \Delta P \cdot A_{\text{total}} = mg$$ ### Core Logic Convert velocity limits to SI units: v_1 = 180mathrm~km/h = 180 times frac518 = 50mathrm~ms^-1$$v_1 = 180\mathrm{~km/h} = 180 \times \frac{5}{18} = 50\mathrm{~ms}^{-1}$$ v_2 = 252mathrm~km/h = 252 times frac518 = 70mathrm~ms^-1$$v_2 = 252\mathrm{~km/h} = 252 \times \frac{5}{18} = 70\mathrm{~ms}^{-1}$$ Total effective wing area layout (2 wings): A_texttotal = 2 times 40 = 80mathrm~m^2$$A_{\text{total}} = 2 \times 40 = 80\mathrm{~m}^2$$ ### Step 1: Calculate Mass Balance Substitute these values into the dynamic lift equation: mg = frac12 rho (v_2^2 - v_1^2) A_texttotal$$mg = \frac{1}{2} \rho (v_2^2 - v_1^2) A_{\text{total}}$$ m(10) = frac12 times 1 times (70^2 - 50^2) times 80$$m(10) = \frac{1}{2} \times 1 \times (70^2 - 50^2) \times 80$$ 10m = 40 times (4900 - 2500) = 40 times 2400 = 96000$$10m = 40 \times (4900 - 2500) = 40 \times 2400 = 96000$$ m = 9600mathrm~kg$$m = 9600\mathrm{~kg}$$ ### Pattern Recognition Remember to multiply individual wing areas by 2 for standard multi-wing lift structures (A_texttotal = 2A$A_{\text{total}} = 2A$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Related Formula The work done in changing the surface area of a liquid is given by: W = T cdot Delta A$$W = T \cdot \Delta A$$ where: * T$T$ is the surface tension of the liquid. * Delta A = A_f - A_i$\Delta A = A_f - A_i$ is the change in the total surface area. ### Core Logic Since the total volume remains constant during splitting: V_i = V_f$V_i = V_f$ frac43pi R^3 = 27 times frac43pi r^3$$\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3$$ R^3 = 27r^3 implies r = fracR3$$R^3 = 27r^3 \implies r = \frac{R}{3}$$ ### Step 1: Calculate the Change in Surface Area Initial surface area of the single drop: A_i = 4pi R^2$A_i = 4\pi R^2$ Final surface area of 27$27$ small drops: A_f = 27 times (4pi r^2) = 27 times 4pi left(fracR3right)^2$$A_f = 27 \times (4\pi r^2) = 27 \times 4\pi \left(\frac{R}{3}\right)^2$$ A_f = 27 times 4pi fracR^29 = 12pi R^2$$A_f = 27 \times 4\pi \frac{R^2}{9} = 12\pi R^2$$ Change in surface area: Delta A = A_f - A_i = 12pi R^2 - 4pi R^2 = 8pi R^2$$\Delta A = A_f - A_i = 12\pi R^2 - 4\pi R^2 = 8\pi R^2$$ ### Step 2: Calculate Work Done Substituting the change in area into the work done formula: W = T cdot Delta A = 8pi R^2 T$$W = T \cdot \Delta A = 8\pi R^2 T$$ ### Pattern Recognition For splitting a large drop of radius R$R$ into n$n$ identical small drops, the change in surface area is given by Delta A = 4pi R^2 (n^1/3 - 1)$\Delta A = 4\pi R^2 (n^{1/3} - 1)$. Substituting n = 27$n = 27$ gives Delta A = 4pi R^2 (3 - 1) = 8pi R^2$\Delta A = 4\pi R^2 (3 - 1) = 8\pi R^2$, leading immediately to 8pi R^2 T$8\pi R^2 T$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Core Logic Statement (I): Liquids have much stronger intermolecular forces compared to gases, leading to significantly higher viscosity in liquids than in gases. Thus, Statement I is incorrect. Statement (II): The presence of insoluble impurities (like soap or detergents) disrupts the cohesive forces between liquid molecules at the surface, which decreases the surface tension. Thus, Statement II is correct. ### Pattern Recognition Viscosity in liquids decreases with temperature, whereas in gases it increases with temperature due to molecular collisions. Insoluble impurities act as surface-active agents that lower surface tension cohesive stability. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Related Formula The height of capillary rise (h$h$) is given by: h = frac2T cos thetarho g r$$h = \frac{2T \cos \theta}{\rho g r}$$ where, T$T$ = surface tension of the liquid theta$\theta$ = angle of contact rho$\rho$ = density of the liquid r$r$ = radius of the capillary tube ### Core Logic As the temperature of water increases, its intermolecular cohesive forces decrease. This leads to a decrease in surface tension (T$T$). Since h propto T$h \propto T$ (assuming rho$\rho$ and theta$\theta$ remain relatively constant), height of capillary rise decreases with increase in temperature. ### Step 1: Evaluate Statement I and II Because T_texthot lt T_textcold$T_{\text{hot}} \lt T_{\text{cold}}$, it follows that h_texthot lt h_textcold$h_{\text{hot}} \lt h_{\text{cold}}$. * **Statement I** states: the height of capillary rise will be smaller in hot water. This is **True**. * **Statement II** states: the height of capillary rise will be smaller in cold water. This is **False**. ### Pattern Recognition Remember the key physical dependence: **Temperature Up implies$\implies$ Surface Tension Down implies$\implies$ Capillary Rise Down**. This basic trend of cohesive/adhesive properties versus thermal energy frequently appears in competitive conceptual physical chemistry/fluid physics questions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Related Formula From Bernoulli's Principle (ignoring small height differences across the wing thickness): P_1 + frac12 rho v_1^2 = P_2 + frac12 rho v_2^2$$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$$ Delta P = P_2 - P_1 = frac12 rho (v_1^2 - v_2^2)$$\Delta P = P_2 - P_1 = \frac{1}{2} \rho (v_1^2 - v_2^2)$$ Net aerodynamic lift force (F$F$) acting on wing area A$A$ is: F = Delta P cdot A = frac12 rho (v_1^2 - v_2^2) A$$F = \Delta P \cdot A = \frac{1}{2} \rho (v_1^2 - v_2^2) A$$ ### Core Logic Given values: * Flow speed on upper surface (v_1$v_1$) = 70 mathrm~ms^-1$70 \mathrm{~ms}^{-1}$ * Flow speed on lower surface (v_2$v_2$) = 65 mathrm~ms^-1$65 \mathrm{~ms}^{-1}$ * Wing area (A$A$) = 2 mathrm~m^2$2 \mathrm{~m}^2$ * Density of air (rho$\rho$) = 1.2 mathrm~kgcdot m^-3$1.2 \mathrm{~kg\cdot m}^{-3}$ ### Step 1: Compute Lift Force Substituting these metrics directly into the dynamic lift equation: F = frac12 times 1.2 times left(70^2 - 65^2right) times 2$$F = \frac{1}{2} \times 1.2 \times \left(70^2 - 65^2\right) \times 2$$ Notice that the factors frac12$\frac{1}{2}$ and 2$2$ cancel out perfectly: F = 1.2 times left(4900 - 4225right)$$F = 1.2 \times \left(4900 - 4225\right)$$ F = 1.2 times 675 = 810 mathrm~N$$F = 1.2 \times 675 = 810 \mathrm{~N}$$ Therefore, the net lift force on the wing is 810 mathrm~N$810 \mathrm{~N}$. ### Pattern Recognition Use the algebraic difference of squares shortcut a^2 - b^2 = (a-b)(a+b)$a^2 - b^2 = (a-b)(a+b)$ to solve velocity square differences quickly: 70^2 - 65^2 = (70-65)(70+65) = 5 times 135 = 675$70^2 - 65^2 = (70-65)(70+65) = 5 \times 135 = 675$, bypassing squaring heavy multi-digit calculations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids