### Related Formula 1. Surface Energy: U = T cdot A = T cdot (4pi R^2)$U = T \cdot A = T \cdot (4\pi R^2)$ 2. Conservation of Volume during coalescence of drops: 2 times left(frac43pi r^3right) = frac43pi R^3$2 \times \left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3$ ### Core Logic When two drops of radius r$r$ coalesce into a single larger drop of radius R$R$, volume is conserved: R^3 = 2r^3 implies R = 2^1/3 r$R^3 = 2r^3 \implies R = 2^{1/3} r$ - Initial surface area of the two separate drops: A_i = 2 times 4pi r^2 = 8pi r^2$A_i = 2 \times 4\pi r^2 = 8\pi r^2$ - Final surface area of the combined single drop: A_f = 4pi R^2 = 4pi (2^1/3 r)^2 = 4pi r^2 2^2/3$A_f = 4\pi R^2 = 4\pi (2^{1/3} r)^2 = 4\pi r^2 2^{2/3}$ - Surface energy released: Delta E = U_i - U_f = T(A_i - A_f)$\Delta E = U_i - U_f = T(A_i - A_f)$ Delta E = T left( 8pi r^2 - 4pi r^2 2^2/3 right) = 4pi r^2 T left[ 2 - 2^2/3 right]$\Delta E = T \left( 8\pi r^2 - 4\pi r^2 2^{2/3} \right) = 4\pi r^2 T \left[ 2 - 2^{2/3} \right]$ ### Pattern Recognition Sees: Coalescence of N$N$ identical drops. Trap: Forgetting to conserve volume first, or confusing initial and final surface areas. Shortcut: Energy released when N$N$ drops coalesce into one big drop is: Delta E = 4pi r^2 T left[ N - N^2/3 right]$\Delta E = 4\pi r^2 T \left[ N - N^{2/3} \right]$ Here, substituting N = 2$N = 2$ directly gives 4pi r^2 T left[ 2 - 2^2/3 right]$4\pi r^2 T \left[ 2 - 2^{2/3} \right]$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Related Formula P = P_0 + rho g h$P = P_0 + \rho g h$ F = Delta P cdot A$F = \Delta P \cdot A$ ### Core Logic Let's analyze the pressure acting on the window at depth h = 3mathrm~m$h = 3\mathrm{~m}$ from both sides: - Water side (density rho_w = 1.0 times 10^3mathrm~kg/m^3$\rho_w = 1.0 \times 10^3\mathrm{~kg/m}^3$): P_w = P_0 + rho_w g h$P_w = P_0 + \rho_w g h$ - Denser liquid side (density rho_l = 1.5 times 10^3mathrm~kg/m^3$\rho_l = 1.5 \times 10^3\mathrm{~kg/m}^3$): P_l = P_0 + rho_l g h$P_l = P_0 + \rho_l g h$ The net pressure difference pushing on the window is: Delta P = P_l - P_w = (rho_l - rho_w) g h$\Delta P = P_l - P_w = (\rho_l - \rho_w) g h$ Delta P = (1.5 times 10^3 - 1.0 times 10^3) times 10 times 3 = 500 times 30 = 15000mathrm~N/m^2$\Delta P = (1.5 \times 10^3 - 1.0 \times 10^3) \times 10 \times 3 = 500 \times 30 = 15000\mathrm{~N/m}^2$ The window has area A = 100mathrm~cm^2 = 100 times 10^-4mathrm~m^2 = 10^-2mathrm~m^2$A = 100\mathrm{~cm}^2 = 100 \times 10^{-4}\mathrm{~m}^2 = 10^{-2}\mathrm{~m}^2$. To keep the door from opening, we must apply a force balancing the pressure difference: F = Delta P cdot A = 15000 times 10^-2 = 150mathrm~N$F = \Delta P \cdot A = 15000 \times 10^{-2} = 150\mathrm{~N}$ ### Step 1: Final Conclusion The required force is $ ### Step 1: Final Conclusion The required force is $150\mathrm{~N}. ### Pattern Recognition For a vertical interface between two fluids, the net pressure difference is simply given by $. ### Pattern Recognition For a vertical interface between two fluids, the net pressure difference is simply given by $\Delta P = \Delta \rho \cdot g h. The ambient atmospheric pressure $. The ambient atmospheric pressure $P_0$ cancels out since it acts on both sides of the window. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Related Formula The gravitational potential energy U$U$ of a liquid column of mass m$m$ and height h$h$ is evaluated relative to its bottom by placing its total mass at its center of mass (h/2$h/2$): U = m g left(frach2right) = (rho A h) g left(frach2right) = frac12 rho A g h^2$U = m g \left(\frac{h}{2}\right) = (\rho A h) g \left(\frac{h}{2}\right) = \frac{1}{2} \rho A g h^2$ Work done by the force of gravity (W$W$) equals the negative change in potential energy: W = -Delta U = U_i - U_f$W = -\Delta U = U_i - U_f$ ### Core Logic Since the vessels are identical and connected at the bottom, water flows from the higher column to the lower one until their final heights equalize at: h_f = frac10 + 62 = 8mathrm~m$h_f = \frac{10 + 6}{2} = 8\mathrm{~m}$ ### Step 1: Calculate Initial Potential Energy (U_i$U_i$) Let the reference level U=0$U=0$ be at the bottom: U_i = U_1 + U_2 = frac12 rho A g h_1^2 + frac12 rho A g h_2^2$U_i = U_1 + U_2 = \frac{1}{2} \rho A g h_1^2 + \frac{1}{2} \rho A g h_2^2$ U_i = frac12 rho A g left(10^2 + 6^2right) = frac12 rho A g (100 + 36) = 68 rho A g$U_i = \frac{1}{2} \rho A g \left(10^2 + 6^2\right) = \frac{1}{2} \rho A g (100 + 36) = 68 \rho A g$

### Step 2: Calculate Final Potential Energy (U_f$U_f$) Both vessels equalize to h_f = 8mathrm~m$h_f = 8\mathrm{~m}$: U_f = 2 times left[ frac12 rho A g h_f^2 right] = rho A g (8^2) = 64 rho A g$U_f = 2 \times \left[ \frac{1}{2} \rho A g h_f^2 \right] = \rho A g (8^2) = 64 \rho A g$ ### Step 3: Work Done by Gravity (W$W$) W = U_i - U_f = 68 rho A g - 64 rho A g = 4 rho A g$W = U_i - U_f = 68 \rho A g - 64 \rho A g = 4 \rho A g$ Substitute the given values ( ho = 10^3mathrm~kg/m^3$ ho = 10^3\mathrm{~kg/m}^3$, A = 2mathrm~m^2$A = 2\mathrm{~m}^2$, g = 10mathrm~m/s^2$g = 10\mathrm{~m/s}^2$): W = 4 times 10^3 times 2 times 10 = 8 times 10^4mathrm~J$W = 4 \times 10^3 \times 2 \times 10 = 8 \times 10^4\mathrm{~J}$ ### Pattern Recognition For leveling liquids between two identical connected columns, the shift in center of mass always simplifies. The loss in potential energy is given by $ ### Pattern Recognition For leveling liquids between two identical connected columns, the shift in center of mass always simplifies. The loss in potential energy is given by $\Delta U = \frac{1}{4} \rho A g (h_1 - h_2)^2. Applying this directly: $. Applying this directly: $Delta U = frac14 times 10^3 times 2 times 10 times (10 - 6)^2 = 5000 times 16 = 8 times 10^4mathrm~J$\Delta U = \frac{1}{4} \times 10^3 \times 2 \times 10 \times (10 - 6)^2 = 5000 \times 16 = 8 \times 10^4\mathrm{~J}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Related Formula Terminal velocity v_t$v_t$ of a spherical body falling through a viscous fluid is given by Stokes' Law: v_t = frac29 fracr^2 g (rho_textbody - rho_textfluid)eta$v_t = \frac{2}{9} \frac{r^2 g (\rho_{\text{body}} - \rho_{\text{fluid}})}{\eta}$ Hence, the viscosity coefficient eta$\eta$ is: eta = frac29 fracr^2 g (rho_textbody - rho_textfluid)v_t$\eta = \frac{2}{9} \frac{r^2 g (\rho_{\text{body}} - \rho_{\text{fluid}})}{v_t}$ ### Core Logic Given parameters: - Diameter d = 3.6mathrm~mm Rightarrow$d = 3.6\mathrm{~mm} \Rightarrow$ radius r = 1.8mathrm~mm = 1.8 times 10^-3mathrm~m$r = 1.8\mathrm{~mm} = 1.8 \times 10^{-3}\mathrm{~m}$ - Terminal velocity v_t = 2.45 times 10^-2mathrm~m/s$v_t = 2.45 \times 10^{-2}\mathrm{~m/s}$ - Liquid density rho_textfluid = 925mathrm~kg/m^3$\rho_{\text{fluid}} = 925\mathrm{~kg/m}^3$ - Steel density rho_textbody = 7825mathrm~kg/m^3$\rho_{\text{body}} = 7825\mathrm{~kg/m}^3$ - Acceleration due to gravity g = 9.8mathrm~m/s^2$g = 9.8\mathrm{~m/s}^2$ ### Step 1: Substitute parameters into formula eta = frac29 times frac(1.8 times 10^-3)^2 times 9.8 times (7825 - 925)2.45 times 10^-2$\eta = \frac{2}{9} \times \frac{(1.8 \times 10^{-3})^2 \times 9.8 \times (7825 - 925)}{2.45 \times 10^{-2}}$ eta = frac29 times frac3.24 times 10^-6 times 9.8 times 69002.45 times 10^-2$\eta = \frac{2}{9} \times \frac{3.24 \times 10^{-6} \times 9.8 \times 6900}{2.45 \times 10^{-2}}$ ### Step 2: Solve the numeric calculation eta = frac2 times 0.36 times 10^-6 times 9.8 times 69002.45 times 10^-2$\eta = \frac{2 \times 0.36 \times 10^{-6} \times 9.8 \times 6900}{2.45 \times 10^{-2}}$ eta = frac0.72 times 9.8 times 6900 times 10^-62.45 times 10^-2$\eta = \frac{0.72 \times 9.8 \times 6900 \times 10^{-6}}{2.45 \times 10^{-2}}$ eta = frac48700.8 times 10^-62.45 times 10^-2 = frac0.04870080.0245 approx 1.99mathrm~Pacdot s$\eta = \frac{48700.8 \times 10^{-6}}{2.45 \times 10^{-2}} = \frac{0.0487008}{0.0245} \approx 1.99\mathrm{~Pa\cdot s}$ ### Pattern Recognition Be careful when converting diameter to radius ($ ### Pattern Recognition Be careful when converting diameter to radius ($r = d/2$). Always ensure all numerical parameters are converted cleanly to SI base units (meters, kilograms, seconds) before applying Stokes' formula. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Related Formula Excess pressure inside a soap bubble in air (which has two liquid-gas interfaces) is given by: Delta P = frac4TR Rightarrow R propto frac1Delta P$\Delta P = \frac{4T}{R} \Rightarrow R \propto \frac{1}{\Delta P}$ The volume V$V$ of a spherical bubble of radius R$R$ is: V = frac43pi R^3 Rightarrow V propto R^3 propto left(frac1Delta Pright)^3$V = \frac{4}{3}\pi R^3 \Rightarrow V \propto R^3 \propto \left(\frac{1}{\Delta P}\right)^3$ ### Core Logic Given state: Delta P_A = frac12 Delta P_B Rightarrow fracDelta P_BDelta P_A = 2$\Delta P_A = \frac{1}{2} \Delta P_B \Rightarrow \frac{\Delta P_B}{\Delta P_A} = 2$ ### Step 1: Calculate the ratio of radii fracR_AR_B = fracDelta P_BDelta P_A = 2$\frac{R_A}{R_B} = \frac{\Delta P_B}{\Delta P_A} = 2$ ### Step 2: Calculate volume scaling parameter (n$n$) V_A = n V_B Rightarrow n = fracV_AV_B = left(fracR_AR_Bright)^3 = (2)^3 = 8$V_A = n V_B \Rightarrow n = \frac{V_A}{V_B} = \left(\frac{R_A}{R_B}\right)^3 = (2)^3 = 8$ ### Pattern Recognition Volume scale factors depend on the cube of linear scale factors (V propto R^3$V \propto R^3$). Since radius is inversely proportional to excess pressure, a halving of excess pressure leads to doubling of radius, scaling volume by 2^3 = 8$2^3 = 8$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids