The excess pressure inside a soap bubble A in air is half the excess pressure inside another soap bubble B in air. If the volume of the bubble A is n$n$ times the volume of the bubble B, then the value of n$n$ is ________.

### Related Formula A_1 V_1 = A_2 V_2$$A_1 V_1 = A_2 V_2$$ P_1 + frac12rho V_1^2 = P_2 + frac12rho V_2^2$$P_1 + \frac{1}{2}\rho V_1^2 = P_2 + \frac{1}{2}\rho V_2^2$$ P_1 - P_2 = rho g h$$P_1 - P_2 = \rho g h$$ ### Core Logic From the continuity equation between A and B: A_AV_A = A_BV_B implies 6V_A = 3V_B implies V_B = 2V_A$$A_{A}V_{A} = A_{B}V_{B} \implies 6V_{A} = 3V_{B} \implies V_{B} = 2V_{A}$$ Applying Bernoulli's equation between A and B for a horizontal pipe: P_A + frac12rho V_A^2 = P_B + frac12rho V_B^2$$P_{A} + \frac{1}{2}\rho V_{A}^{2} = P_{B} + \frac{1}{2}\rho V_{B}^{2}$$ P_A - P_B = frac12rho(V_B^2 - V_A^2)$$P_{A} - P_{B} = \frac{1}{2}\rho(V_{B}^{2} - V_{A}^{2})$$ Since the height difference is h = 5text cm = 0.05text m$h = 5\text{ cm} = 0.05\text{ m}$, the pressure difference is rho gh$\rho gh$. rho g times 0.05 = frac12rho( (2V_A)^2 - V_A^2 )$$\rho g \times 0.05 = \frac{1}{2}\rho( (2V_A)^2 - V_A^2 )$$ g times 0.05 = frac12(3V_A^2)$$g \times 0.05 = \frac{1}{2}(3V_{A}^{2})$$ ### Step 1: Calculate Velocity and Volume Flow Rate Solving for V_A$V_A$: V_A = sqrtfrac2 times 10 times 0.053 = sqrtfrac13 = frac1sqrt3text m/s$$V_{A} = \sqrt{\frac{2 \times 10 \times 0.05}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}\text{ m/s}$$ V_A = frac100sqrt3text cm/s$$V_A = \frac{100}{\sqrt{3}}\text{ cm/s}$$ Volume flow rate Q = A_A V_A$Q = A_{A} V_{A}$: Q = 6text cm^2 times frac100sqrt3text cm/s = frac600sqrt3 = 200sqrt3text cm^3text/s$$Q = 6\text{ cm}^2 \times \frac{100}{\sqrt{3}}\text{ cm/s} = \frac{600}{\sqrt{3}} = 200\sqrt{3}\text{ cm}^{3}\text{/s}$$ ### Pattern Recognition In a horizontal Venturi meter, substituting V_2 = V_1 (A_1/A_2)$V_2 = V_1 (A_1/A_2)$ directly into rho g h = frac12rho (V_2^2 - V_1^2)$\rho g h = \frac{1}{2}\rho (V_2^2 - V_1^2)$ is the standard path. Working in CGS vs MKS units requires careful tracking; convert V_A$V_A$ to cm/s before multiplying by A_A$A_A$ in cm². ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### 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