### Related Formula The absolute pressure inside an air bubble submerged in a liquid is given by : P_textin = P_0 + ho gh + frac2TR$P_{\text{in}} = P_{0} + ho gh + \frac{2T}{R}$ where P_0$P_0$ is atmospheric pressure, ho$ ho$ is the liquid density, g$g$ is gravity, h$h$ is depth, T$T$ is surface tension, and R$R$ is the radius. ### Core Logic We are given that the difference between the inside pressure and atmospheric pressure is 2100text N/m^2$2100\text{ N/m}^2$: P_textin - P_0 = ho gh + frac2TR = 2100$P_{\text{in}} - P_{0} = ho gh + \frac{2T}{R} = 2100$ ### Step 1: Numerical Evaluation Substitute the given values into the relation : ho gh = 1000 times 10 times 0.20 = 2000text N/m^2$ ho gh = 1000 \times 10 \times 0.20 = 2000\text{ N/m}^2$ Now find the excess pressure from surface tension: frac2TR = 2100 - 2000 = 100text N/m^2$\frac{2T}{R} = 2100 - 2000 = 100\text{ N/m}^2$ T = frac100 times R2 = 50 times (0.1 times 10^-2) = 0.05text N/m$T = \frac{100 \times R}{2} = 50 \times (0.1 \times 10^{-2}) = 0.05\text{ N/m}$ ### Pattern Recognition Total inside pressure accounts for both the hydrostatic pressure of the fluid column ( ho gh$ ho gh$) and the spherical geometry boundary constraint pressure (frac2TR$\frac{2T}{R}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Related Formula The work done W$W$ in expanding surface area with surface tension S$S$ is given by: W = S cdot Delta A$W = S \cdot \Delta A$ ### Core Logic By volume conservation during splitting : frac43pi R^3 = n left(frac43pi r^3 ight) implies r = fracRn^1/3 $\frac{4}{3}\pi R^{3} = n \left(\frac{4}{3}\pi r^{3} ight) \implies r = \frac{R}{n^{1/3}} $ Total change in area gives work expression[cite: 636, 637]: W = S left(n cdot 4pi r^2 - 4pi R^2 ight) = 4pi R^2S left(n^1/3 - 1 ight)$W = S \left(n \cdot 4\pi r^{2} - 4\pi R^{2} ight) = 4\pi R^{2}S \left(n^{1/3} - 1 ight)$ ### Step 1: Set up Proportions For n = 27$n = 27$ drops [cite: 53, 644]: W = 4pi R^2S (27^1/3 - 1) = 4pi R^2S (3 - 1) = 2(4pi R^2S) = 10text J$W = 4\pi R^{2}S (27^{1/3} - 1) = 4\pi R^{2}S (3 - 1) = 2(4\pi R^{2}S) = 10\text{ J}$ 4pi R^2S = 5text J$4\pi R^{2}S = 5\text{ J}$ For n = 64$n = 64$ drops [cite: 53, 644]: W' = 4pi R^2S (64^1/3 - 1) = 4pi R^2S (4 - 1) = 3(4pi R^2S)$W' = 4\pi R^{2}S (64^{1/3} - 1) = 4\pi R^{2}S (4 - 1) = 3(4\pi R^{2}S)$ W' = 3 times 5 = 15text J$W' = 3 \times 5 = 15\text{ J}$ ### Pattern Recognition Work scales scaling-wise linearly with the key multiplier index (n^1/3 - 1)$(n^{1/3} - 1)$. Taking ratios between targets directly avoids calculations of constants. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Related Formula By the law of flotation, the weight of a floating body must exactly balance the buoyant force exerted by the displaced fluid volume: M cdot g = rho_textfluid cdot V_textsubmerged cdot g$M \cdot g = \rho_{\text{fluid}} \cdot V_{\text{submerged}} \cdot g$ ### Core Logic Given parameters: * Mass of the cube, M = 400 text g = 0.4 text kg$M = 400 \text{ g} = 0.4 \text{ kg}$ * Total volume of the cube, V_texttotal = (10 text cm)^3 = 1000 text cm^3 = 10^-3 text m^3$V_{\text{total}} = (10 \text{ cm})^3 = 1000 \text{ cm}^3 = 10^{-3} \text{ m}^3$ * Density of water, rho_textwater = 1000 text kg/m^3$\rho_{\text{water}} = 1000 \text{ kg/m}^3$ Equating weight to buoyant force to find the submerged volume V_d$V_d$ : 0.4 = 1000 times V_textsubmerged$0.4 = 1000 \times V_{\text{submerged}}$ V_textsubmerged = frac0.41000 = 4 times 10^-4 text m^3 = 400 text cm^3 quad text$V_{\text{submerged}} = \frac{0.4}{1000} = 4 \times 10^{-4} \text{ m}^3 = 400 \text{ cm}^3 \quad \text{}$ Calculate the volume remaining outside the water surface : V_textoutside = V_texttotal - V_textsubmerged$V_{\text{outside}} = V_{\text{total}} - V_{\text{submerged}}$ V_textoutside = 1000 text cm^3 - 400 text cm^3 = 600 text cm^3 quad text$V_{\text{outside}} = 1000 \text{ cm}^3 - 400 \text{ cm}^3 = 600 \text{ cm}^3 \quad \text{}$ ### Pattern Recognition The fraction of a floating body's volume that is submerged equals the ratio of the body's density to the fluid's density: fracV_textsubmergedV_texttotal = fracrho_textbodyrho_textfluid$\frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{body}}}{\rho_{\text{fluid}}}$. Here, the cube's effective density is 0.4 text g/cm^3$0.4 \text{ g/cm}^3$, meaning 40\%$40\%$ is submerged and 60\%$60\%$ stays outside. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids