### Related Formula B = -fracDelta PDelta V / V implies V = B frac-Delta VDelta P$B = -\frac{\Delta P}{\Delta V / V} \implies V = B \frac{-\Delta V}{\Delta P}$ where, B$B$ = Bulk modulus of liquid Delta P$\Delta P$ = change in pressure Delta V$\Delta V$ = change in volume V$V$ = initial volume ### Core Logic Given parameters: - Initial pressure, P_i = 1mathrm~atm$P_i = 1\mathrm{~atm}$ - Final pressure, P_f = 5mathrm~atm$P_f = 5\mathrm{~atm}$ - Change in pressure, Delta P = P_f - P_i = 4mathrm~atm = 4 times 10^5mathrm~Pa$\Delta P = P_f - P_i = 4\mathrm{~atm} = 4 \times 10^5\mathrm{~Pa}$ - Change in volume, Delta V = -0.8mathrm~cm^3 = -0.8 times 10^-6mathrm~m^3$\Delta V = -0.8\mathrm{~cm}^{3} = -0.8 \times 10^{-6}\mathrm{~m}^{3}$ - Bulk modulus, B = 2mathrm~GPa = 2 times 10^9mathrm~Pa$B = 2\mathrm{~GPa} = 2 \times 10^9\mathrm{~Pa}$ ### Step 1: Compute Initial Volume Substitute the parameters into the formula: V = 2 times 10^9 times frac0.8 times 10^-64 times 10^5$V = 2 \times 10^9 \times \frac{0.8 \times 10^{-6}}{4 \times 10^5}$ V = frac1.6 times 10^34 times 10^5 = 0.4 times 10^-2 = 4 times 10^-3mathrm~m^3$V = \frac{1.6 \times 10^3}{4 \times 10^5} = 0.4 \times 10^{-2} = 4 \times 10^{-3}\mathrm{~m}^{3}$ Convert cubic meters to litres: 1mathrm~m^3 = 1000mathrm~litres$1\mathrm{~m}^3 = 1000\mathrm{~litres}$ V = 4 times 10^-3 times 1000 = 4mathrm~litres$V = 4 \times 10^{-3} \times 1000 = 4\mathrm{~litres}$ Thus, the initial volume was 4mathrm~litres$4\mathrm{~litres}$. ### Pattern Recognition Sees: Bulk modulus definition calculation. Trap: Watch out for unit conversions: 1mathrm~GPa = 10^9mathrm~Pa$1\mathrm{~GPa} = 10^9\mathrm{~Pa}$, and 1mathrm~cm^3 = 10^-6mathrm~m^3$1\mathrm{~cm}^3 = 10^{-6}\mathrm{~m}^3$. Finally, express the answer in Litres, where 1mathrm~L = 10^-3mathrm~m^3$1\mathrm{~L} = 10^{-3}\mathrm{~m}^3$. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Related Formula mathrmV_mathrmT = frac29 mathrmR^2 fracmathrmgeta (mathrmd - rho)$\mathrm{V}_{\mathrm{T}} = \frac{2}{9} \mathrm{R}^{2} \frac{\mathrm{g}}{\eta} (\mathrm{d} - \rho)$ ### Core Logic Let us check each statement sequentially: Statement A: Since mathrmV_mathrmT propto mathrmR^2$\mathrm{V}_{\mathrm{T}} \propto \mathrm{R}^2$, the graph between terminal velocity mathrmV$\mathrm{V}$ and radius mathrmR$\mathrm{R}$ is a quadratic curve, i.e., a parabola. (True) Statement B: Different diameters imply different radii, hence terminal velocity will change. It is not constant. (False) Statement C: Viscosity of a liquid is heavily temperature-dependent, hence mathrmV_mathrmT$\mathrm{V}_{\mathrm{T}}$ changes with temperature. (True) Statement D: Knowing the parameters allows calculation of liquid density rho$\rho$. (True) Statement E: eta$\eta$ is a material property of the fluid and does not change based on the initial dropping speed of the ball. (False) ### Step 1: Final Conclusion Statements A, C, and D are correct, which matches option (2). ### Pattern Recognition Terminal velocity problems rely entirely on parsing the proportional relations hidden in Stokes' Law formulation. Note that viscosity coefficient is an intrinsic characteristic parameter independent of kinematics. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Core Logic Let's audit each fluid mechanics assertion: Statement A: Incorrect. Surface tension arises because surface molecules possess higher potential energy compared to interior bulk molecules due to net cohesive forces pulling inward. Statement B: Incorrect. With rising temperatures, cohesive forces in liquids weaken, causing viscosity to decrease. Statement C: Correct. In gases, viscosity is governed by molecular collisions. Higher temperature leads to increased thermal activity and momentum exchange, increasing viscosity. Statement D: Correct. Critical velocity and turbulence parameters are completely defined by the dimensionless Reynolds number. Statement E: Correct. If streamlines intersected, a fluid particle at that spatial node would have two distinct velocity directions, violating steady-state constraints. ### Step 1: Final Conclusion Statements C, D, and E are correct, leading to option (2). ### Pattern Recognition Contrast Liquid vs Gas viscosity trends under temperature increments: Liquids down (intermolecular bounds weaken), Gases up (random thermal diffusion and collision rates escalate). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Related Formula K = fraccostheta_mathrmAcostheta_mathrmB$K = \frac{\cos\theta_{\mathrm{A}}}{\cos\theta_{\mathrm{B}}}$ * Concave meniscus: theta < 90^circ implies costheta > 0$\theta < 90^\circ implies \cos\theta > 0$ * Convex meniscus: theta > 90^circ implies costheta < 0$\theta > 90^\circ implies \cos\theta < 0$ ### Core Logic For K$K$ to be negative, costheta_mathrmA$\cos\theta_{\mathrm{A}}$ and costheta_mathrmB$\cos\theta_{\mathrm{B}}$ must have opposite mathematical signs. Looking at option (3): * Liquid A has a concave meniscus implies theta_mathrmA < 90^circ implies costheta_mathrmA > 0$implies \theta_{\mathrm{A}} < 90^\circ implies \cos\theta_{\mathrm{A}} > 0$ (positive). * Liquid B has a convex meniscus implies theta_mathrmB > 90^circ implies costheta_mathrmB < 0$implies \theta_{\mathrm{B}} > 90^\circ implies \cos\theta_{\mathrm{B}} < 0$ (negative). This makes the ratio K = fractextpositivetextnegative < 0$K = \frac{\text{positive}}{\text{negative}} < 0$ (negative), which accurately validates the statement. ### Pattern Recognition Meniscus geometry directly correlates with the sign of costheta$\cos\theta$: acute contact angle = concave (wetting fluid), obtuse contact angle = convex (non-wetting fluid). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Related Formula h = frac2Tcosthetarho g r$h = \frac{2T\cos\theta}{\rho g r}$ [cite: 694] ### Core Logic First, calculate the vertical height h$h$ using CGS units: [cite: 694] * T = 70\ textdyn/cm$T = 70\ \text{dyn/cm}$ [cite: 83] * costheta = cos 0^circ = 1$\cos\theta = \cos 0^{\circ} = 1$ [cite: 83] * rho = 1\ textg/cm^3$\rho = 1\ \text{g/cm}^3$ [cite: 694] * g = 980\ textcm/s^2$g = 980\ \text{cm/s}^2$ [cite: 694] * r = 0.1\ textmm = 10^-2\ textcm$r = 0.1\ \text{mm} = 10^{-2}\ \text{cm}$ [cite: 83, 694] h = frac2 times 70 times 11 times 980 times 10^-2 = frac1409.8 = frac1007\ textcm$h = \frac{2 \times 70 \times 1}{1 \times 980 \times 10^{-2}} = \frac{140}{9.8} = \frac{100}{7}\ \text{cm}$ [cite: 694] Since the tube is inclined at 30^circ$30^{\circ}$ to the vertical, the angle it makes with the horizontal is 60^circ$60^{\circ}$[cite: 83, 694]. The relationship between vertical height h$h$ and slant length l$l$ is: [cite: 694] sin 60^circ = frachl implies l = frachsin 60^circ = frac2hsqrt3$\sin 60^{\circ} = \frac{h}{l} \implies l = \frac{h}{\sin 60^{\circ}} = \frac{2h}{\sqrt{3}}$ [cite: 694] l = frac1007 times frac2sqrt3 = frac2007sqrt3 approx 16.49\ textcm$l = \frac{100}{7} \times \frac{2}{\sqrt{3}} = \frac{200}{7\sqrt{3}} \approx 16.49\ \text{cm}$ [cite: 696, 697, 698] Checking option values, frac825 = 16.4\ textcm$\frac{82}{5} = 16.4\ \text{cm}$, which is closest[cite: 85, 698]. ### Pattern Recognition When a capillary tube is tilted, the vertical height of the fluid column remains constant to maintain hydrostatic pressure balance[cite: 694]. Thus, the length of the liquid along the slant always scales as l = frachcosalpha$l = \frac{h}{\cos\alpha}$ where alpha$\alpha$ is the tilt angle relative to the vertical line[cite: 83, 694]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids