Given below are two statements: Statement I : Pressure of fluid is exerted only on a solid surface in contact as the fluid-pressure does not exist everywhere in a still fluid. Statement II: Excess potential energy of the molecules on the surface of a liquid, when compared to interior, results in surface tension. In the light of the above statements, choose the correct answer from the options given below:

### 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 h = frac2Tcosthetarho g r$$h = \frac{2T\cos\theta}{\rho g r}$$ ### Core Logic Since we are assessing surface tension T$T$ as a property of the liquids, the question intends to compare liquids that are identical (implying T_1 = T_2$T_1 = T_2$ and same density).

By Jurin's Law, for identical liquids, the height of capillary rise is inversely proportional to the radius of the tube: h propto frac1r$$h \propto \frac{1}{r}$$ ### Step 1: Final Conclusion Given r_1 > r_2$r_1 > r_2$, the inverse proportionality directly implies that the liquid will rise less in the wider tube. Therefore, h_1 < h_2$h_1 < h_2$ and naturally T_1 = T_2$T_1 = T_2$ for the intrinsic property. ### Pattern Recognition Wider tubes (r$r$ large) cause lower capillary rises (h$h$ small). h propto 1/r$h \propto 1/r$ is a foundational inverse relationship in capillary action. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Related Formula v_T = frac2r^2 g9eta (sigma - rho)$$v_T = \frac{2r^2 g}{9\eta} (\sigma - \rho)$$ ### Core Logic For balls of the same material falling through the same fluid, all terms in the terminal velocity equation except the radius r$r$ are constant. Therefore, v_T propto r^2$v_T \propto r^2$. ### Step 1: Setting up the Ratio frac(v_T)_1(v_T)_2 = left(fracr_1r_2right)^2$$\frac{(v_T)_1}{(v_T)_2} = \left(\frac{r_1}{r_2}\right)^2$$ Substituting the given values: r_1 = 6 text mm$r_1 = 6 \text{ mm}$, (v_T)_1 = 20 text cm/s$(v_T)_1 = 20 \text{ cm/s}$ r_2 = 3 text mm$r_2 = 3 \text{ mm}$ ### Step 2: Final Conclusion frac20(v_T)_2 = left(frac63right)^2$$\frac{20}{(v_T)_2} = \left(\frac{6}{3}\right)^2$$ frac20(v_T)_2 = 2^2 = 4$$\frac{20}{(v_T)_2} = 2^2 = 4$$ (v_T)_2 = frac204 = 5 text cm/s$$(v_T)_2 = \frac{20}{4} = 5 \text{ cm/s}$$ ### Pattern Recognition Terminal velocity is directly proportional to the square of the radius. Halving the radius (6 to 3$6 \to 3$) will drop the terminal velocity by a factor of 1/4$1/4$. ### 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