Grade 11 : Physics - Fluids (Set 4)

Explanation:Gauge pressure at the bottom is the sum of hydrostatic pressures of both liquid columns:
$$P = \rho_w g h_1 + \rho_m g h_2$$
$$P = (1000 \times 10 \times 0.20) + (13600 \times 10 \times 0.10)$$
$$P = 2000 + 13600 = 15600\text{ Pa} = 15.6\text{ kPa}$$

Explanation:At the horizontal level of the interface, hydrostatic pressures in both arms must be equal:
$$P_{\text{left}} = P_{\text{right}} \implies \rho_{\text{oil}} g h_{\text{oil}} = \rho_w g h_w$$
$$h_w = \frac{\rho_{\text{oil}}}{\rho_w} h_{\text{oil}} = \frac{800}{1000} \times 15 = 12\text{ cm}$$

Explanation:By Pascal's principle, pressure is transmitted equally throughout the fluid:
$$\frac{F_A}{A_A} = \frac{F_B}{A_B} \implies F_B = F_A \left(\frac{r_B}{r_A}\right)^2$$
$$F_B = 50 \times \left(\frac{10}{2}\right)^2 = 50 \times 25 = 1250\text{ N}$$

Explanation:For a floating body in equilibrium, weight equals buoyant force:
$$V_{\text{total}} \rho_{\text{wood}} g = V_{\text{sub}} \rho_w g$$
$$\frac{V_{\text{sub}}}{V_{\text{total}}} = \frac{\rho_{\text{wood}}}{\rho_w} = \frac{600}{1000} = 0.60 = 60\%$$

Explanation:The forces acting on the sphere are tension $T$, buoyant force $F_b$, and gravity $W$:
$$W = m g = 4 \times 10 = 40\text{ N}$$
$$F_b = \rho_{\text{oil}} V g = 800 \times (5 \times 10^{-4}) \times 10 = 4\text{ N}$$
$$T = W - F_b = 40 - 4 = 36\text{ N}$$

Explanation:The surface of an accelerating liquid aligns perpendicular to the effective gravity vector:
$$\tan\theta = \frac{a}{g} = \frac{10}{10} = 1$$
$$\theta = \arctan(1) = 45^\circ$$

Explanation:Applying the equation of continuity for incompressible flow:
$$A_1 v_1 = A_2 v_2 \implies d_1^2 v_1 = d_2^2 v_2$$
$$v_2 = v_1 \left(\frac{d_1}{d_2}\right)^2 = 3 \times \left(\frac{4}{2}\right)^2 = 3 \times 4 = 12\text{ m/s}$$

Explanation:First, find $v_B$: $v_B = v_A \frac{A_A}{A_B} = 2 \times \frac{10}{5} = 4m/s$.
Next, apply Bernoulli's equation for horizontal flow:
$$P_A + \frac{1}{2}\rho v_A^2 = P_B + \frac{1}{2}\rho v_B^2$$
$$P_B = 200000 + \frac{1}{2}(1000)(2^2 - 4^2) = 200000 - 6000 = 194000\text{ Pa} = 194\text{ kPa}$$

Explanation:By Torricelli's law, the speed of efflux from an orifice at depth $h$ is:
$$v = \sqrt{2g h} = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10\text{ m/s}$$

Explanation:Depth of hole from surface is $(H-h)$. Horizontal range is given by:
$$R = 2\sqrt{h(H-h)}$$
$$R = 2\sqrt{0.2 \times (1.0 - 0.2)} = 2\sqrt{0.2 \times 0.8} = 2\sqrt{0.16} = 2 \times 0.4 = 0.8\text{ m}$$

Explanation:$$\Delta P = \frac{1}{2}\rho(v_2^2 - v_1^2)$$
Since $A_1 = 2A_2 \implies v_2 = 2v_1$, we have $v_2^2 - v_1^2 = 3v_1^2$.
$$1500 = \frac{1}{2}(1000)(3v_1^2) = 1500 v_1^2 \implies v_1 = 1\text{ m/s}$$
$$Q = A_1 v_1 = 20 \times 10^{-4}\text{ m}^2 \times 1\text{ m/s} = 2.0 \times 10^{-3}\text{ m}^3\text{/s} = 2.0\text{ L/s}$$

Explanation:Terminal velocity is proportional to the square of the radius:
$$v_t = \frac{2 r^2 (\rho - \sigma) g}{9 \eta} \implies v_t \propto r^2$$
Doubling the radius increases the terminal velocity by a factor of $2^2 = 4$, yielding $4v_t$.

Explanation:Substitute parameters into Stokes' terminal velocity formula:
$$v_t = \frac{2 r^2 (\rho - \sigma) g}{9 \eta}$$
$$v_t = \frac{2 \times (10^{-3})^2 \times (2800 - 1000) \times 10}{9 \times 0.1} = \frac{0.036}{0.9} = 0.04\text{ m/s}$$

Explanation:Reynolds number calculation:
$$R_e = \frac{\rho v d}{\eta} = \frac{1000 \times 0.05 \times 0.02}{1.0 \times 10^{-3}} = 1000$$
Since $R_e < 2000$, the fluid motion is smooth and laminar.

Explanation:Volume conservation gives $r = R / 8^{1/3} = R / 2$.
Increase in surface area:
$$\Delta A = 8(4\pi r^2) - 4\pi R^2 = 8 \times 4\pi(R/2)^2 - 4\pi R^2 = 4\pi R^2$$
$$W = T \Delta A = 0.465 \times 4\pi \times (2 \times 10^{-3})^2 \approx 2.34 \times 10^{-5}\text{ J}$$

Explanation:Equating both excess pressure formulas:
$$\frac{4T}{R_1} = \frac{2T}{R_2} \implies \frac{R_1}{R_2} = \frac{4T}{2T} = 2$$
Hence, the ratio $R_1 : R_2 = 2:1$.

Explanation:Using Jurin's law formula for capillary rise:
$$h = \frac{2 T \cos\theta}{r \rho g}$$
$$h = \frac{2 \times 0.07 \times 1}{(5 \times 10^{-4}) \times 1000 \times 10} = \frac{0.14}{5} = 0.028\text{ m} = 2.8\text{ cm}$$

Explanation:Water does NOT overflow from an insufficient capillary tube. According to Jurin's law $h R = h' R'$, as the height reduces to half ($l = 4cm$), the radius of curvature of the meniscus doubles ($R' = 2R$), flattening out to maintain hydrostatic equilibrium.

Explanation:Effective acceleration inside upward accelerating elevator is $g_{\text{eff}} = g + a = \frac{5}{4}g$.
Equating hydrostatic pressure to constant atmospheric pressure:
$$P_{\text{atm}} = \rho g h = \rho \left(\frac{5}{4}g\right) h'$$
$$h' = \frac{4}{5} h = 0.8 \times 76.0\text{ cm} = 60.8\text{ cm}$$

Explanation:Efflux velocity at siphon outlet:
$$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 1.25} = 5\text{ m/s}$$
Volumetric flow rate:
$$Q = A v = (2 \times 10^{-4}\text{ m}^2) \times 5\text{ m/s} = 10^{-3}\text{ m}^3\text{/s} = 1.0\text{ L/s}$$