Class 11 Physics | Mechanical Properties of Fluids MCQs with Answers – Part 5

Explanation: Buoyant force is \(F_B=\rho_f gV_{\text{disp}}\). For a fully immersed block of fixed volume in a uniform liquid, \(V_{\text{disp}}\) remains equal to the block volume. The fluid density \(\rho_f\) and \(g\) also remain unchanged. Although absolute pressure increases with depth, the pressure difference between the lower and upper parts of the same block remains the same for the same vertical size. Buoyant force depends on displaced fluid weight, not on the absolute pressure value at the block’s centre.

Explanation: \( \textbf{Volume conservation:} \) The original drop volume is divided into \(8\) equal smaller drops. Since volume of a sphere is proportional to \(R^3\): \[ 8r^3=R^3 \] Hence: \[ r=\frac{R}{2} \] Excess pressure in a liquid drop is: \[ \Delta P=\frac{2T}{R} \] For each smaller drop: \[ \Delta P'=\frac{2T}{r} \] Substitute \(r=\frac{R}{2}\): \[ \Delta P'=\frac{2T}{R/2}=\frac{4T}{R} \] Since the original value was \(\frac{2T}{R}\): \[ \Delta P'=2\Delta P \] \( \textbf{Final answer:} \) The excess pressure becomes \(2\Delta P\).

Explanation: Capillary rise is \(h=\frac{2T\cos\theta}{\rho gr}\). If \(T\), \(\theta\), \(r\), and \(g\) are fixed, then \(h\propto\frac{1}{\rho}\). A denser liquid has a heavier column for the same height and radius. The same upward surface-tension component can therefore support a smaller height. Density appears in the denominator because capillary rise is balanced against the weight of the liquid column.

Explanation: The sphere is falling downward and speeding up, so its acceleration is downward. Therefore, the net force must be downward. Its weight \(W\) acts downward, while buoyant force \(F_B\) and viscous drag \(F_v\) act upward. For a downward net force, \(W\) must be greater than \(F_B+F_v\). At terminal velocity, the equality \(W=F_B+F_v\) would hold instead.

Explanation: Formula choice in fluid mechanics depends on the physical situation. The relation \(P=\rho gh\) applies to gauge pressure at depth in a static liquid. The relation \(F_B=\rho_f gV_{\text{disp}}\) applies to buoyant force due to displaced fluid. The relation \(Q=Av\) applies to volume flow rate in fluid motion. Looking only at symbols can be misleading because the same letters may appear in different contexts. The physical meaning of the situation should guide the formula choice.

Explanation: Pressure is defined as normal force per unit area, so its unit is \(\text{N m}^{-2}\) or \(\text{Pa}\). Viscosity describes internal friction that opposes relative motion between fluid layers, and its coefficient has unit \(\text{Pa s}\). Surface tension is force per unit length along a liquid surface and has unit \(\text{N m}^{-1}\). These three ideas describe different physical effects in fluids. Keeping their definitions and units separate prevents mixing hydrostatics, viscous flow, and surface phenomena.

Explanation: \( \textbf{Mass continuity relation:} \) \[ \rho_1A_1v_1=\rho_2A_2v_2 \] Here: \[ \rho_1=\rho,\quad A_1=A,\quad v_1=v \] At section Q: \[ \rho_2=2\rho,\quad A_2=A \] Substitute into continuity: \[ \rho Av=(2\rho)Av_2 \] Cancel \(\rho A\): \[ v=2v_2 \] Therefore: \[ v_2=\frac{v}{2} \] \( \textbf{Final answer:} \) The speed at Q is \(\frac{v}{2}\).

Explanation: \( \textbf{Given:} \) \(P_P=1.50\times10^{5}\,\text{Pa}\), \(v_P=4.0\,\text{m s}^{-1}\), \(v_Q=2.0\,\text{m s}^{-1}\), \(h_Q-h_P=5.0\,\text{m}\), \(\rho=1000\,\text{kg m}^{-3}\), and \(g=10\,\text{m s}^{-2}\). Bernoulli’s equation gives: \[ P_P+\frac{1}{2}\rho v_P^2+\rho gh_P=P_Q+\frac{1}{2}\rho v_Q^2+\rho gh_Q \] Rearranging for \(P_Q\): \[ P_Q=P_P+\frac{1}{2}\rho(v_P^2-v_Q^2)-\rho g(h_Q-h_P) \] Substitute: \[ P_Q=1.50\times10^{5}+\frac{1}{2}(1000)(4.0^2-2.0^2)-(1000)(10)(5.0) \] Speed-square difference: \[ 4.0^2-2.0^2=16-4=12 \] Kinetic contribution: \[ \frac{1}{2}(1000)(12)=6000\,\text{Pa} \] Height contribution: \[ (1000)(10)(5.0)=5.0\times10^{4}\,\text{Pa} \] Therefore: \[ P_Q=1.50\times10^{5}+6.0\times10^{3}-5.0\times10^{4} \] \[ P_Q=1.06\times10^{5}\,\text{Pa} \] \( \textbf{Final answer:} \) The pressure at Q is \(1.06\times10^{5}\,\text{Pa}\).

Explanation: \( \textbf{Given:} \) \(A_1=4A_2\), \(P_1-P_2=7.5\times10^{3}\,\text{Pa}\), and \(\rho=1000\,\text{kg m}^{-3}\). From continuity: \[ A_1v_1=A_2v_2 \] Since \(A_1=4A_2\): \[ 4A_2v_1=A_2v_2 \] Therefore: \[ v_2=4v_1 \] For horizontal ideal flow: \[ P_1-P_2=\frac{1}{2}\rho(v_2^2-v_1^2) \] Substitute \(v_2=4v_1\): \[ 7.5\times10^{3}=\frac{1}{2}(1000)(16v_1^2-v_1^2) \] \[ 7.5\times10^{3}=500(15v_1^2) \] \[ 7.5\times10^{3}=7500v_1^2 \] Hence: \[ v_1=1.0\,\text{m s}^{-1} \] Then: \[ v_2=4v_1=4.0\,\text{m s}^{-1} \] \( \textbf{Final answer:} \) The speed at the throat is \(4.0\,\text{m s}^{-1}\).

Explanation: \( \textbf{Efflux speed:} \) \[ v=\sqrt{2gh} \] The height of the hole above the ground is: \[ y=H-h \] Time of fall: \[ t=\sqrt{\frac{2(H-h)}{g}} \] Horizontal range: \[ R=vt \] Substitute: \[ R=\sqrt{2gh}\sqrt{\frac{2(H-h)}{g}} \] Simplifying: \[ R=2\sqrt{h(H-h)} \] For fixed \(H\), the product \(h(H-h)\) is maximum when: \[ h=H-h \] Hence: \[ h=\frac{H}{2} \] \( \textbf{Final answer:} \) The range is maximum when the hole is halfway below the free surface.

Explanation: For a hole at depth \(h\), the height above ground is \(H-h\). The range is: \[ R_1=2\sqrt{h(H-h)} \] For another hole at depth \(H-h\), its height above ground is \(h\). Its range is: \[ R_2=2\sqrt{(H-h)h} \] These two expressions are the same: \[ R_1=R_2 \] The two holes are complementary in depth and height above ground. \( \textbf{Final answer:} \) The horizontal ranges are equal.

Explanation: A soap film has two liquid-air surfaces. Surface tension \(T\) acts along the movable side on each surface. For one surface, the force contribution is \(Tl\). Since the film has two surfaces, the total force is: \[ F=2Tl \] This factor of \(2\) is similar in origin to the factor difference between a liquid drop and a soap bubble in excess-pressure formulas. The length \(l\) is the length of the movable side in contact with the film.

Explanation: \( \textbf{Given:} \) \(T=0.030\,\text{N m}^{-1}\) and \(l=0.10\,\text{m}\). A soap film has two surfaces. The holding force is: \[ F=2Tl \] Substitute: \[ F=2(0.030)(0.10) \] Multiply: \[ F=0.0060\,\text{N} \] Therefore: \[ F=6.0\times10^{-3}\,\text{N} \] The factor \(2\) is needed because both sides of the thin film contribute to the pull. \( \textbf{Final answer:} \) The required force is \(6.0\times10^{-3}\,\text{N}\).

Explanation: \( \textbf{Soap bubble surface count:} \) A soap bubble has two surfaces. Surface area for one spherical surface of radius \(R\) is: \[ 4\pi R^2 \] Total surface area of the bubble initially: \[ A_i=2(4\pi R^2)=8\pi R^2 \] Final radius is \(2R\), so total final area is: \[ A_f=2\{4\pi(2R)^2\} \] \[ A_f=2(16\pi R^2)=32\pi R^2 \] Increase in area: \[ \Delta A=A_f-A_i=32\pi R^2-8\pi R^2 \] \[ \Delta A=24\pi R^2 \] Increase in surface energy: \[ \Delta U=T\Delta A \] Therefore: \[ \Delta U=24\pi R^2T \] \( \textbf{Final answer:} \) The increase in surface energy is \(24\pi R^2T\).

Explanation: For a soap bubble, excess pressure is: \[ \Delta P=\frac{4T}{R} \] The smaller bubble of radius \(R\) has: \[ \Delta P_1=\frac{4T}{R} \] The larger bubble of radius \(2R\) has: \[ \Delta P_2=\frac{4T}{2R}=\frac{2T}{R} \] The smaller bubble has greater internal pressure. When connected, air tends to move from higher pressure to lower pressure. Therefore, air flows from the smaller bubble toward the larger bubble.

Explanation: The calculated capillary rise is the height the liquid would reach if the tube were sufficiently long. If the available tube height above the outside liquid level is smaller than that calculated rise, the liquid can reach the top of the tube. It may overflow or wet the upper end depending on the arrangement. The liquid is not limited to an arbitrary smaller height such as \(2.0\,\text{cm}\). The situation shows that the formula predicts an equilibrium height only when the tube can contain that height.

Explanation: The capillary height is: \[ h=\frac{2T\cos\theta}{\rho gr} \] For water in clean glass, the angle of contact is acute, so \(\cos\theta\) is positive and \(h\) represents rise. For mercury in clean glass, the angle of contact is obtuse, so \(\cos\theta\) is negative and \(h\) represents depression. Both liquids have density and surface tension. The sign of the capillary effect is controlled by the contact angle through \(\cos\theta\).

Explanation: \( \textbf{Poiseuille law:} \) \[ Q=\frac{\pi \Delta P r^4}{8\eta l} \] With \(\Delta P\) and \(\eta\) fixed: \[ Q\propto\frac{r^4}{l} \] Radius is doubled: \[ r'=2r \] Therefore: \[ (r')^4=(2r)^4=16r^4 \] Length is doubled: \[ l'=2l \] New flow-rate ratio: \[ \frac{Q'}{Q}=\frac{16}{2}=8 \] \( \textbf{Final answer:} \) The new flow rate is \(8Q\).

Explanation: Stokes’ law gives the viscous drag on a small sphere moving slowly through a viscous fluid. At terminal velocity, the speed in Stokes’ law is the terminal speed \(v_t\). Therefore, the viscous drag is \(F_v=6\pi\eta rv_t\). The total upward force at terminal velocity includes both buoyant force and viscous drag, while the downward force is weight. The expression \(\rho_f gV_{\text{disp}}\) gives buoyant force, not viscous drag.

Explanation: Terminal velocity means the sphere moves with constant velocity. So, acceleration is zero. If acceleration is zero, the net force is also zero. The downward force is the weight of the sphere. Weight: \(W=0.090\,\text{N}\) The upward forces are buoyant force and viscous drag. Buoyant force: \(F_B=0.030\,\text{N}\) Let the viscous drag be \(F_v\). At terminal velocity: Downward force = Upward forces So: \(W=F_B+F_v\) Rearrange: \(F_v=W-F_B\) Substitute the values: \(F_v=0.090-0.030\) \(F_v=0.060\,\text{N}\) This is the viscous drag when the sphere has reached terminal velocity. \( \textbf{Final answer:} \) The viscous drag is \(0.060\,\text{N}\).