Mix Examples-Fluid Mechanics and Surface Tension Questions in English

(B) Let $M$ be the mass of the meter scale and $m$ be the mass of the body. The center of mass of the uniform meter scale is at the $50 \ cm$ mark. The pivot $P$ is at the $20 \ cm$ mark.
Case $1$: The body is at the $10 \ cm$ mark. The distance of the body from the pivot is $20 \ cm - 10 \ cm = 10 \ cm$. The distance of the center of mass of the scale from the pivot is $50 \ cm - 20 \ cm = 30 \ cm$. For rotational equilibrium,the net torque about the pivot must be zero:
$Mg \times 30 = mg \times 10$
$30M = 10m \Rightarrow m = 3M$
Case $2$: The body is immersed in water and shifted to the $8 \ cm$ mark. The distance of the body from the pivot is $20 \ cm - 8 \ cm = 12 \ cm$. The buoyant force $F_b$ acts upwards at the $8 \ cm$ mark. The weight of the body $mg$ acts downwards at the $8 \ cm$ mark. The effective downward force at the $8 \ cm$ mark is $(mg - F_b)$.
For equilibrium:
$Mg \times 30 = (mg - F_b) \times 12$
$30Mg = 12mg - 12F_b$
Since $F_b = V \rho_w g = (m / \rho) \rho_w g = m g / \rho_r$,where $\rho_r$ is the specific gravity:
$30Mg = 12mg - 12(mg / \rho_r)$
Substitute $m = 3M$:
$30Mg = 12(3M)g - 12(3M)g / \rho_r$
$30 = 36 - 36 / \rho_r$
$36 / \rho_r = 6$
$\rho_r = 36 / 6 = 6$

(C) Let $P_B$ be the pressure at point $B$ at the bottom of the vertical column of liquid of height $h$. Since the top of the column is open to the atmosphere,$P_B = P_0 + \rho gh$,where $P_0$ is the atmospheric pressure.
Now,consider the horizontal segment of the tube rotating with angular velocity $\omega$ about the vertical axis passing through $A$. Point $A$ is at a distance $r = 0$ from the axis,and point $B$ is at a distance $r = R$ from the axis.
The pressure variation in a rotating fluid is given by $dP = \rho \omega^2 r dr$. Integrating from $A$ to $B$:
$P_B - P_A = \int_0^R \rho \omega^2 r dr = \frac{\rho \omega^2 R^2}{2}$.
Therefore,$P_A = P_B - \frac{\rho \omega^2 R^2}{2}$.
Substituting the value of $P_B$,we get $P_A = P_0 + \rho gh - \frac{\rho \omega^2 R^2}{2}$.

(A) When a razor-blade floats on water,it is supported by the buoyant force (Archimedes' principle) and the upward force due to surface tension.
Since the density of the blade is greater than that of water,it would normally sink.
However,surface tension creates a 'dimple' in the water surface,effectively increasing the volume of water displaced beyond the actual volume of the blade.
When the glass is shaken,the surface tension effect is broken,and the blade sinks.
Statement $A$ is incorrect because,while floating,the effective volume of displaced water (including the surface tension effect) is equal to the volume required to balance the weight of the blade.
Statement $B$ is correct because the blade displaces less water when submerged than when it was floating supported by surface tension.
Statement $C$ is correct as it describes the mechanism of support.
Statement $D$ is correct as it explains the equilibrium condition for the floating blade.

(B) Let the cross-sectional area be $A$ and the density of water be $\rho$. The total length of the water column is $L = 4H$. The mass of the water is $m = \rho A L = 4\rho AH$.
Initially,the left column is at height $H/2$ above the mean level and the right column is at $H/2$ below the mean level. The potential energy of the system relative to the mean level is $U_i = \rho A (H/2) g (H/4) + \rho A (H/2) g (-H/4) = 0$ (considering the center of mass of each column).
More simply,we can consider the potential energy change of the displaced water. $A$ volume $V = A(H/2)$ of water moves from the left side to the right side. The center of mass of this volume drops by a distance $H/2$. The change in potential energy is $\Delta U = -(\rho A (H/2)) g (H/2) = -\rho A g H^2 / 4$.
By the law of conservation of energy,the loss in potential energy equals the gain in kinetic energy:
$|\Delta U| = \frac{1}{2} m v^2$
$\rho A g H^2 / 4 = \frac{1}{2} (4 \rho A H) v^2$
$\rho A g H^2 / 4 = 2 \rho A H v^2$
$v^2 = \frac{gH}{8}$
$v = \sqrt{\frac{gH}{8}}$

(B) Let $L$ be the total length of the rod,$A$ be its cross-sectional area,and $\rho_w$ and $\rho_r$ be the densities of water and the rod respectively. Given $\rho_w = \frac{9}{5} \rho_r$. Let $x$ be the submerged length of the rod.
The weight of the rod acts at its center of mass,which is at a distance $L/2$ from the pivot. The buoyant force $F_B$ acts at the center of the submerged part,which is at a distance $(L - x/2)$ from the pivot.
For small angular displacement $\theta$,the torque due to gravity is $\tau_g = (\rho_r A L g) \frac{L}{2} \sin \theta$.
The torque due to the buoyant force is $\tau_B = F_B (L - x/2) \sin \theta$,where $F_B = \rho_w (A x) g$.
Equilibrium is unstable when the net restoring torque becomes zero or negative. The condition for stability is that the restoring torque (gravity) must be greater than or equal to the destabilizing torque (buoyancy).
$(\rho_r A L g) \frac{L}{2} \sin \theta = (\rho_w A x g) (L - x/2) \sin \theta$
$\rho_r L^2 / 2 = \rho_w x (L - x/2)$
Substitute $\rho_w = \frac{9}{5} \rho_r$:
$\rho_r L^2 / 2 = \frac{9}{5} \rho_r x (L - x/2)$
$L^2 / 2 = \frac{9}{5} (Lx - x^2/2)$
$5L^2 = 18Lx - 9x^2$
$9x^2 - 18Lx + 5L^2 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{18L \pm \sqrt{324L^2 - 180L^2}}{18} = \frac{18L \pm \sqrt{144L^2}}{18} = \frac{18L \pm 12L}{18}$
$x = \frac{30L}{18} = \frac{5}{3}L$ (not possible as $x < L$) or $x = \frac{6L}{18} = \frac{L}{3}$.
Thus,$x = L/3$,which means $L/x = 3$.

(A) The cylinder has a radius $r = 4 \ cm = 0.04 \ m$ and height $h = 10 \ cm = 0.1 \ m$. The cross-sectional area $A = \pi r^2 = \pi (0.04)^2 = 0.0016 \pi \ m^2$.
$1$. Force exerted by oil: The oil exerts pressure on the side walls of the cylinder. The net horizontal force is zero due to symmetry. However,the vertical force exerted by the oil is the buoyant force,which is $F_{oil} = \rho_{oil} V_{sub, oil} g$. Here,$\rho_{oil} = 0.5 \times 1000 = 500 \ kg/m^3$ and $V_{sub, oil} = A \times h_{oil} = 0.0016 \pi \times 0.04 = 0.000064 \pi \ m^3$. Thus,$F_{oil} = 500 \times 0.000064 \pi \times 10 = 0.32 \pi \ N$. Therefore,statement $A$ is incorrect.
$2$. Force exerted by water: The water exerts a buoyant force $F_{water} = \rho_{water} V_{sub, water} g$. The submerged height in water is $10 - 2 - 4 = 4 \ cm = 0.04 \ m$. So,$V_{sub, water} = A \times 0.04 = 0.0016 \pi \times 0.04 = 0.000064 \pi \ m^3$. Thus,$F_{water} = 1000 \times 0.000064 \pi \times 10 = 0.64 \pi \ N$. Note: The pressure at the bottom of the cylinder is $P = P_{oil} + P_{water} = (\rho_{oil} g h_{oil} + \rho_{water} g h_{water})$. The total upward force is $F_{total} = P \times A = (500 \times 10 \times 0.04 + 1000 \times 10 \times 0.04) \times 0.0016 \pi = (200 + 400) \times 0.0016 \pi = 0.96 \pi \ N$. This is the net force exerted by the liquids on the bottom surface. Statement $B$ is correct.
$3$. Total force: $F_{total} = F_{oil} + F_{water} = 0.32 \pi + 0.64 \pi = 0.96 \pi \ N$ (buoyant forces). Statement $C$ is incorrect as well,but $A$ is the most fundamentally incorrect statement regarding the net force on the cylinder.

(A) The velocity head is given by $\frac{v^2}{2g}$.
Given pressure head $h = 40\, cm$ of $Hg$.
Converting $Hg$ head to water head: $h_{water} = h_{Hg} \times \frac{\rho_{Hg}}{\rho_{water}} = 40\, cm \times 13.6 = 544\, cm = 5.44\, m$.
Equating velocity head to pressure head: $\frac{v^2}{2g} = 5.44\, m$.
$v^2 = 2 \times 9.8 \times 5.44 = 106.624$.
$v = \sqrt{106.624} \approx 10.32\, m/s$.

(B) The height difference $h$ of a liquid in a rotating cylindrical vessel is given by the formula $h = \frac{\omega^2 r^2}{2g}$.
Here,the angular velocity $\omega = 2\pi f$,where $f = 2\,rev/s$.
So,$\omega = 2 \times \pi \times 2 = 4\pi\,rad/s$.
The radius $r = 0.05\,m$.
Using $g = 10\,m/s^2$ and $\pi^2 = 10$:
$h = \frac{(4\pi)^2 \times (0.05)^2}{2 \times 10} = \frac{16 \times \pi^2 \times 0.0025}{20}$.
Substituting $\pi^2 = 10$:
$h = \frac{16 \times 10 \times 0.0025}{20} = \frac{160 \times 0.0025}{20} = 8 \times 0.0025 = 0.02\,m$.
Converting to centimeters: $0.02\,m = 2\,cm$.

(C) The acceleration of the container sliding down a smooth inclined plane is $a = g \sin \theta$. Given $\theta = 30^o$,$a = g \sin 30^o = g/2$.
In the non-inertial frame of the container,a liquid particle on the surface experiences a pseudo-force $ma$ directed up the incline and gravity $mg$ directed downwards.
The effective acceleration $\vec{g}_{eff} = \vec{g} - \vec{a}$.
The angle $\alpha$ that the liquid surface makes with the horizontal is given by the formula $\tan \alpha = \frac{a \cos \theta}{g - a \sin \theta}$.
Substituting $a = g \sin \theta$:
$\tan \alpha = \frac{(g \sin \theta) \cos \theta}{g - (g \sin \theta) \sin \theta} = \frac{g \sin \theta \cos \theta}{g(1 - \sin^2 \theta)} = \frac{\sin \theta \cos \theta}{\cos^2 \theta} = \tan \theta$.
Thus,$\alpha = \theta = 30^o$.

(D) Let the depth of the lake be $h$. The pressure at the bottom of the lake is $P_1 = P_{atm} + \rho gh$,where $P_{atm} = \rho g(10)$.
Thus,$P_1 = \rho g(10 + h)$.
The pressure at the surface is $P_2 = P_{atm} = \rho g(10)$.
Using Boyle's Law,$P_1 V_1 = P_2 V_2$,assuming temperature is constant.
$\rho g(10 + h) \cdot \frac{4}{3} \pi r^3 = \rho g(10) \cdot \frac{4}{3} \pi \left( \frac{5r}{4} \right)^3$.
$(10 + h) = 10 \cdot \frac{125}{64}$.
$10 + h = \frac{1250}{64} = 19.53$.
$h = 19.53 - 10 = 9.53 \ m$.
Rounding to the nearest value,the depth is $9.5 \ m$.

(B) When small droplets coalesce to form a bigger drop,the change in surface area is $\Delta A = n(4\pi r^2) - 4\pi R^2$. Since the volume is conserved,$n(\frac{4}{3}\pi r^3) = \frac{4}{3}\pi R^3$,so $n = \frac{R^3}{r^3}$.
The energy released is $\Delta E = T \times \Delta A = T(n 4\pi r^2 - 4\pi R^2) = 4\pi T (\frac{R^3}{r} - R^2) = 4\pi R^3 T (\frac{1}{r} - \frac{1}{R})$.
According to the problem,this energy is converted into the kinetic energy of the big drop: $\frac{1}{2} M v^2 = \Delta E$.
Here,$M = \rho \times \text{Volume} = \rho (\frac{4}{3}\pi R^3)$.
Substituting the values: $\frac{1}{2} (\frac{4}{3}\pi R^3 \rho) v^2 = 4\pi R^3 T (\frac{1}{r} - \frac{1}{R})$.
Simplifying: $\frac{2}{3} \pi R^3 \rho v^2 = 4\pi R^3 T (\frac{1}{r} - \frac{1}{R})$.
$v^2 = \frac{4 \times 3}{2} \frac{T}{\rho} (\frac{1}{r} - \frac{1}{R}) = \frac{6T}{\rho} (\frac{1}{r} - \frac{1}{R})$.
Thus,$v = {\left[ {\frac{{6T}}{\rho }\left( {\frac{1}{r} - \frac{1}{R}} \right)} \right]^{\frac{1}{2}}}$.

(B) Let the mass flow rate be $\frac{dm}{dt} = \rho A v$. The force exerted on the mesh is equal to the rate of change of momentum of the liquid.
For the $50\%$ of liquid passing through: $\Delta p = 0$,so $F_1 = 0$.
For the $25\%$ of liquid that loses all momentum: $\Delta p = (0.25 \frac{dm}{dt})v - 0 = 0.25 \rho A v^2$.
For the $25\%$ of liquid that bounces back: $\Delta p = (0.25 \frac{dm}{dt})v - (0.25 \frac{dm}{dt})(-v) = 0.5 \rho A v^2$.
The total force $F = F_1 + F_2 + F_3 = 0 + 0.25 \rho A v^2 + 0.5 \rho A v^2 = 0.75 \rho A v^2 = \frac{3}{4} \rho A v^2$.
The pressure $P = \frac{F}{A} = \frac{3}{4} \rho v^2$.

(A) The shape of the free surface of a liquid in a rotating cylinder is a parabola given by the equation $y = \frac{\omega^2 r^2}{2g}$,where $y$ is the height difference between the center and the wall at radius $r$.
Given:
Radius $r = 5 \, cm = 0.05 \, m$
Rotational frequency $f = 2 \, rev/s$
Angular velocity $\omega = 2 \pi f = 2 \pi \times 2 = 4 \pi \, rad/s$
Acceleration due to gravity $g \approx 10 \, m/s^2$
Substituting the values:
$y = \frac{(4 \pi)^2 \times (0.05)^2}{2 \times 10}$
$y = \frac{16 \times \pi^2 \times 0.0025}{20}$
Using $\pi^2 \approx 10$:
$y = \frac{16 \times 10 \times 0.0025}{20} = \frac{0.4}{20} = 0.02 \, m = 2 \, cm$.

(D) The work done by gravity is equal to the decrease in the potential energy of the liquid system.
Initial potential energy $U_i = (m_1 g \frac{h_1}{2}) + (m_2 g \frac{h_2}{2}) = (A h_1 d g \frac{h_1}{2}) + (A h_2 d g \frac{h_2}{2}) = \frac{Adg}{2}(h_1^2 + h_2^2)$.
When connected, the final height in both vessels will be $h_{avg} = \frac{h_1 + h_2}{2}$.
Final potential energy $U_f = 2 \times (A h_{avg} d g \frac{h_{avg}}{2}) = A d g h_{avg}^2 = Adg (\frac{h_1 + h_2}{2})^2$.
Work done $W = U_i - U_f = \frac{Adg}{2}(h_1^2 + h_2^2) - Adg \frac{(h_1 + h_2)^2}{4}$.
$W = \frac{Adg}{4} [2(h_1^2 + h_2^2) - (h_1^2 + h_2^2 + 2h_1h_2)] = \frac{Adg}{4} (h_1^2 + h_2^2 - 2h_1h_2) = \frac{1}{4} Adg (h_1 - h_2)^2$.

(A) When a vessel containing a liquid moves with a constant velocity (constant speed in a straight line),its acceleration is zero $(a = 0)$.
In the frame of reference of the vessel,there is no pseudo force acting on the liquid because the frame is inertial.
Since the only force acting on the liquid particles is gravity (acting downwards),the free surface of the liquid must remain perpendicular to the effective acceleration due to gravity.
Therefore,the surface of the liquid remains horizontal,just as it would be if the vessel were at rest.
Diagram $(i)$ shows a horizontal liquid surface,which is consistent with this condition.
Hence,option $A$ is correct.

(B) The vapour pressure $(V.P.)$ of a liquid is a characteristic property that depends solely on the temperature of the liquid and the nature of the liquid itself.
It is independent of the surface area of the liquid exposed to the vapour phase or the presence of any inert objects (like a glass plate) floating on the surface.
Since the temperature of the system remains constant,the vapour pressure will remain the same regardless of whether the glass plate is present or removed.

(B) Statement $A$ is true. The excess pressure inside a liquid drop is given by $P = \frac{2T}{R}$,where $T$ is the surface tension and $R$ is the radius of the drop. Since $P \propto \frac{1}{R}$,the excess pressure is inversely proportional to the radius. Therefore,a smaller drop has higher excess pressure than a larger drop.
Statement $B$ is false. According to Bernoulli's principle,for an aeroplane wing,the shape is designed such that air flows faster over the top surface than the bottom surface. Higher speed results in lower pressure on the top surface and higher pressure on the bottom surface,which creates the lift force. Thus,the pressure is actually less on the upper surface and more on the bottom surface.

(B) The $Assertion$ is correct because a thin stainless steel needle can float on a still water surface due to the surface tension of water.
The $Reason$ is also a correct statement in physics,as an object floats when the buoyant force equals its weight (Archimedes' Principle).
However,the $Reason$ is not the correct explanation for the $Assertion$. The needle floats not because of buoyancy,but because the surface tension force acts upwards to balance the weight of the needle.
Therefore,both are correct,but the $Reason$ does not explain the $Assertion$.

(A) Decreases: The surface tension of a liquid is inversely proportional to temperature.
$(b)$ Increases; decreases: Viscosity of gases increases with temperature due to increased molecular collisions,while viscosity of liquids decreases with temperature due to decreased cohesive forces.
$(c)$ Shear strain; rate of shear strain: For solids,the shearing force is proportional to shear strain (Hooke's Law),whereas for fluids,it is proportional to the rate of shear strain (Newton's Law of Viscosity).
$(d)$ Conservation of mass: The increase in flow speed at a constriction is primarily a consequence of the continuity equation,which is based on the conservation of mass.
$(e)$ Greater: For a model in a wind tunnel,the characteristic length is smaller,so to maintain the same Reynolds number as the actual plane,the flow speed must be greater.

(N/A) According to Bernoulli's principle,as the velocity of a fluid increases,its pressure decreases. Blowing over the paper increases the air velocity above it,reducing the pressure above compared to below,which creates an upward lift to keep the paper horizontal.
$(b)$ According to the equation of continuity,$A_1v_1 = A_2v_2$ (Area $\times$ Velocity = Constant). When we partially close the tap with our fingers,the cross-sectional area of the opening decreases,causing the velocity of the water to increase significantly,resulting in fast jets.
$(c)$ The flow rate is governed by the equation of continuity. The needle has a very small cross-sectional area compared to the syringe barrel. Even a small change in the needle's diameter significantly affects the flow velocity,making it a more dominant factor in controlling the flow rate than the thumb pressure.
$(d)$ As fluid exits the hole with high velocity,it carries momentum. According to the law of conservation of momentum,to keep the total momentum of the system zero,the vessel must experience an equal and opposite momentum,resulting in a backward thrust.

(D) The base area of the tank is $A = 1.0 \; m^{2}$.
The area of the hinged door is $a = 20 \; cm^{2} = 20 \times 10^{-4} \; m^{2}$.
The density of water is $\rho_{1} = 10^{3} \; kg/m^{3}$.
The density of acid is $\rho_{2} = 1.7 \times 10^{3} \; kg/m^{3}$.
The height of both liquid columns is $h = 4.0 \; m$.
The pressure exerted by the water column at the bottom is $P_{1} = h \rho_{1} g = 4 \times 10^{3} \times 9.8 = 3.92 \times 10^{4} \; Pa$.
The pressure exerted by the acid column at the bottom is $P_{2} = h \rho_{2} g = 4 \times 1.7 \times 10^{3} \times 9.8 = 6.664 \times 10^{4} \; Pa$.
The pressure difference between the two columns is $\Delta P = P_{2} - P_{1} = (6.664 - 3.92) \times 10^{4} = 2.744 \times 10^{4} \; Pa$.
The force required to keep the door closed is $F = \Delta P \times a = 2.744 \times 10^{4} \times 20 \times 10^{-4} = 54.88 \; N$.

(N/A) Let $P$ be the pressure at height $y$. The change in pressure $dP$ for a small change in height $dy$ is $dP = -\rho g dy$.
Since the atmosphere is isothermal,$PV = nRT$,which implies $P = \frac{\rho RT}{M}$,where $M$ is the molar mass of air.
Thus,$\rho = \frac{PM}{RT}$. Substituting this into the pressure equation: $dP = -\frac{PM}{RT} g dy$,or $\frac{dP}{P} = -\frac{Mg}{RT} dy$.
Integrating from $y=0$ (where $P=P_{0}$) to $y$ (where $P=P$): $\ln(\frac{P}{P_{0}}) = -\frac{Mg}{RT} y$.
So,$P = P_{0} e^{-y/y_{0}}$,where $y_{0} = \frac{RT}{Mg}$. Since $\rho \propto P$ at constant temperature,$\rho = \rho_{0} e^{-y/y_{0}}$.
$(b)$ The balloon rises until its density $\rho$ equals the density of the surrounding air.
Total mass of the balloon system $M_{total} = m_{payload} + m_{He} = 400 + (1425 \times 0.18) = 400 + 256.5 = 656.5 \; kg$.
Density of the balloon $\rho = \frac{M_{total}}{V} = \frac{656.5}{1425} \approx 0.4607 \; kg \, m^{-3}$.
Using the law of atmospheres: $\rho = \rho_{0} e^{-y/y_{0}} \implies 0.4607 = 1.25 e^{-y/8000}$.
$\ln(\frac{0.4607}{1.25}) = -\frac{y}{8000} \implies \ln(0.36856) = -\frac{y}{8000}$.
$-0.998 \approx -\frac{y}{8000} \implies y \approx 8000 \; m = 8 \; km$.

(N/A) Liquid and gas are both classified as fluids. The fundamental differences are as follows:
$1$. Liquids are generally considered incompressible,whereas gases are highly compressible.
$2$. $A$ liquid has a fixed volume and a free surface,while a gas occupies the entire volume of its container and has no free surface.
The common property between them is that both can flow easily and offer very little resistance to shear stress.

(N/A) $(1)$ Fluid Statics: In this branch of fluid mechanics,the forces and pressures acting on a stationary fluid are studied.
$(2)$ Fluid Dynamics: In this branch of fluid mechanics,the motion of fluid and properties related to it as a result of forces acting on the fluid are studied.

(N/A) Fluid mechanics is the branch of physics that deals with the behavior of fluids (liquids and gases) at rest and in motion. It is broadly divided into two main branches:
$1$. Fluid Statics: This is the study of fluids at rest. It focuses on the properties of fluids in equilibrium,such as pressure distribution and buoyancy,where there is no relative motion between fluid particles.
$2$. Fluid Dynamics: This is the study of fluids in motion. It deals with the forces acting on fluids and the resulting flow patterns,including concepts like viscosity,turbulence,and Bernoulli's principle.

(N/A) Dynamic lift is the force that acts on a body,such as an airplane wing,a hydrofoil,or a spinning ball,by virtue of its motion through a fluid.
It occurs when a body moves through a fluid (like air or water) and experiences a pressure difference due to its shape,orientation,or rotation (the Magnus effect).
In many sports such as cricket,tennis,baseball,or golf,dynamic lift is observed when a spinning ball deviates from its expected parabolic trajectory due to the interaction between its spin and the surrounding air.

(B) Skating is possible on ice because of the phenomenon of 'Regelation'.
When a skater stands on ice,the pressure exerted by the skates on the ice surface is very high due to the small surface area of the blade.
According to the phase diagram of water,an increase in pressure lowers the melting point of ice.
As a result,the ice directly beneath the skate blade melts into a thin layer of water,which acts as a lubricant.
This reduces the friction significantly,allowing the skater to glide smoothly over the surface.
Once the pressure is removed,the water refreezes into ice.

(B) Liquids can flow because the intermolecular forces between their molecules are weaker than those in solids.
This allows the molecules to slide over one another,providing the property of fluidity.
Unlike solids,where molecules are held in a fixed position,liquid molecules have enough kinetic energy to overcome the weak attractive forces,enabling them to take the shape of their container.

(N/A) Turbulence dissipates kinetic energy,usually in the form of heat.
Racing cars and planes are engineered to precision in order to minimize turbulence.
On the other hand,turbulence is sometimes desirable. Turbulence promotes mixing and increases the rates of transfer of mass,momentum,and energy.
The blades of a kitchen mixer induce turbulent flow,which helps in preparing thick milkshakes and beating eggs into a uniform texture.

(N/A) Turbulence is a complex fluid motion characterized by chaotic changes in pressure and flow velocity. While often considered a nuisance in engineering due to increased drag,it has several practical applications:
$1$. Mixing: Turbulence is highly effective at mixing fluids,such as in chemical reactors or in the atmosphere,where it helps disperse pollutants.
$2$. Heat Transfer: Turbulent flow significantly enhances heat transfer rates between a fluid and a solid surface,which is essential in heat exchangers,radiators,and cooling systems.
$3$. Combustion: In internal combustion engines and gas turbines,turbulence is used to mix fuel and air rapidly,ensuring efficient and complete combustion.
$4$. Aerodynamics: In some cases,such as on the surface of a golf ball,controlled turbulence (via dimples) is used to reduce drag by delaying flow separation.

(D) $(i)$ When cohesive force $ > $ adhesive force, the liquid does not wet the surface. The angle of contact is obtuse $( > 90^{\circ})$ and the meniscus is convex.
$(ii)$ When cohesive force $ < $ adhesive force, the liquid wets the surface. The angle of contact is acute $( < 90^{\circ})$ and the meniscus is concave.
$(iii)$ Due to surface tension, the pressure on the concave side is always greater than the pressure on the convex side. Therefore, a large pressure is exerted on the concave side of the liquid surface.

(N/A) spinning cylinder moves in a straight path along the direction of its rotational axis. Because of this symmetry,the Magnus effect (which causes lateral force on a spinning object moving through a fluid) is effectively neutralized or balanced,preventing the bullet from deviating from its intended trajectory. For this reason,bullets are designed to be cylindrical.

(A) $(i)$ False. In an incompressible flow,the density $\rho$ of the fluid remains constant throughout the flow.
$(ii)$ False. $A$ non-viscous fluid has a coefficient of viscosity $\eta = 0$. From the Reynolds number formula $R_{e} = \frac{\rho v D}{\eta}$,as $\eta \to 0$,$R_{e} \to \infty$. Such a flow is typically turbulent and cannot be a streamline flow.
$(iii)$ True. By definition,in a steady flow,the velocity of the fluid particles at any given point in space does not change with time.

(N/A) $(i)$ True. The Bernoulli equation is derived from the work-energy theorem,representing the conservation of energy for an incompressible,non-viscous,and steady flow of fluid.
$(ii)$ False. Raincoats are made of hydrophobic materials to repel water. The angle of contact between water and a hydrophobic surface is an obtuse angle (greater than $90^{\circ}$),which prevents the water from wetting the fabric.
$(iii)$ False. Viscosity is actually the measure of the internal friction or resistance to flow between adjacent layers of a fluid moving at different velocities. It is the presence of internal friction,not its absence,that defines viscosity.

(N/A) $(i)$ False. If the weight of the body is greater than the buoyant force,the body will sink in the liquid.
$(ii)$ True. Due to the viscosity of water and the friction between the water and the riverbed,the velocity of water is lowest at the bottom and highest near the surface.
$(iii)$ False. The angle of contact generally decreases as the temperature of the liquid increases because the surface tension of the liquid decreases with an increase in temperature.

(N/A) $(i)$ In steady flow,the path of a particle (line of flow) and the streamline coincide.
$(ii)$ According to Torricelli's law,the velocity of efflux is $v = \sqrt{2gh}$,where $g$ is the acceleration due to gravity.
$(iii)$ Since $1 \ Pa = 1 \ N/m^{2} = 10^{5} \ dyne / 10^{4} \ cm^{2} = 10 \ dyne/cm^{2}$.
$(iv)$ Velocity gradient $= \frac{dv}{dx} = \frac{6 \ cm/s}{0.1 \ mm} = \frac{6 \ cm/s}{0.01 \ cm} = 600 \ s^{-1}$.

(A) $(i)$ At the critical temperature,the interface between liquid and vapor disappears,so the surface tension is $0$.
$(ii)$ Bernoulli's equation is derived from the work-energy theorem,which is based on the law of conservation of energy.
$(iii)$ When a large drop splits into smaller drops,the total surface area increases. Since surface energy is proportional to surface area $(U = T \cdot A)$,the energy is absorbed (or the system requires energy),but in the context of surface tension problems,splitting a drop increases the surface energy,meaning energy is absorbed. However,if the question implies the change in energy,it is positive (absorbed).
$(iv)$ According to Bernoulli's principle,high wind speed over the roof creates low pressure,resulting in an upward lift force.

(A) Rain drops falling through the atmosphere reach a terminal velocity because the viscous drag force of the air balances the gravitational force. This is due to the property of viscosity. Thus,$(a-ii)$.
$(b)$ Clouds float at a height because their density is less than the density of the surrounding air,allowing them to remain suspended due to buoyancy. Thus,$(b-iii)$.
Therefore,the correct match is $(a-ii), (b-iii)$.

(A) The surface of a rotating liquid takes the shape of a paraboloid defined by the equation $z = \frac{\omega^2 r^2}{2g}$.
Here,the diameter of the vessel is $10 \, cm$,so the radius $R = 5 \, cm$.
The difference in height $h$ between the center $(r=0)$ and the side $(r=R)$ is given by:
$h = \frac{\omega^2 R^2}{2g}$
Substituting $R = 5 \, cm$:
$h = \frac{\omega^2 (5)^2}{2g} = \frac{25 \omega^2}{2g}$.

(C) The initial potential energy of the system is given by the sum of the potential energies of the liquid in both vessels:
$U_{i} = (dSx_{1})g \cdot \frac{x_{1}}{2} + (dSx_{2})g \cdot \frac{x_{2}}{2} = \frac{dSg}{2}(x_{1}^{2} + x_{2}^{2})$
By the principle of conservation of volume,the total volume of liquid remains constant:
$Sx_{1} + Sx_{2} = S(x_{f} + x_{f}) = 2Sx_{f}$
$x_{f} = \frac{x_{1} + x_{2}}{2}$
The final potential energy of the system is:
$U_{f} = 2 \times (dSx_{f})g \cdot \frac{x_{f}}{2} = dSgx_{f}^{2} = dSg \left( \frac{x_{1} + x_{2}}{2} \right)^{2}$
The change in potential energy is $\Delta U = U_{f} - U_{i}$:
$\Delta U = dSg \left[ \left( \frac{x_{1} + x_{2}}{2} \right)^{2} - \frac{x_{1}^{2} + x_{2}^{2}}{2} \right]$
$\Delta U = dSg \left[ \frac{x_{1}^{2} + x_{2}^{2} + 2x_{1}x_{2}}{4} - \frac{2x_{1}^{2} + 2x_{2}^{2}}{4} \right]$
$\Delta U = dSg \left[ \frac{2x_{1}x_{2} - x_{1}^{2} - x_{2}^{2}}{4} \right] = -\frac{dSg}{4}(x_{1} - x_{2})^{2}$
The magnitude of the change in energy is $\frac{1}{4} gdS(x_{2} - x_{1})^{2}$.

(B) The initial heights are $h_{1} = 1.0\,m$ and $h_{2} = 1.5\,m$. The cross-sectional area is $A = 16\,cm^{2} = 16 \times 10^{-4}\,m^{2}$.
When interconnected,the final height $h$ in both vessels will be the average of the initial heights: $h = \frac{h_{1} + h_{2}}{2} = \frac{1.0 + 1.5}{2} = 1.25\,m$.
The work done by gravity is equal to the decrease in potential energy: $W = U_{i} - U_{f}$.
Initial potential energy $U_{i} = (m_{1}g \frac{h_{1}}{2}) + (m_{2}g \frac{h_{2}}{2}) = \rho A h_{1} g \frac{h_{1}}{2} + \rho A h_{2} g \frac{h_{2}}{2} = \frac{\rho A g}{2} (h_{1}^{2} + h_{2}^{2})$.
Final potential energy $U_{f} = (m_{1}+m_{2})g \frac{h}{2} = (2 \rho A h) g \frac{h}{2} = \rho A g h^{2} = \rho A g (\frac{h_{1}+h_{2}}{2})^{2}$.
$W = \frac{\rho A g}{2} [h_{1}^{2} + h_{2}^{2} - 2(\frac{h_{1}+h_{2}}{2})^{2}] = \frac{\rho A g}{4} [2h_{1}^{2} + 2h_{2}^{2} - (h_{1}+h_{2})^{2}] = \frac{\rho A g}{4} (h_{1}-h_{2})^{2}$.
Substituting the values: $W = \frac{10^{3} \times 16 \times 10^{-4} \times 10}{4} (1.5 - 1.0)^{2} = \frac{16}{4} (0.5)^{2} = 4 \times 0.25 = 1\,J$.

(B) Let a bubble of radius $R$ expand such that its surface moves with speed $v$. Consider a spherical shell of radius $x$ $(x \ge R)$ and thickness $dx$ in the water.
Due to the incompressibility of water,the volume flow rate across any spherical surface must be constant.
At the bubble surface (radius $R$),the flow rate is $4 \pi R^{2} v$.
At a distance $x$ from the center,the flow rate is $4 \pi x^{2} v_{x}$,where $v_{x}$ is the velocity of the water at distance $x$.
Equating these,$4 \pi x^{2} v_{x} = 4 \pi R^{2} v$,which gives $v_{x} = \frac{R^{2} v}{x^{2}}$.
The mass of the spherical shell of thickness $dx$ is $dm = \rho (4 \pi x^{2} dx)$.
The kinetic energy $dK$ of this shell is $dK = \frac{1}{2} dm v_{x}^{2} = \frac{1}{2} (4 \pi x^{2} \rho dx) \left( \frac{R^{2} v}{x^{2}} \right)^{2}$.
Simplifying,$dK = 2 \pi \rho R^{4} v^{2} \frac{dx}{x^{2}}$.
The total kinetic energy $K$ is the integral from $x = R$ to $x = \infty$:
$K = \int_{R}^{\infty} 2 \pi \rho R^{4} v^{2} \frac{dx}{x^{2}} = 2 \pi \rho R^{4} v^{2} \left[ -\frac{1}{x} \right]_{R}^{\infty} = 2 \pi \rho R^{4} v^{2} \left( 0 - (-\frac{1}{R}) \right) = 2 \pi \rho R^{3} v^{2}$.

(A) The correct answer is $A$.
When we exert pressure on snow,the solid snow melts into liquid because the melting point of ice decreases with an increase in pressure. This phenomenon is a consequence of the anomalous behaviour of water,where water is less dense in its solid form (ice) than in its liquid form.
Because liquid water occupies less volume than the equivalent mass of ice,applying pressure shifts the equilibrium towards the liquid phase. When we release the pressure,the melted water refreezes,binding the snow crystals together into a ball.

(B) For equilibrium,the lift force must equal the weight of the aircraft:
$mg = \frac{1}{2} \rho v^2 A$
Given:
$m = 7500 \, kg$
$v = 8 \, km/s = 8000 \, m/s$
$A = 30 \, m^2$
$\rho(h) = 1.2 e^{-h/10} \, kg/m^3$ (where $h$ is in $km$)
$g \approx 9.8 \, m/s^2$
Substituting the values into the equilibrium equation:
$7500 \times 9.8 = \frac{1}{2} \times (1.2 e^{-h/10}) \times (8000)^2 \times 30$
$73500 = 0.6 \times e^{-h/10} \times 64 \times 10^6 \times 30$
$73500 = 1152 \times 10^6 \times e^{-h/10}$
$e^{-h/10} = \frac{73500}{1152 \times 10^6} \approx 6.38 \times 10^{-5}$
Taking the natural logarithm on both sides:
$-h/10 = \ln(6.38 \times 10^{-5})$
$-h/10 \approx -9.66$
$h \approx 96.6 \, km$
This value falls within the range of $75-100 \, km$.
Therefore,the correct option is $(B)$.

(D) The correct answer is $D$.
An ideal fluid is defined by the following properties:
$1$. It is non-viscous (i.e.,it has zero viscosity).
$2$. It is incompressible (i.e.,its density remains constant).
$3$. Its flow is steady (streamline).
$4$. Its flow is irrotational.
Since an ideal fluid is non-viscous,the statement that it is viscous is incorrect.

(D) Let the initial radius of the air bubble at the bottom be $r_1 = r$.
The final radius at the surface is $r_2 = r + 200\% \text{ of } r = r + 2r = 3r$.
Let the depth of the lake be $h$. The atmospheric pressure is given as $P_{atm} = \rho g H$.
The pressure at the bottom of the lake is $P_1 = P_{atm} + \rho g h = \rho g H + \rho g h = \rho g(H + h)$.
The pressure at the surface is $P_2 = P_{atm} = \rho g H$.
Assuming the temperature remains constant,we use Boyle's Law: $P_1 V_1 = P_2 V_2$.
Since the volume of a sphere is $V = \frac{4}{3} \pi r^3$,we have:
$\rho g(H + h) \times \frac{4}{3} \pi r^3 = \rho g H \times \frac{4}{3} \pi (3r)^3$
$(H + h) r^3 = H \times 27r^3$
$H + h = 27H$
$h = 26H$.
Thus,the depth of the lake is $26H$.

(D) When a vessel containing liquid is accelerated horizontally with an acceleration $a$,the free surface of the liquid tilts at an angle $\theta$ with the horizontal.
The effective acceleration acting on a fluid particle is the vector sum of the acceleration due to gravity $g$ (acting downwards) and the pseudo-acceleration $a$ (acting backwards in the frame of the vessel).
The angle $\theta$ that the free surface makes with the horizontal is given by the relation:
$\tan \theta = \frac{a}{g}$
Given:
$a = 19.6 \, m/s^2$
$g = 9.8 \, m/s^2$
Substituting the values:
$\tan \theta = \frac{19.6}{9.8} = 2$
To express this in terms of $\sin \theta$,we use the trigonometric identity $\sin \theta = \frac{\tan \theta}{\sqrt{1 + \tan^2 \theta}}$:
$\sin \theta = \frac{2}{\sqrt{1 + 2^2}} = \frac{2}{\sqrt{1 + 4}} = \frac{2}{\sqrt{5}}$
Therefore,$\theta = \sin^{-1}\left[\frac{2}{\sqrt{5}}\right]$.

(D) According to Boyle's Law, for a constant temperature, the product of pressure and volume remains constant $(PV = \text{constant})$.
At depth $h$, the total pressure on the bubble is $P_1 = P + \rho g h$, and its volume is $V_1 = V_0$.
At the surface, the pressure is $P_2 = P$, and let the volume be $V_2$.
Applying the equation $P_1 V_1 = P_2 V_2$:
$(P + \rho g h) V_0 = P V_2$
$V_2 = V_0 \frac{P + \rho g h}{P}$
$V_2 = V_0 (1 + \frac{\rho g h}{P})$.

(A) Given:
Initial volume $V_1 = 1\,cm^3$
Depth $h = 40\,m$
Atmospheric pressure $P_0 = 1 \times 10^5\,Pa$
Density of water $\rho = 1000\,kg/m^3$
Acceleration due to gravity $g = 10\,m/s^2$
Temperature is constant.
Pressure at the bottom of the lake is given by:
$P_1 = P_0 + \rho gh$
$P_1 = 1 \times 10^5 + (1000 \times 10 \times 40) = 1 \times 10^5 + 4 \times 10^5 = 5 \times 10^5\,Pa$
Since the temperature is constant,we use Boyle's Law:
$P_1 V_1 = P_0 V_2$
$5 \times 10^5 \times 1 = 1 \times 10^5 \times V_2$
$V_2 = 5\,cm^3$
Therefore,the volume of the air bubble when it reaches the surface is $5\,cm^3$.