NCERT Solutions · Class 11 Physics Mechanical Properties of Fluids

(a) Blood pressure in humans is greater at the feet than at the brain because gravity adds the weight of the intervening column of blood to whatever pressure the heart supplies.Blood forms continuous fluid columns connecting the heart to every part of the body, so the hydrostatic pressure law for a fluid at rest applies along each column: \[P = P_0 + \rho g h \] where \(\displaystyle P_0 \) is the pressure at a reference point, \(\displaystyle \rho \) is the density of blood, \(\displaystyle g \) is the acceleration due to gravity, and \(\displaystyle h \) is the vertical height of fluid between the reference point and the point of measurement, measured downward from the reference.Take the heart as the reference point. The feet lie a height \(\displaystyle h_1 \) below the heart, so the pressure there is boosted by the weight of the blood column sitting above them: \[P_{\text{feet}} = P_{\text{heart}} + \rho g h_1 \]The brain lies a height \(\displaystyle h_2 \) above the heart, so gravity now works against the pump, and the pressure there is reduced by the weight of that column: \[P_{\text{brain}} = P_{\text{heart}} - \rho g h_2 \]Since \(\displaystyle h_1 \) and \(\displaystyle h_2 \) are both positive lengths, \(\displaystyle P_{\text{feet}} > P_{\text{heart}} > P_{\text{brain}} \). One heart, pumping at one pressure, produces a higher reading at the feet and a lower one at the brain purely because of the height of blood between the heart and each point — the same reasoning used for a pressure gauge placed at different heights in any static fluid, not a special property of blood.(b) Atmospheric pressure falls to half its sea-level value within about $\displaystyle 6$ km, while the atmosphere itself extends past $\displaystyle 100$ km, because pressure decreases exponentially with height, not linearly.If air had one fixed density \(\displaystyle \rho_0 \) all the way up, pressure would fall linearly, \(\displaystyle P = P_0 - \rho_0 g h \), reaching zero at a height \(\displaystyle H = P_0/(\rho_0 g) \). But air is a gas, not an incompressible liquid: its density depends on the very pressure squeezing it, through the ideal gas relation \(\displaystyle \rho = PM/RT \) (\(\displaystyle M \) = molar mass of air, \(\displaystyle R \) = gas constant, \(\displaystyle T \) = temperature). Climb higher, pressure drops, so density drops too, so the weight added per extra metre of height keeps shrinking. Combining the hydrostatic equation \(\displaystyle dP = -\rho g\, dh \) with this density–pressure link gives an exponential fall-off instead of a linear one: \[P(h) = P_0 \, e^{-h/H}, \qquad H = \frac{RT}{Mg} \]Using \(\displaystyle T \approx 288\ \text{K} \), \(\displaystyle M = 0.029\ \text{kg mol}^{-1} \), \(\displaystyle R = 8.31\ \text{J mol}^{-1}\text{K}^{-1} \), \(\displaystyle g = 9.8\ \text{m s}^{-2} \): \[H = \frac{8.31 \times 288}{0.029 \times 9.8} \approx 8.4 \times 10^{3}\ \text{m} \]Pressure halves when \(\displaystyle e^{-h/H} = \tfrac12 \), i.e. at \[h = H\ln 2 \approx 8.4 \times 0.693 \approx 5.8\ \text{km} \approx 6\ \text{km} \] — exactly the height quoted in the question. Because the fall-off is exponential rather than linear, pressure keeps approaching zero without ever formally reaching it: it halves in the first $\displaystyle 6$ km, halves again in the next $\displaystyle 6$ km, and so on, so a very thin but non-zero atmosphere is still present $\displaystyle 100$ km up even though its pressure there is a tiny fraction of a percent of the sea-level value. A linear model, by contrast, would force the atmosphere to end abruptly at a single finite height — which is not what is observed.(c) Hydrostatic pressure is a scalar because, at a given point in a fluid at rest, it pushes with the same magnitude on a surface of any orientation — it has no single direction of its own to serve as a vector's component along.Pressure is defined through \[\vec{F} = P\,\vec{A} \] where \(\displaystyle \vec{A} \) is the area vector, whose direction is taken along the outward normal to the surface. A fluid at rest cannot sustain a shear (sideways) force — if it could, the fluid would flow until that shear vanished — so the force it exerts on any imagined surface inside it must always point along the normal to that surface, i.e. \(\displaystyle \vec{F} \) is always parallel to \(\displaystyle \vec{A} \), whatever direction \(\displaystyle \vec{A} \) happens to have. Tilt the surface, and both \(\displaystyle \vec{F} \) and \(\displaystyle \vec{A} \) rotate together by the same amount, so their ratio \[P = \frac{F}{A} \] stays unchanged. Because \(\displaystyle P \) is this fixed ratio between two vectors that are always aligned, it carries no direction of its own — it simply tells you how strongly the fluid pushes normal to any surface placed at that point, in every direction at once (Pascal's law). That is what makes pressure a scalar, even though it is defined as force divided by area: the force \(\displaystyle \vec{F} = P\vec{A} \) on a specific surface is a vector, but the direction there comes entirely from the surface's own area vector \(\displaystyle \vec{A} \), not from \(\displaystyle P \).Answer: (a) The feet lie below the heart, so the weight of the blood column between heart and feet adds to the pressure there, while the brain, being above the heart, has this weight subtracted, giving \(\displaystyle P_{\text{feet}} > P_{\text{brain}} \). (b) Atmospheric pressure falls off exponentially, \(\displaystyle P = P_0 e^{-h/H} \) with scale height \(\displaystyle H \approx 8.4\ \text{km} \), because air's density (and so its weight per unit height) itself falls with pressure; this halves the pressure in about \(\displaystyle H\ln 2 \approx 6\ \text{km} \) while never reaching exactly zero, so the atmosphere is still present past $\displaystyle 100$ km. (c) Hydrostatic pressure is a scalar because a static fluid can exert force only normal to any surface at a point, so \(\displaystyle P = F/A \) is a fixed ratio between two always-parallel vectors and carries no direction of its own.

Angle of contact and wetting both come from a three-way tug-of-war between surface tensions — solid-air, solid-liquid, and liquid-air — not from surface tension alone.At the line where a liquid surface meets a solid, three interfacial tensions pull on that line: \(\displaystyle S_{sa} \) (solid-air), \(\displaystyle S_{sl} \) (solid-liquid), and \(\displaystyle S_{la} \) (liquid-air, the ordinary surface tension of the liquid). Balancing the forces along the solid surface at the contact line gives Young's equation \[S_{sa} = S_{sl} + S_{la}\cos\theta \quad\Rightarrow\quad \cos\theta = \frac{S_{sa}-S_{sl}}{S_{la}} \] where \(\displaystyle \theta \) is the angle of contact. Everything below follows from this one relation.(a) Why mercury-glass is obtuse and water-glass is acuteThe angle of contact is decided by which force wins: the cohesive force between the liquid's own molecules, or the adhesive force between the liquid and the solid.In mercury, the metallic bonding between mercury atoms (cohesion) is far stronger than the attraction between mercury atoms and glass (adhesion). Mercury "resents" contact with glass, so the mercury-glass interfacial tension \(\displaystyle S_{sl} \) is large — larger than \(\displaystyle S_{sa} \). Then \(\displaystyle S_{sa}-S_{sl} < 0 \), so \(\displaystyle \cos\theta < 0 \) and \(\displaystyle \theta \) is obtuse.In water, the adhesive force between water molecules and the glass surface (hydrogen bonding with the oxide surface) is comparable to or stronger than the cohesive force between water molecules themselves. Contact is energetically cheap, so \(\displaystyle S_{sl} \) is small, \(\displaystyle S_{sa}-S_{sl} > 0 \), \(\displaystyle \cos\theta > 0 \), and \(\displaystyle \theta \) is acute. This is the same reason a liquid climbs a capillary tube of one material but is depressed in another — it is a property of the pair of surfaces, not of the liquid alone.(b) Why water spreads and mercury beads upA liquid spreads over a solid only if doing so lowers the total surface energy — that is, only if \(\displaystyle S_{sa} > S_{sl} + S_{la} \) (a positive "spreading coefficient"). Because \(\displaystyle S_{sl} \) is small for water-glass (from part a), this condition is met: replacing the higher-energy glass-air surface with a water film lowers the system's energy, so water spreads out.For mercury, \(\displaystyle S_{sl} \) is large enough that spreading would raise the total surface energy instead of lowering it. The system minimizes energy the other way — by pulling mercury away from the glass into drops, which cuts down the area of the costly mercury-glass contact while mercury's own surface tension pulls each drop into the shape (nearly spherical, see part e) that has the least surface area for its volume. That is why mercury beads instead of wetting the glass.(c) Why surface tension does not depend on the area of the surfaceSurface tension \(\displaystyle T = F/l \) (force per unit length across any line drawn in the surface) comes entirely from short-range cohesive forces between molecules in a layer only a few molecular diameters thick. That force per unit length is the same at every point of the surface, set only by the nature of the liquid and its temperature — it does not care whether the total surface next to it is large or small, exactly as the pressure at a point inside a fluid does not depend on how much fluid there is altogether. So \(\displaystyle T \) is an intensive property of the liquid, unlike the total surface energy \(\displaystyle T \times A \), which obviously does grow with area \(\displaystyle A \).(d) Why a detergent solution needs a small angle of contactAdding detergent lowers the solid-liquid interfacial tension \(\displaystyle S_{sl} \) (its molecules present a hydrophilic end to the water and a hydrophobic end toward the dirt/fibre, reducing the energy cost of that contact). From \(\displaystyle \cos\theta = (S_{sa}-S_{sl})/S_{la} \), a smaller \(\displaystyle S_{sl} \) makes the numerator larger, so \(\displaystyle \cos\theta \) increases and \(\displaystyle \theta \) decreases. This is deliberate: a liquid with a large contact angle would simply bead up on the surface of cloth or bead over the mouth of a narrow crack in dirt, never entering it. A small \(\displaystyle \theta \) lets the soapy water wet and penetrate the fine spaces between fibres, get in under trapped dirt and grease, and lift it away — which is precisely the job a detergent is designed to do.(e) Why a free drop is sphericalSurface tension makes a liquid surface behave like a stretched membrane that always tries to shrink, because the surface carries energy \(\displaystyle T \times A \): more area means more stored energy. With no external force (no gravity, no contact with a wall) to fight against, the drop settles into whatever shape minimizes this surface energy for the volume of liquid it has — and among all three-dimensional shapes enclosing a given volume, the sphere has the least possible surface area. So a drop under no external forces pulls itself into a sphere.Answer: (a) Because mercury's cohesion overwhelms its adhesion to glass (large \(\displaystyle S_{sl}\)) giving \(\displaystyle \cos\theta<0\), while water's adhesion to glass is strong (small \(\displaystyle S_{sl}\)) giving \(\displaystyle \cos\theta>0\) — from \(\displaystyle \cos\theta=(S_{sa}-S_{sl})/S_{la}\). (b) Water spreads because \(\displaystyle S_{sa}>S_{sl}+S_{la}\) for it (lowers total surface energy); mercury beads up because that inequality fails for it. (c) Surface tension is a force per unit length set by short-range intermolecular cohesion, so it is an intensive property independent of how much surface exists. (d) Detergent lowers \(\displaystyle S_{sl}\), raising \(\displaystyle \cos\theta\) and shrinking \(\displaystyle \theta\), so the solution can wet and penetrate narrow spaces between fibres/dirt. (e) With no external force, a drop minimizes its surface energy \(\displaystyle T\times A\) for its fixed volume, and the sphere is the shape of least surface area for a given volume.

Each blank turns on one physical mechanism — cohesion, momentum transfer, elastic vs. flowing response, mass conservation, or the Reynolds number. Work each mechanism through rather than memorising the word.(a) Surface tension: decreases with temperature. Surface tension exists because molecules at a liquid's surface are pulled inward by cohesive forces from the molecules below them, and it takes energy to bring a molecule to the surface against that pull. Raising the temperature gives molecules more kinetic energy, so they sit farther apart on average and the cohesive force between them weakens. A weaker cohesive force means less energy is needed to maintain the surface, so surface tension decreases as temperature rises. (This is also why hot water wets and cleans better than cold water — its lower surface tension lets it spread into narrow gaps.)(b) Viscosity of gases: increases with temperature. Viscosity of liquids: decreases with temperature. This is the pair students most often get backwards, because the microscopic cause is opposite in the two cases. (c) Shearing force: proportional to shear strain for solids; proportional to the rate of shear strain for fluids. A solid with a rigidity modulus \(\displaystyle \eta \) obeys Hooke's law in shear: it deforms by a fixed, definite angle (the shear strain \(\displaystyle \theta \)) for a given shearing stress, and it holds that deformed shape in static equilibrium as long as the stress is applied: \[\text{shearing stress} = \eta \times (\text{shear strain}) \] A fluid, by definition, cannot sustain a shear stress in static equilibrium — it simply keeps flowing (deforming) as long as any shearing force acts, however small. What resists the motion is not how far it has deformed but how fast it is deforming — the velocity gradient between adjacent layers. Newton's law of viscosity captures this: \[\text{shearing stress} = \eta \left(\frac{dv}{dx}\right) \] where \(\displaystyle dv/dx \) is the rate of shear strain. So the shearing force is proportional to shear strain for a solid, and to the rate of shear strain for a fluid — this distinction (a fixed deformation vs. a continuously growing one) is exactly what separates an elastic solid from a viscous fluid.(d) The increase in flow speed at a constriction follows conservation of mass. It is tempting to credit this to Bernoulli's principle, since Bernoulli's equation is what most people reach for with pipe-flow problems, but that is not the mechanism responsible for the speed change itself. For steady, incompressible flow, the mass entering a tube per second must equal the mass leaving it per second — this is exactly the statement of conservation of mass, and it gives the continuity equation: \[A_1 v_1 = A_2 v_2 \] where \(\displaystyle A \) is the cross-sectional area and \(\displaystyle v \) is the flow speed. When the tube narrows (\(\displaystyle A_2 < A_1 \)) at a constriction, this equation forces \(\displaystyle v_2 > v_1 \) — the speed must increase simply to keep the same mass moving through the smaller opening every second. Bernoulli's principle comes in only afterward: it uses energy conservation to say that this now-larger speed must be accompanied by a drop in pressure at the constriction. So the speed increase is a consequence of mass conservation; Bernoulli's principle explains the resulting pressure drop, not the speed increase.(e) Turbulence in the wind-tunnel model sets in at a greater speed than in the actual plane. Whether a flow is smooth (laminar) or turbulent is governed by the Reynolds number, \[R_e = \frac{\rho v d}{\eta} \] where \(\displaystyle \rho \) and \(\displaystyle \eta \) are the density and viscosity of the fluid (air, essentially the same in the tunnel as around the real aircraft), \(\displaystyle v \) is the flow speed, and \(\displaystyle d \) is a characteristic length of the body. Turbulence sets in once \(\displaystyle R_e \) crosses a critical value that is the same for the model and the full-size plane (same shape, same fluid). A wind-tunnel model is built smaller than the real plane, so its characteristic length \(\displaystyle d \) is smaller. To reach the same critical \(\displaystyle R_e \) with \(\displaystyle \rho \) and \(\displaystyle \eta \) unchanged, a smaller \(\displaystyle d \) must be compensated by a greater flow speed \(\displaystyle v \). So turbulence appears in the model only at a higher air speed than it would around the actual, full-sized plane.Answer: (a) decreases; (b) increases (gases), decreases (liquids); (c) shear strain (solids), rate of shear strain (fluids); (d) conservation of mass; (e) greater.

(a) Blow over the top, not under, and the sagging edge rises — that is Bernoulli's principle in action.Bernoulli's equation, applied along a horizontal streamline where height does not change, is \[P + \frac{1}{2}\rho v^{2} = \text{constant} \] where \(\displaystyle P\) is the pressure in the moving air, \(\displaystyle \rho\) its density and \(\displaystyle v\) its speed. Wherever a fluid moves faster, its pressure is lower.Hold a sheet of paper by one edge and the far edge droops under gravity. Blow a stream of air across the top surface: the air there is now moving fast, so by Bernoulli's equation the pressure just above the paper drops below atmospheric pressure \(\displaystyle P_0\). The air trapped underneath is essentially still, so it stays at the full \(\displaystyle P_0\). The resulting pressure difference, \(\displaystyle P_0 - P_{\text{top}} > 0\), pushes upward and lifts the sagging end until the sheet is horizontal.Blowing under the sheet does the opposite: it is now the air below that speeds up, so the pressure below drops instead — the higher pressure ends up on top, pressing the paper further down, not lifting it.(b) Squeezing the tap's opening between your fingers forces the water through a much narrower gap, and a narrower gap means a faster jet.For an incompressible fluid the equation of continuity requires the volume flowing per second to be the same at every cross-section: \[A_1 v_1 = A_2 v_2 \] Here \(\displaystyle A_1\) is the full bore of the tap (speed \(\displaystyle v_1\)) and \(\displaystyle A_2\) is the tiny opening left between your fingers (\(\displaystyle A_2 \ll A_1\)). Rearranging, \[v_2 = v_1 \frac{A_1}{A_2} \] Because \(\displaystyle A_2\) is very small, \(\displaystyle v_2\) becomes very large — the same amount of water must now pass through a much smaller opening every second, so it does so at high speed, emerging as thin, fast jets.(c) The needle's bore, not the thumb's push, is what really sets the injection rate — because flow rate is far more sensitive to radius than to pressure.Flow of a viscous fluid through a narrow tube like a needle follows Poiseuille's law: \[Q = \frac{\pi r^{4}\,\Delta P}{8\,\eta\, l} \] where \(\displaystyle Q\) is the volume flowing per second, \(\displaystyle r\) the needle's internal radius, \(\displaystyle \Delta P\) the pressure difference the thumb creates, \(\displaystyle \eta\) the liquid's viscosity, and \(\displaystyle l\) the needle's length.Notice the two dependences: \(\displaystyle Q\) rises only in direct (first-power) proportion to \(\displaystyle \Delta P\) — doubling the force on the plunger only doubles the flow — but \(\displaystyle Q\) rises with the fourth power of \(\displaystyle r\). Halving the needle's radius, for instance, cuts the flow rate to \(\displaystyle \left(\tfrac12\right)^4 = \tfrac{1}{16}\) of its earlier value. A small, deliberate choice of needle gauge therefore changes the flow rate far more drastically than any change in thumb pressure a person can realistically apply — which is exactly why doctors pick needle size to control how fast an injection is delivered, rather than relying on how hard they press.(d) The vessel recoils backward because the jet carries forward momentum away with it — Newton's third law for a fluid, exactly as in a rocket.Before the hole opens, the fluid-and-vessel system has zero total momentum. Once fluid escapes, it leaves through the hole with a speed \(\displaystyle v\) (given by Torricelli's law, \(\displaystyle v = \sqrt{2gh}\), \(\displaystyle h\) being the height of fluid above the hole), carrying away momentum \(\displaystyle v\,dm\) in a short interval \(\displaystyle dt\). Since no external horizontal force acts on the vessel-plus-fluid system, the total momentum must remain zero — so the vessel is given an equal and opposite momentum, \(\displaystyle -v\,dm\), i.e. it recoils backward. Dividing by \(\displaystyle dt\) gives the recoil force, \[F = v\,\frac{dm}{dt} \] which is precisely the backward thrust felt on the vessel — the same mechanism that propels a rocket by ejecting exhaust gas.(e) Spin drags the nearby air along with the ball, making airflow faster on one side than the other — the resulting pressure difference (the Magnus effect) pushes the ball off a plain parabola.As the ball moves and spins, friction drags a thin layer of air around with its spinning surface. On the side where the surface is moving in the same sense as the oncoming air (relative to the ball), the air is sped up further; on the opposite side, the surface motion opposes the airflow and slows it down. By Bernoulli's principle, pressure is lower where the air is faster: \[P_{\text{fast side}} < P_{\text{slow side}} \] This pressure difference produces a net sideways force (the Magnus force), directed from the high-pressure side toward the low-pressure side, perpendicular to both the ball's velocity and its spin axis. Added to the constant downward pull of gravity, this sideways force bends the ball's path out of the single vertical plane a plain projectile would follow, so instead of the simple parabola of an unspun ball, a spinning cricket ball follows a curved, twisted trajectory — the swing that bowlers rely on.Answer: (a) Blowing over the top speeds up that air, lowering its pressure by Bernoulli's principle; the still air underneath, at full atmospheric pressure, pushes the sagging paper up level — blowing underneath would do the reverse. (b) By the continuity equation \(\displaystyle A_1v_1 = A_2v_2\), squeezing the tap's opening between the fingers sharply increases the outflow speed, producing fast jets. (c) Poiseuille's law gives flow rate \(\displaystyle Q \propto r^4\) but only \(\displaystyle Q \propto \Delta P\), so the needle's bore controls the injection rate far more sensitively than thumb pressure. (d) By conservation of momentum, the forward momentum carried off by the escaping jet is balanced by an equal and opposite backward thrust on the vessel, exactly as in a rocket. (e) Spin makes the airflow faster on one side of the ball than the other, so Bernoulli's principle creates a sideways Magnus force that curves the ball's path away from a plain parabola.

Pressure is force spread over an area, and here almost the girl's whole weight lands on one tiny circular heel — that concentration is what makes the number so large.The heel presses on the floor with a force equal to the girl's weight (she is standing still, so the floor pushes back up with a normal force equal and opposite to her weight — this is the force transmitted through the heel).Step $\displaystyle 1$: Find the force (weight).Weight is a force, not the same thing as mass — mass is $\displaystyle 50$ kg, but the pressing force is \(\displaystyle mg \).\[F = mg = 50 \text{ kg} \times 9.8 \text{ m s}^{-2} = 490 \text{ N} \]Step $\displaystyle 2$: Find the area of contact.The heel is circular with diameter \(\displaystyle d = 1.0 \) cm, so the radius is\[r = \frac{d}{2} = 0.50 \text{ cm} = 0.50 \times 10^{-2} \text{ m} = 5.0 \times 10^{-3} \text{ m} \]The area of a circle is \(\displaystyle A = \pi r^2 \), so\[A = \pi \left(5.0 \times 10^{-3} \text{ m}\right)^2 = \pi \times 25 \times 10^{-6} \text{ m}^2 = 7.85 \times 10^{-5} \text{ m}^2 \]Step $\displaystyle 3$: Apply the definition of pressure.Pressure is force per unit area:\[P = \frac{F}{A} \]Substituting the values found above:\[P = \frac{490 \text{ N}}{7.85 \times 10^{-5} \text{ m}^2} = 6.24 \times 10^{6} \text{ N m}^{-2} \]The data (mass $\displaystyle 50$ kg, diameter $\displaystyle 1.0$ cm) are given to two significant figures, so the result is stated to that precision: \(\displaystyle P \approx 6.2 \times 10^{6} \text{ Pa} \), commonly quoted as \(\displaystyle 6.24 \times 10^{6} \) Pa.This is roughly $\displaystyle 60$ times atmospheric pressure (\(\displaystyle 1.0 \times 10^{5}\) Pa) — concentrating the same weight onto a needle-thin heel rather than a flat sole multiplies the pressure enormously, since pressure scales inversely with area while the force (her weight) stays fixed.Answer: The heel exerts a pressure of about \(\displaystyle 6.24 \times 10^{6}\) Pa (N m\(\displaystyle ^{-2}\)) on the floor.

Pressure at the base of a fluid column depends on the fluid's density, not on how tall the column looks — a lighter liquid needs a taller column to balance the same atmospheric pressure.A barometer works because the atmosphere pushes down on the open pool of liquid, and this pressure is balanced by the weight of the liquid column standing in the tube. For a fluid column of height \(\displaystyle h \) and density \(\displaystyle \rho \), the pressure it exerts at its base is\[P = \rho g h \]where \(\displaystyle \rho \) is the density of the liquid, \(\displaystyle g \) is the acceleration due to gravity, and \(\displaystyle h \) is the height of the column. Rearranging for the height:\[h = \frac{P}{\rho g} \]Normal atmospheric pressure is the quantity being measured, and its standard value is\[P = 1.013 \times 10^{5} \ \text{Pa} \]For French wine, the density given is\[\rho = 984 \ \text{kg m}^{-3} \]Taking \(\displaystyle g = 9.8 \ \text{m s}^{-2} \), substitute into the formula:\[h = \frac{1.013 \times 10^{5} \ \text{Pa}}{(984 \ \text{kg m}^{-3})(9.8 \ \text{m s}^{-2})} \]First find the denominator:\[984 \times 9.8 = 9643.2 \ \text{kg m}^{-2}\text{s}^{-2} \]Then divide:\[h = \frac{1.013 \times 10^{5}}{9643.2} \ \text{m} = 10.505... \ \text{m} \]The density of wine ($\displaystyle 984$ kg m\(\displaystyle ^{-3}\)) has three significant figures, so the answer is rounded to three significant figures at the end, not before:\[h \approx 10.5 \ \text{m} \]This is why Pascal's wine barometer needed a column roughly fourteen times taller than Torricelli's mercury one (mercury's density, $\displaystyle 13600$ kg m\(\displaystyle ^{-3}\), is about $\displaystyle 13.8$ times that of wine) — the same atmospheric pressure has to be balanced, and a less dense liquid can only do that by standing much taller. A barometer built with wine would have needed a tube well over $\displaystyle 10$ m long, which is why mercury barometers stayed the practical choice.Answer: h ≈ $\displaystyle 10.5$ m

The pressure the ocean exerts is just fluid pressure, \(\displaystyle P = h\rho g\) — compare that number to \(\displaystyle 10^{9}\,\text{Pa}\), not the other way round.At a depth \(\displaystyle h\) below the surface of a liquid of density \(\displaystyle \rho\), the pressure in excess of atmospheric pressure (the gauge pressure) is \[P = h\rho g \] where \(\displaystyle h\) is the depth, \(\displaystyle \rho\) is the density of the liquid, and \(\displaystyle g\) is the acceleration due to gravity. This follows from balancing the weight of the fluid column above that depth against the force on the area below it — it does not depend on the shape of the container or the structure sitting in the fluid, only on how far down you are.Taking sea water to have density \(\displaystyle \rho = 1.0\times10^{3}\,\text{kg m}^{-3}\), depth \(\displaystyle h = 3\,\text{km} = 3\times10^{3}\,\text{m}\), and \(\displaystyle g = 9.8\,\text{m s}^{-2}\): \[P = (3\times10^{3}\,\text{m})(1.0\times10^{3}\,\text{kg m}^{-3})(9.8\,\text{m s}^{-2}) \] \[P = 2.94\times10^{7}\,\text{Pa} \]Rounded to two significant figures (matching the two-figure precision of \(\displaystyle \rho\) and \(\displaystyle g\)), this is \[P \approx 2.9\times10^{7}\,\text{Pa}. \]Now compare this to the structure's rated maximum stress: \[\frac{P_{\text{max structure}}}{P_{\text{ocean}}} = \frac{10^{9}\,\text{Pa}}{2.9\times10^{7}\,\text{Pa}} \approx 34. \]The pressure the ocean actually exerts at $\displaystyle 3$ km depth is more than thirty times smaller than the maximum stress \(\displaystyle 10^{9}\,\text{Pa}\) the structure is built to withstand. The easy mistake here is to think a "vertical structure withstanding stress" problem needs something about the structure's own weight or the oil well's internal pressure — it doesn't; the question only asks whether the surrounding water crushes it, and that is set entirely by depth and density through \(\displaystyle h\rho g\).Since the actual pressure \(\displaystyle 2.9\times10^{7}\,\text{Pa}\) is well below the structure's limit of \(\displaystyle 10^{9}\,\text{Pa}\), the structure is safe to place on top of an oil well at this depth.Answer: Yes — the pressure at $\displaystyle 3$ km depth is about \(\displaystyle 2.9\times10^{7}\,\text{Pa}\), well below the structure's \(\displaystyle 10^{9}\,\text{Pa}\) limit, so the structure is suitable.

The maximum pressure on the small piston equals the maximum load's weight divided by the area of the piston that carries that load — Pascal's principle transmits this pressure equally to every part of the enclosed fluid, so the smaller piston must be able to withstand the same pressure as the larger one.In a hydraulic lift, a small piston pushes down on a fluid, and by Pascal's law that pressure is transmitted undiminished to the large piston, which lifts the car. The pressure the large piston bears when it is holding up the maximum load is therefore also the maximum pressure the small piston must be able to generate (and withstand).Step $\displaystyle 1$: Find the maximum weight (force) the large piston must support.Weight, not mass, is what creates a force — this is the point people slip on: mass \(\displaystyle m \) is a property of the car (in kilograms), but the load pushing down on the piston is the weight \(\displaystyle F = mg \) (in newtons).\[m = 3000 \text{ kg}, \qquad g = 9.8 \text{ m s}^{-2} \]\[F = mg = 3000 \text{ kg} \times 9.8 \text{ m s}^{-2} = 29400 \text{ N} \]Step $\displaystyle 2$: Convert the piston's area to SI units.\[A = 425 \text{ cm}^2 = 425 \times 10^{-4} \text{ m}^2 = 4.25 \times 10^{-2} \text{ m}^2 \]Step $\displaystyle 3$: Apply the definition of pressure, \(\displaystyle P = F/A \).Here \(\displaystyle P \) is the pressure on the piston, \(\displaystyle F \) is the force (weight of the car) it must bear, and \(\displaystyle A \) is the cross-sectional area over which that force acts.\[P = \frac{F}{A} = \frac{29400 \text{ N}}{4.25 \times 10^{-2} \text{ m}^2} \]\[P = 6.91765 \times 10^{5} \text{ N m}^{-2} \]Step $\displaystyle 4$: Round to the correct number of significant figures.The area, $\displaystyle 425$ cm², is given to $\displaystyle 3$ significant figures, so the pressure should also be quoted to $\displaystyle 3$ significant figures.\[P \approx 6.92 \times 10^{5} \text{ Pa} \]This is the pressure the fluid carries throughout the system by Pascal's principle — so it is also the maximum pressure the smaller piston has to bear, even though its own area is much smaller than the large piston's. (This is exactly why a hydraulic lift works: the same pressure acting on a much smaller area lets you apply a much smaller force at the small piston to support the same large force at the big piston — pressure, not force, is what's shared.)Answer: The smaller piston must withstand a maximum pressure of \(\displaystyle 6.92 \times 10^{5} \text{ Pa} \) (about \(\displaystyle 6.92 \times 10^{5} \text{ N m}^{-2} \)).

Same level in a connected fluid means same pressure — that is what "mercury columns are in level" is telling you.The two arms of the U-tube are connected through the mercury at the bottom. Since the mercury surface is at the same height in both arms, the two points where water meets mercury and where spirit meets mercury lie at the same level inside one continuous, static fluid (mercury). For a fluid in equilibrium, pressure is the same at all points on a horizontal level within it, so the pressure pushing down on the mercury from the water side must equal the pressure pushing down on the mercury from the spirit side.Setting up the pressure balanceLet \(\displaystyle \rho_w \) = density of water, \(\displaystyle h_w = 10.0\ \text{cm} \) = height of the water column, \(\displaystyle \rho_s \) = density of spirit, \(\displaystyle h_s = 12.5\ \text{cm} \) = height of the spirit column, \(\displaystyle P_0 \) = atmospheric pressure acting on both open surfaces, and \(\displaystyle g \) = acceleration due to gravity.The pressure at the mercury interface on the water side is \[P_0 + \rho_w g h_w \] and the pressure at the mercury interface on the spirit side is \[P_0 + \rho_s g h_s . \]Because both interfaces sit at the same level in the connected mercury, these two pressures must be equal: \[P_0 + \rho_w g h_w = P_0 + \rho_s g h_s . \]Solving for the density ratioThe atmospheric pressure \(\displaystyle P_0 \) and \(\displaystyle g \) cancel from both sides (this is the step that makes the problem simple — you never need the actual value of \(\displaystyle g \) or of atmospheric pressure): \[\rho_w h_w = \rho_s h_s . \]So \[\frac{\rho_s}{\rho_w} = \frac{h_w}{h_s}. \]The quantity \(\displaystyle \rho_s/\rho_w \) is exactly the specific gravity of spirit — the ratio of its density to the density of water. Substituting the given heights: \[\frac{\rho_s}{\rho_w} = \frac{10.0\ \text{cm}}{12.5\ \text{cm}} = 0.800. \]A quick sanity check on the physics: spirit's specific gravity comes out less than $\displaystyle 1$, meaning spirit is lighter than water — which is exactly why it takes a taller column of spirit ($\displaystyle 12.5$ cm) than of water ($\displaystyle 10.0$ cm) to push down on the mercury with the same pressure.Significant figuresBoth given heights carry three significant figures ($\displaystyle 10.0$ cm and $\displaystyle 12.5$ cm), so the ratio is reported to three significant figures: 0.800. There is no rounding to do mid-calculation since this is a single division of the given data.Answer: The specific gravity of spirit is $\displaystyle 0.800$ (its density is $\displaystyle 0.800$ times that of water, i.e. \(\displaystyle 800\ \text{kg/m}^3\)).

When the liquid columns on the two sides of a U-tube don't push down with equal pressure, the connecting mercury shifts until they do — the size of that shift is the "difference in levels" being asked for.Setting up the new column heightsWater and spirit sit on mercury in the two arms of the tube. From the earlier balanced state, the columns were $\displaystyle 10.0$ cm of water and $\displaystyle 12.5$ cm of spirit, which gave the specific gravity of spirit as \[\frac{10.0}{12.5} = 0.800 \] Now $\displaystyle 15.0$ cm more of each liquid is poured into its own arm, so the new column heights are\[h_w = 10.0 + 15.0 = 25.0\ \text{cm (water)}, \qquad h_s = 12.5 + 15.0 = 27.5\ \text{cm (spirit)} \]Why the mercury levels are no longer equalBy Pascal's law, the pressure a liquid column exerts at its base is \(\displaystyle \rho g h \), where \(\displaystyle \rho \) is the liquid's density and \(\displaystyle h \) its height. When the mercury levels were level (equal height in both arms), the two liquid columns had to exert equal pressure at the mercury surface: \[\rho_w g h_w = \rho_s g h_s \] Check this with the new heights, working in specific-gravity units (density relative to water, so water's specific gravity is $\displaystyle 1.00$) — this is a shortcut that is valid only because \(\displaystyle g\) and the tube's cross-section cancel out of every term: \[\rho_w h_w = 1.00 \times 25.0 = 25.0, \qquad \rho_s h_s = 0.800 \times 27.5 = 22.0 \] These are no longer equal. The water column now presses down harder ($\displaystyle 25.0$ vs. $\displaystyle 22.0$, in these units) than the spirit column does. Since the two arms are joined only through the mercury, mercury must be pushed down slightly on the water side and rise by the same amount on the spirit side, until the pressures balance again at a common level.Balancing the pressures with a mercury-height differenceLet \(\displaystyle \delta \) be the amount by which the mercury surface in the spirit arm sits higher than the mercury surface in the water arm — this \(\displaystyle \delta \) is exactly the "difference in levels" the question asks for. Compare pressure at the level of the lower (water-side) mercury surface: Setting these equal (same horizontal level in a connected fluid at rest must carry the same pressure): \[\rho_w g h_w = \rho_s g h_s + \rho_{Hg}\, g\, \delta \] \[\delta = \frac{\rho_w h_w - \rho_s h_s}{\rho_{Hg}} \]This is the step most people skip — mercury doesn't move because "it feels like it should," it moves by exactly enough to re-equalize pressure at a single shared level. Writing that balance explicitly is what turns "mercury shifts a bit" into a number.Substituting the specific gravities (\(\displaystyle \rho_{Hg} = 13.6\) relative to water): \[\delta = \frac{25.0 - 22.0}{13.6} = \frac{3.0}{13.6}\ \text{cm} = 0.2206\ldots\ \text{cm} \]The value came out positive, confirming the assumed direction: mercury really is higher on the spirit side, not the water side.Significant figuresThe subtraction \(\displaystyle 25.0 - 22.0 = 3.0\) keeps only one decimal place (both inputs are known to one decimal place), so the numerator carries just $\displaystyle 2$ significant figures — even though every measured length in the problem has 3. A quotient can be no more precise than its least precise factor, so the final answer is rounded to $\displaystyle 2$ significant figures, not the $\displaystyle 3$ the raw division suggests: \[\delta \approx 0.22\ \text{cm} \]Answer: The mercury level in the spirit arm stands about $\displaystyle 0.22$ cm higher than in the water arm.