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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/physics/ocr/a/density-and-pressure/flashcards","label":"Density and pressure","subjectCode":"a_level_physics__ocr_a","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"8e612e6c-5547-4e5d-88d4-97762fa87db2","title":"Density and pressure Flashcards","topicId":"7f1d7e75-3107-570a-87f7-19fc7e585860","topicSlug":"density-and-pressure","cards":[{"id":"3b2cd0fe-8b90-4297-b8e5-7f0f30f8dcf7","cardType":"basic","points":1,"front":"State the definition of density.","back":"Mass per unit volume of a substance.","mcqOptions":[{"text":"Mass per unit volume of a substance.","correct":true,"explanation":"Density describes how much mass is contained within each unit of volume, so it is calculated using $\\rho = \\frac{m}{V}$."},{"text":"Volume per unit mass of a substance.","correct":false,"explanation":"Reversing mass and volume gives specific volume, the reciprocal of density. Density is mass per unit volume, expressed by $\\rho = \\frac{m}{V}$."},{"text":"Weight per unit volume of a substance.","correct":false,"explanation":"Using weight instead of mass defines specific weight, which depends on gravitational field strength. Density is mass per unit volume and does not depend on gravity."},{"text":"Mass per unit area of a substance.","correct":false,"explanation":"Dividing mass by area gives surface mass density rather than ordinary three-dimensional density. Density is mass divided by the volume occupied by the substance."}]},{"id":"09f81fdb-852f-4e5e-b807-369ab9f5f184","cardType":"cloze","points":null,"front":"To convert density from $\\text{g}\\text{ cm}^{-3}$ to $\\text{kg}\\text{ m}^{-3}$, you must multiply the value by $\\text{[...]}$.","back":"To convert density from $\\text{g}\\text{ cm}^{-3}$ to $\\text{kg}\\text{ m}^{-3}$, you must multiply the value by $\\boxed{1000}$.","mcqOptions":[{"text":"$$1000$","correct":true,"explanation":"One gram is $10^{-3}$ kg while one cubic centimetre is $10^{-6}$ m$^3$, so $1\\text{ g cm}^{-3}=1000\\text{ kg m}^{-3}$."},{"text":"$$0.001$","correct":false,"explanation":"This applies the mass conversion alone in the wrong overall direction and ignores the cubic volume conversion. The density value must be multiplied by $1000$."},{"text":"$$100$","correct":false,"explanation":"This treats the centimetre-to-metre conversion as though volume involved only two powers of length. Accounting for cubic centimetres gives a conversion factor of $1000$."},{"text":"$$1000000$","correct":false,"explanation":"This uses the $10^6$ conversion from cm$^3$ to m$^3$ without also converting grams to kilograms. Combining both conversions gives a factor of $1000$."}]},{"id":"9b7d7b2b-d802-4e29-8625-b06dd6376e23","cardType":"basic","points":2,"front":"How is the volume of an irregular solid measured in a density practical?","back":"Measure the volume of water displaced when the object is fully submerged in a displacement (Eureka) can.","mcqOptions":[{"text":"Measure the volume of water displaced when the object is fully submerged in a displacement (Eureka) can.","correct":true,"explanation":"A fully submerged object displaces a volume of water equal to its own volume. Collecting and measuring the overflow therefore gives the irregular solid’s volume."},{"text":"Measure the volume of water remaining when the object is fully submerged in a displacement (Eureka) can.","correct":false,"explanation":"The water remaining depends on the can’s initial contents and capacity, not just the object. The object’s volume equals the volume of displaced overflow collected."},{"text":"Measure the volume of water absorbed when the object is fully submerged in a displacement (Eureka) can.","correct":false,"explanation":"Absorption is not the basis of displacement and could introduce experimental error for porous objects. The required measurement is the volume of water displaced by full submersion."},{"text":"Measure the volume of water evaporated when the object is fully submerged in a displacement (Eureka) can.","correct":false,"explanation":"Evaporation is unrelated to the solid’s volume and is normally negligible during the experiment. The solid’s volume is found by measuring the displaced water."}]},{"id":"4b889d00-0607-4afd-92c6-0cbe262fa7d3","cardType":"basic","points":1,"front":"Define pressure as it relates to a surface.","back":"The normal force exerted per unit cross-sectional area.","mcqOptions":[{"text":"The normal force exerted per unit cross-sectional area.","correct":true,"explanation":"Pressure measures the force acting perpendicular to a surface divided by the area over which that force acts, so $p=\\frac{F_\\perp}{A}$."},{"text":"The tangential force exerted per unit cross-sectional area.","correct":false,"explanation":"A tangential force produces shear stress rather than pressure. Pressure is determined by the component of force acting normally, or perpendicularly, to the surface."},{"text":"The normal force exerted per unit cross-sectional volume.","correct":false,"explanation":"Pressure is distributed over a surface, so force must be divided by area rather than volume. Its SI units are therefore N m$^{-2}$."},{"text":"The normal force multiplied by the cross-sectional area.","correct":false,"explanation":"Multiplying by area reverses the pressure relationship. Pressure is normal force divided by cross-sectional area, while force can be recovered using $F=pA$."}]},{"id":"efd1eed0-bf13-45d7-9e97-5ce9c8324e57","cardType":"cloze","points":null,"front":"The SI unit of pressure is the Pascal ($\\text{Pa}$), which is equivalent to $\\text{[...]}$.","back":"The SI unit of pressure is the Pascal ($\\text{Pa}$), which is equivalent to $\\boxed{1\\text{ N}\\text{ m}^{-2}}$.","mcqOptions":[{"text":"$$1\\text{ N}\\text{ m}^{-2}$","correct":true,"explanation":"Pressure is force divided by area, so one Pascal equals one newton acting normally over one square metre: $1\\text{ Pa}=1\\text{ N m}^{-2}$."},{"text":"$$1\\text{ N}\\text{ m}^{-1}$","correct":false,"explanation":"Using metres rather than square metres treats the denominator as a length instead of an area. A Pascal is $1\\text{ N m}^{-2}$."},{"text":"$$1\\text{ N}\\text{ m}^{2}$","correct":false,"explanation":"The positive exponent corresponds to multiplying force by area, whereas pressure divides force by area. Therefore one Pascal is $1\\text{ N m}^{-2}$."},{"text":"$$1\\text{ N}\\text{ m}^{-3}$","correct":false,"explanation":"The cubic metre denominator incorrectly treats pressure as force per unit volume. Pressure acts over a surface, making the correct equivalent unit $1\\text{ N m}^{-2}$."}]},{"id":"b3904383-d047-4aac-b9bb-cf17e9395c74","cardType":"basic","points":2,"front":"If a force $F$ is applied at an angle $\\theta$ to the normal of a surface, how is pressure calculated?","back":"Resolve the force to find the perpendicular component ($F\\cos\\theta$) before dividing by area ($p = \\frac{F\\cos\\theta}{A}$).","mcqOptions":[{"text":"Resolve the force to find the perpendicular component ($F\\cos\\theta$) before dividing by area ($p = \\frac{F\\cos\\theta}{A}$).","correct":true,"explanation":"Because $\\theta$ is measured from the normal, the perpendicular component is adjacent to the angle and equals $F\\cos\\theta$. Pressure is this component divided by area."},{"text":"Resolve the force to find the perpendicular component ($F\\sin\\theta$) before dividing by area ($p = \\frac{F\\sin\\theta}{A}$).","correct":false,"explanation":"Using sine would be appropriate if the angle were measured from the surface. Since $\\theta$ is measured from the normal, the perpendicular component is $F\\cos\\theta$."},{"text":"Resolve the force to find the parallel component ($F\\sin\\theta$) before dividing by area ($p = \\frac{F\\sin\\theta}{A}$).","correct":false,"explanation":"The parallel component contributes to shear stress, not pressure. Pressure uses the perpendicular component, which is $F\\cos\\theta$ because the angle is measured from the normal."},{"text":"Use the force as the perpendicular component ($F$) before dividing by area ($p = \\frac{F}{A}$).","correct":false,"explanation":"Treating the whole force as perpendicular ignores its angle to the normal. Only $F\\cos\\theta$ acts normally, so the pressure is $\\frac{F\\cos\\theta}{A}$."}]},{"id":"353640ca-f461-4d60-b8e8-5a57f1c206d7","cardType":"cloze","points":null,"front":"The pressure $p$ due to a column of fluid is given by the formula $\\text{[...]}$.","back":"The pressure $p$ due to a column of fluid is given by the formula $\\boxed{p = h\\rho g}$.","mcqOptions":[{"text":"$$p = h\\rho g$","correct":true,"explanation":"Hydrostatic pressure equals fluid density multiplied by vertical depth and gravitational field strength. It therefore increases proportionally with $h$, $\\rho$, and $g$."},{"text":"$$p = \\frac{\\rho g}{h}$","correct":false,"explanation":"Placing depth in the denominator wrongly predicts that pressure decreases deeper in a fluid. Hydrostatic pressure increases with depth and is given by $p=h\\rho g$."},{"text":"$$p = \\frac{h\\rho}{g}$","correct":false,"explanation":"Dividing by gravitational field strength wrongly predicts lower pressure in a stronger gravitational field. The fluid’s weight produces pressure, so the correct formula is $p=h\\rho g$."},{"text":"$$p = \\rho A h g$","correct":false,"explanation":"Including area gives the weight of the fluid column, not its pressure. Dividing that weight by area cancels $A$, leaving $p=h\\rho g$."}]},{"id":"cc9ffc32-2995-48e0-9139-0eb9e56668c5","cardType":"basic","points":1,"front":"In the derivation of $p = h\\rho g$, what is the expression for the mass of the fluid column in terms of density $\\rho$, area $A$, and height $h$?","back":"$$m = \\rho A h$","mcqOptions":[{"text":"$$m = \\rho A h$","correct":true,"explanation":"The column’s volume is its cross-sectional area multiplied by its height, $V=Ah$. Since $m=\\rho V$, its mass is $m=\\rho Ah$."},{"text":"$$m = \\frac{\\rho A}{h}$","correct":false,"explanation":"Dividing by height wrongly makes the mass decrease as the column becomes taller. Since the volume is $Ah$, the mass is $m=\\rho Ah$."},{"text":"$$m = \\frac{\\rho h}{A}$","correct":false,"explanation":"Dividing by area wrongly predicts that a wider fluid column has less mass. Area and height multiply to give volume, so $m=\\rho Ah$."},{"text":"$$m = \\frac{A h}{\\rho}$","correct":false,"explanation":"Dividing volume by density reverses the relationship $\\rho=\\frac{m}{V}$. Rearranging correctly gives $m=\\rho V=\\rho Ah$."}]},{"id":"4d9bc17a-f714-4aad-a8c9-d840506f7054","cardType":"basic","points":1,"front":"In the derivation of $p = h\\rho g$, what is the expression for the weight ($W$) of the fluid column?","back":"$$W = \\rho A h g$","mcqOptions":[{"text":"$$W = \\rho A h g$","correct":true,"explanation":"The column has volume $Ah$, so its mass is $\\rho Ah$. Multiplying this mass by gravitational field strength $g$ gives the weight $W = \\rho Ahg$."},{"text":"$$W = \\rho A h / g$","correct":false,"explanation":"Dividing by $g$ confuses the conversion from mass to weight. The fluid mass is $\\rho Ah$, which must be multiplied by $g$, giving $W = \\rho Ahg$."},{"text":"$$W = \\rho A g / h$","correct":false,"explanation":"Placing $h$ in the denominator incorrectly implies that a taller fluid column weighs less. Its volume is $Ah$, so its weight is $W = \\rho Ahg$."},{"text":"$$W = \\rho h g / A$","correct":false,"explanation":"Dividing by area incorrectly makes a wider fluid column weigh less. Since the column’s volume is $Ah$, its weight is $W = \\rho Ahg$."}]},{"id":"0bc9da19-c858-4a8e-bf91-14775661424c","cardType":"basic","points":2,"front":"Why is the pressure at the base of a container independent of its shape or surface area?","back":"Pressure only depends on depth ($h$), fluid density ($\\rho$), and gravitational field strength ($g$).","mcqOptions":[{"text":"Pressure only depends on depth ($h$), fluid density ($\\rho$), and gravitational field strength ($g$).","correct":true,"explanation":"Hydrostatic pressure is given by $p = h\\rho g$. Container shape and area do not appear because any increase in fluid weight is balanced by the corresponding increase in base area."},{"text":"Pressure only depends on volume ($V$), fluid density ($\\rho$), and gravitational field strength ($g$).","correct":false,"explanation":"Using total volume overlooks that differently shaped containers can hold different volumes above the same base depth. Hydrostatic pressure depends on $h$, $\\rho$, and $g$, not total volume."},{"text":"Pressure only depends on depth ($h$), fluid mass ($m$), and surface area ($A$).","correct":false,"explanation":"This treats pressure as total weight divided by area without accounting for container geometry. In a static fluid, base pressure is determined by $p = h\\rho g$."},{"text":"Pressure only depends on surface area ($A$), fluid density ($\\rho$), and gravitational field strength ($g$).","correct":false,"explanation":"Area affects both the fluid column’s weight and the area over which that force acts, so it cancels. The remaining dependence is on depth, density, and gravitational field strength."}]},{"id":"32403dec-d603-4417-b9d3-40cbddb44f7e","cardType":"cloze","points":null,"front":"When a container is open to the environment, the total pressure at depth $h$ is the sum of the hydrostatic pressure and [ ].","back":"When a container is open to the environment, the total pressure at depth $h$ is the sum of the hydrostatic pressure and atmospheric pressure.","mcqOptions":[{"text":"atmospheric pressure","correct":true,"explanation":"An open surface is exposed to the atmosphere, so the absolute pressure at depth is the atmospheric pressure plus the hydrostatic increase $h\\rho g$."},{"text":"gauge pressure","correct":false,"explanation":"Gauge pressure is the hydrostatic pressure measured relative to atmospheric pressure, not an additional contribution. Total absolute pressure is hydrostatic pressure plus atmospheric pressure."},{"text":"vapour pressure","correct":false,"explanation":"Vapour pressure concerns equilibrium between a liquid and its vapour and is not normally added to hydrostatic pressure. An open container receives atmospheric pressure at its surface."},{"text":"osmotic pressure","correct":false,"explanation":"Osmotic pressure arises from concentration differences across a partially permeable membrane, not from exposure to air. The external contribution in an open container is atmospheric pressure."}]},{"id":"164ebc8b-679a-4445-a643-6d6c94584eca","cardType":"basic","points":2,"front":"What is the physical origin of upthrust on a submerged object?","back":"The pressure difference between the upper and lower surfaces; pressure at the bottom is higher due to greater depth.","mcqOptions":[{"text":"The pressure difference between the upper and lower surfaces; pressure at the bottom is higher due to greater depth.","correct":true,"explanation":"Fluid pressure increases with depth, so the lower surface experiences a larger upward force than the downward force on the upper surface. Their resultant is the upthrust."},{"text":"The pressure difference between the side surfaces; pressure on one side is higher due to greater area.","correct":false,"explanation":"At equal depths, pressure does not depend on surface area, and horizontal forces on opposite sides cancel. Upthrust arises because the lower surface is at greater depth than the upper surface."},{"text":"The object's weight acting through its centre; weight at the bottom is higher due to greater gravity.","correct":false,"explanation":"Weight acts downward and therefore cannot be the physical origin of an upward buoyant force. Upthrust results from greater fluid pressure on the object’s lower surface."},{"text":"The fluid's surface tension around the object; tension at the bottom is higher due to greater depth.","correct":false,"explanation":"Surface tension can support objects at a liquid surface but does not explain ordinary upthrust on a submerged object. Upthrust comes from the vertical pressure gradient in the fluid."}]},{"id":"ce51bb6a-fff1-47f0-b07f-f06a2ad7143e","cardType":"basic","points":1,"front":"State Archimedes' Principle.","back":"Upthrust is equal to the weight of the fluid displaced by the object.","mcqOptions":[{"text":"Upthrust is equal to the weight of the fluid displaced by the object.","correct":true,"explanation":"Archimedes' Principle states that the resultant upward force on an immersed object equals the weight of the fluid occupying the volume displaced by that object."},{"text":"Upthrust is equal to the mass of the fluid displaced by the object.","correct":false,"explanation":"Mass and force are different quantities, so displaced mass cannot equal upthrust. Multiplying the displaced fluid’s mass by $g$ gives its weight, which equals the upthrust."},{"text":"Upthrust is equal to the volume of the fluid displaced by the object.","correct":false,"explanation":"Volume alone is not a force and does not account for fluid density or gravity. The displaced volume determines a fluid weight $V\\rho g$, which equals the upthrust."},{"text":"Upthrust is equal to the weight of the object immersed in the fluid.","correct":false,"explanation":"Upthrust equals the object’s weight only when the object floats or remains suspended in equilibrium. Archimedes' Principle instead equates upthrust with the displaced fluid’s weight."}]},{"id":"b345c343-3309-4b16-8e53-773060b369f7","cardType":"cloze","points":null,"front":"The formula for upthrust $U$ is $\\text{[...]}$.","back":"The formula for upthrust $U$ is $\\boxed{U = V_{\\text{sub}} \\rho_{\\text{fluid}} g}$.","mcqOptions":[{"text":"$$U = V_{\\text{sub}} \\rho_{\\text{fluid}} g$","correct":true,"explanation":"The displaced fluid has volume $V_{\\text{sub}}$ and mass $V_{\\text{sub}}\\rho_{\\text{fluid}}$. Its weight, and therefore the upthrust, is $V_{\\text{sub}}\\rho_{\\text{fluid}}g$."},{"text":"$$U = V_{\\text{obj}} \\rho_{\\text{fluid}} g$","correct":false,"explanation":"Using the object’s total volume assumes it is completely submerged. Upthrust depends only on displaced volume, so the correct volume is $V_{\\text{sub}}$."},{"text":"$$U = V_{\\text{sub}} \\rho_{\\text{obj}} g$","correct":false,"explanation":"Using the object’s density calculates the weight of the submerged portion of the object, not the displaced fluid. Upthrust uses the fluid density $\\rho_{\\text{fluid}}$."},{"text":"$$U = V_{\\text{sub}} \\rho_{\\text{fluid}} / g$","correct":false,"explanation":"Dividing by $g$ is the wrong mass-to-weight conversion and also gives incorrect units. The displaced fluid’s mass must be multiplied by $g$ to obtain upthrust."}]},{"id":"963788a0-00b4-4649-b467-a5f578116097","cardType":"basic","points":1,"front":"What is the condition for an object to float in equilibrium?","back":"The object's weight must equal the upthrust ($W = U$).","mcqOptions":[{"text":"The object's weight must equal the upthrust ($W = U$).","correct":true,"explanation":"Floating equilibrium requires zero resultant vertical force. The downward weight must therefore be exactly balanced by the upward buoyant force, giving $W = U$."},{"text":"The object's weight must exceed the upthrust ($W > U$).","correct":false,"explanation":"If weight exceeds upthrust, there is a downward resultant force and the object accelerates downward rather than remaining in equilibrium. Equilibrium requires $W = U$."},{"text":"The object's weight must be less than the upthrust ($W < U$).","correct":false,"explanation":"If upthrust exceeds weight, there is an upward resultant force and the object rises until the displaced volume decreases. Floating equilibrium requires $W = U$."},{"text":"The object's density must equal the fluid density ($\\rho_{\\text{obj}} = \\rho_{\\text{fluid}}$).","correct":false,"explanation":"Equal densities describe neutral suspension when the object is fully submerged, not ordinary floating with partial submersion. The universal equilibrium condition is equality of weight and upthrust."}]},{"id":"b0560b79-6042-414e-802c-7eb2d7c8e960","cardType":"cloze","points":null,"front":"An object will sink if its [ ] is greater than the density of the fluid.","back":"An object will sink if its average density is greater than the density of the fluid.","mcqOptions":[{"text":"average density","correct":true,"explanation":"Sinking depends on the object’s total mass divided by its total volume. If this average density exceeds the fluid density, even complete submersion cannot provide enough upthrust."},{"text":"maximum density","correct":false,"explanation":"A dense region does not determine the behaviour of the whole object; hollow or less-dense regions may lower its overall density. Sinking is determined by average density."},{"text":"core density","correct":false,"explanation":"A dense core can be surrounded by a sufficiently large low-density region, allowing the complete object to float. The relevant comparison uses the object’s average density."},{"text":"material density","correct":false,"explanation":"This ignores hollow spaces and trapped air, which can make an object’s overall density lower than that of its constituent material. The sinking criterion uses average density."}]},{"id":"8921f777-bd4d-4ed3-b721-28fd1b1579f4","cardType":"basic","points":1,"front":"Why must the term \"fluid\" be used instead of \"liquid\" when discussing upthrust?","back":"These principles apply to both liquids and gases.","mcqOptions":[{"text":"These principles apply to both liquids and gases.","correct":true,"explanation":"Upthrust and pressure-gradient principles concern fluids, a category that includes both liquids and gases because both can flow and exert pressure on immersed objects."},{"text":"These principles apply only to liquids with low viscosity.","correct":false,"explanation":"Viscosity affects a fluid’s resistance to flow, but it does not determine whether upthrust principles apply. The term fluid is required because the principles cover both liquids and gases."},{"text":"These principles apply to liquids but not gases.","correct":false,"explanation":"Excluding gases is the misconception behind using “liquid” rather than “fluid”. Gases also exert pressure and produce upthrust, so the principles apply to both liquids and gases."},{"text":"These principles apply only to gases at high pressure.","correct":false,"explanation":"Upthrust is not restricted to highly pressurised gases. Both gases and liquids are fluids and can exert the pressure differences responsible for an upthrust force."}]},{"id":"7e4157ce-f0ed-4993-a845-fa64a71300fb","cardType":"cloze","points":null,"front":"When calculating the volume of a sphere $V = \\frac{4}{3}\\pi r^3$, if the radius is in $\\text{cm}$, the resulting volume in $\\text{m}^3$ will be scaled by a factor of $\\text{[...]}$.","back":"When calculating the volume of a sphere $V = \\frac{4}{3}\\pi r^3$, if the radius is in $\\text{cm}$, the resulting volume in $\\text{m}^3$ will be scaled by a factor of $\\boxed{10^{-6}}$.","mcqOptions":[{"text":"$$10^{-6}$","correct":true,"explanation":"Each centimetre equals $10^{-2}$ metres, and the conversion is cubed for volume: $(10^{-2})^3=10^{-6}$. Therefore, the value in cubic centimetres is multiplied by $10^{-6}$."},{"text":"$$10^{-2}$","correct":false,"explanation":"This applies the centimetre-to-metre conversion only once. Because volume depends on $r^3$, the linear factor $10^{-2}$ must be cubed, giving the correct factor $10^{-6}$."},{"text":"$$10^{-3}$","correct":false,"explanation":"This confuses conversion to litres with conversion to cubic metres. Although $1\\text{ cm}^3=10^{-3}$ litres, it equals $10^{-6}\\text{ m}^3$, so the required factor is $10^{-6}$."},{"text":"$$10^6$","correct":false,"explanation":"This reverses the direction of the unit conversion. Converting cubic metres to cubic centimetres uses $10^6$, whereas converting cubic centimetres to cubic metres requires $10^{-6}$."}]},{"id":"037e0434-20f9-416c-a1d3-50b8900ceea9","cardType":"basic","points":2,"front":"How do fluid particles exert pressure on the walls of a container?","back":"Particles in constant random motion collide with walls, exerting a cumulative normal force over the contact area.","mcqOptions":[{"text":"Particles in constant random motion collide with walls, exerting a cumulative normal force over the contact area.","correct":true,"explanation":"Randomly moving particles repeatedly collide with the container walls. Their many microscopic normal forces combine into a macroscopic force, and force per unit area is the pressure."},{"text":"Particles in steady downward motion collide with walls, exerting a cumulative normal force over the contact area.","correct":false,"explanation":"Gravity does not produce a steady downward molecular motion in a fluid at equilibrium. Pressure arises from particles moving randomly in all directions and colliding with the walls."},{"text":"Particles in constant random motion collide with walls, exerting a cumulative tangential force over the contact area.","correct":false,"explanation":"Tangential force would describe shear rather than pressure. Colliding particles transfer momentum perpendicular to the wall, producing a cumulative normal force distributed over the contact area."},{"text":"Particles in constant random motion collide with walls, exerting a cumulative normal force independent of the contact area.","correct":false,"explanation":"The total normal force is not independent of wall area: at fixed pressure, a larger area receives more collisions and therefore greater force. Pressure is normal force divided by area."}]},{"id":"7422ad3f-c28c-4c97-82e7-1204b60dda16","cardType":"cloze","points":null,"front":"Standard atmospheric pressure at sea level is approximately $\\text{[...]}$ (or $101\\text{ kPa}$).","back":"Standard atmospheric pressure at sea level is approximately $\\boxed{1.01 \\times 10^5\\text{ Pa}}$ (or $101\\text{ kPa}$).","mcqOptions":[{"text":"$$1.01 \\times 10^5\\text{ Pa}$","correct":true,"explanation":"The prefix kilo means $10^3$, so $101\\text{ kPa}=101\\times10^3\\text{ Pa}=1.01\\times10^5\\text{ Pa}$. This is the standard approximate sea-level atmospheric pressure."},{"text":"$$1.01 \\times 10^3\\text{ Pa}$","correct":false,"explanation":"This effectively treats $101\\text{ kPa}$ as $1.01\\text{ kPa}$, losing a factor of 100. Converting correctly gives $101\\times10^3\\text{ Pa}=1.01\\times10^5\\text{ Pa}$."},{"text":"$$1.01 \\times 10^4\\text{ Pa}$","correct":false,"explanation":"This is too small by a factor of ten, as though the stated pressure were $10.1\\text{ kPa}$. Converting $101\\text{ kPa}$ correctly gives $1.01\\times10^5\\text{ Pa}$."},{"text":"$$1.01 \\times 10^6\\text{ Pa}$","correct":false,"explanation":"This is too large by a factor of ten, effectively treating the pressure as $1010\\text{ kPa}$. Since one kilopascal is $10^3$ pascals, the correct value is $1.01\\times10^5\\text{ Pa}$."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"Density and pressure","title":"Density and pressure Flashcards","topbar":{"backHref":"/a-level/physics/ocr/a/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"7f1d7e75-3107-570a-87f7-19fc7e585860","walled":"$L39"}],"$L3a","$L3b"]