Upthrust can be demonstrated by the following experiment:<\/strong><\/p>\nTake an empty can and close its mouth with an airtight stopper. Put it in a tub filled with water. It floats with a large part of it above the surface of water and only a small part of it below the surface of water. Push the can into the water. You can feel an upward force and you find it difficult to push the can further into water. It is noticed that as the can is pushed more and more into the water, more and more force is needed to push the can further into water, until it is completely immersed. When the can is fully inside the water, a definite force is still needed to keep it at rest in that position. Again, if the can is released in this position, it is noticed that the can bounces back to the surface and starts floating again.<\/p>\n
Solution 2S.<\/strong><\/span><\/p>\nBuoyant force on a body due to a liquid acts upwards at the centre of buoyancy.<\/p>\n
Solution 3S.<\/strong><\/span><\/p>\nThe property of a liquid to exert an upward force on a body immersed in it is called buoyancy.<\/p>\n
Solution 4S.<\/strong><\/span><\/p>\nThe upward force exerted on a body by the fluid in which it is submerged is called the upthrust. Its S.I. unit is ‘newton’.<\/p>\n
Solution 5S.<\/strong><\/span><\/p>\nA liquid contained in a vessel exerts pressure at all points and in all directions. The pressure at a point in a liquid is the same in all directions – upwards, downwards and sideways. It increases with the depth inside the liquid.
\n
\nWhen a body is immersed in a liquid, the thrusts acting on the side walls of the body are neutralized as they are equal in magnitude and opposite in direction. However, the magnitudes of pressure on the upper and lower faces are not equal. The difference in pressure on the upper and lower faces cause a net upward force (= pressure x area) or upthrust on the body.
\nIt acts at the centre of buoyancy.<\/p>\n
Solution 6S.<\/strong><\/span><\/p>\nUpthrust due to water on block when fully submerged is more than its weight. Density of water is more than the density of cork; hence, upthrust due to water on the block of cork when fully submerged in water is more than its weight.<\/p>\n
Solution 7S.<\/strong><\/span><\/p>\nA piece of wood if left under water comes to the surface of water because the upthrust on body due to its submerged part is equal to its own weight.<\/p>\n
Solution 8S.<\/strong><\/span><\/p>\nExperiment to show that a body immersed in a liquid appears lighter:
\n
\nTake a solid body and suspend it by a thin thread from the hook of a spring balance as shown in the above figure (a). Note its weight. Above figure (a) shows the weight as 0.67 N.
\nThen, take a can filled with water. Immerse the solid gently into the water while hanging from the hook of the spring balance as shown in figure (b). Note its weight. Above figure (b) shows the weight as 0.40 N.
\nThe reading in this case (b) shall be less than the reading in the case (a), which proves that a body immersed in a liquid appears to be lighter.<\/p>\n
Solution 9S.<\/strong><\/span><\/p>\nThe readings in the spring balance decreases.
\nAs the cylinder is immersed in the jar of water, an upward force acts on it, which is in opposition to the weight component of the cylinder. Hence the cylinder appears to be lighter.<\/p>\n
Solution 10S.<\/strong><\/span><\/p>\nA body shall weigh more in vacuum because in vacuum, i.e. in absence of air, no upthrust will act on the body.<\/p>\n
Solution 11S.<\/strong><\/span><\/p>\nUpthrust on a body depends on the following factors:<\/p>\n
\n- Volume of the body submerged in the liquid or fluid.<\/li>\n
- Density of liquid or fluid in which the body is submerged.<\/li>\n<\/ol>\n
Solution 12S.<\/strong><\/span><\/p>\nLarger the volume of body submerged in liquid, greater is the upthrust acting on it.<\/p>\n
Solution 13S.<\/strong><\/span><\/p>\nA stone falls faster.
\nBecause the volume of stone is less than the volume of bunch of feathers of the same mass, the upthrust due to air on stone is less than that on the bunch of feathers, and hence, the stone falls faster.
\nHowever, in vacuum, both shall fall together because there will be no upthrust.<\/p>\n
Solution 14S.<\/strong><\/span><\/p>\nF2\u00a0<\/sub>> F1<\/sub>; Sea water is denser than river water; therefore, the upthrust due to sea water will be greater than that due to river water at the same level. This shall make the body to appear lighter in the sea water.<\/p>\nSolution 15S.<\/strong><\/span><\/p>\nObservation:\u00a0<\/strong>Volume of a block of wood immersed in glycerine is smaller as compared to the volume of block immersed in water.
\nExplanation:\u00a0<\/strong>Density of glycerine is more than that of water. Hence, glycerine exerts more upthrust on the block of wood than water, causing it to float in glycerine with a smaller volume.<\/p>\nSolution 16S.<\/strong><\/span><\/p>\n<\/p>\n
Solution 17S.<\/strong><\/span><\/p>\n<\/p>\n
Solution 18S.<\/strong><\/span><\/p>\n(a) Both have equal volumes.
\n(b) Bounce back to the surface.
\n(c) More than<\/p>\n
Solution 19S.<\/strong><\/span><\/p>\n<\/p>\n
Consider a cylindrical body PQRS of cross-sectional area A immersed in a liquid of density\u00a0\u03c1\u00a0as shown in the figure above. Let the upper surface PQ of the body is at a depth h1<\/sub>\u00a0while its lower surface RS is at depth h2<\/sub>\u00a0below the free surface of liquid.<\/p>\nAt depth h1<\/sub>, the pressure on the upper surface PQ,
\nP1<\/sub>\u00a0= h1\u00a0<\/sub>\u03c1g.<\/p>\nTherefore, the downward thrust on the upper surface PQ,
\nF1<\/sub>\u00a0= Pressure x Area = h1<\/sub>\u00a0\u03c1gA \u2026\u2026\u2026\u2026\u2026\u2026.(i)<\/p>\nAt depth h2<\/sub>, pressure on the lower surface RS,
\nP2<\/sub>\u00a0= h2<\/sub>\u00a0\u03c1g<\/p>\nTherefore, the upward thrust on the lower surface RS,
\nF2<\/sub>\u00a0= Pressure x Area = h2<\/sub>\u00a0\u03c1gA \u2026\u2026\u2026\u2026\u2026\u2026\u2026(ii)<\/p>\nThe horizontal thrust at various points on the vertical sides of body get balanced because the liquid pressure is the same at all points at the same depth.<\/p>\n
From the above equations (i) and (ii), it is clear that F2<\/sub>\u00a0> F1<\/sub>\u00a0because h2<\/sub>\u00a0> h1<\/sub>\u00a0and therefore, body will experience a net upward force.<\/p>\nResultant upward thrust or buoyant force on the body,<\/p>\n
FB<\/sub>\u00a0= F2<\/sub>\u00a0– F1
\n<\/sub>\u00a0= h2<\/sub>\u00a0\u03c1gA – h1<\/sub>\u00a0\u03c1gA
\n= A (h2<\/sub>\u00a0– h1<\/sub>)\u00a0\u03c1g<\/p>\nHowever, A (h2<\/sub>\u00a0– h1<\/sub>) = V, the volume of the body is submerged in a liquid.
\nTherefore, upthrust FB<\/sub>\u00a0= V \u03c1g.<\/p>\nNow, V\u00a0g = Volume of solid immersed x Density of liquid x Acceleration due to gravity
\n= Volume of liquid displaced x Density of liquid x Acceleration due to gravity
\n= Mass of liquid displaced x Acceleration due to gravity
\n= Weight of the liquid displaced by the submerged part of the body<\/p>\n
Thus, Upthrust FB<\/sub>\u00a0= weight of the liquid displaced by the submerged part of the body\u2026..(iii)<\/p>\nNow, let us take a solid and suspend it by a thin thread from the hook of a spring balance and note its weight.<\/p>\n
Then take a eureka can and fill it with water up to its spout. Arrange a measuring cylinder below the spout of the eureka can as shown. Immerse the solid gently in water. The water displaced by the solid is collected in the measuring cylinder.<\/p>\n
When the water stops dripping through the spout, note the weight of the solid and volume of water collected in the measuring cylinder.<\/p>\n
From the diagram, it is clear that
\nLoss in weight (Weight in air – Weight in water) = Volume of water displaced.
\nOr, Loss in weight = Volume of water displaced x 1 gcm-3<\/sup>\u00a0[Because the density of water = 1 gcm-3<\/sup>]
\nOr, Loss in weight = Weight of water displaced \u2026\u2026\u2026\u2026\u2026(iv)<\/p>\nFrom equations (iii) and (iv),
\nLoss in weight = Upthrust or buoyant force<\/p>\n
Solution 20S.<\/strong><\/span><\/p>\nSince the spheres have the same radius, both will have an equal volume inside water, and hence, the upthrust acted by water on both the spheres will be the same.
\nHence, the required ratio of upthrust acting on two spheres is 1:1.<\/p>\n
Solution 21S.<\/strong><\/span><\/p>\nSphere of iron will sink.<\/p>\n
Density of iron is more than the density of water, so the weight of iron sphere will be more than the upthrust due to water in it; thus, it causes the iron sphere to sink.<\/p>\n
Density of wood is less than the density of water, so the weight of sphere of wood shall be less than the upthrust due to water in it. So, the sphere of wood will float with a volume submerged inside water which is balanced by the upthrust due to water.<\/p>\n
Solution 22S.<\/strong><\/span><\/p>\nThe bodies of average density greater than that of the liquid sink in it. While the bodies of average density equal to or smaller than that of liquid float on it.<\/p>\n
Solution 23S.<\/strong><\/span><\/p>\n(i) The body will float if\u00a0\u03c1\u00a0\u2264\u00a0\u03c1L<\/sub>
\n(ii) The body will sink if \u03c1 > \u03c1L<\/sub><\/p>\nSolution 24S.<\/strong><\/span><\/p>\nIt is easier to lift a heavy stone under water than in air because in water, it experiences an upward buoyant force which balances the actual weight of the stone acting downwards. Thus, due to upthrust there is an apparent loss in the weight of the heavy stone, which makes it lighter in water, and hence easy to lift.<\/p>\n
Solution 25S.<\/strong><\/span><\/p>\nArchimedes’ principle states that when a body is immersed partially or completely in a liquid, it experiences an upthrust, which is equal to the weight of liquid displaced by it.<\/p>\n
Solution 26S.<\/strong><\/span><\/p>\nLet us take a solid and suspend it by a thin thread from the hook of a spring balance and note its weight (Fig a).
\nThen take a eureka can and fill it with water up to its spout. Arrange a measuring cylinder below the spout of the eureka can as shown. Immerse the solid gently in water. The water displaced by the solid gets collected in the measuring cylinder.
\n<\/p>\n
When water stops dripping through the spout, note the weight of the solid and volume of water collected in the measuring cylinder.
\nFrom diagram, it is clear that
\nLoss in weight (Weight in air – weight in water) = 300 gf – 200 gf = 100 gf
\nVolume of water displaced = Volume of solid = 100 cm3
\n<\/sup>Because density of water = 1 gcm-3\u00a0<\/sup>
\nWeight of water displaced = 100 gf = Upthrust or loss in weight
\nThis verifies Archimedes’ principle.<\/p>\nSolution 1M.<\/strong><\/span><\/p>\nTurpentine<\/p>\n
Solution 2M.<\/strong><\/span><\/p>\nN<\/p>\n
Solution 3M.<\/strong><\/span><\/p>\n\u03c1 > \u03c1L<\/sub><\/p>\n <\/p>\n
Solution 1N.<\/strong><\/span><\/p>\n<\/p>\n
Solution 2N.<\/strong><\/span><\/p>\n<\/p>\n
Solution 3N.<\/strong><\/span><\/p>\n<\/p>\n
Solution 4N.<\/strong><\/span><\/p>\n<\/p>\n
Solution 5N.<\/strong><\/span><\/p>\n<\/p>\n
Solution 6N.<\/strong><\/span><\/p>\n<\/p>\n
Solution 7N.<\/strong><\/span><\/p>\n<\/p>\n
Solution 8N.<\/strong><\/span><\/p>\n<\/p>\n
Solution 9N.<\/strong><\/span><\/p>\n<\/p>\n
Solution 10N.<\/strong><\/span><\/p>\n<\/p>\n
Exercise 5(B)<\/strong><\/span><\/p>\nSolution 1S.<\/strong><\/span><\/p>\nThe density of a substance is its mass per unit volume.<\/p>\n
Solution 2S.<\/strong><\/span><\/p>\n(i) The C.G.S. unit of density is gcm-3<\/sup>.
\n(ii) The S.I. unit of density is kgm-3<\/sup>.<\/p>\nSolution 3S.<\/strong><\/span><\/p>\n1 gcm-3<\/sup>\u00a0= 1000 kgm-3<\/sup><\/span><\/p>\nSolution 4S.<\/strong><\/span><\/p>\nIt means the mass of 1 m-3<\/sup>\u00a0of iron is 7800 kg.<\/p>\nSolution 5S.<\/strong><\/span><\/p>\nDensity of water at\u00a04\u00b0C<\/span>\u00a0in S.I. units is 1000 kgm-3<\/sup>.<\/p>\nSolution 6S.<\/strong><\/span><\/p>\n(i) Mass of a metallic body remains unchanged with increase in temperature.
\n(ii) Volume of metallic body increases with an increase in temperature.
\n(iii) Density (= Mass\/volume) of a metallic body decreases with an increase in temperature.<\/p>\n
Solution 7S.<\/strong><\/span><\/p>\nOn heating from 0\u00b0C,\u00a0<\/span>the density of water increases up to\u00a04\u00b0C<\/span>\u00a0and then decreases beyond\u00a04\u00b0C.<\/span><\/p>\nSolution 8S.<\/strong><\/span><\/p>\n(i<\/span>) Volume, (ii) kg m-3<\/sup>, (iii) 1000 and (iv) 1000<\/span><\/p>\nSolution 9S.<\/strong><\/span><\/p>\nThe relative density of a substance is the ratio of density of that substance to the density of water at\u00a04\u00b0C.<\/p>\n
Solution 10S.<\/strong><\/span><\/p>\nRelative density is the ratio of two similar quantities; thus, it has no unit.<\/p>\n
Solution 11S.<\/strong><\/span><\/p>\nDensity of a substance is the ratio of its mass to its volume but R.D. of a substance is the ratio of density of that substance to the density of water at 4\u00b0C.<\/p>\n
Solution 12S.<\/strong><\/span><\/p>\n<\/p>\n
Steps:<\/strong><\/p>\n\n- With the help of a physical balance, find the weight, W1<\/sub>\u00a0of the given solid.<\/li>\n
- Immerse the solid completely in a beaker filled with water such that it does not touch the walls and bottom of beaker, and find the weight W2<\/sub>\u00a0of solid in water.<\/li>\n<\/ol>\n