{"id":33677,"date":"2024-12-18T12:34:35","date_gmt":"2024-12-18T07:04:35","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=33677"},"modified":"2024-12-19T10:11:02","modified_gmt":"2024-12-19T04:41:02","slug":"samacheer-kalvi-11th-physics-guide-chapter-7","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-11th-physics-guide-chapter-7\/","title":{"rendered":"Samacheer Kalvi 11th Physics Guide Chapter 7 Properties of Matter"},"content":{"rendered":"

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Physics Guide<\/a> Pdf Chapter 7 Properties of Matter Text Book Back Questions and Answers, Notes.<\/p>\n

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 7 Properties of Matter<\/h2>\n

11th Physics Guide Properties of Matter Book Back Questions and Answers<\/h3>\n

\"Samacheer<\/p>\n

I. Multiple choice questions:<\/span><\/p>\n

Question 1.
\nConsider two wires X and Y. The radius of wire X is 3 times the radius of Y. If they are stretched by the same load then the stress on Y is:
\n(a) equal to that on X
\n(b) thrice that on X
\n(c) nine times that on X
\n(d) Half that on X
\nAnswer:
\n(c) nine times that on X<\/p>\n

Hint:
\n\"Samacheer<\/p>\n

Question 2.
\nIf a wire is stretched to double of its original length, then the strain in the wire is:
\n(a) 1
\n(b) 2
\n(c) 3
\n(d) 4
\nAnswer:
\n(a) 1<\/p>\n

Hint:
\n\"Samacheer<\/p>\n

Question 3.
\nThe load – elongation graph of three wires of the same material are shown in figure. Which of the following wire is the thickest?
\n\"Samacheer
\n(a) wire 1
\n(b) wire 2
\n(c) wire 3
\n(d) all of them have same thickness
\nAnswer:
\n(a) wire 1<\/p>\n

Hint:
\nFor wire stress is less
\nStress = \\(\\frac { Force }{ Area }\\)
\nAs stress is less Area will be greater.
\n\u2234 Wire 1 is thickest.<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nFor a given material, the rigidity modulus is [ \\(\\frac { 1 }{ 3 }\\) ]3<\/sup> of Young\u2019s modulus. Its Poisson\u2019s ratio:
\n(a) 0
\n(b) 0.25
\n(c) 0.3
\n(d) 0.5
\nAnswer:
\n(d) 0.5<\/p>\n

Hint:
\n\"Samacheer<\/p>\n

Question 5.
\nA small sphere of radius 2 cm falls from rest in a viscous liquid. Heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to: [NEET model 2018]
\n(a) 2\u00b2
\n(b) 2\u00b3
\n(c) 24<\/sup>
\n(d) 25<\/sup>
\nAnswer:
\n(d) 25<\/sup><\/p>\n

Hint:
\n\"Samacheer<\/p>\n

Question 6.
\nTwo wires are made of the same material and have the same volume. The area of cross sections of the first and the second wires are A and 2A respectively. If the length of the first wire is increased by \u0394l on applying a force F, how much force is needed to stretch the second wire by the same amount? [NEET model 2018]
\n(a) 2
\n(b) 4
\n(c) 8
\n(d) 16
\nAnswer:
\n(b) 4<\/p>\n

Question 7.
\nWith an increase in temperature, the viscosity of liquid and gas, respectively will:
\n(a) increase and increase
\n(b) increase and decrease
\n(c) decrease and increase
\n(d) decrease and decrease
\nAnswer:
\n(c) decrease and increase<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\nThe Young\u2019s modulus for a perfect rigid body is:
\n(a) 0
\n(b) 1
\n(c) 0.5
\n(d) infinity
\nAnswer:
\n(d) infinity<\/p>\n

Question 9.
\nWhich of the following is not a scalar?
\n(a) viscosity
\n(b) surface tension
\n(c) pressure
\n(d) stress
\nAnswer:
\n(d) stress<\/p>\n

Question 10.
\nIf the temperature of the wire is increased, then the Young\u2019s modulus will:
\n(a) remain the same
\n(b) decrease
\n(c) increase rapidly
\n(d) increase by very a small amount
\nAnswer:
\n(b) decrease<\/p>\n

Question 11.
\nCopper of fixed volume V is drawn into wire of length l. When this wire is subjected to a constant force F, the extension produce in the wire is \u2206l. If Y represents the Young\u2019 modulus, then which of the following graph is a straight line?
\n(a) \u2206l verses V
\n(b) \u2206l verses Y
\n(c) \u2206l verses F
\n(d) \u2206l verses \\(\\frac { 1 }{ l }\\)
\nAnswer:
\n(c) \u2206l verses F<\/p>\n

Hint:
\nStrain \u221d Stress
\n\u2234 \u2206l \u221d F<\/p>\n

\"Samacheer<\/p>\n

Question 12.
\nA certain number of spherical drops of a liquid of radius R coalesce to form a single drop of radius R and volume V If T is the surface tension of the liquid, then:
\n(a) energy = 4VT(\\(\\frac { 1 }{ r }\\) – \\(\\frac { 1 }{ R }\\)) is released
\n(b) energy = 3 VT(\\(\\frac { 1 }{ r }\\) + \\(\\frac { 1 }{ R }\\)) is absorbed
\n(c) energy = 3VT (\\(\\frac { 1 }{ r }\\) – \\(\\frac { 1 }{ R }\\))is released
\n(d) energy is neither released nor absorbed
\nAnswer:
\n(c) energy = 3VT (\\(\\frac { 1 }{ r }\\) – \\(\\frac { 1 }{ R }\\))is released<\/p>\n

Question 13.
\nThe following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?
\n(a) length = 200 cm, diameter = 0.5 mm
\n(b) length= 200 cm, diameter = 1 mm
\n(c) length 200 cm, diameter = 2 mm
\n(d) length= 200 cm, diameter = 3 m
\nAnswer:
\n(a) length = 200 cm, diameter = 0.5 mm<\/p>\n

Hint:
\n\"Samacheer
\nHence the value of l must be most and that of d must be least.<\/p>\n

Question 14.
\nThe wettability of a surface by a liquid depends primarily on:
\n(a) viscosity
\n(b) surface tension
\n(c) density
\n(d) angle of contact between the surface and the liquid
\nAnswer:
\n(d) angle of contact between the surface and the liquid<\/p>\n

Question 15.
\nIn a horizontal pipe of non-uniform cross section, water flows with a velocity of 1 ms-1<\/sup> at a point where the diameter of the pipe is 20 cm. The velocity of water (ms-1<\/sup>) at a point where the diameter of the pipe 10 cm is:
\n(a) .0025 m\/s
\n(b) .25 m\/s
\n(c) 0.025 m
\n(d) .5 m\/s
\nAnswer:
\n(b) .25 m\/s<\/p>\n

Hint:
\n\"Samacheer<\/p>\n

II. Short Answer Questions:<\/span><\/p>\n

Question 1.
\nDefine stress and strain.
\nAnswer:
\nStress: The force per unit area is called as stress.
\nStress, \u03c3 = \\(\\frac { Force }{ Area }\\) = \\(\\frac { F }{ A }\\)<\/p>\n

Strain: Strain is defined as the ratio of change in size to the original size of an object. It measures the degree of deformation.
\n\"Samacheer<\/p>\n

Question 2.
\nState Hooke\u2019s law of elasticity.
\nAnswer:
\nIt states that for small deformation, the stress is directly proportional to strain.<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nDefine Poisson\u2019s ratio.
\nAnswer:
\nIt is defined as the ratio of relative contraction (lateral strain) , to relative expansion (longitudinal strain).
\n\"Samacheer<\/p>\n

Question 4.
\nExplain elasticity using intermolecular forces.
\nAnswer:
\nElastic behaviour of solid. In a solid, atoms and molecules are arranged in such a way that each molecule is acted upon by the forces due to the neighbouring molecules. When deforming force is applied on a body so that its length increases, then the molecules of the body go far apart.<\/p>\n

Question 5.
\nWhich one of these is more elastic, steel or rubber? Why?
\nAnswer:
\nSteel is more elastic than rubber. If equal stress is applied to both steel and rubber, the steel produces less strain. So Young’s modulus is higher for steel than rubber. Hence steel is more elastic than rubber.<\/p>\n

Question 6.
\nA spring balance shows wrong readings after using for a long time. Why?
\nAnswer:
\nWhen a spring balance has been used for a long time, it develops elastic fatigue, the spring of such a balance takes a longer time to recover its original configuration and therefore it does not give correct measurement.<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nWhat is the effect of temperature on elasticity?
\nAnswer:
\nAs the temperature of the substance increases, its elasticity decreases.<\/p>\n

Question 8.
\nWrite down the expression for the elastic potential energy of a stretched wire.
\nAnswer:
\nW = \\(\\frac { 1 }{ 2 }\\) Fl = Elastic potential energy<\/p>\n

Question 9.
\nState Pascal’s law in fluids.
\nAnswer:
\nIf the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid without being diminished in magnitude.<\/p>\n

Question 10.
\nState Archimedes principle.
\nAnswer:
\nWhen a body is partially or wholly immersed in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it and its upthrust acts through the centre of gravity of the liquid displaced.
\nUpthrust or buoyant force = weight of liquid displaced.<\/p>\n

Question 11.
\nWhat do you mean by upthrust or buoyancy?
\nAnswer:
\nThe upward force exerted by a fluid that opposes the weight of an immersed object in a fluid is called upthrust or buoyant force.<\/p>\n

\"Samacheer<\/p>\n

Question 12.
\nState the law of floatation.
\nAnswer:
\nThe law of floatation states that a body will float in a liquid if the weight of the liquid displaced by the immersed part of the body equals the weight of the body.<\/p>\n

Question 13.
\nDefine coefficient of viscosity of a liquid.
\nAnswer:
\nThe coefficient of viscosity of a liquid is defined as the viscous force acting tangentially per unit area of a liquid layer having a unit velocity gradient in a direction perpendicular to the direction of flow of the liquid.<\/p>\n

Question 14.
\nDistinguish between streamlined flow and turbulent flow.
\nAnswer:
\nStreamlined flow:<\/p>\n

    \n
  • When a liquid flows such that each particle of the liquid passing a point moves along the same path and has the same velocity as its predecessor then flow is said to be streamlined flow.<\/li>\n
  • The velocity of the particles is constant.<\/li>\n
  • The path taken by the particle in this flow is a curve.<\/li>\n<\/ul>\n

    Turbulent flow:<\/p>\n

      \n
    • When the speed of the moving liquid, exceeds the critical speed vc<\/sub>, the motion becomes turbulent.<\/li>\n
    • The velocity changes both in magnitude and direction from particle to particle.<\/li>\n
    • The path taken by the particle in this flow becomes erratic and whirlpool. Like circles.<\/li>\n<\/ul>\n

      Question 15.
      \nWhat is Reynold\u2019s number? Give its significance.
      \nAnswer:
      \nReynold\u2019s number Rc<\/sub> is a critical variable, which decides whether the flow of a fluid through a cylindrical pipe is streamlined or turbulent.
      \nRc<\/sub> = \\(\\frac { \u03c1VD }{ \u03b7 }\\)<\/p>\n

      \"Samacheer<\/p>\n

      Question 16.
      \nDefine terminal velocity.
      \nAnswer:
      \nThe maximum constant velocity acquired by a body while falling freely through a viscous medium is called the terminal velocity VT<\/sub>.<\/p>\n

      Question 17.
      \nWrite down the expression for Stoke’s force and explain the symbols involved in it.
      \nAnswer:
      \nF= 6\u03c0\u03b7rv<\/p>\n

        \n
      • Radius (r ) of the sphere<\/li>\n
      • Velocity (v) of the sphere and<\/li>\n
      • Coefficient of viscosity \u03b7 of the liquid.<\/li>\n<\/ul>\n

        Question 18.
        \nState Bernoulli’s theorem.
        \nAnswer:
        \nAccording to Bernoulli\u2019s theorem, the sum of’ pressure energy, kinetic energy, and potential energy per unit mass of an incompressible. Non-viscous fluid in a streamlined flow remains a constant.<\/p>\n

        Question 19.
        \nWhat are the energies possessed by a liquid? Write down their equations.
        \nAnswer:
        \nA liquid in a steady flow can possess three kinds of energy. They are
        \n(i) Kinetic energy, KE = \\(\\frac { 1 }{ 2 }\\)mv\u00b2
        \n(ii) Potential energy, PE = mgh
        \n(iii) Pressure energy, respectively Ep<\/sub> = PV<\/p>\n

        Question 20.
        \nTwo streamlines cannot cross each other. Why?
        \nAnswer:
        \nIf two streamlines cross each other, there will be two directions of flow at the point of intersection which is impossible.<\/p>\n

        \"Samacheer<\/p>\n

        Question 21.
        \nDefine surface tension of a liquid. Mention its S.l unit and dimension.
        \nAnswer:
        \nSurface tension is defined as the force acting on a unit length of an imaginary line drawn on the free surface of the liquid, the direction of the force being perpendicular to the line so drawn and acting parallel to the surface. The SI unit and dimensions of surface tension are Nm-1<\/sup> and MT-2<\/sup>, respectively.<\/p>\n

        Question 22.
        \nHow is surface tension related to surface energy?
        \nAnswer:
        \nThe surface energy per unit area of a surface is numerically equal to the surface tension.<\/p>\n

        Question 23.
        \nDefine the angle of contact for a given pair of solid and liquid.
        \nAnswer:
        \nThe angle between tangents drawn at the point of contact to the liquid surface and solid surface inside the liquid is called the angle of contact for a pair of solid and liquid. It is denoted by \u03b8.<\/p>\n

        Question 24.
        \nDistinguish between cohesive and adhesive forces.
        \nAnswer:
        \nCohesive – The force between the like molecules which holds the liquid together is called \u2018cohesive force\u2019.<\/p>\n

        Adhesive – When the liquid is in contact with a solid, the molecules of these solid and liquid will experience an attractive force which is called \u2018adhesive force\u2019.<\/p>\n

        Question 25.
        \nWhat are the factors affecting the surface tension of a liquid?
        \nAnswer:<\/p>\n

          \n
        1. The presence of any contamination or impurities.<\/li>\n
        2. The presence of dissolved substances.<\/li>\n
        3. Electrification<\/li>\n
        4. Temperature<\/li>\n<\/ol>\n

          \"Samacheer<\/p>\n

          Question 26.
          \nWhat happens to the pressure inside a soap bubble when air is blown into it?
          \nAnswer:
          \nPressure will slowly increase inside the soap bubble.<\/p>\n

          Question 27.
          \nWhat do you mean by capillarity or capillary action?
          \nAnswer:
          \nIn a liquid whose angle of contact with solid is less than 90\u00b0, suffers capillary rise. On the other hand, in a liquid whose angle of contact is greater than 90\u00b0, suffers capillary fall. The rise or fall of a liquid in a narrow tube is called capillarity or capillary action.<\/p>\n

          Question 28.
          \nA drop of oil placed on the surface of water spreads out. But a drop of water place on oil contracts to a spherical shape. Why?
          \nAnswer:
          \nWhen a drop of water is placed on oil, the cohesive force of water molecules dominates the adhesive force between water and oil molecules. Hence drop of water contracts to a spherical shape.<\/p>\n

          Question 29.
          \nState the principle and usage of Venturimeter.
          \nAnswer:
          \nThis device is used to measure the rate of flow (or say flow speed) of the incompressible fluid flowing through a pipe. It works on the principle of Bernoulli\u2019s theorem.<\/p>\n

          \"Samacheer<\/p>\n

          III. Long Answer Questions:<\/span><\/p>\n

          Question 1.
          \nState Hooke’s law and verify it with the help of an experiment.
          \nAnswer:
          \nHooke\u2019s law states that for a small deformation when the stress and strain are proportional to each other.<\/p>\n

          It can be verified in a simple way by stretching a thin straight wire (stretches like spring) of length L and uniform cross-sectional area A spring is suspended from a fixed point O. A pan and a pointer are attached at the free end of the wire as shown in Figure.<\/p>\n

          Using a vernier scale arrangement the extension produced on the wire is measured. From the experiment, it is known that for a given load, the corresponding stretching force is F and the elongation produced on the wire is \u0394L.
          \n\"Samacheer<\/p>\n

          It is directly proportional to the original length L and inversely proportional to the area of cross-section A. A graph is plotted using F on the X-axis and AL on the Y-axis. This graph is a straight line passing through the origin as shown in Figure.
          \n\"Samacheer
          \nVariation of \u2206L with F
          \n\u2234 \u2206L = (slope)F
          \nMultiplying and dividing by volume,
          \nV = AL,
          \nF(slope) = \\(\\frac { AL }{ AL }\\) \u2206L
          \nRearranging, we get
          \n\"Samacheer
          \ni.e., the stress is proportional to the strain in the elastic limit.<\/p>\n

          Question 2.
          \nExplain the different types of modulus of elasticity.
          \nAnswer:
          \nThere are three types of modulus of elasticity. They are:<\/p>\n

            \n
          1. Young\u2019s modulus<\/li>\n
          2. Bulk modulus<\/li>\n
          3. Rigidity modulus<\/li>\n<\/ol>\n

            Young\u2019s modulus: When a wire is stretched or compressed, then the ratio between tensile stress (or compressive stress) and tensile strain (or compressive strain) is defined as Young\u2019s modulus.
            \n\"Samacheer
            \nSI unit of Young\u2019s modulus is Nm-2<\/sup> or pascal.
            \nBulk modulus: Bulk modulus is defined as the ratio of volume stress to the volume strain.
            \nBulk modulus,
            \n\"Samacheer
            \nThe negative sign in equation (1) means that when pressure is applied to the body, its volume decreases.
            \nThe rigidity modulus or shear modulus:
            \nIt is defined as rigidity modulus or Shear modulus,
            \n\"Samacheer<\/p>\n

            Question 3.
            \nDerive an expression for the elastic energy stored per unit volume of a wire.
            \nAnswer:
            \nWhen a body is stretched, work is done against the restoring force (internal force). This work done is stored in the body in the form of elastic energy. Let us consider a wire whose un-stretch length is L and the area of cross-section is A. Let a force produce an extension l and it is assumed that the elastic limit of the wire has not been exceeded and there is no loss in energy. Then, the work done by the force F is equal to the energy gained by the wire. The work is done in stretching the wire by dl,
            \ndW = F dl
            \nThe total work done in stretching the wire from 0 to l is
            \nW = \\(\\int_{0}^{l} \\mathrm{~F} d l\\) … (1)
            \nFrom Young\u2019s modulus of elasticity,
            \nY = \\(\\frac { F }{ A }\\) x \\(\\frac { L }{ l }\\) \u21d2 F = \\(\\frac { YAl }{ L }\\) … (2)
            \nSubstituting equation (2) in equation (1), we get
            \nW = \\(\\int_{0}^{l} \\frac{\\mathrm{YAl}}{\\mathrm{L}} d l\\)
            \nSince l is the dummy variable in the integration, we can change l to l’ (not in limits), therefore
            \n\"Samacheer<\/p>\n

            Question 4.
            \nDerive an equation for the total pressure at a depth \u2018h’ below the liquid surface.
            \nAnswer:
            \nLet us consider a water sample of a cross-sectional area in the form of a cylinder. Let h1<\/sub> and h2<\/sub> be the depths from the air-water interface to level 1 and level 2 of the cylinder, respectively as shown in Figure (a). Let F1<\/sub> be the force acting downwards on level 1 and F2<\/sub> be the force acting upwards on level 2, such that, F1<\/sub> = P1<\/sub> A and F2<\/sub> = P2<\/sub> A Let the mass of the sample to be m and under equilibrium condition, the total upward force (F2<\/sub>) is balanced by the total downward force (F1<\/sub> + mg), otherwise, the gravitational force will act downward which is being exactly balanced by the difference between the force F2<\/sub> – F1<\/sub>
            \nF2<\/sub> – F1<\/sub> = mg = FG<\/sub> … (1)
            \nWhere m is the mass of the water available in the sample element. Let p be the density of the water then, the mass of water available in the sample element is
            \nm = \u03c1V = \u03c1A(h2<\/sub> – h1<\/sub>])
            \nV= A (h2<\/sub> – h1<\/sub>)
            \n\"Samacheer
            \nA sample of water with base area A in a static fluid with its forces in equilibrium
            \nHence, gravitational force,
            \nFG<\/sub> = \u03c1A(h2<\/sub> – h1<\/sub>)g
            \nOn substituting the value of W in equation (1)
            \nF2<\/sub> = F1<\/sub> + mg
            \n\u21d2 P2<\/sub>A = P1<\/sub>A + \u03c1A(h2<\/sub> – h1<\/sub>)g
            \nCancelling out A on both sides,
            \nP2<\/sub> = P1<\/sub> + \u03c1 (h2<\/sub> – h1<\/sub>)g … (2)
            \n\"Samacheer
            \nIf we choose level 1 at the surface of the liquid (i.e., air-water interface) and level 2 at a depth \u2018h \u2019 below the surface (as shown in Figure), then the value of h1<\/sub> becomes zero (h1<\/sub> = 0) when P1<\/sub> assumes the value of atmospheric pressure (say Pa<\/sub>). In addition, the pressure (P2<\/sub>) at a depth becomes P. Substituting these values in the equation,
            \np2<\/sub> = p1<\/sub> + \u03c1(h2<\/sub> – h1<\/sub>)g
            \nwe get
            \np = pa<\/sub> + \u03c1gh
            \nThis means, the pressure at a depth h is greater than the pressure on the surface of the liquid, where Pa<\/sub> is the atmospheric pressure = 1.013 x 105<\/sup> Pa<\/sub>. If the atmospheric pressure is neglected then
            \np = \u03c1 gh
            \nFor a given liquid, p is fixed and g is also constant, then the pressure due to the fluid column is directly proportional to vertical distance or height of the fluid column.<\/p>\n

            \"Samacheer<\/p>\n

            Question 5.
            \nState and prove Pascal’s law in fluids.
            \nAnswer:
            \n\"Samacheer
            \nPascal\u2019s law states that if the effect of gravity can be neglected then the pressure in a fluid in equilibrium is the same everywhere. Let us consider any two points A and B inside the fluid imagined. A cylinder is such that points A and B lie at the centre of the circular surface at the top and bottom of the cylinder.<\/p>\n

            Let the fluid inside this cylinder be in equilibrium under the action of forces from outside the fluid. The forces acting on the circular, top, and bottom surfaces are perpendicular to the forces acting on the cylindrical surface. Therefore the forces acting on the faces at A and B are equal and opposite and hence add to zero.<\/p>\n

            As the areas of these two faces are equal, the pressure at A = pressure at B. This is the proof of Pascal\u2019s law when the effect of gravity is not taken into account.<\/p>\n

            Question 6.
            \nState and prove Archimedes principle.
            \nAnswer:
            \n\"Samacheer
            \nArchimedes principle states that when a body is partially or wholly immersed in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it, and its upthrust acts through the centre of gravity of the liquid displaced.
            \nProof:
            \nConsider a body of height h lying inside a liquid of density p, at a depth x below the free surface of the liquid. The area of a cross-section of the body is a. The forces on the sides of the body cancel out.
            \n\"Samacheer
            \nPressure at the upper face of the body,
            \nP1<\/sub> = x\u03c1g
            \nPressure at the lower face of the body,
            \nP2<\/sub> = (x + h)p\u03c1g
            \nThrust acting on the upper face of the body is
            \nF1<\/sub> = P1<\/sub>a = x\u03c1ga
            \nacting vertically downwards.
            \nThrust acting on the lower face of the body is
            \nF2<\/sub> = P2<\/sub>a = (x+h)\u03c1ga
            \nacting vertically upwards.
            \nThe resultant force (F2<\/sub> – F1<\/sub>) is acting on the body in the upward direction and is called Upthrust (U).
            \n\u2234 U = F2<\/sub> – F1<\/sub> = (x + h)\u03c1g – x\u03c1ga = ah\u03c1g
            \nBut, ah = V, the volume of the body = volume of liquid displaced.
            \nU = V\u03c1g = Mg
            \n[\u2234 M = V\u03c1 = mass of liquid displaced] i.e., upthrust or buoyant force.
            \n= Weight of liquid displaced.
            \nThis proves the Archimedes principle.<\/p>\n

            Question 7.
            \nDerive the expression for the terminal velocity of a sphere moving in a high viscous fluid using stokes force.
            \nAnswer:
            \nLet us consider a sphere of radius r which falls freely through a highly viscous liquid of coefficient of viscosity \u03b7. Let the density of the material of the sphere be p and the density of the fluid be c.
            \n\"Samacheer
            \nGravitational force acting on the sphere,
            \nFG<\/sub> = mg = \\(\\frac { 4 }{ 3 }\\)\u03c0r\u00b3pg (downward force)
            \nUpthrust, U= \\(\\frac { 4 }{ 3 }\\)\u03c0r\u00b3\u03c3g (upward force)
            \nviscous force F = 6\u03c0\u03b7rvt<\/sub>
            \nAt terminal velocity vt<\/sub>.
            \ndownward force = upward force.
            \n\"Samacheer<\/p>\n

            Question 8.
            \nDerive Poiseuille’s formula for the volume of a liquid flowing per second through a pipe under streamlined flow.
            \nAnswer:
            \nConsider a liquid flowing steadily through a horizontal capillary tube.
            \nLet v = (\\(\\frac { V }{ t }\\)) be the volume of the liquid flowing out per second through a capillary tube.
            \nIt depends on
            \n(i) coefficient of viscosity (\u03b7) of the liquid
            \n(ii) radius of the tube (r).
            \n(iii) the pressure gradient (yj Then,
            \n\"Samacheer
            \nSo, equating the powers of M, L, and T on both sides, we get
            \na + c = 0, – a + b – 2c = 3, and – a – 2c = 1 On solving three equations, we get
            \na = – 1, b = 4, and c = 1
            \nTherefore, equation (1) becomes,
            \nv = k\u03b7-1<\/sup>r(\\(\\frac { p }{ l }\\))1<\/sup>
            \nExperimentally, the value of k is shown to be \\(\\frac { \u03c0 }{ 8 }\\), we have
            \nV = \\(\\frac{\\pi r^{4} \\mathrm{P}}{8 \\eta l}\\)
            \nThe above equation is known as Poiseuille\u2019s equation for the flow of liquid through a narrow tube or a capillary tube.<\/p>\n

            Question 9.
            \nObtain an expression for the excess of pressure inside a
            \n(i) liquid drop
            \n(ii) liquid bubble
            \n(iii) air bubble.
            \nAnswer:
            \n(i) Liquid drop: Let us consider a liquid drop of radius R and the surface tension of the liquid is T.
            \n\"Samacheer
            \nThe various forces acting on the liquid drop are,
            \n(a) Force due to surface tension FT<\/sub>= 2\u03c0RT towards the right
            \n(b) Force due to outside pressure, F\\(\\mathrm{F}_{\\mathrm{P}_{1}}\\) towards right
            \n(c) Force due to inside pressure,
            \n\\(\\mathrm{F}_{\\mathrm{P}_{2}}\\) = P2<\/sub> \u03c0R\u00b2 towards left
            \nAs the drop is in equilibrium,
            \n\\(\\mathrm{F}_{\\mathrm{P}_{2}}\\) = FT<\/sub> + \\(\\mathrm{F}_{\\mathrm{P}_{1}}\\)
            \nP2<\/sub>2\u03c0R\u00b2 = 2\u03c0RT + P1<\/sub>\u03c0R\u00b2
            \n\u21d2 (P2<\/sub> – P1<\/sub>)\u03c0R\u00b2 = 2\u03c0RT
            \nExcess pressure is \u2206P = P2<\/sub> – P1<\/sub> = \\(\\frac { 2T }{ R }\\)<\/p>\n

            (ii) Liquid bubble: A soap bubble of radius R and the surface tension of the soap bubble be T is as shown in Figure. A soap bubble has two liquid surfaces in contact with air, one inside the bubble and the other outside the bubble. Hence, the force on the soap bubble due to surface tension is 2 x 2\u03c0RT. The various forces acting on the soap bubble are,
            \n\"Samacheer
            \n(a) Force due to surface tension
            \nFT<\/sub> = 4\u03c0RT towards right<\/p>\n

            (b) Force due to outside pressure,
            \nFP<\/sub> = P1<\/sub>\u03c0R\u00b2 towards right<\/p>\n

            (c) Force due to inside pressure,
            \n\\(\\mathrm{F}_{\\mathrm{P}_{2}}\\) = P2<\/sub>\u03c0R\u00b2 towards left
            \nAs the bubble is in equilibrium,
            \n\\(\\mathrm{F}_{\\mathrm{P}_{2}}\\) = FT<\/sub> + \\(\\mathrm{F}_{\\mathrm{P}_{1}}\\)
            \nP2<\/sub>\u03c0R\u00b2 = 4\u03c0RT + P1<\/sub>\u03c0R\u00b2
            \n\u21d2 (P2<\/sub> – P1<\/sub>)\u03c0R\u00b2 = 4\u03c0RT
            \nExcess pressure is \u2206P = P2<\/sub> – P1<\/sub> = \\(\\frac { 4T }{ R }\\)<\/p>\n

            (iii) Air bubble: Let us consider an air bubble of radius R inside a liquid having surface tension T as shown in Figure. Let P1<\/sub> and P2<\/sub> be the pressures outside and inside the air bubble, respectively. Now, the excess pressure inside the air bubble is \u2206P = P1<\/sub> – P2<\/sub>.
            \n\"Samacheer
            \nIn order to find the excess pressure inside the air bubble, let us consider the forces acting on the air bubble. For the hemispherical portion of the bubble, considering the forces acting on it, we get,
            \n(a) The force due to surface tension acting towards right around the rim of length 2\u03c0R is FT<\/sub> = 2\u03c0RT<\/p>\n

            (b) The force due to outside pressure P1<\/sub> is to the right acting across a cross-sectional area of \u03c0R\u00b2 is \\(\\mathrm{F}_{\\mathrm{P}_{1}}\\) = P1<\/sub>\u03c0R\u00b2<\/p>\n

            (c) The force due to pressure P2<\/sub> inside the bubble, acting to the left is \\(\\mathrm{F}_{\\mathrm{P}_{2}}\\) = P2<\/sub>\u03c0R\u00b2
            \nAs the air bubble is in equilibrium under the action of these forces, \\(\\mathrm{F}_{\\mathrm{P}_{2}}\\) = FT<\/sub> + \\(\\mathrm{F}_{\\mathrm{P}_{1}}\\).
            \n\"Samacheer<\/p>\n

            Question 10.
            \nWhat is capillarity? Obtain an expression for the surface tension of a liquid by the capillary rise method.
            \nAnswer:
            \nThe rise or fall of a liquid in a narrow tube is called capillarity. Let us consider a capillary tube which is held vertically in a beaker containing water; the water rises in the capillary\u2019 tube to a height h due to surface tension.
            \n\"Samacheer
            \nThe surface tension force FT<\/sub>, acts along the tangent at the point of contact downwards and its reaction force upwards. Surface tension T is resolved into two components.
            \n(i) Horizontal component T sin\u03b8 and
            \n(ii) Vertical component T cos\u03b8 acting upwards, all along the whole circumference of the meniscus.
            \nTotal upward force = (T cos\u03b8) (2\u03c0r)
            \n= 2\u03c0rT. cos\u03b8
            \nWhere \u03b8 is the angle of contact, r is the radius of the tube. Let \u03c1 be the density of water and h be the height to which the liquid rises inside the tube. Then,
            \n\"Samacheer
            \nThe upward force supports the weight of the liquid column above the free surface, therefore,
            \n\"Samacheer
            \nIf the capillary is a very fine tube of the radius (i.e., the radius is very small) then \\(\\frac { r }{ 3 }\\) can be neglected when it is compared to the height h. Therefore,
            \nT = \\(\\frac{r \\rho g h}{2 \\cos \\theta}\\)<\/p>\n

            Question 11.
            \nObtain an equation of continuity for a flow of fluid on the basis of conservation of mass.
            \nAnswer:
            \nLet us consider a pipe AB of varying cross-sectional areas a1<\/sub> and a2<\/sub> such that a1<\/sub> > a2<\/sub>. A non – viscous and incompressible liquid flows steadily through the pipe, with velocities v1<\/sub> and v2<\/sub> in areas a1<\/sub> and a2<\/sub>, respectively as shown in Figure.
            \n\"Samacheer
            \nLet m1<\/sub> be the mass of fluid flowing through section A in time \u2206t, m1<\/sub> = (a1<\/sub>v1<\/sub> \u2206t)\u03c1
            \nLet m2<\/sub> be the mass of fluid flowing through section B in time \u2206t, m2<\/sub> = (a2<\/sub>v2<\/sub> \u2206t)\u03c1
            \nFor an incompressible liquid, mass is conserved m1<\/sub> = m2<\/sub>
            \na1<\/sub>v1<\/sub> \u2206t\u03c1 = a2<\/sub>v2<\/sub>\u2206t\u03c1
            \na1<\/sub>v1<\/sub> = a2<\/sub>v2<\/sub> \u21d2 av = constant
            \nWhich is called the equation of continuity. It is based on the conservation of mass in the flow of fluids. In general, av – constant.<\/p>\n

            Question 12.
            \nState and prove Bernoulli’s theorem for a flow of incompressible, non-viscous, and streamlined flow of fluid.
            \nStatement:
            \nAccording to Bernoulli\u2019s theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, non-viscous fluid in a streamlined flow remains a constant.
            \n\"Samacheer
            \nProof:
            \nLet us consider a flow of liquid through a pipe AB as shown in Figure. Let V be the volume of the liquid when it enters A in a time t which is equal to the volume of the liquid leaving B in the same time. Let aA<\/sub>, vA,<\/sub> and PA<\/sub>. be the area of cross-section of the tube, velocity of the liquid, and pressure exerted by the liquid at A respectively.
            \nLet the force exerted by the liquid at A is
            \nFA<\/sub> = VA<\/sub>
            \nDistance travelled by the liquid in time t is
            \nd = vA<\/sub> t
            \nTherefore, the work done is
            \nW = FA<\/sub>d = PA<\/sub> aA<\/sub> vA<\/sub> t
            \nBut aA<\/sub> vA<\/sub> t = aA<\/sub> d = V, the volume of the liquid entering at A.
            \nThus, the work done is the pressure energy (at A),
            \nW = FA<\/sub> d = PA<\/sub> V
            \nPressure energy per unit mass at pressure energy
            \n\"Samacheer
            \nSince m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is
            \nEPA<\/sub> = PA<\/sub>V = PA<\/sub>V x \\(\\frac { m }{ m }\\) = m \\(\\frac{P_{A}}{\\rho}\\)
            \nPotential energy of the liquid at A,
            \nPA<\/sub> = mg hA<\/sub>,
            \nDue to the flow of liquid, the kinetic energy of the liquid at A,
            \nKEA<\/sub> = \\(\\frac { 1 }{ 2 }\\)mv\u00b2A<\/sub>
            \nTherefore, the total energy due to the flow of liquid at A, EA<\/sub> = EPA<\/sub> + KEA<\/sub> + PEA<\/sub>
            \nEA<\/sub> = \\(m \\frac{\\mathrm{P}_{\\mathrm{A}}}{\\rho}+\\frac{1}{2} m v_{\\mathrm{A}}^{2}+m g h_{\\mathrm{A}}\\)
            \nSimilarly, let aB<\/sub>, vB<\/sub>, and PB<\/sub> be the area of a cross section of the tube, the velocity of the liquid, and pressure exerted by the liquid at B. Calculating the total energy at EB, we get
            \nEB<\/sub> = \\(m \\frac{\\mathrm{P}_{\\mathrm{B}}}{\\rho}+\\frac{1}{2} m v_{\\mathrm{B}}^{2}+m g h_{\\mathrm{B}}\\)
            \nFrom the law of conservation of energy,
            \n\"Samacheer
            \nThus, the above equation can be written as
            \n\\(\\frac { p }{ \u03c1g }\\) + \\(\\frac { 1 }{ 2 }\\)\\(\\frac{v^{2}}{g}\\) + h = constant<\/p>\n

            Question 13.
            \nDescribe the construction and working of venturimeter and obtain an equation for the volume of liquid flowing per second through a wider entry of the tube.
            \nAnswer:
            \nConstruction: It consists of two wider tubes A and A’ (with cross-sectional area A) connected by a narrow tube B (with the cross-sectional area a). A manometer in the form of a U-tube is also attached between the wide and narrow tubes as shown in Figure. The manometer contains a liquid of density \u2018\u03c1m<\/sub>‘.
            \n\"Samacheer
            \nTheory:
            \nLet P1<\/sub> be the pressure of the fluid at the wider region of tube A. Let us assume that the fluid of density \u2018\u03c1\u2019 flows from the pipe with speed \u2018v1<\/sub>\u2019 and into the narrow region, its speed increases to \u2018v1<\/sub>‘.<\/p>\n

            According to Bernoulli\u2019s equation, this increase in speed is accompanied by a decrease in the fluid pressure P2<\/sub> at the narrow region of the tube B. Hence, the pressure difference between tubes A and B is noted by measuring the height difference (\u2206P = P1<\/sub> – P2<\/sub>) between the surfaces of the manometer liquid.
            \nFrom the equation of continuity, we can say that Av1<\/sub> = av2<\/sub> which means that
            \nv2<\/sub> = \\(\\frac { A }{ a }\\)v1<\/sub>
            \nUsing Bernoulli\u2019s equation,
            \n\"Samacheer<\/p>\n

            IV. Numerical Problems:<\/span><\/p>\n

            Question 1.
            \nA capillary of diameter d mm is dipped in water such that the water rises to a height of 30 mm. If the radius of the capillary is made (\\(\\frac { 2 }{ 3 }\\))of its previous value, then compute the height up to which water will rise in the new capillary?
            \nAnswer:
            \nLet diameter of capillary tube = d mm
            \nLet the radius of capillary tube r1<\/sub> = r mm
            \nCapillary rise h1<\/sub> = 30 mm
            \nLet the radius of another capillary tube r2<\/sub> = \\(\\frac { 2 }{ 3 }\\) r
            \nLet the capillary rise of another capillary tube be h2<\/sub>
            \nWe know that
            \n\"Samacheer<\/p>\n

            Question 2.
            \nA cylinder of length 1.5 m and diameter 4 cm is fixed at one end. A tangential force of 4 x 105<\/sup> N is applied at the other end. If the rigidity modulus of the cylinder is 6 x 105<\/sup> Nm-2<\/sup> then, calculate the twist produced in the cylinder.
            \nAnswer:
            \nLength of a cylinder l = 1.5 m
            \nDiameter of a cylinder d = 4 x 10-2<\/sup> m
            \nTangential force Ft<\/sub> = 4 x 105<\/sup> N
            \nRigidity modulus \u03b7R<\/sub> = 6 x 1010<\/sup> Nm-2<\/sup>
            \nTwist produced \u03b8 = ?
            \n\"Samacheer<\/p>\n

            Question 3.
            \nA spherical soap bubble A of radius 2 cm is formed inside another bubble B of radius 4 cm. Show that the radius of a single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is lesser than the radius of both soap bubbles A and B.
            \nAnswer:
            \nExcess of pressure inside the liquid due to surface tension,
            \n\u2206P = \\(\\frac { 2T }{ R }\\)
            \nWhere T – surface tension
            \nIn the case of soap bubbles, the excess pressure inside the soap bubble,
            \n\u2206Pb<\/sub> = \\(\\frac { 4T }{ R }\\)
            \nExcess of pressure of air inside the bigger bubble
            \n\u2206Pbigger<\/sub> = \\(\\frac { 4T }{ 4 }\\) = T
            \nExcess of pressure of air inside the smaller bubble
            \n\u2206Psmaller<\/sub> = \\(\\frac { 4T }{ 2 }\\) = 2T
            \nAir pressure different between the smaller bubble and the atmosphere will be equal to the sum of excess pressure inside the bigger and smaller bubbles.
            \n\"Samacheer
            \nExcess pressure inside a single soap bubble
            \n= \\(\\frac { 4T }{ R }\\) = 4T = T
            \n\u2234 Pressure different of single soap bubble less than the radius of both T < 3T.<\/p>\n

            \"Samacheer<\/p>\n

            Question 4.
            \nA block of Ag of mass x kg hanging from a string is immersed in a liquid of relative density 0.72. If the relative density of Ag is 10 and tension in the string is 37.12 N then compute the mass of the Ag block.
            \nAnswer:
            \nLet the mass of Ag block be x kg,
            \nTension in the string T = 37.12 N.
            \nRelative density of liquid Rliq<\/sub> = 0.72
            \nRelative density of silver RAg<\/sub> = 10
            \nAccording to the principle of flotation
            \nVpg<\/sub> = mg
            \n\u2234 Vp=m … (1)
            \nWeight of Ag block = mg = V\u03c1g = RAg<\/sub>Vg<\/sub>
            \nWeight of Ag block W = 10 V\u03c1g [ \u2235 m = V\u03c1]
            \nForce of buoyancy FB<\/sub> = Rliq<\/sub> (V\u03c1)g = 0.72 V\u03c1g<\/sub>
            \nApparent weight Wapp<\/sub> = 10 V\u03c1g<\/sub> – 0.72 V\u03c1g
            \n= 9.78 V\u03c1g
            \n= 9.78 mg
            \nTension in the string = Apparent weight
            \n37.12= 9.78 mg
            \nMass m = \\(\\frac { 37.12 }{ 9.28 }\\) = 4 kg<\/p>\n

            Question 5.
            \nThe reading of the pressure meter attached with a closed pipe is 5 x 105<\/sup> Nm-2<\/sup>. On opening the valve of the pipe, the reading of the pressure meter is 4.5 x 105 <\/sup>Nm-2<\/sup>. Calculate the speed of the water flowing in the pipe.
            \nAnswer:
            \nInitial pressure P2<\/sub> = 5 x 105<\/sup> Nm-2<\/sup>
            \nFinal pressure P1<\/sub> = 4. 5 x 105<\/sup> Nm-2<\/sup>
            \nInitial velocity V1<\/sub> = 0
            \nFinal velocity v2<\/sub> = v2<\/sub>
            \nDensity of water \u03c1 = 10\u00b3 kg\/m\u00b3
            \nUsing Bernoulli\u2019s theorem, we can write,
            \n\"Samacheer<\/p>\n

            V. Conceptual Questions:<\/span><\/p>\n

            Question 1.
            \nWhy coffee runs up into a sugar lump (a small cube of sugar) when one corner of the sugar lump is held in the liquid?
            \nAnswer:
            \nDip the comer of a sugar cube in coffee, and get the whole cube coffee-flavoured due to \u201ccapillary action\u201d.<\/p>\n

            Question 2.
            \nWhy two holes are made to empty an oil tin?
            \nAnswer:
            \nWhen oil comes out of one hole with high velocity, the pressure in the tin decreases. To have a continuous flow of oil proper pressure is to be maintained inside the tin. To achieve this, atmospheric air has to be entered inside the tin. For this purpose, only another hole is made out. Hence two holes are made to empty an oil tin.<\/p>\n

            Question 3.
            \nWe can cut vegetables easily with a sharp knife as compared to a blunt knife. Why?
            \nAnswer:
            \nThe area of a sharp edge is much less than the area of a blunt edge. For the same total force, the effective force per unit area is more for the sharp edge than the blunt edge. Hence, a sharp knife cuts easily than a blunt knife.<\/p>\n

            \"Samacheer<\/p>\n

            Question 4.
            \nWhy the passengers are advised to remove the ink from their pens while going up in an aeroplane?
            \nAnswer:
            \nWhile going up in an aeroplane the atmospheric pressure decreases with height. When aeroplane is going up, the ink in the pen tends to ooze out to equalise the pressure. This may spoil the clothes of the passengers. So they are advised to remove ink from the pen.<\/p>\n

            Question 5.
            \nWe use a straw to suck soft drinks, why?
            \nAnswer:
            \nWhen we suck through the straw, the pressure inside the straw becomes less than the atmospheric pressure. Due to the pressure difference, the soft drink rises in the straw and we are able to take the soft drink easily.<\/p>\n","protected":false},"excerpt":{"rendered":"

            Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Physics Guide Pdf Chapter 7 Properties of Matter Text Book Back Questions and Answers, Notes. Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 7 Properties of Matter 11th Physics Guide Properties of Matter Book Back Questions and Answers I. Multiple choice questions: Question 1. Consider two wires X …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[6],"tags":[],"class_list":["post-33677","post","type-post","status-publish","format-standard","hentry","category-class-11"],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/33677"}],"collection":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=33677"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/33677\/revisions"}],"predecessor-version":[{"id":41671,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/33677\/revisions\/41671"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=33677"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=33677"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=33677"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}