{"id":10765,"date":"2019-12-09T17:11:11","date_gmt":"2019-12-09T11:41:11","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=10765"},"modified":"2021-07-02T15:43:14","modified_gmt":"2021-07-02T10:13:14","slug":"1st-puc-physics-question-bank-chapter-9","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-physics-question-bank-chapter-9\/","title":{"rendered":"1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids"},"content":{"rendered":"

You can Download Chapter 9 Mechanical Properties of Solids Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids<\/h2>\n

1st PUC Physics Mechanical Properties of Solids Textbook Questions and Answers<\/h3>\n

Question 1.
\nA steel wire of length 4.7 m and cross-sectional 3.0 \u00d7 10-5<\/sup> m2<\/sup> stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 \u00d7 10-5<\/sup> m2<\/sup> under a given load. What is the ratio of the Young\u2019s modulus of steel to that of copper?
\nAnswer:
\nLet the Young\u2019s Modulus of steel and copper be Ys<\/sub> and Yc<\/sub> respectively.
\nLength of steel wire, Ls<\/sub> = 4.7 m
\nLength of copper wire, Lc<\/sub> = 3.5 m
\nArea of cross-section of steel wire,
\nAs<\/sub>= 3 \u00d7 10-5<\/sup> m2<\/sup>
\nArea of cross-section of copper wire,
\nAc<\/sub> = 4 \u00d7 10-5<\/sup> m2<\/sup>
\nSince change in lengths are equal.
\n\u2234 \u2206Ls<\/sub> = \u2206Lc<\/sub> = \u2206 L.
\nAs the load is same, Fs<\/sub> = Fc<\/sub> = F
\nWe have, Y = \\(\\frac{F}{A} \\times \\frac{L}{\\Delta L}\\)
\n\\(\\therefore \\frac{Y_{s}}{Y_{c}}=\\frac{L_{s} A_{c}}{A_{s} L_{c}}=\\frac{4 \\times 4.7 \\times 10^{-5}}{3 \\times 3.5 \\times 10^{-5}}\\) = 1.79
\nThe ratio of Young\u2019s Modulus of steel to that of copper is \u2243 1.8 :1.<\/p>\n

Question 2.
\nFigure shows the strain – stress curve for a given material. What are
\n1. Young\u2019s modulus and
\n2. approximate yield strength for this material?
\n\"1st
\nAnswer:
\n1. F from the given graph,
\nFor a stress of 150 \u00d7 106<\/sup> Nm-2<\/sup>
\nStrain = 0.002
\n\u2234 Young\u2019s modulus (Y) = \\(\\frac{\\text { stress }}{\\text { strain }}\\)
\n\\(=\\frac{150 \\times 10^{6} \\mathrm{Nm}^{-2}}{0.002}\\) = 7.5 \u00d7 1010<\/sup> Nm-2<\/sup><\/p>\n

2. 300 \u00d7 106<\/sup> Nm-2<\/sup> is the maximum stress that the material can sustain without crossing the elastic limit. Thus, the yield strength of the material is 3 \u00d7 108<\/sup> Nm-2<\/sup> or 300 \u00d7 106<\/sup> Nm-2<\/sup>.<\/p>\n

Question 3.
\nThe stress-strain graphs for materials A and B are shown in Figure.
\n\"1st
\nThe graphs are drawn to the same scale.<\/p>\n

    \n
  1. Which of the materials has the greater Young\u2019s modulus?<\/li>\n
  2. Which of the two is the stronger material?<\/li>\n<\/ol>\n

    Answer:<\/p>\n

      \n
    1. Material A has greater Young\u2019s modulus as the slope of the graph of material A is greater than material B.
      \nYoung\u2019s Modulus = Y = Slope = \\(\\frac{\\text { stress }}{\\text { strain }}\\)
      \n\u2234 Material A has greater Young\u2019s Modulus.<\/li>\n
    2. From the graph it can be approximated that the stress at the fracture point for material A is higher than material B, thus material A is stronger.<\/li>\n<\/ol>\n

      \"KSEEB<\/p>\n

      Question 4.
      \nRead the following two statements below carefully and state, with reasons, if it is true or false.
      \n1. The Young\u2019s modulus of rubber is greater than that of steel;
      \n2. The stretching of a coil is determined by its shear modulus.
      \nAnswer:
      \n1. The Young\u2019s modulus of a -material defined as,
      \nY = \\(\\frac{\\text { stress }}{\\text { strain }}\\)
      \nSince, rubber stretches more than steel \u2206 L for rubber is greater than that of steel, therefore ( \\(\\frac{\\Delta L}{L}\\) 0f rubber) > (\\(\\frac{\\Delta L}{L}\\) of steel) Since, Y \\(\\alpha \\frac{1}{\\text { strain }}\\) and, strain = \\(\\frac{\\Delta L}{L}\\)
      \nSince the strain of rubber is greater, its Youngs modulus is less. Hence false,<\/p>\n

      2. The Stretching of a coil depends on the shear modulus of the material because for the coil to be stretched, a shear force has to applied to change the shape of the coil. Hence the statement is true.<\/p>\n

      Question 5.
      \nTwo wires of diameter 0.25 cm. one made of steel and the other made of brass are loaded as shown in Fig.9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of steel and the brass wires.
      \n\"1st
      \nAnswer:
      \n\"1st
      \ngiven, d = 0.25 cm
      \nr = \\(\\frac{\\mathrm{d}}{2}\\) = 0.125 cm.
      \nAs<\/sub> = AB<\/sub> = \u03c0 \u00d7 (0.125 \u00d7 10-2<\/sup>)2<\/sup> m2<\/sup>
      \n= 4.91 \u00d7 10-6<\/sup> m2<\/sup>
      \nAs<\/sub> = AB<\/sub> because the diameters are same.
      \nFs<\/sub> = (4 \u00d7 g + 6 \u00d7 g) N
      \n= (4 \u00d7 9.8 + 6 \u00d7 9.8) N
      \n= 10 \u00d7 9.8 N =98 N
      \nFB<\/sub> = (6 \u00d7 9.8) N = 58.8 N
      \n\u2206Ls<\/sub> = ?
      \n\u2206LB<\/sub> = ?
      \nLs<\/sub> = 1.5 m
      \nLB<\/sub> =1.0m
      \nYs<\/sub> = 2 \u00d7 1011<\/sup> N m-2<\/sup> and
      \nYB<\/sub> = 0.91 \u00d7 1011<\/sup> Nm-2<\/sup>
      \n\u2206Ls<\/sub> = \\(\\frac{F_{s} \\cdot L_{s}}{A_{s} \\cdot Y_{s}}=\\frac{98 \\times 1.5}{4.91 \\times 10^{-6} \\times 2 \\times 10^{11}} \\mathrm{m}\\)
      \n= 1.49 \u00d7 10-4<\/sup>m = 0.149 mm
      \nTo find the elongation of Brass,
      \n\u2206LB<\/sub> = \\(\\frac{F_{B} L_{B}}{A_{B} Y_{B}}=\\frac{(58.8)(1)}{\\left(4.91 \\times 10^{-6}\\right)\\left(0.91 \\times 10^{11}\\right)} \\mathrm{m}\\)
      \n= 1.32 \u00d7 10-4<\/sup>m = 0.132 mm
      \nThe elongation of steel wire is 0.149 mm.
      \nThe elongation of Brass wire is 0.132 mm.<\/p>\n

      Question 6.
      \nThe edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
      \nAnswer:
      \nEdge length, L = 10 cm = 0.1 m
      \nWeight attached = 100 kg
      \nF = (100 \u00d7 9.8) N = 980 N
      \nA = L2<\/sup> = 0.1 \u00d7 0.1 m2<\/sup>
      \n= 0.01 m2<\/sup>
      \n\"1st
      \n= 3.92 \u00d7 10-7<\/sup> m
      \n\u2234 the deflection is 3.92 \u00d7 10-7<\/sup> m<\/p>\n

      Question 7.
      \nFour identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
      \nAnswer:
      \nMass of big structure = 50,000 kg
      \nNet force = (5 \u00d7 104<\/sup> \u00d7 9.8) N
      \n= 4.9 \u00d7 105<\/sup> N = FB<\/sub>
      \nSince it is supported by four cylinders the weight is equally distributed.
      \nForce (F) on each cylinder is
      \nF = \\(\\frac{F_{B}}{4}\\) = 122500 N
      \nYoung\u2019s Modulus = Y = = \\(\\frac{\\text { stress }}{\\text { strain }}=\\frac{F \\times L}{A \\times \\Delta L}\\)
      \n\u2234 Compressional strain of each column is given by, \\(\\frac{\\Delta L}{L}=\\frac{F}{A \\times Y}\\)
      \nHere, F = 122500 N
      \nA = \u03c0(0.6)2<\/sup> – \u03c0(0.3)2<\/sup> = \u03c0(0.36 – 0.09)
      \n= 0.27 \u03c0 = 8.48 \u00d7 10-1<\/sup> m2<\/sup>
      \nY = 2 \u00d7 1011<\/sup> N m-2<\/sup>\u00a0(Standard Values)
      \n\u2234 Strain = \\(\\frac{122500}{(0.27 \\pi)\\left(2 \\times 10^{11}\\right)}\\) = 7.22 \u00d7 10-7<\/sup>
      \nThus the compressional strain per cylinder is 7.22 \u00d7 10-7<\/sup><\/p>\n

      \"KSEEB<\/p>\n

      Question 8.
      \nA piece of copper having a rectangular cross-section of 15.2 mm \u00d7 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
      \nAnswer:
      \nF = 44,500 N
      \nStrain =?
      \nLength = 19.1 mm, Breath = 15.2 mm
      \nArea = A = 19.1 \u00d7 15.2 \u00d7 10-6<\/sup>m2<\/sup>
      \n= 2.98 \u00d7 1 O-4<\/sup> m2<\/sup>
      \nG = 42 \u00d7 109<\/sup> Nm-2<\/sup>
      \nWe have, G = \\(\\frac{F}{A \\times \\text { strain }}\\)
      \n\u2234 Strain = \\(\\frac{F}{G \\times A}\\)
      \n\\(=\\frac{44,500}{2.98 \\times 10^{-4} \\times 42 \\times 10^{9}}\\)
      \n= 3.65 \u00d7 10-3<\/sup><\/p>\n

      Question 9.
      \nA steel cable with a radius of 1.5 cm supports a chair lift at a ski area. If the maximum stress is not exceed 108<\/sup> Nm-2<\/sup>, What is the maximum load the cable can support?
      \nAnswer:
      \nr = radius of cable = 1.5 cm.
      \nA = \u03c0(0.015)2<\/sup> m2<\/sup>
      \nmax stress = 108<\/sup> Nm-2<\/sup>
      \nLet F be the maximum force.
      \nstress = \\(\\frac{F}{A}\\)
      \n\u2234 F = max stress \u00d7 area.
      \n= 108<\/sup> \u00d7 \u03c0 \u00d7 (0.015)2<\/sup>
      \n= 7.065 \u00d7 104<\/sup>N
      \n= 70,650 N
      \n\u2234 The maximum load is 70,650 N<\/p>\n

      Question 10.
      \nA rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of Iron. Determine the ratios of their diameters if each is to have the same tension.
      \nAnswer:
      \nSince the rigid bar is supported symmetrically by three wires each of length 2 m, the extension is same in all the three wires. Thus strain is same.
      \n\\(Y=\\frac{F}{\\pi r^{2} \\times \\text { strain }}\\) = \\(\\frac{4 \\mathrm{F}}{\\pi \\mathrm{d}^{2} \\times \\text { strain }}\\)
      \n\u2234 Y \\(\\alpha \\frac{1}{\\mathrm{d}^{2}}\\)
      \nYi<\/sub> = Young\u2019s modulus of Iron
      \ndi<\/sub> = diameter of iron wire.
      \nYc<\/sub> = Young\u2019s modulus of copper
      \ndc<\/sub> = diameter of copper wire.
      \nYi<\/sub> = 190 \u00d7 109<\/sup> N m-2<\/sup>
      \nYc<\/sub> = 110 \u00d7 109<\/sup> N m-2<\/sup>
      \n\\(\\frac{d_{c}}{d_{i}}=\\sqrt{\\frac{Y_{i}}{Y_{c}}}=\\sqrt{\\frac{19}{11}}=1.31: 1\\)<\/p>\n

      Question 11.
      \nA 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev\/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2<\/sup>. Calculate the elongation of the wire when the mass is at the lowest point of its path.
      \nAnswer:
      \nL = 1 m
      \nW = 2 revolutions\/s
      \nA = 0.065 cm2<\/sup> = 6.5 x 10-6<\/sup> m2<\/sup>
      \nF = mg + mlw2<\/sup>
      \n= 14.5 \u00d7 9.8 +14.5 \u00d7 1 \u00d7 (2)2<\/sup>
      \n= 200.1 N
      \n\u2206L = \\(\\frac{F l}{A Y}\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [\u2235 Y = \\(\\frac{\\mathrm{FI}}{\\mathrm{A} \\Delta \\mathrm{I}}\\)]
      \nYoung\u2019s modulus of steel is 2 \u00d7 1011<\/sup> Nm-2<\/sup>
      \n\u2206L = \\(\\frac{200.1 \\times 1}{6.5 \\times 10^{-6} \\times 2 \\times 10^{11}} \\mathrm{m}\\) = 1.539 \u00d7 10-4<\/sup>m<\/p>\n

      Question 12.
      \nCompute the bulk modulus of water from the following data: Initial Volume = 100.0 litre. Pressure increase = 100.0 atm (1 atm = 1.013 \u00d7 105<\/sup> Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
      \nAnswer:
      \nVF<\/sub> = final volume = 100.5 I
      \n= 100.5 \u00d7 10-3<\/sup> m3<\/sup>
      \nVi<\/sub> = initial volume = 100 \u00d7 10-3<\/sup> m3<\/sup>
      \n\u2206V = VF<\/sub> – Vi<\/sub> =0.5 \u00d7 10-3<\/sup> m3<\/sup>
      \n= 5 \u00d7 10-4<\/sup> m3<\/sup>
      \n\u2206 P = 100 atm = 100 \u00d7 1.013 \u00d7 105<\/sup> Pa
      \n\"1st
      \n= 2.03 \u00d7 109<\/sup> Pa
      \nBulk modulus of air = 1 \u00d7 105<\/sup> Pa
      \n\\(\\frac{\\text { Bulk modulus of water }}{\\text { Bulk modulus of air }}=\\frac{2.026 \\times 10^{9}}{1 \\times 10^{5}}\\)
      \n= 2.026 \u00d7 104<\/sup>
      \nSince air is highly compressible compared to water, the ratio is so high.<\/p>\n

      Question 13.
      \nWhat is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 \u00d7 103<\/sup> kg m-3<\/sup>?
      \nAnswer:
      \nDensity of water at surface
      \n= \\(\\rho_{1}\\) = 1.03 \u00d7 103<\/sup> kgm-3<\/sup>
      \ndensity of water at depth h = \\(\\rho_{2}\\)
      \nPressure at depth h = 80 atm
      \n= 80 \u00d7 1.01 \u00d7 105<\/sup> Pa.
      \nV1<\/sub> = Volume of water of mass m at surface.
      \nV2<\/sub> = Volume of water at depth h.
      \n\"1st
      \n\"1st
      \n\\(\\frac{1}{\\mathrm{B}}\\) Compressibility of water
      \n= 45.8 \u00d7 10-11<\/sup>
      \n\u2234 \\(\\frac{\\Delta V}{V_{1}}\\) =80 \u00d7 1.013 \u00d7 105 \u00d7 45.8 \u00d7 10-11<\/sup>
      \n= 3.71 \u00d7 10-3
      \n<\/sup>Since,
      \n\"1st
      \n= 1.034 \u00d7 103<\/sup> kg m-3<\/sup><\/p>\n

      Question 14.
      \nCompute the fractional change in volume of a glass slab, when subjected to hydraulic pressure of 10 atm.
      \nAnswer:
      \nHydraulic pressure exerted = 10 atm
      \n= 10 \u00d7 1.013 \u00d7 105<\/sup> Pa
      \nB = \\(\\frac{P \\cdot V}{\\Delta V}\\)
      \n\\(\\frac{\\Delta V}{V}\\) is the fractional change in volume
      \n\\(\\frac{\\Delta V}{V}=\\frac{P}{B}=\\frac{10 \\times 1.013 \\times 10^{5}}{37 \\times 10^{9}}\\)
      \n= 2.73 \u00d7 10-5<\/sup>
      \n\u2234 The fractional change in Volume is 2.73 \u00d7 10-5<\/sup>.<\/p>\n

      Question 15.
      \nDetermine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 \u00d7 106<\/sup> Pa.
      \nAnswer:
      \nP = 7 \u00d7 106<\/sup> Pa
      \nI = 10 cm
      \nV = I3<\/sup>= 10-3<\/sup> m3<\/sup>
      \nB = 140 \u00d7 109<\/sup>
      \nB = \\(\\frac{P \\times V}{\\Delta V}\\)
      \n\u2206V = \\(\\frac{P}{B} \\times V=\\frac{7 \\times 10^{6}}{140 \\times 10^{9}} \\times 10^{-3} \\mathrm{m}^{3}\\)
      \n= 5 \u00d7 10-8<\/sup> m3<\/sup>
      \n= 0.05 cm3<\/sup>
      \n\u2234 The contraction is 0.05 cm3<\/sup><\/p>\n

      \"KSEEB<\/p>\n

      Question 16.
      \nHow much should the pressure on a litre of water be changed to compress it by 0.10%
      \nAnswer:
      \nP =?
      \nV = 1L ,
      \nB = 2.2 \u00d7 109<\/sup>Nm-2<\/sup>
      \n\\(\\frac{\\Delta v}{V}\\) \u00d7 100 = 0.1
      \n\u21d2 \\(\\frac{\\Delta v}{V}\\) = \\(\\frac{0.1}{100 \\times 1}\\) = 10-3<\/sup>
      \nSince, B = \\(\\frac{\\mathrm{PV}}{\\Delta \\mathrm{V}}\\)
      \nP = B \\(\\frac{\\Delta v}{V}\\)
      \n= 2.2 \u00d7 109<\/sup> Pa \u00d7 10-3<\/sup>
      \n= 2.2 \u00d7 106<\/sup> Pa.
      \n\u2234 The pressure is required is 2.2 \u00d7 106<\/sup> Pa.<\/p>\n

      1st PUC Physics Mechanical Properties of Solids Additional Exercises Questions and Answers<\/strong><\/p>\n

      Question 1.
      \nAnvils made of single crystals of diamond, with the shape as shown in Fig. are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?
      \n\"1st
      \nAnswer:
      \nDiameter at narrow end
      \n= d =0.5 mm
      \nCompressional Force = F = 50,000 N
      \n\"1st
      \n\u2234 The pressure is 2.546 \u00d7 1011<\/sup> Pa<\/p>\n

      Question 2.
      \nA rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig.9.15. The cross-sectional areas of wires A and B are 1.0 mm2<\/sup> and 2.0 mm2<\/sup> respectively. At what point along the rod should a mass m be suspended in order to produce.<\/p>\n

        \n
      1. equal stresses.<\/li>\n
      2. equal in both steel and aluminium wires.<\/li>\n<\/ol>\n

        \"1st
        \nAnswer:
        \nLet the initial lengths of the 2 wires be \u2018L\u2019
        \nGiven that the cross-sectional area of A
        \n= Aa<\/sub> = 1 mm2<\/sup> = 10-6<\/sup>
        \nCross-sectional area of
        \nB = Ab<\/sub> = 2 mm2<\/sup> = 2 \u00d7 10-6<\/sup>m2<\/sup>
        \nLength of rod = 1.05 m.
        \nYoung\u2019s modulus of wire A
        \n= Ysteel<\/sub> = 2 \u00d7 1011<\/sup> N\/m2<\/sup>
        \nYoung\u2019s modulus of wire B
        \nYAl<\/sub> = 7 \u00d7 1010<\/sup> N\/m2<\/sup>
        \nLet the mass \u2018m\u2019 be suspended at a distance \u2018x\u2019 from the end with wire A attached.<\/p>\n

        1. For equal stress on the two wires :
        \ni.e., stress A = stress B
        \n\"1st
        \nor Fb<\/sub> = 2 Fa<\/sub> …..(1)
        \nwhere, Fa<\/sub> is force exerted on steel wire and F\u03c3<\/sub> is force exerted on aluminium wire. Since the total torque at the point when mass is suspended is zero, we can write
        \nFa<\/sub> \u00d7 (x) = Fb<\/sub> \u00d7 (1.05 – x)
        \n\"1st
        \n\u21d2 3x = 2 \u00d7 1.05 or x = 0.7 m
        \nSo, for equal stress on the wires the mass should be suspended at a distance 0.7m, from wire A.<\/p>\n

        2. For equal strain:
        \nWe know that, Strain = \\(\\frac{\\text { Stress }}{Y}=\\frac{F}{\\text { AY }}\\)
        \nwe require (strain)a<\/sub> = (strain)b<\/sub>
        \n\"1st
        \n\\(\\frac{F_{a}}{F_{b}}\\) = \\(\\frac{10}{7}\\)
        \n……………(3) Since the torque is \u20180\u2019 at the point of suspension equation (2) is valid.
        \n\u21d2 \\(\\frac{F_{a}}{F_{b}}\\) = \\(\\frac{10}{7}\\) = \\(\\frac{1.05-x}{x}\\)
        \n\u21d2 17 x = 7 \u00d7 (1.05)
        \nx = 0.432 m
        \nSo, for equal strain on the wires the mass should be suspended at a distance 0.432 m from wire A.<\/p>\n

        Question 3.
        \nA mild steel wire of length 1.0 m and cross-sectional area 0.50 \u00d7 10-2<\/sup>cm2<\/sup> is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the mid-point.
        \nAnswer:
        \nGiven that length of the wire = 1.0 m
        \nArea of cross-section = 0.50 \u00d7 10-2<\/sup>cm2<\/sup>
        \n= 5.0 \u00d7 10-7<\/sup>m2<\/sup>
        \nMass suspended =100g
        \n= 10-1<\/sup> Kg
        \nYsteel<\/sub> = 2 \u00d7 1011<\/sup> Pa
        \nThe mass causes the steel wire to bend as shown.
        \nSo new length of wire = AX + XB
        \n\"1st
        \nChange in length is, \u2206l = (AX + XB) – AB
        \nNow, AX = \\(\\sqrt{\\mathrm{AM}^{2}+\\mathrm{MX}^{2}}\\) = XB
        \n[ Pythogorus Theorem]
        \n\"1st
        \n\"1st
        \n[Expanding and neglecting higher-order terms]
        \nThe tension in the steel wire is,
        \nT cos \u03b8 + T cos \u03b8 = mg
        \n\"1st
        \n\\(\\sqrt[3]{\\frac{0.5 \\times 0.1 \\times 9.8}{4 \\times 2 \\times 10^{11} \\times 5 \\times 10^{-7}}}\\) = 0.0106 m
        \nSo, the depression at the centre is 0.0106 m or 1.06 cm.<\/p>\n

        Question 4.
        \nTwo strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 \u00d7 107<\/sup> Pa? Assume that each rivet is to carry one-quarter of the load.
        \nAnswer:
        \nGiven diameter of metal strips
        \n= 6.0 \u00d7 10-3<\/sup> m
        \n\u21d2 Area = \\(\\pi\\left(\\frac{\\mathrm{d}}{2}\\right)^{2}=\\frac{\\pi \\times 36 \\times 10^{-6} \\mathrm{m}^{2}}{4}\\)
        \n= 2.827 \u00d7 10-5<\/sup>m2<\/sup>
        \nFor the maximum stress of 6.9 \u00d7 107<\/sup> Pa
        \nthe maximum force is given by
        \nMax Force = \\(\\frac{\\text { Max Force }}{\\text { Area }}\\)
        \n\u21d2 Max force = Max Stress \u00d7 Area
        \n= 6.9 \u00d7 107<\/sup> \u00d7 2.827 \u00d7 10-5<\/sup>
        \n= 1950.63 N
        \nSince there are 4 rivets and each rivet carries 1\/4th<\/sup> of the load,
        \nMax tension that can be exerted by the rivet strip is 4 \u00d7 1.95063 KN
        \n= 7.80252 KN.<\/p>\n

        \"KSEEB<\/p>\n

        Question 5.
        \nThe Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench Is about 1.1 \u00d7 108<\/sup> Pa. A steel ball of Initial volume 0.32 m3<\/sup> is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?
        \nAnswer:
        \nPressure at bottom \u2019of the trench = \u03c1gh = 1.1 \u00d7 108<\/sup> Pa
        \nInitial volume of the ball = 0.32 m3<\/sup>
        \nBulk modulus of steel = 1.6 \u00d7 1011<\/sup> Pa
        \nLet \u2206v the change in volume of Ball when it falls down to the bottom.
        \nWe know that, B = \\(\\frac{P}{(\\Delta V \/ V)}\\)
        \n\u21d2 \u2206v = \\(\\frac{P V}{B}\\)
        \n\\(=\\frac{1.1 \\times 10^{8} \\times 0.32}{1.6 \\times 10^{11}}\\) = 2.2 \u00d7 10-4<\/sup> m3<\/sup>
        \nTherefore, change in volume of the ball on reaching the bottom of the trench is 2.2 \u00d7 10-4<\/sup> m3<\/sup> or 220 cm3<\/sup><\/p>\n

        1st PUC Physics Mechanical Properties of Solids One Mark Questions and Answers<\/h3>\n

        Question 1.
        \nWhat is elasticity?
        \nAnswer:
        \nThe property of the body to restore the original shape or size on removal of the external force is called elasticity.<\/p>\n

        Question 2.
        \nDefine stress.
        \nAnswer:
        \nStress is defined as the deforming force acting on a unit area of the body.<\/p>\n

        Question 3.
        \nWhat is the S.l unit of stress?
        \nAnswer:
        \nNm-2<\/sup><\/p>\n

        Question 4.
        \nDefine strain.
        \nAnswer:
        \nStrain is the ratio of change in dimension of a body due to stress to the original dimension.<\/p>\n

        \"KSEEB<\/p>\n

        Question 5.
        \nDefine Shear Strain.
        \nAnswer:
        \nThe angular tilt between different layers of a body due to tangential force (Shear Force) is called Shear Strain.<\/p>\n

        Question 6.
        \nDefine Modulus of elasticity.
        \nAnswer:
        \nModulus of elasticity is defined as the ratio of stress to strain.<\/p>\n

        Question 7.
        \nState Hooke\u2019s law.
        \nAnswer:
        \nWithin elastic limit, stress is directly proportional to strain.<\/p>\n

        Question 8.
        \nMention the expression for Young\u2019s modulus in the case of a stretched string.
        \nAnswer:
        \nYoung\u2019s modulus in case of a stretched string is Y = \\(\\frac{\\mathrm{FL}}{\\pi r^{2}(\\Delta \\mathrm{L})}\\)
        \nF – Force acting on the wire,
        \nL – Initial length of the wire
        \n\u2206L – change in length of the wire,
        \nr – radius of the wire.<\/p>\n

        Question 9.
        \nWhich is more elastic, rubber or iron?
        \nAnswer:
        \nIron.<\/p>\n

        Question 10.
        \nWhat are anisotropic materials?
        \nAnswer:
        \nThe materials which have different physical properties along different directions are called anisotropic materials.<\/p>\n

        Question 11.
        \nName 2 main groups of solids.
        \nAnswer:
        \nAmorphous and Crystalline.<\/p>\n

        \"KSEEB<\/p>\n

        Question 12.
        \nWhat are glass solids?
        \nAnswer:
        \nGlassy solids are amorphous solids.<\/p>\n

        Question 13.
        \nWhat is the Bulk modulus of an incompressible object?
        \nAnswer:
        \nSince its incompressible strain is 0 and Bulk modulus is infinity.<\/p>\n

        Question 14.
        \nShow graphically stress v\/s strain curve for a Brittle material.
        \nAnswer:
        \n\"1st<\/p>\n

        Question 15.
        \nIf the length of the wire Is made 2 times, what Is the effect on the Increase in length under same load?
        \nAnswer:
        \n\u2206l \u221d l. So, increase in length is also doubled.<\/p>\n

        1st PUC Physics Mechanical Properties of Solids Two Marks Questions and Answers<\/h3>\n

        Question 1.
        \nIf a wire is replaced by another similar wire but of half the diameter. What is the effect on<\/p>\n

          \n
        1. Increase in the length for a given load,<\/li>\n
        2. Max load It can sustain.<\/li>\n<\/ol>\n

          Answer:
          \nWe know that, \\(\\frac{\\Delta L}{L}=\\frac{F}{A Y}\\)
          \n1.
          \n\"1st
          \nThe increase in the length (\u2206L)is 4 times the original wire.<\/p>\n

          2.\u00a0 We know that, F = F=A \\(\\left(\\frac{\\mathrm{Y} \\Delta \\mathrm{L}}{\\mathrm{L}}\\right)\\) Thus F \u221d A and hence maximum load it can sustain becomes 1\/4th <\/sup>of the original wire.<\/p>\n

          Question 2.
          \nWhat is a perfectly elastic and inelastic body?
          \nAnswer:
          \nIf a body completely regains its natural shape or size after the removal of external forces, it is called a perfectly elastic body.
          \nEg: rubber.
          \nIf the body does not regain its original size or shape it is called inelastic body or plastic body.<\/p>\n

          Question 3.
          \nIdentical springs made of steel and aluminium are equally stretched. On which more work will have to be done? Why?
          \nAnswer:
          \nWork done for steel spring is more because steel is more elastic than aluminium, (alternatively: Spring constant of steel is more than that of aluminium).<\/p>\n

          \"KSEEB<\/p>\n

          Question 4.
          \nWhy does a cycle tube burst in summer?
          \nAnswer:
          \nIn summer as the temperature increases, the pressure increases inside the tube This pressure would be enough to cross the breaking point of the tyre making it to burst.<\/p>\n

          Question 5.
          \nWhen the pressure on a sphere increases by 50 atm then its volume decreases by 0.02%. Find Bulk modulus.
          \nAnswer:
          \nP = 50 atm
          \nP = 50 atm \u2245 50 \u00d7 1.01 \u00d7 105<\/sup>
          \n\\(\\frac{\\Delta V}{V}=\\frac{0.02}{100}=2 \\times 10^{-4}\\)
          \nBulk modulus= \\(\\frac{\\mathrm{P}}{(\\Delta \\mathrm{V} \/ \\mathrm{V})}=\\frac{5.05 \\times 10^{6}}{2 \\times 10^{-4}}\\)
          \n= 2.525 \u00d7 1010<\/sup>N\/m2<\/sup><\/p>\n

          Question 6.
          \nWhich is more elastic, rubber or Steel? Why?
          \nAnswer:
          \nFor a given force, rubber extends more than steel, i.e., Steel is more elastic than rubber. This is because for the same amount of extension. Steel requires more force. For the same force \u2206l \u221d Y, and F \u221d Y, for same extension Ysteel<\/sub> > Yrubber.<\/sub><\/p>\n

          Question 7.
          \nWhat is elastic Limit?
          \nAnswer:
          \nElastic limit is the value of applied force beyond which the material (body) does not come back to its original shape when the applied force is reduced.<\/p>\n

          Question 8.
          \nPlot load v\/s extension curve for a substance on the graph show,<\/p>\n

            \n
          1. Yield point<\/li>\n
          2. Breaking point<\/li>\n
          3. Elastic limit<\/li>\n
          4. Crushing point.<\/li>\n<\/ol>\n

            Answer:
            \n\"1st
            \n\"1st<\/p>\n

            Question 9.
            \nThe Young ‘s modulus of a wire of length L and radius ‘r’ is Y. If length is reduced to L\/3 and Radius r\/4. What will be Its ratio of extension?
            \nAnswer:
            \nYoung\u2019s modulus is given by
            \nY = \\(\\frac{F}{A} \\frac{l}{\\Delta}\\) \u21d2 \u2206l = \\(\\frac{F}{A} \\frac{l}{Y}\\)
            \nSo, \u2206l \u221d l and \u2206l \u221d \\(\\frac{1}{A}\\) and A \u221d r2<\/sup>
            \n\"1st<\/p>\n

            Question 10.
            \nThe Young\u2019s modulus of a material is Y. It has an area A and length \u2018L’. Now if a is made \\(\\frac{A}{16}\\) and length L\/2. What is the new Young\u2019s modulus?
            \nAnswer:
            \nEven though the value of Young\u2019s modulus is given by, Y = \\(\\frac{F}{A} \\cdot \\frac{l}{\\Delta l}\\). It is only material dependent. It is constant for the given wire irrespective of its dimension. So, the new value of Young\u2019s modulus is Y.<\/p>\n

            \"KSEEB<\/p>\n

            Question 11.
            \nThe length of a metal is L1<\/sub> when a force is F1<\/sub> is applied on it and its length is L2<\/sub> when applied force is F2<\/sub>. Find the original length of wire.
            \nAnswer:
            \nWe know that Y = \\(\\frac{\\mathrm{F} l}{\\mathrm{A} \\Delta l}\\)
            \nLet original length be L. Then,
            \nL1<\/sub> = L + \u2206l1<\/sub>
            \nL2<\/sub> = L + \u2206l2<\/sub>
            \nSince, Y is a constant,
            \n\"1st<\/p>\n

            Question 12.
            \nExplain the terms Young\u2019s modulus and elastic fatigue.
            \nAnswer:
            \nYoung\u2019s modulus is defined as the ratio of longitudinal stress to longitudinal strain.
            \nY = \\(\\frac{(F \/ A)}{(\\Delta l \/ l)}\\)
            \nElastic Fatigue is the time (delay) required for a material to regain its original position on removing the deforming force.<\/p>\n

            1st PUC Physics Mechanical Properties of Solids Three Marks Questions and Answers<\/h3>\n

            Question 1.
            \nWhen a wire is stretched by a certain force its elongation is \u2018a\u2019. If the second wire having double-radius and 4-time length. Find its elongation.
            \nAnswer:
            \n\"1st
            \n= 4 \u00d7 4 a
            \n\u2206l2<\/sub> = 16 a.<\/p>\n

            Question 2.
            \nDefine stress and strain and derive their units. Write one limitation of Hooke\u2019s Law.
            \nAnswer:
            \nStress is the deforming force experience by the body per unit area.
            \nStress = \\(\\frac{F}{A}\\)
            \nUnits = \\(\\frac{(N)}{\\left(m^{2}\\right)}\\) = Nm-2<\/sup>
            \nStrain is the change in the considered quantity (l,v, A, \u03b8…..) per unit of its original value.
            \nStrain = \\(\\frac{\\Delta x}{x}\\)
            \nIt is unitless.
            \nHooke\u2019s Law is valid only for small deformation i.e., in its limit.<\/p>\n

            1st PUC Physics Mechanical Properties of Solids Five Marks Questions and Answers<\/h3>\n

            Question 1.
            \nThe Stress-Strain graph of a metal wire is shown in figure up to pint E. The wire returns to its original state 0 along EPO when it is gradually unloaded point B corresponds to the fracture of wire.
            \n\"1st<\/p>\n

              \n
            1. Up to what point of the curve Hooke\u2019s Law is obeyed?<\/li>\n
            2. Which point on the curve corresponds to yield point or elastic limit.<\/li>\n
            3. Indicate the Elastic region and plastic region.<\/li>\n
            4. Describe what happens when the wire Is loaded up to the stress corresponding to point A and then unloaded gradually. In particular to the dotted line.<\/li>\n
            5. What is peculiar about portion of curve from C to B? Upto what stress can be applied without causing fracture?<\/li>\n<\/ol>\n

              Answer:<\/p>\n

                \n
              1. Since the graph is linear till point E. Hooke\u2019s Law is obeyed till point E.<\/li>\n
              2. The point E corresponds to the elastic limit because till there the body returns to its original state along EPO.<\/li>\n
              3. Elastic – O to E; Plastic – E to B<\/li>\n
              4. Up to point E an almost linear increase in strain with stress. After point E, if it is loaded till A, the strain increases faster for some increase in stress. But on unloading the body returns along line AO to a permanent strain (change in shape) of O’.<\/li>\n
              5. In the curve C to B, the body undergoes a strain even if it is being unloaded and then fractures. To present fracture the value of stress has to be just less than that of C.<\/li>\n<\/ol>\n

                Question 2.
                \nExplain different types of strain.
                \nAnswer:
                \nThere are 3 types of strain.
                \n1. Longitudinal strain :
                \nIt is defined as the ratio of change in length produced (dl) to the original length (L) of a thin rod.
                \nLongitudinal strain = \\(\\frac{d l}{L}\\)<\/p>\n

                2. Shearing strain :
                \nIt is defined as the ratio of displacement of a surface on which stress is acting to the height of the surface.
                \n\"1st
                \nShearing strain \u03b8 = \\(\\frac{x}{h}\\)
                \n\\(=\\frac{\\text { displacement }}{\\text { heightof the surface }}\\)<\/p>\n

                3. Bulk strain :
                \nIt is defined as the ratio of change in volume (dV) to the original volume (V).
                \nBulk strain = \\(\\frac{\\mathrm{d} \\mathrm{v}}{\\mathrm{v}}\\).<\/p>\n

                \"KSEEB<\/p>\n

                Question 3.
                \nExplain the different types of elastic constant.
                \nAnswer:
                \n1. Young\u2019s modulus (y):
                \nIt is defined as the ratio of longitudinal stress to the longitudinal strain.
                \n\"1st
                \nwhere F – Longitudinal force,
                \nL – original length,
                \ndl – change in length,
                \nA – area of cross-section of the wire.<\/p>\n

                2. Rigidity modulus (\u03b7):
                \nIt is defined as the ratio of shearing stress to the shearing strain.
                \n\"1st<\/p>\n

                where F – tangential force,
                \nA – area of the surface,
                \nh – height of the surface,
                \nx – displacement of the surface<\/p>\n

                3. Bulk modulus (K) :
                \nIt is defined as the ratio of bulk stress to the bulk strain. Bulk modulus is given by,
                \n\"1st
                \nwhere F\/A = P – pressure, V is the original volume and dV is the change in volume.<\/p>\n

                1st PUC Physics Mechanical Properties of Solids Numerical Problems Questions and Answers<\/strong><\/p>\n

                Question 1.
                \nWhat is the percentage increase in length of a wire of diameter 12.5 mm stretched by a weight of 1000 kg \u03c9t ?
                \n[Given Y = 12.5 \u00d7 1011<\/sup> dyne\/sq.cm]
                \nAnswer:
                \nArea = \\(\\frac{\\pi d^{2}}{4}=\\frac{\\pi(1.25)^{2}}{4} \\mathrm{cm}^{2}\\)
                \nF = 1000 kg \u03c9t
                \n= 1000 \u00d7 103<\/sup> \u00d7 980
                \nF = 9.8 \u00d7 108<\/sup> dyne
                \nWe know that,
                \n\"1st
                \n\\(=\\frac{9.8 \\times 4 \\times 10^{-3}}{\\pi \\times(12.5)(1.25)^{2}}\\)
                \n= 0.638 \u00d7 10-3<\/sup>
                \nPercentage increase = 0.064%.<\/p>\n

                Question 2.
                \nA cube of side 3 cm Is subjected to shearing force. The top of cube is sheared through 0.012 cm with respect to bottom face. If cube is made of a material of shear modulus 2.1 \u00d7 1010<\/sup>Nm-2<\/sup> then find<\/p>\n

                  \n
                1. Shearing strain<\/li>\n
                2. Shearing stress<\/li>\n
                3. Force applied.<\/li>\n<\/ol>\n

                  Answer:
                  \n1. Shearing strain
                  \n\\(=\\frac{\\Delta l}{l}=\\frac{0.012 \\mathrm{cm}}{3 \\mathrm{cm}}\\)
                  \n= 0.004<\/p>\n

                  2. Shearing stress
                  \n= Shearing strain \u00d7 Shear modulus
                  \nShearing stress = 0.004 \u00d7 2.1 \u00d7 1010<\/sup>Nm-1<\/sup>
                  \n= 8.4 \u00d7 107<\/sup>Nm-2<\/sup><\/p>\n

                  3. Shearing stress = \\(\\frac{\\text { S.Force }}{\\text { Area }}\\)
                  \nShearing force = Shearing stress \u00d7 Area
                  \n= 8.4 \u00d7 107<\/sup> \u00d7 (3 \u00d7 10-1<\/sup>)2<\/sup>
                  \n= 7.56 \u00d7 104<\/sup> N
                  \n= 75.6 KN.<\/p>\n

                  Question 3.
                  \nA composite wire of materials A and B are of lengths 3.2 m and 2.4 m, and areas 1cm2<\/sup> and 1cm2<\/sup> respectively. If the extension observed is 1.2 mm. Find the force ‘F’. Given YA<\/sub> = 1.6 \u00d7 1011<\/sup> Nm-2<\/sup> and YB<\/sub> = 2.4 \u00d7 101<\/sup> Nm-2<\/sup>
                  \nAnswer:
                  \n\"1st
                  \nWe know that Y = \\(\\frac{F}{A} \\times \\frac{l}{\\Delta L}\\)
                  \nLet the individual extension be \u2206lA<\/sub> and \u2206lB<\/sub>
                  \n\u2206LA<\/sub> = \\(F\\left(\\frac{L_{A}}{A_{A} Y_{A}}\\right)\\); \u2206LB<\/sub> = \\(F\\left(\\frac{L_{B}}{A_{B} Y_{B}}\\right)\\)
                  \nGiven \u2206LA<\/sub> + \u2206LB<\/sub> = 1.2 mn
                  \n\"1st
                  \n= 1.2 \u00d7 10-4<\/sup>
                  \nF(3 \u00d7 10-7<\/sup>) = 1.2 \u00d7 10-4<\/sup>
                  \nF = 400 N<\/p>\n

                  Question 4.
                  \nA cube of 5 cm has its upper face displaced by 0.2 cm by a force of 8 N. Find Shear Modulus, Stress and Strain.
                  \nAnswer:
                  \nShear stress = \\(\\frac{F}{A}=\\frac{8 N}{25 \\times 10^{-4} m^{2}}\\)
                  \n= 3200 Nm-2<\/sup>
                  \nShear Strain = \\(\\frac{\\Delta l}{l}\\)
                  \nBut \u2206l = 0.2 cm
                  \nl = 5 cm
                  \nShear strain = \\(\\frac{0.2 \\mathrm{cm}}{5 \\mathrm{cm}}\\)
                  \n= 0.04
                  \nShear Modulus = \\(\\frac{\\text { Shear Stress }}{\\text { Shear Strain }}\\)
                  \n\\(=\\frac{3200}{0.04}\\)
                  \n= 80 \u00d7 103<\/sup>Nm-2<\/sup> or 80 kPa.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 5.
                  \nA rubber string 10 m long is suspended from a rigid support at one end. Calculate extension of the string due to its own weight. The density of rubber is 1.5 \u00d7 103<\/sup> kg\/m3<\/sup> and Yrubber<\/sub> = 5 \u00d7 106<\/sup>Nm-2<\/sup>.
                  \nAnswer:
                  \nThe weight at any point x from below is, \\(\\frac{\\mathrm{M}}{10}\\) (10 – x) where M is mass of string.
                  \nSo, Force at any point dx is
                  \nF = \\(\\frac{\\mathrm{Mg}}{l}\\) (l – x)dx
                  \nWe know that,
                  \n\"1st
                  \nNet extension =
                  \n\"1st
                  \nSo, Net extension of the rubber due to its own weight = \\(\\frac{1.5 \\times 10^{3} \\times 10 \\times 100}{2 \\times 5 \\times 10^{6}}\\)
                  \n= 0.15 m.<\/p>\n

                  Question 6.
                  \nA metal rod is tied vertically. If the breaking stress is 5.4 \u00d7 109<\/sup> Nm-2<\/sup> and density 5.4 \u00d7 103<\/sup> kg\/m3<\/sup>. Find the maximum length of rod that can be held without breaking.
                  \nAnswer:
                  \nStress = \\(\\frac{F}{A}\\)
                  \n= \\(\\frac{\\mathrm{Mg}}{\\mathrm{A}}\\) [\u2235 \\(\\frac{M}{A L}\\) = \u03c1]
                  \n= \u03c1gL
                  \nIf \u03c1gL > Maximum stress that it can sustain than the rod break.
                  \nSo, \u03c1gL < 5.4 \u00d7 109<\/sup> Nm-2<\/sup>
                  \nl< \\(\\frac{5.4 \\times 10^{9}}{5.4 \\times 10^{3} \\times\\left(10 m s^{-1}\\right)}\\)
                  \nl < 105<\/sup> m
                  \nMaximum length of wire is 105<\/sup> m or 100 km..<\/p>\n

                  Question 7.
                  \nConsider a tapered wire as shown. Find the extension if Y is its Young\u2019s modulus.
                  \n\"1st
                  \nAnswer:
                  \nConsider an element dx which is at a distance x from the end with r, as radius. Tfhe change in length in this portion is
                  \ndl = \\(\\left(\\frac{F}{A}\\right) \\frac{d x}{Y}\\)
                  \nFrom figure we can write radius at a distance x is, rx<\/sub> = r1<\/sub> + xtan\u03b8
                  \n\u21d2 A = \u03c0(rx<\/sub>)2<\/sup>
                  \n= \u03c0 (r1<\/sub> + xtan\u03b8)2<\/sup>
                  \n\"1st
                  \nTotal extension = \\(\\frac{F\\left(r_{2}-r_{1}\\right)}{Y \\cdot r_{1} \\cdot r_{2} \\tan \\theta}\\)
                  \n\\(=\\frac{\\mathrm{FL} \\tan \\theta}{\\mathrm{Y} \\mathrm{r}_{1}-\\mathrm{r}_{2} \\tan \\theta}\\)
                  \nTotal extension = \\(\\frac{F \\cdot L}{Y r_{1} r_{2}}\\)
                  \n[\u2235 tan\u03b8 = \\(\\frac{r_{2}-r_{1}}{L}\\)]<\/p>\n","protected":false},"excerpt":{"rendered":"

                  You can Download Chapter 9 Mechanical Properties of Solids Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids 1st PUC Physics Mechanical Properties of …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10765"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=10765"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10765\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=10765"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=10765"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=10765"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}