{"id":27394,"date":"2020-06-30T15:18:33","date_gmt":"2020-06-30T09:48:33","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=27394"},"modified":"2021-07-02T15:07:10","modified_gmt":"2021-07-02T09:37:10","slug":"2nd-puc-physics-previous-year-question-paper-march-2020","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-physics-previous-year-question-paper-march-2020\/","title":{"rendered":"2nd PUC Physics Previous Year Question Paper March 2020"},"content":{"rendered":"

Students can Download 2nd PUC Physics Previous Year Question Paper March 2020, Karnataka 2nd\u00a0PUC Physics Model Question Papers with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Physics Previous Year Question Paper March 2020<\/h2>\n

Time: 3 Hrs 15 Min
\nMax. Marks: 70<\/p>\n

Instructions:<\/p>\n

    \n
  1. All parts are compulsory.<\/li>\n
  2. Answer without relevant diagram\/figure\/circuit wherever necessary will not carry any marks.<\/li>\n
  3. Numerical problems solved without writing the relevant formulae carry no marks.<\/li>\n<\/ol>\n

    Part – A<\/span><\/p>\n

    I. Answer ALL the following questions: (10 \u00d7 1 = 10)<\/span><\/p>\n

    Question 1.
    \nWrite the SI unit of electric flux.
    \nAnswer:
    \nVm.<\/p>\n

    Question 2.
    \nGraphically represent the variation of resistivity of a semiconductor with absolute temperature.
    \nAnswer:
    \n\"2nd<\/p>\n

    \"2nd<\/p>\n

    Question 3.
    \nGive any one use of electromagnet.
    \nAnswer:
    \nElectromagnets are used in calling bells, electromagnetic cranes to displace huge or iron material.<\/p>\n

    Question 4.
    \nWhat is the significance of Lenz’s law?
    \nAnswer:
    \nLenz’s law is used to find polarity induced emf.<\/p>\n

    Question 5.
    \nHow does capacitive reactance vary with frequency?
    \nAnswer:
    \nXc<\/sub> \u221d \\(\\frac{1}{2 \\pi f}\\), Xc<\/sub> \u221d \\(\\frac{1}{f}\\).<\/p>\n

    Question 6.
    \nArrange the following electromagnetic waves in ascending order of their wavelength. Radio waves, Gamma rays, Infrared waves, X-rays.
    \nAnswer:
    \nGamma rays, X-rays, Infrared waves, Radio waves.<\/p>\n

    Question 7.
    \nWhy does the sky appear blue?
    \nAnswer:
    \nThe sky appears blue because of scattering of light by atmospheric particles.<\/p>\n

    Question 8.
    \nMention a method to increase the resolving power of a microscope.
    \nAnswer:
    \nR.P. = \\(\\frac{2 n \\sin \\theta}{\\lambda}\\)
    \n\u2234 R.P can be increased by placing a medium of refractive index ‘n’ between the object and the objective.<\/p>\n

    Question 9.
    \nWrite the nuclear reaction equation for alpha decay of \"2nd
    \nAnswer:
    \n\"2nd<\/p>\n

    \"2nd<\/p>\n

    Question 10.
    \nDraw the logic symbol of NOR gate.
    \nAnswer:
    \n\"2nd<\/p>\n

    Part – B<\/span><\/p>\n

    II. Answer any FIVE of the following questions: (5 \u00d7 2 = 10)<\/span><\/p>\n

    Question 11.
    \nState and explain Coulomb’s law in electrostatics.
    \nAnswer:
    \nCoulomb\u2019s law:
    \n\u2018The Electrostatic force of attraction or repulsion between two stationary point charges separated in free space or air is directly proportional to the product of magnitudes of charge on them and inversely proportional to the square of the distance between them and the force acts along the line joining the two point charges\u2019.
    \n\"2nd
    \n\"2nd<\/p>\n

    Question 12.
    \nA parallel plate capacitor with air between the plates has capacitance C. What will be the capacitance if<\/p>\n

      \n
    1. The distance between the plates is doubled?<\/li>\n
    2. The space between the plates is filled with a substance of dielectric constant 5?<\/li>\n<\/ol>\n

      Answer:<\/p>\n

        \n
      1. Since C \u221d \\(\\frac{1}{d}\\), capacitance gets halved.<\/li>\n
      2. Since C \u221d \u2208r<\/sub> , capacitance increases 5 times.<\/li>\n<\/ol>\n

        Question 13.
        \nWrite two limitations of Ohm’s law.
        \nAnswer:<\/p>\n

          \n
        1. Ohm’s law cannot be applied for non-ohmic devices such as semiconductors, electrolytes of a cell, discharge tubes, thermionic valve devices.<\/li>\n
        2. Ohm’s law cannot be applied for conductors for varying temperatures, for temperatures near absolute temperature and for higher temperatures.<\/li>\n<\/ol>\n

          \"2nd<\/p>\n

          Question 14.
          \nIn a region, an electric field \\(\\overline{\\mathbf{E}}\\) = 5 \u00d7 103<\/sup> \\(\\hat{\\mathbf{j}}\\) NC-1<\/sup> and a magnetic field of \\(\\overline{\\mathbf{B}}\\) = 0.1 \\(\\hat{\\mathbf{k}}\\) T are applied. A beam of charged particles are projected along X-direction. Find the velocity of charged particles which move undeflected in this crossed fields.
          \nAnswer:
          \n\"2nd<\/p>\n

          Question 15.
          \nDefine \u2018retentivity\u2019 and \u2018coercivity\u2019.
          \nAnswer:
          \n1. Retentivity:
          \nThe property of a material to retain the magnetism even after the removal of the magnetising field is known as retentivity.<\/p>\n

          2. Coercivity:
          \nCoercivity refers to the property of a material which shows the tendency of opposition to the removal of residual magnetism in the material.<\/p>\n

          Question 16.
          \nMention two sources of energy loss in transformer.
          \nAnswer:
          \nSources of energy loss in transformers are due to<\/p>\n

            \n
          1. Copper loss<\/li>\n
          2. Magnetic flux linkage loss<\/li>\n
          3. Hysteresis loss<\/li>\n
          4. Eddy current loss.<\/li>\n<\/ol>\n

            Question 17.
            \nWhat is displacement current? Give the expression for it.
            \nAnswer:
            \nThe current in a capacitor is called displacement current.
            \n\"2nd<\/p>\n

            Question 18.
            \nAn alpha particle, a proton and an electron are moving with equal kinetic energy. Which one of these particles has the longest de Broglie wavelength? Give reason.
            \nAnswer:
            \nFor same K.E., \u03bb \u221d \\(\\frac{1}{\\sqrt{\\mathrm{m}}}\\)
            \n\u2235 m\u03b1<\/sub> > mp<\/sub> > me<\/sub>; \u03bbe<\/sub> > \u03bbp<\/sub> > \u03bb\u03b1<\/sub>
            \nHence electrons have longest wavelength.<\/p>\n

            Part – C<\/span><\/p>\n

            III. Answer any FIVE of the following questions: (5 \u00d7 3 = 15)<\/span><\/p>\n

            Question 19.
            \nEstablish the relation between electric field and electric potential.
            \nAnswer:
            \nConsider a point charge ‘+q’. Let ‘P’ be a point initially at infinity. Let ‘A’ be a point inside the . field region. Let ‘+1C’ be a unit positive charge moved from the point at infinity to point ‘A’. Let ‘A’ be at a distance of ‘r’ from the given point charge. Let ‘B’ be another point at distance of ‘dr’, from’A’towards the point charge +q.
            \n\"2nd
            \nThe work done in moving positive test charge from Ato B, against the force of repulsion is dW = -Fdr. The -ve sign indicates that the work is done against the direction of the force and dr is the displacement of +1C of test charge in the direction opposite to the electric field.
            \nBut electric potential dV = \\(\\frac{d W}{+q_{0}}\\) and electric field
            \nE = \\(\\frac{\\mathrm{F}}{+q_{0}}\\)
            \nWhen q0<\/sub> = +1C then dW = dV, E = F
            \nHence dV = -Edr or E = \\(-\\frac{d \\mathrm{V}}{d r}\\)
            \ni.e., Electric field intensity at a point is the negative potential gradient at that point and electric field intensity is in the direction of decreasing electrostatic potential.<\/p>\n

            \"2nd<\/p>\n

            Question 20.
            \nDerive the expression for the energy stored in a charged capacitor?
            \nAnswer:
            \nLet \u2018C\u2019 be the capacitance of a parallel plate capacitor. By definition C = \\(\\frac{Q}{V}\\). By definition amount of work done to raise the charge by \u2018dq\u2019 is dW = Vdq.
            \ni.e., dW= \\(\\frac{q}{C}\\)dq
            \nwhere \u2018q\u2019 is the instantaneous value of charge on the conductor.
            \nTotal work done in charging the capacitor is given by,
            \n\"2nd
            \nor W = \\(\\frac{1}{2} \\frac{Q^{2}}{C}\\), amount of work done is stored in the form of energy.
            \nTherefore E = \\(\\frac{1}{2}\\left(\\frac{Q^{2}}{C}\\right)\\)
            \nUsing Q = CV, E = \\(\\frac{1}{2}\\)CV2<\/sup>, where \u2018 V is the maximum voltage supplied to a capacitor.<\/p>\n

            Question 21.
            \nGive the principle of cyclotron and draw the neat labelled schematic diagram of cyclotron.<\/p>\n

            Question 22.
            \nMention three properties of diamagnetic materials.<\/p>\n

            Question 23.
            \nArrive at the relation between focal length and radius of curvature of a spherical concave mirror.
            \nAnswer:<\/p>\n

              \n
            1. The distance between the pole and the principle focus of the mirror is called focal length (f).<\/li>\n
            2. f = \\(\\frac{\\mathrm{R}}{2}\\) where R – radius of curvature of a spherical mirror.<\/li>\n<\/ol>\n

              Question 24.
              \nUsing Huygen’s principle, show that the angle of incidence is equal to the angle of reflection when a plane wavefront is reflected by a plane surface.<\/p>\n

              Question 25.
              \nDefine work function. Write Einstein’s photoelectric equation and explain the terms.
              \nAnswer:
              \nWork function:
              \nThe minimum energy which is just enough to liberate electrons from a given surface is called its photo electric work function.
              \nhv = w + \\(\\frac{1}{2}\\)mv2<\/sup>, E = hv
              \nenergy of incidence photon.
              \nWhere W – work function; \\(\\frac{1}{2}\\) mv2<\/sup> = maximum kinetic energy of photo electrons<\/p>\n

              \"2nd<\/p>\n

              Question 26.
              \nGive three differences between intrinsic and extrinsic semiconductors.
              \nAnswer:<\/p>\n\n\n\n\n\n\n
              Intrinsic Semiconductors<\/td>\nExtrinsic Semiconductors<\/td>\n<\/tr>\n
              1. These are semiconductors without doping.<\/td>\nThese are semiconductors with doping.<\/td>\n<\/tr>\n
              2. Number density of electrons = Number density of holes<\/td>\nNumber density of electrons \u2260 Number density of holes.<\/td>\n<\/tr>\n
              3. Resistivity is large<\/td>\nResistivity is less than those of intrinsic semiconductors.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Part – D<\/span><\/p>\n

              IV. Answer any TWO of the following questions: (2 \u00d7 5 = 10)<\/span><\/p>\n

              Question 27.
              \nDerive the expression for conductivity of a material: \u03c3 = \\(\\frac{\\mathbf{n} \\mathrm{e}^{2} \\tau}{\\mathbf{m}}\\); where the terms have their usual meaning.<\/p>\n

              Question 28.
              \nObtain the expression for the force between two straight long parallel conductors carrying current. Hence define ‘ampere’.
              \nAnswer:
              \nX and Y are two long straight parallel conductors carrying currents I1<\/sub> and I2<\/sub> respectively, and placed close to each other. d is the separation between the two conductors and L is the length of the conductors.
              \n\"2nd
              \nThe magnetic field at any point on the conductor Y due to current I1<\/sub> in the conductor X, is given by
              \n\"2nd
              \nActs in a direction perpendicular to the plane containing the two conductors. The conductor Y which carries current I2<\/sub>, experiences a mechanical force due to B1<\/sub> acting on it and this force is given by F1<\/sub> = B1 <\/sub>I2<\/sub> L sin \u03b8.
              \n\"2nd
              \nAccording to Fleming\u2019s left hand rule, the direction of the force F1<\/sub> on X is perpendicular to B2<\/sub> and is towards the conductor Y.<\/p>\n

              The magnetic field at any point on the conductor X due to the current I2<\/sub> in y, is given by
              \n\"2nd
              \nActs on X and opposite to B1<\/sub>. The mechanical force acts on X due to B2<\/sub> is I1<\/sub> L sin \u03b8.
              \n\"2nd
              \nAccording to Fleming\u2019s left hand side rule, the direction of the force F2<\/sub> on X is perpendicular to X and it is towards the conductor y if the current I1<\/sub> is inwards (or away from the conductor Y if the current I2<\/sub> is outwards).<\/p>\n

              The Force F1<\/sub> acting on a certain length of the conductor Y due to the current in the conductor X is equal in magnitude to the force F2<\/sub> acting on the same length of X due to the current in conductor Y.<\/p>\n

              If the two conductors carry the currents in the same direction (parallel currents) then the forces attract each other. If the two conductors carry the currents in the opposite directions (anti parallel currents), then they are found to repel each other because the two forces act away from each other.
              \nThe force per unit length on each conductor is
              \n\"2nd
              \nOne ampere of current can be defined as that constant current which when maintained through each of the two infinitely long straight parallel conductors of negligible area of cross section in the same direction placed 1 metre apart in vacuum, causes an attractive force of 2 \u00d7 10-7<\/sup> N m-1 <\/sup>length on each conductor.<\/p>\n

              \"2nd<\/p>\n

              Question 29.
              \nWith the help of a labelled diagram, derive the expression for instantaneous emf induced in an AC generator.<\/p>\n

              V. Answer any Two of the following questions: (2 \u00d7 5 = 10)<\/span><\/p>\n

              Question 30.
              \nObtain the expression for the fringe width of interference fringes in Young’s Double slit experiment.
              \nAnswer:
              \n\"2nd
              \nLet A and B be two slits separated by a distance \u2018d\u2019. Let \u2018\u03bb\u2019 be the wavelength of light. Let \u2018D\u2019 be the distance between the screen and the double slit.
              \nLet \u2018C\u2019 be a point on BP such that AP \u2248 CP.
              \nPath difference between the two waves reaching \u2018P\u2019 is given by BP – AP = BC = \u03b4.
              \nFrom the \u0394 BFP, BP2<\/sup> = BF2<\/sup> + FP2<\/sup> and from the \u0394 AEP, AP2<\/sup> = AE2<\/sup> + EP2<\/sup>
              \nBP2<\/sup> – AP2<\/sup> = (D2<\/sup> + FP2<\/sup>) – (D2<\/sup> – EP2<\/sup>)
              \ni.e; BP2<\/sup> – AP2<\/sup> = FP2<\/sup> – EP2<\/sup>
              \n\"2nd
              \n(BP – AP) (BP + AP) = 2(2x.\\(\\frac{d}{2}\\))
              \nFor a point ‘p’ close to ‘O’,
              \nBP \u2248 AP = D
              \n(BP – AP) (2D) = 2x.d.
              \nBut (BP – AP) = BC = \u03b4
              \ni.e; \u03b4 = \\(\\frac{x d}{D}\\)
              \nor
              \n\"2nd
              \nFor a constructive interference \u03b4 = n\u03bb.<\/p>\n

              \"2nd
              \nDistance of nth<\/sup> bright fringe from the central bright fringe xn<\/sub> = n\\(\\frac{\\lambda D}{d}\\)
              \nDistance of (n + 1)th<\/sup> bright fringe from the central bright fringe xn+1<\/sub> = (n + 1)\\(\\frac{\\lambda D}{d}\\)
              \nBy definition, Fringe width is the distance between two consecutive bright or dark fringes.
              \nxn+1<\/sub> – xn<\/sub> = \u03b2 = \\(\\frac{\\lambda}{d}\\)(n + 1 – n)
              \ni.e;
              \n\"2nd
              \nFringe width \u03b2 \u221d D, \u03b2 \u221d \u03bb and \u03b2 \u221d \\(\\frac{1}{d}\\).
              \nWe can show that the fringe width between any two dark fringes is also \u03b2 = \\(\\frac{\\lambda D}{d}\\).<\/p>\n

              Question 31.
              \nUsing Bohr’s postulates, derive the expression for the radius of nth<\/sup> stationary orbit of electron in hydrogen atom. Hence write the expression for Bohr radius.
              \nAnswer:
              \nConsider an atom of atomic number Z. The charge on its nucleus is + Ze. Let an electron of mass m and charge \u2018-e\u2019 revolve round the nucleus in a circular orbit.
              \nLet v be its velocity.
              \nThe coulomb\u2019s electrostatic force of attraction between the electron and the nucleus is
              \n\"2nd
              \nThis provides the centripetal force \\(\\frac{m v^{2}}{r}\\) needed for the orbital motion of the electron.
              \n\"2nd
              \nFrom Bohr\u2019s Quantization rule, for the nth<\/sup> orbit,
              \nwe have, mvr = \\(\\frac{n h}{2 \\pi}\\) _______(2)
              \nSquaring equation (2)
              \n\"2nd ______(3)
              \nDividing equation (3) by equation (1)
              \n\"2nd
              \n\"2nd
              \nFor H2<\/sub> atom put Z = 1 and r = \\(\\frac{\\varepsilon_{0} n^{2} h^{2}}{\\pi m e^{2}}\\). For I orbit, put n = 1
              \ni.e; r = \u03b50<\/sub>h2<\/sup>\/ \u03c0me2<\/sup> for the I orbit of hydrogen atom.<\/p>\n

              \"2nd<\/p>\n

              Question 32.
              \nWhat is rectification? Explain the working of a p-n junction diode as a half wave rectifier. Draw the input and output waveforms.
              \nAnswer:
              \nThe process of conversion of ac into pulsating dc is called rectification.
              \nAnswer:
              \nThe circuit of a half wave rectifier using a semiconductor diode is shown in figure below.
              \n\"2nd
              \nThe diode in series with a load resistor is connected to the secondary of a transformer. The AC voltage to be rectified is applied to the primary of the transformer.<\/p>\n

              This produces induced voltage across the secondary of the transformer. During the positive half cycles of transformer output i.e., when A is positive, diode D is forward biased and hence it conducts. During negative half cycles, the diode D is reverse biased and hence does not conduct. Current flows through the diode circuit only during positive half cycles. The current flowing through load resistance RL<\/sub> produces the rectified voltage.<\/p>\n

              VI. Answer any THREE of the following questions: (3 \u00d7 5 = 15)<\/span><\/p>\n

              Question 33.
              \nTwo point charges qA<\/sub> = 5\u03bcC and qB<\/sub> = -5\u03bcC are located at A and B separated by 0.2 m in vacuum.<\/p>\n

                \n
              1. What is the electric field at the midpoint O of the line joining the charges?<\/li>\n
              2. If a negative test charge of magnitude 2nC is placed at O, what is the force experienced by the test charge.<\/li>\n<\/ol>\n

                Answer:
                \n\"2nd
                \nWe know that: Electric field at a point due to a point charge
                \n\"2nd
                \nNet electric field at P = E1<\/sub> + E2<\/sub>
                \n= 2 \u00d7 4.5 \u00d7 106<\/sup> Vm-1<\/sup> along AB.
                \ni.e. E = 9.0 \u00d7 106<\/sup> Vm-1<\/sup> along AB.
                \nForce on q0<\/sub> = -2nC is given by
                \nF = Eq0<\/sub> = – 9 \u00d7 106<\/sup> \u00d7 2 \u00d7 10-9<\/sup> = – 1.8 \u00d7 10-2<\/sup> N along BA.<\/p>\n

                \"2nd<\/p>\n

                Question 34.<\/p>\n

                  \n
                1. Three resistors 3\u2126, 4\u2126 and 12\u2126 are connected in parallel. What is the effective resistance of the combination.<\/li>\n
                2. If the combination is connected to a battery of emf 6V and internal resistance 0.5\u2126, find the current drawn from the battery and terminal potential difference across the battery.<\/li>\n<\/ol>\n

                  Answer:
                  \n1. We know that \\(\\frac{1}{\\mathrm{R}_{\\mathrm{p}}}=\\frac{1}{\\mathrm{R}_{1}}+\\frac{1}{\\mathrm{R}_{2}}+\\frac{1}{\\mathrm{R}_{3}}\\)
                  \n\"2nd<\/p>\n

                  2. We know I =
                  \n\"2nd
                  \nand V = IR = 3 \u00d7 1.5 = 4.5 V<\/p>\n

                  Question 35.
                  \nA series LCR circuit contains a pure inductor of inductance 5.0H, a capacitor of capacitance 20pF and a resistor of resistance 40\u2126.<\/p>\n

                    \n
                  1. Find the resonant frequency of the circuit.<\/li>\n
                  2. Calculate the quality factor (Q-factor) of the circuit.<\/li>\n
                  3. What is the impedance at resonant condition?<\/li>\n<\/ol>\n

                    Solution:
                    \nGiven L = 5H, C = 20 \u00d7 10-6<\/sup> F, R = 40\u2126, f0<\/sub> = ?, Q = ?<\/p>\n

                    1. We know that f0<\/sub> =
                    \n\"2nd<\/p>\n

                    2. \u2234 f0<\/sub> = 15.92Hz<\/p>\n

                    3.
                    \n\"2nd
                    \nImpedance of the circuit at resonance = Z = R = 40\u2126.<\/p>\n

                    \"2nd<\/p>\n

                    Question 36.
                    \nAt what angle should a ray of light be incident on the face of an equilateral prism, so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.5
                    \nAnswer:
                    \n\"2nd
                    \nGiven A = 60\u00b0, i2<\/sub> = 90\u00b0, n = 1.5
                    \nr2<\/sub> = C, i1<\/sub> = ?
                    \nWe know that n = \\(\\frac{1}{\\sin C}\\)
                    \ni.e. sinC = \\(\\frac{1}{1.5}=\\frac{2}{3}\\) = 0.6667
                    \nhence C = 41.8\u00b0.
                    \nFor non-syittmetric refraction A = r1<\/sub> + r2<\/sub>
                    \ni.e. r1<\/sub> = A – r2<\/sub> = 60\u00b0 – 41.8\u00b0= 18.2\u00b0
                    \nsince n = \\(\\frac{\\sin \\mathrm{i}_{1}}{\\sin \\mathrm{r}_{1}}\\)
                    \nsin i1<\/sub> = 1.5 \u00d7 sin 18.2\u00b0 = 1.5 \u00d7 0.3123
                    \ni.e. sin i1<\/sub> = 0.4685
                    \nor
                    \n\"2nd
                    \nAngle of incidence = 27.9\u00b0.<\/p>\n

                    \"2nd<\/p>\n

                    Question 37.
                    \nA copper coin has a mass of 63.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. The coin is entirely made of 29<\/sub>63<\/sup>Cu atoms.
                    \nMass of 29<\/sub>63<\/sup>Cu atom = 62.92960 u
                    \nMass of proton = 1.00727 u
                    \nMass of neutron = 1.00866 u
                    \nAvogadro\u2019s number = 6.022 \u00d7 1023<\/sup>.
                    \nAnswer:
                    \nrest mass of constituents:
                    \nrest mass of 29 protons =29 \u00d7 1.00727 = 29.21083 u
                    \nrest mass of 34 neutrons =34 \u00d7 1.00866 = 34.29444 u
                    \nrest mass of 29 electrons = 29 \u00d7 0.00055 = 0.01595 u
                    \n63.52122 u
                    \nMass defect \u0394m = rest mass of constituents – rest mass of atom
                    \ni.e. \u0394m = 63.52122 – 62.92960 = 0.59162 u
                    \nEnergy equivalent of mass defect
                    \nB.E. = 0.59162 \u00d7 931 MeV = 550.79822 MeV per atom
                    \ni.e. E = (550.79822 \u00d7 106<\/sup> \u00d7 1.6 \u00d7 10-19<\/sup>) \u00d7 (6.023 \u00d7 1023<\/sup>) J.
                    \ni.e. E = 5.307 \u00d7 1013<\/sup> J is required to separate all the nucleous in 63.0 g of copper coin.<\/p>\n","protected":false},"excerpt":{"rendered":"

                    Students can Download 2nd PUC Physics Previous Year Question Paper March 2020, Karnataka 2nd\u00a0PUC Physics Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Physics Previous Year Question Paper March 2020 Time: 3 Hrs 15 Min Max. Marks: 70 …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27394"}],"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=27394"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27394\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=27394"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=27394"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=27394"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}