{"id":17780,"date":"2020-01-04T12:22:25","date_gmt":"2020-01-04T06:52:25","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=17780"},"modified":"2021-07-02T15:31:18","modified_gmt":"2021-07-02T10:01:18","slug":"2nd-puc-physics-model-question-paper-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-physics-model-question-paper-2\/","title":{"rendered":"2nd PUC Physics Model Question Paper 2 with Answers"},"content":{"rendered":"

Students can Download 2nd PUC Physics Model Question Paper 2 with Answers, Karnataka 2nd PUC 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 Model Question Paper 2 with Answers<\/h2>\n

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

General Instructions:<\/p>\n

    \n
  1. All parts are compulsory.<\/li>\n
  2. Answers without relevant diagram\/figure\/circuit wherever necessary will not cany any marks<\/li>\n
  3. Direct answers to the Numerical problems without detailed solutions will not carry any 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 field.
    \nAnswer:
    \nNewton per coulomb (N\/C)<\/p>\n

    Question 2.
    \nWhen will the magnetic force on a movi\u00acng charge be maximum in a mag notified?
    \nAnswer:
    \nF = qVBsin\u03b8 , If \u03b8 = 90\u00b0 (Fm<\/sub> – q\u03c5B)
    \nThe magnetic force is maximum only when moving charge is \u22a5r<\/sup>\u00a0to field.<\/p>\n

    Question 3.
    \nWhere on the Earth\u2019s surface is the magnetic dip zero?
    \nAnswer:
    \nAt equator, dip is zero<\/p>\n

    Question 4.
    \nState Curie\u2019s law in magnetism.
    \nAnswer:
    \nThe magnetic susceptability of a para magnetic substance varies inversely to its absolute temperature(T).
    \ni.e x \u03b1 \\(\\frac{\\mathrm{C}}{\\mathrm{T}}\\) (x = \\(\\frac{\\mathrm{C}}{\\mathrm{T}}\\))<\/p>\n

    \"KSEEB<\/p>\n

    Question 5.
    \nWhat is the significance of Lenz\u2019s law?
    \nAnswer:
    \nLaw of conservation of energy.<\/p>\n

    Question 6.
    \nWrite the formula for Malus law.
    \nAnswer:
    \ni = i0<\/sub>cos2<\/sup>\u03b8
    \nWhere I \u2192 is intensity of the light transmitted by the analyser.
    \nI0<\/sub> \u2192 is intensity of the light incident on the analyser.
    \n\u03b8 \u2192 is angle between the pass axes of the analyser & polariser.<\/p>\n

    Question 7.
    \nWhat is the ratio of the nuclear densities of two nuclei having mass numbers in the ratio 1:3?
    \nAnswer:
    \n\"2nd
    \nNuclear density is independent of mass number & is approximately constant.
    \nHence \\(\\frac{P_{1}}{P_{2}}\\) = 1<\/p>\n

    Question 8.
    \nDefine current amplification factor in a common – emitter mode of transistor.
    \nAnswer:
    \nIt is defined as the ratio of change in collector current ( \u0394<\/sub>Ic\u00a0<\/sub>)to the change in base current (\u0394<\/sub>Ib<\/sub>) at constant collector emitter voltage (VCE<\/sub>) when the transistor is in active state
    \n\"2nd<\/p>\n

    Question 9.
    \nWrite the truth table of NAND gate.
    \nTruth table of NAND gate
    \n\"2nd<\/p>\n

    Question 10.
    \nWhy sky wave propagation is not possi\u00acble for waves having frequency more than 30 MHz?
    \nAnswer:
    \nThe sky wave range is short & it is used in short wave broad cast service.<\/p>\n

    Part – B<\/span><\/p>\n

    II. Answer anyfive ofthe following questions. ( 5 \u00d7 3 = 15 )<\/span><\/p>\n

    Question 11.
    \nSketch the electric lines of force due to a point charge q. If i) q<0 and ii) q>0
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 12.
    \nA galvanometer having a coil of resistance 12 \u2126 gives full scale deflection for a current of 4 mA. How can it be converted into a voltmeter of range 0 to 24V?
    \nAnswer:
    \nGiven G = 12 \u2126, Ig<\/sub> = 4 x 10-3<\/sup>A. V = 24V
    \nWe have,
    \n\\(\\mathrm{R}=\\frac{\\mathrm{V}}{\\mathrm{I}_{\\mathrm{B}}}-\\mathrm{G}\\)
    \n\"2nd
    \nR = 5988\u2126
    \nA resistance of 5988 \u2126 must be connected in series with galvanometer.<\/p>\n

    \"KSEEB<\/p>\n

    Question 13.
    \nDistinguish between paramagnetic and ferromagnetic substances.
    \nAnswer:<\/p>\n\n\n\n\n\n\n
    Paramagnetic substances<\/td>\n\u00a0Ferromagnetic Substances<\/td>\n<\/tr>\n
    1. Paramagnetics are feebly attracted by magnets.<\/td>\n1. They are strongly attracted by magnets.<\/td>\n<\/tr>\n
    2. Magnetic permeability is slightly greater than one i.e. Mr>1<\/td>\n2. Relative permeability is greater than 1000<\/td>\n<\/tr>\n
    3. Magnetic susceptability is low & positive.<\/td>\n3. Magnetic susceptability is +ve & large.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Question 14.
    \nWhat is meant by Self Inductance and Mutual Inductance?
    \nAnswer:<\/p>\n

      \n
    1. The phenomena in which an emf is induced in a coil due to change of current through the same coil is known as self-induction.<\/li>\n
    2. The phenomena in which an emf is induced in one coil due to change of current is the neighbouring coil is called mutual induction<\/li>\n<\/ol>\n

      Question 15.
      \nWhat are electromagnetic waves? Write the expression for the velocity of electro magnetic waves in terms of permittivity and magnetic permeability of free space.
      \nAnswer:
      \nThe waves in which there are a sinusoidal variation of electric & magnetic field vectors at right angles to each other & also right angle to the direction of propagation of the wave is called e.m waves.
      \n\u2234 The expression for velocity of e.m. wave is
      \n\\(C=\\frac{1}{\\sqrt{\\mu_{0} \\varepsilon_{0}}}\\) Where \u00b50<\/sub> \u2192 is permeability of free
      \nspace, \u03b50<\/sub> \u2192 is the permittivity of free space.<\/p>\n

      Question 16.
      \nWrite the relation between the path difference and wavelength of light wave used for constructive and destructive interference of light.
      \nAnswer:
      \na) For constructive interference
      \n\u2234 Path difference = 2n\\(\\frac{\\lambda}{2}\\) = n\u03bb
      \nWheve \u03bb \u2192 is wave length of light used
      \nn \u2192 0, 1, 2,
      \nb) For destructive interference :
      \n\u2234 Path difference = (2n+l)\\(\\frac{\\lambda}{2}\\)
      \nWheve \u03bb \u2192 is wave length of light used n \u2192 0, 1, 2,………<\/p>\n

      Question 17.
      \nDefine :
      \ni) photoelectric work function
      \nii) electron volt (ev)
      \ni) Photoelectric work fraction : the minimum energy required to remove an electron from the metal surface is called work function, i.e. w = hV0<\/sub>
      \nWhere h \u2192 is planck\u2019s constant
      \nV0<\/sub> \u2192 is threshold frequency
      \nii) Electron volt is the kinetic energy gained by an electron when it is accelerated through a potential difference of 1 volt.
      \ni.e. lev = 1.602 \u00d7 10-19<\/sup>J.<\/p>\n

      Question 18.
      \nDraw the block diagram of a AM receiver.
      \nAnswer:
      \n\"2nd
      \nx \u2192 Intermediate frequency stage
      \ny \u2192 power amplifier<\/p>\n

      Part – C<\/span><\/p>\n

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

      Question 19.
      \nDerive an expression for potential energy of a system of two charges in the absence of external electric field.
      \n\"2nd
      \nConsider 2 point charges q1<\/sub> & q2<\/sub> are separated by a distance \u2018r\u2019 are as sho in the fig.
      \n\u2234 The change q1<\/sub> is bringing from \u221e to given point A, No work done i.e. w1<\/sub>= 0 |||ly the charge q2<\/sub> is bringing from \u221e to given point \u2018B\u2019 against us field q2<\/sub>
      \n\u2234work is done it is given by
      \ni.e. w2<\/sub> = vq2<\/sub>
      \n\"2nd<\/p>\n

      Question 20.
      \nArrive at an expression for drift velocity.
      \n\"2nd
      \nController a metalic conductor is connected to a battery
      \nLet \\(\\overrightarrow{\\mathrm{E}} \\rightarrow\\) is E.f. setup inside the conductor
      \nm \u2192 is mass of electom
      \ne \u2192 charge of free elector
      \nVd<\/sub> \u2192 is drift velovity
      \n\u03c4 \u2192 is ralaxation time
      \nthe force experienced by an electron in the field is given by \\(\\overrightarrow{\\mathrm{F}}=-\\mathrm{e} \\overrightarrow{\\mathrm{E}}\\)
      \n-Ve sign shows that divertion of \\(\\overrightarrow{\\mathrm{E}}\\) & \\(\\overrightarrow{\\mathrm{F}}\\) are in apposite each other
      \n\"2nd<\/p>\n

      \"KSEEB<\/p>\n

      Question 21.
      \nState and explain Gauss law in magnetism.
      \nAnswer:
      \nThe net magnetic flux through any closed surface is always zero
      \n\\(\\sum \\overrightarrow{\\mathrm{B}} \\cdot \\overrightarrow{\\mathrm{As}}=0\\)
      \n\"2nd
      \nconsider a closed surface
      \nS in a uniform
      \n\\(\\mathrm{M} . \\mathrm{F} \\cdot \\overrightarrow{\\mathrm{B}}\\) Let \\(\\overrightarrow{\\Delta \\mathrm{S}}\\)
      \nbe a small area element of this surface with \\(\\hat{n}\\)
      \nalong its normal
      \n\"2nd
      \nQuestion 22.
      \nDerive the expression for motional emf induced in a conductor moving in a uniform magnetic field.
      \nAnswer:
      \nConsider a straight metallic rod PQ of length \u2018t\u2019 placed in a uniform M.F. \\(\\overrightarrow{\\mathrm{B}}\\). The rod is moved with a velocity \\(\\overrightarrow{\\mathrm{v}}\\) is a direction \u22a5r<\/sup> to
      \n\\(\\overrightarrow{\\mathrm{B}}\\) Let the rod moved through a distance \u2018x\u2019 in time\u2018t\u2019 them the area covered by the rod is A= 1 xx The magnetic flux linked with the rod is
      \n\u03a6 = B.A \u03a6 = B1x
      \n\u2234 The included emf in the rod is
      \n\"2nd
      \nMotional emf (e – Blv)
      \n\u2234 \\(\\mathrm{v}=\\frac{\\mathrm{dx}}{\\mathrm{dt}}\\)<\/p>\n

      Question 23.
      \nWith a diagram, explain the working of a transformer.
      \n\"2nd
      \nAnswer:
      \nAC voltage is applied to the primary, it creates a varying M.F. & hence a changing magnetic flux in the core. Since the secondary coil is magnetically coupled, due to the mutual induction the changing flux causes induced emf in it. Thus, the power is transferred from primary to secondary.<\/p>\n

      Question 24.
      \nWhat is total internal reflection? Mention two applications of optical fibres.
      \nAnswer:
      \nThe phenomena of complete reflection of light at the interface of two optical media when a ray of light travelling in denser medium is called total internal reflection.<\/p>\n

        \n
      • Optical fibre are used in telecommunication<\/li>\n
      • It is used to measure rate of flow of blood.<\/li>\n<\/ul>\n

        Question 25.
        \nWhat are the matter waves? Write the expression for De – Broglie wavelength of a particle and explain the terms.
        \nAnswer:
        \nThe waves associated with material particles in motion is called matter waves. The expression for de Broglie wavelength is \\(\\lambda=\\frac{\\mathrm{h}}{\\mathrm{mv}}\\)
        \nwhere
        \nh \u2192 is Planck’s constant
        \nm \u2192 is mass of moving particle
        \nv \u2192 is the velocity of moving particle.<\/p>\n

        Question 26.
        \nWrite three differences between n-type and p-type semiconductors.
        \nAnswer:
        \nP – type S.C.<\/p>\n

          \n
        1. It is a semiconductor doped with trivalent impurities.<\/li>\n
        2. Majority charge carriers are holes<\/li>\n
        3. The impurit atom is called acceptor impurity.<\/li>\n<\/ol>\n

          n-type S.c<\/p>\n

            \n
          1. It is a semiconductor doped with pentavalent impurities.<\/li>\n
          2. Majority charge carriers are electron<\/li>\n
          3. The impurity atom is called Sonar impurity<\/li>\n<\/ol>\n

            Part – D<\/span><\/p>\n

            IV. Answer any two of the following questions\u00a0 ( 2 \u00d7 5 = 10 )<\/span><\/p>\n

            Question 27.
            \nDerive an expression for the electric field at a point due to an infinitely long thin charged straight wire using Gauss law.
            \nAnswer:
            \nConsider an infinitely long thin straight wive with uniform linear charge q density \u03bb. Let P be a point at \u22a5r<\/sup> distance r from the wire.
            \n\"2nd
            \nTo calculate the E.F \\(\\overrightarrow{\\mathrm{E}}\\) at P, imagine a cylindrical Gaussian surface.
            \n\u2234 The surface area of the] curved part S = 2\u03c0rl
            \nTotal charge enclosed by the Gaussian surface q = \u03bbl
            \nElectric fliix through the end Surfaces of the cylinder is \u03a6 = 0
            \nElectric flux through the curved Surfaces of the cylinder is \u03a62<\/sub> = Ecos\u03b8.s
            \n\u03a62<\/sub> = EX1X2\u03c0rl
            \nThe total electric flux \u03a6 = \u03a61<\/sub> + \u03a62<\/sub>
            \n\u03a6 = 0 + E2\u03c0rl, \u03a62<\/sub>= 2\u03c0rlE …………… 1
            \nA\/C to Gauss law,
            \n\"2nd
            \nfrom (a) and (2)
            \n\"2nd<\/p>\n

            Question 28.
            \nObtain the bridge balanced condition of Wheatstone\u2019s bridge network by applying Kirchhoff\u2019s rules.
            \nAnswer:
            \nWheatstone\u2019s bridge is a device used to determine unknown resistance.
            \n\"2nd<\/p>\n

            It resistances R1<\/sub>, R2<\/sub>, R3<\/sub> & R4<\/sub> a galvanometer of resistance G & a battery with key. Let I is the math current & splits into branch current I1<\/sub> & I2<\/sub> respectively
            \ni.e. I = I1<\/sub> +I2<\/sub>
            \nAt B, I1<\/sub> = I3<\/sub>+ Ig<\/sub>………….(1)
            \nA + D, I2<\/sub> = 4<\/sub> + Ig<\/sub> ………. (2)
            \nApply KVL to the mesh ABDA
            \nI1<\/sub>R1<\/sub> + ig<\/sub>G – I2<\/sub>R2<\/sub> = o ………….. (3)
            \nto the mesh BCDB
            \nI3<\/sub>R3<\/sub> – I4<\/sub>R4<\/sub>– Ig<\/sub>G = 0 ……………… (4)<\/p>\n

            The wheatstones bridge is said to be balanced
            \nwhen no current flowing throught the galvano
            \nmeter i.e. Ig<\/sub> 0 \u2234 eqn (1) is I1<\/sub> = I3<\/sub>
            \neqn(2)is I2<\/sub> = I4<\/sub>
            \neqn (3) is I1<\/sub>R1<\/sub> – I2<\/sub>R2<\/sub> = 0
            \n\u2234 I1<\/sub>R1<\/sub> = I2<\/sub>R2<\/sub> …………. (5)
            \neqn (4) is I3<\/sub>R3<\/sub> – I2<\/sub>R2<\/sub>
            \n\u2234I3<\/sub>R3<\/sub> = I4<\/sub>R4<\/sub> …………………. (6)
            \n\"2nd<\/p>\n

            \"KSEEB<\/p>\n

            Question 29.
            \nTwo straight parallel conductors are placed at certain distance in free space. The direction of current in both the conductors is same. Find the magnitude and direction of the force between them. Hence define ampere.
            \nAnswer:
            \n\"2nd
            \nConsider two infinitely long straight parallel condu ctors x & y carrying a current I1<\/sub> & I2<\/sub> respectively. Let d is the \u22a5r<\/sup> distance between them.
            \nLet I1<\/sub> is the current flowing through x conductor, it produces a M.F. (B1<\/sub>)
            \n\"2nd
            \nNow the conductor Y carrying current I2<\/sub> in the M.F.B, it experiences a mehanical force of length 1 is
            \n\"2nd
            \nThe direction of force can be obtained using ampere’s left-hand rule.
            \n\u2234 The force bn. two ||le conductors carrying currents in the same direction is attractive. ||ly the force bn two ||le conductors carrying current in opposite direction is repulsive.
            \nDefinition of ampere If I1 = I2 = IA, d=lm them F = 2 \u00d7 10-7<\/sup>N.<\/p>\n

            Ampere is the steady current which when flowing through each of 2 infinitely long straight ||le conductors placed in the air at a distance of lm produces a force of 2 \u00d7 10-7<\/sup> N\/m.<\/p>\n

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

            Question 30.
            \nDerive Lens Maker\u2019s formula for a con vex lens.
            \nAnswer:
            \n\"2nd
            \nConsider a thin convex lens of focal length f & R.I (n) placed in air as shown in fig let R1 <\/sub>& R2<\/sub> \u2192 are the of curvatur of the surfaces ABC & ADC of the respectively.
            \no \u2192 Luminous point object on the principle axis.
            \nA ray op invident at p, after refraction, emerges along QI & meet at I on the principal axis.
            \nImage formution takes place in two stages. (1) Refraction at the surface ABC In the observe of ADC, the refracted ray is meet at I1<\/sup>, then
            \n\"2nd
            \n(2) Refraction at the surface ADC the image I1 acts as a virtual object to form a real image at a distance V, then
            \n\"2nd<\/p>\n

            Question 31.
            \nAssuming the expression for radius of the orbit, derive an expression for total energy of an electron in hydrogen atom.
            \nAnswer:
            \nConsider an electron of mass m, charge, e revoking around the nucleus of radius of r. The charge on the nucleus + Te.
            \n\u2234 T.E = KE + P.E
            \nT. E = K + U ………… (1)
            \nFor circulation
            \nCentripetal force = Electrostate force bn nucleus & electron.
            \n\"2nd
            \n\"2nd
            \nThe Potential energy of the electron is the field o nucleus is
            \n\"2nd<\/p>\n

            \"KSEEB<\/p>\n

            Question 32.
            \nWith the help of circuit diagram, explain the working of NPN transistor as a common emitter amplifier.
            \nAnswer:
            \nThe circuit diagram of a CE amplifier using NPN transistor is as shown m fit. The input circuit is forward biased & the output circuit is reverse biased when the ac input signal to be amplified is fed to the base-emitter circuit. The output voltage V0 varies in accordance with the relation, V0<\/sub>=VCE<\/sub> =Vcc<\/sub> – Ic<\/sub>Rc<\/sub>, These variations is the collector voltage VCE<\/sub> uppers as amplified output.<\/p>\n

            During the -t-ve half cycle of ac input signal the forward base of emitter as junction increase Due to this base current IB increase & hence collector current IC increases. As a result of this ICRC increases output voltage VO this indicates that the +ve half cycle of input ac signal voltage is amplified through -ve half cycle.<\/p>\n

            During the -ve half cycle of ac input signal, the forward bias of emitter-base junction decreases. Due to this base current IB<\/sub> decreases & hence collector current IC<\/sub> decreases. As a result of this ICRC decreases the. output voltage V0<\/sub> is +ve. This indicates that the -ve half cycle of input ac signal voltage is amplified through +ve half cycle.<\/p>\n

            Thus, the weak input signal is amplified & output signal is out of phase with the input signal by 180\u00b0
            \n\"2nd<\/p>\n

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

            Question 33.
            \nCharges 2 \u00b5 C, 4 \u00b5 C and 6 \u00b5 C are placed at the three corners A, B and C respectively of a square ABCD of side x metre. Find, what charge must be placed at the fourth corner so that the total potential at the center of the square is 0
            \nAnswer:
            \n\"2nd
            \nIn the \u0394le ABC
            \nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
            \nAC2<\/sup> = x2<\/sup> + x2<\/sup>
            \nAC2<\/sup> = 2x2<\/sup>
            \n\u2234 AC = \u221a2 x,.m
            \n\"2nd
            \n\"2nd<\/p>\n

            Question 34.
            \nA wire having length 2.0m, diameter 1.0 mm and resistivity 1.963 \u00d7 10-8<\/sup> \u2126 m is connected in series with a battery of emf 3 V and internal resistance 1 Calculate the resistance of the wire and current in the circuit.
            \nAnswer:
            \nGiven,
            \nlength (l) = 2m
            \nDiameter (D) = 1mm = 1 \u00d7 10-3<\/sup> m
            \n\u2234 radius (r) = \\(\\frac{\\mathrm{D}}{2}\\) = 0.5 \u00d7 10-3<\/sup> m
            \nresistivity (f) = 1.963 \u00d7 10-8<\/sup>\u2126m
            \nE = 3V Internal senstarce (r) = 1 \u2126
            \nR=? I = ?
            \nWe have,
            \n\"2nd<\/p>\n

            \"KSEEB<\/p>\n

            Question 35.
            \nAn inductor and a bulb are connected in series to an AC source of 220 V, 50 Hz. A current of 11A flows in the circuit and phase angle between voltage and current is \\(\\frac{\\pi}{4}\\) radians. Calculate the impedance and inductance of the circuit.
            \nAnswer:
            \nGiven,
            \nVrms<\/sub> = 220v, f = 50 Hz
            \nIrms<\/sub> = 11A
            \n\u03a6 = \\(\\frac{\\pi}{4}\\) radian, Z=?, L =?
            \n\"2nd<\/p>\n

            Question 36.
            \nIn Young\u2019s double slit experiment while using a source of light of wavelength
            \n4500 \u00c5, the fringe width is 5mm. 1f the distance between the screen and the plane of the slits \u00a1s reduced to half, what should be the wavelength of light to get fringe 4 mm?
            \nAnswer:
            \nGiven,
            \nwavelength (\u03bb) 45OOA\u00b0= 45OO \u00d7 10-10 <\/sup>m
            \nfringe width (w) = 5mm = 5 \u00d7 10-3 <\/sup>m
            \nD = \\(\\frac { D }{ 2 }\\)
            \nwavelength (\u03bb1<\/sup>) = ?
            \nfringe width (w1<\/sup>) = 4mm = 4 \u00d7 10-3 <\/sup>m
            \nwe have, fringe width (w) = \\(\\frac{\\lambda \\mathrm{D}}{\\mathrm{d}}\\)
            \n5 \u00d7 10-3<\/sup> =4500 \u00d7 1010 <\/sup>\u00d7 \\(\\frac { D }{ d }\\)
            \n\"2nd<\/p>\n

            \"KSEEB<\/p>\n

            Question 37.
            \nThe activity of a radioactive substance is 4700 per minute. Five minutes later the activity is 2700 per minute. Find
            \na) decay constant and
            \nb) half – life of the radioactive substance.
            \nAnswer:
            \nGiven A0<\/sub>=4700 per minute
            \nA = 2700 per minute
            \nt = 5 minute \u03bb = ? Ty2<\/sub> = ?
            \nwe have,
            \n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

            Students can Download 2nd PUC Physics Model Question Paper 2 with Answers, Karnataka 2nd PUC 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 Model Question Paper 2 with Answers Time: 3 Hrs 15 Min Max. Marks: …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/17780"}],"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=17780"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/17780\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=17780"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=17780"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=17780"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}