{"id":18057,"date":"2020-01-06T13:03:18","date_gmt":"2020-01-06T07:33:18","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=18057"},"modified":"2021-07-02T15:30:46","modified_gmt":"2021-07-02T10:00:46","slug":"2nd-puc-physics-previous-year-question-paper-june-2019","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-physics-previous-year-question-paper-june-2019\/","title":{"rendered":"2nd PUC Physics Previous Year Question Paper June 2019"},"content":{"rendered":"

Students can Download 2nd PUC Physics Previous Year Question Paper June 2019, 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 June 2019<\/h2>\n

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

Introductions:<\/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 the following questions in one sentence each:\u00a0 ( 10 \u00d7 1 = 10 )<\/span><\/p>\n

    Question 1.
    \nHow does the Resistance of a Conductor Vary with its length?
    \nAnswer:
    \nDirectly proportional OR As length increases, resistance increases.<\/p>\n

    Question 2.
    \nState Ampere\u2019s circuital law.
    \nAnswer:
    \nThe line integral of magnetic field around a closed path in vacuum is equal to p o times the total current passing the closed path.<\/p>\n

    Question 3.
    \nWhen does the force acting on a charged particle moving in a uniform magnetic field is Maximum?
    \nAnswer:
    \nCharged particle moves perpendicular to field OR \u03b8 = 90\u00b0 where \u03b8 is the angle between direction of magnetic field and velocity.<\/p>\n

    Question 4.
    \nDefine declination?
    \nAnswer:
    \nThe angle between the geographic meridian and magnetic meridian at that place.<\/p>\n

    Question 5.
    \nWhat is Retentivity?
    \nAnswer:
    \nThe property of magnetic material to retain magnetism even in the absence of magnetizing field.<\/p>\n

    Question 6.
    \nWrite any one advantage of Eddy Current.
    \nAnswer:<\/p>\n

      \n
    1. Induction furnace<\/li>\n
    2. Magnetic braking in trains<\/li>\n<\/ol>\n

      \"KSEEB<\/p>\n

      Question 7.
      \nWhat is the rest mass of photon?
      \nAnswer:
      \nZero<\/p>\n

      Question 8.
      \nName the Spectral Series of Hydrogen atom which lies in the Visible region of electro magnetic spectrum.
      \nAnswer:
      \nBalmer series.<\/p>\n

      Question 9.
      \nWhat are isotopes?
      \nAnswer:
      \nThe nuclei which have the same atomic number but different mass numbers.<\/p>\n

      Question 10.
      \nMention one need for modulation.
      \nAnswer:<\/p>\n

        \n
      1. To reduce the size of the antenna<\/li>\n
      2. Effective power radiated by the antenna<\/li>\n
      3. Mixing up of signals from different transmitters<\/li>\n<\/ol>\n

        Part -B<\/span><\/p>\n

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

        Question 11.
        \nWrite any two limitations of Ohm\u2019s law.
        \nAnswer:<\/p>\n

          \n
        1. It is not applicable for metallic conductor at very low and very high temperatures.<\/li>\n
        2. It is not applicable for semiconductors, super conductors, triodes.<\/li>\n<\/ol>\n

          Question 12.
          \nDraw a neat labelled diagram of Cyclotron.
          \nAnswer:
          \nCyclotron is a device to accelerate charged particles or ions to high energies. Cyclotron uses both electric field and magnetic field to increase the energy of charged particles.
          \n\"2nd<\/p>\n

          Question 13.
          \nMention an expression for the magnetic field produced at the center on the axis of a current carrying Solenoid and explain the terms.
          \nAnswer:
          \nB = \u03bco<\/sub>nI
          \nWhere \u03bco<\/sub>\u00a0= Permeability of free space
          \nn = number of turns per unit length of solenoid
          \nI = current through solenoid.<\/p>\n

          Question 14.
          \nState and Explain Gauss\u2019s Law in magnetism.
          \nAnswer:
          \nThe net magnetic flux through any closed surface is alwsays zero
          \n\u03a3\\(\\overrightarrow{\\mathrm{B}}\\overrightarrow{\\Delta \\mathrm{S}}\\) = 0
          \n\"2nd
          \nconsider a closed surface S in a uniform
          \nM.F. \\(\\overrightarrow{\\mathrm{B}}\\)Let \\(\\overrightarrow{\\Delta \\mathrm{S}}\\) be a small area element of this surface with along its normal
          \n\"2nd<\/p>\n

          Question 15.
          \nWhat is a Transformer? Mention any one sources of Energy loss.
          \nAnswer:
          \nA transformer is a device used to change AC voltage by using the principle of mutual induction.<\/p>\n

            \n
          1. Loss due to flux leakage<\/li>\n
          2. Loss due to eddy current<\/li>\n<\/ol>\n

            \"KSEEB<\/p>\n

            Question 16.
            \nWrite any two uses of ultra violet rays.
            \nAnswer:<\/p>\n

              \n
            1. Used in LASIK eye surgery<\/li>\n
            2. UV lamps are used to kill germs in water purifiers<\/li>\n
            3. Disinfection for virus and bacteria.<\/li>\n
            4. To produce photo electric current in Burglar alarm.<\/li>\n<\/ol>\n

              Question 17.
              \nGive the two differences betwene Collector region and Emitter region of a transistor.
              \nAnswer:
              \nCollector Region<\/p>\n

                \n
              1. Moderately doped<\/li>\n
              2. Large in size<\/li>\n
              3. Collects charge carriers<\/li>\n<\/ol>\n

                Emitter region<\/p>\n

                  \n
                1. Heavily doped<\/li>\n
                2. Moderate in size<\/li>\n
                3. Emits charge carriers.<\/li>\n<\/ol>\n

                  Question 18.
                  \nDraw a Block diagram showing the important components in a Generalized Communication System.
                  \nAnswer:
                  \n\"2nd<\/p>\n

                  Part-C<\/span><\/p>\n

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

                  Question 19.
                  \nMention any three properties of
                  \nElectric field lines.<\/p>\n

                    \n
                  1. Field lines are continuous curves without any breaks<\/li>\n
                  2. Two field lines cannot cross each other<\/li>\n
                  3. Electrostatic field lines start at positive charges and end at negative charges<\/li>\n
                  4. They do not form closed loops.<\/li>\n<\/ol>\n

                    Question 20.
                    \nObtain an expression for effective capacitance of two capacitors connected in series. (July -2018, Q.No.20)
                    \nAnswer:
                    \n\"2nd
                    \nLet C1<\/sub>, C2<\/sub> -capacitance of 2 capacitors connected in series
                    \nQ-charge on each capacitor
                    \nV1<\/sub> V2 <\/sub>– pd across C1<\/sub> and C2<\/sub>
                    \nV-Total voltage drop across the combination
                    \nThen V = V1<\/sub> + V2<\/sub> (1) ;
                    \nAlso Q = CV
                    \nV1<\/sub> = \\(\\frac{\\mathrm{Q}}{\\mathrm{C}_{1}}\\) and V2<\/sub> = \\(\\frac{\\mathrm{Q}}{\\mathrm{C}_{2}}\\)
                    \nIf system of capacitors is replaced by a single capacitor of
                    \n\"2nd<\/p>\n

                    Question 21.
                    \nExplain with a circuit diagram how a galvanometer can be converted into an ammeter. (March – 2015, Q.No.20)
                    \nAnswer:
                    \nA galvonometer can be converted into an ammeter by connecting a low resistance in parallel with it.
                    \nLet S be the low resistance connected in parallel with the galvonometer, G be the resistance parallel with the galvonometer, G be the resistance of the Galvonometer, I be the maximum current to be measured end Is be the current required, for full scale deflection in the galvonometer<\/p>\n

                    Here Galvonometer and shunt resistances are connected in parallel.
                    \n\u2234 potential difference across \u2018s\u2019 = potential difference across G
                    \n(I – Ig<\/sub>)s = Ig<\/sub>G
                    \n\"2nd<\/p>\n

                    Question 22.
                    \nWrite any three distinguishing properties between diamagnetic and ferromagnetic materials.
                    \nAnswer:
                    \nRefer March – 2019, Q.No. 20
                    \n\"2nd
                    \nOne ampere is that steady current through each of infinitely long straight parallel conductors of negligible area of cross section separated by a distance of lm in vacuum, then each experience a force of 2 \u00d7 10-7<\/sup> Nm-1<\/sup>.<\/p>\n

                    Question 23.
                    \nDerive the relation between focal length and radius of curvature of a concave mirror.
                    \nAnswer:
                    \nConsider a ray parallel to principal axis of a concave mirror of small aperture incident at M. The ray reflected along MF as shown. Let \u03b8 is the angle of incidences. As the aperture is small, angle \u03b8 is small and are PM is considered as straight segment.
                    \n\"2nd
                    \nFrom the \u0394le<\/sup> MCP
                    \ntan \u03b8 = \\(\\frac{P M}{P C}\\)
                    \ntan \u03b8 = \\(\\frac{P M}{R}\\)
                    \nsince \u03b8 is small tan = \u03b8
                    \n\u2234 \u03b8 = \\(\\frac{P M}{R}\\) ……. 1
                    \nFrom the \u0394le<\/sup> PFM
                    \ntan 2\u03b8 = \\(\\frac{P M}{P F}\\)
                    \ntan 2\u03b8 = \\(\\frac{P M}{F}\\)
                    \nsince \u03b8 is small tan 2\u03b8 = 2\u03b8
                    \n\u22342\u03b8 = \\(\\frac{P M}{F}\\)
                    \n\u03b8 = \\(\\frac{P M}{2F}\\) ……..2
                    \nFrom 1 and 2 \\(\\frac{P M}{F}\\) =\\(\\frac{P M}{2F}\\)
                    \nR = 2f
                    \nf = \\(\\frac{R}{2}\\)<\/p>\n

                    \"KSEEB<\/p>\n

                    Question 24.
                    \nUsing Huygens principle, show that the angle of incidence is equal to angle of reflection during a plane wavefront reflected by a plane surface.
                    \nAnswer:
                    \nDepartmental Model Question Paper – 2, Q.No. 24.
                    \nLaw of reflection of light: Angle of incidence = angle of reflection i = r
                    \n\"2nd
                    \nIn above figure, MN in reflecting surface AB in the incident wave front FC in the reflected wave front. i in the angle of incidence. r in the angle of reflection.<\/p>\n

                    Let \u2018v\u2019 be the speed of the light wave in the medium and \u2018t\u2019 be the time taken by the secondary wave letes to move from the point B to C Then the distance. BC = vt<\/p>\n

                    In order to contruct the reflected wave front, Let us draw a sphere of radium AE vt, CE repersents the tangent drawn from the point e to this sphere, This tangent represents the position of the new reflected wave front.<\/p>\n

                    From the figure it is observed that AE = BC vt. the triangles EAC and BAC are congruent
                    \n\u2234 i = r. This is the Law of reflection.<\/p>\n

                    Question 25.
                    \nName the three types of electron Emission.
                    \nAnswer:
                    \nDepending upon the sources of external energy, electron emission is classified into four groups.<\/p>\n

                      \n
                    • Thermionic emission<\/li>\n
                    • Photoelectric emission<\/li>\n
                    • Field emission<\/li>\n
                    • Secondary emission<\/li>\n<\/ul>\n

                      Question 26.
                      \nWhat is NAND gate? Write its logic Symbol and Truth Table. (March – 2016 Q.No. 16)
                      \nAnswer:
                      \nSymbol
                      \n\"2nd
                      \nWhere A & B are inputs and y is output.
                      \n\"2nd<\/p>\n

                      Part- D<\/span><\/p>\n

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

                      Question 27.
                      \nObtain an expression for the Electric Field on the equatorial line of an Electric dipole.
                      \nAnswer:
                      \nRefer Departmental Model Question Paper – 1, Q.No. 27. (March – 2017, Q.No. 27.)
                      \n\"2nd
                      \nLet a point P be at a distance r from the centre of the dipole (O) Dipole length AB = 2a,
                      \nUsing Pythagoras theorem, the distance of the point P from the charges is,<\/p>\n

                      AP = BP = d = \\(\\sqrt{r^{2}+a^{2}}\\)
                      \nUsing, the expression for electric field due to a charge q at a distance d from it, \\(\\mathrm{E}=\\frac{1}{4 \\pi \\varepsilon_{0}}\\) \\(\\frac{\\mathrm{q}}{\\mathrm{d}^{2}}\\) , The magnitude of the electric field at P due to the charge + q is E+q<\/sub> = \\(\\frac{1}{4 \\pi \\varepsilon_{0}}\\) \\(\\frac{q}{r^{2}+a^{2}}\\) The magnitude of the electric field at P due to the charge ‘-q’ is E -q = \\(\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{\\mathrm{q}}{\\mathrm{r}^{2}+\\mathrm{a}^{2}}\\)<\/p>\n

                      The total field at P, \\(\\overrightarrow{\\mathrm{E}}_{+q}\\) and \\(\\overrightarrow{\\mathrm{E}}_{-\\mathrm{q}}\\) are resolved in to components parallel to \\(\\overrightarrow{\\mathrm{p}}\\) and perpendicular to \\(\\overrightarrow{\\mathrm{p}}\\). The direction of p (or p) is from – q to +q.<\/p>\n

                      The components normal to the dipole axis (E+q<\/sub>sin\u03b8) and (E-q<\/sub>sin\u03b8) cancel out as they are equal in magnitude and opposite in direction.
                      \n\"2nd
                      \n\"2nd<\/p>\n

                      \"KSEEB<\/p>\n

                      Question 28.
                      \nDerive an Expression for the Balancing ‘ condition of wheat stone Bridge.
                      \nAnswer:
                      \nIt is the combination of four resistances P, Q, R and S in the form of quadrilateral ABCD. A galvanometer is connected between junctions B and D. The line BD is called galvanometer arm. A battery is connected between junctions A and C. This Ac is called the battery arm.
                      \n\"2nd
                      \nOn applying Kirchhoff\u2019s loop rule to the loop ABDA, we get
                      \nI1<\/sub>P – Ig<\/sub>G – I2<\/sub>R = o
                      \nBut for balance of network Ig<\/sub> = 0
                      \nI1<\/sub>P – I2<\/sub>R = o
                      \nI1<\/sub>P = I2<\/sub>R………………….. (1)
                      \nOn applying Kirchhoff\u2019s loop rule to the loop BCDB, we get
                      \n(I1<\/sub> – Ig<\/sub>)Q – (I2<\/sub> + Ig<\/sub>)S – Ig<\/sub> G = 0
                      \nI1<\/sub>Q – Ig<\/sub>Q – I2<\/sub> S – Ig<\/sub>S – Ig<\/sub>G = 0
                      \nFor balance of network Ig<\/sub> = 0
                      \nI1<\/sub>Q – I2<\/sub>S = 0
                      \nI1<\/sub>Q = I2<\/sub>S …………..(2)
                      \n\"2nd
                      \nThis is the condition for balance Wheatstone bridge.<\/p>\n

                      Question 29.
                      \nDerive an expression for instantaneous induced emf in an A.C. generator.
                      \n\"2nd
                      \nAnswer:
                      \nIn the above figure
                      \nN and S are poles of a magnet, \u03b8 is the angle between the directions of magnetic field B and area vector A.
                      \nWhen the coil is rotated in the magnetic field, the flux linked with the coil varies. At any’ instant of time \u2018t\u2019, A cos\u03b8 is the component of area vector along the direction B.
                      \nThe magnetic flux linked with the coil at any instant of time \u2018t\u2019 is given by
                      \n\u03a6B<\/sub> = B x component of area vector along the field direction.
                      \nFor 1 turn \u03a6B<\/sub> = BA cos \u03b8
                      \nFor n turns \u03a6B<\/sub> = nAB cos \u03b8
                      \n\u03a6B<\/sub> = nABcos\u03c9t….(1)[\u2235 \u03b8 = \u03c9t]
                      \nWhere \u2018 \u03c9 \u2019 is the angular velocity of the coil at time t.
                      \nFrom the Faraday\u2019s second law,
                      \ne = [Latex]-\\frac{\\mathrm{d} \\phi}{\\mathrm{d} \\mathrm{t}}[\/Latex]
                      \ne = -[Latex]\\frac{\\mathrm{d}}{\\mathrm{dt}}[\/Latex][nAB cos cot] [From(l)\u03a6=nABcos\u03c9t]
                      \ne = (-n AB) [-sin \u03c9 t) \u00d7 \u03c9
                      \ne = nAB\u03c9sin\u03c9t
                      \ne = e0<\/sub> sin( \u03c9 t)
                      \nwhere e0 is the peak value of emf = nAB \u03c9<\/p>\n

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

                      Question 30.
                      \nDerive an expression for the fringe witdth of interference fringes in a double slit experiment. (March – 2014, Q.No. 30)
                      \nAnswer:
                      \nLet A and B be two slits separated by a distance \u2018d\u2019. Let ‘\u03bb’ 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 \u00ab CP Path difference between the two waves reaching \u2018P\u2019 is given by BP – AP = BC =????
                      \nFrom the \u0394 BFP, BP2<\/sup> = BF2<\/sup> + FP2<\/sup> and from the A AEP, AP2<\/sup> = AE2<\/sup> + EP2<\/sup>
                      \nBP2<\/sup> – AP2<\/sup> = (D2<\/sup> + FP2<\/sup>) – (D2<\/sup> – EP2<\/sup>)
                      \ni.e, (BP – AP) (BP – AP) = 2 (2 \u00d7 \\(\\frac{\\mathrm{d}}{2}\\) )
                      \n\"2nd
                      \nFor a point ‘P’ close to ‘O’
                      \nBP \u2248 AP = D
                      \n(BP – AP) (2D) = 2x.d
                      \nBut (BP – AP) = BC = \u03b4
                      \ni,e. = \\(\\frac{\\mathrm{xd}}{D}\\) or x = \\(\\frac{\\delta \\mathrm{d}}{\\mathrm{D}}\\)
                      \nFor a constructive interference = n\u03bb
                      \nDistance of nth<\/sup> bright fringe from the central bright fringe xn = n \\(\\frac{\\lambda \\mathbf{D}}{\\mathrm{d}}\\)
                      \nDistance of (n + l)th bright fringe from the central bright fringe xn+1<\/sub> = (n + 1) \\(\\frac{\\lambda \\mathbf{D}}{\\mathrm{d}}\\)
                      \nBy definition, Fringe width is the distance between two consecutive bright or dark fringes.
                      \n\"2nd
                      \nFringe width \\(\\frac{\\lambda}{\\mathrm{d}}\\) \u03b2 \u221d D, \u03b2 \u221d \u03bb and \u03b2 \u221d \\(\\frac{1}{d}\\)
                      \nWe can show that the fringe width between
                      \nany two dark frings is also \u03b2 = \\(\\frac{\\lambda \\mathrm{D}}{\\mathrm{d}}\\)<\/p>\n

                      Question 31.
                      \nDerive an expression for the energy of an electron in nth\u201d stationary orbit of hydrogen atom by assuming the expression for radius. (June – 2017, Q.No. 31)
                      \nAnswer:
                      \n\"2nd
                      \nConsider an electron of mass m Electron and charge e othit revolving round the nucleus of an atom of atomic number Z in the stationary orbit of radius \u2018r\u2019. Let y be the velocity of the nucleus it also possess kinetic energy by virtue of its motion.
                      \nFor stationary orbit, total energy E = KE + PE …………1
                      \nFrom Rutherford\u2019s atom model,
                      \nFor stationary orbit, centripetal force = Electrostatic force.
                      \n\"2nd
                      \n( \u2235 Electrostatic force, according to Coulomb\u2019s law)
                      \non dividing by 2 on both sides, we get
                      \n\"2nd
                      \nWe have PE = Electric potential at a distance are due to + Ze \u00d7 (-e)
                      \n\"2nd
                      \n\"2nd
                      \nThe radius 0fth permitted orbit of electron is given by
                      \n\"2nd
                      \nSubstituting the value of \u2018r\u2019 in equation (4)
                      \n\"2nd<\/p>\n

                      \"KSEEB<\/p>\n

                      Question 32.
                      \nWhat is Rectification? Describe with a circuit diagram the working of a p-n junction diode as half wave rectifier with input and output waveforms.
                      \nAnswer:
                      \nThe process of converting ac into dc is called rectification.
                      \n\"2nd<\/p>\n

                      The diode D in series with a load resistor RL<\/sub> is connected to the secondary of the transformer. The ac voltage to be rectified is applied to the primary of the transformer. The secondary of a transformer supplies the desired ac voltage across terminals A and B. The output dc voltage is taken across the load RL<\/sub>.
                      \nWorking: During the positive half cycle of input ac voltage, the end A becomes positive with respect to the end B. As a res\u00f9\u00cdt the diode becomes forward biased and conducts the current. the current flows in the direction ADRL<\/sub>BA.
                      \nDuring the negative half cycle of input ac voltage, the end A becomes negative with respect to the end B. As a result the diode becomes reverse biased and does not conducts the current.
                      \nThus the current flows through the diode during positive half cycle only. The current flows through the load RL<\/sub> always in the same direction. Hence dc output is obtained.<\/p>\n

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

                      Question 33.
                      \nABCD is a square of side lm. Charges of +3nC, -5nC and +3nC are placed at the comers A, B and C respectively. Calculate the work done in transferring a charge of 12\u00b5c from D to the point of intersection of the diagonals?
                      \nSolution:
                      \nDiagonal = BD = AC = 1.414m Half of diagonal = AO = BO = CO = 0.707m
                      \nElecric potential due system of charges
                      \n\"2nd
                      \nWork done to transfer 12 \u00b5 C charge from D to 0
                      \nW = q[VD<\/sub> – V0<\/sub>] = 12 \u00d7 10-6<\/sup>[22.176 – 12.7298]
                      \nW= 1.1335 \u00d7 10-4<\/sup> J<\/p>\n

                      Question 34.
                      \nA network of Resistors is connected to a 16 V battery with internal resistance 1 \u2126 as shown in Figure below.
                      \na) Compute the equivalent resistance of the network
                      \nb) Calculate the total current in the circuit
                      \n\"2nd
                      \nSolution:
                      \n\"2nd
                      \n4 ohm and 4 ohm in parallel:
                      \n\"2nd
                      \n12 ohm and 6 ohm in parallel:
                      \n\"2nd
                      \nRpl<\/sub>, 1 ohm and Rp2<\/sub> in series : Equivalent resistance o fhetwork
                      \nReq<\/sub> = Rp1<\/sub> + 1 + Rp2<\/sub> = 2 + l+ 4 = 7 ohm
                      \nTotal current in the circuit:
                      \n\"2nd<\/p>\n

                      Question 35.
                      \nA sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 \u2126, L = 25.48 mH and C = 796 \u03bcF. Find
                      \na) Impedance of the circuit.
                      \nb) The phase difference between the Voltage across the source and the current.
                      \nSolution:
                      \n\"2nd
                      \n\"2nd<\/p>\n

                      Question 36.
                      \nAn object of 3cm is placed 14cm in front of a concave lens of focal length 21 cm. Find the position, Nature and size of the Image formed.
                      \nAnswer:
                      \nh0<\/sub> = 3cm, u = -14cm, f = -21 cm
                      \n\"2nd
                      \nPosition of image; 8.4cm from lens on same side as object
                      \nNature of image : Virtual and erect
                      \n\"2nd
                      \nImage is dimished.<\/p>\n

                      \"KSEEB<\/p>\n

                      Question 37.
                      \nCalculate the binding Energy of an alpha (\u03b1) particle in Mev from the following data.
                      \nMass of Helium Nucleus = 4.00260 u
                      \nMass of neutron = 1.008662 u
                      \nMass of proton = 1.007825 u.
                      \nAnswer:
                      \nAlpha particle : \\(_{2}^{4} \\mathrm{He}\\) Z = 2 and A = 4
                      \nMass defect : \u0394m = [Zmp<\/sub> + (A-Z)mn<\/sub>] – M
                      \n\u0394m = 2 \u00d7 1.007825 + 2 \u00d7 1.008662 – 4.00260
                      \n\u0394 m = 2.01565 + 2.017324 – 4.00260 \u0394 m= 0.030374 u
                      \nBinding energy : Eb<\/sub> = \u0394m \u00d7 931 MeV = 0.030374 \u00d7 931 MeV
                      \nEb<\/sub> = 28.2782 MeV<\/p>\n","protected":false},"excerpt":{"rendered":"

                      Students can Download 2nd PUC Physics Previous Year Question Paper June 2019, 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 June 2019 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\/18057"}],"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=18057"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/18057\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=18057"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=18057"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=18057"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}