{"id":17664,"date":"2020-01-03T17:47:59","date_gmt":"2020-01-03T12:17:59","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=17664"},"modified":"2021-07-02T15:31:21","modified_gmt":"2021-07-02T10:01:21","slug":"2nd-puc-physics-model-question-paper-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-physics-model-question-paper-1\/","title":{"rendered":"2nd PUC Physics Model Question Paper 1 with Answers"},"content":{"rendered":"

Students can Download 2nd PUC Physics Model Question Paper 1 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 1 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 x 1 = 10 )<\/span><\/p>\n

    Question 1.
    \nState Faraday\u2019s law of electromagnetic induction.
    \nAnswer:
    \nI law : Whenever a magnetic flux linked with the coil, a change in emf is induced in it.
    \nThis induced emf lasts as long as the change in magnetic flux continues.<\/p>\n

    Question 2.
    \nWrite an expression for the displacement current.
    \nAnswer:
    \nThe expression for displacement current is given by
    \n\"2nd<\/p>\n

    Question 3.
    \nWhat is an electric dipole ?
    \nAnswer:
    \nA pair of equal & opposite charges are seperated by a small distance is called electric dipole.<\/p>\n

    Question 4.
    \nDraw the circuit symbol of p-n-p transistor.
    \nAnswer:
    \n\"2nd
    \nQuestion 5.
    \nHow can the resolving power of a telescope be increased ?
    \nAnswer:
    \nShorten the wavelength of light is used.<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nIn the following nuclear reaction identify the particle X.
    \nAnswer:
    \nN \u2192 P + e\u0304 + X
    \nIn the above reaction, X – represents<\/p>\n

    Question 7.
    \nDefine magnetisation of a sample.
    \nAnswer:
    \nIt is defined as \u201cthe net magnetic moment developed per unit volume of the material\u201d.<\/p>\n

    Question 8.
    \nHow does the power of a lens vary with its focal length ?
    \nAnswer:
    \nPower of a lens is inversely proportional to its focal length.
    \nie. p = \\(\\frac{1}{f}\\)<\/p>\n

    Question 9.
    \nWhat is cyclotron ?
    \nAnswer:
    \nCyclotron is a device used to accelerate heavy charged particle like proton a – pue etc to high speed.<\/p>\n

    Question 10.
    \nGive the wavelength range of X – rays.
    \nAnswer:
    \n1 x 10-10<\/sup> to 3 x 10-8<\/sup> m.<\/p>\n

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

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

    Question 11.
    \nThe current in a coil of self inductance 5 mH changes from 2.5A to 2.0A in 0.01 second. Calculate the value of self induced emf
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 12.
    \nWhat is a toroid ? Mention an expression for magnetic field at a point inside a toroid ?
    \nAnswer:
    \nA solenoid bent in to the form of a closed ring is called toroid.
    \n\u2234 B = \u00b5r<\/sub>\u00b50<\/sub>nI = \u00b5nI
    \nwhere n \u2192 is number of turns per unit length.
    \nI \u2192 is current through the toroid.<\/p>\n

    Question 13.
    \nWhat are isotopes & isobars ?
    \nAnswer:<\/p>\n

      \n
    • Nuclei are having same atomic number but different mass number are called isotopes.<\/li>\n
    • Nuclei are having same mass number but different atomic number are called isobars.<\/li>\n<\/ul>\n

      \"KSEEB<\/p>\n

      Question 14.
      \nDraw the variation of magnetic field (B) with magnetic intensity (H) when a ferromagnetic material is subjected to a cycle of magnetisation.
      \nAnswer:
      \n\"2nd<\/p>\n

      Question 15.
      \nMention two applications of polaroids.
      \nAnswer:<\/p>\n

        \n
      1. It is used in sunglasses, goggles etc.<\/li>\n
      2. It is used in windows, head lamps of buses, cars, aeroplane etc.<\/li>\n<\/ol>\n

        Question 16.
        \nWrite the logic symbol & truth table of NAND gate.
        \nAnswer:
        \nSymbol
        \n\"2nd
        \nWhere A & B arc inputs and y is output.
        \n\"2nd<\/p>\n

        Question 17.
        \nWrite two properties of eletric field lines.
        \nAnswer:<\/p>\n

          \n
        1. They never intersect each other.<\/li>\n
        2. They normal to surface of the conductor.<\/li>\n<\/ol>\n

          Question 18.
          \nWhat is myopia? how to connect it?
          \nAnswer:
          \nIn this human vision defect a person can see near by object clearly but he can not see distant object clearly.
          \nIt can be corrected by using a concave less of suitable focal length.<\/p>\n

          \"KSEEB<\/p>\n

          Part – C<\/span><\/p>\n

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

          Question 19.
          \nWhat is a transformer ? Mention two sources of energy loss in a transformer.
          \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

            Question 20.
            \nWrite three characteristics of nuclear forces ?
            \nAnswer:<\/p>\n

              \n
            1. Nuclear forces is a strongest force in nature.<\/li>\n
            2. They are short range forces<\/li>\n
            3. They are charge independent<\/li>\n<\/ol>\n

              Question 21.
              \nDerive the expression for energy . stored in a charged capacitor.
              \nAnswer:
              \n\"2nd
              \nCharging a capacitor means transfering of eletron from one plate to other plate of capacitor.
              \n\u2234 work will be done because electrons are moving against the electrostatic force of electric field. This work done is stored as a electrostatic potential energy.<\/p>\n

              Consider a capacitor of capacity is charged from DC source of \u2018 V\u2019 volts work done in adding a small charge dq to a capacitor is given by dw = v. dq<\/p>\n

              \u2234 Total workdone in changing a capacitor to Q is
              \n\"2nd
              \nThis work done is stored and a electrostatic potential energy.
              \n\"2nd<\/p>\n

              Question 22.
              \nWhat is an amplifier? Draw the simple circuit of transistor amplifier in CE mode.
              \nAnswer:
              \nAn amplifier is a device which is used for increasing the strength of a weak input ac signal.
              \n\"2nd<\/p>\n

              Question 23.
              \nMention the types of transmission media.
              \nAnswer:
              \nThe modes of propogation of em waves are :<\/p>\n

                \n
              1. Ground wave propogation<\/li>\n
              2. Sky wave propogation<\/li>\n
              3. Space wave propogation<\/li>\n<\/ol>\n

                \"KSEEB<\/p>\n

                Question 24.
                \nDerive an expression for drift velocity of free electrons in a conductor
                \nAnswer:
                \nConsider a conductor under the influence of electric field E. The force experienced by a free electron in a conductor is given by F = -eE, -ve sign shows that the divertion of F & E are in opposite direction.
                \n\"2nd
                \nSince the average time elapsed between successive collision is called relaxation tome (\u03c4)
                \n\"2nd<\/p>\n

                Question 25.
                \nExplain briefly the coil & magnet experiment to demonstrate electromagnetic induction.
                \nAnswer:
                \n\"2nd
                \nThe expt. arrangement consists of coil containing \u2018n\u2019 no of turns. The free end of the coil is connected to sensitive
                \ngalvanometer.<\/p>\n

                  \n
                • When a magnet is at rest, no deflection in the galvanometer.<\/li>\n
                • When a north pole of magnet is moving towards the coil the galvanometers shown deflention in one direction. ||| ly when a north pole of the magnet is moving away from the coil the galvanometer shows deflection in opposite direction.<\/li>\n
                • The deflection of the galvanometer is large, when magnet is moving towards the coil (or) away from the coil.<\/li>\n<\/ul>\n

                  From the expt it is clear that if there is a relative motion between the magnet coil, an emf is induced across the coil & induced current flows in the coil.<\/p>\n

                  Question 26.
                  \nWrite three properties of ferromagnetic materials.<\/p>\n

                    \n
                  1. Ferromagnetic substance is strongly attracted by a magnet.<\/li>\n
                  2. The relative permeability of a ferromagnetic substance is very large ie. Mr<\/sub> >> 1.<\/li>\n
                  3. The magnetic susceptability of a ferromagnetic substance is high +ve value.<\/li>\n<\/ol>\n

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

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

                    Question 27.
                    \nDeduce the condition for balance of a wheat stone’s bridge using Kirchhoff’s rules.
                    \n\"2nd
                    \nWheatstone\u2019s bridge is a device used to determine unknown resistance. It consists of 4 arms of resistances P, Q, R & S, a galvanometer of resistance \u2018G\u2019 & a battery with plug key. Let \u2018I\u2019 be the main current & splits into branch current I1<\/sub> & I2<\/sub> respectively.<\/p>\n

                    Applying KVL to the mesh ABDA
                    \nI1<\/sub>P + Ig<\/sub>G – I2<\/sub>R = 0 ………..(1)
                    \n|||ly to the mesh BCDB
                    \n(I1 <\/sub>– Ig<\/sub>)Q – (I2 <\/sub>+ Ig<\/sub>)S – Ig<\/sub>G = O …… (2)
                    \nThe wheatstone bridge is said to be balanced when no current flowing through the galvonometer.
                    \ni e Ig<\/sub> = 0
                    \neqn (1) I1<\/sub>P – I2<\/sub> R = 0
                    \nI1<\/sub>P – I2<\/sub>R …………(3)
                    \neqn (2) is I1<\/sub> Q – I1<\/sub> S = 0
                    \nI1<\/sub>Q – I2<\/sub>S ………….. (4)
                    \n\"2nd
                    \nThis is the balanced condition of Wheatstone bridge.<\/p>\n

                    Question 28.
                    \nDerive an expression for the force between two parallel conductors carrying currents hence define ampere.
                    \n\"2nd
                    \nConsider two identical parallel conductors each of length seperated by a distance \u2018r\u2019. Let I1<\/sub> is the current flowing through Ist<\/sup> conductor & I2<\/sub> is the current flowing through IInd<\/sup> conductor.
                    \nThe magnetic field is developed in second conductor due to flow of current in I conductor is given by B1<\/sub> = \\(\\frac{\\mu_{0}}{2 \\pi}-\\frac{I_{1}}{r}\\)
                    \nSince the second condutor carries a current, it experiences a mechanical force due to M.F _ developed in first condustor.
                    \n\"2nd<\/p>\n

                    |||ly the first conductor experiences the same force due to presence of M.F developed by the second conductor, The divertions of there forces are obtained using Fleming left hand rule are as shown in the figure. When a current flows through a parallel conductor in same direction are attracted current flows through a opposite direction they repel each other.<\/p>\n

                    Definition of ampere :<\/span> Ampere is the current flowing through conductor of unit length seperated by a distance of lm, it experiences a force of 2 \u00d7 10-7<\/sup>N.<\/p>\n

                    \"KSEEB<\/p>\n

                    Question 29.
                    \nDerive an expression for eletric field due to an electric dipole at a point on the axial line.
                    \nAnswer:
                    \n\"2nd
                    \nConsider a dipole having dipole length \u2018zl Let \u2018p\u2019 be a point at a distance \u2018r\u2019 from the centre of the dipole. Where electric field intensity is required. Imagine a unit +ve charge is placed at p. An electric field intensity is the resultant of two charges & it is given by.-
                    \n\"2nd
                    \nIf 1 << r i.e dipole is short then l2 is neglected compared to \u03bc2<\/sup> .
                    \n\"2nd<\/p>\n

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

                    Question 30.
                    \nWrite the experimental observations of photo electric effect.
                    \nAnswer:
                    \nThe following one the other
                    \nexperimental results of photoelectric effect<\/p>\n

                      \n
                    1. Photoelectric effect is an instantaneous process.<\/li>\n
                    2. The minimum frequency is required to remove the electron from the metal surface is called threshold frequency. If the frequency of incident light is less than threshold frequency them their is no photoeletric effect.<\/li>\n
                    3. The minimum energy required to remove the electron from the metal surface is called work function.
                      \nie. w = h\u03c50<\/sub><\/li>\n
                    4. The intensity of incident light is directly proportional to photoelectric current.<\/li>\n
                    5. The kinetic energy of emitted electron is directly proportional to frequency of incident light.<\/li>\n
                    6. A -ve potential is applied to anode plate just stop the photoelectric current is called stopping potential.<\/li>\n<\/ol>\n

                      Question 31.
                      \nWhat is rectification ? With relevant circuit diagram & waveforms explain the working of p-n junction diode as a full wave retifier.
                      \nAnswer:
                      \nThe process of conversion of AC in to DC is called rectification.
                      \n\"2nd
                      \nThe p- n junction is used as a full wave rectifier are as shown in the figure.
                      \nAn Ac in put is applied to primary of the transformer in first half cycle of AC, end A of secondary centre taping transformer is +ve & end B is -ve diode D1<\/sub> is in forwand bias & hence current flows but diode D2<\/sub> is reverse bias & hence current does not flow in next half cycle of AC end A of secondany centre taping transformer is -ve & end B is +ve diode D1<\/sub> is reverse bias & hence current does not flow but diode D2<\/sub> is in forward bias & hence current flows this process will continue & hence out put DC signal is obtained across the load resistance.<\/p>\n

                      Question 32.
                      \nDerive an expression for equivalent focal length of two thin lenses kept in contact.
                      \n\"2nd
                      \nConsider two thin lenses of focal length f1<\/sub> &f2<\/sub> are in constant let \u2018o\u2019 be a point object placed on the principle axis. A ray of light from the object after passing through the combination of lenses the refronted light ray meet at I. \u2234 An image is formed at I. Thus the formation of image at I can take splace in 2 stages.<\/p>\n

                      Stage (i) :<\/span> In the absence of lens L2<\/sub> the referacted light ray is meet at I1<\/sup> \u2234 Thin less formula can be written as
                      \n\"2nd
                      \nStage (2) :<\/span> In the absence of lens L,, the refracted light ray is meet at p now it acts as a virtual object & hence read final image is formed at I \u2234 Thin lens formula can be written as
                      \n\"2nd<\/p>\n

                      \"KSEEB<\/p>\n

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

                      Question 33.
                      \nIn young\u2019s double slit experiment, fringes of certain.
                      \nAnswer:
                      \nWidth are produced on the screen kept at a certain distance from the slits. When the screen is moved away from the slits by 0.1m, fringe width increases by 6 \u00d7 10-5<\/sup> m. The seperation between the slits is 1mm. calculate the wavelength of the light used.<\/p>\n

                      Question 34.
                      \nWhen two wapasitors are connected in series & connected across 4 kv line the enevgy stored in the system is 8j. The some capacitors, if connected in parallel across the some line, the energy stored is 36 j. Find the individual capacitances.
                      \nAnswer:
                      \nGiven, v = 4 kv = 4 \u00d7 103<\/sup> V = 4000 volts.
                      \nIn series connection, energy stored
                      \n\"2nd
                      \nIn parallel connection energy stored
                      \n\"2nd<\/p>\n

                      C1<\/sub>C2<\/sub> = (C1 <\/sub>+ C2<\/sub>)10-6 <\/sup>= 4.5 \u00d7 1 \u03bcF
                      \nC1<\/sub>C2<\/sub> = 4.5\u03bcF
                      \nWe have (C1<\/sub> – C2<\/sub>)2<\/sup> = (C1<\/sub> + C2<\/sub>)2<\/sup> – 4C1<\/sub>C2<\/sub>
                      \n= (4.5)2<\/sup> – 4 \u00d7 4.5 = 20.25 – 18 = 2.25 \u03bcf
                      \nC1<\/sub> – C2 <\/sub>– 1.5\u00b5f + C2<\/sub> =4.5 \u03bcf<\/p>\n

                      2C1<\/sub> = 6.0 \u03bcf (\u2234 C1<\/sub> =3(\u03bcf)
                      \nalso, C1<\/sub> + C2<\/sub> =4.5
                      \n<c2<\/sub> = 4.5 – 3 = 1.5\u03bcf><\/p>\n

                      Question 35.
                      \nCalculate the shortest & longest wavelength of balmer series of hydrogen atom (Given r = 1.097 \u00d7 107<\/sup>m-1<\/sup>)
                      \nAnswer:
                      \nThe longest wavelength of balmer series of H – atom is given by
                      \n\"2nd
                      \nThe shortest wavelength of balmer series is given by
                      \n\"2nd<\/p>\n

                      \"KSEEB<\/p>\n

                      Question 36.
                      \nCalculate the resonant frequency & Q- factor of a series L-e-R circuit containing a pure inductor of inductance 4H, capacitor of capacitance 27 \u03bcf & resistor of resistance 8.4 \u2126
                      \nAnswer:
                      \nGiven L = 4H,C= 27\u03bcF,R = 8.4\u2126, f0<\/sub> = ?
                      \nQ = factor = ?
                      \nThe expn for resonant frequency is
                      \n\"2nd<\/p>\n

                      \"KSEEB<\/p>\n

                      Question 37.
                      \na)Three resistors of resistances 2 \u2126, 3 \u2126 & 4 \u2126 are combined in series what is the total resistance of the combination.
                      \nb) If this combination is connected to a battery of emf 10V & negligible internal resistance obtain the potential drop across each resistor.
                      \nAnswer:
                      \nGiven :R1<\/sub> = 2\u2126 , R2<\/sub> = 3\u2126, R3<\/sub> = 4\u2126,
                      \nE = 10V,
                      \na) In series combination effective resistance R3<\/sub> is given by
                      \nRs<\/sub> = R1<\/sub> + R2<\/sub> + R3<\/sub> = 2+ 3 + 4 = 9\u2126<\/p>\n

                      b) Current through the circuit
                      \n\\(I=\\frac{E}{R_{s}}=\\frac{10}{9}\\)
                      \n1 = 1.11 A
                      \n\u2234Potential drop across R1<\/sub> = I , R1<\/sub> 1.11 \u00d7 2 = 2.22 = 2.22 Volts
                      \nPotential drop across R2<\/sub> = I R2<\/sub>
                      \n= 1.11 \u00d7 3 = 3.33 = 3.33 Volts
                      \nPotential drop across R3<\/sub> = I R3<\/sub> =1.11 \u00d7 4 = 4.44 = 4.44 Volts<\/p>\n","protected":false},"excerpt":{"rendered":"

                      Students can Download 2nd PUC Physics Model Question Paper 1 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 1 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":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/17664"}],"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=17664"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/17664\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=17664"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=17664"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=17664"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}