OR<\/strong><\/p>\nWrite Maxwell\u2019s generalization of Ampere\u2019s Circuital Law. Show that in the process of charging\u00a0 a capacitor, the current produced within the plates of the capacitor is \n \nis the electric flux produced during charging of the capacitor plates.<\/p>\n
Question 18. \nA charge is distributed uniformly over a ring of radius V. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge.<\/p>\n
Question 19. \n(i) Write the functions of three segments of a transistor. \n(ii) Draw the circuit diagram for studying the input and output characteristics of n-p-n transistor in common emitter configuration. Using the circuit, explain how input, output characteristics are obtained.<\/p>\n
Question 20. \n(a) Calculate the distance of an object of height h from a concave mirror of radius of curvature 20 cm, so as to obtain a real image of magnification 2. Find the location of image also, \n(b) Using mirror formula, explain why does a convex mirror always produce a virtual image.<\/p>\n
Question 21. \n(i) State Bohr\u2019s quantization condition for defining stationary orbits. How does de-Broglie hypothesis explain the stationary orbits? \n(ii) Find the relation between the three wavelengths \u03bb1<\/sub> \u03bb2<\/sub> and \u03bb3<\/sub>\u00a0from the energy level diagram shown below: \n <\/p>\nQuestion 22. \nTwo cells of emf 1.5 V and 2.0 V having internal resistances 0.2 O and 0.3 Q respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.<\/p>\n
SECTION : D<\/strong><\/p>\nQuestion 23. \nMeeta\u2019s father was driving her to the school. At the traffic signal she noticed that each traffic light was made of many tiny lights instead of a single bulb. When Meeta asked this question to her father, he explained the reason for this. \nAnswer the following questions based on above information:<\/strong><\/p>\n\nWhat answer did Meeta\u2019s father give?<\/li>\n What are the tiny lights in traffic signals called and how do these operate?<\/li>\n What were the values displayed by Meeta and her father?<\/li>\n<\/ul>\nSECTION : E<\/strong><\/p>\nQuestion 24. \n(i) Define the term drift velocity. \n(ii) On the basis of electron drift velocity, derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend?(iii) Why alloys like constantan and manganin are used for making standard resistors?<\/p>\n
OR<\/strong><\/p>\n(i) State the principle of working of a potentiometer. \n(ii) In the following potentiometer circuit AB is a uniform wire of length 1 m and resistance 10 \u03a9. Calculate the potential gradient along the wire and balance length AO (= l). \n <\/p>\n
Question 25. \n(i) An a.c. source of voltage V = V0<\/sub> sin at is connected to a series combination of L, C and R. Use the phasor diagram to obtain expressions for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called? \n(ii) In a series LR circuit XL<\/sub> = R and power factor of the circuit is P1<\/sub>. When capacitor with capacitance C such that XL<\/sub> = Xc<\/sub> is put in series, the power factor becomes P2<\/sub>.Calculate P1<\/sub>\/P2<\/sub>.<\/p>\nOR<\/strong><\/p>\n(i) Write the function of a transformer. State its principle of working with the help of a diagram. Mention various energy losses in this device. \n(ii) The primary coil of an ideal step up transformer has 100 turns and transformation ratio is also 100. The input voltage and power are respectively 220V and 1100 W Calculate: \n(a) number of turns in secondary. \n(b) current in primary. \n(c) voltage across secondary. \n(d) current in secondary. \n(e) power in secondary.<\/p>\n
Question 26. \n(i) In Young\u2019s double slit experiment, deduce the condition for (a) constructive, and (b) destructive interference at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position V on the screen. \n(ii) Compare the interference pattern observed in Young\u2019s double slit experiment with single slit diffraction pattern, pointing out three distinguishing features.<\/p>\n
OR<\/strong><\/p>\n(i) Plot a graph to show variation of the angle of deviation as a function of angle of incidence for light passing through a prism. Derive an expression for refractive index of the prism in terms of angle of minimum deviation and angle of prism. \n(ii) What is dispersion of light? What is its cause? \n(iii) A ray oflight incident nonnally on one face ofa right isosceles prism\u00a0 is totally reflected as shown in. What must be the minimum value\u00a0 of refractive index of glass? Give relevant calculations. \n <\/p>\n
Answers\u00a0<\/strong> \nSECTION : A<\/strong><\/p>\nAnswer 1. \nAccording to Gauss\u2019s law :<\/strong> \n \nFlux depends only on the charge enclosed. \nHence, the electric flux remains constant.<\/p>\nAnswer 2. \nPotential at a distance r from a given point charge Q is given by \n \nrA<\/sub> <\u00a0rA<\/sub> \u21d2\u00a0 \u00a0VA<\/sub> \u00a0>\u00a0\u00a0VB<\/sub> \nSince \nHence ,\u00a0VA\u00a0<\/sub>–\u00a0 VB\u00a0<\/sub>\u00a0is positive<\/p>\nAnswer 3. \nMicrowaves of frequency 1 GHz to 300 GHz bounces from even the smallest aircraft so that they are suitable to avoid getting bombed. Microwaves can penetrate through clouds also.<\/p>\n
Answer 4. \nThe Q factor of series resonance circuit is defined as the ratio of the voltage developed across the inductor or capacitor at resonance to the applied voltage, which is the voltage across R. \n \nIt is dimensionless, hence it has no units.<\/p>\n
Answer 5. \nWhen a current carrying coil is placed in magnetic field then it experiences a torque. \nNIAB = k\u03b1 \n \nN = The number of turns, \nI = Current \nA = Area of the loop, \nB = Magnetic field \nk = Torsional constant of the wire, \n\u03b1 = Angle of deflection<\/p>\n
SECTION : B<\/strong><\/p>\nAnswer 6. \n \nLesser mass, greater slope, i.e. of mass m2<\/sub>.<\/p>\nAnswer 7. \nGain in binding energy for nucleon is about 0.9 MeV. \nBinding energy of the nucleus, B1<\/sub>= 7.6 x 240 = 1824 MeV \nBinding energy of each product nucleus, B2<\/sub> = 8.5 x 120 = 1020 MeV. Then, energy released as the nucleus breaks. \nE = 2B2<\/sub>-B1<\/sub>= 2 x 1020 – 1824 = 216 MeV \n <\/p>\nOR<\/strong><\/p>\n\u0394E = (7.73) – 2(2.23) = 7.73 – 4.46 = 3.27 MeV<\/p>\n
Answer 8. \n(i) Attenuation :<\/strong> \nThe loss of strength of a signal while propagating through a medium is known as attenuation.<\/p>\n(ii) Demodulation :<\/strong> \nThe process of retrieval of information from the carrier wave at the receiver is termed demodulation. This is the reverse process of modulation.<\/p>\nAnswer 9. \nBrewster\u2019s law :<\/strong> \nThe law states that the tangent of the polarising angle of incidence of a transparent medium is equal to its refractive index. The light incident at this angle when reflects back is perfectly polarised. \np = tan ip \n<\/sub>The refractive index of a material depends on the colour or wavelength of light. As the polarising angle depends on refractive index (p = tan ip<\/sub>), so it also depends on wavelength of light.<\/p>\nAnswer 10. \n(i) Existence of threshold frequency : \n<\/strong> According to wave theory, there should not exist any threshold frequency but Einstein\u2019s theory explains the existence of threshold frequency. \n(ii) Dependence of kinetic energy on frequency of incident light :<\/strong> \nAccording to wave theory, the maximum kinetic energy of emitted electrons should depend on intensity of incident light and not on frequency whereas Einstein\u2019s equation explains that it depends on frequency and not on intensity. \n(iii) Instantaneous emission of electrons :<\/strong> \nAccording to wave theory there should be time lag between emission of electrons and incident of light whereas Einstein\u2019s equation explains why there is no time lag between incident of light and emission of electrons.<\/p>\nSECTION : C<\/strong><\/p>\nAnswer 11. \n(a) A charge particle having charge q is moving with velocity V in a magnetic field strength \u2018B\u2019 then the force acting on it is given by the formula F = q (\\(\\vec { v } \\) x \\(\\vec { B } \\))\u00a0 and F = qv B sin 0 (where 0 is the angle between velocity vector and magnetic field). \nDirection of force is given by the cross product of velocity and magnetic field. \n \n(b)\u00a0\u03b1 particle will trace circular path in clockwise direction as it’s deviation will be in the direction (\\(\\vec { v } \\) x \\(\\vec { B } \\)) \ni .e, perpendicular to the velocity of particle. neutron will pass without any deviation as\u00a0magnetic field does not exert neutral particle. \nElectron will trace circular path in anticlockwise direction as its deviation will be in the direction opposite to (\\(\\vec { v } \\) x \\(\\vec { B } \\))\u00a0with a\u00a0<\/span>smaller radius due to large charge\/mass ratio as r = mv\/qB<\/p>\nAnswer 12. \n(i) Let capacitance of X be C1<\/sub> and capacitance of Y be C2<\/sub>. \n \n(ii) \n \n \n(iii) \n <\/p>\nAnswer 13 \n(i)\u00a0 Mutual induction is the phenomenon of production of induced emf in one coil due to change of current in the neighbouring coil. The coil in which the current changes is called primary coil and the coil in which emf is induced is called the secondary coil.<\/p>\n
(ii) \n <\/p>\n
Answer 14. \n(a) \n(i) Size of Antenna :<\/strong> \nThe size of antenna required will be of order of \u03bb\/4. When frequency is small, the height of antenna will be large, so audio frequency signal would be modulated over a high frequency carrier wave. \n(ii) Effective power radiated by an Antenna :<\/strong> \nAs power radiated \u03bc 1\/\u00a0\u03bb2<\/sup> \u2014, hence when frequency is increased then the power radiated will be more. \n(b) Advantage of frequency modulation over amplitude modulation. \n(i) Noise can be reduced \n(ii) Transmission efficiency is more because the amplitude of an FM wave is constant.<\/p>\nAnswer 15. \nMagnetic field produced on the wire (carrying current I2<\/sub>) due to I1<\/sub> will be. \n \n\u21d2\u00a0 Attractive force between wires \nIf l = 1 m, d = I m, I1<\/sub> = I2<\/sub> = I and F = 2 x 10-7\u00a0<\/sup>N \n\u21d2\u00a0 \u00a0I = I A \nSo one ampere is defined as the current, which when maintained in two parallel infinite length conductors, held at a separation of one metre will produce a force of 2 x 10-7\u00a0<\/sup>N\u00a0per metre of each conductor.<\/p>\nAnswer 16. \nReflecting Telescope :<\/strong> \nThe reflecting telescope uses concave mirror as objective. The rays of light coming from distant object are incident on the objective (parabolic reflective). After reflection the rays of light meet at a point where another convex mirror is placed. This mirror focusses light inside the telescope tube. The final image is seen through the eye piece. \n \nThe images produced by the reflecting telescope is very bright and its resolving power is high. \nAdvantages :<\/strong><\/p>\n\nThe resolving power (the ability to observe two object distinctly) is high, due to the large diameter of the objective.<\/li>\n There is no chromatic aberration as the objective is a mirror.<\/li>\n<\/ul>\nAnswer 17. \n \nOscillating charges would give rise to oscillating magnetic field. This oscillating magnetic field, according to Faraday\u2019s Law, will induce an emf, i.e., it produces an oscillating electric field. The electric and magnetic fields are perpendicular to each other and are also perpendicular to the direction of propagation of the wave. E is the envelope of electric intensity vector and B is the envelope of magnetic intensity vector.<\/p>\n
OR<\/strong><\/p>\nCorrection in Amperes Circuital law (Modified Ampere\u2019s law): Maxwell removed the problem of current continuity and inconsistency observed in Ampere\u2019s Circuital law by introducing the concept of displacement current, Displacement current arises due to change\u00a0in electric flux with time and is given by \n \n \nConduction current is because of flow of charges but displacement current is not because of flow of charges but because of change in electric flux.<\/p>\n
Answer 18. \n \nSuppose we have a ring of radius \u2018a\u2019 that carries a uniformly distributed positive charge q. As the total charge q is uniformly distributed, the charge dg on the element dl is \n <\/p>\n
\nThe axial component is dE cos \u03b8 and<\/li>\n The perpendicular component is dE sin \u03b8.<\/li>\n<\/ul>\nSince the perpendicular component of any two diametrically opposite elements are equal and opposite, they cancel out in pairs. Only the axial components will add up to produce the resultant field. \n \n <\/p>\n
Answer 19. \n(i) Three segments of transistor are:<\/strong><\/p>\n\nEmitter<\/li>\n \u00a0Base<\/li>\n Collector<\/li>\n<\/ul>\nEmitter :<\/strong> \nIt is of moderate size and heavily doped, it supplies a large number of majority carriers which flow through the transistor. \nBase :<\/strong> \nIt is very thin and lightly doped and it separates emitter and collector region of transistor and controls the flow of charge carriers. \nCollector :<\/strong> \nThis segment is moderately doped and larger in size as compared to emitter. It collects a major portion of majority carriers supplied by the emitter. \n(ii) \n \nFor input characteristics, base current IB<\/sub> versus base emitter voltage VBE<\/sub> is plotted while collector base voltage VCB<\/sub> is kept constant. VCB<\/sub> is kept large, that is 3V to 20V. Input characteristics for various values of VCB<\/sub> gives almost same curves. \n \nOutput characteristics is obtained by varying Ic<\/sub> with VCE<\/sub> keeping IB<\/sub> constant. Different curves are obtained for different values of IB<\/sub>. \n <\/p>\nAnswer 20. \n(a) Given, \nHeight of object = h0 \n<\/sub>Radius of curvature = 20 cm \nMagnification, m = 2 \nObject distance, u = ? \nImage distance, v = ? \n \n \nHence, it will always form virtual image.<\/p>\nAnswer 21. \n(i) Quantization condition :<\/strong> \nOf all possible circular orbits allowed by the classical theory, the electrons are permitted to circulate only in those orbits in which the angular momentumof \n \nWhere L, m, and v are the angular momentum, mass and speed of the electron, r is the radius of the permitted orbit and n is positive integer called principal quantum number. The above equation is Bohr\u2019s famous quantum condition. When an electron of mass m 4<\/sub> is confined to move on a line of length \u03bb with velocity v, the de-Broglie wavelength X associated with electron is: \n \n(ii) In the given energy level diagram, energy increases from level A to B and then B to C. The wavelength corresponds to lowest energy transition is \u03bb2<\/sub>. Then the wavelength \u03bb1\u00a0<\/sub>corresponds to higher energy transition and wavelength \u03bb3<\/sub>\u00a0corresponds to highest energy transition. \n <\/p>\nAnswer 22. \n <\/p>\n
SECTION : D<\/strong><\/p>\nAnswer 23.<\/p>\n
\nMeeta\u2019s father said that these are LED light which consume less power and have high reliability.<\/li>\n The tiny lights in traffic signals are Light Emitting Diode. These are operated by connecting the pn-junction diode in forward biased condition.<\/li>\n Values displayed by Meeta were a good observer, curious and her father shows awareness for energy conservation, power saving and have knowledge about such electrical devices.<\/li>\n<\/ul>\nSECTION : E<\/strong><\/p>\nAnswer 24. \n(i) Drift velocity is defined as the average velocity which the electrons are drifted towards the positive terminal under the effect of applied electric field. \n \n(ii) We know that the current flowing through the conductor is : \n \nWhere p is the specific resistance or resistivity of the material of the wire. It depends on number of free electron per unit volume and temperature.<\/p>\n
(iii) They are used to make standard resistors because :<\/strong><\/p>\n\nThey have high value of resistivity.<\/li>\n Temperature coefficient of resistance is less.<\/li>\n They are least affected by temperature.<\/li>\n<\/ul>\nOR<\/strong><\/p>\n(i) Principle :<\/strong> \nWhen a constant current flows through a wire of uniform cross sectional area then the composition of the potential drop across any length of the wire is directly proportional to that length. Let V be the potential difference across the portion of the wire of length l whose resistance is R. \n \n(ii) Total resistance of the primary circuit \n15 + 10 = 25 \u03a9,\u00a0 emf = 2V \n <\/p>\nAnswer 25. \n(i) Let a, series LCR circuit is connected to an ac source V. We take the voltage of the source to be \nV = V0<\/sub> sin \u03c9t. \n \nThe AC current in each element is the same at any time, having the same amplitude and phase. It is given by, \nI = I0<\/sub> sin (\u03c9t + \u03a6) \nLet VL<\/sub>, VR<\/sub>, Vc<\/sub> and V represent the voltage across the inductor, resistor, capacitor and the source respectively. \n \n \nCondition :<\/strong> \nThe current will be in phase with the voltage at resonance condition.\u00a0 At resonance condition \n <\/p>\nOR<\/strong><\/p>\n(i) A transformer is an electrical device for converting an alternating current at low voltage into that at high voltage or vice-versa. \n <\/p>\n
\nIf it increases the input ac voltage, it is called step up transformer.<\/li>\n If it decreases the input ac voltage, it is called step down transformer.<\/li>\n<\/ul>\nPrinciple :<\/strong> \nIt works on the principle of mutual induction i.e., When a changing current is passed through one of the two inductively coupled coils, an induced emf is set up in the other coil. \n \nWorking Theory : \n<\/strong> As the AC flows through the primary coil, it generates an alternating magnetic flux in the core which passes through the secondary coil. \nLet\u00a0 \u00a0N1<\/sub>\u00a0= No. of turns in primary coils \nN2<\/sub> = No. of turns in secondary coils \nThis changing flux set up an induced emf in the secondary, also a self induced emf in the primary.If there is no leakage of magnetic flux, then flux linked with each turn of the primary coil will be equal to that linked with each of the secondary coil. According to Faraday\u2019s law of induction. \n \nVarious energy losses in transformer are :<\/strong><\/p>\n\nCopper loss :<\/strong> \nSome energy is lost due to the heating of copper wires used in the primary and secondary windings. This power loss (P = I2<\/sup>R) can be minimised by using thick copper wires of low resistance.<\/li>\nEddy current loss :<\/strong> \nThe alternating magnetic flux induces eddy current in the iron core which leads to some energy loss in the form of heat. This loss can be reduced by using laminated iron core.<\/li>\nHysteresis loss :<\/strong> \nThe alternating current carries the iron core through cycles of magnetisation and demagnetisation. Work done in each of these cycles and is lost as heat. This is called hysteresis loss and can be minimised by using core material having narrow hysteresis loop.<\/li>\nFlux leakage :<\/strong> \nThe magnetic flux produced by the primary may not fully pass through the secondary. Some of the flux may leak into air. This loss can be minimised by winding the primary and secondary coils over one another.<\/li>\n<\/ul>\n(ii) Given , N1<\/sub> = 100, k = 100, V1<\/sub> = 220V, P1<\/sub> = 1100W \n \n <\/p>\nAnswer 26 . \n(i) Let the two waves arising from the slits A and B have the amplitudes \na and b and the phase difference \u03a6. Such that y1<\/sub> = a sin \u03c9t and y2<\/sub> =\u00a0 b sin(\u03c9t +\u03a6). \nThe resultant displacement is given as :<\/strong> \n \n \n(ii) Comparison of interference pattern observed in Young\u2019s double slits and the single slits diffraction: \n <\/p>\nOR<\/strong><\/p>\n(i) Angle of deviation is the angle through which the incident rays is deviated on passing through a prism i.e., angle between the incident ray and emergent ray. It is denoted by \u03b4. \n \n \n(ii) Dispersion of light :<\/strong> \nUpon passing through the prism, the white light is separated into its component colours : red, orange, yellow, green, blue, and violet. The separation of visible light into its different colours is known as dispersion. Dispersion occurs because for different colour of light a transparent medium will have different refractive indices (p). \n(iii) For total internal reflection : \n \n<\/strong><\/p>\n<\/div>\nWe hope the CBSE Sample Papers for Class 12 Physics Paper 1 help you. If you have any query regarding CBSE Sample Papers for Class 12 Physics Paper 1, drop a comment below and we will get back to you at the earliest.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"
CBSE Sample Papers for Class 12 Physics Paper 1 are part of CBSE Sample Papers for Class 12 Physics. Here we have given CBSE Sample Papers for Class 12 Physics Paper 1. CBSE Sample Papers for Class 12 Physics Paper 1 Board CBSE Class XII Subject Physics Sample Paper Set Paper 1 Category CBSE Sample … Read more<\/a><\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6805],"tags":[],"yoast_head":"\nCBSE Sample Papers for Class 12 Physics Paper 1 - CBSE Library<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n