allen physics part 2

ALLENÒ CBSE 99 3. Assertion : Hydrogen atom consists of only one electron but its emission spectrum has many lines. Reason : Only Lyman series is found in the emission spectrum. (i) a (ii) b (iii) c (iv) d 4. Assertion : For the scattering of a-particle at a large angle only nucleus of the atom is responsible. Reason : Nucleus is very heavy is comparison to electrons. (i) a (ii) b (iii) c (iv) d 5. Assertion : Atoms are not electrically neutral. Reason : Number of protons and electrons are different. (i) a (ii) b (iii) c (iv) d 6. What is the value of ionisation energy for a hydrogen atoms. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx 7. Define Impact parameter. www.notesdrive.com 8. What are the maximum number of spectral lines emitted by a hydrogen atoms when it is in the third excited state? 9. What are the kinetic & potential energies of electron in ground state. 10. Name the series of hydrogen spectrum lying in visible region? 11. Calculate the frequency of the photon, which can excite the electron to -3.4 eV from -13.6 eV. 12. What is the ratio of radii of the orbits corresponding to first excited state and ground state in a hydrogen atom? 13. The first line of the Lyman series in the hydrogen spectrum has wavelength 1200 Å. Calculate the wavelength of the second line. 14. A hydrogen atom initially in the ground state absorbs a photon, which excites it to the n=4 level. Determine the wavelength and frequency of photon. 15. The total energy of an electron in the first excited state of hydrogen atom is –3.4eV what is its K.E.? What is its potential energy? RACE SOLUTIONS CHAPTER-12 : ATOMS 1. (iii) 2. (i) 3. (ii) 4. (i) 5. (iv) 6. 13.6 eV 7. Impact parameter is the perpendicular distance of a velocity vector from the central line of nucleus. 8. spectral lines = n(n -1) , Here n = 3, So (3) ´ (2) = 3 , So there are 3 spectral lines. 2 2 9. Kinetic energy = 13.6 eV, Potential energy = –27.2 eV 10. Balmer series. 11. E = –3.4eV – (–13.6eV) = 10.2eV, E = 10.2eV u= 10.2eV 10.2 ´1.6 ´10 –19 = 2.46 ´10+15 Hz 6.63 ´ 10 –34 6.63 ´10–34 = E

100 Physics ALLENÒ 12. rn µ n2 Þ r2 / r1= 4/1 13. For first line, 1 = R æ 1 - 1 ö = 3R ………(1) l1 èç 12 22 ÷ø 4 For second line. 1 = R æ 1 - 1 ö = 8R ………(2) l2 çè 12 32 ÷ø 9 1 3R l1 = 4 Þ l2 = 3 ´ 9 Þ l2 = 27 ´ 1200Å = 1012.5Å 1 8R l1 4 ´ 8 32 l2 9 14. Ground state n1 = 1, n2 = 4 é1 = 1 = 912 Å úùû êë R 1.097 ´10+7 www.notesdrive.com 1 = R æ 1 - 1 ö , 1 = R æ 15 ö Þ l = 16 = 16 ´ 912Å = 972Å l node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docxèç1242ø÷lçè16ø÷15R15 Frequency n = c = 3 ´ 108 = 3´1015 Hz l 972 ´10–10 15. E = -3.4eV; KE = -E K.E. = 3.4eV, P.E. = -2 × K.E. = -6.8eV EXERCISE - V (RACE) CHAPTER-13 : NUCLEI For question numbers 1 to 5 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false and R is also false 1. Assertion : Energy is released in nuclear fission. Reason : Total binding energy of the fission fragments is larger then the total binding energy of the parent. (i) a (ii) b (iii) c (iv) d 2. Assertion : Nucleus may emit negative charged particles. Reason : Nucleus contains negative charge also. (i) a (ii) b (iii) c (iv) d 3. Assertion : Nuclear binding energy per nucleon is in the order 9 Be > 7 Li > 4 He 4 3 2 Reason : Binding energy per nucleon increases linearly with difference in number of neutrons and protons. (i) a (ii) b (iii) c (iv) d E

ALLENÒ CBSE 101 4. Assertion : Isobars are the element having same mass number but different atomic number. Reason : Neutrons and protons are present inside nucleus. (i) a (ii) b (iii) c (iv) d 5. Assertion : Density of all the nucleus is same . Reason : Radius of nucleus is directly proportional to the cube roots of mass number. (i) a (ii) b (iii) c (iv) d 6. Give one example of nuclear fusion. 7. What do you mean by mass defect? 8. Name the absorbing material used to control the reaction rate of neutrons in a nuclear reactor? 9. Find the density of 12 C 6 10. Write down the process involved in atoms bomb. 11. Give the names of two moderators used in nuclear reactor. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx 12. Write the necessary condition required for fusion reaction. www.notesdrive.com 13. Write any two characteristic properties of nuclear force. 14. How is the radius of a nucleus related to its mass number? 15. Select the pairs of isobars from the following nuclei ? 11Na22, 11Na23, 10Ne23. RACE SOLUTIONS CHAPTER - 11 : DUAL NATURE OF RADIATION AND MATTER 1. (i) 2. (iii) 3. (iv) 4. (ii) 5. (i) 6. 4 1H1 ® 4 He + 210 e + 2u + 26.7 MeV 2 7. The difference between total mass of nucleons of nucleus and actual mass of a nucleus is called mass defect. 8. Control rods or cadmium rods. 9. r= M = (12)(1.66 ´10-27 kg) = 2.4 × 1017 kgm–3 4 pR 3 4 p[1.2 ´10 -15 ´ (1.2)1/ 3 ] 3 3 10. Nuclear fission 11. (1) Graphite (2) D2O (heavy water) 12. (i) Nuclear fusion will occur when the kinetic energy of colliding nuclei is enough to overcome the strong electrostatic forces of repulsion between the protons. For this, high temperature is required. (ii) The density of nuclei should also be very high to increase the number of collisions. 13. The following are the two characteristic properties: (i) Nuclear force is a short range force. (ii) Nuclear forces show the saturation effect. 14. The radius if of a nucleus of mass number A is related as if = R0A1/3, where R0 is a constant. 15. 11Na23, 10Ne23. E

102 Physics ALLENÒ EXERCISE - VI (MOCK TEST) CHAPTER-12 : ATOMS 1. The first line of the Lyman series in the hydrogen spectrum has wavelength 1200 Å. Calculate the wavelength of the second line. [1] 2. Define ionization energy. [1] 3. What is the ratio of radii of the orbits corresponding to first excited state and ground state in a hydrogen atom? [1] 4. What is the Bohr's quantization condition of the angular momentum of an electron in the second www.notesdrive.com orbit? [2] node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx 5. The kinetic energy of -particle incident on gold foil is doubled then how does the distance of closest approach change? [2] 6. Calculate the ratio of energies of photons produced due to transition of electron of hydrogen atom from its, [2] (i) Second permitted energy level to the first level, and (ii) Highest permitted energy level to the second permitted level. 7. State the limitations of Bohr's atomic model? [2] 8. When is Ha line in the emission spectrum of hydrogen atom obtained ? Calculate the frequency of the photon emitted during this transition. [3] 9. Find the ratio of maximum wavelength to minimum wavelength for the lines of Balmer series in hydrogen spectrum. [3] 10. The energy of the electron in the ground state of hydrogen atom is –13·6 eV. Find the kinetic energy and potential energy of electron in this state. [3] 11. Write first and second postulates of Bohr's model of atom. Derive the expression for radius of stationary orbit of an electron and its orbital velocity. [5] E

ALLENÒ CBSE 103 EXERCISE - VI (MOCK TEST) CHAPTER-13 : NUCLEI 1. Two nuclei have mass numbers in the ratio 7 : 16. What is the ratio of their nuclear densities? [1] 2. What will be the value of neutron multiplication factor for controlled chain reactions ? [1] 3. What is nuclear force ? [1] 4. Distinguish between isotopes and isobars. Give are example for each of the species? [2] 5. Express 16mg mass into equivalent energy in electron volt. [2] 6. Why is nuclear fusion not possible in laboratory? [2] node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx7. You are given two nuclides 3X7 and 3Y4.[2] (i) Are they the isotopes of the same element? Why? www.notesdrive.com (ii) Which one of the two is likely to be more stable? Give reason. 8. Calculate the energy in fusion reaction : 1H2 + 1H2 ® 2He3 + 0n1 [3] where B.E. of 1H2 = 2.23 MeV and of 2He3 = 7.73 MeV. 9. Differentiate between nuclear fission & fusion. Give an example of each. Which of the above reactions take place in nuclear reactor ? [3] 10. Obtain the binding energy of the nuclei 26Fe56 and 83Bi209 in units of MeV from the following data : m(26Fe56) = 55.934939 a.m.ui m(83Bi209) = 208.980388 a.m.u. (Take 1 a.m.u. = 931.5 MeV) [3] 11. (a) Draw a plot showing the variation of potential energy of a pair of nucleons as a function of their separation. Mark the regions on the graph where the force is (i) attractive, and (ii) repulsive. [5] (b) Write three characteristic features of nuclear force which distinguish it from the Coulomb force. (c) What will be the value of neutron multiplication factor for controlled chain reactions ? E

104 Physics ALLENÒ IMPORTANT NOTES ______________________________________________________________________www.notesdrive.com ______________________________________________________________________ ______________________________________________________________________ node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ E

ALLENÒ CBSE 105 UNIT-IX : ELECTRONIC DEVICES CHAPTER-14 : SEMI CONDUCTOR ELECTRONICS 1. CLASSIFICATION OF CONDUCTORS, SEMICONDUCTOR, INSULATOR Properties Conductor Semiconductor Insulator Resistivity 10–2 – 10–8 Wm 10–5 – 106 Wm 1011 – 1019 Wm 102 – 108 mho/m 10-6 – 105 mho/m 10–19 – 10–11 mho/m Conductivity Positive Negative Negative Temp. Due to free electrons Due to electrons and holes No current Coefficient of resistance (a) Conduction Band Conduction Band Conduction Band Current Energy band diagram node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_Unit-9.docx Forbidden No gap Forbidden Gap E g £3eV ForbiddeEng > 3eV Electronwww.notesdrive.com Energyenergy gap Overlapping regionValence BandValence Band Gap Electron EnergyExampleConductorSemi conductor Electron Energy Valence Band @ 0eV £ 3eV Pt, Al, Cu, Ag Ge, Si, GaAs, GaF2 Insulator > 3eV Wood, plastic, Diamond, Mica 2. CLASSIFICATION OF SEMICONDUCTORS (i) Intrinsic Semiconductor : Current due to electron & hole ne = nh = ni µe = e– mobility J = ne [ ve + vh] (Current density) µh = hole mobility s=1 = en [µe + µh] (Conductivity) ve = e– velocity r µh = hole velocity EC EC Eg Eg EV EV (a) (b) (a) An intrinsic semiconductor (b) At T > 0 K, four thermally at T = 0 K behaves like insulator. generated electron-hole pairs. The filled circles ( ) represent Ex: Ge, Si electrons and empty fields E ( ) represent holes.

106 Physics ALLENÒ (ii) Extrinsic semiconductor (Doped semiconductor) : Conductivity s = e(neµe + nhµh) Current Density J = e (neve + nhvh) EC { (a) N-type 0.01 eV Current mainly due to electron ED Electronwww.notesdrive.com energy EV Eg Pentavalent impurity (P, As, Sb etc.) Donor impurity (ND) ne >> nh J @ e ne ve s=1 @ e ne me T > OK r Energy bands of n-type semiconductor (b) P-type at T > OK. Current mainly due to hole EC Trivalent impurity Eg EV (Ga, B, In, Al) E} A 0.01–0.05 eV acceptor impurity (NA) nh >> ne J @ e nh vh s=1 @ e nh mh T > OK r Energy bands of p-type semiconductor at T > OK. w MASS-ACTION LAW : n2i = ne ´ nh For N-type semiconductor ne ; ND For P-type semiconductor nh ; NA 3. P-N JUNCTION (At equilibrium condition) p –+ n node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_Unit-9.docx hole free electron electric V0 potential distance E

ALLENÒ CBSE 107 Direction of diffusion current : P to N side and drift current : N to P side. If there is no biasing then diffusion current = drift current. So total current is zero. (equilibrium condition) In junction N side is at high potential relative to the P side. This potential difference tends to prevent the movement of electron from the N region into the P region. This potential difference called a barrier potential. 4. FORWARD BIAS AND REVERSE BIAS OF SEMICONDUCTOR DIODE Forward Bias Reverse Bias PN PN VV +– –+ 1 Potential Barrier reduces 1 Potential Barrier increases. 2 Width of depletion layer 2 Width of depletion layer decreases increases. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_Unit-9.docx 3 P-N jn. provide very small 3 P-N jn. provide high www.notesdrive.comresistance resistance 4 Forward current flows in 4 Very small current flows. the circuit 5 Order of forward current 5 Order of current is micro is milli ampere. ampere for Ge or Nano ampere for Si. 6 Current flows mainly due 6 Current flows mainly due to majority carriers. to minority carriers. 7 Forward characteristic 7 Reverse characteristic curves. curve Vr(volt) if break down Reverse (mA) voltage saturation 0 current knee voltage Vf(volt) Ir (mA) 8 Forward Resistance : 8 Reverse Resistance Rf = DVf @ 100W : Rr = DVr @ 106W DIf DIr 9 Order of knee or cut in 9 Breakdown voltage voltage Ge® 0.3 V Ge ® 25 V Si ® 0.7 V Si ® 35 V Special point : Generally Rr = 104 : 1 for Si Rf Rr = 103 : 1 for Ge Rf 5. V-I CHARACTERISTICS OF A P-N JUNCTION DIODE I(mA) 100 0.2 0.4 0.6 0.8 1.0 V(V) 80 60 40 100 80 60 40 20 20 Vbr Typical V-I characteristics of a silicon diode. E

108 Physics ALLENÒ 6. APPLICATION OF JUNCTION DIODE AS A RECTIFIER Rectification : Conversion of AC voltage into DC voltage. (a) Half wave rectifier In this rectifier only half of the input wave is rectified Transformer A X Primary Secondary RL (a) B Y Voltageo acrossmRL Voltage at A INPUT VOLTAGE t OUTPUT VOLTAGE (b) t (a) Half-wave rectifier circuit, (b) Input ac voltage and output voltage waveforms from the rectifier circuit. (b) Centre – Tap Full wave Rectifier Waveformw.atAnotesdrive.c In full wave rectifier we get output voltage during both the positive and negative half cycle. Center-Tap Transformer Diode 1(D1) Center A X Tap B RL Output Diode 2(D2) (a) Y Waveform atwB w t (i) (ii) t node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_Unit-9.docx (b) Output waveform Due to Due to Due to Due to (across RL) D1 D2 D1 D2 t (c) (a) A Full-wave rectifier circuit; (b) Input wave forms given to the diode D1 at A and to the diode D2 at B; (c) Output waveform across the load RLconnected in the full wave rectifier circuit. E

ALLENÒ CBSE 109 NCERT IMPORTANT QUESTIONS CHAPTER - 14 : SEMICONDUCTOR ELECTRONICS 1. Pure Si at 300K has equal electron (ne) and hole (nh) concentrations of 1.5 × 1016 m–3. Dopping by indium increases nh to 3 × 1022 m–3. Calculate ne in the doped Si. Sol. For a doped semi-conductor in thermal equilebrium nenh = ni2 (Law of mass action) ne = n 2 = (1.5 ´1016 )2 = 7.5 × 109 m–3 i 3 ´1022 nh 2. The I-V characteristic of a p-n junction diode is shown in figure. Find the approximate dynamic resistance of the p-n junction when (a) a forward bias of 1 volt is applied, (b) a forward bias of 2 volt is applied node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_Unit-9.docx www.notesdrive.com Sol. (a) The current at 1 volt is 10 mA and at 1.2 volt it is 15 mA. The dynamic resistance in this region is R = DV = 0.2 volt = 40 W Di 5 mA (b) The current at 2 volt is 400 mA and at 2.1 volt it is 800 mA. The dynamic resistance in the region is R= DV = 0.1 volt = 0.25 W . Di 400 mA 3. A potential barrier of 0.50 V exists across a p-n junction. If the depletion region is 5.0 × 10–7 m wide, what is the intensity of the electric field in this region ? Sol. The electric field is E = V/d = 0.50 V = 1.0 ´106 V / m. 5.0 ´10-7 m 4. Explain, with the help of a circuit diagram, how the thickness of depletion layer in a p-njunction diode changes when it is forward biased. In the following circuits which one of thetwo diodes is forward biased and which is reversebiased? Sol. (a) The p-n junction is reverse biased. (b) The p-n junction is forward biased. E

110 Physics ALLENÒ 5. In half wave rectifier, if the input frequency is 50 Hz, what will be the output frequency ? What is the output frequency for full wave rectifier for same input frequency ? Sol. output frequency Half wave rectifier Û 50 Hz full wave rectifier Û 100 Hz 6. What is rectification? Draw the circuit diagram of half wave rectifier and explain its working. Show the input ac voltage and output voltage waveforms from the rectifier circuit. Sol. Rectification :- Conversion of A.C. voltage in to DC voltage Working of half wave rectifier :- The secondary coil of a transformer supplies the desired alternating voltage across terminals A and B. When the voltage at A is in positive cycle, the diode is in forward bias and it conducts current. When A is at negative cycle, the diode is in reverse bias and it does not conduct current and only in the positive half-cycle of alternative current (ac), there is a current through the load resistor RL and we get output voltage. Thus, We get half wave rectified output. So this circuit is called as half-wave rectifier. www.notesdrive.com Input Primary A X node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_Unit-9.docxalternatingRL Output Secondary voltage Y B Half wave rectifier circuit Input t® alternating voltage O Output direct voltage O t® Input and Output voltage waveforms from the rectifier circuit 7. What is extrinsic semiconductor? How many types of these are? Write their names. Explain the processes which are occurred during the formation of a P-N junction. Sol. Extrinsic semiconductor :– When suitable impurity is doped into a pure semiconductor in very small quantity like part per million (ppm) then the conductivity of semiconductor is increases very high. These type of material is called extrinsic semiconductor these are of two types : (1) n- type semiconductor (2) p-type semiconductor E

ALLENÒ CBSE 111 Two important processes which are occured during the formation of p-n junction : (1) Diffusion (2) Drift (1) Diffusion :- During the formation of p-n junction and due to the concentration gradient across p and n sides holes diffuses from p side to n side (p®n) and electrons diffuses, from n-side to p-side (n®p). This motion of charge carrier gives rise to diffusion current across the junction. (2) Drift :- When an electron diffuses from n®p it leaves behind an ionized donor on n-side. This ionized donor (positive charge) is immobile as it is bonded to the surrounding atoms. As the electrons continue to diffuse from n®p. A layer of positive charge (or positive space charge region) on n-side of junction is developed. Similarly when a hole diffuses from p®n due to the concentration gradient it leaves behind an ionized acceptor (negative charge) which is immobile as the holes continue to diffuse from p®n. A layer of negative charge (or negative space charge region) on the p side of Junction is developed. Due to this layer an electron on p-side of the junction moves to n-side and a hole on n-side of junction moves to p-side. The motion of charge carriers due to the electric field is called drift. 8. Distinguish between conductors, insulators and semi-conductors on the basis of band theory of solids. Sol. (1) Conductor :– On the basis of energy band theory, those meterial in which valence band and conduction band are partially filled or valance band and conduction band overlap is known as conductor. There is no forbidden energy gap between the valence band and the conduction band. In case of partially filled valence band, electrons from filled energy levels can move easily to the unfilled higher energy levels and hence conduction of electrons take place. In case of the overlap of valence and conduction bands, electrons move easily from valence band to the conduction band. Thus, large number of electrons are available for the conduction of electricity. Resistivity of conducting materials are from 10–2 to 10–8 ohm- metre. E node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_Unit-9.docx www.notesdrive.com CB (Conduction band) VB (Valence band) (2) Insulator :- On the basis of energy band theory, in insulators, the valence band is completely filled with electrons and the conduction band is empty and both the bands are separated by a forbidden energy gap of about greater than 3eV. At room temperature, the electrons in the valence band cannot go to the conduction band and hence insulator cannot conduct electricity. Therefore, insulator is a bad conductor of electricity. Resistivity of insulators are from 1011 to 1019 ohm-metre. (empty) Conduction band E > 3eV (filled) Valence band E

112 Physics ALLENÒ (3) Semiconductors :- On the basis of energy band theory, semiconductors are those material in which the forbidden energy gap between the filled valence band and the empty conduction band is very small (i.e., < 3ev). At absolute zero temperature (0 K), semiconductor behave as an insulator. But at room temperature, some of the electrons in valence band have sufficient thermal energy to jump to the conduction band and a semiconductor can conduct even at room temperature. Germanium and Silicon are the examples of semiconductors. In case of Silicon, the forbidden energy gap (Eg) at room temperature is about 1.1 eV and for Germanium it is about 0.7 eV. Resistivity of semiconductors are from 10–5 to 106 ohm-metre. www.notesdrive.com Conduction band node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_Unit-9.docx E < 3eV Valence band 9. What is rectification ? Explain the working of a fullwave rectifier. Draw necessary circuit diagram. Sol. Rectification : The process in which alternating current is changed into direct current by p-n junction diode is called rectification. Working : When positive half cycle of input a.c. signal flows through the primary coil, induced emf is set up in the secondary coil due to mutual induction. The direction of induced emf is such that the upper end of the secondary coil becomes positive while the lower end becomes negative. Thus, diode D1 is forward biased and diode D2 is reverse biased, so the current due to diode D1 flows through the circuit. The output voltage which varies in accordance with the input half cycle is obtained across the load resistance (RL). During negative half cycle of input a.c. signal, diode D1 is reverse biased and diode D2 is forward biased. The current due to diode D2 flows through the circuit. The output voltage is obtained across the load resistance (RL). The input and corresponding output voltage are shown in figure. Since both the halves of input a.c. (wave) are rectified, so the junction diode is called a full wave rectifier. E

ALLENÒ CBSE 113 Alternating Centre tap X voltage transformer RL (output) A diode D1 Y diode D2 B Full-wave rectifier circuit Input t alternating voltage node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_Unit-9.docxOutputDue Due Due Due direct to D1 to D2 to D1 to D2 www.notesdrive.comvoltage Input and Output wave form from rectifier circuit t 10. Explain with the help of a diagram the formation of depletion region and barrier potential in a pn junction. Sol. It is clear that, N-type semiconductor has an excess of free electrons and P-type has an excess of holes therefore when both are placed together to form a junction, electrons migrate towards the P-side and holes migrate towards N-side due to concentration gradient. Departure of an electron from the N-side to the P-side leaves a positive donor ion on N-side and likewise hole leaves a negative acceptor ion on the P-side resulting in the formation of depletion layer having width ; 10–7m. P-TYPE N-TYPE hhhh eeee hhhh eeee hhhh eeee hhe e e e eh ehhh ehee hhhh e e eh hhhh eeee Depletion layer ®Ei Depletion layer : It is the layer near the junction in which electrons are absent on n side and holes are absent on p side. Potential barrier : Due to the accumulation of immobile ions near the junction an electric potential difference (Vb) developes b/w n side & p side which acts as a barrier for further diffusion of electrons and holes. Vb = Ei × d (volt) E

114 Physics ALLENÒ PREVIOUS YEARS QUESTION CHAPTER - 14 : SEMICONDUCTOR ELECTRONICS EXERCISE-I TWO MARK QUESTIONS : 1. Explain, with the help of a circuit diagram, the working of a p-n junction diode as a half-wave rectifier. Sol. ++ P1 S1 A RL output P–2 –S2 B www.notesdrive.com Transformer Half-wave rectifier:-Diode conducts during positive half cycle & does not conduct during node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_Unit-9.docx negative half cycle. Hence ac is converted into unidirectional pulsating DC by diode. Voltage at A t Voltage across RL t EXERCISE-III THREE MARK QUESTION : 1. Show on a plot, variation of resistivity of (i) a conductor, and (ii) a typical semiconductor as a function of temperature. Using the expression for the resistivity in terms of number density and relaxation time between the collisions, explain how resistivity in the case of a conductor increases while it decreases in a semiconductor, with the rise of temperature. Sol. Variation of resistivity with temperature (i) For conductor (ii) For semiconductor resistivity r (r) O T(K) O T(K) E

ALLENÒ CBSE 115 Resistivity of material is given by m r = ne2t n Þ Number density t Þ relaxation time On increasing the temperature of conductor relaxation time decreases while number density remains constant. Due to this resistivity of conductor increases. In case of semiconductor, on increasing the temperature relaxation time decreases but number density increases. Increase in the number density is more effective then decrease the relaxation time. That's why resistivity of semiconductor decreases. 2. Distinguish between a conductor and a semi conductor on the basis of energy band diagram. Sol. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_Unit-9.docx Conduction band Conduction band Electron Energy Eg Eg Conduction band www.notesdri EEnnveerrggeyy.comValence bandValence band Valence band Insulator Semiconductor Conductor (Eg > 3eV) (Eg < 3eV) (Eg ~ 0) [Eg = Forbidden Energy gap] 3. (i) Explain with the help of a diagram the formation of depletion region and barrier potential (ii) in a p-n junction. (Refer to NCERT Q. No. 10) 4. (i) (ii) Draw the circuit diagram of a half wave rectifier and explain its working Sol. (i) (Refer to NCERT Q. No. 6) Name two important processes that occur during the formation of a pn junction. Draw the circuit diagram of a full wave rectifier along with the input and output waveforms. Briefly explain how the output voltage/current is unidirectional. (Refer to NCERT Q. No.9) Two important phenomena occur during formation of pn junction are : (1) Diffusion – Due to concentration difference the majority charge carrier from p side (holes) and majority charge carrier (electrons) on n side start diffusing towards their opposite side. (2) Drift – As electron-hole pairs neutralises near the junction they leave behind immobile ions, these ions creates an electric field from n side to p side due to which minority charge carrier start moving (drifting). E

116 Physics ALLENÒ 5. (a) In the following diagram, is the junction diode forward biased or reverse biased ? +5V (b) Draw the circuit diagram of a full wave rectifier and state how it works. (Refer to NCERT Q. No. 9) Sol. (a) Reverse biased as p–side is connected to the negative terminal & n–side to the positive terminal. 6. A student wants to use two p-n junction diodes to convert alternating current into direct current. Draw the labelled circuit diagram she would use and explain how it works. (Refer to NCERT Q. No. 9) EXERCISE - III FIVE MARKS QUESTIONS :- www.notesdrive.com Consider a thin p-type silicon (p-Si) semiconductor wafer. By adding precisely a small quantity of pentavalent impurity, part of the p-Si wafer can be converted into n-Si. There are several node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_Unit-9.docx processes by which a semiconductor can be formed. The wafer now contains p-region and n- region and a metallurgical junction between p-, and n- region. Two important processes occur during the formation of a p-n junction: diffusion and drift. We know that in an n-type semiconductor, the concentration of electrons (number of electrons per unit volume) is more compared to the concentration of holes. Similarly, in a p-type semiconductor, the concentration of holes is more than the concentration of electrons. During the formation of p-n junction, and due to the concentration gradient across p-, and n- sides, holes diffuse from p- side to n-side (p ® n) and electrons diffuse from n-side to p-side (n ® p). This motion of charge carriers gives rise to diffusion current across the junction. (i) How can a p-type semiconductor be converted into n- type semiconductor? (A) adding pentavalent impurity (B) adding trivalent impurity (C) not possible (D) heavy doping Ans. (A) (ii) Which of the following is true about n-type semiconductor? (A) concentration of electrons is less than that of holes. (B) concentration of electrons is more than that of holes. (C) concentration of electrons equal to that of holes. (D) None of the above . Ans. (B) (iii) Which of the following is true about p-type semiconductor? (A) concentration of electrons is less than that of holes. (B) concentration of electrons is more than that of holes. (C) concentration of electrons equal to that of holes. (D) None of these Ans. (A) (iv) Which of the following is the reason about diffusion current? (A) diffusion of holes from p to n (B) diffusion of electrons from n to p (C) both (A) and (B) (D) None of these Ans. (C) (v) What are the processes that occur during formation of a p-n junction? (A) drift (B) diffusion (B) both (A) and (B) (D) None of these Ans. (C) E

ALLENÒ CBSE 117 EXERCISE - IV (RACE) CHAPTER- 14 : SEMICONDUCTOR ELECTRONICS For question numbers 1 to 5 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false and R is also false 1. Assertion : The number of electrons in a p-n silicon semiconductor is less than the number of electrons in a pure silicon semiconductor at room temperature. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_Unit-9.docx Reason : It is due to law of mass action www.notesdrive.com (i) a (ii) b (iii) c (iv) d 2. Assertion :Conductivity of a semiconductor increases on doping. Reason : Doping raises the temperature of semiconductor. (i) a (ii) b (iii) c (iv) d 3. Assertion : Semiconductors do not obey ohm's law. Reason : Electric current is determined by the rate of flow of charge carriers. (i) a (ii) b (iii) c (iv) d 4. Assertion : In solar cell it is essential that photo generation of electrons and holes occurs in depletion region. Reason : The junction field separate electron hole pair effectively. (i) a (ii) b (iii) c (iv) d 5. Assertion : In a semiconductor, the conductivity have a higher mobility than holes. Reason : The electrons experience fewer collisions. (i) a (ii) b (iii) c (iv) d 6. What do you mean by rectification? 7. What happens to the width of depletion layer of a p-n junction when it is forwards biased? 8. At what temeprature would an instrinsic semiconductor behaves like a perfect insulator? 9. What is a hole? What is its physical significance? 10. Name one impurity each, which when added, to pure Si, produces (i) n-type, and (ii) p-type semiconductor. 11. Name two factors on which electrical conductivity of a pure semiconductor at a given temperature depends. 12. What is an ideal diode? E

118 Physics ALLENÒ RACE CHAPTER-14 (SOLUTIONS) 1. (i) 2. (iii) 3. (ii) 4. (i) 5. (iii) 6. The process of conversion of AC voltage/Current to DC voltage/Current. 7. Width of the depletion layer reduces. 8. At 0 K temperature. 9. Hole is the vacancy of electron in valence band. The vacancy with the hole behaves as an apparent free particle with effective positive charge. 10. (i) for n-type, arsenic. (ii) for p-type, Indium. 11. (i) Band gap (ii) Biasing. 12. It is a p-n junction diode which offer zero resistance in forward biasing and infinite resistance in reverse biasing, i.e. current flows through it in one direction only. www.notesdrive.comEXERCISE - V (MOCK TEST) CHAPTER- 14 : SEMICONDUCTOR ELECTRONICS node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_Unit-9.docx 1. In half wave rectifier, if the input frequency is 50 Hz, what will be the output frequency ? [1] 2. Show on a plot, variation of resistivity of a typical semoconductor as a fucntion of temperature. [1] 3. Why is the conductivity of n-type semiconductor greater than that of the p-type semiconductor even when both of these have same level of doping? [2] 4. Draw a pn junction with reverse bias? Which biasing will make the resistance of a p-n-junction high? [2] 5. What is an ideal diode? Draw the output wave form across the load resistor R, if the output waveform is as shown in the figure. [2] 0V –6V 6. What do you mean by depletion region and potential barrier in junction diode? [2] 7. In the following diagrams indicate which of the diodes are forward biased and which are reverse bias? [3] 8. Distinguish between conductors, insulators and semi-conductors on the basis of band theory of solids. [3] 9. What is rectification ? Explain the working of a fullwave rectifier. Draw necessary circuit diagram. [5] E

ALLENÒ CBSE 119 SAMPLE QUESTION PAPER 2021-22 Max. Marks : 35 SUBJECT: PHYSICS (042) TERM-I Time : 90 Minutes General Instructions: (i) The Question Paper contains three sections. (ii) Section A has 25 questions. Attempt any 20 questions. (iii) Section B has 24 questions. Attempt any 20 questions. (iv) Section C has 6 questions. Attempt any 5 questions. (v) All questions carry equal marks. (vi) There is no negative marking. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper SECTION-A www.notesdrive.com 1. Which of the following is NOT the property of equipotential surface? (A) They do not cross each other. (B) The rate of change of potential with distance on them is zero. (C) For a uniform electric field they are concentric spheres. (D) They can be imaginary spheres. 2. Two point charges +8q and –2q are located at x=0 and x=L respectively. The point on x axis at which net electric field is zero due to these charges is :- (A) 8L (B) 4L (C) 2 L (D) L 3. An electric dipole of moment p is placed parallel to the uniform electric field. The amount of work done in rotating the dipole by 90° is :- (A) 2pE (B) pE (C) pE/2 (D) Zero 4. Three capacitors 2µF, 3µF and 6µF are joined in series with each other. The equivalent capacitance is :- (A) 1/2µF (B) 1µF (C) 2µF (D) 11µF 5. Two point charges placed in a medium of dielectric constant 5 are at a distance r between them, experience an electrostatic force ‘F’. The electrostatic force between them in vacuum at the same distance r will be :- (A) 5F (B) F (C) F/2 (D) F/5 6. Which statement is true for Gauss law :- (A) All the charges whether inside or outside the gaussian surface contribute to the electric flux. (B) Electric flux depends upon the geometry of the gaussian surface. (C) Gauss theorem can be applied to non-uniform electric field. (D) The electric field over the gaussian surface remains continuous and uniform at every point. E

120 Physics ALLENÒ 7. A capacitor plates are charged by a battery with ‘V’ volts. After charging battery is disconnected and a dielectric slab with dielectric constant ‘K’ is inserted between its plates, the potential across the plates of a capacitor will become :- (A) Zero (B) V/2 (C) V/K (D) KV 8. The best instrument for accurate measurement of EMF of a cell is :- (A) Potentiometer (B) metre bridge (C) Voltmeter (D) ammeter and voltmeter 9. An electric current is passed through a circuit containing two wires of same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 3:2 and 2:3, then the ratio of the current passing through the wire will be :- (A) 2:3 (B) 3:2 (C) 8:27 (D) 27:8 10. By increasing the temperature, the specific resistance of a conductor and a semiconductor :-www.notesdrive.com (A) increases for both. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper (B) decreases for both. (C) increases for a conductor and decreases for a semiconductor. (D) decreases for a conductor and increases for a semiconductor. 11. We use alloys for making standard resistors because they have :- (A) low temperature coefficient of resistivity and high specific resistance (B) high temperature coefficient of resistivity and low specific resistance (C) low temperature coefficient of resistivity and low specific resistance (D) high temperature coefficient of resistivity and high specific resistance 12. A constant voltage is applied between the two ends of a uniform metallic wire, heat ‘H’ is developed in it. If another wire of the same material, double the radius and twice the length as compared to original wire is used then the heat developed in it will be :- (A) H/2 (B) H (C) 2H (D) 4H 13. If the potential difference V applied across a conductor is increased to 2V with its temperature kept constant, the drift velocity of the free electrons in a conductor will :- (A) remain the same (B) become half of its previous value (C) be double of its initial value (D) become zero. 14. The equivalent resistance between A and B is :- 3W 1A 6W A 8W 30W (A) 3 ohms (B) 5.5 ohms (C) 7.5 ohms (D) 9.5 ohms E

ALLENÒ CBSE 121 15. The SI unit of magnetic field intensity is :- (A) AmN–1 (B) NA–1m–1 (C) NA–2m–2 (D) NA–1m–2 16. The coil of a moving coil galvanometer is wound over a metal frame in order to :- (A) reduce hysteresis (B) increase sensitivity (C) increase moment of inertia (D) provide electromagnetic damping 17. Two wires of the same length are shaped into a square of side 'a' and a circle with radius 'r'. If they carry same current, the ratio of their magnetic moment is :- (A) 2 : p (B) p : 2 (C) p : 4 (D) 4 : p 18. The horizontal component of earth's magnetic field at a place is 3 times the vertical node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper component. The angle of dip at that place is :- www.notesdrive.com (A) p/6 (B) p/3 (C) p/4 (D) 0 19. The small angle between magnetic axis and geographic axis at a place is- (A) Magnetic meridian (B) Geographic meridian (C) Magnetic inclination (D) Magnetic Declination 20. Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon the :- (A) rate at which current change in the two coils (B) relative position and orientation of the coils (C) rate at which voltage induced across two coils (D) currents in the two coils 21. A conducting square loop of side 'L' and resistance 'R' moves in its plane with the uniform velocity 'v' perpendicular to one of its sides. A magnetic induction 'B' constant in time and space pointing perpendicular and into the plane of the loop exists everywhere as shown in the figure. The current induced in the loop is :- (A) BL v/R Clockwise (B) BL v/R Anticlockwise (C) 2BL v/R Anticlockwise (D) Zero 22. The magnetic flux linked with the coil (in Weber) is given by theequation :– Փ = 5t2 + 3t + 16 The induced EMF in the coil at time, t=4 will be- (A) –27 V (B) –43 V (C) –108 V (D) 210 V E

122 Physics ALLENÒ 23. Which of the following graphs represent the variation of current (A) with frequency (f) in an AC circuit containing a pure capacitor ? (A) (B) (C) (D) 24. A 20 volt AC is applied to a circuit consisting of a resistance and a coil with negligible resistance. If the voltage across the resistance is 12 volt, the voltage across the coil is - (A) 16 V (B) 10 V (C) 8 V (D) 6 V 25. The instantaneous values of emf and the current in a series ac circuit are :- E = Eo Sinwt and I= Io sin(wt+p/3) respectively, then it is (A) Necessarily a RL circuit www.notesdrive.com (B) Necessarily a RC circuit node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper (C) Necessarily a LCR circuit (D) Can be RC or LCR circuit SECTION-B This section consists of 24 multiple choice questions with overall choice to attempt any 20 questions. In case more than desirable number of questions are attempted, ONLY first 20 will be considered for evaluation. 26. A cylinder of radius r and length l is placed in an uniform electric field parallel to the axis of the cylinder. The total flux for the surface of the cylinder is given by :- (A) zero (B) p r2 (C) Ep r2 (D) 2 Ep r2 27. Two parallel large thin metal sheets have equal surface densities 26.4 × 10–12 C/m2 of opposite signs. The electric field between these sheets is :- (A) 1.5N/C (B) 1.5 × 10–16 N/C (C) 3 × 10–10N/C (D) 3N/C 28. Consider an uncharged conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then, (A) negative and uniformly distributed over the surface of sphere (B) positive and uniformly distributed over the surface of sphere (C) negative and appears at a point the surface of sphere closest to point charge. (D) Zero 29. Three Charges 2q, –q and –q lie at vertices of a triangle. The value of E and V at centroid of triangle will be- (A) E#0 and V#0 (B) E=0 and V=0 (C) E#0 and V=0 (D) E=0 and V#0 E

ALLENÒ CBSE 123 30. Two parallel plate capacitors X and Y, have the same area of plates and same separation between plates. X has air and Y with dielectric of constant 2, between its plates. They are connected in series to a battery of 12 V. The ratio of electrostatic energy stored in X and Y is :- (A) 4:1 (B) 1:4 (C) 2:1 (D) 1:2 31. Which among the following, is not a cause for power loss in a transformer :- (A) Eddy currents are produced in the soft iron core of a transformer. (B) Electric Flux sharing is not properly done in primary and secondary coils (C) Humming sound produced in the transformers due to magnetostriction. (D) Primary coil is made up of a very thick copper wire. 32. An alternating voltage source of variable angular frequency ‘w’ and fixed amplitude ‘V’ is node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paperconnected in series with a capacitance C and electric bulb of resistance R(inductance zero). When www.notesdrive.com‘w’ is increased :-2 (A) The bulb glows dimmer. (B) The bulb glows brighter. (C) Net impedance of the circuit remains unchanged. (D) Total impedance of the circuit increases. 33. A solid spherical conductor has charge +Q and radius R. It is surrounded by a solid spherical shell with charge –Q, inner radius 2R, and outer radius 3R. Which of the following statements is true? R RR A B C D (A) The electric potential has a maximum magnitude at C and the electric field has a maximum magnitude at A (B) The electric potential has a maximum magnitude at D and the electric field has a maximum magnitude at B. (C) The electric potential at A is zero and the electric field has a maximum magnitude at D. (D) Both the electric potential and electric field achieve a maximum magnitude at B. 34. A battery is connected to the conductor of non-uniform cross section area. The quantities or quantity which remains constant is :- (B) drift speed and electric field (A) electric field only (C) electric field and current (D) current only E

124 Physics ALLENÒ 35. Three resistors having values R1, R2, and R3 are connected in series to a battery. Suppose R1 carries a current of 2.0 A, R2, has a resistance of 3.0 ohms, and R3 dissipates 6.0 watts of power. Then the voltage across R3 is:- (A) 1V (B) 2V (C) 3V (D) 4V 36. A straight line plot showing the terminal potential difference (V) of a cell as a function of current (A) drawn from it, is internal resistance of the cell would be then :- (A) 2.8 ohms (B) 1.4 ohms (C) 1.2 ohms (D) zero 37. A 10 m long wire of uniform cross-section and 20 W resistance is used in a potentiometer. The wire is connected in series with a battery of 5 V along with an external resistance of 480 W. If anwww.notesdrive.com unknown emf E is balanced at 6.0 m length of the wire, then the value of unknown emf is :- node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper (A) 1.2 V (B) 1.02 V (C) 0.2 V (D) 0.12 V 38. The current sensitivity of a galvanometer increases by 20%. If its resistance also increases by 5%, the voltage sensitivity will :- (A) decrease by 1% (B) increased by 5% (C) increased by 10% (D) decrease by 4% 39. Three infinitely long parallel straight current carrying wires A, B and C are kept at equal distance from each other as shown in the figure. The wire C experiences net force F .The net force on wire C, when the current in wire A is reversed will be :- rr (A) Zero (B) F/2 3I I I (C) F (D) 2F AB C 40. In a hydrogen atom the electron moves in an orbit of radius 0.5 Ao making 10 revolutions per second, the magnetic moment associated with the orbital motion of the electron will be :- (A) 2.512 × 10–38 Am2 (B) 1.256 × 10–38 Am2 (C) 0.628 × 10–38 Am2 (D) zero 41. An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 800, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10–3s. Ignoring the variation in magnetic field near the ends of the solenoid, the average back emf induced across the ends of the open switch in the circuit would be (A) zero (B) 3.125 volts (C) 6.54 volts (D) 16.74 volts E

ALLENÒ CBSE 125 42. A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3W, L = 25.48 mH, and C = 796 µF, then the power dissipated at the resonant condition will be : (A) 39.70 kW (B) 26.70 kW (C) 13.35 kW (D) Zero 43. A circular loop of radius 0.3cm lies parallel to much bigger circular of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with the bigger loop is :- (A) 3.3 × 10–11 weber (B) 6 × 10–11 weber (C) 6.6 ×10–9 weber (D) 9.1 × 10–11 weber 44. If both the number of turns and core length of an inductor is doubled keeping other factors constant, then its self-inductance will be :- (A) Unaffected (B) doubled (C) halved (D) quadrupled node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper 45. Given below are two statements labelled as Assertion (A) and Reason (R) www.notesdrive.com Assertion (A): To increase the range of an ammeter, we must connect a suitable high resistance in series to it. Reason (R): The ammeter with increased range should have high resistance. Select the most appropriate answer from the options given below: (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is not the correct explanation of A. (C)A is true but R is false. (D) A is false and R is also false. 46. Given below are two statements labelled as Assertion (A) and Reason (R) Assertion (A): An electron has a high potential energy when it is at a location associated with a more negative value of potential, and a low potential energy when at a location associated with a more positive potential. Reason (R): Electrons move from a region of higher potential to region of lower potential. Select the most appropriate answer from the options given below: (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false and R is also false. 47. Given below are two statements labelled as Assertion (A) and Reason (R) Assertion (A): A magnetic needle free to rotate in a vertical plane, orients itself (with its axis) vertical at the poles of the earth. Reason (R): At the poles of the earth the horizontal component of earth’s magnetic field will be zero. Select the most appropriate answer from the options given below: (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false and R is also false. E

126 Physics ALLENÒ 48. Given below are two statements labelled as Assertion (A) and Reason (R) Assertion(A): A proton and an electron, with same momenta, enter in a magnetic field in a direction at right angles to the lines of the force. The radius of the paths followed by them will be same. Reason(R): Electron has less mass than the proton. Select the most appropriate answer from the options given below: (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is not the correct explanation of A. (C)A is true but R is false. (D) A is false and R is also false. 49. Given below are two statements labelled as Assertion (A) and Reason (R) Assertion (A): On Increasing the current sensitivity of a galvanometer by increasing the number of turns, may not necessarily increase its voltage sensitivity. Reason (R): The resistance of the coil of the galvanometer increases on increasing the number of turns. Select the most appropriate answer from the options given below: (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false and R is also false. www.notesdrive.com SECTION-C node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper This section consists of 6 multiple choice questions with an overall choice to attempt any 5. In case more than desirable number of questions are attempted, ONLY first 5 will be considered for evaluation. Q.50 A small object with charge q and weight mg is attached to one end of a string of length ‘L’ attached to a stationary support. The system is placed in a uniform horizontal electric field ‘E’, as shown in the accompanying figure. In the presence of the field, the string makes a constant angle q with the vertical. The sign and magnitude of q :- qL E (A) positive with magnitude mg/E m (B) positive with magnitude (mg/E)tan q (C) negative with magnitude mg/E tan θ (D) positive with magnitude E tanθ/mg E

ALLENÒ CBSE 127 Q.51 A free electron and a free proton are placed between two oppositely charged parallel plates. Both are closer to the positive plate than the negative plate. +++++++++++ +– ––––––––––– Which of the following statements is true? (I) The force on the proton is greater than the force on the electron. (II) The potential energy of the proton is greater than that of the electron. (III) The potential energy of the proton and the electron is the same. (A) I only (B) II only node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper (C) III and I only (D) II and I only www.notesdrive.com CASE STUDY Read the following paragraph and answers the questions: The large-scale transmission and distribution of electrical energy over long distances is done with the use of transformers. The voltage output of the generator is stepped-up. It is then transmitted over long distances to an area sub-station near the consumers. There the voltage is stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our homes. Q.52 Which of the following statement is true? (A) Energy is created when a transformer steps up the voltage (B) A transformer is designed to convert an AC voltage to DC voltage (C) Step–up transformer increases the power for transmission (D) Step–down transformer decreases the AC voltage Q.53 If the secondary coil has a greater number of turns than the primary, (A) the voltage is stepped-up (Vs >Vp ) and arrangement is called a step-up transformer (B) the voltage is stepped-down (Vs <Vp ) and arrangement is called a step down transformer (C) the current is stepped-up (Is >Ip ) and arrangement is called a step-up transformer (D) the current is stepped-down (Is <Ip ) and arrangement is called a step down transformer E

128 Physics ALLENÒ Q.54 We need to step-up the voltage for power transmission, so that (A) the current is reduced and consequently, the I2R loss is cut down (B) the voltage is increased , the power losses are also increased (C) the power is increased before transmission is done (D) the voltage is decreased so V2/R losses are reduced Q.55 A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns. The number of turns in the secondary in order to get output power at 230 V are (A) 4 (B) 40 (C) 400 (D) 4000 www.notesdrive.com node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper ANSWER-KEY 1. (C) 2. (C) 3. (B) 4. (B) 5. (A) 6. (D) 7. (C) 8. (A) 9. (C) 10. (C) 11. (A) 12. (C) 13. (B) 14. (C) 15. (B) 16. (D) 17. (C) 18. (A) 19. (D) 20. (B) 21. (D) 22. (B) 23. (C) 24. (A) 25. (D) 26. (A) 27. (D) 28. (D) 29. (C) 30. (C) 31. (D) 32. (B) 33. (D) 34. (D) 35. (C) 36. (A) 37. (D) 38. (D) 39. (A) 40. (B) 41. (B) 42. (C) 43. (D) 44. (B) 45. (D) 46. (C) 47. (A) 48. (B) 49. (A) 50. (B) 51. (B) 52. (D) 53. (A) 54. (A) 55. (C) E

ALLENÒ CBSE 129 SAMPLE QUESTION PAPER 2021-22 Max. Marks : 35 SUBJECT: PHYSICS THEORY CLASS-XII TERM-II Time : 2 HOURS node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question PaperGeneral Instructions: www.notesdrive.com (i) There are 12 questions in all. All questions are compulsory. (ii) This question paper has three sections: Section A, Section B and Section C. (iii) Section A contains three questions of two marks each, Section B contains eight questions of three marks each, Section C contains one case study-based question of five marks. (iv) There is no overall choice. However, an internal choice has been provided in one question of two marks and two questions of three marks. You have to attempt only one of the choices in such questions. (v) You may use log tables if necessary but use of calculator is not allowed. SECTION-A Q.1 In a pure semiconductor crystal of Si, if antimony is added then what type of extrinsic semiconductor is obtained. Draw the energy band diagram of this extrinsic semiconductor so formed. Q.2 Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but same orbital angular momentum according to the Bohr model? Justify your answer. OR Explain how does (i) photoelectric current and (ii) kinetic energy of the photoelectrons emitted in a photocell vary if the frequency of incident radiation is doubled, but keeping the intensity same? Show the graphical variation in the above two cases. Q.3 Name the device which converts the change in intensity of illumination to change in electric current flowing through it. Plot I-V characteristics of this device for different intensities. State any two applications of this device. SECTION-B Q.4 Derive an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n – 1). Also show that for large values of n, this frequency equals to classical frequency of revolution of an electron. E

130 Physics ALLENÒ Q.5 Explain with a proper diagram how an ac signal can be converted into dc (pulsating) signal with output frequency as double than the input frequency using pn junction diode. Give its input and output waveforms. Q.6 How long can an electric lamp of 100 W be kept glowing by fusion of 2 kg of deuterium? Take the fusion reaction as 2 H +12 H ®32 He + n + 3.27 MeV 1 Q7. Define wavefront. Draw the shape of refracted wavefront when the plane incident wave undergoes refraction from optically denser medium to rarer medium. Hence prove Snell’s law of refraction. Q8. (a) Draw a ray diagram of compound microscope for the final image formed at least distance (b) of distinct vision? An angular magnification of 30X is desired using an objective of focal length 1.25 cm and (a) an eye piece of focal length 5 cm. How will you set up the compound microscope for the (b) final image formed at least distance of distinct vision? OR Draw a ray diagram of Astronomical Telescope for the final image formed at infinity. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. Find the magnifying power of the telescope for viewing distant objects when (i) the telescope is in normal adjustment, (ii) the final image is formed at the least distance of distinct vision. www.notesdrive.com node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper Q.9 Light of wavelength 2000 Å falls on a metal surface of work function 4.2 eV. (a) What is the kinetic energy (in eV) of the fastest electrons emitted from the surface? (b) What will be the change in the energy of the emitted electrons if the intensity of light with same wavelength is doubled? (c) If the same light falls on another surface of work function 6.5 eV, what will be the energy of emitted electrons? Q.10 The focal length of a convex lens made of glass of refractive index (1.5) is 20 cm. What will be its new focal length when placed in a medium of refractive index 1.25 ? Is focal length positive or negative? What does it signify? Q11. (a) Name the e.m. waves which are suitable for radar systems used in aircraft navigation. Write (b) the range of frequency of these waves. (c) If the Earth did not have atmosphere, would its average surface temperature be higher or (a) lower than what it is now? Explain. An e.m. wave exerts pressure on the surface on which it is incident. Justify. (b) OR \"If the slits in Young's double slit experiment are identical, then intensity at any point on the screen may vary between zero and four times to the intensity due to single slit\". Justify the above statement through a relevant mathematical expression. Draw the intensity distribution as function of phase angle when diffraction of light takes place through coherently illuminated single slit. E

ALLENÒ CBSE 131 Q.12 CASE STUDY: MIRAGE IN DESERTS node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question PaperTo a distant observer, the light appears to be coming from somewhere below the ground. The observer naturally assumes that light is being reflected from the ground, say, by a pool of water www.notesdrive.comnear the tall object. Such inverted images of distant tall objects cause an optical illusion to the observer. This phenomenon is called mirage. This type of mirage is especially common in hot deserts. Based on the above facts, answer the following questions: (a) Which of the following phenomena is prominently involved in the formation of mirage in deserts? (A) Refraction, Total internal Reflection (B) Dispersion and Refraction (C) Dispersion and scattering of light (D) Total internal Reflection and diffraction. (b) A diver at a depth 12 m inside water æ a µw = 4 ö sees the sky in a cone of semi- vertical angle èç 3 ø÷ (A) sin-1 4 (B) tan-1 4 (C) sin-1 3 (D) 90° 3 3 4 (c) In an optical fibre, if n1 and n2 are the refractive indices of the core and cladding, then which among the following, would be a correct equation ? (A) n1 < n2 (B) n1 = n2 (C) n1 << n2 (D) n1 > n2 (d) A diamond is immersed in such a liquid which has its refractive index with respect to air as greater than the refractive index of water with respect to air. Then the critical angle of diamond- liquid interface as compared to critical angle of diamond -water interface will (A) depend on the nature of the liquid only (B) decrease (C) remain the same (D) increase E

132 Physics ALLENÒ (e) The following figure shows a cross-section of a ‘light pipe’ made of a glass fiber of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for the following phenomena to occur. (A) 0 < < 90° (B) 0 < < 60° (C) 0 < < 45° (D) 0 < < 30° SECTION-A 1. As given in the statement antimony is added to pure Si crystal, then a n -type extrinsic semiconductor would be so obtained, Since antimony (Sb) is a pentavalent impurity. Energy level diagram of n-type semiconductor www.notesdrive.com 2. No node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper Because according to Bohr's model, En = 13.6 and electrons having different energies belong to different levels having different n2 values of n. So, their angular momenta will be different, as L = mvr = nh 2p OR (i) The increase in the frequency of incident radiation has no effect on photoelectric current. This is because of incident photon of increased energy cannot eject more than one electron from the metal surface. E

ALLENÒ CBSE 133 (ii) The kinetic energy of the photoelectron becomes more than the double of its original energy. As the work function of the metal is fixed, so incident photon of higher frequency and hence higher energy will impart more energy to the photoelectrons. 3. Photodiodes are used to detect optical signals of different intensities by changing current flowing through them. (a) mA node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper reverse bias www.notesdrive.com I volt I1 I2 I3 4 µA (b) I-V Characteristics of a photodiode Applications of photodiodes: 1. In detection of optical signals. 2. In demodulation of optical signals. 3. In light operated switches. 4. In speed reading of computer punched cards. 5. In electronic counters (any two out of these or any other relevant application) SECTION-B 4. From Bohr’s theory, the frequency f of the radiation emitted when an electron de – excites from level n2 to level n1is given as f = 2p2 mk 2 z 2e 4 é1 - 1 ù h3 êë n12 úû n 2 2 Given n1 = n − 1 , n2 = n , derivation of it f = 2p2 mk 2 z 2e 4 (2n -1) h3 (n -1)2 n2 For large n, 2n − 1 = 2n , n − 1 = n and z = 1 Thus, f = 4p2 mk 2e 4 n3h3 which is same as orbital frequency of electron in nth orbit. f = v = 4p2 mk 2e4 2pr n3h3 E

134 Physics ALLENÒ 5. A junction diode allows current to pass only when it is forward biased. So, if an alternating voltage is applied across a diode the current flows only in that part of the cycle when the diode is forward biased. This property is used to rectify alternating voltages and the circuit used for this purpose is called a rectifier. www.notesdrive.comCircuit Diagram Working with input and output waveforms 6. Number of atoms present in 2 g of deuterium = 6 × 1023 Number of atoms present in 2.0 Kg of deuterium = 6 × 1026 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper Energy released in fusion of 2 deuterium atoms = 3.27 MeV Energy released in fusion of 2.0 Kg of deuterium atoms = 3.27 ´ 6 ´1026 MeV 2 9.81 × 1026 MeV = 15.696 × 1013 J Energy consumed by bulb per sec = 100 J Time for which bulb will glow = 15.696 ´1013 s = 4.97 ´104 year 100 7. A locus of points, which oscillate in phase is called a wavefront. OR A wavefront is defined as a surface of constant phase. Proof n1 sin i = n2 sin r (Derivation) This is the Snell’s law of refraction. E

ALLENÒ CBSE 135 8. Diagram of Compound Microscope for the final image formed at D: mo = 30, fo = 1.25 cm, fe = 5 cm when image is formed at least distance of distinct vision, D = 25cm Angular magnification of eyepiece node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper me = æçè1 + D ö =1+ 25 = 6 www.notesdrive.com fe ø÷ 5 Total Angular magnification, m= mome ⇒ mo =m = 30 =5 me 6 As the objective lens forms the real image, mo = vo = –5 Þ vo = -5uo uo Using lens equation, uo = −1.5 cm, vo = −5 × (−1.5) cm = +7.5 cm Given ve = −D = −25 cm, fe = +5 cm, ue = ? using again lens equation ue = 25 6 Thus, object is to be placed at 1.5 cm from the objective and separation between the two lenses should be L = vo + Iuel = 11.67 cm OR 8. (a) Ray diagram of astronomical telescope when image is formed at infinity. (b)(i) In normal adjustment : Magnifying power. m = fo/fe = (140/5) = 28 (ii) When the final image is formed at the least distance of distinct vision (25 cm) : m = fo çæè1 + fe ÷øö = (28 x 1.2) = 33.6 fe D E

136 Physics ALLENÒ 9. l = 2000 Å = (2000 × 10–10)m Wo = 4.2eV h = 6.63 × 10–34 JS (a) Using Einstein's photoelectric equation K. E. = (6.2 – 4.2) eV = 2.0 eV (b) The energy of the emitted electrons does not depend upon intensity of incident light; hence the energy remains unchanged. (c) For this surface, electrons will not be emitted as the energy of incident light (6.2 eV) is less than the work function (6.5 eV) of the surface. 10. Given aµg = 1.5 Focal length of the given convex lens when it is placed in air is f = +20 cm Refractive index of the given medium with respect to air is aµm = 1.25 New focal length of the given convex lens when placed in a medium is f' www.notesdrive.com ( )1 éæ 1 ö æ 1 öù node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question PaperëêèçR1ø÷çèR2ø÷úû f = aµg -1 + ...(A) ( )1 éæ 1 ö æ 1 öù ëêçè R1 ÷ø çè R2 ø÷ûú f' = mµg -1 + ...(B) Dividing (A) by (B), we get f ' = (a µg -1) = (1.5 -1) = 0.5 = 5 = 2.5 f (mµg -1) (1.2 -1) 0.2 2 f' = 2.5f = (2.5 × 20) cm = +50cm as mµg = µg = 1.5 = 1.2 µm 1.25 New focal length is positive. The significance of the positive sign of the focal length is that given convex lens is still converging in the given medium. 11. (a) Microwaves are suitable for the radar system used in aircraft navigation. Range of frequency of microwaves is 108 Hz to 1011 Hz. (b) If the Earth did not have atmosphere, then there would be absence of greenhouse effect of the atmosphere. Due to this reason, the temperature of the earth would be lower than what it is now. (c) An e.m. wave carries momentum with itself and given by P = Energy of wave(U)/Speed of the wave(c) = U/c when it is incident upon a surface it exerts pressure on it. OR 11. (a) The total intensity at a point where the phase difference is ∅, is given by = 1 + 2 + 2 I1I2 cos∅. Here 1 and 2 are the intensities of two individual sources which are equal. When ∅ is 0, I = 4 1. When ∅ is 90°, I = 0 Thus intensity on the screen varies between 4 1 and 0. E

ALLENÒ CBSE 137 (b) Intensity distribution as function of phase angle, when diffraction of light takes place through coherently illuminated single slit 12. (a) Ans (i) Refraction, Total internal reflection node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper (b) Ans (iii) sin -1 æ 3 ö www.notesdrive.com çè 4 ø÷ aµw = 1 sin C Þ sin C = 1 Þ C = sin -1 æ 1 ö aµw ç aµw ÷ è ø (c) Ans (iv) n1 > n2 The refractive index of the core should be greater than the refractive index of the cladding. (d) Ans (iv) increases Iµd = 1 = µd , wµd = 1 = µd sin | C µ1 sin C ' µw µ1 > µw Thus C > C' (e) Ans. (iii) 0<i < 60º , 1µ2 1 = sin C ' sin C ' = 1.44 = 0.8571 1.68 Þ C' = 59º Total internal reflection will occur if the angle ' > 'c, i.e., if ' > 59o or when r < rmax’ where rmax = 90o – 59o = 31o. Using Snell’s law, sin imax = 1.68 sin rmax or max = 1.68 × max = 1.68 × sin 31° = 1.68 × 0.5150 = 0.8662 ∴ max = 60° Thus all incident rays which make angles in the range 0 < < 60° with the axis of the pipe will suffer total internal reflections in the pipe. E

138 Physics ALLENÒ CBSE 2021 SUBJECT: PHYSICS THEORY (CODE : 55/1/ 4) TERM-I SECTION-A 1. An electric dipole placed in a non-uniform electric field will experience : (A) only a force (B) only a torque (C) both force and torque (D) neither force nor torque. 2. Let N1 be the number of electric field lines going out of an imaginary cube of side a that encloses www.notesdrive.com an isolated point charge 2q and N2 be the corresponding number for an imaginary sphere of node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper radius a that encloses an isolated point charge 3 Then (N1/N2) is : (A) 1/p (B) 2/3 (C) 9/4 (D) neither force nor torque 3. In the circuit given below P ¹ R and the reading of the galvanometer is same with switch S open or closed. Then : PQ S RG +– (A) IQ=IR (B) IR = IG (C) IP = IG (D) IQ=IG 4. Two wires A and B of same material having length in the ratio 1:2 and diameter in the ratio 2:3 are connected in series with a battery. The ratio of the potential differences (VA/VB) across the two wires respectively is : (A) 1/3 (B) 3/4 (C) 4/5 (D) 9/8 5. The moving coil galvanometer G1 and G2 have the following particulars respectively: N1 = 30 , A1 = 3.6 x 10–3m2 , B1 = 0.25 T N2 = 42 , A2 = 1.8 x 10-3m2 , B2 = 0.50 T The spring constant is same for both the galvanometers. The ratio of current sensitivities of G1 and G2 is : (A) 5:7 (B) 7:5 (C) 1:4 (D) 1:1 E

ALLENÒ CBSE 139 6. A current I is flowing through the loop as shown in the figure ( MA = R , MB = 2R ). The magnetic field at the centre of the loop is m times : (A) in to the plane of paper I MA (B) out of the plane of paper (C) out of the plane of paper DI (D) in to the plane of paper C 7. A capacitor and an inductor are connected in two different ac circuits with a bulb glowing in each circuit. The bulb glows more brightly when : (A) the number of turns in the inductor is increased node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper (B) the separation between the plates of the capacitor is increased www.notesdrive.com (C) an iron rod is introduced into the inductor (D) a dielectric is introduced into the gap between the plates of the capacitor. 8. A pure inductor of 318mH and a pure resistor of 75 W are connected in series to an ac source of 50 Hz. The voltage across 75 W resistor is found to be 150V . The source voltage is : (A) 150V (B) 175V (C) 220 V (D) 250V 9. The electric potential at a point on the axis of a short electric dipole , at a distance x from the mid point of the dipole is proportional to : (A) (B) (C) (D) 10. Let F1 be the magnitude of the force between two small spheres, charged to a constant potential In free space and F2 be the magnitude of the force between them in a medium of dielectric constant k. Then (F1/F2) is : (A) (B) k (C) k2 (D) 11. Infinity resistance in a resistance box has : (A) a resistance of 105W (B) a resistance of 107W (C) a resistor of ∞ resistance (D) a gap only 12. A battery of 15V and negligible internal resistance is connected across a 50 W resistor. The amount of energy dissipated as heat to the resistor in one minute is : (A) 122 J (B) 270 J (C) 420 J (D) 720 J 13. Lenz’s law is the consequence of the law of conservation of : (A) energy (B) charge (C) mass (D) momentum 14. The vertical component of earth’s magnetic field at a place is times the horizontal component. √ The angle of dip at that place is : (A) 0o (B) 30o (C) 45o (D) 60o E

140 Physics ALLENÒ 15. A long straight wire in the horizontal plane carries a current of 15A in north to south direction. The magnitude and direction of magnetic field at a point 2.5 m east of the wire respectively are : (A) 1.2 mT , vertically upward (B) 1.2 mT , vertically downward (C) 0.6 mT , vertically upward (D) 0.6 mT , vertically downward 16. The emf induced in a 10H inductor in which current changes from 11A to 2A in 9 x 10–1 s is : (A) 104 N (B) 103N (C) 102N (D) 10 N 17. A charge Q is placed at the centre of the line joining two charges q and The system of the three charges will be in equilibrium, if Q is : (A) + (B) – (C) + (D) – 18. Electric flux of an electric field ⃗ through an area d ⃗ is given by : (A) ⃗ ⃗ (B) ⃗ ⃗ (C) ⃗ . ⃗ (D) ] ⃗. ⃗ e e 19. In a potentiometer experiment, the balancing length with a cell is 120cm. When the cell iswww.notesdrive.com shunted with a 1 Ω resistance, the balancing length becomes 40 cm. The internal resistance of the node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper cell is: (A) 10 Ω (B) 7 Ω (C) 3 Ω (D) 2 Ω 20. An electron is projected with velocity v⃗ along the axis of a current carrying long solenoid. Which of the following statements is true? (A) The path of the electron will be circular about the axis (B) The electron will be accelerated along the axis (C) The path of the electron will be helical (D) The electron will continue to move the same velocity v⃗ along the axis of the solenoid. 21. If the speed v of a charged particle moving in a magnetic field B⃗ (v⃗ is perpendicular to B⃗) is halved, then the radius of its path will : (A) not change (B) become two times (C) become one-fourth (D) become half 22. A metal plate is getting heated. Which one of the following statements is incorrect ? (A) It is placed in a space varying magnetic field that does not vary with time. (B) A direct current is passing through the plate. (C) An alternating current is passing through the plate. (D) It is placed in a time varying magnetic field. 23. In an ac circuit the applied voltage and resultant current are E = E0 sin ωt and I = I0 sin (ωt + π/2) respectively. The average power consumed in the circuit is: (A) E0 I0 (B) (C) (D) zero √ 24. The speed acquired by a free electron when accelerated from rest through a potential difference of 100V is: (A) 6 × 106 m s–1 (B) 3 × 106 m s–1 (C) 4 × 105 m s–1 (D) 2 × 103 m s–1 25. Which of the following is not affected by the presence of a magnetic field? (A) A current carrying conductor (B) A moving charge (C) A stationary charge (D) A rectangular current loop with its plane parallel to the field E

ALLENÒ CBSE 141 SECTION-B This section consists of 24 multiple choice questions with overall choice to attempt any 20 questions. In case more than desirable number of questions are attempted, ONLY first 20 will be considered for evaluation. 26. Two point charged +16 q and –4 q are located at x = 0, and x = L. The location of the point on the x-axis at which the resultant electric field due to these charges is zero, is: (A) 8 L (B) 6 L (C) 4 L (D) 2 L 27. An electric dipole of dipole moment 4 x 10–5 C–m, kept in a uniform electric field of 10–3 NC–1 experiences a torque of 2 x 10-8 Nm. The angle which dipole makes with the electric field is: (A) 30o (B) 45o (C) 60o (D) 90o node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper 28. Three identical charges are placed at x-axis from left to right with adjacent charges separated by www.notesdrive.com a distance d. The magnitude of the force on a charge from its nearest neighbour charge is F. Let î be the unit vector along + x-axis, then the net force on each charge from left to right is : (A) (2Fî , –2Fî , 2Fî ) (B) (Fî , 0, Fî) (C) (–5/4 Fî , 0, +5/4 Fî) (D) ( 2 Fî , 0, 2 Fî ) 29. Two students A and B calculate the charge flowing through a circuit. A concludes that 300 C of charge flows in 1 minute. B concludes that 3.125 x 1019 electrons flow in 1 second. If the current measured in the circuit is 5 A, then the correct calculation is done by: (A) A (B) B (C) both A and B (D) neither A nor B 30. The resistances of two wires having same length and same area of cross-section are 2Ω and 8 Ω respectively. If the resistivity of 2 Ω wire is 2.65 x 10-8 m then the resistivity of 8 Ω wire is (A) 10.60 x 10–4 Ω m (B) 8.32 x 10–8 Ω m (C) 7.61 x 10–8 Ω m (D) 5.45 x 10–8 Ω m 31. In a certain region electric field ⃗ and magnetic field ⃗ are perpendicular to each other. An electron enters the region perpendicular to the direction of both ⃗ and ⃗ and moves undeflected. The speed of the electron is – (A) Er gBr (B) Er ´ Br r r E B (C) Br (D) Er ( )32. A test charge of 1.6 ´10-19 C is moving with a velocity vr = 4ˆi + 3kˆ ms-1 in a magnetic field ( )Br = 3kˆ + 4ˆi T . The force on this test charge is :- (A) 24ˆj N (B) - 24 ˆi N (C) 24 kˆ N (D) 0 (D) V 33. In a series LCR circuit , at resonance the current is equal to - R2 + (XL - XC )2 (A) V (B) V (C) V R XC XL - XC E

142 Physics ALLENÒ 34. The frequency of an ac source for which a 10 mF capacitor has a reactance of 1000 ohm is – (A) 1000 Hz (B) 50 Hz (C) 50 Hz (D) 100 Hz p p p 35. In the given network all capacitors used are identical and each one is of capacitance C .Which of the following is the equivalent capacitance between the points A and B ? (A) 6 C (B) 5 C (C) 3 C (D) 5 C 2 2 6 36. The given figure shown I−V graph of a copper wire of length L and area of cross-section A is shown in figure. The slope of the curve becomes- www.notesdrive.com I node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question PaperV (A) Less if the length of the wire is increased (B) more if the length of the wire is increased (C) more if a wire of steel of same dimension is used (D) more if the temperature of wire is increased 37. When a potential difference V is applied across a conductor at temperature T, the drift velocity of the electrons is proportional to – (A) T (B) T (C) V (D) V 38. To identical thick wires and two identical thin wires ,all of the same material and the same length from a square in three different ways P,Q and R as shown. Due to the current in these loops the magnetic field at the centre of the loop will be zero in case of – (A) P and R only (B) Q and R only (C) P and Q only (D) P ,Q and R 39. A circular coil carrying a certain current produces a magnetic field B0 at its centre. The coil is now rewound so as to have three turns and the same current is passed through it. The new magnetic field at the centre is :- (A) 3B0 (B) B0 (C) B0 (D) 9B0 3 9 40. Which one of the following statements is true (A) An inductor has infinite resistance in a dc circuit. (B) A inductor and a capacitor both cannot conduct in a dc circuit (C) A capacitor can conduct in a dc circuit but not an inductor. (D) An inductor can conduct in a dc circuit but not a capacitor. E

ALLENÒ CBSE 143 41. The magnetic flux linked with a coil is given by f = 5t2 + 3t + 10, where f is in weber and t in second. The induced emf in the coil at t= 5 sec will be :– (A) 53 V (B) 43 V (C)10 V (D) 6 V 42. If a charge is moved against a coulomb force of an electric field, then the - (A) intensity of the electric field increases (B) intensity of the electric field decreases (C) work is done by the electric field (D) work is done by the external source 43. A charge Q is located at the centre of a circle of radius r. The work done in moving a test charge q0 from point A to point B (at opposite ends of diameter AB) so as to complete a semicircle is node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper[k=1] 4pe0 www.notesdrive.com (A) k q 0 Q (B) k q0 Q (C) kq0 Q r (D) zero r2 r 44. A long solenoid carrying current produces a magnetic field B along its axis. if the number of turns in the solenoid is halved and current in it is doubled, the new magnetic field will be (A) B/2 (B) B (C) 2B (D) 4B ASSERTION (A) & REASON (R) Given below are two statements labelled as Assertion (A) and Reason (R) (A) Both (A) and (R) are true and (R) is correct explanation of (A) (B) Both (A) and (R) are true, and (R) is not correct explanation of (A) (C) (A) is true but (R) is false. (D) (A) is false (R) is also false 45. Assertion (A) : A bar magnet experiences a torque when place in a magnetic field. Reason (R) : A bar magnet exerts a torque on itself due to its own magnetic field. 46. Assertion (A): In a Series LCR circuit connected to an ac source, resonance can take place. Reason (R): At resonance XL = XC. 47. Assertion (A): When a charged particle moves with velocity V in a magnetic field B (V ^ B), the force on the particle does no work. Reason (R): The magnetic force is perpendicular to the velocity of the particle. 48. Assertion (A): Induced emf in two coils made of wire of the same length and the same thickness, one of copper and another of aluminium is same. The current in copper coil is more than the aluminium coil. Reason (R): Resistance of aluminium coil is more than that of copper coil. 49. Assertion (A): A transformer is used to increase or decrease ac voltage only. Reason (R): A transformer works on the basis of mutual Induction. E

144 Physics ALLENÒ SECTION-C This section consists of 6 multiple choice questions with an overall choice to attempt any 5. In case more than desirable number of questions are attempted, ONLY first 5 will be considered for evaluation. 50. Two charged spheres A and B having their radii in ratio 1 : 2 are connected together with a conducting wire. the ratio of their surface charge densities (sA/ sB) will be (A) (B) 2 (C) (D) 4 51. A current carrying square loop is suspended a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is F, the net force on the remaining three arms of the loop will be :- (A) 3 ⃗ (B) –3 ⃗ (C) ⃗ (D) - ⃗ CASE-STUDYwww.notesdrive.com A battery is a combination of two or more cells. In the following figure, a single battery is represented in which two cells of emf ε1 and ε2 and internal resistance r1 and r2 respectively are node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper connected. Answer the following Questions: 52. The equivalent emf of this combination is: (A) e1r1 + e2 r2 (B) e1r1 - e2 r2 (C) e1r2 - e2 r1 (D) e1 - e2 r1 + r2 r1 + r2 r1 + r2 (D) ε2 r2 = ε1 r1 53. For terminal B to be negative (C) ε1 r1 > ε2 r2 (A) ε1 r2 > ε2 r1 (B) ε1 r2 < ε2 r1 54. The current in the internal circuit is (A) e1 + e2 (B) e1 - e2 (C) e1 - e2 (D) e1 - e2 r1 + r2 r1 + r2 r1 r2 r2 r1 55. The equivalent internal resistance of the combination is :- (A) r1 + r2 (B) r1 + r2 (C) r1 r2 (D) r1 - r2 r1 r2 r1 + r2 ANSWER-KEY 1. (C) 2. (B) 3. (D) 4. (D) 5. (A) 6. (D) 7. (D) 8. (D) 9. (D) 10. (B) 11. (D) 12. (B) 13. (A) 14. (B) 15. (A) 16. (C) 17. (D) 18. (C) 19. (D) 20. (D) 21. (D) 22. (A) 23. (D) 24. (A) 25. (C) 26. (D) 27. (A) 28. (C) 29. (C) 30. (A) 31. (C) 32. (D) 33. (A) 34. (C) 35. (C) 36. (A) 37. (C) 38. (A) 39. (D) 40. (D) 41. (A) 42. (D) 43. (D) 44. (B) 45. (C) 46. (B) 47. (A) 48. (A) 49. (A) 50. (B) 51. (D) 52. (C) 53. (B) 54. (A) 55. (C) E

ALLENÒ CBSE 145 CBSE QUESTIONS PAPER – 2022 (55-3-2) SUBJECT: PHYSICS TERM-II Time : 2 Hours Max. Marks : 35 General Instructions: Read the following instructions very carefully and strictly follow them : (i) This question paper contains 12 questions. All questions are compulsory. (ii) This question paper is divided into three sections - Section A, B and C. (iii) Section A: Q. Nos. 1 to 3 are of 2 marks each. (iv) Section B : Q. Nos. 4 to 11 are of 3 marks each. (v) Section C : Q. No. 12 is a case study based questions of 5 marks. (vi) There is no overall choice in the question paper. However, internal choice has been provided in some of the questions. Attempt any one of the alternatives in such questions. (vii) Use of log tables is permitted, if necessary, but use of calculator is not permitted. c = 3 × 108 m/s h = 6.63 × 10–34 Js e = 1.6 × 10–19 C m0 = 4p × 10–7 T m A–1 e0 = 8.854 × 10–12 C2 N–1 m–2 1 = 9 × 109 N m2 C–2 4pe0 Mass of electron (me) = 9.1 × 10–31 kg Mass of neutron = 1.675 × 10–27 kg Mass of proton = 1.673 × 10–27 kg Avogadro's number = 6.023 × 1023 per gram mole Boltzmann constant = 1.38 × 10–23 JK–1 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper SECTION-A www.notesdrive.com 1. Explain the formation of the barrier potential in a p-n junction. [2] 2. (a) Draw the circuit diagram of an illuminated photodiode and its I-V characteristics. [1] [1] (b) How can a photodiode be used to measure the light intensity ? [2] 3. (a) (i) Distinguish between isotopes and isobars. (ii) Two nuclei have different mass numbers A1 and A2. Are these nuclei necessarily the isotopes of the same element ? Explain. OR (b) (i) Name the factors on which photoelectric emission from a surface depends. (ii) Define the term 'threshold frequency' for a photosensitive material. E

146 Physics ALLENÒ SECTION-B 4. (a) Calculate the frequency of a photon of energy 6.5 × 10–19 J. (b) Can this photon cause emission of an electron from the surface of Cs of work function 2.14 eV ? If yes, what will be maximum kinetic energy of the photoelectron ? [3] 5. Monochromatic light of wavelength 600 nm is incident from air on a water surface. The refractive index of water is 1.33. Find the (i) wavelength, (ii) frequency and (iii) speed, of reflected and refracted light. [3] 6. (a) Electromagnetic waves of wavelengths l1, l2 and l3 are used in radar systems, in water purifiers and in remote switches of TV, respectively. www.notesdrive.com(i) Identify the electromagnetic waves, and node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper(ii) Write one source of each of them.[3] OR (b) (i) State two conditions for two light sources to be coherent. (ii) Give two points of difference between an interference pattern due to a double - slit and a diffraction pattern due to a single slit. 7. State Bohr's postulate to explain stable orbits in a hydrogen atom. Prove that the speed with which the electron revolves in nth orbit is proportional to (1/n). [3] 8. Draw the energy band diagrams for conductors, semiconductors and insulators. Which band determines the electrical conductivity of a solid ? How is the electrical conductivity of a semiconductor affected with rise in its temperature ? Explain. [3] 9. A narrow beam of protons, each having 4.1 MeV energy is approaching a sheet of lead (Z = 82). Calculate : (i) the speed of a proton in the beam, and (ii) the distance of its closest approach [3] 10. How is the spacing between fringes in a double slit experiment affected if: (i) the slits separation is increased, (ii) the colour of light used is changed from red to blue, (iii) the whole apparatus is submerged in a oil of refractive index 1.2 ? Justify your answer in each case. [3] E

ALLENÒ CBSE 147 11. (a) Write two necessary conditions for total internal reflection. (b) Two prisms ABC and DBC are arranged as shown in figure node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question PaperThe critical angles for the two prisms with respect to air are 41.1° and 45° respectively. Trace the www.notesdrive.compath of the ray through the combination. [3] OR (a) An object is placed in front of a converging lens. Obtain the conditions under which the magnification produced by the lens is (i) negative and (ii) positive. (b) A point object is placed at O in front of a glass sphere as shown in figure. Show the formation of image by the sphere. SECTION-C CASE STUDY 12. A compound microscope consists of two conversing lenses. One of them, of smaller aperture and smaller focal length is called objective and the other of slightly larger aperture and slightly larger focal length is called eye-piece. Both the lenses are fitted in a tube with an arrangement to vary the distance between them. A tiny object is placed in front of the objective at a distance slightly greater than its focal length. The objective produces the image of the object which acts as an object for the eye-piece. The eye piece, in turn produces the final magnified image. [1×5 = 5] I. In a compound microscope the images formed by the objective and the eye-piece are respectively. (A) virtual, real (B) real, virtual (C) virtual, virtual (D) real, real E

148 Physics ALLENÒ II. The magnification due to a compound microscope does not depend upon : (A) The aperture of the objective and the eye-piece (B) The focal length of the objective and the eye-piece (C) The length of the tube (D) The colour of the light used III. Which of the following is not correct in the context of a compound microscope ? (A) Both the lenses are of short focal lengths. (B) The magnifying power increases by decreasing the focal lengths of the two lenses. (C) The distance between the two lenses is more than (fo + fe) (D) The microscope can be used as a telescope interchanging the two lenses.www.notesdrive.com IV. A compound microscope conists of an objective of 10X and an eye-piece of 20X. The node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper magnification due to the microscope would be : (A) 2 (B) 10 (C) 30 (D) 200 V. The focal length of objective and eye-piece of a compound microscope are 1.2 cm and 3.0 cm respectively. The object is placed at a distance of 1.25 cm from the objective. If the final image is formed at infinity, the magnifying power of the microscope would be : (A) 100 (B) 150 (C) 200 (D) 250 E