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1 BLUE PRINT OF QUESTION PAPER 1 UNIT NAME 1 MARK 2MARKS 3MARKS VBQ 4 MARKS 5MARKS TOTAL ELECTROSTATICS 6(2) 4(1) 5(1) 15(4) CURRENT ELECTRICITY MAGNETISM&MAGNETIC EFFECTS OF CURRENT 1(1) 4(2) 6(2) 5(1) 16(6) AC&EMI EMW&OPTICS 3(3) 6(3) 3(1) 5(1) 17(8) DUAL NATURE OF 1(1) 9(3) 10(4) MATTER&ATOMS &NUCLEI SEMICONDUCTOR 12(4) 12(4) DEVICES COMMUNICATION TOTAL 5(5) 10(5) 36(12) 4(1) 15(3) 70(26) *TOTAL MARKS (NO OF QUESTIONS)

2 PHYSICS (CLASS XII) Paper-1 Time : 3 Hrs. Max. Marks : 70 General Instruction : (i) All questions are compulsory. (ii) There are 26 questions in total. Questions 1 to 5 are very short answer type questions and carry one mark each. (iii) Questions 6 to 10 carry two marks each, questions 11 to 22 carry three marks each, question 23 is value based question carry 4 marks and questions 24 to 26 carry five marks each. (iv) There is not overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all three question of five marks each weightage. You have to attempt only one of the choices in such questions. (v) Use of calculators is not permitted. However, you may use log tables if necessary. (vi) you may use the following values of physical constants wherever necessary. c = m/s h = Js e = C μ 0 = 4π 10 7 T ma 1 = N m 2 C 1 m e = kg Mass of Neutrons = kg Mass of proton = kg Q Usi g the o ept of fo e et ee i fi ite lo g pa allel u e t a i g o du to s, defi e o e a pe e of u e t. Q A i o e le s ade of t a spa e t ate ial of ef a ti e i de. is i e sed i ate of ef a ti e i de., ill the le s eha e as a o e si g o a di e gi g le s? Gi e easo. Q To hi h pa t of the ele t o ag eti spe t u does a a e of f e ue Hz elo g? Q Ho does fo al le gth of a le s ha ge he ed light i ide t o it is epla ed iolet light? Gi e easo fo ou a s e. Q The gi e g aph sho s the a iatio of photo ele t i u e t I e sus applied oltage V fo t o diffe e e photose siti e ate ials a d fo t o diffe e t i te sities of the i ide t adiatio s. Ide tif the pai s of u es that o espo d to diffe e t ate ials ut sa e i te sit of i ide t adiatio.

3 Q.State the o ditio hi h ust e satisfied fo t o light sou es to e ohe e t? O E plai t o diffe e es et ee teles ope a d a i os ope. Q Wh is the use of AC oltage is p efe ed o e DC oltage? Gi e t o easo s. Q A o e le s of fo al le gth is pla ed oa iall i o ta t ith a o a e le s of fo al le gth. Dete i e the po e of the o i atio. Will the s ste e o e gi g o di e gi g i atu e? Q A a ete of esista e. a easu e u e t up to. A. i What ust e the alue of shu t esista e to e a le the a ete to easu e u e t up to. A? ii What is the o i ed esista e of the a ete a d the shu t? Q a A e a e is t a elli g i a ediu ith a elo it. D a a sket h sho i g the p opagatio of the e a e, i di ati g the di e tio of the os illati g ele t i a d ag eti fields. Ho a e the ag itudes of the ele t i a d ag eti fields elated to elo it of the e a e? Q Figu e sho s a lo k diag a of a t a s itte ide tif the o es X a d Y a d ite thei fu tio s.. Q E plai, ith the help of a i uit diag a, the o ki g of a photo diode. W ite iefl ho it is used to dete t the opti al sig als. OR A Se i o du to has e ual ele t o hole o e t atio of / o dopi g ith a e tai i pu it, ele t o o e t atio i eases to /. i Ide tif Se i o du to o tai ed afte dopi g. ii Cal ulate e hole o e t atio.

4 Q W ite the t o asi odes of o u i atio. Sho diag a ati all ho a a plitude odulated a e is o tai ed he a odulati g sig al is supe i posed o a a ie a e. Q A pa allel plate apa ito of apa ita e C is ha ged to a pote tial V. It is the o e ted to a othe u ha ged apa ito ha i g the sa e apa ita e. Fi d out the atio of e e g sto ed i the o i ed s ste so that e e g sto ed is sa e as i itiall sto ed i the si gle apa ito? OR A hollo li d i al o of le gth a d a ea of oss se tio is pla ed i a th ee di e sio al oo di ate s ste as sho i the figu e. The ele t i field i the egio is gi e et es. Fi d, he e E is NC a d x is i i Net flu th ough the li de. ii Cha ge e losed the li de. Q The de a o sta t, fo a gi e adioa ti e u lei, has a alue of. da. Afte ho u h ti e ill a gi e sa ple of this adio u lide get edu ed to o l. % of its p ese t u e? Q a Wh photoele t i effe t a ot e e plai ed o the asis of a e atu e of light? Gi e easo s. W ite the asi featu es of photo pi tu e of ele t o ag eti adiatio o hi h Ei stei s photoele t i e uatio is ased. Q A sho t a ag et of ag eti o e t. J/T is pla ed ith its a is at to a u ifo ag eti field. It e pe ie es a to ue of. J. (i) Calculate the magnitude of the magnetic field. (ii) In which orientation will the bar magnet be in stable equilibrium in the magnetic field? Q Output ha a te isti s of a p t a sisto i CE o figu atio is sho i the figu e. Dete i e: i d a i output esista e ii d u e t gai a d iii a u e t gai at a ope ati g poi t V CE = V, he I B = µa.

5 Q D a a g aph sho i g the a iatio of pote tial e e g of a pai of u leo s as a fu tio of thei sepa atio. I di ate the egio s i hi h u lea fo e is i att a ti e, a d ii epulsi e. W ite t o ha a te isti featu es of u lea fo e hi h disti guish it f o the Coulo fo e. Q a I hat a is diff a tio f o ea h slit elated to the i te fe e e patte i a dou le slit e pe i e t. T o a ele gths of sodiu light a d a e used, i tu to stud the diff a tio taki g pla e at a si gle slit of ape tu e. The dista e et ee the slit a d the s ee is.. Cal ulate the sepa atio et ee the positio s of the fi st a i a of the diff a tio patte o tai ed i the t o ases. Q Show mathematically that in A.C. circuit containing only inductance, the current lags behind the voltage by a phase of π/2. Q Dedu e a e p essio fo the ele t i pote tial due to a ele t i dipole at a poi t o its a is. Me tio o e o t asti g featu e of ele t i field of a dipole at a poi t as o pa ed to that due to si gle ha ge. Q While t a elli g a k to his eside e i the a, D. Pathak as aught up i a thu de sto. It e a e e da k. He stopped d i i g the a a d aited fo thu de sto to stop. Sudde l he oti ed a hild alki g alo e o the oad. He asked the o to o e i side the a till the thu de sto stopped. D. Pathak d opped the o at his eside e. The od i sisted that D. Pathak should eet his pa e ts. The pa e ts e p essed thei g atitude to D. Pathak fo his o e fo safet of the hild. A s e the follo i g uestio s ased o the a o e i fo atio : a Wh is it safe to sit i side a a du i g a thu de sto? Whi h t o alues a e displa ed D. Pathak i his a tio? Whi h alues a e efle ted i pa e ts' espo se to D. Pathak? d Gi e a e a ple of si ila a tio o ou pa t i the past f o e e da life.

6 Q24 a) What is the effect on the interference fringes to a Young s double slit experiment when (i) the separation between the two slits is decreased? (ii) the width of a source slit is increased? (iii) the monochromatic source is replaced by a source of white light? Justify your answer in each case. (b) The intensity at the central maxima in Young s double slit experimental set-up is I 0. Show that the intensity at a point where the path difference is λ/3 is I 0 /4. (a) Obtain the conditions for the bright and dark fringes in diffraction pattern due to a single narrow slit illuminated by a monochromatic source. OR Why central maxima is more intense as compared to secondary maxima s? Whe the idth of the slit is ade dou le, ho ould this affe t the size a d i te sit of the e t al diff a tio a d? Justif Q a State the o ki g p i iple of a pote tio ete. With the help of the i uit diag a, e plai ho a pote tio ete is used to o pa e the e f's of t o p i a ells. O tai the e ui ed e p essio used fo o pa i g the e f s. W ite t o possi le auses fo o e sided defle tio i a pote tio ete e pe i e t. OR a State Ki hhoff's ules fo a ele t i et o k. Usi g Ki hhoff's ules, o tai the ala e o ditio i te s of the esista es of fou a s of Wheatsto e idge. I the ete idge e pe i e tal set up, sho i the figu e, the ull poi t D is o tai ed at a dista e of f o e d A of the ete idge i e. If a esista e of is o e ted i se ies ith, ull poi t is o tai ed at AD =. Cal ulate the alues of a d. Q. With the help of a s he ati sket h of a lot o e plai its o ki g p i iple a d e tio its t o li itatio s. OR With the help of a eat a d la eled diag a, e plai its p i iple, o st u tio a d o ki g of a o i g oil gal a o ete, hat is the fu tio of

7 i. U ifo adial ag eti field ii. Soft i o o e, i this de i e SOLUTION OF PAPER 1 1 One ampere of current can be defined as the amount of current which when flowing (in same direction) through two infinitely long parallel wires separated by one meter produces an attractive force of 2x 10-7 N/m. The wires must have negligible circular cross section and they must be placed in vacuum. 2 A biconvex lens acts as a converging lens in air because the refractive index of air is less than that of the material of the lens, the refractive index of water is also less than that of refractive index of the material of the lens. So its nature will not change, it behaves as a converging lens. 3. A wave of frequency 5x10 19 Hz belongs to gamma rays of electromagnetic spectrum. 4 The refractive index of the material of a lens increases with the decrease in wavelength of the incident light. So, focal length will decrease with decrease in wavelength according to formula. Thus, when we replace red light with a violet wavelength decrease and hence the focal length of the lens also decrease.

8 5. Curves 1 and 2 correspond to similar materials while curves 3 and 4 represent different materials, since the value of stopping potential for the pair of curves (1 and 2) & (3 and 4) are the same. For given frequency of the incident radiation the stopping potential is independent of its intensity. So, the pairs of curves (1 and 3) and (2 and 4) correspond to different materials but same intensity of incident radiation. 6 The two sources are said to be coherent if they emit continuous monochromatic waves of same wavelength, either in same or constant phase difference. OR Construction Working Telescope Objective has large focal length and large aperture but eyepiece of short focal length and short aperture. It forms magnified image of distant object. Microscope Objective has very short focal length and short aperture and eyepiece has short focal length and large aperture.( fe > fo) It forms magnified image of a small nearby object. 7 The use of AC voltage preferred over DC voltage because of the following: i) The loss energy in transmitting the AC voltage over long distances with the help of step up transformer is negligible as compared to DC voltage. ii) AC voltage can be stepped up and stepped down as per the requirement using a transformer. 8 We have focal length of convex lens, The focal length of the combination = 1 m = 100 cm. As the focal length is in negative, the system will be diverging in nature. 9. We have, resistance of ammeter, RA = 0.80 ohm and maximum current across ammeter, IA = 1.0 A.

9 So, voltage across ammeter, V =. Let the value of shunt be x. (i) Resistance of ammeter with shunt,. Current through ammeter, I = 5 A. Thus, the shunt resistance is 0.2 ohm. (ii) Combined resistance of the ammeter and the shunt, 10 (a) Given that velocity, V = v î and electric field, E along X-axis and magnetic field, B along Z-axis. The propagation of EM wave is following: (b) Speed of EM wave can be given as the ratio of magnitude of electric field (E0) to the magnitude of magnetic field (B0), Q11 Modulator: Since the frequency range of signal is quiet low and it is associated with very small amount of energy, it dies out very soon if transmitted as such. So, it is modulated by mixing with very high frequency waves called carrier waves. This is done by modulator power. Amplifier: Since the signal gets weaken after

10 travelling through long distances it cannot be transmitted as such. Thus, we use a power amplifier to provide it necessary power before feeding the signal to the transmitting antenna.. Q12 Working of photo diode: A junction diode made from light sensitive semi-conductor is called a photodiode. A photodiode is an electrical device used to detect and convert light into an energy signal through the use of a photo detector. It is a pn-junction whose function is controlled by the light allowed to fall on it. Suppose, the wavelength is such that the energy of a photon, hc/λ, is sufficient to break a valance bond. When such light falls on the junction, new hole-electron pairs are created. The number of charge carriers increases and hence the conductivity of the junction increases. If the junction is connected in some circuit, the current in the circuit is controlled by the intensity of the incident light. i. As ele t o o e t atio i eases it is N t pe se i o du to. ii. = e a a = = / OR =. / Q13 The two basic modes of communication are: i) Point to point ii) Broadcast:

11 Q14 Let q be the charge on the charged capacitor. Energy stored in it is given by: U = q 2 /2C When another uncharged similar capacitor is connected, then net capacitance of the system is given by : C = 2C The charge on the system remains constant. So, the energy store in the system is given by: U = = q 2 /2C = = q 2 /4C Thus the required ratio is given by: u /u = (q 2 /4C)/ (q 2 /2C) = 1/2 OR (i) Given, As the electric field is only along the x-axis, so, flux will pass only through the cross-section of cylinder.

12 (ii) Using Gauss s law: Q15 T / =. /λ =. /. =. da N/ N = / t/t. / = / t/t / = / t/t t/t= => t= T=. = da s Q16 (a) Wave nature of radiation cannot explain the following: (i) The instantaneous ejection of photoelectrons. (ii) The existence of threshold frequency for a metal surface. (iii) The fact that kinetic energy of the emitted electrons is independent of the intensity of light and depends upon its frequency. Thus, the photoelectric effect cannot be explained on the basis of wave nature of light. (b) Photon picture of electromagnetic radiation on which Einstein s photoelectric equation is based on particle nature of light. Its basic features are: (i) In interaction with matter, radiation behaves as if it is made up of particles called photons. (ii) Each photon has energy E (=hν) and momentum p (=hν/c), and speed c, the speed of light. (iii) All photons of light of a particular frequency ν, or wavelength λ, have the same energy E (=hν=hc/λ) and momentum p (=hν/c=h/λ), whatever the intensity of radiation may be. (iv) By increasing the intensity of light of given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of intensity of radiation.

13 (v) Photons are electrically neutral and are not deflected by electric and magnetic fields. (vi) In a photon-particle collision (such as photon-electron collision), the total energy and total momentum are conserved. However, number of photons may not be conserved. Q17 (i) Magnetic moment M = 0.9 J/T τ = J, Ɵ = 30 We know = M B = MB sin Ɵ = 0.9 B sin 30 (ii) Stable equilibrium is position of minimum energy. Since (iii) U = M B cosɵ Where, U is the energy stored or P.E. of the magnet inside magnetic field B. So, when = 0, U = MB is the minimum energy. Thus, he a d a e pa allel to ea h othe a ag et is i sta le e uili iu Q18 (i) Dynamic output resistance is given as: (ii) (iii) Q19 (a)

14 (i) The potential energy is minimum at a distance of r0 = 0.8 fm. At this distance, force between nucleons is zero. For distances larger than 0.8 fm, negative P.E. goes on decreasing. The nuclear forces are attractive. (ii) For distances less than 0.8 fm, negative P.E. decreases to zero and then becomes positive. The nuclear forces are repulsive. (b) Properties of nuclear forces: (i) Nuclear force is always attractive in nature. ii Nu lea fo e is a sho t a ge fo e. Q20 (a) If the width of each slit is comparable to the wavelength of light used, the interference pattern thus obtained in the double-slit experiment is modified by diffraction from each of the two slits. (b) Given that: Wavelength of the light beam, λ1 = 590 nm = m Wavelength of another light beam, λ2 = 596 nm = m Distance of the slits from the screen = D = 1.5 m Distance between the two slits = a = m For the first secondary maxima, OR

15 Spacing between the positions of first secondary maxima of two sodium lines = m Q21 As E0sinωt L

16 Q22. Let P be the axial point at a distance r from the centre of the dipole. Electric potential at point P is V. For a distant point r >> a, The potential due to a dipole (at a large distance) falls off as 1/r 2, but potential due to a single charge falls off as 1/r. Q23 It is safer to be inside a car during thunderstorm because the car acts like a Faraday cage. The metal in the car will shield you from any external electric fields and thus prevent the lightning from traveling within the car. (b) Awareness and Humanity (c) Gratitude and obliged (d) I once came across to a situation where a puppy was struck in the middle of a busy road during rain and was not able go cross due to heavy flow, so I quickly rushed and helped him. Q a i F o the f i ge idth e p essio, With the decrease in separation between two slits, d the fringe width increases. (ii) For interference fringes to be seen condition should be satisfied Where, s = size of the source, S = distance of the source from the plane of two slits. As the source slit width increase, fringe pattern gets less and less sharp. When the source slit is so wide the above condition does not satisfied and the interference pattern disappears.

17 (iii) The interference pattern due to different colour component of white light overlap. The central bright fringes for different colours are at the same position. Therefore central fringes are white. And on the either side of the central whit fringe coloured bands will appear. The fringe closed on either side of central white fringe is red and the farthest will be blue. After a few fringes, no clear fringe pattern is seen. (b) Intensity at a point is given by, I = 4I cos 2 /2 Where, = phase difference I = intensity produced by each one of the individual sources. At central maxima, = 0, I = I0 = 4I At path difference Now, intensity at this point, Hence proved. OR (a) Consider a point P on the screen at which wavelets travelling in a direction making angle θ with CO are brought to focus by the lens. The wavelets from points A and B will have a path difference equal to BN. From the right-angled ΔANB, we have BN = AB sinθ BN = a sinθ (i)

18 Suppose BN = λ and θ = θ1 Then, the above equation gives λ = a sin θ1 Such a point on the screen will be the position of first secondary minimum. If BN = 2λ and θ= θ2, then 2λ = a sin θ2 Such a point on the screen will be the position of second secondary minimum. In general, for n th minimum at point P, If yn is the distance of the n th minimum from the centre of the screen, then from right-angled ΔCOP, we have In case θn is small, sin θn tan θn Equations (iv) and (v) give If and = 1, then from equation (i), we have Such a point on the screen will be the position of the first secondary maximum. Corresponding to path difference, and = 2, the second secondary maximum is produced In general, for the nth maximum at point P,

19 If is the distance of n th maximum from the centre of the screen, then the angular position of the n th maximum is given by, In case θ n is small, sin θ n tan θ n For n = 1 (From eq(vi), small angle approximation, This angle is midway between two of the dark fringes. Divide the slit into three equal parts. If we take the first two third part of the slit the path difference between the two ends would be, The first two third is divided into two halves which have path difference. The contribution due to these two halves is 180 out of phase and gets cancel. Only the remaining one third part of the slit contributes to the intensity at a point between the two minima which will be much weaker than the central maxima. Thus with increasing n the intensity of maxima gets weaker. (b) As the number of point sources increases, their contribution towards intensity also increases. Intensity varies as square of the slit width. Thus, when the width of the slit is made double the original width, intensity will get four times of its original value. Width of central maximum is given by, Where, D = Distance between screen and slit, λ = Wavelength of the light, b = Size of slit. So with the increase in size of slit the width of central maxima decreases. Hence, double the size of the slit would result in half the width of the central maxima Q25 (a) Working principle of Potentiometer: When a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion. Applications of Potentiometer for comparing emf s of two cells:

20 The following Figure shows an application of the potentiometer to compare the emf of two cells of emf E1 and E2 E1, E2 are the emf of the two cells. 1, 2, 3 form a two way key. When 1 and 3 are connected, E1 is connected to the galvanometer (G). Jokey is moved to N1, which is at a distance l1 from A, to find the balancing length. Applying loop rule to AN1G31A, Φ l1 + 0 E1 = 0..(1) Where, Φ is the potential drop per unit length Similarly, for E2 balanced against l2 (AN2), Φ l2 + 0 E2 = 0..(2) From equations (1) and (2),.. (3) Thus we can compare the emf s of any two sources. Generally, one of the cells is chosen as a standard cell whose emf is known to a high degree of accuracy. The emf of the other cell is then calculated from Eq. (3). (b) (i) The emf of the cell connected in main circuit may not be more than the emf of the primary cells whose emfs are to be compared. (ii) The positive ends of all cells are not connected to the same end of the wire. OR (a) Kirchhoff s First Law Junction Rule The algebraic sum of the currents meeting at a point in an electrical circuit is always zero. Let the currents be I1, I2I3, and I4 Convention: Current towards the junction positive Current away from the junction negative I3 + ( I1) + ( I2) + ( I4) = 0

21 Kirchhoff s Second Law Loop Rule: In a closed loop, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistances and current flowing through them. For closed part BACB, E1 E2 = I1R1 + I2 R2 I3R3 For closed part CADC, E2 = I3R3 + I4R4 + I5R5 Wheatstone Bridge: The Wheatstone Bridge is an arrangement of four resistances as shown in the following figure. R1, R2, R3,and R4 are the four resistances. Galvanometer (G) has a current Ig flowing through it at balanced condition, Ig = 0 Applying junction rule at B, I2 = I4 Applying junction rule at D, I1 = I3 Applying loop rule to closed loop ADBA, Applying loop rule to closed loop CBDC, From equations (1) and (2), This is the required balanced condition of Wheatstone Bridge. (b) Considering both the situations and writing them in the form of equations,let R' be the resistance per unit length of the potential meter wire,

22 Q26. a C lot o P i iple A ha ged pa ti le a e a ele ated to e high e e gies aki g it pass th ough a ode ate ele t i field a u e of ti es. This a e do e ith the help of a pe pe di ula ag eti field hi h th o s the ha ged pa ti le i to a i ula otio, the f e ue of hi h does ot depe d o the speed of the pa ti le a d adius of the i ula o it. a k a k Wo ki g Suppose a positi e io, sa a p oto, e te s the gap et ee the t o dees a d fi d dee D to e egati e. It gets a ele ated to a ds D. As it e te s the dee D, it does ot e pe ie e a ele t i field due to shieldi g effe t of the etalli dee. The pe pe di ula ag eti field th o s it i to a i ula path. At the i sta t the p oto o es out of the dee D, it fi ds D positi e a d D egati e. It o gets a ele ated to a ds D. It o es faste th ough D des i i g a se i i ula path. Thus, if the f e ue of the applied oltage is kept e a tl the sa e as the f e ue of e olutio of the p oto, the e e ti e the p oto ea hes the gap et ee the dees, the ele t i field is e e sed a d p oto e ei es a push a d fi all it a ui es e high e e g. This is alled lot o eso a e o ditio. a k Li itatio s. It a ot e used to a ele ate ele t o s. ½ + ½.Neut o s, ei g ele t i all eut al, a ot e a ele ated i a lot o. O

23 a k P i iple Its p i iple is ased o the fa t that he a u e t a i g o du to is pla ed i ag eti field, it e pe ie e a to ue. a k Wo ki g As the field is adial, the pla e of oil al a s e ai s pa allel to the field. Whe a u e t flo s th ough the oil, a to ue a ts o it. It is gi e as =NIBA This to ue defle ts the oil th ough a a gle α. A esto i g to ue is set up i the oil due to the elasti it of the sp i gs su h that o α, he e k is the to sio o sta t of the sp i g. I e uili iu esto i g to ue= defle ti g to ue => Kα=NIBA => α=nba/k Thus the defle tio p odu ed i the gal a o ete oil is p opo tio al to the u e t flo i g th ough it. a k i Fu tio of u ifo adial field It p o ides a i u to ue i espe ti e of o ie tatio ii Fu tio of soft i o o e:it i eases the st e gth of the ag eti field a d he e i ease the se siti it of the gal a o ete. ½ + ½

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