Sample Paper 1 Solutions Class XIE 2023-24 Physics ‘Time: $ Hours General Instructions: 1. There are $3 questions in all. All questions are compulsory 2. This question paper has mn A, Section B, Section C, Section D and Section B, 3. All the sections are compulsory. 4. Section A contains sixteen questions, twelve MCQ and four assertion reasoning based of 1 mark each, Section B contains five questions of 2 marks each, Section C contains Seven questions of 3 marks each, Section D contains two ease study based Max. Marks: 70 questious of 4 marks each aud Section E contains three loug questious of 5 marks each, choices in such questions 6. Use of ealeulators is not allowed 5. There is no overall choice. However, an internal choice has been provided in one question in Section B, one questi Section C, one question in each CBQ in Section D andl all three qu cone of the stions in Section E. You have to attempt 7. You may use the following values of physical constants where ever necessary i.e =3x 10'm/s fim 91% Wg fil, © = 16x WC iv. py = 4x 107 Ta G68 x 10s 8851 1 EN Avogadro's mmber = 6.423 x 10" per gram mole vk SECTION-A Tn Huygen’s wave theory, the loeus of all points oscillating in the same phase is called & (a)ray (b) vibrator (c) wavefront (4) half period zone Ans : (¢) wavefront We know from the Huygen’s wave theory that the locns of all points oscillating in the same phase is called a wavefront In other words, a wavefront is defined as a surface of constant phase If a glass rod is immersed in a liquid of the same refractive index, then it will (a) disappear (€) look longer (b) look bent (4) look shorter Ans : (a) disappear We know that if a glass rod is will immersod in a liquid of the same refractive index as that of the rod, then no rufraction will take place Therefore, the glass rod will disappear. Most of the substance show which of the following magnetic property? (b) paramagnetism (@) both b and € Ans : (a) disnaguetism We know that all substances show diamagnotism, as itis niversal. Since hoth the paramagnetism and ferromagnetisin substances have diamagnetism in weak, therefore they are hard to deteet. When two charged capacitors having capacitance and potential G, Vj and C;, Vs respectively, are joined with the help of a wire, the common potential will be (G+ (o) GM AG +O SHE Gu ( GYLG ree Vi+Vi Ans: ) SHES Given, Capacitance of first capacitor = C, Potential of first capacitor = Vi Capacitance of second eapacitor = C; and Potential of second eapacitor = We know that when two charged capacitors are joined with the help of a wire, then the eommon potential, YtGu CHG o Oa In photoelectric effect the maximum kinetic energy of ‘emitted electron depends on (a) wavelength (b) frequency (c) intensity (4) work function Ans: (a,b) ' plotoclectric equation is given by, K, where, y= work function Hence, © y and Ku, © 4 Hence, Ky 2 1 ad Kus 5. Hence maximnm kinetic energy depends on the wavelength, and frequency The number of Photons of frequency 10 Ha in radiation of 6.624 will he (a) 10” (b) 10! (c) 10” (a) 10 Ans (c) 10” Given:Page 2 E=602) vy = 10H Energy Frequency, We know that, ‘Nuuuber of Photons = Energy ‘Energy of one Photon = 6.02 6.62, Ww Gx 10 XO 1 What the estas of a AOW lap tel fal rine by ncuren of EAP (a) 200 (o) 2000 (©) 3600 (a) 4809 Ans : (c) 3600 Given, Power of lam, P =40W Current A We know that resistance of lamp, P_4 Rate = Hy ~ s000 4 Which of the following doe smutual induetion’? (a) dynamo (6) induction coil Ans (d)clectre heater We know that phenomenon according to which an opposing E.ME. is prodhiced in col as «result of ehonge in enrrent or maguotic ux linked with a neighbouring coil, is called suutual induction Since electric heater is based on heating effect of current. therefore it does not obey the phenomenon of mutual induetion. And dynamo, transformer and induction coils are based on the phionomenon ofthe mutual induetion NEET 45 YEARS PAPERS Downdload Free PDF From NODIA App Search Play Store by NODIA For more details vist : www.nodia injec ‘Add 86058 29866 in Your Class Whatsapp Group Got POF not obey the phenomen (b) transformer (@) electric heater & In a series combination, R= 3009, L = 1000 rad — 5". The impedance of th 9H, C=20F, LCReeireut is, (a) 4000 (b) 5009 (e) 9000 (a) 13000 Ans: (b) 5008 Given, Resistance, Inductance, Capacitance, C= 2nP = 2x 104P Angular velocity 1000 rad — s" We know that reactance of inductor, Xi, =wL = 1000 x09 = 9009 and reactance of eapacitor, Mla 1 Xe = So = i000 (2x10) 500 Sample Paper 1 Solutions CBSE Physics Class 12 Therefore, impedance of LCR-cireuit 2=/R+(%- Xe +/(300)* + (900 = 500)" = ¥'90000 + 760000 / 250000 = 5000 1. An electric dipole consists of positive and negative charge ‘of 4yC cach placed at a distance of 5 mm, The dipole moment is (2x 10°C () 6x 10°C m Ans: (a) 2x 10%C-m (b)4x Wm (a) 8x W*Cm Magnitude of each charge on dipole q=4yC =4x 10°C and distance betwoen the charges, x 10%m 2 We know that, Dipole moment,p = 4x 20 =x 9 x 6x09) =2x10°Cm ML Two solenoids of the same length having number of turns in the ratio of 2:3 are connectod in series, The ratio of magnetic fields at their centres is (a) 2:1 (b) 3:1 (a) 3:2 No, of tums in first solenoid, XN No. of tums in second solenoid, —N We know that no. of turns per unit le where, = length of solenoid We also know that maguetic field at the centre of solenoid, B= pont = mol Bree N Therfore BuBs 1, Assertion : In Young’s experiment, the fringe width for dark fringes is different from that for white fringes. Reason : In Young's double slit experiment the fringes are performed with a source of white ligt, then only black ancl bright fringes are observed. (a) Both Assertion and Reason are correct and Reason is the correct explanation of Asserti (b) Both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion, (} Assertion is correct but Reason is incorrect. (a) Assertion is incorrect but Reason is correct. Ans : (il) Both the Assertion and Reason are incorrect. Tn Young's experiment, fringe width of dark and white fringes are equal. IF white light is used as source, coloured fringes axe observed representing bright band of different colours.Page 3 Sample Paper 1 Solutions 1 Assertion We cannot think of magnet ikl configuration with thre poles Reason : A bar magnet docs exert ator its own fil, (#) Both Assertion and Reason are correct aud Reason isthe correct explanation of Assertion. (b)Both Assertion and Reason aro coreet, but Reason is not the correct explanation of Assertion, (6) Asertion is comoct but Reason is incorrect (4) Asertion is incorrect hut Reason is correct. on itself due to Ans : (d) Assertion is incorrect but Reason is correct Magnet field may be formed with the help of three poles. A. bar magnet does not exert a torque on itself due to its own, field M. Assertion (A) : Diamond behaves like an insulator, Reason (R) : There is a large energy gap between valence band and conduetion band of diamond (a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. (b)Both Assertion and Reason are eorreet, but Reason is not the correct explanation of Assertion (c) Assertion is correct but Reason is incorrect, (a) Assertion is incorrect but Reason is correct. ‘Ans : (a) Both Assertion and Reason are correct and Reason, is the correct explanation of Assertion Ta insulator, the forbidden energy gap is quite large. When ‘leetrie field is applied to such a solid, the electron find it difficult to acquire such a large amount of energy. Thus no electron flow ovens. 18 Assertion : In sories LOR cirenit resonance ean take place. Reason : Resonance takes place if inductance and capacitive resistances are equal and opposite (a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. (b)Both Assertion and Reason are correet, but Reason is not the correct explanation of Assertion (c) Assertion is correct but Reason is incorrect, (a) Assertion is incorrect but Reason is correct. Aus : (a) Both Assertion and Reasou are correct and Reason, is the correct explanation of Assertion In series resonance cireuit, current becomes maximum, Decanse total impeslance becomes zero. In case of LC circuit, ‘Total imptance W® Assertion (A) : Capacity of a parallel plate capacitor increases whan distance between the plates is doereased Reason (R) : Capacitance of capacitor is inversely proportional to distance between them. (9) Both Assertion and Reason are correct andl Reason isthe correet explanation of Assertion, (b)Both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion. (¢) Assertion is correct but Reason is incorrect, (4) Assertion is incorrect but Reason is correct Aus : (c) Assertion is correct but Reason is incorrect, Barth is a good conductor of very large size. When some CBSE Physics Class 12 small charge is given to earth, its potential does not change. Hence potential of earth is assumed to be zero, It is just like sen level which does not alter materially when water is added to it or removed from it. Thus, the potential of all, ‘other bodies are measured with reference to the earth, For this if the connection of a charged body to the ground by a. metallic conductor would cause electrons to flow to that body from ground, the body is at positive potential. Conversely, is also true. In either ease the conductor is neutralized and brought to zero potential. In fact the atmosphere does possess significant electric field, T-JEE 45 YEARS 120 PAPERS Downdload Free PDF From NODIA App. Search Play Store by NODIA For more details visit: www.nodia inijee ‘Add 86056 209609 in Your Class Nhasepp Group to Got POF SECTION-B In which sitmation is there a displacement current bnt no ‘conduction current ? Ans : During charging or discharging thereisadisplacement current but no conduction eurrent betwoen plates of eapacitor. Define conductivity of a material. Give its ST unit. Ans : The reciprocal of the resistivity of a material is called its coutnctivity and is denoted by a. Thus, Rar Conductivity = aaa or =} SI unit of conductivity is ohm-'m or ohm m™ or How does a cirenlar loop carrying current behaves as a. magnet? Ans: The eurrent round in the face of the coll isin anti-clockwise direction, then this behaves like a North pole, whereas when it viewed from other seale, then current round ia it isin clockwise direction necessarily forming South pole of magnet. OY Hence, current loop have both magnetic poles and therefore, behaves like a magnetic dipole The current flowing through @ pure inductance 2 mt is (15 c0s3004) A, What is the (i) rans. and (i) average value of current for a complete cycle Ans: Current flowing through the inductor, I = 15c0s300¢ Comparing withPage T= hsinut Here, peak value of curren, 1=15A () For compete cyclo, rms valus of current a v2 V2 (ii) For complete eycle average value of current is zero fie Io =0 11 State the criteria for the phenomenon of total internal reflection of light to take place, or ‘What is the difference between Magnification and Magnifying power? Ans : Following are the criteria for total internal reflection (i), Light must pass from a denser to @ rarer medi, (ii) Angle of incidence must be greater than critical angle. ‘The difference between Magnification and Magnifying power: Magnification ‘Magnifying Power 1. | Magnification is the ratio| Magnifying power is, ‘of image size to object | the ratio of visual angle am sta hy th ge (3) and visual angle bn th (a). Tie wl of maging poner es toner 2 au (I+ 3) CUET 30 SAMPLE PAPERS Downdload Free PDF From NODIA App Search Play Store by NODIA For more details visit: www.nodia.in/cuet ‘Add 88056 280609 n Your Class Whatsaop Group to Get POF 2, | Therangeofmagnification ix 0 to 400 SECTION-C ‘2 In the Rutherford scattering experiment, the distance of ile is closest approach for an a-particle i dk. Mapa replaced by a proton, then how much kinetic energy in comparison to a-particle will be tequted to have the same distance of closest approach d Aus Distance of closest approach = 2kee Ke Since, forgiven distance of cat appronch Knot energy 27 (atomic aber) ere, k= Where, k= = % Draw energy band diagram of n-typed and p-typed semiconduetor at temperature T> 0K. Mark the donar Sample Paper 1 Solutions CBSE Physics Class 12 and acceptor encrgy level with their energies Ans: The required energy band diagram is shown below. ‘Condaction band Acorptorenegy vel eet oe mot ev Valence band (0) pve Condation bane in04 ev eee Donor ene lve Valence band (0) mtype 2A Two equal balls having equal positive charge q caulombs are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is, inserted between the two? Ans: Form Coulomb's law, electri force between the two charged bodies, in a medium, where, K = dielectric coustant of the medium, K=1 For plastic, K >1 Therefore, after insertion of plastic sheet, the force between the two balls wll reduce For vacuum, % State the Bio-Savart law for the magnetic field due to a ‘current carrying element. Use this law to obtain a formula for maguetic field at the eoutre of a eiccular loop of radius R carrying a steady current 1. Indicate the dircetion of the magnetic field. Ans: According to Bio-Savart's law, the magnctie field due to a current element di carrying current I at a point P at distance r from itis given by, a = fe. Hind where, is permeability of free space and 9, is the angle betwoen di’ and 7. Magnetic field at the centre of a circular eurrent carrying coil: Consider a circular loop of wire of radius r carrying & ‘current J. Consider a current clement a of the loop. The direction of df is along the tangent, so dL 7. From Bio-Savart law, magnetic field at the centre Q due to thisPage 5 Sample Paper 1 Solutions [Magnet Fed atthe Centre of Cirelar Current Loop ‘The magnetic field due to all such current elements wil point into the plane of paper at coutze O as shown by encireled cross ©, Hon the total magnetic field at the centre 0 is, fan = [tel dl = ool ae fanm fie Gee fa dol, a sul 28, When does snell’s law fail in reftaction? Ans Snell's law fails, ff ray of light in a denser medium incident ‘on a rarer medium at an angle more than the eritieal angle for that pair of media, then the ray of light is totally reflected, back into the denser medium instead of going to the rarer smeodiuta as predicted by Suell’s law. Hence in this situation, Snell's law fails in refraction, 1. How the size of @ ancleus is experimentally determined? Write the relation between the radius and mass nmmber of tle melons. Show that the density of nucleus is independent of its mass mnmiber. Ans ‘Tho size of the nuclous is experimentally determined using Rutherford's a-scattering experiment and the distance of closest approach ancl impact parameter. Tho relation betwoon radius and mass number of nucleus is, R= RAM 1.2Fm A= mass mmnber where, R, radius of nucleus Nuclear density,p = Masso nucleus_ ___md___ P= Volmmeof uncleus ~ Tz(R, A") amass of each meleon Hence, P= aa where, m From the above forumila, itis clear that p does not depend ‘on the mass number. 2. Explain why do we need coherent sources to produce Interference of light. jonable mass. Give its Define multiplicat CBSE Physics Class 12 plisical significance. Ans: When two monochromatic waves of intensity , fe ancl phase difference 6 meet at a point, the resultant inte is given by, =Hhth+Whhoso ‘The last term 2y I, /, cos is called interference term. There ae «wo posites Le Hew rem c total intensity at any point ill bo constant, The intensity will be maximum (vf +yh)* at points where cosd is +1. aus sininnn (/,~ (/B)? at points where con =— 1 “The sures in this ease are coherent 2. If cong vats contiamosly with tne, thea the inter ference term averages to ner, The tensity, = 1, +h at every point ie., there will be eral illumination on the obnervation sereen. The two foeos inthe case are Incoherent Hence to observe interference, we need to have two sours withthe sume frequency aud with sable phase diference Sich a pir of sources are called ecierent sourees, ins constant with time, t ‘The multiplication factor of a fssionable mass is defined as th ber of neutrous present at the beginning ‘ofa particular generation to the munber of neutrons present at the beginning of the previous generation, ratio of the ‘Ser of en pret he basing of pes et ure of the growth Thos, The multiplication factor k gives a me: rate of the neutrons in a fissionable mass If k> 1, the chain reaction grows. If k= 1, the chain reaction remains steady. If < 1, the chain reaction gradually dies out. SECTION-D Consider the situation shown in figure, The two sits s, and s, placed symmetrically around the central line are illuminated, by monochromatic light of wavelength a. The separation betwen the slits iso The ligt transmitted hy the sits falls ‘oa screen §; place at a distance D frou the slits. The slits ‘is at the contral line and the slit 5, is ta distance fr S Another sereen S., is placed a further distanee D away’ from 5, (i) Find the path difference if 2=42 (aa oan (haya (a Ams :(b) 4/2 As sai At s:AZasPage 6 NO NEED TO PURCHASE SAMPLE PAPERS Download India’s Best 20 Sample Papers “DOWNLOAD NODIA APP Every School conduct Pre-Board From These Sample Papers (ii) Find the ratio of the maximum to minimum intensity observed on §,, if == 42 (a) (© Aus (c) @t wD 7 s:42~ ADS. Ag at S:az=4Pd oy Hence, maxima at Sas well as S, Resultant intensity at §, (iil) Two coherent point sources s, and S, are separated by a suall distance das shown in figure. The fringes ob- tained on the sereen will be 1. SSN sercen D (a) concentric cireles _(b) points (6) straight lines (a) semi-cireles Ans : (a) concentric eizeles When the screen is placed perpendicular to the line joining the sources, the fringes will be concentric circles. (iv) Tn the case of light waves from two colierent sources 5\ and 5, there will be constructive interference at an arbitrary point P, ifthe path difference 5,P~ 5? is (b)m (a) 19 Ans : (b) 0 Constructive interference occurs when the path difference (S,P— ,P) is an integral multiple of». or §P- SP = nd, where n= 0, 1,2, 8, (©) Two monochromatic light waves of amplitudes 34 and 2A interfering at a point have a phase difference of 60°. The intensity at that point will be proportional to (0) sa (bist (ou (die Ans : (@) 194 Hore, 4)=34, 4, =24 and 6 =00 The resultant amplitude at @ point is ab Ay 2AA, Veayy ear oa aay oa As, Intensity = (Amplitude)® ‘Thorefore, intensity at the same point is 2K GA KA X cost" 0 Sample Paper 1 Solutions CBSE Physics Class 12 Tee 194 CO 3 Gauss’s law and Coulomb's law, although expressed in different forms, are equivalent ways of deseribing the relat betwoen charge and cleetric field in statie conditions. Gans law is when gay is the net charge inside an imaginary closed ‘ure called Gaussian surfce. 9 = $B di sive the electri fus through the Ga face, The two ‘equations hold only when the net charge i in vacuum or ai. Gaussian spherical sitfaces (i) there is only one type of charge in the miverse, then (B+ Electric field, dé + Area vector) (of surface (b) f 8 ai could not be defined (6) f B+ i= if charge is inside (l) f= de=0 if charge is ow O Nee wine + i#0 on an fee Ans: (d) fB+ d=0 if charge is outside, fE+ dF dase ; If there is only one type of charge in the universe then it will produce electric field somehow. Hence Ganss's lar is valid (i) What is the nature of Ganssian surface involved in Gans lnw of electrostatic ? (a) Magnetic (b) Scalar (6) Vector (4) Electrical Ans + (¢) Vector (ii) A charge 10 uC is plied at the centre of a hemisphere of radius f= 10 em as shown. The electric lux through, the hemisphere (in MKS units) is H10pe (0) 20x 10° (6x 10 Ans: (0) 6 x 10° According to Ganss’s theorem, Blectrie flax through the sphere = 2 Flectrc fx throngh the hemisphere = (b)10 x 10° (a2 x16 10x10 SSH 06x W°Nm?Ct = 6x 10°NmeC (iv) ‘The clectrie fx through a closed surface area S enclos- ing charge Q is 9. Ifthe surface area is doubled, then the flux is (a) 26 (yee 0.56 x 10NmeePage 7 Sample Paper 1 Solutions (©) of (oe Ams: (a) 6 ‘As fue is the total number of lines passing thongh the surface, for a given charge, it is always the charge enclosed. Qlén- If area is doubled, the flux remains the same. (0) A Gaus surface endosesadpale The ltr through this surface is @e (yt (ot (A) zero Ans : () zero As net charge on a dipole is Card=o Thns, when a ganssian surface encloses a dipole, as per Ganss’s theorem, electric fux through the surface, NEET 45 YEARS PAPERS Downdload Free PDF From NODIN? Search Play Store by NODI For more details vist : www. en injjee ‘Ad 89056 29966 in Your Class Watsapp Group to Got POF SECTION-E 31 Define mutual inductance between a pair of evils, Derive ‘an expression for the mutual inductance of two long coaxial, solenoids of same length round one over the other. In a closed cirenit of resistance 109, the linked flux varies with time according to relation 6 = 6 — 5t+1. At t= 0.25 second, What is the eurrent (in Ampere) flowing through the cirenit? Ans Mutual Inductance : Let » current of jj ampere flow in the primary coil. Let, due to this eurrent, the magnetic flux, Tinked with each turn of the sceondary evil be 6. If Ny be the umber of turns in the secondary coil, then the nmmber of flux-linkages im the coil will be Nags, This nmmber is proportional to the current j flowing in the primary col, iil TRE a 1b: iy or Nib: = Mig Where, Mis constant, called! the coefficient of mutual induetion or mmtual inductance of the two coils. From the ‘equation, we have, Ii =1, then M = Nad Mutual Inductance of Two Long Coaxial Solenoids Consider two loug solenoids $, and S$) of same length (J, CBSE Physics Class 12 such that solenoid (S,) surrounds solenoid (.) incompletely. 5 Let m, be the number of turus per unit ls he the number of turus per unit length of $1, be eurrent passed throngh solenoid $; and ga be fax linked with Sy due to current fowing through S, 64 [oF 6x = Mah Where MG isthe coefficient of nmtual induetion ofthe two solenoids when current is passed through solenoid ($) an emf is induced in solenoid (). Magnetic fell produeed inside solenoid (Si) on passing current through it Bi = torch Magnetic flux linked with each turn of solenoid (%) will be ‘qual to By times the area of cross-section of solenoid (S)) Magnetic fax linked. with each turn ofthe solenoid (5) = BA. Therefore, total magnetic Oh: Knked with the solenaid ($3) BAX tal pomh x AX nal portal hommAL (1) Similaely, the unt inductances between the two solenoids, when current is passed through solenoid (S;) and induced {is produced in solenoid (8) and is given by Ma =ommal M. M (say) Hence, cooficient of mutual induction hetwoen the two long solenoids M = poonmAl. Wo can rewrit oq. (1) as Mamet Given, We know that, Hee, 12x 0.2545 345=2Volt ‘Now, Current -£ Here, R=09Page 8 Sample Paper 1 Solutions 3-02 T-JEE 45 YEARS 120 PAPERS Downdload Free PDF From NODIA App Search Play Store by NODIA For more details visit : www.nodia.in/jee ‘Add 89056 708689 n Your Class Whatsapp Grup to Get POF the inconsistency in Ampere’s cireuital law. What modification was made my Maxwell in this law? A parallel plate capacitor is charged to 60uC. Due to a radioactive souree, the plate loses charge at the rate of 18x 10°Cs"', What is the magnitude of displacement current? Ans : According to Ampere’ aagneti field Bt nlong any closed loop Cis proportional to the current 1 passing through the closed loop, ie SP bmn () 16K = +) 0= i 4 |_| —4 Battery Battery A parallel plite capacitor being charged hy a battery Tn 1864, Maxwell showed that the Eq. (1) is logically inconsistent, Consider a parallel plate capacitor being charged by a battery, as shown in figure (a). As the charging current T flows, a magnetie field is set up around the eapacitor. Now, the current I flows across the area honnded by loop G. Therefore (2) But the are bounded by C; lies in the region between the capacitor plates, so no current flows across it, Hence, f5-d=o (3) Imagine the loops C; ane Cs to be infinitesimally close to ‘each other, as shown in figure (b). Then we must have, f Bedi f Be dl ‘This result is inconsistent with the Eq, (2) anu (8). ‘There inst be sonne missing term in this law. Maxwell's Modification of Ampere’s Law Maxwell argued that a changing electric field between the capacitor plates must induce a magnetic field. As currents tare the usual sources of magnetic fields, a changing electric field amist be associated with a current, Maxwell called this current as the displacement current: If, A be the arva ofthe eapacitor plates and g be the charge on the plates at any’ instant £ during the charging process, ten the electric field in the gap will be, B=2 CBSE Physics Class 12 Fox, Hence, 4/9) _ 1d = ale) = coat But dg/dt is the rate of change of charge on the capacitor plates. It is called displacement current and is given by, & a This isthe missing term in Ampere’ circuit law, The total arnt rast be the sum of the conduction current, and the displacement current [,. Thus Lth= at ‘Hence, the modified form of the Ampere's law is, f8-d = pf ecMte Given, Charge on capacitor, q = 60uC = 60 x 10°C ‘and rate of losses of charge, Mt rs charging capacitor C, a time varying current I(0) flows through the conducting wire, so on applying Ampere’s cireuital law (for loop A) B+ di = jol() (1) f 4 i Now, we cousider a pot like surface enclosing the positively charged plate and nowhere touches the conducting wire (2) If surface A and B forms a tiffin box and electric field B is passing through the surface (B); constitute an electric flux. o=8/4|=2]4] Hank @) If the charge on the plate in the tiffin box is changing with time, there mst be a current between the plates. From equation (8), we got Sample Paper 1 Solutions CBSE Physics Class 12 1Q_ do, 1= B= 4Fe00) =o his is the missing term in Ampere’s cireuital la. ‘The inconsistency may disappear if displacement current is included betwoon the plates, So, generalised! Ampere’s cirenital law can be given as fia o@l(t) + o6de = oll + noo 1. By considering straight line of V-I characteristic curve between 10 mA to 20:mA; and assuming passing through the origin, draw horizontal and vertical lines from 10mA and 20 mA, we have 1 =10mA, iv mA, osv resistance in forward biasing can be given as, Dynami av Roar (8-0.7)V _ o1v ~(20-T0)mA ~ 10mA O19 = $2 x 10'9 =100 2, At -10V. The V-J characteristic graph is a straight line parallel to the voltage axis and not showing any ‘variation. So, the statie resistance ean be given as, wv