PRE-MEDICAL : ENTHUSIAST COURSE
MAJOR TEST08–01–2013PRE-MEDICAL : ENTHUSIAST COURSEHAVE CONTROL ¾ HAVE PATIENCE ¾ HAVE CONFIDENCE Þ 100% SUCCESSBEWARE OF NEGATIVE MARKING1.If switch is open then null point is 4 m away fromA end. What will be distance of null point fromA end, if switch is closed ?1.(10V, 2 W)3Wtc dqath [kqyh gS rks 'kwU; fo{ksi fcanq dh A fljs ls nwjh4 m gAS ;fn dqath can dj ns rks 'kwU; fo{ksi fcanq dh Afljs ls nwjh gksxh\(10V, 2 W)3W4m4mA5W, 5mBA5W, 5mGGE2.(1) 2.8 V (2) 4V(3) 1.6 V (4) 3 VThe ratio of power gain and resistance gain in case3.of common-base amplifier is :(1) a(2) a2(3) a3Find value of I :E2.(1) 2.8 V (2) 4V(3) 1.6 V (4) 3 VCB izo/kZd esa 'kfDr ykHk rFkk izfrjks/k ykHk ds vuqikrdk eku gksxk :-4(4) a3.(1) a(2) a2I dk eku gS :6WI(3) a3I3W1.5W1.5W3W4.5.6.3W2W4V(1) 0.67 A (2) 0.33 A (3) 1 A(4) 0.5 AThe correct relation for a, b for a transistor is:(1) b 1-aa(2) b (3) a b -1b(4) ab 1(1) Emits light when reverse biased(2) Emits light when forward biased(3) Is made from semiconducting compoundgallium arsenide phosphide4.a1- aResistance of a galvanometer 50W and its current–4range is 10 A. We want to measure maximum4V with this meter. Find value of series resistanceto be connected :(1) 9999 W(2) 19950 W(3) 39950 W(4) 29950 WThe light emitting diode (LED) :-(4) Both (2) & (3)(4) a46W3W4VB5.6.2W(1) 0.67 A (2) 0.33 A (3) 1 A(4) 0.5 AVªkalfLVj esa a o b ds e/; laca/k gksrk gS :(1) b 1-aa(2) b a1- a(3) a b -1b(4) ab 1,d xsYosuksehVj dk izfrjks/k 50W gS rFkk bldh /kkjk dh–4ijkl 10 A. ge bl ehVj ls vf/kdre 4 oksYV foHkoukiuk pkgrs gAS dq.Myh ds Js.kh Øe esa tksM s tkus okysizfrjks/k dk ekuk gksxk :-(1) 9999 W(2) 19950 W(3) 39950 W(4) 29950 WLight Emitting Diode (LED) :(1) i'p ck;l esa izdk'k mRlftZr djrk gS(2) vxz ck;l esa izdk'k mRlftZr djrk gS(3) GaAsP tlS s v/kZpkyd ls curk gS(4) (2) o (3) nksuksaiz R;s d iz ' u dks vtqZ u cudj djks AHYour Target is to secure Good Rank in Pre-Medical 20131/31
MAJOR TEST08–01–2013TARGET : PRE-MEDICAL 2013 (NEET-UG)7.Two concentric metallic shells are given in fig.Inner shell is given 2Q charge and outer thickshell is given 3Q charge. Potential on the centreis :-7.nks ladsfUnz; /kkfRod dks'k fp esa fn[kk;s x;s gAS vkUrfjddks'k dks 2Q vkos'k o cká eksVs dks'k dks 3Q vkos'k fn;kx;k gAS dsUnz ij o S qr foHko gS :-aa3a3a2a2a8KQKQ7KQKQ(2)(3)(4)3a5a4aaThe electrical conductivity of pure germaniumcan be increased by :(1) increasing the temperature(2) doping acceptor impurities(3) doping donor impurities(4) All of aboveA half-ring of radius r has linear charge densityl. The electric potential at the centre of the halfring is :-8.9.l(1) 4e010.l(2) 4p 2 e r08.9.v¼Z oy; ds dsUæ ij foHko gksxkl(1) 4e0ll(3) 4pe r (4) 4e r00Identify the part X in following block diagramof a reciever in AM. :-10.Recieving AntennaX2/31(2)l(4) 4e r0AM ds Reciever unit ds Blockdiagram esa X D;kn'kkZrk gS erOutputOutput(1) Square law device(2) Envelope detector(3) IF Stage(4) RectifierTwo identical metal balls with charges 2Q and–Q are separated by some distance, and exert aforce F on each other. They are joined by aconducting wire, which is then removed. Theforce between balls will be :(1) Fll(2) 4p 2 e r (3) 4pe r00Recieving AntennaAmplifier11.8KQKQ7KQKQ(2)(3)(4)3a5a4aa'kq¼ Ge dh o S qr pkydrk fuEu rjhdksa ls c kbZ tk ldrhgS :(1) rkieku c k dj(2) xzkgh v'kqf¼ Mksi dj ds(3) nkrk vk'kqf¼ Mksi dj ds(4) mijksDr lHkh,d r f T;k dks v¼Z oy; dk js[kh; vkos'k ?kuRo l gAS(1)(1)(4)F8(1)(2)(3)(4)11.Square law deviceEnvelope detectorIF StageRectifiernks loZ le /kkfRo; xs a n ks a ij vkos' k 2Q rFkk –QgS rFkk muds e/; dq N nw j h g S vk Sj ;s ,d nw l js ijF cy vkjks f ir djrh g SA budks ,d pkyd rkj lstks M ns a rFkk fQj vyx dj ns a rks ] vc xs a n ks ds eè;cy gks tk;s x k(1) F(2)F2Your Target is to secure Good Rank in Pre-Medical 2013(3)F4(4)F8H
MAJOR TEST08–01–2013PRE-MEDICAL : ENTHUSIAST COURSE12.13.By what percentage will the transmission rangeof a TV tower be affected when height of toweris increased by 21% :(1) 5%(2) 10%(3) 15%(4) 20%In the given circuitRvoltmeter reads 100 V.AVNow R is connectedparallel to voltmeterthen A reads 2A and12.(1) 5%13.120VV reads 60 V. Find initial reading of ammeter:(1) 2/3 Amp(2) 1 Amp14.15.(3) 3 Amp(4) 1.5 AmpIn following radioactive reaction which elementis formedN14 2He4 1H1 ?7(1) 8O17(2) 6C14(3) 7N14(4) 9Ne18Two condensers 1mF & 2mF in a circuit are joinedas shown in figure. The potential of point P & Qare 3 kV & 1 kV respectively. The potential ofpoint R will be :-(1)16.17.3kV5(2)VCCWhen capacitors are fully charged then sum oftheir stored energy is U1. If the key K is openedand a material of dielectric constant Îr 2 isinserted in each capacitors then sum of their stored(1)H35(2)53U1will be :–U2(3)45(3) 15%RA54V120Vdk ikB ;kad 2A rFkk V dkikB ;kad 60 V gAS vehVj dh izkjafHkd ikB ;kad gksxk :(1) 2/3 ,Eih;j(2) 1 ,Eih;j(3) 3 ,Eih;j(4) 1.5 ,Eih;jfuEufyf[kr ukfHkdh; vfHkfØ;k esa mRiUu rRo gS &N14 2He4 1H1 ?(1) 8O17(2) 6C14(3) 7N14(4) 9Ne18nks la/kkfj 1mF rFkk 2mF fp kuqlkj tqM s gq, gAS fcUnq Po Q ds foHko Øe'k% 3 kV o 1 kV gAS R fcUnq ijfoHko gksxk :-15.3kV5(2)(3) 2 kV16.5kV3(4) None of them B ÅtkZ esa ;fn X, B rFkkvfHkfØ;k X AA dh izfr U;wfDyvkWu cU/ku ÅtkZ Øe'k% 7.4 MeV, 8.2MeV rFkk 8.2 MeV gks rks vfHkfØ;k ls mRiUu ÅtkZ g&S20011090(1) 200 MeV(2) 160 MeV(3) 110 MeV(4) 90 MeVKfp kuqlkj nks ,d lekula/kkfj V oksYV dh cVS jhVCCls lekUrj Øe esa tqM s gaSAla / kkfj iw . kZ :i lsvkosf'kr gks tkus ij muesalaxzfgr dqy ÅtkZ dk ;ksx U1 gAS ;fn dqath K [kksy nhtk;s rFkk izR;sd la/kkfj esa Îr 2 ijko S qrkad dk inkFkZHkjus ij dqy ÅtkZ dk ;ksx U2 gks rks(4)(4) 20%717.K(2) 10%fn;s x;s fp esa oksYVehVj dkikB ;kad 100 V gAS vc ;fn Riz f rjks /k dks oks YVehVj dslekukUrj yxk fn;k tk, rks A(1)As shown in figure twoenergy is U2. Then14.5kV3(3) 2 kV(4) None of themIf binding energies per nucleon of X, B and A are7.4 MeV, 8.2 MeV and 8.2 MeV respectively,then the energy released in the reaction : X200 A110 B90 energy will be(1) 200 MeV(2) 160 MeV(3) 110 MeV(4) 90 MeVidentical capacitors areconnected to a batteryof V volts in parallel.;fn TV Tower dh Å¡pkbZ 21% c k nh tk, rks TV towerdh izsf{kr ijkl (Transmission Range) esa fdrukizfr'kr ifjorZu gksxk :-(1)35(2)53Your Target is to secure Good Rank in Pre-Medical 2013(3)45U1U2dk eku gksxk:–(4)543/31
MAJOR TEST08–01–2013TARGET : PRE-MEDICAL 2013 (NEET-UG)18.19.What is the power output of 92Y235 reactor if it takes30 days to use up 2 kg of fuel and if each fissiongives 188 MeV of usable energy (1) 59 MW(2) 51 104 MW(3) 188 MW(4) None of the aboveA point charge Q is placed at origin if electric fieldrrrrat A and B is E1 and E 2 then E1 - E 218.19.dk eku gksxk :-is :-QO20.21.Y235 fj,DVj dh fuxZr 'kfDr D;k gksxh ;fn ;g 30 fnuesa 2 fdxzk bZ/ku dk mi;ksx djrk g]S rFkk izR;sd fo[k.Muesa eqDr ÅtkZ 188 MeV g&S(1) 59 MW(2) 51 104 MW(3) 188 MW(4) mDr esa ls dksbZ ugha,d fcUnq vkos'k Q ewy fcUnq ij fp kuqlkj fLFkr gS ;fnrrrrfcUnq A rFkk B ij fo qr {ks E1 o E 2 gks rks E1 - E 292QOr A r A rr B B2Q2 pÎ0 r 2(1) Zero(2)Q(3) 2 2pÎ r 20Q(4) 2p Î r 20The nucleus 92X234 decays by emitting 3a and 1b–particle. Final product is(1) 87Y228(2) 84Z228(3) 87Y222(4) 84Z222A circuit is arranged as shown. Then, the currentfrom A to B is :20.(2)Q(3) 2 2pÎ r 20Q(4) 2p Î r 2092X234 ukfHkd 3a o 1b– d.k mRlftZr dj fo?kfVr gksrkgAS vafre mRikn g&S(1) 87Y228(3) 87Y22221.2Q2 pÎ0 r 2(1) Zero(2) 84Z228(4) 84Z222fn;s x;s ifjiFk esa A ls B dh vksj /kkjk gS :AA10W10W15W10V10W15W10V5V5V10WBB22.(1) 500 mA(2) 250 mA(3) –500 mA(4) –250 mATwo elements P and Q each of mass 10–2 kg arepresent in a given sample. The ratio of their atomicweights is 1 : 2 and the half lives are 4s and 8srespectively. The masses of P and Q present after16s will be respectively(1) 6.25 10–4 kg and 2.5 10–3 kg.(2) 2.5 10–3 kg and 6.25 10–4 kg.(3) 6.25 10–4 kg and 12.5 10–4 kg.(4) None of the abovedksbZ Hkh iz'u4/31Key Filling(1) 500 mA(3) –500 mA22.(2) 250 mA(4) –250 mAfdlh izfrn'kZ esa nks inkFkksZ P rFkk Q izR;sd dk 10–2 fdxzknzO;eku mifLFkr gAS buds ijekf.od Hkkjksa dk vuqikr1 : 2 rFkk v¼Z vk;q Øe'k% 4 ls- rFkk 8 ls- gaSA 16 lsi'pkr P o Q ds mifLFkr nzO;eku Øe'k% gksxsa&(1) 6.25 10 fdxzk o 2.5 10 fdxzk(2) 2.5 10 fdxzk o 6.25 10 fdxzk(3) 6.25 10 fdxzk o 12.5 10 fdxzk(4) mijksDr esa ls dksbZ ugha–4–3–3–4–4–4ls xyr ugha gks uk pkfg,AYour Target is to secure Good Rank in Pre-Medical 2013H
MAJOR TEST08–01–2013PRE-MEDICAL : ENTHUSIAST COURSE23.A man's near point is 0.5 m and far point is 3m.Power of spectacles lenses required for -23.(i) reading purposes(ii) seeing distant objects, respectively, are :(1) – 2D and 3D24.(2) 2D and – 3D(2) 2D vkSj – 3D(3) 2D and – 0.33 D(3) 2D vkSj – 0.33 D(4) – 2D and 0.33D(4) – 2D vkSj 0.33DThe radioactive rays can be detected by theiraction on photographic film. A radioactivesample enclosed in an enclosure produces a spoton a photographic film. When a perpendicularmagnetic field is applied to the enclosure, howmany spots are observed on the photographicfilm(1) 125.(2) 2121(3) n 227.(3) 324.(4) 4A light ray is incident perpendicular to one faceof a 90 prism and it totally internally reflectedat the glass - air interface. If the angle ofreflection is 45 , we conclude that the refractiveindex n is :(1) n 26.,d O;fDr dk fudV fcUnq 0.5 m vkSj nwj fcUnq 3mgSA p'esa ds ysa Ulksa dh {kerk gksxh Øe'k% (i) i usa ds fy;s(ii) nwjLFk oLrqvksa dks ns[kus ds fy;s :(1) – 2D vkSj 3D25.jsfM;ks&,sfDVo fdj.kksa dks mudh QksVksxzkfQd fQYe ds Åijvuq f Ø;k ds }kjk la l w f pr fd;k tk ldrk g SA ,djsfM;ks&,sfDVo uewuk tks ,d vkoj.k esa cUn g]S QksVksxkz fQdfQYe ij ,d fcUnq mRiUu djrk gAS tc vkoj.k ij ,dyEcor pqEcdh; {ks yxk;k tkrk g]S rks fdrus fcUnqQksVksxzkfQd IysV ij voyksfdr gksaxs g&S(1) 1(2) 2(3) 3(4) 490 fiz T e ds ,d i " B ij iz d k'k fdj.k yEcor vkifrr gksrh gS vkjS dk¡p & ok;q dh lhek ij iw.kZr%vkUrfjd ijkofrZr gksrh gSA ;fn ijkorZu dks.k 45 g S ] rc ge fu"d"kZ fudkyrs g S ] fd viorZ uxq.kkad n gS :(1) n (2) n 245 45 1(3) n 2(4) n 2The inductive time constant is -1226.(2) n 245 45 (4) n 2izjs f.kd le; fu;rkad (inductive time constant) gksrk g&S(1) LR(2) L/R(1) LR(2) L/R(3)(4) R/L(3)(4) R/L(L R)The plane faces of two identical plano - convexlenses each having a forcal length of 50 cm areplaced against each other to form a usualbioconvex lens. The distance from this lenscombination at which an object must be placedto obtain a real, inverted image which has thesame size as the object, is :-27.(L R)50 cm Qks d l nw j h okys nks leryks Ù ky ys U lks a dhlery lrgs a ,d & nw l js dh vks j j[kdj ,dlkekU; mHk;ks Ù ky ys U l cuk;k x;k g SA bl ys U lds la ; ks t u ls fdl nw j h ij oLrq dks j[kk tkukpkfg;s A ftlls oLrq ds vkdkj dk gh okLrfodrFkk mYVk iz f rfcEc iz k Ir gks r k g S :-(1) 50 cm(2) 25 cm(1) 50 cm(2) 25 cm(3) 100 cm(4) 125 cm(3) 100 cm(4) 125 cmHYour Target is to secure Good Rank in Pre-Medical 20135/31
MAJOR TEST08–01–2013TARGET : PRE-MEDICAL 2013 (NEET-UG)28.The power in ac circuit is given by P ErmsIrms cosf. The value of power factor cos f in LCR circuitat resonance is (1) zero29.30.(2) 1(3) 1/2(4) 1,d ,-lh- ifjiFk] esa 'kfDr dk O;atd gS P ErmsIrms cosf fdlh LCR ifjiFk esa vuqukn ds le; cos f dk ekugS&(1) 'kwU;2The focal lengh of field achromatic combinationof telescope is 90 cm. The dispresive powers oflenses are 0.024 and 0.036 respectively. Theirfocal length will be:(1) 30 cm and 60 cm(2) 45 cm and –90 cm(3) 15 cm and 45 cm(4) 30 cm and – 45 cmA conductor rod AB moves parallel to X-axis ina uniform magnetic field, popinting in the positive Z-direction. The end A of the rod getsY28.29.30.(2) 1(3) 1/2YBVVAX32.33.positively chargednegatively chargedneutralfirst positively charged and then negativelychargedIn Young's double slit experiment carried outwith light of wavelength l 5000Å, thedistance between the slits is 0.2 mm and thescreen is at 200 cm from the plane of slits. Thecentral maximum is at x 0. The thirdmaximum will be at x equal to :(1) 1.67 cm(2) 1.5 cm(3) 0.5 cm(4) 5.0 cmThe hardness of a magnetic material is stated onthe basis of the following property of the material(1) retentivity(2) susceptibility(3) permeability(4) coercivityWhen the angle of incidence on a material is 60 ,the reflected light is completely polarised. Thevelocity of the refracted ray inside the materialis (in ms–1):æ 3 ö8(2) ç 10è 2ø(1) 3 108(3)6/313 108(4) 0.5 108X(1) /kukRed vkosf'kr(1)(2)(3)(4)31.2,d nwjn'khZ ds vo.kZd {ks la;kst u dh Qksdl nwjh90 cm g S A ys U lks a dh fo{ks i .k {kerk,¡ Øe'k%0.024 rFkk 0.036 g S A mudh Qks d l nw f j;k¡gksaxh :(1) 30 cm rFkk 60 cm(2) 45 cm rFkk –90 cm(3) 15 cm rFkk 45 cm(4) 30 cm rFkk – 45 cm,d pkyd NM AB, X-v{k ds lekUrj V osx ls xfrdj jgh gAS bl LFkku ij Z-v{k dh fn'kk esa ,d ,dlekupqEcdh; {ks mifLFkr gAS bl NM dk fljk A gks tk;sxk&BA(4) 1(2) ½.kkRed vkosf'kr(3) mnklhu(4) igys /kukRed] fQj ½.kkRed31.rjaxnS/;Z l 5000Å okys izdk'k }kjk fd;s x;s ;axf}fLyLV iz;ksx esa fLyVksa ds chp nwjh 0.2 mm vkjS fLyVksads ry ls ijnk 200 cm ij g S A ds U nz h ; mfPp"Bx 0 ij gSA rc r rh; vf/kdre x ij gksxk] tks cjkcjgS :(1) 1.67 cm(3) 0.5 cm32.33.(2) 1.5 cm(4) 5.0 cmfdlh pqEcdh; inkFkZ dh pqEcdh; dBksjrk (hardness)inkFkZ ds fuEu xq.k ls vkadh tkrh g&S(1) vo'ks"k pqEcdh;(2) pqEcdh; izo fÙk(3) pqEcdh; ikjxE;rk(4) pqEcdh; fuxzkfgrktc ,d inkFkZ ij vkiru dks.k 60 gksrk gS] rc ijkfrZrizdk'k iw.kZr% /kqzfor gksrk gSA inkFkZ ds vUnj viofrZrinkFkZ dk osx gS (ms esa):–1(1) 3 108æ 3 ö8(2) ç 10è 2ø(3)(4) 0.5 1083 108Your Target is to secure Good Rank in Pre-Medical 2013H
MAJOR TEST08–01–2013PRE-MEDICAL : ENTHUSIAST COURSE34.In the figure A, B and C, are current carryingwires. The direction of resultant force on the wireB will be AB35.36.37.A AB Cr3A D C (1) VA VB VC VD (2) VD VC VB VA(3) VD VC VB VA (4) VD VC VB VAHBr2A(1) in the plane of paper and towards the right(2) perpendicular to plane, out of paper(3) in the plane of paper and towards the left(4) along the direction of currentThe main difference in the phenomenon ofinterference and diffraction is that :(1) diffraction is due to interaction of light fromthe same wavefront whereas interference isthe interaction of (light) waves from twoisolated sources.(2) diffraction is due to interaction of light fromsame wavefront, whereas the interference isthe interaction of two waves derived from thesame source.(3) diffraction is due to interaction of wavesderived from the same source, whereas theinterference is the bending of light from thesame wavefront.(4) diffaction is caused by the reflected waves froma source whereas interference is caused due torefraction of waves from a source.The magnetic field at point O due to currentcarrying wire bent in the shape shown in figureism0 i(1)4 prm0 ir O( p 2)(2)4 prm0 ii(p 1)(3)4 pr(4) zeroFor the isolated charged conductor shown in Fig.the potentials at points A, B, C and D are VA, VB,VC and VD respectively. Then fp esa ist ds ry esa A, B rFkk C, /kkjkokgh] yEcs lhèkspkyd rkj gAS rkj B ij yxus okys ifj.kkeh cy dh fn'kkgS&Cr1A34.r1A2A3Ai "B ds ry esa nk;ha vksji "B ry ds yEcor ckgj dh vksji "B ds ry esa ck;ha vksj/kkjk dh gh fn'kk esaO;frdj.k vkSj foorZu dh ifj?kVukvksa esa eq [; ifjorZu;g gS fd :(1) foorZu] leku rjaxkxz okys izdk'k ds O;frdj.k lsgksrk gS] tcfd O;frdj.k nks Lora izdk'k L ksrksadh rjax ksa dh ijLij fØ;k }kjk gksrk gSA(2) foorZu] leku rjxkxz okys izdk'k ds O;frdj.k lsgksrk gS] tcfd O;frdj.k] leku L ksr ls izk Ir rjax ksadh ijLij fØ;k }kjk gksrk gSA(3) foorZ u] ,d gh L ksr ls iz kIr rja xksa ds O;frdj.kds dkj.k gksr k gS] tcfd O;frdj.k mlh rjaxkxzizdk'k ds eq M us ls gksrk gSA(4) ,d L ksr ls ijkofrZ r rja xksa ds }kjk foorZ u gksrkgS] tcfd O;frdj.k ,d L ksr ls izkIr rjaxks a dsviorZu ds dkj.k gksrk gSA,d /kkjkokgh pkyd rkj dks ,d v/kZo Ùkkdkj Hkkx vkSjnks yEcs lh/ks Hkkxksa esa fp ds vuqlkj eksM k x;k gAS rcfcUnq O ij pqEcdh; {ks dk eku g&S(1)(2)(3)(4)35.36.m0 i4 prm0 i( p 2)(2)4 prm0 i(p 1)(3)4 pr(4) zero(1)37.r Oi,d foyfxr] vkosf'kr pkyd] ftls fp esa fn[kk;k x;kg]S ds fcUnqvksa A, B, C vkjS D ij fo qr foHko Øe'k%V , V , V vkjS V gS rcABCD D A B C (1) VA VB VC VD (2) VD VC VB VA(3) VD VC VB VA (4) VD VC VB VAYour Target is to secure Good Rank in Pre-Medical 20137/31
MAJOR TEST08–01–2013TARGET : PRE-MEDICAL 2013 (NEET-UG)38.According to the figure shown, a currentcarrying wire and a rectangular loop are placedclose to each other, then-39.BA PCDQA PCDQ(1) rkj] vk;rkdkj yw i dh vksj vkdf"kZr gksxk](2) rectangular loop will repel the wire(2) vk;rkdkj ywi] rkj dks izfrdf"kZr djsxk](3) position of wire PQ will remain unchanged(3) rkj PQ dh fLFkfr vifjofrZr jgsxh(4) wire PQ will be displaced along its length(4) rkj PQ yEckbZ dh fn'kk esa foLFkkfir gks tk;sxk]6WABH–1–340.6WA0RrRDRrC9W(1) 20V41.(2) 5VR0rrG7W(3) –5V(4) –10VERr(2)00RrRrE(3)0HE(4)40V,d leku vkosf'kr R f T;k ds xksys ds fy;s fuEu esa lsdkSulk xzkQ fo qr {ks rhozrk vkSj xksys ds dsUnz ls nwjhds e/; lgh lEcU/k n'kkZrk g&SE(3)F5W(1)08WEE(2)EB10VE(1)E5W20V(1) 20V(2) 5V(3) –5V(4) –10VFor a uniformly charged sphere of radius R whichof following shows a correct graph between theelectric field intensity and the distance from thecentre of sphere -0–1–3G7WE540V5WC9W–6F10VD,d fo qr f}/kzqo] nks cjkcj fdUrq foijhr izÏfr ds2 10 C eku ds vkos'kksa tks ,d nwljs ls 3 lseh- nwjhij gaS] ls cuk gqvk gAS ;g f}/kzqo 2 10 U;wVu izfr dwykWeds fo qr {ks esa j[kk gqvk gAS bl f}/kzqo ij yxus okyscy vk?kw.kZ dk vf/kdre eku gks ldrk g]S(1) 12 10 U;wVu ehVj(2) 12 10 U;wVu ehVj(3) 24 10 U;wVu ehVj(4) 24 10 U;wVu ehVj;gk¡ fp esa ,d fo qr ifjiFk fn[kk;k x;k gSA blds fy;sfoHkokUrj V – V gS&39.C8WE5W20V41.B(1) wire will be attracted towards rectangular loopAn electric dipole consisting of two oppositecharges of 2 10–6 C each separated by a distanceof 3 cm is placed in an electric field of 2 105newton per coulomb. The maximum torque on thedipole will be (1) 12 10–1 newton metre(2) 12 10–3 newton metre(3) 24 10–1 newton metre(4) 24 10–3 newton metreFor the circuit shown in fig VC – VE is-40.fp kuqlkj ,d /kkjkokgh rkj vkSj ,d /kkjkokgh vk;rkdkjywi ,d nwljs ds fudV j[ks gq;s gaS] rc&38.Rr(4)0Use stop, look and go method in reading the question8/31Your Target is to secure Good Rank in Pre-Medical 2013H
MAJOR TEST08–01–2013PRE-MEDICAL : ENTHUSIAST COURSE42.The effective resistance (in W) between B and Cof letter A containing resistances as shown in figis -42.v{kj A dh vkÏfr esa tqM s gq;s izfrjks/kksa dk fcUnqvksa BvkjS C ds e/; izHkkoh izfrjks/k (W ek d esa) gS&AA10 W10 W10 WB(1) 60(2) 40(3) 80/3A parallel plate capacitor with(1) 60(4) 160/943.AK1K2d/2d/2C(2) 40(3) 80/3(4) 160/9,d lekUrj IysV la/kkfj ] ftldhizR;sd IysV dk {ks Qy A vkjS IysVksads e/; nwjh d g]S esa nks ijkoS qr inkFkZdh ifê;k¡ fp ds vuqlkj Hkjh gqbZ gaSAbu ifê;ksa dk ijkoS qrkad Øe'k% KvkjS K gAS bl foU;kl dh izHkkoh /kkfjrkgS&AK1K2d/2d/21K2. The capacitance of this arrangement is2(1) 2Îo A(K1 K2)/d(1) 2Îo A(K1 K2)/d(2) 2Îo AK1K2/(K1 K2)d(3) Îo AK1K2/d(4) mijksDr esa ls dksbZ ugha(2) 2Îo AK1K2/(K1 K2)d(3) Îo AK1K2/d(4) None of the above45.10 WCplate area A and separationdielectric slabs of equalthickness as shown in fig. Thedielectric constants are K1 and44.10 W10 WB43.10 W10 W10 W10 WWhen a piece of aluminium wire of finite lengthis drawn through a series of dies to reduce itsdiameter to half its original value, its resistancewill become(1) two times(2) four times(3) eight times(4) sixteen timesA full-wave rectifier circuit along with the output44.45.is shown in figure. The contribution(s) from thediode 1 is(are) :-tc ,d ,Y;qfefu;e ds rkj dks dbZ lk¡pksa esa ls [khap djblds O;kl dks ewy O;kl dk vk/kk dj fn;k tkrk gS rcigys dh rqyuk esa bldk izfrjks/k gks tkrk g]S(1) nks xquk(2) pkj xquk(3) vkB xquk(4) lksyg xqukiw.kZ rjax fn"Vdkjh rFkk mlls izkIr fuxZr fp
TARGET : PRE-MEDICAL 2013 (NEET-UG)08–01–2013 4/31 Y T G R P-M 2013 H MAJOR TEST 18. What is the power output of 92 Y235 reactor if it takes 30 days to use up 2 kg of fuel and if each fission gives 188 MeV of usable energy - (1) 59 MW (2) 51 10 4 MW (3) 188 MW (4) None of the above 1
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