ONLINE TEST PAPER JEE MAIN - 2016 - Motion IIT JEE
DATE : 10-04-2016ONLINE TEST PAPERJEE MAIN - 2016ANSWER KEY WITH 228.129.330.3394 - Rajeev Gandhi Nagar Kota, Ph. No. 93141-87482, 0744-2209671www. motioniitjee.com , email-info@motioniitjee.com1
MATHEMATICS1.C4.xxy Ax y(t) dt (x 1) xy 1A (2, 3, 5) y(t) dt1xxy 2 (1 – t) y(t)dt1BD–1, 3, 2 – 14 2 ,4, 2 2dy2xy x2 (1 – x)ydxdy dxx2 (1 – 3x)x2dx – 5 2, 10 7, 10dy y(1 – x – 2x)dxlogy – –1 2 ˆ – 2 ˆi (4 – 3) ĵ – 5 kAD 22 1– 3logx logCx1xlogy logx3 logC – –5 –8 AD 2 î ĵ 2 k̂5. yx3 1log C –x D2y1–yx3 e xC–2y1dydv dxdxI.F. e sec xdxI.F. (sec tanx)ACoplanar20412 21 22y2 1 – 06.xsec x tan xA xi 552(4 – 4) 4( 2 – 2) 04 – 2 2 2 – 4 0 0, 4 – 2 2 03.dy y2 secx tanxdxy2 vdv v sec x tanxdxxy Ce3x2.C2 xi2 xi – 12.45 5 A xi2 12.4 255P4x 3y – 10 0P1P2 4x 3y 5/2 0 5 – 1 5 – 2 5 – 6 5 – 5 5 – 11 5 4 3 1 6 P1OA P OB2 –10 45 5122Z xi 5 7.14 2.85DA – I 0A2014.(A2 – 2A – I)394 - Rajeev Gandhi Nagar Kota, Ph. No. 93141-87482, 0744-2209671www. motioniitjee.com , email-info@motioniitjee.com2
8.12.AA10 (r2 r) r – r rr 1(0, 2)S10 r r 1 – (r – 1) rr 11S (0, –2)1 2 – 0 2 3 – 2 . 10 11 – 9 10 10 119.Eqn to hyperbola–x2y2 154C12HGFra b c3a b c4 9 a2 4 4 – 1 a2 5 a b cr 2(a b c / a) 3313.C14.C(1 – 1 2 sin2 x)2x 0 x(2 tan x – tan2x)lim (a b c)r 2lim 4x 010.sin3 xx3D56 2 c5 3 6 1 2 3 6 c5 3 2566 25 26 729729.x3 2x3x3 – 2x – 2x 33 44 –22 86––3 3315.D–1/711.Df(x) sin4x cos4x 3 4x 22(0, 1) 7x – y 1 0 3 x 82f(x) 1 – 2sin2x cos2xf(x) 1 –f1(x) 0 –1(sin2 2x)21 2sin2x cos2x – 22 –sin 4x 4x 2 x 421– 2m 1 /77 m1 (–2)(–7)1–713 / 77m 1ˆ ˆ b) (a9/77–mEquation ot incident ray41(x – 0)3838y – 38 41x 41x – 38y 38 063m 9 91 – 13m76m 82m 41/38y–1 394 - Rajeev Gandhi Nagar Kota, Ph. No. 93141-87482, 0744-2209671www. motioniitjee.com , email-info@motioniitjee.com3
16.Da1 2d a1 6d a1 10d a1 14d 724(a1 8d) 72a1 8d 18S17 20.54 217(2a1 16d)2 3hS17 17 18 306.17.18.AA – 5I 7A–1 07A–1 5I – AA3 – 2A2 – 3A IA3 – 5A2 7A3A2 – 10A I3A2 – 15A 2I 5A – 20I5(A – 4I)3h2 – h2 954)2 22h2 (54)2 2h 5421.a B2x 1 1 2x – 12x 1 1 2x – 1 2 2x – 11 4(2x – 1)1 8x – 4x 5/8A2 9aDa 24 25–164262 – 4693 1642 222 a b – 2ba 2b – 26 – 2 022.A2102 4 – 4(1)(–2)b 2I 2I 2 dx4 (1 dx2I 6I 31– xx) xdx (1 (1 1– x1 xx)1– xdx1 x1– xx)2 x 1 x 2tdtt–2t C–2210A–2 [x ] [14 – x]4b 1 319.[(14 – x)2 ]dx 1– x1 x23.Dz3 l – i a3 – 3a2 3aiz3 (1 – 3a2) i(3a – a3)pure real no3a – a3 0z 1 i 3a 3z 2(cos /3 i sin /3)S S 1(1 – z12 )1– z1 – 2121 –1 – i 3 40953 3i394 - Rajeev Gandhi Nagar Kota, Ph. No. 93141-87482, 0744-2209671www. motioniitjee.com , email-info@motioniitjee.com4
24.27.Ax –y 12x y 33x 4x 4/3y n 2n– 46 11n–2n– 4 4 1 , 3 3 4–1 y3(1, –1)13(n 2) (n 1) (n) (n – 1) 11.6.5.4.3.2.1(n 2) (n 1) (n) (n – 1) 11.10.9.8n 928.Eqn to tangent–1y 1 –1 – 1 / 31– 4 /3y 1 (x – 1) tan 3 – tan 4 2tan 2 121 tan tan34–1 (x – 1)–4 / 3–1 / 3 3 –1 2 2 ( 3 – 1) 4 – 2 32 2 1 3 29.DBtan 2 1t 1 2 t2 ( 2 – 1) sin cos tan t2 2t2 1 1 1 18 3 – r 3 3 –2 30.2r6– –232r3 r 12(2) Coeff. of x–4 1 1 1 18 – r –43 3 3 r 151518m nt2 44t2 4t2t2 4t2 4.B1dy2 2 4x – 3dx34x – 3 94x 12x 3y 1 3 4P(3, 4)Eqn ot Normal3(x – 3)22y – 8 –3x 93x 2y – 17 0Normal passes through Pt (1, 7).y–4 –2r6– –43n 4Am 4m2 –1B(1) Coeff. of x–28 2tt1 t –Sin ( 2 1) cos 26.Dfor min A BA B /12tanA tanA 2tan /124y 4 –x 1x 4y 3 025.C 1 C15 2 18C12 (1 / 2)12 18218C15 (1 / 2)15394 - Rajeev Gandhi Nagar Kota, Ph. No. 93141-87482, 0744-2209671www. motioniitjee.com , email-info@motioniitjee.com5
PHYSICS1.Wrong question6.2At least three reading are required2.1Qremain 5 336 103 1680 103 J7.4NT2QB remain T Tw12 v n v 5 v v n v 5 5v 5 hc– 1 5 10–5 53495 10 10–5 240 53 4472 10–5 45 mv .(2)8.3 11 Energy 13.6 2 2 ev n1 n2 2hc3hc– – 3 hc 1 13.6 1 4 hc2 3 13.6 4 3.4 3 10.2 eVAn 5 10–5 10 10–5 240 153 12000 120 mVwith left side at high voltageP.D between top & bottom BH v l.(1)2hc– K2 3K12 n 1 sin2 4K2 5. 5 10–5 5 5 10–5 T3 P.D across wings B v l340335 n–n 5335340n 170f E between n2 xBH B sin 1K1 xS v frequency received directly from toy n v 5 frequency received after reflection from wall4.xw2731680 103 27w2w 1680 10 1.68 1053.xBHga12 an 0.199. 12 7 0.183y (a b) . cFor y to be 1(a b) should be 1& c should be 1394 - Rajeev Gandhi Nagar Kota, Ph. No. 93141-87482, 0744-2209671www. motioniitjee.com , email-info@motioniitjee.com6
10.4The V-I graph of i/p characterstic is13.1 2 G 402400 Rg .(1)2 G 204900 Rg.(2)I4900 Rg40 22400 Rg20V4900 Rg 4800 2Rg100 Rgfrom eqn. (1)This means that increasing the voltagedecreases the resistance almost to zero. Thismeans that increases 1/r to Hence. correct graph is11.12.3Redistribution will stop whenqvB qEE vBx E 2 2 000 x vB 0 6 0vBx 1 0vB 2 – 0vBB F G G 0.02 10–3G 2 10–5 20 uA/div ful scape deflection current is 20 A/div 50 1000 AP 1mAHence, 2 is incorrectV– x––– x–––x2 2 10–5 9800 100 14.BA VVB(2 b) (a G)2 a2 )b a2bF 2 abB(a2 a2 2a a a2 )a22 10100 1000 10.19900 299002Image due to reflectionUf 4 0.520 U f 4 51V 20 cmImage due to refractiondapp 2 abB.2a aa2F (4 bB a)Freq. R F (4 bB a) (4 bB) G21 2500 4050 10Hence, correct option is only 1 is correctFVA VF 2 G 402500 15.dRe al nRnI300 1 20 cm1.52The ratio of r0 to ri is very high for transistorsalmost in the range of 100 to 1000394 - Rajeev Gandhi Nagar Kota, Ph. No. 93141-87482, 0744-2209671www. motioniitjee.com , email-info@motioniitjee.com7
16.418.119.1,2yDimensionally B 1D& AD C(0,0)x co-ordinal of AB –x/4x(i)l xl xX co-ordinal of BC 2 22Centre of mass should below A 80xcm 0let lines mass density so mAB x(ii)2(iv)20.17.2S1S2S2a2 4 4(2)( 1)2 12 2 24d a A sin 1.22 1.22 A 4Amplitude modulated O/P is given byC(m) VC cos (Wet) Ad2 3.53.5 1.3745mv ccos (wc w ) t2mv ccos (wC – W ) t2comparing with the given equationmv c 10210 22 303wc 300 2 fc 300 fc 150 Hzwc w 400 w 400 – 300 100 2 fm 100 fm 50 Hz.(1).(2)1.22 600 10 91.22 A30 10 2 24.4 10–7 2.44 10–6 d 10 9.46 1015 2.44 10–6 23.08 10102.308 1011 m108 km m AC–C – C (meaningful)BBDS1 l l2 – 2 – 1 0x x 2n2 – 2n – 1 0 2 C C (iii) A2 – (BC)2 – (meaningful)D D 2n CCCAD2AD2– – –DBDBDBDCAD(meaningful) x l x x l 4 2 0xm x t x2l2 lx 0422l2 – 2xl – x2 0ACAADA– – – A (meaningful)DDDDD21.4In case of EM wave, the di recti on ofoscillation of electric field and magnetic fieldi s pe rp endi cul ar to t he d i rec ti on ofpropagation. Options A and B are incorrect. Also E B gives the direction of propogation( ĵ k̂) ( ĵ k̂) î î 2 îˆ î i 0ˆ k)( ĵ k̂) (jHence correct option is 4394 - Rajeev Gandhi Nagar Kota, Ph. No. 93141-87482, 0744-2209671www. motioniitjee.com , email-info@motioniitjee.com8
22.27.4v n2 Rt2rv nRta dv nRdtFt mnRP FV (MnR) nRt mn2R2 t23.s Ut –1 5 422 100 – 10 90 w F.d Ma.d 10 5 90 –4500 Jrgrw rgw gr2 n 28.3 RT Mw f29.gr24.Wrong question25.24 8 9Ceq. 8 9Induced current A dB r 2 R dtrC Heat generated 2 30.4F e–2t/T RdtRvt2Rmdv 2 vtdt 2t 2r 4B20 e T 0 2TR26.32232 I Rdt0 r 2B0 TR0 8 C3 18 C3 32C 132 9C32 C 32 9C23 C 32ddE1 r 2B0e t / T r 2B0 –t/T eTRTR 1at22s 50 2 –4v 1dv 50 – 5m/s2dt10F ma 10 5 50 Newtondistance travelled in two seconds is22 2r 4B20 [–e– e–0]2TRmdvR 2 dtvfm ln v – 2r 4B20 2 TR3ln v Rt1f vis constant rwhich means that electric field inside thesphere is constant. This is possible only whenThe given situation tells that is function of1r394 - Rajeev Gandhi Nagar Kota, Ph. No. 93141-87482, 0744-2209671www. motioniitjee.com , email-info@motioniitjee.com9
CHEMISTRY1.(3)2.(3)Rate constant does not depned onconcentration of reactans3.(1)4.(1)Rhumann's purple is a confirmatory test ofprotein. Its also know as ninhydrin test5.(4)6.(4)7.(2) 2 MX2 eq. Maq 2xaq3n 312 1 (n – 1) 2 1 (3 – 1) y 1/2 0.5 23.(3)LiH B2H62 Li[BH4]24.(4)N2 2NTaken4rms 4 3RT283R 2T1 24rms (New) 25.12.(4)13.(1)14.(3)15.(4)lone pair of nitrogen at ninth position'sinvolved in resonance i.e. not available fordonation16.yq26.(4)18.(4)Work(4)(4)H2A OH(–)0.1N 0.04(5 bar) –Pekt (V2 – V1) 27.0.1 v 0.041000V 400 ml(2)28.(1)29.(3)(CH2)2 (CO2H2)(–)OOC —CH2 —CH2 100(–) CH2 CH2 (g) 2 CO2(q) 2e(–)0.3 mol 22.4 6.72 lit.30.(3)kp 25 Ans17. 2 Clit is a free radicalr 4(3)XY(s) xa (5 bar) 14 2 –Pext. Vo2 no2RT –1 bar 1bar – 50 8.3 300 –124.5 Ans.394 - Rajeev Gandhi Nagar Kota, Ph. No. 93141-87482, 0744-2209671www. motioniitjee.com , email-info@motioniitjee.com10
394 - Rajeev Gandhi Nagar Kota, Ph. No. 93141-87482, 0744-2209671 www. motioniitjee.com , email-info@motioniitjee
8. JEE (MAIN) 2015 Candidates who wish to appear in JEE (Advanced) 2015 must write Paper-1 of JEE (Main) 2015 which is likely to be held in the month of April 2015. Candidates may visit the website www.jeemain.nic.in for information about JEE (Main) 2015. 9. ELIGIBILITY CRITERIA FOR APPEARING IN JEE (ADVANCED) 2015
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