CONCEPT RECAPITULATION TEST - III [CONCEPT RECAPITULATION TEST - III]
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Bishnu Singh Pau
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- AITS-CRT-III-PCM(S)-JEE(Main)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com1ANSWERS, HINTS & SOLUTIONSCRT –III(Main)Q. No. PHYSICS CHEMISTRY MATHEMATICS1.A B A2.A D D3.C A B4.A A C5.C D A6.C A C7.A A D8.D B C9.B A B10.C A B11.A B C12.D C B13.D D A14.C B B15.C C C16.A D B17.B C B18.B C A19.B A C20.D B A21.A C C22.C D C23.C D D24.A D A25.C C B26.C C C27.A D C28.C D D29.A C B30.C B BALL INDIA TEST SERIESFIITJEEJEE (Main), 2014From Classroom/Integrated School Programs7 in Top 20, 23 in Top 100, 54 in Top 300, 106 in Top 500 All India Ranks & 2314 Studentsfrom Classroom /Integrated School Programs & 3723 Students from All Programs have been Awarded a Rank in JEE (Advanced), 2013
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- AITS-CRT-III-PCM(S)-JEE(Main)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com2PPhhyyssiiccss PART – I1. V = Es ;E =22smqt222s mVqt = 11.375 Volt.2'sin30 sin60 ' 3 ' 'cos30 vcos30 cos60 (v ' ') cos60 (2v) v = v/ =40 34000.1 3 rad/s30120°v3. for x<0It will oscillate like SHMTime period1mTk For x>0It will perform periodic motion under constant force222ET 2mgHence time period1 2T T T 2m 2ET 2kmg 4200mvev BR0mvBeR2 2a R b R 2 2b aR2b02 22bmvB(b a )e
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- AITS-CRT-III-PCM(S)-JEE(Main)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com35. Here the total time is given TTime spent m the region is( 2 ) Hence total time spent2T2 2 v2 v6. Torque of Internal force is always zero. All other quantities will vary.8. Static friction depends on tendency.9.2 22ML 3MLsin 60 212 24 10. When impulse of external force is zero.11. Balance forces.NRmgm2L cosN()12.II13. y(x, t) =2axtbe v = b15. 2L = m (1/2) =m T2f 9L = n (2/2) =n T2f 4 m = 3n = 3 1 = 316. cos = 1 = 0 L = 1/C17. fB= f f0=00 00v v sin30f fv v sin30 =0 002v f2v v
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- AITS-CRT-III-PCM(S)-JEE(Main)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com418.21mv h2 19. = D/d, is minimum for violet.20.PG PC CGˆ ˆV V V 2xj 2i Hence, for the paritcle,Vx= 2m/s and Vy= 2x = 4tax= 0 ay= 4m/s2x = 2t, y = 2t2 aTotal= 4m/s2(a) (along Y-axis)and trajectory 2 2x xy 2.4 2 whenNa a , a cos = a sin = 45ºSo, Vx= Vy= 2 V =2 22CoNV 4 2R 2 2a 4.cos45 m =4m2
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- AITS-CRT-III-PCM(S)-JEE(Main)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com5CChheemmiissttrryy PART – II2.OOHOOHOOHOOHHOOHTautomeriseOOH3.D HDHDHD3D O DDDDDD4. Talc has empirical formula Mg2(Si2O5)2Mg(OH)2which 2n2 5nSi Ounits of two dimensional sheetsilicates.5.OHOH97oIt has half open book like structure having dihedral angle 97o6.o o 1r r 2 4H 2 46.1 187.8 2 285.8 H N H kJ / mol = - 241.0 kJ/mol-1o2 4rH N H 50.6 kJ / mol Decomposition enthalpy = - 50.6 kJ/mol.9. Radius ratio for MgO is = 0.464 which reflect the coordination number 6 and radius ratio for MgSis 0.353. Which reflect the coordination number 4.
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- AITS-CRT-III-PCM(S)-JEE(Main)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com610.2 3 4H O ClO aq ClO aq 2H aq 2e 3 2 22H aq ClO aq 2e ClO aq H O 3 2 42ClO aq ClO aq ClO aq ocellE 0.33 0.36 0.03 ocellRTE lnKnF (at equilibrium)0.0590.03 logK2 logK = - 1K = 0.13 4 22ClO ClO ClOInitially .1 0 0equilibrium .1 2x x x 22xK.1 2x 221 xx 0.019 M10.1 2x 12. Probability of finding an electron is zero, i.e.20 . 23/22 /201 11 8 12 e 0a16 4 21 8 12 0 On solving for we get = 1, 6, 2 and02ra For = 1;00a2r1 ra 2 002r2; 2 r aa 002r6; 6 r 3aa 13. n-factor of H3PO2is 4/3 for the reaction.14.b bT K m i 2.08 = 0.52 × 1 × ii = 4This implies that the salt on dissociation gives 4 ions. Thus, the salt that gives four ions isK3[Fe(CN)6].
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- AITS-CRT-III-PCM(S)-JEE(Main)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com715.DHOCH3HDHO+CH3HOHHCH3D-D shiftOHDCH3HHOHDCH316. O2SeOOOOHBezilic acid rearrangementOHCOO B C18. It is a non-degradable pollutant.19. Colloidal silver bromide solution is used in photography.20. Stability of carbonates increases down the group. Carbonates of alkali metals are more stablethan those of alkali metals.21.2pxand2pyhave one nodal plane each. In 1s there is no nodal plane and in case of*2pxthereare two nodal planes.22. Le-Chatellier’s principle.25.COOH2SOClCOCl3AlCl / OZn/HgHCl26.4NH Cl Nesseler's reagent Brown colour of iodidemillion's base.
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- AITS-CRT-III-PCM(S)-JEE(Main)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com8MMaatthheemmaattiiccss PART – III1. Using T = S1take mid–point (h, k) then equation of PQ isxh + yk – (h2+ k2) = 0 ….. (1)mx – y + 2m = 0 ….. (2)Compare (1) and (2) eliminate m2. As f(x) is continuous decreasing function on (0, ), so 0 < f(8x) < f(6x) < f(4x)So x x xf 6x f 4xlim 1 lim limf 8x f 8x By squeeze play theorem xf 6xlim 1f 8x3. The given lines will be parallel to lines ax2– 6xy + y2= 0So,2y y6 a 0x x m + m2= 6 ….. (1)Which gives m = –3 or 2and mm2= a ….. (2)From (2) a = –27 or 8Hence sum of all possible value of a = –194. 2x4limx2e lng 2 x lng 2 0form0 x 02g' 2lim12g 22g' 2 x2x1g 2 xe e e2xe 5. 1/ n0n2f t dtI lim1n0form0 Using : L’ Hospital and using Leibnitz rule32n n1 1n fn 1nnI lim lim f2 2 n Put1nhhence h 01 f 0 h f 0 1I lim f ' 02 h 2 6. For xy = 1 2dy 1dxx Slope of normal to xy = 1 for x2> 0 (x 0)Slope of line (given) =22log 1 5a a05 So 1 + 5a – a2> 1
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- AITS-CRT-III-PCM(S)-JEE(Main)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com9a2– 5a < 0 hence a (0, 5)7.2a31 1 2 2 2ah2 2 33 3 abc 2 1 1 3 3R43 4 2 2 2 A11MaBC2313138. Area of the parallelogram outside the circle= Area of parallelogram – Area of sector OPQ= 1 1sin 2 sin f2 2 2 QRPOr = 19. Using n n 1 n 2r r 1 r 2n n n 1C C Cr r r 1 is equal to10. We have |(A–1) adj(B–1) adj(2A–1)| =2 21 1 64AB A =6482 1 4 (By using |A–1| =1Aand |adj B| = |B|n–1where n is order of square matrix)11. This distance between the given parallel lines (h) is15length of the side of triangle is =2h3Area of triangle =2 23 4h h 14 33 5 3 12. The volume of tetrahedron =1 0 01 1 1OA OB OC 1 1 06 6 60 1 1 Area of base = 1 1ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆi j i j k i i k2 2 = 1 1222Hence height =3 volume 3 2 1Area of base 62 13. C = 201ab2 S – a = 7S – b = 13a – b = 6 ….. (1)a2+ b2= 400CbA7D13Ba
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- AITS-CRT-III-PCM(S)-JEE(Main)/14FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com10(a – b)2+ 2ab = 400 ab = 1821ab 912 14. Letˆ ˆ ˆr xi yj zk Comparing x = t2+ t – 2 y = t + 2 z = t2+ 6t + 9Hence z – x = 5t + 11 z – x = 5(y – 2) + 11 x + 5y – z + 1 = 016.a b a b 8 2a b 8 222a b sin 8 228b 16112 b 417. If2 22 2x y1a b , a > bSo x1x2= y1y2= b23d = 493d 7(x1y1)(x2y2)y1y218. f(x) = sec x – cosec xf(x) = 0 at x4Alsof " 04 so f(x) is minimum at x4/ 4 / 20 / 4f sec d cosec d4 =2 1ln2 1 =2ln 2 119. a a0 0cos a xln cota tanx dx ln dxsina cosx = a a a0 0 0lncos a x dx lnsinadx lncosxdx I(a) = – ln (sin a) · aI(1) = – ln (sin a) = ln |cosec 1|20. Let1/ xx 1x 01A lim 22 (1form)=xx 1x 0 x 01 1 2 1lim 2 1 limx 2 2xe e =1ln22e 2
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