29- Solution Report (29)

Paper Code : 1001CT103316001 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE (PHASE : III) ANSWER KEY TEST DATE : 03-07-2016 Test Type : MINOR PART-1 : PHYSICS Test Pattern : JEE-Advanced Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. D D B D B B C,D A,B D or A,D A,D Q. 11 12 13 14 A. A D D C SECTION-IV Q. 1 2 3 4 5 6 A. 4 1 6 5 5 5 PART-2 : CHEMISTRY Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. D A B C D B B,C A,B,D B,D A,B,C,D Q. 11 12 13 14 A. C B A C SECTION-IV Q. 1 2 3 4 5 6 A. 2 8 3 0 2 0 PART-3 : MATHEMATICS Q. 1 2 3 4 5 6 7 8 9 10 SECTION-I A. C A A C B B A,B,C A,B,C A,B,D A,C,D Q. 11 12 13 14 A. B D B B Q. 1 2 3 4 5 6 SECTION-IV A. 7 1 4 4 1 5

Paper Code : 1001CT103316001 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) : LEADER COURSE Test Type : MINOR PHASE : III Test Pattern : JEE-Advanced TEST DATE : 03 - 07 - 2016 PART-1 : PHYSICS SOLUTION SECTION-I 5. Ans. (B) 1. Ans. (D) Sol. Heat required to melt 10 gm ice at hc Sol. KEmax =    eV 0°C = 10 × 80 = 800 cal  800 cal heat have to be absorbed from m gm water at 50°C  12400eV  4.5eV  e(2V) Then, 800 = m × 1 × 50, m = 16 gm 200 10 m = 16 gm 2. Ans. (D) = 6.2 – 4.5 + 2 KEmax = 3.7 eV 6. Ans. (B)  dQ    dQ  Sol. P  mv  h  dt B  dt C  Sol. Momentum is same for neutron & -particle to C to A  m > mneutron kA 2T  TC   kA  TC  T  v > vneutron  – particle L L 7. Ans. (C,D)  TC  3T Sol. d’ = d(1+T) = +ive (coefficient of linear 2 expansion) 3. Ans. (B) r’ = r(1+T) = +ive (coefficient of linear expansion) W 8. Ans. (A,B) Sol. Breaking stress = Sol. Intensity = Nh A After joining in parallel We can double the intensity by taking twice Area = 2A no. of photons or by increasing the frequency to twice its initial value.  Breaking stress = W'  W 2A A 9. Ans. (D or A,D) Sol. Since area is same, so spectral emissive  W' = 2W 4. Ans. (D) power must be same. Sol. At steady state  dE    dE    d S1  d S2  P  dQ  eA T4  04 dt  No. of photons is segment S2 > S1 for same  T = 2100 K area Corporate Office :  CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 HS-1/5 +91-744-5156100 [email protected] www.allen.ac.in

Target : JEE (Main + Advanced) 2017/03-07-2016 10. Ans. (A,D) Sol. hC  1  eV0  1eV Sol. Let thermal conducitivity of A, B & C be 2k,  3k & 4k respectively hC At steady state  2k.A 100    3kA    '  4kA ' 30  2  eV0  3eV 0.40 0.30 0.20 2hC    1  2  4eV On solving  = 60°C ' = 40°C  2hC  6.2eV  11. Ans. (A)    2 12400 Å Sol. m1 = 1260 mm 6.2 m1 = 1260 × 10–4, T1 = 2300 K   = 4000 Å = 400 nm E  4, E1  T1  4 E2  T2     1/4 1/4   4  E1  T1  E2  100  E2   T2 , T2  T1  E1     2. Ans. 1   E2  1/4 I 1 2Reh  E1  2 T2  T1    Sol. F   I.R.h  . IRh C CC  500 1/4 1  = 1 3. Ans. 6  Sol. Let volume of mercury be x cm3 T2 = 2300 × 8000  = 2300 × 2 = 1150 k According to question T2 = 1150 k 12. Ans. (D) (34 + x) (1 + 3vessel T) – x(1 + HgT) = 34 On solving, x = 6  T1m1   T2  4. Ans. 5   Sol. T1m1  T2m2 ; m2  Sol. We have m2  2300 1260 109 = 2520 × 10–9 M mu  LuAudu  4  2 m LAd 6 3 1150 Ratio of maximum stress m2 = 2520 nm  (m u  m )g   A  13. Ans. (D) =  u    mu  1  A 51 5 Sol. The value of Young’s modulus for a material  m  Au   mg    depends only on the type of material under   32 6 consideration.] 14. Ans. (C) Sol. The value of the constant k A , where A is (u )max  5  ( )max 6 5. Ans. 5 the cross-sectional area, and l the length of Sol.   F the material. A SECTION-IV so Fbrass = 3 Fsteel 1. Ans. 4 HS-2/5 1001CT100336001

so Fbrass x = Fsteel (2 – x) Leader Course/Phase-III/03-07-2016 x = 0.5 m  kice. .t  x2 6. Ans. 5 L Sol. dQ  k ice A   A  dx  L dt x  dt  x t [t is the time to form ice of thickness x from –°C starting] ice x  2 4 0°C dx water 3 t t = 9 days  Extra time = 9 – 4 = 5 days   kice t dt  x xdx L 0 0 PART–2 : CHEMISTRY SOLUTION SECTION -I 12. Ans.(B) 1. Ans.(D) 13. Ans. (A) M1V1 + M2V2 = M3V3 2. Ans.(A) 14. Ans. (C) n1 = nNH3 × R1 × R2 × R3 × Y1 × Y2 × Y3 SECTION -IV 630  600  4  2  2  20  50  x 1. Ans. 2 63 4 2 3 100 100 100 2. Ans. 8 xy y CxH4 +  4  O2  xCO2 + 2 H2O 3. Ans.(B) M = V.S  44.8  4M 12x + y = 30 88  2 11.2 11.2 44 y=6 x=2 4. Ans. (C) 3. Ans. 3 5. Ans. (D) 6. Ans. (B) nA nA  MB  n n 7. Ans.(B, C) nB  MA  8. Ans.(A,B,D) n A  8 1 16   23 = 2n 9. Ans. (B, D) B  1  1  4  10. Ans. (A, B, C, D) 11. Ans.(C)  n=3 4. Ans. 0 5. Ans. 29 [OMR Ans. 2] 6. Ans. 0 PB = XB × PT  3 15  9atm 5 1001CT100336001 HS-3/5

Target : JEE (Main + Advanced) 2017/03-07-2016 PART-3 : MATHEMATICS SOLUTION SECTION-I  3  sin C  C  3 1. Ans. (C)   2 2  sin 2 16 0 x 0  sin2 C  3 sin C  3  0 ƒ(x) = I + B, B  0 0 0 2 2 2 16 0 0 0  C 3 2  0  sin C  3 (ƒ(x))n = (I + B)n  sin 2 4  2 4  = In + nC1In–1B + nC2B2In–2 .... + Bn = I + nC1B as B2 = 0  cosC  1  3  5 88 1 nx 0 = 0 1 0 5. Ans. (B) tan( +) – tan – tan 0 0 1 = tan{1 + tan tan( +)} – tan tan + tan – tan = tan 2. Ans. (A) 6. Ans. (B) C1  C1  C3 ,C2  C2  C3 1 0 2  cos3A + cos3B – cos(3A + 3B) =1  0 1   2 sin3A sin3B = 1 + cos3A cos3B – cos3A – cos3B 1        1     1     1 0 2  1  cos3A .1  cos3B  1 1 2 sin 3A sin 3B  1    2 0  tan 3A tan 3B 1 2 2    1 = (1 +  + )2[1 +  –  + 2] = (1 +  + )3  3A 3 B  90 = 125 2 2 3. Ans. (A) adjA = adjB  |A|n–1 = |B|n–1 A + B = 60º  C = 120º |A||B| if n is even 1 12  22  C2 C 7  |A| |B| if n is odd  If n is even adjA = adjB 2 2.1.2  |A|A–1 = |B|B–1  A–1 = B–1 = A = B 7. Ans. (A,B,C) A If n is odd adjA = adjB AM = 17  |A|A–1 = |B|B–1  A–1 = ±B–1  A = B or A = –B CM = DM = 12 17 4. Ans. (C) 642 = 12R2S  8 = 3RS ACD = 1 .24.5  60 13 5 2  8r = 3R sin–1 5 D 17 32 sin A sin B sin C  3R 222 B CM r    60  12 S 25 5 8. Ans. (A,B,C) by x = 2 and x = –2 all three rows become identical therefore (x – 2)2, (x + 2)2 will be factor of ƒ(x)   cos A  B  cos A  B  sin C  3 Also by C1  C1 + C2 + C3 we obtain x2 + 8 as  2 2  2 16 a factor. HS-4/5 1001CT100336001

Leader Course/Phase-III/03-07-2016 9. Ans. (A,B,D) 13. Ans. (B) Coefficient of 14. Ans. (B) (1) x17  x2.x5.x10  1 a 2 1 (2) x16  x 2x 4 x10  5  6 2 1 1 0   a = 4  21 x x5 x10  1  1 3 b b3  1 x5 x10  1   (a, b)  (1, 4), (3, 18)  I case for (a, b) = (1, 4)  1  0 (3) x15  x x4 x10 5  21 x 2x 3 x10 no solution   10 II case for (a, b) = (3, 18) 1 = 2 = 3 = 0 infinite solutions. x2 x5 x8  5  SECTION – IV (4) x  1x..x1..11516 10. Ans. (A,C,D) 1. Ans. 7 A2 = 0  C = A Paragraph for Question 11 & 12 CB   24 34  36 51 11. Ans. (B) 2. Ans. 1 cos C  a2  62  52  9  a  4 2.a.6 16 7 6 b 1 3 c 0  0 0 C 4 7  2  c  0 0 1  0   b = –2, c = 5 3. Ans. 4 64 4. Ans. 4 40C25 36C0 – 40C24 36C1 + ...... – 40C0 36C25 is A5B coeff. of x25 in the product of –(1 – x)40(1 + x)36 or – (1 – x2)36(1 – x)4  cos C  1 1  cosC  5  = – 36 C124C1 36 C114C3   4 C36  C36 12 11 22 42   = 4 CM  2.6.4 . 5  3 2 5. Ans. 1 64 4 2 |A1 | 27  A   1 12. Ans. (D) 27   15 . 5 . 7 . 3  15 7 |adj(3A)| = |9adjA| = 93|A|2 2 222 4  93 1 1 15 7 272 4 r    15  7 6. Ans. 5 S 2 2  1 22  3   12  2 3 R  abc  4.5.6  8 2  2   2 4 15 7 7 LA  1 50  72 16  1 106 4  3    5  3 2 3 2 2 3 26 /3 n=5 LB 1 50  32  36  1 46 1  2 2 1001CT100336001 HS-5/5