Solns. Jee Mains'13

Vidyamandir Classes

ANSWERS - JEE MAINS Test - 1/IITJEE-2014
CHEMISTRY
1 D 11 D 21 C 2 D 12 C 22 B 3 D 13 B 23 D 4 C 14 C 24 B 5 C 15 B 25 B 6 B 16 C 26 B 7 A 17 C 27 B 8 B 18 D 28 A 9 D 19 D 29 B 10 A 20 C 30 D

PHYSICS
31
C

32
C

33
A

34
B

35
CANCEL

36
A

37
C

38
B

39
C

40
C

41
A

42
C

43
D

44
C

45
C

46
C

47
C

48
C

49
C

50
D

51
A

52
D

53
C

54
A

55
C

56
B

57
B

58
D

59
D

60
B

MATHEMATICS
61 D 71 C 81 C 62 B 72 B 82 A 63 C 73 C 83 C 64 B 74 B 84 C or D 65 D 75 B 85 D 66 C 76 D 86 D 67 B 77 A 87 D 68 D 78 B 88 B 69 D 79 C 89 C 70 A 80 D 90 A

VMC/JEE-2014

1

JEE MAINS Test - 1/Solutions

Vidyamandir Classes

Solutions to JEE MAINS Test - 1/IITJEE - 2014
[CHEMISTRY]
1.(D) 2.(D) 3.(D) C 2  C3 bond is of the type sp 2  sp3 . 1-butyne is terminal alkyne. It will give ppt with ammoniacal Cu2Cl2 solution. CH3 NH 2  CHCl3  3KOH  CH 3 NC  3KCl  3H2 O 
Carbylamine test

4.(C)

OH | Tert. Pentyl bromide  CH3  C  CH  CH3  CH3  C  CH 2  CH3   E2 | | CH3 CH3
KCN (alc.)

H / H2 O

5.(C) 6.(B)

Cu+2 ions will get deposited and Zn+2 ions come into the solution but SO 2  will remain the same. 4 Tf  iK f m 10 3.72  i  1.86  74.5 100 1000 3.72  74.5 i  1.5 1.86  100  3.72  i  1.86  100 74.5

7.(A)

Calculate using PV = nRT T1  273K T2  546 K and 3R 2 T3  273K 3R 2  273  3.4 kJ

8.(B) 9.(D)

E  nCV T  1 
 Θ

  T2  T1  

N2O ONN CO2 O=C=O Both these species are isoelectronic as well as isostructural. N3H is isoelectronic but not isostructural with these two. 1 BrO 3 t 
 BrO3

10.(A) As we know 

t



 BrO3

t



1 3

 0.056  0.019 L mol1 s1

11.(D) BF3 being symmetric has net dipole = 0 but not in NF3 12.(C) The rate of effusion depends on area, moles and density as well. 13.(B) For a spontaneous process Stotal  0 (IIIrd law)