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8/6/2019 _ATIIT2011T13P2_4_4_3
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Test - 13 (Paper - II) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2011
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TEST - 13 (Paper - II)
ANSWERS
CHEMISTRY MATHEMATICS PHYSICS
39. (A)
40. (A)
41. (A)
42. (D)
43. (A)
44. (A)
45. (8)
46. (1)
47. (3)
48. (4)
49. (9)
50. (B)
51. (B)
52. (C)
53. (C)
54. (D)
55. (A)
56. A (t)
B (s)
C (q)
D (p)
57. A (q, t)
B (r, t)
C (s, t)
D (p)
20. (B)
21. (C)
22. (A)
23. (B)
24. (B)
25. (B)
26. (1)
27. (9)
28. (0)
29. (5)
30. (3)
31. (B)
32. (A)
33. (D)
34. (D)
35. (C)
36. (B)
37. A (s, t)
B (q)
C (p, q, r, s)
D (q, r, s, t)
38. A (t)
B (p, q, r, s, t)
C (p)
D (q, r, s, t)
1. (A)
2. (B)
3. (C)
4. (B)
5. (D)
6. (A)
7. (5)
8. (3)
9. (8)
10. (7)
11. (4)
12. (C)
13. (B)
14. (D)
15. (A)
16. (B)
17. (D)
18. A (p, q, t)
B (p, s, t)C (p, q, r, t)
D (r, t)
19. A (q, r)
B (p, s)
C (q, r)
D (p, s, t)
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All India Aakash Test Series for IIT-JEE 2011 Test - 13 (Paper - II) (Answers & Hints)
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PART - I (CHEMISTRY)
ANSWERS & HINTS
1. Answer (A)
G H T S =
cell cellP
nFE H T.nF ET
=
cell cell
P
HE T E
nF T
= +
For Ecell
versus T,
cellP
Slope ET
= =
Temperature coefficient
Intercept = HnF
2. Answer (B)
CH CNO + N OH3 2 CH CNO + H O CH CNO3 2 2 3 2
H H
(A)
O NOH(Nitrous acid)
(Nitrolic acid)(P)
NONa
(Red solution)
NaOH
RCNO + N = O2 RCNO + H O2 2
(B)
R
(Pseudo-nitrole)
H HO
R
N=O
Gives blue colour with aq. NaOH
RCNO2 No reaction
(C)
R (Due to absence of H in C)
RHNO2
3. Answer (C)
2
213Cl 3 mole
71= =
NH3
= 8 mole and 1 mole
8NH3
+ 3Cl2
N
2+ 6NH
4Cl
NH3
+ 3Cl2NCl3 + 3HCl
N2
evolved = 28 g
NH4Cl formed = 321 g
NCl3
formed = 120.5 g
HCl formed = 109.5 g
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Test - 13 (Paper - II) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2011
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4. Answer (B)
Aldol condensation and cross-aldol condensation products are formed. ( unsaturated carbonyl compounds)
5. Answer (D)
rms(I)
3R(t 273)v
M
+= , mp(I)
2R(t 273)v
M
+=
rms(II)
3R 546v
2M
= , mp(II)
2R 546v
2M
=
rms(III)
3R(2t 546)v
(M / 2)
+= , mp(II)
2R(2t 546)v
(M / 2)
+=
6. Answer (A)
(I)
H
Me
H
BrH
H
Anti arrangement(Favourable)
Syn arrangement(Not favourable)
(II) Major product is H
OH
Me
H
(III) Walden inversion
7. Answer (5)
Brown ring complex = [Fe(H2O)
5NO].SO
4
8. Answer (3)
Adiabatic slope(Poisson's ratio)
Isothermal slope=
2 2He O CO
5 7 4 28 1x y z 3
3 5 3 9 9 = = = = +
9. Answer (8)
4 alcohols and 3 ethers
Butan-2-ol has both d- and l- form
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All India Aakash Test Series for IIT-JEE 2011 Test - 13 (Paper - II) (Answers & Hints)
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10. Answer (7)
Phenolphthalein end point:
NaOH + HCl NaCl + H2O
NaCl + HCl No reaction
NaHCO3
+ HCl No reaction
Na2CO
3+ HCl NaHCO3 + NaCl
x = 1 + 4 = 5 moles
Methyl orange end-point:
NaHCO3
+ HCl NaCl + H2CO3
y = 3 + 4 = 7 moles
xy 7 57
5 5
= =
11. Answer (4)
Anti fluorite structure Na2O
Na+ at all tetrahedral voids. O 2in CCP.
1st C.N. of O2at a corner = 8 = x
(All 8 Na+ in tetrahedral voids in eight cubes sharing a corner)
2nd C.N. of O2at a corner = 12 = y
(12 O2, each three O2in a cube at adjacent face centres sharing a corner)
y x = 12 8 = 4
12. Answer (C)
13. Answer (B)
14. Answer (D)
Solution of Q. 12 to Q.14
NiCl2
+ 2DMG H C C = N3
H C C = N3Ni
N = C CH3
N = C CH3
O -----------H O
O H -----------O
(Cherry red ppt.)
Ni2+ (3d8) Square planar complex due to strong ligand field
dsp2
No unpaired electrons. So, = 0 B.M.
Two chelate rings are present, each are five membered
15. Answer (A)
The order of reactivity can be explained in terms of
(i) Leaving group ability of X
(ii) Electron donating resonance effect ofX
Order of leaving group of X:
Cl > R Cl > R O > R NH() ()
O
O()
()
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Test - 13 (Paper - II) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2011
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16. Answer (B)
RC
O
OHRCOH RCOH RCO
OH
R O 18..
(+)
H
..
H
OH
..
O18
RH +
:OH
18OR
(+)
H
HRCO +
18H O2
OH
..
(+)
H
RCO R18
O
R
17. Answer (D)
COOH
COOH
P O2 5O
O
O
Dry NH3NH
O
O
0C
Br , NaOH2NBr
O
O
(Succinicacid)
(Succinicanhydride)
(Succinimide) (N-bromosuccinimideor NBS)
18. Answer A(p, q, t), B(p, s, t), C(p, q, r, t), D(r, t)
A.Vapour
pressure
Temperature (T)
Tb
SolutionSolvent
B.Stopping
potential
(V
)S
Wavelength ( )
0s
0
hc hcV
e e=
C.
7
Vol. of base (ml)
(titration of weak acidand weak base)
pH
D.
Rate ofbackwardreaction
Time (t)
19. Answer A(q, r), B(p, s), C(q, r), D(p, s, t)
A. All have planar shape.
All have 2 lone pairs.
B. ( 1)2
= +l
hl l
Total nodes = n 1
C. Sideways destructive overlapping between porbitals (in one plane)
2 nodes+
+
D. Single electron species.
2
2
zE
n
zVn
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All India Aakash Test Series for IIT-JEE 2011 Test - 13 (Paper - II) (Answers & Hints)
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20. Answer (B)
After simplifying k= 1, n= 2
21. Answer (C)
P1
O
P2
z1
z2
By rotation it can be easily seen z2
= iz1
(i)
and if we interchange z1
and z2
then we have z1
= iz2
(ii)
Now, clearly |z1
z2| = |z
1+ z
2| for (i) and (ii)
1 2 1 2 0z z z z + = for (i) and (ii)
2
1
Re 0z
z
=
22. Answer (A)
2
161024[ln ] 6[ ] 2011
1
x xx ex
+ = ++
So, for a solution to exist 216
1
x
x+must be an integer.
Now, 22
1 11
x
x
+
So,2
2
1
x
x+may take values 1, 0, 1 as integral values.
But in all cases R.H.S. is ODD while LHS is divisible by 2.
Hence the equation has no solution.
23. Answer (B)
As the letter boxes are identical so we will be only concerned with the number of letters.
The distribution can be as follows
5
4
3
3
2
2
1
0
1
2
1
2
1
1
0
0
0
1
1
1
1
0
0
0
0
0
1
1
0
0
0
0
0
0
1
54
53
5 23 1
5 32 2
52
1 1
5
10
1( . ) 10
2
1( ) 15
2!
10
1 1
C
C
C C
C C
C
=
=
=
=
=
=
=
Ways of distribution Number of ways
So, total number of ways = 1 + 5 + 10 + 10 + 15 + 10 + 1 = 52
PART - II (MATHEMATICS)
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Test - 13 (Paper - II) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2011
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24. Answer (B)
62
, 1
( )i ji ji j
P x x=