T.I.M.E. JEE Advanced 2013 Solution Paper1

1. MODEL SOLUTIONS TO JEE ADVANCED 2013
Paper I – Code 0
PART I
1 2 3 4 5 6 7 8 9 10
D A D A B B C C A B
11 12 13 14 15
B, D A, C B, C A, D B, D
16 17 18 19 20
5 5 1 4 8
Section I
1.
x = a cos
y = a sin
dx = a sin d
dy = a cos d
 





2/
0
33
a
dcosasina
a
dsinacosa
= 0
2. 9
kA2
3


-----(1)
kA3

= t -----(2)
t = 2 s
3. RTnVp
7
4
1





3
4
n
n
,RTnVp
7
3
2
1
2 





n1 =
A
2
2
A
1
mN3
M
n,
mN2
M

2
3
M
M
n
n
2
1
2
1
 ,
9
8
M
M
V
M
:
V
M
d:d,
9
8
M
M
2
121
21
2
1

4.
Energy conservation
(i)  22
0
2
0 cosmu
2
1
mghmu
2
1
∴  22
0
2
0 cosmu
2
1
mghmu
2
1
Vertical velocity of second particle is also
u0cos
tan =


cosu
cosu
0
0
  = 45
5. pt =

nc
Momentum, p =
c
pth


= 8
93
103
101001030

 
= 1  1017
kg m/s
(a, 0)
(0, a)

h
u0
45 u0 cos

2. 6.  =
2
cos2
0


22
1
2
cos



S = odd multiple of
4

=  
4
1n2


7.
R
5.0
24
1
8
1
f
1



R = 3 m
8.
 =
AY
F
, F same
2
2
R4
R
2
2
1
2










9.
Ans. (A)
10. Each MSD = 0.05 cm
50 VSD = 2.45 cm = 49 MSD
∴ Least count = 05.0
50
1
MSD
50
1

= 0.001 cm
Reading = 5.10 + (24  0.001)
= 5.124 cm
Section II
11. S1 closed, Q1 = 2CV0 
S1 open, S2 closed, Q1 = CV0, Q2 = CV0
S2 open, S3 closed, Q2 = CV0 
12.
S = 4  102
=
6
R

R =

 2
1024
, L = R sin30, R = 2L
R =
QB
M4
, B =
Q3
M50
1024.Q
M4
QR
M4
2




 
13. Number of nodes = 6 A is wrong
4 = 2A sinkx cost
k = 62.8 ,  = m01.0
k
2


L =
2
5
= 0.25 m ∴ B correct
2 A = 0.01 m, C is correct
 = 628,  =


2
= 100 Hz
Fundamental frequency =
5
100
= 20 Hz
D is wrong
14. Vg + V3g = V’2g
V’ = 2 V  just fully immersed
∴ kx = V2g  Vg
x =
k
gR
3
4
k
gV 3



15.
E = 0EE 21 
E1 =
  4
R
3
1
R2
R
3
4
4
1 1
0
2
1
3
0





E2 =
 
 
R
3
1
R2
RR2
3
4
4
1
2
0
3
2
3
0





n=
2
3
3
1
u = 24 v = 8
2, 2R , R F same
R, 
R, 3
Vg
V2g
R
E = 0
E1
E2
21
Y
X
30



  jˆ3iˆ
2
1




  jˆ3iˆ
2
1
60
30 RY
R
+Q O
S
L
XL
jˆ2
iˆ32
30
iˆ4

3. 1 = 42 ∴ Option D correct.
E1 =
  4
R
3
1
R2
R
3
4
4
1 1
0
2
1
3
0





E2 =
 
  25
R8
3
1
R5
R2
3
4
4
1 2
0
2
2
3
0





,
25
32
25
8
4 2
121





Section III
16.
∴ 21 g5g  
5
2
1



17. P  t = 2
mv
2
1
0.5  5 = 2
v2.0
2
1

v = 5
18. h = W + VS
VS = h  W
= mx + C
∴ Slope = h
∴ Ans. 1
19. Required:
 
0
0
N
NN 
= t
0
e1
N
N
1 

 = 3
2/1 102
1
1386
693.0
t
2n



t = 3
102
80

= 0.04
∴ 1  et
= 04.0
e
1
1 ;
Expression e
x
= x1...
2
x
x1
2

∴ e
0.04
 1.04
∴
04.1
1
1
e
1
1 04.0
  1  0.96 = 0.04
∴ Ans: 4%
20.
'mR2mR2
2
MR
2
MR 22
22










4
2
101650
2
mR 22




MR
2
= 6.25  4  102
= 0.25
(4  10) = (4 + 1)’  ’ = 8
1
2
m
1g5 
1g
1g

4. PART II
21 22 23 24 25 26 27 28 29 30
B A A C D B B D B D
31 32 33 34 35
A A B, D A, B, C, D B, C, D
36 37 38 39 40
5 8 2 4 6
Section I
21. Q = [V(H2O)6]
2+
 d
3
3 unpaired electrons  = 3.87 BM
R = [Fe(H2O)6]
2+
 d
6
High spin complex 4 unpaired electrons
 = 4.90 BM
P = [FeF6]
3
 d
5
High spin complex 5 unpaired electrons
 = 5.92 BM
22. 
A
r = 0.414 
x
r
= 0.414  250
= 104 pm
23. These metals occur as Ag2S silver glance or
argentite
Cu2S, Chalcocite
Cu2S.Fe2S3 copper pyrites
PbS  galena
24. C6H12O6 + 6O2  6CO2 + 6H2O
H = 6   400 + 6   300 + 1300
= 2900 kJ mol1
=
180
2900
kJ g1
=  16.11 kJ g1
25. In ammoniacal (and neutral) medium Zn(II) is
precipitated as ZnS (white ppt)
Note : Fe
3+
is reduced to Fe
2+
by H2S and the
latter is precipitated as FeS
26. Adsorption is accompanied by decrease in
enthalpy
27. The rate of SN2 reaction is mainly decided by
steric crowding in the transition state
28. Order of the reaction with respect to P is one
(t75% = 2  t50%)
Order of the reaction with respect to Q is zero







k
Q
k
Q
t 0

dt
dx
= k[P]
1
[Q]
0
29. HNO3 decomposes into NO2 on standing in
presence of light
4HNO3  
h
4NO2 + 2H2O + O2
30. Carbolic acid (phenol) (pKa = 10) is weaker than
carbonic acid (pKa = 6.38)
Section II
31. Rate = k[acid] [ester]
1



)HX(H
)HA(H
C
C
2Rate
1Rate
1
C
100
1 

 =
100
1
Ka = C
2
= 104
32. Hyperconjugation involves
  p(empty) and    conjugation
33. (B) cis trans  cis trans
(C) no isomerism  cis trans
(CoBr2Cl2 is tetrahedral)
(D) ionisation  ionisation

5. 34. (P) involves cyclopropenyl cation, (Q) contains
cyclopentadienyl anion (R) is 2,5-dimethyl pyrrole
and (S) is hydroxytropylium chloride.
35. Dissolution of naphthalene in benzene is
accompanied by an increase in entropy.
i.e., Ssystem = positive
Since the solution is ideal,
H = 0 and Ssurroundings = 0
Section III
36.  
mv
1
HeHe
NeNe
Ne
He
v.m
v.m



But v 
M
T
Ne
He
He
Ne
He
Ne
M
M
T
T
v
v

=
20
4
200
1000

= 1
=
4
20
Ne
He



M = 5
37.
Each nitrogen forms 4 NCoO angles with four
oxygens
38. Structure of (P) is
HOOC
HOOC
O
O
39. Tetrapeptide satisfying the given conditions are
Val  Phe  Gly  Ala
Phe  Val  Gly  Ala
Phe  Gly  Val  Ala
Val  Gly  Phe  Ala
40.
N
H2N
N
NH2
N
NH2
Co
O
O N
N
O
O
O
O
O
O

6. PART III
41 42 43 44 45 46 47 48 49 50
A B C B A D C D A C
51 52 53 54 55
B, D B, C A, D C, D A,C
56 57 58 59 60
5 6 6 5 9
Section I
41. Solving the two equation
A 









ba
c
,
ba
c
B (1, 1)
Then distance from AB
2
= 8
(a +b +c)
2
< 4 ( a + b)
2
(a + b + c)
2
– 4(a+ b)
2
<0
((a + b + c) + 2 (a +b)) ((a + b + c –2 ( a+b))
 a + b + c –2a –2b < 0
(a + b +c +2(a + b )>0)
–a –b + c < 0
a + b – c > 0
Choice (A)
42. y = cosx + sinx
= 




 

4
xsin
2
1
y = |cosx –sinx|
y = cosx – sinx, 




 
4
,0
sinx –cosx, 




 
2
,
4
  dx)xsinx(cosxsinxcos
4
0


+   dx)xcosx(sinxsinxcos
2
4









2
4
4
xdxcos2xdxsin2
0
=     2
4
4
xsin2xcos2 0




= 

















2
2
22
2
2
2
2
22
2
2


= 224
2
4
4 
43. f(x) = x
2
–xsinx –cosx
f’(x) = (2– cosx) x>0 for x > 0
and f’(x) < 0 for x < 0
f(–x) = f(x)
f(0) < 0, 0
2
f 




 
But f() > 0
 f(x) has one root for x > 0
and another root in x < 0
 2 Roots
44. 


m
1k
)1n(nk2
cot
–1
[1+n (n+1)]
= cot
–1
[n
2
+ n +1]
= tan
–1






 1nn
1
2
= tan
–1
 







1nn1
n)1n(
= tan
–1
(n +1) –tan
–1
n
 



23
1n
11
1tan2tan
+ (tan
–1
3–tan
-1
2)+-----------
+ (tan
–1
24 – tan
–1
23)
= tan
–1
24 – tan
–1
(1)
= tan
–1






 
23
25
cot
25
23 1
Choice (B)
45. 






x
y
sec
x
y
dx
dy
y = vx,
dx
dv
xv
dx
dy

v + x
dx
dv
= v + sec (v)
x vsec
dx
dy

4
x
1
cos x
sin x

7. cosv dv =
x
dx
sinv = nCx
sin
x
y
= n (Cx)
x = 1, y =
6

 Cn
2
1

sin xlog
2
1
y
x






Choice (A)
46. f’(x) < 2(f(x)
f’(x) –2f(x) < 0 = –k (say)
Where k >0
f’(x) –2f(x) = –k
k)x(f2
dx
df

f(e
–2x
) =   
cdxke x2
= Ce
2
k x2

x =
2
1
1  e
–1
= Ce
2
k 1

C = 






2
k
1
e
1
Since f > 0 in 





1,
2
1
0e
2
k
1
e
1
2
k







f(x) = x2
Ce
2
k

= x2
e
2
k
1
e
1
2
k







 





















1
2
1
1
2
1
x2
2
e
e
2
k
1
2
1
2
k
dx)x(f
= 












2
k
1
2
1
2
k
1
2
e
4
k
2
1
4
ek
2
e
2
k

= 







2
e
1
2
k
2
1e
<
2
1e 
Since f > 0 in 





1,
2
1
and f 





2
1
= 1
Graph of f is above the x– axis
  
1
2
1 0dx)x(f
Hence,  




 1
2
1
2
1e
,0inlies)x(f
47. k2ji3PR 
k4j3iSQ 
k3j2iPT 
k2ji3yx 
xSQy


k4j3iyx 
k3ji2x 
kj2iy


Volume =  Txyp

= 10
321
121
312


48. Point on the line is (–2, –1, 0)
D. Ratio normal to plane <1, 1, 1>
Equation of the line 
r
plane is




1
z
1
1y
1
2x
Point on the line (–2, –1, )
` Which lies on the plane x + y + z –3 = 0
  = 2
 point is (0, 1, 2)
Another point is (0, –2, 3)
 x =
3
2
y =
3
4
, z=
3
11
3
11
3
4
3
2
2,1,0

3
5
3
7
3
2 
 D . Ratio < (2, –7, 5>
Equation line
5
2z
7
1y
2
x 




Choice (D)
49. Required probability
= 1 – P  DCBA 
1 – )D(P).C(P).B(P).A(P
= 1 –
8
7
44
1
2
1



= 1–
256
235
256
21

Choice (A)
y

P Q
RS
x


8. 50.  lies on |z –z0| = r
 | –z0| = r
( –z0)  0z = r
2
||
2
–   2
0000 rzzzz 
||
2
–   2
2
00 r
2
2r
zz 


||
2
–   1
2
r
zz
2
00  ––––––(1)

1
lies on |z – z0| = 2r

1
=
2


lies on |z –z0| = 2r
r2z02



2
0202
r4zz 


























  2
000022
r4zzzz
11




ie, 2
22
2
22
r4
2
2r
1
2
r11















using (1)
–1 + 2
2
2
2
r41
2
r
2
r



2
r7
2
r 2
2
2


||
2
=
7
1
|| =
7
1
Option (C)
Section II
51. The line  in along the line of S. D between
1 & 2
 the d. r. ‘ s are x, y, z where
x +2y + 2z = 0
2x + 2y + z = 0

2
z
3
y
2
x



 line () is r =  (2i –3j +2k)
where  meets 1,  = 1
So the point is 2i –3j +2k
(3 +2s –2)
2
+ (2s+ 6)
2
+ s
2
= 17
9s
2
+28s +20 = 0
(9s+10) (s +2) = 0
 s =
9
10
or –2
points are (–1, –1, 0) & 





9
8
,
9
7
,
9
7
Ans (B) & (D)
52. f’(x) = sinx + x cosx
f’(n) = (–1)
n
n
f’ 






2
1
n = (–1)
n
f’(n +1) = (–1)
n+1
 ( n +1)
 f’(n) f’ 






2
1
n is +ve
f’(n) f’ (n +1) is –ve
f’ 






2
1
n f’(n+1) is –ve
 one point exists in  1n,n 
& is in 





 1n,
2
1
n
Ans (B, C)
53. Sn = –1
2
–2
2
+ 3
2
+4
2
–5
2
–6
2
+ -----
+ (4n –1)
2
+ (4n)
2
    


n
1r
2222
)r4(1r42r4)3r4(
=  


n
1r
3r84
=
 








 n3
2
1nn
.8.4
= 4n [4n +1]
4n (4n +1) = 1056  n = 8
4n (4n +1) = 1332  n = 9
[(B), (C) do not give integral n]
Ans (A, (D)
54. (A) (N
T
MN)
T
= N
T
M
T
N
= N
T
MN when M is symmetric
 N
T
MN is symmetric
(B) (MN – NM)
T
= NM –MN, is skew symmetric
when M, N are symmetric
(C) (MN)
T
= NM where M, N are symmetric
So MN is not symmetric
(D) (adjM) (adjN) = adj (NM) & not adj (MN)
So statement is not correct
Ans (C), & (D)
55. If 46 n is the perimeter
and x is side of square removed
V = (15n –2x) (8n –2x)x
= 4x
3
–46nx
2
+120n
2
x
V’ = 12x
2
–92nx +120n
2

9. = 0 when x = 5
 120n
2
–460n +300 = 0
 6n
2
–23n +15 = 0
 6n
2
–18n –5n +15 = 0
 (6n –5) (n –3) = 0
 n = 3 or
6
5
Possible side – lengths are 45, 24, 12.5,
3
20
. Ans (A), (C)
Section III
56. The eight vectors can be represented as
(1) 1 1 1
(2) 1 1 –1
(3) 1 –1 1
(4) –1 1 1
(5) –1 –1 –1
(6) –1 –1 1
(7) –1 1 –1
(8) 1 –1 –1
From the eight vectors, 3 vectors may be chosen
in 56C3
8
 ways. Among 56 sets, we need to
find coplanar sets
With pair of vectors 1 and 5, any one of the other
six may be chosen to complete coplanar set.
Likewise for each pair of vectors (2 and 6), (3
and 7), (4 and 8), six coplanar sets can be
chosen.
 6  4 = 24 coplanar sets 
56 – 24 = 32 non- coplanar sets = 2
p
 p = 5
57. Let P(E1) = x
P(E2) = y
P(E3) = z
x (1–y) (1 –z) = 
y (1 –z) (1 –x) = 
z (1 –x) (1 –y) = 
Also (1–x) (1 –y) (1 –z) = p
( –2) p = 
( –2r) p = 2r
[x (1–y) (1–z) –2y (1–z) (1–x)]p
= xy (1–z)
2
(1–x) (1–y)
[x(1–y) –2y(1–x)]p
xy (1–z) (1–x) (1-y)
x(1–y) –2y(1–x) = xy
x –xy –2y +2xy = xy
x +xy –2y = xy
x –2y = 0 –––––(1)
[y (1–z) (1–x) –3z (1–x) (1–y)p =
2yz (1–z) (1–x)
2
(1–y)
[y(1–z) –3z(1–y)]p =    z1y1x1yz2 
y –yz –3z+3zy = 2yz
y–3z+ 2zy = 2yz
y –3z = 0 –––––(2)
 
 
6
3
1
2
y
3
1
y2
z
x
EP
EP
3
1

58.
5
7
C
C
;2
C
C
r
5n
1r
5n
1r
5n
r
5n
 






5
7
1r
5rn
,2
r
6rn





 n –3r = –6
5n –12r = –18
 n = 6
59.
    12241k2
2
1nn


 n (n +1) = 2450 +4k
 (n+50) (n –49) = 4k
n = 49  k = 0
n = 50  k = 25
(The next possible value of k is 103, for n =
53, is not feasible)
 the required integer is 25 –20 = 5
60. Let x = h meet the ellipse at   sin3,cos2 ,
so that cos =
2
n
tangent at the point is 1sin
3
y
cos
2
x

where it meets y = 0, x =
h
4
cos
2


Required area, (h)
2 = 







 cos2
cos
2
sin3
2
1
= 






2
h
h
2
sin32
=
 
4
h
1
h2
h432 22


=
 
h2
h43 2
3
2

(1) =
2
9
and
8
1515.3
4
1
43
2
1 2
3













=
8
545
 936458
5
8
21 