jee-adv grand test solutions (p 2)
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IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
1
Key Answers:
1. d 2. a 3. c 4. b 5. d 6. a 7. 5 8. 4 9. 2 10. 6
11. 3 12. b 13. b 14. b 15. c 16. a 17. b 18. 19. 20. b
21. c 22. c 23. d 24. d 25. c 26. 3 27. 4 28. 7 29. 9 30. 1
31. d 32. b 33. a 34. c 35. a 36. c 37. 38. 39. b 40. a
41. a 42. a 43. d 44. c 45. 7 46. 2 47. 2 48. 4 49. 2 50. c
51. a 52. d 53. c 54. d 55. c 56. 57.
18. A – r, B – p, C – s, D – q; 19. A – q, B – r, C – p, D – s;
37. A – p,q, B – r, C – r,s, D – p,q; 38. A – q, B –r, C – s, D-r;
56. A – p,r, B – p,q, C – s, D – p; 57. A – p,q,s, B – p,q,r, C – p,q,s, D p,q,r;
Solutions:
Chemistry
1. No of lone pair of 0Xe and number of bond pair 5
Hybridisation of 3Xe sp d
Hence shape 3 2XeO F should be trigonal bipyramidal and not octahedral
2. For the desired reaction
1 ,
1
GAgCl e Ag Cl E
F
the needed ,G can be obtained by adding the values of ,G for the recton
' 1 ;Ag e Ag G nFE
; ln ,spAgCl Ag Cl G RT K giving
,G = 4 1 101 9.648 10 (298 )ln 1.56 10mol J mol K
77.10 55.95 21.15kJ kJ kJ
The potential is
21.150.2192
1 96.485E V
3.
4. Alkenes with even number of cumulative double bonds are optically active if both sides are
dissymmetric
5. None of the acids evolve 2H gas with alkali metals
CH3
C2H5
O
OH
OH
OCH3
C2H5 O
-H2O
IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
2
6. Given 0.1 , 18.0HCl HClM M V mL
3 30.125 , ?NaHCO NaHCOM M V
on applying , 3 3HCl HCl NaHCO NaHCOM V M V
30.1 18 0.125 NaHCOV
314.4NaHCOM mL
Thus, 14.4mL of 31.25M NaHCO solution is needed to neutralise 18.0 mL of the 0.100
M HCl solution
7. The destruction reaction for 3O is given as 3 22 3O O
The net rate of production of 3O = 15 1 17.2 10 rate of this reactionmol h
Now
23
2 ;d O
k a xdt
but initially
0x
So
23 2 15 82 27.2 10 2 10
d Ok a k
dt
151 1
2 16
7.2 10
4 10k mol h
1 1 1 12
1818
60 60k mol h mol s
3 1 15 10 .mol s So value of 5x
8. We know that , d d
p RT or M RTM P
given, 3 7752.64 / ;
760d g dm P atm
3 1 10.0821 , 310 273 583R dm atmK mol T K
2.64 0.0821 583 760123.9 /
775M g mol
Atomic mass of 31 /P g mol
Number of P atoms in a molecule
123.94
31M
9. There are 2 bonds in C which is and B is 2 3Na SO
10. Total number of unpaired electrons 6
11. C is an amide which on heating with 2 5P O gives propane nitrole and so C is propanamide
2 5
3 2 2 3 2PO
CH CH CONH CH CH CN
NaO S
S
O
ONa
IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
3
C is formed by the action of 3NH on acid B , so acid B is propanoic acid
3 2 3 3 2 4 3 2 2CH CH COOH NH CH CH COONH CH CH CONH
B C
(iii) Acid B is formed from hydrolysis of A as well as A fumes in moist air, so A is
acid halide. Thus A is
3 2 3 2CH CH COCl HOH CH CH COOH HCl
Also, (A) 3 2 (propanoylchloride)CH CH COCl
(B) 3 2 (Propanoicacid)CH CH COOH
(C) 3 2 2(propanamide)CH CH CONH
(i) 3 2CH CH COCl (ii) 3 2CH COCH Cl (iii) 2 2ClCH CH CHO
12. Na metal is soluble in liquid ammonia. The reaction takes place in a homogeneous solution
phase
13. More highly substituted alkenes are adsorbed on metal catalyst effectively so are not readily
reduce
14. The least stable alkenes has highest heat of hydrogenation
15. Yellow curdy precipitate with Ag confirms the presence of I ion in the compound The
precipitate AgI is insoluble in 4NH OH
16. Compound A gives brown precipitate with 4CuSO which becomes white when 2 2 3Na S O is
added. Thus, compound A should be KI
17. 4 2 2 42CuSO KI CuI K SO
A
2 2 2 22CuI Cu I I
brown
2 2 3 2 2 42 2Na S O I Na S O NaI
white
(sodium tetrahtionate)
18. Magnetic moment, 2n n BM
(where n number of unpaired electrons)
For 2 , 0, 0 0 2 0.00Fe n BM
For 3 , 1, 1 1 2 1.73Ti n BM
IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
4
For 2 , 2, 2 2 2 2.83Ni n BM
For 3 , 5, 5 5 2 5.91Fe n BM
19. , order of reactionndCKC n
dt
or, 1/ time /n
mol lit k mol lit
or, 1 1 1time
n nK mol lit
IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
5
Mathematics
20. Favourable number of ways 12 47 41 1 11C C C
Required probability 51 51
52 51 221
21. For the vector a and b to be inclined at an obtuse angle, we must have
. 0 0,a b for all x
2
2 2log 12 6 log 0c x c x for all 0,x
2 6 12 0cy cy for all y R
Where 2logy x
20 and 36 48 0c c c
(using 2
0 if 0 and 0ax bx c x R a D
4
0 3 4 0 ,03
c and c c c
22. We have 2 2 0 0a b c a b c a b c
Line 0ax by c passes through either of the points 1, 1 and ( 1,1)
23. Let 2 and 3y t z u then the equation becomes , where 1, 2, 3........... 2, 4, 6x y u n x t
and 3, 6, 9u
Hence, the number of solution of this equation
= 2 3coefficient of ......nx in x x x 2 4 6 3 6 9..... ....x x x x x x
= 6 2 2 4 3 6coefficient of 1 ... 1 .... 1 .......nx in x x x x x x x
1 116 2 3coefficient of in 1 1 1nx x x x
24. The given quadratic equation is satisfied by , andx a x b x c Hence the quadratic
equation has three roots, which is only possible if it is an identify, hence it has infinite roots
25. Let z i
z i
2Arg Argz
3arg
4z
IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
6
26. 3 2. 5 7A A A A A I
25 7A A
5 5 7 7A I A
18 35A I
Now 4 3. 18 35A A A A A I
218 35A A
18 5 7 35A I A
55 126A I
5 4. 55 126A A A A A I
255 126A A
55 5 7 126A I A
149 385a and b
2 3 2 149 3 385a b
298 1155 1453
27. The period of
sin12
x
is 2
24
as 24 24
sin sin12 12
x x
12sin 2 sin
12
x x
similarly the period of
tan3
x
is 3 and the period of cos4
x
is 2
8
4
Hence the period of the given function
24,8,3 24LCM of
201 201 24 4824
28.
2 3 97
1
.... 1 1 1 ....1lim
1x
x x x x
x
1 2 3 .... 97
97
1 97 97 49 47532
IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
7
29. If ‘r’ be the radius and ' 'h be the height of the cylinder
Volume 2 3 3 2 32 2 4
3 3 3V r h r r r h r
2 242 .3
3V r h h r v tr r
2 22 4r h rh r r r
2
2 4
4
3
r r h h r r rV
Vr rh r
2
1.5 0.05 2 4 0.01 4 1.5 0.01
41.5 4 1.5
3
0.215
9
0.215 21.5
100 100 2.389%9 9
V
V
2389abcd
30. Let 1 2 3 4, 5, , ,E E E E E and 6E be the events of occurrence of 1, 2, 3, 4, 5 and 6 on the dice
respectively, and let E be the even of getting a sum of numbers equal to 9
1 2 31 1 2 1
, ; ;6 6 6
k k kP E P E P E
4 5 61 1 2 1
; ;6 6 6
k k kP E P E P E
and 1 2
( )9 9
P E
Then, E= 3,6 6,3 4,5 5,4
Hence, 3 6 6 3 4 5 5 4( )P E P E E P E E P E E P E E
6 6 3 4 5 5 43P E P E P E P E P E P E P E P E
= 62 3 2 4 5P E P E P E P E
1, 2 3 4Since , , are independentE E E E
1 1 1 1 22 2
6 6 66
k k k k
212 2
18k k
since, 1 2
9 9P E
IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
8
21 1 22 2 ...............(1)
9 18 9k k
22 2 2 4k k
22 2 2k k
22 2 2k k
22 0k k
2 1 0k k
10
2K
Hence, integral value of k is zero and for 0k from equation (1)
Set of integral value of 0k
Number of integral solutions of k is 1
31. Here 1 0a therefore we make the substitution 2 2 2x x t x Squaring both sides of this
equality and reducing the similar terms, we get
2 22
2
2 2 22 2 2 ;
2 1 2 1
t t tx tx t x dx dt
t t
2 22 2 4 4
1 2 2 12 1 2 1
t t tx x t
t t
substituting into the integral, we get
22 2
22
2 1 2 2 2 2
4 4 2 1 1 2
t t t t t dtI dt
t t t t t
Now let us expand the obtained proper rational fraction into per tail fractions
2
3 2
2 2
1 21 2 2
t t A B D
t tt t t
32. 2 1
dxI
x x x
Since here 1 0,c we can apply the second Euler substitution
2 1 1x x tx 2 2
2
2 12 1 1 ;
1
tt x t x x
t
substituting into I, we can get an integral of a rational fraction
2
22
2 2 2,
1 11
dx t tdt
t t tx x x
IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
9
Now,
2
2
2 2 2
1 9 1 1 11
t t A B D E
t t t t t tt
33. In this case 0 0a and c therefore, nether the first, nor the second Euler substitution is
applicable, But the quadratic 27 10x x has the real roots 2, 5 . Therefore we
use the third Euler substitution
27 10 2 5 2x x x x x t
25 2 ;x x t
2
2
5 2
1
tx
t
34. Ans (c)
35. Ans (a)
36. Let the equation of the circle be 2 2 2 2 0x y gx fy c ………..(1)
The line 1lx my will touch the circle (1) if the length of perpendicular from the centre
,g f of the circle on the line is equal to its radius,
i.e.,
2 2
2 2
1gl mfg f c
l m
2 2 2 21gl mf l m g f c
2 2 2 2 2 2 2 1 0 ( )c f l c g m gl fm gflm ii
But the given condition of tangency is
2 24 5 6 1 0l m l ……………….(iii)
Comparing equations (ii) and (iii), we get 2 24 5, 2 6, 2 ,2 0c f mc g g f c gf
solving we get 0, 3, 4f g c
substituting these values in equation (i), the equation of the circle is 2 2 6 4 0x y x .
any point on the line 1 0x y is ,1 ,t t t R
Chord of the contact generated by this point for the circle is 1 3 4 0tx y t t x or
3 3 4 0t x y x y which are concurrent at point of intersection of the 3 0x y and
3 4 0x y for all values of ‘t’. Hence, lines are concurrent at 1 5
,2 2
Also point 2, 3 lies outside the circle from which two tangents can be drawn
IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
10
37. (A) 2 7z i z i k is ellipse if 7 2 5k i i or k
(B) We have, for zC
2 2 2i z i z z i z
2 2z z i
Thus, minimum value of 2z z i is 2 and it is attained for any z lying on the segment joining
2 .z i
(C) 3 4z z i k is hyperbola, if 3 4 0 5k i k
(D) 3 450
kz i az az b
3 45 2 3 4
az az bkz i
i
This is hyperbola if 1 55
kk
38. 1
.12
P A B P A P B
1 1 1 4 3 1 6 1
3 4 12 12 12 12P A B
1
23/ /1 3
2
P A A B P A P A B
/ ' ' 0
' '
PP B A B
P A B
''/
P A s P B P A BP A B
P B P B
1
1 2121 11 3 3
4
IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
11
Physics
39. The centre of mass will follow the same path.
2
Rmx m mR
or 3 3 100
1502 2
Rx
m
40. 4 4 4
1 2 31 2 3
Pr Pr Pr, ,
8 8 8V V V
l l l
and
4Pr
8V
l
Now 1 2 3V V V V
Substituting the values, we get
1 2 3
1 2 1 2 2 3
l l ll
l l l l l l
1 2 3 6
1 2 1 3 2 3 11
m
41. 1 2( )
1
nR T T
2/3 2/31 2(5.6) (0.7)T T
2 14T T
11
3 9
2 2
3
nR TnRT
But 1
4n
1
9
8RT
42. 2
2 21
1 113.6(1)
2 3
hC
2
2 22
1 113.6(2)
2 4
hC
Dividing
2
1
1 5 / 36
4 3 /16
2 15
121527
oA
xR/2
R
m m m
COM
IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
12
43. 212
2iu v
2 2(2 )
2 10 5t
V Vu
Loss of energy 2 2
2 4
5 5
V VV
% loss 2
2
4 / 5100
V
V = 80%
44. Ratio of magnetic moment and angular momentum is given by
2
M q
L m
Which is a function of q and m only. This can be derived as follows :
2( )( )M iA qf r
2
2( ) ( )2 2
q rq r
and 2( )L I mr
2
2
2
2
rq
M q
L mmr
45. Distance between two cars leaving from station X is,
1
60 106
d
km
Man meets the first car after time,
160 1
60 60 2t h
He will meet the next car after time,
210 1
60 60 12t h
In the remaining half an hour, number of cars he will meet again is, 1/ 2
61/12
n
Total number of cars would be meet on route will be 7.
46. From work energy theorem
netKE W
or f iK K pdt
or 2
2 2
0
1 3
2 2mv t dt
( 2m kg)
or
23
2
02
tv
or 2v m/s
2
s
1
2 F
8 F
V
IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
13
47. 24 cos2
oI I
24 cos2
o oI I
1
cos2 2
or 2 3
or 2 2
3x
or 1 1
.3
d
D
yd
xD
7
4
6 10
3 103
yd
D
32 10 m = 2 mm
48. / ,E
v m s t sm v
7 922 10 2 10
For 200 neutrons generation: T t 200
s 74 10
49. d
dq idtR
Area under i - t graph
d (Area under i - t graph) ( R )
1
(4)(0.1)(10) 22
Wb
50. For rod 2
( 3 )1:
(2 )(16 ) / 3
N l
m l
or 32
3 3
mlN
N
A
3l
N
mg
30oB
IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
14
51. (2 ) (2 )(2 ) 4cN F m a m l ml
32 12 3
43 3
F N ml ml
52. 22 2
2
( cos30 ) (2 sin 30 )
( )(2 ) / 3
m gl N l
m l
2
2
32( 3 / 2)
3 3 83 3
8 3(4 / 3)
mlmgl
g
lml
53. (4)(2)(1) 8V Bvl V
,R C and L all are in parallel. Therefore, ,R CV V and LV all are 8V .
Ldi
V Ldt
or 8 (4)di
dt 2
di
dt A/s
i.e.., rate of exchange of current through L is constant. After 2s current will be 4A.
54. ( ) ( )applied mF rightwards F leftwards
IlB 1 2 3( )I I I lB
3I we have already calculated at 2 which is 4 A.
2 (8) 0Cdq d d
I CVdt dt dt
and 18
42
RVI A
R
(4 4)(1)(4) 32appliedF N
55. . (32)(2) 64 /P F v J s
21
1( )
2
d dip Li Li
dt dt
3(4)(4)(2) 32 / ( )J s i I
2 22 1(4) (2) 32 / ( )P i R J s i I
N
a
2l
2ca lC
F
IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
15
56. (A) Since 0E and 0,B so path will be straight line if velocity is parallel to B . Or path will be
circular if V B . Or path will be helical (with uniform pitch) if V is at some other angle to B .
Hence. (A) – p, r
(B) Since 0E and 0,B so path will be straight line if V B or parabola otherwise.
Hence (B) – p, q
(C) 0E , 0,B E B
Helical path with non-uniform pitch Hence, (C) – s
(D) Straight line path if V B E
Hence. (D) – (p)
57.
F f ma
,a
Fx fR Ia aR
Solve to get 2
( )F MxR If
I MR
B
E
90o
v
B
E
V
F
a
xa
R
kf
,M l
IIT Section
Subject Topic Grand Test – Paper II Date
C + M + P Grand Test – 07 IIT – GT – 07
10th May 2014 I20140510
16
0,f if 1
xMR
0,f if 1
xMR
So friction will act in forward direction.
0F if /x I MR , so friction will act in backward direction.
Also solved to get
2
( )FR x Ra
I MR
A is positive for all cases, hence body will accelerate in forward direction in all cases.
It means will be in clockwise direction. Hence bodies will rotate in clockwise direction.
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