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Mock Test - 3 (Paper - 2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
1/12
1. (1)
2. (6)
3. (3)
4. (4)
5. (3)
6. (6)
7. (9)
8. (1)
9. (A, C)
10. (A, B, C)
11. (A, B)
12. (A, B)
13. (B, C)
14. (A, B, C)
15. (B, D)
16. (B, C)
17. (A, B, C)
18. (A, B)
19. (A, B, C)
20. (A, C)
21. (4)
22. (1)
23. (4)
24. (2)
25. (3)
26. (6)
27. (2)
28. (6)
29. (A, B, C, D)
30. (A, B, D)
31. (A, B, C, D)
32. (A, C)
33. (A, B, C)
34. (B, C, D)
35. (A, B, C, D)
36. (A, B)
37. (A, D)
38. (C)
39. (B, D)
40. (D)
41. (1)
42. (7)
43. (0)
44. (2)
45. (2)
46. (5)
47. (3)
48. (8)
49. (A, B, C)
50. (A, C, D)
51. (A, C)
52. (C, D)
53. (A, C, D)
54. (A, B, C)
55. (A, B, C, D)
56. (A, B, C)
57. (B, C)
58. (D)
59. (A, C )
60. (A, C, D)
ANSWERSTest Date: 28/04/2019
PHYSICS CHEMISTRY MATHEMATICS
All India Aakash Test Series for JEE (Advanced)-2019
MOCK TEST - 3 (Paper-2) - Code-C
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 2) (Code-C) (Hints & Solutions)
2/12
PART - I (PHYSICS)
HINTS & SOLUTIONS
1. Answer (1)
Hint: 0p �
Solution : Combined COM of cylindrical shell and
hemisphere must lie on the centre of hemisphere.
P
2
2
2 22 2
(2 2 )
H RRH R R
R
RH R
2. Answer (6)
Hint: Relative velocity at point of contact is zero.
Solution : 2 mg sin – T – f = 2 ma
T – mg sin – f = ma
2f r mr
r = 2a
3. Answer (3)
Hint: String remains taut.
Solution : a cos 53° = 5 cos 37°
20
3
a
2203 m/s
a
4. Answer (4)
Hint: (ap)
t = 2a
0
Solution : a0 = 2 m/s2
at = 4 m/s2
5. Answer (3)
Hint: 2 1 2 1
( )
v u R
Solution : 1
1.5 1 0.5
30 10v
v1 = + 90
Now,
0
31
1 3 2
(2)( 70) 10v
v = 14 cm
D = 14 + 10
= 24 cm
6. Answer (6)
Hint: All images will form on a circle placed
symmetrically on circumference of circle.
Solution : N1 = 3
N2 = 3
7. Answer (9)
Hint: ind
dILdt
Solution : eq
2
3
LL
Also 1 2
1 2
dI dIL L
dt dt
1 1 2 2L I L I
Also, eq 0
dIL E
dt
0 0
eq
E tI
L
8. Answer (1)
Hint: At the time of maximum extension velocity of
each block is same.
Solution : 2
CM
ga
max
2
2
mF mgX
mK K
Mock Test - 3 (Paper - 2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
3/12
9. Answer (A, C)
Hint: Only x-component of velocity changes due to
collision.
Solution : 2
x x x
D D D
V eV e V
2
2
1 11 1 0 e e
e e
1 1 4 5 1
2 2
e
10. Answer (A, B, C)
Hint: const.P �
and K + U = const.
Solution : vm = 0
2 m/s 1 m/s
1
1 22
2 9
mt
k
1
2
9
mt
k
v2m
= 1 m/s
11. Answer (A, B)
Hint: Use principle of superposition.
Solution : 1 2 3 4
R
y y y y y
yR
3 2910sin cos
8 360kx t
ymax
= 10
12. Answer (A, B)
Hint: mV
RqB
Solution : ( )(3) 6
(2) m( )(5) 5
mD
q
ˆ ˆ ˆ ˆ4 3 3 44 3
5 5
i j i jv
�
ˆ ˆ(1.4 4.8 ) m/sv i j �
13. Answer (B, C)
Hint: ind
Bvl
Solution : v = (v0)
2
0
( )(2 )
12 m L
mv L
v
v = 0
0
0
0 0
3,
3( ) 2
B
v
L
vv v L v
L
2 2
1 2
1–
2V B l l
v = B(2v0)L
14. Answer (A, B, C)
Hint: Vrms
= Irms
Z
Solution : 1
100sin4
I t
V
250sin
4
I t
I = I1 + I
2
150sin4
I t
150(100) cos
42
P = 7500 W
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 2) (Code-C) (Hints & Solutions)
4/12
15. Answer (B, D)
Hint: 1 2 1 2
2 cosRI I I I I
Solution : 2
2P.D
2
d
D
D2d
2dS
22
99.52
d
D
, 2 2
02 1
RI A
02 3 2 2 I
16. Answer (B, C)
Hint: Mass of a nucleus is always less than the sum
of masses of its constituent particles.
Solution : M1 < 10 (m
p + m
n)
20(mn + m
p) – M
2 > 10(m
p + m
n) – M
1
m2 < m
1 + 10 (m
p + m
n)
17. Answer (A, B, C)
18. Answer (A, B)
Hint of Q. Nos. 17 and 18
Apply conservation of angular momentum.
Solution of Q. Nos. 17 and 18
17. Li about P = 0
Applying conservation of angular momentum (COAM)
and linear momentum, vp = v
0, = 0.
18. P is fixed.
COAM about P 2
03
mLmv L mvL ...(1)
Elastic collision v0 = L – v ...(2)
0 03
,2 2
v vv
L
19. Answer (A, B, C)
20. Answer (A, C)
Hint of Q. Nos. 19 and 20
max1
t
Q Q e
For maximum power transfer, Rext
= Rint
Solution of Q. Nos. 19 and 20
MN = 27 also 3 N R
M r
N = 9, M = 3
(27)1
9 3 3
NEI
NrR
M
27 3A
18 2
I
1
1 1(1) W
2 2
P EI
max1
t
q q e
(12)2 2418 1 18 1
t t
C CCE e CE e
241
9 1
t
Cq CE e
PART - II (CHEMISTRY)
21. Answer (4)
Hint : [O] 3
2 2 4 3 2Fe (C O ) 2Fe 6CO 6e
Number of waves formed by an electron in an orbit is
equal to number of that orbit.
Solution : Bond order of +NO = 3
Coordination no. of octahedral void = 6
Benzoic acid in benzene undergoes dimerisation. So
100% dimerisation will give the observed molar mass
equal to twice of actual molar mass of benzoic acid.
i.e. 244
22. Answer (1)
Hint : 2 1
m
K1000Scm mol
C
Solution : 5
2 1
m
4.95 10 10005 Scm mol
0.0099
Mock Test - 3 (Paper - 2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
5/12
m
m
5
500
2
2 5K C (0.0099)
500
= 0.99 × 10–6 1 × 10–6
23. Answer (4)
Hint : Ksp
remain same.
Solution : 2
2AB A 2B
���⇀↽���
Ksp
= [A2+] [B–]2
If [A+2] is increased 8 times, then [B–]2 will decrease
8 times.
So, old
new
[B ][B ]
2 2
24. Answer (2)
Hint : 19 2
2 2
1 2
1 1 hcE 21.8 10 Z
n n
Solution :
34 819 2
8
1 6.626 10 3 1021.8 10 1 Z
4 3.055 10
Z2 = 4
Z = 2
25. Answer (3)
Hint : Apply initial rate law method.
Solution : ∵ 1
2
t is independent of conc. of N.
Order of reaction w.r.t. N is 1.
At pH = 2, [H+] = 10–2M, 1
2
t 10 min
pH = 3, [H+] = 10–3M, 1
2
t 100 min
∵1 n
1
2
t [Reactant] K
1 n2
3
10 10
100 10
1 n1(10)
10
n = 2
26. Answer (6)
Hint : Radius ratio rule.
Solution : r 1.62
0.58r 2.8
AB has NaCl type structure.
27. Answer (2)
Hint : VII is displacement reaction.
Solution : In IV and VII no gaseous product
28. Answer (6)
Hint : P1 = CH
3CH OHC
Br
O
— ——
—
— —
Solution : CH3
P1
NH2
CH3
NH2
NH2
(M = 60)
CH
CH C
NH2
O
Br / 2
KOH,
NH3
(excess)— —
— —
—
—
—
— —
29. Answer (A, B, C, D)
Hint : Shape of molecule depends upon the shape of
orbitals involved in hybridization.
Solution : The transition state formed in SN2
mechanism, is trigonal bipyramidal and its hybridization
is sp2.
30. Answer (A, B, D)
Hint : When CHClBr2 reacts with KOH, it forms :CClBr
carbene due to better leaving behaviour of Br–.
Solution :
+ C
H
CCl
Cl
Cl
Br
: C
Cl
Br
+
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 2) (Code-C) (Hints & Solutions)
6/12
31. Answer (A, B, C, D)
Hint : Both ‘N’ of pyrimidine are identical.
Solution : (B) lone pair of N-3 is not involved in
aromaticity.
(D) All are aromatic according to Huckel’s rule.
32. Answer (A, C)
Hint : Avg. bond order = sum of total number of bonds
between the two atoms in all
resonating structures
total number of resonating
structures
Solution : Bond order in CO3
2– = 1.33
SO3
2– = 1.33
ClO4
– = 1.75
ClO3
– = 1.66
33. Answer (A, B, C)
Hint : Apply the law of equivalence.
Solution : 1 1 2 2
5n f n f
3 in 2nd equivalent point, n
1
and n2 are mole of FeSO
4 and FeC
2O
4 , f
1 = f
2 = 1 are n
factor of Fe+3
1 2
5n 1 n 1
3
1 2
5n n
3
[O] 32 4 2FeC O Fe 2CO 3e
nf = 3
34. Answer (B, C, D)
Hint : Compounds having optical activity are
Resolvable.
Solution : A has plane of symmetry.
(B, C, D) are optically active.
35. Answer (A, B, C, D)
Hint : Disproportionation of K2MnO
4.
Solution :
3K MnO + 2H O2 4 2
2KMnO + MnO + 4KOH4 2
(purple) (brown)
36. Answer (A, B)
Hint : Final temperature is 1 2T T
2
.
Ssurr
= 0 [no heat loss to surrounding]
Solution : f f
1 2
T T
System
T T
msdT msdTS
T T
1 2
f
T T(common)T
2
2f
System1 2
(T )S msln
T T
21 2
System1 2
(T T )S msln
4T T
37. Answer (A, D)
38. Answer (C)
Hint of Q. Nos. 37 and 38
rH B D E of Reac tant B D E of Product
Solution of Q. No. 37 and 38
O D/D O(excess)–
2
24 hrCH
3D C
3C C
O O
H H
C O (bond energy) is more than C C ,
O H (bond energy) is more than C H bond energy,
in CH3
CHO & CH2
CH OH therefore formation
of CH3CHO is more exothermic and more stable than
CH2
CH OH
39. Answer (B, D)
40. Answer (D)
Hint of Q. No. 39 and 40
White ppt C, with B
Black residue G, with excess of B
Scarlet ppt D, with KI
all confirm metal is Hg
Hence, A is HgCl2.
Solution of Q. No. 39 and 40
2HgCl2 + SnCl
2 Hg
2Cl
2 + SnCl
4
White ppt (C)
HgCl2 + SnCl
2 Hg + SnCl
4
excess (G)
HgCl2 + 2KI HgI
2 (scarlet) + 2KCl
Hgl2 + KI K
2 [HgI
4]
(E) Soluble
K2[HgI
4] + NH
4OH [HgOHgNH
2]I
Hg2Cl
2 + 2NH
4OH Hg + Hg(NH
2)Cl
black
Mock Test - 3 (Paper - 2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
7/12
PART - III (MATHEMATICS)
41. Answer (1)
Hint: Family of lines are concurrent at (4, – 8)
Solution : c
b a24
4a – 8b + c = 0
Family of lines ax + by + c = 0 are concurrent
at (4,– 8)
(4, – 8) lies on the parabola y2 = 16 x
2yy = 16 Tmy y
16 8
2
N
ym –
8
Nm (4,–8)–8
( ) – 18
42. Answer (7)
Hint: From : 1
Solution : lim cot20 (sin4 cos8 ) Form : 1x
xx x
�
0
(sin4 cos8 – 1) 0lim Form :
tan2 0x
x xe
x
�
Using L.H. rule
x
x xe e
x
2
20
4cos4 – 8sin8lim
2sec (2 ) �
[�] = [e2] = 7
43. Answer (0)
Hint: 2 – 2 – 1 = 0, 2 – 2 –1 = 0, + = 2,
= –1
Solution : + = 2, = –1
x – 2x – 1 = 0 2 – 2= 1, 2 – 2 = 1
3 2 3 2– 2 1 – 2 1
2 ( )0
1 1 ( 1)( 1)
44. Answer (2)
Hint: Draw graph of f(x) = [sinx] [cosx] and g(x)
= [sinx cosx]
Solution : Let f(x) = [sinx] [cosx], g(x) = [sinx cosx]
y
xO
2
23 2
f x x x( ) = [sin ] [cos ]g x x x( ) = [sin cos ]
y
xO
x = 0, 1 satisfies the given equation
45. Answer (2)
Hint: Solve graphically
Solution : y
xO
y = x cos3
y = x sin
y = 1
y = –1
2
tan2x = sinx cosx
sin2x = sinx cos3x
sinx (sinx – cos3x) = 0
sinx 0
then sinx – cos3x = 0
There are two values of x.
46. Answer (5)
Hint: Use binomial theorem
Solution : x x
f x
x x x x
4 5 20
4 4 3 8 4
(1 ) – (1 )( )
[(1 ) – ( )]
x x x x x x
x x x x x x
4 8 12 16 20 20
4 4 8 12 8 4
1 5 10 10 5 –1–
[1 3 3 – – ]
x x x
x x x
4 8 12
4 8 12
5(1 2 2 )5
1 2 2
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 2) (Code-C) (Hints & Solutions)
8/12
47. Answer (3)
Hint: Multiply N and D by (1 – ei)
1 – eir
i
i e 2– 2 sin2
Solution :
nni n i
i
ef n e
e
(2 ) –1 (2 –1)2
1– sin(2 )( , )
1– sin2
f i
128sin
255 2552568, cos sin256 512 512
sin512
i1
sin cos512 512
sin512
fRe 8, 1256
5 – 2Re 8, 5 – 2 3256
f
48. Answer (8)
Hint: Distance between two ellipses.
Solution : f(,) cos= 9 cos2+ 5cos2– 14 coscos– 16 cos– 16 sinsin+ 20
= (4sin– 2sin)2 – 16sin2– 4sin2+ (5cos– 3
cos)2
= (4 sin– 2sin)2+ (5cos– 3 cos)2
Expression denotes the distance between two
points (5 cos, 4 sin) and (3cos, 2sin) is
The maximum distance between the ellipses
x y x y2 2 2 2
1and 125 16 9 4
Maximum value of expression is (5 + 3).
49. Answer (A, B, C)
Hint: Break the function
Solution : f x x x–1 23 1
( ) sin – 1–2 2
x x–1 23 1
– cos – 1–2 2 2
cos–1 a + cos–1b = ab a b–1 2 2cos ( – 1– 1– )
If a + b 0, –1 a, b 1
ab a b–1 2 22 – cos ( – 1– 1– )
If a + b 0, –1 a, b 1
Put a b x3,
2
x x x3 3
If 0, –1 – ,12 2
x x x–1 2 –1 –13 1 3
cos – 1– cos cos2 2 2
x x–1 –12
cos – sin6 3
f x x x–1 –12 –
( ) – sin sin2 3 6
x x x3 – 3
If 0, – 1 1 –1,2 2
then
x x x–1 2 –1 –13 1 3
cos – 1– 2 – cos – cos2 2 2
x–1
2 – – sin6 2
f x x–1( ) – 2 – sin
2 6 2
x–15
– – sin6
f x dx x
1 1
–1
– 3 – 3
2 2
( ) – sin6
x x x
1–1 2
– 3
2
3– 1 sin 1–6 2
3 3 1– 1 – – –6 2 2 1 3 2
Mock Test - 3 (Paper - 2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
9/12
3 1–1– 3 – 3 –
6 2 2
3 3 12 – –
6 2 2
f x dx x dx
– 3 – 3
2 2–1
–1 –1
–5( ) – sin
6
3 3 1– 2 –
6 2 2
50. Answer (A, C, D)
Hint: x x x x1
sin 0,2 2
Solution : f x x f x x f x x2
1 2 3( ) , ( ) , ( ) sin2
xA x x dx x
131 3
2 21
00
2 1 2 1(– ) – –
3 3 3 3 3
A x x dx x x dx
1
12
2
10
2
( – sin ) sin – )2 2
x x x x
11
3 32
2 2
10
2
2 2 2 2cos – cos –
3 2 2 3
2 1 2 1 2 22 – –
3 32 2 2
2 2 2 2 2 2 2– – (1– 2) ( 2 –1)
3 3 3
2 22 –1 –
3
xA x x dx x
11 32
3
0 0
2sin – – cos –
2 2 3
–1 2 2 1–
3 3
51. Answer (A, C)
Hint: x xcos2 sec2 2 0
Solution : x x x
x x x
3 3
2 2 2
sin sec tan
1– 2sin sec – 2tan
x x x
x
3 2
2
tan tan .tan2
21– tan
(tan2x) tan2x = tan2x (sec2x–1) = tan2x . sec2 x – tan2x
xx x
x
2
2
2tansec – tan2
1– tan
x x xdx xdx xdx C
x x
32
2 2
sin sec 1 2tan 1sec – tan2
2 21– 2sin 1– tan
x x C21 1
– ln 1– tan – ln sec 22 4
x x C2 21
– ln (1– tan ) .sec 24
x xC
x
2
4
1 cos 2 sec 2– ln
4 cos
x x C1
ln cos – ln cos24
xC
x
41 cosln
4 cos2
xC
x
21 (1 cos2 )ln
4 cos2
x xC
x
21 1 1 cos 2 2cos2ln
4 4 cos2
f x x x C1 1 1
( ) ln ln cos2 sec 2 24 4 4
passes through 1 1
0,– ln4 4
f x x x1
( ) ln cos2 sec 2 24
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 2) (Code-C) (Hints & Solutions)
10/12
52. Answer (C, D)
Hint: R1 R
1 – R
2, R
2 R
2 – R
3
Solution :
x x x x x
f x x x x x x
2 2 3 3
2 2 3 3
tan – sin sin – cot cot
( ) cot – cos cos – tan tan
0 0 1
= (tanx – sin2x) (cos2x – tan3x) – (cotx – cos2x) (sin2x –
cot3x)
x xx x x
x x
32
3
sin sin(1– sin cos ) cos –
cos cos
x xx x x
x x
32
3
cos cos– 1– cos sin sin –sin sin
x xx x x x x x
x x
4
4
sin4 cos(1– sin cos ) sin cos – – cos sin
cos4 sin
x xx x
x x
8 8
4 4
cos – sin(1– sin cos )
sin cos
If f(x) = 0, 1 – sinx cosx = 0, cos8x = sin8x
sin2 2 Not possible tan 1x x
x n
4
53. Answer (A, C, D)
Hint: �� �
� �
i r
i r
n nn
n n
–
–
Solution :
�
� �
� �
�
�
i r
i r
i k i jn n
nn n i k i j
1 1(3 – 4 ) – (4 3 )
– 5 5
1 1– (3 – 4 ) – (4 3 )5 5
� � �
� � �
�i j k1 3 4
26 26 26
� �
a b c
1 3 4
Let angle between plane mirror and plane containing
reflected rays is .
� �
rn n1 13
cos ( . ) (4 9)5 26 5 26
54. Answer (A, B, C)
Hint: Let A o B b C c( ), ( ), ( )� � �
Solution :
c b( – )
c bA3
4 –
4
c( )
c bA4
4 – 4
5
b c
A22
3
bA1
2
b( )o( )
A
D C
B
3
2
1
1
1
1
4
b c c bA A A A1 2 1 4
2 8 –13,
6 10
� � � �
������ ������
b c c bA A A A3 2 3 4
11 – 8 –4 –11,
12 20
� � � �
������� �������
A A A A A A A b c1 2 4 1 2 1 4
1 34( )
2 120
������ ������ � �
A A A A A A A b c2 3 4 3 2 3 4
1 33( )
2 120
������� ������� � �
A A A A b c1 2 3 4
67( )
120
� �
ABCD b c( ) � �
55. Answer (A, B, C, D)
Hint: r
R ncos
Solution :
A0
/nr
A0
O
B
Mock Test - 3 (Paper - 2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
11/12
In the triangle OAB, r
n Rcos
n
1 3 1cos ,
2 2 2
n
,3 12
n [3, 12]
56. Answer (A, B, C)
Hint: r n – t
Solution : n
nn
r r
r
C
3
0 put n – r = t
n t
n
Ct n
n t
n–
30 ( – )
t
n
Ct n
n n t nt t
n
3 2 2 30– 3 3 –
t
n
n n n
Ct
tn s n nS
n
3 2
0
– 3 3 –
t n t
n n
n
C Ct t
t n tb
n n–
0 0
–
∵
t
n
n n
Ct
n tns b
n0
–
–
n n
nb s
2
n n n n
nn s n s n S3 21
– 3 32 2
n n n
n nS s
33
–2 4
p = 3, q = 2
57. Answer (B, C)
Hint: Locus of centre of circle is a moving circle with
centre lying on a line segment.
58. Answer (D)
Hint: Locus of centre of circle is a moving circle with
centre lying on a line segment.
Solution of Q.Nos. 57 and 58
S4 = (x – a sin)2 + (y – a cos)2 – 1 = 0
Let C1 (h
1k) h = a sin, k = a cosh2 + k2 = a2
locus of C1(h
1 k) is x2 + y2 = a2
O
y
xA a, 0( )
x + y a 2 2 2
= ( – 1)x + y a2 2 2
=
x + y a + 2 2 2
= ( 1)
a a a2 2 2( 1) – ( –1)
a = 4
S2 = (x – sin–1) + (y – cos–1)2 – 4 = 0
C2 (h, k) h = sin–1, k = cos–1 h k
2
Equation of locus of centre of S2 = 0 is
x y x y, – 02 2 2
y
x
2
2
2
2,( )
2,( )0
x y2
A1 1
2 4 4 42 2
4 ( 2 1) sq.units
m = 4, n = 2
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 2) (Code-C) (Hints & Solutions)
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59. Answer (A, C)
Hint: find n (S)
60. Answer (A, C, D)
Hint: Find n (S)
Solution of Q.Nos. 59 and 60
A5
A4
A3
A6A
7A
8A
9
A2
A1
A10
Number of triangles formed = 10C3 – 3C
3 – 4C
3 = 115
Number of quadrilateral formed
= 3C2 [4C
2 +2 C
2 + 4C
1. 2C
1 + 4C
1.1C
1 + 2C
1.1C
1]
+ 3C1 [1C
1 . 4C
1 . 2C
1 + 1C
2 . 4C
1.
2C2 + 1C
1 + 2C
1 + 4C
1 . 2C
2
+ 4C2. 2C
1 ]
+ 3C0 [1C
1 . 4C
1 . 2C
2 + 4C
2 . 2C
1 + 4C
2 + 2C
2 ]
= 3 × 21 + 3 × 31 + 22 = 178
Number of pentagons formed = 3C2 × 31 + 3C
1 × 22 +
3C0 [ 1C
1 + C
2 . 2C
1 ] = 165
Number of hexagons formed = 3C2 × 22 + 3C
1 [1C
1 4C
2 .
2C2 ] = 84
Number of heptagons formed = 1C1 . 2C
2 . 3C
2 . 4C
2 =
18
n(s) = 115 + 780 + 165 + 84 + 18 = 560
165 33
560 112 P ( ) ,
2
2 2
( ) 3 21 7
( ) 3(7 21 31 22 6) 29
P VP
V P V
��
�����
Mock Test - 3 (Paper - 2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
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1. (1)
2. (9)
3. (6)
4. (3)
5. (4)
6. (3)
7. (6)
8. (1)
9. (B, C)
10. (B, D)
11. (A, B, C)
12. (B, C)
13. (A, B)
14. (A, B)
15. (A, B, C)
16. (A, C)
17. (A, B, C)
18. (A, B)
19. (A, B, C)
20. (A, C)
21. (6)
22. (2)
23. (6)
24. (3)
25. (2)
26. (4)
27. (1)
28. (4)
29. (A, B)
30. (A, B, C, D)
31. (B, C, D)
32. (A, B, C)
33. (A, C)
34. (A, B, C, D)
35. (A, B, D)
36. (A, B, C, D)
37. (A, D)
38. (C)
39. (B, D)
40. (D)
41. (8)
42. (3)
43. (5)
44. (2)
45. (2)
46. (0)
47. (7)
48. (1)
49. (A, B, C)
50. (A, B, C, D)
51. (A, B, C)
52. (A, C, D)
53. (C, D)
54. (A, C)
55. (A, C, D)
56. (A, B, C)
57. (B, C)
58. (D)
59. (A, C )
60. (A, C, D)
ANSWERSTest Date: 28/04/2019
PHYSICS CHEMISTRY MATHEMATICS
All India Aakash Test Series for JEE (Advanced)-2019
MOCK TEST - 3 (Paper-2) - Code-D
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 2) (Code-D) (Hints & Solutions)
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PART - I (PHYSICS)
HINTS & SOLUTIONS
1. Answer (1)
Hint: At the time of maximum extension velocity of
each block is same.
Solution : 2
CM
ga
max
2
2
mF mgX
mK K
2. Answer (9)
Hint: ind
dILdt
Solution : eq
2
3
LL
Also 1 2
1 2
dI dIL L
dt dt
1 1 2 2L I L I
Also, eq 0
dIL E
dt
0 0
eq
E tI
L
3. Answer (6)
Hint: All images will form on a circle placed
symmetrically on circumference of circle.
Solution : N1 = 3
N2 = 3
4. Answer (3)
Hint: 2 1 2 1
( )
v u R
Solution : 1
1.5 1 0.5
30 10v
v1 = + 90
Now,
0
31
1 3 2
(2)( 70) 10v
v = 14 cm
D = 14 + 10
= 24 cm
5. Answer (4)
Hint: (ap)
t = 2a
0
Solution : a0 = 2 m/s2
at = 4 m/s2
6. Answer (3)
Hint: String remains taut.
Solution : a cos 53° = 5 cos 37°
20
3
a
2203 m/s
a
7. Answer (6)
Hint: Relative velocity at point of contact is zero.
Solution : 2 mg sin – T – f = 2 ma
T – mg sin – f = ma
2f r mr r = 2a
8. Answer (1)
Hint: 0p �
Solution : Combined COM of cylindrical shell and
hemisphere must lie on the centre of hemisphere.
P
2
2
2 22 2
(2 2 )
H RRH R R
R
RH R
Mock Test - 3 (Paper - 2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
3/12
9. Answer (B, C)
Hint: Mass of a nucleus is always less than the sum
of masses of its constituent particles.
Solution : M1 < 10 (m
p + m
n)
20(mn + m
p) – M
2 > 10(m
p + m
n) – M
1
m2 < m
1 + 10 (m
p + m
n)
10. Answer (B, D)
Hint: 1 2 1 2
2 cosRI I I I I
Solution : 2
2P.D
2
d
D
D2d
2dS
22
99.52
d
D
, 2 2
02 1
RI A
02 3 2 2 I
11. Answer (A, B, C)
Hint: Vrms
= Irms
Z
Solution : 1
100sin4
I t
V
250sin
4
I t
I = I1 + I
2
150sin4
I t
150(100) cos
42
P = 7500 W
12. Answer (B, C)
Hint: ind
Bvl
Solution : v = (v0)
2
0
( )(2 )
12 m L
mv L
v
v = 0
0
0
0 0
3,
3( ) 2
B
v
L
vv v L v
L
2 2
1 2
1–
2V B l l
v = B(2v0)L
13. Answer (A, B)
Hint: mV
RqB
Solution : ( )(3) 6
(2) m( )(5) 5
mD
q
ˆ ˆ ˆ ˆ4 3 3 44 3
5 5
i j i jv
�
ˆ ˆ(1.4 4.8 ) m/sv i j �
14. Answer (A, B)
Hint: Use principle of superposition.
Solution : 1 2 3 4
R
y y y y y
yR
3 2910sin cos
8 360kx t
ymax
= 10
15. Answer (A, B, C)
Hint: const.P �
and K + U = const.
Solution : vm = 0
2 m/s 1 m/s
1
1 22
2 9
mt
k
1
2
9
mt
k
v2m
= 1 m/s
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 2) (Code-D) (Hints & Solutions)
4/12
16. Answer (A, C)
Hint: Only x-component of velocity changes due to
collision.
Solution : 2
x x x
D D D
V eV e V
2
2
1 11 1 0 e e
e e
1 1 4 5 1
2 2
e
17. Answer (A, B, C)
18. Answer (A, B)
Hint of Q. Nos. 17 and 18
Apply conservation of angular momentum.
Solution of Q. Nos. 17 and 18
17. Li about P = 0
Applying conservation of angular momentum (COAM)
and linear momentum, vp = v
0, = 0.
18. P is fixed.
COAM about P 2
03
mLmv L mvL ...(1)
Elastic collision v0 = L – v ...(2)
0 03
,2 2
v vv
L
19. Answer (A, B, C)
20. Answer (A, C)
Hint of Q. Nos. 19 and 20
max1
t
Q Q e
For maximum power transfer, Rext
= Rint
Solution of Q. Nos. 19 and 20
MN = 27 also 3 N R
M r
N = 9, M = 3
(27)1
9 3 3
NEI
NrR
M
27 3A
18 2
I
1
1 1(1) W
2 2
P EI
max1
t
q q e
(12)2 2418 1 18 1
t t
C CCE e CE e
241
9 1
t
Cq CE e
PART - II (CHEMISTRY)
21. Answer (6)
Hint : P1 = CH
3CH OHC
Br
O
— ——
—
— —
Solution : CH3
P1
NH2
CH3
NH2
NH2
(M = 60)
CH
CH C
NH2
O
Br / 2
KOH,
NH3
(excess)— —
— —
—
—
—
— —
22. Answer (2)
Hint : VII is displacement reaction.
Solution : In IV and VII no gaseous product
23. Answer (6)
Hint : Radius ratio rule.
Solution : r 1.62
0.58r 2.8
AB has NaCl type structure.
24. Answer (3)
Hint : Apply initial rate law method.
Solution : ∵ 1
2
t is independent of conc. of N.
Mock Test - 3 (Paper - 2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
5/12
Order of reaction w.r.t. N is 1.
At pH = 2, [H+] = 10–2M, 1
2
t 10 min
pH = 3, [H+] = 10–3M, 1
2
t 100 min
∵1 n
1
2
t [Reactant] K
1 n2
3
10 10
100 10
1 n1(10)
10
n = 2
25. Answer (2)
Hint : 19 2
2 2
1 2
1 1 hcE 21.8 10 Z
n n
Solution :
34 819 2
8
1 6.626 10 3 1021.8 10 1 Z
4 3.055 10
Z2 = 4
Z = 2
26. Answer (4)
Hint : Ksp
remain same.
Solution : 2
2AB A 2B
���⇀↽���
Ksp
= [A2+] [B–]2
If [A+2] is increased 8 times, then [B–]2 will decrease
8 times.
So, old
new
[B ][B ]
2 2
27. Answer (1)
Hint : 2 1
m
K1000Scm mol
C
Solution : 5
2 1
m
4.95 10 10005 Scm mol
0.0099
m
m
5
500
2
2 5K C (0.0099)
500
= 0.99 × 10–6 1 × 10–6
28. Answer (4)
Hint : [O] 3
2 2 4 3 2Fe (C O ) 2Fe 6CO 6e
Number of waves formed by an electron in an orbit is
equal to number of that orbit.
Solution : Bond order of +NO = 3
Coordination no. of octahedral void = 6
Benzoic acid in benzene undergoes dimerisation. So
100% dimerisation will give the observed molar mass
equal to twice of actual molar mass of benzoic acid.
i.e. 244
29. Answer (A, B)
Hint : Final temperature is 1 2T T
2
.
Ssurr
= 0 [no heat loss to surrounding]
Solution : f f
1 2
T T
System
T T
msdT msdTS
T T
1 2
f
T T(common)T
2
2f
System1 2
(T )S msln
T T
21 2
System1 2
(T T )S msln
4T T
30. Answer (A, B, C, D)
Hint : Disproportionation of K2MnO
4.
Solution :
3K MnO + 2H O2 4 2
2KMnO + MnO + 4KOH4 2
(purple) (brown)
31. Answer (B, C, D)
Hint : Compounds having optical activity are
Resolvable.
Solution : A has plane of symmetry.
(B, C, D) are optically active.
32. Answer (A, B, C)
Hint : Apply the law of equivalence.
Solution : 1 1 2 2
5n f n f
3 in 2nd equivalent point, n
1
and n2 are mole of FeSO
4 and FeC
2O
4 , f
1 = f
2 = 1 are n
factor of Fe+3
1 2
5n 1 n 1
3
1 2
5n n
3
[O] 32 4 2FeC O Fe 2CO 3e
nf = 3
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 2) (Code-D) (Hints & Solutions)
6/12
33. Answer (A, C)
Hint : Avg. bond order = sum of total number of bonds
between the two atoms in all
resonating structures
total number of resonating
structures
Solution : Bond order in CO3
2– = 1.33
SO3
2– = 1.33
ClO4
– = 1.75
ClO3
– = 1.66
34. Answer (A, B, C, D)
Hint : Both ‘N’ of pyrimidine are identical.
Solution : (B) lone pair of N-3 is not involved in
aromaticity.
(D) All are aromatic according to Huckel’s rule.
35. Answer (A, B, D)
Hint : When CHClBr2 reacts with KOH, it forms :CClBr
carbene due to better leaving behaviour of Br–.
Solution :
+ C
H
CCl
Cl
Cl
Br
: C
Cl
Br
+
36. Answer (A, B, C, D)
Hint : Shape of molecule depends upon the shape of
orbitals involved in hybridization.
Solution : The transition state formed in SN2 mechanism,
is trigonal bipyramidal and its hybridization is sp2.
37. Answer (A, D)
38. Answer (C)
Hint of Q. Nos. 37 and 38
rH B D E of Reac tant B D E of Product
Solution of Q. No. 37 and 38
O D/D O(excess)–
2
24 hrCH
3D C
3C C
O O
H H
C O (bond energy) is more than C C ,
O H (bond energy) is more than C H bond energy,
in CH3
CHO & CH2
CH OH therefore formation
of CH3CHO is more exothermic and more stable than
CH2
CH OH
39. Answer (B, D)
40. Answer (D)
Hint of Q. No. 39 and 40
White ppt C, with B
Black residue G, with excess of B
Scarlet ppt D, with KI
all confirm metal is Hg
Hence, A is HgCl2.
Solution of Q. No. 39 and 40
2HgCl2 + SnCl
2 Hg
2Cl
2 + SnCl
4
White ppt (C)
HgCl2 + SnCl
2 Hg + SnCl
4
excess (G)
HgCl2 + 2KI HgI
2 (scarlet) + 2KCl
Hgl2 + KI K
2 [HgI
4]
(E) Soluble
K2[HgI
4] + NH
4OH [HgOHgNH
2]I
Hg2Cl
2 + 2NH
4OH Hg + Hg(NH
2)Cl
black
PART - III (MATHEMATICS)
41. Answer (8)
Hint: Distance between two ellipses.
Solution : f(,) cos= 9 cos2+ 5cos2– 14 coscos– 16 cos– 16 sinsin+ 20
= (4sin– 2sin)2 – 16sin2– 4sin2+ (5cos– 3
cos)2
= (4 sin– 2sin)2+ (5cos– 3 cos)2
Expression denotes the distance between two
points (5 cos, 4 sin) and (3cos, 2sin) is
The maximum distance between the ellipses
x y x y2 2 2 2
1and 125 16 9 4
Maximum value of expression is (5 + 3).
Mock Test - 3 (Paper - 2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
7/12
42. Answer (3)
Hint: Multiply N and D by (1 – ei)
1 – eir
i
i e 2– 2 sin2
Solution :
nni n i
i
ef n e
e
(2 ) –1 (2 –1)2
1– sin(2 )( , )
1– sin2
f i
128sin
255 2552568, cos sin256 512 512
sin512
i1
sin cos512 512
sin512
fRe 8, 1256
5 – 2Re 8, 5 – 2 3256
f
43. Answer (5)
Hint: Use binomial theorem
Solution : x x
f x
x x x x
4 5 20
4 4 3 8 4
(1 ) – (1 )( )
[(1 ) – ( )]
x x x x x x
x x x x x x
4 8 12 16 20 20
4 4 8 12 8 4
1 5 10 10 5 –1–
[1 3 3 – – ]
x x x
x x x
4 8 12
4 8 12
5(1 2 2 )5
1 2 2
44. Answer (2)
Hint: Solve graphically
Solution : y
xO
y = x cos3
y = x sin
y = 1
y = –1
2
tan2x = sinx cosx
sin2x = sinx cos3x
sinx (sinx – cos3x) = 0
sinx 0
then sinx – cos3x = 0
There are two values of x.
45. Answer (2)
Hint: Draw graph of f(x) = [sinx] [cosx] and g(x)
= [sinx cosx]
Solution : Let f(x) = [sinx] [cosx], g(x) = [sinx cosx]
y
xO
2
23 2
f x x x( ) = [sin ] [cos ]g x x x( ) = [sin cos ]
y
xO
x = 0, 1 satisfies the given equation
46. Answer (0)
Hint: 2 – 2 – 1 = 0, 2 – 2 –1 = 0, + = 2,
= –1
Solution : + = 2, = –1
x – 2x – 1 = 0 2 – 2= 1, 2 – 2 = 1
3 2 3 2– 2 1 – 2 1
2 ( )0
1 1 ( 1)( 1)
47. Answer (7)
Hint: From : 1
Solution : lim cot20 (sin4 cos8 ) Form : 1x
xx x
�
0
(sin4 cos8 – 1) 0lim Form :
tan2 0x
x xe
x
�
Using L.H. rule
x
x xe e
x
2
20
4cos4 – 8sin8lim
2sec (2 ) �
[�] = [e2] = 7
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 2) (Code-D) (Hints & Solutions)
8/12
48. Answer (1)
Hint: Family of lines are concurrent at (4, – 8)
Solution : c
b a24
4a – 8b + c = 0
Family of lines ax + by + c = 0 are concurrent
at (4,– 8)
(4, – 8) lies on the parabola y2 = 16 x
2yy = 16 Tmy y
16 8
2
N
ym –
8
Nm (4,–8)–8
( ) – 18
49. Answer (A, B, C)
Hint: r n – t
Solution : n
nn
r r
r
C
3
0 put n – r = t
n t
n
Ct n
n t
n–
30 ( – )
t
n
Ct n
n n t nt t
n
3 2 2 30– 3 3 –
t
n
n n n
Ct
tn s n nS
n
3 2
0
– 3 3 –
t n t
n n
n
C Ct t
t n tb
n n–
0 0
–
∵
t
n
n n
Ct
n tns b
n0
–
–
n n
nb s
2
n n n n
nn s n s n S3 21
– 3 32 2
n n n
n nS s
33
–2 4
p = 3, q = 2
50. Answer (A, B, C, D)
Hint: r
R ncos
Solution :
A0
/nr
A0
O
B
In the triangle OAB, r
n Rcos
n
1 3 1cos ,
2 2 2
n
,3 12
n [3, 12]
51. Answer (A, B, C)
Hint: Let A o B b C c( ), ( ), ( )� � �
Solution :
c b( – )
c bA3
4 –
4
c( )
c bA4
4 – 4
5
b c
A2
2
3
bA1
2
b( )o( )
A
D C
B
3
2
1
1
1
1
4
b c c bA A A A1 2 1 4
2 8 –13,
6 10
� � � �
������ ������
b c c bA A A A3 2 3 4
11 – 8 –4 –11,
12 20
� � � �
������� �������
Mock Test - 3 (Paper - 2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
9/12
A A A A A A A b c1 2 4 1 2 1 4
1 34( )
2 120
������ ������ � �
A A A A A A A b c2 3 4 3 2 3 4
1 33( )
2 120
������� ������� � �
A A A A b c1 2 3 4
67( )
120
� �
ABCD b c( ) � �
52. Answer (A, C, D)
Hint: �� �
� �
i r
i r
n nn
n n
–
–
Solution :
�
� �
� �
�
�
i r
i r
i k i jn n
nn n i k i j
1 1(3 – 4 ) – (4 3 )
– 5 5
1 1– (3 – 4 ) – (4 3 )5 5
� � �
� � �
�i j k1 3 4
26 26 26
� �
a b c
1 3 4
Let angle between plane mirror and plane containing
reflected rays is .
� �
rn n1 13
cos ( . ) (4 9)5 26 5 26
53. Answer (C, D)
Hint: R1 R
1 – R
2, R
2 R
2 – R
3
Solution :
x x x x x
f x x x x x x
2 2 3 3
2 2 3 3
tan – sin sin – cot cot
( ) cot – cos cos – tan tan
0 0 1
= (tanx – sin2x) (cos2x – tan3x) – (cotx – cos2x) (sin2x –
cot3x)
x xx x x
x x
32
3
sin sin(1– sin cos ) cos –
cos cos
x xx x x
x x
32
3
cos cos– 1– cos sin sin –sin sin
x xx x x x x x
x x
4
4
sin4 cos(1– sin cos ) sin cos – – cos sin
cos4 sin
x xx x
x x
8 8
4 4
cos – sin(1– sin cos )
sin cos
If f(x) = 0, 1 – sinx cosx = 0, cos8x = sin8x
sin2 2 Not possible tan 1x x
x n
4
54. Answer (A, C)
Hint: x xcos2 sec2 2 0
Solution : x x x
x x x
3 3
2 2 2
sin sec tan
1– 2sin sec – 2tan
x x x
x
3 2
2
tan tan .tan2
21– tan
(tan2x) tan2x = tan2x (sec2x–1) = tan2x . sec2 x – tan2x
xx x
x
2
2
2tansec – tan2
1– tan
x x xdx xdx xdx C
x x
32
2 2
sin sec 1 2tan 1sec – tan2
2 21– 2sin 1– tan
x x C21 1
– ln 1– tan – ln sec 22 4
x x C2 21
– ln (1– tan ) .sec 24
x xC
x
2
4
1 cos 2 sec 2– ln
4 cos
x x C1
ln cos – ln cos24
xC
x
41 cosln
4 cos2
xC
x
21 (1 cos2 )ln
4 cos2
x xC
x
21 1 1 cos 2 2cos2ln
4 4 cos2
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 2) (Code-D) (Hints & Solutions)
10/12
f x x x C1 1 1
( ) ln ln cos2 sec 2 24 4 4
passes through 1 1
0,– ln4 4
f x x x1
( ) ln cos2 sec 2 24
55. Answer (A, C, D)
Hint: x x x x1
sin 0,2 2
Solution : f x x f x x f x x2
1 2 3( ) , ( ) , ( ) sin2
xA x x dx x
131 3
2 21
00
2 1 2 1(– ) – –
3 3 3 3 3
A x x dx x x dx
1
12
2
10
2
( – sin ) sin – )2 2
x x x x
11
3 32
2 2
10
2
2 2 2 2cos – cos –
3 2 2 3
2 1 2 1 2 22 – –
3 32 2 2
2 2 2 2 2 2 2– – (1– 2) ( 2 –1)
3 3 3
2 22 –1 –
3
xA x x dx x
11 32
3
0 0
2sin – – cos –
2 2 3
–1 2 2 1–
3 3
56. Answer (A, B, C)
Hint: Break the function
Solution : f x x x–1 23 1
( ) sin – 1–2 2
x x–1 23 1
– cos – 1–2 2 2
cos–1 a + cos–1b = ab a b–1 2 2cos ( – 1– 1– )
If a + b 0, –1 a, b 1
ab a b–1 2 22 – cos ( – 1– 1– )
If a + b 0, –1 a, b 1
Put a b x3,
2
x x x3 3
If 0, –1 – ,12 2
x x x–1 2 –1 –13 1 3
cos – 1– cos cos2 2 2
x x–1 –12
cos – sin6 3
f x x x–1 –12 –
( ) – sin sin2 3 6
x x x3 – 3
If 0, – 1 1 –1,2 2
then
x x x–1 2 –1 –13 1 3
cos – 1– 2 – cos – cos2 2 2
x–1
2 – – sin6 2
f x x–1( ) – 2 – sin
2 6 2
x–15
– – sin6
f x dx x
1 1
–1
– 3 – 3
2 2
( ) – sin6
Mock Test - 3 (Paper - 2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019
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x x x
1–1 2
– 3
2
3– 1 sin 1–6 2
3 3 1– 1 – – –6 2 2 1 3 2
3 1–1– 3 – 3 –
6 2 2
3 3 12 – –
6 2 2
f x dx x dx
– 3 – 3
2 2–1
–1 –1
–5( ) – sin
6
3 3 1– 2 –
6 2 2
57. Answer (B, C)
Hint: Locus of centre of circle is a moving circle with
centre lying on a line segment.
58. Answer (D)
Hint: Locus of centre of circle is a moving circle with
centre lying on a line segment.
Solution of Q.Nos. 57 and 58
S4 = (x – a sin)2 + (y – a cos)2 – 1 = 0
Let C1 (h
1k) h = a sin, k = a cosh2 + k2 = a2
locus of C1(h
1 k) is x2 + y2 = a2
O
y
xA a, 0( )
x + y a 2 2 2
= ( – 1)x + y a2 2 2
=
x + y a + 2 2 2
= ( 1)
a a a2 2 2( 1) – ( –1)
a = 4
S2 = (x – sin–1) + (y – cos–1)2 – 4 = 0
C2 (h, k) h = sin–1, k = cos–1 h k
2
Equation of locus of centre of S2 = 0 is
x y x y, – 02 2 2
y
x
2
2
2
2,( )
2,( )0
x y2
A1 1
2 4 4 42 2
4 ( 2 1) sq.units
m = 4, n = 2
59. Answer (A, C)
Hint: find n (S)
60. Answer (A, C, D)
Hint: Find n (S)
Solution of Q.Nos. 59 and 60
A5
A4
A3
A6A
7A
8A
9
A2
A1
A10
Number of triangles formed = 10C3 – 3C
3 – 4C
3 = 115
Number of quadrilateral formed
= 3C2 [4C
2 +2 C
2 + 4C
1. 2C
1 + 4C
1.1C
1 + 2C
1.1C
1]
+ 3C1 [1C
1 . 4C
1 . 2C
1 + 1C
2 . 4C
1.
2C2 + 1C
1 + 2C
1 + 4C
1 . 2C
2
+ 4C2. 2C
1 ]
+ 3C0 [1C
1 . 4C
1 . 2C
2 + 4C
2 . 2C
1 + 4C
2 + 2C
2 ]
= 3 × 21 + 3 × 31 + 22 = 178
Number of pentagons formed = 3C2 × 31 + 3C
1 × 22 +
3C0 [ 1C
1 + C
2 . 2C
1 ] = 165
All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 2) (Code-D) (Hints & Solutions)
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Number of hexagons formed = 3C2 × 22 + 3C
1 [1C
1 4C
2 .
2C2 ] = 84
Number of heptagons formed = 1C1 . 2C
2 . 3C
2 . 4C
2 =
18
n(s) = 115 + 780 + 165 + 84 + 18 = 560
165 33
560 112 P ( ) ,
2
2 2
( ) 3 21 7
( ) 3(7 21 31 22 6) 29
P VP
V P V
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