MOCK TEST - 3 (Paper-2) - Code-C · 2019-06-04 · All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 2) (Code-C) (Hints & Solutions) 2/12 PART - I (PHYSICS)

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


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


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


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)


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


20. Answer (A, C)


max1

t

Q Q e

For maximum power transfer, Rext

= Rint


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


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

+


6/12


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


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.


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)


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.


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


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


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).


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


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


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


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


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( ) � �


Hint: r

R ncos

Solution :

A0

/nr

A0

O

B


11/12

In the triangle OAB, r

n Rcos

n

1 3 1cos ,

2 2 2

n

,3 12

n [3, 12]


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)



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


12/12

59. Answer (A, C)

Hint: find n (S)


Hint: Find n (S)


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

1/12

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)

2/12

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


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


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


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


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


18. Answer (A, B)


Apply conservation of angular momentum.


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


20. Answer (A, C)


max1

t

Q Q e

For maximum power transfer, Rext

= Rint


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.


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


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.


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


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


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

+


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)


rH B D E of Reac tant B D E of Product


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.


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).


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


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


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


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]


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

� � � �

��


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( ) � �


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


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


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


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


11/12

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)



58. Answer (D)




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)


Hint: Find n (S)


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


12/12

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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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MOCK TEST - 3 (Paper-2) - Code-C · 2019-06-04 · All India Aakash Test Series for JEE (Advanced)-2019 Mock Test - 3 (Paper - 2) (Code-C) (Hints & Solutions) 2/12 PART - I (PHYSICS)

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