test - 1a (paper - 1) (code-a) (answers) all india aakash ......test - 1a (paper - 1) (code-a)...

Test - 1A (Paper - 1) (Code-A) (Answers) All India Aakash Test Series for JEE (Advanced)-2019

1/12

PHYSICS

1. (A, C)

2. (A, B, D)

3. (A, C)

4. (B, D)

5. (A, C)

6. (A, C)

7. (15)

8. (30)

9. (16)

10. (32)

11. (64)

12. (27)

13. (72)

14. (20)

15. (C)

16. (C)

17. (A)

18. (A)

Test Date : 05/08/2018

ANSWERS

TEST - 1A (Paper-1) - Code-A

All India Aakash Test Series for JEE (Advanced)-2019

CHEMISTRY

19. (A, B, C)

20. (A, B, C, D)

21. (A, B, D)

22. (A, C, D)

23. (A, B, C, D)

24. (B)

25. (35)

26. (02)

27. (64)

28. (28)

29. (06)

30. (10)

31. (10)

32. (09)

33. (B)

34. (D)

35. (C)

36. (A)

MATHEMATICS

37. (D)

38. (B, C, D)

39. (A, B, D)

40. (A, B, C)

41. (B, C, D)

42. (A, B, C)

43. (15)

44. (33)

45. (03)

46. (08)

47. (21)

48. (02)

49. (00)

50. (04)

51. (C)

52. (B)

53. (A)

54. (D)

All India Aakash Test Series for JEE (Advanced)-2019 Test - 1A (Paper - 1) (Code-A) (Hints & Solutions)

2/12

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (A, C)

Hint : At equilibrium net force on each charge is zero.

Solution :

T

T

2 2

2 2 2

16 642 cos cos

4 cos

q qT k

l l

⎛ ⎞ ⎜ ⎟

⎝ ⎠

2 2

2 2 2

162 sin sin

4 sin

q qT k

l l

⎛ ⎞ ⎜ ⎟

⎝ ⎠

3 1tan

64

1 1tan

4

⎛ ⎞ ⎜ ⎟⎝ ⎠

2. Answer (A, B, D)

Hint : Use concept of field inside cavity.

Field at respective points can be considered

superposition of field due to sphere + sphere of

density (–, cavity) + solution

Solution :

sphere cavityAE E E � � �

0

06

A

RE i

��

0 0 0

17

3 54 54B

R R i RE i

��

3. Answer (A, C)

Hint : Use concept of potential due to uniformly

charged disc.

Solution :

04

xd dxdV

x

2

0/2

04

l

d dx

dV

∫ ∫

2

2

0

2 cos

4

r d

dV

∫

0

rV

00

2

Rr

dU rdr ∫

2 3

0

2

3

RU

4. Answer (B, D)

Hint : Apply Ohm’s law, J = E

Solution :

Current density remains same

1E

1 =

2E

2

Apply gauss

2 1

0

( )ds

E E ds

2 1 0

I I Q⎛ ⎞ ⎜ ⎟ ⎝ ⎠

0

2 1

1 1Q I

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

Test - 1A (Paper - 1) (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019

3/12

5. Answer (A, C)

Hint : Potential difference across capacitor is V0

Solution :

03

kQ kQV

a a

03

4

aVQ

k

Wb = QV

0

6. Answer (A, C)

Hint : Find initial charge and final charge.

Solution :

R2

2V0

C

CV q0 +

– – CV q0

–q

+q

0

02

q CV qV

C C

0

2

CVq

2

0bW CV

2

0

1

2i

U CV

2 2

0 091

2 4 4f

V VU C

⎛ ⎞ ⎜ ⎟

⎝ ⎠

Heat dissipated =

2

0

4

CV

7. Answer (15)

Hint : Current through any section would be same.

Solution :

3x x

r

J = E

J2(r + x)l = constant

32 1 constant

xr x E x

r

⎛ ⎞ ⎜ ⎟⎝ ⎠

0

3 31 1

2 2E r E r

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

E = 0

15

4E

= 3.75 E0

8. Answer (30)

Hint : Find distribution of current in circuit

Solution :

R D

2R x

A

B

Let RAB

= x

2

2

Rxx R

R x

2Rx + x2 = 2Rx + 2R2 + xR

x2 – xR = 2R2

2 23

2 2

R Rx

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x = 2R

12060 V

2BD

V

After every section voltage gets halved

VBC

= 30 V

9. Answer (16)

Hint : Find velocity in the direction of electric field.

Solution :

02

,x x

qVl V t V

m

y

qEV dt

m ∫

qat dt

m ∫

2

2y

q tV a

m

2

22

y

x

q lV a

m V


4/12

2

02 2

qal m

m qV

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

04

y

alV

V

2

3

0

tan4 2

y

x

V al m

V qV

10. Answer (32)

Hint : Prove that force is proportional to negative of

displacement.

Solution :

y

2 20

2

2

q yF

a y

For y << a

2

0

qyF

a

2

02

maT

q

11. Answer (64)

Hint : Based on potential energy of dipole in electric

field.

Solution :

Since there is pure rolling

Wfriction

= 0

K = –U

4

4

2 cosi

p Rd R

∫

22 2

ip R

22 2

fp R

pi

pf

/2

U = pE0cos45° + pE

0cos45°

2

02 2 2R E

= –4R2E0

2 2 2

0

12 4

2MR R E

02

E

m

12. Answer (27)

Hint : Find potential across capacitor before and after

switching.

Solution :

Just before switching

02

3C

VV

After switching

0 0 22 4

19 9

t

CRC

V VV e

At t = 2RC ln2

04

9C

VV

0

0

4

9R

VV V

05

9

V

x = 27

13. Answer (72)

Hint : Apply concept of potential gradient

Solution :

G

E0

Eeqreq

R


5/12

eq 1 2

eq 1 2

E E E

r r r

Eeq

= 8 V

req

= 2

Switch S is open

E = 8 V

Switch S is closed

E = 6 V

2

18MN

14. Answer (20)

Hint : Points C, F, G and D, E, H are at same

potential.

Solution :

R

RR

R

R

R

R

R

R

A B

eq

1 1 1 1 1

3 3 3R r r r r

eq

2

rR

15. Answer (C)

16. Answer (C)

Hint and solution of Q. Nos. 15 and 16

Hint : Apply Gauss theorem to find field.

Solution :

= 01

r

R

⎛ ⎞ ⎜ ⎟⎝ ⎠

Q = 4r2dr

=2

0

0

4 1

Rr

r drR

⎛ ⎞ ⎜ ⎟⎝ ⎠∫

=

3 4

04

3 4

r r

R

⎛ ⎞ ⎜ ⎟

⎝ ⎠

=

3 3

0 04

12 3

R R

3 42

04 4

3 4

r rE r

R

⎛ ⎞ ⎜ ⎟

⎝ ⎠

0 31

3 4

r rE

R

⎛ ⎞ ⎜ ⎟⎝ ⎠

0dE

dr

0 02

03 4

r

R

2

3

Rr

17. Answer (A)

18. Answer (A)

Hint and Solution of Q. Nos. 17 and 18

Hint : For maximum power transfer. External

resistance must be equal to resistance of network.

Solution :

Circuit can be redrawn to

eq

RC

+

–

2

eq 60 20

2 4 4

eq

= 40 V

R = req

= 2

t

CV A Be

= Req

C

= C

t

CC

V A Be

At t = 0

VC = A + B

At t =

VC = A = 20 V

VC = 20 – 50e–t/


6/12

19. Answer (A, B, C)

Hint : If T increases Koverall

decreases (as given)

we know, as T increase, K2 increasebut K

eq may

increase or decrease depending on type of reaction.

Since overall rate decreases on increasing

temperature, Keq

must be dominating over K2 and the

reversible reaction must be exothermic.

As 1

1

KT increase, decrease

K

So, K1 must have been greater than K

–1 as K

–1

increases more.

Solution :

Rate = K2[A][B

2] from R.D.S

1

1 2

2

1

K [B ]

K [B]

Keq

[B]2 = [B2]

Rate = K2K

eq[B]2[A]1

Rate = 2 1

1

K K

K

[B]2[A]1

Let 2 1

1

K KK

K

2 1 1(E E E )

RTK e

E2 + E

1 – E

–1 < 0

E–1

– E1 – E

2 > 0

20. Answer (A, B, C, D)

Hint : For immiscible liquid

2

o o o o

A B A H Op p p 755 p p ⇒

Solution :

o

Ap 755 700 55 torr

Mass of A in distillate = 55 2 = 110 g.

Mass of H2O in distillate = 145 1 = 145 g.

For immiscible liquids, from Dalton’s law

2 2 2

o

A A A

o

H O H O H O

w p M

w p M

A

110 18 700M 173.8 g

145 55

21. Answer (A, B, D)

Hint : Boiling point of A is more than B it means B

is more volatile

Solution :

In graph, upper line represent the vapour line while

lower line show liquid line. In between these line

vapour and liquid phase exist.

22. Answer (A, C, D)

Hint : If more H+ is formed pH decreases, but if more

OH– is formed pH increases.

Solution :

Electrolysis

(P) Cathode

2Cu 2e Cu(s)

Anode

2 2

1H O O 2H 2e

2

pH

(Q) Cathode

Ag 1e Ag(s)

Anode

2 2

1H O O 2H 2e

2

pH

(R) Cathode

2 22H O 2e H 2OH

Anode

2

1Cl Cl 1e

2

pH

PART - II (CHEMISTRY)


7/12

(S) Cathode

2 22H O 2e H 2OH

Anode

2 2

1H O O 2H 2e

2

pH remain same.


Hint : For spontaneous cell

Ecell

> 0

Solution :

[Ag+]cathode

> [Ag+]Anode

If [Cl–], [Ag+]

o o

spCl /AgCl/Ag Ag /AgE E 0.059 logK

so o

Ag /AgE act as cathode

24. Answer (B)

Hint :

2

o

2

0.059 [Ag ]E E log

2 [Zn]

Solution :

2

o 1

2

0.1

V0.059E E log

2 0.01

V

⎛ ⎞⎜ ⎟⎝ ⎠ ⎛ ⎞⎜ ⎟⎝ ⎠

o 2

2

1

V0.059E E log

2 V

o 2

2 1E E , V V

(1 + x) > (1 + y)2

V1 volume of solution of cathode chamber.

V2 volume of solution of Anode chamber.

25. Answer (35)

Hint : x = 3, y = 2, so x3 + y3 = 35

Solution :

2 2 2

1N O(g) N (g) O (g)

2

i

i

P1K ln

t 3P 2P

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 2 2

1N O(g) N (g) O (g)

2

Pi

0 0

Pi – p p

p

2

P = Pi +

p

2

2P – 2Pi = p

2N O

p at t = Pi – (2P – 2P

i)

= 3Pi – 2P

K = i

i

P1ln

t 3P 2P

⎛ ⎞⎜ ⎟⎝ ⎠

26. Answer (02)

Hint : C4-axis

Solution :

C4-axis contains opposite face centre and body centre

C4-axis pass through opposite face centres and body

centre. If 2 face centre atom and one body centre

atom is removed then formula becomes Na3Cl

3.

27. Answer (64)

Hint : H.C.P. contain 20 T.H site

Solution :

In H.C.P., 8 T.V are inside the unit cell while 12 are

at vertical edges (2 on each vertical edge)

T.V. = 1

8 12 123

One edge in H.C.P. contributes 1

3.

28. Answer (28)

Hint : x = +2, y = +3, A = 2

Solution :

For electrical neutrality

total positive charge = total negative charge

41 x y 8

A

= i C R T

9.6 = i 0.1 0.08 300

i = 4


8/12

so y = 3 so possible value of A = 1, 2, 3

from eq.

If A = 1

x + 4 3 = 8 x = –4 (Not possible)

If A = 2

x 2 3 = 8

x = +2

so y x

3A

(possible because odd number)

If A = 3

x + 4 = 8

x = 4

then y x 3 4

4A 4

(Not possible because even)

so possible value of A = 2

hence, A2 + 4xy

= 4 + 4 × 2 × 3

= 28

29. Answer (06)

Hint : Mass

DensityVolume

Solution :

Density = 1.315 103 kg m–3

Mass = ?

Volume = 13 7 3 10–27 m3

Let x molecule of insulin be present in one unit cell.

Total mass of insulin = 3

A

x35.8995 kg/m

N

Total volume = 2.73 10–25 m3

Density = 1.315 103 = A

25

x35.8995

N

2.73 10

x = 6

30. Answer (10)

Hint : pA x

A

Solution :

At sea level

2 2N O

p 0.8 p 0.2

at height H, P = 0.4 atm

Mole fraction of N2 and O

2 remain in ratio of 4 : 1 at

any height.

2

2

N

O

p 0.4 0.8 0.32 atm

p 0.4 0.2 0.08 atm

x = h

0.8

k

x = h

0.32

k

x

2.5x

andy

2.5y

x y2

x y

⎛ ⎞⎜ ⎟ ⎝ ⎠

= 10

31. Answer (10)

Hint : 1

2

k2

k

Solution :

1

2

k[C]

[D] k

[C]2

[D]

so, 2k2 = k

1

32. Answer (09)

Hint : Rate =

1

32

1 2

2 3

4

KK [CH CHO]

2K

⎛ ⎞⎜ ⎟⎝ ⎠

Solution : Rate =

1

32

1 2

2 3

4

KK [CH CHO]

2K

⎛ ⎞⎜ ⎟⎝ ⎠

so,

K =

1

a 2

a

1

E 2E E RT

1RT RT

2 E

RT

4

A eAe A e

2A e

⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠

A =

1

2

1

2

4

AA

2A

⎛ ⎞⎜ ⎟⎝ ⎠


9/12

Rate of reaction:

4

2 3 3

d[CH ]K [CH ][CH CHO]

dt

i

Now, by steady state approximation

3

1 3 2 3 3

d[CH ]0 K [CH CHO] K [CH ][CH CHO]

dt

i

i

2

3 2 4 3K [CH CHO] 2K [CH ]

i i

...(i)

2

2 3 3 3 2

d[CH CHO]0 K [CH ][CH CHO] K [CH CHO]

dt

i

i i

...(ii)

so,

1

2

1

3 3

4

K[CH ] [CH CHO] {from (i) and (ii)}

2K

⎛ ⎞ ⎜ ⎟⎝ ⎠

i

1

2

4 1

2 3 3

4

1

32

1 2

2 3

4

d(CH ) KK [CH CHO] [CH CHO]

dt 2K

KK [CH CHO]

2K

⎡ ⎤ ⎢ ⎥

⎣ ⎦

⎛ ⎞ ⎜ ⎟

⎝ ⎠

so

1

2

1

2

4

KK K

2K

⎛ ⎞ ⎜ ⎟

⎝ ⎠

1

2

1

2

4

AA A

2A

⎛ ⎞ ⎜ ⎟

⎝ ⎠

1

10 26

3

6 3 9

10A 10

2 5 10

10 10 10 .

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

33. Answer (B)

Hint : Basic dye stuff like methylene blue are

positively charged.

Solution :

If water moves towards anode it means water

contains negative charge so colloidal sol must

contain positive charge.

34. Answer (D)

Hint : Greater the magnitude of charge on

coagulating ion, greater is its power.

Solution :

Water moves towards negative electrode it means it

is positively charged and sol is negatively charged.

35. Answer (C)

Hint : d[intermediate]

0dt

Solution :

By steady state approximation

1 1 2 2

d[SH ]0 K [S][AH ] K [A][SH ] K [SH ][H O]

dt

1

2 2 1

K [S][AH ][SH ]

K [H O] K [A]

2 1 2

2 2

1 2 2

K K [S][AH ][H O]d[P]Rate K [H O][SH ]

dt K [A] K [H O]

if K2

>> K–1

[A] Rate = K1[S][AH+]

if K2[H

2O] << K

–1[A] Rate =

1 2 2

1

K K [S][H ][H O]

K K

[AH ] [H ]As

[A] K

⎡ ⎤⎢ ⎥⎣ ⎦

36. Answer (A)

Hint : app H

K K [H ]

Solution :

app H HlogK logK logH logK pH

So, Kapp

= expH

(logK pH)

log Kapp

pH

Kapp

pH

(A)


10/12

PART - III (MATHEMATICS)

37. Answer (D)

Hint : Solve equations

Solution :

2 42∵ x yz , y2zx = 34 and z2xy = 44

xyz = 24

A = {1, 2, 3, 4, 6, 8, 12, 24}.

and B = {2, 3, 4, 11}

Number of possible relations from A to B is 232.

38. Answer (B, C, D)

Hint : det(kA) = kn det(A)

Solution :

∵ |A| = 1

then |–A| = (–1)n|A| = 1 or –1.

and det( ) (det )

det

n

n n

kA k AA

k k .


Hint : f(x) is periodic with period 3

Solution :

∵

3, 3 0

, 0 3( ) 3

3, 3 6

x x

x xf x x r

x x

⎧ ⎪

⎪ ⎨ ⎪

⎪⎩ �

f(x) is periodic with period 3.

Range of f(x) is [0, 3 )

f(x) = | |

4

x has exactly four real solutions in (–6, 6).


Hint : Break the function

Solution :

Here

2

2

2 , 0( )

2 , 0

⎧ ⎨

⎩

x x xf x

x x x

The graph of g(x) is given as

–2 –1 O 1 2 3

–2

–1

1

2

3

2

2

2 , 2 1

1, 1 0( )

0, 0 2

2 , 2 3

x x x

xg x

x

x x x

⎧ ⎪ ⎪ ⎨

⎪⎪ ⎩


Hint : sec 8 < 0

Solution :

l = sec2(sec–1 5 ) + cosec2(cosec–1 10 )

l = 15.

And 1 1| sec 8 | 1 1 cos 8

tan tan 4tan 8 sin 8

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

m = – ( – 4) 4m

lm = 15 4 = 60

(B), (C) and (D) options are correct.


Hint : f(x) 0

Solution :

∵ x2 – 4 0 x2 – 1 > 0

(x2 – 1)f(x) = 2

4 | ( ) |x f x

(x2 – 1) f(x) 0 f(x) 0.

|f(x)| = f(x)

(x2 – 1)f(x) = 2

4 ( )x f x

f(x) =

2

2

4

2

x

x


11/12

Domain of f(x) = (–, –2] [2, )

The graph of f(x) is

–2

2

43. Answer (15)

Hint : Property of co-factors

Solution :

, 3

3 | | 3 5 15.ij ij

i i j

a c A

∑

44. Answer (33)

Hint : sin–1(sin7) = 7 – 2

Solution :

f(x + 2) = f(x – 1)

f(x) = f(x + 3)

Period of f(x) is 3.

and

2

1

2

4 13( 2 sin (sin 7))

3

⎛ ⎞⎡ ⎤ ⎜ ⎟⎢ ⎥⎜ ⎟⎣ ⎦⎝ ⎠

xg x g x

x

g(x + 4) = g(x – 7)

g(x) = g(x + 11)

Period of g(x) is 11.

Period of h(x) is 33.

45. Answer (03)

Hint : Draw graph of functions

Solution :

Draw the graphs of 3cos–1x and 1

{ }xthey intersect

each other at three distinct points.

3 real solution are possible

46. Answer (08)

Hint : Make homogeneous type system.

Solution :

(x4 – 1) + (y4 – 1) – (z4 – 1) = 0

(x4 – 1) + (y4 – 1) + (z4 – 1) = 0

(x4 – 1) – (y4 – 1) + (z4 – 1) = 0

2 2

1 2 3

1 1 1

1 1 1 1 1 3 0

1 1

0

4 4 41 1 1 0

1, 1, 1

x y z

x y z

⇒

Number of solutions is 8.

47. Answer (21)

Hint : trace of 1

11 22 33

1 1 1( ) ...

nA

a a a

Solution :

∵

1 1 1

3 2 2

3

0

0 0

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

a b c

A b c

c

then

1 2

1

1

3

2

3

1

10

10 0

k ka

A kb

c

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

3

1

1

1trace of ( )

10n

n

A

⎛ ⎞⎜ ⎟⎝ ⎠∑

1 1 1 1 1

1 2 3 4 5

1(tr( ) tr( ) tr( ) tr( ) tr( ))

10

1(5 2 4 8 3 18 2 32 50)

10

21

A A A A A


12/12

48. Answer (02)

Hint : 2 26

1 23

x

Solution :

For domain of log(15–x)

(x – 2) x – 2 > 0, 15 – x > 0

x (2, 15) – {14} ...(1)

For domain of 1 2 26

cos then3

⎡ ⎤⎢ ⎥⎣ ⎦

x

2 261 2

3

3 2 26 6

2316

2

x

x

x

Domain of f(x) = 23

,15 {14}2

⎡ ⎞ ⎟⎢⎣ ⎠

49. Answer (00)

Hint : Find domains

Solution :

1

1 1, 0

2 2f

D⎛ ⎤ ⎜ ⎥⎝ ⎦

2[ 3, 2]

fD

50. Answer (04)

Hint : Periodic function, draw graph

Solution :

f(x) = sin–1|cosx| + cos–1|sinx|

Here f(x) is periodic function of period .

f(x) =

1 1

1 1

sin (cos ) cos (sin ), 02

sin (cos ) cos (sin ),2

x x x

x x x

⎧ ⎪⎪⎨ ⎪ ⎪⎩

2 , 02

( )

2 ,2

x x

f x

x x

⎧ ⎪⎪ ⎨ ⎪ ⎪⎩

Range of the function is [0, ]

� � �

51. Answer (C)

Hint : (A + B)(A–1 + B–1) = I

Solution :

∵ (A + B)–1 = A–1 + B–1

(A + B)(A–1 + B–1) = I

I + AB–1 + BA–1 = O ...(1)

multiply eq. (1) with matrix A; we get

A + AB–1A + B = O ...(2)

multiply eq. (1) with matrix B; we get

B + A + BA–1B = O ...(3)

from eq. (2) and (3): AB–1A = BA–1B

52. Answer (B)

Hint : AB – A – B + I = I

Solution :

∵ A + B = AB AB – A – B + I = I

(I – A)(I – B) = I

(I – B)(I – A) = I ∵ (inverse are commutative)

I – B – A + BA = I

A + B = BA

AB = A + B = BA

53. Answer (A)

Hint : tan–1x + tan–1y = + 1

tan1

⎛ ⎞⎜ ⎟⎝ ⎠

x y

xy

,

x > 0, y > 0, xy > 1

Solution :

Here x1 = 2 and x

2 = 10

then use these to get the solution

54. Answer (D)

Hint : cos–1x is decreasing function

Solution :

Here x1 = 2 and x

2 = 10

then use to get the solution

Test - 1A (Paper - 1) (Code-B) (Answers) All India Aakash Test Series for JEE (Advanced)-2019

1/12

PHYSICS

1. (A, C)

2. (A, C)

3. (B, D)

4. (A, C)

5. (A, B, D)

6. (A, C)

7. (20)

8. (72)

9. (27)

10. (64)

11. (32)

12. (16)

13. (30)

14. (15)

15. (C)

16. (C)

17. (A)

18. (A)

Test Date : 05/08/2018

ANSWERS

TEST - 1A (Paper-1) - Code-B

All India Aakash Test Series for JEE (Advanced)-2019

CHEMISTRY

19. (B)

20. (A, B, C, D)

21. (A, C, D)

22. (A, B, D)

23. (A, B, C, D)

24. (A, B, C)

25. (09)

26. (10)

27. (10)

28. (06)

29. (28)

30. (64)

31. (02)

32. (35)

33. (B)

34. (D)

35. (C)

36. (A)

MATHEMATICS

37. (A, B, C)

38. (B, C, D)

39. (A, B, C)

40. (A, B, D)

41. (B, C, D)

42. (D)

43. (04)

44. (00)

45. (02)

46. (21)

47. (08)

48. (03)

49. (33)

50. (15)

51. (C)

52. (B)

53. (A)

54. (D)

All India Aakash Test Series for JEE (Advanced)-2019 Test - 1A (Paper - 1) (Code-B) (Hints & Solutions)

2/12

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (A, C)

Hint : Find initial charge and final charge.

Solution :

R2

2V0

C

CV q0 +

– – CV q0

–q

+q

0

02

q CV qV

C C

0

2

CVq

2

0bW CV

2

0

1

2i

U CV

2 2

0 091

2 4 4f

V VU C

⎛ ⎞ ⎜ ⎟

⎝ ⎠

Heat dissipated =

2

0

4

CV

2. Answer (A, C)

Hint : Potential difference across capacitor is V0

Solution :

03

kQ kQV

a a

03

4

aVQ

k

Wb = QV

0

3. Answer (B, D)

Hint : Apply Ohm’s law, J = E

Solution :

Current density remains same

1E

1 =

2E

2

Apply gauss

2 1

0

( )ds

E E ds

2 1 0

I I Q⎛ ⎞ ⎜ ⎟ ⎝ ⎠

0

2 1

1 1Q I

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

4. Answer (A, C)

Hint : Use concept of potential due to uniformly

charged disc.

Solution :

04

xd dxdV

x

2

0/2

04

l

d dx

dV

∫ ∫

2

2

0

2 cos

4

r d

dV

∫

0

rV

00

2

Rr

dU rdr ∫

2 3

0

2

3

RU

5. Answer (A, B, D)

Hint : Use concept of field inside cavity.

Field at respective points can be considered

superposition of field due to sphere + sphere of

density (–, cavity) + solution

Test - 1A (Paper - 1) (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019

3/12

Solution :

sphere cavityAE E E � � �

0

06

A

RE i

��

0 0 0

17

3 54 54B

R R i RE i

��

6. Answer (A, C)

Hint : At equilibrium net force on each charge is zero.

Solution :

T

T

2 2

2 2 2

16 642 cos cos

4 cos

q qT k

l l

⎛ ⎞ ⎜ ⎟

⎝ ⎠

2 2

2 2 2

162 sin sin

4 sin

q qT k

l l

⎛ ⎞ ⎜ ⎟

⎝ ⎠

3 1tan

64

1 1tan

4

⎛ ⎞ ⎜ ⎟⎝ ⎠

7. Answer (20)

Hint : Points C, F, G and D, E, H are at same

potential.

Solution :

R

RR

R

R

R

R

R

R

A B

eq

1 1 1 1 1

3 3 3R r r r r

eq

2

rR

8. Answer (72)

Hint : Apply concept of potential gradient

Solution :

G

E0

Eeqreq

R

eq 1 2

eq 1 2

E E E

r r r

Eeq

= 8 V

req

= 2

Switch S is open

E = 8 V

Switch S is closed

E = 6 V

2

18MN

9. Answer (27)

Hint : Find potential across capacitor before and after

switching.

Solution :

Just before switching

02

3C

VV

After switching

0 0 22 4

19 9

t

CRC

V VV e

At t = 2RC ln2


4/12

04

9C

VV

0

0

4

9R

VV V

05

9

V

x = 27

10. Answer (64)

Hint : Based on potential energy of dipole in electric

field.

Solution :

Since there is pure rolling

Wfriction

= 0

K = –U

4

4

2 cosi

p Rd R

∫

22 2

ip R

22 2

fp R

pi

pf

/2

U = pE0cos45° + pE

0cos45°

2

02 2 2R E

= –4R2E0

2 2 2

0

12 4

2MR R E

02

E

m

11. Answer (32)

Hint : Prove that force is proportional to negative of

displacement.

Solution :

y

2 20

2

2

q yF

a y

For y << a

2

0

qyF

a

2

02

maT

q

12. Answer (16)

Hint : Find velocity in the direction of electric field.

Solution :

02

,x x

qVl V t V

m

y

qEV dt

m ∫

qat dt

m ∫

2

2y

q tV a

m

2

22

y

x

q lV a

m V

2

02 2

qal m

m qV

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

04

y

alV

V

2

3

0

tan4 2

y

x

V al m

V qV

13. Answer (30)

Hint : Find distribution of current in circuit

Solution :

R D

2R x

A

B


5/12

Let RAB

= x

2

2

Rxx R

R x

2Rx + x2 = 2Rx + 2R2 + xR

x2 – xR = 2R2

2 23

2 2

R Rx

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x = 2R

12060 V

2BD

V

After every section voltage gets halved

VBC

= 30 V

14. Answer (15)

Hint : Current through any section would be same.

Solution :

3x x

r

J = E

J2(r + x)l = constant

32 1 constant

xr x E x

r

⎛ ⎞ ⎜ ⎟⎝ ⎠

0

3 31 1

2 2E r E r

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

E = 0

15

4E

= 3.75 E0

15. Answer (C)

16. Answer (C)

Hint and solution of Q. Nos. 15 and 16

Hint : Apply Gauss theorem to find field.

Solution :

= 01

r

R

⎛ ⎞ ⎜ ⎟⎝ ⎠

Q = 4r2dr

=2

0

0

4 1

Rr

r drR

⎛ ⎞ ⎜ ⎟⎝ ⎠∫

=

3 4

04

3 4

r r

R

⎛ ⎞ ⎜ ⎟

⎝ ⎠

=

3 3

0 04

12 3

R R

3 42

04 4

3 4

r rE r

R

⎛ ⎞ ⎜ ⎟

⎝ ⎠

0 31

3 4

r rE

R

⎛ ⎞ ⎜ ⎟⎝ ⎠

0dE

dr

0 02

03 4

r

R

2

3

Rr

17. Answer (A)

18. Answer (A)

Hint and Solution of Q. Nos. 17 and 18

Hint : For maximum power transfer. External

resistance must be equal to resistance of network.

Solution :

Circuit can be redrawn to

eq

RC

+

–

2

eq 60 20

2 4 4

eq

= 40 V

R = req

= 2

t

CV A Be

= Req

C

= C

t

CC

V A Be

At t = 0

VC = A + B

At t =

VC = A = 20 V

VC = 20 – 50e–t/


6/12

19. Answer (B)

Hint :

2

o

2

0.059 [Ag ]E E log

2 [Zn]

Solution :

2

o 1

2

0.1

V0.059E E log

2 0.01

V

⎛ ⎞⎜ ⎟⎝ ⎠ ⎛ ⎞⎜ ⎟⎝ ⎠

o 2

2

1

V0.059E E log

2 V

o 2

2 1E E , V V

(1 + x) > (1 + y)2

V1 volume of solution of cathode chamber.

V2 volume of solution of Anode chamber.


Hint : For spontaneous cell

Ecell

> 0

Solution :

[Ag+]cathode

> [Ag+]Anode

If [Cl–], [Ag+]

o o

spCl /AgCl/Ag Ag /AgE E 0.059 logK

so o

Ag /AgE act as cathode

21. Answer (A, C, D)

Hint : If more H+ is formed pH decreases, but if more

OH– is formed pH increases.

Solution :

Electrolysis

(P) Cathode

2Cu 2e Cu(s)

Anode

2 2

1H O O 2H 2e

2

pH

(Q) Cathode

Ag 1e Ag(s)

Anode

2 2

1H O O 2H 2e

2

pH

(R) Cathode

2 22H O 2e H 2OH

Anode

2

1Cl Cl 1e

2

pH

(S) Cathode

2 22H O 2e H 2OH

Anode

2 2

1H O O 2H 2e

2

pH remain same.


Hint : Boiling point of A is more than B it means B

is more volatile

Solution :

In graph, upper line represent the vapour line while

lower line show liquid line. In between these line

vapour and liquid phase exist.


Hint : For immiscible liquid

2

o o o o

A B A H Op p p 755 p p ⇒

Solution :

o

Ap 755 700 55 torr

Mass of A in distillate = 55 2 = 110 g.

Mass of H2O in distillate = 145 1 = 145 g.

For immiscible liquids, from Dalton’s law

2 2 2

o

A A A

o

H O H O H O

w p M

w p M

A

110 18 700M 173.8 g

145 55

PART - II (CHEMISTRY)


7/12


Hint : If T increases Koverall

decreases (as given)

we know, as T increase, K2 increasebut K

eq may

increase or decrease depending on type of reaction.

Since overall rate decreases on increasing

temperature, Keq

must be dominating over K2 and the

reversible reaction must be exothermic.

As 1

1

KT increase, decrease

K

So, K1 must have been greater than K

–1 as K

–1

increases more.

Solution :

Rate = K2[A][B

2] from R.D.S

1

1 2

2

1

K [B ]

K [B]

Keq

[B]2 = [B2]

Rate = K2K

eq[B]2[A]1

Rate = 2 1

1

K K

K

[B]2[A]1

Let 2 1

1

K KK

K

2 1 1(E E E )

RTK e

E2 + E

1 – E

–1 < 0

E–1

– E1 – E

2 > 0

25. Answer (09)

Hint : Rate =

1

32

1 2

2 3

4

KK [CH CHO]

2K

⎛ ⎞⎜ ⎟⎝ ⎠

Solution :

Rate =

1

32

1 2

2 3

4

KK [CH CHO]

2K

⎛ ⎞⎜ ⎟⎝ ⎠

so,

K =

1

a 2

a

1

E 2E E RT

1RT RT

2 E

RT

4

A eAe A e

2A e

⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠

A =

1

2

1

2

4

AA

2A

⎛ ⎞⎜ ⎟⎝ ⎠

Rate of reaction:

4

2 3 3

d[CH ]K [CH ][CH CHO]

dt

i

Now, by steady state approximation

3

1 3 2 3 3

d[CH ]0 K [CH CHO] K [CH ][CH CHO]

dt

i

i

2

3 2 4 3K [CH CHO] 2K [CH ]

i i

...(i)

2

2 3 3 3 2

d[CH CHO]0 K [CH ][CH CHO] K [CH CHO]

dt

i

i i

...(ii)

so,

1

2

1

3 3

4

K[CH ] [CH CHO] {from (i) and (ii)}

2K

⎛ ⎞ ⎜ ⎟⎝ ⎠

i

1

2

4 1

2 3 3

4

1

32

1 2

2 3

4

d(CH ) KK [CH CHO] [CH CHO]

dt 2K

KK [CH CHO]

2K

⎡ ⎤ ⎢ ⎥

⎣ ⎦

⎛ ⎞ ⎜ ⎟

⎝ ⎠

so

1

2

1

2

4

KK K

2K

⎛ ⎞ ⎜ ⎟

⎝ ⎠

1

2

1

2

4

AA A

2A

⎛ ⎞ ⎜ ⎟

⎝ ⎠

1

10 26

3

6 3 9

10A 10

2 5 10

10 10 10 .

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

26. Answer (10)

Hint : 1

2

k2

k

Solution :

1

2

k[C]

[D] k

[C]2

[D]

so, 2k2 = k

1


8/12

27. Answer (10)

Hint : pA x

A

Solution :

At sea level

2 2N O

p 0.8, p 0.2

at height H, P = 0.4 atm

Mole fraction of N2 and O

2 remain in ratio of 4 : 1 at

any height.

2

2

N

O

p 0.4 0.8 0.32 atm

p 0.4 0.2 0.08 atm

x = h

0.8

k

x = h

0.32

k

x

2.5x

andy

2.5y

x y2

x y

⎛ ⎞⎜ ⎟ ⎝ ⎠

= 10

28. Answer (06)

Hint : Mass

DensityVolume

Solution :

Density = 1.315 103 kg m–3

Mass = ?

Volume = 13 7 3 10–27 m3

Let x molecule of insulin be present in one unit cell.

Total mass of insulin = 3

A

x35.8995 kg/m

N

Total volume = 2.73 10–25 m3

Density = 1.315 103 = A

25

x35.8995

N

2.73 10

x = 6

29. Answer (28)

Hint : x = +2, y = +3, A = 2

Solution :

For electrical neutrality

total positive charge = total negative charge

41 x y 8

A

= i C R T

9.6 = i 0.1 0.08 300

i = 4

so y = 3 so possible value of A = 1, 2, 3

from eq.

If A = 1

x + 4 3 = 8 x = –4 (Not possible)

If A = 2

x 2 3 = 8

x = +2

so y x

3A

(possible because odd number)

If A = 3

x + 4 = 8

x = 4

then y x 3 4

4A 4

(Not possible because even)

so possible value of A = 2

hence, A2 + 4xy

= 4 + 4 × 2 × 3

= 28

30. Answer (64)

Hint : H.C.P. contain 20 T.H site

Solution :

In H.C.P., 8 T.V are inside the unit cell while 12 are

at vertical edges (2 on each vertical edge)

T.V. = 1

8 12 123

One edge in H.C.P. contributes 1

3.


9/12

31. Answer (02)

Hint : C4-axis

Solution :

C4-axis contains opposite face centre and body centre

C4-axis pass through opposite face centres and body

centre. If 2 face centre atom and one body centre

atom is removed then formula becomes Na3Cl

3.

32. Answer (35)

Hint : x = 3, y = 2, so x3 + y3 = 35

Solution :

2 2 2

1N O(g) N (g) O (g)

2

i

i

P1K ln

t 3P 2P

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 2 2

1N O(g) N (g) O (g)

2

Pi

0 0

Pi – p p

p

2

P = Pi +

p

2

2P – 2Pi = p

2N O

p at t = Pi – (2P – 2P

i)

= 3Pi – 2P

K = i

i

P1ln

t 3P 2P

⎛ ⎞⎜ ⎟⎝ ⎠

33. Answer (B)

Hint : Basic dye stuff like methylene blue are

positively charged.

Solution :

If water moves towards anode it means water

contains negative charge so colloidal sol must

contain positive charge.

34. Answer (D)

Hint : Greater the magnitude of charge on

coagulating ion, greater is its power.

Solution :

Water moves towards negative electrode it means it

is positively charged and sol is negatively charged.

35. Answer (C)

Hint : d[intermediate]

0dt

Solution :

By steady state approximation

1 1 2 2

d[SH ]0 K [S][AH ] K [A][SH ] K [SH ][H O]

dt

1

2 2 1

K [S][AH ][SH ]

K [H O] K [A]

2 1 2

2 2

1 2 2

K K [S][AH ][H O]d[P]Rate K [H O][SH ]

dt K [A] K [H O]

if K2

>> K–1

[A] Rate = K1[S][AH+]

if K2[H

2O] << K

–1[A] Rate =

1 2 2

1

K K [S][H ][H O]

K K

[AH ] [H ]As

[A] K

⎡ ⎤⎢ ⎥⎣ ⎦

36. Answer (A)

Hint : app H

K K [H ]

Solution :

app H HlogK logK logH logK pH

So, Kapp

= expH

(logK pH)

log K app

pH

Kapp

pH

(A)


10/12

PART - III (MATHEMATICS)


Hint : f(x) 0

Solution :

∵ x2 – 4 0 x2 – 1 > 0

(x2 – 1)f(x) = 2

4 | ( ) |x f x

(x2 – 1) f(x) 0 f(x) 0.

|f(x)| = f(x)

(x2 – 1)f(x) = 2

4 ( )x f x

f(x) =

2

2

4

2

x

x

Domain of f(x) = (–, –2] [2, )

The graph of f(x) is

–2

2


Hint : sec 8 < 0

Solution :

l = sec2(sec–1 5 ) + cosec2(cosec–1 10 )

l = 15.

And 1 1| sec 8 | 1 1 cos 8

tan tan 4tan 8 sin 8

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

m = – ( – 4) 4m

lm = 15 4 = 60

(B), (C) and (D) options are correct.


Hint : Break the function

Solution :

Here

2

2

2 , 0( )

2 , 0

⎧ ⎨

⎩

x x xf x

x x x

The graph of g(x) is given as

–2 –1 O 1 2 3

–2

–1

1

2

3

2

2

2 , 2 1

1, 1 0( )

0, 0 2

2 , 2 3

x x x

xg x

x

x x x

⎧ ⎪ ⎪ ⎨

⎪⎪ ⎩


Hint : f(x) is periodic with period 3

Solution :

∵

3, 3 0

, 0 3( ) 3

3, 3 6

x x

x xf x x r

x x

⎧ ⎪

⎪ ⎨ ⎪

⎪⎩ �

f(x) is periodic with period 3.

Range of f(x) is [0, 3 )

f(x) = | |

4

x has exactly four real solutions in (–6, 6).


Hint : det(kA) = kn det(A)

Solution :

∵ |A| = 1


11/12

then |–A| = (–1)n|A| = 1 or –1.

and det( ) (det )

det

n

n n

kA k AA

k k .

42. Answer (D)

Hint : Solve equations

Solution :

2 42∵ x yz , y2zx = 34 and z2xy = 44

xyz = 24

A = {1, 2, 3, 4, 6, 8, 12, 24}.

and B = {2, 3, 4, 11}

Number of possible relations from A to B is 232.

43. Answer (04)

Hint : Periodic function, draw graph

Solution :

f(x) = sin–1|cosx| + cos–1|sinx|

Here f(x) is periodic function of period .

f(x) =

1 1

1 1

sin (cos ) cos (sin ), 02

sin (cos ) cos (sin ),2

x x x

x x x

⎧ ⎪⎪⎨ ⎪ ⎪⎩

2 , 02

( )

2 ,2

x x

f x

x x

⎧ ⎪⎪ ⎨ ⎪ ⎪⎩

Range of the function is [0, ]

44. Answer (00)

Hint : Find domains

Solution :

1

1 1, 0

2 2f

D⎛ ⎤ ⎜ ⎥⎝ ⎦

2[ 3, 2]

fD

45. Answer (02)

Hint : 2 26

1 23

x

Solution :

For domain of log(15–x)

(x – 2) x – 2 > 0, 15 – x > 0

x (2, 15) – {14} ...(1)

For domain of 1 2 26

cos then3

⎡ ⎤⎢ ⎥⎣ ⎦

x

2 261 2

3

3 2 26 6

2316

2

x

x

x

Domain of f(x) = 23

,15 {14}2

⎡ ⎞ ⎟⎢⎣ ⎠

46. Answer (21)

Hint : trace of 1

11 22 33

1 1 1( ) ...

nA

a a a

Solution :

∵

1 1 1

3 2 2

3

0

0 0

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

a b c

A b c

c

then

1 2

1

1

3

2

3

1

10

10 0

k ka

A kb

c

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

3

1

1

1trace of ( )

10n

n

A

⎛ ⎞⎜ ⎟⎝ ⎠∑

1 1 1 1 1

1 2 3 4 5

1(tr( ) tr( ) tr( ) tr( ) tr( ))

10

1(5 2 4 8 3 18 2 32 50)

10

21

A A A A A

47. Answer (08)

Hint : Make homogeneous type system.

Solution :

(x4 – 1) + (y4 – 1) – (z4 – 1) = 0

(x4 – 1) + (y4 – 1) + (z4 – 1) = 0

(x4 – 1) – (y4 – 1) + (z4 – 1) = 0


12/12

2 2

1 2 3

1 1 1

1 1 1 1 1 3 0

1 1

0

4 4 41 1 1 0

1, 1, 1

x y z

x y z

⇒

Number of solutions is 8.

48. Answer (03)

Hint : Draw graph of functions

Solution :

Draw the graphs of 3cos–1x and 1

{ }xthey intersect

each other at three distinct points.

3 real solution are possible

49. Answer (33)

Hint : sin–1(sin7) = 7 – 2

Solution :

f(x + 2) = f(x – 1)

f(x) = f(x + 3)

Period of f(x) is 3.

and

2

1

2

4 13( 2 sin (sin 7))

3

⎛ ⎞⎡ ⎤ ⎜ ⎟⎢ ⎥⎜ ⎟⎣ ⎦⎝ ⎠

xg x g x

x

g(x + 4) = g(x – 7)

g(x) = g(x + 11)

Period of g(x) is 11.

Period of h(x) is 33.

50. Answer (15)

Hint : Property of co-factors

Solution :

, 3

3 | | 3 5 15.ij ij

i i j

a c A

∑

� � �

51. Answer (C)

Hint : (A + B)(A–1 + B–1) = I

Solution :

∵ (A + B)–1 = A–1 + B–1

(A + B)(A–1 + B–1) = I

I + AB–1 + BA–1 = O ...(1)

multiply eq. (1) with matrix A; we get

A + AB–1A + B = O ...(2)

multiply eq. (1) with matrix B; we get

B + A + BA–1B = O ...(3)

from eq. (2) and (3): AB–1A = BA–1B

52. Answer (B)

Hint : AB – A – B + I = I

Solution :

∵ A + B = AB AB – A – B + I = I

(I – A)(I – B) = I

(I – B)(I – A) = I ∵ (inverse are commutative)

I – B – A + BA = I

A + B = BA

AB = A + B = BA

53. Answer (A)

Hint : tan–1x + tan–1y = + 1

tan1

⎛ ⎞⎜ ⎟⎝ ⎠

x y

xy

,

x > 0, y > 0, xy > 1

Solution :

Here x1 = 2 and x

2 = 10

then use these to get the solution

54. Answer (D)

Hint : cos–1x is decreasing function

Solution :

Here x1 = 2 and x

2 = 10

then use to get the solution

test - 1a (paper - 1) (code-a) (answers) all india aakash ......test - 1a (paper - 1) (code-a)...

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