Test - 2A (Paper - 2) (Code-E) (Answers) All India Aakash Test … · 2019-06-04 · All India...

Test - 2A (Paper - 2) (Code-E) (Answers) All India Aakash Test Series for JEE (Advanced)-2019

PHYSICS

1. (A, C)

2. (A, B, C)

3. (B, C)

4. (A, B, D)

5. (A, C)

6. (A)

7. (C)

8. (D)

9. (C)

10. (B)

11. (A)

12. (C)

13. (C)

14. (A)

15. (C)

16. A (R, T)

B(R, T)

C(Q, S)

D(P, R)

17. A(R, S)

B (T)

C(P, S)

D(Q, S)

18. (05)

19. (09)

20. (02)

CHEMISTRY

21. (A, C)

22. (A, B, C, D)

23. (A, B, C)

24. (B, D)

25. (A, C)

26. (D)

27. (A)

28. (D)

29. (B)

30. (C)

31. (C)

32. (C)

33. (C)

34. (B)

35. (A)

36. A(P, S)

B(P, S)

C(P, Q, T)

D(P, S)

37. A(P, Q)

B(R, S)

C(T)

D(P)

38. (04)

39. (09)

40. (07)

MATHEMATICS

41. (A, C)

42. (C, D)

43. (B, C, D)

44. (B, C, D)

45. (A, C)

46. (A)

47. (C)

48. (A)

49. (C)

50. (D)

51. (D)

52. (A)

53. (A)

54. (B)

55. (C)

56. A(Q, S)

B(Q, S, T)

C(P, R)

D(Q, R, T)

57. A(P)

B(Q)

C(P, T)

D(P, Q, R)

58. (04)

59. (08)

60. (08)

Test Date : 25/11/2018

ANSWERS

TEST - 2A (Paper-2) - Code-E

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

1/9

All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper - 2) (Code-E) (Hints & Solutions)

2/9

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (A, C)

Hint: x

y f tv

⎛ ⎞ ⎜ ⎟⎝ ⎠

v is wave speed

Solution:

2

0.8

4 – 42

yx

t

⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

v = 2 m/s

m

0.80.2 m

4 y

2. Answer (A, B, C)

Hint: PV2 = C

Solution:

3 3– –

2 2 –1 2 2

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

pr

R R R RC R

TC

V

0

22

TT

⎛ ⎞ ⎜ ⎟⎝ ⎠

dU = 0

3(decrement)

4RT

01

2 2

TRQ

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

.... rejected by gas

Q = U + W

W = Q + U 01

1 2

TR ⎛ ⎞ ⎜ ⎟⎝ ⎠

0

2

RTW

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

3. Answer (B, C)

Hint: Req

= 20

402 A

20 I

Solution:

1 2

1A 1 A

4 2 2 I I

I I=

Radius of V3 = 20 volt

4. Answer (A, B, D)

Hint: Q = CV

0⎛ ⎞ ⎜ ⎟

⎝ ⎠

KAC

d

Solution:

1

11–⎛ ⎞ ⎜ ⎟⎝ ⎠

Q QK

0

4

KbL

d

0 0

0

11–

4

⎛ ⎞ ⎜ ⎟⎝ ⎠

Lb VV

K d

0

2

2

3

bL K

Qd

0 0

0

11–

4

V b LV

d

⎛ ⎞ ⎜ ⎟⎝ ⎠

0

1 2

VE E

d

5. Answer (A, C)

Hint: 1 2

1 2

r

rE E E

r r r

Solution:

If current in ammeter is i2 and current in voltmeter is i

1

then

10 = 11i1 + i

2and 80 = 10i

2 + i

2

it gives i1 =

20

109 A and i

2 =

870

109 A

Voltage across voltmeter V = 20 10 200

volt109 109

V = 200

volt109

and current in ammeter = 870

109 A

6. Answer (A)

7. Answer (C)

8. Answer (D)

Hint & Solution of Question Nos. 6 to 8

Hint: (VA – V

B) = 4(V

P – V

Q)

2

12

⎛ ⎞⎜ ⎟⎝ ⎠

RR

2

4 4

Test - 2A (Paper - 2) (Code-E) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019

3/9

Solution:

Rr = 2 + 2 = 4

9. Answer (C)

10. Answer (B)

11. Answer (A)

Solution of Question Nos. 9 to 11

P = 5

0

0

10 (2 )Kx

P xA

dw = 5 3

010 8 10 2PA dx x dx

∫ ∫

w =

0.12

0

2 1800 2 800 164 J

2 10 200

xx

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

Tf =

4 40

4 40 0

(20 1) 10 (32 10 ) 200

20 10 24 10

f f

PV T

PV

Tf = 280 K

T = 80 K

And U = NCv T =

0 0

0

380

2

⎛ ⎞ ⎜ ⎟

⎝ ⎠

PVR

RT

U =

4 420 10 24 10 3

80 288J200 2

RR

12. Answer (C)

Hint: T2 > T

1

U = nCv T

Solution:

TB > T

A

Q = U + W

Q1 = U

1 + W

1

Q2 = U

2 + W

2

W2 >W

1

13. Answer (C)

Hint: For maximum intensity path difference (x) = n

Solution:

x = 20 – 12 = 8

m

= 8 m

14. Answer (A)

Hint: ·V E r

��

Solution:

0 0

0 0

ˆ ˆ16 123 3

E OO i j⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

��

0

0

ˆ ˆ ˆ ˆ– 16 12 · – – 53

P QV V i j i j

⎛ ⎞ ⎜ ⎟⎝ ⎠

0 0

0 0

7616 60

3 3

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

15. Answer (C)

Hint: E inside shell is zero

Solution:

Vp =

2 2

kQ kQ kQ

R R R

16. Answer A(R, T); B(R, T); C(Q, S), D(P, R)

Hint : Principle of superposition

Solution :

14 2 sin 100 – 2y t x

24 2 cos 100 – 2y t x

(y1 + y

2) = y

R

8sin 100 – 24

t x⎛ ⎞ ⎜ ⎟

⎝ ⎠

yR

= (y1 + y

2 + y

3)

17. Answer A(R, S); B(T), C(P, S); D(Q, S)

Hint : VA – V

B =

B

AE dl∫

��

Solution :

(VA – V

B) =

2

10

4

∫

r

r

drr

= 2

0 1

ln4

r

r

⎛ ⎞⎜ ⎟ ⎝ ⎠

18. Answer (05)

Hint: 1 2

1 2

r

eq

E E E

r r r in parallel combination

Solution:

1 2

1 2

r

r

E E E

r r r

Er = 5 unit and r

eq =

2

3

1

5 152 A

2 175

3

I


4/9

PART - II (CHEMISTRY)

19. Answer (09)

Hint: If PVn = C C

Pr = C

V +

1

R

nSolution:

PT = constant PV1/2 = C

5 9–12 2– 1

2

⎛ ⎞ ⎜ ⎟⎝ ⎠

pr

R R RC

n = 9

20. Answer (02)

Hint: B.F. =1 2

| |f f

Solution:

T2 = T

1 + 4

Hence, f2 = 404 Hz

21. Answer (A, C)

Hint : To find the total spin, we can add spin of each

atom.

Solution :

Ortho hydrogen � para hydrogen

Total spin of ortho hydrogen is 1 and para hydrogen is 0.

At lower temperature para hydrogen is present in

greater concentration.

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

Hint : Dut to inert pair effect, Bi+5 is not stable.

Solution :

Melting point of Bi is low due to metallic behaviour.

23. Answer (A, B, C)

Hint : Factual.

Cu2S

Roasting

n limited i supply airof

Cu2S + Cu

2O

Heating in

absence of air

Cu + SO2

24. Answer (B, D)

Hint : M–L bond is stronger if L is a -acid ligand.

Solution :

Due to Pi-acid behaviour of –NO2, Co–Cl* bond would

be weaker than Co–Cl and the attack would take place

from the back side of the Co–Cl* bond, hence

pentagonal bipyramidal intermediate is formed.

25. Answer (A, C)

Hint : Surface tension first decreases and then

increases with the increase in concentration of the

surfactant.

Solution :

Correct curve for surface tension is

C

26. Answer (D)

Hint : Fact

Solution :

Observations from the curve.

27. Answer (A)

Hint : As given in paragraph, conversion of h.s. to l.s.

shrink the metal ligand bond.

Solution :

For l.s., 0 increase as r decreases.

Curve ‘1’ will be correct.

28. Answer (D)

Hint : Field strength of H2O, that does not cause

pairing in Mn+2.

Solution :

H2O is WFL for Mn2+ therefore, Mn2+ in H

2O form h.s.

complex and CFSE is zero for this system.

29. Answer (B)

Hint : Factual.

Solution :

FeSO + H O Fe(OH)SO 4 2 2 4

yellow


5/9

30. Answer (C)

Hint : Factual.

Solution :

FeSO + NaOH Fe(OH) 4 2

Fe(OH)

Brown 3

White green

31. Answer (C)

Hint : Factual.

Solution :

Mn(NO ) + NaOH Mn(OH) 3 2 2

MnMnO

(Black)

3

White

32. Answer (C)

Hint : Stability of isotope of P.

Solution :

White P is less stable.

33. Answer (C)

Hint : Properties of fullerene.

Solution :

All C atoms are sp2 in fullerene.

34. Answer (B)

Hint : Properties of group-13.

Solution : B > Al > Tl > In > Ga

35. Answer (A)

Hint : Properties of sodium.

Solution: Na forms peroxide.

CsH is the most ionic.

36. Answer A(P, S); B(P, S); C(P, Q, T); D(P, S)

Hint : Reaction of metals with HNO3.

Solution :

R = NH4NO

3.

P1 = N

2O

P2 = N

2O

P3 = NO

2

P4 = N

2O

4

P5 = NO

37. Answer A(P, Q); B(R, S); C(T); D(P)

Hint :

Factual

38. Answer (04)

Hint : Kdesorption

a–E /RT

= A e

Solution :

Life time (y) = a

E /RT

1

Ae

= 10–13

15 1000 3exp

25 300

⎛ ⎞⎜ ⎟⎝ ⎠

= 10 –13 × 400

y = 4 × 10–11 s

39. Answer (09)

Hint : Packing type of lattice.

Solution :

Cu, Ag, Au all have CCP structure, CN = 12

x + y + z = 12 + 12 + 12 = 36

40. Answer (07)

Hint : Electronic configuration of lanthanoids.

Solution :

x = 1

y = 1

z = 7


6/9

PART - III (MATHEMATICS)

41. Answer (A, C)

Hint : Find domain

Solution : Domain of f(x) is x = 1 f(x) = 0

Domain of g(x) is x = –1 g(x) = 3

2

42. Answer (C, D)

Hint : Concept of director circle.

Solution :

O

y

(4, 3)x

y

(3, 4)

xO

y

( , )h k

C

Ox

Locus of C is circle x2 + y2 = a2 + b2 = 25

Hence C moves in a circle of radius 5

Distance covered by C in the motion is equal to

1 14 35 tan tan

3 4

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

1 75tan

24

⎛ ⎞ ⎜ ⎟⎝ ⎠

43. Answer (B, C, D)

Hint : Parallel lines are neither concurrent nor forms

a triangle.

Solution : Lines are concurrent if = 0

2

1 1 1

0( 1) 7 5

2 2 5 0

n n

n n

2 3

2

10 25 5 10 2 2 5 5

7 2 14 0

⇒

n n n n n n

n n

n3 – 4n

2 + 5n – 6 = 0

(n – 3)(n2 – n + 2) = 0 n = 3

but for n = 3, lines become parallel

so these are concurrent for no real value of n.


Hint : Equation of normal y = mx – 2am – am3

Solution : Equation of normal to y2 = 4x

is y = mx – 2m – m3 ...(1)

Equation of normal to y2 = (x – k) is

3

2 4

m my mx km ...(2)

Equations (1) & (2) represent same normal

3

3

1 2

1

2 4

m m m

m m mkm

3m2 = 4k – 6 4k – 6 0

3

2k

45. Answer (A, C)

Hint : 1 1

sin cos2

x x

Solution : Domain of f(x) is [–3, 3]

maximum value of f(x) is

3 397

at 38 4

x

and minimum value of f(x) is

3 311

at 38 4

x

46. Answer (A)

Hint : 1

2r (distance between parallel lines)


7/9

47. Answer (C)

Hint : L4 are concurrent at (12, 12)

48. Answer (A)

Hint : For cyclic quadrilateral L3 and L

4 are parallel

Solution of Q. Nos. 46 to 48

A(2, 7) 4 – 3 + 13 = 0x y

4 – 3 – 37 = 0x yB(10, 1)

(6, 4)

(3, 0

)

(9, 8)

3 + 4 – 34 = 0x y

Centre of circle is either (3, 0) or (9, 8)

+ = 3 or + = 17

(x + y – 24) + (x – y) = 0 are concurrent at (12, 12)

y

x

3 + 4 – 34 = 0x y

4 –

3 +

13 = 0

x

y

4 –

3 – 3

7 =

0

x

y(12, 12)

(2, 7)

(10, 1)

O

1

1m

should not belong to 1 11,

2 2

⎡ ⎤⎢ ⎥⎣ ⎦

= –2, –1, 0, 1

For cyclic quadrilateral

1 3 1

1 4 7

⇒

C (9, 8), r = 5

(x – 9)2 + (y – 8)2 = 52

49. Answer (C)

Hint : Take parametric point on hyperbola.

50. Answer (D)

Hint : Use parametric equation of line.

51. Answer (D)

Hint : 2

( 2) and 4y m x xym

intersect at

exactly one point.


Let a variable point on xy = 4 is 2

2 ,P tt

⎛ ⎞⎜ ⎟⎝ ⎠

Equation of tangent is 2

2 8x y t

t

4x

tyt

4(4 , 0), 0,A t B

t

⎛ ⎞⎜ ⎟⎝ ⎠

Circumcentre is midpoint of AB

2

2 ,h t kt

hk = 4 locus of (h, k) is xy = 4

2e R(2, 0)

r(2, 0)R

x

T

y

S(2

+ co

s,

sin

)

r

r

(2 + rcos)(rsin) = 4

r2sincos + (2sin)r – 4 = 0

1 2

4

sin cosr r

RSRT = 8|cosec2|

(RSRT)min

= 8

y2 = 8(x + 2), xy = 4

y = m(x + 2) + 2

m

is also tangent to xy = 4

242 xmx m

m

⎛ ⎞ ⎜ ⎟⎝ ⎠

has equal roots

D = 0

22

4 ( 4) 12 m mm

m

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

Equation of common tangent is y = –x – 4

2m – 3c = –2 + 12 = 10


8/9

52. Answer (A)

Hint : Find mirror image of (1 + t2, 2t) about line

Solution :

Let variable point on the parabola is P(1+ t2, 2t) and

its reflection in the given line is (h, k)

2 2(1 ) 2 2(1 2 2)

1 1 2

h t k t t t

( 1)( 1) 2

12(1 )

h t t kt

k t h

⇒

2

2 22

hh k

Locus of (h, k) is 4x – 4y = x2 A + B = 0

53. Answer (A)

Hint : Image of C2 in the line, C

2, C

3 are collinear

Solution :

P(7, 5)

(–7, 3)

(1, –3) 3 + 4 – 16 = 0x y

Image of C2(1, –3) in the line 3x + 4y – 16 = 0 is

P(7, 5)

C1C

2 + C

2C

3 + C

3C

1 is minimum

C1, C

3 and P lie on same line

C3(0, 4)

3 150 5 2C C

Radius of 1

3 2C

Radius of 3

2 2C

Equation of C3 is x2 + (y – 4)2 = 8

x2 + y2 – 8y + 8 = 0 a + b + c = 0

54. Answer (B)

Hint : a = A(t 2 + 1) and b = A

2

2

(1 )t

t

Solution :

4A = 1

2

1

8A

a = A(t2+1), b =

2

2

(1 )A t

t

1 1 1

8 a b A

55. Answer (C)

Hint : f(x) = 1 – xn

Solution :

f(x) = xn + 1 or 1 – xn

f(x) = 1 – x4

f(3) = –80

56. Answer A(Q, S); B(Q, S, T); C(P, R); D(Q, R, T)

Hint : For increasing fnf (x) > 0

Solution :

f1(x) is many-one and into

f2(x) is many-one, into and even

f3(x) is one- one and onto

f4(x) is many-one, onto and even

57. Answer A(P); B(Q); C(P, T); D(P, Q, R)

Hint : Let f(x) = ax2 + bx + c

Solution :

Let f(x) = ax2 + bx + c

f (–1) = 0

f(1) = 4 and f(–1) = 2

a + b + c = 4

a – b + c = 2

–2a + b = 0

Solving, we get

1 51, ,

2 2b a c

21 5( )

2 2f x x x

21

[( 1) 4]2

x

1 1cos 1

( )f x

⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

[1 + sgn(f(x))] = 2

1 1tan 0, 1

2f x

⎡ ⎤⎛ ⎞ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

1 22cot 1, 2, 3

( )f x

⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦


9/9

58. Answer (04)

Hint : Let circle (x – r)2 + y2 = r2

Solution :

( , 0 )r

y

xr

Let the equation of circle be

(x – r)2 + y2 = r2

solving with y2 = 8x

x = 0 and x = 2r – 8

2r – 8 0

rmax

= 4

59. Answer (08)

Hint : Let 10 cos , 2sinP

Solution :

Let 10 cos , 2sinP

10 4 3

10 5e

2

2

1 2

320 (given)2 10 cos

5PF PF

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 25 1cos , sin

6 6

2 2

2 10 2 10

1 510sin 4cos10 4

6 6

d

2 2

60. Answer (08)

Hint : Tangent to circle 2

2(1 )y mx m

Solution :

Tangent to the circle is 2

2( 1)y mx m ...(1)

It passes through focus (2, 2) of xy = 2

22 2 2( 1)m m

m2 – 4m + 1 = 0

Line (1) intersects hyperbola xy = 2 at the points

given by 22

2( 1)mx m

x

or2 2

2( 1) 2 0mx m x

2

1 2 1 2

2( 1) 2,

mx x x x

m m

2

1 2

16( )x x

m

Also (y1 – y

2)2 = 16m

Length of focal chord 8

2 8mm

⎛ ⎞ ⎜ ⎟⎝ ⎠

28(1 )

2m

m

8 42 8

m

m

� � �

Test - 2A (Paper - 2) (Code-F) (Answers) All India Aakash Test Series for JEE (Advanced)-2019

PHYSICS

1. (A, C)

2. (A, B, D)

3. (B, C)

4. (A, B, C)

5. (A, C)

6. (A)

7. (C)

8. (D)

9. (C)

10. (B)

11. (A)

12. (C)

13. (A)

14. (C)

15. (C)

16. A(R, S)

B (T)

C(P, S)

D(Q, S)

17. A (R, T)

B(R, T)

C(Q, S)

D(P, R)

18. (02)

19. (09)

20. (05)

CHEMISTRY

21. (A, C)

22. (B, D)

23. (A, B, C)

24. (A, B, C, D)

25. (A, C)

26. (D)

27. (A)

28. (D)

29. (B)

30. (C)

31. (C)

32. (A)

33. (B)

34. (C)

35. (C)

36. A(P, Q)

B(R, S)

C(T)

D(P)

37. A(P, S)

B(P, S)

C(P, Q, T)

D(P, S)

38. (07)

39. (09)

40. (04)

MATHEMATICS

41. (A, C)

42. (B, C, D)

43. (B, C, D)

44. (C, D)

45. (A, C)

46. (A)

47. (C)

48. (A)

49. (C)

50. (D)

51. (D)

52. (C)

53. (B)

54. (A)

55. (A)

56. A(P)

B(Q)

C(P, T)

D(P, Q, R)

57. A(Q, S)

B(Q, S, T)

C(P, R)

D(Q, R, T)

58. (08)

59. (08)

60. (04)

Test Date : 25/11/2018

ANSWERS

TEST - 2A (Paper-2) - Code-F

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

1/9

All India Aakash Test Series for JEE (Advanced)-2019 Test - 2A (Paper - 2) (Code-F) (Hints & Solutions)

2/9

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (A, C)

Hint: 1 2

1 2

r

rE E E

r r r

Solution:

If current in ammeter is i2 and current in voltmeter is i

1

then

10 = 11i1 + i

2and 80 = 10i

2 + i

2

it gives i1 =

20

109 A and i

2 =

870

109 A

Voltage across voltmeter V = 20 10 200

volt109 109

V = 200

volt109

and current in ammeter = 870

109 A

2. Answer (A, B, D)

Hint: Q = CV

0⎛ ⎞ ⎜ ⎟

⎝ ⎠

KAC

d

Solution:

1

11–⎛ ⎞ ⎜ ⎟⎝ ⎠

Q QK

0

4

KbL

d

0 0

0

11–

4

⎛ ⎞ ⎜ ⎟⎝ ⎠

Lb VV

K d

0

2

2

3

bL K

Qd

0 0

0

11–

4

V b LV

d

⎛ ⎞ ⎜ ⎟⎝ ⎠

0

1 2

VE E

d

3. Answer (B, C)

Hint: Req

= 20

402 A

20 I

Solution:

1 2

1A 1 A

4 2 2 I I

I I=

Radius of V3 = 20 volt

4. Answer (A, B, C)

Hint: PV2 = C

Solution:

3 3– –

2 2 –1 2 2

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

pr

R R R RC R

TC

V

0

22

TT

⎛ ⎞ ⎜ ⎟⎝ ⎠

dU = 0

3(decrement)

4RT

01

2 2

TRQ

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

.... rejected by gas

Q = U + W

W = Q + U 01

1 2

TR ⎛ ⎞ ⎜ ⎟⎝ ⎠

0

2

RTW

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

5. Answer (A, C)

Hint: x

y f tv

⎛ ⎞ ⎜ ⎟⎝ ⎠

v is wave speed

Solution:

2

0.8

4 – 42

yx

t

⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

v = 2 m/s

m

0.80.2 m

4 y

6. Answer (A)

7. Answer (C)

8. Answer (D)

Hint & Solution of Question Nos. 6 to 8

Hint: (VA – V

B) = 4(V

P – V

Q)

2

12

⎛ ⎞⎜ ⎟⎝ ⎠

RR

2

4 4

Test - 2A (Paper - 2) (Code-F) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019

3/9

Solution:

Rr = 2 + 2 = 4

9. Answer (C)

10. Answer (B)

11. Answer (A)

Solution of Question Nos. 9 to 11

P = 5

0

0

10 (2 )Kx

P xA

dw = 5 3

010 8 10 2PA dx x dx

∫ ∫

w =

0.12

0

2 1800 2 800 164 J

2 10 200

xx

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

Tf =

4 40

4 40 0

(20 1) 10 (32 10 ) 200

20 10 24 10

f f

PV T

PV

Tf = 280 K

T = 80 K

And U = NCv T =

0 0

0

380

2

⎛ ⎞ ⎜ ⎟

⎝ ⎠

PVR

RT

U =

4 420 10 24 10 3

80 288J200 2

RR

12. Answer (C)

Hint: E inside shell is zero

Solution:

Vp =

2 2

kQ kQ kQ

R R R

13. Answer (A)

Hint: ·V E r

��

Solution:

0 0

0 0

ˆ ˆ16 123 3

E OO i j⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

��

0

0

ˆ ˆ ˆ ˆ– 16 12 · – – 53

P QV V i j i j

⎛ ⎞ ⎜ ⎟⎝ ⎠

0 0

0 0

7616 60

3 3

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

14. Answer (C)

Hint: For maximum intensity path difference (x) = nSolution:

x = 20 – 12 = 8

m

= 8 m

15. Answer (C)

Hint: T2 > T

1

U = nCv T

Solution:

TB > T

A

Q = U + W

Q1 = U

1 + W

1

Q2 = U

2 + W

2

W2 >W

1

16. Answer A(R, S); B(T), C(P, S); D(Q, S)

Hint : VA – V

B =

B

AE dl∫

��

Solution :

(VA – V

B) =

2

10

4

∫

r

r

drr

= 2

0 1

ln4

r

r

⎛ ⎞⎜ ⎟ ⎝ ⎠

17. Answer A(R, T); B(R, T); C(Q, S), D(P, R)

Hint : Principle of superposition

Solution :

14 2 sin 100 – 2y t x

24 2 cos 100 – 2y t x

(y1 + y

2) = y

R

8sin 100 – 24

t x⎛ ⎞ ⎜ ⎟

⎝ ⎠

yR

= (y1 + y

2 + y

3)

18. Answer (02)

Hint: B.F. =1 2

| |f f

Solution:

T2 = T

1 + 4

Hence, f2 = 404 Hz

19. Answer (09)

Hint: If PVn = C CPr

= CV +

1

R

nSolution:

PT = constant PV1/2 = C


4/9

PART - II (CHEMISTRY)

21. Answer (A, C)

Hint : Surface tension first decreases and then

increases with the increase in concentration of the

surfactant.

Solution :

Correct curve for surface tension is

C

22. Answer (B, D)

Hint : M–L bond is stronger if L is a -acid ligand.

Solution :

Due to Pi-acid behaviour of –NO2, Co–Cl* bond would

be weaker than Co–Cl and the attack would take place

from the back side of the Co–Cl* bond, hence

pentagonal bipyramidal intermediate is formed.

23. Answer (A, B, C)

Hint : Factual.

Cu2S

Roasting

n limited i supply airof

Cu2S + Cu

2O

Heating in

absence of air

Cu + SO2

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

Hint : Dut to inert pair effect, Bi+5 is not stable.

Solution :

Melting point of Bi is low due to metallic behaviour.

25. Answer (A, C)

Hint : To find the total spin, we can add spin of each

atom.

Solution :

Ortho hydrogen � para hydrogen

Total spin of ortho hydrogen is 1 and para hydrogen is 0.

At lower temperature para hydrogen is present in

greater concentration.

26. Answer (D)

Hint : Fact

Solution :

Observations from the curve.

27. Answer (A)

Hint : As given in paragraph, conversion of h.s. to l.s.

shrink the metal ligand bond.

Solution :

For l.s., 0 increase as r decreases.

Curve ‘1’ will be correct.

28. Answer (D)

Hint : Field strength of H2O, that does not cause

pairing in Mn+2.

Solution :

H2O is WFL for Mn2+ therefore, Mn2+ in H

2O form h.s.

complex and CFSE is zero for this system.

29. Answer (B)

Hint : Factual.

5 9–12 2– 1

2

⎛ ⎞ ⎜ ⎟⎝ ⎠

pr

R R RC

n = 9

20. Answer (05)

Hint: 1 2

1 2

r

eq

E E E

r r r in parallel combination

Solution:

1 2

1 2

r

r

E E E

r r r

Er = 5 unit and r

eq =

2

3

1

5 152 A

2 175

3

I


5/9

Solution :

FeSO + H O Fe(OH)SO 4 2 2 4

yellow

30. Answer (C)

Hint : Factual.

Solution :

FeSO + NaOH Fe(OH) 4 2

Fe(OH)

Brown 3

White green

31. Answer (C)

Hint : Factual.

Solution :

Mn(NO ) + NaOH Mn(OH) 3 2 2

MnMnO

(Black)

3

White

32. Answer (A)

Hint : Properties of sodium.

Solution: Na forms peroxide.

CsH is the most ionic.

33. Answer (B)

Hint : Properties of group-13.

Solution : B > Al > Tl > In > Ga

34. Answer (C)

Hint : Properties of fullerene.

Solution :

All C atoms are sp2 in fullerene.

35. Answer (C)

Hint : Stability of isotope of P.

Solution :

White P is less stable.

36. Answer A(P, Q); B(R, S); C(T); D(P)

Hint :

Factual

37. Answer A(P, S); B(P, S); C(P, Q, T); D(P, S)

Hint : Reaction of metals with HNO3.

Solution :

R = NH4NO

3.

P1 = N

2O

P2 = N

2O

P3 = NO

2

P4 = N

2O

4

P5 = NO

38. Answer (07)

Hint : Electronic configuration of lanthanoids.

Solution :

x = 1

y = 1

z = 7

39. Answer (09)

Hint : Packing type of lattice.

Solution :

Cu, Ag, Au all have CCP structure, CN = 12

x + y + z = 12 + 12 + 12 = 36

40. Answer (04)

Hint : Kdesorption

a–E /RT

= A e

Solution :

Life time (y) = a

E /RT

1

Ae

= 10–13

15 1000 3exp

25 300

⎛ ⎞⎜ ⎟⎝ ⎠

= 10 –13 × 400

y = 4 × 10–11 s


6/9

PART - III (MATHEMATICS)

41. Answer (A, C)

Hint : 1 1

sin cos2

x x

Solution : Domain of f(x) is [–3, 3]

maximum value of f(x) is

3 397

at 38 4

x

and minimum value of f(x) is

3 311

at 38 4

x


Hint : Equation of normal y = mx – 2am – am3

Solution : Equation of normal to y2 = 4x

is y = mx – 2m – m3 ...(1)

Equation of normal to y2 = (x – k) is

3

2 4

m my mx km ...(2)

Equations (1) & (2) represent same normal

3

3

1 2

1

2 4

m m m

m m mkm

3m2 = 4k – 6 4k – 6 0

3

2k


Hint : Parallel lines are neither concurrent nor forms

a triangle.

Solution : Lines are concurrent if = 0

2

1 1 1

0( 1) 7 5

2 2 5 0

n n

n n

2 3

2

10 25 5 10 2 2 5 5

7 2 14 0

⇒

n n n n n n

n n

n3 – 4n2 + 5n – 6 = 0

(n – 3)(n2 – n + 2) = 0 n = 3

but for n = 3, lines become parallel

so these are concurrent for no real value of n.

44. Answer (C, D)

Hint : Concept of director circle.

Solution :

O

y

(4, 3)x

y

(3, 4)

xO

y

( , )h k

C

Ox

Locus of C is circle x2 + y2 = a2 + b2 = 25

Hence C moves in a circle of radius 5

Distance covered by C in the motion is equal to

1 14 35 tan tan

3 4

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

1 75tan

24

⎛ ⎞ ⎜ ⎟⎝ ⎠

45. Answer (A, C)

Hint : Find domain

Solution : Domain of f(x) is x = 1 f(x) = 0

Domain of g(x) is x = –1 g(x) = 3

2

46. Answer (A)

Hint : 1

2r (distance between parallel lines)


7/9

47. Answer (C)

Hint : L4 are concurrent at (12, 12)

48. Answer (A)

Hint : For cyclic quadrilateral L3 and L

4 are parallel


A(2, 7) 4 – 3 + 13 = 0x y

4 – 3 – 37 = 0x yB(10, 1)

(6, 4)

(3, 0

)

(9, 8)

3 + 4 – 34 = 0x y

Centre of circle is either (3, 0) or (9, 8)

+ = 3 or + = 17

(x + y – 24) + (x – y) = 0 are concurrent at (12, 12)

y

x

3 + 4 – 34 = 0x y

4 –

3 +

13 = 0

x

y

4 –

3 – 3

7 =

0

x

y(12, 12)

(2, 7)

(10, 1)

O

1

1m

should not belong to 1 11,

2 2

⎡ ⎤⎢ ⎥⎣ ⎦

= –2, –1, 0, 1

For cyclic quadrilateral

1 3 1

1 4 7

⇒

C (9, 8), r = 5

(x – 9)2 + (y – 8)2 = 52

49. Answer (C)

Hint : Take parametric point on hyperbola.

50. Answer (D)

Hint : Use parametric equation of line.

51. Answer (D)

Hint : 2

( 2) and 4y m x xym

intersect at

exactly one point.


Let a variable point on xy = 4 is 2

2 ,P tt

⎛ ⎞⎜ ⎟⎝ ⎠

Equation of tangent is 2

2 8x y t

t

4x

tyt

4(4 , 0), 0,A t B

t

⎛ ⎞⎜ ⎟⎝ ⎠

Circumcentre is midpoint of AB

2

2 ,h t kt

hk = 4 locus of (h, k) is xy = 4

2e R(2, 0)

r(2, 0)R

x

T

y

S(2

+ co

s,

sin

)

r

r

(2 + rcos)(rsin) = 4

r2sincos + (2sin)r – 4 = 0

1 2

4

sin cosr r

RSRT = 8|cosec2|

(RSRT)min

= 8

y2 = 8(x + 2), xy = 4

y = m(x + 2) + 2

m

is also tangent to xy = 4

242 xmx m

m

⎛ ⎞ ⎜ ⎟⎝ ⎠

has equal roots

D = 0

22

4 ( 4) 12 m mm

m

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

Equation of common tangent is y = –x – 4

2m – 3c = –2 + 12 = 10


8/9

52. Answer (C)

Hint : f(x) = 1 – xn

Solution :

f(x) = xn + 1 or 1 – xn

f(x) = 1 – x4

f(3) = –80

53. Answer (B)

Hint : a = A(t 2 + 1) and b = A

2

2

(1 )t

t

Solution :

4A = 1

2

1

8A

a = A(t2+1), b =

2

2

(1 )A t

t

1 1 1

8 a b A

54. Answer (A)

Hint : Image of C2 in the line, C

2, C

3 are collinear

Solution :

P(7, 5)

(–7, 3)

(1, –3) 3 + 4 – 16 = 0x y

Image of C2(1, –3) in the line 3x + 4y – 16 = 0 is

P(7, 5)

C1C

2 + C

2C

3 + C

3C

1 is minimum

C1, C

3 and P lie on same line

C3(0, 4)

3 150 5 2C C

Radius of 1

3 2C

Radius of 3

2 2C

Equation of C3 is x2 + (y – 4)2 = 8

x2 + y2 – 8y + 8 = 0 a + b + c = 0

55. Answer (A)

Hint : Find mirror image of (1 + t2, 2t) about line

Solution :

Let variable point on the parabola is P(1+ t2, 2t) and

its reflection in the given line is (h, k)

2 2(1 ) 2 2(1 2 2)

1 1 2

h t k t t t

( 1)( 1) 2

12(1 )

h t t kt

k t h

⇒

2

2 22

hh k

Locus of (h, k) is 4x – 4y = x2 A + B = 0

56. Answer A(P); B(Q); C(P, T); D(P, Q, R)

Hint : Let f(x) = ax2 + bx + c

Solution :

Let f(x) = ax2 + bx + c

f (–1) = 0

f(1) = 4 and f(–1) = 2

a + b + c = 4

a – b + c = 2

–2a + b = 0

Solving, we get

1 51, ,

2 2b a c

21 5( )

2 2f x x x

21

[( 1) 4]2

x

1 1cos 1

( )f x

⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

[1 + sgn(f(x))] = 2

1 1tan 0, 1

2f x

⎡ ⎤⎛ ⎞ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

1 22cot 1, 2, 3

( )f x

⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

57. Answer A(Q, S); B(Q, S, T); C(P, R); D(Q, R, T)

Hint : For increasing fnf (x) > 0

Solution :

f1(x) is many-one and into

f2(x) is many-one, into and even

f3(x) is one- one and onto

f4(x) is many-one, onto and even


9/9

58. Answer (08)

Hint : Tangent to circle 2

2(1 )y mx m

Solution :

Tangent to the circle is 2

2( 1)y mx m ...(1)

It passes through focus (2, 2) of xy = 2

22 2 2( 1)m m

m2 – 4m + 1 = 0

Line (1) intersects hyperbola xy = 2 at the points

given by 22

2( 1)mx m

x

or2 2

2( 1) 2 0mx m x

2

1 2 1 2

2( 1) 2,

mx x x x

m m

2

1 2

16( )x x

m

Also (y1 – y

2)2 = 16m

Length of focal chord 8

2 8mm

⎛ ⎞ ⎜ ⎟⎝ ⎠

28(1 )

2m

m

8 42 8

m

m

59. Answer (08)

Hint : Let 10 cos , 2sinP

Solution :

Let 10 cos , 2sinP

� � �

10 4 3

10 5e

2

2

1 2

320 (given)2 10 cos

5PF PF

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 25 1cos , sin

6 6

2 2

2 10 2 10

1 510sin 4cos10 4

6 6

d

2 2

60. Answer (04)

Hint : Let circle (x – r)2 + y2 = r2

Solution :

( , 0 )r

y

xr

Let the equation of circle be

(x – r)2 + y2 = r2

solving with y2 = 8x

x = 0 and x = 2r – 8

2r – 8 0

rmax

= 4

Test - 2A (Paper - 2) (Code-E) (Answers) All India Aakash Test … · 2019-06-04 · All India...

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Transcript of Test - 2A (Paper - 2) (Code-E) (Answers) All India Aakash Test … · 2019-06-04 · All India...