Aiats Jee Advanced 2015 Test-1

7/18/2019 Aiats Jee Advanced 2015 Test-1

http://slidepdf.com/reader/full/aiats-jee-advanced-2015-test-1 1/8

Test - 1 (Paper - II) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2015

1/8

TEST - 1 (Paper - II)

PHYSICS

1. (B)

2. (A)

3. (B)

4. (D)

5. (C)

6. (B)

7. (A, B, C, D)

8. (A, B, C)

9. (A, B, C, D)

10. (A, D)

11. (A, C)

12. (B)

13. (C)

14. (B)

15. (A)

16. (A)

17. (C)

18. (4)

19. (5)

20. (6)

CHEMISTRY

21. (B)

22. (B)

23. (C)

24. (D)

25. (B)

26. (C)

27. (A, D)

28. (B, C, D)

29. (A, C)

30. (C)

31. (A, D)

32. (B)

33. (A)

34. (C)

35. (A)

36. (B)

37. (A)

38. (5)

39. (8)

40. (4)

MATHEMATICS

41. (B)

42. (D)

43. (B)

44. (B)

45. (A)

46. (B)

47. (A, B, C, D)

48. (A, D)

49. (A, C, D)

50. (A, B)

51. (A, B, C, D)

52. (A)

53. (C)

54. (B)

55. (B)

56. (A)

57. (A)

58. (6)

59. (3)

60. (1)

ANSWERS



All India Aakash Test Series for JEE (Advanced)-2015 Test - 1 (Paper - II) (Answers & Hints)

2/8

1. Answer (B)

120°mg

T

In equilibrium when right spring is cut

2T cos60° = mg T = mg ...(i)

F nett

=2 2

2 cos120mg T mg T

a = g

2. Answer (A)

N = 2 mg cos ...(i)

For wedge, N = 5mg + 2mg cos230°

35 2

4mg mg

3 135

2 2mg mg mg

N

2 c o

s

m g

2 s i n

3 0 °

m g

Also, 2mg cos30°sin30° = N

Wedge

5mg

N

N

m g

= 2

c o s

f

3 0 °

PART - I (PHYSICS)

ANSWERS & HINTS

3 1 132

2 2 2mg mg

3

13

3. Answer (B)

For block,

2

2 –2

22 – 2 0 – 4 –4 ms B

dH d H H t t a

dt dt ⇒ ⇒

aB cos60° = a

W cos30°

60°

aW

aB

60°

aW =

cos60·cos30

Ba

–21 2 44 ms

2 3 3

4. Answer (D)

N

E

i

j

– –2 2

PQ P Q

v v v i j v v

...(i)

2 2– – –

2 2

QR Q R

v v v i j v v ...(ii)

– – – – RP R P R Q P Q

v v v v v v v

= 2 2– –

2 2 2 2

v v v v i j i j

=3

2 2

v v i j




3/8

5. Answer (C)

x = u cos × t ...(i)

21sin –

2y u t gt ...(ii)

2

2 2 2 21

2 x y gt u t

...(iii)

2 4 2

2 2 2 22

4 2

g t gt x y y u t

36 + 64 + 25t 4 + 80t 2 = 625t 2

25t 4 – 545t 2 + 100 = 0

2 2 2 2

1 2 1 2

545 100,

25 25t t t t

2 2

1 2 1 2 1 2

545– – 2 – 2 2

25t t t t t t t

4.2 s

6. Answer (B)

2

rel cos 37 – sin 37v v at v t

2

4 310 – 5 10

5 5t v

rel0

d v

dt

t = 1.6 s

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

For vertical motion, 21–

2

h ut gt ...(i)

1 2

1

2h g t t

For projectile motion,21

sin · –2

h u t gt ...(ii)

1 2

1

2h g t t

1 2 1 2

2 sin 2 cos 1, ,

2

u u t t R g t t

g g

8. Answer (A, B, C)

(i) Time of crossing is minimum when v u and

t min

=d

v

(ii) When v > u , the swimmer can reach, directly

opposite the point of start, t crossing

= 2 2–

d

v u

u

v

(iii) For v < u , drift cannot be zero

u

v v s


F = kt k a t

m ...(i)

2

2

k v t

m

...(ii)

2

2

2

k mv a

m k

...(iii)

a t , v t 2, v a2

10. Answer (A, D)

In (A), For m, 2mg – mg = ma1

...(i)

2

3 3

2 2 2 –In (B), ,

2 2

In (C), – and 2 –

m m m m g T g a

m m m m

T mg ma mg T ma

1 2 3

1 2 3

Solving , ,3 2

84 , , 3

3

g ga g a a

T mg T mg T mg

11. Answer (A, C)

For bead, 1 1

7–

8 8

mg g mg ma a ⇒ ...(i)

For mass block, 2 2

3–

4 4

mg g mg ma a ⇒

For bead w.r.t. string, arel

= a1+ 2 × a

2




4/8

=7 3 7 12 19

8 2 8 8

g g g g g

2 8 16

19 19

l l t

g g

12. Answer (B)

1 2 1 1 2 2

1 2

sin 37 – cos 37m m g m m g a

m m

1 1 2 2

1 2

sin 37 – ·cos 37m m

gm m

3 4

10 – 0.2 2 0.1 15 5 3

3 410 – 0.5

5 5 3

3 0.410 –

5 3

= 4.67 ms–2

13. Answer (C)

m1g sin37° + f –

1m

1g cos 37° = m

1a f = 0.53 N

14. Answer (B)

A

B C a

1

a2

a1

2a32a

3

a3

Let acceleration of A, B and C are a1 , a

2 and a

3 ,

respectively.

Clearly, a2 = a

3...(i)

mg – T = ma1

...(ii)

mg – 2T = ma2

...(iii)

Pulley constraints1 3

2

2–

2

a aa

...(iv)

From (iv), a1 + 2a

3 = –2a

2 a

1 + 2a

2 = –2a

2

a1 = 4a

2...(v)

From (ii) and (iii), 1

2

2 – 2 2

2 –

mg T ma

T mg ma

1

4

9a g

1

2

4 1

4 9 4 9

a g ga

15. Answer (A)

16. Answer (A)

17. Answer (C)

18. Answer (4)

2 2

1

2

1– 10 20 –5 2– 2 s

2

2 2 s

h ut gt t t t

t

⇒ ⇒

t 1+ t

2 = 4 s

19. Answer (5)

20. Answer (6)

f 2 =

2m

2g = 0.5 × 2 × 10 = 10 N,

Tension T = 15 – 10 = 5 N

For m1, T = F

2 + f

1 f

1 = 1 N

21. Answer (B)

Equivalent weight of 2–

4SO ion

96

2 = 48

Total equivalents of sulphates of two metals

2 2 7

12 48 32 48 120

Equivalents of BaSO4

7

120

Mass of BaSO4 7 233 6.8 g

120 2

PART - II (CHEMISTRY)

22. Answer (B)

Let the normalities of HCl and H2SO

4 be N

1 and N

2

respectively.

25(N1 + N

2) = 10 × 2.0 = 20

N1 + N

2 = 0.8

H2SO

4 + BaCl

2 BaSO

4 + 2HCl

2

0.699 100010 N

233

2

; N2 = 0.60

N1 = 0.20




5/8

23. Answer (C)

FeCl3 + 3NaOH Fe(OH)

3 + 3NaCl

Moles of Fe(OH)3

1.425

107

Molarity of FeCl3

1.425

0.1 107

Normality of FeCl3

1.425 30.40 N

0.1 107

24. Answer (D)

Kinetic energy of proton = eV

2p pm v 2 eV ; p p p

m v 2m eV

Similarly, for particle m v 4m eV

p2m eV VV

4m e 8

(∵ m = 4mp)

25. Answer (B)

Energy of photon when electron returns from 6th orbit

to 3rd orbit of Li2+ 1 1

13.6 9 – eV9 36

3

13.6 eV4

Electron in 2nd orbit of He+ ion absorbs this energyand jumps to nth orbit

2

3 1 113.6 13.6 4 –

4 4 n

n = 4

Energy of electron in 4th orbit of He+

4

–13.6 –3.4 eV16

26. Answer (C)

Fact

27. Answer (A, D)

104.5% oleum means 100 g oleum reacts with 4.5 g

H2O.

10 g oleum reacts with 0.45 g H2O

SO3 + H

2O H

2SO

4

Mass of SO3 in 10 g oleum

0.4580 2.0 g

18

10 g oleum is mixed with 0.18 g H2O

Mass of SO3 that combines with

2

0.18 80H O 0.80 g18

Mass of free SO3 left in 10.18 g solution = 1.2 g

Percentage of free SO3

1.2 10011.79%

10.18

Mass of H2SO

4 in 10.18 g solution

0.18 988 8.98 g18

28. Answer (B, C, D)

Let the volume of CO2 in the given mixture be x ml

CO2(g) + C(s) 2CO(g)

Volume of CO after the reaction = 125 – x + 2x = 205

x = 80 ml

Volume of CO in the mixture = 45 ml

Mole percent of CO45 100

36%125

Mole fraction of CO2 80

0.64125

29. Answer (A, C)

Since the number of photons absorbed and

transmitted is same, the electron moves from 1st orbit

to 2nd orbit only. The energy of the photon absorbed is

given by

E = (10.2)Z2 eV

If Z = 2, E = 40.8 eV and if Z = 3, E = 91.8 eV

30. Answer (C)

Electronic configuration of Mn is 1s22s22 p63s23 p63d 54s2

l + m = 0 if l = 0 and m = 0 i.e., all s-orbitals

l = 1 and m = –1 i.e., one of p-orbitals

l = 2 and m = –2 i.e., one of d -orbitals

Total number of electrons having l + m = 0 is equal to 13.

31. Answer (A, D)

Fact

32. Answer (B)

For particle (X),X

x x

h

m v

For particle (Y),Y

y y

h

m v

x xY X

y y

m v 2.416 Å

m v 0.3 0.5

33. Answer (A)

1 n

v Z

2 2

1 1

n

n

2

1

n 3n




6/8

34. Answer (C)

Fact

35. Answer (A)

Fact

36. Answer (B)37. Answer (A)

38. Answer (5)

K4[Fe(CN)

6] + 6H

2SO

4 + 6H

2O 2K

2SO

4 + FeSO

4

+ 3(NH4)2SO

4 + 6CO

Millimoles of K4[Fe(CN)

6] =

1

6 × millimoles of CO

33.6x 0.05

6 22.4

x = 5 ml

39. Answer (8)

(COOH)2 + 2NaOH (COONa)

2 + 2H

2O

NaOH + HCl NaCl + H2O

Milliequivalents of oxalic acid = Milliequivalents of

NaOH – Milliequivalents of HCl

1. 89 2 100040 0.80 – 0.25 x

126

On solving, x = 8 ml

40. Answer (4)

2

2

ZKE

n

He

H

(KE) 4

(KE) 1

PART - III (MATHEMATICS)

41. Answer (B)

(4, 4)(–4, 4)

(4, –4)(–4, –4)

|z – (4 + 4i )| 4

|iz – (–4 + 4i )| 4

|–z – (–4 – 4i )| 4

|–iz – (4 – 4i )| 4

A = 8 × 8 – × 42

= 16(4 – )

42. Answer (D)

2

2 2 2

1 1 1 1 1 1

– – –( – ) ( – ) ( – ) x y y z z x x y y z z x

1 1 1– 2

( – )( – ) ( – )( – ) ( – )( – )y z z x x y z x x y y z

=

21 1 1 – – –

– 2– – – ( – )( – )( – )

x y y z z x

x y y z z x x y y z z x

=

2

1 1 1

– – – x y y z z x

Hence ture for all x R , y R , z R except the

case when x = y or y = z or z = x.

43. Answer (B)

Clearly all the terms of x 28 to x –28 are possible

So, total number of terms are = 28 + 28 + 1 = 57

44. Answer (B)

(72)10 = (70 + 2)10

10 32 2 · 8

27 7

⇒ will be remainder

45. Answer (A)

Coefficient of x 49 is =

2 2 501 2

0 1 49

2 ..... 50 C C C

C C C

50

1

– (51– ) –22100

r

r r

∑

46. Answer (B)

[ x 7 + (1 + x 5)]20

= 20C 0(1 + x 5)20 ( x 7)0 + 20C

1(1 + x 5)19( x 7) + 20C

2(1 + x 5)18

( x 7)2 + ........

Coefficient of x 17 = 20C 1 × 19C

2

20 19 18

2

= 3420




7/8


S = 20C 7 – 20C

8 + 20C

9 – 20C

10 + ....... – 20C

20

We know

(1 + x )20 = 20C 0 + 20C

1x + 20C

2 x 2 + ....... + 20C

20 x 20

(1 + x )–1 = 1 – x + x 2 – x 3 + x 4 – ......

(1 + x )19 = x 13[20C 13

– 20C 12

+ ....... – 20C 0] + .....

20C 7 – 20C

8 + ....... – 20C

20= 19C

13

19 18 17 16 15 14

6!

48. Answer (A, D)

2 2

2

1

| |

z z z z E

z z

Let z = x + iy

2 2

2 2 2

.

| |

x y x iy x iy z z z z E

z x y

2 2 2 2

2 2

x x y y x y y i E

x y

y ( x 2 + y 2) – y = 0

either y = 0 or x 2 + y 2 = 1

49. Answer (A, C, D)

S = 0.100C 02 + 100C

12 + 2.100C

22 + ..... + 100.100 2

100C

S = 100.100 2

100C + 99.100 2

99C + ................ + 100C

12

___________________________________________________________________________

2S = 100 [100C 02 + 100C

12 + 100C

22 + ..... + 100 2

100C ]

S = 50.200C 100

50. Answer (A, B)


3 + 2 = 1

1– 32 1– 3

2

⇒

= 1,1

5

= 11

5and rejected

1

5

= –1

The polynomial will become

( x + 1)2( x – 1)3 = ( x 2 – 1)2( x – 1)

= ( x 4 – 2 x 2 + 1)( x – 1)

= x 5 – x 4 – 2 x 3 + 2 x 2 + x – 1

52. Answer (A)

2

0

1

nn n r

r

x C x

∑

On integrating between 0 to 1

21201 2

0

2 1.....

2 3 21 21

C C C C

... (1)

On integrating between –1 to 0

201 2

0

1.....

2 3 21 21

C C C C

... (2)

(1) + (2), we get

201931 2 1

.....2 4 20 21

C C C

53. Answer (C)

1

1

0 0 0

1 1 1

1 1

n n n

n n n

r r r

r r r

r C C C

r n

∑ ∑ ∑

112 2 1

1

n n

n

Since,

8

0

2 2 1

1 6

n

n

r

r

r C

r

∑

1 8

2 1 2 12 5

1 6

n

nn

n

⇒

54. Answer (B)

55. Answer (B)

Solution of Q. No. 54 and 55

x

y

(–10, 0) (–3, 0) (+5, 0)

(0, 4)

| +3| + | –3| = 10z z

(+3, 0)

– 4 4– 4Re Re 0

4 4 4

z z z

z z z




8/8

put z = x + iy

x 2 + y 2 = 4

– 255

– 1

z

z

put z = x + iy

x 2 +y 2 52 , Clearly B C has no intersection points

area of ( A C ) = area ( A) = 20

area of (C D) = 25 – 20 = 5

area( A) = 20

area(C ) = 25

20 + 5 + 20 – 25

= K

K = 20

56. Answer (A)

(P)

64

64 821

3 5

r r

r r t C

For a rational term r must be a multiple of 8,

hence number of multiples of 8 in 0 to 64 = 9

(Q) By using properties of Binomial coefficient

2 2 2

69 69 69 69 70 70

3 1 3 31r r r r r r

C C C C C C ⇒

r 2 = 3r or r 2 + 3r – 70 = 0

r = 3, 7

(R) 26.3

2 3

x r

x

r = 5

(S) 262 = 4(1 + 7)20 = 4 + 7k ,

Hence remainder is 4

57. Answer (A)

58. Answer (6)

a0

( x – 2)0 + a1

( x – 2)1 +.........+a10

( x – 2)10

= b0( x – 3)0 + b

1( x – 3)1 +........+b

29( x – 3)29

comparing both the side

b5( x – 3)5 = ( x – 2)5 + ( x – 2)6+........+( x – 2)10

= ( x – 3 +1)5 + ( x – 3 + 1)6 +..........+( x – 3 + 1)10

b5 = 5C

5 + 6C

5 + 7C

5+.......+10C

5

b5 = 6C

6 + 6C

5 + 7C

5+.......+10C

5

= 11C 6

Then the value of = 6

59. Answer (3)

1 2 3

18 2

81 9r r r ... (1)

1 2 2 3 3 1

36 4

81 9

r r r r r r

... (2)

1 2 3

8

81r r r ... (3)

From (2) and (3)

1 2 3

1 1 1 4 / 9 9

8 / 81 2r r r

2

2

3 9 2

2 3r

r

⇒

1 3 1 3

4 4,

27 9r r r r

1 2

2 2,

3 9r r

2 2 2

3 9 3

1 6 4 2 1 18 36 6

8 2 / 3 2 / 9 2 / 3 8 2

= 3

60. Answer (1)

5 3 322 1z z z z z

5 3 322 1z z z z z

3 2 22 1z z z z

2 2 2 2

2 2 2 x y x iy x y xyi

3/ 2

2 2

1

x y

(where z = x + iy )

2

3/ 22 2

12 2y x

x y

and y = 0

3

12 x

x

3 1

2 x x

only possibility is z > 0 and 4 1

2 x

1

4

1

10

2z i

,

1

4

2

1– 0

2z i

= +

Documents

Aiats Jee Advanced 2015 Test-1