jee advanced full test ii paper 1 answers

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ANSWERS, HINTS & SOLUTIONS

FULL TEST – II PAPER-1

QQ..NNOO PHYSICS CHEMISTRY MATHEMATICS

1. A A C

2. C A D

3. B A A

4. C A B

5. C A B

6. A A C

7. A, B, C, D A, B, C, D A, C, D

8. A, B A, C, D A, C

9. C, D A, B, C A, B

10. B, C A, D A, D

11. B B C

12. B A B

13. B A A

14. C C A

15. A B C

16. B C D

17. A C C

18. A B B

19. C D C

20. D C D

21. C D B

22. D B D

23. C A A

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AITS-FT-II (Paper-1)-PCM(Sol)-JEE(Advanced)/14


2

PPhhyyssiiccss PART – I

2. (v1cos)t = (v2cos)t = 30 3 . . . (1)

(v1sin)t 21 gt2

+h = (v2sin)t 21 gt2

. . . (2)

from (1) and (2) h = 60 m

3. = tan 2

2 2

1 g2 u cos

= tan 2

21 g 1 tan2 2g

tan2 4 tan + 41 = 0

(, )

u 2gu

If particle will not hit the target. (b2 4ac) < 0

16 4 41 < 0

4 > 3

4. At B, mg sin = 2Bmv

r . . . (1)

Using energy considerations

2B

1 mv2

= mgr(cos sin ) . . . (2)

B

A

Smooth N

vB

From (1) and (2) mg sin = 2mg(cos sin ) 2cos = 3 sin 7. Here the position y on the screen will correspond to maxima.

n Dyd

30,000n

when n = 1, 2, 3, 4, …………..

8. a = kx2

da 2kxdx

tan 60 = 2k 3 1k2

a 2x

2

2dv xv

dx 2

2 3

1

v x C2 6

at x = 0, v = 3 m/s

1

9C2

3

2 xv 93

hence a = 1.5 and v 3 9



3

9. Capillary height 2T coshr g

h = 59.6 mm

here 59.6 mm is greater than the protruding part hence water will rise in the capillary of insufficient height 25 mm.

Now, 2TR 0.6hrg

mm

10. 10. Fluid particles passing through the bend are in circular

motion. The centripetal acceleration is provided by the variation in pressure. In the section as shown

1 2 CP A P A ma 1 2P P applying Bernoulli theorem at 1 and 2

i.e., 2 21 1 1 1 2 2 2

1 1P v gh P v gh2 2

2 21 1 1 2 2

1 1P v P v2 2

as 1 2h h

1 2P P 1 2v v

1

2

maC

P1A

P2A

15. Agm sin = 5 10 sin 37 = 5 10 35

= 30 N

max AB C Bf (m m )gcos 48 N hence f = 30 N

16. C3m gsin 2 10 12N5

max c4f m gcos 0.1 2 10 1.6 N5

hence f = 1.6 N 17. A C BT f f m gsin37 = 91.6 N

T

fC

fa mBg sin 37

18. Optical path difference between the beans arriving at P 2 1x ( ) dsin for maxima 2 1( ) dsin n

2 11sin n ( )d

1 nsin 2 140



4

19. sin 1

n1 2 1 140

or 20 n 60 Hence number of maxima = 60 – 20 = 40 20. At C, phase difference

2 12 ( )

= 80

Hence for maximum intensity will appear at C. Now for minimum intensity at C

( 1)t2

tt 5002( 1)

nm



5

CChheemmiissttrryy PART – II

3. Neither N nor F contain d-orbitals. Further in N2F4 N—N bond is shorter than in N2H4 due to

s-character (Bent’s Rule). 4. Due to poor shielding effect of 3d electron the size of Br ion nearly same as that of Cl ion. 5.

C8H7OClN2O

287 m Molar mass of C8H7OCl = 8 12 + 7 1 + 16 + 35.5 = 154.5 g Molar mass of N2O = 2 14 + 16 = 44g According to Graham’s Law of diffusion

8 72

8 7 2

C H OClN O

C H OCl N O

Mr 154.5 3.5 1.87 : 1r M 44

2N O

1.87d 2872.87

= 187th row from N2O side

8 7C H OCl

1.0d 2872.87

= 100th row from weeping gas side

Therefore, the spectator from the side of N2O in the 187th row will be laughing and weeping simultaneously Alternatively, the spectator from the side of weeping gas in 100th row will laugh and weep.

6. OH attacks at the more reactive (C=O) group, (containing more EWG or less EDG) Et is more

EDG than Me (due to +I effect here no H.C.). Therefore OH attack (C=O) with (Me) group.

Me

Et

O

OMe

Et

C

OH

O

O

OH

OH

3H O

Me

Et

C

OH

O

OH

10. 4 3 2 2NH NO aq N O H O 2 2 2 2NH OH HNO N O 2H O Hence, (A) and (D) are the correct answers.

13. Velocity = 6 Z2.18 10n

14. (i)

3BrOE M/ 5

(ii) 3BrOE M/ 6

ratio = 6/5



6

17.

NH2

Me

Me

OH

2HNO

AN

+Me

Me

OHN

2N Me

Me

OH

12

3

2 4C C bond

breaks Ring exp ansion

O

Me

Me

H

1 2

34

H

Me

Me

O (B)

3RCO HBaeyer Villiger

O

O

Me

Me

(C)(Cyclic ester)

(Lactone)

4

2

i LiAlHii H O

OH

Me

MeOH

(D)

18. 3 3 3 32 2

0.8 15 35 0.1512 5.25

2KIO Pb NO Pb IO 2KNO

1 mol of Pb(NO3) reacts with 2 mol of KIO3 5.25 m mol of Pb(NO3)2 reacts = 2 5.25 = 10.5 m mol m mol of KIO3 left = 12 – 10.5 = 1.5 Hence Pb(NO3)2 is the limiting reagent

19. 3

1.5 m molIO 0.03 M15 35 mL

20. Due to common ion 3IO left in the solution the solubility 3 2Pb IO decreases.

S =

13 4sp

2

K 2.7 10 1090.03

= 0.30 10–9 = 3.0 10–10

21. pH range of all these indicators lies in between 2-12. Hence, all are suitable indicators. 22. Since curve is given for strong acid and strong base hence, pH at the end point is 7.



7

MMaatthheemmaattiiccss PART – III

1. y = ax2 + bx + c, vertex is (4, 2)

4 = –a2

b, b = –8a, 2

a4bac4 2

c = 2 + a4

b2

= 2 + 16a

Now = abc = –8a2(2 + 16a) = –16(a2 + 8a3)

dad

= –16(2a + 24a2) < 0 a [1, 3]

|max = –144, |min = –3600 Difference = 3456 2. x – [x] = {x} x – [x + 1] = {x} – 1

4

2

dxxf = 1.121.6 = 3

1

1 2 3 4-1-2

3.

x

0

2

0x taxsinxdttlim = xsinx

tadtt

lim

x

0

2

0x

=

xcos1xaxlim

2

0x

= xa.

2xsin2

xlim2

2

0x

=

a2

= 1 (given)

a = 4 4. (3)6 = 729 < 900 and (3)7 = 2187 > 900 5. Now by property of triangles

RQ = 12

BC = a2

Similarly PQ = c2

, PR = b2

Area of ABC = abc4R

A

R Q

B P C

That of PQR =

a b c2 2 24R'

Also area of ABC = 4 (area of PQR)

abc abc R 232R' 16R R' 1



8

6. Clearly 2 solutions

B

y = 2x –3

y =logx 3

7. f(x) = cos (|x| + [x]) f(0) =1, f(0 – 0) = cos (–1) = –1

f(1/2) = cos

21

= 0, f

21cos0

21

= 0

02

cos021f

for x (0, 1) f(x) = cos x for x (–1, 0) f(x) = cos (–x – 1) = –cos x 8. OP = 5 2 sec

OP1 = 5 2 cosec

PP1P2 = 2sin

100

( | PP1P2)min = 100 = /4

OP =10 P = (10, 0), (–10, 0) P2

P1

P

O

9. Let P(2, –1) goes 2 units along x + y = 1 upto A and 5 units along x – 2y = 4 up to B

Slope of PA = –1 = tan 135°, slope of PB = 21

= tan

sin = 51

, cos = 5

2

A (x1 + r cos 135°, y1 = r sin 135°) = (2 + 2x 21

, –1 + 2 2

1) = (2 – 2, 2 – 1)

B (x1 + r cos , y1 + r sin ) = (2 + 5 5

2, –1 +

55

) = (25 + 2, 5 – 1)

10. Slope of tangent = ab

b1a1



9

2

2

t1

4t4

dtdxdtdy

dxdy

< 0

So, 0ab

ab

> 0

12. 2

1 P 1 xA A BPP xB B

1

x2 + x – 1 0

x 5 12 or x 1 5

2

x is positive

x 5 12

13. 2 2 2 2 2a b c da b c d

44

22 16 e8 e16 4

64 + e2 – 16e 64 – 4e2 5e2 – 16e 0

0 e 165

14. Let A(a, P(a)), B(b, P(b)), then slope of AB = P(a) = P(b) from LMVT c (a, b) Where P(c) = slope of AB 15. Given QT = QA = 1

Let PQ = x, then 2PT x 1 Then TQP and APO are similar triangles

Then 2

x 1OT OAx 1

22

2(x 1)1 x x 1 8x 1

x = 53

A

P

Q

T

O

16. From above, OA = 2 and AQ = 1 coordinate of Q (2, 1) equation is (x – 2)2 + (y – 1)2 = 1 17. ORQ is a right angled triangle. Then O and R are the extremities of the diameter then the

coordinates of R(2 3, 2)

Equation of circle (x 0)(x 2 3) (y 0)(y 2) 0 2 2x y 2 3x 2y 0



10

18. Perpendicular tangents intersect on the director circle of hyperbola and director circle of rectangular hyperbola is a point circle. Hence centre of hyperbola is (1, 1) and equation of asymptotes are (x – 1) = 0 and y – 1 = 0

19. Equation of hyperbola is xy – x – y + 1 + = 0 It passes through (3, 2) hence = –2 Equation of hyperbola is xy = x + y + 1 20. From the centre of hyperbola we can draw two real normals to the rectangular hyperbola 21. Taking point (r cos , r sin ), we get

r2 = 3 2

3 2 sin 24

rmax = 3 23 2

, rmin = 1

Max exist when 24 2 8

22. Required equation is 2

2x y 13 23 2

23. Centre of circle be origin and its radius is the length of semi minor axis Hence area =

jee advanced full test ii paper 1 answers

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