sol aits 2013 ft i jeea paper 1

AITS-FT-I-(Paper-1)-PCM(S)-JEE(Advanced)/13

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1

ANSWERS, HINTS & SOLUTIONS

FULL TEST –I (Paper-1)

Q. No. PHYSICS CHEMISTRY MATHEMATICS 1. C A D

2. C A A 3. A B B

4. C C A

5. D B D 6. C C B 7. A D B 8. A C B 9. C, D A, C, D A, C 10. A, C A, C A, B 11. A, C A, B, C B, C, D 12. A, B, C, D A, B A, C 13. A B B

14. A A C

15. C B D

16. B D B

17. D A B 18. A D A

1.

(A) → (s) (B) → (p) (C) → (p) (D) → (q, r)

(A) → (p, r, t) (B) → (q) (C) → (r, s) (D) → (r, s)

(A) → (p, q, r, s) (B) → (p) (C) → (q, s) (D)→ (p, r)

2.

(A) → (r) (B) → (q) (C) → (q) (D) → (s)

(A) → (p, q, s) (B) → (p, s) (C) → (r, s) (D) → (q, t)

(A) → (s) (B) → (t) (C) → (q) (D)→ (s)

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2

PPhhyyssiiccss PART – I

SECTION – A

1. E2 = 2 2 2

R L C RV (V V ) V 0+ − = + As, E = VR ∴ VR = V = 220 V

I = E 220 2.2R 100

= = amp

3. Equivalent circuit can be drawn as

Equivalent capacitance = 79 C30

.

C2C

CC

8/3C

CC

C

C

C

C

C

C

2C⇒

C

C

CC

2/3C 2C

2C

C

19C/11⇒⇒

4. This is simply discharging of a capacitor. ∴ t / CR

0q Q e−= where C = 4πε0r 5. When the circuit will reach to steady state the inductor will given zero resistance. Hence the entire

current will only pass through it.

6. X = H 2H 3H2 2+

=

8. Here V ∝ 1/r r → distance between Sun and Planet.

9. 2 2(y h) x h− + + =

2 2

dy x dx 0dt dtx h

+ =+

2 2

dy x dxdt dtx h

= −+

Ady 3 vdt 5

= −

y

x=3cm

h=4c

m

A

Bh = 3 m

h =

4 m

B A3| u | v5

= …(i)

2 2

2A2 2 2

d y hv3dt (x h ) 2

=+

2B A 3

16a v(5)

=



3

2B A

16a v125

= …(ii)

10. Mg – T = Ma …(i) T = ma …(ii) Solving (i) and (ii)

Mga(M m)

=+

FBD of man Mg – N = Ma

MmgN(M m)

=+

N

a

Mg

11. 2 mT 3.14qBπ

= = s.



4

CChheemmiissttrryy PART – II

SECTION – A

1.

Et ( ) 31 O→

O

H

O

Et( )

( )2 4

4

2 Ag O/NH OH2 NaBH→ H+

→O

O

Et

O

OH

OH

Et

2.

C

O

C

O

O

5PCl→

C

C

O

O

Cl

Cl4LiAlH

CH2OH

COO

PCC←

CH2OH

CH2OH

CHO

CHO

OH−

←

3.

C

C

C

C

C

H OH

H OH

OH H

H OH

H

CH2OH

O

1

2

3

4

5

6

and

C

C

C

C

C

OH H

H OH

OH H

H OH

H

CH2OH

O

1

2

3

4

5

6

Cyclic forms of diastereomers that differ only in the configuration of C1, are known as anomers.

The configuration differ only at carbon no. 1. 4. For equimolar solutions, n1 = n2 1x 0.5 and 0.5∴ = o

B B BP x P 0.5 160 80 mm= × = × = o

T T TP x P 0.5 60 30 mm= × = × = TotalP 80 30 110 mm= + =

Mole fraction of toluene in vapour phase = T

Total

P 30 0.27P 110

= =

5. Entropy is a state function, i.e. the change entropy depends upon the initial and final states of the

system and not on how that change is brought about.



5

7.

o0.6∆

o0.4− ∆d orbitals in symmetricalfield of ligands−

4high spin d configurationin an octahedral field

CFSE = 3(- 0.4) o∆ + 0.6 o∆ = - 1.2 o∆ + 0.6 o∆ = -0.6 o∆ 8. Given ( ) 13

sp AgBrK 5 10−= ×

3 3KBr AgNO AgBr KNO+ → + Given, [AgNO3] = 0.05 M 3Ag NO 0.05M+ − ⇒ = =

( )spAg Br K AgBr+ − = 130.05 Br 5 10− − ⇒ × = ×

13

115 10Br 1 10 M0.05

−− −× ⇒ = = ×

[ ] 11Br KBr 1 10 M− − ∴ = = ×

11 9wt. of KBr 1 10 120 1.20 10 gm / litre− −= × × = ×

10.

BrBr

H H

Plane of symmetry

11. (A) ( ) ( )

( )

2anh.ZnCl2 2conc.HClPh CH OH Ph CH Cl Zn OH Cl immediate

white turbidity

− − → − − +

( ) ( )Lucas reagent3 2 3 2CH CH OH CH CH Cl Zn OH Cl slow− − → − − +

(B) ( ) ( )( )

Lucas2 2reagentPh CH OH PhCH Cl Zn OH Cl immediate

White turbidity

− − → +

LucasreagentPh OH No reaction under normal condition− →

(C) ( ) ( ) ( ) ( )( )

Lucas3 3reagent2 2

CH CH OH CH CH Cl Zn OH Cl Moderate rate

White turbidity

− → − +

( ) ( )( )

Lucas3 2 2 2 3 2 2 2reagentCH CH CH CH OH CH CH CH CH Cl Zn OH Cl Slow

White turbidity

− − − − → − − − − +



6

12. 2 3 2 2H S O H O S O+ → + + 3 4 2PbS 4O PbSO 4O+ → + 2 3 33SO O 3SO+ → 2 3 4 23SnCl 6HCl O 3SnCl 3H O+ + → + 13. 2 2Zn Fe Fe Zn+ ++ → +

oeq

0.0591E E logKn

= −

Given, E = 0.2905

o 0.0591 0.010.2905 E log2 0.001

⇒ = −

⇒ 0.02905 + 0.0295 log10 = 0.2905 + 0.0295 = 0.32

o

eq0.0591E logK

2∴ =

eq0.05910.32 logK

2⇒ =

0.320.0295

eqK 10⇒ =

14. For 1 mole of O2

2 22

2O 2O i.e. 3O3

− − → ×

Or 43

mol of Al to change into Al+3 ions.

n = 4 Thus, G nEF∆ = −

G 827000E 2.14 voltnF 4 96500∆ +

⇒ = − = =×

15. ocell eq eq

0.059 1.1 2E logK 37.2881 logK2 0.059

×= ⇒ = =

37

eqK 1.9 10⇒ = × 16. Copper sulphate when reacts with hypophosphorus acid, gets reduced to cuprous hydride. 4 3 2 2 2 2 3 4 2 44CuSO 3H PO 6H O 2Cu H 3H PO 4H SO+ + → + +

17. ( ) ( )

( ) ( ) ( )4 2 2 2 32 2

3Ba OH 2P 6H O 3Ba H PO 2PH

J L K

+ + → +

( )( ) ( ) ( )

2 2 2 4 4 3 22Ba H PO H SO BaSO 2H PO

L N Mgives apple

green colour inthe flame

+ → +



7

18. H4P2O7 (pyrophosphoric acid) is a tetrabasic acid, i.e. 4-hydroxyl groups are present. It is prepared by removing one water molecule from two molecules of orthophosphoric acid. Each phosphorus atom lies in same tetrahedral environment, so called isopolyacid.

+OH P

O

OH

OH P

O

OH

OHOH 2H O−→ OH P

O

OH

P

O

OH

OHO

SECTION-B 1. (A) CH3 CH

Br

COOH ( )( )1 OH2 H

−

+→ CH3 CH

OH

COOH

(B) CH3 CH

Br

CH2 COOH ( )( )1 OH2 H

−

+→ CH3 CH CH COOH

(C)

H2C

Br

CH2 CH2 C

O

OH OH−

→ H2C

Br

CH2 CH2 C

O

OO

C

O

(D)

H2C

Br

CH2 CH2 CH2 C OH

O

OH−

→O

C

O

Reactions A, C, and D show SN2 reactions. 2. In case of (A), compound given in (A) exchanges – H with – D and becomes racemic. In case (B), compound in (B) reacts through a carbocation and gives racemic mixture. In case of (C), no bond attached to the stereocentre breaks in reaction of (C), so retention. In case of (D), compound in (D) gives Hoffmann elimination and it is an example of anti-

elimination. Product does not have any stereogenic centre.

( )d, form−CH3

D

C2H5

OH( )Reaction B gives

CH3

D

C2H5

COOH( )Reaction C gives

( )Reaction D gives

CH3 CH3

( )No stereocentre



8

MMaatthheemmaattiiccss PART – III

SECTION – A

1. We must have f(a2 + 2a + 5) > f(a + 11) ⇒ a2 + 2a + 5 < a + 11 ⇒ a ∈ (–3, 2) ∴ a can take values –2, –1, 0, 1 2. ∆ABC is equilateral ⇒ a = b = c 3. AC2 = AB2 + BC2 ….. (1) BD2 = BC2 + CD2 ….. (2) CA2 = CD2 + DA2 ….. (3) DB2 = DA2 + AB2 ….. (4) Equation (1) + (3) and (2) + (4) gives AC = BD So, equation (1) and (2) give AB = CD and equation (2) and (3) give BC = AD So ∆′s CBD and ADB are congruent ⇒ ∠CBD = ∠ ADB = 90º – ∠ABD ⇒ ∠CBD + ∠ABD = ∠CBA ⇒ DB lies in the plane of ∆ABC, so the points are coplanar 4. Let a – d = 3t; then bc = (d + zt)(d + t) = d2 + 3td + 2t2

Also a 3t1d d= + , so 2 23thk d 1 d t 3td bc

d = + = + <

⇒ bc > hk 5. Clearly P ≡ (1, 2) Equation of tangent at P is 4x(1) + y(2) = 8 ⇒ A ≡ (2, 0) and B ≡ (0, 4) Similarly normal at P is 2x – 4y = 6

⇒ A '( 3, 0)− and 3B' 0, 2

B P

A′

B′

A

∴ Area 0 4 1

1 5BPB' 1 2 12 4

30 12

= = and Area APA ' 5=

∴ Ratio = 4 6. As sin x + sin y ≥ cos α cos x ∀ x ∈ R

Let x2π

= − ⇒ sin y ≥ 1 ⇒ sin y = 1

⇒ 1 + sin x ≥ cos α cos x ⇒ cos α cos x – sin x ≤ 1

⇒ 2cos 1 1α + = ⇒ cos α = 0 ⇒ sin y + cos α = 1 + 0 = 1



9

7. 1 1

0 0

sin x x 2I dx dx3x x

= < =∫ ∫

⇒ 2I3

< and 1 1

1/ 2

0 0

cos xJ x dx 2x

−= < =∫ ∫

⇒ J < 2

8. Equation of tangent 2x t y 2ct+ = ⇒ T ≡ (2ct, 0) and T′ ≡ 2c0, t

Equation of normal ( )3 4t x ty c t 1− = − ; 31N c t , 0t

≡ −

; 31N 0, c tt

≡ −

⇒ 3cNT ct;t

= + 3 cN'T ' ctt

= +

⇒ 31 c cct2 tt

∆ = + ⇒ ( )4

2 41 2t

c 1 t=

∆ + and 31 c' ct ct

2 t∆ = +

⇒ ( )2 4

1 2' c 1 t=

∆ + ⇒ 2

1 1 2' c

+ =∆ ∆

9. We have 2 2 22sin cos x 2sin sin2x2 2π π =

⇒ cos2x = sin 2x ⇒ cos x(cos x – 2 sin x) = 0 ⇒ 1 – 2 tan x = 0 as cos x ≠ 0

⇒ tan x = 12

and cos 2x = 35

10. Clearly, the point lies on 7x – y = 5 Also, centre of the circle must lie on the bisectors of the

lines x + y + 13 = 0 and 7x – y – 5 = 0 given by

x y 13 7x y 52 50

+ + − −= ± ⇒ x – 3y = 35 and 3x + y + 15 = 0

Let (h, k) be the centre of the circle, then h – 3k = 35 ….. (1)

C(h, k)

A

B

x + y + 13 = 0

7x – y – 5 = 0

P(–1, –12)

and 3h + k = 15 ….. (2)

Clearly CB is perpendicular BP ⇒ k 2 7 1h 1−

× = −−

⇒ h + 7k – 15 = 0 ….. (3)

On solving, we get centres as C1 ≡ (29 – 2) and C2 ≡ (–6, 3) ⇒ 2

1r 800= and 21r 50= ⇒ smaller circle has radius = 50

Therefore area of quadrilateral ACBP = 12 50 2002 × ×

sq. units

11. Let 2f(x) ax bx c= + + and f(1) = a + b + c < 0 ⇒ f(x) < 0 for all x ∈ R ⇒ f(–2) < 0 ⇒ 4a – 2b + c <

0 Again f(3) + f(2) < 0 ⇒ 13a + 5b + 2c < 0 Also f(–5) < 0 ⇒ 25a – 5b + c < 0 ⇒ 5b – 25a – c > 0



10

12. ( )

( ) 2

(x x) x 0 ; x 0f(x)

x x x 2x ; x 0

− − = <= + = ≥

⇒ f is continuous and differentiable

Again 0 ; x 0

f '(x)4x ; x 0

<= ≥

⇒ f′(x) is continuous but not differentiable at x = 0 13. Taking B as origin, BC as x–axis and A as (h, k) and

C(4, 0)

We have D ≡ (2, 0) and E ≡ h 4 k, 2 2+

⇒ K2 + (h + 4)(h – 2) = 0 ….. (1) Also AC = 3 ⇒ (h – 4)2 + k2 = 9 ….. (2)

And area = 1 4 K 2K2× × =

y

E

B (0, 0)

A(h, k)

C(4, 0)D x

Solving (1) and (2), we get 3h2

= and 11K2

=

∴ ∆ABC = 11

14. Let AD be angle bisector ⇒ AB BDAC DC

= ⇒ c = 2b

Now b + c > a ⇒ b + c > 6 ⇒ b + 2b > 6 ⇒ b > 2

Again 2 2

2b 4b 6 1

4b+ −

< ⇒ b < 6 ⇒ b ∈ (2, 6) and consequently c ∈ (4, 12)

15. BE 3EC 4

= , take BE = 3K and EC = 4K

⇒ A must be (3K, 6)

Now equation of line BA is 6y x3K

=

⇒ Ky – 2x = 0 Since perpendicular from c on AB = 7

y

E

B

A(3K, 6)

C(7K, 0)E x

y′

x′

⇒ 2

0 14K 7K 4

−=

+ ⇒ 2 4K

3=

Now ( ) ( )2 22c 6 0 3K 0= − + − = 36 + 9K2 = 36 + 12 ⇒ c 48 4 3= =

16–18. Let 1

0

12 yf(y)dy a=∫ and ( )1

2

0

20 y f(y) dy 4 b+ =∫

⇒ 2f(x) ax bx= + ⇒ f(y) = ay2 + by

⇒ ( ) ( )1 1

2 2 2 2

0 0

f(x) 12x y ay by dy 20x y ay by dy 4x= + + + +∫ ∫ = 2 a b a b12x 20x 4x4 3 5 4

+ + + +

⇒ a = 3a + 4b ⇒ a = –2b ….. (1) and 4a + 5b + 4 = b ⇒ a + b = –1 ….. (2) ⇒ a = –2, b = 1 ∴ 2f(x) 2x x= − +



11

SECTION – B

1. Clearly we have 2 ; x [ 2, 1)

f(x) 1 ; x [ 1, 0)x ; x [0, 2]

− ∈ −= − ∈ − ∈

and g(x) = sec x; x ∈ R – (2x + 1)2π

We have ( )2 ; sec x [ 2, 1)

fog x 1 ; sec x [ 1, 0)secx ; secx [0, 2]

− ∈ − −= − ∈ − ∈

∴ ( )

{ }4 2 2 42 ; x , , , 3 3 3 3

fog x 1 ; x ,

secx ; x , 3 3

π π π π − ∈ − − ∪ − −π π = − = −π π π π ∈ −

Limit of fog exist at x = –π, π, –1, points of discontinuity of fog are –π, π and points of

differentiability of fog are 51, 6π

−

Again

{ }

{ }

sec( 2) ; x [ 2, 1)2

g(f(x)) sec( 1) ; x [ 1, 0)

secx ; x [0, 2]2

π− ∈ − − − −

= − ∈ − π ∈ −

Limit of g(f(x)) does not exist at x = –1 2. (A) Let Ei ; i = 1, 2, 3, 4 represents the events that the bag contains I white balls: Clearly

( )i1P E4

= . Let W be the event that the ball drawn is white then,

( ) ( )ii

W 1 1 2 3 4 5P W P E PE 4 4 4 4 4 8

= ⋅ = + + + = ∑

Now ( ) ( )4 44 P E P W /EE 1/ 4 2 pPW P(W) 5 / 8 5 15

= = = =

⇒ p = 6

(B) Required number of ways = ( ) ( ) ( )12 12 2 2 12 3 3 12 12 12

1 2 1 2 3 1 2 12 1 2C C C 2 C C C 2 C ..... C C 2 C+ + ⋅ + + ⋅ + + + ⋅ =

( ) ( )12 12 12 12 12 2 12 3 12 121 2 3 2 2 2 3 2 2 2C 2 C 3 C ..... 12 C 2 C C C C ..... C C+ ⋅ + ⋅ + + ⋅ + ⋅ + ⋅ + + ⋅

= 12 12

12 10 11 10r r 2

r 1 r 2

r C 12 C 12 2 12 11 2−= =

⋅ + = × + × ×∑ ∑

= 13 × 210 × 12 = 13 × 210 × m ⇒ m = 12

(C) 5x 5y 5z x y z52 x 2 y 2 z 2 x 2 y 2 z

+ + = + + − − − − − −

= x 2 2 y 2 2 z 2 2 1 1 15 5 3 22 x 2 y 2 z 2 x 2 y 2 z

− + − + − + + + = − + + + − − − − − −

Now apply A.M ≥ H.M on 2 – x, 2– y and 2 – z



12

⇒ 2 x 2 y 2 z 31 1 13

2 x 2 y 2 z

− + − + −≥

+ +− − −

⇒ 1 1 1 92 x 2 y 2 z 5

+ + ≥− − −

Hence 5x 5y 5z 95 3 2 32 x 2 y 2 z 5

+ + ≥ − + ⋅ = − − −

∴ Least value = 3

(D) ( )n 12 212 11 2 11 2

K K 1 K 1K 1 K 1

22!12 K C C 12 C 1211! 11!− −

= =

⋅ ⋅ = = ⋅∑ ∑

= 1221 19 17.....312 2 611!

⋅ ⋅⋅ − ⋅ ⇒ p = 6

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