Concept Recap Test Mains 3 Sol

AITS-CRT(Set-III)-PCM(S)-JEE(Main)/13

JEE (Main)-2013

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ANSWERS, HINTS & SOLUTIONS CRT (Set-III)

ALL INDIA TEST SERIES

From Long Term Classroom Programs and Medium / Short Classroom Program 4 in Top 10, 10 in Top 20, 43 in Top 100, 75 in Top 200, 159 in Top 500 Ranks & 3542 t o t a l s e l e c t i o n s i n I I T - J E E 2 0 1 2

1

Q. No. 1.

PHYSICS A

CHEMISTRY C

MATHEMATICS A

2.

A

C

B

3.

B

B

A

4.

D

C

C

5.

A

D

B

6.

C

B

A

7.

A

A

A

8.

B

B

C

9.

C

C

D

10.

C

C

A

11.

C

C

B

12.

D

A

A

13.

B

B

C

14.

B

D

B

15.

A

B

B

16.

C

C

D

17.

A

B

C

18.

D

D

A

19.

A

C

A

20.

C

D

D

21.

C

A

D

22.

C

D

A

23.

C

A

C

24.

C

C

25.

D

B C

C

26.

C

A

A

27.

D

B

B

28.

C

B

D

29.

C

C

D

30.

A

A

B

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2

AITS-CRT(Set-III)-PCM(S)-JEE(Main)/13

Physics

PART – I 2

 T2   R 2   =  T  1   R1 

3

1.

Use Kepler’s 3rd law : 

2.

At closed end, Maximum pressure = P0 + ∆P0.

3.

Angular momentum of the comet is conserved.

4.

Z = R2 + X 2 where X = X L − X C

6.

V2 H = I Rt = t R 1 ∴ Hα if V = constant R 2

Current along 4Ω, 6Ω, 12Ω are 6I, 4I, 2I Total current = 12 I

3 × 12I 5 2 Heat through 4Ω = (6I) × 4Ωt = H1 Also current through 2Ω =

2

3  × 12I  × 2Ωt = H 2 5 

Heat through 2Ω = 

H 2 < H1 7.

As the capacitors are uncharged, the initial current

=

V ; R eq

Req = (1.5 + 0.5) Ω = 2 Ω. 8.

EStored =

9.

V=

10.

1 2 di Li ; and emf = L dt 2

2000m S S = 2500 = nλ, number of waves = = = t λ 0.8 m

e S ∆t = 0.2 H ∆I LI2 LI2 E1 = 1 , E 2 = 2 2 2 L 2 2 I2 − I1 = 7.5 J ∆E = 2 Since, the pendulum started with no kinetic energy, conservation of energy implies that the potential energy at Qmax must be equal to the original potential energy, i.e., the vertical position will be same. ∴ L cos α = A + (L – A ) cos θ L cos α − A ⇒ cos θ = L−A L cos α − A   ⇒ θ = cos−1    L−A  L=

(

11.

)

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3 12.

hr = constant for capillarity h = 6 cm for a vertical tube

∴

∴ A=

13.

AITS-CRT(Set-III)-PCM(S)-JEE(Main)/13

h = 12 cm cos 60o

Acceleration =

V(g + a)(ρ − d) Vd

= 20 m / s2 14.

Least count = 1 MSD – 1 VSD In this question (M – 1) MSD ≡ M VSD  M − 1 ∴ 1VSD =   MSD  M  Now, LC = IMSD −

(M − 1) M

MSD

15

M−M+1 1 MSD = MSD M M b ⇒ LC = . M Use Newton’s Law of cooling.

16.

For an adiabatic process, TVγ–1 = const; and U (ideal gas) ∝ T.

18.

The radius of curvature ρ =

=

a = g sing θ - µ g cos θ a = g [ sin θ - 0.1 x cos θ] acceleration will remain +ve upto x = 10 tan θ.

19.

20.

v2 , aN

ω=

2π T

θ sin g m

θ

B N

θ AN = R sin 2 ωt AN = R sin 2 ωt AB = 2R sin 2 ∴ average velocity =

θ os gc m µ

O

θ/2

A

R

AB t

2R ω 2π 3ω sin 2π 2 3ω

=

3 3ωR . 2π

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4

AITS-CRT(Set-III)-PCM(S)-JEE(Main)/13

21. 23.

G G B is parallel to v .

e T 2 3 2 Now T ∝ r and r ∝ n ⇒ T ∝ n3 1 ∴ i∝ 3 n ∴ the current will increase 8 times. ∴ (C) i=

3

25.

26.

T1  n1  =   =8 T2  n2  n1 = 2. ∴ n2

Smin = i1 + i 2 − A for Smin ⇒ i1 = i2 30 = 2I1 − 60 i1 = 45°

λD λD 6 ×10−7 × 40 ×10−2 ⇒ d= = = 20 ×10−4 m = 0.2 cm . −5 d β 12 ×10

27.

β=

28.

Let the mass of wood and concrete be M and m respectively. Then

or or ∴ 29.

M m + = (M + m) 0.5 2.5 1   1   M − 1 = m  1 −   0.5   2.5   1.5   3 M = m  = m   2.5  5 M 3 = m 5

Longitudinal stress = Y × longitudinal strain. Here, longitudinal stress = 0 Then, longitudinal strain = 0.

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5

Chemistry 4.

AITS-CRT(Set-III)-PCM(S)-JEE(Main)/13

PART – II

H3BO3 + Na2CO3 → Na2B4O7 + H2O + CO2 ↑ H3BO3 + NaCl → Na2B4O7 + HCl (↑ remains in solution) + H2O

5.

HCl does not escape from the system as gas like CO2 because of higher magnitude of enthalpy of solution (it is –ve and hence, released) otherwise, like CO2, it too would have escaped the system as gas, thereby driving the reaction forward. OH OH acetylation  →

NH2

9. 10.

NHCOCH3 Paracetamol Being a linear polymer of, natural rubber becomes soft at high temperature (62°C) and brittle at low temperature (< 10°C). It is, therefore, not used in making footwear for polar region. +3 and +4 states are shown by Ce in aqueous solution Dolomite (MgCO3.CaCO3) produces slag, which can be further used in building materials.

11.

2KMnO4 + 3H2SO4 + 5H2O2 → K2SO4 + 2MnSO4 + 8H2O + 5O2

6.

a+b+c 2+3+5 = =5 2 2

=

Meq of ( COOH)2 .2H2O

12.

Noxalic

13.

 6.3  3  126 / 2  × 10   = = 0.4 250 for neutralisation NoxalicVoxalic = NNaOHVNaOH ⇒ 0.4 × 10 – NNaOH × 40 ⇒ NNaOH = 0.1 N C would be sp3 hybridisation O would be sp3 hybridised If the molar volume is very large, the effect of attractive forces as well as molecular size is almost nil, both the pressure correction and volume correction can be neglected. Therefore the gas behaves ideally i.e. PV = nRT The gas in Van der Waals equation is characterised by Van der Waal’s coefficient that are dependent on the identity of gas but are independent of temperature or the effect of temperature or the effect of temperature on these constants is neglected.

14.

acid dihydrate

V(mL ) of ( COO )2 2H2O solution

In case of a real gas, the molecules strike the walls of the containers with less force because they are attracted by other molecules, therefore pressure is always less than pressure in an ideal gas, where molecules collide with a larger impact. 15.

NH2

N2Cl NaNO +HCl

2   →

F HBF

4 →

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6

AITS-CRT(Set-III)-PCM(S)-JEE(Main)/13

16. 17.

SN2 reaction takes place. ⇒ inversion of configuration OH

+K 2CO3 →

CH2OH O

OCH3

+

CH3 − I →

CH2OH

18. 19.

Lack of H-bonding in ether decreases boiling point. In the depresión of freezing point experiment (i) it is found that vapour pressure of solution is less than that of pure solvent. (ii) Only solvent molecules solidify as freezing point. Not applicable for solute particles. SUM = (2) + (3) + (4) = 9 O

20. 21.

22.

C

is meta directing group.

+ − M( s ) → Mn( aq ) + ne

for electrodes oxidation potential 2.303RT 0 EMn + / M = EM log10 Mn+  − n+ /M nF ∴ (1) and (3) correct SUM = 1 + 3 = 4 CH3 O C

O

+

CH3 NaOH / I2

CH  →

C O

B

A

O + CHI3 ↓ ( yellow )

23. OH C

C

+

H → CH 

H

C

C

CH →

C

H

C

CH

H O H

H

Tollen' s Test ←

CH CH CH

Tautomerise ZZZZZZZZ X YZZZZZZZ Z

O

24.

C H

C

C

H

OH

2+ 2− ZZZ X PbCO3 YZZ Z Pb + CO3

(x + y)

x

ZZZ X MgCO3 YZZ Z Mg

2+

y

+ CO32−

(x + y)

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7

∴

K sp PbCO3

x y

=

K sp MgCO3

AITS-CRT(Set-III)-PCM(S)-JEE(Main)/13

x 1.5 × 10 −15 = y 1× 10 −15

x = 1.5 y

x(x + y) = 1.5 × 10-15

1.5y (1.5y + y) = 1.5 × 10-15  1.5 × 10 −15  y=   3.75 

25. 26.

1 2

= 2 × 10 −8

∴ x = 1.5 × 2 × 10−8 = 3 × 10−8 M CsCl has body centered cubic unit cell. Each ion in this structure has a coordination no. of 8. Products Cl CHO +

H

O

O

O

O

C

C

C

C

H

Br

3 chiral C atoms Total number of stereoisomer = 2n = 23 = 8 27.

sp3

sp3d

Cl

sp2

Cl Cl

N

(a)

H

H

P

H

Cl Cl

B

Cl

Cl Cl O (b)

O

+

N

O

N O

sp

H +

O

N H

sp2

H H sp3

sp2

(c)

S O

O

sp2 S (d) H

H angular

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8

AITS-CRT(Set-III)-PCM(S)-JEE(Main)/13

28. H

O H3C

C

OH

CH2 CH3

C

C

CH2 CH3

CN above and below OH H3C

C

OH C2 H5

LiAlH4

←  H3C

CH2NH2

29.

C CN

CH2 CH3 (R as well as S) racemic mixture

O C

O NH

CH3

O C

OH + HN

CH3

COO + H2N

CH3

C

OH  →

18 1. H+ O CH3 ← 18  2. CH3 O H

(C) 30.

meq. Of NaOH = 10 × 0.1 = 1 meq. Of H2SO4 = 10 × 0.05 = 0.5 meq. Of NaOH > meq of H2SO4 ⇒ [OH–] > [H+] ⇒ pOH < 7

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9

Mathematics

AITS-CRT(Set-III)-PCM(S)-JEE(Main)/13

PART – III

1.

D = (2x2 + 4) − 2(−4x − 20) = 2x2 + 8x − 42 = 0 ⇒ α + β = −4.

2.

Total into = 23 − 2 = 6]

3.

G V = 2iˆ − 3ˆj + kˆ × ˆi + 2ˆj − 5kˆ

4.

f(x) = x(9 − x2) ; − 3 ≤ x ≤ 3

5.

Use L’ Hospital’s Rule twice

7.

Number of rectangles = n+1C2.n+1C2 =

8.

f ′( x) = −

(

) (

3

( x + 1)

4

)

n2 ( n + 1)

2

4

− 3 + cos x < 0

Hence f(x) is always decreasing, also as x → ∞, f(x) → −∞ and as x → −∞, f(x) → + ∞ Hence one positive and one negative root 9.

p(x) = ax2 + bx + c p(x) = Q1x + 4 = Q2(x + 1) + 3 = Q3(x − 1) + 1 ∴ p(0) = c = 4 …(1) p(−1) = a − b + 4 = 3 ⇒b−a=1 p(1) = a + b + 4 = 1 ⇒b+a=3 ∴ b = −1 ; a = −2 ∴ p(x) = 2x2 − x + 4 ∴ p(2) = − 8 − 2 + 4 = −6.

10.

As f(x) has continuous derivative ∀ x ∈ [a, b] ⇒ f(x) is continuous and as f(a) and f(b) have opposite signs ∃ at least one value a < x = c < b such that f(c) = 0 also since f′(x) ≠ 0 ⇒ f(x) > 0 or f′(x) < 0 throughout ⇒ function is increasing or decreasing ⇒ will have only one possible root.

11.

x2 + y2 = 1 let x = cosθ and y = sinθ so (x + y)2 can be written in the form of π  (cosθ + sinθ)2 = 2sin2  + θ  4  maximum value = 2.

12.

Put x = tanθ π /2

I=

∫ 1 + ( tan θ )a 0

13.

Let

dθ

( cos θ ) ∫ ( sin θ )a + ( cos θ )a dθ 0

π /2

=

a

⇒I=

π 4

1 = x so that as r → 0 r

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10

AITS-CRT(Set-III)-PCM(S)-JEE(Main)/13

cosax − cosbx.coscx 1 cosax − cosbx ⋅ coscx = lim → x 0 sinbx sincx bc x2 ⋅ ⋅ bc ⋅ x 2 bx cx use L’hospital’s to get (C). lim

x →0

14.

f′*(x) = 2x2 + c f′(−2) = 1 = 8 + c ⇒ c = −7 now f′(x) = 2x2 − 7 2 f(x) = x 3 − 7x + d 3 26 f(−2) = 0 ⇒ d = − 3 1 ∴ f(x) = [2x2 − 21x − 26] 3 45 = −15. f(1) = − 3

15.

If x ∈ 1st quadrant they y = 4 If x ∈ 2nd quadrant they y = −2 If x ∈ 3rd quadrant they y = 0 If x ∈ 4th quadrant they y = −2

y min = −2

16.

Degree of f(x) = n ; degree of f′(x) = n − 1 degree of f″(x) = (n − 2) hence n = (n − 1) + (n − 2) = 2n − 3 ∴n=3 hence f(x) s = ax3 + bx2 + cx + d, (a ≠ 0) f′(x) = 3ax2 + 2bx + c f″(x) = 6ax + 2b ∴ ax3 + bx2 + cx + d = (3ax2 + 2bx + c) (6ax + 2b) 1 ∴ 18a2 = a ⇒ a = . 18

17.

z1z1 = 1; z2 z2 = 1; z3 z3 = 1 given z1 + z2 + z3 = 0 ⇒ z1 + z2 + z3 = 0 hence (z1 + z2 + z3)2 = 0 1 1 1 ∑ z12 + 2z1z2 z3  z + z + z  = 0 hence 2 3  1

18.

∴

∑ z12 = 0

∴

( x + 2)2 = 3  y + 

The other conic is,

∑ z12 + 2z1z2 z3 [ z1 + z2 + z3 ] = 0

13  ⇒ latus rectum = 3 3 

( x − 3) 72

2

( x + 2) ( 7 / 2 )2

2

+

= 1 which is an ellipse

2b2 2.49 7 = = a 4.7 2 7 1 ∴ positive difference − 3 = 2 2 Latus rectum =

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11

19.

G G G Given V + V1 = V2 G 2 G G 2 V1 = V2 − V

( ) (

G G also V ∧ V1 = 2α G G and V ∧ V2 = 2α G G G also s V = V1 = V2 = λ say

;

)

hence, λ2 = 2λ2 − 2λ2 cosa ⇒ cosα ⇒ α = 2nπ ±

1 − cosa −

20.

AITS-CRT(Set-III)-PCM(S)-JEE(Main)/13

1 2

π 3

sin2 ( a / 2 )

cos2 ( a / 2 )

=

2cosa 1 + cosa

a sin2   2 k = 2 ; w = 1 ; p = 1 ⇒ k2 + w2 + p2 = 4 + 1 + 1 = 6

21.

Let x = 2 2 cosθ, y = 2 2 sinθ x − y = 2 2 (cosθ − sinθ) ∴ (x − y)max = 4.

22.

u(x) = 7v(x) ⇒ u′(x) = 7v′(x) ⇒ p = 7(given)  u(x)  u(x) p+q 7+0 = =1 again =7 ⇒   = 0 ⇒ q = 0 ; now  p−q 7−0 v (x)  v ( x) 

23.

f(x) = ax3 + bx2 + cx + d d2 y b is possible point inflexion also since changes sign as we move from left to right of x=− 3a dx 2 b it is a point of inflexion. x=− 3a

24.

y2 = 4ax, 4a = 2p > 0 and t1t2 = −1 4a2 t t ratio = 2 21 22 = −4 a t1 t 2

25.

lim

1

n→∞

( x − 2)

(dividing Nr and Dr by 3n)

n

3n

+3−

1 n n

for lim to be equal to n →∞

1 1  x −2 ; lim → 0 (which is true) and lim   → 0. n→∞  3  3 n→∞ n

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12

AITS-CRT(Set-III)-PCM(S)-JEE(Main)/13

26.

27.

28.

For f to be one-one f′(x) > 0 and f′(x) < 0 for all x clearly f is continuous at x = 0 and f(0) = −1 for x ≤ 0, f′(x) = 2(x + m) if m > 0 then nothing definite can be said about f′(x) hence m < 0 but m ≠ 0 as for x > 0, f is constant and ∀ m < 0, f′(x) < 0, ∀ x > 0

P

Q

2       2  3 3  1 + i 3    = 8 × 3(−ω)3 = −24.  2         

Let nC151 = q.nC150 where q ≠ 1 n! q.n! = (151)! (n − 151)! (150 )! (n − 150 )! 1 q ⇒ n − 150 = q.151 or n = 151q + 150 = 151 n − 150 smallest n will be when q = 2 n = 302 + 150 n = 452 ⇒ sum of the digit = 11.

29.

Coefficient of A in nth term = 8 + (n − 1) (−2) Coefficient of B in nth term = 2 + (n − 1) (−1) 10 − 2n = 2(3 − n) ⇒ 10 = 6 which is absurd

30.

F( x) =

3x + 2

dx ; let x − 9 = t2 ⇒ dx = 2t dt

x −9  3 t2 + 9 + 2  .2t  dt = 2∫ 29 + 3t 2 dt = 2 29t + t 3  ∴ F(x) = ∫    t   3/2 F ( x ) = 2 29 x − 9 + ( x − 9 )  + C   Given F(10) = 60 = 2[29 + 1] + C ⇒ C = 0 3/2 ∴ F ( x ) = 2 29 x − 9 + ( x − 9 )    F(13) = 2[29 × 2 + 4 × 2] = 4 × 33 = 132.

(

)

(

)

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