Concept Recap Test Mains 3 Sol

ANSWERS, HINTS & SOLUTIONS

CRT (Set-III)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. A C 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 _B C 25. D C C 26. C A A 27. D B B 28. C B D 29. C C D 30. A A B

ALL INDIA TEST SERIES

FIITJEE

JEE (Main)-2013

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

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PART – I

1. Use Kepler’s 3rd law :

2 3 2 2 1 1

T

R

T

R









=

















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

3. Angular momentum of the comet is conserved.

4. Z = R2+X2 where X =X_L −X_C 6. 2 2

V

H I Rt

t

R

=

=

1

H

R

∴

α

if V = constant

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

Also current through

2

3

12I

5

Ω = ×

Heat through

4

Ω =

(6I)

2

× Ω =

4 t H

₁

Heat through 2 2

3

2

12I

2 t H

5





Ω =

_

×

_

× Ω =





2 1

H

<

H

7. As the capacitors are uncharged, the initial current

eq

V

;

R

=

Req = (1.5 + 0.5) Ω = 2 Ω. 8. EStored =

1

Li

2

2

; and emf = di L dt 9. V S n , t = = λ number of waves =S λ= 2000m 2500 0.8 m = = 10. _L eS t I ∆ = ∆ = 0.2 H 1 2 LI LI E 2 2 1 2 E , 2 2 = =

(

)

L = I I J 2 2 2 2 1 ∆E − =7.5

11. 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 cos L α − ⇒ θ = − A A 1 Lcos cos−  α −  ⇒ θ = _ A_

12. hr = constant for capillarity

∴ h = 6 cm for a vertical tube

o h 12 cm cos60 ∴ =A = 13. Acceleration = V(g a)( d) Vd + ρ − 2 20 m / s = 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

)

MSD M − − M M 1 1 MSD MSD M M − + = = b LC M ⇒ = .

15 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 =

2 N

v

a

, 19. 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. mg sin θ µ θ mg co s θ 20. 2 T π ω = AN R sin 2 θ = t AN R sin 2 ω = t AB 2R sin 2 ω = A O B N R θ/2 AB average velocity = t ∴ 2R 2 3 sin 2 2 3 ω π ω π ω 3 3 R 2 ω = π .

21. BG is parallel to vG. 23.

i

e

T

=

Now

T

2

∝

r

3 and

r

∝

n

2

⇒

T

∝

n

3 3

1

i

n

∴

∝

∴

the current will increase 8 times.

∴

(C) 25. 1 2

T

T

= 3 1 2

n

n













= 8

∴

1 2

n

n

= 2.

26.

S

_min

= + −

i

₁

i

₂

A

for Smin ⇒ i1 = i2

30 = 2I1 − 60 i1 = 45° 27. 7 2 4 5

D

D

6 10

40 10

d

20 10 m 0.2cm

d

12 10

− − − −

λ

λ

×

× ×

β =

⇒

=

=

=

×

=

β

×

.

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

M m (M m) 0.5 2.5+ = + or M 1 1 m 1 1 0.5 2.5  _{− =}  ₋          or M m 1.5 m 3 2.5 5     = _{ }= _{ }     ∴ M 3 m =5

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

C

C

h

h

e

e

m

m

i

i

s

s

t

t

r

r

y

y

PART – II

4. H3BO3 + Na2CO3 → Na2B4O7 + H2O + CO2 ↑

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

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.

5. OH NH₂ acetylation → OH NHCOCH₃ Paracetamol

6. 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.

9. +3 and +4 states are shown by Ce in aqueous solution

10. Dolomite (MgCO3.CaCO3) produces slag, which can be further used in building materials.

11. 2KMnO4 + 3H2SO4 + 5H2O2 → K2SO4 + 2MnSO4 + 8H2O + 5O2 a b c 2 3 5 5 2 2 + + ₌ + + ₌ 12.

(

)

( )

(

)

2 2 oxalic acid dihydrate

2 mL 2 Meq of COOH .2H O N V of COO 2H O solution = 3 6.3 ₁₀ 126 / 2 _0.4 250  _{ ×}     = = for neutralisation

NoxalicVoxalic = NNaOHVNaOH

⇒ 0.4 × 10 – NNaOH × 40

⇒ NNaOH = 0.1 N

13. C would be sp3 hybridisation O would be sp3 hybridised

14. 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.

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 2 NaNO +HCl → N₂Cl 4 HBF → F

16. SN2 reaction takes place. ⇒ inversion of configuration 17. OH CH₂OH 2 3 K CO + → O CH₂OH 3 CH I + − → OCH₃

18. Lack of H-bonding in ether decreases boiling point. 19. 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

20. C

O

is meta directing group. 21. M_{( )s} →M_{( )}n_aq+ +ne−

for electrodes oxidation potential

n 0n ₁₀ n M / M M / M 2.303RT E E log M nF + = + − _ +_ ∴ (1) and (3) correct SUM = 1 + 3 = 4 22. C O CH₃ CH O A B

+

NaOH / I2→ _{C O} CH₃ O

(

)

3 CHI yellow + ↓ 23. C OH H C CH H+→ C H C CH → C H C CH O H H C H C C OH H Tautomerise ZZZZZZZZX YZZZZZZZZ CH CH CH O Tollen's Test← 24.

(

)

2 2 3 3 PbCO Pb CO x x y +₊ − + ZZZX YZZZ

(

)

2 2 3 3 MgCO Mg CO y x y +₊ − + ZZZX YZZZ

sp 3 sp 3 K PbCO x K MgCO y ∴ = 15 15 x 1.5 10 y 1 10 − − × = × x = 1.5 y x(x + y) = 1.5 × 10-15 1.5y (1.5y + y) = 1.5 × 10-15 1 15 2 8 1.5 10 y 2 10 3.75 − −  ×  =_{ } = ×   8 8 x 1.5 2 10− 3 10 M− ∴ = × × = ×

25. CsCl has body centered cubic unit cell. Each ion in this structure has a coordination no. of 8. 26. Products Cl Br CHO _{H C C C C H} O O O O + 3 chiral C atoms

Total number of stereoisomer = 2n = 23 = 8 27. N H _H H sp3 P Cl Cl Cl _Cl Cl sp3_d B Cl Cl Cl sp2 (a) N+ O O sp N O O O sp2 N+ H H H H sp3 (b) S O O sp2 (c) S H H sp2 angular (d)

28. C H3 C CH2 CH3 O H C C CH2 CH3 OH CN above and below

C

H3 C CH2 CH3

OH

CN (R as well as S) racemic mixture

4 LiAlH ← C C2H5 OH C H3 CH₂NH₂ 29. C NH O CH3 OH→ C O OH HN CH3 COO H2N CH3 (C) 18 3 1. H 2. CH O H + ← C O O18 CH₃ + + 30. meq. Of NaOH = 10 × 0.1 = 1 meq. Of H2SO4 = 10 × 0.05 = 0.5

meq. Of NaOH > meq of H2SO4

⇒ [OH–_{] > [H}+_]

M

M

a

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t

t

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PART – III

1. D = (2x2 _{+ 4) − 2(−4x − 20) = 2x}2 _{+ 8x − 42 = 0 ⇒ α + β = −4.} 2. Total into = 23 − 2 = 6] 3. VG =

(

2i 3 j kˆ− ˆ ˆ+

) (

× +ˆi 2j 5kˆ− ˆ

)

4. f(x) = x(9 − x2_{) ; − 3 ≤ x ≤ 3}

5. Use L’ Hospital’s Rule twice 7. Number of rectangles = n+1_C 2.n+1C2 =

(

)

2 2 n n 1 4 + 8.

( )

(

)

4 3 f x 3 cos x 0 x 1 ′ = − − + < +

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 _{+ y}2 _{= 1}

let x = cosθ and y = sinθ so (x + y)2 _{can be written in the form of}

(cosθ + sinθ)2 _{= 2sin}2

4 π  _{+ θ}     maximum value = 2. 12. Put x = tanθ I =

(

)

(

)

(

) (

)

a /2 /2 a a a 0 0 cos d _d

1 tan sin cos

π _θ π _θ = θ + θ θ + θ

∫

∫

⇒ I = 4 π 13. Let 1 x r = so that as r → 0

2

x 0 ₂ x 0

cosax cosbx.coscx 1 cosax cosbx coscx

lim lim sinbx sincx _{bc x} bc x bx cx → → − ₌ − ⋅ ⋅ ⋅ ⋅

use L’hospital’s to get (C). 14. f′*(x) = 2x2 _{+ c} f′(−2) = 1 = 8 + c ⇒ c = −7 now f′(x) = 2x2 _{− 7} f(x) = 2_x3 _{7x d} 3 − + f(−2) = 0 ⇒ d = 26 3 − ∴ f(x) = 1 3 [2x 2 _{− 21x − 26]} f(1) = 45 3 − = −15. 15. If x ∈ 1st _{quadrant they y = 4} If x ∈ 2nd _{quadrant they y = −2}

If x ∈ 3rd _{quadrant they y = 0 y min = −2}

If x ∈ 4th _{quadrant they y = −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 _{+ bx}2 _{+ cx + d = (3ax2 + 2bx + c) (6ax + 2b)} ∴ 18a2 _{= a ⇒ a =} 1 18. 17. z z_{1 1}=1; z z_{2 2} =1; z z_{3 3} = 1 given z1 + z2 + z3 = 0 ⇒ z1+z2+z3 = 0 hence (z1 + z2 + z3)2 = 0 2 1 1 2 3 1 2 3 1 1 1 z 2z z z 0 z z z   + _ + + _=  

∑

hence 2

[

]

1 1 2 3 1 2 3 z +2z z z z +z +z =0

∑

∴ 2 1 z =0

∑

18. ∴

(

)

2 13 x 2 3 y 3   + = _ + _   ⇒ latus rectum = 3

The other conic is,

(

) (

)

(

)

2 2 2 2 x 3 x 2 1 7 7 / 2 − + + = which is an ellipse Latus rectum = 2b2 2.49 7 a = 4.7 = 2 7 1

19. Given V VG +G₁=VG₂ ; also V VG ∧ G₁= α2

( ) (

2

)

2

1 2

VG = VG −VG and V VG∧ G₂ = α2

also sVG = VG₁ = VG₂ = λ say hence, λ2 = 2λ2 − 2λ2 cosa ⇒ cosα 1

2 ⇒ α = 2nπ ± 3 π 20.

(

)

(

)

2 2 2 sin a / 2 1 cosa cos a / 2 2cosa a 1 cosa sin 2 − − = +       k = 2 ; w = 1 ; p = 1 ⇒ k2 _{+ w}2 _{+ p}2 _{= 4 + 1 + 1 = 6}

21. Let x =2 2cosθ, y =2 2sinθ x − y =2 2(cosθ − sinθ)

∴ (x − y)max = 4.

22. u(x) = 7v(x) ⇒ u′(x) = 7v′(x) ⇒ p = 7(given)

again

( )

( )

u x 7 v x = ⇒

( )

( )

u x 0 v x   =       ⇒ q = 0 ; now p q 7 0 1 p q 7 0 + ₌ + ₌ − − 23. f(x) = ax3 + bx2 + cx + d b x 3a

= − is possible point inflexion also since

2 2

d y

dx changes sign as we move from left to right of

b x 3a = − it is a point of inflexion. 24. y2 = 4ax, 4a = 2p > 0 and t1t2 = −1 ratio = 2 1 2 2 2 2 1 2 4a t t 4 a t t = − 25.

(

)

n n n 1 lim x 2 ₃ 1 n 3 →∞ ₋ + − (dividing Nr and Dr by 3n) for nlim→∞ to be equal to 1 3 ; n 1 lim n

→∞ → 0 (which is true) and

n n x 2 lim 3 →∞ −       → 0.

26. 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 27. 2 3 1 i 3 2 3 2  _{  + }   _{  }       _{  }    = 8 × 3(−ω)3 _{= −24.} 28. Let nC151 = q.nC150 where q ≠ 1

( ) (

151 ! n 151 !n!−

) (

= 150 ! n 150 !

) (

q.n!−

)

1 q 151 n 150= − ⇒ n − 150 = q.151 or n = 151q + 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 2dx x 9 + = − ; let x − 9 = t 2 _{⇒ dx = 2t dt} ∴

( )

(

)

(

)

2 2 3 3 t 9 2 F x .2t dt 2 29 3t dt 2 29t t t  _{+ +}      = = + = _ + _    

∫

∫

( )

(

)

3/2 F x =2 29 x 9_ − + x 9− _+C Given F(10) = 60 = 2[29 + 1] + C ⇒ C = 0 ∴ F x

( )

=2 29 x 9_ − +

(

x 9−

)

3/2_ F(13) = 2[29 × 2 + 4 × 2] = 4 × 33 = 132.