IIT-JEE (PAPER - I)





 →

^NH2OH,H2SO4 oxime formation followed by Beckmann rearrangement.

O||

C –NH–CH3

N-methyl benzamide

So, B is

CH| 3O C= and

A is

3 3 CH CHC| = C| 3.[D] (CH3)3CCH2CH2OH

Heat , O H

H , O Cr K

2 7 2 2



 →⁺

(CH3)3

) A ( 2 CCH COOH





 →

^SOCl2 (CH3)3 CCH2COCl (B)

(CH3)2NH

(CH3)3

) C ( 2 CCH CO||

N(CH3)2

(CH3)3

) D

CC H( 2CH2N(CH3)2 HO ether , LiAlH

2 4





←

4.[C] Attack of nucleophile is a rate determining step

F3C O

O O

I II CH3

I is more electron deficient and facilitates a faster attack.

5.[A] As water introduces, water dissolves HCl(g) and a press drop is produced Liquid level in the capillary rises.

6.[D]

7.[B] The energy of AOs depends on the

(n + l) values n + l value of (n – 1) d = n + 1 ; n + l value of ns = n

(n + l) value of (n + 1)d = n + 3 ; n + l value of nf

= (n + 3)

But due to lower value of principle quantum no.

energy of nf < (n + 1) d

∴ energy of (n + 2) s < nf.

8.[A] A, B and C are magnesium ,aluminium and silicon. Magnesium form ionic oxide, MgO ; Aluminium forms amphoteric oxide ,Al2O3 and silicon forms a giant molecule SiO2.

9.[ B,C,D] Due to resonance cyclohexatriene cation is aromatic which causes it's stability.

10.[ A,B,D] Catalyst lower the activation energy of forward reaction & backward direction keeping the same enthalpy of reaction.

11.[ A,B,C] Greater the value of (IE – ∆_eg H) greater is the electronegativity E.N. of

P = (1680 + 340)/(4.18) (125) ^~– 4 12.[A,B,C] H = E + PV

dP T

dH



 



 =

dP T

dE



 



 + P

dP T

dV



 



 + V

For liquid, dP T

dH



 



 =

dV T

dE



 





dP T

dV



 



 +

P dP T

dV



 



 + V

For incompressible liquid, dP T

dV



 



 _~

– 0.

∴ dP T

dH



 



 _~

– V

SOLUTION FOR MOCK TEST

IIT-JEE (PAPER - I)

For ideal gas,

13.[A] F(Monoester) Molecular weight = 186 No Br2 reaction ⇒ Saturated Two oxygen ⇒ 2 × 16 = 32 No. of CH2 = (186 – 32)/14 = 11

Hence, molecular formula of saturated monoester F, is C11H22O2

Hydrolysis

G H Optically active,

soluble in NaOH ⇒ Acid We have, Ag⁺ salt  →^Br2 ± J Hunsdiecker reaction;

radical intermediate, so racemic mixture (J contains one carbon less than G)

Optically active, (Alcohols are not soluble in NaOH) +ve iodoform test, it suggests

HO–

CH| 3–R

CH On warning with H2SO4

(dehydration) I (no diastereomers means no geometrical isomers) It suggests same alkyl group on one of the doubly bonded carbon atoms

H II.NaBr TsCl

.I optically active ⇒ J

It suggests G contains one more carbon atom than H.

Hence, molecular formula of ester F, is No cis-trans isomers

CH3

14.[C] I exists as diastereomers and H is optically active.

So, H is HO–

15.[D] Since H is optically active and gives negative iodoform test, so H is

or 0.1 v α + 1 = 2 + 2 × 10^–2 × v

Column Matching

19. [A] → r,s,t; [B] → p,r,s; [C] → s; [D] → q,t For CH3CH2CH2NH2^NaNO²^/^HCl→

product the

not intermediate; simplyDiazoniumCH3CH2CHion2isN2

on heating doesn't give alkene.

20. [A] → q,r; [B] → p,r; [C] → r,s; [D] → p,t

1.[A] Let us first count the number of elements in F.

Total number of functions from A to B is 3⁴ = 81.

The number of functions which do not contain x(y) [z] in its range is 2⁴.

∴ the number of functions which contain exactly two elements in the range is 3 . 2⁴ = 48.

The number of functions which contain exactly one element in its range is 3.

Thus, the number of onto functions from A to B is 81 – 48 + 3 = 36

[using principle of inclusion exclusion]

n (F) = 36.

Let f ∈ F. We now count the number of ways in which f ^–1(x) consists of single element.

We can choose preimage of x in 4 ways. The remaining 3 elements can be mapped onto {y, z}

is 2³ – 2 = 6 ways.

∴ f ^–1 (x) will consists of exactly one element in 4 × 6 = 24 ways.

Thus, the probability of the required event is 24/36 = 2/3

2.[A] Let E1 denote the event that the letter came from TATANAGAR and E2 the event that the letter came from CALCUTTA. Let A denote the event that the two consecutive alphabets visible on the envelope are TA. We have P(E1) = 1/2, P(E2) = 1/2, P(A / E1) = 2/8, P (A / E2) = 1/7. Therefore, by Bayes' theorem we have

P(E2 / A) =

3.[D] Required probability = 1 – P (all the letters are put in correct envelops)

The number of the ways of putting the letters in the envelops = ⁴P4 = 4!

The number of ways of putting letters in correct envelops = 1

In the present case

7.[D] Integrating by parts, the given integral is equal to x tan^–1 equation reduces to

dX homogeneous equation, so putting Y = υX, we get XdX

Equations of the given circles can be written as

(x – 3)² + y² = 3² (1)

and (x + 1)² + y² = 1² (2)

Equation of any tangent to circle (2) is

(x + 1) cos θ + y sin θ = 1 (3) This will be a tangent to circle (1) also if

θ common tangent (3) becomes

x + 1 = 1 or x = 0 (4) When cos θ = –1/2, we have sin θ = ± 3/2, and

the equations of the common tangents are –2

1(x + 1) ± 2

3 y = 1 ⇒ x – 3y+ 3 = 0 (5)

and x + 3y+ 3 = 0 (6)

12.[A, B, C, D]

Let F(x) =

∫

0^{x2 −}₊ ⁺t 2

e 2

4 t 5

t dt

⇒ F′(x) = ₂

x 2 4

e 2

4 x 5 x

+ +

− .2x

So from F′(x) = 0, we get x = 0 or x² =

2 16 25 5± −

= 2 3 5±

= 4, 1 Hence x = 0, ±2, ±1.

13.[B]

D C

–1 1

x = 4 x = –4

y = –4

y = +4

A B

Shaded region is S0. Area of S0

= 4 × 2 – 2

1 π (1)² = 8 – π/2

14.[D] y ∈ [0, 4], x ∈ [–1, 1]

m (t) = cost

lines y = 2x + 0.4 lies inside the region so

⇒ t ∈ [0, 1]

t² + (2t + 0.4)² – 1 ≥ 0 ⇒ t ∈ [0.28, 1]

15.[B] (Slope)max. = (cos t)max = cos (0.28) and point is (π, 1)

π

−

− x

1

y = cos (0.28)

16.[C]

S C

S C

2 1 =

2 1

r r =

1 3

x = 3 1 ) 3 ( 1 ) 1 ( 3

−

−

− =

2

−6

= – 3

17.[D] tangents = y = ± 3 x + RT

tan30º =

3

RT ⇒ RT = 3

18.[A]

• 2

C1

C2

60º 1

(h + 1)² + k² = (1 + 2)² (circle) Column Matching

19. [A] → r; [B] → p,r; [C] → s; [D] → r

(A)

∑

= 10

0 r

20C = r ²⁰C0 + ²⁰C1 +……+ ²⁰C10

But, ²⁰C0 + ²⁰C1 +……+ ²⁰C20 = 2²⁰

Also, ²⁰C20 = 1 = ²⁰C0, ²⁰C19 = ²⁰C1, ²⁰C18 = ²⁰C2

etc.

∴ given sum = (²⁰C0 + ²⁰C1 +……+ ²⁰C20) – (²⁰C11 +…..+ ²⁰C20)

2²⁰ + ²⁰C10– (²⁰C10 + ²⁰C9 + ……+ ²⁰C0)

∴ 2 (²⁰C0 + ²⁰C1 +…..+ ²⁰C10)

= 2²⁰ + ²⁰C10

(B)

∑

= 100

0 r

100C (x –3)r ^100–r 2^r

= ((x–3) +2)¹⁰⁰ = (x –1)¹⁰⁰ = (1 –x)¹⁰⁰

∑

=

−

100

0 r

r r

100C ( x) =

∑

=

−

−

100

0 r

r r 100 r r( 1) C x )

1 (

∴ Coeff. of x⁵³ = (–1)⁵³¹⁰⁰C53 = – ¹⁰⁰C53

(C) We have

(1+ x)¹⁰ = ¹⁰C0 + ¹⁰C1 x +¹⁰C2x² +……+ ¹⁰C10 x¹⁰ ....(1) Also (1–x)¹⁰ = ¹⁰C0 – ¹⁰C1x + ¹⁰C2x² +…….

…..+ ¹⁰C10x¹⁰ ....(2) Multiplying, we get

(1 –x²)¹⁰ = (¹⁰C0 + ¹⁰C1 x + ¹⁰C2x² +……

….+ ¹⁰C10x¹⁰) × (¹⁰C0 –¹⁰C1x + ¹⁰C2x²+…

…...+ ¹⁰C10 x¹⁰⁾

Equating the coefficients of x¹⁰, we get

10C5 (–1)⁵ =¹⁰C010C10 –¹⁰C1 10C9 + ¹⁰C2 10C8 +…

….+ ¹⁰C1010C0

⇒ – ¹⁰C5 = (¹⁰C0)² – (¹⁰C1)² + (¹⁰C2)² +……

…+ (¹⁰C10)²

(D) ⁹⁵C4 +

∑

= 5 − 0 j

j 3 100 C

= ⁹⁵C4 + ⁹⁹C3 + ⁹⁸C3 + ⁹⁷C3 + ⁹⁶C3 +⁹⁵C3

= (⁹⁵C4 + ⁹⁵C3) +⁹⁶C3 + ⁹⁷C3 + ⁹⁸C3+ ⁹⁹C3

= (⁹⁶C4 + ⁹⁶C3) + ⁹⁷C3 +⁹⁸C3 +⁹⁹C3

= (⁹⁷C4 +⁹⁷C3) +⁹⁸C3 + ⁹⁹C3

= (⁹⁸C4 + ⁹⁸C3) + ⁹⁹C3

= ⁹⁹C4 + ⁹⁹C3

= ¹⁰⁰C4

20. [A] → p,q; [B] → q,t; [C] → q; [D] → s (A) Given lines intersect if

1 2 1 1

5 4 4 3 1 2

λ

λ

−

−

−

= 0

⇒ λ = 0, – 1

(B) xlim 4x →∞

















+ + +

+

− +

−

2 x

1 1 x

2 x

1 1 x

tan ¹ =2 = y² + 4y + 5

⇒ y = –1, – 3 (C) y² – ax (– x – y) = 0

⇒ for perpendicular lines a + b = 0

⇒ 1 + a = 0 ⇒ a = – 1 (D) (a ×b ) ×a = (jˆ–kˆ) ×a

⇒ (a .a )b – (a .b ) a = (jˆ–kˆ) ×a on solving,

we get b = iˆ

PHYSICS

1.[B] T = m1r1ω12also T = m2 2 2 2 2 2r g +ω

∴ m1r1ω12 = m2 2 22 2 ( 2r ) g + ω or, 0.1 × ω₁² =

2

1 _{2 2}

) 10 ( ) 10

( +

0.1 ω12 = 1

ω₁ = 10 rad/s v1 = r1ω₁ = 10 m/s 2.[A] As f = µN = mg

or, µmlω² = mg ⇒ ω = µl

g

3.[C] At terminal velocity net force is zero.

6πη(r₁ + r2) VT +

34 π (r₁³ + r23) ρg =

34 π (r₁³ + r23) σg 4.[C] As supporting plane is lowered slowly

∴ N = mg – kx

5.[D] Net force acting on container due to liquid coming out from the holes is given by

F = ρA _^ × − × _^ 4 g H 4 2

H g 3

2 = ρgAH towads left

∴ F = f = ρgAH towards right.

Now, τ_F = ρgAH × 2

H into the plane of paper.

τf = ρgAH × 2

H out of plane of papers

∴ τF = τf hence τN = 0

6.[C] TA = R n

V P

A A

A and TB = R n

V P

B B B

Given, PA = PB , VA = VB and nA = 2nB

∴ TA = 2 T_B

Now,

A B B A B A

M M T T V

V = × = 2

7.[A] Equivalent circuit

2, 2'

V0

ε2 C2V0

C1V0

ε1

1 1' 2 2'

3 3'

Middle plate 11' 33'

Total charge on 2 & 2' plate = V d

A d

A

2 0 2 1 0

1 



 



εε +ε ε

σ = ε₀V _



 



 ε

ε +

2 2 1 1

d

d

8.[A]

A

30 B

15 30ºC

15º

O AI

120º 30º C 45º

R B O

l

Sine rule

º 45 sin

R = º 30 sin

l

l

R = 2

º 30 sin

º 45 sin =

9.[A,B] As aT = aN

∴ ds vdv =

R v²

− which can also

be written as dt dv = –

R v²

Integrating the above equations answer is obtained.

10.[A,C] (A) ρ_BLAg = ρ × 4

LAg + 2ρ × 4 LAg

(B) F_B = ρ × 4

LAg + 2ρ × 4

L 3 Ag

FB = 4

7 ρLAg; a = m

mg F_B−

11.[B,D]

c 6V

3Ω b 8V

a

10

13V

•

11

← eq. ckt.

Using Kirchoff's Law Solve the circuit.

12.[A,B,C]

V

CV –CV

If battery is disconnected and plate are pulled apart, then charge will remain constant

E = A 0 2

Q

∈ × 2 =

0 Q

∈ Α

∴ E remain same (A) is correct work is done against attractive force



 →





 

← ⁻

+ Felc. Fext by Fext. (B) is correct.

U = 2 1 CV²

V = constant [as battery is connected]

C = d

0A

∈

as d increase

C decrease ∴ U decrease option (C) is correct.

13.[C] As springs are in parallel

∴ a =

) m m (

x ) k k ( mass

F

2 1

2 1 net

+

= +

14.[A] Frictional force on m2 will act in direction of displacement if

k2x > m2 a 15.[A] as k2A – µm₂g = m2amax

∴ k₂A – µm₂g = m2 



 



 + +

2 1

2 1

m m

k

k A

Solve to get answer. ] 16.[C] Induced emf across OP =

2

B 2 2

1 



 

 ω l =

8 Bωl²

(i) current = R 8 Bωl²

… (i)

Torque on the rod = 2

∫

^/2

0

dx x Bi

l

= 4

Bil² …(ii)

∴ 32R

B dt d 12

Ml²× ω=− ²ωl⁴ [substituting τ = I α]

– 8RM

B 3

d ^{2 2}

0

= l ω

∫

^ω ω

ω

∫

^t

0

dt Solving this eq. & eq. (i)

i =

R 8 Bω₀l²

e^–αt

17.[B] θ =

∫

^∞

0

dt i

18.[A] Heat generated = I ²₀ 21 ω Column Matching

19. [A] → p; [B] → p; [C] → r; [D] → q As cube is floating ρsALg = ρLAxg

∴ x = _



 



 ρ

ρ L

s L

20. [A] → r; [B] → p; [C] → s; [D] → q S = 2

1 × 2 × 16 = 16 m

| Wg | = mg S = WN = m(g + a) cos²θ. S Wf = m(g + a) sin²θ . S

CHEMISTRY

1.[C]

NH2

NOBr

NH2 +

Br^– Br

+ Enantiomer

2.[B] ^CH2=C

COOH Ph

H2/Pd CH3–CH

COOH

Ph (resolvable) 'A'

'B'

CH=CH–COOH H2/Pd

Optically inacitve CH2–CH2–COOH

3.[B] Distribution of electrons in the MO's in He2 is σ_1s² σ∗_1s² . He2 is unstable

Distribution of electrons in the MOS in H2 is σ_1s² . H2 is stable.

4.[B] After mixing total moles of A^–

= 100 × 0.2 × 10^–3 + 100 × 0.3 × 10^–3 = 100 × 10^–3 × 0.5 moles

After mixing total moles of HA

= 100 × 0.1 × 10^–3 + 100 × 0.2 × 10^–3

= 100 × 0.3 × 10^–3 moles After mixing resulting pH = 5 + log

3 5

5.[A,B,C] ^CH2–CH*

NH2

COOH HO

HO

It contains 2 phenolic hydrogens and a carboxylic acidic hydrogens

CH2–HC NH2

COO^– HO

HO

+

Zwitter ion 6.[A,B,C]

Grignard reagent reacting with acyl halide usually gives 3º alcohol.

7.[B,C,D] The order of H-bond energies

F – H ^…….. F^– > F – H ^…….. O > F – H ^…….. F >

O – H ^…….. O > O – H ^…….. F > N – H ^…….. N 8.[C,D]

9.[C,D]

RHE reaction : Hg2Cl2(s) + 2e → 2Hg(l) + 2Cl^– LHE reaction : 2Ag(s) + 2Cl^– → 2AgCl(s) + 2e Net reaction : Hg2Cl2 (s) + 2Ag(s)

→ 2Hg (l) + 2AgCl(s) In case of same concentration of Cl^– ions in the two half cells, Ecell is independent on the concentration of Cl^–. Other substances are either pure solids or liquids, which have unit activities irrespective of their amounts.

Column Matching :

10. [A] → p, r, s ; [B] → q;

[C] → p, q,,s ; [D] → q 11. [A] → p, q,t ; [B] → p, r, s;

[C] → p, q, r; [D] → p, q, r Numerical Response type questions :

12. [4]

E W =

96500 Q or

n / 4 . 106

977 .

2 =

96500 60 60 1 3× × ×

∴ n = 4 13. [8]

O

) reductionKishner Wolff (

glycol / KOH

NH NH2 2

−





 →



(W)

) Ozonolysis (

O H / O_{3 2}





 →



O HO O

HO

(X)

SOLUTION FOR MOCK TEST

PAPER - II

In document Xtraedge April 2010 (Page 89-97)