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Table of Contents
                            AIEEE CHAPTERWISE PHYSICS
	AIEEE CHAPTERWISE PHYSICS
		Chapter _1 Units and Measuremen
		Chapter _ 2 Description of Moti
		Chapter _ 3 Description of Moti
		Chapter _ 4  Law of Motion (11-
		Chapter _ 5  Work Power and Ene
		Chapter _ 6  Rotatioonal Motion
		Chapter _ 7 Gravitation (29-32)
		Chapter _ 8  Properties of Matt
		Chapter _ 9  Oscillation (37-42
		Chapter _ 10 Waves (43-46)
		Chapter _ 11
		Chapter _ 12 Transfer of Heat (
		Chapter _ 13  Electrostatics (5
		Chapter _ 14  Current Electrici
		Chapter _ 15  Thermal and chemi
		Chapter _ 16  Magnetic effect o
		Chapter _ 17  Magnetism (81-84)
		Chapter _ 18  Electromagnetic I
		Chapter _ 19  Ray optics (91-94
		Chapter _ 20  Wave optics (95-9
		Chapter _ 21  Electromagnetic w
		Chapter _ 22  Electrons and Pho
		Chapter _ 23  Atoms, molecules
		Chapter _ 24  Solids and semico
		Chapter _ 25  Practical Physics
	Solved Paper_2009
	Solved Paper_2010
	Solved Paper_2011
chemistry part
	Chapter_1_Atoms, Molecules and
	Chapter_2_States of Matter (03-
	Chapter_3_Atomic Structure, Che
	Chapter_4_Solutions (15-20)
	Chapter_5_Energetics (21-26)
	Chapter_6_Equilibrium (27-34)
	Chapter_7_Redox Reactions and E
	Chapter_8_Kinetics (41-44)
	Chapter_9_Nuclear Chemistry (45
	Chapter_10_Surface Chemistry (4
	Chapter_11_Metallurgy (51-52)
	Chapter_12_Periodic Properties
	Chapter_13_Chemistry of Non-met
	Chapter_14_Chemistry of Metals
	Chapter_15_Chemistry of Represe
	Chapter_16_Transition Metals in
	Chapter_17_Coordination Compoun
	Chapter_18_General Organic Chem
	Chapter_19_Hydrocarbons (87-92)
	Chapter_20_Halides and Hydroxy
	Chapter_21_Aldehydes, Ketones,
	Chapter_22_Nitrogen Containing
	Chapter_23_Polymers (107-108)
	Chapter_24_Biomolecules (109-11
	Chapter_25_Chemistry in Action
	Solved Paper 2009
	Solved Paper_2010
	Solved Paper 2011
mathematics book
	mathematics book
		Chapter _ 1 Set, relations and
		Chapter _ 2 Complex numbers
		Chapter _ 3 Matrices and determ
		Chapter _ 4 Quadratic equations
		Chapter _ 5 Permutation and com
		Chapter _ 6 Mathematical induct
		Chapter _ 7 Binomial Theorem
		Chapter _ 8 Sequences and serie
		Chapter _ 9 Differential calcul
		Chapter _ 10 Integral Calculus
		Chapter _ 11 Differential equat
		Chapter _ 12 Two dimensional ge
		Chapter _ 13 Three dimensional
		Chapter _ 14 Vector algebra
		Chapter _ 15 Statistics
		Chapter _ 16 Probability
		Chapter _ 17 Trigonometry
		Chapter _ 18 Mathematical Logic
		Chapter _ 19 Statics
		Chapter _ 20 Dynamics
	Solved Paper_2009
	Solved Paper_2010
	Solved Paper_2011
	2012 physics
	2012 chemistry
	2012  MATHS
                        
Document Text Contents
Page 1

1

Page 2

MTG Books
(A unit of MTG Learning Media (P) Ltd.)

New Delhi | Gurgaon

Page 220

CHEMISTRY  65 

1.  (a) : 4KO 2 + 2CO 2 → 2K 2 CO 3 + 3O 2 
2.  (c)  : The  negatively  charged  colloidal  particles  of 

impurities get neutralised by the Al 3+  ions and settle 
down and pure water can be decanted off. 

3.  (d) : Be forms water soluble BeSO 4 , water insoluble 
Be(OH) 2  and BeO. Be(OH) 2  is insoluble  in NaOH 
giving sodium beryllate Na 2 BeO 2 . 

4.  (b) : The precipitate of mercuric  iodide dissolves 
in excess of potassium iodide forming a complex, 
K 2 HgI 4 . 

HgI 2 + 2KI → K 2 HgI 4 
HgI 2 on heating liberates I 2 gas. 

HgI 2  Hg   +   I 2 
violet vapours 

5.  (d) : When heated at red heat, AgNO 3 decomposes 
to metallic silver. 

2AgNO 3 → 2Ag + 2NO 2 + O 2 
6.  (a) :Mercury is such a metal which exists as liquid 

at room temperature. 

7.  (c) : Mg 3 N 2  + 6H 2 O → 3Mg(OH) 2 + 2NH 3 
8.  (a) :Beryllium has the valency +2 while aluminium 

exhibits its valency as +3. 

9.  (b) : Al 2 Cl 6 + 12H 2 O  2[Al(H 2 O) 6 ] 3+ + 6Cl – 

10.  (b) :  Grey  tin  is very brittle  and  easily  crumbles 
down to a powder  in very cold climates. 

Grey tin  White tin 
(cubic)  (tetragonal) 

The change of white tin to grey tin is accompanied 

by increase in volume. This is called tin disease or 
tin plague. 

11.  (c) : CN –  ions act both as reducing agent as well 
as good complexing agent. 

12.  (b)  :  Calcium  carbide  is  ionic  carbide  having 

[  C      C  ] . .  . .  2– . 

[  C      C  ] . .  . .  2– Ca 2+ 
1 σ 

2 π 

13.  (b) :Aluminium chloride in aqueous solution exists 
as ion pair. 
2AlCl 3 + aq.→ [AlCl 2 (H 2 O) 4 ] + ( ) + [AlCl 4 (H 2 O) 2 ] – ( ) 
The crystallisation of AlCl 3  from aqueous solution, 
therefore,  yields  an  ionic  solid  of  composition 
[AlCl 2 (H 2 O) 4 ] +  [AlCl 4 (H 2 O) 2 ] – ∙ H 2 O.  This 
compound decomposes at about 190°C to give the 
non­ionic dimer Al 2 Cl 6 . 

[AlCl 2 (H 2 O) 4 ] +  [AlCl 4 (H 2 O) 2 ] –  ∙ H 2 O  190°C 
heat 

Al 2 Cl 6 + H 2 O 

14.  (c) : The elements of group 14 show an oxidation 
state of +4 and +2. The  compounds showing  an 
oxidation state of +4 are covalent compound and 
have tetrahedral structures. . SnCl 4 , PbCl 4 , SiCl 4 , 
etc. whereas those which show +2 oxidation state 
are ionic in nature and behave as reducing agent.

. SnCl 2 , PbCl 2 , etc. 
Further as we move down the group, the tendency 
of the element to form covalent compound decreases 
but the tendency to form ionic compound increases.

JEE MAIN 200

Page 221

CHEMISTRY  67 

1.  In case of nitrogen, NCl 3  is possible but not NCl 5 
while in case of phosphorus, PCl 3  as well as PCl 5 
are possible. It is due to 
(a)  availability  of  vacant  d  orbitals  in P but  not 

in N 
(b)  lower electronegativity of P  than N 
(c)  lower  tendency  of  H­bond  formation  in  P 

than N 
(d)  occurrence of P  in  solid while N  in gaseous 

state at room temperature. 
(2002) 

2.  Which one of  the following statements is correct? 
(a)  Manganese salts give a violet borax test in the 

reducing flame. 
(b)  From a mixed precipitate  of AgCl  and AgI, 

ammonia solution dissolves only AgCl. 
(c)  Ferric  ions give  a  deep  green  precipitate  on 

adding potassium ferrocyanide solution. 
(d)  On  boiling  a  solution  having K + ,  Ca 2+  and 

HCO 3 – ions we get a precipitate of K 2 Ca(CO 3 ) 2 . 
(2003) 

3.  Several blocks of magnesium are fixed to the bottom 
of a  ship to 
(a)  keep away the sharks 
(b)  make the ship lighter 
(c)  prevent action of water and salt 
(d)  prevent puncturing by under­sea rocks. 

(2003) 

4.  In curing  cement plasters water  is  sprinkled from 
time to time. This helps in 
(a)  keeping  it cool 
(b)  developing interlocking needle­like crystals of 

hydrated silicates 
(c)  hydrating sand and gravel mixed with cement 

(d)  converting sand into silicic acid. 
(2003) 

5.  The  solubilities  of  carbonates  decrease  down  the 
magnesium group due to a decrease in 
(a)  lattice energies of solids 
(b)  hydration energies of cations 
(c)  inter­ionic  attraction 
(d)  entropy of  solution  formation. 

(2003) 

6.  The substance not likely to contain CaCO 3  is 
(a)  a marble statue 
(b)  calcined gypsum 
(c)  sea shells 
(d)  dolomite. 

(2003) 

7.  Among Al 2 O 3 , SiO 2 , P 2 O 3 and SO 2 the correct order 
of acid strength is 
(a)  SO 2 < P 2 O 3 < SiO 2 < Al 2 O 3 
(b)  SiO 2 < SO 2 < Al 2 O 3 < P 2 O 3 
(c)  Al 2 O 3 < SiO 2 < SO 2 < P 2 O 3 
(d)  Al 2 O 3 < SiO 2 < P 2 O 3 < SO 2 . 

(2004) 

8.  Of the following outer electronic configurations of 
atoms,  the  highest  oxidation  state  is  achieved by 
which one of  them? 
(a)  (n – 1)d 8 ns 2  (b)  (n – 1)d 5 ns 1 
(c)  (n – 1)d 3 ns 2  (d)  (n – 1)d 5 ns 2 . 

(2004) 

9.  Which  of  the  following oxides  is  amphoteric  in 
character? 
(a)  CaO  (b)  CO 2 
(c)  SiO 2  (d)  SnO 2 

(2005) 

15

JEE MAIN 201

Page 439

Solved Paper-2012 39

10. (c) : f : R R, f x x
x

( ) [ ] cos=
−





2 1
2

π

= −






[ ] cosx xπ
π
2

= [x] sin x
Let n be an integer.

lim ( )
x n

f x
→ +

= 0 ,


lim ( )
x n

f x
→ −

= 0

f(n) = 0
f(x) is continuous for every real x.

11. (d) : Statement 1 :
1 + (1 + 2 + 4) + (4 + 6 + 9) + .... + (361 + 380 + 400)
is 8000
Statement 2 : ( ( ) )k k n

k

n
3 3

1

31− − =
=


Statement 1 : T1 = 1, T2 = 7 = 8 – 1,
T3 = 19 = 27 – 8 Tn = n3 – (n – 1)3

Statement 2 is a correct explanation of
statement 1.

12. (c) : Let the equation of the circle is (x – 1)2 +
(y – k)2 = k2

It passes through (2, 3)
1 + 9 + k2 – 6k = k2

⇒ = ⇒ =k
5
3

10
3

diameter

13. (b) : A Au Au=
















=
















=
















1 0 0
2 1 0
3 2 1

1
0
0

0
1
0

1 2, ,

Let u
a
b
c

1 =


















Au1

1
0
0

=
















a = 1, 2a + b = 0
b = – 2, 3a + 2b + c = 0 c = 1

Let u
p
q
r

2 =
















Au2

0
1
0

=
















p = 0, 2p + q = 1 q = 1,
3p + 2q + r = 0 r = – 2

u u1 2

1
2
1

0
1
2

1
1
1

+ = −
















+


















= −




















14. (c) : ( ) ( )3 1 3 12 2+ − −n n

= + +



− −2 3 32 1
2 1 2

3
2 3n n n nC C·( ) ·( ) ....

irrational number.

15. (b) : Number of ways in which one or more balls
can be selected from 10 white, 9 green, 7 black
balls is
= (10 + 1) (9 + 1) (7 + 1) – 1
= 880 – 1 = 879 ways

16. (b) : (x – 1)2 + y2 = 1, r = 1 a = 2
and x2 + (y – 2)2 = 4, r = 2 b = 4



x y
x y

2 2
2 2

16 4
1 4 16+ = ⇒ + =



17. (a) : x y z r


=
+

=


=
1

2
1

3
1

4 1
and

x y k z
r


=


= =

3
1 2 1 2

or 2r1 + 1 = r2 + 3, 3r1 – 1 = 2r2 + k, 4r1 + 1 = r2
2r1 – r2 = 2, and 4r1 – r2 = – 1

– 2r1 = 3 ⇒ =


= −r r1 2
3

2
5and 5

∴ − − = − + ⇒ = − =
9
2

1 10 10
11
2

9
2

k k

18. (a) : f(x) = ln|x| + bx2 + ax, x 0 has extreme values
at x = – 1, x = 2.

⇒ ′ = + +f x
x

bx a( )
1

2


f (– 1) = 0 and f (2) = 0 [Given]

⇒ − − + = ⇒ = −1 2 0
1
4

b a b

and
1
2

4 0
1
2

+ + = ⇒ =b a a

′′ = − + = − − = − +






<f x
x

b
x x

( )
1

2
1 1

2
1 1

2
0

2 2 2

for all x R – {0}
f has a local maximum at x = – 1, x = 2

Statement 1 : f has local maxima at x = –1, x = 2

Statement 2 : a b= = −
1
2

1
4

,

19. (c) : z 1,
z

z

2

1−
is real.

If z is a real number, then z
z

2

1−
is real.

Let z = x + iy

( ) (( ) )
( )

x y xiy x iy

x y

2 2

2 2
2 1

1
− + − −

− +
is real

– y(x2 – y2) + 2xy(x – 1) = 0
y(x2 + y2 – 2x) = 0 y = 0 or x2 + y2 – 2x = 0

JEE MAIN 419

Page 440

40

\ z lies on real axis or on a circle passing through
origin.

20. (c) : The given statement is
‘‘If I become a teacher, then I will open a school’’
Negation of the given statement is
‘‘ I will become a teacher and I will not open a
school’’ (... ~ (p → q) = p ∧ ~ q)

21. (a, d) : g x t dt
x

( ) cos= ∫ 4
0

⇒ =






=g x
t x

x

( )
sin sin4

4
4

40

⇒ + =
+

=g x
x x

( )
sin ( ) sin

π
π4

4
4

4
⇒ g(p) = 0 ⇒ g(x + p) = g(x) + g(p) or g(x) – g(p).

22. (a) : dv
dt

v= − =72 45003 0π πm / min,

v r
dv
dt

r
dr
dt

= ∴ = × ×
4
3

4
3

33 2π π

After 49 min, v = v0 + 49·
dv
dt

= 4500p – 49 × 72p

= 4500p – 3528p = 972p

Now, ⇒ =972
4
3

3π πr ⇒ r3 = 243 × 3 = 36 ⇒ r = 9

∴ − = × ×72 4 81π π
dr
dt

⇒ = − = −
dr
dt

18
81

2
9

Thus, radius decreases at a rate of
2
9

m/min

23. (d) : esinx – e–sinx – 4 = 0
⇒ (esinx)2 – 4esinx – 1 = 0 ⇒ t2 – 4t – 1 = 0

⇒ =
± +

= ±t
4 16 4

2
2 5

i e e x. ., sin = + −


2 5 2 5or (neglected)
ve

sin ln( )x = + >2 5 1 \ No real roots.

24. (d) : X = {1, 2, 3, 4, 5}; Y ⊆ X , Z ⊆ X, Y ∩ Z = f
Number of ways = 35.

25. (a) : x
y2
4

= , x2 = 9y
Area bounded by the parabolas and y = 2

= × −




 =

= × = × =

∫ ∫2 3 2 5

5
3 2

10
3

2 2
20 2

3

0

2

0

2

3 2

y
y

dy ydy

y( )
/

/
.

26. (a) : P3 = Q3, P2Q = Q2P, PQ2 = P2Q
⇒ P(P2 + Q2) = (Q2 + P2)Q
⇒ P(P2 + Q2) = (P2 + Q2)Q
P ≠ Q ⇒ P2 + Q2 is singular.
Hence, |P2 + Q2| = 0

27. (b) : x1, x2, x3, .... xn, A.M. = x , Variance = s2

Statement 2 : A.M. of 2x1, 2x2, ...., 2xn


=

+ + +
=

2
21 2

( ..... )x x x
n

xn

Given A.M. = 4x \ Statement 2 is false.

28. (c) : d p t
dt

p t
( ( ))

. ( )= −0 5 450

2
900

0850

dp
p

dt
tp


= ∫∫ ⇒




=2
900
50

ln
p

t

⇒ = −p et900 50 2· /

If p = 0, then
900
50

2 182= ⇒ =e tt/ ln

29. (a) : y = mx + c ⇒ 2 = m + c
Co-ordinates of P & Q : P(0, c), Q(–c/m, 0)
1
2

× × =| |c
c
m

A

⇒ =

c
m

A
2

2




=
( )2

2

2m
m

A


− +

= ⇒ − + =
m m

m
A

m
m

A
2 4 4

2 2
2

2

dA
dm

= 0


⇒ − = ⇒ = ⇒ = ⇒ = ±
1
2

2
0

1
2

2
4 2

2 2
2

m m
m m

30. (b) : 100 (a + 99d) = 50 (a + 49d)
⇒ a + 149d = 0 i.e., T150 = 0

JEE MAIN 420

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