IIT-JEE – 2007 Paper – 1 Solutions
Physics
Part – I
Section – I
Straight Objective Type
This section contains 9 multiple choice questions numbered 1 to 9. Each question has 4 choices (A), (B),
(C) and (D), out of which only one is correct.
1.

A circuit is connected as shown in the figure with the switch S open. When the switch is closed,
the total amount of charge that flows from Y to X is

3?F

X

6?F
S

3?

6?

Y
9V
(A) 0
(B) 54? C
(C) 27? C
(D) 81? C
Sol. (C)
When ‘S’is open:

3? F

3?

X

Y

6? F

6?

9V
After the capacitors are fully charged, current flows through resistors only.
? q q
?
Total charge on each capacitor is q ? 18 ? F ?
? ? 9?
? 3 6
?
When ‘S’is closed:

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3? F

X

6? F

i
3?

i
Y 6?

9V
? 9 ?
i? ?
? A ? 1A
? 3? 6?
Potential drop across the 3 ? resister = i × 3 = 3 V
? q' ? 3 ? 3 ? c ? 9 ? c
Potential drop across 6 ? resister = 6 × 1 V = 6V
? q" ? 6 ? 6 ? c ? 36 ? c
? Net charge flowing from Y to X
? ? ? 36 ? 18 ? ? ? ? 9 ? ( ? 18)?? ? c ? 27 ? c
2.

In the options given below, let E denote the rest mass energy of a nucleus and n a neutron. The
correct option is
236
97
U ? E 137
(A) E 92
53 I ? E 39 Y ? 2E ? n ?

?
(B) E ?
(C) E ?
(D) E ?
Sol. (A)

3.

236
92
236
92
236
92

? ? ? ? ?
U ? ? E ? I? ? E ? Y ? ? 2E ? n ?
U ? ? E ? Ba ? ? E ? Kr ? ? 2E ? n ?
U ? ? E ? Ba ? ? E ? Kr ? ? 2E ? n ?
137
53

97
39

140
56

94
36

140
56

94
36

E ? parent nuclie ? ? ? E ? constituents ?
A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting
cylinder of larger radius. Both the cylinders are initially electrically neutral.
(A) A potential difference appears between the two cylinders when a charge density is given to
the inner cylinder
(B) A potential difference appears between the two cylinders when a charge density is given to
the outer cylinder
(C) No potential difference appears between the two cylinders when a uniform line charge is kept
along the axis of the cylinders
(D) No potential difference appears between the two cylinders when same charge density is given
to both the cylinders

Sol. (A)
When a inner cylinder is charged, both negative and positive charges are induced on the outer
cylinder.

Using a gaussian enclosing only the inner surface, a radically symmetric electric field exists.
Hence a non-zero potential difference exists.

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When outer cylinder is charged, no charges are induced on inner cylinder and hence no electric
field exists in between. Potential at all places inside the outer cylinder is constant and hence no
potential difference.
Since in (C) and (D) a net charge (line charge) exists inside a gaussian surface as chosen earlier,
electric field and hence potential difference exists.
4.

The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest
wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is
(A) 802 nm
(B) 823 nm
(C) 1882 nm
(D) 1648 nm

Sol. (B)

hc 13.6 (eV)
?
?
n2
? ? ?

6.6 ? 10? 34 ? 3 ? 108 2 100 ? 6.6 ? 3 2
n ?
n (nm)
13.6 ? 1.6
13.6 ? 1.6 ? 10? 19

? 91 n2

? (n ? 1) ? 91 nm, ? (n ? 2) ? 364 nm
? (n ? 3) ? 819 nm
Note that the minimum ? of IR lies close to end of visible range. Hence (C) and (D) cannot
represent the smallest ? of IR.
Hence (B)
5.

In an experiment to determine the focal length (f) of a concave mirror by the u-v method, a
student places the object pin A on the principal axis at a distance x from the pole P. The student
looks at the pin and its inverted image from a distance keeping his/her eye in line with PA. When
the student shifts his/her eye towards left, the image appears to the right of the object pin. Then.
(A) x < f
(B) f < x < 2f
(C) x = 2f
(D) x > 2f

Sol. (B)
From the problem statement it is clear that the image is formed “behind” the object. This means
v ? u . Thus, f ? x ? 2f .
6.

A ray of light traveling in water is incident on its surface open to air. The angle of incidence is ? ,
which is less than the critical angle. Then there will be
(A) only a reflected ray and no refracted ray
(B) only a refraction ray and no reflected ray
(C) a reflected ray and a refracted ray and the angle between them would be less than 180? – 2?
(D) a reflected ray and a refracted ray and the angle between them would be greater than
180? – 2?

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Sol. (C)
n1 sin ? ? n2 sin(90 ? ? ) ;

n2
n1

? ?

?
90 – ?

? 90 ? ? ? ? ?

7.

n1 ? n2

Angle between reflected and refracted ray = ? ? 90 ? ? ? 180 ? 2?

A resistance of 2 ? is connected across one gap of a metre-bridge (the length of the wire is
100 cm) and an unknown resistance, greater than 2 ? , is connected across the other gap. When
these resistances are interchanged, the balance point shifts by 20 cm . Neglecting any
corrections, the unknown resistance is
(A) 3 ?
(B) 4 ?
(C) 5 ?
(D) 6 ?

Sol. (A)

lo

100 – l

2?
X
l 100 ? l
?
2
x
On interchanging 2? and x,
l' 100 ? l'
?
x
2
Using l'? l = 20 cm, we have
100x 200
?
? 20
2? x x? 2
? x ? 3?
8.

Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The
net charge on the sphere is then,
(A) negative and distributed uniformly over the surface of the sphere
(B) negative and appears only at the point on the sphere closest to the point charge
(C) negative and distributed non-uniformly over the entire surface of the sphere
(D) zero

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Sol. (D)
Although there will be a charge distribution on the outer surface of sphere, the net charge is still
zero.
q

+
+
+
9.

––
–
–
–
–
++

Two particles of mass m each are tied at the ends of a light string of length 2a. The whole system
is kept on a frictionless horizontal surface with the string held tight so that each mass is at a
distance ‘a’from the center P (as shown in the figure). Now, the mid-point of the string is pulled
vertically upwards with a small but constant force F. As a result, the particles move towards each
other on the surface. The magnitude of acceleration, when the separation between them
becomes 2x, is

F
m

m

P
a

a

F
a
2m a2 – x 2
F
x
(B)
2
2m a – x 2
(A)

(C)

F x
2m a

(D)

F a2 – x 2
2m
x

Sol. (B)

F

m

m
a

a

F
T
a
?

T
?
x
2T sin ? ? F
T cos ? ? ma x
2 tan ? ?

F
ma x

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ax ?

F
Fx
?
2m tan ? 2m a2 – x 2

ay ?

F
, is very small and hence can be neglected.
2m

? a ? ax

Section – II
Assertion – Reason Type
This section contains 4 questions numbered 10 to 13. Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is
correct.
10.

Statement - 1
The formula connecting u, v and f for a spherical mirror is valid only for mirrors whose sizes are
very small compared to their radii of curvature.
Because
Statement - 2
Laws of reflection are strictly valid for plane surfaces, but not for large spherical surfaces
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for
statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for
Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True

Sol. (C)

S1 ? correct
S2 ? incorect
Laws of reflection is valid for all surfaces.
11.

Statement - 1
A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to
friction between the block and surface after moving through a certain distance. The surface is
now tilted to an angle 30° with the horizontal and the same block is made to go up on the surface
with the same initial velocity v. The decrease in the mechanical energy in the second situation is
smaller than that in the first situation.
Because
Statement - 2
The coefficient of friction between the block and the surface decreases with the increase in the
angle of inclination.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for
statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for
Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True

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Sol. (C)

V=0

u

(1)
1
? E(1) ? ? mv 2
2
0
V=
u

h
(2)
? E(2) ? mgh ?

1
mv 2
2

? E(2) ? ? E(1)
? S1 ? correct

S2 ? incorrect
? doesn’t change, rather force of friction reduces.
12.

Statement - 1
If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic
X-rays do not change.
Because
Statement - 2
When an electron beams strikes the target in an X-ray tube, part of the kinetic energy is
converted into X-ray energy.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for
statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for
Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True

Sol. (B)
Accelerating potential affects ? min and not characteristic X-ray ?
? S1 ? correct

S2 ? correct

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13.

Statement - 1
In an elastic collision between two bodies, the relative speed of the bodies after collision is equal
to the relative speed before the collision.
Because
Statement – 2
In an elastic collision, the linear momentum of the system is conserved.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for
statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for
Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True

Sol. (B) Contextually.
NB:
S2 need not be true in all cases. Counter e.g. ? collision of a ball with a smooth but rigid wall.
Hence S2 ? incorrect (as a generalized statement)
Note: Elastic refers to conservation of kinetic energy only. If S2 is given as correct, S1 is generally
true.

Section – III
Linked Comprehension Type
This section contains 2 paragraphs P14-16 and P 17-19. Based upon each paragraph, 3 multiple choice
Questions have to be answered. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE
is correct.
P14-16 : Paragraph for Questions Nos. 14 to 16.
A fixed thermally conducting cylinder has a radius R and height L0 . The cylinder is open at its bottom and
has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in
the figure. The atmospheric pressure is P0

2R
L
L0

Piston

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14.

The piston is now pulled out slowly and held at a distance 2 L. from the top. The pressure in the
cylinder between its top and the piston will then be
(A) P0

P0
2
P0 Mg
(C)
?
2 ? R2
P
Mg
(D) 0 ?
2 ? R2
(B)

Sol. (A)
Since the top is open to the atmosphere, pressure inside is equal to pressure outside. Since the
piston is moved slowly, the pressure equalizes to that of atmosphere.
15.

While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then
released, to a position where it can stay in equilibrium. In this condition, the distance of the piston
from the top is
? 2P0 ? R 2 ?
(A) ?
? ? 2L ?
2
? ? R P0 ? Mg ?
? P ? R 2 – Mg ?
(B) ? 0
? ? 2L ?
2
? ? R P0 ?
? P ? R 2 ? Mg ?
(C) ? 0
? ? 2L ?
2
? ? R P0 ?
? P0 ? R 2 ?
(D) ?
? ? 2L ?
2
? ? R P0 – Mg ?

Sol. (D)

2R
y

Since the cylinder is thermally conducting, the expansion is isothermal. Let P be the pressure
inside the cylinder at equilibrium.

P0 (2L)? R2 ? P(y)? R2
2P0L
y
Equilibrium ?
? P?

Forces are balanced.

? ? ? Mg ? P0 ? ? R2 ?

? P ?R
?

2

2 P0 L
Mg
? P0 –
y
? R2

? y?

2P0 ? R2L
? R2P0 ? Mg

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16.

The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is
brought below the cylinder and put in a position so that the water surface in the tank is at the
same level as the top of the cylinder as shown in the figure. The density of water is ? . In
equilibrium, the height H of water column in the cylinder satisfies

L0
H
(A) ? g ?Lo – H? ? P0 ?L o – H? ? L 0P0 ? 0
2

(B) ? g ?Lo – H? ? P0 ?L o – H? ? L0P0 ? 0
2

(C) ? g ?Lo – H? ? P0 ?L o – H? ? L0P0 ? 0
2

(D) ? g ?Lo – H? ? P0 ?L o – H? ? L 0P0 ? 0
2

Sol. (C)
Let the air pressure be P (inside the cylinder). Then

P ?L0 ? H? ? R2 ? P0 ?L0 ? ? R2

? L0 ?
? P? ?
? P0
? L0 ? H ?
This pressure is balanced by the hydrostatic pressure. So
? L0 ?
?
? P0 ? ? g ? L0 ? H? ? P0
? L0 ? H ?
? ? g ? L0 ? H? ? P0 ? L0 ? H? ? P0L0 ? 0
2

P17-19 : Paragraph for Questions Nos. 17 to 19
Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I
respectively about the common axis. Disc A is imparted an initial angular velocity 2? using the entire
potential energy of a spring compressed by a distance x1 . Disc B is imparted an angular velocity ? by a
spring having the same spring constant and compressed by a distance x 2 . Both the discs rotate in the
clockwise direction
17.

The ratio x1 / x 2 is
(A) 2
1
(B)
2
(C) 2
1
(D)
2

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Sol. (C)

1 2 1
2
kx1 ? I ? 2? ? ? 2I? 2
2
2
? x12 ?
Also,

18.

4I? 2
k

1 2 1
kx2 ?
? 2I? ? 2 ? I ? 2
2
2

?

x22 ?

2I ?
k

?

x1
?
x2

2

2

When disc B is brought in contact with disc A, they acquire a common angular velocity in time t.
The average frictional torque on one disc by the other during this period is
2I?
(A)
3t
9I?
(B)
2t
9I?
(C)
4t
3I?
(D)
2t

Sol. (A)
If the common angular velocity is ? 0 ,
I ? 2? ? ? 2I ? ? ? ? 3I? ? 0 ?

4
?
3
Now, for disc A,
? 0 ? 2? ? ? t
? ?0 ?

2? ? ? 0 2?
?
t
3t
2I?
? ? ? I? ?
3t
? ? ?

19.

The loss of kinetic energy during the above process is
I? 2
(A)
2
I? 2
(B)
3
I? 2
(C)
4
I? 2
(D)
6

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Sol. (B)

KEi ?

1
2 1
I ? 2? ? ? ? 2I? ? 2 ? 3I? 2
2
2

KEf ?

1
4?
8
I? 2
? 3I? ?? ?? ? I? 2 ? Loss of KE =
2
3
3
? 3 ?

2

Section – IV
Matrix-Match Type
This section contains 3 questions . Each question contains statement given in two columns which have to
be matched. Statement (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in
column II. The answers to these questions have to be appropriately bubbled as illustrated in the following
example.
If the correct matches are A-p, A-s, B-q, B-r, C-p, C-q and D-s then the correct bubbled 4 × 4 matrix
should be as follows
p q r s
A p q r s
B p q r s
C p q r s
D p q r s
20.

Some physical quantities are given in column I and some possible SI units in which these
quantities may be expressed are given in column II. Match the physical quantities in column I
with the units in Column II and indicate your answer by darkening appropriate bubbles in the 4 ×
4 matrix given in the ORS.

(A)

(B)

(C)

(D)

Column I
GMeMs
G – universal gravitational constant,
Me – mass of the earth
Ms – Mass of the Sun

3RT
M
R – universal gas constant
T – absolute temperature
M – molar mass
F2
q2B2
F – force
Q – charge
B – magnetic field
GMe
Re
G – universal gravitational constant,
Me – mass of the earth

Column II

(p)

(volt) (coulomb) (metre)

(q)

? kilogram ??metre ? ? sec ond?

(r)

? metre ? ? sec ond?

(S)

? farad?? volt ? ? kg?

3

2

2

–2

–2

–1

Re – radius of the earth

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Sol.
(A)
(B)
(C)
(D)
21.

? (p), (q)
? (r), (s)
? (r) , (s)
? (r), (s)

Column I gives certain situations in which a straight metallic wire of resistance R is used and
column II gives some resulting effects. Match the statement in column I with the statements in
column II and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in
the ORS.

(A)

(B)

(C)
(D)

Column I
A charged capacitor is connected to
the ends of the wire
The wire is moved perpendicular to
its length with a constant velocity in
a uniform magnetic field
perpendicular to the plane of motion
The wire is placed in a constant
electric field that has direction along
the length of the wire
A battery of constant emf is
connected to the ends of the wire

Column II
(p)

A constant current flows through the wire

(q)

Thermal energy is generated in the wire

(r)

A constant potential difference develops
between the ends of the wire

(S)

Charges of constant magnitude appear
at the ends of the wire

Sol.
(A)
(B)
(C)
(D)
22.

? (q)
? (r), (s),
? (s)
? (p), (q) , (r)

Some laws/ processes are given in Column I. Match these with physical phenomena given in
Column II and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in
the ORS.

(A)
(B)
(C)
(D)

Column I
Transition between two atomic
energy levels
Electron emission from a material
Mosley’s law
Change of photon energy into
kinetic energy of electrons

Column II
(p)

Characteristic X-rays

(q)
(r)

Photoelectric effect
Hydrogen spectrum

(S)

? – decay

Sol.
(A)
(B)
(C)
(D)

?
?
?
?

(p), (r)
(q), (s),
(p)
(q)

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Chemistry
Part – II
Useful data:
Gas Contant, R = 8.314 K JK –1mol–1
1 F = 96500 C
Atomic Numbers:
H = 1, Li = 3, B= 5, C = 6, N = 7, O = 8, F = 9, Na = 11, P = 15, S = 16, Cl = 17, Ar = 18,
K = 19, V = 23, Cr = 24, Mn = 25, Fe = 26, Co= 27, Ni = 28,Cu = 29, Zn = 30, Ge = 32, Br = 35, Ag = 47,
I = 53, Xe= 54, Pt = 78, Hg = 90 , Pb = 82

Section – I
Straight Objective Type
This section contains 9 multiple choice questions numbered 23 to 31. Each question has 4 choices (A),
(B), and (D), out of which only one is correct
23.

The number of structural isomers for C6H14 is
(A) 3
(B) 4
(C) 5
(D) 6

Sol. (C)
Structural isomers of C6H14
CH3 — CH2 — CH2 — CH2 — CH2 — CH3

CH3 — CH2 — CH2 — CH — CH3
|
CH3

CH3 — CH2 — CH — CH2 — CH3
|
CH3

CH3 — CH— CH— CH3
|
CH3

|
CH3

CH3
CH3 — C— CH2 — CH3
CH3
24.

?

?

When 20 g of naphthoic acid ? C11H8 O2 ? is dissolved in 50 g of benzene K f ? 1.72 K kg mol–1 , a
freezing point depression of 2 K is observed. The van’t Hoff factor (i) is
(A) 0.5
(B) 1
(C) 2
(D) 3

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Sol. (A)
i?

25.

? Tf
2 ? 172 ? 50
?
? 0.5
Km 1.72 ? 20 ? 2000

In the following reaction,

O

conc. HNO3
X
conc. H2SO4

N
H

the structure of the major product ‘X’is
(A)

O
NO2
N
H
(B)

O2N

O
N
H

(C)

O
N
H

NO2

(D)

O
O2N

N
H

Sol. (B)

O2N

O
N

C

H
Because

NH

group is activator and p-directing towards electrophilic aromatic substitution.

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26.

The value of log10 K for a reaction A
(Given ? rH

o
298K

? –54.07 kJmol , ? r S
–1

B is
o
298K

? 10 JK –1mol–1 and

R ? 9314JK – mol–1;2.303 ? 8.314 ? 298 ? 5705)
(A) 5
(B) 10
(C) 95
(D) 100

Sol. (B)

? G ? ? 2.303 RTlog K
? G ? ? H ? T? S
? ? 54070 ? 2980
? ? 54070 J
? — 54070 = –2.303 × 8.314 × 298 log K
log K = 10

27.

Among the following, the paramagnetic compound is
(A) Na2 O2
(B) O3
(C) N2O
(D) KO2

Sol. (D)
KO2 is paramagnetic is nature because it contains unpaired electrons.

28.

The percentage of p-character in the orbitals forming P–P bonds in P4 is
(A) 25
(B) 33
(C) 50
(D) 75

Sol. (D)
Phosphorous will show sp3 hybridisation having 75% p-character.

P
P

P
P
29.

The species having bond order different from that in CO is
(A) NO –
(B) NO?
(C) CN–
(D) N2

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Sol. (A)
Bond order of NO? is 2 while bond order of CO is 3.
30.

The reagent(s) for the following conversion,
Br
?
H
Br
is/are
(A) alcoholic KOH
(B) alcoholic KOH followed by NaHH2
(C) aqueous KOH followed by NaNH2

H

(D) Zn / CH3 OH
Sol. (B)
Br

31.

alc. KOH/NaNH2

CH2
CH2

H

Br

C

C

H

Extraction of zinc form zinc blende is achieved by
(A) electrolytic reduction
(B) roasting followed by reduction with carbon
(C) roasting followed by reduction with another metal
(D) roasting followed by self-reduction

Sol. (B)
Roasting followed by reduction with carbon.

Section – II
Assertion – Reason Type
This section contains 4 questions numbered 32 to 35. Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which only one is
correct.
32.

Statement - 1
p-Hydroxybenzoic acid has a lower boiling point than o-hydroxybenzoic acid
Because
Statement - 2
o-Hydroxybenzoic acid has intramolecular hydrogen bonding
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for
statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for
Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True

Sol. (D)
O-hydroxybenzoic acid has intramolecular hydrogen bonding hence has lower boiling point than
p-hydroxybenzoic acid.

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33.

Statement - 1
Boron always forms covalent bond
Because
Statement - 2
The small size of B3 ? favours formation of covalent bond.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for
statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for
Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True

Sol. (A)
Boron always forms covalent bond, because its compound will have zero resultant dipole moment
due to its trigonal plannar shape.
34.

Statement - 1
Micelles are formed by surfactant molecules above the critical micellar concentration (CMC)
Because
Statement - 2
The conductivity of a solution having surfactant molecules decreases sharply at the CMC.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for
statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for
Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True

Sol. (B)
35.

Statement - 1
In water, orthoboric acid behaves as a weak acid monobasic acid.
Because
Statement – 2
In water, orthoboric acid acts as a proton donor.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for
statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for
Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True

Sol. (C)

IIT-JEE – 2007 / Paper - 1 Solutions

Page: .. 18 ..

Section – III
Linked Comprehension Type
This section contains 2 paragraphs C36-38 and C39-41. Based upon each paragraph, 3 multiple choice
Questions have to be answered. Each question has 4 choices (A), (B), (C) and (D) out of which Only one
is correct.

P36-38 : Paragraph for Questions Nos. 36 to 38.
Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules
(approximately 6.023 × 10 23 ) are present in a few grams of any chemical compound varying with their
atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced.
This concept has implications in diverse areas such as analytical chemistry, biochemistry
electrochemistry and radiochemistry. The following example illustrates a typical case, involving
chemical/electrochemical reaction, which requires a clear understanding of the mole concept.
A 4.0 molar aqueous solution NaCl is prepared and 500 mL of this solution is electrolysed. This leads to
the evolution of chlorine gas at one of the electrodes (atomic mass : Na = 23, Hg = 200;1 Faraday =
96500 coulombs).
36.

The total number of moles of chlorine gas evolved is
(A) 0.5
(B) 1.0
(C) 2.0
(D) 3.0

Sol. (B)
37.

If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is
(A) 200
(B) 225
(C) 400
(D) 446

Sol. (D)
In presence of amalgam cathode sodium ion will discharge in place of hydrogen gas due to over
voltage is the form of amalgams.
Weight of amalgam = 2 × (23 + 200) = 446 g
38.

The total charge (coulombs) required for compete electrolysis is
(A) 24125
(B) 48250
(C) 96500
(D) 193000

Sol. (D)
Total charge = 2 × 96500 = 193000 C

IIT-JEE – 2007 / Paper - 1 Solutions

Page: .. 19 ..

P39-41 : Paragraph for Questions Nos. 39 to 41.
The noble gases have closed-shell electronic configuration and are monoatomic gases under normal
conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between
the atoms and the absence of other interatomic interactions.
The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4
and + 6. XeF4 reacts violently with water to give XeO3 . The compounds of xenon exhibit rich
stereochemistry and their geometries can be deduced considering the total number of electron pairs in
the valence shell
39.

Argon is used in arc welding because of its
(A) low reactivity with metal
(B) ability to lower the melting point of metal
(C) flammability
(D) high calorific value

Sol. (A)
Argon may be used to provide an inert atmosphere in welding.
40.

The structure of XeO3 is
(A) linear
(B) planar
(C) pyramidal
(D) T-shaped

Sol. (C)
Pyramidal
41.

XeF4 and XeF6 are expected to be
(A) oxidizing
(B) reducing
(C) unreactive
(D) strong basic

Sol. (A)

IIT-JEE – 2007 / Paper - 1 Solutions

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Section – IV
Matrix-Match Type
This section contains 3 questions . Each question contains statement given in two columns which have to
be matched. Statement (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in
column II. The answers to these questions have to be appropriately bubbled as illustrated in the following
example.
If the correct matches are A-p, A-s, B-q, B-r, C-p, C-q and D-s then the correct bubbled 4 × 4 matrix
should be as follows

A
B
C
D
42.

p
p
p
p
p

q
q
q
q
q

r s
r s
r s
r s
r s

Match the complexes in column I with their properties listed in column II. Indicate your answer
by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.

(A)
(B)
(C)
(D)

Column I
?? Co ?NH3 ?4 ?H2 O ?2 ?? Cl2
?? Pt ? NH3 ?2 Cl2 ??

?? Co ?H 2O ?5 Cl?? Cl
?? Ni ?H 2 O ?6 Cl?? Cl2

Column II
(p)

geometrical isomers

(q)

paramagnetic

(r)

diamagnetic

(s)

metal ion with +2 oxidation state

Sol.
A—
B—
C—
D—
43.

p, A —
p, B —
s, C —
s, D —

q, A — s.
s, B — r
q
q

Match the chemical substances in column I with type of polymers/type of bonds in column II.
Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.

(A)
(B)
(C)
(D)

Column I
Cellulose
nylon-6,6
Protein
Sucrose

IIT-JEE – 2007 / Paper - 1 Solutions

(p)
(q)
(r)
(s)

Column II
natural polymer
synthetic polymer
amide linkage
glycoside linkage

Page: .. 21 ..

Sol.
A—
B—
C—
D—
44.

p, A — s
q, B — r
r, C — p
s

Match gases under specified conditions listed in column I with their properties/laws in column II.
Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.

(A)
(B)
(C)
(D)

Column I
hydrogen gas (P = 200 atm,
T = 273 K)
hydrogen gas (P = 0, T = 273 K)
CO2 (P = 1 atm, T = 273 K)
real gas with very large molar
volume

Column II
(p)

compressibility factor ? 1

(q)
(r)

attractive forces are dominant
PV = nRT

(s)

P(V – nb) = nRT

Sol.
A — p, A — s, B — r, C — q, C — p, D — p, D — q

IIT-JEE – 2007 / Paper - 1 Solutions

Page: .. 22 ..

Mathematics
Section – III
Straight Objective Type
This section contains 9 multiple choice questions numbered 45 to 53. Each question has 4 choices (A),
(B), (C) and (D), out of which only one is correct
45.

A hyperbola, having the transverse axis of length 2sin? , is confocal with the ellipse
3x 2 ? 4y 2 ? 12 . Then its equation is
(A) x 2 cosec 2 ? ? y 2 sec 2 ? ? 1
(B) x 2 sec 2 ? ? y 2cosec 2 ? ? 1
(C) x 2 sin2 ? ? y 2 cos2 ? ? 1
(D) x 2 cos2 ? ? y 2 sin2 ? ? 1

Sol. (A)

x 2 cosec 2 ? ? y 2 sec 2 ? ? 1

a ? sin ? for hyperbola
ae ? 4 ? 3 ? 1
b2 ? sin2 ? (cosec 2? ? 1) ? cos2 ?
x2
sin2 ?

?

y2
cos2 ?

? 1

x 2 cosec 2 ? ? y 2 sec 2 ? ? 1
46.

A man walks a distance of 3 units from the origin towards the north-east (N 45° E) direction. From
there, he walks a distance of 4 units towards the north-west (N 45° W) direction to reach a point
P. Then the position of P in the Argand plane is
(A) 3ei? / 4 ? 4i
(B) ? 3 – 4i ? ei? / 4
(C) ? 4 ? 3i ? ei? / 4
(D) ? 3 ? 4i ? ei? / 4

Sol. (D)

? 3 ? 4i? ei? / 4
Z
5

4

3
0
Z?

4?
??
i? ? tan? 1 ?
4
3?
?
5.e

IIT-JEE – 2007 / Paper - 1 Solutions

Page: .. 23 ..

i tan? 1

4

3 .ei? / 4
? 5.e
? 3 4?
? 5 ? ? i ? .ei? / 4
? 5 5?

? ? 3 ? 4i ? ei? / 4
47.

?

?

The tangent to the curve y ? e x drawn at the point c, ec intersects the line joining the points

? c ? 1, e ? and ? c ? 1, e ?
c? 1

c? 1

(A) on the left of x = c
(B) on the right of x = c
(C) at no point
(D) at all points
Sol. (A)
On the left of x = c

Shortcut: Let c = 0
48.

The number of solutions of the pair of equations
2sin2 ? ? cos 2? ? 0
2cos2 ? ? 3 sin ? ? 0
in the interval ?0, 2? ? is
(A) zero
(B) one
(C) two
(D) four

Sol. (C)
Add two equations

2 ? 1 ? 2sin2 ? ? 3 sin ? ? 0
2sin2 ? ? 3 sin ? ? 1 ? 0
? 2sin ? ? 1?? sin ? ? 1? ? 0

1
,1
2
If sin ? ? 1 , cos ? ? 0 not satisfies second equation.
sin ? ?

1
3
? cos ? ? ?
2
2
? 5?
? ? ,
6 6

? sin ? ?

IIT-JEE – 2007 / Paper - 1 Solutions

Page: .. 24 ..

49.

Let f(x) be differentiable on the interval ? 0, ?

?

t 2 f(x) ? x 2 f(t)
? 1
x
t? x

such that f(1) = 1, and lim
t?

for each x > 0. Then f(x) is
1 2x 2
(A)
?
3x
3
? 1 4x 2
(B)
?
3x
3
?1 2
(C)
?
x x2
1
(D)
x
Sol. (A)

1 2x 2
?
3x
3
lim ?

t 2 f(x) ? x 2 f(t)
? 1
t? x

lim ?

x ? f(x) ? x 2 f ?(t)
? 1
1

t? x

t? x

? 0?
? ?
? 0?

? 2xf(x) ? x 2 f ?(x) ? 1

dy 2
?1
? y?
dx x
x2
? 2?

dx
x

?2

I.F. ? e
? eln x ? x ? 2
Solution is given by
1
yx ? 2 ? ?
dx ? c
x4

?

?

?3

x
? c
3

at x = 1, y = 1 ? c ?
50.

2
3

The number of distinct real values of ? ,for which the vectors ? ? 2 ˆ
i? ˆ
j ? kˆ , ˆ
i ? ? 2 ˆj ? kˆ and
ˆ
i ? ˆj ? ? 2kˆ are coplanar, is
(A) zero
(B) one
(C) two
(D) three

IIT-JEE – 2007 / Paper - 1 Solutions

Page: .. 25 ..

Sol. (C)

??2

1

1

2

1

??

1

1

1 ? 0
?? 2

?

??

?

2
? ? 2? ? 2 ? 2 ? 1 ? 0 ? ? ? ? 2

51.

Let ? , ? be the roots of the equation x 2 ? px ? r ? 0 and

?
, 2? be the roots of the equation
2

x 2 ? qx ? r ? 0 . Then the value of r is
2
? p ? q?? 2q ? p ?
9
2
(B) ? q ? p ?? 2p ? q ?
9
2
(C) ? q ? 2p ?? 2q ? p ?
9
2
(D) ? 2p ? q ?? 2q ? p ?
9

(A)

Sol. (D)
2
? 2p ? q ?? 2q ? p ?
9
? ?? ? p
?? ? r

?
? 2? ? q
2
? ? 4? ? 2q
2q ? p
3
4p ? 2q
;
? ?
3

? ?

r?

2
? 2q ? p ?? 2p ? q?
9

sec 2 x

52.

lim
x?

?
4

?

f(t)dt

2

x2 ?

?2
16

equals

8
f(2)
?
2
(B) f(2)
?
2 ? 1?
(C) f ? ?
? ? 2?
(D) 4f(2)

(A)

IIT-JEE – 2007 / Paper - 1 Solutions

Page: .. 26 ..

Sol. (A)
sec 2 x

?

f(t)dt

lim

?
4

2f(sec 2 x)sec 2 x tan x
?
2x
x?

? lim

2

?2
16
2f(2).2.1 8
?
? f(2)
?
?
2.
4
x?

53.

Sol. (C)

x2 ?

4

One Indian and four American men and their wives are to be seated randomly around a circular
table. Then the conditional probability that the Indian man is seated adjacent to his wife given that
each American man is seated adjacent to his wife is
1
(A)
2
1
(B)
3
2
(C)
5
1
(D)
5

A1A1?

A 2 A ?2 A 3 A 3? A 4 A?4 II?

Total 5!? 24

A1A1? A 2 A ?2

A 3 A? A 4 A?4 II?
3

Favourable 4!? 25

4!? 25
5!? 2

4

?

2
5

Section – II
Assertion – Reason Type
This section contains 4 questions numbered 54 to 57. Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which only one is
correct.

IIT-JEE – 2007 / Paper - 1 Solutions

Page: .. 27 ..

54.

Let the vectors PQ, QR, RS, ST, TU and UP represent the sides of a regular hexagon.
Statement -1: PQ ?

? RS ?

?

ST ? 0

Because
Statement - 2
PQ ? RS ? 0 and PQ ? ST ? 0
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for
statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for
Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True
Sol. (C)
S1 ?

True, S2 ?

False

RS ? ST ? RT
PQ ? RT ? 0 as PQ ? RT

T

S

R

V

P

Q

PQ ? RS ? 0

55.

Let H1, H2 ,...., Hn be mutually exclusive and exhaustive events with P(Hi ) ? 0 , i ? 1, 2, ...,n . Let E
be any other event with 0 ? P(E) ? 1 .
Statement - 1
P(Hi | E) ? P(E | Hi ) . P(Hi ) for i = 1, 2,… ,n.
Because
Statement - 2
n

?

i? 1

P(Hi ) ? 1

(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for
statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for
Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True

IIT-JEE – 2007 / Paper - 1 Solutions

Page: .. 28 ..

Sol. (D)
S1 ?

False, S2 ?

P ?Hi / E ? ? P

True ,

?Hi ? E ?
P(E)

P ?E / Hi? ? P(Hi ) ?

P ?E ? Hi ?
P(Hi

.P(Hi ) ? P ?E ? Hi ?

P ?Hi / E ? ? P ?E / Hi ? .P(Hi )
as P(E) ? 1
56.

Let F(x) be an indefinite integral of sin2 x .
Statement - 1
The function F(x) satisfies F(x ? ? ) ? F(x) for all real x.
Because
Statement - 2
sin2 (x ? ? ) ? sin2 x for all real x.
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for
statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for
Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True

Sol. (D)

S1 ? False
f(x) ?
57.

x sin2x
?
? c
2
4

Tangents are drawn from the point (17, 7) to the circle x 2 ? y ? ? 169
Statement - 1
The tangents are mutually perpendicular.
Because
Statement - 2
The locus of the points from which mutually perpendicular tangents can be drawn to the given
circle is x 2 ? y ? ? 338 .
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for
statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for
Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True

IIT-JEE – 2007 / Paper - 1 Solutions

Page: .. 29 ..

Sol. (A)
S1 ? True, S2 ? True , S2 explains S1

Director circle of x 2 ? y2 ? r 2 is x 2 ? y 2 ? 2r 2

Section – III
Linked Comprehension Type
This section contains 2 paragraphs M58-60 and M61-63. Based upon each paragraph, 3 multiple choice
Questions have to be answered. Each question has 4 choices (A), (B), (C) and (D) out of which Only one
is correct.
M58-60 : Paragraph for Questions Nos. 58 to 60.
Consider the circle x 2 ? y ? ? 9 and the parabola y ? ? 8x .They intersect at P and Q in the first and the
fourth quadrants, respectively.Tangents to the circle at P and Q intersect the x-axis at R and tangents to
the parabola at P and Q intersect the x-axis at S.
58.

The ratio of the areas of the triangles PQS and PQR is
(A) 1: 2
(B) 1 : 2
(C) 1 : 4
(D) 1 : 8

Sol. (C)

(

P 1, 2 2

)

S

(

Q 1, – 2 2

R

)

PR : x ? 2 2y ? 9
QR : x ? 2 2y ? 9
? R ? 9, 0 ?
PS : 2 2y ? 4 ? x ? 1?
QS : ? 2 2y ? 4 ? x ? 1? ? S( ? 1, 0)

1
? 4 2? 2
? PQS 2
?
? 1: 4
? PQR 1
? 4 2? 8
2

IIT-JEE – 2007 / Paper - 1 Solutions

Page: .. 30 ..

59.

The radius of the circumcircle of the triangle PRS is
(A) 5
(B) 3 3
(C) 3 2
(D) 2 3

Sol. (B)

R?
60.

abc
4?

The radius of the incircle of the triangle PQR is
(A) 4
(B) 3
8
(C)
3
(D) 2

Sol. (D)

r?

?
s
M61-63 : Paragraph for Questions Nos. 61 to 63.

Let Vr denote the sum of the first r terms of an arithmetic progression (A.P.) whose first term is r and the

common difference is ? 2r ? 1? .

Let Tr ? Vr ? 1 ? Vr ? 2 and Qr ? Tr ? 1 ? Tr for r = 1, 2,… .
61.

The sum V1 ? V2 ? ... ? Vn is

1
n(n ? 1)(3n2 ? n ? 1)
12
1
(B)
n(n ? 1)(3n2 ? n ? 2)
12
1
(C) n(2n2 ? n ? 1)
2
1
(D) n(2n3 ? 2n ? 3)
3
(A)

Sol. (B)

?

r2 r
1
Vr ? r 2 ?
? ; ? Vr ?
n ? n ? 1? 3n2 ? n ? 2
2 2
12
62.

?

Tr is always
(A) an odd number
(B) an even number
(C) a prime number
(D) a composite number

IIT-JEE – 2007 / Paper - 1 Solutions

Page: .. 31 ..

Sol. (D)
Always composite number
Tr ? ? 3r ? 1?? r ? 1?
63.

Which one of the following is a correct statement?
(A) Q1, Q 2, Q3 ,... are in A.P. with common difference 5
(B) Q1, Q 2 , Q3 ,... are in A.P. with common difference 6
(C) Q1, Q 2, Q3 ,... are in A.P. with common difference 11
(D) Q1 ? Q2 ? Q3 ? ...

Sol. (B)
Q1, Q 2, Q3 ..... are in A.P. with common difference 6

Qr ? 6r ? 5

Section – IV
Matrix-Match Type
This section contains 3 questions . Each question contains statement given in two columns which have to
be matched. Statement (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in
column II. The answers to these questions have to be appropriately bubbled as illustrated in the following
example.
If the correct matches are A-p, A-s, B-q, B-r, C-p, C-q and D-s then the correct bubbled 4 × 4 matrix
should be as follows

p q r s
A p q r s
B p q r s
C p q r s
D p q r s
64.

Consider the following linear equations
ax ? by ? cz ? 0
bx ? cy ? az ? 0

cx ? ay ? bz ? 0
Match the conditions/expressions in Column I with statement in Column II and indicate your
answer by darkening the appropriate bubbles in the 4 × 4 matrix given in the ORS.

(A)
(B)
(C)
(D)

Column I
a ? b ? c ? 0 and
a2 ? b2 ? c 2 ? ab ? bc ? ca
a ? b ? c ? 0 and
a2 ? b2 ? c 2 ? ab ? bc ? ca
a ? b ? c ? 0 and
a2 ? b2 ? c 2 ? ab ? bc ? ca
a ? b ? c ? 0 and
a ? b ? c ? ab ? bc ? ca
2

2

2

IIT-JEE – 2007 / Paper - 1 Solutions

Column II
(p)

the equations represent planes meeting
only at a single point.

(q)

the equations represent the line x = y =z

(r)

the equations represent identical planes.

(s)

the equations represent the whole of the
three dimensional space.

Page: .. 32 ..

Sol.

65.

A ? r, B ?

q, C ? p, D ? s

In the following ? x ? denotes the greatest integer less than or equal to x.
Math the functions in Column I with the properties in Column II and indicate your answer by
darkening the appropriate bubbles in the 4 ×4 matrix given in the ORS
Column I

Column II

(A)

xx

(p)

continuous in (–1, 1)

(B)

x

(q)

differentiable in (–1, 1)

(C)

x ? ? x?

(r)

(D)

x? 1? x? 1

(s)

strictly increasing in (–1, 1)
not differentiable at least at one point in
(–1, 1)

Sol.

A ? p, q, r

–1

1

xx ?

continuous in (–1, 1) differentiable in (–1, 1) strictly increasing in (–1, 1)

(B) ?

p, s

x
x
–1 0

1

x ?
(C) ?

continuous in (–1, 1) and differentiable in (–1, 1)
r, s

–1

x ? ? x? ?
(D) ?

1

strictly increasing in (–1, 1) not differentiable atleast at one point in (–1, 1)

p, q

IIT-JEE – 2007 / Paper - 1 Solutions

Page: .. 33 ..

2x

– 2x
2
–1

1

x? 1? x? 1 ?

66.

A ?

p, q, r

B ?

p, s

C ?
D ?

r, s
p, q

continuous in (— 1, 1) and differentiable in (–1, 1)

Match the integrals in Column I with the values in Column II and indicate your answer by
darkening the appropriate bubbles in the 4 × 4 matrix given in the CRS.
Column I
1

Column II

dx
x2

? 1?

(A)

?1
1

dx

?

(B)

1? x

0
3

dx

?1? x

(C)

2

2

(p)

1
? 2?
log ? ?
2
? 3?

(q)

? 2?
2log ? ?
? 3?

(r)

?
3

(s)

?
2

2
2

?x

(D)

1

dx
1? x

2

Sol.
A? s
1

1

? 1 ? x2 ?
dx

?1

tan

x

?1

?
0

?
2

?

?
2

?1

B? s
1

?

1

dx
1? x

2

? sin

?1

x
0

C? p
3

?1 ? x2
dx

3

?

2

1
1? x
loge
2
1? x 2

1
2
loge
2
3
D? r
?

2

?x
1

dx
2

x ?1

? sec ? 1 x

2
1

? sec ? 1 2 ?

?
3

IIT-JEE – 2007 / Paper - 1 Solutions

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