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IIT JEE 2009 Physics Paper-I Questions and
Answers
PART III:
PHYSICS
Useful
Data:
Planck�s
constant h = 4.1 �
10�14
eV.s
Velocity of light c = 3 �
108
m/s
SECTION �
I
Straight Objective
Type
This
section contains 8 multiple
choice questions. Each question has 4 choices (A), (B), (C) and (D),
out of which ONLY ONE is
correct.
41. Three
concentric metallic spherical shells of radii R, 2R, 3R, are given
charges
Q1,
Q2,
Q3, respectively. It
is
found that the
surface charges given to the shells,
Q1
:
Q2
:
Q3,
is
(A) 1 : 2 :
3
(B)
1 : 3 : 5
(C) 1 : 4 :
9
(D) 1 : 8 :
18.
18
Key.
(B)
Sol.
It is given
that
Q1/4Π
R2 = (
Q1+
Q2)/4Π(2
R)2
=>
Q1
+
Q2
+
Q3
/4Π(3R)2
i.e
Q1
/1
=
Q1
+
Q2
/4
=
Q1
+
Q2
+
Q3
/9
=>
Q1
/1
=
Q2
/3
=
Q3
/5
So, option (B)
is
correct.
42. A block of base 10
cm � 10 cm and height 15 cm is kepton an inclined
plane. The coefficient of
friction
between them
is 3 . The inclination θ of this inclined plane from the
horizontal
plane is
gradually
increased form
0�. Then
(A) at
θ = 30�, the block will start sliding down the
plane
(B) the block
will remain at rest on the plane up to certain θ and then
it will
topple
(C) at
θ = 60�, the block will start sliding down the plane and
continue to do
so at higher
angles
(D) at
θ = 60�, the block will start sliding down the plane and
further
increasing
θ, it will topple at certain
θ.
Key.
(B)
Sol.
At
θ = 30�, the weight W of the block passes through the base
AB, and hence the
block will not
topple;
since
tan
30� = 1/√3<√3
, So it will
Slide
However,
as θ is increased the block will
topple
θ + tan 1.5
=90�
i.e θ
< 60�
So, option (B)
is
correct.
43. A ball is
dropped from a height of 20 m above the surface of water in a
lake. The refractive index of
water
is 4/3. A fish
inside the lake, in the line of fall of the ball, is looking at
the ball. At an instant, when the
ball
is 12.8 m
above the water surface, the fish sees the speed of the ball as [Take
g = 10 m/
s2.]
(A) 9
m/s
(B)
12
m/s
(C) 16
m/s
(D) 21.23 m/s.
Key.
(C)
Sol.
v = √2 *10*
(20-12.8) = 12 m / s
Due to
refraction at the water
surface,
v/�2 =
u/�1
dv/dt =
4/3*12 = 16
m/s
So, option (C) is
correct.
*44. Look at
the drawing given in the figure which has been drawn with ink of
uniform
line
thickness. The mass of ink used to draw each of the two inner circles,
and
&
nbsp;each
of the two
line segments is m. The mass of the ink used to draw the outer circle
is
6m. The
coordinates of the centers of the different parts are outer circle (0,
0), left
inner circle
(�a, a), right inner circle (a, a), vertical line (0, 0) and
horizontal line
(0,
�a). The
y�coordinate of the center of mass of the ink in this drawing
is
(A) a
/10
(B)
a/8
(C)
a/12
(D)
a/3
.Key.
(A)
Sol.
ycm
=
7m *0 +2m* a+ m(- a) /10m =
a/10
So, option (A)
is
correct.
*45. Two small
particles of equal masses start moving in opposite directions
from a
point A in a
horizontal circular orbit. Their tangential velocities are v and
2v,
respectively,
as shown in the figure. Between collisions, the particles move
with
constant
speeds. After making how many elastic collisions, other than that at
A,
these two
particles will again reach the point A
?
(A)
4
(B) 3
(C)
2
(D) 1.
Key.
(C)
Sol.
The first
collision takes place at A1 as the particle moving in anticlockwise
sense covers half the
distance
covered by
other. They exchange velocities and the second collision takes place
at A2. The third
collision
takes place at
A.
So, option (C)
is correct.
46. The figure shows certain
wire segments joined together to form a coplanar
loop.
The loop is
placed in a perpendicular magnetic field in the direction going into
the
plane of the
figure. The magnitude of the field increases with time I1 and I2
are the
currents in
the segments ab and cd.
Then,
(A) I1
> I2
(B) I1
< I2
(C) I1 is in
the direction ba and I2 is in the direction
cd
(D) I1 is in
the direction ab and I2 is in the direction
dc.
Key.
(D)
Sol.
Since
the field is increasing,
the flux region is increasing; the induced
current (by
Lenz's law) should flow
in an
anticlockwise sense around the boundary of the shaded
region.
So, option (D)
is correct.
.47. A disk of
radius a/4 having a uniformly distributed charges 6C is placed
in
the x�y plane
with its center at (�a/2, 0, 0). A rod of length a carrying
a
uniformly
distributed charge 8C on the x�axis from x = a/4 to x =5a/4.
Two
point charges
�7C and 3C are placed at (a/4, �a/4, 0) and (�3a/4, 3a/4,
0),
respectively.
Consider a cubical surface formed by six surfaces x = � a/2,
y
= � a/2, z = �
a/2. The electric flux through this cubical surface
is
(A)-2C/ε
(B)2C/ε
(C)10C/
ε
(D)12C/ε
Key. (A)
Sol.
Net charge = -7+3
+2/ε
= > -2C/ε
So, option (A)
is correct.
*48. The x�t
graph of a particle undergoing simple harmonic motion is
shown
below. The
acceleration of the particle at t = 4/3 s
is
(A)-√3/32 cm/
s2
(B)
-
π2
/32cm/
s2
(c)
π2
/32
cm/
s2
(D)√3/32*
π2
cm/
s2
Key.
(D)
Sol:
Acceleration
acceleration |d2x
/
dt2|t=4/3
=
-√3/32 cm/
s2
So, option (A)
is correct.
SECTION �
II
Multiple Correct
Answer
Type
This section contains
4 multiple correct answer(s) type questions. Each question
has 4 choices (A), (B), (C)
and (D), out of which ONE OR MORE is/are
correct.
*49. If the
resultant of all the external forces acting on a system of particles
is zero, then from an inertial
frame,
(A) linear
momentum of the system does not change in
time
(B) kinetic
energy of the system does not change in
time
(C) angular
momentum of the system does not change in
time
(D) potential
energy of the system does not change in
time.
Key. (A,
C)
Sol.
∑ Fext = 0
=>P= constant
So, option (A)
is correct.
50. A student
performed the experiment of determination of focal length of a
concave mirror by u�v
method
using an
optical bench of length 1.5 meter. The length of the mirror used is 24
cm. The maximum error
in
the location
of the image can be 0.2 cm. The 5 sets of (u, v) values recorded by
the student (in cm) are :
(42,
56), (48, 48),
(60, 40), (66, 33), (78, 39). The data set(s) that cannot come
from experiment and is
(are)
incorrectly
recorded, is (are)
(A) (42,
56)
(B) (48, 48)
(C) (66,
33)
(D) (78, 39).
Key. (C,
D)
Sol.
1/v + 1/u =
1/f
the data set
(66, 33) does not satisfy the mirror
equation.
So, options
(C) and (D) are correct.
51. For the
circuit shown in the figure
:
(A) the
current I through the battery is 7.5
mA
(B) the
potential difference across RL is 18
V
(C) ratio of
powers dissipated in R1 and R2 is
3
(D) if R1 and
R2 are interchanged, magnitude of the power dissipated in RL will
decrease
by a factor of
9.
Key. (A,
D)
Sol.
R
eq
=3.2kΩ
I = 24/3.2 =7.5mA
& I
=
6mA
After
interchanging R1 with R2
R
eq
=48/7Ω
So, I = 3.5
mA
1
LI =
2mA
So, P
L0 / P
L1 =
9
*52. CV and CP denote
the molar specific heat capacitors of a gas at constant
volume and constant
pressure,
respectively.
Then
(A) CP � CV is
larger for a diatomic ideal gas then for a monoatomic ideal
gas
(B) CP + CV is
larger for a diatomic ideal gas then for a monoatonic ideal
gas
(C) CP / CV is
larger for a diatomic ideal gas then for a monoatonic ideal
gas
(D) CP . CV is
larger for a diatomic ideal gas then for a monoatonic ideal
gas.
Key. (B,
D)
Sol. CP and CV
for diatomic is greater than
monoatomic.
So, CP + Cv,
CP . CV is greater for diatomic ideal
gas.
SECTION �
III
Linked Comprehension
Type
This
section contains
2 paragraphs. Based upon each paragraph, 3 multiple choice
questions have to
beanswered. Each question has 4 choices (A), (B), (C) and (D)
out of which ONLY ONE is
correct
Paragraph
for Questions Nos. 53 to
55
Scientists are working hard to develop unclear fusion
reactor. Nuclei of heavy
hydrogen, 1
H2 , known as
deuteron and
denoted by D, can be thought of as a candidate for fusion reactor. The
D�D reaction is 
In the
core of fusion reactor, a gas of heavy
hydrogen is fully ionized into
deuteron 1 H nuclei and
electrons is known as plasma. The nuclei move randomly
in the reactor core and
occasionally come close enough for nuclear fusion to take
place. Usually, the
temperatures in the reactor
core are too high and no material wall can be used to confine the
plasma. Special techniques
are used which confine the
plasma for a time t0 before the particles fly away from the
core. If n is the
density (number/volume) of
deuterons, the product nt0 is called Lawson number. In one of
the criteria, a reactor is
termed
successful if Lawson number is greater than 5 � 1014
s/cm3.
It
may be helpful to use the following : Boltzmann constant k = 8.6 � 10�5
eV/K;
53. In the core of nuclear fusion
reactor, the gas becomes plasma because
of
(A) strong
nuclear force acting between the
deuterons
(B) Coulomb
force acting between the
deuterons
(C) Coulomb
force acting between the deuteron�electron
pairs
(D) the high
temperature maintained inside the reactor
core.
Key.
(D)
Sol.
Plasma state is achieved at high
temperatures.
54. Assume that two deuteron nuclei in
the core of fusion reactor at
temperature T
are moving towards each
other, each
with kinetic energy 1.5 kT, when the separation between them is
large enough to
neglect
Coulomb
potential energy. Also neglect any interaction from other particles in
the core. The
minimum
temperature T
required for them to reach a separation of 4 � 10�15 m is in the
range
(A) 1.0 � 109
K < T < 2.0 � 109 K (B) 2.0 � 109 K <
T < 3.0 � 109
K
(C) 3.0 � 109
K < T < 4.0 � 109 K (D) 4.0 � 109 K <
T < 5.0 � 109
K.
Key.
(A)
Sol
55. Results of calculations for four
different designs of a fusion reactor
using D�D
reaction are given below.
Which of these
is most promising based on Lawson criterion
?
(A) deuteron
density = 2.0 � 1012 cm�3,
confinement time = 5.0 � 10-3
s
(B) deuteron
density = 8.0 � 1014 cm�3,
confinement time = 9.0 � 10-1
s
(C) deuteron
density = 4.0 � 1023 cm�3,
confinement time = 1.0 � 10-11
s
(D) deuteron
density = 1.0 � 1024 cm�3,
confinement time = 4.0 � 10-12
s.
Key.
(B)
Sol.
Lawson no. = nto Out
of given option nto is greater for option
(B).
Paragraph for
Questions Nos. 56 to
58
When a particle is restricted to move
along x�axis between x = 0 and x = a,
where a is of
nanometer dimension, its energy can take only certain specific
values. The allowed energies of the particle moving in
such a restricted region,
correspond to
the formation of standing waves with nodes at its ends x = 0 and x
= a. The wavelength of this standing wave is related to
the linear momentum p of
the particle
according to the de�Broglie relation. The energy
of the
particle of
mass m is related to its linear momentum as E =
p2/2m . Thus, the
energy of the particle can be denoted by a quantum number
n taking values 1, 2,
3, � (n = 1,
called the ground state) corresponding to the number of
loops
in the
standing wave.
Use the model
described above to answer the following three questions for a
particle moving in the line x = 0 to x =a
56. The allowed energy for the particle
for a particular value of n is
proportional
to
(A)
a-2
(B)
a-3/2
(C)
a-1
(D)
a2.
Key.
(A)
57. If the mass of the particle is m =
1.0 � 10�30 kg and a = 6.6 nm,
the
energy of the
particle in its ground
state
is closest
to
(A) 0.8
meV
(B)
8
meV
(C) 80
meV
(D)
800
meV.
Key.
(B)
Sol.
In ground state n = 1
E =
h2
/8ma2 =8
mev
58. The speed of the particle, that can
take discrete values, is proportional
to
(A)
n�3/2
(B)
n-1
(C)
n1/2
(D)
n.
Key.
(D)
Sol.
1/2 mv2
=> v ά n
SECTION �
IV
Matrix Match
Type
| This section contains 2 questions. Each
question contains statements given in
two columns,
which have to be matched. The statements in Column I are labeled A, B,
C
and D, while the statements
in Column II are labelled p, q, r, s and t. Any given
statement
in Column I can have correct
matching with ONE OR MORE statement(s) in Column
II. The appropriate bubbles
corresponding to the answers to these questions have to
be darkened as
illustrated in the following example:
If the correct matches are A � p, s and t; B � q and r; C
� p and q; and D � s and t;
then the correct darkening of
bubbles will look like the following:
|
 |
59. Six point
charges, each of the same magnitude q, are arranged in different
manners as shown in Column II.In each case, a point M and
a line PQ passing
through M are
shown. Let E be the electric field and V be the electric potential
at M (potential at infinity is zero) due to the given
charge distribution when
it is at
rest.Now, the whole system is set into rotation with a constant angular
velocity about the line PQ. Let B themagnetic field at M
and � be the magnetic
moment of the
system in this condition. Assume each rotating charge to be
equivalent to a steady
current.
*60. Column II
shows five system in which two objects the labelled as X and Y. Also in
each case a point P is
shown. Column
I gives some statements about X and / or Y. Match these statements to
the appropriate
system (s)
from Column
II.
