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MEMORY BASED (IIT-JEE - 2006) TEST PAPER
PHYSICS
TM
Section A : Single choice, number of questions 12, 3 marks for each correct answer and – 1 for incorrect answer.
Section B : Multiple choice, number of questions 8, 5 marks for each corect answer and –1 for incorrect answer.
Section C : Compreshension, 4 paragraphs, with 3 question in each paragraph each question has 5 marks for correct
anwers and –2 marks for incorerct
Section D : Fill in blanks , no negative marking, number of questions 4 each question of 6 marks.
Section E : Matching, no negative marking, each question of 6 marks, number of questions 4.
SECTION A :
1.
A solid sphere of radius R has moment of inertia about its geometrical axis. If it is melted into a disc of radius r and
thickness t. If it's moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also
equal to , then the value of r is equal to :
2
(A)
15
2
R
(B)
5
3
R
(C)
15
R
(D)
3
15
R
TM
Ans.
(A)
2.
System shown in figure is in equilibrium and at rest. The spring and string are massless Now the string is cut. The
acceleration of mass 2m and m just after the string is cut will be :
Sol.
(A) g/2 upwards , g downwards
(B) g upwards, g/ 2 downwards
(C) g upwards , 2g downwards
(D) 2g upwards , g downwards
(A)
TM
RESONANCE
Page # 1
After string is cut, FBD of m
a=
mg
= g
m
TM
FBD of 2m
a=
3.
Sol.
3mg  2mg
g
=

2m
2
A massless rod BD is suspended by two identical massless strings AB and CD of equal lengths. A block of mass
‘m’ is suspended point P such that BP is equal to ‘x’, if the fundamental frequency of the left wire is twice the
fundamental frequency of right wire, then the value of x is :
(A) l/5
(A)
f=
n
2
(B) l/4
(C) 4l/5
(D) 3l/4
T

for both f,land  are same
n T = constt.
Let tensions in the strings be T1 andT2 (as shown)
1 T1 = 2 T2

A = 0
B = 0
TM
T1 = 4T2

mg . x = T2 l

mg (  – x) = T1 
T2
x
1
= T =
x
4
1
4x =  – x
5x = 
x =  /5
4.
A point object is placed at distance of 20 cm from a thin planoconvex lens of focal length 15 cm. The plane surface
of the lens is now silvered. The image created by the system is at :
(A) 60 cm to the left of the system.
(C) 12 cm to the left of the system.
(B) 60 cm to the right of the system.
(D) 12 cm to the right of the system.
TM
RESONANCE
Page # 2
Sol.
(C)
By mirror - lens combination formula
1
2
1
= f – f
F
m
L
2
1
1
=
–
15
F

By mirror formula
TM
1
1
1
=
+
u
F
v
2
1
1
=
+
15
 20
v

–

1
2
1
=
–
20
15
v
38
60
V = –12 cm
negative means towards left
=
5.
Half life of a radio active substance 'A' is 4 days. The probability that a nucleus will decay in two half lives is :
(A)
Sol.
1
4
(B)
3
4
(C)
1
2
(D) 1
(B)
N0
4
probability of a nucleus decaying = no. of nuclei decayed total no.
Number of nuclei left after 2 half lives =
=
6.
3 N0 4
3
=
N0
4
The graph between object distance u and image distance v for a lens is given below. The focal length of the lens is:
TM
Sol.
(A) 5 ± 0.1
(B)
(B) 5 ± 0.05
1
1
1
=
–
u
f
v
(C) 0.5 ± 0.1
(D) 0.5 ± 0.05
...(1)
1
1
1
=
–
10
 10
f
f=+5
By differentiate eq. (1)
 F

f
2

 V
V
2

u
u
2
=
1
V
2
v +
1
u2
u
TM
RESONANCE
Page # 3

+
f
5
2
=
1 (0.1)
 10
2
+
1 (0.1)
10 2
0 .2
0.2
 25 =
 0.05
100
4
f = 5 ± 0.05
Ans.
f =
so,
7.
Two bars of radius 'r' and '2r' are kept in contact as shown. An electric current  is passed through the bars. Which
one of the following is correct?
TM
Ans.
8.
Sol.
(A) Heat produced in bar (1) is 2 times the heat produced in bar (2)
(B) Electric field in both halves is equal
(C) Current density across AB is double that of across BC.
(D) Potential difference across AB is 4 times that of across BC.
(D)
 4 2  


g
A student performs an experiment for determination of   2  , l  1m, and he commits an error of l. For T
T


he takes the time of n oscillations with the stop watch of least count T and he commits a human error of 0.1sec.
For which of the following data, the measurement of g will be most accurate ?
(A) L = 0.5, T = 0.1, n = 20
(B) L = 0.5, T = 0.1, n = 50
(C) L = 0.5, T = 0.01, n = 20
(D) L = 0.1, T = 0.05, n = 50
(D)
g
L
T
2
g = L
T
In option (d) error in g is minimum and number of repetition of measurement are maximum.In this case the error in
g is minimum.
9.
A double star system consists of two stars A and B which have time period TA and TB. Radius RA and RB and mass
MA and MB. Choose the correct option.
(A) If TA > TB then RA > RB
(B) If TA > TB then MA > MB
TM
 TA
(C) 
 TB
2

R
   A

 RB
3



(D) TA = TB
Sol.
(D)
For binary star the time period of the two stars are equal.
10.
Find the time constant for the given RC circuits in correct order :
Sol.
R1 = 1 , R2 = 2 , C1 = 4 F, C2 = 2 F
(A) 18,4, 8/9
(B) 18, 8/9, 4
(B)
eq = Req.Ceq
1 = (2 + 1) (2 + 4) = 18 s
(C) 4,18, 8/9
(D) 4,8/9, 18
TM
RESONANCE
Page # 4
11.
2 =
8
2 1 2  4
.
= s
9
2 1 2  4
3 =
2 1
. (4 + 2) = 4 s.
2 1
A biconvex lens of focal lengthf forms a circular image of radius r of sun in focal plane. Then which option is correct:
(A) r2  f
(B) r2  f2
(C) If lower half part is convered by black sheet, then area of the image is equal to r2/2
(D) if f is doubled, intensity will increase
(B)
r = f tan
TM
Sol.
Hence, r2  f2.
12.
The number of circular divisions on the shown screw gauge is 50. It moves 0.5 mm on main scale for one complete
rotation. The diameter of the ball is :
0
Sol.
(A) 2.25 mm
(C)
5
2 25
(B) 2.20 mm
(C) 1.20 mm
(D) 1.25 mm
0 .5
= 0.05 mm
50
Actual measurement
Zero error = 5 
0 .5
– 0.05 mm
50
= 1 mm + 0.25 mm – 0.05 mm = 1.20 mm.
= 2 × 0.5 mm + 25 ×
TM
SECTION B :
13.
The graph between 1/ and stopping potential (V) of three metals having work functions 1,2 and 3 in an experiment of photo–electric effect is plotted as shown in the figure. Which of the following statement(s) is/are correct ?
[Here  is the wavelength of the incident ray].
(A) Ratio of work functions 1 : 2 : 3 = 1 : 2 : 4.
(B) Ratio of work functions 1 : 2 : 3 = 4 : 2 : 1.
(C) tan is directly proportional to hc/e, where h is Planck’s constant and c is the speed of light.
(D) The violet colour light can eject photoelectrons from metals 2 and 3.
TM
RESONANCE
Page # 5
Sol.
(A, C)
hc
  = eV

hc 

e c
For plate 1 :
V=
Plate 2
1
2
= 0.001
= 0.002
hc
hc
1 : 2 : 3 = 1 : 2 : 4
For plate 2, threshold wavelength
Plate 3
3
= 0.004
hc
TM
hc
hc
1000
=  =
=
= 500 nm
0.002 hc
2
2
For plate 3, threshold wavelength
hc
hc
1000
=  =
=
= 250 nm
0.004 hc
4
3
Since violet colour light  is 400 nm, so violet < threshold for plate 2.
So, violet colour light will eject photo–electrons from plate 2 and not from plate 3.
14.
A field line is shown in the figure. This field can not represent
Ans.
(A) Magnetic field
(C) Induced electric field
(B, D)
15.
A solid sphere is in pure rolling motion on an inclined surface having inclination .
Ans.
(A) frictional force acting on sphere is f =  mg cos .
(B) f is dissipative force.
(C) friction will increase its angular velocity and decreases its linear velocity.
(D) If  decreases, friction will decrease.
(C, D)
16.
Ans.
(B) Electrostatic field
(D) Gravitational field
TM
In a dark room with ambient temperature T0, a black body is kept at a temperature T. Keeping the temperature of the
black body constant (at T), sunrays are allowed to fall on the black body through a hole in the roof of the dark room.
Asuming that there is no change in the ambient temperature of the room, which of the following statement(s) is/are
correct ?
(A) The quantity of radiation absorbed by the black body in unit time will increase.
(B) Since emissivity = absorptivity, hence the quantity of radiation emitted by black body in an unit time will
increase.
(C) Black body radiates more energy in unit time in the visible spectrum.
(D) The reflected energy in unit time by the black body remains same.
(A, B, C, D)
TM
RESONANCE
Page # 6
17.
For spherical symmetrical charge distribution, variation of electric potential with distance from centre is given in
diagram. Given that :
q
V = 4 R for r  R0
0 0
and
q
V = 4 r for r  R0.
0
TM
Ans.
Then which option(s) are correct :
(A) Total charge within 2R0 is q.
(B) Total electrostatic energy for r  R0 is zero.
(C) At r = R0 electric field is dicontinuous.
(D) There will be no charge anywhere except at r = R0
(A, B, C, D)
18.
Which of the following statement is correct in the given figure.
Sol.
19.
Ans.
20.
Ans.
(A) net force on the loop is zero
(B) net torque on the loop is zero
(C) loop will rotate clockwise about axis OO’ when seen from O
(D) loop will rotate anticlockwise about OO’ when seen from O
(A, C)
Magnetic force on wire BC would be perpendicular to the plane of the loop along the outward direction and on wire
DA the magnetic force would be along the inward normal, so net force on the wire loop is zero and torque on the loop
would be along the clockwise sence as seen from O.
Function x = Asin2t + B cos2t + C sint cost represents SHM
(A) for any value of A,B and C (except C = 0)
(B) If A = – B,C = 2B, amplitude = B 2
(C) If A = B ; C = 0
(B, D)
(D) If A = B ; C = 2B, amplitude = |B|
TM
A ball moves over a fixed track as shown in the figure. From A to B the ball rolls without slipping. If surface BC is
frictionless and KA, KB and KC are kinetic energies of the ball at A,B and C respectively, then :
(A) hA > hC ; KB > KC
(C) hA = hC ; KB = KC
(A, B)
(B) hA > hC ; KC > KA
(D) hA < hC ; KB > KC
TM
RESONANCE
Page # 7
21.
A plate of mass M of dimensions (a × b) is hinged along one edge. The plate is maintained in horizontal position by
colliding a ball of mass m, per unit area, elastically 100 times per second this ball is striking on the right half shaded
region of the plate as shown in figure. Find the required speed of the ball (ball is colliding in only half part of the plate
as shown). (It is given M = 3 kg, m = 0.01 kg, b = 2 m, a = 1 m, g = 10 m/s2)
TM
Sol.
3 
b
2mv 100  b  = Mg . .
4


2
Mg
Mg
= 300 m
v = 4m.100 . 3
4
= 10 m/s.
22.
A circular disc is kept on a smooth horizontal plane with its plane parallel to horizontal plane. A groove is made in
the disc along the diameter as shown in the figure. The coefficient of friction between block of mass m placed inside
the groove and the surface of the groove is 2/5 and sin = 3/5. The linear acceleration of the disc is 25 m/s2 towards
left. The acceleration of block with respect to the frame of reference of the disc is :
Sol.
N1 = mg
N2 = ma sin370
ma cos 37 0  N2  N1
= 10 m/s2 .
m
Ans.
10 m/s2
a=
23.
Sol.
TM
In an insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Find the final temperature of
the mixture (in kelwin).
Given, Lfusion = 80 cal/g = 336 J/g, Lvaporization = 540 cal/g = 2268 J/g,
Sice = 2100 J/kg K = 0.5 cal/gK and Swater = 4200 J/kg K = 1 cal/gK
Q = 0
Heat lost by steam to convert into 0ºC water
HL = 0.005 × 540 + 0.05 × 10 × 1
= 27 + 5 = 32 kcal
Heat required by ice to change into 0º C water
1
× 20 + 0.45 × 80 = 4.5 + 36.00 = 40.5 kcal
2
Thus, final temperature of mixture is 0ºC = 273 K
Ans. 273 K.
Hg = 0.45 ×
TM
RESONANCE
Page # 8
24.
Sol.
f the wavelength of the nth line of Lyman series is equal to the de-broglie wavelength of electron in initial orbit of
a hydrogen like element (z = 11). Find the value of n.
1
= RZ2


1 
1
mv
mvr
1 

=
=
 (n  1)2  = (h / mv ) =
h
hr


(n  1)h
n 1
=
2hr
2 r
Putting, r =
1
= RZ2

r0 (n  1)2
Z
TM

(n  1)Z
1 
1 

 (n  1)2  = 2r (n  1) 2
0


 (n +1) –
1
1
1
= 2r ZR =
(n  1)
2  11 0.53  10 10  1.1 10 7
0
1000
2  12.1 0.53
= 24.8
 (n + 1) = 25
 n = 24
Ans.
n = 24.
=
SECTION C :
Paragraph 1
Two plane harmonic sound waves are expressed by the equations.
y1(x, t) = A cos (0.5 x – 100 t)
y2(x, t) = A cos (0.46 x – 92 t)
(All parameters are in MKS) :
25.
Sol.
How many times does an observer hear maximum intensity in one second ?
(A) 4
(B) 10
(C) 6
(D) 8
(A)
2f1  100 
f1 = 50
2f2 = 96
f2 = 48
beat frq = 50 – 48
=2
26.
Sol.
TM
What is the speed of the sound ?
(A) 200 m/s
(B) 180 m/s
(A)
(C) 192 m/s
(D) 96 m/s
2
= 0.5 x

2
=4
0 .5
Speed v = f = 50 × 4
= 200
=
27.
Sol.
At x = 0 how many times the amplitude of y1 + y2 is zero in one second at x = 0 ?
(A) 192
(B) 48
(C) 100
(D) 96
(D)
At x = 0,
y = y1 + y2 = 2A cos 96 t cos 4 t
For y = 0, cos 96  t = 0 or cos 4 t = 0
TM
RESONANCE
Page # 9

96 t = (2n + 1) (/2) and 4t = (2m + 1) /2
For 0 < t < 1
1
1
< n < 95.5 and –
< m < 3.5
2
2
Here n and m are integers, therefore net amplitude becomes zero 100 times.
–
Passage 2
A wooden cylinder of diameter 4r, height H and density /3 is kept on a hole of diamete 2r of a tank, filled with liquid
of density  as shown in the figure.
28.
If level of the liquid starts decreasing slowly when the level of liquid is at a height h1 above the cylinder the block
starts moving up. At what value of h1, will the block rise :
TM
(A)
Sol.
4h
9
(B)
5h
9
(C)
5h
3
(D) Remains same
(C)
A1 = r2 = area of base of cylinder in air
A2 = 3r2 = area of base of cylinder in water
A3 = 4r2 = cross-section area of cylinder
Equating the forces, we get
(Pa + gh1)A3 +

gH A3 = (Pa)A1 + [Pa + g(h1 + H)]A2
3
TM
On solving
h1 =
29.
The block in the above question is maintained at the position by external means and the level of liquid is lowered.
The height h2 when this external force reduces to zero is
(A)
Sol.
5
H
3
4h
9
(B)
5h
9
(C) Remains same
(D)
2h
3
(A)
PaA3+

A Hg = PaA3 + gh2
3 3
TM
RESONANCE
Page # 10
TM
h2 = 4H/9.
30.
Sol.
If height h2 of water level is further decreased, then
(A) cylinder will not move up and remains at its original position
(B) for h2 = h/3, cylinder again starts moving up
(C) for h2 = h/4, cylinder again starts moving up
(D) for h2 = h/5 cylinder again starts moving up
(A)
For h2 < 4h/9 cylinder does not move up because further bouyant force decreases while the weight of block remains
same.
Passage 3
Modern trains are based on Maglev technology in which trains are magnetically leviated, which runs its EDS Maglev
system.
There are coils on both sides of wheels. Due to motion of train current induces in the coil of track which levitate it.
This is in accordance with lenz’s law. If trains lower down then due to lenz’s law a repulsive force increases due to
which train gets uplifted and if it goes much high then there is a net downward force disc to gravity.
The advantage of maglev train is that there is no friction between the train and the track, thereby reducing power
consumption and enabling the train to attain very high speeds.
Disadvantage of maglev train is that as it slows down the electromanetic forces decreases and it becomes difficult
to keep it leviated and as it moves forward according to Lenz law there is an electromagnetic drag force.
31.
Ans.
32.
Ans.
33.
Ans.
What is the advantage of this system ?
(A) No friction hence no power consumption
(C) Gravitation force is zero
(A)
(B) No electric power is used
(D) Electrostatic force draws the train
What is the disadvantage of this system ?
(A) Train experiences upward force according to Lenz’s law
(B) Friction force create a drag on the train
(C) Retardation
(D) By Lenz’s law train experience a drag
(D)
Which force causes the train to elevate up ?
(A) Electrostatic force
(C) Magnetic force
(C)
TM
(B) Time varying electric field
(D) Induced electric field
Passage- 4
The capacitor of capcitance C can be charged (with the help of a resistance R) by a voltage source V, by closing
switch S1 while keeping switch S2 open. The capacitor can be connected in series with an inductor ‘L’ by closing
switch S2 and opening S1.
TM
RESONANCE
Page # 11
34.
Sol.
Initially, the capacitor was uncharged. Now, switch S1 is closed and S2 is kept open. If time constant of this circuit
is , then
(A) after time interval , charge on the capacitor is CV/2
(B) after time interval 2, charge on the capacitor is CV(1 – e–2)
(C) the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged
(D) after time interval 2, charge on the capacitor is CV(1 – e–1)
(B)
Q = Q0(1 – e–t/)
Q = CV(1 – e–t/) after time interval 2.
TM
35.
After the capacitor gets fully charged, S1 is opened and S2 is closed so that the inductor is connected in series with
the capacitor. Then,
(A) at t = 0, energy stored in the circuit is purely in the form of magnetic energy
(B) at any time t > 0, current in the circuit is in the same direction
(C) at t > 0, there is no exchange of energy between the inductor and capacitor
(D) at any time t > 0, instantaneous current in the circuit may V
Sol.
(D)
q = Q0 cos t
i=–

36.
C
L
dq
= Q0  sin t
dt
imax = C V = V
C
L
If the total charge stored in the LC circuit is Q0, then for t  0

t 

(A) the charge on the capacitor is Q = Q0 cos  2 

LC 


t 

(B) the charge on the capacitor is Q = Q0 cos  2 

LC 

(C) the charge on the capacitor is Q = –LC
(D) the charge on the capacitor is Q = –
d2Q
dt 2
1
d2 Q
LC
dt 2
Sol.
(C)
37.
Match the following for the given process :
Sol.
(A) Process J  K
(P) w > 0
(B) Process K  L
(Q) w < 0
(C) Process L  M
(R) Q > 0
(D) Process M  J
(S) Q < 0
(A)  (S), (B)  (P) and (R), (C)  (R), (D)  (Q) and (S).
TM
TM
RESONANCE
Page # 12
38.
Sol.
39.
Sol.
40.
Sol.
Match the following
Column 1
Column 2
(a) Fission
(P)
Matter - energy
(b) Fusion
(Q)
In atoms of high atomic number only
(c)  - decay
(R)
In atoms of low atomic number only
(d) Exothermic nuclear reaction (S)
Involves weak nuclear forces
(A) (a)  (P) and (Q), (b)  (P) and (R),
(c)  (P) and (S),
(d)  (P)
(B) (a)  (R) and (Q), (b)  (Q) and (R),
(c)  (P) and (S),
(d)  (P)
(C) (a)  (S) and (Q), (b)  (P) and (R),
(c)  (P) and (S),
(d)  (P)
(D) (a)  (P) and (R),
(b)  (S) and (R),
(c)  (P) and (S),
(d)  (P)
Match the following
Column 1
(A) Dielectric ring uniformly charged
(B) Dielectric ring uniformly charged rotating
with constant angular velocity 
(C) Constant current in ring i0
(D) A ring carrying current i = i0 cost
(A)  (P) ; (B)  (P), (Q), (S) ; (C)  (Q), (S) ;
TM
Column 2
(P) Time independent electrostatic field out of system
(Q) Magnetic field
(R) Induced electric field
(S) Magnetic moment
(D)  (Q), (R), (S).
A simple telescope used to view distant objects has eyepiece and objective lens of focal lengths fe and f0, respectively. Then :
Column 1
Column 2
(A) Intensity of light receieved by lens
(P) Radius of aperture (R)
(B) Angular magnification
(Q) Dispersion of lens
(C) Length of telescope
(R) focal length f0, fe
(D) Sharpness of image
(S) spherical aberration
(A)  (P), (B)  (R), (C)  (R), (D)  (P), (Q), (S).
TM
TM
RESONANCE
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