PHYSICS STUDY MATERIALS

7/25/19

NEET PHYSICS

1. A spring of force constant k is cut into lengths of
ratio 1 : 2 : 3. They are connected in series and
the new force constant is k'. Then they are connected
in parallel and force constant is k¢¢ . Then k' : k¢¢
is : -
(1) 1 : 9 (2) 1 : 11
(3) 1 : 14 (4) 1 : 16
Ans. (2)
Sol. Length of the spring segments = l, l, l
6 3 2
As we know μ
l
1
K
so spring constants for spring segments will be
K1 = 6K, K2 = 3K, K3 = 2K
so in parallel combination
K" = K1 + K2 + K3 = 11K
in series combination
K' = K (As it will become original spring)
so K' : K" = 1 : 11
2. The ratio of resolving powers of an optical
microscope for two wavelengths l1 = 4000 Å and
l2 = 6000 Å is :-
(1) 9 : 4 (2) 3 : 2
(3) 16 : 81 (4) 8 : 27
Ans. (2)
Sol. Resolving power μ
l
1
l
=
l
1 2
2 1
RP
RP
6000Å 3
4000Å 2
= =
3. The two nearest harmonics of a tube closed at one
end and open at other end are 220 Hz and
260 Hz. What is the fundamental frequency of the
system?
(1) 20 Hz (2) 30 Hz
(3) 40 Hz (4) 10 Hz
Ans. (1)
Sol. Difference between any two consecutive frequencies
of COP = = -
l
2v
260 220
4
= 40 Hz
Þ =
l
v
20Hz
4
So fundamental frequency = 20 Hz
NEET(UG)–2017 TEST PAPER WITH ANSWER & SOLUTIONS
(HELD ON SUNDAY 07th MAY, 2017)
4. Consider a drop of rain water having mass 1 g falling
from a height of 1 km. It hits the ground with a speed
of 50 m/s. Take 'g' constant with a value 10 m/s2.
The work done by the (i) gravitational force and the
(ii) resistive force of air is :-
(1) (i) 1.25 J (ii) – 8.25 J
(2) (i) 100 J (ii) 8.75 J
(3) (i) 10 J (ii) – 8.75 J
(4) (i) – 10 J (ii) – 8.25 J
Ans. (3)
Sol. Work done by the gravity (Wg) = mgh
= 10–3 × 10 × 103
= 10 J
By work–energy theorem = Wg + Wres = DKE
10 + Wres = - ´ ´ 3 2 1
10 (50)
2
Wres = –8.75 J
5. A physical quantity of the dimensions of length that
can be formed out of c, G and pe
2
0
e
4 is [c is velocity
of light, G is universal constant of gravitation and
e is charge] :-
(1)
é ù
êë pe úû
2 1/2
2
0
e
c G
4 (2)
é ù
êë pe úû
2 1/2
2
0
1 e
c G4
(3) pe
2
0
1 e
G
c 4 (4)
é ù
êë pe úû
2 1/2
2
0
1 e
G
c 4
Ans. (4)
Sol. [L] = [c]a [G]b
é ù
ê ú
ë pe û
2 c
0
e
4
[L] = [LT–1]a [M–1L3T–2]béë - ùû
ML3T 2 c
[L] = La+3b+3c M–b+c T–a–2b–2c
a + 3b + 3c =1
–b + c = 0
a + 2b + 2c = 0
On solving,
a = – 2 , b =
1
2
, c =
1
2
\ L =
1
2 2
2
0
1 e
G.
c 4
é ù
ê ú
êë p e úû
2
NEET(UG)-2017
6. Two rods A and B of different materials are welded
together as shown in figure. Their thermal
conductivities are K1 and K2. The thermal
conductivity of the composite rod will be :-
T1 T2
d
(1)
+ 1 2 3(K K )
2
(2) K1 + K2
(3) 2 (K1 + K2)
(4)
+ 1 2 K K
2
Ans. (4)
Sol. In parallel = +
eq 1 2
1 1 1
R R R
= +
l l l
eq 1 2 K (2A) K A K A
+
= 1 2
eq
K K
K
2
7. A capacitor is charged by a battery. The battery is
removed and another identical uncharged capacitor
is connected in parallel. The total electrostatic
energy of resulting system :-
(1) Decreases by a factor of 2
(2) Remains the same
(3) Increases by a factor of 2
(4) Increases by a factor of 4
Ans. (1)
Sol. = 2
i
1
U CV
2
= éêë ùúû =
2
f i
1 V 1
U [2C] U
2 2 2
Decrease by a factor of 2
8. In a common emitter transistor amplifier the audio
signal voltage across the collector is 3V. The
resistance of collector is 3 kW. If current gain is 100
and the base resistance is 2 kW, the voltage and
power gain of the amplifier is :-
(1) 15 and 200
(2) 150 and 15000
(3) 20 and 2000
(4) 200 and 1000
Ans. (2)
Sol.
W
=b = ´ =
W
C
V
B
R 3k
A 100 150
R 2k
Power gain = bAV = 100 × 150 = 15000
9. Thermodynamic processes are indicated in the
following diagram :
f
f
f
700k
500k
300k
f
P
I
IV
III
II
V
i
Match the following
Column-1 Column-2
P. Process I a. Adiabatic
Q. Process II b. Isobaric
R. Process III c. Isochoric
S. Process IV d. Isothermal
(1) P ® c, Q ® a, R ® d, S ® b
(2) P ® c, Q ® d, R ® b, S ® a
(3) P ® d, Q ® b, R ® a, S ® c
(4) P ® a, Q ® c, R ® d, S ® b
Ans. (1)
Sol. Process (1) ® volume constant ® Isochroic
Process (2) ® adiabatic
Process (3) ® Temperature constant ® Isothermal
Process (4) ® Pressure constant ® Isobaric
10. Suppose the charge of a proton and an electron
differ slightly. One of them is – e, the other is
(e + De). If the net of electrostatic force and
gravitational force between two hydrogen atoms
placed at a distance d (much greater than atomic
size) apart is zero, then De is of the order of
[Given mass of hydrogen mh = 1.67 × 10–27 kg]
(1) 10–23 C (2) 10–37 C
(3) 10–47 C (4) 10–20 C
Ans. (2)
Sol.
´ D
=
2 2
2 2
K ( e) Gm
r r
D =
G
e m
K
-
- ´
= ´
´
11
27
9
6.67 10
1.67 10
9 10
C
= 1.436 × 10–37 C
11. The resistance of a wire is 'R' ohm. If it is melted
and stretched to 'n' times its original length, its new
resistance will be :-
(1)
R
n
(2) n2R (3) 2
R
n
(4) nR
Ans. (2)
Sol.
2
R R 2
A volume
=rl!= rl Þ μl
Þ R2 = n2R1
12. The given electrical network is equivalent to :
AB
Y
(1) OR gate (2) NOR gate
(3) NOT gate (4) AND gate
Ans. (2)
Sol. A
B y1 y2 y
NOR NOR NOT
= + 1 y A B
y2=y1+y1=y1=A+B=A+B
y=y2=A+B
NOR GATE
13. The de-Broglie wavelength of a neutron in thermal
equilibrium with heavy water at a temperature T
(Kelvin) and mass m, is :-
(1)
h
3mkT
(2)
2h
3mkT
(3)
2h
mkT
(4)
h
mkT
Ans. (1)
Sol. Kinetic energy of thermal neutron with equilibrium
is
3
KT
2
h h h
mv 2m K.E 3
2m KT
2
l = = =
æ ö
ç ÷
è ø
h
3 mKT
=
14. Which one of the following represents forward bias
diode ?
(1) – 4V R –3V
(2) – 2V R +2V
(3) 3V R 5V
(4) 0V R –2V
Ans. (4)
Sol.
R
V2 V1
In forward bias V1 > V2
Þ only
0V –2V
is in forward bias
15. A long solenoid of diameter 0.1 m has 2 × 104 turns
per meter. At the centre of the solenoid, a coil of
100 turns and radius 0.01 m is placed with its axis
coinciding with the solenoid axis. The current in the
solenoid reduces at a constant rate to 0A from
4 A in 0.05 s. If the resistance of the coil is 10p2W.
the total charge flowing through the coil during this
time is :-
(1) 16 mC (2) 32 mC
(3) 16 p mC (4) 32 p mC
Ans. (2)
Sol.
=éêëæçèDDfö÷ø× ùúûD
1
q t
t R
=êëém p DD úûù D
2
0
i 1
q nNr t
t R
- - =êéëp ´ ´ ´ ´ ´ p ´ ´æçè ö÷øúùû p ´
7 4 2 2
2
4 1
q 4 10 2 10 100 (10 ) 0.05
.05 10
q = 32 µC

16. Preeti reached the metro station and found that the
escalator was not working. She walked up the
stationary escalator in time t1. On other days, if she
remains stationary on the moving escalator, then the
escalator takes her up in time t2. The time taken
by her to walk up on the moving escalator will be
(1) -
1 2
2 1
t t
t t (2) +
1 2
2 1
t t
t t
(3) t1 – t2 (4)
+ 1 2 t t
2
Ans. (2)
Sol. V1 ® velocity of Preeti
V2 ® velocity of escalator
l ® distance
1 2
t
V V
=
+
l 1 2
1 2
1 2
t t
t t
t t
= =
+ +
l
l l
17. Young's double slit experment is first performed in
air and then in a medium other than air. It is found
that 8th bright fringe in the medium lies where 5th
dark fringe lies in air. The refractive index of the
medium is nearly :-
(1) 1.59 (2) 1.69 (3) 1.78 (4) 1.25
Ans. (3)
Sol. (y8)Bright, medium = (y5)Dark, air
l æ öl
ç ÷
è ø
8 mD= 2(5) -1 D
d 2 d
l l
m
8 D 9 D
=
d 2 d Þ m =
16
=1.78
9
18. A beam of light from a source L is incident normally
on a plane mirror fixed at a certain distance x from
the source. The beam is reflected back as a spot
on a scale placed just above the source I. When the
mirror is rotated through a small angle q, the spot
of the light is found to move through a distance y
on the scale. The angle q is given by :-
(1)
y
x
(2)
x
2y (3)
x
y (4)
y
2x
Ans. (4)
Sol.
light spot
Source
(L)
y
2q
x
q
2q =
y
x
; q =
y
2x
19. If q1 and q2 be the apparent angles of dip observed
in two vertical planes at right angles to each other,
then the true angle of dip q is given by :-
(1) tan2q = tan2q1 + tan2q2
(2) cot2q = cot2q1 – cot2q2
(3) tan2q = tan2q1 – tan2q2
(4) cot2q = cot2q1 + cot2q2
Ans. (4)
Sol. tanq1 =
q
a
tan
cos
Þ tanq2 =
q q
=
a a
tan tan
cos(90 – ) sin
Þ sin2a + cos2a = 1
Þ cot2q2 + cot2q1 = cot2q
20. Tow cars moving in opposite directions approach
each other with speed of 22 m/s and 16.5 m/s
respectively. The driver of the first car blows a horn
having a a frequency 400 Hz. The frequency heard
by the driver of the second car is [velocity of sound
340 m/s] :-
(1) 361 Hz (2) 411 Hz
(3) 448 Hz (4) 350 Hz
Ans. (3)
Sol. A
v = 22 m/s s
f = 400 Hz 0
B
v = 16.5 m/s o
As we know for given condition
æ + ö
= çè - ÷ø
observer
app 0
source
v v
f f
v v
æ + ö
= çè - ÷ø
340 16.5
400
340 22
fapp = 448 Hz

21. Two blocks A and B of masses 3 m and m
respectively are connected by a massless and
inextensible string. The whole system is suspended
by a massless spring as shown in figure. The
magnitudes of acceleration of A and B immediately
after the string is cut, are respectively :-
B
A 3m
m
(1)
g
,g
3
(2) g,g (3)
g g
,
3 3
(4) g
g
3
Ans. (1)
Sol. Before cutting the strip :-
A
3mg T
4mg
B
mg
T
\ T = mg
After cutting the strip :-
A
3mg
4mg
B
mg
-
= = A
4mg 3mg g
a
3m 3
= = B
mg
a g
m
22. A thin prism having refracting angle 10° is made
of glass of refractive index 1.42. This prism is
combined with another thin prism of glass of
refractive index 1.7. This combination produces
dispersion without deviation. The refracting angle
of second prism should be :-
(1) 6° (2) 8° (3) 10° (4) 4°
Ans. (1)
Sol. For dispersion without deviation
d1 = d2
A1(m1 – 1) = A2 (m2 – 1)
10(1.42 – 1) = A2 (1.7 – 1)
A2 = 6°
23. The acceleration due to gravity at a height 1 km
above the earth is the same as at a depth d below
the surface of earth. Then :-
(1) d = 1 km
(2) d =
3
2
km
(3) d = 2 km
(4) d =
1
2
km
Ans. (3)
Sol. Q gh = gd
æç - ö÷= æç - ö÷
è ø è ø
2h d
g 1 g 1
R R
d = 2h = 2 km
24. A potentiometer is an accurate and versatile device
to make electrical measurements of E.M.F. because
the method involves :-
(1) Potential gradients
(2) A condition of no current flow through the
galvanometer
(3) A combination of cells, galvanometer and
resistances
(4) Cells
Ans. (2)
Sol. In zero deflection condition, potentiometer draws
no current.
25. A spherical black body with a radius of 12 cm
radiates 450 watt power at 500 K. If the radius were
halved and the temperature doubled, the power
radiated in watt would be :-
(1) 450 (2) 1000
(3) 1800 (4) 225
Ans. (3)
Sol. P μ r2 T4
Þ
æ ö æ ö
= ç ÷ ç ÷
è ø è ø
2 4
1 1 1
2 2 2
P r T
P r T
P2 = 1800 watt
ALLEN
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NEET(UG)-2017
26. Figure shows a circuit that contains three identical
resistors with resistance R = 9.0 W each, two
identical inductors with inductance L = 2.0 mH
each, and an ideal battery with emf e = 18 V. The
current 'i' through the battery just after the switch
closed is,...... :-
R R
R
+ L
– e
L C
(1) 0.2 A (2) 2 A
(3) 0 ampere (4) 2 mA
Ans. (Bonus)
Sol. at t = 0
e
R
R
R
i i1
i1
1
18
i 2A
R 9
e
= = =
\ Current through the battery is
i = 2i1 = 2 × 2 = 4A (Bonus)
OR
According to question language :
Capacitor is not mentioned so i = 2 A
27. Radioactive material 'A' has decay constant '8 l' and
material 'B' has decay constant 'l'. Initially they have
same number of nuclei. After what time, the ratio
of number of nuclei of material 'B' to that 'A' will be
1
e
?
(1)
l
1
7
(2)
l
1
8
(3)
l
1
9
(4)
l
1
Ans. (1)
Sol. lA = 8 l, lB = l
Þ NB = NA
e
Þ N0 e–lt =
8 t
N0e
e
- l
Þ –lt = –8lt – 1 Þ 7lt = –1 Þ t = -
l
1
7
Best answer is t = l
1
7
28. The diagrams below show regions of equipotentials:-
20V 40V
10V 30V
A B
20V 40V
10V 30V
A B
10V 30V
20V 40V
A B
20V
40V
A B
30V
10V
(a) (b) (c) (d)
A positive charge is moved from A to B in each
diagram.
(1) In all the four cases the work done is the same
(2) Minimum work is required to move q in figure (a)
(3) Maximum work is required to move q in figure (b)
(4) Maximum work is required to move q in figure (c)
Ans. (1)
Sol. W = qDV
as DV is same in all conditions, work will be same.
29. Two astronauts are floating in gravitational free
space after having lost contact with their spaceship.
The two will :-
(1) Move towards each other.
(2) Move away from each other.
(3) Will become stationary
(4) Keep floating at the same distance between them.
Ans. (1)
Sol. Astronauts move towards each other under mutual
gravitational force.
30. The x and y coordinates of the particle at any time
are x = 5t – 2t2 and y = 10t respectively, where
x and y are in meters and t in seconds. The
acceleration of the particle at t = 2s is :-
(1) 5 m/s2 (2) – 4 m/s2
(3) – 8 m/s2 (4) 0
Ans. (2)
Sol. vx = 5 – 4t, vy = 10
ax = –4, ay = 0
r = +
x y
a aˆi aˆj
ar = -4ˆi m/s2
ALLEN
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CODE - X
31. One end of string of length l is connected to a
particle of mass 'm' and the other end is connected
to a small peg on a smooth horizontal table. If the
particle moves in circle with speed 'v' the net force
on the particle (directed towards centre) will be
(T represents the tension in the string) :-
(1) +
mv2
T
l
(2) -
mv2
T
l
(3) Zero (4) T
Ans. (4)
Sol. Net force on the particle in uniform circular motion
is centripetal force, which is provided by the tension
in string.
32. A particle executes linear simple harmonic motion
with an amplitude of 3 cm. When the particle is at
2 cm from the mean position, the magnitude of its
velocity is equal to that of its acceleration. Then its
time period in seconds is :-
(1)
p
5
2
(2)
4p
5
(3)
2p
3
(4)
p
5
Ans. (2)
Sol. Amplitude A = 3 cm
When particle is at x = 2 cm ,
its |velocity| = |acceleration|
i.e., w A2- x2 = w2x Þ w =
A2- x2
x
T =
2p
w
= 2p
æç ö÷= p
è ø
2 4
5 5
33. Two Polaroids P1 and P2 are placed with their axis
perpendicular to each other. Unpolarised light I0 is
incident on P1 . A third polaroid P3 is kept in between
P1 and P2 such that its axis makes an angle 45° with
that of P1 . The intensity of transmitted light through
P2 is :-
(1) I0
4
(2) I0
8
(3) I0
16
(4) I0
2
Ans. (2)
Sol.
45°
I2
I1
I0
I3
P1
P3 P2
I1 = I0
2
I2 = I0 2 I0
cos 45°=
2 4
I3 = I0 2 I0
cos 45°=
4 8
34. The bulk modulus of a spherical object is 'B'. If it
is subjected to uniform pressure 'p', the fractional
decrease in radius is :-
(1)
B
3p (2)
3p
B
(3)
p
3B
(4)
p
B
Ans. (3)
Sol. P
B
V
V
D
=
D
-
, D D
=
V 3 R
V R
P
B
3 R
R
D
=
- D Þ
D
- = R P
R 3B
(DP = P)
35. In an electromagnetic wave in free space the root mean
square value of the electric field is Erms = 6V/m. The
peak value of the magnetic field is :-
(1) 2.83 × 10–8 T (2) 0.70 × 10–8 T
(3) 4.23 × 10–8 T (4) 1.41 × 10–8 T
Ans. (1)
Sol. E0 = CB0
= 0
rms
E
E
2
Þ Erms 2 = CB0
Þ
´
= =
´
rms
0 8
E 2 6 2
B
C 3 10
= 2.83 × 10–8 T
36. A rope is wound around a hollow cylinder of mass
3 kg and radius 40 cm. What is the angular
acceleration of the cylinder if the rope is pulled with
a force of 30 N ?
(1) 0.25 rad/s2 (2) 25 rad/s2
(3) 5 m/s2 (4) 25 m/s2
Ans. (2)
Sol. t = Ia
RF = mR2a
30 N
F 30 2
25 rad/s
mR 3 40
100
a = = =
´
ALLEN
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37. Two discs of same moment of inertia rotating about
their regular axis passing through centre and
perpendicular to the plane of disc with angular
velocities w1 and w2 . They are brought into contact
face to face coinciding the axis of rotation. The
expression for loss of energy during this process is:-
(1) (w - w )2
1 2
1
I
4
(2) (w - w )2
I 1 2
(3) (w - w )2
1 2
I
8
(4) (w + w )2
1 2
1
I
2
Ans. (1)
Sol. COAM : 1 w + w
w + w = w Þ w = 2
1 2 I I 2I
2
(K.E.)i = w2+ w2
1 2
1 1
I I
2 2
= ´ w2 = æçw + w ö÷
è ø
2
1 2
f
1
(K.E.) 2I I
2 2
Loss in K.E. = (K.E.)i – (K.E)f = w - w 2
1 2
I
( )
4
38. The photoelectric threshold wavelength of silver is
3250 × 10–10m. The velocity of the electron ejected
from a silver surface by ultraviolet light of
wavelength 2536 × 10–10 m is :-
(Given h = 4.14 × 10–15 eVs and c = 3 × 108 ms–1)
(1) » 0.6 × 106 ms–1 (2) » 61 × 103 ms–1
(3) » 0.3 × 106 ms–1 (4) » 6 × 105 ms–1
Ans. (1 or 4)
Sol. l0 = 3250 Å
l = 2536 Å
= éêël - l ùúû
2
0
1 1 1
mv hc
2
v =
éêël - l0ùúû
2hc 1 1
m
=
19
31
2 12400 1.6 10 714
9.1 10 2536 3250
-
-
´ ´ ´ é ù
´ êë ´ úû
= 0.6 × 106 m/s = 6 × 105 m/s
39. A 250-Turn rectangular coil of length 2.1 cm and
width 1.25 cm carries a current of 85 µA and subjected
to magnetic field of strength 0.85 T. Work done for
rotating the coil by 180º against the torque is:-
(1) 4.55 µJ (2) 2.3 µJ
(3) 1.15 µ J (4) 9.1 µ J
Ans. (4)
Sol. Work = MB[cos q1 – cos q2]
Work = MB[cos 0 – cos 180°]
W = NiAB[1 – (–1)]
W ; 9.1 µJ
40. The ratio of wavelengths of the last line of Balmer
series and the last line of Lyman series is :-
(1) 1 (2) 4 (3) 0.5 (4) 2
Ans. (2)
Sol. For last line of Balmer : n1 = 2 & n2 = ¥
l = êéêë - úùúû= éêë - ¥ ùúû
2 2
2 2 2 2
B 1 2
1 1 1 1 1
RZ R(1)
n n 2
lB =
4
R
...(1)
For last line of Lyman series : n1 = 1 & n2 = ¥
l = êéêë - úùúû= éêë - ¥ ùúû
2 2
2 2 2 2
L 1 2
1 1 1 1 1
RZ R(1)
n n 1
lL = 1/R ...(2)
l
= =
l
B
L
(4/ R)
4
(1/ R)
41. A carnot engine having an efficiency of
1
10
as heat
engine, is used as a refrigerator. If the work done on
the system is 10 J, the amount of energy absorbed
from the reservoir at lower temperature is :-
(1) 90 J (2) 99 J
(3) 100 J (4) 1 J
Ans. (1)
Sol. Q2 1
W
- h
b= =
h
Þ 2 Q 1 0.1
9 0.1
-
=
ÞQ2 = 9 × 10 = 90 J
ALLEN
9
CODE - X
42. A gas mixture consists of 2 moles of O2 and 4 moles
of Ar at temperature T. Neglecting all vibrational
modes, the total internal energy of the system is :-
(1) 15 RT (2) 9 RT (3) 11 RT (4) 4 RT
Ans. (3)
Sol. U =
f
nRT
2
Utotal = +
5 3
(2)RT (4)RT
2 2
Utotal = 11RT
43. An arrangement of three parallel straight wires
placed perpendicular to plane of paper carrying
same current 'I along the same direction is shown
in fig. Magnitude of force per unit length on the
middle wire 'B' is given by :-
B
A
C
d
d
90°
(1)
m
p
2
0 2 i
d
(2)
m
p
2
0 2 i
d
(3)
m
p
2
0i
2 d
(4)
m
p
2
0i
2 d
Ans. (3)
Sol. 0 1 2 i i
F
2 d
m
=
p
= force per unit length
( ) 2
0 0
1
i i i
F
2 d 2 d
m m
= =
p p
= F2
F1 [due to wire A]
F2 [due to wire C]
= 2+ 2
net 1 2 F F F =
m
p
2
0i
2 d
44. A U tube with both ends open to the atmosphere,
is partially filled with water. Oil, which is immiscible
with water, is poured into one side until it stands at
a distance of 10 mm above the water level on the
other side. Meanwhile the water rises by 65 mm
from its original level (see diagram). The density of
the oil is :-
65 mm
65 mm
D
E
F
Pa Pa
Oil
B
10 mm
Final water level
Initial water level
C
A
Water
(1) 425 kg m–3 (2) 800 kg m–3
(3) 928 kg m–3 (4) 650 kg m–3
Ans. (3)
Sol. r0g × 140 × 10–3 = rwg × 130 × 10–3
r = ´ 3» 928 3
0
130
10 kg/m
140
45. Which of the following statements are correct ?
(a) Centre of mass of a body always coincides with
the centre of gravity of the body
(b) Central of mass of a body is the point at which
the total gravitational torque on the body is zero
(c) A couple on a body produce both translational
and rotation motion in a body
(d) Mechanical advantage greater than one means
that small effort can be used to lift a large load
(1) (a) and (b) (2) (b) and (c)
(3) (c) and (d) (4) (b) and (d)
Ans. (4)
Sol. Centre of mass may lie on centre of gravity net
torque of gravitational pull is zero about centre of
mass.
Mechanical advantage =
Load
1
Effort
>
Þ Load > Effort

PAPER-I-PHYSICS & CHEMISTRY VERSION - A1
KEAM-ENGINEERING: PAPER-I-QUESTION WITH ANSWER KEY 1
1. A person observes that the full length of a train subtends an angle of 15 degrees. If the distance
between the train and the person is 3 km, the length of the train, calculated using parallax
method, in meters is
(A) 45 (B) 45π (C) 250 π (D) 250 π (E) 450
Ans : C
2. In a measurement, the random error
(A) Can be decreased by increasing the number of readings and averaging them
(B) Can be decreased by changing the person who takes the reading
(C) Can be decreased by using new instrument
(D) Can be decreased by using a different method in taking the reading
(E) Can never be decreased
Ans : A
3. In order to measure the period of a single pendulum using a stop clock, a student repeated the
experiment for 10 times and noted down the time period for each experiment as 5.1, 5.0, 4.9, 4.9,
5.1, 5.0, 4.9, 5.1, 5.0, 4.9 s. The correct way of expressing the result for the period is
(A) 4.99 s (B) 5.0 s (C) 5.00s (D) 4.9 s (E) 5.1 s
Ans :D
4. The following figure gives the movement of an object. Select the correct statement from the
given choice
(A) The total distance travelled by the object is 975 m
(B) The maximum acceleration of the object is 2m/s2
(C) The maximum declaration happend between 25th and 35th seconds
(D) The object was at rest between 10th and 15th seconds
(E) At 40th second, the object was decelerating
Ans : A
5. Two object, P and Q, travelling in the same direction starts from rest. While the object P starts at
time t = 0 and object Q starts later at t = 30 min. The object P has an acceleration of 40km/h2.
To catch P at a distance of 20 km, the acceleration of Q should be
(A) 40 km/h2 (B) 80 km/h2 (C) 100 km/h2 (D) 120 km/h2 (E) 160 km/h2
Ans :E
PAPER-I-PHYSICS & CHEMISTRY VERSION - A1
KEAM-ENGINEERING: PAPER-I-QUESTION WITH ANSWER KEY 2
6. A train of length L move with a constant speed Vt. A person at the back of the train fires a bullet
at time t = 0 towards a target which is at a distance of D (at time t =0) from the front of the train
(on the same direction of motion). Another person at the front of the train fires another bullet at
time t = T towards the same target. Both bullets reach the target at the same time. Assuming the
speed of the bullets, Vb, are same, the length of the train is
(A) T× (Vb + 2Vt) (B) T× (Vb + Vt) (C) 2× T × (Vb + 2Vt)
(D) 2× T × (Vb – 2Vt) (D) T × (Vb – 2Vt)
Ans :B
7. From the ground, a projectile is fired at an angle of 60 degrees to the horizontal with a speed of
20 m/s. Take acceleration due to gravity as 10 m/s2. The horizontal range of the projectile is
(A) 10 3 m (B) 20 m (C) 20 3 m (D) 40 3 m (E) 400 3 m
Ans :C
8. A person from a truck, moving with a constant speed of 60 km/h, throws a ball upwards with a
speed of 60 km/h. Neglecting the effect of rotation of Earth choose the correct answer from the
given choice
(A) The person cannot catch the ball when it comes down since the truck is moving
(B) The person can catch the ball when it comes down, if the truck is stopped immediately after
throwing the ball
(C) The person can catch the ball when it comes down, if the truck moves with speed less than 60
km/h but does not stop
(D) The person can catch the ball when it comes down, if the truck moves with speed more than
60km/h
(E) The person can catch the ball when it comes down, if the truck continues to move with a
constant speed of 60 km/h
Ans :E
9. A body of mass 2m moving with velocity v makes a head on elastic collision with another body
of mass m which is initially at rest. Loss of kinetic energy of the colliding body ) (mass 2m) is
(A) 1/9 of its initial kinetic energy (B) 1/6 of its initial kinetic energy
(C) ¼ of its initial kinetic energy (D) ½ of its initial kinetic energy
(D) 8/9 of its initial kinetic energy
Ans :E
10. Displacement, x (in meters), of a body of mass 1 kg as a function of time, t, on a horizontal
smooth surface is give as x = 2t2. The work done in the first one second by the external force is
(A) 1 J (B) 2 J (C) 4 J (D) 8 J (E) 16 J
Ans : D
11. A massless spring of length l and spring constant k is placed vertically on a table. A all of mass
m is just kept on top of the spring. The maximum velocity of the ball is
(A) g m
k
(B) g 2m
k
(C) 2g m
k
(D)
2
g m
k
(E)
2
g m
k
Ans : A
12. Under the action of a constant force, a particle is experiencing a constant acceleration. The power
is
(A) Zero (B) Positive constant (C) Negative constant
(D) Increasing uniformly with time (E) Decreasing uniformly with time
Ans :D
PAPER-I-PHYSICS & CHEMISTRY VERSION - A1
KEAM-ENGINEERING: PAPER-I-QUESTION WITH ANSWER KEY 3
13. A copper wire with a cross-section area of 2 × 10-6 m2
has a free electron density equal to
5 × 1022 /cm3. If this wire carries a current of 16A, the drift velocity of the electron is
(A) 1 m/s (B) 0.1 m/s (C) 0.01 m/s
(D) 0.001 m/s (E) 0.0001 m/s
Ans :D
14. The resistance of the tungsten wire in the light bulb, which is rated at 120V/75 W and powred
by a 120 V direct-current supply, is
(A) 0.37 Ω (B) 1.2 Ω (C) 2.66 Ω (D) 192 Ω (D) 9 × 103 Ω
Ans :D
15. The value of the currents I1, I2, and I3 flowing through the circuit given below is
(A) I1 = – 3A, I2 = 2A, I3 = –1A (B) I1 = 2A, I2 = –3A, I3 = –1A
(C) I1 = 3A, I2 = –1A, I3 = –2A (D) I1 = 1A, I2 = –3A, I3 = –2A
(E) ) I1 = 2A, I2 = –1 A, I3 = –3A
Ans :B
16. A silver wire has temperature coefficient of resistivity 4× 10-3/oC and its resistance at 20oC is
10 Ω . Neglecting any change in dimensions due to the change in temperature, its resistance at
40oC is
(A)0.8Ω (B) 1.8 Ω (C) 9.2 Ω (D) 10.8 Ω (E) 11.6 Ω
Ans :D
17. A change Q placed at the center of a metallic spherical shell with inner and outer radii R1 and R2
respectively. The normal component of the electric field at any point on the Gaussian surface
with radius between R1 and R2 will be
(A) Zero (B) 2
4 1
Q
π R
(C) 2
4 2
Q
π R
(D) 2
4 ( 1 2)
Q
π R −R
(E) 2
4 ( 2 1)
Q
π R − R
Ans : A
18. A sphere of radius R has a uniform volume charge density, ρ . The magnitude of electric filed at
a distance r from the centre of the sphere, where r > R, is
(A) 2
0
ρ
4πε r
(B)
2
2
0
ρR
ε r
(C)
3
2
0
ρR
ε r
(D)
3
2
0
ρR
3ε r
(E)
2
2
0
ρR
4ε r
Ans :D
PAPER-I-PHYSICS & CHEMISTRY VERSION - A1
KEAM-ENGINEERING: PAPER-I-QUESTION WITH ANSWER KEY 4
19. Five equal point charges with charge Q = 10 nC are located at x = 2, 4, 5, 10 and 20 m. If
9
ε 0 = [10− / 36π ] F/m, then the potential at the origin (x=0) is
(A) 9.9 V (B) 11.1V (C) 90 V (D) 99 V (E) 111 V
Ans :D
20. Two infinitely long parallel plates of equal areas, 6cm2, are separated by a distance of 1 cm.
While one of the plates has a charge of + 10 nC and the other has - 10nC. The magnitude of the
electric field between the plates, if
9
0
10
36
ε
π
−
= F/m is
(A) 0.6π kV/m (B) 6π kV/m (C)600π kV/m
(D) 60π V/m (E)6π V/m
Ans :C
21. A proton moves with a speed of 5.0 × 106 m/s along the x-axis. It enters a region where there is
a magnetic field of magnitude 2.0 Tesla directed at an angle of 30o to the x-axis and lying in the
xy plane. The magnitude of the magnetic force on the proton is
(A) 0.8 × 10-13 N (B) 1.6 × 10-13 N (C) 8.0 × 10-13 N
(D) 8.0 × 10-13 N (E) 16 × 10-13 N
Ans :D
22. A long straight wire of radius R carries a steady current, I0, uniformly distributed throughout the
cross-section of the wire. The magnetic field at a radial distance r from the centre of the wire, in
the region r> R, is
(A) μ0I0
2π r
(B) μ0I0
2π R
(C)
2
μ0I0R
2π r
(D)
2
μ0I0
2
r
π R
(E)
2
0 0
2
μ I
2
r
π R
Ans : A
23. If the cyclotron oscillator frequency is 16 MHz, then what should be the operating magnetic field
for accelerating the proton of mass 1.6710-27 kg?
(A)0.334π T (B) 3.34 π T (C) 33.4 π T
(D)334 π T (E)3340 π T
Ans : A
24. The speed of light is vacuum is equal to
(A) μ0ε0 (B) 2 2
μ0ε0 (C) μ0ε0 (D)
0 0
1
μ ε
(E)
0 0
1
μ ε
Ans :E
25. A comet orbits around Sun in an elliptical orbit. Which of the following quantities remains
constant during the course of its motion?
(A) Linear velocity (B) Angular velocity (C) Angular momentum
(C) Kinetic energy (E) Potential energy
Ans :C
26. Consider a satellite moving in a circular orbit around Earth. If K and V denote its kinetic energy
and potential energy respectively then (Choose the convention where V = 0 as r → ∞ )
(A) K = V (b) K = 2V (C) V = 2K (D) K = – 2V (D) V = – 2 K
Ans :E
PAPER-I-PHYSICS & CHEMISTRY VERSION - A1
KEAM-ENGINEERING: PAPER-I-QUESTION WITH ANSWER KEY 5
27. Assuming the mass of Earth to be ten times the mass of Mars and its radius to be twice the radius
of Mars and the acceleration due to gravity on the surface of Earth to be 10 m/s2, the acceleration
due to gravity on the surface of Mars is given by
(A) 0.2 m/s2 (B) 0.4 m./2 (C) 2 m/s2 (D) 4 m/s2 (E) 5 m/s2
Ans : D
28. The semi-major axis of the orbit of Saturn is approximately nine times that of Earth. The time
period of revolution of Saturn is approximately equal to
(A) 81 years (B) 27 years (C) 729 years
(D) 3 81 years (E) 9 years
Ans :B
29. A particle of mass 3 kg, attached to a spring with force constant 48 N/m execute simple
harmonic motion on a frictionless horizontal surface. The time period of oscillation of the
particle, in seconds, is
(A) π /4 (B) π /2 (C)2 π (D)8 π (E) π /8
Ans :B
30. The position and velocity of a particle executing simple harmonic motion at t = 0 are given by 3
cm and 8 cm/s respectively. If the angular frequency of the particle is 2 rad/s then the amplitude
of oscillation, in centimeters, is
(A) 3 (B) 4 (C) 5 (D) 6 (E) 8
Ans :C
31. A simple harmonic motion is represented by x(t)=sin2ωt− 2cos2ωt. The angular frequency of
oscillation is given by
(A) ω (B) 2 ω (C) 4 ω (D) ω /2 (E) ω /4
Ans :B
32. A transverse wave is propagating on a stretched string shoes mass per unit length is 32 g/m. The
tension on the string is 80 N. The speed of the wave in the string is
(A) 5/2 m/s (B) 5/ 2 m/s (C) 2/5 m/s (D) 2/5 m/s (E)50 m/s
Ans :E
33. Consider the propagating of sound (with velocity 330 m/s) in a pipe of length 1.5 m with one
end closed and the other open. The frequency associated with the fundamental mode is
(A) 11 Hz (B) 55Hz (C) 110Hz (D) 165 Hz (E) 275 Hz
Ans :B
34. A standing wave propagating with velocity 300 m/s in an open pipe of length 4 m has four nodes.
The frequency of the wave is
(A) 75 Hz (B) 100 Hz (C) 150 Hz (D) 300 Hz (E) 600 Hz
Ans : C
35 Consider a vehicle emitting sound wave of frequency 700 Hz moving towards an observer at a
speed 22 m/s. Assuming the observer as well as the medium to be at rest ad velocity of sound in
the medium to be 330 m/s, the frequency of sound as measured by the observeris
(A) 2525/4 Hz (B) 1960/3 Hz (C) 2240/3 Hz
(D) 750 Hz (E) 5625/7 Hz
Ans :D
PAPER-I-PHYSICS & CHEMISTRY VERSION - A1
KEAM-ENGINEERING: PAPER-I-QUESTION WITH ANSWER KEY 6
36. The x-t plot shown in the figure below describes the motion of the particle, along x-axis, between
two positions A and B. The particle passes through two intermediate points P1 and P2 as shown in
the figure
(A) The instantaneous velocity is positive at P1 and negative at P2
(B) The instantaneous velocity is negative at both P1 and P2
(C) The instantaneous velocity is negative at P1 and positive at P2
(D) The instantaneous velocity is positive at both P1 and P2
(E) The instantaneous velocity is always positive
Ans : A
37. A ball falls from a table top with initial horizontal speed V0. In the absence of air resitance,
which of the following statement is correct
(A) The vertical component of the acceleration changes with time
(B) The horizontal component of the velocity does not change with time
(C) The horizontal component of the acceleration is non zero and finite
(D) The time taken by the ball to touch the ground depends on V0.
(E) The vertical component of the acceleration varies with time
Ans : B
38. A man of mass 60 kg climbed down using an elevator. The elevator had an acceleration 4 ms-2.
If the acceleration due to gravity is 10 ms-2, the main apparent weight on his way down is
(A) 60 N (B) 240 N (C) 360 N (D) 840 N (D) 3600N
Ans :C
39. A uniform rod of length of 1 m and mass of 2 kg is attached to a side support at O as shown in the
figure. The rod is at equilibrium due to upward force T acting at P. Assume the acceleration due
to gravity as 10 m/s2. The value of T is
(A) 0
(B) 2N
(C) 5 N
(C) 5 N
(D) 10 N
(E) 20 N
Ans :D
PAPER-I-PHYSICS & CHEMISTRY VERSION - A1
KEAM-ENGINEERING: PAPER-I-QUESTION WITH ANSWER KEY 7
40. A capillary tube of radius 0.5 mm is immersed in a breaker of mercury. The level inside the tube
is 0.8 cm below the resonance and angle of contact is 120o. What is the surface tension of
mercury if the mass density of mercury is ρ=13.6 ×103kgm−3and acceleration due to gravity is
g = 10 m/s2?
(A) 0.225N/m (B) 0.544 N/m (C) 0.285 N/m
(D) 0.375 N/m (E) 0.425 N/m
Ans :B
41. Which of the following statements related to stress – strain relation is correct
(A) Stress is linearly proportional to strain irrespective of the magnitude of the strain
(B) Stress is linearly proportional to strain above the yield point
(C) Stress is linearly proportional to strain for stress much smaller than at the yield point
(D) Stress – strain curve is same for all materials
(E) Stress is inversely proportional to strain
Ans : C
42. The lower edge of a square slab of side 50 cm and thickness 20 cm is rigidly fixed to the base of a
table. A tangential force of 30 N is applied to the slab. If the shear moduli of the material is
4×1010N/m2 , then displacement of the upper edge, in meters, is
(A) 4×10−12 (B) 4×10−10 (C) 6×10−10 (D) 6×10−12 (E) 8×10−10
Ans : C
43. Initially a beaker had 100g of water at temperature 90oC. Later another 600g of water at
temperature 20oC was poured into the beaker. The temperature, T, of the water after mixing is
(A) 20oC (B) 30oC (C) 45oC (D) 55oC (E)
90oC
Ans : B
44. Match the following
I) Isothermal process 1) ΔQ = 0
II) Isobaric process 2) ΔV = 0
III) Isochoric process 3) ΔP=0
IV) Adiabatic process 4) ΔT=0
(A) I – 4, II – 3, III – 2, IV – 1 (B) I – 3, II – 2, III – 1, IV – 4
(C) I – 1, II – 2, III – 3, IV – 4 (D) I – 4, II – 2, III – 3, IV – 1
(E) I – 1, II – 4, III – 2, IV – 3
Ans : A
45. For an ideal gas, the specific heat at constant pressure Cp is greater than the specific heat at
constant volume Cν . This is because
(A) There is a finite work done by the gas on its environment when its temperature is increased
while the pressure remains constant
(B) There is a finite work done by the gas on its environment when its pressure is increased
while the volume remains constant
(C) There is a finite work done by the gas on its environment when its pressure is increased
while the temperature remains constant
(D) The pressure of the gas remains constant when its temperature remains constant
(E) The internal energy of the gas at constant pressure is more than at constant volume
Ans : A
PAPER-I-PHYSICS & CHEMISTRY VERSION - A1
KEAM-ENGINEERING: PAPER-I-QUESTION WITH ANSWER KEY 8
46. Which of the following statements is correct?
(A) Light waves are transverse but sound waves and waves on strings are longitudinal
(B) Sound waves and waves on a string are transverse but light waves are longitudinal
(C) Light waves and waves on a string are transverse but sound waves are longitudinal
(D) Light waves and waves are transverse but waves on strings are longitudinal
(E) Light waves, sound waves and waves on a string are all longitudinal
Ans : C
47. In Young’s double slit experiment, if the separation between the slits is halved, and the distance
between the slits and the screen is doubled, then the fringe width compared to the unchanged one
will be
(A) Unchanged (B) Halved (C) Doubled
(D) Quadrupled (D) Fringes will disappear
Ans : D
48. The phase velocity of a wave described by the equation ψ = ψ0 sin (kx + ωt + π / 2) is
(A) x/t (B) 0 ψ /ω (C) ω / k (D) π / 2k (E) 0 ψ
Ans : C
49. The direction of propagation of electromagnetic wave is along
(A) Electric field vector, E
􀁇
(B) Magnetic field vector, B 􀁇
(C) E.B
􀁇 􀁇
(D) E×B
􀁇 􀁇
(E) B× E
􀁇 􀁇
Ans : D
50. Assume that a radio station is about 200 km away from your location and the station operates at
972 kHz. How long does it take for an electromagnetic signal to travel from the station to you and
how many wave crests doe it send out per second
(A) 666 μs and 9.72×105 crests per second
(B) 666 μs and 972×105 crests per second
(C) 555 μs and 97.2 ×107 crests per second
(D) 555 μs and 0.972×105 crests per second
(E) 444 μs and 9×106 crests per second
Ans : A
51. What wavelength must electromagnetic radiation have if a photon in the beam has the same
momentum as an electron moving with a speed 1.1×105m / s (Planck’s constant = 6.6 ×10−34 Js ,
rest mass of electron = 9×10−31kg?
(A) 2/3 nm (B) 20/3 nm (C) 4/3 nm (D) 40/3 nm (E) 3/20 nm
Ans : B
52. The electron field portion of an electromagnetic wave is given by (all variables in SI units)
E = 10−4sin(6×105t−0.01x) . The frequency (f) and the speed (ν) of electromagnetic wave are
(A) f = 30 / π kHz and ν = 1.5×107m / s (B) f = 90 / π kHz and ν = 6.0 ×107m / s
(C) f = 300 / π kHz and ν = 6.0 ×107m / s (D) f = 600 / π kHz and ν = 7.5×107m / s
(E) f = 900 / π kHz and ν = 8.0 ×107m / s
Ans : C
PAPER-I-PHYSICS & CHEMISTRY VERSION - A1
KEAM-ENGINEERING: PAPER-I-QUESTION WITH ANSWER KEY 9
53. Huygens’ wave theory of light cannot explain
(A) Diffraction phenomena (B) Interference phenomena (C) Photoelectric effect
(D) Polarization of light (E) Propagation of light
Ans : C
54. An electron, a neutron and an alpha particle have same kinetic energy and their de – Broglie
wavelengths are λe,λn andλα respectively. Which statement is correct about their de – Broglie
wavelengths?
(A) λe > λn > λα (B) λe < λn > λα (C) λe < λn < λα
(D) λe > λn < λα (E) λe = λn < λα
Ans : A
55. It takes 4.6 eV to remove one of the least tightly bound electrons from a metal surface. When
monochromatic photons strike the metal surface, electrons having kinetic energy from zero to
2.2 eV are ejected. What is the energy of the incident photons?
(A) 2.4 eV (B) 2.2 eV (C) 6.8 eV (D) 4.6 eV (E) 5.8 eV
Ans : C
56. If copper and silicon pieces are heated, the resistance of
(A) each will increase (B) each will decrease
(C) copper will increase and silicon will decrease
(D) copper will decrease and silicon will increase
(E) both does not change
Ans : C
57. In an insulator, band gap of the order of
(A) 0.1 eV (B) 1 eV (C) 5 eV (D) 100 eV (E) 1 MeV
Ans : C
58. For a P – N junction diode
(A) Forward current is in mA and reverse current is in μA
(B) Forward current is in μA and reverse current is in mA
(C) Both forward and reverse currents are in μA
(D) Both forward and reverse currents are in mA
(E) No current flows in any direction
Ans : A
59. For a Zener diode
(A) both p and n regions are heavily doped
(B) p region is heavily doped but n region is lightly doped
(C) n region is heavily doped but p region is lightly doped
(D) both p and n regions are lightly doped
(E) depletion region is very thick
Ans : B
60. Speech signal is in the range of
(A) 3700 to 7000
oA
wavelength (B) 20 Hz to 20 kHz frequency (C) 300 to 3100 Hz frequency
(D) 540 to 1600 kHz frequency (E) 88 to 108 MHz frequency
Ans : C
PAPER-I-PHYSICS & CHEMISTRY VERSION - A1
KEAM-ENGINEERING: PAPER-I-QUESTION WITH ANSWER KEY 10
61. Wavelength of the wave with 30 MHz frequency is
(A) 1 cm (B) 10 cm (C) 100 cm (D) 1000 cm (E) 10000 cm
Ans : D
62. To transmit a signal of frequency, m ω , with a carrier frequency, c ω , in AM transmission, the
bandwidth of the filter and amplifier is
(A) m ω (B) m 2ω (C) c ω (D) c m ω − ω (E) c m ω + ω
Ans : B
63. If a magnet is dropped through a vertical hollow copper tube then
(A) the time taken to reach the ground is longer than the time taken if the tube was made out of
plastic
(B) the magnet will get attracted and stick to the copper tube
(C) the time taken to reach the ground is longer than the time taken if the tube was made out of
stainless steel
(D) the time taken to reach the ground does not depend on the radius of the copper tube
(E) the magnet will be repelled away by the tube
Ans : D
64. Consider a circular wire loop of radius R spinning about a diametrical chord which is
perpendicular to a uniform magnetic field 0
(B = B kˆ )
􀁇
(A) The magnitude of the induced EMF in the loop is maximum when the plane of the loop is
perpendicular to B
􀁇
(B) Flux through the loop is minimum when the plane of the loop is perpendicular to B
􀁇
(C) The direction of induced current remains same during the spinning motion of the loop
(D) EMF induced will be the same for a larger radius of the loop in the same field
(E) No EMF will be induced since magnetic field is constant
Ans : A
65. An electric motor when loaded has an effective resistance of 30Ω and an inductive reactance of
40Ω . If the motor is powered by a source with maximum voltage of 420 V, the maximum current
is
(A) 6A (B) 8.4 A (C) 10 A (D) 12 A (E) 13A
Ans : B
66. Which of the following particle when bombards on 65Cu will turn into 66Cu
(A) Proton (B) Neutron (C) Electron (D) Alpha particle (E)
Deutron
Ans : B
67. CO– ion moving with kinetic energy of 20 keV dissociates into O– and C which move along the
parent ion direction. Assuming no energy is released during dissociation, the kinetic energy of the
daughters O (K.E) − and (K.E)C are related as
(A) O C (K.E) (K.E) − = (B) O C (K.E) / (K.E) 16 / 12 − =
(C) O C (K.E) / (K.E) 12 / 16 − = (D) O C (K.E) / (K.E) 16 / 28 − =
(E) O C (K.E) / (K.E) 28 /16 − =
Ans : C
PAPER-I-PHYSICS & CHEMISTRY VERSION - A1
KEAM-ENGINEERING: PAPER-I-QUESTION WITH ANSWER KEY 11
68. If the rms value of sinusoidal input to a full wave rectifier is 0 V / 2 then the rms value of the
rectifier’s output is
(A) 0 V / 2 (B) 2
0 V / 2 (C) 2
0 V /2 (D) 2
0 2V (E) 2
0 2V
Ans : A
69. Eight grams of Cu66 undergoes radioactive decay and after 15 minutes only 1g remains. The half
– life, in minutes, is then
(A) 15 ln (2)/ln (8) (B) 15 ln(8)/ln(2) (C) 15/8 (D) 8/15 (E) 15
ln(2)
Ans : A
70. For a light nuclei, which of the following relation between the atomic number (Z) and mass
number (A) is valid
(A) A = Z/2 (B) Z = A (C) Z = A/2 (D) Z = A2 (E) A = Z2
Ans : C
71. A wheel rotating at 12 rev/s is brought to rest in 6s. The average angular deceleration in rad/s2 of
the wheel during this process is
(A) 4π (B) 4 (C) 72 (D) 1/ π (E) π
Ans : A
72. A torque of 1 N.m is applied to a wheel which is at rest. After 2 seconds the angular momentum
in kg. m2/s is
(A) 0.5 (B) 1 (C) 2 (D) 4 (E) 3
Ans :C
73. Uncertainty principle is valid for
(A) Proton (B) Methane (C) Both (A) and (B)
(D) 1μm sized platinum particles (E) 1μm sized NaCl particles
Ans : A
74. The energy of an electron in the 3S orbital (excited state) of H – atom is
(A) – 1.5 eV (B) – 13.6 eV (C) – 3.4 eV (D) – 4.53 eV (E) 4.53 eV
Ans : A
75. Among the following, the molecule that will have the highest dipole movement is
(A) H2 (B) HI (C) HBr (D) HCl (E) HF
Ans :E
76. Which of the following pair have identical bond order?
(A) CN– and NO+ (B) CN– and 2 O− (C) CN– and
CN+
(D) NO+ and 2 O− (E) 2 O− and CN+
Ans : A
77. A gas will approach ideal behavior at
(A) Low temperature and low pressure (B) Low temperature and high pressure
(C) High temperature and low pressure (D) High temperature and high pressure
(E) Low volume and high pressure
Ans :C
PAPER-I-PHYSICS & CHEMISTRY VERSION - A1
KEAM-ENGINEERING: PAPER-I-QUESTION WITH ANSWER KEY 12
78. Pressure of ideal and real gases at 0K are
(A) > 0 and 0 (B) < 0 and 0 (C) 0 and 0
(D) > 0 and > 0 (E) 0 and > 0
Ans :E
79. For the process A (l, 0.05 atm, 32oC) → A(g, 0.05 atm, 32oC)
The correct set of themrodynamic parameters is
(A) ΔG = 0andΔS= −ve (B) ΔG=0andΔS=+ve
(C) ΔG = +veandΔS= 0 (D) ΔG = −veand ΔS = 0 (E) ΔG = 0andΔS= 0
Ans :B
80. Mixing of N2 and H2 form an ideal gas mixture at room temperature in a container. For this
process, which of the following statement is true?
(A) surrounding system ΔH = 0,ΔS = 0,ΔS = 0andΔG = −ve
(B) surrounding system ΔH = 0,ΔS = 0,ΔS > 0andΔG = −ve
(C) surrounding system ΔH > 0,ΔS = 0,ΔS > 0andΔG = −ve
(D) surrounding system ΔH<0>0,ΔS <0and p="" ve="">(E) surrounding system ΔH=0,ΔS =0,ΔS <0and p="" ve="">Ans : D
81. Which of the following is not true about a catalyst?
(A) Mechanism of the reaction in presence and absence of catalyst could be different
(B) Enthalpy of the reaction does not change with catalysts
(C) Catalyst enhances both forward and backward reaction at equal rate
(D) Catalyst participates in the reaction, but not consumed in the process
(E) Use of catalyst cannot change the order of the reaction
Ans :E
82. In the In K vs. 1
T
plot of a chemical process having ΔS0 > 0 and ΔH0 <0 is="" p="" slope="" the="">proportional to (where K is equilibrium constant)
(A) - ΔH0 (B) ΔH0 (C) ΔS0 (D) - ΔS0 (E) ΔG0
Ans :B
83. For the process
3
2
A→ B , at 298 K, ΔG0 is 163 kJ mol-1. The composition of the reaction mixture is [B] = 1 and
[A] = 10000. Predict the direction of the reaction and the relation between reaction quotient (Q)
and the equilibrium constant (K)
(A) Forward direction because Q > K (B) Reverse direction because Q > K
(C) Forward direction because Q < K (D) Reverse direction because Q < K
(E) It is at equilibrium as Q = K
Ans : C
84. Solubility product (Ksp) of saturated PbCl2 in water is 1.8 × 10-4 mol3 dm-9. What is the
concentration of Pb2+ in the solution?
(A) (0.45 × 10-4)1/3 mol dm-3 (B) (1.8 × 10-4)1/3 mol dm-3 (C) (0.9 × 10-4)1/3 mol dm-3
(D) (2.0 × 10-4)1/3 mol dm-3 (E) (2.45 × 10-4)1/3 mol dm-3
Ans : A
PAPER-I-PHYSICS & CHEMISTRY VERSION - A1
KEAM-ENGINEERING: PAPER-I-QUESTION WITH ANSWER KEY 13
85. The freezing point of equimolal aqueous solutions will be highest for
(A) C6H5NH3Cl (B) AgNO3 (C) Ca(NO3)2
(D) La(NO3)3 (E) D-fructose
Ans :E
86. The molality of the 3M solution of methanol if the density of the solution is 0.9 g cm-3 is
(A) 3.73 (B) 3.0 (C) 3.33 (D) 3.1 (E) 3.2
Ans : A
87. Consider a fuel cell supplied with 1 mole of H2 gas and 10 moles of O2 gas. If fuel cell is
operated at 96.5 mA current, how long will it deliver power?
(Assume 1 F = 96500 C/mole of electrons)
(A) 1 × 106 s (B) 0.5× 106 s (C) 2× 106 s (D) 4× 106 s (E) 5 × 106 s
Ans :C
88. Consider the equilibrium obtained by electrically connecting zinc-amalgam (Zn(Hg) and HgO
electrodes in mercury cell,
An(Hg) + HgO(s) 􀁕 ZnO(s) + Hg(l)
Under this equilibrium, what is the relation between the potential of the Zn(Hg) and HgO
electrodes measured against the standard hydrogen electrode?
(A) Zn(Hg) electrode potential is equal to HgO electrode potential
(B) Zn(Hg) electrode potential is more than HgO electrode potential
(C) HgO electrode potential is more than Zn(Hg) electrode
(D) Cell voltage at above said equilibrium is 1.35 V
(E) Both (C) and (D)
Ans : A
89. 10 g of MgCO3 decomposes on heating to 0.1 mole CO2 and 4 g MgO. The percent purity of
MgCO3 is
(A) 24 % (B) 44% (C) 54% (D) 74% (E) 84%
Ans :E
90. The compound Na2CO3 i x H2O has 50% H2O by mass. The value of “x” is
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8
Ans :C
91. Hybridisation of carbon in 3 CH−
(A) sp2 (B) sp3 (C) sp3d (D) sp3d2 (E) sp2d3
Ans :B
*92. The common features among CO, CN- and 2 NO+ are
(A) Bond order three and isoelectronic (B) Bond order three and weak field ligands
(C) Bond order two and π -acceptors (D) Bond order three and π -donors
(E) Isoelectronic and strong field ligands
Ans : A
93. Which of the following is covalent?
(A) NaCl (B) KCl (C) BeCl2 (D) MgCl2 (E) CaCl2
Ans :C
PAPER-I-PHYSICS & CHEMISTRY VERSION - A1
KEAM-ENGINEERING: PAPER-I-QUESTION WITH ANSWER KEY 14
94. One mole of an unknown compound was treated with excess water and resulted in the evolution
of two moles of a readily combustible gas. The resulting solution was treated with CO2 and
resulted in the formation of white turbidity. The unknown compound is
(A) Ca (B) CaH2 (C) Ca(OH)2 (D) Ca(NO3)2 (E) CaSO4
Ans :B
*95. When potassium is reacted with water, which compound(s) is (are) formed preferentially?
(A) K2O (B) KO2 (C) Both K2O and KO2 (D) K2O2 (E) K2O3
Ans :B
96. Purification of aluminium by electrolytic refining is called
(A) Hall’s process (B) Froth flotation process (C) Bayer’s process
(D) Hoop’s process (E) Serpeck’s process
Ans :D
97. Select the most appropriate statement In BF3
(A) All the bonds are completely ionic
(B) The B-F bond is partially ionic
(C) B-F bond has partial double bond character
(D) Bond energy and bond length data indicates single bond character of the B-F bond
(E) All the bonds are covalent
Ans :E
98. The inert gas found most abundant in the atmosphere is
(A) He (B) Ne (C) Ar (D) Kr (E) Xe
Ans : C
99. When MnO2 is fused with KOH and KNO2, a coloured compound is formed. Choose the right
compound with the appropriate colour
(A) K2MnO4, green (B) KMnO4, purple (C) Mn2O3, brown
(D) Mn3O4, black (E) MnO2, black
Ans : A
100. Identify the case(s) where there is change in oxidation number
(A) Acidified solution of CrO4
2-
(B) SO2 gas bubbled through an acidic solution Cr2O7
2-
(C) Alkaline solution of Cr2O7
2- (D) Ammoniacal solution of CrO4
2-
(E) Aqueous solution of CrO2Cl2 in NaOH
Ans :B
101. Water gas is produced by
(A) Passing steam over red hot coke (B) Passing steam and air over red hot
coke
(C) Burning coke in excess air (D) Burning coke in limited supply of air
(E) Both (A) and (B)
Ans : A
102. The volume of oxygen liberated at STP from 15 ml of 20 volume H2O2 is
(A) 100 mL (B) 150 mL (C) 200 mL (D) 250 mL (E) 300 mL
Ans :E
PAPER-I-PHYSICS & CHEMISTRY VERSION - A1
KEAM-ENGINEERING: PAPER-I-QUESTION WITH ANSWER KEY 15
103. Corundum is _________ mineral of aluminium.
(A) Silicate (B) Oxide (C) Double salt (D) Sulphate (E) Nitrate
Ans :B
104. The solution which does not produce precipitate when treated with aqueous K2CO3 is
(A) BaCL2 (B) CaBr2 (C) MgCl2 (D) Na2SO4 (E)
Pb(N)3)2
Ans :D
105. If the boiling point difference between the two liquids is not much, the _______ method is used to
separate them
(A) Simple distillation (B) Distillation under reduced pressure
(C) Steam distillation (D) Fractional distillation
(E) Differential extraction
Ans :D
106. Lassaigne’s test (with silver nitrate) is commonly used to detect halogens such as chlorine,
bromine and iodine but not useful to detect fluorin because the product AgF formed is
(A) Volatile (B) Reactive (C) Explosive (D) Soluble in water
(E) A liquid
Ans :D
107. Protein is a polymer made of
(A) Carbohydrates (B) Aminoacids (C) Nucleic acids
(D) Carboxylic acids (E) Polycyclic aromatics
Ans :B
108. The letter ‘D’ in D-carbohydrates represents
(A) Dextrorotation (B) Direct synthesis (C) Configuration
(D) Mutarotation (E) Optical activity
Ans :C
109. Phenol is a highly corrosive substance, but its 0.2 per cent solution is used as
(A) Antibiotic (B) Antiseptic (C) Disinfectant
(D) Antihistamine (E) Antacid
Ans :B
110. Name of the following reaction is
(A) Reimer-Tiemann (B) Kolbe-Schmitt (C) Cannizzaro
(D) Gattermann (E) Gattermann-Koch
Ans :B
PAPER-I-
111. X
C
(A
(C
(D
A
112. T
(A
(D
A
113. T
(A
(D
A
114. T(
I
(A
(D
A
115. C
(A
(C
(E
A
116. T
Cd
i
(A
(D
A
-PHYSICS & C
KEA
and Y
C6H5 −CO2H
A) SOCl2 and
C) SOCl2 and
D) SOCl2 and
Ans : A
The reaction o
A) 3 HC CH
•
−
D) 3 HC−CH
Ans :B
The major prod
A)
D)
Ans :B
The correct inc
II), 3 – chlor
A) I < II < III
D) III < I < II
Ans :C
Cycloheptatrie
A) Non-benze
C) Benzenoid
E) Non-benze
Ans :B
The correct or
CH3CH2CH2C
isplacement i
A) I < II < III
D) II < IV < I
Ans : B
CHEMISTRY
AM-ENGINEE
in the
+ X ⎯h⎯eat⎯→
d C6H5CHO
d C6H5CH3
d C6H5CH2Cl
f propene wit
3 H −CH
2 2 H CH
•
−
duct P formed
creasing order
robutyric acid
< IV
< IV
enyl cation is
enoid and non
d and non-arom
enoid and anti
der of increa
H2Br (II),
is
< IV
< III
Y
ERING: PAPE
below reac
6 5 C H −COCl
th HBr in pre
d in the follow
r of the acid s
d (III) and 2, 2
(B) III <
(E) IV <
n-aromatic
matic
i-aromatic
sing reactivit
(CH3)2CClCH
(B) III <
(E) I < I
ER-I-QUESTI
ction are
H2,Pd/BaSO4
quinoline l ⎯⎯⎯⎯⎯→
(B) (
(D) (COCl)2
sence of pero
(B) 3 H C C•
−
(E) None of
wing reaction
(B)
(E)
strength of ac
2-dichlorobut
II < IV < I
III < II < I
(B) N
ty of the follo
H2CH3 (III)
I < IV < II
III < II < IV
ON WITH AN
---------- a
⎯→Y
COCl)2 and C
and C6H5CH
oxide proceed
2 CH−CH Br
the above
n is
cids, butyric a
tyric acid (IV)
(C
Non-benzeno
(D) Benz
owing alkyl h
and CH3C
(C
NSWER KEY
and ---------
C6H5CH3
H2 OH
ds through the
(C)
H
(C)
acid (I), 2 – ch
) is
C) I < III < II
id and aroma
zenoid and aro
halides, CH3C
CH2CH2Cl (I
C) III < I < II
VERSION
Y
----- respec
e intermediate
3
Br
HC CH C
•
− −
hlorobutyric a
I < IV
atic
omatic
CH2CH(Br)CH
IV) towards
I < IV
N - A1
16
tively
e
2 C H
acid
H3 (I),
SN2
PAPER-I-
117. T
(A
(D
A
118. T
be
(A
D
(D
A
119. T
(A
A
120. T
(A
(C
A
-PHYSICS & C
KEA
The strongest b
A) Amide ion
D) Ammonia
Ans : A
The condensa
enzaldehyde
A) Benzalacet
Dibenzylidene
D) Benzoic ac
Ans :C
The product Y
A)
Ans :C
The product fo
A)
C)
Ans : A
CHEMISTRY
AM-ENGINEE
base among th
n
ation reactio
in presence o
tophenone
eacetone
cid and acetic
for the below
(B)
ormed in the f
Y
ERING: PAPE
he following
(B) Hyd
(D) Ani
n between
of dilute alkali
c acid (E) Only
w reaction is
)
following reac
ER-I-QUESTI
is
droxide ion
iline
one equival
i leads to the
(B) Benzylid
y benzoic aci
(C)
ction is
(D)
ON WITH AN
(C
lent of acet
formation of
deneacetone
d
(D)
(B)
NSWER KEY
C) Trimethyl
one and tw
(C)
(E)
VERSION
Y
amine
wo equivalen
(E)
N - A1
17
nts of

7/24/19

Engineering Physics questions and Answers

MODULE IV

1. Define Black Body?

When heat radiations are incident on a surface, a part of it is reflected from the surface, another part is absorbed by it and the remaining part is transmitted through it.

ie r+a+t=100% or r+a+t=1

r→ Amount of heat energy reflected

a→ Amount of heat energy absorbed

t→ Amount of heat energy transmitted

A perfect blackbody is that which absorbs all heat radiations incident on it. There is neither reflection nor transmission of heat energy from that body.

r+a+t=1, generally for a perfect Black body, a=100%, ie r=0, t=0

When black body is heated it emits all the heat radiations absorbed. The thermal radiations depend only on the temperature of the body

2. What is a Ferry’s black body? Explain the black body spectrum? (CU 2006)

It is a double walled Sphere with a very fine hole ‘O’ at one end a projection P on its opposite inner side. The inner surface is perfectly coated with lamp black. The projection ‘P’ prevents the radiations from direct reflection. The inner space between the walls is evacuated

When heat radiations are incident through the hole ‘O’ they undergo multiple reflections until they are completely absorbed by the inner surface. Now it become a perfect black body absorber. When this body is heated to a temperature it emits full radiations through the hole. The internal radiations depend only on the temperature of the body

Distribution Of Energy In The Spectrum Of A Black Body

The distribution of energy of radiations from a black body was first analyzed by Lammer and Pringsheim. The heat radiations passed through carbon tube to slit “s’(they emitted from Carbon tube).The beam falls on a Concave mirror(reflector). The II ’l beam is now passed through a flours per prism which disperses it. The dispersed beam is focused to a Bolometer which measures the intensity (energy) of thermal radiations. By rotating the prism the energies corresponding to different wave at a particular temperature are measured it. The measurements are made with thermal radiations of different temperature. A graph is plotted with the wavelength

on x axis and the intensity of relation of radiation E

on y axis

reaches maximum value E_m

From graph, the energy is not distributed uniformly over the spectrum. At a particular temperature the intensity of radiation E

(energy) at first increases with wavelength

and then decreases with further increases of wavelength. The particular wavelength corresponding to the maximum intensity of emission is taken as

m. Increases of temperature produces increase in energy of all wavelengths. When temperature of black body increases Emission of maximum energy at a particular

also gets increased. As temperature increases the wavelength ‘

(particular

for maximum energy emission) corresponding to the maximum emission of energy gets shifted towards the shorter wavelength side ie

_m gets decreased. This is represented in dotted lines

∝

1/T

mT= a constant .It is called Wien’s law

The area between the curve and the wavelength axis gives the total energy emitted per sec per unit area of the blackbody at the temperature. It is found that this area (ie energy) directly proportioned to the fourth power of its absolute temperature

ie E∝T⁴

it is called Stefan’s Law

3. What are matter waves and De Broglie waves? (KU MAY 2005)

In the phenomenon of interference and diffraction light behaves as a wave while in photoelectric effect and Compton effect it shows particle nature. Thus light has a dual nature. De Broglie hypothesis says that every moving matter exhibit wave like properties under suitable conditions. The wave associated with a particle is called a matter wave.

4. Give an expression for of Broglie wavelength of an electron

Consider a photon of mass m and momentum ‘p’ and frequency ‘

’.

Momentum is given by (p=mc)

De Broglie wavelength

Wavelength of Electrons

Consider an electron of mass ‘m’ and charge e subjected to a potential difference of V volts. If ‘V’ is the velocity acquired by the electron

then ½ mv² = eV

Substituting values, of h, m and e, we have

Wavelength of electron wave is inversely proportional to the square root of the accelerating potential. The wavelength of de Broglie waves associated with electrons accelerated through a potential difference of 100 volt in vacuum is 1.23 A⁰. This value is of the order of the wavelength of x- rays. Davisson and Germer proved experimentally the wave nature of electrons by electron diffraction experiment.

5. What are wave Packets?

A wave that is confined to a small region space in the vicinity of the particle is called a wave packet. It is an envelope of a number of waves super imposed.

by de Broglie hypothesis

We can use wave equations to describe the small particles. This is the significance of de Broglie equations.ie the character of matter waves can be explained with the idea of wave functions.

6. Explain Heisenberg’s uncertainty principle? (KU MAY 2006, CU 2009,2006,2005)

By classical mechanics the position and momentum of a particle can be determined simultaneously with accuracy. In Newtonian mechanics every particle has a fixed position in space and has a definite momentum at any time. According to uncertainty or the principle of indeterminacy. “It is impossible to have an accurate measurement of the position and momentum of particles simultaneously”. The product of the uncertainty in the measurement of position of the particle at a certain instant and the uncertainty in the measurement of the momentum of the particle is of the order of the Planck’s constant ’h’.. If ∆x is the uncertainty (error) in the measurement of the position of particle along x coordinate and ∆Px is the uncertainty (error) in the measurement of its momentum, then

∆_x∙∆P_x=

Similarly

ie if

is small.,

is large and vice versa.

=0,

= infinity

Also

= infinity

∆E.∆t=

An electron exists in an excited state only for a short interval of time. Thus ∆t is small, ∆E must be large

The typical value of ∆t=10^-8seconds.

Also E=h

∆E=h∆

∴ ∆

but ∆E=

This is the irreducible limit to the accuracy with which we can determine the frequency of radiation emitted by an atom which remains in the excited state for about ∆t=10^-8 seconds.

7. What are ultrasonic waves? Mention their properties. (CU 2009, 2008)

Ultrasonic wave means acoustic wanes whose frequency is greater than 20 kH_z. This is a very powerful tool in non-destructive testing of materials, structures and products in engineering.

Ultrasonics

Acoustics is a branch of physics that deals with the study of mechanical waves which gives rise to sound. Sound waves are longitudinal mechanical waves. They can propagate in solids, liquids and gases. Sound waves in the frequency range which can stimulate human ear and brain to the sensation of hearing (ie 20 HZ to 20 KHZ) is called audible range.

Below audible range - infrasonics →eg: earth quake, elephant sound, etc

Above audible range → ultrasonics -eg: elastic vibrations produced by quartz crystal, sound produced by bats etc

Properties of Ultrasonics:-

1. They cannot travel through vacuum

2. They are high energetic waves

3. The speed of ultrasonic waves in a thin rod or crystal is given by

Y→ Young’s modulus of the material

ρ→ density of the rod

Its speed in liquid is given by

K→ Bulk modulus

ρ→ its density

In gas,.

P→ pressure of the gas

ρ→ its density

4. Speed of ultrasonic waves depends on frequency, greater frequency higher speed.

5. They can be reflected, refracted and diffracted like light waves.

6. When ultrasonic waves travel through a medium, they are scattered and a part of energy is absorbed by the medium. This loss of energy by scattering and absorption is called attenuation.

7. They produces heating effect when pass through a medium. A part of energy absorbed by the medium reappears as heat energy

8. When ultrasonic waves pass through a liquid, stationary waves are produced by the reflection. As a result the density of liquid layer is varied from layer to layer. This behaves like a plane diffracting grating which can diffract light waves.

9. Stirring effect

Intense ultrasonic beam produces vigorous agitation in certain low viscous liquids. They produce bubbles due to the disruptive effect.

8. Define piezoelectric effect? What are all the methods available to produce Ultrasonic waves? Explain piezoelectric method. (CU 2005, 2008, 2009)

When certain crystals like quartz tourmaline etc are subjected to stress or pressure along certain axis, a potential difference is developed across the perpendicular axis. This is called Piezo electric effect, discovered by J-Curie and P- Curie in 1880. The converse of this effect is also possible when a potential difference is applied between the two opposite faces of a crystal than stress or strain is induced in the other two opposite faces, ie the crystal is set into vibrations. If the frequency of elastic oscillations coincides with the natural frequency of the crystal, the vibrations will have large amplitude. The crystals which exhibit Piezo- electric effect are called Piezo electric crystals.

E: Quartz, tourmaline, Rochelle selt etc.

`There are 3 main methods available to produce ultrasonic waves

1. Mechanical generator

2. Piezo electric effect method

3. Magneto striction method

Piezoelectric generator Construction

It is based on the converge of Piezo electric effect

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It consists a quartz crystal placed in between the parallel metal plates and it is connected parallel to a tank circuit. The inductance coil L₁ and L₂ is centrally tapped. The coil and the variable capacitor constitutes the tank circuit. One end of the tank circuit is connected to the base of the transistor. The electric power required to operate the transistor is supplied from a D.C source V_CC through R.F.choke. R.F Choke prevents a.c. from flowing to dc source. R_E is a resistor connected to emitter and it regulates the potential of the emitter. C_E is an A.C bypass capacitor which by passes ac alone. R₁ and R₂ resistors provide proper voltage to the transistor. The coupling capacitor C_inprevents the dc flowing to the transistor.

WORKING

When key is switched on high frequency ac is produced from the tank circuit. When this ac is applied across a part of opposite faces of the quartz crystal, it begins to vibrate according to the principle of Piezo electric effect. The variable capacitor is tuned so that the applied frequency becomes equal to the natural frequency of the crystal. Hence tuned resonance takes place and the crystal vibrates with the maximum amplitude. At this condition ultrasonic waves are produced from the sides of the crystal. Ultrasonic waves up to 15 MH_Z can be produced using this method.

The frequency of the circuit is given by

If L= L₁+L₂then

Natural frequency of the crystal is given

Y→ young’s modulus of the crystal

l → thickness of the crystal

ρ → its density

9. Define magnetostriction. Explain this method to produce ultrasonics? (CU 2005, 2011)

When a ferromagnetic material in the form of a bar is subjected to an alternating magnetic field with its length parallel to the magnetic field as shown in fig., the bar undergoes alternate contractions and expansions at a frequency equal to the frequency of the applied magnetic field. This is known as magnetostriction effect.

Ferromagnetic materials which are used for the production of ultrasonic waves are called magnetostriction materials.

PRODUCTION OF ULTRASONICS BY MAGNETOSTRICTION

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The transistorized version of the magnetostriction oscillator is shown in the figure. R is a ferromagnetic rod clamped at the middle. A coil is wound on R near one end and insulated from it. The coil L₁ and variable capacitor constitution the tank circuit. The frequency of the tank circuit is equal to the fundamental frequency of longitudinal vibrations of the rod R. The tank circuit is connected to the collector of the transistor T. The other end of the rod carrier another coil L₂ wound on it. L₂ is connected to the base of the Transistor through the coupling capacitor Cin. The coils are wound loosely on the rod so that rod can vibrate freely

WORKING

When the key k is closed a current flows through the coil L₁. Then a magnetic field is produced. This magnetic field changes the length of the rod due to magnetostriction. This magnetic field changes the length of the rod due to changes the magnetic flux through L₂ thus inducing an emf in it. The coil L₂ is connected between the base and the emitter of the transistor. Hence the emf across it will increase the forward bias.

∴ The collector current increases. This provides a +ve feedback which is required to sustain the oscillations of the tank circuit. The voltage across the Resistors R₁ and R₂ provide proper biasing of the transistor. C_E is ac bypass capacitor. The Magnetostriction method can be used to produce ultrasonic of frequency range 5KHZ to 60 KHZ.

The applied frequency is given by,

Natural frequency of the crystal,

Y → young’s modulus of the crystal

ρ→ its density

Disadvantages of magnetostriction method.

Magnetostriction method is somewhat expensive. Low range frequency ultrasonic can be produced alone up to 60 KHZ. Efficiency affected by eddy current loss and hypothesis loss

These are the disadvantages of this method

10. Mention the Applications of ultrasonics? (CU 2009)

1).Non Destructive Testing of materials (NDT)

It is a method of testing of materials without any destruction in the materials. The present condition and quality of the materials can be examined using NDT without destroying their properties. Ultrasonics can be used to detect the imperfections like the flaws, cracks, breakings, cavity, air pockets, discontinuities etc in material like metals. Any defects in weldings and castings can be exactly located with the help of ultrasonic. Aircraft have to undergo such ultrasonic testings before their flight.

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A flaw or a crack in metal block produces a change in medium which causes reflection of waves. When ultrasonic waves pass through a flaw it is partially reflected from it and partially transmitted through it. The transmitted beam is again reflected from the bottom side of the metal block. When ultrasonic beam is incident on the interface between two media it is partially reflected and partially transmitted. The intensity of the reflected and the transmitted beam is decided by the acoustic impedences of the media. This is the principle of ultrasonic testing. The intensity of ultrasonic wave from the cavity examined using a CRO

→intensity shown by CRO if the object having no cavity or breakings

→ Intensity graph shown by the CRO suppose the object having cavity or breakings etc

2) SONAR

In this method, high frequency ultrasonic waves are used to find the distance and direction of submarines, depth of sea, depth of rocks in the sea shoal of fish in the sea etc. The ultrasonic waves are generated by the Piezo electric method using a quartz crystal placed in between two metal plates. The same quartz crystal is used to detect the ultrasonic waves.

Surface of sea transducer

Bed of sea

These waves are transmitted towards the bed of the sea and get reflected back from the bed in the form of echos. These reflected echos are received by the quartz crystal and they are amplified and fed to the CRO. The time taken by the ultrasonic waves for the to and fro travel is measured

d→ depth sea

t → time taken for to and fro travel

v→ velocity of ultrasonic waves

3) Ultrasonic waves are very effective in cleaning material surfaces. It will agitate dust and impurities on surfaces and remove them.

4) Ultrasonic drilling and welding are very effective in drilling holes and to weld soft metals and plastics.

5) Ultrasonic waves can accelerate chemical reactions. The technique is used by chemical industry to reduce reaction times.

6) They are used for Sterline milk, water etc.

7) Unicellular organisms can be destroyed when exposed to Ultrasonics.

8) Diagnosis is mostly based on Ultrasonic Scanning and imaging.

9) The ultrasonic waves can be for directional signaling on account of their high frequency. It can be concentrated into a sharp beam and can be used for signaling in a particular direction.

10) We can use ultrasonic waves to find the velocity of sound in gases and liquids.

11. Define reverberation & reverberation time? (CU 2008)

The sound produced in a room or a hall suffers multiple reflections from various objects like the walls, ceiling, floor, furniture etc in the hall. The sound appears to remain for a long time after the source of sound is stopped. This persistence of sound even after the sources of sound is stopped is called reverberation.

The reverberation time (T) is defined as the time taken for a sound to decreases in intensity to10^-6 of its original intensity, the time being reckoned from the instant when the source of sound is cut off. The time of reverberation is also defined as the time required for the intensity of sound to decrease by 60db from the moment when the source is cut off. The reverberation time is highly significant in the design of halls and auditoriums.

12. Define absorption coefficient?

The absorption coefficient (∝) of the surface of a material is defined as the ratio of the sound energy absorbed by the surface to the total energy incident in it. Since an open window absorbs the whole amount of sound energy incident on it, the absorption coefficient for it is unity. The unit of absorption is called Sabine. It is the sound energy absorbed by on square Foot of an open window. For unit area of various surfaces, the absorption coefficient is expressed in terms of equivalent area of open window. The equivalent absorbing area A for a surface having total area S and absorption coefficient ∝ is given by

A=∝S

13. Define

(i) Matter waves

(ii) Wave packets

i) Matter waves

ii) Wave packets

According to de Broglie hypothesis a wave is associated with a moving particle. Hence a particle can be represented by a wave confined a space. A plane wave cannot be used to represent ot since it extends to infinity. A wave that is confined to a small region space in the vicinity of the particle is called a wave packet. It is an envelope of a number of waves super imposed

14. Explain the physical concept of wave function (CU 2011)

The quantity with which quantum mechanics is concerned is the wave function ψ of a particle. The quantity that undergoes periodic changes of a body is called wave function ψ. It is in general a complex valued function and itself has no physical interpretation. The square of the absolute magnitude / ψ

or ψ

dxdydz is proportional to the probability of finding the particle in the small volume element dxdydz about the point x, y, z. we can obtain all the physical properties of the system if we know the wave function.

The wave function should fulfill certain requirements.

Since ψ

dxdydz is proportional to the probability of finding the particle with in the volume element, the integral

dxdydz must be finite if the particle is somewhere there. If

dxdydz is zero, the particle doesn’t exists and if it is infinity, the particle is everywhere simultaneously.
Since the probability of finding the particle in the volume element is a surety, then

dxdydz must be equal to 1. The wave function satisfying above condition is called normalised wave function.

The requirements of wave function

1. The wave function ψ must be continuous and single valued everywhere

and

must also be continuous and single valued everywhere

can be normalized.

15. Derive Schrodinger’s Wave Equation for a free Particle and time Dependent Equation (CU 2010)

Schrodinger’s equation is the basic expression used in quantum mechanics. This cannot be derived from elementary rules. We can derive it by considering the plane wave equation and combining with Einstein’s equation for quantum of energy and de Broglie’s’ expression for wavelength.

A particle in motion is associated with a wave function that contains the information about the motion. A plane progressive wave that propagates along X - direction is given by

Ψ = A

‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑ (1)

Where k is the wave vector given by

and

is the angular frequency.

Einstein’s formula for photon energy is

E= h v=

where h =

de – Broglie’s expression for matter wavelength is

p =

= hk

Using the above expression, equation (1) becomes

Ψ = A

Ie ψ = A

On partial differentiation of Ψ with respect to x, twice, we get

Differentiating with respect to time,

These are equivalent to

ψ = -

‑‑‑‑‑‑‑‑‑‑‑‑‑ (4)

= - i

( - i

) ψ ‑‑‑‑‑‑‑‑‑‑(4a)

From 3 and 4, p = i

This is called space operator.

= i

‑‑‑‑‑‑‑‑‑‑‑‑ (5)

‑‑‑‑‑‑‑‑‑‑‑ (5a)

This energy is called time operator.

For a free particle total energy is given by

E =

, since V = 0

ie E

‑‑‑‑‑‑‑‑‑‑ (6)

using equation (4) and (5), Equation (6) becomes

= iħ

‑‑‑‑‑‑‑‑‑‑‑‑ (7)

This is Schrodinger’s equation for a free particle in one dimension. In three dimension it becomes for a free particle, as

ψ = iħ

‑‑‑‑‑‑‑‑(8)

Where

Is called Lapalcian operator

If the particle is moving under a potential V (r, t) then equation (8) becomes

= iħ

This is Schrodinger’s time dependent equation

16. Derive the time independent Schrodinger wave equation or steady state equation. (CU 2010)

According to Schrodinger, de- Broglie’s wavelength holds good for any particle moving in any field of force with potential energy v.

Then total energy E = kinetic energy + potential Energy

E =

+ V

= 2 m (E – V)

or p = [ 2m (E – V)

The wave equation in Cartesian co-ordinate system can be written as

‑‑‑‑‑‑‑‑‑‑‑‑ (9)

Where

‑‑‑‑‑‑‑‑‑‑ (10)

‘u’ is the velocity of motion and

(x y z t) represents the amplitude of the wae associated with the particle

From (10), we get

= -

So Equation (9) becomes

ψ ‑‑‑‑‑‑‑‑‑‑‑‑‑ (11)

So we have

ψ or

ψ +

ψ = 0 ‑‑‑‑‑‑‑‑‑‑‑‑‑ (12)

Equation (12) is a general equation which is independent of time. Let us now introduce the concept of de-Broglie wavelength.

Substituting the wavelength is equation (12)

ψ +

[

ψ = 0

ψ +

ψ = 0

ψ +

ψ = 0………………. (13)

This represents Schrodinger’s time – independent wave equation.

Equation 13 shows that the wave function ψ is a function of coordinates also. V is a function of coordinates.

For the case of a free particle (V=0), the Schrodinger equation becomes

ψ +

ψ = 0

E is the energy having definite values and it also has to be satisfied with certain boundary conditions. The discrete values of E are called Eigen values and the correspondence wave functions are called Eigen functions.

17. Define Expectation values

In quantum mechanics each dynamic variables is represented by an operator which acts on a wave function to give a new wave function.

The expectation value of an operator ‘A’ representing a dynamic variables, denoted by , is defined as

Consider a large number of identical systems. They are in the same state of wave function ψ before measurements. Expectation value is the mean or average value of the results obtained by the measurements. Dynamic quantities like position, momentum, energy etc. are called observables. In quantum mechanics each observable is represented by an operator. When an operator is acting on a wave function we get a new wave function.

< A >=

Where d r is the differential volume element. If ψ is a normalized wave function, then

= 1, then

< A >=

Expectation value depends on the state of the system before measurement. It is the mean value of he results obtained by performing the measurement on a large number of identical systems each of which v was in the same state

before measurement. To emphasise the fact, one may write the expectation value as

18. Derive the time independent Schrodinger equation and solutions

In time dependent Schrodinger equation, the potential energy of a moving particle is a function of time and position also. In certain cases potential energy does not depend explicitly on time. Then we get time independent Schrodinger equation for such particle or system.

The wave function in this case can be expressed as a product of two functions.

(r), a function of position only and f(t), a function of time only

Then

and

= f(t)

putting this on time dependent Schrodinger equation,

(r, t)

f(t)

f(t) = iħ

Dividing throughout by ψ (r) f(r)

……… (14)

In the above equation, the variables are separated.

LHS is a function of position only and the RHS is a function of time only. Therefore each side must be equal to a common constant called separation constant.

= E

dt.

On integration

Log f =

+ constant.

Or f = C

‑‑‑‑‑‑‑‑‑‑‑ (15)

Equating the LHS of equation 14

= E

= E – V or

( E – V)

= 0 ‑‑‑‑‑‑‑‑‑ (16)

The above equation is the Schrodinger time independent wave equation.

This is applicable to problem with potential energy independent of time. This will be applicable to steady state or stationary state problem.

19. Solve the Schrodinger equation for a particle confined in a one-dimensional box of length L. Draw the first few energy levels and the corresponding eigenfunctions. (CU 2011)

Consider the motion of a particle of mass m confined to move between two walls of infinite height at x = 0 and x = L. The width of the box is L. Since there is no interaction between the particle and box, the potential energy of the particle is taken to be zero. It is very clear that potential energy of the particle is taken to be zero. It is very clear that potential energy does not depend on time and we are considering only time independent. Schrodinger equation for the solution. The problem is one dimensional and the Schrodinger equation is.

(E – V)ψ = 0

Since V = 0, the equation becomes

ψ = 0 or

ψ = 0 where

‑‑‑‑‑‑‑‑‑‑‑‑‑ (17)

The solution of the above differential equation is of the form

Ψ = A sin kx + B coskx‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑( 18)

The solution has to be well behaved wave function. Since the particle is inside a box of infinite height, it is impossible to find the particle outside the box ie ψ must be zero for all points outside the box

Ψ = 0 for x < 0 and

Ψ = 0 for x > L

This is possible only if

= 0 at x = 0 and x = L as demanded by continuity condition.

V = 0

X = 0 X = L X

Applying first condition on equation 18, we get

0 = A sin 0 + B cos 0

ie B = 0

So solution reduces to

Ψ = A sin Kx‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑ (19)

Using the next condition ψ = 0 at x = L, we get

0 = A sin kL

There are two possibilities. Either A = 0 or sin kL = 0

A cannot be zero, since the wave function cannot exist. So we have the other possibility

Sin kL = 0

ie kL = n

‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑ (21)

If n = 0, then ψ = 0 for all values if x. This is ruled out. Therefore n is a non-zero integer. We have

on using equation (17)

. ie

Since ħ =

as the energy corresponding to n the different values of energy for n are called Eigen values. Since n is restricted, the particle cannot have any value of energy, but restricted to certain values. The quantity n is called quantum number.

Eigen Function

By applying the normalization condition, ie ∫

ψ d r= 1 we can find the normalized wave function. ∫

dr =1

So we have

A sin

dx = 1

On integration and applying limits we get

= 1 or A=

So the normalized wave function is

(x)=

sin

where n = , 2, 3 ‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑ (22)

For the ground state n = 1, the wave function is given bt

sin

Similarly the wave function for the first two excited states are given by

sin

and

sin

These wave functions associated with different energy are called Eigen wave functions.

=0
X=L

wave function of first three energy levels

20. How will determine the velocity of ultrasonic waves in a liquid by ultrasonic diffractometer?

(CU 2008)

Velocity and wavelength of ultrasonics using ultrasonic diffractometer

When a quartz crystal Q placed between two metal plates is a liquid is set into vibrations using an R.F. Oscillator, ultrasonics are produced. When these ultrasonics are reflected by a reflector, longitudinal stationary waves are produced in the liquid. As a result alternate nodal pulses and antinodal pulses are formed. At nodal planes, the layers are crowded together (compressions or condensations) and density is maximum. At antinodal planes, the layers are separated (rarefactions) and density is minimum. This setup of nodal planes and antinodal planes behaves like slits and opaque spaces of a plane grating. Such an arrangement is called acoustic grating. Using this acoustic grating, velocity 'V’ and wavelength

of ultrasonics can be determined.

A parallel beam of monochromatic light from a sodium vapour lamp is collimated and is allowed to is observed fall normally on this acoustic grating. Diffraction takes place and the diffracted beam through the telescope of a spectrometer. On either side of the central maximum various orders of principal maxima are obtained. If θ is the angle of diffraction for a principal maximum, then

d sinθ = n

———> (1) where'd' is the distance between two consecutive nodal planes or two consecutive anti nodal planes, n is the order of spectrum and

is the wavelength of monochromatic light d can be calculated from this grating equation. But

d =

_a/2 or

_a= 2d---------- >(2) where

_a is

the wavelength of ultrasonic wave in the liquid.

But V=

. Where

is the frequency of oscillations of the crystal and V is the velocity of ultrasonic wave in the liquid.

DESCRIPTION

Ultrasonic diffractometer mainly consists of a quartz crystal Q placed between two metal plates provided with connecting leads. The quartz crystal setup is clamped inside on one face of an optically plane rectangular glass cell filled with kerosene or CCl₄so that the crystal is immersed completely in liquid. The cell is placed on the prism table of a spectrometer and it is illuminated by a beam of monochromatic light from a sodium vapour lamp. The quartz crystal can be subjected to vibrations from an R.F.oscilIator.

PROCEDURE

The initial adjustments of a spectrometer are made. The rectangular cell containing liquid is placed on the prism table perpendicular to the collimator. The quartz crystal Q together with metal plates and connection leads is clamped inside the liquid on another side of cell. A narrow beam of monochromatic light from collimator is allowed to fall normally on the cell. The direct image is observed through the telescope. Now the R.F. oscillator is switched on and the frequency of R.F. oscillator is varied from 2 to 5 MHz . The crystal is subjected to these oscillation and it begins to vibrate resonance with the oscillator. As a result ultrasonic waves are produced in liquid. These waves get reflected from the opposite side of the cell producing longitudinal stationary waves in the form of compressions and rarefactions. This arrangement behaves like a grating and as a result ultrasonic waves are diffracted. On either side of central maximum different orders of principal maxima are obtained. The telescope is turned so that the cross-wire coincides with the spectral line of first order on one side of the central maximum. The main scale reading and vernier scale reading are noted. Now the telescope is turned to the other side so that the cross wire coincides with the spectral line of first order. Main scale reading and vernier scale reading are noted. From these two sets of readings 2θ for first order and hence θ are calculated. The distance'd' between two consecutive nodes or antinodes is found out from the' grating equation dsin θ =n

where

the wavelength of sodium light is known (5893Å)

_a the wavelength of ultrasonics in liquid can be found out from

_a = 2d. Knowing the frequency of oscillator, velocity V of ultrasonics in liquid is calculated from the equation, V =

_a.

21. Derive Sabine’s formula for reverberation time? (CU2008, 2011)

Sabine derived an expression for the reverberation time (T) on the basis of the following assumptions.

· The distribution of sound energy and the intensity of sound is uniform inside an enclosure.

· The dissipation of energy in air is negligible.

· The absorption coefficient of any surface is independent of the intensity of sound.

· The phenomenon of interference and formation of stationery waves are supposed to be absent or non- existent and

· The rate of emission of sound energy is constant.

Consider a hall of volume V. Let a source of sound emit sound. The sound energy spreads out in all directions. The sound waves (energy) are partially reflected and absorbed by various objects in the hall. After some times, a steady state is reached between the energy emitted and the energy dissipated. Then the energy density (energy/ unit volume) becomes uniform throughout the hall.

Let the source of sound be cut off at =0. Let E₀ be the energy density at this instant. The energy density decreases exponent all with time. Let E be the energy density after secs. Then

E= E₀e^-Avt/4V ------(1)

A= the total energy absorbed

V= the velocity of sound

V= the volume of the hall

For a given frequency of sound, the intensity of sound is proportional to the energy. Hence if I₀ is the intensity at t=0 and I is the intensity after t secs

I=I₀e^-Avt/4v

When t= T, the reverberation time,

(From definition of reverberation time)
Then from equations (2) and(3)

Taking logarithms,

=2.303×log₁₀10⁶

=2.303×6×1

Taking

=340m/s at root temperature

This is Sabine’s formula for reverberations time

The total energy absorbed by various surfaces

22. Define the terms

(i) Nano science and Nanotechnology

(ii) Nano materials (CU 2010)

(iii) Nano clusters

(iv) Fullerenes

(i) Nano Science and Nanotechnology

Nanotechnology is the study of the control of matter on an atomic and molecular scale. Generally nanotechnology deals with structure of the size 100 nanometer or smaller, and involves developing materials or devices within that size. Nanotechnology is very diverse, ranging from novel extension of conventional device physics, to completely new approach based upon molecular self assembly, to developing new materials with dimensions on the nanoscale, even to speculation on whether we can directly control matter on the atomic scale. One nanometer (nm) is one billionth or 10^-9, of a meter.

(ii) Nano materials

Nano materials are the materials of a size one billion or 10^-9, of a meter. Nanotechnology is very diverse, ranging from novel extensions of conventional device physics, to completely new approach based up on molecular assembly, to develop new materials with dimensions on the nano scale. Molecules of the dimension 0.1nm. Most of the atoms are on the surface of the clusters. The size of the particle is less than the critical characteristics length of the electron to conduct, they exhibit different properties.

(iii) Nanoclusters

Clusters belong to a new category of materials. Their size is in between bulk materials and their atoms or molecules. Their properties are fundamentally different from those of discrete molecules and bulk solids. They are systems of bound atoms or molecule existing as an intermediate form of matter with properties that lie between those of atoms and bulk materials. Depending on the constituent units they are called either atomic or molecular clusters J Clusters include species only in the gas phase or in the condensed phase or both. They can have either a net charge (ionic clusters) or no charge at all (neutral clusters). The atoms or molecules constitute clusters are bound by forces which may be metallic, covalent, ionic hydrogen bonded or van der. Waal's in character and can up to a few thousand atoms.

iv) Fullerenes

Fullerenes are molecular forms of carbon which are distinctly different from the extended carbon forms known for millennia. There are numerous forms all of which are spheroidal in structures. In Chemistry there is no other molecule formed by the same atom which is as big as fullerenes. A carbon molecule with chemical formula C₆₀ containing 60 carbon atoms in the shape of a soccer ball had been predicted in 1970. An experiment was carried out in Rice university in which a graphite disc was heated by a high intensity laser beam that produces a hot vapour of carbon. A burst of helium gas then swept the vapour of carbon out through an opening where the beam expands. The expansion cooled the atoms and they condensed into clusters. This cooled clusters beam was then narrowed by a skimmer and fed into a mass spectrometer to measure the mass of the molecule in the clusters. A mass number of 720 that would consist of 60 carbon atoms, each of mass 12 was seen. This was evidence of a C₆₀ molecule. This was named after the architect Buckminister Fuller. The name Buckminister fullerene was shortened to fullerene.

23. Mention the applications of Nanotechnology?

Nano medicine: Medical research field has exploited the unique properties of nano materials for various applications, such as cell imaging and treating cancer. This field is called nano medicine. Nanotechnology is also used for diagnostic purposes. The drug consumption and side-effects can be reduced significantly by a targeted or personalized medicine and hence the treating expense is lowered. This is done with the help of nanotechnology by depositing the medicine in the morbid region only and in the correct doze. Nanotechnology can help to reproduce or repair damaged tissue.

Energy: The applications of nanotechnology in the field of energy are

1. by reduction of energy consumption

2. By increasing the efficiency of energy production.

3. By the use of more environmentally friendly energy systems and

4. By recycling batteries

Nanotechnological approaches like LEDS or Quantum Caged Atoms (QCAs) could lead to a strong reduction in energy consumptions. The efficiency of internal combustion engines could improve combustion by designing specific catalysts with maximized surface area. An eg. for an environmentally friendly form of energy is the use of fuel cells powered by hydrogen which is produced by renewable energies. The use of rechargeable batteries with higher rate of recharging using nanomaterials could be helpful for battery disposal problem

Chemistry and Environment: Chemical catalysis from nano particles is highly beneficial due to its extremely large surface to volume ratio. The application ranges from fuel cell to photo catalytic devices.

Filtration: Nanofiltration helps waste water treatment, air purification and energy storage devices.

Information and communication: Memory storage devices such Nino-RAM is developed using carbon nanotubes based crossbar memory.

Food: New consumer products created through nanotechnology are coming to market now. There are nearly a thousand known or claimed nanoproducts. The eg. of food products are canola cooking oil and a tea called nanotea etc.

Textiles: The use of engineered nanofibers already makes cloths stain –repellent and wrinkle-free. These textiles can be washed less frequently and at lower temperatures.

24. Explain the properties & applications of carbon Nano tubes?

Carbon nano tubes are ultimate high strength carbon fibers. They have a high strength to weight ratio. This value is 100 times that a steel. They are highly resistant to chemical attack. It is difficult to them. As a result temperature is not a limitation in practical applications of nano tubes. Surface are of nano tubes is higher than that of graphite. Nano tubes have a high thermal conductivity exhibit a striking telescope property.

In multi walled Nano tubes, multiple concentric nano tubes Presley rested within on another exhibit a striking telescoping property. This is one of the first example of molecular nanotechnology; the precisely positioning of atoms to create useful machines. Thus property has been utilized to create world’s smallest rotational motor.

Because of the symmetry and unique electronic structure of graphine, Nano tubes can be metallic with 1000 times’ greater conductivity then metals or moderate semiconductors. Because of this property CNT are referred to as “one dimensional”

Applications of CNT

Nano tubes can be used as very good electrical conductors. An application of nanotubes is the production of CNT based field emission displays. CNT act as electron emitters at lower turn on voltage and high emissivity. Nanotube tips can be used as a nanoprobs, which does not crash frequently. It can be used for making paper batteries. It is a paper thin sheet of cellulose infused with aligned carbon nanotubes. A CNT complex formed by CNT & fullerenes are used as solar cells. CNT have been implemented in nanoelectromechanical system like nanomotors. Nanotubes are used to form alter capacitors.

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