AIPMT Physics 2008
11
1. (a) A block of mass 5 kg is placed on horizontal surface. A pushing force 20N acting on the block as shown in figure. Coefficient of friction between block and surface is 0.2. Calculate frictional force acting on the block and speed of block after 15 second. (g = 10 ms –2 ) 2 0N 45 o B 5 20N 0.2 5 (g = 10 / 2 ) 2 0N 45 o B (b) In van-der Wall's gas equation 2 a P [V b] RT V , what are dimensions of van-der Wall's constant "a" ? a P [V b] RT V , a Sol. (a) 45 20cos45 20sin45 o 20 5g f N 20sin45 o 20cos45 o 20 cos 45° = 20 10 2 14.1 2 and 20 sin 45° = 20 10 2 14.1 2 Normal reaction on the block N = 5g + 20 cos 45° = 50 + 14.1 = 64.1 newton Frictional force on block is f L = N = 0.2 [5g + 20 sin 45°] = [50 + 14.1] = 12.8 newton Acceleration of block 20cos45 f 14.1 12.8 1.3 a 0.26 5 5 5 = 0.3 ms –2 after 15 sec speed of block v = u + at = 0.26 × 15 = 3.9 ms –1 . (b) 2 a [P] V [a] = [P] [V 2 ] = ML –1 T –2 × L 6 = M 1 L 5 T –2 PHYSICS ww.examrace.com 5 5 mr mr mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mra mr mr mr mr mr mr mr mr mr mr mr mr mr mr mr mr mr mr mr mr mr mr r r r r r r r 2 2 a a [V b] R [V b] R 2 2 2 a a a V V am am m am am a a www P [V P [V a a a V V xa exa a exa xa exa xa ww w ww. w. ww ww ww ww ww ww ww ww ww w w ww ww 45 45 ww s45 s45 sin45 sin45 o o 20 20 20c 20c 20 cos 20 cos N N
description
GOOD
Transcript of AIPMT Physics 2008
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1. (a) A block of mass 5 kg is placed on horizontal surface. A pushing force 20N acting on the block as shown
in figure. Coefficient of friction between block and surface is 0.2. Calculate frictional force acting on the
block and speed of block after 15 second. (g = 10 ms2)
20N
45o B
5 20N
0.2 5
(g = 10 / 2)
20N
45o B
(b) In van-der Wall's gas equation 2a
P [V b] RTV
, what are dimensions of van-der Wall's constant "a" ?
aP [V b] RT
V, a
Sol. (a) 4520cos45
20sin45o
20
5gf
N
20sin45o
20cos45o
20 cos 45 = 20
10 2 14.12
and 20 sin 45 = 20
10 2 14.12
Normal reaction on the block N = 5g + 20 cos 45 = 50 + 14.1 = 64.1 newton
Frictional force on block is fL = N = 0.2 [5g + 20 sin 45] = [50 + 14.1] = 12.8 newton
Acceleration of block 20cos45 f 14.1 12.8 1.3
a 0.265 5 5
= 0.3 ms2
after 15 sec speed of block v = u + at = 0.26 15 = 3.9 ms1.
(b) 2a
[P]V
[a] = [P] [V2] = ML1T2 L6 = M1 L5 T2
PHYSICS
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in figure. Coefficient of friction between block and surface is 0.2. Calculate frictional force acting on the
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in figure. Coefficient of friction between block and surface is 0.2. Calculate frictional force acting on the
5
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5
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2
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2
a
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aP [V b] RT
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P [V b] RTP [V b] RT
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P [V b] RT2P [V b] RT2
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2P [V b] RT2a
P [V b] RTa
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aP [V b] RT
aV
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VP [V b] RT
VP [V b] RT
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P [V b] RTV
P [V b] RT
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a
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a
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P [V b] RT
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P [V b] RTP [V b] RT
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P [V b] RTP [V b] RT
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P [V b] RTa
P [V b] RTa
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aP [V b] RT
aV
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VP [V b] RT
VP [V b] RT
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P [V b] RTV
P [V b] RT
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45
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45
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20cos45
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20cos45
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20sin45
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20sin45o
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o
20
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20
20cos45
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20cos45
20 cos 45 = www.
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20 cos 45 =
Normal reaction on the block N = 5g + 20 cos 45 = 50 + 14.1 = 64.1 newtonwww.
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Normal reaction on the block N = 5g + 20 cos 45 = 50 + 14.1 = 64.1 newton
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2. (a) If three vector A , B and C are such that A B C and their magnitudes are in ratio 5 : 4 : 3 respectively
then find angle between vectors A and C .
A , B C A B C 5 : 4 : 3
A C
(b) Two particles start their motion from same point with initial velocities 4 ms1 and 2 ms1 and accelerations
are 1 ms2 and 2 ms2 respectively. If both reach at final position simultaneously, then determ ne thelength of path travelled by them.
4 2
1 2 2 2
(c) A body of mass 10 kg is released from a 20 m heigh tower, after falling th ough the 20 m distance
body acquires a velocity of 10 ms1. Calculate the work done by the push of the air on the body ?
10 20 20 10 /
Sol. (a) A B C r A C B
(A C) . ( A C) B . B 2 2 2A C 2 A .C B
A2 + C2 2AC cos = B2 [Le angle between A and C = ]
2 2 2 2 2A C B (5) (3) (4) 3cos 53
2AC 2(5)(3) 5
(b) Both particles reach at same position in same time t after travelling same distances
By using 1
s ut at2
For I pa ticle :2
21 ts 4(t) (1) t 4t2 2
...(i)
For II pa cle : 2 21s 2(t) (2) t 2t t2
...(ii)
By equation (i) and (ii) 2
2t4t 2t t2
2t2t 0
2t = 4 s
Subsituting value of t in equation (i) 21
s 4(4) (1) (4) 24m2
(c) Air resistance will oppose the motion and let work done by air resistance = WairWork done by gravity = Wgravity = mgh
Using work energy theorem Wgravity Wair = KE r Wair = mgh 12
mv2
= 10 9.8 20 12
10 (10)2 = 1960 500 = 1460 J
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and accelerations
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and accelerations
respectively. If both reach at final position simultaneously, then determ ne the
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respectively. If both reach at final position simultaneously, then determ ne the
A body of mass 10 kg is released from a 20 m heigh tower, after falling th ough the 20 m distance
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A body of mass 10 kg is released from a 20 m heigh tower, after falling th ough the 20 m distance
. Calculate the work done by the push of the air on the body ?
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. Calculate the work done by the push of the air on the body ?
20
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20
2
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2A C 2 A .C B
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A C 2 A .C B
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[Le angle between
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[Le angle between
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2 2
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2 2A C B (5) (3) (4) 3
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A C B (5) (3) (4) 32 2A C B (5) (3) (4) 32 2
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2 2A C B (5) (3) (4) 32 2
2(5)(3) 5
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2(5)(3) 5
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Both particles reach at same position in same time t after travelling same distances
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Both particles reach at same position in same time t after travelling same distances
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1
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1s ut at
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s ut ats ut at
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s ut at1
s ut at1
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1s ut at
12
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2
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pa ticle :
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pa ticle : s 4(t) (1) t 4t
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s 4(t) (1) t 4t
For
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For II
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II pa cle :
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pa cle :
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By equation (i) and (ii)
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3. (a) A chain of mass m and and length L is placed on a table in such a way that its 1n
th part is hanging
below the edge of table. Calculate the work done to pull the hanging part of chain on the table.
m L 1n
(b) A particle of mass m is connected from a light inextensible string of
12
Alength such that it behave as a simple pendulum. Now string is pulled
to point A making an angle 1 with the vertical and it is released from
the point A, calculate :
(i) velocity of particle at position B, when string makes an angle 2 fr m ver cal.
(ii) tension in the string when particle is at position B, when string makes an angle 2 from vertical.
m
12
A
B
1
A
(i) B 2
(ii) B 2
Sol. (a) Since chain is uniform so depth of ce ter f mass of hanging part from the top of the table = 1 L2 n
Mass of hanging part = m L mL n n
LnCM
L2n
Work done = work done in bringing CM of hanging part on the table = mnanging g hCM
= 2m L mgL
gn 2n 2n
(b) (i) h = (cos 2 cos 1)
at point A and B mechanical energy remain conserved 1
2
2
mgcos 2
T
cos 1
h
mg
cos 2
22 1
1mv mgh v 2gh 2g (cos cos )
2
(ii) Let tension in string at point B is T then for point B 2
2mv
T mgcos
2 2 1 2 1
mT mgcos [2g (cos cos )] mg(3cos 2cos )
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2
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2 fr m ver cal.
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fr m ver cal.
tension in the string when particle is at position B, when string makes an angle
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tension in the string when particle is at position B, when string makes an angle
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1
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1
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2
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2
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Since chain is uniform so depth of ce ter f mass of hanging part from the top of the table =
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Since chain is uniform so depth of ce ter f mass of hanging part from the top of the table =
Mass of hanging part =
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Mass of hanging part =
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m L m
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m L mL n n
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L n n
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Work done = work done in bringing CM of hanging part on the table = m
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Work done = work done in bringing CM of hanging part on the table = m
=
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=
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m L mgL
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m L mgLg
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gm L mgL
gm L mgL
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m L mgLg
m L mgLn 2n
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n 2ng
n 2ng
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gn 2n
g
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(i)
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(i) h =
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h =
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(cos
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(cos
-
4. (a) The weight of a sphere in air is 50g and its weitght in a liquid at temperature 20C is 40 g. When temperature
of liquid incerases to 70C it weight becomes 45g, then find the :
(i) ratio of densities of liquid at given two temperature
(ii) coefficient of cubical expansion of liquid assuming that there is no expansion of the volume of
sphere.
50 20C 40 70C
45
(i)
(ii)
(b) In damped oscillations, the amplitude after 50 oscillations is 0.8 a0, where a0 s the initial amplitude.
Determine amplitude after 150 oscillations.
50 0.8 a0 a0
150
Sol. (a) (i) Let dentsity of liquid at 20C = 1 and dentsit of liquid a 70 C = 2
W apparent =W air V g
at 20C : 40 = 50 V 1g r V 1g = 10
at 70C : 45 = 50 V g V 2g = 5
Dentsity ratio 1
2
10 25 1
(ii) 11
MV and 2
MV r
1 12
1 2
M1
V 1 1
bu2
= 2 so 2 = 1+ r = 1 r1
= 1 1
5070 20/C = 0.02 /C
(b) for damped oscillations amplitude a = a0e bt
Let time period of oscillations = T
for 50 oscillations, time taken = 50 T
0.8a0 = a0eb50 t r e50bT =
45
after 150 oscillations amplitude a = a0eb 50T= a0e
3 b50T = a0(eb50T)3 = a0
3
0644 a
1255
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, where a
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, where a0
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0 s the initial amplitude.
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s the initial amplitude.
dentsit of liquid a 70 C =
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dentsit of liquid a 70 C =
g
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g r
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r V
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V
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1
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1g = 10
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g = 10
45 = 50 V
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45 = 50 V
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g
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g
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V
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V
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10 2
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10 25 1
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5 1
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and
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and
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M
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M
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bu
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bu
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(b) for damped oscillations amplitudewww.
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for damped oscillations amplitude
-
5. (a) Two moles of an ideal gas is taken in a cyclic process abcda. During process ab and cd temperatures are
500K and 300K respectively. Calculate heat absorbed by the system. ( n 2 = 0.69 & R = 8.3 J/mole-K)
abcda ab cd
500K 300K ( n 2 = 0.69 R = 8.3 J/mole-K)
V0 2V0
Internal energy
U
d c
a b
300K
volume ( )
500K
(b) A glass rode having a curved surface at one face as shown in figure below and its e tre of curvature
lies inside the glass rode. Refractive index of glass is 1.5 and radius of cu ved part is R. If a particle
is placed at point P. It forms the real image at point Q. The point O cuts PQ such that OP = 2 OQ
then find out the value of OP.
1 5 R
P Q O, PQ
OP = 2OQ OP
P O Q =1.5
Sol. (a) Processe ab is isothermal expan ion Processe cd is isothermal compression
Processe bc is isoch ric comp ession Processe da is isochoric expansion
Net heat absorbed by he gas Qabcda = Qab + Qbc+ Qcd+ QdaQbc = Qda
Q bcda = Qab+ Qcd = RT1 In0
0
2VV + RT2 In
0
0
V2V
= R (T1 T) ln2 = 2 8.3 0.693 (500 300) = 2300.76 J
(b) L ght is coming from P to O
u = OP = 2OQ and v = OQ
for refraction at curved surface 2 1 2 11.5 1 1.5 1
v u R OQ 2OQ R
r1.5 0.5 0.5OQ OQ R
r2 1
OQ 4ROQ 2R
OP = 2OQ = 2 4R = 8R
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A glass rode having a curved surface at one face as shown in figure below and its e tre of curvature
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A glass rode having a curved surface at one face as shown in figure below and its e tre of curvature
lies inside the glass rode. Refractive index of glass is 1.5 and radius of cu ved part is R. If a particle
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lies inside the glass rode. Refractive index of glass is 1.5 and radius of cu ved part is R. If a particle
is placed at point P. It forms the real image at point Q. The point O cuts PQ such that OP = 2 OQ
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is placed at point P. It forms the real image at point Q. The point O cuts PQ such that OP = 2 OQ
1 5
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1 5
OP = 2OQ OP
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OP = 2OQ OP
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O
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O
Processe ab is isothermal expan ion
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Processe ab is isothermal expan ion
Processe bc is isoch ric comp ession
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Processe bc is isoch ric comp ession
Net heat absorbed by he gas Q
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Net heat absorbed by he gas Q
= Q
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= Qda
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da
Q
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Q bcda
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bcda = Q
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= Qab
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ab+ Q
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+ Q
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(b) L ght is coming from P to Owww.
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L ght is coming from P to O
u = OP = 2OQwww.
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u = OP = 2OQ
-
6. (a) A ring of radius R is uniformaly charged by +Q. Find potential at an axial point which is situated at
distance r from the centre hence deduce the electric field at that point.
R +Q
r
(b) Two short electric dipole have dipole moment 1p and 2p are placed as shown in figure below. Find the
value of torque on 2p due to 1p .
1p 2p 1p
2p
xp1 p2
(c) A conducting coil is bent in the form of equilateral triangle. Leng h of each side is 5 cm and current
flowing through it is 0.2 A. Find the magnetic mom nt f his triangular coil.
5
0.2 A
Sol. (a) Consider a small element of length
RP
r
q
dq= Q2 R
R + r2
2Charge on this small element is q
Potential at point P due t th s small element is
2 2
1 qV
4 r R
Similarly we can take different small elements on the ring and since all these elements are at equal
distance from poi P.
So, p tential due to whole ring at point P is V = V
2 2 2 2 2 2
0 0 0
1 q 1 1 1 Qq
4 4 4r R r R r R
From the symmetry of figure it is clear that direction of electric field at P is along O to P.
Magnitude of electric field at P is E = 10 2 2 2
dV Q d 1dr 4 dr
(R r )=
32 2 2
0
Q 1(R r ) (2r)
4 2
30 2 2 2
1 Qr4
(R r )
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are placed as shown in figure below. Find the
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are placed as shown in figure below. Find the
2
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2p
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2
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2
A conducting coil is bent in the form of equilateral triangle. Leng h of each side is 5 cm and current
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A conducting coil is bent in the form of equilateral triangle. Leng h of each side is 5 cm and current
flowing through it is 0.2 A. Find the magnetic mom nt f his triangular coil.
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flowing through it is 0.2 A. Find the magnetic mom nt f his triangular coil.
Consider a small element of length
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Consider a small element of length
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Charge on this small element is
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Charge on this small element is
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q
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q
Potential at point P due t th s small element is
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Potential at point P due t th s small element is
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2 2
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2 2
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1 q
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1 q4
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4 r R
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r R2 2r R2 2
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2 2r R2 2
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r R
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r R
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Similarly we can take different small elements on the ring and since all these elements are at equal
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Similarly we can take different small elements on the ring and since all these elements are at equal
distance from poi P.
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distance from poi P.
So,
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So, p tential due to whole ring at point P is V =
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p tential due to whole ring at point P is V =
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From the symmetry of figure it is clear that direction of electric field at P is along O to P.www.
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From the symmetry of figure it is clear that direction of electric field at P is along O to P.
-
(b) Let intensity of electric field due to 1p at position of 2p = 1E
Torque on 2p due to 1 2 1p p E
Since angle in between 1E and 2p is 180 So 21 0
(c)a
= 5
10m2
a
I = 0.2Aarea of triangle A = a a sin 6012
each arm of triangle = a
Magnetic moment of loop M = IA 2 2 41 3 5 3
0.2 5 10 5 10 102 2 4
A-m2
7. (a) A cylinder of length 2a cm and radius of cross section r cm is placed in such a way that its axisalong x-axis and its centre is at origin. One face of cylinder is at x = +a cm and ano her face of cylinderis at x = a cm.
a a
E E
y-axis
x axisO
If electric field :
for x > 0 is xE E i N/C fo x < 0 is xE E i N/C
calculate :(i) net outward electric flux through each flat surface.(ii) net outward electric flux throug curved surface.(iii) net charge enclosed inside the cylinder.
2a r x-
x = +a x = a
a a
E E
y-axis
x-axisO
xE E i N/C x > 0
xE E i N/C x < 0
:
(i)
(ii)
(iii)
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www.
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along x-axis and its centre is at origin. One face of cylinder is at x = +a cm and ano her face of cylinder
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along x-axis and its centre is at origin. One face of cylinder is at x = +a cm and ano her face of cylinder
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x axis
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x axis
fo x < 0 is
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fo x < 0 is
net outward electric flux through each flat surface.
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net outward electric flux through each flat surface.
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net outward electric flux throug curved surface.
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net outward electric flux throug curved surface.net charge enclosed inside the cylinder.
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net charge enclosed inside the cylinder.
-
(b) A coil of 500 turns and area 0.04 m2 is placed in a region of transverse magnetic induction of
0.25 Wb/m2. The coil is rotated through 90 about its diameter with a particular angular velocity in
0.1 sec. Coil is connected to a galvanometer in series having a resistance of 25 . Calculate total charge
flow through galvanometer.
500 0.04 2 0.25 / 2
0.1 90
25
Sol. (a)
a
y-axis
O
a
x-axis
E = E a i^E = E a i^
A = A i^ A = A i^
area of cross se tion A = r2
flat surface 1flat surface 2
curved surface
(i) For flat surface 1 :a E.A E i Ai Ea ( r
2) 104 Nm2/C
For flat surface 2 :a
E A ( E i).( Ai) = Ea ( r2) 104 Nm2/C
(ii) For curved surface E and A are perpendicualr so = 0
(iii)0
Q
charged enclosed in cylinder is Q = 0 = 8.85 1012 2Ea ( r2) 104
= (5.5 1015 Ea r2) C
(b) Initial magnetic flux passing through coil is i = N B A
Final magnetic flux passing through coil is f = 0
Change in flux = f i = 0 N B A = N B A
Total charge flowon through coil NBA
qR R
500 0.25 0.040.2C
25
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area of cross se tion A = r
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area of cross se tion A = r
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2
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2area of cross se tion A = r2area of cross se tion A = r
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area of cross se tion A = r2area of cross se tion A = r
a
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a
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E.A E i Ai
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E.A E i Ai E
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Ea
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a (
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(
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r
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r2
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2
a
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a
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aE A ( E i).( Ai)a
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aE A ( E i).( Ai)aE A ( E i).( Ai)
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E A ( E i).( Ai)
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E
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E
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and
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and A
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A
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are perpendicualr so
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are perpendicualr so
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0
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0
Q
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Q
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charged enclosed in cylinder is Q =
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charged enclosed in cylinder is Q =
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(b) Initial magnetic flux passing through coil is www.
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Initial magnetic flux passing through coil is
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Final magnetic flux passing through coil is
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8. (a) An L-C-R series circuit having 220 V a.c. source, inductance 25 mH and resistance 100 . If voltage
across inductor is just double the voltage across resistor then find out frequency of a.c. source.
L-C-R 220 25 mH 100
(b) Find r.m.s. value of voltage given in diagram.
T4
t
VV0
T
(c) A particle and an electron are moving such that the velocity of particle is ree times that of electron.
If ratio of de-Broglie wave length of particle with respect to electron is 1.8 104. Find mass of the
particle (me = 9.1 103 kg).
1.8 10 (me =
9.1 103 )
Sol. (a) VL = 2 VR XL = 2 I R
r L = 2R r2RL
rad s
frequency 31 2R 1 2 100
f 1273.92 2 L 2 3.14 25 10
/s
(b) Root m an square value of voltage Vrms =
T2
0T
0
V dt
dt=
T / 420
0T
0
V dt
dt=
20 2
0 0
TV
V V4T 4 2
(c) de-Broglie wavelength h
mv
Let de-Broglie wavelength : of particle = p
of electron = e
p e e
e p p
m vm v r 1.8 10
4 = 31
e
p e
v9.1 10m 3v r mp =
3127
4
9.1 101.67 10
1.8 3 10kg
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www.
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A particle and an electron are moving such that the velocity of particle is ree times that of electron.
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A particle and an electron are moving such that the velocity of particle is ree times that of electron.
If ratio of de-Broglie wave length of particle with respect to electron is 1.8 10
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If ratio of de-Broglie wave length of particle with respect to electron is 1.8 104
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4If ratio of de-Broglie wave length of particle with respect to electron is 1.8 104If ratio of de-Broglie wave length of particle with respect to electron is 1.8 10
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If ratio of de-Broglie wave length of particle with respect to electron is 1.8 104If ratio of de-Broglie wave length of particle with respect to electron is 1.8 10
1.8 10
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1.8 10
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X
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XL
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L = 2
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= 2 I
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I
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2R
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2RL
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Lrad s
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rad s
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1 2R 1 2 100
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1 2R 1 2 1002 2 L 2 3.14
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2 2 L 2 3.14
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Root m an square value of voltage
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Root m an square value of voltage
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(c) de-Broglie wavelength www.
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de-Broglie wavelength
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9. (a) The graph between the stopping potential and frequency of the incident radiation is shown in figure.
Calculate :
(i) Plank's constant. (ii) Work function.
(i) (ii)
V0
1 5
1.656
(in 110 Hz)14
(in v
olt)
(b) 6C11 undergoes a decay by emitting +. Write the complete equation for this nuclear reaction.
Also calculate the Q-value of reaction.
Given the mass value of : m ( 6C11) = 11.011434 u
m ( 5B11) = 11.009305 u
me = 0.000548 u and
1 u = 931.5 MeV c2
6C11 +
Q-
m ( 6C11) = 11.011434 u
m ( 5B11) = 11.009305 u
me = 0.000548 u
1 u = 931.5 MeV/c2
Sol. (a) (i) By Einstein's photo electric equation
eV0 = h h 0
r19
33 34014
0
eV 1.6 10 1.656 1.6 1.656h 10 6.62 10 J s
( ) 4(5 1) 10
(i ) Work function 0 = h 0 = 6.62 1034 1 1014 = 6.62 1020 J
= 20
19
6.62 10eV 0.414eV
1.6 10
(b) 6C11
5B11 + +1
0 + + Q
Q-value of reaction = mc2
= [m( 6C11) 6me m( 5B
11) + 5me me]c2 = [m( 6C
11) m( 5B11) 2me]c
2
= [11.011434 11.009305 2 0.000548] uc2
= [0.001033] uc2 = 0.001033 931.5 MeV = 0.962 MeV
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. Write the complete equation for this nuclear reaction.
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. Write the complete equation for this nuclear reaction.
) = 11.009305 u
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) = 11.009305 u
and
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and
1 u = 931.5 MeV c
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1 u = 931.5 MeV c2
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2
m (
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m ( 6
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6C
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C11
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11
m (
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m (
By Einstein's photo electric equation
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By Einstein's photo electric equation
eV
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eV0
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0 = h
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= h
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h
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h
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0
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0
r
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r h
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h
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Work function
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10. (a) For given transistor circuit, the base current is 10 A and the collector current is 5.2 mA. Can this transistorcircuit be used as an amplifier ? Your answer must be supported with proper explation.
10 A 5.2 mA
5.5V
R =500kB R =1kC
VBEVCE VCC
(b) For a common emitter amplifier, current gain is 69. If the emitter current s 7 mA then calculate thebase current and collector current.
69 7 mA
Sol. (a)
VBE = 5.5 IB RB = 5.5 10 106 500 103 = 0.5 V
VCE = 5.5 IC RC = 5.5 5.2 103 1 103 = 0.3 V
Hence the emitter-base junction and ase- ollector junction both are forward bias.
Transistor can be used as an amplifier only if emitter-base junction is forward biased and base-collectorjunction is reverse biased
So this circuit can no be u ed as an amplifier.
(b) For CE amplifier :
Cur e t gain = CB
69I
Ir IC = 69 IB
IE = IB + IC = 7 mA
IE = IB + 69 IB r B7mA70
I = 0.1 mA
IC = 69 0.1 = 6.9 mA
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For a common emitter amplifier, current gain is 69. If the emitter current s 7 mA then calculate the
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For a common emitter amplifier, current gain is 69. If the emitter current s 7 mA then calculate the
7 mA
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7 mA
500 10
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500 103
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3 = 0.5 V
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= 0.5 V
3
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3 1 10
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1 103
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3 = 0.3 V
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= 0.3 V
Hence the emitter-base junction and ase- ollector junction both are forward bias.
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Hence the emitter-base junction and ase- ollector junction both are forward bias.
Transistor can be used as an amplifier only if emitter-base junction is forward biased and base-collector
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Transistor can be used as an amplifier only if emitter-base junction is forward biased and base-collectorjunction is reverse biased
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junction is reverse biased
So this circuit can no be u ed as an amplifier.
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So this circuit can no be u ed as an amplifier.
For CE amplifier :
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For CE amplifier :
Cur e t gain =
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Cur e t gain = I
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I
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