Neet 2013 Physics

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TARGET Publicationshttp://www.targetpublications.org NEET UG 2013 Physics Question Paper

TestBooklet

Code WNEET UG 2013, Physics Question Paper

(Date of Examination: 05-05-2013)

Class UnitNo.

Unit name Question No. Total No. ofQuestions

XI NEET

I Physical World and Measurement 1 1

II Scalars and Vectors

III Motion in One Dimension 3 1

IV Laws of Motion 4, 5 2

V Motion in Two Dimensions 2 1

VI Work, Energy and Power 6, 7 2

VII Rotational Motion 8, 9 2

VIII Gravitation 10, 11 2

IX Elasticity 12 1

X ViscosityXI Surface Tension 13 1

XII Heat 14, 15 2

XIII Thermodynamics 16, 17 2

XIV Kinetic Theory of Gases 18, 19 2

XV Oscillations

XVI Wave Mechanics 20, 21, 22 3

Total = 22

XII NEET

I Electrostatics 23, 24 2

II Current Electricity 25, 26, 27 3

III Magnetic effect of electric current 28, 29 2

IV Magnetism 30 1

V Electromagnetic induction and Alternating current 31, 32 2

VI Electromagnetic waves

VII Ray Optics 39, 40 2

VIII Wave Optics 41, 42 2

IX Interference of light

X Diffraction and Polarisation of light

XI Dual nature of matter and Radiation 37, 38 2

XII Atoms and Nuclei 33, 34, 35, 36 4XIII Electronic Devices 43, 44, 45 3

Total = 23

Total = 45


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NEET UG 2013 Physics Question Paper

TestBooklet

Code W TARGET Publicationshttp://www.targetpublications.org1. In an experiment four quantities a, b, c and d are measured with percentage error 1%, 2%, 3% and 4%

respectively. Quantity P is calculated as follows :

P =3 2a b

cd% error in Pis

(A) 14% (B) 10% (C) 7% (D) 4%

1. (A)

1. Given that: P =3 2a b

cd

error contributed by a = 3 a

100a

= 3 1%= 3%

error contributed by b = 2 b

100b

= 2 2%

= 4%

error contributed by c =c

100c

= 3%

error contributed by d =d

100d

= 4%

Percentage error in P is given as,p

100p

=(error contributed by a)+(error contributed by b)+(error contributed by c)+(error contributed by d)

= 3% + 4% + 3% + 4%= 14%

2. The velocity of a projectile at the initial point A is ( ) 2i 3j+ m/s. Its velocity (in m/s) at point B is

(A) 2i 3j (B) 2i + 3j (C) 2i 3j (D) 2i + 3j

2. (C)

2.

Horizontal (X) component remains the same while the vertical (Y) component changes.

Therefore, velocity at B = ( ) 2i 3j m/s

Y

A XB

3j

2i

2i

3j


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TestBooklet

Code W3. A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 seconds, the next 5 seconds

and the next 5 seconds respectively. The relation between h1, h2 and h3 is

(A) h1 = 2h2 = 3h3 (B) h1 = 2h

3= 3

h

5(C) h2 = 3h1 and h3 = 3h2 (D) h1 = h2 = h3

3. (B)3. At point A, u = 0

h1 =12

gt2 = 12

10 25

h1 = 125 mNow, v = u + gt = 0 + 10(5)

v = 50 m/sAt point B, final velocity from A to B = initial velocity at B

h2 = ut + gt2 = 50 5 +

1

2 10 25

Now, h2 = 375 mv = u + gt = 50 + 10(5)

v = 100 m/s

Similarly, At point C, we get,h3 = 625 m

h1 : h2 : h3 = 125 : 375 : 625= 1 : 3 : 5

i.e., h1 =2h

3= 3

h

5

4. Three blocks with masses m, 2m and 3m are connected by strings, as shown in the figure. After an upwardforce F is applied on block m, the masses move upward at constant speed v. What is the net force on theblock of mass 2 m? (g is the acceleration due to gravity)

(A) Zero (B) 2 mg (C) 3 mg (D) 6 mg4. (A)4. Since all three blocks are moving up with a constant speed v, acceleration a is zero.

We know, F = ma

F = 0 [ a = 0] net force is zero.

5. The upper half of an inclined plane of inclination is perfectly smooth while lower half is rough. A blockstarting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of frictionbetween the block and lower half of the plane is given by

(A) =1

tan(B) =

2

tan(C) = 2 tan (D) = tan

5. (C)5. We know that, v2 = u2 + 2as .(1)

Now, initial velocity at midpoint u =L

2g sin2

And final velocity for the lower half = v = 0

m

2 m

3 m

F v

u = 0

t = 5st = 5s

t = 5s

h1

A

B

C

D

h2

h3


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TestBooklet

Code W TARGET Publicationshttp://www.targetpublications.orgAt lower half acceleration = g sin g cos and s =

L

2

From equation (1),

02 2gL

2sin = 2[g sin g cos ]

L

2

2g L2

sin = gL sin gL cos

2gL sin = gL cos

= 2 tan

6. A uniform force of ( ) 3i j+ newton acts on a particle of mass 2 kg. Hence the particle is displaced from

position ( ) 2i k+ metre to position ( ) 4i 3j k + metre. The work done by the force on the particle is

(A) 9 J (B) 6 J (C) 13 J (D) 15 J

6. (A)

6. F

= ( ) 3i j+

S

= 2 1r r

= 2i 3j 2k +

We know,

W = F

S

= ( ) 3i j+ 2i 3j 2k + = 6 + 3 + 0

W = 9 J

7. An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to eachother. The first part of mass 1 kg moves with a speed of 12 ms1 and the second part of mass 2 kg moveswith 8 ms1 speed. If the third part flies off with 4 ms1 speed, then its mass is

(A) 3 kg (B) 5 kg (C) 7 kg (D) 17 kg

7. (B)

7. From law of conservation of momentum,

1 2 3P P P

+ + = 0

Let 1P

and 2P

go off at right angles to each other.

3P

= 2 21 2P P+

m3 4 =2 2(1 12) (2 8) +

= 2 212 16+

= 20

m3 =20

4

m3 = 5 kg

L/2

L/2


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TestBooklet

Code W8. A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to

point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is

(A)3g

2L(B)

g

L(C)

2g

L(D)

2g

3L

8. (A)

8. For the rod PQ,2ML

3 = T

L

2

Now, T = Mg

2ML

3 = Mg

L

2

=3g

2L

9. A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a

maximum height of23v

4gwith respect to the initial position. The object is

(A) Ring (B) Solid sphere (C) Hollow sphere (D) Disc

9. (D)9. Velocity of the small object is given as,

v =2

2

2gh

k1

r+

v2 =2

2

2

2g3v

k4g 1

r

+

1 +2

2

k

r=

3

2

k2 =1

2r2

But k =I

M

I

M= 2

1r

2

I =1

2Mr2 disc

PQ

L

Q

Mg

PL/2


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TestBooklet

Code W TARGET Publicationshttp://www.targetpublications.org10. A body of mass 'm' taken from the earth's surface to the height equal to twice the radius (R) of the earth.

The change in potential energy of body will be

(A) mg2R (B)2

mgR3

(C) 3mgR (D)1

mgR3

10. (B)10. The change in potential energy is given as,

U = Uf Ui

=GMm

R 2R

+

GMm

R

=GMm

R

11

3

=2

3GMm

R

=2

3

2

GMm R

R

= 22 GM

mR3 R

U =2

3mgR

11. Infinite number of bodies, each of mass 2 kg are situated on x-axis at distance 1 m, 2 m, 4 m, 8 m, .....respectively, from the origin. The resulting gravitational potential due to this system at the origin will be

(A) G (B) 8

3G (C)

4

3G (D) 4G

11. (D)11. Gravitational potential is given as,

V =GM

R

V =1 1 1 1

GM ....1 2 4 8

+ + + + +

= G 22 3

1 1 11 ....

2 2 2

+ + + + +

= 2G1

11

2

V = 4G

12. The following four wires are made of the same material. Which of these will have the largest extensionwhen the same tension is applied?(A) Length = 50 cm, diameter = 0.5 mm (B) Length = 100 cm, diameter = 1 mm(C) Length = 200 cm, diameter = 2 mm (D) Length = 300 cm, diameter = 3 mm

12. (A)12. Youngs Modulus for a wire is given as,

Mg LY

L A=

L =Mg L

Y A

L L

A

Now,L

A

is maximum for L= 50cm and diameter = 0.5 mm.

Hence, option (A) is correct.


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TestBooklet

Code W13. The wettability of a surface by a liquid depends primarily on

(A) Viscosity (B) Surface tension(C) Density (D) Angle of contact between the surface and the liquid

13. (D)13. The wettability of a surface of liquid depends on angle of contact between surface and liquid.

14. The molar specific heats of an ideal gas at constant pressure and volume are denoted by Cp and Cvrespectively. If C and R is the universal gas constant, then Cv is equal to

(A)1

1

+

(B)

R

( 1)(C)

( 1)

R

(D) R

14. (B)14. Difference between two molar specific heat is given as,

CP Cv = R

p v

v v v

C C R

C C C =

1 =v

R

C

Cv =R

1

15. A piece of iron is heated in a flame. It first becomes dull red then becomes reddish yellow and finally turnsto white hot. The correct explanation for the above observation is possible by using(A) Stefan's Law(B) Wien's displacement Law(C) Kirchoff's Law(D) Newton's Law of cooling

15. (B)

15. From Wiens displacement law,

m1

T

m T = constant

16. A gas is taken through the cycle A B C A, as shown. What is the net work done by the gas?

(A) 2000 J (B) 1000 J (C) Zero (D) 2000 J16. (B)16. We know,

Work done = Area under P V curve.

3 51 5 10 4 102

=

= 10 102

W = 1000 J

765432

10

A

B

C

2 4 6 8V (103 m3)

P(105 Pa)


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TestBooklet

Code W TARGET Publicationshttp://www.targetpublications.org17. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature.

The ratio of p

v

C

Cfor the gas is

(A)4

3(B) 2 (C)

5

3(D)

3

2

17. (D)17. For an adiabatic process,

P T/1

Given that, P T3

1

= 3

= 3 3

2 = 3

=3

2

18. In the given (V T) diagram, what is the relation between pressures P1 and P2?

(A) P2 = P1 (B) P2 > P1

(C) P2 < P1 (D) Cannot be predicted

18. (C)

18. Assuming the graph for a gas of given mass, we have,

PV = n RT

VT

1P

From the graph,V

T= tan

1

P tan

as angle increases, tan increases and pressure dicreases.

P1 > P2

19. The amount of heat energy required to raise the temperature of 1 g of Helium at NTP, from T1K to T2K is

(A)3

8NakB(T2 T1) (B)

3

2NakB(T2 T1)

(C)3

4NakB(T2 T1) (D)

3

4NakB

2

1

T

T

19. (A)

VP2

P1

T

21


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Code W19. Amount of energy required is given as,

( )2 1f f

E n RT NK T T2 2

= =

E = ( ) ( )A B 2 1f

n N k T T2

Where N = n. NAkB = Boltzmann constant.

E = ( )A B 2 13

n N K T T2

[ f = 3 for He ]

Now,m 1

nM 4

= =

E = 3 12 4

( )A B 2 1N k T T = ( )A B 2 13

N k T T8

20. A wave travelling in the +ve x-direction having displacement along y-direction as 1 m, wavelength 2 m

and frequency of1

Hz is represented by

(A) y = sin(x 2t) (B) y = sin(2x 2t)(C) y = sin(10x 20t) (D) y = sin(2x + 2t)

20. (A)

20. y = a sin (kxt) .(1)

Now, k =2 2

2

=

= 2 .v =1

2 2 =

.

a = 1 m.Substituting these in equation (1),

y = 2sin x 2t2

y = sin [x 2t]

21. If we study the vibration of a pipe open at both ends, then the following statement is not true(A) Open end will be anti-node.(B) Odd harmonics of the fundamental frequency will be generated.(C) All harmonics of the fundamental frequency will be generated.(D) Pressure change will be maximum at both ends.

21. (D)21. The air column in a pipe open at both ends can vibrate in a number of different modes subjected to the

boundary condition that there must be an antinode at the open end.Hence option (A) is correct.The ratio of frequency when pipe is open at both the ends is given as,

: 2 : 3 : 4 : 5

where =v

2L

Both odd as well even i.e., All harmonics are present.Hence, option (B) and (C) are correctPressure variation is minimum at antinode

option (D) is incorrect.


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TestBooklet

Code W TARGET Publicationshttp://www.targetpublications.org22. A source of unknown frequency gives 4 beats/s, when sounded with a source of known frequency 250 Hz.

The second harmonic of the source of unknown frequency gives five beats per second, when sounded with asource of frequency 513 Hz. The unknown frequency is

(A) 254 Hz (B) 246 Hz (C) 240 Hz (D) 260 Hz22. (A)

22. a = 250 4 = 254 Hz or 246 Hzb = 513 5 518 Hz or 508 HzNow, b = 2aWhich is 508 = 2(254)

= 254 Hz

23. Two pith balls carrying equal charges are suspended from a common point by strings of equal length, theequilibrium separation between them is r. Now the strings are rigidly clamped at half the height. Theequilibrium separation between the balls now become

(A)2

1

2

(B)

3

r

2(C)

2r

3

(D)

2r

3

23. (B)23. In the equilibrium position,

T cos = mg

T sin = Fe =2

2

Kq

r

tan =2

2Kqr mg

i.e.,2

2

r / 2 Kq

y mgr =

y =3

2

mgr

2Kq

r y1/3Now, the equilibrium separation for (y/2) is,

r1/3

y

2

( )

1/3

r

2

24. A, B and C are three points in a uniform electric field. The electric potential is

(A) Maximum at A (B) Maximum at B(C) Maximum at C (D) Same at all the three points A, B and C

24. (B)

y

y/2rr

AB

CE

T T cos

T sin

mg

Fe


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Code W24.

dVE

dr

=

i.e., electric field is directed along decreasing potential.

VB > VC > VP Potential is maximum at B.

25. A wire of resistance 4 is stretched to twice its original length. The resistance of stretched wire would be

(A) 2 (B) 4 (C) 8 (D) 16 25. (D)

25. Let R be the resistance and l be the original length.

At constant volume,

Rl2

Resistance of stretched wire is,

R = 4R= 4(4)

R = 16

26. The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 is

(A) 0.2 (B) 0.5 (C) 0.8 (4) 1.0 26. (B)

26. I =E

R r+

0.2 =2.1

10 r+

2 + 0.2r = 2.1

0.2r = 0.1

r = 0.5

27. The resistances of the four arms P, Q, R and S in a Wheatstone's bridge are 10 ohm, 30 ohm, 30 ohm and 90ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If thegalvanometer resistance is 50 ohm, the current drawn from the cell will be

(A) 1.0 A (B) 0.2 A (C) 0.1 A (D) 2.0 A

27. (B)

27. For a Balanced wheatstones Bridge,

P

Q

=R

S

Equivalent resistance = Req =(10 30) (30 90)

(10 30 30 90)

+ +

+ + +

=40 120

160

= 30 Now, Reff= 30 + 5 = 35r

Now, I =eff

V

R=

7

35=

1

5= 0.2 A


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TestBooklet

Code W TARGET Publicationshttp://www.targetpublications.org28. When a proton is released from rest in a room, it starts with an initial acceleration a0 towards west. When it

is projected towards north with a speed v0 it moves with an initial acceleration 3a0 toward west. The electricand magnetic fields in the room are

(A) 0ma

ewest, 0

0

2ma

evup (B) 0

ma

ewest, 0

0

2ma

evdown

(C) 0ma

e east,0

0

3ma

ev up (D)0

ma

e east,0

0

3ma

ev down

28. (B)

28. Electric field =Force

Charge

= 0ma

e(in west direction)

Magnetic force = Fm = 3ma0 ma0= 2ma0 (in west direction)

v

B

is directed towards west.

Since, v

is directed towards north for positive charge, B

is directed vertically down.Now, mF

= q v

B

2ma0 = ev0 B

B = 00

2ma

ev(vertically down)

29. A current loop in a magnetic field(A) Experiences a torque whether the field is uniform or non uniform in all orientations(B) Can be in equilibrium in one orientation(C) Can be in equilibrium in two orientations, both the equilibrium states are unstable(D) Can be in equilibrium in two orientations, one stable while the other is unstable

29. (D)29.

When = 0 (parallel) it is in stable equilibrium.When = 160 (anti-parallel), it is in unstable equilibrium.

30. A bar magnet of length l and magnetic dipole moment Mis bent in the form of an arc as shown in figure.

The new magnetic dipole moment will be

(A) M (B)3

M (C)

2

M (D)

M

2

30. (B)

M

B

B

60 rr


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Code W30. L =

3 r

r =3L

M = m r

= m 3L

=3M

[M = mL]

31. A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced e.m.f. is(A) Once per revolution(B) Twice per revolution(C) Four times per revolution(D) Six times per revolution

31. (B)

31.

e = NAB sinte changes direction twice per revolution.

32. A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulbdecreases when(A) Frequency of the AC source is decreased

(B) Number of turns in the coil is reduced(C) A capacitance of reactance XC = XL is included in the same circuit(D) An iron rod is inserted in the coil

32. (D)32. Impedance is given as,

Z =2 2

L

2 2

R X

R (L 2f )

+

+

If frequency is decreased, impedance decreases.If number of turns decreases, self inductance decreases and thus impedance decreases.At resonance, XC = XL and impedance decreases.When iron rod is inserted, impedance increases and hence current decreases.

33. The condition under which a microwave oven heats up a food item containing water molecules mostefficiently is(A) The frequency of the microwaves must match the resonant frequency of the water molecules(B) The frequency of the microwaves has no relation with natural frequency of water molecules(C) Microwaves are heat waves, so always produce heating(D) Infra-red waves produce heating in a microwave oven

33. (A)33. In presence of microwave, water molecules oscillate in an electric field of microwave resulting in

generation of heat. Amplitude of oscillation will be maximum when frequency of microwave match theresonant frequency of water molecules.

60 rr

r

t

e

+


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Code W TARGET Publicationshttp://www.targetpublications.org34. Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is

(A)5

27(B)

3

23(C)

7

29(D)

9

31

34. (A)34. For Lyman series,

1l

= R(1)22 21 11 2

= 3R4

l =4

3R

For Balmer series,

1

= R(1)2

2 2

1 1

2 3

=5R

36

b =36

5R

l

=4

3R5R

36 =5

3 9 =5

27

35. The half life of a radioactive isotope X is 20 years. It decays to another element Y which is stable. Thetwo elements X and Y were found to be in the ratio 1 : 7 in a sample of a given rock. The age of the rockis estimated to be

(A) 40 years (B) 60 years (C) 80 years (D) 100 years

35. (B)

35.0

N

N=

1

(1 7)+=

1

8=

3

1

(2)

n = 3n 3

1 12 (2)

=

n =t

T

t = 3 20 [Half life of X = T = 20 years] t = 60 years

36. A certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in fusionreaction is 0.02866 u. The energy liberated per u is (given 1 u = 931 MeV)

(A) 2.67 MeV (B) 26.7 MeV (C) 6.675 MeV (D) 13.35 MeV

36. (C)36. Mass defect = m = 0.02866 u

Total energy = E = mc2 = 0.02866 931 MeV

= 26.68 MeV

Energy liberated per nucleon =E

A

=26.68

4

= 6.675 MeV

n = 4

n = 3

n = 1

Longest wavelengthcorresponding to BalmerLongest wavelengthcorresponding to Lyman

n = 2


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Code W37. For photoelectric emission from certain metal the cut-off frequency is . If radiation of frequency 2

impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is theelectron mass)

(A)h

(2m)(B)

h

m(C)

2h

m(D) 2

h

m

37. (C)37. Cut off frequency is given as Work function = hNow, E = K.E. +

2h =1

2mv2 + h

1

2mv2 = 2h h

1

2mv2 = h

v =2h

m

38. The wavelength eof an electron and pof a photon of same energy Eare related by

(A) p2e (B) pe (C) p e (D) p

e

1

38. (A)38. de Broglies wavelength for an electron,

e =h

2mE

i.e., e1

Eor 2

e

1

E.(1)

wavelength of photon is given as,

p =hc

E

i.e., p1

E. (2)

From (1) and (2), we have,2e p

39. A plano-convex lens fits exactly into a planoconcave lens. Their plane surfaces are parallel to each other. Iflenses are made of different materials of refractive indices 1 and 2 and R is the radius ofcurvature of thecurved surface of the lenses, then the focal length of the combination is

(A)1 2

R

2( )+(B)

1 2

R

2( )(C)

1 2

R

( )(D)

2 1

2R

( )

39. (C)

39. Focal length of first lens,1

1

f= (1 1)

1 1

R

= 1

1

R

Focal length of second lens,2

1

f= (2 1)

1 1

R

=

( )1 1

R

So focal length of the combination,


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Code W TARGET Publicationshttp://www.targetpublications.org

1

f=

1

1

f+

2

1

f= 1

1

R

( )2 1

R

1

f= 1 2

R

f = 1 2

R

40. For a normal eye, the cornea of eye provides a converging power of 40 D and the least converging power ofthe eye lens behind the cornea is 20 D. Using this information, the distance between the retina and thecornea - eye lens can be estimated to be(A) 5 cm (B) 2.5 cm (C) 1.67 cm (D) 1.5 cm

40. (C)40. lens + cornea forms an image of distance object at retina.

converging power (40+20) D = 60 DFrom Lens equation,

1 1 60

v 100 =

v =5

3cm

v = 1.67 cm.

41. In Youngs double slit experiment, the slits are 2 mm apart and are illuminated by photons of twowavelengths 1 = 12000 and 2 = 10000 . At what minimum distance from the common central brightfringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with abright fringe from the other?(A) 8 mm (B) 6 mm (C) 4 mm (D) 3 mm

41. (B)

41. 1 1 2 2n D n Dd d =

1

2

n

n= 2

1

=10000A

12000A

=5

6

n1 = 5 and n2 = 6

Therefore, x = 1 1n D

d

=

10

3

5 12000 10 2

2 10

= 6 103 m = 6 mm

42. A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen isplaced at a large distance from the slit. If the speed of the electrons is increased, which of the followingstatements is correct?

(A) Diffraction pattern is not observed on the screen in the case of electrons.(B) The angular width of the central maximum of the diffraction pattern will increase.(C) The angular width of the central maximum will decrease.(D) The angular width of the central maximum will be unaffected.

42. (C)

42. Now, =h

mv

1

v

Therefore, as speed of electron increases, its de-Broglie wavelength decreases.


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Code WAngular width for central maximum is given as,

=2

d

1

v

if speed of electron increases, angular width of central maximum will decreases.

43. In a n-type semiconductor, which of the following statement is true?(A) Electrons are majority carriers and trivalent atoms are dopants(B) Electron are minority carriers and pentavalent atoms are dopants(C) Holes are minority carriers and pentavalent atoms are dopants(D) Holes are majority carriers and trivalent atoms are dopants

43. (C)43. In N-type semiconductors, minority carriers are holes, majority carriers are electrons and pentavalent atoms

are dopants.

44. In a common emitter (CE) amplifier having a voltage gain G, the transistor used has transconductance 0.03mho and current gain 25. If the above transistor is replaced with another one with transconductance 0.02mho and current gain 20, the voltage gain will be

(A)2

3G (B) 1.5 G (C)

1

3G (D)

5

4G

44. (A)

44. Av = oin

R

R

= C

B

I

I o

in

R

R

= c o

in

I R

V

= gm Ro

Av gm.

V1V2

A

A= m1

m2

g

g=

0.03

0.02=

3

2

V2A =2

3V1

A =2

3G.

45. The output (X) of the logic circuit shown in figure will be

(A) X = A.B (B) X = A.B (C) X = A.B (D) X = A B+ 45. (C)45.

A B Y X

0 0 1 0

0 1 1 0

1 0 1 0

1 1 0 1

X = A B = AB

B

AX

y = A.B = A.B

B

A A B

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Neet 2013 Physics