aieee-2011 solution and analysis

8/6/2019 AIEEE-2011 Solution and Analysis

http://slidepdf.com/reader/full/aieee-2011-solution-and-analysis 1/26

12th

60%

11th

40%

1 1 t h

3 0. 0 0%

12th

70.00%

11th

53.33%

12th

46.67%

CA37%

CO10%

TRI3%

ALG

27%

OTH

10%

Vec

13%CA

22.5%

TRI8.75%

ANALYSIS OF 2011 AIEEE PAPERTYPE OF QUESTION SINGLE CHOICE (+4,–1) TOTAL MARKS

PHYSICS 30 120

CHEMISTRY 30 120

MATHEMATICS 30 120

CHAPTER %

Mechanics 16.6

Electrostatics 6.6Optics 10

Heat 13.3

GPM 6.6

SHM/Waves 10

Gravitation 3.3

Modern Physics 10

Semiconductor/Communication 3.3

Capacitor 3.3

CE/MEC/EMI/AC 13.3

Measurement and Error 3.3



CODE (P) PAPER and SOLUTION [CHEMISTRY, PHYSICS AND MATHEMATICS] AIEEE - 01 - May - 2011

[2]

CHEMISTRY

Single Correct Choice Type

This section contains 30 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY

ONE is correct.

01. The presence or absence of hydroxy group on which carbon atom of sugar differentiates RNA and DNA ?

(1) 1st (2) 2nd (3) 3rd (4) 4th

Sol.01. (2)

02. Among the following the maximum covalent character is shown by the compound :

(1) FeCl2

(2) SnCl2

(3) AlCl3

(4) MgCl2

Sol.02. (3)

03. Which of the following statement is wrong ?

(1) The stability of hydrides increases from NH3

to BiH3in group 15 of the periodic table

(2) Nitrogen cannot form d– p bond.

(3) Single N – N bond is weaker than the single P – P bond.

(4) N2O

4has two resonance structures.

Sol.03. (1)

04. Phenol is heated with a solution of mixture of KBr and KBrO3. The major product obtained in the above reaction is :

(1) 2 - Bromophenol (2) 3 - Bromophenol (3) 4 - Bromophenol (4) 2,4,6 -Tribromophenol

Sol.04. (4)

05. A 5.2 molal aqueous solution of methy1 alcohol, CH3OH, is supplied. What is the mole fraction of methyl alcohol in the solution ?

(1) 0.100 (2) 0.190 (3) 0.086 (4) 0.050

Sol. 5. (3)3CH OH

5.20.086

5.2 55.56

.

06. The hybridisation of orbitals of N atom in3 2 4NO , NO , NH are respectively :

(1) sp, sp2, sp3 (2) sp2, sp, sp3 (3) sp, sp3, sp2 (4) sp2, sp3, sp

Sol.06. (2)

07. Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent

it from freezing at – 6°C will be : (Kf for water = 1.86 K kg mol–1, and molar mass of ethylene glycol = 62g mol–1)

(1) 804.32 g (2) 204.30 g (3) 400.00 g (4) 304.60 g

Sol. 7. (1)W

6 1.864 62

4 62 6W 804.32g

1.86




[3]

08. The reduction potential of hydrogen half-cell will be negative if :

(1) p (H2) = 1 atm and [H+] = 2.0 M (2) p (H

2) = 1 atm and [H+] = 1. 0 M

(3) p (H2) = 2 atm and [H+] = 1.0 M (4) p (H

2) = 2 atm and [H+] = 2.0 M

Sol. 8. (3)2H

10 2

P0.0591E log

2 [H ]

for E < O ;2H

2

P1

[H ]

2HP 2atm & [H+] = 1.0 M.

09. Which of the following reagents may be used to distinguish between phenol and benzoic acid ?

(1) Aqueous NaOH (2) Tollen’s reagent (3) Molisch reagent (4) Neutral FeCl3

Sol.09. (4)

10. Tricholoroacetaldehyde was subjected to Cannizzaro’s reaction by using NaOH. The mixture of the products contains sodium

trichloroacetate and another compound. The other compound is :

(1) 2, 2, 2-Trichloroethanol (2) Trichloromethanol (3) 2, 2, 2-Trichloropropanol (4) Chloroform

Sol.10. (1) 2, 2, 2-Trichloroethanol is the answer but chloroform is also the product.

11. Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides ?

(1) Al2O

3< MgO < Na

2O < K

2O (2) MgO < K

2O > Al

2O

3< Na

2O

(2) Na2O < K

2O < MgO < Al

2O

3(4) K

2O < Na

2O < Al

2O

3< MgO

Sol.11. (1)

12. A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emission is at 680 nm, the other is at :

(1) 1035 nm (2) 325 nm (3) 743 nm (4) 518 nm

Sol. 12. (3)1 1 1

355 680

1 1 1

355 680

355 × 680 241400nm nm

680 355 325

= 742.77 nm 743 nm.

13. Which of the following statements regarding sulphur is incorrect ?

(1) S2molecule is paramagnetic

(2) The vapour at 200°C consists mostly of S8rings

(3) At 600° C the gas mainly consists of S2molecules

(4) The oxidation state of sulphur is never less than +4 in its compounds

Sol.13. (4)




[4]

14. The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm3 to a

volume of 100 dm3 at 27°C is

(1) 38.3 J mol–1 K–1 (2) 35.8 J mol–1 K–1 (3) 32.3 J mol–1 K–1 (4) 42.3 J mol–1 K–1

Sol. 14. (1) 2

1

VS nR n

V

S = 2 × 8.314 × 2.303 × log10

100

10

S 38.3 JK–1 mol–1

15. Which of the following facts about the complex [Cr(NH3)

6]Cl

3is wrong ?

(1) The complex involves d2sp3 hybridisation and is octahedral in shape

(2) The complex is paramagnetic

(3) The complex is a outer orbital complex

(4) The complex gives white precipitate with silver nitrate solution

Sol.15. (3)

16. The structure of IF7

is

(1) square pyramid (2) Trigonal bipyramid (3) octahedral (4) pentagonal bipyramid

Sol.16. (4)

17. The rate of a chemical reaction doubles for every 10°C rise of temperature. If the temperature is raised by 50°C, the rate of the

reaction increases by about

(1) 10 times (2) 24 times (3) 32 times (4) 64 times

Sol. 17. (3)

50

510f

i

R2 2 32

R

18. The strongest acid amongst the following compounds is

(1) CH3COOH (2) HCOOH (3) CH

3CH

2CH(Cl)CO

2H (4) ClCH

2CH

2CH

2COOH

Sol.18. (3)

19. Identify the compound that exhibits tautomerism.

(1) 2-Butene (2) Lactic acid (3) 2-Pentanone (4) Phenol

Sol.19. (3) 2-Pentanone is the answer but phenol also exhibits tautomerism.20. A vessel at 1000 K contains CO

2with a pressure of 0.5 atm. Some of the CO

2is converted into CO on the addition of graphite.

If the total pressure at equilibrium is 0.8 atm, the value of K is :

(1) 1.8 atm (2) 3 atm (3) 0.3 atm (4) 0.18 atm

Sol. 20. (1) CO2(g)

+ C(s)

2CO(g)

t = 0 0.5 – 0

At eq. 0.5 – x – 2 x

PT, eq

= 0.8 atm = 0.5 + x

x 0.3

2

1

(2 0.3)K

(0.5 0.3)

= 1.8 atm.




[5]

21. In context of the lanthanoids, which of the following statements is not correct ?

(1) There is a gradual decrease in the radii of the members with increasing atomic number in the series.

(2) All the members exhibit + 3 oxidation state.

(3) Because of similar properties the separation of lanthanoids is not easy.

(4) Availability of 4f electrons results in the formation of compounds in + 4 state for all the members of the series.

Sol.21. (4)

22. ‘a’ and ‘b’ are van der Waals’ constants for gases. Chlorine is more easiliy liquefied than ethane because

(1) a and b for Cl2> a and b for C

2H

6(2) a and b for Cl

2< a and b for C

2H

6

(3) a for Cl2< a for C

2H

6but b for Cl

2> b for C

2H

6(4) a for Cl

2> a for C

2H

6but b for Cl

2< b for C

2H

6

Sol.22. (4) Greater is “a” easier it is to liquify the gas and greater is “ b” , larger is the size of molecule.

23. The magnetic moment (spin only) of [NiCl4]2– is :

(1) 1.82 BM (2) 5.46 BM (3) 2.82 BM (4) 1.41 BM

Sol.23. (3)

24. In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom

of B is missing from one of the face centred points, the formula of the compound is :

(1) A2B (2) AB

2(3) A

2B

3(4) A

2B

5

Sol. 24. (4) A B

18 1

8

1 55

2 2

2 5

Empirical formula = A2B

5

25. The outer electron configuration of Gd (Atomic No. : 64) is :

(1) 4f 3 5d5 6s2 (2) 4f 8 5d0 6s2 (3) 4f 4 5d4 6s2 (4) 4f 7 5d1 6s2

Sol.25. (4)

26. Boron cannot form which one of the following anions ?

(1) 3

6BF (2) BH (3) B(OH) (4)

2BO

Sol.26. (1)

27. Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of :

(1) two ethylenic double bonds (2) a vinyl group

(3) an isopropyl group (4) an acetylenic triple bond

Sol.27. (2)

28. Sodium ethoxide has reacted with ethanoyl chloride. The compound that is produced in the above reaction is

(1) Diethyl ether (2) 2-Butanone (3) Ethyl chloride (4) Ethyl ethanoate

Sol.28. (4)

29. The degree of dissociation () of a weak electrolyte, AxB

yis related to van’t Hoff factor (i) by the expression ;

(1)i 1

(x y 1)

(2)i 1

x y 1

(3)x y 1

i 1

(4)

x y 1

i 1

Sol.29. (1)i 1

(x y 1)

30. Silver Mirror test is given by which one of the following compounds ?

(1) Acetaldehyde (2) Acetone (3) Formaldehyde (4) Benzophenone

Sol.30. (1)




[6]

PHYSICS


This section contains 30 multiple choice questions. Each question has four choices (A), (B), (3)and (4)out of which ONLY

ONE is correct.

31. 100 g of water is heated from 30°C to 50°C. Ignoring the slight expansion of the water, the change in its internal energy is(specific heat of water is 4184 J/kg/K):

(1) 4.2 kJ (2) 8.4 kJ (3) 84 kJ (4) 2.1 kJ

Sol.31. (2)

Q = U + W W 0 U = Q = mc

U = 100 × 10–3 × 4184 × 20 = 8368 U = 8.4 kJ

32. The half life of a radioactive substance is 20 minutes. The approximate time interval (t2– t

1) between the time t

2when

2

3of it

has decayed and time t1

when1

3of it had decayed is:

(1) 7 min (2) 14 min (3) 20 min (4) 28 min

Sol.32. (3)

N = N0e–t

1t

0 0

2N N e

3

2t00

NN e

3

1 2t t2 e e

1 2t t2 e

n2 = (t2– t

1)

2 1 1/ 2

n2t t t

therefor, t2– t

1= 20 min.

33. A mass M, attached to a horizontal spring, executes S.H.M. with amplitude A1. When the mass M passes through its mean

position then a smaller mass m is placed over it and both of them move together with amplitude A2. The ratio of

1

2

A

A

is:

(1)M

M m(2)

M m

M

(3)

1/ 2M

M m

(4)

1/ 2M m

M

Sol. 33. (4)

(A1) M = A

2(M + m) (by conservation of linear momentum)




[7]

1

2

A M m

A M

1

2

k

A M mM m.

A Mk

M

1

2

A M m

A M

34. Energy required for the electron excitation in Li

++

from the first to the third Bohr orbit is :(1) 12.1 eV (2) 36.3 eV (3) 108.8 eV (4) 122.4 eV

Sol. 34. (3)

E = E3– E

1=

2 2z z

13.6 13.69 1

2

2

Zusing, E 13.6 eV

n

=2 1

13.6Z . 19

= 13.6 ×8

9× 9 E = 108.8 eV

35. The transverse displacement y (x, t) of a wave on a string is given by 2 2ax bt 2 abxt

y x, t e

This represents a :

(1) wave moving in +x direction with speeda

b(2) wave moving in –x direction with speed

b

a

(3) standing wave of frequency b (4) standing wave of frequency1

b

Sol. 35. (2)

2 2ax bt 2 a bxt

y x, t e

2ax bt

e

. wave moving in –ve x direction with velocityb

a.

36. A resistor ‘R’ and 2F capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb

that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10

2.5 = 0.4)

(1) 1.3 × 104 (2) 1.7 × 105 (3) 2.7 × 106 (4) 3.3 × 107

Sol.36. (3)

120 V

2 FR

B 120 = 200 (1 – e – t/Rc) 6

5

R 2 1012

1 e20

6

2.5R 0.4 2 10 R = 2.7 × 106




[8]

37. A current I flows in an infinitely long wire with cross section in the form of a semicircular ring of radius R. The magnitude of

the magnetic induction along its axis is :

(1)0

2

I

R

(2)

0

2

I

2 R

(3)

0I

2 R

(4)

0I

4 R

Sol.37. (1)

dB cos

dB sin

dB

/ 2 0

0

IRd

RB 2 sin2 R

0

2B . I

R

38. A Carnot engine operating between temperatures T1

and T2

has efficiency1

6. When T

2is lowered by 62 K, its efficiency

increases to

1

3 . Then T1 and T2 are, respectively :

(1) 372 K and 310 K (2) 372 K and 330 K (3) 330 K and 268 K (4) 310 K and 248 K

Sol.38. (1)

2

1

T1

T (T

2 Lower, TT

1 Upper)

2

1

T11

6 T ;

2 2

1 1 1

T 62 T1 621 1

3 T T T

;

1

1 1 62

3 6 T

1

62 1

T 6 ; T

1= 372 K

2

1

T 1 51

T 6 6

2 1

5 5T T 372 310K

6 6




[9]

39. An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by :

d2.5

dt

where is the instantaneous speed. The time taken by the object, to come to rest, would be :

(1) 1 s (2) 2 s (3) 4 s (4) 8s

Sol.39. (2)

dv2.5 v

dt

1/ 2dv 2.5 v dt

2.5t 2 2sec.

2.5

40. The electrostatic potential inside a charged spherical ball is given by 2ar b where r is the distance from the centre; a, b

are constant. Then the charge density inside the ball is:

(1)0

24 a r (2)0

6 a r (3)0

24 a (4)0

6 a

Sol.40. (4)

2ar b

dE

dr

0

rE 2ar

3

06a

41. A car is fitted with a convex side-view mirror of focal length 20 cm. A second car 2.8 m behind the first car is overtaking the first

car at a relative speed of 15 m/s. The speed of the image of the second car as seen in the mirror of the first one is:

(1)

1

m /s10 (2)

1

m / s15 (3) 10 m/ s (4) 15 m / s

Sol.41. (2)

2 2

I

0

V V f

V u f u

2

IV 20

15 m / s 20 280

2

I

20 1V 15 15 m / s

300 225 15

Ans. 2




[10]

42. If a wire is stretched to make it 0.1 % longer, its resistance will :

(1) increase by 0.05 % (2) increase by 0.2 % (3) decrease by 0.2 % (4) decrease by 0.05 %

Sol.42. (2)

2

RV

2R

R100 2 100

R

R100 2(0.1) 0.2 %

Ans. (2)

43. Three perfect gases at absolute temperatures T1, T

2and T

3are mixed. The masses of molecules are m

1, m

2and m

3and the

number of molecules are n1, n

2and n

3respectively. Assuming no loss of energy, the final temperature of the mixture is:

(1)1 2 3(T T T )

3

(2)

1 1 2 2 3 3

1 2 3

n T n T n T

n n n

(3)

2 2 2

1 1 2 2 3 3

1 1 2 2 3 3

n T n T n T

n T n T n T

(4)

2 2 2 2 2 2

1 1 2 2 3 3

1 1 2 2 3 3

n T n T n T

n T n T n T

Sol.43. (2)

1 1 2 2 3 3 1 2 3 m ix

3 3 3 3

n KT n KT n KT (n n n ) KT2 2 2 2

1 1 2 2 3 3

mix

1 2 3

n T n T n TT

n n n

44. Two identical charged spheres suspended from a common point by two massless strings of length are initially a distance

d(d ) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a

result the charges approach each other with a velocity v. Then as a function of distance x between them,

(1) 1/ 2

v x

(2) 1

v x

(3) 1/ 2

v x(4) v x

Sol.44. (1)

x/2 x/2

x / 2sin tan .




[11]

2

Kqq

x

mg

2

2

2

Tcos mg

kqqTsin

x

kqtan

mgx

2

2

kq x

2mgx

2 3mgq x

k2

2

2

dq dx mg dx 2 q mg2q 3x dq / dt

dt dt k2 dt 3 k2x

1/ 2

3

2

2 dq mg 1v x

3 dt 2 k x

. Here we are assuming velo constant 1/ 2v x

45. Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (surface tension of soap solution

10.03 Nm ) :

(1) 4 mJ (2)0.2 mJ (3) 2 mJ (4) 0.4 mJ

Sol.45. (4)

2 2

2 1W 8 (R R ) . S

48 (25 9) 10 0.03

0.7536 16 10

W 12.05 10

4N 3.84 10 0.38 mJ = 0.4 mJ

46. A fully charged capacitor C with initial charge q0is connected to a coil of self inductance L and t = 0. The time at which the

energy is stored equally between the electric and the magnetic fields is:

(1) LC (2) LC4

(3) 2 LC (4) LC

Sol.46. (2)

Ue= U

B

Tt

8 =

12 LC

8 = LC

4




[12]

47. Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where

the gravitational field is zero is:

(1) zero (2)4Gm

r (3)

6Gm

r (4)

9Gm

r

Sol.47. (4)

P

x

m 4m

r – x

netE 0 when

2

2 2

Gm G.4m 2GM

r xx r x

r xx

2

rx

3

P

Gm G4m Gm 9GmV 3 6

r / 3 2r / 3 r r

.

48. A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the

rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the

angular speed of the disc:

(1) remains unchanged (2)continuously decreases

(3) continuously increases (4) first increases and then decreases

Sol.48. (4)

I constant and I first then then .

49. Let the x-z plane be the boundary between two transparent media. Medium 1 is Z 0 has a refractive index of 2 and medium

2 with z < 0 has a refractive index of 3 . A ray of light in medium 1 given by the vector ˆ ˆ ˆA 6 3i 8 3j 10k

is incident on

the plane of separation. The angle of refraction in medium 2 is:

(1) 30° (2) 45° (3) 60° (4) 75°

Sol.49. (2)

As refractive indices are given in terms of Z 0 and Z 0. So we consider the separating surface as X-Y. So Z-axis will be

normal. Taking angle of incident ray with this normal.

ˆ ˆ ˆ ˆ6 3i 8 3j 10k . k 400 .1. cos cos

sin 3now,

sinr 2

/ 2 3

sinr 2

1sinr

2 r 45




[13]

50. Two particles are executing simple harmonic motion of the same amplitude A and frequency along the x-axis. Their mean

position is separated by distance 0 0X (X A). If the maximum separation between them is 0(X A) , the phase difference

between their motion is :

(1)2

(2)

3

(3)

4

(4)

6

Sol.50. (2)

0x x

1x Asin( t )

2 0x Asin( t) x

Separation between them2 1 0

x x x Asin t A sin( t ) 0

Its maximumvalue should be1

x A sin t sin t

let y sin( t) sin( t ) = y1

+ y2

2y

1y

1 2resul tan t of (y y ) must be 1

So

120

51. Direction :

The question has a paragraph followed by two statement, Statement -1 and Statement-2. Of the given four alternatives after

the statements, choose the one that describes the statements.

A thin air film is formed by putting the convex surface of a plane-convex lens over a plane glass plate. With monochromaticlight, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate)

surface of the film.

Statement -1 : When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of .

Statement -2 : The centre of the interference pattern is dark.

(1) Statement-1 is true, Statement-2 is flase

(2) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of Statement-1.

(3) Statement-1 is true, Statement-2 is true and Statement-2 is not the correct explanation of Statement-1.

(4) Statement-1 is flase, Statement-2 is true.




[14]

Sol.51. (2)

minima

52. A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats . It is moving with speed

v and is suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by :

(1)2( 1)

Mv K2( 1)R

(2)2( 1)

Mv K2 R

(3)2Mv

K2R

(4)

2( 1)Mv K

2R

Sol.52. (4)

P

V

C

C

2

V

1mV n. C . T

2

21 m RmV . . T

2 M 1

2MV . ( 1)T

2R

53. A screw gauge gives the following reading when used to measure the diameter of a wire.

Main scale reading : 0 mm.

Circular scale reading : 52 divisions

Given that 1 mm on main scale corresponds to 100 divisions of the circular scale.

The diameter of wire from the above data is :

(1) 0.52 cm (2) 0.052 cm (3) 0.026 cm (4) 0.005 cm

Sol.53. (2)

Reading = main scale ready + circular scale readiy × least count1

0 52100

0.52 mm = 0.052 cm

54. A boat is moving due east in a region where the earth’s magnetic field is 5.0 × 10–5 NA–1m–1 due north and horizontal. The boat

carries a vertical aerial 2m long. If the speed of the boat is 1.50 ms–1, the magnitude of the induced emf in the wire of aerial is:

(1) 1 mV (2) 0.75 mV (3) 0.50 mV (4) 0.15 mV

Sol.54. (4)

E = HvL

= 5 × 10–5 × 1.5 × 2

= 0.15 mV




[15]

55. This question has Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best

describes the two statements.

Statement-1 : Sky wave signals are used for long distance radio communication. These signals are in general, less stable than

ground wave signals.

Statement-2 : The state of ionosphere varies from hour to hour, day to day season to season.

(1) Statement-1 is true, Statement-2 is false

(2) Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of Statement-1.

(3) Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of Statement-1.

(4) Statement-1 is false, Statement-2 is true.

Sol.55. (2)

Due to variation in composition and position of ionosphere, “fading” of signal takes place due to interference between waves

and change in “skip-distance” sky waves are less stable.

56. A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius

R. Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley,

is

(1)3

g2

(2) g (3)2

g3

(4)g

3

Sol.56. (3)

mg – T = ma ...... (1)m

T

mg

a

TR = I .......(2)

m

m

TR =

2mR a

2 R

;a

for no slippingR

ma

T

2

ma

mg ma

2

3ma

mg2

2

a g3




[16]

57. A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area

around the fountain that gets wet is

(1)

2v

g (2) 2

v

g (3) 2

v

2 g

(4)

2

2

v

g

Sol.57. (2)

max area = 2

maxR

2 sin2R v

g

2

max

vR

g

R

=

2

2vg

= 2v

g

58. This question has Statement -1 and Statement-2. Of the four choices given after the statements, choose the one that best

describes the two statements.

Statement-1 : A metallic surface is irradiated by a monochromatic light of frequency > 0

(the threshold frequency). The

maximum kinetic energy and the stopping potential are Kmax

and V0

respectively. If the requency incident on the suface is

doubled, both the Kmax

and V0

are also doubled.

Statement-2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly

dependent on the frequency of incident light.


(2) Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of Statement-1.

(3) Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of Statement-1.

(4) Statement-1 is false, Statement-2 is true.

Sol.58. (4)

Reason - Kmax

(in eV) = V0=

hW

e

(in eV)

when ‘’ is made ‘n’ time0 0

V nV

59. A pulley of radius 2 m is rotated about its axis by a force F = (20t – 5t2) netwon (where t is measured in second) applied

tangentially. If the moment of inertia of the pulley about it’s axis of rotation is 10 kg m2, the number of rotations made by the

pulley before its direction of motion if reversed, is

(1) less than 3 (2) more than 3 but less than 6

(3) more than 6 but less than 9 (4) more than 9

Sol.59. (2)

= (20t – 5t2) × 2 I = 40t – 10t2 10 = 40t – 10t2 = 4t – t2

2d4t t

dt

0 t

2

0

d (4t t ). dt

2 34t t

2 3




[17]

32 t

2t3

the direction reverses after = 0 = 0 = 2t2 –

3t

3

32 t

2t3

; t = 6 sec.

d

dt

22d t

2tdt 3

6 3

2

0

td 2t . dt

3

63 4

0

2t t

3 12 =

432 6

6 36 4 36 33 12

= 36

36

N2 2

3 < N < 6.

60. Water is flowing continuously from a tap having an internal diameter 8 × 10–3 m. The water velocity as it leaves the tap is

0.4 ms–1. The diameter of the water stream at a distance 2 × 10–1 m below the tap is close to

(1) 35.0 10 m (2) 37.5 10 m (3) 39.6 10 m (4) 33.6 10 m

Sol.60. (4)

2 2

A A B B

1 1P V P gh V

2 2

A B 0 atmP P P P

A

B

h

ref level

VA= 2

B2gh V

and 2 2

B B A AR V R V

BA B

A

VR RV

dia = 2RA

= 8 × 10–3 B

2

B

V

2gh V

= 8 × 10–3 3

2

0.4 0.48 10

4 0.162 10 0.2 0.4

30.44 8 10

33.6 10




[18]

MATHEMATICS


This section contains 30 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of

which ONLY ONE is correct.

61. Let , be real and z be a complex number. If z2 + z + = 0 has two distinct roots on the line Re z = 1, then it is necessary

that :(1) b (0, 1) (2) ( – 1, 0) (3) | | = 1 (4) b (1, )

Sol.61. (4) 2z z 0 1 2z z 2Re(z) 2.1 2

Now. Discriminant D < 0 2 4 0 4 4 0 1 (1, )

62. The value of 1

2

0

8log 1 xdx

1 x

is :

(1) log 2 (2) log28

(3) log2

2

(4) log 2

Sol.62. (1) Let

1

2

0

8log(1 x)I dx1 x

....(i)

Let x = tan t

dx =2

sec tdt

/ 42

2

0

8log(1 tan t)I sec dt

(1 tan t)

/ 4

0

I 8 log(1 tan t) dt

....(ii)

/ 4

0

I 8 log 1 tan t dt4

/ 4

0

1 tan tI 8 log 1 dt

1 tan t

/ 4

0

2I 8 log dt

1 tan t

/ 4

0

I 8 log 2 log(1 tan t) dt

/ 4 /4

0 0

I 8 log 2 dt 8 log(1 tan t) dt

I 8log 2 I4

2I 8log 2.

4

I log 2

63.

2

2

d x

dy equals :

(1)

12

2

d y

dx

(2)

1 32

2

d y dy

dxdx

(3)

22

2

d y dy

dxdx

(4)

32

2

d y dy

dxdx

Sol.63. (4)

2

2

d x d dx

dy dydy

d dx dx

dx dy dy

d 1 dx

dx (dy / dx) dy

2

1 d dy dx.

dx dx dy(dy/dx)

2 2

2

dx d y dx

dy dydx

3 2

2

dx d y

dy dx

3 2

2

dy d y

dx dx




[19]

64. Let I be the purchase value of an equipment and V (t) be the value after it has been used for t years. The value V (t)

depreciates at a rate given by differential equation

dV t

k T t ,dt

where k > 0 is a constant and T is the total life in

years of the equipment. Then the scrap value V (T) of the equipment is :

(1)2 I

Tk

(2)2kT

I2

(3)

2k T t

I2

(4) kTe

Sol.64. (2)dv(t)

k(T t)dt

dv(t) ( kT kt)dt

v(t) = –kT × t +2kt

c2

at t = 0 v (t) = I

I = c

v(t) =2kt

kTt I2

Now scrap value v(T) at t = T

v(T) =2

2 kTkT I

2

2

kTV(T) I

2

65. The coefficient of x7 in the expansion of 6

2 31 x x x is :

(1) 144 (2) – 132 (3) – 144 (4) 132

Sol.65. (3) 2 3 6(1 x x x ) = 6 2 6(1 x) (1 x ) = 2 3 4 5 6[1 6x 15x 20x 15x 6x x ] × 2 4 6 8[1 6x 15x 20x 15x .....]

coeff of 7x 120 300 36 = 156 – 300 = –144

66. For5

x 0, ,2

define x

0

f x t sin t dt Then ƒ has :

(1) local maximum at at and 2 (2) local minimum at and 2(3) local minimum at and local maximum at 2 (4) local maximum at and local minimum at 2.

Sol.66. (4) x

0

sin ƒ x t sin t dt use Newton s leibnitz theorem.

f ' x x sin x

Put f ' x 0

nsin x 0, so

x

y

o 2

52

Thus x , 2

be cause it only depends on graph of sinx.

At point x = .

f ' 0 and f ' 0

So at x = , Function is local maximum

At point x = 2

f ' 2 0 and F' 2 0

So F (x) is local minimum at x = 2

67. The area of the region enclosed by the curves y = x, x = e, y = 1x and the positive x-axis is :

(1) 12 square units (2) 1 square units (3) 3

2 square units (4) 52 square units




[20]

Sol.67. (3) Desined region is as shown is figure

x=e

y=1/x

e1

y = x

Reg. area =

1 e

0 1

1x dx dx

x =

2 1 e0 1

1(x ) | n x |

2

=

1(1 0) (1 0)

2

=

1 31

2 2

68. The lines L1: y – x = 0 and L

2: 2x + y = 0 intersect the line L

3: y + 2 = 0 at P and Q respectivey. The bisector of the acute angle

between L1and L

2intersects L

3at R.

Statement -1 :

The ratio PR : RQ equals 2 2 : 5.

Statement -2 :

In any triangle, bisector of an angle divides the triangle into two similar triangles.

(1) Statement-1 is true, Statement-2 is true ; Statement-2 is a correct explanation for Statement -1.

(2) Statement-2 is true, Statement-2 is true ; Statement-2 is not a correct explanation for Statement-1.

(3) Statement-1 is true, Statement-2 is false.

(4) Statement-1 is false, Statement-2 is ture.

Sol.68. (3) Given lines L1: x – y = 0

L2: 2x + y = 0

L3: y + 2 = 0

Coordinates of P and Q are (–2, –2) and (1, –2) resp.

P R

O

Q

L = 03

L = 02 L = 01

From figure clearly angle bisector divides the app.

side in the ratio of side containing the angle.

Intersection of L1= 0 and L

2= 0 is (0, 0)

Hence

PR : RQ = OP : OQ

PR : RQ8

2 2 : 55

Statement-1 is true, but statement-2 is false

69. The values of p and q for which the function

2

32

sin p 1 x sin x, x 0

x

ƒ x q , x 0

x x x, x 0

x

is continuous for all x in R, are :

(1)1 3

p ,q2 2

(2)5 1

p ,q2 2

(3)3 1

p ,q2 2

(4)1 3

p ,q2 2

Sol.69. (3)x 0 x 0

lim F(x) lim F(x) F(0)

F(0) = ?

x 0x 0

sin(p 1)x sin xlim F(x) lim

x x

= P + 1 + 1 = P + 2

2

3/ 2x 0x 0

x x xlim F(x) lim

x

=

2

2x 0

(x x ) xlim

x x[ x x x]

=

x 0

1lim

1 x 1 =

1

2




[21]

so q = p + 2 =1

2

q =1

2p =

3

2

70. If the angle between the liney 1 z 3

x

2

and the plane x + 2y + 3z = 4 is1 5

cos ,

14

then equals :

(1) 23 (2) 3

2 (3) 25 (4) 5

3

Sol.70. (1)x y 1 z 3

1 2

x + 2y + 3z = 4

2

1 2.2 3sin

1 4 1 4 9

2

5 3

5 14

=1

2

5 3sin

5 14

..........(1)

But 1 5cos

14

1 9sin

14

........(2)

by equation (1) and (2) :

2

5 3 3

145 14

25 3 3 5 25 + 92 + 30 = 9(5 + 2) 30 = 20 = 2/3

71. The domain of the runction 1

ƒ x x x is :

(1) ( – , ) (2) (0, ) (3) ( – , 0) (4) ( – , ) – {0}

Sol.71. (3)1

F(x)| x | x

| x | – x > 0 | x | > x x < 0 x ( ,0) 72. The shortest distance between line y – x = 1 and curve x = y2 is :

(1)3

4(2)

3 2

8(3)

8

3 2(4)

4

3

Sol.72. (2) x – y + 1 = 0

x = y2

D

P2

–2

any pt on

Parabola

P(t2, t)

2t t 1PD

2

PD min =

(1 4)

D4[min ]

442

3

4 2 3 2

8

73. A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases

by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11040

after :

(1) 18 months (2) 19 months (3) 20 months (4) 21 months

Sol.73. (4) 11040 – 600 = 10440




[22]

240 + 280 + .... = 10440 =n

[480 (n 1)40] 104402

n2 [240 20n 20] 10440

2

220n 220n 10440 0. 2n 11n 522 0 2

n 29n 18n 522 0 n (n + 29) – 18(n + 29) = 0

n = 18 Ans 21 months

74. Consider the following statements

P : Suman is brilliant

Q : Suman is rich

R : Suman is honest

The negation of the statement Suman is brilliant and dishonest if and only if Suman is rich can be expressed as :

(1) ~ P ^ Q ~ R (2) ~ Q P^ ~ R (3) ~ Q ~ P ^ R (4) ~ P^ ~ R Q

Sol.74 (2) Simple

75. If 1 is a cube root of units, and 7

1 A B . Then (A, B) equals :

(1) (0, 1) (2) (1, 1) (3) (1, 0) (4) ( – 1, 1)

Sol.75. (2) 7(1 ) A B

Use 21 0 So 1 + = –2

14 A B = 3 4 2( ) . A B = 2 A B

since 21 0

so 2 1

1 A B Thus A = 1 , B = 1

76. If 1 ˆ ˆa 3i k 10

1 ˆ ˆ ˆb 2i 3j 6k ,7

then the value of 2a b . a b a 2b

is :

(1) – 5 (2) – 3 (3) 5 (4) 3

Sol.76. (1) (2a b)

. (a b) a (a b) 2b

(2a b) (a.a)b (b.a)a (2a.b)b (2b.b)a

(2a b).[b 2a]

2 22a.b b 4a 2a.b

2 24a.b b 4a = 0 – 1 – 4 = –5

77. If dy

y 3 0dx

and y (0) = 2, the y (n 2) is equal to :

(1) 7 (2) 5 (3) 13 (4) – 2

Sol.77. (1)dy

y 3dx

dydx

y 3

n(y 3) x nc y 3

n xc

xy 3

ec

Put x = 0 and y = 2

05e c 5

c




[23]

Thus, required equation is y + 3 = 5ex

put x = n2

n2y 3 5e 7

78. Equation of the ellipse whose axes are the axes of coordinates and which passes through the point (– 3, 1) and the

eccentnricity 25 is :

(1) 2 23x 5y 32 0 (2) 2 25x 3y 48 0 (3) 2 23x 5y 15 0 (4) 2 25x 3y 32 0

Sol.78. (1) Let equnation of an ellipse

2 2

2 2

x y1

A B

Since it passes through (–3, 1)

So,2 2

9 11

A B

qB2 + A2 = A2B2

2

2

2

B9 1 B

A

.........(1)

Since

2

2

2B 1 eA

2

2

B 21

A 5 Because e 2 / 5

2

2

B 3

A 5 .......(2)

From euqtion (1)

239 1 B

5

232B

5 .........(3)

and2

2

2

B 32 / 5 32A

1 e 3 / 5 3

Put values of A2 and B2 in

2 2

2 2

x y1

A B

2 2x y1

32 / 3 32 / 5

2 23x 4y 32

79. If the mean deviation about the median of the numbers a, 2a, ......, 50 a is 50, then | a | equals :

(1) 2 (2) 3 (3) 4 (4) 5

Sol.79. (3) a, 2a, ......, 50a

Median =th th1

25 term 26 term2

=1

[25a 26a]2

=51a

2

MD = 50

51a 51a 51a 51aa 2a 3a ..... 50a

2 2 2 250

50

= 1 49a 47a 45a .....a .... 49a 25002 = 1 2 a 3a 5a .... 49a 2500

2 = 25 50a 25002

= 25 × 50a = 5000 5000

a 425 50




[24]

80.

x 2

1 cos 2 x 2lim

x 2

(1) does not exist (2) equals 2 (3) equals 2 (4) equals1

2

Sol.80. (1)x 2

1 cos(x 2)lim

x 2

= x 2

x 21 2 sin

2lim

(x 2)

=x 2

x 21 2 sin

2lim

(x 2)

2

Now use LHL and RHL concept

LHL RHLSo limit does not exists

81. Statement-1 : The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is 9C3.

Statement-2 : The number of ways of choosing any 3 places from 9 different places is 9C3.

(1) Statement-1 is true, Statement-2 is true ; Statement-2 is a correct explanation for Statement-1.(2) Statement-1 is true, Statement-2 is true ; Statement-2 is not a correct explanation for Statemen-1.

(3) Statement-1 is true, Statement-2 is false.

(4) Statement-1 is false, Statement-2 is ture.

Sol.81. (2) Statement 1 No of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is

10 1 94 1 3C C

Statement 2 No of ways of choosing any 3 places from 9 different places is 93C

Statement 1 is correct & statement 2 is also correct but not the correct explaination of 1

82. Let R be the set of real number

Statement-1 :

A = {(x, y) R ×R : y – x is an integer } is an equivalence relation on R.

Statement -2

B = {(x, y) R ×R : x = y for some rational number } is an equivalence relation on R.

(1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

(2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1


(4) Statement-1 is false, Statement-2 is true

Sol.82. (2) A {(x, y) R R : y x is an integer}

(x, x) A x R x x 0 is an integer

A is reflexive

(ii) Let (x, y) AA y – x is an integer

x – y is also an integer

(y, x) A

A is symmetric

(ii i) Let (x,y) A & (y, z) A y – x is an integer & z – y is an integer

(z – y) + (y – x) is an integer z – x is an integer (x,z) R

A is transitive A is an equivalence relation




[25]

B = { (x, y) R R : x = y for some rational number x}

(i) since x = 1 (x) where 1 is rational

B is reflexive

Now clearly (0, 5) B for some 0 [ 0 0 5]

But there is no value of a exist for which (5,0) i.e. (5,0) but (5,0) Hence B is not symmetric hence not an equivalance relation

83. Consider 5 independent Bernoulli’s trials each with probability of sucess p. If the probability of at least one failure is greater

than or equal to31

32, then p lies in the interval :

(1)1 3

,2 4

(2)3 11

,4 12

(3)1

0,2

(4)11

, 112

Sol.83. (3)5 5

5

311 C p

32

5 31p 1

32

5 1p

32

1p

2

1

p 0,2

84. The two circles x2

+ y2

= ax and x2

+ y2

= c2

(c > 0) touch each other if :(1) 2 |a| = c (2) |a| = c (3) a = 2c (4) |a| = 2c

Sol.84. (2) 2 2x y ax & 2 2 2x y c

a/2 a/2C

1

aC ,0

2

2C (0,0)

1

| a |r

2

2r c

| a | | a |

c2 2

c = | a |

85. Let A and B be two symmetric matrices of order 3.Statement-1 :

A(BA) and (AB)A are symmetric matrices.

Statement-2 :

AB is symmetric matrix if matrix multication of A with B is commutative.





Sol.85. (2) Given TA A TB B

T T T(A(BA)) (BA) A = T T T(A B )A = (AB)A = A(BA)

Similarily (AB)A is also symmetric

Statement (1) is true.

Let AB = BA

Now T T T(AB) B A = BA T(AB) AB

Statment 2 is true.

86. If C and D are two events such that C D and P(D) 0, then the correct statement among the following is :

(1) P(C|D) = P(C) (2) P(C|D) P(C) (3) P(C|D) < P(C) (4)P(D)

P(C| D)P(C)




Sol.86. (2)C P(C D)

PD P(D)

=P(C)

P(D) P(D) 1

CP P(C)

D

87. The vectors a

and b

are not perpendicular and c

and d

are two vector satisfying : b c b d

and a ·d 0

. Then the

vector d

is equal to :

(1)b·c

b ca·b

(2)a·c

c ba·b

(3)b·c

b ca·b

(4)a·c

c ba·b

Sol.87. (4) a b 0 ....(i)

also, b c b d

b (c d) 0

....(ii)

a.d 0 ....(iii)

so b (c d)

or d c b

or a.d a.c a.b

a.c

a.b

So,a.c

d c ba.b

or

a.cd c b

a.b

88. Statement-1 The point A(1, 0, 7) is the mirror image of the point B(1, 6, 3) in the line :x y 1 z 2

1 2 3

Statement-2 The line :x y 1 z 3

1 2 2

bisects the line segment joining A(1, 0, 7) and B(1, 6, 3)





Sol.88. (1)x y 1 z 2

1 2 3

(1,3,5)

A(1, 0,7)

B(1,6,3)

(1, 3, 5) satisfies the line and also 1 2 1 2 1 2a a b b c c 0 Statement f is True

A B(1,6,3)

x y 1 z 2

1 2 3

Statement 2 is also true and correct explanation for Statement-1.

89. If A = sin2

x + cos4

x, then for all real x :

(1)3

A 14

(2)13

A 116

(3) 1 A 2 (4)3 13

A4 16

Sol.89. (1) 2 4A sin x cos x

2 4A 1 cos x cos x = 4 2cos x cos x 1 =2 2 21 1 3

(cos x) 2. cos x2 4 4

=

2

2 1 3cos x

2 4

min

3A

4 ; maxA 1

90. The number of values of k for which the linear equations 4x + ky + 2z = 0 ; kx + 4y + z = 0 ; 2x + 2y + z = 0 posses a non

zero solution is :

(1) 3 (2) 2 (3) 1 (4) zero

Sol.90. (2)4 k 2k 4 1 0

2 2 1

4 (2) – k (k – 2) + 2 (2k – 8) = 0 28 k 2k 4k 16 0 2k 6k 8 0 (k – 4) (k – 2) = 0

aieee-2011 solution and analysis

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