CODE-A PHYSICS

1. The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is :

1) 2.5% 2) 3.5% 3) 4.5% 4) 6% Key: (3) Sol: density = M

V

p M r3p V r

= 1.5% + 3 % = 4.5 % 2. All the graphs below are intended to represent the same motion. One of them does it

incorrectly. Pick it up.

1) 2)

3) 4) Key: (2) Sol: Initial and final slopes are zero (so option ‘2’ is wrong) all other are correct 3. Two masses 1 5m kg and 2 10m kg , connected by an inextensible string over a

frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of 2m to stop the motion is :

1) 18.3 kg 2) 27.3 kg 3) 43.3 kg 4) 10.3 kg Key: (2)

Sol:

2

2

m 50N

m m g

For equilibrium. 0.15[10 + m ]10 = 50

10 + m = 50015

M = 23.33 kg 4. A particle is moving in a circular path of radius a under the action of an attractive

potential 22kUr

. Its total energy is :

1) 24ka

2) 22ka

3) Zero 4) 2

32

ka

Key: (3) Sol: 3

du kFdr r

2

3

mv ka a

2

kKE2a

2

kPE2a

TE = 0 5. In a collinear collision, a particle with an initial speed ov strikes a stationary particle of

the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is:

1) 4ov 2) 2 ov 3)

2ov 4)

2ov

Key: (2) Sol: Momentum conservation mv0 = m(v1 – v2) v1 = v2 = v0 ________ (1)

2 2 2 21 2 0 0

1 1 1 1m v v mv mv2 2 2 2

2 2 21 2 0

3v v v2

________ (2)

2 21 2 0v v v

2 2 21 2 1 2 0v v 2v v v ________ (3)

(2) – (3)

20

1 2v2v v2

21 2 04v v v

21 2 1 2 1 2v v v v 4v v

Relative velocity after collision 2 2

0 0v v

02v

6. Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is :

1) 2192

MR 2) 2552

MR 3) 2732

MR 4) 21812

MR

Key: (4) Sol: Use parallel axis theorem

2 2

20

MR MRI 6 6M 2R2 2

2

2 2MR3MR 24MR2

255 MR2

IP = I0 + 7M(3R)2

2 255 MR 63MR2

2181MR2

7. From a uniform circular disc of radius R and mass 9M, a small disc of radius 3R is

removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is :

1) 24 MR 2) 240

9MR 3) 210 MR 4) 237

9MR

Key: (1) Sol: Ifull = Iscooped + Iremaining

2 2 2

rem9M.R MR 4MR I

2 2 9 9

2

2rem

9MR 9 MR I2 18

Irem = 4 MR2 8. A particle is moving with a uniform speed in a circular orbit of radius R in a central force

inversely proportional to the nth power of R. If the period of rotation of the particles is T, then :

1) 3/2T R for any n.

2) 12n

T R

3) 1 /2nT R 4) /2nT R Key: (3) Sol: n

1FR

2

2 n

4 1MRT R

n 1

2T R

9. A solid sphere of radius r made of a soft material of bulk modules K is surrounded by a liquid in a cylindrical container. A massless piston of area of floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of

the sphere, drr

, is :

1) Kamg

2)3Kamg

3) 3mgKa

4) mgKa

Key: (3) Sol: Bulk modulus p

vv

v mgv ak

r mgr 3ak

10. Two moles of an ideal monoatomic gas occupies a volume V at 27o C . The gas expands adiabatically to a volume 2V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

1) (a) 189 K (b) 2.7 kJ 2) (a) 195 K (b) – 2.7 kJ 3) (a) 189 K (b) – 2.7 kJ 4) (a) 195 K (b) 2.7kJ Key: (3) Sol: 1Tv const 300 v2/3 = T. 22/3 v2/3

2/3

300T 189K2

vu nc T

= – 2.7 KJ 11. The mass of a hydrogen molecule is 273.32 10 kg . If 2310 hydrogen molecules strike, per

second, a fixed wall of area 22cm at an angle of 45o to the normal, and rebound elastically with a speed of 310 /m s , then the pressure on the wall is nearly :

1) 3 22.35 10 /N m 2) 3 24.70 10 /N m 3) 2 22.35 10 /N m 4) 2 24.70 10 /N m Key: (1)

Sol: 1 12/

mv mvFP n tA t A

2723

4

2 3.32 10 1000 210

2 10

3 22.35 10 /N m 12. A silver atom in a solid oscillates in simple harmonic motion in some direction with a

frequency of 1210 / sec . What is the force constant of the bonds connecting one atom with the order ? (Mole wt. of silver = 108 and Avagadro number = 23 16.02 10 gm mole )

1) 6.4 N /m 2) 7.1 N/m 3) 2.2 N/m 4) 5.5 N/m Key: (2)

Sol: kwm

1223

1082 106 10

k

k nearly 7.1 N/m 13. A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal

vibrations. The density of granite is 3 32.7 10 /kg m and its Young’s modulus is 109.27 10 Pa . What will be the fundamental frequency of the longitudinal vibrations ?

1) 5 kHz 2) 2.5 kHz 3) 10 kHz 4) 7.5 kHz Key: (1) Sol: 2

2l l

12

Yfl

10

3

1 9.27 10 52 60 /100 2.7 10

kHz

14. Three concentric metal shells A,B and C of respective radii a,b and c (a < b < c) have surface charge densities , and respectively. The potential of shell B is :

1) 2 2

o

a b ca

2) 2 2

o

a b cb

3) 2 2

o

b c ab

4) 2 2

o

b c ac

Key: (2)

Sol: 2 221 4 44

4Bo

b caVb b c

2 2

Bo

a bV cb

15. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20V. If a

dielectric material of dielectric constant 53

K is inserted between the plates, the

magnitude of the included charge will be : 1) 1.2 nC 2) 0.3 nC 3) 2.4 nC 4) 0.9 nC Key: (1)

Sol: 1' 1q qk

11k cvk

5 390 20 1 1.23 5

nc

16. In an a.c. circuit, the instantaneous e.m.f. and current are given by e = 100 sin 30 t

20sin 304

i t

In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively :

1) 50, 10 2) 1000 ,102

3) 50 ,02

4) 50, 0

Key: (2) Sol: cosavg rms rmsp i v

20 100 1000. cos42 2 2

17. Two batteries with e.m.f. 12V and 13V are connected in parallel across a load resistor of 10 . The internal resistances of the two batteries are 1 and 2 respectively. The voltage across the load lies between :

1) 11.6 V and 11.7 V 2) 11.5 V and 11.6 V 3) 11.4 V and 11.5 V 4) 11.7 V and 11.8V

Key: (2)

Sol:

1 2

1 2

1 2

12 6.51

1 1 112

eqr rE

r r

373

37 / 3 3732 / 3 32

ViR

370 11.5632

V i R V

18. An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii , ,e pr r r respectively in a uniform magnetic field B. The relation between , ,e pr r r is :

1) e pr r r 2) e pr r r 3) e pr r r 4) e pr r r Key: (2)

Sol: 2

2mv mk krqB qB q

c pr r r 19. The dipole moment of a circular loop carrying a current I, is m and the magnetic field at

the centre of the loop is 1B . When the dipole moment is doubled by keeping the current

constant, the magnetic field at the centre of the loop is 2B . The ratio 1

2

BB

is :

1) 2 2) 3 3) 2 4) 12

Key: (3) Sol: 2m i a 2 2' 2 2 'm m i a i a

' 2a a

1

2

2 2

2 2

o

o

iB a

iBa

20. For an RLC circuit driven with voltage of amplitude mv and frequency 1o LC

the

current exhibits resonance. The quality factor, Q is given by : 1) oL

R 2) oR

L 3)

o

RC

4) o

CR

Key: (1) Sol: Direct formula, ow LQ

R

21. An EM wave from air enters a medium. The electric fields are 1 01 cos 2 zE E x v t

c

in

air and 2 02 cos 2E E x k z ct

in medium, where the wave number k and frequency v refer to their values in air. The medium is non – magnetic. If

1r and

2r refer to relative

permittivities of air and medium respectively, which of the following options is correct ? 1)

2

1 4r

r

2)

2

1 2r

r

3)

2

1 14

r

r

4)

2

1 12

r

r

Key: (3) Sol: 1

2

2cc

0 0

1c

2

1c

2

1 2

2 1

14

cc

22. Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A. The intensity of light beyond B is found to be 1

2. Now

another identical polarizer C is placed between A and B. The intensity beyond B is now found to be 1

8. The angle between polarizer A and C is :

1) 0o 2) 30o 3) 45o 4) 60o Key: (3) Sol: 20 cos

2II

2 201 2cos cos

2II

40 0 cos8 2I I

1 2

1cos2

045 23. The angular width of the central maximum in a single slit diffraction pattern is 600. The

width of the slit is 1 .m The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.)

1) 25 .m 2) 50 .m 3) 75 .m 4) 100 .m Key: (1) Sol: sin30d

6

6

0.5 10500.5 101

25

mDdm

24. A electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let ,n g be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let n be the wavelength of the emitted photon in the transition from then nth state to the ground state. For large n, (A,B are constants)

1) 2nn

BA

2) n nA B 3) 2 2n nA B 4) 2

n

Key: (1)

Sol: Debroglie wavelength 2

h hp mk

2

213.6 zk

n

n

2

1 1 11

Rn

1

2 2 2

1 1 1 11 1 BAR n R n

25. If the series limit frequency of the Lyman series is ,L then the series limit frequency of the Pfund series is:

1) 25 L 2) 16 L 3) L /16 4) L /25 Key: (4)

Sol: Series limit for Lyman series, 2 2

1 1 11

R R

vf

L

Cf CR v

Series limit for pfund series, 2 2

1 1 15 25

RR

25 25

LvC CRf

26. It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is Pd ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is Pc. The values of Pd and Pc are respectively :

1) (.89, .28) 2) (.28, .89) 3) (0, 0) 4) (0, 1) Key: (1) Sol:

2

4 1 2 8 0.8991 2dp

2

4 1 12 48 0.2813 169cp

27. The reading of the ammeter for a silicon diode in the given circuit is:

1) 0 2) 15mA 3) 11.5mA 4) 13.5 mA Key: (3) Sol: Drop across 200 is 2.3 volt since 0.7 volt is Barrier voltage for diode.

0 2.31 11.5200

mA

28. A telephonic communication service is working at carrier frequency of 10GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz?

1) 32 10 2) 42 10 3) 52 10 4) 62 10 Key: (3)

Sol: =61 10

5kHz

kHz = 52 10

29. In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52cm of the potentiometer wire. If the cell is shunted by a resistance of 5 , a balance is found when the cell is connected across 40cm of the wire. Find the internal resistance of the cell.

1) 1 2) 1.5 3) 2 4) 2.5 Key: (2)

Sol: 1 2

2

l lr Rl

= 52 40540

= 5 12 60 1.5

40 40

30. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10cm. The resistance of their series combination is 1k . How much was the resistance on the left slot before interchanging the resistances ?

1) 990 2) 505 3) 550 4) 910 Key: (3) Sol:

100L

R

R lR l

10110

R

L

R lR l

1L R

R L

R RR R

55l cm

55 1145 9

L

R

RR

911R LR R

1000R LR R

9 1 100011 LR

1000 1120LR

550LR

CHEMISTRY 31. The ratio of mass percent C and H of an organic compound X Y ZC H O is 6 : 1. If one

molecule of the above compound X Y ZC H O contains half as much oxygen as required to burn one molecule of compound X YC H completely to 2CO and 2H O . The empirical formula of compound X Y ZC H O is :

1) 3 6 3C H O 2) 2 4C H O 3) 3 4 2C H O 4) 2 4 3C H O Key: (4)

Sol: 2 2 24 2x yY YH X O XCO H OC

According to given statement

1 22 2

yxZ

2X ; 4Y ; 3Z 32. Which type of ‘defect’ has the presence of cations in the interstitial sites? 1) Schottky defect 2) Vacancy defect 3) Frenkel defect 4) Metal deficiency defect Key: (3) Sol: Frenkel defect is due to presence of cation in interstitial site 33. According to molecular orbital theory, which of the following will not be a viable

molecule? A) 2

2He B) 2He C) 2He D) 22He

Key: (4) Sol: 2

2H is not a viable molecule according to MOT 34. Which of the following lines correctly show the temperature dependence of equilibrium

constant, K, for an exothermic reaction?

1) A and B 2) B and C 3) C and D 4) A and D Key: (1) Sol: A and B have positive slopes for which H is negative 35. The combustion of benzene (1) gives 2CO g and 2 1H O . Given that heat of combustion

of benzene at constant volume is 1 03263.9 25 ;kJ mol at C heat of combustion (in kJ mole-1) of benzene at constant pressure will be : 1 18.314R JK mol

1) 4152.6 2) -452.46 C) 3260 D) -3267.6 Key: (4) Sol:

6 6 2 2 2

15 6 32 g g l

C H O CO H O

15 362 2

H E nRT 333263.9 8.314 10 298

2

3263.9 3.716 3267.676 36. For 1 molal aqueous solution of the following compounds, which one will show the

highest freezing point? 1) 2 3Co H O Cl 2) 2 2 25

.Co H O Cl Cl H O 3) 2 2 24

.2Co H O Cl Cl H O 4) 2 3 23.3Co H O Cl H O

Key: (4) Sol: Freezing point is inversely proportional to number of particles 1

f iT

37. An aqueous solution contains 20.10 M H S and 0.20 M HCl . If the equilibrium constant for lthe formation of HS from 2H S is 71.0 10 and that of 2S from HS ions is 131.2 10 then the concentration of 2S ions in aqueous solution is :

1) 85 10 2) 203 10 3) 216 10 4) 195 10 Key: (2) Sol: 7

2 1 1 10H S HS H Ka 2 13

2 1.2 10HS H S Ka 2

2 2H S H S

2 27 13

2

1 10 1.2 10H S

H S

20.2 0.20.1

S

7 13 1

22

10 1.2 10 102 2 10

S

191.2 104

190.3 10 103 10

2 27 13

2

10 10 1.2H S

H S

38. An aqueous solution contains unknown concentration of 2Ba . When 50 mL of a 1M solution of 2 4Na SO is added, 4BaSO just begins to precipitate. The final volume is 500 mL . The solubility product of 4BaSO is 101 10 . What is the original concentration of 2Ba ?

1) 95 10 M 2) 92 10 M C) 91.1 10 M D) 101.0 10 M Key: (3)

Sol: 2 24 4BaSO Ba SO

1

22 4 4

0.050.05 100.5

2Na SO Na SO

2 24PKS Ba SO

10 2 110 10Ba 2 910Ba

1 29

1 2

500 45010 ?

V VM M

92500 10 450 M

9 92

500 10 1.1 10450

M

39. At 0518 ,C the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr s-1 when 5% had reacted and 0.5 Torr s-1 when 33% had reacted. The order of the reaction is :

1) 2 B) 3 C) 1 D)0 Key: (1) Sol: 1 1

nr k p

1 12 2

2 2

nn r pr k p

r p

1 344.850.5 241.7

n

2 1.42 n 2n 40. How long (approximate) should water be electrolysed by passing through 100 amperes

current so that the oxygen released can completely burn 27.66g of diborane? (Atomic weight of B = 10.8 u)

1) 6.4 hours 2) 0.8 hours 3) 3.2hours 4) 1.6hours Key: (3) Sol: 2 6 2 2 3 23 3B H O B O H O

27.66 127.6

gm mole

2 6 21 3mole of B H requires moles of O required

1

42 2 22 2

mole

FH O O H

21 4mole of O F 3 ? 12 Fmole 12 96500 100 t 11,580 sect 3.2t hr

41. The recommended concentration of fluoride ion in drinking water is up to 1ppm as fluoride ion is required to make teeth enamel harder by converting 3 4 2 2

3 .Ca PO Ca OH to :

1) 2CaF 2) 2 23 .CaF Ca OH

3) 3 4 223 .Ca PO CaF 4) 2 22

3 .Ca OH CaF Key: (3) Sol: 3 4 3 4 22 2. .Ca PO Ca OH Ca PO CaF

42. Which of the following compounds contain(s) no covalent bond(s)? 3 2 2 6 2 4, , , ,KCl PH O B H H SO

1) 2 6 2, ,PKCl B H H 2) 2 4, HKCl SO 3) KCl 4) 2 6,KCl B H Key: (3) Sol: only KCl has ionic bond 43. Which of the following are Lewis acids? 1) 3 3PH and BCl 2) 3 4AlCl and SiCl 3) 3 4PH and SiCl 4) 3 3BCl and AlCl Key: (2 or 4) Sol: 3 3,AlCl BCl are Electron deficient

44. Total number lone pair of electrons in 3I ion is : 1) 3 2) 6 3) 9 4) 12 Key: (3) 0

Sol: I I I .. .. .... .. ..

: ::( )

has 9 lone pairs 45. Which of the following salts is the most basic in aqueous solution? 1) 3Al CN 2) 3CH COOK 3) 3FeCl 4) 3 2Pb CH COO Key: (2) Sol:

3COOCH undergoes anionic hydrolysis and among 3COOKCH has higher pH

46. Hydrogen peroxide oxidizes 4

6Fe CN

to

3

6Fe CN

in acidic medium but reduces

3

6Fe CN

to

4

6Fe CN

in alkaline medium. The other products formed are,

respectively : 1) 2 2 2H O O and H O 2) 2 2 2H O O and H O OH 3) 2 2 2H O and H O O 4) 2 2H O and H O OH Key: (3) Sol: 2 2H O oxidizing properties

4 3

2 2 26 6H O Fe CN Fe CN H O

2 2H O Reducing properties

3 4

2 2 2 26 6H O Fe CN Fe CN H O O

47. The oxidation states of 2 3 6 66 2,Cr in Cr H O Cl Cr C H , and 2 32 2 2

K Cr CN O O NH respectively are :

1) 3, 4 6and 2) 3, 2 4and 3) 3,0 6and 4) 3,0 4and Key: (4) Sol: 2 36

Cr H O Cl 0 3 0x 3x 6 6 2CrCC H 0 0x 0x 2 32 2 2

k G CN O O NH 2 2 4 0 0 0x 4x 48. The compound that does not produce nitrogen gas by the thermal decomposition is : 1) 3 2Ba N 2) 4 2 72NH Cr O 3) 4 2NH NO 4) 4 42NH SO Key: (4) Sol: 3 22Ba NH Ba N 4 2 7 2 32NH Cr O N CrO 4 2 2 2NH NO N H O

49. When metal ‘M’ is treated with NaOH, a white gelatinous precipitate ‘X’ is obtained, which is soluble in excess of NaOH. Compound ‘X’ when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal ‘M’ is :

1) Zn 2) Ca 3) Al 4)Fe Key: (3) Sol: 3Al NaOH Al OH 23 lubNaOH excessAl OH NaAlO So le 2 33

Al OH Al O 50. Consider the following reaction statements: 3 2 3 3 34 3

Co NH Br Br Co NH Br NH

I) Two isomers are produced if the reactant complex ion is a cis-isomer II) Two isomers are produced if the reactant complex ion is a transisomer. III) Only one isomer is produced if the reactant complex ion is a transisomer IV) Only one isomer is produced if the reactant complex ion is a cis-isomer. The correct statements are : 1) (I) and (II) 2) (I) and (III) 3) (III) and (IV) 4) (II) and (IV) Key: (2)

Sol:

M

b

baa

a a

Trans

M

b

aaa

a b

Due to presence of phase of symmetry, III & IV are correct 51. Glucose on prolonged heating with HI gives : 1) n-Hexan 2) 1-Hexene 3) Hexanoic acid 4) 6-iodohexanal Key: (1) Sol:

Pr

4Recos HIolongad

Healing

d PGlu e n Hexan

52. The trans-alkenes are formed by the reduction of alkynes with : 1) 2 4/ ,H Pd C BaSO 2) 4NaBH 3) 3/ .Na liq NH 4) Sn HCl Key: (3) Sol: Birch reduction gives trans alkene with alkyne 53. Which of the following compounds will be suitable for Kjeldhal’s method for nitrogen

estimation?

Key: (2) Sol: Nitrogen in the ring & 2NO groups &

2N Cl fails kjeldah’s test

54. Phenol on treatement with 2CO in the presence of NaOH followed by acidification product. X on treatment with 3 2CH CO O in the presence of catalytic amount of 2 4H SO produceds :

Key: (1)

Sol:

OH

2CO NaOH

OH

COOH

Alicylicociol (x)

(Major)

3 2 /CH CO O H

O-COCH3

COOH

Asprin 55. An alkali is titrated agains an acid with methyl orange an indicator, which of the following

is a correct combination ? Base Acid End point 1) Weak Strong Colourless to pink 2) Strong Strong Pinkish red to yellow 3) Weak Strong Yellow to pinkish

red 4) Strong Strong Pink to colourless

Key: (3) Sol: For methyl orange

N NH

SO3H

N+

CH3

CH3

77 pHpHStrong acid Base Yellow to Pink

N N

S

NCH3

CH3

O

ONa+

O

56. The predominate form of histamine prexent in human blood is (pKa, Histidine = 6.0)

Key: (4) Sol:

Hp of blood = 7.35 Above the Iso-electric point medium, Hence, Ink plasma exist as monocationic

form

57. Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with Br2 to form product B. A and B are respectively :

Key: (3)

Sol:

OH

3

||O

NaOHCl C O CH

O C O

O

CH3

O

O

O CH3

Br

2Br

58. The increasing order of basicity of the following compounds is :

1) a < b < c < d 2) b < a < c < d 3) b < a < d < c 4) d < b < a < c Key: (3) Sol: c > d > a > b 59. The major product formed in the following reaction is :

Key: (4)

Sol: OH

I32 CH I

O

O HI

60. The major product of the following reaction is :

Key: (2)

Sol:

BrNaOMeMeOH

MATHEMATICS

61. Two sets A and B are as under: A a, b R R :|a 5 | 1 and | b 5 | 1 ;

2 2B a,b R R : 4 a 6 9 b 5 36 Then :

(1) B A (2) A B (3) A B = (an empty set) (4) neither A B nor B A Key: (2) Sol: 5 1 5 1a and b

1 5 1 1 5 1a and b

4 6 4 6A a and b

2 24 6 9 5 36a b

2 26 51

9 4a b

A C B

62. Let S x R : x 0 and 2 x 3 x x 6 6 0} . Then S: (1) is an empty set. (2) contains exactly one element (3) contains exactly two element (4) contains exactly four element Key: (3)

Sol: n t 2x t

23 6 02 t t bt

3t 2 6 6 02 6 tt t 2 4 0tt 0 & 4t t x = 16

3t 2 6 6 06 2 tt t

2 8 12 0tt

8 64 482

t

8 4 6, 2

2

X = 4 ; x = 4 & 16 63. If , C, are the distinct roots, of the equation x2 – x + 1 = 0, then 101 + 107 is equal to: (1) – 1 (2) 0 (3) 1 (4) 2 Key: (3)

Sol: 2 1 0xx

2,x w w

1070101 107 2w w

0 214w w

2 1w w

64. If 2x 4 2x 2x2x x 4 2x A Bx x A2x 2x x 4

, then the ordered pair (A, B) is equal to:

(1) (–4, –5) (2) (–4, (3) (3) (–4, 5) ((4) (4, 5) Key: (3) Sol: 1 1 2 3R R RR

1 2 2

5 4 1 4 21 2 4

x xx x x

x x

2 2 1 3 3 1,R R R R RR

1 2 25 4 0 4 0

0 0 4

x xx x

x

25 4 4x x

4 5A B

65. If the system of linear equations

x ky 3z 0

3x ky 2z 02x 4y 3z 0

has a non-zero solution (x, y, z), then 2

xzy

is equal to:

(1) –10 (2) 10 (3) –30 (4) 30 Key: (2)

Sol: 1 33 2 02 4 3

kk

3 8 9 4 3 12 2 0k k k

8 3 5 36 6 0k k k 4 44k 11k 11 3x y p where t = p 3 11 2x y p

52 52px p x

11112 2

p py y

22 2

55 42 102

2

p px z py pp

66. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is :

(1) at least 1000 (2) less than 500 (3) at least 500 but less than 750 (4) at least 750 but less than 1000 Key: (1) Sol: 4 1.3 .4! 10806 CC

67. The sum of the co-efficients of all odd degree terms in the expansion of

5 53 3x x 1 x x 1 , (x > (1) is :

(1) –1 (2) 0 (3) 1 (4) 2 Key: (4)

Sol: 5

3 5 4 3 3 31 21 5 1 5 1x x x c x x c x x 3

2 33 15 x xc 523 3

4 51 5 15 x x c xc

5 33 5 4 3 3 3 2 3

1 2 31 5 1 5 1 5 1x x x c x x c x x c x x 523 34 51 5 15 x x c xc

Adding we have 5 3 3 6 320 1 10 2 12 x x x x xx

5 6 3 7 420 20 10 20 102 x x x x xx 10 2 20 10 2

68. Let a1, a2, a3, …. A49 be in A.P. such that 12

4k 1k 0

a 416

and a9 + a43 = 66. If 2 2 21 2 17a a ..... a 140m , then m is equal to:

(1) 66 (2) 68 (3) 34 (4) 33 Key: (3) Sol: 1 5 9 13 49.... 416a a a aa

1 1 1 14 8 .... 48 416a d a d a da 1 4 1 2 3 .... 12 41612 da

1 1

12 134 416

212 da 1 312 41612 da

1 14162612

da 1 300 79212 da

1 18 42 66d a da

12 376d

376d

1139

4a

2 2 21 2 17.... 140a a ma

2 2 21 12 1 2 3 .... 16 1 2 .... 16 14017 a d d ma

By simplify we get 34m

69. Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 12 + 2.22 + 32 + 2.42 + 52 + 2.62 + …

If B – 2A = 100, then is equal to: (1) 232 (2) 248 (3) 464 (4) 496 Key: (2)

Sol: 2 2 2 2 2 2 21 2 .... 20 2 4 6 .... 20A

2 200 2 440 41 8 4 20 21 416 6 6

B

2 248A B 70. For each t R, let [t] be the greatest integer less than or equal to t. Then

x 0

1 2 15lim x ....x x x

(1) is equal to 0. (2) is equal to 15. (3) is equal to 120. (4) does not exist (in R). Key: (3)

Sol: 0

1 1 2 2 ....xLt x

x x x x

1 2 3 .... 15 1 2 15.....x x x x

x

=120 71. Let |x|S {t R : f x || x . e 1 sin | x | is not differentiable at t}. Then the set S is equal to : (1) ( an empty set) (2) {0} (3) {} (4) {0, } Key: (1) Sol: Check at 0x and

. . 00 &

. . 0L H D

Both at xR H D

differentiable at 0 &x 72. If the curves y2 = 6x, 9x2 + by2 = 16 intersect each other at right angles, then the value of b

is: (1) 6 (2) 7

2 (3) 4 (4) 9

2

Key: (4) Sol: 2 2 26 & 9 16 16x x by

1 2

1dy dydx dx

2b

73. Let 22

1f x xx

and 1g x xx

, x R 1,0,1 . If

f xh x

g x then the local minimum

value of h(x) is: (1) 3 (2) –3 (3) 2 2 (4) 2 2 Key: (4)

Sol: 2

21

1

xxh x

xx

1 tx

x

2 2 2tt t

t th

At 2x local min 22

74. The integral

2 2

5 3 2 3 2 5

sin x cos x dxsin x cos x sin x sin x cos cos x is equal to:

(1) 3

1 C3 1 tan x

(2) 3

1 C3 1 tan x

(3) 3

1 C1 cot x

(4) 3

1 C1 cot x

(where C is a constant of integration) Key: (2)

Sol: 2 2

5 3 2 3 2 5

sin cossin cos sin sin cos cos

x x dxx x x x x x

2 4 2

25 2 3

tan sec sec

tan tan tan 1

x x x dx

x x x

2

2 33

13 1 tan1

t dt cxt

75. The value of 22

x

2

sin x dx1 2

is:

(1) 8 (2)

2 (3) 4 (4)

4

Key: (4)

Sol: 22

2

sin1 2 x

xI dx

2

2

2

2 sinI xdx

4

I

76. Let g(x) = cos x2, f x x and , ( < ) be the roots of the quadratic equation 2 218x 9 x 0 . Then the area (in sq. units) bounded by the curve y = (gof)(x) and the

lines x = , x = and y = 0, is : (1) 1 3 1

2 (2) 1 3 1

2 (3) 1 3 2

2 (4) 1 2 1

2

Key: (1) Sol: cosgof x

2 29 018 xx

,6 3

/3

/6

3 1cos2

x dx

77. Let y = y(x) b the solution of the differential equation dysin x y cos x 4x, x 0,dx

. If

y 02

, then y

6

is equal to:

(1) 249 3

(2) 289 3

(3) 289

(4) 249

Key: (3)

Sol:

cos 4dy y x xdx

six

4cotsin

dy xy xdx x

sin sinin xI e x 2.sin 2 cy x x

2 21 2

2 36 2y

28

9y

78. A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is:

(1) 3x + 2y = 6 (2) 2x + 3y = xy (3) 3x + 2y = xy (4) 3x + 2y = 6xy Key: (3)

Sol: 1x ya b

2 3 1a b

2 3b a ab 2 3y x xy 79. Let the orthocenter and centroid of a triangle be A(– 3, 5) and B(3, 5) respectively. If C is

the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter, is:

(1) 10 (2) 2 10 (3) 532

(4) 3 52

Key: (3)

Sol: O G C

, 3, 3 3,5

2 3 9 62 5 9 2

2 29 3 3 10OC

3 10 532 2

r

80. If the tangent at (1, 7) to the curve x2 = y – 6 touches the circle x2 + y2 + 16x + 12y + c = 0 then the value of c is:

(1) 195 (2) 185 (3) 85 (4) 95 Key: (4)

Sol: 7 62

yx

2 5 2 5 0x y x y

8, 6c 16 6 5 64 36

5c

C = 95 81. Tangent and normal are drawn at P(16, 16) on the parabola y2 = 16x, which intersect the

axis of the parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and CPB, then a value of tan is:

(1) 12

(2) 2 (3) 3 (4) 43

Key: (2)

Sol:

A B

P(16,16)

(-16,0) (24,0)(4,0)

16 24 ,02

c

4,0c

1 2

1 2

21

tan m mm m

82. Tangents are drawn to the hyperbola 2 24 36x y at the points P and Q. If these tangents intersect at the point 0,3T then the area (in sq. units) of PTQ is:

(1) 45 5 (2) 54 3 (3) 60 3 (4) 36 5 Key: (1)

Sol: 2 2

19 36x y

12y

2 12 12 1

9 12 3x

2 5 9x

3 5x

1 2 3 5 15 45 52

A

83. If 1L is the line of intersection of the planes 2 2 3 2 0x y z , 1 0x y z and 2L is the line of intersection of the planes 2 3 0,3 2 1 0x y z x y z , then the distance of the origin from the plane, containing the lines 1L and 2L , is:

(1) 14 2

(2) 13 2

(3) 12 2

(4) 12

Key: (2) Sol: 0y

2 3 2 0x z 2 2 2 0x z 4z 5x

2 2 31 1 1

i j ki j

Plane 5 7 7 4 8 0x y z

35 32162

tandis ce

1 118 3 2

84. The length of the projection of the line segment joining the points 5, 1,4 and 4, 1,3 on the plane, 7x y z is :

(1) 23

(2) 23

(3) 13

(4) 23

Key: (4)

Sol:

A BP Q

5, 1,4 4, 1,3

1 1 2AB

1 1 2 233 2 3 2

cos

2

sin PQ

23

PQ

85. Let u

be a vector coplanar with the vectors ˆˆ ˆ2 3a i j k

and ˆˆb j k

. If u is perpendicular to aand u . 24b

, then 2u is equal to :

(1) 336 (2) 315 (3) 256 (4) 84 Key: (1)

Sol: 2.V h a a b a a b b a

2 14V a b

2. 24 2 . 14V b a b b

22

14 2 336V b a

86. A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is :

(1) 310

(2) 25

(3) 15

(4) 34

Key: (2) Sol: 4 Red 6Blue

2R P BR P RRP

6 4 4 610 12 10 12

48 2120 5

87. If 9

1

5 9ii

x

and 9

2

1

5 45ii

x

, then the standard deviation of the 9 items 1 2 9, ,......,x x x is

: (1) 9 (2) 4 (3) 2 (4) 3 Key: (3)

Sol:

9

19

ixi

2259 9

. ix xi jS D

245 9 36 29 9 9

88. If sum of all the solutions of the equation 18cos . cos .cos 16 6 2

x x x in 0, is

k , then k is equal to : (1) 2

3 (2) 13

9 (3) 8

9 (4) 20

9

Key: (2)

Sol: 2 2 1cos sin 16 2

8cos xx

23 cos 14

8cos xx

3 16 cos 8cos xx

34cos 3cos 12 x x

12

cos3x

2 :3 9n n zx

0, 1n n solution in 0, 13

9sum

89. PQR is a triangular park with 200PQ PR m . A T.V. tower stands at the mid-point of QR . If the angles of elevation of the top of the tower at P, Q and R are respectively 0 045 ,30 and

030 , then the height of the tower (in m) is : (1) 100 (2) 50 (3) 100 3 (4) 50 2 Key: (1)

Sol: Q Dx x R

P

200200

1hpD

pD h 13

hx 3x h

2 2200PD h x 100h

90. The Boolean expression ~ ~p q p q is equivalent to: (1) ~ p (2) p (3) q (4) ~ q

Key: (1) Sol:

P Q ~P p q ~ p q ~ p q ~ ~ p q v p q T F F T F F F F T T T F T T F F T F T F T T T F T F F F

~ ~ ~ p q v p q np