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JEE Main Online test_09-04-19_EveningChemistry

Single Choice Questions :1. The amorphous form of silica is :

(1) quartz (2) tridymite (3) cristobalite (4) kieselguhrAns. (4)Sol. Fact

2. Molal depression constant for a solvent is 4.0 K kg mol–1. The depression in the freezing point of the solventfor 0.03 mol kg–1 solution of K2SO4 is :(Assume complete dissociation of the electrolyte)(1) 0.18 K (2) 0.12 K (3) 0.24 K (4) 0.36 K

Ans. (4)Sol. Tf = iKf × m i for K2SO3 = 3

= 3 × 4 × 0.03Tf = 0.36 K

3. The structures of beryllium chloride in the solid state and vapour, phase, respectively,are :(1) chain and chain (2) chain and dimeric(3) dimeric and chain (4) dimeric and dimeric

Ans. (2)Sol. Vapour form – Dimer

Solid form – Chain

4. The major product of the following reaction is :

2 4

2

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

Ans. (1)Sol. Acid catalysed hydrolysis of ester followed by esterfication to give a six member ring.

5. Increasing order of reactivity of the following compunds for SN1 substitution is :

(1) (B) < (A) < (D) < (C) (2) (B) < (C) < (A) < (D)(3) (A) < (B) < (D) < (C) (4) (B) < (C) < (D) < (A)

Ans. (1)Sol. Reactivity depend upon stability of carbocation.


6. The layer of atmosphere between 10 km to 50 km above the sea level is called as :(1) thermosphere (2) troposphere(3) stratosphere (4) mesosphere

Ans. (3)Sol. Fact

7. Among the following species, the diamagnetic molecule is :(1) NO (2) O2 (3) CO (4) B2

Ans. (3)Sol. CO is diamagnetic.

8. Which one of the following about an electron occupying the 1s orbital in a hydrogen atom is incorrect ?(The Bohr radius is represented by a0)(1) The probability density of finding the electron is maximum at the nucleus.(2) The total energy of the electron is maximum when it is at a distance a0 from the nucleus.(3) The magnitude of potential energy is double that of its kinetic energy on an average.(4) The electron can be found at a distance 2a0 from the nucleus

Ans. (2)Sol.

9. 10 mL of 1mM surfactant solution forms a monolayer covering 0.24 cm2 on a polar substrate. If the polar headis approximated as cube, what is its edge length?(1) 1.0 pm (2) 0.1 nm (3) 2.0 pm (4) 2.0 nm

Ans. (3)Sol. Total area = area covered by single particle × Number of particles

0.24 = 3

2 233

10 10b 6 1010

b2 = 3

3 23

0.24 1010 10 6 10

=

3 2

20

24 10 106 10 10

20

246 10

b2 = 4 × 10–20

b = 2 × 10–10 cm = 2 × 10–12 mb = 2.0 pm

10. What would be the molality of 20% (mass/mass) aqueous solution of KI ?(molar mass of KI = 166 g mol–1)(1) 1.51 (2) 1.48 (3) 1.35 (4) 1.08

Ans. (1)Sol. 20 gm of KI is present in 100 gm solution.

20 gm of KI is present in 80 gm of H2O

20166

mole of KI is present in 80 gm of H2O

molality = moles of solute

weight of solvent in kg = 20

166

801000

= 20 1000

166 80 = 1.51


11. Which of the following compounds is a constitutent of the polymer 2 ?

(1) Formaldehyde (2) N-Methyl urea (3) Ammonia (4) MethylamineAns. (1)

Sol. For given copolymer the main monomer are urea and formaldehyde.

12. The one that is not a carbonate ore is :(1) malachite (2) calamine (3) bauxite (4) siderite

Ans. (3)Sol. Bauxite (Al2O3) is not a carbonate ore.

13. Assertion : For the extraction of iron, haematite ore is used.Reason : Haematite is a carbonate ore of iron.(1) Only the assertion is correct.(2) Both the assertion and reason are correct and the reason is the correct explanation for the assertion.(3) Only the reason is correct.(4) Both the assertion and reason are correct, but the reason is not the correct explanation for the assertion

Ans. (1)Sol. Haematite is an oxide ore.

14. The peptide that gives positive ceric ammonium nitrate and carbylamine tests is :(1) Lys-Asp (2) Gln-Asp (3) Ser-Lys (4) Asp-Gln

Ans. (3)Sol. Dipeptide of Serine-Lysine will respond to both the given test because of presence of free –NH2 & –OH.

15. During compression of a spring the work done is 10 kJ and 2 kJ escaped to the surroundings as heat. Thechange in internal energy, U (in kJ) is :(1) 8 (2) 12 (3) –8 (4) –12

Ans. (1)Sol. According to first law of thermodynamics

U = q + w [q = –2J (heat released, w = +10 (work done on the system)]U = –2 J + 10 JU = +8 J

16. HF has highest boiling point among hydrogen halides, because it has :(1) stongest van der Waals’ interactions(2) lowest ionic character(3) lowest dissociation enthalpy(4) strongest hydrogen bonding

Ans. (4)Sol. H –F ........... H – F ........ H – F

Boiling point Intermolecular forcesHF has hydrogen bond which is strong intermolecular force of attraction.


17. Which of the following potential energy (PE) diagrams represents the SN1 reaction ?

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

Ans. (4)

Sol.

SN1 is two step reaction, where in step (I) formation of carbocation in RDS.

18. p-Hydroxybenzophenone upon reaction with bromine in carbon tetrachloride gives :

(1) (2)

(3) (4)

Ans. (3)

Sol.

Bromination at the activated ring takes place.

19. The maximum number of possible oxidation states of actinoides are shown by :(1) actinium (Ac) and thorium (Th) (2) neptunium (Np) and plutonium (Pu)(3) berkelium (Bk) and californium (Cf) (4) nobelium (No) and lawrencium (Lr)

Ans. (2)Sol. Neptunium shows - +3, +4, +5, +6, +7

Plutonium shows - +3, +4, +5, +6, +7


20. The correct statements among I to III are :(I) Valence bond theory cannot explain the color exhibited by transition metal complexes.(II) Valence bond theory can predict quantitatively the magnetic properties of transtition metal complexes.(III) Valence bond theory cannot distinguish ligands as weak and strong field ones.(1) (I), (II) and (III) (2) (I) and (II) only(3) (I) and (III) only (4) (II) and (III) only

Ans. (3)Sol. (I) Colour in transition series can be explained by the crystal field theory.

(II) Experimental magnetic moment cannot be determined by VBT.(III) Nature of ligand is determined by the ‘Spectrochemical series’.

21. At a given temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas behaviour. Theirequation of state is given as

RTpV b

at T..

Here, b is the van der Waals constant. Which gas will exhibit steepest increase in the plot of Z (compressionfactor) vs p ?(1) Kr (2) Ar (3) Xe (4) Ne

Ans. (3)

Sol. P = RT

V b;

PV Pb1RT RT

; z = Pb1RT

Slope = b

RTSlope b molecular size

22. Consider the given plot of enthalpy of the following reaction between A and B.A + B C + D

Identify the incorrect statement.

(1) Formation of A and B from C has highest enthalpy of activation.(2) C is the thermodynamically stable product.(3) D is kinetically stable product.(4) Activation enthalpy to form C is 5 kJ mol–1 less than to form D.

Ans. (4)Sol. Theory based


23. The correct statements among I to III regarding group 13 element oxides are,(I) Boron trioxide is acidic.(II) Oxides of aluminium ang gallium are amphoteric.(III) Oxides of indium and thallium are basic.(1) (I) and (II) only (2) (I), (II) and (III)(3) (I) and (III) only (4) (II) and (III) only

Ans. (2)Sol. Group 15th oxide : B2O3 Al2O3, Ga2O3 In2O3, Tl2O3

Acidic Amphoteric Basic

24. In the following reaction :

carbonyl compound + MeOH HCl acetal

Rate of the reaction is the highest for :(1) Propanal as substrate and methanol in excess(2) Acetone as substrate and methanol in stoichiometric amount(3) Acetone as substrate and methanol in excess(4) Propanal as substrate and methanol in stoichiometric amount

Ans. (1)Sol. Acetal is form from the reaction of aldehyde with excess of MeOH.

25. The maximum possible denticities of a ligand given below towards a common transition and inner-transitionmetal ion, respectively, are :

(1) 6 and 6 (2) 6 and 8 (3) 8 and 8 (4) 8 and 6Ans. (2)Sol. Maximum coordination number for d-block = 6

Maximum coordination number for f-block = 8

26. Hinsberg’s reagent is :(1) C6H5SO2Cl (2) C6H5COCl (3) (COCl)2 (4) SOCl2

Ans. (1)

Sol.

27. Noradrenaline is a/an :(1) Neurotransmitter (2) Antihistamine (3) Antidepressant (4) Antacid

Ans. (1)Sol. Noradrenaline is a neurotransmitter, function in brain & body as a harmone.


28. The major products A and B for the following reactions are, respectively :

(1)

(2)

(3)

(4)

Ans. (1)Sol. SN2 reaction followed by reduction of ketone and cyanide.

29. In an acid-base titration, 0.1M HCl solution was added to the NaOH solution of unknown strength. Which ofthe following correctly shows the change of pH of the titration mixture in this experiment ?

(A) (B) (C) (D)

(1) (D) (2) (C) (3) (A) (4) (B)Ans. (3)Sol. Theory based

30. A solution of Ni(NO3)2 is electrolysed between platinum electrodes using 0.1 Faraday electricity. How manymole of Ni will be deposited at the cathode ?(1) 0.15 (2) 0.20 (3) 0.05 (4) 0.10

Ans. (3)Sol. From faraday’s first law

w = zQ

w = E 0.1 96500

96500 ; w =

wt

f

M 0.1n

moles × Mwt = wtM 0.12

; moles = 5 × 10–2 moles


JEE Main Online test_09-04-19_EveningPhysics

Single Choice Questions :1. A moving coil galvanometer has a coil with 175 turns and area 1 cm2. It uses a torsion band of torsion

constant 10–6 N-m/rad. The coil is placed in a maganetic field B parallel to its plane. The coil deflects by 1°for a current of 1 mA. The value of B (in Tesla) is approximately :(1) 10–4 (2) 10–3 (3) 10–2 (4) 10–1

Ans. (2)

Sol. =

BM

C= iNAB

B = T10iNAC 3–

2. Two coils 'P' and 'Q' are separated by some distance. When a current of 3 A flows through coil 'P', a magneticflux of 10–3 Wb passes through 'Q'. No current is passed through 'Q'. When no current passes through 'P' anda current of 2 A passes through 'Q', the flux through 'P' is :(1) 3.67 × 10–3 Wb (2) 6.67 × 10–4 Wb (3) 3.67 × 10–4 Wb (4) 6.67 × 10–3 Wb

Ans. (2)

Sol. q = 2/322

20

)xR(2R,i

r2 = 10–3 ; p = 2/322

220

)xr(2r,i

R2

q

p

=

1

2

ii

3–p

2/322

2/322

10)xr()xR(

p = 6.67 × 10–4

3. The physical sizes of the transmitter and receiver antenna in a communication system are :(1) independent of both carrier and modulation frequency(2) inversely proportional to modulation frequency(3) proportional to carrier frequency(4) inversely proportional to carrier frequency

Ans. (4)Sol. The physical size of antenna is inversely proportional to carrier frequency.

4. A string 2.0 m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third harmonicmode. The speed of the wave and its fundamental frequency is :(1) 180 m/s, 80 Hz (2) 320 m/s, 120 Hz (3) 320 m/s, 80 Hz (4) 180 m/s, 120 Hz

Ans. (3)

Sol. 3

2 = 240

3 × 22

= 240 = 320 m/s

fundametal frequency = 2

= 80 Hz


5. A thin smooth rod of length L and mass M is rotating freely with angular speed 0 about an axis perpendicularto the rod and passing through its center. Two beads of mass m and negligible size are at the center of therod initially. The beads are free to slide along the rod. The angular speed of the system , when the beadsreach the opposite ends of the rod, will be :

(1) m3MM 0

(2) m6MM 0

(3) m2M

M 0

(4) mM

M 0

Ans. (2)Sol. Applying conservation of angulajr momentum

0

2

12ML

=

22Lm

12ML 22

= m6MM 0

6. Diameter of the objective lens of a telescope is 250 cm. For light of wavelength 600nm. coming from a distantobject, the limit of resolution of the telescope is close to :(1) 4.5 × 10–7 rad (2) 2.0 × 10–7 rad (3) 3.0 × 10–7 rad (4) 1.5 × 10–7 rad

Ans. (3)

Sol. Limit of resolution = d22.1

= 2.9 × 10–7 rad

7. The specific heats, Cp and CV of a gas of diatomic molecules, A, are given (in units of J mol–1 K–1) by 29 and22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they aretreated as ideal gases, then :(1) A has one vibrational mode and B has two.(2) Both A and B have a vibrational mode each.(3) A has a vibrational mode but B has none.(4) A is rigid but B has a vibrational mode.

Ans. (3)Sol. for A

R =- Cp – Cv = 7

Cv = 2fR

= 22 f = 6.3

Rotation t translation = 5 vibrational = 1For BR = Cp – Cv = 9

Cv = 2fR

f = 942

Rotational + translational = 5 Vibrational = 0


8. A particle of mass 'm' is moving with speed '2' and collides with a mass '2m' moving with speed '' in thesame direction. After collision, the first mass is stopped completely while the second one splits into twoparticles each of mass 'm', which move at angle 45º with respect to the original direction.The speed of each of the moving particle will be :

(1) 2 (2) 2

(3) 22 (4) )22(

Ans. (3)Sol. Lenear momentum conservation

m(2v) + (2m)V = m × 0 + 2V

2

V = 22 V

9. A thin convex lens L (refractive index = 1.5) is placed on a plane mirror M. When a pin is placed at A, suchthat OA = 18 cm, its real inverted image is formed at A itself, as shown in figure. When a liquid of refractiveindex µ is put between the lens and the mirror, The pin has to be moved to A', such that OA' = 27 cm, to getits inverted real image at A' itself. The value of µ will be :

(1) 3 (2) 2 (3) 23

(4) 34

Ans. (4)

Sol. 181

182

21

f1

1

18–1–

f1 1

2

when 1 is filled between lens and mirror

P = 182

– 182

(1 – 1) = 1822–2 1

Feq = – 1–2

18

2 = 6 – 31

1 = 34


10. A particle 'P' is formed due to a completely inelastic collision of particles 'x' and 'y' having de-Broglie wavelengths'x' and 'y' respectively. If x and y were moving in opposite directions, then the de-Broglie wavelength of 'P' is:

(1) yx

yx

(2) |–| yx

yx

(3) x – y (4) x + y

Ans. (2)Sol. by momentum conservation

Px – Py = Pp

pyx

hh–h

p = |–| xy

yx

11. The area of square is 5.29 cm2. The area of 7 such squares taking into account the significant figures is :(1) 37.0 cm2 (2) 37.030 cm2 (3) 37.03 cm2 (4) 37 cm2

Ans. (2)Sol. Vibrant Answer (3)

Total area = A1 + A2 + ...... = 37.03 m2

(Answer should have 2 digits beyond decimal)

12. A wooden block floating in a bucket of water has 54

of its volume submerged. When certain amount of an oil

is poured into the bucket, it is found that the block is just under the oil surface with half of its volume underwater and half in oil. The density of oil relative to that of water is :(1) 0.8 (2) 0.5 (3) 0.7 (4) 0.6

Ans. (4)Sol. In Ist situation

VbPbg = Vs Pw g

54

VV

b

s

finally in equilibrium position.

VbPbg = gP2V

gP2V

wb

0b

2Pb = P0 + Pw

6.053

PP

w

0


13. A wedge of mass M = 4m lies on a frictionless plane. A particle of mass m approaches the wedge with speed. There is no friction between the particle and the plane or between the particle and the wedge. The maximumheight climbed by the particle on the wedge is given by

(1) g52 2

(2) g

2(3) g7

2 2(4) g2

2

Ans. (1)Sol. Applying linear momentum conservation.

mv = (m + 4m) V

V = 5V

– mgh = 21

(m + 4m) V2 – 21

mV2

h = g5V2 2

14. The position vector of a particle changes with time according to the relation j)t20–4(it15)t(r 22

. What

is the magnitude of the acceleration at t = 1?(1) 25 (2) 40 (3) 50 (4) 100

Ans. (3)

Sol. j)t20–4(it15r 22

jt40–it30dtrdv

j40–i30dtvda

|a| = 50

15. A massless spring (k = 800 N/m), attached with a mass (500 g) is completely immersed in 1 kg of water. Thespring is stretched by 2 cm and released so that it starts vibrating. What would be the order of magnitude ofthe change in the temperature of water when the vibrations stop completely ? (Assume that the watercontainer and spring receive negligible heat and specific heat of mass = 400 J/kg K, specific heat ofwater = 4184 J/kg K).(1) 10–3 K (2) 10–5 K (3) 10–1 K (4) 10–4 K

Ans. (2)Sol. Applying conservation of energy

21

kx2 = (m1s1 + m2s2) T

T = 4384

1016 2– = 3.6 × 10–5


16. Four point charges –q, +q, +q and –q are placed on y-axis at y = – 2d, y = – d, y = + d and y = +2d,respectively. The magnitude of the electric field E at a point on the x-axis at x = D, with D>>d, will behave as:

(1) E D1

(2) E 4D1

(3) E 3D1

(4) E 2D1

Ans. (2)Sol. Enet = 2E1 cos1 – 2E2cos2

= )Dd(KQ2

22 cos1 – )Dd4(KQ2

22 . cos2

= 2 2 3/2

2KQD(d D ) – 2/322 )Dd4(

KQD2

= 3DKQD2

2/322/32

1Dd2

1–

1Dd

1

Enet = 2DKQ2

22

Dd2

231–

Dd

23–1 (By binomial approximation)

= 2DKQ2

2

2

2

2

Dd

23–

Dd4

23

Enet = 2DKQ9

2

2

Dd

Enet 4D1

17. A test particle is moving in a circular orbit in the gravitational field produced by a mass density (r) = 2rK

.

Identify the correct relation between the radius R of the particle's orbit and its period T :(1) T/R2 is a constant (2) T/R is a constant(3) TR is a constant (4) T2/R3 is a constant

Ans. (2)

Sol. m = R

0

2drr4

= 4KR

K4

RT

= K42


18. A ve ry long solenoid of radius R is carrying current I(t) = kte–t (k > 0), as a function of time (t 0). counterclockwise current is taken to be positive. A circular conducting coil of radius 2R is placed in the equatorialplane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctlydepicted, as a function of time, by :

(1) (2)

(3) (4)

Ans. (2)Sol. outer = (0 nkte–t) 4R2

= dtd

= – ce–t [1 – t]

i = R

= t–eRC– (1 – t)

at t = 0 i = – veat large t i = +ve

19. A convex lens of focal length 20 cm produces images of the same magnification 2 when an object is kept attwo distances x1 and x2 (x1 > x2) from the lens. The ratio of x1 and x2 is :(1) 5 : 3 (2) 4 : 3 (3) 3 : 1 (4) 2 : 1

Ans. (3)Sol. Object must be placed between lens and focus in one case and beyond focus in other case

1

1 1

v f2

x x f

2

2 2

v f2

x f x

1

2

x3

x


20. The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to abattery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectricmaterial of dielectric constant K is placed between the two plates of the first capacitor. The new potentialdifference of the combined system is :-

(1) nK

V

(2) V (3) )nK(V)1n(

(4) nK

nV

Ans. (3)Sol. When battery is connected total charge on capacitors.

Q = (n + 1) cv

when battery is disconnected total charge on capacitors is conserved.

Ceq = (k + n) C

potential at ABtotal

eq

Q (n 1)vvC (k n)

21. The logic gate equivalent to the given logic circuit is :

(1) OR (2) AND (3) NOR (4) NANDAns. (1)

Sol.

22. A He+ ion is in its first excited state. Its ionization energy is :(1) 13.60 eV (2) 48.36 eV (3) 6.04 eV (4) 54.40 eV

Ans. (1)

Sol.2

2

(13.6)zEn

2

2

(13.6)(2) ( 13.6)eV2


23. In a conductor, if the number of conduction electrons per unit volume is 8.5 × 1028 m–3 and mean free time is25ƒs (femto second), it's approximate resistivity is :(me = 9.1 × 10–31 kg)(1) 10–5 m (2) 10–6 m (3) 10–8 m (4) 10–7 m

Ans. (3)

Sol. = 2ne

m 1.6 × 10–8 m

24. The position of a particle as a function of time t, is given by x(t) = at + bt2 – ct3 where a, b and c areconstants. When the particle attains zero acceleration, then its velocity will be :

(1) c2

ba2

(2) c3

ba2

(3) c

ba2

(4) c4

ba2

Ans. (2)Sol. x = at + bt2 – ct3

= a + 2bt – 3ct2

a = dtdv

= 2b – 6ct = 0 t = c3b

Vt = at C3

b2

25. The resistance of a galvanometer is 50 ohm and the maximum current which can be passed through it is0.002 A. What resistance must be connected to it in order to convert it into an ammeter of range 0 – 0.5 A ?(1) 0.02 ohm (2) 0.2 ohm (3) 0.5 ohm (4) 0.002 ohm

Ans. (2)

Sol.G

G

S(0.5 – 0.002) = 50 × 0.002 S = 0.2


26. Two materials having coefficients of thermal conductivity '3K' and 'K' and thickness 'd' and '3d', respectively,are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are '2' and '1'respectively, (2 > 1). The temperature at the interface is :

12

(1) 101 + 10

9 2 (2) 31 + 3

2 2 (3) 2

12 (4) 6

56

21

Ans. (1)Sol. In steady state heat current will be same in both parts.

2 1( ) ( )3kA kA

d 3d

92 – 9 = – 1

2 1910

27. A metal wire of resistance 3 is elongated to make a uniform wire of double its previous length. This newwire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60° at thecentre, the equivalent resistance between these two points will be :

(1) 25

(2) 5

12(3)

27

(4) 35

Ans. (4)

Sol. Sol. RA

Also, A = constant is doubled so A is halvedR' = 4R = 12 2r =

1R'

R 26

25R'

R 106

1 2eq

1 2

R R 20 5R

R R 12 3


28. Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with aspeed of 20 ms–1 with respect to the ground. If an observer in car A detects a frequency 2000 Hz of the soundcoming from car B, what is the natural frequency of the sound source in car B ?(speed of sound in air = 340 ms–1)(1) 2150 Hz (2) 2250 Hz (3) 2060 Hz (4) 2300 Hz

Ans. (2)

Sol. 2000 = f

20340

20–340

f = 2250 Hz

29. 50 W/m2 energy density of sunlight is normally incident on the surface of a solar panel. Some part ofincidnet energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on 1 m2

surface area will be close to (c = 3 × 108 m/s).(1) 15 × 10–8 N (2) 20 × 10–8 N (3) 35 × 10–8 N (4) 10 × 10–8 N

Ans. (2)

Sol. p = 0.75 CI

+ 2 ×

C

25.0 I

p 0.75 × 8–10350

+ 0.5 × 8–10350

= 8–106125

R = pA = 8–106125

= 20.8 × 108 N

30. Moment of inertia of a body about a given axis is 1.5 kg m2. Initially the body is at rest. In order to producea rotational kinetic energy of 1200 J, the angular acceleration of 20 rad/s2 must be applied about the axis fora duration of :(1) 2 s (2) 2.5 s (3) 3 s (4) 5 s

Ans. (1)

Sol.21

× 1.5 × 2 = 1200

= 40 rad/sec40 = 0 + 20 × tt = 2s


JEE Main Online test_09-04-19_EveningMathematics

Single Choice Questions :1. If p (q r) is false, then the truth values of p, q, r are respectively :

(1) T, T, F (2) F, T, T (3) F, F, F (4) T, F, FAns. (4)Sol. P (q r) F

p is 'T'and q r is F q F and r F

2. The total number of matrices A = 0 2y 12x y –12x –y 1

, (x, y R, x y) for which AATA = 3I3 is :

(1) 2 (2) 6 (3) 3 (4) 4Ans. (4)Sol. ATA = 3I3

0 2x 2x 0 2y 1 3 0 02y y –y 2x y –1 0 3 01 –1 1 2x –y 1 0 0 3

2

2

8x 0 0 3 0 00 6y 0 0 3 00 0 3 0 0 3

8x2 = 3, 6y2 = 3

x = ± 32 2

, y = ±12

3. A rectangle is inscribed in a circle with a diameter lying along the line 3y = x + 7. If the two adjacent verticesof the rectangle are (–8, 5) and (6, 5), then the area of the rectangle (in sq. units) is :(1) 72 (2) 98 (3) 56 (4) 84

Ans. (4)

Sol. 3(a 5)

2

= –1 + 7

a = –1 Sides = 6 & 14

Area = 6 × 14 = 84


4. Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, thesecond row consists of two balls and so on. If 99 more identical balls are added to the total number of ballsused in forming the equilateral triangle, then all these balls can be arranged in a square whose each sidecontains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number ofballs used to form the equilateral triangle is :(1) 262 (2) 225 (3) 190 (4) 157

Ans. (3)

Sol.n(n 1) 99

2

= (n –2)2

n2 + n + 198 = 2n2 – 8n + 8n2 – 9n – 190 = 0n = 19, – 10

Number = n(n 1) 19.20

2 2

= 190

5. Let z C be such that |z| < 1. If = 5 3z5(1– z)

, then :

(1) 5 Re() > 1 (2) 5 Im() < 1 (3) 5 Re() > 4 (4) 4 Im() > 5Ans. (1)Sol. |z| < 1

5w(1 – z) = 5 + 3z

z = 5w – 53 5w

|z| = 5w – 1

3 5w < 1

5|w – 1| < |3 + 5w|

|w – 1| < 3w – –5

6. If the system of equations 2x + 3y – z = 0, x + ky – 2z = 0 and 2x – y + z = 0 has a non-trivial solution

(x, y, z), then x y zy z x + k is equal to :

(1) 34

(2) –4 (3) –14

(4) 12

Ans. (4)

Sol.2 3 –11 k –22 –1 1

= 0 k = 92

2x + 3y – z = 0 ......(1)

x + 92

y – 2z = 0 ......(2)

2x – y + z = 0 ......(3)

y 12 2 &

x 1–y 2 ; Now

zx

= –4 ; x y zy z x + k =

12


7. The value of sin10º sin30º sin50º sin70º is :

(1) 1

32 (2) 1

36 (3) 1

16 (4) 1

18

Ans. (3)Sol. (sin10º sin30º sin70º) sin30º

= 14

(sin30º)2 = 1

16

8. If the function f(x) = a | – x | 1, x 5b | x – | 3,x 5

is continuous at x = 5, then the value of a – b is :

(1) 2

5 – (2) 2– 5 (3)

25 (4)

–25

Ans. (1)Sol. f(x) is continuous at x = 5

a| –5| + 1 = b|5 – | + 3 a(5 – ) + 1 = b(5 – ) + 3 (a – b) (5 – ) = 2

a – b = 2

5 –

9. If f(x) = [x] – x4

, x R, where [x] denotes the greatest integer function, then :

(1) x 4lim

f(x) exists but x 4–lim

f(x) does not exist.

(2) f is continuous at x = 4.

(3) x 4–lim

f(x) exists but x 4lim

f(x) does not exist.

(4) Both x 4–lim

f(x) and x 4lim

f(x) exist but are not equal.

Ans. (2)

Sol.x 4

xlim [x] –4

= 4 – 1 = 3

–x 4

xlim [x] –4

= 3 – 0 = 3

f(4) = 4 – 1 = 3 f(x) is continuous at x = 4


10. If a unit vector a

makes angles /3 with i , /4 with j and (0, ) with k , then a value of is :

(1) 23

(2) 56

(3) 4

(4) 512

Ans. (1)Sol. cos2 + cos2 + cos2 = 1

1 14 2 + cos2 = 1 cos = ±

12

= 2,

3 3

11. If m is chosen in the quadratic equation (m2 + 1)x2 – 3x + (m2 + 1)2 = 0 such that the sum of its roots isgreatest, then the absolute difference of the cubes of its roots is :

(1) 8 3 (2) 10 5 (3) 8 5 (4) 4 3

Ans. (3)

Sol. Sum of roots = 23

1 m if m = 0 then sum of roots has maximum value is 3

x2 – 3x + 1 = 0

+ = 3= 1|3 – 3| = | – |(2 + 2 + )|

= 2 2( ) – 4 . ( ) –

= 5 × 8 = 8 5

12. If the tangent to the parabola y2 = x at a point (, ), ( > 0) is also a tangent to the ellipse, x2 + 2y2 = 1, then is equal to :

(1) 2 2 + 1 (2) 2 2 – 1 (3) 2 + 1 (4) 2 – 1

Ans. (3)

Sol. y() = 12

(x + 2)

2y = x + 2

y = 12 x +

2

m = 12 , c =

2

2

2

21 1

2 24

2

21 1

4 24

2 = 2 + 1


13. If sec x 2e sec x tanx f(x) (sec x tanx sec x) dx = esecxf(x) + C, then a possible choice of f(x) is :

(1) secx + tanx + 12

(2) xsecx + tanx + 12

(3) secx – tanx – 12

(4) secx + xtanx – 12

Ans. (1)

Sol. sec x 2e sec x.tanx.f(x) (sec x. tanx sec x) dx = esecx.f(x) + CDifferentiate with respect to x both side esecx(secx.tanx.f(x) + secx.tanx + sec2x) = esecx(secx.tanx.f(x)) + esecx.f(x) f(x) = sec2x + tanx.secx f(x) = tanx + secx + C

14. The common tangent to the circles x2 + y2 = 4 and x2 + y2 + 6x + 8y – 24 = 0 also passes through the point:(1) (6, –2) (2) (–4, 6) (3) (4, –2) (4) (–6, 4)

Ans. (1)Sol. C1(0, 0), r1 = 2, C2 (3, –4), r2 = 7

C1C2 = |r1 – r2| S1 – S2 = 0 equation of common tangent 6x + 8y – 20 = 0

3x + 4y = 10(6, –2) satisfy it

15. If the sum and product of the first three terms in an A.P. are 33 and 1155, respectively, then a value of its 11th

term is :(1) –36 (2) –25 (3) –35 (4) 25

Ans. (2)Sol. a – d + a + a + d = 33 a = 11

a(a2 – d2) = 1155 121 – d2 = 105 d = ±4If d = 4 then a = 7if d = –4 then a = 15T11 = 7 + 40 = 47


16. The value of the integral 1

–1 2 4

0

xcot (1– x x )dx is :

(1) 2

– loge2 (2) 1–

2 2

loge2 (3) 4

– loge2 (4) 1–

4 2

loge2

Ans. (4)

Sol.1 12 2

1 1 2 1 22 2

0 0

x (x 1)x tan dx x tan (x ) tan (x 1) dx1 x (x 1)

1 11 1 1

0 0

1 tan t tan (t 1)dt 1.tan t dt2

(Applying 'a + b – x')

11120

0

t(tan t)(t) dt1 t

1 10 ln(2) ln24 2 4 2

17. If f : R R is a differentiable function and f(2) = 6, then f (x)

x 26

2t dtlim(x – 2) is :

(1) 12f(2) (2) 24f(2) (3) 0 (4) 2f(2)Ans. (1)

Sol.

f (x)

6x 2

2t dt

lim(x – 2)

=

x 2

2f(x) f '(x)lim1

= 2f(2).f(2) = 12f(2)

18. Two poles standing on a horizontal ground are of heights 5 m and 10 m respectively. The line joining their topsmakes an angle of 15º with the ground. Then the distance (in m) between the poles, is :

(1) 5 (2 3)2

(2) 10( 3 – 1) (3) 5(2 3) (4) 5( 3 1)

Ans. (3)Sol. In ABC

5x

= tan 15º

x = 5

2 3

x = 5 2 3

2 3 2 3

x = 5 2 3


19. The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 + ..... upto 11th term is :(1) 916 (2) 946 (3) 945 (4) 915

Ans. (2)

Sol. S = 11

n 1

n(2n 1)

= 11.12.23 11.122

6 2

= 44.23 – 11.6= 11 (92 – 6)= 11 × 86= 946

20. Two newspapers A and B are published in a city. It is known that 25% of the city population reads A and 20%reads B while 8% reads both A and B. Further, 30% of those who read A but not B look into advertisementsand 40% of those who read B but not A also look into advertisements, while 50% of those who read both Aand B look into advertisements. Then the percentage of the population who look into advertisements is :(1) 13.9 (2) 13.5 (3) 12.8 (4) 13

Ans. (1)Sol. Let population = 100

n(A) = 25n(B) = 20

n(A B) = 17

n(A B) = 12

30 40 5017 12 8

100 100 100

= 51 + 48 = 4 = 13.9

21. The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34, x, 42, 67, 70,

y are 42 and 35 respectively, then yx

is equal to :

(1) 8/3 (2) 7/3 (3) 7/2 (4) 9/4Ans. (2)

Sol.34 x

2

= 35

x = 36

42 = 10 22 26 29 34 36 42 67 70 y

10

y = 84

y 84 7x 36 3


22. The domain of the definition of the function f(x) = 21

4 – x + log10(x3 – x) is :

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

Ans. (2)Sol. domain : 4 – x2 0 x ± 2 ........(1)

& x3 – x > 0 x(x – 1) (x + 1) > 0

x (–1, 0) (1, ) ........(2)(1) (2)

23. Let P be the plane, which contains the line of intersection of the planes, x + y + z – 6 = 0 and 2x + 3y + z +5 = 0 and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to :

(1) 205 5 (2) 17/ 5 (3) 63 5 (4) 11/ 5

Ans. (4)Sol. (x + y + z – 6) + 2x + 3y + z + 5 = 0

( + 2)x + ( + 3)y + ( + 1)z + (5 – 6) = 0 + 1 = 0 = –1 P : x + 2y + 11 = 0

distance = 115

24. The area (in sq. units) of the smaller of the two circles that touch the parabola, y2 = 4x at the point (1, 2) andthe x-axis is :

(1) 4 (3 + 2 ) (2) 8 (2 – 2 ) (3) 8 (3 – 2 2 ) (4) 4(2 – 2 )Ans. (3)Sol. Equation of circle is

(x – 1)2 + (y – 2)2 + (x – y + 1) = 0 x2 + y2 + x( – 2) + y(–4 – ) + (5 + ) = 0 Circle touches x-axis g2 – c = 0

2( – 2)

4

= 5 + = 4 ± 4 2

Radius = –4 –

2

, put & get least value of radius.


25. If the two lines x + (a – 1) y = 1 and 2x + a2y = 1(a R – {0, 1}) are perpendicular, then the distance of theirpoint of intersection from the origin is :

(1) 25 (2)

25 (3)

25 (4)

25

Ans. (2)

Sol. 2–1 –2

a – 1 a

= –1

2 = –a2(a – 1) a3 – a2 + 2 = 0 (a + 1) (a2 – 2a + 2) = 0

D < 0 a = –1 L1 : x – 2y + 1 = 0

L2 : 2x + y – 1 = 0

Now : 0(0, 0) & 1 3P ,5 5

OP = 2 21 3– 0 – 0

5 5

= 1 9 2

25 25 5

26. The area (in sq. units) of the region A = {(x, y) : 2y

2 x y + 4} is :

(1) 533 (2) 18 (3) 30 (4) 16

Ans. (2)Sol. y2 2x & x – y 4

Area = 4 2

2

y(y 4) dy2

16 4 14(4 2) (64 8)2 6

= 6 + 24 – 12

= 18


27. A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is tan–112

. Water

is poured into it at a constant rate of 5 cubic meter per minute. Then the rate (in m/min.), at which the levelof water is rising at the instant when the depth of water in the tank is 10 m; is :(1) 1/15 (2) 1/5 (3) 2/ (4) 1/10

Ans. (2)

Sol. tan = rh

r = h2

15

V = 13

3h4

30º

2dV h dhdt 4 dt

–5 =

100 dh4 dt

;

dh 1dt 5

28. The vertices B and C of a ABC lie on the line, x 2 y – 1 z

3 0 4

such that BC = 5 units. Then the area (in

sq. units) of this triangle, given that the point A(1, –1, 2), is :

(1) 5 17 (2) 34 (3) 6 (4) 2 34

Ans. (2)

Sol. AD(3i 4k) 0

3(3 – 3) + 0 + 4(4 – 2) = 0

= 1725

AD = 51 68ˆ ˆ ˆ– 3 i 2 j – 2 k25 25

= 24 18ˆ ˆ ˆi 2 j k25 25

576 324AD 4625 625

= 900 4625

= 3400 1034625 25

Area of = 1 25 342 5 = 34


29. If cosx dydx – ysinx = 6x, (0 < x <

2

) and y3

= 0, then y6

is equal to :

(1) 2

–2

(2) 2

–2 3

(3) 2

–4 3

(4) 2

2 3

Ans. (2)

Sol. xsecx6xtany–dxdy

xcoseFxdxtan–

Solution is dxxcos·xsecx6)x(cosy or y·cosx = 3x2 + C

30. If some three consecutive coefficients in the binomial expansion of (x + 1)n in powers of x are in the ratio 2 :15 : 70, then the average of these three coefficients is :(1) 964 (2) 227 (3) 625 (4) 232

Ans. (4)

Sol.n

r–1n

r

C 215C

r 2

n – r 1 15

17r = 2n + 2 ......(1)

Nown

rn

r 1

C 1570C

nr

nCr. n – r

r 1 3n – r 14

3n – 17r = 14 ......(2)Solving (1) & (2) : n = 16 & r = 2 16C1, 16C2, 16C3

Now16 16 16

1 2 3C C C 16 120 560 6963 3 3

= 232

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