JEE MAIN-FULL TEST 03 - PHYSICS...JEE MAIN-FULL TEST 03 - PHYSICS A light rod of length , is hanging from the vertical smooth wall of a vehicle moving with acceleration small mass

JEE MAIN-FULL TEST 03

SOLUTIONS OF MOCK TEST

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Q No. - 1

JEE MAIN-FULL TEST 03 - PHYSICS

A swimmer can swim in still water with a speed of . While crossing a river his average speed is

cross the river in the shortest possible time, what is the speed of flow of water ?

(a).

(b).

(c).

(d).

CORRECT OPTION: A

SOLUTION

Avg. speed

Q No. - 2


A car starting from rest is accelerated at constant until it attains a constant speed v. It is then retarded at a

rate until it comes to rest. Considering that the car moves with constant speed for half of the time of total jo

average speed of the car for the journey is

(a).

(b).

(c).

(d). Data insufficient

CORRECT OPTION: B

SOLUTION

Q No. - 3


A smooth ring of mass can slide on a fixed horizontal rod. A string tied to the ring passes over a fixed

carries a block of mass as shown in the figure. At an instant, the string between the ring and the

makes an angle with the rod. The initial acceleration of the ring is

√5m

s3m

s

2m

s

4m

s

6m

s

8m

s

3 =√(Vrt)2 + (Vmrt)2

t⇒ V 2r + 5 = 9

⇒ Vr = 2m

s

v

43v

43v

2

Vavg = =× × v + × v12

t2

12

t

3v

4

P m

Q ( )m2

60 ∘

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(a).

(b).

(c).

(d).

CORRECT OPTION: C

SOLUTION

�..(i)

�.(ii)

Solving (i) and (ii) acceleration of ring

Q No. - 4


A block of mass is hanging over a smooth and light pulley through a light string. The other end of the s

pulled by a constant force . If kinetic energy of the block increases by in . Then

(a). tension in the string is .

2g

32g

62g

9g

3

g − T = am

2

m

2Tcos 60 ∘ =

ma

cos 60 ∘

=2g

g

M

F 20J 1s

Mg

(b). tension in the string is

(c). Work done by the tension on the block is 20 J in 1 sec.

(d). Work done by the force of gravity is 20 J in 1 sec.

CORRECT OPTION: B

SOLUTION

Work done by all the forces on the block equal to change in kinetic energy.

Q No. - 5


A circular ring is fixed in a gravity free space and point of the ring is earthed. Now a magnet is placed alon

the ring at a distance from its centre such that the nearer pole is north pole as shown in figure. A sharp im

applied on the magnet so that it starts to move towards the ring. Then,

(a). Initially magnet experiences an acceleration and then it retards to come to an instantaneous rest.

(b). Magnet starts to oscillate about centr of the ring

(c). Magnet continous to move along the axis with constant velocity

(d). The magnet retards and comes to rest finally.

CORRECT OPTION: D

SOLUTION

N/A

F

Q No. - 6


A block of mass is suspended by means of an ideal spring of force constant from ceiling of a car whic

moving along a circular path of radius with acceleration . The time period of oscillation of the block whe

displeced along the spring, will be

(a).

m k

r a

2π√mg + mak

(b).

(c).

(d).

CORRECT OPTION: C

SOLUTION

No effect of \'a\' and \'g\' on time period of spring pendulum.

Q No. - 7


In the hydrogen atom spectrum and represent wavelengths emitted due to transition from sec

first excited states to the ground state respectively. The value of is

(a).

(b).

(c).

(d).

CORRECT OPTION: A

SOLUTION

√m

k√g2 + a2

2π√m

k

2π√m

√k2 + g2 + a2

λ3 − 1 λ2 − 1λ3 − 1

λ2 − 127

3232

274

99

4

= R( − ) =1λ3 − 1

1

121

328R

9

= R( − ) =1λ2 − 1

1

121

223R

4

⇒ =λ3 − 1

λ2 − 1

27

32

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Q No. - 8


A block of mass 1 kg is pulled along the curve path ACB by a tangential force as shown in figure. The wor

the frictional force when the block moves from A to B is

(a). 5 J

(b). 10 J

(c). 20 J

(d). none of these

CORRECT OPTION: C


SOLUTION

Work done by friction

Q No. - 9


In a capillary tube placed inside the liquid of density in a container, the rise of liquid is h. When block o

is placed on the liquid as shown in figure, liquid in the tube is h\'. If then

(a).

= ∫→F .

→ds = ∫

x

0μmg cos θ

dx

cos θ

= μmg × = 20J

(ρ)

σ < ρ

h' = h

(b).

(c).

(d). insufficient data

CORRECT OPTION: A

SOLUTION

There will be no change.

Q No. - 10


A light rod of length , is hanging from the vertical smooth wall of a vehicle moving with acceleration

small mass attached at it\'s one end is free to to rotate about an axis passing through the other end. The m

velocity given to the mass at it\'s equilibrium possition so that it can complete vertical circular motion is

(a).

(b).

(c).

(d). none of these

CORRECT OPTION: C

h' < h

h' > h

∴ H' = h

L √3g

√5gL

√4gL

√8gL

SOLUTION

Conservation of energy

Q No. - 11


Energy stored in the capacitor in it\'s steady state is

mv2 − m = m + m√3(2l )g12

g(l)

2

g(l)

2

√3

2

v = √8gl

(a).

(b).

(c).

(d). zero

CORRECT OPTION: D

SOLUTION

Potential across capacitor is zero, hence energy stored is zero.

Q No. - 12


The potential difference between two points A and B which are separated by a distance of 1m along the fie

uniform electric field of is

(a). zero

(b). volt

(c). volt

(d). volt

CORRECT OPTION: C

SOLUTION

.

CV

CV

2

QV1

2

10NC − 1

100

10

0.1

VA − VB = ∫E. dx = 10(1)

= 10vo

Q No. - 13


The power factor of a circuit in which a box having unknown electrical devices connected in series with a r

resistance is . The reactance of the box is

(a).

(b).

(c).

(d).

CORRECT OPTION: C

SOLUTION

but cos

.

Q No. - 14


Two points A and B are at distances of \'a\' and \'b\' respectively from an infinite conducting plate having ch

density . The work done in moving charge from A to B is

(a).

(b).

(c).

(d). none of these

CORRECT OPTION: A

SOLUTION

Q No. - 15


A point charge of is placed on the circumference of a non-conducting ring of radius 1m which is rota

constant angular acceleration of . If ring starts it\'s motion at the magnetic field at the cent

ring at sec, is

(a).

(b).

(c).

3Ω3

55Ω

Ω5

34Ω

Ω4

3

Z = √R2 + X2 = √9 + X2

ϕ = =R

Z

3

5X = 4Ω

σ Q

(b − a)Qσ

ε0

Qσ

(b − a)Qσ

(b − a)ε0

ε =σ

ε0

VA − VB = (b) − (a)σ

ε0

σ

ε0

∴ W = Q(VA − VB) = (b − a)Qσ

ε0

0.1C

1rad

sec2t = 0

t = 10

10− 6T

10− 7T

10− 8T

(d).

CORRECT OPTION: B

SOLUTION

Q No. - 16


The wavelength corresponding to maximum spectral of a black body A is . Consider anoth

body B whose surface area is twice of that of A and total radiant energy emitted by B is 16 times that emitt

The wavelength corresponding to maximum spectrum radiancy for B will be

(a).

(b).

(c).

(d).

CORRECT OPTION: D

SOLUTION

107T

ω = 0 + 1 × 10 = 10rad

sec2

∴ V = rω = 1 × 10 = 10m

s

→B = ⇒

∣∣∣

→B

∣∣∣

=μ0

4π

q(→v × →r )

r3μ0qv

4πr2

B = = 10− 7T10− 7 × 0.1 × 10

(1)2

λA = 5000 �

5000(8) �14

2500 �10, 000 �

�5000(8)

14

P = σAT 4

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Since = constant,

Q No. - 17


During an adiabatic process, the density of a gas is found to be proportional to cube of temperature. The d

freedom of gas molecule is

(a). 6

(b). 5

(c). 4

(d). 3

CORRECT OPTION: A

SOLUTION

For adiabatic process, = constant

= constant

= constant

Q No. - 18


Two fixed charge and are located at the point of co-ordinates and respectively

plane. Then all the points in plane where potential is zero lies on a

(a). straight line parallel to x-axis

(b). straight line parallel to y-axis

(c). a circle of radius 4a

(d). circle of radius 2a

CORRECT OPTION: C

SOLUTION

According to equation

⇒ = ( )4

⇒ 16 = ( )4

⇒ TB = TA(8)PA

PB

AB

AA

TB

TA

2

1

TB

TA

14

λmT = = (8)λA

λB

TB

TA

14

⇒ λB = = �λA

(8)14

5000

(8)14

TV γ− 1

T( )γ− 1

m

ρT

ργ− 1

ρ ∝ T ⇒ = 3 ⇒ γ =1

γ− 11

γ − 1

4

3

f = = = 62

γ − 1

2

( − 1)43

−2Q Q ( − 3a, 0) (3a, 0)

(x − y) x − y

P(x, y), A( − 3a, 0), B = (3a, 0)

VPA = , VPB =1

4πε0

−2Q

PA

1

4πε0

Q

PB

( ) + = 014πε0

−2Q

PA

1

4πε0

Q

PB

Q No. - 19


In an circuit shown in the figure, . At time , charge in the capacitor is

decre4asing at a rate of . Choose the correct statements.

(a). maximum charge in the capacitor can be 6C

(b). maximum charge in the capacitor can be 8C

(c). charge in the capacitor will be maximum after time sec

(d). None of these

CORRECT OPTION: A

SOLUTION

Q No. - 20

= ⇒ 4PB2 = PA22

PA

1

PB(x − 5a)2 + y2 = (4a)2

L − C C = 1F , L = 4H t = 0 4C

√5C

s

2 sin− 1( )23

I = √5A

= + Li2 ⇒ qmax = 6Cq2m

2C

q2

2C

1

2


A plank of mass is placed over smooth inclined plane and a sphere is also placed over the plank. Fricti

sufficient between the sphere and the plank to revent slipping. If the plank and the sphere are released fro

frictional force on the sphere is

(a). up the plane

(b). down the plane

(c). zero

(d). horizontal

CORRECT OPTION: C

SOLUTION

N/A

Q No. - 21

M


A disc of mass and radius is placed over a plank of same mass . There is sufficient friction betwee

and the plank to prevent slipping. A force is appliled at the centre of the disc. Choose the correct statem

(a). Acceleration of the plank is

(b). Acceleration of the plank is

(c). Force of friction between disc and plank is

(d). Force of friction between disc and plank is

CORRECT OPTION: A

SOLUTION

m R m

F

F

4mF

2mF

6F

2

f = ma2

α = =τ

I

2f

mR

Q No. - 22


A certain amount of ideal monoatomic gas undergoes, process given by where is the intern

of the gas. The molar specific heat of the gas for the process will be

(a).

(b).

(c).

(d).

CORRECT OPTION: D

⇒ a2 = m, f =F

4

F

4

UV = C12 U

R

23R

5R

2

−R

2

SOLUTION

The process is equivalent to

Compare with

Q No. - 23


An isotropic sound is moving in a circle of radius with a small speed . An observer is hearing this

(See figure). The intensity of the sound heard by will be maximum when the source is at point.

(a). 1

(b). 2

(c). 6

(d). none of these

CORRECT OPTION: A

SOLUTION

Intensity will be highest at the nearest point.

TV = C12

TV x− 1 = C ⇒ x =3

2

⇒ C = + = + = R − 2R = − RR

γ − 1

R

1 − X

R23

R

1 − ( )32

3

2

1

2

A R v B

B


Q No. - 24


A stable species can be obtained through unstable and . The half life period for conversion

that for conversion is . Initially a sample contains unclides of unclides of and no nu

. After how much will the nuclides of species be equal to

(a).

(b).

(c).

(d).

CORRECT OPTION: C

SOLUTION

N/A

Q No. - 25


A certain thermodynamic cycle comprises of two isothermal and adiabatic processes. The highest tempera

obtained in the entire cycle is while the lowest temperature is . When the cycle represented o

curve the sence of the cycle is counter clockwise. The efficiency of the cycle represented on a curv

sence of the cycle is counter clockwise. The efficiency of the cycle is

(a).

(b).

(c).

(d). None of these

CORRECT OPTION: D

C A B A → C

B → C 2T N A, N B

C ( )N2716

2T

3T

4T

5T

600K 300K

P − V

50 %

75 %

25 %


SOLUTION

N/A

Q No. - 26


A man wearing spectales with diverging lenses is viewing a distance object. The image formed on his retin

(a). real and erect.

(b). real and inverted

(c). virtual and erect

(d). virtual and inverted.

CORRECT OPTION: B

SOLUTION

N/A

Q No. - 27


A thin uniform hemispherical bowl of mass and radius is lying on a smooth horizontal surface. A horiz

is now applied perpendicular to the rim of the bowl (see figure). The instantaneous angular acceleration

will be

(a).

(b).

(c).

m R

F

20

3

F

MR10

3

F

MR40

3

F

MR

(d).

CORRECT OPTION: B

SOLUTION

N/A

Q No. - 28


For a stone thrown from a tower of unkown height, the maximum range for a projection peed of is o

a projection angle of . The correcsponding distance between the foot of the tower and the point landin

stone is

(a).

(b).

(c).

(d).

CORRECT OPTION: D

SOLUTION

Q No. - 29


Consider the system shown in the figure. The system is so arranged that both and are in pure trans

a certain instant is moving down with a speed of a speed of then speed of will be

5

3

F

MR

10m

s30 ∘

10m

20m

( )m20√3

( )m10√3

tan θ = ⇒ = = =u2

Rg

u2

g tan θ

100

100 × √3

10

√3

m M

M 1m

s1m

sm

(a).

(b).

(c).

(d).

CORRECT OPTION: C

SOLUTION

N/A

Q No. - 30


Consider the circuit in the adjacent figure. What will be potential difference between and in the steady

(a).

(b).

(c).

(d). zero

CORRECT OPTION: D

SOLUTION

N/A

7m

s

8m

s

1m

s

15m

s

A B

εε

2ε

3

Q No. - 31

JEE MAIN-FULL TEST 03 - CHEMISTRY

For the equilibrium, and

hence, for the equilibrium:

equilibrium constant is:

(a).

(b).

(c).

(d).

CORRECT OPTION: A

SOLUTION

N/A

Q No. - 32


2-Pentyne, on reduction with lithium in liq. yields

(a). pentane

(b). cis-2-pentene

(c). trans-2-pentene

(d). 1-pentene

CORRECT OPTION: C

SOLUTION

N/A

Q No. - 33


Ag+ + 2NH3 ⇔ [Ag(NH3)2]+K1 = 1.8 × 10

7

Ag+ + CI − ⇔ AgCI, K2 = 5.6 × 109 AgCI + 2NH3 ⇔ Ag(NH3)

+2

0.32 × 10− 2

0.31 × 10− 21

1.01 × 1017

1.01 × 10− 17

NH3


Chlordobenzene. A forms only one nitro derivative. Hence is,

(a). Benzyl chloride

(b). p-Chlorotoluene

(c). o-Chlorotoulene

(d). m-Chlorotoulene

CORRECT OPTION: B

SOLUTION

N/A

Q No. - 34


Clasien rearrangement of allyl phenyl ether gives a mixture of

(a). -and p-allyl phenols.

(b). -and -allyl phenols.

(c). -and -allyl phenols.

(d). -and -allyl phenols.

CORRECT OPTION: D

SOLUTION

N/A

Q No. - 35


If in diamond, there is a unit cell fo carbon atoms as fcc and if carbon atom is hybridized, what fraction

ae occupied by carbon atom.

(a). tetrahedral

(b). tetrahedral

(c). octahedral

(d). octahedral

CORRECT OPTION: B

SOLUTION

N/A

Q No. - 36


Out of boiling point (I), entropy (II), (III) and standard e.m.f. of a cell (IV), intensive properties are-

(a).

(b).

(c).

(d). all of these

CORRECT OPTION: C

A(C7H7CI)( i )KMnO4−−−−−−→( ii ) Sodalime

Δ

A

o − , m

o m

m p

o p

sp3

25 %

50 %

25 %

50 %

pH

I, II

I, II, III

I, III, IV

SOLUTION

N/A

Q No. - 37


Which metal liberates with very dilute nitric acid:

(a).

(b).

(c).

(d).

CORRECT OPTION: C

SOLUTION

N/A

Q No. - 38


Which of the following statements are not correct?

(a). The number of and -particles emitted in neptunium series and 7 and 4, respe

(b). The number of and -particles emitted in uranium series are 8 and 7, respective

(c). The number of and -particles emitted in actinium series are 4 and 7, respective

(d). The half-life of a radiactive decay is given as , where is decay constant

CORRECT OPTION: C

SOLUTION

N/A

Q No. - 39


Reactivity of borazole is greater than that of benzene because:

(a). borazole is non-polar compound

(b). borazole is polar compound

(c). borazole has electron in it

(d). of localized electrons in it

CORRECT OPTION: B

SOLUTION

N/A

H2

Zn

Cu

Mn

Hg

α β (.23793 Np →20983 Bi)

α β (.23892 U →20683 Pb)

α β (.23592 U →20783 Bi)

t =12

0.693

λλ


Q No. - 40


When acetone and chloroform are mixed, hydrogen bonding takes place between them. Such a liquid pair

(a). No deviations from Raoult\'s law

(b). Negative deviations from Raoult\'s law

(c). Positive deviations from Raoult\'s law

(d). None

CORRECT OPTION: B

SOLUTION

N/A

Q No. - 41


On heating ammonium dichromate, the gas evolved is

(a). Oxygen

(b). Ammonia

(c). Nitrous oxide

(d). Nitrogen

CORRECT OPTION: D

SOLUTION

N/A

Q No. - 42


When ammonia is added to a cupric salt solution, the deep blue colour, so observed is due to the formatio


(a).

(b).

(c).

(d).

CORRECT OPTION: B

SOLUTION

N/A

Q No. - 43


Arrange the following in order of their decreasing thermal conductivity

(a).

(b).

(c).

(d).

CORRECT OPTION: C

SOLUTION

N/A

Q No. - 44


In a measurement of quantum efficiency of photo-synthesis in green plants, it was found that 10 quanta of

wavelength 6850 were needed to release one molecule of . The average energy storage in this pro

112 kcal/mol evolved. What is the energy conversion efficiency in this experiment? Given

.

(a). 23.5

(b). 26.9

(c). 66.34

(d). 73.1

CORRECT OPTION: B

SOLUTION

Total energy of 10 quanta

Energy stored for process

efficiency

Q No. - 45


The rate of effusion of two gases \'a\' and \'b\' under identical condition of temperature and pressure are in

[Cu(OH)4]2 −

[Cu(NH3)4]2 +

[Cu(OH)2(NH3)2]

[Cu(H2O)4]2 +

AI, Ag, Cu

Cu, Ag, AI

Ag, Cu, AI

AI, Cu, Ag

� O2O2

cal = 4.18J, NA = 6 × 1023, h = 6.64 × 10− 34JS

E = = 2.9 × 10− 19Jhc

λ= 10 × 2.9 × 10− 19J

= = 7.8 × 10− 19J112 × 4.18 × 103

6 × 1023

% = × 1007.8 × 10− 19

29 × 10− 19= 27.9 %

2: 1. What is the ratio of rms velocity of their molecules if and are in the ratio of 2:1 ?

(a).

(b).

(c).

(d).

CORRECT OPTION: C

SOLUTION

As

Q No. - 46


For first order parallel reaction and are 4 and respectively at . If the activation energie

formation of and are respectively and respectively. The Temperature at which

will be obtained in equimolar ratio is:

(a).

Ta Tb

2: 1

√2: 1

2√2: 1

1: √2

= √rarb

2

1

Mb

Ma

Vrms ∝ √T

MVrms = √

3RT

M

= √ = × =Vrms (a )

Vrms ( b )

Ta × MbTb × Ma

2

1

√2

1

2√2

2

k1 k2 2− 1

min 300K

B C 30, 000 33, 314 olJ

m

757.48K

(b).

(c).

(d). None of these

CORRECT OPTION: B

SOLUTION

�(i)

�(ii)

Eqn. (ii) -Eqn.(i)

(For equimolar formation of and

.

Q No. - 47


One mole of was placed in a two litre vessel at a certain temperature. The following equilibrium was

in the vessel.

The equilibrium mixture reacted with 0.2 mole in acidc medium. Hence is:

(a). 0.5

(b). 0.25

(c). 0.125

(d). none of these

CORRECT OPTION: C

SOLUTION

Only will oxidized.

Equivalent of Equivalent of

329.77K

600K

ln( ) = [ − ]k'1k1

E1

R

1

T1

1

T1

ln( ) = [ − ]k'2k2

E2

R

1

T1

1

T2

ln( ) = [ − ]k'2 × k1k2 × k'1

(E2 − E1)

R

1

T1

1

T2

B C, k'2 = k'1 )

ln( ) = ( )( )k1k2

8.314

8.314

T2 − 300

300 × T2

ln 2 = = ( )( )83148.314

T2 − 300

300 × T2T2 = 329.77K

SO3

2SO3(g) ⇔ 2SO2(g) + O2(g)

KMnO4 Kc

2SO3(g)g ⇔ 2SO2(g) + O2(g)

at equilibrium (1 − 2x) (2x) x

SO2

SO2 = KMnO4

2x × 2 = 0.2 × 5

2x = 0.5

Kc = = 0.125[ ]2[ ]0.52

0.252

[ ]20.52


Q No. - 48


A complex of iron and cyanide ions is ionized at 1 molal. If its elevation in boiling point is , th

complex is: (Given

(a).

(b).

(c).

(d).

CORRECT OPTION: B

SOLUTION

It means salt on dissociation gives 4 ions. Thus the salt the gives 4 ions is .

Q No. - 49


Find the standard cell potential involving the cell reaction:

, at

Given:

(a).

(b).

(c).

(d).

CORRECT OPTION: C

100 % 2.08∘

Kb = 0.52∘Cmol− 1kg)

Fe[Fe(CN)6]3K3[Fe(CN)6]K4[Fe(CN)6]Fe3[Fe(CN)6]2

ΔTb = i × kb × m

2.08 = 0.52 × 1 × i

i = 4

K3[Fe(CN)6]

In2 + + Cu+ 2 → In+ 3 + Cu+ 298K

E0C u+

= X1V E0

I n+= X2FV E

0

I n+= X3Vu+ 2

Cn+ 3

In+ 2

I

X1 + X3 − X2X1 + X3 − 2X2

3X1 + X3 − 2X2

X1 + X3 + 2X2


SOLUTION

From Eqn. (i) + (iii) -(ii)

Q No. - 50


is a covalent liquid because

(a). clouds of the ions are weakly polarized to envelop the cation

(b). electron clouds of the ions are strongly polarized to envelop the cation

(c). its molcules are attracted to one another by strong vander Waals\' forces

(d). shown inter-pair effect

CORRECT OPTION: B

SOLUTION

Due to Fajan\'s rule.

Q No. - 51


Boron nitride obtained by heating borazole is:

(a). white solid with a diamoned like structure

(b). slippery white solid with layered structure similar to that of graphite

(c). covalent liquid and is structurally similar to carbon monoxide

(d). low melting solid with rock-salt like structure molecules

CORRECT OPTION: B

SOLUTION

N/A

Q No. - 52


If a metal has low oxygen affinity then purification of metal may be carried out by

(a). Liquation

(b). Distillation

(c). Zone refining

(d). Cupellation

CORRECT OPTION: B

(Cu+ 2 + e− → Cu+ , , E ∘ = X1V , ΔG ∘1 = − FX1, ...(i)),

(In+ 3 + 2e− → In+ 1, E ∘ = X2V , ΔG ∘2 = − 2FX2, . . (ii)),

(In+ 2 + e− → In+ , E ∘ = X2V , ΔG ∘3 = − FX3, ...(iii))

In+ 2 + Cu2 + → In+ 3 _ Cu+ ΔG ∘ = − FE ∘

ΔG ∘ = − F(X1 + X3 − 2X2) = − FE∘

E ∘ = (X1 + X3 − 2X2)

SnCI4

e− CI −

CI −

Sn

SOLUTION

N/A

Q No. - 53


An original salt solution in acidic medium did not give any precipitate on passing gas. Such a solution

obiled, reboiled after dilution 3 times. To such a solutions two drops of conc. were added, then hea

water was added. To this resulting solution, was first added followed by excess of . fina

ppt. was obtained. Hence the cation may be.

(a).

(b).

(c).

(d).

CORRECT OPTION: D

SOLUTION

Green ppt. is

Q No. - 54


Which of the following statement is wrong?

(a). orbital taking part in is

(b). orbital taking part in is

(c). orbitals taking part in are and

(d). orbitals taking part in are and

CORRECT OPTION: B

SOLUTION

N/A

Q No. - 55


Here is:

(a). Glycollic acid

(b). -hydroxypropionic acid

(c). succinic acid

(d). malonic acid

CORRECT OPTION: D

H2S

HNO3

NH4CI NH4OH

AI + 3

Fe+ 2

Fe+ 3

Cr+ 3

−Cr(OH)3

d dsp2 dx2 − y2

d sp3d dyz

d sp3d2 dx2 − y2 dz2

d sp3d3 dxy, dz2 dz2 − y2

H3CCOOHB

−−→ (Y )( i )KCN

−−−−−→( ii )H3O+

(X)

r2P

(X)

α

SOLUTION

Q No. - 56


In order of distinguish between and , which of the following reagents is useful

(a). Heinsberg reagent

(b). p-napthol

(c). Benzene diazonium chloride

(d). None of these

CORRECT OPTION: B

SOLUTION

aromatic amine on diazotisation followed by cuplting with -napthol gives azo dye test.

Q No. - 57


The weakest interparticle forces are present in

(a). thermosetting polymers

(b). thermoplastic polymers

(c). fibres

(d). elastomers

CORRECT OPTION: D

SOLUTION

N/A

H3C −

O

∣ ∣

C − OHHVZ−−→B

BrCH2 −

O

∣ ∣

C − OHKCN−−−→H3O+

(Y )

OH −

O

∣ ∣

C − CH2 − COOH(X ) Malonic acid

r2P

C2H5NH2 C6H5NH2

1∘ β


Q No. - 58


Which of the following statements is false?

(a).

(b).

(c).

(d). in exists in the cage form due to intermolecular hydrogen bonding.

CORRECT OPTION: D

SOLUTION

(A) Viscosity intermolcular attraction

intermolecular bonding tendency

(B) No intramoecular -bonding in the cis-isomer due to loss in planarity as a result of steric and electron

repulsion.

stability symmetry.

(C ) Even though -nitrophenol shows intramolecular -bonding but has greater and effect than

(D) .

No cage lattice structure of .

Q No. - 59


Which among the following compounds will not give effervescence with sodium bicarbonate?

(a).

(b).

(c).

(d). picric acid

CORRECT OPTION: C

SOLUTION

Because phenol is less acidic than .

Q No. - 60


Which of the following is most reactive for reaction?

H3BO3 C2H5OH

α

α H −

H

∴ α

o H −R −I

H3BO3 + 3C2H5OH → B(OC2H5)3 + 3H2O

∴ H3BO3

C6H5COOH

C6H5SO3H

C6H5OH

H2CO3

SN2

(a).

(b).

(c).

(d).

CORRECT OPTION: D

SOLUTION

is less electronegative and therefore more reactive towards .

Q No. - 61

JEE MAIN-FULL TEST 03 - MATHS

Sum of the roots of the equation is

(a). 0

(b). 2

(c). 1

(d). 5

CORRECT OPTION: B

SOLUTION

Given equatio can be written as

or

or

Q No. - 62


if are points on the circle such that and the value of the ex

is least then,

(a). and

(b).

(c). must be real

(d).

CORRECT OPTION: A

S SN2

x2 − 2x + |x − 1| − 5 = 0

( ∣ x − 1)2 + |x − 1| − 6 = 0

t2 + t − 6 = 0

⇒ t = 2 −3 ⇒ |x − 1| = 2

⇒ x − 1 = ± 2 x = 3, − 1

z1, z2, z3, z4 |z| = 1 z1 + z2 + z3 + z4 = 0

|z1 − z2|2 + |z2 − z3|

2 + |z3 − z4|2 + |z4 + z1|

2

z1 + z3 = 0 z2 + z4 = 0

z1 + z2 = z3 + z4

z1, z2, z3, z4

=z1

z3

z2

z4

SOLUTION

Q No. - 63


if and , then the value of is equal to

(a). 13

(b).

(c).

(d).

CORRECT OPTION: A

SOLUTION

Q No. - 64

(|z1 − z2|)2 + (|z2 − z3|)

2 + (|z3 − z4|)2 + (|z4 − z1|)

2

= 2(|z1|2 + |z2|2 + |z3|3 + |z4|2) + |z1 + z3|2 + |z2 + z4|2 > 8 + |z1 + z3|2 + |z2 + z4|2

Tr =r

r4 + 4Sn =

n

∑r= 1

tr 37S5 −7

26

25

261

27

26

Tr = = =r

r4 + 4

r

(r2 + 2)2 − 4r2r

(r2 + 2r + 2)(r2 _ 2r + 2)

= [ − ] = [ − ]14

1

r2 − 2r + 2

1

(r2 + 2r + 2)

1

4

1

(r − 1)2 + 1

1

(r + 1)2 + 1

⇒ Sn = [1 + − − ]1

4

1

2

1

n2 − 1

1

(n + 1)2 + 1

S5 = [1 + − − ] = [ ] = × =1

4

1

2

1

26

1

37

1

4

38 × 37 − 26 × 1

26 × 37

1

4

1380

26 × 37

345

26 × 37

⇒ 37S5 − = = 137

26

338

26



Number of distinct terms in the expansion of is equal to

(a).

(b).

(c).

(d).

CORRECT OPTION: A

SOLUTION

Number of terms number of non-negative integral solution

Q No. - 65


if is symmetric about the lines and then it must be symm

(a). (1,1)

(b).

(c).

(d).

CORRECT OPTION: B

SOLUTION

if a function is symmetric about two mutually perpendicular lines, it must be symmetric about their point of

intersection.

Q No. - 66


if and then,

(a).

(b).

(c).

(d).

CORRECT OPTION: B

SOLUTION

(x + y + z + w)50

.53 C3

51

512

.51 C3

=

y = f(x) 3x + 4y + 1 = 0 4x − 3y − 7 = 0

(1, − 1)

(0, 0)

(1, 0)

√1 − x6 + √1 − y6 = a(x3 − y6) = f(x, y)√dydx

1 − y6

1 − x6

f(xky) =y

x

f(x, y) =x2

y2

f(x, y) =2y2

x2

f(x, y) =y2

x2

− − = a(x2 − y2 )x5

√1 − x6y5

√1 − y6dy

dx

dy

dx

⇒ (ay2 − ) = ax2 +y5

√1 − y6dy

dx

x5

√1 − x6

...(1)

Also, or

Hence, from equation (1),

Q No. - 67


If with standard notations are four co-normal points of the hyperbola then the orthoc

the triangle formed by joining the points is given by

(a).

(b).

(c).

(d).

CORRECT OPTION: B

SOLUTION

For a triangle formed by joining any three co-normal points, orthocentre consides with the fourth point.

Q No. - 68


If be the focus of a parabla and be the focal chord, such that and , then the lengt

rectum of the parabola is

(a). 4

(b). 2

(c). 8

(d). 16

CORRECT OPTION: C

SOLUTION

of and semi latus rectum.

Q No. - 69


if a circle having centre at cut the circles and

orthogonaly, then is equal to

(a). 1

⇒ = ×dy

dx

√1 − y6

√1 − x6

x2(a√1 − x6 + x3)

y2(a√1 − y6 − y3)

= a(x3 − y3)(1 − x6) − (1 − y6)

√1 − x6 − √1 − y6x3 + y3 = a(√1 − y6 − √1 − x6)

⇒ a√1 − y6 − y3 = a√1 − x6 + x3

= ×dy

dx

√1 − y6

√1 − x6x2

y2

t1, t2, t3, t4 xy = x2

t1, t2, t3

(0, 0)

(ct4, )c

t4

(ct1 + t2 + t3 , )c

t1 + t2 + t3

( , ct4)c

t4

S PQ SP = 3 SQ = 6

HM SP SQ = = 4 =2(3)(6)

3 + 6

(α, β) x2 + y2 − 2x − 2y − 7 = 0 x2 + y2 + 4x − 6y∣∣∣

α −∣∣∣

3

4

β

2

(b).

(c).

(d).

CORRECT OPTION: B

SOLUTION

If a circle cuts two circles orthogonally, radical axis of the two sicles pass through the centre of the first circ

Here radical axis of given circles is or

Hence

Q No. - 70


Two circles with radii and . , touch each other externally if be the angle between the d

common tangents, then

(a).

(b).

(c).

(d). none of these

CORRECT OPTION: B

SOLUTION

1

21

40

6x − 4y + 40 3x − 2y + 2 = 0

3α − 2β + 2 = 0 ⇒ α − = −3

4

β

2

1

2

r1 r2 r1 > r2 ≥ 2 θ

θ = sin− 1( )r1 + r2r1 − r2

θ = 2 sin− 1( )r1 − r2r1 + r2

θ = sin− 1( )r1 − r2r1 + r2

Q No. - 71


A unit vector is plane which makes an angle of with the vector and an angle of w

vector is

(a).

(b).

(c).

(d). none of these

CORRECT OPTION: D

SOLUTION

Let �(1)

Also,

�(2)

.(3)

There exists no real values of a and b satisfying (1), (2) and (3)

Hence no such unit vector exists.

sinα =r1 − r2r1 + r2

⇒ θ = 2 sin− 1( )r1 − r2r1 + r2

xy − 45 ∘→i +

→j 60 ∘

→3i −

→4j

→i→i +

→j

√2→i −

→j

√2

→r = a

→i + b

→j ⇒ a2 + b2 = 1

cos 45 ∘ = ⇒ a + b = 1

→r . (

→i +

→j )

∣∣→r ∣∣

∣∣∣

→i +

→j

∣∣∣

cos 60 ∘ = ⇒ 3a − 4b =

→r . (3

→i − 4

→j )

∣∣→r ∣∣

∣∣∣3

→i − 4

→j

∣∣∣

5

2


Q No. - 72


Let �.. . Then the sum of the series

�. Is

(a).

(b).

(c).

(d).

CORRECT OPTION: B

SOLUTION

.... �

On comparison and

Q No. - 73


There are three points (a,x),(b,y) and (x,z) such that the straight lines joining any two of them are not equa

to the corrdinate axes where a,b,c,x,y,z .

if and , then are in.

(a). A.P.

(b). G.P.

(c). H.P.

(1 + x)n = 1 + nx + +n(n − 1)x2

2x, n ∈ R

3 + + + +8

3

80

33240

34

9

27

12

101

(1 + x)n = 3 + + + +8

3

80

33240

34= 1 + nx + +

n(n − 1)x2

2

n = − 3 x = −2

3

∈ R∣∣ ∣ ∣∣

x + a y + b z + c

y + b z + c x + a

z + c x + a y + b

∣∣ ∣ ∣∣

= 0 a + c = − b x, − , zy

2


(d). none of these

CORRECT OPTION: A

SOLUTION

From the given conditions

determinant is a symmetric one. The determinant will be equal to zero is

but (given)

are in A.P.

Q No. - 74


if at is tangent to the parabola , then respective values of a,b,c are

(a).

(b).

(c).

(d). none of these

CORRECT OPTION: A

SOLUTION

For

Tangent at is

Comparing with

Q No. - 75


if : then is increasing on

(a).

(b).

(c).

(d). none of these

CORRECT OPTION: B

SOLUTION

so is increasing in

is increasing on

Q No. - 76


Coordinates of the point on the straight line , which is nearest to the parabola

≠ ± 1, ≠ ± 1, ≠ ± 1 ⇒ x + a ≠ y + b ≠ z +y − x

b − a

z − y

c − b

z − x

c − ax + a + y + b + z + c = 0

a + b + c = 0

⇒ x + y + z = 0

⇒ x + z = 2( − ) ⇒ x, − , zy2

y

2

x = 1, y = 2x y = ax2 + bx + c

, 1,1

2

1

2

1, ,1

2

1

2

, , 11

2

1

2

x = 1, y = a + b + c

(1, a + b + c) (y + a + b + c) = ax + (x + 1) + c1

2

b

2y = 2x, c = a, b = 2(1 − a)

f' (x2 − 4x + 3) > 0, ∀x ∈ (2, 3) f(sinx)

⋃n∈ l

(2nπ, (4n + 1) )π2

⋃n∈ l

((4n − 1) , 2nπ)π2

R

x ∈ (2, 3) ⇒ − 1 < x24x + 3 < 0 f(x) ( − 1, 0)

⇒ f(sinx) ⋃n∈ l

((4n − 1) , 2nπ)π2

x + y = 4 y2 = 4(x − 10

(a).

(b).

(c).

(d). none of these

CORRECT OPTION: A

SOLUTION

Let point on the straight line be , this will be nearest to the parabola if at th

the straight line becomes to the parabola.

let it is normal at

Perpendicular to at is ..(1)

Normal at parabola at is ....(2)

(1) and (2) are same so required point is

Q No. - 77


if be three vectos of magnitude sch that if is the angle between

then is equal to

(a).

(b).

(c).

(d). none of these

CORRECT OPTION: A

SOLUTION

Q No. - 78


if �..then B is equal to

(a). 0

(b). 1

(c). 2

(d). none of these

CORRECT OPTION: A

( , − )172

9

2

(2, 2)

( , )32

5

2

P x + y = 4 (m, 4 − m) ⊥

x − 10 = t2, y = 2t

x + y = 4 (m, 4 − m) y − (4 − m) = (x − m)

(t2 + 10, 2t) y + t(x − 10) = 12t + t3

⇒ t = − 1, m =17

2( , − )17

2

9

2

ā, b̄, c̄ √3, 1, 2 ā × (ā × c̄) + 3b̄ = 0 θ

cos2 θ3

41

21

4

|ā × (ā × c̄)| = ∣∣3b̄∣∣ = 3∣∣b̄∣∣

|ā|. |(ā × c̄)|sin( ) = 3.1 ⇒ 3 = |ā|. (||ā||c̄ ∣ ∣ sin θ)π2

⇒ 3 = 3.2 sin θ ⇒ sin θ = ⇒ cos2 θ =1

2

3

4

△ (x) =

∣∣ ∣ ∣∣

ex sin 2x tanx2

ln(1 + x) cosx sinx

cosx2, ex − 1 sinx2

∣∣ ∣ ∣∣

= A + bx + Cx2 +

SOLUTION

...

put

Q No. - 79


if the function is defined by , then is

(a).

(b).

(c).

(d). does not exist

CORRECT OPTION: A

SOLUTION

Let be the inverse of then

since

so

△ ' (x) =

∣∣ ∣ ∣∣

ex 2 cos 2x 2x sec2 x2

ln(1 + x) cosx sinx


∣∣ ∣ ∣∣

+

∣∣ ∣ ∣∣

ex sin 2x tanx2

−sinx cosx


∣∣ ∣ ∣∣

+

∣∣ ∣ ∣∣

ex sin 2x tanx2

ln(1 + x) cosx sinx

−2xsinx2, ex 2xcosx2

∣∣ ∣ ∣∣

1( 1 +x )

= B + 2Cx +

x = 0, B =

∣∣ ∣∣

1 2 0

0 1 0

1 0 0

∣∣ ∣∣

+

∣∣ ∣∣

1 0 0

1 0 −1

1 0 0

∣∣ ∣∣

+

∣∣ ∣∣

1 0 0

0 1 0

0 1 0

∣∣ ∣∣

= 0

f : [2, ∞) → [1, ∞) f(x) = 3x (x− 2 ) f − 1(x)

1 + √1 + log3 x1 − √1 + log3 x1 + √1 − log3 x

g(x) f f(g(x)) = x

⇒ 3g (x ) ( g (x ) − 2 ) = x

⇒ (g(x))2 − 2g(x) − log3 x = 0

⇒ g(x) = = 1 ± √1 + log3 x2 ± √4 + 4 log3 x

2g : [1, ∞] → [2, ∞]

g(x) = 1 ± √1 + log3 x


Q No. - 80


The number of real solutions of the equations

(a). one

(b). two

(c). zero

(d). infinite

CORRECT OPTION: C

SOLUTION

Since

Since

The given equation has no solution.

Q No. - 81


let then denotes the fractional part of x) is equal to

(a).

(b).

(c).

(d).

CORRECT OPTION: D

SOLUTION

Let

and integer an integer

tan− 1 √x2 − 3x + 2 + cos− 1 √4x − x2 − 3 = π

√x2 − 3x + 2 ≥ ⇒ 0 tan− 1 √x2 − 3x + 2 ≥π

2√4x − x2 − 3 ≥ 0 ⇒ 0 ≤ cos− 1 √4x − x2 − 3 ≤

π

2⇒ 0 ≤ L. H. S. ≤ π ⇒

x = (5√2 + 7)19

x{x}({. }

219

319

0

1

f = (5√2 − 7)19

x − f = ⇒ [x] + {x} − f =


and integer but

So,

Q No. - 82


A flagstaff stands vertically on a pillar, the height of the flagstaff being double the height of the pillar.A man

ground at a distance finds that both the pillar and te flagstaff subtend equal angles at his eyes.the ratio of

of the pillar ad the distance of the man from the pillar, is

(a).

(b).

(c).

(d).

CORRECT OPTION: C

SOLUTION

N/A

Q No. - 83


� then find

(a).

(b).

(c).

(d). none of these

CORRECT OPTION: D

SOLUTION

...

�

so,

Q No. - 84


if lines and third line passing through form triangle of area units t

⇒ {x} − f = −1 < {x} − f < 1 ⇒ {x} = f

x{x} = x. f = 119 = 1

√3: 1

1: 3

1: √3

√3: 2

= + + +1

x

2e

3!

4e

5!

6e

7!∞ ∫

x

0f(y)logy xdy, y > 1

[f(e)]2

2[f( )]

21e

2[f(e2)]92

2

= 2e[ + + +1x

1

3!

2

5!

3

7!∞] = 2eS

S =∞

∑r= 1

=∞

∑r= 1

=∞

∑r= 1

[ − ]r(2r + 1)!

1

2

(2r − 1) − 1

(2r + 1)!

1

2

1

(2r) !

1

(2r + 1)1

= [ − + − +12

1

2!

1

3!

1

4!

1

5!∞]

= e− 1 =1

2

1

2e

⇒ = 2eS = = 11

x

2e

2e

∫1

0f(y)logy xdy = 0

x = y = z, x = =y

2

z

3(1, 1, 1) √6

intersection of third line with second line will be

(a).

(b).

(c).

(d). none of these

CORRECT OPTION: B

SOLUTION

let any point on second line be

So, B is

Q No. - 85


if and be vertices of a triangle whose circumcentre is te origin, then orthocentre is given

(1, 2, 3)

(2, 4, 6)

( , , )43

8

3

12

3

(λ, 2λ, 3λ)

cos θ = sin θ =6

√42

6

√42

△OAB = (OA). Ob sin θ = √3. λ√14 × = √61

2

1

2

6

√42⇒ λ = 2

(2, 4, 6)

A(ā), B(b̄) C(c̄)

(a).

(b).

(c).

(d). none of these

CORRECT OPTION: C

SOLUTION

Centroid triangle will be

Now line joining the orthocentre and the circumcentre is divided by centroid in ratio internally, so ortho

be

Q No. - 86


Let . Then number of roots of the equation in is

(a). 2

(b). 4

(c). 0

(d). infinite

CORRECT OPTION: C

SOLUTION

If we draw the graph of tan x ad cot x, we observes that range of is

So does not have any root.

Q No. - 87


ā + b̄ + c̄

3ā + b̄ + c̄

2ā + b̄ + c̄

ā + b̄ + c̄

32: 1

ā + b̄ + c̄

f(x) = max {tanx, cotx} f(x) =1

√3(0, 2π)

f(x) [ − 1, 0) ∪ [1, ∞)

f(x) =1

√3

if denote the sum of infinity and the sum of terms of the series � such tha

, then the least value of is

(a).

(b).

(c).

(d).

CORRECT OPTION: C

SOLUTION

least value of

Q No. - 88


if , the maximum value of f(x) is

(a).

(b).

(c).

(d). none of these

CORRECT OPTION: A

S Sn n 1 + + + +1

3

1

9

1

27

S − Sn <1

300n

4

5

6

7

S − Sn <1

300

⇒ − < ⇒ [1 − 1 + ] 4501

3n1

300

2

3

1

3n1

450∵ n = 6

f(x) = ∫4

0e | t−x | dt(0 < x < 4)

e4 − 1

2(e2 − 1)e2 − 1


SOLUTION

, so maximum value of is .

Q No. - 89


Let then number of solutions of the inequation is

(a). 4

(b). 2

(c). 1

(d). infinite

CORRECT OPTION: A

SOLUTION

Given inequation can be rewritten as

now and

So, we should have

Q No. - 90


A triangle is inscribed in a circle. The vertices of the triangle divide the circle in to three arcs of length 3,4 a

then area of the triangle is equal to

(a).

(b).

(c).

(d).

CORRECT OPTION: A

SOLUTION

let radius of the circle be

f(x) = ∫x

0e | t−x | dt + ∫

4

x

e | t−x | dt = ∫x

0ex− tdt + ∫

4

x

et−xdt = − ex− t∣∣.x0 + et−x∣∣

x

4= ex +

f(x) = ex − e4 −x = 0 ⇒ x = 4 − ξmpliex = 2

f(0) = f(4) = e4 − 1, f(2) = 2(e2 − 1) f(x) e4 − 1

x, y ∈ [0, 10] (x, y)∣∣∣3

sec2 x− 1√9y2 − 6y + 2 < 1

3sec2 x√y2 + < 1−2y

3

2

9

⇒ 3sec2 x√(y − )

2

+ < 11

3

1

9

3sec2 x > 3 √(y − )

2

+ >1

3

1

9

1

3

sec2 x = 1, y =1

3⇒ x = 0, π, 2π, 3π

9√3(1 + √3)

π2

9√3(√3 − 1)

π2

9√3(1 + √3)

2π2

9√3(√3 − 1)

2π2

arc(AC) = 3, arc(AB) = 4, arc(BC) = 5

r

Now,

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⇒ rα = 3 ⇒ α = , β = , γ =3

r

4

r

5

r

3 + 4 + 5 = 2πr ⇒ r = ⇒ =6

π

1

r

π

6△ABC = △OAC + △OAB + △OBC

= r2(sin( ) + sin( ) + sin( ))12

3

r

4

r

5

r

= . (sin( ) + sin( ) + sin( ))12

36

π2π

2

2π

3

5π

6

= (1 + + ) =18π2

√3

2

1

2

9√3(1 + √3)

π2


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