JEE Solution

8/13/2019 JEE Solution

1/18

Q.No

Question Correct Answer; Detail Solution(optional)

1. Figure below shows a body of massMmoving with the uniform speed on acircular path of radius, R.What is the change in acceleration in going from

1P to 2P

D;

2sin2

aa = 45sin2aR

va

2

22

2. The SI unit of universal gas constant )(R is C; 11

molJKKelvinmole

Joule

nT

PVRnRTPV

3. A particle moves along the sidesAB, BC, CDof a square of side 25mwith a

velocity of 15 .1ms Its average velocity isD; Average velocity =

takenTime

ntdisplacemeTotal

15/75

25 =5m/s

4. The kinetic energy k of a particle moving along a circle of radius R

depends on the distance covered s as 2ask where a is a constant. Theforce acting on the particle ia

B;According to given problem 222

1asmv

m

asv 2

SomR

as

R

vaR

22 2 (i)

Further more asds

dvv

dt

ds

ds

dv

dt

dvat (ii)

(By chain rule)

Which in light of equation (i) i.e.m

asv 2

yields

m

as

m

a

m

asat

222

(iii)

So that222

22 22

m

as

mR

asaaa tR

Hence 2]/[12

Rsm

asa

2]/[12 RsasmaF

C B

AD

P1

P2

R

v


2/18

5. The vector sum of two forces is perpendicular to their vector differences. Inthat case, the force A;Let two vectors be A and B then 0).()( BABA

0.... BBABBBAA

022 BA 22 BA BA

6. A block is kept on an inclined plane of inclination of length l. Thevelocity of particle at the bottom of inclined is (the coefficient of

friction is )

B;Acceleration (a) )cos(sin g ands = l

asv 2 )cos(sin2 gl

7. A body at rest breaks up into 3 parts. If 2 parts having equal masses fly offperpendicularly each after with a velocity of 12m/s, then the velocity of thethird part which has 3 times mass of each part is

D;The momentum of third part will be equal and opposite to the resultant of momentum ofrest two equal parts

let Vis the velocity of third part.

By the conservation of linear momentum

2123 mVm smV /24

8. What is the moment of inertia of a square sheet of side l and mass per unit

area about an axis passing through the centre and perpendicular to itsplane

D;

9. A body falls freely under gravity. Its speed is vwhen it has lost an amount Uof the gravitational energy. Then its mass is

C;U= Loss in gravitational energy = gain in K.E.

So,2

2 2

2

1

v

UmmvU

10 A wire of area of cross-section 2610 m is increased in length by 0.1%. Thetension produced is 1000N. The Young's modulus of wire is

A; 2126

/101.010

1001000mN

Al

FLY

11 Energy required to form a soap bubble of diameter 20 cmwill be (Surfacetension for soap solution is 30 dynes/cm)

D; ergTrE 2400030)10(8 22

m

m

3m

Y

V

X

135

12m/s

12m/s


3/18

12 A spherical ball of radius rand relative density 0.5 is floating in equilibriumin water with half of it immersed in water. The work done in pushing theball down so that whole of it is just immersed in water is : (where is the

density of water)

A;

13 The figure shows a glass tube (linear co-efficient of expansion is )completely filled with a liquid of volume expansion co-efficient . Onheating length of the liquid column does not change. Choose thecorrect relation between and

B;When length of the liquid column remains constant, then the level of liquid moves down

with respect to the container, thus must be less than 3.

Now we can write V= V0(1 + T)

Since V=Al0= [A0(1 + 2T)]l0= V0(1 + 2T)

Hence V0(1 + T) = V0(1 + 2T) = 2.

14 Certain amount of an ideal gas is contained in a closed vessel. The vessel ismoving with a constant velocity v. The molecular mass of gas is M. The rise

in temperature of the gas when the vessel is suddenly stopped is)/( VP CC

B;If mis the total mass of the gas then its kinetic energy 22

1mv

When the vessel is suddenly stopped then total kinetic energy will increase the temperatureof the

gas. Hence TCmv v2

2

1TC

M

mv [As

1

RCv ]

2

2

1

1mvT

R

M

m

R

MvT

2

)1(2

.

15 A gas expands with temperature according to the relation .3/2kTV What is

the work done when the temperature changes by Co30 B; dVV

RTPdVW

Since 3/2kTV dTKTdV 3/13

2

Eliminating ,K we findT

dT

V

dV

3

2

l0

A0


4/18

Hence RRTTRdTT

RTW

T

T20)30(

3

2)(

3

2

3

212

2

1

16 In the following figure, two insulating sheets with thermal resistances Rand

3Ras shown in figure. The temperature is

B;

17 A pendulum has time period Tin air. When it is made to oscillate in water, it

acquired a time period TT 2 . The density of the pendulum bob is equal to

(density of water = 1)

B;The effective acceleration of a bob in water

1gg where and are the

density of water and the bob respectively. Since the period of oscillation

of the bob in air and water are given asg

lT 2 and

g

lT

2

g

g

g

g

T

T )/1(

111

Putting2

1

T

T. We obtain,

11

2

1 = 2

18 Unlike a laboratory sonometer, a stringed instrument is seldom plucked in the

middle. Supposing a sitar string is plucked at about4

1thof its length from the

end. The most prominent harmonic would be

D;When plucked at one fourth it gives two loops, and hence 2 ndharmonic is produced.

19. A point charge is surrounded symmetrically by six identical charges at

distance ras shown in the figure. How much work is done by the forces of

electrostatic repulsion when the point charge qat the centre is removed at

infinity

B;Total potential at the centrer

qV

04

6

Required work doner

qVq

0

2

4

6.

AN

N AN

Q

20C

100C

3R

R

Q


5/18

20. In the circuit as shown in figure the D; 201101101'1 R 45

20'R

Now using ohms law'

25

RRi

4

255.0

R

505.0

254 R 46450R

Current through 20 resistor A1.025

5.2

520

55.0

Potential difference across middle resistor= Potential difference across 20 V21.020

21 A thermo couple develops 40 kelvinV/ . If hot and cold junctions be at 40o

C and 20o

Crespectively then the emf develops by a thermopile using such150 thermo couples in series shall be

D; The temperature difference is 20

o

C = 20 K. So that thermo emf developedVK

K

VE

8002040

Hence total emf = mVV 1201012800150 4

22 In figure shows three long straight wires P, Q and R carrying currentsnormal to the plane of the paper. All three currents have the samemagnitude. Which arrow best shows the direction of the resultant force onthe wireP

C;The forces QF and RF are the forces applies by wires QandRrespectively on the wireP

as shown in figure. Their resultant forceFis best shown by C.

23 The dipole moment of each molecule of a paramagnetic gas is 1.5 10

23 amp m2. The temperature of gas is 27oC and the number of

molecules per unit volume in it is 2 10 26m3. The maximum possible

intensity of magnetisation in the gas will be

A; mAmpV

N

V

MI /103

1

102105.1 32623

P

FRF

FQ Q

PQ

RD

C

B

A

20101025V0.5A

R

q q

q

qq

qr


6/18

24 Two coils have a mutual inductance 0.005H. The current changes in the first

coil according to equation tII sin0 , where AI 100 and = 100

radian/sec. The maximum value of e.m.f. in the second coil is

B; )sin(005.0 0 tidt

d

dt

diMe ti cos005.0 0

510010005.0max e

25 When 100 voltsdc is supplied across a solenoid, a current of 1.0 amperesflows in it. When 100 voltsac is applied across the same coil, the currentdrops to 0.5 ampere. If the frequency of ac source is 50 Hz, then theimpedance and inductance of the solenoid are

A; For dc, 1001

100

i

VR

For ac, 2005.0

100

i

VZ

22 )( LRZ 2222 )50(4)100(200 L

HL 55.0

26 The maximum velocity of an electron emitted by light of wavelength

incident on the surface of a metal of work function , isC;According to Einsteins photoelectric equation

2/1

2 )(2

2

1

m

hcvmv

hc

27 What is the radius of iodine atom (at. no. 53, mass number 126) A;Electronic configuration of iodine is 2, 8, 18, 18, 7,

HereZnmrn

2

9 )10053.0(

Here 5n and ,53Z hence mrn11105.2 .

28 AP-type semiconductor has acceptor levels 57 meVabove the valence band.The maximum wavelength of light required to create a hole is (Plancks

constant h= 34106.6 J-s

C;

hcE

E

hc

193

834

106.11057

103106.6

=217100

29 A ray of light falls on the surface of a spherical glass paper weight makingan angle with the normal and is refracted in the medium at an angle .

The angle of deviation of the emergent ray from the direction of the incidentray

B;From the following ray diagram it is clear that

)()( )(2

30 A thin slice is cut out of a glass cylinder along a plane parallel to its axis.The slice is placed on a flat glass plate as shown. The observed interference

A;The cylindrical surface touches the glass plate along a line parallel to axis of cylinder.The thickness of wedge shaped film increases on both sides of this line. Locus of equal

A

O

B

r r


7/18

fringes from this combination shall be path difference are the lines running parallel to the axis of the cylinder. Hence straightfringes are obtained

31 Which one of the following is not the characteristic of Planck's quantumtheory of radiation

A;Energy is always absorbed or emitted in whole number or multiples of quantum.

32 The molecular orbital configuration of a diatomic molecule is

2

2

22*22*2 2

222211 y

zx

p

ppssss

Its bond order is

A;B.O. 32 4102

ab NN

.

33 Which of the following does not apply to metallic bond D;

34 One mole of 42HN loses 10 mol of electrons to form a new compound Y.

Assuming that all nitrogen appear in the new compound, what is theoxidation state of 2N in Y ? (There is no change in the oxidation state of

hydrogen)

A; eNN a 10222

10)]2(2[2 a

3a .35 Pyrolusite is D; Pyrolusite or Manganese dioxide )( 2MnO is a mineral of manganese.

36 White lead is C;White lead 23 )(.2 OHPbPbCO

37 Which one of the following noble gases is the least polarizable D;Heis least polarizable because of small atomic size.

38 During debromination of meso-dibromobutane, the major compound formedis

D;

CH

CH

CH

CH

CH

BrCH

BrCH

HC

Br ||||

ionDebrominat

butanedibromoMeso3

|

|

3|

2

Trans-2-butene is more stable than its cisisomer.

39 In which of the following processes shape-selective catalysis is occurring

A,C;Shape selective catalyst are zeolites and zeolites are alminosilicates of generalformula OmHSiOAlOM yxnx 222/ ].).()[( . Zeolites are used in conversion of alcohol to

gasoline and in polymerisation of ethylene.

40 A solution containing 32CONa and NaOH requires 300 ml of 0.1 N HCl

using phenolpthalein as an indicator. Methyl orange is then added to theabove titrated solution when a further 25 mlof 0.2 N HCl is required. Theamount ofNaOHpresent in solution is )106,40( 32 CONaNaOH

B;(I) Phenopthalein indicate partial neutralisation of 332 NaHCOCONa

Meq. of 32CONa + Meq. of NaOH = Meq. ofHCl

NVE

W

E

W 10001000

(Suppose gmaCONa 32 ,NaOH= bgm)

1.03001000401000106

ba

.....(1)(II) Methyl orange indicate complete neutralisation


8/18

HCl HCl

2211 VNVN , 1.02.025 mlVV 50so 22 excess

1.0350100040

100053

ba

.....(2)

From (1) and (2) b=1gm.

41 In which of the following arrangements the order is NOT according to theproperty indicated against it

B; ONCB ; When we move from B to O in a periodic table the first ionisationenthalpy increase due to the attraction of nucleous towords the outer most ofelectron.

42 Examine the following three pairs of possible isomers

Now state whether the pairs represent identical compounds ordifferent isomers

D;(ia) and (ib) are identical; (iia) and (iib) are identical, and (iiia) and (iiib) areisomers

Both 1, 2-dichloro benzene

Hence, identical compounds.

Both, 1, 3-dimethyl benzene

Hence, identical compounds.

(iiia) and (iiib) are position isomers.

43 The principal ores of silver are argentite, horn silver and pyrargyrite. Theirformula respectively are

A;

44 Frothfloatation method is successful in separating impurities from oresbecause

D;

45 Which of the following trivalent ion has the largest atomic radii in thelanthanide series

A;Tl+ions are more stable than Tl3+and thus, Tl3+ions change to Tl+ions thereby actingasoxidising agents. Tl3+compounds + 2eTl+compounds

46 Ammonia is a Lewis base. It forms complexes with cations. Which one of

the following cations does not form complex with ammoniaD; Pb because it does not have vacant d-orbitals nor high nuclear charge and it does not

belong to transition series.

COOHHOOC

COOH

Cl

CH3 CH3

Cl

ClCl

COOH

COOH(iiib)

COOH

(iiia)HOOC

Cl

(ib)Cl

CH3

CH3(iia)

CH3

(iib

H3

Cl

Cl

(ia)

(less stable oxidising agent) (more stable oxidising agent)


9/18

47 Which of the following is not true for metal carbonyls D;Metal carbonyls does not show overlapping.

48 Which of the following does not have a metal carbon bond C; 352 )( HOCAl contains bonding through O and thus it does not have metal-carbon bond.

49 What would be the product formed when 1-Bromo-3-chloro cyclobutanereacts with two equivalents of metallic sodium in ether

D;It is the example of Wurtz reaction.

50

BAOHNHOH

PPA

3

2

The product B is

C;

51 Trans-form of polyisoprene is A;Guttapercha rubber is very hard horny material consisting of trans 1, 4 - polyisoprenepolymer

52 Which of the following molecules has three-fold axis of symmetry A; In 3NH molecule, the original appearance is repeated as a result of rotation througho

120 . Such as axis is said to be an axis of three-fold symmetry or a triad axis

53 What does not change on changing temperature A,C;

54 Decomposition of nitrogen pentoxide is known to be a first order reaction 75percent of the oxide had decomposed in the first 24 minutes. At the end ofan hour, after the start of the reaction, the amount of oxide left will be

B;

55 In deriving the kinetic gas equation, use is made of the root mean squarevelocity of the molecules because it is

D;

56 Which alkene is formed from the following reactionButanone23223 PPhCHCHCHCH A; 3223

Butanone2

3|

23 PPhCHCHCHCH

CH

OCCHCH

heptene3methyl,33322

3|

23

POPhCHCHHCH

CH

CCCHCH

57 Which of the following reactions give benzo phenone

B,D;56

563

HCOC

COClHC AlCl

COOHCOOH

O

Br

Cl

2Na2Na+ 2e

Cl

..


10/18

2

56

260

56

CO

HCOCHCOC

C

Cu

o

58 The resistance of 0.01N NaCl solution at Co25 is 200 . Cell constant ofconductivity cell is 1 cm1. The equivalent conductance is

A; 000,101200

11 V

a

l

RVk

1212.105

eqcm

59 The enthalpy of the reaction,

)()(2

1)( 222 gOHgOgH is 1H and that of

)()(2

1)( 222 lOHgOgH is 2H . Then

A;for reaction (i) 5.01 n

for reaction (ii) 5.12 n

So, 21 HH

60 The solubility product of HgSSAgCuS ,, 2 are544431

10,10,10

respectively. The solubilities of these sulphides are in the order

A;

61 1Let a relation R be defined by R= {(4, 5); (1, 4); (4, 6); (7, 6);

(3, 7)} then oRR 1 is

A;We first find ,1R we have )}3,7();7,6();4,6();1,4();4,5{(1 R . We now obtain the

elements of oRR 1 we first pick the element ofR and then of 1R . Since R)5,4( and1

)4,5(

R , we have oRR 1

)4,4(

Similarly, oRRRR 11 )1,1()1,4(,)4,1(

,)4,4()4,6(,)6,4( 11

oRRRR

oRRRR 11

)7,4()7,6(,)6,4(

,)4,7()4,6(,)6,7( 11

oRRRR

oRRRR 11

)7,7()7,6(,)6,7(

,)3,3()3,7(,)7,3( 11

oRRRR

62 Let zand w be two complex numbers such that ,1|| z 1|| w and

2|||| wiziwz . Then z is equal to

C;Let 1||, zibaz 122 ba

and 1||, widcw 122

dc 2|)(||| idciibaiwz

4)()( 22 cbda ......(i)

|)(||| idciibawiz

4)()( 22 cbda ......(ii)

From (i) and (ii), we get 0bc

Either 0b or 0c

If 0b , then 12

a . Then, only possibility is 1a or 1 .63 .If the thp term of an A.P. be q and thq term bep, then its thr term will be B;Given that, qdpaTp )1( ..(i)


11/18

and pdqaTq )1( .. (ii)

From (i) and (ii), we get 1)(

)(

qp

qpd

Putting value of d in equation (i), then 1 qpa

Now, thr term is given by A.P.)1)(1()1()1( rqpdraTr rqp

64 .Let , be the roots of .0)3(2 xx The value of for which22 is minimum, is

C; and3

2)( 222 = 2)3( 2

= 942 from options,

for 9)(,0 022

for 6941)(,1 122

for 5984)(,2 222

for 69129)(,3 322

65 .A library has a copies of one book, b copies of each of two books, c copies of each of three books and single copies of d books. The totalnumber of ways in which these books can be distributed is

B;Total number of books dcba 32

Since there are b copies of each of two books, c copies of each of three booksand single copies of d books. Therefore the total number of arrangements is

32 )!()!(!

!)32(

cba

dcba .

66The coefficient of

3x in the expansion ofx

x

21

)31( 2

will be

C;

67

......!3!2!1

664422 yxyxyx

B; 2222

)1()1( yxyx eeeeS .

68

At what value of ,x will 0

1

1

1

2

2

2

x

x

x

A; 0153419

111

789

151719

69 IfAandBare 33 matrices such that AAB and BBA , then C;

70The value of k, for which 01cossin)sin(cos

2 xxkxx is anidentity, is

B;Given, xxxkxx ,01cossin)sin(cos 2

xxxkxxxx ,01cossinsincos2sincos 22

xxxk ,0sincos)2( 02 k 2k .


12/18

71

In a triangle PQR , .2

R If

2tan

Pand

2tan

Qare the roots of

the equation ).0(02 acbxax then

A;Givena

bQP

2tan

2tan and

a

cQP

2tan

2tan

ora

b

24tan

2tan

,

a

c

24tan.

2tan

a

b

2tan1

2tan1

2tan

a

b

2tan1

12

tan 2

..(i)

Similarly,a

c

2tan1

2tan1

2tan

a

c

2tan1

2tan

2tan

2

......(ii)

By adding (i) and (ii), we geta

c

a

b

2tan1

2tan1

acb bac .

72 The number of real solutions of

21sin)1(tan 211

xxxx is

C;2

1sin)1(tan 211 xxxx

)1(tan 1 xx is defined when

0)1( xx ..(i)

1sin 21 xx is defined when

11)1(0 xx or 0)1(0 xx ..(ii)

Q a R

bcp1

P


13/18

From (i) and (ii), 0)1( xx

or 0x and1.

Hence number of solution is 2.

73 Let PS be the median of the triangle with vertices )1,6(),2,2( QP and

)3,7(R . The equation of the line passing through (1,1) and parallel to PS

is

D;S= midpoint of

1,

2

13

2

31,

2

76QR

m of

9

2

2132

12

PS ,

The required equation is )1(9

21

xy

i.e., 0792 yx .

74 If the pair of straight lines 01 yxxy and the line 032 yax areconcurrent, then a=

D;Given that equation of pair of straight lines 01 yxxy

0)1)(1( yx

01x or 01y

The intersection point is (1, 1).

Lines ,01x 01y and 032 yax are concurrent.

The intersecting point of first two lines satisfy the third line.

Hence, 032 a 1a .

75 The locus of a point which divides the join of )1,1(A and a variable pointP

on the circle 422 yx in the ratio 3 : 2, is

B;Suppose a point on circle is ),( 11 yxB and that which dividesAandBin 3 : 2 isPgiven

by

5

32,

5

32 11 ykx

h

or 13

25x

h

, 1

3

25y

k

As ),( 11 yx lies on circle 422 yx , we get on substituting,

028)(20)(25 22 yxyx .

76 The circle 422 yx cuts the line joining the points )0,1(A and )4,3(B in

two pointsPand Q. Let PA

BPand

QA

BQ. Then and are roots of

the quadratic equation

A;The equation of the line joining A(1, 0) and B(3, 4) is 22 xy . This cuts the circle

422 yx at )2,0( Q and

5

6,

5

8P .

We have5

7,5,53 BPQABQ and

5

3PA


14/18

3

7

5/3

5/7

PA

BP and 3

5

53

QA

BQ

, are roots of the equation 0)(2 xx

i.e., 0)3(3

73

3

72

xx or 02123 2 xx .

77 The angle of intersection of the curves /22 xy and xy sin , is B;The curves /22 xy and xy sin intersect at )0,0( and )1,2/( . Let the gradients of

the tangents to the curves be 1m and 2m respectively. Thenydx

dym

11 and

xdx

dym cos2

At )1,2/( ,

11 m ,

02

cos2

m

Thus

1

)0)(/1(1

0)/1(tan

1cot .

78 The equation of the parabola whose focus is the point (0, 0) and the tangentat the vertex is 01yx is

C;Let focus is )0,0(S andA is the vertex of parabola. Take any point Zsuch thatAS=AZ.

Given tangent at vertex is .01yx

Since directrix is parallel to the tangent at the vertex.

xy

22

O

y = sinx

Y

Q(0,2)

X X

P(8/5,6/5)

B(3,4)Y

A(1,0)O


15/18

Equation of directrix is 0 yx , where is constant.

A is midpoint of SZ, SZ= 2.SA

2222 )1(1

|100|2

)1(1

|00|

2|| i.e. 2

Directrix in this case always lies in II nd quadrant

2

Hence equation of directrix is 02yx

Now,Pbe any point on parabola

SP=PM 22 PMSP

2

22

2

|2|)0()0(

yxyx

.0444222 yxxyyx

79 The vectors b and c are in the direction of north-east and north-westrespectively and |b|=|c|= 4. The magnitude and direction of the vector d= cb, are

B;Clearly, cb , 0. cb

Now, 22 |||| bcdbcd

= 01616.2|||| 22 cbbc

2432|| d and direction of dis west.

80 The co-ordinates of the pointsAandBare (2, 3, 4) (2, 5,4) respectively. If a

pointPmoves so that kPBPA 22 where kis a constant, then the locus of Pis

B; kPBPA 22

])4()3()2[( 222 zyx

kzyx ])4()5()2[( 222

or kzyx 161648 , which is the equation of a plane.

81 The equation of the plane passing through the points )2,3,1( and

perpendicular to planes 522 zyx and 8233 zyx , isA; 0233,022 nmlnml , ,1222 nml we get l, m, n from these equations

and then putting the values in )3()1( ymxl ,0)2( zn we get the required result.

Trick: Checking conversely

xy + 1 = 0

P(x,y)

S (0,0)


16/18

)3(4)1(2 ,08)2(3

So, it passes through given point.

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

So, it is perpendicular to 522 zyx .

,0)3(2)4(3)2(3

So, it is perpendicular to .8233 zyx

82 The value of Nnmxx nmx

,,)(loglim0

isA;

m

n

x

nm

x x

xxx

)(loglim)(loglim

00

Form

1

)1(

0

1)(log

lim

m

n

x mx

xxn

(ByL-Hospital's rule)

m

n

x mx

xn

1

0

)(loglim

Form

12

)2(

0 )(

1)(log)1(

lim

m

n

x xm

xxnn

(ByL-Hospital's rule)

m

n

x xm

xnn

2

2

0

)(log)1(lim

Form

.......................

......................

0)(

!lim

0

mnx xm

n

(Differentiating rN and rD ntimes)83

If

ax

axax

ax

xf

when,1

when,||

)( ,then

B; .1)(,1)(lim,1)(lim

afxfxfaxax

84If tx sin and pty sin , then the value of yp

dx

dyx

dx

ydx 2

2

221 is

equal to

A;Given that tx sin , pty sin

tdt

dxcos , ptp

dt

dycos

2

2

1

1

cos

cos

x

yp

t

ptp

dx

dy

Again differentiate w.r.t.x,

)1(

)2(12

1.1)2.(

12

1.1

2

2

2

2

2

2

2

x

xx

ypdx

dyy

yxp

dx

yd


17/18

2

2

2

2

2

22

1

1

1

1)1(

x

ypx

dx

dy

y

xpy

dx

ydx

dx

dyxyp

dx

ydx 2

2

22 )1(

0)1( 22

22 yp

dx

dyx

dx

ydx -

85 .The radius of the cylinder of maximum volume, which can be inscribed in asphere of radiusRis [AMU 1999] B; If r be the radius and h the height, then from the figure,

22

2

2R

hr

)(4 222 rRh

Now, hrV 2 = 2222 rRr

22

222 )2(

2

1.24

rR

rrrRr

dr

dV

For max. or min., 0dr

dV

22

322 24

rR

rrRr

222 )(2 rrR

22 32 rR Rr3

2 ve

dr

Vd

2

2

.

Hence Vis max., when Rr3

2 .

86

2cossin xx

dxequals D;

2cossin xx

dxI

)14

coscos4

sin.(sin2

xx

dx

)

4

cos(12

1

x

dx

82

2cos12

1

x

dx

A B

R

O

D

CL

r M


18/18

82sin2

2

1

2 x

dx

dx

x

82cosec

22

1 2

c

x

2/1

82cot

22

1

cx

82cot

2

1 .

87 xceb

dxa A; dxecbeae

ecb

dxaxx

x

x 2

Now put ,te x then it reduces to

dttbct

c

ba

bctt

dta

11

)({By partial fraction}

cceb

e

b

ax

x

log .

88

1

0 21

logdx

x

x

C; Put ,sinx we get

ddx

x

x

1

0

2/

02 cos

cos.sinlog

1

log

2log2

sinlog2/

0

d .89

The solution of the differential equation

dx

dyya

dx

dyxy 2 is B;

dx

dyya

dx

dyxy

2 dx

dyaxayy )(2

ax

dx

ayy

dy

)1(

On integrating both sides, we get

caxayy log)log()1log(log

)()1(

axc

ay

y

or yayaxc )1)((

90 In four schools 4321 ,,, BBBB the percentage of girls students is 12, 20, 13, 17

respectively. From a school selected at random, one student is picked up atrandom and it is found that the student is a girl. The probability that theschool selected is ,2B is

B;Favorable number of cases = 20120 C

Sample space = 62162 C

Required probability =31

10

62

20 .

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