01-Offline JEE Main Final Test - 03-04-2016

8/17/2019 01-Offline JEE Main Final Test - 03-04-2016

1/48

Offline JEE Main – 2016

QP Series: E

Max. Marks: 360 Duration: 3 Hours

1. This paper onsists of !0 "uestions #ith 3 parts of $h%sis& 'he(istr% an) Mathe(atis

2. The OM* sheet for 200 "uestions is to +e use)

3. ,se of alulators an) lo- ta+les is prohi+ite)

. Darken the appropriate +u++le usin- a pen in the OM* sheet pro/i)e) to %ou. One entere)&

the ans#er annot +e han-e). n% orretions or (o)ifiations #ill auto(atiall% )ra# a

penalt% of 1 (ark

. o larifiation #ill +e entertaine) )urin- the exa(ination. Dou+ts in the paper an +e

reporte) to the oor)inator after the exa(

6. f the )etails in the OM* 4heet are not fille)& f the OM* sheet is (utilate)& torn& #hite nk

use)& the irles fille) an) srathe)& then the OM* sheet #ill not +e -ra)e)

ll the +est55

,seful Data

t. t.:

N = 14; O = 16; H = 1; S = 32; Cl = 35.5; Mn = 55; Na = 23; C = 12; Ag = 108; K = 39; Fe = 56; Pb = 207

$h%sial 'onstants:

34

6.626 10 Jsh −

= ×,

23 -1

a N 6.022 10 mol= ×,

8 -1

c 2.998 10 ms= ×,

31

em 9.1 10 kg

−

= ×

Copyright © Ace Creative Learning Pvt Ltd.,|| www.ace-online.co.in


2/48


QP Series: E

$h%sis

Multiple 'hoie 7uestions #ith one orret ans#er. orret ans#er arries (arks. #ron-

ans#er arries a penalt% of 1 (ark. 30 x 8 120

. A st!dent "eas!res the ti"e period o# $$ oscillations o# a si"ple pend!l!" #o!r ti"es. %he data set is &$ s &s , &'s and &(s. )# the "ini"!" division in the "eas!ring cloc* is

s, then the reported "ean ti"e sho!ld +e:

92 2 s±(

92 5.0 s±

92 1.8 s±/

92 3 s±

avg90 91 95 92

924

T + + += =

( )avg

2 1 3 01.5

4T ∆

+ + += =

As least co!nt in1s , 2sT ∆ =

Ans:

(. A particle o# "ass " is "oving along the side o# a s0!are side 1

a

2 with a !ni#or" speed

v in the 3-y plane as shown in the 4g!re:

5hich o# the #ollowing state"ents is #alse #or the ang!lar

"o"ent!" L

a+o!t the origin6

ˆ

2

mv L Rk = −r

when the particle is "oving #ro" A to 7.

(

ˆ

2

R L mv a k

= −

r

when the particle is "oving #ro" C to 8.

ˆ

2

R L mv a k

= +

r

when the particle is "oving #or" 7 to C.

/

ˆ2

mv

L R k =

r

when the particle is "oving #ro" 8 to A.

A B→ˆ

2

mv L Rk = −

r

B C →ˆ

2

R L mv a k

= +

r

C D→ˆ

2

R L mv a k

= +

r



3/48


QP Series: E

D A→ ( )ˆ

2

R L mv k

= −

r

Ans: (,/

. A point particle o# "ass

m

"oves along the !ni#or"ly

ro!gh trac* PQ9 as shown in the 4g!re. %he coecient o#

#riction, +etween the particle and the ro!gh trac* e0!als

µ

. %he particle is released, #ro" rest, #ro" the point P

and it co"es to rest at a point 9. %he energies, lost +y

the +all, over the parts, PQ and Q9, o# the trac*, are e0!al to each other, and no energyis lost when particle changes direction #ro" PQ and Q9.

%he val!es o# the coecient o# #riction

µ

and the distance

( ) x QR=

, are respectively

close to:

$.( and ;.' " ( $.( and .' "( $.(& and .' " / (& and ;.' "

PQ QR E E =

( ) ( )cos30mg H mg QR µ µ ° =

2 3QR =

ii

2mgh mg x µ =

2 2 x µ =

1 10.29

2 3 x µ = = =

Ans:

/. A person trying to lose weight +y +!rning #at li#ts a "ass o# $ *g !pto a height o#

" $$$ ti"es. Ass!"e that the potential energy lost each ti"e he lowers the "ass

is dissipated. eciency rate. %a*e/ :

32.45 10 kg−×



4/48


QP Series: E

(

36.45 10 kg−×

39.89 10 kg

−×

/

312.89 10 kg

−×

( ) 41000 10 9.8 1 9.8 10 JW = × × × = ×

(

4output

7input

9.8 10

3.8 10

w

w mη

×= =

×

312.89 10 kgm −= ×

Ans: /

'. A roller is "ade +y ?oining together two cones at their vertices @. )t is *ept on two

rails A7 and C8 which are placed asy""etrically see 4g!re, with its a3is

perpendic!lar to C8 and its centre @ at the centre o# line ?oining A7 and C8 see

4g!re. )t is given a light p!sh so that it starts rolling with its centre @ "oving parallel

to C8 in the direction shown. As it "oves, the roller will tend to :

t!rn right( t!rn le#t go straight/ t!rn le#t and right alternately.

f

is sa"e +!t

x x<

so cloc*wise tor0!e>

anticloc*wise tor0!e %here#ore +ody will t!rn towards le#tAns: (

;. A satellite is revolving in a circ!lar or+it at a height

h

#ro" the earth2s s!r#ace radi!s

o# earth R:h


5/48


QP Series: E

2 gR

(

gR

! 2 gR

/

( )2 1 gR −

GM v

R=

2

GM v

R=

E3tra velocity

( ) 2 1v v gR− = −

Ans: /

. A pend!l!" cloc* loses ( s a day i# the te"perat!re is

40 C °and gains / s a day i#

the te"perat!re is

20 C °. %he te"perat!re at which the cloc* will show correct ti"e,

and the co-ecient o# linear e3pansion

( )α

o# the "aterial o# the pend!l!" sha#t are

respectively:

525 " 1.85 10 !C C α −° = × °

(

460 " 1.85 10 !C C α −° = × °

330 " 1.85 10 !C C α −° = × °

/

255 " 1.85 10 !C C α −° = × °

( )12

T t T ∆ α ∆=

( )1

12 402

t T α = −

D

[ ]1

4 202

t T α = −

D(

25t ⇒ = °

Ans:



6/48


QP Series: E

. An ideal gas !ndergoes a 0!asi-static, reversi+le process in which its "olar heat

capacity C re"ains constant. )# d!ring this process the relation o# press!re P and

vol!"e F is given +y

n

PV = constant, thenn

is given +y


7/48


QP Series: E

00

0

3 pnRT

V pV V

= − +

200

03

p

nRT V p V V = − +D

=orT

to +e "a3i"!"

00

0

2 3 0 pdT

nR V pdV V

= − + =

03

2

V V ⇒ =

S!+stit!ting in

0 094 p V T

nR=

Ans:

$. A particle per#or"s si"ple har"onic "otion with a"plit!de A. )ts speed is tre+led at

the instant that it is at a distance

2

3

A

#ro" e0!ili+ri!" position. %he new a"plit!de o#

the "otion is:

413 A

(

3 A

3 A

/

73 A

PE at

2

3

A

is

2 21 4

2 9

m Aω

HE at

2

3

A

is

2 2 22 21 4 1 5

2 9 2 9

A m Am A

ω ω

− = ÷ ÷

7!t

v

is tre+led. %hen

2 21 9 52

K K m Aω = =

%.E at

2

3

A

is

2 22 2 2 21 4 1 1 495

2 9 2 2 9

m Am A m A

ω ω ω + =

At e3tre"e point#$ %$=

2 2 2 21 49 1

2 9 2m A m Aω ω =



8/48

dx

x


QP Series: E

7 &

3 A =

Ans: /. A !ni#or" string o# length ($ " is s!spended #ro" a rigid s!pport. A short wave p!lse

is introd!ced at its lowest end. )t starts "oving !p the string. %he ti"e ta*en to reach

the s!pport is: ta*e

210 g ms−=

2 2 sπ

(

2 s

2 2 s

/

2 s

T xg v

µ

µ µ = =

dx gx

dt =

dx gx dt =

0 0

Ldx

g dt x

=∫ ∫

2 L g t =

2 202 2 2 s10

Lt g

= = =

Ans: (. %he region +etween two concentric spheres o# radii 1a2 and 1+2 respectively see

4g!re has vol!"e charge density

A

ρ =

, where A is a constant and r is the distance

#ro" the centre. At the centre o# the spheres is a point charge Q. %he val!e o# A s!ch

that the electric 4eld in the region +etween the spheres will +e constant, is:



9/48


QP Series: E

22

Q

aπ

(

( )2 22Q

! aπ −

( )2 22Q

! aπ −

/

2

2Q

aπ

2 2

0

14 4

a

A E Q dx

xπ π

ε

= +

∫

22

0

14

2

a

x E Q Aπ π

ε

= + ÷ ÷

( )2 2 20

12 E Q A aπ π

ε = + −

2 2

2 2 20 0 0

2 2

4 4 4

Q A Aa E

π π

πε πε πε = + −

2

2 20 0 0

2 4 2

A Q Aa E

ε πε ε

= + −

As E

is constant the +rac*eted ter" sho!ld +e Iero2

2 20 0

2

2 4 2

Q a

π πε ε =

22

Q A

aπ =

Ans:

. A co"+ination o# capacitors is set !p as shown in the 4g!re . %he "agnit!de o# the

electric 4eld, d!e to a point charge Q having a charge e0!al to the s!" o# the

charges on the

4 " µ

and

9 " µ

capacitors, at a point distant $ " #ro" it, wo!ld e0!al:



10/48

4'(

24'(

6 )2 )


QP Series: E

(/$ JC ( ;$JC /($ JC / /$JC

24'*# =

9'( 18'*# =

42'*Q =

2 420 N!*

KQ

E = =



11/48

R x2

x


QP Series: E

Ans:

/. %he te"perat!re dependence o# resistances o# C! and !ndoped Si in the

te"perat!re range $$ K /$$ H, is +est descri+ed +y : linear increase #or C!, linear increase #or Si.( linear increase #or C!, e3ponential increase #or Si. linear increase #or C!, e3ponential decrease #or Si./ linear decrease #or C!, linear decrease #or Si.

=or cond!ctor -linear increase=or se"icond!ctor -e3ponential decreaseAns:

'. %wo identical wires A and B, each o# length ‘l’ carry the sa"e c!rrent ). 5ire A is +ent

into a circle o# radi!s R and wire B is +ent to #or" a s0!are o# side 1a2. )#

A B

and

B B

are the val!es o# "agnetic 4eld at the centres o# the circle and s0!are respectively,

then the ratio

A

B

B

B

is:

2

8

π

(

2

16 2

π

2

16

π

/

2

8 2

π

0 0 2

2 2 A

$ $

B R

µ µ π

= = l

0 2 4

42

B

$ B

x

µ

π

= × ÷

02

0 8 28 2

A

B

$

B

$ B

µ π π

µ

π

= =l

l

0 02 2 4 8 2$ $ µ µ

π π = =l l

Ans: /

;.


12/48


QP Series: E

%hese "aterials are !sed

to "a*e "agnets #or electric generators, trans#or"er core and electro"agnet core.

%hen it is proper to !se:

A #or electric generators and trans#or"ers( A #or electro"agnets and 7 #or electric generators. A #or trans#or"ers and 7 #or electric generators

/ 7 #or electro"agnets and trans#or"ers7oth need to have less energy lossAns: /

. An arc la"p re0!ires a direct c!rrent o# $ A at $ F to #!nction. )# it is connected to

a (($ F r"s, '$


13/48


QP Series: E

a ulelig/t ellolig/t aioave x E E E E −∴ > > >

∴8 7 A C

Ans:

&. An o+server loo*s at a distant tree o# height $ " with a telescope o# "agni#ying

power o# ($. %o the o+server the tree appears:

$ ti"es taller. ( $ ti"es nearer

($ ti"es taller / ($ ti"es nearer

Concept o# "agni4cation involves o+?ects appearing nearer not taller

Ans: /

($. %he +o3 o# pin hole ca"era, o# length L, has a hole o# radi!s a. )# is ass!"ed that

when the hole is ill!"inated +y a parallel +ea" o# light o# wavelength

λ

the spread o#

the spot o+tained on the opposite wall o# the ca"era is the s!" o# its geo"etrical

spread and the spread d!e to diBraction. %he spot wo!ld then have its "ini"!" siIe

min sa&!

when:

min2

a and ! L L

λ λ 2 2 = = ÷ ÷

(

min2

a L and ! L

λ λ

2 = = ÷ ÷

min 4a L and ! Lλ λ = =

/

min 4a and ! L L

λ λ 2= =

)# L

is =resnel distance, spread is "ini"!" i#

aL L

λ <

a Lλ ∴ >

a Lλ =



14/48


QP Series: E

min 2 4! a Lλ = =

Ans:

(. 9adiation o# wavelength

λ

is incident on a photo cell. %he #astest e"itted electron

has speed v . )# the wavelength is changed to

3

4

λ

, the speed o# the #astest e"itted

electron will +e:

1

2

3

v 4 > ÷

(

1

2

3

v 4 < ÷

1

2

3

v 4 = ÷

/

1

23

4

v = ÷

21

2

h'W mv

λ = +

214 3 2

h'W mv

λ = +

4

3v v>

Ans:

((.


15/48


QP Series: E

(. )# a,+,c,d are inp!ts to a gate and 3 is its o!tp!t, then, as per #ollowing ti"e graph,

the gate is:

@% ( A8 @9 / A8

=ro" tr!th ta+le, it is @9 gate

Ans:

(/. Choose the correct state"ent: )n a"plit!de "od!lation the a"plit!de o# the high #re0!ency carrier wave is

"ade to vary in proportion to the a"plit!de o# the a!dio signal.( )n a"plit!de "od!lation the #re0!ency o# the high #re0!ency carrier wave is

"ade to vary in proportion to the a"plit!de o# the a!dio signal. )n #re0!ency "od!lation the a"plit!de o# high #re0!ency carrier wave is "ade to

vary in proportion to the a"plit!de o# the a!dio signal./ )n #re0!ency "od!lation the a"plit!de o# the high #re0!ency carrier wave is

"ade to vary in proportion to the #re0!ency o# the a!dio signal.

)n a"plit!de "od!lation, a"plit!de o# caries waveµ

a"plit!de o# a!dio signal

Ans:

('. A screw ga!ge with a pitch o# $.' "" and a circ!lar scale with '$ divisions is !sed to

"eas!re the thic*ness o# a thin sheet o# Al!"ini!". 7e#ore starting the

"eas!re"ent, it is #o!nd that when the two ?aws o# the screw ga!ge are +ro!ght in

contact, the /'th division coincides with the "ain scale and that the Iero o# the "ain

scale is +arely visi+le. 5hat is the thic*ness o# the sheet i# the "ain scale reading is

$.' "" and the ('th division coincides with the "ain scale line6 $.' "" ( $.$ "" $.$ "" / $.'$""

Since Iero o# the "ain scale in not visi+le, )E

in negative

50 45 5 )E = − =

0.50.01mm

50 LC = =



16/48


QP Series: E

5 )C = −

( )TR M*R H*R )C LC = + +

( )0.5 25 5 0.01= + + ×

0.5 0.30 0.80mm= + =

Ans: (

(;. A pipe open at +oth ends has a #!nda"ental #re0!ency # in air. %he pipe is dipped

vertically in water so that hal# o# it is in water. %he #!nda"ental #re0!ency o# the air

col!"n is now:

2

f

(

3

4

f

2 f

/

f

2

v f =

l

42

v f f = =

× l

Ans: /

(. A galvano"eter having a coil resistance o#

100 Ωgives a #!ll scale deNection, when a

c!rrent o# "A is passed thro!gh it. %he val!e o# the resistance, which can convert

this galvano"eter into a""eter giving a #!ll scale deNection #or a c!rrent o# $ A, is:

0.01 Ω(

2 Ω

0.1 Ω

/

3 Ω

0.01 g

g

$ G*

$ $= = Ω

−

Ans:

(. )n an e3peri"ent #or deter"ination o# re#ractive inde3 o# glass o# a pris" +y

$ δ −plot,

it was #o!nd that a ray incident at angle

35°, s!Bers a deviation o#

40°and that it



17/48


QP Series: E

e"erges at angle

79°. )n that case which o# the #ollowing is closest to the "a3i"!"

possi+le val!e o# the re#ractive inde36 .' ( .; . / .

1 235 , 79$ $= ° = °

1 2d $ $ A= + −

35 79 40 74 A = + − = °

minsin2

sin2

A

n A

δ + ÷

= ÷

min5 sin 373 2

δ = + ÷

man

is

5

3

5

3n∴ <

Sinceminδ

is less than

40

,

5 5sin 57 sin 60

3 3n n< ⇒ <

1.446n∴ <

∴nearest val!e is

1.5n =

Ans: (&. )denti#y the se"icond!ctor devices whose characteristics are given +elow, in the

order a, + , c, d.:



18/48


19/48


QP Series: E

'he(istr%



. At $$ H and at", ' "L o# gaseo!s hydrocar+on re0!ires ' "L air

containing ($>2+

+y vol!"e #or co"plete co"+!stion. A#ter co"+!stion the

gases occ!py $"L. Ass!"ing that the water #or"ed is in li0!id #or" and the

vol!"es were "eas!red at the sa"e te"perat!re and press!re, the #or"!la o#

the hydrocar+on is:

3 6

C H

(

3 8C H

4 8

C H

/4 10

C H

Ans: (Sol: 8ata incorrect

Fol. o# hydrocar+on

15="l

Fol. o# o3ygen

20375 75

100= × =

"l

2 2 2

15

154

4 2 x &

& x

& &C H x + xC+ H +

+ ÷

+ + → + ÷

15 754

& x

+ = ÷

54

& x + =

4 20 x &+ =

20

4

& x

−=

)#

3#/is is coect option

8

x

&

= =

)#

4#/is option is not t/ee

4

x

&

= =



20/48


QP Series: E

(. %wo closed +!l+s o# e0!al vol!"e F containing an ideal gas initially at press!re

$ p

and te"perat!re1T

are connected thro!gh a narrow t!+e o# negligi+le vol!"e

as shown in the 4g!re +elow. %he te"perat!re o# one o# the +!l+s is then raised

to2.T

%he 4nal

press!re

f p

is:

1 2

1 2

T T p$

T T

÷+

(

1

1 2

2 T

p$T T

÷+

2

1 2

2 T

p$T T

÷+

/

1 2

1 2

2 T T

p$T T

÷+

Ans:

)nitial no. o# "oles in each +!l+

1

1

PV n

RT = =

5hen te"p o#

1

+!l+ increased #ro"

1T

to

2T

%hen the no. o# "oles o# gas in each +!l+ is

1 2

1 2

f f P V P V n n

RT RT = =

%otal no. o# "oles +e#ore and a#ter is sa"e

1 2 2n n n∴ + =

1 2 1

2 f f $ P V P V PV

RT RT RT + =



21/48


QP Series: E

1 2 1

21 1 $ f

P P

T T T

+ =

2 1

1 2 1

2 $ f P T T P T T T

+ =

[ ]2

1 2

2 $ f

PT P

T T =

+

. A strea" o# electrons #ro" a heated 4la"ent was passed +etween two charged

plates *ept at a potential diBerence F es!. )# e and " are charge and "ass o# an

electron, respectively, then the val!e o#

!h,

5here

,

is wavelength associated

with electron wave is given +y

m-V

'

(

2m-V

m-V

/

2m-V

Ans: /

Sol:

h

mvλ =

Hinetic energy

-V =

21

2mv -V =

2 2mv -V =

2-V v

m=

2

h

-V m

m

λ ∴ =

2

h

m-V λ =

2h

m-V λ

∴ =

/. %he species in which the ato" is in a state o# sp

hy+ridiIation is:



22/48


QP Series: E

2 (++

(

2 (+−

3 (+−

/2 (+

Ans:

Sol: %he str!ct!re o#

2 (++

is

O N O

( *p− hy+ridiIed

'. %he heats o# co"+!stion o# car+on and car+on "ono3ide are

393.5− and

1283.5 ,k. m/0 −−

respectively. %he heat o# #or"ation )n

KJ

o# car+on "ono3ide

per "ole is:

110.5

(

676.5

676.5−

/

110.5−

Ans: /Sol:

2 2 393.5C + C+ H kg + → ∆ = −

(

2 21 283.5

2C+ + C+ H kg + → ∆ = −

9e0!ired E0!ation:

21

2C + C+ H + → ∆ =

( ) ( )1 2 H ∴ − = ∆

( )393.5 283.5 H ∴ ∆ = − − −

110 kg = −



23/48


QP Series: E

;. g gl!cose

( )6 12 6C H +

is added to

178.2 g

water. %he vapo!r press!re o# water

in torr #or this a0!eo!s sol!tion

.;( ;.$ '(.// '&.$

Ans:8ata )ns!cient

)#

373T k =,

760/ P mm=

18 18

180 178.2

/ s

/

P P

P

− ×=

×

7600.010

760

s P − =

760 7.6 s P − =

760 7.6 s P = −

752.4 mm=

. %he e0!ili+ri!" constant at (& H #or a reaction

A B C D+ +R R ST R Ris $$. )# the

initial concentration o# all the #o!r species were U each, then e0!ili+ri!"

concentration o# D

in "ol

1 L−

will +e: $.(( $. ./ .(

Ans: Sol:

100nitial 1 1 1 1

A B C D K

+ +=

R R ST R R

1 100Q = <

∴ =orward 9eaction ta*es place

E0!ations

( ) ( ) ( ) ( )1 1 1 1 x x x x− − + +

[ ] [ ]

[ ] [ ]

C D K

A B=

( ) ( )

( ) ( )

1 1

1 1

x x K

x x

+ +=

− −



24/48


QP Series: E

( )

( )

2

2

1100

1

x

x

+=

−

1101

x x

+∴ =−

10 10 1 x x− = +

9 11 x=

90.818

11 x = =

[ ] 1 D x∴ = +

1 0.818= +

1.818=

. ValvaniIation is applying a coating o#:

P!

(

C

C1

/

)n

Ans: /Sol: =act!al

&. 8eco"position o#2 2 H +

#ollows a 4rst order reaction. )n 4#ty "in!tes the

concentration o#

2 2 H +

decreases #ro"

0.5

to

0.125

U in one s!ch deco"position.

5hen the concentration o#2 2 H +

reaches

0.05 , M

the rate o# #or"ation o#2

+

will

+e:

2 16.93 10 minm/0 − −×

(

4 16.93 10 minm/0 − −×

12.66 min L at *TP

−



25/48


QP Series: E

/

2 11.34 10 minm/0

− −×

Ans: (

Sol:

2.303log

ak t a x= −

2.303 0.5log

50 0.125k =

10.0277mink −=

[ ]1

2 2 R k H +=

( )[ ]2 2 2 2 0.0277 0.05

d H +k H +

dt

− = = ×

2 2 2 22 2 H + H + +→ +

[ ] [ ]2 2 212

d H + d +

dt dt − =

[ ]2 10.0277 0.05

2

d +

dt ∴ = × ×

4 1 16.93 10 m/0 L * − − −= ×

/$. =or a linear plot o# log

! x m

vers!s log p in a =re!ndlich adsorption isother",

which o# the #ollowing state"ents is correct6 * and n are constants 7oth * and Jn appear in the slope ter".( Jn appears as the intercept. @nly Jn appears as the slope./ Log Jn appears as the intercept.

Ans:

Sol:

1

log log log

x

k pm n= + ×

& C mx= +

∴ Slope

1

n=

)ntercept

log k =

/. 5hich o# the #ollowing ato"s has the highest 4rst ioniIation energy6 9+( a H / Sc

Ans: /



26/48


QP Series: E

Sol:

496 (a kJ −

631*' kJ −

/(. 5hich one o# the #ollowing ores is +est concentrated +y #roth Noatation "ethod6 Uagnetite( Siderite Valena/ Ualachite

Ans: Sol: S!lphide ores are concentrated +y #roth Noatation "ethod

/. 5hich one o# the #ollowing state"ents a+o!t water is =ALSE6 5ater is o3idiIed to o3ygen d!ring photosynthesis.( 5ater can act +oth as an acid and as a +ase. %here is e3tensive intra"olec!lar hydrogen +onding in the condensed phase./ )ce #or"ed +y heavy water sin*s in nor"al water.

Ans: Sol: %here will +e e3tensive inter"olec!lar hydrogen +onding in condensed phase

//. %he "ain o3ides #or"ed on co"+!stion o# Li, a and H in e3cess o# air are,

respectively:

2 2 2, L$ + (a + and K+

(2 2 2

, L$+ (a + and K +

2 2 2 2 2

, L$ + (a + and K+

/2 2 2 2, L$ + (a + and K+

Ans: /Sol: Lithi!" #or"s "ono3ideSodi!" #or"s pero3ide "ostlyPotassi!" #or"s s!pero3ide

/'. %he reaction o# Iinc with dil!te and concentrated nitric acid, respectively,

prod!ces:

2 2 ( + and (+

(2

(+ and (+

2

(+ and ( +

/2 2

(+ and ( +

Ans:

Sol:

( )( )3 3 2 22

ilute

4 10 4 5 )n H(+ )n (+ ( + H ++ → + +



27/48


QP Series: E

( )( )3 3 2 22

concentate

4 2 2 )n H(+ )n (+ (+ H ++ → + +

/;. %he pair in which phosphoro!s ato"s have a #or"al o3idation state o#3+ is:

@rthophosphoro!s and pyrophosphoro!s acids( Pyrophosphoro!s and hypophosphoric acids @rthophosphoro!s and hypophosphoric acids/ Pyrophosphoro!s and pyrophosphoric acids

Ans:

Sol: %he #or"!la o# orthophosphor!s acid3 3 H P+=

%he #or"!la o# pyrophosphoro!s acid4 2 5 H P +=

∴

@3idation state o#

3 P = +

/. 5hich o# the #ollowing co"po!nds is "etallic and #erro"agnetic6

2

T$+

(2C+

2V+

/2 Mn+

Ans: (

Sol:2C+ −

#erro"agnetic

2T$+ −8ia"agnetic

2 Mn+ −Anti#erro"agnetic

2V+ −para"agnetic

/. %he pair having the sa"e "agnetic "o"ent is:

. . 24, 25, 26, 27: At (/ C Mn "- C/= = = =

( ) [ ]2 2

2 46 anC H + C/C0

+ −

(

( ) ( )2 2

2 26 6anC H + "- H +

+ +

( ) ( )2 2

2 26 6an Mn H + C H +

+ +

/

[ ] ( ) 22

4 2 6C/C0 and "- H +

+−

Ans: (



28/48


QP Series: E

Sol: %he no. o# !npaired electrons in

2C +

and

2 "- +

is4

( )21 n n= +

( )4 4 2= +

24=

4.89 BM =

/&. 5hich one o# the #ollowing co"ple3es shows optical iso"eris"6

( )3 33C/ (H C0

(( ) 22'$s C/ -n C0 C0

( ) 22tans C/ -n C0 C0

/

( )3 24C/ (H C0 C0

en W ethylenedia"ine

Ans: (

Sol: %he optical iso"ers o# Cis-

( ) 22C/ -n C0 C0

C

Cl

Cl

en

en

C

Cl

Cl

en

en

'$. %he concentration o# N!oride, lead, nitrate and iron in a water sa"ple #ro" an

!ndergro!nd la*e was #o!nd to +e $$$ pp+, /$ pp+, $$ pp" and $.( pp",

respectively. %his water is !ns!ita+le #or drin*ing d!e to high concentration o#: =l!oride( Lead itrate/ )ron

Ans:

Sol: Uore than

40 ppm

gives +l!e +a+y syndro"e disease



29/48


QP Series: E

'. %he distillation techni0!e "ost s!ited #or separating glycerol #ro" spent-lye in

the soap ind!stry is: Si"ple distillation( =ractional distillation

Stea" distillation/ 8istillation !nder red!ced press!re

Ans: /Sol: Vlycerol has a higher +oiling point and it can deco"pose at its +oiling point. So

8istillation !nder red!ced press!re will +e !sed

'(. %he prod!ct o# the reaction given +elow is:

(

/

Ans: (

N!S

!" H2O

K2CO3

OH

Sol:

S!+stit!tion will +e in less sterically hindered allylic position

'. %he a+sol!te con4g!ration o#



30/48


QP Series: E

)s:

( )2 , 3 R *

(

( )2 , 3* R

( )2 , 3* *

/

( )2 , 3 R R

Ans: (Sol: Concept!al

'/. (-chloro-(-"ethylpentane on reaction with sodi!" "etho3ide in "ethanol yields:

All o# these( a and c c only/ a and +

Ans:

Cl

CH3 CH3ONa

CH3OH

#$2%#Sa&'(e)) *"+,-'%

Sol:

Answer can +e all o# these since "a?or prod!ct has not +een "entioned

''. %he reaction o# propene with

( )2 2 H+C0 C0 H ++proceeds thro!gh the inter"ediate:

3 2CH CH CH +H +− − −



31/48


QP Series: E

(3 2CH CH CH C0

+− − −

3 2 CH CH +H CH +− −

/3 2CH CHC0 CH

+− −

Ans: (

CH3 CH CH2HOCl

CH3 CH2 CH2 Cl

#.n'e" /e+0a1%

H

H2O

CH3 CH CH2 Cl

OH

Sol:

';. )n the


32/48


QP Series: E

( Cystine Cysteine/ Uethionine

Ans:

Sol: %he str!ct!re o# Cysteine is

H C

CH2SH

NH2

COOH

'&. 5hich o# the #ollowing is an anionic detergent6 Sodi!" stearate( Sodi!" la!ryl s!lphate Cetyltri"ethyl a""oni!" +ro"ide/ Vlyceryl oleate


=or"!la o# sodi!" la!ryl s!lphate12 25 4C H *+ (a

;$. %he hottest region o# 7!nsen Na"e shown in the 4g!re +elow is:

region ( region ( region / region /




33/48


QP Series: E

Mathe(atis



;. )#

( ) 12 3 0f x f x, xx

+ = ≠ ÷

and

( ) ( ){ }S x R : f x f x ;= ∈ = −

then S :

is an e"pty set. ( contains e3actly one ele"ent.

contains e3actly two ele"ents / contains "ore than two ele"ents.

( ) ( )1

2 3 1 f x f x x

+ = − − − ÷

( ) ( ) ( )1 3 1 6

2 2 4 2 f f x f f x x x x x

⇒ + = ⇒ + = − − − ÷ ÷

E0!ation

( ) ( )1 2−

( ) 2

f x x x

⇒ = −

( ) ( ) f x f x= −

2 2 x x

x x⇒ − = − +

42 x

x⇒ =

2 2 x⇒ = 2 x⇒ = ±

Ans:

;(. A val!e o# #or which

2 3

1 2

isin

isin

+−

is p!rely i"aginary, is :

3

(

6

1 3

4sin−

÷ ÷

/

1 1

3sin

− ÷

( );e 0 2 =

22 6sin 0θ ⇒ − =

2 1sin3

θ ⇒ =

1 1sin

3θ −

⇒ = ± ÷

Ans: /



34/48


QP Series: E

;. %he s!" o# all real val!es o#x

satis#ying the e0!ation

( )2 4 60

25 5

x x

x x+ −

− +

W is

( -/ ; / '

i

24 60 0 x x+ − =

( ) ( )10 6 0 x x⇒ + − =

10 o 6 x⇒ = − ⇒s!" o# the real n!"+ers

10 6 4= − + = −

∴ S!" o# sol!tions

4= −

Ans: (

;/. )#

5

3 2

a bA

− =

and

( )

T

A AdjA AA=then

5a b+is e0!al to:

- ( ' / /

T AA A 3 =

2 225 3a ! !a !⇒ + = +

15 2 0a !− =

2 15, 3

5 2

aa !⇒ = = =

5 5a !⇒ + =

Ans: (;'. %he syste" o# linear e0!ations

0x y z+ − =

0x y z− − =

0x y z+ − = has a non-trival sol!tion #or:

)n4nitely "any val!es o#

( E3actly one val!es o#

E3actly two val!es o#

/ E3actly three val!es o#

1 1

1 1 0

1 1

λ

λ

λ

−− − =

−

( )2

1 0λ λ ⇒ − =



35/48


QP Series: E

0, 1, 1λ λ ⇒ = = −

Ans: /;;. )# all the words with or witho!t "eaning having 4ve letters, #or"ed !sing the letters

o# the word SUALL and arranged as in a dictionaryG then the position o# the word

SUALL is:

/;th ( '&th '(nd

/ 'th

t/4< 3<4 3 1 58

2< 2<

+ + = ÷ ÷

Ans: /

;. )# the n!"+er o# ter"s in the e3pansion o#

2

2 41 0

n

, xx x

− + ≠ ÷

is (, then the s!" o# the

coecients o# all the ter"s in this e3pansion is:

;/ ( ( (/ /

(&

o sol!tion, %here is a"+ig!ity in 0!estion.Ans: -

;. )# the (nd , 'th and &th ter"s o# a non-constant A.P are in V.P., then the co""on ratio

o# this V.P is

8

5

(

4

3

1

/

7

4

Let

, 4 , 8a d a d a d + + +

( ) ( ) ( )2

4 8a d a d a d + = + +

1

8

d

a

⇒ =

11 4

4 481 3

18

a d

a d

+ ×+= = =

+ +

Ans: (



36/48


QP Series: E

;&. )# the s!" o# the 4rst ten ter"s o# the series

2 2 2 2

23 2 11 2 3 4 45 5 5 5

.....4 + + + + + ÷ ÷ ÷ ÷

is

16

5m

then " is e0!al to :

$( ( $ $$

/ &&

2 2 28 12 16 16

... m5 5 5 5

+ + + = ÷ ÷ ÷

2

2 2 2 24 16

2 3 4 ... 11 m5 5 + + + + = ÷

216 1611 1 m m 10125 5

− = ⇒ = ∑

Ans: (

$.Let

( )1

2 2

01 x

xp lim tan x

→ += +

then

logp

is e0!al to :

((

1

2

/

1

4

12 2

0

lim 1 tan x

x

p x+→

= +

is o# the #or"1∞

2

0

1lim 1 tan 1

2 x x

x p - → + + − =

1

2 1log2

p - p= ⇒ =

Ans:

. =or

x R,∈ ( ) 2f x log sin x= −

and

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

then:

g is not diBerentia+le at

0x =

(

( ) ( )0 2g ' cos log=



37/48


QP Series: E

( ) ( )0 2g ' cos log= −

/g is diBerentia+le at

0x =and

( ) ( )0 2g ' sin log= −

( ) ( )( ) ( ) ( )( ) ( ) g x f f x g x f f x f x′ ′ ′= ⇒ = ×

( ) ( )( ) ( )0 0 0 g f f f ′ ′ ′= ×

( ) ( )log 2 sin

coslog 2 sin

x f x x

x

−′ = × −

−

( ) ( ) ( )0 1" log 2 cos log 2 f f ′ ′= − = −

( ) ( )0 cos log2 g ′⇒ =

Ans: (

(. Consider

( ) 1 1

01 2

sinxf x tan , x ,

sinx

− + = ∈ ÷ ÷ ÷− A nor"al to

( )y f x=

at

6x

=

also passes

thro!gh the point:

$,$ (

203

,

÷

06

, ÷ /

04

, ÷

( ) 1 11 cos

1 sin 2tan tan

1 sin1 cos

2

x x

f x x

x

π

π − −

+ − ÷ ÷ + ÷= = ÷ ÷ ÷− − − ÷ ÷ ÷

2

1

2

2cos4 2

tan

2sin4 2

x

x

π

π

−

− ÷ ÷ ÷=

÷ − ÷ ÷ ÷

1tan cot

4 2

xπ − = − ÷ ÷

4 2

xπ = +

)#

6 x

π =

then

3 &

π =

and slope o# nor"al

2= −



38/48

4 x 2 π


QP Series: E

E0!ation

23 6

& xπ π − = − − ÷

22

3 & x π

⇒ + = it is passing thro!gh

20, 3

π

÷

Ans: (

. A wire o# length ( !nits is c!t into two parts which are +ent respectively to #or" a

s0!are o# side W 3 !nits and a circle o# radi!sW r !nits. )# the s!" o# the areas o# the

s0!are and the circle so #or"ed is "ini"!", then:

( )2 4x r= +

(

( )4 x r − =

2x r=/

2x r=

4 2 2 x π + =

2 42

x π

−=

( )2

2 2 2

2

2 4

4

x A x xπ π

π

−= + = +

A

is "on.

( ) ( )2 2 4 40 2 0

4

xdA x

dx π

− −⇒ = ⇒ + =

( )2

2 2 4 x xπ

⇒ = −

2 4 x xπ ⇒ = −

2 1 2 2 1,

4 4 4 x

π π π π π

⇒ = = − = ÷+ + +

2 x ⇒ =

Ans:



39/48


QP Series: E

/. %he integral

( )

12 9

35 3

2 5

1

x xdx

x x

+

+ +∫

is e0!al to :

( )

5

25 3

1

xc

x x

−+

+ +

(

( )

10

25 3

2 1

xc

x x

++ +

( )

5

25 32 1

xc

x x

++ +

/

( )

10

25 32 1

xc

x x

−+

+ +

5here C is an ar+itrary constant.

12 9 3 6

3 35

2 52 5

2 5

2 5

1 11 111

x x x x 3 dx dx

x x x x x

++

= = + ++ + ÷ ÷

∫ ∫

2 5

3 2

2 5 2 5

1 11

1

1 1 1 11 2 1

d x x

C

x x x x

+ + ÷

= − = +

+ + + + ÷ ÷

∫

( )10

25 32 1

x C

x x

= ++ +

Ans: (

'.

( ) ( )1

2

1 2 3 n

nn

n n ... nlim

n→∞

+ + ÷ ÷

is e0!al to :

4

18

e

(

2

27

e

2

9

e

/

3 3 2log −

( )2

1

1log lim log

n

n

n L

n n→∞ =

+= ×∑

( )2

0log 1 x dx= +∫

( ) ( ) 2

0log log 1 x x x x x = − − − +

( ) ( ) 2

01 log 1 x x x= + + −



40/48

( )2, 2

( )2, 0


QP Series: E

3log3 2= −

2

27log

-=

2

27 L

-∴ =

Ans: (

;. %he areain s0 !nits o# the region

( ){ }2 2 22 4 0 0x, y : y x and x y x, x , y≥ + ≤ ≥ ≥

is:

4

3

−

(

8

3

−

4 2

3

−

/

2 2

2 3

−

Solving

2 2 2 24 , 2 x & x & x+ = =

we get

0 o 2 x =

2 2

04 2 A x x x dx = − − ∫

( )2 2

0

4 2 2 x x dx = − − − ∫

230

2 2

20

2 24

3t dt x

−

÷= − − ÷

∫

0

2 1

2

1 84 4sin

2 2 3

t t t

−

−

= − + −

1 80 4

2 2 3

π = − − × − ÷

8

3π = −

s0. !nitsAns: (



41/48


QP Series: E

. )# a c!rve

( )y f x=

passes thro!gh the point

( )1 1, −

and satis4es the diBerential

e0!ation,

( )1y xy dx x dy,+ =

then

1

2f − ÷

is e0!al to :

2

5−

(

4

5−

2

5

/

4

5

2 &dx x& dx xd&+ =

2 0

&dx xd& xdx

&

−⇒ + =

2

02

x xd d

&

⇒ + = ÷ ÷ ÷

2

2

x x'

&⇒ + =

D

( )1, 1−

lies on

11

2'⇒ − + =

12

'⇒ = −

∴ sol is

21

2 2

x x

&+ = −

1 1 1 1

2 2 8 2 x

&= − ⇒ − + = −

1 1 1 5

2 8 2 8 &⇒ = + =

1 4

2 5 f

⇒ − = ÷

Ans: /

. %wo sides o# a rho"+!s are along the lines,

1 0x y− + =and

7 5 0x y− − =. )# its diagonals

intersect at

( )1 2,− −

, then which one o# the #ollowing is a verte3 o# this rho"+!s6



42/48

7 4,

3 3

− − ÷

7 15 x &− = − ( )3, 6−

( )1, 2 1 8,3 3

− ÷

7 5 0 x &− − =

3 x &− =1 0 x &− + =

( ),h k k

'

a

6

( )4, 4 P

( )3, 2−

d

( )2, 3− 5

*


QP Series: E

( )3 9,− −

(

( )3 8,− −

1 8

3 3,

− ÷

/

10 7

3 3,

− − ÷

Ans:

&. %he centres o# those circles which to!ch the circle,

2 28 8 4 0x y x y+ − − − =

, e3ternally

and also to!ch thex axis− , lie on:

a circle ( an ellipse which is not a circle.

a hyper+ola / a para+ola

Let

( ),' h k

+e centre

6CP k = +

( ) ( ) ( )22 2

4 6h & k k ⇒ − + − = +

( ) ( ) ( )22 2

4 6 4h k k ⇒ − = + − −

'∴

lies on a para+olaAns: /

$. )# one o# the dia"eters o# the circle, given +y the e0!ation,

2 2 4 6 12 0x y x y+ − + − =, is a

chord o# a circle S, whose centre is at

( )3 2,−

, then the radi!s o# S is:

5 2

(

5 3

5

/

10

2 225d + =

2 75 ⇒ =

5 3 ⇒ =

Ans: (



43/48


QP Series: E

. Let P +e the point on the para+ola,

28y x=

which is at a "ini"!" distance #ro" the

centre C o# the circle,

( ) 22 6 1x y+ + =

. %hen the e0!ation o# the circle, passing thro!gh

C and having its centre at P is:

2 24 8 12 0x y x y+ − + + =

(

2 24 12 0x y x y+ − + − =

2 22 24 0

4

xx y y+ − + − =

/

2 2 4 9 18 0x y x y+ − + + =

p = point on circle

( )22 , 4t t =

' = centre o# circle

( )0, 6= −

( ) ( )24

4 4 6 f t CP t t = = + +

( ) 0 1 f t t ′ = ⇒ = −

CP ∴

is "in when

1t = −

( )2, 4 P ⇒ = −

∴ E0!ation o# re0!ired circle is

( ) ( )2 2

2 4 8 x &− + + =

i.e.,

2 2 4 8 12 0 x & x &+ − + + =

Ans:

(. %he eccentricity o# the hyper+ola whose length o# the lat!s rect!" is e0!al to and

the length o# its con?!gate a3is is e0!al to hal# o# the distance +etween its #oci, is

4

3

(

4

3

2

3

/

3

222 8 4

!! a

a= ⇒ =

2 2 22 4! a- ! a != ⇒ = +

2 2

3 12a ! a⇒ = =



44/48

( )1, 5, 9 P −

1 5 9 x & 2 − = + = −

m 5 x & 2 − + =

( )9, 15, 1− − −


QP Series: E

212, 48a !⇒ = =

( )2 4 32 212 3

!-

a∴ = = =

Ans:

. %he distance o# the point

( )1 5 9, ,−

#ro" the plane

5x y z− + = "eas!red along the line

x y z= =is

3 10

(

10 3

10

3

/

20

3

∴ 9e0!ired distance

2 2 210 10 10 10 3 PM = = + + =

Ans: (

/. )# the line,

3 2 4

2 1 3

x y z− + += =

− lies in the plane,

9lx my z+ − =, then

2 2l m+

is e0!al to : (; ( ' / (

2 3 00 m− − =D

( )3, 2, 4− −

lies on plane

3 2 5 00 m⇒ − − =D (

Solving and (,1, 10 m= = −

2 2 20 m∴ + =

Ans: /



45/48


QP Series: E

'.Let

a, b

and

c

+e three !nit vectors s!ch that

( ) ( )3

2a b c b c× × = +r r r r r

. )#

b

is not parallel to

c

, then the angle +etween

a

and

b

is :

a

3

4

(2

2

3

/

5

6

1a != =

Let

( ),a ! θ =

( ) ( ) ( )32

a ' ! a ! ' ! '× − × = +

3

2a !⇒ × = −

3cos

2θ ⇒ = −

5

6 6

π π θ π ∴ = − =

Ans: /

;. )# the standard deviation o# the n!"+ers

2 3 11, ,a and

is .', then which o# the

#ollowing is tr!e6

23 26 55 0a a− + = (

23 32 84 0a a− + =

23 34 91 0a a− + =/

23 23 44 0a a− + =

16

4

a x

+=

( )2

49

4 4

$ x x− =∑

( )2

49$ x x⇒ − =∑212 128 336 0a a⇒ − + =

23 32 84 0a a⇒ − + =



46/48


QP Series: E

Ans: (

. Let two #air si3-#aced dice A and 7 +e thrown si"!ltaneo!sly. )#1E

is the event that

die A shows !p #o!r,2

E

is the event the die 7 shows !p two and3

E

is the event that

the s!" o# n!"+ers on +oth dice is odd , then which o# the #ollowing state"ents is

@% tr!e6

1Eand

2Eare independent (

2Eand

3Eare independent

1

E

and3E

are independent. /1 2 3E ,E and E

are independent.

( ) ( ) ( )1 2 1 21 1

,6 36

p E p E p E E = = ∩ =

( )31

2 p E =

1 2 3 E E E ∩ ∩ is an i"possi+le event

Ans: /

. )#

0 2x ≤ <, then the n!"+er o# real val!es o#

x

, which satis#y the e0!ation

2 3 4 0cos x cos x cos x cos x ,+ + + =is:

( ' / &

( ) ( )cos 4 cos 2 cos3 cos 0 x x x x+ + + =

2 cos 3 cos 2 cos 2 cos 0 x x x x⇒ + =

( )2cos cos3 cos 2 0 x x x⇒ + =

54 cos cos cos 0

2 2

x x x⇒ =

5cos 0 o cos 0 o cos 0

2 2

x x x⇒ = = =



47/48

h x60°30° B Ad

10min


QP Series: E

3cos 0 ,

2 2 x x

π π = ⇒ =

{ }cos 02 2 2

x x

x

π

π = ⇒ = ⇒ =

5 5 3 5 7 9cos 0 , , , ,

2 2 2 2 2 2 2

x x π π π π π = ⇒ =

3 7 9, , , ,

5 5 5 5 x

π π π π π

⇒ =

∴ !"+er o# sol!tions

7=

Ans:

&. A "an is wal*ing towards a vertical pillar in a straight path, at a !ni#or" speed. At a

certain point A on the path, he o+serves that the angle o# elevation o# the top o# the

pillar is

30°. A#ter wal*ing #or $ "in!tes #ro" A in the sa"e direction, at a point 7,

he o+serves that the angle o# elevation o# the top o# the pillar is60° . %hen the ti"e

ta*en in "in!tes +y hi", #ro" 7 to reach the pillar, is:

; ( $ ($ / '

cot30 cot60

d h =

° − °

3

2

d h⇒ =

tan60

h

x°

23

h d x⇒ = =

∴ 9e0!ired ti"e

10min 5min

2= =

Ans: /

&$. %he 7oolean E3pression( ) ( )p q q p q∧ ∨ ∨ ∧

: :

is e0!ivalent to :



48/48


QP Series: E

p q∧:(

p q∧

p q∨/

p q∨ :

( ) ( ) P Q Q P Q P Q∩ ∪ ∪ ∩ = ∪

∴ Viven e3pression is e0!ivalent to

P Q∪

Ans:

01-Offline JEE Main Final Test - 03-04-2016

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