2018 jee-main s q p k main... · 2018-04-09 · 2018_jee-main question paper_key & solutions...

2018_Jee-Main Question Paper_Key & Solutions


Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081

www. srichaitanya.net, [email protected]

PAPER-1 CHEMISTRY , MATHEMATICS & PHYSICS

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CHEMISTRY 01. Which of the following salts is the most basic in aqueous solution? . 1) 3CH COOK 2) 3FeCl 3) 3 2( )Pb CH COO 4) 3( )Al CN Key: 1 Sol: Only one anion 3CH COO undergoing hydrolysis. Where as in other salts

3 2Pb CH COO and 3

Al CN , cation is also undergoing hydrolysis.

02. Which of the following compounds will be suitable for Kjeldahl’s method for nitrogen estimaton?

1)

2NH

2)

2NO

3)

2N Cl

4) N Key: 1 Sol: Pyridine, diazonium salts, nitro compounds will not give kjeldahl’s method. 03. Which of the following are Lewis acids? 1) 3 4AlCl andSiCl 2) 3 4PH andSiCl 3) 3 3BCl and AlCl 4) 3 3PH and BCl

Key: 3 Sol: Both 3BCl and 3AlCl are electron difficient molecule.




04. Phenol on treatment with 2CO in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with 3 2

CH CO O in the presence of catalytic amount of 2 4H SO produces:

1)

3CH

O

2COH

O

2)

3CH

O

OC

O

OH

3)

2COH

O2COH3CH

O

4)

3CH

O

2COH

O

Key: 4

Sol:

212

. ,.

CO NaOHH 3 2CHCO O

H

OH OH

COOH COOH

3O COCH




05. An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correct combination?

Base Acid End point 1) Strong Strong Pinkish red to yellow 2) Weak Strong Yellow to pinkish red 3) Strong Strong Pink to colourless 4) Weak Strong Colourless to pink Key: 2 Sol: In basic medium, methyl orange shows yellow colour and in acidic medium shows

pinkish red colour. 06. An aqueous solution contains 0.10 M 2H S and 0.20 M HCl. If the equilibrium

constants for the formation of HS from 2H S is 71 0 10. and that of 2S from

HS ions is 131 2 10. then the concentration of 2S ions in aqueous solution is : 1) 203 10 2) 216 10 3) 195 10 4) 85 10 Key: 1 Sol:

22

22

2 2

H S H HSH S H SH S H S

2 2

1 22

H S

K K KH S

7 13

22

1 10 1.2 10 0.10.2

S

203 10

07. The combustion of benzene (I) gives 2CO (g) and 2H O (I). Given that heat of

combustion of benzene at constant volume is -3263.9 kJ 1mol at 025 C; heat of combustion (in kJ 1mol ) of benzene at constant pressure will be:

1 18 314.R JK mol

1) -452.46 2) 3260 3) -3267.6 4) 4152.6




Key: 3 Sol: 6 6 2 2 27.5 6 3 C H l O g CO g H O l

6 7.5 1.5 H H U nRT

33263.9 1.5 8.314 10 298 13267.6 . kJ mol

08. The compound that does not produce nitrogen gas by the thermal decomposition is: 1) 4 2 72

NH Cr O

2) 4 2NH NO

3) 4 42NH SO

4) 3 2Ba N

Key: 3 Sol: 4 2 7 2 2 3 22

4 NH Cr O N Cr O H O

4 2 2 22 NH NO N H O

4 4 3 4 42 NH SO NH NH HSO

3 223 Ba N N Ba

09. How long(approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane?

(Atomic weight of B=10.8 u) 1) 0.8 hours

2) 3.2 hours

3) 1.6 hours

4) 6.4 hours

Key: 2 Sol: 2 6 2 2 3 23 3 B H g O B O H O

27.66 3 32g g From Faraday’s first law

Mass of gas liberated (m) . .

96500

E c t

8 1003 3296500

t

t in hours 96 96500 3.21

8 100 60 60

hrs




10. Total number of lone pair of electrons in 3I ion is :

1) 6 2) 9 3) 12 4) 3 Key: 2 Sol:

I

I

I

(-)

9 lone pair of electrons. 11. When metal ‘M’ is treated with ,NaOH a white gelatinous precipitate ‘X’ is obtained,

which is soluble in excess of NaOH. Compound ‘X’ when heated strongly gives an oxide

which is used in chromatography as an adsorbent. The metal ‘M’ is :

1) Ca

2) Al

3) Fe

4) Zn.

Key: 2 Sol: Oxide which is used as adsorbent in chromatography is 2 3Al O .

3 3 Al NaOH Al OH Na

gelatinouswhite ppt

23 4 2

lub

Al OH NaOH Na Al OH H O

so le

12. According to molecular orbital theory, which of the following will not be a viable molecule?

1) 2He

2) 2H

3) 22H

4) 22He .




Key: 3 Sol: ion Bond order

2He 0.5

2H 0.5

22H 0

22He 1.

Therefore 22H will not exist.

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

a) 2NH

b)

NH

c)

2NH

NH

d) 3NHCH

1) b a c d

2) b a d c

3) d b a c

4) a b c d .

Key: 2

Sol: ln2NH

, negative inductive effect is present.

Where as 3NHCH is secondary amine. 14. Which type of ‘defect’ has the presence of cations in the interstitial sites? 1) Vacancy defect 2) Frenkel defect 3) Metal deficiency defect 4) Schottky defect Key: 2 Sol: In Frenkel defect, metal cation occupies interstitial site.




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

1) 2 4,KCl H SO 2) KCl 3) 2 6,KCl B H

4) 2 6 3, ,KCl B H PH . Key: 2 Sol: KCl is a ionic compound.

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

and 2 32 2 2K Cr CN O O NH

respectively are :

1) 3, 2, 4and 2) 3,0, 6and 3) 3,0, 4and 4) 3, 4, 6and . Key: 2 Sol: Compound O.s of Cr

2 36

Cr H O Cl +3

6 6 2

Cr C H 0

2 2 2 32 K Cr CN O O NH +6

17. Hydrogen peroxide oxidises 4

6Fe CN

to 3

6Fe CN

in acidic medium but

reduces 3

6Fe CN

to 4

6Fe CN

in alkaline medium. The other products

formed are, respectively.

1) 2 2 2H O O and H O OH

2) 2 2 2H O and H O O

3) 2 2H O and H O OH

4) 2 2 2H O O and H O .

Key: 2

Sol: 34

6 2 2 262 2 2 2Fe CN H O H Fe CN H O

43

6 2 2 2 262 2 2 2Fe CN H O OH Fe CN H O O




18. Glucose on prolonged heating with HI gives : 1) 1-Hexene 2) Hexanoic acid 3) 6-iodohexanal 4) n-Hexane. Key: 4 Sol:

6 12 6 3 2 3 3 2 33 4

HICH O CHCH CH CH CH CH CH

IIodohexane n hexane

19. The predominant form of histamine present in human blood is , 6.0apK Histidine .

1)

N

NH

H

3NH

2)

N

NH

H

2NH

3)

N

N

H

3NH

4)

N

N

H

2NH

Key: 3 Sol:

N

N

H

2NHHistamine

6apK

9apK




In blood (PH7.2). histamine become

N

N

H

3NH

20. The recommended concentration of fluoride ion in drinking water is up to 1 ppm as

fluoride ion is required to make teeth enamel harder by converting

3 4 223 .Ca PO Ca OH to :

1) 2 23 .CaF Ca OH

2) 3 4 223 .Ca PO CaF

3) 223 .Ca OH CaF

4) 2CaF .

Key: 2

Sol: When F ion react with teeth enamel, 3 4 223 .Ca PO Ca OH changes as

fluorapatite 3 4 223 .Ca PO CaF .

21. Consider the following reaction and statements :

3 2 3 3 34 3Co NH Br Br Co NH Br NH

(I) Two isomers are produced if the reactant complex ion is a cis-isomer.

(II) Two isomers are produced if the reactant complex ion is a trans-isomer.

(III) Only one isomer is produced if the reactant complex ion is a trans-isomer.

(IV) Only one isomer is produced if the reactant complex ion is a cis-isomer.

The correct statements are :

1) I and III

2) III and IV

3) II and IV

4) I and II .

Key: 1

Sol: When 3 24Co NH Br

is cis-isomer




Co2HN

2NH

Br

Br

2NH

2HN

Co2HN

2NH

Br

Br2HN

Br

Br

Co2HN

2NH

Br

Br

2NH

Br

Co2HN

2NH

Br

Br

2NH2HN

Br

trans-isomer

Co2HN Br

Br

2NH2HN

Br

fac-isomer only formed

cis-isomer fac-isomer mer-isomer

22. The trans-alkanes are formed by the reduction of alkynes with : 1) 4NaBH

2) 3/ .Na liq NH 3) Sn HCl 4) 2 4/ ,H Pd C BaSO . Key: 2

Sol: Alkyne 3/ .Na liq NH trans-alkene. 23. The ratio of mass percent of C and H of an organic compound X Y ZC H O is 6 : 1. If

one molecule of the above compound X Y ZC H O contains half as much oxygen as

required to burn one molecule of compound X YC H completely to 2CO and 2H O . The

empirical formula of compound X Y ZC H O is :

1) 2 4C H O

2) 3 4 2C H O

3) 2 4 3C H O

4) 3 6 3C H O . Key: 3 Sol: On combustion of X YC H having 6 : 1 ratio of C and H will be

2 4 2 2 23 2 2

6C H O CO H O

atoms of oxygen




3 6 2 2 29 3 329

C H O CO H O

atoms oxygen

2X YC H O will be 2 4 3C H O . 24. Phenol reacts with methyl chloroformate in the presence of NaOH to form product A.

A reacts with 2Br to form product B. A and B are respectively.

1)

O O O O

Br

andO O

2)

O O O O

Br

andO O

3)

OH

Br

and

O

3OCH

OH

O

3OCH

4)

OHBr

and

O

3OCH

OH

O

3OCH

Key: 2 Sol:




OH

Cl C O 3CH

O

OHO O

O O

Br

O

O

2Br(A)

Br(major)

25. The major product of the following reaction is:

Br

NaOMeMeOH

1)

2)

3)

OMe

4)

OMe

Key: 1




Sol: With strong base and bulky groups at carbon leads elimination rather than 2SN substitution

Br

NaOMeMeOH

26. Which of the following lines correctly show the temperature dependence of equilibrium

constant, K, for an exothermic reaction?

A

B

C

D

(0,0)

ln K

1

T K

1) B and C 2) C and D 3) A and D 4) A and B. Key: 4

Sol: 14ln ln

2.303Hkc A

R T

Here .H ve exothermic reaction graph will be

ln K

1

T K

A

B




27. The major product formed in the following reaction is:

O

OHI

Heat

1) I

I

2)

OH

I

3) OH

I

4) OH

OH

Key: 3

Sol: O

OHI

OH

I3CHI 3 2CHCH I




28. An aqueous solution contains an unknown concentration of 2Ba

. When 50 mL of a 1

M solution of 2 4Na SO is added, 4BaSO just begins to precipitate. The final volume is

500 mL. The solubility product of 4BaSO is 101 10 . What is the original

concentration of 2 ?Ba

1) 92 10 M

2) 91.1 10 M

3) 101.0 10 M

4) 95 10 M .

Key: 2 Sol: After addition of 50 mL 1 M 2 4Na SO to the solution having 2Ba ion become

500 mL

24

50 1 0.1500

new SO M .

1042

24

1 10500 ,0.1

spK BaSOIn mL Ba

SO

91 10 M

Now, in 450 mL of initial solution , 9

2 91 10 500 1.1 10450

Ba M

.

29. At 0518 ,C the rate of decomposition of a sample of gaseous acetaldehyde, initially at a

pressure of 363 Torr, was 1.00 Torr 1s when 5% had reacted and 0.5 Torr 1s when

33% had reacted. The order of the reaction is :

1) 3

2) 1

3) 0

4) 2

Key: 4

Sol: 1 1

2 2

nr Pr P




1 1

2 2

log logr Pnr P

Order of the reaction (n) =

1

2

1

2

log

log

rrPP

1log 0.30.5344.8 0.15log243.2

2.

30. For 1 molal aqueous solution of the following compounds, which one will show the

highest freezing point?

1) 2 2 25.Co H O Cl Cl H O

2) 2 2 24.2Co H O Cl Cl H O

3) 2 3 23.3Co H O Cl H O

4) 2 36Co H O Cl .

Key: 3

Sol: Freezing point will be highest, when solute is not undergoing any dissociation

2 3 2.3Co H O Cl H O .




MATHEMATICS 31. The integral

2 2

5 3 2 3 2 2sin x cos x dx

sin x cos xsin x sin x cos x cos x

is equal to

1) 3

1 C3 1 tan x

2) 31 C

1 cot x

3) 31 C

1 cot x

4) 3

1 C3 1 tan x

(Where C is a constant of integration) Key: 1

Sol: Let tan x t

2 6

25 2 3

. sec

1

t xdt

t t t

2 2

2 23 2

1

1 1

t tdt

t t

2 2

23

1

1

t tdt

t

Let 3

11

pt

;

2 2

23

3 113 1

t tdt

t

3

1 1 13 3 1 tan

dp Cx




32. Tangents are drawn to the hyperbola 2 24x y 36 at the points P and Q. If these

tangents intersect at the point T 0,3 then the area (in sq. units) of PTQ is

1)54 3

2) 60 3

3)36 3

4) 45 5 Key: 4

Sol: 2 2

19 36x y

is help

P Q

0,3T

Eq of CC AB is3 1 0 12

36y y

,P Q are 45, 12

area of 1 15 .2 452

TPQ

45 5

33. Tangent and normal are drawn at P 16,16 on the parabola 2y 16x , which intersect

the axis of the parabola at A and B, respectively. If C is the centre of the circle through

the points P, A and B and CPB , then a value of tan is:

1) 2

2) 3

3) 43

4) 12

Key: 1




Sol: 2 16y x

4,0C 24,0

16,16

16,0A B

P

tan 2 (using slopes). 34. Let u

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

and ˆˆb j k

. Ifu is

perpendicular to a and . 24u b and 2u is equal to:

1)315

2)256

3)84

4) 336

Key: 4

Sol: u ka a b

2.u k a b a a b

2 2 2. 24u b k a b a b

4 28 24k

1k

u a a b

2 14a b

2 336u

35. If , C are the distinct roots, of the equation 2 1 0x x , then 101 107 is equal

to:

1)0

2)1

3) 2

4) 1




Key: 2

Sol: 2 1 0x x

2 1 0x x

Root are 2&w w

2,w w

101 107

99 2 105 2. .

2 2 1 2 1

36. Let 2cos ,g x x f x x , , be the roots of the quadratic equation

2 218 9 0x x . Then the area (in sq. units) bounded by the curve y gof x

and the lines ,x x and 0y , is:

1) 1 3 12

2) 1 3 22

3) 1 2 12

4) 1 3 12

Key: 4

Sol: gof x

g x

cos x

area3

3

66

cos sinxdx x

3 12




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

5 5

3 31 1 , 1x x x x x is:

1)0

2)1

3) 2

4) 1

Key: 3

Sol: 5 4 3 3 3 20 0 2 35 5 1 5 1 5C x C x x C x x C x

2 45 3 3 3

0 2 42 5 5 1 5 1C x C x x C x x

25 3 3 30 2 42 5 5 1 5 1C x C x x C x x

5 6 3 6 30 2 2 42 5 5 5 5 2 1C x C x C x C x x x

5 6 3 7 40 2 2 4 4 42 5 5 5 5 2 5 5C x C x C x C x C x C x

0 2 4 42 5 5 5 5C C C C

2 1 10 5 5 2

38. Let 1 2 3 49, , ,...,a a a a be in A.P. Such that12

4 10

416kk

a

and 9 43 66a a . If

2 2 21 2 17.... 140a a a m, then m is equal to:

1)68

2)34

3)33

4) 66

Key: 2

Sol: 1 5 49... 416a a a

9 43 66a a

1 1 1 1 14 8 12 . 48 416a a d a d a d a d




113 4 1 2 3 12 416a d

113 312 416a d

1 24 32a d …. (1)

1 18 42 66a d a d

12 50 66a d

1 25 33a d ….(2)

1d

1 8a

2 2 21 2 17..a a a

2 2 2 28 9 10 ... 24

24 25 44 7 8 15

6 6

25 4 49 7 4 5

5 4 7 5 7 1

20 7 34 140m

34m

39. If 9

1

5 9ii

x

and 9

2

1

5 45ii

x

, then the standard deviation of the 9 items

1 2 9, ,....,x x x is:

1) 4

2) 2

3)3

4) 9

Key: 2

Sol: 9

1

5 9ii

x

69xix

219 ix x




21 5 6 59 ix

21 5 19 ix

9

2

1

1 2 95 59 9 4i i

i

x x

1 45 2 / 9 9 1 49

. 2S D .

40. PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point

of QR. If the angles of elevation of the top of the tower at P,Q and R are respectively

45 ,30 and30 , then the height of the tower (in m) is:

1)50

2)100 3

3)50 2

4)100

Key: 4

Sol:

Q M

P

R

PM h

tan30 h hQM KM

3h QM RM

2 2 2200QM PM

24 40000h

1000h




41. Two sets A and B are as under:

, : 5 1 and 5 1 ;A a b R R a b

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

1) A B

2) A B (an empty set)

3) neither A B nor B A

4) B A

Key: 1

Sol: 2 24 6 9 5 36a b

2 26 5

19 4

a b

Let origin be shifted to 6,5 (i.e. Let 6 , 5a x b y )

2 2

1..... 19 4x y

5 1, 5 1a b 6 5 1, 5 5 1x y

1 1 1x y

1 1 1 1 1x y 2 0, 1 1x y

2,1 0,1

2, 1 0, 1

4 1 16 9 252,1 , 1 19 4 36 36

Similarly 252, 1 1 131

A B




42. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be

selected and arranged in a row on a shelf so that the dictionary is always in the middle.

The number of such arrangements is:

1) less than 500

2) at least 500 but less than 750

3) at least 750 but less than 1000

4) at least 1000

Key: 4

Sol: The number of arrangements

4 16 .3 . 4 1080C C

43. Let 22

1f x xx

and 1 , 1,0,1g x x x Rx

. If

f xh x

g x , then the local

minimum value of h x is:

1) 3

2) 2 2

3)2 2

4) 3

Key: 3

Sol:

2 2f x g xh x

g x g x

1 21h x x

x xx

h x has local mixing 1 0xx

AM GM

1 2 1 22 .1 1x x

x xx xx x

2 2h x




44. For each t R , lest t be the greatest integer less than or equal to t, Then

0

1 2 15lim ....x

xx x x

1) is equal to 15

2) is equal to 120

3) does not exist (in R)

4) is equal to 0.

Key: 2

Sol: 1 1 112 x x

2 2 212 x x

15 15 151x x x

Add: 1 2 .... 15 1 2 15 1 2 15...

x x xx x x x x

1 2 1515 15 ... 15x xx x x

0 0 0

1 2 1515 15 .... 15x x xLt x Lt Lt

x x x

0

1 2 1515 .... 15xLt

x x x

0

1 2 15120 .... 120xLt

x x x

By sandwich theorem

0

1 2 15... 120xLt x

x x x




1 1x x

2 2x x

15 15x x

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

1 2 15... 15xx x x

0

1 2 15 15 16... 1202x

Lt xx x x

45. The value of 22

2

sin1 2x

x dx

is:

1)2

2) 4

3)4

4) 8

Key: 3

Sol: 22

2

sin 11 2x

xA dx

22

2

sin1 2 x

xA dx

( b b

a a

f x dx f a b x dx 1 )




2 22

2

sin 2 sin1 & 2 21 2 1 2

x

x xx xA

2

2

2

2 sinA xdx

12 22 2

A

4

A

46. A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its

colour is observed and this ball along with two additional balls of the same colour are

returned to the bag. If now a ball is drawn at random from the bag, then the

probability that this drawn ball is red, is:

1)25

2)15

3)34

4) 3

10

Key: 1

Sol:

R B

4 6

Req. Prob4 6 6 4

10 12 10 12

2 1 2 15 2 5 2

=25




47. The length of the projection of the line segment joining the points 5, 1,4 and

4, 1,3 on the plane, 7x y z is:

1)23

2)13

3)23

4) 23

Key: 3

Sol:

5, 1,4A 4, 1,3B

1A 1B

7x y z

5 1 4 1

1 1 1 3h k

14 4 11, ,3 3 3

h k

1 14 4 11, ,3 3 3

A

4 1 3 1

1 1 1 3h k

13 2 10, ,3 3 3

h k




1 13 2 10, ,3 3 3

B

2 2 2

1 1 14 13 4 2 11 10 23 3 3 3 3 3 3

A B

.

48. If sum of all solutions of the equation 18cos . cos .cos 1

6 6 2x x x

in

0, is k , then k is equal to:

1)139

2)89

3)209

4) 23

Key: 1

Sol: 1 1

6 6 28 cos x cos xcos x

8 4 16 6

cos x x cos x cos x

238 1 4 14

cos x sin x cos x

238cos 1 cos 4cos 14

x x x

23 4 48 4 1

4cos xcos x cos x

32 8 4 1xcos x cos x cos x

38 6 1cos x cos x

233 3

cos x cos




3 23

x n

2

3 3nx

0x 4

x

1n 2 6 7 53 9 9 9 9

x ,a

7 5 13sin

9 9 9 9

139

k

49. A straight line through a fixed point (2,3) intersects the coordinate axes at distinct

points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus

of R is:

1) 2 3 x y xy

2)3 2 x y xy

3)3 2 6 x y xy

4) 3 2 6 x y

Key: 2

Sol: A straight line through fixed point (2, 3) intersects axes at P, Q. If O is consist of rectangle OPRQ is completed then locus of R is

Let R be 1 1 1, ,0x y P x 1& 0,Q y

O

10, y 1 1,R x y

1,0P x

2,3

equation of PQ is 1 1

1x yx y

1 1

2 3 1x y

But PQ

passes through (R, S) Locus of R is 2 3 1x y or 3 2x y xy




50. Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 2 2 2 2 2 21 2 2 3 2 4 5 2 6 . . . ....

If 2 100 B A , then is equal to : 1) 248

2) 464

3) 496

4) 232

Key: 1

Sol: Sum of 1st 2n terms 22 2 2 2 21 2.2 3 2.4 5 .... 2 2n

2 22 2 2 2 21 2 3 .... 2 2 4 .... 2n n

22 2 1 4 1 4 1 2 1

2 16 6

n n n n n nn n

2 210 21 20 41A B Given 2 100B A

2 220 41 20 21 100

248 .

51. If the curves 2 6y x , 2 29 16 x by intersect each other at right angles, then the value

of b is:

1) 72

2) 4

3)92

4) 6

Key: 3

Sol: 2 6 1y x

132 64

dy dyy Mdx dx

2 2 16ax by




18 2 0dyx bydx

2 18dxby xdy

218 92

dy x xMdx by by

1 2 1M M

1

1 1

3 18 1 1xy by

21 127x by

1 1927 62

x ob x b

52. Let the orthocenter and centroid of a triangle be 3 5A , and 3 3B , respectively. If

C is the circumcenter of this triangle, then the radius of the circle having line segment

AC as diameter, is:

1) 2 10

2) 532

3) 3 5

2

4) 10

Key: 2

Sol: orthocentre and centroid of a triangle be 3,5 , 3,3A B C is circumcentre of triangle.

Then radius of circle having line segment AC as diameter is

Given centroid = B = (3, 3) orthocentre 3,5A

circumcentre 6,2C

2 3 3,5 2 9,9 12,4 6,2A C B C C C

81 9 3 10AC

radius3 10 103

2 2 2AC




53. Let 1 x

S t R : f x x . e sin x is not differentiable at t . Then the set S is

equal to:

1) 0

2)

3) 0 ,

4) (an empty set)

Key: 4

Sol: 1 sinxS t R f x x e x is not differentiable at t

1

0 0 0

1 sin0x

x x

x e xf x ff dt dt

x x

=0

10

1 sin0

x

x x

x e xg x ff dt dt

x x

54. If 24 2 2

2 4 22 2 4

x x xx x x A Bx x A ,x x x

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

1) 4 3 ,

2) 4 5 ,

3) 4 5,

4) 4 5 ,

Key: 2

Sol: 24 2 2

2 2 4 12 2 4

x xx x A Bx x Ax x x

, 4,5A B




1 2 3 1R R R R 1 1 1

5 4 2 4 22 2 4

x x x xx x x

5 4x

1 1 12 4 02 0 4

x xx x

2 25 4 4x x A Bx x A

4A 5B

, 4,5A B

55. The Boolean expression ~ p q ~ p q is equivalent to:

1) p

2) q

3) ~q

4) ~p

Key: 4

Sol: ~ ~ ~ ~ ~p q v p q P q p q

~ ~p q

~ ~p T

~ p

56. If the system of linear equations

3 0 x ky z

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

Has a non-zero solution x,y,z , then 2

xzy

is equal to:

1)10

2)–30

3)30

4) –10




Key: 1

Sol: 1 33 2 02 4 3

kk

11k

11 3 0x y z

3 11 2 0x y z

2 11 3 0x y z

5 1 2x y z

5 2 10

1,2xzyr

57. Let 0 2 3 6 6 0 S x R : x and x x x . Then S:

1) contains exactly one element

2)contains exactly two element

3) contains exactly four element

4) is an empty set.

Key: 2

Sol: 2 3 6 6 0x x x

if 3 2 6 6 6 0x x x x

4 0x x

0 4x x

if 4x if 3x 2 6 6 6 0x x x

8 12 0x x




58. If the tangent at 1 7, to the curve 2 6 x y touches the circle

2 2 16 12 0 x y x y c then the value of c is:

1)185

2)85

3)95

4) 195

Key: 3

Sol: Tangent at 1,7 to the curve

2 6x x is 11 7 62

x y

2 5 0x y

It touching the circle 2 2 16 12 0x y x y c

r d

2 2r d

100 5c

95c

59. Let y y x be the solution of differential equation 4 dysin x ycos x xdx

, 0 x , .

It 02

y . Then

6

y is equal to:

1) 289 3

2) 289

3) 249

4) 249 3




Key: 2

Sol: 4cot

sindy xy xdx x

if =sin x

Given solution 2.sin 2y x x C

20 0

2 4y C

2

rC

2

2sin 22

y x x

2 21 2

2 36 2y

2 21 9

2 18y

21 8

2 18y

89

y

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

1) 13 2

2) 12 2

3) 12

4) 14 2

Key: 1

Sol: point of Intersection of 3 2 3 2 0x y x




1 0x y z

5 8, ,07 7

Drs 1L is 1,1,0

Drs 2L is 3, 5, 7

na phy

5 87 7

1 1 0 03 5 7

x y z

5 87 7 8 07 7

x y z

7 5 7 8 87 0x y

7 7 8 3 0x y z

7 7 8 3 0x y z ….(1)

Distance from (0,0,0) to (1)

3 3 1

49 49 64 9 2 3 2

.




PHYSICS 61. The Angular width of the central maximum in a single slit diffraction pattern is 600.

The width of the slit is 1m. The slit is illuminated by monochromatic plane waves. If

another slit of same width is made near it, young’s fringes can be observed on a screen

placed at a distance 50cm from the slits. If the observed fringe width is 1cm, what is slit

separation Distance? (i.e distance between centers of each slit.)

1) 50m

2) 75 m

3) 100 m

4) 25 m

Key: 4

Sol: 1 ,a m 030

Dd

2

62

50 10 1 101 10 2

Dd

625 10 m

62. An electron from various exited states of hydrogen atom emit radiation to come to the

ground state. Let n , g be the de Broglie wave length of the electron in the nth state

and the ground state respectively. Let n be the wave length of the emitted photon in

the transition from the nth state to the ground state. For large n, (A, B are constants.)

1) n nA B

2) 2 2n nA B

3) 2

n

4) 2nn

BA

Key: 4




Sol: 202 ,n n nr n r n r ;

02n gnr n

2

2

2

2 2

2

1 11

1

n

g

n

n g

n

Rn

R

R

2

2 2

2

2

2

1

nn

n g

g

n

n

R

R

A B

63. The reading of the ammeter for a silicon diode in the given circuit is: 200

3V 1) 15mA

2) 11.5mA

3)13.5mA

4) 0

Key: 2

Sol: 3 0.7

20011.5mA




64. The density of a material in the shape cube is determined by measuring sides of the

cube and its mass. Relative errors in measuring the mass length are respectively 1.5%

and 1% maximum error in determining the density is:

1) 3.5%

2) 4.5%

3) 6%

4) 2.5%

Key: 2

Sol: 3mL

3100 100

1.5 3 4.5

m lm l

65. An electron, a proton and an alpha particle having the same kinetic energy are moving

in circular orbits of radii re, rp, r, respectively in a uniform magnetic field B. The

relation between re, rp, r, is:

1) e Pr r r

2) e Pr r r

3) e Pr r r

4) e Pr r r

Key: 1

Sol:

0 1 41 2

2

e p

mKErqB

mr e H Heq

r r r




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

1) 2 2

[ ]o

a b cb

2) 2 2

[ ]o

b c ab

3) 2 2

[ ]o

b c ac

4) 2 2

[ ]o

a b ca

Key: 1

Sol: 2 2 2

0

1 4 4 44B

a b cVb b c

2 2

0

a b cb

67. Two masses 1 5m kg and 2 10m kg connected by an inextensible string over a Friction less pulley, are moving as shown in the figure. The coefficient of friction of

horizontal surface is 0.15. The weight m that should be put on top of 2m to stop the motion is:

2m

1m

m

1m g

T

T

1) 27.3 kg 2) 43.3 kg 3) 10.3 kg 4) 18.3 kg Key: 1 Sol: 2 1m m g m g

2m

1m

m

12 23.33mm m kg

=27.3kg




68. The particle is moving in a circular path of radius a under the action of an attractive

potential U= - 22kr

its total energy is:

1) 22ka

2) Zero

3) 2

32

ka

4) 24ka

Key: 2

Sol: 2 3 3

22 2

K dU k kU Fr dr r r

2

3mv k

r r

22

12 2

kk mvr

. 0U K E

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

dielectric material of dielectric constant 53

k is inserted between the plates, the

magnitude of the induced charge will be: 1) 0.3 n C

2) 2.4 n C

3) 0.9 n C

4) 1.2 n C

Key: 4

Sol:




0

0

9

9

11

11

5 390 10 20 13 55 290 10 203 51.2

inQ Qk

kc vk

nc

70. A silver atom in a solid oscillates in simple harmonic motion in some direction with a

frequency of 1210 / sec . What is the force constant of the bonds connecting one atom

with the other? (Mole wt. of silver = 108 and Avogadro number

23 16.02 10 gm mole )

1) 7.1 N/m

2) 2.2 N/m

3) 5.5 N/m

4) 6.4 N/m

Key: 1

Sol: 23

123

12 6.02 102 10108 10

23

2 243

12 6.02 104 10108 10

24 3

23

40 10 108 10 126 10

2720 10 7.2 7.1N m N m 71. It is found that if a neutron suffers an elastic collinear collision with deuterium at rest,

fractional loss of its energy is dp ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is cp . The values of dp and cp are respectively.

1) (28, 89) 2) (0, 0) 3) (0, 1) 4) (89, 28) Key: 4 Sol: 2m u m 1 22mu mv mv




2 12 2 2u v v 14 3v

2 12 2 2u v v 21

2 9mu

14 3v 143

v

22

2

1 12 2 9

12

umu mf

mu

119

8 0.899

80 0.89dP

72. The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B1. When the dipole moment is doubled by keeping the

current constant, the magnetic field at the centre of the loop is B2. The ratio 1

2

BB

is:

1) 3

2) 2

3) 12

4) 2 Key: 2 Sol: 2

1 1M R i 22 2M R i

2 1 2 12 2M M R R

01

12u iBR

02

22u iBR

1 2

2 1

2B RB R

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

1) 1.5

2) 2

3) 2.5

4) 1

Key: 1




Sol: 52 5240 40

E R rE RR

r R

52140

rR

1240

rR

12 12 5 1.540 40

r R

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

1) 42 10

2) 52 10

3) 62 10

4) 32 10

Key: 2 Sol: available band width

10 10 10 10100

Hz

Number of channels

9

53

10 2 105 10

75. Unpolarized light of intensity I passes through an ideal polarizer A. Another Identical

polarizer B is paced behind A. The intensity of light beyond B is found to be12

. Now

another identical polarizer C is placed between A and B. The intensity beyond B is now

found to be8I

. The angle between polarizer A and C is:

1) 300

2) 450

3) 600

4) 00

Key: 2

Sol:




0I 02I

0I

02I

A B

CA B

20 cos2I

40 0

0

cos2 8

1cos2

45

I I

76. On interchanging the resistances, the balance point of a meter bridge shifts to the left

by 10cm. The resistance of their series combination is 1k. How much was the resistance on the left slot interchanging the resistances?

1) 505 2) 550 3) 910 4) 990 Key: 2 Sol: 1 2 100l l m 1 2 10l l cm

1

2

5345

l ml m

1 2 1000R R

1 1

2 2

R lR l

11

1 2

1000lRl l

55 1000 550

100




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

is

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

23R

R

1) 2409

MR

2) 210 MR

3) 2379

MR

4) 24 MR Key: 4

Sol: 2

9RM

2

9MR

=M

2 2 29 4

2 9 2 9MR MR M RI

24I MR




78. In collinear collision, a particle with an initial speed 0v strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is:

1) 02 v

2) 0

2v

3) 0

2v

4) 0

4v

Key: 1

Sol: 2 21 122 2f cm resKE m v u v

= 2

201 122 2 2 2 res

v mm v

32f iKE KE

= 2

2 200

1 1 3 122 4 2 2 2 2res

v mm v mv

2 2 20 03resv v v 02resv v

79. An EM wave from air enters a medium. The electric fields are

1 01 cos 2 zE E x v t

c

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

in medium, where the

wave number k and frequency v refer to their values in air. The medium is non-magnetic. If

1r and

2r refer to relative permittivites of air and medium respectively,

which of the fallowing options is correct?

1) 1

2

2r

r

2) 1

2

14

r

r

3) 1

2

12

r

r

4) 1

2

4r

r




Key: 2

Sol: speed of light in medium 2 is 2c

So 0 0

1

r r

cu u

1ru

24r

80. For an RLC circuit driven with voltage of amplitude mv and frequency 1

o LC the

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

1) oRL

2) o

RC

3) o

CR

4) oLR

Key: 4

Sol: Re sonance frequencyQ factor

bandwidth 0L

R

81. All the graphs below are intended to represent the same motion. One of them does it incorrectly, pick it up.

1)

Time

Distance

2)

time

Position

3)

time

velocity

4)

position

velocity




Key: 1 Sol: options 2, 3, 4 represent similar type of motion where as option 1 is different. 82. Two batteries with e.m.f. 12V and 13V are connected in parallel across a load resistor

of10 . The internal resistance of the two batteries is 1 and 2 respectively. The

voltage across the load lies between:

1) 11.5 V and 11.6 V

2) 11.4 V and 11.5 V

3) 11.7 V and 11.8 V

4) 11.6 V and 11.7 V

Key: 1

Sol: 1 112 , 1v r 2 213 , 2v r 10R

12V 13V

1r 2r

10

1 2

1 2

1 2

12 131 2

1 1 1 11 2

r r

r r

24 13 12.33

3

1 2

1 2

23

r rreqr r

10 37 10 37 11.562 3 3210

3

loadv v

83. A particle is moving with a uniform speed in a circular orbit of radius R in a central

force universally proportional to the thn power of R. If the period of rotation of the particle is T, then:

1)1

2n

T R

2) 1 2nT R

3) 2n

T R

4) 3

2T R for any n.

Key: 2

Sol:




122

1

1

c n

n

n

fR

mv v nR R

R

1 11

2 22n nRT R R

v

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

1) 16 L

2) 16

L

3) 25

L

4) 25 L Key: 3 Sol: Limit frequency Lv of Lyman series

1 2211 1,Lhv k n n

1 221 1 5,25Phv k n n

25

LP

vv

85. In an a.c circuit, the instantaneous e.m.f. and current are given by 100sin30e t

20sin 304

i t

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

1) 1000 ,10

2

2) 50 ,0

2

3) 50,0 4) 50,10




Key: 1 Sol:

100sin30

20sin 304

cos100 20 cos

42 21000

2

m m m

e t

i

P I V

W

Wattless current 020 20sin 45 1022

A

86. Two moles of an ideal monotonic gas occupies a volume V at 027 C . The gas expands adiabatically to a volume 2V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

1) (a) 195 K (b) -2.7 kJ 2) (a) 189K (b) -2.7 kJ 3) (a) 195K (b) 2.7 kJ 4) (a) 189K (b) 2.7 kj Key: 2

Sol: 1

Pv Const

Tv Const

T

300

V2V

P

V

213

monoatomic gas 2/3Tv Const

2/32/3300 2v T v 2/3300 1892

T K

vv nC t 2.7kJ

32;2

189 300 111

Rn Cv

T K




87. A solid sphere radius r made of a soft material of bulk modulus K is surrounded by a

liquid in a cylindrical container. A mass less piston of area a floats on the surface of the

liquid, covering entire cross section of cylindrical container. When a mass m is placed

on the surface of the piston to compress the liquid, the fractional decrement in the

radius of the sphere,drr

, is:

1) 3Kamg

2) 3mgKa

3) mgKa

4) Kamg

Key: 2

Sol: Bulk modulus =K, 3v r

v r

P mgK pv av

K

3

3

v mgv aKdr mgr aK

dr mgr Ka




88. A granite rod of 60cm length is clamped at its middle point and is set into longitudinal

vibrations. The density of granite is 3 32.7 10 /kg m and its Young’s modulus is109.27 10 Pa . What will be the fundamental frequency of the longitudinal vibrations?

1) 2.5kHz 2) 10 kHz 3) 7.5kHz 4) 5kHz Key: 4 Sol: 3 22.7 10 /kg m

109.27 10y Pa

0 602

l Cm

102

3

2 3

22

9.27 10 5.85 102.7 10

5.8 10 9.7 10 , 522 6 10

Y

f n kHz

89. The mass of a hydrogen molecule is 273.32 10 .kg if 2310 hydrogen molecules strike,

per second, a fixed wall of area 22cm at an angle of 045 to the normal, and rebound elastically, with a speed of 310 ,m s then the pressure on the wall is nearly:

1) 3 34.70 10 /N m

2) 2 32.35 10 /N m

3) 2 34.70 10 /N m

4) 3 32.35 10 /N m Key: 4 Sol:

2

273.32 10Hm kg

No of Atoms 23/ sec 10hits n , 310 /v m s

2 4 22 2 10A m m , 045

23 27 34

2 cos 12 10 3.32 10 102 2 10

nmvfA




3 2

3 2

2.347 10 /

2.35 10 /

N m

N m

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

P

O

1) 2552

MR

2) 2732

MR

3) 21812

MR

4) 2192

MR

Key: 3 Sol:

2

2

2

2

55 7 32

55 632

1812

pMRI M R

MR

MR

2018 jee-main s q p k main... · 2018-04-09 · 2018_jee-main question paper_key & solutions...

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