Joint Entrance Exam | Mains-2018 - Amazon Web …100p.s3.amazonaws.com/vidyamandir/JEEMAIN2018/jeemain...2. The Answer Sheet is kept inside this Test Booklet. When you are directed

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VMC/JEE Mains-2018 1 JEE Entrance Examination

Joint Entrance Exam | Mains-2018

Paper Code - B 8th April 2018 | 9.30 AM – 12.30 PM

CHMISTRY, MATHEMATICS & PHYSICS

Important Instructions: 1. Immediately fill in the particulars on this page of the Test Booklet with only Black Ball Point Pen

provided in the examination hall.

2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars carefully.

3. The test is of 3 hours duration.

4. The Test Booklet consists of 90 questions. The maximum marks are 360.

5. There are three parts in the question paper A, B, C consisting of Physics, Mathematics and Chemistry having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.

6. Candidate will be awarded marks as stated above in instruction No. 5 for correct response of each

question. 14 (one-fourth) marks of the total marks allotted to the questions (i.e. 1 mark) will be

deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.

7. There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.

8. For writing particulars/marking responses on Side-1 and Side-2 of the Answer Sheet use only Black Ball Point Pen provided in the examination hall.

9. No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc. except the Admit Card inside the examination room/hall.

10. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the bottom of each page and in four pages (Page 20-23) at the end of the booklet.

11. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.

12. The CODE for this Booklet is B. Make sure that the CODE printed on Side-2 of the Answer Sheet is same as that on this Booklet. Also tally the serial number of the Test Booklet and Answer Sheet are the same as that on this booklet. In case of discrepancy, the candidate should immediately report the matter to the Invigilator for replacement of both the Test Booklet and the Answer Sheet.

13. Do not fold or make any stray mark on the Answer Sheet.

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Joint Entrance Exam/IITJEE-2018

PART-A PHYSICS 1. 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) (0, 0) (2) (0, 1) (2) ( 89, 28) (3) ( 28, 89)

1.(3) For collision with deuterium:

1 22mv o mv mv (Conservation of momentum ) ......... (1)

2 1v v v ( 1e ) ......... (2)

By (1) and (2) 1 3vv

2 2

1

2

1 182 2 0.89

1 92

d

mv mvP

mv

For collision with carbon Nucleus

1 20 12mv mv mv (Conservation of momentum ) ......... (1)

2 1v v v ( 1e ) ......... (2)

By (1) and (2)

11113

v v

22

2

1 1 11482 2 13 0.281 169

2

mv m vPc

mv

2. The mass of a hydrogen molecule is 273.32 10 kg. If 2310 hydrogen molecules strike, per second, a fixed wall of area 2 cm2 at an angle of 45° to the normal, and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly:

(1) 2 22.35 10 /N m (2) 2 24.70 10 /N m

(3) 3 22.35 10 /N m (4) 3 24.70 10 /N m 2.(3) Change in momentum of a single molecule.

0 0 22

uP m

Total change in momentum per second

0 0. 2P n P n m u

Pressure 0 2nm uFA A

Substituting values: 3 22.35 10 / .P N m

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3. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless 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) 3mgKa

(2) mgKa

(3) Kamg

(4) 3Kamg

3.(1) PB VV

V PV B

Also, 3V rV r

(As 24V r r )

3

r Pr B

3mgKa

4. Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of 10 . The internal resistances of the two batteries are 1 and 2 respectively. The voltage across the load lies between :

(1) 11.4 V and 11.5 V (2) 11.7 V and 11.8 V (3) 11.6 V and 11.7 V (4) 11.5 V and 11.6 V 4.(4)

Equivalent circuit is

Where,

1 2 21 2 3eqr

1 2

1 2

1 2

12.331 1eq

V Vr r

v

r r

1010AB eq

eqV V

r

= 11.55 volts

5. A particle is moving in a circular path of radius a under the action of an attractive potential 2 .2kUr

Its total

energy is :

(1) Zero (2) 232

ka

(3) 24ka

(4) 22ka

5.(1) 22KUr

3dU KF r rdr r

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2

3K mv

rr (As Force towards center

2mvr

)

K. E. = 22

12 2

KmVr

Total energy = KE + PE 2 22 2

K Kr r

= Zero

6. Two masses 1 5m kg and 2 10m kg, connected by an inextensible string over a frictionless pulley, are

moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of 2m to stop the motion is :

(1) 43.3 kg (2) 10.3 kg (3) 18.3 kg (4) 27.3 kg 6.(4) For 1m to be at rest

5T g

For 2&m m to be at rest

5f T g

( )f N

20.15( )f m m g

23.33m kg

Amongst the options minimum mass that can be kept for no motion is 27.3 kg

7. If the series limit frequency of the Lyman series is ,Lv then the series limit frequency of the Pfund series is ;

(1) /16Lv (2) / 25Lv (3) 25 Lv (4) 16 Lv

7.(2) For Series limit of Lyman : 1 1n and 2n

2 1 11P RcZ

For Series limit of Pfund: 1 5n and 2n

2 1 125 25

LP RcZ

8. Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed

behind A. The intensity of light beyond B is found to be .2I Now another identical polarizer C is placed between

A and B. The intensity beyond B is now found to be .8I The angle between polarizer A and C is :

(1) 45° (2) 60° (3) 0° (4) 30°

8.(1) When an unpolarized light of intensity I passes through a polarizer for the 1st time, intensity of output is 2I

(irrespective of orientation of polarizer) So,

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i.e., polarizers A and B have axes parallel to each other. Now let the axis of C make an angle with A, and with B.

4cos

2 8I I

Solving, 45

9. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let , n g

be the de Broglie wavelength of the electron in the nth state and the ground state respectively. let n be the

wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)

(1) 2 2n nA B (2) 2

n (3) 2nn

BA

(4) n nA B

9 (3). 22 2

1 1 11n

RZn

1

2 21 11n

RZ n

Since n is very large, using binomial

2 21 11n

RZ n

2 2 2

1 1 1n

RZ RZ n

2nn

BA

As 2 2

2 22 12

4n

r n h nn nmZe

10. The reading of the ammeter for a silicon diode in the given circuit is: (1) 11.5 mA

(2) 13.5 mA (3) 0 (4) 15 mA

10 (1).

Voltage across Si diode in forward bias is 0.7 volts. Hence voltage across 200 resister is 3 – 0.7 = 2.3V

2.3200

I = 11.5 mA

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11. An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii , ,e pr r r respectively, in a uniform magnetic field B. The relation between , ,e pr r r is:

(1) e pr r r (2) e pr r r (3) e pr r r (4) e pr r r

11.(4) 2mv mKrqB qB

2 e

em K

reB

2 4

2pm K

reB

2 p

pm K

reB

Comparing (1), (2) and (3) e pr r r

12. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of

dielectric constant 53

K is inserted between the plates, the magnitude of the induced charge will be :

(1) 2.4 nC (2) 0.9 nC (3) 1.2 nC (4) 0.3 nC 12.(3) iq CV

fq KCV

( 1)induced f iq q q K CV 125 1 90 10 20 1.23

nC

13. For an RLC circuit driven with voltage of amplitude mv and frequency 01LC

the current exhibits

resonance. The quality factor, Q is given by :

(1) 0( )RC

(2) 0

CR

(3) 0LR

(4) 0RL

13.(3) Quality factor 02

Q

0LR

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

(1) 52 10 (2) 62 10 (3) 32 10 (4) 42 10

14.(1) Overall bandwidth use for transmission 10% of C

Number of telephonic channel Total bandwidth

Channel bandwidth

95

3

10 10 10100 2 10

5 10

15. A granite rod of 60 cm 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 is 109.27 10 Pa . What will be the fundamental

frequency of the longitudinal vibrations? (1) 10kHz (2) 7.5kHz (3) 5kHz (4) 2.5 kHz

15.(3) 12

c Yfl

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10

31 9.27 10

2 0.6 2.7 10

71 9.27 10

1.2 2.7

34.88 10 Hz 5kHz

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

(1) 2732

MR (2) 21812

MR (3) 2192

MR (4) 2552

MR

16.(2) 2 2

20 6 (2 )

2 2

MR MRI M R 2 21 553 242 2

MR MR

O is the centre of mass of the system. Applying parallel axis theorem between O & P.

2 2 20

557 (3 ) 632PI I M R MR MR 2181

2MR

17. 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

0

b c ab

(2)

2 2

0

b c ac

(3)

2 2

0

a b ca

(4)

2 2

0

a b cb

17.(4) CA BB

kQkQ kQVb b c

2 2 2( )(4 ) ( )(4 ) ( )(4 )a b ck

b b c

2 2

0

1 44

a b cb b

2 2

0

a b cb

18. In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm 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 wire. Find the internal resistance of the cell. (1) 2 (2) 2.5 (3) 1 (4) 1.5 18.(4) Let potential difference per unit length of potentiometer wire be x. In case-I (52)( )x …(i)

In case-II

5

ir

(40)( )ir x

405

r xr

5 405

xr

…(ii)

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From (i) & (ii)

552 405

rx x

52 13 151 5 1 1.540 5 10 10

r r

19. An EM wave from air enters a medium. The electric fields are 1 01 ˆ cos 2 zE E x v tc

in air and

2 02 ˆ 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 permittivities of air and medium respectively, which of

the following options is correct?

(1) 1

2

14

r

r

(2) 1

2

12

r

r

(3) 1

2

4r

r

(4) 1

2

2r

r

19.(1) From wave equations :

In air: 22 , kc

In medium: , 2 kc k k

,

k kc c 2

2

cc

c c

2 2 1 10 0 0 0

1 1 12

r r r r

Medium and air are non-magnetic 1 2

1 ; 1r r

1

2 1 2

1 1 14 4

r

r r r

20. The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 m. 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 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e., distance between the centres of each slit.)

(1) 75 m (2) 100 m (3) 25 m (4) 50 m

20.(3) Angular width of central maxima 2a

; (where a is slit width and is wavelength)

23a

… (i)

In YDSE, fringe width

D

d [where d is slit separation and D is distance of screen from slits)

6

D ad

6D ad

6

21 3.14 102 6 10

d

6100 104

25 m

21. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avagadro number = 6.02 × 1023 gm mole–1)

(1) 2.2 N/m (2) 5.5 N/m (3) 6.4 N/m (4) 7.1 N/m

21.(4) 12 1102

fT

122 10

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2 km

22 3 1223

108 10 2 10 7 16 023 10

k m ..

22. From a uniform circular disc of radius R and mass 9 M, 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 :

(1) 10 MR2 (2) 379

MR2

(3) 4 MR2 (4) 409

MR2

22.(3) 2

2 29 232 2 3

RMM R RI M

2 2 2

29 42 18 9

MR MR MRI MR

23. In a 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) 02v

(2) 02

v (3) 0

4v

(4) 02 v

23.(4)

It is given that final total kinetic energy has increased, so some internal energy of the system must have been

converted into kinetic energy.

2 2 21 2 0

1 1 1mv mv 1.5 mv2 2 2

2 2 21 2 0v v 1.5v ...... (i)

Since, there is no external force, momentum can be conserved 1 2 0mv mv mv

1 2 0v v v ...... (ii)

From (i) & (ii)

2 21 0 1 0v v v 1.5v

2 21 0 1 02v 2v v 0.5v 0

Relative velocity 2 1v v Difference of roots 0D 2 va

24. The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is

1B . When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the

loop is 2.B The ratio 1

2

BB

is :

(1) 2 (2) 12

(3) 2 (4) 3

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24 (1). 201 1 1

1;

2

IB m I r

r

20

2 2 22

;2

IB m I r

r

22221 1

rm

m r

22

12 r

r

2

12

rr

1 2

2 12B r

B r

25. The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is:

(1) 4.5% (2) 6% (3) 2.5% (4) 3.5%

25 (1). 3

M MV L

3M LM L

Maximum % error in density 1.5% 3 1% 4.5%

26. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 .k How much was the resistance on the left slot before interchanging the

resistances? (1) 550 Ω (2) 910 Ω (3) 990 Ω (4) 505Ω

26.(1) Let the resistances in left and right slot be r and 1000 r respectively Initial: (100 ) (1000 )( )r x r x ..................(1)

After interchanging: (1000 )[100 ( 10)] ( 10)r x r x

(1000 )(110 ) ( 10)r x r x ..................(2)

From (1): 100 1000r rx x rx 10r x

From (2): (1000 ) 110 1010 10r rr r

2(1000 )(1100 ) 100r r r r

2 21000 1100 2100 100r r r r 1000 1100 5502000

r

27. In an a.c. circuit, the instantaneous e.m.f. and current are given by 100sin 30e 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) 50 ,02

(2) 50, 0 (3) 50, 10 (4) 1000 ,102

27.(4) cosrms rmsP V I ; 4

100 20 12 2 2

P 1000

2

Wattless current, 20 1sin 102 2rmsI I

28. All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

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(1) (2)

(3) (4) 28.(4) The (1), (2) and (3) graphs can represent the motion of a ball that is thrown in vertically upward direction.

Initially speed decreases, becomes zero and then on the return trip, speed increases. Slope of graph in option (4) does not explain it.

29. Two moles of an ideal monoatomic gas occupies a volume V at 27°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) 189 K (b) – 2.7 kJ (2) (a) 195 K (b) 2.7 kJ (3) (a) 189 K (b) 2.7 kJ (4) (a) 195 K (b) –2.7 kJ

29.(1) For mono atomic gas 53

Using 1 constantTV

2 23 3(300) 2V T V

2

3

300 1892

T K

3 32 8.314 189 300 27682 2

U n R T 2.7 kJ

30. A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then:

(1) 1 / 2nT R (2) / 2nT R

(3) 3/ 2T R for any n (4) 1

2n

T R

30.(1) 1nF

R

2

nk mVF

RR 2

1nkV

mR

(1 )2n

V R

Now 1

2(1 )

2

2n

nR RT R

VR

22H (3) 2

2He (4) 2He

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PART-B MATHEMATICS 31. 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) 85 (2) 95 (3) 195 (4) 185

31.(2) 7 62

yx 2 7 12x y 2 5x y 2 5 0x y

Also, centre of the circle is 8 6 , and the radius is 64 36 c

16 6 5 1005

c

5 100 95c c

32. If 1L is the line of intersection of the planes 2 2 3 2 0 x y z , 1 0 x y z and 2L is the line of

intersection of the planes 2 3 0 x y z , 3 2 1 0 x y z , then the distance of the origin from the

plane, containing the lines 1 2andL L is:

(1) 12 2

(2) 12

(3) 14 2

(4) 13 2

32.(4) 2 2 3 2 0x y z

1 3 2 2 1 3 0x y z

22

1 3 2

2 1 3 2 3 32

So the equation of plane is 7 7 8 3 0x y z

Now, distance from origin equal to 2 2 2

3 13 27 7 8

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

(1) 1 (2) 2 (3) –1 (4) 0

33.(1) 2 1 0x x 21 32

x , (where and 2 are non-real cube roots of unity)

and 2

107101 2 101 214 2 1

34. Tangents are drawn to the hyperbola 2 24 36 x y 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) 60 3 (2) 36 5 (3) 45 5 (4) 54 3 34.(3) Equation of PQ, 4 0 3 36x y

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12y

Area of 1 15 6 5 45 52

TPQ

35. If the curves 2 6y x , 2 29 16 x by intersect each other at right angles, then the value of b is:

(1) 4 (2) 92

(3) 6 (4) 72

35.(2) 2 6yy'

1

6 32

y'y y

1 118 2 0x by y'

1 1 1 12 2

1 1 1 1

18 9 27 2712

x x x xy' bby by by y

21 1

962

y x b

36. 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 2xzy

is equal to:

(1) –30 (2) 30 (3) –10 (4) 10

36.(4) 1 33 2 02 4 3

kk

72

k

3 0x ky z …..(i)

3 2 0x ky z ….(ii)

2 4 3 0x y z …..(iii)

On solving (i) and (ii) 2 5 0x z …..(iv) On solving (iii) and (iv) 4 2y z

2 2

52 10

4

z zxzy z

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

(1) contains exactly two elements. (2) contains exactly four elements. (3) is an empty set. (4) contains exactly one element.

37.(1) 2 3 6 6 0x x x

Case-I: 3x

2 3 6 6 0x x x 4 0x x 0 16x ,

As 9 16x x

Case-II: 3 2 6 6 6 0x x x x 8 12 0x x

6 2 0x x 36 4x ,

As, 3 4x x

There are exactly two elements in the given set.

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

6 6 2

x x x in [0, ] is k , then k is

equal to:

(1) 89

(2) 209

(3) 23

(4) 139

38.(4) 2 2 18cos · cos sin 16 2

x x

23 18cos 1 cos 14 2

x x

28cos 4cos 1 2 14

x x

3cos3 4cos 3cosx x x 2 cos3 1x

1cos32

x

3 0, 3x

3 , 2 , 23 3 3

x Sum = 13

9 .

39. 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) 15

(2) 34

(3) 310

(4) 25

39.(4)

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40. Let 22

1f x xx

and 1g x xx

, 1 0 1x R , , . If

f xh x

g x , then the local minimum value of

h(x) is: (1) 2 2 (2) 2 2 (3) 3 (4) 3

40.(2) Let 1g x x tx

211 0g' xx

20 0t R ; t ,

2

2 22

1 2 2 2f x x x t ,xx

f xh x

g x

2 2 2f x t tg x t t

Let 2h t tt

221h' tt

Local minimum value occurs at 2t

Local minimum value 22 2 2 22

h

41. Two sets A and B are as under:

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

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

Then: (1) A B (an empty set) (2) neither A B nor B A

(3) B A (4) A B 41.(4) Since Set A is, | a 5 | 1 4 < a < 6

and | b 5 | 1 4 < b < 6

Now B is 2 2(a 6) (b 5) 1

9 4

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It can be seen that all vertices of rectangle lie inside the ellipse, therefore A B

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

(1) q (2) ~q (3) ~p (4) p 42.(3) ( ) ( )p q p q

p q ( )p q p q ~ p

T F F F F T F F F F F T F T T F F T F T

43. Tangent and normal are drawn at P(16, 16) on the parabola 2 16y x , 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) 3 (2) 43

(3) 12

(4) 2

43.(4) The equation of tangent at P

1y 16 (x 16)2

A ( 16, 0)

The normal is y 16 2(x 16)

B (24, 0) Since APB

2

AB is the diameter. Center of the circle C (4, 0)

Slope of 1PB 2 m

Slope of 24CP m3

2 1

2 1

m mtan 2

1 m m

44. If 24 2 2

2 4 22 2 4

x x xx x x A Bx x Ax x x

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

(1) 4 5, (2) 4 5, (3) 4 5, (4) 4 3,

44.(1) 2x 4 2x 2x2x x 4 2x (A Bx)(x A)2x 2x x 4

Put x = 0

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34 0 0

0 4 0 A0 0 4

A 4

Put x = 1

23 2 2

2 3 2 (A B)(1 A)2 2 3

3(9 4) 2( 6 4) 2(4 6)

15 20 20 ( 4 B)25

1 ( 4 B)

B = 5

45. 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) 1 (2) 2 (3) 1 (4) 0

45.(2) Let 3 1x y

5 5x y x y

5 5 5 4 5 5 5 5 4 50 1 5 0 1 55 5 C x C x y ....... C y C x C x y ........ C y

5 5 3 2 5 4 5 3 2 5 40 2 4 0 2 42 5 2 5 C x C x y C xy C x C x y C xy

25 3 3 3 5 6 3 6 32 10 1 5 1 2 10 10 5 1 2 x x x x x x x x x x x

5 6 3 7 42 10 10 5 5 10 2 1 10 5 5 2 x x x x x x

46. Let a1, a2, a3, …, a49 be in A.P. such that 12

4 1 9 430

416 and 66kk

a a a

. If 2 2 21 2 17..... 140a a a m ,

then mequal to : (A) 34 (2) 33 (3) 66 (4) 68 46.(1) 1 5 9 49 416a a a ........a 24 32a d ……(i)

9 43 66 25 33a a a d ….(ii)

From (i) and (ii) 1d and 8a

Now, 2 2 21 2 17 140a a ......a m

217

18 1 140

rr m

17

2

17 140 4760 140 34

rr m m m

47. A straight line through a fixed point (2, 3) intersects the coordinates axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is :

(A) 3 2 x y xy (2) 3x + 2y = 6xy

(3) 3x + 2y = 6 (4) 2x + 3y = xy 47.(1) Let, ( , )R h k

(0, )P k

( ,0)Q h

Equation of line would be,

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1x yh k . . . (i)

2 3 1h k

2 3k h hk

Locus of (h, k) is 2 3y x xy

48. The value of 2 2

2

sin1 2x

x dx

is :

(1) 4 (2) 4 (3)

8 (4)

2

48.(2) Given/ 2 2

/ 2

sin1 2x

x dx

2 22sin 2 (sin )( ) ( ) sin

1 2 1 2

x

x xx xf x f x x

/ 22

0

sin x dx

/ 2

2

0

sin4

x dx

49. Let 2( ) cos , ( ) , and , ( )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 , and 0x x y , is :

(1) 1 3 22

(2) 1 2 12

(3) 1 3 12

(4) 1 3 12

49.(3) 2( ) cosg x x

( )f x x

( ( ) ) cosg f x x

Given, 2 218 9 0x x (6 ) (3 ) 0x x

,6 3

x

Area = / 3

/ 6

3 1cos2

x dx

50. For each t R , let [t] be the greatest integer less than or equal to t. Then 0

1 2 15lim .....

xx

x x x

(1) is equal to 120 (2) does not exist (in R) (3) is equal to 0 (4) is equal to 15

50.(1) 0

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

xx x x

0

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

xx x x x x x

= 0 0

1 2 15lim (1 2 3 ........ 15) lim ........x x

xx x x

Now 0 1x x R 120

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51. If 9 9

2

1 1

( 5) 9 and ( 5) 45i ii i

x x

, then the standard deviation of the 9 items x1, x2….. , x9 is :

(1) 2 (2) 3 (3) 9 (4) 4

51.(1) Variance = 245 (1)9 5 1 4

Variance 2

52. The integral 2 2

5 3 2 3 2 5 2sin cos

(sin cos sin sin cos cos )x x dx

x x x x x x is equal to :

(1) 3

11 cot

Cx

(2)

31

1 cotC

x

(3) 31

3(1 tan )C

x

(4) 3

13(1 tan )

Cx

(where C is a constant of integration)

52.(4) 2 2

5 3 2 3 2 5 2sin x cos x dx

(sin x cos x sin x sin x cos x cos x)

2 6

25 2 3

tan x sec xdx

tan x tan x tan x 1

Put tan x t 2 dtsec x

dx

2 2 2

3 2 2 2t (1 t ) dt

(t 1) (t 1)

3t 1 y

2 dy3tdt

21 dy 1 C3 3(y)y

31 C

3(tan x 1)

53. Let | |{ : ( ) | | · ( 1) sin | |xS t R f x x e x is not differentiable at t}. Then the set S is equal to :

(A) (2) 0, (3) (an empty set) (4) {0}

53.(3) Doubtful points for differentiability are 0 and

At x = 0

| || | ( 1) sin | | 0(0 ) limh

h o

h e hfh

( ) ( 1) sinlimh

h o

h e hh

0

sinhlim 1h h

and lim 1 0h

h oe

(0 ) 0 1 0f

| || | ( 1) sin | | 0(0 ) lim

h

h o

h e hfh

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0

( ) ( 1) sinlimh

h

h e hh

0

sinhlim 1h h

and lim 1 0h

h oe

(0 ) 0 1 0f

(0 ) (0 ) 0f f

Similarly ( ) ( ) 0f f

Hence ( )f x is differentiable x R

54. Let y y x be the solution of the differential equation sin cos 4 , 0,dyx y x x xdx

. If 02

y

, then

6y

is equal to :

(1) 289 (2) 24

9 (3) 24

9 3 (4) 28

9 3

54.(1) 4 4dy y cot x x cosec x d y sin x xdxdx

Integrating both sides we get: 22y sin x x c

Also, 02

y

2

2c

2

222

y sin x x

286 9

y

55. Let u be a vector coplanar with the vectors ˆˆ ˆ2 3a i j k and ˆˆb j k

. If u is perpendicular to a and

. 24u b

, then 2u is equal to :

(1) 256 (2) 84 (3) 336 (4) 315

55.(3) ( ) 0;u a b 0 u a and 24.u b

Let ˆ ˆ ˆ ˆ( ) ( )b b a a b u u

2 2 2ˆ ˆ| | ( ) ( )b b a b u

2

2 22ˆ( )ˆ| | ( )

ˆ| |b ub b a

u

2

22 (24)27 ˆ| |u

2| | 336u

56. 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) 13

(2) 23

(3) 23

(4) 23

56.(2) 5 1 41 1 1

x y z

5 1 4 P , ,

P is foot of perpendicular from A to plane 3 8 7

13

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14 4 113 3 3

P , ,

4 1 31 1 1

x y z

4 1 3Q , ,

Q is foot of perpendicular from B to plane 3 6 7

13

13 2 103 3 3

Q , ,

1 4 1 6 23 3 3

PQ

57. 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° and 30°, then the height of the tower (in m) is :

(1) 100 3 (2) 50 2 (3) 100 (4) 50

57.(3) 13

hx

3x h

2 2200 3h h

2 24 (200)h

24 40000h 100h 58. 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) at least 500 but less than 750 (2) at least 750 but less than 1000 (3) at least 1000 (4) less than 500

58.(3) 6 34 1 1 4!C C

6 5 3 24 45 24 10802

59. 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 100B A , then is equal to: (1) 464 (2) 496 (3) 232 (4) 248

59.(4) 2 2 2 2 2 21 2 2 3 2 4 2 20A . . ........ A .

2 2 2 2 2 2 2 21 2 3 4 20 2 4 20......... .......

20 21 41 10 11 2146 6

2870 1540 4410 2870 1540 4410

40 41 81 4 20 21 416 6

B 540 41 41 280 41 820 33620

33620 8820 100 100 24800 248

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60. Let the orthocentre and centroid of a triangle be 3,5A and B(3, 3) respectively. If C is the circumcentre of

this triangle, then the radius of the circle having line segment AC as diameter, is :

(1) 532

(2) 3 52

(3) 10 (4) 2 10

60.(1)

2 3 3

3a

2 12 6a a

2 5 33

b 2 4 2b b

2 2(6 3) 3AC

Diameter 81 9AC 90

Radius 3 10 3 102 2 2

53

2

PART-C CHEMISTRY

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

(1) 9 (2) 12 (3) 3 (4) 6

61.(1) 3I is - 3sp d hybridised

- linear shape

62. Which of the following salts is the most basic in aqueous solution?

(1) 3FeCl (2) 3 2Pb(CH COO) (3) 3Al(CN) (4) 3CH COOK

62.(4) CH3COOK is a salt of a weak acid and a strong base

Most basic 63. 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) (2)

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(3) (4)

63.(1)

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

(a) (b)

(c) (d) (1) (b) < (a) < (d) < (c) (2) (d) < (b) < (a) < (c) (3) (a) < (b) < (c) < (d) (4) (b) < (a) < (c) < (d)

64.(1) Amidines, are stronger organic bases.

65. 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) Weak Strong Yellow to pinkish red (2) Strong Strong Pink to colourless (3) Weak Strong Colourless to pink (4) Strong Strong Pinkish red to yellow

65.(1) Methyl orange is used for titration of strong acid and weak base. 66. The trans-alkenes are formed by the reduction of alkynes with:

(1) 3Na / liq. NH (2) Sn HCl

(3) 2 4H Pd / C, BaSO (4) 4NaBH

66.(1)

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67. The ratio of mass percent of C and H of an organic compound X Y Z(C H O ) is 6 : 1. If one molecule of the above

compound X Y Z(C 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) 3 4 2C H O (2) 2 4 3C H O (3) 3 6 3C H O (4) 2 4C H O

67.(2) x y zC H O has z oxygen atom

x y 2 2 2y y

C H x O x CO H O4 2

O atoms required for combustiony

2 x4

1 y

z 2 x2 4

y

z x4

68. Hydrogen peroxide oxidises 46[Fe(CN) ] to 3

6[Fe(CN) ] in acidic medium but reduces 36[Fe(CN) ] to

46[Fe(CN) ] in alkaline medium. The other products formed are, respectively:

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

(3) 2 2(H O O ) and 2H O (4) 2 2(H O O ) and 2(H O OH )

68.(1) During reduction 2 2 2H O H O

During oxidation 2 2 2H O O

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

(1) (2) (3) (4) 69.(2)

Option (2) is correct [NCERT Class XII Part-II, Page No.-340] 70. How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen

released can completely burn 27.66g of diborane ?

(Atomic weight of B 10.8u)

(1) 3.2 hours (2) 1.6 hours (3) 6.4 hours (4) 0.8 hours

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70.(1) 2 6 2 2 3 2B H 3O B O 3H O

2 627.66

nB H 127.66

2On required = 3

2 2 22H O 2 H O

n-factor for O2 = 4 Number of equivalent = 3 4 12F 12 96500C

i t 12 96500

12 96500

t s100

12 96500

h 3.2 hr100 3600

71. Which of the following lines correctly show the temperature dependence of equilibrium constant, K, for an

exothermic reaction?

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

71.(3) G H T S RT nk H T S

H SnkRT R

Slope is HR

Since H is ve

Slope is positive. 72. At 518° C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was

1.00 Torr s–1 when 5% had reacted and 10.5 Torr s when 33% had reacted. The order of the reaction is :

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

72.(3) nr k A

n1 k 363 0.95 ….(i)

n0.5 k 363 0.67 ….(ii)

From (i) and (ii) n 2

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

73.(3)

Option (3) is correct [NCERT Class XII Part-II, Page No 405]

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74. Consider the following reaction and statements :

3 4 2 3 3 3 3[Co(NH ) Br ] Br [Co(NH ) Br ] NH

(A) Two isomers are produced if the reactant complex ion is a cis-isomer. (B) Two isomers are produced if the reactant complex ion is a trans-isomer. (C) Only one isomer is produced if the reactant complex ion is a trans-isomer. (D) Only one isomer is produced if the reactant complex ion is a cis-isomer. The correct statement are : (1) C and D (2) B and D (3) A and B (4) A and C

74.(4) Case - I

Case - II

Two isomers (fac and mer) are produced if reactant complex ion is a cis isomer.

Only one isomer (fac) is formed if reactant complex ion is a trans isomer. 75. The major product of the following reaction is :

(1) (2)

(3) (4)

75 (4).

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76. Phenol on treatment with CO2 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 H2SO4 produces :

(1) (2) (3) (4) 76 (3).

OH OH

+ CO2NaOH H O3 +

COOH

(X)

OH OCOCH3

+ (CO CO) O3 2

H SO2 4

COOH

Aspirin

COOH

Cat

(X)

+ CH COOH3

77. 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 2Ba ?

(1) 91.1 10 M (2) 101.0 10 M (3) 95 10 M (4) 92 10 M

77.(1) 2 2 104 4 spBaSO (s) Ba (aq) SO (aq) K 10

22 4 4Na SO 2Na SO

Conc. of 24SO in final solution 50 1 0.1M

500

For final solution

2 2 104Ba SO 10 2 9Ba 10 M

i i f fM V M V

9 9C 450 10 500 C 1.1 10 M

78. Which of the following compounds will be suitable for Kjeldahl's method for nitrogen estimation ?

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

78.(4) Kjeldahl method is not applicable to compounds containing nitrogen in nitro (NO2) and azo (N = N –) groups and nitrogen present in the ring (pyridine) as nitrogen of these compounds does not change to ammonium sulphate.

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[NCERT Class XI part II/Page No. 358]

79. 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 and oxide which is used in chromatography as an adsorbent. The metal 'M' is :

(1) Al (2) Fe (3) Zn (4) Ca

79.(1) NaOH23(M) white gel ppt

(X)

Al NaOH Al OH NaAlO

2 33Al OH Al O

2 3Al O is used in chromatography as an absorbent. (Refer NCERT Class XIth/Part-II, Page-352)

80. 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 2S from HS ions is 131.2 10 then the concentration of 2S ions in

aqueous solution is :

(1) 216 10 (2) 195 10 (3) 85 10 (4) 203 10

80.(4) HCl H Cl 0.2M 0.2 M

2H S H HS 71K 10

2HS H S 132K 1.2 10

22H S 2H S 1 2K K K

201.2 10

2 2

2

H SK

H S

H 0.2M , 2H S 0.1

2 2

200.2 S

1.2 100.1

2 20S 3 10 M

81. 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 2 23Ca PO Ca OH to :

(1) 3 4 223Ca PO CaF (2) 3 223 Ca OH CaF

(3) 2CaF (4) 2 23 CaF Ca OH

81.(1) (Refer NCERT Class XIth Part-II, Page-407)

The F ions make the enamel on teeth much harder by converting hydroxyapatite, 3 4 2 23Ca PO Ca OH ,

into much harder fluorapatite i.e. 3 4 223Ca PO CaF

82. The compound that does not produce nitrogen gas by the thermal decomposition is : (1) 4 2NH NO (2) 4 42NH SO (3) 3 2Ba N (4) 4 2 72NH Cr O

82.(2) 4 2 2 2NH NO N 2H O

4 4 3 2 42NH SO NH H SO

3 22Ba N Ba 3N

4 2 7 2 2 2 32NH Cr O N 4H O Cr O

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83. The predominant form of histamine present in human blood is a(pK Histidine = 6.0) :

(1) (2)

(3) (4)

83.(3) NCERT Class XII/Part-II, Page No. 443

84. The oxidation state of Cr in 2 3 6 66 2Cr H O Cl , Cr C H , and 2 2 32 2K Cr CN O O NH

respectively are: (1) 3, 0, and 6 (2) 3, 0 and 4 (3) 3, 4 and 6 (4) 3, 2 and 4

84.(1) 2 36Cr H O Cl

x 0 3 0 x 3

6 6 2Cr C H

x 0 0 x 0

2 2 32 2K Cr CN O O NH

2 x 2 4 2 0 0 x 6

85. Which type of 'defect' has the presence of cations in the interstitial sites ? (1) Frenkel defect (2) Metal deficiency defect (3) Schottky defect (4) Vacancy defect 85.(1) Pressure of cation in interstitial sites is ‘Frenkel’ defect. 86. The combustion of benzene ( ) gives 2CO (g) and 2H O( ). Given that heat of combustion of benzene at

constant volume is 13263.9 kJ mol at 25°C, heat of combustion (in 1kJ mol ) of benzene at constant pressure

will be :

1 1(R 8.314JK mol )

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

86.(2) 6 6 2 2 215C H ( ) O (g) 6CO (g) 3H O( )2

(g)3n2

(g)H U n RT

1.5 8.314 2983263.91000

3267.6 kJ / mol

87. Which of the following are Lewis acids?

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(1) 3PH and 4SiCl (2) 3BCl and 3AlCl

(3) 3PH and 3BCl (4) 3AlCl and 4SiCl

87.(2) 3BCl and 3AlCl are e deficient and thus act as Lewis acid

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

(1) KCl (2) KCl, B2H6 (3) KCl, B2H6, PH3 (4) KCl, H2SO4

88.(1) KCl exist as K and Cl 89. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

(1) 2 4 2 2[Co(H O) Cl ]Cl.2H O (2) 2 3 3 2[Co(H O) Cl ].3H O

(3) 2 6 3[Co(H O) ]Cl (4) 2 5 2 2[Co(H O) Cl]Cl .H O

89.(2) Depression in freezing pt

f fT i K m

Less the value of i,

Higher the value of freezing pt.

For (2) i = 1 (min)

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

(1) 2H (2) 22H (3) 2

2He (4) 2He

90.(2) 22H does not exist as Bond order is zero

Electronic configuration of 2 2 22 1s 1sH : *

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