Jee Mains-2014 Solutions

8/10/2019 Jee Mains-2014 Solutions

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FIITJEE (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 03 Fax: 040-66777004

Note: For the benefit of the students, specially the aspiring ones, the question of JEE2014_MAINS are also given in this booklet.

Keeping the interest of students studying in class XI, the questions based on the topics from class XI have been marked with *,

which can be attempted as a test. For this test the time allocated in Physics, Mathematics and Chemistry are 27 minutes, 28 minutes

and 22 minutes respectively.

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MMAAIINNSSTime: 3 Hours Maximum Marks: 360

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6. Candidates will beawarded marks as stated above in instruction No. 5 for correct response of

each question. (one fourth) marks will be deducted for indicating incorrect response of eachquestion. No deduction from the total score will be made if no response is indicated for anitem in the answer sheet.

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3. 1

Sol. Angular momentum r p

r = Length of string, which is a constant and p is also a constant.Hence magnitude of angular momentum remains constant but direction changes because directions

of r and p change as the particle moves around in circular path.

4. The current voltage relation of diode is given by 1000V/TI e 1 mA , where the applied voltage V is in

volts and the temperature T is in degree Kelvin. If a student makes an error measuring 0.01 V

while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA ?

(1) 0.5 mA (2) 0.05 mA (3) 0.2 mA (4) 0.02 mA

4. 3

Sol.

1000V

T

I e 1

Since T = 300 K

So,10 V

3I e 1 Taking differentials, we get

10

V3

10dI e dV

3

10V

310

I e V .......... 13

10V

3I 5mA e 1 mA 10

V3e 6 mA

Substituting in equation (1)

10

I 6 0.01 0.2mA3

*5. An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the

mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up

by additional 46 cm. What will be length of the air column above mercury in the tube now?

(Atmospheric pressure =76 cm of Hg)

(1) 38 cm (2) 6 cm (3) 16 cm (4) 22 cm

5. 3Sol. Initially 8 cm of air column is at atmosphere pressure, P0

Final pressure, 0P P g 0.54 x .

8cm54cm x

Assuming isothermal conditions

1 1 2 2P V P V

0P 0.08 A P x A


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0 0P 0.08 P g 0.54 x x 76cm 8cm 76cm 54cm x x.

76cm 8cm 22cm x x

Solving we get x 16 cm

6. Match List - I (Electromagnetic wave type) with List - II (Its association/application) and select the

correct option from the choices given below the lists

Lis t I Lis t II

(1) Infrared waves (i) To treat muscular strain

(2) Radio waves (ii) For broadcasting

(3) X rays (iii) To detect fracture of bones

(4) Ultra violet rays (iv) Absorbed by the ozone layer of the atmosphere

(a) (b) (c) (d) (a) (b) (c) (d)

(1) (iii) (ii) (i) (iv) (2) (i) (ii) (iii) (iv)(3) (iv) (iii) (ii) (i) (4) (i) (ii) (iv) (iii)

6. 2

7. Parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a

dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 x 104

V/n the charge density of the positive plate will be close to:

(1) 4 23 10 C/m (2) 4 26 10 C / m (3) 7 26 10 C/m (4) 7 23 10 C/ m

7. 3Sol. Electric field between the plates of capacitor is

0

E

k

Putting 4E 3 10 v / m k = 2.2

We get 7 26 10 c / m

*8. A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to

measure it?

(1) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.

(2) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm.

(3) A meter scale.

(4) A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale

and main scale has 10 divisions in 1 cm.

8. 4Sol. Least count of the used instrument must be = 0.01 cm as per the measured value

The vernier calipers in the given question has a least count of 0.01 cm because1 Main scale division = 0.1 cmand 1 vernier scale division = 0.09 cmand Least count = 1 main scale division 1 vernier scale division.

*9. Four particles, each of mass M and equidistant from each other, move along a circle of radius Runder the action of their mutual gravitational attraction. The speed of each particle is

(1) (1 2 2)GM

R (2)

1(1 2 2 )

2

GM

R (3)

GM

R (4) 2 2

GM

R


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9. 2

Sol. Net gravitational Force = 1 2 3F F F Side of square,

2Ra R 22

2

1 22

GMF F

R 2

2

3 2

GMF

4R

M

AB

CD

M

M M

2F

3F

1F

O

2

1 2 12

GMF F F 2 2

2R

Hence net gravitational.Force on a particle

is2

2

GM 2 1F

2 4R

Gravitational force provides centripetal acceleration for each particle

Hence

2 2

2

GM 2 2 1 Mv

4 RR

2 GM 2 2 1vR 4

1 GM

v 1 2 22 R

10. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of thebuilding will be(1) 12 A (2) 14 A (3) 8 A (4) 10 A

10. 1

Sol. Power = (voltage) (current)Current drawn by 40 W bulb =

40A

220

Current drawn by 100 W bulb =100

A220

Current drawn by 80 W bulb =80

A220

Current drawn by heater =1000

A220

They can be considered to be in parallel, total current drawn from mains

40 100 80 100015 5 5 11.3A

220 220 220 220

Hence minimum capacity of fuse of mains should be 12 A.

*11. A particle moves with simple harmonic motion in a straight line. In first s, after starting from rest ittravels a distance a, and in next s it travels 2a, in same direction, then(1) amplitude of motion is 4a (2) time period of oscillations is 6(3) amplitude of motion is 3a (4) time period of oscillations is 8

11. 2Sol. Let the particle starts at t = 0 from right extreme,

its position after time t is given byx = A cos t

According to the question,A a = A cos


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or (1 cos )A a ..(1)and 3 cos 2A a A or (1 cos 2 ) 3A a (2)dividing these two equations,

1 cos 2 13 cos 1,

1 cos 2

As particle is moving in same direction,

cos 1 is not possible

1

cos2

2

3 3T

or 6T

using equation (1), A = 2a

12. The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 x 103Am1. Thecurrent required to be passed in a solenoid of length 10 cm and number of turns 100, so that themagnet gets demagnetized when inside the solenoid, is(1) 3 A (2) 6 A (3) 30 mA (4) 60 mA

12. 1

Sol.2

100

10 10n

= 1000 per m

H = 3 x 103Am1= ni

,

33 10

1000i

= 3A

13. The forward biased diode connection is

(1) 2 V 4 V (2) 2 V +2V

(3) +2V 2V (4) 3V 3V

13. 3Sol. A diode is forward biased if p side is at higher potential

14. During the propagation of electromagnetic waves in a medium:

(1) Electric energy density is equal to the magnetic energy density.

(2) Both electric and magnetic energy densities are zero.

(3) Electric energy density is double of the magnetic energy density.

(4) Electric energy density is half of the magnetic energy density.

14. 1Sol. During the propagation of electromagnetic waves in a medium, electric energy density is equal to the

magnetic energy density


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15. In the circuit shown here, the point C is kept connected topoint A till the current flowing through the circuit becomesconstant. Afterward, suddenly, point C is disconnected

from point A and connected to point B at time t =0. Ratioof the voltage across resistance and the inductor at t = L/Rwill be equal to:

AC R

LB

(1) 1 (2)1 e

e

(3)

e

1 e (4) 1

15. 1Sol. Voltage across resistance and inductor at any time after C is connected to B will be same as they will

be in parallel.

0R LV V

1R

L

V

V

*16. A mass m is supported by a massless string wound around auniform hollow cylinder of mass m and radius R. If the string doesnot slip on the cylinder, with what acceleration will the mass fall onrelease?

(1)5g

6 (2) g

(3)2g

3 (4)

g

2

m

m

R

16. 4Sol. Let acceleration of the block is a

From FBD of block, mg T ma ........... 1

From FBD of cylinder, TR I

and,

2 aTR mRR

.(2)

a

T

mg

m

(as, a R for no slipping condition)Adding eq, (1) & (2)

mg 2ma

g

a2

T

*17. One mole of diatomic ideal gas undergoes a cyclic process ABC as

shown in figure. The process BC is adiabatic. The temperatures at A, Band C are 400 K, 800 K and 600 K respectively. Choose the correct

statement:

(1) The change in internal energy in the process AB is 350 R.

(2) The change in internal energy in the process BC is 500 R.

(3) The change in internal energy in whole cyclic process is 250 R.

(4) The change in internal energy in the process CA is 700 R.

P

B

AC

V

600K

800K

400K

17. 2

Sol. A B v5

U nC T 1 R 800 400 1000R2


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B C v5

U nC T 1 R 600 800 500R2

C A v5

U nC T 1 R 400 600 500R2 The change in internal energy in whole cyclic process is zero.

*18. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by

the particle, to hit the ground, is n t imes that taken by it to reach the highest point of its path.

The relation between H, u and n is:

(1) 22gH nu n 2 (2) 2gH u n 2 (3) 2 22gH n u (4) 2 2gH n 2 u

18. 1Sol. Let the time taken by the particle to reach the highest point of its path be t0

, 0u

t

g

and the time taken by the particle to hit the ground is 0nu

n tg

20 0

1( ) ( )

2H u n t g n t

21

2

u uH u n g n

g g

2 2 2

2

u n uH n

g g

2 2 22 2gh nu n u 2

2 ( 2)gH nu n

19. A thin convex lens made from crown glass

3

2has focal length f. When it is measured in two

different liquids having refractive indices4 5

and3 3

, it has the focal lengths1 2f and f respectively. The

correct relation between the focal lengths is:

(1)2 1f f and f becomes negative (2) 1 2f and f both become negative

(3) 1 2f f f (4) 1 2f f and f becomes negative

19. 4Sol. Using, Lens maker formula,

2

1 1 2

1 1 11

f R R

We get,

1 2

1 3 1 11

2f R R

(1)

and1 1 2

1 3 3 1 11

2 4f R R

(2)


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and2 1 2

1 3 3 1 11

2 5f R R

(3)

from (1) & (2), 1 4f f from (1) & (3), 2 5f f

*20. Three rods of Copper, Brass and Steel are welded together to form a Y - shaped structure. Area of

cross - section of each rod =4 cm2. End of copper rod is maintained at 100C where as ends of brass

and steel are kept at 0C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms

respectively. The rods are thermally insulated from surroundings except at ends. Thermal

conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of

heat flow through copper rod is:

(1) 4.8 cal / s (2) 6.0 cal / s (3) 1.2 cal / s (4) 2.4 cal / s

20. 1Sol. Let the junction be at 0C.

Thermal resistance of copper rod, Brass rod and steel rod in cgs units is,

25 25, and 25

2 2respectively.

(Thermal resistance isL

kA)

From diagram, 1 2 3i i i where i represent rate of heat flow

100 0 0

25/ 2 25 / 2 25

thermal

As iR

040 C

1100 100 40

i 4.8cal / s25 /2 25 / 2

i1

Brass steel

copper

100C

0C 0C

C

i2 i3

21. A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air

column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

(1) 6 (2) 4 (3) 12 (4) 8

21. 1

Sol. 2n 1

4

2n 1

Speed of sound v f

f 4340

2n 1

340 2n 1f

4

340 2n 1100 2n 1

4 0.85

1n 0; f 100 1 100 Hz

2n 1; f 300 Hz

3n 2; f 500 Hz


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4n 3; f 700 Hz

5n 4; f 900 Hz

6

n 5; f 1100 Hz

7n 6; f 1300 Hz 1250Hz no. of frequencies = 6

*22. There is a circular tube in a vertical plane. Two liquids which do not mixand of densities d1and d2 are filled in the tube. Each liquid subtends90 angle at centre. Radius joining their interface makes an angle

with vertical. Ratio1

2

d

dis:

d2

d1

(1)1 tan

1 tan

(2)1 sin

1 cos

(3)1 sin

1 sin

(4)1 cos

1 cos

22. 1Sol. 1 2 1(1 sin ) [sin cos ] [1 cos ]d gR d gR d gR

1 tan

1 tan

R(1sin )

d2

d2

d1

R(1cos )

R sin

R cos

23. A green light is incident from the water to the airwater interface at the critical angle (). Select thecorrect statement(1) the spectrum of visible light whose frequency is more than that of green light will come out to the

air medium(2) the entire spectrum of visible light will come out of the water at various angles to the normal(3) the entire spectrum of visible light will come out of the water at an angle of 90to the normal(4) the spectrum of visible light whose frequency is less than that of green light will come out to the

air medium23. 4

Sol.1 1

sinC

As is greater for light with greater frequency, C is less for light with greater frequency when green light is falling at its critical angle, for all other higher frequency colours angle of

incidence is greater than their respective critical angle.

24. Hydrogen (1H1), Deuterium(1H

2), singly ionised Helium(2He4)+ and doubly ionised lithium (3Li

6)++allhave one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the

wave lengths of emitted radiation are 1 2 3, , and 4 respectively then approximately which oneof the following is correct?

(1) 1 2 3 44 9 (2) 1 2 3 42 3 4

(3) 1 2 3 44 2 2 (4) 1 2 3 42 2 24. 1


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Sol.2

2 22 1

1 1 1Rz

n n

2

1

z here z is the no. of protons in the nucleus

1

1

(1)K here K is constant

2

1

1K

3

1

4K

4

1

9K

1 2 3 4

4 9

25. The radiation corresponding to 3 2 transition of hydrogen atom falls on a metal surface to producephotoelectrons. These electrons are made to enter a magnetic field of 3 x 104T. If the radius of thelargest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to(1) 0.8 eV (2) 1.6 eV (3) 1.8 eV (4) 1.1 eV

25. 4

Sol.mv

rqB

maxmax

r qBv

m

2max max

12

k mv

2 2 2max

2

1

2

r q Bm

m in Joules

2 2max

2

r qB

m in eV

We know

maxk h

maxh k 2 2

max13.6 13.64 9 2

r qBm

in eV

= 1.889 eV 0.8 eV= 1.1 eV

*26. A block of mass m is placed on a surface with a vertical cross section given by

3

6

xy . If the

coefficient of friction is 0.5, the maximum height above the ground at which the block can be placedwithout slipping is

(1)1

3m (2)

1

2m (3)

1

6m (4)

2

3m

26. 3


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Sol. Let p be a point on the surface.

sinsf mg for block to be stationary

sin Lmg f sin cosmg mg tan slope

for max y= slope

Y

X

P(x,y)fs

mg sin

dy

dx

21 3

2 6

x

x = 1 3

16 6

xy m

*27. When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude F = ax +bx2 where a and b are constants. The work done in stretching the unstretched rubber-band by L is:

(1)

2 3

2 3

aL bL (2)

2 31

2 2 3

aL bL

(3)

2 2aL bL (4) 2 21

2aL bL

27. 1

Sol. W F dx

2

0( )

L

ax bx dx 2 3

02 3

L

x x

a b

=

2 3

2 3

aL bL

*28. On heating water, bubbles being formed at the bottom of the vessel

detatch and rise. Take the bubbles to be spheres of radius R and

making a circular contact of radius r with the bottom of the vessel. If r


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29. 2Sol. At beginning

Axis of the

polaroid

IB

IA

When polaroid is rotated through 30,2' cos 30A AI I 2' cos 60B BI I

as ' 'A BI I

2 2cos 30 cos 60A BI I 2

2

cos 60

cos 30

A

B

I

I

1

3

30. Assume that an electric field 2 E 30 x i

exists in space. Then the potential difference A OV V ,

whereOV is the potential at the origin and AV the potential at x = 2m is:

(1) 80 J (2) 80 J (3) 120 J (4) 120 J

30. 1

Sol.

2

00

A

V V E dr

2

2

0

30x dx 2

2

0

30 x dx 2

3

0

303

x

= 80 V

PART B: MATHEMATICS

31. The image of the linex 1 y 3 z 4

3 1 5

in the plane 2x y + z + 3 = 0 is the line:

(1)x 3 y 5 z 2

3 1 5

(2)x 3 y 5 z 2

3 1 5

(3)x 3 y 5 z 2

3 1 5

(4)

x 3 y 5 z 2

3 1 5

31. 1Sol. Line is parallel to 2x y + z + 3 = 0

Let (h, k, ) is Image of point (1, 3, 4) w.r.t plane

2x y + z + 3 = 0

2 2 3 4 3h 1 k 3 42 1 1 4 1 1

= 2

h 1 = 4 h = 3k = 2 + 3 k = 5= 2 + 4 = 2

(1, 3, 4)


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Equation of image line isx 3 y 5 z 2

3 1 5

32. If the coefficients of x3and x4in the expansion of (1 + ax + bx2) (1 2x)18in powers of x are bothzero, then (a, b) is equal to

(1)251

16,3

(2)251

14,3

(3)272

14,3

(4)272

16,3

32. 4Sol. Coefficient of x

3= 4a.

18C2 2b.

18C1 8 .

18C3= 0

Coefficient of x4= 16.18C4 8a.18C3+ 4b

18C2= 051a 3b = 544 ...... (1)240 32a + 3b = 0 ...... (2)

a = 16; b =272

3

33. If a R and the equation 3 (x [x])2

+ 2 (x [x]) + a2

= 0(where [x] denotes the greatest integer x) has no integral solution, then all possible values of a liein the interval:(1) (1, 0) (0, 1) (2) (1, 2) (3) (2, 1) (4) (, 2) (2, )

33. 1Sol. a2= 3 {x}2 2 {x}

3 {x}2 2 {x} > 0{x} (3 {x} 2) > 0{x} < 0 (or) {x} > 2/3{x} < 0 is not possible {x} (2/3, 1)

a2 2

4 23 2 ,3(1) 2(1)

9 3

(0, 1)a (1, 0) (0, 1)

34. If2

a b b c c a a b c

, then is equal to

(1) 2 (2) 3 (3) 0 (4) 134. 4

Sol. a b b c c a a b b c c a

= a b b c a c c c a b a b b c a c

= 2

a b c a b c a b c

k = 1

35. The variance of first 50 even natural numbers is

(1)833

4 (2) 833 (3) 437 (4)

437

4

35. 2

Sol. Mean =2 4 6 100

x50

= 2 1 2 3 50

50

=

50 51x 51

50

Variance =

22

i2 2ix x x

xn n


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= 2 2 2 22 2 2 2

2 24 1 2 3 502 4 6 100

51 5150 50

= 3434 2601 = 833

36. A bird is sitting on the top of a vertical pole 20m high and its elevation from a point O on the groundis 45. It flies off horizontally straight away from the point O. After one second, the elevation of thebird from O is reduced to 30. Then the speed (in m/s) of the bird is

(1) 40 2 1 (2) 40 3 2 (3) 20 2 (4) 20 3 1 36. 4

Sol. tan 30=20

d 20 d + 20 = 20 3

d = 20 3 1 mts

speed = distance

20 3 1 mts

time

45

30

20 d

20

O

37. The integral 2

0

x x1 4sin 4sin dx

2 2

equals

(1) 4 (2) 2 4 4 33 (3) 4 3 4 (4) 4 3 4

3

37. 4

Sol.2

0 0

x x2sin 1 dx 2sin 1 dx

2 2

=/ 3

0 /3

x x2sin 1 dx 2sin 1 dx

2 2

x

2sin 1 0 for x 0,2 3

x2sin 1 0 for x ,

2 3

=/3

0 /3

x x2 2cos x 4cos x

2 2

=3 4 3

4 42 3 2 3

4 3 4

3

38. The statement ~(p ~q) is(1) equivalent to p q (2) equivalent to ~p q (3) a tautology (4) a fallacy

38. 1Sol. p q ~q p ~q ~(p ~q) p q

T T F F T TT F T T F FF T F T F FF F T F T T


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39. If A is an 3 3 non singular matrix such that AA= AA and B = A1A, then BBequals(1) I + B (2) I (3) B1 (4) (B1)

39. 2Sol. |A| 0

Given AAT= ATAB = A1AT, BT= A(A1)T

BBT= (A

1A

T) A(A

T)1

= A1

(ATA) (A

T)1

= A1

AAT(A

T)1

= I

40. The integral1

xx

11 x e

x

dx is equal to

(1) (x 1)1

xxe

+ c (2)

1x

xxe c

(3) (x + 1)1

xxe

+ c (4)

1x

xxe c

40. 2

Sol.1 1

x xx x

11 x e dx xe c

x

Or put t =1

xxxe

41. If z is a complex number such that |z| 2, then the minimum value of1

z2

(1) is equal to5

2 (2) lies in the interval (1, 2)

(3) is strictly greater than5

2 (4) is strictly greater than

3

2but less than

5

2

41. 2

Sol. Min. of1 1 3

z 22 2 2

(2, 0) (2, 0)1,0

2

42. If g is the inverse of a function f and f(x) =5

1

1 x, then g(x) is equal to

(1) 1 + x5 (2) 5x4 (3)

5

1

1 g(x) (4) 1 + {g(x)}5

42. 4Sol. g(x) = f1(x) fog (x) = x

f(g(x)).g(x) = 1

g(x) = 51

1 g xf (g(x))

43. If , 0, and f(n) = n+ nand

3 1 f(1) 1 f(2)

1 f(1) 1 f(2) 1 f(3)

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

= K (1 )2(1 )2( )2, then K is

equal to


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(1) (2)1

(3) 1 (4) 1

43. 3

Sol. f(1) = + ; f(2) = 2+ 2; f(3) = 3+ 3

=

2 2

2 2 3 3

2 2 3 3 4 4

3 1 1

1 1 1

1 1 1

=

2 2

2 2 2 2

1 1 1 0 1 1

1 0 1 1

1 1

= ( )2(1 )2(1 )2.

44. Let fk(x) =1

k(sinkx + coskx) where x R and k 1. Then f4(x) f6(x) equals

(1) 16

(2) 13

(3) 14

(4) 112

44. 4

Sol. fk(x) =1

k(sin

kx + cos

kx)

= 4 4 6 61 1

sin x cos x sin x cos x4 6

= 2 2 2 21 1

1 2sin xcos x 1 3sin xcos x4 6

= 2 2 2 21 1 1 1 3 2 1

sin xcos x sin xcos x4 2 6 2 12 12

45. Let and be the roots of equation px2+ qx + r = 0, p 0. If p, q, r are in A.P and1 1

= 4, then

the value of | | is

(1)61

9 (2)

2 17

9 (3)

34

9 (4)

2 13

9

45. 4Sol. px

2+ qx + r = 0 p 0

2q = p + r (p, q, r are in A.P)

+ =q

p , =

r

p

q

4 4r

| | = 2

4 q = 4r; p = 9r

2 2

2 2

q 4r q 4pr

pp p

2 2 2

2 2

16r 36r 52r

81r 81r

| | =2 13

9


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46. Let A and B be two events such that 1P A B6

, P(A B) =1

4and

1P A

4 , where A stands

for the complement of the event A. Then the events A and B are

(1) mutually exclusive and independent (2) equally likely but not independent(3) independent but not equally likely (4) independent and equally likely

46. 3

Sol. P(A B) =1 5

16 6

P(A B) =1

4, P(A) =

1 31

4 4

P(B) = P(A B) P(A) + P(A B)

P(B) =5 3 1 5 1 5 3 2 1

6 4 4 6 2 6 6 3

P(A B) =3 1

4 3 = P(A) . P(B) A, B are independent.

P(A) P(B) A, B are not equally likely.

47. If f and g are differentiable functions in [0, 1] satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6, then forsome c ]0, 1[(1) 2f(c) = g(c) (2) 2f(c) = 3g(c) (3) f(c) = g(c) (4) f(c) = 2g(c)

47. 4Sol. f(0) = 2 = g(1), g(0) = 0, f(1) = 6

from LMVT

f(c) = f 1 f 0

41 0

g(c) = g 1 g 0 2 0

21 0 1 0

f(c) = 2 g(c)

48. Let the population of rabbits surviving at a time t be governed by the differential equation

dp(t) 1

dt 2 p(t) 200. If p(0) = 100, then p(t) equals

(1) 400 300 et/2 (2) 300 200 et/2 (3) 600 500 et/2 (4) 400 300 et/2

48. 1

Sol.

dp t 1

p t 200dt 2

I.F =1 t

dt2 2e e

Solution is I.F. p(t) = (I.F)(200) dtet/2p(t) = et/2(200) dtet/2p(t) = 400 et/2+ cPut t = 0 100 = 400 + c c = 300e

t/2p(t) = 400 e

t/2 300

Multiply with et/2p(t) = 400 300 et/2

49. Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle, centred at (0, y), passingthrough origin and touching the circle C externally, then the radius of T is equal to

(1)3

2 (2)

3

2 (3)

1

2 (4)

1

4

49. 4


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Sol. Centre for C is (1, 1)Radius is r1= 1Centre for T is (0, y)

Radius r2= 2 20 y y Touch externally distance between centres = r1+ r2

2

1 y 1 1 y y =1

4 radius =

1

4

50. The area of the region described by A = {(x, y): x2+ y21 and y21 x} is

(1)4

2 3

(2)

4

2 3

(3)

2

2 3

(4)

2

2 3

50. 1

Sol. Area of shaded region = Area of semi circle +1

0

2 1 x dx

= 13/ 22

0

1 x1 42

2 3 / 2 2 3

(0, 1)

(1, 0)

(0, 1)

51. Let a, b, c and d be nonzero numbers. If the point of intersection of the lines 4ax + 2ay + c = 0 and5bx + 2by + d = 0 lies in the fourth quadrant and is equidistant from the two axes then(1) 2bc 3ad = 0 (2) 2bc + 3ad = 0 (3) 3bc 2ad = 0 (4) 3bc + 2ad = 0

51. 3Sol. 4ab x + 2ab x + bc = 0 20abx + 10aby + 5bc = 0

5ab x + 2ab y + ad = 0 20abx + 8aby + 4ad = 0

. .abx + bc ad = 0 2aby + 5bc 4ad = 0

x =bc ad

ab

2ab y = 4ad 5bc y =

4ad 5bc

2ab

y = x bc ad 4ad 5bc

ab 2ab

2ad 2bc = 4ad 5bc. 2ad 3bc = 0

52. Let PS be the median of the triangle with vertices P(2, 2), Q(6, 1) and R(7, 3). The equation of theline passing through (1, 1) and parallel to PS is(1) 4x 7y 11 = 0 (2) 2x + 9y + 7 = 0 (3) 4x + 7y + 3 = 0 (4) 2x 9y 11 = 0

52. 2

Sol. Equation of line parallel to PS and passing through(1, 1) is

y + 1 =1 2

132

2

(x 1)

9(y + 1) = 2 (x 1)2x + 9y + 7 = 0

S13

,12

R (7, 3)Q (6, 1)

P (2, 2)

53. 2

2x 0

sin cos xlim

x

is equal to


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(1)2

(2) 1 (3) (4)

53. 4

Sol. 2 2

2 2x 0 x 0

sin 1 sin x sin sin xlim lim

x x

=

2 2

2 2x 0

sin sin x sin xlim

sin x x

= 1 1 =

54. If X = {4n 3n 1 : n N} and Y = {9(n 1) : n N}, where N is the set of natural numbers, then XY is equal to(1) N (2) Y X (3) X (4) Y

54. 4Sol. X = {(3 + 1)

n 3n 1}

X = {nC01

n3

0+

nC13

11

n 1+

nC23

2+

nC33

3+ ...... +

nCn3

n}

X = 9 {

n

C2+

n

C33 +

n

Cn3

n

+ ..... + 3

n

}Y = {9 (n 1) : n N} X Y X Y = Y

55. The locus of the foot of perpendicular drawn from the centre of the ellipse x2+ 3y2= 6 on anytangent to it is(1) (x

2 y

2)2= 6x

2+ 2y

2(2) (x

2 y

2)2= 6x

2 2y

2

(3) (x2+ y

2)2= 6x

2+ 2y

2(4) (x

2+ y

2)2= 6x

2 2y

2

55. 3

Sol.2 2x y

6 2 = 1

Tangent equation isxcos ysin

a b

= 1

xcos y sin6 2

= 1 ...... (1)

Suppose (h, k) is foot

Tangent slope ish

k

Equation is y k =h

k

(x h)

Ky k2= hx + h2 hx + ky = h2+ k2 ...... (2)(1) and (2) represents same

2 2

cos sin 1

h k6h 2k

cos

2

+ sin2

= 1

2 2

22 2

6x 2y

1x y

(x2

+ y

2

)

2

= 6x

2

+ 2y

2

56. Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the newnumbers are in A.P. Then the common ratio of the G.P is

(1) 2 3 (2) 3 2 (3) 2 3 (4) 2 3 56. 4Sol. a, ar, ar2 ..... G.P

a, 2ar, ar2a + ar

2= 4ar

r2 4r + 1 = 0 r =4 16 4

2


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r =4 2 3

2

r = 2 3

57. If (10)9+ 2(11)1(10)8+ 3(11)2(10)7+ ...... + 10(11)9= k(10)9, then k is equal to

(1)121

10 (2)

441

100 (3) 100 (4) 110

57. 3Sol. S = (10)9+ 2(11)1(10)8+ 3 (11)2(10)7+ ...... + 10 (11)9

11S

10= 11 (10)8+ 2(11)2(10)7+ ...... + (11)10

11S1

10

= ((10)9+ (11) (10)

8+ (11)

2(10)

7+ ......) 11

10

=

10

9 10

111

1010 11

11 110

S

10

= 1010 S = 100 109

58. The angle between the lines whose directions cosines satisfy the equations + m + n = 0 and

2= m2+ n2is

(1)3

(2)

4

(3)

6

(4)

2

58. 1

Sol. = (m + n)

(m + n)2= m2+ n2

mn = 0Case 1:m = 0

= n m = 0.n

n

1 1

m n

0 1

m n

1 0 1

Case 2:n = 0 n = 0.m

= m n m0 1

m n

1 1 0

cos =1 0 0

2 2

cos =

1

2 =

3

59. The slope of the line touching both the parabolas y2= 4x and x2= 32y is

(1)1

2 (2)

3

2 (3)

1

8 (4)

2

3

59. 1


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Sol. y = mx +a

m y = mx +

1

msub in 2ndparabola

x2

= 32

1

mx m

x2+ 32 [mx] +

32

m= 0 = 0

(32)2m

2 4(1)

32

m= 0

m3=1

8 m =

1

2

60. If x = 1 and x = 2 are extreme points of f(x) = log |x| + x2+ x, then

(1) = 6, =1

2 (2) = 6, =

1

2 (3) = 2, =

1

2 (4) = 2, =

1

2

60. 3Sol. f(x) = 2 x 1

x

f(1) = 0 and f(2) = 0

2+ 1 = 0 4 1 02

+ 2 1 = 0 + 8+ 2 = 0+ 8+ 2 = 0

6 3 = 0 =1

2

+1

22

1 = 0 = 2

PART C: CHEMISTRY

*61. Which one of the following properties is not shown by NO ?(1) It combines with oxygen to form nitrogen dioxide(2) Its bond order is 2.5(3) it is diamagnetic in gaseous state(4) It is a neutral oxide

61. 3Sol. NO exists as monomer in gaseous state

In monomeric state NO is paramagnetic

*62. If Z is a compressibility factor, van der Waals equation at low pressure can be written as :

(1)Pb

Z 1RT

(2)Pb

Z 1RT

(3)RT

Z 1Pb

(4)a

Z 1VRT

62. 4Sol. At low pressure Z < 1

Vanderwaals equation

2

aP V b RT

V

at low pressure V is high then b can be neglectedthen equation becomes

2

aP V RT

V


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aPV RT

V

PV a

1RT VRT a

Z 1VRT

63. The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is :(1) Cu (2) Cr (3) Ag (4) Ca

63. 4Sol. More active metal like Ca which is above H in electrochemical series cant be obtained by the

electrolysis of aqueous solutions of its salt.

64. Resistance of 0.2 M solution of an electrolyte is 50 . The specific conductance of the solution is 1.4S m

1. The resistance of 0.5 M solution of the same electrolyte is 280 . The molar conductivity of

0.5M solution of the electrolyte in S m

2

mol

1

is:(1) 5 x 103 (2) 5 x 102 (3) 5 x 104 (4) 5 x 10364. 3

Sol. K Ca

=1

R a

11.4

50 a

70a

For 0.5 M solution

K =1

70280

= 0.25 sm1

3 2 3K 10 m 0.25 10M 0.5

= 5 x 104

.

65. CsCl crystallizes in body centred cubic lattice. If a is its edge length then which of the followingexpressions is correct ?

(1)Cs Cl

3r r a

2 (2) Cs Clr r 3a

(3)Cs Clr r 3a (4)

Cs Cl

3ar r

2

65. 1Sol. For BCC Cs Cl3 a 2 r r

Cs Cl

3ar r

2

66. Consider separate solutions of 0.500 M 2 5C H OH aq , 0.100 M Mg3(PO4)2(aq), 0.250 M KBr(aq) and

0.125 M Na3PO4(aq) at 25C. Which statement is true about these solutions, assuming all salts to bestrong electrolytes ?(1) 0.125 M Na3PO4(aq) has the highest osmotic pressure(2) 0.500 M C2H5OH(aq) has the highest osmotic pressure(3) They all have the same osmotic pressure(4) 0.100 M Mg3(PO4)2(aq) has the highest osmotic pressure.


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66. 3Sol. i CST

C2H5OH i = 1 = 1 x 0.5 ST = 0.5 ST

Mg3(PO4)2 i = 5 = 5 x 0.1 ST = 0.5 STKBr i = 2 = 2 x 0.25 ST = 0.5 STNa3PO4 i = 4 = 4 x 0.125 ST = 0.5 ST

All the electrolytes have same osmotic pressure

*67. In which of the following reactions H2O2acts as a reducing agent ?

(a) 2 2 2H O 2H 2e 2H O (b) 2 2 2H O 2e O 2H

(c) 2 2H O 2e 2OH (d) 2 2 2 2H O 2OH 2e O 2H O

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

67. 2Sol. H2O2is acting as reducing agent means it is undergoing oxidation

i.e., losing electrons

H2O2O2+ 2H++ 2eH2O2+ 2OH

O2+ 2H2O + 2e

68. In 2NS reactions, the correct order of reactivity for the following compounds : CH 3Cl, CH3CH2Cl,

(CH3)2CHCl and (CH3)3CCl is :(1) CH3CH2Cl > CH3Cl > (CH3)2CHCl > (CH3)3CCl(2) (CH3)2CHCl > CH3CH2Cl > CH3Cl > (CH3)3CCl(3) CH3Cl > (CH3)2CHCl > CH3CH2Cl > (CH3)3CCl(4) CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl

68. 4Sol. In 2NS nucleophile attacks the reactant from bask side, and nucleophile can easily attack the

reactant when it is less sterically crowded. With increase in steric hindrance rate of 2NS reaction

decreases.CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl

69. The octahedral complex of a metal ion M3+

with four monodentate ligands L1, L2, L3and L4absorbwavelengths in the region of red, green, yellow and blue, respectively. The increasing order of l igandstrength of the four ligands is :(1) L3< L2< L4< L1 (2) L1< L2< L4< L3 (3) L4 Green > Yellow > RedL4requires more energy so L4becomes strong field ligand in the same way the order isL4> L2> L3> L1

*70. For the estimation of nitrogen, 1.4g of an organic compound was digested by Kjeldahl method and

the evolved ammonia was absorbed in 60 mL ofM

10Sulphuric acid. The unreacted acid required 20

mL ofM

10 sodium hydroxide for complete neutralization. The percentage of nitrogen in the

compound is :


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(1) 3% (2) 5% (3) 6% (4) 10%70. 4Sol. N = M Basicity

No. of milli equivalents of H2SO4reacted=

M60 2 12

10 m. equivalents

Un reacted H2SO4 = m.eq. of NaOH

=1

20 2m.10

equivalents

Then no. of m. equivalents of NH3evolved= 10 m. equivalents

1000 m. equivalents 14gm of N

10 m. equivalents =14

0.14gm100

of N

1.4 gm of organic compound has 0.14 gm of N

Then 100gm of O.C has =0.14 100

1.4

=14

10%1.4

10% of N.

71. The equivalent conductance of NaCl at concentration C and at infinite dilution are C and ,respectively. The correct relationship between Cand is given as :(where the constant B is positive)

(1) C B C (2) C B C

(3) C B C (4) C B C 71. 1Sol. As concentration decreases (with increase in dilution) the equivalent conductivity increases due to

increased mobility of ions. This can be explained by DebyeHuckelonsager equation

C B C

C

C

*72. For the reaction 32 g 2 g1

SO O SO g2

, if KP = KC(RT)

x

where the symbols have usualmeaning then the value of x is : (assuming ideality)

(1)1

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

1

2

72. 4

Sol. n

p cK K RT

n = g.p g.r n n

for 2 2 31

SO g O g SO g2

;1

n2

x = 1/2


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73. In the reaction, 54PClLiAlH Alc.KOH

3CH COOH A B C, the product C is :(1) Ethylene (2) Acetyl chloride (3) Acetaldehyde (4) Acetylene

73. 1

Sol. 54PClLiAlH AlC

3 3 2 3 2 2 2KOHCH COOH H C CH OH CH CH Cl H C CH

74. Sodium phenoxide when heated with CO2 under pressure at 125C yields a product which onacetylation produces C.

ONa

CO2125

5 AtmB

H+

Ac2OC

0

The major product C would be :

(1)

OH

COOCH3

(2)

O COCH3

COOH

(3)

O COCH3

COOH

(4)

OH

COCH3

COCH3 74. 3

Sol.

ONa

CO25 Atm

B

H+

AC2O

0125COO Na

O Na+

COOH

O C

O

CH3

Acetyl

Salicylic acid

(Aspirin)

C

75. On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, theorganic compound formed is :(1) an alkyl cyanide (2) an alkyl isocyanide (3) an alkanol (4) an alkanediol

75. 2Sol.

RNH2+ CHCl3+ 3KOH R N C

(alkyl isocyanide)

3KCl 3H2O

This is carbyl amine reaction and is used in identification of primary amines.

*76. The correct statement for the molecule, CsI3, is :


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(1) It contains Cs3+and Iions. (2) It contains Cs+, Iand lattice I2molecule.

(3) It is a covalent molecule. (4) It contains Cs+and 3I ions.

76. 4

Sol. CsI3is an ionic compound and contains Cs+and 3I

ions.

77. The equation which is balanced and represents the correct product(s) is :

(1) 2 24 excess NaOH

2 26Mg H O EDTA Mg EDTA 6H O

(2) 4 2 2 44CuSO 4KCN K Cu CN K SO

(3) 2 2Li O 2KCl 2LiCl K O

(4) 23 45CoCl NH 5H Co 5NH Cl

77. 4Sol. In general CrII and CoII complexes are substitutionally labile whereas CrIII and CoIII are

substitutionally inert (II). In acidic medium pentammine chlorido cobalt(II) ion dissociates as follows.

23 45

Co NH Cl 5H CO 5 N H Cl

*78. For which of the following molecule significant 0 ?

(a)

Cl

Cl

(b)

CN

CN

(c)

OH

OH

(d)

SH

SH (1) Only (c) (2) (c) and (d) (3) Only (a) (4) (a) and (b)

78. 2Sol.

Cl

Cl

For and

C

C

N

N

OH

OH

SH

SH

net net0 0 netnet

as 'CN' is linear net bond

moments cancel

0 0

79. For the nonstoichiometric reaction 2A + B C + D, the following kinetic data were obtained in

three separate experiments, all at 298 K.


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Initial concentration(A)

Initial concentration(B)

Initial rate o fformation of C

(mol LS

)

0.1 M 0.1 M 1.2 x 103

0.1 M 0.2 M 1.2 x 10

3

0.2 M 0.1 M 2.4 x 103The rate law for the formation of C is :

(1)2dc

k A Bdt

(2)dc

k Adt

(3)dc

k A Bdt

(4)2dc

k A Bdt

79. 2

Sol. Rate (r) =x y

K A B

from (1) and (2)

x y 31

x y 32

K 0.1 0.1r 1.2 10

r 1.2 10K 0.1 0.2

y

11

2

Y = 0

from (1) and (3)

x y 31

x y 33

K 0.1 0.1r 1.2 10

r 2.4 10K 0.2 0.1

y1 1

2 2

Y = 1

Rate equation isdc

K Adt

80. Which series of reactions correctly represents chemical relations related to iron and its compound ?

(1) 2Cl ,heat heat, air Zn

3 2Fe FeCl FeCl Fe

(2)0 0

2O ,heat CO,600 C CO,700 C3 4Fe Fe O FeO Fe

(3) 2 4 2 4 2dil H SO H SO ,O heat4 2 4 3Fe FeSO Fe SO Fe

(4) 2 2 4O ,heat dil H SO heat

4Fe FeO FeSO Fe

80. 2

Sol. 2 3 43Fe 2O Fe O

03 4 2600 C

Fe O CO 3FeO CO

0 2700 CFeO CO Fe CO

*81. Considering the basic strength of amines in aqueous solution, which one has the smallest pKbvalue?(1) (CH3)3N (2) C6H5NH2 (3) (CH3)2NH (4) CH3NH2

81. 3Sol. In polar protic solvent the basic nature of methyl amines is in the order

3 3 2 32 3CH NH CH NH CH N

and C6H5NH2is least basic as lone pair on nitrogen atom is delocalized.So overall (CH3)2NH is more basic.


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82. Which one of the following bases is not present in DNA ?(1) Cytosine (2) Thymine (3) Quinoline (4) Adenine

82. 3

Sol. In DNA the bases present are

(1)

Adenine (A)

N

C

C

N

C

C

N

HC

N

H

NH2

H

(2)

Guanine (G)

NH

C

C

N

C

C

N

C

N

H

O

NH2H

(3)

Cytosine(C)

NC

C

N

HCHC

NH2

H

O

(4)

Thymine (T)

NHC

C

N

CC

O

H

O

CH3

H

*83. The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is :

(1) 5, 1, 1, +1

2 (2) 5, 0, 1, +

1

2 (3) 5, 0, 0, +

1

2 (4) 5, 1, 0, +

1

2

83. 3Sol . The valency electron of rubidium is present in 5s so the set of four quantum numbers are

n = 5, l = 0, m = 0, s = +1

2

*84. The major organic compound formed by the reaction of 1, 1, 1 trichloroethane with silver powder is :(1) 2Butyne (2) 2 Butene (3) Acetylene (4) Ethene

84. 1Sol.

CH3 C

Cl

Cl

Cl 6Ag Cl C

Cl

Cl

CH3 CH3 C C CH3 6AgCl

2-Butyne

85. Given below are the half cell reactions :2Mn 2e Mn ; 0E 1.18V

3 22 Mn e Mn ; 0E 1.51V The Efor 2 33Mn Mn 2Mn will be :(1) 0.33 V ; the reaction will not occur (2) 0.33 V; the reaction will occur(3) 2.69 V ; the reaction will not occur (4) 2.69 V; the reaction will occur

85. 3Sol. Given Mn2++ 2eMn E= 1.18 V (1)

Given 2Mn3+

+ 2e2Mn2+ E= + 1.51 V (2)

So 2Mn2+2Mn3++ 2e E= 1.5 V (3)The required equation is the cell reaction occurring when (1) is taken as cathode and (2) as anode.

2 33Mn Mn 2Mn E= 2.69 VAs Eis ve, G is positive.


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The reaction is non spontaneous.

*86. The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of

number of their molecule is :(1) 1 : 8 (2) 3 : 16 (3) 1 : 4 (4) 7 : 3286. 4Sol. The ratio of masses of oxygen and nitrogen is 1 : 4

The ratio of no. of moles of oxygen and nitrogen is1 4

:32 28

= 7 : 32

87. Which one is classified as a condensation polymer ?(1) Teflon (2) Acrylonitrile (3) Dacron (4) Neoprene

87. 3Sol. Dacron is a condensation polymer.

n CH2OH CH2OH COOHHOOC OCH2 CH2O C

O

CO

O

Dacronn

n

Ethylene glycol

88. Among the following oxoacids, the correct decreasing order of acid strength is :(1) HClO4> HClO3> HClO2> HOCl (2) HClO2> HClO4 > HClO3> HOCl(3) HOCl > HClO2> HClO3> HClO4 (4) HClO4> HOCl > HClO2> HClO3

88. 1Sol. Among oxy acids of chlorine as the oxidation state increases the acidic character increases.

So the acidic strength is in the orderHClO4> HClO3> HClO2> HOCl

89. For complete combustion of ethanol, C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l), the amount of heatproduced as measured in bomb calorimeter, is 1364.47 kJ mol1 at 25C. Assuming ideality theEnthalpy of combustion, CH, for the reaction will be :(R = 8.314 kJ mol1)(1) 1460.50 kJ mol1 (2) 1350.50 kJ mol1(3) 1366.95 kJ mol

1 (4) 1361.95 kJ mol

1

89. 3Sol. In bomb calorimeter the volume is constant so the heat produced is

E = 1364.47 kJ/mole.H = ?n for the given reaction is 1H = E + n RT= 1364.47 + (1) (8.314 x 103) (298)

= 1366.95 kJ/mole

90. The most suitable reagent for the conversion of R CH2 OH R CHO is :(1) CrO3 (2) PCC (Pyridinium Chlorochromate)(3) KMnO4 (4) K2Cr2O7

90. 2Sol. The conversion

RCH2OH RCHO is possible with PCC, as it is a mild oxidising agentBut the remaining reagents CrO3, KMnO4, K2Cr2O7will convert RCH2OH to RCOOH

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JEE MAINS

2014 (Code : H)

32

JEE (MAINS)2014 CODE H

KEY

PHYSICS1. 3 2. 4 3. 1 4. 3

5. 3 6. 2 7. 3 8. 4

9. 2 10. 1 11. 2 12. 1

13. 3 14. 1 15. 1 16. 4

17. 2 18. 1 19. 4 20. 1

21. 1 22. 1 23. 4 24. 1

25. 4 26. 3 27. 1 28. Bonus

29. 2 30. 1

MATHEMATICS31. 1 32. 4 33. 1 34. 4

35. 2 36. 4 37. 4 38. 1

39. 2 40. 2 41. 2 42. 4

43. 3 44. 4 45. 4 46. 3

47. 4 48. 1 49. 4 50. 1

51. 3 52. 2 53. 4 54. 4

55. 3 56. 4 57. 3 58. 1

59. 1 60. 3

CHEMISTRY

61. 3 62. 4 63. 4 64. 3

65. 1 66. 3 67. 2 68. 3

69. 4 70. 4 71. 1 72. 4

73. 1 74. 3 75. 2 76. 4

77. 4 78. 2 79. 2 80. 2

81. C 82. 3 83. 3 84. 1

85. 3 86. 4 87. 3 88. 1

89. 3 90. 2

Jee Mains-2014 Solutions

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