JEE Mains 2014 Paper code G - Amazon Web...

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Vidyamandir Classes VMC/JEE Mains 1 2014 JEE Mains 2014 Paper code G 6 April 2014 | 9:30 AM – 12:30 PM

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JEE Mains 2014

Paper code G 6 April 2014 | 9:30 AM – 12:30 PM

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PART-A MATHEMATICS

1. If 1 and 2x x= − = are extreme points of ( ) 2= + +f x log x x xα β then :

(1) 1

62

,α β= − = − (2) 1

22

,α β= = −

(3) 1

22

,α β= = (4) 1

62

,α β= − =

Ans:

2. The locus of the foot of perpendicular drawn from the centre of the ellipse 2 23 6x y+ = on any tangent

to it is:

(1) ( )22 2 2 26 2x y x y− = − (2) ( )22 2 2 26 2x y x y+ = +

(3) ( )22 2 2 26 2x y x y+ = − (4) ( )22 2 2 26 2x y x y− = +

Ans:

3. Let ( ) ( )1 k kkf x sin x cos x

k= + where and 1x R k∈ ≥ . Then ( ) ( )4 6f x f x− equals :

(1) 1

3 (2)

1

4 (3)

1

12 (4)

1

6

Ans:

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4. If 4 3 1 :nX n n N= − − ∈ and ( ) 9 1 :Y n n N= − ∈ , where N is the set of natural numbers,

then X Y∪ is equal to :

(1) Y X− (2) X (3) Y (4) N

Ans:

5. If A is an 3 3× non-singular matrix such that 1and −= =AA' A' A B A A' , then BB' equals :

(1) I (2) 1B− (3) ( )1− ′B (4) I B+

Ans:

6. The integral

11

1+ + −

∫xxx e dx

xis equal to :

(1)

1xxx e c

++ (2) ( )

1

1xxx e c

++ +

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(3)

1xxx e c

+− + (4) ( )

1

1xxx e c

+− +

Ans:

7. The area of the region described by ( ) 2 2 2: 1 and 1A x,y x y y x= + ≤ ≤ − is :

(1) 4

2 3

π− (2)

2

2 3

π− (3)

2

2 3

π+ (4)

4

2 3

π+

Ans:

8. The image of the line1 3 4

3 1 5

x y z− − −= =

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

(1) 3 5 2

3 1 5

x y z+ − += =

− − (2)

3 5 2

3 1 5

x y z− − −= =

(3) 3 5 2

3 1 5

x y z− + −= =

− − (4)

3 5 2

3 1 5

x y z+ − −= =

Ans:

9. The variance of first 50 even natural numbers is :

(1) 833 (2) 437 (3) 437

4 (4)

833

4

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Ans:

10. If z is a complex number such that 2≥z ,then the minimum value of 1

2z + :

(1) lie in the interval ( )1 2, (2) is strictly greater than5

2

(3) is strictly grater than3

2but less than

5

2 (4) is equal to

5

2

Ans:

11. Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new

numbers are in A.P. Then the common ratio of the G.P. is :

(1) 3 2+ (2) 2 3− (3) 2 3+ (4) 2 3+

Ans:

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12. If the coefficients of 3 4andx x in the expansion of ( ) ( )1821 1 2ax bx x+ + − in power of x are both zero,

then (a, b) is equal to :

(1) 251

143

, (2)

27214

3

, (3)

27216

3,

(4) 251

163

,

Ans:

13. Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4 2 0ax ay c+ + = and

5 2 0bx by d+ + = lies in the fourth quadrant and is equitant from the two axes then:

(1) 2 3 0bc ad+ = (2) 3 2 0bc ad− = (3) 3 2 0bc ad+ = (4) 2 3 0bc ad− =

Ans:

14. If 2

a b b c c a a b cλ× × × =r r rr r r r r

then λ is equal to :

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

Ans:

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15. Let andA B be two events such that ( ) ( )1 1

6 4P A B , P A B= =U I and ( ) 1

4P A ,= where A stands for

the complement of the event A. Then the events andA B are:

(1) Equally likely but not independent (2) Independent but not equally likely

(3) Independent and equally likely (4) Mutually exclusive and independent

Ans:

16. Let PS be the median of the triangle with vertices ( ) ( ) ( )2 2 6 1 7 3, , , and , .P Q R− The equation of the

line passing through ( )1 1, − and parallel to PS is:

(1) 2 9 7 0x y+ + = (2) 4 7 3 0x y+ + = (3) 2 9 11 0x y− − = (4) 4 7 11 0x y− − =

Ans:

17 2

20

sin( cos )limx

x

x

π→

is equal to :

(1) 1 (2) π− (3) π (4) 2

π

Ans:

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18. Let andα β be the roots of equation 2 0, 0.px qx r p+ + = ≠ If p, q, r are in A. P. and 1 1

4α β

+ = ,

then the value of | |α β− is:

(1) 2 17

9 (2)

34

9 (3)

2 13

9 (4)

61

9

Ans:

19. A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is

45o. It flies off horizontally straight away from the point O. After one second, the elevation of the bird

from O is reduced to 30o. Then the speed (in m/s) of the bird is:

(1) 40( 3 2)− (2) 20 2 (3) 20( 3 1)− (4) 40( 2 1)−

Ans:

20. If a R∈ and the equation 2 23( [ ]) 2( [ ]) 0x x x x a− − + − + = (where [x] denotes the greatest integer

x≤ ) has no integral solution, then all possible values of a lie in the interval:

(1) (1, 2) (2) ( )2 1,− −

(3) ( , 2) (2, )−∞ − ∪ ∞ (4) ( 1, 0) (0, 1)− ∪

Ans:

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21. The integral 2

0

1 4sin 4sin2 2

x xdx+ −∫

π equals:

(1) 2

4 4 33

π− − (2) 4 3 4− (3) 4 3 4

3

π− − (4) 4π −

Ans:

22. If f and g are differentiable functions in [0, 1] satisfying f (0) = 2= g(1), g(0) = 0 and f (1) = 6, then for

some ] 0, 1[c∈ :

(1) ( ) ( )2 3f c g c′ ′= (2) ( ) ( )f c g c′ ′=

(3) ( ) ( )2f c g c′ ′= (4) ( ) ( )2 f c g c′ ′=

Ans:

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23. If g is the inverse of a function f and 5

1'( )

1f x

x=

+ , then ( )′g x is equal to :

(1) 45x (2) 5

1

1 ( )g x+ (3) 51 ( )g x+ (4) 51 x+

Ans:

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

(1) 441

100 (2) 100 (3) 110 (4)

121

10

Ans:

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25. If

( ) ( )( ) ( ) ( )( ) ( ) ( )

3 1 1 1 2

, 0, and ( ) and 1 1 1 2 1 3

1 2 1 3 1 4

n n

f f

f n f f f

f f f

+ +

≠ = + + + +

+ + +

α β α β 2 2 2(1 ) (1 ) ( ) ,K α β α β= − − − then K is

equal to :

(1) 1

αβ (2) 1 (3) 1− (4) αβ

Ans:

26. The slope of the line touching both the parabolas 2 24 and 32 :y x x y is= = −

(1) 3

2 (2)

1

8 (3)

2

3 (4)

1

2

Ans:

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27. The statement ~ ( ~ ) :p q is↔

(1) Equivalent to ~ p q↔ (2) A tautology

(3) A fallacy (4) Equivalent to p q↔

Ans:

28. Let the population of rabbits surviving at a time ‘t’ be governed by the differential equation

( ) 1( ) 200.

2

dp tp t

dt= − if p(0) = 100, then p(t) equals:

(1) /2300 200 te−− (2) /2600 500 te− (3) /2400 300 te−− (4) /2400 300 te−

Ans:

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29. Let C be the circle with centre at (1, 1) and radius =1. If T is the circle centred at (0, y), passing through

origin and touching the circle C externally, then the radius of T is equal to:

(1) 3

2 (2)

1

2 (3)

1

4 (4)

3

2

Ans:

30. The angle between the lines whose direction cosines satisfy the equations l + m + n = 0 and

2 2 2l m n= + is :

(1) 4

π (2)

6

π (3)

2

π (4)

3

π

Ans:

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PART-B PHYSICS

31. When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude 2F ax bx= +

where a and b are constants. The work done in stretching the unstretched rubber-band by L is :

(1) 2 31

2 2 3

aL bL +

(2) 2 3aL bL+ (3) ( )2 31

2aL bL+ (4)

2 3

2 3

aL bL+

Ans.(4)

( )2 3

2

0 02 3

L LaL bL

W Fdx ax bx dx= = + = +∫ ∫

32. The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 13 10 Am .−×

The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the

magnet gets demagnetized when inside the solenoid is :

(1) 6A (2) 30 mA (3) 60 mA (4) 3 A

Ans:(4)

3

0

3 10B

A / m n i× = =µ

⇒ i = 3A

33. 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 1 kW.

The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will

be :

(1) 14 A (2) 8 A (3) 10 A (4) 12 A

Ans.(4)

PiV

=

15 40 100 5 80 1 1000

5220 220 220 220

totalI× × ×

= + × + +

= 2500

11 36220

.=

Hence, Imin = 12A.

34. 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) 6 cm (2) 16 cm (3) 22 cm (4) 38 cm

Ans.(2)

0 76P =

76P x= −

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Assuming isothermal condition

( )0 8 54P P x× = × −

⇒ ( )( )76 8 76 54x x× = − −

⇒ 38x cm=

Length of air column above mercury = 54 38 16− = .

35. A bob of mass m attached to an inextensible string of length l is suspended from a vertical support.

The bob rotates in a horizontal circle with an angular speed rad / sω about vertical. About the point of

suspension:

(1) Angular momentum change both in direction and magnitude.

(2) Angular momentum is conserved.

(3) Angular momentum changes in magnitude but not in direction.

(4) Angular momentum changes in direction but not in magnitude.

Ans.(4)

Angular momentum changes direction as shown in figure

36. The current voltage relation of diode is given by ( )1000 1V / TI e mA,= − where the applied voltage V is

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

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

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

Ans.(2)

( )10001

V TI e mA= −

( )10001000 10001V TdI e dV I dV

T T= = +

( )1000

5 1 0 01300

. mA= + ×

= 0.2 mA

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

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

The relation between H, u and n is :

(1) ( ) 22gH n u= − (2) 2 g H = n2u2

(3) ( )2 22g H n u= − (4) ( )22 2g H nu n= −

Ans.(4)

Lr L

r

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21

2H ut gt− = −

Given that u

t n .g

=

2 2

22

nu g n uH u .

g g− = × −

2 2 2

2

nu n uH

g g− = −

⇒ 2 2 22 2gH n u nu= − ⇒ ( )22 2gH nu n= −

38. A thin convex lens made from crown glass 3

2µ =

has focal length f. When it is measured in two

different liquids having refractive indicates 4

3 and

5

3 it has the focal length f1 and f2 respectively.

The correct relation between the focal lengths is :

(1) f1 and f2 both become negative (2) 1 2f f f= <

(3) 1f f> and f2 becomes negative (4) 2f f> and f1 becomes negative

Ans.(3)

2µ µ>

1µ µ<

Hence f2 is negative, f1 is positive.

39. A 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

43 10 V / m,× charge density of the positive plate will be close to :

(1) 4 26 10 C / m× (2) 7 26 10 C / m−×

(3) 7 23 10 C / m−× (4) 4 23 10 C / m×

Ans.(2)

0

QE

AK=

4 120 3 10 8 85 10 2 2

QE K . .

A

−= ∈ = × × × ×

7 26 10 C / m−×

40. In the circuit shown here the point ‘C’ is kept connected to point

‘A’ till the current flowing through the circuit becomes constant.

Afterward, suddenly, point ‘C’ is disconnected form point ‘A’

and connected to point ‘B’ at time t = 0. Ratio of the voltage

across resistance and the inductor at t = L/R will be equal to:

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(1) 1 e

e

− (2)

1

e

e− (3) 1 (4) 1−

Ans.(4)

0 1R

L RL

VV V

V+ = ⇒ = − .

41. Two beams, A and B of plane polarized light with mutually perpendicular planes of polarization are

seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has

zero intensity), a rotation of polariod through 30° makes the two beams appear equally bright. If the

initial intensities of the two beams are IA and IB respectively, then A

B

I

I equals :

(1) 1

3 (2) 3 (3)

3

2 (4) 1

Ans.(1)

3 230 60A BI cos I cos=

2

2

60 1 4 1

3 4 330

A

B

I cos

I cos= = =

42. There is a circular tube in a vertical plane. Two liquids which do

not mix and of densities d1 and d2 are filled in the tube. Each

liquid subtends 90° angle at centre. Radius joining their

interface makes an angle α with vertical. Ratio 1

2

d

d is :

(1) 1

1

sin

cos

αα

+−

(2) 1

1

sin

sin

αα

+−

(3) 1

1

cos

cos

αα

+−

(4) 1

1

tan

tan

αα

+−

Ans.(4)

Balancing torque about centre

1 24 4

m g R sin m gR sinπ π − α = + α

1 12

Rm d A

π= ρ

⇒ 1

2

4

4

sind

dsin

π + α =π + α

= 1

1

tan

tan

+ α− α

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43. The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant

when its temperature is raised by 100 C° is : (For steel Young’s modulus is 11 22 10 N m−× and

coefficient of thermal expansion is 5 11 1 10 ). K− −×

(1) 62 2 10. Pa× (2) 82 2 10. Pa×

(3) 92 2 10. Pa× (4) 72 2 10. Pa×

Ans.(2)

00

F

AYα θ∆ =

ll

⇒ F

YA

α θ= ∆

11 52 10 1 1 10 100P . −= × × × ×

82 2 10. Pa= ×

44. 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 placed

without slipping is :

(1) 1

2m (2)

1

6m (3)

2

3m (4)

1

3m

Ans.(2)

At maximum height block will be in limiting condition.

tanθ µ=

223 1

0 5 16 6

dy x. x y

dxµ= ⇒ = ⇒ = ⇒ =

45. 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 100 C° where as ends of brass

and steel are kept at 0 C.° 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) 6.0 cal/s (2) 1.2 cal/s (3) 2.4 cal/s (4) 4.8 cal/s

Ans.(4)

100 0

B S

CuB S

th thth

th th

C CH

R RR

R T

−=

++

o o

where thRKA

=l

= 4.8 Cal/s

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46. A mass m is supported by a massless string

wound around a uniform hollow cylinder of

mass m and radius R. If the string does not slip

on the cylinder, with what acceleration will the

mass fall on release?

(1) g (2) 2

3

g

(3) 2

g (4)

5

6

g

Ans.(3)

mg T ma− = . . . .(i)

2TR mR α= . . . .(ii)

a R= α . . . .(iii)

Solving equations (i), (ii) and (iii)

2

ga =

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

option from the choices given below the lists:

List I List 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) Ultraviolet rays (iv) Absorbed by the ozone layer of the atmosphere

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

(1) (i) (ii) (iii) (iv)

(2) (iv) (iii) (ii) (i)

(3) (i) (ii) (iv) (iii)

(4) (iii) (ii) (i) (iv)

Ans. (1)

48. The radiation corresponding to 3 2→ transition of hydrogen atom falls on a metal surface to produce

photoelectrons. These electrons are made to enter a magnetic field of 43 10 T−× . If the radius of the

largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to:

(1) 1.6 eV (2) 1.8 eV (3) 1.1 eV (4) 0.8 eV

Ans.(3)

max

mVR , E W K

qB= = +

2 2 2 2 221

2 2 2max

m V q R BK mV

m m= = =

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( )2

219 4

31

101 6 10 3 10

10000 8

2 9 1 10

.

eV . eV.

− −

× × × ×

= =× ×

2 2

1 113 6 1 9

2 3E . . eV

= −

1 9 0 8 1 1maxW E K . . . eV= − = − =

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

(1) Both electric and magnetic energy densities are zero

(2) Electric energy density is double of the magnetic energy density

(3) Electric energy density is half of the magnetic energy density

(4) Electric energy density is equal to the magnetic energy density

Ans.(4)

50. A green light is incident from the water to the air-water interface at the critical angle ( )θ . Select the

correct statement.

(1) The entire spectrum of visible light will come out of the water at various angles to the normal

(2) The entire spectrum of visible light will come out of the water at an angle of 90o to the normal

(3) The spectrum of visible light whose frequency is less than that of green light will come out to

the air medium

(4) The spectrum of visible light whose frequency is more than that of green light will come out to

the air medium

Ans (3)

if g g gv v C C< ⇒ µ < µ ⇒ > .

Hence for gv v i C< ⇒ < So they will refract into air.

51. Four particle, each of mass M and equidistant from each other, move along a circle of radius R under

the action of their mutual gravitational attraction. The speed of each particle is:

(1) ( )11 2 2

2

GM

R+ (2)

GM

R

(3) 2 2GM

R (4) ( )1 2 2

GM

R+

Ans.(1)

2

22

F MVF

R+ =

( )

2 2

2

12

2 2

GM MV

RR

+ =

⇒ ( )12 2 1

2

GMV

R= +

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52. A particle moves with simple harmonic motion in a straight line. In first sτ , after starting from rest it

travels a distance a, and in next sτ it travels 2a, in same direction, then:

(1) time period of oscillations is 6τ (2) amplitude of motion is 3a

(3) time period of oscillations is 8τ (4) amplitude of motion is 4a

Ans.(1)

x Acos tω=

A a Acosωτ− =

3 2A a Acos− = ωτ

22 2 1cos cosωτ ωτ= −

23

2 1A a A a

A A

− − = × −

⇒ 2A a= . Hence 66

TT= ⇒ =τ τ

53. A conductor lies along the z-axis at

1 5 1 5. z . m− ≤ < and carries a fixed current of 10.0

A in za− direction (see figure). For a field

4 0 23 0 10

. xyˆB . e a T

− −= ×r

, find the power required

to move the conductor at constant speed to x = 2.0

m, y = 0m in 35 10 s−× . Assume parallel motion

along the x-axis.

(1) 29.7 W (2) 1.57 W (3) 2.97 W (4) 14.85 W

Ans.(3)

2

0

W F dx Bi l dx= =∫ ∫

=

24 0 2

0

3 10 10 3 . xe dx− −× × × × ∫

= ( )3

0 49 101

0 2

.e.

−−×

( )3

0 4

3

9 101

0 2 5 10

.WP e

t .

−−

×= = −

∆ × ×

= ( )0 49 1 .e−− (Use 2

12

x xe x= + + )

= 2.97 W

54. The forward biased diode connection is :

(1) (2)

(3) (4)

Ans.(2)

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55. Hydrogen ( )11H , Deuterium ( )21 ,H singly ionised Helium ( )42He+

and doubly ionised lithium

( )63Li++ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1.

If the wave lengths of emitted radiation are , ,λ λ λ1 2 3 and λ4 respectively then approximately which

one of the following is correct ?

(1) 1 2 3 42 3 4λ λ λ λ= = = (2) 1 2 3 44 2 2λ λ λ λ= = =

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

Ans.(4)

2

2 2

1 1 1

1 2RZ

= − λ

⇒ 2Z Kλ =

⇒ ( ) ( ) ( ) ( )1 2 3 41 1 4 9λ = λ = λ = λ

56. 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 < < R, and the

surface tension of water is T, value of r just before

bubbles detatch is : (Density of water is wρ )

(1) 2 3 w gRT

ρ (3) 2

3

w gRT

ρ

(3) 2

6

w gRT

ρ (4) 2 w gR

T

ρ

Ans.(None of these)

The bubble will detach when TB F sin= θ

⇒ ( )242

3w

rT R g T R

Rπ = π

⇒ 2 2

3

wTr R gT

=

57. 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) 4 (2) 12 (3) 8 (4) 6

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Ans.(4)

0 1004

Cv Hz= =

l.

Possible frequencies 100 300 500 700 900 1100 300, , , , , ,≡

Hence 6 possible natural oscillations lie below 1250 Hz.

58. Assume that an electric field 230 ˆE x i→

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

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

(1) 80J (2) 120J (3) 120J− (4) 80J−

Ans.(4)

02

0

2

30AV V x dx− = ∫

= 80 V−

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

measure it?

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

(B) A meter scale

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

and main scale has 10 division in 1 cm

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

Ans.(3)

As the student measured length as 3.50 cm hence the least count of the instrument should be .01 cm which is the

LC of option (3).

60. One mole diatomic ideal gas undergoes a cyclic process ABC as shown in the figure. The process BC is

adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct

statement:

(A) The change in internal energy in the process BC is 500R−

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

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

(D) The change in internal energy in the process AB is 350R−

Ans.(1)

( ) ( )51 100 800

2BC vU nC T R

∆ = ∆ = × −

= 500 R−

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PART-C CHEMISTRY

61. Which one is classified as a condensation polymer?

(1) Acrylonitrile (2) Dacron (3) Neoprene (4) Teflon

Ans.(2)

Dacron is a condensation polymer.

62. Which one of the following properties is NOT shown by NO?

(1) It’s bond order is 2.5

(2) It is diamagnetic in gaseous state

(3) It is a neutral oxide

(4) It combines with oxygen to form nitrogen dioxide

Ans.(2)

• Bond order of NO is 2.5

• Paramagnetic (odd electron molecule)

• Neutral oxide

• Combines with O2 to form NO2

2 21

NO O NO2

+ →

63. Sodium phenoxide when heated with CO2 under pressure at 125 Co yields a product which on

acetylation produces C.

The major product C would be :

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

Ans.(2)

[Aspirin]

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64. Given below are the half-cell reactions :

2Mn 2e Mn; E 1.18 V+ −+ → = −o

3 22(Mn e Mn ) ; E 1.51 V+ − ++ → = +o

The Eo for 2 33Mn Mn 2Mn+ +→ + will be:

(1) 0.33V− ; the reaction will occur (2) 2.69 V− ; the reaction will not occur

(3) 2.69 V− ; the reaction will occur (4) 0.33 V− ; the reaction will not occur

Ans.(2)

(a) 2

Mn 2e Mn+ −+ → E 1.18 V° = − : G 2.36 F∆ ° =

(b) 3 2

Mn e Mn+ − ++ → E 1.51V° = + : G 1.51 F∆ ° = −

Required : 2 3

3Mn Mn 2Mn+ +→ + [ 2e− transfer]

⇒ ( ) ( )a 2 b−

( )G 2.36 F 2 1.51 F∆ ° = − −

G 5.38 F∆ ° =

∴ G 5.38

E 2.69 VnF 2F

∆ °° = − = − = − [Non-spontaneous]

65. For complete combustion of ethanol, 2 5 2 2 2C H OH( ) 3O (g) 2CO (g) 3H O( )+ → +l l , the amount

of heat produced as measured in bomb calorimeter, is 11364.47kJ mol at 25 C− o . Assuming ideality the

enthalpy of combustion, cH∆ , for the reaction will be : 1(R 8.314 kJ mol )−=

(1) 11350.50 kJmol−− (2) 11366.95kJmol−−

(3) 11361.95 kJmol−− (4) 11460.50 kJmol−−

Ans.(2)

U 1364. 47KJ /mol∆ = −

gn 1∆ = −

gH U n RT∴ ∆ = ∆ + ∆

1364.47 (8.314) (298)= − −

1366.95KJ / mol= −

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

evolved ammonia was absorbed in 60 mL of M

10 sulphuric acid. The unreacted acid required 20 mL of

M

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

(1) 5% (2) 6% (3) 10% (4) 3%

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Ans.(3)

Millimoles of 2 4H SO taken 1

60 610

= × =

m.moles of NaOH required for black titration = 2

∴ m. moles of 2 4H SO left = 1

∴ m.moles of 2 4H SO consumed = 5.

Each mole 2 4H SO consumes 2 ammonia moles.

∴ 10 m.moles of nitrogen must be present in the sample.

10 14

Mass of nitrogen present in the sample gm 0.14 gm1000

×∴ = =

0.14

% age N 10%1.4

⇒ = =

67. The major organic compound formed by the reaction of 1, 1, 1-trichloroethane with silver powder is:

(1) 2-Butene (2) Acetylene (3) Ethene (4) 2-Butyne

Ans.(4)

3 3 32 Butyne

Cl|

2 CH C Cl 6Ag CH C C CH 6AgCl|Cl

− − + → − ≡ − +

68. The ratio of masses of oxygen and nitrogen in a particular gaseous mixtures is 1 : 4. The ratio of

number of their molecules is:

(1) 3 : 16 (2) 1 : 4 (3) 7 : 32 (4) 1 : 8

Ans.(3)

2

2

O

N

g 1

g 4=

∴ Molar ratio = 2

2

O

N

n

n=

2

2

2 N2

O

O

N

g

g32 28 1 28.

g g 32 4 32 32

28

7 = × = =

69. The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is:

(1) Cr (2) Ag (3) Ca (4) Cu

Ans.(3)

Calcium, being a more reactive metal, cannot be obtained by electrolysis of its salt as H+ will be reduced in

preference to Ca2+.

70. The equivalent conductances of NaCl at concentration C and at infinite dilution are C and ∞λ λ ,

respectively. The correct relationship between C and ∞λ λ is given as: (where the constant B is

positive)

(1) ( )C B C∞λ = λ + (2) ( )C B C∞λ = λ +

(3) ( )C B C∞λ = λ − (4) ( )C B C∞λ = λ −

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Ans.(4)

C B C∞λ = λ −

71. The correct set of four quantum numbers for the valence electron of Rubidium atom (Z = 37) is:

(1) 1

5, 0, 1,2

+ (2) 1

5, 0, 0,2

+

(3) 1

5, 1, 0,2

+ (4) 1

5, 1, 1,2

+

Ans.(2)

Rb atom : Valence electron belongs to 5s orbital

n 5

0

m 0

s 1/ 2

=

=∴

=

=+

l

72. Consider separate solutions of 0.500 M 2 2C H OH(aq) 0.100 M, 3 4 2Mg (PO ) (aq), 0.250 M KBr(aq)

and 0.125 M 3 4Na PO (aq) at 25 C° . Which statement is True about theses solutions, assuming all salts

to be strong electrolytes ?

(1) 2 20.500 M C H OH(aq) has the highest osmotic pressure

(2) They all have the same osmotic pressure

(3) 0.100M 3 4 2Mg (PO ) (aq) has the highest osmotic pressure

(4) 0.125 M Na3PO4 (aq) has the highest osmotic pressure

Ans.(2)

2 50.5M C H OH 0.5 RT⇒ π =

3 4 20.1 M Mg (PO ) iCRT (5 0.1) RT 0.5 RT⇒ π = = × =

0.250 M KBr iCRT (2 0.250)RT 0.5 RT⇒ π = = × =

4 40.125 M Na PO iCRT (4 0.125) RT 0.5 RT⇒ π = = × =

73. The most suitable reagent for the conversion of 2R CH OH R CHO− − → − is :

(1) PCC (Pyridinium Chlorochromate) (2) 4KMnO

(3) K2Cr2O7 (4) 3CrO

Ans.(1) PCC

2R CH OH R CHO− − → −

74. CsCl crystallizes in body-centred cubic lattice. If ‘a’ is its edge length then which of the following

expressions is correct?

(1) Cs Clr r 3a+ −+ = (2)

Cs Clr r 3a+ −+ =

(3) Cs Cl

3ar r

2+ −+ = (4)

Cs Cl

3r r a

2+ −+ =

Ans.(4)

CsCl is BCC .

∴ Cs C

2r 2r 3 a+ −+ =l

Cs C

3r r a

2+ −+ =

l

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75. In which of the following reactions 2 2H O acts 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) b, d (2) a, b (3) c, d (4) a, c

Ans.(1)

(B) 1 2

22 2H O 2e O 2H−

− +− → +

(D) 1 0

2 2 2 2H O 2OH 2e O H O−

− −+ − → +

In both of these reactions H2O2 is acting as a reducing agent as oxygen is getting oxidized.

76. For which of the following molecules, significant 0 ?µ ≠

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

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

Ans.(1)

are “Book – shaped” (Just like that of H2O2) ∴ 0µ ≠

77. On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic

compound formed is :

(1) An alkyl isocyanide (2) An alkanol

(3) An alkanediol (4) An alkyl cyanide

Ans.(1)

2 3(Alkyl isocyanide)

R NH CHCl alc KOH R NC− + + → −

78. In NS 2 reactions, the correct order of reactivity for the following compounds :

3 3 2 3 2 3 3CH Cl, CH CH Cl,(CH ) CHCl and (CH ) CCl is :

(1) 3 2 3 2 3 3 3(CH ) CHCl CH CH Cl CH Cl (CH ) CCl> > >

(2) 3 3 2 3 2 3 3CH Cl (CH ) CHCl CH CH Cl (CH ) CCl> > >

(3) 3 3 2 3 2 3 3CH Cl CH CH Cl (CH ) CHCl (CH ) CCl> > >

(4) 3 2 3 3 2 3 3CH CH Cl CH Cl (CH ) CHCl (CH ) CCl> > >

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Ans.(3)

Reactivity towards NS 2 .

( )3 3 2 3 3 3

3

CH Cl CH CH Cl CH CH Cl CH CCl

|CH

> − − > − − >

79. The octahedral complex of a metal ion 3M + with four monodentate ligands 1 2 3 4L ,L ,L and L absorb

wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand

strength of the four ligands is :

(1) 1 2 4 3L L L L< < < (2) 4 3 2 1L L L L< < <

(3) 1 3 2 4L L L L< < < (4) 3 2 4 1L L L L< < <

Ans.(3)

Greater the absorbed wavelength

⇒ Lesser the energy gap between t2g and eg

⇒ Lesser will be the strength of ligand.

Hence, L1 < L3 < L2 < L4

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

(1) 4 2 4 2 4CuSO 4KCN K [Cu(CN) ] K SO+ → +

(2) 2 2Li O 2KCl 2LiCl K O+ → +

(3) 23 5 4[CoCl(NH ) ] 5H Co 5NH Cl+ + + + −+ → + +

(4) excess NaOH2 4 2

2 6 2[Mg(H O) ] (EDTA) [Mg(EDTA)] 6H O+ − ++ → +

Ans.(3)

Options (1), (2) and (4) are not correct. So, by elimination the best answer will be (3).

81. In the reaction, 4 5LiAIH PCl Alc.KOH

3CH COOH A B C,→ → → the product C is :

(1) Acetyl chloride (2) Acetaldehyde (3) Acetylene (4) Ethylene

Ans.(4)

54 PClLiAlH alc.KOH3 3 2 3 3 2 2CH COOH CH CH OH CH CH Cl CH CH→ − − → − − → =

82. The correct statement for the molecule, 3,CsI is :

(1) it contains Cs , I+ − and lattice I2 molecule (2) it is a covalent molecule

(3) it contains 3Cs and I+ − ions (4) it contains 3Cs and I+ − ions

Ans.(3)

3 3CsI contains Cs & I+ −

83. For the reaction 2(g) 2(g) 3(g)

1SO O SO ,

2+ If x

P CK K (RT)= Where the symbols have usual

meaning then the value of x is : (Assuming ideality)

(1) 1 (2) 1− (3) 1

2−

(4)

1

2

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Ans.(3)

( ) ( ) ( )2 2 3

1SO g O g SO g

2+ →

∴ g

1n

2∆ = −

84. For the non-stoichiometric reaction 2A B C D,+ → + The following kinetic data were obtained in

three separate experiments, all at 298 K.

Initial Concentration (A) Initial Concentration (B) Initial rate of formation of C

(mol − −1 1L S )

0.1 M

0.1 M

0.2 M

0.1 M

0.2 M

0.1 M

31.2 10−×

31.2 10−×

32.4 10−×

The rate law for the formation of C is:

(1) dc

k[A]dt

=

(2) dc

k[A][B]dt

= (3) 2

dck[A] [B]

dt= (4) 2

dck[A][B]

dt=

Ans.(1)

First order w.r.t. A & zero order w.r.t. B

∴ [ ]dcK A

dt=

85. Resistance of 0.2 M solution of an electrolyte is 50Ω . The specific conductance of the solution is

1.4 S 1m− . The resistance of 0.5 M solution of the same electrolyte is 280Ω . The molar conductivity

of 0.5 M solution of the electrolyte in 2 1Sm mol− is :

(1) 25 10× (2) 45 10−× (3) 35 10−× (4) 35 10×

Ans.(2) *G.G =κ

Case1 *1. G 1.4

50=

* 1G 70 m−=

Case 2 *new

1G

280× = κ

1 1new 0.25 Sm 0.0025 Scm− −∴ κ = =

2 11000 0.0025 10005.0 Scm mol

C 0.5

−κ × ×∴ λ = = =

4 2 15 10 S.m . mol− −= ×

86. Among the following oxo-acids, the correct decreasing order of acid strength is :

(1) 2 4 3HClO HClO HClO HOCl> > > (2) 2 3 4HOCl HClO HClO HClO> > >

(3) 4 2 3HClO HOCl HClO HClO> > > (4) 4 3 2HClO HClO HClO HOCl> > >

Ans.(4)

4 3 2HClO HClO HClO HOCl> > >

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87. Which one of the following bases is not present in DNA ?

(1) Thymine (2) Quinoline (3) Adenine (4) Cytosine

Ans.(2)

Quinoline is not included in DNA

Thymine / Guanine / Adenine / Cytosine are the four bases that are present in DNA.

88. Considering the basic strength of amines in aqueous solution , which one has the smallest pKb value ?

(1) 6 5 2C H NH (2) 3 2(CH ) NH

(3) 3 2CH NH (4) 3 3(CH ) N

Ans.(2)

Basic strength in aqueous medium :

3 2 3 2 3 3 2(CH ) NH CH NH (CH ) N Ph NH> > >

89. If Z is a compressibility factor, Van der Waals equation at low pressure can be written as :

(1) Pb

Z 1RT

= +

(2) RT

Z 1Pb

= +

(3) a

Z 1VRT

= − (4)

PbZ 1

RT= −

Ans.(3)

At low pressures : a

Z 1VRT

= −

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

(1) o o

2

4

O , heat CO, 600 C CO, 700 C

3Fe Fe O FeO Fe→ → →

(2) 2 4 2 4 2

4

dil H SO H SO , O heat

2 4 3Fe FeSO Fe (SO ) Fe→ → →

(3) 2 2 4O , heat dil H SO heat

4Fe FeO FeSO Fe→ → →

(4) 2Cl , heat heat, air Zn

3 2Fe FeCl FeCl Fe→ → →

Ans.(1)

2O /Heat CO,600 C CO, 700 C3 4Fe Fe O FeO Fe

° °→ → →