700000712 IITJEEMains2014 CodeE Solution

IITJEE MAIN | PHYSICS, CHEMISTRY AND MATHEMATICS

Paper – 2014

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IITJEE (MAIN) – 2014

(Physics, Chemistry and Mathematics)

Code - E

SOLUTIONS

PART–A : PHYSICS

1. Correct option (1)

The current voltage relation is I = (e 1000V/T – 1) m A …… (1)

1000 V/ T

1000 V/ T

When I 5mA

5 mA = (e 1)

e 6mA

Differentiating equation (1), we get

1000 V/ T

1000 V/ T

1000dI = (e ) dV

T

On substituting the values, e 6 mA ,T= 300K , dV= 0.01

1000= (6 mA) (0.01)

300

=0.2 mA

Therefore the error in the value of current is m A = 0.2 mA.


H is the height of the tower.


Paper – 2014


1

2 2

2

uTime taken by the particle to reach highest point of its path is t

g

We know that v u 2gh

So speed on reaching ground ,v = u 2gh

Now v = u + at ( for particle thrown upwards)

v = -u + gt (For spee

2

2

d of particle reaching to ground)

u 2gh -u + gt

On solving, we get

u u 2gHt =

g

2

1

2

2

Given that time taken by particle to hit ground is n times that

taken by it to reach the highest point of its path.

u u 2gHt = = n t

g

u u 2gH u = n

g g

On solving the equation, we get

2gH = n(n-2)u

3. Correct Option (2)


Paper – 2014


2

2

Acceleration a = R

Now, For the block of mass m, mg -T = ma ---------- equation(1)

Tension T perpendicular distance R I

i e T R mR ......(2)

aT R mR ma R

R

or T = m a ......(3)

Substituting equation

(3) in equation (1), we get

mg -T = ma

mg -ma = ma

mg = 2 ma

ga=

2


3

2 2

2

xGiven y =

6

dy x xOn differentiating 3 =

dx 6 2

dy x

dx 2

Let be the angle at which block start slipping.

Under the condition of limiting equilibrium, we have

tan


Paper – 2014


2

3

where is the coefficient of friction = 0.5

From the figure, we get that

dytan =

dx

Substituting in tan

dy0.5

dx

x0.5 =

2

x = 1

x 1Now, putting value of x in y = ,y

6 6

The maximum horizontal dis

tance from origin where the block can be placed

without slipping is x=1

Hence, vertical height

1y =

6


2

L2

0

L L2

0 0

2

Let the rubber band be stretched by a distance 'x'.

The magnitude of restoring force is F =ax bx

The work done in stretching the rubber-band

dW = F.dx

W (ax bx ) dx

ax dx + bx dx

L a

3

2 3

L + b

2 3

aL bLThe work done in stretching the rubber-band by L is +

2 3


Paper – 2014



Conical pendulum is rotating with an angular velocity ω with the vertical axis.

Torque changes angular momentum

radial

Angular momentum of the pendulum about the suspension point 'O' is

L = m(r v)

r can be resolved into two components, radial components and axial components.

Due to r , L will be axial and due

axial to r , L will be radially outwards.

Here the angular momentum is not conserved.

So magnitude of net angular momentum will be constant |L| = |m(r × v)| but its direction

changes as shown in the figure.


Let the four particles of mass M be placed at the four corners of a square as they are

equidistant from each other as shown in the figure.

The ‘x’ be the distance between the masses and using Pythagoras theorem we can find x as

2 2 2x R R

x 2 R

Under the action of mutual gravitational attraction, they move along the circle of radius R.

The necessary centripetal force of attraction is provided by gravitational force.

Let us consider a single particle acted upon by three other adjacent particles.


Paper – 2014


The two forces acting on that will be the sum of resultant of the force due to adjacent two

masses and that of the third one diagonally opposite to it.

2

diagonally opposite mass

2 2

2

2

2 2

2 2

2

diagonally opposite mass 2

2 2 2

22

MvF' F

R

F' = F F 2FFcos90

F' = 2 F

GM = 2

x

GM GMF' = 2 2

2R2R

GMF

2R

GM GM Mv2

2R R2R

2 2

2

GM 1 1 Mv

R 4 R2

On solving for v, we get

GM 2 4v=

R 4 2

1 GM = (1+2 2 )

2 R


As length is kept constant, we have

11

LStrain = Q

L

Strain = Q

Coefficient of thermal expansion

and Q Temperature change

Thermal stress generated = strain

Pressure =stress = strain

= 2 x 10 Q

11 -5

8

8

= 2 x 10 1.1 x 10 100

= 2.2 10 Pa

Pressure to be applied = 2.2 10 Pa


Paper – 2014



On equating the pressure at A from both L.HS and R.H.S, we get

1 1 2 2 3 2 4 1

1 1 4 1 2 2 3 2

1

1

2

3

3

4

1 1 2 2

h d g h d g h d g h d g

h d g h d g h d g h d g ......(1)

h R R cos(90 )

h (R R sin )

h R sin

h R sin(90 )

h =Rcos

h =R-R cos

Substituting in (1), we get

(R R sin )d g (R-R cos )d g R sin d g Rcos d

1 2

1

2

g

(R cos R sin )d g (Rcos R sin )d g

d cos sin 1 tan =

d cos sin 1 tan

10. Correct option- No option matches the correct answer

When the bubble detaches, force of surface tension will act on complete circumference and

the resultant will be in the downward direction.

The Bubble will detach if


Paper – 2014


Buoyant force ≥ Surface Tension Force

3w

Buoyant force = density (cross sectional area) g

4Buoyant force R g

3

rFrom figure, sin =

R

Surface Tension Force = T dlsin

r = T 2 r

R

On equating both f

3w

42 w

2 w

2 w

orces, we get

4 rR g T 2 r

3 R

2R gr

3T

2 gr R

3T

2 gThe value of r just before bubbles detach is R .

3T


2c B S

C B S

C B S

1 2

Given

A A A 4 cm

L 46 cm L 13 cm L 12 cm

K 0.92 K 0.26 K 0.12 (in CGS units)

KA(T T )Q

l


Paper – 2014


c B S

C C 1 2 S S 1 2B B 1 2

C B S

c

Now,

Q Q Q

K A (T T ) K A (T T )K A (T T ) =

L L L

0.92 4(100 T) 0.26 4(T 0) 0.12 4(T 0) =

46 13 12

200 - 2T = 2T +T

T = 40 C

Hence, rate of heat flow through copper rod is

0.92 4(100 40)Q 4

46

.8 cal/s


Change in internal energy is given as,

fU = nR T

2

For diatomic gas molecule, f = 5

For cyclic process U = 0

Process CA:

5U = 1 R 400 600 U =-500 R

2

Process AB:

5U = 1 R 800 400 U =+1000 R

2

Process BC:

5

U = 1 R 600 800 U =-500 R2



Paper – 2014


1 1 2 2

1 atm

2 atm

2

Let length of the air colunm above mercury in the tube be x.

For the air colunm trapped in tube,

P V P V ......(1)

P P gh g 76

P P g 54 x

= g 76 g 54 x

P = g 22 x

1

2

V Area of cross section length of air colunm

V A 8

V A x

2

2

2

Now substituting all these values in eq(1), we get

76 g 8A g 22 x xA

608 gA 22 gAx x gA (Divide by gA)

x 22x 608 0

x 22x 38 16 0

Solving this, we get

x =16 cm


x Acos t The particle starts from rest

Displacement in time t is = A - x =A -Acos t

For t = :

a =A -Acos

a = A 1-cos ......(1)

For t = 2 + :

2a+a =A -Acos2

3a = A 1-cos2 ......(2)

2 2

Now, by taking ratio of eq(1) and (2), we get

1-cos 1

1-cos2 3

1-cos 1-cos 1

2sin 32 1-cos


Paper – 2014


2

Let cos x

1-x 1-x 1

2 1-x 1 x 32 1-x

1 1

2 1+x 3

2 2x 3

1x cos

2

2

3 T 3

T 6

Hence,time period of the motion is 6 .

From (1);

1a = A 1- A=2a

2

Hence, amplitude of the motion is 2a.

15. Correct option: (3)

2n 1 vf

4L

L 85 cm =0.85 m

For frequencies below 1250 Hz

2n 1 vf 1250 Hz

4L

2n 1 3401250 Hz

4 0.85

2n 1 12.5 Hz

n 6.75

n 6

Number of possible natural oscillations having frequency below 1250 Hz is 6.


Paper – 2014



A

0

2

V 22

V 0

23

A 0

0

A 0

Potential difference dV = -E.dx

E 30x i

dV = -E.dx

dV 30x dx

xV V 30

3

V V 80 J


0

0

0

4 12

7 2

EE

K K

EK

3 10 2.2 8.85 10

6 10 C/m


Power P = VI

Total power:

P = 15 40 5 100 5 80 1 1000

P VI 2500 W

2500 I 11.36 A

220

Hence, minimum capacity of the main fuse should be 12 A.


2

ext0

ext

2

ext0

Work WPower

Time t

W F .dx

F BIL

1P F .dx

t


Paper – 2014


24 0.2x

0

24 0.2x

3 0

20.2x

0

20.2x

0

0.4

1P 3 10 e 10 3 dx

t

1 = 3 10 e 10 3 dx

5 10

=1.8 e dx

e =1.8

0.2

=9 1 e

P 2.97 W


3 1 3 1The coerctivity of 3 10 Am of a magnet implies that magnetic intensity H = 3 10 Am

is required in opposite direction to demagnetise it.

-1

Number of turns per unit length of the solenoid

100n = 1000 turns m

0.1 m

0

0

3

For solenoid

B = H

Now, B = nI

H = nI

H 3 10I = 3 A

n 1000


After changing the switch, the circuit will acts as an L-R discharging circuit.

Applying Kirchhoff’s loop equation in the circuit, we get

R L

R L

R

L

V V 0

V V

V1

V


During the propagation of electromagnetic waves in a medium, the energy density of both

the electric as well as magnetic field is the same.


Paper – 2014



According to the Lens Maker’s formula

21

1 2

1 2 1 2

1 2 1 2

1 1 1n 1

f R R

When crown glass is kept in air, its focal length is f

1 3/2 1 1 3 2 1 11

f 1 R R 2 R R

1 1 1 1 1 1 1 1 Let

f 2 R R 2R R R R

1 1 2 1 2

1 1 2

f 2R ...... (1)

4When crown glass is kept in liquid with n , we get

3

1 3/2 1 1 9 1 11 1

f 4/3 R R 8 R R

1 1 1 1

f 8 R R

1

1

8R

f 8R 4f ...... (2)

2 1 2 1 2

2 1 2

2

5When crown glass is kept in liquid with n , we get

3

1 3/2 1 1 9 1 11 1

f 5/3 R R 10 R R

1 1 1 1 1

f 10 R R 10R

f 10R 5f .

1 2

..... (3)

From (1), (2) and (3), we get

f f , and f becomes negative


We know that, whether the light will refract out to the second medium or totally internally

reflect back in the first medium depends on the critical angle.

Now, critical angle for water to air medium is given as

c

water

1sin

n

We know that, for a light of higher wavelength, i.e., lower frequency the refractive index is

less. This will make c greater for these light rays.

When these light rays travel from water to air, they do not suffer total internal reflection

and refract out in the air medium.


Paper – 2014



The initial intensities of the two beams are IA and IB respectively.

The two beams have mutually perpendicular planes of polarization.

When the Polaroid is rotated through 30, the intensities of both the beams become equal.

If = 30 for beam A, then = 60 for beam B.

Then from Malus’ law, we get

2 2A A A A

2 2B B B B

3I' I cos I cos 30 I

4

1I' I cos I cos 60 I

4

Since I’A = I’B, the ratio is

A B

A B

A

B

I' I'

3 1I I

4 4

I 1 4 1

I 4 3 3


The radius ‘r’ of the path taken by an electron of mass ‘m’ and charge ‘e’ inside a magnetic

field is

mv mv

rqB eB

2 2 2 2 2

2 2 22

2 2 22

m v r e B

r e Bmv

m

1 r e Bmv

2 2m

This energy will be in joules. Hence, energy in eV will be given as

2 2 22

2 2

1 r e B 1mv eV

2 2m e

r eB eV

2m

Therefore, we have

2 2

3 19 4

2

31

302

30

10 10 1.6 10 3 101mv

2 2 9.1 10

1 1.44 10mv 0.79 eV

2 1.82 10

The transition of electron is between 3 → 2 states


Paper – 2014


Thus, we have

2 2

1 1 1 1 5E 13.6 13.6 13.6 1.89 eV

2 3 4 9 36

Thus, the work function of the metal will be

1.89 eV 0.79 eV 1.1 eV


The spectral formula is given as

2

2 2

2

2 2

2

1 1 1Rz

p n

p 1 and n 2

1 1 1Rz

1 2

1 3Rz

4

Now, for Hydrogen 1H1, we have

2

1

1 3 3R 1 R

4 4

For Deuterium 1H2, we have

2

2 1

2 1

1 3 3 1R 1 R

4 4

For Helium 2He4, we have

2

3 1

1 3

1 3 3 4R 2 R 4

4 4

4

For Lithium 3Li6, we have

2

4 1

1 4

1 3 3 9R 3 R 9

4 4

9

Therefore, we have

1 2 3 44 9


A diode is said to be forward biased when the positive side (p) is at a higher potential than

the negative side (n).


Paper – 2014



List I List II

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

(b) Radio waves (ii) For broadcasting

(c) X-rays (iii) To detect fracture of bones

(d) Ultraviolet rays (iv) Absorbed by the ozone layer of the

atmosphere

30. Correct option (B)

The student measures the value as 3.50 cm. This indicates that there is an uncertainty in

the second digit after the decimal point.

Hence, the least count of the instrument should be 0.01 cm = 0.1 mm.

A meter scale has a least count of 1 mm. Hence, the instrument cannot be a meter scale.

Least count of a vernier calliper is calculated as

L.C one main scale division One vernier scale division

Now, for the given vernier calliper there are 10 divisions in 1 cm on the main scale.

Hence, one main scale division is 1 mm

Now, 10 divisions on the vernier scale coincide with 9 divisions on the main scale. Thus, 10

divisions on vernier scale correspond to 9 mm.

Thus, one vernier scale division is 9/10 = 0.9 mm

Hence, the least count of the vernier calliper is

L.C 1 mm 0.9 mm

0.1 mm

0.01 cm

Least count of a screw gauge is

Pitch

L.CNumber of divisions on the circular scale

Hence, for a screw gauge with pitch 1 mm and 100 divisions, we have

1 mm

L.C 0.01 mm100

And, for a screw gauge with pitch 1 mm and 50 divisions, we have

1 mm

L.C 0.02 mm50

Hence, the instrument has to be a vernier calliper.


Paper – 2014


PART–B : CHEMISTRY


Z = 37

Rb is in fifth period.

Electronic configuration: 1s22s22p63s23p63d104s24p65s1

So last electron enters 5s orbital

Hence n = 5, l = 0, m = 0, s = + 1/2 -1/2


Compressibility factor (Z) PVRT

(For one mole of real gas)

van der Waal equation is given by,

aP+ v -b = RT

2V

At low pressure; V – b V

aP+ V = RT

2V

PV + aV

=RT

PV = RT - aV

PVRT

= 1 - a

VRT

Therefore, Z = 1 - a

VRT


In CsCl, Cl- lies at corners of simple cube and Cs+ at the body centre.

Hence, along the body diagonal, Cs+ and Cl- touch each other so,

2r +2r += 3 a- csCI

2

3 ar +r + =- csCI


Paper – 2014



As per the question.

Normality Volume

H2SO4 N5

60mL

NaOH 10

N 20mL

Mass of organic compound is = 1.4 g

n = n + ngeq geq geqH SO NaOH NH42 3

1 60 1 20x = x + ngeq NH5 1000 10 1000 3

6 1= + ngeq NH500 500 3

5 1n = =geq NH 500 1003

1n = n = n =geqmol mol NHN NH 10033

(Mass)N = 14

100= 0.14 g

Percentage of 'N' = 0.141.4

x 100 = 10%


Paper – 2014



For 0.2 M solution,

R = 50 = 1.4 S m–1 = 1.4 × 10–2 S cm–1

1 1

ρ = = Ω cm-2ς 1.4x10

Now, R = l

ρa

l R -2= = 50 x 1.4 x10ς ρ

For 0.5 M solution,

R = 280 = ?

la

= 50 x 1.4 x10-2

l

R = ρa

1 1 l

x x ρ R a

1 -2ς = x 50 x 1.4 x 10

280 1 -2= x 70 x 10

280 = 2.5 x 10-3 S cm-1

ς x 1000Νow, λ = m M

-32.5 x 10 x 1000=

0.5 = 5 S cm2 mol-1

= 5 x 10-4 S m2 mol-1


Paper – 2014



C H OH + 3O 2CO + 3H O2 2 2 25 (l) (g) (g) (l)

Bomb calorimeter gives U of the reaction. Hence, as per the question,

U = - 1364.47 kJ mol-1

ng = -1

H = U + ngRT

1 x 8.314 x 298= -1364.47-

1000 = - 1366.93 kJ mol-1


According to Debye Huckle onsager equation,

λ = λ - A Cc Here, A = B

Therefore, λ = λ -B Cc


π = i CRT πC H OH = 1 x 0.500 x R x T = 0.5 RT2 5 π (PO ) = 5 x 0.100 x R x T = 0.5 RT4Mg3 2 π = 2 x 0.250 x R x T = 0.5 RTKBr π = 4 x 0.125 x R x T = 0.5 RTNa PO43


1SO + O SO2 2 3(g) (g) (g)2 Kp = KC (RT)X

X = Δng = no. of gaseous moles in product – no. of gaseous moles in reactant

1 3 -1= 1- 1+ = 1- =

2 2 2


Paper – 2014



2A + B C + D

Rate of Reaction -1 d[A] d[B]

= - 2 dt dt

d[C] d[D]

= dt dt

Let rate of Reaction = k[A]x[B]y

d[C] yxOr = k[A] [B]dt

Now from table,

1.2 × 10–3 = k [0.1]x[0.1]y ...(i)

1.2 × 10–3 = k [0.1]x[0.2]y ...(ii)

2.4 × 10–3 = k [0.2]x[0.1]y ...(iii)

Dividing equation (i) by (ii)

y-3 x1.2 x 10 k[0.1] [0.1]

= yx-3 k[0.1] [0.2]1.2 x 10

y1

1 =2

y = 0 Now dividing equation (i) by (iii)

y-3 x1.2 x 10 k[0.1] [0.1]

= yx-3 k[0.2] [0.1]2.4 x 10

-1 x1 1

=2 2

x = 1

.d[C] 1 0Hence = k[A] [B]dt


Paper – 2014



Decreasing order of strength of oxoacids:

HClO4> HClO3> HClO2> HOCl

The given acids will get ionised as follows:

HClO4 ⇆ ClO4- + H+



HOCl ⇆ ClO- + H+

The structures of conjugate bases are as follows:

The resonance structure for ClO4

- is as follows:


- is as follows:


- is as follows:

From the above resonating structures, negative charge is more delocalized on ClO4

- due to

resonanace.

The resonanace stability order of conjuhate base is,

ClO4- > ClO3

- > ClO2- > ClO-

Therefore, the acidic strength oreder is,

HClO4 > HClO3 > HClO2 > HOCl


Paper – 2014



Only Ca2+ can not be dischrged as Ca at cathode on electrolysis as in Ca2+ case, H2 gas

gets discharged at cathode.

This is due to the fact that, higher the position of element in the electrochemical series,

more difficult is the reduction of its cations. If Ca2+ is electrolysed then water is reduced

in preference to it. Therefore, Ca2+ can not be reduced electrolytically from an aqueous

solutions.

The chemical equations for cathode are as follows:

In case of Ag2+

At cathode: Ag2+ + e-→ Ag

In case of Ca2+

At cathode: H2O + e-→ ½ H2 + OH-

In case of Cu2+

At cathode: Cu2+ + 2e-→ Cu

In case of Cr3+

At cathode: Cr3+ + 2e-→ Cr


Given:

Ligands L1 L2 L3 L4

Wavelength

(λ) absorbed

region

Red Green Yellow Blue

Now,

Increase in the order of energy of wavelengths absorbed reflect in the greater extent of

crystal-field splitting which in turn reflect into higher field strength of the ligand.

The order of energy of the different regions are as follows:

Red(L1)< Orange< Yellow(L3)< Green(L2)< Blue(L4)< Violet

The ligand which absorb lower energy light has lower strength.

Therefore, the order for the strength of ligand is as follows:

L1<L3< L2< L4


Paper – 2014



Nitric oxide (NO) is paramagnetic in gaseous state. The electronic configuration of NO is

as follows:

σ1s2σ1s*2σ2s

2σ2s *2σ2pz 2σ2pz*2π2px

2 = π2py2 π2px*1

NO is paramagnetic since it has one unpaired electron in its outermost shell.


Reducing agent gets oxidised itself by releasing electrons.

The oxidation number of oxygen in the reactants and product are shown as a power in

brackets in the given following reactions:

(a) H2O2(-1) + 2H+ +2e- → 2H2O(-2)

(b) H2O2(-1) – 2e- →

(0)

2O + 2H+

(c) H2O2(-1) + 2e- →

(-2)-2 O H

(d) H2O2(-1) + 2OH- - 2e- →

(0)

2O + 2H2O

From the oxidation numbers of oxygen in the above reactions, H2O2 acts as reducing

agent in (b) and (d) reactions where oxidation number of oxygen shifts to positive side.


It is a fact that CsI3 contains one Cs+ ion and I3- ions.


The mass ratio of O2 to N2 is given as 1:4.

Let the mass of O2 = x

Hence, the mass of N2 = 4x

Number of moles of O2 = 32

x

Number of moles of N2 = 4

28 7

x x

Now, ratio of moles of O2 to N2 = 32

x:

7

x = 7:32

Therefore, the ratio of number of molecules of O2 to N2 = 7:32


Paper – 2014


48. Correct option: (1) 0

2+ 0

+1.51V -1.18V2+

2+

0

Mn E 1.18V

2( Mn ) E 1.51V

Mn Mn

Therefore, for 3Mn Mn +2

E 1.18V-( 1.51V)

= -2.69 V

2+ -

3+ -

3+

3+

2+

Mn +2e

Mn + e

Mn

Mn

for Mn disproportionation

Since -2.69 V < 0, reaction is non-spontaneous.


The correct option (option 4) is as follows: 2

0 0

O CO CO

Heat 600 C 700 CFeO Fe 3 4Fe Fe O is a correct series of reaction.

2O

Heat 3 4Fe Fe O is a combustion reaction of Fe.

0 0

CO CO

600 C 700 CFeO Fe 3 4Fe O

This series of reactions correspond to the production of Fe by reduction of Fe3O4.

The other options are as follows:

(1) Fe + H2SO4 → FeSO4 + H2,

2FeSO4 + H2SO4 + ½ O2 →Fe2(SO4)3 +H2O

Fe2(SO4)3 Fe2O3(s) + 3SO3 (This reaction is given incorrectly).

(2) Fe 2O

FeO (The products can be Fe2O3 or Fe3O4)

FeO 2 4dil. H SOFeSO4 + H2O

2FeSO4 Fe2O3 + SO2 + SO3

(3) Fe 2Cl

FeCl3 ir no reaction (it cannot give FeCl2)


Paper – 2014



Option (1) Li2O + 2KCl → 2LiCl + K2O

The compound K2O cannot be generated.

Option (2) [CoCl(NH3)5]++5H+→ Co2+ + 5NH4+ + Cl-

This is a correct option. The complex [CoCl(NH3)5]+ decomposes under acidic medium.

All ammine complexes can be decomposed by adding H+.

Option (3) [Mg(H2O)6]2+ + (EDTA)4- excess NaOH [Mg(EDTA)]2+ + 6H2O

The formula of the complex in the product must be [Mg(EDTA)]2-

Option (4) CuSO4 + 4KCN → K2[Cu(CN)4] + K2SO4

It is an incorrect option. The correctly balanced reaction is as follows:

2CuSO4 + 10 KCN → 2 K3[Cu(CN)4] + 2K2SO4 + (CN)2

51. Correct option: ( 2)

In SN2, the order of reactivity depends on the stearic hindrance.

So as the stearic hindrance around the electrophilic carbon increases in the order CH3X

> 10>20>30. Thus the order of reactivity in SN2 reactions is: CH3Cl > (CH3)CH2

– Cl >

(CH3)2CH – Cl > (CH3)3CCl


When aliphatic primary amine is treated with chloroform and ethanolic potassium

hydroxide alkyl isocyanide is obtained.

R – NH2 + CHCl3 R– CH2– NC


Pyridinium chlorochromate is the mild oxidising agent and converts the alcohol to

aldehyde.

R – CH2 - OH R-CHO


2-Butyne


Paper – 2014




In aliphatic amines, the order of basic strength in aqueous solution is as follows:

Since pKb = - log Kb

So (CH3)2NH will have smallest pKb value.


In quinol and thioquinol, the –OH and –SH groups do not cancellise their dipole moment

because they exist in different conformations.

Same as in thioquinol.


Paper – 2014



Dacron is a condensation polymer of ethylene glycol and terepthalic acid.


Quinoline is an alkaloid. DNA contains ATGC bases Adenine, Thymine, Guanine and

Cytocine.


PART–C : MATHEMATICS

61.Correct Option: (2)

n

2 n n n 2

2 3

Let us rewrite the given equation as

X 1 3 3n 1,n N

3 C C ... 3 ,n N expanding by binomial expansion

Divisible by 9

Now let us consider the set Y

Given that

Y= 9 n 1 ,n N

All multiples of 9

Thus,

X Y

X Y=Y


Paper – 2014



1 2 1 2

min

We know that z z z z

1Now consider z+

2

1 1z z+

2 2

1 1z+ 2

2 2

3z

2

Minimum value lies in the interval 1,2


2 2

2 2

2

2

2

2

2

3 x x 2 x x a 0

23 x x a

3

1 13 x a

3 3

1 1 20 x 1 and x

3 3 3

1 40 3 x

3 3

1 1 13 x 1

3 3 3

For non-interval solution 0<a 1 and a 1,0 0,1


Paper – 2014



From the given quadratic equation, we have

q r+ = ;

p p

Given p,q,r in A.P.

2q=p+r....(1)

Also, given that

1 14

4

2

q

p4

r

p

q4

r

q 4r and p= 9r

4

2

2

2 2

q r4

p p

q 4pr

p

16r 36r 2 13

9r 9


Paper – 2014



2 2

2 2 3 3

2 2 3 3 4 4

2

2

2 2

2 2 2

Consider the determinant

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

1+1+1 1+ + 1+ +

1+ + 1+ + 1+ +

1+ + 1+ + 1+ +

1 1 1 1

1 1

1 1 1 1

1 1

Equating this with the given equation,

we ha

ve K=1


1 1

1 1

1 1

1 1

BB' B A A ' ' since B A A '

A A ' A A ' '

A A ' A ' ' A '

A A ' A A '

1 1

1 1

1

1

A A A' A ' since A' A=A A'

A A A ' A '

I A ' A '

A ' A '

I


Paper – 2014



182

2 3 42 18 18 18 18 18

0 1 2 3 4

3

18 3 18 2 18

3 2 1

1 ax bx 1 2x

1 ax bx C C 2x C 2x C 2x C 2x ...

Therefore, coefficient of x

C 2 a C 2 2b C 0

18 17 16 4a 18 178 36b 0

6 2

51 16 8 a 36 17 36b 0

6528 a 36 17 36b 0

51a 3b 54

4

By checking the options it is clear that,

option (2) satisfies the above equation.


9 1 8 2 7 9 9

2 7 3 6 98 10

9 1 8 2 7 9 10

9

Let s= 10 2 11 10 3 11 10 ... 10 11 k 10 ,

11x 11 10 2 11 10 3 11 10 ... 9 11 11

10

Subtracting the above two equations, we have

111 x 10 11 10 11 10 ... 11 11

10

1110 1

101x

10

10

10 Sum of n terms in 11

11 Geometric progression1

10

10 10

109

10

10 10 10

10

11

911

10 111 10x 10 11

10 110

1x 10 11 11

10

1x 10

10

x 10

Given that 10 k 10

Thus,we have

k=100


Paper – 2014



2Let a, ar and ar be three positive integers in a G.P.

Now consider the middle term, which is ar

Given that the middle term is doubled.

Therefore, the new middle term is 2ar,

Hence the three terms, a, 2ar 2

2

2

2

and ar form an AP

4ar=a+ar

4r 1 r

r 4r 1 0

4 2 3r

2

r 2 3 or r 2 3

Since r>1, GP is increasing, we have r 2 3


2

2x 0

22

2 2x 0 x 0

sin cos xWe need to evaluate the expression lim

x

sin 1 sin xsin cos xlim lim

x x

2

2x 0

2

2x 0

2 2

2 2x 0

2 2

2 2x 0 x 0

2 2

2 2x 0 x 0

2

2x 0 x 0

sin sin xlim

x

sin sin xlim sin sin

x

sin sin x sin xlim

sin x x

sin sin x sin xlim lim

sin x x

sin sin x sin xlim lim

sin x x

sin sin x sinxlim lim

xsin x

2

2

0

sin1 1 lim 1


Paper – 2014


71. Correct Option: (2)

Since g(x) is the inverse of f(x),

5

5

1' ( )

'( )

1' ( ) 1

1

1

g f xf x

g f x x

x

Assuming ( )x g y

5

5

'( ) 1 ( )

'( ) 1 ( )

g y g y

g x g x


Consider

f x 2g x h x

Since f and g are continuous and differentiable in 0,1

h x is also continuous and differentiable.

Also we have

h 0 f 0 2g 0 2 0 2

and

h 1 f 1 2g 1 6 2 2 2

Thus, h x satisfies the conditions of Rolles T

heorem in 0,1

Using Rolle's Theorem,

(1) (0) 6 2'( ) 4

1 0 1 0

(1) (0) 2 0'( ) 2

1 0 1 0

f ff c

g gg c

Hence, '( ) 2 '( )f c g c


Paper – 2014



Given,

2( ) log

'( ) 2 1 0 at 1,2

f x x x x

f x x xx

Hence, at 1x ,

2 1 0

2 1 -----Equation 1

And at 2x ,

4 1 02

8 2 -----Equation 2

Solving equations 1 and 2, we get

12,

2


Paper – 2014



Let 1

xxf x e

'( ) f x 1

2

1(1 )

xxe

x

1x+

x

1 1 1x+ x+ x+

x x x

2

1 1 1x+ x+ x+

x x x

2

1Let I= 1+x e dx

x

Rewrite the given integrand as:

1 11+x e =e +x(1 )e

x x

Thus, we have,

1 11+x e dx= e +x(1 )e

x x

1 1 1x+ x+ x+x x x

2

dx

We know that xf'(x)+f(x) dx=xf(x)+c

Thus,

1= e +x(1 )e dx=xe +c

x

I


2

0

0

3

03

3

03

1 4sin 4sin2 2

2sin 12

(1 2sin ) (2sin 1)2 2

4cos 4cos2 2

4 3 34 0 4

3 2 2 3

4 3 43

x xdx

xdx

x xdx dx

x xx x


Paper – 2014



2 2 21 and 1 intersect at 0,1x y y x x

Area of the shaded region =Area of semi-circle plus area bounded by parabola and y-axis

Area of the shaded region = 12

1

(1)

2

ydx

2Since y 1 x y 1 x

Thus, we have,

12

1

12

0

13

2

0

(1)Area (1 )

2

(1)2 (1 ) the parabola is symmetric about x-axis

2

2(1 )( 1)

322

4(0 ( 1))

2 3

4

2 3

x dx

x dx

x


Paper – 2014



Given,

Given equation is:

1'( ) ( ) 200

2

Rewriting the above equation, we have

1'( ) ( ) 200....(1)

2

This is a linear differential equation of the form

dy, where P and Q are functions of x.

dx

From equati

p t p t

p t p t

Py Q

1

2

1on (1), we have, P and Q 200

2

The integrating factor I.F. = e

tPdt

e

Hence,

Pdx Pdx

1 1

2 2 2

2

The solution is of the form,

ye e

Thus, we have,

( ) 200 400

( ) 400

t tt

t

Q dx c

p t e e dt e c

p t ce

Since, p (0) = 100,

=>100 = 400 + c

C = -300

Hence, 2( ) 400 300t

p t e


Paper – 2014



S is the mid-point of QR.

Hence, S = 7 6 3 1 13

, ,12 2 2

Slope of PS = 2 1 2

13 92

2

Equation of the line passing through (1,-1) and parallel to PS is

2( 1) ( 1)

9

9 9 2 2

2 9 7 0

y x

y x

x y


The point of intersection of the lines 4 2 0 and 5 2 0 is given by

4 1

2 2 4 5 8 10

2( )

2

5 4

2

ax ay c bx by d

x

ad bc ad bc ab ab

ad bcx

ab

bc ady

ab

As the point of intersection is in the fourth quadrant, x is positive and y is negative. Also

the distance from the axes is same.

Hence, x = -y

2( ) (5 4 )

2 2

2 2 5 4

3 2 0

ad bc bc ad

ab ab

ad bc bc ad

bc ad


Paper – 2014



Let us rewrite the given equation of the ellipse.

2 2

2 2

2 2

2 2

2 2

2 2

x 3y 6

x 3y 6

6 6 6

x y1

6 2

Now compare the above equation with the

x ygeneral equation of the ellipse 1

a b

a 6;b 2

Equation of a line through (h, k) & perpendicular to line joining it to origin is

2 2

2 2 2 2

2 2

22 2 2

2

( ) ( )

Since the above line is tangent to the ellipse,

it should satisfy the following condition.

C

Substituting the values of C, , and , we have

6 2

hy k x h

k

h h ky x

k k

a m b

a m b

h k h

k k

2

2 2 2 26 2 x y x y


Paper – 2014



2 2

2 2 2

2 2

2 2

2 2

( 1) ( 1) 1

Radius of M=

M touches P externally

(0-1) ( 1) (1 )

1 1 2 1 2

If y > 0,

y 2 2 1 2

4 1

1

4

If y < 0,

y 2 2 1 2

1 2

Which is not possible

1

4

P x y

y

y y

y y y y

y y y

y

y

y y y

y


Paper – 2014



Let the equation of the tangent to 2 14 be y=mx+

my x

The above tangent is also a tangent to 2 32 x y

2

2

2

2

3

3

32

132 mx+

m

3232 0

Since roots are equal, the discriminant is zero.

Thus,

320 32 4 0

32 4

1

8

1

2

x y

x

x mxm

D mm

m

m

m



Paper – 2014


Let B' a,b,c be the image of the point B 1,3,4

The direction ratios of the line BB' is

1 3 4

2 1 1

Any point on the line BB' is of the form

a=2 1, 3 , 4

Let P x,y,z mid point of the line BB'.

Thu

a b c

b c

2 1 1 3 3 4 4s, , ,

2 2 2

1,3 ,42 2

Since the point P lies in the plane, 2 3 0, we have

2 1 3 4 3 02 2

P

P

x y z

2 2 3 4 3 02 2

3 6 0

2

Substituting the value of in

1 3 4

2 1 1

we have

3, 5, 2

Hence the equation of the required line is

3 5 2

3 1 5

a b c

a b c

x y z


Paper – 2014



2 2 2

2 2 2

1 2 1 2 1 2

2 2 2 2 2 2

1 1 1 2 2 2

0..................( )

....................( )

( ) 0

2 ( ) 0

0

So direction ratios are -1, 0, 1 -1, 1, 0

cos

1 0 0cos

2

l m n i

l m n ii

l m l m

m m l

m or l m

and

a a b b c c

a b c a b c

1

22

3


2

LHS = a b b c c a

p b c c a p a b

p b c . c a

p.c b p.b c . c a

a b c b a b c c . c a p a b

a b c b c a 0 a b b 0

a b c b c a a b c

where

where

RHS

2

a b c

1


Paper – 2014



1 1 5( ) ( ) 1

6 6 6

1 1 3( ) ( ) 1

4 4 4

Now, ( ) ( ) ( ) ( )

5 3 1( )

6 4 4

5 3 1 1( )

6 4 4 3

P A B P A B

P A P A

P A B P A P B P A B

P B

P B

Thus, we observe that ( ) ( )P A P B .

Hence, events A and B are not equally likely.

Also, 3 1 1

( ) ( ) ( )4 3 4

P A P B P A B

Thus, events A and B are independent.


Solution:

22

2

22 2 2 22

2 2 2 2

2

2

2

,

2 4 6 .......... 100 2 4 6 ...... 100

50 50

502 100

4(1 2 3 .......... 50 2

50 50

50 (50 1) (2 50 1)

64 5150

50 51 1014 51

50 6

3434 2601

833

ixVariance x

N


Paper – 2014



Solution:

4 4 6 6

4 6

2 2 2 2

2 2 2 2

1( ) sin cos

1 1( ) ( ) sin cos sin cos

4 6

1 11 2sin cos 1 3sin cos

4 6

1 1 1 1sin cos sin cos

4 2 6 2

1 1

4 6

1

12

k k

kf x x xk

f x f x x x x x

x x x x

x x x x


t = 1 second

From the figure, we have

20 20tan 45 1 20

20 1 20tan 30

203

20 3 1

. . 20 3 1 /

o

o

aa a

anda b b

b

i e Speed m s


Paper – 2014



p q q p q ( )p q p q

F F T F T T

F T F T F F

T F T T F F

T T F F T T

Thus, p q is equivalent to p q

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