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.
2. Correct option (3)
H is the height of the tower.
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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)
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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
4. Correct option (1)
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
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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
5. Correct option (3)
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
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6. Correct option (3)
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.
7. Correct option (4)
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.
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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
8. Correct option (1)
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
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9. Correct option (3)
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
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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
11. Correct option (3)
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
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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
12. Correct option (4)
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
13. Correct option (1)
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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
14. Correct option (4)
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
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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.
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16. Correct option (3)
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
17. Correct option (1)
0
0
0
4 12
7 2
EE
K K
EK
3 10 2.2 8.85 10
6 10 C/m
18. Correct option (3)
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.
19. Correct option (2)
2
ext0
ext
2
ext0
Work WPower
Time t
W F .dx
F BIL
1P F .dx
t
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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
20. Correct option (3)
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
21. Correct option (3)
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
22. Correct option (3)
During the propagation of electromagnetic waves in a medium, the energy density of both
the electric as well as magnetic field is the same.
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23. Correct option (2)
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
24. Correct option (2)
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.
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25. Correct option (4)
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
26. Correct option (2)
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
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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
27. Correct option (3)
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
28. Correct option (1)
A diode is said to be forward biased when the positive side (p) is at a higher potential than
the negative side (n).
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29. Correct option (4)
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.
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PART–B : CHEMISTRY
31. Correct option: (1)
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
32. Correct option: (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
33. Correct option (3)
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
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34. Correct option: (2)
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%
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35. Correct option: (1)
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
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36. Correct option: (1)
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
37. Correct option: (3)
According to Debye Huckle onsager equation,
λ = λ - A Cc Here, A = B
Therefore, λ = λ -B Cc
38. Correct option: (1)
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
39. Correct option: (2)
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
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40. Correct option: (4)
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
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41. Correct option: (3)
Decreasing order of strength of oxoacids:
HClO4> HClO3> HClO2> HOCl
The given acids will get ionised as follows:
HClO4 ⇆ ClO4- + H+
HClO3 ⇆ ClO3- + H+
HClO2 ⇆ ClO2- + H+
HOCl ⇆ ClO- + H+
The structures of conjugate bases are as follows:
The resonance structure for ClO4
- is as follows:
The resonance structure for ClO3
- is as follows:
The resonance structure for ClO2
- 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
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42. Correct option: (2)
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
43. Correct option: (2)
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
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44. Correct option: (1)
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.
45. Correct option: (4)
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.
46. Correct option: (2)
It is a fact that CsI3 contains one Cs+ ion and I3- ions.
47. Correct option: (2)
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
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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.
49. Correct option: (4)
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)
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50. Correct option: (2)
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
52. Correct option: (4)
When aliphatic primary amine is treated with chloroform and ethanolic potassium
hydroxide alkyl isocyanide is obtained.
R – NH2 + CHCl3 R– CH2– NC
53. Correct option: (4)
Pyridinium chlorochromate is the mild oxidising agent and converts the alcohol to
aldehyde.
R – CH2 - OH R-CHO
54. Correct option: (3)
2-Butyne
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55. Correct option: (1)
56. Correct option: (1)
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.
57. Correct option: (4)
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.
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58. Correct option: (1)
Dacron is a condensation polymer of ethylene glycol and terepthalic acid.
59. Correct option: (1)
Quinoline is an alkaloid. DNA contains ATGC bases Adenine, Thymine, Guanine and
Cytocine.
60. Correct option: (3)
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
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62.Correct Option: (4)
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
63.Correct Option: (3)
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
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64.Correct Option: (2)
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
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65.Correct Option: (1)
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
66.Correct Option: (4)
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
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67.Correct Option: (2)
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.
68.Correct Option: (1)
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
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69.Correct Option: (2)
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
70.Correct Option: (2)
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
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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
72. Correct Option: (2)
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
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73. Correct Option: (1)
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
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74. Correct Option: (4)
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
75. Correct Option: (2)
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
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76. Correct Option: (3)
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
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77. Correct Option: (3)
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
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78.Correct Option: (4)
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
79. Correct Option: (1)
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
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80. Correct Option: (1)
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
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81. Correct Option: (2)
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
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82. Correct Option: (3)
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
83. Correct Option: (3)
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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
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84. Correct Option: (3)
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
85. Correct Option: (2)
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
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86. Correct Option: (1)
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.
87. Correct Option: (4)
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
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88. Correct Option: (2)
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
89. Correct Option: (2)
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
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90. Correct Option: (3)
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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