3 DISEÑO METODOLÓGICO
3.3 P LAN G ENERAL DE M ARKETING
3.3.2 Procesos y Funcionamiento
3.3.2.1 Objetivos y Estrategias Operativas
× = 0.01 T
6. The charge of the ‘excess’ charge carriers gets balanced by an equal and opposite charge of the ionized cores in the lattice.
7. We have
=
= ∞
=
=
→
∞
−
4 .1 const – 1
2 . 1 const E
and
4 .3 const 2
– 1 1 . 1 const E
2 2 2
2 1 2
2
∴ Ratio = 3 : 1
8. Infrared radiations get readily absorbed by water molecules in most materials. This increases their thermal motion and heats them up.
(i) visible light (ii) Microwaves 9. Focal length do not depend upon the medium in
which the mirror is held. So focal length will not change.
10. As width of central maxima, βc = a λf 2
So, βc ∝ a
1, βc ∝ f & βc ∝ λ thus, (a) width increases with decrease in a (b) width increases with increase in f (c) width decreases with decrease in λ
11. –3nC –2nC
2nC 2 m 1nC r
r = 2
2 = 1 m
Vcentre = r
k [2 + 1 – 2 – 3] × 10–9
= 1
10 9× 9
[–2 × 10–9]
= –18 volt 12. Blue – 6 (A)
Black – 0 (B) yellow – 4 (C)
R = AB × 10C = 60 × 104
∴ i = R
V = 4
10 60
30
× = 0.5 × 10–4 = 0.05 mA
13. Microwave → In microwave oven
Ultraviolet rays → To sterlizing sergical instruments Gamma rays → In medical science.
14. Instantaneous Activity = R = – dt dN = λN
∴ dt dR =
dt
d (λN) = dt λdN
= λ (–λ N) = – λ2 N
= –
2
2 / 1
log 2
T
e N
∴ dt dR ∝
2 2 / 1 ) T (
1
MOCK TEST-1 (SOLUTION)
MOCK TEST– 1 PUBLISHED IN DECEMBER ISSUE
XtraEdge for IIT-JEE 86 JANUARY 2011 15. Energy of a photon of the incident radiation
= λ
This being less than the work function of Mo, there would be no photo-emission from Mo.
16. eVs = hν – w Waves used → space waves
It is both-the height of transmitting antenna as well as the height of receiving antenna that affects the range of the mode of communication.
19. Imax = ( I1 + I2 )2 = ( I + I+δI )2
XtraEdge for IIT-JEE 87 JANUARY 2011 Total B.E. of daughter nucleus
= 7.835 × 231 MeV = 1809.9 MeV Total B.E. of α-particle
= 7.07 × 4 MeV = 28.28 MeV Increase in B.E. after the reaction
= [(180.9 + 28.28) – (1833)] MeV = 5.18 MeV This is the energy released in the reaction, since it assumed to be taken up totally by the α-particle,
2
26. The modulation index (µ) for an AM wave equals the ratio of the peak value of the modulating signal (Am) to the peak value of the carrier wave
27. Postulates of Bohr's Model –(i) In atom, electron moves round the nucleus in stable orbits. Attraction force b/w nucleus and electron given the required centripetal force for electron
mass of electron = m charge on electron = e atomic no. = z radii of orbit = r speed of electron = v
Centripetal force = Attraction force
r
(ii) In stable orbits electron does not emit any radiation. But at jumping from one orbit to another.
It emits or absorbs energy.
This energy equals difference of energy of two energy levels.
∆E = E1 – E2 hν = E2 – E1 h = Plank const.
ν = frequency of radiation
(iii) Electron moves only in those orbits where angular momentum of electrons
mvr =
2π nh
n = principle energy level = 1,2,3, …… ∞ Hydrogen spectrum ––
According to Bohr's theory radii of nth orbit = 22 2 2
4 kme h n
π
velocity of electron v = nh
while electrons jumps from one orbit to another E = E2 – E1
Hydrogen Spectrum Different series –
(1) Lymann - (Ultraviolet) (2) Balmer - (Visible)
(3) Paschen - (Infra Red) (4) Brackett - (Infra Red) (5) Pfund - (Infra Red) (6) Hemfry - (Infra Red)
Shortcomings of Bohr's Model –
XtraEdge for IIT-JEE 88 JANUARY 2011 (1) Applicable only for atoms having one electrons
and not applicable for others.
(2) Nucleous is not stable.
(3) Quantisation of angular momentum is not explained
(4) Orbits of electrons are elliptical, not circular.
(5) Does not explain Stork's and Zeeman's effect.
28. (a) The sources of light, which emit light waves of same frequency with constant phase difference. That is phase difference do not change with time.
Interference pattern is not observed as two physically different light sources can never be coherent.
So, bright fringes are obtained at y = 0,
So, dark fringes are obtained at y = d
So, dark fringes are obtained at y = d
And the distance between two consecutive bright or dark fringe i.e, fringe width, β =
d λD
. OR
(a) In the figure, AB is object placed beyond F perpendicular to the principal axis of a concave lens.
A' B' is the erect, virtual and diminished image.
B
29. Step transfer works on the principle of mutual induction .
Secondary coil Primary coil
Vp = – Np
For step-up transformer,
p
Principle : A.C. Generator works on the principle of E.M.I.
Working : When coil is rotated in the magnetic field (produced in the poles), magnetic flux linking with the coil change.
XtraEdge for IIT-JEE 89 JANUARY 2011
N S
ω
∴ φ = BA cos ωt
∴ ε = – dt Ndφ
= –N dt
d (BA cos ωt)
∴ ε = NB ω sin ωt.
30. U = a
kq2 × 12 + 2
2
a
kq × 12 + 3
2
a
kq × 4
or
Total Electric flux from the surface φ =
∫
→E.ds→=
∫
Edscosθ=
∫
4π∈ 2cosθ0 r
qds
= 4π∈0
q
∫
dscosr2 θ= 4π∈0
q × 4π =
∈0
q
CHEMISTRY
1. Holme signal is mixture of calcium carbide (CaC2) and calcium phosphide (Ca3P2)
2. (a) Copper Pyrite or Chalcopyrite ⇒ CuFeS2 (b) Copper glance ⇒ Cu2S
(c) malachite ⇒ CuCo3. Cu (OH)2 (d) cuprite ⇒ Cu2O
3. Glycine (H2N – CH2 – COOH) is simplest amino acid.
4. t1/2
[A0]
5. Eºcell = 1.229 V
6. (a) The conc. above which associative colloid is formed is called CMC.
(b) Emulsifier stabalise the emulsion → soap or detergents
7. Fluoboric acid
8. CH3 – CH2 – CH – CH2OH + CH3CH2I CH3
9. Structure of N2O5 is – O
O
N – O – N O
O
10. d-block elements are used as catalyst due to two reasons
(A) They show variable oxidation states and hence they are converted to unstable intermediate species while they convert reactants to products this reaction occurs via a path of lower activation energy.
(B) They adsorb reactant molecules on their surface which provide a suitable surface area for the reaction to occur.
11. Small amount of very pure titanium or zirconium metal can be prepared by this method. Impure metal is heated in an evacuated vessel with I2. TiI4 or ZrI4
is formed which. vaporizes leaving behind impurities. The gaseous MI4 is decomposed on a white hot tungsten filament.
⇒ Zr + 2 I2 870 →K ZrI4 2075 →K Zr + 2I2
⇒ Ti + 2I2 →525K TiI4 1675 →K Ti + 2I2 12. Rate constant can be defined as rate of chemical
reaction when conc. of all the reactants become unity Factors on which it depends –
(a) Nature of reactants (b) Temperature (c) Catalyst
13. Ist Method : Given K = 2.303
∴ t1/2 = K
3 . 0 303 .
2 ×
= 2.303 3 . 0 303 .
2 ×
= 0.3 sec.
t90 = 3 10 × t1/2
= 3
10 × 0.3 = 1 sec q
θ ds→ →E
XtraEdge for IIT-JEE 90 JANUARY 2011
II method
K =
t 303 .
2 ln
− x a
a
0 0
2.303 =
t90
303 . 2 ln
10 100
t90 = 1 sec
14. (i) N – Bromo succinimide (ii) sodium acetylide H – C ≡ C–Na+ 15. A = Cresol; B = Benzyl alcohol.
16. (i) 4-Methylhex-2-en-3-ol (ii) 4-chloro-4-phenyl-1-butanol 17. Factual
18. (i) A = ; B =
SO3H
19. The steady decrease in the size of lanthanide ions (M3+) with the increase in atomic number is called Lanthanide contraction.
Cause of Lanthanide Contractions :
As the atomic number increases due to addition of protons nuclear charge increases. And in case of lanthanides electrons are filled in 4f orbital which have poorest screening effect. Thus the increased nuclear charge occurs much on 5d and 6s shell and the electron cloud shrinks. This results in gradual decrease in size of lanthanides with increase in atomic number.
20. (3dNi84s2) =
3d8 4s2 4p
) s 4 d 3 (
2
0
Ni8 + =
3d8 4s0 4p
[NiCl4] 2– = x x x x x x x x Cl¯ Cl¯ Cl¯ Cl¯
sp3
since Cl¯ is weak ligand, thus, it unable to pair electrons and hence it is paramagnetic
[Ni(CN)4]2– = x x x x x x
CN¯ CN¯CN¯
dsp2 x x
CN¯
Since, CN¯ is strong ligand, thus it able to pair electrons and hence it is diamagnetic
21. (i) Buna S –CH2 – CH = CH – CH2 – CH – CH2 – C6H5
( )
n(ii) Buna N –CH2 – CH = CH – CH2 – CH – CH2 – CN
( )
n(iii) Neoprene –CH2 – C = CH – CH2 – Cl
( )
n22. Amino acids can be classified as - (a) Essential – Not produced in body
Ex – Lysine, leucine
(b) Semi Essential – Partially produced in body Ex – Histidine & Arginine
(c) Non essential – Produced in body Ex – Glycine, Alanine
23. Tranquilizers are drugs used to reduce stress. These are of following types :-
(i) Hypnotic Ex – Barbituric acid and it's derivatives (ii) Non-Hypnotic Ex – Chlordiazepoxide,
Meprobamate 24. Eºcell = Eºcathode – Eºanode
= 0.34 – (– 0.25) = +0.59 V ln Kc =
059 . 0
n Eºcell
= 059 . 0
2 × 0.59 = 20 Kc = 1 × 1020
25. (a) Brownian motion : Random movement of colloidal particles in the medium is called brownian movement. Due to brownian motion particles colloide with each other and it imparts stability in the colloidal solution.
(b) Tyndall effect : Scattering of light when rays fall on the colloidal particles is called Tyndall effect.
Due to this scattering of light the path of light in the colloidal solution get illuminated.
Application : colour of sky is blue.
(c) Electrophoresis : Movement of colloidal particles in the presence of electric field is called electrophoresis as charge is present on the colloidal particles and ∴ movement of particles takes place toward opposite pole.
XtraEdge for IIT-JEE 91 JANUARY 2011 26. A is CH3 – C = CH – CH – CH3
CH3
CH3
;
B is CH3 – CH – COOH CH3
;
C is CH3 – C – COOH Br
CH3
;
D is CH3 – C – COOH OH
CH3
27. A is (CH3)3COOH; B is (CH3)C – NH2; C is (CH3)3C – NC;
D and E are (CH3)3C – OH and (CH3)2C = CH2
28. The state of hybridisation of boron in diborane is sp3, In between boron atoms and bridging hydrogens, three centre two electron bond is present.
The structure of diborane is –
x x
x x
x
x
B B
Ht
Ht
Hb
Hb Ht
Ht
Hb = Bridging hydrogens Ht = Terminal hydrogens
29. (i) Rate of rxn gives us how fast reactants are converted into products. Per unit time change in the conc. of reactants or products. Rate constant is equals to rate of rxn when conc. of all the reactants become unity or one.
Rate Rate const.
Depends upon Depend upon - Conc. of reactants - Nature of reactants - Nature of reactants - Temperature - Temperature - Catalyst - Pressure
- Surface area - Catalyst - pH of medium
(ii) m = 62
6 . 222 ×
200
1000 = 17.95 m
d = m M +
1000 MMB
1.072 = M
+
1000 62 95 . 17
1
1.072 = M (0.117) M = 9.107 mol/lit
(iii) Mixture which boils at const temperature is called as azeotropic mixture. There are two types of azeotropes.
(a) maximum boiling point azeotrope – The azeotropic mixture in which boiling point of the mixture is greater than its constituents.
(b) Minimum boiling point azeotrope – The azeotrope in which B.P. of the mixture is less than its constituents
30. (a) (i) 3 a = 4r (ii) 2 a = 4r (iii) a = 2r
(b) The non-stoichiometric point defect responsible for colour in alkali metal is Metal excess defect in which alkali halide when heated in metal vapour results in removal of halogen atom from its position and in place of halogen e– get situated. This results in development of colour in the compound.
↑ ↑ ↑ ↑ (direction of dipoles) Ex [CrO2, Fe, Co, Ni etc]
↑ ↓ ↑ ↓ ( magnetic dipoles) Ex MnO, Mn2O3 etc.
MATHEMATICS
Section A
1. A2 × 2 =
22 21
12 11
a a
a
a aij = 2
3 2i− j
a11 = 2
3 2−
= 2 1; a12 =
2 6 2−
= 2 4= 2
a21 = 2
3 4−
=2 1; a22 =
2 6 4−
= 2 2= 1
A =
1 2 / 1
2 2 / 1
2. 2, 1
6 1 3
2 ⇒ = =
= +
=
− x y
y x
y
x and
5 , 29 3
4 3
2
– ⇒ = =
= +
=
− a b
b a
b a
Hence x = 2, y = 1, a = 3, b = 5
XtraEdge for IIT-JEE 92 JANUARY 2011
Since the given equation contains two arbitrary constants, we shall differentiate it two times and we shall get a differential equation of second order.
Differentiating (i) w.r.t. x, we get
which is the required differential equation of the given family of curves.
8. The vectors ar and br are perpendicular to each other
10. Let the direction ratios of the required line be a, b, c.
Since it is perpendicular to the two given lines.
Therefore,
a + 2b + 3c = 0 … (i) and –3a + 2b + 5c = 0 … (ii)
solving (i) and (ii) by cross-multiplication, we get 4
Thus, the required line passes through (–1, 3, –2) and has direction ratios proportional to 2, –7, 4. So its equation is
2 q = P (two incorrect forecasting) = 2/3 n = 4
Let r be the number of correct forecast P (at least three correct results) = P ( r = 3) + P (r = 4)
XtraEdge for IIT-JEE 93 JANUARY 2011 passes through (1, –6) – 6 = 1 –2 + b ⇒ b = – 5
XtraEdge for IIT-JEE 94 JANUARY 2011
20. The equation of the family of circles of radius r is (x – a)2 + (y – b)2 = r2
where a and b are parameters.
Since equation (i) contains two arbitrary constants, we differentiate it two times w.r.t x and the differential equation will be of second order.
Differentiating (i) w.r.t. x, we get 2(x – a) + 2(y – b)
This is the required differential equation.
OR
So, the coordinates of a general point on this line are (3λ +1, 2λ –1, 5λ + 1)
The equation of the second line is 4
So the coordinates of a general point on this line are (4µ –2; 3µ + 1, –2µ –1)
If the line intersect, then they have a common point.
So, for some values of λ and µ, we must have
XtraEdge for IIT-JEE 95 JANUARY 2011
Hence, the given lines do not intersect.
Section C
(A) required probability = P(A). P(B)+ P(B). P(A)=50 13 (B) required probability
= P(A). P(B) + P(A). P(B)= normal to the plane. Since PQ passes through P and is normal to the given plane, therefore equation of line PQ is
rr= ( iˆ + 3 jˆ + 4 kˆ ) + λ (2 iˆ – jˆ + kˆ )
Since Q lies on line PQ, so let the position vector of Q be ( iˆ +3 jˆ + 4 kˆ ) + λ (2 iˆ – jˆ + kˆ )
XtraEdge for IIT-JEE 96 JANUARY 2011
Thus, the position vector of Q is
( iˆ +3jˆ+4 kˆ ) – 2 (2 iˆ –jˆ+ kˆ ) = –3 iˆ + 5 jˆ +2 kˆ 27. Suppose the dealer buys x fans and y sewing
machines. Since the dealer has space for at most 20 items. Therefore,
x + y ≤ 20
A fan costs Rs.360 and a sewing machine costs Rs.240. Therefore, total cost of x fans and y sewing machines is Rs.(360x +240y). But the dealer has only Rs.5760 to invest . Therefore,
360x + 240y ≤ 5760
Since the dealer can sell all the items that he can buy and the profit on a fan is of Rs.22 and on a sewing machine the profit is of Rs.18. Therefore, total profit on selling x fans.
Let Z denote the total profit. Then, Z = 22x + 18y.
Clearly x, y ≥ 0
Thus, the mathematical formulation of the given problem is
Maximize Z = 22x + 18y S. t. x + y ≤ 20
360 x + 240y ≤ 5760 and x ≥ 0, y ≥ 0
To solve this LPP graphically, we first convert the inequations into equations and draw the corresponding lines. The feasible region of the LPP is shaded in fig. The corner points of the feasible region OA2PB1 are O(0, 0), A2(16, 0), P(8,12) and B1 (0, 20).
These points have been obtained by solving the corresponding intersecting lines, simultaneously.
(0, 0)
The values of the objective function Z at corner points of the feasible region are given in the following table.
Points (x, y) Value of the objective function Z = 22x + 18y
Hence, the dealer should purchase 8 fans and 12 sewing machines to obtain the maximum profit under given conditions.
28. A11 =
The given system of equations can be written
XtraEdge for IIT-JEE 97 JANUARY 2011
The given system of equations can be written as
AX = B where X =