MOCK TEST-3 (SOLUTION)
MOCK TEST– 3 PUBLISHED IN SAME ISSUE
15. X = ? When X and Y are interchanged
Y B X
19. Work-function : Minimum energy needed by a free electron to remove it from metallic surface.
Threshold frequency : Minimum frequency required for photo-electron emission.
Threshold wavelength : Maximum wavelength of incident radiations required for photoelectric emission.
Stopping potential : Minimum potential required across the photo cell to completely stop the photo current.
Number of channels N which can be transmitted simultaneously can be found out by dividing band width by the system with band width of one channel.
i.e.,
(2) Reactance are of two types (a) Capacitive Reactance XC =
C 1 ω
(b) Inductive Reactance XL = ωL
(3) Impendence = Z = R2+(XL–XC)2
(4)
ν XC
ν XL
25. Force = F = Bil For current (i)
i =
R emf =
R
Bvl ⇒ F = R
v B2 l2
26. Cyclotron work on the fact that a positively charged particle can be accelerated to a sufficiently high energy with the help of smaller values of oscillating electric field by making it to cross the same electric field time and again with use of strong magnetic field.
North
target South
B (Magnetic Field) H.F
oscillator
Dees D1
Dees D2
Magnet
27.
+ + + + + + + + + + + + + +
λ
l r
0
qin
ds . E = ∈
∫
rE.2πrl =
0
·
∈ λ l
E =
r 2π∈0
λ
28. Bending of light from sharp edges is called diffraction. Difference between interference and diffraction.
Interference Diffraction (i) Two coherent sources
are necessary
(i) One source is Necessary (ii) All fringes are of
same width
(ii) CM has double width than all other fringes (iii) All bright fringes has
equal intensity
(iii) as order of bright fringes increases, intensity goes down (iv) For bright fringes
path difference = nλ
(iv) For bright fringes, Path difference = (2n – 1)
2 λ. (v) For dark fringes, path
difference = (2n – 1) 2
λ
(v) For dark fringes, path difference
= nλ
Angular width of CM = 3
10
10 5 . 0
10 5000 2 a 2
−
−
×
×
= × λ
= 2 × 10–3 Radian OR
When white light incidents over prism then it divide it into component colours. This phenomenon is called diffraction and angle between end colours is called angular dispersion. Prism deviates violet colour maximum as its refractive index for violet colour is maximum.
ω = y 1
R v
− µ
µ
− µ
∴ ω =
1 5170 . 1
5140 . 1 5318 . 1
−
−
∴ ω = .034
29. white
Types of spectrum Spectrum is of 2 types.
(i) Emission (ii) Absorption.
(i) Emission : - If radiations emitted from a substance is obtained ones screen after passing through the prism then spectrum is called emission spectrum.
It is line emission spectrum for substances at atomic level in gaseous state.
It is continuous spectrum for substances in liquid or solid state.
(ii) Absorption spectrum : - The substance of which absorption spectrum is to be find out is placed in a transparent tube and white light is passed through it now substance absorb radiation corresponding to some particular wavelength and black lines are obtained for these absorbed radiations. These black lines are called absorption spectrum.
OR
Energy of emitted photons from H-atom = 10.2 eV Work – function of metallic surface
= 4000
12400 eV = 3.1 eV
Now According to Einstein's Law of Photo-electric effect '
E = w + K.E.max.
10.2 = 3.1 + K.E.max
∴ K.E.max = 7.1 eV Ans
30. Moving Coil Galvanometer – Principle : When a coil carrying current is placed in a magnetic field, it experiences a Torque τ = NiAB sin α. This Torque is proportional to the current passed through it.
N S
Torsion Head
Mirror Frame
Core Copper coil
Phosphor Bronze
Spring T1
T2
Working : When a current is passed through the coil, the two vertical limbs experience a force normal
to it, equal in magnitude and parallel but opposite in sense. No forces act on the horizontal sides as they are parallel to the field .Since the field is radial, the forces acting on the vertical parts of coil remain always perpendicular to the plane of the coil in all its positions so that the perpendicular distance between the forces is always equal to breadth of the coil. Thus the coil is subjected to a torque whose magnitude is given by
τ = NiAB … (i)
Under the action of this deflecting torque, the coil rotates on its axis. Due to elastic forces, a restoring torque τr is produced in the suspension coil, which is perpendicular to the twist θ produced in the wire.
τr = θ or τr = Cθ … (ii)
where C is restoring torque per unit twist, and it is called the constant of twist for the suspension. As soon as restoring torque becomes equal to the deflecting torque, equilibrium is established. The coil does not rotate further. Let φ be the angle through which the coil rotates till it reaches the equilibrium position.
τ= τr or NiAB = Cφ … (iii) i = φ
NAB
C = K.φ
K = NAB
C is constant for the instrument, called Galvanometer constant.
∴ i ∝ φ .
Thus deflection of coil is proportional to the current passing through it. The deflection can be measured by using a lamp and scale arrangement.
OR
Consider two infinitely long thin conductors carrying currents in opposite directions.
Magnetic field B1 due to I1 at P on conductor CD is given by
B1 = r 2
I1
0
π µ
B I1 D
P B1F2
F1B2
r
A C
I2
Q
The magnetic field B1 is perpendicular to plane of paper and directed inward. This field will produce a force/length F2 on conductor given CD by
F2 = B1I2 = r 2
I I1 2
0
π
µ [Q F = BI l, Here l = 1]
By Fleming's left hand rule direction of F2 is away from conductor AB.
Similarly the current I2 will create a field B2 at Q directed inward which in turn will create force/length F1
F1 = B2I1 = r 2
I I1 2
0π µ
By Fleming's left hand rule, the direction of F1 is away from the conductor CD. Hence the two conductors repel each other.
Definition of Ampere If I1 = I2 = 1A, and r = 1m, then
F = 0 7 2 10 7Nm 1
2 10 4 2
−
− −
× π =
×
= π π µ
Thus one ampere is that current which on flowing through each of the two parallel uniform linear conductors placed in free space at a distance of one meter from each other produces between them a force of 2 × 10–7 N per metre of their lengths.
CHEMISTRY
1. CH3–CH2–CH=CH2 + HCl →
Cl CH CH CH
CH3− 2− | − 3
2. (ii) > (iii) > (i) > (iv)
3. By fehling solution, tollen reagent etc.
4. Amino acid are compound containing amino group and carboxylic group.
Structure of alanine is –
H+3N–CH–COO– CH3
5. Movement of colloidal particles towards opposite terminal takes place which is called as electrophoresis.
6. It is a stoichiometric defect in which an ion (cation) get displaced from its position and get arranged as an interstitial particle it is called Frenkel defect in this defect a vacancy or interstitial defect are formed simultaneously. Due to Frenkel defect.
(i) Density remains same
(ii) Electrical conductivity improved
7. No of particle are different in sugar and NaCl. As dissociation in NaCl takes place.
8. Methyl 3-bromo 2-oxo butanoate
9. (i) Sandmayer reaction. —Cl (ii) Kuchrous reaction
CH ≡ CH → CH2 = CH OH |
CH3 – CHO
10. (i)
Hydrolysis H isocyanide
Ethyl2H5 N C 2HOH
C − = + →+
HCOOH NH
H C
e min a
Ethyl2 5 2+
(ii) NH H
Aniline
N–C–CH3 + HCl H
N-Phenylethanamide + Cl–C–CH3
H
Acetyl chloride
∆ Base
O
11. The given reaction will be as
CH3CH2–C–H+ CH3MgBr H2O
OH –H2O H2SO4
Alc. KOH
(B) O
(A)
CH3CH2–CH–CH3
CH3CH=CH–CH3
Br2
CH3–CH–CH–CH3
Br Br
CH3–C≡C–CH3 (C)
(D)
Structural formula of A : CH CH CO|| H
2
3 − −
Structural formula of B : CH3CH=CH–CH3 Structural formula of C : CH3–CH–CH–CH3
Br Br Structural formula of D : CH3–C≡C–CH3
12.
NC Aniline
(i) ∆
Carbylamine + CH3Cl + 3 KOH
Chloroform NH2
+ 3 KCl + 3 H2O
Phenol
16. (i) Aldol condensation : Two molecules of an aldehyde or a ketone having at least one α-hydrogen atom, condense in the presence of a dilute alkali to give β-hydroxy aldehyde or β-hydroxy ketone.
CH3–C + HCH2CHO
Ethanal Ethanal Aldol
(ii) Trans-Esterification : An ester on reaction with excess of alcohol in the presence of mineral acid forms a new ester.
(i) 1-bromopropane to 2-bromopropane CH3CH2CHBralc. KOH CH3CH = CH2 HBr
20. DNA has a hydrogen bonded double helical structure. The two strands are antiparallel. It can be considered a polymer of nucleotide (Base + sugar + phosphoric acid). In a nucleotide base and phosphate are linked to C1′ and C5 of sugar molecule respectively. Two nucleotide are linked by 3′ – 5′ phosphodiester bonds. Hydrogen bonds are formed between –
Puric bases
Adenine = Thyonine & Guanine ≡ cytosine Pyrimidal bases
P P
S S 5′
3′ 5 4 3 2 1 H bond A = T
S P G S C ≡
P 3′
5′ 21. (a) CH2–CH
Cl H (b) CH2–C
OCOCH3
H CH3
(c) CF2–CF2 H
22. Antacids are drugs used to relieve acidity these can act in any of the following ways –
(a) neutralize acid in stomach Ex – NaHCO3, milk of magnesia (b) stop acid production in stomach Ex – oneprazole, lasoprazole (c) antihistaminic Ex- Ranitidine (zantac)
23. Absorption at the surface is called adsorption. In adsorption the conc. of adsorbate is greater at the surface than in the inner bulk.
Adsorption of gases :
Temperature : As temperature ↑ physisorption continuously ↓ where as with increase in temperature chemisorption first increases than decreases.
Physisorption
T ch
emisorption
T
Pressure : With increases in pressure adsorption of gases increases.
24. (i) Schottky defect : It is a stoichiometric defect in which pair of cations and anions get displaced from their position in such a manner a pair of vacancies are created simultaneously
In this defect
(1) Density ↓ (2) Electrical conductiriy↑
(ii) Interstials : when particles may be an additional particle of same material a foreign particle get arranged in the spaces between regular arrangement of particles.
(iii) F-centres : when crystal of NaCl is heated in the vapours of Na Cl– ions get displaced from its position and diffused into the Na vapour. And NaCl is formed which get deposited at the surface and in place of Cl– ion e– get arranged. This results in colour in the NaCl crystal. Other example LiCl in Li vapours become Pink, KCl in K vapours become violet.
25. (i) Let a0 = 100 M, conc. after 10 min a0 – x = 100 – 20
t 303 .
k= 2 log
x – a
a
0 0
= 10
303 .
2 log 80
100 = 0.02231 min–1 (ii) Let the time for 75% completion = t min x = 75 M
∴ t = k 303 .
2 log
x – a
a
0 0
=
02231 . 0
303 .
2 log 25
100 = 62.15 min
26. (a) Cerium (Ce),
(b) Due to lanthanoide contraction.
(c) In basic medium, K2Cr2O7 converts into K2CrO4
hence its colour changes while in acidic medium K2CrO4 converts into K2Cr2O7
27 (a) 4FeCr2O4+16 NaOH+
) air (O2
7 →
8 Na2CrO4+2Fe2O3 + 8H2O 2Na2CrO4 + H2SO4 →
Na2Cr2O7 + Na2SO4 + H2O Na2Cr2O7 + 2KCl → K2Cr2O7 +2NaCl (b) K2Cr2O7 + 7H2SO4 + 6KI →
Cr2(SO4)3 + 4K2SO4 + 3I2 + 7H2O 2KMnO4+ 3H2SO4 + 5H2S →
2MnSO4 + K2SO4 + 8H2O + 5S
28. (a) Acetylene is first oxidized with 40% H2SO4 in the presence of HgSO4
H – C ≡ C – H + H2O
4 4 2
HgSO
% 1
SO H
%, 40
→
CH3 – CHO
Acetylene Acetaldehyde Acetaldehyde is finally oxidized to acid with air in the presence of manganous acetate catalyst
acid Acetic3 Acetate
Manganous de
AcetaldehyCH3CHO+[O]→CH COOH (b) (i) CH3COOH Ca(OH)2→ (CH3COO)2 Ca
Calcium acetate
∆→
Ca
(HCOO)2 3
de
AcetaldehyCH3CHO+2CaCO (ii) CH3COOH Ca(OH)2→ (CH3COO)2Ca Acetic acid
3
Acetone3COCH3 CaCO
CH +
→
∆
(c) When heated with I2 + NaCO3 Solution, acetone gives yellow crystals of iodoform CH3COCH3 + 3NaOI → CH3I + CH3COONa
Acetone Yellow ppt.
(Iodoform)
Acetic acid does not give iodoform test.
(d) The carbonyl group in – COOH is inert and does not show nucleophilic addition reaction like carbonyl compounds. It is due to resonance stabilization of carboxylate ions.
R – C = O O–
R – C = O– O
Or
(a) (i) Due to smaller + I-effect of one alkyl group in aldehydes as a compared to larger +I-effect of two alkyl groups, the magnitude of positive charge on the carbonyl carbon is more in aldehydes than in ketones.
As a result nucleophilic addition reaction occur more readily in aldehyde than in ketones.
(ii) The boiling points of aldehydes and ketones are lower than corresponding acids and alcohols due to absence of intermolecular hydrogen bonding .
(iii) Aldehydes and ketones undergo a number of addition reactions as both possess the carbonyl functional group which reacts a number of nucleophiles such as HNC, NaHSO3, alcohols, ammonia derivatives and Grignard reagents.
(b) (i) Distinction between acetaldehyde and benzaldehyde: Acetaldehyde and benzaldehyde can be distinguish by Fehling solution.
Acetaldehyde give red coloured precipitate with Fehling solution while benzaldehyde does not.
→
+
+14 24+ 4 34−
Solution Fehling 3CHO 2Cu2 5OH CH
O H O Cu COO
CH 2
. ppt red2
3 −+ +
(ii) Distinction between Propanone and Propanol : Propanone (CH3COCH3) and propanol (CH3CH2CH2OH) can be distinguish by iodoform test. Propanone when warmed with sodium hypoiodide (NaOI) i.e. I2 in NaOH, it gives yellow ppt of idoform while propanol does not respond to iodoform test.
CH COCH + NaOI3 →
opanone
Pr3 3
NaOH 2 COONa CH
I
CH 3
ppt
Yellow3 ↓+ + 29. (a) Net cell reaction
Mg (s) + 2Ag+ (aq) → 2Ag(s) + Mg+2 (aq) According to nernst equation
Ecell = E°cell – n 0591 .
0 log
[ ]
[ ]
22
Ag Mg
+ +
= 0.80 – (– 2.37) – 2 0591 .
0 log –3 2
) 10 1 (
2 . 0
× = 3.32 V
when conc. of Mg+2 is decreased to 0.1, the new emf is Ecell = 3.17 –
2 0591 .
0 log –3 2 ) 10 1 (
1 . 0
× = 3.34 V
(b) (i) Mg > Αl > Zn > Fe > Cu
As we move downwards in the electro chemical series tendency to displace increases.
(ii) K > Mg > Cr > Hg > Ag
30. (i) Cl2 > Br2 > F2 > I2 (due to exceptionally small size of F, F2 has lower bond energy than expected) (ii) HF < HCl < HBr < HI (according to bond length) (iii) NH3 > PH3 > AsH3 > SbH3
(according to density of lonepair on central atom) (iv) H2O > H2Te > H2Se > H2S
(according to strength of intermolecular bonding) (v) HOCl < HClO2 <HClO3 < HClO4
(according to no. of O atoms attached to Cl)
MATHEMATICS
Section A
1. y = A cos (x + B) dx
dy = – A sin (x + B)
2 2
dx y
d = – A cos (x + B)
2 2
dx y
d = – y ⇒ 22 dx
y
d + y = 0
2.
∫
cos2x/4+sin2x/4+2sinx/4cosx/4dxdx x
∫
(cosx/4+sin /4) = 4 sin (x/4) – 4 cos (x/4) + c 3. →a= iˆ–2jˆ+3kˆ; →b = iˆ–3kˆ,⇒ 2→a = 2iˆ–4jˆ+6kˆ
∴ →b × 2→a =
6 4 – 2
3 – 0 1
ˆ ˆ
ˆ j k
i
= – 12iˆ–12ˆj–4kˆ = –4(3iˆ+3ˆj+kˆ)
⇒ |→b × 2→a | = |–4(3iˆ+3ˆj+kˆ)| = 4 32+32+12 = 4 19
4. →a = iˆ +2ˆj–3kˆ, →b = 3iˆ– jˆ+2kˆ ⇒ →a+→b= 4iˆ+ ˆj–kˆ,
→a–→b= 2iˆ+3ˆj–5kˆ
(→a +→b ) . (→a –→b ) = (4iˆ+ ˆj–kˆ).(2iˆ+3jˆ–5kˆ) = 4 × (– 2) + 1 × 3 – 1 × (– 5)
= – 8 + 3 + 5 = 0
⇒ →a +→b is perpendicular to →a –→b . 5. Let the position vectors of A, B, C, D be 6iˆ–7jˆ,
kˆ 4 – jˆ 29 – iˆ
16 , 3jˆ–6kˆ and 2iˆ+5jˆ+10kˆ respectively. Then
AB = OB – OA
= (16iˆ–29jˆ–4kˆ) – (6iˆ–7jˆ) = 10iˆ–22jˆ–4kˆ
AC = (3jˆ–6kˆ) – (6iˆ–7jˆ) = – 6iˆ+10jˆ–6kˆ
AD = (2iˆ+5jˆ+10kˆ) – (6iˆ–7jˆ) = –4iˆ+12jˆ+10kˆ
Now ∆ =
10 12 4 –
6 – 10 6 –
4 – 22 – 10
Operate R1 → R1 + R2 + R3
=
10 12 4 –
6 – 10 6 –
0 0 0
= 0
⇒ The points A, B, C and D are coplanar.
6. Put x = a sin θ
= tan–1
θ θ cos a
sin a
= tan–1 (tan θ) = θ = sin–1
a x
7. g (f(x)) = |5 f(x) – 2|
= | 5 | x | – 2| =
<
≥ 0 x
|, 2 – x 5 –
|
0 x
|, 2 – x 5
|
8.
(
A3×4⋅B4×2)
. C2×3= (X)3×2 . C2×3
= (Y)3×3 ∴ final order = 3 × 3
9. x + 2 = – (2 × –3) ⇒ 3x = 1 ⇒ x = 1/3
10. 2 – 20 = 2x2 – 24 ⇒ 2x2 = 6 ⇒ x = ± 3
Section B
11. Let A and B denote the two events respectively P(A) =
2 5
5 + =
7
5, P(B) = 5 6
6 + =
11 6
P(A) = 1 – 7 5 =
7
2, P( B ) = 1 – 11
6 = 11
5 P(At least one of A and B happens)
= 1 – P(none of A and B happens)
= 1 – P ( A∩ B )
= 1 –
7 5 ×
11 5 =
77 52
OR