0 0
u v ×
ue
D
(ii) When final image is formed at distance D M = –
0 0
u
v
+ f 1 D
4. S1 and S2 are the two desired surfaces.
q S1
S2
5. Superconductors are those material which resistivity is zero below a certain temperature.
6.
~
Induction coil
Conducting Plate I
Sphere I Sphere II Conducting Plate II
Detector
7. Zero
8. High energy X-rays are known as hard X-rays and low energy X-rays are known as soft X-rays. These terms are relative.
9. (1, 3) – (2, 4)
10. The daughter element
(release of energy is accompanied by an increase of B.E)
11. (i) ‘Depletion layer’ width decreases, (ii) Junction field becomes very high
12. The approximate thickness of the film should be of the order of wavelength of the light.
13.
Intensity O I
'θ θ2 θ2
1'
θ θ1 yn
1'
PP1
P2 2'
P
14. d
0 A
∈ = 5 µF ………. (i)
d KA 2
∈0 = 20 µF ……. (ii)
(ii) ÷ (i) 2 K =
5 20
∴ K=8 15.
Uniform magnetic field
Paramagnetic
substance Diamagnetic substance
16. (i) λ = ν c (ii) (Uav)E =
4 1 ∈0 2
E0
MOCK TEST-3 (SOLUTION)
MOCK TEST– 3 PUBLISHED IN SAME ISSUE
17. U = 2 1 Li2 =
2
1 × 2 × 10–3 (5)2 J = 2.5 × 10–2 J
18. XL = ωL = 100 × 30 × 10–3 = 3 Ω R = 4 Ω
V = 200 2 sin 100 t Z = R2+X2L
= 42+32 = 5 Ω
∴ cosφ = Z R =
5 4 = 0.8
19. (i) X-rays are e.m. waves
(ii) X-rays are transverse in nature 20. Reduction factor =
16 1 = 4
2
1 in 4 days.
Hence life = 1 day
∴ For 6 days reduction factor would be 6 2
1 = 64
1
∴ original amount = 4 × 10–3 × 64kg = 0.256 kg 21. A telescope views large objects at large distances; a
microscope views small objects at small distances.
Both need a small field of view. A camera views objects of ordinary sizes at fairly close distances.
Here the field of view is required much more (compare 45° for a camera with about 1° for a microscope objective and something similar for a telescope, a moon subtends about 0.5° at the earth) Thus rays entering a camera lens are far from being paraxial and aberrations will be large and images will be blurred if the apertures are not very small.
For a telescope, on the other hand, the important thing is its ability to resolve distant abjects (i.e., see them as distinct). We have seen that the resolving power increases with increase in aperture.
Therefore, telescopes have as large an aperture as feasible.
22. Q Current in 5 Ω is zero
∴ bridge is balance
∴ R R + 6
6 1 =
20 10
or R
R 6 6+
= 2 1
∴ R = 3 Ω
23. |ε| = L dt di =
dt
L d (t2) = 2 L t
∴ At t = 4, ε = 2 × 5 × 10–3 × 4 = 40 × 10–3 V
24. P Q
B
S a R
b
Let i → current N → No. of turns l → length n = l
N
Applying Ampere's Law along PQRS,
→
∫
→B.dl = µ0 inet∫
→ →Q
P
B l.d +
∫
R→ →Q
B l.d +
∫
S→ →R
B l.d +
∫
P→ →S
B l.d = µ0 inet
B.a + 0 + 0 + 0 = µ0 n (a) i
∴ B=µ0ni
25.
Ig
G
I2 – Ig
Q C A
P
I1 S R
D
I1 + Ig
– +
B
I
I2
+ –
+ –
+ – – +
It is an arrangement of four resistances used for measuring unknown resistance.
Applying KVL in loop ADBA –I1 R + Ig G + I2 P = 0 …. (i)
and in loop DCBD,
– (I1 + Ig) S + (I2 – Ig) Q – Ig G = 0 for balance bridge, Ig = 0
∴ I1 R = I2P …….. (iii) I1 S = I2 Q …….. (iv) (iii) ÷ (iv) we get
S R QP =
26. Diamagnetic substances are feebly magnetised in opposite direction to that of magnetising field Paramagnetic substances are feebly magnetised in the direction of magnetic field.
Ferromagnetic substances are strongly magnetised in the direction of magnetic field.
27. + V –
P N + –
Reverse biasing
In the experimental set, the P and N terminals of a P-N diode are connected to the negative and positive potential point's of a potential divider respectively. As the reverse bias current a feeble current measurable in micro amperes so a micro-ammeter is used to measure it. To plot reverse bias characteristic, we note down reverse currents corresponding to various different reverse voltages on the diode with help of the potential divider.
After obtaining it, the applied voltages are plotted along X-axis and corresponding reverse currents along Y-axis of a graph as shown in the fig.
I (mA)
V + –
Characteristic of a P-N diode
Reverse bias resistance- The ratio of small change in reverse voltage (before break down voltage) to the corresponding change in reverse current for a P-N diode is known as its reverse bias resistance, i.e, Reverse bias resistance
=
current reverse
in change ing Correspond
voltage reverse
in change Small
28. Circuit diagram for drawing the input and output characteristics.
R1
Vcc VCE
+ – Ic
– + mA C E IB
IB
+ – µA VBE VBB
R2
B
Typical shape of the input characteristics.
29.
α β
α Q'
P' B
fe
Q'' E
P''
O
Astronomical telescope
(i) When the final image is formed on the nearest distance of clear vision D
M = – fe
f0
+ D fe 1
(ii) When the final image is formed at infinity M = –
fe
f0
On increasing the aperture of the objective lens the magnifying power of telescope will increase.
30. U =
∫
0qVdq =∫
0qCq dq =C q 2
2
= 2 1 CV2
= 2
1
∈
d
0 A (Ed)2
= 2
1 ∈0E2 (Ad)
∴ Energy density, Ad U =
2 1 ∈0E2 OR
(i) From Gauss' theorem φ =
∈in0
q
∴ qin = φ × ∈0 = – 6 × 103 × 8.85 × 10–12 = –5.31 × 10–8C
(ii) Flux remains the same
+ –
d E
CHEMISTRY
1. H.C.P has highest 74% efficiency
2. When in Fe(OH)3 ppt FeCl3 is added Fe+3 ions are adsorb over the surface of Fe(OH)3 it results in the formation of Fe(OH)3 solution.
3. Lithium tetrahydrido aluminate (III) 4. FeCO3 is siderite ore.
5. Phenol & Formaldehyde
6. Antiseptics are germicides which can be applied on wounds
Ex. Soframycin, Tincture iodine
7. Amino acid in which amino group are more then – COOH group are called basic amino acid.
Lysine ( R is (CH2)3 – NH2 ) 8. CH3F < CH3Cl < CH3Br < CH3I 9. t1/2 = 5730
λ =
2 / 1
693 . 0
t = 1.21 × 10–4 yr–1 λ = t
303 . 2 ln
t 0
a a
t = 4
10 21 . 1
303 . 2
× − ln 8 10
= 1845 yr
10. With the increase in temperature rate constant increases. It is found that with 10 K rise in temperature the rate of reaction become 2 – 3 times.
With the increase in temperature
(1) More no. of collisions occur between the molecules.
(2)Only those molecules which are having minimum sufficient energy to participate in the chemical rxn, reacts with each other and form product
(3) For effective collision activated molecule must collide in the proper orientation
∴ Rate of rxn = P × Z . e−Ea/RT P = Orientation factor Z = No. of collisions
RT Ea
e− / = No. of activated molecules
11. (a) Standard Hydrogen electrode - When H2 gas at 1 atm pr is supplied on Pt sheet dipped in the aqueous solution of an acid having molarity 1M
H2 l atm
Pt
1 M aq acid solution
Following Chemical rxn takes place H2(g, l atm) 2H+ (aq, 1M) + 2e– the potential of this half electrode = 0.0 V (b) Kohlrausch's law states conductivity of a
solution at infinite dilution is equal to sum of molar conductivity of all the ions present in the solution.
0m
Λ = γ⊕ λ⊕º + γ –º λ–º
12. Actinides show much higher oxidation states than Lanthanides because energy difference between 5f, 6d and 7s orbitals is less and hence electrons also participate from 5f orbital also.
13. The seperation of Ag+ and Hg22+ in group – I is carried out by dissolving the precipitate of AgCl in NH4OH, AgCl forms a soluble complex with NH4OH.
Whereas Hg2Cl2 forms a black water insoluble complex.
⇒ AgCl + NH4OH → [Ag(NH3)2]Cl+2H2O (water soluble)
⇒ Hg2Cl2 + NH4OH → Hg(NH2)Cl + Hg + HCl black percipitate + H2O 14. The concentrated ore is heated with excess of air to
remove water and carbon dioxide to remove sulphur and arsenic impurities and to oxidise ferrous to ferric oxide for eg;
2Fe2O3.3H2O → 2Fe2O3 + 3H2O FeCO3 → FeO + CO2 ↑ S + O2 → SO2 ↑ 4As + 3O2 → 2As2O3
4Fe + O2 → 2Fe2O3
15. PHBV has 3hydroxybutanoic acid & 3
hydroxypentanoic acid.
O – CH – CH2 – CO – O – CH – CH2 – CO
CH3 CH2 – CH3
n
16. Purine bases in DNA and RNA are Adenine &
Guanine.
Pyrimidine bases in DNA are Cytosine & Thymine while in RNA are Cytosine & Uracil
17. Detergents are sodium or potassium salts of sulphonic acid.
Cationic detergent – Cetyl trimethyl ammonium bromide
Anionic – Alkyl benzene sulphonate 18. (A) CHCl3 (B) HC ≡ CH
19. (a) Radius of gold r = 0.144 nm F.C.C. 4 r = 2 a
a =
2 r
4 = 2 2r
= 2 × 1.414 × 0.144 = 0.407 nm
Edge length a = 0.407 nm.
(b) (i) When
) p 4 s 4
( Ge2 2 is doped with
) p 5 s 5
( In2 1 13th gr element all the 3e– get bonded and fourth bond of Ge contain only one e– and hence an e– deficient bond or a hole is formed and p type semiconductor is formed.
(ii) Similarly when Si is doped with As 4s24p3 4e–'s get bonded and fifth e– remain unbonded ∴ n-type semiconductor is formed.
20. p = p0AxA + p0BxB
nA = 1 nB = 2 mol.
T1
p = 250 bar 250 = p0A ×
3
1 + p0B× 3
2 ... (1) nA = 2 nB = 2 mol
T2
p = 300 = p0A . 2
1 + p0B × 2
1 ...(2) solving (1) & (2)
0A
p = 450 bar p0B = 150 bar
21. (a) (i) When silica gel is placed is atmosphere saturated with water adsorption of moisture takes place
(ii) CaCl2 adsorbs H2O.
(b) Zeolite is a shape selective catalyst which are metal alumino silicates Mx/n(AlO2)x(SiO2)y.mH2O.
When zeolite is heated pores are generated, these pores are having size 260 – 740 pm which can absorb molecule of definite size. Therefore it is called shape selective catalyst
22. In case of nitrogen family, basic character of hydrides decreases from NH3 to BiH3 because with the increase in size of central element lone pair density about it decreases and tendency of proton (H+) to coordinate with it decreases and hence basic character of hydrides decreases.
23.
O Cr O
O¯
O
Cr O
O¯ O 131º
Cr2O72–
24.
25. 2-Methoxy-2-methylpropane 26.
27. The compound A can be either an aldehyde or a ketone. Since it resists oxidation it must be a ketone. i.e., acetone (CH3COCH3)
The reactions involved are :
CH3 – C – CH3
Reduction O
2[H] CH3 – CH – CH3
OH
(B) 2-Propanol
HBr –H2O
CH3 – CH – CH3
Br
(C) 2-Bromopropane CH3 – C – CH3 + CH3 – CH – CH3 Mg
MgBr O
(B)
(A)
CH3 – C – CH3
OMgBr
CH(CH3)2
H2O
CH3 – C – CH3
OH
CH(CH3)2
2, 3 – Dimethyl-2-butanol
28. (a) The ideal solution is that solution which follows the Raoult's law i.e for ideal solution :-
(i) ∆Hmixing = 0 (ii) ∆Vmixing = 0
Non ideal solution is that solution which does not follow Raoult's law and for which :
(i) ∆Hmixing ≠ 0 (ii) ∆Vmixing = 0
In case of cyclohexane – ethanol a solution with +ve deviation is obtained
In case of a acetone – Chloroform
–ve deviation is obtained Cl
Cl Cl
C H O C
CH3
CH3
+δ –δ +δ
(b) Moles of solute nB = M 30
Moles of H2O (nH2O) = 18 90 = 5
xA =
M / 30 5
5
+ =
M 6
M + pA = p0AxA
pA = p0A. M 6
M
+ ... (1)
When 18g of H2O is added to solution
O H2
x =
M / 30 6
6
+ =
M 5
M + p'A = p0A
M 5
M
+ ... (2)
from (1) & (2) M 6
M 5
+ + =
9 . 2
8 . 2
⇒ M = 23 g/mol
29. (a) Hypophosphorus acid
H OH P
O
H
(b) Pyrophosphoric acid
HO O P
O
OH
OH P
O
OH (c) Dithionic acid
HO — S — S — OH O O
O O (d) Marshall acid
HO — S — O — O — S — OH O O
O O
(e) Hypophosphoric acid
HO P O
OH
OH P
O
OH
30. (a) Since compound (A) is optically inactive and contains nitrogen which gives alcohol with HNO2, it is primary amine. The reactions may be given as CH3CH2CH2CH2NH2HNO →2 CH3CH2CH2CH2OH+N2+H2
O
(A) (B)
1-Aminobutane 1-Butanol CH3CH2CH2CH2OH –H2O
K 440 , SO H24→
CH3CH2CH = CH2
(B) (C)
1-Butene
CH3CH2CH=CH2 HBr → CH3CH2CH–CH3
(C) |
Br
(D) 2-Bromobutane(optically active)
MATHEMATICS
Section A
1. Q (1, 1), (2, 2), (3, 3) ∉ R
⇒ R is not reflexive.
again (1, 2) ∈ R ⇒ (2, 1) ∈ R ⇒ R is symmetric again (1, 2) ∈ R, (2, 1) ∈ R but (1, 1) ∉ R
⇒ R is not transitive.
2. On differentiating 3
2x–1/3 + 3 2y–1/3
dx dy = 0
dx dy = –
3 / 1
3 / 1
x y
3. We have,