Mock Test - 04 - Soluions
Mock Test - 04 - Soluions
Mock Test - 04 - Soluions
Mock Test - 04
ANSWER KEY
[1]
@swar_nava
SECTION – I (PHYSICS)
1. (4) 1 1 1 1 1 1
− = − =
Hint : At Brewster’s angle, refracted light is v u f v ( −30) 20
partially polarised and reflected is fully polarised.
1 1 1 3−2
= − = v = 60 cm
v 20 30 60
2
v
vi = ( v0 )
u
2
60
vi = (10) = 40 cm/s
30
(
sin r = sin 90º − p )
r =90º − p or r + p =90º
Reflected and refracted rays are perpendicular
For refraction at first face
sin 60º
2. (1) = = ….(1)
d sin r1 1
Resolving power of microscope =
1.22 When ray hits normally on silvered face, r2 = 0
1 r1 + r2 = A
R.P
Wavelength r1 + 0 = 37º …(2)
1 1 sin 60º 5 3 5
R1 and R2 = = =
1 2 sin37º 6 2 3
R1 2 450 3
= = =
R2 1 600 4 6. (1)
Real depth
Hint : Apparent depth =
3. (2) Refractive index
Hint: y = n11 = n22
n11 = n22 (for overlapping)
1 n2
=
2 n1
D
1
d = n2 1 n2
=
2
D n1 2 n1
d Let L = thickness of glass slab
Let X = actual distance from first face
n2 1 700 7
= = = X
n1 2 500 5 9= x = 9
n1 = 5
L−x
5=
4. (3)
1 1 1
− = 5 = L − 9
v u f
L =14
2
v L =14 1.5 = 21 cm
vi = v0
u
[2]
@swar_nava
7. (2) 100 1
Current in circuit I = = A
1 2 3
+ + ( 200 + 200 +100) 5
r1 r2 r3
Equivalent EMF = = 1
Voltmeter reading VAB = I R ' = 200 = 40V
1 1 1
+ + 5
r1 r2 r3
and equivalent internal resistance 11. (3)
1 1 1 1 Total mechanical energy remains constant
= = + +
r0 r1 r2 r3 U +K = 0
− ( qQ ) 1 2 − ( Qq ) 1
+ m ( 0) = + mv 2
1 2 40 R 2
( )
40 3R + R 2
2 2
1 2 −qQ Qq
mv = KE = +
2 40 ( 2R ) 40 R
13. (1)
8. (4)
By momentum conservation
Hint: Heat required Q = ms 0 = md .vd + m .v
P t = ms (P = Power of geyser)
m .v 4 v 4v
vd = = =
md ( 210 − 4) 206
ms 1 4200( 42 − 22) 4200 20
P= = = =1400W
t 60 60 14. (3)
Hint : When normal reaction becomes zero block
9. (2) gets separated from ship platform.
A negative charge in electric field will experience
force opposite to direction of electric field.
For induced electric field produced by changing
magnetic field with time forms closed loop.
The electrostatic force between two charges does
At the highest point of ship when motion is
not depend on presence of any other charge. The
downward, acceleration is also along mean
force remains same.
position.
a =2 A (A = amplitude)
10. (3)
Using Kirchhoff’s voltage law for closed loop. When ma mg (N = 0)
300 600 2 A g
R = = 200
900 g
A
2
So block will leave its contact at minimum
g
amplitude A= , at highest position of ship.
2
[3]
@swar_nava
15. (3) u 68
ta = = = 6.8 s
Let frequency of C is n. Beat frequency g 10
n = nA − nB
From t = 6.0 s to t = 6.8 s distance covered
5 21 21
nA = n + n = n = n (n = Frequency of C) = 3.2 m
100 20 20 In next 0.2 second, it falls under gravity
4 24 1 2 1
nB = n − n= n d2 = 0 + g ( 0.2) = 10 ( 0.2) = 5 ( 0.04 )
2
100 25 2 2
Also given nA − nB = 9 Total distance during t = 6 s to t = 7s (7th Second)
21 24 d = 3.2 + 0.2 = 3.4 m
n− n = 9
20 25
21 24 18. (2)
n − = 9 Power of e should be dimensionless
20 25
hc
105 − 96 = M 0 L0T 0
n =9 n =100 Hz x
100
[x] = [hc]
24
nB = 100 = 96Hz [x] = [M.L2T–1] [LT–1] [LT–1] =ML3T–2]
25
19. (3)
16. (3)
F
Given, By second law of motion a a =
M
r = 6.6 × 10–3 m
v = u + at
B = 0.625 T
Force acts for 4s, velocity at end of 4 s
mv2
= qvB v = u + at = 0 + 2 × 4 = 8 m/s
r Now object moves with same velocity due to
mv inertia of motion.
r=
qB
20. (3)
mv = qBr
Given,
= 1.6 × 10–19 × 0.625 × 6.6 × 10–3
T2 (sink) = 500 K
mv = 6.6 10−22 n1 = 50% = 1/2
Therefore, T
n1 = 1 − 2
h T1
=
mv 1 T
=1− 2
6.625 10−34 2 T1
=
6.6 10−22
T1
= 1.0037 10−34+ 22 T2 =
2
= 1.0037 10−12 Now, Temperature of source = T1 (same),
= 0.01 Å n2 = 60%
( )
Temperature of sink required, T2' = ?
17. (1) T2'
u 0.6 = 1 −
ta = , T1
g
T2'
For total distance both upward and downward 0.4 =
T1
distances travelled during 1 second are added.
T2' = 0.4 T1 = 0.4 2T2 = 0.4 1000 = 400 K
T2' = 400 K
[4]
@swar_nava
21. (2) 25. (3)
When no torque acts, angular momentum will IC
= , IB +IC = IE
remain conserved. I B
l = Constant
I
mR2 mR’2 C = 60
0 = 50A
2 2 IC = 50×60×10–6
R’ = R (1 + T) = 3000 × 10– 6 A = 3mA
R20 = R2(1+ T)2 × And IE = IC + IB = (3+50×10–3) mA = 3.05A
0
= (1 + T )
–2
=
(1 + T ) 2 0 26. (1)
Velocity of efflux = 2gh
= 1 + ( −2T ) (Using binomial expansion)
0 v = 2gh , here h = 10 m
(In formula, velocity does not depend on density of
− 1 = −2T
0 the liquid)
v = 2 10 10 = 200 = 14.1 m/s
= −2T
0
27. (2)
W
22. (4) W = PV and P =
According to Malus law, t
I = I0cos2 P = 1.5 m of water = 1.5 × 103 ×10
I = 15×102 N m–2
I1 = 0 V = 60 × 10–6 m3
2
PV 15 102 60 10−6 72
I0 1 I
2
Pav = = =1.08W
I1 = I1 cos2 45 = = 0 t 60
2 2 4 1.1W
1
i.e. intensity becomes times
4 28. (4)
Each planet with respect to sun traces equal area in
23. (4) equal intervals of times. So areal velocity is
l = l0T) constant.
When temperature changes by T By IInd law of Kepler: areal velocity of each planet
l = l0 (1+ T) around sun is constant
l2 – l1 = l2,0 (1 + 2T) – l1,0 (1 +T)
l2 – l1 = (l2,0 – l1,0) (l2,02– l1,0 1) T 29. (2)
Given l2 –l1 = l2,0 – l1,0 Mechanical energy is conserved in conservative
T 0 gravitational field.
U + K = Constant
1l1,0 = 1l2,0
U1 + K1 = U2 + K2
l Fe 1.2 10–5 1 –GmM mgR
Cu = = = + =−
GmM
+ K2
lFe Cu 4.8 10–5 4 R 2 R+h
R
24. (2) h=
2
2T cos 1 –GmM mgR –2 GmM
h= ,h + = + K2
rg r R 2 3 R
r = radius of tube GmM 2 mgR
A = r2 −1 + + = K2
R 3 2
A −GmM mgR
r= + = K2
3R 2
1
h GM
g= 2
A R
h' A 1 gR2 m mgR
= = − + = K2
h 4A 2 3R 2
1 mgR mgR mgR
h ' = 40 = 20 cm K2 = − =
2 2 3 6
[5]
@swar_nava
30. (2) 33. (4)
Ry 1
R = A + B and tan = XL = L and X C =
Rx C
1 1
XC = =
C 2fC
1
XC
f
When frequency increase XC decreases, graph is
rectangular hyperbola. So X2 corresponds to a capacitor
34. (2)
R = A + B = 4iˆ – 2Jˆ + 6Kˆ + (–2Jˆ – 6Kˆ )
1 1
f =
R = 4iˆ – 4 ˆj 2 LC
4 1 1
tan = =1 f1 =
4 2 LC1
= 45° with positive x-axis
1 1
With positive y-axis, angle = 90° + 45° = 135° f2 =
2 LC2
31. (1) f2 C1 C
= =
d |v| f1 C2 KC
at = , Tangential acceleration
dt
25 1
V2 =
ac = Centripetal acceleration 125 K
R
1 1
ac =
= tan 37 5 K
t
1 1
= K = 25
25 K
35. (1)
For uniform acceleration v = u = at
v = u + at
3
ac = t 30 = u + a × 2 = u + 2a ……. (1)
4 60 = u + a × 4 = u + 4a …… (2)
3
V2 = t 12 = 9t Subtracting (1) and (2)
4 30 = 2a a = 15 m s–2
at t = 1s, v2 = 9 × 1 = 9 v = 3 m/s Using in (1), 30 = u + 2× 15 u = 0 ms–1
39. (3)
V Thermal resistance
Hint: E =
x R=
I
, K = Thermal Conductivity
Sol: V = 0.6 V, x = 10–6 m KA
V 0.6 Let T = junction Temperature
E= = = 6 105V / m
x 10−6 From diagram
dH1 dH 2 dH3 dH 4
+ = +
40. (3) dt dt dt dt
Hint: Change equation for y2 to sinusoidal form T1 − T T2 − T T − T4 T − T3
+ = +
Sol: y1 = 4sin 3t + R R R R
3 30 − T 40 − T T − 20 T − 10
+ = +
2
(
y2 = 4 12 + 3
) R R R R
30 – T + 40 – T = T – 20 + T – 10
30 + 40 + 20 + 10 = 4T
1 3
( sin3t ) + .cos3t T=
100
= 25C
( ) ( )
2 2
12 + 3 12 + 3 4
1 3
y2 = 4 2 sin 3t + cos3t 43. (3)
2 2 Hint: Use equation of continuity.
= 8[sin3t.cos60° + cos3t.sin60°] Sol: By continuity equation
y2 = 8[sin(3t + 60°) A1v1 + A2v2 = A3v3 + A4v4
y 4 1 10 + 5 = 8 + A4v4
1 = =
y2 8 2
15 – 8 = A4v4
7 = A4v4
41. (1)
7 = 0.7 × v4
1 T
Hint: Fundamental frequency n = 7
2 v4 =
0.7
Sol:
v = 10 m/s
M r 2 L
= = , T = maxm stress × area
L L
[7]
@swar_nava
44. (4) 46. (1)
u 2 sin 2 u 2 sin 2 Fact based
Hint: R = , Hm =
g 2g
47. (3)
R 4
= Fact based
Hm tan
R 4 12 4 48. (1)
Sol: = =
Hm tan 4 tan Fact based
16 4 4
tan = = sin = 49. (1)
12 3 5
2
Fact based
u 2 sin 2 u2 4
4= 4=
2g 2 g 5 50. (1)
25 25g Fact based
u 2 = 8g =
16 2
25g g
u= = 5
2 2
= 5 5m / s
45. (4)
Hint: vRB = vRG − vBG
Sol:
vRB = vRG
2
+ vBG
2
= 4 2km / h
SECTION – II (CHEMISTRY)
51. (4) 52. (4)
Dalton’s law of partial pressure states that the total All are sweetening agents.
pressure exerted by a mixture of gases is equal to the
53. (1)
sum of the partial pressures exerted by each • Mass of Iron = 40 g
individual gas in the mixture; • Mass of 1.2 atoms of N = 14 × 1.2 = 16.8 g
PH2 = XH2 PT • Mass of 1 × 1023 atoms of C = (12 × 1×1023) / (6.023 ×
1023) = 1.99 g
Lets calculate the mole fraction of hydrogen
• Mass of 1.12 liter of O2 at STP = (32 × 1.2) / 22.4 =
n H2 1.6 g
X H2 =
n gases
54. (4)
XH2 = 0.28/1.31 It comprises four protein chains – two alpha and two
Putting the calculated value of XH2 in Dalton’s law beta chains, wherein each has a ring-like heme group
which contains an iron atom. In CO poisoning, since
relation as: much of the Hb is tied up with CO, O2 transport to
PH2 = (0.28/1.31)×2.35 the tissues is inhibited and it is a correct statement.
Hemoglobin, the oxygen carrier in red blood cells,
PH2 = 0.5 contains the amino acid glutamate at position 6 in
the primary sequence.
[8]
@swar_nava
55. (3) 65. (2)
More the value of ‘a’ more will be intermolecular The pH range of phenolphthalein is 8.2-10.
forces of attraction and so more easily gas can be
liquified; As, value of ‘a’ for SO2 is highest, it can 66. (1)
be easily liquified. A0 – At = kt
69. (3)
Alkane level reduction is possible with clemmensen
reagent
58. (4)
Gammaxene is an insecticide. 70. (1)
m-CPBA converts ketone into ester and base
59. (4) hydrolysis of an ester gives phenol and an acid
1.89 ppm of lithium ions corresponds to 1.89 g of
lithium in 1000000 g of solvent
71. (2)
Thus, 1 kg or 1000 g of solvent will contain 1.89 ×
Reaction of phenyl benzoate with an excess of
10–3 g of lithium
methyl magnesium bromide gives a mixture of 2-
This corresponds to [1.89 × 10–3g]/7 = 2.7 × 10–4 phenyl propan-2-ol and phenol.
mol of lithium
Molality = Moles of Lithium/mass of solvent (in kg) 72. (1)
= 2.7 × 10–4 mol / 1 = 2.7 × 10–4 m Allylic cation undergoes rearrangement to give least
substituted bromide.
60. (2)
Dithionite is S2O42– and dithionate is S2O62– . 73. (2)
Absence of plane, center and improper axis of
61. (3) symmetry.
No free electron in diamond means lower electrical
74. (1)
conductivity but atoms are fixed at a place are
continuously vibrating at their mean position means The monomers of DNA are called nucleotides.
higher thermal conductivity Nucleotides have three components: a base, a sugar
(deoxyribose) and a phosphate residue.
62. (3)
75. (1)
Only s and d orbitals have center of symmetry.
Most natural sugars are D- and most natural amino
acids are L-. One method for determining whether a
63. (3)
molecule is D- or L- by looking at the Fischer
Any species containing odd number of electrons.
projection of a molecule.
64. (3)
76. (4)
Draw the structure of the given compounds and just
The more the number of hyperconjugating structure,
check for the bridging oxygen.
the more is the stability.
@swar_nava
77. (2) 85. (3)
1 3 Apply Hardy-Schulze rule.
2 N2 + 2 H2 → NH3
86. (3)
fH° = –45 kJ mol–1
Number of radial node = 1, n = 3
1 n–l–1=1
N2 → N
2 3–l–1=1
941.8 l=2–1=1
Bond dissociation enthalpy of N2 = = 470.9
2 h
So, orbital angular momentum l ( l + 1)
kJ mol–1 2
3
H2 → H
2 87. (2)
3 In distilled water, there are no ions present except H+
Bond dissociation enthalpy of H–H = (436) = 654
2 and OH– ions, both of which are immensely minute
kJ mol–1 in concentration, which renders their collective
Total enthalpy of atomization = 470.9 + 654 – (–46) conductivity negligible.
= 1170.9 kJ mol–1
88. (4)
Only sulphur of carbon disulphide is having d-
78. (3)
orbital.
Hydroboration-Oxidation is a two step pathway used
to produce alcohols. The reaction proceeds in an
89. (1)
Anti-Markovnikov manner, where the hydrogen
G° = – nFEcell = H° – TS°
(from BH3 or BHR2) attaches to the more substituted
carbon and the boron attaches to the least substituted –TS° = –nFEcell – H°
carbon in the alkene double bond. –(nFEcell + H)
–S° =
T
( )
79. (1)
– 2 96487 4.315 – 825.2 103
Apply CIP Rule. – S =
298
80. (2) S° = 25JK –1
137. (2)
147. (1)
End products of anaerobic respiration in animals
results in lactic acid and energy. Dikaryophase is characteristic featureof
Ascomycetes and Basidiomycetes
138. (2)
148. (2)
Amphibolic pathway is a biochemical pathway
that includes both anabolic and catabolic Selaginella and Salvinia are heterosporous which
processes. produce two kinds of spores {macro (large) micro
(small)}.
In TCA cycle energy is both Consumed and
produced during the process.
149. (4)
139. (1) (1) Subsidiary cells→ Accessory cells
Cellulose is not an alkaloid. (2) Lenticels→ Aerating pores in the back of
plant
(3) Guard cells →Regulate opening and closure
140. (2)
of stomata
Flow of metabolite through metabolic pathways
(4) Bulliform cells→ Rolling in and out of
have definite rate and direction.
leaves
150. (3)
141. (4)
Presence of vessels is a characterise feature of
Genetic material of prokaryotic cell is nucleoid it
angiosperms.
is composed of single circular DNA molecule.
@swar_nava
SECTION – IV (ZOOLOGY)
151. (1) 165. (1)
Aschelminthes are pseudocoelomates. Phosphodiester bond is characteristically found in
Platyhelminthes are acoelomate and Annelida and deoxyribonucleic acid.
Arthropoda are coelomates.
166. (4)
152. (1) Sub-maxillary glands are located on upper jaw.
Saltation - de Vries
Formation of life was preceded by chemical 167. (1)
evolution - Oparin and Haldane Sphincter of oddi – Hepato pancreatic duct
Reproductive fitness - Darwin Cystic duct – Gall bladder
Life comes from pre-existing life - Louis pasteur Hepatic lobule – Glisson’s capsule
Brunner’s gland – Sub-mucosal gland
153. (4)
Respiratory organ as tracheal system is present in 168. (1)
cockroaches. Anxiety and eating spicy food together can lead to
indigestion.
154. (4)
Heart is ventral in position in chordates. 169. (2)
Thoracic cage of man is formed of Ribs, sternum
155. (3) and thoracic vertebrae.
Ichthyophis is a limbless amphibia.
170. (3)
156. (3) Solubility of gases affect the rate of diffusion.
20 (10 pairs) of spiracles are present in Periplanata Solubility of gas rate of diffusion.
americana.
171. (1)
157. (2) 3% of oxygen is carried through dissolved state in
Ligaments connect bone to bone. plasma.
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