Test ID : 627

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nd
LT, XII & CC-I – GRAND TEST – 2 (2 YEAR) (ANSWER KEY) 30.07.2021
Choose the correct answer : 180 x 4 = 720
1. 2) increases In non-uniform electric field, the dipole may
Solution: experience both non-zero torque and
translational force.
A soap bubble has two surfaces. If negative
−Q Q
charge is given to it, then its both surfaces 6. 1) ,
becomes negatively charged. There will be a 2ε0 2ε 0
repulsion between two surfaces consequently Solution:
its radius increases. φ = E.A
2. 4) zero Half of the total flux will go above flat
Solution: surface. Remaining half will pass through flat
as well as curve surface.
The circular path around α-particle acts as
7. 1) 1.1 x 10-10
equipotential surface for another α-particle.
Solution:
Work done in moving a charged particle once
around an equipotential surface is zero. Capacitance of a spherical conductor = 4πε0a
1 1
3. 3) 2 : 1 C= 9
x1 = x10−9
9 x10 9
Solution:
When two conducting spheres are connected = 0.11 x 10-9 F = 1.1 x 10-10F
by a conducting wire, charge will flow from 1
8. 2) (ρ1 + ρ2 )
one sphere to other till both acquire same 2
potential. Solution:
v E r 2 R = R1 + R2
E= ⇒ 1 = 2 = =2 :1 2l ρl ρ l
r E 2 r1 1 ρ 2 = 1 2 + 2 2 ⇒ 2ρ = ρ1 + ρ2
πr πr πr
4. 4) 32 x 10-32J
(ρ1 + ρ2 )
Solution: ρ=
2
1 1 q2 3) 1.1 Ω
W = CV 2 = 9.
2 2 C Solution:
1 (8 x10−18 )2 E 50
= x = 32 x10 −32 J I= ⇒ 4.5 =
2 100 x10 −6
R+r 10 + r
5. 3) a torque as well as translational force 45 + 4.5 r = 50 ⇒ 4.5 r = 5
Solution: 5 1
⇒ r= = = 1.1 Ω
4.5 0.9
10. 3) 15 Ω
Solution:
R BC  100 − 40 
= ⇒ R =10  
10 AC  40 
60
= 10 x =15 Ω
40
 2 Test ID : 627
11. 1) magnetic dipole = M 2 + M 2 + 2MM cos120o
Solution:
A current carrying coil placed in a magnetic  1
= 2M 2 + 2M 2  − 
field behaves like a magnetic dipole.  2
12. 1) zero = 2M 2 − M 2 = M
Solution:
18. 4) 25 x 10-6 H
It can be takes as two sets of parallel wires
Solution:
carrying equal currents in the same direction.
di e
Hence total magnetic field induction at the e=− L ⇒ L=−
centre of diagonal will be zero. dt  
di
 
13. 2) α - particles  dt 
Solution: −3 x10 −4 3 x10 −4 x 0.5
= =
Cyclotron cannot accelerate uncharged  7 − 13  +6
 
particles and higher charged particles like  0.5 
x-rays, γ-rays and β - rays. = 0.25 x 10-4 = 25 x 10-6 H
14. 2) compress 19. 2) 0.2 V
Solution: Solution:
dφ d dB
e = N = N (BA) = NA
dt dt dt
(0.1 − 0)
= 200 x 10 x 10-4 x
Due to flow of current in same direction in 0.1
-4
two adjacent loops, an attractive magnetic = 2000 x 10 = 0.2 V
force will be produced, due to which spring 20. 2) 0.05 mH
will get compressed. Solution:
15. 2) infinity
M = µ0n1n2Al
Solution:
= 4π x 10-7 x 500 x 200 x 4 x 10-4 x 1
I = 16 π x 10-7 x 105 x 10-4
T = 2π , (BH = 0)
MBH = 0.05 x 10-3 H = 0.05 mH
I 21. 3) 4π x 10-2 V
⇒ T = 2π = ∞ (infinity)
M (0) Solution:
1 1
16. 2) 150 kˆ N − m emf induced = BR 2 ω= BR 2 (2πf )
2 2
Solution:
1
τ = M x B = 50 ˆi x (0.5 ˆi + 3j)
ˆ = x 0.2 x (0.1)2 x (2π x 20)
2
= 150 (iˆ x ˆj) =150 kˆ N − m = 4π x 10-2 V
17. 1) M 22. 2) 10 2 A
Solution: Solution:
Here θ = 120o E i   200 x i 0 
Resultant magnetic moment P =  0 0  cos φ =1000 =   cos 60
o

 2   2 
 3 Test ID : 627
1000 = 50 i0 ⇒ i0 = 20 A 27. 2) 50 %
i 20 Solution:
i rms = 0 = =10 2 A
2 2 Intensity of polarized light transmitted from
2) 5 J 1
23. first polarizer = I0 = 50 % I0
Solution: 2
28. 2) light is scattered by droplets in fog
E 0 I0 E2
H = Pt = T= 0 Solution:
2 2Rf
Light is scattered by droplets in fog, so we
100 x100
= =5J (OR) cannot see through fog.
2 x 20 x 50
29. 3) 0.09o
∆H = i2rms RT Solution:
i v 100 5 Angular dispersion = δv - δR = (µv - µR) A
i rms = 0 = 0 = =
2 R 2 20 x 2 2 = (1.523 – 1.514) x 10o
v = v0 sin ωt = 100 sin 100 πt = 0.009 x 10o = 0.09o
ω = 100 π ⇒ 2πf = 100 π 30. 4) all of the above
1 1 Solution:
f = 50 ⇒ T = =
f 50
2
 5  1 25 2
∆H =   x 20 x 50 = 2 x 5 = 5J
 2
24. 2) lags behind the current (i) when a lens is cut along its diameter, radius
Solution: of surfaces do not change. Focal length of
1 each part is equal to original lens.
When, ωL < , emf lags behind the current
ωC (ii) Intensity becomes half
as tan φ is negative. I A 
2

25. 4) a12 + a22 (iii) I α A , 2 =  2 

2
I1  A1 
Solution: 2
π I1  A 2  A2 1 A
Two waves differ in phase by =  ⇒ = ⇒ A2 = 1
2 2I1  A1  A1 2 2

a = a12 + a 22 ⇒ a2 = a12 + a22 (∵ I ∝ a2) (iv) As intensity becoming half, light

transmitting area also becomes half.
I = (a12 + a22) 31. 2) 4 eV
26. 1) decrease
Solution:
Solution: Einstein photo-electric eqn, E = W + KE
Dλ
β= KE = E – W ⇒ KE = hν - W. Comparing this
d equation with y = mx – C, intercept of y-axis
Dλ ' λ represents work function, (C = 4 eV)
⇒ β' = , But λ ' =
d µ 32. 4) 8 x 106 m/s
β Solution:
∴ β'=
µ 1 1
mv12 = 2hν 0 − hν 0 ⇒ mv12 = hν 0 ---- (1)
2 2
 4 Test ID : 627
1 1 38. 2) 6.0 fm
mv 22 = 5hν 0 − hν 0 ⇒ mv 22 = 4hν 0 ----- (2)
2 2 Solution:
1
mv 22 R = R0A1/3
eq(2) 2 4hν 0
⇒ = R Te  125 
1/3
5
eq(1) 1
mv12 hν 0 =  =
2 R Al  27  3
v 22 5 5
⇒ = 4 ⇒ v2 = 2v1 RTe = x R Al = x 3.6
v12 3 3
v2 = 2 x 4 x 106 = 8x 106 m/s = 6 fm
33. 1) 3.33 Ao 39. 2) absorb excess number of thermal neutrons
Solution: Solution:
nh h 2πr Cadmium and Boron rods are used to absorb
mvr = ⇒ =
2π mv n excess number of thermal neutrons.
h 2πr 40. 2) infinite
⇒ λ= ⇒λ=
mv n Solution:
⇒ 2πr = nλ = 2 x 3.14 x 0.53 = 3.33 Ao Applied reverse voltage is greater than VB
34. 3) 3.4 eV (Breakdown) voltage, current is infinite.
Solution: 41. 3) they cannot be used with high voltage
13.6
E n = − 2 eV Solution:
n
Semiconductor devices are suitable for flow
 −13.6  13.6
∴ ∆E = E∝ - E2 = 0 −  2  = = 3.4 voltages.
 2  4 42. 1) 0, 0
35. 2) πr
Solution:
Solution:
nh 2h (0 + 0).(0.0) = 0 . 0 = 1.1 =1
mvr = , For n =2 ⇒ mv =
2π 2πr 43. 4) both I and III
h h Solution:
Now wavelength, λ = = = πr
p  2h  In region I → it acts as closed switch
 
 2πr  In region III → it acts as open switch
36. 3) 0.25 x 107 m-1 44. 2) 6 x 10-8 T
Solution:
Solution:
1  1 1 
Wave number = = R  2 − 2  x12 E E 18
λ 2 ∞  C= ⇒ B= =
B C 3 x108
107
Wave number = = 0.25 x107 m −1 = 6 x 10-8 T
4
45. 2) charge
37. 4) all the above
Solution:
Solution:
Stability of a nucleus can be measured by EM waves are uncharged. So they do not
Avg. binding energy, packing fraction and transport charge. They transport energy,
ratio of no. of neutrons to protons. momentum, information.
 5 Test ID : 627
46. 3) 100 pm 53. 2) D-glucose & L-glucose are diastereomers
Solution : Solution :
For NaCl structure, D-glucose and L-glucose are enantiomers but
a a 580 not diastereomers.
rc + ra = ⇒rc = - ra = - 190 = 100 pm 54. 2) Ni-Cd cell – Primary cell
2 2 2
Solution :
47. 3) 2a : 2 a : 3a
Ni-Cd cell is not a primary cell. It is a
Solution : secondary cell.
a a 3a ln 2
rsc : rfcc : rbcc= : : = 2a: 2a: 3a 55. 2)
2 2 2 4 K
48. 1) A4B3 Solution :
Solution : 0.693 ln 2
For 1st order reaction t1/2 = =
HCP, Z = 6 ⇒ B6 K K
2 2 2 56. 3) R
TV = x 2Z = x 2 x 6 = 8 ⇒ A8
3 3 3 Solution :
Molecular formula is A8B6 ⇒ A4B3 E 1
ln K = ln A - a  
49. 2) 1.5m Al2(SO4)3 R T
Solution : −E a
Slope, = −1 ⇒ Ea = R
1 R
Freezing point ∝
im 57. 3) -0.59 V
Maximum number of particles Solution :
⇒ Maximum ∆Tf pOH = 4 ⇒ pH = 10
⇒ Minimum Tf E H+ /H = -0.059 x pH = -0.059 x 10 = -0.59
2

50. 3) 735 mm 58. 2) 2

Solution : Solution :
10% solution has 10g of urea and 90g of water. n −1
1 t a 
10 1 t1/2 ∝ ⇒ 1 = 2 
RLVP = Xsolute = 60 = 6 = 1 a n −1 t 2  a1 
10 + 90 1 + 5 31 n −1
60 18 6 10  2 ×10−2 
= 
760 − P 1 760 20  4 ×10−2 
= ⇒ 760 − P =
760 31 31 n −1
1 1
P = 735 nm =  ⇒n =2
51. 2) 390.5 2  2
Solution : 59. 3) adsorption, adsorption
Λ 0m(AB) = Λ 0m(AC) + Λ 0m(BD) - Λ 0m(CD) Solution :
Both are applications of adsorption.
= 90.1 + 425.9 – 126.4 60. 4) ZnCO3 – ZnO, Leaching
= 390.5 S cm2 mol-1 Solution :
52. 3) HNO3 + H2O, Minimum boiling point ∆
ZnCO3  Calcination
→ ZnO + CO2
azeotrope
Solution : 61. 4) Zn
HNO3 + H2O, shows negative deviation Solution :
Due to configuration 3d104s2
⇒ maximum boiling point azeotrope
 6 Test ID : 627
62. 3) PtCl4.2NH3
Solution :
[Pt(NH3)2Cl4], non-ionisable and shows GI 70. 3)
M a 2 b4 type ⇒ 2GI
63. 3) Glyptal
Solution : Solution :
Glyptal is a condensation polymer but not a
Both cyanide and ester is reduced to aldehydic
fibre.
function.
64. 4) it has 4 chiral centres in cyclic hemi acetal
structure 71. 2) CH3CHO
Solution : Solution :
Open chain glucose has 4 chiral carbons HCHO not undergo aldol, CH3CHO is more
whereas cyclic hemiacetal has 5 chiral centres. reactive than others.
72. 3) (CH3)2NH
65. 3) Solution :
2° amines more basic than 1° amines aliphatic
Solution : is more basic than aromatic.
Inversion of configuration by SN2 mechanism.
66. 4) PhCH2Br
Solution : 73. 2)
PhCH2Br is more reactive towards both SN1
and SN2 mechanisms then isopropyl bromide.

67. 4) Solution :
Rate of hydrolysis of ester ∝ EWG
Solution : 74. 3) C, D
Solution :
A is , B is . B on anti R − MgX + R '− OH → R − H + R 'OMgX
Base Acid CA CB
addition gives (4). All bronsted bases are Lewis bases.
75. 3) Antiseptic
Solution :
68. 3) , C2H5I

Solution : A is

X is
A is Chloroxylenol used as antiseptic.
76. 2) P2O5/ ∆
69. 2) CH3CHO Solution :
Solution : A is CH3COCl and B is CH3CONH2
X is CH3Cl and Y is CH3CN
CH3CONH2 → P2 O5
∆ , − H2O
CH3CN
 7 Test ID : 627
77. 3) C6H5NC, HCOOH 83. 4)Bleaching action of H2O2 is due to reduction
Solution : Solution :
NH2 NC NH2 Bleaching action of H2O2 is due to oxidation
CHCl3 H3O+ 84. 2) a, c
+ HCOOH
KOH
Z
Solution :
X Y X Down the group reducing nature increases due
78. 2) 5 to decrease in M-H bond energy and correct
Solution : order of boiling points is H2S < H2Se < H2Te
< H2 O
4s23d6 ⇒ Z = 26, given EAN = 36
26 – 0 + 2x = 36 85. 3) H3PO2 < H3PO3 < H3PO4 (acidic strength)
x=5 Solution :
79. 4) magnetic moment Order of acidic strength is H3PO2 > H3PO3 >
Solution : H3PO4
In all isomers of CrCl3.6H2O, weak field Cr+3 86. 2) A-S, B-R, C-Q, D-P
has 3 unpaired electrons. Solution :
80. 3) CO2 (or) CCl4 A) XeO2F2 S) sp3d hybridization
Solution : B) XeOF4 R) square pyramidal
C) XeF4 Q) 2 lp + 4 bp
D) XeF2 P) 1st excited state
87. 3) 1
Solution :
81. 2) between 1 and ∝ in all cases If 100% molecules have same molar mass
Solution : M −w
M n− = M −w ⇒ PDI = =1
For freundlich isotherm, M n−
1 88. 3) A-R, B-S, C-Q, D-P
O< < 1, For all cases
n Solution :
⇒ ∝ > n > 1, For all cases A) Antimalarial R) Chloroquine
82. 2) 1.8 g B) Antiseptic S) Faracine
Solution : C) Antipyretic Q) Paracetamol
W 1000 D) Antimicrobial P) Salvarsan
M= x
MW V
89. 3) A-s, B-r, C-q, D-p
W 1000
0.6 = x ⇒ W1 = 1.8 g Solution :
60 50
A) H3PO2 s) White P4 + Alkali
W 1000
0.3 = x ⇒ W2 = 0.9 g B) H3PO3 r) P2O3 + H2O
60 50
C) H3PO4 q) P4O10 + H2O
WCH3COOH = 1.8 – 0.9 = 0.9 g
D) H4P2O6 p) Red P4 + Alkali
0.9 gm per 0.5 g chreal
90. 4) [PtCl4]-2
⇒ per 1 gram charcoal
Solution :
= 1.8 g
[PtCl4]-2 has square planar structure.
 8 Test ID : 627
91. 3) Bryophytes 121. 3) The pyrimidine complimentary to adenine
92. 3) A – ii, B – iii, C- iv, D – i other than thymine
93. 2) Both the statements are incorrect 122. 4) Extension of DNA in PCR
94. 3) In heterozygous condition characters blend 123. 2) 6
to give intermediate phenotype 124. 2) Recombinant protein
95. 4) more than one option is correct 125. 1) Salmonella typhi
96. 4) It is a dioecious plant 126. 2) i and ii
97. 2) It pushes the developing embryo deep into
127. 4) DNA - DNase
the endosperm
128. 4) Statement – I is incorrect, statement – II is
98. 3) All pollengrains are spherical in shape and
correct
measuring about 25 – 50 µm in diameter
129. 2) A and B chains
99. 3) Orchids, Striga – Thousands of large seeds
130. 3) Retrovirus
100. 4) All the above
131. 2) α-1-antitrypsin
101. 3) Aspergillus niger
132. 3) Vitamin C
102. 3) Reverse transcriptase
133. 3) Repressor
103. 2) sequence before the start codon
134. 3) Introns
104. 2) monozygotic twins
135. 2) IARI and KVIC
105. 1)Taylor and his colleagues – Radioactive
thymidine 136. 4) is the result of first meiotic division
106. 2) All F1 offsprings are heterozygotes 137. 4) Uterus
107. 2) F1 tall plants 138. 2) LH surge induces ovulation
108. 2) sRNA 139. 3) Relaxin, hCG, hPL
109. 4) both 2 and 3 140. 4) oxytocin from the maternal pituitary
110. 2) 4 141. 4) zero
111. 1) Round and yellow 142. 4) both 1 and 2
112. 2) Flower colour – Red / White 143. 3) A - iv, B - iii, C - ii, D - i
113. 3) 5 144. 2) 3
114. 1) Glomus – Pinus 145. 1) World population in 1900 was about 6
115. 3) Many of the microbes in waste water are
billion
non-pathogenic
146. 3) A - iv, B - iii, C - v, D - i
116. 2) Multicellular fungi
147. 2) Neophron, Pavo, Struthio
117. 4) Statement – I is incorrect, statement – II is
148. 3) Klinefelter’s syndrome and Down’s
correct
syndrome
118. 2) Cross hybridization among the selected
parents 149. 4) has all dominant alleles
119. 2) A – iii, B – ii, C – iv, D – i 150. 3) aa, Aa
120. 4) The first transgenic cow, Rosie produced 151. 4) The female offspring inherits both the X
human protein enriched milk i.e. 2.4 g /lit chromosomes from mother only
152. 4) Lamarck - Branching descent
 9 Test ID : 627
153. 3) A - 350 mya, B - 2000 mya, C - 65 mya, 166. 4) All of these
D - 4.5 bya 167. 3) Bamboo plants and pacific salmon fish
154. 3) Natural selection is based on certain 168. 4) N(t+1)= Nt + (B + E) - (D + I)
observations which are not factual 169. 4) lichens and phytoplanktons
155. 3) seed ferns 170. 4) both 1 and 2
156. 4) The first human like being 171. 1) 4 x 1013 kg
157. 2) A - Pteranodon, B - Pelycosaurs 172. 4) i, ii, iii and iv
158. 4) 0.42 173. 2) Electrostatic precipitator
159. 2) the asexual cycle of Plasmodium in human 174. 4) all of these
liver cells 175. 2) Statement I is correct statement II is
160. 4) A - ii, B - iii, C - i, D - iv incorrect
161. 4) both 1 and 2 176. 3) It is an extremely impotent pollutant
162. 4) All of the above 177. 4) N2O, CFCs, CH4, CO2
163. 2) Its receptors present principally in the brain 178. 3) A - v, B - i, C - ii, D - iv
164. 1) A - iii, B - i, C - ii, D - v 179. 2) Statement I is correct statement II is
165. 2) Statement I is correct statement II is incorrect
incorrect 180. 1) Water act - 1985