36 FT - 05

SOLUTIONS
TEST SERIES-NEET [(4+1)(4+1)(+10)]

FULL SYLLABUS TEST - 05 TEST CODE : FT - 05

PART - I : PHYSICS
lD
1. (2) Fringe width b =
d
2V
2. (1)

N cos 60° = Maw Þ N ´ 1 = 16 a w

2
1V Þ N = 32a w ...(i)
For block w.r.t. wedge
Q P
E r - E2 r1 2-2
Enet = 1 2 or Enet = =0
r1 + r2 2 +1
3. (3) t = MB sinq = m × (2l) × B sinq
= 10–4 × 0.1 × 30 sin 30° = 1.5 × 10–4 Nm
4. (2) Given,
Radius of circular wires, r = 20 cm = 0.2 m
Current, i = 2A
Magnetic field due to circular wire X = magnetic field due Balancing vertical forces
to circular wire Y N + maw sin 30° = mg cos 30°
m i Þ N = 8g cos 30° - 8a w sin 30°
\ Bx = By = 0
2r
Þ 32a w = 4 3g - 4a w (Using (i))
4p ´ 10-7
= ´ 2T [Q m0 = 4p × 10–7] 3
2 ´ 0.2 Þ aw = g
Net magnetic field is 9
Along incline plane
4p ´ 10-7 ´ 2 Þ m gsin30° + mawcos30° = mab = 8ab
BN = B2x + B2y = ´ 2T = 2p ´ 10 -6 T
2 ´ 0.2 1 3 3
8´g ´ +8´ ´ g.
= 200p × 10–8 T = 2 × 314 ×10–8 T 2 9 2
Þ ab =
= 628 × 10–8T 8
5. (2) v = f l = f × 2 (l 2 - l1 ) 1 3 3 2g
Þ ab = g ´ + g. =
2 9 2 3
= 480 × 2(0.70 – 0.30) 9. (1) NP = 400, NS = 2000 and VS = 1000 V.
= 384 m/s VP N P VS ´ N P 1000 ´ 400
= of, VP = = = 200V.
6. (3) From formula, drift velocity, Vd = I VS NS NS 2000
neA 10. (1) When a light of single frequency falls on the electrons
I 10 of inner layer of metal, then this electron comes out of the
Þ n= = -6 metal surface after a large number of collisions.
AeVd 5 ´ 10 ´ 1.6 ´ 10 -19 ´ 2 ´ 10 -3
= 625 × 1025 11. (2) A
7. (2) As optical path SB of lower slit is increased, therefore, Y
fringe pattern shifts somewhat downwards. A
8. (4) Let aw be the acceleration of wedge and ab be the B Y = A.B
acceleration of block w.r.t wedge
For the wedge w.r.t. ground
FT - 05 37

u2 2 3 1
A B A Y =
A·B Y Þ sin q1 = ´ sin q2 = ´ =
u1 3 2 2
0 0 1 0 1
æ 1 ö
0 1 1 0 1 So, q1 = sin -1 ç ÷ = 45°
1 0 0 0 0 è 2ø
dQ dT
1 1 0 1 1 22. (4) From formula, = kA
dt dx
12. (3) For stable condition, æ dQ ö
çè ÷ [ML2 T -3 ]
(3iˆ - 4ˆj) + ( 2iˆ - 3jˆ) + F3 = 0 (given) Þk=
dt ø
æ ö
dT
Þ [ k ] =
[L 2
][KL -1
]
= [MLT -3K -1 ]
Þ F3 = -5iˆ + 7ˆj Aç ÷
è dx ø
13. (2) I-V characterstics of solar cell is as shown below. 23. (3) The electric field/emf is induced neither in sides AD
and nor in BC. Unless the metallic square loop is
entering or leaving the magnetic field and the flux linked
with it is changing.
24. (1) In uniform magnetic field
ur r ur ur r r r
14. (4) F = q(v ´ B) So, F ^ v Þ a ^ v .
q1 - q2
15. (3) Electric field between plates given by, E = So, speed and kinetic energy remains same.
2AÎ0
(Here, q1 > q2) 1
25. (2) (i) P = +2.0 Þ f = × 100 cm = + 50 cm
The, the potential difference will be +2.0
q -q q -q æ Î0 A ö (Positive power convex lens)
V = Ed = 1 2 d = 1 2 çQ C = ÷
2AÎ0 2C è d ø (ii) P = P1 + P2 = (+0.25) + (+0.25) = +0.50 D
16. (4) Intensity µ 1/ (distance)2 ; No. of photoelectrons 1 1
f= = × 100 cm = 200 cm
emitted is proportional to intensity of incident light. P + 0.5
17. (4) The work function of a metal depends upon the 1
properties of the metal and the nature of its surface. (iii) P = –2.0 Þ f = × 100 cm = –50 cm
-2.0
18. (4) I + I = 2 MR2 (–ve power concave lens)
I, M (iv) P = P1 + P2 = (–6.0) + (+3.5) = –2.5 D
2 I = 2 MR2
2 1 1
æ Lö f= = × 100 cm = –40 cm
I =Mç ÷ [Q L = pR] I, M P –2.5
è pø
26. (2) Intensity of light transmitted by polariser is half of
19. (1) Using lens formula intensity of unpolarised light.
1 1 1 27. (1) ve = 11.1 km/h
- =
v u f 2GM E
When object is at centre of curvature of lens, Also, ve = 2gRE or ve =
RE
u = – 2f
1 1 1 \ Escape velocity on the surface of moon,
\ - =
v -2f - f ME
2G
For concave lens focal length is negative. 2G M m 81
vm = =
1 -3 -2f Rm RE
Þ = Þv=
v 2f 3 4
Concave lens always forms a virtual and erect image. 2 2
= ve = × 11.1 = 2.46 km/h
Hence, image is not formed at the centre of curvature. 9 9
20. (2) The figure sh ows two C C B.E
28. (4) Mass defect =
independent balanced Wheatstone c2
Bridge connected in parallel each A C C B Mass of nucleus = Mass of proton
having a capacitance C. So, Cnet = C C + mass of neutron – mass defect
CAB = 2C. 29. (4) Displacement = å area = 16 – 8 + 16 – 8 = 16 m
C C
u 2 sin 2 q 2 1 Distance = å |area| = 48 m
21. (3) As, Hmax = Þu µ 2q
2g sin displacement 16m 1
So, = =
1 sin q 2 u1 Distance 48m 3
Þ uµ Þ =
sin q sin q1 u2 30. (1) cd < cp < cf
For diamagnetic substance cd is small negative (10–5)
38 FT - 05

For paramagnetic substances cp is small and positive 38. (3) Maximum particle velocity,
(10–3 to 10–5) vmax = Aw = 3 × 50 = 150 ms–1
For ferromagnetic substanes cf is very large (103 to 105)
31. (2) Volume remains constant 39. (2)
4 3 4
pR = 27 ´ ´ pr3
3 3
3 3 2 R2
R = 27r Þ r =
9
Work done = T.DA = T (27 × 4pr2 – 4pR2)
R2
= 4pT(27 × – R2) = 8pR2T Change in momentum of any one ball
9 r
A Y'
A | D P|= 2 ´ 0.05 ´10 = 1
32. (3) Y r
B
B
r | D P| 1 1000
|Fav |= = = = 200 N
Y ' = A + B. Y = A + B = A + B. Dt 0.005 5
Truth table of the given circuit is given by 40. (1) Balanced wheat stone bridge in circuit so there is no
current in 5W resistor so it can be removed from the circuit.
A B Y' Y
0 0 1 0
0 1 0 1
1 0 0 1
1 1 0 1
33. (1) Comparing the given equation
y = 10–3sin(50t + 2x) with standard equation,
y = a sin(wt – kx) The equivalent resistance will be
Þ wave is moving along –ve x-axis with speed
6 ´ 12
w 50 Req = +2=6W
v= Þ v = = 25 m/sec. 6 + 12
k 2 Now, apply K.V.L, we have
34. (1) Potential at B, VB is maximum V 6
VB > VC > VA I= = = 1A
R eq 6
35. (1)
36. (1) Given : f0 = 1.2 cm; fe = 3.0 cm 41. (3) The electromagnetic waves of all wavelengths travel
with the same speed in space which is equal to velocity of
u0 = 1.25 cm; M¥ = ?
light.
1 1 1
From 1 = 1 - 1 Þ = - 42. (2) (A) ® (3); (B) ® (4); (C) ® (2); (D) ® (1)
f0 v0 u0 1.2 v0 ( -1.25) 43. (1) From the ideal gas equation
1 1 1 P V = nRT
Þ = - Þ v0 = 30 cm
v0 1.2 1.25 V nR æ nR ö 1
Þ = Þ V = çè ÷T Q Slopeof graph µ
Magnification at infinity, T P P ø P
v0 D 30 25 Q (Slope )2 > (Slope )1 \ P2< P1
M¥ = - ´ = ´
u0 f e 1.25 3 44. (1) Dlnet = L1a1DT + L2a2DT
(Q D = 25cm least distance of distinct vision) = 200
(L1 + L2) aDT = (L1a1 + L2a2)DT
Hence the magnifying power of the compound microscope
is 200. L1a1 + L 2a 2
a=
37. (4) When voltage applied is same, then the amount of L1 + L 2
heat dissipated is given by,
45. (2) Dimension of [h] = [ML2T–1]
2
H = V /R 1 [C] = [LT–1]
Þ Hµ
2
V R [G] = [M–1L3T–2]
H= t
R Hence dimension of
Ratio of heat dissipated =
H 2W 4 2
= = é hC 5 ù é ML2T -1 ù × é L5T -5 ù
H 4W 2 1 ê ú=ë û ë û
ê G ú é M -1 L3T -2 ù = [ML2T–2] = energy
ë û ë û
FT - 05 39

PART - II : CHEMISTRY DHC(g) = 81 kJ mol–1

DHD(g) = – 393 kJ mol–1
46. (2) DG = – RT ln k = – nFE° A(g) + 3B(g) ® C(g) + 3D(g)
Þ – RT ln 1 = – nFE° Þ nFE° = 0 Þ E° = 0 DrH = DHprodnet – DHReactant
47. (3) Lucas test is used to distinguish between 1°, 2° and = [3 × (–393) + 81] – [9.7 + 3(– 110)]
3° alcohol on the basis of time taken for turbidity to appear
= – 1098 + 320.3
by using a solution of anhydrous ZnCl2 in conc. HCl.
48. (2) Nitrogen is not able to form pentahalides as it does D r H = -777.7 kJmol -1
not have vacant d-orbitals available for expanding the 60. (3)
octet. Cl
Thus, Assertion is a correct statement.
Nitrogen exhibits an oxidation state of +5 using all five of FeCl3
its valence electrons. + Cl2
D
Thus, Reason is a correct statement. The Reason is not a
correct explanation of the Assertion. (X)

+ R–X
49. (1) -CCl 3 ,- NO 2 and – NH3 are Cl Na/dryether
R

m -directing in nature. (X)

50. (3) Rate1 = k [A]n [B]m; Rate2 = k [2A]n [½B]m The reaction of Aryl halide and Alkyl halide in presence
of sodium metal and dry ether is Wurtz-Fittig reaction.
Rate2 k[2A]n [½B]m 61. (4) The value of equilibrium constant does not depend
\ = = [2]n [½]m = 2n.2–m = 2n–m
Rate1 k[A]n [B]m on initial concentrations of reactants and products.
62. (1)
51. (4)
O O
No. of moles of solute || ||
molality = 63. (4) H 2S2 O8 Þ S2 O82 - Þ O - S- O - O - S- O -
-
Mass of solvent (in kg.) || ||
O O
WB ´ 1000 18.25 ´ 1000
Þ = Thus, there are two peroxide ions and six oxide ions.
M B ´ WA ( in g ) 36.5 ´ 500 = 1 m Þ negative charge = (–1 × 2) + (–2 × 6) = –14
(Q molar mass of HCl = 36.5 g/mol) Positive charge = 2x.
52. (3) A – (p), B – (r), C – (q), D – (s) Þ Total charge = 2x + (–14) = 2x – 14 = –2
53. (2) Þ x = + 6.
54. (2) A – (s), B – (r), C – (q), D – (p) 64. (2) A – (r), B – (s), C – (p), D – (q)
55. (3) The minimum energy required to excite from n = 2 to 65. (4) Atomic radius of gallium is less than that of aluminium.
n = 3 is 66. (4)
(A) Chloroform + Aniline ® (q) Distillation
æ 1 1 ö
DE = E f - E i = -2.18 ´ 10-18 ç - ÷J (B) Benzoic acid + Napthalene ® (p) Crystallisation
2
è3 22 ø (C) Water + Aniline ® (r) Steam distillation
= 1.89 eV [1 J = 6.242 × 1018 eV] » 1.9eV (D) Napthalene + Sodium chloride ® (s) Sublimation
56. (4) SN1 reaction involves carbocation which are planar 67. (3) In Hell - Volhard Zelinsky reaction, when acid reacts
(sp2 hybridised) and thus can be attacked on either face with Br 2 or Cl 2 in presence of red phosphorous a -
of the carbon. hydrogen atom of the acid is replaced by halogen atom of
57. (3) Chlorine has high electron affinity than fluorine. The the acid is replaced by halogen atom HCOOH does not
less negative electron gain enthalpy of fluorine as give HVZ reaction due to absence of a - hydrogen atom.
compared to chlorine is due to very small size of the 68. (4) NaBH4 is mild reducing agent. It selectively
fluorine atom. reduces aldehyde and ketone over alkene.
58. (4) CHO
NaBH 4 OH
Br MgBr D
(i) Mg, dry (ii) D O
¾¾¾® ¾¾¾®
2
CH3 CH3
ether

59. (1) Given, COOCH3 COOCH3

DHA(g) = 9.7 kJ mol–1
DHB(g) = – 110.81 kJ mol–1
40 FT - 05

69. (2) 79. (4)

OH 80. (3) CH3OH + CH3COOH mixture solution is an example
of ideal solution which obey Raoult’s law and DHmix = 0.
CH – OH 81. (1) A – (q), B – (s), C – (p), D – (r)
CH3 CHCl2
82. (2)
Cl 2/hv H2 O Total no. of Bond
Molecule
373 K electron order
(X) H2O O2 16 2
Benzal
Chloride CHO N2 14 3
C2 12 2
B2 10 1
F2 18 1
(i) B2 H6 O2 and C2 have same bond order.
70. (2) R—COOH ¾¾¾¾¾
(ii) H O/H +
®
2
W´F
In the above reaction, when R—COOH is reduced by 83. (2) Equivalent weight =
I´ t
B2H6, followed by acidic hydrolysis;
group changes to —CH2—OH and therefore 0.96 ´ 96500
= = 32.1g ; 31.75g
R—CH2—OH is the main product. 1.2 ´ 40 ´ 60
71. (3) The sequence of reaction is 84. (3) Molar conductance of solution is related to specific
+
conductance as follows :
NaNO + HCl
2
® Ar - N º NCl-
Ar2 - NH 2 ¾¾¾¾¾¾ 1000
(X) Lm = k ´ ....(a)
C
Cu HCl
¾¾¾¾® Ar - Cl + N 2 1000
(Y)
Lm = (6.3 × 10–2 ohm–1 cm–1) ×
(0.1mol / cm3 )
72. (4) 4[H]
R -N C ¾¾¾® R NH - CH 3 = 6.3 × 10–2 × 104 ohm–1cm2 mol–1
alkyl isocyanide secondary amine = 630 ohm–1 cm2 mol–1
73. (2) Total atoms in 1 molecule of C12H22O11 85. (2)
= 12 + 22 + 11 = 45
\ Total atoms in 1 mole of C12H22O11
= 45 × 6.02 × 1023 atoms/mol.
74. (2) DH = DU + DnRT ; for N 2 + 3H 2 ¾ ¾® 2 NH 3
Dng = 2 – 4 = – 2
\ DH = DU - 2 RT or DU = DH + 2 RT \ DU > DH
75. (4) Molecularity is defined as the number of reacting
species taking part in an elementary reaction.
Bond pairs = 5, total lone pairs = 15
76. (2) Reaction of sodium ethoxide with ethyl iodide to
produce diethyl ether is known as Williamson synthesis. 86. (4) Tetrahedral complexes do not show geometrical isom-
It is a nucleophilic substitution reaction and proceeds via erism because due to symmetrical structure, relative posi-
SN2 mechanism. tions of the ligands is same with respect to each other.
77. (1) KMnO4 is a part of the titration reaction so it is an Square planar complexes of the type Ma 3B do not show
internal indicator while phenolphthalein and methyl orange geometrical isomerism because the possible spatial
are added-separately so they are external indicates. arrangements are equivalent.
78. (1) MABCD type square planar complexes show three isomers
which can be obtained by fixing the position of one ligand
HBr
and placing any of the remaining three ligands at the trans
(Anti Markovnikov) Br
Rule position one by one.
(C4H8)
A C A B A B

M M M
+ HBr
Anhydrous D B D C C D
trans cis cis
AlCl3
FT - 05 41

87. (4) Dissociation of electrolytes increases with increase 100. (2) Phloem fibres (bast fibres) are made up of
in temperature: So, Kw = [H+][OH–] also increases. sclerenchymatous cells.
\ pH of the solution decrcases. 101. (4)
CH3 102. (4)
103. (2) A – II, B – IV, C – III, D – I
88. (2) 104. (3) In red algae, sexual reproduction is oogamous and
accompanied by complex post-fertilisation events.
105. (4)
OCrCl2OH +
H 3O 106. (2) The mitochondrial inner membrane forms infoldings
CH CHO known as cristae, which allow large surface area for protein
OCrCl2OH such as cytochrome to function properly and efficiently.
107. (1)
108. (1)
Benzaldehyde
109. (1) Gene mutation or point mutation is the change in
(Z) expression of a gene caused by change in number,
89. (3) k = (mol L–1)1–n time–1. sequence and type of nucleotides.
For given reaction n = 2. 110. (3)
111. (3)
\ k = mol–1 L sec–1 112. (4)
90. (1) Both statement I and II are correct. 113. (4) The given diagram represent T.S. of dicot leaf.
PART - III : BIOLOGY 114. (1) Thorn of Citrus and Bougainvillea is a modified
stem.
91. (2)
115. (4)
92. (3)
116. (2) Synaptonemal complex is formed during zygotene
93. (1) Apical, intercalary and lateral meristems are
stage. The complex is formed by a pair of synapsed
differentiated on the basis of position. Apical meristem is
homologous chromosomes is called bivalent or a tetrad.
situated at the shoot apex and the root apex. Intercalary
These are more clearly visible at pachytene stage.
meristem is present at the base of internodes, e.g., in
117. (1)
grasses or at the base of leaves e.g., in Pinus or at the
118. (3) During lactic acid and alcohol fermentation, not much
base of nodes, e.g., mint. Lateral meristems are present
energy is released, <7% of the energy in glucose is
along the lateral sides of stem and roots.
released.
94. (1)
119. (4) A – II, B – III, C – IV, D – I
95. (1) Growth at cellular level, is principally a consequence
120. (2)
increase in the amount of protoplasm.
121. (4)
96. (3)
122. (2) In leguminosae family, flower is zygomorphic,
97. (1) A – I; B – IV; C – II; D – III
imbricate aestivation and polypetalous.
98. (1)
123. (1) The best stage to observe shape, size and number of
99. (2) Homozygous ‘O’ group Heterozygous ‘A’ group
chromosomes is metaphase.
Parent: 124. (4) The complete oxidation of pyruvate by the stepwise
removal of all the hydrogen atoms, leaving three molecules
of CO2. The passing on of the electrons removed as part
of the hydrogen atoms to molecular O2 with simultaneous
synthesis of ATP.
125. (1) A – I, B – III, C – IV, D – II
126. (2)
127. (3) A – III, B – I, C – IV, D – II
128. (3) 129. (2)
130. (1) P, Q, R and S in the given major pathway of anaerobic
respiration are NAD+, ethanol, lactic acid and PEP
respectively.
Therefore, there are 50% chances that their child will 131. (4) A – III, B – I, C – IV, D – II
have ‘O’ blood group. 132. (1) Statements C and E are incorrect and the statements
A, B and D are correct.
42 FT - 05

133. (4) Antirrhinum plant shows incomplete dominance. Synaptic junctions help in transmission of information
Parents: RR rr through chemicals.
(Red) (White) 151. (3) A – III, B – I, C – II, D – IV
¯ ¯ 152. (4)
153. (1) A – III, B – I, C – II, D – IV
Gametes: 154. (4) During oogenesis, the primary oocyte is diploid; after
the first meiotic division into the secondary oocyte, the
cell becomes haploid.
F1 generation: Rr
155. (1) A – III; B – I; C – IV; D – II
(All pink)
156. (4) A – IV, B – III, C – II, D – I
F2 generation:
157. (4)
R r 158. (4) Exothermic reaction is a type of chemical reaction
which releases energy in the form of heat and light. It is
R RR Rr
(Red) (Pink) opposite to endothermic reaction. Graph shows, energy A
r Rr rr is released in the presence of enzyme and energy B is
(Pink) (White) released in the absence of enzyme.
Phenotypic:- RR : Rr : rr 159. (4)
ratio Red Pink White 160. (2) Both A and R are true but R is NOT the correct
1 : 2 : 1 explanation of A.
Therefore, in F2 generation red, white and pink flowered 161. (2)
plants are obtained. 162. (3) A – III, B – IV, C – I, D – II
134. (3) 163. (4)
135. (3) 164. (2) A – IV; B – III; C – II; D – I
136. (4) Statements A, D and E are correct. 165. (2)
137. (1) Infectious diseases like pneumonia and tetanus have 166. (4) The label I, II, III, IV and V are pubis, femur, tibia,
been controlled to a large extent by the use of vaccines. ilium and sacrum respectively.
138. (2) 167. (4)
139. (2) Acromegaly is caused by tumour in pituitary gland. 168. (1) A – IV, B – III, C – I, D – II
140. (1) 169. (3) Areolar connective tissue contains fibroblasts (cells
141. (2) After the formation of the product in the bioreactors, that produce and secrete fibres), macrophages and mast cells.
it undergoes through separation and purification Inner surface of bronchioles is lined by ciliated epithelium.
processes before a finished product is ready for marketing. Blood is a specialised connective tissue.
These processes are collectively referred to as downstream Tubular parts of nephron are lined by cuboidal epithelium.
processing. 170. (1)
142. (2) A – IV; B – I; C – II; D – III 171. (3)
143. (4) 172. (4) A – III, B – I, C – IV, D – II
144. (1) 173. (2) Statement D is incorrect. Actually, there has been
145. (1) Bowman’s capsule is a double walled cup-like gradual evolution of life forms.
structure. PCT opens into DCT. 174. (1) Ligases join the two DNA molecules
146. (1) A – II, B – I, C – III, D – IV 175. (3)
147. (4) 176. (1) Both Statement I and Statement II are true.
148. (4) A – IV, B – I, C – II, D – III 177. (4) Transgenic models exist for all the given human
149. (4) diseases.
150. (4) Tight junctions provide a barrier which prevents 178. (1)
leakage of fluid across epithelial cells. 179. (1) A – II, B – III, C – I, D – IV
Adhering junctions aid to cement adjacent cells to form a 180. (3) A – II, B – IV, C – I, D – III
sheet.
Gap junctions provide cytoplasmic channels to facilitate
communication between adjacent cells.