SOLUTIONS
TEST SERIES-NEET [(4+1)(4+1)(+10)]
PART TEST - XII/01
PART - I : PHYSICS
1.
8.
q 4 = 6 ´10
D
i
i
Þ Vd =
Þ Vd µ
neA
A
rr
p.r
k pr cos q
(4) V = k
3.
(2) As water droplet is at rest
ur
So, F net = 0
r
3
=
=k
r3
p cos q
r2
4.
5.
6.
4.9 ´ 105
mg
Þ q = 2 × 10–9 C
(3) Work done on equipotential surface is zero.
(4)
(3) For series combination,
Þ
9.
i1 =
Þ
æC ö
CS èç 2 ø÷ 1
=
= = 1: 4
Cp = C + C Þ Cp = 2C Þ
Cp
2C
4
10.
= 9 ´109 ´
(Here, K = 9 × 109)
e2
11.
2
(1.6 ´10-19 )2
(1.6 ´10
-10 2
)
= 9 ´10-9 N
-9
\ Acceleration of electron = F = 9 ´ 10
me 9 ´ 10-31
= 10–9 × 10+31 = 1022 m/s2
6
´ i where i is the current delivered by the battery..
9+ 6
2 ´ 15
= 5A
6
Thus, potential difference across 2W resistor is
V = iR = 5 × 2 = 10V
(1) For conductors
KQ
KQ
; V (r < R ) =
R
R
So, potential inside the surface is same as the potential
on surface. So statement (I) is correct.
ur
s
Just outside the surface of conductor, E =
nˆ
2e 0
n̂ = unit vector perpendicular to surface
So, statement (II) is also correct.
(1) Given : emf e = 2.1 V
I = 0.2 A, R = 10W
Internal resistance r = ?
From formula.
e – Ir = V = IR
2.1 – 0.2r = 0.2 × 10
0.1
2.1 – 0.2 r = 2 or 0.2 r = 0.1 Þ r =
= 0.5 W
0.2
V (r = R ) =
(4) Force of mutual attraction between the electron and
proton is
r
Þ V = 18V
\ i=
C
Þ F = 9 ´ 109 ´
V2
V2
Þ 36 =
R
9
Current through the 9W resistor is i1 =
Cp
r2
(3) We have, P =
V 18
=
= 2A
R
9
The 9W and 6W resistors are in parallel, therefore
CS
Ke 2
B
q 2 = -2 ´10 -8 C
V = k ´ 3 ´ 10 -8 = 9 ´ 10 9 ´ 3 ´ 10 -8 volt
= 27 × 10 = 270 V
For parallel combination,
F=
-8
C q 3 = -3 ´10 C
2
= 1m
2
q
\ Potential at the centre O, V = k
r
é 2 ´ 10-8 -2 ´ 10-8 -3 ´ 10-8 6 ´ 10-8 ù
V = kê
+
+
+
ú
1
1
1
1
êë
úû
1
1 1
= + Þ C =C
S
CS C C
2
7.
= 2m
\ DO = OB = AO = OC =
V=0
C
: PT - XII/01
O
qE
C
C
A
q 1 = 2 ´10 -8 C
0.1 ´ 10 -3 ´ 9.8
C
-8
.
mg
Þ mg = qE Þ q =
E
Þq=
( 2 )2 + ( 2 )2
(1) AC = BD =
(3) i = neAVd
2.
TEST CODE
2
12.
NEET PT - XII/01
(3) Given,
Dipole moment, p = q × l
Resistance at 0°C, R0 = 2 W
q=
Resistance at 80°C, 6.8 W
20.
Using R = R0(1 + aDT)
p 4 ´10-5
=
= 2 ´ 10-3 C = 2mC
0.02
l
R1 = 2W
(4)
where a is the thermal coefficient of resistance
R 3 = 6W R 4 = 8W
\ 6.8 = 2{1 + a(80 – 0)}
X
6.8
– 1 = a ´ 80
Þ
2
R 2 = 4W
3.4 – 1 2.4
=
= 0.03
Þ a=
80
80
Net Resistance =
13.
\ a = 3 × 10–2 °C–1
(3) Flux does not depend on the size and shape of the
close surface, and so, it remains same.
14.
(2) U =
15.
(3) Assertion: In non-polar molecule, centre of +ve
charge coincides with centre of –ve charge, hence net
dipole moment is zero.
Reason: When non-polar material is placed in external
field, centre of charges does not coincide, hence give
non-zero moment in field.
1 é q 2 Qq Qq ù
+
ê +
ú=0
4p Î0 ë a
a
2a û
2q
.
\ Q=(2 + 2)
kp
and Ee = 3 ; \ Ea = 2Ee
r
r
17. (3) Resistance r2 and r 3 are in parallel so potential
difference across r 2 = r3 = V
16.
(2) We have Ea =
\ i 2 r2 = i3 r3 Þ i 2 =
2kp
3
i3r3
r2
... (i)
i
r
\ 3= 2
i1 r2 + r3
18. (4) Charge mobility
V
(m) = d [ Where Vd = drift velocity ]
E
E EA
I(r)
and resistivity (r ) = j = I Þ E = A
21.
=
19.
)
2
m2
Þ m = 1.0
Vs
5 ´ 1.7 ´10 –8
r r r
(4) Torque, t = p ´ E = pE sin q
4 = p × 2 × 105 × sin 30°
4
= 4 ´ 10-5 Cm
or, p =
5
2 ´ 10 ´ sin 30°
1 1
2´ 4
46
+ +6+8 =
+ 14 =
W
2 4
2+4
3
(1) In ohm's law, we check V = IR where I is the corrent
flowing through a resistor and V is the potential
difference across that resistor. Only option (a) fits the
above criteria. Remember that ammeter is connected in
series with resistance and voltmeter parallel with the
resistance.
22. (1) Resistance offer obstruction to current while
conductance allow electric current to flow, so conductance
is the reciprocal of resistance.
dV
= 8x
23. (4) V = 5 + 4x 2 \
...(i)
dx
Force on a charge is
æ dV ö
F = qE = q ç - ÷ = q ( -8x ) ; from (i)
è dx ø
= -2 ´ 10- 6 ´ (–8 ´ 0.5) = 8 ´ 10 -6 N
24.
(2) For the system to be in equilibrium, net force on each
of the charges must be zero. Net force on any of – Q to be
zero, we have
–Q
–Q
2k
Q2
Q 2 kQq
+k 2 - 2 =0
2
2a a 2
Q
\ q=
q
Q
(1 + 2 2)
4
–Q
25.
V VA
Þ m= d = d
E
Ir
(
1
1
+
+ R3 + R 4
R1 R 2
=
ær
ö
Current i1 = i2 + i3 Þ i1 = ç 3 + 1÷ i3
è r2 ø
1.1 ´10 –3 ´ p ´ 5 ´10 –3
Y
26.
a
–Q
(2) Under the effect of applied electric field, the free
electrons get accelerated due to electric field and they
move from lower to higher potential. However, electron
collides with lattice ion during passage and follow curved
path.
(3) We know that potential energy of discrete system of
charges is given by
1 æ q1q 2 q 2 q 3 q 3 q1 ö
ç
÷
+
+
4p Î0 çè r12
r23
r31 ÷ø
According to question,
1 æ q1q 2 q 2 q 3 q 3 q1 ö
+
+
ç
÷
Uinitial =
4p Î0 è 0.3
0.5
0. 4 ø
U=
NEET PT - XII/01
Ufinal =
28.
Q
2
Q ærö
1 4
4pR 2
So,
= q = ´ ç ÷ = 64 ´ =
q èRø
s smaller
16 1
4 pr 2
33. (4) Kirchhoff's first law is based on conservation of charge
and Kirchhoff's second law is based on conservation of
energy.
1 Q
1 æ -2Q ö
34. (1) V = V1 + V2 + V3 =
. +
ç
÷
4p Î0 R 4p Î0 è R ø
1 æ q1q 2 q 2 q 3 q 3 q1 ö
+
+
ç
÷
4p Î0 è 0.3
0.1
0. 4 ø
Uf – Ui =
27.
3
sbigger
1 æ q 2 q 3 q 2q 3 ö
q3
(8q 2 )
çè
÷ø =
4 p Î0 0.1
0.5
4p Î0
(4) Net Power, P
= 15 × 45 + 15 × 100 + 15 × 10 + 2 × 1000
= 15 × 155 + 2000 W
P
Power, P = VI Þ I =
V
15 ´155 + 2000
\ I main =
= 19.66 A » 20 A
220
35.
æ e ö
(4) i = ç
÷
èR+rø
Power delivered to R.
2
æ e ö
P = i2R = ç
÷ R
è R+r ø
dP
=0
P to be maximum,
dR
36.
m
=
9.1 ´ 10-31
1
CV¢2 ´ 2
2
= CV¢2
= 900 × 10–12 × 25 × 102
= 225 × 10–8
= 2.25 × 10–6 J
30. (3)
31. (1) In series
2E
is =
2r + 2
In parallel
E
E
2E
=
r
r
+4
+2
2
as, is = ip
2E
2E
=
2r + 2 r + 4
Þ r = 2W
Þ
E
r
R
E
r
E
r
r
38.
39.
40.
41.
ip
R
(2) From law of conservation of charge qi = q f
Q
4
4
= 64 and, pR 3 = 64 ´ pr 3
3
3
q
r 1
Þ R = 4r Þ =
R 4
Þ 64q = Q Þ
r
r
since angle between E and P is 180°
is
ip =
32.
(3) r =
-19 2
-15
28
ne 2 t 8.5 ´10 ´ (1.6 ´10 ) ´ 25 ´ 10
= 10–8 W-m
37. (3) Potential energy of electric dipole in external electric
field
rr
U = - P.E = -PE cos q = - PE cos180°
2
d éæ e ö ù
êç
or
÷ R ú = 0 or R = r
dR êè R + r ø ú
ë
û
29. (2) By conservation of charge
qi = qf
Þ C × 100 = CV¢ + CV¢
Þ V¢ = 50 V
So, Uf =
1 æ 3Q ö
1 æ 2Q ö
Q
çè ÷ø =
çè
÷ = 2pÎ R
ø
4p Î0 R
4p Î0 R
0
rl
(3) Resistance, R =
A
l l rl 2
[Q Volume (V) = Al.]
R =r ´ =
A l
V
Since resistivity and volume remains constant therefore
% change in resistance
DR 2Dl
=
= 2 ´ (0.5) = 1%
R
l
+
\ U = +PE
On moving towards right electric field strength decrease
\ U decreases.
Net force on electric dipole is towards right and net torque
acting on it is zero so move towards right.
(4)
(3) Capacitance will increase but not 5 times (because
dielectric is not filled completely). Hence, new capacitance
is greater than 100 µF and less than 500 mF.
(4)
(2) Net resistance of the circuit,
Req = 2 W + 1 W + 7 W = 10 W
Net Voltage, Vnet = 10 – 5 = 5 V
Using Kirchhoff’s law
Vnet
5
i = R = 10 = 0.5 A
eq
4
42.
NEET PT - XII/01
R1
(3) R1 + R2 = 1000
Þ R2 = 1000 – R1
On balancing condition
R1(100 – l) = (1000 – R1)l ...(i)
(l )
On Interchanging resistance
balance point shifts left by 10 cm.
On balancing condition
(1000 – R1) (110 – l) = R1 (l – 10)
or, R1 (l – 10) = (1000 – R1) (110 – l)
Dividing eqn (i) by (ii)
1000
\ Efficiency = 1000 + P ´ 100 = 96%.
loss
G
100 – l
44.
Current, i =
100 - l
l
=
G
l - 10 110 - l
Þ (100 – l) (110 – l)
(100 – l + 10)
(l – 10)
= l(l – 10)
=(110 – l)
Þ 11000 – 100l – 110l + l2 = l2 – 10l
Þ 11000 = 200l or, l = 55
Putting the value of ‘l’ in eqn (i)
R1 (100 – 55) = (1000 – R1) 55 Þ R1 (45) = (1000 – R1) 55
Þ R1 (9) = (1000 – R1) 11 Þ 20 R1 = 11000
\ R1 = 550 W
(3) Let P is the observation point at a distance r from
–2q and at (L+r) from +8q.
Given now, net EFI at P = 0
r
\ E1 = EFI (Electric Field Intensity) at P due to +8q
r
E 2 = EFI (Electric Field Intensity) at P due to – 2q
4
1
uur uur
k(8q) k(2q)
\
= 2
E1 = E 2
= 2
\
2
2
+
(L
r)
(r)
(L + r)
r
2
2
4r = (L+r) Þ 2r = L+r
r=L
\ P is at x = L + L = 2L from origin
)dt
0
ö 8 3 3
÷ + (15 - 0 )
ø 3
1 Q
(if r ³ R)
4p Î0 r 2
Electric field at a distance 15 cm from the centre of sphere
will be
E=
E=
9 ´ 109 ´ 3.2 ´ 10 -7
= 0.128 ´ 106 = 1.28 ´ 105 N/C
225 ´ 10 -4
PART - II : CHEMISTRY
51.
(4)
1 æ a ö 1 æ 35 ö
k = ln ç
ln ç
÷=
÷
t è a – x ø 15 è 35 – 9 ø
1 æ 35 ö
ln
15 çè 26 ÷ø
(4) Solutions in which solute - solute and solvent-solvent
interactions are almost similar to solute-solvent
interactions are known as ideal solution.
(2) A – (iv), B – (i), C – (ii), D – (iii)
(3) For following reaction,
2NO (g) + O 2 (g) ® 2NO 2 (g)
1
Wh en the volume of vessel changes to
then
3
concentration of reactant become three times.
The rate of reaction for first order reaction µ concentration
of reactant. So, rate of reaction increases three times.
(1) Molality of solution does not depend on the temperature
as it depends on the moles of solute and mass of solvent.
Hence, both assertion and reason are true.
K ´1000 ´ w 5.12 ´ 1000 ´ 1
(1) By using, M = f
=
DTf ´ WSolvent
0.40 ´ 50
= 256 g mol–1
2.303 RT
0.059 V
0
0
(3) Ecell
=
log K c or Ecell
=
log K c
nF
n
0.059V
=
log1016 = 0.4736 V
2
=
52.
53.
54.
dF K
= [4 - 2q] = 0 Þ q = 2 mC
dq r 2
(3) From the Gauss's law
q
Q
f = in =
Q Q = CV
Î0
Î0
55.
R = 2W, V = 220 V
P 1000
Current, I = =
V
220
2
1000
æ
ö
´2
Ploss = I 2 R = ç
è 220 ø÷
2
8(15)3
Þ q = 2250 + 9000 = 11250 C
3
50. (1) If the charge on a spherical conductor of radius R is
Q, then electric field at distance r from centre is
E=0
(if r Ð R)
r2
For F to be maximum,
46. (2) For metals like copper, at temperature much lower
than 0°C, graph deviates considerably from a straight line.
47. (4) Given : Power, P = 1 kW = 1000 W = Poutput
ò (20t + 8t
Þ q = 10 × (15)2 +
Kq(4 - q)
CV
\f =
Î0
15
15
(2) Let distance between the two divided charges be r.
F=
dq
Þ ò dq = ò idt Þ q =
dt
æ 20t 2 8t 3 ö
æ 152 - 02
+
Þ q = çç
÷÷ = 20 ´ ç
3 ø0
2
è 2
è
R1
From Coulomb’s law, force between two charge,
45.
48. (2) A - (ii); B - (iii); C - (iv); D - (i)
49. (2) Given, i = a0t + bt2
Put a0 = 20 and b = 8, we get i = 20t + 8t2
...(ii)
R2=1000 – R1
43.
Pin = Poutput + Ploss
R2 = 1000 – R1
56.
57.
NEET PT - XII/01
5
58. (2) Higher the value of KH, lower is the solubility of gas,
correct order :
HCHO > CH4 > CO2 > Ar
59. (2) Ionic molar conductivity of H+ is very high and
NH4OH is a weak electrolyte.
60. (2) Depression in freezing point is a colligative property
and depends only on the amount of solute.
DTf = kf .m
1
DT fx
=
DT fy
k f × mx
=
k f × my
1
Mx
1
67. (4) Kohlrausch’s Law states that at infinite dilution, each
ion migrates independently of its co-ion and contributes
to the total equivalent conductance of an electrolyte a
definite share which depends only on its own nature.
From this definition we can see that option (4) is the correct
answer.
68. (3) The rate of formation of SO3 will be equal to rate of
disappearance of SO2. This is because stoichiometry is
same.
69.
425.9 S cm2 mol–1 and L°m (NaA) = 100.5 S cm2 mol–1
My
° + + l° L°m (HA) = l H
A
1
Þ
61.
62.
63.
64.
= l ° + + l°
1 My
=
Þ Mx : My = 1 : 0.25
4 Mx
+ l° - + l °
a=
2a
2
2b
Lm
L 0m
=
Ka(CH3COOH) =
7.8
= 0.02
390
cα 2
cα × cα
=
1– α
c – cα
0.04 ´ (0.02)2
= 1.6 × 10– 5
1– 0.02
pKa = 4.79
Use: pKa + pKb = pKw = 14
Þ pKb = 14 – 4.79 = 9.21
=
=
50
= 0.125
400
1000 g
= 55.55 mole
18 g / mol
nsolute
1
=
nsolute + nwater (1 + 55.55)
= 0.01768 = 0.0177
-1
litre time )
r
71.
(4) Given [A] = 0.01 M
Rate = 2.0 × 10–5 mol L–1 S–1
8r
For a first order reaction;
b
3
2a
2r
Q For (1) and (3), the rate is doubled when conc. of G is
doubled keeping that of H constant i.e., rate µ [G]
\ x=1
From (2) and (3), y = 2
\ Overall order is 3.
(2) a =
L °m (HA)
Mole fraction =
[G]mole [H]mole rate(mole
-
L m (HA)
Moles of 1 kg H2O =
+3
1
Na +
70. (4) 1.00 m i.e., 1 mole of solute is present in 1 kg of water.
® MnO ( OH ) + NH 3
MnO 2 + NH +4 + e - ¾¾
(4) Overall order = sum of orders w.r.t each reactant.
Let the order be x and y for G and H respectively
litre
b
- l°
k (HA) ´1000 5´10-5 ´1000
=
Molarity of HA
0.001
= 50 S cm2 mol–1
litre
a
Cl-
L °m (HA) = 425.9 + 100.5 – 126.4 = 400 S cm2 mol–1
k (HA) = 5 × 10–5 S cm–1
Eab = 50 kJ
(2) Azeotropic mixture have same composition in vapour
and liquid phase.
Interactions between different components is different from
pure components, hence boiling point can be lower or
higher.
(3) At cathode, reduction occurs according to following
reaction.
-1
- l°
= L °m (HCl) + L °m (NaA) - L°m (NaCl)
100 = 150 – Eab
-1
Na +
A
Lm(HA) =
Exp.No.
66.
Cl-
H
(3) As k¢ > k¢¢ , E ¢a < E a¢¢ (Greater the rate constant,
lesser is the activation energy).
(2) DH = Ea f - Eab
+4
65.
° (NaCl)
(3) Given: Lm
= 126.4 S cm2 mol–1, L°m (HCl) =
Rate = k [A] Þ k =
t1/2 =
72.
0.693
2 ´10-3
2.0 ´ 10 -5
= 2 ´ 10 -3
[0.01]
= 347 sec.
(2) p = CRT
w ´ R ´ T 68.4 ´ 0.0821 ´ 273
=
= 4.48 atm
mV
342 ´ 1
(2) The value of DrG depends on n value as per the
equation DrG = –nFEcell
p=
73.
So, assertion statement is correct
Ecell is an intensive property while DrG is an extensive
thermodynamic property
So, reason is correct but not explaining the assertion.
6
NEET PT - XII/01
74. (1) For one molal solution, 1 mol of solute should be
present in 1000 g of solvent. Or we can say that 0.5 mol of
solute should be present in 500g of solvent.
75.
(4) k1 = A × e
–Ea /RT
1
–E
k 2 = A × e a 2 /RT
On dividing equation (ii) by (i)
k
Þ 2 =e
k1
æk
ln ç 2
è k1
76.
So, l°m(CH3COOH) = l°m(CH3COO–) + l°m(H+)
So l°m(CH3COOH)
.....(ii)
= l°m(CH3COOK) +
(E a – Ea )
1
2
RT
10, 000
ö E a1 – E a 2
=
=4
÷=
´ 300
RT
8.314
ø
(2) Cu2+ + 2e– ¾® Cu, DG1° = – 2F(0.34) ...(i)
...(ii)
Cu+ + e– ¾® Cu, DG2° = – F(0.522)
Subtract (ii) from (i)
Cu2+ + e– ¾® Cu+,
DG3° = –F(E0)
°
°
°
\ DG1 – DG2 = DG3
Þ –FE° = –2F(0.34) + F(0.522)
(3) Relative lowering in vapour pressure
=
l°m(AB) = l°m(A+) + l°m(B–)
.....(i)
Þ E° = 0.68 – 0.522 = 0.158 V
77.
82. (1) According to Kohlrausch’s law
83.
Þ 0.222 = 0.7995 -
Rate = k [ A]a [ B]b (K is constant)
Taking ratio of (1) and (2) entry.
Þ
160
nA
760 - p
160 = 1
Þ XA =
=
Þ
nA + nH 2O
760
160 54 4
+
160 18
or
8 ´ 10 -2
79.
80.
81.
3
2
2 o
o
or 184 = ´ 200 + ´ pB or 184 = 120 + pB
5
5
5
or pºB = 160 torr.
(1) Molarity depends on the volume of a solution which
can be changed with change in temperature.
[0.02]b [0.02]a
=
[0.04]a [0.04]b
[2] = [2]a Þ a = 1
= 2 ´ 10 –2 = k [ A]1[ B ]1
k=
10
= antilog 1.2036
a-x
(2) P = PA + PB = xA pAo + xB pBo
where P = total pressure, p = partial pressure
pº = vapour pressure, x = mole fraction
[0.04]b [0.02]a
4 ´ 10 -2 [0.02]a [0.04]b
Order w.r.t [A]
10
= 1.2036
a–x
10
= 0.625 g
or a - x =
15.98
(4) Daniell cell is a type of galvanic cell.
2 ´ 10 -2
=
Order w.r.t [ B] = b = 1
Taking ratio of (2) and (3) entry,
0.693 –1 2.303
10
hr =
log
24
96
a–x
10
= 15.98
a-x
4 ´ 10 -2
...(i)
2 = [2]b Þ [2]1 = [2]b
3
Þ p = ´ 760 = 570 torr..
4
or log
0.0591
1
log
2
[Ag+ ]2
Þ [Ag+] = 10–9.8
Ksp (AgCl) = [Ag+][Cl–] = 10–9.8 × 1
84. (4) A + B ® P
According to rate law expression
\ 54 mL = 54 g water]
k=
0.0591
[H + ]2
log
n
PH 2 ´ [Ag + ]2
Ecell = E°cell -
( p° - p)
= XA
p°
(2)
x y
æx- yö
- = z+ ç
÷
2 2
è 2 ø
(1) The given EMF at those conditions is equal to
standard electrol potential. The cell reaction:
H2(g) + 2Ag+ (aq) ® 2H+ (aq) + 2Ag(s); n = 2
=z+
[Given p° = 760 torr. Let, d water = 1 g/mL
78.
1
1
l° (H SO ) – l° (K SO )
2 m 2 4 2 m 2 4
2 ´ 10-2
0.02 ´ 0.02
2 ´ 10 -2 ´ 104 100
=
= 50
4
2
(2) Effective collision depends on proper orientation and
colliding energy. The colliding energy has to overcome
the energy of activation in order to form the product.
(4) Correct matching for pair (iii) will be
[G (conductance) – siemens or ohm–1(S).]
(3) For the reaction
A ® Product
Given t1/ 2 = 1 hour
For a zero order reaction
=
85.
86.
87.
\ t1 / 2 =
[ A0 ]
2k
NEET PT - XII/01
or k =
[ A0 ]
2 t1/ 2
7
=
2
= 1 mol L–1 hr–1
2 ´1
K f ´ wH
DTf =
MH
Further for a zero order reaction,
19.23 =
dx change in concentration
k=
=
dt
time
88.
89.
90.
91.
MH =
%/atomic mass
Molratio
= mol
C
92.3%
92.3/12 = 7.7
7.7/7.7 = 1
H
7.7%
7.7/7.7 = 7.7
7.7/7.7 = 1
\ Empirical formula = CH
Empirical formula weight = 12 + 1 = 13
95.
Thus, molecular formula = (CH)4 = C4H4
(2) 4A + 3B ® 6C + 9D
Rate of reaction :
-
(2) Molarity is intrinsic property. Hence, it is independent
of amount of solution.
\
r = k[A]α [B]β
...(i)
2r = k[2A]α [B]β
...(ii)
β
3r = k[A] [9B]
Dividing eq. (ii) by eq. (i)
1 d [A ]
1 d [ B] 1 d [ C ] 1 d [ D ]
==
=
4 dt
3 dt
6 dt
9 dt
Rate of reaction =
Now, -
3 = 9β or, 3 = 32β or, 2β = 1 or, β = 1/2
1 3
=
2 2
92. (1) Effective collision is the deciding factor for complex
reactions.
93. (4) Let, volume of solution = 1 L
Mole of NaOH = 2
Mass of NaOH solution = volume × density
= 1000 × 1.28 = 1280 g
Mass of H2O = (1280 – 2 × 40) = 1200 g = 1.2 kg
1 d [B]
= 1´10-2
3 dt
d [B]
= 3 ´10-2
dt
Rate of consumption of B = 3 × 10–2 mol L–1 s–1
\ Consumption of B in 10s
= 3 × 10–2 × 10 = 3 × 10–1
= 0.3 mol L–1 = 30 × 10–2 mol L–1
...(iii)
2 = 2α or α = 1
Dividing eq. (iii) by eq. (i)
96.
or
-
(4)
é MnO 4- + 8H + + 5e- ® Mn 2+ + 4H 2O ù ´ 2
ë
û
E°
MnO-4 /Mn 2 +
\ Order of the reaction = 1 +
Molality (m) =
94.
1 d [C] 1
= ´ 6 ´10-2
6 dt
6
= 1 × 10–2 mol L–1 s–1
A + B ¾¾
®C + D
α
%
Molecular weight
52
Factor (n) = Empirical formula weight = 13 = 4
C
(3)
40 ´ 0.02 ´ 1000
Mc ´ 0.8
40 ´ 0.02 ´ 1000
= 52g
19.23 ´ 0.8
Element
strong electrolyte (KI)
weak electrolyte (H2CO3) ]
1000
Wc
Molar mass of solute
0.50 - 0.25
1=
Þ time = 0.25 hr..
time
(1) For 10K rise in temperature, the rate of reaction nearly
doubles.
(2) Statement I: KI is strong electrolyte, it dissociates
completely. So, lm increases slowly on dilution.
Statement II: In weak electrolyte lm increases, sharply
as the dissociation increases.
lm
´
Mole of NaOH
2
=
= 1.67 m
Mass of solvent (kg) 1.2
(3) Melting point of camphor = 176°C
Melting temperature of camphor solution = 156.7°C
Thus, DTf for camphor solution = 176 – 156.77
= 19.23°C
\ Depression in freezing point
97.
= +1.510V
1
é
°
+
-ù
ê H 2 O ® 2 O 2 + 2H + 2e ú ´ 5 E H2 O/O2 = -1.223V
ë
û
+
2+ 5
2MnO 4 + 6H ® 2Mn + O2 + 3H 2 O
2
E °cell = +1.510V - 1.223V = 0.287 V
(1) A ® Products
Given: Initial conc. [A0] = 0.1 M
Conc. After 5 min [At] = 0.001 M
t = 5 min.
For first order reaction
k=
[A ] 2.303
2.303
æ 0.1 ö
log 0 =
log ç
÷
t
[A t ]
5
è 0.001 ø
k = 0.9212 min–1
8
98.
99.
NEET PT - XII/01
(1) r = K[A]1[B]1
0.1 = K(20)1 (0.5)1
0.40 = K(x)1 (0.5)1
0.80 = K(40)1 (y)1
From (i) and (ii)
x = 80
From (i) and (iii)
y= 2
(2) Arrhenius equation
... (i)
... (ii)
... (iii)
112.
113.
114.
k = Ae- Ea / RT
115.
ln k = ln A + ln e - Ea / RT
116.
E
ln k = ln A - a
R
æ1ö
ç ÷
èT ø
Slope of ln k vs
Ea
1
curve, m = T
R
...... (1)
Ea
R
3
Ea = 5× 10 × 8.314 J/mol
= 41.57 × 103 J/mol = 41.5 kJ/mol
100. (1) A-III, B-I, C-II, D-IV
-5 ´103 = -
PART - III : BOTANY
101. (3) The correct sequence of formation of female
gametophyte of flowering plants is- Meiosis in megaspore
mother cell and formation of megaspore tetrad ®
Functional megaspore undergoes three mitotic divisions,
results in the formation of eight nuclei ® Cell walls are
laid down ® Formation of seven cells with eight nuclei.
102. (4) The endosperm is absent in the seed of pea because
endosperm is completely consumed by the developing
embryo.
103. (2) Xenogamy refers to the transfer to pollen grains from
anthers of one plant to stigma of a different plant which
during pollination brigs genetically different types of pollen
grains to stigma.
104. (1)
105. (4) In majority of aquatic plants, the flowers emerge above
the level of water. These may be pollinated by insects or
wind eg.: Water hyacinth and water lily.
106. (3) Sporopollenin is the most resistant known biological
material. It is resistant to several biological and chemical
decomposition and can be preserved as fossils.
107. (4)
108. (3) The fruits that develop without fertilization are
known as parthenocarpic fruits such as banana.
109. (2)
Polyembryony ® Orange
Parthenocarpy ® Banana
False fruit ® Apple
Embryo Sac ® Female gametophyte
110. (3) Assertion is incorrect and reason is correct.
Pollination by wind is more common among abiotic
pollinations.
111. (2) The correct stages of development of a dicot embryo
in the order of their occurrence are – (iii). Formation of
zygote ® (iv). Formation of globular embryo ® (i).
117.
118.
119.
120.
121.
122.
123.
124.
Formation of heart shaped embryo ® (ii). Formation of
typical dicot embryo.
(2)
(2) Syngamy is the most vital event of sexual
reproduction is perhaps the fusion of gametes, resulting
the formation of a diploid zygote.
(3)
•
Wheat ® Albuminous seed
•
Black pepper ® Perisperm
•
Cashew ® False fruits
•
Citrus ® Polyembryony
(2) In a fertilized ovule n, 2n and 3n conditions occur
respectively in egg, nucellus and endosperm.
(4) Cleistogamy is a self-fertilization that occurs within a
permanently closed flower. In cleistogamous flower, the
anther and stigma lies close to each other. When anther
dehisces in the flower buds, pollen grains come in contact
with the stigma to effect pollination. Thus, cleistogamous
flowers are invariably autogamous as there is no chance of
cross - pollen landing on the stigma. Cleistogamous flowers
produce assured seed set even in the absence of pollinators.
(1) Synergids bear prominent structure called ‘filiform’
apparatus which are finger like projections. This apparatus
is present in upper part of each synergid. This apparatus
is useful for the absorption and transportation of materials
from the nucellus to the embryo sac.
(4) Statement (iv) is not the characteristic of wind
pollinated flowers.
Wind pollination requires such type of the pollen grains
that are light and non-sticky so that they can be
transported in wind currents. They often possess wellexposed stamens so that the pollens are easily dispersed
into wind currents, and large often-feathery stigma to
easily trap air-borne pollen grains. Wind pollinated flowers
often have a single ovule in each ovary and numerous
flowers packed into an inflorescence.
(3)
Non-albuminous seed ® Residual endosperm absent
Albuminous seed ® Residual endosperm present
Perisperm ® Residual nucellus
Pericarp ® Fruit wall
(1) The gynoecium represents the female reproductive
part of the flower that consists of pistil. Each pistil
comprises three parts, i.e., stigma, style and ovary.
(1) Majority of insect pollinated flowers are large-sized.
These flowers are colorful and possess fragrance to attract
the insects.
(1) Tapetum is important for the nutrition an d
development of pollen grains, as well as a source of
precursors for the pollen coat.
(4) A – III; B – IV; C – I; D – II
(3) All the statements are correct about self incompatibility. Self-incompatibility is a general name for
several genetic mechanisms in angiosperms, which
prevent self-fertilization and thus encourage outcrossing
and allogamy. In plants with self - incompatibility, when a
pollen grain produced in a plant reaches a stigma of the
same plant or another plant with a similar genotype, the
process of pollen germination, pollen tube growth, ovule
fertilization, and embryo development is halted at one of
its stages, and consequently no seeds are produced.
NEET PT - XII/01
125. (2) In general the egg apparatus of embryo sac in
angiosperm consists of one egg cell, two synergids, three
antipodal cells, two polar nuclei.
126. (2) In over 60% of angiosperm pollen grain is liberated
at 2 celled stage.
127. (1) Megaspore mother cell (MMC) undergoes meiosis
to form four haploid cells (called megaspores) and the
process of formation is known as megasporogenesis. The
MMC undergoes meiotic division results in the production
of four megaspores. 100 functional megaspores are
produced by 100 MMC, since three out of four megaspores
degenerate in each case.
128. (4)
129. (3)
130. (4) The attachment point of funicle and body of ovule is
known as hilum.
131. (4) Two types of pollination are recognised based on the
destination of pollen grains. When pollen grains are
transferred from an anther to the stigma of the same flower,
the process is called self pollination or autogamy. Cross
pollination is further classified depending on whether the
pollination has occurred between two flowers on the same
plant (geitonogamy) or between two flowers on different
plants (xenogamy).
132. (3) Pollen grains are rich in nutrients therefore used as
food supplements. Athletes and race horses use these
tablets to enhance performance.
133. (2) Endosperm of angiosperms is triploid in nature. As,
one of the male gamete moves towards the egg cell and
fuses with its nucleus results in the formation of a diploid
zygote where as the other male gamete moves towards
the two polar nuclei located in the central cell and fuses
with the polar nuclei results in the formation of a triploid
primary endosperm nucleus.
134. (1)
135. (2) As the anther develops, the microspore mother cells
of the sporogenous tissue undergoes meiotic divisions
to form microspore tetrads. After dehydration, the
microspore tetrad is separated into pollen grains.
136. (3)
137. (3) A – III; B – I; C – IV; D – II
138. (3)
139. (4) Double fertilisation is a characteristic feature of
flowering plants. It involves two fusions in which one
male gamete fuses with egg cell to form zygote and other
male gamete fuses with the diploid secondary nucleus to
produce triploid primary endosperm nucleus.
140. (2) A – IV; B – III; C – II; D – I
141. (4)
142. (3) The pollen grains represent the male gametophytes.
As, the anthers mature and dehydrate, the microspores
dissociate from each other and develop into pollen grains.
So, embryo sac is to ovule as pollen grains is to an anther.
143. (3) Egg apparatus consists of two synergids and one egg
cell lying at the micropylar end. Synergids bear prominent
structure called ‘ filiform’ apparatus which are finger like
projections. Synergids guide the path of pollen tube towards
the egg, help in obtaining nourishment from the outer nucellar
cells and also function as shock absorbers during the
penetration of pollen tube into the embryo sac. Cytoplasm of
egg is inactive, rich in ribosomes, and contains plastids.
144. (2) Aquatic plants such as water hyacinth & Lily, the
flower semerge above the level of water & are pollinated
by insects & wind.
9
145. (1) The majority of angiosperms bear chasmogamous
flowers, which means the flowers expose their mature
anthers and stigma to the pollinating agents. There is
another group of plants which set seeds without exposing
their sex organs. Such flowers are called cleistogamous
and the phenomenon is cleistogamy.
146. (3)
147. (3) Primary endosperm nucleus is formed by the fusion
of diploid secondary nucleus with a male gamete. Therefore,
it is triploid. Zygote is formed by fusion of two gametes
and thus it is diploid. Synergids are the cells of
gametophyte and hence these are haploid.
148. (1) One can study pollen germination in plants like pea
in a lab by using 10% of sugar solution.
149. (1)
150. (4) Cleistogamy is the phenomenon, where flowers never
open and in such flowers, only self-pollination occurs within
the bud (unopened flower). Bisexual flowers which do not
open at all are called cleistogamous. In such flowers, anthers
and stigma lie close to each other. When the anthers dehisce
in the flower buds, pollen grains come in contact with the
stigma and pollination occurs. So, there is no chance of crosspollination in this type of flower. Therefore, both statements
are true.
PART - IV : ZOOLOGY
151. (4) Both Statement I and Statement II are correct.
152. (2) The glandular tissue of each breast is divided into 15-20
mammary lobes containing clusters of cells called alveoli.
153. (4) A - II, B - IV, C - III, D - I
154. (1) 155. (1)
156. (1) Both assertion and reason are correct but reason is
not the correct explanation of assertion.
157. (4) The male sex accessory ducts include rete testis, vasa
efferentia, epididymis and vas deferens.
158. (4)
159. (1) Diaphragms- they cover the cervix blocking the entry
of sperms
Contraceptive pills – inhibit ovulation and implantation
Intra uterine devices – increases phagocytosis of sperm
within uterus
Lactational amenorrhea – absences of menstrual cycle and
ovulation following parturition.
160. (4) Seminal plasma is secreted from male acessory glands
and it is rich in fructose, calcium and certain enzymes.
161. (4)
162. (3) The correct option is (3) because placenta secretes
human chorionic gonadotropin (hCG). Zona pellucida is a
primary egg membrane secreted by the secondary oocyte.
The secretions of bulbourethral glands help in lubrication
of the penis Leydig cells synthesise and secrete testicular
hormones called androgens.
163. (2) The spermatids are transformed into spermatozoa
(sperms) by the process called spermiogenesis. After
spermiogenesis, sperm heads become embedded in the
Sertoli cells, and are finally released from the seminiferous
tubules by the process called spermiation.
164. (3) Both LH and FSH attain a peak level in the middle of
menstrual cycle (about 14th day) resulting in ovulation.
165. (4) Milk produced during the initial few days of lactation is
called colostrum. This contains several antibodies absolutely
essential to develop resistance for the new-born babies.
10
166. (1) In this method, ova from female and sperms from
male are collected and are induced to form zygote under
simulated condition in the lab.
167. (4) According to the World Health Organisation (WHO),
reproductive health means a total well-being in all aspects
of reproduction, i.e., physical, emotional, behavioural and
social.
168. (3) LNG 20 is hormone releasing IUD that releases
levonorgestrel. It is responsible to increase the sperm
phagoeytosis.
Multiload 375, Cu T and Cu 7 are copper releasing IUDs.
169. (1)
170. (3) The acrosome (head of the sperm) contains the
necessary enzymes (hyal uronidase and acrosin) to
penetrate the membrane of the ovum.
171. (4) Gonorrhoea, Syphilis, Genital herpes are sexually
transmitted diseases.
172. (2) Gonorrhoea is a bacterial disease which can be
treated with the help of antibiotics and cured completely.
Rest other diseases are not completely cured as other
mentioned options are viral diseases.
173. (4)
174. (2) By the end of the 12 weeks, most of the major organ
systems are formed.
175. (3) The head of the sperm is known as acrosome that
has enzymes which help sperm to enter an egg.
176. (2) Graafian follicle releases secondary oocyte from the
ovary by the process of ovulation.The fluid filled cavity
of tertiary follicle is known as antrum.
177. (2) Rapid declines in death rate, maternal mortality rate
(MMR) and infant mortality rate (IMR) as well as an increase
in number of people in reproducible age are probable
reasons for the present increase in india’s population.
178. (4)
179. (1) Sexually transmitted diseases (STDs) are infections
that are commonly spread by sex, especially vaginal
intercourse, anal sex or oral sex. The causes of STDs are
bacteria, parasites, yeast, and viruses.
180. (2) Semen, or seminal fluid, is an alkaline fluid that contains
spermatozoa embedded in seminal plasma. Semen is
ejaculated by male reproductive system during orgasm.
181. (1) Both the statements are correct. Vas deferens receives
a duct from seminal vesicle and opens into urethra as the
ejaculatory duct. The cavity of the cervix is called cervical
canal which along with vagina forms birth canal.
182. (1) To get rid of unwanted pregnancies and to prevent
the fatality or harmfulness to the mother or to foetus or
both due to the continuation of pregnancy are the reasons
on the basis of which pregnancy can be terminated.
183. (1) Blastomeres are a type of cell produced by cleavage
(cell division) of the zygote after fertilization and are an
essential part of blastula formation.
184. (4) A - III, B - IV, C - I, D - II
185. (3) The use of condoms during coitus is some of the
simple precautions to avoid contacting STIs like AIDS
and syphilis.
186. (2) The correct matching is:
Vasectomy is a surgical method of contraception
Coitus interruptus is a natural method of contraception
Cervical cap is a barrier method of contraception
Saheli is an oral method of contraception which is a nonsteroidal pill
NEET PT - XII/01
187. (2) Lactational amenorrhoea do not play any role in
phagocytosis of sperms.
188. (4) 189. (2) 190. (2)
191. (3) Oocytes start division and enter into prophase-I of
the meiotic division and get temporarily arrested at that
stage, called primary oocytes.
Oogenesis is initiated during the embryonic development
stage when a couple of million gamete mother cells
(oogonia) are formed within each fetal ovary; no more
oogonia are formed and added after birth.
192. (4) Contraceptive pills have to be taken daily for a period
of 21 days starting preferably within the first five days of
menstrual cycle. It helps in the prevention of conception.
As long as the mother feeds breast the child fully, chances
of conception are almost nil only up to a maximum period
of six months following parturition.
193. (4) Condoms are used as barriers made of thin rubber/
latex sheath used to cover the penis in the male or vagina
and cervix in females. It prevents the deposition of
ejaculated semen into the vagina of the female. It should
be discarded after a single use. It is also a safe guard
against transmission of AIDS and other Sexually
Transmitted Diseases (STDs).
194. (3)
195. (4) Such type of infections or disease that are spreaded
through sexual intercourse are known as veneral diseases
or sexually transmitted disease. These disease are
transmitted by the sharing of infected needles, transfusion
of blood from an infected person or from an infected
mother to foetus.
196. (1) In lactational amenorrhea, pills and emergency
contraceptives methods, there is the involvement of
hormones.
Lactational amenorrhea (absence of menstruation) is
based on the fact that ovulation and therefore the cycle
do not occur during the period of intense lactation
following parturition. Prolactin is the major hormone
responsible for milk production and is present in sufficient
quantities in almost all women to allow the establishment
of normal lactation.
Emergency contraception methods include emergency
contraception pills (ECP), intrauterine device, e.g., LNG20 (Levonor- gestrel) and ulipristal acetate.
CuT and barrier method do not involve any hormonal role.
197. (4) Both Statement I and Statement II are correct.
198. (2) Fully developed foetus and the placenta induce foetal
ejection reflex. Foetal ejection reflex is also called mild
uterine contraction.
199. (2) Such type of infections or disease that are spreaded
through sexual intercourse are known as reproductive tract
infections (RTI) or sexually transmitted disease. These disease are transmitted by the sharing of infected needles,
transfusion of blood from an infected person or from an
infected mother to foetus.
200. (4) Uterus is also known as womb. Oogenesis is the
process of formation of mature female gametes.
Primary oocytes get surrounded by layer of granulosa cells.