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CAREER POINT
[ CODE – A ]
CAREER POINT
AIPMT Exam 2015 (Paper & Solution)
Code – A
Q.1
Date : 25-07-2015
2,3-Dimethyl-2-butene can be prepared by
Ans.
[1]
heating which of the following compounds
Sol.
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with a strong acid?
Exercise Sheet: [Chapter : Periodic Table,
(1) (CH3)2C=CH–CH2–CH3
Ex.3A, Q.No.15, Page no. 187]
(2) (CH3)2CH–CH2–CH=CH2
64Gd
(3) (CH3)2CH–CH–CH=CH2
⇒ [Xe] 4f75d16s2
54
Half filled 'f' subshell is more stable
CH3
(4) (CH3)3C–CH=CH2
Q.3
The formation of the oxide ion, O2– (g),
Ans.
[4]
from oxygen
Sol.
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exothermic and then an endothermic step as
Class notes: [Chapter : GOC-II]
shown below :
CH3
H⊕
⊕
CH3–C=C–CH3
Thus process of formation of O2– in gas
CH3
–H⊕
⊕
CH3–C–CH–CH3
H3C CH3
an
O–(g) + e– → O2–(g) ; ∆fHΘ = +780 kJ mol–1
CH3–C – CH – CH2
CH3
first
O(g) + e– → O–(g), ∆fHΘ = –141 kJ mol–1
CH3
CH3–C–CH=CH2
atom requires
phase is unfavourable even though O2– is
Θ
~CH3
isoelectronic with neon. It is due to the fact
Rearrangement
that :
H3C CH3
(1) oxygen is more electronegative
(2,3-Dimethyl-2-butene)
(2) addition of electron in oxygen results in
larger size of the ion
Q.2
(3) electron
Gadolinium belongs to 4f series. It's atomic
repulsion
outweighs
the
number is 64. Which of the following is the
stability gained by achieving noble gas
correct
configuration
electronic
configuration
of
gadolinium?
(4) O– ion has comparatively smaller size
(1) [Xe] 4f75d16s2
(2) [Xe] 4f65d26s2
(3) [Xe] 4f86d2
(4) [Xe] 4f95s1
than oxygen atom
Ans.
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CAREER POINT
Sol.
[ CODE – A ]
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Sol.
Exercise Sheet: [Chapter : Periodic Table,
Exercise Sheet: [Chapter : Chemical
Ex.3B, Q.No.76, Page no. 194]
equilibrium, Ex.3(A), Q.No.1, Page no. 120]
Second electron gain enthalpy is positive
(energy
absorbed)
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because
If equilibrium constant for following
incoming
reaction is
electron realise more repulsion when enter
in uninegative anion
N 2 + O2
2NO ; K
then
Q.4
The number of structural isomers possible
1
1
N2 + O2
2
2
from the molecular formula C3H9N is :
(1) 2
(2) 3
(3) 4
[3]
Sol.
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Sheet:
[Chapter
K
(4) 5
Ans.
Exercise
NO ; K' =
:
Q.6
Which one of the following pairs of
solution is not an acidic buffer?
GOC-I
(Isomerism), Ex.(7B), Q.No.23, Page no. 79]
(1) H2CO3 and Na2CO3
Given molecular formula = C3H9N
(2) H3PO4 and Na3PO4
1°-amine ⇒ CH3–CH2–CH2–NH2 & CH3–CH–NH2
(3) HClO4 and NaClO4
CH3
2°-amine ⇒ CH3–NH–CH2CH3
(4) CH3COOH and CH3COONa
3°-amine ⇒ CH3–N–CH3
Ans.
[3]
Sol.
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CH3
Total 4 structural isomers are possible.
Exercise
Sheet:
[Chapter
:
Ionic
equilibrium, Ex.3A, Q.No.30, Page no. 156]
Q.5
If the equilibrium constant for
N2(g) + O2(g)
Weak acid and its conjugate salt with
2NO(g) is K,
strong base is an acidic buffer but
according to option Ans is (3), HClO4 is a
the equilibrium constant for
1
1
N2(g) + O2(g)
2
2
(1) K
Ans.
(2) K2
strong acid
NO(g) will be :
(3) K1/2
(4)
∴ HClO4 + NaClO4 is not an acidic buffer
1
K
2
[3]
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CAREER POINT
Q.7
[ CODE – A ]
Aqueous solution of which of the following
Q.9
compounds is the best conductor of electric
CO2 most easily ?
current ?
(1) MgCO3
(2) CaCO3
(1) Ammonia, NH3
(3) K2CO3
(4) Na2CO3
(2) Fructose, C6H12O6
Ans.
[1]
Sol.
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(3) Acetic acid, C2H4O2
Exercise
Sheet:
Thermal stability ∝
Ans.
[4]
Sol.
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Exercise
Sheet:
[Chapter
:
[Chapter
:
φ=
Electro
1
Ionic potential(φ)
Charg e
Size
Chemistry, Ex.1, Q.No.4, Page no. 14]
Thermal stability :
Anhydrous HCl is non conducting but in
K2CO3 > Na2CO3 > CaCO3 > MgCO3
aqueous medium conducting as it provide
ions
s-block,
Q.No.117, Page no. 28]
(4) Hydrochloric acid, HCl
Q.8
On heating which of the following releases
+
∆
Na2CO3 
→
×
–
HCl + H2O → H3O + Cl
∆
K2CO3 
→
×
∆
MgO + CO2
MgCO3 
→
Caprolactam is used for the manufacture of:
(1) Terylene
(2) Nylon-6, 6
(3) Nylon - 6
(4) Teflon
∆
→
CaO + CO2
CaCO3 
Q.10
Strong reducing behaviour of H3PO2 is due
Ans.
[3]
to :
Sol.
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(1) High oxidation state of phosphorus
Exercise Sheet: [Chapter : Polymers,
(2) Presence of two –OH groups and one
P–H bond
Page no. 194]
(3) Presence of one –OH group and two
NH
CH2
O
C=O
H2O
CH2
CH2
∆
P–H bonds
C–(CH2)5–NH
(4) High
n
Nylon-6
gain
enthalpy
of
phosphorus
CH2–CH2
Caprolactum
electron
Ans.
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Sol.
[ CODE – A ]
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Stability
Exercise
Bond order 2.5
Sheet:
[Chapter
:
p-block
O +2 > O2 > O −2 > O 2−2
(Pnicogen family), Ex.1, Q.No.28, Page no. 4]
1.5
1.0
Bond order↑
Bond dissociation energy↑
Stability↑
O
H3PO2 ⇒
2
P
HO
H H
Q.12
H3PO2 is monobasic acid and act reducing
agent due to presence of two P–H bond
The number
maximum in :
of
water
molecules
is
(1) 18 gram of water
H3PO2 + 2Ag2O → H3PO4 + 4Ag
(2) 18 moles of water
(3) 18 molecules of water
Q.11
(4) 1.8 gram of water
Decreasing order of stability of O 2 , O −2 , O +2
and O 22 − is :
(1) O 2 >
O 2+
>
O 22−
>
Ans.
[2]
Sol.
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Exercise Sheet: [Chapter : Atom,
molecules and Chemical Arithmetic, Ex.1,
Q.No.61, Page no. 67]
O −2
(2) O 2− > O 22− > O 2+ > O 2
18 g H2O is 1 mol H2O
(3) O +2 > O 2 > O −2 > O 22−
∴ no. of H2O molecules are 6.02 × 1023
(4) O 22− > O −2 > O 2 > O +2
Ans.
[3]
Sol.
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18 mol H2O has 18 × 6.02 × 1023 molecules
of H2O
1.8 g H2O has 0.1 mol H2O
∴ no. of molecules = 0.1 × 6.02 × 1023
Exercise Sheet: [Chapter : Chemical
= 6.02 × 1022
Bonding, Ex.1, Q.No.114, Page no. 219]
Stability ∝ bond order
O2 ⇒ σ1s 2 σ* 1s 2 σ2s2 σ*2s2 σ2p 2z
Q.13
species are not isostructural?
(1) NH3, PH3
( π2px 2 = π2py 2 ) (π∗2px1 = π*2py1)
(2) XeF4, XeO4
N − Na
Bond order = b
2
Bond order =
6−4
=2
2
In which of the following pairs, both the
(3) SiCl4, PCl +4
(4) diamond, silicon carbide
Ans.
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Sol.
[ CODE – A ]
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Q.15
notes: [Chapter : Chemical Bonding]
XeF4
XeO4
sp3d2
sp3
F
of which of the following compounds will
require the least amount of acidified
KMnO4 for complete oxidation?
O
F
Xe
Xe
F
O
F
Square planar
O O
tetrahedral
Assuming complete ionization, same moles
(1) FeC2O4
(2) Fe(NO2)2
(3) FeSO4
(4) FeSO3
Ans.
[3]
Sol.
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Exercise Sheet: [Chapter : Oxidation
Reduction, Ex.2, Q.No.25, Page no. 93]
Q.14
In the reaction with HCl, an alkene reacts
in accordance with the Markovnikov's
rule,
to
give
a
product
1-chloro-1-
methylcyclohexane. The possible alkene is :
CH2
(1)
CH3
(2)
(A)
KMnO4 + FeC2O4 → Mn+2 + Fe+3 + CO2
v.f = 5
v.f. = 3
let number of reacting moles of FeC2O4 = x
∴ n KMnO 4 =
(B)
3x
5
KMnO4 + Fe(NO2)2 → Mn+2 + Fe+3 + NO 3−
v.f = 5
CH3
(3) (A) and (B)
v.f = 5
similarity if moles of Fe(NO2)2 is x
(4)
5x
=x
5
n KMnO 4 =
Ans.
[3]
Sol.
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Class notes: [Chapter : GOC-II]
CH3
CH2
H3 C
⊕
Cl
ClΘ
H⊕
KMnO4 + FeSO4 → Mn+2 + Fe+3
v.f = 5
v.f = 1
and also the moles of FeSO4 is x
n KMnO 4 =
(HCl)
(A)
x
5
KMnO4 + FeSO3 → Mn+2 + SO 24 − + Fe+3
v.f = 5
CH3
CH3
H3C
⊕
H⊕
v.f = 3
Cl
ClΘ
(HCl)
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and also the moles of FeSO4 is x
n KMnO 4 =
3x
5
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CAREER POINT
Q.16
[ CODE – A ]
Reaction of phenol with chloroform in
(3) Cl2O7 is an anhydride of perchloric acid
presence of dilute sodium hydroxide finally
introduces which one of the following
functional group ?
(1) –CHCl2
(2) –CHO
(3) –CH2Cl
(4) –COOH
(4) O3 molecule is bent
Ans.
[2]
Sol.
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notes]
OF2 is fluoride of oxygen because F is more
Ans.
[2]
electronegative than oxygen
Sol.
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2O
Cl2O7 H

→ 2HClO4
Exercise
Sheet:
[Chapter
:
Oxygen
O3 ⇒
Compounds, Ex.6, Q.No.31]
O
CHO
CHCl3 + KOH
contain 24 electron.
(Salicylaldehyde)
(Phenol)
Q.19
Q.17
(2) 32%
(3) 26%
(2) Hexacyanidoferrate (III) ion
(4) 48%
Ans.
[2]
Sol.
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Exercise Sheet: [Chapter : Solid state,
Ex.1, Q.No.28, Page no. 116]
The name of complex ion, [Fe(CN)6]3– is :
(1) Tricyanoferrate (III) ion
The vacant space in bcc lattice unit cell is :
(1) 23%
O
ONF and O2N– are isoelectronic because both
OH
OH
sp2, bent
O
(3) Hexacyanoiron (III) ion
(4) Hexacyanitoferrate (III) ion
Ans.
[2]
Sol.
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Packing efficiency of bcc unit cell = 68%
vacant space = 100 – 68 = 32%
Exercise Sheet: [Chapter : Coordination
Chemistry, Ex.11(A), Page no. 65]
[Fe(CN)6]3–
Q.18
Which of the statements given below is
x – 6 = –3
incorrect ?
x = +6 – 3
(1) ONF is isoelectronic with O2N–
x = +3
(2) OF2 is an oxide of fluorine
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Hexa cyanido ferrate (III) ion.
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CAREER POINT
Q.20
[ CODE – A ]
If Avogadro number NA, is changed from
6.022 × 10
23
–1
mol
to 6.022 × 10
20
Sol.
–1
mol ,
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Exercise
Sheet:
[Chapter
this would change :
Page no. 99]
(1) the ratio of chemical species to each
Nucleophile are e– rich species
other in a balanced equation
:
GOC-II,
∴ It is Lewis Base.
(2) the ratio of elements to each other in a
compound
(3) the definition of mass in units of grams
Q.22
A gas such as carbon monoxide would be
most likely to obey the ideal gas law at :
(4) the mass of one mole of carbon
(1) high temperatures and high pressures
Ans.
[3]
Sol.
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(2) low temperatures and low pressures
Exercise Sheet: [Chapter Atom, molecules
(3) high temperatures and low pressures
and Chemical Arithemic, Ex.1, Q.No.65,
(4) low temperatures and high pressures
Page no. 68]
1 amu =
1 amu =
1
× mass of 1 atom of C – 12 isotope
12
Ans.
[3]
Sol.
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1
gm = 1.66 × 10–24 gm
6.023
Exercise Sheet: [Chapter : Gaseous state,
Q.No.42, Page no. 111]
If Avogadro number is change then the
definition of mass in unit of gm is changed
Q.21
Ans.
Which of the following statements is not
For real gases

an 2 
P +
 (V – nb) = nRT

V 2 

….(1)
correct for a nucleophile?
At high temperature intramolecular attractive
(1) Nucleophiles attack low e– density sites
forces are neglected i.e.
(2) Nucleophiles are not electron seeking
neglected. At low pressure volume is high
(3) Nucleophile is a Lewis acid
So nb term is neglected.
(4) Ammonia is a nucleophile
Hence from equation (1)
an 2
V2
term is
PV = nRT
[3]
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CAREER POINT
Q.23
Hybridization
[ CODE – A ]
involved
in complex
C + O2 → CO2 ; ∆H = – 393.3 kJ/mol
the
2–
[Ni(CN)4] is : (At. No. Ni = 28)
2
2
2
44 g CO2 then heat evolved is 393.5 kJ when
3
(1) d sp
(2) d sp
(3) dsp2
(4) sp3
35.2 g then heat evolved is
393.5
× 35.2
44
= – 315 kJ
Ans.
[3]
Sol.
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Exercise Sheet: 11A [Chapter : Coordination
Q.25
20.0 g of magnesium carbonate sample
decomposes on hating to give carbon dioxide
chemistry, Q.No.41, Page no.63]
and 8.0 g magnesium oxide. What will be the
[Ni(CN)4]2–
percentage purity of magnesium carbonate in
x–4=–2
the sample ?
x = + 4 –2
(1) 60
(2) 84
x = +2
(3) 75
(4) 96
28Ni
= [Ar] 3d84s2
Ni+2 =[Ar] 3d8
Ans.
[2]
Sol.
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Exercise
Sheet:
2
[Chapter
:
Atom
molecule, Q.No. 29, Page no. 73]
3d
∆
MgCO3 
→
MgO + CO2
CN– is strong ligand so pairing possible
3d
40g MgO obtained from 84 g MgCO3
Then 8g obtained from
4p
4s
2
dsp complex
= 16.8 g MgCO3
∴ % purity =
Q.24
The heat of combustion of carbon to CO2 is
–393.5kJ/mol.
The
heat
released
84
×8
40
16.8
× 100 = 84%
20
upon
formation of 35.2g of CO2 from carbon and
oxygen gas is -
Q.26
What is the mole fraction of the solute in a
1.00 m aqueous solution ?
(1) –630 kJ
(2) –3.15 kJ
(3) –315 kJ
(4) + 315 kJ
(1) 0.0354
(2) 0.0177
(3)0.177
(4) 1.770
Ans.
[3]
Ans.
[2]
Sol.
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Sol.
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Exercise Sheet: 1 [Chapter : Chemical
Exercise Sheet: 3A [Chapter : Solution
thermodynamics, Q.No.112, Page no. 144]
Volumetric Analysis, Q.No.10, Page no. 51]
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CAREER POINT
XA
1000
×
XB
MB
m=
[ CODE – A ]
Q.29
CH3CHOH.COOH,
XA
1000
1=
×
18
(1 – X A )
56.5 XA = 1
1
56.5
∴ XA = 0.0177
which
optically
Ans.
(1) Enantiomers
(2) Mesomers
(3) Diastereomers
(4) Atropisomers
[1]
Sol.
COOH
Q.27
are
active, are called :
1 – XA = 55.5 XA
XA =
Two possible stero-structuers of
The correct statement regarding defects in
H
crystalline solids is
(1) Frenkel defect is a dislocation defect
*
COOH
OH & HO
*
CH3
CH3
[R]
[S]
H
(2) Frenkel defect is found in the halides of
Both are optically active and Enantiomers.
alkaline metals
(3) Schottky defects have no effect on the
density of crystalline solids
Q.30
The following reaction
(4) Frenkel defects decrease the density of
NH2
+ Cl
crystalline solids
Ans.
[1]
Sol.
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O
NaOH
H
N
Exercise Sheet: 1 [Chapter : Solid State,
Q.No. 63, 64, Page no. 118]
O
Frenkel defect it is a dislocation defect
is known by the name Q.28
(1) Acetylation reaction
The stability of +1 oxidation state among Al,
Ga, In and Tl in creases in the sequence -
(2) Schotten-Baumen reaction
(1) Tl < In < Ga < Al (2) In < Tl < Ga < Al
(3) Friedel-Craft's reaction
(3) Ga < In < Al < Tl (4) Al < Ga < In < Tl
Ans.
[3]
Sol.
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(4) Perkin's reaction
Ans.
[2]
Exercise Sheet: Class Notes]
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CAREER POINT
Sol.
[ CODE – A ]
Benzoylatino of aniline is called schotten-
Q.32
Baumen's reaction.
NaOH
(–HCl)
NH2 + Cl–C
O
(1) hydrocyanic acid
(2) sodium hydrogen sulphite
NH–C
O
Q.31
Reaction of a carbonyl compound with one of
the following reagents involves nucleophilic
addition followed by elimination of water.
The reagent is -
The sum of coordination number and
oxidation number of the metal M in the
complex [M(en)2(C2O4)]Cl (where en is
ethylenediamine) is (1) 7
(2) 8
(3) 9
(4) 6
Ans.
[3]
Sol.
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Exercise Sheet: 11B [Chapter : Coordination
chemistry, Q.No.116, Page no. 77]
(3) a Grignard reagent
(4) hydrazine in presence of feebly acidic
solution
Ans.
[4]
Sol.
R
C=O
R/H
+δ – δ
H–NH–NH2
H⊕
(Nucleophilic
Addition)
R
R/H
C
N
NH2
H
OH
(–H2O) H⊕
R
[M (en)2(C2O4)]Cl
R/H
C=N–NH2
(Elimination
reaction)
x+0–2–2–1=0
x=+2+1
x = +3
Q.33
–2
en, and C2O4 are bidentate ligand.
O
CH2–NH2
CH2–NH2
–
C–O
–
C–O
m
Which one of the following esters gets
hydrolysed most easily under alkaline
conditions -
OCOCH3
m
(1)
O
OCOCH3
N
N
(2) Cl
O
OCOCH3
N
N
(3) O2N
O
OCOCH3
N
(4) H3CO
Coordination No = 6
sum of coordination no and oxidation no.
Ans.
[3]
6+3=9
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CAREER POINT
[ CODE – A ]
O
Sol.
Q.35
NO2 is
CH3–C–O
having
best
0.6 × 10–3 mole per second. If the concentration
of A is 5 M, then concentration of B after
leaving group (due to –M effect of nitro
20 minutes is -
gruop) so it is most reactivge towards alkaline
hydrolysis.
Q.34
In an SN1 reaction on chiral centres, there
The rate constant of the reaction A → B is
(1) 0.36 M
(2) 0.72 M
(3) 1.08 M
(4) 3.60 M
Ans.
[2]
Sol.
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Exercise Sheet: 2 [Chapter : Chemical
is -
Kinetics, Q.No. 29, Page no. 194]
(1) 100% retention
Zero order reaction
(2) 100% inversion
x = kt
x = 0.6 × 10–3 × 20 × 60
(3) 100% racemization
= 72 × 10–2
(4) inversion more than retention leading to
x = 0.72 M
partial racemization
A
Ans.
[4]
Initial case
Sol.
Not 100% recemisation
After 20 min. 5 – x
So
R2
*
R1
Cl
SN1
NuΘ
⊕
Q.36
R3
x
[B] = x = 0.72M
What is the pH of the resulting solution when
HCl are mixed ?
R2
R3
0
equal volumes of 0.1 M NaOH and 0.01 M
R3
*
5
B
R2
R1
R1
→
R2
Nu + Nu
*
R3
R1
(1) 7.0
(2) 1.04
(3) 12.65
(4) 2.0
Ans.
[3]
Sol.
Students may find similar question in CP
Exercise Sheet: 1B [Chapter : Ionic
equilibrium, Q.No. 22, Page no. 145]
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11
CAREER POINT
[ CODE – A ]
Cl
Let volume of each solution is V.
Acid
Base
en
Eq. of HCl = 0.01 × V Equivalent of NaOH = 0.1 ×V
Co
Net Equivalent = 0.1V – 0.01V
Cl
= V [0.1 – 0.01]
Trans
= 0.09V
∴ N × 2V = 0.09V
en
Cl
(Basic)
Cl
N
0.09
[OH]Θ N =
= 4.5 × 10–2
2
Cl
Co
N
Co
N
= 2 – 0.65
N
N
N
pOH = 2 – log 4.5
N
Cl
N
d & l of Cis from
= 1.35
∴ pH = 14 – 1.35 = 12.65
Q.38
The variation of the boiling points of the
hydrogen halides is in the order HF > HI >
Q.37
Number of possible isomers for the complex
HBr > HCl. What explains the higher boiling
[Co(en)2Cl2]Cl will be : (en = ethylene
point of hydrogen fluoride ?
diamine)
(1) The bond energy of HF molecules is
(1) 3
(2) 4
greater than in other hydrogen halides
(3) 2
(4) 1
(2) The effect of nuclear shielding is much
Ans.
[1]
reduced in fluorine which polarizes the
Sol.
Students may find similar question in CP
HF molecule
Exercise Sheet: 5 [Chapter : Coordination
(3) The electro negativity of fluorine is much
chemistry, Q.No.30, Page no. 54]
higher than for other elements in the
(i) tota isomer = 3
group
[CO(en)2Cl2]Cl
(4) there is strong hydrogen bonding between
HF molecules
Cis
⇓
d& l
trans
Ans.
[4]
Sol.
Students may find similar question in CP
trans form is optically inactive due to plane of
symmetry
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Exercise Sheet: Class Notes [Chapter :
Chemical Bonding]
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12
CAREER POINT
[ CODE – A ]
Q.40
HF mass ↑
HCl Vanderwall force↑
HBr Bpt↑
HI
presence of air produces (1) benzoic acid
(2) benzaldehyde
Boiling point of HF is maximum because HF
molecules associated by hydrogen bond
δ+
δ–
δ+
δ–
The oxidation of benzene by V2O5 in the
δ–
(3) benzoic anhydride
δ–
(4) maleic anhydride
H–––––F--------H–––––F--------H–––––F
Bpt. HF > HI > HBr > HCl
Ans.
[4]
Sol.
Q.39
What is the mass of the precipitate formed
O
O
when 50 mL of 16.9% solution of AgNO3 is
mixed with 50 mL of 5.8% NaCl solution ?
OH
V2O5
∆
(Ag = 107.3, N = 14, O = 16, Na = 23, Cl = 35.5)
(1) 7g
(2) 14g
(3) 28g
(4) 3.5 g
(Benzene)
Ans.
[1]
Sol.
Students may find similar question in CP
Q.41
Exercise
Sheet:
1
[Chapter
:
Moles n =
+
0.99M × 50
1000
~
− 0.05
O
O
(Maleic anhydride)
Which of the following is not the product of
dehydration of
OH
?
16.9 × 10
= 1.0 M
169.3
5.8
Molarity (M) of NaCl =
× 10 = 0.9914M
58.5
AgNO3
O
(Maleic Acid)
Atom
Molecule, Q.No.76, Page no. 65]
Molarity (M) of AgNO3 =
OH
∆
–H2O
NaCl →AgCl↓ + NaNO3
0.99 × 50
1000
~
− 0.05
Ans.
Sol.
(1)
(2)
(3)
(4)
[4]
⊕
stable carbocation
∴ moles of AgCl formed = 0.05
Mass of AgCl = 0.05 × 142.5 = 7 g
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No rearrangement occur.
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13
CAREER POINT
Q.42
[ CODE – A ]
Method by which Aniline cannot be prepared
Sol.
Luca's reagent can be used for [HCl + ZnCl2]
is -
1º/2º/3º-alcohol But if ZnCl2 is absent than
(1) reduction of nitrobenzene with H2/Pd in
only 3ºalcohol can be used because it is most
reactive due to 3º carbocation.
ethanol
(2) potassium salt of phthalimide treated with
chlorobenzene followed by hydrolysis
Q.44
with aqueous NaOH solution
(3) hydrolysis
of
energy of the listed orbitals in the atom of
phenyisocyanide
with
titanium ?
acidic solution
(At. no. Z = 22)
(4) degradation of benzamide with bromine in
alkaline solution
Ans.
Which is the correct order of increasing
[2]
Sol.
(1) 3s 3p 3d 4s
(2) 3s 3p 4s 3d
(3) 3s 4s 3p 3d
(4) 4s 3s 3p 3d
Ans.
[1]
Sol.
Students may find similar question in CP
Exercise Sheet: 3B [Chapter : Atomic
H2/Pd
NO2
O
C
C
Structure, Q.No. 163, Page no. 43]
NH2
Ti [22], [Ar] 3d24s2
Cl
Θ⊕
NK
Inert halide due
to resonance
X
NaOH
H2O
When electron will fill in 3d then it's energy
X
will becomes less than 4s. So electron will
O
N
C
C–NH2
(2) H2O/H⊕
Br2+ KOH
∆
emit from 4s not from 3d. So energy of 3d
NH2 + HCOOH
NH2
O
will be less then 4s.
Q.45
In the extraction of copper from its sulphide
ore, the metal is finally obtained by the
Q.43
reduction of cuprous oxide with -
Which of the following reaction(s) can be
(1) copper (I) sulphide (2) sulphur dioxide
used for the preparation of alkyl halides ?
ZnCl 2
(I) CH3CH2OH + HCl anh
.
→
(II) CH3CH2OH + HCl →
(III) (CH3)3COH + HCl →
(3) iron(II) sulphide
(4) carbon monoxide
Ans.
[1]
Sol.
Students may find similar question in CP
ZnCl2
(IV) (CH3)2CHOH + HCl anh
.
→
(1) (IV) only
Exercise Sheet: [Chapter : Metallurgy,
Q.No.60, Page no. 98]
(2) (III) and (IV) only
Copper is extracted by self reduction process
(3) (I), (III) and (IV) only
2Cu2O + Cu2S → 6Cu +SO2
(4) (I) and (II) only
Ans.
[3]
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14
CAREER POINT
Q.46
[ CODE – A ]
Root pressure develops due to -
(1) Plasma membrane
(1) Increase in transpiration
(2) Nuclear envelope
(2) Active absorption
(3) Ribosome
(3) Low osmotic potential in soil
(4) Mesosome
(4) Passive absorption
Ans.
[2]
Ans.
[2]
Sol.
Students may find it in CP Sheet:
Sol.
Students may find similar question in CP
[Chapter: Cell biology, Page no. 3]
Nuclear membrane is absent in prokaryotes.
Exercise Sheet: Plant physiology [Chapter:
Plant water relation, Q.No 258, Ex. # 1]
Root pressure is developed in herbaceous
plants due to osmotic active water absorption.
Q.47
Q.49
two separate circulatory pathways?
Which one is a wrong statement?
(1) Brown algae have chlorophyll a and c,
and fucoxanthin
(2) Archegonia are found in Bryophyta,
diversity, Page no. 132
Whale have double circulation which consist
of
(a) Pulmonary circulation
(b) Systemic circulation
Q.50
[Chapter : Plant diversity, Page no. 81]
Mucor is member of zygomycetes and form
aplanospores (Aflagellate spores).
Q.48
(4) Whale
Students may find it in CP Sheet: Animal
gymnosperms
Students may find this in CP Sheet:
(3) Lizard
Sol.
(4) Haploid endosperm is typical feature of
Sol.
(2) Frog
[4]
(3) Mucor has biflagellate zoospores
[3]
(1) Shark
Ans.
Pteridophyta and Gymnosperms
Ans.
Which one of the following animals has
Which of the following structures is not
found in a prokaryotic cell?
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Most animals that live in deep oceanic
waters are (1) detritivores
(2) primary consumers
(3) secondary consumers
(4) tertiary consumers
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15
CAREER POINT
Ans.
[1]
Sol.
In deep oceanic water generally detrivore's
are found.
Q.51
An association of individuals of different
species living in the same habitat and
having functional interactions is -
[ CODE – A ]
Q.53
Axile placentation is present in (1) Argemone
(2) Dianthus
(3) Lemon
(4) Pea
Ans.
[3]
Sol.
Students may find it in CP Sheet:
(1) Population
[Chapter: Structural organization in plant,
(2) Ecological niche
Page no. 104]
(3) Biotic community
Argemone : Parietal
(4) Ecosystem
Dianthus
: Free central
Ans.
[3]
Lemon
: Axile
Sol.
Students may find similar question in CP
Sheet: Ecology [Chapter : Population and
community, Page no. 52]
Pea
: Marginal
Community is group of interacting
populations of different species occupying
same habitat.
Q.52
The oxygen evolved during photosynthesis
comes from water molecules. Which one of
the following pairs of elements is involved
in this reaction?
(1) Magnesium and Chlorine
(2) Manganese and Chlorine
(3) Manganese and Potassium
(4) Magnesium and Molybdenum
Ans.
[2]
Sol.
Students may find similar question in CP
Exercise Sheet: Plant physiology [Chapter
Q.54
In which of the following both pairs have
correct combination?
Gaseous nutrient
(1) cycle
Sedimentary nutrient
cycle
Gaseous nutrient
(2) cycle
Sedimentary nutrient
cycle
Gaseous nutrient
(3) cycle
Sedimentary nutrient
cycle
Gaseous nutrient
(4) cycle
Sedimentary nutrient
cycle
Sulphur and
Phosphorus
Carbon and Nitrogen
Carbon and Nitrogen
Sulphur and
Phosphorus
Carbon and Sulphur
Nitrogen and
Phosphorus
Nitrogen and Sulphur
Carbon and
Phosphorus
: Photosynthesis, Q.No.158, Ex. # 1]
Ans.
[2]
Photolysis of water helped by minerals of
Sol.
Students may find similar question in CP
oxygen evolving complex (OEC) which
Exercise
involves Mangnese and Chlorine.
Ecosystem, Q.No.78, Ex.# 2]
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Sheet:
Ecology
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[Chapter
:
16
CAREER POINT
[ CODE – A ]
Carbon and nitrogen cycles are gaseous
cycles as their reservoir lies in hydrosphere
and atmosphere respectively while sulphur
Ans.
[4]
Sol.
Students may find this in CP Sheet:
[Chapter : Plant diversity, Page no. 83]
and phosphorus cycles are sedimentory cycles
as their reservoir lies in rocks or sediment.
Morels and truffles are edible member of
ascomycetes.
Q.55
In mammalian eye, the 'fovea' is the center
Q.57
of the visual field, where -
Which of the following are not membranebound?
(1) more rods than cones are found
(2) high density of cones occur, but has no
rods
(1) Mesosomes
(2) Vacuoles
(3) Ribosomes
(4) Lysosomes
(3) the optic nerve leaves the eye
Ans.
[3]
(4) only rods are present
Sol.
Students may find it in CP Sheet:
[Chapter: Cell biology, Page no. 28]
Ans.
[2]
Sol.
Students may find it in CP Sheet: Animal
Ribosomes are membrane less cell organelle.
physiology-II, Page no. 271
Fovea centralis have only cone cell so act as
Q.58
partners are adversely affected?
centre of visual field.
Q.56
Choose the wrong statement (1) Yeast is unicellular and useful in
fermentation
(2) Penicillium
is
In which of the following interactions both
multicellular
(1) Mutualism
(2) Competition
(3) Predation
(4) Parasitism
Ans.
[2]
Sol.
Students may find theory in CP Sheet:
and
produces antibiotics
(3) Neurospora is used in the study of
biochemical genetics
Ecology
[Chapter
:
Community
&
population theory, Page no. 64]
Competition is the interaction where fitness
of both interacting species decrease ( – / – )
(4) Morels and truffles are poisonous
mushrooms
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17
CAREER POINT
Q.59
[ CODE – A ]
A colour blind man marries a woman with
Implantation of embryo of site other then
normal sight who has no history of colour
uterus is called Ectopic pregnancy or Tubal
pregnancy
blindness in her family. What is the
probability of their grandson being colour
Q.61
blind?
(1) 0.25
(2) 0.5
(3) 1
(4) Nil
Cellular organelles with membranes are (1) lysosomes,
Golgi
apparatus
and
mitochondria
(2) nuclei, ribosomes and mitochondria
Ans.
[4]
Sol.
Students may find similar question in CP
(3) chromosomes,
ribosomes
and
endoplasmic reticulum
Exercise Sheet: Genetic, Page no. 20
(4) endoplasmic reticulum, ribosomes and
Nil
X
C
C
Y
X
X X
XY
X
C
XY
X X
nuclei
Ans.
[1]
Sol.
Students may find it in CP Sheet:
Son is not affected so grand son will also not
[Chapter: Cell biology]
affected.
Lysosome : single membrane bound,
Golgi body and endoplasmic reticulam :
Q.60
Ectopic pregnancies are referred to as(1) Pregnancies
terminated
due
Membranous, mitochondria and nucleus
double membrane, ribosome membrane less.
to
hormonal imbalance
(2) pregnancies with genetic abnormality
Q.62
(3) Implantation of embryo at site other
than uterus
(4) Implantation of defective embryo in the
uterus
Cell wall is absent in (1) Nostoc
(2) Aspergillus
(3) Funaria
(4) Mycoplasma
Ans.
[4]
Sol.
Students may find this in CP Sheet:
Ans.
[3]
[Chapter : Plant diversity, Page no. 44]
Sol.
Students may find this in class notes
Cell wall is not present in mycoplasma.
[Chapter : Reproductive system]
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CAREER POINT
Q.63
[ CODE – A ]
The term "linkage" was coined by (1) W. Sutton
(2) T.H. Morgan
(3) T. Boveri
(4) G. Mendel
Ans.
[2]
Sol.
Students may find similar question in CP
Sol.
Students may find it in CP Sheet: Animal
physiology-I, Page no. 133
Premolar teeth are absent in human
childhood (milk teeth dentition). Pre molar
teeth are monophyodont.
Exercise Sheet: Genetic, Page no. 15
Q.66
T.H. Morgan
A protoplast is a cell (1) without cell wall
(2) without plasma membrane
Q.64
Which of the following biomolecules does
(3) without nucleus
have a phosphodiester bond?
(4) undergoing division
(1) Nucleic acids in a nucleotide
(2) Fatty acids in a diglyceride
(3) Monosaccharides in a polysaccharide
Ans.
[1]
Sol.
Students may find it in CP Sheet: [Chapter
: Cell biology, Page no. 6]
(4) Amino acids in a polyptide
Ans.
[1]
Sol.
Students may find it in CP Sheet:
= cell – cell wall
[Chapter: Protoplasm, Page no. 18]
Q.65
Protoplasm = cell wall less cell
Q.67
In which group of organisms the cell walls
Phosphodiester bond is present between two
form two thin overlapping shells which fit
nucleotide of a nucleic acid.
together?
The primary dentition in human differs
(1) Slime moulds
(2) Chrysophytes
(3) Euglenoids
(4) Dinoflagellates
from permanent dentition in not having one
of the following type of teeth -
Ans.
(1) Incisors
(2) Canine
(3) Premolars
(4) Molars
Ans.
[2]
Sol.
Students may find this in CP Sheet:
[3]
Kota H.O. : Career Point Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744‐5151200
[Chapter : Plant diversity, Page no. 64]
Overlapping shells are present in Diatoms
member of Chrysophytes.
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19
CAREER POINT
Q.68
[ CODE – A ]
The DNA molecule to which the gene of
Ans.
[3]
interest is integrated for cloning is called -
Sol.
Students may find this in CP Sheet:
(1) Carrier
(2) Transformer
(3) Vector
(4) Template
[Chapter : Reproduction in flowering
plants, Page no. 32]
Coconut water consist of free nuclear
Ans.
[3]
Sol.
Students may find similar question in CP
endosperm with each nucleus been triploid in
nature.
Exercise Sheet: Genetic Page no. 44
Vector.
Q.71
The species confined to a particular region
and not found elsewhere is termed as -
Q.69
Male gametophyte in angiosperms produces
(1) Rate
(2) Keystone
(1) Three sperms
(3) Alien
(4) Endemic
(2) Two sperms and a vegetative cell
Ans.
[4]
(3) Single sperm and a vegetative cell
Sol.
Students may find similar question in CP
Sheet: Ecology [Chapter : Biodiversity,
(4) Single sperm and two vegetative cell
Theory, Page no. 134]
Ans.
[2]
Sol.
Students may find this in CP Sheet:
Species finds only in a particular region are
called as endemic species.
[Chapter : Reproduction in flowering
plants, Page no. 13]
Q.72
Pollen grain
Generative cell
Vegetative cell
Q.70
Metagenesis refers to -
Sperm
(1) Presence of a segmented body and
Sperm
(male
gamete)
Coconut water from a tender coconut is -
parthenogenetic mode of reproduction
(2) Presence of different morphic forms
(3) Alternation
of
generation
between
(1) Degenerated nucellus
asexual and sexual phases of an
(2) Immature embryo
organism
(3) Free nuclear endosperm
(4) Innermost layers of the seed coat
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(4) Occurrence of a drastic change in form
during post-embryonic development
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CAREER POINT
[ CODE – A ]
Eutrophication cause increase in Biochemical
Ans.
[3]
Sol.
Students may find it in CP Sheet: Animal
Oxygen
Demand
(BOD)
and
decrease
Dissolve Oxygen (D.O.), leads to death of
diversity, Page no. 19
fishes.
Metagenesis is alternation of generation
between asexual polyp form and sexual
medusa form of some coelentrates like obelia.
Q.75
The function of the gap junction is to (1) stop substance from leaking across a
Q.73
tissue
The enzyme that is not present in succus
(2) performing
entericus is (1) lipase
(2) maltase
(3) nucleases
(4) nucleosidase
cementing
to
keep
neighbouring cells together
(3) facilitate
communication
between
adjoining cells by connecting the
Ans.
[3]
Sol.
Students may find it in CP Sheet: Animal
small
physiology-I, Page no. 150
molecules.
cytoplasm for rapid transfer of ions,
Lipase, maltase, nucleosidase enzymes are
juice.
and
some
large
(4) separate two cells from ach other
found in succus entericus (intestinal juice)
while Nuclease enzyme in found in pancreatic
molecules
Ans.
[3]
Sol.
Students may find it in CP Sheet: Animal
physiology-I, Page no. 5, (NCERT-XI
Q.74
Eutrophication of water bodies leading to
(Eng.) Page No. 102)
killing of fishes is mainly due to nonavailability of -
Function of gap junction is to facilitate
(1) oxygen
(2) food
communication between adjoining cell by
(3) light
(4) essential minerals
connecting the cytoplasm for rapid transfer of
ions,
Ans.
[1]
Sol.
Students may find similar question in CP
Exercise
Sheet:
Ecology
[Chapter
small
molecule
and
some
large
molecules.
:
Environmental issue, Q.No.28, Ex. # 2]
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CAREER POINT
Q.76
[ CODE – A ]
Match the following list of microbes and
Q.77
their importance Sacharomyces
(1) cerevisiae
(i)
correct sequence -
(ii) Ripening of Swiss
cheese
(3) Trichoderma
polysporum
(iii) Commercial
production of ethanol
(iv) Production of blood
cholesterol lowering
agents
(4) bacterium
(a) Crossing over
Production of
immunosuppressive
agents
(2) Monascus
purpureus
Propioni-
Arrange the following events of meiosis in
(b) Synapsis
(c) Terminalisation of chiasmata
(d) Disappearance of nucleolus
sharmanii
(1) (b), (c), (d), (a)
(2) (b), (a), (d), (c)
(3) (b), (a), (c), (d)
(4) (a), (b), (c), (d)
(b)
(c)
(d)
Ans.
[3]
(1) (iii)
(i)
(iv)
(ii)
Sol.
Students may it in CP Sheet: [Chapter :
(2) (iii)
(iv)
(i)
(ii)
(3) (iv)
(iii)
(ii)
(i)
(a)
Cell division, Page no. 54]
(a) Crossing over : Pachytene
(b) Synapsis : Zygotene
(4) (iv)
(ii)
(i)
(iii)
(c) Terminalization of chiasma : Dikinesis
Ans.
[2]
Sol.
Students may find this in CP Sheet:
(d) Disappearance of nucelolus : end of
Dikinesis
[Chapter : Economic zoology, Page no. 113
They are different events of prophase-I of
& 117]
Meiosis.
(a)
Saccharomyces cerevisiae →
Fermentation → Ethanol production
(b)
Q.78
Monoscus purpureus → statins → blood
became possible with the discovery of -
cholesterol lowering agents
(c)
The cutting of DNA at specific locations
(1) Ligases
Trichoderma polysporum → cyclosporin
A → Production of immuno suppressive
(2) Restriction enzymes
agent
(d)
Propionibacterium
shormanii
→
(3) Probes
Ri
penning of swiss cheese
(4) Selectable markers
Ans.
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CAREER POINT
Sol.
[ CODE – A ]
Students may find similar question in CP
Q.81
Exercise Sheet: Genetic Page no. 42
The body cell in cockroach discharge their
nitrogenous waste in the haemolymph
mainly in the form of -
Q.79
During
biological
nitrogen
(1) Calcium carbonate
fixation,
inactivation of nitrogenase by oxygen
(2) Ammonia
poisoning is prevented by -
(3) Potassium urate
(1) Cytochrome
(2) Leghaemoglobin
(3) Xanthophyll
(4) Carotene
Ans.
[2]
Sol.
Students may find similar question in CP
(4) Urea
Ans.
[3]
Sol.
Students may find it in CP Sheet: Lower
animal, Page no. 89-90
Exercise Sheet: Plant Physiology [Chapter
Body cells in cockroach discharges their
: Plant water relation, Q.No. 345, Ex. # 1]
nitrogenous waste in form potassium urate in
Leghaemoglobin acts as O2 scavenger as it
the haemolymph from where it comes into
combined
malpighian tubule.
with
oxyleghaemoglobin
O2
and
to
regulates
form
O2
concentration in root nodules of leguminous
plants.
Q.82
Filiform apparatus is characteristic feature
of -
Q.80
Grafted kidney may be rejected in a patient
(1) Synergids
due to -
(2) Generative cell
(1) Innate immune response
(3) Nucellar embryo
(2) Humoral immune response
(4) Aleurone cell
(3) Cell-mediated immune response
(4) Passive immune response
Ans.
[3]
Sol.
Students may find similar question in CP
Ans.
[1]
Sol.
Students may find this in CP Sheet:
[Chapter : Reproduction in flowering
plants, Page no. 21]
Exercise Sheet: Immunity and disease
Page no. 124
Synergid of embryo sac consist of filiform
apparatus.
Cell mediated immune response.
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23
CAREER POINT
Q.83
[ CODE – A ]
Acid rain is caused by increase in the
Ans.
[3]
atmospheric concentration of -
Sol.
Students may find it in CP Sheet:
(1) O3 and dust
(2) SO2 and NO2
(3) SO3 and CO
(4) CO2 and CO
[Chapter: Structural organization of plant,
Page no. 98]
Superior ovary or hypogynous flower is
Ans.
[2]
Sol.
Students may find similar question in CP
Exercise
present in china rose, mustard, brinzal, potato,
Sheet:
Ecology
[Chapter
onion and tulip.
:
Environmental issue]
Q.86
Acid rain have H2SO4 and HNO3 in the ratio
Which of the following is not a function of
the skeletal system?
of 7 : 3 due to high concentration of oxides of
(1) Locomotion
sulphur & nitrogen.
(2) Production of erythrocytes
Q.84
The wheat grain has an embryo with one
(3) Storage of minerals
large, shield-shaped cotyledon known as (1) Coleoptile
(2) Epiblast
(3) Coleorrhiza
(4) Scutellum
(4) Production of body heat
Ans.
[4]
Sol.
Students may find this in class notes
Ans.
[4]
[Chapter : Skeletal system]
Sol.
Students may find this in CP Sheet:
Function
[Chapter : Reproduction in flowering
Locomotion, production of erythrocyte (By
plants, Page no. 36]
bone marrow) and storage of minerals.
of
skeletal
system
are
–
Scutellum is the single cotyledon present in
monocot.
Q.87
Golden rice is a genetically modified crop
plant where the incorporated gene is meant
Q.85
Among china rose, mustard, brinjal, potato,
for biosynthesis of -
guava, cucumber, onion and tulip, how
many plants have superior ovary?
(1) Four
(2) Five
(3) Six
(4) Three
Ans.
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(1) Vitamin A
(2) Vitamin B
(3) Vitamin C
(4) Omega 3
[1]
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CAREER POINT
Sol.
[ CODE – A ]
Students may find similar question in CP
Q.90
Exercise Sheet: Genetics, Page no. 50
A pleiotropic gene (1) Controls multiple traits in an individual
Vitamin-A
(2) is expressed only in primitive plants
(3) is a gene evolved during Pliocene
Q.88
Chromatophores take part in (4) controls a trait only in combination
(1) Respiration
(2) Photosynthesis
(3) Growth
(4) Movement
Ans.
[2]
Sol.
Students may find it in CP Sheet:
with anther gene
Ans.
[1]
Sol.
Students may find similar question in CP
Exercise Sheet: Genetics Page no. 23
[Chapter: Cell biology, Page no. 19]
Control multiple trait in a individual.
Chromatophore are membraneous structure
and have photosynthetic pigments. These
pigments are involved in photosynthesis.
Q.91
Human urine is usually acidic because (1) hydrogen ions are actively secreted into
Q.89
the filtrate
Select the wrong statement -
(2) the sodium transporter exchanges one
(1) Mosaic disease in tobacco and AIDS in
hydrogen ion for each sodium ion, in
human being are caused by viruses
peritubular capillaries
(2) The viroids were discovered by D.J.
(3) excreted plasma proteins are acidic
Ivanowski
(4) Potassium
(3) W.M. Stanley showed that viruses
was coined by M.W. Beijerinek
Ans.
[2]
Sol.
Students may find this in CP Sheet:
sodium
exchange
generates acidity
could be crystallized
(4) The term 'contagium vivum fluidum'
and
Ans.
[1]
Sol.
Students may it in CP Sheet: Animal
[Chapter : Plant diversity – Virus, Page no.
189]
physiology-II, Page no. 19
Hydrogen ion are actively secreted into the
filtrate so urine become acidic in nature.
Viroids were discovered by T.O. Diener.
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25
CAREER POINT
Q.92
Auxin can be bioassayed by -
[ CODE – A ]
Q.94
cavities lined with food filtering flagellated
(1) Lettuce hypocotyl elongation
cells and indirect development are the
(2) Avena coleoptile curvature
characteristics of phylum :
(3) Hydroponics
(4) Potometer
Ans.
[2]
Sol.
Students may find similar question in CP
Exercise Sheet: Plant physiology [Chapter:
(2) Coelenterata
(3) Porifera
(4) Mollusca
[3]
Sol.
Students may find it in CP Sheet: Animal
diversity, Page no. 15
In porifera body having meshwork of cells
internal cavities lined with food filtering
flagellated cells 'Choanocyte' indirect
development with larva like parenchymata
and amphiblastula.
Avena coleoptile curvature test is the
bioassay of auxin.
Which of the following events is not
associated with ovulation in human female?
(1) Protozoa
Ans.
Plant growth, Q.No.3, Ex. # 1]
Q.93
Body having meshwork of cells, internal
Q.95
(1) LH surge
Which one of the following hormones is
not involved in sugar metabolism?
(2) Decrease in estradiol
(3) Full development of Graafian follicle
(1) Glucagon
(2) Cortisone
(3) Aldosterone
(4) Insulin
(4) Release of secondary oocyte
Ans.
[3]
Ans.
[2]
Sol.
Students may find it in CP Sheet: Animal
Sol.
Students may find it in CP Sheet: Animal
physiology-II, Page no. 163, (NCERT-XI
physiology-II, Page no. 149, (NCERT-XI
(Eng.) Page No. 335-336)
(Eng.) Page No. 332)
Aldosterone produced by adrenal cortex is
Decrease in levels of estradiol is not
involved in regulating the amount of minerals
associated with ovulation in human female,
is our body fluid, while glucagon, cortisone &
where as LH surge, full development of
insulin are involved in sugar metabolism.
Grafian follicle and release of secondary
oocyte is associated with ovulation.
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26
CAREER POINT
Q.96
[ CODE – A ]
Which of the following diseases is caused
Q.98
child through a technique called GIFT. The
by a protozoan?
(1) Blastomycosis
A childless couple can be assisted to have a
full form of this technique is :
(2) Syphilis
(1) Germ cell internal fallopian transfer
(3) Influenza
(4) Babesiosis
Ans.
[4]
Sol.
Students may find it in Class notes
(2) Gamete inseminated fallopian transfer
(3) Gamete intra fallopian transfer
(4)
[Chapter : Animal diversity
Gamete
internal
fertilization
and
transfer
Babesiosis also called Red fever caused by
Ans.
[3]
Sol.
Students may find it in CP Sheet:
Babesia sp member of class sporozoa of
phylum protozoa.
Reproductive system, Page no. 92
Q.97
GIFT = Gamete Intra Falopian Transfer
Outbreeding is an important strategy of
animal husbandry because it :
Q.99
(1) exposes harmful recessive genes that
A jawless fish, which lays eggs in fresh
water and whose ammocoetes larvae after
are eliminated by selection.
metamorphosis return to the ocean is :
(2) helps in accumulation of superior genes.
(1) Petromyzon
(2) Eptatretus
(3) is useful in producing purelines of
(3) Myxine
(4) Neomyxine
animals.
(4) is useful in overcoming inbreeding
Ans.
[1]
Sol.
Students may find it in CP Sheet: Animal
depression.
diversity, Page no. 105
Ans.
[4]
Petromyzon is jawless marine fish which lays
Sol.
Students may find this in class notes
egg in fresh water and whose ammocoetes
[Chapter : Economic zoology], (NCERT-
larva after metamorphosis return to the ocean.
XI, Page No. 167)
Outbreeding is cross between different breads
&
useful
in
overcoming
Q.100 The structures that help some bacteria to
inbreeding
depression.
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attach to rocks and / or host tissues are :
(1) Holdfast
(2) Rhizoids
(3) Fimbriae
(4) Mesosomes
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CAREER POINT
[ CODE – A ]
Sol.
Ans.
[3]
Sol.
Students may find this in CP Sheet:
Students may find it in CP Sheet:
Reproductive system, Page no. 51
In human female meiosis-II completed after
[Chapter : Plant diversity, Page no. 29]
fertilization.
Bacterial surface consist of fimbriae that help
in attachment of substratum.
Q.103 Which of the following layers in an antral
Q.101 If
you
suspect
major
deficiency
follicle is acellular ?
of
antibodies in a person, to which of the
(1) Zona pellucida
following would you look for confirmatory
(2) Granulosa
evidence ?
(3) Theca interna
(1) Serum globulins
(2) Fibrinogin in plasma
(3) Serum albumins
(4) Stroma
Ans.
[1]
Sol.
Students may find it in CP Sheet:
(4) Haemocytes
Reproductive system, Page no. 16
Ans.
[1]
Zona pellucida = Acellular layer between
Sol.
Students may find similar question in CP
follicular cell and oocyte.
Exercise Sheet: Immunity and disease,
Page no. 125
Q.104 In his classic experiments on pea plants,
Serum globulins.
Mendel did not use :
(1) Flower position
(2) Seed colour
(3) Pod length
(4) Seed shape
Q.102 In human females, meiosis-II is not
completed until?
(1) birth
(2) puberty
Ans.
[3]
Sol.
Students may find similar question in CP
Exercise Sheet: Genetics, Page no. 5
(3) fertilization
Pod length.
(4) uterine implantation
Ans.
[3]
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CAREER POINT
[ CODE – A ]
Q.105 Which one of the following fruits is
(2) one allele dominant on the other
parthenocarpic?
(3) alleles tightly linked on the same
(1) Banana
(2) Brinjal
(3) Apple
(4) Jackfruit
chromosome
(4) alleles that are recessive to each other
Ans.
[1]
Ans.
[1]
Sol.
Students may find it in CP Sheet:
Sol.
Students may find similar question in CP
[Chapter: Structural organization of plant,
Exercise Sheet: Genetics, Page no. 14
Page no. 105]
Parthenocarpy : Formation of fruit without
fertilization e.g. Banana.
Q.108 The chitinous exoskeleton of arthropods is
formed by the polymerization of :
Q.106 In angiosperms, microsporogenesis and
(1) lipoglycans
megasporogenesis :
(2)
(1) occur in ovule
[4]
Sol.
Students may find this in CP Sheet:
chondroitin
(4) N – acetyl glucosamine
Ans.
[4]
Sol.
Students may find it in CP Sheet:
[Chapter: Protoplasm, Page no. 106]
[Chapter : Reproduction in flowering
Chitin is a polymer of N-acteyl glucosamine.
plants, Page no. 4]
Microspore and megaspore are formed by
and
(3) D – glucosamine
(3) form gametes without further divisions
Ans.
sulphate
sulphate
(2) occur in anther
(4) Involve meiosis
keratin
Q.109 The imperfect fungi which are decomposers
meiosis .
of litter and help in mineral cycling belong
to :
Q.107 A gene showing codominance has :
(1) both alleles independently expressed in
(1) Ascomycetes
(2) Deuteromycetes
(3) Basidiomycetes
(4) Phycomycetes
the heterozygote
Ans.
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CAREER POINT
Sol.
[ CODE – A ]
Students may find this in CP Sheet:
Sol.
Students may find it in CP Sheet:
[Chapter: Structural organization of plant,
[Chapter : Plant diversity, Page no. 92]
Page no. 156]
Deuteromycetes lack sexual reproduction and
so are termed fungi imperfectii.
Unisexual flower either have androecium or
gyanoecium only.
Unisexual flowers are present in cucumber.
Q.110 The wings of a bird and the wings of an
insect are :
(1) homologous structures and represent
Q.112 Increase in concentration of the toxicant at
successive trophic levels is known as :
convergent evolution
(1) Biogeochemical cycling
(2) homologous structures and represent
divergent evolution
(2) Biomagnification
(3) analogous structures and represent
(3) Biodeterioration
convergent evolution
(4) phylogenetic structures and represent
(4) Biotransformation
Ans.
[2]
Sol.
Students may find similar question in CP
divergent evolution
Ans.
[3]
Sol.
Students may find it in CP Sheet: Origin
Exercise
Sheet:
Ecology
[Chapter
:
Environmental issues, Ex. 2, Q.No. 6]
and Evolution of life, Page no. 16
Increased concentration of non biodegradable
Wings of birds and the wings of an insect are
pollutants at successive trophic levels of food
similar in function but differ in origin so they
chain is known of Biomagnification.
are analogous structure & analogy shows
convergent evolution.
Q.113 Destruction of the anterior horn cells of the
spinal cord would result in loss of :
Q.111 Flowers are unisexual in :
Ans.
(1) Onion
(2) Pea
(3) Cucumber
(4) China rose
(1) integrating impulses
(2) sensory impulses
(3) voluntary motor impulses
[3]
(4) commissural impulses
Ans.
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CAREER POINT
Sol.
[ CODE – A ]
Students may find it in CP Sheet: Animal
physiology-II, Page no. 229
Ans.
[3]
Sol.
Students may find it in CP Sheet:
Nerves arises from dorsal or anterior root
[Chapter: Cell biology]
ganglia of spinal cord are sensory in nature.
Q.114 Roots play insignificant role in absorption
Q.116 Identify the correct order of organisation of
genetic material from largest to smallest :
of water in :
(1) Wheat
(2) Sunflower
(3) Pistia
(4) Pea
(1) Chromosome,
genome,
nucleotide,
gene
(2) Chromosome, gene, genome, nucleotide
Ans.
[3]
Sol.
Pistia is an aquatic plant and root is generally
(3) Genome,
(4) Genome, chromosome, gene, nucleotide
plant.
Q.115 Match the columns and identify the correct
Ans.
[4]
Sol.
Students may find it in class notes
option.
Protoplasm.
Column I
Column II
(a) Thylakoids
(i) Disc-shaped sacs in Golgi
apparatus
(b) Cristae
(ii) Condensed structure of DNA
(c) Cisternae
(iii) Flat
Genome > Chromosome > Gene > Nucleotide
membranous
sacs
Q.117 Which one of the following hormones
though synthesised elsewhere, is stored and
in
released by the master gland ?
stroma
(a)
nucleotide,
gene
not involved in water absorption in aquatic
(d) Chromatin
chromosome,
(1) Melanocyte stimulating hormone
(iv) Infoldings in mitochondria
(b)
(c)
(d)
(1) (iii)
(iv)
(ii)
(i)
(2) (iv)
(iii)
(i)
(ii)
(3) (iii)
(iv)
(i)
(ii)
(4) (iii)
(i)
(iv)
(ii)
(2) Antidiuretic hormone
(3) Luteinizing hormone
(4) Prolactin
Ans.
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CAREER POINT
Sol.
[ CODE – A ]
Students may find it in CP Sheet: Animal
Q.119 Which of the following joints would allow
physiology-II, Page no. 152, (NCERT-XI
no movement?
(Eng.) Page No. 332)
Antidiuretic
hormone
(Vasopressin)
(1) Ball and Socket joint
is
synthesized by hypothalamic nuclei. It is
(2) Fibrous joint
stored and released by neurohypophysis
(3) Cartilaginous joint
(posterior lobe of pituitary gland).
(4) Synovial joint
Q.118 Read the different components from (a) to
(d) in the list given below and tell the
correct order of the components with
Ans.
[2]
Sol.
Students may find it in CP Sheet: Animal
reference to their arrangement from outer
physiology-I, Page no. 112
side to inner side in a woody dicot stem :
Fibrous joint do not allow movement.
(1) Secondary cortex
(2) Wood
Q.120 Which one of the following is not
(3) Secondary phoem
applicable to RNA?
(4) Phellem
(1) Chargaff's rule
The correct order is :
(2) Complementary base pairing
(1) (d), (c), (a), (b)
(3) 5' phosphoryl and 3' hydroxyl ends
(2) (c), (d), (b), (a)
(4) Heterocyclic nitrogenous bases
(3) (a), (b), (d), (c)
(4) (d), (a), (c), (b)
Ans.
[1]
Ans.
[4]
Sol.
Students may find it in class notes
Sol.
Students may find similar question in CP
Exercise Sheet: [Chapter : Structural
[Chapter : Protoplasm]
Chargaff's principle is not applicable for RNA
organization of plant, Q.No. 137, Page no. 66]
as purine is not equal to pyrimidine.
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CAREER POINT
[ CODE – A ]
Q.121 Doctors use stethoscope to hear the sounds
produced during each cardiac cycle. The
second sound is heard when :
Ans.
[2]
Sol.
Students may find similar question in CP
Exercise
(1) AV node receives signal from SA node
walls
Ecology
[Chapter
:
Population & Community, Ex. 1, Q.No. 78]
(2) AV valves open up
(3) Ventricular
Sheet:
Succession is gradual & Predictable change in
vibrate
due
species composition occurs in a area.
to
gushing in of blood from atria
(4) Semilunar valves close down after the
Q.123 In the following human pedigree, the filled
blood flows into vessels from ventricles
symbols represent the affected individuals.
Identify the type of given pedigree.
Ans.
[4]
Sol.
Students may find it in CP Sheet: Animal
(I)
physiology-II, Page no. 104
(II)
Closure of Bicuspid & tricuspid valve
(III)
produces first heart sound (lubb) while of
semilunar valve produces second heart sound
(IV)
(Dup).
(1) X-linked dominant
(2) Autosomal dominant
Q.122 During ecological succession :
(3) X-linked recessive
(1) the changes lead to a ommunity that is
in
near
equilibrium
with
the
environment and is called pioneer
community
(2) the gradual and predictable change in
(4) Autosomal recessive
Ans.
[4]
Sol.
Students may find similar question in CP
species composition occurs in a given
area
(3) the establishment of a new biotic
community is very fast in its primary
Exercise Sheet: Genetics, Page no. 63]
Autosomal recessive.
Offspring is affected but parent are not
affected
is
characteristic
of
autosomal
recessive.
phase.
(4) the numbers and types of animals
remain constant.
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CAREER POINT
[ CODE – A ]
Q.124 Balbiani rings are sites of :
Q.126 Which of the following are most suitable
indicators
(1) RNA and protein synthesis
(3) Nucleotide synthesis
(4) Polysaccharide synthesis
Ans.
[1]
Sol.
Students may find it in CP Sheet: [Chapter
are
pollution
in
the
(1) Fungi
(2) Lichens
(3) Conifers
(4) Algae
Ans.
[2]
Sol.
Students may find similar question in CP
Exercise
: Cell biology, Page no. 43]
ring
SO2
environment?
(2) Lipid synthesis
Balbiani
of
Sheet:
Ecology
[Chapter
:
Environmental issues, Ex. 1, Q.No. 19]
present
in
polytene
Lichens are SO2 Pollution indicators or they
chromosome & it is the site of RNA &
do not grow in SO2 Polluted habitat.
protein synthesis.
Q.127 Satellite DNA is important because it :
Q.125 Name the pulmonary disease in which
alveolar
surface
area
involvedin
(1) codes for enzymes needed for DNA
gas
replication.
exchange is drastically reduced due to
damage in the alveolar walls.
(2) codes for proteins needed in cell cycle.
(1) Asthma
(3) Shows high degree of polymorphism in
population and also the same degree of
(2) Pleurisy
polymorphism in an individual, which
(3) Emphysema
is heritable from parents to children.
(4) Pneumonia
Ans.
[3]
Sol.
Students may find it in CP Sheet: Animal
physiology-II, Page no. 66
(4) does not code for proteins and is same
in all members of the population.
Ans.
[3]
Sol.
Students may find similar question in CP
Emphysema is pulmonary disease caused by
air pollutant causes break down of alveolar
Exercise Sheet: Genetics, Page no. 51
wall so reduces surface area of gaseous
V.N.T.R. or satellite DNA show high degree
exchange.
of polymorphism is basis of DNA test.
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CAREER POINT
Q.128 Industrial melanism is an example of -
[ CODE – A ]
Q.130 The introduction of t-DNA into plants
involves -
(1) Neo Lamarckism
(1) Allowing the plant roots to stand in
(2) Neo Darwinism
water
(3) Natural selection
(2) Infection of the plant by Agrobacterium
(4) Mutation
Ans.
[3]
Sol.
Students may find it in CP Sheet: Origin &
tumefaciens
(3) Altering the pH of the soil, then heatshocking the plants
Evolution of life, Page no. 32
(4) Exposing the plants to cold for a brief
Industrial melanism shows natural selection.
period
Ans.
[2]
Sol.
Students may find similar question in CP
Q.129 A column of water within xylem vessels of
tall trees does not break under its weight
Sheet: Genetics, Page no. 47
because of -
Agrobacterium tumefaciens bacteria contain
(1) Positive root pressure
t-DNA (transfer DNA)
(2) Dissolved sugars in water
In its Ti plasmid (Tumor inducing plasmid)
(3) Tensile strength of water
This bacterium transfers this t-DNA fragment
into host plant nuclear genome during
(4) Lignification of xylem vessels
Ans.
[3]
Sol.
Students may find similar question in CP
infection.
Q.131 Pick up the wrong statement :
Exercise Sheet: Plant Physiology [Chapter
(1) Nuclear
: Plant water relation theory, Page no. 196]
membrane
is
present
in
Monera
High Tensile strength of water, helps in
(2) Cell wall is absent in Animalia
prevention of breakage of water column
within xylem vessels of high tree.
(3) Protista
have
photosynthetic
and
heterophic modes of nutrition
(4) Some fungi are edible
Ans.
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CAREER POINT
Sol.
[ CODE – A ]
Students may find this in CP Sheet:
Q.134 Which of the following pairs is not
[Chapter : Plant diversity, Page no. 23]
correctly matched?
Monera (Prokaryotes) do not have well
defined
nucleus
as
they
lack
Mode of reproduction
nuclear
membrane and nucleolus.
Q.132 In photosynthesis, the light-independent
reactions take place at :
(1) Stromal matrix
(2) Thylakoid lumen
(3) Photosystem I
(4) Photosystem II
Example
(1) Conidia
Penicillium
(2) Offset
Water hyacinth
(3) Rhizome
Banana
(4) Binary fission
Sargassum
Ans.
[4]
Sol.
Students may find this in CP Sheet:
Ans.
[1]
[Chapter : Plant diversity, Page no. 102]
Sol.
Students may find similar question in CP
Binary fission occurs in unicellular organisms
Exercise Sheet: Plant Physiology [Chapter
whereas sargassum is multicellular brown
algae.
: Photosynthesis theory, Page no. 16]
The light independent reaction or dark
reaction
occurs
in
stromal
matrix
of
Q.135 The UN conference of Parties on climate
chloroplast.
change in the year 2012 was held at :
(1) Warsaw
Q.133 Which of the following immunoglobulins
(2) Durban
does constitute the largest percentage in
(3) Doha
human milk?
(1) IgG
(2) IgD
(3) IgM
(4) IgA
Ans.
[4]
Sol.
Students may find similar question in CP
(4) Lima
Ans.
[3]
Sol.
UN Conference of Parties (COP) 2012 was
held at Qatar, Doha.
Exercise Sheet: Immunity and disease,
Page no. 125
IgA.
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36
CAREER POINT [ CODE – A ]
Q.136 In the spectrum of hydrogen, the ratio of the
Sol.
Students may find similar question in CP
longest wavelength in the Lyman series to the
Exercise Sheet: [Chapter : Photo electric
longest wavelength in the Balmer series is :
effect, Ex-2, Q.No. 28, Page no. 78]
5
(1)
27
(3)
E = 15 keV = 15 × 103 eV
12400
Q E=
eVÅ
λ
12400
λ=
eVÅ
E
12400
λ =
eV Å
15 × 103 eV
λ = .826 Å (λ < 0.01 Å)
4
(2)
9
9
4
(4)
27
5
Ans.
[1]
Sol.
Students may find similar question in :
[Chapter
:
Atomic structure,
similar
question in CP minor test paper ]
⎛ 1
1 ⎞⎟
⎜
⎜ n2 − n2 ⎟
2 ⎠
⎝ 1
Lyman series : n1 = 1, n2 = 2
1 ⎞
1
⎛1
= Rz2 ⎜ 2 − 2 ⎟
λ1
2 ⎠
⎝1
Balmer series :
n1 = 2, n2 = 3
1 ⎞
1
⎛ 1
= Rz2 ⎜ 2 − 2 ⎟
λ2
2
3
⎝
⎠
It is part of X-rays
Q.138 An electron moves on a straight line path
1
= Rz2
λ
λ
eq n (2)
⇒ 1 =
n
λ2
eq (1)
=
XY as shown. The abcd is a coil adjacent to
the path of
electron. What will be the
direction of current, if any, induced in the
...(1)
coil ?
a
...(2)
b
⎛1 1⎞
5
Rz 2 ⎜ − ⎟
9
4
⎝
⎠ = 36
3
1⎞
2⎛
Rz ⎜1 − ⎟
4
⎝ 4⎠
d
c
X
Y
electron
(1) No current induced
5
4
5
×
=
36
3
27
(2) abcd
(3) adcb
Q.137 The energy of the em waves is of the order of
(4) The current will reverse its direction as
15 keV. To which part of the spectrum does it
the electron goes past the coil
belong ?
(1) γ-rays
(3) Infra - red rays
Ans.
[2]
Ans.
[4]
Sol.
Students may find similar question in CP
(2) X-rays
(4) Ultraviolet rays
Exercise
Sheet:
[Chapter
:
EMI,
Ex-3B, Q.No. 47, Page no. 183]
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37
CAREER POINT [ CODE – A ]
When e– is coming towards the loop magnetic
the same level, then the weights added to the
flux of one type increased and when going
steel and brass wires must be in the ratio
away same magnetic flux decreased so
of :
induced current opposite to each other.
(1) 1 :1
(2) 1 :2
(3) 2 :1
(4) 4 :1
Q.139 The cylindrical tube of a spray pump has
radius R, one end of which has n fine holes,
Ans.
[3]
each of radius r. If the speed of the liquid in
Sol.
Students may find similar question in CP
the tube is V, the speed of the ejection of the
liquid through the holes is :
(1)
(3)
V 2R
nr
(2)
VR 2
nr 2
(4)
Exercise Sheet: [Chapter : Elasticity,
Ex-1B, Q.No. 42, Page no. 180]
VR 2
n 2r 2
VR 2
n 3r 2
Ans.
[3]
Sol.
Students may find similar question in CP :
[Class room notes]
Ysteel = 2 Ybrass
Ls = Lb
As = Ab
ΔLs = ΔLb
F
stress
WL
= A =
Y=
ΔL AΔL
strain
L
W=
n.r
2R
YAΔL
∝Y
L
Ws
Y
= s =2:1
Wb
Yb
A1V1 = A2V2
⇒ V2 =
A1V1
πR 2 × V
=
A2
n (πr 2 )
R2
= 2 V
nr
Q.141 A potentiometer
resistance r
wire of length L and a
are connected in series with a
battery of e.m.f. E0 and a resistance r1. An
unknown e.m.f. E is balanced at a length l of
the potentiometer wire. The e.m.f. E will be
Q.140 The Young’s modulus of steel is twice that of
brass. Two wires of same length and of same
given by :
(1)
LE 0 r
(r + r1 )l
(2)
LE 0 r
lr1
(3)
E0r l
(r + r1 ) L
(4)
E 0l
L
area of cross section, one of steel and another
of brass are suspended from the same roof. If
we want the lower ends of the wires to be at
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38
CAREER POINT [ CODE – A ]
Maximum velocity
β = Aω
Ans.
[3]
Sol.
Students may find similar question in CP
Exercise
Sheet:
[Chapter
:
n
Eq (1)
⇒
Eq n (2)
Current
electricity, Ex.-3B, Q.No. 113, Page no.
T = 2π
219]
l
A
P
α
=ω ⇒
β
2π
α
=
T
β
β
α
r
Q.143 If vectors A = cos ωt î + sinωt ĵ
r1
E0
....(2)
and
r
ωt
ωt
ĵ are functions of
B = cos
î + sin
2
2
r
B
time, then the value of t at which they are
E
orthogonal to each other is :
E0
Current in wire AB =
r1 + r
(1) t = 0
⎛ E ⎞ r
Potential gradient = ⎜⎜ 0 ⎟⎟ .
⎝ r1 + r ⎠ L
(3) t =
⎛ E ⎞ r
×l
E = ⎜⎜ 0 ⎟⎟
⎝ r1 + r ⎠ L
∴
Q.142 A particle is executing a simple harmonic
motion. Its maximum acceleration is α and
maximum velocity is β. Then, its time period
of vibration will be:
(1)
2πβ
α
(2)
β2
α2
(3)
α
β
(4)
β2
α
Ans.
[1]
Sol.
Students may find similar question in CP
Exercise Sheet: [Chapter : SHM, Ex-1,
Q.No. 25, Page no. 242]
Maximum acceleration
α = Aω2
π
2ω
(2) t =
π
4ω
(4) t =
π
ω
Ans.
[4]
Sol.
Students may find similar question in CP
[CP Minor test paper ]
For perpendicular vectors
→
→
A . B=0
ωt
ωt ⎤
⎡
⇒ [cos ωt î + sin ωt ĵ ] . ⎢cos î + sin
ĵ = 0
2
2 ⎥⎦
⎣
ωt
ωt
+ sin ωt sin
cos ωt cos
=0
2
2
{Q cos A cos B + sin A sin B = cos (A – B)}
ωt ⎞
⎛
∴ cos ⎜ ωt − ⎟ = 0
2 ⎠
⎝
ωt
=0
⇒ cos
2
ωt
π
⇒
=
2
2
⇒
t=
π
ω
....(1)
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39
CAREER POINT [ CODE – A ]
⎛ 330 ⎞
= 100 ⎜
⎟
⎝ 330 − 9.7 ⎠
Q.144 A source of sound S emitting waves of
frequency 100
Hz and an observer O are
⎛ 330 ⎞
= 100 ⎜
⎟
⎝ 320.3 ⎠
located at some distance from each other. The
source is moving with a speed of 19.4 ms–1 at
= 103.02 Hz
an angle of 60° with the source observer line
as shown in the figure. The observer is at rest.
The apparent frequency observed by the
observer (velocity of sound in air 330 ms–1),
Q.145 An automobile moves on a road with a speed
of 54 km h–1. The radius of its wheels is 0.45
is :
m and the moment of inertia of the wheel
vs
about its axis of rotation is 3 kg m2. If the
vehicle is brought to rest in 15s, the
60°
S
magnitude of average torque transmitted by
O
its brakes to the wheel is :
(1) 97 Hz
(2) 100 Hz
(3) 103 Hz
(1) 2.86 kg m2 s–2
(2) 6.66 kg m2 s–2
(3) 8.58 kg m2 s–2
(4) 10.86 kg m2 s–2
(4) 106 Hz
Ans.
[3]
Sol.
Students may find similar question in CP
[Chapter : Doppler Effect, CP Classroom
Ans.
[2]
Sol.
Students may find similar question in CP
Exercise Sheet: [Chapter : Rotation ,
notes]
Ex- 1, Q.No. 41, Page no. 141]
Vs sin 60º
S
V = 54
Vs
60º
Vs cos 60º
⎛
⎞
v
⎟
n' = n ⎜⎜
⎟
⎝ v − v s cos 60º ⎠
⎛
⎞
⎜
⎟
330
⎟
= 100 ⎜
⎜ 330 − 19.4 × 1 ⎟
⎜
⎟
2⎠
⎝
ω0 =
O
Doppler
line
km
= 15 m/s
h
v
15 rad
=
,
r
.45 s
ω=0
ω = ω0 + αt
0=
15
+ α15
.45
α=–
15
1 rad
=–
.45 × 15
.45 s 2
τ = Iα = –
3
300
=–
= –6.66 kgm2/s2
.45
45
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40
CAREER POINT [ CODE – A ]
Q.146 A rectangular coil of length 0.12 m and width
(1)
C2V 2
2d 2
(2)
C2V 2
2d
(3)
CV 2
2d
(4)
CV 2
d
0.1 m having 50 turns of wire is suspended
vertically in a
uniform magnetic field of
strength 0.2 Weber/m2. The coil carries a
current of 2A. If the plane of the coil is
Ans.
[3]
Sol.
Students may find similar question in CP:
inclined at an angle of 30° with the direction
of the field, the torque required to keep the
[Chapter : Capacitance, Classroom notes]
coil in stable equilibrium will be :
Ans.
(1) 0.12 Nm
(2) 0.15 Nm
(3) 0.20 Nm
(4) 0.24 Nm
Force between plates
of capacitor
F = qE
⎛ q ⎞
⎟
= q⎜⎜
⎟
⎝ 2A ∈0 ⎠
[3]
q2
2A ∈0
∴ q = CV
C2V 2
F=
2A ∈0
F=
Sol.
Students may find similar question in CP
Exercise Sheet: [Chapter : Magnetic field,
Ex. -1, Q.No. 117, Page no. 70]
⎛ A ∈0 ⎞ 2
⎜
⎟CV
d ⎠
⎝
F=
2A ∈0
B
30º
60º
F=
A
τ = NAIB sin θ
= 50 × .012 × 2 × 0.2 × sin 60º
= 0.20 Nm
CV 2
2d
Q.148 Two vessels separately contain two ideal
gases A and B at the same temperature, the
pressure of A being twice that of B. Under
such conditions, the density of A is found to
Q.147 A parallel plate air capacitor has capacity ‘C’,
be 1.5 times the density of B. The ratio of
distance of separation between plates is ‘d’
molecular weight of A and B is :
and potential difference ‘V’ is applied
between the plates. Force of attraction
(1)
1
2
(2)
(3)
3
4
(4) 2
between the plates of the parallel plate air
capacitor is :
Ans.
2
3
[3]
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41
CAREER POINT Sol.
[ CODE – A ]
Students may find similar question in CP
Sol.
[Chapter : KTG, Classroom notes]
Students may find similar question in CP
Exercise Sheet: [Chapter : Gravitation, CP
Minor Test ]
A
B
Force on satellite is only gravitational force,
which will always be towards the centre of
T
T
PA = 2PB
PB
ρA = 1.5 ρB
MA
=?
MB
P RT
ρRT
⇒ M0 =
Q =
ρ M0
P
R, T = const.
ρ
M0 ∝
P
ρ
M
P
⇒ A = A× B
M B ρ B PA
1
= 1.5 ×
2
= 0.75 = 3/4
earth
Q.150 In the given figure, a diode D is connected to
an external resistance R = 100 Ω and an
e.m.f. of 3.5 V. If the barrier potential
developed across the diode is 0.5V, the
current in the circuit will be:
100 Ω
D
R
Q.149 A satellite S is moving in an elliptical orbit
3.5 V
around the earth. The mass of the satellite is
very small compared to the mass of the earth.
(1) 35 mA
(2) 30 mA
Then,
(3) 40 mA
(4) 20 mA
(1) the acceleration of S is always directed
Ans.
[2]
Sol.
Students may find similar question in CP
towards the centre of the earth.
(2) the angular momentum of S about the
centre of the earth changes in direction,
Exercise Sheet: [Chapter : Electronics,
Solved Example, Page no. 179]
but its magnitude remains constant.
i=
(3) the total mechanical energy of S varies
periodically with time
Vnet 3.5 − 0.5
=
R net
100
=
3
Amp.
100
(4) the linear momentum of S remains
constant in magnitude.
Ans.
= 30 mA
[1]
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42
CAREER POINT [ CODE – A ]
Q.151 A remote – sensing satellite of earth revolves
(1) Path of the particle is a circle of radius 4
in a circular orbit at a height of 0.25 × 106 m
meter
r
(2) Acceleration vector is along – R .
above the surface of earth. If earth’s radius is
6.38 × 106 m and g = 9.8 ms–2, then the
orbital speed of the satellite is :
(1) 6.67 km s–1
(2) 7.76 km s–1
(3) 8.56 km s–1
(4) 9.13 km s–1
(3) Magnitude of acceleration vector is
v2
,
R
where v is the velocity of particle
(4) Magnitude of the velocity of particle is
8 meter/ second
Ans.
[2]
Sol.
Students may find similar question in CP
Exercise Sheet: [Chapter : Gravitation ,
Ans.
[4]
Sol.
Students may find similar question in CP
Ex-2, Q.No. 6, Page no. 208]
[Chapter : Motion in plane, Class room
notes]
GMm mv 2
=
r
r2
v=
⎛
v= ⎜
⎜
⎝
=
r
R = 4 sin(2πt )î + 4 cos(2πt ) ĵ
r=R+h
GM
GMR 2
g
=
=
R
r
r
R 2r
⎞
9.8
⎟ × 6.38 × 10 6
6
6 ⎟
.25 × 10 + 6.38 × 10 ⎠
1.47
× 6.38 × 10 6
10 6
3
= 7.76 × 10 m/s = 7.76 km/s
r
Q.152 The position vector of a particle R as a
x = 4 sin 2πt
….(1)
y = 4 cos 2πt
…(2)
x2 + y2 = 42 (sin2 2πt + cos2 2πt)
(1)
x2 + y2 = 42 equation of circle and radius 4m
.
r V2
(−R̂ )
(2) a =
R
V2
R
function of time is given by :
(3) a =
r
R = 4 sin(2πt) î + 4cos(2πt) ĵ
(4) Vx = + 4(cos 2πt) (2π)
Where R is in meters, t is in seconds and î
and ĵ denote unit vectors along x-and ydirections, respectively. Which one of the
following statements is wrong for the motion
and Vy = – 4(sin 2πt) (2π)
V=
=
Vx2 + Vx
2
(8π) 2 (cos 2 2πt + sin 2 2πt )
V = 8π
of particle ?
so 4th option is incorrect.
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43
CAREER POINT [ CODE – A ]
ω0
Q.153 A string is stretched between fixed points
separated by 75.0 cm. It is observed to have
resonant frequencies of 420 Hz and 315 Hz.
m1
m2
P
There are no other resonant frequencies
x
(L – x)
between these two. The lowest resonant
frequency for this string is :
Ans.
Sol.
(1) 105 Hz
(2) 155 Hz
(3) 205 Hz
(4) 10.5 Hz
(1) x =
m2L
m1 + m 2
(2) x =
m1L
m1 + m 2
(3) x =
m1
L
m2
(4) x =
m2
L
m1
[1]
Students may find similar question in CP
Ans.
[1]
Sol.
Students may find similar question in CP
Exercise Sheet: [Chapter : Wave theory,
[Chapter : Rotation, classroom notes]
Ex-1, Q.No. 98, Page no. 41]
I = m1x2 + m2(L – x)2
I = m1x2 + m2L2 + m2x2 – 2m2Lx
315 : 420 = 3 : 4
315 Hz is a 3 harmonic of wire and 420 Hz
dI
= 2m1x + 0 + 2xm2 – 2m2L = 0
dx
is a 4th harmonic.
x(2m1 + 2m2) = 2m2L
If 315 is a 3rd harmonic then 3n fundamental = 315
x=
rd
n fundamental =
315
= 105Hz
3
m2L
m1 + m 2
1
Iω02
2
When I is minimum, work done W =
will also be minimum.
Q.154 Point masses m1 and m2 are placed at the
opposite ends of a rigid rod of length L and
Q.155 At the first minimum adjacent to the central
maximum of a single-slit diffraction pattern,
negligible mass. The rod is to be set rotating
the phase difference between the Huygen’s
about an axis perpendicular to it.
The
wavelet from the edge of the slit and the
position of point P on this rod through which
wavelet from the midpoint of the slit is :
the axis should pass so that the work required
π
radian
8
(2)
(3)
π
radian
2
(4) π radian
to set the rod rotating with angular velocity
ω0 is minimum, is given by :
Ans.
π
radian
4
(1)
[4]
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44
CAREER POINT Sol.
[ CODE – A ]
⇒ 0 î – 12(1 + α) ĵ + 6(1 + α) k̂ = 0
Students may find similar question in CP
[Chapter : Wave optics, Classroom notes]
–12(1 + α) = 0
α = –1
A
B λ/2
For first minima path difference between A
Q.157 Two particles A and B, move with constant
r
r
velocities v1 and v 2 . At the initial moment
r
r
their position vectors are r1 and r2
and C is λ so path difference between A and
respectively. The condition for particles A
B should be λ/2 and
and B for their collision is :
Δ=λ
C
phase difference π
r r
r
r
(1) r1 – r2 = v1 – v 2
r
Q.156 A force F = α î + 3 ĵ + 6k̂ is acting at a point
r r
r r
r –r
v –v
(2) r1 r2 = r 2 r1
r1 – r2
v 2 – v1
r
r = 2î − 6 ĵ − 12k̂ . The value of α for which
r r
r r
(3) r1 ⋅ v1 = r2 ⋅ v 2
angular momentum about origin is conserved
is :
r r
r r
(4) r1 × v1 = r2 × v 2
(1) 1
(2) –1
(3) 2
(4) zero
Ans.
[2]
Sol.
Students may find similar question in CP
Ans.
[2]
Sol.
Students may find similar question in CP
[Chapter : Vector, Classroom notes]
Exercise
Sheet:
[Chapter
:
r
v1t
Rotation,
Solved Example 13, Page no. 134]
Angular momentum
r
L = constant
r
r1
r
v2t
r
R
r
r2
⇒
r
Torque τ = 0
⇒
r
r
r × F =0
r
r r
r r
R = r1 + v1t = r2 + v 2 t
k̂
r r
r
r
r1 − r2 = ( v 2 − v1 ) t
î
2
α
ĵ
− 6 − 12 = 0
3
6
⇒ (–36 + 36) î – (12 + 12α) ĵ + (6 + 6α) k̂ =0
r r
r
r
r1 − r2
( v 2 − v1 ) t
=
r r
r
r
| r1 − r2 |
| v 2 − v1 | t
r r
r
r
r1 − r2
v 2 − v1
r r = r
r
| r1 − r2 | | v 2 − v1 |
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45
CAREER POINT [ CODE – A ]
(1)
σ1σ 2
σ1 + σ 2
(2)
(3)
σ1 + σ 2
2 σ1σ 2
(4)
Q.158 A nucleus of uranium decays at rest into
nuclei of thorium and helium. Then :
(1) The helium nucleus has less kinetic
energy than the thorium nucleus
(2) The helium nucleus has more kinetic
2 σ1σ 2
σ1 + σ 2
σ1 + σ 2
σ1σ 2
Ans.
[2]
Sol.
Students may find similar question in CP
energy than the thorium nucleus
[Chapter : Current Electricity, Classroom
notes]
(3) The helium nucleus has less momentum
than the thorium nucleus
L
ρ1
(4) The helium nucleus has more momentum
than the thorium nucleus.
2L
L
L
= ρ1
+ ρ2
A
A
A
[2]
ρ
Sol.
Students may find similar question in CP
2ρ = ρ1 + ρ2
Exercise
2 1
1
=
+
σ σ1 σ 2
[Chapter
:
Nuclear
Physics, Ex-2, Q.No. 158, Page no. 17]
238
92U
2 σ1 + σ 2
=
σ
σ1σ 2
→ 90Th234 + 2He4
Momentum conservation
r
r
⇒ | p Th | = | p He | = p
Kinetic energy K =
K∝
A
Req = R1 + R2
Ans.
Sheet:
L
ρ2
p2
2m
σ=
2σ1σ 2
σ1 + σ 2
Q.160 Light of wavelength 500 nm is incident on a
1
m
metal with work function 2.28 eV. The de
Broglie wavelength of the emitted electron is
mHe < mTh
:
⇒ KHe > KTh
(1) ≤ 2.8 × 10–12 m
(2) < 2.8 × 10–10 m
Q.159 Two metal wires of identical dimensions are
connected in series. If σ1 and σ2 are the
(3) < 2.8 × 10–9m
conductivities of the metal wires respectively,
(4) ≥ 2.8 × 10–9m
the effective conductivity of the combination
is :
Ans.
[4]
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46
CAREER POINT Sol.
[ CODE – A ]
Students may find similar question in CP
vsound = 952 m/sec
[Chapter : P.E.E., , Classroom notes]
CP = ?
EPh =
1240
eV = 2.48 eV
500
vsound =
K.Emax = EPh – W
γ=
= 2.48 – 2.28
= 0.2 eV
* λe min =
=
12.27
K.E max (eV)
12.27
0.2
γPV
M
C
M 2
Vsound = P
CV
PV
⎛ M⎞ 2
CP = CV ⎜
⎟ v sound
⎝ PV ⎠
Å
⎡
⎤
4 × 10 −3
= 5⎢ −5
(952)2
−3 ⎥
×
×
10
22
.
4
10
⎣
⎦
Å
=
= 27.436 Å
= 27.436 × 10–10 m
20
× (952)2 × 10–5
22.4
= 809, 200 × 10–5 = 8.09 J/moles k
λmin. = 2.7436 × 10 m
–9
λ ≥ λmin
Q.162 A series R-C circuit is connected to an
alternating voltage source. Consider two
Q.161 4.0 g of a gas occupies 22.4 liters at NTP.
situations:
The specific heat capacity of the gas at
(a) When capacitor is air filled.
constant volume is 5.0 JK–1 mol–1. If the
speed of sound in this gas at NTP is 952 ms–1,
(b) When capacitor is mica filled.
then the heat capacity at constant pressure is
Current through resistor is i and voltage
(Take gas constant R = 8.3 JK–1 mol–1)
across capacitor is V then :
(1) 8.5 JK–1mol–1
(1) Va = Vb
(2) 8.0 JK–1 mol–1
(2) Va < Vb
(3) 7.5 JK–1 mol–1
(3) Va > Vb
(4) 7.0 JK–1 mol–1
(4) ia > ib
Ans.
[1]
Ans.
[3]
Sol.
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Sol.
Students may find similar question in CP
Exercise Sheet: [Chapter : KTG, Solved
[Chapter : AC, Classroom notes]
Example 9, Page no. 123]
M = 4 gm
V = 22.4 litre
CV = 5
J
mole K
C
R
~
V = V0 sin ωt
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47
CAREER POINT XC =
[ CODE – A ]
1
2πfC
Ans.
[3]
Sol.
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current in circuit
V
I=
=
Z
or I =
[Chapter : Friction, classroom notes]
V
⎛ 1 ⎞
R2 + ⎜
⎟
⎝ 2πfC ⎠
μs = tan30º =
2
2πfC
S = ut +
×V
4π 2 f 2 C 2 R 2 + 1
4=0+
Voltage drop across capacitor
1
3
= 0.57 ≈ 0.6
1 2
at
2
1
(gsin30º – μkcos 30º)(4)2
2
VC = I × XC
=
VC =
2πfC × V
4π 2 f 2 C 2 R 2 + 1
×
0.5 = 10 ×
1
2πfC
1
3
– μk 10 ×
2
2
5 3μ k = 4.5
V
μk = 0.51
4π 2 f 2 C 2 R 2 + 1
When mica is introduced capacitance will
Q.164 Two stones of masses m and 2 m are whirled
increase, hence voltage across capacitor gets
in horizontal circles, the heavier one in a
decrease.
radius
Q.163 A plank with a box on it at one end is
r
and lighter one in radius r. The
2
tangential speed of lighter stone is n times
gradually raised about the other end. As the
that of the value of heavier stone when they
angle of inclination with the horizontal
experience same centripetal forces. The value
reaches 30°, the box starts to slip and slides
of n is :
4.0 m down the plank in 4.0 s. The
coefficients of static and kinetic friction
(1) 1
(2) 2
between the box and the plank will be,
(3) 3
(4) 4
respectively :
Ans.
[2]
Sol.
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[Chapter : Circular Motion, Minor Test]
mg
θ
v2
(1) 0.4 and 0.3
r/2
(2) 0.6 and 0.6
r
2m
m
(3) 0.6 and 0.5
v1
Equal centripetal force
(4) 0.5 and 0.6
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48
CAREER POINT [ CODE – A ]
Q.166 An ideal gas is compressed to half its initial
mv12 (2m) v 22
=
r
r/2
volume by means of several processes. Which
⇒ v12 = 4v 22
of the process results in the maximum work
⇒ v1 = 2v2
done on the gas ?
Q.165 The
coefficient
of
performance
of
a
refrigerator is 5. If the temperature inside
freezer is –20°C, the temperature of the
(2) 31°C
(3) 41°C
(4) 11°C
(2) Adiabatic
(3) Isobaric
(4) Isochoric
Ans.
[2]
Sol.
Students may find similar question in CP
surroundings to which it rejects heat is :
(1) 21°C
(1) Isothermal
[Chapter
Thermodynamics,
Classroom
notes]
V1 → V1/2
Ans.
[2]
Sol.
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adiabatic
[Chapter : Thermodynamics, Classroom
isothermal
notes]
α=5
T2 = – 20ºC
isobaric
= – 20 + 273
T2 = 253K
V1/2
T1 = ?
α =
Work done = area under curve
T2
T1 − T2
253
⇒5=
T1 − 253
V1
Wadiabatic > Wisothermal > Wisobaric
Q.167 A ball is thrown vertically downwards from a
height of 20 m with an initial velocity v0. It
collides with the ground, loses 50 percent of
⇒ 5T1 – 1265 = 253
its energy in collision and rebounds to the
1518
T1 =
= 303.6K
5
same height. The initial velocity v0 is :
= 30.6ºC ≈ 31ºC
(Take g = 10 ms– 2)
T1 = temp. of surrounding
(1) 10 ms–1
(2) 14 ms–1
(3) 20 ms–1
(4) 28 ms–1
T2 = temp. of source (inside temp.)
Ans.
[3]
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49
CAREER POINT Sol.
[ CODE – A ]
Using conservation of K.E
Students may find similar question in CP
[Chapter
:
Work,
power,
energy,
2
1
1 ⎛v⎞
1
Mv2 + 0 = M ⎜ ⎟ + Mv 22
2
2 ⎝3⎠
2
Classroom notes]
1
⎛
⎞
0. 5⎜ mgh + mv2 ⎟ = mgh
2
⎝
⎠
v2 =
0.25mv2 = 0.5mgh
v=
=
2gh
v2
+ v 22
9
8v 2
9
v 22 =
2 × 10 × 20 =
400 = 20 m/s
v2 =
2 2v
3
Q.168 On a frictionless surface, a block of mass M
moving at speed v collides elastically with
another block of same mass M which is
Q.169 If potential(in volts) in a region is expressed
initially at rest. After collision the first block
as V (x, y, z) = 6xy – y + 2yz, the electric
moves at an angle θ to its initial direction
field (in N/C) at point (1, 1, 0) is :
and has a speed
(
v
. The second block‘s speed
3
(1) – 6î + 9 ĵ + k̂
(
(3) – 6î + 5 ĵ + 2k̂
after the collision is :
(1)
(3)
3
v
2
3
v
4
)
(2)
(4)
2 2
v
3
3
2
(
(2) – 3î + 5 ĵ + 3k̂
)
(
(4) – 2î + 3 ĵ + k̂
)
)
Ans.
[3]
Sol.
Students may find similar question in CP
Exercise Sheet: [Chapter : Electrostatics,
v
Ex-3A, Q.No. 26, Page no. 56]
Ans.
[2]
Sol.
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V = 6xy – y + 2yz
Exercise Sheet: [Chapter : Work, power,
E= −
∂V
∂V ∂ V
î –
ĵ –
k̂
∂x
∂y
∂z
energy , Ex-3B, Q.No. 9, Page no. 101]
E = – 6 yî – (6x – 1) ĵ – 2 yk̂
v/3
At (1, 1, 0)
M
M →v
M → u2 = 0 ≡
θ
E = – 6î – 5 ĵ – 2k̂
M
v2
E = – (6î + 5 ĵ + 2k̂ )
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50
CAREER POINT [ CODE – A ]
Q.170 Two slits in Youngs experiment have widths
Sol.
Students may find similar question in CP
in the ratio 1 : 25. The ratio of intensity at the
Exercise
maxima and minima in the interference
Mechanics, Ex-1A, Q.No. 4, Page no. 215]
pattern,
I max
is :
I min
P=
4
(1)
9
(3)
9
(2)
4
121
49
(4)
=
=
49
121
[Chapter
:
Fluid
W mgh V
=
=
ρgh
t
t
t
5 × 10 −3
× 13.6 × 103 × 10 × 150 × 10–3
60
5 × 13.6 × 150
× 10–3
6
= 1700 × 10–3
P = 1.7 Watt
Ans.
[2]
Sol.
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Exercise Sheet: [Chapter : Wave Optics ,
Q.172 A proton and an alpha particle both enter a
Ex-1B , Q.No. 2, Page no.89]
I max
I min
Sheet:
⎛ I1
⎞
⎜
+1 ⎟
⎜ I
⎟
=⎜ 2
⎟
⎜ I1 − 1 ⎟
⎜ I
⎟
⎝ 2
⎠
region of uniform magnetic field B, moving
at right angles to the field B. If the radius of
2
circular orbits for both the particles is equal
and the kinetic energy acquired by proton is
1MeV, the energy acquired by alpha particle
will be :
2
⎛ 1
⎞
⎜
+1 ⎟
25
⎟ = 36 = 9
= ⎜⎜
⎟
16
4
1
− 1 ⎟⎟
⎜⎜
⎝ 25
⎠
(1) 1 MeV
(2) 4 MeV
(3) 0.5 MeV
(4) 1.5 MeV
Q.171 The heart of a man pumps 5 liters of blood
through the arteries per minute at a pressure
of 150 mm of mercury. If the density of
mercury
be
13.6
×
103
kg/m3
and
g = 10 m/s2, then the power of heart in
watt is :
(1) 1.50
Ans.
[1]
Sol.
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Exercise Sheet: [Chapter : Magnetic field,
Solved example 19 Page no. 35]
Radius in magnetic field
(2) 1.70
R=
(3) 2.35
Ans.
(4) 3.0
mv
=
qB
2mE
qB
[2]
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51
CAREER POINT E=
[ CODE – A ]
V0 = Av × VI = 300
q 2 B2 R 2
2m
Vouput = 300 cos (15t + π/3 + π)
For proton
E1 =
Vouput = 300 cos (15t + 4π/3)
e ×B ×R
2 × mp
2
2
2
Q.174 If dimensions
for α-particle
of critical velocity vc of a
liquid flowing through a tube are expressed as
( 2e ) × B × R
2 × 4m p
2
E2 =
2
2
[ηx ρy rz] , where η, ρ and r are the
coefficient of viscosity of liquid, density of
∴ E1 = E2
liquid and radius of the tube respectively,
then the values of x, y, and z are given by :
(1) 1, 1, 1
Q.173 The input signal given to a CE amplifier
having
a
voltage
gain
of
150
is
(2) 1, –1, –1
π⎞
⎛
Vi = 2 cos ⎜15t + ⎟ . The corresponding
3⎠
⎝
(3) –1, –1, 1
(4) –1, –1, –1
output signal will be :
4π ⎞
⎛
(1) 300 cos ⎜15 t +
⎟
3 ⎠
⎝
π⎞
⎛
(2) 300 cos ⎜15 t + ⎟
3⎠
⎝
Ans.
[2]
Sol.
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Exercise
Sheet:
[Chapter
:
Unit
&
Dimensions, Ex-2, Q.No. 42, Page no. 24]
2π ⎞
⎛
(3) 75 cos ⎜15 t +
⎟
3 ⎠
⎝
5π ⎞
⎛
(4) 2 cos ⎜15 t + ⎟
6 ⎠
⎝
Ans.
Vc = ηxρyrz
M0L1T–1 = (M1L–1T–1)x (M1L–3)y (L1)z
M0L1T–1 = Mx+y L–x–3y +z T–x
–x = – 1
x=1
[1]
... (1)
x+y=0=0
Sol.
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y=–x =–1
Exercise Sheet: [Chapter : Electronics,
–x –3y + z = 1
Q.No. 30, Page no. 191]
–1 – 3(–1) + z = 1
Vi = 2 cos (15t + π/3)
Av =
...... (2)
z=–1
V0
VI
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52
CAREER POINT [ CODE – A ]
Q.175 A circuit contains an ammeter, a battery of 30
Q.176 Water rises to a height ‘h’ in capillary tube. If
V and a resistance 40.8 ohm all connected in
the length of capillary tube above the surface
series. If the ammeter has a coil of resistance
of water is made less than ‘h’, then :
480
ohm
and
a
shunt
of
(1) Water does not rise at all.
20 ohm, the reading in the ammeter will be :
(2) water rises upto the tip of capillary tube
(1) 1 A
and then starts overflowing like a
fountain.
(2) 0.5 A
(3) water rises upto the top of capillary tube
(3) 0.25 A
and stays there without overflowing.
(4) 2 A
(4) water rises upto a point a little below the
Ans.
[2]
Sol.
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top and stays there
[Chapter : Current electricity, Classroom
notes]
Ans.
[3]
Sol.
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[Chapter : Surface tension, Classroom
notes]
RS =20Ω
40.8 Ω
Height of water column > length of tube
RA =480 Ω
⇒ so liquid will be stay there
I
30 V
Q.177 In an astronomical telescope in normal
adjustment a straight black line of length L is
480 × 20
Req = 40.8 +
500
drawn on inside part of objective lens. The
= 40.8 + 19.2
eye-piece forms a real image of this line. The
= 60 Ω
length of the image is I. The magnification of
I=
30
= 0.5 amp
60
the telescope is :
So reading of ammeter = 0.5 amp
Ans.
(1)
L
I
(2)
L
+1
I
(3)
L
–1
I
(4)
L+I
L–I
[1]
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53
CAREER POINT Sol.
[ CODE – A ]
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ρ = ρ0 (1 + YΔT)
[Chapter : Optical Instrument, Classroom
ρ – ρ0 = ρ0 Y ΔT
notes]
fractional change
fO + fe
ρ − ρ0
= YΔT = 5 × 10–4 × 40
ρ0
= 200 × 10–4 = 0.020
fe
fO
Q.179 A
For eye – piece lens
photoelectric
surface
is
illuminated
successively by monochromatic light of
h
f
m=
= I
f + u hO
⇒
fe
I
=
f e + [− (f O + f e )] L
⇒–
⇒
wavelength λ and
in the seconds case is 3 times that in the first
fe
I
=
L
fO
case, the work function of the surface of the
material is :
fO
L
= m.p.
=–
l
fe
(h = Planck’s constant, c = speed of light)
(1)
hc
3λ
(2)
hc
2λ
(3)
hc
λ
(4)
2hc
λ
Q.178 The value of coefficient of volume expansion
Ans.
[2]
Sol.
40°C in its temperature, is :
Students may find similar question in CP
Exercise Sheet: [Chapter : P.E.E , Ex-1,
Q.No. 34, Page no. 75]
(1) 0.010
(2) 0.015
kmax = E – φ
(3) 0.020
(4) 0.025
k=
of glycerin is 5 × 10–4 K–1. The fractional
change in the density of glycerin for a rise of
hc
–φ
λ
[3]
3k =
Sol.
maximum
kinetic energy of the emitted photoelectrons
(–) sign stands for inverted image
Ans.
λ
. If the
2
Students may find similar question in CP
hc
–φ
( λ / 2)
[Chapter : Thermal Expansion, CP Major
On solving
Paper]
φ=
... (1)
.... (2)
hc
2λ
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54
CAREER POINT [ CODE – A ]
Q.180 A beam of light consisting of red, green and
blue colors is incident on a right angled
prism. The refractive index of the material of
the prism for the above red, green and blue
wavelengths are 1.39, 1.44 and 1.47,
respectively.
Hence green and Blue colour will
not emerge out
A
Blue
Green
Red
B
45°
C
The prism will :
(1) separate the red colour part from the
green and blue colours
(2) separate the blue colour part from the red
and green colours
(3) separate all the three colours from one
another
(4) not separate the three colours at all
Ans.
[1]
Sol.
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[Chapter : Ray Optics, Classrooms Notes]
i
i = 45°
∴
for no emergence
i>C
sin i > sin C
sin i >
1
2
µ>
1
µ
>
1
µ
2
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55