01-06-2025

9610WMD801490250004 MD

PHYSICS

1) Two cells each of same e.m.f. but of internal resistance r1 and r2 are joined to form a series circuit
through an external resistance R. Value of R in term of r1 and r2 for which cell 1 has zero p.d. across

it is :-

(1) R = r1 – r2
(2) R = r1 + r2

(3)

(4)

2) Figure shows a balanced wheatstone bridge :

Then which of the following is false ?

(1) There is no flow of current from galvano meter in the balanced condition
(2) If R1 is slightly increased, the current in galvanometer flows from D to B
(3) If R4 is slightly decreased, the current in galvanometer flows from D to B
(4) If R3 is slightly increased, the current in galvanometer flows from B to D

3) The temperature coefficient of resistance of a wire is 0.00125 per degree celcius. At 300 K its
resistance is 1 ohm. The resistance of the wire will be 2 ohms at a temperature :–

(1) 1154 K
(2) 1127 K
(3) 600 K
(4) 1400 K

4) Two cells A and B of emf 1.3 V and 1.5 V respectively are arranged as shown in figure. The
voltmeter reads 1.45 V. The voltmeter is assumed to be ideal. Then :-

(1) r1 = 2r2
(2) r1 = 3r2
(3) r2 = 2r1
(4) r2 = 3r1

5) In the circuit shown in figure, V must be :-

(1) 50 V
(2) 80 V
(3) 100 V
(4) 1290 V

6) Two electric bulbs whose resistance are in the ratio of 1 : 2, are connected in series to a constant
voltage source. The power dissipated in them has the ratio :-

(1) 1 : 2
(2) 1 : 1
(3) 1 : 4
(4) 2 : 1

7) The Galvanometer G reads zero. If the battery A has negligible internal resistance the value of
resistance 'X' will be :-

(1) 200 Ω
(2) 50 Ω
(3) 500 Ω
(4) 100 Ω

8) For the circuit shown in figure, which one of the following equation is correct :-
(1) E1 – i2x + E2 + i1y = 0
(2) E1 – i2x – i1z = 0
(3) E2 + i2y – i2z = 0
(4) –E2 + (i1 + i2)z + i1y = 0

9) A ring is made of a wire having a resistance R0 = 15 Ω . Find the points A and B as shown in figure
at which a current carrying conductor should be connected so that the resistance R of the sub circuit

between these point is equal to Ω :-

(1)

(2)

(3)

(4)

10) In the adjoining circuit, the potential difference across 3 Ω is :

(1) 2 V
(2) 4 V
(3) 8 V
(4) 16 V

11) Equivalent resistance between A and B is:

(1)

(2)

(3)
R
(4) R

12) An infinite ladder network of resistance is constructed with 1Ω and 2Ω resistance. The 6V
battery between A and B has negligible internal resistance. The current that passes through 2Ω

resistance nearest to the battery is

(1) 1A
(2) 1.5 A
(3) 2 A
(4) 2.5 A

13) The resultant capacitance between A and B in the figure is :

(1) 1µF
(2) 10 µF
(3) 50 µF
(4) 1.5

14) Find potential at A and B in the given circuit

(1) 40 V, 0 V
(2) 50 V, 0 V
(3) 60 V, –40 V
(4) 40 V, –10 V

15) A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then
disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the
capacitor, is now inserted in it. Which of the following is incorrect ?

(1) The energy stored in the capacitor decreases K times.

(2)
The change in energy stored is
(3) The charge on the capacitor is not conserved.
(4) The potential difference between the plates decreases K times.

16) The magnetic field due to a small current element at a distance and element carrying
current i is

(1)

(2)

(3)

(4)

17) The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at
a distance of 4 cm from the centre is 54 µT. What will be its value at the centre of the loop

(1) 250 µT
(2) 150 µT
(3) 125 µT
(4) 75 µT

18) A wire carrying current I has the shape as shown adjoining figure. Linear parts of the wire are
very long and parallel to X-axis while semicircular portion of radius R is lying Y-Z plane. Magnetic
field at point O is

(1)

(2)

(3)

(4)

19) A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y
plane and the other in x-z plane. If the current in the loop is i. The resultant magnetic field due to
the two semicircular parts at their common centre is

(1)

(2)

(3)

(4)

20) The distance at which the magnetic field on axis as compared to the magnetic field at the center

of the coil carrying current I and radius R is , would be

(1) R
(2)
(3) 2R
(4)

21) A coil having N turns carry a current I as shown in the figure. The magnetic field intensity at
point P is

(1)

(2)

(3)

(4) Zero

22) A wire carrying current I is shaped as shown. Section AB is a quarter circle of radius r. The

magnetic field is directed

(1) At an angle π/4 to the plane of the paper

(2) Perpendicular to the plane of the paper and directed in to the paper
(3) Along the bisector of the angle ACB towards AB
(4) Along the bisector of the angle ACB away from AB

23) In the figure, what is the magnetic field at the point O

(1)

(2)

(3)
(4)

24) Magnetic field at a distance r from an infinitely long straight conductor carrying a steady current
varies as

(1) 1/r2
(2) 1/r
(3) 1/r3
1/
(4)

25) A proton moves along vertical line and towards observer, the pattern of concentric circular field
lines of magnetic field which is produced due to its motion:

(1) ACW, In horizontal plane

(2) ACW, In vertical plane
(3) CW, In horizontal plane
(4) CW, In vertical plane

26) For an infinite long solid cylindrical current carrying conductor if magnetic field at a point P as
shown in figure is B0, then the magnetic field at point R will be (uniformly distributed current)

(1)

(2)

(3)

(4)

27) The magnetic induction at the centre O in the figure shown is

(1)

(2)

(3)

(4)

28) Two concentric coplanner coils of turns n1 and n2 have radii ratio 2 : 1 respectively. Equal
current in both the coils flows in opposite direction. If net magnetic field is zero at their common
centre then n1 : n2 is :-

(1) 2 : 1
(2) 1 : 2
(3) 1 : 1
(4) 4 : 1

29) Find magnetic field at point P due to current carrying wire as shown in figure.

(1)

(2)

(3)

(4)

30) A and B are two concentric circular conductors of centre O and carrying currents i1 and i2 as
shown in the adjacent figure. If ratio of their radii is 1 : 2 and ratio of the flux densities at O due to A
and B is 1 : 3, then the value of i1/i2 is

(1)

(2)

(3)

(4)

31) PQRS is a square loop made of uniform conducting wire the current enters the loop at P and

leaves at S. Then the magnetic field will be

(1) Maximum at the centre of the loop

(2) Zero at the centre of loop
(3) Zero at all points inside the loop
(4) Zero at all points outside of the loop

32) Two long straight wires are set parallel to each other. Each carries a current i in the same
direction and the separation between them is 2r. Magnetic field midway between them is

(1)
(2)
(3) Zero
(4)
33) The magnetic field intensity at the centre of cubical cage of identical wires of length 'a' due to a

current I flowing as shown in the figure is

(1)

(2)

(3) 0
(4) 1/2a

34) A circular coil of radius R carries an electric current. The magnetic field due the coil at a point
on the axis of the coil located at a distance r from the center of the coil, such that r >> R varies as :-

(1) 1/r
(2) 1/r3/2
(3) 1/r2
(4) 1/r3

35) Magnetic field at origin 'O' due to given current distribution is :-

(1)

(2)

(3)

(4) Zero

36) A charge +3C moves vertically downward with speed 4π m/s. Find magnitude of magnetic field
at point P.
(1)

(2)

(3)

(4)

37) A current i ampere flows in a circular arc of wire whose radius is R, which subtends an angle

radian at its centre. The magnetic induction B at the centre is

(1)

(2)

(3)

(4)

38) A long, straight wire is turned into a loop of radius 10 cm (see figure) If a current of 8 amperes is
passed through the loop, then the value of the magnetic field and its direction at the centre C of the

loop shall be close to

(1) 5.0 × 10–5 Newton/(amp-meter), upward

(2) 3.4 × 10–5 Newton/(amp-meter), upward
(3) 1.6 × 10–5 Newton/(amp-meter), downward
(4) 1.6 × 10–5 Newton/(amp-meter), upward

39) Two straight long conductors AOB and COD are perpendicular to each other and carry currents
i1 and i2. The magnitude of the magnetic induction at a point P at a distance a from the point O in a
direction perpendicular to the plane ACBD is

(1)
(2)

(3)

(4)

40) Circular loop of a wire and a long straight wire carry currents Ic and Ie, respectively as shown in
figure. Assuming that these are placed in the same plane. The magnetic fields will be zero at the

centre of the loop when the separation H is

(1)

(2)

(3)

(4)

41) Two symmetrical current carrying rings are placed perpendicular to each other with common
centre. If magnetic field at the centre due to one coil is B, then find net magnetic field.

(1)
(2) B
(3)
(4)
42) Current 'I' is flowing in a conductor shaped as shown in the figure. The radius of the curved part
is r and the length of straight portion is very large. The value of the magnetic field at the centre O

will be

(1)

(2)

(3)

(4)

43) A steady current I goes through a wire loop PQR having shape of a right angle triangle with PQ

= 3x, PR = 4x and QR = 5x. If the magnitude of the magnetic field at P due to this loop is k ,
find the value of k.

(1) 8
(2) 3
(3) 7
(4) None of these

44) A square loop of side 'a' is made by a current carrying wire. Magnetic field at its vertex 'P' is :-

(1)

(2)

(3)

(4)

45) Magnetic field due to a moving charge be the -

(1)

(2)

(3)

(4)

CHEMISTRY

1) Which of the following is correct about a solution showing positive deviation?

(1) Vapour pressure observed will be the less than that calculated from Raoult's law
(2) Minimum boiling azeotrope will be formed
(3) ΔHmix < 0
(4) ΔVmix < 0

2) Which of the following physical properties is used to determine, the molecular mass of a polymer
solution?

(1) Relative lowering of vapour pressure

(2) Elevation in boiling point
(3) Depression in freezing point
(4) Osmotic pressure

3) In the case of osmosis net flow of solvent molecules move from solution having

(1) Higher vapour pressure to lower vapour pressure

(2) Higher concentration to lower concentration
(3) Lower vapour pressure to higher vapour pressure
(4) Higher osmotic pressure to lower osmotic pressure

4) Which of the following equimolar solution have highest vapour pressure?

(1) Glucose
(2) NaCl
(3) K2SO4
(4) K4Fe(CN)6

5) If α is the degree of dissociation of Na2SO4. The van't Hoff factor used for the calculation of
molecular mass is

(1) 1 + 2α
(2) 1 – 2α
(3) 1 + α
(4) 1 – α

6) An aqueous solution freezes at –0.36°C. Kf and Kb for water are 1.8 and 0.52 K kg mol–1
respectively then value of boiling point of solution at 1 atm pressure is

(1) 101.04°C
(2) 100.104°C
(3) 0.104°C
(4) 100°C

7) The osmotic pressure of decimolar solution of urea at 27°C is

(1) 2.49 bar

(2) 5.0 bar
(3) 3.4 bar
(4) 1.25 bar

8) Charge required for reduction of 0.1 mol Al2O3 into Al :-

(1) 0.3 F
(2) 0.2 F
(3) 0.6 F
(4) 0.4 F

9) If limiting molar conductances of Ca+2 and Cl– ions are 119 and 76.3 Scm2 mol–1 respectively. The
limiting molar conductivity of CaCl2 is :-

(1) 347 S cm2 mol–1

(2) 271.6 S cm2 mol–1
(3) 42.7 S cm2 mol–1
(4) 195.3 S cm2 mol–1

10) Specific conductance of 0.1 M Nitric acid is 6.3 × 10–2 ohm–1 cm–1. The molar conductance of the
solution is :

(1) 630 ohm–1 cm2 mol–1

(2) 315 ohm–1 cm2 mol–1
(3) 100 ohm–1 cm2 mol–1
(4) 6.300 ohm–1 cm2 mol–1

11) The thermodynamic efficiency of cell is given by-

(1)
(2)

(3)

(4) Zero

12) At equilibrium:

(1)
(2) Ecell = 0, ΔG = 0
(3) both are correct
(4) none is correct

13) Statement-I: Conductivity always increases with decrease in the concentration of electrolyte.
Statement-II: Molar conductivity always increases with decrease in the concentration of electrolyte

(1) Both Statement I and Statement II are true

(2) Both Statement I and Statement II are false
(3) Statement I is false but Statement II is true
(4) Statement I is true but Statement II is false

14) Assertion :- Conductivity always decreases with decrease in concentration of weak and strong
electrolyte.
Reason :- On dilution number of ions per unit volume linearly decreases in both weak and strong
electrolytes.

(1) Both Assertion and Reason are true and Reason is the correct explanation of Assertion
(2) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion
(3) Assertion is true but Reason is false
(4) Both Assertion and Reason are false

15) Which of the following solution of KCl has the maximum value of conductivity?

(1) 1M
(2) 0.1M
(3) 0.01M
(4) 0.001M

16) The common features among the species CN–, CO and NO+ are :

(1) bond order three and diamagnetic

(2) bond order three and weak field ligands
(3) paramagnetic and strong field ligands
(4) paramagnetic and π-acceptor ligands

17) CO bond length is longest in :

 +
(1) [Mn(CO)6]
–
(2) [V(CO)6]
(3) Cr(CO)6
2–
(4) [Ti(CO)6]

18) Which of the following ligand does not act as π-acid ligand ?

(1) CN¯
(2) CO
(3) C2H4
(4)

19) What is the oxidation number of Fe in [Fe(H2O)5(NO)]2+ ion ?

(1) +2
(2) +3
(3) +1
(4) 0

20) Which is not true about metal carbonyls ?

(1) Here CO act as a Lewis base as well as Lewis acid

(2) Here metal acts as Lewis base as well as Lewis acid
(3) Here dπ-pπ back bonding takes place
(4) Here pπ-pπ back bonding takes place

21) Find IUPAC name of the hydrate isomer of CrCl3.6H2O, which is having lowest electrical
conductivity.

(1) Hexaaquachromium(III)chloride
(2) Tetraaquadichloridochromium(III)chloridedihydrate
(3) Pentaaquachloridochromium(III)chloridemonohydrate
(4) Triaquatrichloridochromium(III)trihydrate

22) The IUPAC name of Xe[PtF6] is :

(1) Hexafluoroplatiate(VI)xenon
(2) Xenohexafluoroplatinate(V)
(3) Xenohexafluoroplatiate(VI)
(4) Xenoniumhexafluoroplatinum(V)

23) Select the compound having maximum conductivity in aqueous medium.

(1) [Cr(NH3)6]Cl3
(2) [Cr(NH3)5Cl]Cl2
(3) [Cr(NH3)4Cl2]Cl
(4) [Cr(NH3)3Cl3]

24) Select correct statement for Cr.6NH3.Cl3 and Cr.5NH3.Cl3

(1) In both complex compounds secondary valency is satisfied by only NH3

(2) In both complex compounds Cl– are satisfying only primary valency
(3) In both complex compounds primary valency is satisfied by only Cl–
(4) In both complex compounds all Cl– are ionizable

25) Select the correct IUPAC name of Prussian blue :

(1) Iron(II) hexacyanoferrate(II)

(2) Iron(II) hexacyanoferrate(III)
(3) Iron(III) hexacyanoferrate(II)
(4) Iron(III) hexacyanoferrate(III)

26) Which of the following complex is an outer orbital complex ?

2+
(1) [Ni(NH3)6]
4–
(2) [Mn(CN)6]
3+
(3) [Co(NH3)6]
4–
(4) [Fe(CN)6]

27) Octahedral complexes of Ni(II) are always :

(1) Inner orbital and diamagnetic

(2) Outer orbital ad paramagnetic
(3) Inner or outer orbital depending upon the strong or weak field ligand
(4) None of these

28) The magnetic moment of [MnX4]2– is 5.9 BM. The geometry of the complex ion is :
(X = monodentate halide ion)

(1) tetrahedral
(2) square planar
(3) both are possible
(4) none of these

29) The geometry of [Ni(CO)4] and[NiCl2(PPh3)2] are :

(1) both square planar

(2) tetrahedral and square planar respectively
(3) both are tetrahedral
(4) square planar and tetrahedral respectively
30) Which of the following order is correct in spectrochemical series of ligands ?

(1)
(2)
(3)
(4)

31) Which of the following complexes have a maximum number of unpaired electrons ?

(1) [Ni(CO)4]
+
(2) [Co(NH3)4(NO2)2]
–
(3) [Ag(CN)2]
2–
(4) [CuBr4]

32) The structure of K[PtCl3(C2H4)] and hybridisation of Pt respectively are :

(1) square planar, sp2d2

(2) square planar, dsp2
(3) tetrahedral, sp3
3
(4) octahedral, d2sp

33) The crystal field stabilisation energy of [Co(NH3)6]Cl3 is :

(1) –7.2 Δ0
(2) –0.4 Δ0
(3) –2.4 Δ0
(4) –3.6 Δ0

34) Statement-1 :- [Ti(H2O)6]4+ is coloured while [Sc(H2O)6]3+ is colourless.

Statement-2 :- d-d transition is not possible in [Sc(H2O)6]3+.

(1) If statement-1 is true but the statement-2 is false.

(2) If statement-1 is false but statement-2 is true.
(3) If both statements are true and the statement-2 is the correct explanation of statement-1.
(4) If both statements are true but the statement-2 is not the correct explanation of statement-1.

35) Statement-1 :- Transition metal ion forming octahedral complexes undergo sp3d2 or d2sp3
hybridisation.
Statement-2 :- Strong field ligands force the unpaired electrons of central metal ion to pair up
causing d2sp3 hybridisation whereas weak field ligands do not after electronic configuration of the
metal ion undergoes in sp3d2 hybridisation.

(1) If statement-1 is true but the statement-2 is false.

(2) If statement-1 is false but statement-2 is true.
(3) If both statements are true and the statement-2 is the correct explanation of statement-1.
(4) If both statements are true but the statement-2 is not the correct explanation of statement-1.

36) Statement-1 :- CO and CN are referred as π acid ligands.

Statement-2 :- In CO and CN vacant π type orbitals are present.

(1) If statement-1 is true but the statement-2 is false.

(2) If statement-1 is false but statement-2 is true.
(3) If both statements are false.
(4) If both statements are true.

37) Given below are two statements :

Statement-I : Complex [Fe(H2O)5(NO)]SO4 is paramagnetic.
Statement-II : The Fe in Complex [Fe(H2O)5(NO)]SO4 has three unpaired electrons.
In the light of the above statements, choose the most appropriate answer from the options given
below :

(1) Statement-I and Statement-II are correct.

(2) Statement-I is correct but Statement-II is incorrect.
(3) Both Statement-I and II are incorrect.
(4) Statement-I is incorrect but Statement-II is correct.

38) Given below are two statements :

Statement-I : The primary valency are normally ionisable and are satisfied by negative ions.
Statement-II : The secondary valency are non-ionisable are satisfied by neutral molecules or
negative ions.
In the light of the above statements, choose the most appropriate answer from the options given
below :

(1) Statement-I and Statement-II are correct.

(2) Statement-I is correct but Statement-II is incorrect.
(3) Both Statement-I and II are incorrect.
(4) Statement-I is incorrect but Statement-II is correct.

39) Given below are two statements :

Statement-I : Decacarbonyl dimangnese (0) is made up of two square pyramidal [Mn(CO)5] units
joined by a Mn–Mn bond.
Statement-II : Zero oxidation state of central metal cannot be stabilised by CO groups.
In the light of the above statements, choose the most appropriate answer from the options given
below :

(1) Statement-I and Statement-II are correct.

(2) Statement-I is correct but Statement-II is incorrect.
(3) Both Statement-I and II are incorrect.
(4) Statement-I is incorrect but Statement-II is correct.

40) CN– is an ambidentate ligand because

(1) it has multiple bonds

(2) carbon has a negative charge
(3) it form chelate
(4) both C & N can donate lone pair

41) Match the following and choose correct option.

Ligand Property

(A) ethylenediamine (P) bidentate ligand

(B) hydrazene (Q) tridentate ligand

(C) dimethylgyloximato(dmg–) (R) Monodentate ligand

(D) dien (S) can form hydrogen bond

A B C D
(1) P Q R S
(2) P R Q S
(3) P R S Q
(4) P S Q R
(1) 1
(2) 2
(3) 3
(4) 4

42) In the complex ion [Fe(EDTA)]– the coordination number (C.N.) and oxidation number (O.N.) of
central metal ion is :-

(1) C. N.= 6, O. N. = +3
(2) C.N. =1, O. N. = –1
(3) C. N. = 4, O. N. = +2
(4) C. N. = 3, O. N. = +3

43) Which one is a heteroleptic complex ?

(1) Ferrocene
(2) Chromocene
(3) Prussian blue
(4) Zeise's salt

44) Which of the following species is not expected to be ligand :-

(1) NO
+
(2) NH4
(3) NH2CH2CH2NH2
(4) CO

45) Identify the statement which is not correct :-

(1) Coordination compounds are mainly known for transition metals.
(2) Coordination number and oxidation state of central metal atom in a complex is always same.
(3) Ligand donates at least one electron pair to CMA
+
(4) [Co(NH3)4Cl2] is a heteroleptic complex.

BIOLOGY

1) Which of the following is not a bacterial disease?

(1) Typhoid
(2) Tuberculosis
(3) Ringworm
(4) Pneumonia

2) The causative organism of malaria is:

(1) Plasmodium
(2) Entamoeba
(3) Leishmania
(4) Trypanosoma

3) The drug used for treatment of allergy is:

(1) Antibiotics
(2) Antihistamines
(3) Antivirals
(4) Interferons

4) Malaria is transmitted by:

(1) Male Anopheles mosquito

(2) Female Anopheles mosquito
(3) Culex mosquito
(4) Sandfly

5) AIDS is caused by:

(1) Bacteria
(2) Virus
(3) Protozoa
(4) Fungus

6) Initial symptoms of HIV infection resemble:

(1) Diabetes
(2) Cancer
(3) Common cold
(4) Hypertension

7) Heroin is obtained from:

(1) Erythroxylum coca

(2) Cannabis sativa
(3) Papaver somniferum
(4) Nicotiana tabacum

8) HIV targets which immune cells?

(1) B cells
(2) Cytotoxic-T cells
(3) Helper-T cells
(4) Plasma cells

9) Passive immunity is acquired by:

(1) Vaccination
(2) Infection
(3) Injection of antibodies
(4) Exposure to antigens

10) BCG vaccine is used for:

(1) Typhoid
(2) Polio
(3) Tuberculosis
(4) Hepatitis

11) Which is not a mode of HIV transmission?

(1) Blood transfusion

(2) Sexual contact
(3) Mosquito bite
(4) Infected needles

12) Histamine causes:

(1) Dilation of blood vessels

(2) Constriction of bronchi
(3) Increased permeability of capillaries
(4) All of the above
13) Cancer cells differ from normal cells by:

(1) Uncontrolled growth

(2) Ability to undergo apoptosis
(3) Limited life span
(4) Lack of mutations

14) Leukemia is cancer of:

(1) Liver
(2) Bone
(3) Blood-forming tissues
(4) Skin

15) Metastasis refers to:

(1) Cell maturation

(2) Formation of benign tumors
(3) Spread of cancer to distant sites
(4) Cell death

16) Interferons are produced in response to:

(1) Bacterial infection

(2) Allergens
(3) Viral infection
(4) Protozoa

17) Cocaine is a:

(1) Depressant
(2) Stimulant
(3) Narcotic
(4) Hallucinogen

18) Cell-mediated immunity is provided by:

(1) Antibodies
(2) B cells
(3) T cells
(4) Plasma cells

19) ELISA is used to detect:

(1) Malaria
(2) Typhoid
(3) HIV
(4) Tuberculosis

20) Which is a secondary lymphoid organ?

(1) Thymus
(2) Bone marrow
(3) Lymph node
(4) Liver

21) Which component is used in vaccines to produce immunity?

(1) Toxins
(2) Antibodies
(3) Attenuated microbes
(4) Plasma

22) NACO works for :

(1) hepatitis B
(2) Syphilis
(3) AIDS
(4) Cancer

23) Match the following diseases with their causative agents

Column I Column II
(Disease) (Causative Agent)
A. Malaria 1. Salmonella typhi
B. Typhoid 2. Plasmodium falciparum
C. Filariasis 3. Wuchereria bancrofti
D. Pneumonia 4. Streptococcus pneumoniae
Options:
(1) A–2, B–1, C–3, D–4
(2) A–3, B–2, C–1, D–4
(3) A–4, B–3, C–2, D–1
(4) A–1, B–2, C–3, D–4

24) Match the following immune cells with their functions -

Column I Column II
(Immune Cell) (Function)
A. B-lymphocytes 1. Antibody production
B. T-lymphocytes 2. Cell-mediated immunity
C. Macrophages 3. Phagocytosis
D. Mast cells 4. Release of histamine
Options:
(1) A–1, B–2, C–3, D–4
(2) A–2, B–1, C–4, D–3
(3) A–3, B–4, C–2, D–1
(4) A–4, B–3, C–1, D–2

25) Match the following diseases with their mode of transmission -

Column I (Disease) Column II (Mode of Transmission)

A. AIDS 1. Blood transfusion/sexual contact
B. Malaria 2. Bite of infected female Anopheles
C. Common Cold 3. Droplet infection
D. Ascariasis 4. Contaminated food and water
(1) A–3, B–2, C–1, D–4
(2) A–1, B–2, C–3, D–4
(3) A–4, B–1, C–2, D–3
(4) A–2, B–3, C–4, D–1

26) Match the following types of immunity with examples

Column I (Immunity Type) Column II (Example)

A. Active natural immunity 1. Infection recovery
B. Active artificial immunity 2. Vaccination
C. Passive natural immunity 3. Mother's milk antibodies
D. Passive artificial immunity 4. Injection of pre-formed antibodies
(1) A–2, B–1, C–4, D–3
(2) A–3, B–4, C–2, D–1
(3) A–1, B–2, C–3, D–4
(4) A–4, B–3, C–1, D–2

27) Assertion (A) : AIDS is caused by HIV which is a retrovirus.

Reason (R) : Retroviruses have RNA as genetic material and use reverse transcriptase enzyme.

(1) Both A and R are true, and R is the correct explanation of A.

(2) Both A and R are true, but R is not the correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.

28) Assertion (A): Vaccination provides passive immunity.

Reason (R): Passive immunity involves the transfer of antibodies.

(1) Both A and R are true, and R is the correct explanation of A.

(2) Both A and R are true, but R is not the correct explanation of A
(3) A is true but R is false.
(4) Both A and R are false.
29) Which of the following statements about the immune system are correct?
1. B-lymphocytes are responsible for cell-mediated immunity.
2. T-lymphocytes help in the destruction of infected cells.
3. Antibodies are produced by plasma cells derived from B-cells.
4. Vaccines provide active immunity.

(1) 1, 2 and 3 only

(2) 2, 3 and 4 only
(3) 1, 3 and 4 only
(4) 1, 2, 3 and 4

30) Which of the following statements about HIV/AIDS are correct?

1. HIV specifically attacks helper T-cells.
2. AIDS can be diagnosed by ELISA test.
3. HIV has DNA as its genetic material.
4. HIV infection leads to immunodeficiency.

(1) 1, 2 and 3 only

(2) 1, 2 and 4 only
(3) 1, 3 and 4 only
(4) 1, 2, 3 and 4

31) Assertion (A): Lactobacillus is used in the production of curd.

Reason (R): Lactobacillus ferments lactose to lactic acid, which curdles milk proteins.

(1) Both A and R are true, and R is the correct explanation of A.

(2) Both A and R are true, but R is not the correct explanation of A.
(3) A is true, but R is false.
(4) Both A and R are false.

32) Match the column-I with column-II :

Column I Column II
(Microbe) (Product)
A. Saccharomyces cerevisiae 1. Citric acid
B. Trichoderma polysporum 2. Statins
C. Monascus purpureus 3. Ethanol
D. Aspergillus niger 4. Cyclosporin A
(1) A-3, B-4, C-2, D-1
(2) A-4, B-3, C-1, D-2
(3) A-1, B-2, C-3, D-4
(4) A-2, B-1, C-4, D-3

33) Which of the following is used as a biofertilizer in paddy fields?

(1) Azospirillum
(2) Rhizobium
(3) Azolla
(4) Glomus

34) Which of the following microbes is used in the production of vinegar?

(1) Saccharomyces cerevisiae

(2) Acetobacter aceti
(3) Lactobacillus
(4) Trichoderma

35) Match the column-I with column-II :

Column I Column II
(Microbe) (Use)
A. Rhizobium 1. Nitrogen fixation in legume roots
B. Azotobacter 2. Nitrogen fixation in free-living state
C. Azospirillum 3. Nitrogen fixation in cereal crops
D. Frankia 4. Nitrogen fixation in actinorhizal plants
(1) A-1, B-2, C-3, D-4
(2) A-2, B-1, C-4, D-3
(3) A-3, B-4, C-1, D-2
(4) A-4, B-3, C-2, D-1

36) Which of the following is used in the production of antibiotics?

(1) Aspergillus niger

(2) Penicillium
(3) Saccharomyces cerevisiae
(4) Trichoderma

37) Which microbe is used to ripen Swiss cheese and gives it its characteristic holes?

(1) Aspergillus niger

(2) Propionibacterium shermanii
(3) Lactobacillus
(4) Rhizobium

38) Assertion (A): Antibiotics are effective against bacterial infections.

Reason (R): Antibiotics inhibit virus replication inside host cells.

(1) Both A and R are true, and R is the correct explanation of A.

(2) Both A and R are true, but R is not the correct explanation of A.
(3) A is true, but R is false.
(4) A is false, but R is true.

39) Which of the following is used in the production of statins (cholesterol-lowering agents)?
(1) Aspergillus
(2) Monascus purpureus
(3) Trichoderma
(4) Rhizopus stolonifer

40) Match the column-I with column-II :

Column I Column II
(Microbe/Use) (Function/Product)
A. Bacillus thuringiensis 1. Insect pest control
B. Saccharomyces cerevisiae 2. Alcohol production
C. Methanobacterium 3. Biogas production
D. Penicillium notatum 4. Antibiotic production
(1) A–1, B–2, C–3, D–4
(2) A–2, B–1, C–4, D–3
(3) A–3, B–2, C–1, D–4
(4) A–4, B–3, C–2, D–1

41) Which microbe is commonly used as a starter culture in alcohol fermentation?

(1) Rhizopus
(2) Saccharomyces cerevisiae
(3) Penicillium
(4) Bacillus subtilis

42) Identify the correct statements about biogas:

1. Biogas mainly contains methane.
2. It is produced by aerobic decomposition.
3. Cow dung is a common substrate.
4. Methanogens are involved in the process.

(1) 1, 2, 3
(2) 1, 3, 4
(3) 2, 3, 4
(4) 1, 2, 4

43) Which of the following is used in sewage treatment?

(1) Methanobacterium
(2) Monascus
(3) Floc-forming bacteria
(4) Trichoderma

44) What is cyclosporin A used for?

(1) Blood pressure regulation

(2) Immunosuppression in organ transplant
(3) Treating bacterial infections
(4) Killing nematodes

45) Which of the following is a free-living nitrogen fixer?

(1) Rhizobium
(2) Frankia
(3) Azotobacter
(4) Anabaena in Azolla

46) The term “gene” was coined by:

(1) Mendel
(2) Darwin
(3) Bateson
(4) Johanssen

47) In a monohybrid cross, the F2 phenotypic ratio is :

(1) 1 : 2 : 1
(2) 3 : 1
(3) 9 : 3 : 3 : 1
(4) 1 : 1

48) Alleles are :

(1) Different chromosomes

(2) Identical genes
(3) Alternative forms of a gene
(4) Homologous chromosomes

49) Law of Independent Assortment is applicable only when:

(1) Genes are linked

(2) Genes are on the same chromosome
(3) Genes are on different chromosomes
(4) Genes are recessive

50) Which disease is caused due to a single recessive gene located on the X chromosome?

(1) Phenylketonuria
(2) Sickle-cell anaemia
(3) Haemophilia
(4) Down syndrome
51) In incomplete dominance, the F1 phenotype is:

(1) Similar to recessive parent

(2) Similar to dominant parent
(3) Intermediate of both parents
(4) Similar to both parents

52) In codominance:

(1) One allele is dominant

(2) Both alleles express partially
(3) Both alleles express equally
(4) None express

53) The disorder caused by an autosomal recessive gene leading to abnormal haemoglobin is:

(1) Thalassemia
(2) Colour blindness
(3) Turner syndrome
(4) Klinefelter syndrome

54) XO condition in females causes:

(1) Down syndrome

(2) Klinefelter syndrome
(3) Turner syndrome
(4) Superfemale

55) If a man with blood group A marries a woman with blood group B, the possible blood groups of
offspring are:

(1) A, B
(2) A, B, AB
(3) A, B, AB, O
(4) AB only

56) Which of the following is an example of polygenic inheritance?

(1) Colour blindness

(2) Sickle cell anaemia
(3) Skin colour in humans
(4) Thalassemia

57) Number of chromosomes in an individual with Klinefelter syndrome:

(1) 44
(2) 45
(3) 47
(4) 46

58) The gene responsible for sex determination in humans is present on:

(1) Chromosome 1
(2) Autosomes
(3) Y chromosome
(4) Mitochondrial DNA

59) Which disorder shows a non-disjunction of chromosome 21?

(1) Turner syndrome

(2) Down syndrome
(3) Sickle-cell anaemia
(4) Colour blindness

60) If both parents are carriers of thalassemia, what is the chance that their child is affected?

(1) 100%
(2) 25%
(3) 50%
(4) 0%

61) Identify the correct statements:

1. In codominance, both alleles express themselves equally.
2. In incomplete dominance, one allele is fully dominant.
3. Human blood group AB is an example of codominance.
4. Monohybrid F2 phenotypic ratio in incomplete dominance is 1 : 2 : 1.

(1) 1, 3, 4
(2) 2, 3
(3) 1 and 2
(4) All of these

62) Choose the true statements:

1. Sickle-cell anaemia is caused by substitution of glutamic acid by valine.
2. It is inherited in an autosomal dominant fashion.
3. Carriers show symptoms in all conditions.
4. The mutation is in the β-globin chain of haemoglobin.

(1) 1 and 4
(2) 1, 2, 3
(3) 2 and 3
(4) All of these
63) Which statements are false?
1. Turner syndrome individuals have 46 chromosomes.
2. Klinefelter syndrome individuals are female.
3. Haemophilia is X-linked recessive.
4. XO females are sterile.

(1) 1 and 2
(2) 1 and 3
(3) 2 and 4
(4) 3 and 4

64) Select the correct statements regarding sex determination in humans:

1. Females are homogametic.
2. Males produce two types of gametes.
3. The ovum determines the sex of the child.
4. The presence of Y determines maleness.

(1) 1, 2, 4
(2) 2 and 3
(3) 1 and 4
(4) All of these

65) Regarding pedigree analysis, identify correct options:

1. Helps identify inheritance patterns.
2. Used in human genetics.
3. Can identify carriers.
4. Cannot detect sex-linked traits.

(1) 1, 2, 3
(2) 1 and 2
(3) Only 3
(4) All of these

66) Match the inheritance pattern with the example:

A Codominance 1 Skin colour

B Incomplete dominance 2 AB blood group

C Polygenic inheritance 3 Flower colour in Snapdragon

D Sex-linked recessive 4 Haemophilia

(1) A–2, B–3, C–1, D–4
(2) A–3, B–2, C–1, D–4
(3) A–2, B–1, C–4, D–3
(4) A–1, B–2, C–3, D–4

67) Match syndrome with chromosomal condition:

 A Down Syndrome 1 XXY

B Turner Syndrome 2 Trisomy 21

C Klinefelter Syndrome 3 XO

D Normal Male 4 XY
(1) A–2, B–3, C–1, D–4
(2) A–3, B–2, C–1, D–4
(3) A–1, B–2, C–3, D–4
(4) A–2, B–1, C–4, D–3

68) Match the terms with their definitions:

A Gene 1 Physical expression

B Allele 2 Alternate forms of gene

C Phenotype 3 Segment of DNA coding for protein

D Genotype 4 Genetic constitution

(1) A–3, B–2, C–1, D–4
(2) A–2, B–3, C–1, D–4
(3) A–3, B–1, C–2, D–4
(4) A–4, B–2, C–1, D–3

69) Match the scientists with their contributions:

A Mendel 1 Term – Linkage

B Morgan 2 Father of genetics

C Hugo de Vries 3 Chromosomal theory

D Sutton 4 Rediscovered Mendel’s laws

(1) A–2, B–3, C–1, D–4
(2) A–3, B–2, C–4, D–1
(3) A–2, B–1, C–4, D–3
(4) A–2, B–3, C–4, D–1

70) Match the blood groups with possible genotypes:

A Blood group A 1 IAIA or IAi

B Blood group B 2 IBIB or IBi

C Blood group AB 3 IAIB

D Blood group O 4 ii
(1) A–1, B–2, C–3, D–4
(2) A–2, B–1, C–3, D–4
(3) A–1, B–2, C–4, D–3
(4) A–1, B–3, C–2, D–4

71) Assertion (A) : There are four genotypes of gametes RY, Ry, rY and ry each with a frequency of
25 per cent or 1/4th of the total gametes produced in F1 generation of dihybrid cross.
Reason (R) : During gametes formation, segregation of 50 per cent R and 50 per cent r is
independent from the segregation of 50 per cent Y and 50 per cent y.
In the light of the above statement, choose the correct answer from the option given below

(1) Both A and R are true and R is the correct explanation of A.

(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true and R is false.
(4) A is false and R is true.

72) From the statements given below choose the correct option :
A. Dependent pairs segregate independently of each other.
B. Segregate at the time of gamete formation such that only one of each pair is transmitted to a
gamete.
C. Chromosome occurs in pair but genes are unpaired in haploid organisms.
D. Occur in pairs in diploid organisms.
E. Chromosome are paired but genes are unpaired in diploid organisms.
Which statement/s is/are not applicable for both Chromosomes and Genes?

(1) Only C and D

(2) Only A, C, D
(3) Only A
(4) Only A,C, E

73) Match List 1with List 2

List 1 List 2

A Flower position 1 Round

B Seed shape 2 Terminal

C Flower colour 3 Green

D Seed colour 4 Violet

(1) A- 2, B-1, C-4, D-3
(2) A- 4, B-1, C-2, D-3
(3) A- 2, B-3, C-4, D-1
(4) A- 2, B-4, C-1, D-3

74) Assertion: (A) A carrier mother of haemophilia can pass the disorder to her son.
Reason: (R) Haemophilia is an autosomal recessive disorder.

(1) Both A and R are correct; R explains A.

(2) A is correct, R is incorrect.
(3) Both A and R are incorrect.
(4) Both A and R are correct; R does not explain A.

75) Assertion (A) : Males are more frequently affected by X-linked disorders.
Reason (R) : They have only one X chromosome.

(1) Both A and R are correct; R explains A.

(2) Both A and R are correct; R does not explain A.
(3) A is correct; R is incorrect.
(4) Both A and R are incorrect.

76) Which of the following statements are false?

1. Down syndrome is caused by trisomy 13.
2. Klinefelter individuals are male.
3. Turner syndrome individuals have 47 chromosomes.
4. Trisomy 21 causes mental retardation.

(1) 1 and 3 only

(2) 2 and 3 only
(3) 1 and 4 only
(4) All except 2

77) Match diseases with their genetic basis:

A Colour blindness 1 Autosomal recessive

B Thalassemia 2 Trisomy 21

C Down syndrome 3 X-linked recessive

(1) A–3, B–1, C–2
(2) A–2, B–3, C–1
(3) A–1, B–2, C–3
(4) A–3, B–2, C–1

78) Who is regarded as the "Father of Genetics"?

(1) Hugo de Vries

(2) Charles Darwin
(3) Gregor Mendel
(4) T.H. Morgan

79) A test cross is used to determine:

(1) The genotype of a recessive individual.

(2) The genotype of a dominant individual.
(3) The phenotype of F1 generation.
(4) The number of genes involved.
80) Assertion(A): X-linked recessive traits appear more frequently in males.
Reason(R): Males are hemizygous for X-linked genes.

(1) Both A and R are correct, R explains A.

(2) Both A and R are correct, R does not explain A.
(3) A is correct, R is incorrect.
(4) Both A and R are incorrect.

81) Which chromosomal abnormality causes Klinefelter syndrome?

(1) XO
(2) XX
(3) XXY
(4) XYY

82) A couple, both with normal vision, have a son with colour blindness. Which parent is likely the
carrier?

(1) Father
(2) Mother
(3) Both (1) and (2)
(4) None of these

83) A couple has a daughter with Turner syndrome. What is the likely chromosomal composition?

(1) 46, XX
(2) 45, XO
(3) 47, XXX
(4) 46, XY

84) Who proposed the Chromosomal Theory of Inheritance?

(1) Morgan and Muller

(2) Sutton and Boveri
(3) Mendel and Correns
(4) Darwin and Wallace

85) If a pedigree chart shows a trait skipping generations, it most likely represents:

(1) Autosomal dominant inheritance

(2) Autosomal recessive inheritance
(3) X-linked dominant inheritance
(4) Y-linked inheritance

86) Statement Ι : When many genes govern a character it is called polygenic inheritance.
Statement ΙI : In polygenic inheritance, the effect of each allele is additive.
(1) Statement Ι : is true but II is false
(2) Statement Ι : is false but II is true
(3) Both Statement I & II are true
(4) Both Statement I & II are false

87)
In this given pedigree what is the mode of inheritance ?

(1) Autosomal dominant

(2) Autosomal recessive
(3) X-linked dominant
(4) X-linked recessive

88) SRY gene is a : -

(1) A gene present on Y-chromosome

(2) A gene present on X-chromosome
(3) A gene present on X-autosome
(4) A segment of m-RNA

89) Which increasing age, the effect on linkage is :-

(1) Becomes strong

(2) Becomes weak
(3) Terminates
(4) Unchanged

90) Match the following columns and choose the incorrect match ?

I II
(1)
male
I II
(2)
female
I II
(3)
affected individual
I II
(4)
consanguineous mating
 ANSWER KEYS

PHYSICS

Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A. 1 3 2 2 2 1 4 1 1 1 1 2 1 3 3 4 1 2 1 4
Q. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
A. 1 2 3 2 1 1 1 1 2 1 2 3 3 4 4 3 4 2 3 1
Q. 41 42 43 44 45
A. 3 3 3 1 2

CHEMISTRY

Q. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65
A. 2 4 1 1 1 2 1 3 2 1 3 2 3 3 1 1 4 4 3 4
Q. 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85
A. 2 2 1 3 3 1 2 1 3 1 4 2 3 2 3 4 3 1 2 4
Q. 86 87 88 89 90
A. 1 1 4 2 2

BIOLOGY

Q. 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110
A. 3 1 2 2 2 3 3 3 3 3 3 4 1 3 3 3 2 3 3 3
Q. 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130
A. 3 3 1 1 2 3 1 4 2 2 1 1 3 2 1 2 2 3 2 1
Q. 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150
A. 2 2 3 2 3 4 2 3 3 3 3 3 1 3 3 3 3 3 2 2
Q. 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170
A. 1 1 1 1 1 1 1 1 3 1 1 4 1 2 1 1 1 3 2 1
Q. 171 172 173 174 175 176 177 178 179 180
A. 3 2 2 2 2 3 2 1 1 2
 SOLUTIONS

PHYSICS

1)

Explaining : Asking Value of R

Given Condition Potential Difference across 1st Cell is zero.
Concept : This question is based on Key Points :
(1) Discharging of cell, V = ε – I r
(2) KVL : ΣΔV = 0
KCL : ΣIin = ΣIout
Ohm's Law : V = IR

Solution : Current,

P.D. across cell 1 = Ir1 =

For zero pd., the fall of potential should be equal to its e.m.f

i.e.,
i.e., R = r1 – r1
Final Answer : (1) R = r1 – r2

2)

Explaining : Given : A wheat-stone bridge

Ask : direction of current flow in slightly unbalanced Wheat stone Bridge.
Concept : This question is based on key points :

(1) Current B D

(2) Current B D

Solution : If Current B → D

As, and

If R4 is slightly decreased
∴ Direction of Current in galvanometer is from B to D
Final Answer : (3)

3)

Explaining : Given : α = 0.00125 °C–1, R = at T = 300K,

Ask : Value of T when R =
Concept : This question is based on key points :
(1) Temperature dependent resistance
(2) RT = R0 (1 + α ΔT)
Solution : R = R0 (1 + αΔT)
1 = R0[1 + α(27)]

0
2 = R [1 + α(T)] ⇒

T = 854°C
T = 854 + 273 = 1127 K
Final Answer : (2) T = 1127K

4) Explaining : Given : Ideal Voltmeter and equivalent single loop circuit.

Ask : Relation between (r1,r2)
Concept : This question is based on key points :
(1) Ideal Voltmeter Open Circuit
(2) KVL : ΣΔV = 0
Solution : No current will flow through voltmeter. As it is ideal (infinite resistance). Current
Through two batteries

Now, V = E2 – ir2

∴
Solving this equation, we get
r1 = 3r2
Final Answer : (2) r1 = 3r2

5) Explaining : Given : Series and Parallel Combination of Resistance.

Ask : Applied Voltage V
Concept : This question is based on key points :
(1) Req = R1 + R2 + ......

(2)
(3) V = I Req
Solution : 100Ω, 25Ω and 20Ω are in parallel.
Their, net resistance is 10Ω
∴ Rnet = 4Ω + 10Ω + 6Ω = 20Ω
V = i Rnet = 80 V
Final Answer : (2) V = 4 × 20 = 80v

6)

Explaining : Given : Ration of resistance of bulb 1: 2

Ask : Ration of Power dissipated
Concept : This question is based on key points :
(1) In Series current (I) is same.
(2) P1 : P2 = R1 : R2

Solution :
P1 = I2R
P2 = I22R

∴
Final Answer : (1) P1 : P2 = R1 : R2 = 1: 2

7)

Explaining : Given : Reads zero ∴

Ask : Value of resistance 'x'
Concept : This question is based on key points :

(1) If no current flows through G

(2) KVL + KCL

Solution :
If Ig = 0 than VA – VB = 4V
By potential divider rule

⇒ X = 100 Ω
Final Answer : (4) X =

8) Explaining : Given : multiloop circuit

Ask : Correct equation based on KVL
Concept : This question is based on key points :
(1) KVL in a loop ΣΔV = 0
(2)
(3) In this question: option elimination is the key
Solution : For upper loop. by KVL
E1 + E2 = i2x – i1y
⇒ E1 – i2x + E2 + i1y = 0
Final Answer : (1) E1 – i2x + E2 + i1y = 0

9)

Explaining : Given : RAB =

Ask : ,
Concept : This question is based on Key points:

(1) In Parellel: Req =

(2)

Solution : As and R1 + R2 = 15
then R1R2 = 50

Now R2 – R1 = = =5
Hence ⇒ R2 = 10Ω and R1 = 5Ω

Final Answer : (1)

10)

Explaining : Given : Electric circuit with series and parallel combination of resistance.
Ask : Potential Difference across
Concept : This question is based on Key points :

(1) Current Division rule

(2) Series and parallel combination of resistances
(3) Potential difference is same across the element connected in parallel.

Solution : Effective circuit is

ii = (2A) = 1A
V1 = i, (2) = (1) (2) = 2V
Final Answer : (1) V = 2 Volt

11)

Explaining : Given : Electric circuit

Ask : Equivalent resistance between A and B
Concept : This question is based on key point :
Identification of series and Parallel connected resistance.
Solution : Explanation : In this question we have to find the equivalent resistance between
A and B.
Calculation :

Resistors (1) and (2) are in parallel

∴ Re1 = Re2 ..................(1)

Position (5) and (6) are in parallel

∴ Req = R56 = ..................(2)

The circuit is now reduced

Req 2 =
Req =
Conclusion : The correct answer is option (1)

Final Answer : (1)

12) Explaining : Given : Infinite ladder network

Ask : Current Passes through resistance nearest to battery.
Concept : This question is based on key points :
(1) Identification of repeating Pattern
(12 Circuit reduction

(3) Solution of Quadratic Equation

Solution : R = 1 +
2R + R2 = 2 + R + 2R
⇒ R2 – R – 2 = 0

R= = =2

i= = = 3A
V1 = 3 × 1 = 3V
V2 = 6 – 3 = 3V

i2 = = = 1.5A

Final Answer : (2 ) Req = ,

13)

Explaining : Given : Circuit

Ask : Equivalent Capacitance between A and B
Concept : This question is based on Key points :

(1) In series :
(2) In Parallel :
Solution : The given circuit can be redraws as follows
Final Answer : (1) CAB =

14)

Explaining : Given : Capacitive circuit

Ask: Potential at A and B
Concept : This question is based on Key
(1) Q = Ceq V
(2) KVL in Capacitive circuit ΣΔV = 0

(3)

Solution :
Q = Ceq V

Q= = 20 C

VA – 0 = 0 – VB =

and
= 60 V VB = – 40 V
Final Answer : (3) VA = 60V, VB = – 40V

15)

Explaining : Given : Charged isolated capacitor and insertion of dielectric

Ask : Incorrect Option
Concept : This question is based on key points :
(1) Q : Remains same (as capacitor is isolated)
(2) C = KCo
(3) Q =

(4)
Solution : Once the capacitor is charged, its charge will be constant Q = CV
When dielectric slab is inserted

E= ⇒ ENew = Einitial

V = so Vnew = V
Final Answer : (3)

16) Explaining : Vector form of Biot Savart Law.

Concept : This question is based on Biot Savart Law.
Solution :

Final Answer : (4)

17) Explaining : Magnetic field at the axis of a current carrying circular coil.

Concept : This question is based on

Solution :

Final Answer : (1)

18) Explaining : Magnetic field due to combination of wire & arc

Concept : This question is based on .

[where θ in radian]
Solution : → Net magnetic field due to semi infinite wire
→ magnetic field due to semi circular wire

So,
Final Answer : (2)

19) Explaining : The resultant magnetic field due to the two semicircular parts at their
common centre is.

Concept : This question is based on

Solution :
 .
Final Answer : (1)

20) Explaining : The distance at which the magnetic field on axis as compared to the

magnetic field at the center of the coil carrying current I and radius R is , would be

Concept : This question is based on .

Solution :

⇒ x2 + R2 = 4R2 ⇒ x =
Final Answer : (4)

21) Explaining : Magnetic field at the axis at any point due to current carrying circular coil.

Concept : This question is based on .

Solution : Theoretical.
Final Answer : (1)

22) Explaining : Magnetic field due to combination of wires & arc.

Concept : This question is based on use Right Hand Rule.
Solution : Use Right hand rule.
Final Answer : (2)

23) Explaining : Magnetic field due to combination of wire & arc.

Concept : This question is based on

Solution :
Magnetic field due to different parts are
B1 = 0
Final Answer : (3)

24) Explaining : Magnetic field at a distance r from an infinitely long straight conductor
carrying a steady current varies as

Concept : This question is based on

Solution :
Final Answer : (2)

25) Explaining : Magnetic field line due to a moving charge.

Concept : This question is based on field lines.
Solution : Theoretical.
Final Answer : (1)

26) Explaining : Magnetic field due to a solid cylinder.

Concept : This question is based on

Solution :

Final Answer : (1)

27) Explaining : Magnetic field due to semicircular coils.

Concept : This question is based on

Solution : In the following figure, magnetic fields at O due to sections 1, 2, 3 and 4 are
considered as and respectively.

As
So
Final Answer : (1)

28)

Explaining : Magnetic field due to concentric coplanner coils at their common centre.

Concept : This question is based on .

Solution :
Magnetic field at centre is zero
B1 = B2

Given that
(I1 = I2, N1 : N2 = n1 : n2, R1 : R2 = 2 : 1)

⇒ ( I1 = I2) ⇒
Final Answer : (1)

29)

Explaining : Find MF due to finite current carrying wire.

Concept : This question is based on

Solution :

Final Answer : (2)

30) Explaining : The ratio of current due to concentric circular coils.

Concept : This question is based on .

Solution : and
We know that

Final Answer : (1)

31) Explaining : Magnetic field due to symmetric loop.

Concept : This question is based on Bat centre = 0
Solution : By theory
Final Answer : (2)

32) Explaining : Magnetic field between the midpoint of two current carrying parallel wires.

Concept : This question is based on .

Solution : At P

Since
So, Bnet = 0

Final Answer : (3)

33) Explaining : M.F. due to symmetric loop.

Concept : This question is based on Bnet = 0
Solution : For symmetrical current distribution, magnetic field at the centre due to all current
carrying conductors will get cancelled
Hence Bnet = 0
Final Answer : (3)

34) Explaining : A circular coil of radius R carries an electric current. The magnetic field due
the coil at a point on the axis of the coil located at a distance r from the center of the coil, such
that r >> R varies as

Concept : This question is based on .

Solution :
Final Answer : (4)

35) Explaining : M.F. due to combination of wires.

Concept : This question is based on .

Solution : Direction of magnetic field are equal and opposite to each other having same
magnitude. Therefore, net magnetic field is zero.
Final Answer : (4)

36) Explaining : Magnetic field due to moving charge.

Concept : This question is based on

Solution : ⇒ ⇒B= Tesla

Final Answer : (3)

37) Explaining : Magnetic field due to circular arc.

Concept : This question is based on .

Solution :
Final Answer : (4)

38) Explaining : MF due to combination of wire and coil.

Concept : This question is based on

Solution : B at the centre of a coil carrying a current, i is

B due to wire

Given i = 8A; r = 10 × 10–2 m,

Magnetic field at centre C,

 .
Final Answer : (2)

39) Explaining : MF due to combination of wire.

Concept : This question is based on .

Solution : At P :

Final Answer : (3)

40) Explaining : MF due to combination of loop & wire.

Concept : This question is based on

Solution :
Final Answer : (1)

41) Explaining : MF due to combination of arc & wire.

Concept : This question is based on .

Solution : ⇒

⇒ ⇒
Final Answer : (3)

42) Explaining : Find MF due to given configuration.

Concept : This question is based on
Solution :

Due to AC,
Due to AD, A is at the end of the wire, therefore at O,

∴ Total induction =
Final Answer : (3)

43) Explaining : MF due to combination of wire.

Concept : This question is based on .

Solution :
Magnetic field at P due to PQ & PR is zero.
∴ Magnetic field at P due to QR

α = 37° & β = 53°

where,

∴ k = 7.
Final Answer : (3)
 44) Explaining : MF due to combination of wire.

Concept : This question is based on

Solution : Magnetic field due to wire (1) & (2) is zero, because point P lies on the axis of these
wires

Magnetic field due to wire (3)

B3 = =
Magnetic field due to wire (4)

B4 =
Net magnetic field due to wire (3) and (4) :-

Bp = B3 + B4 =
Final Answer : (1)

45) Explaining : M.F due to moving charge.

Concept : This question is based on .

Solution : Theoritical.
Final Answer : (2)

CHEMISTRY

46)

Explaining : Correct statement given about the solution showing +ve deviation.
Concept : This question is based on non ideal solution.
Solution : Solution showing +ve deviation con forms minimum boiling azeotropic mixture.
Final Answer : (2)

47) Explaining : The molecular mass of a polymer solution can be determined by which
physical property
Concept : This question is based on colligative property
Solution : Osmatic pressure is best method to determine the molecular mass of polymer
solution
Final Answer : (4)

48)

Explaining : Net movement of solvent in case of osmosis

Concept : This question is based on osmosis
Solution : Net movement of solvent molecule from high up to low up
Final Answer : (1)

49)

Explaining : Vapour pressure dependance on vant Hoff factor.

Concept : This question is based on abnormal C.P.

Solution :
'i' is minimum for glucose so it has highest up
Final Answer : (1)

50)

Explaining : If α is the degree of dissociation of Na2SO4. The van't Hoff factor used for
calculation of molecular mass
Concept : This question is based on abnormal CP
Solution : i = 1 + (n – 1) α
n = 3 for Na2SO4
i = 1 + (3 – 1) α
i = 1 + 2α
Final Answer : (1)

51) Explaining : Freezing point Kf and Kb of water are given then boiling point of solution at 1
atm pressure is
Concept : This question is based on colligative property

Solution :

Tb = 100°C + 0.104 = 100.104

Final Answer : (2)

52)

Explaining : Osmotic pressure of decimolar solution of urea at 27°C are asked

Concept : This question is based on osmotic pressure
Solution : π = CRT

= × 0.0831 × 300
= 2.49 bar
Final Answer : (1)

53)

Explaining : We need to determine charge required for reduction process

Concept : This question is based on reduction Al2O3 → 2Al + O2

Solution : For each mole of Al2O3, 2 moles of Al are produces. To reduce aluminium from Al+3
to Al it require 3 moles e¯ per mole of aluminium.
Thus for 2 mole of Al we require - 6 moles of e¯.
Since 1 Faraday correspond to 1 mole of electron the charge require for 1 mole of
Al2O3 is 6 Faraday
Then for 0.1 mole = 0.10 × 6 = 0.6 Faraday
Final Answer : (3)

54)

Explaining : The limiting molar conductance of Ca+2 and Cl– are given. Then the limiting
molar conductance of CaCl2 is asked
Concept : This question is based on calculation of limiting molar conductance
Solution : CaCl2 → Ca+2 + 2Cl–
= 119 + 2 × 76.3
= 271.6 s cm2 mol–1
Final Answer : (2)

55)

Explaining : Specific conductance of 0.1 M Nitric acid is given and molar conductance of
solution is asked.
Concept : This question is based on the relation between Λm and κ

Solution : Λm =
Λm = 630 ohm–1 cm2 mol–1
Final Answer : (1)

56)

Explaining : Expression of thermodynamic efficiency of cell

Concept : This question is based on thermodynamic efficiency

Solution :
Final Answer : (3)

57)

Explaining : Expression and value of and ΔG° at equilibrium

Concept : This question is based on keq and ΔG°
Solution : Ecell =
At equilibrium
=0
ΔG = 0
Final Answer : (2)

58)

Explaining : Relation of conductivity and molar conductivity with concentration of electrolyte

Concept : This question is based on dilution effect on electrolytic conductance
Solution : Conductivity decreases with dilution
Final Answer : (3)

59)

Explaining : Assertion states that Conductivity always decrease with decrease in

concentration of weak and strong electrolyte.
Reason state that on dilution number of ions per unit volume linearly decreases in
both weak and strong electrolytes.
Concept : This question is based on factors affecting electrolyte conductance
Solution : Conductivity decreases with decrease in concentration
On dilution number of ion in 1 ml solution decreases.
Final Answer : (3)

60)

Explaining : Conductivity and concentration relation based question

Concept : This question is based on factors affecting electrolyte conductance
Solution : As we know that conductivity decreases on dilution with decrease in concentration
Final Answer : (1)

61) Explaining : Common features among the species CN–, CO and NO+
Concept : This question is based on CN–, CO and NO+ have bond order three and diamagnetic.
Solution : CN–, CO and NO+ have bond order three and diamagnetic.
Final Answer : (1)

62) Explaining : Longest bond length is

Concept : This question is based on bond order

Solution :

Final Answer : (4)

63) Explaining : π-acid ligand is

Concept : This question is based on ligand.
Solution : Due to absence of vacant atomic orbital as well as π* molecular orbital, does
not act as π acid ligand.
Final Answer : (4)

64) Explaining : Oxidation number of Fe

Concept : This question is based on oxidation number.
Solution : [Fe+1(H2O)5(NO)2+] ion
Final Answer : (3)

65) Explaining : Metal carbonyls

Concept : This question is based on synergic bonding.
Solution : CO is called p-acid ligand. In metal carbonyl complexes, there is donation of an
electron pair from carbon to the empty orbital of metal and then simultaneously a back p-
bonding is formed by sideways overlap of a filled orbital on the metal with empty antibonding
orbital of CO.
Final Answer : (4)

66) Explaining : IUPAC name of hydrate isomer of CrCl3.6H2O

Concept : This question is based on IUPAC nomenclature
Solution : [Cr(H2O)4Cl2]Cl.2H2O ; hydrate isomer of complex [Cr(H2O)6]Cl3 has lowest
electrical conductivity, because it gives only two ions on ionisation in aqueous medium.
Final Answer : (2)

67) Explaining : IUPAC nomenclature

Concept : This question is based on IUPAC nomenclature
Solution : The IUPAC name of Xe[PtF6] is Xenon hexafluoroplatinate (V).
Final Answer : (2)

68) Explaining : Secondary and primary valency

Concept : This question is based on Werner's theory
Solution : Conductivity ∝ number of ions in aqueous medium.
[Cr(NH3)6]Cl3 → [Cr(NH3)6]3+ + 3Cl–
[Cr(NH3)5Cl]Cl2 → [Cr(NH3)5Cl]2+ + 2Cl–
[Cr(NH3)4Cl2]Cl → [Cr(NH3)4Cl2] + + Cl–
[Cr(NH3)3Cl3] → Does not ionize
Final Answer : (1)

69) Explaining : Valence bond theory

Concept : This question is based on valence bond theory
Solution : Cr.6NH3.Cl3 → [Cr(NH3)6]Cl3
Secondary valency ‘6’ satisfied by only NH3, all Cl– satisfy only primary valency.
Cr.5NH3.Cl3 →[Cr(NH3)5Cl]Cl2
Secondary valency ‘6’ satisfied by five NH3 and one Cl– while all three Cl– satisfy
primary valency.
Final Answer : (3)

70) Explaining : IUPAC name

Concept : This question is based on IUPAC nomenclature
Solution : Prussian Blue : Fe4[Fe(CN)6]3 or Fe3+[Fe2+(CN)6]4–
Iron(III) hexacyanoferrate(II)
Final Answer : (3)

71) Explaining : Magnetic properties

Concept : This question is based on Crystal Field Theory
Solution : [Ni(NH3)6]2+ → sp3d2 → outer orbital complex.
[Mn(CN)6]4– → d2sp3 → inner orbital complex.
[Co(NH3)6]3+ → d2sp3 → inner orbital complex.
[Fe(CN)6]4– → d2sp3 → inner orbital complex.
Final Answer : (1)

72) Explaining : Magnetic properties

Concept : This question is based on VBT
Solution : Octahedral complexes of Ni(II) are always outer orbital and paramagnetic.
Final Answer : (2)

73) Explaining : Magnetic moment is

Concept : This question is based on CFT
Solution : [MnX4]2–, MM→5.9 BM i.e., 5 unpaired electrons, geometry is tetrahedral.
Final Answer : (1)

74) Explaining : Geometry of complex

Concept : This question is based on CFT
Solution : The geometry of [Ni(CO)4] and [NiCl2(PPh3)2] is tetrahedral.
Final Answer : (3)

75) Explaining : Ligands field strength (spectrochemical series)

Concept : This question is based on CFT
Solution : Correct order in spectrochemical series of ligands is :

As CN– ligand is involved in synergic bonding so CN– is a strong field ligand.

Final Answer : (1)

76) Explaining : Number of unpaired electrons (Magnetic moment)

Concept : This question is based on magnetic moment
Solution : µ = B.M.
Ni(CO)4 ; µeff = 0
[Co(NH3)4(NO2)2]+ ; µeff = 0
[Ag(CN)2] ; µeff = 0
[CuBr4]2–; µeff = 1.732 BM
Final Answer : (4)

77) Explaining : Geometry and hybridisation

Concept : This question is based on VBT
Solution : K[PtCl3(C2H4)] → square planar geometry, dsp2 hybridisation.
Final Answer : (2)

78) Explaining : Crystal field stabilisation energy

Concept : This question is based on CFT
Solution : CFSE for octahedral complex in case of strong field ligand is :
CFSE for [Co(NH3)6]Cl3 is
CFSE =
= –0.4Δ0(6) + 0 + 2P
CFSE = –2.4Δ0 + 2P
Final Answer : (3)

79) Explaining : Colour of complexes

Concept : This question is based on CFT
Solution : Both [Ti(H2O)6]4+ and [Sc(H2O)6]3+ are colourless, due to absence of free electrons in
3d subshell.
Final Answer : (2)

80) Explaining : Hybridisation

Concept : This question is based on VBT
Solution : Transition metal ion forming octahedral complexes undergo sp3d2 or d2sp3
hybridisation. Statement-2 is the correct explanation of statement-1.
Final Answer : (3)

81) Explaining : π acid ligands

Concept : This question is based on synergic bonding
Solution : The ligands having vacant π-type orbitals have a tendency to receive back donated
π-electrons, thus these are called π-acid ligands or π-acceptor ligands, e.g., CO and CN have
lone pair as well as π or π*-orbitals which take part in the formation of π-bond with central
metal atom as observed in case of carbonyls.
Final Answer : (4)

82)

Explaining : Magnetic Properties

Concept : This question is based on CFT
Solution : Statement-l: Complex [Fe(H₂O)5(NO)]SO4 is paramagnetic.
In Fe2+, the d6 configuration with weak field ligands like H2O and NO would result in unpaired
electrons. Since there are unpaired electrons, the complex is paramagnetic. This statement is
correct.
Statement-II: The Fe in Complex [Fe(H₂O)(NO)]SO₄ has three unpaired electrons.
With weak field ligands, the d6 configuration of Fe2+ would typically result in 3 unpaired
 electrons because of the insufficient splitting of the d-orbitals, leading to a high-spin
configuration. This statement is correct.
Thus, both statements are correct, and the correct answer is:
Final Answer : (3)

83)

Explaining : Primary and Secondary valency

Concept : This question is based on Werner's theory
Solution : Statement-I correctly describes primary valencies as ionizable and typically
satisfied by negative ions. (Correct) Statement-II accurately describes secondary valencies
as non-ionizable and satisfied by neutral molecules or negative ions directly coordinated to the
central metal atom. (Correct).
Therefore, both statements are correct according to Werner's theory.
Final Answer : (1)

84) Explaining : Metal Carbonyl

Concept : This question is based on organometallic.
Solution :

A. Statement I : This statement is correct. Decacarbonyl dimanganese (O) indeed has a

structure consisting of two square pyramidal [Mn(CO)5] units linked by an Mn-Mn bond.
B. Statement II : This statement is incorrect. CO ligands are well-known for their ability to
stabilize low oxidation states of metals through л-backbonding. In fact, many metal carbonyls,
including Decacarbonyl dimanganese (O), feature metals in zero or low oxidation states.

Final Answer : (2)

85)

Explaining : CN– is an ambidentate ligand

Concept : This question is based on classification of ligands
Solution : It has two donor site but at a time only one site can donate their electron pair.
Final Answer : (4)

86)

Explaining : Classify the ligands

Concept : This question is based on Classification of ligands
Solution :
(A) Ethylenediamine with two donor site (bidentate)

(B) Hydrazine with only one donor site or site (Monodentate ligand)
(C) Dimethylgyloximato (dmg–) can form hydrogen bond

(D) Dien with three donor site (Tridentate ligand)

Final Answer : (3)

87) Explanation: In this question we have to find out CN and oxidation number of central
metal ion of given compound.
Concept : This question is based on oxidation number.
Solution :
In [Fe(EDTA)]– , the denticity of EDTA = 6
So the co-ordination number = 6
The change of EDTA = – 4
Let oxidation state of Fe is n
⇒ n – 4 = –1
n=+3
CN = 6, oxidation number = +3
Final Answer : (1)

88) Explaining : One is a heteroleptic complex

Concept : This question is based on types of complexes
Solution : Zeise’s salt K[Pt(C2H4)Cl3] has more than one type of ligands, hence it’s hetroleptic.
(1) (C5H5)Fe
(2) (C5H5)2Cr
(3) Fe4[Fe(CN)6]3
 Final Answer : (4)

89) Explaining : The following species is not expected to be ligand

Concept : This question is based on types of ligands
Solution : Ligand donate lone pair to CMA in NH4+ species does not have lone pair to donate.
Final Answer : (2)

90)

Explaining : To identify the false statement among the provided options concerning the
characteristics of coordination compounds.
Concept : This question is based on understanding key definitions and properties of
coordination compounds, including the roles of transition metals and ligands.
Solution : Coordination compounds are primarily formed by transition metals.Ligands donate
electron pairs to the central metal atom.
A heteroleptic complex contains different types of ligands.
However, the coordination number (number of bonds to the metal) and the oxidation state
(formal charge on the metal) are generally not the same.
Final Answer : (2)

BIOLOGY

91)

Explaining :
Concept :
Solution : Ringworm is caused by fungi, not bacteria.
Final Answer : (3)

92)

Explaining :
Concept :
Solution : Malaria is caused by Plasmodium, a protozoan parasite transmitted by the female
Anopheles mosquito.
Final Answer : (1)

93)

Explaining :
Concept :
Solution : Antihistamines counteract histamines released during allergic reactions.
Final Answer : (2)

94) Explaining :
Concept :
Solution : Only female Anopheles transmits the Plasmodium parasite.
Final Answer : (2)

95) Explaining :
Concept :
Solution : HIV (a retrovirus) causes AIDS.
Final Answer : (2)

96) Explaining :
Concept :
Solution : Early symptoms are mild and flu-like.
Final Answer : (3)

97) Explaining :
Concept :
Solution : Heroin (diacetylmorphine) is synthesized from morphine extracted from the opium
poppy.
Final Answer : (3)

98) Explaining :
Concept :
Solution : HIV infects T-helper cells, which leads to weakening the immune system.
Final Answer : (3)

99) Explaining :
Concept :
Solution : Passive immunity involves direct transfer of ready-made antibodies.
Final Answer : (3)

100) Explaining :
Concept :
Solution : BCG (Bacillus Calmette–Guérin) protects against TB.
Final Answer : (3)

101) Explaining :
Concept :
Solution : HIV is not transmitted by insects.
Final Answer : (3)

102) Explaining :
Concept :
Solution : Histamine causes multiple allergic responses.
Final Answer : (4)
103) Explaining :
Concept :
Solution : Cancer cells proliferate without regulation.
Final Answer : (1)

104) Explaining :
Concept :
Solution : Abnormal increase in WBCs.
Final Answer : (3)

105) Explaining :
Concept :
Solution : Characteristic of malignant tumors.
Final Answer : (3)

106) Explaining :
Concept :
Solution : Interferons are antiviral proteins.
Final Answer : (3)

107) Explaining :
Concept :
Solution : It stimulates the central nervous system.
Final Answer : (2)

108) Explaining :
Concept :
Solution : T cells destroy infected host cells directly.
Final Answer : (3)

109) Explaining :
Concept :
Solution : ELISA detects HIV antibodies or antigens.
Final Answer : (3)

110) Explaining :
Concept :
Solution : Secondary lymphoid organs initiate immune responses.
Final Answer : (3)

111) Explaining :
Concept :
Solution : These stimulate the immune system without causing disease.
Final Answer : (3)
 112) Explaining :
Concept :
Solution : National AIDS Control Organization.
Final Answer : (3)

113) Explaining :
Concept :
Solution :

A. Malaria → Plasmodium falciparum

B. Typhoid → Salmonella typhi
C. Filariasis → Wuchereria bancrofti
D. Pneumonia → Streptococcus pneumoniae
E. Final Answer : (1)

114) Explaining :
Concept :
Solution :

A. B-lymphocytes → produce antibodies

B. T-lymphocytes → perform cell-mediated immunity
C. Macrophages → engulf pathogens
D. Mast cells → release histamine (allergic responses)

Final Answer : (1)

115) Explaining :
Concept :
Solution : AIDS → through blood/sexual contact
Malaria → mosquito bite
Common cold → droplet infection (airborne)
Ascariasis → fecal-oral route (contaminated food/water)
Final Answer : (2)

116) Explaining :
Concept :
Solution :

A. Active natural → infection and self-recovery

B. Active artificial → vaccine induces immunity
C. Passive natural → maternal antibodies (milk, placenta)
D. Passive artificial → injection of ready-made antibodies

Final Answer : (3)

117) Explaining :
Concept :
Solution : HIV (Human Immunodeficiency Virus) is a retrovirus with RNA as genetic material.
It uses the reverse transcriptase enzyme to convert RNA into DNA, which integrates into the
 host genome. Thus, both assertion and reason are true, and reason explains the assertion
correctly.
Final Answer : (1)

118) Explaining :
Concept :
Solution : Vaccination induces active immunity, not passive. In active immunity, the host’s
immune system is stimulated to produce antibodies. Passive immunity involves direct transfer
of pre-formed antibodies (e.g., anti-venom).
Final Answer : (4)

119) Explaining :
Concept :
Solution :

A. Statement 1 is incorrect: B-cells are involved in humoral immunity, not cell-mediated.

B. Statement 2 is correct: T-cells attack infected cells directly.
C. Statement 3 is correct: Plasma cells (from B-cells) secrete antibodies.
D. Statement 4 is correct: Vaccines stimulate the immune system → active immunity.

Final Answer : (2)

120) Explaining :
Concept :
Solution :

A. Statement 1: Correct → HIV targets CD4+ T-helper cells.

B. Statement 2: Correct → ELISA is used to detect HIV antibodies.
C. Statement 3: Incorrect → HIV has RNA, not DNA.
D. Statement 4: Correct → leads to Acquired ImmunoDeficiency Syndrome (AIDS).

Final Answer : (2)

121) Explaining :
Concept :
Solution : Lactobacillus bacteria ferment lactose in milk to produce lactic acid, which lowers
the pH and causes milk proteins to coagulate, forming curd.
Final Answer : (1)

122) Explaining :
Concept :
Solution :

A. Saccharomyces cerevisiae is used in the production of ethanol.

B. Trichoderma polysporum produces cyclosporin A, an immunosuppressive agent.
C. Monascus purpureus is used to produce statins, which lower cholesterol levels.
D. Aspergillus niger is used to produce citric acid, a food preservative.

Final Answer : (1)

123)
 Explaining :
Concept :
Solution : Azolla is a water fern that forms a symbiotic relationship with nitrogen-fixing
cyanobacteria, providing a natural source of nitrogen to paddy fields.
Final Answer : (3)

124) Explaining :
Concept :
Solution : Acetobacter aceti is a bacterium that oxidizes ethanol to acetic acid, the main
component of vinegar.
Final Answer : (2)

125) Explaining :
Concept :
Solution :

A. Rhizobium forms symbiotic relationships with legumes for nitrogen fixation.

B. Azotobacter fixes nitrogen in free-living conditions.
C. Azospirillum associates with cereal crops for nitrogen fixation.
D. Frankia forms symbiotic relationships with actinorhizal plants for nitrogen fixation.

Final Answer : (1)

126) Explaining :
Concept :
Solution :
Final Answer : (2)

127) Explaining :
Concept :
Solution : Propionibacterium shermanii produces carbon dioxide during fermentation,
creating holes in Swiss cheese.
Final Answer : (2)

128) Explaining :
Concept :
Solution : Antibiotics are effective only against bacteria, not viruses. So, A is true, R is
false.
Final Answer : (3)

129) Explaining :
Concept :
Solution : Monascus purpureus produces statins which inhibit cholesterol synthesis enzymes.
Final Answer : (2)

130) Explaining :
Concept :
 Solution :
Final Answer : (1)

131) Explaining :
Concept :
Solution : Saccharomyces cerevisiae (yeast) is used to ferment sugars to alcohol.
Final Answer : (2)

132) Explaining :
Concept :
Solution :

A. Biogas contains mostly methane.

B. It is produced by anaerobic (not aerobic) bacteria.
C. Cow dung is used.
D. Methanogens (like Methanobacterium) are key bacteria.

Final Answer : (2)

A.

133) Explaining :
Concept :
Solution : Floc-forming bacteria like Pseudomonas and Bacillus form aggregates in aeration
tanks during secondary sewage treatment.
Final Answer : (3)

134) Explaining :
Concept :
Solution : Cyclosporin A, produced by Trichoderma polysporum, is an immunosuppressant
used in organ transplant surgeries.
Final Answer : (2)

135) Explaining :
Concept :
Solution : Azotobacter is a free-living nitrogen-fixing bacterium, unlike Rhizobium (symbiotic
in legumes).
Final Answer : (3)

136) Explaining :
Concept :
Solution : Wilhelm Johanssen coined "gene" in 1909 for hereditary units.
Final Answer : (4)

137) Explaining :
Concept :
Solution : A monohybrid cross between heterozygotes yields a 3 : 1 dominant to recessive
phenotype ratio.
Final Answer : (2)

138) Explaining :
Concept :
Solution : Alleles are different versions of a gene on the same locus.
Final Answer : (3)

139) Explaining :
Concept :
Solution : This law applies only when genes segregate independently.
Final Answer : (3)

140) Explaining :
Concept :
Solution : X-linked recessive disorder, mostly affects males.
Final Answer : (3)

141) Explaining :
Concept :
Solution : Incomplete dominance results in a blended phenotype.
Final Answer : (3)

142) Explaining :
Concept :
Solution : Codominance shows simultaneous expression (e.g. AB blood group).
Final Answer : (3)

143) Explaining :
Concept :
Solution : Thalassemia results from defective haemoglobin production.
Final Answer : (1)

144) Explaining :
Concept :
Solution : Turner’s syndrome = 45 chromosomes, missing one X.
Final Answer : (3)

145) Explaining :
Concept :
Solution : If both are heterozygous (IAi × IBi), all four blood types are possible.
Final Answer : (3)

146) Explaining :
Concept :
Solution : Polygenic = multiple genes control a trait; skin colour involves 3–4 genes.
Final Answer : (3)

147) Explaining :
Concept :
Solution : XXY condition (male with one extra X) = 47 chromosomes.
Final Answer : (3)

148) Explaining :
Concept :
Solution : The SRY gene on Y chromosome triggers male development.
Final Answer : (3)

149) Explaining :
Concept :
Solution : Caused by trisomy of chromosome 21.
Final Answer : (2)

150) Explaining :
Concept :
Solution : Autosomal recessive disorder — 1/4 chance if both are heterozygous.
Final Answer : (2)

151) Explaining :
Concept :
Solution : In incomplete dominance, neither allele is completely dominant.
Final Answer : (1)

152) Explaining :
Concept :
Solution : It is autosomal recessive, and carriers are usually asymptomatic.
Final Answer : (1)

153) Explaining :
Concept :
Solution : Turner = 45 chromosomes (XO); Klinefelter = XXY, male.
Final Answer : (1)

154) Explaining :
Concept :
Solution : Male gamete (sperm) determines sex — not ovum.
Final Answer : (1)
155) Explaining :
Concept :
Solution : Pedigree is useful in tracing sex-linked traits too.
Final Answer : (1)

156) Explaining :
Concept :
Solution : Codominance = AB, Incomplete = snapdragon, Polygenic = skin, X-linked =
haemophilia.
Final Answer : (1)

157) Explaining :
Concept :
Solution : Down Syndrome - Trisomy 21st chromosome.
Turner Syndrome - AA + XO.
Klinefelter Syndrome - AA + XXY.
Final Answer : (1)

158) Explaining :
Concept :
Solution : Gene - Segment of DNA coding for protein
Allele - Alternate forms of gene
Phenotype - Physical expression
Genotype - Genetic constitution
Final Answer : (1)

159) Explaining :
Concept :
Solution : Mendel - Father of genetics
Morgan - Chromosomal theory
Hugo de Vries - Rediscovered Mendel’s laws
Sutton - Term – Linkage
Final Answer : (3)

160) Explaining :
Concept :
Solution : Blood group A - IAIA or IAi
Blood group B - IBIB or IBi
Blood group AB - IAIB
Blood group O - ii
Final Answer : (1)

161) Explaining :
Concept :
Solution : NCERT page…64
Final Answer : (1)

162) Explaining :
Concept :
Solution : NCERT page…66
Wrong statements after correction.
A. Independent pairs segregate independently of each other
C. Both Chromosome and genes are unpaired in haploid organisms.
E. Both Chromosome and genes are paired in diploid organisms.
Final Answer : (4)

163) Explaining :
Concept :
Solution : (A) Flower position- Axial/terminal.
(B) Seed shape- Round/wrinkled.
(C) Flower colour- Violet/white.
(D) Seed colour- Yellow/green.
Final Answer : (1)

164) Explaining :
Concept :
Solution : Haemophilia is X-linked recessive, not autosomal.
Final Answer : (2)

165) Explaining :
Concept :
Solution : In males, one recessive X-linked allele causes disease.
Final Answer : (1)

166) Explaining :
Concept :
Solution : Down syndrome = trisomy 21, Turner = 45 chromosomes (XO).
Final Answer : (1)

167) Explaining :
Concept :
Solution : Colour blindness - X-linked recessive
Thalassemia - Autosomal recessive
Down syndrome - Trisomy 21
Final Answer : (1)
168) Explaining :
Concept :
Solution : Mendel's work on Pisum sativum laid the foundation of modern genetics.
Final Answer : (3)

169) Explaining :
Concept :
Solution : Test cross = crossing dominant phenotype with homozygous recessive.
Final Answer : (2)

170) Explaining :
Concept :
Solution : Males have only one X → recessive allele expresses directly.
Final Answer : (1)

171) Explaining :
Concept :
Solution : Klinefelter = Male with 47 chromosomes → extra X.
Final Answer : (3)

172) Explaining :
Concept :
Solution : X-linked recessive disorder; mother must be carrier (XcX).
Final Answer : (2)

173) Explaining :
Concept :
Solution : Female missing one X chromosome.
Final Answer : (2)

174) Explaining :
Concept :
Solution : Linked Mendelian inheritance to chromosomes.
Final Answer : (2)

175) Explaining :
Concept :
Solution : Skipping generations is typical of recessive traits.
Final Answer : (2)

176) Explaining :
Concept :
Solution : Fact based question.
Final Answer : (3)
177) Explaining :
Concept :
Solution : It is an autosomal recessive trait then can be transmitted from parents to the
offspring when both the partners are carrier for the gene.
Final Answer : (2)

178) Explaining :
Concept :
Solution : fact based question
Final Answer : (1)

179) Explaining :
Concept :
Solution : With increasing age , frequency of crossing over between genes may decrease,
leading to increase in strength of linkage.
Final Answer : (1)

180) Explaining :
Concept :
Solution : Fact based question.
Final Answer : (2)