08-06-2025

1950CMD303021250046 MD

PHYSICS

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

(1) 1µF
(2) 10 µF
(3) 50 µF
(4) 1.5

2) Three capacitors are connected as shown in fig. Then the charge on capacitor C1 is :-

(1) 6μC
(2) 12μC
(3) 18μC
(4) 24μC

3) Three dielectric are filled in a parallal plate capacitor as shown. What should be the net

capacitance –

(1)

(2)

(3)

(4)
4) A parallel plate capacitor has two layers of dielectric as shown in figure. This capacitor is
connected across a battery. The graph which shows the variation of electric field (E) & distance (x)
from left plate.

(1)

(2)

(3)

(4)

5) There are 50 turns of a wire in every cm length of a long solenoid. If 4 ampere current is flowing
in the solenoid, the approximate value of magnetic field along its axis at an internal point and at one
end will be respectively

(1) 12.6 × 10–3 Weber/m2, 6.3 × 10–3 Weber/m2

(2) 12.6 × 10–3 Weber/m2, 25.1 × 10–3 Weber/m2
(3) 25.1 × 10–3 Weber/m2, 12.6 × 10–3 Weber/m2
(4) 25.1 × 10–5 Weber/m2, 12.6 × 10–5 Weber/m2

6) Find the additional charge that flows from battery if the space between the plate is filled with
dielectric (see figure) :-

(1) 150 μC
(2) 300 μC
(3) 250 μC
(4) 100 μC

7) Find out equivalent capacitance between A and B if each capacitor has capacitance ‘C’.

(1)

(2)

(3)

(4) None

8) Two uncharged capacitors are charged with a battery of E volt. by shifting key from 1 to 2. The

ratio of charges produced on these capacitors will be :–

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

9) Three identical capacitors are given a charge Q each and they are then allowed to discharge
through resistance R1, R2 and R3. Their charges as a function of time shown in the graph below. The
smallest of the three resistance is :-

(1) R3
(2) R2
(3) R1
(4) Cannot be predicted

10) The figure gives the electric potential V as a function of distance through five regions on x-axis.
Which of the following is true for the electric field E in these regions :-

(1) E1 > E2 > E3 > E4

(2) E1 = E3 and E2 < E4
(3) E2 = E4 and E1 < E3
(4) E1 < E2 < E3 < E4

11)

In the above figure magnetic field at point C will be:-

(1)

(2)
(3)

(4)

12) Two coaxial solenoids 1 and 2 of the same length are set so that one is inside the other. The
number of turns per unit length are n1 and n2. The current i1 and i2 are flowing in opposite directions.
The magnetic field inside the inner coil is zero. This is possible when :-

(1) i1 ≠ i2 and n1 = n2
(2) i1 = i2 and n1 ≠ n2
(3) i1 = i2 and n1 = n2
(4) None of these

13) Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out
of the plane of the paper as shown. The variation of magnetic field B along the line XX' is given by :

(1)

(2)

(3)

(4)

14) Current I is flowing in 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)
15) A long thin hollow metallic cylinder of radius 'R' has a current i ampere. The magnetic induction
'B'-away from the axis at a distance r from the axis varies as shown in

(1)

(2)

(3)

(4)

16) Magnetic field at point 'P' due to following current distribution is :-

(1)

(2)

(3)

(4)

17) The dimensions of the magnetic field B is

(1) [MLT–2A–1]
(2) [MT–2A–1]
(3) [ML2TA–2]
(4) [M2T–2A–1]

18) A uniform wire is bent in the form of a circle of radius R. A current I enters at A and leaves at C
as shown in the figure : If the length ABC is half of the length ADC, the magnetic field at the centre

O will be :-

(1) Zero

(2)

(3)

(4)

19) Magnetic field at point O due to straight conductor PQ is.

(1)

(2)

(3)

(4)

20) In the toroid, the number of tums per metre length is 500 and current through it is .
Mangetic field product inside (in weber/ } will be

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

21) A charged particle enters a magnetic field at right angles to the magnetic field. The field exists
for a lenght equal to 1.5 times the radius of the circular path of the particle. The particle will be

deviated from its path by

(1) 90°

(2)

(3) 30°
(4) 180°

22) Find the position of point from wire 'B' where net magnetic field is zero due to following current
distribution

(1)

(2)

(3)

(4)

23) A and B are two concentric circular conductors of center O and carrying currents i1 and i2 as
shown. If ratio of r1 : r2 is 1: 2 and ratio of flux densities at O due to A and B is 1 : 3, then the ratio of

i1 : i2 is

(1)
(2)

(3)

(4)

24) In the given figure magnitude of net magnetic field at O will be :-

(1)

(2)

(3)

(4)

25) Magnetic field at point O will be :-

(1)

(2)

(3)

(4)

26) A point charge q is place at rest near current carring wire then -

(1) Magnetic force acting on wire is zero

(2) Electric force acting on wire is zero
(3) Both (1) & (2)
(4) None
27) A current carrying wire AB of the length 2πR is turned along a circle, as shown in figure. The

magnetic field at the centre O.

(1)

(2)

(3)

(4)

28) Two wires with currents 2 A & 1 A are enclosed in circular loop. Another wire with current 3 A is

situated outside the loop as shown. The around the loop is :-

(1) μ0
(2) 3μ0
(3) 6μ0
(4) 2μ0

29) Rank the value of for the closed paths shown in figure from the smallest to largest.

(1) a, b, c, d
(2) a, c, d, b
(3) a, d, c, b
(4) a, c, b, d

30) Two identical long conducting wires AOB and COD are placed at right angle to each other, with
one above other such that ‘O’ is their common point for the two. The wires carry I1 and I2
currents respectively. Point ‘P’ is lying at distance ‘d’ from ‘O’ along a direction perpendicular to the
plane containing the wires. The magnetic field at the
point ‘P’ will be :-

(1)

(2)

(3)

(4)

31) Match the following

Column-I Column-II
Magnetic field inside a
(A) (p) Not constant
long straight solenoid is
Magnetic field inside a
(B) (q) ZERO
toroidal solenoid is
Magnetic field inside a
conducting hollow pipe
(C) (r) Constant
having current parallel to
its axis
Magnetic field due to
(D) current carrying wire on (s) Maximum
its surface is
(1) A→p, B→s, C→q, D→r
(2) A→s, B→r, C→q,s, D→p
(3) A→r, B→p, C→s, D→(q,r)
(4) A→r, B→p, C→(q,r), D→s

32) A toroid with mean radius r0, diameter 2a has N turns carrying current I. What is the magnetic
field B outside the toroid :-

(1)

(2)

(3)

(4) Zero

33) A charge having q/m equal to 108 C/kg and with velocity 3 × 105 m/s enters into a uniform
magnetic field B = 0.3 tesla at an angle 30° with direction of field. Then radius of curvature will be:-
(1) 0.01 cm
(2) 0.5 cm
(3) 1 cm
(4) 2 cm

34) A 2C charge of mass 50g, is moving in a magnetic field with velocity . Find
out pitch of its path :-

(1) π × 10–2 m
(2) 10–2 m
(3) 2π × 10–2 m
(4) π cm

35) A positive change particle of mass m and change q is projected with velocity v as shown. If
radius of change particle is greater than d then find the time spent by the change particle in

magnetic field.

(1)

(2)

(3)

(4)

36) A rod of length L and mass M0 is bent to form a semicircular ring as shown. The M.I. about X'X is

:-

(1)

(2)

(3)

(4)
37) Three-point masses, each m, are placed at the vertices of an equilateral triangle of side 'a'
moment of inertia of the system about the axis COD which passes through the mass at O and lies in

the plane of the triangle and perpendicular to OA is

(1) ma2

(2)

(3)

(4)

38) One quarter section is cut from a uniform circular disc of radius R. This section has a mass M. It
is made to rotate about a line perpendicular to its plane and passing through the centre of the
original disc. It moment of inertia about the axis of rotation is

(1)

(2)

(3)

(4)

39) M.I. of uniform solid sphere about axis passing through center.
(1)

(2)

(3)

(4)

40) Two spheres of same mass and radius are in contact with each other. If the moment of inertia of
a sphere about its diameter is I, then the moment of inertia of both the spheres about the tangent at
their common point would be

(1) 3I
(2) 7I
(3) 4I
(4) 5I

41) The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing

through the centre O and perpendicular to plate is

(1) I1 + I2
(2) I2 + I3
(3) I1 + I3
(4) All of the above

42) Four solid rigid balls each of mass m and radius r are fixed on a rigid-ring of radius 2r and mass
2m. The system is whirled about ‘O’ as shown. The radius of gyration of the system is :-

(1)
r

(2)
(3)

(4)

43) A thin uniform disc has mass 9m and radius R. A hole of radius R/3 is cut from it as shown in the
figure. The moment of inertia of the remaining part about an axis passing through the centre O of

the disc and perpendicular to the plane of the disc is :

(1) 8mR2
(2) 4mR2

(3)

(4)

44) Four similar point masses (each of mass m) are placed on the circumference of a disc of mass M

and radius R. The M.I. of the system about normal axis through the centre O will be -

(1) MR2 + 4mR2

(2)
MR2 + 4mR2

(3)
MR2 + mR2
(4) None of these

45) Four thin rods of same mass M and same length ℓ form a square as shown in figure. Moment of
inertia of this system about an axis through centre O and perpendicular to its plane is :-

(1)
Mℓ2

(2)
(3)

(4)
Mℓ2

CHEMISTRY

1) Given:
The potential for the cell
cr | cr3+ (0.1 M) || Fe2+ (0.01 M) | Fe, is:

(1) 0.339V
(2) –0.339V
(3) –0.26V
(4) 0.26V

(C1) | Co for this cell, ΔG is negative if :

2+ 2+
2) Co | Co (C2) || Co

(1) C2 > C1
(2) C1 > C2
(3) C1 = C2
(4) unpredictable

3) Standard reduction potential of half-cells are given as follows :

Zn2+ + 2e– → Zn, E° = –0.76V
Fe2+ + 2e– → Fe, E° = –0.44V
What is the emf of following reactions ?
Zn(s) + Fe(aq)2+ → Zn(aq)2+ + Fe(s)

(1) –1.20V
(2) +1.20V
(3) +0.32V
(4) –0.32V

4) Standard emf of 0.59 V of galvanic cell in which 3 mole electron taking part in redox reaction. So
for such reaction find out value of equilibrium constant ?

(1) 1025
(2) 1020
(3) 1015
(4) 1030

5) If pH of hydrogen electrode is 10, then what is the potential of it….

(1) 0.59 V
(2) 0.00 V
(3) –0.59 V
(4) –0.059 V

6) Standard electrode potential of three metals x, y and z are –1.2 V, +0.5 V and –3.0 V. The order of
reducing agent of three metals are….

(1) y > z > x

(2) y > x > z
(3) z > x > y
(4) x > y > z

7) The molar conductances of and CH3COONa at infinite dilution are 126.45, 426.16 and
respectively. The molar conductance of CH3COOH at infinite dilution is

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

8) What happens to the voltage in a galvanic cell when the salt bridge is removed?

(1) The voltage drops to zero.

(2) The voltage remains the same.
(3) The voltage gradually increases.
(4) The voltage rapidly increases.

9) Pick out the incorrect statement

(1) Equivalent conductance increases with dilution

(2) Molar conductance increases with dilution
(3) Specific conductance increases with dilution
(4) Specific resistance increases with dilution

10) When NaCl solution is electrolysed using Pt electrodes then pH of solution

(1) Increases
(2) Decreases
(3) Remains same
(4) Firstly increases and then decreases

11) Electrolysis of aq. CuSO4 causes

(1) An increase in pH
(2) A decrease in pH
(3) Either decrease or increase
(4) None

12) The equilibrium constant for a feasible cell reaction

(1) < 1
(2) 0
(3) = 0
(4) > 1

13) The variation of equivalent conductance of strong electrolyte with concentration is correctly
shown in which figure?

(1)

(2)

(3)
(4)

14) A cell, with cell constant 0.4 cm–1 has the resistance of 40 ohm of a 0.01 M solution of an
electrolyte. Then the molar conductivity in ohm–1 cm2 mol–1 will be :

(1) 104
(2) 103
(3) 102
(4) 101

15) Saturated solution of KNO3 is used to make salt bridge because -

+ –
(1) velocity of K is greater than that of NO 3
– +
(2) velocity of NO 3 is greater than that of K
+ –
(3) velocity of both K and NO 3 are nearly the same
(4) KNO3 is highly soluble in water

16) In a galvanic cell, the electrons flow from :-

(1) Anode to cathode through the solution

(2) Cathode to anode through the solution
(3) Anode to cathode through the external circuit
(4) Cathode to anode through the external circuit

17) Select the correct statement if -

, , ,

(1) is the strongest oxidizing agent and Mg is the strongest reducing agent.
4+
(2) Sn + 2I– → Sn2+ + I2 is a spontaneous reaction
(3) Mg2+ + Sn2+ → Mg + Sn4+ is a spontaneous reaction
(4) Here, weakest oxidizing agent is Sn4+ and weakest reducing agent is Mn2+

18) The standard EMF of Daniel cell is 1.10 volt. The maximum electrical work obtained from the
Daniel cell is :

(1) 212.3 kJ
(2) 175.4 kJ
(3) 106.15 kJ
(4) 53.07 kJ

19) Consider following half-cell reactions :

I. A + e– → A– E° = 0.96 V
II. B– + e– → B2– E° = –0.12 V
III. C+ + e– → C E° = +0.18 V
IV. D2+ + 2e– → D E° = –1.12 V
What combination of two half cells would result in a cell with the largest potential?

(1) I and II
(2) I and III
(3) I and IV
(4) II and IV

20) If is x1, is x2 then will be.

(1) 3x2 – 2x1

(2) x2 – x1
(3) x2 + x1
(4) 2x1 + 3x2

21) The emf of a Daniell cell at 298 K is Ei

Zn|ZnSO4| |CuSO4|Cu
(0.01 M) (1.0 M)
when the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, the emf changed to E2.
What is the relationship between E1 and E2?

(1) E1 = E2
(2) E2 = O ≠ E1
(3) E1 > E2
(4) E1 < E2

22) For half reaction

then Nernst equation for ERP is :-

(1)

(2)

(3)
(4)

23) If the standard reduction potential, Eo for four divalent elements, X, Y, Z, W are –1.46V, –0.36 V,
–0.15 V and –1.24 V respectively, then:

(1) X will replace Z+2 from aqueous solution

(2) Y will replace Z+2 from aqueous solution
(3) W will replace Z+2 from aqueous solution
(4) All the above statements are correct

24) Which of the following will increase the voltage of the cell :-
Sn(s) + 2Ag+(aq) Sn2+ (aq) + 2 Ag (s)-

(1) Increase in the concentration of Sn2+ ions

(2) Increase in the concentration of Ag+ ions
(3) Increase in the size of silver rod
(4) None

25) Calculate the emf of the following concentration cell at 25°C :

Ag(s) | AgNO3 (0.01 M) | | AgNO3 (0.05 M) | Ag(s)

(1) –0.414 V
(2) 0.828 V
(3) 0.414 V
(4) 0.0414 V

26)

Which of the following is most basic?

(1)

(2)

(3) CH3–NH–CHO

(4)

27) Which of the following resonating structure is least stable?

(1)
(2)

(3)

(4)

28) Arrange the following compound in the correct order of pKa ?

(1) III > I > IV > II

(2) III > IV > I > II
(3) II > IV > I > III
(4) III > II > IV > I

29) Which of the following does not show hyperconjugation?

(1)

(2)

(3)

(4) CH3–CH=CH–CH3

30) Correct order of acidic strength ?

(1) I > II > III

(2) II > III > I
(3) III > II > I
(4) I > III > II

31) Which hydrogen is most acidic :-

(1) a
(2) b
(3) c
(4) d

32) Least stable Carbanion is :-

(1)

(2)

(3)

(4)

33) Out of the following series, the one containing only electrophiles is :

(1)

(2)

(3)

(4) AlCl3, NH3, H2O

34) (A) and (B) are respectively

(1)
both

(2)
both

(3)
and

(4)
and

35) ;
Y is :-

(1)
(2)

(3)

(4)

36) The major product of following reaction will be

products

(1)

(2)

(3)

(4)

37) Most reactive compound for electrophilic addition reaction :-

(1)

(2)

(3)
(4)

38) product (B) is :-

(1)

(2)

(3)

(4) None

39) P, [Major] P is :-

(1)

(2)

(3) both are correct

(4) None

40) IUPAC name of is :-

(1) 1,6-Dimethylcyclohexene
(2) 1,2-Dimethylcyclohex-2-ene
(3) 1,2-Dimethylcyclohex-2-ene
(4) 2,3-Dimethylcyclohexene

41)

IUPAC name of is :-

(1) Methyl ethanoate

(2) Aceto ethanoate
(3) Ethanoyl ethanoate
(4) Ethanoic anhydride

42) How many carbon atoms are present in principle carbon chain of following compound ?

(1) 4
(2) 6
(3) 5
(4) 7

43) Match the following column on the basis of given data

Column-
Column-II
I

(i) F2 (a) Strongest reducing agent

(ii) F– (b) Weakest oxidising agent

(iii) Li (c) Weakest reducing agent

(iv) Li+ (d) Strongest oxidising agent

(1) (i) - c, (ii) - d, (iii) - b, (iv) - a
(2) (i) - d, (ii) - c, (iii) - b, (iv) - a
(3) (i) - d, (ii) - c, (iii) - a, (iv) - b
(4) (i) - a, (ii) - c, (iii) - b, (iv) - d

44) Match the following columns :

Column-I Column-II

Electrolysis of salts Nature of solution

using inert electrode after electrolysis

(i) Aqueous NaCl (a) Acidic

(ii) Aqueous AgNO3 (b) Neutral

(iii) Aqueous CuCl2 (c) Basic

(1) (i) - a, (ii) - c, (iii) - b
(2) (i) - b, (ii) - a, (iii) - c
(3) (i) - c, (ii) - a, (iii) - b
(4) (i) - c, (ii) - b, (iii) - a

45) Given :
E°(Cu2+/Cu) = 0.337 V and E°(Sn2+/Sn) = –0.136 V.
Which of the following statements is correct ?

2+
(1) Cu ions can be reduced by H2(g)
(2) Cu can be oxidized by H+
2+
(3) Sn ions can be reduced by H2 (g)
(4) Cu can reduce Sn+2

BIOLOGY

1)

Which of the following pedigree depicts unaffected parents with affected male child?

(1)

(2)

(3)

(4)

2)

Given pedigree chart represent autosomal recessive disease. What is the probability of 5th child
inheriting this disease :-

(1) 25%
(2) 75%
(3) 50%
(4) 100%

3) A colour blind girl is rare because she will be born only when :

(1) Her mother and maternal grand father were colour blind
(2) Her father and maternal grand father were colour blind
(3) Her mother is colour blind and father has normal vision
(4) Parents have normal vision but grand parents were colour blind

4) A woman receives her 'X'-Chromosome from :

(1) Her mother only

(2) Her father only
(3) Both her mother and father
(4) Mitochondria of mother only

5) In a certain taxon of insects, some have 17 chromosomes and the others have 18 chromosome.
The 17 and 18 chromosome-bearing organisms are

(1) Males and females, respectively

(2) Females and males, respectively
(3) All males
(4) All females

6)

The inheritance pattern of a gene over generations among humans is studied by the pedigree
analysis. Character studied in the pedigree analysis is equivalent to

(1) Quantitative trait

(2) Mendelian trait
(3) Polygenic trait
(4) Maternal trait

7) Match the following -

Column-I Column-II

Board plam with characteristic

A Haemophilia (i)
palm crease

B Down's Syndrome (ii) Delayed clotting of blood

Klinefelter's
C (iii) Few feminine characters
syndrome

D Turner's Syndrome (iv) Rudimentary ovaries

(1) (A)-(i), B-(ii), C-(iii), D-(iv)
(2) (A)-(iii), B-(ii), C-(i), D-(iv)
(3) (A)-(ii), B-(i), C-(iv), D-(iii)
(4) (A)-(ii), B-(i), C-(iii), D-(iv)

8) Refer to the given figure which is followed by few statements. Choose the incorrect statement

(1) It shows male heterogamety

(2) Both possess same types of autosomes
(3) The sex of progeny is determined by females
(4) This type of sex determination is different from humans

9) Which one of the following symbols and its representation, used in human pedigree analysis is
correct?

(1)
= unaffected female

(2)
= male affected

(3)
= mating between relatives

(4)
= unaffected male

10) A female suffering from Turner's syndrome is sterile due to

(1) Gynaecomastia
(2) Broad Palm
(3) Presence of Barr body
(4) Rudimentary ovaries

11) Select the odd statement w.r.t. sickle-cell anaemia?

(1) It is an autosome linked recessive trait

(2) The disease is controlled by a single pair of allele HbA and HbS.
(3) Homozygous individuals for HbS appear apparently unaffected by the disease
The defect is caused by the substitution of glutamic acid by valine at the sixth position of beta
(4)
globin chain of haemoglobin molecule

12) Drones do not have (i) and thus cannot have (ii) , but have a (iii) , and can have (iv) ,
The correct words for (i), (ii), (iii) and (iv) respectively are

(1) Grandfather, grandsons, father and sons

(2) Grandsons, sons, father and Grandfather
(3) Sons, grandsons, father and Grandfather
(4) Father, sons, grandfather and grandsons

13) A couple has four daughter's, the probability of fifth child being son is

(1) 50 %
(2) 25%
(3) 75%
(4) 100%

14) An individual shows the following characteristics

(a) Physical, Psychomotor and mental development is retarded
(b) Congenital heart disease
(c) Furrowed tongue
This individual suffers from

(1) Klinefelter’s syndrome

(2) Down’s syndrome
(3) Turner’s syndrome
(4) Thalassemia

15) Choose the correct option with respect to ploidy and chromosomal number of male honey bees ?

(1) 2n = 32
(2) n = 32
(3) n = 16
(4) 2n = 16

16) The weight of fruit in a plant is determined by the number of dominant alleles of a certain
number of genes. If seven weight categories are noticed, how many gene sites would be involved?

(1) Two
(2) Three
(3) Four
(4) Five

17) Assertion : Haemophilia is commonly found in male.

Reason : Haemophilia is X-linked recessive disorder.

(1) If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.
(2) If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.
(3) If Assertion is True but the Reason is False.
(4) If both Assertion & Reason are False.

18) If both parents are affected by thalassemia, what are the chances of pregnancy resulting in
a normal child?
(1) 25%
(2) 75%
(3) 100%
(4) 0%

19) Blue eye colour is recessive to brown eye colour in human. The expected children of a marriage
between a blue-eyed woman and a brown eyed man who had a blue-eyed mother will be-

(1) All black-eyed`

(2) All blue-eyed
(3) All brown-eyed
(4) One blue-eyed and one brown-eyed

20) If parents are carrier for albinism, then what will be the possibility in the children?

(1) Normal, heterozygous and albino

(2) All normal
(3) All heterozygous
(4) All albino

21) Which of the following pedigrees is not possible for autosomal dominant disease?

(1)

(2)

(3)

(4)

22) A classical example of point mutation is

(1) Sickle cell anaemia

(2) Turner's syndrome
(3) Thalassemia
(4) Haemophilia
23) Pedigree symbols A and B indicate

(1) Normal female

(2) Carrier female
(3) Affected female
(4) More than one option is correct

24) The pedigree chart clearly indicating recessive disorder is

(1)

(2)

(3)

(4)

25) Find out the probability of a human couple having 4 sons.

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

26) Which of the following human syndrome occurs due to monosomy:-

(1) Down syndrome

(2) Turner's syndrome
(3) Klinefelter's syndrome
(4) Jacob's syndrome

27) Which of the following disease is not applicable for pedigree analysis:-

(1) Cystic fibrosis

(2) Sickle Cell Anaemia
(3) Thalassemia
(4) AIDS

28) Which of the following statement is correct about the diagram given below:-

(1) This disorder was first described by Marshall Nirenberg.

(2) This genetic disorder is due to the presence of an additional copy of X-chromosome
(3) This genetic disorder is due to the presence of an additional copy of the chromosome number 21
(4) The affected individual is long statured with large rounded head.

29) Down syndrome is due to:-

(1) Sex-linked inheritance

(2) Crossing over
(3) Nondisjunction of chromosome
(4) Linkage

30) All are characters of Down’s syndrome except:

(1) Congenital heart disease

(2) Broad flat face
(3) Small and wrinkled tongue
(4) Many ‘loops’ on finger tips

31) Gynaecomastia is seen in case of:

(1) Down’s syndrome

(2) Klinefelter’s syndrome
(3) Turner’s syndrome
(4) All of these

32) In which of the following disorders, mental retardation is a common feature

(1) Thalassemia and Turner’s syndrome

(2) Phenylketonuria and Down’s syndrome
(3) Haemophilia and phenylketonuria
(4) Cystic fibrosis and Turner’s syndrome

33) Which of the following option have example of primary meristem

(1) Apical meristem, Intrafascicular cambium

(2) Intercalary meristem, cork cambium
(3) Intercalary meristem, interfascicular cambium
(4) Intra fascicular cambium, interfascicular cambium

34) Which of the following statements is not correct ?

(A) Apical meristems are primary meristem while intercalary meristem are secondary
(B) Intrafasicular cambium are primary tissue while interfasicular cambium are secondary meristem
(C) Apical and intercalary meristem both are primary
(D) Collenchymatous cell do not have protoplasm at maturity

(1) A & D
(2) A & B
(3) B & D
(4) C & D

35) Read the following four statements (A-D)

(A) Bast fibre are made up of sclerenchymatous cells.
(B) Bast fibre are absent in primary phloem.
(C) At maturity sieve tube element have developed nucleus.
(D) At maturity bast fibre lose their protoplasm and become dead.
How many of the above statements are correct?

(1) Four
(2) One
(3) Two
(4) Three

36) Read the following statements and find out correct statements.
(I) Bulliform cells are found in dorsiventral leaves and help in photosynthesis.
(II) Bulliform cells in isobilateral leaves help in minimizing water loss.

(1) Only I is correct

(2) Only II is correct
(3) Both I and II are correct
(4) Both I and II are Incorrect

37) Which one of the following statement is correct?

(1) Initiation of lateral roots occurs from pericycle.

(2) Vascular bundles in roots are conjoint.
(3) Dicot root do not have unicellular hairs.
In isobilateral leaf mesophyll has two types of cells- the palisade parenchyma and spongy
(4)
parenchyma.

38) Which of the following is not correct statement about the dicotyledonous stem?

(1) Endodermis is also reffered to as the starch sheath.

(2) Sclerenchymatous layer is present below the epidermis
(3) Vascular bundle is conjoint, open and with endarch protoxylem.
(4) Epidermis may bear trichomes and stomata.

39) Choose the correct statement.

(1) Internal structure also shows adaptation to diverse environment.

(2) Internal morphology of plant and animals are very similar.
(3) The meristem occurs between mature tissues is known as axillary bud.
(4) Apical meristem is secondary meristem.

40) Choose the correct and incorrect statements :-

(a) All the tissues on the inner side of the endodermis forms stele.
(b) Pith is large and well developed in dicot root
(c) Conjuctive tissue is present in both stem and root
(d) In monocot stem, water containing cavities are present within the vascular bundles
(e) Bulliform cells occurs in grasses.

(1) a, b, e-correct and c, d-incorrect

(2) a, b, d, e-correct and c-incorrect
(3) a, d, e-correct, and b, c-incorrect
(4) a, e-correct, and b, c, d-incorrect

41) Read the given statements and give the answer from the options which include only the correct
combinations.
(A) Both monocot and dicot roots have casparian strips.
(B) In stems both collateral open and collateral closed conditions may be present
(C) Bulliform cell may be present in both monocot stem and monocot roots
(D) Vascular cambium and cork cambium both are secondary meristems.

(1) A, C, D
(2) A, B, D
(3) A, B, C
(4) A, B, C and D
42) In dorsiventral leaf, what is true regarding the number of stomata?

(1) Stomata more on adaxial surface

(2) Stomata more on abaxial surface
(3) Stomata absent
(4) Number of stomata equal on both surface

43)

Read the following statements carefully :-

(a) Cambium is present in open vascular bundles
(b) In conjoint type of vascular bundles, the xylem and phloem are situated at different radii.
(c) All tissues on the inner side of the endodermis constitute the stele.
(d) Pith is large and well developed in monocot roots
(e) Sclerenchymatous hypodermis is present in monocot stem.
Choose the correct answer from the options given below.

(1) a, c, d and e
(2) a, b, c, d and e
(3) b, c, d and e
(4) a, b, d and e

44) Following are the anatomical features –

(a) Polyarch condition of vascular bundles.
(b) Metaxylem faces inwards.
(c) Radial vascular bundles.
(d) Innermost cortex layer is thick walled and suberised.
Select correctly matched pair :-

(1) Dicot Root – a, b and d

(2) Monocot Root – a, b, c and d
(3) Dicot stem – a and b
(4) Monocot stem – a, b and c

45) Statement-I :- In dicotyledonous stem (eg. sunflower stem), pericycle is present on the inner
side of endodermis located just overhead to phloem bundles in the form of semilunar patches of
sclerenchyma
Statement-II :- In conjoint vascular bundles phloem usually located on the exterior to xylem.

(1) Both statements I and II are correct.

(2) Both statements I and II are incorrect.
(3) Only Statement I is correct
(4) Only Statement II is correct

46) Forecoming generations are less adaptive than their parental generation due to :-

(1) Natural selection

(2) Mutation
(3) Genetic drift
(4) Adaptation

47) Darwin in his 'Natural Selection Theory' did not believe in any role of which one of the following
in organic evolution :-

(1) Parasites and predators as natural enemies

(2) Survival of the fittest
(3) Struggle for existence
(4) Discontinuous variations

48)

According to Darwin the evolution is:

(1) Slow process

(2) Directional process
(3) Both (1) & (2)
(4) Fast and Random process

49)

Founder effect can be seen in a population because of :

(1) Constant allele frequency

(2) Genetic drift
(3) Absence of gene flow
(4) Absence of natural Selection

50)

Oxygen in atmosphere has been formed by:

(1) Evaporation of water

(2) Photosynthesis of blue green algae
(3) Metabolism of microorganisms
(4) Decaying organisms

51)

Golden age Dinosaurs was during:

(1) Cenozoic era

(2) Palaeozoic era
(3) Archeozoic era
(4) Mesozoic era

52)
Which type of selection is industrial melanism observed in moth, Biston betularia?

(1) Stabilising
(2) Directional
(3) Disruptive
(4) Artificial

53)

Select the correct statement

A. Adaptive ability is the end result of fitness.
B. Fitness is selected by nature.
C. Any population has built in variation.
D. first form of life arose about 2000 MYA.

(1) Only A & B

(2) Only B & C
(3) Only A, B & C
(4) Only C & D

54) Which of the following are example of adaptive radiation ?

(1) Darwin’s finches

(2) Placental mammals
(3) Australian marsupials
(4) All of the above

55) Which of the following statements is /are correct regarding special creation theory.

(1) The first cellular form of life did not possibly originate till about 2000 million years ago
(2) All living species or types of organisms that we see today were created as such
(3) Existing living forms share similarities to varying degrees among themselves
(4) The first form of life arose slowly through evolutionary forces from non-living molecules

56) The two key concepts of Darwinian Theory of Evolution are

(1) Natural selection and over production

(2) Branching descent and natural selection
(3) Branching descent and struggle for existence
(4) Variation and heredity

57) Read the following (A-D) statements:

A. When migration of a section of population to another place and population occurs, gene
frequencies changes.
B. If gene migration happens multiple times there would be a gene flow
C. If loss of gene from a population occurs by chance it is called a genetic drift
D. Natural selection is a process in which heritable variations enabling better survival are enabled to
reproduce greater number of progeny
How many of above statements are true

(1) Two
(2) One
(3) Four
(4) Three

58) Assertion (A) : Hugo de Vries said mutations cause speciation.

Reason (R) : Mutations are slow changes

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

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

59) The original birds were_____, from which the various Darwin's finches arose:

(1) Insectivorous
(2) Cactus eating
(3) Carnivores
(4) Seed eating

60) Single step large mutation leading to speciation is also called :

(1) Founder effect

(2) Saltation
(3) Branching descent
(4) Natural selection

61) An isolated population with equal number of blue-eyed and brown-eyed individuals was
decimated by earthquake only a few brown-eyed individuals survived to form next generation. The
change in gene pool is:

(1) Blocked gene flow

(2) Founder effect
(3) Bottleneck effect
(4) Hardy-Weinberg equilibrium

62) Statement-I : Crocodiles, Birds and Dinosaurs have common ancestor i.e. Thecodonts.
Statement-II : Turtles, Lizards, Snakes and Tuataras have common ancestor i.e Sauropsids.

(1) Both statements are correct

(2) Statement I is correct and statement II is incorrect
(3) Statement I is incorrect and statement II is correct
(4) Both statements are incorrect
63) Identify the dinosaurs A and B:

(1) A-Tyrannosaurus, B-Stegosaurus

(2) A-Triceratops, B-Stegosaurus
(3) A-Tyrannosaurus, B-Brachiosaurus
(4) A-Triceratops, B-Tyrannosaurus

64)
A & B is :-

(1) Cycad, Genetales

(2) Conifers, Cycads
(3) Ginkgos, Seed fern
(4) Seed Fern, Arborescent lycopods

65) Which of the following is the correct sequence of events depicting Darwin's theory for evolution?
I. Natural selection
II. Struggle for existence
III. Variations
IV. Overproduction.

(1) I – II – III – IV
(2) IV – II – III – I
(3) IV – I – II – III
(4) II – III – IV – I

66) Which plants were present but they all fell to form coal deposits slowly-

(1) Monocots
(2) Gymnosperms
(3) Giant ferns
(4) Dicots

67) Hugo de Vries mutations are A while Darwinian variations are B . Options

(1) A small and directional B random and directionless

(2) A small and directional B random and directional
(3) A random and directionless B small and directional
(4) A random and directional B small and directional

68) Which of the following evidences does not favour the Lamarckian concept ?

(1) Absence of limbs in snakes

(2) Presence of Webbed toes in aquatic birds.
(3) Melanization in peppered moth in industrial area
(4) Lack of pigment in cave dwelling animals

69) Which of the following is an example of anthropogenic action ?

(1) Darwin finches

(2) Drug resistance
(3) Australian marsupials
(4) Placental mammals

70) The diagrammatic representation of operation of natural selection is shown. Which of the
following option indicates 'directional selection' ?

(1)

(2)

(3)
(4)

71) Match the followings :-

Column-A Column-B
A. Invertebrates formed i. 65 MYA
B. Jawless fish evolved ii. 200 MYA
C. Dinosaurs disappeared iii. 500 MYA
D. Ichthyosaurs evolved iv. 350 MYA
(1) A-iii, B-iv, C-ii, D-i
(2) A-iii, B-iv, C-i, D-ii
(3) A-iv, B-i, C-ii, D-iii
(4) A-iv, B-iii, C-i, D-ii

72) Which of the following is correct sequence of evolution?

(1) Synapsids → Thecodont → Dinosaur → Birds

(2) Sauropsids → Pelycosaurs → Dinosaur → Crocodile
(3) Synapsids → Pelycosaurs → Therapsids → Mammals
(4) Sauropsids → Synapsids → Pelycosaurs → Dinosaur

73) Lymph capillaries of intestine villi are called as:-

(1) Lymphatics
(2) Portal vein
(3) Thoracic duct
(4) Lacteals

74) When heart muscles are suddenly damaged by an inadequate blood supply, this condition called
as:

(1) Heart failure

(2) Heart attack
(3) Cardiac arrest
(4) Congentive heart failure

75) Ventricles receive maximum blood from atria during:-

(1) Auricular systole

(2) Ventricular systole
(3) Auricular diastole
(4) Joint diastole

76) First heart sound LUB is associated with :

(1) Opening of semilunar valves
(2) Opening of tricuspid and bicuspid valves
(3) Closure of tricuspid and bicuspid valves
(4) Closure of semilunar valves

77) Match the excretory functions of section I with the part of the excretory system in section II.
Choose the correct combination from among the option given :-

Section-I Section - II
(Function) (Parts of excretory systems)

(A) Ultra filtration (1) Henle's loop

(B) Concentration of urine (2) Ureter

(C) Transport of urine (3) Urinary bladder

(D) Storage of urine (4) Malpighian corpuscle

(5) Proximal convoluted tubule

(1) A–4, B–1, C–2, D–3
(2) A–4, B–3, C–2, D–1
(3) A–5, B–4, C–1, D–3
(4) A–5, B–4, C–1, D–2

78) Kidney stones are produced due to :-

(1) Precipitation of cholesterol.

(2) Deposition of sand particles.
(3) Blockage of fat.
(4) Crystallisation of oxalates.

79) In erythroblastosis foetalis, which conditions are observed :-

(1) Severe jaundice only

(2) Severe anaemia only
(3) Severe anaemia and jaundice
(4) Severe anaemia and diarrhoea

80) Which of the following condition are indicative of diabetes mellitus ?

(1) Glucosuria
(2) Proteinuria
(3) Ketonuria
(4) Glucosuria and ketonuria (both)

81) Which of the following statements is incorrect?

(1) In ureotelic organisms, ammonia is not a product of metabolism.

(2) In mammals, some amount of urea may be retained in the kidney matrix to maintain osmolarity.
(3) In fish, kidneys do not play any significant role in the removal of ammonium ions.
(4) Urea and uric acid are less toxic than ammonia.

82) The contractile protein of skeletal muscle involving ATPase activity is :-

(1) Actin
(2) Myosin
(3) Troponin
(4) Tropomyosin

83) Which event does not happen during muscle contraction ?

(1) Actin filaments move towards H-zone

(2) Decreased length of myosin filament
(3) Decreased length of sarcomere
(4) Length of A-band remains unchanged.

84) Match column-I with column-II and choose the correct answer from the options given below :-

Column-I Column-II

(a) Smooth muscle (i) Myoglobin

(b) Tropomyosin (ii) Thin filament

(c) Red muscle (iii) Sutures

(d) Skull (iv) Involuntary

(1) a-iv; b-ii; c-i; d-iii
(2) a-ii; b-iv; c-i; d-iii
(3) a-iv; b-i; c-iii; d-ii
(4) a-iv; b-i; c-ii; d-iii

85) Read the following statements and find out the incorrect statement.
a. The CNS is the site of information processing and control.
b. The somatic neural system relays impulses from the CNS to the involuntary organs and smooth
muscles of the body.
c. The autonomic neural system transmits impulses from the CNS to skeletal muscles.
d. The autonomic neural system is further classified into sympathetic and parasympathetic neural
system.

(1) a and b
(2) b and c
(3) c and d
(4) a and d

86) If the Na+ – K+ pump stops working then the

(1) Na+ and K+ will be in excess in extracellular fluid
(2) Na+ will be in excess in extracellular fluid
(3) K+ will be excess in intracellular fluid
(4) Na+ will be in excess in intracellular fluid

87) In the presence of Ca2+ channel blockers, which of the following will be true ?

(1) Neurotransmitter is released but Na+ channel of post-synaptic neuron will not open
(2) Neurotransmitter is not released but Na+ channel of post-synaptic neuron will open up
(3) Neurotransmitter is released but K+ channel of post-synaptic neuron open up
(4) Neither neurotransmitter is released nor the Na+ channel of post-synaptic neuron open up

88)

How many hormones are steroids in nature :-

Cortisol, Hypothalamic hormone, Pituitary hormone, Testosterone, Estradiol, Progesterone,

Epinephrine.
(1) 4
(2) 5
(3) 6
(4) 3

89) Leydig's cells are stimulated by the pituitary hormone known as

(1) Prolactin
(2) FSH
(3) LH
(4) GH

90) Which hormone act on both pancreas and gall bladder and stimulate secretion of pancreatic
enzymes and bile juice :-

(1) Gastrin
(2) Cholecystokinin
(3) Gastric inhibitory peptide (GIP)
(4) Duocrinin
 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 1 3 1 3 1 1 2 3 2 3 3 1 1 1 2 2 1 2 2
Q. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
A. 4 3 1 2 3 3 1 1 2 4 4 4 2 3 3 1 4 1 2 2
Q. 41 42 43 44 45
A. 4 3 2 2 1

CHEMISTRY

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

BIOLOGY

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

PHYSICS

1)

Concept :
Combine capacitors using series and parallel rules, and simplify the circuit systematically to
find net capacitance between two terminals.
Formula

• Series:
• Parallel:
Calculation:
The given circuit can be redraws as follows

On further solving the network in similar manner equivalent capacitance obtained between A
and B will be 1μF.

Answer Option -1

2)
2(6 – x) + 4 (0 – x) + 2 (6 – x) = 0
12 – 2x – 4x + 12 – 2x = 0
24 = 8x ∴ x = 3V
∴ qC1 = 2 (6 – 3) = 6 μC
3)

Ans is -

4)

Ans -

5) The magnetic field in the solenoid along its axis

(i) At an internal point = μ0ni

(ii) At one end

6)

Ans - 150 μC
7)

8)

9) Slope

0
At t =0, Slope = i =
Q Slope 1 > Slope2 > Slope3
∴ R1 < R2 < R3

10) Electric field in the region 1, 3 and 5 is zero

i.e. E1 = E3 = E5
Slope of the line BC < Slope of the line DE
i.e. E2 < E4

11) Question Explanation:

A current carrying wire is composed of straight and curved. We are asked to calculate the net
magnetic field at point C using contributions from all segments.

concept:
Magnetic field due to current elements Biot-savart law.
Formula:

For circular arcs:

Calculation:
The straight segment along the z-axis contributes a magnetic field in directions

is the semicircular arc in the xy-plane centered at point C contributes along

.

Another straight segment adds .

Combining with the component total field.

Final answer: 3

12) Explanation :
Two coaxial solenoids have same length & different turn densities n1 & n2. They carry currents
i, & i2 opposite directions and the MF inside the inner solenoid is zero. We need to determine
this condition.

Concept :
Superposition of magnetic fields and the magnetic field due to solenoid.

Formula :
Bsolenoid = Honi

Calculation :
For net B = O
B1 → MF due to one solenoid B1 = H0 n1 i1
B2 → MF due to other solenoid B2 = H0 n2 i2
Net B1 – B2 = 0
H0n1i1 – H0n2i2 =
n1i1 = n2 i2
Now we check each option
Option (3)

13)
Towards left of both wires direction of B is downward and at mid point between two wires,
magnetic field is zero.

14) Magnetic field at centre due to th of circular current carrying are is

 ...(i)
Magnetic field at O due to straight wire P is
BP = 0 ...(ii)
Now, magnetic field due to straight wire R is

...(iii)

15) Magnetic field inside the hollow metallic cylinder and magnetic field outside it

16)

17)

Ans - [MT–2A–1]

18)

Zero

19) 1. Problem Explanation: Find the magnetic field at point O due to the straight wire PQ.
2. Concept Used: Biot-Savart Law (for a finite wire).
3. Formulas Used:

We use the absolute value here because direction doesn't matter for the multiple-choice
options, and we've already oriented our angles.
Where:
• μ0 is the permeability of free space
• I is the current
• r is the perpendicular distance from point O to the wire
• θ1 and q2 are the angles made by the lines joining the ends of the wire ( P and Q ) to point O,
measured with respect to the perpendicular to the wire.
4. Calculation:
1. Identify the angles:
• (angle between the line connecting O to P and the perpendicular)
• (angle between the line connecting O to Q and the perpendicular)
2. Apply the formula:

20)

Ans -

21)

Ans - 180°

22)
Suppose P is the point between the conductors where net magnetic field is zero.
So at P | Magnetic field due to conductor
1| = |Magnetic field due to conductor 2|

i.e.,

Hence position from B

23)

B=
i ∝ Br

24)

Barc =

Bst.wire =
Barc ⊥ Bst.wire

0
B =
25)

Ans -

26)

Both (1) & (2)

27)

Ans -

28)
= 2 – 1 = 1A

= μo

29) We shall use where I is the current enclosed by loop

Net current enclosed by path a is zero.
Net current enclosed by path c is 1A.
Net current enclosed by path d is 3A.
Net current enclosed by path b is 5A.

30)

Ans -

31)

By theory

32) Magnetic field outside the volume of toroid is zero.

33)
r=

34)

pitch = Vcosθ × T

= V11 × T = 2 ×

=
= 2π × 10–2 m

35) Deviation as r > d

sinθ =

θ = sin–1 =

36)

For the complete ring, M.I. of the ring about diameter . For the semi-circular ring.

37)

Ans -

38)

Ans -

39)
 Ans -

40)

Ans - 7I

41)

All of the above

42) Moment of inertia of the system

I = 2m(2r)2 + 4x

= 8mr2 + 4

I=
For radius of gyration I = (6m)K2

⇒ 6mK2 = mr2

⇒K=r Hence option (3)

44)

Ans - MR2 + 4mR2

45) Moment of inertia of rod AB about point P and perendicular to the plane =
MI. of rod AB about point

'O' = by using parallel axis theorem but the system consist of four side

of similar type so by the symmetry Isystem = 4

CHEMISTRY
46)

Net cell reaction :

2Cr + 3Fe2+ —→ 2Cr3+ + 3Fe

Ecell =

47)

C1 > C2

48)

+0.32V

49)

Ans - 1030

50)

Ans - –0.59 V

51)

z>x>y

52)

53)

The voltage drops to zero.

54)

Specific conductance increases with dilution

55)

Increases
56) A decrease in PH

57) For feasible cell reaction → ΔG = –ve and acceleration to ΔG° = –2.303 RT log KC
If (KC > 1) → (positive) then ΔG° = (negative)

58) According to onsage equation i.e., on

increasing concentration dissociation of electrolyte decreases or decreases.

59)

Ans - 103

60)

velocity of both K+ and NO–3 are nearly the same

61)

Anode to cathode through the external circuit

62) is the strongest oxidizing agent and Mg is the strongest reducing agent.

63)

212.3 kJ

64)

I and IV

65)

3x2 – 2x1

66)
E1 > E2

67)

68)

All the above statements are correct

69)

Increase in the concentration of Ag+ ions

70)

0.0414 V

71)
lone pair of 'N' is localized.

72) is least stable due to less number of covalent bond (incomplete octet of O) and
+ve charge on more EN atom.

73)
Acidic strength(A.s/ka)→III>I>IV>II
pKa → II > IV>I>III

74) Carbanion does not show hyperconjugation so does not show hyperconjugation.

75)

A.S ∝ stability of C.B

 76)

Conjugate base is stabilised by – M of two group.

77) , 3° least stable.

78)

A) Question explanation: Identify the series containing only electrophiles (electron pair
acceptors).

C) Concept: Electrophiles are electron-deficient species. Nucleophiles donate electron pairs.

D) Answer Explanation:

A. Series 1 & 4 contain nucleophiles (H₂O, NH₃). Incorrect.

B. Series 2 contains nucleophiles (H₂O, R−NH₂). Incorrect.
C. Series 3 contains only electrophiles: BF₃ (incomplete octet), SO₃ (electron-deficient S), and
NO₂⁺ (positive charge). Correct.

E) Final Answer: Series 3 (BF₃, SO₃, NO₂⁺) contains only electrophiles. This corresponds to
option 3.

F) Question Level: Moderate

79)

Ans - and

80)
Ans -

81)

Ans -

82)
[More stable C⊕]

83)

84)

Ans -

85)

1,6-Dimethylcyclohexene
 86)

Ethanoic anhydride

87)

Ans is 6

88)
F2 + 2e⊝ → 2F–
Li+ + e⊝ → Li

reducing power ∝ SOP

oxidising power ∝ SRP

89)

(i) - c, (ii) - a, (iii) - b

90)

Cu2+ ions can be reduced by H2(g)

BIOLOGY

91)

Ans

92)

Ans - 25%

93)

Her father and maternal grand father were colour blind

94)

Both her mother and father

95)

Males and females, respectively

96)

Mendelian trait

97)

(A)-(ii), B-(i), C-(iii), D-(iv)

98)

It shows male heterogamety

99) NCERT Pg no. 88

100)

Rudimentary ovaries

101)

Homozygous individuals for HbS appear apparently unaffected by the disease

102)

Father, sons, grandfather and grandsons

103) Allen Module Pg No. 55

104) NCERT Pg.no. 92

105) NCERT Pg No. 87

106) Allen Module Pg No. 101

107)

If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.

108)
Ans 0%

109)

One blue-eyed and one brown-eyed

110)

Normal, heterozygous and albino

111)

Ans -

112) Ncert Pg No. 88

113)

Carrier female

114)

Ans -
115)

Ans- 1/16

116)

Turner's syndrome

117)

AIDS

118)

This genetic disorder is due to the presence of an additional copy of the chromosome number
21

119)

Nondisjunction of chromosome

120)

NCERT-XII, Pg. # 76

121)

NCERT-XII, Pg. # 76

122)

Phenylketonuria and Down’s syndrome

123)
 NCERT (XIth) Pg. # 85

124)

Ncert page 85 and 86.

125)

Three

126) NCERT Pg. No. 94

127) NCERT XI, Pg. # 90, 91, 93, 94

128) NCERT XI, Pg. # 92, 93

129) NCERT-XI old Pg.# 84, 85

130) NCERT Page No. # 91, 93, 94

131) NCERT XI Page No. 91, 90

132) NCERT XI, Pg. # 93

133)

NCERT 11th Pg. No. 90

134) NCERT-XII, Pg. # 91 & 92

135) Both statements I & II are correct.

above → overhead
outer side - exterior to
In NCERT Page No. 90
The conjoint vascular bundles usually have the phloem located only on the outer side of xylem.

136) Mutations are random changes in the genetic material, and while they can introduce new
genetic variations, they can also lead to harmful or less adaptive traits. In some cases, the
accumulation of mutations over generations can result in a decrease in the population's overall
adaptability, particularly if the mutations are mostly neutral or detrimental rather than
beneficial.
So, the correct answer would be:
 2. Mutation : Mutations introduce genetic changes, but if the mutations are harmful or not
beneficial, they can reduce the overall adaptability of subsequent generations.

137)

A. Charles Darwin's theory of Natural Selection emphasized the gradual accumulation of small,
continuous variations over generations as the key to evolution.
B. He believed that continuous variations (small differences among individuals) were heritable
and could be selected for in nature.These variations contribute to adaptation and survival in a
changing environment.

However, Darwin did not believe that discontinuous variations (sudden, large changes or mutations)
played a major role in evolution.
The correct answer is 4. Discontinuous variations.

138)

NCERT (XII) Pg # 135/146(H) Para : 7.6

139)

Genetic drift change gene frequeancy.

140)

Photosynthesis of blue green algae

141)

Mesozoic era

142)

NCERT XII, Pg. No.135

143)

The correct answer is Option 2: B & C.

A. Adaptive ability is the end result of fitness.

A. False : because Fitness is the end result of the ability to adapt and get selected by nature.

B. Fitness is selected by nature.

A. True: Natural selection is the process where nature "selects" for traits that increase an
organism's fitness. Individuals with higher fitness (i.e., those better suited to their
environment) are more likely to survive and reproduce, passing their advantageous traits to
the next generation.

C. Any population has built-in variation.

 A. True: Genetic variation is essential for evolution. Within any population, there is natural
variation in traits due to genetic mutations and the shuffling of genes during
sexual reproduction. This variation provides the raw material for natural selection to act upon.

D. First form of life arose about 2000 MYA.

A. False: The first forms of life on Earth are estimated to have appeared around 3.5-3.8 billion
years ago, not 2000 million years ago (which is 2 billion years ago).

Therefore, only statements B, and C are correct.

144)

The answer to the question is 4. All of the above.

Adaptive radiation is a rapid evolutionary process in which a single ancestral species
diversifies into multiple species, each adapted to a different ecological niche.

145)

All living species or types of organisms that we see today were created as such

146)

Branching descent and natural selection

147)

Four

148)

A is true, R is false.

149)

Seed eating

150)

Saltation

151)

Bottleneck effect

152)

Both statements are correct

153) NCERT (XII) Pg # 130 Fig:7.2

154)

Seed Fern, Arborescent lycopods

155)

IV – II – III – I

156) NCERT Pg. # (E)-140, (H)-149

157)

NCERT Page no.135

158) Allen module Ex : I, Question No : 75, Page no 134

159) NCERT Pg. # 119

160) NCERT XII, Pg.# 137

161)

A-iii, B-iv, C-i, D-ii

162) NCERT (XII) Page no. 123

163)

Lacteals

164) NCERT XI Pg.# 288

165)

Joint diastole

166) NCERT XIth Pg.#285

167)
NCERT - Pg. No.# 295

168) Kidney stones are produced due to crystallisation of oxalates.

169) NCERT (XI) Pg. # 281; para-1

170) NCERT-XI, Pg. # 298

171)

NCERT Pg. No: 205

172)

NCERT - Pg. No. 221 Para 2 Fig 17.3 (b)

173)

During muscle contraction, the following events occur:

A. Actin filaments move towards the H-zone: This is a key feature of the sliding filament
theory. Actin filaments slide past myosin filaments, pulling them towards the center of
the sarcomere.

B. Decreased length of sarcomere: As the actin and myosin filaments slide past each other,
the distance between the Z-lines (the boundaries of the sarcomere) decreases, resulting
in a shorter sarcomere.

C. Length of A-band remains unchanged: The A-band is the region of the sarcomere that
contains the myosin filaments. During contraction, the A-band does not change in length
because the myosin filaments themselves do not shorten.

Therefore, the length of the myosin filament does not decrease during muscle contraction.

So,the correct answer is 2.

174) NCERT-XI, Pg. # 314, Q.6

175)

b and c

176)
Na+ will be in excess in intracellular fluid

177)

Neither neurotransmitter is released nor the Na+ channel of post-synaptic neuron open up

178) NCERT Pg. # 338, 22.4

179)

ANS - LH

180)

Cholecystokinin