NEET-UG 2023 Question Paper — Set 2 Solutions

As there is no external force acting on COM i.e., vector F_ext =0 and initial velocity of COM is zero, SESY) therefore it will remain at rest i.e., the speed of centre of mass is zero.

Resistivity ρ = 1/σ(conductivity) and σ maximum for silver so silver has the smallest re resistivity.

From Gauss's law, total electric flux through a closed surface

Φ_net = q / ε₀

∴ Total electric flux through the cube

Φ_c = (Q × 10⁻⁶) / ε₀

As charge is placed at the centre of the cube, so flux through each face is same.

∴ Φ = (1 / 6) Φ_c = (1 / 6) × (Q × 10⁻⁶) / ε₀

Force due to magnetic field F_m = qvB sinθ Here θ = 0 ⇒ F_m = 0

Force due to electric field vector F_e = qE vector

Here , vector F_e = -eE vector

Clearly, force due to electric field is opposite to the direction of motion, so its speed will decrease.

At the highest point (B) of a projectile,

v = v₀ cos 60° = 20 cos 60° = 20 × (1 / 2) = 10 m/s

[ ∵ At B, V_y = 0 ]

Elastic potential energy per unit volume

= (1 / 2) × Young's modulus × (Strain)²

= (1 / 2) × 2 × 10¹¹ × [ 1 / 1000 ]²

= (1 / 2) × 2 × 10¹¹ × 10⁻⁶ = 10⁵ J/m³

∵ Resistance R = ρl / A , where A = (π / 4) d²

∴ 81 = (ρ × l) / ((π / 4) d²) , and R' = (ρ × l) / ((π / 4) × (3d)²) = (ρl) / ((π / 4) × 9d²)

∴ R' = R / 9 = 81 / 9 = 9 Ω

As, D. C. voltage is supplied so impedance Z₁ = R = 10 Ω

And Z₂ = √[ R² + (X_c)² ]

= √[ (10)² + [ 1 / (2 × π × 50 × (10³ / π) × 10⁻⁶) ]² ]

⇒ Z₂ = √[ (10)² + (10)² ] = 10√2

∴ Z₂ > Z₁ or Z₁ < Z₂

Total energy of electron in hydrogen atom

E_η = -13.6 / n²

In second excited state = -13.6 / (3)² = -1.51 eV

Therefore 1.51 eV energy required to ionize hydrogen atom from its second excited state.

Gravitational force, F = GM₁M₂ / r²

⇒ [G] = ([F][r]²) / ([M₁][M₂]) = ([MLT⁻²][L²]) / ([M][M])

⇒ [G] = [M⁻¹ L³ T⁻²] Torque T = F · r = [ML² T⁻²] Angular momentum, L = Mvr = [ML² T⁻¹]

To make a p–type extrinsic semiconductor, an intrinsic semiconductor is doped with trivalent intrinsic impurity like Gallium, Indium, Boron, Aluminum, etc.

From stoke's law, Viscous drag acting on sphere Fᵥ = 6πηrv

= 6 × (22 / 7) × 0.8 × (10⁻³ / 2) × 2 = 15.08 × 10⁻³ N

= 15 × 10⁻³ N

Power, P = V_rms I_rms cosϕ In purely resistive circuit voltage and current are in same phase ∴ cosϕ = cos0° = 1 ∴ Power dissipated in maximum.

According to Gauss' law, total electric flux

ϕ = ∮ 𝐄⃗ · d 𝐬⃗ = q_enclosed / ε₀

∴ Electric flux depends on quantity of charges enclosed by surface.

As per Curie law χ_m ∝ 1 / T (Paramagnetic)

Susceptibility χ varies inversely with absolute temperature. Therefore option (c) represents correct χ versus 1 / T graph.

Truth table of given logic circuit is as follows.

Find out put is same as that of and gate out put so given circuit is equivalent to and gate.

The given transistor represents a n–p–n transistor so statement (A) is incorrect Emitter base junction is forward biased. Emitter is of moderate size and heavily doped and central segment is thin and lightly doped. So, statements (B) and (C) are correct.

Deviation produced by a thin prism

δ = (μ - 1)A = (3/2 - 1) 6° = 3°

Angle of emergence, e=6° +3°=9°

Instantaneous power P = 𝐅⃗ ·𝐕⃗

P = [10î + 10ĵ + 20k̂] · [5î - 3ĵ + 6k̂] W

= [50 - 30 + 120] W = 140 W

By Einstein's photoelectric equation KE_max = hf - ϕ₀

or, KE_max = hc / λ - hc / λ₀

Clearly KE_max is independent of intensity of incident radiation.

Mean diameter,

d̄ = [3.33 cm + 3.32 cm + 3.34 cm + 3.33 cm + 3.32 cm] / 5

= 3.328 cm

After rounding off to 2 decimal places, d̄ = 3.33 cm

All pressure, relative density, coefficient of viscosity and surface tension are scalar quantity

Given, position,𝐫⃗(t) = (4t î + 2t² ĵ + 5k̂) m

∴ Velocity, 𝐯⃗ = dr𝐫⃗(t) / dt = 4î + 4t ĵ |_(t=1) m s⁻¹ =

= 4î + 4ĵ m s⁻¹

∴ | 𝐯⃗| = √[4² + 4²] = 4√2 m s⁻¹

And, tan ϕ = v_y / v_x = 4 / 4 = 1

∴ Direction with respect to X-axis is 45°

Work done in isobaric i.e., when pressure is constant (AB) process = P ΔV = 3 × 10² [2] J = 600 J

De Broglie wavelength,

λ = 12.27 / √V Å = 12.27 / √81 Å = 12.27 / 9 Å

∴ λ = 1.36 Å or 0.136 nm

4ᵗʰ overtone for closed organ pipe → 9ᵗʰ harmonic and 3ʳᵈ overtone for open organ pipe → 4ᵗʰ harmonic

∴ 9 × (V / 4l_c) = 4 × (V / 2l_o) ⇒ l_c / l_o = 9 / 8

The equivalent circuit of the given circuit is as follows

∴ C_net = 15 + 5 = 20 μF

We know Refractive index of the medium, μ_med

= √[μ ε] / √[μ₀ ε₀] = √[ (1.5)μ₀ (2)ε₀ ] / √[μ₀ ε₀] = √3

From FBD of the given diagram, F = mg or, I l B = mg

⇒ I = mg / (l B) = (250 × 10⁻³ × 9.8) / ( (2)(0.7) ) = 1750 × 10⁻³ A

= 1.75 A

Since object distance u > 2f so all three medium will form individual real images at different points on the other side of lens ∴ Number of real images formed = 3.

Inductive reactance, X_L = ωL

X_L = 100 π × (1 / π) = 100 Ω

∴ tan ϕ = X_L / R = 100 / 100 = 1 ∴ ϕ = 45°

During elastic collision between two equal masses, the velocity of the two bodies gets interchanged so if one body is at rest, energy transfer will be maximum for M = m

The velocity of the body at infinite distance from the centre of the earth, V' = √[ V² - V_e² ]

= √[ (22.4)² - (11.2)² ]

∴ V' = 11.2√3 km/s

From the relation, sin i_c = 1 / μ and μ ∝ 1 / λ

sin i_c ∝ λ Therefore, for higher value of λ critical angle will be more, so Yellow, Orange and Red will come out to air. The order of wavelength of different colours V < I < B < G < Y < O < R

Lyman series of hydrogen atom appears in Ultraviolet region. Balmer series in visible region and Paschen, Bracket and P–fund series in infrared region.

In uniform circular motion, due to change in direction, velocity 𝐯⃗ and acceleration 𝐚⃗ is not a constant.

Force is equal to rate of change of momentum

∴ F = Δp / Δt

∴ Δp = 2 mv cos30°

= 2 × 1 × (√3 / 2) = √3

∴ F = Δp / Δt = √3 / 0.1 = 10√3 N

Time period of simple pendulum T = 2π √[ l / g_eff ]

∴ √3 = 2π √[ l / g ]

After immersed in liquid Mg_eff = Mg - (Mg / ρ_s) · ρ_l

or, Mg_eff = Mg - (Mg / 4)= (3Mg / 4) ⇒ g_eff = 3g / 4

∴ T₁ = 2π √[ l / (3g/4) ] = ( 2π √[l/g] ) × 2 / √3 = 2T/√3 √3 = 2 s

If object is placed at 2f then its image of equal size is to be obtained on a parallel wall with the help of a convex lens placed between the two walls.

∴ 2f = x ⇒ f = x / 2

Inductive reactance, X_L = ωL and Capacitive reactance, X_C = 1 / ωC At very high frequency X_C = 0₁ , X_L =∞

Therefore the effective impedance of the circuit,

Z = (1 + 2) Ω = 3 Ω

Drift velocity, V_d = i / (neA)

V_d = 10 / ( 10²² × 1.6 × 10⁻¹⁹ × π(10⁻³)² )

V_d = 10 / (1.6π × 10⁻³) = 6.25 / π × 10³ m/s

We know that PV = nRT

P_net = P₁ + P₂ = (n₁ / V)RT + (n₂ / V)RT

= (n₁ + n₂) RT / V

= (0.2 + 0.3) × (8.3 × 200) / (200 × 10⁻⁶)

= 8.3 × 0.5 × 10⁶ = 4.15 × 10⁵ Pa

About fixed axis, torque = I α

α = 100 / 300 = 1 / 3 rad/s²

For constant α ⇒ ω_i = 0 ω_f = ω_i + αt ∴ At t = 3 s

ω_f = 1 / 3 × 3 = 1 rad/s

For case – I ⇒ ε_s = ϕ × 330

⇒ ϕ = εₛ / 330 (V / cm) ....(1)

For case – II ⇒ i × 2 = ϕ i₂

⇒ (εₛ / 3) × 2 = (εₛ/ 330) × i₂ [From equation 1] ⇒ i₂ = 220 cm

Acceleration due to gravity on the earth surface

g = GM / R² = (G · 4/3 π R³ ρ) / R²

g = G · 4/3 πRρ ⇒ ρ = 3g / 4GπR

For r > R, Electric Field E = Kq / r² ....(i)

and potential at surface of sphere V = Kq / R

∴ From equation (i)

V = Er² / R ⇒ E = RV / r²

Displacement current i_d = C (dV / dt)

⇒ 2 × 10⁻³ = 4 × 10⁻⁶ × (dV / dt)

dV / dt = (2 × 10⁻³) / (4 × 10⁻⁶) = 500 V/s

⇒ (F_net)_H = mg sin30° - μmg cos30°

= mg / 2 - (1 / √3) mg · (√3 / 2)

F_net = 0

Maximum value of emf generated by an ac generator ε₀ = NBA_ω = 100 × 0.05 × 1 × 2π = 31.415 V ∴ ε₀ = 31.4 V

Angular momentum of electron moving in nᵗʰ orbit from Bohr's postulate

L = nh / 2π ....(i)

Given, L = 1.5 (h / π) So, n = 3 ....(ii)

Now, energy of electron in 3ʳᵈ orbit

E = (-13.6 × z²) / n² = (-13.6 × 1²) / 3² ≈ -1.5 eV

3, 4–Dimethylhex–3–ene will show geometrical isomerism

Boron trihalides cannot be hydrolysed to produce an ion of the form [B(H₂O)₆]³⁺ because of absence of d orbitals in boron.

Rate = k [A]⁻¹/² [B]³/²

∴ order of reaction = -1/2 + 3/2 = 1

Ca²⁺ = 20 - 2 = 18 e⁻

Ar = 18 e⁻

K⁺ = 19 - 1 = 18 e⁻

Cl⁻ = 17 + 1 = 18 e⁻

So, Ca²⁺, Ar, K⁺ and Cl⁻ are isoelectronic species.

Cheilosis refers to the inflammation and small cracks in one or both corners of the mouth. Cheilosis occurs due to deficiency of vitamin B₂ (Riboflavin)

So, the functional group present in compound "X" is alcohol and Ketone.

PV = nRT

P = (n / V) RT

P = (d / M) RT

PM / RT = d

⇒ d ∝ 1 / T , d ∝ P

So, for maximum density P should be maximum and T should be minimum which is in option 0°C, 4 atm

PV = nRT

P = (n / V) RT

P = (d / M) RT

PM / RT = d

⇒ d ∝ 1 / T , d ∝ P

So, for maximum density P should be maximum and T should be minimum which is in option (A), (B) and (C) only

Fluorine is the most electronegative element. H – F bond has maximum and polarity. Order of dipole moment is

H – F > NH₃ > H₂S > CH₄ (μ = 0)

Λₘ = Λ°ₘ – A√c

• For strong electrolytes, Λₘ increases slowly with dilution.

• The value of 'A' depends on the type of electrolyte.

• All electrolytes of particular type have the same value for 'A'

• e.g. BaCl₂ and Mg(OH)₂ → 2 – 1 electrolytes

• LiAlH₄ and NaBH₄ cannot reduce C = C

• H₂/Pd reduce alkene to alkane

Due to an increase in nuclear charge which accompanies the filling of the inner d-orbitals, there is an increase in ionisation enthalpy along each series of the transition elements, from left to right. However, many small variations occur.

• According to molecular orbital theory, the sigma (σ) molecular orbitals are symmetrical around the bond-axis while pi (π) molecular orbitals are not symmetrical.

• As z–axis is taken as internuclear axis, 2pₓ and 2p_y forms π molecular orbitals, hence these are unsymmetrical around the bond-axis

Tranquilizers and analgesics are neurologically active drugs. Equanil is used in controlling depression and hypertension.

• The wave function is a mathematical function whose value depends coordinates of the electron in an atom and does not carry any physical significance. So, Statement I is true.

• The probability of finding an electron at a point within an atom is proportional to the square of the orbital wave function i.e. |ψ|² at the point. So, Statement II is false.

• Chloroform (b.p. 334 K) and aniline (b.p. 457) are easily separated by the technique of distillation as both the liquids having sufficient difference in their boiling points.

• The fractions of crude oil in petroleum industry is separated using fractional distillation due to less difference in boiling points of liquids.

• Glycerol can be separated from spent–dye using distillation under reduced pressure.

• Aniline is separated by steam distillation from aniline-water mixture.

• Mn₂O₇, SO₂ and TeO₃ are acidic oxides.

• GeO is distinctly acidic in nature.

• NO₂ is acidic in nature.

Relation between Molality (m), Molarity (M) and density (d) is

m = 1000M / (1000 × d - M × M_solute)

M_solute denotes molar mass of solute.

m = (1000 × 1) / ((1000 × 1.25) - (1 × 85)) = 1000 / (1250 - 85) = 1000 / 1165

m = 0.858 m

High density polythene is formed by addition polymerisation of ethene in the presence of catalyst triethyl aluminium and titanium tetrachloride (Ziegler Natta Catalyst) (TiCl₄ and Al(C₂H₅)₃ Cl) So, Statement I is correct. High density polymers are chemically inert. So, Statement II is also correct.

Hydrogenation of Acyl chloride in the presence of catalyst palladium on barium sulphate to form aldehyde is known as Rosenmund reaction.

• (A) NaH → Saline or Ionic hydride

• (B) PH₃ → Electron rich hydride (Elements of group 15-17 form such hydrides)

• (C) GeH₄ → Electron precise (Elements of group 14 form such hydrides)

• (D) LaH₂.₈₇ → Metallic or Non-stoichiometric hydrides

For any value of l, mₗ value will be -l to +l

So for n = 5 and l = 2: mₗ = -2, -1, 0, +1, +2 and mₛ = +1/2 or -1/2

Hence,

the set n = 5, l = 2, mₗ= -2, -1, +1, +2, mₛ = +1/2 is incorrect.

Lithium and Beryllium are the first members of their respective groups (i.e. group 1 and 2 respectively). Due to their exceptionally small atomic and ionic sizes they form compounds with covalent character not ionic. Hence, Assertion statement is wrong. Lithium and Magnesium have similar properties due to diagonal relationship hence, reason statement is correct.

E°_cell = E°_cathode - E°_anode = 1.33 - (-0.44) = 1.33 + 0.44 = +1.77 V

Acidic buffer is a mixture of weak acid and its conjugate base or salt of that weak acid with another strong base in a molar ratio ranging from 1/10 to 10 e.g.,

CH₃COOH + CH₃COONa

(Weak acid) (Salt of weak acid with strong base NaOH)

Amides are not basic in nature because of non availability of lone-pair on nitrogen, protonated amide will be even weaker bases.

van der Waal equation for n mole gas is

(p + an²/V²) (V - nb) = nRT

For n = 1 mol

(p + a/V²) (V - b) = RT

(1) Al⁺ is unstable in solution because SRP of Al⁺/Al is positive hence, it will get reduced in solution.

(2) Tl/Tl⁺ = +0.34 (SOP)

Tl/Tl³⁺ = -1.26 (SOP)

Since, SOP Tl/Tl⁺ > Tl/Tl³⁺

Hence, Tl will get oxidised to Tl⁺ easily than Tl³⁺

(3) Since Al/Al³⁺ has more SOP than Tl/Tl⁺ it will be more electropositive

(4) Since Tl³⁺ + 2e⁻ → Tl⁺; E = (3 × 1.26 + 0.34) / 2 = +2.04 Hence, Tl³⁺ is better oxidising agent not reducing agent.

Since hydrated chlorides and bromides of Ca, Sr and Ba are ionic in nature they will get dehydrated on heating and hydrolysis will not take place, while chloride and bromide of Be and Mg has covalent nature hence, they will get hydrolysed.

K₂PtCl₂ · 2NH₃ is complex compound

KAl(SO₄)₂ · 12H₂O is empirical formula of potash alum which is a double salt K₂SO₄ · Al₂(SO₄)₃ · 24H₂O

For reaction, 3A → 2B

Average rate of reaction = - (1 / 3) (Δ[A] / Δt) = (1 / 2) (Δ[B] / Δt)

So, Δ[B] / Δt = - (2 / 3) (Δ[A] / Δt)

HA ⇌ H⁺ + A⁻

At t = 0: C - -

At eq.: C - Cα, Cα Cα

K = ([H⁺][A⁻]) / [HA] = (Cα)(Cα) / (C(1 - α)) = Cα² / (1 - α) = Cα²

(∵ α is very small) = 0.1 (0.01)² = 1 × 10⁻⁵

• Methylene Blue sol is a positively charged sol.

• Silver sol, Sb₂S₃ sol and congo red sol are negatively charged sol.

KCN is predominately ionic and provides cyanide ions in solution. Although both carbon and nitrogen atoms are in a position to donate electron pairs, the attack takes place mainly through carbon atom and not through nitrogen atom since C – C bond is more stable than C – N bond. However, AgCN is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as the major product

R – X + AgCN → R – NC + AgX

More is the stability of alkene formed, more will be the rate of α–β dehydrohalogenation

Solvate isomerism differ by whether or not a solvent molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice.

• [Cr(H₂O)₆]Cl₃, [Cr(H₂O)₅Cl]Cl₂ · H₂O and [Cr(H₂O)₄Cl₂]Cl · 2H₂O show solvate isomerism.

• Co–ordination number of Cr is 6

In an organic compound when inductive and electromeric effects operate in opposite directions, the electromeric effect predominates.

In alkylarene, sp³ hybridized C – H bonds are present the carbon of which is directly bonded to sp² carbon atoms so, hyperconjugation effect is observed in o-xylene.

I

For a ccp structure,

4r = a√2

∴ a = 4r / √2

a = 2√2r

Photochemical smog is harmful to humans as well as plants so option (c) is incorrect.

Photochemical smog occurs in warm, dry and sunny climate.

As in a ccp unit cell having 'N' number of atoms there are '2N' number of tetrahedral voids, so number of tetrahedral voids formed in a 5 mol of a compound having ccp structure is 5 × 2 × 6.022 × 10²³ = 6.022 × 10²⁴ voids.

A redox couple is defined as having together the oxidised and reduced forms of a substance taking part in an oxidation or reduction half reaction.

2H₂O(g) → 2H₂(g) + O₂(g)

ΔH° = 483.64 kJ

For decomposition of 2 moles of H₂O, enthalpy change = 483.64 kJ For decomposition of 1 mole of H₂O, enthalpy change = 483.64 / 2 kJ = 241.82 kJ

Chrome steel is used for cutting tools and crushing machines while fine dust of aluminium is used in paints and lacquers. Zinc dust is a reducing agent for manufacture of dye–stuffs and paints. So, statements A, B and D are correct. Copper is used for oxidation of primary and secondary alcohols into aldehydes and ketones respectively at high temperature. So, statement (C) is incorrect. For galvanisation of iron, zinc is used to prevent iron from corrosion so, statement (E) is also incorrect. Hence, correct statements are (A), (B) and (D) only.

Lesser is the number of particles, lower is the elevation in boiling point

Ionisation enthalpies of early actinoids are lower than early lanthanoids. This is because 5f orbitals will penetrate less into the inner core of electrons. The 5f electrons, will be more effectively shielded from the nuclear charge than the 4f electrons of corresponding lanthanoids.

Major product formed will be a conjugated alkene.

At anaphase the doubling of chromosomes takes place. During this phase the separation of sister chromatids occurs, since mitotic apparatus is disrupted soon after metaphase therefore option (b) is correct.

(A)-(III), (B)-(II), (C)-(IV), (D)-(I) Chlorophyll a - Blue green

Chlorophyll b - Yellow green

Xanthophyll - Yellow

Carotenoid - Yellow to yellow orange.

Phaeophyceae and Rhodophyceae classes of algae possess pigment fucoxanthin and pigment phycoerythrin, respectively.

Ligation of foreign DNA at BamH I site will result in loss of tetracycline resistance of pBR322.

Mutualism is a type of interaction in which both the organisms are benefitted from each other. Since plants offer rewards to animals in the form of pollen and nectar and the animals facilitate the pollination process so this is an example of mutualism.

In some members of Leguminosae, Solanaceae and Rosaceae sets of families, the pollen grains are viable for months.

Such type of species of the plants that plays a vital role in controlling the relative abundance of other species in a community is known as Keystone species. They play important role in the maintenance of the species and function of an ecosystem.

Diplotene is a fourth stage of the prophase of meiosis. During this stage of the chromosome synapsis dissolves. However, chiasmata remain attached.

(A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Hydrarch succession → Phytoplankton to mesic conditions.

Xerarch succession → Lichens to mesic conditions.

Ecological succession → Gradual change in the species composition.

Secondary succession → Faster and climax reached quickly.

Paul Ehrlich used the "rivet popper hypothesis" and considered Airplane as ecosystem and rivets used to join all parts together is considered as species. Thus he compared rivets in an airplane to species within a genus.

(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Monohybrid cross →3: 1

Dihybrid cross → 9 : 3 : 3 : 1

Incomplete dominance → 1 : 2 : 1

Test cross → 1 : 1

(A)-(IV), (B)-(I), (C)-(II), (D)-(III)

Protein → Peptide bond.

Unsaturated fatty acid → C = C double bonds.

Nucleic acid → Phosphodiester bond.

Polysaccharide → Glycosidic bonds.

Phototropism is the phenomenon which is influenced by auxin and also played a major role in its discovery. Charles Darwin and his son Francis Darwin observed that the coleoptiles of canary grass responded to unilateral illumination by growing towards the light source (phototropism). Auxin was isolated by F.W. Went from tips of coleoptiles of oat seedlings.

When sepals or petals in a whorl just touch one another at the margin, without overlapping, said to be valvate such as in Calotropis.

Both statements are correct.

RuBisCO is the most abundant enzyme in the world. Photorespiration does not occur in C₄ plants. It takes place in C₃ plants.

The correct sequence of tissues in the stellar region of the stem that show secondary growth from pith towards cortex is:

Primary xylem → Secondary xylem → Secondary phloem → Primary phloem.

(A), (B), (D) only

According to the Mass flow hypothesis, the statements A, B and D are correct while statement C is incorrect.

Mass flow is the movement of substances in bulk or en masse from one point to another as a result of pressure differences between the two points. It is a characteristic of mass flow that substances, whether in solution or in suspension, are swept along at the same pace, as in flowing river. Bulk flow can be achieved either through a positive hydrostatic pressure gradient (e.g., a garden hose) or a negative hydrostatic pressure gradient (e.g., suction through a straw).

Fatty acids are connected with the respiratory pathway through Acetyl CoA.

Inulin is a polymer of fructose.

Genotype of heterozygous violet flowers is Ww.

Genotype of homozygous white flower is ww.

According to the cross, the white and violet flowers will be produced in equal proportion that is (20 : 20) out of 40 progenies.

The expected ratio of violet and white flower plant is 1 : 1.

The amount of nutrients such as carbon, nitrogen, potassium and calcium present in the soil at any given time is referred to as Standing state.

Thermostable DNA polymerase used in PCR was isolated from Thermus aquaticus. It remains active during the high temperature induced denaturation of double stranded DNA.

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Pteropsida → Adiantum

Lycopsida → Selaginella

Psilopsida → Psilotum

Sphenopsida → Equisetum

When addition of polluted water with high BOD take place in fresh water lake then the lake gets contaminated and amount of dissolved oxygen in the lake will decrease.

Eutrophication is the natural aging of a lake by nutrient enrichment of its water. In a young lake the water is cold and clear, supporting little life. With time, streams draining into the lake introduce nutrients such as nitrogen and phosphorus, which encourage the growth of aquatic organisms and finally converting into land. Thus nitrates and phosphates flowing from agricultural farms into water bodies are a significant cause of Eutrophication.

Transfer of pollen grain from anther to stigma of another flower of same plant is known as Geitonogamy.

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Auxin → Promotes flowering in pineapple.

Gibberellin → Increase in the length of grape stalks.

Cytokinin → Overcoming apical dominance.

Ethylene → Promotes female flower formation in cucumber.

The TCA cycle starts with the condensation of acetyl group with oxaloacetic acid (OAA) and water to yield citric acid. The reaction is catalysed by the enzyme citrate synthase and a molecule of CoA is released.

Citrate is then isomerised to isocitrate. During this cycle two successive steps of decarboxylation takes place that leads to the formation of α-ketoglutaric acid and then succinyl-CoA. During this cycle, first decarboxylation occurs when isocitrate is converted into α-ketoglutarate and second decarboxylation occurs when α-ketoglutarate is converted into succinyl CoA.

The Calotropis is a type of weed that grows in abandoned fields. The plant produces highly poisonous cardiac glycosides that is why never any cattle or goats browsing on this plant.

Calotropis, commonly called milkweed, produces very toxic chemicals called cardiac glycosides that affect the heart. Hence, cattle avoid browsing this plant.

Repressor protein is the component that binds to the operator region of an operon and prevents RNA polymerase from transcribing the operon.

The last chromosome sequenced in Human Genome Project was Chromosome 1. The sequence of chromosome 1 was completed only in May 2006. This was the last of the 24 human chromosomes - 22 autosomes and X and Y - to be sequenced.

In sickle cell anaemia disorder change of single base pair in the gene for beta globin chain results in change of glutamic acid to valine.

The monocot root has epidermis, cortex, endodermis, pericycle, vascular bundles and pith. There are usually more than six (polyarch) xylem bundles in the monocot root. Pith is large and well developed.

In pea and bean flowers, there are five petals, the largest (standard) overlaps the two lateral petals (wings) which in turn overlap the two smallest anterior petals (keel).

During symport two different molecules move across the membrane in same direction with the help of same carrier.

The deficiency symptoms tend to appear first in the young tissues whenever the elements are relatively immobile and are not transported out of the mature organs, for example, element like sulphur and calcium are a part of the structural component of the cell and hence are not easily released.

(A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Nitrococcus - Conversion of ammonia to nitrate.

Rhizobium - Conversion of atmospheric nitrogen to ammonia.

Thiobacillus - Denitrification.

Nitrobacter - Conversion of nitrite to nitrate.

During the formation of the female gametophyte, first of all the megaspore mother cell (MMC) undergoes meiotic division. In the next step, in a majority of flowering plants, one of the megaspores is functional while the other three degenerate. Then the nucleus of the functional megaspore divides mitotically to form two nuclei which move to the opposite poles, forming the 2-nucleate embryo sac. Two more sequential mitotic nuclear divisions result in the formation of the 4-nucleate and later the 8-nucleate stages of the embryo sac. After the 8-nucleate stage, cell walls are laid down leading to the organisation of the typical female gametophyte or embryo sac. So the correct sequence is the statements of (C), (B), (A), (D), (E).

(A)-(II), (B)-(III), (C)-(I), (D)-(IV)

Kanamycin - Selectable marker

ClaI - Restriction site

Disarmed retroviruses - Deliver genes into animal cell. Kanamycin Rgene. - Antibiotic resistance.

Pteridophytes are the first terrestrial plants to possess vascular tissues - xylem and phloem. All pteridophytes exhibit haplo-diplontic pattern of lifecycle.

The type of meristem that occurs in the mature regions of roots and shoots of many plants, particularly those that produce woody axis and appear later than primary meristem is called the secondary or lateral meristem. They are cylindrical meristems. Fascicular vascular cambium, interfascicular cambium and cork-cambium are examples of lateral meristems.

6 is correct answer.

In the given statements, statement I is incorrect and the statement II is correct. The correct statement is- In rooted plants, transport in xylem (of water and minerals) is essentially unidirectional, from roots to the stems. Organic and mineral nutrients however, undergo multidirectional transport in phloem.

In the given statements, only statement 1 is incorrect and the rest are correct. The correct statement is- Plant growth is generally indeterminate.

The statement that is incorrect about Agrobacterium tumifaciens is statement

. The correct statement is that Agrobacterium tumifaciens is used to deliver gene of interest into a variety of plant i.e., in eukaryotic host cells and not in any prokaryotic cell. All other given statements are correct about Agrobacterium tumifaciens.

Both the statements are correct. The process of copying genetic information from one strand of the DNA into RNA is termed as transcription. A transcription unit in DNA is defined primarily by the three regions in the DNA: (i) A Promoter (ii) The Structural gene (iii) A Terminator

Hershey and Chase used 32P and 35S labelled phages for their experiment to discover whether it was protein or DNA from the viruses that entered the bacteria. They grew some viruses on a medium that contained radioactive phosphorus and some others on medium that contained radioactive sulfur. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. Similarly, viruses grown on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulfur.

In the given statements Cyclic photophosphorylation involves both PS I and PS II is only correct.

The correct statement is- In cyclic photophosphorylation only PS I is functional. In this system the electron is circulated within the photosystem and the phosphorylation occurs due to cyclic flow of electrons. This process takes place in the stroma lamellae.

During anaphase stage of mitosis and anaphase II stage of meiosis the centromere of each chromosome splits and moves towards the opposite poles.

(A)-(III), (B)-(I), (C)-(II), (D)-(IV)

Zygomorphic - Cassia

Hypogynous - Mustard

Perigynous - Plum

Epigynous - Cucumber

In the given statements, only statements (A) and (B) are correct about the hormone and its functions. Statements (C), (D) and (E) are incorrect. The correct statements are- Goitre is a form of hypothyroidism. Glucocorticoids are secreted in Adrenal Cortex. The parathyroid glands secrete a peptide hormone called parathyroid hormone (PTH). The secretion of PTH is regulated by the circulating levels of calcium ions.

Binomial nomenclature has two components - the Generic name and the specific epithet. This naming system was given by Carolus Linnaeus. Thus, X - Generic name, Y - Specific epithet, Z - Carolus Linnaeus.

The sequences of respiration are:- (D) Pulmonary ventilation by which atmospheric air is drawn in and CO₂ rich alveolar air is released out → (A) Diffusion of gases (O₂ and CO₂) across alveolar membrane → (C) Transport of gases by the blood → (B) Diffusion of O₂ and CO₂ between blood and tissues→ (E) Utilisation of O₂ by the cells for catabolic reactions and resultant release of CO₂.

The accelerated rates of species extinctions that the world is facing now are largely due to human activities. The four major causes of loss of biodiversity are- Habitat loss and fragmentation Over-exploitation Alien species invasions Co-extinctions

(A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Terpenoides→ Diterpenes

Lectins → Concanavalin A

Alkaloids → Codeine

Toxins→ Ricin

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Palm bones → Metacarpals

Wrist bones → Carpals

Ankle bones → Tarsals

Digit bones → Phalanges

Gonorrhoea, syphilis, genital herpes, chlamydiasis, genital warts, trichomoniasis, hepatitis-B and HIV infection are sexually transmitted infections. Except for hepatitis-B, genital herpes and HIV infections, other diseases are completely curable if detected early and treated properly.

Restriction enzyme is a type of enzyme that is used to cut the DNA molecule near a specific sequence of bases. So it is also called as molecular scissors.

Prophase I of the first meiotic division is typically longer and more complex. It has been further subdivided into the following five phases based on chromosomal behaviour, i.e., Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis. So the correct sequence of substages of Prophase-I of meiotic division is (D), (A), (B), (E), (C).

Both the statements are correct. RNA being unstable, mutate at a faster rate. Consequently, viruses having RNA genome and having shorter life span mutate and evolve faster. RNA can directly code for the synthesis of proteins, hence can easily express the characters.

Both the statements are correct.

Goblet cells are unicellular glands.

Earwax is the secretion of exocrine gland.

In the question initial population = 250, final population = 2500 and time = 1 year are given.

So change in population = Final population − Initial population

= 2500 − 250 = 2250

So birth rate = 2250/250 × 1 = 9.

Statement II is correct and Statement I is incorrect.

Each primary oocyte gets surrounded by a layer of granulosa cells and is called the primary follicle. The primary follicle gets surrounded by more layers of granulosa cells and a new theca layer and are called secondary follicles. The secondary follicle soon transforms into a tertiary follicle which is characterised by a fluid filled cavity called antrum.

House fly belongs to the family of Muscidae.

Housefly belongs to the family Muscidae, class Insecta and phylum Arthropoda.

Smack is chemically diacetylmorphine which is a white, odourless, bitter crystalline compound. This is obtained by acetylation of morphine.

In the given statements, statements (A), (B) and (D) only are correct while statement (C) is incorrect.

Certain cells do not divide further etc. G₁ phase to enter an inactive stage called quiescent stage (G₀) of the cell cycle. Cells in this stage remain metabolically active but no longer proliferate unless called on to do so depending on the requirement of the organism.

(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

P-wave → Depolarisation of atria.

QRS complex → Depolarisation of ventricles.

T-wave → Repolarisation of ventricles.

End of T-wave → End of systole.

Alkaloids such as morphine, cannabinoids, rubber, essential oils, antibiotics, coloured pigments, scents, gums, spices. These are called secondary metabolites. Lecithin is not a secondary metabolite. It is a phospholipid that is found in cell membrane.

Statement I is incorrect but Statement II is true.

The correct statement is that Intra Cytoplasmic Sperm Injection (ICSI) is the another specialised procedure of in-vitro fertilization to correct infertility cases due to the inability of the male partner to inseminate the female or due to very low sperm counts in the ejaculates, could be corrected by artificial insemination.

A-(II), B-(III), C-(IV), D-(I)

Eosinophils → 2-3%

Lymphocytes → 20-25%

Neutrophils → 60-65%

Monocytes → 6-8%

In the given question only statement (c) is incorrect about chemical structure of insulin. The correct statement is that the mature insulin does not contain C-peptide chain which is removed during maturation of pro-insulin into insulin. Thus it has only two polypeptide chains that are only A and B chains.

In the given question, both assertion and reason are true but reason is not the correct explanation of assertion. Ascending limb of loop of Henle is impermeable to water and allows transport of electrolytes actively or passively. Dilution of filtrate takes place due to efflux of electrolytes in the medullary fluid.

A-(II), B-(III), C-(IV), D-(I)

Typhoid → Salmonella

Elephantiasis → Aschelminthes

Ringworm → Microsporum

Malaria → Protozoan

Insects such as cockroaches excrete nitrogenous wastes as uric acid and thus are called uricotelic.

There are three major regions that make up the brain stem; mid brain, pons and medulla oblongata. Brain stem forms the connections between the brain and spinal cord.

(A)-(III), (B)-(I), (C)-(IV), (D)-(II).

Non-medicated IUDs → Lippes loop

Copper releasing IUDs → Multiload 375

Hormone releasing IUDs → LNG-20

Vaults → Rubber barrier

Lactose is the substrate for the enzyme beta-galactosidase and it regulates switching on and off of the operon. So it is termed as inducer.

The given characteristics are characteristics of the Neanderthal man that has a brain size of 1400cc, lived in near east and central Asia between 1,00,000-40,000 years back. They used hides to protect their body and buried their dead.

The correct sequence of different hormones for their role during gametogenesis is:-(B) Gonadotropin releasing hormone from hypothalamus → (E) Gonadotrophin from anterior pituitary gland → (A) Gonadotropin LH stimulates synthesis and secretion of Androgen → (C) Androgen stimulates spermatogenesis → (D) Gonadotropin FSH helps in the process of spermiogenesis.

Both Statement I and Statement II are correct. In bacteria, the mesosomes are formed by the extensions of plasma membrane. The mesosomes, in bacteria, help in DNA replication and cell wall formation.

(A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Deforestation → Conversion of forested areas to non-forested areas.

Reforestation → Process of restoring a forest that once existed but was removed.

Green-house effect → Responsible for heating earth's surface and atmosphere.

Eutrophication → Natural ageing of lake by nutrient enrichment of its water.

In the given statements about sickle cell anaemia only statement (A) and (C) are correct. Sickle cell anaemia occurs due to change in a single base pair of DNA. This is an example of point mutation. The defect is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the beta globin chain of the haemoglobin molecule. The substitution of amino acid in the globin protein results due to the single base substitution at the sixth codon of the beta globin gene from GAG to GUG.

Statement I is true but Statement II is false. The adult human eye ball is nearly a spherical structure. The wall of the eye ball is composed of three layers. The external layer is composed of a dense connective tissue and is called the sclera. The anterior portion of this layer is called the cornea. The middle layer is known as choroid.

(A)-(II), (B)-(I), (C)-(III), (D)-(IV).

Contractile vacuole → Amoeba

Water vascular system → Asterias

Canal system → Spongilla

Flame cells → Taenia

The correct sequence of the events of Renin-Angiotensin mechanism is:-(C) Fall in glomerular blood pressure' !(A) Activation of J.G cells and release of renin' !(E) Angiotensinogen is converted to Angiotensin I and then to Angiotensin II ' !(B) Angiotensin II activates release of aldosterone ! (D) Reabsorption of Na+ and water from distal convoluted tubule.

Only statement (B), (C) and (D) are correct. Ctenophores reproduce only sexually and fertilization is external. In tapeworm, fertilization is internal but sexes are not separate. Ctenophores are exclusively marine, diploblastic and bioluminescent organisms.

The correct sequential steps regarding absorption of fatty acids and glycerol, in intestine is:-(C) Fatty acid and glycerol are incorporated into small droplets called micelles !(B) Micelles move into intestinal mucosa !(A) Micelles are reformed into small protein coated fat globules called chylomicrons !(E) Chylomicrons are transported into lacteals !(D) Lacteals release the absorbed substances into blood stream.

Both Statement I and Statement II are false. The correct statatements are:- In cockroach forewings (mesothoracic) called tegmina are opaque dark and leathery and cover the hind wings when at rest. The hind wings are transparent, membranous and are used in flight.

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Cytokine barriers ! Interferons

Cellular barriers ! Neutrophils and macrophages Physiological barriers ! Tears and saliva

Physical barriers ! Mucus coating of respiratory tract

According to Hershey and Chase experiment only statement (A) and (B) are corrects. Hershey and Chase grew some viruses on a medium that contained radioactive phosphorus and some others on medium that contained radioactive sulfur. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. Similarly, viruses grown on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.

Only statements (C) and (D) are correct about genetic code. The characteristics of genetic code are:- The codon is triplet. Some amino acids are coded by more than one codon, so the code is degenerate. The codon is read in mRNA in a contiguous fashion. The code is nearly universal. AUG has dual functions. It codes for Methionine (met), and it also act as initiator codon. UAA, UAG, UGA are stop terminator codons.

The correct statements regarding dissolved Oxygen and biochemical oxygen demand are statements (A), (B), (D), and (E) only. BOD is inversely related to dissolved oxygen. Low dissolved oxygen and high BOD lead to loss of aquatic life. High BOD leads to high dissolved oxygen. Both BOD and dissolved oxygen are indicator of health of a water body. Both BOD and dissolved oxygen are affected by amount of organic matter in the water body.

In Multiple Ovulation Embryo Transfer (MOET) Technology, the fertilised eggs at 8–32 cells stages, are recovered non-surgically and transferred to surrogate mothers.

(A)-(IV), (B)-(III), (C)-(II), (D)-(I).

Gene therapy ! Allows correction of a gene defect.

RNA interferance → Cellular defence.

ELISA ! Diagnostic test for AIDS.

Gel Electrophoresis ! Separation of DNA fragments.

Statement I is correct but Statement II is false. The correct statement is :- Parathyroid hormone(PTH) along with TCT plays a significant role in calcium balance in the body.

The correct statements with respect to mechanism of sex determination in Grasshopper are only statements (B) and (D). Male Grasshopper produces two different types of gametes either with or without X chromosome. All eggs bear an additional X chromosome besides the autosomes.

(A)-(II), (B)-(III), (C)-(IV), (D)-(I).

Columnar epithelium → Inner lining of stomach and intestine.

Ciliated epithelium → Inner lining of bronchioles Squamous epithelium → Endothelium

Cuboidal epithelium → Ducts of glands.

Both assertion and reason are true and reason is the correct explanation of assertion. Some organisms possess adaptations that are physiological which allow them to respond quickly to a stressful situation. When a person goes to high altitude experiences altitude sickness. Its symptoms include nausea, fatigue and heart palpitations. This is because in the low atmospheric pressure of high altitudes, the body does not get enough oxygen.

Only statements (A) and (C) are correct about Golgi apparatus. Golgi apparatus is the important site of formation of glycoprotein and glycolipids. It modifies the protein synthesized by ribosomes on ER.