NEET-UG 2023 Question Paper Solutions

Since, eV = hc / λ_min => λ_min = hc / eV ∴ λ_min ∝ 1 / V NOTES: X ray photo of maximum energy is emitted when an electron loses its whole energy during a single collision with the atom

Given, efficiency η = 50 / 100 = 1 / 2 Source temperature, T₁ = 327 + 273 = 600 K Efficiency of Carnot engine η = 1 - T₂ / T₁ => η = 1 - T₂ / 600 => 1 / 2 = 1 - T₂ / 600 T₂ / 600 = 1 / 2 => T₂ = 300 K ∴ Sink temperature, T₂ = 300 - 273 = 27°C

Maximum height, H = u² sin² θ / 2g H = (280)² × (sin² 30) / (2 × 9.8) = 280 × 280 × 0.5 × 0.5 / (2 × 9.8) ∴ H = 1000 m

For resonance frequency f = 1 / (2π √(LC)) => f = 1 / (2 × π × √(10 × 10⁻³ × 1 × 10⁻⁶)) = 10⁴ / 2π = 10⁴ / (2 × 22/7) = 1.591 × 10³ ∴ f = 1.591 kHz

Statement I : Photovoltaic devices convert optical radiation into electricity. II : Zener diode is designed to operate under reverse bias in breakdown region. NOTES: Zener diodes are not damaged by high reverse current.

µ₂ sin i_c = µ₁ => sin i_c = µ₁ / µ₂ µ = c / V ∴ sin i_c = µ₁ / µ₂ = V₂ / V₁ sin i_c = (10x / t₂) / (x / t₁) ∴ i_c = sin⁻¹(10t₁ / t₂)

Using 1 / λ = R [ 1 / n₂² - 1 / n₁² ] For Balmer series, [n₂ = 2, n₁ = ∞] 1 / λ = R [ 1 / 4 - 1 / ∞ ] => λ = 4 / R ... (i) And for bracket series, (n₂ = 4, n₁ = ∞) 1 / λ' = R [ 1 / 16 - 1 / ∞ ] => λ' = 16 / R ... (ii) Dividing eq (i) by (ii)

λ / λ' = (4 / R) / (16 / R) ∴ λ' = 4λ

Since galvanometer does not show any deflection ∴ i_g = 0 From the circuit diagram, (10 - 2) / 400 = 2 / R ∴ R = (2 × 400) / 8 = 100 Ω

Amount of energy required to form a soap bubble = [S × ΔA] × 2 = [0.03 × 4 × π × 4 × 10⁻⁴] × 2 = 3.015 × 10⁻⁴ J

Magnetic energy stored in an inductor U = 1/2 L i² = 1/2 × 4 × 10⁻⁶ × (2)² = 8 × 10⁻⁶ J = 8 µJ

As transfer to be ideal and for ideal transformer P_input = P_output ∴ (V I)_in = 60 => 220 × I = 60 ∴ I = 0.27 A

Given E = 2 × 10⁵ N/C, l = 2 cm and τ = 4 Nm̅ τ̅ = p̅ × E̅ = p E sin θ

4 = p × 2 × 10⁵ × sin 30° => p = 4 × 10⁻⁵ c m ∴ Charge on the dipole, q = p / l = (4 × 10⁻⁵) / 0.02 = 2 mC

Average speed, v_avg = 2 v₁ v₂ / (v₁ + v₂) = (2 × v × 2v) / (v + 2v) = 4v / 3

Longitudinal stress = Internal restoring force / Area = F_ext / Area ∴ Stress = W / A

Since, ϕ_net = ∮_s E . dS = 0 i.e., Net flux through surface is zero, so number of flux lines entering the surface must be equal to the number of flux lines leaving it. NOTES: In the expression ∮_s E . dS = 0 the electric field is the sum of electric field due to charge within the surface and due to charge outside the surface.

Energy of incident radiation = 2.20 eV Work function of Cs, ϕ_Cs = 2.14 eV and ϕ_K = 2.30 eV ϕ_Na = 2.75 eV Since the work function of caesium, ϕ_Cs < ϕ_incident. Hence photons may be emitted from caesium.

RMS speed, v_rms = √(3RT / m) => v_rms ∝ √T v_rms is increased by 3 times, So, final rms speed = v + 3v = 4v T₁ = 273 - 50 = 223 K and T₂ = ? v₁ / v₂ = √(T₁ / T₂) => v / 4v = √(223 / T₂) => 1 / 16 = 223 / T₂ ∴ T₂ = 3568 K or, T₂ = 3568 - 273 = 3295°C

Fundamental harmonic produced by open pipe, f_open pipe = v / 2l And by closed pipe, f_closed pipe = v / 4l ∴ f_o / f_c = (v / 2l) × (4l / v) = 2 / 1 ∴ f_o : f_c = 2 : 1

For Young's double slit experiment, angular fringe width α = λ / d It does not depend on the distance of screen from the slit, so statement I is correct. Angular fringe width ∝ λ So if λ increases, angular separation of fringes increases. Therefore, statement I is true and statement II is false.

For parallel grouping, C_eq = C₁ + C₂ + ..... C_n C₁ = 3 + 3 = 6 µF For series grouping, 1 / C_eq = 1 / C₁ + 1 / C₂ + ..... 1 / C_n C_eq = C₁C₂ / (C₁ + C₂) = 3 × 6 / (3 + 6) = 18 / 9 = 2 µF

Let gravitational field at point Q be zero ∴ Gm / x² = G(9m) / (R - x)² (R - x)² / x² = 9 => x = R / 4 V_Q = -Gm / x - G(9m) / (R - x) V_Q = -Gm / (R / 4) - G(9m) / (3R / 4) = -4Gm / R - 12Gm / R = -16Gm / R

Venturi-meter works on the Bernoulli's principle.

N = N₀ / 2ⁿ => N / N₀= 1 / 2ⁿ => 1 / 16 = 1 / 2ⁿ => 1 / 2⁴ = 1 / 2ⁿ ∴ n = 4 n = t / T₁∴ t = 4 × T₁/₂ = 4 × 20 = 80 minutes

Change in momentum, ΔP vector = vector P_f - vector P_î

vector P_f = m uî

vector P_î = m u (-ĵ) ΔP = m uî - m u (-ĵ) ΔP = m u (î + ĵ)

vector F = ΔP vector / Δt

Direction of change of momentum and direction of force acting on the player will be same i.e., along north-east.

NOTES: Random errors is also called chance error. It occurs due to parameter which are beyond the control of experimenter.

Using the Right hand screw rule, angular acceleration of a body, moving along the circumference of a circle is along the axis of rotation.

Capacitor is used to remove the ac ripple from the rectified output.

Net resistance of the circuit, R_eq = 2 Ω + 1 Ω + 7 Ω = 10 Ω Net Voltage, V_net = 10 - 5 = 5 V i = V_net / R_eq = 5 / 10 = 0.5 A

Capacitive reactance X_C = 1 / ωC = 1 / 2πνC Since ν decreasing, X_C will increase ∴ Current, I = E₀ / X_C will decrease As conduction current = displacement current Therefore displacement current will decrease.

For closed loops Magnetic flux = ∮ vector B . ds vector = zero Magnetic monopole doesn't exist. Hence net magnetic flux through any closed surface is zero.

Given, Amplitude of oscillating electric field component, E₀ = 48 Vm⁻¹ Using, c = E₀ / B₀ Here, B₀ = Amplitude of oscillating magnetic field ∴ B₀ = 48 / (3 × 10⁸) = 16 × 10⁻⁸ = 1.6 × 10⁻⁷ T

Potential energy stored in spring (U) is given by U = 1/2 Kx² Initially U_i = 1/2 K(2)² where x = 2 cm => U_i = 1/2 (K) . (4) = 2 K ... (i) Finally, U_f = 1/2 K(8)² = 1/2 K × 64 = 32 K ... (ii) On dividing (i) by (ii) U_i / U_f = 2K / 32K = 1 / 16 => U_f = 16 U_i

Using third equation of motion between I and II v² - u² = 2as => (u / 3)² = u² - 2a × 24 => 2a(24) = 8u² / 9 ... (i) Using third equation of motion again between II and III

0 = (u / 3)² - 2as => u² / 9 = 2as ... (ii) From equation (i) and (ii) 2a × 24 = 8(2as) => s = 3 cm

Length of wooden block = 24 + 3 = 27 cm

Given, Radius of inner most orbit of hydrogen atom, r₁ = 5.3 × 10⁻¹¹ m Radius of nᵗʰ orbit r_n = n² / Z => r_n ∝ n² ∴ r₁ / r₂ = (n₁ / n₂)² => r₁ / r₂ = (1 / 3)² => r₂ = 9r₁ = 5.3 × 10⁻¹¹ × 9 = 47.7 × 10⁻¹¹ = 4.77 Å

Inductance, L = 50 / π mH Resistance, R = 10 Ω Capacitance, C = 10³ / π × 10⁻⁶ Inductive reactance, X_L = ωL = 2πfL = 2π × 50 × (50 / π) × 10⁻³ = 5 Ω Capacitive reactance, X_C = 1 / ωC = 1 / 2πfC = (1 × π) / (2π × 50 × 10³ × 10⁻⁶) = 10³ / 100 = 10 Ω Impedance Z = √((X_C - X_L)² + R²) = √((10 - 5)² + 10²) => Z = √125 = 5√5 Ω

Position of particle as function of time x = A sin ωt

From figure, at t = 2 s x = 1 Time period, T = 8 s ω = 2π / T = 2π / 8 = π / 4 a = -ω²x = -(π / 4)² × 1 => a = -π² / 16 m/s⁻²

NOTES: In SHM, we cannot apply equation of translatory motion as acceleration is not (a = ω²x) constant.

From the lens maker's formula 1 / f = (µ - 1)(1 / R₁ - 1 / R₂) ∴ 1 / f₁ = (1.6 - 1)(1 / ∞ - 1 / 20) = -0.6 / 20 = -3 / 100 ∴ 1 / f₂ = (1.5 - 1)(1 / 20 - 1 / -20) = 0.5 / 10 = 1 / 20 ∴ 1 / f₃ = (1.6 - 1)(1 / -20 - 1 / ∞) = -0.6 / 10 = -3 / 100 Effective focal length (f_eff) is 1 / f_eff = 1 / f₁ + 1 / f₂ + 1 / f₃ => 1 / f_eff = -3 / 100 + 1 / 20 - 3 / 100 => 1 / f_eff = -1 / 100 => f_eff = -100 cm

Electrostatic potential due to a point charge is given by V = Kq / r ∴ V_net = Kq / (2 × 10⁻²) - Kq / (8 × 10⁻²) => V_net = (Kq × 10²) / 2 (1 - 1 / 4) = (3/8 Kq) × 10² V = 3/8 qK

As the body remains stationary, ∴ ma_max = µ_s mg ⇒ a_max = µ_s g = 0.15 × 10 = 1.5 m/s²

For series combination, Equivalent resistance, R_eq = 10R Current i₁ = E / 10R For parallel combination, Equivalent resistance, R_eq = R / 10 ∴ i₂ = E / (R / 10) = 10E / R ∴ i₂ / i₁ = (10E / R) × (R / E) = 100 => i₂ = 100i₁ According to question, i₂ = ni₁ ∴ n = 100

Magnetic field due to wire I = µ₀i / 4πR

Magnetic field due to wire III = µ₀i / 4πR

Magnetic field due to semiconductor arc II = µ₀i / 4R

Magnetic field due to wire I and III = -µ₀i / 4πR (into page) + µ₀i / 4πR (into page) = µ₀i / 2πR (into page)

B_total = B_arc - B_wires = µ₀i / 4R - µ₀i / 2πR = µ₀i / 4R [ 1 - 2/π ] Pointed away from page.

Given, Resistance at 0°C, R₀ = 2 Ω Resistance at 80°C, R = 6.8 Ω Using R = R₀(1 + αΔT) where α is the temperature coefficient of resistance ∴ 6.8 = 2{1 + α(80 - 0)} => 6.8 / 2 - 1 = α × 80 => α = (3.4 - 1) / 80 = 2.4 / 80 = 0.03 ∴ α = 3 × 10⁻² °C⁻¹

Given, Magnetic Field, B = (2î+3ĵ-4k̂) T Magnetic force acting on a current carrying wire is F = Il vector × B vector = I L Î × (2î+3ĵ-4k̂) = 3IL k̂+ 4IL ĵ Magnitude of force, |F| = √((3IL)² + (4IL)²) = 5IL

We know that For convex lens f₁ > 0, concave lens f₂ < 0 1 / f_net = 1 / f₁ + 1 / f₂ = 1 / f - 1 / f = 0 => f_net = Infinite NOTES: If two thin lenses of same focal length but having opposite nature are placed in contact then combination will behave like a plane glass plate.

(A) Atoms consist of three fundamental particles : Electrons, protons and neutrons

(D) Protons and neutrons present in the nucleus are collectively called as nucleons. So, the correct statements are B, C, E only.

Activation energy is defined as the minimum amount of extra energy absorbed by reactant molecules to achieve threshold energy. Zero activation energy means the reactant molecules have the sufficient energy to form product.

NOTES: According to Arrhenius equation k = A e^(-Eₐ / RT) Zero activation energy means Eₐ = 0 => k = A e⁰ => k = A, where A is the frequency factor. It simply implies that all collisions lead to successful collisions and should results in the formation of product which is practically not possible.

Let the number of atoms of element B = N So, the number of atoms of element A = 1/3 × 2 N ∴ The formula of the compound is A₂/₃ _N B_N = A₂B₃ So, x = 2, y = 3, x + y = 5

On dissolving alkali metal (sodium) in liquid ammonia, a deep blue solution is developed due to ammoniated electron which absorbs energy in visible region of light and imparts blue colour. Due to the presence of unpaired electrons, solution is paramagnetic.

M + (x + y)NH₃ ——> [M(NH₃)ₓ]⁺ + [e(NH₃)_y]⁻

Neoprene is formed by free radical polymerisation of chloroprene.

In case, nitrogen and sulphur both are present in organic compound, sodium thiocyanate is formed. It gives blood red colour and no Prussian blue since there are no free cyanide ions. Na + C + N + S ———> NaSCN Fe³⁺ + SCN⁻ ———> [Fe(SCN)]²⁺ (Blood red)

NOTES:

If sodium fusion is carried out with excess of sodium, the thiocyanate decomposes to yield cyanide and sulphide. These ions now give their usual tests. NaSCN + 2Na ———> NaCN + Na₂S

This reaction is called soda lime decarboxylation

Molar mass of CH₄ = 16 g/mol

Weight of 2 moles of CH₄ = 16 × 2 = 32 g

NOTES: Decarboxylation with soda lime produces alkane containing one carbon less than the carboxylic acid.

Conductivity = conductance × cell constant k = 1/R × l/A l/A = k × R = 0.0210 × 60 = 1.26 cm⁻¹

complexes in which a metal is bound to only one kind of donor groups are called as homoleptic complexes. Potassium trioxalatoaluminate (III). It is a homoleptic complex.

n_m = 2l + 1 => l = (n_m - 1) / 2

The stability of Cu²⁺ (aq) is more than Cu⁺ (aq) is due to the much more negative Δ_hydH° of Cu²⁺ (aq) than Cu⁺ (aq), which more than compensates for second ionisation enthalpy of Cu. Δ_hyd H° of Cu²⁺ (aq) = -2121 kJ mol⁻¹, Δᵢ H°₁ of Cu = +745 kJ mol⁻¹, Δᵢ H°₂ of Cu = +1960 kJ mol⁻¹

As we move down the group, due to poor shielding effect of intervening d and f orbitals, the increased effective nuclear charge holds ns electrons tightly and therefore restricting their participation in bonding. This effect is known as inert pair effect. So, the relative stability of +1 O.S increases for heavier elements. Hence, TlI is more stable than TlI₃. NOTES: Boron and Aluminium do not show the inert pair effect. The inert pair effect is only shown by the elements which have d and f orbitals influencing their outermost s-orbital electrons by having poor shielding.

Combination of N₂ and H₂ to form NH₃ in presence of finely divided Fe is an example of heterogeneous catalysis.

No. of σ bonds = 11, No. of π bonds = 3, No. of lone pair of e⁻ = 1

For molecules like B₂, C₂, N₂ etc. the increasing order of energies of various molecular orbitals is σ1s < σ*1s < σ2s < σ*2s < (π2p_x = π2p_y) < σ2p_z < (π*2p_x = π*2p_y) < σ*2p_z

Helium is used as diluent for oxygen in modern diving apparatus because of its very low solubility in blood.

NOTES: The solubility of he is less than that of nitrogen and it diffuses more rapidly, the elimination of inert gas from blood and tissues would be more rapid if Helium was substituted for nitrogen.

Rate = k[A]²[B] When concentration of A is tripled, [A'] = [3A] New rate, r' = k[A']²[B] = k[3A]²[B] = 9k[A]²[B] => r' = 9r, increases by factor of nine.

This reaction is Clemmensen reduction

Lewis acids are the one which accepts lone pair of electron due to presence of vacant orbital in outermost shell.

For isoelectronic species, as the charge on anion increases, ionic size increases O ——> O²⁻, F ——> F⁻, N ——> N³⁻, Na ——> Na⁺ Ionic size order : N³⁻ > O²⁻ > F⁻ > Na⁺ So, N forms N³⁻ anion with largest ionic size.

Allylic halides are the compound where halogen atom attached to sp³ hybridised carbon atom next to double bond.

A unit formed by the attachment of a base to 1' position of sugar is known as nucleoside. In nucleosides, the sugar carbons are numbered as 1', 2', 3', etc. in order to distinguish these from the bases (Fig. (a)). When nucleoside is linked to phosphoric acid at 5' -position of sugar moiety, we get a nucleotide (Fig. (b)).

Intermolecular forces are the forces of attraction and repulsion between interacting molecules. This term does not include covalent s covalent bond holds atoms of a molecule together. Hence, dipole - dipole forces, dipole - induced dipole forces, hydrogen bonding and dispersion forces are intermolecular forces.

Statement A, B, C are incorrect.

(D) H - H bond dissociation energy is maximum as compared to single bond between two atom of any element.

(E) Hydrogen reduces oxides of metal that are less active than iron.

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

CaCO₃ ———1200 K———> CaO + CO₂

1 mol ——> 1 mol

100 g ——> 44 g

As CaCO₃ is 20% pure, So, mass of pure CaCO₃ = 20 × 20 / 100 = 4 g

So, 100 g CaCO₃ produces 44 g CO₂

∴ 4 g CaCO₃ produces 44 / 100 × 4 g CO₂ = 1.76 g CO₂

The value of ΔᵣG depends on n value as per the equation ΔᵣG = -nFE_cell So, assertion statement is correct E_cell is an intensive property while ΔᵣG is an extensive thermodynamic property So, reason is correct but not explaining the assertion.

Criteria for Huckel's rule: Planarity Complete delocalisation of electrons Presence of (4n + 2)n electrons in ring where π is an integer (=0, 1,2, ..) The compounds which follow Huckel's rule are option (i), (ii), (iv), (vii).

Ammoniacal silver nitrate solution is Tollens’ reagent. Aldehyde upon warming with Tollens’ reagent produces a silver mirror due to formation of silver metal in alkaline medium. Aldehyde is oxidised to corresponding carboxylate anion.

(C) Basic character decreases from V₂O₃ to V₂O₄ to V₂O₅. (D) V₂O₄ dissolves in acids to give VO²⁺.

In other options carbocation formed destabilised by -I effect of -NO₂ group.

The bond between O - CH₂ - is weaker than the bond between O - C₆H₅ because the carbon of phenyl group is sp² hybridised and there is a partial double bond character. Therefore I⁻ ion breaks O - CH₂ - bond.

At 900-1500 K (higher temperature range in the blast furnace) Reaction which take place are:

C + CO₂ ——> 2CO

FeO + CO ——> Fe + CO₂

CaO + SiO₂ ——> CaSiO₃ (Slag formation)

Fe₂O₃ + CO ——> 2FeO + CO₂ at 500-800 K.

Correct relation between change in enthalpy and change in internal energy is ΔH = ΔU + Δn_g RT

Using Ion electron method: Reduction Half reaction: Cr₂O₇²⁻ + 6e⁻ ——> 2Cr³⁺ Oxidation

Half reaction:

SO₃²⁻ ——> SO₄²⁻ + 2e⁻ × 3

Overall reaction: Cr₂O₇²⁻ + 3SO₃²⁻ ——> 2Cr³⁺ + 3SO₄²⁻

To balance 'H' atoms, adding H₂O on LHS Cr₂O₇²⁻ + 3SO₃²⁻ ——> 2Cr³⁺ + 3SO₄²⁻ + 4H₂O

To balance 'H' atoms, adding H⁺ on RHS

Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ ——> 2Cr³⁺ + 3SO₄²⁻ + 4H₂O

∴ a = 1, b = 3, c = 8

Nutrient enriched water bodies support a dense plant population, which kills animal life by depriving it of oxygen and results in subsequent loss of biodiversity. This process is called as eutrophication. ∴ Statement I is incorrect and Statement II is correct.

A + B ⇌ C + D k_eq = [C] [D] / [A] [B] k_eq = (10 × 6) / (2 × 3) = 10 ΔG° = -RT ln k_eq = -2.303 RT log k_eq = -2.303 × 2 × 300 × log 10 = -1381.8 cal

Chelating ligands in general form more stable complexes ∴ The most stable complex is [CoCl₂(en)₂]NO₃

Both the statements are correct because transpiration cools leaf surfaces, sometimes 10 to 15 degrees, by evaporative cooling and forces generated by transpiration can create pressures sufficient to lift a xylem sized column of water up to 130 meters high.

In this method of making competent cells, suitable for plants, cells are bombarded with high velocity micro-particles of gold or tungsten coated with DNA in a method known as biolistics or gene gun.

Both the assertion and reason is correct and reason is the correct explanation of assertion. The predominant stage of life cycle of moss is gametophyte which consist of 2 stages. The first stage is protonema that directly develops from spore.

The unequivocal proof that DNA is the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952). They worked with viruses that infect bacteria called bacteriophages. The bacteriophages attaches to the bacteria and its genetic material then enter the bacterial cell.

Dedifferentiation is a phenomenon by which the living differentiated plant cells, that by now have lost the capacity to divide canregain the capacity of division under certain conditions. In tissue culture experiments, leaf mesophyll cells are put in a culture medium to form callus. This phenomenon may be called as dedifferentiation.

Majority of insect pollinated flowers are large-sized. These flowers are colorful and possess fragrance to attract the insects.

The frequency of recombination is the number of crossovers formed between any two genes. Alfred Sturtevant explained that the recombination frequency between gene pairs is the measure of distance between the genes.

In eukaryotes there are three major types of RNA polymerases are found that are RNA polymerase I, RNA polymerase II and RNA polymerase III. RNA polymerase III transcribes : tRNAs, ScRNA, 5S rRNA and snRNA.

Habitat loss and fragmentation is the most important cause driving animals and plants to extinction.

Statement I is incorrect because endarch and exarch are the terms often used for describing the position of primary xylem in the plant body. Exarch condition is the most common feature of the root system. Hence, statement I is incorrect and II is correct.

Axile placentation occurs in syncarpous pistils. The ovary is partitioned into two or more chambers. Axile placentation is observed in China rose, Petunia and Lemon.

Expressed Sequence Tags or ESTs are genes that are expressed as RNA in the body. ESTs represent portions of expressed genes. They may be represented as either cDNA/mRNA sequence or as the reverse complement of the mRNA, the template strand.

Tassels in the com cob represents stigma and style which wave in the wind to trap pollen grains.

Spraying juvenile conifers with gibberellins hastens their maturity period, thus leading to early seed production.

The given equation represent Net primary productivity which is gross primary productivity minus respiration loss. In the given equation, R represent “Respiratory loss”.

The characteristics specific to family Fabaceae but not found in Solanaceae or Liliaceae is Diadelphous and dithecous anther.

Primary endosperm nucleus is formed by the fusion of diploid secondary nucleus with a male gamete. Therefore, it is triploid. Zygote is formed by fusion of two gametes and thus it is diploid. Synergids are the cells of gametophyte and hence these are haploid.

Ethylene promotes rapid internode/petiole elongation on deep water rice plants.

Late wood is also known as autumn wood. It is a wood in a tree's growth ring which is produced later in the growing season when the growth is not rapid.

Both assertion and reason are correct and reason is the correct explanation of assertion because ATP in glycolysis is used at two steps of conversion that are Glucose→ Glucose-6-phosphate Fructose-6-phosphate → Fructose-1, 6-bisphosphate

The detritus food chain begins with detritus that is dead organic matter. The saprotrophic bacteria and fungi breakdown detritus into simpler inorganic substances by a process called catabolism.

For every CO₂ molecule entering the Calvin cycle, 3 molecules of ATP and 2 of NADPH₂, are required. To make one molecule of glucose, 6 turns of the cycle are required. Thus, ATP and NADPH₂ molecules required for synthesis of one molecule of glucose during Calvin cycle will be: 6*3 = 18 ATP 6*2 = 12 NADPH₂

The phenomenon of pleiotropism refers to phenomenon where a single gene affecting multiple phenotypic expression.

The reaction centre in PS II has an absorption maxima at 680 nm.

Manganese (Mn) is an essential micronutrient that while needed in small amounts, plays a key role in photosynthesis. Mn sparks the photosynthesis process by splitting water after Photosytem II (PSII) fixes light to initiate the conversion of CO₂ and water into carbohydrates.

The historic convention on Biological Diversity, “The Earth Summit” was held in Rio de Janeiro in the year 1992. It called upon all nations to take appropriate measures for conservation of biodiversity and sustainable utilisation of its benefits.

Pteridophytes that produce two types of spores are known as heterosporous. Selaginella and Salvinia are heterosporous pteridophytes.

The correct matching is: M phase or Mitosis is also called equational division. During G₂ phase DNA synthesis stops but cell synthesis RNA, proteins, etc. for next phase.

Quiescent stage is inactive phase in which non-dividing cells enters.

G1 phase is the interval between mitosis and initiation of DNA replication.

The correct matching is: Iron activates catalase enzyme. Zinc is needed in the synthesis of auxin. Boron is required for cell elongation and cell differentiation. Molybdenum is component of nitrogenase and nitrate reductase enzyme.

The correct matching is: Cohesion represents mutual attraction between water molecules. Adhesion represents attraction of water molecules to polar surfaces Surface tension represents water molecules are attracted to each other in the liquid phase more than to water in the gas phase., Guttation represent loss of water in liquid phase.

The correct matching is: (+,+) Mutualism : In this interaction, both the interacting species are benefitted. (+, 0) Commensalism : Only one species is benefitted and the other species remains unharmed (-, 0) Amensalism : Neither species is benefitted. One remains unharmed and the other is harmed. (+, -) Parasitism : One species is benefitted and other is negatively effected.

The correct matching is:

Pyruvate, which is formed by the glycolytic catabolism of carbohydrates in the cytosol, after it enters mitochondrial matrix undergoes oxidative decarboxylation by a complex set of reactions catalyzed by pyruvate dehydrogenase.

The scheme of glycolysis was given by Gustay Embden, Otto Meyrhof and J. Parnas, and is often referred to as the EMP pathway.

In electron transport system, the energy of oxidation-reduction is utilized for the production of proton gradient required for phosphorylation, thus, this process is also called oxidative phosphorylation.

The TCA (tricarboxylic acid 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.

The ribosome consists of 80 different proteins.

The correct sequence of formation of recombinant DNA is:

i. Cutting of DNA at specific location by restriction enzyme

ii. Isolation of desired DNA fragment

iii. Amplification of gene of interest using PCR

iv. Insertion of recombinant DNA into the host cell

NOTES: Recombinant DNA Technology is the joining together of DNA molecules from two different species. The recombinant DNA molecule is inserted into a host organism to produce new genetic combinations that are of value to science, medicine, agriculture, and industry.

Statement II is incorrect as in general, herbivores and plants appear to be more adversely affected by competition than carnivores. Statement I is correct.

A flower is defined as the modification of shoot apical meristem into floral meristem. Internodes do not elongate and the axis gets condensed. The apex produces different kinds of floral appendages laterally at the successive nodes instead of leaves.

Statement A, C and D are incorrect w.r.t. Klinefelter’s syndrome s they are associated with Down’s syndrome.

The required components for chemiosmosis are proton gradient, proton pump, plasma membrane and ATP synthase.

Statement B,C and E are incorrect because: Bark that is formed early in the season is called early or soft bark. Towards the end of the season late or hard bark is formed. Bark is non-technical term that refer to all tissues exterior to vascular cambium. Phellogen is couple of layers thick.

Assertion is correct but reason is false as in gymnosperms the pollen grains are released from the microsporangium they are carried in air currents. They come in constant with the opening of the ovules borne on megasporophylls. The pollen tube carrying the male gametes grows towards archegonia in the ovules and discharge their contents near the mouth of the archegonia.

The endomembrane system includes the nuclear envelope, lysosomes, vesicles, the ER, and Golgi apparatus, as well as the plasma membrane

Amniocentesis can lead to female foeticide as it detects the sex of the foetus which decreases the gender ratio. So, it has a Statutory ban to avoid female.

Both the statement are false because ligaments is an example of dense regular connective tissue. Cartilage is an example of specialized connective tissue.

The correct matching is: P-wave represents the electrical excitation (or depolarisation) of the atria which leads to the contraction of both the atria, QRS complex represents the depolarisation of ventricles which initiates the ventricular contraction. T-wave represents the return of the ventricles from excited to normal state.

the correct matching is: Vasectomy is a surgical method of contraception Coitus interruptus is a natural method of contraception Cervical cap is a barrier method of contraception Saheli is an oral method of contraception which is a non-steroidal pill

Both the statements are correct. Vas deferens receives a duct from seminal vesicle and opens into urethra as the ejaculatory duct, The cavity of the cervix is called cervical canal which along with vagina forms birth canal.

Assertion is true as there are two types of nephrons, i.e., cortical nephrons and juxtamedullary nephrons based on their relative position in the cortex and medulla. Reason is not correct as loop of Henle in juxtamedullary nephrons is very long and runs deep into the medulla.

The correct matching is following:

Gene a codes for enzyme transacetylase.

Gene y codes for enzyme permease.

Gene i codes for repressor protein

Gene z codes for enzyme -galactosidase.

The correct matching will be:

Ringworm is caused by Trichophyton.

Filariasis is caused by Wuchereria bancrofti.

Malaria is caused by Plasmodium species.

Pneumonia is caused by Haemophilus influenzae.

Cytoskeleton in a cell involved in many functions such as mechanical support, motility, maintenance of the shape of the cell

Large amount of nutrients in water promotes growth of algal bloom. Algal bloom increases fish mortality. Eutrophication result due to natural ageing of lakes

The correct matching will be: Heroin belongs to the category of opioids, and it is a depressant

that slows down body functions. Marijuana is known for its effect on the cardiovascular system of the body.

Cocaine interferes with the transport of the neurotransmitter dopamine.

Morphine is used is a sedative and painkiller.

Radial symmetry is found in adults of phylum of ctenophore,

coelenterate and Echinodermata.

In prokaryotes, the negatively charged DNA is held with some positively charged proteins in a region termed as nucleoid. In eukaryotes, the negatively charged DNA is wrapped around

the positively charged histone octamer to form a structure called nucleosome.

The incorrect answer is option (d) as incorrect statement is:

First menstrual cycle that begins at puberty is called menarche. Rest other statements are incorrect.

Mole and flying squirrel are placental mammals. Lemur and wolf are placental mammals.

Bobcat is a placental mammal

Therefore only option Numbat, Spotted cuscus, Flying phalanger is correct.

The correct answer is Serum and Urine analysis

because using conventional

methods of diagnosis like serum and urine analysis, etc, do not help in early diagnosis. Recombinant DNA technology, Polymerase Chain Reaction and Enzyme Linked Immuno-Sorbent

Assay (ELISA) are some of the techniques that serve the purpose of early diagnosis.

The correct matching will be:

Cartilaginous joint is present between the adjacent vertebrae in the vertebral column.

Ball and socket joint is present between between Humerus and Pectoral girdle

Fibrous Joint is present between between flat skull bones Saddle Joint is present between between carpal and metacarpal

of thumb

The correct matching is: A leopard and a lion in a forest/grassland exemplify competition where both the species are competing for the same resources. Fungi and root of a higher plant in mycorrhizae exemplify mutualism where both the species are benefitted. The fungi help the plant in the absorption of essential nutrients from the soil while the plant in turn provides the fungi with energy yielding carbohydrates

A cattle egret and a cattle in a field exemplify commensalism where one species benefits and the other remains unaffected. The egrets always forage close to where cattle are grazing because the cattle, as they move, stir up and flush out insects from the vegetation that otherwise might be difficult for the egrets to find and catch.

Vital capacity is the maximum volume of air a person can breathe in after forced expiration. Vital capacity includes ERV, TV and IRV.

The correct matching is:

(i) Fovea is the point of greatest visual acuity or resolution.

(ii) Iris is the visible coloured portion of the eye that regulates diameter of pupil.

(iii) Blind spot is the point where optic nerve leaves the eye-ball and photoreceptor cells are absent.

(iv) Sclerais the extenal layer of eye formed of dense connective tissue.

Once the undigested and unabsorbed substances enter the caecum, their backflow is prevented by Ileo-caecal valve

Gonorrhoea is a bacterial disease which can be treated with the help of antibiotics and cured completely. Rest other diseases are not completely cured as other mentioned options are viral diseases.

Statement I is incorrect because left-hand represented by first amino acid is N terminal and right-hand by C terminal

Statement II is incorrect because it can particulate matter present in the exhaust from a thermal power plant.

HIV enters into helper T-lymphocytes (TH), replicates and produces progeny viruses. The progeny viruses released into blood attack other helper lymphocyte.

Both the statements are correct. RNA mutate at much higher rate as compare to DNA because RNA has higher mutation rate that DNA because DNA has deoxyribose pentose sugar in it’s configuration whereas RNA has extra oxygen in its ribose sugar which makes it more reactive. As oxygen has got lone pair it will have tendency to react.

Both assertion and reason are correct but reason is not the correct explanation of assertion. Corpus luteum secretes large amount of progesterone which is essential for maintenance of endometrium of uterus. In absence of fertilisation, the corpus luteum degenerates hence the decrease in the level of progesterone hormone will cause disintegration of endometrium leading to menstruation.

The correct matching is:

Cholecystokinin (CCK) acts on both gall bladder and pancreas and stimulates the secretion of bile juice and pancreatic enzymes respectively.

GIP inhibits gastric secretion and motility.

Atrial Natriuretic Factor (ANF) is released from the atrial wall of our heart.

Anti-diuretic hormone (ADH) acts mainly on the kidney and stimulates resorption of water and electrolytes by the distal tubules.

Consanguineous marriage is matrimony between individuals who are closely related. In a clinical sense, marriage between £wo family members who are second cousins or closer qualify as having a consanguineous marriage.

So the correct option is mating between relatives.

The unique mammalian characteristics are presence of hairs, pinna and mammary gland. Presence of tympanic membrane is present in amphibians too and monocondylic skull is present in reptiles and aves.

Basophils are one of three granulocytes, along with neutrophils and eosinophils. Basophils are the smallest in number of granulocytes but the largest in cell size. Neutrophils are most abundant. Monocytes have kidney shape nucleus.

The correct statement regarding chordates are b and c. The chordate characters are presence of closed circulatory system and presence of pharyngeal gill slits. Nerve cord is dorsal, hollow and single. Heart is ventral. They are triploblastic and coelomate.

Thyroid hormone is involved in maintaining water and electrolytes balance, help in process of RBC production and maintaining basal metabolic rate.

The chordate characters are presence of closed circulatory system and presence of pharyngeal gill slits. Nerve cord s dorsal, hollow and single. Heart is ventral. They are triploblastic and coelomate.

The parts of human brain that helps in regulation of sexual behaviour, expression of excitement, pleasure, rage, fear etc. are Limbic system and hypothalamus,

The correct matching will be: Areolar connective tissue contains fibroblasts (cells that produce and secrete fibres), macrophages and mast cells, Inner surface of bronchioles is lined by ciliated epithelium. Blood is a specialised connective tissue. Tubular parts of nephron are lined by cuboidal epithelium.

The correct matching will be: Exponential growth occurs when there is unlimited resource availability condition. Expanding age pyramid reflects growing population where the percent individuals of pre-reproductive age is largest followed by reproductive and post-reproductive age groups. Stable age pyramid shows stable population where the percent individuals of pre-reproductive and reproductive age group are same.

Statement A and E are incorrect because excessive loss of body fluid from the body switches on osmoreceptors and ADH is responsible for increasing in GER.

Statement I is not correct because During GO phase of cell cycle, the cell is metabolically active but do not divide.

Tetrad formation is seen during zygotene stage. Terminalisation of chiasmata takes place during diakinesis, Crossing over takes place between non-sister chromatids of homologus chromosomes.

Inbreeding decreases the productivity of inbred population, after continuous inbreeding is not an advantage of inbreeding

Anal style is present only in male cockroach, so it show sexual dimorphism.

The template strand serves in mRNA synthesis while the other strand is called coding strand as its base sequence is same as that of newly synthesized mRNA. So, the sequence of mRNA will be identical to the given sequence of coding strand except for the presence of uracil in place of thymine in mRNA

Muscle bundles are held together by collagenous connective tissue layer called fascia. Muscle bundles are called fascicles. The portion of the myofibril between two functional unit of contraction called sarcomere.