NEET-UG 2018 Question Paper Solutions

Diameter of the ball
= MSR + CSR × (least count) – zero error
= 0.5 cm + 25 × 0.001 – (–0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm

Let the mass of block is m . It will remain stationary if forces acting on it are in equilibrium, i.e. ma cos θ = mg sin θ ⇒ a = g tan θ.

Here ma = Pseudo force on block, mg = weight.

According to law of conservation of linear momentum,

mv+4m x 0 = 4mv' + 0 ⇒ V' = v/4

Coefficient of restitution,

e = Relative velocity of separation / Relative velocity of approach v/4/v

or, e = 1/4 = 0.25

Angular momentum, remains conserved until the torque acting on sphere remains zero. 𝝉ₑₓ = 0

So, dL/dt = 0

i.e, angular momentum L = constant.

In rolling motion, rotational kinetic energy,

Kₜ = 1/2 Iv²

And, Kₜ + Kᵣ = 1/2 mv² +1 / 2 lω²

= 1/2 mv² + 1/2 (2/5 mr²)(v/r²) = 7 / 10 mv²

∴ Kₜ/(Kₜ + Kᵣ) = (1/2 mv²)/(7/10 mv²) = 5/7

Work done required to bring them to rest ΔW = ΔKE (work-energy theorem)

ΔW = 1/2 Iω² (ΔKEᵣₒₜ = 1/2 Iω²)

or, ΔW ∝ I (for same ω)

I_ Solid sphere = 2/5 MR² , I_disk = 1/2 MR²

I_(ring) = MR² ∴ W_C > W_B > W_A

Speed of the planet will be maximum when its distance from the sun is minimum as mvr =constant.

Point A is perihelion and C is aphelion.

Clearly, V_A > V_B > V_C

So, K_A > K_B > K_C

Δl = (F'/3AY)l ....(ii)

From equation (i) & (ii),

Δl =(F/AY)3l = (F' /3AY)l or, F'=9F

Power = rate of production of heat F.V

= 6πηrV_T · V_T = 6πηrV_T²

(∵F = 6πηrV_Tr Stoke’s formula)

V_T∝ r²

∵ V_T = 2/9 ( r²(ρ − σ) / η( g)

∴ Power ∝ r⁵

From Wien’s law

λₘₐₓ T = constant

i.e., λₘₐₓ₁T₁ = λₘₐₓ₂T₂

⇒ λ₀ T = 3 λ₀/4 T

⇒ T' =4/3 T

Power radiated P ∝ T⁴

So, P₂/P₁ = n = (T'/T)⁴ = (4/3)⁴ = 256/81

Gas is monatomic, so Cₚ = 5/2 R

Given process isobaric

∴ dQ = nCₚ dT

⇒ dQ = (5/2 R) dT

dW = PdV = nRdT

∴ Required ratio = dW/dQ = nRdT / n(5/2 R)dT = 2/5

Efficiency of ideal heat engine, η = (1 − T₂/T₁)

Sink temperature, T₂ = 100°C = 100 + 273 = 373 K

Source temperature, T₁ = 0°C + 0° + 273 = 273 K

Percentage efficiency, % η = (1 − T₂/T₁) × 100

=(1- 273/373)x100 = (100/373) × 100 = 26.8%

Let at temperature T rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere.

V_escape =11200m/s

Also, Vᵣₘₛ = V_ escape = √(3k_BT/m₀₂) = 11200 m/s

Putting value of k_B and mO₂, we get,

T = 8.360 × 10⁴ K

From equation, acceleration, a = 20 m/s², and displacement, y = 5m |a| = ω²y

⇒20= ω²(5)

⇒ ω = 2 rad / s

Time period of pendulum,

T = 2π/ω = 2π/2 = πs

Two successive resonance are produced at 20 cm and 73 cm of column length

∴ λ/2 = (73 − 20) × 10⁻² m

⇒ λ = 2 × (73 − 20) × 10⁻²

Velocity of sound, ν = nλ

= 2 × 320 [73 − 20] × 10⁻² = 339.2 m/s⁻¹

For closed organ pipe, third harmonic

n = (2N − 1)V / 4ℓ= 3V/4ℓ (∵ N = 2)

For open organ pipe, fundamental frequency

n = NV / 2ℓ′ =V/2ℓ (∵ N = 1)

According to question, 3V/4ℓ = V/2ℓ′

⇒ ℓ = 4ℓ′/3 × 2 = 2ℓ /3 = 2 x 20 / 3 =13.33cm

As we know, F = qE =ma

⇒ a = qE/m

h = 1qE/2m (t²)

∴ t= √(2hm/qE)

i.e., time t ∝ √m as ‘q’ is same for electron and proton. Since, electron has smaller mass so it will take smaller time.

Acceleration, a = v-u /t = 6-0/1

= 6 ms⁻²

For t = 0 to t = 1s,

s₁ = 1/2 x 6 (1)² = -3m ...(i)

For t=1s=tot=2s,

S₂ 6.1 -1/2 x 6(l)² = -3 ...(ii)

For t = 2s to t = 3s,

S₃ = 0-1/2 x 6(l)² = -3 (iii)

Total displacement S = S₁ + S₂ +S₃ = 3 m

Average velocity = 3/3 = 1 ms⁻¹

Total distance travelled =9 m

Average speed = 9/3 = 3 ms⁻¹

Electrostatic force between the metal plates

F_plate = Q²/ 2Aᵋ₀

For isolated capacitor Q = constant

Clearly, F is independent of the distance between plates.

Colour code for carbon resistor

0 Black

1 Brown Tolerance: ±5% Gold

2 Red    ±10% Silver

3 Orange  ±20% No colour

4Yellow

5 Green

6 Blue

7 Violet

8 Grey

9 White

(47 ± 4.7)KΩ= 47 × 10³ ± 10%

Yellow -Violet - Orange - Silver

In series grouping equivalent resistance Rₛₑᵣᵢₑₛ = nR

In parallel grouping equivalent resistance Rₚₐᵣₐₗₗₑₗ = R/n

I = E/nR + R ...(i)

10I = E/R/n+R ...(ii)

Dividing eq. (ii) by (i),

10 = (n+1)R/( 1/n +1 )R

Solving we get, n = 10

From figure, for equilibrium,

mg sin 30°=1/B 30°

⇒ I =mg/ℓB tan 30°

= 0.5 x9.8/0.25 x√3 = 11.32A

Current sensitivity of moving coil galvanometer

Iₛ= NBA/C ...(i)

Voltage sensitivity of moving coil galvanometer,

Vₛ= NBA/ CR_G ..(ii)

Dividing eq. (i) by (ii)

Resistance of galvanometer

R_G = Iₛ / Vₛ = 5 ×1 / 20 ×10⁻³ = 5000 / 20 = 250 Ω

Rod gains gravitational potential energy which comes from energy of current source.

From question energy stored in inductor,

U = 25 × 10⁻³ J

Current, I = 60 mA

Energy stored in inductor U = 1/2LI²

25 × 10⁻³ = 1/2 × L × (60 × 10⁻³)²

L = 25 × 2 × 10⁶ x10⁻³ /3600 = 13.89 Hm

Power dissipated in an LCR series circuit connected to an a.c. source emf E

P = Eᵣₘₛ Iᵣₘₛ cos Φ = E²ᵣₘₛR/Z² = E²ᵣₘₛR / R² + (ωL − 1/Cω)²

(10/√2)² × 50 / (50)² + (314 × 20 × 10⁻³ − 1/(314 × 100 × 10⁻⁶))²] Solving we get, P =0.79 W

As we know, 𝐄⃗ × 𝐁⃗= 𝐕⃗

( 𝐄⃗ĵ × 𝐁⃗) =𝐕⃗î (∵Electric field vector is along +y axis)

So, 𝐁⃗ = Bk̂

i.e., direction of magnetic field vector is along +z direction.

For refracting the path, light ray should be normally incident on slivered face.

A = r + o ⇒ r = 30°

Applying Snell’s law at point M,

sin i/sin30° = √2/1

⇒sin i = √2 × 1/2

or, sin r = 1/√2 i.e., i = 45°

For telescope, angular magnification = f₀ / f_E

So, focal length of objective lens should be large.

Angular resolution = D/1.22 λ So, D objective lens of refracting telescope should have large focal length (f₀) and larger diameter D for larger magnification.

Angular width = λ/d

So, 0.20° = λ/2 mm

⇒ λ = 0.20° × 2

Again, 0.21° = λ/d

Now putting the value of λ,

d = 0.20° × 2 mm/0.21°

∴ d = 1.9 mm

When reflected light rays and refracted rays are perpendicular, reflected light is polarised with electric field vector perpendicular to the plane of incidence.

Also, tan i = μ (i (Brewster angle)

Initial de-Broglie wavelength

λ₀ = h/(mV₀) ...(i)

Acceleration of electron

a = eE₀/m (∵ F = ma = eE₀)

Velocity after time ‘t’

V = (V₀ + (eE₀/m(t) )

So, λ = h/mV = h/m(V₀+eE₀/m(t)

= h/mV₀[1+eE₀/mV₀ (t)] = λ₀/[1+eE₀/mV₀ (t) ] ...(ii)

Dividing eqs. (ii) by (i),

de- Broglie wavelength λ = λ₀/[1+ eE₀/mV₀(t)]

In a Bohr orbit of hydrogen atom Kinetic energy,

K = kze²/2rₙ

Total energy, E = −kze²/2rₙ

So, Kinetic energy : total energy = 1 : −1

Number of nuclei remaining, N = 600 − 450 = 150 after time ‘T’

N/N₀ = 150/600 = 1/4

N = N₀e⁻λt ⇒ ℓn(N₀/N) = λt ⇒ℓn (N₀/N ) = λt

⇒ t = 1/ λ ℓn (N₀/N)

2.303 × T₁/₂ / 0.693( log₁₀ ) N₀/N

= 2.303 × 10 / 0.693 (log₁₀ 4)

On heating, number of electron-hole pairs increases, so overall resistance of diode will change. Hence forward biasing and reversed biasing both are changed.

From question, V_BE = = 0,  Vᵢ = 20V

V_CE = 0

V_b = 0 (earthed)

I_B = ?, I𝒸 = ?, β = ?

I𝒸 = (20 − 0)/4 × 10³ = 5 × 10⁻³ = 5 mA

Vᵢ = V_BE + I_BR_B  or, Vi = 0 + I_BR_B

⇒ 20 = I_B × 500 × 10³

⇒ I_B = 20/500 × 10³ = 40 μA

β= I_c /I_b= 25 × 10⁻³ / 40 × 10⁻⁶ = 125

Y=(A.B̅ + A̅ . B)

Short circuited current,

I = nε /nr1 = ε/r

So, I is independent of n and I is constant.

using mirror formula,

1/f = 1/v₁ + 1/u

-1/15 = 1/v₁ + 1/u

⇒ 1/v₁ = 1/-15 + 1/40

∴ v₁ = 24 cm

When object is displaced by 20 cm towards mirror

Now, u₂ = -20

So, 1/f = 1/v₂ + 1/u₂

1/-15 = 1/v₂ + 1/20 ⇒ 1/v₂ = 1/20 - 1/15

∴ v₂ = -60 cm

Therefore image shifts away from mirror by

60 − 24 = 36 cm

Using Einstein's photoelectric equation,

E = W₀ + Kₘₐₓ

When light of frequency, 2v₀ is incident on a metal plate,

h(2v₀) = hv₀ +1/2 mv₁²

hν₀ = ½ mv₁² ..........(i)

When light of frequency, 5ν₀ is incident on a metal plate

h(5ν₀) = hν₀ + ½mv₂²

4hν₀ = ½mv₂² ..........(ii)

Dividing eqn. (i) by (ii),

1/4 = v₁²/v₂²

∴ v₁/v₂ = 1/2

If universal gravitational constant becomes ten times,

G' = 10G

Acceleration due to gravity, g = GM/R²

So, acceleration due to gravity increases.

Using first law of thermodynamics equation,

ΔQ = ΔU + ΔW

⇒54 × 4.18

= ΔU + 1.013 × 10⁵(167.1 × 10⁻⁶ − 0)

(∵ ΔW = PΔV)

⇒ ΔU = 208.7 J

As track is frictionless, so total mechanical energy will remain constant

i.e.,0 + mgh = ½mvL² + 0

Using v² − u² = 2gh, h = v²/2g (∵ u = 0)

For completing the vertical circle, v_L ≥ √5gR

or, h = 5gR/2g = 5/2(R ) = 5/4(D)

(a) Mass of water = 18 × 1 = 18 g

Molecules of water = mole × N_A

= 18/18 N_A = N_A

(b) Molecules of water = mole × Nₐ

= 0.18/18 NA = 10⁻² N_A

(c) Molecules of water = mole × N_A

= 0.00224/0.0224 × N_A = 10⁻³ N_A

(d) Moles of water = 0.00224/22.4 = 10⁻⁴

Molecules of water = mole × N_A = 10⁻⁴ N_A

CO₂ is absorbed by KOH.

So, the remaining product is only CO.

Moles of CO formed from both reactions

= 1/20 + 1/20 = 1/10

Left mass of CO = moles × molar mass

= 1/10 × 28 = 2.8 g

The correct configuration of N' is

Due to poor shielding effect of 3d electrons in Ga, the to mic radii of Ga < AL Thus, the correct order of atomic is B<Ga<Al <In<Tl

The structure of CIF₃ is

The number of lone pair of electrons on central Cl is 2.

sp², sp², sp, sp

CH₂ = CH - C ≡ CH

NO : (σ1s)², (σ*s)², (σ2s)², (σ*2s)², (σ2p𝓏)²,(π2pₓ)², =(π2pᵧ)², (π*2pₓ)¹= (π2pᵧ)⁰,

B.O. = 10 − 5/2 = 2.5

CN⁻ : (σ1s)², (σ*1s)², (σ2s)², (σ*2s)², (π2pₓ)²,=(π2pᵧ)², (σ2p𝓏)²

B.O. = 10 − 4/2 = 3

CN : (σ1s)², (σ*1s)², (σ2s)², (σ*2s)², (π2pₓ)²=(π2pᵧ)², (σ2p𝓏)¹

B.O. = (9 − 4)/2 = 2.5

CN⁺ : (σ1s)², (σ*1s)², (σ2s)², (σ*2s)², (π2pₓ)² = (π2pᵧ)²

B.O. = (8 − 4)/2 = 2

Hence, option CN⁻ should be the right answer.

In real gas equation,

van der Wall constant (a) α forces of attraction.

van der Waal constant ‘a’, signifies intermolecular forces of attraction.

Higher is the value of ‘a’, easier will be the liquefaction of gas.

Let B.E. of X₂, Y₂ and XY are x kJ mol⁻¹, 0.5x kJ mol⁻¹ and x kJ mol⁻¹ respectively.

1/2 X₂ + 1/2 Y₂ → XY ; ΔH = −200 kJ mol⁻¹

ΔH = Σ(B.E.)_Reactants −Σ(B.E.)_product

= [1/2 x + ½(0.5x)] − [1 x (x)]

On solving, x = 800 kJ mol⁻¹.

A₂(g) + B₂(g) ⇌ X₂(g) ΔᵣH = −X kJ;

On increasing pressure equilibrium shifts in a direction where number of moles decreases i.e. forward direction. On decreasing temperature, equilibrium shifts in exothermic direction .e., forward direction.

So, high pressure and low temperature favours maximum formation of product.

Meq. of HCl = 75 × 1/5 × 1 = 15

Meq. of NaOH = 25 × 1/5 × 1 = 5

Meq. of HCl in resulting solution = 10

Molarity of [H⁺] in resulting mixture = 10/100 = 1/10

pH = −log[H⁺] = −log[1/10] = 1.0

Solubility of BaSO₄ = 2.42 × 10⁻³ g L⁻¹

∴ s = (2.42 × 10⁻³)/233 = 1.038 × 10⁻⁵ mol L⁻¹

Kₛₚ= s² = (1.038 × 10⁻⁵)² = 1.08 × 10⁻¹⁰ mol² L⁻²

Calculate E°_cell corresponding to each compound undergoing disproportionation reaction. The reaction for which E°_cell comes out +ve is spontaneous.

HBrO → Br₂ E° = 1.595 V, SRP (cathode)

HBrO → BrO₃⁻ E° = −1.5 V, SOP (anode)

2HBrO ⟶ Br₂ + BrO₃⁻

E°_cell = SRP (cathode) − SRP (anode)

= 1.595 − 1.5

= 0.095 V

E°_cell > 0 ⇒ ΔG° < 0 [spontaneous]

In metals ‘moving down the group, metallic character Al increases, so basic nature increases hence most acidic will be BeO.

BeO <Mg0 <Ca0 <BaO

___________________

increasing basic character

BeH₂ < CaH₂ < BaH₂

Smaller the size of cation, more will be its polarising power.

Hence, BeH₂, will be least ionic.

Boron belongs to 2nd period and it does not have vacant d-orbital.

−I effect increases on increasing electronegativity of atom. So, correct order of −I effect is

−NH₂ < −OR < −F.

Nitrous oxide (N₂O) occurs naturally in environment.

In automobile engine, when fuel is burnt dinitrogen and dioxygen combine to yield NO and NO₂.

Thus, N₂O₅ is the answer

For bcc lattice : Z = 2, a = 4r/√3

For fcc lattice : Z = 4, a = 2√2r

∴ d₂₅°C/d₉₀₀°C = (ZM/N_A ᵃ³)_bcc /(ZM/_Aᵃ³)_fcc

= 2/4 × (2√2r/(4r/√3)³

= 3√3/4√2

(t₁/₂)¹ˢᵗ order = Independent of concentration

(t₁/₂)²ⁿᵈ order ∝ 1/[A]₀

(t₁/₂)_zero = [A]₀/2k

∴ If [A]₀ = doubled, t₁/₂ doubled

According to Hardy Schulze rule, coagulating power of an ion depends on both magnitude and sign of the charge on the ion

Mg has more −ΔG value than alumina. So it will be in the lower part of Ellingham diagram. Metals which have more −ΔG value can reduce those metal oxides which have less −ΔG value.

CrO₄²⁻  Cr⁶⁺  diamagnetic
Cr₂O₇²⁻  Cr⁶⁺  diamagnetic
MnO₄⁻  Mn⁷⁺  diamagnetic
MnO₄²⁻  Mn⁶⁺  paramagnetic

Complexes are respectively [Co(NH₃)₆]Cl₃, [Co(NH₃)₅Cl]Cl₂ and [Co(NH₃)₄Cl₂]Cl.

Ni(28) : Ar]3d⁸ 4s²

∵ CO is a strong field ligand, so unpaired electrons get paired. Hence, configuration would be:

For four 'CO' ligands hybridisation would be sp³ and thus the complex would be diamagnetic and of tetrahedral geometry.

Co³⁺ = [Ar]3d⁶, unpaired e⁻(n) = 4

Spin magnetic moment
= √4(4+2) = √24 B.M.

Cr³⁺ = [Ar]3d³, unpaired e⁻(n) = 3

Spin magnetic moment
= √3(3+2) = √15 B.M.

Fe³⁺ = [Ar]3d⁵, unpaired e⁻(n) = 5

Spin magnetic moment
= √5(5+2) = √35 B.M.

Ni²⁺ = [Ar]3d⁸, unpaired e⁻(n) = 2

Spin magnetic moment
= √2(2+2) = √8 B.M.

[Fe(CO)₅]

EAN = Z − O.N. + 2(C.N.)

= 26 − 0 + 2(5)

= 26 + 10

= 36

Only one central metal atom/ion is present and it follows the EAN rule, so it is mononuclear.

Carboxylic acids have higher boiling points than

aldehydes, ketones and even alcohols of comparable

molecular mass.

This is due to more extensive association through

intermolecular H-bonding

In acidic medium aniline is protonated to form anilinium

ion which is meta-directing.

HOOC–CH₂–NH₂ ⇌

Glycine

⁻OOC–CH₂–⁺NH₃

Zwitter ion

Cross-linked or network polymers are usually formed from bi-functional and tri-functional monomers and contain strong covalent bonds between various linear polymer chains like melamine, bake lite etc.

Amylose and amylopectin are polymers of α-D-glucose, so β-link is not possible. Amylose is linear with 1 → 4 α-linkages whereas amylopectin is branched and has both 1 → 4 and 1 → 6 α-linkages.

So option Amylopectin has 1 -----> 4∝ -linkage and 1 ----. 6 α- linkage should be the correct option.

In metals moving down the group metallic character increases, so basic nature increases; hence most acidic will be BeO.

Sporopollenin cannot be degraded by enzyme: strong acids and alkali, therefore it is helpful in preserving pollen as fossil

Saccharomyces i.e. yeast is an eukaryote (unicellular fungi).

Mycobacterium is a bacterium. Oscillatoria and nostoc

are cyanobacteria.

Diatoms are the chief producers in some oceans and in some seasons as they are the primary producers and the food chain in marine ecosystem depends on it.

Ciliates differs from other protozoans in having two

types of nuclei. B.g., Paramoecium have two types of nuclei i.e. macronucleus & micronucleus.

Polysiphonia is a genus of red algae, where asexual spores and gametes are non-motile or non-flagellated.

In Agaricus (a genus of basidiomycetes), basidio- spores or meiospores are produced exogenously. Neurosa pora (a genus of ascomycetes) produces ascospores as meiospores but endogenously inside the ascus). Alternaria (a genus of deuteromycetes) does not produce sexual spores. Saccharomyces (Unicellular ascomycetes) produces ascospores, endogenously.

Pseudopodia are locomotory and feeding structures in sarcodines

(amoeboid).

Metamorphosis refers to transformation of larva into adult. Animal that perform metamorphosis are said to have indirect development. Metamorphosis includes, in insects, the transformation of a maggot into an adult fly and a caterpillar into a butterfly and, in amphibians, the changing of a tadpole into a frog. In earthworm development is direct which means no larval stage are there and hence no metamorphosis.

Some lateral roots of mangroves become specialised as pneumatophores in saline mud flats. These are lateral roots that grow upward (negative geotropism) for varying distances and function as the site of oxygen intake for the submerged primary root system.

Winged pollen grains are present in Pinus. Each pollen grain has two wing-like structures which enables it to float in air, as an adaptation for dispersal by the wind.

Casparian strip is a band of cell wall material deposited in the radial and transverse walls of the endodermis. Casparian strip is made of suberin and sometimes lignin

Secondary tissues are generated from the growth of a cambium. Vascular cambium gives rise to secondary xylem to the inside, and to secondary phloem to the outside.

Grasses are monocots and monocots usually do not have secondary growth. Palm like monocots have anomalous secondary growth.

Sugar is a carbohydrate. These are polyhydroxy aldehyde, ketone or their derivatives, which means they ‘have carbonyl and hydroxyl groups in its structure.

Diplotene is a stage of meiotic prophase which follows the pachytene and during which the paired homologous chromosomes begin to separate and chiasmata become visible.

Light, temperature and concentration of CO₂ affect opening and closing of stomata. They are not affected by O₂ concentration.

Grass being a monocot, has dumb-bell shaped stomata in their leaves.

Iron is absorbed by plants in the form of ferric ions

Plants require potassium ions (K+) for protein synthesis and for the opening and closing of stomata, which is regulated by proton pumps to make surrounding guard cells either turgid or flaccid.

Green sulphur bacteria do not use H₂O as source of proton, therefore they do not evolve O₂.

The light reactions of photosynthesis take place on the thylakoid membranes of the chloroplast. During light reactions of photosynthesis, light is absorbed and the energy is used to drive electrons from water to generate NADPH and to drive protons across a membrane. These protons return through ATP synthase to make ATP. ATP, NADPH and oxygen are products of light reaction, while NADH is a product of respiration process.

In cellular respiration, NAD+ act as an electron carrier.

In thecodont teeth are present in the sockets of the jaw bone called alveoli. In diphyodont teeth erupts twice, temporary milk or deciduous teeth are replaced by a set of permanent or adult teeth. In heterodont dentition it consists of different types of teeth namely incisors, canine, premolars and molars.

Tidal volume (TV) is volume of air inspired or expired during normal respiration. It is approximately 500 mL. Inspiratory reserve volume (IRV) is additional volume of air a person can inspire by a forceful inspiration. It is around 2500 - 3000 mL. Expiratory reserve volume (ERV) is additional volume of air a person can be expired by a forceful expiration. This averages 1000 - 1100 mL. Residual volume

(RV) is volume of air remaining in lungs even after forceful expiration. This averages 1100 - 1200 mL.

‘Smack’ also called as brown sugar/Heroin is formed by acetylation of morphine. It is obtained from the latex of unripe capsule of Poppy plant (Papaver somniferum).

Silicosis is an occupational disease caused due to excess inhalation of silica dust in the workers involved grinding or stone breaking industries.

Fibrinogen forms fibrin strands during coagulation. These strands forms a network and the meshes of which are occupied by blood cells, this structure - finally forms a clot.

Tricuspid valves are AV valve present between right atrium and right ventricle in heart. Bicuspid valves (mitral valves) are AV valve present between left atrium and left ventricle. Semilunar valves are present at the openings of aortic and pulmonary aorta.

Glycosuria denotes presence of glucose in the urine. Gout is due to deposition of uric acid crystals in the joint. Renal calculi are precipitates of calcium phosphate produced in the pelvis of the kidney. Glomerular nephritis is the inflammatory condition of glomerulus characterised by proteinuria and haematuria.

The sarcoplasmic reticulum to release calcium ions into the muscle interior where they bind to troponin, thus causing tropomyosin to shift from the face of the actin filament to which myosin heads need to bind to produce contraction.

Alzheimer’s disease is a neurodegenerative disorder due to deficiency of neurotransmitter acetylcholine. Rheumatoid arthritis is an autoimmune disorder in which antibodies are produced against the synovial membrane and cartilage. Vitiligo causes white patches on skin also characterized as autoimmune disorder. Psoriasis is a skin disease that causes itchy or sore patches of thick red skin and is also autoimmune.

Osteoporosis has three causes: excess parathyroid hor- ‘mone, advanced age, and lack of estrogen in older females. Estrogen promotes the activity of osteoblast. and inhibits osteoclast.

‘Parathormone promotes mobilisation of calcium from bone into blood. Excessive activity of parathormone causes demineralisation leading to osteoporosis.

The limbic system (emotional motor system) is responsible for the experience and expression of emotion but not movement. It is located in the core of the brain and includes the amygdala, hippocampus and hypothalamus.

Nissl bodies are present in the soma or cell body of a neuron. When observed under electron microscope, they appear to be composed of rough endoplasmic reticulum (RER) and free ribosomes hence help in protein synthesis.

The muscles that move the eyeball are attached to the sclera. Suspensory ligament of lens - a series of fibers that connect the ciliary body of the eye with the lens, holding it in place.

Epinephrine is derived from tyrosine by the removal of carboxyl group. It is a catecholamine.

An offset is a small, virtually complete daughter plant that has been naturally and asexually produced on the mother plant. They are clones, meaning that they, genetically identical to the mother plant. Offset is a vegetative part of a plant, formed by mitosis.

A monocarpic plant flowers and produces seeds only once before dying, e.g. bamboo. A polycarpic plant reproduces sexually more than once in its lifetime, , e.g jackfruit , papaya , mango etc.

Pollen grains can be stored for several years in liquid nitrogen at -196°C (cryopreservation). Cryopreservation is the process of cooling and storing cells, tissues, or org y at very low temperatures; most common -196°C/-321° in liquid nitrogen (N₂) to maintain their viability.

Double fertilization is a unique phenomenon that occurs in angiosperms only.

Double fertilization = Syngamy + Triple fusion

Ovulation is followed by the luteal phase during the corpus luteum secretes oestrogen and progesterone hormones that exerts negative feedback on hypothalamus in turn inhibits the release of FSH and LH. This prevents the development of new follicles during the luteal phase.

Spermiogenesis is conversion of spermatids info: spermatozoa whereas spermiation is the release of the sperms from sertoli cells into the cavity of seminiferous tubule.

The extraembryonic membranes are amnion, chorion allantois and yolk sac. Amnion is derived from mesoderm on the outer side and ectoderm on the inner side. Chorion is formed from trophoectoderm and mesoderm, whereas allantois and yolk sac membrane have mesoderm on outerside and endoderm in inner side.

The placenta s an endocrine gland that is only present during pregnancy. It relases hCG, hPL, progestogens, estro gens, etc. Human chorionic gonadotropic hormone (hCG) stimulates the corpus luteum: during pregnancy to release estrogen and progesterone. Human placental lactogen (hPL) is involved in growth of body of mother and breast. Progesterone maintains pregnancy.

Saheli is world’s first and only oral non steroidal contraceptive pill, dosage-once a week pill. Its functioning is based upon selective estrogen receptor modulation prevents egg from getting implanted.

'Woman acts as a carrier. Both son & daughter X-chromosome. Although only son would be the diseaséd one.

Coding strand and mRNA have the similar nucleotide

‘sequence except, Thymine (T) is replaced by Uracil (U) in mRNA

‘Francois Jacob and Jacque Monod proposed the model of gene regulation known as operon model/lac operon. Alec Jeffreys gave DNA fingerprinting technique. Matthew Meselson and F. Stahl gave semi-conservative DNA replication in E.coli. Alfred Hershey and Martha Chase proved DNA as genetic material not protein.

Semi-conservative DNA replication was first shown in bacterium Escherichia coli by Matthew Meselson and Franklin Stahl.

Unlike eukaryotes in prokaryotes the genes are organised into operon. Operon is a co-ordinated group of genes which are all transcribed together & regulate a metabolic pathway as a unit.

As per mutation theory which was given by Hugo de Vries state that the evolution is a discontinuous phenomenon or saltatory phenomenon/saltation.

Retrovirus is a type of RNA virus that inserts a copy of its genome into the DNA of a host cell that it invades, thus changing the genome of that cell. Retrovirus is commonly used as vector for introducing a DNA fragment in human lymphocyte.

PCR is based on three simple steps required for any

DNA synthesis reaction: (i) denaturation of the template into single strands; (ii) annealing of primers to each original strand for new strand synthesis; and (iii) extension of the new DNA strands from the primers.

Biopiracy refers to the use of bioresdurces by multinational companies and other organisation without proper authorisation from the countries and people concerned with compensatory payment.

Indian Government has setup organisation such as GEAC (Genetic Engineering Appraisal Committee) which will make decisions regarding the validity of GM research and safety of introducing GM-organism for public services.

In 1997, an American company got patent rights on Basmati rice through the US patent and trademark office that was actually been derived from Indian farmer’s varieties. The diversity of rice in India is one of the richest in the world, 27 documented varieties of Basmati are grown in India.

The two extremes in the animal kingdom are endothermic homeotherms and ectothermic poikilotherms. Most mammals, including humans, as well as most birds are endothermic homeotherms, while most fish, invertebrates, reptiles, and amphibians are ectothermic poikilotherms. Chelone (Turtle) belongs to class reptilia which is poikilotherm or cold blooded.

Saprophytes are organisms which live on dead organic matter and help in their decomposition. Parasites is an organisms that live in or on other living plants and animals and dependent on them for their food. Lichens represent a type of symbiotic association of algae

and fungi, in which, both of them dependent on each other

for their food and shelter.

Mycorrhiza is also a type of symbiotic association of fungi and plant roots, e.g., Cycas coralloid root.

Whenever the pre-reproductive individuals or the younger population size is larger than the reproductive group, the population will be an increasing population.

Amensalism /antibiosis, association between organisms of two different species in which one is inhibited or destroyed and the other is unaffected. These are chemicals secreted by one microbial group (eg : Penicillium) which harm other microbes (eg : Staphylococcus). It has no effect on Penicillium or the organism which produces it.

In obligate mutualism, one organism cannot survive without the other. Yucca have an obligate mutualism with a species of moth i.e. Pronuba.

World Ozone day is celebrated on 16th September. 5th June is World Environment Day. 21 April is National Yellow Bat Day. 22 April is National Earth Day.

Joseph Grinnell in 1917 coined the term niche, which he used as largely equivalent to a species habitat. It refers the functional role played by the organism where it lives.

The given data depicts the inverted pyramid of biomass, usually present in aquatic ecosystem. Upright pyramid of biomass and numbers are not possible, as the data depicts primary producer is less than primary consumer and this is less than secondary consumers. Pyramid of energy is always upright.

In-situ conservation is the on-site conservation or the conservation of genetic resources in natural populations of plant or animal species. E.g. Sacred groves.

A primary pollutant is an air pollutant emitted. directly from a source. A secondary pollutant is not directly emitted as such, but forms when other pollutants (primary pollutants) react in the atmosphere. O₃ (ozone) is a secondary pollutant.

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

Chloroftuorocarbons (CFCs) and other halogenated ozone depleting substances (ODS) are ‘mainly responsible for man-made chemical ozone depletion. CFCs rise into the stratosphere where they are eventually broken down by ultraviolet (UV) rays from the Sun. This causes them to release free chlorine which then reacts with oxygen leading to the chemical process of destroying ozone molecules.

Asthma is a common long-term inflammatory disease of the airways of the lungs. It is a difficulty in breathing causing wheezing due to inflammation of bronchi and bronchioles. Emphysema is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased, and is mainly caused by smoking.

Polytene chromosomes are found in salivary glands of insects of order Diptera.

A polyribosome (or polysome) is a complex of an mRNA molecule and two or more ribosomes that act to translate mRNA instructions into polypeptides.

Phospholipid synthesis does not take place in rough endoplasmic reticulum (RER). Smooth endoplasmic reticulum (SER) is involved in lipid synthesis.

The digestive tract of aves has additional chambers in their digestive system as crop and gizzard. crop is concerned with storage of food grains, whereas gizzard is a masticatory organ in birds used to crush food grain.

Males bear a pair of short, thread like anal styles which are absent in females. Anal/ caudal styles arise from 9th abdominal segment in male cockroach.

In renal physiology, ultrafiltration occurs, at the barrier between the blood and the filtrate in the glomerular capsule (Bowman’s capsule) in the kidneys. Concentration of urine refers to water absorption from glomerular filtrate created by counter-current mechanism in Henle’ loop. Urine is carried from kidney to bladder through ureter. Urinary bladder is for storage of urine

Golgi complex after processing, packages them in vesicles, and either stores them for later use or sends them out of the cell. It is also the organelle that builds lysosomes (cell digestion machines).

The nucleolus (plural nucleoli) is a large, distinct, spheroidal subcompartment of the nucleus of eukaryotic cells that is the site of ribosomal RNA (rRNA) synthesis and assembly of ribosomal subunits.

Ribozyme is a catalytic RNA, which thus nucleic acid.

Starch synthesis in pea is controlled by pleiotropic gene. Pleiotropy occurs when one gene influences two or more seemingly unrelated phenotypic traits.

Natality is the birth rate within a population. When compared with the death or mortality rate, the growth or decrease in a population can be determined.

Herbarium - Dried and pressed plant specimen.
• Key - Identification tool of various taxa.
• Museum - Plant and animal specimen are prese

The gymnosperms are a group of seed-producing plants. The name is based on their unenclosed condition of their seeds (called ovules in their unfertilized state).

Sweet potato is a modified adventitious root for storage of food. Rhizomes are underground modified stem. Tap root is primary root directly elongated from the radicle.

Curd has enriched presence of B vitamins specially Vit-B₁₂ which improves its nutritional value than milk.

Parietal or oxyntic cell is a source of HCl and intrinsic factor. HCl converts iron present in diet from ferric to ferrous form so that it can be absorbed easily and used during erythropoiesis. Intrinsic factor is essential for the absorption of vitamin B₁₂ and its deficiency causes
pernicious anaemia.

Coefficient of friction or sliding friction has no dimension.

f = μₛN ⇒ μₛ = f/N = [M⁰L⁰T⁰]

Due to high electronegativity and small size, F forms only one oxoacid, HOF known as fluoric (I) acid.

Oxidative phosphorylation takes place in inner mitochondrial membrane.

IAIO, IBIO – Dominant-recessive relationship

IAIB – Codominance

IA, IB & IO – Three different allelic forms of a gene (multiple allelism)

Divergent evolution demonstrates how species can have common (homologous) anatomical structures, where as eye of octopus, bat and man are examples of analogous rgans showing convergent evolution.

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Bird and bat wings are analogous, as forelimb is homologous. In different vertebrates, bones of forelimbs are similar but their forelimbs are adapted in different way as per their adaptation, show homology.

‘Smack’ also called as brown sugar/Heroin is formed by acetylation of morphine. It is obtained from the latex of unripe capsule of Poppy plant (Papaver somniferum)