NEET-UG 2006 Question Paper Solutions

Dimension of a.t = dimension of velocity a.t = [LT⁻¹]
⇒ [a = LT⁻²]
Dimension of c = dimension of t
(two physical quantity of same dimension can only be added)
So, dimension of c = [T]
Dimension of b / t + c = Dimension of velocity
[b] / [T] = [LT⁻¹] ⇒ [b · T⁻¹] = [LT⁻¹] ⇒ b = [L]
So, answer is [LT⁻²], [L] and [T]

x = 40 + 12t − t³
v = dx / dt = 12 − 3t²
For v = 0; t = √(12 / 3) = 2 sec
So, after 2 seconds velocity becomes zero.
Value of x in 2 sec.
= 40 + 12 × 2 − 2³
= 40 + 24 − 8 = 56 m

Let t₁ & t₂ be the time taken by A and B respectively to reach the ground then from the formula,
h = (1 / 2)gt²,
For first body,
16 = (1 / 2)gt₁²

For second body, 25 = (1 / 2)gt₂²
∴ 16 / 25 = t₁² / t₂² ⇒ t₁ / t₂ = 4 / 5.

Horizontal range for projection angle
(45° − θ) is, R₁ = u² sin 2(45 − θ) / g
Horizontal range projection
angle (45° + θ) is, R₂ = u² sin 2(45 + θ) / g
According to the condition,
⇒ R₁ / R₂ = u² sin 2(45 − θ) / u² sin 2(45 + θ) = sin(90 − 2θ) / sin(90 + 2θ)
⇒ R₁ / R₂ = cos 2θ / cos 2θ = 1 / 1
So, R₁ : R₂ : 1 : 1

Distance covered in one circular loop = 2πr
= 2 × 3.14 × 100 = 628 m
avg Speed = 628 / 62.8 = 10 m/sec
Displacement in one circular loop = 0
avg Velocity = 0 / time = 0

The magnitude of both velocity v₁ and v₂ are equal. Components of velocity along the wall are equal and are in same direction.

But the component of velocity perpendicular to wall are equal but opposite in direction.

So, net change in momentum in perpendicular direction,

ΔP = mv₁ sin 30° - (-mv₂ sin 30°)

⇒ 2mv sin 30° [As v₁ = v₂ = v]

Average force acting on the wall will be given by

⇒ Ft = 2mv sin 30°

⇒ F = 2mv sin30° / t

⇒ F = 2 × 0.5 × 12 × 1 / 2 × 0.25 = 24

F = 24 N

|𝐀⃗ + 𝐁⃗|² = |𝐀⃗ - 𝐁⃗|²
= |𝐀⃗|² + |𝐁⃗|² + 2𝐀⃗· 𝐁⃗ = A² + B² + 2AB cos θ
and, |𝐀⃗ − 𝐁⃗|² = |𝐀⃗|² + |𝐁⃗|² − 2𝐀⃗ · 𝐁⃗
= A² + B² − 2AB cos θ
So, A² + B² + 2AB cos θ
= A² + B² − 2AB cos θ
4AB cos θ = 0 ⇒ cos θ = 0
θ = 90°
So, angle between A & B is 90°.

Acceleration = d²s / dt² = 2 / 3 m/s²
Force acting on the body
= 3 × 2 / 3 = 2 newton
Displacement in 2 secs = 1 / 2 × 2 / 3 × 2² = (4 / 3) m
Work done = 2 × 4 / 3 = (8 / 3) J

Work done against gravity = mg sin θ × d
(d sin θ = 10)
= 2 × 10 × 10
= 200 J
Actual work done = 300 J
Work done against friction = 300 − 200 = 100 J

Rotation will create centrifugal force due to which pressure will be created at the other end. Now, this centrifugal force will depend on distance of the liquid from the axis as centrifugal force = mω²r.

Let us consider a small part dx of the liquid column which is at a distance of x.

Mass of this element will be,

dm = (M / L)dx

So, force exerted on this element will be,

⇒ df = dm.ω²x

As the mass of liquid extends from x = 0 to x = L, the net force on the wall experienced by the mass of the liquid is,

M.I. of a uniform circular disc of radius ‘R’ and mass ‘M’ about an axis passing through C.M. and normal to the disc is
I_C.M. = 1 / 2 MR²
From parallel axis theorem,
I_T = I_C.M. + MR²
= 1 / 2 MR² + MR²
= 3 / 2 MR²

Potential energy at height R, PE = −GMm / 2R
If the mass of a body which is thrown with velocity vₑ so that it goes out of gravitational field from distance R, then
1 / 2 mvₑ² = (GM / 2R).m
⇒ vₑ = √(GM / R)
or, vₑ = √gR
Now, v = √2gR, so, v = √2 vₑ
or, vₑ = v / √2
Comparing it with given equation,
f = 1 / √2

Applying Wien’s displacement law,
λₘT = constant
5000 Å × (1227 + 273)
= (2227 + 273) × λₘ
λₘ = (5000 × 1500) / 1000 = 3000 Å

We know that efficiency of Carnot Engine
η = (T₁ − T₂) / T₁
where, T₁ is temp. of source and T₂ is temp. of sink
∴ 0.40 = (T₁ − 300) / T₁
⇒ T₁ − 300 = 0.40T₁
0.6T₁ = 300 ⇒ T₁ = 300 / .6
= 3000 / 6
= 500 K

Now efficiency to be increased by 50%
∴ 0.60 = (T₁ − 300) / T₁
⇒ T₁ − 300 = 0.6T₁
0.4T₁ = 300
⇒ T₁ = 300 / .4
= (300 × 10) / 4
= 750
Increase in temp. = 750 − 500 = 250 K

PV / T = constant

C_P = 7 / 2 R; C_V = C_P − R
= (7 / 2)R − R = (5 / 2)R

C_P / C_V = (7 / 2)R / (5 / 2)R = 7 / 5

The potential energy of a spring = (1 / 2) kx²
U = (1 / 2) k.(2)² = 4 × (1 / 2) k
For x = 8 cm,
Energy stored = (1 / 2) k.(8)² = 64 × (1 / 2) k
= 64 × U / 4 = 16U

Restoring force, F = kx = Axρg
⇒ k = Aρg
∴ T = 2π √(m / k) = 2π√(m / Aρg)

Sound waves in air are longitudinal and the light waves are transverse.

Speed of a wave is represented by the equation
y(x,t) = A sin (kx − ωt + φ) is v = ω / k
By comparison, ω = - 4π; k = - 0.5π
v = ω / k = 4π / 0.5π = 8 m/sec

The time gap between the initial direct note and the reflected note upto the minimum andibility level is called reverberation time.
Sabine has shown that standard reverberation time for an auditorium is given by the formula
T_R =K(V / αS)
Where, V is volume of the auditorium, S is the surface area. So,
T_R = K.V / α S = 1 (given)
T_R = K·l³ / 6αl²
(Assuming auditorium to be cubic in shape)

= (K / 6α)l
So, T_R ∝ l

If dimension is doubled, reverberation time t will be doubled.
So, New T_R = 2 sec.

Frequencies of sound waves are 330 / 5 and 330 / 5.5 i.e., 66 Hz and 60 Hz
Frequency of beat = 66 − 60 = 6 per second

Flux = 𝐄⃗. 𝐀⃗

𝐄⃗ is electric field vector & 𝐀⃗ is area vector.
Here, angle between 𝐄⃗ & 𝐀⃗ is 90°.
So, 𝐄⃗. 𝐀⃗ = 0 ; Flux = 0

If we increase the distance between the plates its capacity
decreases resulting in higher potential as we know Q = CV. Since Q is constant (battery has been disconnected), on decreasing C, V will increase.

Power = V × I
I = Power/ V = 100 / 125
E.C.E. of chlorine is 0.367 × 10⁻⁶ kg/coulomb
Charge passing in one minute = (100 / 125) × 60
= 48 coulomb
Chlorine precipitated = 0.367 × 10⁻⁶ × 48
= 17.6 × 10⁻⁶ kg

= 17.6 mg

Kirchhoff's law deals with conservation of electrical charge and the second law deals with conservation of energy.

Current in the circuit
= E + E / r₁ + r₂ + R = 2E / (r₁ + r₂) + R
P.D. across first cell = E − ir₁
E −((2E × r₁) / (r₁ + r₂) + R)
Now, E = 2Er₁ / (r₁ + r₂) + R) = 0

⇒ E = 2Er₁ / (r₁ + r₂) + R) ⇒ 2r₁ = r₁ + r₂ + R
R = r₁ − r₂

Power = V.I = I²R
i₂ = √(Power / R) = √(2 / 8) = √(1 / 4) = (1 / 2)A

Potential over 8Ω = Ri₂ = 8 × 1 / 2 = 4V
This is the potential over parallel branch. So,
i₁ = 4 / 4 = 1A
Power of 3Ω = i₁²R = 1 × 1 × 3 = 3W

Current will flow from B to A.

Potential drop over the resistance CA will be more due to higher value of resistance. So potential at A will be less as compared with B. Hence, current will flow from B to A.

Same wire ⇒ Resistance ∝ length ∝ radius.

R₁ : R₂ = 2 : 1

Magnetic field at the center:

B = μ₀I/2r

For equal magnetic fields:

I₁/ r₁ =I ₂ / r₂

Since r₁=2r₂

I₁ / 2I₂

Using V= IR :

V₁ / V ₂ = I₁ / I ₂ R₁ / R₂ = 2 * 2 = 4

V₁ : V ₂ 4 : 1

Force on a particle moving with velocity v in a magnetic field B is
F = q (𝐯⃗and 𝐁⃗)
If angle between 𝐯⃗ & 𝐁⃗ is either zero or 180°, then value of F
will be zero as cross product of 𝐯⃗and 𝐁⃗ will be zero.

So option (angle between 𝐯⃗and 𝐁⃗ can have any value other than zero and 180°) is correct.

Curie temperature is the temperature above which ferromagnetic material becomes paramagnetic material.

Mutual Inductance of two coils
M = √(M₁M₂)
= √(2mH × 8mH)
= 4mH

Power factor, cos φ = R / √[(ωL − 1 / ωC)² + R²]
= 8 / √((31 − 25)² + 8²)
= 8 / √(6² + 8²)
= 8 / 10
= 0.8

When there is change of flux in the core of a transformer due to change in current around it, eddy current is produced. The direction of this current is opposite to the current which produces it, so it will reduce the main current. We laminate the core so that flux is reduced resulting in the reduced
production of eddy current.

We know that frequency of electrical oscillation in L.C.
circuit is
f = (1 / 2π)√LC
Now, L = 2L & C = 4C
f′ = (1 / 2π)√(1 / 2L · 4C) = (1 / 2π) √(1 / LC) × 1 / 2√2
⇒ f′ = 1 / (2√2) × f

From the formula,
1 / f = (1 / f₁) + (1 / f₂) = (1 / 25) - (1 / 25) = 0
Power of combination = 1 / f = 0

In the later case microscope will be focussed for O′. So, it is required to be lifted by distance OO′.

OO′ = real depth of O - apparent depth of O.

= 3 − 3 / 1.5
[ μ = real depth / apparent depth ]
= 3[(1.5 − 1) / 1.5]
= (3 × .5) / 1.5 = 1 cm

When electrons emitted from cathode collide with gas molecules or atoms, they knock out outer electrons and produce positively charged ions. They become part of positive rays.

Applying Einstein's formula for photo-electricity
hν = ϕ + 1 / 2 mv²; hν = ϕ + K
ϕ = hν − K
If we use 2ν frequency then let the kinetic energy becomes K′
So, h . 2ν = ϕ + K′
2hν = hν − K + K′
K′ = hν + K

A photo-cell employs photoelectric effect to convert light energy into photoelectric current.

1 MeV = 10⁶ × 1.6 × 10⁻¹⁹ joule
Momentum of photon,
p = E / c = 1.6 × 10⁻¹³ / 3 × 10⁸

p = (1.6 / 3) × 10⁻²¹ = 16 / 3 × 10⁻²²
p = 5 × 10⁻²² kg m/sec

Energy of ground state = 13.6 eV
Energy of first excited state
= −13.6 / 4 = −3.4 eV
Energy of second excited state
= −13.6 / 9 = −1.5 eV

Difference between ground state and 2nd excited state
= 13.6 − 1.5 = 12.1 eV
So, electron can be excited upto 3rd orbit.
No. of possible transition
1 → 2, 1 → 3, 2 → 3
So, three lines are possible.

We use the formula,
R = R₀A¹ᐟ³

This represents relation between atomic mass and radius of the
nucleus.

For beryllium, R₁ = R₀(9)¹ᐟ³

For germanium, R₂ = R₀(A)¹ᐟ³

R₁ / R₂ = (9)¹ᐟ³ / (A)¹ᐟ³ ⇒  1 / 2 = (9)¹ᐟ³ / (A)¹ᐟ³
⇒ 1 / 8 = 9 / A ⇒ A = 8 × 9 = 72

₁D² → ₂He⁴
Energy released
= 28 − 2 × 2.2
= 23.6 MeV

Let at time t₁ & t₂, number of particles of Nucleus be N₁ & N₂. So,
R₁ = dN₁ / dt = = −λN₁;
R₂ = dN₂ / dt = −λN₂

R₁ / R₂ = λN₁ / λN₂ = N₁ / N₁e_(-λ(t₂−t₁)) = e_λ(t₂−t₁)
= R₁ = R₂e_λ(t₂−t₁)
= R₂e_(−λ(t₁−t₂))

In forward bias, V₁ > V₂ i.e., in figure p-side is at higher potential w.r.t. n-side of p-n junction diode.

ΔI_b = +50 μA, ΔI_c = 5 × 10⁻³ A

β = ΔI_c / ΔI_b

= (5 × 10⁻³) / (50 × 10⁻⁶)

= (5 × 1000) / 50

= 100

On the basis of given graph, following table is possible.

It is the truth table of AND gate.

Weknow that

Δx . Δp ≥ h / 4π

Δx . mΔv > h / 4π ⇒ Δv > h / 4π Δ xm

Δv = 6.626 × 10⁻³⁴ / 4π × 0.1 × 10⁻¹⁰ × 9.11 × 10⁻³¹

= (66 / 4π × 9) × 10⁷ = 5.79 × 10⁶ ms⁻¹

Magnetic quantum no. represents the orientation of atomic orbitals in an atom. For example pₓ, p_y & p_z have orientation along X-axis, Y-axis & Z-axis.

Statement (The canonical structures have no real existence), (Multiple bonds are always shorter than corresponding single bonds), (The electron-deficient molecules can act as Lewis acids) are correct. Statement (Every AB₅ molecule does in fact have square pyramidal structure) is incorrect statement.

AB₅ may have two structures as follows:

In NH₃ the atomic dipole and bond dipole are in the same direction whereas in NF₃ these are in opposite direction so in the former case they are added up whereas in the latter case net result is reduction of dipole moment. It has been shown in the following figure :

In SiCl₄ there is sp³ hybridisation, so the structure is tetrahedral. In SO₄²⁻, PO₄³⁻, NH₄⁺ the structure is tetrahedral with sp³ hybridisation. But in SCl₄, sp³d hybridisation is present, so its shape is different i.e., see saw.

NO₂⁺ will have linear shape as it will have sp hybridisation.

In BF₃, AlF₃ & NF₃ all fluoride atoms are symmetrically oriented with respect to central metal atom but in ClF₃ three fluorine atoms are arranged as follows :

Here two bonds are in equatorial plane & one bond is in axial plane.

Electronic configuration of the molecule according to molecular orbital theory, is

σ1s² σ*1s² σ2s² σ*2s² σ2p_z² (π2pₓ² = 2p_y²)

(π*2pₓ = π2p_y)

Last two electrons are unpaired. So no. of unpaired electron is 2.

We know that
ΔH = ΔE + PΔV
In the reaction, H₂ + Br₂ → 2HBr
this is no change in volume or ΔV = 0 as number of gaseous moles are same.
So, ΔH = ΔE for this reaction

The resonance energy provides extra stability to the benzene molecule so it has to overcome for hydrogenation to take place.

So ΔH = −358.5 − (−150.4) = −208.1 kJ

We know that
ΔG = ΔH − TΔS
When the reaction is in equilibrium, ΔG = 0
0 = ΔH − TΔS
⇒ T = ΔH / ΔS

T = (2 × 30 × 1000) / (2 × 105) = 285.7 K

If ΔG_system = 0 the system has attained equilibrium.
If ΔG < 0, the process is spontaneous
If ΔG > 0, the process does not occur in the forward direction.
If may occur in the backward direction.

The equilibrium constant for the reaction is given by Kₚ = [CO₂] / [CH₄][O₂]² option is incorrect as the value of Kₚ given is wrong.
It should have been
K_P = P_CO₂ / P_CH₄ × (P_O₂)²

For a solution of 10⁻⁸ M HCl, [H⁺] = 10⁻⁸
[H⁺] of water = 10⁻⁷
Total [H⁺] = 10⁻⁷ + 10⁻⁸ = 10 × 10⁻⁸ + 10⁻⁸
10⁻⁸ (10 + 1) = 11 × 10⁻⁸

HNO₂ is a weak acid and NaNO₂ is salt of that weak acid and strong base (NaOH).

Smaller the ion more is its ionic mobility in aqueous solution.
Ionic radii of the given alkali metals is in the order Na⁺ < K⁺ < Rb⁺ < Cs⁺ and the expected ionic mobility will be in the same order
Cs⁺ < Rb⁺ < K⁺ < Na⁺. However due to high degree of solvation
(or hydration) and because of lower size or high charge density,
hydrated ion size follows the same order Cs⁺ < Rb⁺ < K⁺ < Na⁺ and
thus conductivity order is Cs⁺ > Rb⁺ > K⁺ > Na⁺.

Al₂O₃ can be converted to anhydrous AlCl₃ by heating a mixture of Al₂O₃ and carbon in dry Cl₂ gas.

General molecular formula of alkanols is
CₙH₂ₙ ₊₂o, (CₙH₂ₙ₊₁OH)

It is 2,3-dimethylpentanoyl chloride.

Among the three given hybrid orbitals, sp hybrid orbital is most electronegative.

In body centred cubic lattice one molecule of CsBr is within one unit cell.
Atomic mass of unit cell = 133 + 80 = 213 a.m.u.
Volume of cell = a³ = (436.6 × 10⁻¹⁰)³ cm³
Density = (Z × M) / (N_A × a³)
Density = (1 × 213) / (6.02 × 10²³ × (436.6)³ × 10⁻³⁰)
= 213 × 10⁷ / 6.02 × (436.6)³
= 4.25 g/cm³

F-centres

A solution of acetone in ethanol shows positive deviation from Raoult's law. It is because ethanol molecules has strong hydrogen bond. When acetone is added, they hydrogen bonds of ethanol becomes is weaker. Therefore, its volatility and vapour pressure increases.

The thalloid body of slime moulds is made up of multinucleate cell which lacks septa in between and hence it is multinucleated single celled mass called plasmodium.

Trypanosoma, Noctiluca, Monocystis & Giardia are all unicellular protists. unicellular eukaryotes.

Mucor is a fungus and most of the fungi require the optimum temperature of about 15–30°C, good moisture content in atmosphere and not very dark and not very lighted place. So Mucor requires a temperature of about 25°C, humidity about 95% and a shady place to grow fully.

Mad cow disease is actually Bovine Spongiform Encephalopathy or BSE. In this disease cattle in Britain got spongy brain and it ultimately gradual degradation of nervous system. It is caused by some virus like but nucleic acid devoid proteinaceous particles called prions (proteinaceous infectious particle).

Peat moss is used wherever we require to retain water in a long because peat mosses are hygroscopic in nature and they absorb the moisture from the atmosphere and this moisture keep living materials and flowers fresh for a long time.

Conifers Gymnosperms differ from grasses angiosperms (angiosperms) because in gymnosperms the female gametophyte is actually endosperm which is made before fertilisation, while in angiosperm it is a tissue formed by the fertilisation of second male gamete to polar nuclei. Moreover in gymnosperms the endosperm is a haploid tissue while in angiosperms it is triploid.

Jointed legs & chitinous exoskeleton are the common characters found in centipede, cockroach & crab.

Aurelia, Hydra & Sea anemone have the characteristic cnidoblast cells. Cnidoblast is shown in the given figure.

Metameric segmentation means body is divided externally as well as internally. This characteristic is present in annelida (e.g. earth worm) and arthropoda (e.g. cockroach). Their body is divided externally and internally as well.

Mollusca is the second-largest phylum of invertebrate animals after the Arthropoda. The members are known as molluscs or mollusks. They are soft-bodied, bilaterally symmetrical, segmented, coelomate animals; usually shelled having a mantle, ventral foot, anterior head, and a dorsal visceral mass. Loligo, Ptero, Octopus are some examples of Mollusca.

Penguin, Kiwi & Ostrich all belong to class Aves of chordata (i.e. birds) and they do not give birth to young ones, they are oviparous, while Kangaroo, Hedgehog, Dolphin, Loris all belong to class mammalia and are viviparous.

Salamander does not undergo any annual migration as it occurs in Siberian crane. Arctic tern and Salmon mainly for the search of food or for breeding.

In sweet pea, the gynoecium is monocarpellary and bears a single locule along the junction of two fused margins, this is called marginal placentation.

Long filamentous threads of maize are the styles of ovaries. In fact, these are the longest styles in plants.

Single cotyledon of embryo in cereal grain is represented by scutellum. Coleoptile represents the covering of stem. Coleorhiza represents the covering of root.
Notes:
Prophyll is a plant structure resembling a leaf (as a bracteole) or consisting of a modified or rudimentary leaf (as a floral primordium). The coleoptile is a rigid piece of plant tissue that encloses the first few leaves (or plumule) of the corn seedling as the mesocotyl elongates and "pushes" the coleoptile to the soil surface during the emergence process. The coleorhiza is a protective sheath that envelops the radicle in certain plants that is penetrated by the root immediately. It helps in protecting the radicle.

Pineapple (Ananas sativus) is a sorosis fruit developed from a composite inflorescence which is actually a cluster of compactly borne flowers on a common axis.

Aril develops from ovule coat and is the edible part of litchi.
Notes:
An aril (or arillus) is a fleshy covering of certain seeds formed from the funiculus attachment point of the seed. The aril may create a fruit-like structure called a false-fruit and is produced by a few species of gymnosperms, notably the yews and related conifers in the families Taxaceae and Cephalotaxaceae.

In plants, conducting tissue xylem has an important integral cell as xylem vessel which is enucleate. The phloem on the other hand has a row of sieve tubes which are also enucleate at maturity.

Areolar connective tissue has rich supply of nerve fibres and blood vessels. It joins skin epithelia with muscles, nerves & blood vessels. Binding of body parts together is the main function of areolar connective tissue.

Mast cell release histamine which increases the permeability of blood vessels at the site of inflammation so that more blood supply takes place. The site gets warm and red due to this.

A major break through in cell studies came with the development of EM having great resolution power due to which we can see the ultrastructures of cell organelles.
Notes:
There are several different types of electron microscopes, including the transmission electron microscope (TEM), scanning electron microscope (SEM), and reflection electron microscope (REM). The electron microscope uses a beam of electrons and their wave-like characteristics to magnify an object's image, unlike the optical microscope that uses visible light to magnify images. This stream is confined and focused using metal apertures and magnetic lenses into a thin, focused, monochromatic beam.

Eukaryotic cilia are made up of tubulin protein. Cilia have a core of 2 microtubules and around there are 9 doublet microtubules. These are not all hair like structures.
Notes:
Microtubules are polymers of tubulin that form part of the cytoskeleton and provide structure and shape to eukaryotic cells. Microtubules can grow as long as 50 micrometres and are highly dynamic. The outer diameter of a microtubule is between 23 and 27 nm while the inner diameter is between 11 and 15 nm.

The enzymes of electron transport system are present in inner mitochondrial membrane.

2nm is the correct answer

Length of one turn of DNA helix is 34 Å.
1 Å = 10⁻¹ nm
34⁻ = 10⁻¹ × 34 = 34 × 1/10 nm
= 3.4 nm

Coenzyme is an organic substance that enhances the action of an enzyme by binding with the protein molecule. Holoenzyme is a biochemically active compound formed by the combination of an enzyme with a coenzyme. Apoenzyme is the protein part of an enzyme to which the coenzyme attaches to form an active enzyme. Isoenzyme is one of several forms of an enzyme that catalyse the same reaction but differ from each other in such properties as substrate affinity and maximum rates of enzyme substrate reaction.

P -Protein is the supporting factor for the translocation of organic solutes in sieve tubes.

Because iron promotes the formation of chlorophyll and magnesium is an integral part of chlorophyll molecule so in the absence of these nutrients plants show chlorosis & leaves get yellow. Application of iron and magnesium to promote the synthesis of chlorophyll can cure the disease.
Notes:
The lack of iron is one of the more common nutrients associated with chlorosis. Manganese or zinc deficiencies in the plant will also cause chlorosis. The way to separate an iron deficiency from zinc or manganese deficiency is to check what foliage turned chlorotic first .

In photosystem I electrons released by P-700 are accepted by a substance called iron-sulphur complex denoted as A(FeS) which is thus reduced and it passes these high electrons to oxidised ferredoxin and it gets reduced.

Photorespiration is the process which occurs in C₃ plants. In this process, peroxisomes, chloroplast and mitochondria take part. The oxygen consuming reactions occur in peroxisomes and stroma of chloroplast while CO₂ releasing reaction occurs in mitochondria.

Clostridium botulinum is an obligate anaerobe i.e. it normally lives in the absence of oxygen. Facultative anaerobes are those who generally live in oxygen but may live without oxygen in suitable medium. Obligate aerobes can only live in the presence of oxygen while facultative aerobes generally live in oxygen but can also live without oxygen.
Notes:
Botulism is a rare and potentially fatal illness caused by a toxin produced by the bacterium Clostridium botulinum. The disease begins with weakness, blurred vision, feeling tired, and trouble speaking.

12 kcal of energy is present in one molecule of ATP & on oxidation of one mole of glucose into CO₂ and H₂O energy released is 686 kcal. So no. of ATP which can store this energy would be = 57.1 = 57 ATPs.

a-amylase enzyme stimulates the germination of barley seed. Barley seeds are rich in carbohydrate (starch). The starch is hydrolysed by α-amylase to monosaccharide unit at the time of germination of seeds.

When an apical bud is present on a plant, it suppresses the growth of axillary buds, this is called apical dominance. When apical bud is cut off the axillary buds start growing & hedge become dense.

Treated seeds at a low temperature under moist conditions to break dormancy is called stratification.

The amount of gases remaining in the lungs even after a forcible expiration is known as residual volume. Variation in the level of inspiratory and expiratory gases is maintained by the residual volume. So residual air in lungs does not affect the respiratory efficiency.

Transport of CO₂ is much easier than O₂ because of the higher solubility of CO₂ than O₂. Almost 7% of CO₂ is transported in dissolved state in plasma. About 20–25% CO₂ is transported by RBCs. 70% of CO₂ is transported in the form of bicarbonate ions. CO₂ reacts with water present in plasma.
Notes:
Carbonic acid is unstable & gives H⁺ and HCO₃⁻. H⁺ bind with haemoglobin to maintain the pH of blood while HCO₃⁻ remain in the blood and carries the CO₂ to lungs.

Periplaneta (P. americana, the Indian Cockroach) is an insect & insects do not have closed blood circulation. Their blood (called haemolymph) even does not serve for respiration. They do not possess blood vessels in this circulatory system and hence blood is an open system.

Folic acid and cobalamine is the correct answer.

The correct matching will be:
• Protonephridia or flame cells are the excretory structures in platyhelminthes.
• Nephridia are the tubular excretory structures of earthworms (Phretima) and other annelids.
• Single celled organisms like Paramoecium have contractile vacuoles for excretion.
• Urecose glands are present in cockroach.

Those animals that excrete ammonia are called ammonotelic, e.g. aquatic amphibia. Those animals that excrete urea are called ureotelic, e.g. frog, humans.
Those animals that excrete uric acid are called uricotelic, e.g. pigeon, lizards, and cockroach.

A – Adrenal gland – located at the anterior part of kidney, secrete Catecholamines which stimulate glycogen breakdown.

Angiotensinogen is a protein secreted by the liver cells.

Actin and tropomyosin are part of thin filaments of skeletal muscle. Tubulin is present in microtubules. Myosin is muscle protein.

Bowman's glands are found in olfactory epithelium which secrete mucus to keep the olfactory epithelium moist and free from germs.
Notes:
Smell detects volatile chemicals in the inhaled air, and taste detects dissolved chemicals in food and drink. Olfactory receptor cells have long cilia that protrude into a mucus layer that is secreted by Bowman's glands in the olfactory epithelium.

Endocrine glands regulate neural activity as endocrine glands secrete epinephrine & norepinephrine which have their effects over neuron activity.

During spermatogenetic developments spermatids get nourished by nurse or Sertoli cells. The spermatogenesis is under the direct influence of follicle-stimulating hormone (FSH) that is secreted by adenohypophysis of pituitary gland.

Neurohormones are actually secreted by the neurosecretory cells of the hypothalamus. They are circulated to the posterior part of the pituitary gland through the blood & stored there and released when required.

Adrenaline (called Fight-Flight hormone) is responsible for the dilation of blood vessels, increased oxygen consumption by tissues & glycogenesis to prepare the body for emergency reactions under life threatening conditions.

Cortisol is the steroid hormone secreted by adrenal cortex and plays an important role in carbohydrate metabolism. It retards the glucose consumption & hence level of glucose in blood increases & blood pressure increases.

There is no role of sodium in hormonal action. In heart cells cAMP acts as secondary messenger which is made in the response of adrenaline and it stimulates Ca²⁺ ions to come out from the sarcoplasmic reticulum of muscle fibres which causes muscle contraction.
cGMP (Cyclic Guanosine Monophosphate) which acts as secondary messenger works on the action of acetylcholine, increase in flow of Ca²⁺ into muscle fibres & hence causes muscle relaxation.

Epinephrine (adrenaline), norepinephrine (noradrenaline) and acetylcholine are the neurotransmitters. These are released by the nerve fibres to transmit the impulse to the next neuron. Cortisone is not the neurotransmitter.

In dicot plant general arrangement of nuclei in the embryo sac is that 3 nuclei in 3 cells of egg apparatus (one egg cell & 2 synergids) at micropylar end and 2 polar nuclei at almost middle region and 3 nuclei in the form of antipodals at the distal end of micropyle, towards nucellus. So total nuclei = 3 + 2 + 3.

Aleurone layer is the layer formed by the endospermic tissue in the members of family Graminae e.g. wheat, rice, maize etc. Obviously it will be a triploid tissue. So if synergids have 8 no. of chromosomes which are haploid then aleurone-layer must have 24 chromosomes.

Progesterone hormone is secreted by the corpus luteum on ovaries and is responsible to stop ovulation (during pregnancy) and fixes the foetus to the uterine wall, facilitates placentation and controls the foetus development in uterus. If this hormone is withdrawn then fixation of foetus to uterine wall will not take place, formation of placenta will also not take place, ovulation starts from the ovaries and hence in human female menstruation starts.

Polygenic inheritance is the inheritance of traits which are dependent on the number of genes such as the skin colour of human beings, e.g. three genes A, B and C. Genotype AABBCC will have darkest skin colour and genotype aabbcc will have light skin colour. The number of each type of alleles in the genotype would determine the darkness or lightness of the skin of an individual.
Notes:
Polygenic inheritance was discovered in 1909 by the Swedish scientist H. Nilsson-Ehle, who studied the inheritance of grain colour in wheat by the analytical dissociation of the character.

So the ratio of Round seeded with yellow cotyledons : Round seeded with green cotyledons : Wrinkled seeded yellow cotyledons : Wrinkled seeded green cotyledons is 9 : 3 : 3 : 1. So in the F₂ generation, the round seeded character remains with yellow cotyledons & wrinkled seeds will be yellow cotyledons.

A cross of F₁ hybrid with its recessive homozygous parent is called the test cross. It is done to determine the genotype of a given plant. If the given plant has homozygous dominant traits then test cross gives all dominant trait plants but if it is heterozygous dominant then it gives dominant and recessive phenotypes in 1 : 1 ratio.

Phenotype is the appearance of one organism while genotype is the genetic complement it has from its ancestors. These genes only show their effect in phenotype but environmental also plays an important role in this. Hence phenotype is a result of genotype and environmental interaction.

It would make only two types of gametes, these are ABC & AbC.

The genes of cytoplasmic characters occur outside the chromosome, in cytoplasm, they inherit their traits like chromosomal genes do. In a zygote formation, the cytoplasmic comes from the mother's ova and hence the cytoplasmic characters must come from maternal side and this is the reason that cytoplasmic inheritance is also known as maternal inheritance.

Colour blindness is X chromosome linked character. So they'll be having all colour blind sons & carrier daughters.

Cri-du-chat syndrome in humans is caused by the partial deletion of the short arm of chromosome no. 5. This child has abnormalities in his/her facial expressions, mental retardation, they speak and cry like that of cat, heart does not work properly.

Sickle cell anaemia is a biochemical disorder in which shape of RBCs become sickle-shaped due to the defective haemoglobin. Haemoglobin becomes useless for oxygen transport.
Huntington's Chorea is a disease in which atrophy of brain occurs resulting in respiratory irregularities, articulation of speech and irregular limb movements. They both are genetic diseases present in any person since birth hence congenital diseases. .

In sickle cell anaemia, RBCs become sickle shaped

which are not supportive for the growth of malarial parasite

Plasmodium so it provides immunity against malaria disease.

Neomycin is the antibiotic which inhibits the translation of bacterial cell so that it cannot affect the host cell.

Beadle & Tatum postulated the theory of "One gene-one enzyme" in which they stated that in any living cell there are numerous of genes present on chromosomes in a linear fashion. One single gene controls the synthesis of one particular enzyme (or protein) which is responsible for its phenotypic character.

Glutamic Acid is one of the non-essential amino acids, biochemically related to glutamine. The two constitute a substantial fraction of the amino acids in many proteins (10–20% in many cases and up to 45% in some plant proteins). An important metabolic intermediate as well as a neurotransmitter molecule in the central nervous system, glutamic acid finds uses in medicine and biochemical research.
Notes:

Its sodium salt is the food flavour enhancer monosodium glutamate (MSG).

Paleontology is the study of fossils. Ontogeny is the process of development of organism in the embryo stage and phylogeny is known as the history of an organism through ages during evolution.

Homologous organs represent the divergent evolution. These are the organs belonging to different organisms of different species may be for dissimilar function but they all have the same basic structure. On the other hand vestigial organs

show that how a species evolved from its ancestors through the

ages according to the use and disuse of organs
Notes:
These are the organs which were sometime functional in humans (or may be any other organisms) but during the course of evolution their utility gradually declined and now they are present as non-functional and vestigial organs.

Jurassic period of Mesozoic era was about 19–20 crore years ago & lasted for about 5.5–6 crore years. The climate was hot and damp. It is called the age of dinosaurs. Its primitive bird Archaeopteryx evolved from reptiles. Its angiosperm appeared as dicotyledon but gymnosperms were dominant.

Archaeopteryx is a fossil (dead) found from the Mesozoic rocks. It is a connecting link between reptiles and birds.

Infant Methemoglobinemia is also called "blue baby syndrome". It is a condition where a baby's skin turns blue. This happens when there is not enough oxygen in the blood. Methemoglobinemia is a condition that some babies are born with (congenital) or some develop early in life (acquired).

Notes:

The most common cause of blue baby syndrome is water contaminated with nitrates. After a baby drinks formula made with nitrate-rich water, the body converts the nitrates into nitrites. These nitrites bind to the haemoglobin in the body, forming methemoglobin, which is unable to carry oxygen.

Whenever our body gets attacked by some foreign invaders, our body's immune system produces some chemicals to kill or react against the invader. These chemicals are actually made up of carbohydrates and proteins i.e. glycoproteins called antibody.

AIDS is characterized by reduction in the number of helper T-lymphocytes because of HIV infection. It suppresses human immune system due to which any secondary infection may lead to death. T-lymphocytes are main cells of immune system.

In Borderline Personality Disorder (BPD) a person suffers from emotionally unstable personality, unpredictable moods, highly reactive, anxiety and irritability. While in Schizophrenia a person suffers from distorted thoughts, he/she shows some extreme responses, sometimes auditory hallucinations.

The most suitable culture medium for Drosophila melanogaster is ripe banana.
Drosophila melanogaster is commonly called as fruitfly and is often used in genetic and developmental biology researches. Moist bread is a culture medium for the fungus Rhizopus while Agar-agar is used as a tissue culture medium.
Notes:
Ripe banana contains all essential nutrients such as carbohydrates, minerals and vitamins that are required for the growth of Drosophila melanogaster.

Hybrid vigour is also called Heterosis which is measured in terms of size, growth rate, yield and number of other characters. Hybridisation increases vigour and fertility. It is exploited by crossing of two inbred parental lines. It is best maintained in vegetatively reproducing crops because of genetic recombinations.

Crop plants grown in monoculture are highly prone to pests.

Triticale is the 1st man-made cereal which is made by the crossing of wheat (Triticum aestivum) and rye (Secale cereale).

For curing of tea, a bacteria Micrococcus candidans is used. Curing of tea leaves is brought about by the activity of bacteria. It is essentially an oxidation, dry fermentation process, during which water is driven off, the green colour is lost and the leaves assume a tougher texture and undergo chemical changes.
Mycorrhiza is a mutually beneficial association between fungi and the roots of higher plants. Virus and fungi are not involved in the curing process of tea leaves.

Gibberellins and Auxins are known to induce parthenocarpy in plants. If a tomato plant is treated with a low concentration of auxin and gibberellic acid it'll produce fruits without fertilization i.e. parthenocarpic fruits.
Notes:
Vernalized seeds are treated at low temperature for breaking dormancy. Phenyl Mercuric Acetate is an antitranspirant.

Meristem culture is done for the development of virus-free plants. Meristematic tissue cell can be taken either from shoot or root tip.
Notes:
Meristem culture was developed by Morel and Martin in 1952 for virus elimination from Dahlia. Orchid Cymbidium was micropropagated using meristem culture by Morel in 1965.

Escherichia coli is a bacterium found in human colon. On this bacterium scientists have made extensive genetic experiments to make some valuable products like insulin. Another bacterium is Agrobacterium tumefaciens which causes crown gall in plants and is extensively used for genetic experiments.

Golden rice is a transgenic crop of the future with high Vit. A content. Millions of people suffer from Vit. A deficiency which leads to vision impairment.

At an altitude of 5400 metres the low atmospheric pressure of O₂ will be very low, so the solubility of oxygen in the blood will be very less; hence oxygen carried by each RBC will be low. To fulfil the oxygen requirement of the body, blood has to carry more oxygen to the body tissue and this is done by the increased number of RBCs.

Niche indicate the habitat of a particular species and the interaction of that species with the resources present in the habitat. Niche overlap means that two or more species share the resources present in a particular niche.

The formula of exponential growth is dN/dt = rN where dN/dt is the rate of change in population size, r is the biotic potential and N is the population size.

Praying mantis is a good example of camouflage. Camouflage is the natural colouring of an animal which enables it to blend in with its surroundings.
Notes:
A praying mantis, or praying mantid, is the common name for an insect of the order Mantodea. Often mistakenly spelled praying mantis (a tempting mistake, as they are notoriously predatory), they are in fact named for the typical "prayer-like" stance. The word mantis derives from the Greek word mantis for prophet or fortune teller.

Primary production is defined as the amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis. It is expressed in terms of weight (g m⁻²) or energy (kcal m⁻²). The rate of biomass production is called productivity. It is expressed in terms of g m⁻² yr⁻¹ (or kcal m⁻² yr⁻¹) to compare the productivity of different ecosystems. It can be divided into Gross Primary Productivity (GPP) and Net Primary Productivity (NPP). Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilised by plants in respiration. Gross primary productivity minus respiration losses (R), is the occurrence of vertical productivity (NPP).

Stratification is the occurrence of vertical zonation in an ecosystem and indicates the presence of favorable environmental conditions, for e.g., trees occupy top vertical strata or layer of a forest, shrub the second.

The greatest biomass of autotrophs in the oceans is that of free floating microalgae, cyanobacteria and nanoplankton.

Four important functional aspects of the ecosystem are:
(i) Productivity
(ii) Decomposition
(iii) Energy flow
(iv) Nutrient cycling

Fresh weight is not used for the construction of ecological pyramids because the total fresh weight does not change into energy. Hence we can say that fresh weight is not continuous in the trophic levels.

Kaziranga National Park in Assam is famous as a protection environmental park for rhinoceros.

Endangered plant is Benthamidia nicobarica and endangered animal is Red Panda (Ailurus fulgens).

In-situ means keeping endangered species of animals or plants into natural environment and not in the environment that looks like natural but man made, like zoological & botanical gardens. In-situ includes, national parks, sanctuaries and biosphere reserve.

Hotspots are those areas which were rich in biodiversity but now under threat due to direct or indirect interference of human activities. These regions are on the edge to get some of their species extinct due to humans. Western Ghats in India are under threat due to continuous developmental activities and Doon valley is under threat due to continuous mining activities.

Photochemical smog is made by the deposition of dust and carbon particles on the pollutant gases and water vapours. The gases found in photochemical smog are NO₂, PAN, O₃ etc.

Biochemical Oxygen Demand (BOD) is the amount of oxygen required to destroy the organic waste by the bacteria. The more oxygen required for bacteria that means there is more organic waste. The limiting value that is acceptable for BOD is less than 30 ppm (mg/d).

During osmosis, water flows through semipermeable membrane from both sides but the net flow is from solution having lower concentration to higher concentration.

Osmotic pressure = π = cRT ⇒ π = nRT / V

5% solution means
100 ml = 5 g
1000 ml = 50 g / L
For isotonic solution, π₁ = π₂
(10 / 60) × 0.0821 × T = (50 / M) × 0.0821 × T
[V = 1 dm³ = 1 litre]
M = 300 g mol⁻¹

ΔT = K_f m = 5.12 × 1 / 250 × 1000 / 51.2 = 0.4 K

Fe → Fe²⁺ + 2e⁻ [Anode], E = −0.441 V

[Fe³⁺ + e⁻ → Fe²⁺] × 2 [Cathode]: E = +0.771 V

_______________________

Fe + 2Fe³⁺ → 3Fe2+ (overall)

E°_cell = E°_cathode − E°_anode

= 0.771 − (−0.441) = 0.771 + 0.441 = 1.212 V

The cell reaction is as follows:
A → A⁺ + e⁻
B⁺ + e⁻ → B

__________
Adding A + B⁺ → A⁺ B

In the given options, −d[C] / 3dt will not represent the reaction rate. It should not have −ve sign as it is product, since (1 / 3)(d[C] / dt) show the rate of formation of product C, whose value should be positive.

If we write rate of reaction in terms of concentration of NH₃ and H₂, then
Rate of reaction = (1 / 2) (d[NH₃] / dt) = (−1 / 3) (d[H₂] / dt)
So,
d[NH₃] / dt = (−2 / 3) d[H₂] / dt

According to Freundlich adsorption isotherm.
At intermediate pressure, extent of adsorption
x / m = kp¹ᐟⁿ or log (x / m) = log k + (1 / n) log p;
Plot of log(x / m) vs log p is linear with slope = 1 / n.

More the oxidation state of the central atom (metal), more is its acidity. Hence, SeO₂ (O.S. of Se = +4) is acidic. Further for a given O.S., the basic character of the oxides increases with the increasing size of the central atom. Thus Al₂O₃ and Sb₂O₃ are amphoteric and Bi₂O₃ is basic.

Sc³⁺ : 1s², 2s²p⁶, 3s²p⁶d⁰, 4s⁰; no unpaired electron

Cu⁺ : 1s², 2s²p⁶, 3s²p⁶d¹⁰,4s⁰; no unpaired electron

Ni²⁺ : 1s², 2s²p⁶, 3s²p⁶d⁸, 4s⁰; 2 unpaired electrons present

Ti³⁺ : 1s², 2s²p⁶, 2s²p⁶d¹, 4s⁰; 1 unpaired electron present

Co²⁺ : 1s², 2s²p⁶, 3s²p⁶d⁷, 4s⁰; 3 unpaired electrons present

So, from the given option, the only correct combination is Ni²⁺ and Ti³⁺.

Copper sulphate on treatment with excess of KCN forms complex K₃[Cu(CN)₄] or [Cu(CN)₄]³⁻
CuSO₄ + 2KCN → Cu(CN)₂ + K₂SO₄

2Cu(CN)₂ → Cu₂(CN)₂ + (CN)₂

Cu₂(CN)₂ + 3KCN → K₃[Cu(CN)₄] + CuCN

Lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals result in larger no. of oxidation state.

The given compound may have linkage isomerism due to presence of NO₂ group which may be in the form –NO₂ or –ONO. It may have ionisation isomerism due to presence of two ionisable group –NO₂ & –Cl. It may have geometrical isomerism in the form of cis-trans form as follows:

[Co(NH₃)₄Cl(NO₂)]NO₂ &

[Co(NH₃)₄(NO₂)₂]Cl – ionisation isomers.

[Co(NH₃)₄(NO₂)₂]Cl [Co(NH₃)₄(ONO)₂]Cl

– Linkage isomers.

μ = √(n(n + 2))

3.83 = √(n(n + 2)), on solving n = 3

As per magnetic moment, it has three unpaired electron.

Cr³⁺ will have configuration as = [Ar]3d³

So, 3d_xy¹ 3d_yz¹ 3d_xz¹

Ethylene oxide when treated with Grignard reagent gives primary alcohol.

In case of unsymmetrical ethers, the site of cleavage depends on the nature of alkyl group e.g.,

The alkyl halide is always formed from the smaller alkyl group.

Out of given compound only acetaldehyde can form optical active hydroxy acid as it contains one asymmetric carbon atom as marked below :

Aldehydes are more reactive than ketones due to +I effect of –CH₃ group. There are two –CH₃ group in acetone which reduces +ve charge density on carbon atom of carbonyl group. More hindered carbonyl group too becomes less reactive. So, in the given case CH₃CHO is the right choice.

It is an example of Claisen condensation. The product is
acetoacetic ester.

Benzylamine is more basic than aniline. The reason is that in aniline, the lone pair of nitrogen is in conjugation with benzene ring so it is not available readily. On the other hand in Benzylamine, nitrogen is not directly attached with ring so lone pair is not in conjugation with ring.

Pepsin and Trypsin are two enzymes involved in the process (hydrolysis of proteins).

Testosterone and Adrenaline are steroid hormone, Thyroxin is a non-steroid hormone and glucagon is a peptide hormone.

Vitamins are organic substances which does not provide energy but are essential for healthy growth and proper functioning of body. Synthetically vitamins are not synthesized inside human body but they are essential part of our diet.

The given compound is a copolymer of hexamethylene diamine and adipic acid. It is actually Nylon-6,6.

Whenever our body gets attacked by some foreign invadors, our body's immune system produces some chemicals to kill or to react against the invader. These chemicals are actually made up of carbohydrates and proteins i.e. glycoproteins called antibody