NEET-UG 2011 Question Paper — Set 2 Solutions

Average acceleration

<a>= Change in velocity / Total Time.

<a>= |40ĵ - 30î| / 10

<a>= √(4² + (-3)²)

<a>= 5 m/s².

Here, u = 0. We have, v² = u² + 2gh ⇒ v = √(2gh) = √(2 × 10 × 20) = 20 m/s.

For maximum range, the angle of projection, θ = 45°.

∴ R = u²sin(2θ)/g

= (20)²sin(2×45°)/10 = 400×1/10

= 40 m.

Centripetal acceleration a_c = ω²r = (2π/T)²r = (2π/0.2π)² × 5 × 10⁻² = 5 m/s².

Impulse experienced by the body = change in momentum

= MV - (-MV)

= 2MV.

Total mass = (60 + 940) kg = 1000 kg.

Let T be the tension in the supporting cable, then T - 1000g = 1000 × 1

⇒ T = 11000N.

When work is done by the system against a conservative force, its potential energy increases

Power exerted by a force is given by P = F.v. When the body is just above the earth's surface, its velocity is greatest. At this instant, gravitational force is also maximum. Hence, the power exerted by the gravitational force is greatest at the instant just before the body hits the earth.

θ(t) = 2t³ - 6t²

⇒ dθ/dt = 6t² - 12t ⇒ α = d²θ/dt²

= 12t - 12 = 0

∴ t = 1 s

By theorem of parallel axes,

I = I_cm + Md²

I = I₀ + M(L/2)² = I₀ + ML²/4.

Angular momentum is conserved. ∴ L₁ = L₂

⇒ mr₁v₁ = mr₂v₂

⇒r₁v₁ = r₂v₂

⇒ v₁/v₂ = r₂/r₁.

According to question work done by gas is given -150J so that according to it answer will be 150 J of heat has been added to the gas

Change in entropy is given by

ΔS = dQ/T or ΔS = ΔQ/T= mL_f / 273

ΔS= (1000 × 80) / 273 = 293 cal/K.

The given velocity-position graph depicts that the motion of the particle is SHM.

In SHM, at t=0, v=0 and x = x_max.

Only functions given in (A) & (C) represent SHM.

Here, y₁ = a sin(ωt + kx + 0.57)

so, ϕ₁ = 0.57

and y₂ = a cos(ωt + kx)

= a sin[π/2 + (ωt + kx)]

So, ϕ₂ = π/2

Phase difference, Δϕ = ϕ₂ - ϕ₁

= π/2 - 0.57 = 3.14/2 - 0.57

= 1.57 - 0.57 = 1 radian.

We have, v = nλ.

⇒ v ∝ λ (as n remains constant). Thus, as v increases 10 times , λ also increases 10 times.

By Gauss's theorem,

ϕ = Q_in / ε₀.

Thus, the net flux depends only on the charge enclosed by the surface. Hence, there will be no effect on the net flux if the radius of the surface is doubled

Distance of point A from the two +q charges = L.

Distance of point A from the two -q charges = √(L² + (2L)²) = √5 L.

∴ V_A = (Kq/L × 2) - (Kq/√5L × 2)

= (2Kq/L)[1 - 1/√5]

= (1/4πε₀) (2q/L) (1 - 1/√5).

Let the internal resistance of the battery be r. Then the current flowing through the circuit is given by

i = E / (R + r).

In first case,

2 = E / (2 + r). .......(i)

In second case,

0.5 = E / (9 + r). ......(ii)

From (i) & (ii),

4 + 2r = 4.5 + 0.5r

⇒ 1.5r = 0.5 ⇒ r = 1/3 Ω.

We have, e = at + bt².

⇒ de/dt = a + 2bt.

At neutral temperature,

t = -a/2b

∴ de/dt = 0.

𝐯⃗ and 𝐁⃗ are in the same direction, so the magnetic force on the electron becomes zero. But force on the electron due to the electric field is opposite to the direction of velocity

We have, P = V²/R ⇒ 36 = V²/9

⇒ V = 18V.

Current passing through the 9Ω resistor is

i₁ = V/R = 18/9 = 2A.

The 9Ω and 6Ω resistors are in parallel, therefore

i₁ = 6/9+6 × i

where i is the current delivered by the battery.

∴ i = 2×15/6 = 5A

Thus, Potential difference across 2Ω resistor is V = iR = 5 × 2 = 10V.

Leta current i be flowing in the loop ABC in the direction shown in the figure. If the length of each of the

sides AB and BC be x then

|𝐅⃗| =i x B

where B is the magnitude of the magnetic force.

The direction of 𝐅⃗ will be n the direction perpendicular to the plane of the paper and going into it.

By Pythagorus theorem,

AC = √ x² + x² = √2x

∴ Magnitude of force on AC

= i√2× B sin 45°

= i√2xB × 1/√2

= ixB = |𝐅⃗|

The direction of the force on AC is perpendicular to the plane of the paper and going out of it. Hence, force on AC = -𝐅⃗

A → diamagnetic,

B → paramagnetic,

C → ferromagnetic,

D → non-magnetic.

e = -L(di/dt).

During 0 to T/4, di/dt = const.

∴ e = -ve

During T/4 to T/2, di/dt = 0,

∴ e = 0

During T/2 to 3T/4, di/dt = const.

∴ e = +ve

Thus graph given in option (a) represents the variation of induced emf with time.

V_rms = 200√2 / √2 = 200V.

I_rms = V_rms / X_C = (200 / 1) / (100 × 10⁻⁶)

= 2 × 10⁻² = 20mA.

The phase difference φ is given by tan ϕ = X_L / R = 3 / 3 = 1.

→ ϕ = π/4

E.M. wave always propagates in a direction perpendicular to both electric and magnetic fields. So, electric and magnetic fields should be along + X and + Y—

directions respectively.

The decreasing order of the wavelengths is as given below: microwave, infrared, ultraviolet, gamma rays.

Difference between apparent and real depth of a pond is due to the refraction of light, not due to the total internal reflection. Other three phenomena are due to the total internal reflection.

R = 20 cm

h₀ = 2

u = - 30 cm

We have, 1/f = (μ-1)(1/R₁ - 1/R₂)

= (3/2-1)(1/20 - (-1/20))

⇒ 1/f = (3/2-1) × 2/20

∴ f = 20 cm

1/f = 1/v - 1/u ⇒1/20 = 1/v + 1/ 30

1/v = 1/20 - 1/30 = 10/600

v = 60 cm

m = hᵢ / h₀ = v/u

⇒ hᵢ = v/u× h₀ = 60 / 30 × 2

= -4 cm.

So, image is inverted

λ ∝ 1/√V

⇒ λ₁/λ₂ = √(V₂/V₁) = √(100keV/25keV) = 2

⇒ λ₂ = λ₁/2

In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by increasing the potential difference between the anode and filament.

For occurrence of photoelectric effect, the incident light should have frequency more than a certain minimum which is called the threshold frequency (ν₀).

The maximum kinetic energy of emitted electrons is given by

K.E. = ϕ - ϕ₀

K.E₁ = 1 eV - 0.5 eV = 0.5 eV.

K.E₂ = 2.5 eV - 0.5 eV = 2 eV

∴ K.E₁/K.E₂ = 0.5eV/ 2eV =1/4

⇒ v₁/v₂ = √(1/4) = 1/2

The stopping potential is numerically equal to the maximum kinetic energy.

For first line of Lyman series of hydrogen,

hc/λ₁ = Rhc(1/1² - 1/2²).

For second line of Balmer series of hydrogen like ion,

hc/λ₂ = Z²Rhc(1/2² - 1/4²)

By question, λ₁ = λ₂,

⇒ (1/1 - 1/2) = Z²(1/4 - 1/16)

or Z = 2

E = mc²

m = E/c²

So, Mass decay per second

dm/dt = 1/c² dE/dt = 1 / c²(Power in watt)

= 1/(3×10⁸)² × 1000 × 10³

and mass decay per hour = dm/dt ×60 × 60

= 1/(3×10⁸)² × 10⁶ × 3600

= 4 × 10⁻⁸ kg

= 40 microgram.

Momentum Mu = E/c = hv/c. Recoil energy = 1/2 Mu²

= 1/2 M²u²/M = 1 / 2M (hv/c)²

= h²v² / 2Mc².

When the coulomb repulsion between the nuclei is overcome then nuclear fusion reaction takes place. This is possible when the temperature is too high.

Let number of atoms in X = Nₓ

Number of atoms in Y = N_y

By question

Nₓ / N_y = 1 / 15

∴ Part of Nₓ = 1/16 (Nₓ + N_y)

=1/2⁴(Nₓ + N_y)

So, total 4 half-lives are passed, so age of rock is 4 × 50 = 200 years.

When ᵐₙX emits one α-particle, its atomic mass decreases by 4 units and atomic number by 2. Therefore, the new nucleus becomes ᵐ⁻⁴ₙ₋₂Y. But as it emits two β particles, its atomic number increases by 2. Thus the resulting nucleus is ᵐ⁻⁴ₙX.

In forward biasing of the p-n junction, the positive terminal of the battery is connected to p-side and the negative terminal of the battery is connected to n-side. The depletion region becomes thin.

An ideal junction diode when forward biased offers zero resistance. Voltage drop across such a junction diode is zero. In reverse bias diode offers infinite resistance and voltage drop across it is equal to voltage applied.

When small amount of antimony (pentavalent) is added to germanium crystal then crystal becomes n-type semi conductor. Therefore, there will be more free electrons than holes in the semiconductor.

The current gain

β = ΔI_C / ΔI_B = 10mA / 200µA

= (10 × 10³) / 200 = 50.

The energy stored by a capacitor U = 1/2 C V² .....(i)

V is the p.d. between two plates of the capacitor potential difference

V = E.d

The capacitance of the parallel plate capacitor

C = Aε₀/d

substituting the value of C in equation (i)

U = 1/2 (Aε₀/d)(Ed)² = 1/2 ε₀ E² A d

Given E₁ = 25 eV E₂ = 50 eV

E₁ = hc/λ₁ and E₂ = hc/λ₂.

∴ E₁/E₂ = λ₂/λ₁

∴ λ₂/λ₁ = 25/50 = 1/2

∴ λ₁ = 2λ₂.

When n = 6, the sequence of filling of electrons: 6s → 4f → 5d → 6p or ns → (n-2)f → (n-1)d → np

Total no. of atomic orbitals in a shell = n². Given n = 4, hence number of atomic orbitals in 4th shell will be 16.

NO₂⁻, H = 1/2[5+0+1-0] = 3= (sp²)

NO₃⁻, H = 1/2[5+0+1-0] = 3 =(sp²)

NH₂⁻, H = 1/2[5+2+1-0] = 4 =(sp³)

NH₄⁺, H = 1/2[5+4+0-1] = 4 (sp³)

SCN⁻, H =1/2[4+0+0+0] = 2 = (sp)

∴ NO₂⁻ and NO₃⁻ have the same hybridization.

.

O₂ (16) : KK (σ 2s)² (σ* 2s)² (σ 2p_z)² (π 2p_x)² (π 2p_y)², (π*₂ p_y)¹ = (π*₂p_x)¹, B.O. : 2

O₂⁺ (15) : Remove one electron from π*2p_y from O₂, B.O. : 2.5

O₂⁻ (17) : KK (σ 2s)² (σ* 2s)² (σ 2p_z)² (π 2p_x)² (π 2p_y)², (π* 2p_x)² (π* 2p_y)¹, B.O. : 1.5

O₂²⁻ (18) : KK (σ 2s)² (σ* 2s)² (σ 2p_z)² (π 2p_x)² (π 2p_y)², (π* 2p_x)² (π* 2p_y)²

Since, the bond length decreases as the bond order increases, hence, O₂⁺ have least bond length.

r_A / r_B = √(M_B / M_A)

(V / 20) / (V / 10) = √(M_B / 49)

⇒ 1/2 = √(M_B / 49)

M_B = 1/4 × 49 = 12.25u

Average velocity v = √(8RT/πM)

i.e., v ∝ √T

∴ V₂/V₁ = √(2T/T) = 1.41

Given n_CO = ⁿN₂

P_CO + P_N₂ = 1 atm

Partial pressure of a gas

= mole fraction of gas × total pressure

∴ P_N₂ = (ⁿN₂ / (ⁿCO + ⁿN₂)) × 1 = (ⁿN₂ / ²ⁿN₂) × 1 = 1/2 = 0.5 atm.

For adiabatic process ΔQ = 0.

In free expansion, external pressure is zero. Thus, work done w = 0 and ΔE = w = 0. If ΔE = 0, then ΔT = 0.

Given

4H(g) → 2H₂(g); ΔH = -869.6 kJ.

or 2H₂(g) → 4H(g); ΔH = 869.6 kJ. H₂(g) → 2H(g); ΔH = 869.6 / 2

= 434.8 kJ.

Given ΔH = 30 kJ mol⁻¹ T = 273 + 27 = 300 K

ΔS_T = ΔH_T / T

= (3 × 10⁴) / 300 J mol⁻¹

= 100 J mol⁻¹ K⁻¹

For the reaction N₂(g) + O₂(g) ⇌ 2NO(g)

K₁ = [NO]² / ([N₂][O₂])

For the reaction 2NO(g) + O₂(g) ⇌ 2NO₂(g)

K₂ = [NO₂]² / ([NO]²[O₂])

For the reaction NO₂(g) ⇌ 1/2 N₂(g) + O₂(g)

K = [N₂]¹ᐟ²[O₂] / [NO₂]

Hence, K = √(1 / (K₁ × K₂))

K = ([N₂]¹ᐟ²[O₂]¹ᐟ² / [NO]) × ([NO][O₂]¹ᐟ² / [NO₂]) = [N₂]¹ᐟ²[O₂] / [NO₂]

BF₃ behaves as a Lewis acid because it is an electron-deficient species

Given : [NH₃] = 0.3 M, [NH₄⁺] = 0.2 M,

K_b = 1.8 × 10⁻⁵

pOH = pK_b + log ([salt] / [base]) [pK_b = -log K_b; pK_b = -log 1.8 × 10⁻⁵]

∴ pK_b = 4.74

pOH = 4.74 + log(0.2 / 0.3) = 4.74 + 0.3010 - 0.4771 = 4.56

pH = 14 - 4.56 = 9.436

The active ingredient in bleaching powder for bleaching action is Ca(OCl)₂.

.

Given wt of compound taken (w)=0.35g

Volume of nitrogen collected (V)= 55 mL.

Room temperature (tK)=300K

Atmospheric pressure (P) =715 mm

Aq. tension (p) = 15mm

Calculation -

Volume of N₂ at NTP= ((P - p) × V/ t) × 273 / 760mL

=(( 715 - 15) × 55 / 300) ×273 / 760

=46.098 mL

% of nitrogen

=28 x voL.of N₂ at NTP ×100 /

22400 × wt of organic compound

= 28 × 46.098 × 100 / 22400 × 0.35 =16.46%

Conc. HNO₃ decomposes NaCN and Na₂S to avoid their interference.

NaCN + HNO₃ → NaNO₃ + HCN↑

Na₂S + 2HNO₃ → 2NaNO₃ + H₂S↑

Due to +M effect of — OH group and hyperconjugation of —CH, group.

The growth of fishes gets hindered if the concentration of (D.O.) is below 6 ppm.

Given K_f = - 1.86° cm⁻¹, mass of solute = 5.00 g, mass of solvent = 45.0 g

ΔT_f = i × K_f . m

3.82 = i × 1.86 × (5 × 1000) / (142 × 45) (Molecular mass of Na₂SO₄ = 142)

∴ i = 2.63

If a compound dissociates in solvent i > 1 and on association i < 1.

As the value of standard reduction potential decreases, the reducing power increases i.e.,

Z (-3.0) > X (-1.2) > Y (+0.5).

.

The couple for which SRP value is high, will act as cathode and the other couple will act as an anode.

E°_cell = E°_cathode - E°_anode = 0.15 V - (-0.74 V) = +0.89 V

Standard Gibbs free energy is given as

ΔG° = -nE°F

If E°_cell < 0 i.e. -ve

ΔG° > 0

Further ΔG° = -RT ln K_eq

∴ ΔG° > 0 when K_eq < 1

Order of reaction can be zero, whole number, or fractional.

x/m = f(P) at constant T and

x/m = f(T) at constant T

Thus, it can be stated that for constant value of x/m, pressure and temperature can take different values or P = f(T) at constant (x/m).

But x/m ≠ P × T

Pig iron or cast iron contains 3 – 5% carbon and varying amounts of Mn, Si, P and S which makes the iron hard and brittle.

Zr and Ti are purified by van Arkel method.

Zr(s) + 2I₂(g) ⟶ ZrI₄(g)

ZrI₄(g) ── On the hot filament ──> Zr(s) + 2I₂(g)

.

The green colour appears due to the formation of Cr³⁺ ion

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

Interchange of ligand between the complexes give isomers e.g. [Co(NH₃)₆] [Cr(CN)₆] is an isomer of [Co(CN)₆] [Cr(NH₃)₆]

Complexes of the type [M_ABCD] may exist in three isomeric forms.

Similarly, [Pt(Py)(NH₃)BrCl] may exist in three isomeric form in which

M = Pt, A = Py, B = NH₃, C = Br, D = Cl.

Ni²⁺ = 3d⁸ 4s⁰

Since, the coordination number of Ni in this complex is 4, the configuration of Ni²⁺ at first sight shows that the complex is paramagnetic with two unpaired electron. However, experiments show that the complex is diamagnetic. This is possible when the 3d electrons rearrange against the Hund's rule as shown below. This is in accordance with the fact that the ligand involved here is strong i.e., CN⁻ ion. Ni²⁺ (after rearrangement)

Hence, now dsp² hybridization involving one 3d, one 4s and two 4p orbitals, takes place leading to four dsp² hybrid orbitals, each of which accepts four electron pairs from CN⁻ ion forming [Ni(CN)₄]²⁻ ion.

[Ni(CN)₄]²⁻

Thus, the complex is diamagnetic as it has no unpaired electron.

Minimum paramagnetic behaviour. =[Co(H₂O)₆]²⁺

Clemmensen reduction is

Clemmensen reduction is

When a nitro compound is reduced with a neutral reducing agent (Zn dust + NH₄Cl), the corresponding hydroxyl amine is formed.

All reducing sugars shows mutarotation.

Polyesters are condensation polymers of a diabasic acid and a diol. e.g., Terylene.

Diphenyl hydramine, also known as Benadryl, is an antihistamine.

For a first order reaction

Total time T = no. of half lives (n) × half life (t₁/₂)

N / N₀ = (1/2)ⁿ

where n = no. of half lives

Give N₀ (original amount) = 1.28 mg/l

N (amount of substance left after time T) = 0.04 mgL⁻¹

∴ 0.04 / 1.28 = (1/2)ⁿ ; 1 / 32 = (1/2)ⁿ ; (1/2)⁵ = (1/2)ⁿ

n = 5

T = 5 × 138 = 690

1.00 molal aqueous solution contain mole in 1000 g water

n_solute = 1 mol

w_solvent = 1000 g

n_solvent = 1000 / 18 = 55.56 mol

x_solute = 1 / (1 + 55.56) = 0.0177

Tiger and tigris both are from same genus with a

particular taxonomic category.

Eubacteria are prokaryotic but eubacteria are

enclosed by plasma membrane like eukaryotic cells.

Methanogens are archacbacteria, abundant in cattle

yard and paddy fields.

Methanogens produce methane as a metabolic byproduct in hypoxic conditions. They are prokaryotic and belong to the domain of archaea.

E.coliis a prokaryotic gram negative bacterium

Archegoniophore is the female sex organ of bryophytes (Marchantia) and pteridophytes. Its neck

region is made up of 4-6 vertical rows of cells.

Unlike bryophytes and pteridophytes, in gymnosperms the male and female gametophytes do not have an independent free-living existence. They remain with in the sporangia retained on the sporophytes.

The gametophyte of ‘bryophytes are smaller but have large sex organs.

In nature insects have the highest number of species 1t (30 million). Fungi have 100,000, angiosperms have 260000 and birds have 10,000 species.

A male shark possesses a pair of claspers which are inserted into a female shark’s cloaca (an opening on the underside of the body) at the time of mating. Claspers are located on the inner edge of the pelvic fins near the male's cloaca. The function of claspers is to introduce sperm into a female shark's body for the purpose of fertilizing her eggs. Female sharks do not have claspers.

Chondrichthyes are the cartilaginous fish with a flexible skeleton made of cartilage rather than bone.

The characteristics and phylum of Preropus is accurate.

When a flower can be divided into two similar halves only in one particular vertical plane, it is zygomorphic, e.g.

pea, gulmohar, bean, etc.

Tulips are most popular and wildly grown flowers.

Tulips are bulbs, possessing modified stems and leaves.

Aleurone cells are the outer cell layer of the endo sperm, usually only one cell thick in wheat and the only endosperm tissue alive at maturity. The cell layers of this layer are responsible for the de-novo synthesis of enzyme needed during germination. The chromosome number is 63 of a plant with 42 chromosome in its root tip cells.

Floral formula of chilli is

The ovary is half-inferior in flowers of peach.

Ground tissue includes all tissues except epidermis and vascular bundles. The ground tissue comprises the -bulk of the primary plant body. Parenchyma, collenchyma and sclerenchyma cells are common in the ground tissue,

Guard cells differ from epidermal cells in having chloroplast. The cell wall of guard cells are not uniform, -

inner walls are thicker than the outer walls, epidermal are uniformly thin.

Smut is a disease of cereals, corn, grasses and

sorghum caused by many species of fungi.

Phellem, phellogen and phelloderm are collectively called periderm.

Ciliated epithelium is usually found in the respiratory tract and fallopian tubes. Ciliated epithelium is a thin tissue that has hair-like structures on it. These hairs, called cilia, move back and forth to help move particles out of our body. We find ciliated epithelial tissue in our respiratory tract and in the fallopian tubes of women. Ciliated epithelium performs the function of moving particles or fluid over the epithelial surface in such structures as the trachea, bronchial tubes, and nasal cavities. It often occurs in the vicinity of mucus-secreting goblet cells.

Cockroach intake the food after grounding it by its

mandibles and gizzard.

The intestine of an earthworm has a peculiar fold of tissue that forms a pocket in the intestine that is visible when viewing a cross section. This space, which runs almost the whole length of the body dorsally, is known as the typhlosole and it serves to increase the surface area of the intestine for absorption.

Golgi apparatus is the important site for formation of glycoprotein and glycolipid

Peptide synthesis takes place in ribosome of the cell.

Under electron microscope the nucleosomes appear as beads on a string in chromosome, due to a short length of DNA wrapped around a core of histone proteins.

Lecithin is a fat like substance (a phospholipid), which is a part of plasma membrane

In the given curve, the relation between temperature and enzyme activity is shown.

In mitosis, chromosomes move to the equator and get aligned along equatorial plate in metaphase

Calcium is not remobilized from the leaves to the fruits, like potassium, phosphorus and sulphur. It occurs abundantly in a non-exchangeable form such as anorthite (CaAl₂Si₂O₈).

A prokaryotic autotrophic nitrogen fixing symbiont is found in Cycas.

The nitrifying bacteria bring about transformations of ammonia to nitrite and of nitrite to nitrate.

Leghaemoglobin is the haemoglobin like red pigment found in root nodules of legumes and reported to function as an oxygen-carrying pigment in symbiotic nitrogen fixation.

CAM plants are most common in arid environments.

Being able to keep stomata closed during the hottest and driest part of the day reduces the loss of water through evapotranspiration.

The coughing would have been due to improper movement of epiglottis.

Rennin is an enzyme that is essential for the protein

digestion. It is active in acidic medium and inactive in

normal gastric juice of adult.

In the given figure the exchange of respiratory gases occurs in alveolar cavity.

A large portion of oxygen is left unused in the human

‘blood even after its uptake by the body tissues. This O, acts as a reserve during muscular exercise.

Fibrinogen, the protein of blood plasma is converted into insoluble protein fibrin during the clotting process. The fibrinogen free fluid, obtained after removal of the clot, called blood serum, is plasma minus fibrinogen.

Bundle of His’ are a typical cardiac muscle fibres, connecting the atria with ventricle.

Arteries are the vessels which break up into capillaries which reunite to form one visceral organ.

190/110 mmHg may harm vital organs like brain and kidney. This is called hypertension which can give rise to increased heart rate and palpitation.

Organisms which excrete nitrogenous wastes in the form of urea are called ureotelic organisms whereas the organisms which excrete nitrogenous wastes in the form of uric acid are called uricotelic organisms. Uric acid is the least poisonous and the least soluble in water, relative to Ammonia and Urea.

Renal pelvis is the innermost portion of kidney. So, collecting ducts are not a part of renal pelvis.

The pituitary gland is located in a bony cavity called sella tursica and is attached to hypothalamus by a stalk. It is divided anatomically into an adenohypophysis and a neurohypophysis. The Iatter is also called pars nervosa or posterior pituitary. The latter is also called pars nervosa or posterior pituitary. It stores and releases two hormone

called oxytocin and vasopressin, which are actually synthesized by the hypothalamus and are transported axonally to neurohypophysis. Vasopressin acts mainly at the kidney and stimulates resorption of water and electrolytes by the distal tubules in the nephron and

thereby reduced loss of water through urine (divresis). Hence, it is also called as anti-divretic hormone (ADH).

Podocytes are highly specialized cells with a complex

cytoarchitecture plays a major role in establishing the

selective permeability of glomerular filtration barrier.

When some one drinks lots of water, kidney release of ADH is suppressed.

When a neuron is in resting state i.e., not conducting any impulse, the axonal membrane is comparatively more permeable to K⁺ ion and nearly impermeable to Na⁺ ions.

Rhodopsin contained in the rod type of photoreceptor cells of human eye is a derivative of vitamin A.

Anterior pituitary secretes growth hormone. Ovary secretes oestrogen hormone. Alpha cells of islets of Langerhans secrete glucagon.

The axillary buds of the potato tuber are called “eyes”

in common language. They are found at the nodes of the stem tuber.

A synergid cell wall forms a highly thickened structure called the filiform apparatus at the micropylar end consisting of numerous finger like projections into synergid cytoplasm. These synergid cells are necessary for pollen tube guidance in ovule.

Cleistogamy is autogamous pollination. When pollination and fertilization occur in unopened flower bud, it is known as cleistogamy. It ensures self-pollination and prevents crosspollination.

Wind pollination is common in grasses. Grasses produce large amount of pollen which by the help of wind reach to opposite sex for reproduction.

Some fleshy fruits such as mango, plum etc. usually have a single hard stone that encloses a seed, called drupe.

Nucellar polyembryony is reported in a Citrus species.

If the vasa efferentia in the human reproductive system get blocked the gametes will not be transformed from testes to epididymis.

The testes are present in the scrotum which maintains the scrotal temperature below 2°C of the internal body temperature.

In the given figure

III - marked portion is Infundibulum IV - marked portion is Fimbriae

V - marked portion is Cervix

Intra uterine device (IUD) is a method of contraception in India. The IUD is inserted in the woman’s uterus through the cervix.

Medical termination of pregnancy is considered safe up to twelve weeks of pregnancy.

Heterosis or hybrid vigor occurs when two unrelated individuals or lines are crossed, the performance of F₁ hybrid, which is often superior to both the parents.

TypeO

XO type of sex chromosomes determine male sex in grasshopper

Mutation can be induced by gamma radiation.

The sample of a portion of DNA strand shown in the figure is palindromic sequence of base pairs.

23S rRNA acts as a catalystina bammal cell.

Increasing brain capacity

Saliva of infected female Anopheles mosquito

Headlouse is an obligate ectoparasite of human scalp and as well as laying egg on human hair.

Skin has a metabolic function to prevent non- resident bacteria from developing.

When HIV replicates rapidly in helper T- lymphocytes and damages large number of these

ELISA is an fundamental tool of clinical immunology and is used as an initial screen for HIV detection.

rice

‘Himgiri’ developed by hybridisation and selection for disease resistance against rust pathogens is a variety of wheat. It is resistant to leaf/stripe rust and hill bunt.

Saccharomyces

Glomus

Agrobacterium

EcoRI is an endonuclease enzyme isolated from strains of E.coli and a part of restriction modified system. So co part stands for coli.

Gel electrophoresis

molasses

RNA interference has been used to develop plants resistant to nematodes. RNA interference is a double stranded RNA (dsRNA) induced for gene sllencm, phenomenon

purify enzymes

In the world maximum number of existing transgenic animals is mice.

tropical rainforests

Some species make permanent burrows deep into the soil to escape high temperature or sunlight and some cold blooded animals often like to bask in the sun to warm up their body.

Secondary succession refers to the regrowth of a habitat in the area where disruptive event has occurred and eliminated the existing, above ground plant life of the natural habitat. So, it occurs on a deforested site.

Declining population

Out of total incident solar radiation, about 50% ofi forms Photosynthetically Active Radiation (PAR).

A standing crop is the quantity or total weight or energy content of the organism, which are in a particular location at a particular time.

It is inverted in shape

In India maximum genetic diversity is seen in rice. Total of 23 variety of irrigated ecology and 8 varieties of rainfed ecology has been invented in India.

germplasm

biological process

Marsh gas is another term for methanc produced by plants decomposmg under water.

fresh water lakes

The Bhopal gas tragedy occured on the night of December 2-3, 1984 at the Union carbide India Limited pesticide plant in Bhopal, M.P. A leak of methyl isocyanate gas and other chemicals from the plant resulted in the exposure of thousands of people.

The correct acronym of IUCN is International Union for Conservation of Nature and Natural resources.

CO₂ and N₂O are major green house gases

Mango

In this case, dehydration is governed by Saytzeff's rule, according to which hydrogen is preferentially eliminated from the carbon atom with fewer number of hydrogen atoms i.e., poor becomes poorer. Thus, 2-methyl-but-2-ene is the major product.

This reaction is governed by Markownikoff's rule according to which when an unsymmetrical reagent e.g. HBr adds to an unsymmetrical alkene, then the negative part of the reagent is added to that carbon atom of the double bond which bears the least number of hydrogen atom. Thus, in above case 2-methyl-2-bromobutane will be the major product.

Work done = area under F-d graph = [2 × (7 - 3)] + [1/2 × 2 × (12 - 7)] = 8 + 5 = 13J.

(ii), (iv) and (iii)

The given compound is

hv/c is correct answer.

Melting point of metal halides decreases as the sjze

of the halogen increases. The correct order is

CaF₂.>CaCl₂ >CaBr₂>CaI₂

Bond length order is

C–C < C=C < C–O < C–H

1.10A° 1.34A° 1.40A° 1.54A°

Nucleophilic substitution reactions are the most common reactions of alkyl halides.