NEET-UG 2008 Question Paper Solutions

Energy density = Energy / Volume

⇒ [ML² T⁻²] / [L³]= [ML⁻¹ T⁻²]

Refractive index and dielectric constant have no dimensions.

Young’s Modulus = F / A × 𝑙 / Δ𝑙

⇒ [MLT⁻²] / [L²] . [L] / [L]= [ML⁻¹T⁻²]·

Magnetic field, B = F / iℓ = [MLT⁻²] / [A][L]

= [MT⁻²A⁻¹]

Given, error in the measurement of radius of a sphere Δr / r × 100 = 2%

Volume of the sphere V = (4 / 3) πr³

∴ Error in the volume ΔV / V × 100

= ±3 . Δr / r × 100 = 3 × 2% = ±6%

Distance travelled in the nth second is given by Sₙ = u + (a / 2) (2n − 1)

Given u = 0, a = (4 / 3) ms⁻², n = 3

∴ Sₙ = 0 + ((4 / 3 × 2) (2 × 3 − 1)) = 4 / 6 × 5 = (10 / 3) m

The slope of the graph (ds / dt) is maximum at C and hence the instantaneous velocity is maximum at C.

Instantaneous velocity,

Initial velocity, u = 10 ms⁻¹

Final velocity, v = 20 ms⁻¹

Distance, s = 135 m

Let, acceleration = a

Using the formula, v² = u² + 2as

a = v²- u² / 2s = ((20)² - (10)² / 2 ×135) = 400 -100 / 2×135

or, a = 300 / 2×135 = 150 / 135 = (10 / 9) ms⁻²

Now, using the relation, v = u + at

t = v - u / a = (20 - 10 / (10 / 9)) = 10 / 10 × 9 sec= 9 sec.

The momentum along a-axis remains unchanged

Clearly, change in momentum along x-axis = mvcos⁡θ − mvcos⁡θ=0

Momentum changed only in vertical direction or y-axis.

So, ΔP = P_vertical

⇒ P_final = P_initial

= mvsin⁡θ − (−mvsin⁡θ) = 2mvsin⁡θ

= 2mv × sin⁡ 45°

= 2mv × 1 / √2 = √2mv

Hence, resultant change in momentum = √2mv

F= d(Mv) / dt = M(dv / dt) + v (dM / dt)

∴ v is constant, 

∴ F = v (dM / dt) But (dM / dt) = Mkg/s ∴F=vM newton.

The Components of 1 N and 2 N forces along + x-axis

=1 cos 60° + 2 sin 30°

=1× 1 / 2 + 2× 1 / 2 = 1 / 2 + 1 = 3 / 2 = 1.5N

The component of 4 N force along –x-axis = 4 sin 30°

= 4 × 1 / 2 = 2 N.

Therefore, if a force of 0.5 N is applied along +x-axis, the resultant force along x-axis will become zero and the resultant force will be obtained only along y-axis.

Given, h = 60 m, g = 10 ms⁻²,

Rate of flow of water = 15 kg/s

∴ Power of the falling water

= 15 kgs⁻¹ × 10 ms⁻² × 60 m = 9000 watt.

Loss in energy due to friction

= 9000 × 10 / 100 = 900 watt.

∴ Power generated by the turbine

= (9000 – 900) watt = 8100 watt = 8.1 kW

Let the initial velocity of the shell be v, then by the conservation of momentum mv = Mv′

where v′ = velocity of gun.

∴ v′ = (m / M) v

Now, total K.E. = (1 / 2)mv² + (1 / 2)Mv′²

= (1 / 2)mv² + ((1 / 2) M) (m / M)² v² = (1 / 2)mv² [1 + m / M]

= (1 / 2 × 0.2) (1 + 0.2 / 4) v² = (0.1 × 1.05) v²

But total K.E. =1.05 kJ =1.05 × 10³ J

∴ 1.05 × 10³ = 0.1 × 1.05 × v²

⇒ v² = 1.05 × 10³ / 0.1 × 1.05 = 10⁴

∴ v = 10² = 100 ms⁻¹

∵ I = MK² ∴ K = √(I / M)

I_ring =MR² and I_disc = (1 / 2) MR²

K₁ / K₂ = √(I₁ / I₂) = √(MR² / (MR²/2)) = √(2:1)

Mass of each part = M / 2

Length of each part = L / 2

Total M.Ⅰ. = Sum of M.Ⅰ.s of both parts 

= (M / 2) (L / 2)² × 1 / 3 + (M / 2) (L / 2)² × 1 / 3

=2× M / 2 × L² / 4 ×1 / 3 = ML² / 12

For the riders to experience weightlessness at the top of the hill, the weight of the rider must be balanced by the centripetal force.

i.e., mg = mv² / R

⇒ v= √gR = √10×20 = 14.1 ms⁻¹

Hence, the speed of the car should be between 14 ms⁻¹ an 15 ms⁻¹

For different temperature scales, we have

x − L.F.P / U.F.P. − L.F.P. =constant

Where L.F.P ⇒ Lower Fixed Point

U.F.P ⇒ Upper Fixed Point

Where x is the measurement at that scale. Here, if C and W be the measurements on Celsius and W scale then,

C - 0 / 100 - 0 = W - 39 / 239 - 39 (∴ C = 39°C)

⇒ W = (39 × 200 / 100) + 39 = 78 + 39 = 117° W

Heat required to raise the temperature of 1 kg water from 20°C to 100°C is given by

Q = msΔθ = (1 × 4200 × 80) J

Power of kettle (P) = VI = (220 × 4) W

Time taken=Q / P = ((1 × 4200 × 80) / (220 × 4))

= 381.81 sec =6.36 min

In a cyclic process, the initial state coincides with the final state. Hence, the change in internal energy is zero, as it depends only on the initial and final states.

Let the mass of the gas be m.

At a fixed temperature and pressure, volume is fixed.

Density of the gas, ρ = m / V

Now ρ / P = m / nRT ⇒ m / nRT = x (By question)

⇒ xT = constant ⇒ x₁T₁= x₂T₂

⇒ x₂ ⇒ x₁T₁ / T₂ = (283 / 383)x

[given T₁= 283K]

[T₂     = 383K] 

Maximum acceleration of a particle in the simple harmonic motion is directly proportional to the square of angular frequency

i.e., a ∝ ω²

∴ a₁ / a₂ = ω₁² / ω₂² = (100)² / (1000)² = 1 / 100

⇒ a₁ : a₂ = 1 : 10².

We have x = a sin (ωt + π / 6)

∴ Velocity, v = dx / dt = aω cos (ωt + π / 6)

Maximum velocity = aω

According to question,

aω / 2 = aω cos (ωt + π / 6)

or, cos (ωt + π / 6) = 1 / 2 = cos 60° or cos π / 3

⇒ ωt + π / 6 = π / 3

ωt = π / 3 − π / 6 or, ωt = π / 6

or, (2π / T) · t = π / 6 ⇒ t = T / 12

y=0.25sin⁡(10πx−2πt)

Comparing this equation with the standard wave equation

y=a sin⁡(kx−ωt)

We get, k=10π

⇒ 2π / λ = 10π ⇒ λ = 0.2 m

And ω = 2π or, 2πν = 2π ⇒ ν = 1 Hz

The sign inside the bracket is negative, hence the wave travels in +ve x-direction.

By the symmetry of the figure, the electric fields at O due to the portions AC and BD are equal in magnitude and opposite in direction. So, they cancel each other.

Similarly, the field at O due to CD and AKB are equal in magnitude but opposite in direction. Therefore, the electric field at the centre due to the charge on the part ACDB is E along OK.

Given that, V = Q × 10¹¹ volts

Electric potential at point is given by

V = 1 / 4π ε₀ . Q / r or, Q × 10¹¹ = 1 / 4π ε₀ . Q / r

⇒ r = 1 / 4π ε₀ × 10⁻¹¹ m.

As |E| = |V| / r = (Q × 10¹¹ / (1 / 4π ε₀ ).10⁻¹¹)

= 4π ε₀ Q × 10²² volt m⁻¹.

The energy stored by a capacitor

U = (1 / 2) CV² ...........(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) Aε₀E²d

 Resistance of a wire is given by R= ρ (𝑙 / a)

If the length is increased by 10 % then new

length 𝑙′ = 𝑙 + (1 / 10) = (11 / 10)𝑙

In that case, area of cross section of wire would decrease by 10 %

∴ New area of cross section

A′= A - A / 10 = (9 / 10) A

∴ R′ = ρ (ℓ′ / A′) = ρ ((11 / 10)𝑙 / (9 / 10)) A

R′ = (1 1 / 9) ρ (𝑙 / R)              R′=1.21R

The new resistance increases by 1.21 times. The specific resistance (resistivity) remains unchanged as it depends on the nature of the material of the wire.

Bacterial leaf blight of rice is caused by a species Xanthomonas. Mature rice plants are infected by these bacteria, lesion begin as water-soaked stripes on the leaf blades and eventually increase in length and width, becoming yellow to grayish-white until the entire leaf dries up.

A domain of prokaryotic organisms containing the archaeobacteria including the methanogens, which produce methane; the thermoacidophilic bacteria, which live in extremely hot and acidic environments & and the halophilic bacteria, which can only function at high salt concentrations, are abundant in the world’s oceans.

Thermococcus, Methanococcus and Methanobacterium exemplify archaebacteria that contain protein homologous to eukaryotic core histones.

Male and female gametophytes of Cedrus do not have free-living independent existence. Cedrus belongs to conifer.

Salvinia is an heterosporous aquatic fern with both annual and perennial species. The heterospores have two types of spores — microspores and megaspores.

The important features distinguishing Gnetum from Cycas and Pinus and showing affinities with angiosperms are the presence of vessel elements and absence of archegonia.

Ascaris is a common parasite found in the large intestine of man. It is worldwide in distribution. The number of worms may be 500 or more in a single host. Ascaris is characterized by the presence of neither true coelom nor metamerism.

Characteristic of phylum Annelida is pseudocoelomate. Pseudocoelomate is any invertebrate animal whose body cavity is a pseudocoel, a cavity between the gut and the outer body wall derived from a persistent blastocoel, rather than a true coelom. Pseudocoelomate animals include the Rotifera and Nematoda.

Mollusca mostly oviparous and a few viviparous. The development may be direct or indirect with trochophore, velliger, and glochidium.

Scorpion, spider, and cockroach have a ventral solid central nervous system. All three belong to phylum Arthropoda.

— Spider belongs to class Arachnida

— Scorpion belongs to class Arachnida

— Cockroach belongs to class Insecta

Cockroach belongs to phylum Arthropoda. It has 10 pairs of Spiracles (2 pairs on thorax and 8 pairs on abdomen).

Replum is present in the ovary of flower of mustard. A false septun called replum develops between the two parietal placentae in mustard and other members of family

Brassicaceae. The ovary becomes bilocular.

The length of different internodes in a culm of sugarcane is variable because of intercalary meristem. Intercalary meristem is not a part of apical meristem. It occurs in the internodes of grasses (sugarcane) between leaf nodes and enables longitudinal growth of the stem.

Vascular tissues in flowering plants develop from plerome. Plerome is a central core of primary meristem which gives rise to all cells of the stele from the pericycle inward.

Pollen exine is resistant to enzyme action. The hard outer layer called the exine is made up of sporopollenin which is one of the most resistant organic material known.

It can withstand high temperatures and strong acids and alkali. No enzymes that degrade sporopollenin are so far known.

Earthworms have no skeleton but during burrowing, anterior end becomes turgid and acts as a hydraulic skeleton. It is due to coelomic fluid. The coelom is filled with an alkaline milky coelomic fluid.

Keeping in view of the ‘Fluid mosaic model’ for the structure of cell membrane, the movement of lipids and proteins from one lipid monolayer to the other, lipid can rarely flip-flop, protein can not. The fluid mosaic model is the most important and widely accepted latest model for plasma membrane was given by Singer and Nicolson in 1972. According to them it resembles “protein icebergs in a sea of lipids”.

Cellulose is a major component of cell wall of Pythium. Pythium is a genus of parasitic Oomycete. Because this group of organisms were once classified as fungi, they are sometimes still treated as such. Pythium, like others in the family Pythiaceae, are usually characterized by their production of coenocytic hyphae, hyphae without separations. Oogonia generally contain a single oospore. Antheridia contain an elongated and club shaped antheridium.

The two subunits of ribosome remain united at a critical ion level of magnesium. The presence of magnesium and its amount plays an important role in the appearance and structure of the ribosomes.

Vacuole in a plant cell is membrane bound and contains water and excretory substance. Vacuole is a space within the cytoplasm of living cell that is filled with air,

water or other liquid, cell sap or food particles. In plant cells, there is usually one large vacuole bounded by a single layered membrane (tonoplast or vacuole membrane);

animal cells usually have several small vacuoles.

Malonate is a powerful inhibitor of cellular respiration, because it binds to the active site of succinate dehydrogenase in the citric acid cycle but does not react. It thereby competes with succinate, since it does not have the -CH₂,-CH₂,- group (as in succinate which is required for dehydrogenation). For the oxidative phosphorylation reaction, malonate is an inhibitor for complex II which again contains succinate dehydrogenase.

Rupture and fractionation do not usually occur in the water column in vessel/ tracheids during the ascent of sap because of cohesion and tension.

Carbohydrates are commonly found as starch in plant storage organs. It is a chemically non-reactive and osmotically inactive polysaccharide with a molecular weight. Carbohydrates perform a vital role in living organisms. Starch and other polysaccharides serve as energy storage in plants, particularly in seeds, tubers, etc. which provide a major energy source for animals, including humans.

Nitrogen fixation in root nodules of Alnus is brought about by Frankia. It is a symbiotic nitrogen-fixing bacteria in root nodules of several non-legume plants like Casuarina (Australian Pine), Myrica and Alnus (Alder).

Electrons from excited chlorophyll molecule of photosystem II are accepted first by quinone. Photosystem II is a photosynthetic pigment system along with some electron carriers that are located in the appressed part of the grana thylakoids. Photosystem II has chlorophyll a, b and carotenoids. Other components of PS II are pheophytin, plastoquinone (PQ), cytochrome complex and blue coloured copper containing plastocyanin.

In leaves of C₄ plants, malic acid synthesis during CO₂ fixation occurs in mesophyll cells. The mesophyll cells perform C₄ cycle and the cells of bundle sheath perform C₃ cycle.

Glycolysis takes place in all body cells and is of two types—

(a) Anaerobic glycolysis—From glycogen or glucose to lactic acid in muscles.

(b) Aerobic glycolysis—From glycogen or glucose to pyruvic acid (all cells of body).

Chemiosmotic theory was postulated by the British biochemist Peter Mitchell (1920–22) to explain the formation of ATP in the mitochondrial electron transport chain. As electrons are transferred along the electron carrier system in the inner mitochondrial membrane, hydrogen ions (protons) are actively transported into the space between the inner and outer mitochondrial membranes, which thus contains a higher concentration of protons than the matrix. This creates an electrochemical gradient across the inner membrane, through which protons move back into the matrix.

In germinating seeds, fatty acids are degraded exclusively in the glyoxysomes.

Glyoxylate cycle occurs in tissues rich in fats, such as those of germinating seeds; the enzymes involved in the cycle, which have not been found in mammals, are contained in organelles called glyoxysomes.

Senescence is an active developmental cellular process in the growth and functioning of a flowering plant, indicated in leaf abscission. Senescence is the changes that occur in an organism between maturity and death, i.e. ageing.

Characteristically there is a deterioration in functioning as the cells become less efficient in maintaining and replacing vital cell components.

Photoperiodism was first studied by Garner and Allard (1920). They observed that “Maryland Mammoth” variety of tobacco could be made to flower in summer by reducing the light hours with artificial darkening. It could be made to remain vegetative in winter by providing extra light.

Small intestine- Starch disaccharide (maltose)

Small intestine is the portion of the alimentary canal between the stomach and large intestine. It is subdivided into the duodenum, jejunum and ileum. It plays an essential

role in the final digestion and absorption of food.

If the secretion of parietal cells of gastric glands is blocked with an inhibitor, in the absence of HCI secretion, inactive pepsinogen is not converted into the active enzyme pepsin.

Basophil is a type of white blood cell (leucocyte) that has a lobed nucleus surrounded by granular cytoplasm. Basophils are produced continually by stem cells in the red bone marrow & move about in an amoeboid fashion. Like mast cells, they produce histamine and heparin as part of the body’s defences at the site of an infection or injury.

The most active phagocytic white blood cells are neutrophils and monocytes which modulate the innate immune response and phagocytose infectious agents.

Neutrophil has a lobed nucleus and granular cytoplasm. Neutrophils engulf bacteria and release various substances such as lysozyme and oxidising agents.

Monocyte is the largest form of WBC in vertebrates. It has a kidney-shaped nucleus and is actively phagocytic, ingesting bacteria and cell debris.

The haemoglobin of a human foetus has a higher affinity for oxygen than that of an adult. Foetal haemoglobin binds to oxygen more strongly than adult haemoglobin, enabling the transfer of oxygen from mother to foetus prenatally.

In humans, blood passes from the post caval to the diastolic right atrium of heart due to suction pull.

In humans total number of ribs are Fourteen (14). These are of three types:

(1) True ribs – These are 7 in number.

(2) False ribs – These are 2 in number.

(3) Floating ribs – These are 4 in number.

Total sum of physio-electrochemical changes that takes place along the length of nerve fibre is known as nerve impulse.

Change in potential due to stimulation of nerve fibre is called action potential. During propagation of nerve impulse, Na⁺ enters inside so (+ve) charge is formed inside the membrane. K⁺ ions come out.

The diagrammatic cross section of a single loop of human cochlea represents three different parts—

A - Perilymph

B - Tectorial membrane

C - Endolymph

Cochlea arises from sacculus. It is spirally coiled duct. It is also known as Lagena.

It is connected with sacculus by duct of reuniens.

Rod cells is a type of light sensitive receptor cell present in the retinas of vertebrates. Rod contains the pigment rhodopsin and are essential for vision in dim light.

Cone cells is a type of light sensitive receptor cell, found in retinas of all diurnal vertebrates. Cones are specialized to transmit information about colour and are responsible for the visual acuity of the eye. They function best in bright light.

Here E > ER / R + r, hence the lengths 110 cm and 100 cm are interchanged.

Without being short-circuited through R, only the battery E is balanced.

E = V / L × 𝑙₁ = V / L ×110      ...(i)

When R is connected across E, Ri = V / L× 𝑙₂

or, R (E / R+r) = V / L × 100

Diving (i) by (ii), we get 

R + r / R = 110 / 100

or, 100 R +100r = 110 R

or, 10R = 100 r

∴ r = 10R / 100 = ((10 × 10) / 100) (∴R = 10Ω)

⇒ r = 1 Ω

Clearly, 2Ω, 4Ω and (1+5) Ω resistors are in parallel.

Hence, potential difference is same across each of them.

∴ Ⅰ₁ × 2 = Ⅰ₂ × 4 = Ⅰ₃ × 6

Given Ⅰ₁ = 3A (∴ Ⅰ₁ × 2 = Ⅰ₃ × 6 )

Given Ⅰ₁= 3A

∴ Ⅰ₁ × 2 = Ⅰ₃ × 6 provides

Ⅰ₃ = (Ⅰ₁ × 2 / 6) = (3 × 2 / 6) = 1A.

Now, the potential across the 5Ω resistor is 

V = Ⅰ₃ × 5 = 1 × 5 = 5V

the power dissipated in the 5Ω resistor

P = V² / R = 5² / 5 = 5watt.

When a charged particle enters a transverse magnetic field it traverse a circular path. Its kinetic energy remains constant.

According to the figure the magnitude of force on the segment QM is F₃ —F₁, and PM is F₂.

Therefore, the magnitude of the force on segment PQ is √(F₃-F₁)² + F₂²

Total internal resistance = (50 +2950)Ω = 3000Ω

Emf of the cell, ε= 3V

∴ Current = ε / R = 3 / 3000 = 1×10⁻³A = 1.0mA

∴ Current for full scale deflection of 30 division is 1.0mA

∴ Current for a deflection of 20 divisions,

Ⅰ = (20 / 30 ×1) mA or Ⅰ = (2 / 3)mA

Let the resistance be x Ω. Then 

x= ε / Ⅰ = (3V / (2/3×10⁻³A)) = (3 × 3 × 10³ / 2) Ω = 4500Ω

But the resistance of the galvanometer is 50Ω.

∴ Resistance to be added 

= (4500-50)Ω= 4450Ω

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

Here, B=1 / π (Wb/m²)

θ = 60°

Area normal to the plane of the disc

=πr² cos 60° = πr² / 2

Flux = B × normal area 

= 0.2 × 0.2 / 2 = 0.02Wb

Total number of turns in the solenoid, N= 500

Current , Ⅰ = 2A.

Magnetic flux linked with each turn

= 4×10⁻³Wb

As, Φ = LⅠ or NΦ= LⅠ ⇒ L = NΦ / 1

= (500 × 4 × 10⁻³ / 2) henry = 1 H.

The average power in the circuit over one cycle of a.c. is given by

Pₐᵥ = eᵣₘₛ ×iᵣₘₛ ×cosΦ

= E₀ / √2 × Ⅰ₀ /√2 × cos Φ = (E₀ Ⅰ₀ / 2) cosΦ

The velocity of electromagnetic radiation in a medium of permittivity ε₀ and permeability μ₀ is equal to 1/√(μ₀ ε₀) ·

The focal length of the combination 

1 / f = 1 / f₁+ 1 / f₂

∴ Power of the combinations,

P= f₁+f₂ / f₁f₂ (∵ P=1 / f)

We have, |v / u| = size of image / size of object

or, size of image = |v / u| × size of object

= (10⁻¹ / 1.5 ×10¹¹) × (1.39 × 10⁹)

= 0.92 × 10⁻³ m = 9.2 × 10⁻⁴m

∴ Diameter of the sun’s image = 9.2 ×10⁻⁴m

The resultant intensity of two periodic waves at a point is given by

I = I₁ + I₂ + 2√(I₁I₂) . cosϕ.

Resultant intensity is maximum if

cosϕ = −1

i.e. Iₘₐₓ = I₁ + I₂ + 2√(I₁I₂)

Resultant intensity is minimum if

cosϕ = +1

i.e., Iₘᵢₙ = I₁ + I₂ − 2√(I₁I₂)

Therefore, the sum of the maximum and minimum intensities is

Iₘₐₓ + Iₘᵢₙ

= I₁ + I₂ + 2√(I₁I₂) + I₁ + I₂ − 2√(I₁I₂)

= 2(I₁ + I₂)

In discharge tube, collisions between the charged particles emitted from the cathode and atoms of the gas results to the coloured glow in the tube.

Wavelength of particle

(λ₁)= h / mv = (h / (1×10⁻⁶)× v)

where v is the velocity of the particle.

Wave length of electron,

(λ₂) = (h / (9.1 ×10⁻³¹) × (3×10⁶))

but  λ₁ = λ₂ (Given)

∴ (h / (1×10⁻⁶) × v) = (h / (9.1 ×10⁻³¹) × (3×10⁶))

⇒ v = 9.1 ×10⁻³¹ × 3×10⁶ / 10⁻⁶ = 2.73 ×10⁻¹⁸ ms⁻¹

Work function, Φ₀= 6.2eV.

Stopping Potential V₀ = 5V.

As, eV₀ = hv - Φ₀

or eV₀ + Φ₀ =hc / λ

or λ = (hc / eV₀ + Φ₀)

λ = (6.62 ×10⁻³⁴ Js × 3 × 10⁸ms⁻¹ / 1.6 × 10⁻¹⁹ × 5J + 6.2 × 1.6 × 10⁻¹⁹J)

= (6.62 × 10⁻³⁴ × 3 × 10⁸ / 1.6 × 10⁻¹⁹ (5+6.2)) m

= (19.86 ×10⁻²⁶ /17.92 ×10⁻¹⁹) = 1.10 × 10⁻⁷ m.

When the electron is in first excited state

(n = 2), the excitation energy is given by 

ΔE = E₂ - E₁

We have, Eₙ = (-13.6 / n²) eV

∴ E₂ = (-13.6 / 2²) eV = -3.4eV

Given E₁ = -13.6eV

 ∴ ΔE = (-3.4) - (-13.6)= 10.2 eV.

Required ratio of nuclear densities = ρ₁ / ρ₂

= ((M₁/V₁) / (M₂/V₂)) = M₁/M₂ × V₂/V₁ = 1 / 3 × ((4/3)πR₂³ / (4/3)πR₁³

= 1 / 3 × (R₂ / R₁)³ = 1 / 3 × (R₀M₂¹ᐟ³ / R₀M₁¹ᐟ³)³ [∴ R = R₀M¹ᐟ³ ]

= 1 / 3 × (M₂ / M₁) = 1 / 3 × ( 3 / 1)

= 1 : 1

Mass defect = ZMₚ + (A − Z)Mₙ − M(A, Z)

or, B.E. / c² = ZMₚ + (A − Z)Mₙ − M(A, Z)

∴ M(A, Z) = ZMₚ + (A − Z)Mₙ − B.E. / c²

Let the required time be t. Then

N₁ = N₀e_(−λ₁t) ; N₂ = N₀e_(−λ₂t)

Where

N₁ = number of nuclei of X₁ after time t

N₂ = number of nuclei of X₂ after time t

N₀ = initial number of nuclei of X₁ and X₂ each

Now, N₁ / N₂ = ((N₀e_(−λ₁t) / N₀e_(−λ₂t)) Here N₁ / N₂ = 1 / e

λ₁ = 5λ ; λ₂ = λ

∴ 1 / e = (e_(−5λt) / e_(−λt)) ⇒ e⁻¹ = e_(−4λt) ⇒ 4λt = 1

∴ t = 1 / 4λ

E_g = 2.0 eV = 2 × 1.6 × 10⁻¹⁹ J

∴ E_g = hν

∴ ν = E_g / h

= ((2 × 1.6 × 10⁻¹⁹ J) / (6.62 × 10⁻³⁴Js))

= 0.4833 × 10¹⁵ s⁻¹ = 4.833 × 10¹⁴ Hz ≃ 5×10¹⁴ Hz

Negative feedback is applied to reduce the output voltage of an amplifier. If there is no negative feedback, the value of output voltage could be very high. In the options given, the maximum value of voltage gain is 100. Hence it is the correct option.

Atomic radius is the distance between two nearest neighbours. For fcc crystal

Atomic radius = (1 / 2√2) × lattice parameter

or r = (a / (2√2))

r = (3.6 / (2√2)) Å = 1.27 Å

Energy given by the cell E = CV²

Here, C = capacitance of condenser = (Aε₀ / d)

V = potential difference across the plates = Ed

∴ E = (Aε₀ / d) (Ed)² = A ε₀ E² d

Potential difference across PM

V = 4 × 1 = 4 volt (i.e., across 4 Ω)

Equivalent resistance of lower side arm,

R_eq = 1 + ((0.5 × 0.5) / (0.5 + 0.5)) = 1.25 Ω

Now the circuit can be shown as

Let current I flows in lower branch, so

1.25 I = 4 V

or I = 4 / 1.25 = 3.2 A

Therefore, 3.2 A current flows in 1 Ω resistance, hence potential difference between M and N

V′ = 3.2 × 1 = 3.2 volt

Writing the equation of combustion of propane (C₃H₈), we get

From the above equation we find that we need 5 L of oxygen at N.T.P. to completely burn 1 L of propane at N.T.P.

If we change the conditions for both the gases from N.T.P. to same conditions of temperature and pressure, the same results are obtained. i.e. 5 L is the correct answer.

From above equation, 2 moles of MnO₄ required to oxidise 5 moles of oxalate.

Thus, number of moles of MnO₄⁻ required to oxidise one mole of oxalate = (2 × 2 / 5) = 0.4 moles.

∴ Volume occupied by 1 gram water = 1 cm³

or Volume occupied by

(6.023 × 10²³ / 18) molecules of water = 1 cm³

[∴ 1 g water = (1 / 18) moles of water]

Thus, volume occupied by 1 molecule of water

= (1 × 18 / 6.023 × 10²³) cm³ = 3.0 × 10⁻²³ cm³

i.e. the correct answer is option (3.0 × 10⁻²³ cm³).

Δp = mΔv

Substituting the given values of Δx and m, we get

1×10⁻¹⁸ gcms⁻¹

= 9×10⁻²⁸ g × Δv

or Δv = (1×10⁻¹⁸ / 9×10⁻²⁸)

= 1.1 × 10⁹ cm s⁻¹ ≃ 1×10⁹ cm s⁻¹

Writing the equation for the reaction, we get

From this equation we find 223 g of PbO reacts 0with 73 g of HCl to form 278 g of PbCl₂.

If we carry out the reaction between 3.2 g HCl and 6.5 g PbO.

Amount of PbO that reacts with 3.2 g HCl = (223 / 73) × 3.2 = 9.77 g.

Since amount of PbO present is only 6.5 g so PbO is the limiting reagent.

Amount of PbCl₂ formed by 6.5 g of PbO = (278 / 223) × 6.5 g

Number of moles of PbCl₂ formed

= (278 / 223) × (6.5 / 278) moles = 0.029 moles.

We know Δp·Δx ≥ h / 4π

or m·Δv·Δx = h / 4π  

[∴ Δp = mΔv]

Since Δp = Δx (given)

∴ Δp·Δp = h / 4π or mΔv mΔv = h / 4π

or (Δv)² = h / 4πm²

∴ Δv = √(h / 4πm²) = 1 / 2m √(h / π)

The shape of ozone molecules is

In it we find 2σ and 1π bond, i.e., the correct answer is 2 sigma and 1 pi bond

From the structure of three species we can determine the number of lone pair electrons on the central atom (i.e. N atom) and thus the bond angle.

We know that higher the number of lone pairs of electron on the central atom, greater is the lp–lp repulsion between Nitrogen and Oxygen atoms. Thus smaller is the bond angle.

The correct order of bond angle is

NO₂⁻ < NO₂ < NO₂⁺

We know that q (heat) and work (w) are not state functions but (q + w) is a state functions. H – TS (i.e. G) is also a state function.

Cornea transplant in humans is almost never rejected. This is because it has no blood supply. Cornea is a transparent layer of tissue, continuous with the sclerotic, that forms the front part of the vertebrate eye, over the iris and lens.

A peptide hormone secreted by the parathyroid gland in response to low levels of calcium in the blood. It acts to maintain normal blood levels of calcium by increasing the number of osteoclasts, which break down the bone matrix and release calcium into the blood. It also increases the reabsorption of calcium and magnesium ions in the kidney tubules, so that their concentration is maintained in the blood.

Parathyroid and adrenal are the endocrine glands. Adrenal glands are the pairs of endocrine glands situated immediately above the kidneys. Hence they are also known as suprarenal glands.

Parathyroid glands are the two pairs of endocrine glands situated behind or embedded within the thyroid gland in higher vertebrates. They produce parathyroid hormone, which controls the amount of calcium in the blood.

Unisexuality of flowers prevents autogamy, but not geitonogamy. In self fertilisation, the male and female gametes are derived from the same individual. Among plants, self fertilization also called autogamy is common in many cultivated species, e.g., wheat and oats.

However, self fertilization is a form of inbreeding and does not allow for the mixing of genetic material; if it occurs over a number of generations it will result in offspring being less vigorous and productive than those resulting from cross fertilization.

Filiform apparatus helps in the entry of pollen tube into a synergid in ovule.

Filiform apparatus is form of finger like projection comprising a core of micro fibrils enclosed in a sheath. The filiform apparatus resembles transfer cells meant for short distance movement of metabolites. The filiform apparatus is responsible for the absorption of food from the nucleus.

In pea, endosperm is consumed by developing embryo in the seed. The endosperm is completely absorbed by the growing embryo and the food reserve gets stored in the cotyledons. Such seeds are called non endospermic or exalbuminous.

Dry indehiscent single-seeded fruit formed from bicarpellary syncarpous inferior ovary is cypsela. Cypsela is also calied inferior, false or pseudocarpic achene, the thin fruit wall (developed from pericarp and thalamus)is attached to the seed at one point but the fruits develops from an inferior, unilocular and uniovuled ovary, e.g., sunflower, marigold. Some cypsela develop pappus for dispersal e.g. Soncus, Taraxacum.

The fleshy receptacle of syconus of fig encloses a number of Achenes. Achene is the thin dry pericarp is free from seed except at one point. The fruit develops from a monocarpellary pistil having superior unilocular and ovuled ovary.

The fruit of pomegranate is chambered, developed from inferior ovary and has seeds with succulent testa. The fruit is Balausta. It is a special type of false or pseudocarpic

berry in which the syncarpous pistil consists of two rows of fused carpels, one above the other.

In human female, the periodic discharge of blood, mucus and cellular debris from uterine wall from non pregnant women of sexual maturity is known as menstrual

cycle. Few hours before the start of menstruation, the spiral arterioles constrict one by one resulting into the blanched mucosa. The mucosa shrinks and the death of the blood deprived tissues takes place. Blood clotting doesn’t occur due to presence of fibrinolysin.

An organizer is a signal aligning center that directs the development of embryo or part of embryo, e.g. in amphibians, it is dorsal tip of blastophore.

A. The pill — Prevents ovulation

B. Condom — Prevents sperm reaching cervix

C. Vasectomy — Semen contains nosperms

D. Copper‑T — Prevent implantation

Statements i and iii are correct.

• Medical Termination of Pregnancy (MTP) during first trimester is generally safe.

• Intrauterine device like copper‑T are effective contraceptives.

Klinefelter’s syndrome is a genetic disorder affecting men in which an individual gains an extra X chromosome, so that the usual karyotype of XY is replaced by XXY. Symptoms of Klinefelter, syndrome named after U.S. physician H.P. Klinefelter, include female characteristics (such as breast enlargement).

Haploids describes a nucleus cell or organism with a single set of unpaired chromosomes. The haploid number is designated as X. Reproductive cells, formed as a result of meiosis are haploid. Fusion of two such cells restores the normal diploid number. Therefore, haploids are more suitable for mutation studies than the diploids. This is because all mutations, whether dominant or recessive are expressed in haploids.

Purine is an organic nitrogenous base sparingly soluble in water, that gives rise to a group of biologically important derivatives, notably adenine and guanine, which occur in nucleotides and nucleic acids (DNA and RNA).

GCU indicates alanine but GUU indicates valine. Stop codons are AUG, UGA and UAA.

AUG is the most common start codon which codes for methionine.

UUA indicates leucine but UCA indicates serine.

Polysome (Polyribosome) is a complex formed by several ribosomes attached to a single mRNA molecule in the process of translation.

In the DNA molecule, there are two strands which run anti‑parallel, one in 5’–3’ direction and the other in 3’–5’ direction. The two strands are held together by hydrogen bonds between their bases. Adenine (A), a purine of one chain is exactly opposite to thymine (T), a pyrimidine of the other chain. Similarly, cytosine (C), a pyrimidine lies opposite to guanine (G), a purine. This allows a sort of lock & key arrangement between large sized purine & small sized pyrimidine. It is strengthened by the appearance of hydrogen bonds between the two.

The theory of the continuity of the germplasm published by August Weismann (1834–1914) in 1886. It proposes that the contents of the reproductive cells (sperms and ova) are passed on unchanged from one generation to the next, unaffected by any changes undergone by the rest of the body. It thus rules out any possibility of the inheritance of acquired characteristics, and has become fundamental to Neo‑Darwinian theory.

Darwin’s Finches are an excellent example of adaptive radiation. Adaptive radiation is the occurrence of the diversification of a single or small groups of species into a large

number or descendents that occupy various ecological niches. This is an evolutionary process driven by natural selection.

Thorn of Bougainvillea and tendril of Cucurbita are example of homologous organs.

Nephridia of earthworm and malphigian tubules of cockroach belongs to excretory organs. Earthworm has a well‑developed excretory system which is composed of large number of minute, coiled and glandular segmentally arranged excretory tubules called the nephridia. Malphigian tubules of cockroach are extremely fine yellowish unbranched thread‑like structure present at the function of midgut and hindgut.

Coacervate is an aggregate of macromolecules such as proteins, lipids and membrane, and contain enzymes that are capable of converting substances such as glucose into more complex molecules, such as starch. Coacervate droplets arise spontaneously under appropriate conditions and may have been the prebiological systems from which living organisms originated.

Amoebiasis : Use only sterilized food and water

Diphtheria : DPT Vaccine

Cholera : Use oral rehydration therapy

Syphilis : Treponema pallidum

The saliva in the mouth and the tears from the eyes belong to physiological barriers.

Hashish causes after‑thought perceptions and hallucinations. Hashish is a preparation of Cannabis composed of the compressed trichomes collected from the Cannabis plant. Psychoactive effects vary between types of Hashish but are usually the same as those of other Cannabis preparations such as marijuana. Hashish is generally prohibited to the same extent as all other forms of cannabis. It is consumed in much the same way as Cannabis buds, used by itself in miniature smoking pipes, vapourized, hot‑knifed, or smoked in joints mixed with tobacco, Cannabis buds or other herbs.

Jatropha is being tried in India as a biofuel substitute for fossil fuels. Biofuels are the fuels of biological origin. They consist of renewable form of energy.

Energy problems can be solved up to a great extent by efficient and proper used of these biofuels.

Use of only disease free seeds and use of disease resistant varieties are the most important control measures that could be taken to successfully grow chickpea in an

area where bacterial blight disease is common.

Paecilomyces is a cosmopolitan filamentous fungus which inhabits the soil, decaying plants, and food products. Some species of Paecilomyces are isolated from insect. The telomorphs of Paecilomyces are classified in the genera Cilromocleista, Talaromyces. Paecilomyces is usually considered as a contaminant but may also cause

infections in humans and animals.

Trichoderma harzianum has proved a useful microorganism for biological control of soil borne plant pathogens. Trichoderma harzianum is a fungus that is also used as a fungicide. It is used for foliar application, seed treatment and soil treatment for suppression of various disease causing fungal pathogens. Commercial biotechnological products such as 3 Tac have been useful for treatment of Botrytis, Fusarium, Penicillium sp. It is also used for manufacturing enzymes.

Modern detergent contain enzymes preparations of alkaliphiles. Alkaliphiles live in alkaline environments like soda lakes, or alkaline soil. The pH level in alkaline substances is from about 9 to 11 on the pH scale. The alkaliphiles keep an alkaline level of about pH 9 inside the cells, but the environment has a higher level of alkalinity. Alkaliphiles can keep the right pH level by pumping in hydrogen ions in out at the right rate.

Human insulin is being commercially produced from a transgenic species of Escherichia coli. E. coli is a bacterium that is commonly found in the lower intestine of warm blooded animals. The bacteria can also be grown easily and its genetics are comparatively simple and easily manipulated, making it one of the best studied prokaryotic model organisms, and an important species in biotechnology.

Downy mildew of grapes disease is linked to the discovery of Bordeaux mixture as a popular fungicide. Bordeaux mixture was discovered by Millardet in France in 1882. Bordeaux mixture is prepared by dissolving 40g of copper sulphate and 40 g of calcium hydroxide in 5 litres of water. Bordeaux mixture is used primarily as a fungicide, it was first used to control downy mildew disease of grape‑vine caused by a fungus, Plasmopara viticola.

The linking of antibiotic resistance gene with the plasmid vector became possible with DNA ligase. DNA ligase is an enzyme that is able to join together two portions of DNA and therefore plays an important role in DNA repair. DNA ligase is also used in recombinant DNA technology as it ensures that the foreign DNA is bound to the plasmid into which it is incorporated.

Gel electrophoresis is a technique far separation of DNA fragments according to their size. DNA is negatively charged so in gel tank when electricity is passed, DNA move towards positive electrode.

An RNA molecule whose base sequence is complementary to that of the RNA transcript of a gene, i.e. the sense RNA, such as a messenger RNA (mRNA). Hence, an antisense RNA can undergo base pairing with its complementary mRNA sequence. This blocks gene expression, either by preventing access for ribosome to translate the mRNA or by triggering degradation of the double stranded RNA by ribonuclease

enzymes. Antisense RNA and DNA both have therapeutic potential for modifying the activity of disease causing genes.

Cry endotoxin obtained from Bacillus thuringiensis are effective against bollworms. A bollworm is a common term for any larva of a moth that attacks the fruiting bodies

of certain crops, especially cotton.

A modern hospital can make use of variety of sophisticated instruments and equipment for accurate diagnosis and treatment of diseases. Three main categories of instruments and equipments used are diagnostic, imaging and therapeutic. From the above given statements, statement 1 & 3 are correct.

• During open heart surgery blood is circulated in the heart lung machine.

• Computerised Axial Tomography (CAT) shows detailed internal structure as seen in a section of body.

Golden rice is a transgenic food crop which may help in solving the problem of night blindness in developing countries. Golden rice or miracle rice is rich in vitamin A or β‑carotene and iron.

Characteristics of certain desert animals such as kangaroo, rat are

• They feed on dry seeds and do not require drinking water.

• They excrete very concentrated urine and do not use water to regulate body temperature.

‘A’ is more recent and shows slight reduction in growth rate.

About 70% of total global carbon is found in oceans. This oceanic reservoir regulates the amount of carbon dioxide in the atmosphere. Atmosphere contains only about one percent of total global carbon.

Low moisture content will lead to decrease in number of decomposers because decomposers need an optimum moisture for their growth and functioning.

Food chain is the transfer of energy from green plants (Primary producers), through a sequence of organisms occupies in a food chain is known as its trophic level. Therefore, statements b and c are correct.

Species richness refers to the number of species per unit area. Species evenness is the relative abundance with which each species is represented in an area. Thus variation in the number of species, kinds of species as well as the number of individuals per species lead to diversity. Area(d) and number of habitats (12) shows the maximum species diversity.

Quercus species is a dominant component in temperate - deciduous forests. The common name of Quercus is cork oak belongs to family fabaceae. Cork used as a bottle stoppers, soles for shoes, insulating material and for manufacture of linoleum.

Lesser inter-specific competition is not observe in biodiversity hotspots, levels of species richness and high degree of endemism. Initially 25 biodiversity hotspots

were identified but subsequently nine more have of biodiversity hotspots in the world to 34. These hotspots are also regions of accelerated habitat loss.

World Summit on sustainable development (2002) was held in Johannesburg, South Africa, 190 countries pledged their commitment to achieve by 2010, a significant in reduction the current rate of biodiversity loss at global, regional and local levels.

The Central Pollution Control Board (CPCB), statutory organisation, was constituted in September 1974 under the Water (Prevention and Control of Pollution) Act 1974. Further, CPCB was entrusted with the powers and functions under Air (Prevention and Control of Pollution) Act, 1981. It serves as a field formation and also provides technical services to the Ministry of Environment and Forest, of the provisions of the Environment (Protection) Act, 1986. Principal functions of the CPCB, as spelt out in the Water (Prevention and Control of Pollution) Act, 1974, and the Air (Prevention and Control of Pollution) Act, 1981.

(i) To promote cleanliness of streams and wells in different areas of the states by prevention, control and abatement of water pollution.

(ii) To improve the quality of air and to prevent, control or abate air pollution in the country.

Lots of urea and phosphate fertilizer were used in the crops in the vicinity and the lake water turned green and stinky. Due to this, lake near a village suffered heavy mortality of fishes within a few days.

CFC 14%, Methane 20% is the correct percentage of the two (out of the total 4) greenhouse gases that contribute to the total global warming. A regular assessment of abundance of greenhouse gases and their impact on global environment is being made by IPCC (Inter‑Governmental Panel on Climate Change). The various greenhouse gases are CO₂ (warming effect 60%), CH₄ (effect 20%), chlorofluorocarbons or CFCs (14%) and nitrous oxide (N₂O, 6%). Others of minor significance are water vapours and ozone.

In human adult females, oxytocin is a hormone released by the pituitary gland (neurohypophysis), that causes contraction of uterine muscles during labour.

Vital capacity of lungs is the largest possible expiration after largest possible inspiration that is greatest. Volume of air can be exchanged in single respiration or amount of air breathed in and out with greatest possible efforts.

Vital capacity = TV + IRV + ERV

Total lung capacity = VC + RV

So, we can say that VC = Total lung capacity - Residual volume

C4plants are photosynthetically more efficient than C3 plants because C4 plant contain two types of chloroplast i.e., bundle sheath chloroplast and mesophyll chloroplast. So such plants operate a dicarboxylic acid cycle in addition to Calvin

cycle. CO₂, acceptor molecule (PEP) is present in large bundle sheath cell which has higher efficiency in picking up CO₂,.

Thus, photosynthesis continues even at low CO₂, concentration and the rate of photorespiration is also negligible.

Coacervates and microspheres reproduced by asexual methods such as budding, fission and fragmentation.

Bacterial leaf blight of rice is caused by Xanthomonasoryzae a bacterium which is gram- negative, aerobic, capsulated, and motile with a single polar flagellum. Primary

infection is carried through the infected seeds. The entry of the pathogen occurs through wounds and stomata. The symptoms of the disease are the appearance of linear, yellow to straw coloured stripes, usually on both the edges of the leaf. As the disease progresses, the drying and twisting of the leaf tip occurs. The most destructive phase of the disease is the ‘Ikresek’ or wilt resulting from early systematic infection.

At any constant temperature the K.E. of gaseous molecules remains same.

The reaction for formation of HCl can be written as

H₂ + Cl₂ → 2HCl

H–H + Cl–Cl → 2(H–Cl)

Substituting the given values, we get enthalpy of formation of

ΔH_f(HCl) = [(B.E)_H–H + (B.E)_Cl–Cl] – [2 × B.E_(H–Cl)]

= 434 + 242 – 2 × (431) = –186 kJ

∴ Enthalpy of formation for

1 mol HCl = (–186 / 2) kJ = –93 kJ

For the reaction:

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

The reaction given is an example of decomposition reaction and we know that decomposition reactions are endothermic in nature, i.e. ΔH > 0.

Further

Δn = (1 + 1) – 1 = +1

Hence more number of molecules are present in products which shows more randomness i.e. ΔS > 0 (ΔS is positive).

For this reaction K_eq is given by

K = [Fe³⁺][OH⁻]³ / [Fe(OH)₃]

= (Fe³⁺)(OH⁻)³  [∴ solid = 1]

If (OH⁻) is decreased by 1 / 4 times, then for given reaction, equilibrium constant to remain constant, we have to increase the concentration of (Fe³⁺) by a factor of (4³) i.e. 4 × 4 × 4 = 64. Thus, option (64 times) is correct.

For the reaction 

             2 AB₂(g)⇌2 AB (g) + B₂(g)

initially   2              0             0

 at equi   2(1 -x)          2x             x

The partial pressure at equilibrium are calculated on the basis of total numbers of moles at equilibrium.

Total number of moles

= 2 (1- x) + 2x + x = (2 + x)

∴ P_AB₂ = (2(1 - x) / (2 + x)) × P where P is the total pressure.

P_AB = (2x / (2 + x)) × P, P_B₂ = (x / (2+x)) ×P

Since x is very small so can be neglected in denominator Thus, we get

P_AB₂ = (1 - x) × P                         P_AB = x × P

P_B₂ = x / 2 ×P

Now, K_P = ((P_AB)² (P_B₂) / (P_AB₂)²) = ((x)² × P²·P× x / 2 / (1-x)² × P²)

= (x³·P³ / 2 × 1 × P²)                                [∵ 1 - x ≃1]

= x³·P / 2 or x³ = (2·Kp / P) or x = (2K_P/ P)¹ᐟ³

Given reaction are

x⇌ Y+Z   ...(i)

and A ⇌ 2B  ...(ii)

Let the total pressure for reaction (i) and (ii) be P₁ and P₂ respectively, then

Kp₁/Kp₂ = 9/1    (given)

After dissociation 

                       x⇌Y+Z

                    (1-α)  α  α

At equilibrium

[Let 1 mole of X dissociate with α as degree of dissociation]

Total number of moles = 1-α + α+ α = (1+α)

Thus Pₓ = (1-α/1+α)· P₁; P_y= (α/1+α) P₁;

P_z = (α/1+α)·P₁

∴Kp₁= (α/1+α)·P₁×α/(1+α)·

P₁/(1-α/1+α)·P₁                   ....(i)

Similarly for                A⇌2B

At equilibrium           (1-α) 2α

We have,

Kp₂ = (2αP₂/1+α)² / (1-α/1+α) P₂    ....(ii)

Diving (i) by (ii), we get

Kp₁/Kp₂ = α²·P₁/4α²·P₂ or Kp₁/Kp₂ = 1/4 · P₁/P₂

or 9= 1/4 · P₁/P₂                 [∴ Kp₁/Kp₂ = 9/1]

or P₁/P₂ = 36/1 or P₁: P₂ = 36:1

[H₃O]⁺ for a solution having pH = 3 is given by

[H₃O]⁺ = 1 × 10⁻³ moles/litre [∴ [H₃O]⁺ = 10⁻ᵖᴴ]

Similarly for solution having pH = 4,

[H₃O]⁺ = 1 × 10⁻⁴ moles/litre and for pH = 5

[H₃O]⁺ = 1 × 10⁻⁵ moles/litre

Let the volume of each solution in the mixture be 1 L ,then total volume of mixture solution

(1+1+1) L = 3 L

Total [H₃O]⁺ ion present in mixture solution

= (10⁻³ + 10⁻⁴ + 10⁻⁵) moles

Then [H₃O]⁺ ions concentration of mixture solution

= ((10⁻³ + 10⁻⁴ + 10⁻⁵) / 3) M = (0.00111 / 3) M

= 0.00037 M = 3.7 × 10⁻⁴ M.

The highest pH will be recorded by the most basic solution. The basic nature of hydroxides of alkaline earth metals increase as we move from Mg to Ba thus the solution of BaCl₂ in water will be most basic and so it will have highest pH.

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 thus 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, the hydrated ion size follows the same order Cs⁺ > Rb⁺ > K⁺ > Na⁺ and thus, conductivity order is Cs⁺ > Rb⁺ > K⁺ > Na⁺.

The stability of alkali metal hydrides decreases from Li to Cs. It is due to the fact that M–H bonds become weaker with increase in size of alkali metals as we move down the group from Li to Cs. Thus the order of stability of hydrides is

LiH > NaH > KH > RbH > CsH

C–1 is sp hybridized (C≡C)

C–3 is sp³ hybridized (C–C)

C–5 is sp² hybridized (C=C)

Correct sequence: sp, sp³, sp²

The amount of s‑character in various hybrid orbitals is as follows.

sp = 50%, sp² = 33%, sp³ = 25%

Therefore, s‑character of the C–H bond in acetylene (sp) is greater than that of the C–H bond in alkene (sp² hybridized), which in turn has greater s‑character of the C–H bond in alkanes. Thus, owing to a high s‑character of the C–H bond in alkynes, the electrons constituting this bond are more strongly held by the carbon nucleus with the result the hydrogen present on such a carbon atom can be easily removed as a proton.

The acidic nature of three types of C–H bonds follows the order

≡C–H > =C–H > –C–H

Further, as we know that conjugate base of a strong acid is a weak base, hence the correct order of basicity is:

As 's' character increases, stability of carbanion increases. Electron donating groups decrease stability of carbanion. Hence, correct stability order:

The formed 2° carbocation will rearrange to a more stable 3° carbocation as :

Out of the given compounds the most reactive towards nucleophilic attack is

Green chemistry are the reaction, which reduces the use of hazardous chemicals.

Percentage of occupied space in a bcc is 68%.

∴ Percentage of free space in a bcc is

(100 – 68) = 32%

Following generalization can be easily derived for various types of lattice arrangements in cubic cells between the edge length (a) of the cell and r the radius of the sphere.

For simple cubic : a = 2r or r = a / 2

For body centred cubic:

a = (4 / √3)r or r = (√3 / 4)a

For face centred cubic:

a = 2√2r or r = (1 / 2√2)a

Thus the ratio of radii of spheres will be simple: bcc : fcc

= a / 2 :(√3 / 4)a :(1 / 2√2)a

Kohlrausch’s Law states that at infinite dilution, each ion migrates independently of its co‑ion and contributes to the total equivalent conductance of an electrolyte a definite share which depends only on its own nature.

From this definition we can see that option (infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte.) is the correct answer.

The semiconductors formed by the introduction of impurity atoms containing one elecron less than the parent atoms of insulators are termed as p-type semiconductors.

Therefore, silicon containing 14 electrons has to be doped with boron containing 13 electrons to give a p-type semi-conductor.

From the given data we find Fe³⁺

(E°_(Fe³⁺/ Fe²⁺) = + 0.77V) is strongest oxidizing agent.

Writing the equation for pentane–oxygen fuel cell at respective electrodes and over all reaction, we get

At Anode:

C₅H₁₂ + 10H₂O → 5CO₂ + 32H⁺ + 32e⁻

(pentane)

At Cathode:

8O₂ + 32H⁺ + 32e⁻ → 16H₂O

__________________________

Overall: C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O

Calculation of ΔG° for the above reaction

ΔG° = [5×(−394.4) + 6×(−237.2)] − [−8.2]

= −1972.0 − 1423.2 + 8.2 = −3387.0 kJ

= −3387000 Joules

From the overall equation we find n = 32

Using the relation ΔG° = −nFE°_cell and

substituting values we get

−3387000 = −32 × 96500 × E°_cell (F = 96500 C)

or E°_cell = 3387000 / (32×96500) = 3387000 / 3088000

= (3387 / 3088) V = 1.0968 V

This reaction is autocatalyzed and involves complex calculation for concentration terms.

We can look at the above results in a simple way to find the dependence of reaction rate (i.e. rate of disappearance of Br₂).

From data (1) and (2) in which concentration of CH₃COCH₃ and H⁺ remain unchanged and only the concentration of Br₂ is doubled, there is no change in rate of reaction. It means the rate of reaction is independent of concentration of Br₂.

Again from (2) and (3) in which (CH₃COCH₃) and (Br₂) remain constant but H⁺ increases from 0.05 M to 0.10 i.e. doubled, the rate of reaction changes from 5.7×10⁻⁵ to 1.2×10⁻⁴ (or 12×10⁻⁵), thus it also becomes almost doubled. It shows that rate of reaction is directly proportional to [H⁺]. From (3) and (4), the rate should have doubled due to increase in conc of [H⁺] from 0.10 M to 0.20 M but the rate has changed from 1.2×10⁻⁴ to 3.1×10⁻⁴. This is due to change in concentration of CH₃COCH₃ from 0.30 M to 0.40 M. Thus, the rate is directly proportional to [CH₃COCH₃].

Rate = k[CH₃COCH₃]¹[Br₂]⁰[H⁺]¹

= k[CH₃COCH₃][H⁺]

Given: k₁ = 10¹⁶ e_(−2000/T) and k₂ = 10¹⁵ .e_(-1000/T)

When k₁ = k₂ at temperature T, we have

10¹⁶. e_(-2000/T) = 10¹⁵. e_(−1000/T)

or 10 × 10¹⁵ e_(−2000/T) = 10¹⁵ e_(−1000/T)

or 10 .e_(−2000/T) = e_(−1000/T)

or ln 10 - (2000/T) = − (1000/T)

or ln 10 = 2000/T - 1000/T

or 2.303 log 10 = 1000/T

or 2.303 × 1 × T = 1000 [∴ log 10 = 1]

or T = (1000 / 2.303) K

Ti ; Z (22) is 1s²2s²2p⁶3s²3p⁶4s²3d²

V ; Z (23) is 1s²2s²2p⁶3s²3p⁶4s²3d³

Cr ; Z (24) is 1s²2s²2p⁶3s²3p⁶3d⁵4s¹

Mn ; Z (25) is 1s²2s²2p⁶3s²3d⁵4s²

The Second electron in all the cases (except Cr) is taken out from 4s orbital and for Cr it is taken from half‑filled 3d orbital. The force required for removal of second electron will be more for Mn than (except for Cr) due to having more positive charge. Based on this, we find the correct order Mn > V > Ti.

i.e. Cr > Mn > V > Ti.

S_N2 mechanism is followed in case of primary and secondary halides i.e., S_N2 reaction is favoured by small groups on the carbon atom attached to halogens so reactivity order is:

In the molecule

the number of stereoisomers is given by sum of geometrical isomers (because of presence of C = C) and optical isomers (because of presence of chiral carbon atom).

Number of geometrical isomers = 2 (one C = C is present).

Number of optical isomers = 2 (one chiral carbon atom).

Total number of stereoisomers = 2 + 2 = 4

A strong base (OH⁻) abstract an α — H atom from ketone which is acidic in nature.

Thus, two molecules of acetophenone condense to form a β‑hydroxy ketone which gets dehydrated in the presence of acid upon heating to form α, β‑unsaturated compound.

The more the basic character of the leaving group, the lesser is the reactivity. The basic character follows the order:

NH₂⁻ > OR⁻ > RCOO⁻ > Cl⁻

Hence, the relative reactivities of acyl compounds towards nucleophilic substitution follow the order Acyl halides > Acid anhydride > Ester > Amide.

Thyroxine is an amino hormone.

More the number of unpaired electrons present in a complex, more is its paramagnetic behaviour.

Out of the given ligands water, ammonia, cyanide and oxalate, maximum splitting will occur in case of cyanide (CN⁻), i.e. the magnitude of Δ₀ will be maximum in case of [Co(CN)₆]³⁻.

Diamond has ZnS type structure. So, number of atoms in a unit cell of diamond is 8.

He₂⁺ < O₂⁻ < NO < C₂²⁻

A = = = T

G ≡ ≡ ≡ C

Diazotisation followed by electrophilic substitution at para position.

Paecilomyces is a cosmopolitan filamentous fungus which inhabits the soil, decaying plants, and food prod-ucts. Some species of Paecilomyces are isolated from insect. The telomorphs of Paecilomyces are classified in the genera Cilromocleista, Talaromyces. Paecilomyces is usually considered as a contaminant but may also cause infections in humans and animals.

Trichoderma harzianum has proved a useful microor- 18 ganism for biological control of soil borne plant pathogens. Trichoderma harzianum is a fungus that is also used as a fun-gicide. It is used for foliar application, seed treatment and soil treatment for suppression of various disease causing fungal pathogens. Commercial biotechnological products such as 3 Tac have been useful for treatment of Botrytis, Fusarium, Penicillium sp. It is also used for manufacturing enzymes.

Modern detergent contain enzymes preparations of alkaliphiles. Alkaliphiles live in alkaline environments like soda lakes, or alkaline soil. The pH level in alkaline substances is from about 9 to 11 on the pH scale. The al-kaliphiles keep an alkaline level of about pH 9 inside the cells, but the environment has a higher level of alkalinity. Alkaliphiles can keep the right pH level by pumping in hydrogen ions in out at the right rate.