NEET-UG 2010

Initially body is at rest.

$\therefore$ $h=\frac{1}{2}g{t}^2$

$\Rightarrow$ $g=\frac{2h}{t^2}$

Maximum percentage error,

$\frac{\Delta g}{g}\times 100=\left[\frac{\Delta h}{h}+2\frac{\Delta t}{t}\right]\times 100$

Given that
$\frac{\Delta h}{h}\times 100$ = e₁

and $\frac{\Delta t}{t}\times 100$ = e₂

$\therefore$ $\frac{\Delta g}{g}\times 100$ = e₁ + 2e₂

$x=a\sin \omega t$ or $\frac{x}{a}=\sin \omega t$    ....(1)
$y=a\cos \omega t$ or $\frac{y}{a}=\cos \omega t$    ....(2)

Squaring and adding, we get
$\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$    ($\therefore$ ${\cos }^2\omega t+{\sin }^2\omega t=1$)
or $x^2+y^2=a^2$

This is the equation of a circle. Hence particle follows a circular path.

Let v be velocity of a projectile at maximum height H.


v = ucos$\theta$

According to given problem, $v=\frac{u}{2}$

$\therefore$ $\frac{u}{2}=u\cos \theta \Rightarrow \cos \theta =\frac{1}{2}\Rightarrow \theta =60^{\circ }$

Power delivered in time T is
P = F·V = MaV

or $P=MV\frac{dV}{dT}\Rightarrow PdT=MVdV$

$\Rightarrow PT=\frac{M{V}^2}{2}$ or $P=\frac{1}{2}\frac{M{V}^2}{T}$

Time taken to reach the bottom of inclined plane.

$t=\sqrt{\frac{2l\left(1+\frac{K^2}{R^2}\right)}{g\sin \theta }}$

Here, $l$ is length of incline plane

For solid cylinder $K^2=\frac{R^2}{2}$

For hollow cylinder K² = R²

Hence, solid cylinder will reach the bottom first.

As no external torque is acting about the axis, angular momentum of system remains conserved.

I₁${\omega }_1$ = I₂${\omega }_2$

$\Rightarrow {\omega }_2=\frac{I_1{\omega }_1}{\left(M+2m\right)r^2}=\frac{M\omega }{\left(M+2m\right)}$

Mass of the disc = 9M Mass of removed portion of disc = M The moment of inertia of the complete disc about an axis passing through its centre O and perpendicular to its plane is $I_1=\frac{9}{2}M{R}^2$

Now, the moment of inertia of the disc with removed portion

$I_2=\frac{1}{2}M{\left(\frac{R}{3}\right)}^2=\frac{1}{18}M{R}^2$

Therefore, moment of inertia of the remaining portion of disc about O is

$I=I_1-I_2=\frac{9}{2}M{R}^2-\frac{M{R}^2}{18}=\frac{40M{R}^2}{9}$

The acceleration due to gravity at a depth d below surface of earth is

$g^{\prime }=\frac{GM}{R^2}\left(1-\frac{d}{R}\right)=g\left(1-\frac{d}{R}\right)$

g' = 0 at d = R.

i.e., acceleration due to gravity is zero at the centre of earth.
Thus, the variation in value g with r is

For, r > R,

$g^{\prime }=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}=\frac{g{R}^2}{r^2}\Rightarrow g^{\prime }\propto \frac{1}{r^2}$

Here, R + h = r

For r < R, $g^{\prime }=g\left(1-\frac{d}{R}\right)=\frac{gr}{R}$

Here, $R-d=r\Rightarrow g^{\prime }\propto r$

Therefore, the variation of g with distance from centre of the earth will be as shown in the figure.

Centre of gravity of a body is the point at which the weight of the body acts and the radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the CG of the body to the axis.

Total energy of the orbiting satellite of mass m having orbital radius r is

$E=-\frac{GMm}{2r}$

where M is the mass of the planet.

Additional kinetic energy required to transfer the satellite from a circular orbit of radius R₁ to another radius R₂ is E₂−E₁

$=-\frac{GMm}{2R_E}-\left(-\frac{GMm}{2R_1}\right)=-\frac{GMm}{2R_2}+\frac{GMm}{2R_1}$

$=\frac{GMm}{2}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Ideal gas equation, for an adiabatic process is

$P{V}^{\gamma }=$ constant $\Rightarrow$ $P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

For monoatomic gas $\gamma =\frac{5}{3}$

$\therefore$ $P_1{V}_1^{5/3}=P_2{\left(\frac{V_1}{8}\right)}^{5/3}$

$\Rightarrow$ P₂ = P₁ × (2)⁵ = 32 P₁.

C_v = molar specific heat of the ideal gas at constant volume

C_p = molar specific heat of the ideal gas at constant pressure,

C_p' = MC_p and C_v’ = MC_v

Also ${C_p}^{\prime }-{C_v}^{\prime }=R$

MC_p – MC_v = R

C_p – C_v = R/M

Electric field inside a charged conductor is always zero.

Electric field between two parallel plates placed in vacuum is given by

$E=\frac{\sigma }{{\epsilon }_0}$

In a medium of dielectric constant K, $E^{\prime }=\frac{\sigma }{{\epsilon }_0}$

For kerosene oil K > 1 $\Rightarrow$ E' < E

Magnetic moment of the loop.

M = NIA = 2000 × 2 × 1.5 × 10^–4 = 0.6 J/T

Torque $\tau$ = MBsin30°

$=0.6\times 5\times 10^{-2}\times \frac{1}{2}=1.5\times 10^{-2}Nm$

Force due to electric field acts along the direction of the electric field but force due to the magnetic field acts along a direction perpendicular to both the velocity of the charged particle and the magnetic field. Hence both statements (2) and (3) are true. In statement (2), magnetic force is zero, so, electric force will keep the particle continue to move in horizontal direction. In statement (3), both electric and magnetic forces will be opposite to each other. If their magnitudes will be equal then the particle will continue horizontal motion.

Magnetic fields due to the two parts at their common centre are respectively,

$B_y=\frac{{\mu }_{0}i}{4R}$ and $B_z=\frac{{\mu }_{0}i}{4R}$


Resultant field = $\sqrt{B_y^2+B_z^2}$

= $\sqrt{{\left(\frac{{\mu }_{0}i}{4R}\right)}^2+{\left(\frac{{\mu }_{0}i}{4R}\right)}^2}$

$=\sqrt{2}\frac{{\mu }_{0}i}{4R}=\frac{{\mu }_{0}i}{2\sqrt{2}R}$

Magnetic field due to bar magnets exerts force on moving charges only. Since the charge is at rest, zero force acts on it.

The magnetic moment of a diamagnetic atom is equal to zero.

q = q₀ cos$\omega$t

$\Rightarrow$ cos$\omega$t = $\frac{q}{q_0}=\frac{C{V}_2}{C{V}_1}=\frac{V_2}{V_1}$

Current through the inductor

I = $\frac{dq}{dt}=\frac{d}{dt}\left(q_0\cos \omega t\right)$ = - q₀$\omega$ sin$\omega$t

$\therefore$ |I| = $C{V}_1\frac{1}{\sqrt{LC}}{\left[1-{\cos }^2\omega t\right]}^{1/2}$

= $V_1\sqrt{\frac{C}{L}}{\left[1-\left(\frac{V_2}{V_1}\right)\right]}^{1/2}$

= ${\left(\frac{C\left(V_1^2-V_2^2\right)}{L}\right)}^{\frac{1}{2}}$

As given, $E=10\cos (10^{7}t+kx)\hat{j}V/m$

here amplitude E₀ = 10 V/m and $\omega$ = 10⁷ rad/s

As c = $\nu \lambda =\frac{\omega \lambda }{2\pi }$

$\Rightarrow$ $\lambda$ = $\frac{2\pi c}{\omega }$ = $\frac{2\pi \times 3\times {10}^8}{{10}^7}$ = 188.4 m

Also, c = $\frac{\omega }{k}$

$\Rightarrow$ k = $\frac{\omega }{c}=\frac{{10}^7}{3\times {10}^8}$ = 0.033

The wave is propagating along y direction.

Angle of prism, A = r₁ + r₂

For minimum deviation r₁ = r₂ = r

$\therefore$ A = 2r

Given, A = 60°

Hence, r = $\frac{A}{2}$ = $\frac{60^{\circ }}{2}$ = 30^o

Refractive index for medium M₁ is

$\mu$₁ = $\frac{3\times {10}^8}{1.5\times {10}^8}$ = 2

Refractive index for medium M₂ is

$\mu$₂ = $\frac{3\times {10}^8}{2.0\times {10}^8}$ = $\frac{3}{2}$

For total internal reflection, sini $\ge$ sinC

But sinC = $\frac{{\mu }_2}{{\mu }_1}$

sini $\ge$ $\frac{{\mu }_2}{{\mu }_1}$ $\ge$ $\frac{3/2}{2}$

$\Rightarrow$ i $\ge$ sin^$-$1 $$\left(\frac{3}{4}\right)$$

A₁ = $\lambda$N₁ at time t₁

A₂ = $\lambda$N₂ at time t₂

Therefore, number of nuclei decayed during time interval (t₁ – t₂ ) is

N₁ - N₂ = $\frac{\left[A_1-A_2\right]}{\lambda }$

₁H² + ₁H² $$\to$$ ₂He⁴ + $\Delta$E

The binding energy per nucleon of a deuteron = 1.1 MeV

$\therefore$ Total binding energy = 2 × 1.1 = 2.2 MeV

The binding energy per nucleon of a helium nuclei = 7 MeV

$\therefore$ Total binding energy = 4 × 7 = 28 MeV

Hence, energy released

$\Delta$E = (28 – 2 × 2.2) = 23.6 MeV

The number of photoelectrons ejected is directly proportional to the intensity of incident light. Maximum kinetic energy is independent of intensity of incident light but depends upon the frequency of light.
Hence option (b) is correct

E = h$\nu$ = E₃ - E₁

= $-\frac{13.6}{3^2}-\left(-\frac{13.6}{1^2}\right)$

= 12.1 eV

Therefore, stopping potential

eV₀ = $h\nu$ - $\varphi$

= (12.1 – 5.1) eV

V₀ = 7 V

For transistor action, the base region must be very thin and lightly doped. Also, the emitter-base junction is forward biased and base-collector junction is reverse biased.

It is clear from given logic circuit, that out put Y is low when both the inputs are high, otherwise it is high. Thus logic circuit is NAND gate.

According to the de-Broglie equation, $\lambda$ = $\frac{h}{mv}$

$\therefore$ $\lambda$ = $\frac{6.6\times {10}^{-34}}{0.66\times 100}=1\times 10^{-35}m$

PV = nRT

$\Rightarrow$ $P=\frac{nRT}{V}$ = $\frac{w}{m}\frac{RT}{V}$

= $\frac{6}{16.05}\times \frac{8.314\times 402}{0.03}$

= 41648 Pa

	2A(g)	+	B(g)	⇌	3C(g)	+	D(g)
Initial mole	1		1		0		0
At equilibrium	1 - (2 $\times$ 0.25) = 0.5		1 - 0.25 = 0.75		3 $\times$ 0.25 = 0.75		0.25

Equilibrium constant, K = $\frac{{\left[C\right]}^3\left[D\right]}{{\left[A\right]}^2\left[B\right]}$

$\therefore$ K = $\frac{{\left[0.75\right]}^3\left[0.25\right]}{{\left[0.5\right]}^2\left[0.75\right]}$

Given

Fe₂O_3(s) + 3CO_(g) $\to$ 2Fe_(s) + 3CO_2(g); $\Delta$H = $-$ 26.8 kJ .....(1)

FeO_(s) + CO(g) $\to$  Fe_(s) + CO_2(g); $\Delta$H = $-$ 16.5 kJ .....(2)

Fe₂O_3(s) + CO_(g) $\to$  2FeO_(s) + CO_2(g), $\Delta$H = ? ....(3)

Equation (3) can be calculated as :

(1) - 2(2)

$\therefore$ $\Delta$H = –26.8 + 33.0 = +6.2 kJ

We know, from Gibb's equation,

$\Delta$G = $\Delta$H – T$\Delta$S

When $\Delta$G = 0, $\Delta$H = T$\Delta$S

$\therefore$ $T=\frac{\Delta H}{\Delta S}$ = $\frac{40.63\times {10}^3}{108.8}$ = 373.4 K

Since the ideal gas expands spontaneously into vacuum,, P_ext = 0.

$\therefore$ Work done is also zero.

When K_p > Q, rate of forward reaction > rate of backward reaction.

$\therefore$ Reaction is spontaneous.

When $\Delta$G^o < RT ln Q, $\Delta$G^o is positive, reverse reaction is feasible, thus reaction is non spontaneous.

When K_p = Q, rate of forward reaction > rate of backward reaction.

$\therefore$ Reaction is in equilibrium.

When T$\Delta$S > $\Delta$H, $\Delta$G will be negative only when $\Delta$H = +ve.

$\therefore$ Reaction is spontaneous and endothermic.

EMF of a cell = Reduction potential of cathode – Reduction potential of anode

= Reduction potential of cathode + Oxidation potential of anode

= Oxidation potential of anode – Oxidation potential of cathode.

Equivalent conductance of an electrolyte at infinite dilution is given by the sum of equivalent conductances of the respective ions at infinite dilution.

The value of rate constant can be increased by increasing the temperature and is independent of the initial concerntration of the reactants.

The ratio in cation and anion size will be maximum when the cation is largest size and the anion is smallest size. So among the given species, Cs⁺ has maximum size among given cations and F^- has smallest size among the given anions, So Cs/F has highest r_c/r_a

Atomic radius decreases on moving from left to right in a period. so order of sizes for Cl, P and Mg is Cl < P < Mg. Down the group size increases. So overall order is : Cl < P < Mg < Ca.

No. of electron pairs = No. of atoms bonded to it + 1/2[Gp. no. of central atom - Valency of central atom $\pm$ No. of electrons]

For NO₃^- = 3 + 1/2[5 - 6 + 1] = 3 (sp² hybridisation)

For H₃O⁺ = 3 + 1/2[6 - 3 - 1] = 4 (sp³ hybridisation)

For neutral molecules

No. of electron pairs = No. of atoms bonded to it + 1/2[Gp. no. of central atom - Valency of central atom]

$\therefore$For CH₄, no. of e^- pairs = 4 + 1/2[4 - 4] = 4 (sp³ hybridisation)

For SF₄, no. of e^- pairs = 4 + 1/2[6 - 4] = 5 (sp³d hybridisation)

and For ions,

No. of electron pairs = No. of atoms bonded to it + 1/2[Gp. no. of central atom - Valency of central atom $\pm$ No. of electrons]

$\therefore$For BF₄^-, no. of e^- pairs = 4 + 1/2[3 - 4 + 1] = 4 (sp³ hybridisation)

For NH₄⁺, no. of e^- pairs = 4 + 1/2[5 - 4 - 1] = 4 (sp³ hybridisation)

A $\frac{\text{buildrel}\Delta }{⟶}$ colourless gas + residue

Residue + H₂O $\to$ B $\frac{\text{buildrel}excessC{O}_2}{⟶}$ C $\frac{\text{buildrel}\Delta }{⟶}$ A

This is possible only when ‘A’ is CaCO₃. The reactions are as follow :

Heavy water is used for slowing down the speed of neutrons in nuclear reactors, hence used as moderator. Boiling point of heavy water is greater (374.42 K) than that of ordinary water (373 K), hence heavy water is more associated. Dielectric constant of ordinary water is greater than that of heavy water, hence ordinary water is a better solvent.

The common stable oxidation state of all the lanthanoids is +3. The oxidation state of +2 and +4 are also exhibited by some of the elements. These oxidation states are only stable in those cases where stable 4f⁰ , 4f⁷ or 4f¹⁴ configurations are achieved.

(1) Sulphuric acid     (iv) Contact process

(2) Steel                     (ii) Bessemer’s process

(3) Sodium hydroxide    (iii) Leblanc process

(4) Ammonia                 (i) Haber’s process

H₃$\stackrel{+}{O}$ has a lone pair of electrons on oxygen atom, thus it is not an electrophile. Also the octet is complete

The conformation (d) is most stable because of intermolecular H-bonding.

If a molecule contains both carbon-carbon double or triple bonds, those two are treated as per in seeking the lowest number combination. However, if the sum of numbers turns out to be the same starting from either of the carbon chain, then lowest number is given to the C = C double bond.

The iodoform test is exhibited by ethyl alcohol, acetaldehyde, acetone, methyl ketones, those alcohols which possess
CH₃CH(OH)— group, acetophenone, $\alpha$-hydroxypropionic acid, keto acid, 2-aminoalkanes, etc.

Glycerol when treated with excess HI produces 2–iodopropane.

The intermediate is carbocation which is destabilised by C = O group (present on - carbon to the –OH group) in the first three cases. In (c), –hydrogen is more acidic which can be removed as water. Moreover, the positive charge on the intermediate carbocation is relatively away from the C = O group.

Under alkaline conditions of the reagent, fructose gets converted into a mixture of glucose and mannose (Lobry de Bruyn van Ekenstein rearrangement) both of which contain the –CHO group and hence reduce Tollens’ reagent to give silver mirror test.

A – Rough endoplasmic reticulum
B – Smooth endoplasmic reticulum
C – Nucleus
D – Cytoplasm

Enzymes are mostly proteins but some are RNA (ribozymes). No lipid working as enzymes are known.

In Pistia (water lettuce) vegetative propagation occurs by offset where one internode long runners grows horizontally along the soil surface and gives rise to new plants either from axillary or terminal buds.

A – offset of water hyacinth (Eichhornia)
B – Antheridiophore of Marchantia
C – Antipodals of the mature embryo sac
D – Oogonium of Chara

The genus Glomus form endomycorrhiza, a symbiotic associations with plants. The fungal symbiont in these associations absorbs phosphorus from soil and passes it to the plant. Plants having such associations show other benefits also, such as resistance to root-borne pathogens, tolerance to salinity and drought, and an overall increase in plant growth and development.
Nostoc is a blue green algae, Agaricus is a basidiomycetes, Rhizobium is a eubacteria.

A – Denitrification

B – Ammonification

C – Plants

D – Animals

Plasmid and bacteriophage are used as vectors in genetic engineering. Plasmid is an autonomously replicating circular extra chromosomal DNA found in bacteria. They can be transferred from cell to cell in a bacterial colony. This characteristic is being used in biotechnology for transferring desirable gene into target gene of the host. Bacteriophage is a bacterial virus which can infect it, quickly multiply within and destroy (lyse) their host (virus) cells. During infection bacteriophages inject their DNA into these cells. The injected DNA selectively replicate and are expressed in the host that results in a multiplication of phages that ultimately burst out of the cell (by lysis). This ability of transferring DNA from the phage genome to specific host during infection process gave scientists the idea that specially designed phage vectors could be used for gene cloning.

The Indian rhinoceros is a endemic of north-east region of India. Kaziranga National Park (Assam) is famous for rhinoceros.

Chemosynthetic autotrophic bacteria oxidise various inorganic substances such as nitrates, nitrites and ammonia and use the released energy for their ATP production.

They play a great role in recycling nutrients like nitrogen, phosphorous, iron and sulphur.

Black stem rust is caused by Puccinia graminis tritici. The genus Puccinia includes 700 species, which cause rust diseases of many economic plants such as wheat, barley, oats, etc. It is called a rust because of the reddish brown colour of the spores that are found chiefly upon the surface of the host leaves and stems. P. graminis is heteroecious, i.e., requiring two hosts, wheat and barberry for the completion of normal life cycle.

According to the nature of the spores, the life cycle of the P. graminis is divided into five stages. It is during, teleuto stage, the teliospore (or teleutospores) produce dark brown to black pustules on the surface of stems and leaves of the wheat that results into ‘black stem rust of wheat’.

In cotton, china rose and lady’sfinger margins of sepals or petals overlap that of the next one this mode of arrangement (aestivation) is called twisted.

Opuntia is a xerophytic plant which lives in dry habitat. The plant has fleshy organs where water and mucilage are stored. The stem is modified into flat green structure, therefore, Opuntia is also called as phylloclades.

Flowers in pea have diadelphous stamens. The floral formula for Solanaceae is

The plants belonging to the Family Fabaceae such as soybean, pea, sem, moong, gram, etc have the floral formula

A – Fixation of CO₂ by PEP carboxylase
B – Decarboxylation
C – Regeneration

Z scheme involves both PSI and PSII to transfer electron excited by light starting from PSII uphill to the acceptor, down to the electron transport chain to PSI, which further comprise of excitation of electrons, transfer to another acceptor and finally down hill to NADP⁺ causing reduction of it to NADPH + H+. Stroma lamella contains PSI only.

In kranz anatomy, the mesophyll is undifferentiated and its cells occur in concentric layers around vascular bundles. Vascular bundles are surrounded by large sized bundle sheath cells which are arranged in a wreath-like manner in one to several layers. C4 plants, both monocots and dicots, such as sugarcane, maize, sorghum have kranz anatomy in leaf.

The given pedigree chart shows that both the daughters received the gene from the parents, while son may be normal or affected. It shows that the female parent is heterozygous.

The given situation is an example of incomplete dominance where phenotype found in F1 generation do not resemble either of the two parents. The genotype of the two plants used for cross will be The incomplete dominance of dominant allele (here ‘R’) over recessive allele (here ‘r’) could be due to mutations (insertion, deletion, substitution or inversion of nucleotides). The mutant allele generally produces a faulty or no product. This modification in the product may lead to incomplete dominance of the (unmodified) wild type dominant allele.

Test cross is performed to determine the genotype of F₂ plant. In a typical test cross an organism showing dominant phenotype and whose genotype is to be determined is crossed with one that is homozygous recessive for the allele being investigated, instead of self-crossing. The progenies of such a cross can easily be analysed to predict the genotype of the test organism.

The three alleles I^A, I^B and i of gene I in ABO blood group system can produce six different genotypes and four different phenotypes as shown below :

Insulin gene is found in every body cell but is not expressed in all cells. It is nucleosome which consists of a core of eight histones. DNA is composed of nucleotides. Centriole is found in animal cells, which produces aster during cell division.

Unlike in prokaryotes where transcription and translation take place in the same compartment, in eukaryotes primary transcript is first processed in the nucleus and then transported outside of the nucleus. Since the primary transcripts of the eukaryotes contains both the expressing genes (exons) and nonexpressing genes (introns), it undergoes splicing of introns and later capping and tailing at 5'-end and 3'-end respectively.

Lac operon is the operon that regulates lactose metabolism in the bacterium Escherichia coli. Its form wast first postulated in 1961 by Francois Jacob and Jacques Monod to explain the control of β- galactosidase synthesis and for this work, they were awarded Nobel Prize. Lac operon system or inducible operon system consists of regulator gene (i) promoter gene (z, y and a), repressor protein and inducer.

The phosphodiester bonds is formed between the phosphate group, which is connected to carbon 5' of the sugar residue of one nucleotide and carbon 3' of the sugar residue of the next nucleotide.

Insulin is now being commercially produced by genetic engineering. Insulin consists of two short polypeptide chains: chain A and chain B, that are linked together by disulphide bonds. Insulin, in mammal is synthesised as a prohormone which contains an extra stretch called the C-peptide. During maturation this
C-peptide is removed. The production of insulin could only have been commercially possible if somehow the maturation process of C-peptide been skipped.
This problem was solved in 1988 by Eli Lilly, an American company which prepared functionable insulin from two DNA sequences corresponding to A and B chains of human insulin and introduced them in plasmids of E.coli to produce insulin chains. In this way, chains A and B were produced separately which was extracted, combined by creating disulfide bonds to get human insulin.

Predation is an interaction between members of two species in which members of one species capture, kill and eat up members of other species. Host is a term which is specifically related to parasitism. Amensalism is an interspecies interaction in which one species is harmed whereas the other one is unaffected. Parasitic organism can live both over the surface of their host or inside their body.

When sewage mixes with water body, microorganisms present in it biodegrade organic matter of sewage using oxygen. This results into a sharp decline in dissolved oxygen which may cause mortality of aquatic creatures. Gradually, however, dissolved oxygen increases in concentration with the completion of biodegradation of sewage matter.

Monoecious plants have separate male and female flowers on the same plant. Pinus have both the male and female cones or strobili on the same tree.

Ethylene promotes root growth and root hair formation. In low concentration ethylene is used for initiation of roots and also of lateral roots.

Indole-3-acetic acid (also called auxin) is a phytohormone which is generally produced by the growing apices of the stems and roots, from where they migrate to the regions of their action. It is observed that the growing apical bud inhibits the growth of the lateral (axillary) bud (apical dominance).

Since apical meristem is the site of auxin synthesis, it is the physiological effect of the auxin which results in the phenomenon of apical dominance. When shoot tips is removed it usually results in the growth of lateral buds. This phenomenon is widely applied in tea plantations and hedge-making because as in tea plantation and industries, the apical bud is plucked for tea processing which results in more lateral buds thus enhancing plantation and further industrial purposes

The representation of forest ecosystem in pyramid of numbers is always upright because higher trophic level comprising of tertiary consumers is generally smaller than that of the lower trophic levels (i.e., secondary consumer, than primary consumer and primary producer).
The pyramid of number is inverted in case of single tree producer which can provide nourishment to several herbivores such as birds which can further support larger population of ectoparasites.

Histopathological study is the invasive technique. Radiography and CT involves X-rays which are harmful.
In MRI strong magnetic fields and non-ionising radiations are used to detect any physiological changes in the concerned tissue. Hence it is safe for detection of cancers.

Plasmodium is a tiny protozoan which is responsible for malaria in the human. In malaria the patient experiences high fever which periodically rises and also experiences recurring chills with fever. Such symptoms are seen because when the RBCs carrying Plasmodium (one of the stage in the life cycle of the parasite) ruptures it releases a toxic substance called haemozoin which is chiefly responsible for the chill and high fever recurring every three to four days.

The haemoglobin content per 100 ml of blood in a normal healthy human adult is actually found in the range of 12 - 16 grams. Therefore, the correct answer is Option D: 12 - 16 mg. However, it's important to note that there seems to be a mistake in the unit provided in the options. The correct unit should be grams (g) and not milligrams (mg), as haemoglobin content in blood is measured in grams per deciliter (g/dL). So, the correct expression reflecting normal haemoglobin levels should be 12 - 16 g/dL, not mg.

In general, average ranges for haemoglobin levels vary by age and sex, typically:

• For men: 13.8 to 17.2 grams per deciliter


• For women: 12.1 to 15.1 grams per deciliter

These levels can be affected by various factors including diet, physical activity, and medical conditions. It's essential for individuals to have their haemoglobin levels within the normal range as it's crucial for effectively transporting oxygen throughout the body. Deviations from the normal range can indicate different medical conditions that may require further investigation.

The term angina means chest pain. In this disease enough oxygen does not reach the heart muscles. The patient experiences heart pain usually in front of the chest.

AB blood group person have both antigens A and B, but do not have antibodies in the plasma. Due to the presence of both the antigens, AB blood group person cannot donate blood to anyone. However, the person can receive blood from any blood group as it has no antibodies. Thus, AB blood group is a universal recepient.

The fastest distribution of an injectable material or medicine, with minimal risk, is achieved by injecting it into the veins, making Option C the correct answer. Here’s why:

• Veins: When medicine is injected into the venous system (intravenously), it is quickly carried throughout the body by the bloodstream. The advantage of venous injection (IV injection) is the rapid onset of effects as the medicine is directly entered into the circulatory system, bypassing the need for absorption. This method also allows for precise control over the dosage and is used for substances that might be irritants or not effective if administered by other routes.
• Arteries: Injecting a substance directly into an artery is far more dangerous and is rarely done outside of specialized medical procedures. The arterial system is under much higher pressure than the venous system, which increases the risk of complications, including arterial damage and the potential for creating embolisms that could cause strokes or other serious issues.
• Muscles: Intramuscular injections are absorbed more slowly than intravenous injections because the blood flow to muscle tissues is less than to veins. This method is used for certain types of medication that need to be absorbed over a longer period. It's safer than arterial injection but slower in effect compared to venous injection.
• Lymph Vessels: The lymphatic system is not designed for the fast distribution of substances throughout the body. It has a much slower flow than the circulatory system and is involved in managing the immune response and transporting waste and other materials out of the bodily tissues. Injection into the lymphatic system is not a common practice for administering medications for systemic effect.

In summary, intravenous injection (Option C) is the preferred route for the immediate and safest distribution of injectable substances throughout the body.

Cartilage is avascular, as the blood vessels innervate only perichondrium through which nutrition diffuses into cartilage cells. Cornea is also avascular.

Hypothalamus is the region of the forebrain in the floor of the third ventricle, linked with the thalamus above and the pituitary gland below. It contains several important centres controlling body temperature, thirst, hunger, and eating, water balance and sexual function. It is also closely connected with emotional activity and sleep and functions as a centre for the integration of hormonal and autonomic nervous activity through its control of the pituitary secretions.

Jaundice is a disorder in which skin and eyes turn yellow due to the deposition of bile pigment. This happens when bile made in the liver fails to reach the intestine due to obstruction of the bile duct. As a result, the bile is absorbed into the blood instead of going to the duodenum and cause yellowing of eyes and skin.

Parietal or oxyntic cells secrete HCl (due to which pH of stomach is highly acidic) and intrinsic factor. Parietal glands also secrete pepsinogen to which hydrochloric acid acts to convert it into pepsin. Pepsin in return causes digestion of proteins. If parietal cells become non-functional it will directly affect protein digestion.

Process of parturition is induced by both nervous system and hormones secreted by the endocrine glands of the mother. The signals for child birth (parturition) originate from the fully developed fetus and placenta which induce mild uterine contractions called fetal ejection reflex. This causes quick release of oxytocin from the maternal posterior lobe of pituitary gland which induces labour pains. Prostoglandins, progesterone and estrogen also play a role. Prolactin is the hormone which induces lactation and has no role in parturition.

The male accessory glands include paired seminal vesicles, a prostate and paired bulbourethral glands. Secretions of these glands constitute the seminal plasma which is rich in fructose, calcium and certain enzymes. The secretions of bulbourethral glands also helps in the lubrication of the penis.

Implantation in endometrial uterine wall takes place at blastocyst stage of embryonic development. Before implantation, the blastomeres of early blastocyst get arranged into an outer layer called trophoblast and an inner group of cells attached to trophoblast called inner cell mass. It is the trophoblast layer through which blastocyst gets attached to the endometrium and the inner cell mass gets differentiated as the embryo.

Nephridia is the excretory organ of the earthworm. Earthworms have three types of nephridial structures called as septal, integumentary and pharyngeal nephridia. These three nephridial structures are present on different positions in the body and also vary in structures.

Septal and pharyngeal nephridia are both enteronephric i.e., nitrogen products are expelled in gut. Integumentary nephridia is exonephric i.e., nitrogen waste products are directly discharged outside.

Animals belonging to Phylum Chordata are fundamentally characterised by the presence of a notochord, a dorsal hollow nerve cord and paired pharyngeal gill slits. Crocodile, penguin, whale and dogfish are all chordates.

All of them have gill slits or have had it during embryonic development. Thus, paired gill slits are present in these animal at some stage of life.

The brain capacity gradually increased from early human ancestors to modern man. Homo habilis had 650 - 800cc brain capacity which increased around 900cc in Homo erectus. The Neanderthal man (Homo neanderthalensis) had 1400cc brain capacity which evolved to around 1450cc in Homo sapiens.