NEET-UG 2001

Plank constant (h) = $\frac{E\lambda }{c}$ = $\frac{\left[M{L}^2{T}^{-2}\right]\left[L\right]}{\left[L{T}^{-1}\right]}$ = [ML²T^-1]

Angular momentum (L) = I$\times$$\omega$ = [ML²][T^-1] = [ML²T^-1]

At height $\frac{h}{2}$, speed is = 10 m/s

and at height h, speed = 0

Using formula,

v² = u² - 2gh

0 = (10)² - 2(10)$\frac{h}{2}$

$\Rightarrow$ h = $\frac{100}{10}$ = 10 m

$\omega =\frac{2\pi }{t}$. As t is same

$\therefore$ $\frac{{\omega }_1}{{\omega }_2}=1$

$\left|\overrightarrow{A}+\overrightarrow{B}\right|=\left|\overrightarrow{A}\right|+\left|\overrightarrow{B}\right|$ if $\overrightarrow{A}\parallel \overrightarrow{B}$. $\therefore$ $\theta$ = 0^o

Apply conservation of linear momentum. Total momentum before explosion = total momentum after explosion

$0=\frac{m}{5}{v}_1\hat{i}+\frac{m}{5}{v}_2\hat{j}+\frac{3m}{5}{\overrightarrow{v}}_3$

$\frac{3m}{5}{\overrightarrow{v}}_3=-\frac{m}{5}\left[v_1\hat{i}+v_2\hat{j}\right]$

${\overrightarrow{v}}_3=\frac{-v_1}{3}\hat{i}-\frac{v_2}{3}\hat{j}$
     $\therefore$ $v_1=v_2=30$ m/sec

${\overrightarrow{v}}_3=-10\hat{i}-10\hat{j};v_3=10\sqrt{2}$ m/sec

Net force experienced = $\frac{Total\mathrm{Impulse}}{Timetaken}$

$=\frac{m\Delta v}{t}=0.15\times \frac{20}{0.1}=30N$

Maximum friction force = $\mu$mg

= .6 × 1 × 9.8 = 5.88 N

But here required friction force

= ma = 1 × 5 = 5 N

Load W = Mg = 75 × 10 = 750 N

Effort (P) = 250 N

$\therefore$ Mechanical advantage
$=\frac{load}{effort}=\frac{W}{P}=\frac{750}{250}=3$

Velocity ratio
$=\frac{distancetravelledbyeffort}{distancetravelledbyload}=4$

Efficiency $(\eta )=\frac{Mechanicaladvantage}{Velocityratio}$

$=\left(3/4\right)\times 100=75\%$

Kinetic energy of the ball = K and angle of projection ($\theta$) = 45^o

Velocity of the ball at the highest point = v cos $\theta$

$=v\cos 45^o=\frac{v}{\sqrt{2}}$

Therefore kinetic energy of the ball

$\Rightarrow$ $\frac{1}{2}m\times \left(\frac{v}{\sqrt{2}}\right)=\frac{1}{4}m{v}^2=\frac{K}{2}$

Energy = $\frac{1}{2}K{x}^2=\frac{1}{2}\frac{F^2}{K}$

$\frac{K_A}{K_B}=2$

$\therefore$ $\frac{E_A}{E_B}=\frac{1}{2}\Rightarrow E_B=2E_A$

Drop in P.E. = maximum K.E.

mg(2 – 0.75) = 1mv²

$\Rightarrow$ v = $\sqrt{2g\left(1.25\right)}=5m/s$

Using $v^2=\frac{2gh}{1+\frac{h}{R}}$ and given h = R

$\therefore$ $v=\sqrt{gR}=\sqrt{\frac{GM}{R}}$

Heat flow rate $\frac{dQ}{dt}=\frac{KA\left(T_1-T_2\right)}{L}=Q$

When linear dimensions are double.

$A_1\propto r_1^2,L_1=L$

$A_2\propto 4r_1^2,L_2=2L\Rightarrow Q_2=2Q_1$

Efficiency is maximum in Carnot engine which is an ideal engine.

$\because \eta =\frac{400-300}{400}\times 100\%=25\%$

$\therefore$ efficiency 26% is impossible for his heat engine.

Energy = $\frac{1}{2}m{\omega }^2{a}^2=\frac{1}{2}k{a}^2$

Resultant amplitude = 2a (1 + cos$\varphi$) = a

$\therefore$ $\left(1+\cos \varphi \right)=1/2;$

$cos\varphi =-\frac{1}{2};\varphi =\frac{2\pi }{3}$

$n=\frac{1}{2l}\sqrt{\frac{T}{\pi r^2\rho }}$

${\rho }_1^{{}^{\prime }}=\frac{\rho }{2};T^{\prime }=2T$

D' = 2D or r' = 2r

$n^{\prime }=\frac{1}{2l}\sqrt{\frac{2T}{\pi {\left(2r\right)}^2\frac{\rho }{2}}}$

After solving, $n^{\prime }=\frac{1}{2l}\sqrt{\frac{2T}{\pi r^{2}l}}=n$

Comparing the given equation with general equation,
$y=a\sin 2\pi \left(\frac{t}{T}-\frac{x}{\lambda }\right)$, we get

$T=\frac{2\pi }{100}$ and $\lambda =20\pi$

$\therefore v=\upsilon \lambda =\frac{100}{2\pi }\times 20\pi =1000m/s$

For complete cube $\varphi =\frac{Q}{{\epsilon }_0}\times 10^{-6}$

For each face $\varphi =\frac{1}{6}\frac{Q}{{\epsilon }_0}\times 10^{-6}$

When an electric dipole is placed in a uniform electrical field $\overrightarrow{E}$, the torque on the dipole is given by

$\overrightarrow{\tau }=\overrightarrow{p}\times \overrightarrow{E}$

For metals specific resistance decrease with decrease in temperature whereas for semiconductors specific resistance increases with decrease in temperature.

Potential gradient = Potential fall per unit length. In this case resistance of unit length.

$R=\frac{\rho l}{A}=\frac{{10}^{-7}\times 1}{{10}^{-6}}=10^{-1}\Omega$

Potential fall across R is

V = I.R = 0.1 $\times$ 10^-1 = 0.01volt/m.

$=10^{-2}$ volt/m

Here, P = Q = S = R = 10 $\Omega$

since $\frac{P}{Q}=\frac{S}{R}=1$, so it is balanced Wheatstone bridge and no current pass through galvanometer.

$\therefore$ Equivalent resistance is given by

R = $\frac{\left(10+10\right)\left(10+10\right)}{\left(10+10\right)+\left(10+10\right)}$

$=\frac{20\times 20}{40}=10\Omega$

Energy stored per unit volume

$=\frac{1}{2}{\epsilon }_0{E}^2=\frac{1}{2}{\epsilon }_0{\left(\frac{V}{d}\right)}^2=\frac{1}{2}{\epsilon }_0\frac{V^2}{d^2}$
   $\left(\because E=\frac{V}{d}\right)$

The frequency of revolution of charged particle in a perpendicular magnetic field is

$f=\frac{1}{T}=\frac{1}{\frac{2\pi r}{v}}=\frac{v}{2\pi r}$

$=\frac{v}{2\pi }\times \frac{eB}{mv}=\frac{eB}{2\pi m}$

Magnetic moment M = niA

Tangent galvanometer is used to measure current.

Diamagnetism oes not depened on the temperature

Given, $I=t^2{e}^{-t}$

So, $\frac{dI}{dt}=2t{e}^{-t}-t^2{e}^{-t}$

$\left|\epsilon \right|=L\frac{dI}{dt}$

here emf is zero when $\frac{dI}{dt}$ = 0

$\therefore$ $2t{e}^{-t}-t^2{e}^{-t}$ = 0

$\Rightarrow$ $2t{e}^{-t}=t^2{e}^{-t}$

$\Rightarrow$ $t{e}^{-t}\left(t-2\right)$ = 0

As t $\ne$ $\infty$, t $\ne$ 0

$\therefore$ t = 2 sec

Reactance X = $\frac{1}{\omega C}=\frac{1}{2\pi fC}$

$\therefore$ X $\propto$ $\frac{1}{fC}$

$\therefore$ $\frac{X_1}{X}=\frac{fC}{f_1{C}_1}$ = $\frac{fC}{2f\times 2C}$ = $\frac{1}{4}$

$\Rightarrow$ X₁ = $\frac{X}{4}$

Quality factor Q = $\frac{\omega L}{R}$

As ${\omega }^2=\frac{1}{LC}$

$\therefore$ Q = $\frac{1}{\omega RC}$

The ozone layer absorbs the harmful ultraviolet rays coming from sun.

Optical fibre are based on total internal reflection.

$\theta ={\sin }^{-1}\left(\frac{1}{\mu }\right)$ = ${\sin }^{-1}\left(\frac{3}{5}\right)$

$\Rightarrow$ $\sin \theta =\frac{3}{5}$

$\therefore$ tan$\theta$ = $\frac{3}{4}$ = $\frac{r}{4}$

$\Rightarrow$ r = 3 m

Frequency remains same.

Energy released in nuclear fission is due to some mass is converted into energy.

Given : Mass of neutron = M_n
Mass of proton = M_p;
Atomic mass of the element = M ;
Number of neutrons in the element = N and number of protons in the element = Z.

We know that the atomic mass (M) of any stable nucleus is always less than the sum of the masses of the constituent particles.

Therefore, M $<$ [NM_n + ZM_p].

X is a neutrino, when b-particle is emitted.

2dsin$\theta$ = n$\lambda$

$\because$ $-1\le \sin \theta \le 1$

Therefore $\lambda$_max = 2d

$\Rightarrow$ $\lambda$_max = 2 $\times$ 2.8 $\times$ 10^$-$8

= 5.6 $\times$ 10^$-$8 m

N = N₀${\left(\frac{1}{2}\right)}^n$

$\Rightarrow$ 1 = 256${\left(\frac{1}{2}\right)}^n$

$\Rightarrow$ n = 8

Time for 8 half life = 100 hours

${}_5^{10}$ B + ${}_0^1$ n $\to$ ${}_2^4$ He + ${}_3^7$ Li

$\alpha$-rays contain Helium nuclei which contains 2 unit of positive charge.

As E_n = $\frac{-2{\pi }^{2}m{K}^2{Z}^2{e}^4}{n^2{h}^2}$

For helium Z = 2. Hence requisite answer is 4E_n.

As Intensity becomes 4 times. So number of electrons emitted per second also becomes 4 times.

photo electric effect shows particle nature of light.

As eE = evB

$\therefore$ v = $\frac{\left|\overrightarrow{E}\right|}{\left|\overrightarrow{B}\right|}$

D₁ is reverse biased and D₂ is forward biased.

$\therefore$ I = $\frac{5}{30+20}$ = $\frac{5}{50}$ A

$\frac{I_C}{I_E}$ = 0.98 = $\alpha$

$\therefore$ Current gain, $\beta$ = $\frac{I_C}{I_B}$ = $\frac{\alpha }{1-\alpha }$ = 49

This truth table represents NAND gate.

Molarity = $\frac{1.17\times 1000}{36.5\times 1}=32.05$

Specific volume ( vol. of 1 g) cylindrical virus particle

= 6.02 $\times$ 10^-2 cc/g

Radius of virus, r = 7 Å = 7 $\times$ 10^-8 cm

Volume of virus = $\pi r^{2}l$

= $\frac{22}{7}\times (7\times 10^{-8})\times 10\times 10^{-8}$

= 154 $\times$ 10^-23 cc

wt. of one virus particle = $\frac{Volume}{SpecificVolume}$

$\Rightarrow$ $\frac{1.54\times {10}^{-23}}{6.02\times {10}^{-2}}g$

$\therefore$ Molecular wt of virus = wt. of N_A particle

= $\frac{1.54\times {10}^{-23}}{6.02\times {10}^{-2}}\times 6.02\times 10^{-23}g/mol$

= 15400 g/mol = 15.4 kg/mol

n = 3, l = 2, m = +2

It shows one of the five d-orbitals(3d)

For $\pi$ overlap, the lobes of the atomic orbitals are perpendicularto the line joining the nuclei.

Hence only sidewise overlapping takes place.

When water pressure increases in the pressure cooker, water boils at lower temperature and the beans in pressure cooker are cooked earlier.

CH₃COOH ⇌ CH₃COO^– + H⁺

K_c = $\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$

$\Rightarrow$ [CH₃COOH] = $\frac{3.4\times {10}^{-4}\times 3.4\times {10}^{-4}}{1.7\times {10}^{-5}}$

= 6.8 × 10^–3

Strong base has higher tendency to accept the proton. Increasing order of base and hence the order of accepting tendency of proton is

R $-$ NH₂ > NH₃ > HS^$$-$$ > I^$$-$$

M₂S ⇌ 2M⁺ + S^2–

K_sp = [M⁺]²[S^2–] = (2s)²(s) = 4s ³

K_sp = 4(3.5 × 10^–6)³ = 1.7 × 10^–16

In polyprotic acids the loss of second proton occurs much less readily than the first. Usually the K_a values for successive loss of protons from these acids differ by at least a factor of 10^–3.

$\therefore$ K_a1 > K_a2

The osmotic pressure of sea water is 25 atm at 15°C. When pressure greater than 26 atm is applied on sea water separated by a rigid semipermeable membrane. Pure water is obtained. This is also called desalination of sea water.

PbO₂ → PbO

Pb has +4 oxidation state In PbO₂.

Pb has +2 oxidation state In PbO.

Here $\Delta$G is negative that is why reaction is spontaneous and Pb⁴⁺ reduces Pb²⁺ thus, Pb²⁺ is most probable oxidation state of Pb.

SnO₂ $\to$ SnO;

Sn has +4 oxidation state In SnO₂.

Sn has +2 oxidation state In SnO.

$\Delta$G is positive that is why reaction is not spontaneous thus, +4 oxidation state of Sn is more stable as it does not change to +2 oxidation state spontaneously.

2H₂O_2(l) $\to$ 2H₂O_(l) + O_2(g), $\Delta$H_r = ?

H_2(g) + O_2(g) → H₂O_2(l), $\Delta$H = – 188 kJ mol ...(1)

H_2(g) + O_2(g) → H₂O_(l), $\Delta$H = –286 kJ/mol ....(2)

(1) - (2)

H₂O_2(l) → H₂O_(l) + $\frac{1}{2}$O_2(g)

$\Delta$H = –286 – (– 188) = –98 kJ mol^–1

Multiplying this equation by 2 we get the required equation

2H₂O_2(l) → 2H²O_(l) + O_2(g)

$\Delta$H = 2 (– 98) kJ/mol = – 196 kJ/mol

As, $\Delta$E = q + W

Also, W = – p$\Delta$V

As $\Delta$V = 0

So, W = 0

$\Rightarrow$ $\Delta$E = q

Now, q = 500 J

Thus, $\Delta$E = q = 500 J

CH₄ + $\frac{1}{2}$ O₂ $\to$ CH₃OH

Given that for this reaction, $\Delta$H_r = -ve

CH₄ + 2O₂ → CO₂ + 2H₂O,    $\Delta$H = x ....(1)

CH₃OH + $\frac{3}{2}$ O₂ → CO₂ + 2H₂O,    $\Delta$H = y ....(2)

Eqn. (1) – Eqn. (2)

CH₄ + $\frac{1}{2}$ O₂ $\to$ CH₃OH

$\Delta$H_r = x - y = -ve

$\therefore$ x < y

The metals having higher negative values of their electrode potential can displace metals having lower values from their salt solutions.

For the reaction;

2N₂O₅ $\to$ 4NO₂ + O₂

This is a first order reaction.

$\therefore$ rate = k [N₂O₅] ;

[N₂O₅] = $\frac{rate}{k}$

= $\frac{1.02\times {10}^{-4}}{3.4\times {10}^{-5}}$

= 3

From Arrhenius equation, k = Ae ^–E_a/RT

The activation energy of the reaction in the presence of enzyme is different from E_a obtained in laboratory.

At the end of 24 hrs. activity = 0.01 M

Half life = 6 hrs.

In 24 hrs. there are 24/6 = 4 half life.

Activity of substance after n half-life = $\frac{A}{2^n}$

$\therefore$ $\frac{A}{2^4}=0.01$

$\Rightarrow$ A = 0.16

hcp is a closed packed arrangement in which the unit cell is hexagonal and coordination number is 12.

Adsorption is the property of a substance to hold or to concentrate liquid or dissolved substances upon its surface. Solids adsorb greater amount of substance at lower temperature and in general adsorption decreases with increasing temperature.

Be – 1s²2s²;

B – 1s²2s²2p¹;

C – 1s²2s²2p²;

N – 1s²2s²2p³;

O – 1s²2s²2p⁴.

IP increases along the period. But IP of Be > B. Further IP of O < N because atoms with fully or partly filled orbitals are most stable and hence have high ionisation energy.

As in case of boron, 2p¹ electron have to be removed to get B⁺(1s² 2s²) from B(1s² 2s² 2p¹), while in case of Be (1s²2s²) electron have to be removed to get Be⁺ (1s²2s¹). p electron can be removed more easily than s electron so the energy required to remove electron will be less in case of boron.

As X and Y both are electronegative elements thus both attracts the electron density from H thus, electron density on H decreases and on X it increases.

Bond angle is maximum in NH₄⁺ tetrahedral molecule with bond angle 109°.

Compounds with same shape and same hybridisation are known as isostructural.

XeF₂, IF₂^- $\to$ both are sp³ hybridised linear molecules.

For strong $\pi$-bonding, p$\pi$ - p$\pi$ bonding should be strong. In case of P, due to larger size as compared to N-atom, p$\pi$ - p$\pi$ bonding is not so strong.

The most convenient method to protect the bottom of the ship made of iron is white tin plating preventing the build up of barnacles.

In lanthanides the basic character of hydroxides decreases as the ionic radius decreases.

La³⁺ ions are larger than Li³⁺ so it easily gives OH^– ion thus La(OH)₃ is more basic than Li(OH)₃.

In [Cr(SCN)₂(NH₃)₄]⁺ it shows linakge, geometrical and optical isomerism.

Coordination number of nickel in [Ni(C₂O₄)₃]^4$-$ is 6 because C₂O₄^2– is a bidentate ligand.

In Ni(CO)₄^-, Ni in zero oxidation state

Ni(0) = [Ar] 3d⁸4s²


It has no unpaired electron so acts as diamagnetic and has tetrahedral structure.

[Co(CO)₅NH₃]²⁺ : In this complex, Co-atom is attached with NH₃ through $\sigma$ bonding and with CO with dative $\pi$-bond.

Ionic conductance increases with increasing the number of ions, produced after decomposition.

In K₄[Fe(CN)₆], it produces maximum 5 ions thus show more conductivity.

K₄[Fe(CN)₆] ⇌ 4K⁺ + [Fe(CN)₆]^4–

Organic compounds exhibit the property of enantiomerism (optical isomerism) only when their molecules are chiral. Most chiral compounds have a chiral centre, which is an atom bonded to four different atoms or groups.
2,3-Dimethylpentane has one chiral C-atom and do not have any symmetric element.

Presence of electron withdrawing groups (– NO₂, –X, – CN) increase the acidity of phenols while the presence of electron releasing groups (– NH₂, – CH₃) decrease the acidity of phenols. This explains the following order of acidity.

p-nitrophenol > phenol > p-cresol.

In steam distillation of toluene, the pressure of toluene in vapour is less than pressure of barometer, because it is carried out when a solid or liquid is insoluble in water and is volatile with steam but the impurities are non-volatile.

Chlorination of n-butane takes place via free radical formation.

2C₄H₉Cl+2Na $\frac{\text{buildrel}Ethanol}{⟶}$ C₄H₉ - C₄H₉ + 2NaCl

In C₄H₉ - C₄H₉ i.e., (H₃C)₃C—C(CH₃)₃ there are primary hydrogens only.

Thus, only one chloro derivative is possible.

C₂H₅OH + 4I₂ + NaOH $\to$

CHI₃ + NaI + HCOONa + H₂O

Iodoform is a pale yellow solid which crystallises in hexagonal plates.

Tollen's reagent is used to detect aldehydic group. Tollen's reagent is an ammonical solution of silver nitrate. When aldehyde is added to Tollen's reagent, silver oxide is reduced to metallic silver which deposits as mirror.

RCHO + Ag₂O $\to$ RCOOH + 2Ag

Glucose reduces Fehling solution because glucose has free –CHO group which is readily oxidised.

The Hofmann bromamide reaction is as :

RCONH₂ + Br₂ + KOH $\to$ RNH₂

The mechanism is as follows :

Terylene is the condensation polymer of ethylene glycol and terephthalic acid.

Diastase is an enzyme. (The protein produced by living systems which acts as a biological catalyst). Enzymes are characterised by the name ending with -ase.

Due to resonance C — N bond in protein acquires double bond character and is smaller than usual C — N bond.

Starch is also know as amylum which occurs in all green plants. A molecule of starch (C₆H₁₀O₅)_n is built of a large number of $\alpha$-glucose ring joined through oxygen-atom.

Guanine G is linked to cytosine C by three hydrogen bonds and adenine A is linked with thymine T by two hydrogen bonds. It is abbreviate as : G ≡ C and A = T

Middle lamella is the first formed layer, present between the two adjacent cells. It is situated outside the primary cell wall. It is made up of calcium and magnesium pectate.

Phytochrome is a plant pigment that can detect the presence or absence of light and is involved in regulating many processes that are linked to day length (photoperiod), such as seed germination and initiation of flowering. It consists of a light-detecting portion, called a chromophore, linked to a small protein and exists in two interconvertible forms with different physical properties, particularly in the ability to bind to membranes.

Nitrogen fixation involves conversion of atmospheric nitrogen to ammonia. It is done with the help of nitrogenase enzyme which occurs inside thick walled heterocysts of the blue green algae. These provide suitable anaerobic environment for nitrogenase activity even in aerobic conditions.

Presence of more than one embryo inside the seed is called polyembryony. It is more common in gymnosperms than angiosperms. In angiosperms, it is generally present as an unusual feature in few cases like Citrus, mango etc.
In Citrus many embryos are formed from the structures outside the embryo (like nucellus). This is commonly called adventive polyembryony. In Citrus upto 10 nucellar embryos are formed.

Grass is a monocot plant. Primary sporogenous cell gives rise to microspore mother cells or pollen mother cells. Each MMC on reduction division gives rise to 4 microspores or pollens and this formation of microspores or pollens is called microsporogenesis. Karyokinesis is of successive type. The successive type of cytokinesis is common in monocots. Here both meiotic I and II nuclear divisions are followed by wall formation and it leads to isobilateral tetrad.

Vallisneria exhibits hydrophily and Salvia is pollinated by honey bee. In bottle brush pollination is carried out by birds.

Azolla and blue green algae- Anabaena form biofertilizer in rice fields.

Erucic acid is a monounsaturated omega-9 fatty acid, denoted 22 : 1 $\omega$-9. It is prevalent in rapeseed, wallflower seed, and mustard seed, making up 40 to 50 percent of their oils. Erucic acid is also known as cis13-docosenoic acid and the trans isomer is known as brassidic acid.

Inulin is a water soluble fructosan. It is a common reserve food in members of family compositae. Pectin is a mucopolysaccharides and occur in plant cell walls. At the time of fruit ripening wall of pectins hydrolyse to give constituent sugars.

Rate of salt absorption increases when temperature increases but to a certain limit as salt absorption is inhibited at higher temperature because certain enzymes are not active at higher temperature and carriers are not synthesised. These carriers are required for active transport of salts from outer space in inner space.

Rate of diffusion of ions and molecules increases at enhanced temperature due to their increased kinetic activity. Thus passive salt absorption will increase.

Phloem is involved in transport of organic food. According to mass flow hypothesis organic substances flow in the form of solution in sieve elements due to development of an osmotically generated pressure gradient.

Tropical plants have evolved C₄ cycle to overcome photorespiration.

Azolla performs nitrogen fixation through its colonies of Anabaena azollae, which have the capacity to fix atmospheric nitrogen.

Leghaemoglobin is a red respiratory pigment found in the root nodules of leguminous plant if Rhizobium is present. Soybean is a legume plant so it contains leghaemoglobin in its root nodules.

Cytochrome is an Fe containing porphyrin pigment. These are electron transferring proteins often regarded as enzymes.

A plasmid is a DNA molecule separate from the chromosomal DNA and capable of autonomous replication. In many cases, it is typically circular and double-stranded. It usually occurs in bacteria and is sometimes found in eukaryotic organisms. The size of plasmids varies from 1 to over 400 kilobase pairs (kbp). There may be one copy, for large plasmids, to hundreds of copies of the same plasmid in a single cell. The term plasmid was first introduced by the American molecular biologist Joshua Lederberg in 1952.

The enzyme that cuts DNA at specific places is known as a restriction endonuclease. Among the options provided, the one that fits this description is :

Option C : E.coli restriction endonuclease I

Restriction endonucleases, also known as restriction enzymes, are enzymes that cut DNA at or near specific recognition nucleotide sequences known as restriction sites. E.coli restriction endonuclease I is an example of such an enzyme, derived from the bacterium Escherichia coli (E. coli). These enzymes are essential tools in molecular biology, particularly in processes like gene cloning, as they allow for precise cutting and manipulation of DNA.

This combination allows to select only streptomycin resistant strains and non resistant strains would die on medium containing antibiotic streptomycin.

A mutant strain of T4 -bacteriophage, RII, fails to lyse the E.coli but when two strains R-IIX and R-IIY are mixed then they lyse the E.coli because both strains have different cistrons.

A taxonomic hierarchy is the sequence of arrangement of taxonomic categories in a descending order during the classification of an organism.

There are seven obligate categories - kingdom, division , class, order, family, genus and species. Species is the lowest category while kingdom is the highest category. The number of common characters is maximum in case of organisms placed in the lowest category.

Number of common characters decreases with the rise in category. Species are the smallest group of individuals which can be recognized by ordinary methods as groups and which are consistently and persistently different from other groups because their characters are less general.

The fungal hyphae secrete enzymes which convert insoluble complex food material in the substratum to the soluble ones. The hyphae wall of intracellular hyphae come in contact with the host protoplasm and obtain food by direct diffusion.

Cauliflower mosaic virus contain dsDNA. It is circular and shows semidiscontinuous type of replication.

Archaebacteria are the most ancient and halogenous group of bacteria and are called living fossils.

Using Gram stain, developed by Danish physician, Christian Gram in 1884, two kinds of bacteria were noted - those species of bacteria that are decolorized by alcohol are called gram negative and those that retain the stain are called gram positive.

This property of bacteria is related with the structure and compositional differences between the walls of gram positive and gram negative forms. In the cell wall of Gram +ve bacteria, both horizontal and vertical peptide linkages are present, due to which mesh is dense and hence the stain does not come out. Further outer layer of cell wall of Gram +ve bacteria is made of teichoic acid.

In the cell wall of Gram –ve bacteria, either horizontal or vertical peptide linkage are present, due to which mesh is loose and hence stain comes out. Further outermost layer of cell wall of Gram –ve bacteria is made of lipopolysaccharides.

Cyanobacteria are a group of photosynthetic bacteria that are capable of performing oxygenic photosynthesis. During this process, they use sunlight to convert carbon dioxide and water into organic compounds, while producing oxygen as a byproduct. This oxygen production is a key characteristic of oxygenic photosynthesis, and is what allows cyanobacteria to release oxygen into the atmosphere.

In addition to oxygenic photosynthesis, many species of cyanobacteria also have the ability to fix atmospheric nitrogen into a form that can be used by other organisms. This process is facilitated by an enzyme called nitrogenase, which converts nitrogen gas into ammonia, a compound that can be taken up by plants and other organisms. Cyanobacteria are the primary nitrogen fixers in many ecosystems, and they play a critical role in maintaining the balance of nutrients in many environments.

So, to summarize, cyanobacteria are oxygenic because they produce oxygen during photosynthesis, and they are capable of fixing nitrogen because they contain the enzyme nitrogenase. Option A, "Oxygenic with nitrogenase," is the correct answer.

Malvaceae is also known as cotton family or mallow family. The plants of this family are cosmopolitan in distribution, although more common in tropical (warm) regions. Gossypium (cotton) is an important genera of this family. Sunflower belongs to Family Compositae. Wheat belong to Family Poaceae. Catechu belongs to Family Mimosaceae.

Banana is a berry. It develops from monocarpellary or multicarpellary syncarpous ovary. Epicarp makes the rind of the fruit, mesocarp is fleshy and endocarp is thin and membranous. The edible portion of banana is endocarp and less developed mesocarp.