AIPMT 2011 Chemistry Question Paper with Answer and Solution

(A) Nitrifying bacteria are chemoautotrophic organisms that play a crucial role in the nitrogen cycle.
They perform the process of nitrification,which involves the oxidation of ammonia $(NH_3)$ into nitrites $(NO_2^-)$ and subsequently into nitrates $(NO_3^-)$.
Examples include $Nitrosomonas$ (which oxidizes ammonia to nitrites) and $Nitrobacter$ (which oxidizes nitrites to nitrates).
Therefore,the correct description is that they oxidize ammonia to nitrates.

(B) Leghaemoglobin is an oxygen scavenger found in the root nodules of leguminous plants.
The enzyme nitrogenase,which is responsible for biological nitrogen fixation,is highly sensitive to molecular oxygen $(O_2)$.
Leghaemoglobin binds to $O_2$ and maintains a low concentration of free oxygen in the nodules,thereby creating an anaerobic environment that protects the nitrogenase enzyme from oxidative damage and allows it to function efficiently.

(C) Essential mineral elements are those that are absolutely necessary for the growth,development,and reproduction of plants.
According to the criteria established by Arnon and Stout,an element is considered essential if it is required for the plant's life cycle,its deficiency cannot be replaced by another element,and it is directly involved in plant metabolism.
$Iron$ $(Fe)$,$Manganese$ $(Mn)$,and $Phosphorus$ $(P)$ are well-known essential macronutrients or micronutrients.
$Cadmium$ $(Cd)$ is a toxic heavy metal and is not required for any physiological process in plants; in fact,it is harmful even in small concentrations.
Therefore,$Cadmium$ is not an essential mineral element.

(B) $C_4$ plants exhibit a special type of leaf anatomy known as Kranz anatomy.
In this anatomy,the bundle sheath cells are arranged in several layers around the vascular bundles.
These cells are characterized by having thick walls that are impervious to gaseous exchange,no intercellular spaces,and a large number of chloroplasts to facilitate the $C_4$ cycle (Calvin cycle occurs in these cells).

(A) Pancreatic juice contains inactive enzymes known as proenzymes or zymogens,which include $trypsinogen$,$chymotrypsinogen$,and $procarboxypeptidase$.
$Trypsinogen$ is secreted by the pancreas into the duodenum.
In the duodenum,$trypsinogen$ is activated into its active form,$trypsin$,by an enzyme called $enterokinase$ (also known as $enteropeptidase$),which is secreted by the intestinal mucosa.
Therefore,$trypsinogen$ is a constituent of pancreatic juice,while $trypsin$ is the activated form,and $enterokinase$ is an intestinal enzyme.

(B) In the given diagram of the alveolus:
$A$ represents the Alveolar cavity,which is the primary site for the exchange of respiratory gases ($O_2$ and $CO_2$) between the air in the lungs and the blood.
$B$ represents a Red blood cell (erythrocyte),which is primarily responsible for the transport of $O_2$ and $CO_2$.
$C$ represents the pulmonary artery/arteriole bringing deoxygenated blood.
$D$ represents the capillary wall,which along with the alveolar wall,forms the respiratory membrane for gas exchange.
Comparing the options,option $B$ correctly identifies $A$ as the alveolar cavity and its function as the main site of gas exchange.

(B) The Bundle of His (also known as the atrioventricular bundle) is a collection of heart muscle cells specialized for electrical conduction.
It transmits the electrical impulses from the atrioventricular node $(AVN)$ to the ventricles of the heart.
Therefore,it is an essential component of the human heart's conducting system.

(D) The $24$ hour (diurnal) rhythm of our body,such as the sleep-wake cycle,is regulated by the hormone Melatonin.
Melatonin is secreted by the pineal gland located on the dorsal side of the forebrain.
It plays a very important role in the regulation of a $24$-hour (diurnal) rhythm of our body.
It also helps in maintaining the normal rhythms of sleep-wake cycle,body temperature,and influences metabolism,pigmentation,the menstrual cycle,as well as our defense capability.

(A) The plasma membrane of eubacteria is structurally and functionally similar to that of eukaryotic cells. Both consist of a phospholipid bilayer with embedded proteins.
In contrast,eubacteria lack a membrane-bound nucleus (they have a nucleoid instead).
While both have ribosomes,eukaryotic ribosomes are $80S$ (composed of $60S$ and $40S$ subunits),whereas eubacterial ribosomes are $70S$ (composed of $50S$ and $30S$ subunits).
The cell wall of eubacteria is made of peptidoglycan,which is absent in eukaryotic cells.

(D) In the plant kingdom,the life cycle involves an alternation of generations between a haploid gametophyte and a diploid sporophyte.
In Bryophytes (e.g.,$Polytrichum$,$Marchantia$) and Pteridophytes (e.g.,$Adiantum$),the gametophyte is an independent,free-living generation.
In Gymnosperms (e.g.,$Pinus$) and Angiosperms,the gametophyte is highly reduced and depends on the sporophyte for nutrition and support.
Therefore,in $Pinus$,the gametophyte is not an independent,free-living generation.

(A) In angiosperms,the megaspore mother cell undergoes meiosis to produce four haploid megaspores. Out of these four,three degenerate,and only one remains functional. This functional megaspore undergoes three successive mitotic divisions to form the female gametophyte,which is known as the embryo sac.

(B) In plant anatomy,the ground tissue system consists of all tissues except the epidermis and the vascular bundles. It includes simple tissues such as parenchyma,collenchyma,and sclerenchyma. In primary stems and roots,the ground tissue consists of the cortex,pericycle,pith,and medullary rays.

(C) During the metaphase stage of cell division,the chromosomes align at the equatorial plate.
Each chromosome consists of two sister chromatids held together at the centromere.
On the surface of the centromere,there are small disc-shaped structures called kinetochores.
The spindle fibres attach to these kinetochores to facilitate the movement and segregation of chromosomes during anaphase.
Therefore,the correct answer is $C$ (kinetochores).

(B) The voltage across the Zener diode remains constant at $15 \, V$ when it is in the breakdown region.
The current through the load resistor $(R_L = 1 \, k\Omega)$ is:
$i_0 = \frac{V_z}{R_L} = \frac{15 \, V}{1000 \, \Omega} = 0.015 \, A = 15 \, mA$
The total current $(i)$ supplied by the source is determined by the voltage drop across the series resistor $(R = 250 \, \Omega)$:
$i = \frac{V_{in} - V_z}{R} = \frac{20 \, V - 15 \, V}{250 \, \Omega} = \frac{5 \, V}{250 \, \Omega} = 0.02 \, A = 20 \, mA$
Applying Kirchhoff's Current Law at the junction,the current through the Zener diode $(i_z)$ is:
$i_z = i - i_0 = 20 \, mA - 15 \, mA = 5 \, mA$

(B) Given:
$(1) \ Cu^{2+}_{(aq)} + e^- \to Cu^{+}_{(aq)}$; $\Delta G_1 = -1 \times 0.15 \times F$
$(2) \ Cu^{+}_{(aq)} + e^- \to Cu_{(s)}$; $\Delta G_2 = -1 \times 0.50 \times F$
We need to find $E^\circ$ for the reaction:
$(3) \ Cu^{2+}_{(aq)} + 2e^- \to Cu_{(s)}$; $\Delta G_3 = -2 \times F \times E^\circ_3$
Since reaction $(3) = (1) + (2)$,we have:
$\Delta G_3 = \Delta G_1 + \Delta G_2$
$-2 \times F \times E^\circ_3 = (-1 \times 0.15 \times F) + (-1 \times 0.50 \times F)$
$-2 \times E^\circ_3 = -0.15 - 0.50$
$-2 \times E^\circ_3 = -0.65$
$E^\circ_3 = \frac{0.65}{2} = 0.325 \ V$

(C) Let $N_P(t)$ and $N_Q(t)$ be the number of nuclei of $P$ and $Q$ at time $t$.
Given: $N_P(0) = 4N_0$,$T_{1/2, P} = 1$ min,$N_Q(0) = N_0$,$T_{1/2, Q} = 2$ min.
The decay equations are: $N_P(t) = 4N_0 \cdot (1/2)^{t/1}$ and $N_Q(t) = N_0 \cdot (1/2)^{t/2}$.
We need to find $t$ when $N_P(t) = N_Q(t)$:
$4N_0 \cdot (1/2)^t = N_0 \cdot (1/2)^{t/2}$
$4 = (1/2)^{t/2} / (1/2)^t = (1/2)^{-t/2} = 2^{t/2}$
$2^2 = 2^{t/2} \implies t/2 = 2 \implies t = 4$ minutes.
At $t = 4$ min:
$N_P(4) = 4N_0 \cdot (1/2)^4 = 4N_0 / 16 = N_0/4$.
$N_Q(4) = N_0 \cdot (1/2)^{4/2} = N_0 / 4$.
The number of nuclei of $R$ formed is the total number of decayed nuclei of $P$ and $Q$:
$N_R = (N_P(0) - N_P(4)) + (N_Q(0) - N_Q(4))$
$N_R = (4N_0 - N_0/4) + (N_0 - N_0/4) = 15N_0/4 + 3N_0/4 = 18N_0/4 = 9N_0/2$.

(C) The rate of reaction $(ROR)$ is defined as:
$ROR = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$
Given the rate expressions:
$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$
$\frac{d[NO_2]}{dt} = k'[N_2O_5]$
$\frac{d[O_2]}{dt} = k''[N_2O_5]$
Substituting these into the $ROR$ expression:
$\frac{1}{2} k[N_2O_5] = \frac{1}{4} k'[N_2O_5] = k''[N_2O_5]$
Comparing $k$ and $k'$:
$\frac{k}{2} = \frac{k'}{4} \implies k' = 2k$
Comparing $k$ and $k''$:
$\frac{k}{2} = k'' \implies k'' = \frac{k}{2}$

(A) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is $\vec E \times \vec B$.
Given that the wave propagates along the $+z-$ axis (unit vector $\hat k$),we must have $\hat E \times \hat B = \hat k$.
Testing option $A$: $\hat i \times \hat j = \hat k$. This satisfies the condition.
Therefore,the electric field is along the $x-$ axis and the magnetic field is along the $y-$ axis.

(D) The average velocity $(V_{av})$ of a gaseous molecule is given by the formula $V_{av} = \sqrt{\frac{8RT}{\pi M}}$.
From this expression,it is clear that $V_{av} \propto \sqrt{T}$.
If the temperature is doubled,the new temperature $T' = 2T$.
The new average velocity $V_{av}'$ will be proportional to $\sqrt{2T}$.
Therefore,$V_{av}' = \sqrt{2} \times V_{av}$.
Since $\sqrt{2} \approx 1.414$,the average velocity increases by a factor of $1.4$.

(D) According to the Freundlich adsorption isotherm,the relationship between the amount of gas adsorbed $(x)$ and the mass of the adsorbent $(m)$ at a constant temperature is given by $\frac{x}{m} = k p^{1/n}$.
Here,$k$ and $n$ are constants,$x$ is the amount of gas adsorbed,and $m$ is the amount of solid adsorbent.
Adsorption isotherms and isobars describe the variation of $\frac{x}{m}$ with pressure at constant temperature or with temperature at constant pressure,respectively.
The relation $\frac{x}{m} = P \times T$ is incorrect because the adsorption capacity $\frac{x}{m}$ does not follow a simple product relationship with pressure and temperature.

(B) Fat-soluble vitamins are those that are similar to oil and do not dissolve in water.
Vitamin $A, D, E,$ and $K$ are the fat-soluble vitamins.
Vitamin $B$ complex is not a fat-soluble vitamin because it is water-soluble.
It consists of eight vitamins: $B_1, B_2, B_3, B_5, B_6, B_7, B_9,$ and $B_{12}$.
Since it is water-soluble,our body does not store it,and therefore,it must be consumed daily.

(D) The number of molecules is directly proportional to the number of moles $(n = \text{mass} / \text{molar mass})$.
For $A$: $n(SO_2) = 64 \ g / 64 \ g \ mol^{-1} = 1 \ mol$.
For $B$: $n(CO_2) = 44 \ g / 44 \ g \ mol^{-1} = 1 \ mol$.
For $C$: $n(O_3) = 48 \ g / 48 \ g \ mol^{-1} = 1 \ mol$.
For $D$: $n(H_2) = 8 \ g / 2 \ g \ mol^{-1} = 4 \ mol$.
Since $H_2$ has the highest number of moles $(4 \ mol)$,it contains the maximum number of molecules.

(A) Given the angular position: $\theta(t) = 2t^3 - 6t^2$.
First,find the angular velocity $\omega$ by differentiating $\theta$ with respect to time $t$: $\omega = \frac{d\theta}{dt} = 6t^2 - 12t$.
Next,find the angular acceleration $\alpha$ by differentiating $\omega$ with respect to time $t$: $\alpha = \frac{d\omega}{dt} = 12t - 12$.
The torque $\tau$ is related to angular acceleration by the equation $\tau = I\alpha$,where $I$ is the moment of inertia.
For the torque to be zero,the angular acceleration $\alpha$ must be zero (assuming $I \neq 0$).
Setting $\alpha = 0$: $12t - 12 = 0$.
Solving for $t$: $12t = 12$,which gives $t = 1 \text{ s}$.

(C) Lewis base is a substance that can donate a lone pair of electrons.
$H_2O$,$NH_3$,and $OH^{-}$ all possess lone pairs of electrons on their central atoms,allowing them to act as Lewis bases.
$BF_3$ has an incomplete octet (only $6$ electrons around the central $B$ atom) and acts as an electron-pair acceptor,making it a Lewis acid.
Therefore,$BF_3$ is the least likely to behave as a Lewis base.

(B) Let $N$ be the amount of radioactive isotope $X$ remaining and $N_0$ be the initial amount of $X$.
The amount of stable element $Y$ formed is $N_0 - N$.
Given the ratio of $X$ to $Y$ is $1:15$,we have $\frac{N}{N_0 - N} = \frac{1}{15}$.
Adding $1$ to both sides or rearranging gives $\frac{N_0 - N}{N} = 15$,so $\frac{N_0}{N} - 1 = 15$,which implies $\frac{N_0}{N} = 16$.
We know the radioactive decay law is $\frac{N}{N_0} = (\frac{1}{2})^{t/T_{1/2}}$,where $T_{1/2} = 50$ years.
Thus,$\frac{N_0}{N} = 2^{t/T_{1/2}} = 16$.
Since $16 = 2^4$,we have $\frac{t}{T_{1/2}} = 4$.
Therefore,$t = 4 \times 50 = 200$ years.

(B) The voltage across the Zener diode is constant at $15 \, V$ because it is in the breakdown region.
First,calculate the current through the load resistor $(R_L = 1 \, k\Omega)$:
$I_L = \frac{V_Z}{R_L} = \frac{15 \, V}{1 \times 10^3 \, \Omega} = 15 \, mA$.
Next,calculate the total current $(I)$ flowing through the series resistor $(R_S = 250 \, \Omega)$:
The voltage drop across the series resistor is $V_S = V_{in} - V_Z = 20 \, V - 15 \, V = 5 \, V$.
$I = \frac{V_S}{R_S} = \frac{5 \, V}{250 \, \Omega} = 0.02 \, A = 20 \, mA$.
Finally,apply Kirchhoff's Current Law at the node:
$I = I_Z + I_L$
$I_Z = I - I_L = 20 \, mA - 15 \, mA = 5 \, mA$.
Therefore,the current through the Zener diode is $5 \, mA$.

(B) The energy of a radiation is given by the formula $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.
Since $h$ and $c$ are constants,$E \propto \frac{1}{\lambda}$.
Therefore,$\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.
Given $E_1 = 25 \ eV$ and $E_2 = 50 \ eV$,we have:
$\frac{25}{50} = \frac{\lambda_2}{\lambda_1}$
$\frac{1}{2} = \frac{\lambda_2}{\lambda_1}$
$\lambda_1 = 2\lambda_2$.

(B) In the Davisson and Germer experiment,the electrons are accelerated by an electric field created by the potential difference $V$ applied between the filament and the anode.
According to the work-energy theorem,the kinetic energy $K$ gained by an electron of charge $e$ is given by $K = eV = \frac{1}{2}mv^2$.
From this relation,the velocity $v$ is given by $v = \sqrt{\frac{2eV}{m}}$.
Since $v \propto \sqrt{V}$,the velocity of the electrons increases as the potential difference $V$ between the anode and the filament is increased.

(D) In a nuclear fusion reaction,two lighter nuclei combine to form a heavier nucleus.
Since all nuclei are positively charged,there exists a strong electrostatic (Coulomb) repulsion between them.
To overcome this repulsive force and bring the nuclei close enough for the strong nuclear force to take over,they must possess very high kinetic energy.
This high kinetic energy is achieved at extremely high temperatures (on the order of $10^7$ to $10^8 \ K$),which is why fusion reactions are also known as thermonuclear reactions.

(B) $1$. The input voltage is $V_{in} = 20\,V$ and the series resistance is $R_s = 250\,\Omega$.
$2$. The Zener diode is in parallel with the load resistor $R_L = 1\,k\Omega = 1000\,\Omega$. Since the Zener diode is in breakdown,the voltage across the load resistor is $V_L = V_Z = 15\,V$.
$3$. The current through the load resistor is $I_L = \frac{V_L}{R_L} = \frac{15\,V}{1000\,\Omega} = 0.015\,A = 15\,mA$.
$4$. The total current supplied by the source through the series resistor $R_s$ is $I_s = \frac{V_{in} - V_Z}{R_s} = \frac{20\,V - 15\,V}{250\,\Omega} = \frac{5\,V}{250\,\Omega} = 0.02\,A = 20\,mA$.
$5$. Applying Kirchhoff's current law at the node,the current through the Zener diode is $I_Z = I_s - I_L = 20\,mA - 15\,mA = 5\,mA$.

(A) The input voltage is $V_{in} = 20\, V$. The series resistance is $R_s = 250\, \Omega$. The Zener breakdown voltage is $V_z = 15\, V$. The load resistance is $R_L = 1\, k\Omega = 1000\, \Omega$.
$1$. Calculate the total current $(I)$ flowing through the series resistor $R_s$:
$I = \frac{V_{in} - V_z}{R_s} = \frac{20\, V - 15\, V}{250\, \Omega} = \frac{5\, V}{250\, \Omega} = 0.02\, A = 20\, mA$.
$2$. Calculate the current $(I_L)$ flowing through the load resistor $R_L$:
$I_L = \frac{V_z}{R_L} = \frac{15\, V}{1000\, \Omega} = 0.015\, A = 15\, mA$.
$3$. The current through the Zener diode $(I_z)$ is given by:
$I_z = I - I_L = 20\, mA - 15\, mA = 5\, mA$.

(C) The given complex is of the type $[Mabcd]$,where $M$ is the central metal atom $(Pt)$ and $a, b, c, d$ are $4$ different ligands $(py, NH_3, Br, Cl)$ attached to it.
Square planar complexes of the type $[Mabcd]$ exhibit $3$ geometrical isomers.
These isomers are formed by fixing one ligand and changing the positions of the other three ligands relative to it,as shown in the provided image.

(A) Given the angular position $\theta(t) = 2t^3 - 6t^2$.
First,find the angular velocity $\omega$ by differentiating $\theta(t)$ with respect to time $t$:
$\omega = \frac{d\theta}{dt} = \frac{d}{dt}(2t^3 - 6t^2) = 6t^2 - 12t$.
Next,find the angular acceleration $\alpha$ by differentiating $\omega$ with respect to time $t$:
$\alpha = \frac{d\omega}{dt} = \frac{d}{dt}(6t^2 - 12t) = 12t - 12$.
The torque $\tau$ is given by $\tau = I\alpha$,where $I$ is the moment of inertia.
For the torque to be zero,$\tau = 0$,which implies $I\alpha = 0$. Since $I \neq 0$,we must have $\alpha = 0$.
Setting $\alpha = 12t - 12 = 0$,we get $12t = 12$,which results in $t = 1 \text{ sec}$.

(B) Given:
$(1) \ Cu^{2+}_{(aq)} + e^- \to Cu^{+}_{(aq)} \quad \Delta G_1^{\circ} = -1 \times 0.15 \times F$
$(2) \ Cu^{+}_{(aq)} + e^- \to Cu_{(s)} \quad \Delta G_2^{\circ} = -1 \times 0.50 \times F$
Adding $(1)$ and $(2)$ gives the overall reaction:
$Cu^{2+}_{(aq)} + 2e^- \to Cu_{(s)} \quad \Delta G_3^{\circ} = -2 \times F \times E^{\circ}_{Cu^{2+}/Cu}$
Since $\Delta G_3^{\circ} = \Delta G_1^{\circ} + \Delta G_2^{\circ}$,we have:
$-2 \times F \times E^{\circ}_{Cu^{2+}/Cu} = (-1 \times 0.15 \times F) + (-1 \times 0.50 \times F)$
$-2 \times E^{\circ}_{Cu^{2+}/Cu} = -0.65$
$E^{\circ}_{Cu^{2+}/Cu} = \frac{0.65}{2} = 0.325 \, V$.

(B) The total number of orbitals in a given energy level $n$ is given by the formula $n^2$.
For the fourth energy level,$n = 4$.
Therefore,the total number of orbitals = $n^2 = 4^2 = 16$.
Thus,the correct option is $B$.

(B) The reaction of nitrobenzene with $Sn + HCl$ is a standard reduction reaction.
Nitrobenzene $(C_6H_5NO_2)$ undergoes reduction in the presence of a metal-acid catalyst like $Sn + HCl$ to form aniline $(C_6H_5NH_2)$.
The chemical equation is:
$C_6H_5NO_2 + 6[H] \xrightarrow{Sn+HCl} C_6H_5NH_2 + 2H_2O$
Therefore,the product obtained is aniline.

(A) The box is dropped on the moving belt,so its initial velocity relative to the belt is $u_{rel} = 2\, m/s$.
The frictional force acting on the box provides an acceleration $a = \mu g = 0.5 \times 10 = 5\, m/s^2$.
This acceleration acts in the direction of the belt's motion to bring the box to rest relative to the belt.
Using the kinematic equation $v_{rel}^2 = u_{rel}^2 - 2 a s$,where $v_{rel} = 0$ (when the box stops relative to the belt):
$0 = (2)^2 - 2(5)s$
$10s = 4$
$s = 0.4\, m$.

(A) According to the principle of conservation of linear momentum,the total momentum before the collision equals the total momentum after the collision.
Initial momentum of mass $m$ is $\vec{p}_1 = m v \hat{i}$.
Initial momentum of mass $3m$ is $\vec{p}_2 = 3m (2v \hat{j}) = 6m v \hat{j}$.
Total initial momentum $\vec{p}_{total} = m v \hat{i} + 6m v \hat{j}$.
Let the final velocity of the combined mass $(m + 3m = 4m)$ be $\vec{v}_c$.
By conservation of momentum: $4m \vec{v}_c = m v \hat{i} + 6m v \hat{j}$.
Dividing by $4m$,we get $\vec{v}_c = \frac{m v}{4m} \hat{i} + \frac{6m v}{4m} \hat{j}$.
Therefore,$\vec{v}_c = \frac{1}{4} v \hat{i} + \frac{3}{2} v \hat{j}$.

(C) Vitamins are classified into two groups based on their solubility: fat-soluble and water-soluble.
Fat-soluble vitamins include Vitamin $A$,$D$,$E$,and $K$.
Water-soluble vitamins include Vitamin $B$ complex and Vitamin $C$.
Therefore,Vitamin $C$ is not a fat-soluble vitamin; it is water-soluble.

(B) Let $N$ be the amount of radioactive isotope $X$ remaining and $N_0$ be the initial amount.
The amount of stable element $Y$ formed is $N_0 - N$.
Given the ratio $\frac{N}{N_0 - N} = \frac{1}{15}$.
Adding $1$ to both sides: $\frac{N}{N_0 - N} + 1 = \frac{1}{15} + 1 \implies \frac{N + N_0 - N}{N_0 - N} = \frac{16}{15} \implies \frac{N_0}{N_0 - N} = \frac{16}{15}$.
Taking the reciprocal: $\frac{N}{N_0} = \frac{1}{16}$.
We know the radioactive decay law: $\frac{N}{N_0} = (\frac{1}{2})^n$,where $n$ is the number of half-lives.
$\frac{1}{16} = (\frac{1}{2})^4$,so $n = 4$.
The age of the rock $t = n \times t_{1/2} = 4 \times 50 \, years = 200 \, years$.

(A) The direction of propagation of an electromagnetic wave is given by the direction of the vector $\vec{E} \times \vec{B}$.
Given that the wave propagates along the $+z-$ axis,we must have $\vec{E} \times \vec{B} = \hat{k}$.
Since the electric field $\vec{E}$ and magnetic field $\vec{B}$ must be perpendicular to each other and to the direction of propagation,they must lie in the $xy-$ plane.
Checking the options:
For option $A$: $\vec{E} = E_0 \hat{i}$ and $\vec{B} = B_0 \hat{j}$.
Then $\vec{E} \times \vec{B} = (E_0 \hat{i}) \times (B_0 \hat{j}) = E_0 B_0 (\hat{i} \times \hat{j}) = E_0 B_0 \hat{k}$.
This matches the direction of propagation $+z$.
Therefore,option $A$ is correct.

(B) The fundamental frequency $n$ of a stretched wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Since the wires are identical,$n \propto \sqrt{T}$.
Differentiating this relation,we get $\frac{dn}{n} = \frac{1}{2} \frac{dT}{T}$.
For small changes,we can write $\frac{\Delta n}{n} = \frac{1}{2} \frac{\Delta T}{T}$.
Here,the beat frequency is $\Delta n = 6\, Hz$ and the original frequency is $n = 600\, Hz$.
Substituting these values,we get $\frac{\Delta T}{T} = 2 \left( \frac{\Delta n}{n} \right)$.
$\frac{\Delta T}{T} = 2 \left( \frac{6}{600} \right) = 2 \times 0.01 = 0.02$.

(D) The rays are converging towards a point $15 \, cm$ behind the lens,so the image distance $v = +15 \, cm$.
If the lens is removed,the rays meet $5 \, cm$ closer to the lens,meaning the original convergence point was at $15 - 5 = 10 \, cm$ from the lens position. Since the rays were converging towards this point,the object distance $u = +10 \, cm$.
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Substituting the values: $\frac{1}{15} - \frac{1}{10} = \frac{1}{f}$
$\frac{2 - 3}{30} = \frac{1}{f}$
$-\frac{1}{30} = \frac{1}{f}$
$f = -30 \, cm$

(D) For the first line of the Lyman series of hydrogen $(n_1=1, n_2=2)$:
$\frac{1}{\lambda_1} = R(1)^2 \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R \left(1 - \frac{1}{4}\right) = \frac{3R}{4}$
For the second line of the Balmer series of a hydrogen-like ion $(n_1=2, n_2=4)$:
$\frac{1}{\lambda_2} = R Z^2 \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = R Z^2 \left(\frac{1}{4} - \frac{1}{16}\right) = R Z^2 \left(\frac{3}{16}\right)$
Given that $\lambda_1 = \lambda_2$,we have $\frac{1}{\lambda_1} = \frac{1}{\lambda_2}$:
$\frac{3R}{4} = \frac{3RZ^2}{16}$
Dividing both sides by $\frac{3R}{4}$:
$1 = \frac{Z^2}{4}$
$Z^2 = 4$
$Z = 2$

(C) The number of molecules is calculated as: $\text{Number of molecules} = \text{Moles} \times N_A$.
For $A$: $\text{Moles of } CO_2 = \frac{44 \ g}{44 \ g/mol} = 1 \ mol$. Number of molecules = $1 \times N_A$.
For $B$: $\text{Moles of } O_3 = \frac{48 \ g}{48 \ g/mol} = 1 \ mol$. Number of molecules = $1 \times N_A$.
For $C$: $\text{Moles of } H_2 = \frac{8 \ g}{2 \ g/mol} = 4 \ mol$. Number of molecules = $4 \times N_A$.
For $D$: $\text{Moles of } SO_2 = \frac{64 \ g}{64 \ g/mol} = 1 \ mol$. Number of molecules = $1 \times N_A$.
Comparing the values,$8 \ g \ H_2$ contains the maximum number of molecules $(4 \times N_A)$.

(A) All tissues except epidermis and vascular bundles constitute the ground tissue or fundamental tissue. It consists of simple tissues such as parenchyma,collenchyma,and sclerenchyma. Ground tissue includes cortex,pericycle,and medullary rays. In leaves,the ground tissue consists of mesophyll.

(B) The endomembrane system consists of organelles whose functions are coordinated. These include the endoplasmic reticulum $(ER)$,Golgi complex,lysosomes,and vacuoles.
Peroxisomes are not part of the endomembrane system because their functions are not coordinated with the aforementioned components.

(A) Leghaemoglobin is an oxygen scavenger found in the root nodules of leguminous plants.
It protects the nitrogen-fixing enzyme,nitrogenase,from the damaging effects of oxygen,as nitrogenase is highly sensitive to molecular oxygen.

(A) Elements that are structural components of the cell,such as $Calcium$,are not remobilized from older tissues to younger tissues.
Because these elements are immobile,the deficiency symptoms of $Calcium$ appear first in the young tissues.
In contrast,elements like $Potassium$,$Sulphur$,and $Phosphorus$ are highly mobile and are readily transported from older tissues to younger,developing tissues.

(D) The epiglottis is a thin,leaf-like flap of elastic cartilage covered with a mucous membrane,located at the entrance of the larynx (voice box).
Its primary function is to act as a lid that closes the glottis during swallowing,preventing food or liquid from entering the trachea (windpipe).
When food is swallowed,the epiglottis moves downward to cover the opening of the larynx.
If the epiglottis fails to close properly,food particles may enter the larynx,triggering a cough reflex to expel the foreign material and protect the airway.