AIPMT 2010 Chemistry Question Paper with Answer and Solution

(A) The protein coat that surrounds the genetic material ($DNA$ or $RNA$) of a virus is called the $Capsid$.
This coat is composed of smaller protein subunits known as $capsomeres$.
$Virion$ refers to the complete,infectious virus particle.
$Nucleoprotein$ refers to the complex of nucleic acid and protein.
$Core$ refers to the central part of the virus containing the genome.

(A) Triploblastic animals are those that possess three germ layers: ectoderm,mesoderm,and endoderm during embryonic development.
$1$. Flatworms (Phylum $Platyhelminthes$) are the first group of animals to exhibit a triploblastic body organization.
$2$. Sponges (Phylum $Porifera$) are diploblastic or lack true tissue organization.
$3$. Ctenophores (Phylum $Ctenophora$) are diploblastic.
$4$. Corals (Phylum $Cnidaria$) are diploblastic.
Therefore,the correct answer is Flatworms.

(C) Heartwood,also known as $duramen$,is the central,darker,and harder part of the secondary xylem.
It consists of dead elements with highly lignified walls,which makes it non-conducting.
In contrast,sapwood,or $alburnum$,is the peripheral,lighter-colored part of the secondary xylem that is involved in the conduction of water and minerals.
Heartwood is resistant to microbial and insect attacks due to the deposition of organic compounds like tannins,resins,oils,gums,and aromatic substances in the vessels,which block them (tyloses).
Therefore,the primary difference is that heartwood consists of dead and non-conducting elements.

(B) The fallopian tube (oviduct) consists of three main parts: Infundibulum,Ampulla,and Isthmus.
$1$. The $Infundibulum$ is the funnel-shaped part closest to the ovary,which possesses finger-like projections called fimbriae that help in the collection of the ovum after ovulation.
$2$. The $Ampulla$ is the wider part of the oviduct where fertilization usually takes place.
$3$. The $Isthmus$ is the last and narrow part of the oviduct that joins the uterus.
Therefore,the part closest to the ovary is the $Infundibulum$.

(D) Goblet cells are specialized epithelial cells found in the mucosal lining of the digestive tract.
They are responsible for the secretion of mucus.
Mucus acts as a lubricant,which protects the intestinal wall from mechanical injury and chemical damage (like acid).
It also facilitates the smooth movement of food (chyme) through the intestinal lumen.
Therefore,if goblet cells are non-functional,the lubrication of the intestinal tract is lost,leading to difficulty in the movement of food.

(C) An electrophile is an electron-deficient species that can accept an electron pair.
$Cl^{\oplus}$,$BH_3$,and $NO_2^{\oplus}$ are electron-deficient and act as electrophiles.
In $H_3O^{\oplus}$,the oxygen atom has a complete octet (eight electrons in its valence shell) and a positive charge,but it cannot accept further electrons. Therefore,it is not electrophilic in nature.

(B) Strong electrolytes are completely ionised at all concentrations.
On increasing dilution,the number of ions remains the same because they are already fully dissociated.
However,the inter-ionic attractions decrease,which leads to an increase in the ionic mobility of the ions.
Therefore,the equivalent conductance increases primarily due to the increase in ionic mobility.

(B) In an $LCR$ series circuit,the voltage across the inductor is $V_L$ (reading of $V_1$) and across the capacitor is $V_C$ (reading of $V_2$).
Given $V_1 = V_L = 300 \, V$ and $V_2 = V_C = 300 \, V$.
Since $V_L = V_C$,the circuit is in a state of electrical resonance.
At resonance,the net impedance $Z$ of the circuit is equal to the resistance $R$,so $Z = R = 100 \, \Omega$.
The total voltage applied is $V = 220 \, V$.
The current $I$ in the circuit is given by $I = \frac{V}{Z} = \frac{220 \, V}{100 \, \Omega} = 2.2 \, A$.
This is the reading of the ammeter $A$.
At resonance,the total voltage $V$ is equal to the voltage across the resistor $V_R$ (reading of $V_3$).
Therefore,$V_3 = V_R = V = 220 \, V$.
Thus,the reading of $V_3$ is $220 \, V$ and the reading of $A$ is $2.2 \, A$.

(A) For $H_4P_2O_5$: $4(+1) + 2x + 5(-2) = 0$ $\Rightarrow 2x = +6$ $\Rightarrow x = +3$.
For $H_4P_2O_6$: $4(+1) + 2x + 6(-2) = 0$ $\Rightarrow 2x = +8$ $\Rightarrow x = +4$.
For $H_4P_2O_7$: $4(+1) + 2x + 7(-2) = 0$ $\Rightarrow 2x = +10$ $\Rightarrow x = +5$.
Thus,the oxidation states are $+3, +4, +5$.

(B) complex ion absorbs visible light if it has unpaired electrons,which allows for $d-d$ transitions.
In $[Ni(CN)_4]^{2-}$,$Ni$ is in the $+2$ oxidation state ($3d^8$ configuration).
Since $CN^-$ is a strong field ligand,it causes pairing of electrons,resulting in a square planar geometry with zero unpaired electrons.
Therefore,it does not undergo $d-d$ transitions and is not expected to absorb visible light.

(C) Valium is a tranquilizer drug.
It is mainly used to treat anxiety,insomnia,and symptoms of acute alcohol withdrawal.
It acts by enhancing the effects of gamma-aminobutyric acid $(GABA)$ in the brain.

(D) The structure of $P_4O_{10}$ consists of a tetrahedral arrangement of four phosphorus atoms.
Each phosphorus atom is bonded to three other phosphorus atoms through oxygen bridges ($P-O-P$ bonds).
There are $6$ such bridging oxygen atoms in the structure.
Additionally,each phosphorus atom is bonded to one terminal oxygen atom via a double bond $(P=O)$,totaling $4$ terminal oxygen atoms.
Thus,the total number of bridging oxygen atoms is $6$.

(B) complex ion absorbs visible light if it contains unpaired electrons,which allow for $d-d$ transitions.
$1$. $[Ni(H_2O)_6]^{2+}$: $Ni^{2+}$ is $3d^8$,which has $2$ unpaired electrons.
$2$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons to form a $dsp^2$ square planar complex with $0$ unpaired electrons.
$3$. $[Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ is $3d^3$,which has $3$ unpaired electrons.
$4$. $[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$,which has $4$ unpaired electrons.
Since $[Ni(CN)_4]^{2-}$ has no unpaired electrons,it does not undergo $d-d$ transitions and is not expected to absorb visible light.

(D) The focal length $f$ of a lens depends on the refractive index of the material and the radii of curvature of its surfaces. Covering a part of the lens does not change these parameters,so the focal length remains $f$.
Intensity $I$ is proportional to the area of the aperture $A = \pi (d/2)^2 = \pi d^2/4$.
When the central part with diameter $d/2$ is covered,the radius of the covered part is $d/4$. The area of the covered part is $A_{covered} = \pi (d/4)^2 = \pi d^2/16 = A/4$.
The remaining area of the lens is $A' = A - A/4 = 3A/4$.
Since intensity $I$ is proportional to the area,the new intensity $I'$ is $I' = I \times (A'/A) = I \times (3/4) = 3I/4$.

(B) In an $LCR$ series circuit,the voltage across the inductor $V_L$ and the capacitor $V_C$ are in opposite phases.
Given $V_1 = V_L = 300 \, V$ and $V_2 = V_C = 300 \, V$.
Since $V_L = V_C$,the circuit is in a state of resonance.
In resonance,the net voltage across the $LC$ combination is $V_{LC} = V_L - V_C = 300 - 300 = 0 \, V$.
The total applied voltage $V$ is equal to the voltage across the resistor $V_R$.
Therefore,$V_3 = V_R = V = 220 \, V$.
The current in the circuit is given by $I = \frac{V_R}{R} = \frac{220 \, V}{100 \, \Omega} = 2.2 \, A$.
Thus,the reading of voltmeter $V_3$ is $220 \, V$ and the reading of ammeter $A$ is $2.2 \, A$.

(C) In a gravity-free space,there is no external force acting on the system (man + stone). Therefore,the center of mass of the system remains at rest.
Let the initial position of the center of mass be $Y_{cm} = 10\, m$ from the floor.
Let the man move upwards by a distance $x$ and the stone move downwards by a distance $y$ when the stone hits the floor.
The stone travels $10\, m$ downwards,so $y = 10\, m$.
Since the center of mass remains stationary,the displacement of the center of mass is zero: $m_1 \Delta y_1 + m_2 \Delta y_2 = 0$.
Here,$m_1 = 50\, kg$,$\Delta y_1 = x$ (upwards),$m_2 = 0.5\, kg$,$\Delta y_2 = -10\, m$ (downwards).
$50(x) + 0.5(-10) = 0$.
$50x = 5$.
$x = 0.1\, m$.
The new height of the man above the floor is $10 + x = 10 + 0.1 = 10.1\, m$.

(C) Magnetic field,$B = 0.025 \, T$.
Radius of the loop,$r = 2 \, cm = 2 \times 10^{-2} \, m$.
Constant rate at which the radius of the loop shrinks,$\frac{dr}{dt} = 1 \times 10^{-3} \, m/s$.
Magnetic flux linked with the loop is $\phi = B A \cos \theta = B(\pi r^2) \cos 0^{\circ} = B \pi r^2$.
The magnitude of the induced $e.m.f.$ is $|\varepsilon| = \frac{d\phi}{dt} = \frac{d}{dt}(B \pi r^2) = B \pi (2r) \frac{dr}{dt}$.
Substituting the values:
$|\varepsilon| = 0.025 \times \pi \times 2 \times (2 \times 10^{-2}) \times (1 \times 10^{-3})$.
$|\varepsilon| = 0.025 \times \pi \times 4 \times 10^{-5} = 0.1 \times 10^{-5} \times \pi = 1 \times 10^{-6} \pi \, V$.
$|\varepsilon| = \pi \, \mu V$.

(C) In an $LCR$ series circuit,the voltage across the inductor $V_L$ and the capacitor $V_C$ are $180^{\circ}$ out of phase.
Given $V_1 = V_L = 300 \, V$ and $V_2 = V_C = 300 \, V$.
The net voltage across the $LC$ combination is $V_{LC} = |V_L - V_C| = |300 - 300| = 0 \, V$.
The total supply voltage $V$ is given by $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
Since $V = 220 \, V$ and $(V_L - V_C) = 0$,we have $V = V_R = 220 \, V$.
Thus,the reading of voltmeter $V_3$ (which measures voltage across resistor $R$) is $V_3 = 220 \, V$.
The current $I$ in the circuit is given by $I = \frac{V_R}{R} = \frac{220 \, V}{100 \, \Omega} = 2.2 \, A$.
Therefore,the reading of ammeter $A$ is $2.2 \, A$.

(C) The number of moles of the gas is $0.1 \ mol$.
Since the gas is triatomic,each molecule contains $3$ atoms.
The number of molecules $= \text{moles} \times N_A = 0.1 \times 6.02 \times 10^{23} = 6.02 \times 10^{22} \ \text{molecules}$.
The number of atoms $= \text{number of molecules} \times 3 = 6.02 \times 10^{22} \times 3 = 1.806 \times 10^{23} \ \text{atoms}$.

(C) When a current-carrying loop is placed in a uniform magnetic field,the net magnetic force acting on the entire closed loop is zero.
Let $\vec{F}_{net}$ be the total force on the loop,which is given by $\vec{F}_{net} = \vec{F}_{arm1} + \vec{F}_{remaining} = 0$.
Given that the force on one arm is $\vec{F}$,we have $\vec{F} + \vec{F}_{remaining} = 0$.
Therefore,the net force on the remaining three arms is $\vec{F}_{remaining} = -\vec{F}$.

(A) In a body-centred cubic $(BCC)$ lattice,the atoms touch along the body diagonal.
The body diagonal length is given by $a \sqrt{3}$.
The distance between two oppositely charged ions (nearest neighbour distance) is half of the body diagonal.
Distance $= \frac{a \sqrt{3}}{2}$.
Given $a = 387 \ pm$.
Distance $= \frac{387 \times 1.732}{2} \approx 335 \ pm$.

(B) The structure of $P_4O_{10}$ consists of four phosphorus atoms at the corners of a tetrahedron. Each phosphorus atom is bonded to one terminal oxygen atom via a double bond $(P=O)$. The remaining six oxygen atoms act as bridging atoms,each connecting two phosphorus atoms ($P-O-P$ linkages). Therefore,there are $6$ bridging oxygen atoms in the $P_4O_{10}$ molecule.

(D) In series,the equivalent capacitance is $C_S = \frac{C_1}{n_1}$.
Energy stored,$E_S = \frac{1}{2} C_S V_S^2 = \frac{1}{2} \left(\frac{C_1}{n_1}\right) (4V)^2 = \frac{8 C_1 V^2}{n_1}$.
In parallel,the equivalent capacitance is $C_P = n_2 C_2$.
Energy stored,$E_P = \frac{1}{2} C_P V_P^2 = \frac{1}{2} (n_2 C_2) (V)^2 = \frac{n_2 C_2 V^2}{2}$.
Given that $E_S = E_P$,we have:
$\frac{8 C_1 V^2}{n_1} = \frac{n_2 C_2 V^2}{2}$.
Solving for $C_2$:
$C_2 = \frac{16 C_1}{n_1 n_2}$.

(D) For the coin to revolve with the record without slipping,the required centripetal force must be provided by the static frictional force.
The centripetal force required is $F_c = mR\omega^2$.
The maximum static frictional force available is $f_{s,max} = \mu N = \mu mg$.
For the coin to remain stationary relative to the record,the centripetal force must be less than or equal to the maximum static friction:
$mR\omega^2 \leq \mu mg$
Dividing both sides by $m\omega^2$,we get:
$R \leq \frac{\mu g}{\omega^2}$

(B) Strong electrolytes are completely ionised at all concentrations.
On increasing dilution,the number of ions remains the same,but the inter-ionic attractions decrease,which leads to an increase in the ionic mobility of the ions.
Therefore,the equivalent conductance increases.

(C) The circuit is an $LCR$ series circuit connected to an $AC$ source of $220 \, V$.
Given: $V_L = V_1 = 300 \, V$ and $V_C = V_2 = 300 \, V$.
Since $V_L = V_C$,the circuit is in resonance.
In a series $LCR$ circuit at resonance,the net voltage across the $LC$ combination is zero $(V_{LC} = V_L - V_C = 300 - 300 = 0 \, V)$.
Therefore,the entire source voltage appears across the resistor $R$.
Thus,the reading of voltmeter $V_3$ is $V_R = 220 \, V$.
The current $I$ in the circuit is given by $I = \frac{V_R}{R} = \frac{220 \, V}{100 \, \Omega} = 2.2 \, A$.
So,the reading of voltmeter $V_3$ is $220 \, V$ and the reading of ammeter $A$ is $2.2 \, A$.

(D) For the coin to revolve with the record without slipping,the required centripetal force must be provided by the static frictional force.
The centripetal force required for circular motion is $F_c = m \omega^2 r$.
The maximum static frictional force available is $f_{s, \text{max}} = \mu N = \mu m g$.
For the coin to stay in place,the static friction must be greater than or equal to the required centripetal force:
$f_{s, \text{max}} \ge F_c$
Substituting the expressions:
$\mu m g \ge m \omega^2 r$
Dividing both sides by $m \omega^2$ (assuming $m$ and $\omega$ are non-zero):
$r \le \frac{\mu g}{\omega^2}$

(C) The particle starts from rest,so its initial velocity $u = 0$.
After time $T$,the final velocity is $V$.
The work done on the particle is equal to the change in its kinetic energy,according to the work-energy theorem.
$W = \Delta K = \frac{1}{2} M V^{2} - \frac{1}{2} M u^{2} = \frac{1}{2} M V^{2} - 0 = \frac{1}{2} M V^{2}$.
The average power delivered to the particle in time $T$ is defined as the work done divided by the time taken.
$P_{\text{avg}} = \frac{W}{T} = \frac{\frac{1}{2} M V^{2}}{T} = \frac{1}{2} \frac{M V^{2}}{T}$.

(D) For the series combination of $n_1$ capacitors of capacitance $C_1$ connected to a $4V$ source:
The equivalent capacitance is $C_s = \frac{C_1}{n_1}$.
The energy stored is $U_s = \frac{1}{2} C_s (4V)^2 = \frac{1}{2} \left( \frac{C_1}{n_1} \right) (16V^2) = \frac{8 C_1 V^2}{n_1}$.
For the parallel combination of $n_2$ capacitors of capacitance $C_2$ connected to a $V$ source:
The equivalent capacitance is $C_p = n_2 C_2$.
The energy stored is $U_p = \frac{1}{2} C_p V^2 = \frac{1}{2} (n_2 C_2) V^2$.
Given that $U_s = U_p$:
$\frac{8 C_1 V^2}{n_1} = \frac{1}{2} n_2 C_2 V^2$.
Solving for $C_2$:
$C_2 = \frac{16 C_1}{n_1 n_2}$.

(B) The electric field between two parallel plates is given by $E = \frac{\sigma}{\epsilon_0 K}$,where $\sigma$ is the surface charge density,$\epsilon_0$ is the permittivity of free space,and $K$ is the dielectric constant of the medium.
When the plates are dipped in kerosene oil,the medium between the plates changes from air (or vacuum) to kerosene oil.
The dielectric constant $K$ of kerosene oil is greater than $1$ $(K > 1)$.
Since the charge $Q$ on the plates remains constant,the surface charge density $\sigma = \frac{Q}{A}$ also remains constant.
Therefore,the new electric field $E'$ becomes $E' = \frac{E}{K}$.
Since $K > 1$,the electric field $E'$ will be less than the original electric field $E$. Thus,the electric field decreases.

(D) The focal length $f$ of a lens depends on the refractive index of the material and the radii of curvature of its surfaces. Covering the central part of the lens does not change these parameters,so the focal length remains $f' = f$.
Intensity $I$ is proportional to the area of the aperture $A$. The initial area is $A = \pi (d/2)^2 = \pi d^2 / 4$.
The area of the covered central part is $A_{covered} = \pi (d/4)^2 = \pi d^2 / 16$.
The remaining area is $A' = A - A_{covered} = \frac{\pi d^2}{4} - \frac{\pi d^2}{16} = \frac{3\pi d^2}{16}$.
Since $I \propto A$,the new intensity $I'$ is given by $I' = I \times (A'/A) = I \times (3/4) = 3I/4$.

(A) The distance $h$ covered by a body in free fall starting from rest in time $t$ is given by the equation $h = \frac{1}{2} gt^2$.
Rearranging this formula to solve for $g$,we get $g = \frac{2h}{t^2}$.
According to the rules of error propagation,for a quantity $g = k \cdot h^a \cdot t^b$,the relative error is given by $\frac{\Delta g}{g} = a \frac{\Delta h}{h} + |b| \frac{\Delta t}{t}$.
Here,$a = 1$ and $b = -2$. Therefore,the maximum relative error is $\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2 \frac{\Delta t}{t}$.
Multiplying by $100$ to express this as a percentage error,we get $\frac{\Delta g}{g} \times 100 = \left( \frac{\Delta h}{h} \times 100 \right) + 2 \left( \frac{\Delta t}{t} \times 100 \right)$.
Given that the percentage error in distance is $e_1$ and in time is $e_2$,the percentage error in $g$ is $e_1 + 2e_2$.

(D) The period of oscillation for a mass-spring system is given by the formula $T = 2\pi \sqrt{\frac{m}{k}}$,where $m$ is the mass and $k$ is the spring constant.
Initially,the mass is $M$,so $T = 2\pi \sqrt{\frac{M}{k}}$.
When another mass $M$ is added,the total mass becomes $M' = M + M = 2M$.
The new period of oscillation $T'$ will be $T' = 2\pi \sqrt{\frac{2M}{k}}$.
Dividing the new period by the initial period: $\frac{T'}{T} = \frac{2\pi \sqrt{\frac{2M}{k}}}{2\pi \sqrt{\frac{M}{k}}} = \sqrt{\frac{2M}{M}} = \sqrt{2}$.
Therefore,the new period is $T' = \sqrt{2}T$.

(B) In basal placentation,the placenta develops at the base of the ovary and a single ovule is attached to it. The ovary is typically unilocular. Examples include sunflower and marigold.

(D) Tracheids are elongated,angular,dead cells with hard,lignified,wide lumens and narrow end walls.
The walls of tracheids possess different types of thickenings,and the unthickened areas (pits) of their walls allow the rapid movement of water from one tracheid to another.
Tracheids are the characteristic cell types of xylem tissues in gymnosperms and pteridophytes,where they serve as the chief elements for water conduction.

(A) Plasmodesmata are microscopic channels that traverse the cell walls of plant cells and some algal cells.
They act as an effective transport pathway between two adjacent cells.
These structures facilitate the transport of molecules and communication between the cytoplasm of neighboring cells.

(D) Thylakoid: These are flattened sac-like structures present inside the chloroplast,involved in light-dependent reactions of photosynthesis.
Endoplasmic reticulum $(ER)$: This is a network of membranes involved in protein and lipid synthesis.
Plasmalemma: This is another term for the plasma membrane,which acts as the outer boundary of the cell.
Cytoskeleton: This is an elaborate network of filamentous proteinaceous structures (such as microtubules,microfilaments,and intermediate filaments) present in the cytoplasm that provides structural support,maintains cell shape,and facilitates cell motility.

(B) Essential elements required by plants in relatively large amounts are called macronutrients,such as $C, H, O, N, P, K, Ca, S, Mg.$
Essential elements required by plants in very small amounts are called micronutrients or trace elements,such as $Fe, Mn, Zn, Cu, B, Mo, Cl$ and $Ni.$
Magnesium $(Mg)$ is a macronutrient because it is a constituent of the chlorophyll molecule and is required in significant quantities for plant growth and development.
Therefore,$Mg$ is not a micronutrient.

(D) Phototropic movement is the result of the uneven distribution of $Auxin$.
When a plant is exposed to light from one side,$Auxin$ migrates to the shaded side of the stem.
This higher concentration of $Auxin$ on the shaded side causes cells to elongate more rapidly than those on the illuminated side,leading to the bending or curvature of the plant towards the light source.

(D) Goblet cells are specialized cells found in the columnar epithelium of the intestinal mucosa. They are responsible for the secretion of mucin,a glycoprotein that forms mucus when hydrated. Mucus acts as a lubricant,protecting the intestinal lining and facilitating the smooth movement of food (chyme) through the digestive tract. If goblet cells become non-functional,the lack of mucus will lead to friction and difficulty in the passage of food,thereby adversely affecting the smooth movement of food downwards the intestine.

(D) In humans,the principal nitrogenous excretory compound (i.e.,urea) is synthesized in the liver via the ornithine cycle. Urea is then transported through the blood and eliminated mostly through the kidneys as an excretory product.

(C) The adrenal gland is composed of two distinct parts: the adrenal cortex and the adrenal medulla.
Adrenaline (epinephrine) and noradrenaline (norepinephrine) are catecholamine hormones secreted by the adrenal medulla.
In contrast,the adrenal cortex secretes steroid hormones such as aldosterone (a mineralocorticoid),cortisol (a glucocorticoid),and sex steroids like androstenedione and dehydroepiandrosterone.
Since adrenaline is produced in the adrenal medulla,an injury to the adrenal cortex will not affect its secretion.

(A) In the pancreatic islets (Islets of Langerhans),$\alpha$-cells secrete glucagon,whereas $\beta$-cells secrete insulin.
Therefore,the pair 'Glucagon - Beta cells' is incorrectly matched because glucagon is secreted by $\alpha$-cells,not $\beta$-cells.
Somatostatin is correctly secreted by $\delta$-cells.
Corpus luteum secretes relaxin during the later stages of pregnancy.
Deficiency or dysfunction of insulin leads to diabetes mellitus.

(B) Vegetative propagation is a form of asexual reproduction in plants.
In aquatic plants like $Pistia$ and $Eichhornia$,vegetative propagation occurs through a specialized structure known as an $Offset$.
An $Offset$ is a lateral branch with short internodes and each node bears a rosette of leaves and a tuft of roots.
Therefore,the correct option is $B$.

(B) Biofortification is the process of breeding crops to increase their nutritional value. This includes increasing the levels of vitamins,minerals,proteins,and healthy fats. It is a practical and cost-effective method to improve public health by addressing hidden hunger or micronutrient deficiencies.

(B) Biofortification is the process of breeding crops to increase their nutritional value. This includes increasing the levels of vitamins,minerals,and proteins to improve public health and combat malnutrition. Somatic hybridization involves the fusion of protoplasts from two different plant species. Micropropagation is a technique for rapid vegetative multiplication of plants using tissue culture. Biomagnification refers to the increasing concentration of toxic substances in the tissues of organisms at successively higher levels in a food chain.

(B) Vegetative propagation in aquatic plants like $Pistia$ and $Eichhornia$ (water hyacinth) occurs through a specialized structure called an $Offset$.
An $Offset$ is a short,thick,horizontal branch that produces a rosette of leaves above and a tuft of roots below at the node.
Therefore,the correct option is $B$.

(B) The part of the fallopian tube closer to the ovary is the funnel-shaped $Infundibulum$.
It possesses finger-like projections called $fimbriae$,which help in the collection of the ovum after ovulation.

(A) Seminal plasma is composed of the fluid and sperms from the vas deferens (about $10 \%$ of the total),fluid from the seminal vesicles (almost $60 \%$ ),fluid from the prostate gland (about $30 \%$ ),and a small amount of secretions from the bulbourethral glands.
It is rich in fructose,calcium,citrate ions,phosphate ions,a clotting enzyme,profibrinolysin,inositol,prostaglandins,and several proteins.
Fructose serves as the primary energy source for the sperm,while calcium and other ions help in maintaining the motility and viability of the sperm.

(B) In the mammalian ovum,the process of meiosis occurs during the maturation phase.
Initially,the nucleus shifts towards the animal pole and undergoes meiosis-$I$.
The second meiotic division remains arrested at the metaphase-$II$ stage.
It is only completed after the penetration of the ovum by a sperm during the process of fertilization.
This division results in an unequal cytoplasmic cleavage,forming a large cell called the ootid (which contains almost all the cytoplasm) and a very small cell called the second polar body.