AIPMT 2009 Chemistry Question Paper with Answer and Solution

(C) Cyclic photophosphorylation involves only Photosystem $I$ $(PSI)$.
In this process,the excited electrons are cycled back to the reaction center $(P700)$ through an electron transport chain.
As the electrons move through the electron transport system,energy is released which is used to pump protons across the membrane,creating a proton gradient.
This gradient drives the synthesis of $ATP$ from $ADP$ and inorganic phosphate $(Pi)$ via $ATP$ synthase.
Unlike non-cyclic photophosphorylation,cyclic photophosphorylation does not involve the photolysis of water (so no $O_2$ is released) and does not reduce $NADP^+$ to $NADPH$.

(A) The pinna (external ear) is supported by cartilage,specifically elastic cartilage.
Elastic cartilage provides both strength and flexibility to the structure.
This type of cartilage is also found in the tip of the nose,the epiglottis,and the Eustachian tubes.
Therefore,the correct option is $A$.

(D) The correct answer is $D$.
$Frankia$ is a symbiotic nitrogen-fixing bacterium that forms root nodules in non-leguminous plants like $Alnus$.
$Azolla$ is a water fern that has a symbiotic relationship with the cyanobacterium $Anabaena$ $azollae$,but $Frankia$ is a classic example of a symbiotic nitrogen-fixing actinomycete.
$Glomus$ is a genus of fungi that forms mycorrhizal associations (not nitrogen fixation).
$Azotobacter$ is a free-living nitrogen-fixing bacterium.

(D) Allergies are the exaggerated response of the immune system to certain antigens present in the environment.
Common symptoms of allergies include sneezing,watery eyes,running nose,and difficulty in breathing.
These symptoms are caused by the release of chemicals like histamine and serotonin from the mast cells.
To quickly reduce the symptoms of allergy,drugs like antihistamines,adrenaline,and steroids are used.
Therefore,these medications are specifically administered to treat allergic reactions.

(D) Let the charges on the shells be $q_1, q_2,$ and $q_3$ respectively.
$q_1 = 4\pi a^2\sigma, q_2 = -4\pi b^2\sigma, q_3 = 4\pi c^2\sigma$.
The potential at the surface of shell $a$ is $V_A = \frac{kq_1}{a} + \frac{kq_2}{b} + \frac{kq_3}{c} = k4\pi\sigma(a - b + c)$.
The potential at the surface of shell $b$ is $V_B = \frac{kq_1}{b} + \frac{kq_2}{b} + \frac{kq_3}{c} = k4\pi\sigma(\frac{a^2}{b} - b + c)$.
The potential at the surface of shell $c$ is $V_C = \frac{kq_1}{c} + \frac{kq_2}{c} + \frac{kq_3}{c} = \frac{k4\pi\sigma}{c}(a^2 - b^2 + c^2)$.
Given $c = a + b$,then $c^2 = (a + b)^2 = a^2 + b^2 + 2ab$,so $a^2 - b^2 + c^2 = a^2 - b^2 + a^2 + b^2 + 2ab = 2a^2 + 2ab = 2a(a + b) = 2ac$.
Substituting this into $V_C$: $V_C = \frac{k4\pi\sigma}{c}(2ac) = k8\pi\sigma a$.
Also,$V_A = k4\pi\sigma(a - b + a + b) = k4\pi\sigma(2a) = k8\pi\sigma a$.
Thus,$V_A = V_C \neq V_B$.

(A) According to the first law of thermodynamics,the heat energy supplied to a system is given by $dQ = dU + dW$,where $dU$ is the change in internal energy and $dW$ is the work done by the system.
Given:
Heat absorbed $dQ = 2 \, kcal = 2000 \, cal$.
Since $1 \, cal = 4.2 \, J$,we have $dQ = 2000 \times 4.2 \, J = 8400 \, J$.
Work done $dW = 500 \, J$.
Substituting these values into the equation:
$8400 \, J = dU + 500 \, J$
$dU = 8400 \, J - 500 \, J$
$dU = 7900 \, J$.
Therefore,the change in internal energy is $7900 \, J$.

(B) When a charged particle of mass $m$ and charge $q$ moves in a uniform magnetic field $B$ with speed $v$ perpendicular to the field,it experiences a magnetic Lorentz force that acts as a centripetal force.
The magnetic force is given by $F = qvB$.
The centripetal force required for circular motion is $F = \frac{mv^2}{R}$.
Equating the two forces: $qvB = \frac{mv^2}{R}$.
Solving for the radius $R$: $R = \frac{mv}{qB}$.
The angular velocity $\omega$ is given by $\omega = \frac{v}{R}$.
Substituting $R$: $\omega = \frac{v}{(mv/qB)} = \frac{qB}{m}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
Substituting $\omega$: $T = \frac{2\pi m}{qB}$.
Since $T$ depends only on the mass $m$,charge $q$,and magnetic field $B$,it is independent of both the radius $R$ and the speed $v$.

(c) : The floral formula of tobacco is (image). It belongs to the family Solanaceae. The flower is actinomorphic, bisexual, 5 sepals gamosepalous, $5$ gamopetalous corolla, $5$ epipetalous stamens and $2$ carpels syncarpous havingsuperior ovary.

(D) $DDT$ causes biomagnification because it is a non-biodegradable pollutant.
These substances are lipo-soluble (fat-soluble),which allows them to accumulate in the fatty tissues of organisms.
As they are not metabolized or excreted easily,their concentration increases at each successive trophic level in the food chain,a phenomenon known as biomagnification.

(B) The balanced chemical equation for the reaction is: $2 H_2(g) + O_2(g) \rightarrow 2 H_2 O(l)$.
Calculate the number of moles for the reactants:
$n(H_2) = \frac{10 \ g}{2 \ g/mol} = 5 \ mol$.
$n(O_2) = \frac{64 \ g}{32 \ g/mol} = 2 \ mol$.
According to the stoichiometry,$1 \ mol$ of $O_2$ reacts with $2 \ mol$ of $H_2$ to produce $2 \ mol$ of $H_2 O$.
Here,$O_2$ is the limiting reagent because it will be consumed completely first ($2 \ mol$ of $O_2$ requires $4 \ mol$ of $H_2$,and we have $5 \ mol$ of $H_2$).
Since $1 \ mol$ of $O_2$ produces $2 \ mol$ of $H_2 O$,then $2 \ mol$ of $O_2$ will produce $2 \times 2 = 4 \ mol$ of $H_2 O$.

(D) For a given azimuthal quantum number $l$,the number of orbitals in a subshell is given by $(2l + 1)$.
Since each orbital can accommodate a maximum of $2$ electrons,the total number of electrons in a subshell is $2 \times (2l + 1) = 4l + 2$.

(B) For an electron in an atom,the quantum numbers must satisfy the following rules:
$1$. $n$ is a positive integer $(1, 2, 3, \dots)$.
$2$. $l$ can have values from $0$ to $(n - 1)$.
$3$. $m$ can have values from $-l$ to $+l$ (including $0$).
$4$. $s$ can be $+1/2$ or $-1/2$.
In option $B$,$n = 3$ and $l = 2$. The allowed values for $m$ are $-2, -1, 0, +1, +2$. Since $m = -3$ is outside this range,this arrangement is not permissible.

(D) Ionisation energy generally increases across a period due to a decrease in atomic size and effective nuclear charge.
Comparing the configurations:
$A: [Ne] 3s^2 3p^2$ (Period $3$,Group $14$)
$B: [Ar] 3d^{10} 4s^2 4p^3$ (Period $4$,Group $15$)
$C: [Ne] 3s^2 3p^1$ (Period $3$,Group $13$)
$D: [Ne] 3s^2 3p^3$ (Period $3$,Group $15$)
Elements in Period $3$ have smaller atomic radii than those in Period $4$,so elements in Period $3$ generally have higher ionisation energies than those in Period $4$.
Among the Period $3$ elements $(A, C, D)$,the configuration $[Ne] 3s^2 3p^3$ corresponds to a half-filled $p$-orbital,which is exceptionally stable.
Therefore,$[Ne] 3s^2 3p^3$ has the highest ionisation energy.

(D) When an $H$ atom is directly bonded to a highly electronegative atom like $N$,$O$,or $F$,a strong dipole is created,leading to the formation of intermolecular hydrogen bonding.
In liquid $CH_3OH$ (methanol),the $H$ atom is bonded to an $O$ atom,which allows for the formation of intermolecular hydrogen bonds between adjacent molecules.
To convert liquid $CH_3OH$ to a gas,these intermolecular hydrogen bonds must be overcome,as they are the dominant attractive forces holding the liquid molecules together.

(A) The molecular orbital configuration for $N_2$ ($14$ electrons) is: $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p_x)^2(\pi 2p_y)^2(\sigma 2p_z)^2$.
Bond order $(B.O.)$ $= \frac{10-4}{2} = 3$.
For $N_2^-$ ($15$ electrons): $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p_x)^2(\pi 2p_y)^2(\sigma 2p_z)^2(\pi^* 2p_x)^1$.
$B.O. = \frac{10-5}{2} = 2.5$.
For $N_2^{2-}$ ($16$ electrons): $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p_x)^2(\pi 2p_y)^2(\sigma 2p_z)^2(\pi^* 2p_x)^1(\pi^* 2p_y)^1$.
$B.O. = \frac{10-6}{2} = 2$.
Comparing the bond orders: $2 < 2.5 < 3$.
Therefore,the increasing order is $N_2^{2-} < N_2^- < N_2$.

(C) For $sp^2$ hybridization,the steric number should be $3$.
In $BF_3$ molecule,the central atom $B$ is bonded to $3$ atoms and has $0$ lone pairs,so steric number $= 3 + 0 = 3$. Hence,it is $sp^2$ hybridized.
In $NO_2^-$ ion,the central atom $N$ is bonded to $2$ oxygen atoms and has $1$ lone pair,so steric number $= 2 + 1 = 3$. Hence,it is $sp^2$ hybridized.
In $NH_2^-$ ion,the central atom $N$ is bonded to $2$ hydrogen atoms and has $2$ lone pairs,so steric number $= 2 + 2 = 4$. Hence,it is $sp^3$ hybridized.
In $H_2O$ molecule,the central atom $O$ is bonded to $2$ hydrogen atoms and has $2$ lone pairs,so steric number $= 2 + 2 = 4$. Hence,it is $sp^3$ hybridized.
Thus,in $BF_3$ and $NO_2^-$,the central atom is $sp^2$ hybridised.

(D) Energy absorbed by molecule $(A_2) = 4.4 \times 10^{-19} \ J$
Bond energy per molecule $(A_2) = 4.0 \times 10^{-19} \ J$
Excess energy available as kinetic energy $(K.E.) = \text{Energy absorbed} - \text{Bond energy}$
$K.E. = 4.4 \times 10^{-19} - 4.0 \times 10^{-19} = 0.4 \times 10^{-19} \ J$
Since the molecule is $(A_2)$,it contains $2$ atoms.
$K.E.$ per atom $= \frac{0.4 \times 10^{-19} \ J}{2} = 0.2 \times 10^{-19} \ J = 2 \times 10^{-20} \ J$
Thus,the kinetic energy per atom is $2 \times 10^{-20} \ J$.

(B) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative $(\Delta G < 0)$.
Given: $\Delta H = 170 \ kJ = 170000 \ J$ and $\Delta S = 170 \ J \ K^{-1}$.
Using the relation $\Delta G = \Delta H - T \Delta S$,for spontaneity:
$\Delta H - T \Delta S < 0$
$\Delta H < T \Delta S$
$T > \frac{\Delta H}{\Delta S}$
$T > \frac{170000 \ J}{170 \ J \ K^{-1}}$
$T > 1000 \ K$.
Among the given options,the reaction becomes spontaneous at temperatures above $1000 \ K$,and $1110 \ K$ is the only value satisfying this condition.

(B) The enthalpy of reaction $(\Delta H)$ is calculated using the formula:
$\Delta H = \sum \text{Bond Energy (Reactants)} - \sum \text{Bond Energy (Products)}$
For the reaction: $CH_2=CH_2 + H-H \to CH_3-CH_3$
Reactants have: $1 \times (C=C)$,$4 \times (C-H)$,and $1 \times (H-H)$ bonds.
Products have: $1 \times (C-C)$ and $6 \times (C-H)$ bonds.
$\Delta H = [BE(C=C) + 4 \times BE(C-H) + BE(H-H)] - [BE(C-C) + 6 \times BE(C-H)]$
$\Delta H = BE(C=C) + BE(H-H) - BE(C-C) - 2 \times BE(C-H)$
$\Delta H = 606.10 + 431.37 - 336.49 - 2(410.50)$
$\Delta H = 1037.47 - 336.49 - 821.00$
$\Delta H = -120.02 \approx -120 \text{ kJ mol}^{-1}$.

(C) The dissociation of acetic acid is given by:
$CH_3COOH \rightleftharpoons CH_3COO^{-} + H^{+}$,$K_{a1} = 1.5 \times 10^{-5}$
The dissociation of $HCN$ is given by:
$HCN \rightleftharpoons H^{+} + CN^{-}$,$K_{a2} = 4.5 \times 10^{-10}$
For the reaction $CN^{-} + CH_3COOH \rightleftharpoons HCN + CH_3COO^{-}$,the equilibrium constant $K$ is calculated as:
$K = \frac{K_{a1}}{K_{a2}}$
Substituting the values:
$K = \frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}} = \frac{1.5}{4.5} \times 10^{5} = \frac{1}{3} \times 10^{5} = 0.333 \times 10^{5} = 3.33 \times 10^{4}$
Rounding to the nearest provided option,$K \approx 3.0 \times 10^{4}$.

(D) Lewis acid is defined as an electron-pair acceptor.
In $(CH_3)_3B$,the central boron atom has only $6$ electrons in its valence shell,meaning it has an incomplete octet.
Therefore,it can accept a lone pair of electrons to complete its octet,acting as a Lewis acid.
In contrast,$(CH_3)_2O$,$(CH_3)_3P$,and $(CH_3)_3N$ all have central atoms with complete octets and at least one lone pair of electrons,which allows them to act as Lewis bases (electron-pair donors).

(D) The ionization constant of the weak base $NH_{4}OH$ is given as $K_{b} = 1.77 \times 10^{-5}$.
Ammonium chloride $(NH_{4}Cl)$ is a salt of a weak base $(NH_{4}OH)$ and a strong acid $(HCl)$.
The hydrolysis constant $(K_{h})$ for such a salt is given by the formula:
$K_{h} = \frac{K_{w}}{K_{b}}$
Where $K_{w}$ is the ionic product of water,which is $1.0 \times 10^{-14}$ at $298 \ K$.
Substituting the values:
$K_{h} = \frac{1.0 \times 10^{-14}}{1.77 \times 10^{-5}}$
$K_{h} \approx 0.56497 \times 10^{-9} = 5.65 \times 10^{-10}$.

(D) Number of millimoles of $HCl = 20.0 \times 0.050 = 1.0 \ mmol$.
Number of millimoles of $Ba(OH)_2 = 30.0 \times 0.10 = 3.0 \ mmol$.
Since $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^{-}$,$3.0 \ mmol$ of $Ba(OH)_2$ provides $6.0 \ mmol$ of $OH^{-}$ ions.
Number of $H^{+}$ ions from $HCl = 1.0 \ mmol$.
Remaining $OH^{-}$ moles after neutralization $= 6.0 - 1.0 = 5.0 \ mmol$.
Total volume $= 20.0 + 30.0 = 50.0 \ mL$.
$[OH^{-}] = \frac{5.0 \ mmol}{50.0 \ mL} = 0.10 \ M$.

(D) $(i)$ Sum of oxidation states of all atoms $=$ charge of ion.
$(ii)$ Oxidation number of oxygen $= -2$.
Let the oxidation state of $P$ in $PO_{4}^{3-}$ be $x$:
$x + 4(-2) = -3 \implies x - 8 = -3 \implies x = +5$.
Let the oxidation state of $S$ in $SO_{4}^{2-}$ be $y$:
$y + 4(-2) = -2 \implies y - 8 = -2 \implies y = +6$.
Let the oxidation state of $Cr$ in $Cr_{2}O_{7}^{2-}$ be $z$:
$2z + 7(-2) = -2 \implies 2z - 14 = -2 \implies 2z = 12 \implies z = +6$.
Thus,the oxidation states of $P, S$ and $Cr$ are $+5, +6$ and $+6$ respectively.

(C) According to Fajans' rule,the covalent character is directly proportional to the polarizability of the anion.
As the size of the anion increases,its polarizability increases,which leads to an increase in the covalent character of the bond.
The order of the size of halide ions is $F^- < Cl^- < Br^- < I^-$.
Therefore,the order of covalent character in alkali metal halides is $MI > MBr > MCl > MF$.

(A) Sodium hydroxide,$NaOH$,is a strong base. It reacts with acidic and amphoteric oxides but does not react with basic oxides.
$CaO$ is a basic oxide (alkaline earth metal oxide).
$SiO_2$ is an acidic oxide.
$BeO$ and $B_2O_3$ are amphoteric oxides.
Since $CaO$ is a basic oxide,it will not react with the base $NaOH$.

(A) The stability of the $+1$ oxidation state increases down the group $13$ elements as $Al < Ga < In < Tl$.
This is due to the inert pair effect,where the $ns^2$ electrons become increasingly reluctant to participate in bond formation due to poor shielding by $d$ and $f$ orbitals.
As we move down the group,the stability of the $+1$ oxidation state increases,making $Tl^+$ the most stable among the given ions.

(D) The correct option is $(d)$.
In the given hydrocarbon structure:
$1.$ $C_2$ is involved in a triple bond $(C \equiv C)$,so it is $sp$ hybridized.
$2.$ $C_3$ is bonded to four other atoms ($C_2, C_4, H$,and the $C$ of the $CH_3$ group) via single bonds,so it is $sp^3$ hybridized.
$3.$ $C_5$ is involved in a double bond $(C=C)$,so it is $sp^2$ hybridized.
$4.$ $C_6$ is bonded to four other atoms ($C_5, C_7$,and the $C$ atoms of two $CH_3$ groups) via single bonds,so it is $sp^3$ hybridized.
Thus,the sequence for $C_2, C_3, C_5$ and $C_6$ is $sp, sp^3, sp^2, sp^3$.

(D) Compounds that exhibit geometrical isomerism must have a restricted rotation,such as a double bond $(C=C)$,where each carbon atom of the double bond is attached to two different groups.
For $2-$Butene $(CH_3-CH=CH-CH_3)$,each double-bonded carbon is attached to a hydrogen atom and a methyl group. Since these groups are different,it exhibits $cis-trans$ isomerism.
$1.$ $cis-2-$Butene: Both methyl groups are on the same side of the double bond.
$2.$ $trans-2-$Butene: Both methyl groups are on opposite sides of the double bond.

(D) The reaction of benzene with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ is known as Friedel-Crafts alkylation.
In this reaction,the methyl group $(-CH_3)$ replaces a hydrogen atom on the benzene ring to form toluene $(C_6H_5CH_3)$.
The chemical equation is: $C_6H_6 + CH_3Cl \xrightarrow{Anhy. AlCl_3} C_6H_5CH_3 + HCl$.

(D) Degree of dissociation,$\alpha = \frac{\Lambda^{c}}{\Lambda^{\infty}}$
Given,$\Lambda^{c} = 8.0 \, \text{mho} \, \text{cm}^2$ and $\Lambda^{\infty} = 400 \, \text{mho} \, \text{cm}^2$.
$\alpha = \frac{8.0}{400} = 0.02 = 2 \times 10^{-2}$.
For a weak monobasic acid,the dissociation constant $K_{a}$ is given by Ostwald's dilution law:
$K_{a} = \frac{C \alpha^{2}}{1 - \alpha}$.
Since $\alpha$ is very small $(1 - \alpha \approx 1)$,the formula simplifies to $K_{a} \approx C \alpha^{2}$.
Given concentration $C = \frac{1}{32} \, M$.
$K_{a} = \frac{1}{32} \times (2 \times 10^{-2})^{2} = \frac{1}{32} \times 4 \times 10^{-4} = \frac{1}{8} \times 10^{-4} = 0.125 \times 10^{-4} = 1.25 \times 10^{-5}$.

(B) According to the Bronsted-Lowry theory,an acid is a proton donor and a base is a proton acceptor.
In the nitrating mixture,conc. $H_2SO_4$ acts as a strong acid and donates a proton to conc. $HNO_3$.
The reaction is: $HNO_3 + H_2SO_4 \rightarrow H_2NO_3^+ + HSO_4^-$.
Here,$HNO_3$ accepts a proton from $H_2SO_4$,therefore $HNO_3$ acts as a base.
Subsequently,$H_2NO_3^+$ decomposes to form the electrophile $NO_2^+$: $H_2NO_3^+ \rightarrow NO_2^+ + H_2O$.

(A) Fajan's rule states that the covalent character increases with the decrease in the size of the cation and increases with the increase in the size of the anion.
Since the size of the anion increases down the group,the order of ionic size is $I^{-} > Br^{-} > Cl^{-} > F^{-}$.
According to Fajan's rule,the covalent character increases as the size of the anion increases.
Therefore,the order of covalent character is $MI > MBr > MCl > MF$.

(A) The yellow color of the stool in infants is primarily due to the presence of bile pigments,specifically bilirubin and biliverdin.
Even though the infant's diet consists solely of white mother's milk,the liver secretes bile juice into the digestive tract.
Bile pigments are metabolic waste products resulting from the breakdown of old red blood cells.
These pigments are excreted into the intestine along with bile juice and are modified by intestinal bacteria,giving the stool its characteristic yellow color.

(B) In humans,the process of digestion begins in the mouth.
Salivary amylase (ptyalin) acts on starch in the mouth,breaking it down into maltose.
Therefore,starch is partially digested before reaching the stomach.
Fats are not digested in the mouth; they undergo emulsification and digestion primarily in the small intestine.
Cellulose is a complex carbohydrate (polysaccharide) that humans lack the enzymes to digest.
Thus,both fat and cellulose reach the stomach in a completely undigested state.

(C) The $Hypothalamus$ is a vital part of the diencephalon in the human brain.
It contains a number of centers which control body temperature,urge for eating and drinking.
It also contains several groups of neurosecretory cells,which secrete hormones called hypothalamic hormones.
Therefore,the correct option is $C$.

(A) plant is called monoecious when both male and female reproductive structures (cones or flowers) are present on the same individual plant.
In the given options,$Pinus$ is a monoecious plant because it bears both male and female cones on the same tree.
$Cycas$ is dioecious (male and female plants are separate).
$Papaya$ is dioecious.
$Marchantia$ is a liverwort where male and female sex organs are produced on separate thalli (dioecious).

(C) The process of micturition (urination) is initiated by the stretching of the urinary bladder wall as it fills with urine.
Stretch receptors present in the wall of the bladder send signals to the Central Nervous System $(CNS)$ when the bladder is stretched.
In response,the $CNS$ sends motor messages to initiate the contraction of smooth muscles of the bladder and the simultaneous relaxation of the urethral sphincter,causing the release of urine.
If the stretch receptors are removed,the brain will not receive the signal that the bladder is full.
Consequently,the reflex arc required for micturition will not be triggered,and micturition will not occur,leading to the bladder becoming over-distended.

(D) Biopesticides are biological agents used to control pests.
$1$. $Bacillus$ $thuringiensis$ $(Bt)$ is a well-known bacterium used as a biopesticide to control insect larvae.
$2$. $Trichoderma$ $harzianum$ is a free-living fungus used as a biocontrol agent against several plant pathogens.
$3$. $Nuclear$ $Polyhedrosis$ $Virus$ $(NPV)$ belongs to the genus $Baculovirus$ and is used as a narrow-spectrum insecticidal agent.
$4$. $Xanthomonas$ $campestris$ is a bacterium that causes black rot disease in cruciferous plants,making it a plant pathogen rather than a biopesticide.
Therefore,the correct option is $D$.

(D) Biomagnification refers to the increase in concentration of a toxic substance at successive trophic levels.
$DDT$ (Dichlorodiphenyltrichloroethane) is a persistent organic pollutant.
It is highly lipid-soluble (fat-soluble),which means it is not easily excreted by the body.
Instead,it gets stored in the adipose (fat) tissues of organisms.
As it moves up the food chain,the concentration of $DDT$ increases significantly in the tissues of higher trophic level organisms,leading to biomagnification.

(D) The torque $\vec \tau$ is defined as the cross product of the position vector $\vec r$ and the force vector $\vec F$,given by $\vec \tau = \vec r \times \vec F$.
By the definition of the cross product,the resulting vector $\vec \tau$ is perpendicular to both $\vec r$ and $\vec F$.
Since the dot product of two perpendicular vectors is zero,we have $\vec r \cdot \vec \tau = 0$ and $\vec F \cdot \vec \tau = 0$.

(D) The balanced chemical equation is: $2H_2(g) + O_2(g) \to 2H_2O(l)$.
Calculate the moles of reactants:
$n(H_2) = \frac{10 \ g}{2 \ g/mol} = 5 \ mol$.
$n(O_2) = \frac{64 \ g}{32 \ g/mol} = 2 \ mol$.
Identify the Limiting Reagent $(L.R.)$:
According to the stoichiometry,$2 \ mol$ of $H_2$ requires $1 \ mol$ of $O_2$.
So,$5 \ mol$ of $H_2$ would require $2.5 \ mol$ of $O_2$.
Since we only have $2 \ mol$ of $O_2$,$O_2$ is the $L.R.$
Calculate the product:
Since $1 \ mol$ of $O_2$ produces $2 \ mol$ of $H_2O$,$2 \ mol$ of $O_2$ will produce $2 \times 2 = 4 \ mol$ of $H_2O$.

(C) For a body-centred cubic $(BCC)$ crystal,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by the formula: $\sqrt{3}a = 4r$.
Given that the edge length $a = 351 \ pm$.
Substituting the values: $r = \frac{\sqrt{3} \times a}{4} = \frac{1.732 \times 351}{4}$.
$r = \frac{607.932}{4} = 151.98 \ pm \approx 151.8 \ pm$.

(D) The initial kinetic energy of the body is given by $K_i = \frac{1}{2} m u^2 = \frac{1}{2} \times 1 \times (20)^2 = 200\, J$.
If there were no air friction,the maximum height $H$ reached by the body would be determined by the conservation of energy: $m g H = K_i$.
$1 \times 10 \times H = 200 \implies H = 20\, m$.
However,the body only reaches a height of $h' = 18\, m$ due to energy loss from air friction.
The potential energy at this height is $U_f = m g h' = 1 \times 10 \times 18 = 180\, J$.
The energy lost due to air friction is the difference between the initial kinetic energy and the final potential energy: $\Delta E = K_i - U_f = 200\, J - 180\, J = 20\, J$.

(B) Applying Kirchhoff's Voltage Law $(KVL)$ to the upper loop containing the resistor $R$ and the cell with emf ${\varepsilon _1}$ and internal resistance ${r_1}$:
Starting from the positive terminal of the cell ${\varepsilon _1}$ and moving in the direction of the current:
$1$. The potential drop across the cell is ${\varepsilon _1}$.
$2$. The current flowing through the resistor $R$ is $(i_1 + i_2)$,so the potential drop is $-(i_1 + i_2)R$.
$3$. The current flowing through the internal resistance $r_1$ is $i_1$,so the potential drop is $-i_1r_1$.
Summing these potential changes to zero gives: ${\varepsilon _1} - (i_1 + i_2)R - i_1r_1 = 0$.

(A) Thyroxine hormone contains iodine. The thyroid gland produces two primary hormones - thyroxine (also referred to as $T4$) and tri-iodothyronine (also referred to as $T3$). The numbers $3$ and $4$ refer to the number of atoms of iodine in the hormones. Iodine is essential for the production of thyroid hormones. Iodine is found in most foods,especially seafood.

(A) To convert a galvanometer into an ammeter,a shunt resistance $S$ must be connected in parallel with the galvanometer.
The formula for the shunt resistance is given by $I_{g} \cdot G = (I - I_{g}) \cdot S$,where $I_{g}$ is the full-scale deflection current,$G$ is the galvanometer resistance,$I$ is the maximum current to be measured,and $S$ is the shunt resistance.
Rearranging for $S$,we get $S = \frac{I_{g} \cdot G}{I - I_{g}}$.
Given values are $I_{g} = 1.0\,A$,$G = 60\,\Omega$,and $I = 5.0\,A$.
Substituting these values: $S = \frac{1.0 \times 60}{5.0 - 1.0} = \frac{60}{4} = 15\,\Omega$.
Thus,a resistance of $15\,\Omega$ must be connected in parallel.

(C) An isochoric process is defined as a process in which the volume of the system remains constant. In such a process,the pressure and temperature may change,but the volume does not. Therefore,the statement that pressure remains constant in an isochoric process is incorrect.

(D) For $PO_4^{3-}$: Let the oxidation number of $P$ be $x$. Then $x + 4(-2) = -3$,which gives $x - 8 = -3$,so $x = +5$.
For $SO_4^{2-}$: Let the oxidation number of $S$ be $x$. Then $x + 4(-2) = -2$,which gives $x - 8 = -2$,so $x = +6$.
For $Cr_2O_7^{2-}$: Let the oxidation number of $Cr$ be $x$. Then $2x + 7(-2) = -2$,which gives $2x - 14 = -2$,so $2x = 12$,and $x = +6$.
Thus,the oxidation numbers are $+5, +6, +6$.

(D) Ciliated epithelium is characterized by the presence of cilia on the free surface of the cells.
These cilia help in moving particles or mucus in a specific direction over the epithelial surface.
This tissue is specifically found on the inner surface of hollow organs such as the bronchioles (part of the respiratory tract) and the fallopian tubes (oviducts) to facilitate the movement of mucus and ova,respectively.

(C) The hydrolysis of dimethyl dichlorosilane,$(CH_3)_2SiCl_2$,produces a dihydroxy derivative,$(CH_3)_2Si(OH)_2$.
This dihydroxy derivative undergoes condensation polymerisation to form a straight chain silicone polymer,as shown in the reaction:
$n(CH_3)_2SiCl_2 + 2nH_2O \rightarrow n(CH_3)_2Si(OH)_2 + 2nHCl$
$n(CH_3)_2Si(OH)_2 \rightarrow -[O-Si(CH_3)_2-O]_n- + nH_2O$

(A) For a body-centred cubic $(bcc)$ crystal, the relationship between the edge length $a$ and the atomic radius $r$ is given by $a \sqrt{3} = 4r$.
Given, edge length $a = 351 \ pm$.
Thus, the atomic radius $r = \frac{a \sqrt{3}}{4}$.
Substituting the values, $r = \frac{351 \times 1.732}{4} = 151.98 \ pm$.
Rounding to the nearest option, the value is approximately $151.8 \ pm$.

(D) In a face-centred cubic $(fcc)$ lattice, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $r = \frac{\sqrt{2}a}{4}$.
Given, $a = 361 \text{ pm}$.
Substituting the value of $a$ in the formula:
$r = \frac{\sqrt{2} \times 361}{4} \approx \frac{1.414 \times 361}{4} \approx 127.6 \text{ pm}$.
Rounding to the nearest integer, the radius is $128 \text{ pm}$.

(D) The depression in freezing point is given by the formula: $\Delta T_f = i \cdot K_f \cdot m$
Here,$\Delta T_f = 0 - (-0.00732) = 0.00732 \ ^oC$,$K_f = 1.86 \ ^oC/m$,and $m = 0.0020 \ m$.
Calculating the van't Hoff factor $(i)$:
$i = \frac{\Delta T_f}{K_f \cdot m} = \frac{0.00732}{1.86 \times 0.0020} = \frac{0.00732}{0.00372} \approx 1.968 \approx 2$.
The ionic compound $[Co(NH_3)_5(NO_2)]Cl$ dissociates in water as: $[Co(NH_3)_5(NO_2)]Cl \rightarrow [Co(NH_3)_5(NO_2)]^+ + Cl^-$.
Thus,$1 \ mol$ of the compound produces $2 \ mol$ of ions.

(D) The standard Gibbs free energy change is given by $\Delta G^{o} = -nFE^{o}$.
For reaction $(i): Cu^{2+} + 2e^{-} \rightarrow Cu$,$\Delta G_{1}^{o} = -2 \times F \times 0.337 = -0.674F$.
For reaction $(ii): Cu^{2+} + e^{-} \rightarrow Cu^{+}$,$E^{o} = 0.153 \, V$. We need the reverse reaction: $Cu^{+} \rightarrow Cu^{2+} + e^{-}$,where $E^{o} = -0.153 \, V$.
Thus,for $Cu^{+} \rightarrow Cu^{2+} + e^{-}$,$\Delta G_{2}^{o} = -1 \times F \times (-0.153) = 0.153F$.
Adding the two reactions:
$(Cu^{2+} + 2e^{-}$ $\rightarrow Cu) + (Cu^{+}$ $\rightarrow Cu^{2+} + e^{-})$ $\rightarrow (Cu^{+} + e^{-}$ $\rightarrow Cu)$.
$\Delta G_{total}^{o} = \Delta G_{1}^{o} + \Delta G_{2}^{o} = -0.674F + 0.153F = -0.521F$.
For the target reaction $Cu^{+} + e^{-} \rightarrow Cu$,$n=1$,so $\Delta G^{o} = -1 \times F \times E^{o}$.
$-0.521F = -1 \times F \times E^{o} \implies E^{o} = 0.52 \, V$.

(A) The reduction reaction for aluminium is: $Al^{3+} + 3e^- \rightarrow Al(s)$.
The equivalent mass of $Al$ is $E = \frac{\text{Atomic mass}}{\text{Valency factor}} = \frac{27}{3} = 9 \ g \ mol^{-1}$.
Using Faraday's law of electrolysis: $W = \frac{E \times I \times t}{96500}$.
Given: $I = 4.0 \times 10^4 \ A$,$t = 6 \ hours = 6 \times 3600 \ s = 21600 \ s$.
$W = \frac{9 \times 4.0 \times 10^4 \times 21600}{96500} \approx 8.05 \times 10^4 \ g \approx 8.1 \times 10^4 \ g$.

(D) For the reaction,$N_2 + 3H_2 \rightarrow 2NH_3,$ the rate of reaction is given by:
Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
Given that $\frac{d[NH_3]}{dt} = 2 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}.$
Equating the terms for $H_2$ and $NH_3$:
$-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
$-\frac{d[H_2]}{dt} = \frac{3}{2} \times \frac{d[NH_3]}{dt}$
$-\frac{d[H_2]}{dt} = \frac{3}{2} \times (2 \times 10^{-4}) = 3 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}.$

(D) For the reaction $BrO_{3(aq)}^{-} + 5Br_{(aq)}^{-} + 6H^{+} \to 3Br_{2(l)} + 3H_2O_{(l)}$,the rate of reaction is given by:
Rate $= -\frac{d[BrO_3^-]}{dt} = -\frac{1}{5} \frac{d[Br^-]}{dt} = -\frac{1}{6} \frac{d[H^+]}{dt} = \frac{1}{3} \frac{d[Br_2]}{dt} = \frac{1}{3} \frac{d[H_2O]}{dt}$.
We need to relate the rate of appearance of $Br_2$ to the rate of disappearance of $Br^-$.
From the expression: $\frac{1}{3} \frac{d[Br_2]}{dt} = -\frac{1}{5} \frac{d[Br^-]}{dt}$.
Multiplying both sides by $3$,we get:
$\frac{d[Br_2]}{dt} = -\frac{3}{5} \frac{d[Br^-]}{dt}$.

(B) Given: Half-life period $(t_{1/2})$ = $1386 \ s$.
For a first-order reaction,the relationship between half-life and rate constant $(k)$ is given by:
$t_{1/2} = \frac{0.693}{k}$
Rearranging for $k$:
$k = \frac{0.693}{t_{1/2}}$
Substituting the value:
$k = \frac{0.693}{1386} \ s^{-1}$
$k = 0.0005 \ s^{-1}$
$k = 5.0 \times 10^{-4} \ s^{-1} = 0.5 \times 10^{-3} \ s^{-1}$.

(A) For the reaction,$A + B \longrightarrow$ Products.
Let the rate law be: $\text{Rate} = k[A]^x[B]^y$.
From observation $(i)$,doubling $[A]$ doubles the rate: $2 \times \text{Rate} = k[2A]^x[B]^y \implies 2 = 2^x \implies x = 1$.
From observation $(ii)$,doubling both $[A]$ and $[B]$ increases the rate by a factor of $8$: $8 \times \text{Rate} = k[2A]^1[2B]^y$.
Dividing this by the original rate law: $8 = 2^1 \times 2^y \implies 8 = 2 \times 2^y \implies 4 = 2^y \implies y = 2$.
Thus,the rate law is: $\text{Rate} = k[A][B]^2$.

(D) The strength of an oxidising agent is determined by its ability to gain electrons,which is related to its reduction potential.
Fluorine $(F_2)$ has the highest electronegativity and the highest standard reduction potential among the halogens.
As we move down the group,the electronegativity decreases,making it harder to gain electrons.
Therefore,$F_2$ is the strongest oxidising agent as it is reduced most readily to the $F^-$ ion.

(B) The number of oxidation states exhibited by a transition element depends on the number of valence electrons available for bonding.
$1$. For $3d^54s^1$ (Chromium),the total number of valence electrons is $6$,so it can show oxidation states up to $+6$.
$2$. For $3d^54s^2$ (Manganese),the total number of valence electrons is $7$,so it can show oxidation states up to $+7$.
$3$. For $3d^24s^2$ (Titanium),the total number of valence electrons is $4$,so it can show oxidation states up to $+4$.
$4$. For $3d^34s^2$ (Vanadium),the total number of valence electrons is $5$,so it can show oxidation states up to $+5$.
Thus,the element with the configuration $3d^54s^2$ exhibits the largest number of oxidation states.

(B) To determine if a species is colourless,we check for the presence of unpaired electrons in the $d$-orbitals. If the $d$-orbital is either completely empty $(d^0)$ or completely filled $(d^{10})$,the species is colourless due to the absence of $d-d$ transitions.
$1$. $TiF_6^{2-}$: $Ti$ is in $+4$ oxidation state. Electronic configuration: $Ti^{4+} = [Ar] 3d^0$. Since it has no $d$-electrons,it is colourless.
$2$. $CoF_6^{3-}$: $Co$ is in $+3$ oxidation state. Electronic configuration: $Co^{3+} = [Ar] 3d^6$. It has unpaired electrons,so it is coloured.
$3$. $Cu_2Cl_2$: $Cu$ is in $+1$ oxidation state. Electronic configuration: $Cu^{+} = [Ar] 3d^{10}$. Since the $d$-orbital is fully filled,it is colourless.
$4$. $NiCl_4^{2-}$: $Ni$ is in $+2$ oxidation state. Electronic configuration: $Ni^{2+} = [Ar] 3d^8$. It has unpaired electrons,so it is coloured.
Therefore,$TiF_6^{2-}$ and $Cu_2Cl_2$ are the colourless species.

(A) Optical isomerism is exhibited by only those complexes in which elements of symmetry are absent.
Octahedral complexes of the types $[M(aa)_3]$,$[M(aa)x_2y_2]$,and $[M(aa)_2x_2]$ lack elements of symmetry and thus exhibit optical isomerism,where $aa$ represents a bidentate ligand,$x$ or $y$ represent monodentate ligands,and $M$ represents the central metal ion.
Complexes of the type $[MA_3B_3]$,such as $[Co(NH_3)_3Cl_3]^0$,possess planes of symmetry and therefore do not exhibit optical isomerism.
Conversely,$[Co(en)_3]^{3+}$ (type $[M(aa)_3]$),$[Co(en)_2Cl_2]^+$ (type $[M(aa)_2x_2]$),and $[Co(en)Cl_2(NH_3)_2]^+$ (type $[M(aa)x_2y_2]$) all exhibit optical isomerism.

(B) complex ion absorbs visible light if it contains unpaired electrons in its $d$-orbitals,allowing for $d-d$ transitions.
$1$. In $[Ti(en)_2(NH_3)_2]^{4+}$,$Ti$ is in the $+4$ oxidation state. $Ti^{4+} = [Ar] 3d^0$. No $d$-electrons,so no $d-d$ transition.
$2$. In $[Cr(NH_3)_6]^{3+}$,$Cr$ is in the $+3$ oxidation state. $Cr^{3+} = [Ar] 3d^3$. It has $3$ unpaired electrons,allowing for $d-d$ transitions.
$3$. In $[Zn(NH_3)_6]^{2+}$,$Zn$ is in the $+2$ oxidation state. $Zn^{2+} = [Ar] 3d^{10}$. Fully filled $d$-orbitals,no $d-d$ transition.
$4$. In $[Sc(H_2O)_3(NH_3)_3]^{3+}$,$Sc$ is in the $+3$ oxidation state. $Sc^{3+} = [Ar] 3d^0$. No $d$-electrons,so no $d-d$ transition.
Therefore,only $[Cr(NH_3)_6]^{3+}$ is expected to absorb visible light.

(D) $KOH \rightarrow K^{+} + OH^{-}$
$RX + OH^{-} \rightarrow R-OH + X^{-}$
$OH^{-}$ is a stronger nucleophile than the halide ion $(X^{-})$,so it easily replaces the weaker nucleophile. Nucleophiles are species that are either negatively charged or possess lone pairs of electrons,such as $OH^{-}$,$\ddot{N}H_{3}$,etc.

(C) $1$. $CH_3CH_2OH \xrightarrow{PBr_3} CH_3CH_2Br \ (X)$
$2$. $CH_3CH_2Br \xrightarrow{alc. KOH} CH_2=CH_2 \ (Y) \text{ (Dehydrohalogenation)}$
$3$. $CH_2=CH_2$ $\xrightarrow{H_2SO_4} CH_3CH_2OSO_3H$ $\xrightarrow{H_2O, \text{heat}} CH_3CH_2OH \ (Z) \text{ (Hydration of ethene)}$
Thus,the product $Z$ is ethanol.

(C) Periodic acid $(HIO_4)$ is a selective oxidizing agent used for the oxidative cleavage of vicinal diols ($1,2$-diols).
Ethylene glycol $(HO-CH_2-CH_2-OH)$ is a vicinal diol that reacts with $HIO_4$ to produce two molecules of formaldehyde $(HCHO)$.
$HO-CH_2-CH_2-OH + HIO_4 \rightarrow 2HCHO + HIO_3 + H_2O$.

(B) Step $1$: Phenol reacts with $Zn$ dust to form benzene $(X)$.
$C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO$
Step $2$: Benzene undergoes Friedel-Crafts alkylation with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ to form toluene $(Y)$.
$C_6H_6 + CH_3Cl \xrightarrow{Anhyd.\,AlCl_3} C_6H_5CH_3 + HCl$
Step $3$: Toluene $(Y)$ is oxidized by alkaline $KMnO_4$ to form benzoic acid $(Z)$.
$C_6H_5CH_3 \xrightarrow{Alkaline\,KMnO_4} C_6H_5COOH$
Therefore,the product $Z$ is benzoic acid.

(A) The reaction of trichloroacetaldehyde (chloral) with chlorobenzene in the presence of concentrated sulphuric acid $(H_2SO_4)$ is a condensation reaction.
In this reaction,one molecule of chloral reacts with two molecules of chlorobenzene to form $1,1,1-$trichloro$-2,2-$bis($p-$chlorophenyl)ethane,which is commonly known as $DDT$.
The chemical equation is:
$CCl_3CHO + 2C_6H_5Cl \xrightarrow{Conc. H_2SO_4} (ClC_6H_4)_2CHCCl_3 + H_2O$
Thus,the product formed is $DDT$.

(C) The reaction of propionic acid $(CH_3-CH_2-COOH)$ with $Br_2$ in the presence of red phosphorus is known as the $Hell-Volhard-Zelinsky$ $(HVZ)$ reaction.
This reaction specifically substitutes the $\alpha$-hydrogen atoms with bromine atoms.
Propionic acid has two $\alpha$-hydrogens on the carbon atom adjacent to the $-COOH$ group.
Since a dibromo product is formed,both $\alpha$-hydrogens are replaced by bromine,resulting in the structure $CH_3-C(Br)_2-COOH$.

(A) $N$-methylaniline is a secondary amine. When it reacts with nitrous acid $(HNO_2)$,generated in situ by the reaction of $NaNO_2$ and $HCl$,it undergoes $N$-nitrosation to form $N$-nitroso-$N$-methylaniline. The reaction is as follows:
$C_6H_5NHCH_3 + NaNO_2 + HCl \rightarrow C_6H_5N(NO)CH_3 + NaCl + H_2O$
This product is a yellow oily liquid.

(B) The segment of $DNA$ which acts as the instrumental manual for the synthesis of the protein is a $gene$.
Every protein in a cell has a corresponding $gene$ that contains the information required for its synthesis.

(C) Thyroxine is $3, 5, 3', 5'$-tetraiodothyronine. It is secreted by the follicular cells of the thyroid gland. The structure is shown below:
$HO-C_6H_2I_2-O-C_6H_2I_2-CH(NH_2)COOH$
Thyroxine stimulates the consumption of oxygen and thus,the metabolism of all cells or tissues in the body.

(A) Neoprene is a polymer of chloroprene ($2$-chloro-$1,3$-butadiene). Its correct repeating unit is $[ - CH_2 - C(Cl) = CH - CH_2 - ]_n$. The structure provided in option $(A)$ is incorrect because it contains an extra $-CH_2-$ group in the main chain.

(D) Tranquilizers are chemical substances that reduce mental tension and anxiety.
They are often referred to as psychotherapeutic drugs.
$Equanil$,$Valium$,and $Serotonin$ are common examples of tranquilizers.
Therefore,$Equanil$ is the correct answer.