AIPMT 2007 Chemistry Question Paper with Answer and Solution

(C) The $DNA$ double helix model proposed by Watson and Crick states that the two polynucleotide chains run in opposite directions.
One chain runs in the $5' \rightarrow 3'$ direction,while the other runs in the $3' \rightarrow 5'$ direction.
This orientation is known as antiparallel.

(B) Living organisms exhibit several characteristics such as growth,reproduction,metabolism,cellular organization,and consciousness.
Among these,consciousness (the ability to sense their environment and respond to environmental stimuli) is considered the defining property of living organisms.
While growth and reproduction are characteristics of living beings,they are not defining properties because non-living objects can also grow (e.g.,mountains,boulders) and some living organisms (e.g.,sterile worker bees,infertile human couples) cannot reproduce.
Therefore,interaction with the environment and the resulting progressive evolution is a fundamental trait that distinguishes living organisms from non-living matter.

(D) The biological organization of living organisms begins at the sub-microscopic molecular level. Atoms combine to form molecules,which then form macromolecules,organelles,and eventually cells. Therefore,the fundamental basis of life's organization starts at the molecular level.

(D) Living organisms exhibit several defining characteristics such as metabolism,cellular organization,and consciousness (response to stimuli).
While growth and reproduction are characteristics of living organisms,they are not defining properties because non-living things can also show growth (e.g.,accumulation of material) and some living organisms (like sterile worker bees or mules) cannot reproduce.
However,the ability to sense the environment and respond to stimuli (consciousness) is a defining property of all living organisms.
Among the given options,'Response to touch' (a form of consciousness/stimuli response) is the most appropriate characteristic that distinguishes living beings from non-living things,as non-living objects do not possess consciousness.

(D) The biological organization of living organisms begins at the sub-microscopic molecular level. Atoms combine to form molecules,which then form macromolecules,organelles,and eventually cells. Therefore,the fundamental basis of life's organization starts at the molecular level.

(D) 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 $B$ is water-soluble.

(A) Given:
Angular acceleration,$\alpha = 3.0\, rad/s^2$
Initial angular velocity,$\omega_i = 2.00\, rad/s$
Time,$t = 2\, s$
Using the kinematic equation for rotational motion:
$\theta = \omega_i t + \frac{1}{2} \alpha t^2$
Substituting the values:
$\theta = (2.00)(2) + \frac{1}{2}(3.0)(2)^2$
$\theta = 4 + \frac{1}{2}(3.0)(4)$
$\theta = 4 + 6 = 10\, rad$
Thus,the wheel has rotated through an angle of $10\, radians$.

(B) The satellite of mass $m$ is moving in a circular orbit of radius $r.$
$\therefore$ Kinetic energy of the satellite,$K = \frac{GMm}{2r} \dots (i)$
Potential energy of the satellite,$U = -\frac{GMm}{r} \dots (ii)$
Orbital speed of the satellite,$v = \sqrt{\frac{GM}{r}} \dots (iii)$
Time period of the satellite,$T = 2\pi \sqrt{\frac{r^3}{GM}} \dots (iv)$
Given $m_{S_1} = 4m_{S_2}$.
Since $M$ and $r$ are the same for both satellites $S_1$ and $S_2$:
From equation $(iii)$,the orbital speed $v$ is independent of the mass of the satellite. Therefore,both satellites move with the same speed.
From equation $(i)$,$K \propto m$,so $K_{S_1} = 4K_{S_2}$.
From equation $(ii)$,$U \propto m$,so $U_{S_1} = 4U_{S_2}$.
From equation $(iv)$,$T$ is independent of the mass of the satellite,so $T_{S_1} = T_{S_2}$.
Thus,statement $(b)$ is correct.

(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 $v$.

(D) The ferromagnetic property of a material is due to the alignment of atomic magnetic moments in the same direction.
When the temperature of a ferromagnetic material is increased beyond its Curie temperature $(T_c)$,the thermal agitation becomes strong enough to disrupt the long-range order of the magnetic moments.
As a result,the material loses its spontaneous magnetization and transitions into a paramagnetic state.
Therefore,nickel will show paramagnetism above its Curie temperature.

(C) : Calcium $(Ca)$ is an essential macronutrient for plant growth.
Macronutrients are essential elements required by plants in quantities greater than $1 \ mg/g$ of dry matter.
Calcium is used as calcium pectinate for the formation of the middle lamella in the cell wall,is involved in lipid metabolism,cell division,and cell enlargement,helps in the translocation of carbohydrates,and activates various enzymes in plants.
In contrast,$Zn$ (Zinc),$Cu$ (Copper),and $Mn$ (Manganese) are classified as micronutrients because they are required in very small amounts.

(A) : Serum globulins are proteins found in blood serum that contain most of the antibodies (immunoglobulins) of the blood.
Serum globulin electrophoresis is a laboratory test used to examine specific proteins in the blood.
Globulins are categorized into alpha,beta,and gamma globulins.
Since antibodies are primarily gamma globulins,a deficiency in these proteins serves as confirmatory evidence for an antibody deficiency.

(D) The correct answer is $D$.
Net primary productivity $(NPP)$ is defined as the total amount of organic matter stored by producers per unit area per unit time.
Gross primary productivity $(GPP)$ is the total organic matter synthesized by producers through photosynthesis per unit area per unit time.
$NPP = GPP - \text{Respiration losses}$.
Tropical rainforests are located in equatorial and sub-equatorial regions, which receive abundant sunlight and rainfall throughout the year.
Due to these favorable climatic conditions, these ecosystems exhibit the highest biological diversity and the highest annual net primary productivity compared to other ecosystem types.

(D) The weighted average atomic mass is calculated as:
$\text{Average atomic mass} = (\% \text{ abundance of isotope } 1 \times \text{mass of isotope } 1) + (\% \text{ abundance of isotope } 2 \times \text{mass of isotope } 2) + (\% \text{ abundance of isotope } 3 \times \text{mass of isotope } 3)$
$\text{Average atomic mass} = (0.90 \times 200) + (0.08 \times 199) + (0.02 \times 202)$
$\text{Average atomic mass} = 180.00 + 15.92 + 4.04 = 199.96 \ amu$
Rounding to the nearest whole number,the value is $200 \ amu$.

(B) The rules for quantum numbers are: $n > 0$,$0 \le l < n$,$-l \le m \le +l$,and $s = \pm 1/2$.
$(i)$ $(3, 2, 1, +1/2)$ is possible $(n=3, l=2, m=1)$.
$(ii)$ $(2, 2, 1, +1/2)$ is impossible because $l$ must be less than $n$ $(l < n)$.
$(iii)$ $(4, 3, -2, -1/2)$ is possible $(n=4, l=3, m=-2)$.
$(iv)$ $(1, 0, -1, -1/2)$ is impossible because $m$ must be between $-l$ and $+l$ (here $l=0$,so $m$ must be $0$).
$(v)$ $(3, 2, 3, +1/2)$ is impossible because $m$ must be between $-l$ and $+l$ (here $l=2$,so $m$ can only be $-2, -1, 0, 1, 2$).
Thus,sets $(ii), (iv)$ and $(v)$ are not possible.

(A) All the given species ($Ca^{2+}$,$K^{+}$,$Ar$,$Cl^{-}$,$S^{2-}$) are isoelectronic,meaning they all contain $18$ electrons.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number,$Z$) increases.
The atomic numbers are: $Ca$ $(Z=20)$,$K$ $(Z=19)$,$Ar$ $(Z=18)$,$Cl$ $(Z=17)$,$S$ $(Z=16)$.
Since the nuclear charge follows the order $Ca^{2+} > K^{+} > Ar > Cl^{-} > S^{2-}$,the ionic size follows the inverse order:
$Ca^{2+} < K^{+} < Ar < Cl^{-} < S^{2-}$.

(B) Ionisation enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
As the principal quantum number $(n)$ increases,the distance of the valence electron from the nucleus increases,leading to a decrease in the effective nuclear charge experienced by the electron.
Option $B$ $(1s^2 \ 2s^2 \ 2p^5 \ 3s^1)$ represents an atom with its valence electron in the $n=3$ shell,which is further from the nucleus compared to the electrons in the $n=2$ shell in the other options.
Therefore,the electron in the $3s$ orbital is held least tightly,resulting in the lowest ionisation enthalpy.

(A) Proton affinity is directly related to the basicity of the species.
In a period,as electronegativity increases,the ability of the atom to hold its lone pair increases,thus decreasing basicity. Comparing $NH_{2}^{-}$ and $F^{-}$,nitrogen is less electronegative than fluorine,so $NH_{2}^{-}$ is a stronger base.
In a group,as atomic size increases,the charge density decreases,which reduces the basicity. Comparing $F^{-}$ and $I^{-}$,$F^{-}$ is a stronger base.
Comparing $F^{-}$ and $HS^{-}$,$F^{-}$ is a stronger base because oxygen (in $OH^{-}$) is more electronegative than sulfur (in $HS^{-}$),and $F^{-}$ is smaller than $HS^{-}$.
Therefore,$NH_{2}^{-}$ has the highest proton affinity among the given species.

(C) To be isostructural,the species must have the same hybridization and the same geometry.
$(a)$ $SO_{3}^{2-}$ has $sp^{3}$ hybridization (pyramidal) and $NO_{3}^{-}$ has $sp^{2}$ hybridization (trigonal planar).
$(b)$ $BF_{3}$ has $sp^{2}$ hybridization (trigonal planar) and $NF_{3}$ has $sp^{3}$ hybridization (pyramidal).
$(c)$ $BrO_{3}^{-}$ has $sp^{3}$ hybridization with one lone pair (pyramidal) and $XeO_{3}$ has $sp^{3}$ hybridization with one lone pair (pyramidal). Thus,they are isostructural.
$(d)$ $SF_{4}$ has $sp^{3}d$ hybridization (see-saw) and $XeF_{4}$ has $sp^{3}d^{2}$ hybridization (square planar).

(C) Bond length is inversely proportional to bond order. Higher bond order results in shorter bond length.
$1.$ In $CO$, the bond order is $3$ $(:C \equiv O:^+)$, so the bond length is shortest $(112.8 \ pm)$.
$2.$ In $CO_2$, the bond order is $2$ $(O=C=O)$, so the bond length is intermediate $(122 \ pm)$.
$3.$ In $CO_3^{2-}$, the carbon atom is $sp^2$ hybridized and exhibits resonance. The bond order is $1.33$, which is the lowest, resulting in the longest bond length $(136 \ pm)$.
Therefore, the correct order of increasing bond length is $CO < CO_2 < CO_3^{2-}$.

(D) The standard enthalpy change of formation of a compound is the enthalpy change which occurs when one mole of the compound is formed from its elements in their standard states.
The equation representing the standard enthalpy of formation of $H_2O_{(l)}$ is the reaction where $1 \ mol$ of $H_2O_{(l)}$ is formed from its constituent elements,$H_{2_{(g)}}$ and $O_{2_{(g)}}$.
This is given by reaction $(ii)$: $H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_2 \ kJ \ mol^{-1}$.
Therefore,the enthalpy of formation of $H_2O_{(l)}$ is $-X_2 \ kJ \ mol^{-1}$.

(B) The formation reaction for $HCl$ is: $\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightarrow HCl(g)$.
The enthalpy of formation is given by: $\Delta H_f = [\sum \text{Bond Energy of Reactants}] - [\sum \text{Bond Energy of Products}]$.
Substituting the given values: $-90 = [\frac{1}{2} \times BE(H-H) + \frac{1}{2} \times BE(Cl-Cl)] - [BE(H-Cl)]$.
$-90 = [\frac{1}{2} \times 430 + \frac{1}{2} \times 240] - BE(H-Cl)$.
$-90 = [215 + 120] - BE(H-Cl)$.
$-90 = 335 - BE(H-Cl)$.
$BE(H-Cl) = 335 + 90 = 425 \ kJ \ mol^{-1}$.

(A) Given reactions:
$(1) \ N_2 + 3H_2 \rightleftharpoons 2NH_3 \quad K_1$
$(2) \ N_2 + O_2 \rightleftharpoons 2NO \quad K_2$
$(3) \ H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O \quad K_3$
Target reaction:
$(4) \ 2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O \quad K$
To obtain equation $(4)$,we perform the operation: $(2) + 3 \times (3) - (1)$.
Applying the rules of equilibrium constants:
$K = \frac{K_2 \times (K_3)^3}{K_1} = \frac{K_2 K_3^3}{K_1}$

(A) The dissociation of the weak acid is given by: $HA \rightleftharpoons H^{+} + A^{-}$
At equilibrium,let the concentration of $H^{+}$ be $x$. Then $[H^{+}] = [A^{-}] = x$ and $[HA] = 0.1 - x \approx 0.1$ (since $K_a$ is very small).
$K_a = \frac{[H^{+}][A^{-}]}{[HA]} = \frac{x^{2}}{0.1} = 1.00 \times 10^{-5}$
$x^{2} = 1.00 \times 10^{-6}$
$x = 1.00 \times 10^{-3} \ M$
Percentage of dissociation $\alpha = \frac{x}{C} \times 100 = \frac{1.00 \times 10^{-3}}{0.100} \times 100 = 1 \%$

(A) Given the concentration of hydronium ions: $[H_3O^+] = 1 \times 10^{-10} \, M$.
First,calculate the $pH$ of the solution:
$pH = -\log[H_3O^+] = -\log[10^{-10}] = 10$.
At $25 \, ^\circ C$,the relationship between $pH$ and $pOH$ is:
$pH + pOH = 14$.
Substituting the value of $pH$:
$10 + pOH = 14$.
Therefore,$pOH = 14 - 10 = 4$.

(A) The thermal stability of metal carbonates depends on the electropositive nature of the metal.
As the electropositive character of the metal increases,the thermal stability of its carbonate increases.
For Group $2$ elements,the electropositive character increases down the group $(Be < Mg < Ca)$. Thus,the thermal stability order is $BeCO_3 < MgCO_3 < CaCO_3$.
Group $1$ metals are more electropositive than Group $2$ metals,so $K_2CO_3$ is more stable than the alkaline earth metal carbonates.
Therefore,the correct order is $BeCO_3 < MgCO_3 < CaCO_3 < K_2CO_3$.

(A) The hydration energy of sulphates decreases as we move down the group $II$ in the periodic table.
$Mg^{2+}$ is the smallest ion among the group $II$ alkaline earth metal ions,which results in a very high charge density.
Due to its small size,the hydration energy of $Mg^{2+}$ is significantly high,making it greater than its lattice energy.
As a result,$MgSO_4$ is highly soluble in water,whereas the solubility of other sulphates like $SrSO_4$ and $BaSO_4$ decreases due to their lattice energy being higher than their hydration energy.

(A) The $ns^2$ electrons of the outermost shell in group $14$ elements may not participate in bonding due to the inert pair effect.
This effect increases down the group,making the lower oxidation state $(+2)$ more stable for heavier elements.
For $Sn$ (tin),the $+4$ oxidation state is more stable.
For $Pb$ (lead),the $+2$ oxidation state is more stable due to the significant inert pair effect.
Therefore,the most characteristic oxidation states for lead and tin are $+2$ and $+4$ respectively.

(B) In chain silicates,each $SiO_4^{4-}$ tetrahedron shares two oxygen atoms with other tetrahedra to form a long chain.
This results in the general formula $(SiO_3^{2-})_n$ for pyroxenes.
Another type of chain silicate is the double chain silicate (amphiboles),which has the formula $(Si_4O_{11}^{6-})_n$.

(B) To determine the $R$-configuration,we assign priorities to the groups attached to the chiral carbon based on the Cahn-Ingold-Prelog $(CIP)$ rules: $Cl(1) > C_2H_5(2) > CH_3(3) > H(4)$.
In the Fischer projection,the configuration is $R$ if the sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is clockwise,provided the lowest priority group $(H)$ is on a vertical bond.
Analyzing the structures,the configuration $R$ corresponds to the arrangement where the priority sequence is clockwise. Based on the provided image,the third structure shows the $R$-configuration where $Cl$ is on the left,$C_2H_5$ is on top,$CH_3$ is on the right,and $H$ is at the bottom. This corresponds to the structure $C_2H_5 - C(Cl)(CH_3) - H$ (Option $B$).

(C) The reactivity towards electrophilic aromatic substitution depends on the electron density of the benzene ring.
$1$. Phenol $(C_6H_5OH)$ has an $-OH$ group,which is a strong activating group due to its strong $+M$ (mesomeric) effect,making it the most reactive.
$2$. Toluene $(C_6H_5CH_3)$ has a $-CH_3$ group,which is an activating group due to $+I$ (inductive) and hyperconjugation effects,making it more reactive than benzene.
$3$. Benzene $(C_6H_6)$ is the reference compound.
$4$. Chlorobenzene $(C_6H_5Cl)$ has a $-Cl$ group,which is deactivating due to its strong $-I$ effect,although it is ortho/para-directing due to the $+M$ effect. Overall,it is less reactive than benzene.
Thus,the correct order of decreasing reactivity is: $(iv) > (ii) > (i) > (iii)$.

(C) In the reaction of $1-$butyne with $HCl$,Markovnikov addition occurs to form $2-$chloro-$1-$butene $(B)$.
Subsequent addition of $HI$ to $B$ also follows Markovnikov's rule,where the iodine atom attaches to the carbon already containing the chlorine atom,resulting in $2-$chloro-$2-$iodobutane $(C)$.
Reaction:
$CH_3-CH_2-C \equiv CH + HCl \rightarrow CH_3-CH_2-CCl=CH_2 (B)$
$CH_3-CH_2-CCl=CH_2 + HI \rightarrow CH_3-CH_2-C(I)(Cl)-CH_3 (C)$

(D) $2-$Methyl$-2-$butene (molecular formula $C_5H_{10}$) yields acetone on ozonolysis.
$3-$Methyl$-1-$butene $(CH_3-CH(CH_3)-CH=CH_2)$ $\xrightarrow[(ii) Zn/H_2O]{(i) O_3}$ $2-$Methylpropanal $(CH_3-CH(CH_3)-CHO)$ + Formaldehyde $(CH_2O)$
Cyclopentane $(C_5H_{10})$ $\xrightarrow[(ii) Zn/H_2O]{(i) O_3}$ $\text{No reaction}$
$2-$Methyl$-1-$butene $(CH_3-CH_2-C(CH_3)=CH_2)$ $\xrightarrow[(ii) Zn/H_2O]{(i) O_3}$ Butanone $(CH_3-CH_2-CO-CH_3)$ + Formaldehyde $(CH_2O)$
$2-$Methyl$-2-$butene $(CH_3-C(CH_3)=CH-CH_3)$ $\xrightarrow[(ii) Zn/H_2O]{(i) O_3}$ Acetone $(CH_3-CO-CH_3)$ + Acetaldehyde $(CH_3CHO)$

(C) First,calculate the molarity $(M_1)$ of the concentrated acid:
$M_1 = \frac{\text{density} \times 10 \times \% \text{ by mass}}{\text{molar mass}} = \frac{1.80 \times 10 \times 98}{98} = 18 \ M$.
Now,use the dilution formula $M_1 V_1 = M_2 V_2$:
$18 \times V_1 = 0.1 \times 1000 \ mL$.
$V_1 = \frac{100}{18} \approx 5.55 \ mL$.

(D) The acidic strength of oxyacids increases with an increase in the oxidation state of the central atom.
The oxidation states of chlorine in the given oxyacids are:
$HOCl$ $(+1)$,
$HOClO$ $(+3)$,
$HOClO_2$ $(+5)$,
$HOClO_3$ $(+7)$.
Since the oxidation state increases from $HOCl$ to $HOClO_3$,the acidic strength increases in the order: $HOCl < HOClO < HOClO_2 < HOClO_3$.

(C) compound that does not rotate plane polarised light is optically inactive.
If a compound is thought to be chiral but shows no optical rotation,it could be a racemic mixture (an equimolar mixture of two enantiomers where the rotation of one is cancelled by the other,i.e.,externally compensated) or it could be a meso compound (which contains chiral centers but is achiral due to an internal plane of symmetry,i.e.,internally compensated).
Therefore,the observation of no rotation does not definitively prove the compound is meso; it could also be a racemic mixture.

(D) Let us analyze each option:
$1$. Floating ribs in humans: There are $2$ pairs ($4$ total) of floating ribs (the $11^{th}$ and $12^{th}$ pairs). However,the question asks for the total number of items,and usually,in biological contexts,we refer to pairs or specific counts. Let's re-evaluate.
$2$. Amino acids found in proteins: There are $20$ standard amino acids found in proteins,not $16$.
$3$. Types of diabetes: There are two main types of diabetes mellitus ($Type-1$ and $Type-2$),not $3$.
$4$. Cervical vertebrae in humans: All mammals,including humans,have exactly $7$ cervical vertebrae. This is a biologically constant and correct fact.
Therefore,option $D$ is the correct statement.

(A) Slime moulds are saprophytic protists. The body moves along decaying twigs and leaves engulfing organic material. Under suitable conditions,they form an aggregation called $Plasmodium$ which may grow and spread over several feet. Under unfavourable conditions,the $Plasmodium$ differentiates and forms fruiting bodies bearing spores at their tips. Therefore,$Plasmodium$ is a well-known example of a slime mould. $Thiobacillus$ is a bacterium,$Anabaena$ is a cyanobacterium,and $Rhizopus$ is a fungus (bread mould).

(D) The classification of algae into major groups (Chlorophyceae,Phaeophyceae,and Rhodophyceae) is primarily based on the types of photosynthetic pigments present in their cells. These pigments determine the characteristic color of the algae and are essential for their photosynthetic activity. While other factors like stored food and cell wall composition are also used,the primary diagnostic character for classification is the pigment composition.

(A) Let the allele for yellow seeds be $Y$ and the allele for green seeds be $y$.
Since yellow is dominant over green,the genotype for a heterozygous yellow-seeded plant is $Yy$.
The genotype for a green-seeded plant is $yy$ (as it is a recessive trait).
When we perform a test cross between $Yy$ and $yy$:
$Yy \times yy \rightarrow Yy, Yy, yy, yy$.
The resulting offspring are $50\%$ yellow-seeded $(Yy)$ and $50\%$ green-seeded $(yy)$.
Therefore,the ratio of yellow to green-seeded plants is $1:1$,which is equivalent to $50:50$.

(B) Sulphide ores are generally concentrated by the froth floatation process. However,$Argentite$ $(Ag_2S)$ is an exception.
Because silver is a precious metal,a significant amount of ore can be lost during the froth floatation process as it may remain at the bottom of the tank.
Therefore,$Argentite$ is concentrated by chemical leaching using a dilute solution of sodium cyanide $(NaCN)$ in the presence of air $(O_2)$,which forms a soluble complex: $4Ag_2S + 8NaCN + 2H_2O + O_2 \rightarrow 8Na[Ag(CN)_2] + 4NaOH + 2S$.
Thus,option $B$ is correct.

(C) The induced $emf$ in the primary coil is given by Faraday's law: $E_P = \frac{d\phi}{dt}$.
Substituting the given flux $\phi = \phi_0 + 4t$,we get $E_P = \frac{d}{dt}(\phi_0 + 4t) = 4 \, V$.
For an ideal transformer,the ratio of the secondary voltage $E_S$ to the primary voltage $E_P$ is equal to the ratio of the number of turns in the secondary coil $N_S$ to the number of turns in the primary coil $N_P$.
$\frac{E_S}{E_P} = \frac{N_S}{N_P}$.
Given $N_P = 50$ and $N_S = 1500$,we have $\frac{E_S}{4} = \frac{1500}{50}$.
$\frac{E_S}{4} = 30$.
Therefore,$E_S = 30 \times 4 = 120 \, V$.

(C) Given that the concentration of hydronium ions is $[H_3O^+] = 1 \times 10^{-10} \, M$.
At $25\,^{\circ}C$,the ionic product of water is $K_w = [H_3O^+][OH^-] = 10^{-14}$.
Therefore,the concentration of hydroxide ions is $[OH^-] = \frac{K_w}{[H_3O^+]} = \frac{10^{-14}}{10^{-10}} = 10^{-4} \, M$.
The $pOH$ is calculated as $pOH = -\log[OH^-] = -\log(10^{-4}) = 4$.

(A) The work done in moving a charge $Q$ from point $C$ to point $D$ is given by $W = Q(V_D - V_C)$.
Point $C$ is the midpoint of $AB$ (distance $2L$),so $AC = L$ and $CB = L$. The potential at $C$ due to charges $+q$ at $A$ and $-q$ at $B$ is $V_C = \frac{1}{4\pi \epsilon_0} (\frac{q}{L} - \frac{q}{L}) = 0$.
Point $D$ is on the semicircle with diameter $CD$. Since $C$ is the midpoint of $AB$,$D$ is at a distance $3L$ from $A$ and $L$ from $B$. The potential at $D$ is $V_D = \frac{1}{4\pi \epsilon_0} (\frac{q}{3L} - \frac{q}{L}) = \frac{1}{4\pi \epsilon_0} (\frac{q - 3q}{3L}) = -\frac{2q}{12\pi \epsilon_0 L} = -\frac{q}{6\pi \epsilon_0 L}$.
Therefore,$W = Q(V_D - V_C) = Q(-\frac{q}{6\pi \epsilon_0 L} - 0) = -\frac{qQ}{6\pi \epsilon_0 L}$.

(C) For a weak acid $HA$,the dissociation equilibrium is $HA \rightleftharpoons H^+ + A^-$.
The dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2 / (1 - \alpha)$.
Since the acid is weak,$\alpha \ll 1$,so we approximate $K_a \approx C \alpha^2$.
Given $K_a = 1.00 \times 10^{-5}$ and $C = 0.100 \ M$.
$1.00 \times 10^{-5} = 0.100 \times \alpha^2$.
$\alpha^2 = 1.00 \times 10^{-4}$.
$\alpha = 0.01$.
The percentage of dissociation is $\alpha \times 100 = 0.01 \times 100 = 1 \%$.

(B) The stability of transition metal ions in aqueous solution depends on the electronic configuration and crystal field stabilization energy.
For $Cr^{3+}$ $(3d^3)$,the configuration is $t_{2g}^3$ in an octahedral field,which is a half-filled $t_{2g}$ subshell,providing extra stability.
$Mn^{3+}$ $(3d^4)$ is less stable due to the Jahn-Teller distortion.
$V^{3+}$ $(3d^2)$ and $Ti^{3+}$ $(3d^1)$ do not have the extra stability associated with the $t_{2g}^3$ configuration.
Therefore,$Cr^{3+}$ is the most stable ion in aqueous solution.

(B) The orbital velocity of a satellite is given by $V_0 = \sqrt{\frac{GM}{r}}$. Since both satellites are in the same orbit,they have the same orbital radius $r$. Thus,both move with the same speed.
Potential energy is given by $U = -\frac{GM_e m}{r}$. Since the masses $m$ of the satellites are different,their potential energies are different.
Kinetic energy is given by $K = \frac{1}{2}mv^2$. Since the masses are different,their kinetic energies are different.
The time period is given by $T = 2\pi \sqrt{\frac{r^3}{GM}}$. Since $r$ is the same for both,their time periods are equal.
Therefore,the statement that $S_1$ and $S_2$ are moving with the same speed is true.

(A) The solubility of an ionic compound occurs when the hydration energy $(H.E.)$ is greater than the lattice energy $(L.E.)$.
For alkaline earth metal sulfates,the solubility decreases down the group as the size of the cation increases.
Therefore,the solubility order is $MgSO_4 > CaSO_4 > SrSO_4 > BaSO_4 > RaSO_4$.
Among the given options,$MgSO_4$ has the highest solubility,which implies that its hydration energy is significantly higher than its lattice energy compared to the others.

(B) The net work done on the ball is the sum of the work done by the gravitational force and the work done by the spring force.
$1$. Work done by gravity $(W_g)$: The total vertical displacement of the ball is $(h + d)$ downwards. Since the force of gravity acts downwards,$W_g = mg(h + d)$.
$2$. Work done by the spring $(W_s)$: The spring force acts upwards while the displacement is downwards. The work done by a spring is given by $-\frac{1}{2}Kd^2$.
$3$. Net work done $(W_{net})$: $W_{net} = W_g + W_s = mg(h + d) - \frac{1}{2}Kd^2$.

(B) The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$.
In an octahedral field,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.
For $Cr^{3+}$,the three electrons occupy the $t_{2g}$ orbitals,resulting in a half-filled $t_{2g}^3$ configuration,which provides extra stability in aqueous solution.

(A) Nucleophilicity is the ability of a species to donate an electron pair to an electrophile.
In a polar protic solvent,nucleophilicity increases down the group because the smaller ions are more heavily solvated,which hinders their ability to act as nucleophiles.
Therefore,the order of nucleophilicity for halide ions is $Cl^{-} < Br^{-} < I^{-}$.

(D) Freons or chlorofluorocarbons $(CFCs)$ are responsible for the depletion of the ozone layer in the upper strata of the atmosphere.
They are used as propellants,aerosol spray caps,refrigerants,and fire-fighting reagents.
They are stable and chemically inert compounds that absorb $UV$ radiation and break down,liberating free atomic chlorine.
This chlorine atom causes the decomposition of ozone through a free radical reaction,resulting in the depletion of the ozone layer.
Common freons include $Freon-11$ $(CFCl_3)$ and $Freon-12$ $(CF_2Cl_2)$.
The reaction mechanism is as follows:
$Cl^{\bullet} + O_3 \longrightarrow ClO^{\bullet} + O_2$
$ClO^{\bullet} + O_3 \longrightarrow Cl^{\bullet} + 2O_2$

(A) For a simple cubic unit cell,the number of atoms per unit cell $(Z)$ is $1$.
The relationship between the edge length $(a)$ and the atomic radius $(r)$ is $a = 2r$ or $r = \frac{a}{2}$.
The volume of one atom is $V_{atom} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (\frac{a}{2})^3 = \frac{4}{3} \pi \frac{a^3}{8} = \frac{\pi a^3}{6}$.
The packing fraction is defined as the ratio of the volume of atoms to the total volume of the unit cell $(a^3)$:
$\text{Packing Fraction} = \frac{Z \times V_{atom}}{a^3} = \frac{1 \times \frac{\pi a^3}{6}}{a^3} = \frac{\pi}{6}$.

(B) Doping $NaCl$ with $10^{-4} \ mol \%$ of $SrCl_2$ means that $100 \ mol$ of $NaCl$ contains $10^{-4} \ mol$ of $SrCl_2.$
Therefore,$1 \ mol$ of $NaCl$ contains $10^{-4} / 100 = 10^{-6} \ mol$ of $SrCl_2.$
Each $Sr^{2+}$ ion replaces two $Na^+$ ions to maintain electrical neutrality,creating one cation vacancy per $Sr^{2+}$ ion introduced.
Concentration of cation vacancies $= (10^{-6} \ mol \ SrCl_2 / mol \ NaCl) \times (N_A \ ions / mol) \times (1 \ \text{vacancy} / Sr^{2+} \ \text{ion}).$
$= 10^{-6} \times 6.02 \times 10^{23} \ \text{vacancies} \ mol^{-1} = 6.02 \times 10^{17} \ mol^{-1}.$

(B) The depression in freezing point is given by the formula: $\Delta T_{f} = i \times K_{f} \times m$.
For the dissociation of weak acid $HX \rightleftharpoons H^{+} + X^{-}$,the van't Hoff factor $i$ is calculated as $i = 1 + \alpha$,where $\alpha$ is the degree of dissociation.
Given $\alpha = 20\% = 0.2$,so $i = 1 + 0.2 = 1.2$.
Given molality $m = 0.5\, mol\, kg^{-1}$ and $K_{f} = 1.86\, K\, kg\, mol^{-1}$.
Substituting these values: $\Delta T_{f} = 1.2 \times 1.86 \times 0.5 = 1.116\, K \approx 1.12\, K$.

(C) For a cell reaction in equilibrium at $298 \ K$,the relationship between the standard cell potential and the equilibrium constant is given by:
$E_{cell}^{\circ} = \frac{0.0591}{n} \log K_{C}$
where $K_{C}$ is the equilibrium constant and $n$ is the number of electrons transferred in the balanced redox reaction.
For the reaction $Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$,the number of electrons transferred is $n = 2$.
Given $E_{cell}^{\circ} = 0.46 \ V$,we substitute the values into the equation:
$0.46 = \frac{0.0591}{2} \log K_{C}$
$\log K_{C} = \frac{0.46 \times 2}{0.0591} = \frac{0.92}{0.0591} \approx 15.567$
$K_{C} = 10^{15.567} \approx 3.69 \times 10^{15} \approx 4.0 \times 10^{15}$

(B) The efficiency of a fuel cell $(\phi)$ is defined as the ratio of the useful work done (Gibbs free energy change,$\Delta G$) to the total heat energy available from the combustion of the fuel (Enthalpy change,$\Delta H$).
Efficiency $(\phi) = \frac{\Delta G}{\Delta H} \times 100$.
Ideally,fuel cells are expected to have an efficiency of $100 \%$.

(A) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
Given that $60\%$ is completed in $60 \ min$,$[A]_t = 100 - 60 = 40$.
$k = \frac{2.303}{60} \log \frac{100}{40} = \frac{2.303}{60} \log 2.5$.
Using $\log 2.5 = \log(10/4) = \log 10 - \log 4 = 1 - 0.60 = 0.40$.
$k = \frac{2.303 \times 0.40}{60} = \frac{0.9212}{60} \ min^{-1}$.
For $50\%$ completion $(t_{1/2})$,$t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693 \times 60}{0.9212} \approx 45.12 \ min$.
Thus,the reaction is completed in approximately $45 \ min$.

(C) For a first-order reaction,the rate constant $k$ is given by the equation:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
At half-life $(t_{1/2})$,the concentration of the reactant $[A]_t = \frac{[A]_0}{2}$.
Substituting this into the equation:
$t_{1/2} = \frac{2.303}{k} \log \frac{[A]_0}{[A]_0 / 2}$
$t_{1/2} = \frac{2.303}{k} \log 2$
Since $2.303 \log 2 = \ln 2$,we get:
$t_{1/2} = \frac{\ln 2}{k}$
The half-life of a first-order reaction is independent of the initial concentration of the reactant.

(D) The rate law for the reaction is given as $Rate = k[H_2]^1[ICl]^1$.
For Mechanism $A$,which is a single-step elementary reaction,the rate law would be $Rate = k[H_2][ICl]^2$. This does not match the given rate law.
For Mechanism $B$,the rate-determining step is the slow step: $H_{2(g)} + ICl_{(g)} \rightarrow HCl_{(g)} + HI_{(g)}$.
The rate law for this elementary step is $Rate = k[H_2][ICl]$.
This matches the given rate law,which is first order with respect to both $H_2$ and $ICl$. Therefore,only Mechanism $B$ is consistent with the given information.

(A) The main assumptions of the Langmuir adsorption isotherm are:
$(i)$ The surface of the solid is homogeneous,meaning all adsorption sites are equivalent in their ability to adsorb the particles.
$(ii)$ Adsorption is limited to the formation of a unimolecular layer (monolayer).
$(iii)$ There are no lateral interactions between the adsorbed molecules.
$(iv)$ The heat of adsorption is constant and independent of the surface coverage.

(D) Froth Floatation Process is generally used for the concentration of sulphide ores based on the differential wetting properties of ore and gangue particles with oil and water.
However,$Argentite$ $(Ag_{2}S)$ is an exception as it is concentrated by chemical leaching (MacArthur-Forrest cyanide process).
The chemical reactions involved are:
$Ag_{2}S + 4NaCN \rightarrow 2Na[Ag(CN)_{2}] + Na_{2}S$
Silver is subsequently recovered by displacement using $Zn$:
$2Na[Ag(CN)_{2}] + Zn \rightarrow Na_{2}[Zn(CN)_{4}] + 2Ag$

(D) The standard free energies of formation $(\Delta G_f^o)$ of most metal sulphides are greater than those of $CS_2$ and $H_2S$.
This means that carbon and hydrogen cannot effectively reduce metal sulphides to their respective metals.
However,the standard free energies of formation of metal oxides are significantly lower than those of $SO_2$.
Therefore,the oxidation of metal sulphides to metal oxides is thermodynamically favourable.
Thus,roasting the sulphide ore to an oxide before reduction is a necessary step in the extraction process.
Consequently,the statement that carbon and hydrogen are suitable reducing agents for metal sulphides is false.

(D) The stability of ions in aqueous solution depends on their electronic configuration and hydration energy.
For $Ti^{3+}$ $(3d^1)$,$V^{3+}$ $(3d^2)$,$Cr^{3+}$ $(3d^3)$,and $Mn^{3+}$ $(3d^4)$,the $Cr^{3+}$ ion is particularly stable due to its half-filled $t_{2g}$ subshell $(t_{2g}^3 e_g^0)$ in an octahedral field.
$Mn^{3+}$ is unstable in aqueous solution because it easily reduces to $Mn^{2+}$ $(3d^5)$,which has a stable half-filled $d$-orbital configuration.
Therefore,$Cr^{3+}$ is the most stable among the given options in aqueous solution.

(B) The regular decrease in the radii of lanthanide ions from $La^{3+}$ to $Lu^{3+}$ is known as lanthanide contraction.
It is due to the poor shielding effect of $4f$ electrons,which allows the increased nuclear charge to pull the electrons closer to the nucleus.
As a result of lanthanide contraction,the atomic radii of elements of the $4d$ and $5d$ series become very similar,leading to similarities in their chemical properties.
Therefore,the statement that $4d$ and $5d$ series have no similarities is incorrect.

(B) Enantiomorphs (or enantiomers) are non-superimposable mirror images of each other.
For a coordination complex to exhibit optical isomerism (enantiomerism),it must lack a plane of symmetry and a center of inversion.
In the complex $[Co(en)_2Cl_2]^+$,the $cis$-isomer lacks a plane of symmetry and thus exists as a pair of enantiomorphs ($d$ and $l$ forms).
The $trans$-isomer of $[Co(en)_2Cl_2]^+$ possesses a plane of symmetry and is optically inactive.
Other options like $[Cr(NH_3)_6][Co(CN)_6]$,$[Pt(NH_3)_4][PtCl_6]$,and $[Co(NH_3)_4Cl_2]NO_2$ do not exhibit optical isomerism due to the presence of planes of symmetry in their structures.

(B) Paramagnetic behaviour is directly proportional to the number of unpaired electrons present in the metal ion.
$H_2O$ is a weak field ligand,so it does not cause pairing of electrons in these $3d$ metal ions.
$1$. For $[Cr(H_2O)_6]^{2+}$: $Cr^{2+}$ is $3d^4$,having $4$ unpaired electrons.
$2$. For $[Mn(H_2O)_6]^{2+}$: $Mn^{2+}$ is $3d^5$,having $5$ unpaired electrons.
$3$. For $[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$,having $4$ unpaired electrons.
$4$. For $[Ni(H_2O)_6]^{2+}$: $Ni^{2+}$ is $3d^8$,having $2$ unpaired electrons.
Since $[Ni(H_2O)_6]^{2+}$ has the minimum number of unpaired electrons $(2)$,it exhibits the minimum paramagnetic behaviour.

(C) The reaction of an ether with $HI$ follows the $S_N2$ mechanism when the alkyl groups are primary or secondary.
The iodide ion $(I^-)$ attacks the less sterically hindered alkyl group.
In isobutyl ethyl ether $(CH_3-CH(CH_3)-CH_2-O-CH_2-CH_3)$,the ethyl group is less sterically hindered than the isobutyl group.
Therefore,the $I^-$ ion attacks the ethyl group,resulting in the formation of isobutyl alcohol $(CH_3-CH(CH_3)-CH_2OH)$ and ethyl iodide $(CH_3CH_2I)$.

(D) The reduction of aldehydes or ketones to alkanes using zinc amalgam $(Zn(Hg))$ and concentrated hydrochloric acid $(HCl)$ is known as the Clemmensen reduction.
In this reaction,the carbonyl group $(>C=O)$ is reduced to a methylene group $(-CH_2-)$.

(A) Aldehydes that do not possess $\alpha$-hydrogen atoms,such as benzaldehyde $(C_6H_5CHO)$,undergo self-oxidation and reduction (disproportionation) when treated with concentrated alkali ($50\%$ aqueous $NaOH$).
This reaction is known as the Cannizzaro reaction.
It yields the corresponding alcohol (benzyl alcohol) and the salt of the corresponding acid (sodium benzoate).
The reaction is: $2C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa$.

(A) Aldol condensation involves the reaction between two molecules of an aldehyde or a ketone containing at least one $\alpha-$hydrogen atom in the presence of a dilute base.
This reaction results in the formation of a $\beta-$hydroxy aldehyde (aldol) or a $\beta-$hydroxy ketone (ketol).
These products can further undergo dehydration upon heating to form $\alpha, \beta-$unsaturated carbonyl compounds.
Therefore,the primary product formed in the initial step of Aldol condensation is a $\beta-$hydroxy aldehyde or a $\beta-$hydroxy ketone.

(C) The reactivity of acid chlorides towards hydrolysis depends on the magnitude of the positive charge on the carbonyl carbon atom.
Electron-withdrawing groups $(EWG)$ increase the positive charge on the carbonyl carbon,thereby increasing reactivity.
Electron-donating groups $(EDG)$ decrease the positive charge on the carbonyl carbon,thereby decreasing reactivity.
The substituents at the para-position are:
$(i)$ $-H$ (no effect)
$(ii)$ $-NO_2$ (strong $EWG$)
$(iii)$ $-CH_3$ ($EDG$ via hyperconjugation)
$(iv)$ $-CHO$ $(EWG)$
Comparing the strength of $EWG$,$-NO_2$ is a stronger electron-withdrawing group than $-CHO$.
Thus,the order of reactivity is: $(ii) > (iv) > (i) > (iii)$.

(C) The acidity of carboxylic acids is influenced by the inductive effect of the substituent attached to the alpha-carbon.
Electron-withdrawing groups (EWGs) increase the acidity by stabilizing the carboxylate anion through the $-I$ effect.
The strength of the $-I$ effect depends on the electronegativity of the halogen atom: $F > Cl > Br$.
Therefore,the acidity order is $FCH_2COOH > ClCH_2COOH > BrCH_2COOH > CH_3COOH$.

(A) $CH_{3}NC + 4[H] \xrightarrow{LiAlH_{4}} CH_{3}NHCH_{3}$
$CH_{3}NC$ (Methyl isocyanide) on reduction with lithium aluminium hydride $(LiAlH_{4})$ yields $CH_{3}NHCH_{3}$ (Dimethylamine),which is a secondary $(2^{\circ})$ amine.
In contrast,reduction of alkyl cyanides $(RCN)$ with $LiAlH_{4}$ yields primary $(1^{\circ})$ amines $(RCH_{2}NH_{2})$.
Reduction of acetamide $(CH_{3}CONH_{2})$ yields ethylamine $(CH_{3}CH_{2}NH_{2})$,a primary amine.
Reduction of nitroethane $(CH_{3}CH_{2}NO_{2})$ yields ethylamine $(CH_{3}CH_{2}NH_{2})$,a primary amine.

(C) $RNA$ and $DNA$ molecules contain ribose and deoxyribose sugars, respectively.
Both sugars possess multiple chiral centers.
The chirality of these nucleic acids is primarily attributed to the presence of the $D$-sugar component (specifically $D$-ribose in $RNA$ and $D$-2-deoxyribose in $DNA$).

(D) Vitamins are classified into two categories based on their solubility:
$(I)$ Fat-soluble vitamins: $A, D, E, K$.
$(II)$ Water-soluble vitamins: Vitamin $B$ complex and Vitamin $C$.
Therefore,Vitamin $B$ is water-soluble.

(D) Condensation Polymerization: This is a process of polymer formation by the reaction of smaller molecules to form bigger ones,with the elimination of small molecules such as $H_{2}O$ or methanol.
For example: In the formation of $Nylon-6,6$,small molecules of hexamethylenediamine and adipic acid react,resulting in the removal of a water molecule for each condensation step between the two monomers.