AIEEE 2003 Chemistry Question Paper with Answer and Solution

(A) The work done $W$ in rotating a magnetic needle from an angle $\theta_1$ to $\theta_2$ in a magnetic field $B$ is given by $W = MB(\cos \theta_1 - \cos \theta_2)$.
Given $\theta_1 = 0^{\circ}$ and $\theta_2 = 60^{\circ}$,we have:
$W = MB(\cos 0^{\circ} - \cos 60^{\circ}) = MB(1 - 0.5) = 0.5 MB = \frac{MB}{2}$.
Thus,$MB = 2W$.
The torque $\tau$ required to maintain the needle at an angle $\theta = 60^{\circ}$ is given by $\tau = MB \sin \theta$.
Substituting the values,$\tau = MB \sin 60^{\circ} = (2W) \times \frac{\sqrt{3}}{2} = \sqrt{3} W$.

(C) The electromagnetic spectrum is ordered by wavelength. Among the given options,$\gamma$-rays (gamma rays) are electromagnetic waves with the highest frequency and,consequently,the shortest wavelength. $X$-rays have longer wavelengths than $\gamma$-rays. $\alpha$-rays and $\beta$-rays are particles (helium nuclei and electrons,respectively),not electromagnetic radiation,and thus do not have a wavelength in the same context as photons.

(C) Optical fibres operate on the principle of total internal reflection of light. Because they transmit signals as light pulses rather than electrical currents,they are immune to electromagnetic interference from external sources. Therefore,the statement that they are subject to electromagnetic interference is false.

(C) The balanced chemical equation for the reduction of boron trichloride by hydrogen is:
$2BCl_3(g) + 3H_2(g) \to 2B(s) + 6HCl(g)$
First,calculate the number of moles of Boron $(B)$ produced:
$n(B) = \frac{\text{mass}}{\text{atomic mass}} = \frac{21.6 \ g}{10.8 \ g/mol} = 2 \ mol$
From the stoichiometry of the reaction,$2 \ mol$ of $B$ are produced by $3 \ mol$ of $H_2$ gas.
Since $2 \ mol$ of $B$ are produced,the moles of $H_2$ required is $3 \ mol$.
At $STP$ ($273 \ K$ and $1 \ atm$),$1 \ mol$ of any ideal gas occupies $22.4 \ L$.
Therefore,the volume of $H_2$ required is:
$V = 3 \ mol \times 22.4 \ L/mol = 67.2 \ L$

(B) Isoelectronic species are those that have the same number of electrons.
For $N^{3-}$,the number of electrons is $7 + 3 = 10$.
For $F^{-}$,the number of electrons is $9 + 1 = 10$.
For $Na^{+}$,the number of electrons is $11 - 1 = 10$.
Since all three species have $10$ electrons,they are isoelectronic.

(B) The Balmer series corresponds to electronic transitions ending at the $n = 2$ energy level.
Lines in the series are ordered by increasing energy and decreasing wavelength starting from the red end.
The first line (red end) corresponds to the transition $n = 3 \to n = 2$.
The second line corresponds to the transition $n = 4 \to n = 2$.
The third line corresponds to the transition $n = 5 \to n = 2$.

(C) The equilibrium constant $K_c$ for the reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ is given by the expression:
$K_c = \frac{[NO_2]^2}{[N_2O_4]}$
Given:
$[NO_2] = 1.2 \times 10^{-2} \ mol \ L^{-1}$
$[N_2O_4] = 4.8 \times 10^{-2} \ mol \ L^{-1}$
Substituting the values:
$K_c = \frac{(1.2 \times 10^{-2})^2}{4.8 \times 10^{-2}} = \frac{1.44 \times 10^{-4}}{4.8 \times 10^{-2}}$
$K_c = 0.3 \times 10^{-2} = 3 \times 10^{-3} \ mol \ L^{-1}$

(C) The given reaction is $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ with $\Delta H^\circ = -198 \ kJ$.
Since $\Delta H^\circ < 0$,the reaction is exothermic.
According to Le Chatelier's principle,for an exothermic reaction,a decrease in temperature favors the forward reaction.
Regarding pressure,the number of moles of gaseous products is $2$,while the number of moles of gaseous reactants is $2 + 1 = 3$.
Since the number of moles decreases in the forward direction $(3 \rightarrow 2)$,an increase in pressure favors the forward reaction.
Therefore,lowering the temperature and increasing the pressure are the favorable conditions.

(C) Proton affinity is directly related to the basicity of a substance.
Among the given compounds,$NH_3$ is the strongest Lewis base because the lone pair on the nitrogen atom is more available for donation compared to the lone pairs on $O$,$S$,or $P$.
Therefore,$NH_3$ has the highest affinity for hydrogen ions $(H^+)$.

(A) The $Br\text{ø}nsted-Lowry$ theory defines acids as proton $(H^+)$ donors and bases as proton acceptors.
$H_2SO_4$ and $HNO_3$ act as proton donors,fitting the $Br\text{ø}nsted-Lowry$ definition.
$AlCl_3$ and $SO_2$ do not contain ionizable protons and act as Lewis acids (electron pair acceptors),which is explained by the Lewis theory,not the $Br\text{ø}nsted-Lowry$ theory.
Since both $AlCl_3$ and $SO_2$ are listed as options,and the question asks for an example not covered by the theory,both are technically correct; however,$AlCl_3$ is the classic example of a Lewis acid.

(A) The dissociation of the salt is given by: $AB_2(s) \rightleftharpoons A^{2+}(aq) + 2B^{-}(aq)$
Let the solubility be $s = 1.0 \times 10^{-5} \ mol \ L^{-1}$.
The concentrations of the ions are $[A^{2+}] = s$ and $[B^{-}] = 2s$.
The solubility product constant $K_{sp}$ is defined as: $K_{sp} = [A^{2+}][B^{-}]^2$
Substituting the values: $K_{sp} = (s)(2s)^2 = 4s^3$
Calculating the value: $K_{sp} = 4 \times (1.0 \times 10^{-5})^3 = 4 \times 10^{-15}$

(A) During a thunderstorm,high energy lightning causes atmospheric nitrogen and oxygen to react to form nitrogen oxides ($NO$ and $NO_2$).
These oxides dissolve in rainwater to form nitric acid $(HNO_3)$ and nitrous acid $(HNO_2)$.
The formation of these acids increases the concentration of $H^+$ ions in the rainwater.
Consequently,the acidity of the rainwater increases,which leads to a decrease in the $pH$ value compared to normal rainwater.

(C) Statement $(C)$ is incorrect.
For a very dilute acid solution like $1 \times 10^{-8} \ M \ HCl$,the contribution of $H^+$ ions from water cannot be neglected.
The total $[H^+] = 10^{-8} + 10^{-7} \approx 1.1 \times 10^{-7} \ M$.
Therefore,$pH = -\log(1.1 \times 10^{-7}) \approx 6.96$,which is slightly acidic,not $8$ (which is basic).

(C) According to Hess's Law,the enthalpy change for a chemical reaction is a state function.
This means it depends only on the initial and final states of the system and is independent of the path taken or the nature of intermediate reaction steps.

(D) The relationship between standard Gibbs free energy change $(\Delta G^o)$ and the equilibrium constant $(K_c)$ is derived from the equation: $\Delta G = \Delta G^o + RT \ln Q$.
At equilibrium,the reaction quotient $Q = K_c$ and the change in Gibbs free energy $\Delta G = 0$.
Substituting these values into the equation gives: $0 = \Delta G^o + RT \ln K_c$.
Rearranging this,we get: $\Delta G^o = -RT \ln K_c$ or $-\Delta G^o = RT \ln K_c$.

(B) For any spontaneous (irreversible) process in an isolated system,the change in entropy is positive,i.e.,$(dS)_{V,E} > 0$.
For a process occurring at constant temperature $(T)$ and pressure $(P)$ involving only pressure-volume work,the spontaneity criterion is given by the change in Gibbs free energy,which must be negative,i.e.,$(dG)_{T,P} < 0$.
Therefore,the correct criteria are $(dS)_{V,E} > 0$ and $(dG)_{T,P} < 0$.

(D) The reaction is: $H_2C=CH_{2(g)} + H_{2(g)} \to H_3C-CH_{3(g)}$
$\Delta H = \sum \text{Bond energy of reactants} - \sum \text{Bond energy of products}$
$\text{Reactants: } (4 \times C-H) + (1 \times C=C) + (1 \times H-H) = (4 \times 414) + 615 + 435 = 1656 + 615 + 435 = 2706 \, kJ \, mol^{-1}$
$\text{Products: } (6 \times C-H) + (1 \times C-C) = (6 \times 414) + 347 = 2484 + 347 = 2831 \, kJ \, mol^{-1}$
$\Delta H = 2706 - 2831 = -125 \, kJ$

(D) The thermal decomposition of metal nitrates depends on the position of the metal in the electrochemical series.
Nitrates of noble metals like silver $(Ag)$ and mercury $(Hg)$ decompose upon heating to yield the corresponding metal,nitrogen dioxide $(NO_2)$,and oxygen $(O_2)$.
The reaction for silver nitrate is:
$2AgNO_3 \xrightarrow{\Delta} 2Ag + 2NO_2 + O_2$
Other nitrates like those of iron,copper,or manganese typically decompose to form their respective metal oxides.

(C) The modern periodic law states that the physical and chemical properties of elements are a periodic function of their atomic numbers $(Z)$.
In the modern periodic table,elements are arranged in the order of increasing atomic number.
This arrangement explains the periodicity in properties,as elements with similar electronic configurations recur at regular intervals.

(B) The solubility of alkaline earth metal carbonates decreases down the group.
This is because the hydration energy of the cation decreases more rapidly than the lattice energy as the size of the cation increases down the group.
Since the hydration energy is not sufficient to overcome the lattice energy,the solubility decreases.
Therefore,the correct option is $(b)$.

(B) The reaction between potassium chromate $(K_2CrO_4)$ and dilute nitric acid $(HNO_3)$ is an acid-base equilibrium reaction where the chromate ion $(CrO_4^{2-})$ is converted into the dichromate ion $(Cr_2O_7^{2-})$.
The balanced chemical equation is:
$2K_2CrO_4 + 2HNO_3 \rightarrow K_2Cr_2O_7 + 2KNO_3 + H_2O$
In ionic form:
$2CrO_4^{2-} + 2H^{+} \rightarrow Cr_2O_7^{2-} + H_2O$
This is not a redox reaction; the oxidation state of chromium remains $+6$ in both $CrO_4^{2-}$ and $Cr_2O_7^{2-}$.

(C) Concentrated $HCl$ is a highly volatile liquid that releases $HCl$ gas.
$HCl$ gas has a very strong affinity for water vapor present in the atmosphere.
When $HCl$ gas molecules come in contact with atmospheric moisture,they absorb the water vapor to form tiny droplets of concentrated hydrochloric acid solution.
These microscopic droplets scatter light and appear as a white cloud or fumes in the air.

(B) . Marble statue contains $CaCO_3$.
$B$. Calcined gypsum is $CaSO_4 \cdot \frac{1}{2}H_2O$ (or $CaSO_4 \cdot 2H_2O$ for gypsum),which does not contain $CaCO_3$.
$C$. Sea shells are primarily composed of $CaCO_3$.
$D$. Dolomite is a double carbonate of calcium and magnesium,$CaCO_3 \cdot MgCO_3$.

(B) The red solid is $HgI_2$ (mercuric iodide).
$HgI_2$ is insoluble in water but dissolves in a solution of $KI$ due to the formation of a soluble complex: $HgI_2 + 2KI \to K_2[HgI_4]$.
Upon heating,$HgI_2$ undergoes thermal decomposition: $HgI_2 \to Hg + I_2$.
The $I_2$ vapours appear as violet-coloured fumes,and metallic mercury $(Hg)$ droplets condense on the cooler parts of the test tube.

(D) The structure is $CH_3-CO-CH(CH_3)_2$.
The longest carbon chain containing the ketone functional group consists of $4$ carbon atoms.
Numbering the chain from the end that gives the ketone group the lowest possible locant,we start from the left: $C_1$ is $CH_3$,$C_2$ is $CO$,$C_3$ is $CH(CH_3)$,and $C_4$ is $CH_3$.
There is a methyl substituent at the $C_3$ position.
Therefore,the $IUPAC$ name is $3-$methylbutan$-2-$one or $3-$methyl$-2-$butanone.

(B) The correct answer is $(b)$.
In neopentane $(CH_3-C(CH_3)_2-CH_3)$,all $12$ hydrogen atoms are chemically equivalent because the molecule is highly symmetrical.
Therefore,the replacement of any of these hydrogen atoms by a chlorine atom results in the formation of only one unique monosubstituted product,which is $1-$chloro$-2,2-$dimethylpropane.

(D) The conversion of an alkene to an alkane is a hydrogenation reaction.
$But-1-ene$ $(CH_3CH_2CH=CH_2)$ reacts with hydrogen gas in the presence of a metal catalyst like $Pd$,$Pt$,or $Ni$ to form $butane$ $(CH_3CH_2CH_2CH_3)$.
This process is known as catalytic hydrogenation.
Therefore,the correct reagent is $Pd / H_2$.

(B) In a pressure cooker,the pressure inside is higher than the atmospheric pressure.
According to the principle of elevation of boiling point,as the pressure increases,the boiling point of water increases.
This allows the food to cook at a higher temperature,which significantly reduces the cooking time.

(B) The electronic configurations of the neutral atoms are:
$V: [Ar] 3d^3 4s^2$
$Cr: [Ar] 3d^5 4s^1$
$Mn: [Ar] 3d^5 4s^2$
$Fe: [Ar] 3d^6 4s^2$
To find the second ionization enthalpy,we remove one electron to form the $+1$ ion and then remove the second electron:
$V^+: [Ar] 3d^3 4s^1 \rightarrow V^{2+}: [Ar] 3d^3$
$Cr^+: [Ar] 3d^5 \rightarrow Cr^{2+}: [Ar] 3d^4$
$Mn^+: [Ar] 3d^5 4s^1 \rightarrow Mn^{2+}: [Ar] 3d^5$
$Fe^+: [Ar] 3d^6 4s^1 \rightarrow Fe^{2+}: [Ar] 3d^6$
Among these,$Cr^+$ has a stable half-filled $d^5$ configuration. Removing an electron from this stable configuration requires a very high amount of energy. Therefore,$Cr$ has the highest second ionization enthalpy.

(B) $LiAlH_4$ is a strong reducing agent that reduces carboxylic acids to primary alcohols but does not reduce isolated carbon-carbon double bonds.
Therefore,$CH_2=CH-COOH \xrightarrow{LiAlH_4} CH_2=CH-CH_2OH$.

(D) Given $|z\omega| = 1$,which implies $|z||\omega| = 1$ ... $(i)$
Also,$\text{arg}(z) - \text{arg}(\omega) = \frac{\pi}{2}$,which implies $\text{arg}(\frac{z}{\omega}) = \frac{\pi}{2}$.
This means $\frac{z}{\omega} = |\frac{z}{\omega}| e^{i\pi/2} = |\frac{z}{\omega}| i$. Since $|\frac{z}{\omega}| = \frac{|z|}{|\omega|}$,we have $\frac{z}{\omega} = \frac{|z|}{|\omega|} i$ ... $(ii)$
From $(i)$,$|z| = \frac{1}{|\omega|}$. Substituting this into $(ii)$,we get $\frac{z}{\omega} = \frac{1}{|\omega|^2} i$.
However,a simpler approach: $\frac{z}{\omega} = i \implies z = i\omega$.
Taking the conjugate,$\bar{z} = \bar{i}\bar{\omega} = -i\bar{\omega}$.
Then $\bar{z}\omega = (-i\bar{\omega})\omega = -i(\bar{\omega}\omega) = -i|\omega|^2$.
Since $|z\omega| = 1$ and $\frac{z}{\omega} = i$,we have $|z| = |i\omega| = |\omega|$,so $|\omega|^2 = |z||\omega| = 1$.
Therefore,$\bar{z}\omega = -i(1) = -i$.

(B) The given expression is $^nC_{r+1} + ^nC_{r-1} + 2 \times ^nC_r$.
We can rewrite $2 \times ^nC_r$ as $^nC_r + ^nC_r$.
So,the expression becomes $(^nC_{r+1} + ^nC_r) + (^nC_r + ^nC_{r-1})$.
Using the Pascal's identity formula $^nC_r + ^nC_{r-1} = ^{n+1}C_r$,we get:
$(^nC_{r+1} + ^nC_r) + (^nC_r + ^nC_{r-1}) = ^{n+1}C_{r+1} + ^{n+1}C_r$.
Applying the same identity again:
$^{n+1}C_{r+1} + ^{n+1}C_r = ^{n+2}C_{r+1}$.
Thus,the correct option is $B$.

(B) The student must choose $10$ questions out of $13$. The first $5$ questions are in one group and the remaining $8$ questions are in another group.
He must choose at least $4$ from the first $5$ questions. The possible cases are:
Case $1$: Choose $4$ from the first $5$ and $6$ from the remaining $8$.
Number of ways $= {^5}C_4 \times {^8}C_6 = 5 \times 28 = 140$.
Case $2$: Choose $5$ from the first $5$ and $5$ from the remaining $8$.
Number of ways $= {^5}C_5 \times {^8}C_5 = 1 \times 56 = 56$.
Total number of ways $= 140 + 56 = 196$.

(C) Let $D$ be the midpoint of $BC$. The median through $A$ is the vector $\overrightarrow{AD}$.
Since $D$ is the midpoint of $BC$,we have $\overrightarrow{AD} = \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AC})$.
Substituting the given vectors:
$\overrightarrow{AD} = \frac{1}{2}((3\hat{i} + 4\hat{k}) + (5\hat{i} - 2\hat{j} + 4\hat{k}))$
$\overrightarrow{AD} = \frac{1}{2}(8\hat{i} - 2\hat{j} + 8\hat{k}) = 4\hat{i} - \hat{j} + 4\hat{k}$.
The length of the median is the magnitude of $\overrightarrow{AD}$:
$|\overrightarrow{AD}| = \sqrt{4^2 + (-1)^2 + 4^2} = \sqrt{16 + 1 + 16} = \sqrt{33}$.

(B) The general equation of a parabola with its axis as the $x$-axis is given by $y^2 = 4a(x - h)$,where $a$ and $h$ are arbitrary constants.
Since there are two arbitrary constants,we differentiate the equation twice.
First derivative: $2y y_1 = 4a$,which simplifies to $y y_1 = 2a$.
Second derivative: $y_1^2 + y y_2 = 0$.
In the differential equation $y_2 y + y_1^2 = 0$,the highest order derivative is $y_2$,so the order is $2$.
The power of the highest order derivative $y_2$ is $1$,so the degree is $1$.
Therefore,the degree is $1$ and the order is $2$.

(A) Given $n = 15$,$\sum x = 170$,and $\sum x^2 = 2830$.
When the incorrect observation $20$ is replaced by $30$,the new sum $\sum x'$ is:
$\sum x' = 170 - 20 + 30 = 180$.
The new sum of squares $\sum x'^2$ is:
$\sum x'^2 = 2830 - (20)^2 + (30)^2 = 2830 - 400 + 900 = 3330$.
The variance is given by the formula $\sigma^2 = \frac{1}{n} \sum x'^2 - \left( \frac{\sum x'}{n} \right)^2$.
Substituting the values:
$\sigma^2 = \frac{3330}{15} - \left( \frac{180}{15} \right)^2$
$\sigma^2 = 222 - (12)^2$
$\sigma^2 = 222 - 144 = 78$.

(C) The hormone secretin is produced by the $S$-cells located in the mucosal lining of the duodenum. When acidic chyme from the stomach enters the duodenum,it triggers the release of secretin into the bloodstream. This hormone then stimulates the pancreas to secrete bicarbonate-rich pancreatic juice to neutralize the acidity.

(B) Ionization potential is the energy required to remove the most loosely bound electron from an isolated gaseous atom.
As we move down a group in the periodic table,the atomic size increases due to the addition of new shells.
As the atomic size increases,the distance between the nucleus and the valence electron increases,resulting in a weaker electrostatic force of attraction.
Therefore,it becomes easier to remove the valence electron,leading to a decrease in ionization potential.
Among the given elements,$_7^{14}N$ (Nitrogen),$_8^{16}O$ (Oxygen),and $_{18}^{40}Ar$ (Argon) are non-metals or noble gases with relatively small atomic radii.
$_{55}^{133}Cs$ (Cesium) is an alkali metal located at the bottom of the periodic table (Group $1$,Period $6$),which has the largest atomic radius among the options.
Thus,$_{55}^{133}Cs$ has the lowest ionization potential.

(B) The electronic configurations of the given elements are:
$V (Z=23): [Ar] 3d^3 4s^2$
$Cr (Z=24): [Ar] 3d^5 4s^1$
$Mn (Z=25): [Ar] 3d^5 4s^2$
$Fe (Z=26): [Ar] 3d^6 4s^2$
To find the second ionization enthalpy,we remove one electron from each to form the $M^+$ ion:
$V^+: [Ar] 3d^3 4s^1$
$Cr^+: [Ar] 3d^5$
$Mn^+: [Ar] 3d^5 4s^1$
$Fe^+: [Ar] 3d^6 4s^1$
The second ionization enthalpy involves removing an electron from the $M^+$ ion. For $Cr^+$,the configuration is $[Ar] 3d^5$,which is a stable half-filled $d$-orbital configuration. Removing an electron from this stable state requires a significantly higher amount of energy compared to the others. Therefore,$Cr$ has the highest second ionization enthalpy.

(B) The electronic configurations are as follows:
$V (Z=23): [Ar] 3d^3 4s^2$
$Cr (Z=24): [Ar] 3d^5 4s^1$
$Mn (Z=25): [Ar] 3d^5 4s^2$
$Fe (Z=26): [Ar] 3d^6 4s^2$
To find the second ionization enthalpy,we remove one electron from each:
$V^+: [Ar] 3d^3 4s^1$
$Cr^+: [Ar] 3d^5 4s^0$
$Mn^+: [Ar] 3d^5 4s^1$
$Fe^+: [Ar] 3d^6 4s^1$
Removing the second electron from $Cr^+$ involves removing an electron from the stable half-filled $3d^5$ subshell,which requires a very high amount of energy.
Therefore,$Cr$ has the highest second ionization enthalpy.

(A) Let the oxidation state of $Ni$ be $x$.
The oxidation state of $K$ is $+1$ and the oxidation state of $CN^-$ is $-1$.
The sum of oxidation states in a neutral complex is $0$.
$4(+1) + x + 4(-1) = 0$
$4 + x - 4 = 0$
$x = 0$.
Therefore,the oxidation state of $Ni$ in $K_4[Ni(CN)_4]$ is $0$.

(B) Let $\vec{n_1}$ and $\vec{n_2}$ be the normal vectors to the faces $OAB$ and $ABC$ respectively.
For face $OAB$,the normal vector is $\vec{n_1} = \vec{OA} \times \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(3-2) + \hat{k}(1-4) = 5\hat{i} - \hat{j} - 3\hat{k}$.
For face $ABC$,the vectors are $\vec{AB} = (2-1, 1-2, 3-1) = (1, -1, 2)$ and $\vec{AC} = (-1-1, 1-2, 2-1) = (-2, -1, 1)$.
The normal vector is $\vec{n_2} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -2 & -1 & 1 \end{vmatrix} = \hat{i}(-1+2) - \hat{j}(1+4) + \hat{k}(-1-2) = \hat{i} - 5\hat{j} - 3\hat{k}$.
If $\theta$ is the angle between the faces $OAB$ and $ABC$,then $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
$\vec{n_1} \cdot \vec{n_2} = (5)(1) + (-1)(-5) + (-3)(-3) = 5 + 5 + 9 = 19$.
$|\vec{n_1}| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{25 + 1 + 9} = \sqrt{35}$.
$|\vec{n_2}| = \sqrt{1^2 + (-5)^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$.
Therefore,$\cos \theta = \frac{19}{\sqrt{35} \cdot \sqrt{35}} = \frac{19}{35}$.
$\theta = \cos^{-1}\left(\frac{19}{35}\right)$.

(D) For the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,we rewrite it as $\frac{x^2}{144/25} - \frac{y^2}{81/25} = 1$.
Here,$a^2 = \frac{144}{25}$ and $b^2 = \frac{81}{25}$.
The eccentricity $e_h$ is given by $e_h^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{81/25}{144/25} = 1 + \frac{81}{144} = \frac{225}{144}$.
So,$e_h = \frac{15}{12} = \frac{5}{4}$.
The foci of the hyperbola are $(\pm a_h e_h, 0) = (\pm \frac{12}{5} \times \frac{5}{4}, 0) = (\pm 3, 0)$.
For the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$,the foci are $(\pm ae, 0)$ where $a^2 = 16$.
Since the foci are the same,$ae = 3$,so $e = \frac{3}{4}$.
For an ellipse,$e^2 = 1 - \frac{b^2}{a^2}$,so $(\frac{3}{4})^2 = 1 - \frac{b^2}{16}$.
$\frac{9}{16} = 1 - \frac{b^2}{16} \implies \frac{b^2}{16} = 1 - \frac{9}{16} = \frac{7}{16}$.
Therefore,$b^2 = 7$.

(D) Given the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$ with $a > 0$.
First,find the derivative: $f'(x) = 6x^2 - 18ax + 12a^2$.
Set $f'(x) = 0$ to find critical points: $6(x^2 - 3ax + 2a^2) = 0$,which factors to $6(x - a)(x - 2a) = 0$.
Thus,the critical points are $x = a$ and $x = 2a$.
Find the second derivative: $f''(x) = 12x - 18a$.
At $x = a$,$f''(a) = 12a - 18a = -6a < 0$ (since $a > 0$),so $x = a$ is a point of local maximum. Thus,$p = a$.
At $x = 2a$,$f''(2a) = 12(2a) - 18a = 6a > 0$,so $x = 2a$ is a point of local minimum. Thus,$q = 2a$.
Given the condition $p^2 = q$,we substitute the values: $a^2 = 2a$.
Since $a > 0$,we can divide by $a$ to get $a = 2$.

(B) The orbital angular momentum of an electron is given by the formula: $\text{Angular momentum} = \sqrt{l(l + 1)} \frac{h}{2\pi}$.
For an $s$ orbital,the azimuthal quantum number $l = 0$.
Substituting $l = 0$ into the formula: $\text{Angular momentum} = \sqrt{0(0 + 1)} \frac{h}{2\pi} = 0$.
Therefore,the orbital angular momentum of an electron in an $s$ orbital is zero.

(B) For the system of linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero:
$\left|\begin{array}{lll}1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c\end{array}\right| = 0$
Applying row operations $R_{2} \rightarrow R_{2} - R_{1}$ and $R_{3} \rightarrow R_{3} - R_{1}$:
$\left|\begin{array}{ccc}1 & 2a & a \\ 0 & 3b - 2a & b - a \\ 0 & 4c - 2a & c - a\end{array}\right| = 0$
Expanding along the first column:
$(3b - 2a)(c - a) - (4c - 2a)(b - a) = 0$
Expanding the terms:
$(3bc - 3ab - 2ac + 2a^{2}) - (4bc - 4ab - 2ac + 2a^{2}) = 0$
Simplifying the expression:
$3bc - 3ab - 2ac + 2a^{2} - 4bc + 4ab + 2ac - 2a^{2} = 0$
$-bc + ab = 0$
$ab = bc$
Wait,let us re-evaluate the expansion:
$3bc - 3ab - 2ac + 2a^{2} - (4bc - 4ab - 2ac + 2a^{2}) = 0$
$3bc - 3ab - 2ac + 2a^{2} - 4bc + 4ab + 2ac - 2a^{2} = 0$
$-bc + ab = 0 \Rightarrow ab = bc$.
Actually,let's re-calculate:
$(3b-2a)(c-a) = 3bc - 3ab - 2ac + 2a^2$
$(4c-2a)(b-a) = 4bc - 4ac - 2ab + 2a^2$
Subtracting: $(3bc - 3ab - 2ac + 2a^2) - (4bc - 4ac - 2ab + 2a^2) = 0$
$-bc - ab + 2ac = 0$
$2ac = ab + bc$
Dividing by $abc$:
$\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$
This implies that $a, b, c$ are in $H.P.$

(A) The Curie temperature $(T_{C})$ is the temperature at which certain materials lose their permanent magnetic properties and transition to induced magnetism.
Specifically,when a ferromagnetic material is heated above its Curie temperature,the thermal agitation overcomes the alignment of magnetic dipoles.
As a result,the material loses its spontaneous magnetization and behaves as a paramagnetic material.
Therefore,above the Curie temperature,a ferromagnetic material becomes paramagnetic.

(A) $C_6H_5I$ is an aryl halide where the $C-I$ bond has partial double bond character due to resonance,making it resistant to nucleophilic substitution under these conditions. Thus,it does not release $I^-$ ions to form a yellow precipitate of $AgI$ with $AgNO_3$.
$C_6H_5CH_2I$ is a primary alkyl halide (benzyl iodide) which undergoes nucleophilic substitution with $NaOH$ to form $C_6H_5CH_2OH$ and $NaI$. The $NaI$ dissociates to provide $I^-$ ions,which react with $AgNO_3$ to form a yellow precipitate of $AgI$.
Since $B$ gave a yellow precipitate,$B$ must be $C_6H_5CH_2I$,and $A$ must be $C_6H_5I$.
Therefore,the statement '$A$ was $C_6H_5I$' is true.

(A) The reaction of the complex with $AgNO_3$ to yield $2$ moles of $AgCl_{(s)}$ indicates that there are $2$ chloride ions $(Cl^-)$ present outside the coordination sphere (ionizable chloride ions).
When $[Co(NH_3)_5Cl]Cl_2$ dissolves in water,it dissociates as follows:
$[Co(NH_3)_5Cl]Cl_2 \rightarrow [Co(NH_3)_5Cl]^{2+} + 2Cl^-$
This results in a total of $1 + 2 = 3$ moles of ions per mole of the complex,which matches the given information.
Therefore,the correct structure is $[Co(NH_3)_5Cl]Cl_2$.

(D) Let the $9$ distinct observations in ascending order be $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9$.
Since the number of observations $n = 9$ is odd,the median is the $\left(\frac{n+1}{2}\right)^{th}$ observation,which is the $5^{th}$ observation,$x_5$.
Given,$x_5 = 20.5$.
When the largest $4$ observations $(x_6, x_7, x_8, x_9)$ are increased by $2$,the order of the observations remains unchanged because $x_5 < x_6 < x_7 < x_8 < x_9$ and the new values $x_6+2, x_7+2, x_8+2, x_9+2$ are still greater than $x_5$.
The new set of observations is $x_1, x_2, x_3, x_4, x_5, x_6+2, x_7+2, x_8+2, x_9+2$.
The median of the new set is still the $5^{th}$ observation,which is $x_5 = 20.5$.
Thus,the median remains the same.

(A) According to Faraday's law of electrolysis,the mass deposited or dissolved is given by $m = ZIt$.
Since the current $I$ and time $t$ are constant for both electrodes in the series circuit,we have the ratio: $\frac{m_{Cu}}{m_{Zn}} = \frac{Z_{Cu}}{Z_{Zn}}$.
Given $m_{Zn} = 0.13 \ g$,$Z_{Zn} = 32.5$,and $Z_{Cu} = 31.5$.
Substituting the values: $m_{Cu} = m_{Zn} \times \frac{Z_{Cu}}{Z_{Zn}} = 0.13 \times \frac{31.5}{32.5} = 0.126 \ g$.
Therefore,the increase in the mass of the $Cu$ pole is $0.126 \ g$.

(C) . Magnesium acts as a sacrificial anode because it is more reactive than iron. It undergoes oxidation instead of iron,thereby protecting the iron hull of the ship from corrosion caused by water and salt.

(B) Glass is considered a supercooled liquid because it possesses a disordered,amorphous structure similar to a liquid,but it has a very high viscosity that prevents it from flowing at room temperature.
Supercooling is a process of lowering the temperature of a liquid or a gas below its freezing point without it becoming a solid.

(C) When phosphine $(PH_3)$ is mixed with chlorine gas $(Cl_2)$,it reacts vigorously to form phosphorus pentachloride $(PCl_5)$ and hydrogen chloride $(HCl)$.
The balanced chemical equation is: $PH_3 + 4Cl_2 \to PCl_5 + 3HCl$.
This reaction is highly exothermic,meaning the mixture warms up.

(A) The balanced chemical equation for the reaction is: $Ba(OH)_2 + 2HCl \rightarrow BaCl_2 + 2H_2O$.
According to the stoichiometry,$1 \ mole$ of $Ba(OH)_2$ reacts with $2 \ moles$ of $HCl$.
Using the molarity equation: $\frac{M_1 V_1}{n_1} = \frac{M_2 V_2}{n_2}$,where $M_1$ and $V_1$ are for $Ba(OH)_2$ and $M_2$ and $V_2$ are for $HCl$.
Given: $V_1 = 25 \ mL$,$M_2 = 0.1 \ M$,$V_2 = 35 \ mL$,$n_1 = 1$,$n_2 = 2$.
$\frac{M_1 \times 25}{1} = \frac{0.1 \times 35}{2}$.
$M_1 = \frac{0.1 \times 35}{25 \times 2} = \frac{3.5}{50} = 0.07 \ M$.

(A) For an ideal solution,the following conditions must be satisfied:-
$1. \Delta H_{mix} = 0$ (Enthalpy of mixing is zero)
$2. \Delta V_{mix} = 0$ (Volume of mixing is zero)
$3. \Delta S_{mix} > 0$ (Entropy of mixing is positive)
$4. \Delta G_{mix} < 0$ (Gibbs free energy of mixing is negative)
Since $\Delta H_{mix} = 0$ is a defining characteristic of an ideal solution,option $A$ is correct.

(A) The molar mass of $NaCl$ is $23 + 35.5 = 58.5 \ g/mol$.
Number of formula units in $1.00 \ g$ of $NaCl = \frac{1.00 \ g}{58.5 \ g/mol} \times 6.022 \times 10^{23} \text{ formula units/mol} = 1.029 \times 10^{22}$ formula units.
Since one unit cell of $NaCl$ (fcc structure) contains $4$ formula units of $NaCl$,the number of unit cells is $\frac{1.029 \times 10^{22}}{4} = 2.57 \times 10^{21}$ unit cells.

(C) The initial radionuclide is $_{90}^{234}Th$.
$1$. First $\beta$-decay: $_{90}^{234}Th \rightarrow {}_{91}^{234}X + {}_{-1}^{0}e$.
$2$. Second $\beta$-decay: $_{91}^{234}X \rightarrow {}_{92}^{234}Y + {}_{-1}^{0}e$.
$3$. One $\alpha$-decay: $_{92}^{234}Y \rightarrow {}_{90}^{230}Z + {}_{2}^{4}He$.
Thus,the resulting radionuclide has an atomic number of $90$ and a mass number of $230$.

(A) The number of half-lives $(n)$ is calculated as: $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{18 \ h}{3 \ h} = 6$.
Using the radioactive decay formula: $N_t = N_o \times (1/2)^n$.
Substituting the values: $N_t = 256 \times (1/2)^6$.
$N_t = 256 \times \frac{1}{64} = 4 \ g$.
Therefore,the remaining mass is $4.0 \ g$.

(C) The rate law for the reaction is given by: $R = k[NO]^2[O_2]$.
When the volume is reduced to half,the concentration of each reactant doubles because $C = \frac{n}{V}$.
Let the new concentrations be $[NO]' = 2[NO]$ and $[O_2]' = 2[O_2]$.
The new rate $R'$ is: $R' = k(2[NO])^2(2[O_2])$.
$R' = k \times 4[NO]^2 \times 2[O_2] = 8 \times k[NO]^2[O_2]$.
Therefore,$R' = 8R$. The rate of reaction increases to eight times its initial value.

(C) The given equation $k = A e^{-E_a/RT}$ is known as the Arrhenius equation.
In this equation:
$k$ is the rate constant.
$A$ is the Arrhenius factor or frequency factor.
$E_a$ is the activation energy.
$R$ is the universal gas constant.
$T$ is the temperature in Kelvin.
Therefore,the statement '$E_a$ is energy of activation' is correct.

(A) The initial rate is given by $R = k[A]^n[B]^m$.
When the concentration of $A$ is doubled $([A]' = 2[A])$ and the concentration of $B$ is halved $([B]' = \frac{B}{2})$,the new rate $R'$ is:
$R' = k[2A]^n[\frac{B}{2}]^m$
$R' = k \cdot 2^n [A]^n \cdot 2^{-m} [B]^m$
$R' = k[A]^n[B]^m \cdot 2^{n - m}$
$R' = R \cdot 2^{n - m}$
Therefore,the ratio of the new rate to the earlier rate is:
$\frac{R'}{R} = 2^{n - m}$.

(B) The reduction reaction at the cathode is: $Ag^{+} + e^{-} \rightarrow Ag(s)$.
According to Faraday's first law of electrolysis,$1 \ \text{mole}$ of electrons $(96500 \ C)$ deposits $1 \ \text{mole}$ of silver $(108 \ g)$.
Therefore,the mass of silver deposited by $9650 \ C$ of charge is:
$\text{Mass} = \frac{108 \ g}{96500 \ C} \times 9650 \ C = 10.8 \ g$.

(C) The Nernst equation for the cell reaction is given by:
$E_{cell} = E_{cell}^o - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
Here,$n = 2$ (number of electrons transferred).
Substituting the given values:
$E_{cell} = 1.10 - \frac{0.0591}{2} \log \frac{1}{0.1}$
$E_{cell} = 1.10 - 0.02955 \times \log(10)$
Since $\log(10) = 1$,we get:
$E_{cell} = 1.10 - 0.02955 = 1.07045 \ V \approx 1.07 \ V$.

(A) The reducing power of a metal is inversely proportional to its standard reduction potential.
Lower reduction potential indicates a greater tendency to lose electrons,hence higher reducing power.
Given reduction potentials: $E^{\circ}_{A} = +0.5 \ V$,$E^{\circ}_{B} = -3.0 \ V$,$E^{\circ}_{C} = -1.2 \ V$.
Comparing the values: $-3.0 \ V < -1.2 \ V < +0.5 \ V$.
Therefore,the order of reducing power is $B > C > A$.

(D) The relationship between standard emf $(E^o)$ and equilibrium constant $(K)$ is given by the equation: $\log \ K = \frac{nFE^o}{2.303 \ RT}$.
Given: $n = 2$,$E^o = 0.295 \ V$,$T = 298 \ K$,$F = 96500 \ C \ mol^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values: $\log \ K = \frac{2 \times 96500 \times 0.295}{2.303 \times 8.314 \times 298}$.
$\log \ K = \frac{56935}{5705.8} \approx 9.978$.
Since $\log \ K \approx 10$,we get $K = 10^{10}$.

(B) Physical adsorption is an exothermic process. According to Le Chatelier's Principle,for an exothermic process,an increase in temperature favors the reverse process (desorption). Therefore,physical adsorption decreases with an increase in temperature.
For a process to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative. Since $\Delta G = \Delta H - T\Delta S$,and adsorption involves a decrease in entropy $(\Delta S < 0)$,the process can only be spontaneous if $\Delta H$ is negative (exothermic) and the temperature $T$ is low enough such that $|\Delta H| > |T\Delta S|$.

(C) The phenomenon where there is a regular decrease in the atomic and ionic radii of lanthanoids with an increase in atomic number is known as lanthanide contraction. This is a characteristic feature of the $f-$ block elements.

(B) The correct statement is that from a mixed precipitate of $AgCl$ and $AgI$,ammonia solution dissolves only $AgCl$.
$AgCl$ is soluble in aqueous ammonia due to the formation of the complex $[Ag(NH_3)_2]Cl$,whereas $AgI$ is much less soluble and does not dissolve significantly in dilute ammonia.

(B) Ammonia acts as a ligand because it donates its lone pair of electrons to $Cu^{2+}$ ions to form the $[Cu(NH_3)_4]^{2+}$ complex.
This reaction occurs in a basic medium.
In an acidic medium,the lone pair of electrons on the nitrogen atom of ammonia is donated to $H^+$ ions (protons) to form ammonium ions $(NH_4^+)$.
Since $NH_4^+$ does not possess a lone pair,it cannot act as a ligand,and thus the complex cannot form.
The reaction is: $Cu^{2+}_{(aq)} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}$
In the presence of acid: $NH_3 + H^+ \rightarrow NH_4^+$

(B) Let the oxidation state of $Ni$ be $x$.
In $K_4[Ni(CN)_4]$,the oxidation state of $K$ is $+1$ and $CN$ is $-1$.
The sum of oxidation states of all atoms in a neutral complex is $0$.
$4(+1) + x + 4(-1) = 0$
$4 + x - 4 = 0$
$x = 0$
Therefore,the oxidation state of $Ni$ is $0$.

(A) The complex $Co(NH_3)_5Cl_3$ reacts with $2$ moles of $AgNO_3$ to form $2$ moles of $AgCl_{(s)}$,which indicates that there are $2$ ionizable $Cl^-$ ions outside the coordination sphere.
Therefore,the formula is $[Co(NH_3)_5Cl]Cl_2$.
Upon dissolution in water,it dissociates as: $[Co(NH_3)_5Cl]Cl_2 \rightleftharpoons [Co(NH_3)_5Cl]^{2+} + 2Cl^-$.
This produces $1$ complex cation and $2$ chloride anions,totaling $3$ moles of ions per mole of the complex.

(A) $C_6H_5CH_2I$ is an alkyl halide where the iodine is attached to an $sp^3$ hybridized carbon,making it reactive towards nucleophilic substitution with $NaOH$ to produce $I^-$ ions.
$C_6H_5I$ is an aryl halide where the iodine is attached to an $sp^2$ hybridized carbon,making it inert towards nucleophilic substitution under these conditions.
When $AgNO_3$ is added to the acidified solution,$I^-$ ions react to form a yellow precipitate of $AgI$.
Since $B$ gave a yellow precipitate,$B$ must be $C_6H_5CH_2I$.
Therefore,$A$ must be $C_6H_5I$.

(A) The dehydration of alcohols to alkenes in the presence of concentrated $H_2SO_4$ follows an $E1$ mechanism.
Step $1$: The initiation step involves the protonation of the alcohol molecule by the acid catalyst $(H_2SO_4)$ to form a protonated alcohol (alkyloxonium ion).
$CH_3CH_2OH + H^+ \rightarrow CH_3CH_2OH_2^+$
This step makes the $-OH$ group a better leaving group (as $H_2O$).
Therefore,the correct initiation step is the protonation of the alcohol molecule.

(D) Due to inter-molecular hydrogen bonding in alcohols,the boiling point of alcohols is much higher than that of ethers.
Because ethers do not form inter-molecular hydrogen bonds,they have weaker intermolecular forces,making them more volatile.

(B) The general formula $C_nH_{2n}O_2$ represents compounds with a degree of unsaturation (double bond equivalent) of $1$.
For carboxylic acids $(R-COOH)$,the general formula is $C_nH_{2n}O_2$.
For example,acetic acid $(CH_3COOH)$ has $n=2$,giving $C_2H_4O_2$.
Diketones,diols,and dialdehydes generally have different general formulas or require different degrees of unsaturation.
Thus,the correct option is $B$.

(C) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom.
$1$. The alkyl group $(CH_3-)$ is an electron-donating group that exhibits the $+I$ effect,which increases the electron density on the nitrogen atom.
$2$. Ammonia $(NH_3)$ has no alkyl groups,so it is the least basic among the three.
$3$. Methylamine $(CH_3NH_2)$ has one methyl group,while dimethylamine $((CH_3)_2NH)$ has two methyl groups.
$4$. Since $(CH_3)_2NH$ has two electron-donating groups,it has higher electron density on the nitrogen atom compared to $CH_3NH_2$.
Therefore,the correct order of increasing basicity is $NH_3 < CH_3NH_2 < (CH_3)_2NH$.

(C) Cellulose is a linear polysaccharide consisting of a large number of $D$-glucose units joined by $\beta$-glycosidic linkages.
Upon complete hydrolysis,cellulose yields only $D$-glucose molecules.

(C) Nylon threads are made up of polyamide.
Some common examples are nylon $6$ and nylon $6,6$.
Polyamides are polymers which contain repeating amide,$-CONH-$ linkages.