AP EAMCET 2020 Chemistry Question Paper with Answer and Solution

(D) For a sparingly soluble salt of molecular formula $A_2 X_3$,if solubility in pure water is $y \ M$ at a given temperature:
$A_2 X_3 \rightleftharpoons 2 A^{3+} + 3 X^{2-}$
Initial concentrations: $0, 0$
Equilibrium concentrations: $2y \ M, 3y \ M$
Solubility product,$K_{sp} = [A^{3+}]^2 [X^{2-}]^3$
$K_{sp} = (2y)^2 \times (3y)^3$
$K_{sp} = 4y^2 \times 27y^3 = 108y^5 \ M^5$

(A) When equal volumes are mixed,the concentration of each ion is halved. Let the initial concentrations be $[Ca^{2+}]_i$ and $[F^{-}]_i$. The new concentrations are $[Ca^{2+}] = [Ca^{2+}]_i / 2$ and $[F^{-}] = [F^{-}]_i / 2$.
Precipitation occurs if the ionic product $Q_{sp} = [Ca^{2+}][F^{-}]^2 > K_{sp} = 1.6 \times 10^{-10}$.
For option $A$: $[Ca^{2+}] = 0.5 \times 10^{-2}$,$[F^{-}] = 0.5 \times 10^{-5}$. $Q_{sp} = (0.5 \times 10^{-2})(0.5 \times 10^{-5})^2 = 0.5 \times 0.25 \times 10^{-12} = 1.25 \times 10^{-13} < 1.6 \times 10^{-10}$. Precipitation will not occur.
For option $B$: $[Ca^{2+}] = 0.5 \times 10^{-3}$,$[F^{-}] = 0.5 \times 10^{-3}$. $Q_{sp} = (0.5 \times 10^{-3})(0.5 \times 10^{-3})^2 = 0.125 \times 10^{-9} = 1.25 \times 10^{-10} < 1.6 \times 10^{-10}$. Precipitation will not occur.
For option $C$: $[Ca^{2+}] = 0.5 \times 10^{-4}$,$[F^{-}] = 0.5 \times 10^{-2}$. $Q_{sp} = (0.5 \times 10^{-4})(0.5 \times 10^{-2})^2 = 0.5 \times 0.25 \times 10^{-8} = 1.25 \times 10^{-9} > 1.6 \times 10^{-10}$. Precipitation occurs.
For option $D$: $[Ca^{2+}] = 0.5 \times 10^{-2}$,$[F^{-}] = 0.5 \times 10^{-3}$. $Q_{sp} = (0.5 \times 10^{-2})(0.5 \times 10^{-3})^2 = 0.5 \times 0.25 \times 10^{-8} = 1.25 \times 10^{-9} > 1.6 \times 10^{-10}$. Precipitation occurs.
Note: Both $A$ and $B$ do not precipitate. Given the original question structure,$A$ is the most dilute case.

(B) The dissolution of $AgBr$ is represented by the equilibrium: $AgBr(s) \rightleftharpoons Ag^{+}(aq) + Br^{-}(aq)$.
The solubility product expression is $K_{sp} = [Ag^{+}][Br^{-}] = 5.0 \times 10^{-13}$.
In the presence of $0.1 \ M$ $NaBr$,the concentration of $Br^{-}$ ions is dominated by the strong electrolyte $NaBr$,so $[Br^{-}] \approx 0.1 \ M$.
Let the solubility of $AgBr$ be $S$. Then $[Ag^{+}] = S$.
Substituting these values into the $K_{sp}$ expression: $S \times 0.1 = 5.0 \times 10^{-13}$.
Solving for $S$: $S = (5.0 \times 10^{-13}) / 0.1 = 5.0 \times 10^{-12} \ M$.
Therefore,the solubility of $AgBr$ in $0.1 \ M$ $NaBr$ is $5.0 \times 10^{-12} \ M$.

(D) Let $S$ be the solubility of $CaSO_4$ in $mol/L$.
$CaSO_4(s) \rightleftharpoons Ca^{2+}(aq) + SO_4^{2-}(aq)$
$K_{sp} = [Ca^{2+}][SO_4^{2-}] = S^2 = 9 \times 10^{-6}$
$S = \sqrt{9 \times 10^{-6}} = 3 \times 10^{-3} \ M$ (or $mol/L$).
The molar mass of $CaSO_4 = 40 + 32 + (4 \times 16) = 136 \ g/mol$.
Solubility in $g/L = S \times \text{Molar Mass} = 3 \times 10^{-3} \ mol/L \times 136 \ g/mol = 0.408 \ g/L$.
This means $0.408 \ g$ of $CaSO_4$ dissolves in $1 \ L$ of water.
Therefore,the volume required to dissolve $1 \ g$ of $CaSO_4 = \frac{1 \ g}{0.408 \ g/L} \approx 2.45 \ L$.
Hence,the correct option is $D$.

(D) Consider the block of mass $m$ in the frame of the wedge. Since the block does not slip,it is in equilibrium in the frame of the wedge.
The forces acting on the block are:
$1$. Gravitational force $m g$ acting vertically downwards.
$2$. Normal reaction $R$ exerted by the wedge,acting perpendicular to the inclined surface.
$3$. Pseudo force $m a$ acting horizontally in the direction opposite to the acceleration of the wedge.
Resolving the forces along the inclined plane:
$m g \sin \theta = m a \cos \theta$
$\Rightarrow a = g \tan \theta$ ... $(i)$
Resolving the forces perpendicular to the inclined plane:
$R = m g \cos \theta + m a \sin \theta$
Substituting the value of $a$ from equation $(i)$:
$R = m g \cos \theta + m (g \tan \theta) \sin \theta$
$R = m g \cos \theta + m g \frac{\sin^2 \theta}{\cos \theta}$
$R = \frac{m g \cos^2 \theta + m g \sin^2 \theta}{\cos \theta}$
$R = \frac{m g (\cos^2 \theta + \sin^2 \theta)}{\cos \theta}$
$R = \frac{m g}{\cos \theta} = m g \sec \theta$

(C) Given: Mass $m = 5 \ kg$,initial velocity $u = 30 \hat{i} + 40 \hat{j} \ ms^{-1}$,and force $F = -(\hat{i} + 5 \hat{j}) \ N$.
We focus on the $y$-component of motion.
Initial $y$-velocity $u_y = 40 \ ms^{-1}$ and $y$-component of force $F_y = -5 \ N$.
Using Newton's second law,the acceleration along the $y$-axis is $a_y = \frac{F_y}{m} = \frac{-5 \ N}{5 \ kg} = -1 \ ms^{-2}$.
We want to find the time $t$ when the final $y$-velocity $v_y = 0$.
Using the first equation of motion $v_y = u_y + a_y t$:
$0 = 40 + (-1)t$
$t = 40 \ s$.

(A) The particle moves in a horizontal circle on a smooth table. The forces acting on the particle are:
$1$. The gravitational force $(mg)$ acting downwards.
$2$. The normal reaction $(N)$ from the table acting upwards.
$3$. The tension $(T)$ in the string acting towards the centre of the circular path.
Since the particle is moving in a horizontal circle,there is no vertical acceleration,so $N = mg$.
The horizontal force acting towards the centre is the tension $T$,which provides the necessary centripetal force for the circular motion. Therefore,the net force on the particle directed towards the centre is equal to $T$.

(C) $BF_3$ is used as a catalyst in several industrial processes due to its strong Lewis acid nature.
$sp^2$-hybridized boron in $BF_3$ is electron-deficient ($6 \ e^-$ in valence shell,$2 \ e^-$ less than the octet) and acts as a strong Lewis acid.

(A) $BF_3 \xrightarrow{LiAlH_4} B_2H_6 (X)$
$3B_2H_6 + 6NH_3 \xrightarrow{\Delta} 2B_3N_3H_6 (Y) + 12H_2$
Compound $X$ is $B_2H_6$ (Diborane).
Compound $Y$ is $B_3N_3H_6$ (Borazine),which is known as inorganic benzene.

(A) When boron trichloride $(BCl_3)$ reacts with water,it undergoes hydrolysis to form boric acid $(H_3BO_3)$ and hydrochloric acid $(HCl)$.
The balanced chemical equation is:
$BCl_3 + 3H_2O \longrightarrow H_3BO_3 + 3HCl$

(C) In $B_2H_6$,there are $4$ terminal $B-H$ bonds which are $2$-centre-$2$-electron $(2c-2e)$ bonds.
There are $2$ bridged hydrogen atoms which form $2$ bridge $B-H-B$ bonds.
Each bridge $B-H-B$ bond is a $3$-centre-$2$-electron $(3c-2e)$ bond.
Therefore,there are only $2$ such $3c-2e$ bonds,not $4$.
Thus,statement $(C)$ is incorrect.

(A) Boron halides $(BX_3)$ have a central boron atom bonded to three halogen atoms.
In these molecules,boron has only $6$ electrons in its valence shell (an incomplete octet).
Due to this electron deficiency,they act as Lewis acids by accepting a lone pair of electrons from a Lewis base to complete their octet.

(A) When carbon combines with oxygen,it can form four distinct oxides or oxocarbons:
$1$. Carbon monoxide: $CO$
$2$. Carbon dioxide: $CO_2$
$3$. Carbon suboxide: $C_3O_2$
$4$. Mellitic anhydride: $C_{12}O_9$
Thus,there are $4$ such compounds.

(A) $CCl_4$ cannot undergo hydrolysis because the $C$ atom,being a member of the second period,does not have vacant $d$-orbitals in its valence shell.
Therefore,it cannot accommodate the lone pair donated by the oxygen atom of the water molecule.
In contrast,$SiCl_4$,$VCl_4$,and $TiCl_4$ possess vacant $d$-orbitals,allowing them to undergo hydrolysis.
Hence,the correct option is $A$.

(B) In group $14$,as we move down the group,the atomic size increases and the bond energy of the element-element bond decreases.
This leads to a decrease in the tendency to show catenation.
The order of catenation is: $C >> Si > Ge \approx Sn$.
Lead $(Pb)$ does not show catenation due to its large size and weak bond strength.
Thus,the correct option is $B$.

(B) $1$. In $N(SiH_3)_3$,the nitrogen atom has a lone pair which is donated into the vacant $3d$-orbital of the silicon atom,forming a $p\pi-d\pi$ back-bonding. This makes the nitrogen atom $sp^2$ hybridized,resulting in a $planar$ geometry.
$2$. In $Me_3N$ (trimethylamine),the carbon atom does not have vacant $d$-orbitals,so no back-bonding occurs. The nitrogen atom remains $sp^3$ hybridized with a lone pair,resulting in a $pyramidal$ geometry.
$3$. In $(SiH_3)_3P$,the phosphorus atom is larger and has a lower electronegativity compared to nitrogen. The lone pair on phosphorus is stereochemically active and does not participate in significant back-bonding with silicon. Thus,it retains a $pyramidal$ geometry.

(C) In pyrosilicate $(Si_2O_7)^{6-}$,two silicate units $(SiO_4)^{4-}$ share one oxygen atom. The structure is represented as $(Si_2O_7)^{6-} = O_3Si-O-SiO_3$.

(A) $(CH_3)_3SiCl$ is used during the polymerisation of organosilicones because it acts as a chain terminator.
When added to the reaction mixture,it reacts with the terminal $-OH$ groups of the growing siloxane chain,forming a stable $(CH_3)_3Si-O-$ group at the end.
This prevents further condensation and thus allows the control of the chain length of the organosilicone polymers.
Hence,the correct option is $A$.

(D) Silicones are synthetic organosilicon polymers with the general formula $(R_2SiO)_n$.
They possess a hydrophobic (water-repelling) nature due to the presence of non-polar organic groups attached to the silicon atoms.
They are widely used in medical applications because they are biocompatible.
However,silicones are excellent electrical insulators,not conductors,due to the strong $Si-O$ bonds and the absence of free electrons or ions for charge transport.

(D) Among the dioxides of group-$14$ elements ($CO_2$,$SiO_2$,$GeO_2$,$SnO_2$,and $PbO_2$),$PbO_2$ is weakly basic in nature.
Due to the inert pair effect,the $+2$ oxidation state of $Pb$ is more stable than the $+4$ oxidation state.
Therefore,$PbO_2$ (where $Pb$ is in $+4$ state) has a strong tendency to get reduced to $Pb^{2+}$ $(PbO)$,which makes it a powerful oxidizing agent.

(D) The nitrogen atom in the $NO_3^{-}$ ion has a valence shell configuration of $2s^2 2p^3$.
In the nitrate ion,nitrogen forms three sigma bonds (one with each oxygen atom) and one pi bond (delocalized over the three oxygen atoms).
This means nitrogen uses all its $5$ valence electrons to form $4$ bonds ($3$ single bonds and $1$ double bond equivalent in resonance).
Therefore,the number of bond pairs of electrons on the nitrogen atom is $4$.
Since all valence electrons are involved in bonding,there are $0$ lone pairs of electrons on the nitrogen atom.
Thus,the correct answer is $4,0$.

(C) The structure of $P_4O_{10}$ consists of a $P_4$ tetrahedral core where each edge is bridged by an oxygen atom ($P-O-P$ linkages).
Additionally,each phosphorus atom is bonded to a terminal oxygen atom via a double bond $(P=O)$.
Counting these,there are $12$ $P-O$ single bonds and $4$ $P=O$ double bonds.
Thus,the correct option is $C$.

(D) $O_3$ acts as a strong oxidizing agent.
$KI$ is oxidized to $I_2$.
$FeSO_4$ is oxidized to $Fe_2(SO_4)_3$.
$K_2MnO_4$ is oxidized to $KMnO_4$.
In $KMnO_4$,the oxidation state of $Mn$ is $+VII$,which is the maximum possible oxidation state for $Mn$.
Therefore,$KMnO_4$ cannot be further oxidized by $O_3$.
Hence,the correct option is $D$.

(A) The balanced chemical equation is: $AgNO_{3(aq)} + NaCl_{(aq)} \rightarrow AgCl_{(s)} + NaNO_{3(aq)}$
Step $1$: Calculate the mass of reactants.
Mass of $AgNO_3 = 25 \ mL \times 0.35 = 8.75 \ g$.
Moles of $AgNO_3 = \frac{8.75 \ g}{169.87 \ g/mol} \approx 0.0515 \ mol$.
Mass of $NaCl = 25 \ mL \times 0.116 = 2.9 \ g$.
Moles of $NaCl = \frac{2.9 \ g}{58.44 \ g/mol} \approx 0.0496 \ mol$.
Step $2$: Identify the limiting reagent.
Since the stoichiometry is $1:1$,$NaCl$ is the limiting reagent $(0.0496 \ mol < 0.0515 \ mol)$.
Step $3$: Calculate the mass of $AgCl$ formed.
Moles of $AgCl$ formed = Moles of $NaCl = 0.0496 \ mol$.
Mass of $AgCl = 0.0496 \ mol \times 143.32 \ g/mol \approx 7.1 \ g$.
Rounding to the nearest given option,the answer is $7 \ g$.

(C) reducing agent is a substance that can be oxidized,meaning its central atom must be able to increase its oxidation state.
In $SO_2$,the oxidation state of $S$ is $+4$,and its maximum oxidation state is $+6$,so it can be oxidized.
In $NO_2$,the oxidation state of $N$ is $+4$,and its maximum oxidation state is $+5$,so it can be oxidized.
In $CO_2$,the oxidation state of $C$ is $+4$,which is its maximum group oxidation state (Group $14$). Therefore,it cannot be oxidized further and cannot act as a reducing agent.
In $ClO_2$,the oxidation state of $Cl$ is $+4$,and its maximum oxidation state is $+7$,so it can be oxidized.

(B) In $SF_6$,the sulphur atom is $sp^3d^2$ hybridized,resulting in an octahedral geometry.
Six fluorine atoms surround the central sulphur atom,creating a highly crowded environment.
This steric hindrance prevents the attack of any nucleophile or electrophile on the sulphur atom,making $SF_6$ kinetically inert.

(B) As we move down the group in the periodic table,the atomic size of the halogen increases significantly.
This increase in size leads to a decrease in the bond dissociation enthalpy of the $H-X$ bond.
Consequently,the bond becomes weaker and it becomes easier to release the $H^+$ ion.
Therefore,the acidic strength increases down the group.
The correct order is $HF < HCl < HBr < HI$.

(C) The formation of molecular orbitals is described by the Linear Combination of Atomic Orbitals $(LCAO)$ method.
Bonding molecular orbitals are formed by the constructive interference of atomic orbitals,which corresponds to the addition of their wave functions.
Therefore,the wave function for a bonding molecular orbital is given by $\psi_{bonding} = \psi_A + \psi_B$.
Conversely,antibonding molecular orbitals are formed by the destructive interference,represented by the subtraction of wave functions,$\psi_{antibonding} = \psi_A - \psi_B$.

(D) We have the expression: $\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}$
Rearranging the terms: $(\tan 81^{\circ}+\tan 9^{\circ}) - (\tan 63^{\circ}+\tan 27^{\circ})$
Using the identity $\tan(90^{\circ}-\theta) = \cot \theta$:
$= (\cot 9^{\circ}+\tan 9^{\circ}) - (\cot 27^{\circ}+\tan 27^{\circ})$
Using $\cot \theta + \tan \theta = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin 2\theta}$:
$= \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$
$= 2 \left( \frac{\sin 54^{\circ} - \sin 18^{\circ}}{\sin 54^{\circ} \sin 18^{\circ}} \right)$
Using the identity $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:
$= 2 \left( \frac{2 \cos 36^{\circ} \sin 18^{\circ}}{\sin 54^{\circ} \sin 18^{\circ}} \right)$
Since $\sin 54^{\circ} = \cos 36^{\circ}$:
$= 2 \left( \frac{2 \cos 36^{\circ} \sin 18^{\circ}}{\cos 36^{\circ} \sin 18^{\circ}} \right) = 2 \times 2 = 4$

(B) homogeneous equation of second degree in $x$ and $y$,given by $ax^2 + 2hxy + by^2 = 0$,represents a pair of straight lines passing through the origin,provided that $h^2 \geq ab$.

(A) The pair of straight lines $A x^2+2 H x y+B y^2=0$ passes through the origin. For these lines to form an equilateral triangle with the line $a x+b y+c=0$,the angle between the pair of lines must be $60^{\circ}$ or $\frac{\pi}{3}$.
The angle $\theta$ between the lines $A x^2+2 H x y+B y^2=0$ is given by $\tan \theta = \frac{2 \sqrt{H^2-A B}}{|A+B|}$.
Setting $\theta = \frac{\pi}{3}$,we have $\tan \frac{\pi}{3} = \sqrt{3}$.
So,$\sqrt{3} = \frac{2 \sqrt{H^2-A B}}{|A+B|}$.
Squaring both sides,we get $3 = \frac{4(H^2-A B)}{(A+B)^2}$.
$3(A+B)^2 = 4(H^2-A B)$.
$3(A^2+2 A B+B^2) = 4 H^2-4 A B$.
$3 A^2+6 A B+3 B^2 = 4 H^2-4 A B$.
$3 A^2+10 A B+3 B^2 = 4 H^2$.
Factoring the left side: $3 A^2+9 A B+A B+3 B^2 = 4 H^2$.
$3 A(A+3 B)+B(A+3 B) = 4 H^2$.
$(A+3 B)(3 A+B) = 4 H^2$.

(B) It is given that one of the lines $2x^2 - xy + by^2 = 0$ passes through the point $(-4, -2)$.
Substituting the point $(-4, -2)$ into the equation:
$2(-4)^2 - (-4)(-2) + b(-2)^2 = 0$
$2(16) - (8) + b(4) = 0$
$32 - 8 + 4b = 0$
$24 + 4b = 0$
$4b = -24$
$b = -6$
Therefore,$b^2 = (-6)^2 = 36$.
Hence,option $B$ is correct.

(C) The general equation of a pair of straight lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Comparing this with $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$,we get $a = 2$,$h = \frac{5}{2}$,and $b = 3$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.
Substituting the values: $\tan \theta = \left| \frac{2\sqrt{(\frac{5}{2})^2 - (2)(3)}}{2 + 3} \right|$.
$\tan \theta = \left| \frac{2\sqrt{\frac{25}{4} - 6}}{5} \right| = \left| \frac{2\sqrt{\frac{1}{4}}}{5} \right| = \left| \frac{2 \times \frac{1}{2}}{5} \right| = \frac{1}{5}$.
Since the angle between the lines is $\tan^{-1}(k)$,we have $\tan \theta = |k| = \frac{1}{5}$,which implies $k = \pm \frac{1}{5}$.

(D) The equation of the angle bisectors for the pair of lines $a x^2+2 h x y+b y^2=0$ is given by $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$.
For the first pair $x^2-2 p x y-y^2=0$,we have $a=1, b=-1, h=-p$.
The angle bisectors are $\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p}$,which simplifies to $\frac{x^2-y^2}{2}=\frac{x y}{-p}$,or $x^2-y^2+\frac{2 x y}{p}=0$.
Given that this pair of bisectors is $x^2-2 q x y-y^2=0$,we compare the coefficients:
Comparing $x^2+\frac{2}{p} x y-y^2=0$ with $x^2-2 q x y-y^2=0$,we get $\frac{2}{p}=-2 q$.
Therefore,$p q=-1$.

(B) Given the equation of the pair of straight lines is $6x^2 - 5xy + y^2 = 0$.
Dividing by $x^2$ (assuming $x \neq 0$),we get:
$\left(\frac{y}{x}\right)^2 - 5\left(\frac{y}{x}\right) + 6 = 0$.
Let $m = \frac{y}{x}$,then $m^2 - 5m + 6 = 0$.
Factoring the quadratic equation: $(m - 3)(m - 2) = 0$.
So,the slopes are $m_1 = \tan \alpha = 3$ and $m_2 = \tan \beta = 2$.
Using the formula for the tangent of the difference of two angles:
$\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$.
Substituting the values:
$\tan(\alpha - \beta) = \frac{3 - 2}{1 + (3)(2)} = \frac{1}{1 + 6} = \frac{1}{7}$.

(A) The general equation of a pair of straight lines is given by $ax^2 + 2hxy + by^2 = 0$. The lines are perpendicular if the sum of the coefficients of $x^2$ and $y^2$ is zero,i.e.,$a + b = 0$.
For option $A$: $2 x^2 = y(x + 2 y) \Rightarrow 2 x^2 - xy - 2 y^2 = 0$.
Here,$a = 2$ and $b = -2$.
Since $a + b = 2 + (-2) = 0$,the lines represented by this equation are perpendicular.

(B) The pair of straight lines passing through the origin is given by the equation $ax^2 + 2hxy + by^2 = 0$.
The angle $\theta$ between these lines is given by the formula $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.
When the lines are perpendicular,the angle $\theta = 90^{\circ}$.
Since $\tan 90^{\circ}$ is undefined (approaches $\infty$),the denominator must be zero.
Therefore,$a + b = 0$.

(A) The acute angle $(\theta)$ between the lines represented by the homogeneous equation $ax^2+2hxy+by^2=0$ is given by the formula:
$\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$
If the lines are at right angles (perpendicular),then $\theta = 90^\circ$.
Since $\tan 90^\circ$ is undefined,the denominator must be zero.
Therefore,$a+b=0$.

(C) The given equation of the pair of straight lines is $x^2+4xy+y^2=0$.
Comparing this with the general equation $ax^2+2hxy+by^2=0$,we get $a=1$,$2h=4 \Rightarrow h=2$,and $b=1$.
The angle $\theta$ between the pair of straight lines is given by the formula:
$\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Substituting the values,we get:
$\tan \theta = \left| \frac{2\sqrt{2^2-(1)(1)}}{1+1} \right| = \frac{2\sqrt{4-1}}{2} = \sqrt{3}$.
Since $\tan \theta = \sqrt{3}$,we have $\theta = 60^{\circ}$.

(B) The given equation of the pair of straight lines is $3dx^2 - 5xy + (d^2 - 2)y^2 = 0$.
For a general equation of a pair of lines $Ax^2 + 2Hxy + By^2 = 0$,the lines are perpendicular if the sum of the coefficients of $x^2$ and $y^2$ is zero,i.e.,$A + B = 0$.
Here,$A = 3d$ and $B = d^2 - 2$.
Setting $A + B = 0$,we get $3d + d^2 - 2 = 0$,which is $d^2 + 3d - 2 = 0$.
Using the quadratic formula $d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we find $d = \frac{-3 \pm \sqrt{3^2 - 4(1)(-2)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 8}}{2} = \frac{-3 \pm \sqrt{17}}{2}$.
Since there are two distinct real values for $d$,the condition is satisfied for $2$ values of $d$.
Thus,option $B$ is correct.

(C) The equation of the angle bisectors of the pair of lines $ax^2+2hxy+by^2=0$ is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.
For the pair $x^2-2pxy-y^2=0$,we have $a=1, h=-p, b=-1$.
The angle bisectors are $\frac{x^2-y^2}{1-(-1)} = \frac{xy}{-p}$,which simplifies to $\frac{x^2-y^2}{2} = \frac{xy}{-p}$,or $x^2-y^2 = -\frac{2}{p}xy$,which is $x^2 + \frac{2}{p}xy - y^2 = 0$.
It is given that this pair of bisectors is the same as the pair $x^2-2qxy-y^2=0$.
Comparing the coefficients of $xy$,we get $\frac{2}{p} = -2q$.
This implies $pq = -1$,or $pq+1=0$.
Thus,option $C$ is correct.

(A) Let the two fixed lines be $L_1$ and $L_2$ intersecting at a point $P$.
The lines that are equally inclined to $L_1$ and $L_2$ are the angle bisectors of the angles formed by $L_1$ and $L_2$.
The angle bisectors of any two intersecting lines are always perpendicular to each other because the angle between the internal and external bisectors is always $90^{\circ}$.
Thus,Statement-$II$ is true and it correctly explains why the two lines described in Statement-$I$ must be perpendicular.
Therefore,both statements are true and Statement-$II$ is the correct explanation of Statement-$I$.

(C) The general equation of a second-degree curve is $ax^2+2hxy+by^2+2gx+2fy+c=0$.
For the equation $lx^2+3xy-2y^2-5x+5y+k=0$ to represent a pair of perpendicular lines,the sum of the coefficients of $x^2$ and $y^2$ must be zero,i.e.,$a+b=0$.
Here,$a=l$ and $b=-2$.
So,$l+(-2)=0 \Rightarrow l=2$.
Now the equation becomes $2x^2+3xy-2y^2-5x+5y+k=0$.
For this to represent a pair of lines,the determinant condition $abc+2fgh-af^2-bg^2-ch^2=0$ must be satisfied.
Here,$a=2, b=-2, c=k, h=\frac{3}{2}, g=-\frac{5}{2}, f=\frac{5}{2}$.
Substituting these values: $(2)(-2)(k) + 2(\frac{5}{2})(-\frac{5}{2})(\frac{3}{2}) - 2(\frac{5}{2})^2 - (-2)(-\frac{5}{2})^2 - k(\frac{3}{2})^2 = 0$.
$-4k - \frac{75}{4} - \frac{50}{4} + \frac{50}{4} - \frac{9k}{4} = 0$.
$-4k - \frac{9k}{4} - \frac{75}{4} = 0$.
$-\frac{25k}{4} = \frac{75}{4} \Rightarrow k=-3$.
Thus,$l=2$ and $k=-3$.

(B) The equation of the pair of angle bisectors for the pair of straight lines $a x^2+2 h x y+b y^2=0$ is given by $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$.
For the pair $x^2-2 p x y-y^2=0$,the bisectors are given by $\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p}$.
This simplifies to $\frac{x^2-y^2}{2}=\frac{x y}{-p}$,which implies $-p(x^2-y^2)=2 x y$,or $p x^2+2 x y-p y^2=0$.
Dividing by $p$,we get $x^2+\frac{2}{p} x y-y^2=0$.
Since this is the pair of bisectors,it must be identical to the given pair $x^2-2 q x y-y^2=0$.
Comparing the coefficients of $x y$,we have $-2 q = \frac{2}{p}$.
Thus,$-2 p q = 2$,which gives $p q = -1$.

(D) The general equation of a pair of straight lines passing through the origin is given by $ax^2 + 2hxy + by^2 = 0$.
The equation of the pair of angle bisectors for these lines is given by the formula:
$\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
Comparing this with the given options,option $D$ is the correct answer.

(A) The given equation is $x^2+4xy-y^2=0$. Comparing this with $ax^2+2hxy+by^2=0$,we get $a=1, h=2, b=-1$.
The equation of the angle bisectors is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.
Substituting the values,we get $\frac{x^2-y^2}{1-(-1)} = \frac{xy}{2}$.
$\frac{x^2-y^2}{2} = \frac{xy}{2} \implies x^2-xy-y^2=0$.
Since $y=mx$ is a bisector,substitute $y=mx$ into the equation: $x^2-x(mx)-(mx)^2=0$.
$x^2(1-m-m^2)=0$.
For non-trivial lines,$m^2+m-1=0$.
Using the quadratic formula,$m = \frac{-1 \pm \sqrt{1^2-4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.
Therefore,$2m = -1 \pm \sqrt{5}$.

(B) The given equation is $x^2+2 \sqrt{2} xy+2y^2+4x+4 \sqrt{2}y+1=0$.
This can be written as $(x+\sqrt{2}y)^2 + 4(x+\sqrt{2}y) + 1 = 0$.
Let $t = x+\sqrt{2}y$. Then the equation becomes $t^2 + 4t + 1 = 0$.
Solving for $t$ using the quadratic formula,$t = \frac{-4 \pm \sqrt{16-4}}{2} = -2 \pm \sqrt{3}$.
Thus,the two parallel lines are $x+\sqrt{2}y + 2 - \sqrt{3} = 0$ and $x+\sqrt{2}y + 2 + \sqrt{3} = 0$.
The distance between two parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is given by $d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.
Here,$A=1, B=\sqrt{2}, C_1=2-\sqrt{3}, C_2=2+\sqrt{3}$.
$d = \frac{|(2-\sqrt{3})-(2+\sqrt{3})|}{\sqrt{1^2+(\sqrt{2})^2}} = \frac{|-2\sqrt{3}|}{\sqrt{3}} = \frac{2\sqrt{3}}{\sqrt{3}} = 2 \text{ units}$.
Therefore,the distance between the lines is $2 \text{ units}$.

(A) Given circle is $x^2+y^2-6x-4y-12=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=-2, c=-12$.
Centre $(h, k) = (-g, -f) = (3, 2)$.
Radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-3)^2+(-2)^2-(-12)} = \sqrt{9+4+12} = \sqrt{25} = 5$.
Since the tangents are parallel to the $X$-axis,their equations are of the form $y=k$.
The perpendicular distance from the centre $(3, 2)$ to the line $y-k=0$ must be equal to the radius $r=5$.
$\frac{|2-k|}{\sqrt{0^2+1^2}} = 5 \Rightarrow |2-k| = 5$.
This gives $2-k = 5$ or $2-k = -5$.
So,$k = -3$ or $k = 7$.
The equations of the lines are $y=-3$ and $y=7$,which can be written as $(y+3)=0$ and $(y-7)=0$.
The combined equation of the pair of lines is $(y+3)(y-7) = 0$.
$y^2 - 7y + 3y - 21 = 0 \Rightarrow y^2 - 4y - 21 = 0$.

(C) The given equation is $x^2(\sec^2 \theta - \sin^2 \theta) - 2xy \tan \theta + y^2 \sin^2 \theta = 0$.
Comparing this with the general form $Ax^2 + 2Hxy + By^2 = 0$,we have:
$A = \sec^2 \theta - \sin^2 \theta$,$2H = -2 \tan \theta \implies H = -\tan \theta$,and $B = \sin^2 \theta$.
Let $m_1$ and $m_2$ be the slopes of the lines. Then $m_1 + m_2 = -\frac{2H}{B} = \frac{2 \tan \theta}{\sin^2 \theta}$ and $m_1 m_2 = \frac{A}{B} = \frac{\sec^2 \theta - \sin^2 \theta}{\sin^2 \theta} = \sec^2 \theta \csc^2 \theta - 1$.
The square of the difference of the slopes is $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$.
$(m_1 - m_2)^2 = \left(\frac{2 \tan \theta}{\sin^2 \theta}\right)^2 - 4\left(\frac{\sec^2 \theta - \sin^2 \theta}{\sin^2 \theta}\right)$.
$(m_1 - m_2)^2 = \frac{4 \sin^2 \theta}{\cos^2 \theta \sin^4 \theta} - \frac{4(\sec^2 \theta - \sin^2 \theta)}{\sin^2 \theta} = \frac{4}{\cos^2 \theta \sin^2 \theta} - \frac{4}{\cos^2 \theta \sin^2 \theta} + 4 = 4$.

(A) The given circle is $x^2+y^2-6x-4y-12=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=-2, c=-12$.
Centre $(h, k) = (-g, -f) = (3, 2)$.
Radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-3)^2+(-2)^2-(-12)} = \sqrt{9+4+12} = \sqrt{25} = 5$.
Since the tangents are parallel to the $X$-axis,they are of the form $y=k$.
The perpendicular distance from the centre $(3, 2)$ to the line $y-k=0$ must be equal to the radius $r=5$.
$\frac{|2-k|}{\sqrt{0^2+1^2}} = 5 \Rightarrow |2-k| = 5$.
This gives $2-k = 5$ or $2-k = -5$.
So,$k = -3$ or $k = 7$.
The equations of the lines are $y=-3$ and $y=7$,which can be written as $(y+3)=0$ and $(y-7)=0$.
The combined equation is $(y+3)(y-7) = y^2-7y+3y-21 = y^2-4y-21=0$.
Thus,the correct option is $A$.

(A) When a crystal lattice contains extra cations in the interstitial sites to maintain electrical neutrality,it is known as a metal excess defect.
This is a type of non-stoichiometric defect.

(A) Piezoelectric materials are materials that have the ability to generate an internal electrical charge from applied mechanical stress.
Quartz $(SiO_2)$ is a well-known example of a piezoelectric material.
Hence,option $(A)$ is correct.

(A) Bactericidal antibiotics are those that kill bacteria. Examples include $Penicillin$,$Aminoglycosides$,and $Ofloxacin$.
Bacteriostatic antibiotics are those that inhibit the growth of bacteria. Examples include $Tetracycline$,$Chloramphenicol$,and $Erythromycin$.
Therefore,$Penicillin$ is a bactericidal antibiotic.

(C) Relative Lowering of Vapour Pressure $(RLVP)$ is given by the formula: $\frac{p^{\circ} - p}{p^{\circ}} = \chi_A$
At $373 \ K$,the vapour pressure of pure water $(p^{\circ})$ is $760 \ Torr$.
Given mass of solute '$A$' = $160 \ g$,molar mass of '$A$' = $160 \ g \ mol^{-1}$.
Number of moles of '$A$' $(n_A)$ = $\frac{160 \ g}{160 \ g \ mol^{-1}} = 1 \ mol$.
Given volume of water = $54 \ mL$,density of water $\approx 1 \ g \ mL^{-1}$,so mass of water = $54 \ g$.
Molar mass of water $(H_2O)$ = $18 \ g \ mol^{-1}$.
Number of moles of water $(n_{H_2O})$ = $\frac{54 \ g}{18 \ g \ mol^{-1}} = 3 \ mol$.
Mole fraction of solute '$A$' $(\chi_A)$ = $\frac{n_A}{n_A + n_{H_2O}} = \frac{1}{1 + 3} = \frac{1}{4}$.
Using the formula: $\frac{760 - p}{760} = \frac{1}{4}$.
$760 - p = \frac{760}{4} = 190$.
$p = 760 - 190 = 570 \ Torr$.

(B) When a non-volatile solute like salt is added to water in a sealed vessel,the number of solvent molecules at the surface decreases,which reduces the rate of evaporation. This leads to a decrease in the vapour pressure of the solution compared to pure water. This phenomenon is known as the lowering of vapour pressure,given by $\Delta p = p^{\circ} - p$,where $\Delta p \propto \chi_B$ ($p^{\circ}$ is the vapour pressure of pure water,$p$ is the vapour pressure of the solution,and $\chi_B$ is the mole fraction of the solute).

(A) According to Raoult's law,the relative lowering of vapour pressure is directly proportional to the mole fraction of the solute.
Therefore,mole fraction is the unit used to relate the concentration of a solution with its vapour pressure.

(D) Two solutions having the same osmotic pressure $(\pi)$ at the same temperature are called isotonic solutions.
Osmotic pressure is given by $\pi = i \times C \times R \times T$,where $i$ is the van't Hoff factor.
For $0.1 \ M$ $NaCl$,$i = 2$ (as it dissociates into $Na^+$ and $Cl^-$),so $\pi = 2 \times 0.1 \times R \times T = 0.2 \ RT$.
For $0.1 \ M$ $KNO_3$,$i = 2$ (as it dissociates into $K^+$ and $NO_3^-$),so $\pi = 2 \times 0.1 \times R \times T = 0.2 \ RT$.
Since both solutions have the same osmotic pressure,they are isotonic.

(D) The correct matching of the methods with their respective colligative properties is:
$A$. Beckmann method $\rightarrow$ $3$. Depression in $F$.$P$.
$B$. Ostwald-Walker method $\rightarrow$ $4$. Relative lowering of vapour pressure
$C$. Berkeley-Hartley method $\rightarrow$ $1$. Osmotic pressure
$D$. Landsberger method $\rightarrow$ $2$. Elevation in $B$.$P$.
Therefore,the correct order is $A-3, B-4, C-1, D-2$.

(C) The depression in freezing point is a colligative property,which depends on the number of particles in the solution.
Glucose is a non-electrolyte and does not dissociate in water,so its van't Hoff factor $(i)$ is $1$.
$MgCl_2$ is a strong electrolyte that dissociates as $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$,producing $3$ ions per formula unit,so its van't Hoff factor $(i)$ is $3$.
Since the concentration of both solutions is $0.01 \ M$,the depression in freezing point for $MgCl_2$ will be approximately $3$ times that of the glucose solution.
Therefore,the correct option is $(C)$.

(D) The freezing point depression is a colligative property,which depends on the van't Hoff factor $(i)$,representing the number of particles produced per formula unit in solution.
Urea is a non-electrolyte,so $i = 1$.
$NaCl$ dissociates as $NaCl \rightarrow Na^+ + Cl^-$,so $i = 2$.
$Na_2SO_4$ dissociates as $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$,so $i = 3$.
$Al_2(SO_4)_3$ dissociates as $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$,so $i = 5$.
Since the freezing point depression $\Delta T_f = i \times K_f \times m$,the solution with the highest value of $i$ will have the maximum depression in freezing point,resulting in the lowest freezing point.
Thus,$1 \ M \ Al_2(SO_4)_3$ has the lowest freezing point. Hence,the correct option is $D$.

(A) Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
Step $1$: Calculate the number of moles of $NaOH$.
Moles of $NaOH = \frac{\text{mass}}{\text{molar mass}} = \frac{0.4 \ g}{40 \ g \ mol^{-1}} = 0.01 \ mol$.
Step $2$: Convert the volume of the solution to liters.
Volume $= 500 \ mL = 0.5 \ L$.
Step $3$: Calculate Molarity.
$M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.01 \ mol}{0.5 \ L} = 0.02 \ M$.

(C) Initial mass of solution $= 100 \ g$.
Mass of sucrose $= 40 \ g$ and mass of water $= 60 \ g$.
After heating,the mass of sucrose remains constant at $40 \ g$,but the concentration becomes $50 \%$.
Let the new mass of the solution be $m$.
$50 \% \text{ of } m = 40 \ g$.
$m = \frac{40}{0.50} = 80 \ g$.
Mass of water lost $= 100 \ g - 80 \ g = 20 \ g$.

(A) Schiff bases are formed by the reaction of a primary amine $(R-NH_2)$ with a carbonyl compound,such as an aldehyde or a ketone.
In this reaction,the amine attacks the carbonyl carbon to form a hemiaminal intermediate,which subsequently undergoes dehydration to produce an imine,commonly known as a Schiff base.
Therefore,$A$ represents an aldehyde or a ketone.

(D) Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
Step $1$: Calculate the number of moles of $NaOH$.
Moles of $NaOH = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{10 \ g}{40 \ g \ mol^{-1}} = 0.25 \ mol$.
Step $2$: Convert the volume of the solution to liters.
Volume $= 500 \ mL = 0.5 \ L$.
Step $3$: Calculate Molarity.
Molarity $= \frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{0.25 \ mol}{0.5 \ L} = 0.5 \ M$.

(C) The relationship between molarity $(M)$,molality $(m)$,and density $(d)$ is given by the formula:
$m = \frac{1000 \times M}{1000 \times d - M \times M_B}$
Rearranging for density $(d)$:
$d = \frac{M}{m} + \frac{M \times M_B}{1000} = \frac{1.44}{1.5} + \frac{1.44 \times 166}{1000}$
$d = 0.96 + 0.23904 = 1.19904 \ g \ mL^{-1} \approx 1.20 \ g \ mL^{-1}$
Here,$M = 1.44 \ mol \ L^{-1}$,$m = 1.5 \ mol \ kg^{-1}$,and $M_B (KI) = 166 \ g \ mol^{-1}$.

(B) $PPM$ (Parts Per Million) is defined as the mass of the solute in $mg$ per $kg$ of the solution.
Given,mass of toothpaste $= 500 \ g = 0.5 \ kg$.
Mass of fluorine $= 0.2 \ g = 200 \ mg$.
Concentration in $ppm = \frac{\text{mass of solute in } mg}{\text{mass of solution in } kg} = \frac{200 \ mg}{0.5 \ kg} = 400 \ ppm$.
$400 \ ppm$ can be written as $4 \times 10^2 \ ppm$.
Therefore,option $(B)$ is correct.

(B) Given: Molality $(m) = 5 \ mol / kg$,Density $(d) = 1.3 \ g / mL$,Molar mass of urea $(M_w) = 60.06 \ g / mol$.
$5$ molal solution means $5$ moles of urea are present in $1000 \ g$ of solvent (water).
Mass of urea = $5 \ mol \times 60.06 \ g / mol = 300.3 \ g$.
Total mass of solution = Mass of solute + Mass of solvent = $300.3 \ g + 1000 \ g = 1300.3 \ g$.
Volume of solution = $\frac{\text{Mass}}{\text{Density}} = \frac{1300.3 \ g}{1.3 \ g / mL} \approx 1000.23 \ mL = 1.00023 \ L$.
Molarity $(M) = \frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{5 \ mol}{1.00023 \ L} \approx 4.9988 \ M$.
Since $4.9988 \ M$ is approximately $5 \ M$,the correct option is $(B)$.

(B) Mass $\%$ of solute = $\frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$
Mass of solute = $1 \ g$
Mass of solvent = $18 \ g$
Mass of solution = $1 \ g + 18 \ g = 19 \ g$
$\therefore$ Mass $\%$ of solute = $\frac{1}{19} \times 100 \approx 5.26 \%$
Rounding to the nearest whole number,we get $5 \%$.
Hence,the correct option is $(B)$.

(C) Given,mass of $HCl = 18.25 \ g$.
Molar mass of $HCl = 36.5 \ g/mol$.
Mass of water $= 500 \ g = 0.5 \ kg$.
Number of moles of $HCl = \frac{\text{Mass of } HCl}{\text{Molar mass of } HCl} = \frac{18.25}{36.5} = 0.5 \ mol$.
Molality $= \frac{\text{Number of moles of solute}}{\text{Mass of solvent in } kg}$.
Molality $= \frac{0.5 \ mol}{0.5 \ kg} = 1 \ m$.
Hence,the molality of the solution is $1 \ m$,thus the correct option is $C$.

(C) Mole fraction is a unit of concentration,defined as the ratio of the number of moles of a specific component to the total number of moles of all components in the solution.
Since it is a ratio of two quantities with the same unit (moles),the units cancel out,making mole fraction a unitless quantity.
The sum of the mole fractions of all components in a solution is always equal to $1$.
Therefore,the correct option is $(C)$.

(D) The half-life of a reaction is the time required for the reactant concentration to decrease to one-half its initial value.
For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
From the graph,at $t = 30 \ minutes$,the concentration of reactant $[A]_t$ equals the concentration of product $[B_2]$.
Reaction: $A_5 \rightarrow 5 B_2$
Initial: $a \quad 0$
At $t = 30 \ min$: $a-x \quad 5x$
Given $a-x = 5x$,so $a = 6x$.
Substituting into the rate equation:
$k = \frac{2.303}{30} \log \frac{a}{a-x} = \frac{2.303}{30} \log \frac{6x}{x} = \frac{2.303}{30} \log 6 \approx \frac{2.303 \times 0.778}{30} \approx 0.0597 \ min^{-1}$.
The half-life is $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0597} \approx 116 \ minutes$.
Given the options,$114 \ minutes$ is the closest value.

(A) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Given: Molality $(m)$ = $0.25 \ mol/kg$,Mass of solution = $2.5 \ kg$.
Let the mass of urea be $w \ g$. Molar mass of urea $(NH_2CONH_2)$ = $60 \ g/mol$.
Moles of urea $(n)$ = $\frac{w}{60}$.
Mass of solvent = Mass of solution - Mass of solute = $(2500 - w) \ g = \frac{2500 - w}{1000} \ kg$.
Using the formula: $m = \frac{n}{Mass \ of \ solvent \ (kg)}$.
$0.25 = \frac{w/60}{(2500 - w)/1000}$.
$0.25 = \frac{w}{60} \times \frac{1000}{2500 - w}$.
$0.25 = \frac{1000w}{60(2500 - w)}$.
$15(2500 - w) = 1000w$.
$37500 - 15w = 1000w$.
$1015w = 37500$.
$w \approx 36.94 \ g \approx 37 \ g$.

(A) Let $R_1, R_2$,and $R_3$ be the rates of three reactions of first,second,and third-order,respectively,and $k$ be the rate constant for all three reactions.
The rate laws are:
$R_1 = k[A]^1$
$R_2 = k[A]^2$
$R_3 = k[A]^3$
where $[A]$ is the concentration of reactant $A$ in moles per litre.
Given that $[A] = 1$,we have:
$R_1 = k(1)^1 = k$
$R_2 = k(1)^2 = k$
$R_3 = k(1)^3 = k$
Therefore,$R_1 = R_2 = R_3$.

(B) $N$-ethylbenzene sulphonamide $(C_6H_5SO_2NHC_2H_5)$ is soluble in alkali because the hydrogen atom attached to the nitrogen atom is acidic due to the strong electron-withdrawing effect of the sulphonyl group $(-SO_2-)$.
This allows it to form a water-soluble salt with alkali.
$N,N$-diethylbenzene sulphonamide does not have an acidic hydrogen atom attached to the nitrogen,so it is insoluble in alkali.
Tertiary amines do not react with acid chlorides or Hinsberg's reagent $(C_6H_5SO_2Cl)$ because they lack an acidic hydrogen atom on the nitrogen.

(C) The 'spin only' magnetic moment $(\mu_{so})$ is calculated using the number of unpaired electrons $(n)$ in an atom or ion. The formula is given by:
$\mu_{so} = \sqrt{n(n+2)} \ BM$
where $BM$ stands for Bohr Magneton.
Hence,the correct option is $(C)$.

(D) Chemisorption is characterized by high specificity,high enthalpy of adsorption (typically $80-400 \ kJ \ mol^{-1}$),and it is an irreversible process.
Chemisorption results in the formation of a unimolecular layer on the surface of the adsorbent,not a multimolecular layer.
Therefore,option $(d)$ is not a characteristic of chemisorption.

(A) Initial surface area of the cube $= 6 \times (10 \ cm)^2 = 600 \ cm^2$.
When the cube is chopped lengthwise into $5$ equal pieces,each piece is a cuboid with dimensions $2 \ cm \times 10 \ cm \times 10 \ cm$.
Surface area of one such cuboid $= 2 \times (2 \times 10 + 10 \times 10 + 10 \times 2) = 2 \times (20 + 100 + 20) = 280 \ cm^2$.
Total surface area of $5$ such cuboids $= 5 \times 280 \ cm^2 = 1400 \ cm^2$.
Since adsorption power is directly proportional to the surface area,the ratio of final to initial adsorption power $= \frac{1400}{600} = \frac{14}{6} = 2.33$.
Thus,the adsorption power increases by $2.33$ times.

(B) The electronic configuration of $Sc^{3+}$ is $[Ar] 3d^0$.
Since there are no unpaired electrons in the $d$-orbital,$d-d$ transitions are not possible,making the ion colourless.
In contrast,$Ti^{3+}$ $(3d^1)$,$V^{4+}$ $(3d^1)$,and $Cr^{4+}$ $(3d^2)$ contain unpaired electrons,allowing for $d-d$ transitions,which result in the exhibition of colour.

(B) The adsorption theory of catalysis explains the mechanism of $Heterogeneous \ catalysis$.
According to this theory,the reactant molecules are adsorbed on the surface of the solid catalyst,which increases the concentration of reactants on the surface and facilitates the formation of an intermediate,thereby lowering the activation energy of the reaction.

(A) According to the Freundlich adsorption isotherm,the relationship is given by $\frac{x}{m} = K p^{\frac{1}{n}}$.
Taking the logarithm on both sides,we get $\log \left( \frac{x}{m} \right) = \log K + \frac{1}{n} \log p$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \left( \frac{x}{m} \right)$,$x = \log p$,$m = \frac{1}{n}$,and $c = \log K$.
Therefore,the slope of the straight line is $\frac{1}{n}$.

(C) The Freundlich adsorption isotherm represents the variation in the amount of adsorbate $(x/m)$ adsorbed on the surface of an adsorbent with the change in pressure $(p)$ at a constant temperature.
Physical adsorption is an exothermic process. According to Le Chatelier's principle,as the temperature increases,the extent of adsorption decreases.
In the given graph,for a constant pressure,the amount of adsorption $(x/m)$ follows the order: $T_1 > T_2 > T_3 > T_4$.
Since adsorption decreases with an increase in temperature,the corresponding temperatures must follow the order: $T_1 < T_2 < T_3 < T_4$.
Comparing the given values,the order $195 \ K < 244 \ K < 298 \ K < 303 \ K$ matches the trend shown in the graph for $T_1, T_2, T_3, T_4$ respectively.
Therefore,the correct option is $(C)$.

(B) Arsenic sulphide solution $(As_2S_3)$ is a hydrophobic sol,meaning it has little or no affinity for the dispersion medium $(H_2O)$.
Conversely,solutions of gum,starch,and protein are hydrophilic in nature because they are macromolecules (natural polymers) that have a strong affinity for the dispersion medium.

(C) The colloidal particles have a tendency to preferentially adsorb a particular type of ions from the solution.
$A$ colloidal particle usually adsorbs those ions which are in excess and are common to its own lattice.
This preferential adsorption of a particular type of ions imparts a particular type of charge to colloidal particles.
Hence,the correct option is $(C)$.

(B) Glycine is an amino acid with an isoelectric point $(pI)$ of $5.97$. At this $pH$,glycine exists as a zwitterion,meaning it has no net charge and does not move towards any electrode in an electric field.
When a base is added,the $pH$ of the solution increases above the $pI$ $(pH > 5.97)$.
In a basic medium,the amino acid loses a proton from the $-NH_3^+$ group,resulting in a net negative charge on the glycine molecule.
Since the colloidal particles become negatively charged,they will move towards the anode under the influence of an electric field.
The movement of negatively charged colloidal particles towards the anode is known as $Anaphoresis$.

(B) Sulphapyridine is a sulfonamide antibiotic. Its chemical structure consists of a benzene ring substituted with an amino group $(-NH_2)$ at the para position and a sulfonamide group $(-SO_2NH-)$ attached to a pyridine ring. The correct structure is shown in option $B$.