MHT CET 2011 Chemistry Question Paper with Answer and Solution

(C) The relationship between linear speed $v$ and angular velocity $\omega$ for a body moving in a circle of radius $r$ is given by the formula $v = r\omega$.
Rearranging this formula to solve for angular velocity,we get $\omega = v/r$.
Since both the speed $v$ and the radius $r$ are constant,the angular velocity $\omega$ is also constant.

(D) According to the law of conservation of mechanical energy,the kinetic energy given to the sphere at the lowest point must be equal to the potential energy gained at the height of the suspension point.
Let $m$ be the mass of the sphere and $v$ be the initial horizontal velocity.
At the lowest point,the kinetic energy is $K.E. = \frac{1}{2}mv^2$.
At the height of the suspension point,the potential energy is $P.E. = mgl$.
Equating the two:
$\frac{1}{2}mv^2 = mgl$
Solving for $v$:
$v^2 = 2gl$
$v = \sqrt{2gl}$

(C) The formula for escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$.
Let $M_e$ and $R_e$ be the mass and radius of the earth,and $M_m$ and $R_m$ be the mass and radius of the moon.
Given: $M_e = 81 M_m$ and $R_e = 3.5 R_m$.
The ratio of escape velocity on earth $(v_e)$ to that on the moon $(v_m)$ is given by:
$\frac{v_e}{v_m} = \sqrt{\frac{M_e}{R_e} \cdot \frac{R_m}{M_m}} = \sqrt{\frac{M_e}{M_m} \cdot \frac{R_m}{R_e}}$
Substituting the given values:
$\frac{v_e}{v_m} = \sqrt{81 \cdot \frac{1}{3.5}} = \sqrt{\frac{81}{3.5}} = \sqrt{23.14} \approx 4.81$.

(D) Young's modulus $Y$ is given by $Y = \frac{F L}{A \Delta l}$,where $A = \pi r^2$.
Thus,the force $F$ is given by $F = \frac{Y A \Delta l}{L} = \frac{Y (\pi r^2) \Delta l}{L}$.
Since the material is the same,$Y$ is constant. Since the increase in length $\Delta l$ is the same for both wires,we have $F \propto \frac{r^2}{L}$.
Therefore,the ratio of the forces is $\frac{F_A}{F_B} = \left( \frac{r_A}{r_B} \right)^2 \times \left( \frac{L_B}{L_A} \right)$.
Given $\frac{L_A}{L_B} = \frac{1}{2}$ and the ratio of diameters is $2:1$,so $\frac{r_A}{r_B} = \frac{2}{1}$.
Substituting these values: $\frac{F_A}{F_B} = (2)^2 \times (2) = 4 \times 2 = 8$.
Thus,the ratio is $8:1$.

(B) Using the ideal gas equation $PV = nRT = (m/M)RT$,where $m$ is the mass and $M$ is the molar mass. Since $V$ and $M$ are constant,we have $P \propto mT$.
Let the initial state be $(P_1, m_1, T_1)$ and the final state be $(P_2, m_2, T_2)$.
Given: $P_1 = 10 \ atm$,$T_1 = 27 + 273 = 300 \ K$,$m_1 = m$.
After removing half the mass: $m_2 = m/2$.
Final temperature: $T_2 = 87 + 273 = 360 \ K$.
Using the ratio: $\frac{P_2}{P_1} = \frac{m_2}{m_1} \times \frac{T_2}{T_1}$
$\frac{P_2}{10} = \frac{m/2}{m} \times \frac{360}{300}$
$\frac{P_2}{10} = \frac{1}{2} \times \frac{6}{5}$
$\frac{P_2}{10} = \frac{3}{5}$
$P_2 = 10 \times \frac{3}{5} = 6 \ atm$.

(A) According to Stefan-Boltzmann Law,the rate of heat radiation per unit area $E$ is proportional to the fourth power of the absolute temperature $T$,i.e.,$E \propto T^4$.
Given:
$T_1 = 227^{\circ}C = 227 + 273 = 500 \ K$
$E_1 = 5 \ cal/cm^2-sec$
$T_2 = 727^{\circ}C = 727 + 273 = 1000 \ K$
Using the ratio formula:
$\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4$
$\frac{E_2}{5} = \left( \frac{1000}{500} \right)^4$
$\frac{E_2}{5} = (2)^4 = 16$
$E_2 = 16 \times 5 = 80 \ cal/cm^2-sec$.
Thus,the correct option is $A$.

(D) The displacement of a particle in simple harmonic motion starting from the mean position is given by $y = A \sin(\omega t)$.
Here,$A$ is the amplitude,$\omega = \frac{2\pi}{T}$ is the angular frequency,and $T$ is the time period.
We want to find the time $t$ when the displacement $y = \frac{A}{2}$.
Substituting the values into the equation: $\frac{A}{2} = A \sin\left(\frac{2\pi t}{T}\right)$.
Dividing both sides by $A$: $\frac{1}{2} = \sin\left(\frac{2\pi t}{T}\right)$.
Since $\sin(30^\circ) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$,we have $\frac{2\pi t}{T} = \frac{\pi}{6}$.
Solving for $t$: $t = \frac{T}{12}$.

(A) The work done $W$ in stretching a spring by an extension $x$ is given by the formula $W = \frac{1}{2} k x^2$.
Since the extension $x$ is the same for both springs,the work done is directly proportional to the spring constant $k$ $(W \propto k)$.
Given that $K_A > K_B$,it follows that $W_A > W_B$.
Therefore,more work is required to stretch spring $A$.

(A) The standard equation for a stationary wave is given by $y = 2a \sin \left( \frac{2 \pi x}{\lambda} \right) \cos (\omega t)$.
Comparing the given equation $y = 4 \sin \left( \frac{\pi x}{15} \right) \cos (96 \pi t)$ with the standard form,we get:
$\frac{2 \pi}{\lambda} = \frac{\pi}{15}$.
Solving for $\lambda$:
$\lambda = 30$.
The distance between a node and the next consecutive antinode is given by $\frac{\lambda}{4}$.
Therefore,the distance $= \frac{30}{4} = 7.5$.

(A) The speed of sound in a gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$. As the temperature $T$ increases,the speed of sound $v$ increases.
For an organ pipe of length $L$,the frequency $n$ is given by $n = \frac{v}{\lambda}$,where $\lambda$ depends on the length of the pipe and the mode of vibration.
Since the length of the pipe $L$ remains effectively constant (ignoring thermal expansion),the wavelength $\lambda$ remains unchanged.
Therefore,as $v$ increases,the frequency $n$ must also increase.

(A) For a closed resonance tube,the resonance lengths are given by $l_1 = \frac{\lambda}{4}$ and $l_2 = \frac{3\lambda}{4}$.
The difference between the two resonance lengths is $l_2 - l_1 = \frac{\lambda}{2}$.
Given $l_1 = 16 \ cm = 0.16 \ m$ and $l_2 = 49 \ cm = 0.49 \ m$.
Thus,$\frac{\lambda}{2} = 0.49 \ m - 0.16 \ m = 0.33 \ m$,which implies $\lambda = 0.66 \ m$.
Using the wave equation $v = n\lambda$,where $v$ is the velocity of sound and $n$ is the frequency:
$n = \frac{v}{\lambda} = \frac{330 \ m/s}{0.66 \ m} = 500 \ Hz$.

(D) The momentum $p$ of a charged particle accelerated from rest through a potential difference $V$ is given by $p = \sqrt{2mK}$,where $K = qV$ is the kinetic energy.
Thus,$p = \sqrt{2mqV}$.
Since $V$ is the same for both particles,the ratio of momenta is $\frac{p_e}{p_\alpha} = \sqrt{\frac{m_e q_e}{m_\alpha q_\alpha}}$.
For an electron,$q_e = e$. For an $\alpha$-particle,$q_\alpha = 2e$.
Substituting these values,we get $\frac{p_e}{p_\alpha} = \sqrt{\frac{m_e \cdot e}{m_\alpha \cdot 2e}} = \sqrt{\frac{m_e}{2m_\alpha}}$.

(D) The momentum $p$ of a charged particle accelerated from rest through a potential difference $V$ is given by $p = \sqrt{2mqV}$,where $m$ is the mass and $q$ is the charge of the particle.
Since $V$ is the same for both particles,$p \propto \sqrt{mq}$.
For an electron,$m = m_e$ and $q = e$.
For an $\alpha$-particle,$m = m_\alpha$ and $q = 2e$.
Therefore,the ratio of the momenta is $\frac{p_e}{p_\alpha} = \sqrt{\frac{m_e \cdot e}{m_\alpha \cdot 2e}} = \sqrt{\frac{m_e}{2m_\alpha}}$.

(B) Hess's law of constant heat summation states that the total enthalpy change for a chemical reaction is the same,whether the reaction takes place in one step or in several steps.
Therefore,the heat of reaction depends only on the initial state of the reactants and the final state of the products.

(C) When $3$ moles of benzaldehyde $(C_6H_5CHO)$ react with $2$ moles of ammonia $(NH_3)$,they undergo a condensation reaction to form hydrobenzamide $( (C_6H_5CH=N)_2CH-C_6H_5 )$ and release $3$ moles of water $(H_2O)$.
The chemical equation is:
$3C_6H_5CHO + 2NH_3 \rightarrow (C_6H_5CH=N)_2CH-C_6H_5 + 3H_2O$
Thus,the correct option is $(C)$.

(B) The bisector of the angle between the positive directions of the axes is the line $y = x$.
Since this line is one of the lines represented by the pair $ax^2 + 2hxy + by^2 = 0$,it must satisfy the equation.
Substituting $y = x$ into the equation,we get:
$ax^2 + 2hx(x) + b(x)^2 = 0$
$ax^2 + 2hx^2 + bx^2 = 0$
$x^2(a + 2h + b) = 0$
Since this holds for all $x$,we must have $a + 2h + b = 0$,which simplifies to $a + b = -2h$.

(A) The coordinates of the ends of the latus rectum of the parabola $y^2 = 4ax$ are $(a, 2a)$ and $(a, -2a)$.
The equation of the normal at $(x_1, y_1)$ to $y^2 = 4ax$ is given by $y - y_1 = -\frac{y_1}{2a}(x - x_1)$.
For the point $(a, 2a)$,the normal is $y - 2a = -\frac{2a}{2a}(x - a) \implies y - 2a = -(x - a) \implies x + y - 3a = 0$ $(i)$.
For the point $(a, -2a)$,the normal is $y - (-2a) = -\frac{-2a}{2a}(x - a) \implies y + 2a = 1(x - a) \implies x - y - 3a = 0$ $(ii)$.
The combined equation is $(x + y - 3a)(x - y - 3a) = 0$.
This simplifies to $((x - 3a) + y)((x - 3a) - y) = 0 \implies (x - 3a)^2 - y^2 = 0$.
Expanding this,we get $x^2 - 6ax + 9a^2 - y^2 = 0$,which is $x^2 - y^2 - 6ax + 9a^2 = 0$.

(B) The inverse of a conditional statement $P \Rightarrow Q$ is defined as $\sim P \Rightarrow \sim Q$.
For the given proposition $(p \wedge \sim q) \Rightarrow r$,we identify $P = (p \wedge \sim q)$ and $Q = r$.
The inverse is $\sim (p \wedge \sim q) \Rightarrow \sim r$.
Using De Morgan's Law,$\sim (p \wedge \sim q)$ is equivalent to $\sim p \vee \sim (\sim q)$,which simplifies to $\sim p \vee q$.
Therefore,the inverse is $(\sim p \vee q) \Rightarrow \sim r$.

(C) To determine the nature of the proposition $(p$ $\Rightarrow \;\sim p) \wedge (\sim p$ $\Rightarrow p)$,we construct a truth table:

$p$	$\sim p$	$p \Rightarrow \;\sim p$	$\sim p \Rightarrow p$	$(p$ $\Rightarrow \;\sim p) \wedge (\sim p$ $\Rightarrow p)$
$T$	$F$	$F$	$T$	$F$
$F$	$T$	$T$	$F$	$F$

Since the truth value of the proposition is $F$ for all possible values of $p$,it is a contradiction.

(C) In option $A$,a coordinate (dative) bond is formed between the lone pair of the Oxygen atom and the empty $1s$ orbital of the $H^+$ ion.
In option $B$,the Boron atom has an empty $2p$ orbital after the formation of $BF_3$,so it accepts a lone pair from a Fluoride ion $(F^-)$ to form $BF_4^-$.
In option $C$,$HF_2^-$ involves a strong Hydrogen bond between $F^-$ and $HF$,not a coordinate bond.
In option $D$,the Nitrogen atom in $NH_3$ has one lone pair,which it shares with an $H^+$ ion to form a coordinate bond in $NH_4^+$.
Therefore,$HF_2^-$ is the correct answer.

(C) Step $1$: Reaction of ethyne with excess $HCl$ gives $1,1-$dichloroethane $(A)$.
$CH \equiv CH + 2HCl \rightarrow CH_3-CHCl_2 (A)$
Step $2$: Hydrolysis of $1,1-$dichloroethane $(A)$ followed by heating gives ethanal $(B)$.
$CH_3-CHCl_2 + H_2O \xrightarrow{\Delta} CH_3CHO (B) + 2HCl$
Step $3$: Reduction of ethanal $(B)$ with $Na-Hg / H_2O$ gives ethanol $(C)$.
$CH_3CHO + 2[H] \xrightarrow{Na-Hg / H_2O} CH_3CH_2OH (C)$
Thus,compound $C$ is ethanol.

(D) The transition $M_{(g)}^{2+} \longrightarrow M_{(g)}^{3+}$ involves the maximum amount of energy.
As the positive charge on the ion increases,the effective nuclear charge per electron increases.
This results in a stronger electrostatic attraction between the nucleus and the remaining electrons.
Therefore,the energy required to remove the $3^{rd}$ electron (third ionization energy) is significantly higher than the energy required for the removal of the first or second electron.

(C) coordinate bond (or dative bond) is formed when one atom donates a lone pair of electrons to another atom that needs them to complete its octet.
$H_{3}O^{+}$ contains a coordinate bond between $O$ and $H^{+}$.
$BF_{4}^{-}$ contains a coordinate bond between $F^{-}$ and $BF_{3}$.
$NH_{4}^{+}$ contains a coordinate bond between $N$ and $H^{+}$.
$HF_{2}^{-}$ is formed by hydrogen bonding between $F^{-}$ and $HF$,represented as $[F-H...F]^{-}$. It does not contain a coordinate bond.

(D) In $SnCl_{2}$,the central atom $Sn$ has a valence shell configuration of $5s^{2} 5p^{2}$. It forms two covalent bonds with two $Cl$ atoms,utilizing two $5p$ electrons. This leaves one lone pair in the $5s$ orbital. The total number of electron domains around $Sn$ is $3$ ($2$ bond pairs + $1$ lone pair),which corresponds to $sp^{2}$ hybridization. Due to the presence of one lone pair,the geometry is bent or $V$-shaped.

(A) Antibiotics are classified as bactericidal or bacteriostatic based on their action.
Bactericidal antibiotics kill the microorganisms in the body.
Examples of bactericidal antibiotics include $Penicillin$,$Ofloxacin$,and $Aminoglycosides$.
Bacteriostatic antibiotics inhibit the growth of microorganisms.
Examples of bacteriostatic antibiotics include $Chloramphenicol$,$Erythromycin$,and $Tetracycline$.
Therefore,$Ofloxacin$ is a bactericidal antibiotic.

(C) $1$. Identify the parent chain: The compound consists of two cyclobutane rings connected by a single bond,forming a $1, 1'-$bicyclobutane system.
$2$. Numbering: Number the rings such that the substituents get the lowest possible locants. The connection point is $1$ and $1'$.
$3$. Substituents: There is a bromine atom at position $3$ and a chlorine atom at position $3'$.
$4$. Alphabetical order: $B$ (bromo) comes before $C$ (chloro).
$5$. Final Name: $3-$bromo$-3'-$chloro$-1, 1'-$bicyclobutane.

(D) Electrophiles are electron-loving chemical species. These may be either positively charged or electrically neutral species with an incomplete octet or vacant orbitals.
$H_{2}O$,$NH_{3}$,and $C_{2}H_{5}OH$ are nucleophiles because they possess lone pairs of electrons.
$SO_{3}$ is an electrophile because the sulfur atom is bonded to three highly electronegative oxygen atoms,making it electron-deficient. It can be represented as a resonance hybrid of the following structures:
$O=S(=O)-O^{-} \leftrightarrow -O-S(=O)=O \leftrightarrow -O-S(=O)=O$ (with a positive charge on $S$ in each resonance structure).
Due to the partial positive charge on the $S$ atom,it acts as an electrophile.

(B) The central carbon atom of a free radical contains $7$ electrons.
Free radicals are paramagnetic due to the presence of an unpaired electron and are formed by homolysis of covalent bonds either by heat or by light.

(D) The reaction of benzene with dichloromethane $(CH_2Cl_2)$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction.
Since benzene is present in excess,the reaction proceeds as follows:
$2C_6H_6 + CH_2Cl_2 \xrightarrow{\text{Anhy. } AlCl_3} C_6H_5-CH_2-C_6H_5 + 2HCl$
The product formed is diphenylmethane.

(A) $Neo$-pentane ($2,2$-dimethylpropane) has all $12$ hydrogen atoms equivalent. Therefore,its monochlorination yields only one product,$1$-chloro-$2,2$-dimethylpropane.
$n$-butane gives two isomers ($1$-chlorobutane and $2$-chlorobutane).
$Iso$-butane gives two isomers ($1$-chloro-$2$-methylpropane and $2$-chloro-$2$-methylpropane).
$Iso$-pentane gives four isomers.
Thus,$Neo$-pentane does not give different isomers on monochlorination.

(D) Given,$pH = 6$.
Therefore,the initial concentration of $[H^{+}] = 10^{-6} \ M$.
After diluting the solution $1000$ times,the new concentration of $[H^{+}]$ from the acid is $[H^{+}]_{acid} = \frac{10^{-6}}{1000} = 10^{-9} \ M$.
Since this concentration is very low,the contribution of $[H^{+}]$ from the auto-ionization of water $(10^{-7} \ M)$ cannot be neglected.
Total $[H^{+}] = [H^{+}]_{acid} + [H^{+}]_{water} = 10^{-9} + 10^{-7} \ M$.
Total $[H^{+}] = 10^{-7} (0.01 + 1) = 1.01 \times 10^{-7} \ M$.
$pH = -\log(1.01 \times 10^{-7}) = 7 - \log(1.01) \approx 7 - 0.0043 = 6.9957$.
Thus,the $pH$ of the final solution is approximately $6.99$.

(C) $KNO_3$ is a salt formed from a strong acid $(HNO_3)$ and a strong base $(KOH)$.
Salts of strong acids and strong bases do not undergo hydrolysis because their constituent ions do not react with water to change the $pH$ of the solution.
Therefore,$KNO_3$ will not undergo hydrolysis.

(A) The dissociation of $Ag_{2}CrO_{4}$ is given by: $Ag_{2}CrO_{4} \rightleftharpoons 2 Ag^{+} + CrO_{4}^{2-}$.
Let the solubility be $s$. Then $[Ag^{+}] = 2s$ and $[CrO_{4}^{2-}] = s$.
The solubility product expression is $K_{sp} = [Ag^{+}]^{2} [CrO_{4}^{2-}]$.
Substituting the values: $K_{sp} = (2s)^{2} (s) = 4s^{3}$.
Given $K_{sp} = 32 \times 10^{-12}$,we have $4s^{3} = 32 \times 10^{-12}$.
$s^{3} = 8 \times 10^{-12}$.
$s = (8 \times 10^{-12})^{1/3} = 2 \times 10^{-4} \ M$.
Since $[CrO_{4}^{2-}] = s$,the concentration is $2 \times 10^{-4} \ M$.

(C) In both silica $(SiO_{2})$ and silicates,the fundamental building block is the orthosilicate unit,which is the tetrahedral ($SiO_{4})^{4-}$ anion.
In this structure,one silicon atom is bonded to four oxygen atoms arranged at the corners of a tetrahedron.

(C) When $Zn$ reacts with an excess of $NaOH$ solution,it forms sodium zincate and releases hydrogen gas.
The reaction is:
$Zn + 2NaOH + 2H_2O \longrightarrow Na_2[Zn(OH)_4] + H_2$
This can also be represented as:
$Zn + 2NaOH \longrightarrow Na_2ZnO_2 + H_2$
Thus,the product obtained is sodium zincate $(Na_2ZnO_2)$.

(B) The atomic mass unit $(amu)$,also known as the unified atomic mass unit $(u)$,is defined as $1/12$ of the mass of a carbon-$12$ atom.
$1 \ amu = 1.66057 \times 10^{-27} \ kg$.
Since $1 \ kg = 10^3 \ g$,we convert the value to grams:
$1 \ amu = 1.66057 \times 10^{-27} \ kg \times 10^3 \ g/kg = 1.66057 \times 10^{-24} \ g$.
Therefore,the correct value is $1.66 \times 10^{-24} \ g$.

(A) The normality $(N)$ of a solution is related to its molarity $(M)$ by the formula: $N = M \times \text{n-factor}$.
For $H_2SO_4$, the n-factor is $2$ (since it is a dibasic acid).
Therefore, the normality of $2 \,M \,H_2SO_4$ is $2 \times 2 = 4 \,N$.
Using the dilution equation $N_1 V_1 = N_2 V_2$:
$4 \,N \times V_1 = 0.2 \,N \times 100 \,mL$.
$V_1 = \frac{0.2 \times 100}{4} \,mL$.
$V_1 = 5 \,mL$.

(C) According to Heisenberg's uncertainty principle: $\Delta x \cdot m \Delta v \geq \frac{h}{4 \pi}$.
Given velocity $v = 300 \ ms^{-1}$ and accuracy is $0.001 \ \%$,so uncertainty in velocity $\Delta v = 300 \times \frac{0.001}{100} = 3 \times 10^{-3} \ ms^{-1}$.
Substituting the values: $\Delta x = \frac{h}{4 \pi m \Delta v} = \frac{6.63 \times 10^{-34}}{4 \times 3.1416 \times 9.1 \times 10^{-31} \times 3 \times 10^{-3}}$.
$\Delta x = \frac{6.63 \times 10^{-34}}{34.21 \times 10^{-33}} \approx 0.1938 \times 10^{-1} \ m = 1.938 \times 10^{-2} \ m$.
The closest value is $1.92 \times 10^{-2} \ m$.

(C) The combustion reaction is: $C(s) + O_{2}(g) \longrightarrow CO_{2}(g)$
The enthalpy change for the formation of $1 \ mol$ $(44 \ g)$ of $CO_{2}$ is $\Delta H = -393.5 \ kJ / mol$.
Heat released for $44 \ g$ of $CO_{2} = 393.5 \ kJ$.
Heat released for $1 \ g$ of $CO_{2} = \frac{393.5}{44} \ kJ$.
Heat released for $35.2 \ g$ of $CO_{2} = \frac{393.5}{44} \times 35.2 \ kJ = 314.8 \ kJ \approx 315 \ kJ$.
Since heat is released,the value is $-315 \ kJ$.

(B) According to Hess's law,the total enthalpy change of a reaction is the same whether the reaction occurs in one step or in several steps. Therefore,the heat of reaction depends only on the initial and final states of the reactants and products,not on the path taken.

(B) The reaction between a strong acid $(HNO_3)$ and a strong base $(KOH)$ is represented as: $H^+ + OH^- \rightarrow H_2O$; $\Delta H = -57.0 \ kJ \ mol^{-1}$.
Since $0.2 \ mol$ of $KOH$ is the limiting reagent,it will react with $0.2 \ mol$ of $HNO_3$ to produce $0.2 \ mol$ of $H_2O$.
The heat evolved is calculated as: $\text{Heat} = \Delta H \times \text{moles of water formed} = 57.0 \ kJ \ mol^{-1} \times 0.2 \ mol = 11.4 \ kJ$.

(D) The equation of the given circle is $x^{2}+y^{2}+4x-6y+9 \sin^{2} \alpha + 13 \cos^{2} \alpha = 0$.
Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get center $A(-2, 3)$ and radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{4+9-(9 \sin^{2} \alpha + 13 \cos^{2} \alpha)} = \sqrt{13 - 9 \sin^{2} \alpha - 13 \cos^{2} \alpha} = \sqrt{13(1-\cos^{2} \alpha) - 9 \sin^{2} \alpha} = \sqrt{13 \sin^{2} \alpha - 9 \sin^{2} \alpha} = \sqrt{4 \sin^{2} \alpha} = 2 \sin \alpha$.
Let $P(h, k)$ be the point. In $\Delta ABP$,where $B$ is the point of contact,$\angle APB = \alpha$ and $\angle ABP = 90^{\circ}$.
Thus,$\sin \alpha = \frac{AB}{AP} = \frac{r}{AP} = \frac{2 \sin \alpha}{\sqrt{(h+2)^{2}+(k-3)^{2}}}$.
$\Rightarrow \sqrt{(h+2)^{2}+(k-3)^{2}} = 2$.
Squaring both sides,$(h+2)^{2}+(k-3)^{2} = 4$.
$h^{2}+4h+4+k^{2}-6k+9 = 4$.
$h^{2}+k^{2}+4h-6k+9 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^{2}+y^{2}+4x-6y+9=0$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For a doubly ionized lithium atom $(Li^{2+})$, the atomic number $Z = 3$.
In the ground state, the principal quantum number $n = 1$.
Substituting these values into the formula: $E_1 = -13.6 \times \frac{3^2}{1^2} \ eV = -13.6 \times 9 \ eV = -122.4 \ eV$.
The energy required to remove the electron (ionization energy) is the energy needed to bring the electron from the ground state to infinity $(E_{\infty} = 0)$.
Therefore, Ionization Energy $= E_{\infty} - E_1 = 0 - (-122.4 \ eV) = +122.4 \ eV$.

(B) The surface energy $E$ of a soap bubble is given by the product of surface tension $T$ and the total surface area. Since a soap bubble has two surfaces (inner and outer),the total surface area is $2 \times 4 \pi R^{2} = 8 \pi R^{2}$.
Initial surface energy $E_{i} = T \times (2 \times 4 \pi R^{2}) = 8 \pi R^{2} T$.
When the radius is doubled,the new radius becomes $R' = 2R$.
The final surface energy $E_{f} = T \times (2 \times 4 \pi (2R)^{2}) = T \times (2 \times 4 \pi \times 4R^{2}) = 32 \pi R^{2} T$.
The work done $W$ is the change in surface energy:
$W = E_{f} - E_{i} = 32 \pi R^{2} T - 8 \pi R^{2} T = 24 \pi R^{2} T$.

(B) To prepare $3-$methylpentan$-3-$ol,we need a Grignard reagent to attack an ester. The structure of $3-$methylpentan$-3-$ol is $CH_3CH_2-C(OH)(CH_3)-CH_2CH_3$.
This tertiary alcohol can be formed by the reaction of an ester with two equivalents of a Grignard reagent.
Specifically,reacting ethyl propanoate $(CH_3CH_2COOC_2H_5)$ with two equivalents of methyl magnesium bromide $(CH_3MgBr)$ yields $3-$methylpentan$-3-$ol.
The reaction proceeds as follows:
$CH_3CH_2COOC_2H_5 + CH_3MgBr \rightarrow CH_3CH_2COCH_3 + C_2H_5OMgBr$
$CH_3CH_2COCH_3 + CH_3MgBr \rightarrow CH_3CH_2C(OMgBr)(CH_3)_2$
After hydrolysis: $CH_3CH_2C(OH)(CH_3)_2$ (which is $2-$methylbutan$-2-$ol). Wait,let us re-evaluate.
To get $3-$methylpentan$-3-$ol $(CH_3CH_2-C(OH)(CH_3)-CH_2CH_3)$,we need to react ethyl ethanoate $(CH_3COOC_2H_5)$ with two equivalents of ethyl magnesium bromide $(C_2H_5MgBr)$.
$CH_3COOC_2H_5 + 2C_2H_5MgBr \rightarrow CH_3C(OH)(C_2H_5)_2$.
Thus,the correct option is $B$.

(D) The reaction of alcohols with Lucas reagent (conc. $HCl +$ anhydrous $ZnCl_2$) is used to distinguish between primary,secondary,and tertiary alcohols.
Tertiary alcohols react most rapidly and give immediate turbidity at room temperature due to the formation of stable carbocations.
Among the given options,$2-$methylpropan$-2-$ol is a tertiary alcohol,while the others are primary or secondary alcohols.
Therefore,$2-$methylpropan$-2-$ol gives immediate turbidity.

(C) The rate of $S_{N}2$ reactions is significantly enhanced in polar aprotic solvents.
Polar aprotic solvents like $DMSO$ (Dimethyl sulfoxide) do not form hydrogen bonds with the nucleophile,leaving it 'naked' and more reactive.
In contrast,polar protic solvents like $CH_{3}OH$ and $H_{2}O$ solvate the nucleophile through hydrogen bonding,which decreases its nucleophilicity and slows down the $S_{N}2$ reaction.

(C) The reaction of an alcohol $(R-OH)$ with diazomethane $(CH_2N_2)$ in the presence of a catalyst like $HBF_4$ results in the formation of a methyl ether $(R-OCH_3)$ and the release of nitrogen gas $(N_2)$.
For the reaction of propan$-1-$ol $(CH_3CH_2CH_2OH)$ with diazomethane $(CH_2N_2)$:
$CH_3CH_2CH_2OH + CH_2N_2 \xrightarrow{HBF_4} CH_3CH_2CH_2OCH_3 + N_2$
The product formed is $1-$methoxypropane.

(D) Acetone $(CH_3COCH_3)$ undergoes aldol condensation in the presence of a base like $Ba(OH)_2$ to form $4-$hydroxy$-4-$methylpentan$-2-$one (diacetone alcohol).
$2CH_3COCH_3 \xrightarrow{Ba(OH)_2} (CH_3)_2C(OH)CH_2COCH_3$
Upon dehydration (heating with acid),the $4-$hydroxy$-4-$methylpentan$-2-$one loses a water molecule to form $4-$methylpent$-3-$en$-2-$one (mesityl oxide).
$(CH_3)_2C(OH)CH_2COCH_3 \xrightarrow{\Delta, H^+} (CH_3)_2C=CHCOCH_3 + H_2O$
Thus,the final product is $4-$methylpent$-3-$en$-2-$one.

(C) The reaction of benzaldehyde with ammonia is a condensation reaction. Three molecules of benzaldehyde $(3C_6H_5CHO)$ react with two molecules of ammonia $(2NH_3)$ to produce hydrobenzamide $((C_6H_5CH)_3N_2)$ and three molecules of water $(3H_2O)$.
The balanced chemical equation is:
$3C_6H_5CHO + 2NH_3 \rightarrow (C_6H_5CH)_3N_2 + 3H_2O$

(D) In aniline,$N$-methylaniline,and o-toluidine,the lone pair of electrons on the nitrogen atom is in conjugation with the benzene ring,which decreases its availability for protonation,making them weaker bases.
In benzylamine $(C_6H_5CH_2NH_2)$,the lone pair on the nitrogen atom is not in conjugation with the benzene ring because of the presence of a $-CH_2-$ group between the benzene ring and the nitrogen atom.
Therefore,the lone pair is more available for protonation,making benzylamine the strongest base among the given compounds.

(C) Step $1$: Reaction of isopropylamine with $HNO_2$ gives propan-$2$-ol $(A)$.
$CH_3-CH(NH_2)-CH_3 + HNO_2 \rightarrow CH_3-CH(OH)-CH_3 + N_2 + H_2O$
Step $2$: Oxidation of propan-$2$-ol $(A)$ gives acetone $(B)$.
$CH_3-CH(OH)-CH_3 \xrightarrow{[O]} CH_3-CO-CH_3$
Step $3$: Reaction of acetone $(B)$ with methylmagnesium iodide $(CH_3MgI)$ followed by hydrolysis gives $2$-methylpropan-$2$-ol $(C)$.
$CH_3-CO-CH_3 + CH_3MgI$ $\rightarrow (CH_3)_3C-OMgI$ $\xrightarrow{H_2O/H^+} (CH_3)_3C-OH$
Thus,the final product $C$ is $2$-methylpropan-$2$-ol.

(A) Aldehydes and ketones react with hydroxylamine $(NH_2OH)$ to form oximes,which is a condensation reaction.
Self-condensation (Aldol condensation) requires the presence of at least one $\alpha$-hydrogen atom.
Methanal $(HCHO)$ does not contain any $\alpha$-hydrogen atom,so it cannot undergo self-aldol condensation.
Propanal $(CH_3CH_2CHO)$,Acetone $(CH_3COCH_3)$,and Ethanal $(CH_3CHO)$ all contain $\alpha$-hydrogen atoms and thus undergo self-condensation.
Therefore,the correct answer is Methanal.

(B) $Br_{2}$ water is a mild oxidizing agent.
Glucose $(CHO(CHOH)_{4}CH_{2}OH)$ on oxidation with $Br_{2}$ water selectively oxidizes the aldehyde group $(-CHO)$ to a carboxylic acid group $(-COOH)$,resulting in the formation of gluconic acid $(COOH(CHOH)_{4}CH_{2}OH)$.
The reaction is:
$CHO(CHOH)_{4}CH_{2}OH + [O] \xrightarrow{Br_{2}, H_{2}O} COOH(CHOH)_{4}CH_{2}OH$

(C) Haemoglobin is a conjugated protein consisting of a protein part called globin and a non-protein prosthetic group called haem. The haem group contains an $Fe^{2+}$ ion coordinated to a porphyrin ring.

(B) Malonic acid $(CH_{2}(COOH)_{2})$ undergoes decarboxylation upon heating to produce acetic acid $(CH_{3}COOH)$ and carbon dioxide $(CO_{2})$.
The reaction is as follows:
$CH_{2}(COOH)_{2} \xrightarrow{\Delta} CH_{3}COOH + CO_{2}$

(A) The rate of a chemical reaction depends upon the slowest step in a multistep mechanism,which is known as the rate-determining step.
Therefore,the overall rate of a multistep reaction is equal to the rate of the slowest step.

(C) For a first order reaction,the rate law is given by $-\frac{d[A]}{dt} = k[A]$.
Integrating this expression from time $t=0$ (where $[A] = a$) to time $t$ (where $[A] = a-x$),we get:
$k = \frac{2.303}{t} \log \frac{a}{a-x}$
where '$a$' is the initial concentration and '$a-x$' is the concentration at time '$t$'.

(A) The nuclear reaction is: $_{90}Th^{228}$ $\xrightarrow{-4 \alpha} _{82}Pb^{212}$ $\xrightarrow{-1 \beta} _{83}Bi^{212}$.
The daughter element is $_{83}Bi^{212}$.
The number of neutrons is calculated as: $\text{Mass number} - \text{Atomic number}$.
$\text{Number of neutrons} = 212 - 83 = 129$.

(C) Radioactive disintegration follows first-order kinetics.
The rate law is given by $k = \frac{2.303}{t} \log \frac{N_0}{N}$.
Here,$k$ is the decay constant (or disintegration constant).
Since the unit of time $t$ is in the denominator,the unit of $k$ is $\text{time}^{-1}$ (e.g.,$\text{s}^{-1}$,$\text{min}^{-1}$,or $\text{year}^{-1}$).

(B) The electronic configuration of $Ce$ $(Z=58)$ is $[Xe] 4f^1 5d^1 6s^2$.
Cerium $(Ce)$ exhibits $+3$ and $+4$ oxidation states.
The $+3$ state is the most stable,while the $+4$ state is also common due to the attainment of a stable noble gas configuration $([Xe])$ upon losing four electrons.

(D) The given cell reaction is: $Fe^{2+} + Sn \longrightarrow Fe + Sn^{2+}$
In this reaction,$Fe^{2+}$ is reduced to $Fe$ (cathode) and $Sn$ is oxidized to $Sn^{2+}$ (anode).
The standard reduction potentials are given as: $E^o_{Fe^{2+}/Fe} = -0.44 \ V$ and $E^o_{Sn^{2+}/Sn} = -0.14 \ V$.
The standard emf of the cell $(E^o_{cell})$ is calculated as:
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
$E^o_{cell} = E^o_{Fe^{2+}/Fe} - E^o_{Sn^{2+}/Sn}$
$E^o_{cell} = -0.44 \ V - (-0.14 \ V)$
$E^o_{cell} = -0.44 + 0.14 = -0.30 \ V$

(B) According to Faraday's first law of electrolysis,the mass of the substance deposited $(w)$ is directly proportional to the quantity of electricity $(Q)$ passed through the electrolyte.
$w \propto Q$
$w = Z \cdot Q$
Since $Q = I \times t$ and $Z = \frac{E}{96500}$ (where $E$ is the equivalent weight and $96500$ is Faraday's constant),
$w = \frac{E \times I \times t}{96500}$.

(B) Metamers are isomers that have the same molecular formula but differ in the nature of the alkyl groups attached to the same polyvalent functional group (in this case,the secondary amine group,$-NH-$).
For the molecular formula $C_4H_{11}N$,the secondary amines ($2^{\circ}$ amines) are:
$1$) $CH_3CH_2-NH-CH_2CH_3$ (Diethylamine)
$2$) $CH_3-NH-CH_2CH_2CH_3$ (Methylpropylamine)
$3$) $CH_3-NH-CH(CH_3)_2$ (Methylisopropylamine)
Thus,there are $3$ possible metamers.

(B) The thermal decomposition of $TiI_{4}$ is a key step in the van Arkel-de Boer process used for the refining of titanium metal.
Upon heating,$TiI_{4}$ decomposes into pure titanium metal and iodine vapor according to the reaction:
$TiI_{4} \xrightarrow{\Delta} Ti + 2I_{2}$

(C) Step $1$: Reaction of $2,2-$dichloropropane with aqueous $KOH$ gives a gem-diol intermediate,which loses a water molecule to form acetone $(CH_3COCH_3)$ as product $A$.
Step $2$: Clemmensen reduction of acetone $(CH_3COCH_3)$ using $Zn(Hg)/HCl$ reduces the carbonyl group to a methylene group $(-CH_2-)$,yielding propane $(CH_3CH_2CH_3)$ as product $B$.
Therefore,the correct option is $C$.

(B) The order $HI < HBr < HCl < HF$ represents the increasing trend for thermal stability,ionic character,and dipole moment.
However,the reducing power of hydrogen halides increases as the bond dissociation enthalpy decreases,which follows the order $HF < HCl < HBr < HI$.
Therefore,the property that does not correspond to the given order is reducing power.

(C) Rayon is a regenerated cellulose fiber that is highly absorbent,capable of holding over $90 \%$ of its own mass in water. Due to its non-adherent properties,it is commonly used in medical dressings as it does not stick to wounds.

(D) High density polythene (or plastic) is prepared by heating ethylene at about $330-350 \ K$,under a pressure of $1-2 \ atm$ and in the presence of Ziegler-Natta catalyst.
$n \ CH_2=CH_2 \xrightarrow[\text{Ziegler-Natta catalyst}]{330-350 \ K, 1-2 \ atm} (-CH_2-CH_2-)_n$
High density polythene.

(C) Phenols react with neutral $FeCl_{3}$ solution to form a complex,which results in a characteristic blue,violet,or green colouration. This is a standard test for the presence of the phenolic group.

(A) An oxyacid of phosphorus that contains $P-H$ bonds can act as a reducing agent.
$H_{3}PO_{3}$ (phosphorous acid) contains one $P-H$ bond,which makes it a reducing agent.
In contrast,$H_{3}PO_{4}$,$H_{2}P_{2}O_{6}$,and $H_{4}P_{2}O_{7}$ do not contain any $P-H$ bonds and therefore do not act as reducing agents.

(B) Given: Volume of solution $= 1500 \ cm^{3}$,Density of solution $= 1.052 \ g/cm^{3}$,Mass of urea (solute) $= 18 \ g$,Molar mass of urea $= 60 \ g/mol$.
Mass of solution $= \text{Volume} \times \text{Density} = 1500 \ cm^{3} \times 1.052 \ g/cm^{3} = 1578 \ g$.
Mass of solvent (water) $= \text{Mass of solution} - \text{Mass of solute} = 1578 \ g - 18 \ g = 1560 \ g = 1.560 \ kg$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in } kg} = \frac{18 / 60}{1.560} = \frac{0.3}{1.560} \approx 0.192 \ m$.

(A) The relative lowering of vapour pressure is given by the formula: $\frac{p^{\circ}-p_{s}}{p^{\circ}} = \frac{n_{2}}{n_{1}+n_{2}} = \frac{\frac{w_{2}}{M_{2}}}{\frac{w_{1}}{M_{1}}+\frac{w_{2}}{M_{2}}}$
Given: Mass of cane sugar $(w_{2})$ = $34.2 \ g$,Molar mass of cane sugar $(M_{2})$ = $342 \ g/mol$,Mass of water $(w_{1})$ = $180 \ g$,Molar mass of water $(M_{1})$ = $18 \ g/mol$.
Calculating moles:
$n_{2} = \frac{34.2}{342} = 0.1 \ mol$
$n_{1} = \frac{180}{18} = 10 \ mol$
Relative lowering of vapour pressure = $\frac{0.1}{10+0.1} = \frac{0.1}{10.1} \approx 0.0099$.