TS EAMCET 2002 Chemistry Question Paper with Answer and Solution

(D) Given the equation $\left| z + \frac{i}{2} \right|^2 = \left| z - \frac{i}{2} \right|^2$.
Substitute $z = x + iy$:
$\left| x + iy + \frac{i}{2} \right|^2 = \left| x + iy - \frac{i}{2} \right|^2$
$\left| x + i(y + \frac{1}{2}) \right|^2 = \left| x + i(y - \frac{1}{2}) \right|^2$
Using the property $|a + ib|^2 = a^2 + b^2$:
$x^2 + (y + \frac{1}{2})^2 = x^2 + (y - \frac{1}{2})^2$
$x^2 + y^2 + y + \frac{1}{4} = x^2 + y^2 - y + \frac{1}{4}$
$y = -y$
$2y = 0$
$y = 0$
This represents the $x$-axis.

(B) Let $f(x) = x^5 - 6x^2 - 4x + 5$.
According to Descartes' Rule of Signs,the number of positive real roots is at most the number of sign changes in the coefficients of $f(x)$.
The coefficients are $(1, 0, 0, -6, -4, 5)$.
The sign changes occur from $1$ to $-6$ and from $-4$ to $5$. Thus,there are $2$ sign changes,meaning $f(x)$ has at most $2$ positive real roots.
Now,consider $f(-x) = (-x)^5 - 6(-x)^2 - 4(-x) + 5 = -x^5 - 6x^2 + 4x + 5$.
The coefficients are $(-1, 0, 0, -6, 4, 5)$.
The sign changes occur from $-6$ to $4$. Thus,there is $1$ sign change,meaning $f(x)$ has at most $1$ negative real root.
Therefore,the maximum possible number of real roots is $2 + 1 = 3$.

(A) Given that $A$ and $B$ are square matrices of order $3$.
Since $A$ is a non-singular matrix,its determinant $|A| \neq 0$,which implies that the inverse matrix $A^{-1}$ exists.
Given the equation $AB = O$.
Pre-multiplying both sides by $A^{-1}$,we get:
$A^{-1}(AB) = A^{-1}O$
$(A^{-1}A)B = O$
$IB = O$
$B = O$
Therefore,$B$ is a null matrix.

(A) Let the point $P = (-2, 6)$ and $Q = (4, 2)$. The line $L = 0$ is the perpendicular bisector of the segment $PQ$.
The midpoint of $PQ$ is $\left( \frac{-2 + 4}{2}, \frac{6 + 2}{2} \right) = (1, 4)$.
The slope of $PQ$ is $m_{PQ} = \frac{2 - 6}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3}$.
Since line $L$ is perpendicular to $PQ$,its slope $m_L$ is the negative reciprocal of $m_{PQ}$,so $m_L = \frac{3}{2}$.
The equation of line $L$ passing through $(1, 4)$ with slope $\frac{3}{2}$ is given by:
$y - 4 = \frac{3}{2}(x - 1)$
$2(y - 4) = 3(x - 1)$
$2y - 8 = 3x - 3$
$3x - 2y + 5 = 0$.

(A) Given points are $A(1, x, 3)$,$B(3, 4, 7)$,and $C(y, -2, -5)$.
Since the points are collinear,the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ must be parallel,i.e.,$\overrightarrow{AB} = \lambda \overrightarrow{BC}$ for some scalar $\lambda$.
$\overrightarrow{AB} = (3-1)\hat{i} + (4-x)\hat{j} + (7-3)\hat{k} = 2\hat{i} + (4-x)\hat{j} + 4\hat{k}$.
$\overrightarrow{BC} = (y-3)\hat{i} + (-2-4)\hat{j} + (-5-7)\hat{k} = (y-3)\hat{i} - 6\hat{j} - 12\hat{k}$.
Equating the components of $\overrightarrow{AB} = \lambda \overrightarrow{BC}$:
$2 = \lambda(y-3)$ ... $(i)$
$4-x = -6\lambda$ ... $(ii)$
$4 = -12\lambda$ ... $(iii)$
From $(iii)$,$\lambda = \frac{4}{-12} = -\frac{1}{3}$.
Substituting $\lambda = -\frac{1}{3}$ into $(i)$: $2 = -\frac{1}{3}(y-3) \Rightarrow -6 = y-3 \Rightarrow y = -3$.
Substituting $\lambda = -\frac{1}{3}$ into $(ii)$: $4-x = -6(-\frac{1}{3}) \Rightarrow 4-x = 2 \Rightarrow x = 2$.
Thus,$(x, y) = (2, -3)$.

(A) Given $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$.
Using the identity $\sin^4 x = \sin^2 x (1 - \cos^2 x)$ and $\cos^4 x = \cos^2 x (1 - \sin^2 x)$,we substitute these into the expression:
$f(x) = \frac{\cos^2 x + \sin^2 x (1 - \cos^2 x)}{\sin^2 x + \cos^2 x (1 - \sin^2 x)}$
Expanding the terms:
$f(x) = \frac{\cos^2 x + \sin^2 x - \sin^2 x \cos^2 x}{\sin^2 x + \cos^2 x - \cos^2 x \sin^2 x}$
Since $\sin^2 x + \cos^2 x = 1$,the numerator and denominator become identical:
$f(x) = \frac{1 - \sin^2 x \cos^2 x}{1 - \sin^2 x \cos^2 x} = 1$
Therefore,for any $x \in R$,$f(x) = 1$. Thus,$f(2002) = 1$.

(A) The given differential equation is $\frac{dy}{dx} - y = x^2$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = x^2$.
The Integrating Factor $(I.F.)$ is given by $e^{\int P dx} = e^{\int -1 dx} = e^{-x}$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.
Substituting the values,$y e^{-x} = \int x^2 e^{-x} dx + c$.
Using integration by parts,$\int x^2 e^{-x} dx = x^2(-e^{-x}) - \int 2x(-e^{-x}) dx = -x^2 e^{-x} + 2 \int x e^{-x} dx$.
$= -x^2 e^{-x} + 2[x(-e^{-x}) - \int 1(-e^{-x}) dx] = -x^2 e^{-x} - 2x e^{-x} - 2e^{-x} + c$.
Thus,$y e^{-x} = -e^{-x}(x^2 + 2x + 2) + c$.
Multiplying both sides by $e^x$,we get $y = -(x^2 + 2x + 2) + ce^x$,which simplifies to $y + x^2 + 2x + 2 = ce^x$.

(A) The formula for the centre of mass of a system of particles is given by $X_{CM} = \frac{\sum m_i x_i}{\sum m_i}$.
Here,the masses are $m_i = i \cdot m$ and their positions are $x_i = i \cdot l$ for $i = 1, 2, \dots, n$.
Substituting these values:
$X_{CM} = \frac{m(l) + 2m(2l) + 3m(3l) + \dots + nm(nl)}{m + 2m + 3m + \dots + nm}$
$X_{CM} = \frac{ml(1^2 + 2^2 + 3^2 + \dots + n^2)}{m(1 + 2 + 3 + \dots + n)}$
Using the standard summation formulas $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{i=1}^n i = \frac{n(n+1)}{2}$:
$X_{CM} = \frac{l \cdot \frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}$
$X_{CM} = l \cdot \frac{n(n+1)(2n+1)}{6} \cdot \frac{2}{n(n+1)}$
$X_{CM} = \frac{(2n + 1)l}{3}$

(A) The position of the centre of mass $X_{CM}$ is given by the formula:
$X_{CM} = \frac{\sum m_i x_i}{\sum m_i}$
Substituting the given values:
$X_{CM} = \frac{m(l) + 2m(2l) + 3m(3l) + \dots + nm(nl)}{m + 2m + 3m + \dots + nm}$
Factoring out $m$ and $l$:
$X_{CM} = \frac{ml(1^2 + 2^2 + 3^2 + \dots + n^2)}{m(1 + 2 + 3 + \dots + n)}$
Using the standard summation formulas $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$:
$X_{CM} = \frac{l \cdot \frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}$
Simplifying the expression:
$X_{CM} = l \cdot \frac{n(n+1)(2n+1)}{6} \cdot \frac{2}{n(n+1)}$
$X_{CM} = \frac{(2n+1)l}{3}$

(D) Step $1$: Oxidation of ethanol $(C_2H_5OH)$ with $KMnO_4 / H^{+}$ yields acetic acid $(CH_3COOH)$ as $X$.
$C_2H_5OH + [O] \xrightarrow{KMnO_4 / H^{+}} CH_3COOH (X) + H_2O$
Step $2$: Esterification of acetic acid $(X)$ with ethanol $(Y)$ in the presence of conc. $H_2SO_4$ yields ethyl acetate $(CH_3COOC_2H_5)$.
$CH_3COOH + C_2H_5OH (Y) \xrightarrow{H_2SO_4 / \Delta} CH_3COOC_2H_5 + H_2O$

(A) When diethyl ether reacts with cold $HI$,the cleavage of the $C-O$ bond occurs.
Since the reaction is carried out with cold $HI$,only one molecule of $HI$ reacts,leading to the formation of ethyl alcohol and ethyl iodide.
The reaction is as follows:
$C_2H_5-O-C_2H_5 + HI \rightarrow C_2H_5OH + C_2H_5I$

(D) Alkaline formaldehyde solution reacts with $H_2O_2$ as follows:
$2HCHO + H_2O_2 \rightarrow 2HCOOH + H_2 \uparrow$
In this reaction,formaldehyde $(HCHO)$ is oxidized to formic acid $(HCOOH)$ and hydrogen gas $(H_2)$ is liberated.

(A) The reaction of acetone $(CH_3COCH_3)$ in the presence of a base like $Ba(OH)_2$ is an aldol condensation reaction.
In this reaction,two molecules of acetone undergo condensation to form diacetone alcohol.
The mechanism involves the formation of an enolate ion from one molecule of acetone,which then attacks the carbonyl carbon of another acetone molecule.
The product $X$ is $4-hydroxy-4-methylpentan-2-one$,which has the structure $CH_3-C(OH)(CH_3)-CH_2-CO-CH_3$.

(A) The oxidation of acetaldehyde $(CH_3CHO)$ to acetic acid $(CH_3COOH)$ is industrially carried out using manganese acetate as a catalyst in the presence of air or oxygen.
The reaction is: $2CH_3CHO + O_2 \xrightarrow{\text{Manganese acetate}} 2CH_3COOH$.

(B) The charge $Q$ on a capacitor is given by the formula $Q = C \times V$,where $C$ is the capacitance and $V$ is the potential difference.
Given $C = 1000 \ \mu F = 1000 \times 10^{-6} \ F$ and $V = 20 \ V$.
Total charge $Q = 1000 \times 10^{-6} \ F \times 20 \ V = 20,000 \ \mu C$.
The rate of charging is $I = 200 \ \mu C/s$.
The time $t$ required is given by $t = \frac{Q}{I}$.
$t = \frac{20,000 \ \mu C}{200 \ \mu C/s} = 100 \ s$.

(A) Initial capacitance $C = 100 \mu F = 100 \times 10^{-6} \text{ F}$ and voltage $V = 50 \text{ V}$.
Initial energy stored $E_1 = \frac{1}{2} C V^2 = \frac{1}{2} \times 100 \times 10^{-6} \times (50)^2 = 50 \times 10^{-6} \times 2500 = 0.125 \text{ J} = 125 \times 10^{-3} \text{ J}$.
When the distance $d$ is doubled,the new capacitance $C' = \frac{\epsilon_0 A}{2d} = \frac{C}{2} = 50 \mu F = 50 \times 10^{-6} \text{ F}$.
Since the battery remains connected,the voltage $V$ remains constant at $50 \text{ V}$.
Final energy stored $E_2 = \frac{1}{2} C' V^2 = \frac{1}{2} \times 50 \times 10^{-6} \times (50)^2 = 25 \times 10^{-6} \times 2500 = 0.0625 \text{ J} = 62.5 \times 10^{-3} \text{ J}$.
The charge on the capacitor changes from $Q_1 = CV$ to $Q_2 = C'V$. The change in charge is $\Delta Q = (C' - C)V = (50 - 100) \times 10^{-6} \times 50 = -50 \times 10^{-6} \times 50 = -2500 \times 10^{-6} \text{ C}$.
The work done by the battery is $W = \Delta Q \times V = (-2500 \times 10^{-6}) \times 50 = -125000 \times 10^{-6} = -0.125 \text{ J} = -125 \times 10^{-3} \text{ J}$.
The negative sign indicates that energy is returned to the battery. The magnitude of energy change supplied by the battery is $125 \times 10^{-3} \text{ J}$ (returned). However,the question asks for the additional energy supplied. Since the energy of the capacitor decreases,the battery absorbs energy. The change in energy of the capacitor is $E_2 - E_1 = 62.5 \times 10^{-3} - 125 \times 10^{-3} = -62.5 \times 10^{-3} \text{ J}$. The battery supplies $-62.5 \times 10^{-3} \text{ J}$,which is $\frac{125}{2} \times 10^{-3} \text{ J}$ removed.

(C) Acetylene $(HC \equiv CH)$ undergoes oxidation in the presence of alkaline potassium permanganate $(KMnO_4 / KOH)$ at $25^{\circ}C$ to form oxalic acid $(HOOC-COOH)$.
The chemical reaction is:
$HC \equiv CH + 4[O] \xrightarrow{KMnO_4 / KOH, 25^{\circ}C} HOOC-COOH$.

(C) When oxalic acid $(COOH)_2$ is heated with concentrated $H_2SO_4$,it undergoes dehydration to produce a mixture of carbon monoxide $(CO)$ and carbon dioxide $(CO_2)$ gases along with water:
$(COOH)_2 \xrightarrow{conc. H_2SO_4} CO + CO_2 + H_2O$
When this gaseous mixture is passed through caustic potash $(KOH)$,the acidic gas $CO_2$ is absorbed,while $CO$ (a neutral gas) passes through.
The reaction between $CO_2$ and $KOH$ is:
$CO_2 + 2KOH \rightarrow K_2CO_3 + H_2O$
Thus,the product formed is potassium carbonate $(K_2CO_3)$.

(C) Aspirin (acetylsalicylic acid) is prepared by the acetylation of the phenolic $-OH$ group of salicylic acid using acetic anhydride in the presence of an acid catalyst like concentrated $H_2SO_4$.
The reaction is:
$\text{Salicylic acid} + (CH_3CO)_2O \xrightarrow{\text{Conc. } H_2SO_4} \text{Acetylsalicylic acid (Aspirin)} + CH_3COOH$
Therefore,the correct reagent is $(CH_3CO)_2O / \text{Conc. } H_2SO_4$.

(B) The reaction of acetic acid $(CH_3COOH)$ with ammonia $(NH_3)$ forms ammonium acetate $(CH_3COONH_4)$ as an intermediate product $X$.
$CH_3COOH + NH_3 \longrightarrow CH_3COONH_4$ $(X)$
On heating,ammonium acetate undergoes dehydration to form acetamide $(CH_3CONH_2)$ as product $Y$ and water $(H_2O)$.
$CH_3COONH_4 \stackrel{\Delta}{\longrightarrow} CH_3CONH_2$ $(Y)$ $+ H_2O$
Thus,$X$ is $CH_3COONH_4$ and $Y$ is $CH_3CONH_2$.

(B) The conjugate base of $NH_4^{+}$ is formed by the removal of a proton $(H^{+})$.
$NH_4^{+} \rightarrow NH_3 + H^{+}$.
In $NH_3$,the central nitrogen atom has $3$ bond pairs and $1$ lone pair.
According to $VSEPR$ theory,the steric number is $3 + 1 = 4$,which corresponds to $sp^3$ hybridization.

(C) To determine the hybridization of $Xe$ in its fluorides,we use the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $XeF_2$: $H = \frac{1}{2}(8 + 2) = 5$,which corresponds to $sp^3d$ hybridization.
For $XeF_4$: $H = \frac{1}{2}(8 + 4) = 6$,which corresponds to $sp^3d^2$ hybridization.
Thus,the correct pair is $XeF_2, sp^3d$.

(B) An ionic bond is formed by the electrostatic attraction between oppositely charged ions,while a covalent bond is formed by the sharing of electrons between atoms.
In $K_2SO_4$,the potassium ions $(K^+)$ and the sulfate ion $(SO_4^{2-})$ are held together by ionic bonds.
Within the sulfate ion $(SO_4^{2-})$,the sulfur and oxygen atoms are held together by covalent bonds.
Therefore,$K_2SO_4$ contains both ionic and covalent bonds.
$CH_2Cl_2$,$BeCl_2$,and $SO_2$ contain only covalent bonds.

(C) At equilibrium,the rate of the forward reaction is equal to the rate of the backward reaction.
Given that the rate of the forward reaction at equilibrium is $0.02 \ mol \ L^{-1} \ min^{-1}$.
Therefore,the rate of the backward reaction at equilibrium is also $0.02 \ mol \ L^{-1} \ min^{-1}$.

(C) For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate expression is given by: $\text{Rate} = -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Given that $-\frac{d[N_2]}{dt} = 0.02 \ mol \ L^{-1} \ s^{-1}$.
Equating the terms: $-\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt}$.
Therefore,$-\frac{d[H_2]}{dt} = 3 \times (-\frac{d[N_2]}{dt}) = 3 \times 0.02 \ mol \ L^{-1} \ s^{-1} = 0.06 \ mol \ L^{-1} \ s^{-1}$.

(D) Radioactive disintegration follows first-order kinetics.
The rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{a}{a-x}$.
Given $a = 10 \ g$,$a-x = 1 \ g$,and $t = 2.303 \ \text{min}$.
Substituting the values: $k = \frac{2.303}{2.303} \log \frac{10}{1} = 1 \times \log(10) = 1 \ \text{min}^{-1}$.
The half-life $t_{1/2}$ is calculated as $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{1} = 0.693 \ \text{min}$.

(C) Electronegativity increases across a period from left to right and decreases down a group.
In the periodic table,$O$ (Oxygen) is in group $16$ and period $2$,$N$ (Nitrogen) is in group $15$ and period $2$,and $P$ (Phosphorus) is in group $15$ and period $3$.
Comparing these,$O$ has the highest electronegativity due to its position at the far right of period $2$.
$N$ has a higher electronegativity than $P$ because $N$ is in period $2$ while $P$ is in period $3$ (electronegativity decreases down a group).
Therefore,the correct order is $O > N > P$.

(C) Resistance of each piece $= \frac{R}{20} \Omega$.
Equivalent resistance of $10$ such pieces connected in series is $R_1 = 10 \times \frac{R}{20} = \frac{R}{2}$.
Similarly,equivalent resistance of $10$ such pieces connected in parallel is $\frac{1}{R_2} = \frac{1}{R/20} + \dots (10 \text{ times}) = \frac{10}{R/20} = \frac{200}{R}$,so $R_2 = \frac{R}{200}$.
Now,$R_1$ and $R_2$ are connected in series,so the total resistance is $R_{eq} = R_1 + R_2 = \frac{R}{2} + \frac{R}{200} = \frac{100R + R}{200} = \frac{101R}{200}$.

(B) When a wire is stretched,its volume remains constant. Resistance $R = \rho \frac{l}{A} = \rho \frac{l^2}{V}$. Since $\rho$ and $V$ are constant,$R \propto l^2$.
Given initial resistance $R_1 = 3 \Omega$ and length $l_1 = l$. After stretching,$l_2 = 2l$.
$\frac{R_2}{R_1} = \left(\frac{l_2}{l_1}\right)^2 = \left(\frac{2l}{l}\right)^2 = 4$.
So,$R_2 = 4 \times R_1 = 4 \times 3 = 12 \Omega$.
The wire of resistance $12 \Omega$ is bent into an equilateral triangle. Each side will have a resistance of $R_{side} = \frac{12 \Omega}{3} = 4 \Omega$.
Let the vertices be $A, B, C$. The sides are $AB, BC, CA$ each of $4 \Omega$.
To find the effective resistance between the ends of any side (e.g.,across $B$ and $C$),the side $BC$ $(4 \Omega)$ is in parallel with the series combination of $AB$ and $AC$ $(4 \Omega + 4 \Omega = 8 \Omega)$.
$R_{eq} = \frac{R_{BC} \times (R_{AB} + R_{AC})}{R_{BC} + (R_{AB} + R_{AC})} = \frac{4 \times 8}{4 + 8} = \frac{32}{12} = \frac{8}{3} \Omega$.

(C) Given:
Resistance of the galvanometer,$G = 100 \ \Omega$
Full-scale deflection current,$I_g = 100 \ \mu A = 100 \times 10^{-6} \ A = 0.1 \times 10^{-3} \ A$
Required total current,$I = 1 \ mA = 1 \times 10^{-3} \ A$
To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with it.
The formula for shunt resistance is given by:
$S = \frac{I_g G}{I - I_g}$
Substituting the values:
$S = \frac{(0.1 \times 10^{-3}) \times 100}{1 \times 10^{-3} - 0.1 \times 10^{-3}}$
$S = \frac{0.1 \times 10^{-1}}{0.9 \times 10^{-3}}$
$S = \frac{0.01}{0.0009} = \frac{100}{9} \ \Omega$
Thus,the required shunt resistance is $\frac{100}{9} \ \Omega$.

(C) In a potentiometer experiment, the balancing length $l_1$ is proportional to the electromotive force $(E)$ of the cell: $E = k l_1$, where $k$ is the potential gradient.
When an external resistance $R$ is connected in parallel, the terminal potential difference $V$ is given by $V = k l_2$, where $l_2$ is the new balancing length.
Given $l_1 = 560 \, cm$ and the change in length is $60 \, cm$, the new balancing length $l_2 = 560 \, cm - 60 \, cm = 500 \, cm$.
The formula for internal resistance $r$ is $r = R \left( \frac{E}{V} - 1 \right) = R \left( \frac{l_1}{l_2} - 1 \right) = R \left( \frac{l_1 - l_2}{l_2} \right)$.
Substituting the values: $r = 10 \left( \frac{560 - 500}{500} \right) = 10 \left( \frac{60}{500} \right) = 10 \times 0.12 = 1.2 \, \Omega$.

(A) The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
For $Cu^{2+}$ ion,two electrons are removed (one from $4s$ and one from $3d$),resulting in $[Ar] 3d^9$.
In the $3d^9$ configuration,there is $1$ unpaired electron $(n = 1)$.
The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n = 1$,we get $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

(C) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.
For a proton,$\lambda_0 = \frac{h}{\sqrt{2m_p q_p V}}$.
For an $\alpha$-particle,the mass $m_{\alpha} = 4m_p$ and the charge $q_{\alpha} = 2q_p$.
Substituting these into the formula for the $\alpha$-particle:
$\lambda_{\alpha} = \frac{h}{\sqrt{2(4m_p)(2q_p)V}} = \frac{h}{\sqrt{8(2m_p q_p V)}} = \frac{1}{\sqrt{8}} \left( \frac{h}{\sqrt{2m_p q_p V}} \right)$.
Since $\sqrt{8} = 2\sqrt{2}$,we get $\lambda_{\alpha} = \frac{\lambda_0}{2\sqrt{2}}$.

(D) Let the work function of the metal be $W$.
According to Einstein's photoelectric equation,the maximum kinetic energy $(KE)_{\max }$ is given by $(KE)_{\max } = E - W$,where $E$ is the energy of the incident photon.
For the first photon,$E_1 = 2W$. Thus,$(KE_1)_{\max } = 2W - W = W$.
For the second photon,$E_2 = 3W$. Thus,$(KE_2)_{\max } = 3W - W = 2W$.
Since $(KE)_{\max } = \frac{1}{2}mv^2$,we have the ratio of kinetic energies:
$\frac{(KE_1)_{\max }}{(KE_2)_{\max }} = \frac{W}{2W} = \frac{1}{2}$.
Substituting the velocity terms:
$\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{1}{2} \implies \frac{v_1^2}{v_2^2} = \frac{1}{2}$.
Taking the square root on both sides:
$\frac{v_1}{v_2} = \frac{1}{\sqrt{2}}$.
Therefore,the ratio of maximum velocities is $1: \sqrt{2}$.

(B) The formula for the mass deposited is $w = z \cdot i \cdot t$,where $w$ is the mass,$z$ is the electrochemical equivalent,$i$ is the current,and $t$ is the time.
Rearranging for $z$,we get $z = \frac{w}{i \times t}$.
Substituting the given values: $z = \frac{0.066 \ g}{2 \ A \times 100 \ s} = \frac{0.066}{200} \ g \ C^{-1}$.
Thus,$z = 3.3 \times 10^{-4} \ g \ C^{-1}$.

(C) In Down's process,molten sodium chloride $(NaCl)$ is electrolyzed.
At the anode,oxidation of chloride ions $(Cl^{-})$ occurs to produce chlorine gas $(Cl_2)$.
The reaction is: $2 Cl^{-} \longrightarrow Cl_2 + 2 e^{-}$

(B) The reduction reaction for the copper electrode is: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$.
Using the Nernst equation at $25^{\circ} C$ $(298 \ K)$:
$E_{Cu^{2+}/Cu} = E^{\circ}_{Cu^{2+}/Cu} - \frac{0.0591}{n} \log \frac{1}{[Cu^{2+}]}$
$E_{Cu^{2+}/Cu} = E^{\circ}_{Cu^{2+}/Cu} + \frac{0.0591}{2} \log [Cu^{2+}]$
Substituting the given values:
$E_{Cu^{2+}/Cu} = 0.34 + \frac{0.0591}{2} \log (10^{-2})$
$E_{Cu^{2+}/Cu} = 0.34 + \frac{0.0591}{2} \times (-2)$
$E_{Cu^{2+}/Cu} = 0.34 - 0.0591$
$E_{Cu^{2+}/Cu} = 0.2809 \ V$

(A) Consider a small element of length $dr$ at a distance $r$ from the axis of rotation.
The induced emf $de$ across this small element is given by $de = B v dr$,where $v = r \omega$.
So,$de = B (r \omega) dr$.
To find the total induced emf $e$ between the center and one end of the rod (length $L/2$),we integrate from $r = 0$ to $r = L/2$:
$e = \int_{0}^{L/2} B \omega r dr = B \omega \left[ \frac{r^2}{2} \right]_{0}^{L/2} = B \omega \frac{(L/2)^2}{2} = \frac{B \omega L^2}{8}$.
Since the rod rotates about its center,the emf induced in each half of the rod is $\frac{B \omega L^2}{8}$.
Because the two halves are connected in parallel with respect to the center,the total emf between the two ends is the sum of the emf induced in each half: $e_{total} = \frac{B \omega L^2}{8} + \frac{B \omega L^2}{8} = \frac{B \omega L^2}{4}$.

(D) For the Compton effect,the shift in wavelength is given by the relation:
$\Delta \lambda = \lambda_2 - \lambda_1 = \frac{h}{m_0 c}(1 - \cos \theta)$
Here,$\lambda_1$ is the wavelength of incident radiation,$\lambda_2$ is the wavelength of scattered radiation,and $\frac{h}{m_0 c}$ is the Compton wavelength,which is approximately $0.024 \ \mathring{A}$.
Given: $\lambda_2 = 0.22 \ \mathring{A}$ and $\theta = 60^{\circ}$.
Substituting the values:
$0.22 - \lambda_1 = 0.024(1 - \cos 60^{\circ})$
$0.22 - \lambda_1 = 0.024(1 - 0.5)$
$0.22 - \lambda_1 = 0.024 \times 0.5$
$0.22 - \lambda_1 = 0.012$
$\lambda_1 = 0.22 - 0.012 = 0.208 \ \mathring{A}$.

(A) Given:
Mass $m = 1 ~g = 10^{-3} ~kg$
Charge $q = 10^{-8} ~C$
Potential at $P$,$V_P = 600 ~V$
Potential at $Q$,$V_Q = 0 ~V$
Velocity at $Q$,$v_Q = 20 ~cm/s = 0.2 ~m/s$
According to the work-energy theorem,the work done by the electric field is equal to the change in kinetic energy:
$W = \Delta KE = KE_Q - KE_P$
$q(V_P - V_Q) = \frac{1}{2} m v_Q^2 - \frac{1}{2} m v_P^2$
Substituting the values:
$10^{-8} (600 - 0) = \frac{1}{2} (10^{-3}) (0.2)^2 - \frac{1}{2} (10^{-3}) v_P^2$
$600 \times 10^{-8} = 0.5 \times 10^{-3} \times 0.04 - 0.5 \times 10^{-3} v_P^2$
$6 \times 10^{-6} = 2 \times 10^{-5} - 0.5 \times 10^{-3} v_P^2$
$0.5 \times 10^{-3} v_P^2 = 20 \times 10^{-6} - 6 \times 10^{-6}$
$0.5 \times 10^{-3} v_P^2 = 14 \times 10^{-6}$
$v_P^2 = \frac{14 \times 10^{-6}}{0.5 \times 10^{-3}} = 28 \times 10^{-3} = 0.028$
$v_P = \sqrt{0.028} ~m/s$

(A) Fluorosis is a condition caused by the excessive intake of fluoride. In the body,excess fluoride reacts with calcium $(Ca)$ present in bones and teeth to form calcium fluoride $(CaF_2)$,which leads to the hardening of bones and mottling of teeth.
The chemical reaction is: $Ca + F_2 \rightarrow CaF_2$ (Fluorosis disease).

(D) secondary $(2^{\circ})$ alcohol is an alcohol where the hydroxyl group $(-OH)$ is attached to a carbon atom that is bonded to two other carbon atoms.
$1.$ $2-$methyl$-2-$propanol: The $-OH$ group is attached to a tertiary carbon. It is a tertiary alcohol.
$2.$ $1-$propanol: The $-OH$ group is attached to a primary carbon. It is a primary alcohol.
$3.$ $1-$butanol: The $-OH$ group is attached to a primary carbon. It is a primary alcohol.
$4.$ $2-$pentanol: The structure is $CH_3-CH(OH)-CH_2-CH_2-CH_3$. The $-OH$ group is attached to the $C2$ carbon,which is bonded to two other carbon atoms ($C1$ and $C3$). Thus,it is a secondary alcohol.

(B) tertiary amine ($3^{\circ}$ amine) has the general formula $R_3N$,where $R$ represents an alkyl group.
For a tertiary amine,the minimum number of carbon atoms required is $3$ (e.g.,trimethylamine,$(CH_3)_3N$).
Calculating the molecular formula for trimethylamine: $C_3H_9N$.
$C_2H_7N$ corresponds to ethylamine $(1^{\circ})$ or dimethylamine $(2^{\circ})$.
$CH_5N$ corresponds to methylamine $(1^{\circ})$.
$CH_3N$ is not a stable saturated amine.
Therefore,$C_3H_9N$ is the molecular formula of a tertiary amine.

(B) According to Kepler's second law (law of areas),the areal velocity $(A)$ is defined as the rate at which the area is swept out by the position vector of the planet.
Mathematically,the areal velocity is given by the relation:
$A = \frac{dA}{dt} = \frac{L}{2M}$
where:
$L$ is the angular momentum of the planet,
$M$ is the mass of the planet.
Rearranging this formula to solve for the angular momentum $(L)$:
$L = 2 M A$
Therefore,the correct option is $B$.

(D) The Wurtz reaction for the preparation of ethane is given by the equation:
$2CH_3I + 2Na \xrightarrow{\text{dry ether}} C_2H_6 + 2NaI$
According to the stoichiometry of the reaction,$2$ moles of methyl iodide $(CH_3I)$ are required to produce $1$ mole of ethane $(C_2H_6)$.
The molar mass of $CH_3I$ is calculated as: $12 (C) + 3 (H) + 127 (I) = 142 \ g/mol$.
Therefore,the mass of $2$ moles of $CH_3I$ required is: $2 \times 142 \ g = 284 \ g$.

(A) The chemical formula of $chloroethane$ is $CH_3CH_2Cl$.
In this molecule,the carbon atom attached to the chlorine atom (halogen) is bonded to two hydrogen atoms.
Since the carbon atom bearing the halogen is attached to only one other carbon atom,it is classified as a primary $(1^{\circ})$ alkyl halide.

(A) The given reaction is $A$ $\xrightarrow{HBr} C_2H_5Br$ $\xrightarrow{B} A$.
If $A = C_2H_4$ (ethene) and $B = \text{alcoholic } KOH / \Delta$:
$1$. $C_2H_4 + HBr \rightarrow C_2H_5Br$ (Electrophilic addition reaction).
$2$. $C_2H_5Br + \text{alcoholic } KOH \xrightarrow{\Delta} C_2H_4 + KBr + H_2O$ (Dehydrohalogenation reaction).
Thus,$A$ is $C_2H_4$ and $B$ is alcoholic $KOH / \Delta$.

(A) The conversion of $1, 2-$dibromoethane to ethene is a dehalogenation reaction.
This reaction involves the removal of two bromine atoms from adjacent carbon atoms using zinc dust in the presence of alcohol under heating $(\Delta)$.
The chemical equation is: $BrCH_2-CH_2Br + Zn \xrightarrow{\text{alcohol}, \Delta} CH_2=CH_2 + ZnBr_2$.
Thus,the correct reaction conditions are $Zn$,alcohol,$\Delta$.

(D) The basic character of a molecule depends on the availability of the lone pair of electrons on the central atom for donation.
In $PH_3$,the lone pair is present in an orbital with high $s$-character,making it less available for donation.
In $P(CH_3)_3$,the three methyl groups are electron-donating groups ($+I$ effect),which increase the electron density on the phosphorus atom,making the lone pair more available for donation.
Therefore,$P(CH_3)_3$ is a stronger base than $PH_3$.

(D) When heavy water $(D_2O)$ reacts with magnesium nitride $(Mg_3N_2)$,the deuterium atoms replace the hydrogen atoms in the reaction products.
The chemical reaction is:
$Mg_3N_2 + 6 D_2O \longrightarrow 3 Mg(OD)_2 + 2 ND_3$
Thus,the products formed are magnesium deuteroxide $(Mg(OD)_2)$ and deuteroammonia $(ND_3)$.

(C) For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate of reaction is given by:
Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Given that $-\frac{d[N_2]}{dt} = 0.02 \ mol \ L^{-1} \ s^{-1}$.
Equating the terms for $N_2$ and $H_2$:
$-\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt}$.
Therefore,$-\frac{d[H_2]}{dt} = 3 \times (-\frac{d[N_2]}{dt})$.
$-\frac{d[H_2]}{dt} = 3 \times 0.02 = 0.06 \ mol \ L^{-1} \ s^{-1}$.

(A) The basicity of phosphines depends on the availability of the lone pair on the phosphorus atom.
In $P(CH_3)_3$,the three methyl groups exert a $+I$ (inductive) effect,which increases the electron density on the phosphorus atom.
This makes the lone pair more available for donation compared to $PH_3$,where no such electron-donating groups are present.
Therefore,the correct order of basic character is $P(CH_3)_3 > PH_3$.

(C) Acetophenone is prepared by the Friedel-Crafts acylation of benzene.
In this reaction,benzene $(C_6H_6)$ reacts with acetyl chloride $(CH_3COCl)$ in the presence of an anhydrous Lewis acid catalyst,anhydrous $AlCl_3$.
The reaction is:
$C_6H_6 + CH_3COCl \xrightarrow{\text{anhydrous } AlCl_3} C_6H_5COCH_3 + HCl$
Thus,the chemicals required are $C_6H_6$ $(A)$,$CH_3COCl$ $(C)$,and anhydrous $AlCl_3$ $(D)$.

(B) The bond energy depends on the bond length. As the size of the central atom increases down the group $(N < P < As)$,the bond length increases,and the bond energy decreases.
The order of bond energies is $N-H > P-H > As-H$.
The values are $389 \ kJ \ mol^{-1}$ for $N-H$,$318 \ kJ \ mol^{-1}$ for $P-H$,and $247 \ kJ \ mol^{-1}$ for $As-H$.
Therefore,the correct order is $389, 318, 247$.

(D) The reaction sequence is as follows:
$1$. Oxidation of ethanol $(C_2H_5OH)$ with $KMnO_4 / H^+$ yields acetic acid $(CH_3COOH)$ as the final product $X$.
$2$. The reaction of acetic acid $(CH_3COOH)$ with ethanol $(C_2H_5OH)$ in the presence of concentrated $H_2SO_4$ is an esterification reaction,which produces ethyl acetate $(CH_3COOC_2H_5)$.
$3$. Thus,$X = CH_3COOH$ and $Y = C_2H_5OH$.

(A) When diethyl ether reacts with cold $HI$,it undergoes cleavage to form ethyl alcohol and ethyl iodide as follows:
$C_2H_5-O-C_2H_5 + HI \rightarrow C_2H_5OH + C_2H_5I$

(A) The reaction $2 CH_3COCH_3 \stackrel{Ba(OH)_2}{\longrightarrow} X$ is an aldol condensation reaction of acetone.
In this reaction,two molecules of acetone undergo condensation in the presence of a base like $Ba(OH)_2$ to form $4-hydroxy-4-methylpentan-2-one$.
The reaction mechanism involves the formation of an enolate ion from one acetone molecule,which then attacks the carbonyl carbon of another acetone molecule.
The product $X$ is $CH_3-C(OH)(CH_3)-CH_2-CO-CH_3$.

(A) The oxidation of acetaldehyde $(CH_3CHO)$ to acetic acid $(CH_3COOH)$ is an industrial process.
This reaction is typically catalyzed by manganese acetate $(Mn(CH_3COO)_2)$ in the presence of air or oxygen.
The reaction is: $2CH_3CHO + O_2 \xrightarrow{Mn(CH_3COO)_2} 2CH_3COOH$.

(C) The reaction of oxalic acid with concentrated $H_2SO_4$ is as follows:
$(COOH)_2 \xrightarrow{conc. H_2SO_4} CO + CO_2 + H_2O$
The gaseous mixture consists of $CO$ and $CO_2$.
When this mixture is passed through caustic potash $(KOH)$,$CO_2$ (an acidic gas) is absorbed,while $CO$ (a neutral gas) passes through.
The reaction of $CO_2$ with $KOH$ is:
$CO_2 + 2KOH \rightarrow K_2CO_3 + H_2O$
Thus,the product formed is potassium carbonate $(K_2CO_3)$.

(C) The preparation of aspirin (acetylsalicylic acid) from salicylic acid involves the acetylation of the phenolic $-OH$ group.
This reaction is carried out by reacting salicylic acid with acetic anhydride in the presence of an acid catalyst like concentrated $H_2SO_4$.
The reaction is: $\text{Salicylic acid} + (CH_3CO)_2O \xrightarrow{Conc. H_2SO_4} \text{Aspirin} + CH_3COOH$.
Therefore,the correct reagent is $(CH_3CO)_2O / \text{Conc. } H_2SO_4$.

(B) The reaction of acetic acid $(CH_3 COOH)$ with ammonia $(NH_3)$ forms ammonium acetate $(CH_3 COONH_4)$ as an intermediate $(X)$.
$CH_3 COOH + NH_3 \longrightarrow CH_3 COONH_4$ $(X)$
On heating,ammonium acetate undergoes dehydration to form acetamide $(CH_3 CONH_2)$ as the final product $(Y)$.
$CH_3 COONH_4 \stackrel{\Delta}{\longrightarrow} CH_3 CONH_2$ $(Y)$ $+ H_2 O$
Therefore,$X$ is $CH_3 COONH_4$ and $Y$ is $CH_3 CONH_2$.

(C) The hybridization of the central atom $Xe$ is determined by the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $XeF_2$: $H = \frac{1}{2}(8 + 2) = 5$,which corresponds to $sp^3d$ hybridization.
For $XeF_4$: $H = \frac{1}{2}(8 + 4) = 6$,which corresponds to $sp^3d^2$ hybridization.
Therefore,the correct pair is $XeF_2, sp^3d$.

(D) Radioactive disintegration follows first-order kinetics.
$k = \frac{2.303}{t} \log \frac{a}{a-x}$
Given $a = 10 \ g$,$(a-x) = 1 \ g$,and $t = 2.303 \ \text{min}$.
$k = \frac{2.303}{2.303} \log \frac{10}{1} = 1 \ \text{min}^{-1}$.
The half-life is given by $t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693}{1} = 0.693 \ \text{min}$.

(A) The atomic number of $Cu$ is $29$. The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
For $Cu^{2+}$ ion,two electrons are removed (one from $4s$ and one from $3d$),resulting in the configuration $[Ar] 3d^9$.
In the $3d^9$ configuration,there is $1$ unpaired electron $(n = 1)$.
The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$.
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM}$.

(B) The formula for the mass deposited is $w = z \cdot i \cdot t$,where $w$ is the mass,$z$ is the electrochemical equivalent,$i$ is the current,and $t$ is the time.
Given: $w = 0.066 \ g$,$i = 2 \ A$,$t = 100 \ s$.
Rearranging the formula to solve for $z$: $z = \frac{w}{i \times t}$.
Substituting the values: $z = \frac{0.066}{2 \times 100} = \frac{0.066}{200} = 0.00033 \ g \ C^{-1}$.
Therefore,$z = 3.3 \times 10^{-4} \ g \ C^{-1}$.

(B) The reduction half-reaction for the copper electrode is: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$.
Using the Nernst equation at $25^{\circ} C$ $(298 \ K)$:
$E_{Cu^{2+}/Cu} = E^{\circ}_{Cu^{2+}/Cu} - \frac{0.0591}{n} \log \frac{1}{[Cu^{2+}]}$
Given $E^{\circ} = +0.34 \ V$,$n = 2$,and $[Cu^{2+}] = 0.01 \ M = 10^{-2} \ M$.
$E_{Cu^{2+}/Cu} = 0.34 - \frac{0.0591}{2} \log \frac{1}{10^{-2}}$
$E_{Cu^{2+}/Cu} = 0.34 - \frac{0.0591}{2} \log(10^2)$
$E_{Cu^{2+}/Cu} = 0.34 - \frac{0.0591}{2} \times 2$
$E_{Cu^{2+}/Cu} = 0.34 - 0.0591 = 0.2809 \ V$.

(D) secondary $(2^{\circ})$ alcohol is one in which the hydroxyl group $(-OH)$ is attached to a carbon atom that is bonded to two other carbon atoms.
$1$. $2$-methyl-$2$-propanol: $(CH_3)_3COH$ is a tertiary $(3^{\circ})$ alcohol.
$2$. $1$-propanol: $CH_3CH_2CH_2OH$ is a primary $(1^{\circ})$ alcohol.
$3$. $1$-butanol: $CH_3CH_2CH_2CH_2OH$ is a primary $(1^{\circ})$ alcohol.
$4$. $2$-pentanol: $CH_3CH(OH)CH_2CH_2CH_3$ is a secondary $(2^{\circ})$ alcohol,as the $-OH$ group is attached to the $C2$ atom,which is bonded to $C1$ and $C3$.

(B) tertiary amine has the general formula $R_3N$,where $R$ is an alkyl group.
For a tertiary amine,the minimum number of carbon atoms required is $3$ (e.g.,trimethylamine,$(CH_3)_3N$).
Calculating the molecular formula for $(CH_3)_3N$:
Total carbons = $3$,Total hydrogens = $3 \times 3 = 9$,Total nitrogens = $1$.
Thus,the molecular formula is $C_3H_9N$.

(A) The chemical formula of chloroethane is $CH_3CH_2Cl$.
In this molecule,the carbon atom attached to the chlorine atom (halogen) is bonded to two hydrogen atoms.
Since this carbon atom is attached to only one other carbon atom,it is a primary $(1^{\circ})$ carbon.
Therefore,chloroethane is a primary alkyl halide.
Thus,the correct answer is: Two,primary.

(D) Acetophenone $(C_6H_5COCH_3)$ is prepared by the Friedel-Crafts acylation of benzene.
In this reaction,benzene $(C_6H_6)$ reacts with acetyl chloride $(CH_3COCl)$ in the presence of an anhydrous Lewis acid catalyst,such as anhydrous $AlCl_3$.
The chemical equation is:
$C_6H_6 + CH_3COCl \xrightarrow{\text{anhydrous } AlCl_3} C_6H_5COCH_3 + HCl$
Therefore,the chemicals required are benzene $(A)$,acetyl chloride $(C)$,and anhydrous $AlCl_3$ $(D)$.

(D) The basic character of a molecule depends on the availability of the lone pair of electrons on the central atom for donation.
In $PH_3$,the lone pair is present in an orbital with high $s$-character,making it less available for donation.
In $P(CH_3)_3$,the three methyl groups are electron-donating ($+I$ effect),which increases the electron density on the phosphorus atom,making the lone pair more available for donation.
Therefore,$P(CH_3)_3$ is a stronger base than $PH_3$.

(A) The thermite reaction involves the reduction of ferric oxide $(Fe_2O_3)$ by aluminium $(Al)$ powder.
The balanced chemical equation is:
$Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3$
From the stoichiometry of the reaction,$1$ mole of $Fe_2O_3$ reacts with $2$ moles of $Al$.
However,the question refers to the mass ratio or parts by weight commonly used in the thermite process.
In the standard thermite mixture,$3$ parts of ferric oxide $(Fe_2O_3)$ are mixed with $1$ part of aluminium powder $(Al)$ by weight.
Therefore,$X = 3$ and $Y = 1$.

(A) Bauxite powder reacts with coke and nitrogen as follows:
$Al_2O_3 + 3C + N_2 \xrightarrow{2075 \ K} 2AlN (X) + 3CO$
When $AlN$ reacts with water,it undergoes hydrolysis:
$AlN + 3H_2O \longrightarrow Al(OH)_3 + NH_3 \uparrow$
Thus,the gas formed is ammonia $(NH_3)$.

(B) When ammonia reacts with excess chlorine,the reaction is as follows:
$NH_3 + 3Cl_2 \longrightarrow NCl_3 + 3HCl$
In this reaction,$NCl_3$ (nitrogen trichloride) is formed as an explosive product along with $HCl$ (hydrogen chloride).

(D) When ammonia reacts with excess chlorine,the reaction is as follows:
$NH_3 + 3Cl_2 \longrightarrow NCl_3 + 3HCl$
In this reaction,nitrogen trichloride $(NCl_3)$ and hydrogen chloride $(HCl)$ are formed as the products.

(D) $1$. The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $A$ is the mass number. Option $A$ is incorrect because it states $A^{1/2}$.
$2$. Isobars are atoms with the same mass number but different atomic numbers. $_{7}N^{15}$ and $_{8}O^{16}$ have different mass numbers ($15$ and $16$),so they are not isobars. Option $B$ is incorrect.
$3$. The thorium series ($4n$ series) ends at the stable isotope $_{82}Pb^{208}$. Option $C$ is incorrect as it mentions $_{83}Bi^{209}$.
$4$. Magic numbers for nucleons are $2, 8, 20, 28, 50, 82, 126$. For $_{20}Ca^{40}$,the number of protons is $20$ and the number of neutrons is $40 - 20 = 20$. Since both $20$ and $20$ are magic numbers,option $D$ is correct.

(C) Physical adsorption (physisorption) is characterized by weak van der Waals forces between the adsorbate and the adsorbent.
Because these forces are weak,the activation energy required for physical adsorption is very low,not high.
Therefore,the statement that 'Activation energy of physical adsorption is very high' is incorrect.
In contrast,chemical adsorption (chemisorption) typically requires a high activation energy due to the formation of chemical bonds.

(D) The bond energy depends on the bond length. As the size of the central atom increases,the bond length increases,and the bond energy decreases.
The order of atomic size is $N < P < As$.
Therefore,the order of bond energy is $N-H > P-H > As-H$.
The bond energy values are $N-H = 389 \ kJ \ mol^{-1}$,$P-H = 318 \ kJ \ mol^{-1}$,and $As-H = 247 \ kJ \ mol^{-1}$.
Thus,the bond energies for $P-H$,$As-H$,and $N-H$ are $318, 247, 389$ respectively.