TS EAMCET 2008 Chemistry Question Paper with Answer and Solution

(D) Given curves are $y^2=4x+4$ $(i)$ and $y^2=36(9-x)$ (ii).
To find the intersection points,equate the two expressions for $y^2$:
$4x+4 = 324-36x$
$40x = 320 \Rightarrow x = 8$.
Substituting $x=8$ into $(i)$,$y^2 = 4(8)+4 = 36 \Rightarrow y = \pm 6$.
So,the intersection points are $(8,6)$ and $(8,-6)$.
Differentiating $(i)$ with respect to $x$: $2y \frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{2}{y}$.
Differentiating (ii) with respect to $x$: $2y \frac{dy}{dx} = -36 \Rightarrow \frac{dy}{dx} = \frac{-18}{y}$.
At point $(8,6)$:
$m_1 = \frac{2}{6} = \frac{1}{3}$ and $m_2 = \frac{-18}{6} = -3$.
Since $m_1 \times m_2 = \frac{1}{3} \times (-3) = -1$,the tangents are perpendicular.
Therefore,the angle between the curves is $90^{\circ}$ or $\frac{\pi}{2}$.

(C) Given curve is $y^4=ax^3$.
On differentiating with respect to $x$,we get:
$4y^3 \frac{dy}{dx} = 3ax^2$.
At the point $(a, a)$,the slope of the tangent is:
$\frac{dy}{dx} = \frac{3a(a)^2}{4(a)^3} = \frac{3a^3}{4a^3} = \frac{3}{4}$.
The slope of the normal is the negative reciprocal of the slope of the tangent:
$m_{\text{normal}} = -\frac{1}{3/4} = -\frac{4}{3}$.
The equation of the normal at $(a, a)$ is given by:
$y - a = -\frac{4}{3}(x - a)$.
Multiplying by $3$:
$3y - 3a = -4x + 4a$.
Rearranging the terms:
$4x + 3y = 7a$.

(D) Given the integral: $\int e^x(1+x) \cdot \sec^2(x e^x) \, dx = f(x) + C$.
Let $t = x e^x$.
Then,differentiating with respect to $x$ using the product rule: $\frac{dt}{dx} = e^x + x e^x = e^x(1+x)$.
Thus,$dt = e^x(1+x) \, dx$.
Substituting these into the integral,we get: $\int \sec^2(t) \, dt$.
The integral of $\sec^2(t)$ is $\tan(t) + C$.
Substituting back $t = x e^x$,we get $\tan(x e^x) + C$.
Comparing this with $f(x) + C$,we find $f(x) = \tan(x e^x)$.

(C) Let $I = \int_{-\pi / 2}^{\pi / 2} \sin |x| \, dx$.
Since $f(x) = \sin |x|$ is an even function because $f(-x) = \sin |-x| = \sin |x| = f(x)$,we can use the property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$.
Thus,$I = 2 \int_{0}^{\pi / 2} \sin |x| \, dx$.
For $x \in [0, \pi / 2]$,$|x| = x$,so $I = 2 \int_{0}^{\pi / 2} \sin x \, dx$.
Evaluating the integral: $I = 2 [-\cos x]_{0}^{\pi / 2}$.
$I = 2 [-\cos(\pi / 2) - (-\cos 0)]$.
$I = 2 [0 - (-1)] = 2(1) = 2$.

(D) Let $I = \int_0^1 x^{3/2} \sqrt{1-x} \, dx$.
Substitute $x = \sin^2 \theta$,then $dx = 2 \sin \theta \cos \theta \, d\theta$.
When $x=0, \theta=0$ and when $x=1, \theta=\frac{\pi}{2}$.
$I = \int_0^{\pi/2} (\sin^2 \theta)^{3/2} \sqrt{1-\sin^2 \theta} \cdot (2 \sin \theta \cos \theta) \, d\theta$
$I = \int_0^{\pi/2} \sin^3 \theta \cdot \cos \theta \cdot 2 \sin \theta \cos \theta \, d\theta$
$I = 2 \int_0^{\pi/2} \sin^4 \theta \cos^2 \theta \, d\theta$.
Using Wallis' formula $\int_0^{\pi/2} \sin^m \theta \cos^n \theta \, d\theta = \frac{(m-1)(m-3)\dots(1) \cdot (n-1)(n-3)\dots(1)}{(m+n)(m+n-2)\dots(2)} \cdot \frac{\pi}{2}$ (if both $m, n$ are even):
$I = 2 \left[ \frac{(4-1)(4-3) \cdot (2-1)}{(4+2)(4+2-2)(4+2-4)} \cdot \frac{\pi}{2} \right]$
$I = 2 \left[ \frac{3 \cdot 1 \cdot 1}{6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} \right] = 2 \left[ \frac{3}{48} \cdot \frac{\pi}{2} \right] = 2 \left[ \frac{1}{16} \cdot \frac{\pi}{2} \right] = \frac{\pi}{16}$.

(B) The given differential equation is $x y^2 d y = (x^3 + y^3) d x$.
This can be rewritten as $\frac{d y}{d x} = \frac{x^3 + y^3}{x y^2}$.
This is a homogeneous differential equation. Let $y = v x$,then $\frac{d y}{d x} = v + x \frac{d v}{d x}$.
Substituting these into the equation:
$v + x \frac{d v}{d x} = \frac{x^3 + v^3 x^3}{x(v x)^2} = \frac{x^3(1 + v^3)}{x^3 v^2} = \frac{1 + v^3}{v^2}$.
$x \frac{d v}{d x} = \frac{1 + v^3}{v^2} - v = \frac{1 + v^3 - v^3}{v^2} = \frac{1}{v^2}$.
Separating the variables,we get $v^2 d v = \frac{1}{x} d x$.
Integrating both sides: $\int v^2 d v = \int \frac{1}{x} d x$.
$\frac{v^3}{3} = \log |x| + C$,where $C = \log c$.
$\frac{v^3}{3} = \log |x| + \log c = \log |c x|$.
Substituting $v = \frac{y}{x}$,we get $\frac{1}{3} (\frac{y}{x})^3 = \log |c x|$.
$\frac{y^3}{3 x^3} = \log |c x| \Rightarrow y^3 = 3 x^3 \log |c x|$.

(B) The given linear differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = -\tan x$ and $Q = e^x \sec x$.
First,we find the integrating factor $(IF)$:
$IF = e^{\int P dx} = e^{\int -\tan x dx} = e^{\ln(\cos x)} = \cos x$.
The general solution is given by $y \cdot (IF) = \int Q \cdot (IF) dx + c$.
Substituting the values:
$y \cos x = \int (e^x \sec x) \cdot \cos x dx + c$.
Since $\sec x \cdot \cos x = 1$,the equation simplifies to:
$y \cos x = \int e^x dx + c$.
Integrating $e^x$,we get:
$y \cos x = e^x + c$.

(C) Let $\overrightarrow{B} = \hat{i} - \hat{j}$.
To find the component of vector $\overrightarrow{A}$ along the direction of $\overrightarrow{B}$,we calculate the scalar projection of $\overrightarrow{A}$ onto the unit vector of $\overrightarrow{B}$.
The unit vector along $\overrightarrow{B}$ is $\hat{u}_B = \frac{\overrightarrow{B}}{|\overrightarrow{B}|} = \frac{\hat{i} - \hat{j}}{\sqrt{1^2 + (-1)^2}} = \frac{\hat{i} - \hat{j}}{\sqrt{2}}$.
The component of $\overrightarrow{A}$ along $\overrightarrow{B}$ is given by the dot product $\overrightarrow{A} \cdot \hat{u}_B$.
$\overrightarrow{A} \cdot \hat{u}_B = (a_x \hat{i} + a_y \hat{j} + a_z \hat{k}) \cdot \left( \frac{\hat{i} - \hat{j}}{\sqrt{2}} \right)$.
$= \frac{1}{\sqrt{2}} (a_x(1) + a_y(-1) + a_z(0)) = \frac{a_x - a_y}{\sqrt{2}}$.

(C) Let the position vectors be $\vec{a} = \overrightarrow{OA} = 2\hat{i}-\hat{j}+\hat{k}$,$\vec{b} = \overrightarrow{OB} = \hat{i}-3\hat{j}-5\hat{k}$,and $\vec{c} = \overrightarrow{OC} = 3\hat{i}-4\hat{j}-4\hat{k}$.
To find $\cos A$,we consider the triangle $ABC$. The vectors forming the sides are $\vec{AB} = \vec{b} - \vec{a} = (1-2)\hat{i} + (-3+1)\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}$ and $\vec{AC} = \vec{c} - \vec{a} = (3-2)\hat{i} + (-4+1)\hat{j} + (-4-1)\hat{k} = \hat{i} - 3\hat{j} - 5\hat{k}$.
The angle $A$ is the angle between vectors $\vec{AB}$ and $\vec{AC}$.
$\cos A = \frac{\vec{AB} \cdot \vec{AC}}{|\vec{AB}| |\vec{AC}|}$.
$\vec{AB} \cdot \vec{AC} = (-1)(1) + (-2)(-3) + (-6)(-5) = -1 + 6 + 30 = 35$.
$|\vec{AB}| = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$.
$|\vec{AC}| = \sqrt{1^2 + (-3)^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$.
$\cos A = \frac{35}{\sqrt{41} \sqrt{35}} = \sqrt{\frac{35}{41}}$.
Therefore,$\cos^2 A = \frac{35}{41}$.

(A) Let the points be $A(60, 3)$,$B(40, -8)$,and $C(a, -52)$.
Since the points are collinear,the area of the triangle formed by them is zero,or the slope of $AB$ must be equal to the slope of $BC$.
Slope of $AB = \frac{-8-3}{40-60} = \frac{-11}{-20} = \frac{11}{20}$.
Slope of $BC = \frac{-52-(-8)}{a-40} = \frac{-44}{a-40}$.
Equating the slopes: $\frac{11}{20} = \frac{-44}{a-40}$.
$11(a-40) = 20(-44)$.
$11a - 440 = -880$.
$11a = -440$.
$a = -40$.

(A) Let $\overrightarrow{p} = \overrightarrow{a}$ and $\overrightarrow{q} = \overrightarrow{b}$ be the position vectors of points $P$ and $Q$ respectively.
Given that $\overrightarrow{PR} = 5 \overrightarrow{PQ}$.
We know that $\overrightarrow{PR} = \overrightarrow{r} - \overrightarrow{p}$ and $\overrightarrow{PQ} = \overrightarrow{q} - \overrightarrow{p}$.
Substituting these into the given equation:
$\overrightarrow{r} - \overrightarrow{p} = 5(\overrightarrow{q} - \overrightarrow{p})$
$\overrightarrow{r} - \overrightarrow{a} = 5(\overrightarrow{b} - \overrightarrow{a})$
$\overrightarrow{r} = \overrightarrow{a} + 5\overrightarrow{b} - 5\overrightarrow{a}$
$\overrightarrow{r} = 5\overrightarrow{b} - 4\overrightarrow{a}$
Thus,the position vector of $R$ is $5\overrightarrow{b} - 4\overrightarrow{a}$.

(C) Let the vertices of the triangle be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.
Given the midpoints of $AB, BC$,and $CA$ are $(l, 0, 0), (0, m, 0)$,and $(0, 0, n)$ respectively.
Using the midpoint formula:
$\frac{x_1+x_2}{2} = l, \frac{y_1+y_2}{2} = 0, \frac{z_1+z_2}{2} = 0 \implies x_1+x_2=2l, y_1+y_2=0, z_1+z_2=0$
$\frac{x_2+x_3}{2} = 0, \frac{y_2+y_3}{2} = m, \frac{z_2+z_3}{2} = 0 \implies x_2+x_3=0, y_2+y_3=2m, z_2+z_3=0$
$\frac{x_3+x_1}{2} = 0, \frac{y_3+y_1}{2} = 0, \frac{z_3+z_1}{2} = n \implies x_3+x_1=0, y_3+y_1=0, z_3+z_1=2n$
Solving these systems of equations:
For $x$: $x_1+x_2=2l, x_2+x_3=0, x_3+x_1=0 \implies x_1=l, x_2=l, x_3=-l$
For $y$: $y_1+y_2=0, y_2+y_3=2m, y_3+y_1=0 \implies y_1=-m, y_2=m, y_3=m$
For $z$: $z_1+z_2=0, z_2+z_3=0, z_3+z_1=2n \implies z_1=n, z_2=-n, z_3=n$
Thus,$A(l, -m, n), B(l, m, -n), C(-l, m, n)$.
Now,calculate the squared side lengths:
$AB^2 = (l-l)^2 + (m-(-m))^2 + (-n-n)^2 = 0 + (2m)^2 + (-2n)^2 = 4m^2 + 4n^2$
$BC^2 = (-l-l)^2 + (m-m)^2 + (n-(-n))^2 = (-2l)^2 + 0 + (2n)^2 = 4l^2 + 4n^2$
$CA^2 = (l-(-l))^2 + (-m-m)^2 + (n-n)^2 = (2l)^2 + (-2m)^2 + 0 = 4l^2 + 4m^2$
Summing these: $AB^2+BC^2+CA^2 = (4m^2+4n^2) + (4l^2+4n^2) + (4l^2+4m^2) = 8(l^2+m^2+n^2)$.
Therefore,$\frac{AB^2+BC^2+CA^2}{l^2+m^2+n^2} = \frac{8(l^2+m^2+n^2)}{l^2+m^2+n^2} = 8$.

(C) Given the direction cosines of the two lines as $(l_1, m_1, n_1) = \left(\frac{\sqrt{3}}{4}, \frac{1}{4}, \frac{\sqrt{3}}{2}\right)$ and $(l_2, m_2, n_2) = \left(\frac{\sqrt{3}}{4}, \frac{1}{4}, \frac{-\sqrt{3}}{2}\right)$.
The cosine of the angle $\theta$ between two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ is given by $\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.
Substituting the values:
$\cos \theta = \left| \left(\frac{\sqrt{3}}{4} \times \frac{\sqrt{3}}{4}\right) + \left(\frac{1}{4} \times \frac{1}{4}\right) + \left(\frac{\sqrt{3}}{2} \times \frac{-\sqrt{3}}{2}\right) \right|$
$\cos \theta = \left| \frac{3}{16} + \frac{1}{16} - \frac{3}{4} \right|$
$\cos \theta = \left| \frac{4}{16} - \frac{12}{16} \right| = \left| -\frac{8}{16} \right| = \left| -\frac{1}{2} \right| = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = \frac{\pi}{3}$.

(A) The sample space $S$ for throwing two dice has $6 \times 6 = 36$ outcomes.
$p_k = P(E_k) = \frac{|E_k|}{36}$.
For $k=1$: $E_1 = \{(1, 1)\}$,so $p_1 = \frac{1}{36}$.
For $k=2$: $E_2 = \{(1, 2), (2, 1)\}$,so $p_2 = \frac{2}{36}$.
For $k=4$: $E_4 = \{(1, 4), (4, 1), (2, 2)\}$,so $p_4 = \frac{3}{36}$.
For $k=6$: $E_6 = \{(1, 6), (6, 1), (2, 3), (3, 2)\}$,so $p_6 = \frac{4}{36}$.
For $k=30$: $E_{30} = \{(5, 6), (6, 5)\}$,so $p_{30} = \frac{2}{36}$.
For $k=11$: $E_{11} = \emptyset$,so $p_{11} = 0$.
For $k=36$: $E_{36} = \{(6, 6)\}$,so $p_{36} = \frac{1}{36}$.
Comparing the values: $p_1 = \frac{1}{36}$,$p_{30} = \frac{2}{36}$,$p_4 = \frac{3}{36}$,$p_6 = \frac{4}{36}$.
Thus,$p_1 < p_{30} < p_4 < p_6$ is correct.

(B) Let $R$ be the event of drawing a red ball. The contents of the boxes are:
$B_1: 1R, 2W \implies P(R|B_1) = \frac{1}{3}$
$B_2: 2R, 3W \implies P(R|B_2) = \frac{2}{5}$
$B_3: 3R, 4W \implies P(R|B_3) = \frac{3}{7}$
Using Bayes' Theorem,the probability that the ball came from box $B_2$ given that it is red is:
$P(B_2|R) = \frac{P(B_2)P(R|B_2)}{P(B_1)P(R|B_1) + P(B_2)P(R|B_2) + P(B_3)P(R|B_3)}$
Substituting the values:
$P(B_2|R) = \frac{\frac{1}{3} \times \frac{2}{5}}{\frac{1}{2} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{5} + \frac{1}{6} \times \frac{3}{7}}$
$P(B_2|R) = \frac{\frac{2}{15}}{\frac{1}{6} + \frac{2}{15} + \frac{1}{14}}$
Finding a common denominator for the denominator $(210)$:
$P(B_2|R) = \frac{\frac{2}{15}}{\frac{35 + 28 + 15}{210}} = \frac{2}{15} \times \frac{210}{78} = \frac{280}{1170} = \frac{14}{39}$

(C) For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
Given that $P(X=1) = P(X=2)$,we have:
$\frac{e^{-\lambda} \lambda^1}{1!} = \frac{e^{-\lambda} \lambda^2}{2!}$
Dividing both sides by $e^{-\lambda} \lambda$ (assuming $\lambda \neq 0$):
$1 = \frac{\lambda}{2}$
$\lambda = 2$
Now,we need to find $P(X=4)$:
$P(X=4) = \frac{e^{-\lambda} \lambda^4}{4!} = \frac{e^{-2} (2)^4}{4!}$
$P(X=4) = \frac{e^{-2} \times 16}{24} = \frac{2}{3 e^2}$.

(A) For a probability distribution,the sum of all probabilities must be equal to $1$.
Therefore,$\sum P(X = x) = 1$.
$\frac{1}{10} + k + \frac{1}{5} + 2k + \frac{3}{10} + k = 1$
Combining the constant terms:
$(\frac{1}{10} + \frac{2}{10} + \frac{3}{10}) + (k + 2k + k) = 1$
$\frac{6}{10} + 4k = 1$
$4k = 1 - \frac{6}{10}$
$4k = \frac{10 - 6}{10}$
$4k = \frac{4}{10}$
$k = \frac{4}{10 \times 4}$
$k = \frac{1}{10}$

(D) The terminal velocity $v_T$ of a spherical drop is given by $v_T = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta}$,where $r$ is the radius of the drop.
Thus,$v_T \propto r^2$.
Given the ratio of terminal velocities is $\frac{v_{T_1}}{v_{T_2}} = \frac{9}{4}$.
Since $\frac{v_{T_1}}{v_{T_2}} = \left(\frac{r_1}{r_2}\right)^2$,we have $\frac{r_1}{r_2} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.
The volume $V$ of a spherical drop is given by $V = \frac{4}{3} \pi r^3$,so $V \propto r^3$.
The ratio of their volumes is $\frac{V_1}{V_2} = \left(\frac{r_1}{r_2}\right)^3 = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$.

(C) According to Lewis,compounds that can accept a lone pair of electrons are called Lewis acids.
Boron halides,such as $BX_3$,have only $6$ electrons in the valence shell of the boron atom.
Due to this electron deficiency,they can accept a lone pair of electrons from a donor to complete their octet,thus behaving as Lewis acids.

(D) Orthosilicic acid $(H_4SiO_4)$,on heating at $1000^{\circ}C$,loses two water molecules to form silica $(SiO_2)$ as product $A$.
$H_4SiO_4 \xrightarrow[{-2H_2O}]{1000^{\circ}C} SiO_2 (A)$
Silica $(SiO_2)$ on reduction with carbon at high temperature gives carborundum $(SiC)$ as product $B$ and carbon monoxide $(CO)$.
$SiO_2 + 3C \xrightarrow{\Delta} SiC (B) + 2CO$
Therefore,$B$ is carborundum.

(B) The chemical formula of peroxodisulphuric acid (Marshall's acid) is $H_2S_2O_8$.
Its structure consists of two $SO_3$ groups linked by a peroxide linkage $(-O-O-)$.
In the structure:
- There are $2$ $S=O$ double bonds per sulfur atom,contributing $2$ $\pi$ bonds each,totaling $4$ $\pi$ bonds.
- Counting the $\sigma$ bonds: $4$ $S=O$ bonds,$2$ $S-OH$ bonds,$2$ $O-H$ bonds,$2$ $S-O$ bonds (to peroxide oxygen),and $1$ $O-O$ bond.
- Total $\sigma$ bonds = $4 + 2 + 2 + 2 + 1 = 11$.
- Total $\pi$ bonds = $4$.
Thus,the number of $\sigma$ and $\pi$ bonds are $11$ and $4$ respectively.

(D) An oxidising agent is a substance that undergoes reduction or facilitates the oxidation of another reactant (e.g.,by removing hydrogen).
In all three given reactions,the oxidation state of chlorine decreases from $0$ in $Cl_2$ to $-1$ in $HCl$,meaning chlorine is reduced.
$(i)$ $CH_3CH_2OH + Cl_2 \longrightarrow CH_3CHO + HCl$: Chlorine removes hydrogen from ethanol,oxidizing it to acetaldehyde.
(ii) $CH_3CHO + Cl_2 \longrightarrow CCl_3CHO + HCl$: Chlorine replaces hydrogen atoms in acetaldehyde,acting as an oxidant.
(iii) $CH_4 + Cl_2 \stackrel{hv}{\longrightarrow} CH_3Cl + HCl$: Chlorine removes hydrogen from methane,oxidizing it to chloromethane.
Since chlorine is reduced in all these reactions,it acts as an oxidising agent in all of them.

(B) The oxidizing power of halogens decreases down the group as the reduction potential decreases.
Fluorine $(F_2)$ is the strongest oxidizing agent,while Iodine $(I_2)$ is the weakest.
$A$ halogen with a higher reduction potential can displace a halide ion with a lower reduction potential from its salt solution.
Since the reduction potential of $Cl_2$ is lower than that of $F_2$,$Cl_2$ cannot oxidize $F^-$ to $F_2$.
Therefore,the reaction $Cl_2 + 2F^- \longrightarrow 2Cl^- + F_2$ is not spontaneous and does not occur.

(B) $(i)$ Superoxides contain the $O_2^-$ ion,which has one unpaired electron,making them paramagnetic. This statement is correct.
(ii) As we move down the group,the size of the metal ion increases,which decreases the lattice energy and increases the solubility/dissociation of hydroxides,thus increasing basic strength. This statement is correct.
(iii) Conductivity in aqueous solutions depends on ionic mobility. As we move down the group,the size of the hydrated ion decreases (due to less hydration),leading to higher ionic mobility and higher conductivity. Thus,conductivity increases down the group. This statement is incorrect.
(iv) The basic nature of carbonates is due to anionic hydrolysis (hydrolysis of the $CO_3^{2-}$ ion),not cationic hydrolysis. This statement is incorrect.
Therefore,only statements $(i)$ and $(ii)$ are correct.

(A) $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$. $100 \ g \ CaCO_3$ gives $22.4 \ L \ CO_2$ at $STP$. So,$10 \ g \ CaCO_3$ gives $2.24 \ L \ CO_2$. Thus,$A-(iv)$.
$(B)$ $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$. $106 \ g \ Na_2CO_3$ gives $22.4 \ L \ CO_2$. So,$1.06 \ g \ Na_2CO_3$ gives $0.224 \ L \ CO_2$. Thus,$B-(i)$.
$(C)$ $C + O_2 \rightarrow CO_2$. $12 \ g \ C$ gives $22.4 \ L \ CO_2$. So,$2.4 \ g \ C$ gives $(22.4 \times 2.4) / 12 = 4.48 \ L \ CO_2$. Thus,$C-(ii)$.
$(D)$ $2CO + O_2 \rightarrow 2CO_2$. $56 \ g \ CO$ gives $2 \times 22.4 \ L \ CO_2 = 44.8 \ L \ CO_2$. So,$0.56 \ g \ CO$ gives $(44.8 \times 0.56) / 56 = 0.448 \ L \ CO_2$. Thus,$D-(iii)$.
Therefore,the correct match is $A-(iv), B-(i), C-(ii), D-(iii)$.

(C) The kinetic energy $(KE)$ of an ideal gas is given by the formula $KE = nRT$, where $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature in Kelvin.
Given:
For helium: $n_{He} = 0.3 \text{ mol}$
For argon: $n_{Ar} = 0.4 \text{ mol}$, $T_{Ar} = 400 \text{ K}$
According to the problem, $KE_{He} = KE_{Ar}$.
Substituting the values:
$0.3 \times R \times T = 0.4 \times R \times 400$
Dividing both sides by $R$:
$0.3 \times T = 160$
$T = \frac{160}{0.3} = 533.33 \text{ K} \approx 533 \text{ K}$.

(B) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{313.52 Z^2}{n^2} \ kcal \ mol^{-1}$.
For a hydrogen atom,$Z = 1$.
The Lyman series corresponds to transitions ending at $n_1 = 1$.
The $H_\alpha$ line in the Lyman series corresponds to the transition from $n_2 = 2$ to $n_1 = 1$.
Energy in the $n_1 = 1$ orbit: $E_1 = -\frac{313.52 \times (1)^2}{(1)^2} = -313.52 \ kcal \ mol^{-1} \approx -313.6 \ kcal \ mol^{-1}$.
Energy in the $n_2 = 2$ orbit: $E_2 = -\frac{313.52 \times (1)^2}{(2)^2} = -\frac{313.52}{4} = -78.38 \ kcal \ mol^{-1} \approx -78.4 \ kcal \ mol^{-1}$.
Thus,the energies are $-313.6 \ kcal \ mol^{-1}$ and $-78.4 \ kcal \ mol^{-1}$.

(A) Given,velocity of particle $A$ $(v_A)$ = $0.05 \ ms^{-1}$.
Velocity of particle $B$ $(v_B)$ = $0.02 \ ms^{-1}$.
Let the mass of particle $A$ $(m_A)$ = $m$.
Then,the mass of particle $B$ $(m_B)$ = $5m$.
The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
For particle $A$,$\lambda_A = \frac{h}{m_A v_A} = \frac{h}{m \times 0.05}$.
For particle $B$,$\lambda_B = \frac{h}{m_B v_B} = \frac{h}{5m \times 0.02} = \frac{h}{0.1m}$.
Taking the ratio $\frac{\lambda_A}{\lambda_B} = \frac{h}{m \times 0.05} \times \frac{0.1m}{h} = \frac{0.1}{0.05} = \frac{2}{1}$.
Thus,the ratio is $2: 1$.

(A) To find the enthalpy change for the reaction $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$,we use Hess's Law by adding the given equations:
$(i) \ H_2O_{(g)} + C_{(s)} \longrightarrow CO_{(g)} + H_{2(g)} ; \Delta H_1 = +131 \ kJ$
$(ii) \ CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H_2 = -282 \ kJ$
$(iii) \ H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(g)} ; \Delta H_3 = -242 \ kJ$
Adding equations $(i)$,$(ii)$,and $(iii)$:
$(H_2O_{(g)} + C_{(s)} + CO_{(g)} + \frac{1}{2} O_{2(g)} + H_{2(g)} + \frac{1}{2} O_{2(g)})$ $\longrightarrow (CO_{(g)} + H_{2(g)} + CO_{2(g)} + H_2O_{(g)})$
Canceling common species on both sides,we get:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
The total enthalpy change is $\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = 131 + (-282) + (-242) = -393 \ kJ$.