TS EAMCET 2003 Chemistry Question Paper with Answer and Solution

(D) For total internal reflection to occur at the hypotenuse,the angle of incidence $i$ must be greater than or equal to the critical angle $C$.
Here,the ray travels parallel to the base. From the geometry of the right-angled prism,the angle of incidence $i$ at the hypotenuse is $(90^{\circ} - \theta)$,where $\theta$ is the base angle.
For total internal reflection,$i \geq C$,where $\sin C = \frac{1}{\mu}$.
Thus,the condition is $i \geq \sin^{-1}\left(\frac{1}{\mu}\right)$.
Substituting $i = 90^{\circ} - \theta$,we get $90^{\circ} - \theta \geq \sin^{-1}\left(\frac{1}{\mu}\right)$.
Taking the sine of both sides: $\sin(90^{\circ} - \theta) \geq \frac{1}{\mu}$.
This simplifies to $\cos \theta \geq \frac{1}{\mu}$.
Therefore,the maximum value of the base angle $\theta$ is $\cos^{-1}\left(\frac{1}{\mu}\right)$.

(C) The refractive index of the material of the prism is given by the formula:
$\mu = \frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}$
Given that the angle of minimum deviation $\delta_m = A$,we substitute this into the formula:
$\mu = \frac{\sin \left(\frac{A+A}{2}\right)}{\sin \frac{A}{2}}$
$\mu = \frac{\sin A}{\sin \frac{A}{2}}$
Using the trigonometric identity $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$,we get:
$\mu = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}$
$\mu = 2 \cos \frac{A}{2}$
$\cos \frac{A}{2} = \frac{\mu}{2}$
$\frac{A}{2} = \cos ^{-1}\left(\frac{\mu}{2}\right)$
$A = 2 \cos ^{-1}\left(\frac{\mu}{2}\right)$

(C) The oxidation state of $Xe$ in $XeO_3$ is calculated as follows:
$x + 3(-2) = 0 \implies x = +6$.
$XeO_3$ has a trigonal pyramidal geometry due to $sp^3$ hybridization with one lone pair on the $Xe$ atom.
Due to the presence of the lone pair,the bond angle is compressed from the ideal tetrahedral angle of $109.5^{\circ}$ to approximately $103^{\circ}$.

(A) Sodium metal is manufactured by the electrolysis of fused sodium chloride mixed with $KCl$ and $KF$.
On electrolysis:
At iron cathode: $Na^{+} + e^{-} \longrightarrow Na_{(s)}$ (metallic sodium)
At graphite anode: $2Cl^{-} \longrightarrow Cl_{2(g)} + 2e^{-}$
$NaCl$ melts at $800^{\circ}C$. It is difficult to attain and maintain this melting point. So,$KCl$ and $KF$ are added to lower the melting point of the mixture to about $600^{\circ}C$. $KCl$ and $KF$ are not electrolyzed under the voltage conditions used for sodium.

(C) In the Castner process,the extraction of sodium metal involves the electrolysis of molten sodium hydroxide $(NaOH)$.
The dissociation of sodium hydroxide is represented as: $4 NaOH \longrightarrow 4 Na^{+} + 4 OH^{-}$.
At the anode,oxidation occurs,where hydroxide ions $(OH^{-})$ lose electrons to form water and oxygen gas:
$4 OH^{-} \longrightarrow 2 H_2 O + O_2 + 4 e^{-}$.

(A) When an aqueous solution of magnesium bicarbonate is boiled,it undergoes thermal decomposition to form magnesium oxide,water,and carbon dioxide gas.
The chemical equation is:
$Mg(HCO_3)_2(aq) \xrightarrow{\Delta} MgO(s) + H_2O(l) + 2CO_2(g)$

(C) When $Na_2O$ and $CaO$ are dissolved in water,they form $NaOH$ and $Ca(OH)_2$ respectively.
$Na_2O + H_2O \longrightarrow 2NaOH$
$CaO + H_2O \longrightarrow Ca(OH)_2$
When this mixture is saturated with excess $CO_2$,the hydroxides react to form bicarbonates:
$2NaOH + 2CO_2 \longrightarrow 2NaHCO_3$
$Ca(OH)_2 + 2CO_2 \longrightarrow Ca(HCO_3)_2$
Since $NaHCO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$,the solution is basic due to the hydrolysis of the bicarbonate ion $(HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^-)$.

(A) In an $n-p-n$ transistor,the emitter is $n$-type,the base is $p$-type,and the collector is $n$-type.
When used as an amplifier,the base-emitter junction is forward-biased and the collector-base junction is reverse-biased.
Due to the forward bias,electrons are injected from the emitter into the base.
Because the base is very thin and lightly doped,most of these electrons diffuse through the base into the collector region.
The collector is kept at a positive potential relative to the base to attract these electrons.
Therefore,electrons move from the base to the collector.

(B) The standard equation of a parabola with focus at $(0,0)$ and axis along the $X$-axis is $(y-0)^2 = 4a(x+a)$,where $a$ is a parameter.
This simplifies to $y^2 = 4ax + 4a^2$.
Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $a = \frac{y}{2} \frac{dy}{dx}$.
Substituting this value of $a$ into the original equation:
$y^2 = 4\left(\frac{y}{2} \frac{dy}{dx}\right)x + 4\left(\frac{y}{2} \frac{dy}{dx}\right)^2$.
$y^2 = 2xy \frac{dy}{dx} + 4\left(\frac{y^2}{4} \left(\frac{dy}{dx}\right)^2\right)$.
$y^2 = 2xy \frac{dy}{dx} + y^2 \left(\frac{dy}{dx}\right)^2$.
Dividing by $y$ (assuming $y \neq 0$):
$y = 2x \frac{dy}{dx} + y \left(\frac{dy}{dx}\right)^2$.
Rearranging the terms gives $-y \left(\frac{dy}{dx}\right)^2 = 2x \frac{dy}{dx} - y$.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x \log x^2 + x}{\sin y + y \cos y}$
By separating the variables,we get: $(\sin y + y \cos y) dy = (x \log x^2 + x) dx$
Integrating both sides: $\int (\sin y + y \cos y) dy = \int (x \log x^2 + x) dx$
For the left side,using integration by parts on $\int y \cos y dy$: $\int \sin y dy + (y \sin y - \int \sin y dy) = y \sin y$
For the right side,$\int (2x \log x + x) dx = 2 \int x \log x dx + \int x dx$
Using integration by parts for $\int x \log x dx = \frac{x^2}{2} \log x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \log x - \frac{x^2}{4}$
Thus,$\int (x \log x^2 + x) dx = 2(\frac{x^2}{2} \log x - \frac{x^2}{4}) + \frac{x^2}{2} + C = x^2 \log x - \frac{x^2}{2} + \frac{x^2}{2} + C = x^2 \log x + C$
Therefore,the solution is $y \sin y = x^2 \log x + C$.

(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.
Since $D, E, F$ are mid-points of $AB, AC, BC$ respectively,their position vectors are:
$\vec{d} = \frac{\vec{a} + \vec{b}}{2}$,$\vec{e} = \frac{\vec{a} + \vec{c}}{2}$,$\vec{f} = \frac{\vec{b} + \vec{c}}{2}$.
Now,we calculate the vectors $\overrightarrow{BE}$ and $\overrightarrow{AF}$:
$\overrightarrow{BE} = \vec{e} - \vec{b} = \frac{\vec{a} + \vec{c}}{2} - \vec{b} = \frac{\vec{a} + \vec{c} - 2\vec{b}}{2}$
$\overrightarrow{AF} = \vec{f} - \vec{a} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2}$
Adding these two vectors:
$\overrightarrow{BE} + \overrightarrow{AF} = \frac{\vec{a} + \vec{c} - 2\vec{b} + \vec{b} + \vec{c} - 2\vec{a}}{2} = \frac{2\vec{c} - \vec{a} - \vec{b}}{2} = \vec{c} - \frac{\vec{a} + \vec{b}}{2}$
Since $\vec{d} = \frac{\vec{a} + \vec{b}}{2}$,we have $\overrightarrow{BE} + \overrightarrow{AF} = \vec{c} - \vec{d} = \overrightarrow{DC}$.

(B) The given vector equation is $\overrightarrow{r}=(1-p-q) \overrightarrow{a}+p \overrightarrow{b}+q \overrightarrow{c}$.
This can be rewritten as $\overrightarrow{r} = \overrightarrow{a} + p(\overrightarrow{b} - \overrightarrow{a}) + q(\overrightarrow{c} - \overrightarrow{a})$.
Let $\overrightarrow{u} = \overrightarrow{b} - \overrightarrow{a}$ and $\overrightarrow{v} = \overrightarrow{c} - \overrightarrow{a}$.
Then the equation becomes $\overrightarrow{r} = \overrightarrow{a} + p\overrightarrow{u} + q\overrightarrow{v}$.
Since $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are non-coplanar,the vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ are non-collinear.
An equation of the form $\overrightarrow{r} = \overrightarrow{a} + p\overrightarrow{u} + q\overrightarrow{v}$ represents a plane passing through the point with position vector $\overrightarrow{a}$ and parallel to the vectors $\overrightarrow{u}$ and $\overrightarrow{v}$.

(A) Given that $\overrightarrow{a} = \overrightarrow{b} + \overrightarrow{c}$.
Taking the dot product of both sides with themselves:
$(\overrightarrow{a}) \cdot (\overrightarrow{a}) = (\overrightarrow{b} + \overrightarrow{c}) \cdot (\overrightarrow{b} + \overrightarrow{c})$
$|\overrightarrow{a}|^2 = |\overrightarrow{b}|^2 + |\overrightarrow{c}|^2 + 2(\overrightarrow{b} \cdot \overrightarrow{c})$
Since the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $\frac{\pi}{2}$,their dot product $\overrightarrow{b} \cdot \overrightarrow{c} = |\overrightarrow{b}||\overrightarrow{c}| \cos(\frac{\pi}{2}) = 0$.
Therefore,$a^2 = b^2 + c^2$.

(B) The vector area of a triangle with vertices having position vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ is given by the formula:
$\text{Vector Area} = \frac{1}{2} (\overrightarrow{AB} \times \overrightarrow{AC})$
Since $\overrightarrow{AB} = \overrightarrow{b} - \overrightarrow{a}$ and $\overrightarrow{AC} = \overrightarrow{c} - \overrightarrow{a}$,we have:
$\text{Vector Area} = \frac{1}{2} ((\overrightarrow{b} - \overrightarrow{a}) \times (\overrightarrow{c} - \overrightarrow{a}))$
$= \frac{1}{2} (\overrightarrow{b} \times \overrightarrow{c} - \overrightarrow{b} \times \overrightarrow{a} - \overrightarrow{a} \times \overrightarrow{c} + \overrightarrow{a} \times \overrightarrow{a})$
Since $\overrightarrow{a} \times \overrightarrow{a} = 0$,$-\overrightarrow{b} \times \overrightarrow{a} = \overrightarrow{a} \times \overrightarrow{b}$,and $-\overrightarrow{a} \times \overrightarrow{c} = \overrightarrow{c} \times \overrightarrow{a}$,we get:
$\text{Vector Area} = \frac{1}{2} (\overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} + \overrightarrow{c} \times \overrightarrow{a})$

(A) Given vectors are $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$,$\overrightarrow{b}=\hat{i}+\hat{j}$,and $\overrightarrow{c}=\hat{i}$.
First,calculate the cross product $\overrightarrow{a} \times \overrightarrow{b}$:
$\overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0-1) - \hat{j}(0-1) + \hat{k}(1-1) = -\hat{i} + \hat{j}$.
Now,calculate $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}$:
$(-\hat{i} + \hat{j}) \times \hat{i} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ 1 & 0 & 0 \end{vmatrix} = \hat{k}(0-1) = -\hat{k}$.
We are given $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} = \lambda \overrightarrow{a} + \mu \overrightarrow{b}$.
Substituting the vectors:
$-\hat{k} = \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} + \hat{j})$
$-\hat{k} = (\lambda + \mu)\hat{i} + (\lambda + \mu)\hat{j} + \lambda\hat{k}$.
Comparing the coefficients of $\hat{i}$,$\hat{j}$,and $\hat{k}$:
$1$) $\lambda + \mu = 0$
$2$) $\lambda + \mu = 0$
$3$) $\lambda = -1$
From equation $(1)$,since $\lambda = -1$,we have $-1 + \mu = 0$,which implies $\mu = 1$.
Therefore,$\lambda + \mu = -1 + 1 = 0$.

(D) Given equations are:
$3lm - 4ln + mn = 0$ ... $(i)$
$l + 2m + 3n = 0$ ... (ii)
From (ii),$l = -(2m + 3n)$.
Substituting this in $(i)$:
$-3(2m + 3n)m + 4(2m + 3n)n + mn = 0$
$-6m^2 - 9mn + 8mn + 12n^2 + mn = 0$
$-6m^2 + 12n^2 = 0$
$m^2 = 2n^2 \Rightarrow m = \pm \sqrt{2}n$.
Case $1$: $m = \sqrt{2}n$. Then $l = -(2\sqrt{2} + 3)n$.
Direction ratios $(l_1, m_1, n_1) = (-(3 + 2\sqrt{2}), \sqrt{2}, 1)$.
Case $2$: $m = -\sqrt{2}n$. Then $l = -(-2\sqrt{2} + 3)n = (2\sqrt{2} - 3)n$.
Direction ratios $(l_2, m_2, n_2) = (2\sqrt{2} - 3, -\sqrt{2}, 1)$.
For angle $\theta$ between lines:
$\cos \theta = \frac{l_1l_2 + m_1m_2 + n_1n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}}$
Numerator: $(-(3 + 2\sqrt{2}))(2\sqrt{2} - 3) + (\sqrt{2})(-\sqrt{2}) + (1)(1)$
$= -((2\sqrt{2})^2 - 3^2) - 2 + 1 = -(8 - 9) - 1 = 1 - 1 = 0$.
Since the numerator is $0$,$\cos \theta = 0$,which implies $\theta = \frac{\pi}{2}$.

(C) Let the $XOZ$-plane divide the line segment joining the points $A(2, 3, 1)$ and $B(6, 7, 1)$ in the ratio $m: n$.
Using the section formula,the coordinates of the point of division are given by:
$\left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}, \frac{m z_2 + n z_1}{m + n}\right)$
Substituting the given values:
$\left(\frac{6m + 2n}{m + n}, \frac{7m + 3n}{m + n}, \frac{m + n}{m + n}\right)$
Since the point lies on the $XOZ$-plane,its $y$-coordinate must be $0$.
Therefore,$\frac{7m + 3n}{m + n} = 0$.
$7m + 3n = 0$
$7m = -3n$
$\frac{m}{n} = -\frac{3}{7}$.
Thus,the ratio is $-3: 7$.

(A) The intercept form of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $X, Y, Z$ axes respectively.
Given that the plane makes intercepts $4$ on the $X$-axis $(a=4)$ and $3$ on the $Z$-axis $(c=3)$.
Since the plane is parallel to the $Y$-axis,the intercept on the $Y$-axis is infinite,which means $\frac{1}{b} = 0$.
Substituting these values into the intercept form equation:
$\frac{x}{4} + \frac{y}{\infty} + \frac{z}{3} = 1$
$\frac{x}{4} + 0 + \frac{z}{3} = 1$
Multiplying the entire equation by $12$ to clear the denominators:
$3x + 4z = 12$.

(B) Let the equation of the plane passing through $(1, 1, 1)$ be $a(x - 1) + b(y - 1) + c(z - 1) = 0$ $\dots (i)$.
Since the plane passes through $(1, -1, -1)$,we have $a(1 - 1) + b(-1 - 1) + c(-1 - 1) = 0$,which simplifies to $-2b - 2c = 0$,or $b + c = 0$ $\dots (ii)$.
The plane is perpendicular to $2x - y + z + 5 = 0$,so their normal vectors are perpendicular. Thus,$2a - b + c = 0$ $\dots (iii)$.
From $(ii)$,we have $b = -c$. Substituting this into $(iii)$,we get $2a - (-c) + c = 0$,which simplifies to $2a + 2c = 0$,or $a = -c$.
Let $c = -1$,then $a = 1$ and $b = 1$.
Substituting these values into $(i)$,we get $1(x - 1) + 1(y - 1) - 1(z - 1) = 0$.
$x - 1 + y - 1 - z + 1 = 0$.
$x + y - z - 1 = 0$.

(D) We know that for any two events $A$ and $B$,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Given $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3$,we have:
$P(A) + P(B) = P(A \cup B) + P(A \cap B) = 0.8 + 0.3 = 1.1$.
We know that $P(\bar{A}) = 1 - P(A)$ and $P(\bar{B}) = 1 - P(B)$.
Therefore,$P(\bar{A}) + P(\bar{B}) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$.
Substituting the value of $P(A) + P(B)$:
$P(\bar{A}) + P(\bar{B}) = 2 - 1.1 = 0.9$.

(B) The probability of getting a head in a single toss is $p = \frac{1}{2}$,and the probability of getting a tail is $q = 1 - p = \frac{1}{2}$.
The probability of getting no heads in $n$ tosses is $P(\text{no head}) = q^n = (\frac{1}{2})^n$.
The probability of getting at least one head is $P(\text{at least one head}) = 1 - P(\text{no head}) = 1 - (\frac{1}{2})^n$.
Given that $1 - (\frac{1}{2})^n > 0.8$,we have:
$1 - 0.8 > (\frac{1}{2})^n$
$0.2 > (\frac{1}{2})^n$
$\frac{1}{5} > \frac{1}{2^n}$
$2^n > 5$.
For $n = 2$,$2^2 = 4 < 5$.
For $n = 3$,$2^3 = 8 > 5$.
Thus,the least value of $n$ is $3$.

(C) According to the Clausius-Clapeyron equation,the relationship between vapour pressure $(p)$ and absolute temperature $(T)$ is given by:
$\log p = -\frac{\Delta H_{vap}}{2.303 R} \cdot \frac{1}{T} + C$
Here,$\Delta H_{vap}$ is the enthalpy of vaporization,$R$ is the gas constant,and $C$ is an integration constant.
This equation is in the form of a straight line $y = mx + c$,where $y = \log p$,$x = \frac{1}{T}$,and the slope $m = -\frac{\Delta H_{vap}}{2.303 R}$.
Since the slope is negative,the graph of $\log p$ versus $\frac{1}{T}$ is a straight line with a negative slope.

(C) The balanced chemical equation for the combustion of sulphur is:
$S(s) + O_2(g) \longrightarrow SO_2(g)$
From the stoichiometry of the reaction,$1 \ mol$ of $S$ reacts with $1 \ mol$ of $O_2$.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Therefore,$1 \ mol$ of $S$ requires $22.4 \ L$ of $O_2$.
For $1.5 \ mol$ of $S$,the volume of $O_2$ required is:
$V = 1.5 \ mol \times 22.4 \ L/mol = 33.6 \ L$.

(C) Given velocities: $C_1 = 100 \ ms^{-1}, C_2 = 200 \ ms^{-1}, C_3 = 500 \ ms^{-1}$.
$rms$ velocity is defined as $C_{rms} = \sqrt{\frac{C_1^2 + C_2^2 + C_3^2}{n}}$.
Substituting the values: $C_{rms} = \sqrt{\frac{100^2 + 200^2 + 500^2}{3}}$.
$C_{rms} = \sqrt{\frac{10000 + 40000 + 250000}{3}} = \sqrt{\frac{300000}{3}}$.
$C_{rms} = \sqrt{100000} = 100 \sqrt{10} \ ms^{-1}$.

(B) The mass defect is given as $3.32 \times 10^{-26} \ g$.
First,convert the mass defect from grams to atomic mass units $(amu)$ using the conversion factor $1 \ amu = 1.66 \times 10^{-24} \ g$.
$\text{Mass defect in } amu = \frac{3.32 \times 10^{-26} \ g}{1.66 \times 10^{-24} \ g/amu} = 0.02 \ amu$.
The binding energy is calculated by multiplying the mass defect in $amu$ by $931 \ MeV/amu$.
$\text{Binding energy} = 0.02 \ amu \times 931 \ MeV/amu = 18.62 \ MeV$.

(C) The kinetic energy $(KE)$ of an electron in an orbit is given by $KE = \frac{1}{2}mv^2$.
From the electrostatic force balance,the centripetal force is provided by the Coulombic attraction: $\frac{mv^2}{r} = \frac{e^2}{r^2}$,which implies $mv^2 = \frac{e^2}{r}$.
Substituting this into the $KE$ expression: $KE = \frac{1}{2} \times \frac{e^2}{r} = \frac{e^2}{2r}$.
The potential energy $(PE)$ of the electron is given by $PE = \frac{-e^2}{r}$.
The total energy $(E)$ is the sum of kinetic and potential energy: $E = KE + PE = \frac{e^2}{2r} - \frac{e^2}{r} = \frac{-e^2}{2r}$.

(A) The energy $E$ of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Given: $\lambda = 2000 \ \mathring{A} = 2000 \times 10^{-8} \ cm = 2 \times 10^{-5} \ cm$.
Speed of light $c = 3 \times 10^{10} \ cm/s$.
Planck's constant $h = 6.626 \times 10^{-27} \ erg \ s$.
Substituting the values: $E = \frac{6.626 \times 10^{-27} \ erg \ s \times 3 \times 10^{10} \ cm/s}{2 \times 10^{-5} \ cm}$.
$E = \frac{19.878 \times 10^{-17}}{2 \times 10^{-5}} \ erg = 9.939 \times 10^{-12} \ erg \approx 9.94 \times 10^{-12} \ erg$.

(B) The Freundlich adsorption isotherm is given by the empirical relation:
$\frac{x}{m} = K p^{1/n}$
where:
$x$ is the mass of the adsorbate,
$m$ is the mass of the adsorbent,
$p$ is the pressure,
$K$ and $n$ are constants which depend on the nature of the adsorbent and the gas at a particular temperature.

(C) The energy radiated per unit time is given by the Stefan-Boltzmann law: $E = \sigma A T^4$,where $\sigma$ is the Stefan-Boltzmann constant and $A = 4\pi R^2$ is the surface area of the star.
For the sun: $E_{\text{sun}} = \sigma (4\pi R_{\text{sun}}^2) T_{\text{sun}}^4$.
For the star $A$: $E_{\text{star}} = \sigma (4\pi R_{\text{star}}^2) T_{\text{star}}^4$.
Given that $E_{\text{star}} = 10000 E_{\text{sun}}$,we have:
$\sigma (4\pi R_{\text{star}}^2) T_{\text{star}}^4 = 10000 \times \sigma (4\pi R_{\text{sun}}^2) T_{\text{sun}}^4$.
Canceling $\sigma$ and $4\pi$ from both sides:
$R_{\text{star}}^2 T_{\text{star}}^4 = 10000 R_{\text{sun}}^2 T_{\text{sun}}^4$.
Rearranging to find the ratio of radii:
$\left(\frac{R_{\text{star}}}{R_{\text{sun}}}\right)^2 = 10000 \left(\frac{T_{\text{sun}}}{T_{\text{star}}}\right)^4$.
Substituting the given values $T_{\text{sun}} = 6000 \ K$ and $T_{\text{star}} = 2000 \ K$:
$\left(\frac{R_{\text{star}}}{R_{\text{sun}}}\right)^2 = 10000 \left(\frac{6000}{2000}\right)^4 = 10000 \times (3)^4 = 10000 \times 81 = 810000$.
Taking the square root of both sides:
$\frac{R_{\text{star}}}{R_{\text{sun}}} = \sqrt{810000} = 900$.
Thus,the ratio $R_{\text{star}} : R_{\text{sun}} = 900 : 1$.

(B) Density at $0^{\circ} C$,$\rho_0 = 1.0127$.
Density at $100^{\circ} C$,$\rho_{100} = 1$.
Coefficient of real expansion of liquid,$\gamma_{\text{real}} = \frac{\rho_0 - \rho_{100}}{\rho_{100} \times \Delta t} = \frac{1.0127 - 1}{1 \times 100} = 1.27 \times 10^{-4} /^{\circ} C$.
Coefficient of volume expansion of glass,$\gamma_g = 3 \alpha = 3 \times 9 \times 10^{-6} = 2.7 \times 10^{-5} = 0.27 \times 10^{-4} /^{\circ} C$.
Coefficient of apparent expansion,$\gamma_{\text{app}} = \gamma_{\text{real}} - \gamma_g = 1.27 \times 10^{-4} - 0.27 \times 10^{-4} = 1 \times 10^{-4} /^{\circ} C$.
The mass expelled is given by $\Delta m = m_0 \times \gamma_{\text{app}} \times \Delta t$ (approximately,for small changes) or using the relation $\frac{m_1}{m_2} = 1 + \gamma_{\text{app}} \Delta t$.
Here,$m_1 = 300 \ g$.
$\frac{300}{m_2} = 1 + (10^{-4} \times 100) = 1 + 0.01 = 1.01$.
$m_2 = \frac{300}{1.01}$.
Mass expelled $= m_1 - m_2 = 300 - \frac{300}{1.01} = 300 \left(1 - \frac{1}{1.01}\right) = 300 \left(\frac{0.01}{1.01}\right) = \frac{3}{1.01} \ g$.

(D) Statement $(A)$ is false because Duddell's thermo galvanometer is designed to measure both alternating current $(AC)$ and direct current $(DC)$ by utilizing the heating effect of the current.
Statement $(B)$ is true because a thermopile is a highly sensitive device consisting of several thermocouples connected in series,which allows it to detect very small temperature differences,typically on the order of $10^{-3} {}^{\circ}C$.

(C) The combustion reaction of methane is: $CH_4 + 2O_2 \longrightarrow CO_2 + 2H_2O$.
The molar mass of methane $(CH_4)$ is $12 + (4 \times 1) = 16 \ g \ mol^{-1}$.
Given that the heat evolved for $10 \ g$ of $CH_4$ is $560 \ kJ$,the enthalpy change $\Delta H$ for $10 \ g$ is $-560 \ kJ$.
To find the heat of combustion per mole $(16 \ g)$,we use the unitary method:
Heat of combustion $= \frac{-560 \ kJ}{10 \ g} \times 16 \ g \ mol^{-1} = -896 \ kJ \ mol^{-1}$.

(B) Given wavelength $\lambda = 6000 \text{ } \mathring{A} = 6 \times 10^{-7} \text{ m}$.
For dark fringes in Young's double slit experiment, the path difference $\Delta x$ is given by $\Delta x = (2n - 1) \frac{\lambda}{2}$ for $n = 1, 2, 3, \dots$ where $n$ represents the order of the dark fringe.
For the third dark fringe, we take $n = 3$.
Substituting the values: $\Delta x = (2 \times 3 - 1) \frac{6 \times 10^{-7}}{2} \text{ m}$.
$\Delta x = 5 \times 3 \times 10^{-7} \text{ m} = 15 \times 10^{-7} \text{ m}$.
$\Delta x = 1.5 \times 10^{-6} \text{ m} = 1.5 \text{ } \mu\text{m}$.

(A) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.
Initial frequency $n_1 = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
New length $l' = l - 0.40l = 0.6l$.
New tension $T' = T + 0.44T = 1.44T$.
New frequency $n_2 = \frac{1}{2l'} \sqrt{\frac{T'}{m}}$.
Taking the ratio of the final to initial frequency: $\frac{n_2}{n_1} = \frac{l}{l'} \sqrt{\frac{T'}{T}}$.
Substituting the values: $\frac{n_2}{n_1} = \frac{l}{0.6l} \sqrt{\frac{1.44T}{T}} = \frac{1}{0.6} \times \sqrt{1.44} = \frac{1.2}{0.6} = 2$.
Therefore,the ratio $n_2 : n_1 = 2 : 1$.

(B) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $m = \pi r^2 \rho$ is the mass per unit length.
Thus,$n = \frac{1}{2lr} \sqrt{\frac{T}{\pi \rho}}$.
The frequency of the $p$-th overtone is $(p+1)n$.
For string $A$,the first overtone is $2n_A = \frac{2}{2l_A r_A} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{l_A r_A} \sqrt{\frac{T}{\pi \rho}}$.
For string $B$,the second overtone is $3n_B = \frac{3}{2l_B r_B} \sqrt{\frac{T}{\pi \rho}}$.
Given $2n_A = 3n_B$ and $r_A = 2r_B$:
$\frac{1}{l_A r_A} = \frac{3}{2l_B r_B}$
$\frac{l_A}{l_B} = \frac{2 r_B}{3 r_A} = \frac{2 r_B}{3(2 r_B)} = \frac{2}{6} = \frac{1}{3}$.
Therefore,the ratio of the lengths $l_A : l_B = 1 : 3$.

(C) For statement $A$: The kinetic energy $KE$ of a body starting from rest under a constant force $F$ is given by $KE = \frac{1}{2}mv^2$. Since $v = at$,we have $KE = \frac{1}{2}m(at)^2 = \frac{1}{2}ma^2t^2$.
The rate of change of kinetic energy is $\frac{d(KE)}{dt} = \frac{d}{dt}(\frac{1}{2}ma^2t^2) = ma^2t$.
Since $m$ and $a$ are constants,$\frac{d(KE)}{dt} \propto t$. Thus,the rate of change of kinetic energy varies linearly with time. Statement $A$ is correct.
For statement $B$: $A$ body at rest is in equilibrium only if the net force acting on it is zero. $A$ body can be momentarily at rest (e.g.,a ball thrown vertically upward at its highest point) while being acted upon by a non-zero net force (gravity). Thus,it is not necessarily in equilibrium. Statement $B$ is wrong.

(D) For two particles to collide,their position vectors at time $t$ must be equal: $r_1(t) = r_2(t)$.
Given $r_1(t) = r_1 + v_1 t$ and $r_2(t) = r_2 + v_2 t$,the condition for collision is $r_1 + v_1 t = r_2 + v_2 t$.
Rearranging,we get $r_1 - r_2 = (v_2 - v_1) t$.
Substitute the given values:
$r_1 - r_2 = (3 \hat{i} + 5 \hat{j}) - (-5 \hat{i} - 3 \hat{j}) = 8 \hat{i} + 8 \hat{j}$.
$v_2 - v_1 = (a \hat{i} + 7 \hat{j}) - (4 \hat{i} + 3 \hat{j}) = (a - 4) \hat{i} + 4 \hat{j}$.
Given $t = 2 \ s$,substitute into the equation:
$8 \hat{i} + 8 \hat{j} = ((a - 4) \hat{i} + 4 \hat{j}) \times 2$.
Dividing by $2$:
$4 \hat{i} + 4 \hat{j} = (a - 4) \hat{i} + 4 \hat{j}$.
Comparing the coefficients of $\hat{i}$:
$a - 4 = 4 \implies a = 8$.

(A) The anhydride of sulphuric acid $(H_2SO_4)$ is sulphur trioxide $(SO_3)$.
In the structure of $SO_3$,there are three $S=O$ double bonds.
Each double bond consists of one $\sigma$-bond and one $\pi$-bond.
Therefore,the total number of $\sigma$-bonds is $3$ and the total number of $\pi$-bonds is $3$.

(D) The ground state electronic configuration of $Cl$ $(Z=17)$ is $[Ne] 3s^2 3p_x^2 3p_y^2 3p_z^1$.
In the first excited state,one electron from $3p$ moves to $3d$,giving $3$ unpaired electrons.
In the second excited state,one electron from $3s$ moves to $3d$,giving $5$ unpaired electrons.
In the third excited state,one more electron from $3p$ moves to $3d$,resulting in $7$ unpaired electrons $(3s^1, 3p_x^1, 3p_y^1, 3p_z^1, 3d^3)$.
These $7$ unpaired electrons react with $7$ fluorine atoms to form $ClF_7$.
The hybridization is $sp^3d^3$,which corresponds to a pentagonal bipyramidal geometry.

(B) When $SO_3$ is dissolved in heavy water $(D_2O)$,it forms deuterated sulphuric acid $(D_2SO_4)$ as the compound $X$.
The chemical reaction is: $SO_3 + D_2O \longrightarrow D_2SO_4$ $(X)$.
In $D_2SO_4$,the central sulphur atom is bonded to two $OD$ groups and two oxygen atoms via double bonds.
The steric number of sulphur is calculated as: $\text{Number of sigma bonds} + \text{Number of lone pairs} = 4 + 0 = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridisation.

(A) In $H_2O$,the oxygen atom is $sp^3$ hybridized with two lone pairs and two bond pairs,resulting in an angular (bent) shape.
$NH_4^+$ is $sp^3$ hybridized and tetrahedral.
$CH_4$ is $sp^3$ hybridized and tetrahedral.
Therefore,the correct set is $H_2O, sp^3$,angular.

(C) The dipole moment of a purely ionic bond is calculated as $\mu_{\text{theoretical}} = q \times d$.
Given: bond length $d = 1.25 \ \mathring{A} = 1.25 \times 10^{-8} \ cm$.
The charge of an electron $q = 4.8 \times 10^{-10} \ esu$.
$\mu_{\text{theoretical}} = 1.25 \times 10^{-8} \ cm \times 4.8 \times 10^{-10} \ esu = 6.0 \times 10^{-18} \ esu \ cm = 6.0 \ D$.
Given experimental dipole moment $\mu_{\text{experimental}} = 1.0 \ D$.
Percent ionic character $= \frac{\mu_{\text{experimental}}}{\mu_{\text{theoretical}}} \times 100 = \frac{1.0}{6.0} \times 100 = 16.66 \%$.

(C) The reaction is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
Initially: $1 \ mol \ N_2, 3 \ mol \ H_2, 0 \ mol \ NH_3$.
At equilibrium: $(1-x) \ mol \ N_2, (3-3x) \ mol \ H_2, 2x \ mol \ NH_3$.
Given that at equilibrium,moles of $N_2 = 0.6$.
So,$1-x = 0.6 \implies x = 0.4$.
Now,calculate moles of each component at equilibrium:
Moles of $N_2 = 0.6 \ mol$.
Moles of $H_2 = 3 - 3(0.4) = 3 - 1.2 = 1.8 \ mol$.
Moles of $NH_3 = 2(0.4) = 0.8 \ mol$.
Total moles at equilibrium = $0.6 + 1.8 + 0.8 = 3.2 \ mol$.

(D) For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate of reaction is expressed as:
$Rate = -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
Equating the terms for $NH_3$ and $H_2$:
$-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
Multiplying both sides by $6$:
$-2 \frac{d[H_2]}{dt} = 3 \frac{d[NH_3]}{dt}$
Rearranging gives $3 \frac{d[NH_3]}{dt} = -2 \frac{d[H_2]}{dt}$.

(A) The stability of oxidation states in Group $14$ elements is governed by the inert pair effect.
As we move down the group from $Si$ to $Pb$,the stability of the $+2$ oxidation state increases,while the stability of the $+4$ oxidation state decreases.
For Lead $(Pb)$,which is at the bottom of the group,the $+2$ oxidation state is significantly more stable than the $+4$ oxidation state due to the inert pair effect.
Therefore,$Pb^{2+} > Pb^{4+}$ is the correct order of stability.

(B) Let the remainder when $p(x)$ is divided by $x^2-a^2$ be $R(x) = mx + n$,where $x^2-a^2 = (x-a)(x+a)$.
By the Remainder Theorem,$p(a) = -a$ and $p(-a) = a$.
Since $p(x) = (x^2-a^2)q(x) + mx + n$,we have:
$p(a) = ma + n = -a$ ... $(i)$
$p(-a) = -ma + n = a$ ... (ii)
Adding $(i)$ and (ii),we get $2n = 0$,so $n = 0$.
Subtracting (ii) from $(i)$,we get $2ma = -2a$,so $m = -1$.
Thus,the remainder $R(x) = -1(x) + 0 = -x$.

(A) The concentration of dissolved oxygen $(D.O.)$ in water is a key indicator of water quality.
According to environmental standards,water is considered to be polluted if the $D.O.$ content is less than $5 \ ppm$.
If the $D.O.$ value is below $5 \ ppm$,it indicates that the water is not suitable for the survival of aquatic life.

(A) Let $P = \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15}$.
Note that $\frac{14 \pi}{15} = \pi - \frac{\pi}{15}$,so $\cos \frac{14 \pi}{15} = -\cos \frac{\pi}{15}$.
Thus,$P = -\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15}$.
Using the formula $\cos \theta \cos 2 \theta \cos 4 \theta \dots \cos 2^{n-1} \theta = \frac{\sin 2^n \theta}{2^n \sin \theta}$ with $\theta = \frac{\pi}{15}$ and $n = 4$:
$P = -\left( \frac{\sin(2^4 \cdot \frac{\pi}{15})}{2^4 \sin \frac{\pi}{15}} \right) = -\frac{\sin \frac{16 \pi}{15}}{16 \sin \frac{\pi}{15}}$.
Since $\sin \frac{16 \pi}{15} = \sin(\pi + \frac{\pi}{15}) = -\sin \frac{\pi}{15}$,
$P = -\left( \frac{-\sin \frac{\pi}{15}}{16 \sin \frac{\pi}{15}} \right) = \frac{1}{16}$.

(B) We know that for any $x \in [-1, 1]$,the identity $\cos ^{-1} x + \sin ^{-1} x = \frac{\pi}{2}$ holds true.
Given the expression $\cos \left[\cos ^{-1}\left(-\frac{1}{7}\right)+\sin ^{-1}\left(-\frac{1}{7}\right)\right]$,we substitute $x = -\frac{1}{7}$.
This simplifies to $\cos \left( \frac{\pi}{2} \right)$.
Since $\cos \left( \frac{\pi}{2} \right) = 0$,the value of the expression is $0$.

(C) Given the equation: $(5+4 \cos \theta)(2 \cos \theta+1)=0$.
Since $5+4 \cos \theta$ can never be $0$ (because $-1 \le \cos \theta \le 1$,so $1 \le 5+4 \cos \theta \le 9$),we must have $2 \cos \theta + 1 = 0$.
$\Rightarrow \cos \theta = -\frac{1}{2}$.
In the interval $[0, 2 \pi]$,$\cos \theta$ is negative in the second and third quadrants.
$\cos \theta = \cos \left(\pi - \frac{\pi}{3}\right) = \cos \left(\frac{2 \pi}{3}\right)$ and $\cos \theta = \cos \left(\pi + \frac{\pi}{3}\right) = \cos \left(\frac{4 \pi}{3}\right)$.
Thus,the solution set is $\left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}$.