AP EAMCET 2010 Chemistry Question Paper with Answer and Solution

(C) White tin,an allotropic form of tin,is an example of a tetragonal crystal system.
For a tetragonal system,the unit cell parameters are $a = b \neq c$ and $\alpha = \beta = \gamma = 90^{\circ}$.
The reasoning provided states $a=b=c$ and $\alpha=\beta=\gamma \neq 90^{\circ}$,which is incorrect.
Therefore,Assertion $(A)$ is true,but Reason $(R)$ is false.

(D) Given,the radius ratio of anion to cation $= 10 : 9.3$.
$\therefore$ The radius ratio of cation to anion $= 9.3 : 10 = 0.93$.
When the radius ratio of cation to anion lies between $0.732$ and $1.00$,the coordination number is $8$.
Thus,the coordination number of the cation in the crystal is $8$.

(D) When a non-volatile solute is added to a volatile solvent,the solute particles occupy some of the surface area of the solvent.
This reduces the number of solvent molecules available at the surface for vaporization,leading to a decrease in vapour pressure.
As the concentration of the non-volatile solute increases,the vapour pressure of the solution decreases.
Given concentrations of $X$ are:
$0.01 \text{ mol L}^{-1} < 0.10 \text{ mol L}^{-1} < 0.25 \text{ mol L}^{-1}$
Correspondingly,the vapour pressures are:
$p_3 > p_1 > p_2$
This can be rewritten as:
$p_2 < p_1 < p_3$

(A) The dissociation reaction is: $BaCl_{2(aq)} \rightarrow Ba^{2+} + 2Cl^-$.
Let the initial moles be $1$.
Degree of dissociation,$\alpha = 0.8$.

Species	$BaCl_2$	$Ba^{2+}$	$Cl^-$
Initial moles	$1$	$0$	$0$
Moles at equilibrium	$1 - \alpha$	$\alpha$	$2\alpha$

Total moles at equilibrium $= (1 - \alpha) + \alpha + 2\alpha = 1 + 2\alpha$.
Van't Hoff factor $i = 1 + 2\alpha$.
Given $\alpha = 0.8$,so $i = 1 + 2(0.8) = 1 + 1.6 = 2.6$.

(C) The combustion reaction of graphite is given by:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
From the stoichiometry of the reaction,$1 \text{ mole}$ of $C$ produces $1 \text{ mole}$ of $CO_2$.
Since $1 \text{ mole}$ of any substance contains $6.022 \times 10^{23}$ molecules:
$1 \text{ mole of } C \text{ produces } 6.022 \times 10^{23} \text{ molecules of } CO_2$.
Therefore,$0.1 \text{ mole}$ of graphite will produce:
$0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22} \text{ molecules of } CO_2$.

(B) From Graham's law of diffusion,$\frac{r_{CH_4}}{r_X} = \sqrt{\frac{M_X}{M_{CH_4}}}$.
Given that $r_{CH_4} = 2 \cdot r_X$,we have $2 = \sqrt{\frac{M_X}{16}}$.
Squaring both sides,$4 = \frac{M_X}{16}$,which gives $M_X = 64 \ g/mol$.
The number of moles of gas $X$ in $32 \ g$ is $n = \frac{32 \ g}{64 \ g/mol} = 0.5 \ mol$.
The number of molecules is $n \times N = 0.5 \times N = \frac{N}{2}$.

(C) The energy of a photon emitted during an electronic transition is given by $\Delta E = E_{n_2} - E_{n_1} = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV}$.
Since $\Delta E = \frac{hc}{\lambda}$,the wavelength $\lambda$ is inversely proportional to the energy difference $\Delta E$.
To obtain the lowest wavelength,we need the highest energy transition.
Comparing the transitions:
$(A)$ $n_2=\infty$ to $n_1=2$: $\Delta E \propto (1/4 - 0) = 0.25$
$(B)$ $n_2=4$ to $n_1=3$: $\Delta E \propto (1/9 - 1/16) = 0.0486$
$(C)$ $n_2=2$ to $n_1=1$: $\Delta E \propto (1/1 - 1/4) = 0.75$
$(D)$ $n_2=5$ to $n_1=3$: $\Delta E \propto (1/9 - 1/25) = 0.0711$
The transition $n_2=2$ to $n_1=1$ has the largest energy difference,therefore it emits radiation of the lowest wavelength.

(C) For a wave function to be physically acceptable or well-behaved,it must satisfy the following Born's conditions:
$1$. $\psi$ must be finite everywhere.
$2$. $\psi$ must be single-valued.
$3$. $\psi$ must be continuous and its first derivative must also be continuous.
Therefore,the condition that $\psi$ must be infinite is incorrect.

(B) Soap molecules contain a hydrophobic (water-repelling) tail and a hydrophilic (water-attracting) head.
When soap is added to water containing grease or dirt,the hydrophobic tails dissolve into the grease,while the hydrophilic heads remain in the water.
This arrangement forms a spherical structure where the grease is trapped inside the hydrophobic core,and the hydrophilic heads point outward towards the water.
This aggregate of soap and dirt particles is known as a micelle.
When the surface is rinsed with water,the micelle is washed away,effectively removing the dirt.

(B) Let the initial length of each rod be $l$. In the equilateral triangle $ABC$,$D$ is the mid-point of $AB$,so $AD = l/2$. The length of the altitude $DC$ is given by $DC^2 = AC^2 - AD^2 = l^2 - (l/2)^2 = 3l^2/4$.
After a small change in temperature $\Delta t$,the new lengths become $l' = l(1 + \alpha \Delta t)$.
The new length of $AC$ is $l_{AC}' = l(1 + \alpha_2 \Delta t)$ and the new length of $AD$ is $l_{AD}' = (l/2)(1 + \alpha_1 \Delta t)$.
Since $DC$ remains constant,$DC^2 = (l_{AC}')^2 - (l_{AD}')^2$.
$3l^2/4 = [l(1 + \alpha_2 \Delta t)]^2 - [(l/2)(1 + \alpha_1 \Delta t)]^2$.
$3l^2/4 = l^2(1 + 2\alpha_2 \Delta t + \alpha_2^2 \Delta t^2) - (l^2/4)(1 + 2\alpha_1 \Delta t + \alpha_1^2 \Delta t^2)$.
Neglecting higher-order terms $\alpha^2 \Delta t^2$,we get:
$3l^2/4 = l^2 + 2l^2 \alpha_2 \Delta t - l^2/4 - (l^2/4)(2\alpha_1 \Delta t)$.
$3l^2/4 = 3l^2/4 + 2l^2 \alpha_2 \Delta t - (l^2/2)\alpha_1 \Delta t$.
$0 = 2\alpha_2 \Delta t - (1/2)\alpha_1 \Delta t$.
$2\alpha_2 = \alpha_1/2 \implies \alpha_1 = 4\alpha_2$.

(B) Let $L_0$ be the initial length of each strip before heating. Let $d$ be the total thickness of the bimetallic strip.
After heating,the length of the brass strip is $L_B = L_0(1 + \alpha_B \Delta T) = (R + d) \theta$.
The length of the copper strip is $L_C = L_0(1 + \alpha_C \Delta T) = R \theta$.
Dividing the two equations,we get $\frac{R + d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T}$.
This simplifies to $1 + \frac{d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T}$.
$\frac{d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T} - 1 = \frac{1 + \alpha_B \Delta T - 1 - \alpha_C \Delta T}{1 + \alpha_C \Delta T} = \frac{(\alpha_B - \alpha_C) \Delta T}{1 + \alpha_C \Delta T}$.
Since $\alpha \Delta T \ll 1$,we can approximate $1 + \alpha_C \Delta T \approx 1$.
Thus,$R \approx \frac{d}{(\alpha_B - \alpha_C) \Delta T}$.
Therefore,$R \propto \frac{1}{\Delta T}$.

(B) Let the temperature of the junction $B$ be $\theta$.
Since the rods $BC$ and $BD$ are connected to the same temperature $80^{\circ}C$ at their ends $C$ and $D$,they act as parallel thermal resistances.
Let $R$ be the thermal resistance of each rod. The equivalent resistance of the parallel combination of rods $BC$ and $BD$ is $R_{eq} = \frac{R \times R}{R + R} = \frac{R}{2}$.
According to the principle of steady-state heat flow,the heat current flowing through rod $AB$ must equal the sum of the heat currents flowing through rods $BC$ and $BD$.
Using the formula for heat current $\frac{Q}{t} = \frac{\Delta T}{R}$,we have:
$\frac{\theta - 20}{R} = \frac{80 - \theta}{R/2}$
$\theta - 20 = 2(80 - \theta)$
$\theta - 20 = 160 - 2\theta$
$3\theta = 180$
$\theta = 60^{\circ}C$.

(B) The process is given on a $p-T$ diagram. For an ideal gas, $p V = \mu R T$, which implies $p = \frac{\mu R}{V} T$.
Since $p \propto T$ for a constant volume, the lines passing through the origin in a $p-T$ diagram represent isochoric (constant volume) processes.
In the given figure, $A B$ and $C D$ are straight lines passing through the origin, so these are isochoric processes.
Work done in an isochoric process is zero. Thus, $W_{A B} = 0$ and $W_{C D} = 0$.
For the isobaric processes $B C$ and $D A$, the work done is $W = p \Delta V = \mu R \Delta T$.
For process $B C$ (at constant pressure $p_2$): $W_{B C} = \mu R (T_C - T_B) = 3 \times R \times (2400 - 800) = 3 R \times 1600 = 4800 R$.
For process $D A$ (at constant pressure $p_1$): $W_{D A} = \mu R (T_A - T_D) = 3 \times R \times (400 - 1200) = 3 R \times (-800) = -2400 R$.
The total work done in the cycle is $W = W_{A B} + W_{B C} + W_{C D} + W_{D A} = 0 + 4800 R + 0 - 2400 R = 2400 R$.
Substituting $R = 8.314 \, J / mol K$: $W = 2400 \times 8.314 = 19953.6 \, J \approx 20 \, kJ$.

(D) $1$. In the $p-V$ diagram,the isothermal expansion from $A$ to $B$ follows the curve $AB$. The adiabatic compression from $B$ to $C$ follows the curve $BC$.
$2$. From the graph,it is clear that the final pressure $p_2$ at point $C$ is greater than the initial pressure $p_1$ at point $A$ $(p_2 > p_1)$.
$3$. The net work done $W$ by the gas in a cyclic process is equal to the area enclosed by the cycle $ABCA$. Since the process is counter-clockwise (expansion $A \to B$ followed by compression $B \to C$),the work done by the gas is negative $(W < 0)$.
$4$. Alternatively,the area under the adiabatic curve $BC$ (which is the work done on the gas during compression) is greater than the area under the isothermal curve $AB$ (which is the work done by the gas during expansion). Thus,the net work done by the gas is negative.
$5$. Therefore,$p_2 > p_1$ and $W < 0$.

(A) From the first law of thermodynamics,$\Delta U = Q + W$.
Since heat is provided to the system,$Q = +50 \ J$.
Since work is done on the system,$W = +10 \ J$.
Therefore,the change in internal energy is $\Delta U = 50 \ J + 10 \ J = 60 \ J$.

(B) According to the principle of homogeneity of dimensions,the dimensions of each term in a physical equation must be the same.
Given the equation $F = at + bt^2$,where $F$ is force and $t$ is time.
The dimension of force $F$ is $[MLT^{-2}]$.
For the first term: $[at] = [F]$
$[a] = [F] / [t] = [MLT^{-2}] / [T] = [MLT^{-3}]$.
For the second term: $[bt^2] = [F]$
$[b] = [F] / [t^2] = [MLT^{-2}] / [T^2] = [MLT^{-4}]$.
Thus,the dimensions of $a$ and $b$ are $[MLT^{-3}]$ and $[MLT^{-4}]$ respectively.

(C) The ratio of intensities of two coherent sources is given as $\frac{I_1}{I_2} = \frac{64}{1}$.
Let $I_1 = 64k$ and $I_2 = k$,where $k$ is a constant.
The ratio of maximum intensity to minimum intensity in an interference pattern is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2}$.
Substituting the values:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{64k} + \sqrt{k})^2}{(\sqrt{64k} - \sqrt{k})^2} = \frac{(8\sqrt{k} + \sqrt{k})^2}{(8\sqrt{k} - \sqrt{k})^2}$.
$\frac{I_{\max}}{I_{\min}} = \frac{(9\sqrt{k})^2}{(7\sqrt{k})^2} = \frac{81k}{49k} = \frac{81}{49}$.

(D) The frequency of a closed organ pipe vibrating in its first harmonic is $n_1 = \frac{v_1}{4 l_1}$.
The frequency of an open organ pipe vibrating in its third harmonic is $n_3 = \frac{3 v_2}{2 l_2}$.
Since both pipes are in resonance with the same tuning fork,their frequencies are equal: $n_1 = n_3$.
Therefore,$\frac{v_1}{4 l_1} = \frac{3 v_2}{2 l_2}$,which simplifies to $\frac{l_1}{l_2} = \frac{1}{6} \left( \frac{v_1}{v_2} \right)$.
The speed of sound in a gas is given by $v = \sqrt{\frac{B}{\rho}}$,where $B$ is the bulk modulus (reciprocal of compressibility) and $\rho$ is the density.
Given that the compressibility is equal in both pipes,$B_1 = B_2 = B$.
Thus,$\frac{v_1}{v_2} = \sqrt{\frac{B/\rho_1}{B/\rho_2}} = \sqrt{\frac{\rho_2}{\rho_1}}$.
Substituting this into the ratio of lengths: $\frac{l_1}{l_2} = \frac{1}{6} \sqrt{\frac{\rho_2}{\rho_1}}$.

(D) The fundamental frequency $n$ of a sonometer wire of length $l$ is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
Since $T$ and $m$ are constant for the wire,$n \propto \frac{1}{l}$,which implies $l \propto \frac{1}{n}$.
Given the ratio of frequencies $n_1 : n_2 : n_3 = 1 : 3 : 4$,the ratio of the lengths of the segments must be $l_1 : l_2 : l_3 = \frac{1}{1} : \frac{1}{3} : \frac{1}{4}$.
To simplify this ratio,multiply by the least common multiple of $1, 3, 4$,which is $12$: $l_1 : l_2 : l_3 = 12 : 4 : 3$.
The sum of the parts is $12 + 4 + 3 = 19$.
The total length is $114 ~cm$.
Thus,$l_1 = \frac{12}{19} \times 114 = 12 \times 6 = 72 ~cm$.
$l_2 = \frac{4}{19} \times 114 = 4 \times 6 = 24 ~cm$.
$l_3 = \frac{3}{19} \times 114 = 3 \times 6 = 18 ~cm$.
Therefore,the lengths are $72 ~cm, 24 ~cm, 18 ~cm$.

(C) Let the mass of the ball be $m$. When the ball is at a height $h = 10 ~m$,its velocity is $v_0$. The kinetic energy at this point is $K_1 = \frac{1}{2} m v_0^2$.
Just before hitting the ground,the ball falls an additional $10 ~m$. By the work-energy theorem,the kinetic energy just before impact is $K_{impact} = K_1 + mgh = \frac{1}{2} m v_0^2 + mg(10)$.
After the collision,the ball loses $50 \%$ of its energy,so the kinetic energy just after impact is $K_{after} = 0.5 \times K_{impact} = 0.5 \times (\frac{1}{2} m v_0^2 + 10mg)$.
The ball rises back to a height of $10 ~m$,meaning its potential energy at the peak is $mgh = 10mg$. Since the kinetic energy is zero at the peak,by conservation of energy,$K_{after} = 10mg$.
Equating the two: $0.5 \times (\frac{1}{2} m v_0^2 + 10mg) = 10mg$.
$\frac{1}{2} m v_0^2 + 10mg = 20mg$.
$\frac{1}{2} m v_0^2 = 10mg$.
$v_0^2 = 20g$.
Using $g = 10 ~m/s^2$,$v_0^2 = 200$,so $v_0 = \sqrt{200} \approx 14.14 ~m/s$. Given the options,$14 ~m/s$ is the correct choice.

(C) When a ball falls from a height $h$,the height reached after the $n^{th}$ rebound is given by $h_n = h e^{2n}$,where $e$ is the coefficient of restitution.
The total distance $H$ covered by the ball before it comes to rest is the sum of the initial fall and the subsequent upward and downward paths for each rebound:
$H = h + 2h_1 + 2h_2 + 2h_3 + \dots$
$H = h + 2(h e^2) + 2(h e^4) + 2(h e^6) + \dots$
$H = h + 2h(e^2 + e^4 + e^6 + \dots)$
The term in the bracket is an infinite geometric progression with first term $a = e^2$ and common ratio $r = e^2$. The sum is $S = \frac{a}{1-r} = \frac{e^2}{1-e^2}$.
Substituting this back into the expression for $H$:
$H = h + 2h \left( \frac{e^2}{1-e^2} \right)$
$H = h \left( 1 + \frac{2e^2}{1-e^2} \right)$
$H = h \left( \frac{1 - e^2 + 2e^2}{1-e^2} \right)$
$H = h \left( \frac{1 + e^2}{1-e^2} \right)$

(A) Let the mass of the smaller piece be $m$ and the mass of the larger piece be $4m$. The total mass is $5m$.
According to the law of conservation of linear momentum,the initial momentum equals the final momentum:
$P_{initial} = P_{final}$
$5m(40 \hat{i} + 50 \hat{j} - 25 \hat{k}) = m(200 \hat{i} + 70 \hat{j} + 15 \hat{k}) + 4m(\vec{v}_{large})$
Dividing by $m$:
$5(40 \hat{i} + 50 \hat{j} - 25 \hat{k}) = (200 \hat{i} + 70 \hat{j} + 15 \hat{k}) + 4\vec{v}_{large}$
$(200 \hat{i} + 250 \hat{j} - 125 \hat{k}) - (200 \hat{i} + 70 \hat{j} + 15 \hat{k}) = 4\vec{v}_{large}$
$0 \hat{i} + 180 \hat{j} - 140 \hat{k} = 4\vec{v}_{large}$
$\vec{v}_{large} = \frac{180 \hat{j} - 140 \hat{k}}{4} = 45 \hat{j} - 35 \hat{k} \text{ m/s}$.

(A) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{V - M}{2}$,where $V$ is the number of valence electrons of the central atom and $M$ is the number of monovalent atoms bonded to it.
$A. \ NH_3$: Central atom $N$ has $5$ valence electrons. It is bonded to $3$ $H$ atoms. $\text{Lone pairs} = \frac{5-3}{2} = 1$ (Option $5$).
$B. \ H_2O$: Central atom $O$ has $6$ valence electrons. It is bonded to $2$ $H$ atoms. $\text{Lone pairs} = \frac{6-2}{2} = 2$ (Option $1$).
$C. \ XeF_2$: Central atom $Xe$ has $8$ valence electrons. It is bonded to $2$ $F$ atoms. $\text{Lone pairs} = \frac{8-2}{2} = 3$ (Option $2$).
$D. \ CH_4$: Central atom $C$ has $4$ valence electrons. It is bonded to $4$ $H$ atoms. $\text{Lone pairs} = \frac{4-4}{2} = 0$ (Option $3$).
Therefore,the correct matching is $A-5, B-1, C-2, D-3$.

(D) For the reaction: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$,the equilibrium constant is $K = \frac{[HI]^2}{[H_2][I_2]}$ ...$(i)$
For the reaction: $HI_{(g)} \rightleftharpoons \frac{1}{2} H_{2(g)} + \frac{1}{2} I_{2(g)}$,the equilibrium constant is $K' = \frac{[H_2]^{1/2} [I_2]^{1/2}}{[HI]}$ ...$(ii)$
Comparing $(ii)$ with $(i)$,we can see that $K' = \sqrt{\frac{1}{K}} = \frac{1}{\sqrt{K}}$.

(C) Elements with half-filled or completely-filled orbitals,$i.e.$,having stable electronic configurations,have very low negative values of electron affinity.
Since the electron affinity of $B$ $(-60 \ kJ \ mol^{-1})$ is the lowest among the given values,it corresponds to the most stable electronic configuration,which is $3s^2 3p^3$ (half-filled $p$-orbital).

(C) Given equation: $x^3-ax^2+ax-1=0$.
We can factorize the equation as: $(x^3-1) - ax(x-1) = 0$.
$(x-1)(x^2+x+1) - ax(x-1) = 0$.
$(x-1)(x^2+x+1-ax) = 0$.
Since $\alpha \neq 1$ is a root,$\alpha$ must satisfy $x^2+(1-a)x+1=0$.
Thus,$\alpha^2+(1-a)\alpha+1=0$.
If $x = \frac{1}{\alpha}$ is a root,then $(\frac{1}{\alpha})^3 - a(\frac{1}{\alpha})^2 + a(\frac{1}{\alpha}) - 1 = 0$.
Multiplying by $\alpha^3$,we get $1 - a\alpha + a\alpha^2 - \alpha^3 = 0$,which is $-(x^3-ax^2+ax-1) = 0$ evaluated at $x=\alpha$.
Since $\alpha$ is a root,$\frac{1}{\alpha}$ is also a root of the equation.

(A) Given the cubic equation $x^3-b x^2+c x-d=0$.
Let the roots of this equation in geometric progression be $\frac{a}{r}, a, ar$.
From the relationship between roots and coefficients:
$1$. Sum of roots: $\frac{a}{r} + a + ar = b \Rightarrow a(\frac{1}{r} + 1 + r) = b$ ... $(i)$
$2$. Sum of roots taken two at a time: $\frac{a}{r} \cdot a + a \cdot ar + ar \cdot \frac{a}{r} = c \Rightarrow a^2(\frac{1}{r} + 1 + r) = c$ ... (ii)
$3$. Product of roots: $\frac{a}{r} \cdot a \cdot ar = d \Rightarrow a^3 = d$ ... (iii)
Dividing equation (ii) by equation $(i)$:
$\frac{a^2(\frac{1}{r} + 1 + r)}{a(\frac{1}{r} + 1 + r)} = \frac{c}{b} \Rightarrow a = \frac{c}{b}$.
Substituting $a = \frac{c}{b}$ into equation (iii):
$(\frac{c}{b})^3 = d$ $\Rightarrow \frac{c^3}{b^3} = d$ $\Rightarrow c^3 = b^3 d$.

(B) Given equation $x^3-6x^2+11x-6=0$ has the roots $\alpha, \beta, \gamma$.
By factoring,$(x-1)(x-2)(x-3)=0$,so the roots are $1, 2, 3$.
Let $\alpha=1, \beta=2, \gamma=3$.
Then,$a = \alpha^2+\beta^2+\gamma^2 = 1^2+2^2+3^2 = 1+4+9 = 14$.
$b = \alpha\beta+\beta\gamma+\gamma\alpha = (1)(2)+(2)(3)+(3)(1) = 2+6+3 = 11$.
$c = (\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) = (1+2)(2+3)(3+1) = 3 \times 5 \times 4 = 60$.
Comparing the values,$11 < 14 < 60$,which implies $b < a < c$.

(D) Given $z=1+i\sqrt{3}$.
We know that for a complex number $z=x+iy$,the argument $\operatorname{Arg} z$ is $\tan^{-1}(\frac{y}{x})$.
Here,$x=1$ and $y=\sqrt{3}$.
Since $x>0$ and $y>0$,$z$ lies in the first quadrant.
$\operatorname{Arg} z = \tan^{-1}(\frac{\sqrt{3}}{1}) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.
Now,the conjugate of $z$ is $\bar{z} = 1-i\sqrt{3}$.
Here,$x=1$ and $y=-\sqrt{3}$.
Since $x>0$ and $y < 0$,$\bar{z}$ lies in the fourth quadrant.
$\operatorname{Arg} \bar{z} = \tan^{-1}(\frac{-\sqrt{3}}{1}) = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$.
Therefore,$|\operatorname{Arg} z|+|\operatorname{Arg} \bar{z}| = |\frac{\pi}{3}| + |-\frac{\pi}{3}| = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}$.

(A) The acceptable level of carbon monoxide gas $(CO)$ in the atmosphere is approximately $9 \ ppm$.

(A) Given that $\cos (x-y), \cos x, \cos (x+y)$ are in harmonic progression $(HP)$.
Therefore,$\cos x = \frac{2 \cos (x-y) \cos (x+y)}{\cos (x+y) + \cos (x-y)}$.
Using the identity $\cos (A-B) \cos (A+B) = \cos ^2 A - \sin ^2 B = \cos ^2 A - (1 - \cos ^2 B) = \cos ^2 A + \cos ^2 B - 1$,we have:
$\cos x = \frac{2(\cos ^2 x + \cos ^2 y - 1)}{2 \cos x \cos y}$.
$\cos ^2 x \cos y = \cos ^2 x + \cos ^2 y - 1$.
Rearranging the terms:
$\cos ^2 x \cos y - \cos ^2 x = \cos ^2 y - 1$.
$\cos ^2 x (\cos y - 1) = - (1 - \cos ^2 y)$.
$\cos ^2 x (\cos y - 1) = - \sin ^2 y$.
Since $1 - \cos ^2 y = \sin ^2 y$,we have:
$\cos ^2 x (1 - \cos y) = (1 - \cos y)(1 + \cos y)$.
Given $\cos x \neq \cos y$,we can assume $\cos y \neq 1$ (otherwise $\cos x = 1$,which contradicts the distinct numbers condition).
Dividing both sides by $(1 - \cos y)$:
$\cos ^2 x = 1 + \cos y$.

(B) The equation of a line that makes equal intercepts $a$ on the positive $x$ and $y$ axes is given by $\frac{x}{a} + \frac{y}{a} = 1$,which simplifies to $x + y = a$ (where $a > 0$).
The perpendicular distance of this line from the origin $(0, 0)$ is given as $1$ unit.
Using the distance formula $\left| \frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}} \right| = d$,we have:
$\left| \frac{0 + 0 - a}{\sqrt{1^2 + 1^2}} \right| = 1$
$\left| \frac{-a}{\sqrt{2}} \right| = 1$
Since $a > 0$,we get $a = \sqrt{2}$.
Thus,the equation of the line is $x + y = \sqrt{2}$ ... $(i)$.
The second line is given as $y = 2x + 3 + \sqrt{2}$,which can be rewritten as $2x - y = -3 - \sqrt{2}$ ... (ii).
To find the intersection point $(x_0, y_0)$,we solve equations $(i)$ and (ii):
Adding $(i)$ and (ii):
$(x + y) + (2x - y) = \sqrt{2} + (-3 - \sqrt{2})$
$3x = -3 \implies x_0 = -1$.
Substituting $x_0 = -1$ into $(i)$:
$-1 + y_0 = \sqrt{2} \implies y_0 = \sqrt{2} + 1$.
Finally,we calculate $2x_0 + y_0$:
$2(-1) + (\sqrt{2} + 1) = -2 + \sqrt{2} + 1 = \sqrt{2} - 1$.

(A) To find the image of the line $x+y-2=0$ with respect to the $y$-axis,we replace $x$ with $-x$ in the equation.
Substituting $x = -x$ into the given equation $x+y-2=0$,we get:
$(-x)+y-2=0$
$-x+y-2=0$
Multiplying by $-1$,we obtain:
$x-y+2=0$
Thus,the image of the line is $x-y+2=0$.

(C) The given line is $4x - 2y = 1$,which can be written as $2y = 4x - 1$ or $y = 2x - 1/2$. The slope of this line is $m_1 = 2$.
Since line $L$ is perpendicular to this line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
Thus,$2 \times m_2 = -1$,which gives $m_2 = -1/2$.
The equation of line $L$ with slope $-1/2$ can be written as $x + 2y + \lambda = 0$.
This line intersects the coordinate axes at points $(-\lambda, 0)$ and $(0, -\lambda/2)$.
The area of the triangle formed by the line with the coordinate axes is given by $\frac{1}{2} \times |\text{base}| \times |\text{height}| = 4$.
$\frac{1}{2} \times |-\lambda| \times |-\lambda/2| = 4$.
$\frac{\lambda^2}{4} = 4 \implies \lambda^2 = 16 \implies \lambda = \pm 4$.
Substituting $\lambda = 4$ into the equation $x + 2y + \lambda = 0$,we get $x + 2y + 4 = 0$,which is equivalent to $2x + 4y + 8 = 0$.
Substituting $\lambda = -4$ into the equation $x + 2y + \lambda = 0$,we get $x + 2y - 4 = 0$,which is equivalent to $2x + 4y - 8 = 0$.

(A) Let the image of the point $P(4, -13)$ be $P'(x_1, y_1)$.
Since $Q$ is the midpoint of $PP'$,its coordinates are $Q = \left(\frac{x_1 + 4}{2}, \frac{y_1 - 13}{2}\right)$.
Since $Q$ lies on the line $5x + y + 6 = 0$,we have:
$5\left(\frac{x_1 + 4}{2}\right) + \left(\frac{y_1 - 13}{2}\right) + 6 = 0$
$5x_1 + 20 + y_1 - 13 + 12 = 0$
$5x_1 + y_1 + 19 = 0$ $\ldots$ $(i)$
Since the line $PP'$ is perpendicular to the line $5x + y + 6 = 0$ (which has slope $-5$),the slope of $PP'$ is $\frac{1}{5}$.
Thus,$\frac{y_1 - (-13)}{x_1 - 4} = \frac{1}{5}$
$5(y_1 + 13) = x_1 - 4$
$x_1 - 5y_1 - 69 = 0$ $\ldots$ (ii)
Solving equations $(i)$ and (ii):
Multiply $(i)$ by $5$: $25x_1 + 5y_1 + 95 = 0$
Add to (ii): $(25x_1 + 5y_1 + 95) + (x_1 - 5y_1 - 69) = 0$
$26x_1 + 26 = 0 \Rightarrow x_1 = -1$
Substitute $x_1 = -1$ into $(i)$: $5(-1) + y_1 + 19 = 0 \Rightarrow y_1 = -14$.
Thus,the image is $(-1, -14)$.

(B) The given equation of the pair of straight lines is $3x^2-11xy+10y^2-7x+13y+k=0$.
Comparing this with the general equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get:
$a=3, h=-\frac{11}{2}, b=10, g=-\frac{7}{2}, f=\frac{13}{2}$.
The point of intersection $(x, y)$ of the pair of lines is given by the formula:
$x = \frac{hf-bg}{ab-h^2}$ and $y = \frac{gh-af}{ab-h^2}$.
Calculating the denominator: $ab-h^2 = 3(10) - (-\frac{11}{2})^2 = 30 - \frac{121}{4} = \frac{120-121}{4} = -\frac{1}{4}$.
Calculating $x$: $x = \frac{(-\frac{11}{2})(\frac{13}{2}) - (10)(-\frac{7}{2})}{-\frac{1}{4}} = \frac{-\frac{143}{4} + 35}{-\frac{1}{4}} = \frac{\frac{-143+140}{4}}{-\frac{1}{4}} = \frac{-3/4}{-1/4} = 3$.
Calculating $y$: $y = \frac{(-\frac{7}{2})(-\frac{11}{2}) - (3)(\frac{13}{2})}{-\frac{1}{4}} = \frac{\frac{77}{4} - \frac{39}{2}}{-\frac{1}{4}} = \frac{\frac{77-78}{4}}{-\frac{1}{4}} = \frac{-1/4}{-1/4} = 1$.
Thus,the point of intersection is $(3, 1)$.

(A) The equation of the pair of lines passing through the origin and the intersection of $x^2+y^2=4$ and $x+y=a$ is obtained by homogenizing the circle equation using the line equation:
$x^2+y^2=4 \left(\frac{x+y}{a}\right)^2$
$a^2(x^2+y^2)=4(x^2+y^2+2xy)$
$(a^2-4)x^2 - 8xy + (a^2-4)y^2 = 0$
Since the lines are perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(a^2-4) + (a^2-4) = 0$
$2(a^2-4) = 0$
$a^2 = 4$
Since $a>0$,we have $a=2$.

(D) The given equation is $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$.
We can rewrite this as $2(4x^2 - 12xy + 9y^2) - 3(2x - 3y) - 5 = 0$.
This simplifies to $2(2x - 3y)^2 - 3(2x - 3y) - 5 = 0$.
Let $t = 2x - 3y$. Then the equation becomes $2t^2 - 3t - 5 = 0$.
Factoring the quadratic,we get $(2t - 5)(t + 1) = 0$.
So,$2x - 3y = 5/2$ and $2x - 3y = -1$.
These are two parallel lines of the form $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$,where $A = 2, B = -3$.
Rewriting the lines: $2x - 3y - 2.5 = 0$ and $2x - 3y + 1 = 0$.
The distance $d$ between parallel lines is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
$d = \frac{|-2.5 - 1|}{\sqrt{2^2 + (-3)^2}} = \frac{|-3.5|}{\sqrt{4 + 9}} = \frac{3.5}{\sqrt{13}} = \frac{7/2}{\sqrt{13}} = \frac{7}{2\sqrt{13}}$.

(B) Enantiomers are stereoisomers that are non-superimposable mirror images of each other.
They possess identical physical properties such as boiling point,density,and bond lengths in an achiral environment.
However,they differ in their interaction with plane-polarized light,which is measured as specific rotation.
Therefore,the two enantiomers of secondary butyl chloride differ in specific rotation.

(A) Enantiomers are non-superimposable mirror images of each other. For a molecule with multiple chiral centers,the enantiomer is formed by inverting the configuration at every chiral center.
For $2,3-$butanediol,the chiral centers are at positions $2$ and $3$.
The enantiomer of the $(2R, 3R)$ isomer is the $(2S, 3S)$ isomer.
Therefore,the pair $(2R, 3R)$ and $(2S, 3S)$ represents enantiomers.

(A) The Diels-Alder reaction requires a conjugated diene and a dienophile (an alkene or alkyne).
In option $A$,the reactant $CH_2=CH-CH_2-CH=CH_2$ is $penta-1,4-diene$.
This is an isolated diene,not a conjugated diene.
Therefore,it cannot participate in the Diels-Alder reaction.

(D) When an acidified solution of $TiO_2$ is treated with $H_2O_2$,an intense yellow-orange colour is obtained due to the formation of pertitanic acid,$H_2TiO_4$.
The chemical reaction is:
$TiO_2 + H_2O_2 \xrightarrow{H_2SO_4} H_2TiO_4$
This reaction is used for the detection of both $Ti(IV)$ and $H_2O_2$.

(D) Given the recurrence relation $f(x)=f(x-2)+f(x-3)$ with initial values $f(0)=0, f(1)=1, f(2)=2$.
We calculate the values step by step:
For $x=3: f(3)=f(1)+f(0)=1+0=1$.
For $x=4: f(4)=f(2)+f(1)=2+1=3$.
For $x=5: f(5)=f(3)+f(2)=1+2=3$.
For $x=6: f(6)=f(4)+f(3)=3+1=4$.
For $x=7: f(7)=f(5)+f(4)=3+3=6$.
For $x=8: f(8)=f(6)+f(5)=4+3=7$.
For $x=9: f(9)=f(7)+f(6)=6+4=10$.
Thus,$f(9)=10$.

(D) Given $u=\sin ^{-1}\left(\frac{x^4+y^4}{x+y}\right)$.
Let $v=\sin u=\frac{x^4+y^4}{x+y}$.
Here,$v$ is a homogeneous function of $x$ and $y$ with degree $n = 4 - 1 = 3$.
By Euler's theorem for homogeneous functions,we have $x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = n v$.
Substituting $v = \sin u$ and $n = 3$,we get $x \frac{\partial}{\partial x}(\sin u) + y \frac{\partial}{\partial y}(\sin u) = 3 \sin u$.
Applying the chain rule,$x \cos u \frac{\partial u}{\partial x} + y \cos u \frac{\partial u}{\partial y} = 3 \sin u$.
Dividing both sides by $\cos u$,we get $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 3 \frac{\sin u}{\cos u} = 3 \tan u$.

(A) Given: $pH = 5.0$ and concentration $C = 0.01 \ M = 10^{-2} \ M$.
$[H^{+}] = 10^{-pH} = 10^{-5} \ M$.
For a weak acid,the dissociation constant $K_a$ is given by the formula $[H^{+}] = \sqrt{K_a \cdot C}$.
Squaring both sides,we get $[H^{+}]^2 = K_a \cdot C$.
Therefore,$K_a = \frac{[H^{+}]^2}{C} = \frac{(10^{-5})^2}{10^{-2}} = \frac{10^{-10}}{10^{-2}} = 10^{-8}$.
Thus,$[H^{+}] = 1 \times 10^{-5} \ M$ and $K_a = 1 \times 10^{-8}$.

(C) Given,$y = \cos^{-1}\left(\frac{a^2-x^2}{a^2+x^2}\right) + \sin^{-1}\left(\frac{2ax}{a^2+x^2}\right)$.
Substitute $x = a \tan \theta$,which implies $\theta = \tan^{-1}\left(\frac{x}{a}\right)$.
Then,$y = \cos^{-1}\left(\frac{a^2 - a^2 \tan^2 \theta}{a^2 + a^2 \tan^2 \theta}\right) + \sin^{-1}\left(\frac{2a^2 \tan \theta}{a^2 + a^2 \tan^2 \theta}\right)$.
Simplifying the terms inside the brackets:
$y = \cos^{-1}\left(\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}\right) + \sin^{-1}\left(\frac{2 \tan \theta}{1 + \tan^2 \theta}\right)$.
Using trigonometric identities $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$ and $\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$,we get:
$y = \cos^{-1}(\cos 2\theta) + \sin^{-1}(\sin 2\theta)$.
$y = 2\theta + 2\theta = 4\theta$.
Substituting back $\theta = \tan^{-1}\left(\frac{x}{a}\right)$,we have $y = 4 \tan^{-1}\left(\frac{x}{a}\right)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 4 \cdot \frac{1}{1 + (x/a)^2} \cdot \frac{1}{a} = 4 \cdot \frac{a^2}{a^2 + x^2} \cdot \frac{1}{a} = \frac{4a}{a^2 + x^2}$.

(A) We are given the integral $I = \int(1-\cos x) \operatorname{cosec}^2 x \, dx$.
Using the trigonometric identities $1-\cos x = 2 \sin^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we have:
$I = \int \left(2 \sin^2 \frac{x}{2}\right) \cdot \frac{1}{\sin^2 x} \, dx$
$I = \int \frac{2 \sin^2 \frac{x}{2}}{(2 \sin \frac{x}{2} \cos \frac{x}{2})^2} \, dx$
$I = \int \frac{2 \sin^2 \frac{x}{2}}{4 \sin^2 \frac{x}{2} \cos^2 \frac{x}{2}} \, dx$
$I = \frac{1}{2} \int \frac{1}{\cos^2 \frac{x}{2}} \, dx$
$I = \frac{1}{2} \int \sec^2 \frac{x}{2} \, dx$
Integrating $\sec^2 \frac{x}{2}$ gives $2 \tan \frac{x}{2}$:
$I = \frac{1}{2} \cdot (2 \tan \frac{x}{2}) + c = \tan \frac{x}{2} + c$.
Comparing this with $f(x) + c$,we get $f(x) = \tan \frac{x}{2}$.

(C) Given $I_n = \int_0^{\pi / 4} \tan^n x \, dx$.
Consider the sum $I_r + I_{r+2} = \int_0^{\pi / 4} \tan^r x \, dx + \int_0^{\pi / 4} \tan^{r+2} x \, dx$.
$I_r + I_{r+2} = \int_0^{\pi / 4} \tan^r x (1 + \tan^2 x) \, dx$.
Since $1 + \tan^2 x = \sec^2 x$,we have $I_r + I_{r+2} = \int_0^{\pi / 4} \tan^r x \sec^2 x \, dx$.
Let $t = \tan x$,then $dt = \sec^2 x \, dx$. When $x=0, t=0$ and when $x=\pi/4, t=1$.
Thus,$I_r + I_{r+2} = \int_0^1 t^r \, dt = \left[ \frac{t^{r+1}}{r+1} \right]_0^1 = \frac{1}{r+1}$.
For $r=2$,$I_2 + I_4 = \frac{1}{3}$.
For $r=3$,$I_3 + I_5 = \frac{1}{4}$.
For $r=4$,$I_4 + I_6 = \frac{1}{5}$.
The sequence is $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots$,which are in harmonic progression $(HP)$ because their reciprocals $3, 4, 5, \ldots$ are in arithmetic progression.

(D) The given series is $\log _4 2 - \log _8 2 + \log _{16} 2 - \dots$
Using the property $\log _b a = \frac{1}{\log _a b}$,we get:
$= \frac{1}{\log _2 4} - \frac{1}{\log _2 8} + \frac{1}{\log _2 16} - \dots$
$= \frac{1}{\log _2(2^2)} - \frac{1}{\log _2(2^3)} + \frac{1}{\log _2(2^4)} - \dots$
$= \frac{1}{2 \log _2 2} - \frac{1}{3 \log _2 2} + \frac{1}{4 \log _2 2} - \dots$
Since $\log _2 2 = 1$,the expression becomes:
$= \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots$
We know the expansion $\log _e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$
For $x = 1$,$\log _e(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$
Rearranging the terms,we get $\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \log _e 2$.

(A) Given,the condition is $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ ...$(i)$
Since $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are unit vectors,we have $|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1$.
Let the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ be $\theta$.
From equation $(i)$,we can write:
$\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}$
Squaring both sides,we get:
$(\overrightarrow{a}+\overrightarrow{b})^2=(-\overrightarrow{c})^2$
$|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2(\overrightarrow{a} \cdot \overrightarrow{b})=|\overrightarrow{c}|^2$
Substituting the values $|\overrightarrow{a}|=1, |\overrightarrow{b}|=1, |\overrightarrow{c}|=1$:
$1^2+1^2+2(\overrightarrow{a} \cdot \overrightarrow{b})=1^2$
$2+2(\overrightarrow{a} \cdot \overrightarrow{b})=1$
$2(\overrightarrow{a} \cdot \overrightarrow{b})=-1$
$\overrightarrow{a} \cdot \overrightarrow{b}=-\frac{1}{2}$
Using the definition of the dot product,$\overrightarrow{a} \cdot \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}| \cos \theta$:
$1 \cdot 1 \cdot \cos \theta = -\frac{1}{2}$
$\cos \theta = -\frac{1}{2}$
Since $\cos \theta = -\frac{1}{2}$,the angle $\theta = \frac{2 \pi}{3}$.