AP EAMCET 2009 Chemistry Question Paper with Answer and Solution

(B) Let $I = \int_0^\pi \frac{1}{1+\sin x} \, dx$.
Using the identity $\sin x = \frac{2 \tan(x/2)}{1+\tan^2(x/2)}$,we get:
$I = \int_0^\pi \frac{1}{1 + \frac{2 \tan(x/2)}{1+\tan^2(x/2)}} \, dx = \int_0^\pi \frac{1+\tan^2(x/2)}{(1+\tan(x/2))^2} \, dx$.
Since $1+\tan^2(x/2) = \sec^2(x/2)$,we have:
$I = \int_0^\pi \frac{\sec^2(x/2)}{(1+\tan(x/2))^2} \, dx$.
Let $t = \tan(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) \, dx$,which implies $\sec^2(x/2) \, dx = 2 \, dt$.
As $x \to 0$,$t \to 0$. As $x \to \pi$,$t \to \infty$.
Thus,$I = \int_0^\infty \frac{2 \, dt}{(1+t)^2} = 2 \left[ -\frac{1}{1+t} \right]_0^\infty$.
$I = 2 [0 - (-1)] = 2 \times 1 = 2$.

(D) The region is bounded by $y=\sin x$,$y=\cos x$ and the $x$-axis for $0 \leq x \leq \frac{\pi}{2}$.
For $0 \leq x \leq \frac{\pi}{4}$,the region is bounded by $y=\sin x$ and the $x$-axis. Thus,the area $A_1$ is given by:
$A_1 = \int_0^{\pi/4} \sin x \, dx = [-\cos x]_0^{\pi/4} = -(\cos \frac{\pi}{4} - \cos 0) = -(\frac{1}{\sqrt{2}} - 1) = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}}$.
For $\frac{\pi}{4} \leq x \leq \frac{\pi}{2}$,the region is bounded by $y=\cos x$ and the $x$-axis. Thus,the area $A_2$ is given by:
$A_2 = \int_{\pi/4}^{\pi/2} \cos x \, dx = [\sin x]_{\pi/4}^{\pi/2} = \sin \frac{\pi}{2} - \sin \frac{\pi}{4} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}}$.
Comparing the two areas,we get $A_1 : A_2 = \frac{\sqrt{2}-1}{\sqrt{2}} : \frac{\sqrt{2}-1}{\sqrt{2}} = 1 : 1$.

(D) For an achromatic doublet, the condition to eliminate chromatic aberration is that the sum of the product of the power $(P)$ and the dispersive power $(\omega)$ of the two lenses must be zero.
Mathematically, this is expressed as: $P_1 \omega_1 + P_2 \omega_2 = 0$.
Therefore, the sum of the product of their powers and dispersive powers must be equal to zero.

(C) Superoxides are species containing the $O_2^-$ ion,where oxygen has an oxidation state of $-\frac{1}{2}$.
These are typically formed by larger alkali metals such as $K, Rb$ and $Cs$,resulting in $KO_2, RbO_2$ and $CsO_2$.
For salts containing large anions like $O_2^-$,the lattice energy increases as the size of the cation increases down the group.
Since higher lattice energy contributes to greater thermodynamic stability of the ionic crystal,the stability of superoxides increases from $K$ to $Cs$.
Therefore,the assertion $(A)$ is true,but the reason $(R)$ is false because stability increases due to an increase in lattice energy,not a decrease.

(C) In circuit $A$,both $p-n$ junction diodes are in forward bias. Therefore,current flows through both branches.
The equivalent resistance $R_A$ is given by:
$\frac{1}{R_A} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \implies R_A = 2 \Omega$
Using Ohm's law,$V = I_A R_A$:
$8 = I_A \times 2 \implies I_A = 4 \text{ A}$
In circuit $B$,the upper diode is in forward bias,but the lower diode is in reverse bias. Therefore,no current flows through the lower branch.
The resistance of the circuit is $R_B = 4 \Omega$.
Using Ohm's law,$V = I_B R_B$:
$8 = I_B \times 4 \implies I_B = 2 \text{ A}$
Thus,the currents are $4 \text{ A}$ and $2 \text{ A}$ respectively.

(A) The magnitude of a vector $\overrightarrow{a} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $|\overrightarrow{a}| = \sqrt{x^2 + y^2 + z^2}$.
Calculating the magnitudes:
$m_1 = |\overrightarrow{a}_1| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \approx 2.45$
$m_2 = |\overrightarrow{a}_2| = \sqrt{3^2 + (-4)^2 + (-4)^2} = \sqrt{9 + 16 + 16} = \sqrt{41} \approx 6.40$
$m_3 = |\overrightarrow{a}_3| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \approx 1.73$
$m_4 = |\overrightarrow{a}_4| = \sqrt{(-1)^2 + 3^2 + 1^2} = \sqrt{1 + 9 + 1} = \sqrt{11} \approx 3.32$
Comparing the values: $\sqrt{3} < \sqrt{6} < \sqrt{11} < \sqrt{41}$,which implies $m_3 < m_1 < m_4 < m_2$.

(A) The angle $\theta$ between two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\cos \theta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|}$.
Given that the angle $\theta > 90^{\circ}$,we know that $\cos \theta < 0$.
Since the magnitudes $|\overrightarrow{a}|$ and $|\overrightarrow{b}|$ are always positive,the condition $\cos \theta < 0$ implies that the dot product $\overrightarrow{a} \cdot \overrightarrow{b} < 0$.
Calculating the dot product: $\overrightarrow{a} \cdot \overrightarrow{b} = (\lambda)(\lambda) + (-7)(1) + (3)(2\lambda) = \lambda^2 - 7 + 6\lambda$.
Setting the dot product to be less than zero: $\lambda^2 + 6\lambda - 7 < 0$.
Factoring the quadratic expression: $(\lambda + 7)(\lambda - 1) < 0$.
The inequality holds when $\lambda$ lies between the roots of the quadratic equation,which are $-7$ and $1$.
Therefore,the solution is $-7 < \lambda < 1$.

(A) Given the expression: $\overrightarrow{AB}+2\overrightarrow{AD}+\overrightarrow{BC}-2\overrightarrow{DC}$.
Using the triangle law of vector addition,$\overrightarrow{AB}+\overrightarrow{BC} = \overrightarrow{AC}$.
So,the expression becomes: $\overrightarrow{AC}+2\overrightarrow{AD}-2\overrightarrow{DC}$.
Since $\overrightarrow{AD}+\overrightarrow{DC} = \overrightarrow{AC}$,we have $\overrightarrow{AD} = \overrightarrow{AC}-\overrightarrow{DC}$.
Substituting this: $\overrightarrow{AC}+2(\overrightarrow{AC}-\overrightarrow{DC})-2\overrightarrow{DC} = 3\overrightarrow{AC}-4\overrightarrow{DC}$.
Given $Q$ is the midpoint of $AC$,$\overrightarrow{AC} = 2\overrightarrow{QC}$.
Given $P$ divides $DC$ in ratio $1:2$,$\overrightarrow{DP} = \frac{1}{3}\overrightarrow{DC}$ and $\overrightarrow{PC} = \frac{2}{3}\overrightarrow{DC}$,so $\overrightarrow{DC} = \frac{3}{2}\overrightarrow{PC}$.
Substituting these: $3(2\overrightarrow{QC})-4(\frac{3}{2}\overrightarrow{PC}) = 6\overrightarrow{QC}-6\overrightarrow{PC} = 6(\overrightarrow{QC}+\overrightarrow{CP}) = 6\overrightarrow{QP}$.
Since $\overrightarrow{QP} = -\overrightarrow{PQ}$,we have $6\overrightarrow{QP} = -6\overrightarrow{PQ}$.
Thus,$k\overrightarrow{PQ} = -6\overrightarrow{PQ}$,which implies $k = -6$.

(D) Let the vertices of the triangle be $A = (1, 0, 0)$,$B = (0, 1, 0)$,and $C = (0, 0, 1)$.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$,we calculate the lengths of the sides:
$AB = \sqrt{(0-1)^2 + (1-0)^2 + (0-0)^2} = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{1+1} = \sqrt{2}$.
$BC = \sqrt{(0-0)^2 + (0-1)^2 + (1-0)^2} = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{1+1} = \sqrt{2}$.
$CA = \sqrt{(1-0)^2 + (0-0)^2 + (0-1)^2} = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2}$.
The perimeter of the triangle is the sum of the lengths of its sides:
$\text{Perimeter} = AB + BC + CA = \sqrt{2} + \sqrt{2} + \sqrt{2} = 3\sqrt{2}$.

(A) The given equation of the sphere is $x^2+y^2+z^2-12x-4y-3z=0$.
Comparing this with the general equation of a sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$,we get $2u=-12$,$2v=-4$,and $2w=-3$.
Thus,$u=-6$,$v=-2$,and $w=-\frac{3}{2}$.
The radius of the sphere is given by the formula $r = \sqrt{u^2+v^2+w^2-d}$.
Substituting the values,we get $r = \sqrt{(-6)^2+(-2)^2+(-\frac{3}{2})^2-0}$.
$r = \sqrt{36+4+\frac{9}{4}} = \sqrt{40+\frac{9}{4}} = \sqrt{\frac{160+9}{4}} = \sqrt{\frac{169}{4}}$.
Therefore,$r = \frac{13}{2}$.

(C) Given,$l^2+m^2-n^2=0$ $(i)$ and $l+m+n=0$ $(ii)$.
From $(ii)$,$n=-(l+m)$. Substituting this into $(i)$:
$l^2+m^2=(-(l+m))^2 = l^2+m^2+2lm$.
This implies $2lm=0$,so $l=0$ or $m=0$.
Case $1$: If $l=0$,then $n=-m$. Since $l^2+m^2+n^2=1$,we have $0^2+m^2+(-m)^2=1 \Rightarrow 2m^2=1 \Rightarrow m=\pm\frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ and $(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.
Case $2$: If $m=0$,then $n=-l$. Since $l^2+m^2+n^2=1$,we have $l^2+0^2+(-l)^2=1 \Rightarrow 2l^2=1 \Rightarrow l=\pm\frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$ and $(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$.
Let the direction vectors of the two lines be $\vec{a} = (0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ and $\vec{b} = (\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$.
The angle $\theta$ between the lines is given by $\cos \theta = |l_1l_2 + m_1m_2 + n_1n_2|$.
$\cos \theta = |(0)(\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}})(0) + (-\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})| = |0 + 0 + \frac{1}{2}| = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

(D) Given the equations for the direction cosines $(l, m, n)$ of two lines:
$l+m+n=0 \implies n = -(l+m)$
Substitute $n$ into the second equation $l^2-5m^2+n^2=0$:
$l^2-5m^2+(-l-m)^2 = 0$
$l^2-5m^2+l^2+2lm+m^2 = 0$
$2l^2+2lm-4m^2 = 0$
$l^2+lm-2m^2 = 0$
$(l+2m)(l-m) = 0$
This gives two cases:
Case $1$: $l=m$. Then $n = -(l+m) = -2l$. The direction ratios are $(l, l, -2l)$,which simplifies to $(1, 1, -2)$. The direction cosines are $(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}})$.
Case $2$: $l=-2m$. Then $n = -(-2m+m) = m$. The direction ratios are $(-2m, m, m)$,which simplifies to $(-2, 1, 1)$. The direction cosines are $(-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})$.
Let the two lines have direction cosines $(l_1, m_1, n_1) = (\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}})$ and $(l_2, m_2, n_2) = (-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})$.
The cosine of the angle $\theta$ between them is given by:
$\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$
$\cos \theta = |(\frac{1}{\sqrt{6}})(-\frac{2}{\sqrt{6}}) + (\frac{1}{\sqrt{6}})(\frac{1}{\sqrt{6}}) + (-\frac{2}{\sqrt{6}})(\frac{1}{\sqrt{6}})|$
$\cos \theta = |-\frac{2}{6} + \frac{1}{6} - \frac{2}{6}| = |-\frac{3}{6}| = \frac{1}{2}$
Since $\cos \theta = \frac{1}{2}$,we have $\theta = 60^{\circ} = \frac{\pi}{3}$.

(C) Given,$l+m+n=0 \implies l = -m-n$ and $l^2+m^2-n^2=0$.
Substituting $l = -m-n$ into the second equation:
$(-m-n)^2 + m^2 - n^2 = 0$
$m^2 + 2mn + n^2 + m^2 - n^2 = 0$
$2m^2 + 2mn = 0$
$2m(m+n) = 0$.
This gives two cases:
Case $1$: If $m=0$,then $l = -n$. The direction ratios are $(-n, 0, n)$,which simplifies to $(-1, 0, 1)$. Let $\vec{v_1} = (-1, 0, 1)$.
Case $2$: If $m+n=0$,then $m = -n$. Substituting into $l = -m-n$,we get $l = -(-n)-n = 0$. The direction ratios are $(0, -n, n)$,which simplifies to $(0, -1, 1)$. Let $\vec{v_2} = (0, -1, 1)$.
The angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.
$\vec{v_1} \cdot \vec{v_2} = (-1)(0) + (0)(-1) + (1)(1) = 1$.
$|\vec{v_1}| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{2}$.
$|\vec{v_2}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$.
$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(B) First,calculate the vectors $2 \overrightarrow{a}-\overrightarrow{c}$ and $\overrightarrow{a}+\overrightarrow{b}$.
$2 \overrightarrow{a}-\overrightarrow{c} = 2(-\hat{i}+\hat{j}+2 \hat{k}) - (-2 \hat{i}+\hat{j}+3 \hat{k}) = (-2\hat{i}+2\hat{j}+4\hat{k}) + (2\hat{i}-\hat{j}-3\hat{k}) = \hat{j}+\hat{k}$.
$\overrightarrow{a}+\overrightarrow{b} = (-\hat{i}+\hat{j}+2 \hat{k}) + (2 \hat{i}-\hat{j}-\hat{k}) = \hat{i}+\hat{k}$.
Let $\theta$ be the angle between these two vectors.
The cosine of the angle is given by $\cos \theta = \frac{(\hat{j}+\hat{k}) \cdot (\hat{i}+\hat{k})}{|\hat{j}+\hat{k}| |\hat{i}+\hat{k}|}$.
Calculating the dot product: $(\hat{j}+\hat{k}) \cdot (\hat{i}+\hat{k}) = (0)(1) + (1)(0) + (1)(1) = 1$.
Calculating the magnitudes: $|\hat{j}+\hat{k}| = \sqrt{0^2+1^2+1^2} = \sqrt{2}$ and $|\hat{i}+\hat{k}| = \sqrt{1^2+0^2+1^2} = \sqrt{2}$.
Thus,$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(C) The formula for the image $(x, y, z)$ of a point $(x_1, y_1, z_1)$ in the plane $ax + by + cz + d = 0$ is given by:
$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \frac{-2(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}$
Given the point $(3, 2, 1)$ and the plane $2x - y + 3z - 7 = 0$,we have $a = 2, b = -1, c = 3, d = -7$.
Substituting these values into the formula:
$\frac{x - 3}{2} = \frac{y - 2}{-1} = \frac{z - 1}{3} = \frac{-2(2(3) - 1(2) + 3(1) - 7)}{2^2 + (-1)^2 + 3^2}$
$\frac{x - 3}{2} = \frac{y - 2}{-1} = \frac{z - 1}{3} = \frac{-2(6 - 2 + 3 - 7)}{4 + 1 + 9}$
$\frac{x - 3}{2} = \frac{y - 2}{-1} = \frac{z - 1}{3} = \frac{-2(0)}{14} = 0$
Setting each part equal to $0$:
$x - 3 = 0 \Rightarrow x = 3$
$y - 2 = 0 \Rightarrow y = 2$
$z - 1 = 0 \Rightarrow z = 1$
Thus,the image of the point is $(3, 2, 1)$,which means the point lies on the plane.

(C) Given,$P(\bar{A} \cup \bar{B}) = P(\overline{A \cap B}) = \frac{7}{10}$.
Since,$P(A \cap B) + P(\overline{A \cap B}) = 1$.
$\Rightarrow P(A \cap B) = 1 - \frac{7}{10} = \frac{3}{10}$.
Also,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$\Rightarrow \frac{4}{5} = P(A) + \frac{2}{5} - \frac{3}{10}$.
$\Rightarrow P(A) = \frac{4}{5} - \frac{2}{5} + \frac{3}{10}$.
$\Rightarrow P(A) = \frac{2}{5} + \frac{3}{10} = \frac{4+3}{10} = \frac{7}{10}$.

(D) The given quadratic equation is $x^2 + 4x + c = 0$.
For the roots to be real,the discriminant $D$ must be greater than or equal to $0$.
$D = b^2 - 4ac \geq 0$
Substituting $a = 1$,$b = 4$,we get:
$4^2 - 4(1)(c) \geq 0$
$16 - 4c \geq 0$
$16 \geq 4c$
$c \leq 4$.
Since $c$ is chosen from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$,the possible values for $c$ are $\{1, 2, 3, 4\}$.
There are $4$ favorable outcomes out of $9$ total outcomes.
Therefore,the required probability is $\frac{4}{9}$.

(B) The density formula is given by $d = \frac{Z M}{N_A a^3}$,where $Z$ is the number of atoms per unit cell,$M$ is the molar mass,$N_A$ is Avogadro's constant $(6.022 \times 10^{23} \ mol^{-1})$,and $a$ is the edge length.
Rearranging for $Z$: $Z = \frac{d N_A a^3}{M}$.
Given: $d = 8.92 \ g \ cm^{-3}$,$M = 63.55 \ g \ mol^{-1}$,$a = 362 \ pm = 362 \times 10^{-10} \ cm$.
Substituting the values: $Z = \frac{8.92 \times 6.022 \times 10^{23} \times (362 \times 10^{-10})^3}{63.55}$.
$Z \approx \frac{8.92 \times 6.022 \times 10^{23} \times 47.458 \times 10^{-24}}{63.55} \approx 4.01$.
Since $Z = 4$,the unit cell is face-centred cubic $(FCC)$.

(A) The freezing point of a substance is defined as the temperature at which the solid and liquid phases of the substance exist in equilibrium,meaning their vapour pressures are equal.
In the context of the depression in freezing point experiment,the equilibrium is established between the $ \text{liquid solvent} $ and the $ \text{solid solvent} $.
When a non-volatile solute is added to a solvent,the vapour pressure of the solution decreases,which leads to a lower temperature required to reach the equilibrium between the solid solvent and the liquid solution.

(C) Mass of $Cd = 0.9 \ g$.
Mass of $Cl_2 = 1.5 \ g - 0.9 \ g = 0.6 \ g$.
Atomic weight of $Cl = 35.5 \ g/mol$,so mass of $Cl_2 = 2 \times 35.5 = 71 \ g/mol$.
In $CdCl_2$,$0.6 \ g$ of $Cl_2$ corresponds to $71 \ g/mol$.
Therefore,$0.9 \ g$ of $Cd$ corresponds to atomic weight $X$ of $Cd$.
Using the ratio: $\frac{\text{Mass of } Cd}{\text{Mass of } Cl_2} = \frac{\text{Atomic weight of } Cd}{\text{Atomic weight of } Cl_2}$.
$\frac{0.9}{0.6} = \frac{X}{71}$.
$X = \frac{0.9 \times 71}{0.6} = 1.5 \times 71 = 106.5 \ g/mol$.

(B) $1$. Determine the empirical formula of $A$:
$C = 85.71 \% / 12 = 7.14$; $H = 14.29 \% / 1 = 14.29$.
Ratio $C:H = 7.14 : 14.29 = 1 : 2$. Empirical formula is $CH_2$.
$2$. Determine the molecular formula of $A$:
Molecular weight $= 2 \times \text{vapour density} = 2 \times 14 = 28$.
$n = 28 / 14 = 2$. Molecular formula is $(CH_2)_2 = C_2H_4$ (ethene).
$3$. Reaction sequence:
$A$ is $CH_2=CH_2$.
$CH_2=CH_2 + HOCl \rightarrow HO-CH_2-CH_2-Cl$ ($B$ is $2-$chloroethanol).
$HO-CH_2-CH_2-Cl + KCN \rightarrow HO-CH_2-CH_2-CN + KCl$.
$HO-CH_2-CH_2-CN + H_3O^+ \rightarrow HO-CH_2-CH_2-COOH$ ($C$ is $3-$hydroxypropanoic acid).
Therefore,the correct option is $B$.

(A) Perhydrol is a $30\% \ w/v$ $H_2O_2$ solution,which is known as $100$ volume $H_2O_2$.
This means $1 \ mL$ of perhydrol produces $100 \ mL$ of $O_2$ at $\text{STP}$.
The chemical reaction for the conversion is:
$2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$
From the stoichiometry,$2 \ L$ of $SO_2$ requires $1 \ L$ (or $1000 \ mL$) of $O_2$ for complete conversion.
Therefore,the volume of perhydrol required $= \frac{1000 \ mL}{100} = 10 \ mL$.

(B) The average kinetic energy per molecule of an ideal gas is given by the formula: $KE_{avg} = \frac{3}{2} k T$
Where $k$ is the Boltzmann constant $(k = \frac{R}{N_A})$ and $T$ is the temperature in Kelvin.
Given $T = 27^{\circ} C = 27 + 273 = 300 \ K$.
Substituting the values:
$KE_{avg} = \frac{3}{2} \times \frac{8.314 \ J \ K^{-1} \ mol^{-1}}{6.022 \times 10^{23} \ mol^{-1}} \times 300 \ K$
$KE_{avg} = 1.5 \times 1.38 \times 10^{-23} \times 300 \ J$
$KE_{avg} \approx 6.21 \times 10^{-21} \ J \ molecule^{-1}$

(A) The kinetic energy $(KE)$ of $1 \ mol$ of electrons is $6.023 \times 10^4 \ J$.
Therefore,the $KE$ of $1$ electron is $\frac{6.023 \times 10^4 \ J/mol}{6.023 \times 10^{23} \ mol^{-1}} = 1.0 \times 10^{-19} \ J$.
The energy of the incident photon $(E)$ is given by $E = \frac{hc}{\lambda}$.
$E = \frac{6.626 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m/s}{600 \times 10^{-9} \ m} = 3.313 \times 10^{-19} \ J$.
According to the photoelectric effect equation,$E = \Phi + KE$,where $\Phi$ is the work function (threshold energy).
$\Phi = E - KE = 3.313 \times 10^{-19} \ J - 1.0 \times 10^{-19} \ J = 2.313 \times 10^{-19} \ J$.

(A) From de-Broglie's equation,$\lambda = \frac{h}{mv}$.
Since $KE = \frac{1}{2}mv^2 = \frac{p^2}{2m}$ and $p = \frac{h}{\lambda}$,we have $KE = \frac{h^2}{2m\lambda^2}$.
Thus,$KE \propto \frac{1}{\lambda^2}$.
Therefore,$\frac{K_1}{K_2} = \left( \frac{\lambda_2}{\lambda_1} \right)^2$.
Given $\frac{\lambda_1}{\lambda_2} = \frac{3}{5}$,we have $\frac{\lambda_2}{\lambda_1} = \frac{5}{3}$.
So,$\frac{K_1}{K_2} = \left( \frac{5}{3} \right)^2 = \frac{25}{9}$.
The ratio of kinetic energy is $25: 9$.

(B) $As_2S_3$ is a negatively charged sol.
According to the Hardy-Schulze rule,the coagulating power of an ion is directly proportional to its valency.
Since $As_2S_3$ is a negative sol,it is coagulated by the addition of cations.
The valencies of the cations in the given options are:
$K^+$ $(KCl)$ = $+1$
$Al^{3+}$ $(AlCl_3)$ = $+3$
$Mg^{2+}$ $(MgSO_4)$ = $+2$
$K^+$ $(K_3Fe(CN)_6)$ = $+1$
Since $Al^{3+}$ has the highest valency $(+3)$,it is the most effective in causing coagulation.

(A) In the steady state,the heat current flowing into the junction must equal the heat current flowing out of the junction.
Let $T$ be the temperature of the junction.
The heat current $H$ is given by $H = \frac{KA(T_1 - T_2)}{l}$.
Since the dimensions (area $A$ and length $l$) are the same for all rods,the heat current is proportional to the thermal conductivity $K$.
According to the figure,the heat flows from the $100^{\circ} C$ source to the junction,and then from the junction to the $50^{\circ} C$ and $0^{\circ} C$ sinks.
Applying the principle of conservation of energy (Kirchhoff's law for heat current):
$H_{in} = H_{out1} + H_{out2}$
$\frac{3KA(100 - T)}{l} = \frac{2KA(T - 50)}{l} + \frac{KA(T - 0)}{l}$
Canceling $\frac{KA}{l}$ from both sides:
$3(100 - T) = 2(T - 50) + T$
$300 - 3T = 2T - 100 + T$
$300 - 3T = 3T - 100$
$6T = 400$
$T = \frac{400}{6} = \frac{200}{3}^{\circ} C$.

(C) The apparent weight loss is equal to the weight of the displaced liquid (Archimedes' principle). Let $V_{30}$ and $V_{40}$ be the volumes of the metal at $30^{\circ} C$ and $40^{\circ} C$ respectively.
At $30^{\circ} C$: Loss in weight $= 45 - 25 = 20 \ g$.
$V_{30} = \frac{\text{Loss in weight}}{\rho_{30}} = \frac{20 \ g}{1.5 \ g/cm^3} = 13.33 \ cm^3$.
At $40^{\circ} C$: Loss in weight $= 45 - 27 = 18 \ g$.
$V_{40} = \frac{\text{Loss in weight}}{\rho_{40}} = \frac{18 \ g}{1.25 \ g/cm^3} = 14.40 \ cm^3$.
Using the volume expansion formula: $V_{40} = V_{30}(1 + \gamma \Delta T)$,where $\Delta T = 10^{\circ} C$.
$\gamma = \frac{V_{40} - V_{30}}{V_{30} \Delta T} = \frac{14.40 - 13.33}{13.33 \times 10} = \frac{1.07}{133.3} \approx 8.027 \times 10^{-3} /^{\circ} C$.
The coefficient of linear expansion $\alpha = \frac{\gamma}{3} = \frac{8.027 \times 10^{-3}}{3} \approx 2.67 \times 10^{-3} /^{\circ} C$.
Rounding to the nearest option,we get $2.6 \times 10^{-3} /^{\circ} C$.

(D) For an ideal monoatomic gas,the molar heat capacity at constant pressure is $C_P = \frac{5}{2}R$ and at constant volume is $C_V = \frac{3}{2}R$.
For cylinder $A$ (isobaric process),the heat supplied is $Q = n C_P \Delta T_A$.
Given $n$ is the same,$Q = n \times \frac{5}{2}R \times 42 = 105 nR$.
For cylinder $B$ (isochoric process),the heat supplied is $Q = n C_V \Delta T_B$.
Since the heat supplied is the same,$n C_P \Delta T_A = n C_V \Delta T_B$.
$\frac{5}{2}R \times 42 = \frac{3}{2}R \times \Delta T_B$.
$5 \times 42 = 3 \times \Delta T_B$.
$\Delta T_B = \frac{210}{3} = 70 ~K$.

(B) For a cyclic process,the change in internal energy over the complete cycle is zero,i.e.,$\sum \Delta U = 0$.
From the first law of thermodynamics,$\Delta U = Q - W$.
Calculating $\Delta U$ for each state:
$\Delta U_1 = Q_1 - W_1 = 6000 - 2500 = 3500 \text{ J}$
$\Delta U_2 = Q_2 - W_2 = -5500 - (-1000) = -4500 \text{ J}$
$\Delta U_3 = Q_3 - W_3 = -3000 - (-1200) = -1800 \text{ J}$
$\Delta U_4 = Q_4 - W_4 = 3500 - x$
Since $\sum \Delta U = 0$:
$3500 - 4500 - 1800 + 3500 - x = 0$
$700 - x = 0 \implies x = 700 \text{ J}$.
Net work done $W_{\text{net}} = W_1 + W_2 + W_3 + W_4 = 2500 - 1000 - 1200 + 700 = 1000 \text{ J}$.
Total heat absorbed $Q_{\text{in}} = Q_1 + Q_4 = 6000 + 3500 = 9500 \text{ J}$.
Efficiency $\eta = \frac{W_{\text{net}}}{Q_{\text{in}}} \times 100 = \frac{1000}{9500} \times 100 \approx 10.5 \%$.

(B) We need to find $\Delta H^{\circ}$ for the reaction: $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$.
Given equations:
$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2} H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$
$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$
$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$
To get the target reaction,perform the operation: $2 \times (A) + \frac{1}{2} \times (C) - (B)$.
$2 \times [Na_{(s)} + H_2O_{(l)}$ $\longrightarrow NaOH_{(s)} + \frac{1}{2} H_{2(g)}] \implies 2Na_{(s)} + 2H_2O_{(l)}$ $\longrightarrow 2NaOH_{(s)} + H_{2(g)} \quad \Delta H^{\circ} = 2 \times (-146) = -292 \ kJ$
$\frac{1}{2} \times [2Na_2O_{(s)} + 2H_{2(g)}$ $\longrightarrow 4Na_{(s)} + 2H_2O_{(l)}] \implies Na_2O_{(s)} + H_{2(g)}$ $\longrightarrow 2Na_{(s)} + H_2O_{(l)} \quad \Delta H^{\circ} = \frac{259}{2} = 129.5 \ kJ$
$-(B) \implies 2NaOH_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)} + H_2O_{(l)} \quad \Delta H^{\circ} = -418 \ kJ$
Adding these: $(2Na_{(s)} + 2H_2O_{(l)} + Na_2O_{(s)} + H_{2(g)} + 2NaOH_{(s)} + SO_{3(g)})$ $\longrightarrow (2NaOH_{(s)} + H_{2(g)} + 2Na_{(s)} + H_2O_{(l)} + Na_2SO_{4(s)} + H_2O_{(l)})$
Simplifying gives: $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$.
$\Delta H^{\circ} = -292 + 129.5 - 418 = -580.5 \ kJ \approx -581 \ kJ$.

(C) The enthalpy of formation of $H$ atoms is given as $\Delta H_f(H) = 218 \ kJ/mol$.
This corresponds to the reaction: $\frac{1}{2} H_2(g) \longrightarrow H(g) ; \Delta H = 218 \ kJ/mol$.
For the dissociation of one mole of $H_2$ molecules into $H$ atoms: $H_2(g) \longrightarrow 2H(g) ; \Delta H = 2 \times 218 = 436 \ kJ/mol$.
To convert this energy into $kcal/mol$,we use the conversion factor $1 \ kcal = 4.18 \ kJ$.
Bond energy $= \frac{436 \ kJ/mol}{4.18 \ kJ/kcal} \approx 104.3 \ kcal/mol$.
Thus,the $H-H$ bond energy is approximately $104 \ kcal/mol$.

(D) Given the wave equation: $y = a \sin (bt - cx)$.
In the expression $y = a \sin (\theta)$,the argument of the trigonometric function $\theta = (bt - cx)$ must be dimensionless.
Since $bt$ and $cx$ are added/subtracted,they must have the same dimensions as the argument of the sine function,which is dimensionless.
$(a)$ $\frac{y}{a}$ is the ratio of two lengths,so it is dimensionless.
$(b)$ $bt$ is the argument of the sine function,so it is dimensionless.
$(c)$ $cx$ is the argument of the sine function,so it is dimensionless.
$(d)$ The dimension of $b$ is $[T^{-1}]$ and the dimension of $c$ is $[L^{-1}]$.
Therefore,the dimension of $\frac{b}{c} = \frac{[T^{-1}]}{[L^{-1}]} = [LT^{-1}]$,which represents the dimension of velocity.
Thus,$\frac{b}{c}$ is a quantity with dimensions.

(A) The position of the $n^{\text{th}}$ bright fringe (maximum) from the central maximum in Young's double slit experiment is given by the formula:
$y_n = \frac{n \lambda D}{d}$
where $D$ is the distance between the screen and the slits,and $d$ is the distance between the two slits.
For the first case,with wavelength $\lambda_1$ and $n_1 = 10$:
$y_1 = \frac{10 \lambda_1 D}{d}$
For the second case,with wavelength $\lambda_2$ and $n_2 = 5$:
$y_2 = \frac{5 \lambda_2 D}{d}$
Now,calculating the ratio $\left(\frac{y_1}{y_2}\right)$:
$\frac{y_1}{y_2} = \frac{\frac{10 \lambda_1 D}{d}}{\frac{5 \lambda_2 D}{d}}$
$\frac{y_1}{y_2} = \frac{10 \lambda_1}{5 \lambda_2}$
$\frac{y_1}{y_2} = \frac{2 \lambda_1}{\lambda_2}$

(B) Let the frequency of the sources be $n = 680 \, Hz$ and the speed of sound be $v = 340 \, ms^{-1}$.
The listener moves from $A$ towards $B$ with velocity $u$.
As the listener moves away from source $A$, the apparent frequency $n^{\prime}$ heard from $A$ is:
$n^{\prime} = n \left( \frac{v - u}{v} \right) = 680 \left( \frac{340 - u}{340} \right) = 2(340 - u)$
As the listener moves towards source $B$, the apparent frequency $n^{\prime \prime}$ heard from $B$ is:
$n^{\prime \prime} = n \left( \frac{v + u}{v} \right) = 680 \left( \frac{340 + u}{340} \right) = 2(340 + u)$
The beat frequency is the difference between the two apparent frequencies:
$n^{\prime \prime} - n^{\prime} = 10$
$2(340 + u) - 2(340 - u) = 10$
$680 + 2u - 680 + 2u = 10$
$4u = 10$
$u = 2.5 \, ms^{-1}$

(C) The phenomenon of interference occurs between two waves that have the same frequency and a constant phase difference.
Looking at the given waves:
$(i)$ $y_1 = a \sin(\omega t + \phi_1)$ has angular frequency $\omega$.
(ii) $y_2 = a \sin(2\omega t)$ has angular frequency $2\omega$.
(iii) $y_3 = d' \sin(\omega t + \phi_2)$ has angular frequency $\omega$.
(iv) $y_4 = d' \sin(3\omega t + \phi)$ has angular frequency $3\omega$.
Since waves $(i)$ and (iii) have the same angular frequency $\omega$,their superposition will result in interference.
Therefore,the correct option is $(C)$.

(B) The fundamental frequency of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
Given,$n_1 = 600 \text{ Hz}$.
When the tension in one wire is increased to $T'$,its new frequency becomes $n_2 = 606 \text{ Hz}$ to produce $6 \text{ beats per second}$ $(n_2 - n_1 = 606 - 600 = 6 \text{ Hz})$.
We have $\frac{n_2}{n_1} = \sqrt{\frac{T'}{T}}$.
Substituting the values: $\frac{606}{600} = \sqrt{\frac{T'}{T}}$.
$1.01 = \sqrt{\frac{T'}{T}}$.
Squaring both sides: $\frac{T'}{T} = (1.01)^2 = 1.0201 \approx 1.02$.
The fractional increase in tension is $\frac{\Delta T}{T} = \frac{T' - T}{T} = \frac{T'}{T} - 1 = 1.02 - 1 = 0.02$.

(C) The power $P$ required to move a fluid through a pipe is given by the work done per unit time. For a horizontal pipe,the work done is primarily to overcome the kinetic energy of the fluid.
The kinetic energy $K$ of a mass $m$ moving with velocity $v$ is $K = \frac{1}{2}mv^2$.
The power $P$ is the rate of change of kinetic energy: $P = \frac{dK}{dt} = \frac{1}{2} v^2 \frac{dm}{dt}$.
Let the mass flow rate be $\frac{dm}{dt} = \dot{m}$. Since the pipe is the same,the cross-sectional area $A$ is constant. The flow rate (volume per unit time) is $Q = Av$. Thus,the mass flow rate $\dot{m} = \rho A v$,where $\rho$ is the density of water.
Substituting $\dot{m}$ into the power equation: $P = \frac{1}{2} v^2 (\rho A v) = \frac{1}{2} \rho A v^3$.
Since $Q = Av$,we have $v = Q/A$. Substituting this into the power equation: $P = \frac{1}{2} \rho A (Q/A)^3 = \frac{\rho}{2A^2} Q^3$.
This shows that $P \propto Q^3$.
Given that the initial flow rate is $Q_0$ and the final flow rate is $Q_1 = n Q_0$,the ratio of powers is:
$\frac{P_1}{P_0} = \left( \frac{Q_1}{Q_0} \right)^3 = n^3$.
Therefore,the ratio $P_1 : P_0 = n^3 : 1$.

(A) $1$. The Wacker process converts ethene $(\text{CH}_2=\text{CH}_2)$ to ethanal $(\text{CH}_3\text{CHO})$,so $B$ is $\text{CH}_3\text{CHO}$.
$2$. Ozonolysis of an alkene $A$ yields $B$ (ethanal). Since $A \xrightarrow{\text{O}_3} 2 \text{CH}_3\text{CHO}$,$A$ must be but$-2-$ene $(\text{CH}_3\text{CH}=\text{CHCH}_3)$.
$3$. The alkyne is reduced to $A$ (cis-but$-2-$ene) using Lindlar's catalyst. Therefore,the starting alkyne is but$-2-$yne $(\text{CH}_3\text{C} \equiv \text{CCH}_3)$.

(B) Sulphuric anhydride is $SO_3$. In its structure,the central sulfur atom is $sp^2$ hybridized.
It forms three $\sigma$ bonds with three oxygen atoms.
Out of the three $\pi$ bonds,one is a $p \pi-p \pi$ bond (formed by the overlap of $p$-orbitals of $S$ and $O$) and two are $p \pi-d \pi$ bonds (formed by the overlap of $d$-orbitals of $S$ and $p$-orbitals of $O$).
Therefore,the molecule contains $3 \sigma$,$1 p \pi-p \pi$,and $2 p \pi-d \pi$ bonds.

(C) From the definition of dipole moment,$\mu = \delta \times d$,where $\delta$ is the magnitude of the electric charge and $d$ is the bond length.
Therefore,$\delta = \frac{\mu}{d}$.
The ratio of the fraction of electric charge for $HCl$ and $HI$ is given by:
$\frac{\delta_{HCl}}{\delta_{HI}} = \frac{\mu_{HCl}}{d_{HCl}} \times \frac{d_{HI}}{\mu_{HI}}$
Substituting the given values:
$\frac{\delta_{HCl}}{\delta_{HI}} = \frac{1.03 \times 1.6}{1.3 \times 0.38} = \frac{1.648}{0.494} \approx 3.33: 1$.
Thus,the ratio is approximately $3.3: 1$.

(C) For the reaction: $N_{2(g)} + 2 O_{2(g)} \rightleftharpoons 2 NO_{2(g)}$,the equilibrium constant is $K_1 = 100$.
$K_1 = \frac{[NO_2]^2}{[N_2][O_2]^2} = 100$ ... $(i)$
For the target reaction: $NO_{2(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + O_{2(g)}$,the equilibrium constant is $K_2$.
$K_2 = \frac{[N_2]^{1/2} [O_2]}{[NO_2]}$ ... $(ii)$
Comparing $(i)$ and $(ii)$,we observe that $K_2 = \sqrt{\frac{1}{K_1}}$.
$K_2 = \sqrt{\frac{1}{100}} = \frac{1}{10} = 0.1$.

(C) The first ionisation energy $(IE)$ generally increases across a period from left to right. For the second period elements,the expected order is $Li < Be < B < C < N < O < F < Ne$.
However,due to the stability of fully-filled and half-filled orbitals,the actual order for $B, Be, N,$ and $O$ is $B < Be < O < N$.
The lower $IE$ of $B$ $(1s^2, 2s^2 2p^1)$ compared to $Be$ $(1s^2, 2s^2)$ is because the electron is removed from a $2p$ orbital in $B$,which is easier than removing an electron from the stable,fully-filled $2s$ orbital in $Be$.
The lower $IE$ of $O$ $(1s^2, 2s^2 2p^4)$ compared to $N$ $(1s^2, 2s^2 2p^3)$ is because $N$ has a stable,half-filled $2p$ subshell,making it harder to remove an electron compared to $O$.

(C) Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2-2x+4=0$.
From the relation between roots and coefficients,we have:
$\alpha+\beta = 2$ $(i)$
$\alpha\beta = 4$ $(ii)$
We can write $x^2-2x+4=0$ as $x^2-2x+1+3=0$,which implies $(x-1)^2 = -3$.
Thus,$x-1 = \pm \sqrt{3}i$,so $x = 1 \pm \sqrt{3}i$.
Let $\alpha = 1+\sqrt{3}i = 2(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 2e^{i\pi/3}$ and $\beta = 1-\sqrt{3}i = 2(\frac{1}{2} - i\frac{\sqrt{3}}{2}) = 2e^{-i\pi/3}$.
Then,$\alpha^6 = (2e^{i\pi/3})^6 = 2^6 e^{i2\pi} = 64(1) = 64$.
Similarly,$\beta^6 = (2e^{-i\pi/3})^6 = 2^6 e^{-i2\pi} = 64(1) = 64$.
Therefore,$\alpha^6+\beta^6 = 64+64 = 128$.

(D) Total number of points is $3p$.
To form a triangle,we need to select $3$ points that are not collinear.
The total number of ways to select $3$ points from $3p$ points is $^{3p}C_3$.
Since $p$ points are collinear on each of the $3$ lines,we must subtract the sets of $3$ points that lie on the same line.
Number of triangles $= ^{3p}C_3 - 3 \times (^pC_3)$.
$= \frac{3p(3p-1)(3p-2)}{6} - 3 \times \frac{p(p-1)(p-2)}{6}$.
$= \frac{p(3p-1)(3p-2) - p(p-1)(p-2)}{2}$.
$= \frac{p}{2} [ (9p^2 - 9p + 2) - (p^2 - 3p + 2) ]$.
$= \frac{p}{2} [ 8p^2 - 6p ] = p(4p^2 - 3p) = p^2(4p-3)$.

(A) The thermosphere is the fourth layer of the Earth's atmosphere and is located above the mesosphere.
In this region,the air is very thin.
The thermosphere includes the ionosphere,which is a region filled with charged particles.
Due to high temperatures and solar radiation,molecules undergo ionization to form species such as $O_2^{+}$,$O^{+}$,and $NO^{+}$.

(B) We have $\tan (x-y) \tan y = \frac{\sin (x-y) \sin y}{\cos (x-y) \cos y}$.
Multiplying numerator and denominator by $2$,we get $\frac{2 \sin (x-y) \sin y}{2 \cos (x-y) \cos y}$.
Using the product-to-sum formulas $2 \sin A \sin B = \cos (A-B) - \cos (A+B)$ and $2 \cos A \cos B = \cos (A-B) + \cos (A+B)$:
$\tan (x-y) \tan y = \frac{\cos (x-2y) - \cos x}{\cos (x-2y) + \cos x}$.
Dividing numerator and denominator by $\cos (x-2y)$:
$= \frac{1 - \frac{\cos x}{\cos (x-2y)}}{1 + \frac{\cos x}{\cos (x-2y)}}$.
Substituting $\lambda = \frac{\cos x}{\cos (x-2y)}$,we get $\frac{1-\lambda}{1+\lambda}$.

(C) Given,$\sinh ^{-1} 2 + \sinh ^{-1} 3 = x$.
Taking $\cosh$ on both sides,$\cosh x = \cosh(\sinh ^{-1} 2 + \sinh ^{-1} 3)$.
Using the identity $\cosh(A + B) = \cosh A \cosh B + \sinh A \sinh B$,we get:
$\cosh x = \cosh(\sinh ^{-1} 2) \cosh(\sinh ^{-1} 3) + \sinh(\sinh ^{-1} 2) \sinh(\sinh ^{-1} 3)$.
Since $\cosh(\sinh ^{-1} y) = \sqrt{1 + y^2}$,we have:
$\cosh(\sinh ^{-1} 2) = \sqrt{1 + 2^2} = \sqrt{5}$ and $\cosh(\sinh ^{-1} 3) = \sqrt{1 + 3^2} = \sqrt{10}$.
Substituting these values:
$\cosh x = (\sqrt{5})(\sqrt{10}) + (2)(3) = \sqrt{50} + 6$.
To match the options,we write this as $\frac{2(6 + \sqrt{50})}{2} = \frac{1}{2}(12 + 2 \sqrt{50})$.

(C) Given equation: $\sin^2 x - \cos 2x = 2 - \sin 2x$
Using identities $\cos 2x = 1 - 2\sin^2 x$ and $\sin 2x = 2\sin x \cos x$:
$\sin^2 x - (1 - 2\sin^2 x) = 2 - 2\sin x \cos x$
$3\sin^2 x - 1 = 2 - 2\sin x \cos x$
Alternatively,using $\sin^2 x = 1 - \cos^2 x$ and $\cos 2x = 2\cos^2 x - 1$:
$(1 - \cos^2 x) - (2\cos^2 x - 1) = 2 - 2\sin x \cos x$
$2 - 3\cos^2 x = 2 - 2\sin x \cos x$
$-3\cos^2 x + 2\sin x \cos x = 0$
$\cos x (2\sin x - 3\cos x) = 0$
Since $3\cos x \neq 2\sin x$,we have $2\sin x - 3\cos x \neq 0$.
Therefore,$\cos x = 0$.
The general solution for $\cos x = 0$ is $x = (2n + 1) \frac{\pi}{2}$,which can be written as $x = (4n \pm 1) \frac{\pi}{2}$ for $n \in \mathbb{Z}$.

(A) The given equation is $x^2+y^2=r^2$.
When the axes are rotated through an angle $\theta = 36^{\circ}$,the transformation equations are:
$x = X \cos 36^{\circ} - Y \sin 36^{\circ}$
$y = X \sin 36^{\circ} + Y \cos 36^{\circ}$
Substituting these into the original equation:
$(X \cos 36^{\circ} - Y \sin 36^{\circ})^2 + (X \sin 36^{\circ} + Y \cos 36^{\circ})^2 = r^2$
Expanding the terms:
$X^2 \cos^2 36^{\circ} + Y^2 \sin^2 36^{\circ} - 2XY \sin 36^{\circ} \cos 36^{\circ} + X^2 \sin^2 36^{\circ} + Y^2 \cos^2 36^{\circ} + 2XY \sin 36^{\circ} \cos 36^{\circ} = r^2$
Simplifying:
$X^2(\cos^2 36^{\circ} + \sin^2 36^{\circ}) + Y^2(\sin^2 36^{\circ} + \cos^2 36^{\circ}) = r^2$
Since $\cos^2 \theta + \sin^2 \theta = 1$,we get:
$X^2(1) + Y^2(1) = r^2$
$X^2 + Y^2 = r^2$