TS EAMCET 2011 Chemistry Question Paper with Answer and Solution

(B) Potassium crystallizes in a $BCC$ system.
Number of moles of potassium $= \frac{39 \ g}{39 \ g/mol} = 1 \ mol$.
Total number of atoms $= 1 \ mol \times N = N \ \text{atoms}$.
In a $BCC$ unit cell,the number of atoms per unit cell is $2$.
Therefore,the number of unit cells $= \frac{\text{Total number of atoms}}{2} = \frac{N}{2}$.

(A) Lowering of vapour pressure is a colligative property,which depends on the number of particles of the solute (van't Hoff factor,$i$).
For $0.1 \ M$ aqueous solutions,the relative lowering of vapour pressure is proportional to the number of ions produced upon dissociation.
$BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$; $i = 3$
$NaCl \rightarrow Na^+ + Cl^-$; $i = 2$
$Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$; $i = 5$
Thus,the ratio of the lowering of vapour pressure for $BaCl_2 : NaCl : Al_2(SO_4)_3$ is $3 : 2 : 5$.

(D) The decomposition reaction of $NaHCO_3$ is: $2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2$.
Let the mass of $NaHCO_3$ be $x \ g$ and $Na_2CO_3$ be $y \ g$.
Given $x + y = 19$.
Moles of $CO_2$ produced = $\frac{1.12 \ L}{22.4 \ L/mol} = 0.05 \ mol$.
From the reaction,$2 \ mol$ of $NaHCO_3$ produces $1 \ mol$ of $CO_2$.
So,moles of $NaHCO_3 = 2 \times 0.05 = 0.1 \ mol$.
Mass of $NaHCO_3 = 0.1 \ mol \times 84 \ g/mol = 8.4 \ g$.
Mass of $Na_2CO_3$ in the original mixture = $19 \ g - 8.4 \ g = 10.6 \ g$.

(C) Real gases deviate from ideal behaviour due to intermolecular forces and the finite volume of gas molecules.
At $High \text{ temperature}$,the kinetic energy of molecules is high,making intermolecular forces negligible.
At $Low \text{ pressure}$,the volume occupied by the gas molecules becomes negligible compared to the total volume of the container.
Therefore,real gases approach ideal gas behaviour under conditions of $High \text{ temperature}$ and $Low \text{ pressure}$.

(A) According to Bohr's postulate,the angular momentum $(L)$ of an electron in a hydrogen atom is quantized and is given by the formula: $L = \frac{n h}{2 \pi}$,where $n$ is an integer $(n = 1, 2, 3, \dots)$.
This can be rewritten as $L = n \times (0.5 \frac{h}{\pi})$.
For the given options:
$A) 1.25 \frac{h}{\pi} = 2.5 \times \frac{h}{2 \pi}$ (Not an integer multiple of $\frac{h}{2 \pi}$)
$B) 1 \frac{h}{\pi} = 2 \times \frac{h}{2 \pi}$ ($n = 2$,permitted)
$C) 1.5 \frac{h}{\pi} = 3 \times \frac{h}{2 \pi}$ ($n = 3$,permitted)
$D) 0.5 \frac{h}{\pi} = 1 \times \frac{h}{2 \pi}$ ($n = 1$,permitted)
Therefore,$1.25 \frac{h}{\pi}$ is not a permitted value.

(D) The relationship between frequency $(v)$,speed of light $(c)$,and wavelength $(\lambda)$ is given by $v = \frac{c}{\lambda}$.
Given: $c = 3 \times 10^8 \ m/s$ and $\lambda = 600 \ nm = 600 \times 10^{-9} \ m$.
Substituting the values: $v = \frac{3 \times 10^8}{600 \times 10^{-9}} = \frac{3 \times 10^8}{6 \times 10^{-7}} = 0.5 \times 10^{15} = 5.0 \times 10^{14} \ Hz$.

(D) The spin-only magnetic moment $(\mu)$ is calculated using the formula: $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Electronic configurations and unpaired electrons $(n)$:
$I. Fe^{2+}: [Ar] 3d^6 \rightarrow n = 4$
$II. Ti^{2+}: [Ar] 3d^2 \rightarrow n = 2$
$III. Cu^{2+}: [Ar] 3d^9 \rightarrow n = 1$
$IV. V^{2+}: [Ar] 3d^3 \rightarrow n = 3$
The magnetic moment increases with the number of unpaired electrons.
Comparing $n$ values: $1 (III) < 2 (II) < 3 (IV) < 4 (I)$.
Therefore,the increasing order is $III < II < IV < I$.

(A) The Freundlich adsorption isotherm is given by the equation:
$\frac{x}{m} = K p^{1/n}$
Taking the logarithm on both sides:
$\log \frac{x}{m} = \log K + \frac{1}{n} \log p$
This equation is in the form of a straight line equation $y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log p$,slope $m = \frac{1}{n}$,and intercept $c = \log K$.
Therefore,plotting a graph between $\log \frac{x}{m}$ and $\log p$ yields a straight line.

(C) In a steady state,the rate of heat flow $(H)$ through both slabs connected in series must be equal.
$H = \frac{k_1 A (T_1 - T_x)}{L} = \frac{k_2 A (T_x - T_2)}{L}$
Since the thickness $(L)$ and cross-sectional area $(A)$ are the same for both slabs,we have:
$k_1 (12 - x) = k_2 (x - 0)$
Given $k_1 = \frac{k_2}{2}$,we substitute this into the equation:
$\frac{k_2}{2} (12 - x) = k_2 x$
$12 - x = 2x$
$3x = 12 \implies x = 4^{\circ} C$
The temperature difference across slab $A$ is $(12 - x) = 12 - 4 = 8^{\circ} C$.

(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature in Kelvin.
Given $\eta = \frac{1}{6}$,so $1 - \frac{T_2}{T_1} = \frac{1}{6} \implies \frac{T_2}{T_1} = \frac{5}{6} \implies T_1 = 1.2 T_2$.
When the sink temperature is reduced by $62^{\circ} C$,the new sink temperature is $T_2' = T_2 - 62$. The new efficiency is doubled,so $\eta' = 2 \times \frac{1}{6} = \frac{1}{3}$.
Thus,$1 - \frac{T_2 - 62}{T_1} = \frac{1}{3} \implies \frac{T_2 - 62}{T_1} = \frac{2}{3}$.
Substituting $T_1 = 1.2 T_2$ into the equation: $\frac{T_2 - 62}{1.2 T_2} = \frac{2}{3}$.
$3(T_2 - 62) = 2.4 T_2 \implies 3 T_2 - 186 = 2.4 T_2 \implies 0.6 T_2 = 186 \implies T_2 = 310 \ K$.
Converting to Celsius: $T_2 = 310 - 273 = 37^{\circ} C$.
Then $T_1 = 1.2 \times 310 = 372 \ K$. Converting to Celsius: $T_1 = 372 - 273 = 99^{\circ} C$.
Therefore,the temperatures are $99^{\circ} C$ and $37^{\circ} C$.

(D) For an adiabatic process,the relation between pressure $p$ and volume $V$ is given by $pV^\gamma = k$,where $\gamma = C_p / C_V$ is the adiabatic index. $(i)$
Given that $p \propto T^3$,we can write $p = k' T^3$,where $k'$ is a constant.
Using the ideal gas equation $pV = nRT$,we have $T = \frac{pV}{nR}$.
Substituting this into the given relation:
$p = k' \left( \frac{pV}{nR} \right)^3$
$p = \left( \frac{k'}{n^3 R^3} \right) p^3 V^3$
$1 = \left( \frac{k'}{n^3 R^3} \right) p^2 V^3$
$p^2 V^3 = \text{constant}$
$p V^{3/2} = \text{constant}$ (ii)
Comparing equation $(i)$ and (ii),we get $\gamma = 3/2$.
Therefore,the value of $C_p / C_V$ is $3/2$.

(D) The enthalpy change of a reaction,$\Delta_r H$,is related to the activation energies of the forward reaction $(E_f)$ and the backward reaction $(E_b)$ by the equation: $\Delta_r H = E_f - E_b$.
For an exothermic reaction,the enthalpy change is negative,i.e.,$\Delta_r H < 0$.
Substituting this into the equation,we get $E_f - E_b < 0$,which implies $E_f < E_b$.

(D) The entropy change for a phase transition is given by the formula $\Delta S = \frac{q_{rev}}{T}$.
For the melting of ice,the heat absorbed is $q = m \times L_f$,where $m = 27.3 \ g$ and $L_f = 330 \ J g^{-1}$.
$q = 27.3 \ g \times 330 \ J g^{-1} = 9009 \ J$.
The melting temperature of ice is $T = 0^{\circ} C = 273 \ K$.
Therefore,$\Delta S = \frac{9009 \ J}{273 \ K} = 33 \ J K^{-1}$.

(C) The expression $\frac{1}{2} \mu_0 H^2$ represents the energy density of a magnetic field.
Energy density is defined as energy per unit volume.
The dimensional formula for energy is $[ML^2 T^{-2}]$ and for volume is $[L^3]$.
Therefore,the dimensional formula for energy density is $\frac{[ML^2 T^{-2}]}{[L^3]} = [ML^{-1} T^{-2}]$.
Alternatively,using the dimensions of $\mu_0$ and $H$:
$[\mu_0] = [MLT^{-2} A^{-2}]$ and $[H] = [AL^{-1}]$.
Substituting these into the expression: $[\mu_0 H^2] = [MLT^{-2} A^{-2}] \times [AL^{-1}]^2 = [MLT^{-2} A^{-2}] \times [A^2 L^{-2}] = [ML^{-1} T^{-2}]$.

(B) The physical quantities and their corresponding $SI$ units are as follows:
$1$. Magnetic field intensity $(H)$: The unit is Ampere per meter $(A \cdot m^{-1})$.
$2$. Magnetic flux $(\phi)$: The unit is Weber $(Wb)$.
$3$. Magnetic pole strength $(m)$: The unit is Ampere-meter $(A \cdot m)$.
$4$. Magnetic induction $(B)$: The unit is Weber per square meter $(Wb \cdot m^{-2})$,which is equivalent to Tesla $(T)$.
Matching these with the lists:
$(A)$ Magnetic field intensity $\rightarrow$ (iv) $A \cdot m^{-1}$
$(B)$ Magnetic flux $\rightarrow$ $(i)$ $Wb$
$(C)$ Magnetic pole strength $\rightarrow$ (iii) $A \cdot m$
$(D)$ Magnetic induction $\rightarrow$ (ii) $Wb \cdot m^{-2}$
Therefore,the correct matching is $(A)-(iv), (B)-(i), (C)-(iii), (D)-(ii)$. The correct option is $(b)$.

(C) The resultant intensity $I_R$ at any point in an interference pattern is given by $I_R = I_{\max} \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.
Given that $I_R = 75 \% \text{ of } I_{\max} = 0.75 I_{\max} = \frac{3}{4} I_{\max}$.
Substituting this into the formula: $\frac{3}{4} I_{\max} = I_{\max} \cos^2(\frac{\phi}{2})$.
$\cos^2(\frac{\phi}{2}) = \frac{3}{4}$.
Taking the square root on both sides: $\cos(\frac{\phi}{2}) = \frac{\sqrt{3}}{2}$.
We know that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,so $\frac{\phi}{2} = \frac{\pi}{6}$.
Therefore,the phase difference $\phi = \frac{\pi}{3}$.

(A) Let $v$ be the velocity of sound in air and $f_0$ be the frequency of the third note.
The frequencies of the two sound notes are $f_1 = \frac{v}{\lambda_1} = \frac{v}{40/195} = \frac{195v}{40}$ and $f_2 = \frac{v}{\lambda_2} = \frac{v}{40/193} = \frac{193v}{40}$.
Since each note produces $9$ beats per second with the third note,we have:
$|f_1 - f_0| = 9$ and $|f_2 - f_0| = 9$.
This implies $f_1 - f_0 = 9$ and $f_0 - f_2 = 9$ (since $f_1 > f_2$).
Adding these two equations:
$(f_1 - f_0) + (f_0 - f_2) = 9 + 9$
$f_1 - f_2 = 18$
Substituting the expressions for $f_1$ and $f_2$:
$\frac{195v}{40} - \frac{193v}{40} = 18$
$\frac{2v}{40} = 18$
$\frac{v}{20} = 18$
$v = 360 ~m/s$.

(B) Let the centre of the bigger disc of radius $2R$ be at the origin $(0,0)$.
Let $M$ be the mass of the bigger disc. The mass per unit area is $\sigma = \frac{M}{\pi(2R)^2} = \frac{M}{4\pi R^2}$.
The mass of the removed smaller disc of radius $R$ is $m = \sigma \cdot \pi R^2 = \frac{M}{4\pi R^2} \cdot \pi R^2 = \frac{M}{4}$.
The centre of mass of the bigger disc is at $x_1 = 0$. The centre of mass of the removed smaller disc is at $x_2 = R$ (since the circumferences touch).
The centre of mass of the remaining part is given by:
$x_{CM} = \frac{M x_1 - m x_2}{M - m}$
$x_{CM} = \frac{M(0) - (M/4)(R)}{M - M/4} = \frac{-MR/4}{3M/4} = -\frac{R}{3}$.
The distance from the centre of the bigger disc is $|x_{CM}| = \frac{R}{3}$.
Comparing this with $\alpha R$,we get $\alpha = \frac{1}{3}$.

(B) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{1}{2} (V - N - C)$,where $V$ is the number of valence electrons of the central atom,$N$ is the number of monovalent atoms bonded to it,and $C$ is the charge on the species.
$1$. For $XeF_2$: $V=8, N=2, C=0$. $\text{Lone pairs} = \frac{1}{2}(8-2) = 3$.
$2$. For $ICl_2^{-}$: $V=7, N=2, C=1$. $\text{Lone pairs} = \frac{1}{2}(7-2+1) = 3$.
Both $XeF_2$ and $ICl_2^{-}$ have $3$ lone pairs on their respective central atoms.

(B) The formal charge $(FC)$ is calculated as: $FC = (\text{Total valence electrons}) - (\text{Non-bonding electrons}) - \frac{1}{2}(\text{Bonding electrons})$.
For $N_{(1)}$ (terminal nitrogen with $4$ non-bonding electrons and $2$ bonding electrons from a double bond): $FC = 5 - 4 - \frac{1}{2}(4) = -1$.
For $N_{(2)}$ (central nitrogen with $0$ non-bonding electrons and $8$ bonding electrons from two double bonds): $FC = 5 - 0 - \frac{1}{2}(8) = +1$.
For $O$ (oxygen with $4$ non-bonding electrons and $4$ bonding electrons from a double bond): $FC = 6 - 4 - \frac{1}{2}(4) = 0$.
Thus,the formal charges are $-1, +1, 0$.

(D) In $XeF_6$,the central atom $Xe$ has $8$ valence electrons. It forms $6$ bonds with $F$ atoms,leaving $2$ electrons as one lone pair.
Steric number = (Number of sigma bonds) + (Number of lone pairs) = $6 + 1 = 7$.
$A$ steric number of $7$ corresponds to $sp^3d^3$ hybridisation.
Thus,$XeF_6$ has $sp^3d^3$ hybridisation and $1$ lone pair of electrons.

(D) Graphite has a two-dimensional sheet-like structure.
In graphite,each carbon atom is bonded to three other carbon atoms in the same plane,forming hexagonal rings.
Therefore,each carbon atom in graphite is $sp^2$ hybridized.
In contrast,in diamond,each carbon atom is bonded to four other carbon atoms in a tetrahedral arrangement,making it $sp^3$ hybridized.

(A) For the reaction $2AB \rightleftharpoons A_2 + B_2$,the equilibrium constant is $K_c = 49$.
When the coefficients of a balanced chemical equation are multiplied by a factor $n$,the new equilibrium constant $K'_c$ is given by $K'_c = (K_c)^n$.
In this case,the original equation is multiplied by $n = 1/2$.
Therefore,$K'_c = (49)^{1/2} = \sqrt{49} = 7$.

(C) The electronic configurations of $Na^+$,$Ne^+$,$Mg^+$,and $Al^+$ (after the removal of the first electron) are as follows:
$Na^+: 1s^2, 2s^2, 2p^6$ (Stable noble gas configuration)
$Ne^+: 1s^2, 2s^2, 2p^5$
$Mg^+: 1s^2, 2s^2, 2p^6, 3s^1$
$Al^+: 1s^2, 2s^2, 2p^6, 3s^2$
Ionisation energy depends on the stability of the electronic configuration,effective nuclear charge,and atomic size.
$Na^+$ has a stable noble gas configuration $(2p^6)$,making the removal of a second electron extremely difficult,resulting in the highest $IE_2$.
$Mg^+$ has a $3s^1$ configuration,which is relatively easy to remove compared to the others.
$Al^+$ has a $3s^2$ configuration,which is more stable than $Mg^+$.
$Ne^+$ has a $2p^5$ configuration,which is more stable than $Al^+$.
Thus,the correct order of $IE_2$ is $Mg < Al < Ne < Na$.

(D) Given that $\tan A$ and $\tan B$ are the roots of the quadratic equation $x^2-px+q=0$.
From the properties of roots,we have:
$\tan A + \tan B = p$
$\tan A \tan B = q$
Using the formula for $\tan(A+B)$:
$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{p}{1-q}$
Now,consider a right-angled triangle where the opposite side is $p$ and the adjacent side is $(1-q)$. The hypotenuse is $\sqrt{p^2 + (1-q)^2}$.
Therefore,$\sin(A+B) = \frac{p}{\sqrt{p^2 + (1-q)^2}}$.
Squaring both sides,we get:
$\sin^2(A+B) = \frac{p^2}{p^2 + (1-q)^2}$.

(D) Given,$\frac{x^2+x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$
Multiplying both sides by $(x-1)(x-2)(x-3)$,we get:
$x^2+x+1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$
To find $A$,put $x=1$:
$1^2+1+1 = A(1-2)(1-3) \Rightarrow 3 = A(-1)(-2) \Rightarrow 3 = 2A \Rightarrow A = \frac{3}{2}$
To find $C$,put $x=3$:
$3^2+3+1 = C(3-1)(3-2) \Rightarrow 13 = C(2)(1) \Rightarrow 13 = 2C \Rightarrow C = \frac{13}{2}$
Therefore,$A+C = \frac{3}{2} + \frac{13}{2} = \frac{16}{2} = 8$

(B) Given,${ }^{15}P_8 = A + 8 \cdot { }^{14}P_7$
We know that ${ }^{n}P_r = \frac{n!}{(n-r)!}$.
So,$\frac{15!}{(15-8)!} = A + 8 \cdot \frac{14!}{(14-7)!}$
$\frac{15!}{7!} = A + 8 \cdot \frac{14!}{7!}$
$A = \frac{15!}{7!} - 8 \cdot \frac{14!}{7!}$
$A = \frac{15 \cdot 14!}{7!} - 8 \cdot \frac{14!}{7!}$
$A = \frac{14!}{7!} (15 - 8)$
$A = \frac{14!}{7!} (7)$
$A = \frac{14!}{6!} = { }^{14}P_8$

(B) five-digit number is divisible by $5$ if its unit digit is $0$ or $5$.
Case $I$: When $0$ is in the unit place.
The remaining $4$ places can be filled by the remaining $5$ digits $(1, 2, 3, 4, 5)$ in $^5P_4 = 5! = 120$ ways.
Case $II$: When $5$ is in the unit place.
The first place (ten-thousands place) cannot be $0$. Thus,the first place can be filled by any of the $4$ digits $(1, 2, 3, 4)$.
The remaining $3$ places can be filled by the remaining $4$ digits (including $0$) in $^4P_3 = 4 \times 3 \times 2 = 24$ ways.
So,the number of ways for Case $II$ is $4 \times 24 = 96$.
Total number of ways $= 120 + 96 = 216$.

(D) Given the inequality: ${ }^{n-1} C_3+{ }^{n-1} C_4>{ }^n C_3$
Using the Pascal's identity ${ }^{n} C_r+{ }^{n} C_{r-1}={ }^{n+1} C_r$,we have ${ }^{n-1} C_3+{ }^{n-1} C_4={ }^n C_4$.
Substituting this into the inequality: ${ }^n C_4>{ }^n C_3$
Using the formula ${ }^n C_r = \frac{n!}{r!(n-r)!}$,we get:
$\frac{n!}{4!(n-4)!} > \frac{n!}{3!(n-3)!}$
$\frac{1}{4(n-4)!} > \frac{1}{(n-3)(n-4)!}$
$\frac{1}{4} > \frac{1}{n-3}$
$n-3 > 4$
$n > 7$
Since $n$ must be an integer,the minimum value of $n$ is $8$.

(B) The total number of balls is $2n$. We draw $n$ pairs of balls one by one.
For the first pair,the number of ways to choose one white and one black ball is $\binom{n}{1} \times \binom{n}{1} = n^2$.
For the second pair,the number of ways is $\binom{n-1}{1} \times \binom{n-1}{1} = (n-1)^2$.
Continuing this until the last pair,the total number of ways is $(n^2) \times (n-1)^2 \times \dots \times 1^2 = (n!)^2$.
Given that $(n!)^2 = 14400$.
Taking the square root on both sides,$n! = \sqrt{14400} = 120$.
Since $5! = 120$,we have $n = 5$.

(C) The Bhopal gas tragedy,which occurred in $1984$,was caused by the accidental release of Methyl isocyanate $(CH_3NCO)$ gas.
Methyl isocyanate is commonly abbreviated as $MIC$.

(C) We know that $60^{\circ} 1^{\prime} = 60^{\circ} + 1^{\prime} = 60^{\circ} + \left(\frac{1}{60}\right)^{\circ}$.
Since $1^{\circ} = \alpha$ radians,then $1^{\prime} = \left(\frac{1}{60}\right)^{\circ} = \frac{\alpha}{60}$ radians.
Using the approximation $\cos(A + B) \approx \cos A - B \sin A$ for small $B$ (where $B$ is in radians):
$\cos(60^{\circ} + 1^{\prime}) \approx \cos(60^{\circ}) - \left(\frac{\alpha}{60}\right) \sin(60^{\circ})$.
Substituting the values $\cos(60^{\circ}) = \frac{1}{2}$ and $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$:
$\cos(60^{\circ} 1^{\prime}) \approx \frac{1}{2} - \left(\frac{\alpha}{60}\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{1}{2} - \frac{\alpha \sqrt{3}}{120}$.

(C) Given,$f(x) = \sin^6 x + \cos^6 x$.
We can write this as $f(x) = (\sin^2 x)^3 + (\cos^2 x)^3$.
Using the identity $a^3 + b^3 = (a+b)(a^2 + b^2 - ab)$,we get:
$f(x) = (\sin^2 x + \cos^2 x)(\sin^4 x + \cos^4 x - \sin^2 x \cos^2 x)$.
Since $\sin^2 x + \cos^2 x = 1$,this simplifies to:
$f(x) = \sin^4 x + \cos^4 x - \sin^2 x \cos^2 x$.
Adding and subtracting $2\sin^2 x \cos^2 x$,we get:
$f(x) = (\sin^2 x + \cos^2 x)^2 - 3\sin^2 x \cos^2 x = 1 - 3\sin^2 x \cos^2 x$.
Multiplying and dividing by $4$,we get $f(x) = 1 - \frac{3}{4}(2\sin x \cos x)^2 = 1 - \frac{3}{4}\sin^2(2x)$.
Since $0 \leq \sin^2(2x) \leq 1$,we have:
$1 - \frac{3}{4}(1) \leq f(x) \leq 1 - \frac{3}{4}(0)$.
$\frac{1}{4} \leq f(x) \leq 1$.
Thus,$f(x) \in \left[\frac{1}{4}, 1\right]$.

(D) Given expression: $32 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5A}{2}\right)$
$= 16 \left[ 2 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5A}{2}\right) \right]$
Using the formula $2 \sin X \sin Y = \cos(X-Y) - \cos(X+Y)$:
$= 16 \left[ \cos \left( \frac{5A}{2} - \frac{A}{2} \right) - \cos \left( \frac{5A}{2} + \frac{A}{2} \right) \right]$
$= 16 [ \cos 2A - \cos 3A ]$
$= 16 [ (2 \cos^2 A - 1) - (4 \cos^3 A - 3 \cos A) ]$
Substitute $\cos A = \frac{3}{4}$:
$= 16 \left[ 2 \left( \frac{3}{4} \right)^2 - 1 - 4 \left( \frac{3}{4} \right)^3 + 3 \left( \frac{3}{4} \right) \right]$
$= 16 \left[ 2 \left( \frac{9}{16} \right) - 1 - 4 \left( \frac{27}{64} \right) + \frac{9}{4} \right]$
$= 16 \left[ \frac{9}{8} - 1 - \frac{27}{16} + \frac{9}{4} \right]$
$= 16 \left[ \frac{18 - 16 - 27 + 36}{16} \right]$
$= 18 - 16 - 27 + 36 = 11$

(B) Given trigonometric equations are $\tan \theta = -1$ and $\cos \theta = \frac{1}{\sqrt{2}}$.
Since $\tan \theta$ is negative and $\cos \theta$ is positive,$\theta$ must lie in the fourth quadrant.
The general solution for $\tan \theta = -1$ is $\theta = n \pi + \frac{3 \pi}{4}$.
The general solution for $\cos \theta = \frac{1}{\sqrt{2}}$ is $\theta = 2 n \pi \pm \frac{\pi}{4}$.
For $\theta$ to satisfy both,we look for the angle in the fourth quadrant where $\cos \theta = \frac{1}{\sqrt{2}}$ and $\tan \theta = -1$,which is $\theta = 2 n \pi + \frac{7 \pi}{4}$ (or $2 n \pi - \frac{\pi}{4}$).

(A) The slope of line $AB$ is $m = \frac{1-0}{3-2} = 1$.
Since $\tan \theta = 1$,the angle that $AB$ makes with the positive $x$-axis is $\theta = 45^{\circ}$.
When the line is rotated by $45^{\circ}$ in the anti-clockwise direction about point $A$,the new angle with the positive $x$-axis becomes $45^{\circ} + 45^{\circ} = 90^{\circ}$.
The length of the segment $AB$ is $r = \sqrt{(3-2)^2 + (1-0)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.
Let the new position of $B$ be $C(x, y)$. Since the length remains constant,$AC = AB = \sqrt{2}$.
The coordinates of $C$ are given by $(x_A + r \cos \phi, y_A + r \sin \phi)$,where $\phi = 90^{\circ}$.
$x = 2 + \sqrt{2} \cos 90^{\circ} = 2 + \sqrt{2}(0) = 2$.
$y = 0 + \sqrt{2} \sin 90^{\circ} = 0 + \sqrt{2}(1) = \sqrt{2}$.
Thus,the new coordinates are $(2, \sqrt{2})$.

(C) The quadrilateral is bounded by the lines $x = 0$ ($y$-axis),$y = 0$ ($x$-axis),$2x + y = 2$,and $x + y = 5$.
For a point $(x, y)$ to be inside the quadrilateral with natural number coordinates $(x \geq 1, y \geq 1)$,it must satisfy:
$1) \ x + y < 5$
$2) \ 2x + y > 2$
Since $x, y \geq 1$,the condition $2x + y > 2$ is always satisfied for all natural numbers $(x, y)$.
Now we check the condition $x + y < 5$ for natural numbers:
If $x = 1$,then $1 + y < 5 \implies y < 4$. Possible values for $y$ are $1, 2, 3$. Points: $(1, 1), (1, 2), (1, 3)$.
If $x = 2$,then $2 + y < 5 \implies y < 3$. Possible values for $y$ are $1, 2$. Points: $(2, 1), (2, 2)$.
If $x = 3$,then $3 + y < 5 \implies y < 2$. Possible value for $y$ is $1$. Point: $(3, 1)$.
If $x = 4$,then $4 + y < 5 \implies y < 1$. No natural number $y$ exists.
Total points are $(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)$.
There are $6$ such points.

(C) The equation of the line perpendicular to $x + 3y = 7$ is of the form $3x - y + \lambda = 0$.
Since this line passes through $(3, 8)$,we substitute the coordinates:
$3(3) - 8 + \lambda = 0$ $\Rightarrow 9 - 8 + \lambda = 0$ $\Rightarrow \lambda = -1$.
Thus,the equation of the perpendicular line is $3x - y - 1 = 0$.
The point of intersection of $x + 3y = 7$ and $3x - y - 1 = 0$ is the foot of the perpendicular. Solving these equations:
From the second equation,$y = 3x - 1$. Substituting into the first: $x + 3(3x - 1) = 7$ $\Rightarrow x + 9x - 3 = 7$ $\Rightarrow 10x = 10$ $\Rightarrow x = 1$.
Then $y = 3(1) - 1 = 2$. So,the foot of the perpendicular is $(1, 2)$.
Let the image of the point $(3, 8)$ be $(x_1, y_1)$. Since $(1, 2)$ is the midpoint of the segment joining $(3, 8)$ and $(x_1, y_1)$:
$\frac{3 + x_1}{2} = 1$ $\Rightarrow 3 + x_1 = 2$ $\Rightarrow x_1 = -1$.
$\frac{8 + y_1}{2} = 2$ $\Rightarrow 8 + y_1 = 4$ $\Rightarrow y_1 = -4$.
Therefore,the image of the point is $(-1, -4)$.

(C) Let $P(x, y)$ be any point on the locus. According to the problem,the sum of the distances from $P$ to $F_1(0, 2)$ and $F_2(0, -2)$ is $6$.
$\sqrt{x^2 + (y - 2)^2} + \sqrt{x^2 + (y + 2)^2} = 6$
$\sqrt{x^2 + (y - 2)^2} = 6 - \sqrt{x^2 + (y + 2)^2}$
Squaring both sides:
$x^2 + (y - 2)^2 = 36 + x^2 + (y + 2)^2 - 12\sqrt{x^2 + (y + 2)^2}$
$x^2 + y^2 - 4y + 4 = 36 + x^2 + y^2 + 4y + 4 - 12\sqrt{x^2 + (y + 2)^2}$
$-8y - 36 = -12\sqrt{x^2 + (y + 2)^2}$
$2y + 9 = 3\sqrt{x^2 + (y + 2)^2}$
Squaring again:
$(2y + 9)^2 = 9(x^2 + y^2 + 4y + 4)$
$4y^2 + 36y + 81 = 9x^2 + 9y^2 + 36y + 36$
$9x^2 + 5y^2 = 45$

(D) In cycloalkanes,the heat of combustion per $CH_2$ group for cyclohexane $(657.9 \ kJ \ mol^{-1})$ is the lowest,which indicates it has the least ring strain and is the most stable cycloalkane.
Cyclopropane and cyclobutane possess significant angle strain and torsional strain,making them less stable and more reactive compared to cyclohexane.
Therefore,both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.

(B) compound exhibits enantiomerism if it contains at least one chiral center (an asymmetric carbon atom,which is bonded to four different groups).
In $CH_3-CH(Br)-CH_2-CH_3$ ($2$-bromobutane),the second carbon atom is bonded to four different groups: $-H$,$-CH_3$,$-Br$,and $-CH_2CH_3$.
Since it has a chiral center,it is optically active and exhibits enantiomerism.
The other options do not contain any chiral carbon atoms.

(A) The percentage of sulphur is calculated using the formula:
$\text{Percentage of } S = \frac{32 \times \text{weight of } BaSO_4 \times 100}{233 \times \text{weight of compound}}$
Given:
$\text{Weight of } BaSO_4 = 0.233 \ g$
$\text{Weight of compound} = 0.16 \ g$
Substituting the values:
$\text{Percentage of } S = \frac{32 \times 0.233 \times 100}{233 \times 0.16}$
$\text{Percentage of } S = \frac{32 \times 0.233 \times 100}{233 \times 0.16} = \frac{32 \times 0.233 \times 100}{233 \times 0.16} = \frac{32 \times 100}{1000 \times 0.16} = \frac{3200}{160} = 20 \%$

(C) The reaction of benzene $(C_6H_6)$ with ozone $(O_3)$ is an ozonolysis reaction.
Benzene reacts with three molecules of ozone to form benzene triozonide $(X)$.
Benzene triozonide on reductive hydrolysis with $Zn / H_2O$ yields three molecules of glyoxal $(Y)$ $(CHO-CHO)$.
Therefore,$X$ is triozonide and $Y$ is glyoxal.

(D) Both hydrogen $(H_2)$ and deuterium $(D_2)$ molecules have the same bond length of $74 \ pm$.
However, they differ in their physical and chemical properties such as bond energy, melting point, and boiling point due to the difference in their isotopic masses.
Based on the data:
- Bond length: $74 \ pm$ (Same)
- Bond energy: $436 \ kJ \ mol^{-1}$ for $H_2$ vs $443.3 \ kJ \ mol^{-1}$ for $D_2$
- Melting point: $0.00 \ ^\circ C$ for $H_2$ vs $3.81 \ ^\circ C$ for $D_2$
- Boiling point: $100.0 \ ^\circ C$ for $H_2$ vs $101.42 \ ^\circ C$ for $D_2$
Therefore, the bond length is the same.

(C) The given function is $f(x) = |x| + |\sin x|$.
To find the left hand derivative $(LHD)$ at $x=0$,we use the definition:
$LHD = \lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h}$
Since $f(0) = |0| + |\sin 0| = 0$,we have:
$LHD = \lim_{h \to 0^+} \frac{|-h| + |\sin(-h)|}{-h}$
For small $h > 0$,$|-h| = h$ and $|\sin(-h)| = |-\sin h| = \sin h$ (since $\sin h > 0$ for $h \in (0, \pi/2)$).
$LHD = \lim_{h \to 0^+} \frac{h + \sin h}{-h}$
$LHD = \lim_{h \to 0^+} -\left( \frac{h}{h} + \frac{\sin h}{h} \right)$
$LHD = -(1 + 1) = -2$.

(D) $CH_3COOH \rightleftharpoons H^{+} + CH_3COO^{-}$
$K_a = \frac{[H^{+}][CH_3COO^{-}]}{[CH_3COOH]} = \frac{[H^{+}]^2}{[CH_3COOH]}$
$[H^{+}] = \sqrt{K_a \times [CH_3COOH]}$
$[H^{+}] = \sqrt{2 \times 10^{-5} \times 0.05}$
$[H^{+}] = \sqrt{10^{-6}} = 10^{-3} \ M$
$pH = -\log[H^{+}]$
$pH = -\log(10^{-3}) = 3$

(B) Let $\lambda$ be the linear mass density of the chain. The total mass of the chain is $M = \lambda L$.
Let $l'$ be the length of the chain hanging over the edge. The length of the chain on the table is $(L - l')$.
The mass of the hanging part is $m_h = \lambda l'$,and the mass of the part on the table is $m_t = \lambda (L - l')$.
The force pulling the chain down is the weight of the hanging part: $F_g = m_h g = \lambda l' g$.
The maximum static frictional force acting on the part on the table is $f_{max} = \mu N = \mu m_t g = \mu \lambda (L - l') g$.
For the chain to be on the verge of sliding,the pulling force must equal the maximum static friction:
$\lambda l' g = \mu \lambda (L - l') g$
$l' = \mu (L - l')$
$l' = \mu L - \mu l'$
$l' (1 + \mu) = \mu L$
$l' = \frac{\mu L}{(1 + \mu)}$

(D) Let the maximum tension in the rope be $T = 60 ~kg-wt = 60g ~N$. The two boys have masses $m_1 = 20 ~kg$ and $m_2 = 30 ~kg$. Let their accelerations be $a_1$ and $a_2$ respectively.
The total force exerted on the rope is the sum of the forces exerted by each boy: $T = m_1(g + a_1) + m_2(g + a_2)$.
Substituting the given values: $60g = 20(g + a_1) + 30(g + a_2)$.
$60g = 20g + 20a_1 + 30g + 30a_2$.
$60g = 50g + 20a_1 + 30a_2$.
$10g = 20a_1 + 30a_2$.
Dividing by $10$,we get $g = 2a_1 + 3a_2$.
To find the ratio of maximum accelerations,we consider the individual limits. If boy $1$ climbs with maximum acceleration $a_1$,then $a_2 = 0$,so $g = 2a_1 \Rightarrow a_1 = g/2$. If boy $2$ climbs with maximum acceleration $a_2$,then $a_1 = 0$,so $g = 3a_2 \Rightarrow a_2 = g/3$.
The ratio of their maximum possible accelerations is $a_1 : a_2 = (g/2) : (g/3) = 3 : 2$.

(C) Given the distance function $s = t^2 - 2t + 5$.
Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(t^2 - 2t + 5) = 2t - 2$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(2t - 2) = 2$.
Thus,the acceleration of the particle is $2$ units.

(A) Let $I = \int \frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx$.
We know that $\sin^8 x - \cos^8 x = (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)$.
Also,$\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1 - 2 \sin^2 x \cos^2 x$.
Substituting this into the integral:
$I = \int \frac{(\sin^4 x - \cos^4 x)(1 - 2 \sin^2 x \cos^2 x)}{1 - 2 \sin^2 x \cos^2 x} dx$.
$I = \int (\sin^4 x - \cos^4 x) dx$.
Since $\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) = -\cos 2x$.
$I = \int -\cos 2x dx = -\frac{\sin 2x}{2} + B$.
Comparing this with $I = A \sin 2x + B$,we get $A = -\frac{1}{2}$.