AP EAMCET 2014 Chemistry Question Paper with Answer and Solution

(D) In $KO_2$,the potassium ion is $K^{+}$ and the superoxide ion is $O_2^{-}$.
$K^{+}$ has a noble gas configuration and is diamagnetic.
The superoxide ion $O_2^{-}$ has $17$ electrons.
Its molecular orbital electronic configuration is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$.
Due to the presence of $1$ unpaired electron in the antibonding $\pi^*$ molecular orbital,$O_2^{-}$ is paramagnetic.

(B) The output characteristics of a transistor in common-emitter configuration are defined as the variation of collector current $(I_C)$ with collector-emitter voltage $(V_{C E})$ while keeping the base current $(I_B)$ constant.
This is represented by the graph of $I_C$ versus $V_{C E}$ for different fixed values of $I_B$.

(D) Given:
Intrinsic carrier concentration,$n_i = 1.6 \times 10^{16} / m^3$
Donor concentration,$n_e \approx N_D = 4.8 \times 10^{20} / m^3$
According to the law of mass action for semiconductors,the product of electron and hole concentrations is equal to the square of the intrinsic carrier concentration:
$n_e \times n_h = n_i^2$
Substituting the given values:
$4.8 \times 10^{20} \times n_h = (1.6 \times 10^{16})^2$
$4.8 \times 10^{20} \times n_h = 2.56 \times 10^{32}$
$n_h = \frac{2.56 \times 10^{32}}{4.8 \times 10^{20}}$
$n_h = 0.533 \times 10^{12} / m^3 = 5.33 \times 10^{11} / m^3$
Thus,the concentration of holes is approximately $5.3 \times 10^{11} / m^3$.

(C) The equation of the family of parabolas with vertex at $(0, -1)$ and axis along the $Y$-axis is given by:
$x^2 = 4a(y + 1)$ --- $(i)$
Differentiating both sides with respect to $x$,we get:
$2x = 4a y^{\prime}$
$a = \frac{2x}{4y^{\prime}} = \frac{x}{2y^{\prime}}$
Substituting the value of $a$ into equation $(i)$:
$x^2 = 4 \left( \frac{x}{2y^{\prime}} \right) (y + 1)$
$x^2 = \frac{2x}{y^{\prime}} (y + 1)$
Assuming $x \neq 0$,we divide by $x$:
$x = \frac{2(y + 1)}{y^{\prime}}$
$x y^{\prime} = 2y + 2$
$x y^{\prime} - 2y - 2 = 0$

(A) The given differential equation is $\cos y + (x \sin y - 1) \frac{dy}{dx} = 0$.
Rearranging the terms,we get $(x \sin y - 1) \frac{dy}{dx} = -\cos y$.
Taking the reciprocal,$\frac{dx}{dy} = -\frac{x \sin y - 1}{\cos y} = -x \tan y + \sec y$.
This can be written as $\frac{dx}{dy} + x \tan y = \sec y$.
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = \tan y$ and $Q = \sec y$.
The integrating factor $(IF)$ is $e^{\int P dy} = e^{\int \tan y dy} = e^{\ln |\sec y|} = \sec y$.
The general solution is given by $x(IF) = \int Q(IF) dy + C$.
Substituting the values,$x \sec y = \int \sec y \cdot \sec y dy + C$.
$x \sec y = \int \sec^2 y dy + C$.
$x \sec y = \tan y + C$.

(A) The given lines are in the form $\vec{r} = \vec{a} + \lambda \vec{b}$.
Comparing the given equations,the direction vectors are $\vec{b_1} = \hat{i} + 4 \hat{j} + 3 \hat{k}$ and $\vec{b_2} = \hat{i} + 2 \hat{j} - 3 \hat{k}$.
The angle $\theta$ between two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$ is given by $\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.
Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(1) + (4)(2) + (3)(-3) = 1 + 8 - 9 = 0$.
Since the dot product is $0$,the angle $\theta$ is $\frac{\pi}{2}$.

(A) Given that $\vec{a}+3\vec{b}$ is collinear with $\vec{c}$.
Therefore,$\vec{a}+3\vec{b} = \lambda\vec{c}$ for some scalar $\lambda$.
This implies $\vec{a}+3\vec{b}-\lambda\vec{c} = 0$ $(i)$.
Also,$3\vec{b}+2\vec{c}$ is collinear with $\vec{a}$.
Therefore,$3\vec{b}+2\vec{c} = \mu\vec{a}$ for some scalar $\mu$.
This implies $3\vec{b}+2\vec{c}-\mu\vec{a} = 0$ $(ii)$.
From $(i)$,we have $3\vec{b} = \lambda\vec{c} - \vec{a}$.
Substituting this into $(ii)$:
$(\lambda\vec{c} - \vec{a}) + 2\vec{c} - \mu\vec{a} = 0$
$(\lambda+2)\vec{c} - (1+\mu)\vec{a} = 0$.
Since $\vec{a}$ and $\vec{c}$ are non-collinear,their coefficients must be zero.
Thus,$\lambda+2 = 0 \implies \lambda = -2$ and $1+\mu = 0 \implies \mu = -1$.
Substituting $\lambda = -2$ into $(i)$:
$\vec{a}+3\vec{b} - (-2)\vec{c} = 0$
$\vec{a}+3\vec{b}+2\vec{c} = 0$.

(C) Given,$|\vec{a}|=2, |\vec{b}|=3$ and $|\vec{c}|=4$.
We know that $|\vec{x}-\vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2(\vec{x} \cdot \vec{y})$.
Therefore,$|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2 = (|\vec{a}|^2+|\vec{b}|^2-2\vec{a} \cdot \vec{b}) + (|\vec{b}|^2+|\vec{c}|^2-2\vec{b} \cdot \vec{c}) + (|\vec{c}|^2+|\vec{a}|^2-2\vec{c} \cdot \vec{a})$
$= 2(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2) - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.
We also know that $|\vec{a}+\vec{b}+\vec{c}|^2 \geq 0$,which implies $|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq 0$.
Thus,$-2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \leq |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2$.
Substituting this into our expression:
$|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2 \leq 2(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2) + (|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2) = 3(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2)$.
Substituting the given magnitudes:
$3(2^2+3^2+4^2) = 3(4+9+16) = 3(29) = 87$.
Thus,the upper bound is $87$.

(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) Let $S$ be the sample space of outcomes where the sum of the two dice is $7$. The possible outcomes are:
$S = \{(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)\}$
Thus,the total number of outcomes $n(S) = 6$.
Let $E$ be the event that the number $3$ appears at least once.
The outcomes in $E$ are:
$E = \{(3, 4), (4, 3)\}$
Thus,the number of favorable outcomes $n(E) = 2$.
The required probability is $P(E) = \frac{n(E)}{n(S)} = \frac{2}{6} = \frac{1}{3}$.

(A) Let $S$ denote success (pass) and $F$ denote failure (fail). The probability of passing the first test is $P(S_1) = p$,so $P(F_1) = 1-p$.
For subsequent tests,$P(S_{n+1} | S_n) = p$ and $P(S_{n+1} | F_n) = \frac{p}{2}$.
The candidate is selected if they pass at least two tests. The possible favorable outcomes are:
$1$. Pass $I$,Pass $II$,Fail $III$: $p \times p \times (1-p) = p^2(1-p)$
$2$. Pass $I$,Fail $II$,Pass $III$: $p \times (1-p) \times \frac{p}{2} = \frac{p^2(1-p)}{2}$
$3$. Fail $I$,Pass $II$,Pass $III$: $(1-p) \times \frac{p}{2} \times p = \frac{p^2(1-p)}{2}$
$4$. Pass $I$,Pass $II$,Pass $III$: $p \times p \times p = p^3$
Summing these probabilities:
$P = p^2(1-p) + \frac{p^2(1-p)}{2} + \frac{p^2(1-p)}{2} + p^3$
$P = p^2(1-p) + p^2(1-p) + p^3$
$P = 2p^2 - 2p^3 + p^3 = 2p^2 - p^3 = p^2(2-p)$.

(D) Given that $A, B$ and $C$ are mutually exclusive and exhaustive events,we have $P(A) + P(B) + P(C) = 1$.
Given $P(B) = \frac{3}{2} P(A)$ and $P(C) = \frac{1}{2} P(B) = \frac{1}{2} \times \frac{3}{2} P(A) = \frac{3}{4} P(A)$.
Substituting these into the sum equation:
$P(A) + \frac{3}{2} P(A) + \frac{3}{4} P(A) = 1$
$P(A) \left( 1 + \frac{6}{4} + \frac{3}{4} \right) = 1$
$P(A) \left( \frac{4+6+3}{4} \right) = 1$
$\frac{13}{4} P(A) = 1 \implies P(A) = \frac{4}{13}$.
Now,$P(C) = \frac{3}{4} P(A) = \frac{3}{4} \times \frac{4}{13} = \frac{3}{13}$.
Since $A$ and $C$ are mutually exclusive,$P(A \cup C) = P(A) + P(C)$.
$P(A \cup C) = \frac{4}{13} + \frac{3}{13} = \frac{7}{13}$.

(B) For a probability distribution,the sum of probabilities is $1$.
$\Sigma P(X=x) = k + 2k + 3k + 2k + k = 9k = 1 \implies k = \frac{1}{9}$.
The mean $E(X) = \Sigma x_i P(x_i) = (1 \times k) + (2 \times 2k) + (3 \times 3k) + (4 \times 2k) + (5 \times k) = k + 4k + 9k + 8k + 5k = 27k$.
Substituting $k = \frac{1}{9}$,$E(X) = 27 \times \frac{1}{9} = 3$.
The expected value of $X^2$ is $E(X^2) = \Sigma x_i^2 P(x_i) = (1^2 \times k) + (2^2 \times 2k) + (3^2 \times 3k) + (4^2 \times 2k) + (5^2 \times k) = k + 8k + 27k + 32k + 25k = 93k$.
Substituting $k = \frac{1}{9}$,$E(X^2) = 93 \times \frac{1}{9} = \frac{93}{9} = \frac{31}{3}$.
Variance $Var(X) = E(X^2) - [E(X)]^2 = \frac{31}{3} - (3)^2 = \frac{31}{3} - 9 = \frac{31-27}{3} = \frac{4}{3}$.

(C) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute.
$n_{\text{urea}} = 0.1 \ mol$
$n_{\text{water}} = \frac{180 \ g}{18 \ g/mol} = 10 \ mol$
Mole fraction of urea $(x_2)$ = $\frac{0.1}{0.1 + 10} = \frac{0.1}{10.1} \approx 0.0099$
Using the formula $\frac{p^{\circ} - p_s}{p^{\circ}} = x_2$:
$\frac{24 - p_s}{24} = \frac{0.1}{10.1}$
$24 - p_s = 24 \times 0.0099 = 0.2376$
$p_s = 24 - 0.2376 = 23.7624 \ mm \ Hg \approx 23.76 \ mm \ Hg$

(A) For isotonic solutions,the osmotic pressures are equal,so $\frac{W_1}{V_1 M_1} = \frac{W_2}{V_2 M_2}$.
Given that the solutions are $1 \%$ and $5 \%$,we have $\frac{W_1}{V_1} = 1 \ g/100 \ mL$ and $\frac{W_2}{V_2} = 5 \ g/100 \ mL$.
Let $M_1$ be the molar mass of solute $X$ and $M_2 = 342 \ g \ mol^{-1}$ be the molar mass of cane sugar.
Substituting the values: $\frac{1}{M_1} = \frac{5}{342}$.
Solving for $M_1$: $M_1 = \frac{342}{5} = 68.4 \ g \ mol^{-1}$.

(A) Given,percentage of oxygen $= 40 \%$.
Percentage of metal $= 100 - 40 = 60 \%$.
Valency of metal $= 2$.
Since the valency of the metal is $2$ and the valency of oxygen is $2$,the formula of the oxide is $MO$.
Let the atomic weight of the metal be $M$.
Percentage of oxygen $= \frac{16}{M + 16} \times 100 = 40$.
$1600 = 40(M + 16)$.
$1600 = 40M + 640$.
$40M = 960$.
$M = 24$.

(C) Given,$\frac{r_X}{r_Y} = \frac{1}{5}$ and $\frac{r_Y}{r_Z} = \frac{1}{6}$.
Multiplying the two ratios:
$\frac{r_X}{r_Y} \times \frac{r_Y}{r_Z} = \frac{1}{5} \times \frac{1}{6} = \frac{1}{30}$.
Thus,$\frac{r_X}{r_Z} = \frac{1}{30}$.
Therefore,the ratio of rates of diffusion of $Z$ and $X$ is $\frac{r_Z}{r_X} = \frac{30}{1}$,which is $30:1$.

(D) For $4d$ orbital,the principal quantum number $n = 4$ and the azimuthal quantum number $l = 2$.
Number of angular nodes $= l = 2$.
Number of radial nodes $= n - l - 1 = 4 - 2 - 1 = 1$.
Therefore,the number of angular and radial nodes are $2$ and $1$ respectively.

(D) The order of increasing energy is determined using the $(n+l)$ rule. If two orbitals have the same $(n+l)$ value,the orbital with the lower value of $n$ has lower energy.
$(i)$ For $n=4, l=1$,$(n+l) = 4+1 = 5$.
$(ii)$ For $n=4, l=0$,$(n+l) = 4+0 = 4$.
$(iii)$ For $n=3, l=2$,$(n+l) = 3+2 = 5$.
$(iv)$ For $n=3, l=1$,$(n+l) = 3+1 = 4$.
Comparing the values: $(iv)$ and $(ii)$ have $(n+l) = 4$,where $(iv)$ has a lower $n$ value $(3 < 4)$. Thus,energy of $(iv)$ < $(ii)$.
$(i)$ and $(iii)$ have $(n+l) = 5$,where $(iii)$ has a lower $n$ value $(3 < 4)$. Thus,energy of $(iii)$ < $(i)$.
Combining these,the correct order of increasing energy is $(iv) < (ii) < (iii) < (i)$.

(A) van der Waals' forces are responsible for physisorption,not chemisorption. Therefore,$Assertion (A)$ is false.
Chemisorption involves the formation of chemical bonds,which requires activation energy. Thus,high temperature is favourable for chemisorption. Therefore,$Reason (R)$ is true.
Hence,the correct option is $A$ is false,but $R$ is true.

(C) An agonist is a chemical substance that binds to a receptor and activates it to produce a biological response. It mimics the action of natural chemical messengers.

(B) According to the principle of calorimetry,Heat lost by steam = Heat gained by water and calorimeter.
Let $m$ be the mass of steam condensed in grams.
Heat lost by steam = $m \times L + m \times C_w \times (T_{steam} - T_{final})$
$= m \times 540 + m \times 1 \times (100 - 90) = 550m \ cal$.
Heat gained by water and calorimeter = $(m_{water} + m_{eq}) \times C_w \times (T_{final} - T_{initial})$
$= (1000 \ g + 200 \ g) \times 1 \times (90 - 9) = 1200 \times 81 = 97200 \ cal$.
Equating the two: $550m = 97200$.
$m = \frac{97200}{550} \approx 176.7 \ g$.
Converting to $kg$,$m \approx 0.1767 \ kg$,which is approximately $0.18 \ kg$.

(D) According to Stefan-Boltzmann Law,the power $P$ radiated by a black body is given by $P = \sigma A T^4$.
Given:
$\sigma = 5.67 \times 10^{-8} ~W m^{-2} K^{-4}$
$A = 200 ~mm^2 = 200 \times 10^{-6} ~m^2$
$T = 727^{\circ} C = 727 + 273 = 1000 ~K$
Substituting these values into the formula:
$P = (5.67 \times 10^{-8}) \times (200 \times 10^{-6}) \times (1000)^4$
$P = 5.67 \times 10^{-8} \times 2 \times 10^{-4} \times 10^{12}$
$P = 5.67 \times 2 \times 10^{-12} \times 10^{12}$
$P = 11.34 ~J/s$.

(C) The change in internal energy $\Delta U$ for an ideal gas in an adiabatic process is given by the formula:
$\Delta U = n C_v \Delta T$
Since $C_v = \frac{R}{\gamma - 1}$, we have:
$\Delta U = n \frac{R}{\gamma - 1} (T_2 - T_1)$
Given values:
$n = 5 \, mol$
$T_1 = 273 \, K$ (at $STP$)
$T_2 = 673 \, K$
$R = 8.3 \, J/mol-K$
$\gamma = 1.4$
Substituting these values into the equation:
$\Delta U = 5 \times \frac{8.3}{1.4 - 1} \times (673 - 273)$
$\Delta U = 5 \times \frac{8.3}{0.4} \times 400$
$\Delta U = 5 \times 8.3 \times 1000$
$\Delta U = 41500 \, J$
Converting to kilo joules:
$\Delta U = 41.50 \, kJ$

(B) The relationship between Gibbs free energy,enthalpy,and entropy is given by the equation: $\Delta G = \Delta H - T \Delta S$
Given that $\Delta G = 0$,the equation becomes: $0 = \Delta H - T \Delta S$
Rearranging for $T$,we get: $T = \frac{\Delta H}{\Delta S}$
Convert $\Delta H$ from $kJ \ mol^{-1}$ to $J \ mol^{-1}$: $\Delta H = -20.5 \times 10^3 \ J \ mol^{-1}$
Substitute the values: $T = \frac{-20.5 \times 10^3 \ J \ mol^{-1}}{-50.0 \ J \ K^{-1} \ mol^{-1}} = 410 \ K$

(C) Let the length of the pipe be $l$ and the speed of sound be $v$.
The fundamental frequency of the closed pipe is $f_c = \frac{v}{4l}$.
The $3^{rd}$ harmonic of the closed pipe is $f_{3c} = 3 \times f_c = \frac{3v}{4l}$.
The fundamental frequency of the open pipe is $f_o = \frac{v}{2l}$.
According to the problem,the fundamental frequency of the open pipe is less than the $3^{rd}$ harmonic of the closed pipe by $55 ~Hz$:
$f_{3c} - f_o = 55 ~Hz$
$\frac{3v}{4l} - \frac{v}{2l} = 55$
$\frac{3v - 2v}{4l} = 55$
$\frac{v}{4l} = 55 ~Hz$.
Since the fundamental frequency of the closed pipe is $f_c = \frac{v}{4l}$,the value is $55 ~Hz$.

(B) Given that the wheel starts from rest,the initial angular velocity $\omega_0 = 0$. The angular displacement $\theta$ under constant angular acceleration $\alpha$ is given by $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$.
For the first time interval $t$,the angle rotated is $\theta_1 = 15^{\circ}$.
$15^{\circ} = 0 + \frac{1}{2} \alpha t^2 \implies \frac{1}{2} \alpha t^2 = 15^{\circ} \quad \dots(i)$
For the total time interval $(t + 2t) = 3t$,the total angular displacement $\theta_{total}$ is:
$\theta_{total} = \frac{1}{2} \alpha (3t)^2 = \frac{1}{2} \alpha (9t^2) = 9 \left( \frac{1}{2} \alpha t^2 \right)$.
Substituting the value from equation $(i)$:
$\theta_{total} = 9 \times 15^{\circ} = 135^{\circ}$.
The increase in angle in the next $2t \ s$ is $\Delta \theta = \theta_{total} - \theta_1 = 135^{\circ} - 15^{\circ} = 120^{\circ}$.

(D) Let the mass of the shell be $M$. At the highest point,the velocity of the shell is $v_x = u \cos \theta$. The shell breaks into two equal parts of mass $m = M/2$.
Let the velocity of the first part (which retraces its path) be $v_1 = -u \cos \theta$. By the law of conservation of linear momentum:
$M(u \cos \theta) = m v_1 + m v_2$
$M(u \cos \theta) = (M/2)(-u \cos \theta) + (M/2)v_2$
$u \cos \theta = -0.5 u \cos \theta + 0.5 v_2$
$1.5 u \cos \theta = 0.5 v_2$
$v_2 = 3 u \cos \theta$
The kinetic energy of the first part is $E_1 = \frac{1}{2} (M/2) (u \cos \theta)^2 = \frac{1}{4} M u^2 \cos^2 \theta$.
The kinetic energy of the second part is $E_2 = \frac{1}{2} (M/2) (3 u \cos \theta)^2 = \frac{1}{2} (M/2) (9 u^2 \cos^2 \theta) = \frac{9}{4} M u^2 \cos^2 \theta$.
Comparing $E_1$ and $E_2$:
$E_2 = 9 \times (\frac{1}{4} M u^2 \cos^2 \theta) = 9 E_1$.

(A) Concentrated $H_2SO_4$ is a powerful dehydrating agent.
When it reacts with sugar $(C_{12}H_{22}O_{11})$,it removes water molecules,leaving behind a black residue of carbon.
This process is known as the charring of sugar.
$C_{12}H_{22}O_{11} \xrightarrow{\text{conc. } H_2SO_4} 12C + 11H_2O$.

(C) The reaction given is a decarboxylation reaction,also known as the soda-lime decarboxylation.
When a sodium salt of a carboxylic acid is heated with soda lime $(NaOH + CaO)$,it loses a molecule of $CO_2$ to form an alkane with one carbon atom less than the parent carboxylic acid.
$CH_3-CH_2-CO_2^{\ominus} Na^{\oplus} + NaOH \xrightarrow{CaO, \Delta} CH_3-CH_3 + Na_2CO_3$
Here,sodium propanoate ($3$ carbons) reacts to form ethane ($2$ carbons).
Therefore,$Z$ is ethane.

(A) The bond length is defined as the average distance between the centers of the nuclei of two bonded atoms in a molecule. It depends on factors such as atomic size, hybridization, and electronegativity difference.
As the atomic size increases, the bond length generally increases.
Comparing the covalent radii of the atoms involved:
$C-C$ bond length is approximately $154 \ pm$.
$C-H$ bond length is approximately $109 \ pm$.
$C-N$ bond length is approximately $147 \ pm$.
$C-O$ bond length is approximately $143 \ pm$.
Therefore, the $C-C$ bond has the longest bond distance among the given options.

(C) Ethyne $(C_2H_2)$: The carbon atom is $sp$ hybridized with $2$ $\sigma$-bonds and $0$ lone pairs,resulting in a linear geometry.
Methane $(CH_4)$: The carbon atom is $sp^3$ hybridized with $4$ $\sigma$-bonds and $0$ lone pairs,resulting in a tetrahedral geometry.
Therefore,the shapes are linear and tetrahedral.

(B) hydrogen bond is a special type of dipole-dipole attraction between a hydrogen atom covalently bonded to a highly electronegative atom and another electronegative atom with a lone pair.
In an $HF$ molecule,there is a significant electronegativity difference between $H$ and $F$,which creates a permanent dipole.
Consequently,the interaction between $HF$ molecules is a form of dipole-dipole interaction.
In the gaseous state,several $HF$ molecules polymerize through these $H$-bonding interactions.

(A) The reaction is $A + B \rightleftharpoons C + D$.
Initial moles at $t = 0$: $A = 1 \ mol$,$B = 1 \ mol$,$C = 0 \ mol$,$D = 0 \ mol$.
At equilibrium,$40 \%$ of $B$ has reacted,so $0.4 \ mol$ of $B$ is consumed.
Moles at equilibrium: $A = (1 - 0.4) = 0.6 \ mol$,$B = (1 - 0.4) = 0.6 \ mol$,$C = 0.4 \ mol$,$D = 0.4 \ mol$.
Since the volume is $10 \ L$,the concentrations are $[A] = 0.6/10 = 0.06 \ M$,$[B] = 0.06 \ M$,$[C] = 0.04 \ M$,$[D] = 0.04 \ M$.
$K_C = \frac{[C][D]}{[A][B]} = \frac{0.04 \times 0.04}{0.06 \times 0.06} = \frac{0.0016}{0.0036} = \frac{16}{36} = \frac{4}{9} \approx 0.44$.

(D) For group $13$ elements,electronegativity $(EN)$ decreases from $B$ to $Al$ due to the increase in atomic size.
From $Al$ to $Tl$,the electronegativity increases gradually due to the poor shielding effect of $d$ and $f$ orbitals,which increases the effective nuclear charge.
The values of $EN$ (on the Pauling scale) are: $B (2.0)$,$Al (1.5)$,$Ga (1.6)$,$In (1.7)$,and $Tl (1.8)$.
Thus,the correct variation shows a decrease from $B$ to $Al$ followed by an increase from $Al$ to $Tl$.

(C) To determine the number of electrons in the valence shell,we calculate the total number of bonding and lone pairs around the central atom.
In $SCl_2$,the central atom is $S$ (Sulfur).
Sulfur has $6$ valence electrons.
It forms $2$ single bonds with $Cl$ atoms,using $2$ electrons.
Number of lone pairs on $S = \frac{6 - 2}{2} = 2$.
Total electron pairs around $S = 2 \text{ (bonding pairs)} + 2 \text{ (lone pairs)} = 4 \text{ pairs}$.
Total electrons in the valence shell $= 4 \times 2 = 8$ electrons.
Thus,$SCl_2$ follows the octet rule.

(A) In the periodic table,as we move from left to right across a period,the effective nuclear charge increases while the number of shells remains the same.
This results in a stronger attraction between the nucleus and the valence electrons,causing the atomic radius to decrease.
All the given elements $(Na, Mg, Al, Si, S)$ belong to the $3^{rd}$ period.
Following the trend of decreasing atomic radius from left to right,the order is $Na > Mg > Al > Si > S$.
Therefore,the increasing order is $S < Si < Al < Mg < Na$.

(B) Given: $L_1 = 108 \ mH$,$N_1 = 600$ turns,$N_2 = 500$ turns,and $L_2 = ?$
For a circular coil,the self-inductance $L$ is proportional to the square of the number of turns $N^2$,given by $L = \frac{\mu_0 \pi N^2 r}{2}$.
Since the radius $r$ is the same for both coils,we have the relation:
$\frac{L_1}{L_2} = \left(\frac{N_1}{N_2}\right)^2$
Substituting the given values:
$\frac{108}{L_2} = \left(\frac{600}{500}\right)^2$
$\frac{108}{L_2} = \left(\frac{6}{5}\right)^2 = \frac{36}{25}$
$L_2 = \frac{108 \times 25}{36}$
$L_2 = 3 \times 25 = 75 \ mH$
Thus,the self-inductance of the coil is $75 \ mH$.

(A) Let $\alpha, \beta, \gamma$ be the roots of $x^3-2x^2+10x-8=0$.
From Vieta's formulas:
$\alpha+\beta+\gamma = 2$
$\alpha\beta+\beta\gamma+\gamma\alpha = 10$
$\alpha\beta\gamma = 8$
We want the equation with roots $\alpha^2, \beta^2, \gamma^2$.
Sum of new roots: $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (2)^2 - 2(10) = 4 - 20 = -16$.
Sum of roots taken two at a time: $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha+\beta+\gamma) = (10)^2 - 2(8)(2) = 100 - 32 = 68$.
Product of new roots: $\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = (8)^2 = 64$.
The required cubic equation is $x^3 - (\text{sum of roots})x^2 + (\text{sum of roots taken two at a time})x - (\text{product of roots}) = 0$.
Substituting the values: $x^3 - (-16)x^2 + 68x - 64 = 0$.
Thus,$x^3+16x^2+68x-64=0$.

(D) Given $Z_r = \cos \left(\frac{\pi}{2^r}\right) + i \sin \left(\frac{\pi}{2^r}\right) = e^{i \frac{\pi}{2^r}}$.
Therefore,the product $P = Z_1 Z_2 Z_3 \ldots \infty = e^{i \frac{\pi}{2^1}} \cdot e^{i \frac{\pi}{2^2}} \cdot e^{i \frac{\pi}{2^3}} \ldots$
Using the property of exponents,$P = e^{i \left( \frac{\pi}{2} + \frac{\pi}{4} + \frac{\pi}{8} + \ldots \right)}$.
The exponent is a geometric series with first term $a = \frac{\pi}{2}$ and common ratio $r = \frac{1}{2}$.
The sum of the infinite geometric series is $S = \frac{a}{1 - r} = \frac{\pi/2}{1 - 1/2} = \frac{\pi/2}{1/2} = \pi$.
Thus,$P = e^{i \pi} = \cos \pi + i \sin \pi$.
Since $\cos \pi = -1$ and $\sin \pi = 0$,we get $P = -1 + i(0) = -1$.

(A) Given: $x=p+q$,$y=p \omega+q \omega^2$,and $z=p \omega^2+q \omega$.
We need to calculate $xyz = (p+q)(p \omega+q \omega^2)(p \omega^2+q \omega)$.
First,multiply $y$ and $z$:
$yz = (p \omega+q \omega^2)(p \omega^2+q \omega) = p^2 \omega^3 + pq \omega^2 + pq \omega^4 + q^2 \omega^3$.
Since $\omega^3=1$ and $\omega^4=\omega$,we have:
$yz = p^2(1) + pq(\omega^2+\omega) + q^2(1) = p^2 + pq(\omega^2+\omega) + q^2$.
Using the property $1+\omega+\omega^2=0$,we know $\omega+\omega^2=-1$.
So,$yz = p^2 - pq + q^2$.
Now,$xyz = (p+q)(p^2 - pq + q^2)$.
Using the algebraic identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$,we get:
$xyz = p^3+q^3$.

(A) Chlorofluorocarbons (CFCs),also known as freons,are stable,non-toxic,and non-reactive compounds used in refrigeration and air conditioning.
Due to their long atmospheric lifetime,they eventually reach the stratosphere.
In the stratosphere,they are decomposed by high-energy $UV$ radiation,leading to the formation of chlorine free radicals.
The reaction for dichlorodifluoromethane $(CF_2Cl_2)$ is:
$CF_2Cl_2 \stackrel{hv}{\longrightarrow} \dot{C}F_2Cl + \dot{Cl}$
Thus,the products $X$ and $Y$ are the free radicals $\dot{C}F_2Cl$ and $\dot{Cl}$.

(D) Given,$\cos x = \tan y$,$\cot y = \tan z$ and $\cot z = \tan x$.
From the given equations,we have $\tan y = \cos x$ and $\cot y = \frac{1}{\tan y} = \frac{1}{\cos x}$.
Since $\cot y = \tan z$,we have $\tan z = \frac{1}{\cos x}$.
Then $\cot z = \frac{1}{\tan z} = \cos x$.
Given $\cot z = \tan x$,we get $\tan x = \cos x$.
$\Rightarrow \frac{\sin x}{\cos x} = \cos x$
$\Rightarrow \sin x = \cos^2 x$
$\Rightarrow \sin x = 1 - \sin^2 x$
$\Rightarrow \sin^2 x + \sin x - 1 = 0$.
Using the quadratic formula $\sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $\sin x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.
Since $\sin x$ must be positive in the range where $\tan x$ and $\cos x$ are defined as equal,we take the positive root: $\sin x = \frac{\sqrt{5}-1}{2}$.

(C) We use the logarithmic forms of inverse hyperbolic functions:
$\sec h^{-1} x = \log _e\left(\frac{1+\sqrt{1-x^2}}{x}\right)$ and $\operatorname{cosec} h^{-1} x = \log _e\left(\frac{1+\sqrt{1+x^2}}{x}\right)$.
Substituting $x = \frac{1}{2}$ in $\sec h^{-1} x$:
$\sec h^{-1}\left(\frac{1}{2}\right) = \log _e\left(\frac{1+\sqrt{1-\frac{1}{4}}}{\frac{1}{2}}\right) = \log _e\left(\frac{1+\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) = \log _e(2+\sqrt{3})$.
Substituting $x = \frac{3}{4}$ in $\operatorname{cosec} h^{-1} x$:
$\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right) = \log _e\left(\frac{1+\sqrt{1+\frac{9}{16}}}{\frac{3}{4}}\right) = \log _e\left(\frac{1+\frac{5}{4}}{\frac{3}{4}}\right) = \log _e\left(\frac{\frac{9}{4}}{\frac{3}{4}}\right) = \log _e(3)$.
Therefore,$\sec h^{-1}\left(\frac{1}{2}\right) - \operatorname{cosec} h^{-1}\left(\frac{3}{4}\right) = \log _e(2+\sqrt{3}) - \log _e(3) = \log _e\left(\frac{2+\sqrt{3}}{3}\right)$.

(D) Given,$\cos x + \cos y = \frac{3}{2}$ and $\sin x + \sin y = \frac{3}{4}$.
Using sum-to-product formulas:
$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = \frac{3}{2}$ $(1)$
$2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = \frac{3}{4}$ $(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)} = \frac{3/4}{3/2}$
$\tan \left(\frac{x+y}{2}\right) = \frac{1}{2}$
Using the identity $\sin(x+y) = \frac{2 \tan \left(\frac{x+y}{2}\right)}{1 + \tan^2 \left(\frac{x+y}{2}\right)}$:
$\sin(x+y) = \frac{2 \times \frac{1}{2}}{1 + (\frac{1}{2})^2} = \frac{1}{1 + \frac{1}{4}} = \frac{1}{\frac{5}{4}} = \frac{4}{5}$.

(D) The given pair of lines is $x^2-3xy+y^2=0$. This represents two lines passing through the origin $(0,0)$.
Let the lines be $y=m_1x$ and $y=m_2x$. Then $m_1+m_2=3$ and $m_1m_2=1$.
The area of the triangle formed by the lines $ax^2+2hxy+by^2=0$ and $lx+my+n=0$ is given by the formula:
$\text{Area} = \frac{n^2\sqrt{h^2-ab}}{|am^2-2hlm+bl^2|}$.
Here,$a=1, h=-\frac{3}{2}, b=1, l=1, m=1, n=1$.
Substituting these values:
$\text{Area} = \frac{1^2\sqrt{(-\frac{3}{2})^2-(1)(1)}}{|(1)(1)^2-2(-\frac{3}{2})(1)(1)+(1)(1)^2|}$.
$\text{Area} = \frac{\sqrt{\frac{9}{4}-1}}{|1+3+1|} = \frac{\sqrt{\frac{5}{4}}}{5} = \frac{\frac{\sqrt{5}}{2}}{5} = \frac{\sqrt{5}}{10} = \frac{1}{2\sqrt{5}}$.

(B) The given equation is $x^2 + \alpha y^2 + 2 \beta y - a^2 = 0$.
Comparing this with the general equation of a pair of lines $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$,we get:
$A = 1, B = \alpha, H = 0, G = 0, F = \beta, C = -a^2$.
For the lines to be perpendicular,the condition is $A + B = 0$.
$1 + \alpha = 0 \Rightarrow \alpha = -1$.
For the equation to represent a pair of lines,the determinant condition must be satisfied:
$ABC + 2FGH - AF^2 - BG^2 - CH^2 = 0$.
Substituting the values:
$(1)(\alpha)(-a^2) + 2(\beta)(0)(0) - (1)(\beta)^2 - (\alpha)(0)^2 - (-a^2)(0)^2 = 0$.
$-\alpha a^2 - \beta^2 = 0$.
Since $\alpha = -1$,we have:
$-(-1)a^2 - \beta^2 = 0
$ $\Rightarrow a^2 - \beta^2 = 0
$ $\Rightarrow \beta^2 = a^2
$ $\Rightarrow \beta = a$ (taking the positive root as per options).

(C) The relative strengths of the four fundamental forces in nature,taking the strongest (strong nuclear force) as $1$,are as follows:
$1$. Strong nuclear force: Relative strength = $1$ (Matches with $f$)
$2$. Electromagnetic force: Relative strength = $10^{-2}$ (Matches with $e$)
$3$. Weak nuclear force: Relative strength = $10^{-13}$ (Matches with $h$)
$4$. Gravitational force: Relative strength = $10^{-39}$ (Matches with $i$)
Therefore,the correct matching is: $A-f, B-h, C-e, D-i$.