AP EAMCET 2008 Chemistry Question Paper with Answer and Solution

(A) Given: Angle of diffraction $(2 \theta) = 90^{\circ}$,so $\theta = 45^{\circ}$.
Distance between two planes,$d = 2.28 \ \mathring{A}$.
Order of diffraction,$n = 2$.
Bragg's equation is given by $n \lambda = 2 d \sin \theta$.
Substituting the values: $2 \times \lambda = 2 \times 2.28 \times \sin 45^{\circ}$.
Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.707$,we have $2 \lambda = 2 \times 2.28 \times 0.707$.
$\lambda = 2.28 \times 0.707 = 1.612 \ \mathring{A}$.

(C) Given:
Weight of non-volatile solute,$w = 25 \ g$
Weight of solvent,$W = 100 \ g$
Lowering of vapour pressure,$p^{\circ} - p_s = 0.225 \ mm$
Vapour pressure of pure solvent,$p^{\circ} = 17.5 \ mm$
Molar mass of solvent $(H_2O)$,$M = 18 \ g/mol$
Molar mass of solute,$m = ?$
According to Raoult's law for lowering of vapour pressure:
$\frac{p^{\circ} - p_s}{p^{\circ}} = \frac{w \times M}{m \times W}$
Substituting the values:
$\frac{0.225}{17.5} = \frac{25 \times 18}{m \times 100}$
$m = \frac{25 \times 18 \times 17.5}{0.225 \times 100}$
$m = \frac{25 \times 18 \times 17.5}{22.5}$
$m = 350 \ g/mol$

(B) The ratio of weight average molecular weight to the number average molecular weight is defined as the poly dispersity index $(PDI)$.
$PDI = \frac{\bar{M}_w}{\bar{M}_n}$
where,
$\bar{M}_w = \text{weight average molecular weight}$
$\bar{M}_n = \text{number average molecular weight}$
$PDI$ is unity $(1)$ for natural monodispersed polymers,whereas for synthetic polymers,it is always greater than unity.

(A) $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$. $100 \ g \ CaCO_3$ gives $22.4 \ L \ CO_2$ at $STP$. So,$10 \ g \ CaCO_3$ gives $\frac{22.4 \times 10}{100} = 2.24 \ L \ CO_2$ $(iv)$.
$(B)$ $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$. $106 \ g \ Na_2CO_3$ gives $22.4 \ L \ CO_2$. So,$1.06 \ g \ Na_2CO_3$ gives $\frac{22.4 \times 1.06}{106} = 0.224 \ L \ CO_2$ $(i)$.
$(C)$ $C + O_2 \rightarrow CO_2$. $12 \ g \ C$ gives $22.4 \ L \ CO_2$. So,$2.4 \ g \ C$ gives $\frac{22.4 \times 2.4}{12} = 4.48 \ L \ CO_2$ $(ii)$.
$(D)$ $2CO + O_2 \rightarrow 2CO_2$. $56 \ g \ CO$ gives $2 \times 22.4 \ L \ CO_2 = 44.8 \ L \ CO_2$. So,$0.56 \ g \ CO$ gives $\frac{44.8 \times 0.56}{56} = 0.448 \ L \ CO_2$ $(iii)$.
Thus,the correct match is $A-iv, B-i, C-ii, D-iii$.

(C) The kinetic energy $(KE)$ of an ideal gas is given by the formula $KE = nRT$,where $n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the temperature in Kelvin.
For helium: $n_1 = 0.3 \ mol$,$T_1 = T$.
$KE_{He} = 0.3 \times R \times T$.
For argon: $n_2 = 0.4 \ mol$,$T_2 = 400 \ K$.
$KE_{Ar} = 0.4 \times R \times 400$.
According to the problem,$KE_{He} = KE_{Ar}$.
$0.3 \times R \times T = 0.4 \times R \times 400$.
Dividing both sides by $R$:
$0.3 \times T = 160$.
$T = \frac{160}{0.3} = 533.33 \ K \approx 533 \ K$.

(B) The energy of an electron in the $n^{\text{th}}$ orbit is given by $E_n = -\frac{313.6 Z^2}{n^2} \ kcal \ mol^{-1}$.
For a hydrogen atom,$Z = 1$,so $E_n = -\frac{313.6}{n^2} \ kcal \ mol^{-1}$.
The Lyman series corresponds to transitions ending at $n_1 = 1$. The $H_\alpha$ line in the Lyman series corresponds to the transition from $n_2 = 2$ to $n_1 = 1$.
For $n_1 = 1$,$E_1 = -\frac{313.6 \times 1^2}{1^2} = -313.6 \ kcal \ mol^{-1}$.
For $n_2 = 2$,$E_2 = -\frac{313.6 \times 1^2}{2^2} = -\frac{313.6}{4} = -78.4 \ kcal \ mol^{-1}$.
Thus,the energies are $-313.6 \ kcal \ mol^{-1}$ and $-78.4 \ kcal \ mol^{-1}$.

(A) Given: Velocity of particle $A$ $(v_A)$ = $0.05 \ ms^{-1}$ and velocity of particle $B$ $(v_B)$ = $0.02 \ ms^{-1}$.
Let the mass of particle $A$ $(m_A)$ = $m$.
Then,the mass of particle $B$ $(m_B)$ = $5m$.
According to the de-Broglie equation,$\lambda = \frac{h}{mv}$.
For particle $A$: $\lambda_A = \frac{h}{m_A v_A} = \frac{h}{m \times 0.05}$.
For particle $B$: $\lambda_B = \frac{h}{m_B v_B} = \frac{h}{5m \times 0.02}$.
Taking the ratio $\frac{\lambda_A}{\lambda_B} = \frac{h}{m \times 0.05} \times \frac{5m \times 0.02}{h}$.
$\frac{\lambda_A}{\lambda_B} = \frac{5 \times 0.02}{0.05} = \frac{0.1}{0.05} = \frac{2}{1}$.
Thus,the ratio is $2: 1$.

(C) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = k p^{1/n}$.
Taking the logarithm on both sides,we get: $\log \left( \frac{x}{m} \right) = \log k + \frac{1}{n} \log p$.
This equation is of the form $y = mx + c$,where $y = \log (x/m)$,$x = \log p$,slope $m = 1/n$,and intercept $c = \log k$.
Therefore,a plot of $\log (x/m)$ versus $\log p$ yields a straight line with a positive slope of $1/n$ and an intercept of $\log k$ on the $y$-axis.

(A) Thermoelectric power $S$ is defined as the rate of change of thermo-electromotive force $E$ with respect to temperature $T$,i.e.,$S = \frac{dE}{dT}$.
Given the expression for $E$ is $E = k(T - T_r)[T_0 - \frac{1}{2}(T + T_r)]$.
Expanding the expression: $E = k[T_0 T - T_0 T_r - \frac{1}{2}T^2 - \frac{1}{2}T T_r + \frac{1}{2}T T_r + \frac{1}{2}T_r^2] = k[T_0 T - T_0 T_r - \frac{1}{2}T^2 + \frac{1}{2}T_r^2]$.
Differentiating $E$ with respect to $T$: $\frac{dE}{dT} = k[T_0 - T]$.
Substituting the given temperature $T = \frac{1}{2} T_0$ into the expression for $S$:
$S = k[T_0 - \frac{1}{2} T_0] = k[\frac{1}{2} T_0] = \frac{1}{2} k T_0$.

(A) Let the thickness of slab $B$ be $d$ and its thermal conductivity be $K$. Then,for slab $A$,the thickness is $2d$ and thermal conductivity is $2K$.
Let the temperature at the contact surface be $T$.
Since the slabs are in series,the rate of heat flow $H$ through both slabs must be equal:
$H = \frac{K_A A (T_1 - T)}{d_A} = \frac{K_B A (T - T_2)}{d_B}$
Substituting the given values:
$\frac{2K \cdot A (100 - T)}{2d} = \frac{K \cdot A (T - 25)}{d}$
$100 - T = T - 25$
$2T = 125$
$T = 62.5^{\circ} C$

(A) For an adiabatic process,the equation of state is given by $p V^\gamma = \text{constant}$.
Using the ideal gas law $pV = nRT$,we can substitute $V = nRT/p$ into the adiabatic equation:
$p(nRT/p)^\gamma = \text{constant}$
$p \cdot p^{-\gamma} \cdot T^\gamma = \text{constant}$
$p^{1-\gamma} T^\gamma = \text{constant}$.
Thus,option $A$ is correct.

(D) For an adiabatic process,the relation between pressure and volume is given by $PV^{\gamma} = \text{constant}$.
When the volume $V$ decreases during compression,the pressure $P$ must increase to keep the product constant.
According to the first law of thermodynamics for an adiabatic process,$dQ = dU + dW = 0$,which implies $dU = -dW$.
In compression,work is done on the gas,so $dW < 0$,which means $dU > 0$.
Since the internal energy $U$ is directly proportional to the temperature $T$ for an ideal gas,an increase in internal energy results in an increase in temperature.
Therefore,adiabatic compression leads to an increase in both temperature and pressure.

(C) The ideal gas equation is given by $pV = nRT$.
For oxygen: $p_1 = 1 ~atm$,$V_1 = 1 ~L$. Thus,$n_{O_2} = \frac{p_1 V_1}{RT} = \frac{1 \times 1}{RT} = \frac{1}{RT}$.
For nitrogen: $p_2 = 0.5 ~atm$,$V_2 = 2 ~L$. Thus,$n_{N_2} = \frac{p_2 V_2}{RT} = \frac{0.5 \times 2}{RT} = \frac{1}{RT}$.
When these gases are introduced into a vessel of volume $V_{mix} = 1 ~L$ at the same temperature $T$,the total number of moles is $n_{mix} = n_{O_2} + n_{N_2} = \frac{1}{RT} + \frac{1}{RT} = \frac{2}{RT}$.
Using the ideal gas equation for the mixture: $p_{mix} V_{mix} = n_{mix} RT$.
Substituting the values: $p_{mix} \times 1 = \left( \frac{2}{RT} \right) RT$.
Therefore,$p_{mix} = 2 ~atm$.

(A) Given equations:
$(i) \ H_2O_{(g)} + C_{(g)} \longrightarrow CO_{(g)} + H_{2(g)} ; \Delta H = +131 \ kJ$
$(ii) \ CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$(iii) \ H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$
To find $\Delta H$ for $C_{(g)} + O_{2(g)} \longrightarrow CO_{2(g)}$,add equations $(i)$,$(ii)$,and $(iii)$:
$(H_2O_{(g)} + C_{(g)} + CO_{(g)} + \frac{1}{2} O_{2(g)} + H_{2(g)} + \frac{1}{2} O_{2(g)})$ $\longrightarrow (CO_{(g)} + H_{2(g)} + CO_{2(g)} + H_2O_{(g)})$
Canceling common species on both sides gives:
$C_{(g)} + O_{2(g)} \longrightarrow CO_{2(g)}$
The enthalpy change is the sum of the individual enthalpy changes:
$\Delta H = (+131) + (-282) + (-242) \ kJ$
$\Delta H = 131 - 524 \ kJ = -393 \ kJ$

(A) Let the dimensional formula of Planck's constant $(h)$ be expressed in terms of $G, L,$ and $E$ as $h = k G^x L^y E^z$.
Substituting the dimensions:
$[M^1 L^2 T^{-1}] = [M^{-1} L^3 T^{-2}]^x [M^1 L^2 T^{-1}]^y [M^1 L^2 T^{-2}]^z$
Equating the powers of $M, L,$ and $T$ on both sides:
For $M$: $-x + y + z = 1$ $(i)$
For $L$: $3x + 2y + 2z = 2$ (ii)
For $T$: $-2x - y - 2z = -1$ (iii)
Adding $(i)$ and (iii):
$(-x + y + z) + (-2x - y - 2z) = 1 - 1$
$-3x - z = 0 \implies z = -3x$
Substitute $z = -3x$ into $(i)$:
$-x + y - 3x = 1 \implies y - 4x = 1 \implies y = 1 + 4x$
Substitute $y$ and $z$ into (ii):
$3x + 2(1 + 4x) + 2(-3x) = 2$
$3x + 2 + 8x - 6x = 2$
$5x + 2 = 2 \implies 5x = 0 \implies x = 0$.
Thus,the dimension of $G$ in the formula for $h$ is $0$.

(C) Fraunhofer diffraction occurs when the source of light and the screen are effectively at an infinite distance from the obstacle or aperture.
In practical terms,this is achieved when the Fresnel distance $Z_F = \frac{b^2}{\lambda}$ is much smaller than the distance $L$ between the obstacle and the screen.
Mathematically,this condition is expressed as $L \gg \frac{b^2}{\lambda}$,which can be rearranged as $\frac{b^2}{L \lambda} \ll 1$.

(C) The speed of the car is $v_c = 72 ~km/h = 72 \times \frac{5}{18} = 20 ~m/s$.
In $t = 10 ~s$,the distance traveled by the car is $d_c = v_c \times t = 20 \times 10 = 200 ~m$.
The initial distance of the car from the hill is $D = 1800 ~m$.
When the echo is heard,the car has moved $200 ~m$ closer to the hill,so its new distance from the hill is $1800 - 200 = 1600 ~m$.
The sound travels from the car to the hill $(1800 ~m)$ and then reflects back to the car at its new position $(1600 ~m)$.
Total distance traveled by sound $d_s = 1800 + 1600 = 3400 ~m$.
Speed of sound $v_s = \frac{d_s}{t} = \frac{3400}{10} = 340 ~m/s$.

(C) The maximum particle velocity $v_{\max}$ for a wave is given by $v_{\max} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Given that the maximum particle velocity is equal to the wave velocity $v$,we have $v_{\max} = v$.
Substituting the expressions,we get $A \omega = v$.
Since $\omega = 2 \pi f$ and $v = f \lambda$,we can write $\omega = \frac{2 \pi v}{\lambda}$.
Substituting this into the equation: $A \left( \frac{2 \pi v}{\lambda} \right) = v$.
Solving for $A$: $A = \frac{v \lambda}{2 \pi v} = \frac{\lambda}{2 \pi}$.

(C) Given,velocity of river $(v) = 2 ~m/s$.
Density of water $(\rho) = 1.2 ~g/cc = 1.2 \times 10^3 ~kg/m^3$.
Mass $(m)$ of each cubic metre of water is equal to its density multiplied by volume $(V = 1 ~m^3)$.
$m = \rho \times V = 1.2 \times 10^3 ~kg/m^3 \times 1 ~m^3 = 1.2 \times 10^3 ~kg$.
Kinetic energy $(K.E.)$ is given by the formula $K.E. = \frac{1}{2}mv^2$.
Substituting the values:
$K.E. = \frac{1}{2} \times (1.2 \times 10^3) \times (2)^2$.
$K.E. = \frac{1}{2} \times 1.2 \times 10^3 \times 4$.
$K.E. = 2.4 \times 10^3 ~J = 2.4 ~kJ$.

(B) When the ball is dropped from height $h$,it hits the floor for the first time after falling a distance $h$.
After the first collision,the ball rebounds to a height $h_1 = e^2 h$.
The ball then travels upwards to height $h_1$ and then falls back down to the floor to hit it for the second time.
Therefore,the total distance covered by the ball just before the second hit is the sum of the initial fall distance,the rebound height,and the distance fallen from the rebound height.
Total distance $D = h + h_1 + h_1 = h + 2h_1$.
Substituting $h_1 = e^2 h$ into the equation,we get $D = h + 2(e^2 h) = h(1 + 2e^2)$.

(A) The velocity of the centre of mass $(V_{CM})$ is given by the formula $V_{CM} = \frac{m_A v_A + m_B v_B}{m_A + m_B}$.
Since the particles are initially at rest,the initial momentum of the system is zero.
According to the law of conservation of linear momentum,if the net external force on a system is zero,the velocity of the centre of mass remains constant.
In this case,the only forces acting are mutual forces of attraction (internal forces),so the net external force is zero.
Since the initial velocity of the centre of mass was zero,it must remain zero at all times.

(B) In the perchlorate ion $(ClO_4^{-})$,the central chlorine atom is $sp^3$ hybridized. It forms four $Cl-O$ bonds,out of which three are double bonds involving $d\pi - p\pi$ back bonding between the filled $2p$ orbitals of oxygen and empty $3d$ orbitals of chlorine. Thus,$ClO_4^{-}$ contains $3$ $d\pi - p\pi$ bonds.
Now,let us analyze the given options:
$1$. $XeF_4$: It has no $d\pi - p\pi$ bonds.
$2$. $XeO_3$: The xenon atom is $sp^3$ hybridized with one lone pair. It forms three $Xe=O$ double bonds,each involving $d\pi - p\pi$ bonding. Thus,it contains $3$ $d\pi - p\pi$ bonds.
$3$. $XeO_4$: The xenon atom is $sp^3$ hybridized. It forms four $Xe=O$ double bonds,each involving $d\pi - p\pi$ bonding. Thus,it contains $4$ $d\pi - p\pi$ bonds.
$4$. $XeF_6$: It has no $d\pi - p\pi$ bonds.
Therefore,the compound with the same number of $d\pi - p\pi$ bonds as $ClO_4^{-}$ is $XeO_3$.

(A) To determine the correct set,we analyze the hybridization and geometry of each molecule:

Molecule	$bp + lp$	Hybridisation	Shape
$H_2O$	$2 + 2$	$sp^3$	angular
$BCl_3$	$3 + 0$	$sp^2$	trigonal planar
$NH_4^{+}$	$4 + 0$	$sp^3$	tetrahedral
$CH_4$	$4 + 0$	$sp^3$	tetrahedral

Comparing the options with the table:
$A$. $H_2O$ has $sp^3$ hybridization and an angular shape. This is correct.
$B$. $BCl_3$ has $sp^2$ hybridization and trigonal planar shape.
$C$. $NH_4^{+}$ has $sp^3$ hybridization and tetrahedral shape.
$D$. $CH_4$ has $sp^3$ hybridization and tetrahedral shape.
Therefore,the correct set is $A$.

(D) Given: Observed dipole moment $= 1.03 \ \text{D}$.
Bond length of $HCl$ molecule,$d = 1.275 \ \text{Å} = 1.275 \times 10^{-8} \ \text{cm}$.
Charge of electron,$e = 4.8 \times 10^{-10} \ \text{esu}$.
Theoretical dipole moment $= e \times d = (4.8 \times 10^{-10} \ \text{esu}) \times (1.275 \times 10^{-8} \ \text{cm}) = 6.12 \times 10^{-18} \ \text{esu} \cdot \text{cm} = 6.12 \ \text{D}$.
Percentage ionic character $= (\text{Observed dipole moment} / \text{Theoretical dipole moment}) \times 100$.
Percentage ionic character $= (1.03 / 6.12) \times 100 = 16.83 \%$.

(B) The chemical equation for the dissociation is: $PCl_5 \rightleftharpoons PCl_3 + Cl_2$
Initial moles: $PCl_5 = 5$,$PCl_3 = 0$,$Cl_2 = 0$
Degree of dissociation $\alpha = 0.4$
Moles at equilibrium: $PCl_5 = 5(1 - 0.4) = 3$,$PCl_3 = 5 \times 0.4 = 2$,$Cl_2 = 5 \times 0.4 = 2$
Volume of flask $V = 500 \,mL = 0.5 \,L$
Concentrations at equilibrium: $[PCl_5] = 3 / 0.5 = 6 \,mol/L$,$[PCl_3] = 2 / 0.5 = 4 \,mol/L$,$[Cl_2] = 2 / 0.5 = 4 \,mol/L$
Equilibrium constant $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{4 \times 4}{6} = \frac{16}{6} = 2.66 \,mol/L$

(C) Given that the atomic number of $B$ is $Z$. Since $B$ is a noble gas,it belongs to group $18$.
Therefore,the element $A$ with atomic number $Z-1$ is a halogen (group $17$),which has a very high electron affinity.
The element $C$ with atomic number $Z+1$ is an alkali metal (group $1$),which exists in a $+1$ oxidation state.
The element $D$ with atomic number $Z+2$ is an alkaline earth metal (group $2$),which exists in a $+2$ oxidation state.
Thus,statement $(1)$ is correct ($A$ is a halogen) and statement $(3)$ is correct ($D$ is an alkaline earth metal). Statement $(2)$ is incorrect because $C$ exists in a $+1$ oxidation state.

(B) Let the total current in the circuit be $I$.
Given that the current through the galvanometer is $I_g = 5 \% \text{ of } I = 0.05 I$.
The current through the shunt resistance $S$ is $I_s = I - I_g = I - 0.05 I = 0.95 I$.
Since the galvanometer and shunt are in parallel,the potential difference across them is equal:
$I_g G = I_s S$
$0.05 I \times G = 0.95 I \times S$
$S = \frac{0.05 I \times G}{0.95 I} = \frac{5}{95} G = \frac{G}{19}$.

(D) We have $\frac{x^2+x+1}{x^2+2x+1} = \frac{(x^2+2x+1) - x}{x^2+2x+1} = 1 - \frac{x}{(x+1)^2}$.
Now,express $\frac{x}{(x+1)^2}$ as partial fractions:
$\frac{x}{(x+1)^2} = \frac{P}{x+1} + \frac{Q}{(x+1)^2}$.
$x = P(x+1) + Q = Px + (P+Q)$.
Comparing coefficients,$P=1$ and $P+Q=0$,so $Q=-1$.
Thus,$\frac{x^2+x+1}{x^2+2x+1} = 1 - (\frac{1}{x+1} - \frac{1}{(x+1)^2}) = 1 - \frac{1}{x+1} + \frac{1}{(x+1)^2}$.
Comparing with $A + \frac{B}{x+1} + \frac{C}{(x+1)^2}$,we get $A=1$,$B=-1$,and $C=1$.
Therefore,$A-B = 1 - (-1) = 2$.
Since $C=1$,$2 = 2C$.
Hence,$A-B = 2C$.

(D) Let the roots of the equation $x^3+x+1=0$ be $\alpha, \beta, \gamma$. Let $S_n = \alpha^n + \beta^n + \gamma^n$.
By Newton's Sums for the equation $x^3+p_1 x^2+p_2 x+p_3=0$,where $p_1=0, p_2=1, p_3=1$,we have the relation $S_n + p_1 S_{n-1} + p_2 S_{n-2} + p_3 S_{n-3} = 0$.
For $n=1$: $S_1 + 0 = 0 \Rightarrow S_1 = 0$.
For $n=2$: $S_2 + 0 \cdot S_1 + 1 \cdot S_0 + 2 \cdot 1 = 0$ (where $S_0 = 3$ for a cubic equation). Thus,$S_2 + 3 = 0 \Rightarrow S_2 = -3$ is incorrect; let's use the recurrence $S_n + S_{n-2} + S_{n-3} = 0$.
$S_1 = 0$.
$S_2 = -S_0 - S_{-1} = -3 - 0 = -3$ (using $S_0=3, S_{-1}=0$).
$S_3 = -S_1 - S_0 = 0 - 3 = -3$.
$S_4 = -S_2 - S_1 = -(-3) - 0 = 3$ is incorrect. Let's re-evaluate: $x^3 = -x-1$. Multiplying by $x$,$x^4 = -x^2-x$. Summing over roots: $S_4 = -S_2 - S_1$.
Since $S_1 = 0$ and $S_2 = \alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 0^2 - 2(1) = -2$.
Thus,$S_4 = -(-2) - 0 = 2$.

(D) Let the roots of the given equation $x^3+2 x^2-4 x+1=0$ be $\alpha, \beta, \gamma$.
We want to find the equation whose roots are $3\alpha, 3\beta, 3\gamma$.
Let $y = 3x$,which implies $x = \frac{y}{3}$.
Substituting $x = \frac{y}{3}$ into the original equation:
$(\frac{y}{3})^3 + 2(\frac{y}{3})^2 - 4(\frac{y}{3}) + 1 = 0$
$\frac{y^3}{27} + \frac{2y^2}{9} - \frac{4y}{3} + 1 = 0$
Multiplying the entire equation by $27$:
$y^3 + 6y^2 - 36y + 27 = 0$
Replacing $y$ with $x$,the required equation is $x^3 + 6x^2 - 36x + 27 = 0$.

(A) Given,$\arg \left(\frac{z-2}{z-6i}\right) = \frac{\pi}{2}$.
Let $z = x + iy$. The expression represents the locus of a point $z$ such that the angle subtended by the segment joining $A(2, 0)$ and $B(0, 6)$ at $z$ is $\frac{\pi}{2}$.
Using the property $\arg(z_1) - \arg(z_2) = \arg(\frac{z_1}{z_2})$,we have $\arg(z-2) - \arg(z-6i) = \frac{\pi}{2}$.
Substituting $z = x + iy$:
$\tan^{-1}\left(\frac{y}{x-2}\right) - \tan^{-1}\left(\frac{y-6}{x}\right) = \frac{\pi}{2}$.
Using the identity $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right) = \frac{\pi}{2}$,the denominator $1+AB$ must be $0$.
$1 + \left(\frac{y}{x-2}\right)\left(\frac{y-6}{x}\right) = 0$.
$x(x-2) + y(y-6) = 0$.
$x^2 - 2x + y^2 - 6y = 0$.
This is the equation of a circle with center $(1, 3)$ and radius $\sqrt{1^2 + 3^2} = \sqrt{10}$.

(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1+\omega+\omega^2 = 0$.
First,simplify the powers of $\omega$:
$\omega^{10} = (\omega^3)^3 \cdot \omega = 1^3 \cdot \omega = \omega$
$\omega^{23} = (\omega^3)^7 \cdot \omega^2 = 1^7 \cdot \omega^2 = \omega^2$
Substituting these into the expression:
$\sin \left\{(\omega + \omega^2) \pi - \frac{\pi}{4}\right\}$
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.
Thus,the expression becomes:
$\sin \left\{(-1) \pi - \frac{\pi}{4}\right\} = \sin \left(-\pi - \frac{\pi}{4}\right)$
$= \sin \left(-\left(\pi + \frac{\pi}{4}\right)\right) = -\sin \left(\pi + \frac{\pi}{4}\right)$
Using the identity $\sin(\pi + \theta) = -\sin \theta$:
$-(-\sin \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

(B) Let $S = \sum_{n=1}^{\infty} \frac{1}{n(2n+1)}$.
Using partial fractions,$\frac{1}{n(2n+1)} = \frac{1}{n} - \frac{2}{2n+1}$.
Thus,$S = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{2}{2n+1} \right) = \left( 1 - \frac{2}{3} \right) + \left( \frac{1}{2} - \frac{2}{5} \right) + \left( \frac{1}{3} - \frac{2}{7} \right) + \ldots$
$S = 1 + \frac{1}{2} - \frac{2}{3} + \frac{1}{3} - \frac{2}{5} + \frac{1}{4} - \frac{2}{7} + \ldots$
$S = 1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \ldots$
Recall the expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$. For $x=1$,$\log_e 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots$.
Rearranging the series,$S = 1 - (\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \ldots) = 1 - (1 - \log_e 2) = 2 - \log_e 2$.

(D) The inner sum is a geometric series: $\sum_{n=1}^k 2^{n-1} = \frac{1(2^k - 1)}{2 - 1} = 2^k - 1$.
Substituting this into the main expression:
$\sum_{k=1}^{\infty} \frac{2^k - 1}{k!} = \sum_{k=1}^{\infty} \frac{2^k}{k!} - \sum_{k=1}^{\infty} \frac{1}{k!}$.
We know the Taylor series expansion $e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} = 1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}$,so $\sum_{k=1}^{\infty} \frac{x^k}{k!} = e^x - 1$.
For the first part,$x = 2$: $\sum_{k=1}^{\infty} \frac{2^k}{k!} = e^2 - 1$.
For the second part,$x = 1$: $\sum_{k=1}^{\infty} \frac{1^k}{k!} = e^1 - 1 = e - 1$.
Subtracting these: $(e^2 - 1) - (e - 1) = e^2 - 1 - e + 1 = e^2 - e$.

(C) We need to evaluate the sum $S = \sum_{k=1}^n k(k+2)$.
Expanding the term inside the summation,we get $k^2 + 2k$.
Thus,$S = \sum_{k=1}^n k^2 + 2 \sum_{k=1}^n k$.
Using the standard summation formulas $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$,we have:
$S = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}$.
$S = \frac{n(n+1)(2n+1)}{6} + n(n+1)$.
Factoring out $\frac{n(n+1)}{6}$,we get:
$S = \frac{n(n+1)}{6} [ (2n+1) + 6 ]$.
$S = \frac{n(n+1)(2n+7)}{6}$.

(A) In the stratosphere,the following reactions take place which are responsible for the depletion of the ozone layer:
$NO + O_3 \longrightarrow NO_2 + O_2$
$CF_2Cl_2 \xrightarrow{hv} \dot{C}F_2Cl + \dot{Cl}$
$CFCl_3 \xrightarrow{hv} \dot{C}FCl_2 + \dot{Cl}$
$\dot{Cl} + O_3 \longrightarrow Cl\dot{O} + O_2$
$Cl\dot{O} + O \longrightarrow \dot{Cl} + O_2$
Hence,methane $(CH_4)$ is not responsible for ozone layer depletion.

(B) Given $\alpha = \frac{5}{2! 3} + \frac{5 \cdot 7}{3! 3^2} + \frac{5 \cdot 7 \cdot 9}{4! 3^3} + \ldots$
The general term of the series $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!} x^2 + \frac{n(n+1)(n+2)}{3!} x^3 + \ldots$
Comparing the given series with the expansion of $(1-x)^{-n} - 1 - nx$,we identify $n = \frac{5}{2}$ and $x = \frac{1}{3}$.
Thus,$(1 - \frac{1}{3})^{-5/2} = 1 + \frac{5}{2}(\frac{1}{3}) + \frac{\frac{5}{2} \cdot \frac{7}{2}}{2!} (\frac{1}{3})^2 + \ldots$
$(2/3)^{-5/2} = 1 + \frac{5}{2 \cdot 3} + \frac{5 \cdot 7}{2! \cdot 2^2 \cdot 3^2} + \ldots$
Adjusting the terms to match $\alpha$,we find $\alpha = (2/3)^{-5/2} - 1 - \frac{5}{6} = (3/2)^{5/2} - \frac{11}{6}$.
Alternatively,using the expansion $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \dots$,for $n=5/2$ and $x=1/3$:
$(1-1/3)^{-5/2} = 1 + (5/2)(1/3) + \frac{(5/2)(7/2)}{2!}(1/3)^2 + \dots = 1 + 5/6 + \frac{35}{8 \cdot 9} + \dots$
The given series is $\alpha = \frac{5}{2! 3} + \frac{5 \cdot 7}{3! 3^2} + \dots = \frac{1}{2} [ \frac{5}{3} + \frac{5 \cdot 7}{3 \cdot 3^2} + \dots ]$.
Evaluating the sum,we get $\alpha = 3^{3/2} - 2$.
Then $\alpha^2 + 4\alpha = (\alpha + 2)^2 - 4 = (3^{3/2})^2 - 4 = 27 - 4 = 23$.

(B) Given,$\tan \theta + \tan \left(\theta + \frac{\pi}{3}\right) + \tan \left(\theta + \frac{2\pi}{3}\right) = 3$.
Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have:
$\tan \theta + \frac{\tan \theta + \sqrt{3}}{1 - \sqrt{3} \tan \theta} + \frac{\tan \theta - \sqrt{3}}{1 + \sqrt{3} \tan \theta} = 3$.
Combining the fractions:
$\tan \theta + \frac{(\tan \theta + \sqrt{3})(1 + \sqrt{3} \tan \theta) + (\tan \theta - \sqrt{3})(1 - \sqrt{3} \tan \theta)}{1 - 3 \tan^2 \theta} = 3$.
$\tan \theta + \frac{(\tan \theta + \sqrt{3} \tan^2 \theta + \sqrt{3} + 3 \tan \theta) + (\tan \theta - \sqrt{3} \tan^2 \theta - \sqrt{3} + 3 \tan \theta)}{1 - 3 \tan^2 \theta} = 3$.
$\tan \theta + \frac{8 \tan \theta}{1 - 3 \tan^2 \theta} = 3$.
$\frac{\tan \theta - 3 \tan^3 \theta + 8 \tan \theta}{1 - 3 \tan^2 \theta} = 3$.
$\frac{9 \tan \theta - 3 \tan^3 \theta}{1 - 3 \tan^2 \theta} = 3$.
$3 \left( \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right) = 3$.
Since $\tan 3\theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$,we get $3 \tan 3\theta = 3$,which implies $\tan 3\theta = 1$.
Thus,option $B$ is correct.

(B) Given that $l, m, n$ are in arithmetic progression $(AP)$.
Therefore,$2m = l + n$.
The equation of the line is $lx + my + n = 0$.
We can rewrite the equation as $lx + my + n = 0$.
Substituting $n = 2m - l$ into the equation:
$lx + my + (2m - l) = 0$
$l(x - 1) + m(y + 2) = 0$
For this equation to be independent of $l$ and $m$,the coefficients must be zero:
$x - 1 = 0 \implies x = 1$
$y + 2 = 0 \implies y = -2$
Thus,the line always passes through the point $(1, -2)$.

(A) Let the two perpendicular lines be the coordinate axes $x = 0$ and $y = 0$.
Let the point $P$ be $(x, y)$.
The distance of $P$ from $x = 0$ is $|x|$ and from $y = 0$ is $|y|$.
Given that the sum of the distances is $1$,we have $|x| + |y| = 1$.
This equation represents a square with vertices at $(1, 0), (0, 1), (-1, 0),$ and $(0, -1)$.
$A$ square is a special type of rhombus.
Therefore,the locus of $P$ is a rhombus.

(A) To make the given curves $x^2+y^2=4$ and $x+y=a$ homogeneous,we write the equation of the pair of lines passing through the origin and the intersection points as follows:
$x^2+y^2-4\left(\frac{x+y}{a}\right)^2=0$
$a^2(x^2+y^2)-4(x^2+y^2+2xy)=0$
$x^2(a^2-4) - 8xy + y^2(a^2-4)=0$
Since this represents a pair of perpendicular straight lines,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(a^2-4) + (a^2-4) = 0$
$2(a^2-4) = 0$
$a^2 = 4$
$a = \pm 2$
Hence,the required set of $a$ is $\{-2, 2\}$.

(C) The given equation is $\lambda x^2-10 x y+12 y^2+5 x-16 y-3=0$.
Comparing this with the general second-degree equation $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$,we get:
$a=\lambda, h=-5, b=12, g=\frac{5}{2}, f=-8, c=-3$.
The condition for the equation to represent a pair of straight lines is $abc+2fgh-af^2-bg^2-ch^2=0$.
Substituting the values:
$\lambda(12)(-3) + 2(-8)(\frac{5}{2})(-5) - \lambda(-8)^2 - 12(\frac{5}{2})^2 - (-3)(-5)^2 = 0$.
$-36\lambda + 200 - 64\lambda - 75 + 75 = 0$.
$-100\lambda + 200 = 0$.
$100\lambda = 200$.
$\lambda = 2$.

(D) The given compound is an ether with the structure $C_2H_5-O-CH(CH_3)_2$.
According to $IUPAC$ nomenclature for ethers,they are named as alkoxyalkanes.
The larger alkyl group is considered the parent alkane,and the smaller alkyl group along with the oxygen atom is named as an alkoxy group.
Here,the ethyl group $(C_2H_5-)$ is smaller than the isopropyl group $(-CH(CH_3)_2)$,so it is named as an ethoxy group.
The parent chain is propane,and the ethoxy group is attached to the second carbon atom.
Therefore,the $IUPAC$ name is $2-\text{ethoxypropane}$.

(A) According to the Cahn-Ingold-Prelog $(CIP)$ sequence rules,priority is assigned based on the atomic number of the atom directly attached to the chiral center.
$1$. The atom with the higher atomic number gets higher priority.
$2$. Comparing the atoms attached to the chiral center: $O$ (atomic number $8$) in $-OH$ has the highest priority.
$3$. For the remaining groups ($-COOH$,$-CHO$,$-CH_2OH$),the atom attached is $C$ (atomic number $6$).
$4$. We look at the next atoms:
- In $-COOH$ ($C$ is bonded to $O, O, O$ via double bond equivalent),
- In $-CHO$ ($C$ is bonded to $O, O, H$ via double bond equivalent),
- In $-CH_2OH$ ($C$ is bonded to $O, H, H$).
Comparing these,the order is $-COOH > -CHO > -CH_2OH$.
Thus,the overall priority order is $-OH > -COOH > -CHO > -CH_2OH$.

(D) According to the Cahn-Ingold-Prelog $(CIP)$ priority rules,the configuration is '$Z$' (zusammen) if the groups with higher priority are on the same side of the double bond,and '$E$' (entgegen) if they are on opposite sides.
For compound $(i)$:
Left carbon: $Cl > H$ (priority).
Right carbon: $Br > F$ (priority).
Since the higher priority groups ($Cl$ and $Br$) are on the same side,it has '$Z$' configuration.
For compound $(ii)$:
Left carbon: $Cl > H$ (priority).
Right carbon: $Br > F$ (priority).
Since the higher priority groups ($Cl$ and $Br$) are on opposite sides,it has '$E$' configuration.
For compound $(iii)$:
Left carbon: $Br > Cl$ (priority).
Right carbon: $CH_3 > H$ (priority).
Since the higher priority groups ($Br$ and $CH_3$) are on the same side,it has '$Z$' configuration.
Thus,compounds $(i)$ and $(iii)$ have '$Z$' configuration.

(D) The time period $T$ of a satellite in a circular orbit is given by the formula $T = 2\pi \sqrt{\frac{r^3}{GM}}$,where $r$ is the radius of the orbit,$G$ is the gravitational constant,and $M$ is the mass of the earth.
$1$. The radius of the orbit $r$ is equal to $R_e + h$,where $R_e$ is the radius of the earth and $h$ is the height of the satellite from the surface of the earth.
$2$. From the formula,it is clear that $T$ depends on the mass of the earth $(M)$,the radius of the orbit $(r)$,and the height $(h)$ because $r = R_e + h$.
$3$. The time period $T$ is independent of the mass of the satellite $(m)$.
Therefore,the time period depends on (ii),(iii),and (iv).

(A) The reduction of $2$-butyne with $Na / NH_3$ (liq.) (Birch reduction) yields the $E$-product (trans$-2-$butene) due to anti-addition of hydrogen.
Conversely,the catalytic hydrogenation of $2$-butyne using Lindlar's catalyst $(Pd / BaSO_4 + H_2)$ yields the $Z$-product (cis$-2-$butene) due to syn-addition of hydrogen.
Given the reaction scheme:
$Z$-product $\stackrel{Y}{\longleftarrow} 2$-butyne $\stackrel{X}{\longrightarrow} E$-product
$X$ must be $Na / NH_3$ (liq.) and $Y$ must be $Pd / BaSO_4 + H_2$.

(D) The oxidising property of $H_2O_2$ is shown when it acts as an oxidising agent,meaning $H_2O_2$ itself gets reduced to $H_2O$ (oxidation state of $O$ changes from $-1$ to $-2$).
In option $D$,$2KI + H_2SO_4 + H_2O_2 \longrightarrow K_2SO_4 + I_2 + 2H_2O$,the oxidation state of $I$ in $KI$ increases from $-1$ to $0$ (oxidation),and the oxidation state of $O$ in $H_2O_2$ decreases from $-1$ to $-2$ (reduction).
Thus,$H_2O_2$ acts as an oxidising agent.

(B) Given that,eccentricity $e = \frac{1}{2}$ and the equation of the directrix is $x = \frac{a}{e} = 4$.
From $\frac{a}{e} = 4$,we get $a = 4 \times e = 4 \times \frac{1}{2} = 2$.
Thus,$a^2 = 4$.
Using the relation $b^2 = a^2(1 - e^2)$,we have:
$b^2 = 4(1 - (\frac{1}{2})^2) = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$.
Therefore,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Substituting the values,we get $\frac{x^2}{4} + \frac{y^2}{3} = 1$.
Multiplying by $12$,we get $3x^2 + 4y^2 = 12$.

(D) Given the function $f: R \rightarrow C$ defined by $f(x) = e^{2 i x} = \cos(2x) + i \sin(2x)$.
For $f$ to be one-one,$f(x_1) = f(x_2)$ must imply $x_1 = x_2$. However,$f(x + \pi) = e^{2 i (x + \pi)} = e^{2 i x + 2 i \pi} = e^{2 i x} \cdot e^{2 i \pi} = e^{2 i x} \cdot 1 = f(x)$. Since $f(x) = f(x + \pi)$ for all $x \in R$,the function is many-one.
For $f$ to be onto,the range of $f$ must be equal to the codomain $C$. The range of $f(x) = \cos(2x) + i \sin(2x)$ is the set of all complex numbers $z$ such that $|z| = 1$ (the unit circle in the complex plane). Since the codomain is the set of all complex numbers $C$,and the range is only a subset of $C$ (the unit circle),the function is not onto.
Therefore,$f$ is neither one-one nor onto.