AP EAMCET 2007 Chemistry Question Paper with Answer and Solution

(A) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a double convex lens,the radii of curvature are $R_1 = R$ and $R_2 = -R$.
Given $f = 5 ~cm$ and $\mu = 1.5$.
Substituting these values into the formula:
$\frac{1}{5} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right)$
$\frac{1}{5} = 0.5 \times \left( \frac{1}{R} + \frac{1}{R} \right)$
$\frac{1}{5} = 0.5 \times \frac{2}{R}$
$\frac{1}{5} = \frac{1}{R}$
Therefore,$R = 5 ~cm$.

(A) In an $n$-type semiconductor,the majority charge carriers are electrons,which are provided by donor impurity atoms.
These donor energy levels are located just below the conduction band.
As the temperature increases,these electrons are excited into the conduction band.
Consequently,the Fermi energy level shifts upward and lies in the forbidden energy gap,but closer to the conduction band.

(A) Given equation: $xy = ae^x + be^{-x}$ $(1)$
Differentiating with respect to $x$ using the product rule on the left side:
$x \frac{dy}{dx} + y = ae^x - be^{-x}$ $(2)$
Differentiating again with respect to $x$:
$x \frac{d^2 y}{dx^2} + \frac{dy}{dx} + \frac{dy}{dx} = ae^x + be^{-x}$
$x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} = ae^x + be^{-x}$
From equation $(1)$,we know that $ae^x + be^{-x} = xy$. Substituting this into the equation:
$x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} = xy$
Rearranging the terms,we get:
$x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} - xy = 0$

(B) Given the homogeneous differential equation: $\frac{dy}{dx} = \frac{y^2}{xy - x^2}$.
Divide numerator and denominator by $x^2$: $\frac{dy}{dx} = \frac{(y/x)^2}{(y/x) - 1}$.
Let $v = y/x$,then $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{v^2}{v - 1}$.
$x\frac{dv}{dx} = \frac{v^2}{v - 1} - v = \frac{v^2 - v^2 + v}{v - 1} = \frac{v}{v - 1}$.
Separating variables: $\frac{v - 1}{v} dv = \frac{dx}{x}$.
$(1 - \frac{1}{v}) dv = \frac{dx}{x}$.
Integrating both sides: $\int (1 - \frac{1}{v}) dv = \int \frac{dx}{x}$.
$v - \ln|v| = \ln|x| + C$.
$v = \ln|x| + \ln|v| + C = \ln|xv| + C = \ln|y| + C$.
Since $v = y/x$,we have $y/x = \ln|y| + C$.
Taking exponential on both sides: $e^{y/x} = e^{\ln|y| + C} = e^C \cdot y$.
Let $e^C = k$,then $e^{y/x} = ky$.

(C) Total ways to choose $4$ numbers from $40$ is $^{40}C_4 = \frac{40 \times 39 \times 38 \times 37}{4 \times 3 \times 2 \times 1} = 91390$.
Number of ways to choose $4$ consecutive numbers is $37$ (i.e.,$\{1,2,3,4\}, \{2,3,4,5\}, \ldots, \{37,38,39,40\}$).
Probability that the $4$ numbers are consecutive is $P(C) = \frac{37}{91390} = \frac{1}{2470}$.
Probability that they are not consecutive is $1 - P(C) = 1 - \frac{1}{2470} = \frac{2469}{2470}$.

(D) Total number of balls $= 6 + 4 = 10$.
Number of ways to select $2$ balls out of $10 = {}^{10}C_2 = \frac{10 \times 9}{2 \times 1} = 45$.
Number of ways to select $2$ white balls from $6 = {}^{6}C_2 = \frac{6 \times 5}{2 \times 1} = 15$.
Number of ways to select $2$ black balls from $4 = {}^{4}C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
Total favorable outcomes $= 15 + 6 = 21$.
$\therefore$ The required probability $= \frac{21}{45} = \frac{7}{15}$.

(D) Given that $A$ and $B$ are mutually exclusive events,so $A \cap B = \phi$,which implies $P(A \cap B) = 0$.
By the definition of conditional probability,$P(A \mid \bar{B}) = \frac{P(A \cap \bar{B})}{P(\bar{B})}$.
Since $A = (A \cap B) \cup (A \cap \bar{B})$ and these are disjoint,$P(A) = P(A \cap B) + P(A \cap \bar{B})$.
Thus,$P(A \cap \bar{B}) = P(A) - P(A \cap B) = P(A) - 0 = P(A)$.
Also,$P(\bar{B}) = 1 - P(B)$.
Therefore,$P(A \mid \bar{B}) = \frac{P(A)}{1 - P(B)}$.

(B) For a binomial distribution,the mean is given by $np = 4$ and the variance is $\sigma^2 = npq = (\sqrt{3})^2 = 3$.
Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{3}{4}$,which implies $q = \frac{3}{4}$.
Since $p + q = 1$,we have $p = 1 - \frac{3}{4} = \frac{1}{4}$.
Substituting $p = \frac{1}{4}$ into $np = 4$,we get $n \times \frac{1}{4} = 4$,so $n = 16$.
We need to find $P(X \geq 1)$. Using the complement rule,$P(X \geq 1) = 1 - P(X = 0)$.
For a binomial distribution,$P(X = k) = \binom{n}{k} p^k q^{n-k}$.
Thus,$P(X = 0) = \binom{16}{0} (\frac{1}{4})^0 (\frac{3}{4})^{16} = (\frac{3}{4})^{16}$.
Therefore,$P(X \geq 1) = 1 - (\frac{3}{4})^{16}$.

(C) By definition,$1$ mole of any substance contains $6.022 \times 10^{23}$ constituent particles (Avogadro's number). Thus,equal moles of different substances contain the same number of constituent particles. So,Assertion $(A)$ is true.
However,the number of particles in a given weight depends on the molar mass of the substance $(n = \frac{w}{M})$. Since different substances have different molar masses,equal weights of different substances will contain different numbers of moles and,consequently,different numbers of constituent particles. So,Reason $(R)$ is false.

(A) Number of moles of ethyl alcohol $(C_2H_5OH)$ $= \frac{138 \ g}{46 \ g/mol} = 3 \ mol$.
Number of moles of water $(H_2O)$ $= \frac{72 \ g}{18 \ g/mol} = 4 \ mol$.
Mole fraction of alcohol $(X_{C_2H_5OH})$ $= \frac{3}{3+4} = \frac{3}{7}$.
Mole fraction of water $(X_{H_2O})$ $= \frac{4}{3+4} = \frac{4}{7}$.
The ratio of mole fraction of alcohol to water is $\frac{X_{C_2H_5OH}}{X_{H_2O}} = \frac{3/7}{4/7} = \frac{3}{4}$.

(C) According to Charles's Law,at constant pressure,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 2 \ L$,$T_1 = 273 \ K$ (at $STP$),$V_2 = 4 \ L$ (doubled volume).
Substituting the values: $\frac{2}{273} = \frac{4}{T_2}$.
$T_2 = \frac{273 \times 4}{2} = 546 \ K$.
Converting to Celsius: $546 - 273 = 273^{\circ} C$.

(C) For the Lyman series,the Rydberg formula is given by: $\frac{1}{\lambda} = R \left[ \frac{1}{1^2} - \frac{1}{n_2^2} \right]$.
Given $\lambda = \frac{16}{15 R}$,so $\frac{1}{\lambda} = \frac{15 R}{16}$.
Substituting this into the formula: $\frac{15 R}{16} = R \left[ 1 - \frac{1}{n_2^2} \right]$.
Dividing both sides by $R$: $\frac{15}{16} = 1 - \frac{1}{n_2^2}$.
Rearranging the terms: $\frac{1}{n_2^2} = 1 - \frac{15}{16} = \frac{1}{16}$.
Therefore,$n_2^2 = 16$,which gives $n_2 = 4$.

(C) For $N$ shell,the principal quantum number $(n) = 4$.
The number of sub-levels (sub-shells) is equal to $n$,which is $4$.
The number of orbitals is given by $n^2 = 4^2 = 16$.
The number of electrons is given by $2n^2 = 2 \times 4^2 = 32$.
Therefore,the values are $4, 16, 32$.

(D) Enzymes are biological catalysts. They are highly specific in nature and catalyze various biological reactions within living organisms.

(C) catalyst increases the rate of a reaction by providing an alternative pathway with lower activation energy.
Therefore,the assertion $(A)$ is true because the rate of reaction increases.
However,the reason $(R)$ is false because the activation energy of the reaction decreases in the presence of a catalyst,not increases.

(D) According to Stefan-Boltzmann law,the radiant energy $E$ emitted by a black body is directly proportional to the fourth power of its absolute temperature $T$.
Mathematically,$E \propto T^4$.
Let $E_1 = E$ at temperature $T_1 = T$.
Let $E_2$ be the radiant energy at temperature $T_2 = \frac{T}{2}$.
Using the ratio formula: $\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4$.
Substituting the values: $\frac{E_2}{E} = \left(\frac{T/2}{T}\right)^4 = \left(\frac{1}{2}\right)^4$.
Therefore,$E_2 = \frac{E}{16}$.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
Taking the differential,the fractional change in time period is $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l} = \frac{1}{2} \alpha \Delta \theta$.
Given $\alpha = 12 \times 10^{-6} /{ }^{\circ} C$ and $\Delta \theta = 40^{\circ} C - 20^{\circ} C = 20^{\circ} C$.
Substituting the values: $\frac{\Delta T}{T} = \frac{1}{2} \times 12 \times 10^{-6} \times 20 = 120 \times 10^{-6} = 1.2 \times 10^{-4}$.
The time lost or gained per day is $\Delta T = \frac{\Delta T}{T} \times T_{day}$.
Since $T_{day} = 24 \times 60 \times 60 = 86400 \ s$,we have $\Delta T = 1.2 \times 10^{-4} \times 86400 = 10.368 \ s$.
Rounding to the nearest value,the clock loses $10.3 \ s/\text{day}$.

(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,$n C_P \Delta T_A = n C_V \Delta T_B$.
$\frac{5}{2}R \times 42 = \frac{3}{2}R \times \Delta T_B$.
$5 \times 42 = 3 \times \Delta T_B$.
$\Delta T_B = \frac{210}{3} = 70 ~K$.

(D) For cylinder $B$ (fixed piston,isochoric process): $\Delta Q = n C_v \Delta T_B = n (\frac{3R}{2}) \Delta T_B$.
For cylinder $A$ (free piston,isobaric process): $\Delta Q = n C_p \Delta T_A = n (\frac{5R}{2}) \Delta T_A$.
Since the heat supplied $\Delta Q$ is the same for both: $n (\frac{3R}{2}) \Delta T_B = n (\frac{5R}{2}) \Delta T_A$.
Given $\Delta T_A = 42 ~K$,we have: $3 \Delta T_B = 5 \times 42$.
$\Delta T_B = \frac{210}{3} = 70 ~K$.

(B) The relationship between the temperature of inversion $(T_i)$,neutral temperature $(T_n)$,and the temperature of the cold junction $(T_0)$ in a thermocouple is given by the formula:
$T_n = \frac{T_i + T_0}{2}$
Rearranging this formula to solve for the temperature of inversion $(T_i)$:
$T_i = 2T_n - T_0$
Given values are:
$T_0 = 10^{\circ} C$
$T_n = 270^{\circ} C$
Substituting these values into the equation:
$T_i = 2 \times 270^{\circ} C - 10^{\circ} C$
$T_i = 540^{\circ} C - 10^{\circ} C$
$T_i = 530^{\circ} C$

(B) In an adiabatic process,the system is thermally insulated from its surroundings,meaning $Q = 0$.
According to the first law of thermodynamics,$\Delta U = Q - W$.
Since $Q = 0$,we have $\Delta U = -W$.
In an expansion process,the gas does work on the surroundings,so $W > 0$.
Therefore,$\Delta U = -W < 0$,which means the internal energy of the system decreases.
Since the internal energy of an ideal gas is directly proportional to its temperature $(U \propto T)$,a decrease in internal energy results in a decrease in the temperature of the system.

(A) The bond dissociation energies are approximately as follows:
$H-H \approx 436 \ kJ/mol$
$C-H \approx 413 \ kJ/mol$
$C-C \approx 348 \ kJ/mol$
Therefore,the decreasing order is $H-H > C-H > C-C$.

(C) The formation reaction of ethylene is: $2C_{(graphite)} + 2H_{2(g)} \rightarrow C_2H_{4(g)}$
Using Hess's Law,the enthalpy of formation is given by: $\Delta H_f = 2 \times \Delta H_I + 2 \times \Delta H_{II} - \Delta H_{III}$
Substituting the given values:
$\Delta H_f = 2(-393.5 \ kJ) + 2(-286.2 \ kJ) - (-1410.8 \ kJ)$
$\Delta H_f = -787.0 \ kJ - 572.4 \ kJ + 1410.8 \ kJ$
$\Delta H_f = 51.4 \ kJ$

(D) $(1)$ Planck's constant: $E = h\nu \implies [h] = [E]/[\nu] = [ML^2 T^{-2}] / [T^{-1}] = [ML^2 T^{-1}]$. This matches (iii).
$(2)$ Gravitational constant: $F = G(m_1 m_2)/r^2 \implies [G] = [Fr^2]/[M^2] = [MLT^{-2}][L^2]/[M^2] = [M^{-1} L^3 T^{-2}]$. This matches (iv).
$(3)$ Bulk modulus: $B = \text{Stress} / \text{Strain} = [ML^{-1} T^{-2}] / [1] = [ML^{-1} T^{-2}]$. This matches $(i)$.
$(4)$ Coefficient of viscosity: $F = \eta A (dv/dx) \implies [\eta] = [F] / ([A][dv/dx]) = [MLT^{-2}] / ([L^2][LT^{-1}/L]) = [ML^{-1} T^{-1}]$. This matches (ii).
Therefore,the correct matching is $(1)$-(iii),$(2)$-(iv),$(3)$-$(i)$,$(4)$-(ii). Hence,option $(d)$ is correct.

(B) The Huygen eyepiece is designed to minimize aberrations. It satisfies the condition for achromatism (elimination of chromatic aberration) by placing the two plano-convex lenses at a distance equal to half the sum of their focal lengths. It also satisfies the condition for minimum spherical aberration.

(B) The velocity of the source $v_s$ is given by $v_s = r \omega$.
Given $r = 2 \ m$ and $\omega = 15 \ rad/s$,we have $v_s = 2 \times 15 = 30 \ m/s$.
The listener is at the center,so the source moves towards the listener when its velocity vector points directly at the center.
The highest frequency $f'$ heard by a stationary listener is given by the Doppler effect formula for a source moving towards the listener:
$f' = f \left( \frac{v}{v - v_s} \right)$
Substituting the given values:
$f' = 540 \left( \frac{330}{330 - 30} \right) = 540 \left( \frac{330}{300} \right) = 540 \times 1.1 = 594 \ Hz$.

(A) The fundamental frequency $f$ of a vibrating wire is given by the formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since the wire is the same,$L$ and $\mu$ are constant,so $f \propto \sqrt{T}$.
Therefore,we can write the ratio: $\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}}$.
Given $f_1 = 450 ~Hz$,$T_1 = 9 ~kg-wt$,and $f_2 = 900 ~Hz$.
Substituting the values: $\frac{900}{450} = \sqrt{\frac{T_2}{9}}$.
$2 = \sqrt{\frac{T_2}{9}}$.
Squaring both sides: $4 = \frac{T_2}{9}$.
$T_2 = 4 \times 9 = 36 ~kg-wt$.

(C) For the water not to spill out of the bucket at the highest point,the velocity at the highest point must be at least $\sqrt{gR}$.
Using the principle of conservation of energy between the lowest point $(v_L)$ and the highest point $(v_H)$:
$\frac{1}{2}mv_L^2 = \frac{1}{2}mv_H^2 + mg(2R)$
$v_L^2 = v_H^2 + 4gR$
Since $v_H = \sqrt{gR}$,we have $v_L^2 = gR + 4gR = 5gR$.
$v_L = \sqrt{5gR}$
Substituting the given values $g = 10 ~m/s^2$ and $R = 0.5 ~m$:
$v_L = \sqrt{5 \times 10 \times 0.5} = \sqrt{25} = 5 ~m/s$.

(A) When the block strikes the spring,the kinetic energy of the block is converted into the potential energy of the spring. By the law of conservation of energy:
$\frac{1}{2} m v^2 = \frac{1}{2} k x^2$
where $x$ is the maximum compression in the spring.
$x = \sqrt{\frac{m v^2}{k}} = \sqrt{\frac{25 \times (3)^2}{100}} = \sqrt{\frac{225}{100}} = \frac{15}{10} = 1.5 \ m$.
When the block returns to the original position,the potential energy stored in the spring is completely converted back into the kinetic energy of the block. Since the surface is smooth (no friction),energy is conserved. The magnitude of the velocity remains the same,but the direction is reversed.
Thus,the velocity of the block is $v = -3 \ ms^{-1}$.

(A) $1$. Pyridinium chlorochromate $(PCC)$ is a mild oxidizing agent that oxidizes primary alcohols to aldehydes and secondary alcohols to ketones.
$2$. The compound '$Y$' gives a positive iodoform test (reacts with $I_2$ and alkali to form triiodomethane,$CHI_3$),which indicates that '$Y$' must contain a $CH_3CO-$ group or be ethanol $(C_2H_5OH)$ which oxidizes to acetaldehyde $(CH_3CHO)$.
$3$. Ethanol $(C_2H_5OH)$ on oxidation with $PCC$ gives acetaldehyde $(CH_3CHO)$.
$4$. Acetaldehyde $(CH_3CHO)$ reacts with $I_2$ and $NaOH$ to give $CHI_3$ (iodoform).
$5$. The reaction sequence is:
$C_2H_5OH + [O] \xrightarrow{PCC \text{ in } CH_2Cl_2} CH_3CHO$
$CH_3CHO + 4NaOH + 3I_2 \rightarrow CHI_3 + HCOONa + 3H_2O + 3NaI$
$6$. Thus,'$X$' is $C_2H_5OH$.

(D) The force exerted by the bullets on the rifle is equal to the rate of change of momentum of the bullets.
Given:
Mass of rifle,$M = 20 ~kg$
Number of bullets per second,$n = 4$
Mass of each bullet,$m = 35 \times 10^{-3} ~kg$
Velocity of each bullet,$v = 400 ~ms^{-1}$
The force exerted by the bullets on the rifle (recoil force) is given by:
$F_{recoil} = n \times (m \times v)$
$F_{recoil} = 4 \times (35 \times 10^{-3} ~kg) \times (400 ~ms^{-1})$
$F_{recoil} = 4 \times 35 \times 0.4 = 56 ~N$
To prevent the rifle from moving backwards,an equal and opposite force must be applied on the rifle.
Therefore,the required force is $56 ~N$ in the forward direction.

(C) In the $CO_3^{2-}$ ion,the central $C$ atom undergoes $sp^2$ hybridization,resulting in a trigonal planar geometry.
In contrast,$BF_4^{-}$,$NH_4^{+}$,and $SO_4^{2-}$ all involve $sp^3$ hybridization of the central atom,which gives them a tetrahedral structure.

(D) The fluorides of alkaline earth metals,except for $BeF_2$,are generally insoluble in water due to their high lattice energy compared to their hydration energy.
Among the given options,$NaF$ and $KF$ are alkali metal fluorides and are soluble in water.
$BeF_2$ is covalent and soluble in water.
$MgF_2$ is an alkaline earth metal fluoride that is insoluble in water.

(C) For the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$,the equilibrium constant is $K_1 = 5 \times 10^{-2}$.
For the reaction $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$,we observe that the reaction is the reverse of the first reaction multiplied by a factor of $2$.
Therefore,the new equilibrium constant $K_2$ is given by $K_2 = \frac{1}{K_1^2}$.
Substituting the value of $K_1$:
$K_2 = \frac{1}{(5 \times 10^{-2})^2} = \frac{1}{25 \times 10^{-4}} = \frac{10^4}{25} = 400 = 4 \times 10^2$.

(B) Non-metallic oxides are generally acidic in nature. Carbon is a non-metal belonging to group $14$ ($IVA$ group).
$C + O_2 \longrightarrow CO_2$
$CO_2 + H_2O \longrightarrow H_2CO_3$ (Carbonic acid)
Since $CO_2$ is a gas and forms an acidic solution,the element belongs to group $IV$ $(IVA)$.

(C) Given that $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$.
Then,$\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
The roots of the new equation are $\gamma = \frac{1-\alpha}{\alpha} = \frac{1}{\alpha}-1$ and $\delta = \frac{1-\beta}{\beta} = \frac{1}{\beta}-1$.
The new equation is $x^2 - (\gamma+\delta)x + \gamma\delta = 0$.
Sum of roots: $\gamma+\delta = (\frac{1}{\alpha}-1) + (\frac{1}{\beta}-1) = \frac{\alpha+\beta}{\alpha\beta} - 2 = \frac{-b/a}{c/a} - 2 = -\frac{b}{c} - 2 = -\frac{b+2c}{c}$.
Product of roots: $\gamma\delta = (\frac{1}{\alpha}-1)(\frac{1}{\beta}-1) = \frac{1}{\alpha\beta} - (\frac{1}{\alpha}+\frac{1}{\beta}) + 1 = \frac{1}{\alpha\beta} - \frac{\alpha+\beta}{\alpha\beta} + 1 = \frac{a}{c} - \frac{-b/a}{c/a} + 1 = \frac{a}{c} + \frac{b}{c} + 1 = \frac{a+b+c}{c}$.
The equation is $x^2 - (-\frac{b+2c}{c})x + \frac{a+b+c}{c} = 0$.
Multiplying by $c$,we get $cx^2 + (b+2c)x + (a+b+c) = 0$.
Comparing with $px^2+qx+r=0$,we find $r = a+b+c$.

(B) Given,$\frac{3x}{(x-a)(x-b)} = \frac{2}{x-a} + \frac{1}{x-b}$
Multiplying both sides by $(x-a)(x-b)$,we get:
$3x = 2(x-b) + 1(x-a)$
$3x = 2x - 2b + x - a$
$3x = 3x - (a + 2b)$
Comparing the constant terms on both sides:
$0 = -(a + 2b)$
$a + 2b = 0$
$a = -2b$
Therefore,$\frac{a}{b} = -2$,which means $a:b = -2:1$.

(D) First,arrange the $8$ men around a circular table. The number of ways to arrange $n$ objects in a circle is $(n-1)!$. So,the number of ways to arrange $8$ men is $(8-1)! = 7!$.
This arrangement creates $8$ gaps between the men.
To ensure no two women sit together,we must place the $4$ women in these $8$ gaps.
The number of ways to choose and arrange $4$ women in $8$ gaps is given by the permutation formula ${}^{8}P_{4}$.
Therefore,the total number of ways is $7! \times {}^{8}P_{4}$.

(A) The number of diagonals in a polygon of $n$ sides is given by the formula $\frac{n(n-3)}{2}$.
Given that the number of diagonals is $275$,we have:
$\frac{n(n-3)}{2} = 275$
$n(n-3) = 550$
$n^2 - 3n - 550 = 0$
Factoring the quadratic equation:
$n^2 - 25n + 22n - 550 = 0$
$n(n - 25) + 22(n - 25) = 0$
$(n - 25)(n + 22) = 0$
This gives $n = 25$ or $n = -22$.
Since the number of sides $n$ must be positive,we have $n = 25$.

(A) Given,$5 \tan \theta = 4$.
Dividing the numerator and denominator of the expression by $\cos \theta$,we get:
$\frac{5 \sin \theta - 3 \cos \theta}{\sin \theta + 2 \cos \theta} = \frac{5 \frac{\sin \theta}{\cos \theta} - 3}{\frac{\sin \theta}{\cos \theta} + 2}$
Since $\tan \theta = \frac{4}{5}$,substitute this value into the expression:
$= \frac{5(\frac{4}{5}) - 3}{(\frac{4}{5}) + 2}$
$= \frac{4 - 3}{\frac{4 + 10}{5}}$
$= \frac{1}{\frac{14}{5}} = \frac{5}{14}$

(A) Given,$\sin A + \sin B = \sqrt{3}(\cos B - \cos A)$.
Rearranging the terms,we get $\sin A + \sqrt{3} \cos A = \sqrt{3} \cos B - \sin B$.
Dividing both sides by $2$,we get $\frac{1}{2} \sin A + \frac{\sqrt{3}}{2} \cos A = \frac{\sqrt{3}}{2} \cos B - \frac{1}{2} \sin B$.
Using the identity $\sin(x+y) = \sin x \cos y + \cos x \sin y$ and $\sin(x-y) = \sin x \cos y - \cos x \sin y$,we have:
$\sin(A + \frac{\pi}{3}) = \sin(\frac{\pi}{3} - B)$.
This implies $A + \frac{\pi}{3} = \frac{\pi}{3} - B$,which simplifies to $A = -B$ or $A + B = 0$.
Now,we need to find the value of $\sin 3A + \sin 3B$.
Since $A = -B$,then $3A = -3B$.
Therefore,$\sin 3A + \sin 3B = \sin(-3B) + \sin 3B = -\sin 3B + \sin 3B = 0$.

(D) Given,$\cos (A-B)=3/5$ and $\tan A \tan B=2$.
We know that $\tan A \tan B = \frac{\sin A \sin B}{\cos A \cos B} = 2$.
Applying componendo and dividendo:
$\frac{\cos A \cos B + \sin A \sin B}{\cos A \cos B - \sin A \sin B} = \frac{1+2}{1-2}$.
This simplifies to $\frac{\cos (A-B)}{\cos (A+B)} = \frac{3}{-1}$.
Substituting $\cos (A-B) = 3/5$:
$\frac{3/5}{\cos (A+B)} = -3$.
$\cos (A+B) = \frac{3/5}{-3} = -1/5$.

(B) The area of a triangle with vertices in polar coordinates $(r_1, \theta_1)$,$(r_2, \theta_2)$,and $(r_3, \theta_3)$ is given by $\frac{1}{2} |r_1 r_2 \sin(\theta_2 - \theta_1) + r_2 r_3 \sin(\theta_3 - \theta_2) + r_3 r_1 \sin(\theta_1 - \theta_3)|$.
Substituting the given values $(1, 0)$,$(2, \frac{\pi}{3})$,and $(3, \frac{2\pi}{3})$:
Area $= \frac{1}{2} |(1)(2) \sin(\frac{\pi}{3} - 0) + (2)(3) \sin(\frac{2\pi}{3} - \frac{\pi}{3}) + (3)(1) \sin(0 - \frac{2\pi}{3})|$
$= \frac{1}{2} |2 \sin(\frac{\pi}{3}) + 6 \sin(\frac{\pi}{3}) + 3 \sin(-\frac{2\pi}{3})|$
$= \frac{1}{2} |2(\frac{\sqrt{3}}{2}) + 6(\frac{\sqrt{3}}{2}) + 3(-\frac{\sqrt{3}}{2})|$
$= \frac{1}{2} |\sqrt{3} + 3\sqrt{3} - \frac{3\sqrt{3}}{2}|$
$= \frac{1}{2} |4\sqrt{3} - 1.5\sqrt{3}| = \frac{1}{2} |2.5\sqrt{3}| = \frac{5\sqrt{3}}{4} \text{ sq units}$.

(D) The given equation of the pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$.
Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=1, b=-35, g=-2, f=22, c=-12$.
The point of intersection $(x_0, y_0)$ of the pair of lines is given by the formula:
$x_0 = \frac{hf-bg}{ab-h^2} = \frac{(1)(22)-(-35)(-2)}{(1)(-35)-(1)^2} = \frac{22-70}{-35-1} = \frac{-48}{-36} = \frac{4}{3}$
$y_0 = \frac{gh-af}{ab-h^2} = \frac{(-2)(1)-(1)(22)}{(1)(-35)-(1)^2} = \frac{-2-22}{-36} = \frac{-24}{-36} = \frac{2}{3}$
Since the lines are concurrent,the point $(\frac{4}{3}, \frac{2}{3})$ must satisfy the line $5x+\lambda y-8=0$.
$5(\frac{4}{3}) + \lambda(\frac{2}{3}) - 8 = 0$
$\frac{20}{3} + \frac{2\lambda}{3} - 8 = 0$
Multiply by $3$: $20 + 2\lambda - 24 = 0$
$2\lambda - 4 = 0$
$2\lambda = 4$
$\lambda = 2$

(A) $1$. Identify the longest carbon chain containing the principal functional group $(-OH)$ and the double bond $(C=C)$. The chain has $6$ carbon atoms,so the parent alkane is hexane.
$2$. Number the chain from the end that gives the lowest possible locant to the principal functional group $(-OH)$. Numbering from right to left,the $-OH$ group is at position $2$.
$3$. The double bond starts at carbon $3$,so it is a hex-$3$-ene derivative.
$4$. There is a methyl group $(-CH_3)$ at position $5$.
$5$. Combining these,the $IUPAC$ name is $5$-methylhex-$3$-en-$2$-ol.

(A) Feldspar is a group of rock-forming tectosilicate minerals that make up about $41\%$ of the Earth's continental crust. The general chemical formula for potassium feldspar (orthoclase) is $KAlSi_3O_8$.

(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
Let $M_e$ and $R_e$ be the mass and radius of the Earth,and $M_p$ and $R_p$ be the mass and radius of the planet.
Given: $M_p = \frac{M_e}{2}$ and $R_p = \frac{R_e}{4}$.
The escape velocity on the planet $v_p$ is given by $v_p = \sqrt{\frac{2GM_p}{R_p}}$.
Taking the ratio of escape velocities: $\frac{v_p}{v_e} = \sqrt{\frac{M_p}{M_e} \times \frac{R_e}{R_p}}$.
Substituting the values: $\frac{v_p}{v_e} = \sqrt{\frac{1/2}{1} \times \frac{1}{1/4}} = \sqrt{\frac{1}{2} \times 4} = \sqrt{2}$.
Therefore,$v_p = v_e \times \sqrt{2} = 11 \times 1.414 = 15.55 \ km \ s^{-1}$.

(B) Step $1$: The reaction of ethene $(CH_2=CH_2)$ with $HCl$ in the presence of anhydrous $AlCl_3$ yields chloroethane $(CH_3CH_2Cl)$ as product $A$.
$CH_2=CH_2 + HCl \rightarrow CH_3CH_2Cl$ $(A)$
Step $2$: The reduction of alkyl halides using $Zn-Cu$ couple in the presence of ethanol $(C_2H_5OH)$ is a standard method to obtain alkanes.
$CH_3CH_2Cl + 2[H] \xrightarrow{Zn-Cu / C_2H_5OH} CH_3CH_3$ $(B)$ $+ HCl$
Thus,$B$ is ethane $(C_2H_6)$.

(C) $ZnH_2$ is an example of an interstitial (metallic) hydride.
$NH_3$,$CH_4$,and $H_2O$ are examples of covalent (molecular) hydrides.

(A) The solubility of an ionic compound like $NaCl$ depends on the dielectric constant of the solvent.
Higher dielectric constant leads to better solvation of ions,which increases solubility.
The dielectric constant of ordinary water $(H_2O)$ is approximately $81$,while that of heavy water $(D_2O)$ is approximately $80$.
Since the dielectric constant of ordinary water is higher,$NaCl$ is more soluble in ordinary water and less soluble in heavy water.
Therefore,both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.