TS EAMCET 2002 Chemistry Question Paper with Answer and Solution

(A) The reaction of iron sulphide with oxygen is: $2FeS + 3O_2 \rightarrow 2FeO + 2SO_2$ $(A)$.
When $SO_2$ $(A)$ is dissolved in water,it forms sulphurous acid: $SO_2 + H_2O \rightarrow H_2SO_3$.
Sulphurous acid $(H_2SO_3)$ is a diprotic acid,meaning it has two replaceable hydrogen atoms.
Therefore,the basicity of $H_2SO_3$ is $2$.

(C) Option $A$ is correct because Bronsted-Lowry theory defines acids as proton donors,but $BCl_3$ is a Lewis acid that accepts electron pairs,not protons.
Option $B$ is incorrect because for $0.01 \ M \ NaOH$,$[OH^-] = 10^{-2} \ M$,so $pOH = 2$ and $pH = 14 - 2 = 12$.
Option $C$ is correct as the ionic product of water $(K_w)$ at $25^{\circ} C$ is $1.0 \times 10^{-14} \ mol^2 \ L^{-2}$.
Option $D$ is correct as the definition of $pH$ is $pH = -\log [H^+]$.
Note: In many contexts,multiple statements may be technically correct. However,$C$ and $D$ are fundamental definitions. Given the standard format,$C$ is a specific constant value.

(A) Step $1$: Calculate the total moles of $HCl$ in the mixture.
$n_1 = M_1 \times V_1 = 0.2 \ M \times 0.075 \ L = 0.015 \ mol$
$n_2 = M_2 \times V_2 = 1 \ M \times 0.025 \ L = 0.025 \ mol$
Total moles $n = n_1 + n_2 = 0.015 + 0.025 = 0.040 \ mol$
Step $2$: Calculate the total volume of the final solution.
$V_{total} = 75 \ mL + 25 \ mL + 300 \ mL = 400 \ mL = 0.4 \ L$
Step $3$: Calculate the final molarity of $HCl$.
$M_{final} = \frac{n}{V_{total}} = \frac{0.040 \ mol}{0.4 \ L} = 0.1 \ M$
Step $4$: Calculate the $pH$.
$[H^+] = 0.1 \ M = 10^{-1} \ M$
$pH = -\log_{10}[H^+] = -\log_{10}(10^{-1}) = 1$

(C) Given: Coefficient of friction $\mu = 0.5$. The ratio of net force $F$ to normal reaction $R$ is $\frac{F}{R} = \frac{1}{2}$.
The net force $F$ acting down the plane is given by $F = mg \sin \theta - f$,where $f$ is the frictional force.
The normal reaction $R$ is $mg \cos \theta$.
The frictional force $f$ is given by $f = \mu R = \mu mg \cos \theta$.
Substituting these into the ratio:
$\frac{mg \sin \theta - \mu mg \cos \theta}{mg \cos \theta} = \frac{1}{2}$
Dividing the numerator by the denominator:
$\tan \theta - \mu = \frac{1}{2}$
Given $\mu = 0.5$,we have:
$\tan \theta - 0.5 = 0.5$
$\tan \theta = 1$
Therefore,$\theta = 45^{\circ}$.

(C) Let the body leave the surface of the hemisphere at point $P$. At this point,the radius vector makes an angle $\theta$ with the vertical.
The forces acting on the body at point $P$ are the gravitational force $mg$ (acting downwards) and the normal reaction $R$ (acting radially outwards).
The component of gravitational force towards the center is $mg \cos \theta$.
The net centripetal force required for circular motion is provided by the difference between the radial component of gravity and the normal reaction:
$mg \cos \theta - R = \frac{mv^2}{r}$
When the body leaves the surface,the normal reaction $R$ becomes zero.
Therefore,$mg \cos \theta = \frac{mv^2}{r}$.
Substituting the given values $v = 5 \ m/s$,$r = 5 \ m$,and $g = 10 \ m/s^2$:
$\cos \theta = \frac{v^2}{rg} = \frac{5^2}{5 \times 10} = \frac{25}{50} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = 60^{\circ}$.

(C) Given: radius $r = 0.4 \text{ \AA} = 0.4 \times 10^{-10} \text{ m}$, speed $v = 10^6 \text{ m/s}$, charge $q = 1.6 \times 10^{-19} \text{ C}$.
The moving electron constitutes an electric current $i = \frac{q}{T}$, where $T$ is the time period.
$T = \frac{2\pi r}{v}$, so $i = \frac{qv}{2\pi r}$.
Substituting the values:
$i = \frac{1.6 \times 10^{-19} \times 10^6}{2\pi \times 0.4 \times 10^{-10}} = \frac{1.6 \times 10^{-13}}{0.8\pi \times 10^{-10}} = \frac{2 \times 10^{-3}}{\pi} \text{ A}$.
The magnetic field $B$ at the centre of a circular loop is given by $B = \frac{\mu_0 i}{2r}$.
$B = \frac{4\pi \times 10^{-7} \times (2 \times 10^{-3} / \pi)}{2 \times 0.4 \times 10^{-10}}$.
$B = \frac{4 \times 2 \times 10^{-10}}{0.8 \times 10^{-10}} = \frac{8}{0.8} = 10 \text{ T}$.

(D) Statement $(A)$ is wrong because domain theory is used to explain ferromagnetism,not paramagnetism.
Statement $(B)$ is correct because the magnetic susceptibility of a diamagnetic substance is independent of temperature,as it arises from the orbital motion of electrons which is not significantly affected by thermal agitation.

(C) Given the formula for the physical quantity $X = \frac{A^2 B}{C^{1/3} D^3}$.
Using the principle of propagation of errors,the relative error in $X$ is given by:
$\frac{\Delta X}{X} = 2 \frac{\Delta A}{A} + \frac{\Delta B}{B} + \frac{1}{3} \frac{\Delta C}{C} + 3 \frac{\Delta D}{D}$.
Now,we calculate the contribution of each term to the percentage error:
Contribution of $A = 2 \times (2\%) = 4\%$.
Contribution of $B = 1 \times (2\%) = 2\%$.
Contribution of $C = \frac{1}{3} \times (4\%) = 1.33\%$.
Contribution of $D = 3 \times (5\%) = 15\%$.
Comparing these values,the minimum contribution to the percentage error in $X$ is from $C$ $(1.33\%)$.

(D) To compare the numbers $\sqrt[3]{4}, \sqrt[4]{5}, \sqrt[4]{7}, \sqrt[3]{8}$,we convert them to exponents: $4^{1/3}, 5^{1/4}, 7^{1/4}, 8^{1/3}$.
Find the least common multiple $(LCM)$ of the denominators $3$ and $4$,which is $12$.
Raise each number to the power of $12$:
$(4^{1/3})^{12} = 4^4 = 256$
$(5^{1/4})^{12} = 5^3 = 125$
$(7^{1/4})^{12} = 7^3 = 343$
$(8^{1/3})^{12} = 8^4 = 4096$
Comparing the values $256, 125, 343, 4096$,the smallest value is $125$,which corresponds to $\sqrt[4]{5}$.

(B) Given the equation: $6^x = 7^{x+4}$
Taking logarithm on both sides:
$\log(6^x) = \log(7^{x+4})$
$x \log 6 = (x+4) \log 7$
$x(\log 2 + \log 3) = x \log 7 + 4 \log 7$
$x(\log 2 + \log 3) - x \log 7 = 4 \log 7$
$x(\log 2 + \log 3 - \log 7) = 4 \log 7$
Substituting the given values $\log 2=a, \log 3=b, \log 7=c$:
$x(a+b-c) = 4c$
$x = \frac{4c}{a+b-c}$

(B) We know that the Taylor series expansion for the exponential function is $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \ldots$
Let $u = x \log _e a$.
Then the given series becomes $1 + (x \log _e a) + \frac{(x \log _e a)^2}{2!} + \frac{(x \log _e a)^3}{3!} + \ldots$
This is equivalent to $e^{x \log _e a}$.
Using the property of logarithms $n \log _e m = \log _e m^n$,we have $x \log _e a = \log _e a^x$.
Therefore,$e^{\log _e a^x} = a^x$.

(D) The general term $T_n$ of the series is given by $T_n = \frac{1+2+2^2+\ldots+2^{n-1}}{n !}$.
Using the sum of a geometric progression,$T_n = \frac{2^n-1}{n !} = \frac{2^n}{n !} - \frac{1}{n !}$.
The sum of the series is $S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{2^n}{n !} - \sum_{n=1}^{\infty} \frac{1}{n !}$.
We know that $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n !} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots$.
Thus,$\sum_{n=1}^{\infty} \frac{2^n}{n !} = e^2 - 1$ and $\sum_{n=1}^{\infty} \frac{1}{n !} = e - 1$.
Therefore,$S = (e^2 - 1) - (e - 1) = e^2 - e$.

(A) Let $\alpha$ be the common root of the two equations.
Then,$\alpha^2 + a\alpha + b = 0$ and $\alpha^2 + b\alpha + a = 0$.
Subtracting the two equations: $(\alpha^2 + a\alpha + b) - (\alpha^2 + b\alpha + a) = 0$.
$(a-b)\alpha + (b-a) = 0$.
$(a-b)\alpha = (a-b)$.
Since $a \neq b$,we can divide by $(a-b)$ to get $\alpha = 1$.
Substituting $\alpha = 1$ into the first equation: $1^2 + a(1) + b = 0$.
$1 + a + b = 0$.
Therefore,$a + b = -1$.

(C) Given that $3$ is a root of the equation $x^2+kx-24=0$.
Substituting $x=3$ into the equation:
$(3)^2 + k(3) - 24 = 0$
$9 + 3k - 24 = 0$
$3k - 15 = 0$
$3k = 15 \Rightarrow k = 5$.
Now,we check the options by substituting $k=5$ and $x=3$:
For option $C$: $x^2-kx+6=0$
Substituting $x=3$ and $k=5$:
$(3)^2 - (5)(3) + 6 = 9 - 15 + 6 = 0$.
Since the equation is satisfied,$3$ is a root of $x^2-kx+6=0$.

(D) Given the equation $x^4-x^2+x-1=0$.
Let $\alpha = \frac{1+\sqrt{3}i}{2}$ be a root.
Since the coefficients are real,the conjugate $\beta = \frac{1-\sqrt{3}i}{2}$ must also be a root.
Sum of roots $\alpha+\beta = \frac{1+\sqrt{3}i}{2} + \frac{1-\sqrt{3}i}{2} = 1$.
Product of roots $\alpha\beta = \left(\frac{1+\sqrt{3}i}{2}\right)\left(\frac{1-\sqrt{3}i}{2}\right) = \frac{1+3}{4} = 1$.
The quadratic factor corresponding to these roots is $x^2-(\alpha+\beta)x+\alpha\beta = x^2-x+1 = 0$.
Dividing $x^4-x^2+x-1$ by $x^2-x+1$,we get:
$(x^2-x+1)(x^2+x-1) = 0$.
The real roots are obtained from $x^2+x-1=0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $x = \frac{-1 \pm \sqrt{1+4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.
Note: The original question statement implies the roots are $1, -1$,but the polynomial $x^4-x^2+x-1$ factors as $(x-1)(x+1)(x^2+1)$ is incorrect. Let us re-evaluate: $x^4-x^2+x-1 = x^2(x^2-1) + (x-1) = (x-1)(x^2(x+1)+1) = (x-1)(x^3+x^2+1)$.
Actually,$x^4-x^2+x-1 = (x^2-x+1)(x^2+x-1)$.
The real roots are $\frac{-1 \pm \sqrt{5}}{2}$.

(A) Let $z = \frac{3+2i \sin \theta}{1-2i \sin \theta}$.
To make $z$ a real number,the imaginary part must be zero.
Multiply the numerator and denominator by the conjugate of the denominator $(1+2i \sin \theta)$:
$z = \frac{(3+2i \sin \theta)(1+2i \sin \theta)}{(1-2i \sin \theta)(1+2i \sin \theta)}$
$z = \frac{3 + 6i \sin \theta + 2i \sin \theta + 4i^2 \sin^2 \theta}{1 + 4 \sin^2 \theta}$
Since $i^2 = -1$,we have:
$z = \frac{(3 - 4 \sin^2 \theta) + i(8 \sin \theta)}{1 + 4 \sin^2 \theta}$
For $z$ to be real,the imaginary part must be $0$:
$\frac{8 \sin \theta}{1 + 4 \sin^2 \theta} = 0$
This implies $\sin \theta = 0$.
Given $0 < \theta < 2\pi$,the only solution is $\theta = \pi$.

(A) We know that the definition of the hyperbolic sine function is $\sinh(z) = \frac{e^z - e^{-z}}{2}$.
Substituting $z = ix$,we get:
$\sinh(ix) = \frac{e^{ix} - e^{-ix}}{2}$.
Using Euler's formula,$\sin x = \frac{e^{ix} - e^{-ix}}{2i}$,which implies $\frac{e^{ix} - e^{-ix}}{2} = i \sin x$.
Therefore,$\sinh(ix) = i \sin x$.

(C) The set of available odd digits is $\{1, 3, 5, 7, 9\}$.
Total number of $5$-digit numbers formed using these $5$ distinct digits is $5! = 120$.
$A$ number is divisible by $5$ if its last digit is $5$.
If the last digit is fixed as $5$,the remaining $4$ positions can be filled by the remaining $4$ digits in $4!$ ways.
Number of $5$-digit numbers divisible by $5 = 4! = 24$.
Therefore,the number of $5$-digit numbers not divisible by $5 = 5! - 4! = 120 - 24 = 96$.

(C) The velocity of efflux of the liquid from the hole at the bottom is given by Torricelli's law: $v = \sqrt{2gd}$.
The vertical height through which the water falls is $h$.
The time taken $t$ for the water to reach the ground is given by the equation of motion $h = \frac{1}{2}gt^2$,which gives $t = \sqrt{\frac{2h}{g}}$.
The horizontal range $R$ is the product of the horizontal velocity and the time of flight:
$R = v \times t$
$R = \sqrt{2gd} \times \sqrt{\frac{2h}{g}}$
$R = \sqrt{2gd \times \frac{2h}{g}}$
$R = \sqrt{4dh}$
Squaring both sides,we get $R^2 = 4dh$.
Therefore,$d = \frac{R^2}{4h}$.

(D) Given:
Inner diameter $d = 0.28 \ mm = 0.28 \times 10^{-3} \ m$.
Radius $r = d/2 = 0.14 \times 10^{-3} \ m$.
Surface tension $T = 0.07 \ N/m$.
For water,the contact angle $\theta = 0^\circ$,so $\cos \theta = 1$.
Due to surface tension,the capillary rise $h$ is given by $h = \frac{2T}{r \rho g}$.
The pressure difference across the meniscus is $\Delta P = \frac{2T}{r}$.
To keep the water level in the tube the same as in the vessel,we must apply an external pressure equal to the capillary pressure difference.
$\Delta P = \frac{2 \times 0.07}{0.14 \times 10^{-3}} = \frac{0.14}{0.14 \times 10^{-3}} = 10^3 \ N/m^2$.
Since the water surface in the tube is depressed by this pressure relative to the atmospheric pressure,the total pressure required to maintain the level is the sum of atmospheric pressure and the capillary pressure:
$P_{total} = P_{atm} + \Delta P = 10^5 + 10^3 = 100 \times 10^3 + 1 \times 10^3 = 101 \times 10^3 \ N/m^2$.

(D) The horizontal displacement is given by $x = u_x t = 36t$,so the horizontal component of velocity is $u_x = 36 \ m/s$.
The vertical displacement is given by $y = u_y t - \frac{1}{2} g t^2 = 48t - 4.9t^2$,so the vertical component of velocity is $u_y = 48 \ m/s$.
The initial velocity $u$ is given by the magnitude of the velocity vector: $u = \sqrt{u_x^2 + u_y^2}$.
Substituting the values: $u = \sqrt{36^2 + 48^2} = \sqrt{1296 + 2304} = \sqrt{3600} = 60 \ m/s$.

(A) Given: Mass of neutron $m_n = 1.00893 \ amu$,mass of proton $m_p = 1.00813 \ amu$,and mass of deuteron $m_d = 2.01473 \ amu$.
$A$ deuteron nucleus $(^2_1H)$ consists of one proton and one neutron.
The mass defect $\Delta m$ is calculated as:
$\Delta m = (m_n + m_p) - m_d$
$\Delta m = (1.00893 + 1.00813) - 2.01473$
$\Delta m = 2.01706 - 2.01473 = 0.00233 \ amu$.
The packing fraction is defined as the ratio of mass defect to the mass number $(A)$.
For deuteron,$A = 2$.
$\text{Packing fraction} = \frac{\Delta m}{A} = \frac{0.00233}{2} = 0.001165 = 11.65 \times 10^{-4}$.

(C) The nuclear force acting between proton-neutron $(p-n)$,proton-proton $(p-p)$,and neutron-neutron $(n-n)$ are approximately equal and charge independent. Therefore,statement $A$ is wrong.
The neutron reproduction factor $k$ (often denoted as $k$ instead of $A$) represents the ratio of neutrons produced in one generation to the number of neutrons in the preceding generation. If $k > 1$,the chain reaction is supercritical and the fission rate increases,meaning the reaction is in an accelerating state. Therefore,statement $B$ is correct.

(D) The time period of an oscillating magnet is given by $T = 2\pi \sqrt{\frac{I}{MH}}$.
Here,$I$ is the moment of inertia and $M$ is the magnetic moment.
For a rod of mass $m$ and length $l$,$M = m_p l$ (where $m_p$ is pole strength) and $I = \frac{m l^2}{12}$.
When the rod is broken into three equal parts,the new length is $l' = \frac{l}{3}$ and the new mass is $m' = \frac{m}{3}$.
The new magnetic moment is $M' = m_p \times \frac{l}{3} = \frac{M}{3}$.
The new moment of inertia is $I' = \frac{m' (l')^2}{12} = \frac{(m/3) (l/3)^2}{12} = \frac{m l^2}{12 \times 3 \times 9} = \frac{I}{27}$.
Wait,let's re-evaluate: $I = \frac{1}{12} m l^2$. For a part of mass $m/3$ and length $l/3$,$I' = \frac{1}{12} (m/3) (l/3)^2 = \frac{1}{12} \frac{m l^2}{27} = \frac{I}{27}$.
Then $T' = 2\pi \sqrt{\frac{I'}{M' H}} = 2\pi \sqrt{\frac{I/27}{(M/3) H}} = 2\pi \sqrt{\frac{I}{9MH}} = \frac{T}{3} = \frac{4}{3} \ s$.
Correction: The standard formula for a magnet oscillating is $T = 2\pi \sqrt{\frac{I}{MH}}$. Given the options,the intended calculation assumes $I \propto l^3$ and $M \propto l$,leading to $T \propto \sqrt{l^2} \propto l$. Thus $T' = T/3$. However,checking the provided solution logic: $I' = I/9$ is used. If $I' = I/9$ and $M' = M/3$,then $T' = T/\sqrt{3} = 4/\sqrt{3} \ s$.

(A) For a body of mass $m$ executing $SHM$ under a force $F = -kx$,the time period is $T = 2\pi \sqrt{\frac{m}{k}}$.
Since $F = ma = -kx$,the force constant $k$ is proportional to $\frac{1}{T^2}$ (as $k = m\omega^2 = m(\frac{2\pi}{T})^2$).
Thus,$k_1 \propto \frac{1}{T_1^2}$ and $k_2 \propto \frac{1}{T_2^2}$.
When both forces act in the same direction,the effective force constant is $k_{eff} = k_1 + k_2$.
The new time period $T$ is given by $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$.
Substituting the values: $\frac{1}{T^2} = (\frac{5}{4})^2 + (\frac{5}{3})^2 = \frac{25}{16} + \frac{25}{9}$.
$\frac{1}{T^2} = 25 \times (\frac{9+16}{144}) = 25 \times \frac{25}{144} = \frac{625}{144}$.
Taking the square root: $\frac{1}{T} = \frac{25}{12}$,which gives $T = \frac{12}{25} \ s$.

(C) Given equation: $4 v^2 = 25 - x^2$.
Divide by $4$: $v^2 = \frac{25}{4} - \frac{x^2}{4}$.
Comparing this with the standard equation of simple harmonic motion $v^2 = \omega^2 (A^2 - x^2)$,we get $v^2 = \frac{1}{4} (25 - x^2)$.
Here,$\omega^2 = \frac{1}{4}$,which implies $\omega = \sqrt{\frac{1}{4}} = \frac{1}{2} \text{ rad/s}$.
The time period $T$ is given by $T = \frac{2 \pi}{\omega}$.
Substituting the value of $\omega$: $T = \frac{2 \pi}{1/2} = 4 \pi \text{ s}$.

(A) The thermite reaction involves the reduction of ferric oxide $(Fe_2O_3)$ by aluminium $(Al)$ powder. The balanced chemical equation is:
$Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3$
According to the stoichiometry of the reaction,$1$ mole of $Fe_2O_3$ reacts with $2$ moles of $Al$.
However,in the thermite mixture used for welding,the ratio by mass is typically $3$ parts of ferric oxide to $1$ part of aluminium powder.
Therefore,$X = 3$ and $Y = 1$.

(A) The reaction of bauxite $(Al_2O_3)$ with coke $(C)$ and nitrogen $(N_2)$ at $2075 \ K$ is given by:
$Al_2O_3 + 3C + N_2 \xrightarrow{2075 \ K} 2AlN + 3CO$
Here,$X$ is aluminum nitride $(AlN)$.
When $AlN$ reacts with water,it produces ammonia gas $(NH_3)$:
$AlN + 3H_2O \longrightarrow Al(OH)_3 + NH_3 \uparrow$

(B) When ammonia reacts with excess chlorine,nitrogen trichloride $(NCl_3)$ and hydrogen chloride $(HCl)$ are formed.
The balanced chemical equation is:
$NH_3 + 3Cl_2 \text{ (excess)} \longrightarrow NCl_3 + 3HCl$

(D) When ammonia reacts with excess chlorine,the reaction is as follows:
$NH_3 + 3Cl_2 \longrightarrow NCl_3 + 3HCl$
Thus,the products obtained are nitrogen trichloride $(NCl_3)$ and hydrogen chloride $(HCl)$.

(A) The balanced chemical equation for the reaction of fluorine with hot concentrated $KOH$ is:
$2 F_2 + 4 KOH \longrightarrow 4 KF + 2 H_2O + O_2$
From the stoichiometry of the reaction,the molar ratio of $KF : H_2O : O_2$ is $4 : 2 : 1$.
Dividing by $2$,we get the ratio $2 : 1 : 0.5$.

(A) The number of ways to arrange $n$ distinct objects in a circle is $(n-1)!$.
For a necklace,the clockwise and counter-clockwise arrangements are considered identical,so the number of ways is $\frac{(n-1)!}{2}$.
Here,$n = 8$.
Number of ways = $\frac{(8-1)!}{2} = \frac{7!}{2} = \frac{5040}{2} = 2520$.

(A) Let the old coordinates be $(x, y)$ and the new coordinates be $(x', y')$. The transformation equations for rotation of axes by an angle $\theta$ are:
$x = x' \cos \theta - y' \sin \theta$
$y = x' \sin \theta + y' \cos \theta$
Given $\theta = 45^{\circ}$,$x' = \sqrt{2}$,and $y' = 4$.
Substituting these values:
$x = \sqrt{2} \cos 45^{\circ} - 4 \sin 45^{\circ}$
$x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \right) - 4 \left( \frac{1}{\sqrt{2}} \right) = 1 - \frac{4}{\sqrt{2}} = 1 - 2\sqrt{2}$
$y = \sqrt{2} \sin 45^{\circ} + 4 \cos 45^{\circ}$
$y = \sqrt{2} \left( \frac{1}{\sqrt{2}} \right) + 4 \left( \frac{1}{\sqrt{2}} \right) = 1 + \frac{4}{\sqrt{2}} = 1 + 2\sqrt{2}$
Thus,the coordinates in the old system are $(1 - 2\sqrt{2}, 1 + 2\sqrt{2})$.

(D) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since the circle passes through $(0,0)$,we have $c = 0$.
Since it passes through $(2,0)$,we have $2^2 + 0^2 + 2g(2) + 2f(0) + 0 = 0$,which gives $4 + 4g = 0$,so $g = -1$.
Since it passes through $(0,-2)$,we have $0^2 + (-2)^2 + 2g(0) + 2f(-2) + 0 = 0$,which gives $4 - 4f = 0$,so $f = 1$.
The equation of the circle is $x^2 + y^2 - 2x + 2y = 0$.
Since the point $(k,-2)$ lies on the circle,we substitute $x = k$ and $y = -2$:
$k^2 + (-2)^2 - 2(k) + 2(-2) = 0$
$k^2 + 4 - 2k - 4 = 0$
$k^2 - 2k = 0$
$k(k - 2) = 0$.
Since the points must be distinct,$k$ cannot be $0$ (as that would make the point $(0,-2)$,which is already given).
Therefore,$k = 2$.

(B) The given line is $2x - 3y + 7 = 0$.
Any line perpendicular to the given line is of the form $3x + 2y + k = 0$.
To find the intercepts,set $y = 0$ to get $3x = -k$,so $x = -k/3$.
Set $x = 0$ to get $2y = -k$,so $y = -k/2$.
The area of the triangle formed with the coordinate axes is given by $\frac{1}{2} |x| |y| = 3$.
$\frac{1}{2} |-\frac{k}{3}| |-\frac{k}{2}| = 3$.
$\frac{k^2}{12} = 3$ $\Rightarrow k^2 = 36$ $\Rightarrow k = \pm 6$.
Substituting $k$ back into the equation,we get $3x + 2y = \pm 6$.

(C) The equation of the pair of angle bisectors of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
Since the coordinate axes are the bisectors,their equation is $xy = 0$.
Comparing $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ with $xy = 0$,we observe that the $x^2$ and $y^2$ terms must vanish.
This implies $a - b = 0$,which contradicts the given condition $a \neq b$.
However,if we look at the structure $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$,for this to represent $xy = 0$,the coefficient of $xy$ in the bisector equation must be non-zero,and the $x^2 - y^2$ term must be zero.
Actually,the condition for the coordinate axes to be the bisectors is $a + b = 0$ and $h \neq 0$.

(D) The given equation is $x^2(\cos \theta - \sin \theta) + 2xy \cos \theta + y^2(\cos \theta + \sin \theta) = 0$.
Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get:
$a = \cos \theta - \sin \theta$,$h = \cos \theta$,and $b = \cos \theta + \sin \theta$.
The formula for the acute angle $\alpha$ between the pair of lines is given by $\tan \alpha = \left| \frac{2 \sqrt{h^2 - ab}}{a + b} \right|$.
Substituting the values:
$h^2 - ab = \cos^2 \theta - (\cos \theta - \sin \theta)(\cos \theta + \sin \theta) = \cos^2 \theta - (\cos^2 \theta - \sin^2 \theta) = \sin^2 \theta$.
$a + b = (\cos \theta - \sin \theta) + (\cos \theta + \sin \theta) = 2 \cos \theta$.
Thus,$\tan \alpha = \left| \frac{2 \sqrt{\sin^2 \theta}}{2 \cos \theta} \right| = \left| \frac{2 \sin \theta}{2 \cos \theta} \right| = |\tan \theta|$.
Since $2 \theta$ is acute,$\theta$ is also acute,so $\alpha = \theta$.

(B) The given pair of straight lines is $xy-x-y+1=0$.
Factoring the expression: $x(y-1)-1(y-1)=0$,which gives $(x-1)(y-1)=0$.
This represents two lines: $x=1$ and $y=1$.
The point of intersection of these two lines is $(1, 1)$.
Since the lines are concurrent with the line $ax+2y-3a=0$,the point $(1, 1)$ must satisfy the equation of the line $ax+2y-3a=0$.
Substituting $x=1$ and $y=1$ into the equation: $a(1)+2(1)-3a=0$.
$a+2-3a=0$.
$-2a+2=0$.
$2a=2$.
$a=1$.

(B) The given pair of lines is $xy-x-y+1=0$.
This can be factored as $x(y-1)-1(y-1)=0$,which gives $(x-1)(y-1)=0$.
The two lines are $x-1=0$ and $y-1=0$.
The point of intersection of these two lines is $(1, 1)$.
Since the lines are concurrent with the line $ax+2y-3a=0$,the point $(1, 1)$ must satisfy the equation of the line $ax+2y-3a=0$.
Substituting $x=1$ and $y=1$ into the equation,we get:
$a(1)+2(1)-3a=0$
$a+2-3a=0$
$-2a+2=0$
$2a=2$
$a=1$

(B) Since the circle touches both coordinate axes in the third quadrant,its center must be at a distance of $5$ units from both axes.
In the third quadrant,both $x$ and $y$ coordinates are negative.
Therefore,the center of the circle is $(-5, -5)$ and the radius is $r = 5$.
The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting $h = -5$,$k = -5$,and $r = 5$,we get:
$(x - (-5))^2 + (y - (-5))^2 = 5^2$
$(x+5)^2 + (y+5)^2 = 25$.

(B) Let the circle have radius $r$. Since the circle lies in the first quadrant and touches both coordinate axes,its center is $(r, r)$.
The equation of the circle is $(x - r)^2 + (y - r)^2 = r^2$.
The circle touches the line $4x + 3y - 12 = 0$. The perpendicular distance from the center $(r, r)$ to the line must be equal to the radius $r$:
$\frac{|4r + 3r - 12|}{\sqrt{4^2 + 3^2}} = r$
$\frac{|7r - 12|}{5} = r$
$|7r - 12| = 5r$
This gives two cases:
$1) 7r - 12 = 5r$ $\Rightarrow 2r = 12$ $\Rightarrow r = 6$
$2) 7r - 12 = -5r$ $\Rightarrow 12r = 12$ $\Rightarrow r = 1$
The question asks for the radius of the larger circle,which is $r = 6$.

(D) Let the line be $\frac{x}{a} + \frac{y}{b} = 1$. Since the distance from the origin $(0,0)$ to the line is $c$,we have $\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} = c$,which implies $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$.
Since the circle passes through $O(0,0)$,$A(a,0)$,and $B(0,b)$,the circle is $x^2 + y^2 - ax - by = 0$.
The centre of this circle is $(h, k) = (\frac{a}{2}, \frac{b}{2})$.
Thus,$a = 2h$ and $b = 2k$.
Substituting these into the equation $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$,we get $\frac{1}{(2h)^2} + \frac{1}{(2k)^2} = \frac{1}{c^2}$.
$\frac{1}{4h^2} + \frac{1}{4k^2} = \frac{1}{c^2} \Rightarrow \frac{h^2+k^2}{4h^2k^2} = \frac{1}{c^2}$.
Wait,the standard approach for the locus of the midpoint of the hypotenuse of a right triangle with fixed legs is different. Let the centre be $(x, y)$. Since the triangle $OAB$ is a right-angled triangle at $O$,the centre of the circle is the midpoint of the hypotenuse $AB$. Thus $A = (2x, 0)$ and $B = (0, 2y)$.
The line equation is $\frac{X}{2x} + \frac{Y}{2y} = 1$. The distance from origin is $c = \frac{|-1|}{\sqrt{(\frac{1}{2x})^2 + (\frac{1}{2y})^2}} = \frac{1}{\sqrt{\frac{1}{4x^2} + \frac{1}{4y^2}}}$.
$c^2 = \frac{1}{\frac{1}{4x^2} + \frac{1}{4y^2}} = \frac{4x^2y^2}{x^2+y^2}$. This does not match the options. Let's re-evaluate.
Actually,the distance from origin to line $Ax+By+C=0$ is $c = \frac{|C|}{\sqrt{A^2+B^2}}$. For $\frac{x}{a} + \frac{y}{b} = 1$,$c = \frac{1}{\sqrt{1/a^2+1/b^2}}$. The centre is $(a/2, b/2)$. The locus is $x^2+y^2 = (a/2)^2 + (b/2)^2 = \frac{a^2+b^2}{4}$. Using $1/c^2 = 1/a^2 + 1/b^2 = \frac{a^2+b^2}{a^2b^2}$,we have $a^2+b^2 = \frac{a^2b^2}{c^2}$.
Given the options,the intended answer is $x^2+y^2 = c^2$ is incorrect. Let's check $x^2+y^2 = c^2/4$ or similar. Actually,if $c$ is the distance,the locus is $1/x^2 + 1/y^2 = 4/c^2$.

(C) Given that the focus is $S(3,0)$ and the directrix is $x+3=0$. Let $P(x, y)$ be any point on the parabola.
By the definition of a parabola,the distance from $P$ to the focus $S$ is equal to the perpendicular distance from $P$ to the directrix.
$SP^2 = PM^2$
$(x-3)^2 + (y-0)^2 = (x+3)^2$
$y^2 = (x+3)^2 - (x-3)^2$
Using the identity $a^2 - b^2 = (a+b)(a-b)$:
$y^2 = (x+3+x-3)(x+3-x+3)$
$y^2 = (2x)(6)$
$y^2 = 12x$

(C) Let the focal chord pass through the focus $S(a, 0)$ and connect points $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$ on the parabola $y^2 = 4ax$.
The equation of the chord $PQ$ is $y(t_1 + t_2) = 2x + 2at_1t_2$.
Since it passes through $(a, 0)$,we have $0 = 2a + 2at_1t_2$,which implies $t_1t_2 = -1$.
Let $(x_1, y_1)$ be the pole of this chord with respect to the parabola $y^2 = 4ax$.
The equation of the polar of $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$,or $yy_1 - 2ax = 2ax_1$.
Comparing this with the chord equation $y(t_1 + t_2) - 2x = 2at_1t_2$,we get:
$\frac{t_1 + t_2}{y_1} = \frac{-2}{-2a} = \frac{2at_1t_2}{2ax_1}$
From $\frac{-2}{-2a} = \frac{2at_1t_2}{2ax_1}$,we get $x_1 = at_1t_2$.
Since $t_1t_2 = -1$,we have $x_1 = -a$.
Thus,the locus of the pole is $x = -a$,which is the directrix of the parabola.

(C) The general term in the expansion of $\left(x^2+\frac{k}{x}\right)^5$ is given by:
$T_{r+1} = { }^5 C_r (x^2)^{5-r} \left(\frac{k}{x}\right)^r$
$T_{r+1} = { }^5 C_r x^{10-2r} \cdot k^r \cdot x^{-r} = { }^5 C_r k^r x^{10-3r}$
For the coefficient of $x$,we set the exponent of $x$ to $1$:
$10 - 3r = 1$
$3r = 9 \implies r = 3$
Now,substitute $r=3$ into the coefficient expression:
Coefficient of $x = { }^5 C_3 k^3 = 10 k^3$
Given that the coefficient is $270$:
$10 k^3 = 270$
$k^3 = 27$
$k = 3$

(B) The $p^{th}$ term in the expansion of $(1+x)^n$ is $T_p = { }^n C_{p-1} x^{p-1}$,so its coefficient is $p = { }^n C_{p-1}$.
The $(p+1)^{th}$ term is $T_{p+1} = { }^n C_p x^p$,so its coefficient is $q = { }^n C_p$.
We know the property of binomial coefficients: $\frac{{ }^n C_r}{{ }^n C_{r-1}} = \frac{n-r+1}{r}$.
Here,$\frac{q}{p} = \frac{{ }^n C_p}{{ }^n C_{p-1}} = \frac{n-p+1}{p}$.
Thus,$q = \frac{p(n-p+1)}{p} = n-p+1$.
Rearranging the terms,we get $p+q = n+1$.

(C) To find the sum of the coefficients in the expansion of a polynomial $P(x)$,we substitute $x = 1$ into the expression.
Given the expression $(1+x+x^2)^n$.
Substituting $x = 1$:
$(1 + 1 + 1^2)^n = (1 + 1 + 1)^n = 3^n$.
Therefore,the sum of the coefficients is $3^n$.

(A) The equation of the ellipse is $5x^2 + 9y^2 = 45$,which can be written as $\frac{x^2}{9} + \frac{y^2}{5} = 1$.
Here,$a^2 = 9$ and $b^2 = 5$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.
The equation of the hyperbola is $5x^2 - 4y^2 = 45$,which can be written as $\frac{x^2}{9} - \frac{y^2}{45/4} = 1$.
Here,$a^2 = 9$ and $b^2 = \frac{45}{4}$.
The eccentricity $e^{\prime}$ is given by $e^{\prime} = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{45/4}{9}} = \sqrt{1 + \frac{5}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.
Therefore,$ee^{\prime} = \frac{2}{3} \times \frac{3}{2} = 1$.

(A) The equation of the line is $x+4y-4=0$,which is in the form $lx+my+n=0$,where $l=1, m=4, n=-4$.
The equation of the ellipse is $x^2+4y^2=4$,which can be written as $\frac{x^2}{4}+\frac{y^2}{1}=1$.
Here,$a^2=4$ and $b^2=1$.
The formula for the pole $(x_1, y_1)$ of the line $lx+my+n=0$ with respect to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is given by:
$x_1 = \frac{-a^2l}{n}$ and $y_1 = \frac{-b^2m}{n}$.
Substituting the values:
$x_1 = \frac{-(4)(1)}{-4} = 1$
$y_1 = \frac{-(1)(4)}{-4} = 1$
Thus,the pole is $(1,1)$.

(C) Given the polar equation: $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$
Multiply by $8r$:
$8 = r + 3r \cos \theta$
Since $x = r \cos \theta$ and $r = \sqrt{x^2 + y^2}$,we have:
$r = 8 - 3x$
Squaring both sides:
$r^2 = (8 - 3x)^2$
$x^2 + y^2 = 64 - 48x + 9x^2$
Rearranging the terms:
$8x^2 - y^2 - 48x + 64 = 0$
This is of the form $Ax^2 + By^2 + Dx + Ey + F = 0$ where the coefficients of $x^2$ and $y^2$ have opposite signs ($8$ and $-1$).
Therefore,it represents a hyperbola.