TS EAMCET 2001 Chemistry Question Paper with Answer and Solution

(D) The magnet is placed along $AB$ with its center at $O$. The length of the magnet is $2l = 10 \text{ cm}$,so $l = 5 \text{ cm} = 0.05 \text{ m}$. The magnetic moment $M = 1 \text{ Am}^2$.
Point $C$ lies on the perpendicular bisector of the magnet,which corresponds to the equatorial position.
The distance $r$ of point $C$ from the center $O$ of the magnet is the height of the equilateral triangle with side $a = 10 \text{ cm} = 0.1 \text{ m}$.
$r = \sqrt{a^2 - (a/2)^2} = \sqrt{0.1^2 - 0.05^2} = \sqrt{0.01 - 0.0025} = \sqrt{0.0075} = \sqrt{75} \times 10^{-2} \text{ m} = 5\sqrt{3} \times 10^{-2} \text{ m}$.
The magnetic field $B$ at the equatorial position is given by $B = \frac{\mu_0}{4\pi} \frac{M}{(r^2 + l^2)^{3/2}}$.
Since $r^2 + l^2 = (5\sqrt{3} \times 10^{-2})^2 + (5 \times 10^{-2})^2 = (75 + 25) \times 10^{-4} = 100 \times 10^{-4} = 10^{-2} \text{ m}^2$.
Thus,$(r^2 + l^2)^{3/2} = (10^{-2})^{3/2} = 10^{-3} \text{ m}^3$.
Substituting the values: $B = 10^{-7} \times \frac{1}{10^{-3}} = 10^{-4} \text{ T}$.

(C) Let $x = \frac{\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}}}{\sqrt{8+\sqrt{28}}-\sqrt{8-\sqrt{28}}}$.
Rationalizing the denominator:
$x = \frac{(\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}})^2}{(\sqrt{8+\sqrt{28}})^2-(\sqrt{8-\sqrt{28}})^2}$
$x = \frac{(8+\sqrt{28}) + (8-\sqrt{28}) + 2\sqrt{(8+\sqrt{28})(8-\sqrt{28})}}{(8+\sqrt{28}) - (8-\sqrt{28})}$
$x = \frac{16 + 2\sqrt{64-28}}{2\sqrt{28}}$
$x = \frac{16 + 2\sqrt{36}}{2\sqrt{4 \times 7}}$
$x = \frac{16 + 2(6)}{2(2\sqrt{7})}$
$x = \frac{16 + 12}{4\sqrt{7}} = \frac{28}{4\sqrt{7}} = \frac{7}{\sqrt{7}} = \sqrt{7}$.

(D) Given that $\alpha$ and $\beta$ are the roots of the equation $x^2+bx+c=0$.
From the relation between roots and coefficients,we have $\alpha+\beta = -b$.
Also,$\alpha+h$ and $\beta+h$ are the roots of the equation $x^2+qx+r=0$.
Therefore,the sum of the roots is $(\alpha+h) + (\beta+h) = -q$.
This simplifies to $(\alpha+\beta) + 2h = -q$.
Substituting $\alpha+\beta = -b$,we get $-b + 2h = -q$.
Rearranging the terms,$2h = b-q$.
Thus,$h = \frac{1}{2}(b-q)$.

(D) Let $y = (x-\alpha)(x-\beta)$.
Expanding the expression,we get $y = x^2 - (\alpha+\beta)x + \alpha\beta$.
To find the minimum value,we can complete the square:
$y = x^2 - (\alpha+\beta)x + \left(\frac{\alpha+\beta}{2}\right)^2 - \left(\frac{\alpha+\beta}{2}\right)^2 + \alpha\beta$.
$y = \left(x - \frac{\alpha+\beta}{2}\right)^2 - \frac{(\alpha+\beta)^2 - 4\alpha\beta}{4}$.
$y = \left(x - \frac{\alpha+\beta}{2}\right)^2 - \frac{\alpha^2 + 2\alpha\beta + \beta^2 - 4\alpha\beta}{4}$.
$y = \left(x - \frac{\alpha+\beta}{2}\right)^2 - \frac{(\alpha-\beta)^2}{4}$.
Since the square term $\left(x - \frac{\alpha+\beta}{2}\right)^2 \ge 0$,the minimum value occurs when $x = \frac{\alpha+\beta}{2}$.
Thus,the minimum value is $-\frac{1}{4}(\alpha-\beta)^2$.

(A) Given the equation: $\frac{x-4}{x^2-5x-2k} = \frac{2}{x-2} - \frac{1}{x+k}$
Simplifying the right side: $\frac{2(x+k) - (x-2)}{(x-2)(x+k)} = \frac{2x + 2k - x + 2}{(x-2)(x+k)} = \frac{x + 2k + 2}{x^2 + kx - 2x - 2k} = \frac{x + 2k + 2}{x^2 + (k-2)x - 2k}$
Comparing the denominators: $x^2 - 5x - 2k = x^2 + (k-2)x - 2k$
This implies $k-2 = -5$,so $k = -3$
Comparing the numerators: $x-4 = x + 2k + 2$
Substituting $k = -3$: $x-4 = x + 2(-3) + 2 = x - 6 + 2 = x - 4$
Both sides match,therefore $k = -3$.

(B) The given equation is $x^3-6x^2+6x-5=0$.
To shift the roots by $h$,we replace $x$ with $x+h$.
The new equation becomes $(x+h)^3-6(x+h)^2+6(x+h)-5=0$.
Expanding the terms:
$(x^3+3x^2h+3xh^2+h^3) - 6(x^2+2xh+h^2) + 6(x+h) - 5 = 0$.
Grouping the terms by powers of $x$:
$x^3 + (3h-6)x^2 + (3h^2-12h+6)x + (h^3-6h^2+6h-5) = 0$.
For the $x^2$ term to be absent,its coefficient must be zero:
$3h-6 = 0$.
Therefore,$h = 2$.

(D) Given equation: $x^3-14x^2+56x-64=0$.
Let $f(x) = x^3-14x^2+56x-64$.
Testing $x=2$: $f(2) = 8 - 14(4) + 56(2) - 64 = 8 - 56 + 112 - 64 = 0$.
Since $x=2$ is a root,$(x-2)$ is a factor.
Dividing the polynomial by $(x-2)$,we get $x^2(x-2) - 12x(x-2) + 32(x-2) = 0$.
$(x-2)(x^2-12x+32) = 0$.
Factoring the quadratic: $(x-2)(x-4)(x-8) = 0$.
The roots are $2, 4, 8$.
Since $\frac{4}{2} = 2$ and $\frac{8}{4} = 2$,the common ratio is constant,so the roots are in $GP$.

(B) Given the polynomial equation $P(x) = x^4-2x^3+2x-1=0$.
Since $1$ is a root of order $3$,$(x-1)^3$ must be a factor of $P(x)$.
Expanding $(x-1)^3 = x^3-3x^2+3x-1$.
Dividing $x^4-2x^3+2x-1$ by $(x-1)^3$:
$x^4-2x^3+2x-1 = (x-1)^3(x+1)$.
Setting the factors to zero,we get $(x-1)^3 = 0$ and $x+1 = 0$.
Thus,the roots are $1, 1, 1$ and $-1$.
The other root is $-1$.

(A) Since the coefficients of the biquadratic equation are assumed to be rational,the irrational and complex roots must occur in conjugate pairs. Thus,the roots are $1+i, 1-i, 1-\sqrt{2}, 1+\sqrt{2}$.
For the roots $1+i$ and $1-i$:
Sum $= (1+i) + (1-i) = 2$
Product $= (1+i)(1-i) = 1^2 - i^2 = 1+1 = 2$
The quadratic factor is $x^2 - 2x + 2 = 0$.
For the roots $1-\sqrt{2}$ and $1+\sqrt{2}$:
Sum $= (1-\sqrt{2}) + (1+\sqrt{2}) = 2$
Product $= (1-\sqrt{2})(1+\sqrt{2}) = 1^2 - (\sqrt{2})^2 = 1-2 = -1$
The quadratic factor is $x^2 - 2x - 1 = 0$.
The biquadratic equation is the product of these two factors:
$(x^2-2x+2)(x^2-2x-1) = 0$
$x^2(x^2-2x-1) - 2x(x^2-2x-1) + 2(x^2-2x-1) = 0$
$x^4 - 2x^3 - x^2 - 2x^3 + 4x^2 + 2x + 2x^2 - 4x - 2 = 0$
$x^4 - 4x^3 + 5x^2 - 2x - 2 = 0$.

(B) According to Hooke's Law,the extension of an elastic string is proportional to the applied tension. Let $L_0$ be the natural length of the string and $k$ be the force constant.
The length $L$ under tension $T$ is given by $L = L_0 + \frac{T}{k}$.
For $T_1 = 4 \ N$,$L_1 = a = L_0 + \frac{4}{k}$ --- $(1)$
For $T_2 = 5 \ N$,$L_2 = b = L_0 + \frac{5}{k}$ --- $(2)$
Subtracting $(1)$ from $(2)$: $b - a = \frac{5}{k} - \frac{4}{k} = \frac{1}{k} \Rightarrow k = \frac{1}{b-a}$.
Substituting $k$ into $(1)$: $L_0 = a - 4(b-a) = a - 4b + 4a = 5a - 4b$.
Now,for $T_3 = 9 \ N$,the length $x$ is:
$x = L_0 + \frac{9}{k} = (5a - 4b) + 9(b - a)$
$x = 5a - 4b + 9b - 9a$
$x = 5b - 4a$.

(D) Given: Initial velocity $u = 20 \ m/s$,angle of projection $\theta = 45^{\circ}$,and $g = 10 \ m/s^2$.
The standard equation of the trajectory of a projectile is given by $h = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Comparing this with the given equation $h = Ax - Bx^2$,we get:
$A = \tan \theta = \tan 45^{\circ} = 1$.
$B = \frac{g}{2u^2 \cos^2 \theta} = \frac{10}{2 \times (20)^2 \times (\cos 45^{\circ})^2} = \frac{10}{2 \times 400 \times (1/\sqrt{2})^2} = \frac{10}{800 \times 1/2} = \frac{10}{400} = \frac{1}{40}$.
Now,the ratio $A:B$ is calculated as:
$\frac{A}{B} = \frac{1}{1/40} = 40$.
Therefore,the ratio $A:B$ is $40:1$.

(A) According to the law of conservation of linear momentum,the initial momentum is zero,so the magnitudes of the momenta of the two fragments must be equal:
$m_1 v_1 = m_2 v_2$
$\frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{1}{3}$
Assuming the density $\rho$ of the nucleus is constant,the mass $m$ is proportional to the volume $V = \frac{4}{3} \pi R^3$,where $R$ is the radius.
$\frac{m_1}{m_2} = \frac{R_1^3}{R_2^3}$
Equating the two expressions for the mass ratio:
$\frac{R_1^3}{R_2^3} = \frac{v_2}{v_1} = \frac{1}{3}$
Taking the cube root on both sides:
$\frac{R_1}{R_2} = \left(\frac{1}{3}\right)^{1/3} = 1 : 3^{1/3}$

(B) The potential energy $U$ of a body executing simple harmonic motion $(SHM)$ is given by $U = \frac{1}{2} k r^2$,where $k = m \omega^2$ is the force constant and $r$ is the displacement.
Given:
$E_1 = \frac{1}{2} k x^2 \implies x = \sqrt{\frac{2 E_1}{k}}$
$E_2 = \frac{1}{2} k y^2 \implies y = \sqrt{\frac{2 E_2}{k}}$
At displacement $(x+y)$,the potential energy $E$ is:
$E = \frac{1}{2} k (x+y)^2$
Taking the square root of both sides:
$\sqrt{E} = \sqrt{\frac{1}{2} k} (x+y)$
Substituting the expressions for $x$ and $y$:
$\sqrt{E} = \sqrt{\frac{1}{2} k} \left( \sqrt{\frac{2 E_1}{k}} + \sqrt{\frac{2 E_2}{k}} \right)$
$\sqrt{E} = \sqrt{\frac{1}{2} k} \cdot \sqrt{\frac{2}{k}} (\sqrt{E_1} + \sqrt{E_2})$
$\sqrt{E} = \sqrt{\frac{1}{2} \cdot 2} (\sqrt{E_1} + \sqrt{E_2})$
$\sqrt{E} = \sqrt{E_1} + \sqrt{E_2}$

(B) Let the amplitude be $A$. The angular frequencies are $\omega_1 = \frac{2\pi}{T_1} = \frac{2\pi}{3}$ and $\omega_2 = \frac{2\pi}{T_2} = \frac{2\pi}{6} = \frac{\pi}{3}$.
Since they start from the origin,their displacements are $x_1 = A \sin(\omega_1 t)$ and $x_2 = A \sin(\omega_2 t)$.
When they meet,$x_1 = x_2$,so $\sin(\omega_1 t) = \sin(\omega_2 t)$.
This implies $\omega_1 t = \pi - \omega_2 t$ (for the first meeting point),so $t = \frac{\pi}{\omega_1 + \omega_2} = \frac{\pi}{\frac{2\pi}{3} + \frac{\pi}{3}} = 1 \ s$.
The velocity of a particle in $SHM$ is $v = A\omega \cos(\omega t)$.
At $t = 1 \ s$,$v_P = A(\frac{2\pi}{3}) \cos(\frac{2\pi}{3} \cdot 1) = A(\frac{2\pi}{3})(-\frac{1}{2}) = -\frac{A\pi}{3}$.
At $t = 1 \ s$,$v_Q = A(\frac{\pi}{3}) \cos(\frac{\pi}{3} \cdot 1) = A(\frac{\pi}{3})(\frac{1}{2}) = \frac{A\pi}{6}$.
The ratio of the magnitudes of velocities is $|v_P| : |v_Q| = \frac{A\pi}{3} : \frac{A\pi}{6} = 2 : 1$.

(C) Borazole,also known as borazine,is an inorganic compound with the chemical formula $B_3N_3H_6$.
It is often referred to as 'inorganic benzene' because it is isoelectronic and structurally similar to benzene $(C_6H_6)$.

(A) The tendency for catenation is directly proportional to the bond dissociation energy of the element-element bond.
As we move down the group from $C$ to $Ge$,the atomic size increases,which leads to a decrease in the overlap of orbitals and consequently a decrease in bond strength.
The bond dissociation energies for $C-C$,$Si-Si$,and $Ge-Ge$ are approximately $348 \ kJ \ mol^{-1}$,$180 \ kJ \ mol^{-1}$,and $167 \ kJ \ mol^{-1}$ respectively.
Therefore,the correct order of bond energies matches the order of catenation tendency.

(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 $(HCl)$.

(D) Fluorine $(F_2)$ is the most electronegative element and acts as a strong oxidizing agent. When passed through hot concentrated $KOH$ solution,it oxidizes water to oxygen.
The chemical reaction is:
$2F_2 + 4KOH \longrightarrow 4KF + 2H_2O + O_2$.

(C) When chlorine is passed through an aqueous solution of sodium thiosulfate (hypo),it acts as an oxidizing agent. The reaction is as follows:
$Na_2S_2O_3 + H_2O + Cl_2 \longrightarrow Na_2SO_4 + 2HCl + S \downarrow$
Thus,the products formed are sodium sulfate $(Na_2SO_4)$,hydrochloric acid $(HCl)$,and sulfur $(S)$.

(A) The general binomial expansion for $(1-x)^{-n}$ is given by $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \frac{n(n+1)(n+2)}{3!}x^3 + \ldots$
For $n = \frac{1}{2}$,we have $(1-x)^{-1/2} = 1 + \frac{1}{2}x + \frac{\frac{1}{2}(\frac{3}{2})}{2!}x^2 + \frac{\frac{1}{2}(\frac{3}{2})(\frac{5}{2})}{3!}x^3 + \ldots$
$(1-x)^{-1/2} = 1 + \frac{1}{2}x + \frac{1 \cdot 3}{2^2 \cdot 2!}x^2 + \frac{1 \cdot 3 \cdot 5}{2^3 \cdot 3!}x^3 + \ldots$
$(1-x)^{-1/2} = 1 + \frac{x}{2} + \frac{1 \cdot 3}{8}x^2 + \frac{1 \cdot 3 \cdot 5}{48}x^3 + \ldots$
Comparing this with the given series $1 + \frac{1}{4} + \frac{1 \cdot 3}{32} + \ldots$,we set $\frac{x}{2} = \frac{1}{4}$,which gives $x = \frac{1}{2}$.
Substituting $x = \frac{1}{2}$ into the expansion:
$(1 - \frac{1}{2})^{-1/2} = (\frac{1}{2})^{-1/2} = \sqrt{2}$.
Thus,the sum of the series is $\sqrt{2}$.

(A) We know that $\sin 5 \theta = 5 \sin \theta - 20 \sin ^3 \theta + 16 \sin ^5 \theta$.
Dividing by $\sin \theta$ (assuming $\sin \theta \neq 0$):
$\frac{\sin 5 \theta}{\sin \theta} = 5 - 20 \sin ^2 \theta + 16 \sin ^4 \theta$.
Using the identity $\sin ^2 \theta = 1 - \cos ^2 \theta$:
$= 5 - 20(1 - \cos ^2 \theta) + 16(1 - \cos ^2 \theta)^2$
$= 5 - 20 + 20 \cos ^2 \theta + 16(1 - 2 \cos ^2 \theta + \cos ^4 \theta)$
$= -15 + 20 \cos ^2 \theta + 16 - 32 \cos ^2 \theta + 16 \cos ^4 \theta$
$= 16 \cos ^4 \theta - 12 \cos ^2 \theta + 1$.

(A) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$A + C = 180^{\circ}$ and $B + D = 180^{\circ}$.
This implies $C = 180^{\circ} - A$ and $D = 180^{\circ} - B$.
Now,consider the expression $\cos A + \cos B + \cos C + \cos D$.
Substituting the values of $C$ and $D$:
$\cos A + \cos B + \cos(180^{\circ} - A) + \cos(180^{\circ} - B)$
Since $\cos(180^{\circ} - \theta) = -\cos \theta$,we have:
$\cos A + \cos B - \cos A - \cos B = 0$.

(A) Given that $\tan \theta + \cot \theta = 2$.
We know that $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$.
Substituting these values,we get $\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = 2$.
Taking the common denominator,$\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = 2$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\frac{1}{\sin \theta \cos \theta} = 2$,which implies $\sin \theta \cos \theta = \frac{1}{2}$.
Multiplying both sides by $2$,we get $2 \sin \theta \cos \theta = 1$,which is $\sin 2 \theta = 1$.
Thus,$2 \theta = \frac{\pi}{2}$,so $\theta = \frac{\pi}{4}$.
Therefore,$\sin \theta = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

(A) Given $\operatorname{cosec} \theta = \frac{p+q}{p-q}$.
We know $\sin \theta = \frac{p-q}{p+q}$.
Using the formula $\sin \theta = \frac{2 \tan(\theta/2)}{1 + \tan^2(\theta/2)}$,we have $\frac{2 \tan(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{p-q}{p+q}$.
Inverting both sides: $\frac{1 + \tan^2(\theta/2)}{2 \tan(\theta/2)} = \frac{p+q}{p-q}$.
Applying Componendo and Dividendo:
$\frac{1 + \tan^2(\theta/2) + 2 \tan(\theta/2)}{1 + \tan^2(\theta/2) - 2 \tan(\theta/2)} = \frac{(p+q) + (p-q)}{(p+q) - (p-q)}$.
$\frac{(1 + \tan(\theta/2))^2}{(1 - \tan(\theta/2))^2} = \frac{2p}{2q} = \frac{p}{q}$.
Taking the square root: $\frac{1 + \tan(\theta/2)}{1 - \tan(\theta/2)} = \sqrt{\frac{p}{q}}$.
Since $\tan(\pi/4) = 1$,this is $\tan(\pi/4 + \theta/2) = \sqrt{\frac{p}{q}}$.
Therefore,$\cot(\pi/4 + \theta/2) = \frac{1}{\tan(\pi/4 + \theta/2)} = \sqrt{\frac{q}{p}}$.

(A) We use the trigonometric identity $\cos^2 A - \sin^2 B = \cos(A+B) \cos(A-B)$.
Here,$A = \frac{\pi}{6} + \theta$ and $B = \frac{\pi}{6} - \theta$.
Substituting these values:
$\cos^2\left(\frac{\pi}{6}+\theta\right)-\sin^2\left(\frac{\pi}{6}-\theta\right) = \cos\left(\frac{\pi}{6}+\theta + \frac{\pi}{6}-\theta\right) \cos\left(\frac{\pi}{6}+\theta - (\frac{\pi}{6}-\theta)\right)$
$= \cos\left(\frac{2\pi}{6}\right) \cos(2\theta)$
$= \cos\left(\frac{\pi}{3}\right) \cos(2\theta)$
Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$,the expression becomes $\frac{1}{2} \cos 2\theta$.

(D) We have,$\sqrt{3} \sin x + \cos x = 4$.
Dividing both sides by $2$,we get: $\frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x = 2$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we can write this as: $\sin(x + \frac{\pi}{6}) = 2$.
Since the range of the sine function is $[-1, 1]$,the value $\sin(x + \frac{\pi}{6}) = 2$ is impossible.
Therefore,the equation has no solution.

(C) The lines $x=1$ and $y=1$ intersect at $B(1, 1)$.
The line $x+y=1$ intersects $x=1$ at $A(1, 0)$ and $y=1$ at $C(0, 1)$.
The vertices of the triangle are $A(1, 0)$,$B(1, 1)$,and $C(0, 1)$.
The lengths of the sides are:
$a = BC = \sqrt{(1-0)^2 + (1-1)^2} = 1$
$b = CA = \sqrt{(1-0)^2 + (0-1)^2} = \sqrt{1+1} = \sqrt{2}$
$c = AB = \sqrt{(1-1)^2 + (1-0)^2} = 1$
The incentre $(I_x, I_y)$ is given by:
$I_x = \frac{ax_1 + bx_2 + cx_3}{a+b+c} = \frac{1(1) + \sqrt{2}(1) + 1(0)}{1+\sqrt{2}+1} = \frac{1+\sqrt{2}}{2+\sqrt{2}} = \frac{1+\sqrt{2}}{\sqrt{2}(\sqrt{2}+1)} = \frac{1}{\sqrt{2}}$
$I_y = \frac{ay_1 + by_2 + cy_3}{a+b+c} = \frac{1(0) + \sqrt{2}(1) + 1(1)}{1+\sqrt{2}+1} = \frac{1+\sqrt{2}}{2+\sqrt{2}} = \frac{1}{\sqrt{2}}$
Thus,the incentre is $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.

(C) The given equation is $(a+2b)x + (a-b)y + (a+5b) = 0$.
Rearranging the terms by grouping $a$ and $b$:
$a(x + y + 1) + b(2x - y + 5) = 0$.
For this equation to hold for all values of $a$ and $b$,the coefficients of $a$ and $b$ must be zero independently:
$x + y + 1 = 0$ --- $(1)$
$2x - y + 5 = 0$ --- $(2)$
Adding equations $(1)$ and $(2)$:
$(x + y + 1) + (2x - y + 5) = 0$
$3x + 6 = 0$
$3x = -6$
$x = -2$.
Substituting $x = -2$ into equation $(1)$:
$-2 + y + 1 = 0$
$y - 1 = 0$
$y = 1$.
Thus,the line passes through the fixed point $(-2, 1)$.

(A) The given lines are $x+3y=10$ and $6x^2+xy-y^2=0$.
Factorizing the second equation: $6x^2+3xy-2xy-y^2=0$ $\Rightarrow 3x(2x+y)-y(2x+y)=0$ $\Rightarrow (3x-y)(2x+y)=0$.
So,the lines are $3x-y=0$ and $2x+y=0$.
Let the vertices of the triangle be $A, B, C$.
Solving $2x+y=0$ and $3x-y=0$ gives $A(0,0)$.
Solving $x+3y=10$ and $3x-y=0$ gives $B(1,3)$.
Solving $x+3y=10$ and $2x+y=0$ gives $C(-2,4)$.
The altitude from $A(0,0)$ to $BC$ $(x+3y=10)$ has slope $3$. Equation: $y-0=3(x-0) \Rightarrow 3x-y=0$.
The altitude from $B(1,3)$ to $AC$ $(2x+y=0)$ has slope $1/2$. Equation: $y-3=\frac{1}{2}(x-1) \Rightarrow x-2y+5=0$.
Solving $3x-y=0$ and $x-2y+5=0$: $y=3x$ $\Rightarrow x-2(3x)+5=0$ $\Rightarrow -5x=-5$ $\Rightarrow x=1, y=3$.
The orthocentre is $(1,3)$.

(B) The given pair of straight lines is $ax^2 + 2hxy + by^2 = 0$.
The lines bisecting the angle between the coordinate axes are $y = x$ and $y = -x$.
If $y = x$ is one of the lines,then substituting $y = x$ into the equation gives:
$ax^2 + 2h(x)(x) + b(x)^2 = 0$
$ax^2 + 2hx^2 + bx^2 = 0$
$(a + 2h + b)x^2 = 0$
This implies $a + b + 2h = 0$,or $a + b = -2h$.
Squaring both sides,we get $(a + b)^2 = (-2h)^2 = 4h^2$.
Similarly,if $y = -x$ is one of the lines,substituting $y = -x$ gives:
$ax^2 + 2h(x)(-x) + b(-x)^2 = 0$
$ax^2 - 2hx^2 + bx^2 = 0$
$(a - 2h + b)x^2 = 0$
This implies $a + b = 2h$,which also leads to $(a + b)^2 = 4h^2$ upon squaring.
Thus,the condition is $(a + b)^2 = 4h^2$.

(B) Given the pair of straight lines $ax^2 + 2hxy + by^2 = 0$.
Let the slopes of the lines be $m_1$ and $m_2$.
Given $m_1 = 2m_2$ ...$(i)$
We know that $m_1 + m_2 = -\frac{2h}{b}$ ...(ii)
and $m_1 m_2 = \frac{a}{b}$ ...(iii)
Substituting $(i)$ into (ii): $2m_2 + m_2 = -\frac{2h}{b} \implies 3m_2 = -\frac{2h}{b} \implies m_2 = -\frac{2h}{3b}$ ...(iv)
Substituting $(i)$ into (iii): $2m_2 \cdot m_2 = \frac{a}{b} \implies 2m_2^2 = \frac{a}{b}$ ...$(v)$
Substituting (iv) into $(v)$: $2(-\frac{2h}{3b})^2 = \frac{a}{b} \implies 2 \cdot \frac{4h^2}{9b^2} = \frac{a}{b} \implies \frac{8h^2}{9b^2} = \frac{a}{b} \implies 8h^2 = 9ab$.

(B) Let the point on the circle $x^2+y^2=p^2$ be $(x_1, y_1)$.
Then $x_1^2+y_1^2=p^2$.
The equation of the polar of $(x_1, y_1)$ with respect to the circle $x^2+y^2=q^2$ is $x x_1+y y_1=q^2$.
This line touches the circle $x^2+y^2=r^2$.
The perpendicular distance from the center $(0, 0)$ to the line $x x_1+y y_1-q^2=0$ is equal to the radius $r$.
So,$\frac{|0(x_1)+0(y_1)-q^2|}{\sqrt{x_1^2+y_1^2}} = r$.
This simplifies to $\frac{q^2}{\sqrt{p^2}} = r$,which gives $\frac{q^2}{p} = r$.
Therefore,$q^2 = pr$,which implies that $p, q, r$ are in $GP$.

(C) The given circles are $S_1: x^2+y^2+2x-2y+2=0$ and $S_2: x^2+y^2-\frac{2}{5}x-\frac{16}{5}y+\frac{13}{5}=0$.
The family of co-axial circles is given by $S_1 + \lambda(S_1 - S_2) = 0$ or $S_1 + k S_2 = 0$.
The limiting points are the centers of the point circles in the system,which occur when the radius $r = 0$.
For a circle $x^2+y^2+2gx+2fy+c=0$,the radius is $r = \sqrt{g^2+f^2-c}$. Setting $r=0$ gives $g^2+f^2=c$.
The radical axis is $S_1 - S_2 = 0$ $\Rightarrow (2 + \frac{2}{5})x + (-2 + \frac{16}{5})y + (2 - \frac{13}{5}) = 0$ $\Rightarrow \frac{12}{5}x + \frac{6}{5}y - \frac{3}{5} = 0$ $\Rightarrow 4x + 2y - 1 = 0$.
Any circle in the system is $S_1 + \lambda(4x + 2y - 1) = 0 \Rightarrow x^2+y^2+(2+4\lambda)x+(-2+2\lambda)y+(2-\lambda)=0$.
For limiting points,$g^2+f^2=c \Rightarrow (1+2\lambda)^2 + (-1+\lambda)^2 = 2-\lambda$.
$1+4\lambda+4\lambda^2 + 1-2\lambda+\lambda^2 = 2-\lambda \Rightarrow 5\lambda^2+3\lambda=0$.
Thus,$\lambda=0$ or $\lambda=-\frac{3}{5}$.
For $\lambda=0$,the center is $(-1, 1)$.
For $\lambda=-\frac{3}{5}$,the center is $(-(1+2(-\frac{3}{5})), -(-1+(-\frac{3}{5}))) = (-1+\frac{6}{5}, 1+\frac{3}{5}) = (\frac{1}{5}, \frac{8}{5})$.
The limiting points are $(-1, 1)$ and $(\frac{1}{5}, \frac{8}{5})$.

(C) The given equation of the parabola is $y^2+8x-2y+17=0$.
Rearranging the terms to complete the square for $y$:
$y^2-2y+1=-8x-17+1$
$(y-1)^2 = -8x-16$
$(y-1)^2 = -8(x+2)$
Comparing this with the standard form $(y-k)^2 = -4a(x-h)$,we get $4a = 8$.
Thus,the length of the latus rectum is $8$.

(B) The given equation of the parabola is $y^2=4x$.
For a point $P(x_1, y_1)$ on the parabola $y^2=4ax$,the normal at $P$ is given by $y-y_1 = -\frac{y_1}{2a}(x-x_1)$.
Here,$a=1$ and $P=(1,2)$,so the slope of the normal is $m = -\frac{2}{2(1)} = -1$.
The equation of the normal at $P(1,2)$ is $y-2 = -1(x-1)$,which simplifies to $x+y=3$,or $x=3-y$.
Substituting $x=3-y$ into the parabola equation $y^2=4x$:
$y^2 = 4(3-y)$
$y^2 = 12-4y$
$y^2+4y-12=0$
$(y-2)(y+6)=0$
This gives $y=2$ (which corresponds to point $P$) and $y=-6$.
For $y=-6$,$x = 3-(-6) = 9$.
Therefore,the coordinates of point $Q$ are $(9,-6)$.

(D) We have,$\frac{(1-3x)^2}{(1-2x)} = (1 - 6x + 9x^2)(1 - 2x)^{-1}$.
Using the binomial expansion $(1-2x)^{-1} = 1 + (2x) + (2x)^2 + (2x)^3 + (2x)^4 + \dots = 1 + 2x + 4x^2 + 8x^3 + 16x^4 + \dots$.
Now,multiply $(1 - 6x + 9x^2)$ by $(1 + 2x + 4x^2 + 8x^3 + 16x^4 + \dots)$.
The coefficient of $x^4$ is obtained by:
$1 \times (16x^4) - 6x \times (8x^3) + 9x^2 \times (4x^2) = 16x^4 - 48x^4 + 36x^4$.
Summing these coefficients: $16 - 48 + 36 = 4$.

(C) The given equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we get $a^2 = 16$ and $b^2 = 9$.
The formula for eccentricity $e$ is $e = \sqrt{1 - \frac{b^2}{a^2}}$.
Substituting the values,we get $e = \sqrt{1 - \frac{9}{16}}$.
$e = \sqrt{\frac{16 - 9}{16}} = \sqrt{\frac{7}{16}}$.
Therefore,$e = \frac{\sqrt{7}}{4}$.

(B) The given equation is $16 x^2+y^2+8 x y-74 x-78 y+212=0$.
Comparing this with the general second-degree equation $a x^2+b y^2+2 h x y+2 g x+2 f y+c=0$,we get:
$a=16, b=1, h=4, g=-37, f=-39, c=212$.
Now,we calculate the discriminant $h^2 - ab$:
$h^2 - ab = (4)^2 - (16)(1) = 16 - 16 = 0$.
Since $h^2 - ab = 0$,the equation represents a parabola.

(B) Given the polar equation: $\frac{l}{r} = 2 \sin^2 \frac{\theta}{2}$.
Using the trigonometric identity $2 \sin^2 \frac{\theta}{2} = 1 - \cos \theta$,we get:
$\frac{l}{r} = 1 - \cos \theta$
$l = r(1 - \cos \theta)$
$l = r - r \cos \theta$
Since $x = r \cos \theta$ and $r = \sqrt{x^2 + y^2}$,we substitute these into the equation:
$l = \sqrt{x^2 + y^2} - x$
$l + x = \sqrt{x^2 + y^2}$
Squaring both sides:
$(l + x)^2 = x^2 + y^2$
$l^2 + 2lx + x^2 = x^2 + y^2$
$y^2 = 2lx + l^2$
$y^2 = 2l(x + \frac{l}{2})$
This is the standard form of a parabola $Y^2 = 4AX$ with vertex at $(-\frac{l}{2}, 0)$.

(C) The equation of the hyperbola is $x^2 - y^2 = 8$.
The asymptotes of the hyperbola $x^2 - y^2 = a^2$ are given by $x^2 - y^2 = 0$,which implies $x = \pm y$,or $x + y = 0$ and $x - y = 0$.
Let $P(x, y)$ be any point on the hyperbola.
The length of the perpendicular from $P(x, y)$ to the line $x - y = 0$ is $d_1 = \frac{|x - y|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y|}{\sqrt{2}}$.
The length of the perpendicular from $P(x, y)$ to the line $x + y = 0$ is $d_2 = \frac{|x + y|}{\sqrt{1^2 + 1^2}} = \frac{|x + y|}{\sqrt{2}}$.
The product of the lengths of the perpendiculars is $d_1 \times d_2 = \frac{|x - y|}{\sqrt{2}} \times \frac{|x + y|}{\sqrt{2}} = \frac{|x^2 - y^2|}{2}$.
Since the point $(x, y)$ lies on the hyperbola $x^2 - y^2 = 8$,we have $d_1 \times d_2 = \frac{8}{2} = 4$.

(B) We know that $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x} = 1$.
Given limit is $\lim _{x \rightarrow 0} \frac{\sin x \sin ^{-1} x}{x^2}$.
This can be rewritten as $\lim _{x \rightarrow 0} \left(\frac{\sin x}{x}\right) \left(\frac{\sin ^{-1} x}{x}\right)$.
Applying the limit properties,we get $1 \times 1 = 1$.

(A) We evaluate the limit $\lim _{x \rightarrow \infty} \left(\frac{x+a}{x+b}\right)^{x}$.
This is of the form $1^{\infty}$.
We can rewrite the expression as:
$\lim _{x \rightarrow \infty} \left(\frac{x+b+a-b}{x+b}\right)^{x} = \lim _{x \rightarrow \infty} \left(1 + \frac{a-b}{x+b}\right)^{x}$.
Using the standard limit formula $\lim _{x \rightarrow \infty} (1 + \frac{k}{x})^x = e^k$,we have:
$\lim _{x \rightarrow \infty} \left(1 + \frac{a-b}{x+b}\right)^{x} = \lim _{x \rightarrow \infty} \left[\left(1 + \frac{a-b}{x+b}\right)^{\frac{x+b}{a-b}}\right]^{\frac{x(a-b)}{x+b}}$.
Since $\lim _{x \rightarrow \infty} \frac{x(a-b)}{x+b} = a-b$,the limit is $e^{a-b}$.

(B) Given limit: $L = \lim _{x \rightarrow 0} \frac{x(10^x - 1)}{1 - \cos x}$.
Since this is a $\frac{0}{0}$ form,we apply $L$-Hospital's rule:
$L = \lim _{x \rightarrow 0} \frac{x \cdot 10^x \ln 10 + (10^x - 1)}{\sin x}$.
Again,this is a $\frac{0}{0}$ form,so we apply $L$-Hospital's rule again:
$L = \lim _{x \rightarrow 0} \frac{x \cdot 10^x (\ln 10)^2 + 10^x \ln 10 + 10^x \ln 10}{\cos x}$.
Substituting $x = 0$:
$L = \frac{0 \cdot 10^0 (\ln 10)^2 + 10^0 \ln 10 + 10^0 \ln 10}{\cos 0} = \frac{0 + \ln 10 + \ln 10}{1} = 2 \ln 10 = 2 \log_{e} 10$.
Note: In the context of the options provided,$\log 10$ refers to $\ln 10$ (natural logarithm).

(D) Styrene $(C_6H_5CH=CH_2)$ is prepared from benzene $(C_6H_6)$ in two steps:
$1$. Benzene reacts with ethylene $(CH_2=CH_2)$ in the presence of $AlCl_3$ to form ethylbenzene $(C_6H_5CH_2CH_3)$.
$2$. Ethylbenzene is then dehydrogenated at high temperature (approx. $700 \ ^\circ C$) using a catalyst (like iron oxide or $Al_2O_3$) to produce styrene and hydrogen gas $(H_2)$.
Thus,benzene is the starting material used in the preparation of styrene.

(B) Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these into the first term:
$\frac{\cos C+\cos A}{c+a} = \frac{2 \cos \frac{C+A}{2} \cos \frac{C-A}{2}}{2R(\sin C+\sin A)} = \frac{2 \cos \frac{C+A}{2} \cos \frac{C-A}{2}}{2R(2 \sin \frac{C+A}{2} \cos \frac{C-A}{2})} = \frac{1}{2R} \cot \frac{C+A}{2}$.
Since $A+B+C = \pi$,$\frac{C+A}{2} = \frac{\pi}{2} - \frac{B}{2}$,so $\cot \frac{C+A}{2} = \tan \frac{B}{2}$.
Thus,the first term is $\frac{1}{2R} \tan \frac{B}{2}$.
For the second term:
$\frac{\cos B}{b} = \frac{\cos B}{2R \sin B} = \frac{1}{2R} \cot B = \frac{1}{2R} \left( \frac{1-\tan^2 \frac{B}{2}}{2 \tan \frac{B}{2}} \right)$.
Adding both terms:
$\frac{1}{2R} \left( \tan \frac{B}{2} + \frac{1-\tan^2 \frac{B}{2}}{2 \tan \frac{B}{2}} \right) = \frac{1}{2R} \left( \frac{2 \tan^2 \frac{B}{2} + 1 - \tan^2 \frac{B}{2}}{2 \tan \frac{B}{2}} \right) = \frac{1}{2R} \left( \frac{1+\tan^2 \frac{B}{2}}{2 \tan \frac{B}{2}} \right) = \frac{1}{2R \sin B} = \frac{1}{b}$.

(D) Given,$\frac{a}{b^2-c^2} + \frac{c}{b^2-a^2} = 0$.
Using the sine rule $a = 2R \sin A$,$b = 2R \sin B$,$c = 2R \sin C$:
$\frac{2R \sin A}{4R^2(\sin^2 B - \sin^2 C)} + \frac{2R \sin C}{4R^2(\sin^2 B - \sin^2 A)} = 0$
$\frac{\sin A}{\sin(B+C)\sin(B-C)} + \frac{\sin C}{\sin(B+A)\sin(B-A)} = 0$
Since $A+B+C = \pi$,$\sin(B+C) = \sin A$ and $\sin(B+A) = \sin C$:
$\frac{\sin A}{\sin A \sin(B-C)} + \frac{\sin C}{\sin C \sin(B-A)} = 0$
$\frac{1}{\sin(B-C)} + \frac{1}{\sin(B-A)} = 0$
$\sin(B-A) = -\sin(B-C) = \sin(C-B)$
$B-A = C-B$ or $B-A = \pi - (C-B)$
$2B = A+C$. Since $A+B+C = \pi$,$A+C = \pi - B$.
$2B = \pi - B$ $\Rightarrow 3B = \pi$ $\Rightarrow B = \frac{\pi}{3}$.

(B) We know that $a = 2R \sin A$ and $c = 2R \sin C$. Substituting these into the expression:
$a^2 \sin 2C + c^2 \sin 2A = (2R \sin A)^2 (2 \sin C \cos C) + (2R \sin C)^2 (2 \sin A \cos A)$
$= 8R^2 \sin^2 A \sin C \cos C + 8R^2 \sin^2 C \sin A \cos A$
$= 8R^2 \sin A \sin C [\sin A \cos C + \cos A \sin C]$
$= 8R^2 \sin A \sin C \sin(A + C)$
Since $A + B + C = 180^{\circ}$,$\sin(A + C) = \sin B$.
$= 8R^2 \sin A \sin B \sin C$
Using the formula for the area of a triangle $\Delta = \frac{abc}{4R}$,we have $abc = 4R\Delta$. Also,$\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$.
$= 8R^2 \left( \frac{a}{2R} \right) \left( \frac{b}{2R} \right) \left( \frac{c}{2R} \right) = \frac{abc}{R} = \frac{4R\Delta}{R} = 4\Delta$.

(D) Let $h$ be the height of the tower $AB$ and $x$ be the length of the shadow when the Sun's altitude is $45^{\circ}$.
In $\triangle ABD$,$\tan 45^{\circ} = \frac{AB}{AD} = \frac{h}{x}$. Since $\tan 45^{\circ} = 1$,we have $h = x$.
In $\triangle ABC$,the shadow length is $AC = AD + DC = x + 60$. The Sun's altitude is $30^{\circ}$.
$\tan 30^{\circ} = \frac{AB}{AC} = \frac{h}{x + 60}$.
Substituting $x = h$,we get $\frac{1}{\sqrt{3}} = \frac{h}{h + 60}$.
$h + 60 = h\sqrt{3}$.
$60 = h(\sqrt{3} - 1)$.
$h = \frac{60}{\sqrt{3} - 1}$.
Rationalizing the denominator: $h = \frac{60(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{60(\sqrt{3} + 1)}{3 - 1} = \frac{60(\sqrt{3} + 1)}{2} = 30(\sqrt{3} + 1) \ m$.

(B) square matrix is called a scalar matrix if all its diagonal elements are equal to a constant $k$ and all non-diagonal elements are zero.
Given that $a_{ij} = 0$ for $i \neq j$ and $a_{ij} = k$ for $i = j$,this definition perfectly matches the definition of a scalar matrix.
Therefore,the correct option is $B$.