TS EAMCET 2003 Chemistry Question Paper with Answer and Solution

(C) Given $A+B+C=270^{\circ}$,so $A+B = 270^{\circ}-C$.
Using the identity $\cos 2A + \cos 2B = 2 \cos(A+B) \cos(A-B)$:
$\cos 2A + \cos 2B + \cos 2C = 2 \cos(A+B) \cos(A-B) + (2 \cos^2 C - 1)$
$= 2 \cos(270^{\circ}-C) \cos(A-B) + 2 \cos^2 C - 1$
$= 2(-\sin C) \cos(A-B) + 2 \cos^2 C - 1$
$= -2 \sin C \cos(A-B) + 2(1-\sin^2 C) - 1$
$= 1 - 2 \sin C [\cos(A-B) + \sin C]$
Since $\sin C = \sin(270^{\circ}-(A+B)) = -\cos(A+B)$:
$= 1 - 2 \sin C [\cos(A-B) - \cos(A+B)]$
$= 1 - 2 \sin C [2 \sin A \sin B]$
$= 1 - 4 \sin A \sin B \sin C$.

(A) Expand each term using the formula $\sin(x-y) = \sin x \cos y - \cos x \sin y$:
$\cos \alpha (\sin \beta \cos \gamma - \cos \beta \sin \gamma) + \cos \beta (\sin \gamma \cos \alpha - \cos \gamma \sin \alpha) + \cos \gamma (\sin \alpha \cos \beta - \cos \alpha \sin \beta)$
$= \cos \alpha \sin \beta \cos \gamma - \cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha - \cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta - \cos \gamma \cos \alpha \sin \beta$
Observe that all terms cancel out:
$(\cos \alpha \sin \beta \cos \gamma - \cos \gamma \cos \alpha \sin \beta) + (-\cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha) + (-\cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta) = 0 + 0 + 0 = 0$

(D) We know that $\sin 6 \theta = \sin 3(2 \theta)$.
Using the identity $\sin 3A = 3 \sin A - 4 \sin^3 A$,we get:
$\sin 6 \theta = 3 \sin 2 \theta - 4 \sin^3 2 \theta$.
Substitute $\sin 2 \theta = 2 \sin \theta \cos \theta$:
$\sin 6 \theta = 3(2 \sin \theta \cos \theta) - 4(2 \sin \theta \cos \theta)^3$
$= 6 \sin \theta \cos \theta - 4(8 \sin^3 \theta \cos^3 \theta)$
$= 6 \sin \theta \cos \theta - 32 \sin^3 \theta \cos^3 \theta$
$= 6 \sin \theta \cos \theta - 32 \sin \theta \cos^3 \theta (1 - \cos^2 \theta)$
$= 6 \sin \theta \cos \theta - 32 \sin \theta \cos^3 \theta + 32 \sin \theta \cos^5 \theta$
$= 32 \cos^5 \theta \sin \theta - 32 \cos^3 \theta \sin \theta + 3(2 \sin \theta \cos \theta)$
$= 32 \cos^5 \theta \sin \theta - 32 \cos^3 \theta \sin \theta + 3 \sin 2 \theta$.
Comparing this with the given equation $\sin 6 \theta = 32 \cos^5 \theta \sin \theta - 32 \cos^3 \theta \sin \theta + 3x$,we find $3x = 3 \sin 2 \theta$,which implies $x = \sin 2 \theta$.

(D) Let the coordinates of $P$ be $(x, y)$.
Since $P$ is equidistant from $A, B, C$,we have $PA^2 = PB^2$ and $PB^2 = PC^2$.
From $PA^2 = PB^2$:
$(x-1)^2 + (y-3)^2 = (x+3)^2 + (y-5)^2$
$x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 + 6x + 9 + y^2 - 10y + 25$
$-2x - 6y + 10 = 6x - 10y + 34$
$8x - 4y + 24 = 0 \Rightarrow 2x - y + 6 = 0$ ... $(i)$
From $PB^2 = PC^2$:
$(x+3)^2 + (y-5)^2 = (x-5)^2 + (y+1)^2$
$x^2 + 6x + 9 + y^2 - 10y + 25 = x^2 - 10x + 25 + y^2 + 2y + 1$
$16x - 12y + 8 = 0 \Rightarrow 4x - 3y + 2 = 0$ ... (ii)
Solving $(i)$ and (ii):
Multiply $(i)$ by $2$: $4x - 2y + 12 = 0$ ... (iii)
Subtract (ii) from (iii): $(-2y - (-3y)) + (12 - 2) = 0$ $\Rightarrow y + 10 = 0$ $\Rightarrow y = -10$
Substitute $y = -10$ into $(i)$: $2x - (-10) + 6 = 0$ $\Rightarrow 2x + 16 = 0$ $\Rightarrow x = -8$
Thus,$P$ is $(-8, -10)$.
$PA = \sqrt{(-8-1)^2 + (-10-3)^2} = \sqrt{(-9)^2 + (-13)^2} = \sqrt{81 + 169} = \sqrt{250} = 5 \sqrt{10}$.

(D) Let $(x, y)$ be the coordinates in the original system and $(x', y')$ be the coordinates in the new system after rotation by an angle $\theta = 135^{\circ}$.
The transformation equations are:
$x' = x \cos \theta + y \sin \theta$
$y' = -x \sin \theta + y \cos \theta$
Given $(x', y') = (4, -3)$ and $\theta = 135^{\circ}$,we have $\cos 135^{\circ} = -\frac{1}{\sqrt{2}}$ and $\sin 135^{\circ} = \frac{1}{\sqrt{2}}$.
Substituting these values:
$4 = x(-\frac{1}{\sqrt{2}}) + y(\frac{1}{\sqrt{2}}) \implies -x + y = 4\sqrt{2} \quad (i)$
$-3 = -x(\frac{1}{\sqrt{2}}) + y(-\frac{1}{\sqrt{2}}) \implies -x - y = -3\sqrt{2} \implies x + y = 3\sqrt{2} \quad (ii)$
Adding $(i)$ and $(ii)$:
$2y = 7\sqrt{2} \implies y = \frac{7\sqrt{2}}{2} = \frac{7}{\sqrt{2}}$
Subtracting $(i)$ from $(ii)$:
$2x = -\sqrt{2} \implies x = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$
Thus,the original coordinates are $(x, y) = \left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$.

(A) The given equation is $\sqrt{3} \sin \theta + 2 \cos \theta = \frac{4}{r}$.
Multiplying by $r$,we get $\sqrt{3} r \sin \theta + 2 r \cos \theta = 4$.
Using polar coordinates $x = r \cos \theta$ and $y = r \sin \theta$,the equation becomes $\sqrt{3} y + 2 x = 4$,or $2x + \sqrt{3}y - 4 = 0$.
The slope of this line is $m_1 = -\frac{2}{\sqrt{3}}$.
$A$ line perpendicular to this will have a slope $m_2 = \frac{\sqrt{3}}{2}$.
The equation of a line with slope $m_2$ passing through the point $(x_1, y_1) = (-1, \frac{\pi}{2})$ in polar form is tricky,so we convert the point to Cartesian coordinates: $x = r \cos \theta = -1 \cos(\frac{\pi}{2}) = 0$ and $y = r \sin \theta = -1 \sin(\frac{\pi}{2}) = -1$.
So the point is $(0, -1)$.
The equation of the line is $y - (-1) = \frac{\sqrt{3}}{2}(x - 0)$,which simplifies to $2y + 2 = \sqrt{3}x$,or $\sqrt{3}x - 2y = 2$.
Substituting $x = r \cos \theta$ and $y = r \sin \theta$,we get $\sqrt{3} r \cos \theta - 2 r \sin \theta = 2$.

(C) Since the given three lines are concurrent,the determinant of their coefficients must be zero:
$\left|\begin{array}{rrr} 4 & 3 & -1 \\ 1 & -1 & 5 \\ k & 5 & -3 \end{array}\right| = 0$
Expanding the determinant along the first row:
$4((-1)(-3) - (5)(5)) - 3((1)(-3) - (5)(k)) - 1((1)(5) - (-1)(k)) = 0$
$4(3 - 25) - 3(-3 - 5k) - 1(5 + k) = 0$
$4(-22) + 9 + 15k - 5 - k = 0$
$-88 + 9 - 5 + 14k = 0$
$-84 + 14k = 0$
$14k = 84$
$k = 6$

(D) The pair of lines is given by $A x^2+2 H x y+B y^2=0$.
Since the triangle formed by these lines and $a x+b y+c=0$ is equilateral,the angle between the lines of the pair must be $60^{\circ}$.
The angle $\theta$ between the lines $A x^2+2 H x y+B y^2=0$ is given by $\tan \theta = \frac{2\sqrt{H^2-A B}}{|A+B|}$.
For an equilateral triangle,$\theta = 60^{\circ}$,so $\tan 60^{\circ} = \sqrt{3} = \frac{2\sqrt{H^2-A B}}{|A+B|}$.
Squaring both sides,we get $3 = \frac{4(H^2-A B)}{(A+B)^2}$.
$3(A+B)^2 = 4(H^2-A B)$.
$3(A^2+B^2+2 A B) = 4 H^2-4 A B$.
$3 A^2+3 B^2+6 A B = 4 H^2-4 A B$.
$3 A^2+10 A B+3 B^2 = 4 H^2$.
Factoring the left side,we get $(3 A+B)(A+3 B) = 4 H^2$.

(A) The given equations are $\lambda^2 x^2 - m^2 y^2 \mp n(\lambda x + m y) = 0$.
Factoring these,we get:
$(\lambda x - m y)(\lambda x + m y) - n(\lambda x + m y) = 0 \implies (\lambda x + m y)(\lambda x - m y - n) = 0$
$(\lambda x - m y)(\lambda x + m y) + n(\lambda x + m y) = 0 \implies (\lambda x + m y)(\lambda x - m y + n) = 0$
This represents two pairs of parallel lines:
Pair $1$: $\lambda x + m y = 0$ and $\lambda x + m y + n = 0$
Pair $2$: $\lambda x - m y = 0$ and $\lambda x - m y - n = 0$
The area of the parallelogram formed by lines $a_1 x + b_1 y + c_1 = 0, a_1 x + b_1 y + c_2 = 0$ and $a_2 x + b_2 y + d_1 = 0, a_2 x + b_2 y + d_2 = 0$ is given by $\left| \frac{(c_1 - c_2)(d_1 - d_2)}{a_1 b_2 - a_2 b_1} \right|$.
Here,$c_1 = 0, c_2 = n$ and $d_1 = 0, d_2 = -n$.
$a_1 = \lambda, b_1 = m$ and $a_2 = \lambda, b_2 = -m$.
Area $= \left| \frac{(0 - n)(0 - (-n))}{\lambda(-m) - \lambda(m)} \right| = \left| \frac{-n^2}{-2\lambda m} \right| = \frac{n^2}{2|\lambda m|}$ sq units.

(A) The structure of $2, 3-$dimethylhexane is: $CH_3-CH(CH_3)-CH(CH_3)-CH_2-CH_2-CH_3$.
$1$. Tertiary $(3^{\circ})$ carbon atoms: These are carbon atoms bonded to three other carbon atoms. In this structure,the carbons at positions $2$ and $3$ of the hexane chain are tertiary. Total $= 2$.
$2$. Secondary $(2^{\circ})$ carbon atoms: These are carbon atoms bonded to two other carbon atoms. In this structure,the carbons at positions $4$ and $5$ of the hexane chain are secondary. Total $= 2$.
$3$. Primary $(1^{\circ})$ carbon atoms: These are carbon atoms bonded to only one other carbon atom. In this structure,the two methyl groups attached at positions $2$ and $3$,and the terminal carbons at positions $1$ and $6$ are primary. Total $= 4$.
Thus,the number of tertiary,secondary,and primary carbon atoms are $2, 2, 4$ respectively. However,looking at the options provided,the correct sequence is $2, 2, 4$. Since $2, 2, 4$ is not explicitly listed as an option,we evaluate the provided options. Option $A$ is $2, 2, 4$.

(D) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$T^2 \propto R^3$.
Let $T_1$ be the time period for radius $R_1 = R$ and $T_2$ be the time period for radius $R_2 = 1.02 R$.
Then,$\frac{T_2}{T_1} = \left( \frac{R_2}{R_1} \right)^{3/2} = (1.02)^{3/2}$.
Using the binomial approximation $(1 + x)^n \approx 1 + nx$ for small $x$:
$\frac{T_2}{T_1} = (1 + 0.02)^{3/2} \approx 1 + \frac{3}{2} \times 0.02 = 1 + 0.03 = 1.03$.
Thus,$T_2 = 1.03 T_1$.
The percentage difference is given by $\frac{T_2 - T_1}{T_1} \times 100 = \frac{1.03 T_1 - T_1}{T_1} \times 100 = 0.03 \times 100 = 3\%$.

(B) The reaction of propene with $HBr$ in the presence of peroxide (Kharasch effect) proceeds via a free radical mechanism.
In this mechanism,the bromine radical $(\dot{Br})$ attacks the double bond to form a more stable secondary free radical intermediate $(CH_3-\dot{C}H-CH_2Br)$.
This secondary free radical then abstracts a hydrogen atom from $HBr$ to form the final product,$1$-bromopropane $(CH_3-CH_2-CH_2Br)$.

(A) When acetylene $(C_2H_2)$ is passed through a red hot iron tube,it undergoes cyclic polymerization to form benzene $(C_6H_6)$,which is compound $X$.
$3C_2H_2 \xrightarrow{\text{red hot iron tube}} C_6H_6 (X)$
Option $A$ represents the reduction of phenol using zinc dust,which is a standard laboratory method to produce benzene:
$C_6H_5OH + Zn \xrightarrow{\text{distillation}} C_6H_6 + ZnO$
Thus,reaction $A$ yields $X$ as the major product.

(A) From the given reactions:
$1$. $X + HCl \rightarrow C_2H_5Cl$ (Addition reaction indicates $X$ is an alkene,$C_2H_4$).
$2$. $Y + HCl \rightarrow C_2H_5Cl$ (Substitution reaction indicates $Y$ is an alcohol,$C_2H_5OH$).
$3$. The conversion of $Y$ $(C_2H_5OH)$ to $X$ $(C_2H_4)$ is a dehydration reaction.
$4$. Dehydration of ethanol to ethene occurs by heating with $Al_2O_3$ at $350^{\circ}C$.
$\underset{(Y)}{C_2H_5OH} \xrightarrow[350^{\circ}C]{Al_2O_3} \underset{(X)}{C_2H_4} + H_2O$

(A) Given the system of equations:
$2x + y - z = 7$ $(i)$
$x - 3y + 2z = 1$ $(ii)$
$x + 4y - 3z = 5$ $(iii)$
To find the number of solutions,we check the consistency of the system.
Let $D$ be the determinant of the coefficient matrix:
$D = \begin{vmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & -3 \end{vmatrix} = 2(9 - 8) - 1(-3 - 2) - 1(4 + 3) = 2(1) - 1(-5) - 1(7) = 2 + 5 - 7 = 0$.
Since $D = 0$,the system is either inconsistent or has infinitely many solutions.
Let us check the consistency using the augmented matrix or by eliminating variables.
From $(i) + (ii) \times 2$: $(2x + y - z) + 2(x - 3y + 2z) = 7 + 2(1) \implies 4x - 5y + 3z = 9$.
Alternatively,consider $(ii) + (iii)$: $(x - 3y + 2z) + (x + 4y - 3z) = 1 + 5 \implies 2x + y - z = 6$.
Comparing this with equation $(i)$,which is $2x + y - z = 7$,we see a contradiction $(6 = 7)$.
Therefore,the system is inconsistent and has $0$ solutions.

(B) Given $f(x) = \frac{x-1}{2x^2-7x+5}$ for $x \neq 1$.
Factoring the denominator: $2x^2-7x+5 = 2x^2-2x-5x+5 = 2x(x-1)-5(x-1) = (2x-5)(x-1)$.
Thus,for $x \neq 1$,$f(x) = \frac{x-1}{(2x-5)(x-1)} = \frac{1}{2x-5}$.
The function is $f(x) = \begin{cases} \frac{1}{2x-5}, & x \neq 1 \\ -\frac{1}{3}, & x=1 \end{cases}$.
By the definition of the derivative,$f'(1) = \lim_{h \to 0} \frac{f(1+h)-f(1)}{h}$.
$f'(1) = \lim_{h \to 0} \frac{\frac{1}{2(1+h)-5} - (-\frac{1}{3})}{h} = \lim_{h \to 0} \frac{\frac{1}{2h-3} + \frac{1}{3}}{h}$.
$f'(1) = \lim_{h \to 0} \frac{3 + (2h-3)}{3h(2h-3)} = \lim_{h \to 0} \frac{2h}{3h(2h-3)} = \lim_{h \to 0} \frac{2}{3(2h-3)}$.
Substituting $h=0$,we get $f'(1) = \frac{2}{3(-3)} = -\frac{2}{9}$.

(B) Given the function $f(x) = \frac{x}{1+|x|}$.
To find $f^{\prime}(0)$,we check the left-hand derivative and right-hand derivative at $x = 0$.
For $x > 0$,$|x| = x$,so $f(x) = \frac{x}{1+x}$. Then $f^{\prime}(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} = \frac{1}{(1+x)^2}$. Thus,$f^{\prime}(0^+) = \lim_{x \to 0^+} \frac{1}{(1+x)^2} = 1$.
For $x < 0$,$|x| = -x$,so $f(x) = \frac{x}{1-x}$. Then $f^{\prime}(x) = \frac{(1-x)(1) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2}$. Thus,$f^{\prime}(0^-) = \lim_{x \to 0^-} \frac{1}{(1-x)^2} = 1$.
Since $f^{\prime}(0^+) = f^{\prime}(0^-) = 1$,the derivative $f^{\prime}(0)$ exists and is equal to $1$.

(C) Given,$u(x, y) = y \log x + x \log y$.
First,we find the partial derivatives $u_x$ and $u_y$:
$u_x = \frac{\partial}{\partial x}(y \log x + x \log y) = \frac{y}{x} + \log y$.
$u_y = \frac{\partial}{\partial y}(y \log x + x \log y) = \log x + \frac{x}{y}$.
Now,consider the expression $E = u_x u_y - u_x \log x - u_y \log y + \log x \log y$.
We can factor this expression as:
$E = u_x(u_y - \log x) - \log y(u_y - \log x) = (u_x - \log y)(u_y - \log x)$.
Substituting the values of $u_x$ and $u_y$:
$u_x - \log y = (\frac{y}{x} + \log y) - \log y = \frac{y}{x}$.
$u_y - \log x = (\log x + \frac{x}{y}) - \log x = \frac{x}{y}$.
Therefore,$E = (\frac{y}{x}) \times (\frac{x}{y}) = 1$.

(C) The auto-ionization of water,$H_2O_{(l)} \rightleftharpoons H^+_{(aq)} + OH^-_{(aq)}$,is an endothermic process $(\Delta H > 0)$.
According to Le Chatelier's principle,an increase in temperature shifts the equilibrium to the right,increasing the value of the ionic product $(K_w)$.
Conversely,a decrease in temperature shifts the equilibrium to the left,decreasing the value of $K_w$.
Since the temperature is decreased from $35^{\circ} C$ to $10^{\circ} C$,the value of $K_w$ must be less than $1.96 \times 10^{-14}$.
Among the given options,$2.95 \times 10^{-15}$ is the only value smaller than $1.96 \times 10^{-14}$.

(B) Given: Number of moles $n = 5$,initial temperature $T_1 = 100^{\circ} C$,final temperature $T_2 = 120^{\circ} C$,change in internal energy $\Delta U = 80 \ J$.
The change in temperature is $\Delta T = T_2 - T_1 = 120^{\circ} C - 100^{\circ} C = 20 \ K$.
The change in internal energy for a gas at constant volume is given by the formula $\Delta U = C_V \Delta T$,where $C_V$ is the total heat capacity of the gas at constant volume.
Rearranging the formula to solve for $C_V$: $C_V = \frac{\Delta U}{\Delta T}$.
Substituting the values: $C_V = \frac{80 \ J}{20 \ K} = 4 \ J/K$.
Therefore,the total heat capacity of the gas at constant volume is $4 \ J/K$.

(A) Let the horizontal acceleration given to the inclined plane be $a$. To keep the object stationary relative to the plane,the pseudo force $ma$ acting on the object in the horizontal direction must balance the component of gravity acting down the plane.
Resolving the forces parallel to the inclined plane:
The component of the pseudo force $ma$ along the plane is $ma \cos \theta$.
The component of gravity $mg$ along the plane is $mg \sin \theta$.
For the object to remain stationary relative to the plane,these forces must be equal: $ma \cos \theta = mg \sin \theta$.
Thus,$a = g \tan \theta$.
Given $\sin \theta = \frac{1}{l}$,we can find $\tan \theta$. In a right-angled triangle with opposite side $1$ and hypotenuse $l$,the adjacent side is $\sqrt{l^2 - 1}$.
Therefore,$\tan \theta = \frac{1}{\sqrt{l^2 - 1}}$.
Substituting this into the expression for $a$,we get $a = \frac{g}{\sqrt{l^2 - 1}}$.

(A) Given that,the rate of change of volume is $\frac{dV}{dt} = 30 \ ft^3 / min$ and the radius is $r = 15 \ ft$.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Differentiating both sides with respect to time $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Substituting the given values: $30 = 4 \pi (15)^2 \frac{dr}{dt}$.
$30 = 4 \pi (225) \frac{dr}{dt} = 900 \pi \frac{dr}{dt}$.
Therefore,$\frac{dr}{dt} = \frac{30}{900 \pi} = \frac{1}{30 \pi} \ ft / min$.

(B) Given integral is $I = \int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$.
We can rewrite the numerator as: $1+x+\sqrt{x(1+x)} = \sqrt{1+x} \cdot \sqrt{1+x} + \sqrt{x} \cdot \sqrt{1+x}$.
Factoring out $\sqrt{1+x}$,we get: $\sqrt{1+x}(\sqrt{1+x} + \sqrt{x})$.
Substituting this back into the integral: $I = \int \frac{\sqrt{1+x}(\sqrt{1+x} + \sqrt{x})}{\sqrt{x}+\sqrt{1+x}} d x$.
The term $(\sqrt{x}+\sqrt{1+x})$ cancels out from the numerator and denominator.
Thus,$I = \int \sqrt{1+x} d x$.
Using the power rule $\int u^n du = \frac{u^{n+1}}{n+1} + C$,we get: $I = \frac{(1+x)^{3/2}}{3/2} + C = \frac{2}{3}(1+x)^{3/2} + C$.

(C) Given the interval $[2,6]$ is divided into $n=4$ equal sub-intervals.
The step size is $h = \frac{6-2}{4} = 1$.
Let $f(x) = \frac{1}{x^2-x}$.
The values of $x_i$ and $y_i = f(x_i)$ are:
$x_0 = 2, y_0 = \frac{1}{2^2-2} = \frac{1}{2} = 0.5$
$x_1 = 3, y_1 = \frac{1}{3^2-3} = \frac{1}{6} \approx 0.1667$
$x_2 = 4, y_2 = \frac{1}{4^2-4} = \frac{1}{12} \approx 0.0833$
$x_3 = 5, y_3 = \frac{1}{5^2-5} = \frac{1}{20} = 0.05$
$x_4 = 6, y_4 = \frac{1}{6^2-6} = \frac{1}{30} \approx 0.0333$
By Simpson's rule:
$\int_2^6 f(x) dx \approx \frac{h}{3} [y_0 + y_4 + 4(y_1 + y_3) + 2y_2]$
$\approx \frac{1}{3} [\frac{1}{2} + \frac{1}{30} + 4(\frac{1}{6} + \frac{1}{20}) + 2(\frac{1}{12})]$
$\approx \frac{1}{3} [\frac{16}{30} + 4(\frac{10+3}{60}) + \frac{1}{6}]$
$\approx \frac{1}{3} [\frac{16}{30} + \frac{52}{60} + \frac{5}{30}] = \frac{1}{3} [\frac{32+52+10}{60}] = \frac{94}{180} = \frac{47}{90} \approx 0.5222$.

(A) Given the differential equation: $y^2 dx + (x^2 - xy + y^2) dy = 0$.
Rearranging,we get: $\frac{dy}{dx} = \frac{-y^2}{x^2 - xy + y^2}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{-(vx)^2}{x^2 - x(vx) + (vx)^2} = \frac{-v^2}{1 - v + v^2}$.
$x \frac{dv}{dx} = \frac{-v^2}{v^2 - v + 1} - v = \frac{-v^2 - v(v^2 - v + 1)}{v^2 - v + 1} = \frac{-v^2 - v^3 + v^2 - v}{v^2 - v + 1} = \frac{-(v^3 + v)}{v^2 - v + 1}$.
Separating variables: $\frac{v^2 - v + 1}{v^3 + v} dv = -\frac{1}{x} dx$.
Using partial fractions: $\frac{v^2 - v + 1}{v(v^2 + 1)} = \frac{A}{v} + \frac{Bv + D}{v^2 + 1}$.
Solving gives $A = 1, B = 0, D = -1$.
So,$\left(\frac{1}{v} - \frac{1}{v^2 + 1}\right) dv = -\frac{1}{x} dx$.
Integrating both sides: $\log|v| - \tan^{-1}(v) = -\log|x| + C$.
$\log|v| + \log|x| = \tan^{-1}(v) + C$.
$\log|vx| = \tan^{-1}(v) + C$.
Since $y = vx$,we have $\log|y| = \tan^{-1}\left(\frac{y}{x}\right) + C$,or $\tan^{-1}\left(\frac{y}{x}\right) = \log y + C$.

(D) We have the equations for the direction ratios $(l, m, n)$ as:
$3lm - 4ln + mn = 0$ $\ldots$ $(i)$
$l + 2m + 3n = 0$ $\ldots$ (ii)
From (ii),we get $l = -(2m + 3n)$.
Substituting this into $(i)$:
$-3(2m + 3n)m + 4(2m + 3n)n + mn = 0$
$-6m^2 - 9mn + 8mn + 12n^2 + mn = 0$
$-6m^2 + 12n^2 = 0 \Rightarrow m^2 = 2n^2 \Rightarrow m = \pm \sqrt{2}n$.
Case $1$: $m = \sqrt{2}n$. Then $l = -(2\sqrt{2} + 3)n$. The direction ratios are $(l_1, m_1, n_1) = (-(3 + 2\sqrt{2}), \sqrt{2}, 1)$.
Case $2$: $m = -\sqrt{2}n$. Then $l = -(-2\sqrt{2} + 3)n = (2\sqrt{2} - 3)n$. The direction ratios are $(l_2, m_2, n_2) = (-(3 - 2\sqrt{2}), -\sqrt{2}, 1)$.
For the angle $\theta$ between the lines:
$\cos \theta = \frac{l_1l_2 + m_1m_2 + n_1n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}}$
The numerator is $(3 + 2\sqrt{2})(3 - 2\sqrt{2}) + (\sqrt{2})(-\sqrt{2}) + (1)(1) = (9 - 8) - 2 + 1 = 1 - 2 + 1 = 0$.
Since the numerator is $0$,$\cos \theta = 0$,which implies $\theta = \frac{\pi}{2}$.

(D) Let $P(X=0) = p_0, P(X=1) = p_1, P(X=2) = p_2, P(X=3) = p_3$.
Given that the mean is $1.3$,we have:
$0 \cdot p_0 + 1 \cdot p_1 + 2 \cdot p_2 + 3 \cdot p_3 = 1.3$
Substituting $p_2 = 0.3$ and $p_3 = 2 p_1$:
$0 + p_1 + 2(0.3) + 3(2 p_1) = 1.3$
$p_1 + 0.6 + 6 p_1 = 1.3$
$7 p_1 = 0.7 \Rightarrow p_1 = 0.1$
Now,$p_3 = 2 p_1 = 2(0.1) = 0.2$.
Since the sum of probabilities is $1$:
$p_0 + p_1 + p_2 + p_3 = 1$
$p_0 + 0.1 + 0.3 + 0.2 = 1$
$p_0 + 0.6 = 1 \Rightarrow p_0 = 0.4$
Thus,$P(X=0) = 0.4$.

(A) The probability mass function of a Poisson distribution is given by $P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$,where $\lambda$ is the mean of the distribution.
Given the condition $P(X=2) = 3 P(X=3)$.
Substituting the formula:
$\frac{\lambda^2 e^{-\lambda}}{2!} = 3 \cdot \frac{\lambda^3 e^{-\lambda}}{3!}$
Dividing both sides by $e^{-\lambda}$ (since $e^{-\lambda} \neq 0$) and $\lambda^2$ (assuming $\lambda > 0$):
$\frac{1}{2} = 3 \cdot \frac{\lambda}{3 \times 2 \times 1}$
$\frac{1}{2} = \frac{3\lambda}{6}$
$\frac{1}{2} = \frac{\lambda}{2}$
$\lambda = 1$.
Thus,the mean of the Poisson distribution is $1$.

(C) Let $E_1$ be the event of selecting bag $X$ and $E_2$ be the event of selecting bag $Y$. Since the bag is selected at random,$P(E_1) = P(E_2) = \frac{1}{2}$.
Let $W$ be the event of drawing a white ball.
The probability of drawing a white ball from bag $X$ is $P(W|E_1) = \frac{2}{2+3} = \frac{2}{5}$.
The probability of drawing a white ball from bag $Y$ is $P(W|E_2) = \frac{4}{4+2} = \frac{4}{6} = \frac{2}{3}$.
Using the law of total probability,the probability of drawing a white ball is $P(W) = P(E_1) \cdot P(W|E_1) + P(E_2) \cdot P(W|E_2)$.
$P(W) = \left(\frac{1}{2} \cdot \frac{2}{5}\right) + \left(\frac{1}{2} \cdot \frac{2}{3}\right) = \frac{1}{5} + \frac{1}{3} = \frac{3+5}{15} = \frac{8}{15}$.

(B) The rate of flow of water through a capillary tube is given by Poiseuille's equation: $V = \frac{\pi p r^4}{8 \eta l}$.
This can be rewritten as a pressure drop: $p = V \left( \frac{8 \eta l}{\pi r^4} \right) = V R_h$,where $R_h = \frac{8 \eta l}{\pi r^4}$ is the hydraulic resistance.
For the first tube,$R_1 = \frac{8 \eta l}{\pi r^4}$.
For the second tube,$R_2 = \frac{8 \eta l}{\pi (r/2)^4} = \frac{8 \eta l}{\pi r^4 / 16} = 16 R_1$.
In a series combination,the total pressure difference $p$ is the sum of pressure drops: $p = p_1 + p_2 = V' R_1 + V' R_2$,where $V'$ is the new flow rate.
Since $p = V R_1$,we have $V R_1 = V' (R_1 + 16 R_1) = V' (17 R_1)$.
Therefore,$V' = \frac{V}{17}$.

(D) The excess pressure inside a soap bubble of radius $r$ is given by $p = \frac{4T}{r}$,where $T$ is the surface tension.
For the two initial bubbles,the excess pressures are $p_1 = \frac{4T}{r_1}$ and $p_2 = \frac{4T}{r_2}$.
The volumes of these bubbles are $V_1 = \frac{4}{3}\pi r_1^3$ and $V_2 = \frac{4}{3}\pi r_2^3$.
Let the radius of the resulting bubble be $R$. The excess pressure inside it is $p = \frac{4T}{R}$ and its volume is $V = \frac{4}{3}\pi R^3$.
Under isothermal conditions,the total number of moles of gas remains constant. Since $PV = nRT$,and $T$ is constant,the product $PV$ is conserved (assuming the external pressure is zero in vacuum).
Thus,$PV = p_1V_1 + p_2V_2$.
Substituting the values: $\frac{4T}{R} \left( \frac{4}{3}\pi R^3 \right) = \frac{4T}{r_1} \left( \frac{4}{3}\pi r_1^3 \right) + \frac{4T}{r_2} \left( \frac{4}{3}\pi r_2^3 \right)$.
Simplifying this,we get $R^2 = r_1^2 + r_2^2$.
Therefore,$R = \sqrt{r_1^2 + r_2^2}$.

(D) Given mass defect,$\Delta m = 0.3 ~g = 0.3 \times 10^{-3} ~kg = 3 \times 10^{-4} ~kg$.
Velocity of light,$c = 3 \times 10^8 ~m/s$.
The energy liberated is given by Einstein's mass-energy equivalence formula,$E = \Delta m c^2$.
$E = (3 \times 10^{-4} ~kg) \times (3 \times 10^8 ~m/s)^2$.
$E = 3 \times 10^{-4} \times 9 \times 10^{16} ~J = 27 \times 10^{12} ~J$.
To convert Joules to kilowatt-hour $(kWh)$,we divide by $3.6 \times 10^6 ~J/kWh$.
$E = \frac{27 \times 10^{12}}{3.6 \times 10^6} ~kWh = 7.5 \times 10^6 ~kWh$.

(C) The refractive index of the prism material is given by the formula: $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.
Given that the angle of minimum deviation $\delta_m = A$,we substitute this into the formula:
$\mu = \frac{\sin((A + A)/2)}{\sin(A/2)}$
$\mu = \frac{\sin(2A/2)}{\sin(A/2)} = \frac{\sin A}{\sin(A/2)}$.
Using the trigonometric identity $\sin A = 2 \sin(A/2) \cos(A/2)$,we get:
$\mu = \frac{2 \sin(A/2) \cos(A/2)}{\sin(A/2)}$.
Simplifying the expression,we have $\mu = 2 \cos(A/2)$.
Therefore,$\cos(A/2) = \mu/2$.
Taking the inverse cosine on both sides,$A/2 = \cos^{-1}(\mu/2)$.
Thus,$A = 2 \cos^{-1}(\mu/2)$.

(C) When an aqueous solution of magnesium bicarbonate is boiled,it undergoes thermal decomposition to form magnesium hydroxide,carbon dioxide,and water.
The balanced chemical equation is:
$Mg(HCO_3)_{2(aq)} \xrightarrow{\Delta} Mg(OH)_{2(s)} + 2CO_{2(g)} + 2H_2O_{(l)}$

(C) $1$. When $Na_2O$ and $CaO$ are dissolved in water,they form their respective hydroxides: $Na_2O + H_2O \rightarrow 2NaOH$ and $CaO + H_2O \rightarrow Ca(OH)_2$.
$2$. When this solution is saturated with excess $CO_2$,the hydroxides react to form bicarbonates: $NaOH + CO_2 \rightarrow NaHCO_3$ and $Ca(OH)_2 + 2CO_2 \rightarrow Ca(HCO_3)_2$.
$3$. Both $NaHCO_3$ and $Ca(HCO_3)_2$ are soluble in water.
$4$. The resulting solution contains these bicarbonates,which make the solution slightly acidic due to the presence of excess dissolved $CO_2$ (forming carbonic acid) and the nature of the bicarbonate salts.

(C) The balanced chemical equation for the combustion of sulphur is: $S(s) + O_2(g) \longrightarrow SO_2(g)$.
From the stoichiometry of the reaction,$1 \ mol$ of $S$ reacts with $1 \ mol$ of $O_2$.
At $STP$,$1 \ mol$ of any gas occupies a volume of $22.4 \ L$.
Therefore,$1 \ mol$ of $S$ requires $22.4 \ L$ of $O_2$.
For $1.5 \ mol$ of $S$,the volume of $O_2$ required is: $1.5 \ mol \times 22.4 \ L/mol = 33.6 \ L$.

(C) Given velocities: $C_1 = 100 \ ms^{-1}, C_2 = 200 \ ms^{-1}, C_3 = 500 \ ms^{-1}$.
Root mean square (rms) velocity is given by the formula: $C_{rms} = \sqrt{\frac{C_1^2 + C_2^2 + C_3^2}{n}}$.
Substituting the values: $C_{rms} = \sqrt{\frac{100^2 + 200^2 + 500^2}{3}}$.
$C_{rms} = \sqrt{\frac{10000 + 40000 + 250000}{3}}$.
$C_{rms} = \sqrt{\frac{300000}{3}} = \sqrt{100000}$.
$C_{rms} = 100 \sqrt{10} \ ms^{-1}$.

(C) The kinetic energy $(KE)$ of an electron in an orbit is given by $KE = \frac{1}{2} mv^2$.
From the electrostatic force balance,$\frac{mv^2}{r} = \frac{e^2}{r^2}$,which implies $mv^2 = \frac{e^2}{r}$.
Substituting this into the kinetic energy expression,we get $KE = \frac{1}{2} \frac{e^2}{r}$.
The potential energy $(PE)$ of the electron is $PE = \frac{-e^2}{r}$.
The total energy $(E)$ is the sum of kinetic and potential energy: $E = KE + PE = \frac{1}{2} \frac{e^2}{r} - \frac{e^2}{r} = \frac{-e^2}{2r}$.

(A) The energy $E$ of a photon is given by $E = h \nu = \frac{hc}{\lambda}$.
Given: $\lambda = 2000 \ \text{Å} = 2000 \times 10^{-8} \ \text{cm} = 2 \times 10^{-5} \ \text{cm}$.
Speed of light $c = 3 \times 10^{10} \ \text{cm/s}$.
Planck's constant $h = 6.626 \times 10^{-27} \ \text{erg} \cdot \text{s}$.
Frequency $\nu = \frac{c}{\lambda} = \frac{3 \times 10^{10}}{2 \times 10^{-5}} = 1.5 \times 10^{15} \ \text{s}^{-1}$.
Energy $E = h \nu = (6.626 \times 10^{-27} \ \text{erg} \cdot \text{s}) \times (1.5 \times 10^{15} \ \text{s}^{-1}) \approx 9.94 \times 10^{-12} \ \text{erg}$.

(C) The combustion reaction of methane is: $CH_4(g) + 2O_2(g) \longrightarrow CO_2(g) + 2H_2O(l)$.
Given that $10 \ g$ of $CH_4$ releases $560 \ kJ$ of heat,the enthalpy change $\Delta H$ for $10 \ g$ is $-560 \ kJ$.
The molar mass of methane $(CH_4)$ is $12 + (4 \times 1) = 16 \ g \ mol^{-1}$.
Heat of combustion is defined as the heat evolved per mole of substance.
For $10 \ g$ of $CH_4$,$\Delta H = -560 \ kJ$.
For $1 \ g$ of $CH_4$,$\Delta H = \frac{-560}{10} \ kJ$.
For $16 \ g$ $(1 \ mole)$ of $CH_4$,$\Delta H = \frac{-560}{10} \times 16 = -896 \ kJ \ mol^{-1}$.