AP EAMCET 2005 Chemistry Question Paper with Answer and Solution

(D) Given the differential equation: $x^2 y - x^3 \frac{dy}{dx} = y^4 \cos x$.
Dividing both sides by $x^3 y^4$,we get:
$\frac{x^2 y}{x^3 y^4} - \frac{x^3}{x^3 y^4} \frac{dy}{dx} = \frac{y^4 \cos x}{x^3 y^4}$
$\Rightarrow \frac{1}{x y^3} - \frac{1}{y^4} \frac{dy}{dx} = \frac{\cos x}{x^3}$
$\Rightarrow -\frac{1}{y^4} \frac{dy}{dx} + \frac{1}{x} y^{-3} = \frac{\cos x}{x^3}$
Let $v = y^{-3}$. Then $\frac{dv}{dx} = -3 y^{-4} \frac{dy}{dx}$,which implies $-\frac{1}{y^4} \frac{dy}{dx} = \frac{1}{3} \frac{dv}{dx}$.
Substituting this into the equation:
$\frac{1}{3} \frac{dv}{dx} + \frac{1}{x} v = \frac{\cos x}{x^3}$
Multiplying by $3$: $\frac{dv}{dx} + \frac{3}{x} v = \frac{3 \cos x}{x^3}$.
This is a linear differential equation of the form $\frac{dv}{dx} + P(x)v = Q(x)$,where $P(x) = \frac{3}{x}$ and $Q(x) = \frac{3 \cos x}{x^3}$.
Integrating factor ($I$.$F$.) $= e^{\int P(x) dx} = e^{\int \frac{3}{x} dx} = e^{3 \ln x} = x^3$.
The solution is $v \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + c$.
$v \cdot x^3 = \int \frac{3 \cos x}{x^3} \cdot x^3 dx + c$
$v x^3 = 3 \int \cos x dx + c$
$v x^3 = 3 \sin x + c$.
Since $v = y^{-3}$,we have $x^3 y^{-3} = 3 \sin x + c$.

(A) We evaluate each expression:
$(A)$ The scalar triple product is defined as $[\mathbf{a} \mathbf{b} \mathbf{c}] = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$,which matches $3$.
$(B)$ Using the vector triple product formula $(\mathbf{x} \times \mathbf{y}) \times \mathbf{z} = (\mathbf{x} \cdot \mathbf{z})\mathbf{y} - (\mathbf{y} \cdot \mathbf{z})\mathbf{x}$,we have $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b} = (\mathbf{c} \cdot \mathbf{b})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c} = (\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$,which matches $5$.
$(C)$ Using the vector triple product formula $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$,which matches $2$.
$(D)$ The dot product is defined as $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos(\theta)$,where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$,which matches $1$.
Thus,the correct matching is $A-3, B-5, C-2, D-1$.

(A) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$,which implies $a \cdot a = 1$ and $b \cdot b = 1$.
Expanding the expression $(a+b) \times (a \times b)$ using the distributive property of the cross product:
$(a+b) \times (a \times b) = a \times (a \times b) + b \times (a \times b)$.
Using the vector triple product formula $A \times (B \times C) = (A \cdot C)B - (A \cdot B)C$:
$a \times (a \times b) = (a \cdot b)a - (a \cdot a)b = (a \cdot b)a - b$.
$b \times (a \times b) = (b \cdot b)a - (b \cdot a)b = a - (a \cdot b)b$.
Adding these results:
$(a \cdot b)a - b + a - (a \cdot b)b = a(a \cdot b + 1) - b(1 + a \cdot b) = (a - b)(a \cdot b + 1)$.
Wait,re-evaluating the expansion:
$a \times (a \times b) + b \times (a \times b) = (a \cdot b)a - (a \cdot a)b + (b \cdot b)a - (b \cdot a)b = (a \cdot b)a - b + a - (a \cdot b)b = a(1 + a \cdot b) - b(1 + a \cdot b) = (a - b)(1 + a \cdot b)$.
Thus,the vector $(a+b) \times (a \times b)$ is parallel to the vector $(a - b)$.

(C) Let the given points be $A(1, -2, -3)$ and $B(2, 0, 0)$.
The vector equation of the line passing through $A$ and $B$ is given by $\vec{r} = \vec{a} + t(\vec{b} - \vec{a})$,where $\vec{a} = \hat{i} - 2\hat{j} - 3\hat{k}$ and $\vec{b} = 2\hat{i}$.
The direction vector is $\vec{v} = \vec{b} - \vec{a} = (2-1)\hat{i} + (0 - (-2))\hat{j} + (0 - (-3))\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k}$.
The line equation is $\vec{r} = (1\hat{i} - 2\hat{j} - 3\hat{k}) + t(\hat{i} + 2\hat{j} + 3\hat{k})$.
This gives the parametric equations: $x = 1 + t$,$y = -2 + 2t$,$z = -3 + 3t$.
For option $(C) (0, -4, -6)$,we set $x = 0 \implies 1 + t = 0 \implies t = -1$.
Substituting $t = -1$ into $y$ and $z$: $y = -2 + 2(-1) = -4$ and $z = -3 + 3(-1) = -6$.
Since the point $(0, -4, -6)$ satisfies the line equation,it is collinear.

(B) Let $P(B) = p$. Given that $P(A) = 2P(B)$,we have $P(A) = 2p$. Since $P(A) + P(B) = 1$,we get $2p + p = 1$,so $3p = 1$,which means $p = \frac{1}{3}$. Thus,$P(B) = \frac{1}{3}$ and $P(A) = \frac{2}{3}$.
Let $R$ be the event of drawing a red ball.
The probability of drawing a red ball from Box $A$ is $P(R|A) = \frac{3}{2+3} = \frac{3}{5}$.
The probability of drawing a red ball from Box $B$ is $P(R|B) = \frac{4}{3+4} = \frac{4}{7}$.
Using Bayes' Theorem,the probability that the ball came from Box $B$ given that it is red is:
$P(B|R) = \frac{P(B) \cdot P(R|B)}{P(A) \cdot P(R|A) + P(B) \cdot P(R|B)}$
$P(B|R) = \frac{\frac{1}{3} \cdot \frac{4}{7}}{\frac{2}{3} \cdot \frac{3}{5} + \frac{1}{3} \cdot \frac{4}{7}}$
$P(B|R) = \frac{\frac{4}{21}}{\frac{6}{15} + \frac{4}{21}} = \frac{\frac{4}{21}}{\frac{2}{5} + \frac{4}{21}}$
$P(B|R) = \frac{\frac{4}{21}}{\frac{42 + 20}{105}} = \frac{\frac{4}{21}}{\frac{62}{105}} = \frac{4}{21} \cdot \frac{105}{62} = \frac{4 \cdot 5}{62} = \frac{20}{62} = \frac{10}{31}$.

(C) Let $E$ be the event of getting a head from the coin.
$P(E) = \frac{1}{2}$.
Let $F$ be the event of getting an odd number $(1, 3, 5)$ from the die.
$P(F) = \frac{3}{6} = \frac{1}{2}$.
Since the events $E$ and $F$ are independent,the probability of both occurring is given by the product of their individual probabilities:
$P(E \cap F) = P(E) \times P(F)$.
$P(E \cap F) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

(C) Given that $n=6$ and $P(X=2)=9 P(X=4)$.
Using the binomial probability formula $P(X=k) = {^nC_k} p^k q^{n-k}$:
${^6C_2} p^2 q^4 = 9 \cdot {^6C_4} p^4 q^2$
Since ${^6C_2} = 15$ and ${^6C_4} = 15$,we have:
$15 p^2 q^4 = 9 \cdot 15 p^4 q^2$
$q^2 = 9 p^2$
$q = 3p$ (since $p, q > 0$).
We know that $p+q=1$,so $p + 3p = 1 \Rightarrow 4p = 1 \Rightarrow p = \frac{1}{4}$.
Then $q = 1 - \frac{1}{4} = \frac{3}{4}$.
The variance of a binomial distribution is given by $npq$:
$\text{Variance} = 6 \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{18}{16} = \frac{9}{8}$.

(B) Given that $P(X=k) = \frac{(k+1)a}{3^k}$ for $k \in \{0, 1, 2, \ldots, \infty\}$.
We know that the sum of all probabilities in a probability distribution is $1$,i.e.,$\sum_{k=0}^{\infty} P(X=k) = 1$.
Let $S = \sum_{k=0}^{\infty} \frac{(k+1)a}{3^k} = a \left( 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \ldots \infty \right) = 1$.
Let $S = a \left( 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \ldots \right)$.
Multiplying by $\frac{1}{3}$,we get $\frac{1}{3}S = a \left( \frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \ldots \right)$.
Subtracting the two equations:
$S - \frac{1}{3}S = a \left( 1 + (\frac{2}{3} - \frac{1}{3}) + (\frac{3}{9} - \frac{2}{9}) + (\frac{4}{27} - \frac{3}{27}) + \ldots \right)$
$\frac{2}{3}S = a \left( 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots \right)$.
The term in the bracket is an infinite geometric series with first term $1$ and common ratio $r = \frac{1}{3}$.
Sum $= \frac{a}{1-r} = \frac{1}{1 - 1/3} = \frac{1}{2/3} = \frac{3}{2}$.
So,$\frac{2}{3}S = a \left( \frac{3}{2} \right) = \frac{3a}{2}$.
Thus,$S = \frac{3a}{2} \times \frac{3}{2} = \frac{9a}{4}$.
Since $S = 1$,we have $\frac{9a}{4} = 1$,which implies $a = \frac{4}{9}$.

(A) The terminal velocity $v$ of a sphere falling through a viscous liquid is given by $v = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}$.
Since the material of the sphere is the same,the density $\rho$ is constant,so $v \propto r^2$.
Given the mass $M = \frac{4}{3} \pi r^3 \rho$,we have $r \propto M^{1/3}$.
Substituting this into the proportionality,we get $v \propto (M^{1/3})^2 = M^{2/3}$.
Therefore,$\frac{v_1}{v_2} = \left(\frac{M_1}{M_2}\right)^{2/3}$.
Given $M_1 = 20 \times 10^{-3} \ kg$,$v_1 = 0.5 \ ms^{-1}$,and $M_2 = 54 \times 10^{-2} \ kg = 540 \times 10^{-3} \ kg$.
$\frac{0.5}{v_2} = \left(\frac{20 \times 10^{-3}}{540 \times 10^{-3}}\right)^{2/3} = \left(\frac{1}{27}\right)^{2/3} = \left(\left(\frac{1}{3}\right)^3\right)^{2/3} = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$.
Thus,$v_2 = 0.5 \times 9 = 4.5 \ ms^{-1}$.

(D) The reaction between sodium thiosulphate $(Na_2S_2O_3)$ and iodine $(I_2)$ is given by:
$2Na_2S_2O_3 + I_2 \rightarrow Na_2S_4O_6 + 2NaI$
In this reaction,sodium thiosulphate is oxidised to sodium tetrathionate $(Na_2S_4O_6)$.
Therefore,option $A$ is correct.
Sodium thiosulphate is highly soluble in water,so option $B$ is correct.
Ozone reacts with alkenes to form ozonides,which is a standard test for unsaturation,so option $C$ is correct.
Option $D$ states that sodium thiosulphate reacts with iodine to form sodium sulphate,which is incorrect as it forms sodium tetrathionate.

(B) When sodium is heated in air at $300^{\circ} C$,it forms sodium peroxide $(X = Na_2O_2)$.
$2Na + O_2 \xrightarrow{300^{\circ} C} Na_2O_2$
Sodium peroxide reacts with $CO_2$ to form sodium carbonate and oxygen gas $(Y = O_2)$.
$2Na_2O_2 + 2CO_2 \rightarrow 2Na_2CO_3 + O_2$
Therefore,$Y$ is $O_2$.

(B) The reactions are as follows:
$1$. $2Mg + CO_2 \longrightarrow 2MgO + C$ ($MgO$ is formed).
$2$. $Mg + 2HNO_3 \text{ (dil.)} \longrightarrow Mg(NO_3)_2 + H_2$ ($MgO$ is not formed; $Mg(NO_3)_2$ is formed).
$3$. $2Mg + 2NO \xrightarrow{\Delta} 2MgO + N_2$ ($MgO$ is formed).
$4$. $3Mg + B_2O_3 \longrightarrow 3MgO + 2B$ ($MgO$ is formed).
Therefore,the correct option is $B$.

(D) The thermal decomposition of calcium carbonate is given by the reaction:
$CaCO_{3(s)} \xrightarrow{\Delta} CaO_{(s)} + CO_2(g)$
From the stoichiometry of the reaction:
$1 \text{ mole of } CaCO_3 (100 \text{ g}) \text{ produces } 1 \text{ mole of } CaO (56 \text{ g})$.
Since $56 \text{ g}$ of $CaO$ is obtained from $100 \text{ g}$ of $CaCO_3$,
$28 \text{ g}$ of $CaO$ is obtained from:
$x = \frac{100 \times 28}{56} = 50 \text{ g}$.
Therefore,the value of $x$ is $50$.

(A) Let the weight of both gases $A$ and $B$ be $w \text{ g}$.
Given molecular weight ratio $M_A : M_B = 1 : 4$. Let $M_A = M$ and $M_B = 4M$.
Moles of $A$ $(n_A)$ $= \frac{w}{M}$.
Moles of $B$ $(n_B)$ $= \frac{w}{4M}$.
Mole ratio $n_A : n_B = \frac{w}{M} : \frac{w}{4M} = 4 : 1$.
Partial pressure of $B$ $(p_B)$ $= \text{Mole fraction of } B \times P_{\text{total}}$.
$p_B = \frac{n_B}{n_A + n_B} \times P = \frac{1}{4 + 1} \times P = \frac{P}{5} \text{ atm}$.

(C) According to Bohr's postulate,the angular momentum of an electron in an orbit is quantized as $mvr = \frac{nh}{2\pi}$.
From the de-Broglie relation,the wavelength is given by $\lambda = \frac{h}{mv}$,which implies $mv = \frac{h}{\lambda}$.
Substituting $mv$ into the angular momentum equation: $r = \frac{n}{2\pi} \times \frac{h}{mv} = \frac{n\lambda}{2\pi}$.
The circumference of the orbit is $C = 2\pi r$.
Substituting the expression for $r$: $C = 2\pi \times \frac{n\lambda}{2\pi} = n\lambda$.
For the fourth orbit,$n = 4$,therefore the circumference is $4\lambda$.

(C) The electronic configurations of the elements are as follows:
For $X$ $(Z=19)$: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$. The $M$-shell $(n=3)$ contains $2+6 = 8$ electrons.
For $Y$ $(Z=21)$: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2$. The $M$-shell $(n=3)$ contains $2+6+1 = 9$ electrons.
For $Z$ $(Z=25)$: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2$. The $M$-shell $(n=3)$ contains $2+6+5 = 13$ electrons.
Comparing the number of electrons in the $M$-shell: $Z (13) > Y (9) > X (8)$.
Thus,the correct order is $Z > Y > X$.

(C) The absolute value of enthalpy $(H)$ cannot be determined experimentally. Only the change in enthalpy $(\Delta H)$ during a process can be measured. Therefore,statement $C$ is incorrect.