AP EAMCET 2006 Chemistry Question Paper with Answer and Solution

(D) Given the interval $[0, 6]$ is divided into $n = 6$ equal parts.
The width of each sub-interval is $h = \frac{b-a}{n} = \frac{6-0}{6} = 1$.
The values of $f(x) = x^3$ at $x = 0, 1, 2, 3, 4, 5, 6$ are:
$y_0 = f(0) = 0^3 = 0$
$y_1 = f(1) = 1^3 = 1$
$y_2 = f(2) = 2^3 = 8$
$y_3 = f(3) = 3^3 = 27$
$y_4 = f(4) = 4^3 = 64$
$y_5 = f(5) = 5^3 = 125$
$y_6 = f(6) = 6^3 = 216$
By the trapezoidal rule:
$\int_0^6 x^3 dx \approx \frac{h}{2} \{y_0 + y_6 + 2(y_1 + y_2 + y_3 + y_4 + y_5)\}$
$= \frac{1}{2} \{0 + 216 + 2(1 + 8 + 27 + 64 + 125)\}$
$= \frac{1}{2} \{216 + 2(225)\}$
$= \frac{1}{2} \{216 + 450\}$
$= \frac{666}{2} = 333$.

(A) Given equation is $x^y = y^x$.
Taking natural logarithm on both sides,we get $y \log x = x \log y$.
Differentiating both sides with respect to $x$ using the product rule:
$y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx} = x \cdot \frac{1}{y} \cdot \frac{dy}{dx} + \log y$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx} \left( \log x - \frac{x}{y} \right) = \log y - \frac{y}{x}$.
$\frac{dy}{dx} \left( \frac{y \log x - x}{y} \right) = \frac{x \log y - y}{x}$.
Multiplying both sides by $x$ and rearranging:
$x \left( \frac{y \log x - x}{y} \right) \frac{dy}{dx} = x \log y - y$.
Multiplying by $-1$ to match the required form:
$x \left( \frac{x - y \log x}{y} \right) \frac{dy}{dx} = y - x \log y$.
Therefore,$x(x - y \log x) \frac{dy}{dx} = y(y - x \log y)$.

(A) Given differential equation is $(x^2+y^2) dx = 2xy dy$.
Rearranging,we get $\frac{dy}{dx} = \frac{x^2+y^2}{2xy}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} = \frac{x^2(1+v^2)}{2x^2v} = \frac{1+v^2}{2v}$.
$x \frac{dv}{dx} = \frac{1+v^2}{2v} - v = \frac{1+v^2-2v^2}{2v} = \frac{1-v^2}{2v}$.
Separating variables: $\frac{2v}{1-v^2} dv = \frac{1}{x} dx$.
Integrating both sides: $\int \frac{2v}{1-v^2} dv = \int \frac{1}{x} dx$.
Let $1-v^2 = t$,then $-2v dv = dt$.
$-\int \frac{dt}{t} = \log|x| + \log|c|$.
$-\log|1-v^2| = \log|cx|$.
$\log|\frac{1}{1-v^2}| = \log|cx| \Rightarrow \frac{1}{1-v^2} = cx$.
Substituting $v = \frac{y}{x}$: $\frac{1}{1-(y^2/x^2)} = cx \Rightarrow \frac{x^2}{x^2-y^2} = cx$.
$x = c(x^2-y^2)$.

(B) Using the binomial expansion,we have $4^n = (1+3)^n$.
By the binomial theorem,$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots + x^n$.
Substituting $x=3$,we get $4^n = 1 + 3n + \frac{n(n-1)}{2!} \cdot 3^2 + \dots + 3^n$.
Rearranging the terms,we get $4^n - 3n - 1 = \frac{n(n-1)}{2} \cdot 9 + \dots + 3^n$.
Since every term on the right side contains a factor of $9$ (or a higher power of $3$ which is divisible by $9$),the expression $4^n - 3n - 1$ is divisible by $9$ for all integers $n \geq 2$.
For $n=1$,$4^1 - 3(1) - 1 = 0$,which is divisible by $9$ (as $0 = 9 \times 0$).
Thus,$4^n - 3n - 1$ is divisible by $9$ for all $n \geq 1$.

(D) Let $\overrightarrow{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$.
Given $\overrightarrow{a} \cdot \hat{i} = 1$,we have $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot \hat{i} = 1$,which implies $a_1 = 1$.
Next,$\overrightarrow{a} \cdot (2 \hat{i} + \hat{j}) = 1$,so $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot (2 \hat{i} + \hat{j}) = 1$.
This gives $2a_1 + a_2 = 1$. Substituting $a_1 = 1$,we get $2(1) + a_2 = 1$,so $a_2 = -1$.
Finally,$\overrightarrow{a} \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$,so $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$.
This gives $a_1 + a_2 + 3a_3 = 1$. Substituting $a_1 = 1$ and $a_2 = -1$,we get $1 - 1 + 3a_3 = 1$,which implies $3a_3 = 1$,so $a_3 = \frac{1}{3}$.
Thus,$\overrightarrow{a} = \hat{i} - \hat{j} + \frac{1}{3} \hat{k} = \frac{1}{3}(3 \hat{i} - 3 \hat{j} + \hat{k})$.

(B) The position vector of a point on the line joining two points with position vectors $\vec{a}$ and $\vec{b}$ is given by the section formula $\vec{r} = (1-t)\vec{a} + t\vec{b}$ for some scalar $t$.
If we consider the midpoint (where $t = 0.5$),the position vector is:
$\vec{r} = \frac{(\hat{i}+\hat{j}-\hat{k}) + (\hat{i}-\hat{j}+\hat{k})}{2}$
$= \frac{2\hat{i} + 0\hat{j} + 0\hat{k}}{2}$
$= \hat{i}$
Thus,the position vector of the midpoint is $\hat{i}$.

(C) Given that $A$ and $B$ are independent events,$A$ and $B^c$ are also independent events.
$P(B) = \frac{2}{7}$,so $P(B^c) = 1 - P(B) = 1 - \frac{2}{7} = \frac{5}{7}$.
We are given $P(A \cup B^c) = 0.8$.
Using the formula $P(A \cup B^c) = P(A) + P(B^c) - P(A \cap B^c)$ and the independence property $P(A \cap B^c) = P(A) \cdot P(B^c)$:
$0.8 = P(A) + \frac{5}{7} - P(A) \cdot \frac{5}{7}$
$0.8 - \frac{5}{7} = P(A) \cdot (1 - \frac{5}{7})$
$\frac{5.6 - 5}{7} = P(A) \cdot \frac{2}{7}$
$\frac{0.6}{7} = P(A) \cdot \frac{2}{7}$
$P(A) = \frac{0.6}{2} = 0.3$.

(D) Let the radii of the two soap bubbles be $a$ and $b$ respectively,and the radius of the resulting single larger bubble be $c$.
The excess pressure inside a soap bubble is $\Delta P = \frac{4 T}{r}$,where $T$ is surface tension and $r$ is the radius.
The internal pressure of the bubbles are $P_a = P + \frac{4 T}{a}$,$P_b = P + \frac{4 T}{b}$,and $P_c = P + \frac{4 T}{c}$.
Assuming isothermal conditions and conservation of the number of moles of air,$P_a V_a + P_b V_b = P_c V_c$.
Substituting the volumes $V_a = \frac{4}{3} \pi a^3$,$V_b = \frac{4}{3} \pi b^3$,and $V_c = \frac{4}{3} \pi c^3$,we get:
$(P + \frac{4 T}{a})(\frac{4}{3} \pi a^3) + (P + \frac{4 T}{b})(\frac{4}{3} \pi b^3) = (P + \frac{4 T}{c})(\frac{4}{3} \pi c^3)$.
Simplifying,$P(a^3 + b^3 - c^3) + 4 T(a^2 + b^2 - c^2) = 0$.
The change in volume is $V = V_c - (V_a + V_b) = \frac{4}{3} \pi (c^3 - a^3 - b^3)$,so $a^3 + b^3 - c^3 = -\frac{3 V}{4 \pi}$.
The change in surface area is $A = 4 \pi c^2 - (4 \pi a^2 + 4 \pi b^2) = 4 \pi (c^2 - a^2 - b^2)$,so $a^2 + b^2 - c^2 = -\frac{A}{4 \pi}$.
Substituting these into the equation: $P(-\frac{3 V}{4 \pi}) + 4 T(-\frac{A}{4 \pi}) = 0$.
Multiplying by $-4 \pi$,we get $3 P V + 4 T A = 0$.

(C) Ductile materials are those that can be drawn into thin wires. This property is due to the fact that they undergo a large plastic deformation before breaking.
In the stress-strain curve of ductile metals,the region between the elastic limit and the breaking point (fracture point) is very large,not small. This large plastic region allows the material to be stretched significantly without breaking.
Therefore,the Assertion $(A)$ is true,but the Reason $(R)$ is false.

(A) Silica $(SiO_2)$ is used for making optical instruments due to its high transparency and thermal stability.

(C) The balanced chemical equation for the reaction between oxalic acid $(H_2C_2O_4)$ and acidified $KMnO_4$ is:
$2MnO_4^- + 5H_2C_2O_4 + 6H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$
From the stoichiometry,$5$ moles of $H_2C_2O_4$ react with $2$ moles of $MnO_4^-$.
Using the relation $n_{H_2C_2O_4} / 5 = n_{MnO_4^-} / 2$:
$(40 \times 10^{-3} \times x) / 5 = (16 \times 10^{-3} \times 0.05) / 2$
$8 \times 10^{-3} \times x = 4 \times 10^{-4}$
$x = (4 \times 10^{-4}) / (8 \times 10^{-3}) = 0.05 \ M$
Oxalic acid is a diprotic acid: $H_2C_2O_4 \rightarrow 2H^+ + C_2O_4^{2-}$.
Assuming complete dissociation,$[H^+] = 2 \times [H_2C_2O_4] = 2 \times 0.05 = 0.1 \ M$.
$pH = -\log[H^+] = -\log(0.1) = 1$.

(A) $1$. In option $A$,the reaction is $AlCl_3 + 3NaOH \longrightarrow Al(OH)_3(s) + 3NaCl(aq)$. This reaction produces a precipitate of aluminum hydroxide and aqueous sodium chloride; no gas is evolved.
$2$. In option $B$,white phosphorus reacts with $NaOH$ to produce phosphine gas $(PH_3)$.
$3$. In option $C$,aluminum reacts with $NaOH$ to produce sodium aluminate and hydrogen gas $(H_2)$.
$4$. In option $D$,zinc reacts with $NaOH$ to produce sodium zincate and hydrogen gas $(H_2)$.
Therefore,the reaction that does not liberate a gaseous product is $A$.

(D) The chemical reaction for the formation of ozone is: $3O_2(g) \rightarrow 2O_3(g)$.
Let the initial volume of $O_2$ be $100 \ L$.
Let $x \ L$ of $O_2$ be converted into ozone.
According to the stoichiometry,$3 \ L$ of $O_2$ produces $2 \ L$ of $O_3$.
Therefore,$x \ L$ of $O_2$ will produce $\frac{2}{3}x \ L$ of $O_3$.
The remaining volume of $O_2$ is $(100 - x) \ L$.
The volume of $O_3$ formed is $\frac{2}{3}x \ L$.
Given that the volumes of $O_2$ and $O_3$ become equal:
$100 - x = \frac{2}{3}x$
$100 = x + \frac{2}{3}x = \frac{5}{3}x$
$x = \frac{100 \times 3}{5} = 60 \ L$.
The volume of ozone formed is $\frac{2}{3}x = \frac{2}{3} \times 60 = 40 \ L$.

(A) According to the ideal gas equation,$PV = nRT$,which can be rearranged as $V = (\frac{nR}{P})T$.
This is the equation of a straight line $V = mT$,where the slope $m = \frac{nR}{P}$.
Since $n$ and $R$ are constants,the slope is inversely proportional to pressure $(m \propto \frac{1}{P})$.
From the given graph,the slope of the lines follows the order: $m_2 > m_3 > m_1$.
Therefore,the order of pressures is $P_1 > P_3 > P_2$.

(A) Given: $E = 3 \times 10^{-12} \ erg$,$h = 6.62 \times 10^{-27} \ erg \cdot s$,$c = 3 \times 10^{10} \ cm/s$.
Using the formula $E = \frac{hc}{\lambda}$,we get $\lambda = \frac{hc}{E}$.
$\lambda = \frac{6.62 \times 10^{-27} \times 3 \times 10^{10}}{3 \times 10^{-12}} \ cm$.
$\lambda = 6.62 \times 10^{-5} \ cm$.
Since $1 \ cm = 10^7 \ nm$,$\lambda = 6.62 \times 10^{-5} \times 10^7 \ nm = 6.62 \times 10^2 \ nm = 662 \ nm$.

(A) According to Heisenberg's uncertainty principle: $\Delta x \cdot m \cdot \Delta v = \frac{h}{4 \pi}$.
Let the mass of particle $A$ be $m_A = m$ and the mass of particle $B$ be $m_B = 5m$.
The uncertainties in velocities are $\Delta v_A = 0.05 \ m \ s^{-1}$ and $\Delta v_B = 0.02 \ m \ s^{-1}$.
For particle $A$: $\Delta x_A \cdot m \cdot 0.05 = \frac{h}{4 \pi} \dots (i)$.
For particle $B$: $\Delta x_B \cdot 5m \cdot 0.02 = \frac{h}{4 \pi} \dots (ii)$.
Equating $(i)$ and $(ii)$ since both equal $\frac{h}{4 \pi}$:
$\Delta x_A \cdot m \cdot 0.05 = \Delta x_B \cdot 5m \cdot 0.02$.
$\Delta x_A \cdot 0.05 = \Delta x_B \cdot 0.10$.
$\frac{\Delta x_A}{\Delta x_B} = \frac{0.10}{0.05} = 2$.

(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by $\Delta H = \Delta E + \Delta n_g RT$.
For $\Delta H \neq \Delta E$,the change in the number of gaseous moles $(\Delta n_g)$ must not be equal to $0$.
$(A)$ $\Delta n_g = 1 - 1 = 0$.
$(B)$ $\Delta n_g = 2 - (1 + 1) = 0$.
$(C)$ $\Delta n_g = 2 - (1 + 1) = 0$.
$(D)$ $\Delta n_g = 1 - (1 + 0.5) = -0.5$.
Since $\Delta n_g \neq 0$ for option $D$,$\Delta H \neq \Delta E$ for this reaction.

(D) The lattice energy $(U)$ of an ionic crystal is defined as the energy released when gaseous ions combine to form one mole of the solid ionic crystal.
According to the Born-Landé equation,the lattice energy is inversely proportional to the inter-ionic distance $(r_0)$.
Since the inter-ionic distance $r_0$ is the sum of the ionic radii of the cation $(r_c)$ and the anion $(r_a)$,we have $r_0 = r_c + r_a$.
Therefore,the lattice energy $U$ is proportional to $\frac{1}{(r_c + r_a)}$.