AP EAMCET 2001 Chemistry Question Paper with Answer and Solution

(C) In heterogeneous catalysis,the reactants and the catalyst exist in different phases.
In the reaction $2 CO_{(g)} + O_{2(g)} \xrightarrow{Pt_{(s)}} 2 CO_{2(g)}$,the reactants ($CO$ and $O_2$) are in the gaseous phase,while the catalyst $(Pt)$ is in the solid phase.
Since they are in different phases,this is an example of heterogeneous catalysis.

(D) The heat absorbed due to the Thomson effect is given by the formula: $H = \sigma q \Delta T$,where $\sigma$ is the Thomson coefficient,$q$ is the charge,and $\Delta T$ is the temperature difference.
Given:
$\sigma = 10 \mu V/K = 10 \times 10^{-6} \text{ } V/K$
$q = 10 \text{ } C$
$\Delta T = 60^{\circ}C - 50^{\circ}C = 10 \text{ } K$
Substituting the values:
$H = (10 \times 10^{-6} \text{ } V/K) \times (10 \text{ } C) \times (10 \text{ } K)$
$H = 1000 \times 10^{-6} \text{ } J$
$H = 10^{-3} \text{ } J = 1 \text{ } mJ$.

(C) For an adiabatic process,the relationship between pressure $P$ and density $d$ is given by $P \propto d^{\gamma}$.
Given $\gamma = \frac{7}{5}$ and $\frac{d^{\prime}}{d} = 32$.
We have the relation $\frac{P^{\prime}}{P} = \left(\frac{d^{\prime}}{d}\right)^{\gamma}$.
Substituting the given values:
$\frac{P^{\prime}}{P} = (32)^{7/5}$.
Since $32 = 2^5$,we have:
$\frac{P^{\prime}}{P} = (2^5)^{7/5} = 2^7$.
Calculating the value:
$2^7 = 128$.
Therefore,the ratio $\frac{P^{\prime}}{P} = 128$.

(D) The bond dissociation energy of halogen molecules generally decreases down the group due to the increase in atomic size and bond length.
However,$F_2$ is an exception due to inter-electronic repulsion between lone pairs.
The correct order for the bond dissociation energies is $Cl_2 > Br_2 > F_2 > I_2$.
Among the given options,the order $Cl_2 > Br_2 > I_2$ is correct.

(C) The formula for unit conversion is $n_2 = n_1 \left[ \left( \frac{M_1}{M_2} \right)^a \left( \frac{L_1}{L_2} \right)^b \left( \frac{T_1}{T_2} \right)^c \right]$.
Dimensions of force are $[M L T^{-2}]$,so $a=1, b=1, c=-2$.
Given $n_1 = 100$,$M_1 = 1 \text{ g}$,$L_1 = 1 \text{ cm}$,$T_1 = 1 \text{ s}$.
New system: $M_2 = 1 \text{ kg} = 1000 \text{ g}$,$L_2 = 1 \text{ m} = 100 \text{ cm}$,$T_2 = 1 \text{ min} = 60 \text{ s}$.
Substituting these values:
$n_2 = 100 \left[ \left( \frac{1 \text{ g}}{1000 \text{ g}} \right)^1 \left( \frac{1 \text{ cm}}{100 \text{ cm}} \right)^1 \left( \frac{1 \text{ s}}{60 \text{ s}} \right)^{-2} \right]$
$n_2 = 100 \left[ \frac{1}{1000} \times \frac{1}{100} \times (60)^2 \right]$
$n_2 = 100 \times \frac{1}{1000} \times \frac{1}{100} \times 3600$
$n_2 = 3.6$.

(C) Given the ratio of amplitudes of two light waves is $\frac{a_1}{a_2} = \frac{3}{2}$.
Let $a_1 = 3k$ and $a_2 = 2k$,where $k$ is a constant.
The intensity $I$ is proportional to the square of the amplitude,$I \propto a^2$.
The ratio of maximum intensity to minimum intensity in an interference pattern is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2}$.
Substituting the values:
$\frac{I_{\max}}{I_{\min}} = \frac{(3k + 2k)^2}{(3k - 2k)^2} = \frac{(5k)^2}{(k)^2} = \frac{25k^2}{k^2} = \frac{25}{1}$.
Thus,the intensity ratio is $25: 1$.

(A) Given wavelengths are $\lambda_1 = 5 ~m$ and $\lambda_2 = 6 ~m$.
Let the velocity of sound be $v$.
The frequencies of the two waves are $n_1 = \frac{v}{\lambda_1} = \frac{v}{5}$ and $n_2 = \frac{v}{\lambda_2} = \frac{v}{6}$.
The number of beats produced in $3 ~s$ is $30$,so the beat frequency is $f_{beat} = \frac{30}{3} = 10 ~Hz$.
The beat frequency is the difference between the two frequencies: $n_1 - n_2 = 10$.
Substituting the values: $\frac{v}{5} - \frac{v}{6} = 10$.
Taking the common denominator: $v \left( \frac{6 - 5}{30} \right) = 10$.
$v \left( \frac{1}{30} \right) = 10$.
Therefore,$v = 300 ~m/s$.

(D) The frequency $n$ of a stretched string is given by the formula: $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.
From the formula,we have the relation: $n \propto \frac{\sqrt{T}}{l}$.
Let the initial frequency be $n_1 = n$ and the final frequency be $n_2 = 2n$.
Let the initial length be $l_1 = l$ and the final length be $l_2 = \frac{3}{4}l$.
Using the ratio: $\frac{n_1}{n_2} = \frac{l_2}{l_1} \sqrt{\frac{T_1}{T_2}}$.
Substituting the values: $\frac{n}{2n} = \frac{\frac{3}{4}l}{l} \sqrt{\frac{T_1}{T_2}}$.
$\frac{1}{2} = \frac{3}{4} \sqrt{\frac{T_1}{T_2}}$.
$\sqrt{\frac{T_1}{T_2}} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}$.
Squaring both sides: $\frac{T_1}{T_2} = \frac{4}{9}$.
Therefore,$T_2 = \frac{9}{4} T_1$.
The tension must be changed by a factor of $\frac{9}{4}$.

(C) Mass of the bullet,$m = 10 \ g = 10 \times 10^{-3} \ kg$.
Velocity of the bullet,$v = 300 \ m/s$.
Kinetic energy of the bullet,$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times (10 \times 10^{-3}) \times (300)^2 = 0.005 \times 90000 = 450 \ J$.
Heat produced,$Q = 50\% \text{ of } KE = 0.5 \times 450 = 225 \ J$.
The heat absorbed by the bullet raises its temperature,given by $Q = mc\Delta T$.
Here,$c = 150 \ J/kg \cdot ^{\circ}C$.
$225 = (10 \times 10^{-3}) \times 150 \times \Delta T$.
$225 = 1.5 \times \Delta T$.
$\Delta T = \frac{225}{1.5} = 150^{\circ}C$.

(B) Given:
Mass $m = 2.05 \times 10^6 \ kg$
Initial velocity $v_1 = 5 \ m/s$
Final velocity $v_2 = 25 \ m/s$
Time $t = 5 \ minutes = 5 \times 60 = 300 \ s$
Power is defined as the rate of work done,which is equal to the rate of change of kinetic energy:
$P = \frac{W}{t} = \frac{\Delta KE}{t}$
$P = \frac{\frac{1}{2} m (v_2^2 - v_1^2)}{t}$
Substituting the values:
$P = \frac{1}{2} \times \frac{2.05 \times 10^6 \times (25^2 - 5^2)}{300}$
$P = \frac{1}{2} \times \frac{2.05 \times 10^6 \times (625 - 25)}{300}$
$P = \frac{1}{2} \times \frac{2.05 \times 10^6 \times 600}{300}$
$P = \frac{1}{2} \times 2.05 \times 10^6 \times 2$
$P = 2.05 \times 10^6 \ W = 2.05 \ MW$

(D) Given: Mass $m = 6 ~kg$,Displacement $s = \frac{t^2}{4} ~m$.
Velocity $v = \frac{ds}{dt} = \frac{d}{dt}(\frac{t^2}{4}) = \frac{2t}{4} = \frac{t}{2} ~m/s$.
Acceleration $a = \frac{dv}{dt} = \frac{d}{dt}(\frac{t}{2}) = \frac{1}{2} ~m/s^2$.
Force $F = m \times a = 6 \times \frac{1}{2} = 3 ~N$.
At $t = 2 ~s$,displacement $s = \frac{(2)^2}{4} = \frac{4}{4} = 1 ~m$.
Work done $W = F \times s = 3 ~N \times 1 ~m = 3 ~J$.

(B) Step $1$: Reaction of acetic acid with metallic sodium:
$2CH_3COOH + 2Na \rightarrow 2CH_3COONa + H_2 \uparrow$
Here,$X$ is sodium acetate $(CH_3COONa)$.
Step $2$: Decarboxylation of sodium acetate with sodalime $(NaOH + CaO)$:
$CH_3COONa + NaOH \xrightarrow{CaO, \Delta} CH_4 + Na_2CO_3$
Here,$Y$ is methane $(CH_4)$.

(A) Given masses: $m_1 = 20 \ g$,$m_2 = 30 \ g$,$m_3 = 50 \ g$.
Velocities: $\vec{v}_1 = 10 \hat{i} \ m/s$,$\vec{v}_2 = 10 \hat{j} \ m/s$,$\vec{v}_3 = 10 \hat{k} \ m/s$.
The velocity of the centre of mass is given by:
$\vec{v}_{cm} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2 + m_3 \vec{v}_3}{m_1 + m_2 + m_3}$
Substituting the values:
$\vec{v}_{cm} = \frac{20(10 \hat{i}) + 30(10 \hat{j}) + 50(10 \hat{k})}{20 + 30 + 50}$
$\vec{v}_{cm} = \frac{200 \hat{i} + 300 \hat{j} + 500 \hat{k}}{100}$
$\vec{v}_{cm} = 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$

(C) $NH_3$ has $sp^3$ hybridisation with a trigonal pyramidal shape.
$BeCl_2$ is linear and $SO_2$ is bent ($V$-shaped).
$SF_6$ has $sp^3d^2$ hybridisation,resulting in an octahedral geometry where all $F-S-F$ bond angles are $90^{\circ}$.
$CO_2$ is a linear molecule with a net dipole moment of zero.

(D) The central atom $Xe$ has $8$ valence electrons.
In $XeF_4$,$Xe$ forms $4$ single bonds with $4$ fluorine atoms.
Number of electrons used for bonding $= 4$.
Number of remaining valence electrons $= 8 - 4 = 4$.
Number of lone pairs $= \frac{4}{2} = 2$.
Thus,there are $2$ lone pairs on $Xe$ in $XeF_4$.

(B) The equilibrium constant $K_c$ is related to the forward rate constant $k_f$ and the backward rate constant $k_b$ by the expression: $K_c = \frac{k_f}{k_b}$.
Given $K_c = 81$ and $k_f = 162 \ L \ mol^{-1} \ s^{-1}$.
Substituting the values: $81 = \frac{162}{k_b}$.
Solving for $k_b$: $k_b = \frac{162}{81} = 2 \ L \ mol^{-1} \ s^{-1}$.

(C) The electronic configuration of the element with atomic number $Z = 12$ is $1s^2, 2s^2, 2p^6, 3s^2$.
Since the valence shell is the $3^{rd}$ shell,the period is $3$.
There are $2$ electrons in the valence shell $(3s^2)$,which corresponds to group $II A$ (or group $2$ in the modern periodic table).

(C) According to the Mulliken scale,the electronegativity $(EN)$ of an element is defined as the arithmetic mean of its ionisation potential $(IP)$ and electron affinity $(EA)$.
Mathematically,this is expressed as: $EN = \frac{IP + EA}{2}$.

(A) The elements $A, B$ and $C$ are Lithium $(Li)$,Sodium $(Na)$ and Potassium $(K)$ respectively,which belong to Group $1$ of the periodic table.
As we move down the group,the atomic size increases due to the addition of new shells.
Ionization potential is inversely proportional to the atomic size.
Therefore,as the atomic size increases from $A$ to $C$,the energy required to remove the valence electron decreases.
The correct order of first ionization potential is $A > B > C$.

(D) The given series is a geometric progression where the first term $a = 1$ and the common ratio $r = (\cos \theta + i \sin \theta) = e^{i \theta}$.
The $n^{th}$ term of a geometric progression is given by $T_n = a \cdot r^{n-1}$.
For the $10^{th}$ term,$n = 10$,so $T_{10} = 1 \cdot (e^{i \theta})^{10-1} = e^{i 9 \theta}$.
Given $\theta = \frac{\pi}{6}$,we substitute the value:
$T_{10} = e^{i 9 (\frac{\pi}{6})} = e^{i \frac{3\pi}{2}}$.
Using Euler's formula $e^{i \phi} = \cos \phi + i \sin \phi$:
$T_{10} = \cos(\frac{3\pi}{2}) + i \sin(\frac{3\pi}{2}) = 0 + i(-1) = -i$.

(C) To form a $5$-digit number using the digits $0, 2, 4, 6, 8$ without repetition,the first digit cannot be $0$.
Total permutations of $5$ digits taken all at once is $^5P_5 = 5! = 120$.
Numbers starting with $0$ are formed by arranging the remaining $4$ digits in the last $4$ positions,which is $^4P_4 = 4! = 24$.
Therefore,the number of $5$-digit numbers is $120 - 24 = 96$.

(D) The depletion of the ozone layer is primarily caused by the release of $Chloro-fluoro \ carbons$ $(CFCs)$.
These compounds,when released into the atmosphere,reach the stratosphere where they are broken down by ultraviolet radiation to release chlorine atoms.
These chlorine atoms then catalyze the destruction of ozone $(O_3)$ molecules into oxygen $(O_2)$.

(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:
$\sin x \cos \frac{\pi}{6} + \cos x \sin \frac{\pi}{6} = 2$.
$\sin(x + \frac{\pi}{6}) = 2$.
Since the range of the sine function is $[-1, 1]$,the value $2$ is not possible.
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 $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.

(D) The given lines are $L_1: x+y-4=0$,$L_2: x-y+2=0$,and $L_3: y-2=0$.
These three lines form a triangle.
$A$ circle that touches all three sides of a triangle is called an incircle or an excircle.
For any triangle,there is exactly $1$ incircle and $3$ excircles.
Therefore,there are $1+3=4$ circles that touch all three given lines.

(A) Given lines are $2x + 3y = 6$ and $2x + 3y = 8$.
These lines meet the $X$-axis at $A(3, 0)$ and $B(4, 0)$ respectively.
The line $L$ passes through $(2, 2)$ and cuts the $X$-axis at $C(x_1, 0)$,such that the $x$-coordinates of $A, B,$ and $C$ are in arithmetic progression,i.e.,$3, 4, x_1$.
Since they are in arithmetic progression,$2 \times 4 = 3 + x_1$,which gives $x_1 = 8 - 3 = 5$.
So,the coordinates of $C$ are $(5, 0)$.
$A$ line $L$ passing through the points $(2, 2)$ and $(5, 0)$ is given by:
$y - 0 = \frac{0 - 2}{5 - 2}(x - 5)$
$y = \frac{-2}{3}(x - 5)$
$3y = -2x + 10$
$2x + 3y = 10$

(D) The $IUPAC$ name $2$-methyl-$2$-butene indicates a $4$-carbon chain (butene) with a double bond at the $2$-nd position and a methyl group at the $2$-nd position.
The structure is $CH_3-CH=C(CH_3)-CH_3$.
Therefore,the correct option is $D$.

(D) Functional isomers are compounds that have the same molecular formula but possess different functional groups.
In option $(D)$,the first compound is $CH_3CH_2CH_2OH$ (propan$-1-$ol),which contains an alcohol $(-OH)$ functional group.
The second compound is $CH_3OCH_2CH_3$ (methoxyethane),which contains an ether $(-O-)$ functional group.
Since they have the same molecular formula $(C_3H_8O)$ but different functional groups,they are functional isomers.

(A) Let the two parts be $m_1 = xM$ and $m_2 = (1-x)M$.
According to Newton's law of gravitation,the force $F$ between them at a distance $r$ is given by $F = G \frac{m_1 m_2}{r^2}$.
Substituting the values,we get $F = \frac{G}{r^2} (xM)(1-x)M = \frac{GM^2}{r^2} (x - x^2)$.
For the force $F$ to be maximum,the derivative of $F$ with respect to $x$ must be zero: $\frac{dF}{dx} = 0$.
$\frac{d}{dx} [\frac{GM^2}{r^2} (x - x^2)] = 0$.
Since $\frac{GM^2}{r^2}$ is a constant,we have $\frac{d}{dx} (x - x^2) = 0$.
$1 - 2x = 0$.
$2x = 1$,which gives $x = 1/2$.

(D) $1$. The reaction of $C_2 H_2$ (acetylene) with a red hot iron tube at $500^{\circ}C$ is a cyclic polymerization reaction that produces $C_6 H_6$ (benzene). Therefore,$A = C_6 H_6$.
$2$. The reaction of benzene $(A)$ with a mixture of concentrated $HNO_3$ and concentrated $H_2 SO_4$ at $70^{\circ}C$ is a nitration reaction,which produces $C_6 H_5 NO_2$ (nitrobenzene). Therefore,$B = C_6 H_5 NO_2$.
$3$. The reduction of nitrobenzene $(B)$ with $LiAlH_4$ can lead to the formation of azobenzene $(C_6 H_5-N=N-C_6 H_5)$ under specific conditions.
$4$. Thus,$A = C_6 H_6$ and $B = C_6 H_5 NO_2$.

(C) The bond dissociation energy depends on the bond length. As the size of the halogen atom increases from $F$ to $Br$,the bond length increases,which leads to a decrease in the bond dissociation energy.
The atomic size order is $F < Cl < Br$.
Therefore,the bond length order is $H-F < H-Cl < H-Br$.
Consequently,the bond dissociation energy order is $HF > HCl > HBr$.

(A) reducing agent is a substance that reduces another species and itself gets oxidized. In the reaction $PbO_{2(s)} + H_2O_{2(aq)} \longrightarrow PbO_{(s)} + H_2O_{(l)} + O_{2(g)}$,the oxidation state of oxygen in $H_2O_2$ changes from $-1$ to $0$ (in $O_2$),which indicates oxidation. Therefore,$H_2O_2$ acts as a reducing agent in this reaction. In the other options $(B, C, D)$,$H_2O_2$ acts as an oxidizing agent because it gets reduced to $H_2O$ (oxygen oxidation state changes from $-1$ to $-2$).

(A) The $Calgon$ process is used for the removal of hardness of water. $Calgon$ is the trade name for sodium hexametaphosphate,$Na_2[Na_4(PO_3)_6]$,which removes $Ca^{2+}$ and $Mg^{2+}$ ions by forming soluble complexes.

(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 $x-y=0$ 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 lies on the hyperbola $x^2-y^2=8$,we substitute the value: $\frac{8}{2} = 4$.

(A) Since $f(x)$ is continuous at $x = 5$,we have $f(5) = \lim_{x \rightarrow 5} f(x)$.
First,we factor the numerator and denominator:
$f(x) = \frac{x^2-10x+25}{x^2-7x+10} = \frac{(x-5)^2}{(x-5)(x-2)}$.
For $x \neq 5$,we can simplify the expression by canceling the common factor $(x-5)$:
$f(x) = \frac{x-5}{x-2}$.
Now,calculate the limit as $x \rightarrow 5$:
$f(5) = \lim_{x \rightarrow 5} \frac{x-5}{x-2} = \frac{5-5}{5-2} = \frac{0}{3} = 0$.

(C) Given that,$u=e^{x^2-y^2}$.
First,we find the partial derivative of $u$ with respect to $x$:
$u_x = \frac{\partial}{\partial x}(e^{x^2-y^2}) = e^{x^2-y^2}(2x)$.
Multiplying by $y$,we get:
$y u_x = 2xy e^{x^2-y^2} \quad (i)$.
Next,we find the partial derivative of $u$ with respect to $y$:
$u_y = \frac{\partial}{\partial y}(e^{x^2-y^2}) = e^{x^2-y^2}(-2y)$.
Multiplying by $x$,we get:
$x u_y = -2xy e^{x^2-y^2} \quad (ii)$.
Adding equations $(i)$ and $(ii)$:
$y u_x + x u_y = 2xy e^{x^2-y^2} - 2xy e^{x^2-y^2} = 0$.

(B) Let $y = \sin^{-1}(3x - 4x^3)$.
Substitute $x = \sin \theta$,which implies $\theta = \sin^{-1} x$.
Using the trigonometric identity $\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$,we get:
$y = \sin^{-1}(\sin 3\theta) = 3\theta$.
Substituting back the value of $\theta$:
$y = 3\sin^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\sin^{-1} x) = 3 \cdot \frac{1}{\sqrt{1-x^2}} = \frac{3}{\sqrt{1-x^2}}$.

(C) The given function is $u(x, y) = x y^2 \tan^{-1}\left(\frac{y}{x}\right)$.
This is a homogeneous function of degree $n = 1 + 2 = 3$ because $u(tx, ty) = (tx)(ty)^2 \tan^{-1}\left(\frac{ty}{tx}\right) = t^3 x y^2 \tan^{-1}\left(\frac{y}{x}\right) = t^3 u(x, y)$.
According to Euler's Theorem on homogeneous functions,if $u$ is a homogeneous function of degree $n$ in $x$ and $y$,then $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n u$.
Here,$n = 3$.
Therefore,$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 3 u$.

(A) buffer solution is formed by mixing a weak acid and its salt with a strong base,or a weak base and its salt with a strong acid.
When $1 \ M \ CH_3COOH$ and $0.5 \ M \ NaOH$ are mixed in equal volumes,the reaction is:
$CH_3COOH + NaOH \longrightarrow CH_3COONa + H_2O$.
Since the concentration of $CH_3COOH$ $(1 \ M)$ is double that of $NaOH$ $(0.5 \ M)$,after the reaction,$0.5 \ M$ of $CH_3COOH$ remains unreacted,and $0.5 \ M$ of $CH_3COONa$ is formed.
This mixture of a weak acid $(CH_3COOH)$ and its salt $(CH_3COONa)$ acts as an acidic buffer.

(A) Using the Henderson-Hasselbalch equation for an acidic buffer:
$pH = pK_a + \log \frac{[\text{salt}]}{[\text{acid}]}$
Given $pH = 5.8$ and $pK_a = 4.8$:
$5.8 = 4.8 + \log \frac{[\text{salt}]}{[\text{acid}]}$
$\log \frac{[\text{salt}]}{[\text{acid}]} = 5.8 - 4.8 = 1.0$
Taking the antilog on both sides:
$\frac{[\text{salt}]}{[\text{acid}]} = 10^1 = 10$
Therefore,the ratio $\frac{[\text{acid}]}{[\text{salt}]} = \frac{1}{10} = 0.1$.

(D) Let the total length of the cylinder be $L$. Initially,the piston is at the middle,so the length of each side is $l = L/2$.
According to Charles's law,at constant pressure,$V \propto T$,which implies $l \propto T$ for a cylinder of uniform cross-section.
Let the initial temperature be $T_1 = 0^{\circ} C = 273 ~K$ and the final temperature of the heated side be $T_2 = 100^{\circ} C = 373 ~K$.
When the piston moves by $5 ~cm$,the length of the heated side becomes $(l + 5)$ and the length of the other side becomes $(l - 5)$.
Applying the ratio: $\frac{l+5}{l-5} = \frac{373}{273}$.
Using componendo and dividendo: $\frac{(l+5) + (l-5)}{(l+5) - (l-5)} = \frac{373 + 273}{373 - 273}$.
$\frac{2l}{10} = \frac{646}{100}$.
$2l = 64.6 ~cm$.
Since the total length of the cylinder is $L = 2l$,the total length is $64.6 ~cm$.

(C) When a particle is projected up an inclined plane,both the component of gravity acting down the plane and the frictional force act in the direction opposite to the motion.
The force acting down the plane is $F = mg \sin \theta + f_k$,where $f_k = \mu N = \mu mg \cos \theta$.
Using Newton's second law,$ma = -(mg \sin \theta + \mu mg \cos \theta)$.
The magnitude of the retardation (deceleration) is $a = g(\sin \theta + \mu \cos \theta)$.
Given $\theta = 45^{\circ}$ and $\mu = 0.5 = \frac{1}{2}$.
Substituting these values: $a = g(\sin 45^{\circ} + 0.5 \cos 45^{\circ})$.
$a = g\left(\frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}}\right)$.
$a = g\left(\frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}}\right) = g\left(\frac{2+1}{2\sqrt{2}}\right) = \frac{3g}{2\sqrt{2}}$.

(D) Given: Weight $W = 64 ~N$,coefficient of static friction $\mu_s = 0.6$,coefficient of kinetic friction $\mu_k = 0.4$.
To start the motion,the applied force $F$ must be equal to the limiting friction: $F = \mu_s N = \mu_s W$.
Substituting the values: $F = 0.6 \times 64 ~N = 38.4 ~N$.
Once the body is in motion,the kinetic friction acting on it is $f_k = \mu_k N = \mu_k W$.
Substituting the values: $f_k = 0.4 \times 64 ~N = 25.6 ~N$.
The net force acting on the body is $F_{net} = F - f_k = (\mu_s - \mu_k) W$.
Using Newton's second law,$F_{net} = ma$,where $m = \frac{W}{g}$.
So,$(\mu_s - \mu_k) W = \frac{W}{g} \times a$.
$a = (\mu_s - \mu_k) g$.
$a = (0.6 - 0.4) g = 0.2 ~g$.

(D) Given that $y = \cos(\sin x)$.
First derivative $y_1 = \frac{dy}{dx} = -\sin(\sin x) \cdot \cos x$.
Second derivative $y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx}[-\sin(\sin x) \cdot \cos x]$.
Using the product rule: $y_2 = -[\cos(\sin x) \cdot \cos x \cdot \cos x + \sin(\sin x) \cdot (-\sin x)]$.
$y_2 = -\cos(\sin x) \cos^2 x + \sin(\sin x) \sin x$.
Now,calculate $y_1 \sin x + y_2 \cos x$:
$y_1 \sin x + y_2 \cos x = [-\sin(\sin x) \cos x] \sin x + [-\cos(\sin x) \cos^2 x + \sin(\sin x) \sin x] \cos x$.
$= -\sin(\sin x) \sin x \cos x - \cos(\sin x) \cos^3 x + \sin(\sin x) \sin x \cos x$.
$= -\cos(\sin x) \cos^3 x$.
Since $y = \cos(\sin x)$,the expression becomes $-y \cos^3 x$.

(C) Given $f(x) = \frac{x^2}{x+a}$.
Using the quotient rule,$f^{\prime}(x) = \frac{(x+a)(2x) - x^2(1)}{(x+a)^2} = \frac{2x^2 + 2ax - x^2}{(x+a)^2} = \frac{x^2 + 2ax}{(x+a)^2}$.
Now,find $f^{\prime \prime}(x)$ by differentiating $f^{\prime}(x)$ again:
$f^{\prime \prime}(x) = \frac{(x+a)^2(2x + 2a) - (x^2 + 2ax)(2(x+a))}{(x+a)^4}$
$f^{\prime \prime}(x) = \frac{(x+a)(2x + 2a) - 2(x^2 + 2ax)}{(x+a)^3} = \frac{2x^2 + 4ax + 2a^2 - 2x^2 - 4ax}{(x+a)^3} = \frac{2a^2}{(x+a)^3}$.
Substituting $x = a$:
$f^{\prime \prime}(a) = \frac{2a^2}{(a+a)^3} = \frac{2a^2}{(2a)^3} = \frac{2a^2}{8a^3} = \frac{1}{4a}$.

(A) Given the function: $y = A \cos n x + B \sin n x$.
First,differentiate with respect to $x$:
$y_1 = \frac{dy}{dx} = -A n \sin n x + B n \cos n x$.
Now,differentiate again with respect to $x$ to find the second derivative $y_2$:
$y_2 = \frac{d^2y}{dx^2} = -A n^2 \cos n x - B n^2 \sin n x$.
Factor out $-n^2$:
$y_2 = -n^2 (A \cos n x + B \sin n x)$.
Since $y = A \cos n x + B \sin n x$,we can substitute $y$ into the equation:
$y_2 = -n^2 y$.
Rearranging the terms gives:
$y_2 + n^2 y = 0$.

(D) We use the integration by parts formula: $\int u v d x = u \int v d x - \int (u' \int v d x) d x$.
Let $u = (x+1)^2$ and $v = e^x$.
Then $u' = 2(x+1)$ and $\int v d x = e^x$.
Applying the formula:
$\int(x+1)^2 e^x d x = (x+1)^2 e^x - \int 2(x+1) e^x d x$.
Now,apply integration by parts again to $\int (x+1) e^x d x$:
$\int (x+1) e^x d x = (x+1) e^x - \int 1 \cdot e^x d x = (x+1) e^x - e^x$.
Substituting this back:
$= (x+1)^2 e^x - 2[(x+1) e^x - e^x] + C$
$= (x^2+2x+1) e^x - 2x e^x - 2 e^x + 2 e^x + C$
$= (x^2+2x+1-2x) e^x + C$
$= (x^2+1) e^x + C$.

(A) The length of the sub-tangent is given by the formula $y \cdot \frac{dx}{dy}$.
Given that the sub-tangent is double the abscissa $(x)$,we have the differential equation:
$y \cdot \frac{dx}{dy} = 2x$
Rearranging the terms to separate the variables:
$\frac{dx}{x} = 2 \frac{dy}{y}$
Integrating both sides:
$\int \frac{1}{x} dx = 2 \int \frac{1}{y} dy$
$\ln|x| = 2 \ln|y| + \ln|C|$
Using logarithmic properties:
$\ln|x| = \ln|y^2| + \ln|C|$
$\ln|x| = \ln|C y^2|$
Taking the exponential of both sides,we get:
$x = C y^2$

(B) Let the quantity be $X = \frac{L}{T^2}$.
The relative error in $X$ is given by $\frac{\Delta X}{X} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.
Given that the percentage error in $L$ is $\frac{\Delta L}{L} \times 100 = 3 \%$ and the percentage error in $T$ is $\frac{\Delta T}{T} \times 100 = 2 \%$.
The maximum percentage error in $X$ is $\frac{\Delta X}{X} \times 100 = \left( \frac{\Delta L}{L} \times 100 \right) + 2 \left( \frac{\Delta T}{T} \times 100 \right)$.
Substituting the given values: $\text{Percentage Error} = 3 \% + 2 \times (2 \%) = 3 \% + 4 \% = 7 \%$.

(B) Let the side of the square be $a$.
The length of the diagonal $d$ of a square with side $a$ is given by $d = a\sqrt{2}$.
The distance between the two given points $(1, -2, 3)$ and $(2, -3, 5)$ is the length of the diagonal $d$.
$d = \sqrt{(2-1)^2 + (-3 - (-2))^2 + (5-3)^2}$
$d = \sqrt{(1)^2 + (-1)^2 + (2)^2}$
$d = \sqrt{1 + 1 + 4} = \sqrt{6}$
Since $d = a\sqrt{2}$,we have:
$a\sqrt{2} = \sqrt{6}$
$a = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}$
Therefore,the length of the side of the square is $\sqrt{3}$.