AP EAMCET 2004 Chemistry Question Paper with Answer and Solution

(A) The kinetic energy $(KE)$ of $n$ moles of an ideal gas is given by $KE = \frac{3}{2} nRT$.
For $16 \ g$ of methane ($CH_4$,molar mass = $16 \ g/mol$),moles $n_1 = \frac{16}{16} = 1 \ mol$.
For $32 \ g$ of oxygen ($O_2$,molar mass = $32 \ g/mol$),moles $n_2 = \frac{32}{32} = 1 \ mol$.
Since both have $1 \ mol$ at the same temperature $(300 \ K)$,their kinetic energies are equal. Thus,$(A)$ is true.
Reason $(R)$ states that at constant temperature,kinetic energy of one mole of all gases is equal,which is true $(KE = \frac{3}{2} RT)$.
Therefore,$(R)$ is the correct explanation of $(A)$.

(A) The electronic configurations are as follows:
$K$ $(Z=19)$: $1s^2, 2s^2 2p^6, 3s^2 3p^6, 4s^1$. Electrons in $M$ shell $(n=3)$ = $8$.
$Mn$ $(Z=25)$: $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^5, 4s^2$. Electrons in $M$ shell $(n=3)$ = $13$.
$Ni$ $(Z=28)$: $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^8, 4s^2$. Electrons in $M$ shell $(n=3)$ = $16$.
$Sc$ $(Z=21)$: $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^1, 4s^2$. Electrons in $M$ shell $(n=3)$ = $9$.
Thus,$K$ has the least number of electrons in its $M$ shell.

(A) thermocouple consists of two dissimilar metal wires joined at two junctions.
When one junction is kept at a constant temperature and the other is exposed to a changing temperature,a thermoelectric electromotive force $(EMF)$ is generated.
Because the junction of a thermocouple is very small,its thermal capacity is extremely low.
Thermal capacity is defined as the product of mass and specific heat capacity $(C = mc)$.
$A$ low thermal capacity means the junction can gain or lose heat very quickly to reach thermal equilibrium with the surroundings.
Therefore,it can respond rapidly to changes in temperature,making it suitable for measuring rapidly changing temperatures.
Thus,$(A)$ is true,$(R)$ is true,and $(R)$ is the correct explanation of $(A)$.

(C) The rate of cooling of a body is given by the formula: $\frac{d\theta}{dt} = \frac{\sigma A(T^4 - T_0^4)}{ms}$.
Here,$m = 34.38 \times 10^{-3} \ kg$,$A = 19.2 \times 10^{-4} \ m^2$,$T = 400 \ K$,$T_0 = 300 \ K$,and $\frac{d\theta}{dt} = 0.04 \ K/s$.
Rearranging for specific heat $s$: $s = \frac{\sigma A(T^4 - T_0^4)}{m(\frac{d\theta}{dt})}$.
Substituting the values:
$s = \frac{(5.73 \times 10^{-8}) \times (19.2 \times 10^{-4}) \times (400^4 - 300^4)}{(34.38 \times 10^{-3}) \times 0.04}$.
$s = \frac{(5.73 \times 10^{-8}) \times (19.2 \times 10^{-4}) \times (256 \times 10^8 - 81 \times 10^8)}{34.38 \times 10^{-3} \times 0.04}$.
$s = \frac{5.73 \times 19.2 \times 10^{-12} \times 175 \times 10^8}{34.38 \times 10^{-3} \times 0.04}$.
$s = \frac{19249.92 \times 10^{-4}}{1.3752 \times 10^{-3}} \approx 1400 \ J \ kg^{-1} \ K^{-1}$.

(D) For an ideal monoatomic gas,the internal energy is given by $U = n C_V T$,where $C_V = \frac{3}{2} R$.
When two gases are mixed in an adiabatic container,the total internal energy is conserved: $U_{mix} = U_1 + U_2$.
$(n_1 + n_2) C_V T_{mix} = n_1 C_V T_1 + n_2 C_V T_2$.
Since the gases are both monoatomic,$C_V$ cancels out:
$T_{mix} = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$.
Substituting the given values:
$T_{mix} = \frac{4(400) + 2(700)}{4 + 2}$.
$T_{mix} = \frac{1600 + 1400}{6} = \frac{3000}{6} = 500 ~K$.

(D) For an adiabatic process,the relationship between pressure $P$ and density $\rho$ is given by $P \propto \rho^{\gamma}$.
Therefore,we can write the ratio as $\frac{P^{\prime}}{P} = \left( \frac{d^{\prime}}{d} \right)^{\gamma}$.
Given $\frac{d^{\prime}}{d} = 32$ and $\gamma = \frac{7}{5}$.
Substituting these values,we get $\frac{P^{\prime}}{P} = (32)^{7/5}$.
Since $32 = 2^5$,we have $\frac{P^{\prime}}{P} = (2^5)^{7/5} = 2^7$.
Calculating $2^7$,we get $2^7 = 128$.
Thus,the ratio $\frac{P^{\prime}}{P} = 128$.

(B) For an ideal gas,the pressure $P$ is given by $P = \frac{nRT}{V} = \frac{m}{M} \frac{RT}{V}$,where $m$ is the mass,$M$ is the molar mass,$R$ is the gas constant,and $T$ is the temperature.
Initially,for vessel $A$,$P_{A,i} = \frac{m_A RT}{MV}$. After isothermal expansion to volume $2V$,$P_{A,f} = \frac{m_A RT}{M(2V)}$.
The change in pressure for vessel $A$ is $\Delta P = |P_{A,f} - P_{A,i}| = \frac{m_A RT}{2MV}$.
Similarly,for vessel $B$,the change in pressure is $1.5 \Delta P = |P_{B,f} - P_{B,i}| = \frac{m_B RT}{2MV}$.
Dividing the two equations:
$\frac{1.5 \Delta P}{\Delta P} = \frac{\frac{m_B RT}{2MV}}{\frac{m_A RT}{2MV}}$
$1.5 = \frac{m_B}{m_A}$
$\frac{3}{2} = \frac{m_B}{m_A}$
$3 m_A = 2 m_B$.

(B) An endothermic reaction is a process that absorbs heat from its surroundings,resulting in a positive enthalpy change $(\Delta H > 0)$.
In option $B$,$180.8 \ kJ$ of heat is added to the reactants,indicating that energy is absorbed to form the products.
Therefore,the reaction $N_{2(g)} O_{2(g)} 180.8 \ kJ \longrightarrow 2 NO_{(g)}$ is endothermic.

(C) The molecule $CH_4$ contains four $C-H$ bonds.
To break one mole of $CH_4$ into gaseous carbon and hydrogen atoms,all four $C-H$ bonds must be broken.
The total energy required is $4 \times 416 \ kJ \ mol^{-1} = 1664 \ kJ \ mol^{-1}$.
Therefore,the correct thermochemical equation is $CH_{4(g)} + 1664 \ kJ \longrightarrow C_{(g)} + 4H_{(g)}$.

(D) In the equation $x(t) = \frac{v_0}{A}(1 - e^{-At})$,the exponent of the exponential function must be dimensionless. Therefore,the dimension of $At$ must be $1$.
$[A][T] = [M^0 L^0 T^0] \implies [A] = [T^{-1}] = [M^0 L^0 T^{-1}]$.
Since the term $(1 - e^{-At})$ is dimensionless,the dimension of $x$ must be equal to the dimension of $\frac{v_0}{A}$.
$[x] = \frac{[v_0]}{[A]} \implies [L] = \frac{[v_0]}{[T^{-1}]} \implies [v_0] = [L][T^{-1}] = [M^0 LT^{-1}]$.
Thus,the dimensions of $v_0$ are $[M^0 LT^{-1}]$ and the dimensions of $A$ are $[M^0 L^0 T^{-1}]$.

(A) Statement $A$ is true: Fresnel diffraction is characterized by the source and/or the screen being at a finite distance from the diffracting aperture or obstacle.
Statement $B$ is true: $X$-ray diffraction (a form of diffraction) is a standard technique used to determine the molecular structure of complex biological molecules,including the helical structure of nucleic acids ($DNA$/$RNA$).

(D) Let the frequency of the third note be $n$ and the velocity of sound be $v$. The frequency of a note is given by $f = \frac{v}{\lambda}$.
For the first note,frequency $f_1 = \frac{v}{36/195} = \frac{195v}{36}$.
For the second note,frequency $f_2 = \frac{v}{36/193} = \frac{193v}{36}$.
Since both produce $10$ beats per second with the third note,we have:
$\frac{195v}{36} - n = 10$ ...$(i)$
$n - \frac{193v}{36} = 10$ ...(ii)
Adding equations $(i)$ and (ii):
$\frac{195v}{36} - \frac{193v}{36} = 20$
$\frac{2v}{36} = 20$
$\frac{v}{18} = 20$
$v = 360 ~m/s$.

(D) The frequency of a sonometer wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$. Since $f$ and $\mu$ are constant,$l \propto \sqrt{T}$.
Therefore,$\frac{l_{\text{air}}}{l_{\text{water}}} = \sqrt{\frac{T_{\text{air}}}{T_{\text{water}}}}$.
In air,the tension $T_{\text{air}} = mg$. When immersed in water,the buoyant force $F_B$ acts upwards,so $T_{\text{water}} = mg - F_B$.
Given specific gravity $\sigma = 8$,the density of iron $\rho = 8 \rho_w$. The buoyant force $F_B = V \rho_w g = \frac{m}{\rho} \rho_w g = \frac{m}{8 \rho_w} \rho_w g = \frac{mg}{8}$.
Thus,$T_{\text{water}} = mg - \frac{mg}{8} = \frac{7}{8} mg = \frac{7}{8} T_{\text{air}}$.
Substituting this into the ratio: $\frac{l_{\text{air}}}{l_{\text{water}}} = \sqrt{\frac{T_{\text{air}}}{\frac{7}{8} T_{\text{air}}}} = \sqrt{\frac{8}{7}}$.
Given $l_{\text{air}} = \frac{1}{\sqrt{7}} \times 1 \ m$,we have $l_{\text{water}} = l_{\text{air}} \times \sqrt{\frac{7}{8}} = \frac{1}{\sqrt{7}} \times \sqrt{\frac{7}{8}} = \frac{1}{\sqrt{8}} \ m$.

(C) The work done by the force on the block is equal to the change in its kinetic energy.
Since the block starts from rest at $x = 0$,the kinetic energy $KE$ at any position $x$ is given by the work done by the force $F$ from $x = 0$ to $x$.
$KE = \int_{0}^{x} F dx = \int_{0}^{x} (9 - x^2) dx = 9x - \frac{x^3}{3}$.
To find the maximum kinetic energy,we find where the force becomes zero:
$F = 9 - x^2 = 0 \implies x^2 = 9 \implies x = 3 ~m$.
At $x = 3 ~m$,the kinetic energy is:
$KE_{max} = \int_{0}^{3} (9 - x^2) dx = [9x - \frac{x^3}{3}]_{0}^{3} = (9(3) - \frac{3^3}{3}) - 0 = 27 - 9 = 18 ~J$.

(B) The velocity of the centre of mass is given by $\overrightarrow{v}_{CM} = \frac{m_1 \overrightarrow{v}_1 + m_2 \overrightarrow{v}_2}{m_1 + m_2}$.
Since $m_1 = m_2 = m$,we have $\overrightarrow{v}_{CM} = \frac{\overrightarrow{v}_1 + \overrightarrow{v}_2}{2} = \frac{4 \hat{i} + 4 \hat{j}}{2} = (2 \hat{i} + 2 \hat{j}) \text{ ms}^{-1}$.
The acceleration of the centre of mass is $\overrightarrow{a}_{CM} = \frac{m_1 \overrightarrow{a}_1 + m_2 \overrightarrow{a}_2}{m_1 + m_2}$.
Given $\overrightarrow{a}_1 = (5 \hat{i} + 5 \hat{j}) \text{ ms}^{-2}$ and $\overrightarrow{a}_2 = 0$,we get $\overrightarrow{a}_{CM} = \frac{\overrightarrow{a}_1 + 0}{2} = (2.5 \hat{i} + 2.5 \hat{j}) \text{ ms}^{-2}$.
Since the acceleration $\overrightarrow{a}_{CM}$ is constant and the initial velocity $\overrightarrow{v}_{CM}$ is not parallel to the acceleration $\overrightarrow{a}_{CM}$,the path of the centre of mass is a parabola.

(A) The hybridization of carbon atoms in the given molecules is as follows:
$A$. Ethane $(CH_3-CH_3)$: Both carbon atoms are $sp^3$ hybridized. Thus,$A-3$.
$B$. Ethylene $(CH_2=CH_2)$: Both carbon atoms are $sp^2$ hybridized. Thus,$B-4$.
$C$. Acetylene $(CH \equiv CH)$: Both carbon atoms are $sp$ hybridized. Thus,$C-1$.
$D$. Benzene $(C_6H_6)$: All six carbon atoms in the ring are $sp^2$ hybridized. Thus,$D-2$.
Therefore,the correct matching is $A-3, B-4, C-1, D-2$.

(C) Cation formation involves the loss of an electron from a neutral atom. $M \rightarrow M^+ + e^-$.
This process requires energy,known as ionisation energy.
Therefore,a lower ionisation potential (or ionisation energy) makes it easier for an atom to lose an electron and form a cation.

(C) Ionic compounds consist of ions that are held together by strong electrostatic forces of attraction.
In the solid state,these ions are fixed in a lattice and cannot move,making them poor conductors.
However,in the molten state or when dissolved in water,the ions become free to move,which allows them to conduct electricity.
Therefore,the statement that ionic compounds do not conduct electricity in molten or aqueous solutions is incorrect.

(B) For a general reaction,the equilibrium constant is defined as $K_c = \frac{[\text{Product}]}{[\text{Reactant}]}$.
If $[\text{Product}] > [\text{Reactant}]$,then the ratio $K_c$ must be greater than $1$.
Comparing the given values:
$A: K = 0.001 < 1$
$B: K = 10 > 1$
$C: K = 0.005 < 1$
$D: K = 0.01 < 1$
Therefore,in reaction $B$,the concentration of the product is higher than the concentration of the reactant.

(D) Since $(x-2)$ is a common factor of the expressions $x^2+ax+b$ and $x^2+cx+d$,we have:
$2^2 + 2a + b = 0 \Rightarrow 4 + 2a + b = 0$ ... $(i)$
$2^2 + 2c + d = 0 \Rightarrow 4 + 2c + d = 0$ ... $(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$(4 + 2a + b) - (4 + 2c + d) = 0$
$2a + b - 2c - d = 0$
$b - d = 2c - 2a$
$b - d = 2(c - a)$
Therefore,$\frac{b-d}{c-a} = 2$.

(A) Let $y = \sqrt{42+\sqrt{42+\sqrt{42+\ldots}}}$.
Since the expression is infinite,we can write:
$y = \sqrt{42+y}$.
Squaring both sides,we get:
$y^2 = 42 + y$.
Rearranging the terms to form a quadratic equation:
$y^2 - y - 42 = 0$.
Factoring the quadratic equation:
$(y - 7)(y + 6) = 0$.
This gives two possible values for $y$: $y = 7$ or $y = -6$.
Since the square root of a positive number must be positive,$y$ cannot be $-6$.
Therefore,the value is $y = 7$.

(A) Given that $f(x)$ is a polynomial with rational coefficients.
If a complex number $a+bi$ is a root,its conjugate $a-bi$ must also be a root.
Thus,$1+2i$ and $1-2i$ are roots.
If a surd $a+\sqrt{b}$ is a root,its conjugate $a-\sqrt{b}$ must also be a root.
Thus,$2-\sqrt{3}$ and $2+\sqrt{3}$ are roots.
Additionally,$5$ is given as a root.
Therefore,the roots are $1+2i, 1-2i, 2-\sqrt{3}, 2+\sqrt{3},$ and $5$.
Counting these,we have $5$ distinct roots.
Since the polynomial has at least these $5$ roots,the least value of $n$ is $5$.

(A) Given that $x_n = \cos \frac{\pi}{2^n} + i \sin \frac{\pi}{2^n} = e^{i \frac{\pi}{2^n}}$.
We need to evaluate the infinite product $\prod_{n=1}^{\infty} x_n = \prod_{n=1}^{\infty} e^{i \frac{\pi}{2^n}}$.
Using the property of exponents,$\prod_{n=1}^{\infty} e^{i \frac{\pi}{2^n}} = e^{i \sum_{n=1}^{\infty} \frac{\pi}{2^n}}$.
The exponent is a geometric series: $\sum_{n=1}^{\infty} \frac{\pi}{2^n} = \pi \left( \frac{1/2}{1 - 1/2} \right) = \pi \left( \frac{1/2}{1/2} \right) = \pi$.
Therefore,the product is $e^{i \pi}$.
Using Euler's formula,$e^{i \pi} = \cos \pi + i \sin \pi = -1 + 0i = -1$.

(A) The given series is an infinite geometric progression with the first term $a = 1$ and common ratio $r = \frac{2i}{3}$.
Since $|r| = |\frac{2i}{3}| = \frac{2}{3} < 1$,the sum $S$ is given by $S = \frac{a}{1-r}$.
$S = \frac{1}{1 - \frac{2i}{3}} = \frac{1}{\frac{3-2i}{3}} = \frac{3}{3-2i}$.
To simplify,multiply the numerator and denominator by the conjugate $(3+2i)$:
$S = \frac{3(3+2i)}{(3-2i)(3+2i)} = \frac{9+6i}{3^2 - (2i)^2} = \frac{9+6i}{9 - (-4)} = \frac{9+6i}{13}$.

(D) There are $10$ speakers in total. The condition is that $S_1$ must speak after $S_2$.
In any arrangement of $10$ speakers,there are only two possibilities for the relative order of $S_1$ and $S_2$: either $S_1$ speaks before $S_2$ or $S_2$ speaks before $S_1$.
Since these two cases are equally likely,exactly half of the total permutations $(10!)$ will satisfy the condition that $S_2$ speaks before $S_1$ (which is equivalent to $S_1$ speaking after $S_2$).
Therefore,the required number of ways is $\frac{10!}{2}$.

(C) We are given the limit: $\lim_{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$.
Since $x$ is independent of $k$,we can take it out of the summation: $x \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \left(\frac{k}{n}\right)^2$.
Using the definition of a definite integral as the limit of a Riemann sum,$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n f\left(\frac{k}{n}\right) = \int_0^1 f(t) dt$.
Here,$f(t) = t^2$,so the expression becomes: $x \int_0^1 t^2 dt$.
Evaluating the integral: $x \left[ \frac{t^3}{3} \right]_0^1 = x \left( \frac{1}{3} - 0 \right) = \frac{x}{3}$.

(C) Let $y = x \log _e a$. The given series is $y + \frac{y^3}{3!} + \frac{y^5}{5!} + \dots$
We know the Taylor series expansion for the hyperbolic sine function is $\sinh(y) = y + \frac{y^3}{3!} + \frac{y^5}{5!} + \dots$
Substituting $y = x \log _e a$ back into the series,we get $\sinh(x \log _e a)$.
Thus,the value of the series is $\sinh(x \log _e a)$.

(A) We know that the sum of the first $k$ cubes is $\sum_{i=1}^k i^3 = \left(\frac{k(k+1)}{2}\right)^2$ and the sum of the first $k$ odd numbers is $\sum_{i=1}^k (2i-1) = k^2$.
Substituting these into the expression:
$\sum_{k=1}^5 \frac{\left(\frac{k(k+1)}{2}\right)^2}{k^2} = \sum_{k=1}^5 \frac{k^2(k+1)^2}{4k^2} = \sum_{k=1}^5 \frac{(k+1)^2}{4}$.
Expanding the sum for $k=1$ to $5$:
$= \frac{1}{4} [2^2 + 3^2 + 4^2 + 5^2 + 6^2] = \frac{1}{4} [4 + 9 + 16 + 25 + 36] = \frac{90}{4} = 22.5$.

(B) The atmosphere is divided into layers based on altitude from the Earth's surface:
$1$. Troposphere: $0-10 \ km$
$2$. Stratosphere: $10-50 \ km$
$3$. Mesosphere: $50-85 \ km$
$4$. Thermosphere: $85-500 \ km$
Comparing the altitudes,the order is Thermosphere $(85-500 \ km)$ > Mesosphere $(50-85 \ km)$ > Troposphere $(0-10 \ km)$.
Therefore,the decreasing order of altitude is $III, II, I$.

(D) The given equation is $x^2(\cos^2 \theta - 1) - xy \sin^2 \theta + y^2 \sin^2 \theta = 0$.
Since $\cos^2 \theta - 1 = -\sin^2 \theta$,the equation becomes:
$-\sin^2 \theta x^2 - \sin^2 \theta xy + \sin^2 \theta y^2 = 0$.
Dividing by $-\sin^2 \theta$ (assuming $\sin \theta \neq 0$),we get:
$x^2 + xy - y^2 = 0$.
Comparing this with the general homogeneous equation $ax^2 + 2hxy + by^2 = 0$,we have $a = 1$,$2h = 1$,and $b = -1$.
The condition for the lines to be perpendicular is $a + b = 0$.
Here,$a + b = 1 + (-1) = 0$.
Therefore,the lines are perpendicular to each other,and the angle between them is $\frac{\pi}{2}$.

(B) The equation of the circle is $x^2+y^2=25$.
Two points $(x_1, y_1)$ and $(x_2, y_2)$ are conjugate with respect to the circle $x^2+y^2=r^2$ if $x_1x_2 + y_1y_2 = r^2$.
Here,$(x_1, y_1) = (1, a)$ and $(x_2, y_2) = (b, 2)$,and $r^2 = 25$.
Substituting these values into the condition,we get:
$(1)(b) + (a)(2) = 25$
$b + 2a = 25$
We need to find the value of $4a + 2b$.
Multiplying the equation $b + 2a = 25$ by $2$,we get:
$2(b + 2a) = 2(25)$
$2b + 4a = 50$
Thus,$4a + 2b = 50$.

(B) To determine the structure of $4$-bromo-$3$-methylbut-$1$-ene,we analyze the $IUPAC$ name:
$1$. The parent chain is 'but$-1-$ene',which means a four-carbon chain with a double bond at the first position: $CH_2=CH-CH_2-CH_3$.
$2$. The substituent '$3$-methyl' indicates a methyl group $(-CH_3)$ attached to the third carbon atom.
$3$. The substituent '$4$-bromo' indicates a bromine atom $(-Br)$ attached to the fourth carbon atom.
$4$. Combining these,the structure is $CH_2=CH-CH(CH_3)-CH_2Br$.
Thus,the correct option is $B$.

(B) The de-bromination of vicinal dihalides like $1, 2$-dibromoethane is carried out using zinc $(Zn)$ dust in the presence of alcohol (usually ethanol) upon heating.
This reaction is a type of elimination reaction where the two bromine atoms are removed to form an alkene.
The chemical equation is:
$BrCH_2-CH_2Br + Zn \xrightarrow{\text{alcohol}, \Delta} CH_2=CH_2 + ZnBr_2$
Thus,the correct metal is $Zn$.

(B) The given reaction is the reaction of ethylene dibromide with alcoholic $KOH$ to form acetylene.
In this reaction,two molecules of $HBr$ are removed from the vicinal dihalide,which is known as dehydrobromination (a type of dehydrohalogenation).

(A) In the presence of a Lewis acid (like $AlCl_3$),benzene undergoes an electrophilic substitution reaction with an alkyl halide. This reaction is known as Friedel-Crafts alkylation.
The reaction is as follows:
$C_6H_6 + C_2H_5Cl \xrightarrow{AlCl_3} C_6H_5-C_2H_5 + HCl$
The product formed is ethylbenzene,which has the molecular formula $C_8H_{10}$.

(A) Permutit,also known as zeolite,is a hydrated sodium aluminosilicate with the general formula $Na_2Al_2Si_2O_8 \cdot xH_2O$.
It is used to remove the hardness of water by ion exchange.
During the process,the $Na^{+}$ ions in the zeolite are replaced by $Ca^{2+}$ and $Mg^{2+}$ ions present in hard water,making the water soft.
When all the $Na^{+}$ ions are replaced by $Ca^{2+}$ or $Mg^{2+}$ ions,the permutit becomes exhausted.
Therefore,exhausted permutit does not contain $Na^{+}$ ions.

(C) Given function is $f(x, y) = 2(x-y)^2 - x^4 - y^4$.
First,we find the partial derivatives:
$f_x = \frac{\partial}{\partial x} [2(x-y)^2 - x^4 - y^4] = 4(x-y) - 4x^3$.
$f_y = \frac{\partial}{\partial y} [2(x-y)^2 - x^4 - y^4] = -4(x-y) - 4y^3$.
Now,we find the second-order partial derivatives:
$f_{xx} = \frac{\partial}{\partial x} [4(x-y) - 4x^3] = 4 - 12x^2$.
$f_{yy} = \frac{\partial}{\partial y} [-4(x-y) - 4y^3] = 4 - 12y^2$.
$f_{xy} = \frac{\partial}{\partial y} [4(x-y) - 4x^3] = -4$.
Evaluating these at $(0,0)$:
$f_{xx}(0,0) = 4 - 12(0)^2 = 4$.
$f_{yy}(0,0) = 4 - 12(0)^2 = 4$.
$f_{xy}(0,0) = -4$.
Finally,calculate the expression:
$f_{xx}f_{yy} - f_{xy}^2 = (4)(4) - (-4)^2 = 16 - 16 = 0$.

(C) Given that,$\frac{x+1}{(2x-1)(3x+1)}=\frac{A}{(2x-1)}+\frac{B}{(3x+1)}$
Multiplying both sides by $(2x-1)(3x+1)$,we get:
$x+1 = A(3x+1) + B(2x-1)$
$x+1 = x(3A+2B) + (A-B)$
Comparing the coefficients of $x$ and the constant terms on both sides,we get:
$3A+2B = 1$ ...$(i)$
$A-B = 1$ ...$(ii)$
From equation $(ii)$,$A = B+1$. Substituting this into equation $(i)$:
$3(B+1) + 2B = 1$
$3B+3+2B = 1$
$5B = -2 \Rightarrow B = -\frac{2}{5}$
Now,$A = -\frac{2}{5} + 1 = \frac{3}{5}$
We need to find $16A+9B$:
$16\left(\frac{3}{5}\right) + 9\left(-\frac{2}{5}\right) = \frac{48}{5} - \frac{18}{5} = \frac{30}{5} = 6$

(D) Statement $A$ is incorrect. When a person walks,they push the ground backward with their foot. According to Newton's third law,the ground exerts an equal and opposite force on the person in the forward direction. Thus,the frictional force acts in the direction of motion.
Statement $B$ is correct. For a cycle,the rear wheel is driven by the chain,so the ground exerts a forward frictional force on it. However,the front wheel is a rolling wheel that is not driven; it experiences a backward frictional force from the ground to oppose its rotation and motion.

(C) Given $y = \sin^{-1} x$.
Differentiating with respect to $x$,we get:
$\frac{d y}{d x} = \frac{1}{\sqrt{1 - x^2}}$.
Squaring both sides:
$\left(\frac{d y}{d x}\right)^2 = \frac{1}{1 - x^2}$.
$(1 - x^2) \left(\frac{d y}{d x}\right)^2 = 1$.
Differentiating again with respect to $x$:
$(1 - x^2) \cdot 2 \left(\frac{d y}{d x}\right) \cdot \frac{d^2 y}{d x^2} + \left(\frac{d y}{d x}\right)^2 \cdot (-2x) = 0$.
Dividing by $2 \frac{d y}{d x}$ (assuming $\frac{d y}{d x} \neq 0$):
$(1 - x^2) \frac{d^2 y}{d x^2} - x \frac{d y}{d x} = 0$.
Therefore,$(1 - x^2) \frac{d^2 y}{d x^2} = x \frac{d y}{d x}$.

(A) Let $I = \int_0^2 \frac{2x-2}{2x-x^2} dx$.
Notice that the integrand is improper at the boundaries $x=0$ and $x=2$ because the denominator $2x-x^2 = x(2-x)$ becomes zero at these points.
We evaluate the improper integral as a limit: $I = \lim_{\epsilon \to 0^+} \int_{\epsilon}^{2-\epsilon} \frac{2x-2}{2x-x^2} dx$.
Let $f(x) = 2x-x^2$,then $f'(x) = 2-2x = -(2x-2)$.
Thus,$\int \frac{2x-2}{2x-x^2} dx = -\int \frac{f'(x)}{f(x)} dx = -\ln|2x-x^2| + C$.
Evaluating the definite integral:
$I = \lim_{\epsilon \to 0^+} [-\ln|2x-x^2|]_{\epsilon}^{2-\epsilon} = \lim_{\epsilon \to 0^+} [-\ln|2(2-\epsilon)-(2-\epsilon)^2| - (-\ln|2\epsilon-\epsilon^2|)]$.
$I = \lim_{\epsilon \to 0^+} [\ln|2\epsilon-\epsilon^2| - \ln|4-2\epsilon-(4-4\epsilon+\epsilon^2)|] = \lim_{\epsilon \to 0^+} [\ln|2\epsilon-\epsilon^2| - \ln|2\epsilon-\epsilon^2|] = 0$.
Since the limits of the logarithmic terms cancel out symmetrically,the principal value of the integral is $0$.

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta$.
Define $f(\theta) = \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right)$.
Now,check for parity:
$f(-\theta) = \log \left(\frac{2-\sin(-\theta)}{2+\sin(-\theta)}\right) = \log \left(\frac{2+\sin \theta}{2-\sin \theta}\right)$.
Using the property $\log(\frac{a}{b}) = -\log(\frac{b}{a})$,we get:
$f(-\theta) = -\log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) = -f(\theta)$.
Since $f(-\theta) = -f(\theta)$,the function $f(\theta)$ is an odd function.
By the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) dx = 0$.
Therefore,$I = 0$.

(A) Given that,$y = A e^x + B e^{2x} + C e^{3x}$ ... $(i)$
Differentiating with respect to $x$,we get:
$y^{\prime} = A e^x + 2B e^{2x} + 3C e^{3x}$ ... (ii)
Differentiating again with respect to $x$,we get:
$y^{\prime \prime} = A e^x + 4B e^{2x} + 9C e^{3x}$ ... (iii)
Differentiating again with respect to $x$,we get:
$y^{\prime \prime \prime} = A e^x + 8B e^{2x} + 27C e^{3x}$ ... (iv)
The given equation is a linear combination of $e^x, e^{2x}, e^{3x}$. The characteristic equation is $(m-1)(m-2)(m-3) = 0$.
Expanding this,we get:
$(m^2 - 3m + 2)(m - 3) = 0$
$m^3 - 3m^2 - 3m^2 + 9m + 2m - 6 = 0$
$m^3 - 6m^2 + 11m - 6 = 0$
Replacing $m^k$ with $y^{(k)}$,we get the differential equation:
$y^{\prime \prime \prime} - 6y^{\prime \prime} + 11y^{\prime} - 6y = 0$.

(C) Given that $d(n)$ represents the number of positive divisors of $n$.
For a prime number $P$,the number of divisors of $P^7$ is $7+1 = 8$.
So,$d(P^7) = 8$.
Next,we find $d(d(P^7)) = d(8)$. Since $8 = 2^3$,the number of divisors is $3+1 = 4$.
So,$d(d(P^7)) = 4$.
Finally,we find $d(d(d(P^7))) = d(4)$. Since $4 = 2^2$,the number of divisors is $2+1 = 3$.
Therefore,$d(d(d(P^7))) = 3$.

(D) Given that,$\log _{27}(\log _3 x) = \frac{1}{3}$.
Using the definition of logarithm,$\log _a b = c \implies b = a^c$,we get:
$\log _3 x = (27)^{1/3}$.
Since $27 = 3^3$,we have $(27)^{1/3} = (3^3)^{1/3} = 3^1 = 3$.
So,$\log _3 x = 3$.
Again,using the definition of logarithm,$x = 3^3$.
Therefore,$x = 27$.

(A) Let $\vec{a} = 3 \hat{i} + 3 \hat{j} + \sqrt{3} \hat{k}$,$\vec{b} = \hat{i} + \hat{k}$,and $\vec{c} = \sqrt{3} \hat{i} + \sqrt{3} \hat{j} + \lambda \hat{k}$.
Since these vectors are coplanar,their scalar triple product must be zero,i.e.,$\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$.
This is equivalent to the determinant of the matrix formed by their components being zero:
$\begin{vmatrix} 3 & 3 & \sqrt{3} \\ 1 & 0 & 1 \\ \sqrt{3} & \sqrt{3} & \lambda \end{vmatrix} = 0$
Expanding the determinant along the first row:
$3(0 - \sqrt{3}) - 3(\lambda - \sqrt{3}) + \sqrt{3}(\sqrt{3} - 0) = 0$
$-3\sqrt{3} - 3\lambda + 3\sqrt{3} + 3 = 0$
$-3\lambda + 3 = 0$
$3\lambda = 3$
$\lambda = 1$

(C) Let $\vec{a} = \hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = \lambda\hat{i} - 4\hat{j} + \hat{k}$.
Since the vectors $\vec{a}$ and $\vec{b}$ are orthogonal,their dot product must be zero,i.e.,$\vec{a} \cdot \vec{b} = 0$.
Calculating the dot product:
$(\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (\lambda\hat{i} - 4\hat{j} + \hat{k}) = 0$
$(1)(\lambda) + (3)(-4) + (4)(1) = 0$
$\lambda - 12 + 4 = 0$
$\lambda - 8 = 0$
$\lambda = 8$.

(A) Let the sides of the parallelogram be represented by vectors $\vec{a} = \hat{i}+2 \hat{j}+3 \hat{k}$ and $\vec{b} = 3 \hat{i}+2 \hat{j}+\hat{k}$.
The diagonals of the parallelogram are given by $\vec{d_1} = \vec{a} + \vec{b}$ and $\vec{d_2} = \vec{a} - \vec{b}$.
Calculating $\vec{d_1} = (\hat{i}+2 \hat{j}+3 \hat{k}) + (3 \hat{i}+2 \hat{j}+\hat{k}) = 4\hat{i} + 4\hat{j} + 4\hat{k} = 4(\hat{i} + \hat{j} + \hat{k})$.
The unit vector parallel to $\vec{d_1}$ is $\frac{\vec{d_1}}{|\vec{d_1}|} = \frac{4(\hat{i} + \hat{j} + \hat{k})}{\sqrt{4^2 + 4^2 + 4^2}} = \frac{4(\hat{i} + \hat{j} + \hat{k})}{\sqrt{16+16+16}} = \frac{4(\hat{i} + \hat{j} + \hat{k})}{\sqrt{48}} = \frac{4(\hat{i} + \hat{j} + \hat{k})}{4\sqrt{3}} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.
Calculating $\vec{d_2} = (\hat{i}+2 \hat{j}+3 \hat{k}) - (3 \hat{i}+2 \hat{j}+\hat{k}) = -2\hat{i} + 0\hat{j} + 2\hat{k} = -2\hat{i} + 2\hat{k}$.
The unit vector parallel to $\vec{d_2}$ is $\frac{-2\hat{i} + 2\hat{k}}{\sqrt{(-2)^2 + 2^2}} = \frac{-2\hat{i} + 2\hat{k}}{\sqrt{8}} = \frac{-2\hat{i} + 2\hat{k}}{2\sqrt{2}} = \frac{-\hat{i} + \hat{k}}{\sqrt{2}}$.
Comparing with the given options,the correct unit vector is $\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.

(D) The given equation of the plane is $3x - 2y - z = 18$.
Dividing by $18$ to convert it into intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$:
$\frac{3x}{18} - \frac{2y}{18} - \frac{z}{18} = 1$
$\frac{x}{6} + \frac{y}{-9} + \frac{z}{-18} = 1$.
Thus,the plane meets the coordinate axes at points $A(6, 0, 0)$,$B(0, -9, 0)$,and $C(0, 0, -18)$.
The centroid $(G)$ of $\triangle ABC$ with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.
$G = (\frac{6+0+0}{3}, \frac{0-9+0}{3}, \frac{0+0-18}{3}) = (2, -3, -6)$.

(B) Let the origin be $O(0,0,0)$ and the foot of the perpendicular from the origin to the plane be $P(2,-1,3)$.
Since $OP$ is perpendicular to the plane,the vector $\vec{OP}$ is the normal vector to the plane.
The direction ratios of the normal vector $\vec{OP}$ are $(2-0, -1-0, 3-0) = (2, -1, 3)$.
The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector $(a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.
Substituting the point $P(2, -1, 3)$ and the normal vector $(2, -1, 3)$ into the equation:
$2(x-2) - 1(y-(-1)) + 3(z-3) = 0$
$2(x-2) - 1(y+1) + 3(z-3) = 0$
$2x - 4 - y - 1 + 3z - 9 = 0$
$2x - y + 3z - 14 = 0$
Thus,the equation of the plane is $2x - y + 3z - 14 = 0$.