TS EAMCET 2012 Chemistry Question Paper with Answer and Solution

(C) Insulin is a peptide hormone secreted by the pancreas.
It is transported to different body parts via the bloodstream.
Insulin acts by binding to specific receptors located on the surface of the target cells.
Therefore,its site of action is the $plasma \ membrane$.

(B) In jelly,the dispersed phase is $liquid$ and the dispersion medium is $solid$.
Thus,it is a colloidal solution of $liquid$ in $solid$.

(D) According to Wien's displacement law, $\lambda_m T = \text{constant}$, so $T \propto \frac{1}{\lambda_m}$.
Given $\lambda_2 = \frac{\lambda_1}{2}$, therefore $T_2 = 2 T_1$.
According to Stefan-Boltzmann law, the power radiated by a black body is $P = \sigma A T^4$, which implies $P \propto T^4$.
Thus, $\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4 = (2)^4 = 16$.
Therefore, $P_2 = 16 P_1 = 16 P$.

(B) The moment of inertia $I$ of a body is given by $I = Mk^2$,where $M$ is the mass and $k$ is the radius of gyration.
Since mass $M$ remains constant,we have $I \propto k^2$.
Taking the natural logarithm on both sides: $\ln I = \ln M + 2 \ln k$.
Differentiating both sides,we get $\frac{dI}{I} = 2 \frac{dk}{k}$.
For thermal expansion,the radius of gyration $k$ changes with temperature as $k' = k(1 + \alpha \Delta T)$,which implies $\frac{\Delta k}{k} = \alpha \Delta T$.
Substituting this into the expression for the change in moment of inertia:
$\frac{\Delta I}{I} = 2 \left( \frac{\Delta k}{k} \right) = 2 \alpha \Delta T$.

(D) The coefficient of real expansion of a liquid $(\gamma_r)$ is constant and is given by the sum of the coefficient of apparent expansion $(\gamma_a)$ and the coefficient of volume expansion of the container $(\gamma_v = 3\alpha$, where $\alpha$ is the coefficient of linear expansion).
Thus, $\gamma_r = \gamma_{a1} + 3\alpha_1 = \gamma_{a2} + 3\alpha_2$.
Given:
$\gamma_{a1} = 6 \times 10^{-6} /{ }^{\circ} C$ (for copper vessel)
$\alpha_1 = 18 \times 10^{-6} /{ }^{\circ} C$ (for copper)
$\gamma_{a2} = 24 \times 10^{-6} /{ }^{\circ} C$ (for steel vessel)
Substituting the values:
$6 \times 10^{-6} + 3(18 \times 10^{-6}) = 24 \times 10^{-6} + 3\alpha_2$
$6 \times 10^{-6} + 54 \times 10^{-6} = 24 \times 10^{-6} + 3\alpha_2$
$60 \times 10^{-6} = 24 \times 10^{-6} + 3\alpha_2$
$3\alpha_2 = 36 \times 10^{-6} /{ }^{\circ} C$
$\alpha_2 = 12 \times 10^{-6} /{ }^{\circ} C$.

(C) Given the condition $V \propto T^{2/3}$,we can write $V = c T^{2/3}$,which implies $T \propto V^{3/2}$.
Using the ideal gas law $PV = nRT$,we substitute $T \propto V^{3/2}$ to get $PV \propto V^{3/2}$,which simplifies to $P \propto V^{1/2}$.
This represents a polytropic process $PV^x = \text{constant}$,where $x = -1/2$.
The work done in a polytropic process is given by $W = \frac{nR \Delta T}{1-x}$.
Here,$n = 1 \ mol$,$\Delta T = 30 \ K$,and $x = -1/2$.
Substituting the values: $W = \frac{1 \times 8.314 \times 30}{1 - (-1/2)} = \frac{249.42}{1.5} = 166.28 \ J$.
Rounding to the nearest option,we get $166.2 \ J$.

(C) For an adiabatic process,the relation between temperature $T$ and pressure $p$ is given by $T^\gamma p^{1-\gamma} = \text{constant}$.
Taking the natural logarithm on both sides: $\gamma \ln T + (1-\gamma) \ln p = \text{constant}$.
Differentiating both sides: $\gamma \frac{\Delta T}{T} + (1-\gamma) \frac{\Delta p}{p} = 0$.
Rearranging for $\Delta T$: $\frac{\Delta T}{T} = \frac{\gamma - 1}{\gamma} \frac{\Delta p}{p}$.
Given: $T = 273 \text{ K}$ (at $NTP$),$\gamma = 1.5$,$\Delta p = 0.001 \text{ dyne}/cm^2$,$p = 1.013 \times 10^6 \text{ dyne}/cm^2$.
Substituting the values: $\Delta T = 273 \times \left( \frac{1.5 - 1}{1.5} \right) \times \frac{0.001}{1.013 \times 10^6}$.
$\Delta T = 273 \times \frac{0.5}{1.5} \times 9.87 \times 10^{-10}$.
$\Delta T = 273 \times \frac{1}{3} \times 9.87 \times 10^{-10} \approx 8.97 \times 10^{-8} \text{ K}$.

(A) The formula for heat supplied at constant pressure is given by $q = n \times C_p \times \Delta T$.
Given:
Heat supplied $(q)$ = $1 \ kJ = 1000 \ J$.
Molar mass of water $(H_2O)$ = $18 \ g \ mol^{-1}$.
Number of moles $(n)$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{100 \ g}{18 \ g \ mol^{-1}} = \frac{100}{18} \ mol$.
Molar heat capacity $(C_p)$ = $75 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values in the formula:
$1000 = \frac{100}{18} \times 75 \times \Delta T$.
$\Delta T = \frac{1000 \times 18}{100 \times 75} = \frac{180}{75} = 2.4 \ K$.

(A) As the size of the halogen atom increases from $Cl$ to $I$,the $A-A$ bond length increases.
Generally,as the bond length increases,the bond dissociation energy decreases.
Therefore,the bond energy follows the order: $Cl_2 > Br_2 > I_2$.

(C) From the figure,the path difference between the waves reaching point $P$ from $S_1$ and $S_2$ is given by $\Delta x = S_1 P - S_2 P$.
For the first bright fringe,the path difference must be equal to the wavelength $\lambda$ (since they are in phase).
From the geometry of the setup,the path difference can be approximated as $\Delta x = d \cos \theta$,where $d = 2 \lambda$ is the separation between the sources.
Thus,$2 \lambda \cos \theta = \lambda$.
$\cos \theta = \frac{1}{2}$,which implies $\theta = 60^{\circ}$.
From the right-angled triangle formed by the screen,$\tan \theta = \frac{OP}{D} = \frac{x}{D}$.
Since $\theta = 60^{\circ}$,$\tan 60^{\circ} = \sqrt{3}$.
Therefore,$\frac{x}{D} = \sqrt{3}$,which gives $x = \sqrt{3} D$.

(D) The speed of a transverse wave in a rope at a distance $x$ from the free end is given by $v = \sqrt{gx}$.
Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \sqrt{gx}$.
Rearranging the terms,we get $dt = \frac{dx}{\sqrt{gx}}$.
Integrating from $x=0$ to $x=l$,the total time $t$ is:
$t = \int_{0}^{l} \frac{dx}{\sqrt{gx}} = \frac{1}{\sqrt{g}} [2\sqrt{x}]_{0}^{l} = 2\sqrt{\frac{l}{g}}$.
Substituting the given values $l = 2.45 ~m$ and $g = 9.8 ~m/s^2$:
$t = 2 \sqrt{\frac{2.45}{9.8}} = 2 \sqrt{\frac{1}{4}} = 2 \times 0.5 = 1 ~s$.

(B) The frequency of a vibrating wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$. Since tension $T$ and mass per unit length $\mu$ are constant, we have $n \propto \frac{1}{l}$, which implies $n l = \text{constant}$.
Let $n$ be the frequency of the tuning fork.
At $l_1 = 100 \, cm$, the frequency of the wire is $n_1 = n + 8$ (since the wire is shorter, its frequency is higher).
At $l_2 = 101 \, cm$, the frequency of the wire is $n_2 = n - 8$.
Using $n_1 l_1 = n_2 l_2$:
$(n + 8) \times 100 = (n - 8) \times 101$
$100n + 800 = 101n - 808$
$101n - 100n = 800 + 808$
$n = 1608 \, Hz$.

(A) The initial kinetic energy of the vehicle is $K = \frac{p^2}{2M}$.
When the engine is switched off,the only force acting on the vehicle to stop it is the kinetic friction force $f_k = \mu_k M g$.
According to the work-energy theorem,the work done by the friction force is equal to the change in kinetic energy.
$W = \Delta K$
$-f_k \cdot s = 0 - K$
$\mu_k M g s = \frac{p^2}{2M}$
Solving for $s$ (the distance travelled):
$s = \frac{p^2}{2 M^2 \mu_k g}$.

(C) Power $P$ is defined as the rate of change of kinetic energy: $P = \frac{dK}{dt} = \frac{d}{dt} (\frac{1}{2} m v^2) = m v \frac{dv}{dt}$.
Since $a = \frac{dv}{dt}$,we have $P = m v a = m v (v \frac{dv}{dx}) = m v^2 \frac{dv}{dx}$.
Rearranging the terms: $v^2 dv = \frac{P}{m} dx$.
Integrating both sides: $\int_0^v v^2 dv = \int_0^x \frac{P}{m} dx$.
$\frac{v^3}{3} = \frac{P x}{m}$.
Thus,$v^3 = \frac{3 P x}{m}$.
Since $P$,$m$,and $x$ (at a certain distance) are related,we find $v^3 \propto \frac{P}{m}$.

(D) Statement $(A)$ is incorrect because the position of the centre of mass is a physical property of the system and is independent of the choice of the coordinate system.
Statement $(B)$ is correct because the motion of the centre of mass is governed by the net external force acting on the system,i.e.,$\vec{F}_{ext} = M\vec{a}_{cm}$.
Statement $(C)$ is correct because internal forces cancel each other out in pairs according to Newton's third law,so they do not affect the motion of the centre of mass.
Statement $(D)$ is incorrect because internal forces cannot change the state of motion of the centre of mass.
Therefore,statements $(A)$ and $(D)$ are wrong.

(C) Initial kinetic energy of ball $A$ is $K = p^2 / (2m)$,so $p = \sqrt{2mK}$.
Initial velocity of ball $A$ is $u_1 = p/m = \sqrt{2K/m}$.
After collision,ball $A$ moves in the negative $x$-direction with kinetic energy $K' = K/9$.
Let $v_1$ be the final velocity of ball $A$. Then $K' = 1/2 m v_1^2 = K/9$,which gives $v_1 = \sqrt{2K/(9m)} = (1/3) \sqrt{2K/m} = p/(3m)$.
Since ball $A$ moves in the negative $x$-direction,its final velocity is $v_1 = -p/(3m)$.
By the law of conservation of linear momentum: $p_{initial} = p_{final}$.
$p = m v_1 + p_B$,where $p_B$ is the final momentum of ball $B$.
$p = m(-p/(3m)) + p_B$.
$p = -p/3 + p_B$.
$p_B = p + p/3 = 4p/3$.

(D) The formal charge $(FC)$ is calculated using the formula: $FC = V - L - \frac{1}{2} B$,where $V$ is the number of valence electrons,$L$ is the number of lone pair electrons,and $B$ is the number of bonding electrons.
For the $CO_2$ molecule with the structure $\ddot{O}=C=\ddot{O}$:
For the $O$ atom: $V = 6$,$L = 4$,$B = 4$. Thus,$FC = 6 - 4 - \frac{1}{2}(4) = 0$.
For the $C$ atom: $V = 4$,$L = 0$,$B = 8$. Thus,$FC = 4 - 0 - \frac{1}{2}(8) = 0$.
Therefore,the formal charges of $C$ and $O$ atoms are $0$ and $0$,respectively.

(C) The molecular orbital electronic configuration of $O_2$ ($16$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^1, \pi^* 2p_y^1$.
Bonding molecular orbitals are those without an asterisk $(*)$.
The bonding orbitals are $\sigma 1s, \sigma 2s, \sigma 2p_z, \pi 2p_x, \pi 2p_y$.
Each of these $5$ orbitals contains $2$ electrons,totaling $10$ bonding electrons.
Therefore,the number of bonding electron pairs is $10 / 2 = 5$.

(D) The correct answer is $D$.
$A$: $LiOH$ is a weaker base than $NaOH$ because the $Li-O$ bond is stronger than the $Na-O$ bond due to the smaller size of $Li^+$,making it harder to release $OH^-$ ions. This is correct.
$B$: Salts of $Be$ undergo hydrolysis because $Be^{2+}$ has a high charge density and small size,which polarizes water molecules. This is correct.
$C$: Calcium bicarbonate,$Ca(HCO_3)_2$,exists only in aqueous solution and is soluble in water. This is correct.
$D$: Hydrolysis of beryllium carbide $(Be_2C)$ produces methane $(CH_4)$,not acetylene $(C_2H_2)$. The reaction is: $Be_2C + 4H_2O \rightarrow CH_4 + 2Be(OH)_2$. Thus,this statement is incorrect.

(B) The variation of the viscosity coefficient $(\eta)$ of liquids with temperature $(T)$ is described by the Andrade equation,which is given by the expression: $\eta = A e^{E / R T}$.
Here,$A$ is a constant,$E$ is the activation energy for viscous flow,$R$ is the universal gas constant,and $T$ is the absolute temperature.
As temperature increases,the viscosity of liquids decreases,which is consistent with the exponential form $\eta = A e^{E / R T}$.

(B) The work function is the minimum energy required to remove an electron from the surface of a metal. As the atomic size increases down a group,the ionization energy decreases,meaning the work function also decreases.
$Li$,$Na$,$K$,and $Rb$ belong to Group $1$ of the periodic table.
The order of their atomic sizes is $Li < Na < K < Rb$.
Therefore,the order of their work functions is $Li > Na > K > Rb$.
Given values: $Li = 2.41 \ eV$,$Na = 2.30 \ eV$,and $Rb = 2.09 \ eV$.
The work function of $K$ must lie between $Na$ $(2.30 \ eV)$ and $Rb$ $(2.09 \ eV)$.
Among the given options,$2.20 \ eV$ is the only value that falls within the range $2.09 \ eV < K < 2.30 \ eV$.

(A) Given,$(a+ib)^2=(c+id)^2(x+iy)$
Taking the modulus on both sides:
$|(a+ib)^2| = |(c+id)^2(x+iy)|$
$|a+ib|^2 = |c+id|^2 |x+iy|$
Since $|z|^2 = a^2+b^2$,we have:
$a^2+b^2 = (c^2+d^2) \sqrt{x^2+y^2}$
Substituting the given values $a^2+b^2=4$ and $c^2+d^2=2$:
$4 = 2 \sqrt{x^2+y^2}$
$\sqrt{x^2+y^2} = 2$
Squaring both sides:
$x^2+y^2 = 4$

(D) Given that $\alpha$ is a non-real root of $x^6-1=0$.
Since $\alpha^6=1$ and $\alpha \neq 1$ (as $\alpha$ is non-real),we have the sum of the roots of the geometric series:
$1+\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5 = \frac{\alpha^6-1}{\alpha-1} = 0$.
This implies $\alpha^2+\alpha^3+\alpha^4+\alpha^5 = -(1+\alpha)$.
Therefore,the expression becomes:
$\frac{\alpha^2+\alpha^3+\alpha^4+\alpha^5}{\alpha+1} = \frac{-(1+\alpha)}{\alpha+1} = -1$.

(C) Since $a, b, c$ are in geometric progression,we have $b = ar$ and $c = ar^2$.
Substituting these into the line equation $ax + by + c = 0$,we get $ax + ary + ar^2 = 0$.
Dividing by $a$ (assuming $a \neq 0$),we get $x + ry + r^2 = 0$,or $x = -ry - r^2$.
Substituting this into the curve equation $x + 2y^2 = 0$,we get $(-ry - r^2) + 2y^2 = 0$.
Rearranging gives the quadratic equation $2y^2 - ry - r^2 = 0$.
The sum of the ordinates ($y$-coordinates) is the sum of the roots of this quadratic equation,which is given by $-\frac{\text{coefficient of } y}{\text{coefficient of } y^2} = -(\frac{-r}{2}) = \frac{r}{2}$.

(C) Acid rain is primarily caused by the emission of oxides of nitrogen and sulphur into the atmosphere.
These gases react with water vapor in the atmosphere to form strong acids,namely nitric acid $(HNO_3)$ and sulphuric acid $(H_2SO_4)$,which then fall as acid rain.
The chemical reactions are as follows:
$2NO_2 + H_2O \rightarrow HNO_3 + HNO_2$
$2SO_2 + O_2 + 2H_2O \rightarrow 2H_2SO_4$
Thus,the pair of gases responsible for acid rain is $NO_2$ and $SO_2$.

(B) We know that for any positive real numbers $a$ and $b$,the Arithmetic Mean is greater than or equal to the Geometric Mean,i.e.,$\frac{a+b}{2} \geq \sqrt{ab}$.
Let $a = 27 \tan ^2 \theta$ and $b = 3 \cot ^2 \theta$.
Then,$\frac{27 \tan ^2 \theta + 3 \cot ^2 \theta}{2} \geq \sqrt{27 \tan ^2 \theta \cdot 3 \cot ^2 \theta}$.
$\frac{27 \tan ^2 \theta + 3 \cot ^2 \theta}{2} \geq \sqrt{81 \tan ^2 \theta \cdot \cot ^2 \theta}$.
Since $\tan \theta \cdot \cot \theta = 1$,we have $\frac{27 \tan ^2 \theta + 3 \cot ^2 \theta}{2} \geq \sqrt{81 \cdot 1}$.
$\frac{27 \tan ^2 \theta + 3 \cot ^2 \theta}{2} \geq 9$.
$27 \tan ^2 \theta + 3 \cot ^2 \theta \geq 18$.
Thus,the minimum value is $18$.

(B) The given equation is $(x^2+y^2) \cos^2 \theta = (x \cos \theta + y \sin \theta)^2$.
Expanding the right side:
$(x^2+y^2) \cos^2 \theta = x^2 \cos^2 \theta + y^2 \sin^2 \theta + 2xy \sin \theta \cos \theta$.
Rearranging the terms to the form $Ax^2 + 2Hxy + By^2 = 0$:
$x^2(\cos^2 \theta - \cos^2 \theta) + 2xy(\sin \theta \cos \theta) + y^2(\sin^2 \theta - \cos^2 \theta) = 0$.
$0x^2 + 2xy(\sin \theta \cos \theta) - y^2(\cos^2 \theta - \sin^2 \theta) = 0$.
For a pair of lines to be perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero $(A + B = 0)$.
Here,$A = 0$ and $B = -(\cos^2 \theta - \sin^2 \theta) = -\cos(2\theta)$.
Setting $A + B = 0$:
$0 - \cos(2\theta) = 0 \Rightarrow \cos(2\theta) = 0$.
Thus,$2\theta = \frac{\pi}{2} \Rightarrow \theta = \frac{\pi}{4}$.

(D) The given pair of lines is $8x^2 - 6xy + y^2 = 0$.
Factoring the quadratic equation: $8x^2 - 4xy - 2xy + y^2 = 0$ $\Rightarrow 4x(2x - y) - y(2x - y) = 0$ $\Rightarrow (4x - y)(2x - y) = 0$.
Thus,the two lines are $y = 4x$ and $y = 2x$.
The third line is $2x + 3y = a$.
The intersection points are:
$1$. Intersection of $y = 4x$ and $y = 2x$ is $O(0, 0)$.
$2$. Intersection of $y = 4x$ and $2x + 3y = a$: $2x + 3(4x) = a$ $\Rightarrow 14x = a$ $\Rightarrow x = a/14, y = 4a/14 = 2a/7$. Point $A(a/14, 2a/7)$.
$3$. Intersection of $y = 2x$ and $2x + 3y = a$: $2x + 3(2x) = a$ $\Rightarrow 8x = a$ $\Rightarrow x = a/8, y = 2a/8 = a/4$. Point $B(a/8, a/4)$.
The area of the triangle with vertices $(0, 0), (x_1, y_1), (x_2, y_2)$ is $\frac{1}{2} |x_1y_2 - x_2y_1|$.
Area $= \frac{1}{2} |(a/14)(a/4) - (a/8)(2a/7)| = \frac{1}{2} |a^2/56 - 2a^2/56| = \frac{1}{2} |-a^2/56| = a^2/112$.
Given area is $7$,so $a^2/112 = 7$ $\Rightarrow a^2 = 784$ $\Rightarrow a = 28$ (assuming $a > 0$ for area).

(D) Conformational isomers are stereoisomers that can be interconverted simply by rotation about a single bond. These are also known as $rotamers$ or $conformers$,and the phenomenon is known as $rotamerism$.

$(A)$. Acetaldehyde $(CH_3CHO)$ and vinyl alcohol $(CH_2=CH-OH)$ are tautomers,as they exist in dynamic equilibrium involving the migration of an $\alpha$-hydrogen atom.
$B$. Eclipsed and staggered ethane are different spatial arrangements of the same molecule,known as conformational isomers.
$C$. $(+)$$2$-butanol and $(-)$$2$-butanol are non-superimposable mirror images of each other,hence they are enantiomers.
$D$. Methyl-$n$-propylamine $(CH_3-NH-CH_2CH_2CH_3)$ and diethylamine $(CH_3CH_2-NH-CH_2CH_3)$ have different alkyl groups attached to the same nitrogen atom,which defines them as metamers.

(C) The chlorination of ethane is a classic example of a free radical substitution reaction. It proceeds through three main stages:
Step $I$: Initiation step,where chlorine molecules undergo homolytic fission in the presence of light $(hv)$ to form chlorine free radicals:
$Cl_2 \stackrel{hv}{\longrightarrow} 2 Cl^{\bullet}$
Step $II$: Propagation step,where the chlorine radical reacts with ethane to form an ethyl radical,which then reacts with another chlorine molecule to form chloroethane and regenerate a chlorine radical:
$CH_3CH_3 + Cl^{\bullet} \longrightarrow CH_3CH_2^{\bullet} + HCl$
$CH_3CH_2^{\bullet} + Cl_2 \longrightarrow CH_3CH_2Cl + Cl^{\bullet}$
Step $III$: Termination step,where free radicals combine to end the chain reaction:
$CH_3CH_2^{\bullet} + Cl^{\bullet} \longrightarrow CH_3CH_2Cl$
$Cl^{\bullet} + Cl^{\bullet} \longrightarrow Cl_2$
$2 CH_3CH_2^{\bullet} \longrightarrow CH_3CH_2CH_2CH_3$

(D) In benzene,all the $C-C$ bond lengths are equal to $1.39 \ Å$ due to resonance. Therefore,the statement that benzene has two different bond lengths of $1.54 \ Å$ and $1.34 \ Å$ is incorrect.

(B) $H_2O_2$ reacts with disodium hydrogen phosphate $(Na_2HPO_4)$ to form an addition product.
The reaction is as follows:
$H_2O_2 + Na_2HPO_4 \rightarrow Na_2HPO_4 \cdot H_2O_2$
This is a stable addition product.

(A) Given,$x^4+y^4=t^2+\frac{1}{t^2}$.
We know that $(t+\frac{1}{t})^2 = t^2 + \frac{1}{t^2} + 2$.
Thus,$t^2 + \frac{1}{t^2} = (t+\frac{1}{t})^2 - 2$.
Substituting the given values,we get $x^4+y^4 = (x^2+y^2)^2 - 2$.
Expanding the right side: $x^4+y^4 = x^4 + y^4 + 2x^2y^2 - 2$.
Subtracting $x^4+y^4$ from both sides,we get $0 = 2x^2y^2 - 2$,which implies $x^2y^2 = 1$.
Therefore,$y^2 = \frac{1}{x^2} = x^{-2}$.
Differentiating both sides with respect to $x$: $\frac{d}{dx}(y^2) = \frac{d}{dx}(x^{-2})$.
$2y \frac{dy}{dx} = -2x^{-3} = -\frac{2}{x^3}$.
Dividing by $2$,we get $y \frac{dy}{dx} = -\frac{1}{x^3}$.
Multiplying both sides by $x^3$,we get $x^3 y \frac{dy}{dx} = -1$.

(A) Given,$x^m y^n = (x+y)^{m+n}$.
Taking the natural logarithm on both sides,we get:
$m \ln x + n \ln y = (m+n) \ln (x+y)$.
Differentiating both sides with respect to $x$:
$\frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = \frac{m+n}{x+y} \left(1 + \frac{dy}{dx}\right)$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{m}{x} - \frac{m+n}{x+y} = \frac{dy}{dx} \left( \frac{m+n}{x+y} - \frac{n}{y} \right)$.
Simplifying both sides:
$\frac{m(x+y) - x(m+n)}{x(x+y)} = \frac{dy}{dx} \left( \frac{y(m+n) - n(x+y)}{y(x+y)} \right)$.
$\frac{mx + my - mx - nx}{x(x+y)} = \frac{dy}{dx} \left( \frac{my + ny - nx - ny}{y(x+y)} \right)$.
$\frac{my - nx}{x(x+y)} = \frac{dy}{dx} \left( \frac{my - nx}{y(x+y)} \right)$.
Since $my - nx \neq 0$ (as $x, y$ are variables),we can cancel the terms:
$\frac{1}{x} = \frac{dy}{dx} \left( \frac{1}{y} \right)$.
Therefore,$\frac{dy}{dx} = \frac{y}{x}$.

(C) $Cl^{-}$ is a Lewis base,not a Lewis acid,because its octet is complete and it has a lone pair of electrons to donate.
For $1.0 \times 10^{-8} \ M \ HCl$,the contribution of $H^{+}$ from water must be considered,resulting in a $pH$ slightly less than $7$ (approx $6.98$).
The ionic product of water $(K_w)$ at $25^{\circ} C$ is $1.0 \times 10^{-14} \ mol^2 \ L^{-2}$.
$AlCl_{3}$ acts as a Lewis acid due to an incomplete octet,which is not explained by the Bronsted-Lowry theory.

(B) Given that the rate of change of volume is $\frac{dV}{dt} = 2 \pi \text{ cm}^3/\text{s}$.
We know that the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Differentiating both sides with respect to $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Given $V = 288 \pi$,we find the radius $r$ as:
$288 \pi = \frac{4}{3} \pi r^3 \Rightarrow 216 = r^3 \Rightarrow r = 6 \text{ cm}$.
Substituting the values into the derivative equation:
$2 \pi = 4 \pi (6)^2 \frac{dr}{dt}$
$2 \pi = 4 \pi (36) \frac{dr}{dt}$
$2 \pi = 144 \pi \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{2 \pi}{144 \pi} = \frac{1}{72} \text{ cm/s}$.

(A) Let $I = \int \frac{dx}{\sqrt{x-x^2}}$.
We can rewrite the denominator as $\sqrt{x(1-x)} = \sqrt{x} \sqrt{1-x}$.
So,$I = \int \frac{dx}{\sqrt{x} \sqrt{1-x}}$.
Let $\sqrt{x} = \sin \theta$. Then $x = \sin^2 \theta$,which implies $dx = 2 \sin \theta \cos \theta \, d\theta$.
Substituting these into the integral:
$I = \int \frac{2 \sin \theta \cos \theta \, d\theta}{\sin \theta \sqrt{1-\sin^2 \theta}}$
$I = \int \frac{2 \sin \theta \cos \theta \, d\theta}{\sin \theta \cos \theta}$
$I = \int 2 \, d\theta = 2\theta + C$.
Since $\sin \theta = \sqrt{x}$,we have $\theta = \sin^{-1} \sqrt{x}$.
Therefore,$I = 2 \sin^{-1} \sqrt{x} + C$.

(D) Let $I = \int \sec^2 x \operatorname{cosec}^4 x \, dx$.
We can write $\sec^2 x = \frac{1}{\cos^2 x}$ and $\operatorname{cosec}^4 x = \frac{1}{\sin^4 x}$.
So,$I = \int \frac{1}{\sin^4 x \cos^2 x} \, dx$.
Using the identity $1 = \sin^2 x + \cos^2 x$,we have:
$I = \int \frac{\sin^2 x + \cos^2 x}{\sin^4 x \cos^2 x} \, dx = \int \frac{\sin^2 x}{\sin^4 x \cos^2 x} \, dx + \int \frac{\cos^2 x}{\sin^4 x \cos^2 x} \, dx$.
$I = \int \frac{1}{\sin^2 x \cos^2 x} \, dx + \int \frac{1}{\sin^4 x} \, dx$.
$I = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} \, dx + \int \operatorname{cosec}^4 x \, dx$.
$I = \int (\sec^2 x + \operatorname{cosec}^2 x) \, dx + \int \operatorname{cosec}^2 x (1 + \cot^2 x) \, dx$.
$I = \tan x - \cot x + \int \operatorname{cosec}^2 x \, dx + \int \operatorname{cosec}^2 x \cot^2 x \, dx$.
$I = \tan x - \cot x - \cot x + \int \cot^2 x \operatorname{cosec}^2 x \, dx$.
Let $u = \cot x$,then $du = -\operatorname{cosec}^2 x \, dx$.
$I = \tan x - 2 \cot x - \int u^2 \, du = \tan x - 2 \cot x - \frac{u^3}{3} + C$.
$I = -\frac{1}{3} \cot^3 x + \tan x - 2 \cot x + C$.
Comparing this with the given expression $-\frac{1}{3} \cot^3 x + k \tan x - 2 \cot x + C$,we get $k = 1$.

(A) The work done $W$ in rotating a magnetic dipole of magnetic moment $M$ in a magnetic field $B$ from an angle $\theta_1 = 0^{\circ}$ to $\theta_2 = 60^{\circ}$ is given by:
$W = MB(1 - \cos 60^{\circ}) = MB(1 - 0.5) = \frac{MB}{2}$.
From this,we find $MB = 2W$.
The torque $\tau$ required to maintain the needle at $\theta = 60^{\circ}$ is given by:
$\tau = MB \sin 60^{\circ}$.
Substituting $MB = 2W$ and $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$:
$\tau = (2W) \times \frac{\sqrt{3}}{2} = \sqrt{3} W$.

(C) Let the point of intersection be $P$. Any point on line $l_1$ is given by $(1+t, -6+2t, 2+t)$.
Any point on line $l_2$ is given by $(2u, 4+u, 1+2u)$.
For the lines to intersect,these points must be equal:
$1+t = 2u$ $(i)$
$-6+2t = 4+u$ (ii)
$2+t = 1+2u$ (iii)
From $(i)$,$t = 2u-1$. Substituting this into (ii):
$-6+2(2u-1) = 4+u$
$-6+4u-2 = 4+u$
$3u = 12 \Rightarrow u = 4$.
Substituting $u=4$ into $t=2u-1$,we get $t=2(4)-1 = 7$.
Check with (iii): $2+7 = 9$ and $1+2(4) = 9$. The values are consistent.
The point of intersection is $(1+7, -6+2(7), 2+7) = (8, 8, 9)$.

(B) Let the direction cosines of the line be $(l, m, n)$. Since the line is equally inclined to the coordinate axes,the angles made by the line with the $x, y,$ and $z$-axes are equal. Let this angle be $\alpha$.
Thus,$l = \cos \alpha, m = \cos \alpha, n = \cos \alpha$.
We know that for any line,the sum of the squares of the direction cosines is $1$,i.e.,$l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$.
$3 \cos^2 \alpha = 1$.
$\cos^2 \alpha = \frac{1}{3}$.
$\cos \alpha = \pm \frac{1}{\sqrt{3}}$.
Since the angle $\alpha$ is the angle with the $y$-axis,$\cos \alpha = \frac{1}{\sqrt{3}}$ (taking the positive value for the acute angle).
Therefore,$\alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

(A) Let the points be $A(1,0,0)$,$B(0,1,0)$,and $C(1,1,1)$.
Calculate the distances between the points:
$AB = \sqrt{(0-1)^2 + (1-0)^2 + 0^2} = \sqrt{1+1} = \sqrt{2}$
$BC = \sqrt{(1-0)^2 + (1-1)^2 + (1-0)^2} = \sqrt{1+0+1} = \sqrt{2}$
$CA = \sqrt{(1-1)^2 + (1-0)^2 + (1-0)^2} = \sqrt{0+1+1} = \sqrt{2}$
Since $AB = BC = CA = \sqrt{2}$,$\triangle ABC$ is an equilateral triangle.
The sphere with the smallest radius passing through these points has its center at the centroid of $\triangle ABC$.
Center $C' = \left(\frac{1+0+1}{3}, \frac{0+1+1}{3}, \frac{0+0+1}{3}\right) = \left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)$.
The radius $R$ is the distance from $C'$ to $A(1,0,0)$:
$R^2 = \left(1-\frac{2}{3}\right)^2 + \left(0-\frac{2}{3}\right)^2 + \left(0-\frac{1}{3}\right)^2 = \left(\frac{1}{3}\right)^2 + \left(-\frac{2}{3}\right)^2 + \left(-\frac{1}{3}\right)^2 = \frac{1}{9} + \frac{4}{9} + \frac{1}{9} = \frac{6}{9} = \frac{2}{3}$.
The equation of the sphere is $(x-\frac{2}{3})^2 + (y-\frac{2}{3})^2 + (z-\frac{1}{3})^2 = \frac{2}{3}$.
Expanding this: $x^2 - \frac{4x}{3} + \frac{4}{9} + y^2 - \frac{4y}{3} + \frac{4}{9} + z^2 - \frac{2z}{3} + \frac{1}{9} = \frac{6}{9}$.
$x^2 + y^2 + z^2 - \frac{4x}{3} - \frac{4y}{3} - \frac{2z}{3} + \frac{9}{9} = \frac{6}{9}$.
$x^2 + y^2 + z^2 - \frac{4x}{3} - \frac{4y}{3} - \frac{2z}{3} + 1 = \frac{2}{3}$.
Multiplying by $3$: $3(x^2+y^2+z^2) - 4x - 4y - 2z + 3 = 2$.
$3(x^2+y^2+z^2) - 4x - 4y - 2z + 1 = 0$.

(A) Let $X$ be the number of tails in $100$ tosses. $X$ follows a binomial distribution with $n = 100$ and $p = \frac{1}{2}$.
We need to find the probability that $X$ is odd,i.e.,$P(X \in \{1, 3, 5, \ldots, 99\})$.
The probability is given by $\sum_{k \in \{1, 3, \ldots, 99\}} \binom{100}{k} p^k q^{n-k}$.
Since $p = q = \frac{1}{2}$,this becomes $\sum_{k \in \{1, 3, \ldots, 99\}} \binom{100}{k} (\frac{1}{2})^{100} = \frac{1}{2^{100}} \sum_{k \in \{1, 3, \ldots, 99\}} \binom{100}{k}$.
We know that the sum of odd-indexed binomial coefficients $\binom{n}{1} + \binom{n}{3} + \ldots = 2^{n-1}$.
For $n = 100$,the sum is $2^{100-1} = 2^{99}$.
Therefore,the required probability is $\frac{1}{2^{100}} \times 2^{99} = \frac{2^{99}}{2^{100}} = \frac{1}{2}$.

(A) The probability mass function for a binomial distribution is given by $P(X=r) = { }^n C_r p^r q^{n-r}$,where $q = 1-p$.
We are given the ratio $\frac{P(X=r)}{P(X=n-r)}$.
Substituting the formula:
$\frac{P(X=r)}{P(X=n-r)} = \frac{{ }^n C_r p^r q^{n-r}}{{ }^n C_{n-r} p^{n-r} q^r}$.
Since ${ }^n C_r = { }^n C_{n-r}$,the expression simplifies to:
$\frac{P(X=r)}{P(X=n-r)} = \frac{p^r q^{n-r}}{p^{n-r} q^r} = \left(\frac{p}{q}\right)^{r - (n-r)} = \left(\frac{p}{q}\right)^{2r-n} = \left(\frac{q}{p}\right)^{n-2r}$.
For this expression to be independent of $n$,the base must be $1$ (or the exponent must be $0$,but it must hold for all $r$).
Thus,$\frac{q}{p} = 1$,which implies $q = p$.
Since $p + q = 1$,we have $p + p = 1$,which gives $2p = 1$.
Therefore,$p = \frac{1}{2}$.

(A) Given: Tension $F = 22 ~N$,Area $A = 0.02 ~cm^2 = 0.02 \times 10^{-4} ~m^2$,Young's modulus $Y = 1.1 \times 10^{11} ~N/m^2$,Poisson's ratio $\sigma = 0.32$.
Longitudinal strain is given by $\frac{\Delta l}{l} = \frac{F}{AY} = \frac{22}{0.02 \times 10^{-4} \times 1.1 \times 10^{11}} = \frac{22}{2.2 \times 10^5} = 10^{-4}$.
Poisson's ratio is defined as $\sigma = - \frac{\Delta r/r}{\Delta l/l}$. The lateral strain is $\frac{\Delta r}{r} = -\sigma \frac{\Delta l}{l}$.
The area $A = \pi r^2$,so $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.
Substituting the lateral strain,$\frac{\Delta A}{A} = -2 \sigma \frac{\Delta l}{l}$.
The magnitude of decrease in area is $\Delta A = 2 \sigma A \frac{\Delta l}{l}$.
$\Delta A = 2 \times 0.32 \times 0.02 ~cm^2 \times 10^{-4} = 0.64 \times 0.02 \times 10^{-4} ~cm^2 = 0.0128 \times 10^{-4} ~cm^2 = 1.28 \times 10^{-6} ~cm^2$.

(B) The reaction of $BCl_3$ with $H_2$ in the presence of a $Cu-Al$ catalyst at $450^{\circ}C$ produces diborane $(B_2H_6)$ as product $X$:
$2BCl_3 + 6H_2 \xrightarrow{Cu-Al, 450^{\circ}C} B_2H_6 + 6HCl$
Diborane $(B_2H_6)$ undergoes methylation with $CH_3Cl$ to form tetramethyldiborane,which is $Z$:
$B_2H_6 + 4CH_3Cl \rightarrow (CH_3)_4B_2H_2 + 4HCl$
Thus,$Z$ is $(CH_3)_4B_2H_2$.

(C) The initial oxidation state of $N$ in $N_2H_4$ is calculated as: $2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$.
Since $1 \ mole$ of $N_2H_4$ loses $10 \ moles$ of electrons,the total increase in oxidation number is $10$.
Let the oxidation state of $N$ in $Z$ be $y$. Since there are $2$ nitrogen atoms,the total oxidation state of $N$ in $Z$ is $2y$.
The change in oxidation state is $2y - (-4) = +10$.
$2y + 4 = 10 \implies 2y = 6 \implies y = +3$.
Therefore,the oxidation state of nitrogen in $Z$ is $+3$.

(C) The reduction of $NaOH$ to $Na$ is a highly endothermic and difficult process because $Na$ is a very strong reducing agent. Among the given options,$C$ (carbon) is capable of reducing $NaOH$ to $Na$ at high temperatures. The reaction is: $2NaOH + 2C \rightarrow 2Na + 2CO + H_2$.

(A) The $Spin$ quantum number $(s)$ explains the fine structure of spectral lines,such as the doublets observed in the spectra of $H$ and alkali metals,and the doublets and triplets observed in the spectra of alkaline earth metals,due to the electron spin angular momentum.