TS EAMCET 2011 Chemistry Question Paper with Answer and Solution

(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \frac{\phi \left( \frac{y}{x} \right)}{\phi' \left( \frac{y}{x} \right)}$.
Substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the differential equation,we get:
$v + x \frac{dv}{dx} = v + \frac{\phi(v)}{\phi'(v)}$.
Subtracting $v$ from both sides,we have:
$x \frac{dv}{dx} = \frac{\phi(v)}{\phi'(v)}$.
Separating the variables,we get:
$\frac{\phi'(v)}{\phi(v)} dv = \frac{dx}{x}$.
Integrating both sides:
$\int \frac{\phi'(v)}{\phi(v)} dv = \int \frac{dx}{x}$.
This yields $\log |\phi(v)| = \log |x| + \log |k|$,where $\log |k|$ is the constant of integration.
Using the property of logarithms,$\log |\phi(v)| = \log |kx|$.
Therefore,$\phi(v) = kx$.
Substituting $v = \frac{y}{x}$ back,we obtain the solution:
$\phi \left( \frac{y}{x} \right) = kx$.

(A) In a series $L-R$ circuit,the current $i$ at any time $t$ is given by $i = i_0(1 - e^{-Rt/L})$,where $i_0 = V/R$.
At $t = 0$,the current $i = i_0(1 - e^0) = i_0(1 - 1) = 0$.
The potential difference across the resistor is $V_R = iR$. Since $i = 0$ at $t = 0$,$V_R = 0 ~V$.
The potential difference across the inductor is $V_L = L(di/dt)$. According to Kirchhoff's voltage law,$V = V_R + V_L$. At $t = 0$,$V = 0 + V_L$,so $V_L = 25 ~V$.
Therefore,the potential difference across the resistor is $0 ~V$ and across the inductor is $25 ~V$.

(D) The reduction of nitrobenzene with zinc in an alkaline medium (e.g.,$Zn/NaOH$) yields hydrazobenzene $(C_{12}H_{12}N_2)$ as the product $X$.
Structure of hydrazobenzene: $C_6H_5-NH-NH-C_6H_5$.
Calculation of bonds:
$1$. Each phenyl ring $(C_6H_5)$ contains $6$ $C-C$ bonds (including $3$ $\pi$ bonds),$5$ $C-H$ bonds,and $1$ $C-N$ bond.
$2$. Total $\sigma$ bonds = $(6 \times 2) + (5 \times 2) + (1 \times 2) + (N-N \sigma) + (2 \times N-H \sigma) = 12 + 10 + 2 + 1 + 2 = 27$.
$3$. Total $\pi$ bonds = $3$ (in first ring) + $3$ (in second ring) = $6$.
Thus,$X$ contains $27 \sigma$ and $6 \pi$ bonds.

(B) At the isoelectric point,the concentration of the neutral dipolar ion (Zwitter ion) is maximum,which leads to minimum intermolecular electrostatic repulsion and minimum interaction with water molecules. Consequently,$\alpha$-amino acids have minimum solubility at their isoelectric point. Therefore,the statement that they have maximum solubility at the isoelectric point is incorrect.

(A) Let the initial capacitance of each condenser be $C = \frac{A \varepsilon_0}{d}$.
For condenser $M$,the space is filled with a dielectric of constant $K = 8$. The new capacitance is $C_M = K C = 8C$.
For condenser $N$,a copper plate of thickness $t = d/2$ is introduced. The new capacitance is $C_N = \frac{\varepsilon_0 A}{d - t} = \frac{\varepsilon_0 A}{d - d/2} = \frac{\varepsilon_0 A}{d/2} = 2 \left( \frac{\varepsilon_0 A}{d} \right) = 2C$.
Since the capacitors are connected in series,the charge $Q$ on both is the same.
The potential difference $V$ is given by $V = Q/C$,so $V \propto 1/C$.
Therefore,the ratio of potential differences is $V_M : V_N = \frac{1}{C_M} : \frac{1}{C_N} = \frac{1}{8C} : \frac{1}{2C} = \frac{1}{8} : \frac{1}{2} = 2 : 8 = 1 : 4$.

(B) Let $V$ be the potential difference across the capacitor when it is fully charged. The energy stored in the capacitor is given by $U = \frac{1}{2} C V^2$.
When the capacitor is fully discharged through the resistance wire,the entire stored energy is dissipated as heat $\Delta H$ in the block.
Since the system is thermally isolated,the heat gained by the block is equal to the energy lost by the capacitor: $\Delta H = U = \frac{1}{2} C V^2$.
The heat gained by the block is also given by the formula $\Delta H = m s \Delta T$,where $m$ is the mass,$s$ is the specific heat,and $\Delta T$ is the rise in temperature.
Equating the two expressions for $\Delta H$: $\frac{1}{2} C V^2 = m s \Delta T$.
Solving for $V$: $V^2 = \frac{2 m s \Delta T}{C}$,which gives $V = \left(\frac{2 m s \Delta T}{C}\right)^{1 / 2}$.

(B) Let the centre of the bigger disc be the origin $O(0,0)$.
Let the mass of the original disc be $M$ and its radius be $2R$.
The mass of the removed smaller disc of radius $R$ is $m = \frac{\pi R^2}{\pi (2R)^2} M = \frac{M}{4}$.
The centre of the removed disc is at a distance $R$ from the origin along the $x$-axis,so its coordinates are $(R, 0)$.
The centre of mass of the remaining part $(x_{CM})$ is given by the formula:
$x_{CM} = \frac{M_1 x_1 - m x_2}{M_1 - m}$
Here,$M_1 = M$,$x_1 = 0$,$m = M/4$,and $x_2 = R$.
$x_{CM} = \frac{M(0) - (M/4)(R)}{M - M/4} = \frac{-MR/4}{3M/4} = -\frac{R}{3}$.
The distance from the centre is $|x_{CM}| = \frac{R}{3}$.
Comparing this with $\alpha R$,we get $\alpha = \frac{1}{3}$.

(A) The velocity of the ball just before the first collision is $v_0 = \sqrt{2gh_0}$.
After the first collision,the velocity is $v_1 = ev_0$.
After the second collision,the velocity is $v_2 = ev_1 = e^2v_0$.
Following this pattern,after $n$ collisions,the velocity of the ball is $v_n = e^n v_0$.
The height $h_n$ reached after the $n$th collision is given by $h_n = \frac{v_n^2}{2g}$.
Substituting $v_n = e^n v_0$,we get $h_n = \frac{(e^n v_0)^2}{2g} = e^{2n} \frac{v_0^2}{2g}$.
Since $h_0 = \frac{v_0^2}{2g}$,we have $h_n = e^{2n} h_0$.
Therefore,$e^{2n} = \frac{h_n}{h_0}$,which implies $e = \left[\frac{h_n}{h_0}\right]^{1/2n}$.

(C) The structure of pyrophosphoric acid $(H_4P_2O_7)$ contains a $P-O-P$ linkage,four $P-OH$ groups,and two $P=O$ bonds.
Counting the bonds:
- There are $4$ $P-OH$ sigma bonds.
- There are $4$ $O-H$ sigma bonds.
- There is $1$ $P-O-P$ sigma bond (two $P-O$ bonds).
- There are $2$ $P=O$ sigma bonds.
- Total sigma bonds = $4 + 4 + 2 + 2 = 12$.
- There are $2$ $P=O$ pi bonds.
Thus,it contains $12$ $\sigma$ and $2$ $\pi$-bonds.

(B) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{1}{2} (V - N - C)$,where $V$ is the number of valence electrons,$N$ is the number of monovalent atoms,and $C$ is the charge.
$A) PCl_5$: $P$ has $5$ valence electrons. Lone pairs = $\frac{1}{2}(5 - 5) = 0$. $BrF_5$: $Br$ has $7$ valence electrons. Lone pairs = $\frac{1}{2}(7 - 5) = 1$.
$B) XeF_2$: $Xe$ has $8$ valence electrons. Lone pairs = $\frac{1}{2}(8 - 2) = 3$. $ICl$: $I$ has $7$ valence electrons. Lone pairs = $\frac{1}{2}(7 - 1) = 3$.
$C) XeF_4$: $Xe$ has $8$ valence electrons. Lone pairs = $\frac{1}{2}(8 - 4) = 2$. $ClO_4^{-}$: $Cl$ has $7$ valence electrons. Lone pairs = $\frac{1}{2}(7 - 0 + 1) = 4$ (Oxygen is divalent,so $N=0$).
$D) SCl_4$: $S$ has $6$ valence electrons. Lone pairs = $\frac{1}{2}(6 - 4) = 1$. $CH_4$: $C$ has $4$ valence electrons. Lone pairs = $\frac{1}{2}(4 - 4) = 0$.
Thus,both $XeF_2$ and $ICl$ have $3$ lone pairs on their central atoms.

(B) The formula for formal charge is:
$\text{Formal charge} = [\text{Total number of valence electrons in the free atom}] - [\text{Total number of non-bonding (lone pair) electrons}] - \frac{1}{2} [\text{Total number of bonding (shared) electrons}]$
For the end $N$ atom marked $(1)$:
$\text{Formal charge} = 5 - 4 - \frac{1}{2}(4) = 5 - 4 - 2 = -1$
For the central $N$ atom marked $(2)$:
$\text{Formal charge} = 5 - 0 - \frac{1}{2}(8) = 5 - 4 = +1$
For the end $O$ atom:
$\text{Formal charge} = 6 - 4 - \frac{1}{2}(4) = 6 - 4 - 2 = 0$
Thus,the formal charges are $-1, +1, 0$ for $N_{(1)}, N_{(2)}$ and $O$ respectively.

(D) The central atom $Xe$ has $8$ valence electrons. In $XeF_6$,it forms $6$ bond pairs with $F$ atoms and has $1$ lone pair.
Steric number = (Number of bond pairs) + (Number of lone pairs) = $6 + 1 = 7$.
$A$ steric number of $7$ corresponds to $sp^3d^3$ hybridisation.
Thus,$XeF_6$ has $sp^3d^3$ hybridisation and $1$ lone pair of electrons.

(D) Graphite has a two-dimensional sheet-like structure.
The adjacent layers of the sheet are held together by weak van der Waals' forces.
In graphite,each carbon atom is present in an $sp^2$ hybridized state.
In contrast,in diamond,each carbon atom is $sp^3$ hybridized.

(A) For the reaction $2 AB \rightleftharpoons A_2 + B_2$,the equilibrium constant is $K_c = \frac{[A_2][B_2]}{[AB]^2} = 49$.
For the reaction $AB \rightleftharpoons \frac{1}{2} A_2 + \frac{1}{2} B_2$,the equilibrium constant is $K_c' = \frac{[A_2]^{1/2} [B_2]^{1/2}}{[AB]}$.
This is equal to $\sqrt{K_c}$.
Therefore,$K_c' = \sqrt{49} = 7$.

(D) Alitame is an artificial sweetening agent. It is $2000$ times sweeter than cane sugar. Its structure is as follows:
$HOOC-CH_2-CH(NH_2)-CONH-CH(CH_3)-CONH-C(CH_3)_2-S-C(CH_3)_2-CH_2-$ (Note: The structure provided in the image represents Alitame).

(C) The electronic configurations of the ions formed after the removal of the first electron are:
$Na^{+} (1s^{2} 2s^{2} 2p^{6})$
$Ne^{+} (1s^{2} 2s^{2} 2p^{5})$
$Mg^{+} (1s^{2} 2s^{2} 2p^{6} 3s^{1})$
$Al^{+} (1s^{2} 2s^{2} 2p^{6} 3s^{2})$
The second ionisation potential $(IE_{2})$ is the energy required to remove an electron from these unipositive ions.
$Na^{+}$ has a stable noble gas configuration $(2s^{2} 2p^{6})$,making it extremely difficult to remove another electron,thus it has the highest $IE_{2}$.
$Mg^{+}$ has a single electron in the $3s$ orbital,which is relatively easy to remove.
$Al^{+}$ has a stable $3s^{2}$ configuration,making its $IE_{2}$ higher than $Mg^{+}$.
$Ne^{+}$ has a $2p^{5}$ configuration,which is less stable than $Na^{+}$ but requires significant energy.
Comparing these,the correct order of $IE_{2}$ is $Mg < Al < Ne < Na$.

(A) The sensitivity of a galvanometer is defined as the deflection per unit current,$S = \theta / I$.
When a shunt $S_h$ is connected in parallel with a galvanometer of resistance $G$,the new sensitivity $S'$ is given by the formula $S' = S \times \frac{S_h}{G + S_h}$.
Given: $S = 60 \text{ div/A}$,$S' = 10 \text{ div/A}$,and $G = 20 \ \Omega$.
Substituting these values into the formula: $10 = 60 \times \frac{S_h}{20 + S_h}$.
Dividing both sides by $10$: $1 = 6 \times \frac{S_h}{20 + S_h}$.
$20 + S_h = 6 S_h$.
$5 S_h = 20$.
$S_h = 4 \ \Omega$.
Therefore,the value of the shunt used is $4 \ \Omega$.

(A) The sensitivity of a galvanometer is defined as the deflection per unit current,given by $S_g = \frac{\theta}{i_g}$.
When a shunt $S$ is connected in parallel with a galvanometer of resistance $G$,the current $i_g$ flowing through the galvanometer is given by $i_g = i \left( \frac{S}{G+S} \right)$,where $i$ is the total current.
The new sensitivity $S'$ is given by $S' = \frac{\theta}{i} = \frac{\theta}{i_g} \cdot \frac{i_g}{i} = S_g \cdot \left( \frac{S}{G+S} \right)$.
Given $S_g = 60 \text{ division/A}$ and $S' = 10 \text{ division/A}$.
Substituting the values: $10 = 60 \cdot \left( \frac{S}{20+S} \right)$.
$\frac{10}{60} = \frac{S}{20+S} \Rightarrow \frac{1}{6} = \frac{S}{20+S}$.
$20 + S = 6S \Rightarrow 5S = 20$.
$S = 4 \ \Omega$.

(B) Since the ammeter reading is zero,no current flows through the right branch of the circuit.
Let $i_1$ be the current flowing through the left loop.
Applying Kirchhoff's voltage law to the left loop:
$12 - 500 i_1 - R i_1 = 0$ --- $(i)$
Since the ammeter reading is zero,the potential difference across the resistor $R$ must be equal to the electromotive force of the battery in the right branch.
Thus,the voltage across $R$ is $V_R = R i_1 = 2 \text{ V}$.
Substituting $R i_1 = 2$ into equation $(i)$:
$12 - 500 i_1 - 2 = 0$
$10 = 500 i_1$
$i_1 = \frac{10}{500} = \frac{1}{50} \text{ A}$.
Now,using $R i_1 = 2$:
$R \times \frac{1}{50} = 2$
$R = 100 \Omega$.

(D) According to Kirchhoff's Current Law $(KCL)$,the sum of currents entering a junction equals the sum of currents leaving it.
At junction $A$:
The incoming currents are $3 \ A$ and $2 \ A$. Let the current flowing from $A$ to $B$ be $i_1$.
$i_1 = 3 \ A + 2 \ A = 5 \ A$.
At junction $B$:
The incoming current is $i_1 = 5 \ A$. The outgoing currents are $2 \ A$ and the current flowing from $B$ to $C$,let it be $i_2$.
$i_1 = 2 \ A + i_2 \implies 5 \ A = 2 \ A + i_2 \implies i_2 = 3 \ A$.
At junction $C$:
The incoming currents are $i_2 = 3 \ A$ and $1 \ A$. The outgoing current is $i$.
$i = i_2 + 1 \ A = 3 \ A + 1 \ A = 4 \ A$.
Therefore,the value of current $i$ is $4 \ A$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of emitted photoelectrons is given by $K_{max} = E - \phi_0$,where $E$ is the energy of the incident photon and $\phi_0$ is the work function of the metal.
For the first photon: $\frac{1}{2} m v_1^2 = E_1 - \phi_0 = 2.5 \ eV - 1.5 \ eV = 1.0 \ eV$ $(i)$
For the second photon: $\frac{1}{2} m v_2^2 = E_2 - \phi_0 = 3.5 \ eV - 1.5 \ eV = 2.0 \ eV$ (ii)
Dividing equation $(i)$ by equation (ii),we get:
$\frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \frac{1.0 \ eV}{2.0 \ eV}$
$\frac{v_1^2}{v_2^2} = \frac{1}{2}$
Taking the square root on both sides:
$\frac{v_1}{v_2} = \frac{1}{\sqrt{2}}$
Thus,the ratio of the maximum velocities is $1 : \sqrt{2}$.

(A) According to Kohlrausch's law of independent migration of ions:
$\Lambda_{\infty} (NH_4Cl) = \Lambda_{\infty} (NH_4^+) + \Lambda_{\infty} (Cl^-) = 130 \ \Omega^{-1} \ cm^2 \ equiv^{-1} \ (i)$
$\Lambda_{\infty} (NaOH) = \Lambda_{\infty} (Na^+) + \Lambda_{\infty} (OH^-) = 217 \ \Omega^{-1} \ cm^2 \ equiv^{-1} \ (ii)$
$\Lambda_{\infty} (NaCl) = \Lambda_{\infty} (Na^+) + \Lambda_{\infty} (Cl^-) = 109 \ \Omega^{-1} \ cm^2 \ equiv^{-1} \ (iii)$
To find $\Lambda_{\infty} (NH_4OH)$,we perform the operation: $(i) + (ii) - (iii)$
$\Lambda_{\infty} (NH_4OH) = \Lambda_{\infty} (NH_4^+) + \Lambda_{\infty} (Cl^-) + \Lambda_{\infty} (Na^+) + \Lambda_{\infty} (OH^-) - (\Lambda_{\infty} (Na^+) + \Lambda_{\infty} (Cl^-))$
$\Lambda_{\infty} (NH_4OH) = \Lambda_{\infty} (NH_4^+) + \Lambda_{\infty} (OH^-)$
$\Lambda_{\infty} (NH_4OH) = 130 + 217 - 109 = 238 \ \Omega^{-1} \ cm^2 \ equiv^{-1}$

(C) Methyl isocyanate $(MIC)$ gas was responsible for the Bhopal gas tragedy that occurred in $1984$. $MIC$ is a highly toxic chemical intermediate used in the production of carbamate pesticides.

(C) Nitration is an electrophilic aromatic substitution reaction. The rate of this reaction depends on the electron density of the benzene ring.
Groups with a $+R$ (resonance) effect increase the electron density of the ring,making it more reactive towards electrophiles. The $-OH$ group in phenol $(I)$ has a strong $+R$ effect,significantly increasing the electron density.
Groups with a $-R$ effect decrease the electron density of the ring,making it less reactive. The $-NO_2$ group in nitrobenzene $(II)$ has a strong $-R$ effect,significantly decreasing the electron density.
Benzene $(III)$ has no substituent.
Therefore,the order of reactivity is: Phenol $(I)$ $>$ Benzene $(III)$ $>$ Nitrobenzene $(II)$.

(D) The stability of cycloalkanes is determined by the total strain present in the ring,which includes angle strain and torsional strain.
Cyclohexane exists in a chair conformation that is essentially free of both angle strain and torsional strain,making it the most stable cycloalkane.
Cyclopropane and cyclobutane possess significant angle strain due to the deviation of bond angles from the ideal tetrahedral angle of $109.5^{\circ}$,as well as torsional strain due to eclipsed hydrogen atoms.
Therefore,both the assertion and the reason are true,and the reason correctly explains why cyclohexane is more stable than the smaller cycloalkanes.

(B) compound exhibits enantiomerism if it contains a chiral center (an asymmetric carbon atom bonded to four different groups).
In $CH_3-CH(Br)-CH_2-CH_3$ ($2$-bromobutane),the carbon atom at position $2$ is bonded to four different groups: $-H$,$-CH_3$,$-Br$,and $-CH_2CH_3$.
Since it has a chiral center,it is optically active and exhibits enantiomerism.

(A) The molar mass of $BaSO_4$ is $233 \ g/mol$.
The atomic mass of sulphur $(S)$ is $32 \ g/mol$.
The formula for the percentage of sulphur is:
$\text{Percentage of } S = \frac{32 \times \text{mass of } BaSO_4 \times 100}{233 \times \text{mass of organic compound}}$.
Substituting the given values:
$\text{Percentage of } S = \frac{32 \times 0.233 \times 100}{233 \times 0.16}$.
Since $0.233 / 233 = 0.001$,the expression becomes:
$\frac{32 \times 0.001 \times 100}{0.16} = \frac{3.2}{0.16} = 20 \%$.
Thus,the percentage of sulphur is $20 \%$.

(A) Nickel oxide $(NiO)$ is reduced by water gas $(CO + H_2)$ to obtain the metal nickel $(Ni)$.
$2NiO + CO + H_2 \rightarrow 2Ni + CO_2 + H_2O$
Other oxides are reduced by different reducing agents:
$ZnO + C \xrightarrow{\Delta} Zn + CO$
$WO_3 + 3H_2 \xrightarrow{\Delta} W + 3H_2O$
$Fe_2O_3 + 3CO \xrightarrow{\Delta} 2Fe + 3CO_2$

(D) The gravitational force acting on an astronaut inside an orbiting spaceship is $F = G \frac{Mm}{r^2}$. This force is finite and non-zero. However,the astronaut is in a state of free fall along with the spaceship. In the non-inertial frame of the spaceship,the gravitational force is balanced by the centrifugal force,resulting in a net force of zero (weightlessness). Therefore,the astronaut does not 'experience' any weight or gravitational force. Thus,Assertion $(A)$ is false.
Reason $(R)$ is a standard physical fact: the gravitational force between the Earth and the spaceship provides the necessary centripetal force for orbital motion. Thus,Reason $(R)$ is true.

(A) In the presence of air and light,chloroform $(CHCl_3)$ undergoes oxidation to form carbonyl chloride,commonly known as phosgene $(COCl_2)$,which is a highly toxic gas.
The chemical reaction is:
$2CHCl_3 + O_2 \xrightarrow{\text{Air and light}} 2COCl_2 + 2HCl$

(A) The Wolff-Kishner reduction involves the reduction of carbonyl compounds (aldehydes or ketones) to alkanes.
In this reaction,the carbonyl compound is heated with hydrazine $(NH_2NH_2)$ and a strong base like potassium hydroxide $(KOH)$ in a high-boiling solvent such as ethylene glycol.
The reaction proceeds via the formation of a hydrazone intermediate,which then undergoes base-catalyzed decomposition to yield the corresponding alkane and nitrogen gas $(N_2)$.

(C) The reaction of benzene $(C_6H_6)$ with ozone $(O_3)$ is an ozonolysis reaction.
Benzene reacts with three molecules of ozone to form benzene triozonide $(X)$.
Benzene triozonide on reductive hydrolysis with $Zn / H_2O$ yields three molecules of glyoxal $(Y)$ $(CHO-CHO)$.
Thus,$X$ is triozonide and $Y$ is glyoxal.

(D) Both hydrogen $(H_2)$ and deuterium $(D_2)$ molecules have the same bond length of $74 \ pm$.
However,they differ in their physical and chemical properties such as bond energy,melting point,and boiling point due to the difference in their isotopic masses.

(D) For a weak acid,the concentration of $H^+$ ions is given by $[H^+] = \sqrt{K_a \times C}$.
Given $K_a = 2 \times 10^{-5}$ and $C = 0.05 \ M$.
$[H^+] = \sqrt{2 \times 10^{-5} \times 0.05} = \sqrt{10^{-6}} = 10^{-3} \ M$.
$pH = -\log[H^+] = -\log(10^{-3}) = 3$.

(C) The strength of an acid is directly proportional to its acid dissociation constant $(K_a)$ and inversely proportional to its $pK_a$ value.
Mathematically,$pK_a = -\log(K_a)$.
Therefore,a stronger acid possesses a higher $K_a$ value and a smaller $pK_a$ value.
Comparing the given values $(4.76, 4.19, 0.23, 3.41)$,the smallest value is $0.23$.
Thus,the strongest carboxylic acid among them has a $pK_a$ value of $0.23$.

(C) The total kinetic energy of the system consists of translational kinetic energy and rotational kinetic energy.
Total mass of the system $M = m + m = 2m$.
Translational kinetic energy $KE_{trans} = \frac{1}{2} M v^2 = \frac{1}{2} (2m) v^2 = m v^2$.
Since the sphere is hollow and filled with liquid,the liquid does not rotate with the sphere (it remains stationary relative to the ground). Only the hollow shell rotates.
The moment of inertia of a thin hollow sphere is $I = \frac{2}{3} m R^2$.
Rotational kinetic energy $KE_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{2}{3} m R^2) (\frac{v}{R})^2 = \frac{1}{3} m v^2$.
Total kinetic energy $KE_{total} = KE_{trans} + KE_{rot} = m v^2 + \frac{1}{3} m v^2 = \frac{4}{3} m v^2$.

(C) The pendulum bob is in equilibrium under the action of three forces: the tension $T$ in the string,the horizontal force $F = 2 ~N$,and the weight $W = 1 ~N$ acting vertically downwards.
For equilibrium,the net force in both horizontal and vertical directions must be zero.
Resolving the tension $T$ into components:
Horizontal component: $T \sin \theta = F = 2 ~N$ $(i)$
Vertical component: $T \cos \theta = W = 1 ~N$ (ii)
Squaring and adding equations $(i)$ and (ii):
$(T \sin \theta)^2 + (T \cos \theta)^2 = F^2 + W^2$
$T^2 (\sin^2 \theta + \cos^2 \theta) = F^2 + W^2$
$T^2 = F^2 + W^2$
$T = \sqrt{F^2 + W^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} ~N$.

(B) The correct matches are as follows:
$(A)$ Fleming's left hand rule is used to determine the direction of the force on a current-carrying conductor in a magnetic field. Thus,$(A)$-(iii).
$(B)$ Right hand thumb rule is used to determine the direction of magnetic field lines around a current-carrying conductor. Thus,$(B)$-(iv).
$(C)$ Biot-Savart law is used to calculate the magnitude and direction of magnetic induction due to a current element. Thus,$(C)$-(ii).
$(D)$ Fleming's right hand rule is used to determine the direction of the induced current in a conductor moving in a magnetic field. Thus,$(D)$-$(i)$.
Therefore,the correct matching is $(A)$-(iii),$(B)$-(iv),$(C)$-(ii),$(D)$-$(i)$,which corresponds to option $(b)$.

(C) Initial pole strength $= m$ and magnetic moment $= M = m \times (2l)$.
When a magnet is cut $n$ times parallel to its axis,the pole strength of each piece becomes $\frac{m}{n+1}$. Here,$n=5$,so the new pole strength $m' = \frac{m}{5+1} = \frac{m}{6}$.
When a magnet is cut $k$ times perpendicular to its axis,the length of each piece becomes $\frac{2l}{k+1}$. Here,$k=3$,so the new length $2l' = \frac{2l}{3+1} = \frac{2l}{4}$.
The magnetic moment of each piece is $M' = m' \times (2l') = \left(\frac{m}{6}\right) \times \left(\frac{2l}{4}\right) = \frac{m \times 2l}{24} = \frac{M}{24}$.
Thus,the pole strength is $\frac{m}{6}$ and the magnetic moment is $\frac{M}{24}$.

(A) Let the common root be $\alpha$. Then $\alpha^3+a\alpha+1=0$ and $\alpha^4+a\alpha^2+1=0$.
From the first equation,$a\alpha = -\alpha^3-1$,so $a = -\alpha^2 - \frac{1}{\alpha}$.
Substitute $a$ into the second equation: $\alpha^4 + (-\alpha^2 - \frac{1}{\alpha})\alpha^2 + 1 = 0$.
$\alpha^4 - \alpha^4 - \alpha + 1 = 0$.
This gives $\alpha = 1$.
Substituting $\alpha = 1$ into $x^3+ax+1=0$,we get $1^3+a(1)+1=0$,which implies $1+a+1=0$,so $a = -2$.

(D) The given quadratic expression is $y = ax^2 + bx + c$.
Since the discriminant $D = b^2 - 4ac = 0$,the quadratic equation $ax^2 + bx + c = 0$ has two equal real roots.
This implies that the parabola $y = ax^2 + bx + c$ touches the $x$-axis at exactly one point.
Since $a > 0$,the parabola opens upwards.
Therefore,the curve touches the $x$-axis and lies above it.

(D) We know that $(1+i)^2 = 1 + i^2 + 2i = 2i$ and $(1-i)^2 = 1 + i^2 - 2i = -2i$.
Given expression: $E = \frac{(1+i)^{2011}}{(1-i)^{2009}} = \frac{(1+i)^{2009} \cdot (1+i)^2}{(1-i)^{2009}}$.
$E = \left(\frac{1+i}{1-i}\right)^{2009} \cdot (1+i)^2$.
Since $\frac{1+i}{1-i} = \frac{(1+i)^2}{1^2 - i^2} = \frac{2i}{2} = i$,
$E = (i)^{2009} \cdot (2i)$.
$E = (i^{2008} \cdot i) \cdot 2i = (1 \cdot i) \cdot 2i = 2i^2 = 2(-1) = -2$.

(B) Given,$z = a - \frac{i}{2}$.
Substitute $z$ into the expression $|i + z|^2 - |i - z|^2$:
$|i + (a - \frac{i}{2})|^2 - |i - (a - \frac{i}{2})|^2$
$= |a + \frac{i}{2}|^2 - |-a + \frac{3i}{2}|^2$
$= (a^2 + (\frac{1}{2})^2) - ((-a)^2 + (\frac{3}{2})^2)$
$= a^2 + \frac{1}{4} - (a^2 + \frac{9}{4})$
$= \frac{1}{4} - \frac{9}{4} = -\frac{8}{4} = -2$.

(A) Let $z = x + iy$.
Given $\arg \left(\frac{z-2}{z+2}\right) = \frac{\pi}{3}$.
Using the property $\arg \left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$,we have:
$\arg(z-2) - \arg(z+2) = \frac{\pi}{3}$
$\arg((x-2) + iy) - \arg((x+2) + iy) = \frac{\pi}{3}$
$\tan^{-1}\left(\frac{y}{x-2}\right) - \tan^{-1}\left(\frac{y}{x+2}\right) = \frac{\pi}{3}$
Applying the formula $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$:
$\tan^{-1}\left[\frac{\frac{y}{x-2} - \frac{y}{x+2}}{1 + \left(\frac{y}{x-2}\right)\left(\frac{y}{x+2}\right)}\right] = \frac{\pi}{3}$
$\frac{\frac{y(x+2) - y(x-2)}{x^2-4}}{\frac{x^2-4+y^2}{x^2-4}} = \tan\left(\frac{\pi}{3}\right)$
$\frac{4y}{x^2+y^2-4} = \sqrt{3}$
$x^2 + y^2 - \frac{4}{\sqrt{3}}y - 4 = 0$
This is the equation of a circle.

(A) The volume flow rate (quantity of water per second) is given by $Q = A v$,where $A$ is the area of the hole and $v$ is the velocity of efflux.
According to Torricelli's law,the velocity of efflux at a depth $h$ is $v = \sqrt{2gh}$.
For the square hole: $A_1 = L^2$ and $v_1 = \sqrt{2gy}$.
Thus,$Q_1 = L^2 \sqrt{2gy}$.
For the circular hole: $A_2 = \pi R^2$ and $v_2 = \sqrt{2g(4y)} = 2\sqrt{2gy}$.
Thus,$Q_2 = \pi R^2 (2\sqrt{2gy})$.
Given that $Q_1 = Q_2$,we have:
$L^2 \sqrt{2gy} = \pi R^2 (2\sqrt{2gy})$
$L^2 = 2\pi R^2$
$R^2 = \frac{L^2}{2\pi}$
$R = \frac{L}{\sqrt{2\pi}}$.

(B) Let the total length of the cylindrical vessel be $L$. Initially,the piston is at the midpoint,so the initial length of the gas column is $V_1 \propto L/2$ at $T_1 = 0^{\circ} C = 273 ~K$.
When heated to $T_2 = 100^{\circ} C = 373 ~K$,the piston moves by $5 ~cm$,so the new length of the gas column is $V_2 \propto (L/2 + 5)$.
Since the pressure remains constant (as the piston is weightless and moves freely),we apply Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Substituting the values: $\frac{L/2}{273} = \frac{L/2 + 5}{373}$.
Cross-multiplying: $373(L/2) = 273(L/2 + 5)$.
$373(L/2) = 273(L/2) + 1365$.
$100(L/2) = 1365$.
$L/2 = 13.65 ~cm$.
Therefore,$L = 27.3 ~cm$.

(B) The tension $T$ in the wire when two masses $m_1 = 1 \ kg$ and $m_2 = 2 \ kg$ are connected over a pulley is given by $T = \frac{2 m_1 m_2 g}{m_1 + m_2}$.
Substituting the values: $T = \frac{2 \times 1 \times 2 \times 10}{1 + 2} = \frac{40}{3} \ N$.
Breaking stress is defined as $\sigma = \frac{T}{A}$,where $A = \pi r^2$ is the cross-sectional area.
Given $\sigma = \frac{40}{3 \pi} \times 10^6 \ N m^{-2}$.
Equating the two: $\frac{40}{3 \pi} \times 10^6 = \frac{40/3}{\pi r^2}$.
Simplifying: $10^6 = \frac{1}{r^2} \Rightarrow r^2 = 10^{-6} \ m^2$.
Therefore,$r = 10^{-3} \ m = 1 \ mm$.

(C) Let the police party catch the thief after time $t$.
Distance traveled by the police party in time $t$ is $d_p = v t$.
Distance traveled by the thief in time $t$ starting from rest with acceleration $a$ is $d_t = x + \frac{1}{2} a t^2$.
For the police to catch the thief,the distance covered by the police must be greater than or equal to the distance covered by the thief:
$v t \geq x + \frac{1}{2} a t^2$
Rearranging the terms,we get:
$\frac{1}{2} a t^2 - v t + x \leq 0$
For this quadratic inequality in $t$ to have real solutions,the discriminant $D$ must be greater than or equal to zero:
$D = (-v)^2 - 4(\frac{1}{2} a)(x) \geq 0$
$v^2 - 2 a x \geq 0$
$v^2 \geq 2 a x$

(A) The energy released per fission of one nucleus is $E_1 = 200 \text{ MeV}$.
First,convert this energy into Joules $(J)$:
$E_1 = 200 \times 10^6 \text{ eV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.
We need to find the number of nuclei $N$ required to release a total energy $E_{\text{total}} = 1000 \text{ J}$.
The formula is $N = \frac{E_{\text{total}}}{E_1}$.
Substituting the values:
$N = \frac{1000}{3.2 \times 10^{-11}} = \frac{1000}{3.2} \times 10^{11} = 312.5 \times 10^{11} = 3.125 \times 10^{13}$.
Therefore,the number of nuclei required is $3.125 \times 10^{13}$.

(A) Energy released in the fission of one nucleus is $E_1 = 200 \text{ MeV}$.
Converting this energy into Joules:
$E_1 = 200 \times 1.6 \times 10^{-13} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.
We need to find the number of nuclei $n$ required to release a total energy $E_{total} = 1000 \text{ J}$.
The relationship is $E_{total} = n \times E_1$.
Therefore,$n = \frac{E_{total}}{E_1} = \frac{1000}{3.2 \times 10^{-11}}$.
$n = \frac{1000}{3.2} \times 10^{11} = 312.5 \times 10^{11} = 3.125 \times 10^{13}$ nuclei.

(C) For hydrogen electrode: $E = E^0 - 0.059 \ pH = 0 - 0.059 \times 10 = -0.591 \ V$. Thus,$(A)$ matches with $(III)$.
$(B)$ The standard reduction potential of $Cu^{2+} | Cu$ is $E^0_{Cu^{2+}/Cu} = 0.337 \ V$. Thus,$(B)$ matches with $(IV)$.
$(C)$ The standard oxidation potential of $Zn | Zn^{2+}$ is $E^0_{Zn/Zn^{2+}} = 0.76 \ V$. Thus,$(C)$ matches with $(I)$.
$(D)$ The value of $\frac{2.303 RT}{F}$ at $298 \ K$ is $\frac{2.303 \times 8.314 \times 298}{96500} \approx 0.059$. Thus,$(D)$ matches with $(II)$.
Therefore,the correct matching is $A-III, B-IV, C-I, D-II$,which corresponds to option $(c)$.

(A) When moist chlorine gas reacts with hypo $(Na_2S_2O_3)$,it acts as an oxidizing agent. The reaction produces sodium sulphate,hydrochloric acid,and colloidal sulphur.
The balanced chemical equation is:
$Na_2S_2O_3 + H_2O + Cl_2 \rightarrow Na_2SO_4 + 2HCl + S$

(D) The reduction of nitrobenzene with zinc in an alkaline medium (e.g.,$Zn/NaOH$) produces hydrazobenzene $(C_{12}H_{12}N_2)$ as the final product $X$.
Structure of hydrazobenzene: $Ph-NH-NH-Ph$.
Each phenyl ring $(C_6H_5)$ contains $6 \pi$ bonds (from the aromatic ring) and $12 \sigma$ bonds ($6$ $C-C$ and $6$ $C-H$).
For two phenyl rings,we have $12 \pi$ bonds and $24 \sigma$ bonds.
Additionally,the central $N-N$ linkage has $1 \sigma$ bond,and the two $N-H$ bonds contribute $2 \sigma$ bonds.
Total $\sigma$ bonds = $24 + 1 + 2 = 27$.
Total $\pi$ bonds = $3 + 3 = 6$ (from the two benzene rings).
Thus,$X$ contains $27 \sigma$ and $6 \pi$ bonds.

(B) At the isoelectric point,the net charge on an amino acid is zero,which results in minimum intermolecular electrostatic repulsion and minimum interaction with water molecules. Consequently,amino acids have the least solubility in water at their isoelectric point. Therefore,the statement that they have maximum solubility at their isoelectric point is incorrect.

(C) The structure of pyrophosphoric acid $(H_4P_2O_7)$ consists of two $P=O$ bonds,four $P-OH$ bonds,and one $P-O-P$ bridge.
Counting the bonds:
- Each $P=O$ bond contains $1$ $\sigma$ and $1$ $\pi$-bond (Total: $2$ $\sigma$,$2$ $\pi$).
- Each $P-OH$ bond contains $1$ $\sigma$-bond (Total: $4$ $\sigma$).
- The $P-O-P$ bridge contains $2$ $\sigma$-bonds.
- Each $O-H$ bond contains $1$ $\sigma$-bond (Total: $4$ $\sigma$).
Total $\sigma$-bonds = $2 + 4 + 2 + 4 = 12$.
Total $\pi$-bonds = $2$.
Therefore,the correct answer is $12, 2$.

(D) An artificial sweetening agent is a food additive that provides a sweet taste like that of sugar but with significantly less food energy.
Among the given options,$Alitame$ is an artificial sweetening agent.
It is approximately $2000$ times sweeter than cane sugar.
Its chemical structure is represented as follows:
$HOOC-CH_2-CH(NH_2)-CONH-CH(CH_3)-CONH-C(CH_3)_2-CH_2-S-C(CH_3)_2$ (or as shown in the provided image).

(C) For hydrogen electrode: $E = E^0 - 0.0591 \ pH = 0 - 0.0591 \times 10 = -0.591 \ V$. Thus,$A-III$.
$(B)$ Standard reduction potential of $Cu^{2+}|Cu$ is $0.337 \ V$. Thus,$B-IV$.
$(C)$ Standard oxidation potential of $Zn|Zn^{2+}$ is $0.76 \ V$. Thus,$C-I$.
$(D)$ The value of $\frac{2.303RT}{F}$ at $298 \ K$ is $0.059$. Thus,$D-II$.
The correct matching is $A-III, B-IV, C-I, D-II$,which corresponds to option $(c)$.

(A) According to Kohlrausch's law of independent migration of ions:
$\Lambda_{\infty}(NH_4Cl) = \Lambda_{\infty}(NH_4^+) + \Lambda_{\infty}(Cl^-) = 130 \ ohm^{-1} \ cm^2 \ equiv^{-1}$ $(i)$
$\Lambda_{\infty}(NaOH) = \Lambda_{\infty}(Na^+) + \Lambda_{\infty}(OH^-) = 217 \ ohm^{-1} \ cm^2 \ equiv^{-1}$ $(ii)$
$\Lambda_{\infty}(NaCl) = \Lambda_{\infty}(Na^+) + \Lambda_{\infty}(Cl^-) = 109 \ ohm^{-1} \ cm^2 \ equiv^{-1}$ $(iii)$
To find $\Lambda_{\infty}(NH_4OH)$,we perform the operation: $(i) + (ii) - (iii)$
$\Lambda_{\infty}(NH_4OH) = \Lambda_{\infty}(NH_4^+) + \Lambda_{\infty}(OH^-) = \Lambda_{\infty}(NH_4Cl) + \Lambda_{\infty}(NaOH) - \Lambda_{\infty}(NaCl)$
$\Lambda_{\infty}(NH_4OH) = 130 + 217 - 109 = 238 \ ohm^{-1} \ cm^2 \ equiv^{-1}$

(C) Nitration is an electrophilic aromatic substitution reaction. The rate of this reaction depends on the electron density of the aromatic ring.
Groups that increase electron density (electron-donating groups) activate the ring,while groups that decrease electron density (electron-withdrawing groups) deactivate the ring.
In phenol $(I)$,the $-OH$ group exerts a strong $+R$ (resonance) effect,which significantly increases the electron density of the ring,making it highly reactive.
In benzene $(III)$,there is no substituent to alter the electron density.
In nitrobenzene $(II)$,the $-NO_2$ group exerts a strong $-R$ effect,which significantly decreases the electron density of the ring,making it the least reactive.
Therefore,the order of reactivity is $(I) > (III) > (II)$.

(A) Nickel oxide $(NiO)$ is reduced by water gas $(CO + H_2)$ to obtain the metal. The other oxides are reduced by different reducing agents as shown below:
$ZnO + C \xrightarrow{\Delta} Zn + CO$
$WO_3 + 3H_2 \xrightarrow{\Delta} W + 3H_2O$
$Fe_2O_3 + 3CO \xrightarrow{\Delta} 2Fe + 3CO_2$
$2NiO + \underbrace{CO + H_2}_{\text{Water gas}} \longrightarrow 2Ni + CO_2 + H_2O$

(A) In the presence of air and light,chloroform $(CHCl_3)$ undergoes oxidation to form carbonyl chloride,commonly known as phosgene $(COCl_2)$,which is a highly toxic gas.
The chemical reaction is:
$CHCl_3 + \frac{1}{2}O_2 \xrightarrow{\text{Air and light}} COCl_2 + HCl$

(A) The Wolff-Kishner reduction is a chemical reaction used to convert carbonyl groups (aldehydes or ketones) into methylene groups $(-CH_2-)$.
It involves heating the carbonyl compound with hydrazine $(NH_2NH_2)$ and a strong base like potassium hydroxide $(KOH)$ in a high-boiling solvent such as ethylene glycol.
The reaction proceeds via the formation of a hydrazone intermediate,which then undergoes base-catalyzed decomposition to release nitrogen gas $(N_2)$ and form the corresponding alkane.

(C) The acidity of a substance is inversely proportional to its $pK_a$ value.
Mathematically,$pK_a = -\log(K_a)$.
$A$ stronger acid has a higher $K_a$ value,which corresponds to a smaller $pK_a$ value.
Comparing the given values: $0.23 < 3.41 < 4.19 < 4.76$.
Therefore,the smallest $pK_a$ value of $0.23$ corresponds to the strongest carboxylic acid.

(B) Polymers which control various life processes in plants and animals are called biopolymers.
Cellulose,Insulin,$DNA$,starch,and proteins are examples of biopolymers.
Nylon-$6$ is a synthetic polymer,not a biopolymer.

(A) Hypo is sodium thiosulfate,$Na_2S_2O_3 \cdot 5H_2O$. When moist chlorine gas reacts with sodium thiosulfate,it acts as an oxidizing agent. The reaction is:
$Na_2S_2O_3 + 5H_2O + 4Cl_2 \rightarrow 2NaHSO_4 + 8HCl$.
However,in the context of standard chemistry problems regarding the reaction of hypo with chlorine,it is often represented as:
$Na_2S_2O_3 + H_2O + Cl_2 \rightarrow Na_2SO_4 + S + 2HCl$.
Thus,the products formed are $Na_2SO_4, S, \text{and } HCl$.

(B) Potassium crystallizes in a $BCC$ system.
Number of moles of potassium $= \frac{39 \ g}{39 \ g/mol} = 1 \ mol$.
$1 \ mol$ of atoms contains $N$ atoms.
In a $BCC$ unit cell,the number of atoms per unit cell is $2$.
Therefore,the number of unit cells $= \frac{\text{Total number of atoms}}{\text{Atoms per unit cell}} = \frac{N}{2}$.

(A) The lowering of vapour pressure is a colligative property,which is directly proportional to the van't Hoff factor $(i)$ for the same molar concentration of solute.
$BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$; $i = 3$
$NaCl \rightarrow Na^+ + Cl^-$; $i = 2$
$Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$; $i = 5$
Since the concentration is the same $(0.1 \ M)$ for all,the ratio of the lowering of vapour pressure is equal to the ratio of their van't Hoff factors.
Therefore,the ratio is $3 : 2 : 5$.

(D) The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For the given ions:
$(I) Fe^{2+} ([Ar] 3d^6)$: Number of unpaired electrons $(n)$ = $4$.
$(II) Ti^{2+} ([Ar] 3d^2)$: Number of unpaired electrons $(n)$ = $2$.
$(III) Cu^{2+} ([Ar] 3d^9)$: Number of unpaired electrons $(n)$ = $1$.
$(IV) V^{2+} ([Ar] 3d^3)$: Number of unpaired electrons $(n)$ = $3$.
Comparing the number of unpaired electrons: $1 (III) < 2 (II) < 3 (IV) < 4 (I)$.
Therefore,the increasing order of magnetic moment is $III < II < IV < I$.

(A) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = K p^{1/n}$.
Taking the logarithm on both sides,we get: $\log \frac{x}{m} = \log K + \frac{1}{n} \log p$.
This equation is in the form of a straight line $y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log p$,slope $m = \frac{1}{n}$,and intercept $c = \log K$.
Therefore,plotting a graph between $\log \frac{x}{m}$ and $\log p$ yields a straight line.

(D) The activation energy for the forward reaction $(E_f)$ and the activation energy for the backward reaction $(E_b)$ are related to the enthalpy of the reaction $(\Delta_r H)$ by the equation: $\Delta_r H = E_f - E_b$.
For an exothermic reaction,the enthalpy change is negative,meaning $\Delta_r H < 0$.
Substituting this into the equation,we get $E_f - E_b < 0$,which implies $E_f < E_b$.