TS EAMCET 2010 Chemistry Question Paper with Answer and Solution

(C) The conversion of $O$-acylated phenol to $C$-acylated phenol in the presence of a Lewis acid like $AlCl_3$ is known as the Fries rearrangement.
In this reaction,the acyl group $(RCO-)$ migrates from the oxygen atom of the phenolic ester to the ortho or para position of the benzene ring.
Since the atoms or groups within the molecule rearrange to form a new isomer,this process is classified as a molecular rearrangement reaction.

(B) The reaction of diethyl ether with carbon monoxide in the presence of a Lewis acid catalyst like $BF_3$ at high pressure and temperature is a carbonylation reaction.
The reaction proceeds as follows:
$C_2H_5-O-C_2H_5 + CO \xrightarrow[500 \ atm]{BF_3 / 150^{\circ}C} C_2H_5COOC_2H_5$
The product $(A)$ formed is ethyl propionate.

(A) Alcohols containing the $CH_3CH(OH)-$ group and carbonyl compounds containing the $CH_3CO-$ group give a yellow precipitate with iodine and $NaOH$ solution. This reaction is known as the iodoform test.
Thus,$CH_3CHO$,due to the presence of the $CH_3CO-$ group,gives a yellow precipitate with $I_2$ and $NaOH$.
The reaction is: $CH_3CHO + 3I_2 + 4NaOH \longrightarrow CHI_3 + HCOONa + 3NaI + 3H_2O$.

(A) The reduction of nitrobenzene $(C_6H_5NO_2)$ with $Zn$ dust and alcoholic $KOH$ is a classic reaction in organic chemistry.
This reaction proceeds through several intermediates (nitrosobenzene,azoxybenzene,azobenzene) and finally yields hydrazobenzene $(C_6H_5NH-NHC_6H_5)$ as the major product.

(B) According to Chargaff's rules,the base composition of $DNA$ varies between species.
In human beings,the ratio of $(A+T)/(G+C)$ is approximately $1.52$.

(B) The potential across a discharging capacitor is given by $V = V_0 e^{-t/RC}$.
According to the problem,the potential falls to half its original value,so $V = V_0/2$.
Substituting this into the equation: $V_0/2 = V_0 e^{-t/RC}$.
This simplifies to $1/2 = e^{-t/RC}$,or $e^{t/RC} = 2$.
Taking the natural logarithm on both sides: $t/RC = \ln(2)$.
We know that $\ln(2) = 2.3026 \times \log_{10}(2)$.
Given $\log_{10}(2) = 0.3010$,so $\ln(2) = 2.3026 \times 0.3010 \approx 0.693$.
Now,calculate the time constant $RC$:
$R = 10 M\Omega = 10 \times 10^6 \Omega = 10^7 \Omega$.
$C = 0.1 \mu F = 0.1 \times 10^{-6} F = 10^{-7} F$.
$RC = 10^7 \times 10^{-7} = 1 ~s$.
Therefore,$t = RC \times 0.693 = 1 \times 0.693 = 0.693 ~s$.

(B) When capacitors are connected in series,the charge on each capacitor is the same as the total charge on the combination.
Given,$q = 80 \mu C$ for both capacitors.
The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.
First,we find the value of $C$ using the equivalent capacitance formula: $C_{\text{eq}} = \frac{q}{V} = \frac{80 \mu C}{120 \ V} = \frac{2}{3} \mu F$.
For series connection,$\frac{1}{C_{\text{eq}}} = \frac{1}{1} + \frac{1}{C}$.
$\frac{3}{2} = 1 + \frac{1}{C} \implies \frac{1}{C} = \frac{1}{2} \implies C = 2 \mu F$.
Now,the energy stored in the capacitor $C$ is $U = \frac{q^2}{2C} = \frac{(80 \mu C)^2}{2 \times 2 \mu F} = \frac{6400 \times 10^{-12}}{4 \times 10^{-6}} \ J = 1600 \times 10^{-6} \ J = 1600 \ \mu J$.

(D) The electric field $E$ between two parallel plates is given by $E = \frac{V}{d}$.
Given,potential difference $V = 10^4 ~V$ and separation $d = 0.5 ~cm = 0.5 \times 10^{-2} ~m$.
The force $F$ on an electron of charge $e = 1.6 \times 10^{-19} ~C$ is $F = eE = \frac{eV}{d}$.
Substituting the values:
$F = \frac{1.6 \times 10^{-19} \times 10^4}{0.5 \times 10^{-2}}$
$F = \frac{1.6 \times 10^{-15}}{0.5 \times 10^{-2}}$
$F = 3.2 \times 10^{-13} ~N$.

(C) The velocity of the centre of mass $(v_{CM})$ is given by the formula: $v_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$.
Substituting the given values: $v_{CM} = \frac{4 \times 5 \hat{i} + 2 \times 10 \hat{i}}{4 + 2}$.
$v_{CM} = \frac{20 \hat{i} + 20 \hat{i}}{6} = \frac{40 \hat{i}}{6} = \frac{20}{3} \hat{i} \ m/s$.
The kinetic energy of the centre of mass is given by $K = \frac{1}{2} M v_{CM}^2$,where $M = m_1 + m_2 = 6 \ kg$.
$K = \frac{1}{2} \times 6 \times (\frac{20}{3})^2$.
$K = 3 \times \frac{400}{9} = \frac{400}{3} \ J$.

(C) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{V - M}{2}$,where $V$ is the number of valence electrons of the central atom and $M$ is the number of monovalent atoms bonded to it.
For $A$. $NH_3$: Central atom $N$ has $5$ valence electrons and $3$ $H$ atoms attached. $\text{Lone pairs} = \frac{5-3}{2} = 1$. Thus,$A-5$.
For $B$. $H_2O$: Central atom $O$ has $6$ valence electrons and $2$ $H$ atoms attached. $\text{Lone pairs} = \frac{6-2}{2} = 2$. Thus,$B-1$.
For $C$. $XeF_2$: Central atom $Xe$ has $8$ valence electrons and $2$ $F$ atoms attached. $\text{Lone pairs} = \frac{8-2}{2} = 3$. Thus,$C-2$.
For $D$. $CH_4$: Central atom $C$ has $4$ valence electrons and $4$ $H$ atoms attached. $\text{Lone pairs} = \frac{4-4}{2} = 0$. Thus,$D-3$.
Therefore,the correct match is $A-5, B-1, C-2, D-3$.

(D) For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$,the equilibrium constant is $K = \frac{[HI]^2}{[H_2][I_2]}$ ...$(i)$
For the reaction $HI_{(g)} \rightleftharpoons \frac{1}{2} H_{2(g)} + \frac{1}{2} I_{2(g)}$,the equilibrium constant is $K' = \frac{[H_2]^{1/2} [I_2]^{1/2}}{[HI]}$
Comparing this with equation $(i)$,we see that $K' = \sqrt{\frac{1}{K}} = \frac{1}{\sqrt{K}}$.

(B) The Arrhenius equation is given by:
$k = A e^{-E_a / RT}$
Taking the natural logarithm $(\ln)$ on both sides:
$\ln k = \ln A - \frac{E_a}{RT}$
Rearranging this into the linear equation form $y = mx + c$:
$\ln k = (-\frac{E_a}{R}) (\frac{1}{T}) + \ln A$
Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln k$,$x = \frac{1}{T}$,and $c = \ln A$,the slope $(m)$ is equal to $-\frac{E_a}{R}$.

(B) Among the given drugs,$ibuprofen$ is a non-narcotic (i.e.,not habit-forming) analgesic.
$Note$: $Diazepam$ is a hypnotic and sedative drug. $Formalin$ and $terpineol$ have antiseptic properties.

(C) Elements with half-filled or completely-filled orbitals possess stable electronic configurations and exhibit very low negative values of electron affinity.
Given the electron affinity values,element $B$ has the lowest magnitude $(-60 \ kJ \ mol^{-1})$,which indicates it possesses a stable half-filled $p$-orbital configuration.
Therefore,the outer electronic configuration of element $B$ is $3s^2 3p^3$.

(B) Given: Area $A = 0.3 ~m^2$,charge density $n = 2 \times 10^{25} / m^3$,and charge $q = 3t^2 + 5t + 2$.
First,find the current $i$ by differentiating $q$ with respect to time $t$:
$i = \frac{dq}{dt} = \frac{d}{dt}(3t^2 + 5t + 2) = 6t + 5$.
At $t = 2 ~s$,the current is $i = 6(2) + 5 = 17 ~A$.
The relationship between current and drift velocity $v_d$ is given by $i = neAv_d$.
Rearranging for $v_d$:
$v_d = \frac{i}{neA}$.
Substituting the values $(e = 1.6 \times 10^{-19} ~C)$:
$v_d = \frac{17}{2 \times 10^{25} \times 1.6 \times 10^{-19} \times 0.3}$.
$v_d = \frac{17}{0.96 \times 10^6} = 17.708 \times 10^{-6} ~m/s = 1.77 \times 10^{-5} ~m/s$.

(A) First,calculate the equivalent resistance of the parallel combination of $6 \Omega$ and $12 \Omega$ resistors:
$R_p = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4 \Omega$
Next,calculate the total resistance of the circuit,as the parallel combination is in series with the $6 \Omega$ resistor:
$R_{eq} = R_p + 6 \Omega = 4 \Omega + 6 \Omega = 10 \Omega$
Now,calculate the total current flowing through the circuit using Ohm's law $(I = V / R)$:
$I = \frac{10 \text{ V}}{10 \Omega} = 1 \text{ A}$
The potential difference across the parallel combination $(V_p)$ is the current multiplied by the equivalent parallel resistance:
$V_p = I \times R_p = 1 \text{ A} \times 4 \Omega = 4 \text{ V}$
Since resistors in parallel have the same potential difference across them,the potential difference across the $12 \Omega$ resistor is $4 \text{ V}$.

(A) . $Cu^{+} = [Ar] 3d^{10}$ (No unpaired electrons,so it is colourless). $Zn^{2+} = [Ar] 3d^{10}$ (No unpaired electrons,so it is colourless).
$B$. $Cu^{2+} = [Ar] 3d^9$ (One unpaired electron is present,so it is coloured). $Zn^{2+}$ is colourless.
$C$. The colour of $Fe^{2+}$ is green while that of $Fe^{3+}$ is yellow/brown.
$D$. $MnO_4^{-}$ contains no unpaired electrons,however,it is coloured due to charge transfer.
Thus,only the statement given in option $A$ is correct.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = h v - h v_0$,where $v$ is the frequency of incident light and $v_0$ is the threshold frequency.
For frequency $v_1$,$K_1 = h(v_1 - v_0)$.
For frequency $v_2$,$K_2 = h(v_2 - v_0)$.
Given the ratio of kinetic energies is $\frac{K_1}{K_2} = \frac{1}{n}$.
Substituting the expressions,we get $\frac{h(v_1 - v_0)}{h(v_2 - v_0)} = \frac{1}{n}$.
$\frac{v_1 - v_0}{v_2 - v_0} = \frac{1}{n}$.
Cross-multiplying gives $n(v_1 - v_0) = v_2 - v_0$.
$n v_1 - n v_0 = v_2 - v_0$.
$n v_1 - v_2 = n v_0 - v_0$.
$n v_1 - v_2 = v_0(n - 1)$.
Therefore,$v_0 = \frac{n v_1 - v_2}{n - 1}$.

(C) According to Kohlrausch's law of independent migration of ions,the equivalent conductance at infinite dilution for benzoic acid $(C_6H_5COOH)$ can be calculated as follows:
$\Lambda^{\infty}_{C_6H_5COOH} = \Lambda^{\infty}_{C_6H_5COONa} + \Lambda^{\infty}_{HCl} - \Lambda^{\infty}_{NaCl}$
Given values:
$\Lambda^{\infty}_{C_6H_5COONa} = 240 \ \Omega^{-1} \ cm^2 \ \text{equiv}^{-1}$
$\Lambda^{\infty}_{HCl} = 349 \ \Omega^{-1} \ cm^2 \ \text{equiv}^{-1}$
$\Lambda^{\infty}_{NaCl} = 229 \ \Omega^{-1} \ cm^2 \ \text{equiv}^{-1}$
Substituting the values:
$\Lambda^{\infty}_{C_6H_5COOH} = 240 + 349 - 229 = 360 \ \Omega^{-1} \ cm^2 \ \text{equiv}^{-1}$

(C) Equivalent conductivity,$\wedge_{eq} = \frac{\kappa \times 1000}{C}$,where $\kappa$ is the specific conductance and $C$ is the concentration in $g \ \text{equiv}/L$.
Since specific conductance $\kappa = \frac{1}{R}$,where $R$ is the specific resistance.
Substituting the value of $\kappa$ in the formula:
$\wedge_{eq} = \frac{1}{R} \times \frac{1000}{C} = \frac{1000}{R C}$.

(A) The time constant of an $LR$ circuit is given by $\tau = \frac{L}{R}$.
Initially,$\tau_1 = \frac{L}{R} = 3 \times 10^{-3} \text{ s}$ ...$(i)$
When a $90 \Omega$ resistor is added in series,the new resistance is $(R + 90) \Omega$.
The new time constant is $\tau_2 = \frac{L}{R + 90} = 0.5 \times 10^{-3} \text{ s}$ ...(ii)
Dividing equation $(i)$ by (ii):
$\frac{3 \times 10^{-3}}{0.5 \times 10^{-3}} = \frac{L/R}{L/(R + 90)}$
$6 = \frac{R + 90}{R}$
$6R = R + 90$
$5R = 90 \implies R = 18 \Omega$
Substituting $R = 18 \Omega$ into equation $(i)$:
$L = 3 \times 10^{-3} \times 18 = 54 \times 10^{-3} \text{ H} = 54 \text{ mH}$.
Thus,the inductance is $54 \text{ mH}$ and the resistance is $18 \Omega$.

(D) The thermo emf is given by $E = aT + bT^2$.
At the neutral temperature $(T_n)$,the emf is maximum,and at the inversion temperature $(T_i)$,the emf becomes zero.
Setting $E = 0$ for the inversion temperature:
$0 = aT_i + bT_i^2$
$T_i(a + bT_i) = 0$
Since $T_i \neq 0$,we have $a + bT_i = 0$,which gives $T_i = -\frac{a}{b}$.
Given $\frac{a}{b} = -200^{\circ}C$,so $T_i = -(-200^{\circ}C) = 200^{\circ}C$.
This $T_i$ is the temperature relative to the cold junction being at $0^{\circ}C$.
Since the cold junction is kept at $T_c = 30^{\circ}C$,the actual inversion temperature is $T_{inv} = T_i + T_c = 200^{\circ}C + 30^{\circ}C = 230^{\circ}C$.
Converting to Kelvin: $T(K) = 273 + 230 = 503 \ K$.
Wait,re-evaluating the standard relation: The inversion temperature $T_i$ is defined such that $T_i - T_c = T_c - T_n$,where $T_n$ is the neutral temperature. The equation $E = aT + bT^2$ implies $T_n = -a/2b = 100^{\circ}C$. If $T_c = 30^{\circ}C$,then $T_i = 2T_n - T_c = 2(100) - 30 = 170^{\circ}C$.
Converting to Kelvin: $T = 273 + 170 = 443 \ K$.

(A) The deflection $y$ of a charged particle moving with velocity $u$ in a uniform electric field $E$ over a distance $L$ is given by $y = \frac{1}{2} a t^2 = \frac{1}{2} (\frac{qE}{m}) (\frac{L}{u})^2 = \frac{qEL^2}{2mu^2}$.
Since momentum $p = mu$, we have $u = p/m$. Substituting this, $y = \frac{qEL^2}{2m(p/m)^2} = \frac{qEL^2m}{2p^2}$.
Given that $E, L,$ and $p$ are constant for all particles, we have $y \propto qm$.
For a proton $(p)$, deuteron $(d)$, and $\alpha$-particle $(\alpha)$:
Charge ratio: $q_p : q_d : q_\alpha = 1 : 1 : 2$.
Mass ratio: $m_p : m_d : m_\alpha = 1 : 2 : 4$.
Therefore, the ratio of deflections is $y_p : y_d : y_\alpha = (q_p m_p) : (q_d m_d) : (q_\alpha m_\alpha) = (1 \times 1) : (1 \times 2) : (2 \times 4) = 1 : 2 : 8$.

(A) The acceptable level of carbon monoxide gas $(CO)$ in the atmosphere is approximately $9 \ ppm$.

(B) The $-OH$ group is ortho/para directing and increases the electron density of the benzene ring,making it highly reactive towards electrophilic substitution.
Thus,ortho/para substitution in phenol by an electrophile is very facile.

(B) Enantiomers are stereoisomers that are non-superimposable mirror images of each other.
They possess identical physical properties such as boiling point,density,and bond lengths in an achiral environment.
However,they differ in their interaction with plane-polarized light,which is measured as specific rotation.
Therefore,the two enantiomers of secondary butyl chloride differ in specific rotation.

(A) Enantiomers are non-superimposable mirror images of each other.
For a molecule with multiple chiral centers,the enantiomer is formed by inverting the configuration at every chiral center.
Thus,the enantiomer of $(2R, 3R)$-butanediol is $(2S, 3S)$-butanediol.
Therefore,the pair $(2R, 3R)$ and $(2S, 3S)$ represents enantiomers.

(C) The minerals and their chemical formulas are as follows:

Mineral	Formula
Galena	$PbS$
Cerussite	$PbCO_3$
Cassiterite	$SnO_2$
Anglesite	$PbSO_4$

From the table,it is clear that $SnO_2$ (Cassiterite) is the mineral for tin $(Sn)$.

(C) Roasting is a metallurgical process in which sulphide ores are heated in the presence of excess air to convert them into their corresponding metal oxides.
In the given options,the reaction $2ZnS + 3O_2 \longrightarrow 2ZnO + 2SO_2$ represents the roasting of zinc blende $(ZnS)$.

(B) The initial energy of the satellite on the surface of the earth is $E_1 = -\frac{GMm}{R}$.
When the satellite is in a circular orbit of radius $r = 7R$,its total energy is $E_2 = -\frac{GMm}{2r} = -\frac{GMm}{2(7R)} = -\frac{GMm}{14R}$.
The minimum energy required to be spent by the launching vehicle is the difference between the final energy and the initial energy:
$\Delta E = E_2 - E_1$
$\Delta E = -\frac{GMm}{14R} - \left(-\frac{GMm}{R}\right)$
$\Delta E = -\frac{GMm}{14R} + \frac{14GMm}{14R}$
$\Delta E = \frac{13GMm}{14R}$.

(C) $1$. $CH_3 Cl + KCN \longrightarrow CH_3 CN (A) + KCl$
(Nucleophilic substitution of $Cl^{-}$ by $CN^{-}$ gives methyl cyanide or acetonitrile).
$2$. $CH_3 CN + 2H_2 O \xrightarrow{H_3 O^{\oplus}} CH_3 COOH (B) + NH_3$
(Acidic hydrolysis of nitrile gives ethanoic acid).
$3$. $CH_3 COOH + C_2 H_5 OH \xrightarrow{H^{+}, \Delta} CH_3 COOC_2 H_5 (C) + H_2 O$
(Esterification of ethanoic acid with ethanol gives ethyl ethanoate).
Thus,$A = CH_3 CN, B = CH_3 CO_2 H, C = CH_3 CO_2 C_2 H_5$.

(A) The Diels-Alder reaction requires a conjugated diene and a dienophile (an alkene or alkyne).
In option $A$,$1,4$-pentadiene is an isolated diene,not a conjugated diene.
Conjugated dienes have alternating single and double bonds (e.g.,$CH_2=CH-CH=CH_2$).
Since $1,4$-pentadiene lacks the necessary conjugated system,it cannot participate in the Diels-Alder reaction.

(C) As the size of the central atom (order of size $O < S < Se < Te$) increases,the $H-A$ (where $A$ is the central atom) bond length increases.
This leads to a decrease in the $H-A$ bond dissociation energy.
Consequently,$H_2Te$ releases a proton $(H^+)$ more readily than the others.
Therefore,$H_2Te$ is the most acidic among the given compounds.
$H_2Te + H_2O \longrightarrow H_3O^+ + HTe^-$

(D) When an acidified solution of $TiO_2$ is treated with $H_2O_2$,an intense yellow-orange colour is obtained due to the formation of pertitanic acid,$H_2TiO_4$.
$TiO_2 + H_2O_2 \xrightarrow{H_2SO_4} H_2TiO_4$
This reaction is used for the detection of both $Ti(IV)$ and $H_2O_2$.

(A) Given,$pH$ of $0.01 \ M \ CH_3COOH$ solution $= 5.0$.
Concentration,$C = 0.01 \ M = 10^{-2} \ M$.
We know that $[H^{+}] = 10^{-pH} = 10^{-5} \ M$.
For a weak acid,the dissociation constant $K_a$ is given by the formula:
$[H^{+}] = \sqrt{K_a \cdot C} \implies [H^{+}]^2 = K_a \cdot C$.
Therefore,$K_a = \frac{[H^{+}]^2}{C} = \frac{(10^{-5})^2}{10^{-2}} = \frac{10^{-10}}{10^{-2}} = 10^{-8}$.
Thus,the values are $[H^{+}] = 1 \times 10^{-5} \ M$ and $K_a = 1 \times 10^{-8}$.

(A) For a smooth inclined plane,the acceleration is $a_s = g \sin \theta$. The time taken to cover distance $s$ is $t_s = \sqrt{\frac{2s}{g \sin \theta}}$.
For a rough inclined plane,the acceleration is $a_r = g(\sin \theta - \mu \cos \theta)$. The time taken is $t_r = \sqrt{\frac{2s}{g(\sin \theta - \mu \cos \theta)}}$.
Given $t_r = n t_s$,we have $\frac{t_r}{t_s} = n$.
Squaring both sides: $\frac{t_r^2}{t_s^2} = n^2 \implies \frac{\sin \theta}{\sin \theta - \mu \cos \theta} = n^2$.
Rearranging: $\sin \theta = n^2 \sin \theta - n^2 \mu \cos \theta$.
$n^2 \mu \cos \theta = (n^2 - 1) \sin \theta$.
$\mu = \frac{n^2 - 1}{n^2} \tan \theta$.
Since $\theta = 45^{\circ}$,$\tan 45^{\circ} = 1$.
Therefore,$\mu = 1 - \frac{1}{n^2}$.

(C) The kinetic energy $K$ is given by $K = \frac{1}{2} m v^2$.
Given $K = 45.5 \text{ eV} = 45.5 \times 1.6 \times 10^{-19} \text{ J}$.
$v^2 = \frac{2K}{m} = \frac{2 \times 45.5 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$.
$v^2 = \frac{145.6 \times 10^{-19}}{9.1 \times 10^{-31}} = 16 \times 10^{12} \text{ m}^2 \text{s}^{-2}$.
$v = 4 \times 10^6 \text{ m s}^{-1}$.
For an undeflected beam in crossed fields,the electric force equals the magnetic force: $qE = qvB$,which implies $v = \frac{E}{B}$.
$B = \frac{E}{v} = \frac{1 \times 10^3}{4 \times 10^6} = 0.25 \times 10^{-3} = 2.5 \times 10^{-4} \text{ Wb m}^{-2}$.

(B) The frequency of vibration in a vibration magnetometer is given by $n = \frac{1}{2\pi} \sqrt{\frac{MB}{I}}$.
For the combination of two magnets,the effective magnetic moment $M_{eff}$ and moment of inertia $I_{eff} = I_1 + I_2$ are used.
When like poles are tied together,$M_{eff} = M_1 + M_2$,so $n_1 = \frac{1}{2\pi} \sqrt{\frac{(M_1+M_2)B}{I_1+I_2}} = 6 ~Hz$.
When unlike poles are tied together,$M_{eff} = M_1 - M_2$,so $n_2 = \frac{1}{2\pi} \sqrt{\frac{(M_1-M_2)B}{I_1+I_2}} = 2 ~Hz$.
Taking the ratio of the two frequencies:
$\frac{n_1}{n_2} = \sqrt{\frac{M_1+M_2}{M_1-M_2}} = \frac{6}{2} = 3$.
Squaring both sides:
$\frac{M_1+M_2}{M_1-M_2} = 9$.
$M_1 + M_2 = 9M_1 - 9M_2$.
$10M_2 = 8M_1$.
$\frac{M_1}{M_2} = \frac{10}{8} = \frac{5}{4}$.
Thus,the ratio $M_1: M_2$ is $5: 4$.

(A) The formula for the magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 I}{2R}$.
Given values are:
Permeability of free space $\mu_0 = 4 \pi \times 10^{-7} ~T \cdot m/A$
Current $I = 0.9 ~A$
Radius $R = 5 ~cm = 5 \times 10^{-2} ~m$
Substituting these values into the formula:
$B = \frac{4 \pi \times 10^{-7} \times 0.9}{2 \times 5 \times 10^{-2}}$
$B = \frac{3.6 \pi \times 10^{-7}}{10 \times 10^{-2}}$
$B = \frac{3.6 \pi \times 10^{-7}}{10^{-1}}$
$B = 3.6 \pi \times 10^{-6} ~T = 36 \pi \times 10^{-7} ~T$.

(D) When a magnetic needle is placed in two mutually perpendicular magnetic fields $B_1$ and $B_2$,it aligns itself along the resultant magnetic field direction.
Let $B_1 = 1 ~T$ and $B_2 = \sqrt{3} ~T$.
The deflection angle $\theta$ of the needle from the direction of $B_1$ is given by the tangent law:
$\tan \theta = \frac{B_2}{B_1}$
Substituting the given values:
$\tan \theta = \frac{\sqrt{3}}{1} = \sqrt{3}$
Since $\tan 60^{\circ} = \sqrt{3}$,we have:
$\theta = 60^{\circ}$

(B) Let the magnitudes of the two vectors be $|\overrightarrow{A}| = |\overrightarrow{B}| = a$.
The resultant vector $\overrightarrow{R} = \overrightarrow{A} + \overrightarrow{B}$ has a magnitude $R = \sqrt{a^2 + a^2 + 2a^2 \cos \theta} = \sqrt{2a^2(1 + \cos \theta)} = \sqrt{2a^2(2 \cos^2 \frac{\theta}{2})} = 2a \cos \frac{\theta}{2}$.
The angle $\alpha$ that the resultant $\overrightarrow{R}$ makes with vector $\overrightarrow{A}$ is given by $\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta}$.
Substituting $A = B = a$,we get $\tan \alpha = \frac{a \sin \theta}{a + a \cos \theta} = \frac{\sin \theta}{1 + \cos \theta} = \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}} = \tan \frac{\theta}{2}$.
Therefore,$\alpha = \frac{\theta}{2}$.
Since the magnitudes are equal,the resultant bisects the angle between the two vectors.

(B) Given the quadratic equation $x^2+(\sqrt{3}+2) x+(\sqrt{3}-1)=0$.
By Vieta's formulas,the sum of roots $m_1+m_2 = -(\sqrt{3}+2)$ and the product of roots $m_1 m_2 = \sqrt{3}-1$.
The difference of the roots is $|m_1-m_2| = \sqrt{(m_1+m_2)^2 - 4m_1 m_2} = \sqrt{(\sqrt{3}+2)^2 - 4(\sqrt{3}-1)} = \sqrt{3+4+4\sqrt{3}-4\sqrt{3}+4} = \sqrt{11}$.
The vertices of the triangle formed by $y=m_1 x, y=m_2 x$ and $y=c$ are $(0,0), (c/m_1, c)$ and $(c/m_2, c)$.
The area of the triangle is $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| = \frac{1}{2} |0(c-c) + \frac{c}{m_1}(c-0) + \frac{c}{m_2}(0-c)| = \frac{1}{2} c^2 |\frac{1}{m_1} - \frac{1}{m_2}| = \frac{1}{2} c^2 \frac{|m_2-m_1|}{|m_1 m_2|}$.
Substituting the values,Area $= \frac{1}{2} c^2 \frac{\sqrt{11}}{\sqrt{3}-1} = \frac{1}{2} c^2 \frac{\sqrt{11}(\sqrt{3}+1)}{3-1} = \frac{\sqrt{33}+\sqrt{11}}{4} c^2$.

(C) Let $z = \sqrt{3}+i$.
Converting to polar form: $r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$.
$\theta = \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.
So,$z = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$.
Then $\bar{z} = 2(\cos \frac{\pi}{6} - i \sin \frac{\pi}{6})$.
Using De Moivre's theorem:
$z^7 + \bar{z}^7 = 2^7(\cos \frac{7\pi}{6} + i \sin \frac{7\pi}{6}) + 2^7(\cos \frac{7\pi}{6} - i \sin \frac{7\pi}{6})$.
$= 2^7 \times 2 \cos \frac{7\pi}{6} = 2^8 \cos(\pi + \frac{\pi}{6})$.
$= -2^8 \cos \frac{\pi}{6} = -256 \times \frac{\sqrt{3}}{2} = -128 \sqrt{3}$.

(B) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
We need to evaluate the expression: $(x+1)(x+\omega)(x-\omega-1)$.
First,rewrite the third term: $(x - (\omega + 1))$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega + 1 = -\omega^2$.
So the expression becomes $(x+1)(x+\omega)(x - (-\omega^2)) = (x+1)(x+\omega)(x+\omega^2)$.
Now,multiply $(x+\omega)(x+\omega^2) = x^2 + \omega^2 x + \omega x + \omega^3 = x^2 + x(\omega + \omega^2) + 1$.
Since $\omega + \omega^2 = -1$ and $\omega^3 = 1$,this simplifies to $x^2 - x + 1$.
Finally,multiply by $(x+1)$: $(x+1)(x^2 - x + 1) = x^3 + 1$.

(C) The correct matches are as follows:
$A$. Hooke's law: Within the elastic limit,stress is directly proportional to strain. Thus,$A-3$.
$B$. Shearing strain: It is defined as the angle in radians through which a plane perpendicular to the fixed surface of a cubical body gets turned under the effect of a tangential force. It is also known as tangential strain. Thus,$B-1$.
$C$. Bulk strain: For a solid,the volumetric strain (bulk strain) is equal to $3$ times the linear strain. Thus,$C-4$.
$D$. Elastic fatigue: The temporary loss of elastic properties due to the action of repeated alternating deforming forces is called elastic fatigue. Thus,$D-2$.
Therefore,the correct matching is $A-3, B-1, C-4, D-2$,which corresponds to option $C$.

(B) The total time is $2 \ min \ 20 \ s = 2 \times 60 \ s + 20 \ s = 140 \ s$.
In $40 \ s$,the athlete completes $1$ round.
In $140 \ s$,the number of rounds completed is $\frac{140}{40} = 3.5$ rounds.
After $3$ complete rounds,the athlete returns to the starting point,so the displacement is $0$.
After the remaining $0.5$ round,the athlete reaches the diametrically opposite point of the starting position.
The displacement is equal to the diameter of the circular track,which is $2 R$.

(D) The total energy of a particle executing $SHM$ is given by $E = \frac{1}{2} m \omega^2 A^2$,where $m$ is the mass,$\omega$ is the angular frequency,and $A$ is the amplitude.
For the first particle,$x_1 = 4 \sin \left(10 t + \frac{\pi}{6}\right)$,the angular frequency $\omega_1 = 10 \text{ rad/s}$ and amplitude $A_1 = 4 \text{ units}$.
Thus,$E_1 = \frac{1}{2} m (10)^2 (4)^2 = \frac{1}{2} m (100)(16) = 800m$.
For the second particle,$x_2 = 5 \cos (\omega t)$,the angular frequency is $\omega$ and amplitude $A_2 = 5 \text{ units}$.
Thus,$E_2 = \frac{1}{2} m \omega^2 (5)^2 = \frac{25}{2} m \omega^2$.
Given that the energies are equal,$E_1 = E_2$.
$800m = \frac{25}{2} m \omega^2$.
$1600 = 25 \omega^2$.
$\omega^2 = \frac{1600}{25} = 64$.
$\omega = 8 \text{ units}$.

(B) Diborane reacts with ammonia to give different products under different reaction conditions:
$1$. At low temperature: $B_2H_6 + 2NH_3 \rightarrow B_2H_6 \cdot 2NH_3$ (ionic salt $[BH_2(NH_3)_2]^+[BH_4]^-$).
$2$. At high temperature: $3B_2H_6 + 6NH_3 \rightarrow 2B_3N_3H_6$ (Borazine,also known as inorganic benzene).
$3$. At very high temperature: $B_2H_6 + 2NH_3 \rightarrow 2(BN)_n + 6H_2$ (Boron nitride,which resembles graphite).
Therefore,$B_{12}H_{12}$ is not formed in the reaction between diborane and ammonia.

(B) The thermal decomposition of ammonium nitrate $(NH_4NO_3)$ at $250^{\circ}C$ yields nitrous oxide $(N_2O)$,also known as laughing gas,and water vapor.
The balanced chemical equation is:
$NH_4NO_3 \xrightarrow{\Delta, 250^{\circ}C} N_2O + 2H_2O$

(D) mixture of helium and oxygen is used in the treatment of asthma.
Because helium has a low density,this mixture flows easily through restricted respiratory passages,making it easier for patients to breathe.

(C) The conversion of $O$-acylated phenol to $C$-acylated phenol in the presence of a Lewis acid like $AlCl_3$ is known as the Fries rearrangement.
In this reaction,the acyl group $(RCO-)$ migrates from the oxygen atom of the phenolic ester to the ortho or para position of the benzene ring.
Since the atoms or groups within the same molecule rearrange to form a new isomer,this process is classified as a molecular rearrangement reaction.

(B) The reaction of diethyl ether with carbon monoxide in the presence of a Lewis acid catalyst like $BF_3$ at high pressure and temperature is a carbonylation reaction.
$C_2H_5-O-C_2H_5 + CO \xrightarrow[500 \text{ atm}]{BF_3 / 150^{\circ}C} C_2H_5COOC_2H_5$
The product $(A)$ formed is ethyl propionate.

(A) Alcohols containing the $CH_3CH(OH)-$ group and carbonyl compounds containing the $CH_3CO-$ group give a yellow precipitate with iodine and $NaOH$ solution. This reaction is known as the iodoform test.
Thus,$CH_3CHO$ gives a yellow precipitate with $I_2$ and $NaOH$ due to the presence of the $CH_3CO-$ group.
The reaction is: $CH_3CHO + 3I_2 + 4NaOH \longrightarrow CHI_3 + HCOONa + 3NaI + 3H_2O$.

(A) The reduction of nitrobenzene $(C_6H_5NO_2)$ with $Zn$ dust and alcoholic $KOH$ is a specific chemical reaction that proceeds through several intermediates to finally yield hydrazobenzene $(C_6H_5NH-NHC_6H_5)$.
This reaction is a classic example of the reduction of nitro compounds in a basic medium.

(B) The base composition of human $DNA$ varies,but the average ratio of $(A+T)/(G+C)$ in human beings is approximately $1.52$.

(B) The Arrhenius equation is given by:
$k = A e^{-E_a / RT}$
Taking the natural logarithm on both sides:
$\ln k = \ln A - \frac{E_a}{RT}$
Rearranging this into the linear equation form $y = mx + c$:
$\ln k = -\frac{E_a}{R} \left( \frac{1}{T} \right) + \ln A$
Comparing this with $y = mx + c$,where $y = \ln k$,$x = \frac{1}{T}$,and $c = \ln A$,the slope $m$ is equal to $-\frac{E_a}{R}$.

(B) Among the given drugs,$ibuprofen$ is a non-narcotic (i.e.,not habit-forming) analgesic.
Note: $Diazepam$ is a hypnotic and sedative drug.
$Formalin$ and $terpineol$ possess antiseptic properties.

(A) $Cu^{+} = [Ar] 3d^{10}$ (No unpaired electrons,so it is colourless).
$Zn^{2+} = [Ar] 3d^{10}$ (No unpaired electrons,so it is colourless).
Therefore,aqueous solutions of $Cu^{+}$ and $Zn^{2+}$ are colourless.
$Cu^{2+} = [Ar] 3d^9$ (One unpaired electron is present,so it is coloured).
$Fe^{3+}$ is yellow/brown in aqueous solution.
$MnO_4^{-}$ is purple due to charge transfer,despite having no unpaired electrons.
Thus,the statement in option $(A)$ is correct.

(C) According to Kohlrausch's law of independent migration of ions,the equivalent conductance at infinite dilution for benzoic acid $(C_6H_5COOH)$ can be calculated as follows:
$\wedge_{C_6H_5COOH}^{\infty} = \wedge_{C_6H_5COONa}^{\infty} + \wedge_{HCl}^{\infty} - \wedge_{NaCl}^{\infty}$
Given values:
$\wedge_{C_6H_5COONa}^{\infty} = 240 \ \Omega^{-1} \ cm^2 \ equiv^{-1}$
$\wedge_{HCl}^{\infty} = 349 \ \Omega^{-1} \ cm^2 \ equiv^{-1}$
$\wedge_{NaCl}^{\infty} = 229 \ \Omega^{-1} \ cm^2 \ equiv^{-1}$
Substituting the values:
$\wedge_{C_6H_5COOH}^{\infty} = 240 + 349 - 229$
$= 589 - 229$
$= 360 \ \Omega^{-1} \ cm^2 \ equiv^{-1}$

(C) Equivalent conductivity,$\wedge_{eq} = \frac{\kappa \times 1000}{C}$,where $\kappa$ is specific conductance.
Since specific conductance $\kappa = \frac{1}{R}$ (where $R$ is specific resistance,often denoted as resistivity $\rho$ in some contexts,but here $R$ represents the specific resistance value).
Substituting the value of $\kappa$ into the formula:
$\wedge_{eq} = \frac{1}{R} \times \frac{1000}{C} = \frac{1000}{R C}$.

(B) The $-OH$ group is an electron-donating group by resonance effect,which makes the benzene ring highly activated towards electrophilic substitution.
It is ortho/para directing in nature.
Therefore,ortho/para substitution in phenol by an electrophile is very facile compared to the other options,which contain electron-withdrawing groups.

(C) The minerals and their formulas are as follows:

Mineral	Formula
Galena	$PbS$
Cerussite	$PbCO_3$
Cassiterite	$SnO_2$
Anglesite	$PbSO_4$

From the table,it is clear that $SnO_2$ (Cassiterite) is the mineral for tin $(Sn)$.
Therefore,the correct option is $C$.

(C) Roasting is a metallurgical process in which sulphide ores are heated in the presence of excess air to convert them into their corresponding metal oxides.
In the given options,the reaction $2ZnS + 3O_2 \longrightarrow 2ZnO + 2SO_2$ represents the roasting of zinc sulphide $(ZnS)$ ore to form zinc oxide $(ZnO)$.

(C) The reaction sequence is as follows:
$1$. $CH_3 Cl + KCN \rightarrow CH_3 CN (A) + KCl$. Here,$A$ is methyl cyanide (acetonitrile).
$2$. $CH_3 CN + 2H_2 O \xrightarrow{H_3 O^{\oplus}} CH_3 COOH (B) + NH_3$. Here,$B$ is ethanoic acid (acetic acid).
$3$. $CH_3 COOH + C_2 H_5 OH \xrightarrow{H^{+}} CH_3 COOC_2 H_5 (C) + H_2 O$. This is an esterification reaction,where $C$ is ethyl ethanoate (ethyl acetate).
Thus,the correct sequence is $A = CH_3 CN, B = CH_3 CO_2 H, C = CH_3 CO_2 C_2 H_5$.

(C) As the size of the central atom increases (order of size $O < S < Se < Te$),the $H-A$ (where $A$ is the central atom) bond length increases.
Consequently,the $H-A$ bond dissociation energy decreases.
Therefore,$H_2Te$ releases a proton $(H^+)$ most readily,making it the most acidic among the given compounds.
The acidic strength order is $H_2O < H_2S < H_2Se < H_2Te$.

(B) The thermal decomposition of ammonium nitrate $(NH_4NO_3)$ at $250^{\circ}C$ yields nitrous oxide $(N_2O)$ and water vapor.
$NH_4NO_3 \xrightarrow{\Delta, 250^{\circ}C} N_2O + 2H_2O$
Thus,the correct oxide formed is $N_2O$.

(A) The polydispersity index $(PDI)$ is defined as the ratio of weight average molecular weight to number average molecular weight.
$PDI = \frac{\bar{M}_w}{\bar{M}_n}$
Given:
$\bar{M}_w = 60000$
$\bar{M}_n = 40000$
$PDI = \frac{60000}{40000} = 1.5$
Since $1.5 > 1$,the polydispersity index of the polymer is $>1$.

(B) When chlorine gas reacts with hot and concentrated sodium hydroxide $(NaOH)$,it undergoes a disproportionation reaction to form sodium chloride $(NaCl)$ and sodium chlorate $(NaClO_3)$.
The balanced chemical equation is:
$3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O$
Therefore,apart from sodium chloride,sodium chlorate $(NaClO_3)$ is formed.

(C) White tin,an allotropic form of tin,is an example of a tetragonal system.
For a tetragonal system,the unit cell parameters are defined as $a=b \neq c$ and $\alpha=\beta=\gamma=90^{\circ}$.
The given reasoning states $a=b=c$ and $\alpha=\beta=\gamma \neq 90^{\circ}$,which describes a rhombohedral system,not a tetragonal one.
Therefore,$Assertion (A)$ is true but $Reasoning (R)$ is false.

(D) Given,the radius ratio of anion to cation $= 10 : 9.3$.
$\therefore$ The radius ratio of cation to anion $= \frac{9.3}{10} = 0.93$.
When the radius ratio of cation to anion lies between $0.732$ and $1.00$,the coordination number is $8$.
Thus,the coordination number of the cation in the crystal is $8$.

(D) When a non-volatile solute is added to a volatile solvent,the solute particles occupy some of the surface area of the solvent.
This reduces the number of solvent molecules available at the surface for vaporization,leading to a decrease in vapour pressure.
As the concentration of the non-volatile solute increases,the vapour pressure of the solution decreases.
Given the concentrations of $X$ as $0.01 < 0.10 < 0.25$,the corresponding vapour pressures follow the order $p_3 > p_1 > p_2$.
Thus,the correct order is $p_2 < p_1 < p_3$.

(A) The dissociation reaction for $BaCl_2$ is: $BaCl_{2(aq)} \rightarrow Ba^{2+}_{(aq)} + 2Cl^-_{(aq)}$
Initially,we have $1$ mole of $BaCl_2$.
Given the degree of dissociation $\alpha = 80\% = 0.8$.
At equilibrium,the moles are:
$BaCl_2 = 1 - \alpha = 1 - 0.8 = 0.2$
$Ba^{2+} = \alpha = 0.8$
$Cl^- = 2\alpha = 2 \times 0.8 = 1.6$
Total moles at equilibrium = $0.2 + 0.8 + 1.6 = 2.6$
The van't Hoff factor $i$ is defined as the ratio of total moles after dissociation to the initial moles:
$i = \frac{2.6}{1} = 2.6$

(B) Soap molecules contain a hydrophobic (water-repelling) tail and a hydrophilic (water-attracting) head.
When soap is added to water,the hydrophobic tails dissolve in the grease or dirt particles,while the hydrophilic heads remain in the water.
This arrangement results in the formation of spherical clusters known as micelles.
Thus,a micelle is an aggregate of soap molecules surrounding a dirt or grease particle.