TS EAMCET 2007 Chemistry Question Paper with Answer and Solution

(B) The formula for the orthogonal projection of vector $\vec{b}$ on vector $\vec{a}$ is given by $\text{proj}_{\vec{a}} \vec{b} = \frac{(\vec{b} \cdot \vec{a})}{|\vec{a}|^2} \vec{a}$.
Given $\vec{a} = \hat{i} - \hat{j} - \hat{k}$,we have $|\vec{a}|^2 = (1)^2 + (-1)^2 + (-1)^2 = 1 + 1 + 1 = 3$.
Given $\vec{b} = \lambda\hat{i} - 3\hat{j} + \hat{k}$,the dot product is $\vec{b} \cdot \vec{a} = (\lambda)(1) + (-3)(-1) + (1)(-1) = \lambda + 3 - 1 = \lambda + 2$.
Substituting these into the projection formula:
$\frac{(\lambda + 2)}{3} (\hat{i} - \hat{j} - \hat{k}) = \frac{4}{3} (\hat{i} - \hat{j} - \hat{k})$.
Comparing the coefficients of the vectors,we get $\frac{\lambda + 2}{3} = \frac{4}{3}$.
$\lambda + 2 = 4$.
$\lambda = 2$.

(B) The formula for the orthogonal projection of vector $\vec{b}$ on vector $\vec{a}$ is given by $\frac{(\vec{b} \cdot \vec{a})\vec{a}}{|\vec{a}|^2}$.
Given that the projection is $\frac{4}{3}(\hat{i} - \hat{j} - \hat{k})$,we have:
$\frac{(\vec{b} \cdot \vec{a})\vec{a}}{|\vec{a}|^2} = \frac{4}{3}(\hat{i} - \hat{j} - \hat{k})$
First,calculate $|\vec{a}|^2 = (1)^2 + (-1)^2 + (-1)^2 = 1 + 1 + 1 = 3$.
Next,calculate the dot product $\vec{b} \cdot \vec{a} = (\lambda \hat{i} - 3\hat{j} + \hat{k}) \cdot (\hat{i} - \hat{j} - \hat{k}) = \lambda(1) + (-3)(-1) + (1)(-1) = \lambda + 3 - 1 = \lambda + 2$.
Substituting these into the projection formula:
$\frac{(\lambda + 2)(\hat{i} - \hat{j} - \hat{k})}{3} = \frac{4}{3}(\hat{i} - \hat{j} - \hat{k})$
Comparing the coefficients,we get $\frac{\lambda + 2}{3} = \frac{4}{3}$.
$\lambda + 2 = 4 \Rightarrow \lambda = 2$.

(D) Step $1$: Reduction of acetic acid $(CH_3COOH)$ with $LiAlH_4$ gives ethanol $(C_2H_5OH)$ as product '$A$'.
$CH_3COOH + 4[H] \xrightarrow{LiAlH_4} CH_3CH_2OH (A) + H_2O$
Step $2$: Ethanol $(A)$ reacts with acetic acid in the presence of acid catalyst $(H_3O^+)$ to form ethyl acetate $(CH_3COOC_2H_5)$ as product '$B$' via esterification reaction.
$CH_3CH_2OH (A) + CH_3COOH \xrightarrow{H_3O^+} CH_3COOC_2H_5 (B) + H_2O$
Thus,'$A$' is $C_2H_5OH$ and '$B$' is $CH_3COOC_2H_5$.

(A) $1$. Pyridinium chlorochromate $(PCC)$ is a mild oxidizing agent that oxidizes primary alcohols to aldehydes and secondary alcohols to ketones.
$2$. The reaction of compound '$Y$' with $I_2$ and alkali (iodoform test) indicates that '$Y$' must be a methyl ketone or acetaldehyde $(CH_3CHO)$.
$3$. If '$X$' is ethanol $(C_2H_5OH)$,its oxidation with $PCC$ yields acetaldehyde $(CH_3CHO)$,which is compound '$Y$'.
$4$. Acetaldehyde $(CH_3CHO)$ gives a positive iodoform test with $I_2$ and $NaOH$ to form triiodomethane $(CHI_3)$.
$5$. The reaction sequence is:
$C_2H_5OH + [O] \xrightarrow{PCC, CH_2Cl_2} CH_3CHO$
$CH_3CHO + 3I_2 + 4NaOH \rightarrow CHI_3 + HCOONa + 3NaI + 3H_2O$
Thus,'$X$' is $C_2H_5OH$.

(C) The natural frequency of an $L-C$ circuit is given by $f = \frac{1}{2 \pi \sqrt{LC}}$.
This implies $f \propto \frac{1}{\sqrt{C}}$.
When the capacitor $C$ is replaced by a capacitor with a dielectric medium of constant $K$,the new capacitance becomes $C' = KC$.
The new frequency $f'$ is $f - 25 \text{ kHz} = 125 \text{ kHz} - 25 \text{ kHz} = 100 \text{ kHz}$.
Using the ratio $\frac{f'}{f} = \sqrt{\frac{C}{C'}} = \sqrt{\frac{C}{KC}} = \frac{1}{\sqrt{K}}$.
Substituting the values: $\frac{100}{125} = \frac{1}{\sqrt{K}}$.
$\frac{4}{5} = \frac{1}{\sqrt{K}} \implies \sqrt{K} = \frac{5}{4} = 1.25$.
Therefore,$K = (1.25)^2 = 1.5625 \approx 1.56$.

(D) The reduction of nitrobenzene with $Zn$ dust and aqueous $NH_4Cl$ is a controlled reduction process.
This reaction yields $N$-phenyl hydroxylamine as the major product.
The chemical equation is:
$C_6H_5NO_2 + 4[H] \xrightarrow{Zn, NH_4Cl} C_6H_5NHOH + H_2O$

(D) burning candle emits a continuous spectrum because it contains a wide range of wavelengths.
Sodium vapour emits a characteristic line spectrum due to electronic transitions.
$A$ Bunsen flame often shows a band spectrum due to molecular emissions.
Dark lines in the solar spectrum (Fraunhofer lines) are caused by the absorption of specific wavelengths by gases in the solar atmosphere,forming an absorption spectrum.
Therefore,the correct matching is $1-(ii), 2-(i), 3-(iii), 4-(iv)$.

(C) Given,$m_1 = 6 \ kg, m_2 = 4 \ kg$.
$\overrightarrow{v}_1 = 5 \hat{i} - 2 \hat{j} + 10 \hat{k}, \quad \overrightarrow{v}_2 = 10 \hat{i} - 2 \hat{j} + 5 \hat{k}$.
The velocity of the centre of mass is given by the formula:
$\overrightarrow{v}_{cm} = \frac{m_1 \overrightarrow{v}_1 + m_2 \overrightarrow{v}_2}{m_1 + m_2}$.
Substituting the values:
$\overrightarrow{v}_{cm} = \frac{6(5 \hat{i} - 2 \hat{j} + 10 \hat{k}) + 4(10 \hat{i} - 2 \hat{j} + 5 \hat{k})}{6 + 4}$.
$\overrightarrow{v}_{cm} = \frac{(30 \hat{i} - 12 \hat{j} + 60 \hat{k}) + (40 \hat{i} - 8 \hat{j} + 20 \hat{k})}{10}$.
$\overrightarrow{v}_{cm} = \frac{70 \hat{i} - 20 \hat{j} + 80 \hat{k}}{10}$.
$\overrightarrow{v}_{cm} = 7 \hat{i} - 2 \hat{j} + 8 \hat{k}$.

(C) In the $CO_3^{2-}$ ion,the central $C$-atom undergoes $sp^2$-hybridisation.
It possesses a triangular planar structure.
Conversely,$BF_4^{-}$,$NH_4^{+}$,and $SO_4^{2-}$ all exhibit a tetrahedral geometry due to $sp^3$-hybridisation of the central atom.

(D) The fluorides of alkaline earth metals,except for $BeF_2$,are generally insoluble in water due to their high lattice energy compared to their hydration energy.
$BeF_2$ is soluble in water due to the high hydration energy of the small $Be^{2+}$ ion.
Among the given options,$MgF_2$ is insoluble in water.

(A) For the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$,the equilibrium constant is $K_1 = 5 \times 10^{-2}$.
For the reaction $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$,the reaction is the reverse of the first reaction multiplied by $2$.
Therefore,the new equilibrium constant $K_2$ is given by $K_2 = \frac{1}{K_1^2}$.
$K_2 = \frac{1}{(5 \times 10^{-2})^2} = \frac{1}{25 \times 10^{-4}} = \frac{10^4}{25} = 400 \ atm$.

(B) An element that forms a gaseous oxide which dissolves in water to form an acidic solution is typically a non-metal,such as carbon.
Carbon belongs to group $IV$ $(14)$ of the periodic table.
The reaction is as follows:
$C + O_2 \longrightarrow CO_2$
$CO_2 + H_2O \longrightarrow H_2CO_3$ (carbonic acid)
Since $H_2CO_3$ is an acid,the element belongs to group $IV$.

(C) The resistance $R$ of a wire is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.
Given: $R = 4 \Omega$,$\rho = 2.2 \times 10^{-8} \Omega \cdot m$,and diameter $d = 1.4 \text{ mm} = 1.4 \times 10^{-3} \text{ m}$.
The radius $r = \frac{d}{2} = 0.7 \times 10^{-3} \text{ m}$.
The cross-sectional area $A = \pi r^2 = \pi (0.7 \times 10^{-3})^2 \text{ m}^2$.
Rearranging the formula for length: $l = \frac{R A}{\rho}$.
Substituting the values: $l = \frac{4 \times \pi \times (0.7 \times 10^{-3})^2}{2.2 \times 10^{-8}}$.
$l = \frac{4 \times 3.14159 \times 0.49 \times 10^{-6}}{2.2 \times 10^{-8}} \approx \frac{6.1575 \times 10^{-6}}{2.2 \times 10^{-8}} \approx 2.7988 \times 10^2 \approx 280 \text{ m}$.

(C) In a metre bridge,the balance condition is given by the formula $\frac{P}{Q} = \frac{l}{100-l}$,where $P$ is the resistance in the left gap,$Q$ is the resistance in the right gap,and $l$ is the balance point distance from the left end in $cm$.
Given the ratio $\frac{P}{Q} = \frac{2}{3}$.
Substituting the values into the formula:
$\frac{2}{3} = \frac{l}{100-l}$
Cross-multiplying gives:
$2(100 - l) = 3l$
$200 - 2l = 3l$
$200 = 5l$
$l = \frac{200}{5} = 40 ~cm$.
Therefore,the balance point from the left is $40 ~cm$.

(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$Cr^{2+} (Z=24): [Ar] 3d^4$. Number of unpaired electrons $(n) = 4$.
$Fe^{2+} (Z=26): [Ar] 3d^6$. Number of unpaired electrons $(n) = 4$.
Since both $Cr^{2+}$ and $Fe^{2+}$ have $4$ unpaired electrons,they have the same calculated value of magnetic moment.

(A) In the Sun,the primary source of energy is the proton-proton cycle.
In this cycle,hydrogen nuclei (protons) undergo a series of nuclear fusion reactions to form a helium nucleus.
During this fusion process,a small amount of mass is converted into a large amount of energy,as described by Einstein's mass-energy equivalence principle,$E = \Delta mc^2$.

(C) According to Einstein's photoelectric equation: $e V_0 = h \nu - h \nu_0$
Where $V_0$ is the stopping potential,$\nu$ is the frequency of incident light,and $\nu_0$ is the threshold frequency.
Rearranging for $\nu_0$: $\nu_0 = \nu - \frac{e V_0}{h} = \frac{c}{\lambda} - \frac{e V_0}{h}$
Given: $\lambda = 2 \times 10^{-7} \ m$,$V_0 = 2.5 \ V$,$e = 1.6 \times 10^{-19} \ C$,$h = 6.6 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$.
Calculating frequency of incident light: $\nu = \frac{3 \times 10^8}{2 \times 10^{-7}} = 1.5 \times 10^{15} \ Hz$.
Calculating the work function term: $\frac{e V_0}{h} = \frac{1.6 \times 10^{-19} \times 2.5}{6.6 \times 10^{-34}} \approx 0.606 \times 10^{15} \ Hz$.
Therefore,$\nu_0 = 1.5 \times 10^{15} - 0.606 \times 10^{15} = 0.894 \times 10^{15} \ Hz \approx 9.0 \times 10^{14} \ Hz$.

(C) The stopping potential $V_0$ is given by the Einstein's photoelectric equation: $e V_0 = K_{\text{max}} = \frac{hc}{\lambda} - \phi$.
Given: $h = 6.67 \times 10^{-34} \text{ J-s}$, $c = 3 \times 10^8 \text{ m/s}$, $\lambda = 2000 \text{ \AA} = 2 \times 10^{-7} \text{ m}$, and $\phi = 5 \text{ eV} = 5 \times 1.6 \times 10^{-19} \text{ J} = 8 \times 10^{-19} \text{ J}$.
Calculating the energy of the incident photon: $E = \frac{hc}{\lambda} = \frac{6.67 \times 10^{-34} \times 3 \times 10^8}{2 \times 10^{-7}} = 10.005 \times 10^{-19} \text{ J}$.
Substituting into the equation: $e V_0 = 10.005 \times 10^{-19} \text{ J} - 8 \times 10^{-19} \text{ J} = 2.005 \times 10^{-19} \text{ J}$.
Therefore, $V_0 = \frac{2.005 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ C}} \approx 1.25 \text{ V}$.

(A) The standard $emf$ of the cell is calculated using the formula: $E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}$.
In the given cell representation $Cu_{(s)}|Cu^{2+}_{(aq)}||Ag^{+}_{(aq)}|Ag_{(s)}$,$Ag$ is the cathode and $Cu$ is the anode.
Substituting the given values: $E_{\text{cell}}^{\circ} = 0.80 \ V - 0.34 \ V$.
Therefore,$E_{\text{cell}}^{\circ} = +0.46 \ V$.

(A) Let the two unit negative charges be $q_1 = -1 \ C$ and $q_2 = -1 \ C$,separated by a distance $d$.
For the system to be in equilibrium,the net force on each charge must be zero.
Consider the equilibrium of charge $q_1$ at point $A$. The force exerted by $q$ at $B$ and $q_2$ at $C$ on $q_1$ must sum to zero:
$F_{AB} + F_{AC} = 0$
$\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q}{(d/2)^2} + \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{d^2} = 0$
Dividing by $\frac{q_1}{4 \pi \varepsilon_0}$,we get:
$\frac{q}{(d/2)^2} + \frac{q_2}{d^2} = 0$
$\frac{q}{d^2/4} + \frac{-1}{d^2} = 0$
$\frac{4q}{d^2} = \frac{1}{d^2}$
$4q = 1$
$q = 0.25 \ C$

(B) The charges are located at $x=0$ $(q/2)$,$x=a$ $(-q)$,and $x=2a$ $(q/2)$.
The point $P$ is at a distance $r$ from the charge $-q$ at $x=a$. Since $P$ is on the $x$-axis to the right of the charges,its coordinate is $x_P = a + r$.
The distances of point $P$ from the three charges are:
$1$. From $q/2$ at $x=0$: $d_1 = (a+r) - 0 = r+a$
$2$. From $-q$ at $x=a$: $d_2 = (a+r) - a = r$
$3$. From $q/2$ at $x=2a$: $d_3 = (a+r) - 2a = r-a$
The total electric potential $V$ at point $P$ is the sum of potentials due to individual charges:
$V = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q/2}{r+a} - \frac{q}{r} + \frac{q/2}{r-a} \right]$
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{1}{r+a} - \frac{2}{r} + \frac{1}{r-a} \right]$
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{r(r-a) - 2(r^2-a^2) + r(r+a)}{r(r^2-a^2)} \right]$
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{r^2 - ar - 2r^2 + 2a^2 + r^2 + ar}{r(r^2-a^2)} \right]$
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{2a^2}{r(r^2-a^2)} \right]$
Since $r \gg a$,we can approximate $r^2 - a^2 \approx r^2$:
$V \approx \frac{q}{8 \pi \varepsilon_0} \cdot \frac{2a^2}{r^3} = \frac{q a^2}{4 \pi \varepsilon_0 r^3}$

(B) The electric potential $V$ at a point due to a system of charges is the algebraic sum of the potentials due to individual charges.
The positions of the charges are $x_1 = 0$,$x_2 = a$,and $x_3 = 2a$. The point $P$ is at $x = a+r$.
The distances of point $P$ from the charges are:
$r_1 = (a+r) - 0 = r+a$
$r_2 = (a+r) - a = r$
$r_3 = (a+r) - 2a = r-a$
The total potential $V_P$ is:
$V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q/2}{r+a} - \frac{q}{r} + \frac{q/2}{r-a} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{1}{2(r+a)} - \frac{1}{r} + \frac{1}{2(r-a)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{r(r-a) - 2(r^2-a^2) + r(r+a)}{2r(r^2-a^2)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{r^2 - ar - 2r^2 + 2a^2 + r^2 + ar}{2r(r^2-a^2)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{2a^2}{2r(r^2-a^2)} \right] = \frac{q a^2}{4 \pi \varepsilon_0 r(r^2-a^2)}$
Since $a \ll r$,we have $r^2 - a^2 \approx r^2$.
Therefore,$V_P = \frac{q a^2}{4 \pi \varepsilon_0 r^3}$.

(D) The Bhopal gas tragedy of $1984$ was caused by the leakage of methyl isocyanate $(CH_3NCO)$ gas.

(A) $1$. Identify the longest carbon chain containing the principal functional group $(-OH)$ and the double bond $(C=C)$. The chain has $6$ carbon atoms,so the parent alkane is hexane.
$2$. Number the chain starting from the end that gives the lowest locant to the principal functional group $(-OH)$. Numbering from right to left gives the $-OH$ group at position $2$.
$3$. The double bond is at position $3$,and the methyl substituent is at position $5$.
$4$. Combining these,the $IUPAC$ name is $5-$methylhex$-3-$en$-2-$ol.

(D) Salol is phenyl salicylate. Its chemical structure consists of a benzene ring substituted with a hydroxyl group $(-OH)$ at the ortho position and an ester group ($-COOR$,specifically phenyl ester) at the adjacent position. Therefore,the functional groups present are $-OH$ and $-COOR$.

(A) Feldspar is a group of rock-forming tectosilicate minerals that make up about $41\%$ of the Earth's continental crust. The general chemical formula for potassium feldspar is $KAlSi_3O_8$.

(B) Step $1$: The reaction of ethene $(CH_2=CH_2)$ with $HCl$ in the presence of anhydrous $AlCl_3$ gives ethyl chloride $(CH_3CH_2Cl)$ as product $A$.
$CH_2=CH_2 + HCl \rightarrow CH_3CH_2Cl$ $(A)$
Step $2$: The reduction of ethyl chloride $(A)$ with $Zn-Cu$ couple in the presence of ethanol $(C_2H_5OH)$ is a dehalogenation/reduction reaction that yields ethane $(C_2H_6)$ as product $B$.
$CH_3CH_2Cl + 2[H] \xrightarrow{Zn-Cu, C_2H_5OH} CH_3CH_3$ $(B)$ $+ HCl$
Thus,$B$ is $C_2H_6$.

(C) $ZnH_2$ is an example of an interstitial (or metallic) hydride,whereas $NH_3$,$CH_4$,and $H_2O$ are examples of covalent (or molecular) hydrides.

(A) $I$. The reducing property of $VA$ group hydrides increases from $NH_3$ to $BiH_3$ due to the decrease in bond dissociation enthalpy $(NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3)$.
$II$. The tendency to donate a lone pair (basic strength) decreases from $NH_3$ to $BiH_3$ because the electron density on the central atom decreases as the size of the central atom increases.
$III$. Thermal stability of $VA$ group hydrides decreases from $NH_3$ to $BiH_3$ as the $M-H$ bond length increases and bond strength decreases $(NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3)$.
$IV$. Bond angle of $VA$ group hydrides decreases from $NH_3$ to $BiH_3$ due to the decrease in electronegativity of the central atom,which leads to more $p$-character in the bonding orbitals $(NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3)$.
All statements $I, II, III,$ and $IV$ are correct.

(A) The solubility of an ionic compound like $NaCl$ depends on the dielectric constant of the solvent. Higher dielectric constant leads to better solvation of ions,which increases solubility.
The dielectric constant of ordinary water $(H_2O)$ is approximately $81$,while that of heavy water $(D_2O)$ is approximately $80$.
Since the dielectric constant of ordinary water is higher,$NaCl$ is more soluble in ordinary water than in heavy water.
Therefore,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(D) conjugate acid-base pair differs by only one proton $(H^+)$.
$1.$ $HPO_3^{2-}$ and $PO_3^{3-}$ differ by one $H^+$,so they form a conjugate pair.
$2.$ $H_2PO_4^{-}$ and $HPO_4^{2-}$ differ by one $H^+$,so they form a conjugate pair.
$3.$ $H_3PO_4$ and $H_2PO_4^{-}$ differ by one $H^+$,so they form a conjugate pair.
$4.$ $H_2PO_4^{-}$ and $PO_3^{3-}$ differ by two protons $(2H^+)$,therefore they do not form a conjugate acid-base pair.

(B) $NH_4Cl$ is a salt of a weak base $(NH_4OH)$ and a strong acid $(HCl)$.
In water,it undergoes hydrolysis to form $NH_4OH$ (weakly ionized) and $H_3O^{+}$ ions (strongly ionized),which makes the solution acidic.
The reaction is as follows:
$NH_4^{+} + H_2O \rightleftharpoons NH_4OH + H^{+}$

(C) Let the mass of the man be $m$. The weight of the man is $w = mg$.
The forces acting on the man are his weight $mg$ acting downwards and the frictional force $F$ acting upwards.
According to Newton's second law of motion,the net force acting on the man is equal to the product of his mass and acceleration.
Since the man is sliding down with an acceleration $a = g/4$,we have:
$mg - F = ma$
Substituting $a = g/4$:
$mg - F = m(g/4)$
$mg - F = mg/4$
$F = mg - mg/4$
$F = \frac{3mg}{4}$
Since $w = mg$,the frictional force is $F = \frac{3w}{4}$.

(A) When a magnet is displaced by a very small angle $\theta$,the restoring torque acting on the magnet is given by $\tau = -M B_H \sin \theta$.
The negative sign indicates that the torque is restoring in nature.
Since the torque is also given by $\tau = I \alpha$,where $\alpha$ is the angular acceleration,and for small angles $\sin \theta \approx \theta$,we have:
$I \alpha = -M B_H \theta$.
Taking the magnitude of angular acceleration:
$\alpha = \frac{M B_H \theta}{I}$.

(A) For wire $A$ bent into a circle of radius $r$,the circumference is $2 \pi r = 40 \ cm$. Thus,$r = \frac{40}{2 \pi} \ cm$. The magnetic field at the center is $B_1 = \frac{\mu_0 i_1}{2 r}$.
For wire $B$ bent into an arc of radius $r$,the length of the arc is $l = r \theta = 30 \ cm$. Thus,the angle subtended is $\theta = \frac{30}{r} = \frac{30}{40 / 2 \pi} = \frac{3}{2} \pi$ radians. The magnetic field at the center is $B_2 = \frac{\mu_0 i_2 \theta}{4 \pi r}$.
Given $B_1 = B_2$,we have $\frac{\mu_0 i_1}{2 r} = \frac{\mu_0 i_2 \theta}{4 \pi r}$.
Substituting $\theta = \frac{3}{2} \pi$,we get $\frac{i_1}{2} = \frac{i_2 (3/2 \pi)}{4 \pi} = \frac{3 i_2}{8}$.
Therefore,$\frac{i_1}{i_2} = \frac{3}{8} \times 2 = \frac{3}{4}$.

(C) When a charged particle moves perpendicularly to a magnetic field,it follows a circular path.
The radius $r$ of the circular path is given by the formula:
$r = \frac{mv}{qB}$
Given values:
$m = 9 \times 10^{-31} \ kg$
$v = 1.6 \times 10^7 \ m/s$
$q = e = 1.6 \times 10^{-19} \ C$
$B = 0.1 \ T$
Substituting these values into the formula:
$r = \frac{9 \times 10^{-31} \times 1.6 \times 10^7}{1.6 \times 10^{-19} \times 0.1}$
$r = \frac{9 \times 1.6 \times 10^{-24}}{0.16 \times 10^{-19}}$
$r = \frac{14.4 \times 10^{-24}}{1.6 \times 10^{-20}}$
$r = 9 \times 10^{-4} \ m$

(C) When a charged particle moves perpendicular to a uniform magnetic field,it undergoes uniform circular motion.
In uniform circular motion,the acceleration (centripetal acceleration) is always directed towards the center of the circle,while the velocity vector is directed along the tangent to the path.
Therefore,the velocity vector $\overrightarrow{v}$ and the acceleration vector $\overrightarrow{a}$ are always perpendicular to each other.
Given,$\overrightarrow{v} = 2 \hat{i} + c \hat{j}$ and $\overrightarrow{a} = 3 \hat{i} + 4 \hat{j}$.
Since $\overrightarrow{v} \perp \overrightarrow{a}$,their dot product must be zero:
$\overrightarrow{v} \cdot \overrightarrow{a} = 0$
$(2 \hat{i} + c \hat{j}) \cdot (3 \hat{i} + 4 \hat{j}) = 0$
$(2)(3) + (c)(4) = 0$
$6 + 4c = 0$
$4c = -6$
$c = -1.5$

(D) Given inequalities are:
$x^2 - 3x - 10 < 0$
$(x - 5)(x + 2) < 0$
This implies $x \in (-2, 5)$.
Also,$10x - x^2 - 16 > 0$
Multiplying by $-1$,we get $x^2 - 10x + 16 < 0$
$(x - 2)(x - 8) < 0$
This implies $x \in (2, 8)$.
For both inequalities to hold simultaneously,we take the intersection:
$x \in (-2, 5) \cap (2, 8) = (2, 5)$.

(B) We simplify the expression from the innermost radical outward:
$\sqrt{14-6 \sqrt{5}} = \sqrt{9+5-2 \times 3 \times \sqrt{5}} = \sqrt{(3-\sqrt{5})^2} = 3-\sqrt{5}$
Substituting this back into the expression:
$\sqrt{2+\sqrt{5}-\sqrt{6-3 \sqrt{5} + (3-\sqrt{5})}} = \sqrt{2+\sqrt{5}-\sqrt{9-4 \sqrt{5}}}$
Now,simplify $\sqrt{9-4 \sqrt{5}}$:
$\sqrt{9-2 \times 2 \times \sqrt{5}} = \sqrt{(\sqrt{5}-2)^2} = \sqrt{5}-2$
Substituting this back:
$\sqrt{2+\sqrt{5}-(\sqrt{5}-2)} = \sqrt{2+\sqrt{5}-\sqrt{5}+2} = \sqrt{4} = 2$

(C) The given series is $\frac{1}{2}-\frac{1}{2 \cdot 2^2}+\frac{1}{3 \cdot 2^3}-\frac{1}{4 \cdot 2^4}+\ldots$
We know the logarithmic expansion $\log _e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$ for $|x| < 1$.
Comparing the given series with the expansion,we can write it as $\frac{1}{2} - \frac{(1/2)^2}{2} + \frac{(1/2)^3}{3} - \frac{(1/2)^4}{4} + \ldots$
By substituting $x = \frac{1}{2}$ in the expansion,we get:
$\log _e(1 + \frac{1}{2}) = \log _e(\frac{3}{2})$.

(B) Given that $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$.
Then,$\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
The roots of the equation $px^2+qx+r=0$ are $x_1 = \frac{1-\alpha}{\alpha} = \frac{1}{\alpha}-1$ and $x_2 = \frac{1-\beta}{\beta} = \frac{1}{\beta}-1$.
The equation with roots $x_1$ and $x_2$ is given by $x^2 - (x_1+x_2)x + x_1x_2 = 0$.
Sum of roots: $x_1+x_2 = (\frac{1}{\alpha}-1) + (\frac{1}{\beta}-1) = \frac{\alpha+\beta}{\alpha\beta} - 2 = \frac{-b/a}{c/a} - 2 = -\frac{b}{c} - 2 = -\frac{b+2c}{c}$.
Product of roots: $x_1x_2 = (\frac{1}{\alpha}-1)(\frac{1}{\beta}-1) = \frac{1}{\alpha\beta} - (\frac{1}{\alpha}+\frac{1}{\beta}) + 1 = \frac{1}{\alpha\beta} - \frac{\alpha+\beta}{\alpha\beta} + 1 = \frac{a}{c} - (\frac{-b/a}{c/a}) + 1 = \frac{a}{c} + \frac{b}{c} + 1 = \frac{a+b+c}{c}$.
Substituting these into the quadratic equation: $x^2 - (-\frac{b+2c}{c})x + \frac{a+b+c}{c} = 0$.
Multiplying by $c$: $cx^2 + (b+2c)x + (a+b+c) = 0$.
Comparing this with $px^2+qx+r=0$,we get $r = a+b+c$.

(C) Given that $1, 2, 3, 4$ are the roots of the equation $x^4+ax^3+bx^2+cx+d=0$.
Therefore,we can write the polynomial as:
$(x-1)(x-2)(x-3)(x-4) = x^4+ax^3+bx^2+cx+d$
Expanding the left side:
$(x^2-3x+2)(x^2-7x+12) = x^4-7x^3+12x^2-3x^3+21x^2-36x+2x^2-14x+24$
$= x^4-10x^3+35x^2-50x+24$
Comparing the coefficients with $x^4+ax^3+bx^2+cx+d=0$,we get:
$a=-10, b=35, c=-50, d=24$
Now,calculate $a+2b+c$:
$a+2b+c = -10 + 2(35) + (-50)$
$= -10 + 70 - 50 = 10$

(A) Given the cubic equation $x^3-2x^2+3x-4=0$,the roots are $\alpha, \beta, \gamma$.
By Vieta's formulas:
$\alpha+\beta+\gamma = 2$
$\alpha\beta+\beta\gamma+\gamma\alpha = 3$
$\alpha\beta\gamma = 4$
We need to find the value of $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$.
Using the identity $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$,let $a=\alpha\beta, b=\beta\gamma, c=\gamma\alpha$.
Then $(\alpha\beta+\beta\gamma+\gamma\alpha)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 2(\alpha\beta^2\gamma + \beta\gamma^2\alpha + \gamma\alpha^2\beta)$.
This simplifies to $(\alpha\beta+\beta\gamma+\gamma\alpha)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 2\alpha\beta\gamma(\alpha+\beta+\gamma)$.
Substituting the known values:
$(3)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 2(4)(2)$.
$9 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 16$.
$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = 9 - 16 = -7$.

(A) Given,$\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^n=1$.
We can write the complex number in polar form as $\cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$.
So,$(e^{i\pi/6})^n = 1$.
$e^{in\pi/6} = e^{i(2k\pi)}$ for some integer $k$.
Equating the arguments,$\frac{n\pi}{6} = 2k\pi$.
$n = 12k$.
For the smallest positive integer $n$,we take $k=1$,which gives $n=12$.

(B) Given,$a = \frac{1-i \sqrt{3}}{2} = \frac{1}{2} - i \frac{\sqrt{3}}{2}$.
Then,$\bar{a} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$.
$(i) \ a \bar{a} = |a|^2 = \left(\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$. Matches $(D)$.
$(ii) \ \arg \left(\frac{1}{\bar{a}}\right) = \arg(a) = \tan^{-1}\left(\frac{-\sqrt{3}/2}{1/2}\right) = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$. Matches $(A)$.
$(iii) \ a - \bar{a} = \left(\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) - \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = -i \sqrt{3}$. Matches $(B)$.
$(iv) \ \frac{4}{3a} = \frac{4}{3} \cdot \frac{1}{a} = \frac{4}{3} \cdot \frac{\bar{a}}{|a|^2} = \frac{4}{3} \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = \frac{2}{3} + i \frac{2\sqrt{3}}{3} = \frac{2}{3} + i \frac{2}{\sqrt{3}}$.
Thus,$\operatorname{Im}\left(\frac{4}{3a}\right) = \frac{2}{\sqrt{3}}$. Matches $(F)$.
The correct sequence is $(i)-D, (ii)-A, (iii)-B, (iv)-F$.

(A) Given,$\left|\frac{z-2i}{z+2i}\right|=1$
$\Rightarrow |z-2i| = |z+2i|$
Substituting $z=x+iy$:
$|x+i(y-2)| = |x+i(y+2)|$
Squaring both sides:
$x^2 + (y-2)^2 = x^2 + (y+2)^2$
$x^2 + y^2 - 4y + 4 = x^2 + y^2 + 4y + 4$
$-4y = 4y$
$8y = 0$
$y = 0$
This represents the $x$-axis.

(C) According to Bernoulli's equation for a horizontal pipe,the sum of pressure energy and kinetic energy per unit volume remains constant:
$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$
Given:
$P_1 = 50 \ kPa = 50 \times 10^3 \ Pa$
$v_1 = 1 \ m/s$
$v_2 = 2 \ m/s$
Density of water $\rho = 1000 \ kg/m^3 = 10^3 \ kg/m^3$
Rearranging the equation to solve for $P_2$:
$P_2 = P_1 + \frac{1}{2} \rho (v_1^2 - v_2^2)$
$P_2 = 50 \times 10^3 + \frac{1}{2} \times 10^3 \times (1^2 - 2^2)$
$P_2 = 50 \times 10^3 + 0.5 \times 10^3 \times (1 - 4)$
$P_2 = 50 \times 10^3 - 1.5 \times 10^3$
$P_2 = 48.5 \times 10^3 \ Pa = 48.5 \ kPa$

(D) liquid does not wet a solid surface if the angle of contact is obtuse,i.e.,greater than $90^{\circ}$.
In this scenario,the cohesive forces between the liquid molecules are stronger than the adhesive forces between the liquid and the solid surface.
Consequently,the liquid tends to minimize contact with the surface,resulting in the surface remaining dry.

(A) Poisson's ratio $(\sigma) = \frac{\text{lateral strain}}{\text{longitudinal strain}}$.
Given: lateral strain $= 0.01 \times 10^{-3}$,$\sigma = 0.4$.
Longitudinal strain $(\frac{\Delta L}{L}) = \frac{\text{lateral strain}}{\sigma} = \frac{0.01 \times 10^{-3}}{0.4} = 0.025 \times 10^{-3} = 2.5 \times 10^{-5}$.
Young's modulus $(Y) = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} = \frac{F}{A} \times \frac{L}{\Delta L}$.
Given: $F = 100 \ N$,$A = 0.025 \ m^2$,$\frac{\Delta L}{L} = 2.5 \times 10^{-5}$.
$Y = \frac{100}{0.025 \times 2.5 \times 10^{-5}} = \frac{100}{0.0625 \times 10^{-5}} = \frac{100}{6.25 \times 10^{-7}} = 16 \times 10^7 \ N/m^2 = 1.6 \times 10^8 \ N/m^2$.

(C) The initial horizontal velocity of the object is $u_x = u \cos \theta$. Given $\sin \theta = 3/5$,then $\cos \theta = 4/5$. So,$u_x = 100 \times (4/5) = 80 \ m/s$.
The horizontal range $R$ of the projectile is $R = \frac{u^2 \sin 2\theta}{g} = \frac{2 u^2 \sin \theta \cos \theta}{g} = \frac{2 \times 100^2 \times (3/5) \times (4/5)}{10} = 960 \ m$.
The horizontal distance to the highest point is $R/2 = 480 \ m$.
At the highest point,the momentum is conserved. The initial momentum is $P_i = (2m) u_x = (2m)(80) = 160m$.
After the explosion,the first piece $(m)$ comes to rest $(v_1 = 0)$,so $P_f = m(0) + m(v_2) = m v_2$.
Equating $P_i = P_f$,we get $160m = m v_2$,so $v_2 = 160 \ m/s$.
The second piece travels from the highest point with horizontal velocity $160 \ m/s$ for the remaining time of flight $t = \frac{u \sin \theta}{g} = \frac{100 \times (3/5)}{10} = 6 \ s$.
The additional horizontal distance covered by the second piece is $d = v_2 \times t = 160 \times 6 = 960 \ m$.
The total distance from the point of projection is $R/2 + d = 480 + 960 = 1440 \ m$.

(C) The given reactions represent the haloform reaction process for the preparation of chloroform $(CHCl_3)$:
$1$. Bleaching powder reacts with water to produce chlorine gas $(X = Cl_2)$:
$CaOCl_2 + H_2O \longrightarrow Ca(OH)_2 + Cl_2$
$2$. Chlorine reacts with acetaldehyde $(CH_3CHO)$ to form chloral $(Y = CCl_3CHO)$:
$CH_3CHO + 3Cl_2 \longrightarrow CCl_3CHO + 3HCl$
$3$. Chloral reacts with calcium hydroxide to produce chloroform:
$2CCl_3CHO + Ca(OH)_2 \longrightarrow 2CHCl_3 + (HCOO)_2Ca$
Thus,'$Y$' is $CCl_3CHO$ (chloral).

(C) In diethyl ether $(CH_3CH_2-O-CH_2CH_3)$,the oxygen atom is bonded to two carbon atoms and has two lone pairs of electrons.
According to the steric number rule,the steric number is calculated as: $\text{Steric Number} = \text{Number of sigma bonds} + \text{Number of lone pairs} = 2 + 2 = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridisation.

(D) The initial radioactive substance is $_{88}^{228}X$.
Each $\alpha$ emission decreases the atomic number by $2$ and the mass number by $4$.
Each $\beta$ emission increases the atomic number by $1$ and does not change the mass number.
After emitting $3$ $\alpha$ and $3$ $\beta$ particles,the atomic number of $Y$ becomes: $88 - (3 \times 2) + (3 \times 1) = 88 - 6 + 3 = 85$.
The element with atomic number $85$ is Astatine $(At)$,which belongs to group $VIIA$ (Halogens) of the periodic table.

(C) The reactions are as follows:

Reactants	Products
$A$. $C_2H_5Cl + \text{moist } Ag_2O$	$III$. $CH_3CH_2OH$ (Substitution)
$B$. $C_2H_5Cl + \text{aqueous ethanolic } AgCN$	$IV$. $CH_3CH_2NC$ (Isocyanide formation)
$C$. $C_2H_5Cl + \text{aqueous ethanolic } AgNO_2$	$I$. $CH_3CH_2ONO$ (Nitrite formation)
$D$. $C_2H_5Cl + \text{ethanolic } KOH$	$II$. $C_2H_4$ (Elimination/Dehydrohalogenation)

Therefore,the correct match is $A-III, B-IV, C-I, D-II$.

(B) The correct answer is $B$.
Gold sol is a lyophobic sol,not a lyophilic sol.
Milk is a natural emulsion of fat in water.
Physical adsorption (physisorption) is exothermic and decreases with an increase in temperature.
Chemical adsorption (chemisorption) is specific and forms a unilayer (monolayer) on the surface.

(D) Step $1$: Reduction of acetic acid $(CH_3COOH)$ with $LiAlH_4$ is a strong reducing agent that reduces carboxylic acids to primary alcohols.
$CH_3COOH + 4[H] \xrightarrow{LiAlH_4} CH_3CH_2OH + H_2O$
Thus,$A$ is ethanol $(C_2H_5OH)$.
Step $2$: Esterification reaction between ethanol $(A)$ and acetic acid $(CH_3COOH)$ in the presence of an acid catalyst $(H_3O^+)$ yields an ester.
$CH_3CH_2OH + CH_3COOH \xrightarrow{H_3O^+} CH_3COOC_2H_5 + H_2O$
Thus,$B$ is ethyl acetate $(CH_3COOC_2H_5)$.
Therefore,$A = C_2H_5OH$ and $B = CH_3COOC_2H_5$.

(C) The reaction sequence represents the haloform reaction for the preparation of chloroform $(CHCl_3)$:
$1$. Bleaching powder reacts with water to produce chlorine gas $(X)$:
$CaOCl_2 + H_2O \longrightarrow Ca(OH)_2 + Cl_2 (X)$
$2$. Acetaldehyde reacts with chlorine to form chloral $(Y)$:
$CH_3CHO + 3Cl_2 \longrightarrow CCl_3CHO (Y) + 3HCl$
$3$. Chloral reacts with calcium hydroxide to produce chloroform:
$2CCl_3CHO + Ca(OH)_2 \longrightarrow 2CHCl_3 + (HCOO)_2Ca$
Thus,'$Y$' is $CCl_3CHO$ (chloral).

(D) The reduction of nitrobenzene with zinc dust $(Zn)$ and aqueous ammonium chloride $(NH_4Cl)$ is a controlled reduction process.
This reaction yields $N$-phenylhydroxylamine as the major product.
The chemical equation is:
$C_6H_5NO_2 + 4[H] \xrightarrow{Zn, NH_4Cl} C_6H_5NHOH + H_2O$

(C) The chemical formula of diethyl ether is $C_2H_5-O-C_2H_5$.
In this molecule,the oxygen atom is bonded to two carbon atoms and possesses two lone pairs of electrons.
The steric number of the oxygen atom is calculated as: $\text{Steric Number} = \text{Number of sigma bonds} + \text{Number of lone pairs} = 2 + 2 = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridisation.

(D) The initial element is ${}_{88}X^{228}$ belonging to group $IIA$ (atomic number $Z = 88$).
When an element emits one $\alpha$-particle,its atomic number decreases by $2$ and mass number by $4$.
When an element emits one $\beta$-particle,its atomic number increases by $1$ and mass number remains unchanged.
For the emission of $3 \alpha$ and $3 \beta$ particles:
Change in atomic number $(Z)$ = $(3 \times -2) + (3 \times 1) = -6 + 3 = -3$.
New atomic number of $Y$ = $88 - 3 = 85$.
An element with atomic number $88$ is Radium $(Ra)$,which is in group $IIA$ (alkaline earth metals).
Group $IIA$ elements have a valence shell configuration of $ns^2$.
Atomic number $88$ is in the $7^{th}$ period. The element with atomic number $85$ is Astatine $(At)$.
Astatine $(Z=85)$ belongs to the halogen family,which is group $VIIA$ (or group $17$).

(C) The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$Cr^{2+} (Z=24): [Ar] 3d^4 4s^0$; $n = 4$ unpaired electrons.
$Fe^{2+} (Z=26): [Ar] 3d^6 4s^0$; $n = 4$ unpaired electrons.
Since $Cr^{2+}$ and $Fe^{2+}$ have the same number of unpaired electrons $(n=4)$,they have the same calculated value of magnetic moment.

(A) The cell reaction is: $Cu_{(s)} + 2Ag^+_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$.
For the given cell,the cathode is $Ag^+/Ag$ and the anode is $Cu^{2+}/Cu$.
The standard cell potential is calculated as:
$E^0_{cell} = E^0_{cathode} - E^0_{anode}$
$E^0_{cell} = E^0_{Ag^+/Ag} - E^0_{Cu^{2+}/Cu}$
$E^0_{cell} = 0.80 \ V - 0.34 \ V = +0.46 \ V$.

(D) The Bhopal gas tragedy of $1984$ was caused by the leakage of methyl isocyanate $(CH_3NCO)$.

(D) Salol is phenyl salicylate. Its chemical structure consists of a benzene ring substituted with a hydroxyl group $(-OH)$ at the ortho position and an ester group ($-COOC_6H_5$,which is a type of $-COOR$) at the adjacent position.
Therefore,the functional groups present in salol are $-OH$ (hydroxyl) and $-COOR$ (ester).

(A) $C_2H_5Cl + \text{moist } Ag_2O \rightarrow C_2H_5OH + AgCl$. Thus,$A-iii$.
$(B)$ $C_2H_5Cl + AgCN \rightarrow C_2H_5NC + AgCl$ (Isocyanide is the major product). Thus,$B-iv$.
$(C)$ $C_2H_5Cl + AgNO_2 \rightarrow C_2H_5ONO + AgCl$ (Alkyl nitrite is the major product). Thus,$C-i$.
$(D)$ $C_2H_5Cl + \text{ethanolic } KOH \rightarrow C_2H_4 + KCl + H_2O$ (Dehydrohalogenation). Thus,$D-ii$.
Therefore,the correct match is $A-iii, B-iv, C-i, D-ii$.

(A) The reducing property of the hydrides of $VA$ group increases from $NH_3$ to $BiH_3$ due to the decrease in bond dissociation enthalpy: $NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$.
The tendency to donate a lone pair (basic strength) decreases from $NH_3$ to $BiH_3$ because the electron density on the central atom decreases as the size of the atom increases.
Thermal stability of $VA$ group hydrides decreases from $NH_3$ to $BiH_3$ as the $M-H$ bond strength decreases: $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$.
Bond angle of $VA$ group hydrides decreases from $NH_3$ to $BiH_3$ due to the decrease in electronegativity of the central atom: $NH_3 (107.8^{\circ}) > PH_3 (93.6^{\circ}) > AsH_3 (91.8^{\circ}) > SbH_3 (91.3^{\circ}) > BiH_3 (90^{\circ})$.
Thus,all statements $I, II, III,$ and $IV$ are correct.

(C) Metals are good conductors of electricity due to the presence of free electrons.
Electrical conductivity of semiconductors increases with an increase in temperature.
Silicon doped with boron (Group $13$ element) creates a $p$-type semiconductor.
Silicon doped with arsenic (Group $15$ element) creates an $n$-type semiconductor.

(C) When sodium thiosulphate solution is added to $AgBr$,a complex compound called sodium argentothiosulphate is formed,which is soluble in water.
The chemical reaction is:
$AgBr + 2Na_2S_2O_3 \longrightarrow Na_3[Ag(S_2O_3)_2] + NaBr$
The product formed is $Na_3[Ag(S_2O_3)_2]$.

(D) Perchloric acid $(HClO_4)$ is an oxoacid of chlorine where the oxidation state of chlorine is $+7$. It does not contain a peroxy linkage ($-O-O-$ bond).
In contrast,perphosphoric acid $(H_3PO_5)$,pernitric acid $(HNO_4)$,and perdisulphuric acid $(H_2S_2O_8)$ all contain at least one peroxy linkage ($-O-O-$ bond) in their structures.

(D) Krypton is used in miner's cap lamps because it provides a high-intensity light and has a long filament life.

(C) $PHBV$ (Polyhydroxy butyrate-co-$\beta$-hydroxy valerate) is a biodegradable polymer.
It is obtained by the copolymerization of $3$-hydroxybutanoic acid and $3$-hydroxypentanoic acid.

(B) Gold sol is a lyophobic sol,not a lyophilic sol. Therefore,option $B$ is incorrect.

(D) Enzymes are biological catalysts. They are highly specific in nature and catalyze various biological reactions within living organisms.

(C) catalyst increases the rate of a reaction by providing an alternative pathway with lower activation energy.
Therefore,the assertion $(A)$ is true,but the reason $(R)$ is false because the activation energy decreases,not increases.