AP EAMCET 2012 Chemistry Question Paper with Answer and Solution

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

(B) Let the point $P$ divide the line segment $QR$ in the ratio $\lambda : 1$.
The coordinates of $P$ are given by the section formula as $\left( \frac{5\lambda + 2}{\lambda + 1}, \frac{2\lambda + 2}{\lambda + 1}, \frac{-2\lambda + 1}{\lambda + 1} \right)$.
Given that the $x$-coordinate of $P$ is $4$,we have:
$\frac{5\lambda + 2}{\lambda + 1} = 4$
Solving for $\lambda$:
$5\lambda + 2 = 4(\lambda + 1)$
$5\lambda + 2 = 4\lambda + 4$
$\lambda = 2$
Now,substitute $\lambda = 2$ into the expression for the $z$-coordinate of $P$:
$z = \frac{-2\lambda + 1}{\lambda + 1} = \frac{-2(2) + 1}{2 + 1} = \frac{-4 + 1}{3} = \frac{-3}{3} = -1$.
Thus,the $z$-coordinate of $P$ is $-1$.

(B) Let the following events be defined:
$A_1$: The student knows the answer.
$A_2$: The student does not know the answer (guesses).
$E$: The student gets the correct answer.
Given:
$P(A_1) = 0.9 = \frac{9}{10}$
$P(A_2) = 1 - 0.9 = 0.1 = \frac{1}{10}$
$P(E|A_1) = 1$ (If he knows,he answers correctly with certainty).
$P(E|A_2) = \frac{1}{4}$ (If he guesses,there is $1$ correct option out of $4$).
We need to find the probability that he was guessing given that he got the correct answer,i.e.,$P(A_2|E)$.
Using Bayes' Theorem:
$P(A_2|E) = \frac{P(A_2)P(E|A_2)}{P(A_1)P(E|A_1) + P(A_2)P(E|A_2)}$
$P(A_2|E) = \frac{(\frac{1}{10}) \times (\frac{1}{4})}{(\frac{9}{10}) \times (1) + (\frac{1}{10}) \times (\frac{1}{4})}$
$P(A_2|E) = \frac{\frac{1}{40}}{\frac{9}{10} + \frac{1}{40}} = \frac{\frac{1}{40}}{\frac{36+1}{40}} = \frac{1}{37}$.

(B) The given reaction is the Reimer-Tiemann reaction.
In this reaction,phenol reacts with chloroform $(CHCl_3)$ in the presence of an aqueous base like sodium hydroxide $(NaOH)$ to form an ortho-substituted product,which is salicylaldehyde ($o$-hydroxybenzaldehyde).
The mechanism involves the formation of a dichlorocarbene $(:CCl_2)$ intermediate,which acts as an electrophile and attacks the phenol ring at the ortho position.

(A) The compound $A$ $(C_3H_6O)$ reacts with $I_2 / NaOH$ (iodoform test),which indicates it is a methyl ketone. Thus,$A$ is acetone $(CH_3COCH_3)$.
$1$. Reaction with $I_2 / NaOH$ (Iodoform reaction):
$CH_3COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 (B) + CH_3COONa + 3NaI + 3H_2O$
Here,$B$ is iodoform $(CHI_3)$.
$2$. Reaction with $Zn-Hg / HCl$ (Clemmensen reduction):
$CH_3COCH_3 + 4[H] \xrightarrow{Zn-Hg / HCl} CH_3CH_2CH_3 (C) + H_2O$
Here,$C$ is propane $(CH_3CH_2CH_3)$.

(A) $1$. The $-NO_2$ group is a deactivating and meta-directing group. Therefore,electrophilic aromatic substitution of nitrobenzene with $Cl_2$ in the presence of $Fe$ (Lewis acid) yields $m\text{-chloronitrobenzene}$ as the major product $(A)$.
$2$. The reduction of nitrobenzene with $LiAlH_4$ is a complex process. While $LiAlH_4$ is a strong reducing agent,under specific conditions,it can reduce nitrobenzene to azobenzene $(C_6H_5-N=N-C_6H_5)$ $(B)$.

(D) The formula for formal charge $(FC)$ is: $FC = V - lp - \frac{1}{2} bp$
Where $V$ is the number of valence electrons,$lp$ is the number of lone pair electrons,and $bp$ is the number of bonding electrons.
For the structure $\ddot{O}=C=\ddot{O}$:
For $O$ atom: $V = 6$,$lp = 4$,$bp = 4$. So,$FC = 6 - 4 - \frac{1}{2}(4) = 0$.
For $C$ atom: $V = 4$,$lp = 0$,$bp = 8$. So,$FC = 4 - 0 - \frac{1}{2}(8) = 0$.
Thus,the formal charges of $C$ and $O$ are $0$ and $0$ respectively.

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

(D) . $LiOH$ is a weaker base than $NaOH$ because $Li^+$ has a higher charge density than $Na^+$,which increases the covalent character of the $Li-OH$ bond,making it less ionic and thus a weaker base. This statement is correct.
$B$. Salts of $Be$ undergo hydrolysis due to the high charge density and small size of $Be^{2+}$ ion,which polarizes the water molecules. This statement is correct.
$C$. $Ca(HCO_3)_2$ is soluble in water. This statement is correct.
$D$. Hydrolysis of beryllium carbide $(Be_2C)$ gives methane $(CH_4)$,not acetylene $(C_2H_2)$. The reaction is: $Be_2C + 4H_2O \rightarrow CH_4 + 2Be(OH)_2$. Therefore,this statement is incorrect.

(D) Phosphorus $(P)$ is a non-metal and typically forms acidic oxides such as $P_4O_6$ and $P_4O_{10}$.
$Al_2O_3$,$SnO$,and $Sb_2O_3$ are known to be amphoteric oxides,meaning they can react with both acids and bases.

(B) As the atomic size increases down a group,the ionization energy and the work function decrease because the outermost electron is held less tightly by the nucleus.
$Li$,$Na$,$K$,and $Rb$ belong to Group $1$ of the periodic table.
The order of their atomic sizes is $Li < Na < K < Rb$.
Therefore,the order of their work functions is $Li > Na > K > Rb$.
Given values: $Li = 2.41 \ eV$,$Na = 2.30 \ eV$,and $Rb = 2.09 \ eV$.
The work function of $K$ must lie between the values of $Na$ $(2.30 \ eV)$ and $Rb$ $(2.09 \ eV)$.
Among the given options,$2.20 \ eV$ is the only value that falls in the range $2.09 \ eV < K < 2.30 \ eV$.

(B) When the complex is dissolved in water,it dissociates to give $3$ ions in total.
This indicates that there are $2$ chloride ions $(Cl^-)$ outside the coordination sphere and one complex cation.
Thus,the formula of the complex is $[Co(NH_3)_5Cl]Cl_2$.
Dissociation: $[Co(NH_3)_5Cl]Cl_2 \rightarrow [Co(NH_3)_5Cl]^{2+} + 2Cl^-$
Total ions = $1$ (complex ion) + $2$ (chloride ions) = $3$ ions.

(C) For a balanced Wheatstone bridge,the condition is $\frac{A}{B} = \frac{C}{D}$.
Case $1$: When $C = 100 \Omega$,the bridge is balanced,so $\frac{A}{B} = \frac{100}{D}$ ... $(i)$
Case $2$: When $A$ and $B$ are interchanged,the new resistances in the arms are $B$ and $A$ respectively. The bridge balances for $C = 121 \Omega$,so $\frac{B}{A} = \frac{121}{D}$ ... (ii)
From equation $(i)$,we have $\frac{A}{B} = \frac{100}{D}$. Therefore,$\frac{B}{A} = \frac{D}{100}$.
Substituting this into equation (ii),we get $\frac{D}{100} = \frac{121}{D}$.
$D^2 = 100 \times 121 = 12100$.
$D = \sqrt{12100} = 110 \Omega$.

(C) For a balanced Wheatstone bridge,the ratio of resistances is given by $\frac{A}{B} = \frac{C}{D}$.
Let $D = x$. Initially,$\frac{A}{B} = \frac{100}{x}$.
When $A$ and $B$ are interchanged,the new condition is $\frac{B}{A} = \frac{121}{x}$.
From the first equation,$\frac{B}{A} = \frac{x}{100}$.
Equating the two expressions for $\frac{B}{A}$,we get $\frac{x}{100} = \frac{121}{x}$.
$x^2 = 121 \times 100 = 12100$.
$x = \sqrt{12100} = 110 \ \Omega$.
Thus,the value of $D$ is $110 \ \Omega$.

(A) The de-Broglie wavelength $\lambda$ for a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.
For the proton $(p)$: $\lambda_p = \frac{h}{\sqrt{2m_p q_p V}}$.
For the $\alpha$-particle $(\alpha)$: $\lambda_\alpha = \frac{h}{\sqrt{2m_\alpha q_\alpha V_\alpha}}$.
Given $\lambda_p = \lambda_\alpha$, we have $\sqrt{2m_p q_p V} = \sqrt{2m_\alpha q_\alpha V_\alpha}$.
Squaring both sides: $m_p q_p V = m_\alpha q_\alpha V_\alpha$.
We know that $m_\alpha = 4m_p$ and $q_\alpha = 2q_p$.
Substituting these values: $m_p q_p V = (4m_p)(2q_p) V_\alpha$.
$m_p q_p V = 8 m_p q_p V_\alpha$.
Therefore, $V_\alpha = \frac{V}{8}$.

(D) $E_{\text{cell}}^{\circ} = E_{Cu^{2+} / Cu}^{\circ} - E_{Zn^{2+} / Zn}^{\circ} = +0.34 - (-0.76) \ V = 1.1 \ V$
Using the Nernst equation: $E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
For the reaction $Cu^{2+} + Zn \longrightarrow Zn^{2+} + Cu$,$n = 2$.
$E_{\text{cell}} = 1.1 - \frac{0.059}{2} \log \frac{0.1}{0.01}$
$E_{\text{cell}} = 1.1 - 0.0295 \times \log(10)$
$E_{\text{cell}} = 1.1 - 0.0295 \times 1 = 1.0705 \ V \approx 1.07 \ V$

(A) The reduction reaction for $Al^{3+}$ is: $Al^{3+} + 3e^{-} \longrightarrow Al(s)$.
From the stoichiometry,$1 \ mol$ of $Al$ $(27 \ g)$ requires $3 \ moles$ of electrons.
To deposit $36 \ g$ of $Al$,the number of moles of electrons required is:
$\text{Moles of } e^{-} = \frac{3 \ mol \ e^{-}}{27 \ g} \times 36 \ g = 4 \ mol$.

(D) The induced electromotive force (emf) in a moving conductor is given by the formula:
$e = B v l \sin \theta$
Where:
$B = 0.1 ~Wb/m^2$ (Magnetic field induction)
$v = 10 ~m/s$ (Velocity of the conductor)
$l = 4 ~m$ (Length of the conductor)
$\theta = 30^{\circ}$ (Angle between the conductor and the magnetic field)
Substituting the values into the formula:
$e = 0.1 \times 10 \times 4 \times \sin(30^{\circ})$
$e = 0.1 \times 10 \times 4 \times 0.5$
$e = 1 \times 4 \times 0.5$
$e = 2 ~V$
Thus,the induced emf is $2 ~V$.

(D) The total electromotive force $(emf)$ produced in a thermocouple is determined by the Seebeck effect,which depends on the nature of the metals used (Seebeck coefficients),the temperatures of the hot and cold junctions,and the Thomson coefficients of the metals. It is an instantaneous phenomenon related to the temperature gradient and material properties. Therefore,the total $emf$ does not depend on the duration of time for which the current is passed through the thermocouple.

(B) The induced emf $(e)$ in the secondary coil is given by the formula: $e = M \frac{di}{dt}$.
Given:
Mutual inductance $(M) = 92 \times 10^{-6} \, H$
Rate of change of current $(\frac{di}{dt}) = \frac{25 \, A}{1 \, ms} = \frac{25 \, A}{1 \times 10^{-3} \, s} = 25 \times 10^{3} \, A/s$.
Substituting these values into the formula:
$e = (92 \times 10^{-6} \, H) \times (25 \times 10^{3} \, A/s)$
$e = 92 \times 25 \times 10^{-3} \, V$
$e = 2300 \times 10^{-3} \, V$
$e = 2.3 \, V$.

(A) In the absence of gravity,the only force acting on each sphere is the electrostatic repulsive force exerted by the other sphere.
Since the spheres are held in a horizontal position,the distance between the two spheres is $2L$.
According to Coulomb's law,the electrostatic force $F$ between the two charges is given by:
$F = \frac{1}{4 \pi \varepsilon_0} \frac{Q \cdot Q}{(2L)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{4L^2} = \frac{Q^2}{16 \pi \varepsilon_0 L^2}$.
Since the spheres are in equilibrium,the tension $T$ in each string must balance this electrostatic force.
Therefore,$T = F = \frac{Q^2}{16 \pi \varepsilon_0 L^2}$.

(C) Oxides of nitrogen and sulphur are responsible for acid rain.
$NO_2 + H_2O \rightarrow HNO_3$
$SO_2$ $\xrightarrow{[O]} SO_3 + H_2O$ $\rightarrow H_2SO_4$
Due to the presence of $HNO_3$ and $H_2SO_4$ (strong acids),rainwater becomes acidic,and this phenomenon is called acid rain.

(D) The chlorine atom $(Cl)$ in chlorobenzene exhibits both $-I$ (inductive) and $+M$ (mesomeric) effects.
Due to the $-I$ effect,it is ring deactivating.
Due to the $+M$ effect,it is ortho/para directing.
Therefore,the statement that $Cl$ is meta directing is incorrect.

(D) Conformational isomers are also known as $rotamers$ and the isomerism as $rotamerism$. These are different spatial arrangements of atoms in a molecule that can be interconverted by rotation about single bonds.

(B) The correct matches are as follows:
$A$. Acetaldehyde $(CH_3CHO)$ and vinyl alcohol $(CH_2=CH-OH)$ are tautomers.
$B$. Eclipsed and staggered ethane are two conformations of ethane,hence they are conformational isomers.
$C$. $(+)2$-butanol and $(-)2$-butanol are non-superimposable mirror images of each other,hence they are enantiomers.
$D$. Methyl-$n$-propylamine $(CH_3-NH-CH_2CH_2CH_3)$ and diethylamine $((C_2H_5)_2NH)$ have different alkyl groups attached to the same functional group $(-NH-)$,hence they are metamers.
Therefore,the correct matching is $A-2, B-4, C-1, D-5$.

(C) The chlorination of ethane is a classic example of a free radical substitution reaction. It proceeds through a chain mechanism involving three main steps:
$1.$ Initiation: The homolytic cleavage of $Cl_2$ by ultraviolet light $(h\nu)$ to form chlorine free radicals.
$Cl_2 \xrightarrow{h\nu} 2Cl^{\bullet}$
$2.$ Propagation: The chlorine radical abstracts a hydrogen atom from ethane to form an ethyl radical $(CH_3CH_2^{\bullet})$,which then reacts with another $Cl_2$ molecule to form ethyl chloride and regenerate a chlorine radical.
$CH_3CH_3 + Cl^{\bullet} \rightarrow CH_3CH_2^{\bullet} + HCl$
$CH_3CH_2^{\bullet} + Cl_2 \rightarrow CH_3CH_2Cl + Cl^{\bullet}$
$3.$ Termination: The combination of any two free radicals to end the chain reaction.
$CH_3CH_2^{\bullet} + Cl^{\bullet} \rightarrow CH_3CH_2Cl$
$Cl^{\bullet} + Cl^{\bullet} \rightarrow Cl_2$
$2CH_3CH_2^{\bullet} \rightarrow CH_3CH_2CH_2CH_3$

(D) In benzene,all the $C-C$ bond lengths are equal to $1.39 \mathring{A}$ due to resonance.
Therefore,the statement that benzene has two different carbon-carbon bond lengths is incorrect.

(B) $H_2O_2$ reacts with disodium hydrogen phosphate $(Na_2HPO_4)$ to form an addition product known as sodium phosphate perhydrate.
The chemical reaction is:
$H_2O_2 + Na_2HPO_4 \rightarrow Na_2HPO_4 \cdot H_2O_2$

(C) Lewis acid is an electron pair acceptor,while a Lewis base is an electron pair donor.
$Cl^{-}$ has a complete octet and possesses lone pairs of electrons,which it can donate; therefore,it acts as a Lewis base,not a Lewis acid.
Thus,the statement '$Cl^{-}$ is a Lewis acid' is incorrect.

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

(D) The time period of a vibrating magnet is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$.
This implies $T \propto \frac{1}{\sqrt{B_H}}$.
Given at the first place,$n_1 = 30 \text{ oscillations/min} = 0.5 \text{ oscillations/s}$.
Therefore,the time period $T_1 = \frac{1}{n_1} = \frac{1}{0.5} = 2 ~s$.
At the second place,the magnetic field is doubled,so $(B_H)_2 = 2(B_H)_1$.
The ratio of time periods is $\frac{T_2}{T_1} = \sqrt{\frac{(B_H)_1}{(B_H)_2}} = \sqrt{\frac{(B_H)_1}{2(B_H)_1}} = \frac{1}{\sqrt{2}}$.
Thus,$T_2 = T_1 \times \frac{1}{\sqrt{2}} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} ~s$.

(A) Let the two forces be $F_1$ and $F_2$,where $F_1$ is the smaller force.
Given: $F_1 + F_2 = 16 ~N$,so $F_2 = 16 - F_1$.
Let the resultant force be $R = 8 ~N$.
Since the resultant is perpendicular to the smaller force $F_1$,we have the relation $R^2 + F_1^2 = F_2^2$.
Substituting the values: $8^2 + F_1^2 = (16 - F_1)^2$.
$64 + F_1^2 = 256 + F_1^2 - 32 F_1$.
$32 F_1 = 256 - 64$.
$32 F_1 = 192$.
$F_1 = 6 ~N$.
Then,$F_2 = 16 - 6 = 10 ~N$.
Thus,the forces are $6 ~N$ and $10 ~N$.

(C) The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.
Squaring both sides,we get $T^2 = 4\pi^2 \frac{l}{g}$,which implies $g = 4\pi^2 \frac{l}{T^2}$.
The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T}{T}$.
Given: $l = 1.01 \ m$,$\Delta l = 0.01 \ m$,$t_{total} = 63 \ s$,$\Delta t = 3 \ s$.
The time period $T = \frac{t_{total}}{30} = \frac{63}{30} = 2.1 \ s$.
The error in time period $\Delta T = \frac{\Delta t}{30} = \frac{3}{30} = 0.1 \ s$.
Substituting these values into the error formula:
$\frac{\Delta g}{g} = \frac{0.01}{1.01} + 2 \left( \frac{0.1}{2.1} \right) \approx 0.0099 + 0.0952 \approx 0.1051$.
Converting to percentage: $\frac{\Delta g}{g} \times 100 \% \approx 10.51 \% \approx 10 \%$.

(B) Given $\alpha, \beta, \gamma$ are roots of $x^3+p x^2+q x+r=0$.
By Vieta's formulas: $\alpha+\beta+\gamma = -p$,$\alpha\beta+\beta\gamma+\gamma\alpha = q$,and $\alpha\beta\gamma = -r$.
Let the new roots be $y_1 = \alpha(\beta+\gamma)$,$y_2 = \beta(\gamma+\alpha)$,and $y_3 = \gamma(\alpha+\beta)$.
Note that $\alpha(\beta+\gamma) = \alpha(\beta+\gamma+\alpha) - \alpha^2 = -p\alpha - \alpha^2$.
Since $\alpha$ is a root,$\alpha^3+p\alpha^2+q\alpha+r=0$,so $\alpha^3+p\alpha^2 = -q\alpha-r$.
Thus,$y_1 = -p\alpha - \alpha^2$.
Alternatively,$y_1 = \alpha\beta+\alpha\gamma = q - \beta\gamma = q - (\frac{-r}{\alpha}) = q + \frac{r}{\alpha}$.
So,$\alpha = \frac{r}{y-q}$.
Substituting this into the original equation: $(\frac{r}{y-q})^3 + p(\frac{r}{y-q})^2 + q(\frac{r}{y-q}) + r = 0$.
Dividing by $r$ (assuming $r \neq 0$): $\frac{r^2}{(y-q)^3} + \frac{pr}{(y-q)^2} + \frac{q}{y-q} + 1 = 0$.
Multiplying by $(y-q)^3$: $r^2 + pr(y-q) + q(y-q)^2 + (y-q)^3 = 0$.
Expanding: $r^2 + pry - prq + q(y^2 - 2qy + q^2) + (y^3 - 3y^2q + 3yq^2 - q^3) = 0$.
$y^3 + (q-3q)y^2 + (3q^2 - 2q^2 + pr)y + (r^2 - prq + q^3 - q^3) = 0$.
$y^3 - 2qy^2 + (q^2 + pr)y + (r^2 - prq) = 0$.
The coefficient of $y$ is $q^2+pr$.

(D) Given that $\alpha$ is a non-real root of $x^6-1=0$.
Since $x^6-1 = (x-1)(x^5+x^4+x^3+x^2+x+1) = 0$,and $\alpha \neq 1$ (as it is a non-real root),we have:
$\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1 = 0$ --- $(i)$
We need to evaluate $\frac{\alpha^2+\alpha^3+\alpha^4+\alpha^5}{\alpha+1}$.
Factorizing the numerator:
$\alpha^2+\alpha^3+\alpha^4+\alpha^5 = \alpha^2(1+\alpha) + \alpha^4(1+\alpha) = (1+\alpha)(\alpha^2+\alpha^4)$
Therefore,the expression becomes:
$\frac{(1+\alpha)(\alpha^2+\alpha^4)}{\alpha+1} = \alpha^2+\alpha^4$
From equation $(i)$,$\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1 = 0$.
This can be written as $(1+\alpha) + \alpha^2(1+\alpha) + \alpha^4(1+\alpha) = 0$.
$(1+\alpha)(1+\alpha^2+\alpha^4) = 0$.
Since $\alpha \neq -1$ (because $(-1)^6-1 \neq 0$),we must have $1+\alpha^2+\alpha^4 = 0$.
Thus,$\alpha^2+\alpha^4 = -1$.

(B) committee of $12$ members is to be formed such that women are in majority. The total number of women is $9$ and men is $8$.
Case $I$: $9$ women and $3$ men
$\text{Number of ways} = {^9C_9} \times {^8C_3} = 1 \times \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$
Case $II$: $8$ women and $4$ men
$\text{Number of ways} = {^9C_8} \times {^8C_4} = 9 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 9 \times 70 = 630$
Case $III$: $7$ women and $5$ men
$\text{Number of ways} = {^9C_7} \times {^8C_5} = \frac{9 \times 8}{2 \times 1} \times \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 36 \times 56 = 2016$
$\text{Total number of ways} = 56 + 630 + 2016 = 2702$

(C) The student must choose $10$ questions out of $13$. The first $6$ questions form one group,and the remaining $7$ questions form another group. The student must choose at least $5$ questions from the first $6$.
Case $I$: Choosing $5$ questions from the first $6$ and $5$ questions from the remaining $7$.
Number of ways $= {}^{6}C_{5} \times {}^{7}C_{5} = 6 \times 21 = 126$.
Case $II$: Choosing $6$ questions from the first $6$ and $4$ questions from the remaining $7$.
Number of ways $= {}^{6}C_{6} \times {}^{7}C_{4} = 1 \times 35 = 35$.
Total number of ways $= 126 + 35 = 161$.

(C) For a drop floating in equilibrium,the downward gravitational force must be balanced by the upward buoyant force and the upward surface tension force.
Weight of the drop = $\frac{4}{3} \pi r^3 d g$.
Buoyant force (half immersed) = $\frac{2}{3} \pi r^3 \rho g$.
Surface tension force acting along the circumference of the contact circle = $T \times 2 \pi r$.
Equating the forces: $\frac{4}{3} \pi r^3 d g = \frac{2}{3} \pi r^3 \rho g + 2 \pi r T$.
Dividing by $\pi r$: $\frac{4}{3} r^2 d g = \frac{2}{3} r^2 \rho g + 2 T$.
Rearranging terms: $r^2 g (\frac{4}{3} d - \frac{2}{3} \rho) = 2 T$.
$r^2 g (\frac{2}{3}) (2 d - \rho) = 2 T$.
$r^2 = \frac{3 T}{g(2 d - \rho)}$.
Therefore,$r = \sqrt{\frac{3 T}{g(2 d - \rho)}}$.

(B) The half-life $(T_{1/2})$ of the radioactive element is $10 \ h$.
The total time elapsed $(t)$ is $40 \ h$.
The number of half-lives $(n)$ is calculated as $n = \frac{t}{T_{1/2}} = \frac{40 \ h}{10 \ h} = 4$.
The fraction of the initial radioactivity remaining is given by the formula $\frac{N}{N_0} = (\frac{1}{2})^n$.
Substituting the value of $n$,we get $\frac{N}{N_0} = (\frac{1}{2})^4 = \frac{1}{16}$.

(C) The number of atoms decaying per second is given by the activity $A = \lambda N$.
First,calculate the decay constant $\lambda = \frac{0.693}{T_{1/2}}$.
Half-life $T_{1/2} = 1620 \text{ years} = 1620 \times 365 \times 24 \times 3600 \text{ seconds} \approx 5.11 \times 10^{10} \text{ s}$.
The number of atoms $N$ in $1 \ g$ of $Ra^{226}$ is $N = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{1}{226} \times 6.023 \times 10^{23} \approx 2.665 \times 10^{21} \text{ atoms}$.
Now,calculate activity $A = \lambda N = \left( \frac{0.693}{5.11 \times 10^{10}} \right) \times 2.665 \times 10^{21} \approx 3.61 \times 10^{10} \text{ decays/s}$.

(C) The displacement of a particle in simple harmonic motion starting from the mean position is given by $y(t) = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \ rad/s$.
At $t=1 \ s$,the distance from the mean position is $y_1 = A \sin(\frac{\pi}{4} \times 1) = A \times \frac{1}{\sqrt{2}}$.
Since the particle starts from the mean position,the distance travelled in the first second is $d_1 = y_1 = \frac{A}{\sqrt{2}}$.
At $t=2 \ s$,the displacement is $y_2 = A \sin(\frac{\pi}{4} \times 2) = A \sin(\frac{\pi}{2}) = A$.
The distance travelled in the second second is $d_2 = y_2 - y_1 = A - \frac{A}{\sqrt{2}} = A(1 - \frac{1}{\sqrt{2}})$.
The ratio of the distances is $\frac{d_1}{d_2} = \frac{A/\sqrt{2}}{A(1 - 1/\sqrt{2})} = \frac{1/\sqrt{2}}{(\sqrt{2}-1)/\sqrt{2}} = \frac{1}{\sqrt{2}-1}$.

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

(C) When silicon is doped with an impurity having a valency less than $4$,a $p$-type semiconductor is obtained.
The valency of $As$,$P$,and $Sb$ is $+5$ (Group $15$ elements),while the valency of $In$ is $+3$ (Group $13$ element).
Thus,$In$ is the impurity which,when added to $Si$,creates a $p$-type semiconductor.

(B) The treatment of cellulose with concentrated $HNO_3$ (nitric acid) in the presence of concentrated $H_2SO_4$ results in the formation of cellulose nitrate.
Cellulose nitrate is a significant industrial chemical used in the manufacture of explosives,such as gun cotton,as well as in lacquers,paints,and certain medicinal applications.

(C) In the Haber's process for the synthesis of ammonia from $N_2$ and $H_2$,$Fe$ is used as a catalyst and $Mo$ acts as a promoter (activator).
The chemical equation is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons[Mo]{Fe} 2NH_{3(g)} + 22.4 \ kcal$.

(A) The structures of the given oxoacids of sulphur are as follows:
$1$. Pyrosulphurous acid $(H_2S_2O_5)$: Contains two $S$ atoms with oxidation states $+3$ and $+5$.
$2$. Hyposulphurous acid $(H_2S_2O_4)$: Contains two $S$ atoms with oxidation states $+3$ and $+3$.
$3$. Pyrosulphuric acid $(H_2S_2O_7)$: Contains two $S$ atoms with oxidation states $+6$ and $+6$.
$4$. Peroxodisulphuric acid $(H_2S_2O_8)$: Contains two $S$ atoms with oxidation states $+6$ and $+6$.
Thus,pyrosulphurous acid is the only oxoacid among the options that contains two sulphur atoms in different oxidation states.

(A) As the atomic size of noble gases increases from $He$ to $Xe$,the magnitude of the interatomic van der Waals' forces increases.
Stronger van der Waals' forces require more energy to overcome,which leads to an increase in the boiling point of the noble gases from $He$ to $Xe$.
Therefore,both $(A)$ and $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(C) The number of diagonals in a polygon with $n$ sides is given by the formula: $\frac{n(n-3)}{2}$.
Given that the number of diagonals is $170$,we have:
$\frac{n(n-3)}{2} = 170$
$n(n-3) = 340$
$n^2 - 3n - 340 = 0$
Solving the quadratic equation:
$n^2 - 20n + 17n - 340 = 0$
$n(n - 20) + 17(n - 20) = 0$
$(n - 20)(n + 17) = 0$
Since $n$ must be positive,$n = 20$.

(B) The given series is $S = 1 + \frac{1}{3 \cdot 2^2} + \frac{1}{5 \cdot 2^4} + \frac{1}{7 \cdot 2^6} + \ldots$
We know the logarithmic expansion $\log _e \left( \frac{1+x}{1-x} \right) = 2 \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots \right)$ for $|x| < 1$.
Multiplying the series by $2$,we get $2S = 2 + \frac{2}{3 \cdot 2^2} + \frac{2}{5 \cdot 2^4} + \ldots = 2 + \frac{1}{3 \cdot 2} + \frac{1}{5 \cdot 2^3} + \ldots$
This is not quite right. Let us rewrite the series as $S = 2 \left( \frac{1}{2} + \frac{(1/2)^3}{3 \cdot 2} + \ldots \right)$ is incorrect.
Let $x = 1/2$. The expansion is $2(x + x^3/3 + x^5/5 + \ldots) = \log _e \left( \frac{1+x}{1-x} \right)$.
Our series is $1 + \frac{1}{3 \cdot 2^2} + \frac{1}{5 \cdot 2^4} + \ldots = \frac{1}{x} (x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots)$ where $x = 1/2$.
So,$S = \frac{1}{x} \cdot \frac{1}{2} \log _e \left( \frac{1+x}{1-x} \right) = \frac{1}{1/2} \cdot \frac{1}{2} \log _e \left( \frac{1+1/2}{1-1/2} \right) = \log _e \left( \frac{3/2}{1/2} \right) = \log _e 3$.

(C) Given the equation: $\tan x + \tan \left(x + \frac{\pi}{3}\right) + \tan \left(x + \frac{2\pi}{3}\right) = 3$.
Using the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have:
$\tan x + \frac{\tan x + \sqrt{3}}{1 - \sqrt{3} \tan x} + \frac{\tan x - \sqrt{3}}{1 + \sqrt{3} \tan x} = 3$.
Combining the last two terms:
$\tan x + \frac{(\tan x + \sqrt{3})(1 + \sqrt{3} \tan x) + (\tan x - \sqrt{3})(1 - \sqrt{3} \tan x)}{1 - 3 \tan^2 x} = 3$.
Simplifying the numerator:
$(\tan x + \sqrt{3} \tan^2 x + \sqrt{3} + 3 \tan x) + (\tan x - \sqrt{3} \tan^2 x - \sqrt{3} + 3 \tan x) = 8 \tan x$.
So,$\tan x + \frac{8 \tan x}{1 - 3 \tan^2 x} = 3$.
$\frac{\tan x(1 - 3 \tan^2 x) + 8 \tan x}{1 - 3 \tan^2 x} = 3$.
$\frac{\tan x - 3 \tan^3 x + 8 \tan x}{1 - 3 \tan^2 x} = 3$.
$\frac{9 \tan x - 3 \tan^3 x}{1 - 3 \tan^2 x} = 3$.
Dividing both sides by $3$:
$\frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x} = 1$.
Since $\tan 3x = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x}$,we get $\tan 3x = 1$.

(B) The reaction of an alkyl aryl ether with $HI$ involves the cleavage of the $C-O$ bond. In the case of benzyl phenyl ether $(C_6H_5CH_2-O-C_6H_5)$,the $C-O$ bond between the benzyl carbon and oxygen is cleaved because the benzyl carbocation is stabilized by resonance. The oxygen atom remains attached to the phenyl ring,forming phenol $(C_6H_5OH)$,while the benzyl group forms benzyl iodide $(C_6H_5CH_2I)$. Therefore,$A = C_6H_5CH_2I$ and $B = C_6H_5OH$.

(A) The reaction of acetaldehyde $(CH_3CHO)$ with semicarbazide $(H_2N-NH-CO-NH_2)$ is a nucleophilic addition-elimination reaction.
In this reaction,the carbonyl oxygen of the aldehyde is replaced by the nitrogen of the semicarbazide,resulting in the formation of a semicarbazone.
The reaction is as follows:
$CH_3CHO + H_2N-NH-CO-NH_2 \rightarrow CH_3CH=N-NH-CO-NH_2 + H_2O$
The product formed is acetaldehyde semicarbazone,which corresponds to the structure $H_3C-CH=N-NH-CO-NH_2$.

(C) For a first order reaction,the integrated rate equation is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
Given that $93.75 \%$ of the reaction is complete,the remaining concentration is $[A]_t = [A]_0 - 0.9375[A]_0 = 0.0625[A]_0$.
Substituting this into the rate equation for $t = x$:
$k = \frac{2.303}{x} \log \frac{[A]_0}{0.0625[A]_0} = \frac{2.303}{x} \log(16) = \frac{2.303}{x} \log(2^4) = \frac{2.303 \times 4 \times \log(2)}{x} = \frac{4 \times 0.693}{x}$.
We know that the half-life $t_{1/2}$ is given by $k = \frac{0.693}{t_{1/2}}$.
Equating the two expressions for $k$:
$\frac{0.693}{t_{1/2}} = \frac{4 \times 0.693}{x}$.
Therefore,$t_{1/2} = \frac{x}{4}$ minutes.

(A) For a first order reaction,the rate constant $k$ is given by the formula:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$
Given values are:
$t = 5 \ min$
$[A]_0 = 0.6 \ mol \ L^{-1}$
$[A] = 0.2 \ mol \ L^{-1}$
Substituting the values in the formula:
$k = \frac{2.303}{5} \log \frac{0.6}{0.2}$
$k = \frac{2.303}{5} \log 3$
Using $\log 3 = 0.4771$:
$k = \frac{2.303 \times 0.4771}{5}$
$k = \frac{1.0988}{5} = 0.21976 \ min^{-1} \approx 0.219 \ min^{-1}$

(A) $4-$hydroxyacetanilide is commonly known as paracetamol.
It acts as an antipyretic (used to lower body temperature) and an analgesic (used to relieve pain).

(B) The given reaction is the Reimer-Tiemann reaction. In this reaction,phenol reacts with chloroform $(CHCl_3)$ in the presence of an aqueous base like sodium hydroxide $(NaOH)$ to form an ortho-substituted product,which is salicylaldehyde ($o$-hydroxybenzaldehyde). The mechanism involves the formation of a dichlorocarbene $(:CCl_2)$ intermediate,which acts as an electrophile and attacks the phenol ring.

(A) The reaction between acetaldehyde $(CH_3CHO)$ and semicarbazide $(H_2N-NHCONH_2)$ is a nucleophilic addition-elimination reaction.
In this reaction,the carbonyl oxygen of the acetaldehyde is replaced by the nitrogen atom of the semicarbazide,resulting in the formation of a semicarbazone derivative.
The reaction is as follows:
$CH_3CHO + H_2N-NHCONH_2 \rightarrow CH_3CH=N-NHCONH_2 + H_2O$
The product formed is acetaldehyde semicarbazone,which corresponds to the structure $H_3C-CH=N-NH-C(=O)-NH_2$.

(A) The compound $A$ is $CH_3COCH_3$ (acetone),which has the molecular formula $C_3H_6O$.
$1$. Reaction with $I_2/NaOH$ (Iodoform test): Acetone reacts with $I_2$ in the presence of $NaOH$ to give a yellow precipitate of iodoform $(CHI_3)$.
$CH_3COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 + CH_3COONa + 3NaI + 3H_2O$. Here,$B$ is $CHI_3$.
$2$. Reaction with $Zn-Hg/HCl$ (Clemmensen reduction): Acetone undergoes reduction to form propane $(CH_3CH_2CH_3)$.
$CH_3COCH_3 \xrightarrow{Zn-Hg/HCl} CH_3CH_2CH_3$. Here,$C$ is $CH_3CH_2CH_3$.

(A) $1$. The $-NO_2$ group is a deactivating and meta-directing group. Therefore,when nitrobenzene reacts with $Cl_2$ in the presence of $Fe$ (a Lewis acid catalyst),electrophilic aromatic substitution occurs at the meta position to form $m$-chloronitrobenzene $(A)$.
$2$. The reduction of nitrobenzene with $LiAlH_4$ is a complex process. While $LiAlH_4$ is a strong reducing agent,its reaction with nitrobenzene typically leads to the formation of azobenzene $(C_6H_5-N=N-C_6H_5)$ as the major product $(B)$.

(A) $4$-hydroxyacetanilide is commonly known as $Paracetamol$.
It acts as an $Antipyretic$ (used to reduce body temperature) and as an $Analgesic$ (used to relieve pain).

(D) Phosphorus $(P)$ is a non-metal and typically forms acidic oxides such as $P_2O_3$ and $P_4O_{10}$.
$Al_2O_3$ (Aluminum oxide),$SnO$ (Tin oxide),and $Sb_2O_3$ (Antimony oxide) are known to be amphoteric,meaning they can react with both acids and bases.
Therefore,$P$ cannot form an amphoteric oxide.

(B) When the complex is dissolved in water,it dissociates to give $3$ ions. This indicates that there are $2$ $Cl^{-}$ ions present outside the coordination sphere as counter ions.
Therefore,the formula of the complex must be $[Co(NH_3)_5Cl]Cl_2$.
The dissociation reaction is:
$[Co(NH_3)_5Cl]Cl_2 \xrightarrow{H_2O} [Co(NH_3)_5Cl]^{2+} + 2Cl^{-}$

(D) The standard cell potential is calculated as: $E_{\text{cell}}^{\circ} = E_{Cu^{2+} / Cu}^{\circ} - E_{Zn^{2+} / Zn}^{\circ} = 0.34 - (-0.76) = 1.1 \ V$.
Using the $Nernst$ equation: $E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
For the reaction $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$,the number of electrons transferred is $n = 2$.
Substituting the values: $E_{\text{cell}} = 1.1 - \frac{0.059}{2} \log \frac{0.1}{0.01} = 1.1 - 0.0295 \times \log(10)$.
Since $\log(10) = 1$,we get $E_{\text{cell}} = 1.1 - 0.0295 = 1.0705 \ V \approx 1.07 \ V$.

(A) The reduction reaction for $Al^{3+}$ is:
$Al^{3+} + 3e^{-} \longrightarrow Al$
From the stoichiometry,$1 \ mol$ of $Al$ $(27 \ g)$ requires $3 \ mol$ of electrons.
Therefore,the number of moles of electrons required for $36 \ g$ of $Al$ is:
$\text{Moles of electrons} = \frac{3 \ mol \ e^{-}}{27 \ g \ Al} \times 36 \ g \ Al = 4 \ mol \ e^{-}$

(D) In chlorobenzene,the chlorine atom exhibits both $-I$ (inductive) and $+M$ (mesomeric) effects.
Due to the $+M$ effect,the electron density increases at the ortho and para positions,making it ortho/para directing.
However,because of the strong $-I$ effect,the overall electron density of the benzene ring decreases,making it ring deactivating.
Therefore,the statement that $Cl$ is meta directing is incorrect.

(C) When silicon $(Si)$ is doped with an impurity having a valency less than $4$,a $p$-type semiconductor is obtained.
Silicon belongs to group $14$ and has a valency of $4$.
Elements like $As$ (Arsenic),$P$ (Phosphorus),and $Sb$ (Antimony) belong to group $15$ and have a valency of $5$,which results in $n$-type semiconductors.
$In$ (Indium) belongs to group $13$ and has a valency of $3$.
Therefore,doping $Si$ with $In$ creates electron holes,resulting in a $p$-type semiconductor.

(B) Cellulose is treated with concentrated $HNO_3$ in the presence of concentrated $H_2SO_4$ to produce cellulose nitrate.
Cellulose nitrate is an important substance used in the manufacture of explosives (gun cotton),lacquers,and plastics.

(C) In the Haber's process for the synthesis of ammonia from $N_2$ and $H_2$,$Fe$ is used as a catalyst and $Mo$ acts as a promoter (activator).
The chemical equation is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons[Mo]{Fe} 2NH_{3(g)} + 22.4 \text{ kcal}$.

(A) The structures of the given oxoacids of sulphur are as follows:
$1$. Pyrosulphurous acid $(H_2S_2O_5)$: Contains two $S$ atoms with oxidation states $+3$ and $+5$.
$2$. Hyposulphurous acid $(H_2S_2O_4)$: Contains two $S$ atoms with oxidation states $+3$ and $+3$.
$3$. Pyrosulphuric acid $(H_2S_2O_7)$: Contains two $S$ atoms with oxidation states $+6$ and $+6$.
$4$. Peroxodisulphuric acid $(H_2S_2O_8)$: Contains two $S$ atoms with oxidation states $+6$ and $+6$.
Thus,pyrosulphurous acid is the only oxoacid among the options where the two sulphur atoms have different oxidation states.

(A) Noble gases are monoatomic and held together by weak van der Waals' forces.
As the atomic size increases from $He$ to $Xe$,the magnitude of van der Waals' forces increases due to an increase in the number of electrons and polarizability.
Consequently,more energy is required to overcome these forces,leading to an increase in the boiling point.
Therefore,both $(A)$ and $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(B) Neoprene is a synthetic rubber formed by the polymerization of chloroprene.
Chloroprene is chemically known as $2-$chlorobuta$-1,3-$diene.
The reaction is as follows:
$n \ CH_2=C(Cl)-CH=CH_2 \xrightarrow{\text{Polymerization}} [-CH_2-C(Cl)=CH-CH_2-]_n$

(D) $Ni$ anode is used in the electrolytic extraction of $Na$ by Castner's process. In this process,molten $NaOH$ is electrolyzed using a $Ni$ anode and an iron cathode.

(C) The formula for relative lowering of vapour pressure is given by: $\frac{p^{\circ}-p_s}{p^{\circ}} = \frac{n_A}{n_A+n_B} = \frac{\frac{w_A}{m_A}}{\frac{w_A}{m_A}+\frac{w_B}{m_B}}$
Where $w_A$ and $m_A$ are the mass and molar mass of the solute,and $w_B$ and $m_B$ are the mass and molar mass of the solvent (water).
Given: $\frac{p^{\circ}-p_s}{p^{\circ}} = 0.02$,$m_A = 60 \ g/mol$,$w_B = 90 \ g$,$m_B = 18 \ g/mol$.
Substituting the values: $0.02 = \frac{\frac{w_A}{60}}{\frac{w_A}{60} + \frac{90}{18}}$
$0.02 = \frac{\frac{w_A}{60}}{\frac{w_A}{60} + 5}$
$0.02 \times (\frac{w_A}{60} + 5) = \frac{w_A}{60}$
$0.02 \times \frac{w_A}{60} + 0.1 = \frac{w_A}{60}$
$0.1 = \frac{w_A}{60} - 0.02 \times \frac{w_A}{60}$
$0.1 = \frac{w_A}{60} \times (1 - 0.02) = \frac{w_A}{60} \times 0.98$
$w_A = \frac{0.1 \times 60}{0.98} = \frac{6}{0.98} \approx 6.12 \ g$. Since the closest option is $6 \ g$,the correct answer is $C$.

(C) $BaCl_2$ is an electrolyte that dissociates in water as $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$.
Since the number of particles increases due to dissociation,the observed colligative property (elevation in boiling point) is higher than the theoretical value.
Because the molar mass is inversely proportional to the colligative property,the experimental molar mass is found to be less than the calculated (theoretical) molar mass.

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

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

(A) As the atomic size increases down the group,the $A-A$ bond length increases.
Since bond energy is inversely proportional to bond length,the bond energy decreases as the size of the halogen atom increases.
Therefore,the order of bond energy for the given halogens is $Cl_2 > Br_2 > I_2$.