MHT CET 2020 Chemistry Question Paper with Answer and Solution

(C) At the mean position,the velocity of the mass $M$ is maximum,given by $v_{max} = \omega_1 A_1 = \sqrt{\frac{k}{M}} A_1$.
When mass $m$ is placed on $M$,the total mass becomes $(M+m)$.
Since the collision is perfectly inelastic and occurs at the mean position,the momentum is conserved.
$M v_{max} = (M+m) v'_{max}$.
So,$v'_{max} = \frac{M}{M+m} v_{max}$.
The new angular frequency is $\omega_2 = \sqrt{\frac{k}{M+m}}$.
Since $v'_{max} = \omega_2 A_2$,we have $\frac{M}{M+m} (\omega_1 A_1) = \omega_2 A_2$.
Substituting the values: $\frac{M}{M+m} \sqrt{\frac{k}{M}} A_1 = \sqrt{\frac{k}{M+m}} A_2$.
$\frac{A_1}{A_2} = \frac{M+m}{M} \sqrt{\frac{k}{M+m}} \sqrt{\frac{M}{k}} = \frac{M+m}{M} \sqrt{\frac{M}{M+m}} = \sqrt{\frac{M+m}{M}} = \left(\frac{M+m}{M}\right)^{\frac{1}{2}}$.

(C) When the mass $M$ passes through the mean position,its velocity $V$ is maximum,given by $V = \omega_{1} A_{1} = \sqrt{\frac{k}{M}} A_{1}$.
When mass $m$ is placed on $M$,the momentum is conserved during the collision (as the impulsive force acts internally).
$M V = (M + m) V^{\prime}$,where $V^{\prime}$ is the new velocity at the mean position.
$V^{\prime} = \frac{M}{M+m} V$.
The new amplitude $A_{2}$ is determined by the new angular frequency $\omega_{2} = \sqrt{\frac{k}{M+m}}$.
Since $V^{\prime} = \omega_{2} A_{2}$,we have $A_{2} = \frac{V^{\prime}}{\omega_{2}} = \frac{M V}{(M+m) \sqrt{\frac{k}{M+m}}} = \frac{M \sqrt{\frac{k}{M}} A_{1}}{(M+m) \sqrt{\frac{k}{M+m}}} = \frac{M}{\sqrt{M}} \cdot \frac{\sqrt{M+m}}{M+m} A_{1} = \sqrt{\frac{M}{M+m}} A_{1}$.
Therefore,$\frac{A_{1}}{A_{2}} = \sqrt{\frac{M+m}{M}} = \left(\frac{M+m}{M}\right)^{\frac{1}{2}}$.

(D) The critical angle $c$ is given by the formula $\sin c = \frac{1}{\mu}$,where $\mu$ is the refractive index of the denser medium with respect to the rarer medium.
For the critical angle to be minimum,$\sin c$ must be minimum,which implies $\mu$ must be maximum.
Refractive indices are: $a\mu_w = 1.33$,$a\mu_g = 1.50$.
For light traveling from glass to air,$\mu = \frac{a\mu_g}{a\mu_a} = \frac{1.5}{1} = 1.5$.
For light traveling from water to glass,$\mu = \frac{a\mu_g}{a\mu_w} = \frac{1.5}{1.33} \approx 1.125$.
For light traveling from glass to water,$\mu = \frac{a\mu_w}{a\mu_g} = \frac{1.33}{1.5} < 1$ (Total internal reflection is not possible).
Comparing the values of $\mu$ for which total internal reflection is possible,the value $\mu = 1.5$ (glass to air) is the maximum.
Therefore,the critical angle is minimum for light traveling from glass to air.

(A) For a real image formed by a convex lens,the magnification $m$ is given by $m = -n$.
Since $m = \frac{v}{u}$,we have $\frac{v}{u} = -n$,which implies $u = -\frac{v}{n}$.
Using the lens formula $\frac{1}{F} = \frac{1}{v} - \frac{1}{u}$,we substitute $u$:
$\frac{1}{F} = \frac{1}{v} - \frac{1}{(-v/n)} = \frac{1}{v} + \frac{n}{v} = \frac{1+n}{v}$.
Therefore,the image distance $v = F(n+1)$.

(B) The magnifying power $m$ of an astronomical telescope in normal adjustment is given by the formula $m = \frac{f_o}{f_e}$,where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.
If the focal length of the eyepiece is doubled,the new focal length $f_e' = 2f_e$.
The new magnifying power $m'$ becomes $m' = \frac{f_o}{f_e'} = \frac{f_o}{2f_e} = \frac{1}{2} \left( \frac{f_o}{f_e} \right) = \frac{m}{2}$.
Therefore,the magnifying power becomes half of its original value.

(C) The rotational kinetic energy $K$ is given by $K = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.
Given that the kinetic energies are equal,$K_1 = K_2$.
Therefore,$\frac{L_1^2}{2I_1} = \frac{L_2^2}{2I_2}$.
Substituting $I_1 = I$ and $I_2 = 2I$:
$\frac{L_1^2}{2I} = \frac{L_2^2}{2(2I)}$
$\frac{L_1^2}{I} = \frac{L_2^2}{2I}$
$\frac{L_1^2}{L_2^2} = \frac{I}{2I} = \frac{1}{2}$
Taking the square root on both sides:
$\frac{L_1}{L_2} = \frac{1}{\sqrt{2}}$
Thus,the ratio of their angular momenta is $1: \sqrt{2}$.

(C) The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
Given that the frequency $f$ is halved,the angular velocity $\omega = 2\pi f$ is also halved,so $\omega_2 = \frac{\omega_1}{2}$.
Given that the kinetic energy is doubled,$K_2 = 2K_1$.
Using the ratio: $\frac{K_2}{K_1} = \frac{\frac{1}{2} I_2 \omega_2^2}{\frac{1}{2} I_1 \omega_1^2} \implies 2 = \frac{I_2}{I_1} \left( \frac{1}{2} \right)^2 \implies 2 = \frac{I_2}{I_1} \times \frac{1}{4} \implies \frac{I_2}{I_1} = 8$.
Angular momentum is given by $L = I\omega$.
The new angular momentum $L_2$ is $\frac{L_2}{L_1} = \frac{I_2 \omega_2}{I_1 \omega_1} = \left( \frac{I_2}{I_1} \right) \left( \frac{\omega_2}{\omega_1} \right) = 8 \times \frac{1}{2} = 4$.
Therefore,$L_2 = 4L$.

(D) The kinetic energy of rotation $K$ is related to angular momentum $L$ and moment of inertia $I$ by the formula $K = \frac{L^2}{2I}$.
Rearranging for angular momentum,we get $L = \sqrt{2KI}$.
Since the kinetic energies $K$ are equal for both bodies,we have $L \propto \sqrt{I}$.
Therefore,the ratio of the angular momenta of the two bodies is $\frac{L_1}{L_2} = \sqrt{\frac{I_1}{I_2}}$.
Given $I_1 = I$ and $I_2 = 2I$,we substitute these values:
$\frac{L_1}{L_2} = \sqrt{\frac{I}{2I}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Thus,the ratio is $1: \sqrt{2}$.

(D) Let the mass per unit length be $\lambda$. The mass of a ring is $M = \lambda \times (2\pi r)$.
For the first ring of radius $R$,mass $M_1 = \lambda (2\pi R)$. The moment of inertia is $I_1 = M_1 R^2 = \lambda (2\pi R) R^2 = 2\pi \lambda R^3$.
For the second ring of radius $nR$,mass $M_2 = \lambda (2\pi nR) = n M_1$. The moment of inertia is $I_2 = M_2 (nR)^2 = (n M_1) (n^2 R^2) = n^3 M_1 R^2 = n^3 I_1$.
Given the ratio $\frac{I_1}{I_2} = \frac{1}{8}$.
Substituting $I_2 = n^3 I_1$,we get $\frac{I_1}{n^3 I_1} = \frac{1}{n^3} = \frac{1}{8}$.
Therefore,$n^3 = 8$,which gives $n = 2$.

(D) The moment of inertia of the original disc of mass $M$ and radius $R$ about an axis passing through its center and perpendicular to its plane is $I_{\text{total}} = \frac{1}{2} MR^2$.
The area of the original disc is $A = \pi R^2$. The diameter of the hole is $R$,so its radius is $r = R/2$. The area of the hole is $A_{\text{hole}} = \pi (R/2)^2 = \frac{\pi R^2}{4} = \frac{A}{4}$.
Since the mass is proportional to the area,the mass of the removed part is $M_{\text{removed}} = \frac{M}{4}$.
The center of the hole is at a distance $d = R/2$ from the center of the original disc. Using the parallel axis theorem,the moment of inertia of the removed disc about the axis passing through the center of the original disc is:
$I_{\text{removed}} = I_{\text{cm}} + M_{\text{removed}} d^2$
$I_{\text{removed}} = \left[ \frac{1}{2} M_{\text{removed}} r^2 \right] + M_{\text{removed}} (R/2)^2$
$I_{\text{removed}} = \left[ \frac{1}{2} \left( \frac{M}{4} \right) \left( \frac{R}{2} \right)^2 \right] + \left( \frac{M}{4} \right) \left( \frac{R}{2} \right)^2$
$I_{\text{removed}} = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{MR^2 + 2MR^2}{32} = \frac{3MR^2}{32}$.
The moment of inertia of the remaining part is:
$I_{\text{remaining}} = I_{\text{total}} - I_{\text{removed}}$
$I_{\text{remaining}} = \frac{1}{2} MR^2 - \frac{3}{32} MR^2 = \frac{16 MR^2 - 3 MR^2}{32} = \frac{13 MR^2}{32}$.

(A) At the highest point of the vertical circle, the condition to prevent water from falling is that the gravitational force must be provided by the centripetal force at minimum velocity.
$mr\omega^2 = mg$
$\omega^2 = \frac{g}{r}$
$\omega = \sqrt{\frac{g}{r}}$
Since angular frequency $\omega = 2\pi f$, where $f$ is the frequency of revolution:
$2\pi f = \sqrt{\frac{g}{r}}$
$f = \frac{1}{2\pi} \sqrt{\frac{g}{r}}$

(A) The given circuit consists of a $NOR$ gate followed by a $NOT$ gate.
$1$. The output of the $NOR$ gate is $\overline{A+B}$.
$2$. This output is then passed through a $NOT$ gate (inverter),which performs the complement operation.
$3$. The final output $Y$ is the complement of the $NOR$ output: $Y = \overline{(\overline{A+B})} = A+B$.
$4$. The Boolean expression $A+B$ corresponds to the $OR$ gate.
Therefore,the combination of a $NOR$ gate and a $NOT$ gate acts as an $OR$ gate.

(B) The circuit consists of two $NOT$ gates ($G1$ and $G2$) followed by a $NOR$ gate. The inputs to the $NOT$ gates are $A$ and $B$. Thus,the outputs of the $NOT$ gates are $C = \overline{A}$ and $D = \overline{B}$. These are fed into a $NOR$ gate. The output $Y$ of the $NOR$ gate is given by $Y = \overline{C+D}$. Substituting the values of $C$ and $D$,we get $Y = \overline{\overline{A} + \overline{B}}$. By De Morgan's theorem,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$. This is the Boolean expression for an $AND$ gate. The truth table is as follows:

$A$	$B$	$C=\overline{A}$	$D=\overline{B}$	$Y=\overline{C+D}$
$0$	$0$	$1$	$1$	$0$
$0$	$1$	$1$	$0$	$0$
$1$	$0$	$0$	$1$	$0$
$1$	$1$	$0$	$0$	$1$

(A) The given circuit consists of a $NOR$ gate followed by a $NOT$ gate.
$1$. The output of the $NOR$ gate with inputs $A$ and $B$ is $C = \overline{A+B}$.
$2$. This output $C$ is then passed through a $NOT$ gate.
$3$. The final output $Y$ is the complement of $C$,so $Y = \overline{C} = \overline{\overline{A+B}}$.
$4$. By the law of double negation,$\overline{\overline{X}} = X$,therefore $Y = A+B$.
$5$. $A$ gate with the Boolean expression $Y = A+B$ is an $OR$ gate.

(A) The emissive power $P$ of a body is given by the Stefan-Boltzmann law: $P = \sigma e T^{4}$,where $\sigma$ is the Stefan-Boltzmann constant,$e$ is the emissivity,and $T$ is the absolute temperature.
Given that the emissive power of both spheres is the same,we have $P_{1} = P_{2}$.
Therefore,$\sigma e_{1} T_{1}^{4} = \sigma e_{2} T_{2}^{4}$.
This simplifies to $\frac{T_{1}^{4}}{T_{2}^{4}} = \frac{e_{2}}{e_{1}}$.
Given the ratio of emissivity $e_{1} : e_{2} = 1 : 4$,we have $\frac{e_{2}}{e_{1}} = \frac{4}{1} = 4$.
Substituting this into the equation: $\frac{T_{1}^{4}}{T_{2}^{4}} = 4$.
Taking the fourth root on both sides: $\frac{T_{1}}{T_{2}} = (4)^{1/4} = (2^{2})^{1/4} = 2^{1/2} = \sqrt{2}$.
Thus,the ratio $T_{1} : T_{2} = \sqrt{2} : 1$.

(A) The angular width of the central maximum for a single slit diffraction is given by $\theta = \frac{2 \lambda}{a}$,where $\lambda$ is the wavelength of light and $a$ is the width of the slit.
From this formula,it is clear that the angular width $\theta$ is inversely proportional to the slit width $a$ $(\theta \propto \frac{1}{a})$.
Therefore,as the slit width $a$ increases,the angular width $\theta$ of the central maximum decreases.

(A) The position of the $n^{\text{th}}$ bright fringe is given by $y_{n} = \frac{n \lambda_{1} D}{d}$. For the $5^{\text{th}}$ bright band,$y_{5} = \frac{5 \lambda_{1} D}{d}$.
The position of the $m^{\text{th}}$ dark fringe is given by $y'_{m} = \frac{(2m - 1) \lambda_{2} D}{2d}$. For the $6^{\text{th}}$ dark band,$y'_{6} = \frac{(2 \times 6 - 1) \lambda_{2} D}{2d} = \frac{11 \lambda_{2} D}{2d}$.
Since the bands coincide,$y_{5} = y'_{6}$.
$\frac{5 \lambda_{1} D}{d} = \frac{11 \lambda_{2} D}{2d}$.
Simplifying,$5 \lambda_{1} = \frac{11 \lambda_{2}}{2}$.
Therefore,$\frac{\lambda_{1}}{\lambda_{2}} = \frac{11}{10}$,which implies $\frac{\lambda_{2}}{\lambda_{1}} = \frac{10}{11}$.

(B) The linear width of the central maximum in a single-slit diffraction pattern is given by $w = \frac{2 \lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance from the slit to the screen,and $d$ is the slit width.
The angular width of the central maximum is defined as the angle subtended by the central maximum at the slit,given by $\theta = \frac{w}{D} = \frac{2 \lambda}{d}$.
From this formula,it is clear that the angular width depends only on the wavelength $\lambda$ and the slit width $d$. It does not depend on the distance $D$ between the slit and the screen.

(A) The general equation for a wave is $Y = A \sin(\omega t)$,where $\omega = 2 \pi f$.
Given $Y_1 = 0.25 \sin 316 t$,we have $\omega_1 = 316 \text{ rad/s}$.
Thus,$2 \pi f_1 = 316 \implies f_1 = \frac{316}{2 \pi} = \frac{158}{\pi} \text{ Hz}$.
Given $Y_2 = 0.25 \sin 310 t$,we have $\omega_2 = 310 \text{ rad/s}$.
Thus,$2 \pi f_2 = 310 \implies f_2 = \frac{310}{2 \pi} = \frac{155}{\pi} \text{ Hz}$.
The beat frequency is the difference between the two frequencies:
$\text{Beat frequency} = f_1 - f_2 = \frac{158}{\pi} - \frac{155}{\pi} = \frac{3}{\pi} \text{ Hz}$.

(D) The general equation for a wave is $Y = A \sin(\omega t)$,where $\omega = 2 \pi n$ and $n$ is the frequency in Hz.
For the first wave,$Y_{1} = 0.25 \sin(316 t)$,so $\omega_{1} = 316 \text{ rad/s}$.
The frequency $n_{1} = \frac{\omega_{1}}{2 \pi} = \frac{316}{2 \pi} \text{ Hz}$.
For the second wave,$Y_{2} = 0.25 \sin(310 t)$,so $\omega_{2} = 310 \text{ rad/s}$.
The frequency $n_{2} = \frac{\omega_{2}}{2 \pi} = \frac{310}{2 \pi} \text{ Hz}$.
The number of beats produced per second is the difference in frequencies: $n_{beat} = |n_{1} - n_{2}|$.
$n_{beat} = \frac{316}{2 \pi} - \frac{310}{2 \pi} = \frac{316 - 310}{2 \pi} = \frac{6}{2 \pi} = \frac{3}{\pi} \text{ beats/s}$.

(B) Let the length of the pipe be $\ell$ and the velocity of sound be $V$.
For an open organ pipe,the fundamental frequency is $f_{open} = \frac{V}{2\ell}$.
When one end is closed,the pipe becomes a closed organ pipe. The harmonics of a closed organ pipe are odd multiples of the fundamental frequency $f_{closed} = \frac{V}{4\ell}$.
The third harmonic of a closed organ pipe is $f_{3, closed} = 3 \times \frac{V}{4\ell} = \frac{3V}{4\ell}$.
According to the problem,$f_{3, closed} - f_{open} = 50 ~Hz$.
Substituting the expressions: $\frac{3V}{4\ell} - \frac{V}{2\ell} = 50$.
$\frac{3V - 2V}{4\ell} = 50 \implies \frac{V}{4\ell} = 50 ~Hz$.
We need to find the fundamental frequency of the open pipe,$f_{open} = \frac{V}{2\ell} = 2 \times \frac{V}{4\ell} = 2 \times 50 = 100 ~Hz$.

(A) The standard equation for a stationary wave is given by $y = A \sin(kx) \cos(\omega t)$.
Comparing this with the given equation $y = 10 \sin \frac{\pi x}{4} \cos 20 \pi t$,we find the wave number $k = \frac{\pi}{4} \ cm^{-1}$.
The wave number $k$ is related to the wavelength $\lambda$ by the formula $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$: $\frac{\pi}{4} = \frac{2\pi}{\lambda}$.
Solving for $\lambda$,we get $\lambda = 8 \ cm$.
The distance between two consecutive nodes in a stationary wave is equal to $\frac{\lambda}{2}$.
Therefore,the distance $= \frac{8 \ cm}{2} = 4 \ cm$.

(A) The angular frequency $\omega$ is given by $\omega = 2\pi f$.
Given $f = 50 ~Hz$,we have $\omega = 2 \times \pi \times 50 = 100\pi ~rad/s$.
The phase change $\Delta \phi$ for a time interval $\Delta t$ is given by the formula $\Delta \phi = \omega \Delta t$.
Substituting the given values,$\Delta \phi = (100\pi) \times (0.01) = \pi ~rad$.

(B) The wavelength $\lambda$ is given by $\lambda = \frac{v}{f} = \frac{350}{50} = 7 ~m$.
The distance $x$ covered by the wave in time $t = 0.01 ~s$ is $x = v \times t = 350 \times 0.01 = 3.5 ~m$.
The phase change $\Delta \phi$ is given by the formula $\Delta \phi = \frac{2\pi}{\lambda} \times x$.
Substituting the values,$\Delta \phi = \frac{2\pi}{7} \times 3.5 = \frac{2\pi}{7} \times \frac{7}{2} = \pi ~rad$.

(A) Given: Conductivity $(k)$ = $0.0112 \ \Omega^{-1} cm^{-1}$ and Resistance $(R)$ = $55.0 \ \Omega$.
The formula for cell constant $(G^*)$ is: $G^* = k \times R$.
Substituting the values: $G^* = 0.0112 \ \Omega^{-1} cm^{-1} \times 55.0 \ \Omega$.
Therefore,$G^* = 0.616 \ cm^{-1}$.

(B) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{1000 \times \kappa}{C}$
Here,conductivity $(\kappa)$ = $2 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$ and concentration $(C)$ = $0.08 \ M$.
Substituting the values: $\Lambda_m = \frac{1000 \times 2 \times 10^{-2}}{0.08} = \frac{20}{0.08} = 250 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.

(B) The conductivity $(k)$ of an electrolytic solution is defined as the product of the conductance $(G)$ and the cell constant $(G^*)$.
Since conductance $(G)$ is the reciprocal of electrical resistance $(R)$,we have $G = \frac{1}{R}$.
Substituting this into the formula for conductivity,we get $k = G \cdot G^* = \frac{1}{R} \cdot G^* = \frac{G^*}{R}$.
Therefore,the correct relation is $k = \frac{G^*}{R}$.

(A) Given: Molar conductivity $(\Lambda_m)$ = $124.3 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$,Conductivity $(\kappa)$ = $1.243 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$.
Formula: $\Lambda_m = \frac{1000 \times \kappa}{C}$,where $C$ is the concentration in $mol \ L^{-1}$.
Rearranging for $C$: $C = \frac{1000 \times \kappa}{\Lambda_m}$.
Calculation: $C = \frac{1000 \times 1.243 \times 10^{-4}}{124.3} = \frac{0.1243}{124.3} = 0.001 \ mol \ L^{-1}$.
Therefore,the correct option is $A$.

(D) Conductivity $(k) = \frac{1}{\text{resistivity}} = \frac{1}{2.5 \times 10^{-3} \ \Omega \ cm} = 400 \ \Omega^{-1} \ cm^{-1}$.
Molar conductivity $(\Lambda_{m}) = \frac{1000 \times k}{C}$.
Substituting the values: $\Lambda_{m} = \frac{1000 \times 400}{0.8} = \frac{400000}{0.8} = 5 \times 10^{5} \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.
Thus,the correct option is $D$.

(A) The cell constant $(G^*)$ is defined as the ratio of the distance between the electrodes $(\ell)$ to the area of cross-section $(a)$.
$G^* = \frac{\ell}{a}$
Given: $\ell = 0.98 \ cm$,$a = 1.96 \ cm^{2}$.
$G^* = \frac{0.98 \ cm}{1.96 \ cm^{2}} = 0.5 \ cm^{-1}$.
Therefore,the correct option is $A$.

(D) Given: Concentration $C = 0.01 \ M$,Molar conductivity $\Lambda_m = 400.0 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.
Formula: $\Lambda_m = \frac{1000 \times \kappa}{C}$,where $\kappa$ is the conductivity.
Rearranging for $\kappa$: $\kappa = \frac{\Lambda_m \times C}{1000}$.
Calculation: $\kappa = \frac{400.0 \times 0.01}{1000} = \frac{4}{1000} = 4.0 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$.

(D) Conductivity $(\kappa)$ is defined as the reciprocal of resistivity $(\rho)$.
$\kappa = \frac{1}{\rho}$

(B) Freons are chlorofluoro derivatives of methane and ethane. $CCl_2F_2$ (Dichloro difluoromethane),$CCl_3F$ (Trichloro fluoromethane),and $CHClF_2$ (Chloro difluoromethane) are common examples of freons. Diphenyl $(C_6H_5-C_6H_5)$ is an aromatic hydrocarbon and is not a freon.

(B) The structure of benzylamine is $C_6H_5CH_2NH_2$.
In the $IUPAC$ nomenclature,the parent chain is methane,which is substituted by a phenyl group and an amino group.
Therefore,the $IUPAC$ name is phenylmethanamine.

(C) The given compound is an ether: $CH_3-CH_2-O-CH_2-CH(CH_3)_2$.
To name it according to $IUPAC$ rules,we identify the longest carbon chain attached to the oxygen atom.
The longest chain has $3$ carbon atoms,which is a propane chain.
The ethoxy group $(-OCH_2CH_3)$ is attached to the first carbon of this chain.
There is a methyl group $(-CH_3)$ attached to the second carbon of the propane chain.
Thus,the $IUPAC$ name is $1-$ethoxy$-2-$methylpropane.

(B) The chemical formula for $n-hexadecyl$ trimethyl ammonium chloride is $[CH_3(CH_2)_{15}N(CH_3)_3]^+Cl^-$.
In this structure,there are $3$ methyl groups attached to the nitrogen atom (trimethyl part).
Additionally,there is $1$ methyl group at the end of the $n-hexadecyl$ chain.
Total number of methyl groups = $3 + 1 = 4$.

(B) The structure of $3-$chloropropyl ethyl ether is $CH_3-CH_2-O-CH_2-CH_2-CH_2-Cl$.
In $IUPAC$ nomenclature,the ether group is named as an alkoxy substituent.
The longest carbon chain containing the ether oxygen is a propane chain.
The ethoxy group $(-OCH_2CH_3)$ is attached to the $1^{st}$ carbon,and the chlorine atom is attached to the $3^{rd}$ carbon.
Therefore,the $IUPAC$ name is $1-$ethoxy$-3-$chloropropane,which is equivalent to $3-$chloro$-1-$ethoxypropane.

(C) The structure of mesityl oxide is $(CH_3)_2C=CHCOCH_3$.
To name this compound according to $IUPAC$ rules:
$1$. Identify the longest carbon chain containing the functional group (ketone) and the double bond. The chain has $5$ carbon atoms,so the parent alkane is pentane.
$2$. Number the chain starting from the end closer to the ketone group: $C_1$ is the methyl carbon,$C_2$ is the carbonyl carbon,$C_3$ is the $CH$ group,$C_4$ is the $C$ group,and $C_5$ is the methyl group attached to $C_4$.
$3$. The ketone is at position $2$,and the double bond starts at position $3$. There is a methyl group at position $4$.
$4$. Combining these,the name is $4-$methylpent$-3-$en$-2-$one.

(B) The structure of allyl chloride is $CH_{2}=CH-CH_{2}Cl$.
Counting the atoms in the structure:
There are $3$ carbon atoms $(C_{3})$.
There are $5$ hydrogen atoms $(H_{5})$.
There is $1$ chlorine atom $(Cl)$.
Therefore,the molecular formula is $C_{3}H_{5}Cl$.

(B) An allylic alcohol is one where the $-OH$ group is attached to a carbon atom adjacent to a carbon-carbon double bond.
For an allylic alcohol to be secondary,the carbon atom bearing the $-OH$ group must be attached to two other carbon atoms.
In $CH_2=CH-CH(OH)-CH_3$ (but$-3-$en$-2-$ol),the $-OH$ group is on the $C-2$ carbon,which is adjacent to the double bond $(C-3=C-4)$ and is attached to two other carbon atoms ($C-1$ and $C-3$).
Thus,but$-3-$en$-2-$ol is an allylic secondary alcohol.

(A) Neohexyl chloride is the common name for $3,3-$dimethylbutyl chloride,which has the structure $(CH_{3})_{3}C-CH_{2}-CH_{2}-Cl$.
Option $A$ represents $3,3-$dimethylbutyl chloride (neohexyl chloride).
Option $B$ represents $n-$hexyl chloride.
Option $C$ represents $5-$methylhexyl chloride.
Option $D$ represents $3,3-$dimethyl$-2-$chlorobutane.

(C) Mandelonitrile is the cyanohydrin of benzaldehyde.
Its chemical formula is $C_6H_5CH(OH)CN$.
It is also known as $\alpha$-hydroxyphenylacetonitrile or benzaldehyde cyanohydrin.

(B) An allylic alcohol is an alcohol in which the $-OH$ group is attached to a $sp^3$ hybridized carbon atom adjacent to a carbon-carbon double bond $(C=C)$.
In $But-3-en-2-ol$ $(CH_2=CH-CH(OH)-CH_3)$, the $-OH$ group is attached to the carbon atom adjacent to the double bond, which satisfies the definition of an allylic alcohol.

(A) The structure of $1-$chloro$-2,2-$dimethylpropane is $(CH_3)_3C-CH_2Cl$.
In this molecule,the carbon atom attached to the chlorine is a primary carbon,which is bonded to a quaternary carbon atom (a carbon bonded to four other carbons).
This structural arrangement,where a central carbon is bonded to three methyl groups and one $-CH_2Cl$ group,is commonly referred to as the $neo-$pentyl group.
Therefore,the common name for this compound is $neo-$pentyl chloride.

(C) The chemical formula of salicylaldehyde is $C_7H_6O_2$.
It consists of a benzene ring substituted with an aldehyde group $(-CHO)$ and a hydroxyl group $(-OH)$ at the ortho position.
Looking at the structure,there is one oxygen atom in the aldehyde group and one oxygen atom in the hydroxyl group.
Therefore,the total number of oxygen atoms present in salicylaldehyde is $1 + 1 = 2$.

(A) The correct answer is $A$ (Ether).
An ether functional group has the general structure $R-O-R'$,where $R$ and $R'$ are alkyl or aryl groups.
It does not contain a carbonyl group $(>C=O)$.
In contrast,esters $(R-CO-OR')$,amides $(R-CO-NH_2)$,and acyl halides $(R-CO-X)$ all contain the carbonyl group $(>C=O)$.

(C) benzylic alcohol is one where the hydroxyl group $(-OH)$ is attached to a carbon atom that is directly bonded to a benzene ring.
In a tertiary $(3^{\circ})$ alcohol,the carbon atom bearing the $-OH$ group is attached to three other carbon atoms.
Let's analyze the structures:
$A$: Phenyl methanol $(C_6H_5CH_2OH)$ - The carbon attached to $-OH$ is bonded to one carbon (the benzene ring),making it a primary $(1^{\circ})$ alcohol.
$B$: $1$-phenyl ethanol $(C_6H_5CH(OH)CH_3)$ - The carbon attached to $-OH$ is bonded to two carbons,making it a secondary $(2^{\circ})$ alcohol.
$C$: $2$-phenyl propan-$2$-ol $(C_6H_5C(CH_3)_2OH)$ - The carbon attached to $-OH$ is bonded to three carbons (two methyl groups and one phenyl group),making it a tertiary $(3^{\circ})$ alcohol.
$D$: $1$-phenyl propan-$2$-ol $(C_6H_5CH_2CH(OH)CH_3)$ - This is not a benzylic alcohol because the $-OH$ is not on the carbon directly attached to the benzene ring.
Therefore,the correct option is $C$.

(A) Hydroquinone is a phenolic compound with two hydroxyl $(-OH)$ groups attached to the benzene ring at the para positions ($1$ and $4$ positions).
Therefore,the $IUPAC$ name of hydroquinone is Benzene-$1,4$-diol.

(D) tricarboxylic acid is an organic compound that contains three carboxyl groups $(-COOH)$.
$1$. Valeric acid $(CH_3(CH_2)_3COOH)$ is a monocarboxylic acid.
$2$. Oxalic acid $(HOOC-COOH)$ is a dicarboxylic acid.
$3$. Caproic acid $(CH_3(CH_2)_4COOH)$ is a monocarboxylic acid.
$4$. Citric acid $(HOOC-CH_2-C(OH)(COOH)-CH_2-COOH)$ is a tricarboxylic acid.
Therefore,the correct option is $D$.

(D) Butylated hydroxytoluene $(BHT)$ has the structure $2,6-di-tert-butyl-4-methylphenol$.
Each $tert-butyl$ group $(-C(CH_3)_3)$ contains $3$ methyl groups.
There are two $tert-butyl$ groups,so $2 \times 3 = 6$ methyl groups.
Additionally,there is one methyl group attached to the para position of the benzene ring.
Total number of methyl groups = $6 + 1 = 7$.

(A) Glyceraldehyde is an aldotriose with the structural formula $HOCH_{2}-CH(OH)-CHO$.
Counting the atoms in the structure: there are $3$ carbon atoms,$6$ hydrogen atoms,and $3$ oxygen atoms.
Therefore,the molecular formula is $C_{3}H_{6}O_{3}$.

(D) $CFC$s (Chlorofluorocarbons) are widely used as refrigerants in refrigerators and air conditioning systems.
$Dichlorodifluoromethane$ $(CCl_2F_2)$,also known as $Freon-12$,is a common $CFC$ used for this purpose.

(B) In the carbonyl group,the $C$ atom is $sp^2$ hybridised and forms three sigma bonds.
The $C-C-O$ bond angle is approximately $120^{\circ}$.
The $C=O$ bond is stronger than the $C=C$ bond in alkenes because oxygen has higher electronegativity than carbon,which results in partial ionic character in the $C=O$ bond,making it stronger.

(D) Glycolic acid,also known as hydroxyacetic acid,has the chemical formula $HO-CH_2-COOH$.
It consists of a carboxylic acid group $(-COOH)$ attached to a hydroxymethyl group $(-CH_2OH)$.

(A) The acidic strength of aromatic carboxylic acids is influenced by the electronic effect of the substituent group attached to the benzene ring.
Electron-withdrawing groups $(EWG)$ like $-Cl$,$-CN$,and $-NO_{2}$ stabilize the carboxylate anion through inductive and/or resonance effects,thereby increasing the acidic strength.
Conversely,electron-donating groups $(EDG)$ like $-NH_{2}$ destabilize the carboxylate anion by increasing electron density,which decreases the acidic strength of the aromatic carboxylic acid.
Therefore,$-NH_{2}$ is the group that decreases the acidic strength.

(C) The molecular formula $C_4H_9X$ corresponds to the alkyl group derived from butane $(C_4H_{10})$.
There are two isomeric alkanes with the formula $C_4H_{10}$: $n$-butane $(CH_3-CH_2-CH_2-CH_3)$ and isobutane $(CH_3-CH(CH_3)-CH_3)$.
For $n$-butane,there are two non-equivalent sets of hydrogen atoms,leading to two isomers: $1$-halobutane and $2$-halobutane.
For isobutane,there are two non-equivalent sets of hydrogen atoms,leading to two isomers: $1$-halo-$2$-methylpropane and $2$-halo-$2$-methylpropane.
Thus,the total number of possible monohalogen derivatives is $2 + 2 = 4$.

(B) The boiling point of amines depends on the extent of intermolecular $H$-bonding.
$1^{\circ}$-amines (primary amines) have two $H$-atoms attached to the nitrogen atom,allowing for extensive intermolecular $H$-bonding.
$n$-butylamine $(CH_3CH_2CH_2CH_2NH_2)$ is a $1^{\circ}$-amine,whereas diethylamine is a $2^{\circ}$-amine and ethyldimethylamine is a $3^{\circ}$-amine.
$tert$-butylamine is also a $1^{\circ}$-amine,but its branched structure reduces the surface area for van der Waals forces compared to the straight-chain $n$-butylamine.
Therefore,$n$-butylamine has the highest boiling point.

(B) primary amine has the general structure $R-NH_2$. For the molecular formula $C_4H_{11}N$,the alkyl group $R$ must be $C_4H_9$.
The possible isomers for the butyl group $(C_4H_9-)$ are:
$1$. $n$-butylamine: $CH_3CH_2CH_2CH_2NH_2$
$2$. Isobutylamine: $(CH_3)_2CHCH_2NH_2$
$3$. sec-butylamine: $CH_3CH_2CH(NH_2)CH_3$
$4$. tert-butylamine: $(CH_3)_3CNH_2$
Thus,there are $4$ possible primary amines.

(B) The correct answer is $Chromatography$.
Adsorption chromatography is based on the principle of selective adsorption of components from a mixture onto a stationary phase.
It involves a stationary phase (solid) and a mobile phase (liquid or gas).
The separation occurs due to the difference in the extent to which different components are adsorbed on the stationary phase.

(A) The dry distillation of calcium salts of carboxylic acids is a standard method for the preparation of ketones.
For calcium propionate,$(CH_3CH_2COO)_2Ca$,the reaction proceeds as follows:
$(CH_3CH_2COO)_2Ca \xrightarrow{\text{Dry Distillation}} CaCO_3 + CH_3CH_2COCH_2CH_3$
The product formed is $CH_3CH_2COCH_2CH_3$,which is diethyl ketone.
Diethyl ketone is also known as pentan-$3$-one.

(D) In gas chromatography,a carrier gas is an inert gas used to transport the sample through the column.
Commonly used carrier gases include helium $(He)$,nitrogen $(N_2)$,hydrogen $(H_2)$,and argon $(Ar)$.
Helium and nitrogen are the most frequently used gases,and helium is particularly preferred when using a capillary column due to its efficiency.

(A) The mineral of aluminium among the given options is $Diaspore$.
$Diaspore$ is an aluminium oxide hydroxide mineral with the chemical formula $\alpha-AlO(OH)$.
$Limonite$ is an iron ore,$Azurite$ is a copper mineral,and $Chalcopyrite$ is a copper iron sulfide mineral.

(A) Calamine is an ore of zinc.
It is zinc carbonate with the chemical formula $ZnCO_{3}$.

(D) Pyrolusite is a mineral consisting essentially of manganese dioxide $(MnO_{2})$.
It is the most important ore of manganese.

(D) The manufacturing process of $K_{2}Cr_{2}O_{7}$ from chromite ore $(FeCr_{2}O_{4})$ involves three main steps:
$1$. Conversion of chromite ore into sodium chromate:
$4FeCr_{2}O_{4} + 8Na_{2}CO_{3} + 7O_{2} \longrightarrow 2Fe_{2}O_{3} + 8CO_{2} + 8Na_{2}CrO_{4}$
$2$. Conversion of sodium chromate into sodium dichromate:
$2Na_{2}CrO_{4} + H_{2}SO_{4} \longrightarrow Na_{2}Cr_{2}O_{7} + Na_{2}SO_{4} + H_{2}O$
$3$. Conversion of sodium dichromate to potassium dichromate:
$Na_{2}Cr_{2}O_{7} + 2KCl \longrightarrow K_{2}Cr_{2}O_{7} + 2NaCl$
In the first step,$Na_{2}CrO_{4}$ (sodium chromate) is produced. This $Na_{2}CrO_{4}$ is then used as a reactant in the second step to produce $Na_{2}Cr_{2}O_{7}$.

(D) .
Duralumin is an alloy,which is a trade name for the earliest types of age-hardenable aluminum alloys.
It is composed of approximately $90 \% \ Al$,$4 \% \ Cu$,$1 \% \ Mg$,and $0.5 \% - 1 \% \ Mn$.
It is known for being a very hard and lightweight alloy.
Therefore,option $D$ is correct.

(C) The correct answer is $C$ (Carnotite).
Carnotite is a mineral that contains vanadium. Its chemical formula is approximately $K_2(UO_2)_2(VO_4)_2 \cdot 3H_2O$.
It contains approximately $53 \% \text{ uranium}$,$12 \% \text{ vanadium}$,and trace quantities of radium.
It is a radioactive,bright-yellow,soft,and earthy mineral that serves as a significant source of uranium and vanadium.

(D) The chemical formula for chalcopyrite is $CuFeS_{2}$.
It is a copper iron sulphide mineral.
Therefore,the elements present in chalcopyrite are $Cu$,$Fe$,and $S$.

(C) The chemical name of cryolite is sodium hexafluoroaluminate and its chemical formula is $Na_{3}AlF_{6}$.
It is a halide mineral that contains aluminium.
Magnesite is $MgCO_{3}$,Haematite is $Fe_{2}O_{3}$,and Siderite is $FeCO_{3}$.
Cryolite is used as a solvent for bauxite in the electrolytic production of aluminium.

(B) Cassiterite $(SnO_2)$ is the principal ore of tin.
It often contains magnetic impurities of wolframite,which is a mixture of iron tungstate $(FeWO_4)$ and manganese tungstate $(MnWO_4)$.
These magnetic impurities are separated from the non-magnetic $SnO_2$ ore using electromagnetic separation.

(B) Azurite is a soft,deep-blue copper mineral with the chemical formula $Cu_3(CO_3)_2(OH)_2$.
It is produced by the weathering of copper ore deposits.
Carnotite is an ore of uranium,Pyrolusite is an ore of manganese $(MnO_2)$,and Chromite is an ore of chromium $(FeCr_2O_4)$.

(D) The correct answer is $Galena$.
$Galena$ $(PbS)$ is a sulphide ore.
The froth flotation process is specifically used for the concentration of sulphide ores because sulphide ores are preferentially wetted by oil,while gangue particles are wetted by water.
Since $Galena$ is lead$(II)$ sulphide,it is concentrated using this method.

(A) The zone refining process is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.
- This process is used to obtain metals of very high purity.
- Elements like $Germanium$,$Silicon$,$Boron$,$Gallium$,and $Indium$ are purified using this method.
- Therefore,$Germanium$ is the correct answer.

(D) The $Van \ Arkel$ method is used for the refining of $Zr$ and $Ti$.
The $Van \ Arkel$ process involves heating the impure metal with iodine to form a volatile complex.
This complex,upon decomposition at high temperatures,yields the pure metal.
For $Zirconium$:
$Zr \text{ (impure)} + 2I_2 \rightarrow ZrI_4$
$ZrI_4 \rightarrow Zr \text{ (pure)} + 2I_2 \text{ (at } 1800 \ K)$
For $Titanium$:
$Ti \text{ (impure)} + 2I_2 \rightarrow TiI_4$
$TiI_4 \rightarrow Ti \text{ (pure)} + 2I_2 \text{ (at } 1800 \ K)$