MHT CET 2023 Chemistry Question Paper with Answer and Solution

(B) The particle moves with a constant speed $v$ in a circular path. Let the initial velocity be $\vec{v}_i = v\hat{i}$.
After moving to the diametrically opposite point,the velocity becomes $\vec{v}_f = -v\hat{i}$.
The change in momentum is $\Delta \vec{p} = m\vec{v}_f - m\vec{v}_i = m(-v\hat{i}) - m(v\hat{i}) = -2mv\hat{i}$.
The magnitude of the change in momentum is $|\Delta \vec{p}| = 2mv$.
Since the speed $v$ is uniform,the kinetic energy $K = (1/2)mv^2$ remains constant throughout the motion.
Therefore,the change in kinetic energy is zero.
Thus,the correct option is $B$.

(A) According to the First Law of Thermodynamics $(FLOT)$: $\Delta Q = \Delta U + \Delta W$.
Here,$\Delta Q$ is the heat added to the system,$\Delta U$ is the change in internal energy,and $\Delta W$ is the work done by the gas.
Given: Pressure $P = 50 \ N/m^2$,initial volume $V_i = 10 \ m^3$,final volume $V_f = 4 \ m^3$,and heat added $\Delta Q = 100 \ J$.
The work done by the gas is $\Delta W = P \Delta V = P(V_f - V_i) = 50 \times (4 - 10) = 50 \times (-6) = -300 \ J$.
Substituting these values into the $FLOT$ equation: $100 = \Delta U + (-300)$.
Therefore,$\Delta U = 100 + 300 = 400 \ J$.
Since $\Delta U$ is positive,the internal energy increases by $400 \ J$.

(C) The coefficient of performance $(K)$ of an ideal refrigerator is given by the formula: $K = \frac{T_2}{T_1 - T_2}$,where $T_2$ is the temperature of the freezer and $T_1$ is the temperature of the surroundings.
Given: $T_2 = -13^{\circ}C = 273 - 13 = 260K$ and $K = 5$.
Substituting the values into the formula: $5 = \frac{260}{T_1 - 260}$.
Multiplying both sides by $(T_1 - 260)$: $5(T_1 - 260) = 260$.
$5T_1 - 1300 = 260$.
$5T_1 = 1560$.
$T_1 = \frac{1560}{5} = 312K$.
Converting back to Celsius: $T_1 = 312 - 273 = 39^{\circ}C$.

(A) Given: Charge $q = 3\, C$,Force $F = 3000\, N$,Distance $d = 1\, cm = 10^{-2}\, m$.
First,calculate the electric field intensity $E$ using the formula $F = qE$:
$E = \frac{F}{q} = \frac{3000}{3} = 1000\, N/C$.
The potential difference $V$ between two points in a uniform electric field is given by $V = E \times d$:
$V = 1000 \times 10^{-2} = 10\, V$.
Therefore,the potential difference is $10\, V$.

(C) The initial capacitance of the air-filled capacitor is $C = \frac{\varepsilon_0 A}{d} = 10\,\mu F$.
When the area is divided into two equal halves $(A_1 = A_2 = A/2)$ and filled with dielectrics,the two parts act as capacitors in parallel.
The equivalent capacitance $C_{eq}$ is given by $C_{eq} = C_1 + C_2$.
Here,$C_1 = \frac{k_1 \varepsilon_0 (A/2)}{d} = k_1 \times \frac{C}{2} = 2 \times \frac{10}{2} = 10\,\mu F$.
And $C_2 = \frac{k_2 \varepsilon_0 (A/2)}{d} = k_2 \times \frac{C}{2} = 4 \times \frac{10}{2} = 20\,\mu F$.
Therefore,$C_{eq} = 10 + 20 = 30\,\mu F$.

(C) Initial energy of the system is $U_i = \frac{1}{2}CV_1^2 + \frac{1}{2}CV_2^2$.
When the capacitors are connected in parallel,the common potential is $V = \frac{CV_1 + CV_2}{2C} = \frac{V_1 + V_2}{2}$.
Final energy of the system is $U_f = \frac{1}{2}(2C)V^2 = C \left( \frac{V_1 + V_2}{2} \right)^2 = \frac{1}{4}C(V_1 + V_2)^2$.
The decrease in energy is $\Delta U = U_i - U_f = \frac{1}{2}C(V_1^2 + V_2^2) - \frac{1}{4}C(V_1 + V_2)^2$.
Simplifying this,$\Delta U = \frac{1}{4}C [2V_1^2 + 2V_2^2 - (V_1^2 + V_2^2 + 2V_1V_2)] = \frac{1}{4}C(V_1^2 + V_2^2 - 2V_1V_2) = \frac{1}{4}C(V_1 - V_2)^2$.

(A) The average power consumed in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$,where $\phi$ is the phase difference between voltage and current.
If the current is wattless,it means the average power consumed in the circuit is zero $(P = 0)$.
Substituting $P = 0$ into the power formula: $0 = V_{rms} I_{rms} \cos \phi$.
Since $V_{rms}$ and $I_{rms}$ are not zero,we must have $\cos \phi = 0$.
This implies that $\phi = 90^o$ (or $\pi/2$ radians).
Therefore,the phase difference between virtual voltage and virtual current is $90^o$.

(C) When the reverse bias voltage applied to a $PN-$junction diode is increased to a sufficiently large value,the electric field across the junction becomes strong enough to break the covalent bonds of the semiconductor atoms. This leads to the generation of a large number of charge carriers,causing a sudden and sharp increase in the reverse current. This phenomenon is known as the breakdown of the $PN-$junction diode.

(D) The distance between the first dark fringes on either side of the central bright fringe is equal to the width of the central maximum.
The formula for the width of the central maximum in a single-slit diffraction pattern is given by $w = \frac{2\lambda D}{d}$.
Given:
$\lambda = 600\, nm = 600 \times 10^{-9}\, m$
$d = 1\, mm = 1 \times 10^{-3}\, m$
$D = 2\, m$
Substituting the values:
$w = \frac{2 \times 600 \times 10^{-9} \times 2}{1 \times 10^{-3}}$
$w = \frac{2400 \times 10^{-9}}{10^{-3}}$
$w = 2400 \times 10^{-6}\, m = 2.4 \times 10^{-3}\, m = 2.4\, mm$.
Therefore, the correct option is $D$.

(D) Given the constraint $x + 2y = 8$,we can express $y$ in terms of $x$ as $y = \frac{8 - x}{2}$.
Let the function to be maximized be $f(x) = xy$. Substituting the expression for $y$,we get $f(x) = x \left( \frac{8 - x}{2} \right) = 4x - \frac{x^2}{2}$.
To find the critical points,we calculate the first derivative: $f'(x) = \frac{d}{dx} (4x - \frac{x^2}{2}) = 4 - x$.
Setting $f'(x) = 0$,we get $4 - x = 0$,which implies $x = 4$.
Substituting $x = 4$ into the constraint equation,we find $y = \frac{8 - 4}{2} = 2$.
To verify that this is a maximum,we check the second derivative: $f''(x) = -1$. Since $f''(x) < 0$,the function attains a local maximum at $x = 4$.
The maximum value is $f(4) = 4 \times 2 = 8$.

(A) Given the curve $y = a\sqrt{x} + bx$.
Since the curve passes through $(1, 2)$,we have $2 = a(1)^{1/2} + b(1)$,which implies $a + b = 2$ ... $(i)$.
The area bounded by the curve,the line $x = 4$,and the $x$-axis is given by $\int_{0}^{4} (a\sqrt{x} + bx) dx = 8$.
Evaluating the integral: $\left[ \frac{2a}{3}x^{3/2} + \frac{b}{2}x^2 \right]_{0}^{4} = 8$.
Substituting the limits: $\frac{2a}{3}(8) + \frac{b}{2}(16) = 8$.
This simplifies to $\frac{16a}{3} + 8b = 8$.
Dividing by $8$,we get $\frac{2a}{3} + b = 1$,or $2a + 3b = 3$ ... $(ii)$.
From $(i)$,$b = 2 - a$. Substituting into $(ii)$: $2a + 3(2 - a) = 3$.
$2a + 6 - 3a = 3$,which gives $-a = -3$,so $a = 3$.
Then $b = 2 - 3 = -1$.
Thus,$a = 3$ and $b = -1$.

(A) The equation of a circle passing through the origin with its center on the $y$-axis is given by ${x^2} + {(y - a)^2} = {a^2}$,where $(0, a)$ is the center.
This simplifies to ${x^2} + {y^2} - 2ay = 0$.
Differentiating both sides with respect to $x$,we get $2x + 2y\frac{{dy}}{{dx}} - 2a\frac{{dy}}{{dx}} = 0$.
From this,we find $a\frac{{dy}}{{dx}} = x + y\frac{{dy}}{{dx}}$,so $a = \frac{{x + y\frac{{dy}}{{dx}}}}{\frac{{dy}}{{dx}}} = x\frac{{dx}}{{dy}} + y$.
Substituting the value of $2a$ into the original equation ${x^2} + {y^2} - 2ay = 0$:
${x^2} + {y^2} - 2y(x\frac{{dx}}{{dy}} + y) = 0$.
${x^2} + {y^2} - 2xy\frac{{dx}}{{dy}} - 2{y^2} = 0$.
${x^2} - {y^2} - 2xy\frac{{dx}}{{dy}} = 0$.
Multiplying by $\frac{{dy}}{{dx}}$,we get $({x^2} - {y^2})\frac{{dy}}{{dx}} - 2xy = 0$.

(A) यदि तीन सदिश $\vec{a}, \vec{b}, \vec{c}$ समतलीय हैं,तो उनका अदिश त्रिक गुणनफल (Scalar Triple Product) शून्य होता है,अर्थात $[\vec{a} \vec{b} \vec{c}] = 0$
सारणिक रूप में:
$\begin{vmatrix} p & 1 & 1 \\ 1 & q & 1 \\ 1 & 1 & r \end{vmatrix} = 0$
$R_1 \to R_1 - R_2$ और $R_2 \to R_2 - R_3$ का उपयोग करने पर:
$\begin{vmatrix} p-1 & 1-q & 0 \\ 0 & q-1 & 1-r \\ 1 & 1 & r \end{vmatrix} = 0$
विस्तार करने पर:
$(p-1)[(q-1)r - (1-r)] - (1-q)[0 - (1-r)] = 0$
$(p-1)(qr - q - 1 + r) + (1-q)(1-r) = 0$
$(p-1)(qr - q - 1 + r) + (1 - r - q + qr) = 0$
$(p-1)(qr - q - 1 + r) + 1(qr - q - 1 + r) = 0$
$(p-1+1)(qr - q - 1 + r) = 0$
$p(qr - q - 1 + r) = 0$
$pqr - pq - p + pr = 0$
हमें दिया गया है कि $p \neq 1, q \neq 1, r \neq 1$।
सारणिक को हल करने पर: $pqr - (p+q+r) + 2 = 0$
अतः,$pqr - (p+q+r) = -2$।

(A) Given $\bar{a} \times (\bar{b} \times \bar{c}) = \frac{\bar{b} + \bar{c}}{\sqrt{2}}$.
Using the vector triple product formula $\bar{a} \times (\bar{b} \times \bar{c}) = (\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c}$.
So,$(\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c} = \frac{1}{\sqrt{2}}\bar{b} + \frac{1}{\sqrt{2}}\bar{c}$.
Rearranging terms: $(\bar{a} \cdot \bar{c} - \frac{1}{\sqrt{2}})\bar{b} - (\bar{a} \cdot \bar{b} + \frac{1}{\sqrt{2}})\bar{c} = 0$.
Since $\bar{b}$ and $\bar{c}$ are non-coplanar (and thus linearly independent),their coefficients must be zero.
Therefore,$\bar{a} \cdot \bar{c} = \frac{1}{\sqrt{2}}$ and $\bar{a} \cdot \bar{b} = -\frac{1}{\sqrt{2}}$.
Since $\bar{a}$ and $\bar{b}$ are unit vectors,$|\bar{a}| = 1$ and $|\bar{b}| = 1$.
The angle $\theta$ between $\bar{a}$ and $\bar{b}$ is given by $\cos \theta = \frac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|} = \frac{-1/\sqrt{2}}{1 \times 1} = -\frac{1}{\sqrt{2}}$.
Thus,$\theta = \cos^{-1}(-\frac{1}{\sqrt{2}}) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

(B) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by: $eV_s = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function of the metal.
For wavelength $\lambda$,$e(4.8) = \frac{hc}{\lambda} - \phi$ --- $(1)$
For wavelength $2\lambda$,$e(1.6) = \frac{hc}{2\lambda} - \phi$ --- $(2)$
Subtracting equation $(2)$ from $(1)$:
$e(4.8 - 1.6) = \frac{hc}{\lambda} - \frac{hc}{2\lambda}$
$3.2e = \frac{hc}{2\lambda}$ --- $(3)$
From $(1)$,$\phi = \frac{hc}{\lambda} - 4.8e$. Since $\frac{hc}{\lambda} = 2 \times 3.2e = 6.4e$ (from eq $3$),
$\phi = 6.4e - 4.8e = 1.6e$.
The threshold wavelength $\lambda_0$ is given by $\phi = \frac{hc}{\lambda_0}$.
Since $\phi = 1.6e$ and $\frac{hc}{\lambda} = 6.4e$,we have $\frac{hc}{\lambda_0} = \frac{1}{4} \times \frac{hc}{\lambda}$.
Therefore,$\lambda_0 = 4\lambda$.

(D) Given $\vec{a} = \frac{1}{\sqrt{10}}(3\hat{i} + \hat{k})$ and $\vec{b} = \frac{1}{7}(2\hat{i} + 3\hat{j} - 6\hat{k})$.
First,note that $|\vec{a}|^2 = \frac{1}{10}(3^2 + 1^2) = 1$ and $|\vec{b}|^2 = \frac{1}{49}(2^2 + 3^2 + (-6)^2) = \frac{4+9+36}{49} = 1$.
Also,$\vec{a} \cdot \vec{b} = \frac{1}{7\sqrt{10}}(3(2) + 0(3) + 1(-6)) = 0$.
Now,expand the expression: $(2\vec{a} - \vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} + 2\vec{b})]$.
Using the vector triple product formula $\vec{u} \times (\vec{v} + \vec{w}) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w}$,we have:
$(\vec{a} \times \vec{b}) \times (\vec{a} + 2\vec{b}) = (\vec{a} \times \vec{b}) \times \vec{a} + 2((\vec{a} \times \vec{b}) \times \vec{b})$.
Using $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$,we get:
$(\vec{a} \times \vec{b}) \times \vec{a} = -(\vec{a} \times (\vec{a} \times \vec{b})) = -((\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}) = - (0\vec{a} - 1\vec{b}) = \vec{b}$.
$2((\vec{a} \times \vec{b}) \times \vec{b}) = 2((\vec{a} \cdot \vec{b})\vec{b} - (\vec{b} \cdot \vec{b})\vec{a}) = 2(0\vec{b} - 1\vec{a}) = -2\vec{a}$.
So,the expression becomes $(2\vec{a} - \vec{b}) \cdot (\vec{b} - 2\vec{a})$.
$= -(2\vec{a} - \vec{b}) \cdot (2\vec{a} - \vec{b}) = -|2\vec{a} - \vec{b}|^2$.
$= -(4|\vec{a}|^2 + |\vec{b}|^2 - 4\vec{a} \cdot \vec{b}) = -(4(1) + 1 - 4(0)) = -5$.

(B) Let the slopes of $PQ$ and $PR$ be $m_1$ and $m_2$. Since $\triangle PQR$ is a right-angled isosceles triangle at $P$,the lines $PQ$ and $PR$ are perpendicular,so $m_1 m_2 = -1$. Also,the angle between $PQ$ and $QR$ is $45^\circ$. The slope of $QR$ is $m = -2$. Using the formula $\tan 45^\circ = |\frac{m_1 - (-2)}{1 + m_1(-2)}| = 1$,we get $|\frac{m_1 + 2}{1 - 2m_1}| = 1$. This gives $m_1 + 2 = 1 - 2m_1$ or $m_1 + 2 = -(1 - 2m_1)$. Solving $3m_1 = -1$ gives $m_1 = -1/3$,and $m_1 + 2 = -1 + 2m_1$ gives $m_1 = 3$. Thus,the slopes are $3$ and $-1/3$. The equations of the lines passing through $P(2, 1)$ are $y - 1 = 3(x - 2)$ and $y - 1 = -1/3(x - 2)$. Rewriting these as $3x - y - 5 = 0$ and $x + 3y - 5 = 0$. The combined equation is $(3x - y - 5)(x + 3y - 5) = 0$. Expanding this: $3x^2 + 9xy - 15x - xy - 3y^2 + 5y - 5x - 15y + 25 = 0$,which simplifies to $3x^2 - 3y^2 + 8xy - 20x - 10y + 25 = 0$.

(B) The two springs on the left side,each having a spring constant of $2k$,are connected in series. Their equivalent spring constant $k_{eq1}$ is given by:
$\frac{1}{k_{eq1}} = \frac{1}{2k} + \frac{1}{2k} = \frac{2}{2k} = \frac{1}{k} \implies k_{eq1} = k$.
The two springs on the right side of mass $M$ are connected in parallel. Their effective spring constant $k_{eq2}$ is:
$k_{eq2} = k + 2k = 3k$.
Now,the mass $M$ is connected to these two equivalent systems ($k_{eq1}$ and $k_{eq2}$) in parallel. Therefore,the net equivalent spring constant $k_{net}$ of the system is:
$k_{net} = k_{eq1} + k_{eq2} = k + 3k = 4k$.
The frequency of oscillation $f$ for a spring-mass system is given by:
$f = \frac{1}{2\pi}\sqrt{\frac{k_{net}}{M}} = \frac{1}{2\pi}\sqrt{\frac{4k}{M}}$.

(A) Let the two spheres be $P$ and $Q$. The axis of rotation passes through the centre of sphere $Q$ and is perpendicular to the rod.
The moment of inertia of the system is $I_{\text{net}} = I_Q + I_P + I_{\text{rod}}$.
$1$. For sphere $Q$: The axis passes through its centre. The moment of inertia of a solid sphere about its central axis is $I_Q = \frac{2}{5} M r^2$,where $r = \frac{R}{2}$.
$I_Q = \frac{2}{5} M \left(\frac{R}{2}\right)^2 = \frac{2}{5} M \left(\frac{R^2}{4}\right) = \frac{1}{10} MR^2$.
$2$. For sphere $P$: The axis is at a distance $d = 2R$ from its centre. Using the parallel axis theorem,$I_P = I_{\text{cm}} + Md^2$.
$I_P = \frac{2}{5} M \left(\frac{R}{2}\right)^2 + M(2R)^2 = \frac{1}{10} MR^2 + 4MR^2 = \frac{41}{10} MR^2$.
$3$. For the rod: Since the rod is massless,$I_{\text{rod}} = 0$.
Total moment of inertia $I_{\text{net}} = I_Q + I_P = \frac{1}{10} MR^2 + \frac{41}{10} MR^2 = \frac{42}{10} MR^2 = \frac{21}{5} MR^2$.

(B) Let the slopes of $PQ$ and $PR$ be $m_1$ and $m_2$. Since $\triangle PQR$ is a right-angled isosceles triangle at $P$,the lines $PQ$ and $PR$ are perpendicular,so $m_1 m_2 = -1$. Let $m_1 = m$,then $m_2 = -\frac{1}{m}$.
Since $\triangle PQR$ is isosceles with $\angle P = 90^\circ$,we have $\angle PQR = \angle PRQ = 45^\circ$.
The slope of line $QR$ is $m_{QR} = -2$.
The angle between $PQ$ (slope $m$) and $QR$ (slope $-2$) is $45^\circ$:
$\tan 45^\circ = \left| \frac{m - (-2)}{1 + m(-2)} \right| = 1$
$\left| \frac{m+2}{1-2m} \right| = 1 \Rightarrow (m+2)^2 = (1-2m)^2$
$m^2 + 4m + 4 = 1 - 4m + 4m^2 \Rightarrow 3m^2 - 8m - 3 = 0$
$(3m+1)(m-3) = 0 \Rightarrow m = 3, -\frac{1}{3}$.
The equations of lines $PQ$ and $PR$ passing through $P(2, 1)$ are $y-1 = 3(x-2)$ and $y-1 = -\frac{1}{3}(x-2)$.
These can be written as $3x - y - 5 = 0$ and $x + 3y - 5 = 0$.
The joint equation is $(3x - y - 5)(x + 3y - 5) = 0$.
$3x^2 + 9xy - 15x - xy - 3y^2 + 5y - 5x - 15y + 25 = 0$
$3x^2 - 3y^2 + 8xy - 20x - 10y + 25 = 0$.

(D) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = \text{constant}$.
Thus,$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
For a monoatomic gas,the adiabatic index $\gamma = 5/3$,so $\gamma - 1 = 2/3$.
The volume of the gas in a cylinder of cross-sectional area $A$ is $V = A \times L$,where $L$ is the length of the gas column.
Substituting $V_1 = A L_1$ and $V_2 = A L_2$ into the adiabatic equation:
$T_1 (A L_1)^{2/3} = T_2 (A L_2)^{2/3}$.
Dividing both sides by $T_2 (A L_1)^{2/3}$,we get:
$\frac{T_1}{T_2} = \frac{(A L_2)^{2/3}}{(A L_1)^{2/3}} = (\frac{L_2}{L_1})^{2/3}$.

(C) Let the observations be $x_1, x_2, \dots, x_{20}$ with mean $\bar{x}$.
Given,variance $\sigma^2 = \frac{1}{20} \sum_{i=1}^{20} (x_i - \bar{x})^2 = 5$.
When each observation is multiplied by a constant $k=2$,the new observations are $y_i = 2x_i$.
The new mean is $\bar{y} = 2\bar{x}$.
The new variance is $\sigma_{new}^2 = \frac{1}{20} \sum_{i=1}^{20} (y_i - \bar{y})^2$.
Substituting $y_i = 2x_i$ and $\bar{y} = 2\bar{x}$:
$\sigma_{new}^2 = \frac{1}{20} \sum_{i=1}^{20} (2x_i - 2\bar{x})^2 = \frac{1}{20} \sum_{i=1}^{20} 4(x_i - \bar{x})^2$.
$\sigma_{new}^2 = 4 \times \left[ \frac{1}{20} \sum_{i=1}^{20} (x_i - \bar{x})^2 \right] = 4 \times 5 = 20$.

(B) $1$. The two springs on the left with force constants $2k$ and $2k$ are in series. Their equivalent spring constant $k_{L}$ is given by $\frac{1}{k_{L}} = \frac{1}{2k} + \frac{1}{2k} = \frac{2}{2k} = \frac{1}{k}$,so $k_{L} = k$.
$2$. The two springs on the right with force constants $k$ and $2k$ are in parallel. Their equivalent spring constant $k_{R}$ is given by $k_{R} = k + 2k = 3k$.
$3$. The left and right combinations are effectively in parallel with respect to the mass $M$. Thus,the total effective spring constant $k_{eff} = k_{L} + k_{R} = k + 3k = 4k$.
$4$. The frequency of oscillation $f$ is given by $f = \frac{1}{2 \pi} \sqrt{\frac{k_{eff}}{M}} = \frac{1}{2 \pi} \sqrt{\frac{4k}{M}}$.

(D) The magnetic field $B$ produced by the larger loop (radius $R_1$) at its centre is given by $B = \frac{\mu_0 i_1}{2 R_1}$.
Since $R_1 >> R_2$,we can assume this magnetic field is approximately uniform over the area of the smaller loop (radius $R_2$).
The magnetic flux $\phi_2$ passing through the smaller loop is $\phi_2 = B \cdot A_2$,where $A_2 = \pi R_2^2$ is the area of the smaller loop.
Substituting the values,we get $\phi_2 = \left( \frac{\mu_0 i_1}{2 R_1} \right) (\pi R_2^2)$.
By definition,the mutual inductance $M$ is given by $M = \frac{\phi_2}{i_1}$.
Therefore,$M = \frac{\mu_0 \pi R_2^2}{2 R_1}$.
From this expression,it is clear that $M \propto \frac{R_2^2}{R_1}$.

(D) According to Einstein's photoelectric equation,the stopping potential $V$ is given by $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.
For the first case: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$ --- $(i)$
For the second case: $e(\frac{V}{4}) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$ --- $(ii)$
Multiplying equation $(ii)$ by $4$,we get: $eV = \frac{2hc}{\lambda} - \frac{4hc}{\lambda_0}$ --- $(iii)$
Equating $(i)$ and $(iii)$: $\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{2hc}{\lambda} - \frac{4hc}{\lambda_0}$
Rearranging the terms: $\frac{3hc}{\lambda_0} = \frac{hc}{\lambda}$
Therefore,$\lambda_0 = 3\lambda$.

(D) According to Einstein's photoelectric equation,the stopping potential $V$ is given by: $eV = \frac{hc}{\lambda} - \Phi$,where $\Phi = \frac{hc}{\lambda_0}$ is the work function.
For wavelength $\lambda$: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$ --- $(i)$
For wavelength $2\lambda$: $e(\frac{V}{4}) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$ --- $(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$4 = \frac{\frac{hc}{\lambda} - \frac{hc}{\lambda_0}}{\frac{hc}{2\lambda} - \frac{hc}{\lambda_0}} = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$
$4(\frac{1}{2\lambda} - \frac{1}{\lambda_0}) = \frac{1}{\lambda} - \frac{1}{\lambda_0}$
$\frac{2}{\lambda} - \frac{4}{\lambda_0} = \frac{1}{\lambda} - \frac{1}{\lambda_0}$
$\frac{1}{\lambda} = \frac{3}{\lambda_0}$
$\lambda_0 = 3\lambda$.

(C) The initial energy of the combined system is $U_i = \frac{1}{2}CV_1^2 + \frac{1}{2}CV_2^2$.
When the capacitors are connected in parallel,the common potential $V$ is given by $V = \frac{CV_1 + CV_2}{C + C} = \frac{V_1 + V_2}{2}$.
The final energy of the system is $U_f = \frac{1}{2}(2C)V^2 = C \left( \frac{V_1 + V_2}{2} \right)^2 = \frac{C}{4}(V_1 + V_2)^2$.
The decrease in energy is $\Delta U = U_i - U_f = \frac{1}{2}C(V_1^2 + V_2^2) - \frac{1}{4}C(V_1 + V_2)^2$.
Simplifying this,$\Delta U = \frac{C}{4} [2V_1^2 + 2V_2^2 - (V_1^2 + V_2^2 + 2V_1V_2)] = \frac{C}{4}(V_1^2 + V_2^2 - 2V_1V_2) = \frac{1}{4}C(V_1 - V_2)^2$.

(B) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by:
$\frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) = V_0$
For the first case,with wavelength $\lambda$ and stopping potential $4.8 \ V$:
$\frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) = 4.8$ --- $(i)$
For the second case,with wavelength $2\lambda$ and stopping potential $1.6 \ V$:
$\frac{hc}{e} \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right) = 1.6$ --- $(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}} = \frac{4.8}{1.6} = 3$
$\frac{1}{\lambda} - \frac{1}{\lambda_0} = 3 \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right)$
$\frac{1}{\lambda} - \frac{1}{\lambda_0} = \frac{1.5}{\lambda} - \frac{3}{\lambda_0}$
$\frac{3}{\lambda_0} - \frac{1}{\lambda_0} = \frac{1.5}{\lambda} - \frac{1}{\lambda}$
$\frac{2}{\lambda_0} = \frac{0.5}{\lambda} = \frac{1}{2\lambda}$
$\lambda_0 = 4\lambda$

(B) An electromagnet must be easily magnetized and demagnetized by the application and removal of an external magnetic field.
To achieve this,the material must have low retentivity so that it does not retain magnetism after the current is switched off.
Additionally,it must have low coercivity so that it can be easily demagnetized.
Therefore,the correct choice is low retentivity and low coercivity.

(D) The total induced $emf$ $(E)$ in a coil of $N$ turns due to mutual inductance $M$ is given by Faraday's law: $E = M \frac{dI}{dt}$.
Given that the current changes from $I$ to $0$ in time $t$,the magnitude of the rate of change of current is $\frac{dI}{dt} = \frac{I}{t}$.
Therefore,the total induced $emf$ in the coil is $E = M \frac{I}{t}$.
Since the question asks for the $emf$ induced per turn,we divide the total induced $emf$ by the number of turns $N$.
However,note that the total flux linkage is $N\phi = MI$. The induced $emf$ in the entire coil is $E = \frac{d(N\phi)}{dt} = M \frac{dI}{dt} = \frac{MI}{t}$.
The $emf$ induced per turn is $\frac{E}{N} = \frac{MI}{Nt}$ is incorrect based on standard definitions; rather,the $emf$ induced in the whole coil is $E = \frac{MI}{t}$. If the question asks for $emf$ per turn,it is $\frac{E}{N} = \frac{MI}{Nt}$.
Wait,re-evaluating: The total $emf$ induced in the secondary coil is $E = M \frac{dI}{dt} = \frac{MI}{t}$. This $emf$ is distributed across all $N$ turns. Thus,$emf$ per turn is $\frac{E}{N} = \frac{MI}{Nt}$.

(D) Grignard reagents $(RMgX)$ react with compounds containing active hydrogen atoms (like water,alcohols,or amines) to form the corresponding hydrocarbon.
The reaction is as follows:
$n CH_3 MgI + n H_2 O \rightarrow n CH_4 + n Mg(OH)I$
Here,$CH_3 MgI$ (Methyl magnesium iodide) reacts with $H_2 O$ to produce $CH_4$ (Methane) and $Mg(OH)I$ (Hydroxy magnesium iodide).

(C) The Cannizzaro reaction is a redox reaction in which two molecules of an aldehyde (lacking an $\alpha$-hydrogen atom) react in the presence of a strong base to produce a primary alcohol and a carboxylic acid salt. Since the same aldehyde species is simultaneously oxidized and reduced,it is classified as a disproportionation reaction.

(B) Acylation reaction is characteristic of primary and secondary amines because they possess at least one replaceable hydrogen atom attached to the nitrogen atom.
$N$-Methylaniline $(C_6H_5NHCH_3)$ is a secondary amine,which contains one $N-H$ bond,allowing it to undergo acylation.
Ethyldimethylamine,$N,N$-Dimethylmethanamine,and $N,N$-Dimethylaniline are all tertiary amines and lack the necessary $N-H$ bond to undergo acylation.

(C) salt turns red litmus blue if its aqueous solution is basic.
$NH_4NO_3$ is a salt of a strong acid $(HNO_3)$ and a weak base $(NH_4OH)$,making it acidic.
$NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,making it acidic.
$NH_4F$ is a salt of a weak acid $(HF)$ and a weak base $(NH_4OH)$. Since $K_a$ of $HF$ $(6.8 \times 10^{-4})$ is greater than $K_b$ of $NH_4OH$ $(1.8 \times 10^{-5})$,the solution is acidic.
$NH_4CN$ is a salt of a weak acid $HCN$ $(K_a = 4.0 \times 10^{-10})$ and a weak base $NH_4OH$ $(K_b = 1.8 \times 10^{-5})$.
Since $K_b > K_a$,the aqueous solution of $NH_4CN$ is basic and turns red litmus blue.

(B) The reaction sequence is as follows:
$CH_3 Br \xrightarrow{KCN} CH_3 CN (A)$
$CH_3 CN \xrightarrow{Na/C_2H_5OH} CH_3-CH_2-NH_2 (B)$
The product '$B$' is $CH_3-CH_2-NH_2$,which is Ethylamine.

(B) Cellulose is a linear polysaccharide composed of $D$-glucose units.
These units are joined together by $\beta-1,4$-glycosidic linkages.

(D) Maltose is a disaccharide composed of two $\alpha-D-glucose$ units.
These units are linked together by a glycosidic bond formed between the $C-1$ of one glucose molecule and the $C-4$ of the other.
Therefore,the linkage is an $\alpha-1,4-glycosidic$ linkage.

(D) To determine which molecule does not contain nitrogen,let us examine the chemical structures of each:
$1$. Pyrrole $(C_4H_5N)$: Contains a nitrogen atom in a five-membered ring.
$2$. Piperidine $(C_5H_{11}N)$: Contains a nitrogen atom in a six-membered saturated ring.
$3$. Pyridine $(C_5H_5N)$: Contains a nitrogen atom in a six-membered aromatic ring.
$4$. Pyran $(C_5H_6O)$: Contains an oxygen atom in a six-membered ring,but no nitrogen atom.
Therefore,Pyran is the molecule that does not contain nitrogen.

(D) Monocarboxylic acids contain only one $-COOH$ group,while dicarboxylic acids contain two $-COOH$ groups.
$A$. Malonic acid $(HOOC-CH_2-COOH)$ and succinic acid $(HOOC-(CH_2)_2-COOH)$ are dicarboxylic acids.
$B$. Adipic acid $(HOOC-(CH_2)_4-COOH)$ is a dicarboxylic acid.
$C$. Butyric acid $(CH_3(CH_2)_2COOH)$ and caproic acid $(CH_3(CH_2)_4COOH)$ both contain only one $-COOH$ group,making them a pair of monocarboxylic acids.

(C) The reaction of carboxylic acids with phosphorus pentachloride $(PCl_5)$ yields acid chlorides,phosphorus oxychloride $(POCl_3)$,and hydrogen chloride $(HCl)$.
The chemical equation is:
$C_6H_5COOH + PCl_5 \xrightarrow{\Delta} C_6H_5COCl + POCl_3 + HCl$
Thus,the reagent used is $PCl_5$.

(D) In $BeF_2$,the central atom $Be$ has only $4$ electrons in its valence shell after forming two covalent bonds with $F$ atoms.
Since the number of electrons is less than $8$,it is an electron-deficient molecule and does not obey the octet rule.

(A) The formula for formal charge is: $\text{Formal Charge} = \text{Total valence electrons} - \text{Non-bonding electrons} - \frac{1}{2} \times \text{Bonding electrons}$.
For the carbon atom $(C)$ in the $CO_2$ molecule:
- Total valence electrons $(VE)$ = $4$
- Non-bonding electrons $(NE)$ = $0$
- Bonding electrons $(BE)$ = $8$ (four bonds,each with two electrons)
$\text{Formal Charge on } C = 4 - 0 - \frac{8}{2} = 4 - 4 = 0$.

(B) The number of lone pairs on the central atom can be calculated using the formula: $\text{Lone pairs} = \frac{1}{2} [V - B]$,where $V$ is the number of valence electrons of the central atom and $B$ is the number of bonding electrons (or bonds formed).
For $SF_6$: The central atom Sulfur $(S)$ has $6$ valence electrons and forms $6$ bonds with Fluorine atoms. $\text{Lone pairs} = \frac{1}{2} [6 - 6] = 0$.
For $SO_2$: Sulfur has $6$ valence electrons,forms $4$ bonds with Oxygen,leaving $1$ lone pair.
For $NH_3$: Nitrogen has $5$ valence electrons,forms $3$ bonds with Hydrogen,leaving $1$ lone pair.
For $SF_4$: Sulfur has $6$ valence electrons,forms $4$ bonds with Fluorine,leaving $1$ lone pair.
Thus,$SF_6$ has no lone pair on the central atom.

(B) In $CH_4$,the carbon atom is $sp^3$ hybridised.
In $C_2H_2$ (acetylene),the carbon atoms are $sp$ hybridised due to the presence of a triple bond.
In $H_2O$,the oxygen atom is $sp^3$ hybridised.
In $NH_3$,the nitrogen atom is $sp^3$ hybridised.
Therefore,$C_2H_2$ is the molecule that does not involve $sp^3$ hybridisation.

(A) In $AB_4E$ type molecules,the central atom $A$ has $4$ bond pairs and $1$ lone pair.
The total number of electron pairs is $5$,which corresponds to a trigonal bipyramidal electron geometry.
To minimize electron-electron repulsions,the lone pair occupies an equatorial position,resulting in a see-saw molecular shape.
Example: $SF_4$.

(B) The structure of $2-$Methylbut$-2-$ene is $CH_3-C(CH_3)=CH-CH_3$.
In this molecule,the carbon atoms at positions $1$,$3$,and the methyl group attached to position $2$ are $sp^3$ hybridized.
The carbon atoms at positions $2$ and $3$ are $sp^2$ hybridized because they are involved in the double bond.
Thus,there are $3$ $sp^3$ hybridized carbon atoms in one molecule of $2-$Methylbut$-2-$ene.
Therefore,in one mole of $2-$Methylbut$-2-$ene,there are $3$ moles of $sp^3$ hybrid carbon atoms.

(A) In the $PCl_5$ molecule,the central phosphorus atom $(P)$ is bonded to five chlorine atoms $(Cl)$.
According to $VSEPR$ theory,the molecule has $5$ bonding pairs and $0$ lone pairs of electrons around the central atom.
This corresponds to $sp^3d$ hybridization,which results in a trigonal bipyramidal geometry.

(C) The given molecule is $HO-CH_2-CH_2-CH_2-CH(CH_3)-CH(CH_3)_2$.
Expanding the structure: $HO-CH_2-CH_2-CH_2-CH(CH_3)-CH(CH_3)_2$.
All carbon atoms in this molecule are bonded to four other atoms via single bonds,meaning they are all $sp^3$ hybridized.
Counting the carbon atoms: $3$ (from $(CH_2)_3$) + $1$ (from $CH$) + $1$ (from $CH_3$) + $1$ (from $CH$) + $2$ (from $(CH_3)_2$) = $8$ carbon atoms.
Therefore,there are $8$ $sp^3$ hybridized carbon atoms.

(B) The total number of electrons in the $NO$ molecule is $7 + 8 = 15$.
According to Molecular Orbital Theory,the electronic configuration of $NO$ is:
$\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$.
There is $1$ unpaired electron in the antibonding $\pi^* 2p_x$ orbital.
Therefore,the number of unpaired electrons is $1$.

(C) $Li$ in $LiCl$ has less than eight electrons in its valence shell. It has only two electrons shared in the bond. Hence,it has an incomplete octet.
$SF_6$,$PCl_5$,and $H_2SO_4$ are examples of molecules with an expanded octet,where the central atom has more than eight electrons.

(C) The elevation in boiling point $(\Delta T_b)$ is a colligative property,which is directly proportional to the van't Hoff factor $(i)$ and the molarity $(M)$ of the solution: $\Delta T_b = i \times K_b \times M$.
For non-electrolytes like urea and glucose,$i = 1$.
For $KNO_3$,$i = 2$ $(K^{+} + NO_3^{-})$.
For $Na_2SO_4$,$i = 3$ $(2Na^{+} + SO_4^{2-})$.
Calculating the effective concentration $(i \times M)$:
$A: 1 \times 0.015 = 0.015 \ M$
$B: 2 \times 0.01 = 0.02 \ M$
$C: 3 \times 0.01 = 0.03 \ M$
$D: 1 \times 0.015 = 0.015 \ M$
Since $0.01 \ M \ Na_2SO_4$ has the highest effective concentration of solute particles,it will exhibit the highest boiling point.

(D) Alcohols are generally neutral or very weakly acidic,whereas phenols are more acidic than alcohols.
Electron-withdrawing groups,such as the nitro group $(-NO_2)$,stabilize the phenoxide ion through $-I$ and $-M$ effects,thereby increasing the acidity of the phenol.
Since $p-$nitrophenol contains a strong electron-withdrawing $-NO_2$ group at the para position,it exhibits the highest acidic strength among the given compounds.

(C) In a tertiary benzylic alcohol,the $-OH$ group is bonded to a $sp^3$ hybridized tertiary carbon atom,which is directly attached to an aromatic ring.
In option $C$,the structure is $C_6H_5-C(CH_3)_2-OH$. Here,the carbon atom attached to the $-OH$ group is bonded to one phenyl group and two methyl groups,making it a tertiary carbon atom attached to an aromatic ring. Thus,it is a tertiary benzylic alcohol.

(C) Methanal $(HCHO)$ reacts with Grignard reagent $(RMgX)$ to form primary alcohols $(R-CH_2-OH)$.
Propan-$2$-$ol$ is a secondary alcohol,which is formed by the reaction of Grignard reagent with aldehydes other than methanal (e.g.,ethanal).
Therefore,propan-$2$-$ol$ cannot be prepared by the action of Grignard reagent on methanal.

(D) The reaction of a Grignard reagent $(C_2H_5MgBr)$ with a ketone $(A)$ followed by acid hydrolysis $(H_2O/H^+)$ yields a tertiary alcohol.
Given the product is $\text{3-Methylpentan-3-ol}$,which has the structure $CH_3-CH_2-C(OH)(CH_3)-CH_2-CH_3$.
The Grignard reagent $C_2H_5MgBr$ provides an ethyl group $(C_2H_5)$.
Removing the ethyl group from the tertiary carbon of the product leaves a ketone with a methyl group and a propyl group attached to the carbonyl carbon,which is $CH_3-CH_2-C(=O)-CH_3$,known as $\text{Butanone}$.
Thus,the compound $A$ is $\text{Butanone}$.

(C) In the reaction of alcohols with phosphorus trichloride $(PCl_3)$,the $C-O$ bond of the alcohol is broken to form an alkyl chloride.
The chemical equation is: $3 R-OH + PCl_3 \longrightarrow 3 R-Cl + H_3PO_3$.
In contrast,reactions with carboxylic acids,acid anhydrides,and acid chlorides typically involve the cleavage of the $O-H$ bond of the alcohol.

(B) When vapours of tertiary alcohols like $2-$methylpropan$-2-$ol are passed over hot copper at $573 \ K$,they undergo dehydration rather than dehydrogenation.
This reaction results in the formation of an alkene.
The reaction is: $(CH_3)_3C-OH \xrightarrow{Cu, 573 \ K} CH_3-C(CH_3)=CH_2 + H_2O$.
The product formed is $2-$methylpropene.

(D) The reactivity of alcohols with haloacids $(HX)$ follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
This is because the reaction proceeds via the formation of a carbocation intermediate,and the stability of the carbocation follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
Analyzing the given structures:
$(A)$ $CH_3CH_2C(CH_3)(OH)CH_3$ is a $3^{\circ}$ alcohol.
$(B)$ $CH_3CH_2CH(OH)CH_2CH_3$ is a $2^{\circ}$ alcohol.
$(C)$ $CH_3CH_2CH_2CH(OH)CH_3$ is a $2^{\circ}$ alcohol.
$(D)$ $CH_3CH_2C(OH)(CH_3)CH_2CH_3$ is a $3^{\circ}$ alcohol.
Both $(A)$ and $(D)$ are tertiary alcohols. However,$(D)$ is a more substituted tertiary alcohol ($3$-methylpentan$-3-$ol) compared to $(A)$ ($2$-methylbutan$-2-$ol),making the carbocation formed from $(D)$ slightly more stable due to increased hyperconjugation and inductive effects. Thus,$(D)$ reacts fastest.

(B) The reaction sequence is a Wurtz-Fittig type coupling reaction.
Step $1$: $A \xrightarrow{SOCl_2, \Delta} B$. This is the conversion of an alcohol to an alkyl chloride. Propan$-2-$ol $(CH_3CH(OH)CH_3)$ reacts with $SOCl_2$ to form $2-$chloropropane $(CH_3CHClCH_3)$ as $B$.
Step $2$: $B + C_2H_5Cl \xrightarrow{Na/\text{ether}} \text{2-Methylbutane}$. This is a Wurtz reaction where $2-$chloropropane $(CH_3CHClCH_3)$ reacts with chloroethane $(C_2H_5Cl)$ in the presence of sodium and dry ether to form $2-$methylbutane $(CH_3CH(CH_3)CH_2CH_3)$.
Therefore,substrate $A$ is Propan$-2-$ol.

(B) When anisole $(C_6H_5OCH_3)$ is heated with dilute sulfuric acid $(H_2SO_4)$ under pressure,it undergoes acid-catalyzed hydrolysis.
The reaction is as follows:
$C_6H_5OCH_3 + H_2O \xrightarrow{H^+, \Delta} C_6H_5OH + CH_3OH$
Thus,the products obtained are phenol $(C_6H_5OH)$ and methanol $(CH_3OH)$.

(C) When phenol is treated with concentrated nitric acid in the presence of concentrated $H_2SO_4$,it undergoes electrophilic aromatic substitution to form $2,4,6-\text{trinitrophenol}$,which is commonly known as picric acid.
This reaction involves the nitration of the phenol ring at all available ortho and para positions due to the strong activating effect of the $-\text{OH}$ group.

(C) When phenol $(C_6H_5OH)$ is heated with zinc dust,it undergoes a reduction reaction.
Zinc dust acts as a reducing agent and removes the oxygen atom from the phenol group.
The reaction is as follows:
$C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO$
Thus,phenol is converted into benzene.

(B) When phenol is treated with bromine water,it undergoes electrophilic substitution to form a white precipitate of $2,4,6$-tribromophenol.
The reaction is as follows:
$C_6H_5OH + 3Br_2 \xrightarrow{H_2O} C_6H_2Br_3OH + 3HBr$
Thus,water is the solvent used.

(C) When phenol reacts with dilute nitric acid $(HNO_3)$ at low temperature $(298 \ K)$,it undergoes electrophilic aromatic substitution to form a mixture of $ortho$-nitrophenol and $para$-nitrophenol.
$C_6H_5OH + dil. HNO_3 \rightarrow o-Nitrophenol + p-Nitrophenol$

(A) The reaction of an alkyl aryl ether like ethoxybenzene $(C_6H_5OC_2H_5)$ with hot and concentrated $HI$ involves the cleavage of the $C-O$ bond between the alkyl group and the oxygen atom.
This occurs because the $C-O$ bond between the phenyl ring and oxygen has partial double bond character due to resonance,making it stronger and harder to break.
Therefore,the $HI$ attacks the alkyl group,resulting in the formation of phenol $(C_6H_5OH)$ and ethyl iodide $(C_2H_5I)$.

(C) Williamson's synthesis involves the reaction of an alkyl halide with a sodium alkoxide to form an ether.
This reaction proceeds via an $S_N2$ mechanism.
Aryl halides,such as $C_6H_5-Cl$,do not undergo $S_N2$ reactions easily because the $C-Cl$ bond has partial double bond character due to resonance and the carbon atom is $sp^2$ hybridized,making it less susceptible to nucleophilic attack.
Therefore,$C_6H_5-Cl$ does not undergo Williamson's synthesis.

(A) $Crotonyl$ alcohol is represented by the formula $CH_3CH=CHCH_2OH$.
In this molecule,the $-OH$ group is attached to a carbon atom that is $sp^3$ hybridized and is adjacent to a carbon-carbon double bond $(C=C)$.
This structural feature defines an allylic alcohol.

(B) In vinylic alcohol,the $-OH$ group is attached to a $sp^2$ hybridized carbon atom which is part of a carbon-carbon double bond.
In $but-2-en-2-ol$ $(CH_3-C(OH)=CH-CH_3)$,the $-OH$ group is directly attached to the $sp^2$ hybridized carbon of the double bond.
Therefore,$but-2-en-2-ol$ is a vinylic alcohol.

(D) Dihydric phenols are compounds that contain two hydroxyl $(-OH)$ groups attached to a benzene ring.
Based on the provided structures:
$1$. Catechol: Contains two $-OH$ groups (dihydric).
$2$. Resorcinol: Contains two $-OH$ groups (dihydric).
$3$. Quinol: Contains two $-OH$ groups (dihydric).
$4$. Phloroglucinol: Contains three $-OH$ groups (trihydric).
$5$. Pyrogallol: Contains three $-OH$ groups (trihydric).
Therefore,the pair consisting of two dihydric phenols is Catechol and Quinol.

(C) An allylic alcohol is one in which the $-OH$ group is attached to a carbon atom adjacent to a carbon-carbon double bond $(C=C-C-OH)$.
$A$. $H_2C=CH-CH_2-OH$: Primary allylic alcohol ($-OH$ is on a primary carbon).
$B$. $CH_3-CH=CH-CH_2-OH$: Primary allylic alcohol ($-OH$ is on a primary carbon).
$C$. $H_2C=CH-CH(OH)-CH_3$: Secondary allylic alcohol ($-OH$ is on a secondary carbon).
$D$. $H_2C=CH-C(OH)(CH_3)_2$: Tertiary allylic alcohol ($-OH$ is on a tertiary carbon).
Therefore,the correct option is $C$.

(A) The Rosenmund reduction is a hydrogenation process where an acyl chloride $(R-COCl)$ is reduced to an aldehyde $(R-CHO)$ using hydrogen gas in the presence of a poisoned palladium catalyst,specifically palladium on barium sulfate $(Pd-BaSO_4)$.
The reaction is represented as:
$R-COCl + H_2 \xrightarrow{Pd-BaSO_4} R-CHO + HCl$
Therefore,option $A$ represents the Rosenmund reduction.

(C) Rosenmund reduction involves the hydrogenation of acid chlorides to aldehydes using $H_2$ gas in the presence of palladium supported on barium sulfate $(Pd-BaSO_4)$.
In this reaction,benzoyl chloride $(C_6H_5COCl)$ is reduced to benzaldehyde $(C_6H_5CHO)$.
The chemical equation is:
$C_6H_5COCl + H_2 \xrightarrow{Pd-BaSO_4} C_6H_5CHO + HCl$
Therefore,the correct product is benzaldehyde.

(B) $p-$Nitrophenol has the highest melting point among the given compounds.
This is due to the presence of strong intermolecular hydrogen bonding in $p-$nitrophenol,which leads to the association of molecules.
In contrast,$o-$nitrophenol exhibits intramolecular hydrogen bonding,which reduces the intermolecular forces of attraction,resulting in a lower melting point.

(A) The reaction of an organic compound '$A$' with $PCl_3$ to form an alkyl chloride $(C_3H_7Cl)$ suggests that '$A$' is a primary alcohol,$CH_3-CH_2-CH_2-OH$ (Propan$-1-$ol).
Reaction $1$: $CH_3-CH_2-CH_2-OH + PCl_3 \rightarrow CH_3-CH_2-CH_2-Cl + H_3PO_3$.
Reaction $2$: Oxidation of a primary alcohol with $PCC$ (Pyridinium chlorochromate) yields an aldehyde.
$CH_3-CH_2-CH_2-OH \xrightarrow{PCC} CH_3-CH_2-CHO$ (Propanal,$C_3H_6O$).
Thus,'$A$' is Propan$-1-$ol.

(B) In the Etard reaction,the methyl group of toluene (methylbenzene) is oxidized using $CrO_2Cl_2$ (chromyl chloride) in the presence of $CS_2$ (carbon disulfide) as a solvent.
This process forms a chromium complex intermediate,which upon acid hydrolysis yields benzaldehyde.
Therefore,the reagent used is chromyl chloride.

(C) The reaction of toluene with chromyl chloride $(CrO_2Cl_2)$ in the presence of carbon disulfide $(CS_2)$ followed by acid hydrolysis is known as the $Etard$ reaction.
In this reaction,the methyl group of toluene is oxidized to an aldehyde group,resulting in the formation of benzaldehyde $(C_6H_5CHO)$.

(B) The given reaction is the Etard reaction.
In this reaction,Toluene is oxidized by chromyl chloride $(CrO_2Cl_2)$ in the presence of carbon disulfide $(CS_2)$ to form a brown chromium complex $(A)$.
This complex $(A)$ on hydrolysis with water $(H_3O^+)$ gives Benzaldehyde $(B)$.
The reaction is:
$C_6H_5CH_3 + 2CrO_2Cl_2$ $\xrightarrow{CS_2} C_6H_5CH(OCrOHCl_2)_2$ $\xrightarrow{H_3O^+} C_6H_5CHO$
Thus,the product '$B$' is Benzaldehyde.

(C) The reaction of acyl chlorides with dialkyl cadmium reagents is a standard method for the preparation of ketones.
Given the reaction: $2CH_3COCl + (CH_3)_2Cd \rightarrow 2CH_3COCH_3 + CdCl_2$.
Here,$CH_3COCl$ is $Ethanoyl$ chloride,which reacts with dimethyl cadmium to form $Propanone$ $(CH_3COCH_3)$.

(A) The formation of aldol from aldehyde is known as an aldol condensation reaction.
In this reaction,two molecules of an aldehyde or ketone containing at least one $\alpha$-hydrogen atom react in the presence of a dilute alkali to form $\beta$-hydroxy aldehyde (aldol) or $\beta$-hydroxy ketone (ketol).
Since the reaction involves the combination of two molecules with the elimination of a water molecule (in the subsequent step),it is classified as a condensation reaction.

(C) The given reaction is the Wolff-Kishner reduction.
In the first step,ethyl phenyl ketone $(C_6H_5-CO-CH_2CH_3)$ reacts with hydrazine $(H_2N-NH_2)$ to form a hydrazone $(A)$.
In the second step,the hydrazone is treated with a strong base $(KOH)$ in a high-boiling solvent like ethylene glycol $(HO-(CH_2)_2-OH)$ and heated $(\Delta)$.
This process reduces the carbonyl group $(C=O)$ to a methylene group $(CH_2)$.
Thus,the ethyl phenyl ketone $(C_6H_5-CO-CH_2CH_3)$ is reduced to $n$-propyl benzene $(C_6H_5-CH_2-CH_2CH_3)$.

(D) The reaction of $2$ molecules of propanone in the presence of $Ba(OH)_2$ is an aldol condensation reaction.
Step $1$: Two molecules of propanone undergo aldol condensation to form $4-$hydroxy$-4-$methylpentan$-2-$one (product $A$).
$2 CH_3COCH_3 \xrightarrow{Ba(OH)_2} (CH_3)_2C(OH)CH_2COCH_3$ (Product $A$)
Step $2$: Upon heating $( \Delta )$ and dehydration $(-H_2O)$,product $A$ loses a water molecule to form an $\alpha, \beta-$unsaturated ketone,which is $4-$methylpent$-3-$en$-2-$one (product $B$).
$(CH_3)_2C(OH)CH_2COCH_3 \xrightarrow{\Delta, -H_2O} (CH_3)_2C=CHCOCH_3$ (Product $B$)
Thus,the correct product $B$ is $4-$methylpent$-3-$en$-2-$one.

(C) When an alcoholic solution of an aldehyde is treated with a few drops of Schiff's reagent,a pink,red,or magenta colour appears.

(A) The reaction between methanal $(HCHO)$ and benzaldehyde $(C_6H_5CHO)$ in the presence of concentrated $NaOH$ is a cross-Cannizzaro reaction.
In this reaction,the more reactive aldehyde (methanal) undergoes oxidation to form a salt of carboxylic acid (sodium methanoate),which upon acidification $(H_3O^+)$ yields methanoic acid $(HCOOH)$.
The less reactive aldehyde (benzaldehyde) undergoes reduction to form the corresponding alcohol,phenyl methanol $(C_6H_5CH_2OH)$.

(B) Ammoniacal silver nitrate is known as Tollens' reagent.
It acts as a mild oxidizing agent and specifically oxidizes aldehydes to their corresponding carboxylate anions.
During this process,the $Ag^+$ ions in the reagent are reduced to metallic silver,which forms a silver mirror or a greyish-black precipitate.
Among the given options,$Ethanal$ $(CH_3CHO)$ is an aldehyde,while the others are an alcohol,an ether,and a carboxylic acid,respectively.
Therefore,only $Ethanal$ reacts with Tollens' reagent to form a silver precipitate.

(B) Acylation reaction occurs with primary and secondary amines because they possess at least one hydrogen atom attached to the nitrogen atom,which can be replaced by an acyl group.
$N$-Methylaniline $(C_6H_5NHCH_3)$ is a secondary amine and thus undergoes acylation.
Ethyldimethylamine,$N,N$-Dimethylmethanamine,and $N,N$-Dimethylaniline are all tertiary amines,which lack a hydrogen atom on the nitrogen and therefore do not undergo acylation.

(B) Step $1$: $CH_3Br$ reacts with $KCN$ to form methyl cyanide $(CH_3CN)$ as product $A$.
$CH_3Br + KCN \rightarrow CH_3CN + KBr$
Step $2$: Methyl cyanide $(CH_3CN)$ undergoes reduction with $Na / C_2H_5OH$ (Mendius reduction) to form ethyl amine $(CH_3CH_2NH_2)$ as product $B$.
$CH_3CN + 4[H] \xrightarrow{Na / C_2H_5OH} CH_3CH_2NH_2$
Therefore,the product $B$ is ethyl amine.

(D) When aniline $(C_6H_5NH_2)$ reacts with nitrous acid $(HNO_2)$,which is generated in situ by the reaction of $NaNO_2$ and $HCl$ at a low temperature of $273-278 \ K$,it undergoes a diazotization reaction.
This reaction converts the amino group $(-NH_2)$ into a diazonium group $(-N_2^+Cl^-)$,resulting in the formation of benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$.

(B) The Gatterman-Koch formylation of an arene involves the treatment of benzene or a substituted benzene with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ under high pressure.
This reaction yields benzaldehyde or a substituted benzaldehyde.
The reaction is catalyzed by anhydrous aluminium chloride $(AlCl_3)$ or cuprous chloride $(CuCl)$.
Therefore,the correct reagent is $CO, HCl / AlCl_3$ (anhydrous).

(B) $LiAlH_4$ is a strong reducing agent that reduces amides to corresponding amines.
The reaction is: $CH_3CONH_2 \xrightarrow[\text{ether}]{LiAlH_4} CH_3CH_2NH_2$ (Ethanamine).
Thus,the reactant '$A$' is acetamide $(CH_3CONH_2)$.

(A) The reaction is Hofmann elimination.
Substrate $A$ is diethyldimethyl ammonium halide,which reacts with moist $Ag_2O$ (forming $AgOH$) to produce the quaternary ammonium hydroxide.
$(CH_3CH_2)_2N^+(CH_3)_2X^- + AgOH \rightarrow (CH_3CH_2)_2N^+(CH_3)_2OH^- + AgX$
Upon heating,the quaternary ammonium hydroxide undergoes Hofmann elimination to form the alkene and the tertiary amine.
$(CH_3CH_2)_2N^+(CH_3)_2OH^- \xrightarrow{\Delta} CH_3CH_2N(CH_3)_2 + CH_2=CH_2 + H_2O$

(D) The reaction proceeds as follows:
$1$. $Aniline$ reacts with $NaNO_2 + HCl$ at $273 \ K$ to form $Benzenediazonium \ chloride$ $(A)$.
$C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273 \ K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$
$2$. $Benzenediazonium \ chloride$ $(A)$ on heating with water $(H_2O, \Delta)$ undergoes hydrolysis to form $Phenol$ $(B)$ and releases $N_2$ gas.
$C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2 \uparrow + HCl$
Therefore,product $B$ is $Phenol$.

(B) Arylamines are generally weaker bases than ammonia and aliphatic amines due to the resonance effect,which delocalizes the lone pair of electrons on the nitrogen atom into the benzene ring.
Aliphatic amines are stronger bases than ammonia due to the electron-donating inductive effect ($+I$ effect) of the alkyl groups.
Among aliphatic amines,$(CH_3)_2NH$ ($2^{\circ}$ amine) is a stronger base than $CH_3NH_2$ ($1^{\circ}$ amine) in the gas phase or non-polar solvents due to the greater inductive effect of two methyl groups.
Therefore,the correct decreasing order of basic strength is: $(CH_3)_2NH > CH_3NH_2 > NH_3 > C_6H_5NH_2$.

(C) The reaction involves the exhaustive methylation of an amine using excess methyl iodide $(CH_3 I)$.
$C_2 H_5 NH_2 + 3CH_3 I \xrightarrow{\Delta} C_2 H_5(CH_3)_3 N^+ I^- + 2HI$
This reaction is known as the exhaustive alkylation (or Hofmann's exhaustive methylation) of amines,where a primary amine reacts with excess alkyl halide to form a quaternary ammonium salt.

(D) Step $1$: Reaction of $CH_3Br$ with $KCN$ gives methyl cyanide $(CH_3CN)$ as $A$.
$CH_3Br + KCN \rightarrow CH_3CN + KBr$
Step $2$: Reduction of methyl cyanide $(CH_3CN)$ with $Na / C_2H_5OH$ (Mendius reduction) yields ethylamine $(CH_3CH_2NH_2)$ as $B$.
$CH_3CN + 4[H] \xrightarrow{Na / C_2H_5OH} CH_3CH_2NH_2$

(B) $p$-Toluidine is $4$-methylaniline.
Its structure consists of a benzene ring with a methyl group $(-CH_3)$ at the $1$-position and an amino group $(-NH_2)$ at the $4$-position (para position).
The molecular formula is derived by counting the atoms: $6$ carbons in the ring + $1$ carbon in the methyl group = $C_7$.
$4$ hydrogens on the ring + $3$ hydrogens in the methyl group + $2$ hydrogens in the amino group = $H_9$.
$1$ nitrogen atom = $N$.
Thus,the molecular formula is $C_7H_9N$.

(D) Cyclohexylamine consists of a cyclohexane ring $(C_6H_{11})$ attached to an amino group $(-NH_2)$.
Combining these,we get $C_6H_{11} + NH_2 = C_6H_{13}N$.
Therefore,the molecular formula of cyclohexylamine is $C_6H_{13}N$.

(A) The reaction of primary amines with chloroform $(CHCl_3)$ in the presence of an alcoholic base $(KOH)$ is known as the carbylamine reaction.
This reaction produces isocyanides (carbylamines),which have a characteristic foul smell.
Since only primary amines undergo this reaction,among the given options,$CH_3CH_2NH_2$ (Ethanamine) is a primary amine,while the others are secondary or tertiary amines.

(A) The specific rotation of $D-(+)$-glucose in aqueous solution is $+52.7^{\circ}$.
This value is obtained after the process of mutarotation,where the equilibrium mixture of $\alpha-D$-glucose $(+112^{\circ})$ and $\beta-D$-glucose $(+19^{\circ})$ is formed.

(D) The ring structure of glucose is formed by the intramolecular hemiacetal formation.
This involves the reaction between the aldehyde group at $C_1$ and the hydroxyl group at $C_5$ to form a six-membered pyranose ring.

(D) Glucose reacts with acetic anhydride to form glucose pentaacetate. The reaction is as follows:
$CHO-(CHOH)_4-CH_2OH + 5(CH_3CO)_2O \rightarrow CHO-(CHOCOCH_3)_4-CH_2OCOCH_3 + 5CH_3COOH$
This reaction confirms the presence of five hydroxyl $(-OH)$ groups in the glucose molecule.