MHT CET 2021 Chemistry Question Paper with Answer and Solution

(A) Given that the rate of change of the radius is $\frac{dr}{dt} = 2 \text{ cm/sec}$.
We need to find the rate of change of the area $A$ when the radius $r = 10 \text{ cm}$.
The area of a circle is given by $A = \pi r^2$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$.
Substituting the given values $r = 10 \text{ cm}$ and $\frac{dr}{dt} = 2 \text{ cm/sec}$:
$\frac{dA}{dt} = 2 \pi \times 10 \times 2 = 40 \pi \text{ cm}^2/\text{sec}$.

(D) Given the function $f(x) = \log(1+x) - \frac{2x}{2+x}$. The domain of the function requires $1+x > 0$,so $x > -1$.
First,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx} [\log(1+x)] - \frac{d}{dx} \left[ \frac{2x}{2+x} \right]$
$f'(x) = \frac{1}{1+x} - \left[ \frac{(2+x)(2) - (2x)(1)}{(2+x)^2} \right]$
$f'(x) = \frac{1}{1+x} - \frac{4}{(2+x)^2}$
Simplifying the expression:
$f'(x) = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} = \frac{4 + 4x + x^2 - 4 - 4x}{(1+x)(2+x)^2} = \frac{x^2}{(1+x)(2+x)^2}$
For the function to be increasing,we require $f'(x) > 0$.
Since $x^2 > 0$ and $(2+x)^2 > 0$ for all $x > -1$ (excluding $x=0$),the condition $f'(x) > 0$ is satisfied when $1+x > 0$,which implies $x > -1$.
Therefore,the function is increasing on $(-1, \infty)$.

(A) We have $f(x)=\frac{1-x+x^2}{1+x+x^2}$.
To find the critical points,we differentiate $f(x)$ with respect to $x$ using the quotient rule:
$f^{\prime}(x)=\frac{(1+x+x^2)(2x-1)-(1-x+x^2)(2x+1)}{(1+x+x^2)^2}$.
Expanding the numerator:
$f^{\prime}(x)=\frac{(2x^3+x^2+x-1)-(2x^3-x^2+x+1)}{(1+x+x^2)^2} = \frac{2x^2-2}{(1+x+x^2)^2}$.
Setting $f^{\prime}(x)=0$ gives $2(x^2-1)=0$,which implies $x=1$ or $x=-1$.
Evaluating the function at these points:
For $x=1$,$f(1)=\frac{1-1+1}{1+1+1} = \frac{1}{3}$.
For $x=-1$,$f(-1)=\frac{1-(-1)+(-1)^2}{1+(-1)+(-1)^2} = \frac{3}{1} = 3$.
Comparing these values,the minimum value of $f(x)$ is $\frac{1}{3}$.

(D) To find the area bounded by the parabola $y=x^2$ and the line $y=x$,we first determine their points of intersection by setting $x^2 = x$.
This gives $x^2 - x = 0$,which implies $x(x-1) = 0$.
Thus,the points of intersection are $x=0$ and $x=1$.
For $x \in [0, 1]$,the line $y=x$ lies above the parabola $y=x^2$.
The required area $A$ is given by the integral:
$A = \int_0^1 (x - x^2) dx$
$A = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1$
$A = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - (0 - 0)$
$A = \frac{1}{2} - \frac{1}{3} = \frac{3-2}{6} = \frac{1}{6}$ sq. units.

(B) To find the area bounded by the curve $y=2x-x^2$ and the $X$-axis,we first find the intersection points by setting $y=0$:
$0 = 2x - x^2$
$x(2-x) = 0$
So,$x=0$ and $x=2$.
The required area $A$ is given by the integral of the curve from $x=0$ to $x=2$:
$A = \int_{0}^{2} (2x - x^2) dx$
$A = \left[ \frac{2x^2}{2} - \frac{x^3}{3} \right]_{0}^{2}$
$A = \left[ x^2 - \frac{x^3}{3} \right]_{0}^{2}$
$A = (2^2 - \frac{2^3}{3}) - (0^2 - \frac{0^3}{3})$
$A = (4 - \frac{8}{3}) - 0$
$A = \frac{12-8}{3} = \frac{4}{3}$ sq. units.

(D) Let $I = \int_0^1 \tan ^{-1}\left(\frac{2x-1}{1+x-x^2}\right) dx$.
Using the property $\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right)$,we can rewrite the integrand:
$\frac{2x-1}{1+x-x^2} = \frac{x - (1-x)}{1 + x(1-x)}$.
Thus,$\tan^{-1}\left(\frac{x-(1-x)}{1+x(1-x)}\right) = \tan^{-1}(x) - \tan^{-1}(1-x)$.
So,$I = \int_0^1 \tan^{-1}(x) dx - \int_0^1 \tan^{-1}(1-x) dx$.
Let $J = \int_0^1 \tan^{-1}(1-x) dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get $J = \int_0^1 \tan^{-1}(1-(1-x)) dx = \int_0^1 \tan^{-1}(x) dx$.
Therefore,$I = \int_0^1 \tan^{-1}(x) dx - \int_0^1 \tan^{-1}(x) dx = 0$.

(A) The implication $p \rightarrow (\sim p \vee q)$ is false only when the antecedent is true and the consequent is false.
Thus,$p \equiv T$ and $(\sim p \vee q) \equiv F$.
Since $p \equiv T$,then $\sim p \equiv F$.
Substituting this into the second condition: $F \vee q \equiv F$.
For a disjunction to be false,both components must be false,so $q \equiv F$.
Therefore,the truth values are $p \equiv T$ and $q \equiv F$.

(A) Let the given expression be $E = [(p \wedge \sim q) \vee q] \vee (\sim p \wedge q)$.
Using the distributive law,$(p \wedge \sim q) \vee q = (p \vee q) \wedge (\sim q \vee q)$.
Since $(\sim q \vee q) = T$ (a tautology),we have $(p \vee q) \wedge T = p \vee q$.
Now,the expression becomes $E = (p \vee q) \vee (\sim p \wedge q)$.
Using the distributive law again,$(p \vee q) \vee (\sim p \wedge q) = (p \vee q \vee \sim p) \wedge (p \vee q \vee q)$.
Since $(p \vee \sim p) = T$,we have $(T \vee q) \wedge (p \vee q) = T \wedge (p \vee q) = p \vee q$.
Thus,the expression is equivalent to $p \vee q$.

(D) Let the slopes of the lines be $m_1$ and $m_2$. The equation $ax^2+2hxy+by^2=0$ represents two lines passing through the origin.
We have the sum of slopes $m_1+m_2 = -\frac{2h}{b}$ and the product of slopes $m_1m_2 = \frac{a}{b}$.
Given the ratio of slopes is $m_1:m_2 = 5:3$,let $m_1 = 5k$ and $m_2 = 3k$.
Then $m_1+m_2 = 8k = -\frac{2h}{b} \Rightarrow k = -\frac{h}{4b}$.
And $m_1m_2 = 15k^2 = \frac{a}{b}$.
Substituting $k$ in the product equation: $15(-\frac{h}{4b})^2 = \frac{a}{b}$.
$15 \times \frac{h^2}{16b^2} = \frac{a}{b}$.
$\frac{15h^2}{16b^2} = \frac{a}{b} \Rightarrow \frac{h^2}{ab} = \frac{16}{15}$.
Thus,the ratio $h^2:ab$ is $16:15$.

(C) Given that $\sin (y+z-x), \sin (z+x-y), \sin (x+y-z)$ are in $AP$.
Therefore,$2 \sin (z+x-y) = \sin (y+z-x) + \sin (x+y-z)$.
Using the sum-to-product formula $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$,we get:
$2 \sin (z+x-y) = 2 \sin \frac{(y+z-x) + (x+y-z)}{2} \cos \frac{(y+z-x) - (x+y-z)}{2}$
$2 \sin (z+x-y) = 2 \sin y \cos (z-x)$.
Since $\sin (z+x-y) = \sin (z-(y-x)) = \sin z \cos (y-x) - \cos z \sin (y-x)$,this approach is complex.
Alternatively,using $\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$:
$\sin (z+x-y) - \sin (y+z-x) = \sin (x+y-z) - \sin (z+x-y)$
$2 \cos z \sin (x-y) = 2 \cos x \sin (y-z)$
$\cos z (\sin x \cos y - \cos x \sin y) = \cos x (\sin y \cos z - \cos y \sin z)$
$\cos z \sin x \cos y - \cos z \cos x \sin y = \cos x \sin y \cos z - \cos x \cos y \sin z$
$2 \cos z \cos x \sin y = \cos x \cos y \sin z + \sin x \cos y \cos z$
Dividing by $\cos x \cos y \cos z$:
$2 \tan y = \tan z + \tan x$.

(D) Let the initial population be $P$ and the rate of increase be $8 \%$ per year.
The differential equation representing the growth is $\frac{dP}{dt} = \frac{8}{100} P = 0.08 P$.
Separating the variables,we get $\frac{dP}{P} = 0.08 dt$.
Integrating both sides,we have $\int \frac{dP}{P} = \int 0.08 dt$,which gives $\log P = 0.08 t + C$.
At $t = 0$,the population is $P_0$,so $C = \log P_0$.
Thus,$\log P = 0.08 t + \log P_0$,or $\log \left( \frac{P}{P_0} \right) = 0.08 t$.
For the population to double,$P = 2 P_0$,so $\log 2 = 0.08 t$.
Given $\log 2 = 0.6912$,we have $0.6912 = 0.08 t$.
Therefore,$t = \frac{0.6912}{0.08} = 8.64 \text{ years}$.

(A) Given that $\bar{a}, \bar{b}, \bar{c}$ are mutually perpendicular vectors,we have $\bar{a} \cdot \bar{b} = 0, \bar{b} \cdot \bar{c} = 0, \bar{c} \cdot \bar{a} = 0$ and magnitudes $|\bar{a}|=1, |\bar{b}|=2, |\bar{c}|=3$.
The scalar triple product is defined as $[\bar{a}+\bar{b}+\bar{c} \quad \bar{b}-\bar{a} \quad \bar{c}] = (\bar{a}+\bar{b}+\bar{c}) \cdot ((\bar{b}-\bar{a}) \times \bar{c})$.
Expanding the cross product: $(\bar{b}-\bar{a}) \times \bar{c} = (\bar{b} \times \bar{c}) - (\bar{a} \times \bar{c})$.
Now,the dot product becomes: $(\bar{a}+\bar{b}+\bar{c}) \cdot ((\bar{b} \times \bar{c}) - (\bar{a} \times \bar{c}))$.
Since $\bar{a}, \bar{b}, \bar{c}$ are mutually perpendicular,$\bar{b} \times \bar{c}$ is parallel to $\bar{a}$ and $\bar{a} \times \bar{c}$ is parallel to $\bar{b}$.
Specifically,$\bar{a} \cdot (\bar{b} \times \bar{c}) = [\bar{a} \bar{b} \bar{c}]$ and $\bar{b} \cdot (\bar{a} \times \bar{c}) = -[\bar{a} \bar{b} \bar{c}]$.
The expression simplifies to: $\bar{a} \cdot (\bar{b} \times \bar{c}) - \bar{a} \cdot (\bar{a} \times \bar{c}) + \bar{b} \cdot (\bar{b} \times \bar{c}) - \bar{b} \cdot (\bar{a} \times \bar{c}) + \bar{c} \cdot (\bar{b} \times \bar{c}) - \bar{c} \cdot (\bar{a} \times \bar{c})$.
Terms like $\bar{a} \cdot (\bar{a} \times \bar{c}) = 0$ and $\bar{c} \cdot (\bar{b} \times \bar{c}) = 0$.
Thus,we are left with $\bar{a} \cdot (\bar{b} \times \bar{c}) - \bar{b} \cdot (\bar{a} \times \bar{c}) = [\bar{a} \bar{b} \bar{c}] - (-[\bar{a} \bar{b} \bar{c}]) = 2[\bar{a} \bar{b} \bar{c}]$.
Since they are mutually perpendicular,$[\bar{a} \bar{b} \bar{c}] = |\bar{a}| |\bar{b}| |\bar{c}| = 1 \times 2 \times 3 = 6$.
Therefore,$2 \times 6 = 12$.

(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors,we have $|\vec{a}|=1, |\vec{b}|=2, |\vec{c}|=3$ and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$.
The scalar triple product is defined as $[\vec{x} \quad \vec{y} \quad \vec{z}] = (\vec{x} \times \vec{y}) \cdot \vec{z}$.
We need to evaluate $[\vec{a}+\vec{b}+\vec{c} \quad \vec{b}-\vec{a} \quad \vec{c}] = ((\vec{a}+\vec{b}+\vec{c}) \times (\vec{b}-\vec{a})) \cdot \vec{c}$.
Expanding the cross product: $(\vec{a}+\vec{b}+\vec{c}) \times (\vec{b}-\vec{a}) = \vec{a} \times \vec{b} - \vec{a} \times \vec{a} + \vec{b} \times \vec{b} - \vec{b} \times \vec{a} + \vec{c} \times \vec{b} - \vec{c} \times \vec{a}$.
Since $\vec{x} \times \vec{x} = 0$ and $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$,we get $\vec{a} \times \vec{b} + \vec{a} \times \vec{b} + \vec{c} \times \vec{b} - \vec{c} \times \vec{a} = 2(\vec{a} \times \vec{b}) + \vec{c} \times \vec{b} - \vec{c} \times \vec{a}$.
Now,taking the dot product with $\vec{c}$: $(2(\vec{a} \times \vec{b}) + \vec{c} \times \vec{b} - \vec{c} \times \vec{a}) \cdot \vec{c} = 2(\vec{a} \times \vec{b}) \cdot \vec{c} + (\vec{c} \times \vec{b}) \cdot \vec{c} - (\vec{c} \times \vec{a}) \cdot \vec{c}$.
Since the scalar triple product is zero if any two vectors are the same,$(\vec{c} \times \vec{b}) \cdot \vec{c} = 0$ and $(\vec{c} \times \vec{a}) \cdot \vec{c} = 0$.
Thus,the expression simplifies to $2(\vec{a} \times \vec{b}) \cdot \vec{c} = 2[\vec{a} \quad \vec{b} \quad \vec{c}]$.
Since $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular,$[\vec{a} \quad \vec{b} \quad \vec{c}] = |\vec{a}| |\vec{b}| |\vec{c}| = 1 \times 2 \times 3 = 6$.
Therefore,the result is $2 \times 6 = 12$.

(C) Concept: The energy $U$ stored in a capacitor is given by the formula:
$U = \frac{q^2}{2C}$
where $q$ is the charge and $C$ is the capacitance.
Let the original charge be $q$. The initial energy is $U = \frac{q^2}{2C}$.
When the charge is increased by $\Delta q = 2 \ C$,the new charge becomes $(q + 2)$.
The new energy $U'$ is given as $U' = U + 21\% \text{ of } U = 1.21U$.
Substituting the formula for energy:
$1.21 \left( \frac{q^2}{2C} \right) = \frac{(q + 2)^2}{2C}$
$1.21 q^2 = (q + 2)^2$
Taking the square root on both sides:
$\sqrt{1.21} q = q + 2$
$1.1q = q + 2$
$0.1q = 2$
$q = \frac{2}{0.1} = 20 \ C$.
Therefore,the original charge on the capacitor is $20 \ C$.

(C) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.
Let the original charge be $q$. The original energy is $U = \frac{q^2}{2C}$.
When the charge is increased by $2 \ C$,the new charge becomes $(q + 2) \ C$.
The new energy $U'$ is $21 \%$ more than the original energy,so $U' = U + 0.21U = 1.21U$.
Substituting the expressions for energy: $\frac{(q + 2)^2}{2C} = 1.21 \times \frac{q^2}{2C}$.
Canceling $2C$ from both sides,we get $(q + 2)^2 = 1.21q^2$.
Taking the square root of both sides: $q + 2 = 1.1q$.
Rearranging the terms: $1.1q - q = 2$,which gives $0.1q = 2$.
Therefore,$q = \frac{2}{0.1} = 20 \ C$.

(B) Let the total current $I = 2.1 \text{ A}$ enter at point $P$ and leave at point $R$. Let the current through the branch $PQ$ be $i$. Then the current through the branch $QR$ is also $i$ (assuming the galvanometer branch $QS$ has zero current,which is true if the bridge is balanced,but here we must check the potentials).
Applying Kirchhoff's Voltage Law $(KVL)$ to the loops or using the potential divider principle:
The potential difference between $P$ and $R$ is the same for both parallel paths $PQR$ and $PSR$.
For path $PQR$,the total resistance is $R_1 = 5 \Omega + 1 \Omega = 6 \Omega$.
For path $PSR$,the total resistance is $R_2 = 12.5 \Omega + 2.5 \Omega = 15 \Omega$.
Let $i_1$ be the current through path $PQR$ and $i_2$ be the current through path $PSR$.
Since $i_1 + i_2 = 2.1 \text{ A}$ and $i_1 R_1 = i_2 R_2$:
$i_1 (6) = i_2 (15) \Rightarrow i_2 = \frac{6}{15} i_1 = 0.4 i_1$.
Substituting into the total current equation:
$i_1 + 0.4 i_1 = 2.1 \text{ A} \Rightarrow 1.4 i_1 = 2.1 \text{ A} \Rightarrow i_1 = \frac{2.1}{1.4} = 1.5 \text{ A}$.
However,looking at the provided diagram,the current $i$ is flowing through the $1 \Omega$ resistor. The current $i_1$ calculated above is the current through the entire branch $PQR$,which includes the $1 \Omega$ resistor. Thus,the current through the $1 \Omega$ resistor is $1.5 \text{ A}$.

(D) The magnetic field produced by the larger coil of radius '$R$' at its center is given by $B = \frac{\mu_0 I}{2R}$.
Since the smaller coil of radius '$r$' is placed at the center and $R \gg r$,we can assume the magnetic field is uniform over the area of the smaller coil.
The magnetic flux $\phi$ passing through the smaller coil is $\phi = B \times A$,where $A = \pi r^2$ is the area of the smaller coil.
$\phi = \left( \frac{\mu_0 I}{2R} \right) \times (\pi r^2) = \frac{\mu_0 \pi r^2 I}{2R}$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{I}$.
Substituting the expression for $\phi$,we get $M = \frac{\mu_0 \pi r^2}{2R}$.
Therefore,the mutual inductance $M$ is proportional to $\frac{r^2}{R}$.

(B) The frequency of a simple pendulum is given by $n = \frac{1}{2\pi} \sqrt{\frac{g}{\ell}}$.
Since $n \propto \sqrt{g}$,the ratio of the new frequency $n'$ to the original frequency $n$ is $\frac{n'}{n} = \sqrt{\frac{g'}{g}}$.
The acceleration due to gravity at a depth $d$ is given by $g' = g(1 - \frac{d}{R})$.
Substituting $d = \frac{R}{2}$,we get $g' = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.
Therefore,$\frac{n'}{n} = \sqrt{\frac{g/2}{g}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Thus,the new frequency is $n' = \frac{n}{\sqrt{2}}$.

(B) The kinetic energy per mole of an ideal gas is given by the formula $E = \frac{3}{2} R T$.
Here,$E$ is the kinetic energy per mole,$R$ is the universal gas constant,and $T$ is the absolute temperature.
To find the universal gas constant $R$,we rearrange the equation:
$R = \frac{2 E}{3 T}$.
Therefore,the correct option is $B$.

(C) When the lift is stationary,the reading of the spring balance is equal to the weight of the bag,$W = mg = 49 ~N$.
Taking $g = 9.8 ~m/s^2$,the mass of the bag is $m = \frac{49}{9.8} = 5 ~kg$.
When the lift moves downward with an acceleration $a = 5 ~m/s^2$,the apparent weight $W'$ is given by the formula $W' = m(g - a)$.
Substituting the values,we get $W' = 5 \times (9.8 - 5) = 5 \times 4.8 = 24 ~N$.
Therefore,the reading of the spring balance will be $24 ~N$.

(C) The force per unit length between two parallel conductors carrying currents $I_1$ and $I_2$ separated by distance $d$ is given by $F = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
Since the currents are in the same direction,the force is attractive. Let the initial force be $F = k \frac{I_1 I_2}{d}$,where $k = \frac{\mu_0}{2 \pi}$.
In the new scenario,the current in one conductor becomes $2I_1$ (or $2I_2$) and the direction is reversed,making the force repulsive (negative sign).
The new distance is $d' = 3d$.
The new force $F'$ is given by $F' = -k \frac{(2I_1) I_2}{3d} = -\frac{2}{3} \left( k \frac{I_1 I_2}{d} \right) = -\frac{2}{3} F$.

(C) The angular momentum $L$ of a particle of mass $m$ moving in a circular orbit of radius $r$ with angular speed $\omega$ is given by $L = I\omega = (mr^2)\omega$.
The magnetic moment $M$ associated with the circulating charge is given by $M = IA$,where $I$ is the equivalent current and $A$ is the area of the orbit.
The current $I = \frac{q}{T} = \frac{q}{2\pi/\omega} = \frac{q\omega}{2\pi}$.
The area $A = \pi r^2$.
Thus,$M = \left(\frac{q\omega}{2\pi}\right)(\pi r^2) = \frac{1}{2}q\omega r^2$.
The ratio of the magnetic moment to the angular momentum is $\frac{M}{L} = \frac{\frac{1}{2}q\omega r^2}{mr^2\omega} = \frac{q}{2m}$.
Therefore,the ratio depends only on the charge '$q$' and the mass '$m$' of the particle.

(C) Let $T$ be the surface tension of mercury.
Total surface energy before coalescence $(E_1)$: Since there are two drops of radius $R$,$E_1 = 2 \times (4 \pi R^2 T) = 8 \pi R^2 T$.
Let the radius of the new large drop be $R'$. By conservation of volume: $\frac{4}{3} \pi R'^3 = 2 \times \frac{4}{3} \pi R^3$,which gives $R'^3 = 2R^3$,or $R' = 2^{1/3} R$.
Total surface energy after coalescence $(E_2)$: $E_2 = 4 \pi R'^2 T = 4 \pi (2^{1/3} R)^2 T = 4 \pi (2^{2/3}) R^2 T$.
The ratio of surface energies is $\frac{E_1}{E_2} = \frac{8 \pi R^2 T}{4 \pi (2^{2/3}) R^2 T} = \frac{2}{2^{2/3}} = 2^{1 - 2/3} = 2^{1/3}$.
Thus,the ratio is $2^{1/3} : 1$.

(C) The height of water rise in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
From this,we see that $h \propto \frac{1}{r}$.
The cross-sectional area $A$ is given by $A = \pi r^2$,which implies $r \propto \sqrt{A}$.
Substituting this into the proportionality for $h$,we get $h \propto \frac{1}{\sqrt{A}}$.
Let the initial height be $h_1 = h$ and initial area be $A_1 = A$. Let the new height be $h_2$ and new area be $A_2 = \frac{A}{9}$.
Using the ratio: $\frac{h_2}{h_1} = \sqrt{\frac{A_1}{A_2}}$.
Substituting the values: $\frac{h_2}{h} = \sqrt{\frac{A}{A/9}} = \sqrt{9} = 3$.
Therefore,$h_2 = 3h$.

(B) The height $h$ to which water rises in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$,which implies $h \propto \frac{1}{r}$.
If the radius changes from $r_1 = r$ to $r_2 = \frac{r}{3}$,the new height $h_2$ becomes $h_2 = h_1 \times \frac{r_1}{r_2} = h \times \frac{r}{r/3} = 3h$.
The mass of water in the capillary is given by $m = V \rho = (\pi r^2 h) \rho$.
For the first tube,$m_1 = \pi r^2 h \rho = m$.
For the second tube,$m_2 = \pi (r_2)^2 h_2 \rho = \pi (\frac{r}{3})^2 (3h) \rho$.
$m_2 = \pi (\frac{r^2}{9}) (3h) \rho = \frac{1}{3} \pi r^2 h \rho$.
Therefore,$m_2 = \frac{m}{3}$.

(A) For body $B$,the distance traveled is $x = V t$.
For body $A$,starting from rest,the distance traveled is $x = \frac{1}{2} a t^2$.
Since they start from the same point and meet at time $t$,their displacements must be equal:
$\frac{1}{2} a t^2 = V t$
Dividing both sides by $t$ (assuming $t \neq 0$):
$\frac{1}{2} a t = V$
$t = \frac{2V}{a}$

(A) Given: Initial velocity $u = 15 ~m/s$,acceleration $a = -0.3 ~m/s^2$,time $t = 60 ~s$.
First,we calculate the time taken to stop: $v = u + at \implies 0 = 15 - 0.3t \implies t = 50 ~s$.
Since the driver applies brakes for $60 ~s$,the vehicle stops at $t = 50 ~s$ and remains at rest for the remaining $10 ~s$.
Distance covered in $50 ~s$: $s = ut + \frac{1}{2}at^2 = 15(50) + \frac{1}{2}(-0.3)(50)^2 = 750 - 375 = 375 ~m$.
Distance from the traffic light $= 400 ~m - 375 ~m = 25 ~m$.

(B) The centripetal acceleration $a$ of a particle moving in a circular path of radius $R$ with speed $v$ is given by the formula: $a = \frac{v^2}{R}$.
Since the radius $R$ of the circular path remains constant,the acceleration is directly proportional to the square of the speed: $a \propto v^2$.
Let the initial speed be $v_1 = v$ and the initial acceleration be $a_1 = a = \frac{v^2}{R}$.
When the speed is doubled,the new speed is $v_2 = 2v$.
The new acceleration $a_2$ is given by: $a_2 = \frac{(2v)^2}{R} = \frac{4v^2}{R} = 4a$.
Therefore,the ratio of the acceleration after the change to the acceleration before the change is: $\frac{a_2}{a_1} = \frac{4a}{a} = \frac{4}{1}$.
Thus,the ratio is $4:1$.

(A) The frequency of the $S.H.M.$ is $f = 5 \ Hz$.
Angular frequency $\omega = 2\pi f = 2\pi \times 5 = 10\pi \ rad/s$.
At the highest point of oscillation,the spring is unstretched,which means the extension at the equilibrium position is equal to the amplitude $A$ of the oscillation.
Thus,$A = \Delta l = \frac{mg}{k}$.
Since $\omega^2 = \frac{k}{m}$,we have $\frac{g}{\omega^2} = \frac{mg}{k} = A$.
Substituting the values: $A = \frac{10}{(10\pi)^2} = \frac{10}{100\pi^2} = \frac{1}{10\pi^2} \ m$.
The maximum speed is given by $V_{\max} = A\omega$.
$V_{\max} = \left(\frac{1}{10\pi^2}\right) \times (10\pi) = \frac{1}{\pi} \ m/s$.

(B) Given: Moment of inertia $I = 1.2 ~kg \cdot m^2$,Rotational Kinetic Energy $K.E. = 1500 ~J$,Angular acceleration $\alpha = 25 ~rad/s^2$,Initial angular velocity $\omega_0 = 0$.
Step $1$: Calculate the final angular velocity $\omega$ using the rotational kinetic energy formula:
$K.E. = \frac{1}{2} I \omega^2$
$1500 = \frac{1}{2} \times 1.2 \times \omega^2$
$1500 = 0.6 \times \omega^2$
$\omega^2 = \frac{1500}{0.6} = 2500$
$\omega = 50 ~rad/s$.
Step $2$: Calculate the time $t$ using the kinematic equation for rotation:
$\omega = \omega_0 + \alpha t$
$50 = 0 + 25 \times t$
$t = \frac{50}{25} = 2 ~s$.

(D) The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.
For a solid sphere rotating about its diameter,the moment of inertia is $I_{\text{sphere}} = \frac{2}{5} M R^2$.
Thus,$K_{\text{sphere}} = \frac{1}{2} \left( \frac{2}{5} M R^2 \right) \omega_{\text{sphere}}^2 = \frac{1}{5} M R^2 \omega_{\text{sphere}}^2$.
For a solid cylinder rotating about its geometrical axis,the moment of inertia is $I_{\text{cylinder}} = \frac{1}{2} M R^2$.
Given that $\omega_{\text{cylinder}} = 2 \omega_{\text{sphere}}$,the kinetic energy is $K_{\text{cylinder}} = \frac{1}{2} \left( \frac{1}{2} M R^2 \right) (2 \omega_{\text{sphere}})^2 = \frac{1}{4} M R^2 (4 \omega_{\text{sphere}}^2) = M R^2 \omega_{\text{sphere}}^2$.
The ratio is $\frac{K_{\text{sphere}}}{K_{\text{cylinder}}} = \frac{\frac{1}{5} M R^2 \omega_{\text{sphere}}^2}{M R^2 \omega_{\text{sphere}}^2} = \frac{1}{5}$.

(A) The moment of inertia $I$ of a body in terms of its mass $m$ and radius of gyration $K$ is given by $I = mK^2$.
Angular momentum $L$ is related to moment of inertia $I$ and angular velocity $\omega$ by the formula $L = I\omega$.
Substituting the expression for $I$,we get $L = (mK^2)\omega$.
Rearranging to solve for angular velocity $\omega$,we get $\omega = \frac{L}{mK^2}$.

(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:

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

(C) perfectly black body is defined as an ideal body that absorbs all incident radiation of any wavelength.
According to Kirchhoff's law of thermal radiation,for an arbitrary body in thermal equilibrium,the emissivity $(e)$ is equal to its absorptivity $(a)$.
Since a perfectly black body has an absorptivity of $a = 1$,its emissivity must also be $e = 1$.
Therefore,the coefficient of emission for a perfectly black body is unity.

(B) The relationship between temperature in Kelvin $(T_K)$ and temperature in Celsius $(T_C)$ is given by the formula:
$T_K = T_C + 273$
Given $T_C = -197^{\circ} C$.
Substituting the value:
$T_K = -197 + 273$
$T_K = 76 ~K$

(A) For an adiabatic process,the relation between pressure and volume is given by $P_1 V_1^\gamma = P_2 V_2^\gamma$.
Given that the final volume $V_2 = V_1 / 8$,so the ratio $V_1 / V_2 = 8$.
The adiabatic index for a monoatomic gas is $\gamma = 5/3$.
Substituting these values into the equation:
$\frac{P_2}{P_1} = \left( \frac{V_1}{V_2} \right)^\gamma = (8)^{5/3}$.
Since $8 = 2^3$,we have $(2^3)^{5/3} = 2^5 = 32$.
Therefore,the ratio of the final pressure to the initial pressure is $32$.

(D) We know that for an ideal gas,the relationship between specific heats is given by Mayer's relation: $C_p - C_v = R$.
Given that the ratio of specific heats is $\gamma = \frac{C_p}{C_v}$,we can write $C_p = \gamma C_v$.
Substituting this into Mayer's relation: $\gamma C_v - C_v = R$.
Factoring out $C_v$,we get $C_v(\gamma - 1) = R$.
Therefore,$C_v = \frac{R}{\gamma - 1}$.

(B) Given:
Slit separation $d = 1 ~mm = 10^{-3} ~m$
Distance between slit and screen $D = 1.33 ~m$
Refractive index of water $\mu = 1.33$
Wavelength of light in air $\lambda = 6300 ~Å = 6.3 \times 10^{-7} ~m$
The wavelength of light in water is given by $\lambda_w = \frac{\lambda}{\mu} = \frac{6.3 \times 10^{-7}}{1.33} ~m$.
The fringe width $X$ is given by the formula:
$X = \frac{\lambda_w D}{d}$
Substituting the values:
$X = \frac{(6.3 \times 10^{-7} / 1.33) \times 1.33}{10^{-3}}$
$X = \frac{6.3 \times 10^{-7}}{10^{-3}} = 6.3 \times 10^{-4} ~m$.
Thus,the fringe width is $6.3 \times 10^{-4} ~m$.

(B) The resultant intensity $I$ at any point in Young's double slit experiment is given by the formula $I = I_{max} \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.
Given that $I_0$ is the maximum intensity,we have $I = I_0 \cos^2(\frac{\phi}{2})$.
The relationship between phase difference $\phi$ and path difference $\Delta x$ is $\phi = \frac{2\pi}{\lambda} \Delta x$.
Given the path difference $\Delta x = \frac{\lambda}{6}$,the phase difference is $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$.
Substituting this into the intensity formula:
$\frac{I}{I_0} = \cos^2(\frac{\pi/3}{2}) = \cos^2(\frac{\pi}{6})$.
Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,we get $\frac{I}{I_0} = (\frac{\sqrt{3}}{2})^2 = \frac{3}{4}$.

(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:
$\text{Beat frequency} = f_1 - f_2 = \frac{158}{\pi} - \frac{155}{\pi} = \frac{3}{\pi} \text{ Hz}$.

(A) The reaction of benzene diazonium chloride with cuprous chloride $(CuCl)$ in the presence of hydrochloric acid $(HCl)$ to form chlorobenzene is a classic example of the Sandmeyer reaction.
In this reaction,the diazonium group $(-N_2^+Cl^-)$ is replaced by a chlorine atom.

(A) The given reaction is a Wurtz reaction: $2R-Cl + 2Na \xrightarrow{\text{dry ether}} R-R + 2NaCl$.
To find '$A$',we split the product $3,4-$diethyl$-3,4-$dimethylhexane at the bond between $C3$ and $C4$.
The product is $CH_3CH_2-C(CH_3)(CH_2CH_3)-C(CH_3)(CH_2CH_3)-CH_2CH_3$.
Splitting this at the central $C-C$ bond gives two identical fragments: $CH_3CH_2-C(CH_3)(CH_2CH_3)-$.
Replacing the radical with a chlorine atom gives $CH_3CH_2-C(Cl)(CH_3)-CH_2CH_3$,which is $3-$chloro$-3-$methylpentane.

(C) The reaction that converts the carbonyl group $(>C=O)$ of aldehydes and ketones into a methylene group $(>CH_2)$ using zinc amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$ is known as the Clemmensen reduction.
The general reaction is:
$R-CO-R' \xrightarrow{Zn-Hg/HCl} R-CH_2-R'$
Therefore,the correct option is $C$.

(A) The Wurtz reaction involves the coupling of two alkyl halide molecules in the presence of sodium metal to form a symmetrical alkane.
To obtain $2,2,3,3-$tetramethylbutane,we must identify the alkyl group that,when doubled,forms this structure.
The structure of $2,2,3,3-$tetramethylbutane is $(CH_3)_3C-C(CH_3)_3$.
Splitting this molecule at the central $C-C$ bond gives two $(CH_3)_3C-$ groups.
Therefore,the required alkyl halide is $tert-$butyl bromide,$(CH_3)_3C-Br$.

(C) The reaction of two molecules of aryl halides (like chlorobenzene) with sodium metal in the presence of dry ether to form diaryl compounds (like biphenyl) is known as the Fittig reaction.
The chemical equation is:
$2C_6H_5Cl + 2Na \xrightarrow{\text{dry ether}} C_6H_5-C_6H_5 + 2NaCl$
Therefore,the correct option is $C$.

(B) The conversion of alkyl chlorides or alkyl bromides to alkyl iodides is known as the Finkelstein reaction.
In this reaction,the alkyl halide is treated with sodium iodide $(NaI)$ in the presence of dry acetone.
The reaction is: $R-X + NaI \xrightarrow{\text{dry acetone}} R-I + NaX$ (where $X = Cl, Br$).

(A) The dehydration of $2-$Methylhexan$-3-$ol with concentrated $H_2SO_4$ proceeds via the formation of a carbocation intermediate.
Upon elimination of a proton,multiple alkenes can be formed.
According to the Saytzeff rule,the more substituted alkene is the major product.
$2-$Methylhexan$-3-$ol undergoes dehydration to form $2-$Methylhex$-2-$ene as the major product because it is a trisubstituted alkene,which is more stable than the disubstituted $2-$Methylhex$-3-$ene.

(A) Step $1$: Dehydration of $Propan-1-ol$ with $Al_2O_3$ at $623 \ K$ gives propene $(A)$ as the product.
$CH_3-CH_2-CH_2-OH \xrightarrow[623 \ K]{Al_2O_3} CH_3-CH=CH_2 (A)$
Step $2$: Addition of $conc. H_2SO_4$ to propene follows $Markovnikov's$ rule to form isopropyl hydrogen sulphate,which upon hydrolysis with water $(H_2O, \Delta)$ yields $Propan-2-ol$ $(B)$.
$CH_3-CH=CH_2$ $\xrightarrow{conc. H_2SO_4} CH_3-CH(OSO_3H)-CH_3$ $\xrightarrow{H_2O, \Delta} CH_3-CH(OH)-CH_3 (B)$
Thus,the final product $B$ is $Propan-2-ol$.

(A) Tertiary butyl bromide $(CH_3)_3C-Br$ is a tertiary alkyl halide. When it reacts with an alcoholic solution of $NH_3$,it undergoes a dehydrohalogenation reaction ($\alpha, \beta-$elimination).
In this reaction,the base (ammonia) abstracts a proton from the $\beta-$carbon,leading to the elimination of $HBr$ and the formation of an alkene.
The reaction is: $(CH_3)_3C-Br \xrightarrow{alc. NH_3} CH_3-C(CH_3)=CH_2 + HBr$.
The product formed is $2-$methylpropene.

(B) Step $1$: Dehydration of $CH_3-CH_2-CH_2-OH$ (Propan$-1-$ol) with $Al_2 O_3$ at $623 \ K$ gives $A$,which is Propene $(CH_3-CH=CH_2)$.
Step $2$: Propene reacts with conc. $H_2 SO_4$ followed by hydrolysis $(H_2 O)$ to undergo hydration according to Markovnikov's rule.
Step $3$: The reaction follows the mechanism: $CH_3-CH=CH_2 + H_2 SO_4$ $\rightarrow CH_3-CH(OSO_3 H)-CH_3$ $\xrightarrow{H_2 O} CH_3-CH(OH)-CH_3$ (Propan$-2-$ol).
Thus,the product $B$ is Propan$-2-$ol.

(D) The given reaction is a Friedel-Crafts acylation reaction.
In this reaction,chlorobenzene reacts with an acylating agent in the presence of an anhydrous Lewis acid catalyst like $AlCl_3$.
The reaction is:
$\text{Chlorobenzene} + CH_3COCl$ $\xrightarrow[anhydrous]{AlCl_3} 2-\text{chloroacetophenone} + 4-\text{chloroacetophenone}$
Here,the reactant $(A)$ is acetyl chloride $(CH_3COCl)$.

(B) Chlorobenzene undergoes electrophilic aromatic substitution (nitration) when treated with a mixture of concentrated $HNO_3$ and $H_2SO_4$.
Since the chlorine atom is ortho/para-directing,the reaction yields a mixture of $2-$chloronitrobenzene and $4-$chloronitrobenzene.

(C) The reaction of $Toluene$ with $Chromyl chloride$ $(CrO_2Cl_2)$ in the presence of $Carbon disulfide$ $(CS_2)$ followed by acid hydrolysis $(H_3O^+)$ to form $Benzaldehyde$ is known as the $Etard reaction$.

(D) When ethyl benzene is treated with strong oxidizing agents like $HNO_3$ or $KMnO_4$,the alkyl side chain attached to the benzene ring is oxidized to a carboxylic acid group $(-COOH)$.
Thus,ethyl benzene undergoes strong oxidation to form benzoic acid.

(A) $N_2O_5$ is a non-metallic oxide of nitrogen in a high oxidation state $(+5)$,which makes it acidic in nature.
$NO$ is a neutral oxide.
$Na_2O$ is a metallic oxide,which makes it basic in nature.
$CO$ is a neutral oxide.

(A) Selenium exists in two allotropic forms: red (non-metallic) and grey (metallic).

(C) $SO_2$ is used as a food preservative,as an antichlor,and in the manufacture of $H_2SO_4$ (via the contact process).
However,$SO_2$ does not form oleum with concentrated $H_2SO_4$.
Oleum $(H_2S_2O_7)$ is formed by the reaction of sulfur trioxide $(SO_3)$ with concentrated $H_2SO_4$.

(C) Mustard gas,also known as sulfur mustard,has the chemical formula $S(CH_2CH_2Cl)_2$.
It is a potent vesicant chemical warfare agent that causes severe blistering and burning of the skin,eyes,and respiratory tract.

(C) At room temperature,$F_2$ and $Cl_2$ are gases,$Br_2$ is a liquid,and $I_2$ and $At$ are solids. Therefore,the correct answer is $Bromine$.

(C) $IF_3$ (Iodine trifluoride) is an interhalogen compound.
It is a yellow solid at room temperature,whereas $ClF$,$IF_7$,and $ClF_3$ are colourless gases.

(C) The acidic strength of hydrogen halides depends on the bond dissociation enthalpy of the $H-X$ bond.
As we move down the group from $F$ to $I$,the atomic size increases,which leads to a decrease in the bond dissociation enthalpy.
Consequently,the ease of releasing an $H^{+}$ ion increases down the group.
Therefore,the acidic strength increases in the order: $HF < HCl < HBr < HI$.
Thus,$HF$ has the lowest acidic strength.

(C) The oxidation states of $Cl$ in the given acids are as follows:
$1$. Hypochlorous acid $(HClO)$: $+1$
$2$. Chlorous acid $(HClO_2)$: $+3$
$3$. Chloric acid $(HClO_3)$: $+5$
$4$. Perchloric acid $(HClO_4)$: $+7$
Therefore,the $Cl$ atom in Perchloric acid $(HClO_4)$ has the highest oxidation state of $+7$.

(C) Ionization enthalpy decreases on moving down the group,and $Xe$ has the lowest ionization enthalpy among the given noble gases.
Due to this low ionization energy,$Xe$ can easily form stable crystalline fluorides such as $XeF_2$,$XeF_4$,and $XeF_6$ upon reaction with fluorine.

(C) Argon is used in the welding of specialty alloys as well as in the welding of automobile frames,mufflers,and other automotive parts. It is called a shield gas because it does not react with the gases and metals in the vicinity of the welding process. It merely takes up space and prevents unwanted reactions from occurring due to reactive gases such as $N_2$ and $O_2$.

(C) The filament in an electric bulb is made using tungsten,which would burn quickly when it comes in contact with oxygen in the air.
This is the reason why unreactive gases such as $Ar$ and $N_2$ are used for filling electric bulbs.
These gases are chemically inactive and prevent the oxidation of the tungsten filament,thereby increasing the lifespan of the bulb.

(A) Semisynthetic polymers,such as cellulose acetate,are used in the preparation of cinema films.

(C) The polymerization of tetrafluoroethylene to produce polytetrafluoroethylene $(PTFE)$ is carried out by heating or exposing tetrafluoroethylene to a free radical initiator such as ammonium persulphate or hydrogen peroxide under high pressure.

(A) $\varepsilon$-Caprolactam does not undergo vinyl polymerization.
It undergoes ring-opening polymerization to form Nylon-$6$.

(C) $Nylon-6$ is prepared by heating caprolactam with water at a high temperature.
In this process,the ring of caprolactam opens to form aminocaproic acid,which then undergoes polymerization to form $Nylon-6$.
The reaction is as follows:
$\text{Caprolactam}$ $\xrightarrow{H_2O, \text{Heat}} H_2N-(CH_2)_5-COOH$ $\xrightarrow{\text{Heat}} (-NH-(CH_2)_5-CO-)_n \text{ (Nylon-6)}$

(C) Polyacrylamide is a synthetic polymer that forms a cross-linked gel structure when polymerized. This gel is widely used in gel electrophoresis to separate biological molecules like proteins and nucleic acids based on their size and charge.

(C) $Buna-N$ is a synthetic rubber copolymer of $1,3-butadiene$ and $acrylonitrile$. It is highly resistant to the action of petrol,lubricating oil,and organic solvents. Due to these properties,it is primarily used in the manufacture of oil seals,tank linings,and adhesives.

(A) $Polyvinyl$ $chloride$ $(PVC)$ is a polymer formed by the polymerization of $vinyl$ $chloride$ monomer. The chemical formula for $vinyl$ $chloride$ is $CH_2=CHCl$. The polymerization reaction is as follows:
$n(CH_2=CHCl) \xrightarrow{\text{catalyst}} -[CH_2-CHCl]_n-$

(A) homopolymer is a polymer formed from only one type of monomer unit.
$Teflon$ is a homopolymer of tetrafluoroethene $(CF_2=CF_2)$.
$Nylon$ $2-$nylon $6$,$PHBV$,and $Bakelite$ are copolymers because they are formed from two or more different types of monomer units.

(C) In the process of vulcanization,sulfur $(S)$ is added to the rubber to form crosslinks between polymer chains,which improves the mechanical properties of the rubber. Therefore,$S$ forms crosslinks in the vulcanization of $SBR$ rubber.

(D) Neoprene is a synthetic rubber formed by the free-radical polymerization of chloroprene ($2$-chloro-$1,3$-butadiene).
It is an addition polymer,not a condensation polymer.
It is highly resistant to petroleum,vegetable oils,and light.
Therefore,the statement that it is a condensation polymer is incorrect.

(D) The polymer obtained from $\varepsilon$-caprolactam is Nylon-$6$.
When $\varepsilon$-caprolactam is heated with water at high temperature,it undergoes ring-opening polymerization to form Nylon-$6$.

(C) copolymer is a polymer formed from two or more different types of monomer units.
Buna-$S$ is a copolymer formed by the polymerization of $1,3-$butadiene and styrene.

(B) $PVC$ (Polyvinyl chloride) is widely used in the manufacturing of floor tiles,pipes,and various construction materials due to its durability and chemical resistance.

(C) Buna-$S$ is a copolymer of $1,3-$butadiene and styrene.
It is obtained by the polymerization of $1,3-$butadiene and styrene in the ratio of $3:1$ in the presence of a catalyst (often sodium or free radical initiator).
The reaction is represented as:
$nCH_2=CH-CH=CH_2 + nC_6H_5CH=CH_2 \xrightarrow{\text{Polymerization}} (-CH_2-CH=CH-CH_2-CH(C_6H_5)-CH_2-)_n$

(B) $SBR$ (Styrene-Butadiene Rubber) is a synthetic rubber used extensively in the manufacture of automobile tyres due to its high abrasion resistance and durability.

(B) $LDPE$ (Low Density Polyethylene) is used to obtain shopping bags because of its flexibility and toughness.

(C) cyclic amide is also known as a lactam.
$\varepsilon$-caprolactam is a seven-membered cyclic amide that is used as a monomer for the preparation of Nylon-$6$.
Therefore,the correct option is $C$.

(A) $HDP$ stands for high-density polyethylene. It is a linear polymer with high density due to close packing.
$HDP$ is obtained by the polymerization of ethene in the presence of a Ziegler-Natta catalyst,which is a combination of triethylaluminium and titanium tetrachloride,at a temperature of $333 \ K$ to $343 \ K$ and a pressure of $6-7 \ atm$.

(B) Teflon is a polymer formed by the polymerization of tetrafluoroethylene $(CF_2=CF_2)$.
The reaction is carried out by heating tetrafluoroethylene with a free radical or persulphate catalyst at high pressure.
The chemical equation is: $nCF_2=CF_2 \xrightarrow[(NH_4)_2S_2O_8]{\text{Heat, pressure}} -[CF_2-CF_2]_n-$
Thus,the monomer used is tetrafluoroethylene.

(B) $Terylene$ (also known as $Dacron$) is a polyester fiber.
It is prepared by the condensation polymerization of $Ethylene \ glycol$ $(HO-CH_2-CH_2-OH)$ and $Terephthalic \ acid$ $(HOOC-C_6H_4-COOH)$.
The reaction is as follows:
$n(HO-CH_2-CH_2-OH) + n(HOOC-C_6H_4-COOH) \rightarrow -[O-CH_2-CH_2-O-CO-C_6H_4-CO]_n- + 2nH_2O$
Therefore,the correct monomers are $Ethylene \ glycol$ and $Terephthalic \ acid$.

(D) Melamine-formaldehyde resin is formed by the co-polymerisation of melamine and formaldehyde.
It is a classic example of a cross-linked polymer,where the polymer chains are connected by covalent bonds to form a three-dimensional network structure.

(B) Ozone acts as a reducing agent in reactions where it reduces peroxides to oxides,while itself being reduced to oxygen.
In the reaction $BaO_{2(s)} + O_{3(g)} \rightarrow BaO_{(s)} + 2 O_{2(g)}$,ozone reduces barium peroxide $(BaO_2)$ to barium oxide $(BaO)$.

(D) There are $14$ distinct Bravais lattices that can be formed within the $7$ crystal systems.

(C) $X$-ray crystallography is used to identify the molecular and atomic structure of a crystal.
The crystal diffracts the incident $X$-ray beam.
By measuring the intensities and angles of these diffracted beams,the molecular structure of the crystal can be evaluated.
The instrument used for this purpose is known as an $X$-ray diffractometer.

(D) Two or more substances having the same crystal structure are called isomorphous.
They show the same atomic ratio (Iso- Same,Morphous- Form).
Examples include $NaF$ and $MgO$ ($1:1$ ratio),and $NaNO_3$ and $CaCO_3$ ($1:1:3$ ratio).
$NaCl$ and $KCl$ have similar properties but different crystal structures.