MHT CET 2008 Chemistry Question Paper with Answer and Solution

(C) An open window acts as a perfectly black body because any radiation entering the window is absorbed by the interior of the room and does not reflect back.
For a perfectly black body,the absorption coefficient $(a)$ is defined as the ratio of the energy absorbed to the total energy incident on the body.
Since a black body absorbs all incident radiation,the absorption coefficient is $a = 1$.

(B) Option $(a)$ is incorrect because during clear nights,objects on the surface of the Earth radiate heat,causing the temperature near the ground to fall.
Option $(c)$ is incorrect because the total energy radiated by a black body per unit time per unit area is given by Stefan's law,$E \propto T^4$,not $T^2$.
For option $(d)$,the energy radiated per second is given by $P = A \varepsilon \sigma T^4 = 4 \pi r^2 \varepsilon \sigma T^4$.
Calculating the ratio: $\frac{P_1}{P_2} = \left( \frac{r_1}{r_2} \right)^2 \left( \frac{T_1}{T_2} \right)^4 = \left( \frac{1}{4} \right)^2 \left( \frac{4000}{2000} \right)^4 = \frac{1}{16} \times 16 = 1$.
Since $P_1 = P_2$,option $(d)$ is incorrect.
Option $(b)$ is correct because Newton's law of cooling is an approximate form of Stefan's law of radiation and is valid for small temperature differences,typically observed in natural convection.

(C) Metals reflect incident light due to the oscillations of free electrons under the influence of the electric field of the incident light wave.
As the temperature increases,the random motion of free electrons increases,which leads to more frequent collisions with the lattice ions.
This increase in collisions results in an increase in electrical resistance,and consequently,the electrical conductivity of the metal decreases.
Therefore,the bonding in such a solid is metallic.

(A) compound is chiral if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
$1-$chloropentane: $CH_2(Cl)-CH_2-CH_2-CH_2-CH_3$. The carbon atom bonded to chlorine is attached to two identical hydrogen atoms,so it is achiral.
$2-$chloropentane: $CH_3-CH(Cl)-CH_2-CH_2-CH_3$. The $C-2$ atom is bonded to $-H, -Cl, -CH_3, \text{ and } -CH_2CH_2CH_3$. Since all four groups are different,it is chiral.
$1-$chloro$-2-$methylpentane: $CH_2(Cl)-CH(CH_3)-CH_2-CH_2-CH_3$. The $C-2$ atom is bonded to $-H, -CH_3, -CH_2Cl, \text{ and } -CH_2CH_2CH_3$. Since all four groups are different,it is chiral.
$3-$chloro$-2-$methylpentane: $CH_3-CH(CH_3)-CH(Cl)-CH_2-CH_3$. The $C-3$ atom is bonded to $-H, -Cl, -CH(CH_3)_2, \text{ and } -CH_2CH_3$. Since all four groups are different,it is chiral.
Therefore,$1-$chloropentane is the only achiral compound.

(A) Molarity is defined as the number of moles of solute per liter of solution $(M = \frac{n}{V(L)})$.
Since volume $(V)$ is temperature-dependent,molarity changes with temperature.
Molality,mole fraction,and weight fraction are based on mass,which is independent of temperature.

(B) Chloramine-$T$ is an antiseptic.
Antiseptics are chemical substances that either kill or prevent the growth of microorganisms and can be applied to living tissues.
Examples include $0.2$ percent solution of phenol,$KMnO_4$,and Chloramine-$T$.

(C) The general equation of a pair of straight lines is given by $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$.
Comparing this with the given equation $12x^2 + 7xy + ay^2 + 13x - y + 3 = 0$,we get $A = 12$ and $B = a$.
For a pair of lines to be perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero,i.e.,$A + B = 0$.
Substituting the values,we get $12 + a = 0$.
Therefore,$a = -12$.

(C) Let the one end of the diameter be $A = (1, 1)$.
Since the other end $B$ lies on the line $x + y = 3$,we can represent $B$ as $(t, 3 - t)$.
The centre $(h, k)$ of the circle is the midpoint of the diameter $AB$.
Thus,$h = \frac{1 + t}{2}$ and $k = \frac{1 + (3 - t)}{2} = \frac{4 - t}{2}$.
From these equations,we have $t = 2h - 1$ and $t = 4 - 2k$.
Equating the two expressions for $t$:
$2h - 1 = 4 - 2k$
$2h + 2k = 5$
Replacing $(h, k)$ with $(x, y)$,the locus of the centre is $2x + 2y = 5$.

(A) Given the circle equation $x^2 + y^2 = 13$.
Let the point on the circle be $(2, y')$.
Substituting $x = 2$ into the circle equation:
$2^2 + (y')^2 = 13$
$4 + (y')^2 = 13$
$(y')^2 = 9$
$y' = \pm 3$.
So,the points are $(2, 3)$ and $(2, -3)$.
The equation of the tangent to the circle $x^2 + y^2 = r^2$ at point $(x_1, y_1)$ is given by $xx_1 + yy_1 = r^2$.
For point $(2, 3)$: $2x + 3y = 13$.
For point $(2, -3)$: $2x - 3y = 13$.
Thus,the equations of the tangents are $2x + 3y = 13$ and $2x - 3y = 13$.

(B) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.
Let the point on the parabola be $(h, k)$.
Given that the ordinate is twice the abscissa,we have $k = 2h$.
Since the point lies on the parabola,$k^2 = 16h$.
Substituting $k = 2h$ into the equation,we get $(2h)^2 = 16h$,which simplifies to $4h^2 = 16h$.
This gives $4h^2 - 16h = 0$,or $4h(h - 4) = 0$.
Thus,$h = 0$ or $h = 4$.
If $h = 0$,then $k = 0$. The point is $(0, 0)$. The focal distance is $h + a = 0 + 4 = 4$.
If $h = 4$,then $k = 8$. The point is $(4, 8)$. The focal distance is $h + a = 4 + 4 = 8$.
Since $4$ is not among the options,the required focal distance is $8$.

(B) For the parabola $y^2 = 8x$,we have $4a = 8$,so $a = 2$.
Any tangent to the parabola is given by $y = mx + \frac{a}{m}$,which simplifies to $mx - y + \frac{2}{m} = 0$.
If this line is a tangent to the circle $x^2 + y^2 = 2$,the perpendicular distance from the center $(0, 0)$ to the line must equal the radius $r = \sqrt{2}$.
Thus,$\frac{|2/m|}{\sqrt{m^2 + 1}} = \sqrt{2}$.
Squaring both sides,we get $\frac{4}{m^2} = 2(m^2 + 1)$,which leads to $2 = m^2(m^2 + 1)$.
$m^4 + m^2 - 2 = 0$.
Factoring gives $(m^2 + 2)(m^2 - 1) = 0$.
Since $m^2$ must be positive,$m^2 = 1$,so $m = \pm 1$.
Substituting $m = 1$ into the tangent equation,we get $y = x + \frac{2}{1} = x + 2$.
Substituting $m = -1$,we get $y = -x - 2$.

(C) Let the vertices of the triangle be $O(0)$,$A(a + b)$,and $B(a - b)$.
The area of the triangle is given by $\frac{1}{2} |\vec{OA} \times \vec{OB}|$.
Substituting the vectors,we get $\text{Area} = \frac{1}{2} |(a + b) \times (a - b)|$.
Expanding the cross product: $(a + b) \times (a - b) = a \times a - a \times b + b \times a - b \times b$.
Since $a \times a = 0$ and $b \times b = 0$,and $b \times a = -(a \times b)$,we have:
$(a + b) \times (a - b) = 0 - (a \times b) - (a \times b) - 0 = -2(a \times b) = 2(b \times a)$.
Thus,$\text{Area} = \frac{1}{2} |2(b \times a)| = |b \times a| = |b| |a| \sin(90^\circ)$.
Given $|a| = 2$ and $|b| = 3$,and the vectors are perpendicular $(\theta = 90^\circ)$:
$\text{Area} = 3 \times 2 \times 1 = 6$.

(D) Let the points be $P_1 = 6a - 4b + 4c$,$P_2 = -4c$,$P_3 = -a - 2b - 3c$,and $P_4 = a + 2b - 5c$.
The equation of the line passing through $P_1$ and $P_2$ is $r = P_1 + m(P_2 - P_1) = 6a - 4b + 4c + m(-6a - 4b - 8c)$ ..... $(i)$.
The equation of the line passing through $P_3$ and $P_4$ is $r = P_3 + n(P_4 - P_3) = -a - 2b - 3c + n(2a + 4b - 2c)$ ..... $(ii)$.
For the point of intersection,the equations $(i)$ and $(ii)$ must yield the same position vector $r$.
Equating the coefficients of $a, b,$ and $c$ from $(i)$ and $(ii)$:
$6 - 6m = -1 + 2n \implies 6m + 2n = 7$
$-4 - 4m = -2 + 4n \implies 4m + 4n = -2 \implies 2m + 2n = -1$
$4 - 8m = -3 - 2c \implies 8m - 2n = 7$
Subtracting the second equation from the first: $(6m + 2n) - (2m + 2n) = 7 - (-1) \implies 4m = 8 \implies m = 2$.
Substituting $m = 2$ into $2m + 2n = -1$: $4 + 2n = -1 \implies 2n = -5 \implies n = -2.5$.
Checking these values in the third equation: $8(2) - 2(-2.5) = 16 + 5 = 21 \neq 7$.
Since the system of equations is inconsistent,the lines do not intersect.
Therefore,the correct option is $(d)$.

(C) Let $a, b, c$ be the direction ratios of the required line.
Since the line lies in both planes,its direction vector is perpendicular to the normals of both planes,$(3, 2, 1)$ and $(1, 1, -2)$.
Thus,$3a + 2b + c = 0$ and $a + b - 2c = 0$.
Using cross product,$\frac{a}{(-4 - 1)} = \frac{b}{(1 + 6)} = \frac{c}{(3 - 2)}$,which gives $\frac{a}{-5} = \frac{b}{7} = \frac{c}{1}$.
To find a point on the line,set $z = 0$ in the given equations: $3x + 2y = 5$ and $x + y = 3$.
Solving these,we get $x = -1$ and $y = 4$.
Thus,the point is $(-1, 4, 0)$.
The symmetric equation is $\frac{x - (-1)}{-5} = \frac{y - 4}{7} = \frac{z - 0}{1}$,which simplifies to $\frac{x + 1}{-5} = \frac{y - 4}{7} = \frac{z - 0}{1}$.

(C) The equation of a plane passing through the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ is given by $a(x + 1) + b(y - 3) + c(z + 2) = 0$,where $-3a + 2b + c = 0$ (since the normal vector is perpendicular to the line direction vector $(-3, 2, 1)$).
Since the plane passes through the point $(0, 7, -7)$,we substitute these coordinates into the plane equation:
$a(0 + 1) + b(7 - 3) + c(-7 + 2) = 0$
$a + 4b - 5c = 0$.
Now we have the system of equations:
$1) -3a + 2b + c = 0$
$2) a + 4b - 5c = 0$
Using cross-multiplication to find the ratios of $a, b, c$:
$\frac{a}{(2)(-5) - (1)(4)} = \frac{b}{(1)(1) - (-3)(-5)} = \frac{c}{(-3)(4) - (2)(1)}$
$\frac{a}{-10 - 4} = \frac{b}{1 - 15} = \frac{c}{-12 - 2}$
$\frac{a}{-14} = \frac{b}{-14} = \frac{c}{-14}$
This simplifies to $a = b = c$. Let $a = b = c = 1$.
The equation of the plane is $1(x + 1) + 1(y - 3) + 1(z + 2) = 0$,which simplifies to $x + y + z = 0$.

(C) We are given $f(x) = \sqrt{\frac{x - \sin x}{x + \cos^2 x}}$.
To find $\lim_{x \to \infty} f(x)$,we divide the numerator and denominator inside the square root by $x$:
$\lim_{x \to \infty} \sqrt{\frac{x - \sin x}{x + \cos^2 x}} = \lim_{x \to \infty} \sqrt{\frac{1 - \frac{\sin x}{x}}{1 + \frac{\cos^2 x}{x}}}$
As $x \to \infty$,we know that $\frac{\sin x}{x} \to 0$ and $\frac{\cos^2 x}{x} \to 0$ because $-1 \le \sin x \le 1$ and $0 \le \cos^2 x \le 1$.
Therefore,the limit becomes $\sqrt{\frac{1 - 0}{1 + 0}} = \sqrt{1} = 1$.

(C) For the function $f(x)$ to be continuous at $x = 0$,the following condition must be satisfied:
$f(0) = \lim_{x \to 0} f(x)$
Given $f(0) = k$,we need to evaluate the limit:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} \left( x \sin \frac{1}{x} \right)$
We know that for all $x \ne 0$,the sine function is bounded:
$-1 \le \sin \frac{1}{x} \le 1$
Multiplying by $x$ (where $x > 0$):
$-x \le x \sin \frac{1}{x} \le x$
Applying the Squeeze Theorem as $x \to 0$:
$\lim_{x \to 0} (-x) = 0$ and $\lim_{x \to 0} (x) = 0$
Therefore,$\lim_{x \to 0} x \sin \frac{1}{x} = 0$.
Since $f(0) = k$,we have $k = 0$.

(A) Given equation: $x^3 + y^3 - 3axy = 0$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) - 3a \frac{d}{dx}(xy) = 0$
Using the chain rule and product rule:
$3x^2 + 3y^2 \frac{dy}{dx} - 3a \left( x \frac{dy}{dx} + y \right) = 0$
Divide the entire equation by $3$:
$x^2 + y^2 \frac{dy}{dx} - ax \frac{dy}{dx} - ay = 0$
Group the terms containing $\frac{dy}{dx}$:
$\frac{dy}{dx}(y^2 - ax) = ay - x^2$
Therefore,$\frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax}$.

(C) Given the velocity $v = \frac{ds}{dt} = 6t - \frac{t^2}{6}$.
To find the distance $s$,we integrate the velocity with respect to time $t$:
$s = \int (6t - \frac{t^2}{6}) dt = 3t^2 - \frac{t^3}{18} + C$.
Given that $s = 0$ at $t = 0$,we substitute these values to find the constant $C$:
$0 = 3(0)^2 - \frac{0^3}{18} + C \implies C = 0$.
Thus,the expression for distance is $s = 3t^2 - \frac{t^3}{18}$.
To find the distance traveled in $3$ seconds,we substitute $t = 3$:
$s = 3(3)^2 - \frac{3^3}{18} = 3(9) - \frac{27}{18} = 27 - \frac{3}{2} = \frac{54 - 3}{2} = \frac{51}{2}$.

(B) Let $f(x) = x^3 - 12x^2 + 36x + 17$.
To find the critical points,we calculate the derivative $f'(x)$:
$f'(x) = 3x^2 - 24x + 36$.
Setting $f'(x) = 0$:
$3(x^2 - 8x + 12) = 0$
$3(x - 2)(x - 6) = 0$
So,the critical points are $x = 2$ and $x = 6$.
Now,we evaluate the function $f(x)$ at the critical points and the endpoints of the interval $[1, 10]$:
$f(1) = (1)^3 - 12(1)^2 + 36(1) + 17 = 1 - 12 + 36 + 17 = 42$.
$f(2) = (2)^3 - 12(2)^2 + 36(2) + 17 = 8 - 48 + 72 + 17 = 49$.
$f(6) = (6)^3 - 12(6)^2 + 36(6) + 17 = 216 - 432 + 216 + 17 = 17$.
$f(10) = (10)^3 - 12(10)^2 + 36(10) + 17 = 1000 - 1200 + 360 + 17 = 177$.
Comparing these values,the maximum value is $177$ at $x = 10$.

(D) Given differential equation is $\sqrt{\frac{dy}{dx}} - 4\frac{dy}{dx} - 7x = 0$.
Rearranging the terms,we get $\sqrt{\frac{dy}{dx}} = 4\frac{dy}{dx} + 7x$.
Squaring both sides to remove the radical,we obtain $\frac{dy}{dx} = (4\frac{dy}{dx} + 7x)^2$.
Expanding the right side,we get $\frac{dy}{dx} = 16(\frac{dy}{dx})^2 + 49x^2 + 56x\frac{dy}{dx}$.
The highest derivative present is $\frac{dy}{dx}$,so the order is $1$.
The highest power of the highest derivative after making the equation a polynomial in derivatives is $2$,so the degree is $2$.
Thus,the order is $1$ and the degree is $2$.

(C) Let $x - y = v$. Then,differentiating with respect to $x$,we get $1 - \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dv}{dx}$.
Substituting this into the given differential equation: $1 - \frac{dv}{dx} = \frac{v + 3}{2v + 5}$.
Rearranging the terms: $\frac{dv}{dx} = 1 - \frac{v + 3}{2v + 5} = \frac{2v + 5 - v - 3}{2v + 5} = \frac{v + 2}{2v + 5}$.
Separating the variables: $\int \frac{2v + 5}{v + 2} dv = \int dx$.
Simplifying the integrand: $\int \left( 2 + \frac{1}{v + 2} \right) dv = \int dx$.
Integrating both sides: $2v + \log|v + 2| = x + c$.
Substituting $v = x - y$ back into the equation: $2(x - y) + \log|x - y + 2| = x + c$.

(A) The given differential equation is $(3xy + y^2)dx + (x^2 + xy)dy = 0$.
This can be written as $\frac{dy}{dx} = - \frac{3xy + y^2}{x^2 + xy}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = - \frac{3x^2v + x^2v^2}{x^2 + x^2v} = - \frac{v^2 + 3v}{v + 1}$.
$x\frac{dv}{dx} = - \frac{v^2 + 3v}{v + 1} - v = - \frac{v^2 + 3v + v^2 + v}{v + 1} = - \frac{2v^2 + 4v}{v + 1} = - \frac{2v(v + 2)}{v + 1}$.
Separating the variables: $\frac{v + 1}{v(v + 2)} dv = - \frac{2}{x} dx$.
Using partial fractions: $\frac{v + 1}{v(v + 2)} = \frac{A}{v} + \frac{B}{v + 2} \implies v + 1 = A(v + 2) + Bv$.
For $v = 0$,$A = 1/2$. For $v = -2$,$B = 1/2$.
So,$\int (\frac{1}{2v} + \frac{1}{2(v + 2)}) dv = - \int \frac{2}{x} dx$.
$\frac{1}{2} \ln|v| + \frac{1}{2} \ln|v + 2| = - 2 \ln|x| + C$.
$\ln|v(v + 2)| = - 4 \ln|x| + C' \implies \ln|v(v + 2)x^4| = C'$.
$v(v + 2)x^4 = c^2$. Substituting $v = y/x$: $\frac{y}{x}(\frac{y}{x} + 2)x^4 = c^2$.
$x^2(y^2 + 2xy) = c^2$.

(B) Let $p$ be the statement: "It rains".
Let $q$ be the statement: "$I$ shall go to school".
The given conditional statement is $p \Rightarrow q$.
The negation of a conditional statement $p \Rightarrow q$ is given by $\sim (p \Rightarrow q) \equiv p \wedge \sim q$.
Here,$p$ is "It rains" and $\sim q$ is "$I$ shall not go to school".
Therefore,the negation is "It rains and $I$ shall not go to school".

(D) Let the initial angular velocity be $\omega_0$ and the final angular velocity be $\omega = \frac{\omega_0}{2}$. The angular displacement for $36$ rotations is $\theta_1 = 36 \times 2\pi = 72\pi \text{ rad}$.
Using the equation of motion $\omega^2 = \omega_0^2 + 2\alpha\theta$:
$(\frac{\omega_0}{2})^2 = \omega_0^2 + 2\alpha(72\pi)$
$\frac{\omega_0^2}{4} - \omega_0^2 = 144\pi\alpha$
$-\frac{3\omega_0^2}{4} = 144\pi\alpha \implies \alpha = -\frac{3\omega_0^2}{576\pi} = -\frac{\omega_0^2}{192\pi}$.
Now,for the second part,the fan starts with $\omega_i = \frac{\omega_0}{2}$ and comes to rest $(\omega_f = 0)$:
$0^2 = (\frac{\omega_0}{2})^2 + 2\alpha\theta_2$
$0 = \frac{\omega_0^2}{4} + 2(-\frac{\omega_0^2}{192\pi})\theta_2$
$\frac{\omega_0^2}{4} = \frac{\omega_0^2}{96\pi}\theta_2$
$\theta_2 = \frac{96\pi}{4} = 24\pi \text{ rad}$.
Number of rotations = $\frac{\theta_2}{2\pi} = \frac{24\pi}{2\pi} = 12$ rotations.

(C) chiral compound is one that contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
$1-$chloropentane is $CH_3-CH_2-CH_2-CH_2-CH_2Cl$. Here,the carbon atom attached to the chlorine atom is bonded to two hydrogen atoms,which are identical. Therefore,it is achiral.
$1-$chloro$-2-$methylpentane has a chiral center at $C-2$.
$2-$chloropentane has a chiral center at $C-2$.
$3-$chloro$-2-$methylpentane has a chiral center at $C-3$.

(A) compound is chiral if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
$1$. $1$-chloropentane: $CH_3-CH_2-CH_2-CH_2-CH_2Cl$. The $C_1$ atom is bonded to two hydrogen atoms,so it is achiral.
$2$. $2$-chloropentane: $CH_3-CH(Cl)-CH_2-CH_2-CH_3$. The $C_2$ atom is bonded to $-H, -Cl, -CH_3, -CH_2CH_2CH_3$. It is chiral.
$3$. $1$-chloro-$2$-methylpentane: $CH_2(Cl)-CH(CH_3)-CH_2-CH_2-CH_3$. The $C_2$ atom is bonded to $-H, -CH_3, -CH_2Cl, -CH_2CH_2CH_3$. It is chiral.
$4$. $3$-chloro-$2$-methylpentane: $CH_3-CH(CH_3)-CH(Cl)-CH_2-CH_3$. The $C_3$ atom is bonded to $-H, -Cl, -CH_2CH_3, -CH(CH_3)_2$. It is chiral.
Therefore,$1$-chloropentane is the only achiral compound.

(A) molecule is chiral if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
$1-$Chloropentane $(CH_3-CH_2-CH_2-CH_2-CH_2Cl)$: The carbon atom attached to the chlorine is bonded to two hydrogen atoms. Since it is not bonded to four different groups,it is achiral.
$2-$Chloropentane $(CH_3-CHCl-CH_2-CH_2-CH_3)$: The $C-2$ atom is bonded to $-H, -Cl, -CH_3,$ and $-CH_2CH_2CH_3$. It is chiral.
$1-$Chloro$-2-$methylpentane $(CH_3-CH_2-CH_2-CH(CH_3)-CH_2Cl)$: The $C-2$ atom is bonded to $-H, -CH_3, -CH_2CH_2CH_3,$ and $-CH_2Cl$. It is chiral.
$3-$Chloro$-2-$methylpentane $(CH_3-CH(CH_3)-CHCl-CH_2-CH_3)$: The $C-3$ atom is bonded to $-H, -Cl, -CH_2CH_3,$ and $-CH(CH_3)_2$. It is chiral.
Therefore,$1-$chloropentane is not chiral.

(D) When the surface area of a liquid is increased,molecules from the interior of the liquid rise to the surface.
As these molecules reach the surface,work is done against the cohesive force.
This work is stored in the molecules in the form of potential energy.
Thus,the potential energy of the molecules lying on the surface is greater than that of the molecules in the interior of the liquid.

(D) The energy of the incident photon must be at least equal to the band gap energy $(E_g)$ for the material to absorb it.
Given $E_g = 2.0\, eV$.
The relationship between energy and frequency is given by $E = h\nu$,where $h$ is Planck's constant $(6.63 \times 10^{-34}\, J\cdot s)$ and $\nu$ is the frequency.
First,convert the energy from $eV$ to Joules:
$E = 2.0 \times 1.6 \times 10^{-19}\, J = 3.2 \times 10^{-19}\, J$.
Now,calculate the minimum frequency $\nu$:
$\nu = \frac{E}{h} = \frac{3.2 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 0.4826 \times 10^{15}\, Hz$.
$\nu \approx 4.83 \times 10^{14}\, Hz$.
Rounding to the nearest given option,we get $\nu \approx 5 \times 10^{14}\, Hz$.

(C) $1$. Identify the longest carbon chain containing the ether functional group. The structure is $CH_3-CH(OC_3H_7)-C_3H_7$.
$2$. The longest chain has $5$ carbon atoms,which is a pentane chain.
$3$. The alkoxy group $-OC_3H_7$ (propoxy group) is attached to the second carbon atom of the pentane chain.
$4$. Numbering the chain from the end closer to the substituent gives the position $2$ for the propoxy group.
$5$. Therefore,the $IUPAC$ name is $2-$propoxy pentane.

(C) molecule is chiral if it contains at least one chiral center (an asymmetric carbon atom bonded to four different groups).
$1-$chloropentane $(CH_3-CH_2-CH_2-CH_2-CH_2Cl)$ has no chiral carbon atom because the $C_1$ carbon is bonded to two identical hydrogen atoms.
$2-$chloropentane,$1-$chloro$-2-$methylpentane,and $3-$chloro$-2-$methylpentane all contain at least one chiral carbon atom.
Therefore,$1-$chloropentane is not chiral.

(D) Grignard reagents like methyl magnesium iodide $(CH_3MgI)$ act as strong bases and react with compounds containing acidic hydrogen atoms.
Terminal alkynes,which have a hydrogen atom attached to a triply bonded carbon atom $(R-C \equiv C-H)$,possess acidic hydrogen.
Among the given options,$CH_3CH_2CH_2C \equiv CH$ is a terminal alkyne.
The reaction is:
$CH_3CH_2CH_2C \equiv CH + CH_3MgI \rightarrow CH_3CH_2CH_2C \equiv CMgI + CH_4 \uparrow$
Therefore,option $(D)$ is correct.

(C) species that acts as both a Bronsted acid and a base is called an amphoteric species. It must be able to donate a proton $(H^+)$ and accept a proton $(H^+)$.
$HPO_4^{2-}$ can accept a proton to form $H_2PO_4^-$ (acting as a base) and donate a proton to form $PO_4^{3-}$ (acting as an acid).
$HPO_4^{2-} + H_2O \rightleftharpoons H_2PO_4^- + OH^-$
$HPO_4^{2-} + H_2O \rightleftharpoons PO_4^{3-} + H_3O^+$
$H_2PO_2^-$ is the conjugate base of $H_3PO_2$ (a monobasic acid) and cannot donate any more protons.
$HPO_3^{2-}$ is the conjugate base of $H_2PO_3^-$ (derived from the dibasic acid $H_3PO_3$) and cannot donate any more protons.

(C) For a weak base,the concentration of hydroxyl ions is given by $[OH^{-}] = \sqrt{K_{b} \times C}$.
$[OH^{-}] = \sqrt{1.0 \times 10^{-5} \times 0.1} = \sqrt{1.0 \times 10^{-6}} = 1.0 \times 10^{-3} \ M$.
Using the ionic product of water,$K_{w} = [H^{+}] [OH^{-}] = 1.0 \times 10^{-14}$.
$[H^{+}] = \frac{K_{w}}{[OH^{-}]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-3}} = 1.0 \times 10^{-11} \ M$.
The $pH$ is calculated as $pH = -\log [H^{+}]$.
$pH = -\log (1.0 \times 10^{-11}) = 11$.

(A) The dissociation of $Ca(OH)_{2}$ in water is represented as:
$Ca(OH)_{2}(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^{-}(aq)$
Let the solubility be $s \ mol \ L^{-1}$.
At equilibrium,$[Ca^{2+}] = s$ and $[OH^{-}] = 2s$.
The solubility product expression is:
$K_{sp} = [Ca^{2+}][OH^{-}]^{2}$
Substituting the values:
$K_{sp} = (s)(2s)^{2} = s \times 4s^{2} = 4s^{3}$

(A) Isobars are atoms of different elements that have the same mass number but different atomic numbers.
For ${ }_{20} Ca^{40}$,the mass number is $40$ and the atomic number is $20$.
Among the options,${ }_{18} Ar^{40}$ has the same mass number $(40)$ but a different atomic number $(18)$.
Therefore,${ }_{18} Ar^{40}$ is an isobar of ${ }_{20} Ca^{40}$.

(B) Given equations:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}; \Delta H = r$ ...$(I)$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)}; \Delta H = s$ ...$(II)$
We need to find the heat of formation of $CO$,which corresponds to the reaction:
$C_{(s)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{(g)}; \Delta H = ?$
Subtracting equation $(II)$ from equation $(I)$:
$(C_{(s)} + O_{2(g)}) - (CO_{(g)} + \frac{1}{2} O_{2(g)}) \longrightarrow CO_{2(g)} - CO_{2(g)}$
$C_{(s)} + \frac{1}{2} O_{2(g)} - CO_{(g)} \longrightarrow 0$
$C_{(s)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{(g)}$
The enthalpy change for this reaction is $\Delta H = r - s$.

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$.
Here,$\Delta n_g$ is the change in the number of moles of gaseous species,calculated as: $\Delta n_g = \sum n_{g, \text{products}} - \sum n_{g, \text{reactants}}$.
For the reaction $C_2H_5OH_{(l)} + 3O_{2(g)} \longrightarrow 2CO_{2(g)} + 3H_2O_{(l)}$,the gaseous moles are:
Products: $2 \text{ moles of } CO_2$.
Reactants: $3 \text{ moles of } O_2$.
$\Delta n_g = 2 - 3 = -1$.
Substituting this value into the equation,we get: $\Delta H = \Delta E + (-1)RT = \Delta E - RT$.

(C) Given the curve equation is $y^{2}=2(x-3)$ ...$(i)$
On differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 2$,which implies $\frac{dy}{dx} = \frac{1}{y}$.
The slope of the tangent at any point $(x, y)$ is $m_{T} = \frac{1}{y}$.
The slope of the normal at $(x, y)$ is $m_{N} = -\frac{1}{m_{T}} = -y$.
The given line is $y - 2x + 1 = 0$,which can be written as $y = 2x - 1$. The slope of this line is $m = 2$.
Since the normal is parallel to the line,their slopes must be equal:
$-y = 2 \Rightarrow y = -2$.
Substituting $y = -2$ into the curve equation $(i)$:
$(-2)^{2} = 2(x - 3)$
$4 = 2(x - 3)$
$2 = x - 3 \Rightarrow x = 5$.
Thus,the required point is $(5, -2)$.

(D) In the given circuit,there are two parallel branches.
Each branch consists of two switches connected in series.
The upper branch has switches $\sim p$ and $q$ in series,which corresponds to the expression $(\sim p \wedge q)$.
The lower branch has switches $p$ and $\sim q$ in series,which corresponds to the expression $(p \wedge \sim q)$.
Since the two branches are connected in parallel,the total Boolean polynomial is the disjunction of the two expressions: $(\sim p \wedge q) \vee (p \wedge \sim q)$.

(D) The current $I$ flowing through the galvanometer circuit is given by Ohm's law: $I = \frac{V}{R_{total}}$.
Initially,the total resistance is $R_{total,1} = 50 \Omega + 2950 \Omega = 3000 \Omega$.
The current for $30$ divisions is $I_1 = \frac{3 \text{ V}}{3000 \Omega} = 10^{-3} \text{ A}$.
Since the deflection is proportional to the current,the current per division is $\frac{10^{-3} \text{ A}}{30}$.
To obtain a deflection of $20$ divisions,the required current is $I_2 = 20 \times \left( \frac{10^{-3}}{30} \right) = \frac{2}{3} \times 10^{-3} \text{ A}$.
Let the new total resistance be $R_{total,2}$. Then $I_2 = \frac{3 \text{ V}}{R_{total,2}}$.
$\frac{2}{3} \times 10^{-3} = \frac{3}{R_{total,2}} \implies R_{total,2} = \frac{3 \times 3}{2 \times 10^{-3}} = 4500 \Omega$.
The total resistance in the circuit is $R_{total,2} = R_G + R_{series}$.
$4500 \Omega = 50 \Omega + R_{series} \implies R_{series} = 4450 \Omega$.

(B) Let $i_{a}$ be the current flowing through the ammeter and $i$ be the total current. So,a current $(i - i_{a})$ will flow through the shunt resistance.
The potential difference across the ammeter and the shunt resistance is the same because they are connected in parallel.
Therefore,$i_{a} \times R = (i - i_{a}) \times S$
Rearranging for $S$,we get:
$S = \frac{i_{a} R}{i - i_{a}}$
Given values are $i_{a} = 100 \text{ A}$,$i = 750 \text{ A}$,and $R = 13 \Omega$.
Substituting these values into the formula:
$S = \frac{100 \times 13}{750 - 100}$
$S = \frac{1300}{650}$
$S = 2 \Omega$
Thus,the value of the shunt resistance is $2 \Omega$.

(A) At the top of the hill, the forces acting on the rider are the gravitational force $Mg$ (downwards) and the normal reaction $N$ (upwards).
The net centripetal force is provided by the difference between these forces:
$Mg - N = M \frac{v^2}{R}$
For the condition of "weightlessness", the normal reaction $N$ must be zero $(N = 0)$.
Substituting $N = 0$ into the equation:
$Mg = M \frac{v^2}{R}$
$g = \frac{v^2}{R}$
$v = \sqrt{Rg}$
Given $R = 20 \, m$ and taking $g = 10 \, m/s^2$:
$v = \sqrt{20 \times 10} = \sqrt{200} \approx 14.14 \, m/s$
Therefore, the speed of the car at the top of the hill is between $14 \, m/s$ and $15 \, m/s$.

(B) The time period of oscillation of a magnet is given by $T = 2 \pi \sqrt{\frac{I}{MH}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $H$ is the horizontal component of the Earth's magnetic field.
For the original magnet of mass $m$ and length $L$,$I = \frac{mL^2}{12}$ and $M = q_m L$ (where $q_m$ is pole strength).
When the magnet is cut into three equal parts along its length,each part has mass $m' = m/3$,length $L' = L/3$,and pole strength $q_m' = q_m$.
When these three parts are stacked together with like poles aligned,the new moment of inertia $I'$ is the sum of the moments of inertia of the three parts: $I' = 3 \times \frac{m'(L')^2}{12} = 3 \times \frac{(m/3)(L/3)^2}{12} = \frac{mL^2}{12 \times 9} = \frac{I}{9}$.
The new magnetic moment $M'$ is the sum of the magnetic moments of the three parts: $M' = 3 \times (q_m \times L/3) = q_m L = M$.
The new time period $T'$ is $T' = 2 \pi \sqrt{\frac{I'}{M'H}} = 2 \pi \sqrt{\frac{I/9}{MH}} = \frac{1}{3} \times 2 \pi \sqrt{\frac{I}{MH}} = \frac{T}{3}$.
Given $T = 2 ~s$,therefore $T' = 2/3 ~s$.

(B) The maximum acceleration of a particle in simple harmonic motion is given by the formula $a_{\max} = \omega^{2} A$,where $\omega$ is the angular frequency and $A$ is the amplitude.
Given that the amplitudes $A$ are the same for both motions,the ratio of their maximum accelerations is:
$\frac{(a_{\max})_{1}}{(a_{\max})_{2}} = \frac{\omega_{1}^{2} A}{\omega_{2}^{2} A} = \frac{\omega_{1}^{2}}{\omega_{2}^{2}}$
Substituting the given values $\omega_{1} = 100 \ rad \ s^{-1}$ and $\omega_{2} = 1000 \ rad \ s^{-1}$:
$\frac{(a_{\max})_{1}}{(a_{\max})_{2}} = \frac{(100)^{2}}{(1000)^{2}} = \left(\frac{100}{1000}\right)^{2} = \left(\frac{1}{10}\right)^{2} = \frac{1}{100} = 1: 10^{2}$

(C) Let the radii of the thin spherical shell and the solid sphere be $R_{1}$ and $R_{2}$ respectively.
The moment of inertia of the thin spherical shell about its diameter is given by $I_{1} = \frac{2}{3} M R_{1}^{2}$.
The moment of inertia of the solid sphere about its diameter is given by $I_{2} = \frac{2}{5} M R_{2}^{2}$.
Given that the masses $M$ and the moments of inertia $I$ are equal for both bodies,we have $I_{1} = I_{2}$.
Therefore,$\frac{2}{3} M R_{1}^{2} = \frac{2}{5} M R_{2}^{2}$.
Canceling $M$ and the factor $2$ from both sides,we get $\frac{R_{1}^{2}}{3} = \frac{R_{2}^{2}}{5}$.
Rearranging the terms,we get $\frac{R_{1}^{2}}{R_{2}^{2}} = \frac{3}{5}$.
Taking the square root on both sides,we get $\frac{R_{1}}{R_{2}} = \sqrt{\frac{3}{5}} = \frac{\sqrt{3}}{\sqrt{5}}$.
Thus,the ratio of their radii is $\sqrt{3}: \sqrt{5}$.

(B) $p-n$ photodiode is a semiconductor device that generates current when illuminated by light. For a photon to be absorbed by the semiconductor material, its energy $E$ must be at least equal to the band gap energy $E_g$.
Given, $E_g = 2.0 \text{ eV}$.
The relationship between energy and frequency is $E = h\nu$, where $h$ is Planck's constant $(6.63 \times 10^{-34} \text{ J} \cdot \text{s})$ and $\nu$ is the frequency.
Converting the band gap energy into Joules: $E_g = 2.0 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-19} \text{ J}$.
Setting $h\nu = E_g$, we get $\nu = \frac{E_g}{h} = \frac{3.2 \times 10^{-19} \text{ J}}{6.63 \times 10^{-34} \text{ J} \cdot \text{s}}$.
Calculating the value: $\nu \approx 0.482 \times 10^{15} \text{ Hz} = 4.82 \times 10^{14} \text{ Hz}$.
Rounding to the nearest option, we get $\nu \approx 5 \times 10^{14} \text{ Hz}$.

(A) The boiling point of alcohols depends on the extent of intermolecular hydrogen bonding and the surface area of the molecule.
As the branching increases from primary $(1^{\circ})$ to secondary $(2^{\circ})$ to tertiary $(3^{\circ})$ alcohols,the surface area of the molecule decreases.
$A$ smaller surface area leads to weaker van der Waals forces of attraction.
Additionally,steric hindrance in tertiary alcohols reduces the effectiveness of intermolecular hydrogen bonding.
Therefore,the boiling point decreases in the order: $1^{\circ} > 2^{\circ} > 3^{\circ}$.

(C) The mixture of anhydrous $ZnCl_2$ and conc. $HCl$ is known as Lucas reagent.
Lucas test is used to distinguish between primary,secondary,and tertiary alcohols.
Tertiary alcohols react immediately with Lucas reagent to produce turbidity due to the formation of alkyl chlorides.
Secondary alcohols give turbidity within $5-10 \ \text{min}$,while primary alcohols do not produce turbidity at room temperature.
Among the given options,$2-\text{hydroxy}-2-\text{methylpropane}$ is a $3^{\circ}$ alcohol,which is the most reactive and produces turbidity immediately.

(C) In the given reaction,the bromine atom $(-Br)$ in tert-butyl bromide is replaced by a hydroxyl group $(-OH)$ from water.
Since the incoming group is a nucleophile $(-OH^-)$,this is a nucleophilic substitution reaction.
Specifically,it follows an $S_N1$ mechanism because the substrate is a tertiary alkyl halide.

(C) The reaction of an alkyl halide $(R-X)$ with dry silver oxide $(Ag_2O)$ upon heating is a method for the preparation of ethers (Williamson ether synthesis variant).
The chemical equation is:
$2 R-X + Ag_2O \xrightarrow{\Delta} R-O-R + 2 AgX$
Comparing this with the given reaction $2 A + \text{dry silver oxide} \xrightarrow{\Delta} \text{ether} + 2 \operatorname{Ag} X$,we can conclude that $A$ represents an alkyl halide $(R-X)$.

(C) The reaction of a Grignard reagent $(RMgX)$ with a carbonyl compound followed by hydrolysis yields an alcohol.
To obtain $\text{tert-pentyl alcohol}$ $(CH_3CH_2C(CH_3)_2OH)$,we need to react a ketone with a Grignard reagent.
Here,the Grignard reagent is $C_2H_5MgI$ (ethylmagnesium iodide).
When $C_2H_5MgI$ reacts with $A$ (acetone,$CH_3COCH_3$),the nucleophilic ethyl group attacks the carbonyl carbon of acetone.
$CH_3COCH_3 + C_2H_5MgI \rightarrow CH_3C(OMgI)(CH_3)(C_2H_5)$.
Upon hydrolysis with $H_2O/H^+$,this forms $CH_3C(OH)(CH_3)(C_2H_5)$,which is $2-$methylbutan$-2-$ol,commonly known as $\text{tert-pentyl alcohol}$.
Therefore,compound $A$ is acetone.

(D) The ninhydrin test is a chemical test used to detect ammonia or primary and secondary amines. In biochemistry,it is specifically used to detect proteins and amino acids,which produce a characteristic blue or purple color.
Benedict's solution is used to detect reducing sugars (monosaccharides and some disaccharides).
Since proteins are not reducing sugars,they give a negative Benedict's test.
Therefore,a protein is the compound that gives a positive ninhydrin test and a negative Benedict's solution test.

(A) The reaction of phenol with $CHCl_3$ and $KOH$ is the Reimer-Tiemann reaction,which yields salicylaldehyde ($2$-hydroxybenzaldehyde) as the intermediate $X$.
Salicylaldehyde contains an aldehyde group but no $\alpha$-hydrogen atoms.
When treated with $50 \%$ $KOH$ solution,it undergoes the Cannizzaro reaction,a disproportionation reaction.
One molecule of salicylaldehyde is reduced to the corresponding alcohol ($2$-hydroxybenzyl alcohol),and another molecule is oxidized to the corresponding carboxylic acid salt (potassium $2-$hydroxybenzoate).

(D) The iodoform test is given by compounds that contain either the $CH_{3}CO-$ group or the $CH_{3}CH(OH)-$ group.
$2-$pentanone $(CH_{3}COCH_{2}CH_{2}CH_{3})$ contains the $CH_{3}CO-$ group.
Ethanal $(CH_{3}CHO)$ contains the $CH_{3}CO-$ group.
Ethanol $(CH_{3}CH_{2}OH)$ contains the $CH_{3}CH(OH)-$ group.
However,$3-$pentanone $(CH_{3}CH_{2}COCH_{2}CH_{3})$ does not contain either of these groups,so it does not give the iodoform test.

(A) $1$. Ester $(A)$ is $C_6H_5COOC_2H_5$ (Ethyl benzoate).
$2$. Reaction with excess $CH_3MgBr$ followed by acid workup yields a tertiary alcohol: $C_6H_5C(OH)(CH_3)_2$.
$3$. Dehydration of this alcohol with $H_2SO_4$ gives the olefin $(B)$,which is $C_6H_5-C(CH_3)=CH_2$ ($2$-phenylpropene).
$4$. Ozonolysis of $(B)$ $(C_6H_5-C(CH_3)=CH_2)$ yields acetophenone $(C_6H_5COCH_3)$,which has the formula $C_8H_8O$.
$5$. Acetophenone contains a methyl ketone group $(-COCH_3)$,which gives a positive iodoform test.
$6$. Therefore,the ester $(A)$ is $C_6H_5COOC_2H_5$.

(A) In the crossed aldol condensation between acetaldehyde $(CH_{3}CHO)$ and formaldehyde $(HCHO)$,the base $(OH^{-})$ abstracts an $\alpha$-hydrogen atom from the acetaldehyde molecule because the $\alpha$-hydrogens are acidic due to the electron-withdrawing effect of the carbonyl group.
This abstraction results in the formation of an enolate ion (carbanion) intermediate,which is represented as $: \overline{C}H_{2}CHO$.

(B) The basic strength of amines depends on the availability of the lone pair of electrons on the nitrogen atom.
Alkyl groups are electron-releasing groups that exert a positive inductive effect $(+I)$,which increases the electron density on the nitrogen atom,thereby increasing the basicity.
In $NH_3$ (ammonia),there is no alkyl group attached to the nitrogen atom.
Therefore,$NH_3$ is the weakest base among the given options.

(C) Amides react with bromine and caustic soda to give their corresponding primary amines. This reaction is known as Hofmann's bromamide degradation reaction.
The chemical equation is:
$CH_{3}CONH_{2} + Br_{2} + 4KOH \xrightarrow{343K} CH_{3}NH_{2} + 2KBr + K_{2}CO_{3} + 2H_{2}O$
Thus,acetamide $(CH_{3}CONH_{2})$ gives methanamine $(CH_{3}NH_{2})$.

(D) Benzyl amine $(C_6H_5CH_2NH_2)$ is a primary $(1^{\circ})$ amine.
Primary amines react with chloroform $(CHCl_3)$ and ethanolic potassium hydroxide $(KOH)$ to form isocyanides (carbylamines),which have a foul smell.
This reaction is known as the carbylamine reaction.
The chemical equation is:
$C_6H_5CH_2NH_2 + CHCl_3 + 3KOH (alc.) \rightarrow C_6H_5CH_2NC + 3KCl + 3H_2O$
Thus,the product obtained is benzyl isocyanide.

(D) The reaction of glucose with phenylhydrazine proceeds in three steps:
$1$. Glucose reacts with one molecule of phenylhydrazine to form phenylhydrazone.
$2$. The phenylhydrazone then reacts with a second molecule of phenylhydrazine,which acts as an oxidizing agent,to form an intermediate keto-imine.
$3$. Finally,this intermediate reacts with a third molecule of phenylhydrazine to form the stable osazone.
Thus,a total of $3$ molecules of phenylhydrazine are required for the formation of osazone from one molecule of glucose.

(C) The reaction sequence is as follows:
$1$. The first step is the Hell-Volhard-Zelinsky $(HVZ)$ reaction,where propanoic acid reacts with $Cl_2$ in the presence of red phosphorus to form $\alpha$-chloropropanoic acid $(X)$:
$CH_3-CH_2-COOH \xrightarrow[\text{red } P]{Cl_2} CH_3-CHCl-COOH \text{ (} X \text{)}$
$2$. The second step is a dehydrohalogenation reaction using alcoholic $KOH$,which removes $HCl$ from the $\alpha$-carbon and $\beta$-carbon to form acrylic acid $(Y)$:
$CH_3-CHCl-COOH \xrightarrow{\text{Alc. KOH}} CH_2=CH-COOH \text{ (} Y \text{)}$
Therefore,the end product $Y$ is acrylic acid $(CH_2=CH-COOH)$.

(B) For a general reaction $aA + bB \longrightarrow cC + dD$,the rate of reaction is given by: $\text{Rate} = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$.
For the reaction $H_{2} + I_{2} \longrightarrow 2 HI$,the stoichiometric coefficients are $1, 1,$ and $2$ respectively.
Therefore,the differential rate law is: $\text{Rate} = -\frac{d[H_{2}]}{dt} = -\frac{d[I_{2}]}{dt} = \frac{1}{2} \frac{d[HI]}{dt}$.

(B) The rate law expression is given as: $\text{Rate} = \frac{+d[C]}{dt} = k[A]^1[B]^1$.
The order of reaction with respect to $A$ is $1$.
The order of reaction with respect to $B$ is $1$.
The total order of the reaction is the sum of the exponents of the concentration terms in the rate law: $1 + 1 = 2$.

(A) For a first order reaction,the half-life period $(t_{1/2})$ is given by the formula: $t_{1/2} = \frac{0.693}{k}$.
Given the rate constant $k = 1.155 \times 10^{-3} \ s^{-1}$.
Substituting the value of $k$ in the formula:
$t_{1/2} = \frac{0.693}{1.155 \times 10^{-3}} \ s$.
$t_{1/2} = \frac{0.693}{1.155} \times 10^{3} \ s$.
$t_{1/2} = 0.6 \times 1000 \ s$.
$t_{1/2} = 600 \ s$.

(C) $\gamma$-radiations are electromagnetic waves with no mass and no charge.
Due to their high energy and neutral nature,they have the highest penetrating power compared to charged particles like $\alpha$-particles,protons,or positrons.

(D) The number of half-lives $(n)$ is calculated as: $n = \frac{T}{t_{1/2}} = \frac{12 \ days}{3 \ days} = 4$.
Using the radioactive decay formula: $N = N_0 \times (1/2)^n$.
Given $N = 3 \ g$ and $n = 4$:
$3 = N_0 \times (1/2)^4$.
$3 = N_0 \times (1/16)$.
$N_0 = 3 \times 16 = 48 \ g$.

(D) The half-life period $(t_{1/2})$ for a reaction of order $n$ is given by the relation: $t_{1/2} \propto \frac{1}{[A]_0^{n-1}}$,where $[A]_0$ is the initial concentration.
For a first order reaction,$n = 1$.
Substituting $n = 1$ in the expression: $t_{1/2} \propto \frac{1}{[A]_0^{1-1}} = \frac{1}{[A]_0^0} = \text{constant}$.
Therefore,for a first order reaction,the half-life period is independent of the initial concentration.

(A) The nuclear reaction is: ${}^{238}_{92}U \longrightarrow {}^{206}_{82}Pb + m({}^{4}_{2}He) + n({}^{0}_{-1}e)$.
Comparing the mass numbers on both sides: $238 = 206 + 4m$ $\Rightarrow 4m = 32$ $\Rightarrow m = 8$.
Comparing the atomic numbers on both sides: $92 = 82 + 2m - n$.
Substituting $m = 8$: $92 = 82 + 2(8) - n$ $\Rightarrow 92 = 82 + 16 - n$ $\Rightarrow 92 = 98 - n$ $\Rightarrow n = 6$.
Therefore,$8$ $\alpha$-particles and $6$ $\beta$-particles are emitted.

(B) Tranquilisers are drugs used to relieve mental ailments.
These are also known as psychotherapeutic drugs as they act on the central nervous system.

(A) Aspirin (acetylsalicylic acid) is prepared by the acetylation of salicylic acid.
Salicylic acid is chemically known as $o-$hydroxybenzoic acid.
The reaction involves the treatment of salicylic acid with acetic anhydride in the presence of an acid catalyst.

(B) Antiseptics are chemical substances that either kill or prevent the growth of microorganisms on living tissues. $Chloramine-T$ is a well-known antiseptic used to treat wounds and infections. Other examples include $Dettol$,$Savlon$,$Acriflavin$,$Boric \ acid$,$Phenol$,$Iodoform$,$KMnO_4$,and $Methylene \ blue$.

(A) The electronic configuration of $Cerium$ ($Ce$,$Z=58$) is $[Xe] 4f^1 5d^1 6s^2$.
Due to the loss of three electrons,it forms a stable $+3$ oxidation state.
It also exhibits a $+4$ oxidation state because the removal of four electrons results in a stable noble gas configuration $([Xe])$.

(D) $Sc^{3+}$ has an electronic configuration of $[Ar] 3d^0$.
Since there are no electrons in the $d$-orbital,$d-d$ transitions are not possible,making the ion colourless.

(D) Transition elements are defined as elements which have incompletely filled $d$-orbitals in their ground state or in any of their oxidation states.
Generally,the $d$-block elements are referred to as transition elements.
The general electronic configuration of these elements is represented as $(n-1) d^{1-10}, n s^{1-2}$.

(D) Lanthanides and actinides are both considered inner transition elements and commonly exhibit a $+3$ oxidation state.
However,a key point of dissimilarity is their radioactivity.
All actinides are radioactive in nature,whereas among lanthanides,only promethium $(Pm)$ is radioactive,and the rest are non-radioactive.

(A) The reducing power of a substance is inversely proportional to its standard reduction potential $(E^{\circ}_{red})$.
$A$ more negative $E^{\circ}_{red}$ value indicates a greater tendency to undergo oxidation (loss of electrons),making the species a stronger reducing agent.
Comparing the given values:
$E^{\circ}_{Zn^{2+}/Zn} = -0.762 \ V$
$E^{\circ}_{Cr^{3+}/Cr} = -0.74 \ V$
$E^{\circ}_{H^{+}/H_2} = 0.00 \ V$
$E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.77 \ V$
Since $Zn_{(s)}$ has the most negative standard reduction potential $(-0.762 \ V)$,it is the strongest reducing agent among the given options.

(A) The given reaction is a disproportionation reaction: $2 Cu^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + Cu_{(s)}$.
This can be split into two half-reactions:
$1$. Oxidation: $Cu^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + e^{-}$,where $E^{\circ}_{ox} = -E^{\circ}_{Cu^{2+}/Cu^{+}} = -0.15 \ V$.
$2$. Reduction: $Cu^{+}_{(aq)} + e^{-} \rightarrow Cu_{(s)}$,where $E^{\circ}_{red} = E^{\circ}_{Cu^{+}/Cu}$.
To find $E^{\circ}_{Cu^{+}/Cu}$,we use the relation: $\Delta G^{\circ}_{Cu^{2+}/Cu} = \Delta G^{\circ}_{Cu^{2+}/Cu^{+}} + \Delta G^{\circ}_{Cu^{+}/Cu}$.
$-(2)F(E^{\circ}_{Cu^{2+}/Cu}) = -(1)F(E^{\circ}_{Cu^{2+}/Cu^{+}}) + -(1)F(E^{\circ}_{Cu^{+}/Cu})$.
$2(0.34) = 0.15 + E^{\circ}_{Cu^{+}/Cu} \implies E^{\circ}_{Cu^{+}/Cu} = 0.68 - 0.15 = 0.53 \ V$.
Now,$E^{\circ}_{cell} = E^{\circ}_{ox} + E^{\circ}_{red} = -0.15 \ V + 0.53 \ V = +0.38 \ V$.

(C) For a spontaneous cell reaction,the Gibbs free energy change $(\Delta G)$ must be negative.
According to the relation $\Delta G = -nFE^{\circ}_{cell}$,for $\Delta G$ to be negative,$E^{\circ}_{cell}$ must be positive.
Thus,both the condition of $\Delta G^{\circ} < 0$ and $E^{\circ}_{cell} > 0$ indicate spontaneity.
However,in the context of standard thermodynamic criteria for spontaneity,$\Delta G^{\circ}$ being negative is the fundamental condition.

(C) The isomeric ethers for the molecular formula $C_4H_{10}O$ are:
$1$. Diethyl ether: $CH_3CH_2OCH_2CH_3$
$2$. Methyl propyl ether: $CH_3OCH_2CH_2CH_3$
$3$. Methyl isopropyl ether: $CH_3OCH(CH_3)_2$
Hence,there are $3$ isomeric ethers possible.

(B) Zone refining is a technique used to produce metals of very high purity. It is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal. This method is particularly effective for obtaining ultra-pure samples of metals like $Zn$,$Ge$,$Si$,$B$,$Ga$,and $In$.

(C) Calamine is the carbonate ore of zinc,represented by the chemical formula $ZnCO_{3}$.

(C) Rayon is a semi-synthetic polymer obtained from wood pulp,which is why it is classified as a regenerated fibre. It is also commonly known as artificial silk because its texture and appearance resemble natural silk.

(C) $n HOCH_2-CH_2OH + n HOOC-C_6H_4-COOH \rightarrow HOCH_2CH_2O[OC-C_6H_4-CO-O-CH_2-CH_2-O]_{n-1} OC-C_6H_4-COOH + n H_2O$
Terylene (also known as $Dacron$ or $Terene$) is a condensation polymer formed by the reaction between ethylene glycol and terephthalic acid. It is widely used as a synthetic fibre.

(A) Molarity is defined as the number of moles of solute per unit volume of solution $(M = \frac{n}{V})$.
Since volume $(V)$ of a solution changes with temperature due to thermal expansion or contraction,molarity is temperature-dependent.
In contrast,molality,mole fraction,and weight fraction are based on mass,which remains constant regardless of temperature changes.

(A) Isotonic solutions have the same molar concentration of solute particles in the solution.
We calculate the molar concentration of particles (van't Hoff factor $i \times M$):
For $A$ (glucose): $1 \times 0.1 \ M = 0.1 \ M$.
For $B$ $(NaCl)$: $2 \times 0.05 \ M = 0.1 \ M$.
For $C$ $(BaCl_{2})$: $3 \times 0.05 \ M = 0.15 \ M$.
For $D$ $(AlCl_{3})$: $4 \times 0.1 \ M = 0.4 \ M$.
Since solutions $A$ and $B$ have the same concentration of particles $(0.1 \ M)$,they are isotonic.

(C) Given,vapour pressure of pure benzene,$p^{\circ} = 640 \ mm \ Hg$.
Vapour pressure of the solution,$p = 600 \ mm \ Hg$.
Weight of solute,$w = 2.175 \ g$.
Weight of solvent (benzene),$W = 39.08 \ g$.
Molecular weight of benzene,$M = 78 \ g/mol$.
Let the molecular weight of the solute be $m$.
According to Raoult's law for non-volatile solutes:
$\frac{p^{\circ} - p}{p^{\circ}} = \frac{w \times M}{m \times W}$
Substituting the values:
$\frac{640 - 600}{640} = \frac{2.175 \times 78}{m \times 39.08}$
$\frac{40}{640} = \frac{169.65}{m \times 39.08}$
$\frac{1}{16} = \frac{169.65}{m \times 39.08}$
$m = \frac{16 \times 169.65}{39.08} \approx 69.46 \approx 69.60 \ g/mol$.

(A) The elevation in boiling point is a colligative property,which depends on the van't Hoff factor $(i)$,representing the number of particles produced upon dissociation.
For $0.1 \ M \ FeCl_3$,$i = 4$ $(Fe^{3+} + 3Cl^-)$.
For $0.1 \ M \ BaCl_2$,$i = 3$ $(Ba^{2+} + 2Cl^-)$.
For $0.1 \ M \ NaCl$,$i = 2$ $(Na^+ + Cl^-)$.
For $0.1 \ M \ \text{urea}$,$i = 1$ (non-electrolyte).
Since $0.1 \ M \ FeCl_3$ produces the maximum number of ions,it exhibits the highest elevation in boiling point and thus the highest boiling point.