MHT CET 2023 Chemistry Question Paper with Answer and Solution

(B) An isothermal process is defined as a process in which the temperature of the system remains constant $(dT = 0)$.
For an ideal gas,the internal energy $(U)$ is a function of temperature only $(U = f(T))$. Therefore,if the temperature is constant,the internal energy remains constant $(dU = 0)$.
However,for real gases or condensed phases,enthalpy $(H = U + PV)$ depends on pressure as well. Even if temperature is constant,enthalpy may change if the pressure changes.
Thus,the statement that the enthalpy of the system remains constant is not universally true for all isothermal processes,making it the false statement.

(C) The change in the number of moles of gaseous species is given by $\Delta n_g = (n_{products, g} - n_{reactants, g})$.
For the reaction: $NH_2CN_{(g)} + \frac{3}{2} O_{2_{(g)}} \longrightarrow N_{2_{(g)}} + CO_{2_{(g)}} + H_2O_{(l)}$
$\Delta n_g = (1 + 1) - (1 + 1.5) = 2 - 2.5 = -0.5 \ mol$.
Given $\Delta U = -740.5 \ kJ$,$T = 25 + 273 = 298 \ K$,and $R = 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.
Using the relation $\Delta H = \Delta U + \Delta n_g RT$:
$\Delta H = -740.5 \ kJ + (-0.5 \ mol) \times (8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}) \times 298 \ K$.
$\Delta H = -740.5 \ kJ - 1.2388 \ kJ = -741.7388 \ kJ \ mol^{-1}$.
Rounding to one decimal place,we get $\Delta H \approx -741.7 \ kJ \ mol^{-1}$.

(C) The work done in a $PV$ system is given by the formula $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$.
Initial volume $V_1 = 150 \ mL + 150 \ mL = 300 \ mL = 0.3 \ dm^3$.
Final volume $V_2 = 150 \ mL = 0.15 \ dm^3$.
External pressure $P_{ext} = 1 \ bar$.
Substituting the values: $W = -1 \ bar \times (0.15 \ dm^3 - 0.3 \ dm^3) = -1 \times (-0.15) \ dm^3 \ bar = 0.15 \ dm^3 \ bar$.
Since $1 \ dm^3 \ bar = 100 \ J$,the work done is $0.15 \times 100 \ J = 15.00 \ J$.

(B) Thermodynamic properties are classified as either state functions or path functions.
State functions depend only on the initial and final states of the system,such as $Internal \ energy$ $(U)$,$Enthalpy$ $(H)$,and $Entropy$ $(S)$.
Path functions depend on the path taken to reach the final state,such as $Work$ $(w)$ and $Heat$ $(q)$.
Therefore,$Work$ is a path function.

(D) The relationship between enthalpy change $(\Delta H)$,internal energy change $(\Delta U)$,and work done $(W)$ is given by: $\Delta H = \Delta U + P \Delta V$.
Since work done by the system during expansion is $W = -P \Delta V$,we have $P \Delta V = -W$.
Therefore,$\Delta H = \Delta U - W$.
Given:
Internal energy change $\Delta U = +432 \ J$.
Work done by the gas $W = 200 \ J$.
Since the gas performs work,the work done on the system is $-200 \ J$.
Substituting the values: $\Delta H = 432 \ J - (-200 \ J) = 432 \ J + 200 \ J = 632 \ J$.

(C) The relationship between enthalpy of solution,lattice enthalpy,and hydration enthalpy is given by:
$\Delta_{soln} H = \Delta_{lattice} H + \Delta_{hyd} H$
Rearranging the formula to solve for hydration enthalpy:
$\Delta_{hyd} H = \Delta_{soln} H - \Delta_{lattice} H$
Substituting the given values:
$\Delta_{hyd} H = 4 \ kJ \ mol^{-1} - 790 \ kJ \ mol^{-1}$
$\Delta_{hyd} H = -786 \ kJ \ mol^{-1}$

(D) The chemical equation for the formation of ethane is:
$2 C_{(s)} + 3 H_{2(g)} \longrightarrow C_2 H_{6(g)} ; \Delta_{f} H^{\circ} = ?$
The molar mass of ethane $(C_2 H_6)$ is $(2 \times 12) + (6 \times 1) = 30 \ g \ mol^{-1}$.
The number of moles in $3 \ g$ of $C_2 H_6$ is $n = \frac{3 \ g}{30 \ g \ mol^{-1}} = 0.1 \ mol$.
Given that $8.84 \ kJ$ of heat is liberated for $0.1 \ mol$ of $C_2 H_6$,the heat liberated for $1 \ mol$ is:
$\Delta_{f} H^{\circ} = \frac{-8.84 \ kJ}{0.1 \ mol} = -88.4 \ kJ \ mol^{-1}$.
Since heat is liberated,the enthalpy change is negative.

(D) The enthalpy of solution $(\Delta_{\text{soln}} H)$ is given by the sum of the lattice enthalpy $(\Delta_{L} H)$ and the hydration enthalpy $(\Delta_{\text{hyd}} H)$.
$\Delta_{\text{soln}} H = \Delta_{L} H + \Delta_{\text{hyd}} H$
Given: $\Delta_{L} H = 699 \ kJ \ mol^{-1}$ and $\Delta_{\text{hyd}} H = -681.8 \ kJ \ mol^{-1}$.
Substituting the values:
$\Delta_{\text{soln}} H = 699 \ kJ \ mol^{-1} + (-681.8 \ kJ \ mol^{-1})$
$\Delta_{\text{soln}} H = 17.2 \ kJ \ mol^{-1}$.

(A) For the reaction: $2 H_2 + O_2 \rightarrow 2 H_2 O$ (which contains $4 \times O-H$ bonds).
$\Delta_{r} H = [2 \times BE(H-H) + BE(O=O)] - [4 \times BE(O-H)]$
$-571 = [2 \times 435 + 498] - 4 \times BE(O-H)$
$-571 = [870 + 498] - 4 \times BE(O-H)$
$-571 = 1368 - 4 \times BE(O-H)$
$4 \times BE(O-H) = 1368 + 571$
$4 \times BE(O-H) = 1939$
$BE(O-H) = \frac{1939}{4} \approx 484.75 \ kJ/mol$
Thus,the average bond energy is approximately $484 \ kJ/mol$.

(A) The electron gain enthalpy is the energy change when an electron is added to a neutral gaseous atom.
Noble gases like $Ne$ have a stable $ns^2 np^6$ electronic configuration.
Adding an electron to $Ne$ requires energy to place it in the next higher energy shell,making the process endothermic.
Therefore,$Ne$ has a positive electron gain enthalpy.

(A) The entropy change of the surroundings is given by $\Delta S_{surr} = \frac{-\Delta H}{T}$.
Given $\Delta H = 28.1 \ kJ \ mol^{-1} = 28100 \ J \ mol^{-1}$ and $T = 300 \ K$.
$\Delta S_{surr} = \frac{-28100 \ J \ mol^{-1}}{300 \ K} = -93.67 \ J \ K^{-1} \ mol^{-1}$.
The total entropy change is $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}$.
$\Delta S_{total} = 108.7 \ J \ K^{-1} \ mol^{-1} + (-93.67 \ J \ K^{-1} \ mol^{-1}) = 15.03 \ J \ K^{-1} \ mol^{-1}$.
Rounding to one decimal place,we get $15.1 \ J \ K^{-1} \ mol^{-1}$.

(B) The entropy change of the surrounding is given by the formula: $\Delta S_{surr} = -\frac{\Delta H_{sys}}{T}$.
Given,$\Delta H_{sys} = +7 \ kJ \ mol^{-1} = +7000 \ J \ mol^{-1}$ and $T = 300 \ K$.
Substituting the values: $\Delta S_{surr} = -\frac{7000 \ J \ mol^{-1}}{300 \ K} = -23.33 \ J \ K^{-1} \ mol^{-1}$.
Thus,the correct option is $B$.

(D) Given:
$\Delta H = 7 \ kJ$
$\Delta S = 24.8 \ J \ K^{-1} = 24.8 \times 10^{-3} \ kJ \ K^{-1}$
$T = 300 \ K$
The Gibbs free energy change is given by the formula:
$\Delta G = \Delta H - T \Delta S$
Substituting the values:
$\Delta G = 7 \ kJ - (300 \ K \times 24.8 \times 10^{-3} \ kJ \ K^{-1})$
$\Delta G = 7 \ kJ - 7.44 \ kJ$
$\Delta G = -0.44 \ kJ$

(A) The given differential equation is $\frac{dy}{dx} = \frac{y}{x} + \sec \left(\frac{y}{x}\right)$.
This is a homogeneous differential equation. Let $y = vx$,so $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = v + \sec(v)$.
This simplifies to $x \frac{dv}{dx} = \sec(v)$,which can be written as $\cos(v) dv = \frac{1}{x} dx$.
Integrating both sides: $\int \cos(v) dv = \int \frac{1}{x} dx$,which gives $\sin(v) = \log(x) + c$.
Substituting $v = \frac{y}{x}$ back,we get $\sin \left(\frac{y}{x}\right) = \log(x) + c$.
The curve passes through $\left(1, \frac{\pi}{6}\right)$,so $\sin \left(\frac{\pi/6}{1}\right) = \log(1) + c$.
Since $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\log(1) = 0$,we have $c = \frac{1}{2}$.
Thus,the equation of the curve is $\sin \left(\frac{y}{x}\right) = \log(x) + \frac{1}{2}$.

(A) Given curves are $y^2=6x$ $(i)$ and $9x^2+by^2=16$ (ii).
Differentiating $(i)$ with respect to $x$: $2y \frac{dy}{dx} = 6 \Rightarrow \frac{dy}{dx} = \frac{3}{y}$.
Differentiating (ii) with respect to $x$: $18x + 2by \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{9x}{by}$.
Since the curves intersect at right angles,the product of their slopes at the point of intersection $(x, y)$ must be $-1$.
$\left(\frac{3}{y}\right) \times \left(-\frac{9x}{by}\right) = -1$.
$\Rightarrow \frac{27x}{by^2} = 1 \Rightarrow by^2 = 27x$.
Substitute $y^2=6x$ from $(i)$ into this equation: $b(6x) = 27x$.
Assuming $x \neq 0$,we get $6b = 27 \Rightarrow b = \frac{27}{6} = \frac{9}{2}$.

(B) Let height and breadth of the sheet be $y \ m$ and $x \ m$ respectively.
Since the area is $18 \ m^2$,we have $x y = 18$.
Converting to $cm$,$x y = 180000 \ cm^2$,so $y = \frac{180000}{x}$.
The margins are $75 \ cm$ at top and bottom (total $150 \ cm = 1.5 \ m$) and $50 \ cm$ at each side (total $100 \ cm = 1 \ m$).
The area available for printing is $A = (y - 1.5)(x - 1)$.
Substituting $y = \frac{18}{x}$,we get $A = (\frac{18}{x} - 1.5)(x - 1) = 18 - \frac{18}{x} - 1.5x + 1.5 = 19.5 - 1.5x - \frac{18}{x}$.
To maximize $A$,find $\frac{dA}{dx} = -1.5 + \frac{18}{x^2} = 0$.
$\frac{18}{x^2} = 1.5 \Rightarrow x^2 = \frac{18}{1.5} = 12$.
$x = \sqrt{12} = 2 \sqrt{3} \ m$.
Then $y = \frac{18}{2 \sqrt{3}} = \frac{9}{\sqrt{3}} = 3 \sqrt{3} \ m$.
Thus,the dimensions are height $3 \sqrt{3} \ m$ and breadth $2 \sqrt{3} \ m$.

(A) First,we determine the set $S$ by solving the inequality $x^2+30 \leq 11x$.
$x^2-11x+30 \leq 0$
$(x-5)(x-6) \leq 0$
Thus,$S = [5, 6]$.
Next,we find the derivative of $f(x) = 3x^3-18x^2+27x-40$:
$f'(x) = 9x^2-36x+27 = 9(x^2-4x+3) = 9(x-1)(x-3)$.
For $x \in [5, 6]$,both $(x-1)$ and $(x-3)$ are positive,so $f'(x) > 0$.
Since $f'(x) > 0$ on the interval $[5, 6]$,the function $f(x)$ is strictly increasing on this interval.
Therefore,the maximum value occurs at the right endpoint $x=6$.
$f(6) = 3(6)^3 - 18(6)^2 + 27(6) - 40$
$f(6) = 3(216) - 18(36) + 162 - 40$
$f(6) = 648 - 648 + 162 - 40 = 122$.

(D) Since the function is continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$,the left-hand limit must equal the right-hand limit at these points.
At $x = \frac{\pi}{4}$:
$\lim_{x \to \frac{\pi}{4}^-} (x + a \sqrt{2} \sin x) = \lim_{x \to \frac{\pi}{4}^+} (2x \cot x + b)$
$\frac{\pi}{4} + a \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 2(\frac{\pi}{4}) \cdot 1 + b$
$\frac{\pi}{4} + a = \frac{\pi}{2} + b \implies a - b = \frac{\pi}{4} \quad (i)$
At $x = \frac{\pi}{2}$:
$\lim_{x \to \frac{\pi}{2}^-} (2x \cot x + b) = \lim_{x \to \frac{\pi}{2}^+} (a \cos 2x - b \sin x)$
Using $\lim_{x \to \frac{\pi}{2}} x \cot x = \lim_{x \to \frac{\pi}{2}} x \frac{\cos x}{\sin x} = 0$:
$0 + b = a \cos(\pi) - b \sin(\frac{\pi}{2})$
$b = -a - b \implies a + 2b = 0 \quad (ii)$
From $(ii)$,$a = -2b$. Substituting into $(i)$:
$-2b - b = \frac{\pi}{4} \implies -3b = \frac{\pi}{4} \implies b = -\frac{\pi}{12}$
Then $a = -2(-\frac{\pi}{12}) = \frac{\pi}{6}$.
Thus,$a = \frac{\pi}{6}$ and $b = -\frac{\pi}{12}$.

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\cos^{\frac{2}{3}} x \sin^{\frac{4}{3}} x} \, dx$
Divide numerator and denominator by $\cos^{\frac{4}{3}} x$:
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\frac{1}{\cos^{\frac{4}{3}} x}}{\frac{\cos^{\frac{2}{3}} x \sin^{\frac{4}{3}} x}{\cos^{\frac{4}{3}} x}} \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sec^2 x}{\tan^{\frac{4}{3}} x} \, dx$
Put $\tan x = t$,then $\sec^2 x \, dx = dt$.
When $x = \frac{\pi}{6}$,$t = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.
When $x = \frac{\pi}{3}$,$t = \tan(\frac{\pi}{3}) = \sqrt{3}$.
$I = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} t^{-\frac{4}{3}} \, dt = \left[ \frac{t^{-\frac{4}{3} + 1}}{-\frac{4}{3} + 1} \right]_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} = \left[ \frac{t^{-\frac{1}{3}}}{-\frac{1}{3}} \right]_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} = -3 \left[ t^{-\frac{1}{3}} \right]_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}$
$I = -3 \left( (\sqrt{3})^{-\frac{1}{3}} - (\frac{1}{\sqrt{3}})^{-\frac{1}{3}} \right) = -3 \left( (3^{\frac{1}{2}})^{-\frac{1}{3}} - (3^{-\frac{1}{2}})^{-\frac{1}{3}} \right)$
$I = -3 \left( 3^{-\frac{1}{6}} - 3^{\frac{1}{6}} \right) = 3 \cdot 3^{\frac{1}{6}} - 3 \cdot 3^{-\frac{1}{6}} = 3^{1+\frac{1}{6}} - 3^{1-\frac{1}{6}} = 3^{\frac{7}{6}} - 3^{\frac{5}{6}}$

(A) The circle passes through the origin $(0,0)$ and its centre lies on the $Y$-axis. Let the centre be $(0, k)$. Since it passes through the origin,the radius is $k$.
The equation of the circle is $(x-0)^2 + (y-k)^2 = k^2$.
Expanding this,we get $x^2 + y^2 - 2yk + k^2 = k^2$,which simplifies to $x^2 + y^2 = 2yk$.
From this,we find $k = \frac{x^2+y^2}{2y}$.
Differentiating $x^2 + y^2 = 2yk$ with respect to $x$:
$2x + 2y \frac{dy}{dx} = 2k \frac{dy}{dx}$.
Substituting $k = \frac{x^2+y^2}{2y}$ into the equation:
$2x + 2y \frac{dy}{dx} = 2 \left( \frac{x^2+y^2}{2y} \right) \frac{dy}{dx}$.
$2x + 2y \frac{dy}{dx} = \left( \frac{x^2+y^2}{y} \right) \frac{dy}{dx}$.
Multiplying by $y$: $2xy + 2y^2 \frac{dy}{dx} = (x^2+y^2) \frac{dy}{dx}$.
Rearranging terms: $2xy = (x^2+y^2-2y^2) \frac{dy}{dx}$.
$2xy = (x^2-y^2) \frac{dy}{dx}$.
Thus,$(x^2-y^2) \frac{dy}{dx} - 2xy = 0$.

(D) Given the differential equation: $(2+\sin x) \frac{dy}{dx} + (y+1) \cos x = 0$
Rearranging the terms to separate the variables:
$\frac{dy}{y+1} = -\frac{\cos x}{2+\sin x} dx$
Integrating both sides:
$\int \frac{dy}{y+1} = -\int \frac{\cos x}{2+\sin x} dx$
$\ln(y+1) = -\ln(2+\sin x) + C$
$\ln(y+1) + \ln(2+\sin x) = C$
$\ln((y+1)(2+\sin x)) = C$
$(y+1)(2+\sin x) = e^C = K$
Given $y(0) = 1$,substitute $x=0$ and $y=1$:
$(1+1)(2+\sin 0) = K \Rightarrow 2(2+0) = K \Rightarrow K = 4$
So,$(y+1)(2+\sin x) = 4$
For $x = \frac{\pi}{2}$:
$(y+1)(2+\sin \frac{\pi}{2}) = 4$
$(y+1)(2+1) = 4$
$3(y+1) = 4$
$y+1 = \frac{4}{3}$
$y = \frac{4}{3} - 1 = \frac{1}{3}$

(A) Given the equation: $\log(x+y) = 2xy$ ... $(i)$
At $x = 0$,substituting into $(i)$ gives: $\log(0+y) = 2(0)y \implies \log(y) = 0 \implies y = e^0 = 1$.
Now,differentiate both sides of $(i)$ with respect to $x$:
$\frac{d}{dx}(\log(x+y)) = \frac{d}{dx}(2xy)$
$\frac{1}{x+y} \cdot (1 + y') = 2(y + xy')$
Substitute $x = 0$ and $y = 1$ into the differentiated equation:
$\frac{1}{0+1} \cdot (1 + y'(0)) = 2(1 + 0 \cdot y'(0))$
$1 \cdot (1 + y'(0)) = 2(1)$
$1 + y'(0) = 2$
$y'(0) = 2 - 1 = 1$
Thus,the value of $y'(0)$ is $1$.

(A) Given equation: $\log(x+y)=2xy$
Differentiating both sides with respect to $x$:
$\frac{1}{x+y} \cdot (1 + y') = 2y + 2xy'$
At $x=0$,the equation becomes $\log(0+y) = 2(0)y$,which implies $\log(y) = 0$,so $y = e^0 = 1$.
Substituting $x=0$ and $y=1$ into the differentiated equation:
$\frac{1}{0+1} \cdot (1 + y'(0)) = 2(1) + 2(0)y'(0)$
$1 \cdot (1 + y'(0)) = 2 + 0$
$1 + y'(0) = 2$
$y'(0) = 2 - 1 = 1$

(C) Given $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+6$. $(i)$
Differentiating with respect to $x$,we get $f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2)$. (ii)
Differentiating again,we get $f^{\prime \prime}(x)=6 x+2 f^{\prime}(1)$. (iii)
Substituting $x=1$ in (ii),we get $f^{\prime}(1)=3(1)^2+2(1) f^{\prime}(1)+f^{\prime \prime}(2) \Rightarrow f^{\prime}(1)+f^{\prime \prime}(2)=-3$. (iv)
Substituting $x=2$ in (iii),we get $f^{\prime \prime}(2)=6(2)+2 f^{\prime}(1) \Rightarrow f^{\prime \prime}(2)=12+2 f^{\prime}(1)$. $(v)$
Substituting $(v)$ into (iv),we get $f^{\prime}(1)+12+2 f^{\prime}(1)=-3 \Rightarrow 3 f^{\prime}(1)=-15 \Rightarrow f^{\prime}(1)=-5$.
From $(v)$,$f^{\prime \prime}(2)=12+2(-5)=2$.
Now,$f(2)=2^3+2^2(-5)+2(2)+6 = 8-20+4+6 = -2$.

(B) $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+6$
$\therefore f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2)$ ...$(i)$
$\therefore f^{\prime \prime}(x)=6 x+2 f^{\prime}(1)$ ...(ii)
Substituting $x=1$ in $(i)$,we get
$f^{\prime}(1)=3(1)^2+2(1) f^{\prime}(1)+f^{\prime \prime}(2)$
$\Rightarrow f^{\prime}(1)+f^{\prime \prime}(2)=-3$ ...(iii)
Substituting $x=2$ in (ii),we get
$f^{\prime \prime}(2)=6(2)+2 f^{\prime}(1)$
$\Rightarrow f^{\prime \prime}(2)=12+2 f^{\prime}(1)$ ...(iv)
From (iii) and (iv),we get
$f^{\prime}(1)+12+2 f^{\prime}(1)=-3$
$\Rightarrow 3 f^{\prime}(1)=-15$
$\Rightarrow f^{\prime}(1)=-5$
From (iii),$-5+f^{\prime \prime}(2)=-3$
$\Rightarrow f^{\prime \prime}(2)=2$
$\therefore f(2) = 2^3+2^2(-5)+2(2)+6$
$= 8-20+4+6$
$= -2$

(B) Given that $f^{\prime}(x)=f(x)$ for all $x \in \mathbb{R}$.
This is a first-order linear differential equation: $\frac{f^{\prime}(x)}{f(x)}=1$.
Integrating both sides with respect to $x$,we get $\ln|f(x)| = x + C$.
Thus,$f(x) = e^{x+C} = k e^x$,where $k = e^C$ is a constant.
Given $f(1)=2$,we have $k e^1 = 2$,which implies $k = \frac{2}{e}$.
So,$f(x) = \frac{2}{e} \cdot e^x = 2 e^{x-1}$.
Now,$h(x) = f(f(x))$.
Using the chain rule,$h^{\prime}(x) = f^{\prime}(f(x)) \cdot f^{\prime}(x)$.
Since $f^{\prime}(x) = f(x)$,we have $h^{\prime}(x) = f(f(x)) \cdot f(x)$.
At $x=1$,$h^{\prime}(1) = f(f(1)) \cdot f(1)$.
Since $f(1)=2$,$h^{\prime}(1) = f(2) \cdot 2$.
Using $f(x) = 2 e^{x-1}$,we find $f(2) = 2 e^{2-1} = 2e$.
Therefore,$h^{\prime}(1) = (2e) \cdot 2 = 4e$.

(C) We are given $f(x) = \int \frac{\sqrt{x}}{(1+x)^2} \, dx$.
We need to evaluate $f(3) - f(1) = \int_1^3 \frac{\sqrt{x}}{(1+x)^2} \, dx$.
Let $\sqrt{x} = \tan \theta$,then $x = \tan^2 \theta$ and $dx = 2 \tan \theta \sec^2 \theta \, d\theta$.
When $x = 1$,$\tan \theta = 1 \Rightarrow \theta = \frac{\pi}{4}$.
When $x = 3$,$\tan \theta = \sqrt{3} \Rightarrow \theta = \frac{\pi}{3}$.
Substituting these into the integral:
$I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\tan \theta \cdot 2 \tan \theta \sec^2 \theta}{(1 + \tan^2 \theta)^2} \, d\theta = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{2 \tan^2 \theta \sec^2 \theta}{\sec^4 \theta} \, d\theta = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} 2 \sin^2 \theta \, d\theta$.
Using the identity $2 \sin^2 \theta = 1 - \cos 2\theta$:
$I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (1 - \cos 2\theta) \, d\theta = [\theta - \frac{\sin 2\theta}{2}]_{\frac{\pi}{4}}^{\frac{\pi}{3}}$.
$I = (\frac{\pi}{3} - \frac{\sin(2\pi/3)}{2}) - (\frac{\pi}{4} - \frac{\sin(\pi/2)}{2}) = (\frac{\pi}{3} - \frac{\sqrt{3}}{4}) - (\frac{\pi}{4} - \frac{1}{2})$.
$I = \frac{\pi}{12} - \frac{\sqrt{3}}{4} + \frac{1}{2}$.

(D) Let $y = \log \left(t+\sqrt{1+t^2}\right)$.
Then,differentiating with respect to $t$,we get $dy = \frac{1}{t+\sqrt{1+t^2}} \left(1 + \frac{2t}{2\sqrt{1+t^2}}\right) dt$.
Simplifying the term inside the parenthesis: $1 + \frac{t}{\sqrt{1+t^2}} = \frac{\sqrt{1+t^2}+t}{\sqrt{1+t^2}}$.
Thus,$dy = \frac{1}{t+\sqrt{1+t^2}} \cdot \frac{\sqrt{1+t^2}+t}{\sqrt{1+t^2}} dt = \frac{1}{\sqrt{1+t^2}} dt$.
Substituting this into the integral: $\int y \, dy = \frac{y^2}{2} + c$.
Comparing this with the given expression $\frac{1}{2}[g(t)]^2 + c$,we identify $g(t) = y = \log \left(t+\sqrt{1+t^2}\right)$.
Therefore,$g(2) = \log \left(2+\sqrt{1+2^2}\right) = \log (2+\sqrt{5})$.

(D) Given that $\tan ^{-1} a+\tan ^{-1} b+\tan ^{-1} c=\pi$.
Using the formula $\tan ^{-1} a+\tan ^{-1} b+\tan ^{-1} c = \tan ^{-1} \left( \frac{a+b+c-abc}{1-ab-bc-ca} \right) = \pi$.
Taking $\tan$ on both sides,we get $\frac{a+b+c-abc}{1-ab-bc-ca} = \tan \pi$.
Since $\tan \pi = 0$,we have $\frac{a+b+c-abc}{1-ab-bc-ca} = 0$.
This implies $a+b+c-abc = 0$.
Therefore,$a+b+c=abc$.

(A) Let $\sin^{-1} x = \theta$,which implies $x = \sin \theta$. Since $0 \leq x \leq \frac{1}{2}$,we have $0 \leq \theta \leq \frac{\pi}{6}$.
Then $\sqrt{1-x^2} = \cos \theta$.
The expression becomes $\tan \left[ \sin^{-1} \left( \frac{\sin \theta}{\sqrt{2}} + \frac{\cos \theta}{\sqrt{2}} \right) - \theta \right]$.
Using the identity $\sin \theta + \cos \theta = \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right)$,we get:
$\frac{\sin \theta + \cos \theta}{\sqrt{2}} = \sin \left( \theta + \frac{\pi}{4} \right)$.
Substituting this into the expression:
$= \tan \left[ \sin^{-1} \left( \sin \left( \theta + \frac{\pi}{4} \right) \right) - \theta \right]$.
Since $0 \leq \theta \leq \frac{\pi}{6}$,then $\frac{\pi}{4} \leq \theta + \frac{\pi}{4} \leq \frac{\pi}{6} + \frac{\pi}{4} = \frac{5\pi}{12}$.
Since this range is within $[-\frac{\pi}{2}, \frac{\pi}{2}]$,$\sin^{-1}(\sin(\theta + \frac{\pi}{4})) = \theta + \frac{\pi}{4}$.
Thus,the expression simplifies to $\tan \left( \theta + \frac{\pi}{4} - \theta \right) = \tan \frac{\pi}{4} = 1$.

(D) Let $p$ : The payment will be made.
Let $q$ : The work is finished in time.
The given statement is a biconditional statement: $p \iff q$,which is equivalent to $(p \rightarrow q) \wedge (q \rightarrow p)$.
The negation of $p \iff q$ is $\sim(p \iff q)$,which is equivalent to $(p \wedge \sim q) \vee (q \wedge \sim p)$.
This translates to: "The payment is made and the work is not finished in time,or the work is finished in time and the payment is not made."
Therefore,option $(D)$ is correct.

(A) Given expression: $[\sim(\sim p \vee q) \vee (p \wedge r)] \wedge (\sim q \wedge r)$
Applying De Morgan's law: $[(p \wedge \sim q) \vee (p \wedge r)] \wedge (\sim q \wedge r)$
Applying Distributive law: $[p \wedge (\sim q \vee r)] \wedge (\sim q \wedge r)$
Applying Associative law: $p \wedge [(\sim q \vee r) \wedge (\sim q \wedge r)]$
Since $(\sim q \vee r) \wedge (\sim q \wedge r) = (\sim q \wedge r)$,the expression becomes: $p \wedge (\sim q \wedge r)$
Rearranging using Commutative law: $(p \wedge r) \wedge \sim q$

(C) Given expression: $[(\sim(\sim p \vee q)) \vee (p \wedge r)] \wedge (\sim q \wedge r)$
Using De Morgan's law,$\sim(\sim p \vee q) \equiv (p \wedge \sim q)$.
So,the expression becomes: $[(p \wedge \sim q) \vee (p \wedge r)] \wedge (\sim q \wedge r)$
Applying the Distributive law: $[p \wedge (\sim q \vee r)] \wedge (\sim q \wedge r)$
Using the Associative and Commutative laws: $p \wedge [(\sim q \vee r) \wedge (\sim q \wedge r)]$
Since $(\sim q \vee r) \wedge (\sim q \wedge r) \equiv (\sim q \wedge r)$ by the Absorption law:
The expression simplifies to: $p \wedge (\sim q \wedge r)$
Which is equivalent to: $(p \wedge r) \wedge \sim q$

(B) The number of diagonals in a polygon with $n$ sides is given by the formula $\frac{n(n-3)}{2}$.
Given that the number of diagonals is $54$,we have:
$\frac{n(n-3)}{2} = 54$
$n(n-3) = 108$
$n^2 - 3n - 108 = 0$
Factoring the quadratic equation:
$(n - 12)(n + 9) = 0$
Since the number of sides $n$ must be a positive integer,we have $n = 12$.

(D) Let the observations be $x_1, x_2, \dots, x_n$ with variance $\sigma^2 = 5$.
When each observation is multiplied by a constant $k$,the new variance $\sigma'^2$ is given by $\sigma'^2 = k^2 \times \sigma^2$.
Here,$k = 2$ and $\sigma^2 = 5$.
Therefore,the new variance $= 2^2 \times 5 = 4 \times 5 = 20$.

(C) The equation of the plane passing through $(2,1,0)$,$(4,1,1)$,and $(5,0,1)$ is given by the determinant equation:
$\begin{vmatrix} x-2 & y-1 & z-0 \\ 4-2 & 1-1 & 1-0 \\ 5-2 & 0-1 & 1-0 \end{vmatrix} = 0$
$\Rightarrow \begin{vmatrix} x-2 & y-1 & z \\ 2 & 0 & 1 \\ 3 & -1 & 1 \end{vmatrix} = 0$
Expanding the determinant: $(x-2)(0+1) - (y-1)(2-3) + z(-2-0) = 0$
$\Rightarrow (x-2) + (y-1) - 2z = 0 \Rightarrow x+y-2z = 3$.
Let $R'(x, y, z)$ be the image of $R(2,1,6)$ with respect to the plane $x+y-2z-3=0$.
The formula for the image $(x, y, z)$ of a point $(x_1, y_1, z_1)$ in the plane $ax+by+cz+d=0$ is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -2 \frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$.
Substituting the values: $\frac{x-2}{1} = \frac{y-1}{1} = \frac{z-6}{-2} = -2 \frac{2+1-2(6)-3}{1^2+1^2+(-2)^2} = -2 \frac{3-12-3}{6} = -2 \frac{-12}{6} = 4$.
Thus,$x-2 = 4 \Rightarrow x=6$,$y-1 = 4 \Rightarrow y=5$,and $z-6 = -8 \Rightarrow z=-2$.
Therefore,the image $R'$ is $(6, 5, -2)$.

(A) Let $\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}=\lambda$.
Then $x = \lambda+3, y = 3\lambda-1, z = 6-\lambda$.
Let $\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}=\mu$.
Then $x = 7\mu-5, y = -6\mu+2, z = 4\mu+3$.
Since the lines intersect,we equate the coordinates:
$\lambda+3 = 7\mu-5 \Rightarrow \lambda - 7\mu = -8$ $(i)$
$3\lambda-1 = -6\mu+2 \Rightarrow 3\lambda + 6\mu = 3 \Rightarrow \lambda + 2\mu = 1$ (ii)
Subtracting $(i)$ from (ii): $9\mu = 9 \Rightarrow \mu = 1$.
Substituting $\mu=1$ in (ii): $\lambda + 2(1) = 1 \Rightarrow \lambda = -1$.
Substituting $\lambda = -1$ into the first line equations: $x = -1+3 = 2, y = 3(-1)-1 = -4, z = 6-(-1) = 7$.
Thus,the point of intersection is $R(2, -4, 7)$.
The reflection of a point $(x, y, z)$ in the $xy$-plane is $(x, y, -z)$.
Therefore,the reflection of $R(2, -4, 7)$ in the $xy$-plane is $(2, -4, -7)$.

(B) The equation of a plane passing through $(1, 1, 1)$ is $a(x-1) + b(y-1) + c(z-1) = 0$.
Since the plane is parallel to the lines with direction ratios $(1, 0, -1)$ and $(-1, 1, 0)$,the normal vector $(a, b, c)$ must be perpendicular to both direction vectors.
Thus,$a(1) + b(0) + c(-1) = 0 \Rightarrow a = c$ and $a(-1) + b(1) + c(0) = 0 \Rightarrow a = b$.
Therefore,$a = b = c$. Setting $a = b = c = 1$,the equation of the plane becomes $1(x-1) + 1(y-1) + 1(z-1) = 0$,which simplifies to $x + y + z = 3$.
Dividing by $3$,we get the intercept form $\frac{x}{3} + \frac{y}{3} + \frac{z}{3} = 1$.
The intercepts are $A(3, 0, 0)$,$B(0, 3, 0)$,and $C(0, 0, 3)$.
The volume of the tetrahedron $OABC$ is given by $V = \frac{1}{6} |x_A y_B z_C| = \frac{1}{6} |3 \times 3 \times 3| = \frac{27}{6} = \frac{9}{2} \text{ cubic units}$.

(C) Let $\theta$ be the angle between $\overline{AB}$ and $\overline{AD}$.
$\cos \theta = \frac{\overline{AB} \cdot \overline{AD}}{|\overline{AB}||\overline{AD}|} = \frac{(2 \hat{i}+10 \hat{j}+11 \hat{k}) \cdot(-\hat{i}+2 \hat{j}+2 \hat{k})}{\sqrt{4+100+121} \sqrt{1+4+4}} = \frac{-2+20+22}{\sqrt{225} \sqrt{9}} = \frac{40}{15 \times 3} = \frac{40}{45} = \frac{8}{9}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin \theta = \sqrt{1 - (\frac{8}{9})^2} = \sqrt{\frac{81-64}{81}} = \frac{\sqrt{17}}{9}$.
Let $\alpha$ be the angle of rotation of $AD$ such that $AD^{\prime} \perp AB$. The angle between $AB$ and $AD^{\prime}$ is $90^{\circ}$.
Thus,$\alpha + \theta = 90^{\circ}$,which implies $\alpha = 90^{\circ} - \theta$.
Therefore,$\cos \alpha = \cos(90^{\circ} - \theta) = \sin \theta$.
Substituting the value of $\sin \theta$,we get $\cos \alpha = \frac{\sqrt{17}}{9}$.

(A) Using the vector triple product formula,$\overline{a} \times (\overline{b} \times \overline{c}) = (\overline{a} \cdot \overline{c}) \overline{b} - (\overline{a} \cdot \overline{b}) \overline{c}$.
Given $\overline{a} \times (\overline{b} \times \overline{c}) = \frac{\overline{b} + \overline{c}}{\sqrt{2}}$,we have $(\overline{a} \cdot \overline{c}) \overline{b} - (\overline{a} \cdot \overline{b}) \overline{c} = \frac{1}{\sqrt{2}} \overline{b} + \frac{1}{\sqrt{2}} \overline{c}$.
Rearranging the terms,we get $(\overline{a} \cdot \overline{c} - \frac{1}{\sqrt{2}}) \overline{b} - (\overline{a} \cdot \overline{b} + \frac{1}{\sqrt{2}}) \overline{c} = 0$.
Since $\overline{a}, \overline{b}, \overline{c}$ are non-coplanar,$\overline{b}$ and $\overline{c}$ are linearly independent.
Therefore,the coefficients must be zero: $\overline{a} \cdot \overline{b} + \frac{1}{\sqrt{2}} = 0$.
This implies $\overline{a} \cdot \overline{b} = -\frac{1}{\sqrt{2}}$.
Since $\overline{a}$ and $\overline{b}$ are unit vectors,$|\overline{a}| |\overline{b}| \cos \theta = -\frac{1}{\sqrt{2}}$,where $\theta$ is the angle between $\overline{a}$ and $\overline{b}$.
Thus,$\cos \theta = -\frac{1}{\sqrt{2}}$,which gives $\theta = \frac{3 \pi}{4}$.

(D) Let the given expression be $E = (2 \bar{a}-\bar{b}) \cdot ((\bar{a} \times \bar{b}) \times (\bar{a}+2 \bar{b}))$.
Using the vector triple product identity $(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{v} \cdot \vec{w}) \vec{u}$,we have:
$(\bar{a} \times \bar{b}) \times (\bar{a}+2 \bar{b}) = (\bar{a} \cdot (\bar{a}+2 \bar{b})) \bar{b} - (\bar{b} \cdot (\bar{a}+2 \bar{b})) \bar{a}$.
Since $\bar{a}$ and $\bar{b}$ are unit vectors,$\bar{a} \cdot \bar{a} = 1$ and $\bar{b} \cdot \bar{b} = 1$.
Also,$\bar{a} \cdot \bar{b} = \frac{1}{7\sqrt{10}} (3 \times 2 + 0 \times 3 + 1 \times (-6)) = 0$.
Thus,$(\bar{a} \cdot (\bar{a}+2 \bar{b})) = \bar{a} \cdot \bar{a} + 2(\bar{a} \cdot \bar{b}) = 1 + 0 = 1$.
And $(\bar{b} \cdot (\bar{a}+2 \bar{b})) = \bar{b} \cdot \bar{a} + 2(\bar{b} \cdot \bar{b}) = 0 + 2(1) = 2$.
So,$(\bar{a} \times \bar{b}) \times (\bar{a}+2 \bar{b}) = 1 \bar{b} - 2 \bar{a} = - (2 \bar{a} - \bar{b})$.
Now,$E = (2 \bar{a}-\bar{b}) \cdot (-(2 \bar{a}-\bar{b})) = - |2 \bar{a}-\bar{b}|^2$.
$|2 \bar{a}-\bar{b}|^2 = (2 \bar{a}-\bar{b}) \cdot (2 \bar{a}-\bar{b}) = 4|\bar{a}|^2 + |\bar{b}|^2 - 4(\bar{a} \cdot \bar{b}) = 4(1) + 1 - 4(0) = 5$.
Therefore,$E = -5$.

(B) Given that $\bar{a}$ and $\bar{b}$ are unit vectors,so $|\bar{a}| = 1$ and $|\bar{b}| = 1$.
Since $(\bar{a}+2\bar{b})$ and $(5\bar{a}-4\bar{b})$ are perpendicular,their dot product is zero:
$(\bar{a}+2\bar{b}) \cdot (5\bar{a}-4\bar{b}) = 0$
$5|\bar{a}|^2 - 4(\bar{a} \cdot \bar{b}) + 10(\bar{a} \cdot \bar{b}) - 8|\bar{b}|^2 = 0$
$5(1)^2 + 6(\bar{a} \cdot \bar{b}) - 8(1)^2 = 0$
$5 + 6(\bar{a} \cdot \bar{b}) - 8 = 0$
$6(\bar{a} \cdot \bar{b}) = 3$
$\bar{a} \cdot \bar{b} = \frac{3}{6} = \frac{1}{2}$
Since $\bar{a} \cdot \bar{b} = |\bar{a}||\bar{b}| \cos \theta$,we have:
$(1)(1) \cos \theta = \frac{1}{2}$
$\cos \theta = \frac{1}{2}$
$\theta = \frac{\pi}{3}$

(D) First,we calculate the magnitudes of $\overline{a}$ and $\overline{b}$:
$|\overline{a}| = \frac{1}{\sqrt{10}} \sqrt{3^2 + 0^2 + 1^2} = \frac{\sqrt{10}}{\sqrt{10}} = 1$
$|\overline{b}| = \frac{1}{7} \sqrt{2^2 + 3^2 + (-6)^2} = \frac{1}{7} \sqrt{4 + 9 + 36} = \frac{7}{7} = 1$
Next,we calculate the dot product $\overline{a} \cdot \overline{b}$:
$\overline{a} \cdot \overline{b} = \frac{1}{7\sqrt{10}} (3 \times 2 + 0 \times 3 + 1 \times (-6)) = \frac{1}{7\sqrt{10}} (6 + 0 - 6) = 0$
Since $\overline{a} \cdot \overline{b} = 0$,the vectors are perpendicular.
Now,consider the expression $E = (2 \overline{a}-\overline{b}) \cdot [(\overline{a} \times \overline{b}) \times (\overline{a}+2 \overline{b})]$.
Using the vector triple product identity $(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$:
$(\overline{a} \times \overline{b}) \times (\overline{a}+2 \overline{b}) = (\overline{a} \cdot (\overline{a}+2 \overline{b}))\overline{b} - (\overline{b} \cdot (\overline{a}+2 \overline{b}))\overline{a}$
$= (\overline{a} \cdot \overline{a} + 2(\overline{a} \cdot \overline{b}))\overline{b} - ((\overline{b} \cdot \overline{a}) + 2(\overline{b} \cdot \overline{b}))\overline{a}$
$= (1 + 0)\overline{b} - (0 + 2(1))\overline{a} = \overline{b} - 2\overline{a}$
Now substitute this back into the expression:
$E = (2 \overline{a}-\overline{b}) \cdot (\overline{b} - 2\overline{a})$
$= -(2 \overline{a}-\overline{b}) \cdot (2 \overline{a}-\overline{b}) = -|2 \overline{a}-\overline{b}|^2$
$= -(4|\overline{a}|^2 + |\overline{b}|^2 - 4(\overline{a} \cdot \overline{b}))$
$= -(4(1) + 1 - 4(0)) = -5$.

(B) Let $\theta$ be the angle between $\overline{AB}$ and $\overline{AD}$.
$\cos \theta = \frac{\overline{AB} \cdot \overline{AD}}{|\overline{AB}||\overline{AD}|} = \frac{(2 \hat{i}+10 \hat{j}+11 \hat{k}) \cdot(-\hat{i}+2 \hat{j}+2 \hat{k})}{\sqrt{4+100+121} \sqrt{1+4+4}} = \frac{-2+20+22}{15 \times 3} = \frac{40}{45} = \frac{8}{9}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin \theta = \sqrt{1 - (\frac{8}{9})^2} = \sqrt{\frac{81-64}{81}} = \frac{\sqrt{17}}{9}$.
Let $\alpha$ be the angle of rotation of $AD$ such that the new vector $AD^{\prime}$ is perpendicular to $AB$.
The angle between $AB$ and $AD^{\prime}$ is $90^{\circ}$.
Since $AD$ is rotated by $\alpha$ towards $AB$,the angle between $AB$ and $AD$ is $\theta = \alpha + 90^{\circ}$ (or $\alpha = \theta - 90^{\circ}$ depending on direction,but here $AD^{\prime}$ is perpendicular to $AB$,so the angle between $AB$ and $AD$ is $\theta = 90^{\circ} - \alpha$ if rotating towards $AB$).
Thus,$\alpha = 90^{\circ} - \theta$.
$\cos \alpha = \cos(90^{\circ} - \theta) = \sin \theta = \frac{\sqrt{17}}{9}$.

(A) In an $LCR$ series circuit,the phase difference $\phi$ between voltage and current is given by $\tan \phi = \frac{X_L - X_C}{R}$.
Given $\phi = 45^{\circ}$,so $\tan 45^{\circ} = 1$.
Substituting the values of $X_L = 2 \pi fL$ and $X_C = \frac{1}{2 \pi fC}$:
$1 = \frac{2 \pi fL - \frac{1}{2 \pi fC}}{R}$
$R = 2 \pi fL - \frac{1}{2 \pi fC}$
$R = \frac{(2 \pi f)^2 LC - 1}{2 \pi fC}$
$2 \pi fCR = 4 \pi^2 f^2 LC - 1$
$4 \pi^2 f^2 LC = 1 + 2 \pi fCR$
$L = \frac{1 + 2 \pi fCR}{4 \pi^2 f^2 C}$.

(A) At resonance,the inductive reactance equals the capacitive reactance $(X_L = X_C)$.
Given: $R = 2 \Omega$,$L = 100 \times 10^{-6} \text{ H}$,$C = 400 \times 10^{-12} \text{ F}$,$e_{rms} = 0.1 \text{ V}$.
The impedance at resonance is $Z = R = 2 \Omega$.
The current in the circuit is $I_{rms} = \frac{e_{rms}}{Z} = \frac{0.1}{2} = 0.05 \text{ A}$.
The resonant angular frequency is $\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{100 \times 10^{-6} \times 400 \times 10^{-12}}} = \frac{1}{\sqrt{4 \times 10^{-14}}} = \frac{1}{2 \times 10^{-7}} = 0.5 \times 10^7 = 5 \times 10^6 \text{ rad/s}$.
The voltage drop across the inductor is $V_L = I_{rms} \times X_L = I_{rms} \times L\omega$.
$V_L = 0.05 \times (100 \times 10^{-6}) \times (5 \times 10^6) = 0.05 \times 10^{-4} \times 5 \times 10^6 = 0.05 \times 500 = 25 \text{ V}$.

(B) The phase angle $\phi$ in a series $LCR$ circuit is given by $\tan \phi = \frac{X_L - X_C}{R}$.
Given that the voltage leads the current by $45^{\circ}$,we have $\phi = 45^{\circ}$.
Therefore,$\tan 45^{\circ} = \frac{X_L - X_C}{R} = 1$,which implies $X_L - X_C = R$.
Substituting the expressions for inductive reactance $X_L = 2 \pi f L$ and capacitive reactance $X_C = \frac{1}{2 \pi f C}$,we get:
$2 \pi f L - \frac{1}{2 \pi f C} = R$
$2 \pi f L = R + \frac{1}{2 \pi f C}$
$2 \pi f L = \frac{2 \pi f C R + 1}{2 \pi f C}$
$L = \frac{1 + 2 \pi f C R}{4 \pi^2 f^2 C}$
Thus,the correct option is $(B)$.

(D) The frequency $f$ of the emitted radiation is given by the Rydberg formula:
$f = c \cdot \frac{1}{\lambda} = cR \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$
Here,the electron jumps from $n_2 = 3$ to $n_1 = 2$.
Substituting the values:
$f = cR \left[ \frac{1}{2^2} - \frac{1}{3^2} \right]$
$f = cR \left[ \frac{1}{4} - \frac{1}{9} \right]$
$f = cR \left[ \frac{9 - 4}{36} \right]$
$f = \frac{5RC}{36}$
Thus,the correct option is $D$.

(C) Initial energy of the system is $U_1 = \frac{1}{2} CV_1^2 + \frac{1}{2} CV_2^2 = \frac{C}{2}(V_1^2 + V_2^2)$.
When the capacitors are connected in parallel,the common potential is $V = \frac{V_1 + V_2}{2}$.
The final energy of the system is $U_2 = \frac{1}{2}(C + C)V^2 = C \left(\frac{V_1 + V_2}{2}\right)^2 = \frac{C}{4}(V_1 + V_2)^2$.
The decrease in energy is $\Delta U = U_1 - U_2$.
$\Delta U = \frac{C}{2}(V_1^2 + V_2^2) - \frac{C}{4}(V_1 + V_2)^2$.
$\Delta U = \frac{C}{4} [2(V_1^2 + V_2^2) - (V_1^2 + V_2^2 + 2V_1V_2)]$.
$\Delta U = \frac{C}{4} [V_1^2 + V_2^2 - 2V_1V_2] = \frac{1}{4} C(V_1 - V_2)^2$.

(B) Let the total current in the circuit be $I$ and the current through the galvanometer be $I_g$.
Given that $I_g = 4\% \text{ of } I = 0.04 I$.
Therefore,the ratio $\frac{I}{I_g} = \frac{1}{0.04} = 25$.
The shunt resistance $S$ connected in parallel to a galvanometer of resistance $G$ is given by the formula:
$S = \frac{I_g \times G}{I - I_g}$.
Dividing the numerator and denominator by $I_g$,we get:
$S = \frac{G}{\frac{I}{I_g} - 1}$.
Substituting the value $\frac{I}{I_g} = 25$:
$S = \frac{G}{25 - 1} = \frac{G}{24}$.

(D) The solubility of most salts increases with an increase in temperature. However,for certain salts like $Na_2SO_4$ (specifically the decahydrate form,$Na_2SO_4 \cdot 10H_2O$),the solubility decreases as the temperature increases beyond a certain point due to the transition to the anhydrous form.

(D) Polymorphism is the ability of a solid material to exist in more than one form or crystal structure.
Since these forms have different crystal structures,their physical properties like crystal shape,density,and melting point are different.
Therefore,the statement that the crystal shape of polymorphic substances is identical is false.

(B) Galena is a naturally occurring mineral form of lead$(II)$ sulfide.
Its chemical formula is $PbS$.

(B) reaction intermediate is a substance that is produced in one step of a reaction mechanism and consumed in a subsequent step.
In the given mechanism:
Step $(i)$: $2 SO_{2(g)} + 2 NO_{2(g)} \rightarrow 2 SO_{3(g)} + 2 NO_{(g)}$
Step $(ii)$: $2 NO_{(g)} + O_{2(g)} \rightarrow 2 NO_{2(g)}$
Here,$NO_{(g)}$ is produced in step $(i)$ and consumed in step $(ii)$.
$NO_{2(g)}$ acts as a catalyst because it is consumed in step $(i)$ and regenerated in step $(ii)$.
Therefore,$NO_{(g)}$ is the reaction intermediate.

(B) Among the given elements,$Se$ (Selenium),$I$ (Iodine),and $S$ (Sulfur) are solids at room temperature.
$Br$ (Bromine) is a non-metallic element that exists in a liquid state at room temperature.

(D) The acidic character of the hydrides of group $16$ elements increases down the group as the bond dissociation enthalpy decreases.
Therefore,the order of acidic strength is $H_2O < H_2S < H_2Se < H_2Te$.
Thus,$H_2Te$ is the most acidic molecule among the given options.

(B) Among the given elements,$S$ (Sulfur) exhibits the highest tendency for catenation due to its moderate size and electronic configuration.
This property allows it to form a large number of allotropic forms,such as $S_8$,$S_6$,and other cyclic structures.

(B) The thermal stability of the hydrides of group $16$ elements depends on the bond dissociation energy of the $E-H$ bond.
As the size of the central atom increases down the group $(O < S < Se < Te)$,the bond length increases and the bond dissociation energy decreases.
Therefore,the thermal stability decreases in the order: $H_2O > H_2S > H_2Se > H_2Te$.
Thus,$H_2Te$ has the lowest thermal stability.

(A) The melting and boiling points of elements of group $16$ increase with an increase in atomic number due to an increase in the magnitude of van der Waals forces.
Therefore,the boiling point of sulfur is higher than that of oxygen.
Thus,the statement that the boiling point of sulfur is lower than oxygen is $FALSE$.

(B) In the contact process for the industrial production of sulphuric acid,$SO_2$ and $O_2$ react to form $SO_3$ in the presence of a catalyst.
Historically,platinised asbestos was used as the catalyst.
Currently,vanadium pentoxide $(V_2O_5)$ is more commonly used,but among the given options,$Platinum$ is the correct metal catalyst.

(D) In industry,sulphur dioxide is produced as a byproduct by roasting sulphide ores like zinc sulphide and iron pyrites in air.
$2 ZnS_{(s)} + 3 O_{2(g)} \xrightarrow{\Delta} 2 ZnO_{(s)} + 2 SO_{2(g)}$
$4 FeS_{2(s)} + 11 O_{2(g)} \xrightarrow{\Delta} 2 Fe_2O_{3(s)} + 8 SO_{2(g)}$

(D) The acid strength of oxoacids of chlorine increases with the increase in the oxidation state of the central chlorine atom.
The increasing order of acid strength is: $HOCl < HClO_2 < HClO_3 < HClO_4$.
In $HClO_4$,the chlorine atom is in the $+7$ oxidation state,which makes it the most stable conjugate base due to the maximum dispersal of negative charge.
Therefore,perchloric acid $(HClO_4)$ is the strongest acid among the given oxoacids of chlorine.

(C) The acidic strength of hydrohalic acids $(HX)$ depends on the bond dissociation enthalpy of the $H-X$ bond.
As the size of the halogen atom increases from $F$ to $I$,the bond length increases and the bond dissociation enthalpy decreases.
Therefore,the acidic strength increases in the order: $HF < HCl < HBr < HI$.
Hence,the weakest halogen acid is $HF$.

(A) The chalcogen family consists of elements in Group $16$ of the periodic table,which are Oxygen $(O)$,Sulfur $(S)$,Selenium $(Se)$,Tellurium $(Te)$,and Polonium $(Po)$.
Astatine $(At)$ belongs to Group $17$,which is the halogen family.
Therefore,$At$ does not belong to the chalcogen family.

(A) Xenon $(Xe)$ has a large atomic size and lower ionization enthalpy compared to $He$,$Ne$,$Ar$,and $Kr$.
Due to its lower ionization energy,it can easily form compounds with highly electronegative elements like fluorine and oxygen.
Consequently,xenon exhibits the highest number of different oxidation states (e.g.,$+2, +4, +6, +8$) among the noble gases.

(A) homopolymer is a polymer formed from only one type of monomer unit.
Polyacrylonitrile is formed by the polymerization of acrylonitrile $(CH_2=CH-CN)$ monomers only,making it a homopolymer.
Glyptal is a copolymer formed from ethylene glycol and phthalic acid.
Polycarbonate is a copolymer formed from bisphenol-$A$ and phosgene.
Buna-$S$ is a copolymer formed from $1,3$-butadiene and styrene.

(B) Elastomers are polymers in which the polymer chains are held together by the weakest intermolecular forces. These weak forces permit the polymer to be stretched. $A$ few crosslinks are introduced in between the chains,which help the polymer retract to its original position after the force is released. Examples include Buna-$S$,Buna-$N$,and natural rubber. Nylon $6,6$ is a fiber,Terylene is a fiber,and Polythene is a thermoplastic polymer.

(A) Linear polymers consist of long and straight chains. $High \ density \ polythene$ $(HDPE)$ is a linear polymer formed by addition polymerization of ethene under specific conditions (low pressure and use of Ziegler-Natta catalyst).
$Low \ density \ polythene$ $(LDPE)$ is a branched-chain polymer.
$Bakelite$ and $Melamine$ are cross-linked or network polymers.

(A) Biodegradable polymers are those that can be decomposed by microorganisms.
Nylon $2$-nylon $6$ is a polyamide copolymer of glycine $(H_2N-CH_2-COOH)$ and amino caproic acid $(H_2N-(CH_2)_5-COOH)$.
It is a well-known biodegradable polymer used in medical sutures and other applications.
Other options like Terylene,Nylon $6$,and Nylon $6,6$ are synthetic polymers that are not biodegradable.

(D) Thermocol is a trade name for expanded polystyrene $(EPS)$.
It is prepared by the polymerization of the monomer $Styrene$ $(C_6H_5CH=CH_2)$.
Therefore,the correct monomer used for the preparation of thermocol is $Styrene$.

(D) Terylene is a copolymer formed by the polymerization of two different monomers,ethylene glycol and terephthalic acid.
Natural rubber,polypropene,and $PVC$ (polyvinyl chloride) are homopolymers because they are formed from a single type of monomer unit.

(D) $PHBV$ is a copolymer of two bifunctional $\beta$-hydroxy carboxylic acids,namely,$3$-hydroxybutanoic acid and $3$-hydroxypentanoic acid.
$SBR$ (Styrene-Butadiene Rubber) is a copolymer of styrene and $1,3$-butadiene.
Both $SBR$ and $PHBV$ are examples of copolymers.

(A) In the process of vulcanization of rubber,accelerators are added to speed up the crosslinking process. $Zinc \ butyl \ xanthate$ is a commonly used accelerator for this purpose.

(A) Thermosetting polymers are cross-linked or heavily branched molecules that undergo extensive cross-linking during molding,which makes them hard and infusible.
$A$. Urea formaldehyde resin is a classic example of a thermosetting polymer.
$B$. Polythene is a thermoplastic polymer.
$C$. Polystyrene is a thermoplastic polymer.
$D$. Polyvinyls (like $PVC$) are thermoplastic polymers.
Therefore,the correct option is $A$.

(A) Elastomers are polymers that have weak intermolecular forces of attraction,which allow the polymer to be stretched.
$Neoprene$ is a synthetic rubber and is classified as an elastomer.
$Terylene$ is a polyester (fiber),$Polystyrene$ is a thermoplastic,and $Bakelite$ is a thermosetting polymer.

(D) The monomer used in the preparation of Teflon is tetrafluoroethylene,$(CF_2=CF_2)$.
Polymerization of tetrafluoroethylene under high pressure in the presence of a peroxide or persulphate catalyst yields Teflon,which is $(CF_2-CF_2)_n$.

(B) High density polythene $(HDP)$ is obtained by the polymerization of ethene in the presence of a Zieglar-Natta catalyst at a temperature of $333 \ K$ to $343 \ K$ and a pressure of $6-7 \ atm$.
Low density polythene $(LDP)$ requires high pressure $(1000-2000 \ atm)$.
Therefore,the statement that $HDP$ needs high pressure is incorrect.
Its melting point (in the range of $144-150^{\circ}C$) is higher than that of $LDP$ (melting point $110^{\circ}C$).

(B) Natural rubber is a high molecular mass linear polymer of isoprene ($2$-methylbuta-$1,3$-diene).

(A) $LDP$ (Low Density Polyethylene) is amorphous in nature due to its highly branched structure. $HDP$ (High Density Polyethylene) is crystalline due to its linear structure. Therefore,the property of being crystalline is exhibited by $HDP$,not $LDP$.

(A) Due to its high tensile strength and luster,Nylon $6$ fibres are used for the manufacturing of tyre cords.

(C) $LDP$ (Low Density Polyethylene) is prepared by the polymerization of ethylene under high pressure $(1000-2000 \ atm)$ and temperature $(350-570 \ K)$ in the presence of traces of $O_2$ or peroxide as an initiator.
Therefore,the statement that it needs low pressure of $6-7 \ atm$ is incorrect.

(D) Teflon is a polymer of tetrafluoroethylene.
The chemical formula of the monomer tetrafluoroethylene is $CF_2=CF_2$,which is represented as $C_2F_4$.

(D) Natural rubber is a high molecular mass linear polymer of $2$-methyl-$1,3$-butadiene,also known as isoprene.
It consists of various chains held together by weak van der Waals forces and has a coiled structure.
Chloroprene is the monomer used for the preparation of Neoprene,not natural rubber.
Therefore,the statement that natural rubber is obtained from chloroprene is incorrect.

(B) The classification of polymers based on their structure is as follows:
$1$. High density polyethylene $(HDPE)$: Linear chain polymer.
$2$. Low density polyethylene $(LDPE)$: Branched chain polymer.
$3$. Bakelite: Cross-linked or network polymer.
$4$. Melamine: Cross-linked or network polymer.
Therefore,the correct answer is $B$.

(C) Novolac is a linear polymer formed by the condensation polymerization of $Phenol$ and $Formaldehyde$ in the presence of an acid catalyst.
It is a type of phenol-formaldehyde resin.

(C) Melamine-formaldehyde resin is a thermosetting polymer formed by the condensation polymerization of melamine and formaldehyde.
It is highly durable,heat-resistant,and shatterproof,which makes it an ideal material for manufacturing unbreakable plastic dinnerware,such as plates,bowls,and cups.

(B) The copolymerization of $CH_2=CH-CN$ (acrylonitrile) and $CH_2=CH-CH=CH_2$ ($1$,$3$-butadiene) results in the formation of Buna-$N$ (also known as nitrile rubber or $NBR$).
This is a synthetic rubber used in oil-resistant applications.

(B) $HDP$ stands for High Density Polyethene. It is a tough,hard,and chemically inert polymer. It is primarily used in the manufacture of containers like buckets,dustbins,bottles,and pipes,as well as for the manufacture of toys.

(C) Polymers are classified based on intermolecular forces into elastomers,fibres,thermoplastics,and thermosetting polymers.
Fibres are thread-forming solids that possess high tensile strength and high modulus.
These characteristics are due to strong intermolecular forces like hydrogen bonding or dipole-dipole interactions.
Examples of fibres include polyamides (nylon-$6,6$) and polyesters (Terylene).
Vulcanized rubber,Buna-$S$,and Polystyrene are classified as elastomers or thermoplastics,which do not exhibit the strong intermolecular forces characteristic of fibres.

(A) The dissolution of $Na_2SO_4$ in water is an exothermic process.
When a substance dissolves in water via an exothermic process,its solubility decreases as the temperature increases.
Therefore,the solubility of $Na_2SO_4$ in water decreases with an increase in temperature.

(D) $Cu^{+}$ has a $3d^{10}$ electronic configuration.
Due to the absence of unpaired electrons and a completely filled $d$-subshell,$d-d$ transitions are not possible,resulting in a colorless aqueous solution.
In contrast,$Fe^{3+}$ $(3d^5)$,$Fe^{2+}$ $(3d^6)$,and $Cu^{2+}$ $(3d^9)$ have unpaired electrons and exhibit color.

(D) The ionic character of metal halides follows the order $MF > MCl > MBr > MI$,where $M$ is a monovalent metal.
According to Fajan's rule,the smaller the size of the anion,the greater is the ionic character of the bond.
Since the fluoride ion $(F^-)$ is the smallest among the halide ions,it forms metal halides with the highest ionic character.

(A) The molality $(m)$ of the solution is calculated as: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
Given mass of glucose $= 36 \ g$,molar mass $= 180 \ g/mol$.
Moles of glucose $= \frac{36}{180} = 0.2 \ mol$.
Assuming the density of the solution is approximately $1 \ g/mL$,$1 \ dm^3$ (or $1 \ L$) of solution contains $1000 \ g$ of water,which is $1 \ kg$.
Thus,$m = 0.2 \ mol / 1 \ kg = 0.2 \ m$.
The elevation in boiling point is given by $\Delta T_{b} = K_{b} \times m = K_{b} \times 0.2 = 0.2 K_{b}$.
The boiling point of the solution $= 100 + \Delta T_{b} = 100 + 0.2 K_{b}$.

(D) Crystalline solids are anisotropic in nature,meaning they exhibit different physical properties (such as refractive index) when measured in different directions. Therefore,the statement that they are isotropic is incorrect.

(D) Diamond is a crystalline solid because its constituent particles are arranged in a regular,repeating pattern throughout the entire structure.
In contrast,glass,plastic,and rubber are amorphous solids because they lack a long-range ordered arrangement of particles.

(A) There are $7$ primitive crystal systems in nature.
These $7$ crystal systems can be further classified into $14$ types of Bravais lattices based on the arrangement of lattice points within the unit cell.

(A) There are $7$ distinct crystal systems in crystallography. These $7$ crystal systems give rise to $14$ possible Bravais lattices (also known as space lattices) based on the arrangement of lattice points within the unit cells. Therefore,the total number of crystal systems is $7$.

(A) The number of atoms in one mole of a metal is $6.022 \times 10^{23}$.
In a simple cubic unit cell,the number of atoms per unit cell is $1$.
Therefore,the number of unit cells $= \frac{\text{Total number of atoms}}{\text{Number of atoms per unit cell}} = \frac{6.022 \times 10^{23}}{1} = 6.022 \times 10^{23}$.

(D) Silver $(Ag)$ crystallizes in a face-centered cubic $(fcc)$ lattice structure.
In an $fcc$ unit cell,the packing efficiency is calculated as $74.0 \%$.

(C) The number of atoms of $A$ per unit cell (at corners) = $8 \times (1/8) = 1$.
The number of atoms of $B$ per unit cell (at face centres) = $6 \times (1/2) = 3$.
Therefore,the ratio of atoms $A:B = 1:3$.
Hence,the formula of the compound is $AB_3$.