AP EAMCET 2020 Chemistry Question Paper with Answer and Solution

(D) light year is a unit of distance,not time. It is defined as the distance that light travels in a vacuum in one Julian year.
$1 \text{ light year} = 9.46 \times 10^{15} \text{ m}$.
Leap year,microsecond,and lunar month are all units used to measure time intervals.

(C) In the given graph, the $t$-axis represents time and the $s$-axis represents position (displacement).
For any graph, the slope is given by $\frac{dy}{dx}$. Here, the slope is $\frac{dt}{ds} = \frac{1}{v}$, where $v$ is the velocity.
At point $D$, the position $s$ returns to $0$ at some time $t > 0$. Since the object starts from the origin ($s=0$ at $t=0$) and returns to the origin ($s=0$ at $t=t_D$), the net displacement is $\Delta s = s_{final} - s_{initial} = 0 - 0 = 0$.
Average velocity is defined as $\text{Average velocity} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{0}{t_D} = 0$.
Therefore, option $C$ is correct.

(C) The work done by a force is given by the formula $W = \vec{F} \cdot \vec{s} = FS \cos \theta$,where $\theta$ is the angle between the force vector and the displacement vector.
In uniform circular motion,the centripetal force acts towards the center of the circle,while the displacement (which is along the tangent to the path) is perpendicular to the radius.
Therefore,the angle between the force and the displacement is $\theta = 90^\circ$.
Since $\cos 90^\circ = 0$,the work done $W = FS \cos 90^\circ = 0$.

(C) The formula for work done is $W = F \cdot s \cdot \cos(\theta)$,where $F$ is the force applied,$s$ is the displacement,and $\theta$ is the angle between the force and displacement vectors.
Since the man fails to displace the wall,the displacement $s = 0$.
Therefore,$W = F \cdot 0 \cdot \cos(\theta) = 0$.
Thus,the man does no work at all.

(A) Power is defined as the rate of doing work.
$P = \frac{dW}{dt}$
Since work done $dW = F \cdot dx$,we can write:
$P = \frac{F \cdot dx}{dt}$
Since velocity $v = \frac{dx}{dt}$,substituting this into the equation gives:
$P = F \times v$
Therefore,the power is the product of force and velocity.

(C) The area swept by a planet in a small time interval $dt$ is given by $dA = \frac{1}{2} r \cdot (r d\theta)$.
Therefore,the rate of area swept is $\frac{dA}{dt} = \frac{1}{2} r \left(r \frac{d\theta}{dt}\right)$.
We know that the angular momentum is $\vec{L} = m r^2 \frac{d\vec{\theta}}{dt}$.
Substituting this into the area rate equation,we get $\frac{dA}{dt} = \frac{1}{2m} \left(m r^2 \frac{d\theta}{dt}\right) = \frac{L}{2m}$.
According to Kepler's $2^{nd}$ law,a planet sweeps out equal areas in equal intervals of time,meaning $\frac{dA}{dt} = \text{constant}$.
Since $m$ is constant,$L$ must also be constant.
Therefore,Kepler's $2^{nd}$ law is a statement of the conservation of angular momentum.

(D) The formula for the extension in length $(\Delta L)$ of a wire is given by $\Delta L = \frac{FL}{AY}$,where $F$ is the applied force,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
Since $A = \pi r^2$,where $r$ is the radius of the wire,we have $\Delta L = \frac{FL}{\pi r^2 Y}$.
Given that $F$,$L$,and $Y$ are constant,the extension is inversely proportional to the square of the radius: $\Delta L \propto \frac{1}{r^2}$.
Since the diameter $D = 2r$,the relationship also holds for the diameter: $\Delta L \propto \frac{1}{D^2}$.
Let $\Delta L_1 = 2.4\, cm$ and $D_1 = 0.4\, mm$. If the diameter is doubled,$D_2 = 2D_1$.
Then,$\frac{\Delta L_2}{\Delta L_1} = \left( \frac{D_1}{D_2} \right)^2 = \left( \frac{D_1}{2D_1} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Therefore,$\Delta L_2 = \frac{\Delta L_1}{4} = \frac{2.4\, cm}{4} = 0.6\, cm$.

(A) According to the Doppler effect,when a source moves towards a stationary observer,the apparent frequency $n'$ is given by the formula:
$n' = n \left( \frac{v}{v - v_S} \right)$
where $n$ is the real frequency,$v$ is the speed of sound,and $v_S$ is the speed of the source.
Given that the source is moving with $1/10$ of the speed of sound,we have $v_S = v/10$.
Substituting this into the formula:
$n' = n \left( \frac{v}{v - v/10} \right)$
$n' = n \left( \frac{v}{9v/10} \right)$
$n' = n \left( \frac{10}{9} \right)$
Therefore,the ratio of apparent to real frequency is $\frac{n'}{n} = \frac{10}{9}$.

(D) An equipotential surface is a surface where the electric potential is the same at every point.
By definition,the work done $W$ in moving a charge $q$ from one point to another is given by $W = q(V_2 - V_1)$.
Since the surface is equipotential,the potential at both points is equal,i.e.,$V_1 = V_2$.
Therefore,$V_2 - V_1 = 0$.
Consequently,the work done $W = q(0) = 0$.
Thus,no work is done in moving a charge over an equipotential surface.

(A) The magnetic force $F$ on a charged particle moving in a magnetic field is given by the formula: $F = qvB \sin \theta$.
Given:
Charge of a proton $q = 1.6 \times 10^{-19} \, C$,
Velocity $v = 2 \times 10^7 \, m/s$,
Magnetic flux density $B = 1.5 \, Wb/m^2$,
Angle $\theta = 30^\circ$.
Substituting these values into the formula:
$F = (1.6 \times 10^{-19}) \times (2 \times 10^7) \times 1.5 \times \sin(30^\circ)$
$F = (1.6 \times 10^{-19}) \times (2 \times 10^7) \times 1.5 \times 0.5$
$F = 1.6 \times 10^{-19} \times 10^7 \times 1.5$
$F = 2.4 \times 10^{-12} \, N$.

(D) For a hollow cylindrical conductor carrying a uniformly distributed current $i$,the magnetic field $B$ at a distance $r$ (where $R < r < 2R$) from the axis is given by Ampere's Law:
$B = \frac{\mu_0 I_{enclosed}}{2\pi r}$
Here,the current $I_{enclosed}$ within the radius $r$ is the fraction of the total current $i$ proportional to the cross-sectional area:
$I_{enclosed} = i \left( \frac{\pi r^2 - \pi R^2}{\pi (2R)^2 - \pi R^2} \right) = i \left( \frac{r^2 - R^2}{4R^2 - R^2} \right) = i \left( \frac{r^2 - R^2}{3R^2} \right)$
Substituting $r = \frac{3R}{2}$:
$I_{enclosed} = i \left( \frac{(\frac{3R}{2})^2 - R^2}{3R^2} \right) = i \left( \frac{\frac{9R^2}{4} - R^2}{3R^2} \right) = i \left( \frac{\frac{5R^2}{4}}{3R^2} \right) = \frac{5i}{12}$
Now,calculate the magnetic field $B$:
$B = \frac{\mu_0}{2\pi r} \left( \frac{5i}{12} \right) = \frac{\mu_0}{2\pi (\frac{3R}{2})} \left( \frac{5i}{12} \right) = \frac{\mu_0}{3\pi R} \left( \frac{5i}{12} \right) = \frac{5\mu_0 i}{36\pi R}$

(A) The magnetic moment $M$ of the bar magnet is given by the product of pole strength $m$ and length $l$: $M = m \times l$.
Given $m = 10^{-3} \, Wb$ and $l = 10 \, cm = 0.1 \, m$,we have $M = 10^{-3} \times 0.1 = 10^{-4} \, A \cdot m^2$.
The torque $\tau$ acting on a magnetic dipole in a magnetic field $B$ is given by $\tau = MB \sin \theta$.
Here,$B = 4\pi \times 10^{-3} \, T$ and $\theta = 30^{\circ}$.
Substituting the values: $\tau = (10^{-4}) \times (4\pi \times 10^{-3}) \times \sin(30^{\circ})$.
$\tau = 4\pi \times 10^{-7} \times 0.5$.
$\tau = 2\pi \times 10^{-7} \, N \cdot m$.

(C) The magnetic moment of a single bar magnet is given by $M = m \times l$,where $m$ is the pole strength and $l$ is the length of the magnet.
Since the two magnets are identical and placed at a right angle to each other,their individual magnetic moment vectors $\vec{M_1}$ and $\vec{M_2}$ are also at a right angle to each other.
The magnitude of each magnetic moment is $M = ml$.
The net magnetic moment of the system is the vector sum of the two individual magnetic moments:
$M_{net} = \sqrt{M_1^2 + M_2^2 + 2M_1M_2 \cos(90^\circ)}$
Since $\cos(90^\circ) = 0$,we have:
$M_{net} = \sqrt{M^2 + M^2} = \sqrt{2M^2} = \sqrt{2}M$
Substituting $M = ml$,we get:
$M_{net} = \sqrt{2}\,ml$.

(B) The magnetic moment $M$ of a circular loop carrying current $i$ is given by the formula $M = iA$,where $A$ is the area of the loop.
For a circular loop of radius $R$,the area is $A = \pi R^2$.
If the length of the wire is $L$,then the circumference of the loop is $L = 2\pi R$,which implies $R = \frac{L}{2\pi}$.
Substituting the value of $R$ into the area formula,we get $A = \pi \left(\frac{L}{2\pi}\right)^2 = \frac{L^2}{4\pi}$.
Now,substituting $A$ into the magnetic moment formula: $M = i \left(\frac{L^2}{4\pi}\right)$.
Rearranging the equation to solve for $L$: $L^2 = \frac{4\pi M}{i}$.
Therefore,the length of the wire is $L = \sqrt{\frac{4\pi M}{i}}$.

(A) The horizontal component of the Earth's magnetic field $(B_H)$ is given by the formula $B_H = B \cos \phi$,where $B$ is the total magnetic field and $\phi$ is the angle of dip.
Assuming the total magnetic field $B$ is the same at both locations:
For the first place,$(B_H)_1 = B \cos 30^o = B \frac{\sqrt{3}}{2}$.
For the second place,$(B_H)_2 = B \cos 45^o = B \frac{1}{\sqrt{2}}$.
The ratio of the horizontal components is $\frac{(B_H)_1}{(B_H)_2} = \frac{B \cos 30^o}{B \cos 45^o} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{2} \times \sqrt{2} = \frac{\sqrt{3}}{\sqrt{2}}$.
Thus,the ratio is $\sqrt{3} : \sqrt{2}$.

(D) Lenz's law states that the direction of the induced current is such that it opposes the change in magnetic flux that produced it. If the induced current were to assist the change,it would lead to an infinite increase in energy,which violates the law of conservation of energy. Therefore,Lenz's law is a direct consequence of the law of conservation of energy.

(C) For an ideal transformer,the relationship between turns,voltage,and current is given by: $\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{i_s}{i_p}$.
Given that it is a step-down transformer,the primary coil must have more turns than the secondary coil. Thus,$N_p = 5000$ and $N_s = 500$.
Given values: $V_p = 2200 \, V$,$i_p = 4 \, A$.
Using the ratio: $\frac{5000}{500} = \frac{2200}{V_s} = \frac{i_s}{4}$.
Calculating voltage $V_s$: $\frac{5000}{500} = \frac{2200}{V_s} \Rightarrow 10 = \frac{2200}{V_s} \Rightarrow V_s = 220 \, V$.
Calculating current $i_s$: $\frac{5000}{500} = \frac{i_s}{4} \Rightarrow 10 = \frac{i_s}{4} \Rightarrow i_s = 40 \, A$.
Therefore,the current is $40 \, A$ and the potential difference is $220 \, V$.

(B) The induced electromotive force $(EMF)$ in a moving conductor is given by $\varepsilon = B \cdot l_{eff} \cdot V$,where $l_{eff}$ is the effective length of the conductor perpendicular to the velocity vector and the magnetic field.
For a wire moving in a uniform magnetic field,the effective length is the straight-line distance between the two points of contact on the rails.
Even if the wire $AB$ is replaced by a semicircular wire,the distance between the points of contact on the rails remains the same as the original straight wire $AB$.
Since the velocity $V$,the magnetic field $B$,and the effective length $l_{eff}$ remain unchanged,the induced $EMF$ $\varepsilon$ remains the same.
Consequently,the magnitude of the induced current $I = \varepsilon / R$ also remains the same.

(A) The speed of light in a vacuum is $c = 3 \times 10^8 \text{ m/s}$.
The speed of light in the medium $(v)$ is given by the product of frequency $(\nu)$ and wavelength $(\lambda)$: $v = \nu \lambda$.
Substituting the given values: $v = (4 \times 10^{14} \text{ Hz}) \times (5 \times 10^{-7} \text{ m}) = 20 \times 10^7 \text{ m/s} = 2 \times 10^8 \text{ m/s}$.
The refractive index $(\mu)$ is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium: $\mu = \frac{c}{v}$.
$\mu = \frac{3 \times 10^8 \text{ m/s}}{2 \times 10^8 \text{ m/s}} = 1.5$.

(A) Cis-trans isomerism (geometrical isomerism) is exhibited by compounds that have restricted rotation around a double bond,where each carbon atom of the double bond is attached to two different groups.
$A$. $2-$butene $(CH_3-CH=CH-CH_3)$ satisfies this condition as each carbon of the $C=C$ bond is attached to a hydrogen atom and a methyl group,allowing for cis and trans configurations.
$B$. $2-$butyne has a triple bond,which is linear and does not allow for cis-trans isomerism.
$C$. $2-$butanol and $D$. Butanone do not contain a $C=C$ double bond with the required substitution pattern.
Therefore,the correct answer is $2-$butene.

(C) Some colloidal particles develop electrical charges due to the dissociation or ionization of the surface molecules.
The charge on the colloidal particles is primarily caused by the selective adsorption of ions from the dispersion medium (solution) onto the surface of the particles.

(D) Glucose on reaction with bromine water ($Br_2$ water) acts as a mild oxidizing agent,which oxidizes the aldehyde group $(-CHO)$ of glucose to a carboxylic acid group $(-COOH)$,resulting in the formation of gluconic acid.
$C_6H_{12}O_6 + Br_2 + H_2O \rightarrow CH_2OH(CHOH)_4COOH + 2HBr$

(A) Given equation is $x^2 - 5|x| + 6 = 0$.
Since $|x|^2 = x^2$,we can write the equation as $|x|^2 - 5|x| + 6 = 0$.
Let $|x| = t$,where $t \ge 0$. Then the equation becomes $t^2 - 5t + 6 = 0$.
Factoring the quadratic: $(t - 2)(t - 3) = 0$.
This gives $t = 2$ or $t = 3$.
Case $1$: $|x| = 2 \implies x = 2, -2$.
Case $2$: $|x| = 3 \implies x = 3, -3$.
Thus,the solutions are $x \in \{2, -2, 3, -3\}$.
There are $4$ solutions in total.

(B) Given $f(x) = \left| \begin{array}{ccc} x - 3 & 2(x^2 - 9) & 3(x^3 - 27) \\ x - 5 & 2(x^2 - 25) & 4(x^3 - 125) \\ 1 & 2 & 3 \end{array} \right|$.
Observe that if $x = 3$,the first row becomes $(3-3, 2(9-9), 3(27-27)) = (0, 0, 0)$. Since one row is zero,$f(3) = 0$.
Observe that if $x = 5$,the second row becomes $(5-5, 2(25-25), 4(125-125)) = (0, 0, 0)$. Since one row is zero,$f(5) = 0$.
We need to calculate $f(1)f(3) + f(3)f(5) + f(5)f(1)$.
Substituting the values $f(3) = 0$ and $f(5) = 0$ into the expression:
$f(1) \times 0 + 0 \times 0 + 0 \times f(1) = 0 + 0 + 0 = 0$.
Since $f(3) = 0$,the result is $f(3)$.

(C) The given equation is a homogeneous second-degree equation of the form $ax^2 + 2hxy + by^2 = 0$.
Comparing $4x^2 - 24xy + 11y^2 = 0$ with the general form,we have $a = 4$,$2h = -24$,and $b = 11$.
Since the equation is homogeneous and has no constant term,it represents a pair of straight lines passing through the origin.
To factorize: $4x^2 - 22xy - 2xy + 11y^2 = 0$
$2x(2x - 11y) - y(2x - 11y) = 0$
$(2x - y)(2x - 11y) = 0$
Thus,the lines are $2x - y = 0$ and $2x - 11y = 0$,both of which pass through the origin $(0, 0)$.

(C) Given centre $(h, k) = (2, -3)$ and circumference $= 10\pi$.
Since the circumference of a circle is given by $2\pi r$,we have $2\pi r = 10\pi$,which implies $r = 5$.
The standard equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting the values,we get $(x - 2)^2 + (y - (-3))^2 = 5^2$.
$(x - 2)^2 + (y + 3)^2 = 25$.
Expanding the squares: $(x^2 - 4x + 4) + (y^2 + 6y + 9) = 25$.
$x^2 + y^2 - 4x + 6y + 13 = 25$.
$x^2 + y^2 - 4x + 6y - 12 = 0$.

(C) The function is defined as $f(x) = |x| + \frac{|x|}{x}$.
First,consider the term $|x|$. The function $g(x) = |x|$ is continuous for all real numbers,including $x = 0$.
Second,consider the term $h(x) = \frac{|x|}{x}$.
For $x > 0$,$h(x) = \frac{x}{x} = 1$.
For $x < 0$,$h(x) = \frac{-x}{x} = -1$.
Since the left-hand limit $\lim_{x \to 0^-} h(x) = -1$ and the right-hand limit $\lim_{x \to 0^+} h(x) = 1$ are not equal,the function $h(x)$ is discontinuous at $x = 0$.
Since the sum of a continuous function and a discontinuous function is discontinuous,$f(x) = |x| + \frac{|x|}{x}$ is discontinuous at $x = 0$ due to the term $\frac{|x|}{x}$.

(C) To find the derivative of the product $(1 + x^2)\tan^{-1}x$,we use the product rule: $\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx}$.
Let $u = (1 + x^2)$ and $v = \tan^{-1}x$.
Then $\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = \frac{1}{1 + x^2}$.
Applying the product rule:
$\frac{d}{dx}[(1 + x^2)\tan^{-1}x] = (1 + x^2) \cdot \frac{d}{dx}(\tan^{-1}x) + \tan^{-1}x \cdot \frac{d}{dx}(1 + x^2)$
$= (1 + x^2) \cdot \frac{1}{1 + x^2} + \tan^{-1}x \cdot (2x)$
$= 1 + 2x \tan^{-1}x$.
Thus,the correct option is $C$.

(D) Given $y = \frac{e^x \log x}{x^2}$.
Using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$,where $u = e^x \log x$ and $v = x^2$:
$\frac{dy}{dx} = \frac{x^2 \frac{d}{dx}(e^x \log x) - e^x \log x \frac{d}{dx}(x^2)}{(x^2)^2}$
$\frac{dy}{dx} = \frac{x^2 (e^x \log x + e^x \cdot \frac{1}{x}) - e^x \log x (2x)}{x^4}$
$\frac{dy}{dx} = \frac{x^2 e^x \log x + x e^x - 2x e^x \log x}{x^4}$
$\frac{dy}{dx} = \frac{x e^x (x \log x + 1 - 2 \log x)}{x^4}$
$\frac{dy}{dx} = \frac{e^x [1 + (x - 2) \log x]}{x^3}$.

(A) Given that the range $R$ is twice the maximum height $H$,so $R = 2H$.
We know the formulas for range $R = \frac{v^2 \sin(2\theta)}{g} = \frac{2v^2 \sin\theta \cos\theta}{g}$ and maximum height $H = \frac{v^2 \sin^2\theta}{2g}$.
Substituting these into the given condition $R = 2H$:
$\frac{2v^2 \sin\theta \cos\theta}{g} = 2 \left( \frac{v^2 \sin^2\theta}{2g} \right)$
$\frac{2v^2 \sin\theta \cos\theta}{g} = \frac{v^2 \sin^2\theta}{g}$
$2 \cos\theta = \sin\theta$
$\tan\theta = 2$.
From the right-angled triangle with opposite side $2$ and adjacent side $1$,the hypotenuse is $\sqrt{2^2 + 1^2} = \sqrt{5}$.
Thus,$\sin\theta = \frac{2}{\sqrt{5}}$ and $\cos\theta = \frac{1}{\sqrt{5}}$.
Now,substitute these into the range formula:
$R = \frac{2v^2 \sin\theta \cos\theta}{g} = \frac{2v^2}{g} \left( \frac{2}{\sqrt{5}} \right) \left( \frac{1}{\sqrt{5}} \right) = \frac{4v^2}{5g}$.

(C) The equation of a line passing through $(2, 3)$ with slope $m$ is $y - 3 = m(x - 2)$,which simplifies to $y = mx + (3 - 2m)$.
The $x$-intercept $(A)$ is found by setting $y = 0$: $0 = mx + 3 - 2m \Rightarrow x = \frac{2m - 3}{m}$.
The $y$-intercept $(B)$ is found by setting $x = 0$: $y = 3 - 2m$.
The area of the triangle formed with the coordinate axes is $\frac{1}{2} |x_{intercept}| |y_{intercept}| = 12$.
$\frac{1}{2} |\frac{2m - 3}{m}| |3 - 2m| = 12$.
Since $(3 - 2m)^2 = (2m - 3)^2$,we have $\frac{(2m - 3)^2}{|m|} = 24$.
Case $1$: $m > 0$,then $(2m - 3)^2 = 24m$ $\Rightarrow 4m^2 - 12m + 9 = 24m$ $\Rightarrow 4m^2 - 36m + 9 = 0$. The discriminant $D = (-36)^2 - 4(4)(9) = 1296 - 144 = 1152 > 0$,so there are $2$ distinct positive values for $m$.
Case $2$: $m < 0$,then $(2m - 3)^2 = -24m$ $\Rightarrow 4m^2 - 12m + 9 = -24m$ $\Rightarrow 4m^2 + 12m + 9 = 0$ $\Rightarrow (2m + 3)^2 = 0$. This gives $m = -3/2$,which is $1$ negative value.
Total number of possible lines = $2 + 1 = 3$.

(D) In $XeF_4$,the geometry is square planar with all $Xe-F$ bonds being equal due to symmetry.
In $BF_4^-$ and $SiF_4$,the geometry is tetrahedral,where all four bonds are equivalent.
In $SF_4$,the sulfur atom undergoes $sp^3d$ hybridization with one lone pair,resulting in a see-saw geometry.
Due to the presence of the lone pair,the axial and equatorial $S-F$ bonds have different bond lengths and bond angles,making them unequal.

(D) Lenz's law is a direct consequence of the law of conservation of $ENERGY$.
According to Lenz's law,the direction of the induced current is such that it always opposes the cause (change in magnetic flux) that produces it.
If the induced current were to assist the cause,it would lead to an increase in magnetic flux,which would further increase the current,violating the law of conservation of $ENERGY$.
Therefore,to overcome the opposing force,external work must be done. This mechanical work is converted into electrical energy,confirming that Lenz's law is consistent with the law of conservation of $ENERGY$.

(D) In the given graph,the $t$-axis (time) is on the vertical axis and the $s$-axis (displacement) is on the horizontal axis.
For a physical motion,time $t$ must always increase. In this graph,as the object moves from $O$ to $A$,$B$,$C$,and $D$,the value of $t$ increases. However,at point $D$,the graph shows that for a further increase in displacement $s$,the time $t$ would have to decrease,which is physically impossible.
Furthermore,for any time $t$,there cannot be two different positions $s$ (or for any position $s$,there cannot be two different times $t$ in a way that violates causality). Specifically,in this $t$-$s$ graph,the slope represents $dt/ds = 1/v$. At point $D$,the curve turns back,implying that time starts decreasing,which is impossible.
Therefore,the graph shown is impossible.

(C) In a $V-I$ graph,the slope represents the resistance $R$.
For temperature $T_1$,the line makes an angle $\theta$ with the $I$-axis. Thus,$R_1 = \tan \theta$. Since $R \propto T$,we have $R_1 = k T_1 = \tan \theta$.
For temperature $T_2$,the line makes an angle $\theta$ with the $V$-axis. The angle with the $I$-axis is $(90^\circ - \theta)$. Thus,$R_2 = \tan(90^\circ - \theta) = \cot \theta$.
Since $R \propto T$,we have $R_2 = k T_2 = \cot \theta$.
Now,consider $(T_2 - T_1) \propto (R_2 - R_1)$.
$(T_2 - T_1) \propto (\cot \theta - \tan \theta)$.
Using trigonometric identities: $\cot \theta - \tan \theta = \frac{\cos \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta} = \frac{\cos^2 \theta - \sin^2 \theta}{\sin \theta \cos \theta} = \frac{\cos 2\theta}{\frac{1}{2} \sin 2\theta} = 2 \cot 2\theta$.
Therefore,$(T_2 - T_1) \propto \cot 2\theta$.

(A) The change in internal energy $\Delta U$ for an ideal gas is given by $\Delta U = n C_V \Delta T$.
For a diatomic gas,the molar heat capacity at constant volume is $C_V = \frac{R}{\gamma - 1}$,where $\gamma = 1.4 = 7/5$.
Using the ideal gas law $PV = nRT$,we have $\Delta T = \frac{P_B V_B - P_A V_A}{nR}$.
Substituting this into the internal energy formula:
$\Delta U = n \left( \frac{R}{\gamma - 1} \right) \left( \frac{P_B V_B - P_A V_A}{nR} \right) = \frac{P_B V_B - P_A V_A}{\gamma - 1}$.
From the graph,at point $A$: $P_A = 5 \times 10^3 \, Pa$,$V_A = 4 \, m^3$.
At point $B$: $P_B = 2 \times 10^3 \, Pa$,$V_B = 6 \, m^3$.
Calculating the product $PV$ at each point:
$P_A V_A = (5 \times 10^3) \times 4 = 20 \times 10^3 \, J$.
$P_B V_B = (2 \times 10^3) \times 6 = 12 \times 10^3 \, J$.
Now,$\Delta U = \frac{12 \times 10^3 - 20 \times 10^3}{7/5 - 1} = \frac{-8 \times 10^3}{2/5} = -8 \times 10^3 \times \frac{5}{2} = -20 \times 10^3 \, J = -20 \, kJ$.

(B) The motional electromotive force $(EMF)$ induced in a conductor moving in a magnetic field is given by $\varepsilon = \int (\vec{v} \times \vec{B}) \cdot d\vec{l}$.
For a conductor moving with constant velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$,this simplifies to $\varepsilon = \vec{v} \times \vec{B} \cdot \vec{L}_{eff}$,where $\vec{L}_{eff}$ is the effective length vector connecting the two ends of the conductor.
In both cases (straight wire $AB$ and a semicircular wire connecting points $A$ and $B$),the effective length vector $\vec{L}_{eff}$ is the same,as it is the straight-line distance between the points of contact on the rails.
Since the velocity $V$,magnetic field $B$,and effective length $L_{eff}$ remain unchanged,the induced $EMF$ $\varepsilon = B V L_{eff}$ remains the same.
Consequently,the induced current $I = \varepsilon / R$ also remains the same.

(A) To determine if all bonds are equal,we examine the molecular geometry and hybridization:
$1$. $SiF_4$: $sp^3$ hybridization,tetrahedral geometry. All $Si-F$ bonds are equivalent.
$2$. $XeF_4$: $sp^3d^2$ hybridization,square planar geometry. All $Xe-F$ bonds are equivalent.
$3$. $BF_4^-$: $sp^3$ hybridization,tetrahedral geometry. All $B-F$ bonds are equivalent.
$4$. $ClF_3$: $sp^3d$ hybridization,$T$-shaped geometry. Due to the presence of two lone pairs in the equatorial positions,the axial $Cl-F$ bonds are longer than the equatorial $Cl-F$ bond. Thus,all bonds are not equal.
Therefore,the correct option is $A$.

(B) The induced electromotive force $(emf)$ in a moving conductor is given by $e = B l v$,where $B$ is the magnetic field,$l$ is the length of the conductor perpendicular to the velocity,and $v$ is the velocity.
In this setup,the effective length $l$ of the wire that cuts the magnetic field lines is the distance between the two rails (the chord length of the wire).
When the straight wire $AB$ is replaced by a semicircular wire with the same chord length,the effective length $l$ remains unchanged because the distance between the points of contact on the rails remains the same.
Since $e = B l v$,and $B, l,$ and $v$ remain constant,the induced $emf$ remains the same.
Consequently,the induced current $I = e / R$ also remains the same,assuming the resistance $R$ of the circuit does not change.

(D) Glucose $(CHO-(CHOH)_4-CH_2OH)$ is a mild reducing agent.
When it reacts with bromine water $(Br_2/H_2O)$,which is a mild oxidizing agent,the aldehyde group $(-CHO)$ is oxidized to a carboxylic acid group $(-COOH)$.
The resulting product is gluconic acid $(COOH-(CHOH)_4-CH_2OH)$.

(A) For an ideal gas,the change in internal energy is given by $\Delta U = n C_V \Delta T$.
Since $C_V = \frac{R}{\gamma - 1}$ and $PV = nRT$,we have $\Delta U = \frac{n R \Delta T}{\gamma - 1} = \frac{P_B V_B - P_A V_A}{\gamma - 1}$.
For a diatomic gas,the adiabatic index $\gamma = 1.4 = \frac{7}{5}$,so $\gamma - 1 = 0.4 = \frac{2}{5}$.
From the graph,at point $A$: $P_A = 5 \times 10^3 \, Pa$,$V_A = 4 \, m^3$.
At point $B$: $P_B = 2 \times 10^3 \, Pa$,$V_B = 6 \, m^3$.
Calculating the product $PV$ at each point:
$P_A V_A = (5 \times 10^3) \times 4 = 20 \times 10^3 \, J$.
$P_B V_B = (2 \times 10^3) \times 6 = 12 \times 10^3 \, J$.
Substituting these values into the formula:
$\Delta U = \frac{12 \times 10^3 - 20 \times 10^3}{2/5} = \frac{-8 \times 10^3}{0.4} = -20 \times 10^3 \, J = -20 \, kJ$.

(D) The correct option is $D$.
Archimedes' upward thrust (buoyant force) is defined as the upward force exerted by a fluid that opposes the weight of an immersed object. This force arises specifically due to the pressure gradient in a fluid column,which is caused by gravity $(P = \rho gh)$.
If there were no gravity,there would be no pressure variation with depth in the fluid. Consequently,the net upward force (buoyant force) on an object would be zero.
Viscosity,surface tension,and pressure (in a compressed fluid) are intrinsic properties or states that do not depend on the presence of gravity.

(A) The formula for the apparent frequency $f$ when the source moves towards a stationary observer is given by $f = f_0 \left( \frac{v}{v - v_s} \right)$,where $f_0$ is the real frequency,$v$ is the speed of sound,and $v_s$ is the speed of the source.
Given that the source speed $v_s = v/10$.
Substituting this into the formula:
$f = f_0 \left( \frac{v}{v - v/10} \right)$
$f = f_0 \left( \frac{v}{9v/10} \right)$
$f = f_0 \left( \frac{10}{9} \right)$
Therefore,the ratio of apparent frequency to real frequency is $f/f_0 = 10/9$.

(A) The speed of light in a medium $(v)$ is given by the product of frequency $(f)$ and wavelength $(\lambda)$:
$v = f \times \lambda$
Given $f = 4 \times 10^{14} ~Hz$ and $\lambda = 5 \times 10^{-7} ~m$.
$v = (4 \times 10^{14}) \times (5 \times 10^{-7}) = 20 \times 10^7 ~m/s = 2 \times 10^8 ~m/s$.
The refractive index $(\mu)$ is defined as the ratio of the speed of light in a vacuum $(c)$ to the speed of light in the medium $(v)$:
$\mu = \frac{c}{v}$
Taking $c = 3 \times 10^8 ~m/s$:
$\mu = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$.

(B) $AlCl_3$ is a covalent compound due to the high polarizing power of the $Al^{3+}$ ion (Fajans' rule).
Because it is covalent,it is soluble in organic solvents like ether.
$LiCl$ and $NaCl$ are ionic compounds and are insoluble in ether.
Therefore,only $AlCl_3$ is extracted into ether.

(C) transformer works on the principle of mutual induction.
Mutual induction is a phenomenon where a change in current in one coil induces an electromotive force $(emf)$ in an adjacent coil due to the change in magnetic flux linked with it.

(D) transformer is a device that transfers electrical energy between two or more circuits through electromagnetic induction.
It consists of two coils,the primary and the secondary,wound on a common magnetic core.
When an alternating current flows through the primary coil,it creates a changing magnetic flux in the core.
This changing flux links with the secondary coil and induces an electromotive force $(EMF)$ in it,which is the phenomenon of mutual induction.
Therefore,a transformer works on the principle of mutual induction.

(D) Essential amino acids are those that cannot be synthesized by the human body and must be obtained through the diet. $Lysine$,$Histidine$,and $Valine$ are essential amino acids.
$Tyrosine$ is a non-essential amino acid because it can be synthesized by the body from the essential amino acid $Phenylalanine$.

(A) In a hydroxyl ion,$OH^{\ominus}$,the oxygen atom has $6$ valence electrons.
It shares $1$ electron with the hydrogen atom to form a covalent bond,and it gains $1$ electron due to the negative charge.
This results in $8$ electrons around the oxygen atom,which corresponds to $4$ electron pairs.
Out of these $4$ pairs,$1$ pair is involved in the $O-H$ covalent bond,and the remaining $3$ pairs are lone pairs on the oxygen atom.

(B) To determine the hybridization $(H)$,we use the formula: $H = \frac{1}{2}(V + M - C + A)$
Where $V$ = number of valence electrons of the central atom,$M$ = number of monovalent atoms,$C$ = cationic charge,and $A$ = anionic charge.
$1$. For $NO_2^{+}$:
$H = \frac{1}{2}(5 + 0 - 1 + 0) = 2$,which corresponds to $sp$ hybridization.
$2$. For $NO_3^{-}$:
$H = \frac{1}{2}(5 + 0 - 0 + 1) = 3$,which corresponds to $sp^2$ hybridization.
$3$. For $NH_4^{+}$:
$H = \frac{1}{2}(5 + 4 - 1 + 0) = 4$,which corresponds to $sp^3$ hybridization.
Thus,the hybridization types are $sp, sp^2, sp^3$ respectively.

(D) Essential amino acids are those that cannot be synthesized by the human body and must be obtained through the diet.
Valine,Leucine,and Lysine are essential amino acids.
Tyrosine is a non-essential amino acid because it can be synthesized in the body from phenylalanine.

(B) Secondary alcohols like $CH_3-CH(OH)-CH_2-CH_3$ (butan$-2-$ol) on oxidation with strong oxidizing agents like alkaline $KMnO_4$ undergo oxidation to form ketones.
Specifically,butan$-2-$ol is oxidized to butanone $(CH_3-CO-CH_2-CH_3)$.
Primary alcohols like butan$-1-$ol $(CH_3-CH_2-CH_2-CH_2OH)$ are oxidized to carboxylic acids (butanoic acid) under these conditions.
Hence,the correct option is $(B)$.

(D) Cumene (isopropylbenzene) is oxidised in the presence of air to form cumene hydroperoxide.
On treating with dilute acid,it undergoes rearrangement to form phenol and acetone (propan$-2-$one) as a by-product.

(D) Iron $(Fe)$ is used as a catalyst in Haber's process.
The Haber process,also called the Haber-Bosch process,is the industrial implementation of the reaction of nitrogen gas $(N_2)$ and hydrogen gas $(H_2)$.
The chemical equation is: $N_2(g) + 3H_2(g) \xrightarrow{Fe, 450^{\circ}C, 250 \ atm} 2NH_3(g)$.
It is the main industrial procedure to produce ammonia $(NH_3)$ and the catalyst used is iron with a suitable promoter like $K_2O$,$CaO$,$SiO_2$,and $Al_2O_3$.
Hence,Iron is used in the Haber process as a cheap catalyst because it allows reaching a reasonable yield in an acceptable time.

(A) When an iron rod is exposed to a moist atmosphere,it undergoes corrosion,commonly known as rusting.
This process involves the formation of hydrated ferric oxide $(Fe_2O_3 \cdot xH_2O)$ on the surface of the iron.
Since this involves the formation of chemical bonds between the iron surface and the oxygen/water molecules,it is classified as a surface phenomenon known as chemisorption.

(B) The reaction between a ketone (cyclobutanone) and semicarbazide $(H_2N-NH-CO-NH_2)$ is a nucleophilic addition-elimination reaction.
In this reaction,the carbonyl oxygen of the ketone is replaced by the nitrogen atom of the semicarbazide,resulting in the formation of a semicarbazone.
The general structure of a semicarbazone is $R_2C=N-NH-CO-NH_2$.
For cyclobutanone,the structure is a cyclobutane ring double-bonded to a nitrogen atom,which is attached to an $NH$ group,which is attached to a carbonyl group,which is attached to an $NH_2$ group.
This matches the structure shown in option $B$.

(C) Formaldehyde is $HCHO$. $A$ hemiacetal is formed by the addition of one molecule of an alcohol to an aldehyde. The reaction of formaldehyde with methanol $(CH_3OH)$ forms methyl hemiacetal of formaldehyde.
The reaction is: $HCHO + CH_3OH \rightleftharpoons H_2C(OH)(OCH_3)$.
The structure consists of a central carbon atom bonded to two hydrogen atoms,one hydroxyl group $(-OH)$,and one methoxy group $(-OCH_3)$.
This corresponds to the structure $H_2C(OH)(OCH_3)$.
Hence,the correct option is $(C)$.

(B) In a cross aldol condensation,the nucleophile (enolate) attacks the carbonyl carbon of the electrophile.
Here,$CH_3CH_2CH_2CHO$ (butanal) acts as the nucleophile,forming an enolate at the $\alpha$-carbon: $CH_3CH_2CH^-CHO$.
This enolate attacks the carbonyl carbon of $CH_3CH_2CHO$ (propanal) as the electrophile.
The reaction is: $CH_3CH_2CHO + CH_3CH_2CH_2CHO \rightarrow CH_3CH_2-CH(OH)-CH(CH_2CH_3)-CHO$.
The product formed is $2-$ethyl$-3-$hydroxypentanal.
Hence,the correct option is $B$.

(C) Aldol condensation occurs in aldehydes or ketones that possess at least one $\alpha$-hydrogen atom.
$1$. Methanal $(HCHO)$ has no $\alpha$-carbon,hence no $\alpha$-hydrogen.
$2$. Butan-$1$-ol is an alcohol,not an aldehyde or ketone.
$3$. $2$-methyl pentanal $(CH_3CH_2CH_2CH(CH_3)CHO)$ has one $\alpha$-hydrogen atom at the $C-2$ position,so it undergoes aldol condensation.
$4$. $2,2$-dimethyl pentanal $(CH_3CH_2CH_2C(CH_3)_2CHO)$ has no $\alpha$-hydrogen atom because the $C-2$ carbon is fully substituted with two methyl groups and the aldehyde group.
Therefore,the correct option is $C$.

(D) Due to the presence of bulky aryl groups,the nucleophilic attack on the $sp^2$ carbon of the $>C=O$ group experiences steric hindrance,decreasing the reaction rate.
$(B)$ Fehling's solution is a mild oxidizing agent and cannot oxidize benzaldehyde.
$(C)$ Ethanal contains $\alpha$-hydrogens which are acidic due to the electron-withdrawing effect of the carbonyl group,allowing for enolization.
$(D)$ The $-NO_2$ group at the $p$-position exerts a strong $-R$ (resonance) and $-I$ (inductive) effect,which increases the partial positive charge on the carbonyl carbon,making it more susceptible to nucleophilic attack. Thus,$p$-nitrobenzaldehyde is more reactive than benzaldehyde. Therefore,statement $D$ is incorrect.

(D) Formaldehyde $(HCHO)$ reacts with ammonia $(NH_3)$ to produce hexamethylenetetramine,which is commonly known as urotropine.
The balanced chemical equation is:
$6 HCHO + 4 NH_3 \rightarrow (CH_2)_6N_4 + 6 H_2O$

(D) Ethanal $(CH_3CHO)$ and propanone $(CH_3COCH_3)$ both react with a Grignard reagent $(RMgX)$ to form alcohols.
Specifically,ethanal reacts with a Grignard reagent to form a secondary alcohol (after hydrolysis),while propanone reacts with a Grignard reagent to form a tertiary alcohol (after hydrolysis).
In contrast,Tollen's reagent,Fehling's reagent,and Schiff's reagent are specific tests for aldehydes and do not react with ketones like propanone.
Therefore,the correct option is $(D)$.

(B) The reaction of an aromatic nitro compound $(Ar-NO_2)$ with $Zn / NH_4Cl$ is a selective reduction that produces an aromatic hydroxylamine $(Ar-NHOH)$.
$Ar-NO_2 + Zn / NH_4Cl \rightarrow Ar-NHOH + ZnO$
Aromatic hydroxylamines are strong reducing agents and can reduce Tollen's reagent (ammoniacal silver nitrate) to metallic silver $(Ag)$,which appears as a black precipitate.
$Ar-NHOH + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow Ar-NO + 2Ag(s) + 4NH_3 + 2H_2O$
Therefore,the functional group present in compound "$A$" is the nitro group $(-NO_2)$.

(D) The boiling points of amines depend on the extent of intermolecular hydrogen bonding.
Primary amines $(R-NH_2)$ have two hydrogen atoms available for hydrogen bonding,secondary amines $(R_2NH)$ have one,and tertiary amines $(R_3N)$ have none.
Therefore,the extent of hydrogen bonding follows the order: primary > secondary > tertiary.
Compound $2$ $(CH_3CH_2CH_2NH_2)$ is a primary amine,compound $3$ $(CH_3CH_2NHCH_3)$ is a secondary amine,and compound $1$ $((CH_3)_3N)$ is a tertiary amine.
Thus,the order of boiling points is $2 > 3 > 1$.

(B) The reaction of aniline with nitrous acid $(HNO_2)$ at low temperature $(273-278 \ K)$ is known as the diazotization reaction.
In this reaction,aniline reacts with $HNO_2$ (generated in situ from $NaNO_2$ and $HCl$) to form benzene diazonium chloride.
The chemical equation is:
$C_6H_5NH_2 + HNO_2 + HCl \xrightarrow{273-278 \ K} C_6H_5N_2^+Cl^- + 2H_2O$.
Thus,the product $A$ is benzene diazonium salt.

(A) According to Lewis theory,a base is a species that donates a lone pair of electrons.
In aniline,the lone pair on the nitrogen atom is involved in resonance with the benzene ring,making it less available for donation compared to the lone pair in $NH_3$.
Therefore,$NH_3$ is a stronger base than aniline.
Statement $(A)$ claims aniline is a stronger base than ammonia,which is false.
Statement $(B)$ is true because methylamine is an aliphatic amine with an electron-donating alkyl group,making it more basic than aniline.
Statement $(C)$ is true because a weaker base has a higher $pK_b$ value.
Statement $(D)$ is true because aniline reacts with bromine water to form $2,4,6$-tribromoaniline,which appears as a white precipitate.
Thus,the false statement is $(A)$.

(C) To prepare a $1^{\circ}$-amine from an alkyl halide with the simultaneous addition of one $-CH_2-$ group to the carbon chain,$KCN$ is used as the reagent.
The reaction proceeds as follows:
$1$. Nucleophilic substitution: $R-X + KCN \rightarrow R-CN + KX$
$2$. Reduction: $R-CN + 4[H] \rightarrow R-CH_2-NH_2$
$KCN$ provides the cyanide ion $(CN^-)$,which acts as a nucleophile to extend the carbon chain by one carbon atom,and subsequent reduction of the nitrile group yields the $1^{\circ}$-amine.

(C) Auto-reduction (or self-reduction) is a process used for metals like $Pb$,$Hg$,and $Cu$ where the sulfide ore is partially roasted to form oxide,which then reacts with the remaining sulfide to yield the metal.
For example,$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$ followed by $2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$.
Titanium $(Ti)$ is a highly reactive metal and cannot be extracted by auto-reduction. It is typically extracted using the Kroll process,which involves the reduction of $TiCl_4$ with magnesium $(Mg)$ or sodium $(Na)$.

(C) Tollen's reagent is $(AgNO_3 + NH_4OH)$ and it provides a basic medium. Fructose is a keto-hexose,yet it can reduce Tollen's reagent. This is because,in a basic medium,it undergoes enolisation and rearrangement (Lobry de-Bruyn-van Ekenstein transformation) to form an aldose (like $D$-glucose or $D$-mannose) which contains an aldehyde group. The equilibrium involves the formation of an enediol intermediate.

(B) Oligosaccharides are formed by the linkage of two or more monosaccharide units through $O$-glycosidic bonds.
This linkage involves the formation of a $C-O-C$ bond between two sugar units,which is chemically classified as an ether linkage.
Thus,the correct option is $(B)$.

(D) Glucose contains an aldehyde group $(-CHO)$,but it does not give all the characteristic reactions of aldehydes due to the formation of a cyclic hemiacetal structure.
Glucose reacts with $NH_2OH$ to form an oxime.
Glucose reacts with $HCN$ to form a cyanohydrin.
Glucose reacts with $Br_2 / H_2O$ (mild oxidizing agent) to form gluconic acid.
However,glucose does not react with $NaHSO_3$ (sodium bisulphite) to form a hydrogen sulphite addition product,which is a characteristic reaction of most aldehydes and ketones.

(D) When glucose reacts with bromine water $(Br_2/H_2O)$,it undergoes mild oxidation of the aldehyde group to a carboxylic acid group,resulting in the formation of gluconic acid.
The reaction is as follows:
$CH_2OH(CHOH)_4CHO + [O] \xrightarrow{Br_2/H_2O} CH_2OH(CHOH)_4COOH$
Therefore,the correct option is $(D)$.

(A) When glucose $(CHO(CHOH)_4CH_2OH)$ is heated with concentrated $HI$ for a prolonged period,the reduction of all hydroxyl groups and the aldehyde group occurs.
This reaction results in the formation of $n-$hexane $(CH_3(CH_2)_4CH_3)$,which indicates that all six carbon atoms in glucose are linked in a straight chain.

(B) Glucose + Fructose $\longrightarrow$ Sucrose.
Sucrose is a disaccharide,which is made of $\alpha-D$-glucose and $\beta-D$-fructose joined together by a $1, 2$-glycosidic linkage.

(D) Proteins are polymers of amino acids linked by peptide bonds.
Upon complete hydrolysis,proteins yield only $\alpha$-amino acids.
An $\alpha$-amino acid has the general structure $R-CH(NH_2)-COOH$,where the amino group is attached to the $\alpha$-carbon atom.

(D) Watson and Crick proposed that the $DNA$ molecule is in the form of a double helix with two polynucleotide chains coiled around one another in a spiral.
This is known as $B-DNA$ or biological $DNA$.
It has $10$ nucleotide pairs per turn of the double helix and is right-handed with a diameter of $20 \mathring{A}$.
One complete turn of the double helix is $34 \mathring{A}$.
The distance between adjacent stacks or nucleotides is $3.4 \mathring{A}$.
Hence,option $(D)$ is correct.

(C) nucleotide is the basic structural unit of nucleic acids ($DNA$ and $RNA$). It consists of three main components:
$1$. $A$ nitrogenous base (purine or pyrimidine) attached to the $1'$ carbon of the sugar.
$2$. $A$ pentose sugar (ribose in $RNA$ or $2'$-deoxyribose in $DNA$).
$3$. $A$ phosphate group attached to the $5'$ carbon of the sugar.
Looking at the provided structures:
- The structure must show the phosphate group attached to the $5'-CH_2$ group.
- The nitrogenous base must be attached to the $1'$ carbon.
- The $3'$ carbon must have a hydroxyl $(-OH)$ group.
Based on the standard structural representation of a nucleotide,the correct structure is shown in option $(C)$.

(B) Vitamin $C$,also known as ascorbic acid,is a water-soluble vitamin found in various foods and used as a dietary supplement.
It is essential for the synthesis of collagen and acts as an antioxidant.
Hence,the correct option is $(B)$.

(A) The acidic strength depends on the stability of the conjugate base formed after the loss of a proton.
$(i)$ is $4$-ethylcyclohexanecarboxylic acid (aliphatic).
$(ii)$ is $4$-methoxybenzoic acid (aromatic with $+R$ effect of $-OCH_3$ group).
$(iii)$ is $4$-nitrobenzoic acid (aromatic with $-R$ effect of $-NO_2$ group).
$1$. Aromatic carboxylic acids are generally stronger than aliphatic carboxylic acids due to the resonance stabilization of the carboxylate anion by the benzene ring.
$2$. The $-NO_2$ group in $(iii)$ exerts a strong $-R$ (electron-withdrawing) effect,which stabilizes the carboxylate anion and significantly increases acidity.
$3$. The $-OCH_3$ group in $(ii)$ exerts a $+R$ (electron-donating) effect,which destabilizes the carboxylate anion and decreases acidity compared to benzoic acid,but it is still more acidic than the aliphatic acid $(i)$ due to the aromatic ring.
Therefore,the order of acidic strength is $(iii) > (ii) > (i)$.

(B) The reaction proceeds as follows:
$CH_3CHO + HCN \rightarrow CH_3CH(OH)CN$ (Acetaldehyde cyanohydrin)
$CH_3CH(OH)CN + 2H_2O + H^+ \rightarrow CH_3CH(OH)COOH + NH_4^+$
Acetaldehyde on reaction with hydrogen cyanide $(HCN)$ gives a cyanohydrin,which on hydrolysis yields lactic acid $(CH_3CH(OH)COOH)$.
Therefore,compound '$A$' is acetaldehyde. Hence,the correct option is $(B)$.

(D) The main constituent of vinegar is acetic acid $\left( CH_3COOH \right)$.
It contains $5-8 \%$ acetic acid by volume.
Vinegar is mainly used in culinary arts.

(C) Due to the $+R$ effect (resonance effect) of the $Cl$ atom with the benzene ring,the $C-Cl$ bond in chlorobenzene acquires partial double bond character.
This partial double bond character makes the $C-Cl$ bond in chlorobenzene shorter and stronger than the $C-Cl$ bond in methyl chloride $(CH_3-Cl)$,which exhibits only a single bond character.

(C) The acidic strength of substituted benzoic acids depends on the electronic effects of the substituents attached to the benzene ring.
$1.$ Electron-donating groups $(EDG)$ like $-OCH_3$ (methoxy group) decrease the acidity by destabilizing the carboxylate anion. Thus,$4-$methoxybenzoic acid $(IV)$ is the weakest acid.
$2.$ Electron-withdrawing groups $(EWG)$ like $-NO_2$ (nitro group) increase the acidity by stabilizing the carboxylate anion through $-I$ and $-M$ effects.
$3.$ Benzoic acid $(I)$ has no substituent. $4-$nitrobenzoic acid $(II)$ has one $-NO_2$ group,and $3, 4-$dinitrobenzoic acid $(III)$ has two $-NO_2$ groups.
$4.$ Since $III$ has two electron-withdrawing groups,it is more acidic than $II$.
Therefore,the increasing order of acidic strength is: $IV < I < II < III$.

(B) substance is diamagnetic if all its electrons are paired (i.e.,no unpaired electrons,$n = 0$).
$NO$ has $15$ valence electrons (odd),so it is paramagnetic.
$NO_2$ has $17$ valence electrons (odd),so it is paramagnetic.
$N_2O$ has $16$ valence electrons (even),all paired,so it is diamagnetic.
$N_2O_3$ has $30$ valence electrons (even),all paired,so it is diamagnetic.
$N_2O_4$ has $34$ valence electrons (even),all paired,so it is diamagnetic.
$N_2O_5$ has $40$ valence electrons (even),all paired,so it is diamagnetic.
Therefore,the diamagnetic compounds are $N_2O, N_2O_3, N_2O_4, N_2O_5$.

(A) Molecular solids are composed of molecules held together by weak van der Waals' forces or hydrogen bonds.
Because of these weak intermolecular forces,molecular solids are generally soft and have low melting points.
Many molecular solids are also volatile in nature.
Therefore,the correct option is $A$.

(D) Carboxylic acids form dimers in the vapor phase or in non-polar solvents due to intermolecular hydrogen bonding.
In this structure,two carboxylic acid molecules are linked by two hydrogen bonds.
Counting the atoms involved in the ring:
$1$. Carbon atom of the first carboxyl group
$2$. Oxygen atom of the first carboxyl group
$3$. Hydrogen atom involved in the first hydrogen bond
$4$. Oxygen atom of the second carboxyl group
$5$. Carbon atom of the second carboxyl group
$6$. Oxygen atom of the second carboxyl group
$7$. Hydrogen atom involved in the second hydrogen bond
$8$. Oxygen atom of the first carboxyl group
Thus,it forms an $8-$membered ring.

(C) Intermolecular $H$-bonding occurs in molecules where a hydrogen atom is covalently bonded to a highly electronegative atom $(N, O, F)$ and is attracted to another electronegative atom in a neighboring molecule.
Acetamide $(CH_3CONH_2)$ contains an $N-H$ bond,which allows it to act as a hydrogen bond donor,and a $C=O$ group,which acts as a hydrogen bond acceptor. This enables the formation of intermolecular $H$-bonds between the $N-H$ of one molecule and the $C=O$ of another.
Ethyl acetate,methyl formate,and acetic anhydride lack $N-H$ or $O-H$ bonds,so they cannot form intermolecular $H$-bonds.
Thus,the correct option is $(C)$.

(B) The rate of reaction is given by the expression: $-\frac{1}{3} \frac{d[X]}{dt} = \frac{1}{2} \frac{d[Y]}{dt} = \frac{d[Z]}{dt}$
Given that the rate of disappearance of $X$ is $-\frac{d[X]}{dt} = 7.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
From the rate expression,we have: $\frac{1}{2} \frac{d[Y]}{dt} = -\frac{1}{3} \frac{d[X]}{dt}$
Therefore,the rate of formation of $Y$ is: $\frac{d[Y]}{dt} = \frac{2}{3} \times (-\frac{d[X]}{dt}) = \frac{2}{3} \times 7.2 \times 10^{-3} = 4.8 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

(A) Given: $\Delta t = 25 \ min = 25 \times 60 \ s = 1500 \ s$.
Change in concentration $\Delta [R] = [R]_f - [R]_i = 0.02 - 0.03 = -0.01 \ mol \ L^{-1}$.
Rate $= -\frac{\Delta [R]}{\Delta t} = -\frac{-0.01}{1500} \ mol \ L^{-1} \ s^{-1}$.
Rate $= \frac{0.01}{1500} = \frac{1}{150000} = 6.667 \times 10^{-6} \ mol \ L^{-1} \ s^{-1}$.

(A) For the given reaction: $5Br^-{_{\text{(aq)}}} + BrO_3^-{_{\text{(aq)}}} + 6H^+{_{\text{(aq)}}} \rightarrow 3Br_{2\text{(aq)}} + 3H_2O_{\text{(l)}}$
The rate of reaction is defined as the rate of disappearance of reactants divided by their stoichiometric coefficients.
Rate $= -\frac{1}{5} \frac{\Delta[Br^{-}]}{\Delta t} = -\frac{\Delta[BrO_3^{-}]}{\Delta t} = -\frac{1}{6} \frac{\Delta[H^{+}]}{\Delta t} = \frac{1}{3} \frac{\Delta[Br_2]}{\Delta t} = \frac{1}{3} \frac{\Delta[H_2O]}{\Delta t}$
Given that $-\frac{\Delta[Br^{-}]}{\Delta t} = x \ mol \ L^{-1} \ min^{-1}$.
Substituting this into the rate expression:
Rate $= \frac{1}{5} \times (-\frac{\Delta[Br^{-}]}{\Delta t}) = \frac{1}{5} \times x = \frac{x}{5} \ mol \ L^{-1} \ min^{-1}$.

(A) For a first-$order$ reaction,the rate equation is $R = R_0 e^{-kt}$.
Taking the natural logarithm on both sides: $\ln R = \ln R_0 - kt$.
This can be rearranged as $\ln(R_0/R) = kt$.
Converting to base $10$: $\log(R_0/R) = \frac{kt}{2.303}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log(R_0/R)$ and $x = t$,the slope is $m = \frac{k}{2.303}$,which is positive.
Therefore,plotting $\log(R_0/R)$ vs time gives a straight line with a positive slope.

(D) For a zero-order reaction,the integrated rate law is given by $[A] = -kt + [A]_0$.
This equation is in the form of a linear equation $y = mx + c$,where $y = [A]$,$x = t$,$m = -k$ (slope),and $c = [A]_0$ ($y$-intercept).
Therefore,a plot of $[A]$ versus time $(t)$ results in a straight line for a zero-order reaction.
Hence,the correct option is $(D)$.

(A) For a first-order reaction,$t_{1/2} = \frac{0.693}{k}$,so $k \propto \frac{1}{t_{1/2}}$.
Given: $T_1 = 300 \ K, t_{1/2}(1) = 20 \ s$ and $T_2 = 350 \ K, t_{1/2}(2) = 2 \ s$.
Therefore,$\frac{k_2}{k_1} = \frac{t_{1/2}(1)}{t_{1/2}(2)} = \frac{20}{2} = 10$.
Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 \ R} [\frac{T_2 - T_1}{T_1 T_2}]$.
Substituting the values: $\log(10) = \frac{E_a}{2.303 \times 8.314 \times 10^{-3}} [\frac{350 - 300}{350 \times 300}]$.
$1 = \frac{E_a}{2.303 \times 8.314 \times 10^{-3}} [\frac{50}{105000}]$.
$E_a = \frac{2.303 \times 8.314 \times 10^{-3} \times 105000}{50} \approx 40.2 \ kJ \ mol^{-1}$.

(C) Antibiotics have either a cidal (killing) effect or a static (inhibitory) effect on microbes.
Bactericidal antibiotics include: $Penicillin$,$aminoglycosides$,$ofloxacin$.
Bacteriostatic antibiotics include: $Erythromycin$,$tetracycline$,$chloramphenicol$.
Therefore,the correct pair for bactericidal and bacteriostatic antibiotics is $Penicillin$ and $erythromycin$ (or $tetracycline$).
Looking at the options,option $B$ provides $erythromycin$ and $tetracycline$,which are both bacteriostatic. However,checking the standard classification,$Penicillin$ is bactericidal and $Erythromycin$ is bacteriostatic. Given the options provided,$B$ is the most appropriate choice as it lists two bacteriostatic agents,but the question asks for one of each. Re-evaluating the options,$A$ ($Penicillin$ - bactericidal,$Ofloxacin$ - bactericidal) is incorrect. $C$ ($Penicillin$ - bactericidal,$Chloramphenicol$ - bacteriostatic) is the correct match.

(B) Sulphapyridine is a well-known sulphonamide antibiotic. Its chemical structure consists of a $p$-aminobenzenesulphonamide group attached to a pyridine ring at the nitrogen atom of the sulphonamide group. The correct structure is represented by option $B$.

(A) Neurologically active drugs are those that affect the message transfer mechanism from nerve to receptor. $Analgesics$ and $Tranquilizers$ are examples of neurologically active drugs. Therefore,$Analgesic$ is the correct answer.

(C) Tincture of iodine is a common antiseptic solution.
It typically consists of $2-3 \%$ elemental iodine,along with potassium iodide or sodium iodide,dissolved in a mixture of ethanol and water.
The presence of alcohol is a defining characteristic of tincture solutions.

(B) The number of moles of solute is calculated using the formula: $n = M \times V$ (where $M$ is molarity and $V$ is volume in liters).
For $1000 \ mL$ of $1 \ M \ NaOH$,$n = 1 \ M \times 1 \ L = 1 \ mol$.
Checking the options:
$(1)$ $0.2 \ M \times 0.2 \ L = 0.04 \ mol$.
$(2)$ $1 \ M \times 0.2 \ L = 0.2 \ mol$ and $0.2 \ M \times 1 \ L = 0.2 \ mol$. Since $0.2 = 0.2$,this statement is true.
$(3)$ $0.2 \ M \times 0.1 \ L = 0.02 \ mol$.
$(4)$ $0.2 \ M \times 2 \ L = 0.4 \ mol$.

(D) Sodium lauryl sulphate,$CH_3(CH_2)_{10}CH_2-OSO_3^{\ominus}Na^{\oplus}$,is an anionic detergent.
It is the sodium salt of sulphonated lauryl alcohol,$CH_3(CH_2)_{10}CH_2-OH$.
Sodium stearate is a soap,not a detergent.

(D) Bithionol is added to soaps to impart antiseptic properties and to reduce the odour produced by bacterial decomposition of organic matter on skin.
Bithionol is an aromatic compound,which contains sulphur and is used in soaps.
Hence,the correct option is $D$.