AIEEE 2005 Chemistry Question Paper with Answer and Solution

(C) Polyamides are polymers containing amide linkages $(-CONH-)$ in their backbone.
Nylon-$66$ is a condensation polymer formed by the reaction of hexamethylenediamine and adipic acid,which results in the formation of amide bonds.
Bakelite is a phenol-formaldehyde resin.
Terylene is a polyester.
Teflon is a polymer of tetrafluoroethene.

(B) In a Young's double slit experiment, the condition for constructive or destructive interference at any point $P(x, y)$ on the screen is that the path difference between the waves from the two slits $S_1$ and $S_2$ must be constant.
Mathematically, this is expressed as $|PS_1 - PS_2| = \text{constant}$.
By definition, the locus of a point such that the difference of its distances from two fixed points (the foci, which are the slits $S_1$ and $S_2$) is constant, is a hyperbola.
Therefore, the interference fringes formed on the screen are hyperbolic in shape.

(C) Aqueous solution of $AlCl_3$ undergoes hydrolysis to form $Al(OH)_3$ and $HCl$:
$AlCl_3 + 3 H_2O \rightarrow Al(OH)_3 + 3 HCl$
Upon heating the solution to dryness,the volatile $HCl$ gas escapes,and the remaining $Al(OH)_3$ undergoes thermal decomposition to form aluminium oxide $(Al_2O_3)$:
$2 Al(OH)_3 \xrightarrow{\Delta} Al_2O_3 + 3 H_2O$
Therefore,the final product obtained is $Al_2O_3$.

(C) The heat exchanged in a process is given by $Q = \int T \, dS$,which represents the area under the curve in a $T-S$ diagram.
$1$. Heat absorbed $(Q_1)$ from $A$ to $B$: The process $A \rightarrow B$ is a straight line. The area under this line is a trapezoid with parallel sides $2T_0$ and $T_0$ and height $(2S_0 - S_0) = S_0$.
$Q_1 = \text{Area under } AB = \frac{1}{2} (2T_0 + T_0) (2S_0 - S_0) = \frac{3}{2} T_0 S_0$.
$2$. Heat rejected $(Q_2)$ from $B$ to $C$: The process $B \rightarrow C$ is at constant temperature $T_0$. The area under this line is a rectangle.
$Q_2 = \text{Area under } BC = T_0 (2S_0 - S_0) = T_0 S_0$.
$3$. Efficiency $(\eta)$: The efficiency of a heat engine is given by $\eta = 1 - \frac{Q_2}{Q_1}$.
$\eta = 1 - \frac{T_0 S_0}{\frac{3}{2} T_0 S_0} = 1 - \frac{2}{3} = \frac{1}{3} \approx 0.33$.

(A) The number of moles of Helium $(He)$ is $n_1 = \frac{16}{4} = 4 \ mol$. Helium is a monoatomic gas,so its degrees of freedom $f_1 = 3$ and $C_{v1} = \frac{3}{2}R$.
The number of moles of Hydrogen $(H_2)$ is $n_2 = \frac{16}{2} = 8 \ mol$. Hydrogen is a diatomic gas,so its degrees of freedom $f_2 = 5$ and $C_{v2} = \frac{5}{2}R$.
The equivalent $C_v$ of the mixture is given by $C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{4(\frac{3}{2}R) + 8(\frac{5}{2}R)}{4 + 8} = \frac{6R + 20R}{12} = \frac{26R}{12} = \frac{13}{6}R$.
The equivalent $C_p$ of the mixture is given by $C_{p,mix} = C_{v,mix} + R = \frac{13}{6}R + R = \frac{19}{6}R$.
The ratio $\gamma = \frac{C_{p,mix}}{C_{v,mix}} = \frac{19/6 R}{13/6 R} = \frac{19}{13} \approx 1.46$.

(B) Given: Initial velocity $u = 100\, m/s$,Final velocity $v = 0\, m/s$,Coefficient of kinetic friction $\mu_k = 0.5$,Acceleration due to gravity $g = 10\, m/s^2$.
The retardation force acting on the car is $F = \mu_k N = \mu_k mg$.
Using Newton's second law,the retardation $a = \frac{F}{m} = \mu_k g = 0.5 \times 10 = 5\, m/s^2$.
Using the kinematic equation $v^2 - u^2 = 2as$:
$0^2 - (100)^2 = 2(-5)s$
$-10000 = -10s$
$s = \frac{10000}{10} = 1000\, m$.
Therefore,the distance at which the car can be stopped is $1000\, m$.

(A) The line $y = mx + c$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ if $c^2 = a^2m^2 - b^2$.
Given the line $y = \alpha x + \beta$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we substitute $m = \alpha$ and $c = \beta$ into the condition.
Thus,$\beta^2 = a^2\alpha^2 - b^2$.
Replacing $(\alpha, \beta)$ with $(x, y)$ to find the locus,we get $y^2 = a^2x^2 - b^2$.
Rearranging the terms,we have $a^2x^2 - y^2 = b^2$,or $\frac{x^2}{(b/a)^2} - \frac{y^2}{b^2} = 1$.
This equation represents a hyperbola.

(D) The number of moles for helium $(He)$ is $n_1 = \frac{16\,g}{4\,g/mol} = 4\,mol$. Helium is a monoatomic gas,so $f_1 = 3$,$C_{v1} = \frac{3R}{2}$,and $C_{p1} = \frac{5R}{2}$.
The number of moles for oxygen $(O_2)$ is $n_2 = \frac{16\,g}{32\,g/mol} = 0.5\,mol$. Oxygen is a diatomic gas,so $f_2 = 5$,$C_{v2} = \frac{5R}{2}$,and $C_{p2} = \frac{7R}{2}$.
The ratio of specific heats for the mixture is given by $\gamma_{mix} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}}$.
Substituting the values:
$\gamma_{mix} = \frac{4 \times (\frac{5R}{2}) + 0.5 \times (\frac{7R}{2})}{4 \times (\frac{3R}{2}) + 0.5 \times (\frac{5R}{2})}$
$\gamma_{mix} = \frac{10R + 1.75R}{6R + 1.25R} = \frac{11.75R}{7.25R} = \frac{1175}{725} \approx 1.62$.

(A) According to the law of conservation of linear momentum: $m_{1} \vec{u}_{1} + m_{2} \vec{u}_{2} = m_{1} \vec{v}_{1} + m_{2} \vec{v}_{2}$.
Let the initial velocity of the first mass be $\vec{u}_{1} = v \hat{i}$ and the second mass be at rest,$\vec{u}_{2} = 0$.
After the collision,the first mass moves with velocity $\vec{v}_{1} = \frac{v}{\sqrt{3}} \hat{j}$ (perpendicular to the initial direction).
Substituting these values into the momentum equation:
$m(v \hat{i}) + m(0) = m(\frac{v}{\sqrt{3}} \hat{j}) + m \vec{v}_{2}$.
Dividing by $m$,we get:
$\vec{v}_{2} = v \hat{i} - \frac{v}{\sqrt{3}} \hat{j}$.
The speed of the second mass is the magnitude of $\vec{v}_{2}$:
$|\vec{v}_{2}| = \sqrt{v^{2} + (-\frac{v}{\sqrt{3}})^{2}} = \sqrt{v^{2} + \frac{v^{2}}{3}} = \sqrt{\frac{4v^{2}}{3}} = \frac{2}{\sqrt{3}}v$.

(B) The internal resistance $r$ of a cell is given by the formula:
$r = R \left[ \frac{l_1}{l_2} - 1 \right]$
Where:
$R = 2\,\Omega$ (external shunt resistance)
$l_1 = 240\,cm$ (balancing length without shunt)
$l_2 = 120\,cm$ (balancing length with shunt)
Substituting the values into the formula:
$r = 2 \left[ \frac{240}{120} - 1 \right]$
$r = 2 [2 - 1]$
$r = 2 \times 1 = 2\,\Omega$
Thus,the internal resistance of the cell is $2\,\Omega$.

(C) The condition for the resolution of two point objects by an optical system (like the human eye) is given by the Rayleigh criterion: $\theta = \frac{1.22 \lambda}{d}$.
Here,$\theta$ is the angular separation,$\lambda$ is the wavelength of light,and $d$ is the diameter of the pupil.
Also,for small angles,the angular separation $\theta$ can be expressed as $\theta = \frac{y}{D}$,where $y$ is the linear separation between the dots and $D$ is the distance from the eye.
Equating the two expressions: $\frac{y}{D} = \frac{1.22 \lambda}{d}$.
Rearranging to solve for $D$: $D = \frac{y \cdot d}{1.22 \cdot \lambda}$.
Given values: $y = 1 \, mm = 1 \times 10^{-3} \, m$,$d = 3 \, mm = 3 \times 10^{-3} \, m$,and $\lambda = 500 \, nm = 500 \times 10^{-9} \, m$.
Substituting these values: $D = \frac{1 \times 10^{-3} \times 3 \times 10^{-3}}{1.22 \times 500 \times 10^{-9}}$.
$D = \frac{3 \times 10^{-6}}{610 \times 10^{-9}} = \frac{3000}{610} \approx 4.918 \, m$.
Rounding to the nearest integer,we get $D \approx 5 \, m$.

(B) Let the initial velocity be $V$ and the constant retardation be $a$.
Using the equation of motion $v^2 = u^2 + 2as$,where $v = V/2$,$u = V$,and $s = 3 \ cm$:
$(V/2)^2 = V^2 - 2a(3) \implies V^2/4 = V^2 - 6a \implies 6a = 3V^2/4 \implies a = V^2/8$.
Now,let the total distance covered be $S$. When the bullet comes to rest,final velocity $v = 0$:
$0^2 = V^2 - 2aS \implies V^2 = 2(V^2/8)S \implies V^2 = (V^2/4)S \implies S = 4 \ cm$.
The additional distance covered is $S - 3 \ cm = 4 \ cm - 3 \ cm = 1 \ cm$.

(C) polyamide is a polymer containing the amide linkage $(-CONH-)$ in its chain.
$Nylon-6,6$ is a polyamide. It is formed by the condensation polymerisation of hexamethylenediamine $(H_2N(CH_2)_6NH_2)$ and adipic acid $(HOOC(CH_2)_4COOH)$.
Bakelite is a phenol-formaldehyde resin.
Terylene is a polyester formed by the condensation of ethylene glycol and terephthalic acid.
Teflon is a polymer of tetrafluoroethylene $(CF_2=CF_2)$.

(A) The energy of an emitted photon is given by $\Delta E = E_{initial} - E_{final}$.
For emission,the electron must move from a higher energy level to a lower energy level.
Transition $I$ is an absorption transition (upward arrow).
Transitions $II$,$III$,and $IV$ are emission transitions (downward arrows).
Comparing the energy gaps:
Transition $II$: $n=4$ to $n=3$
Transition $III$: $n=2$ to $n=1$
Transition $IV$: $n=4$ to $n=2$
The energy difference between levels is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$ (for Hydrogen-like atoms).
The energy gap $\Delta E$ increases as we move to lower $n$ values.
The gap between $n=2$ and $n=1$ (Transition $III$) is the largest among the emission transitions shown.
Therefore,transition $III$ corresponds to the emission of a photon with the most energy.

(A) In Young's double slit experiment,the path difference between the two waves reaching a point on the screen is given by $\Delta x = d \sin \theta \approx d \tan \theta = d(y/D)$.
For bright fringes,the path difference is $n\lambda$,so $d(y/D) = n\lambda$,which implies $y = (n\lambda D)/d$.
Since $y$ is constant for a given order $n$,the fringes appear as straight lines parallel to the slits on the screen.
Therefore,the shape of the interference fringes is a straight line.

(B) Given that the galvanometer $G$ reads zero,no current flows through the branch containing $E_2$ and the galvanometer.
Therefore,the current $I$ flows only through the loop containing $E_1$,the $500 \, \Omega$ resistor,and the resistor $X$.
Using Ohm's law for the circuit:
$I = \frac{E_1}{500 + X}$
The potential difference across the resistor $X$ must be equal to the emf of the battery $E_2$ for the galvanometer to read zero:
$V_X = I \cdot X = E_2$
Substituting the value of $I$:
$\left( \frac{E_1}{500 + X} \right) \cdot X = E_2$
Given $E_1 = 12 \, V$ and $E_2 = 2 \, V$:
$\frac{12 \cdot X}{500 + X} = 2$
$12X = 2(500 + X)$
$12X = 1000 + 2X$
$10X = 1000$
$X = 100 \, \Omega$

(C) Let the point $P$ be at a distance $x$ from the origin $(x=0)$.
The electric field due to $+8q$ at $P$ is $E_1 = \frac{k(8q)}{x^2}$ (directed away from the origin).
The electric field due to $-2q$ at $P$ is $E_2 = \frac{k(2q)}{(x-L)^2}$ (directed towards the origin).
For the net electric field to be zero at $P$,the magnitudes must be equal: $E_1 = E_2$.
$\frac{k(8q)}{x^2} = \frac{k(2q)}{(x-L)^2}$
$\frac{4}{x^2} = \frac{1}{(x-L)^2}$
Taking the square root on both sides:
$\frac{2}{x} = \frac{1}{x-L}$
$2(x-L) = x$
$2x - 2L = x$
$x = 2L$
Thus,the point is located at $x = 2L$.

(C) When unpolarized light of intensity $I_0$ passes through a polarizing sheet (polaroid),the transmitted light becomes plane-polarized.
According to Malus's Law and the properties of polarizers,the intensity of the transmitted light is $I_t = \frac{1}{2} I_0$.
The intensity of the light that does not get transmitted (absorbed or reflected) is the difference between the incident intensity and the transmitted intensity.
$I_{not\ transmitted} = I_0 - I_t = I_0 - \frac{1}{2} I_0 = \frac{1}{2} I_0$.

(A) The melting point of ionic compounds is primarily determined by their lattice energy.
Lattice energy is inversely proportional to the interionic distance $(U \propto \frac{1}{r_+ + r_-})$.
Among the given alkali metal chlorides,$LiCl$ has the smallest cation size,but it exhibits significant covalent character due to Fajan's rule,which lowers its melting point.
$NaCl$ has a high lattice energy and a stable crystal structure,resulting in the highest melting point among the options provided.

(A) The van't Hoff equation is given by: $\ln K_{eq} = -\frac{\Delta H^{\circ}}{R} (\frac{1}{T}) + \frac{\Delta S^{\circ}}{R}$.
Comparing this with the equation of a straight line $y = mx + c$,we have $y = \ln K_{eq}$ and $x = 1/T$.
The slope of the line is $m = -\frac{\Delta H^{\circ}}{R}$.
From the given graph,the slope is positive (as $\ln K_{eq}$ increases with $1/T$).
Therefore,$-\frac{\Delta H^{\circ}}{R} > 0$,which implies $\Delta H^{\circ} < 0$.
$A$ negative enthalpy change $(\Delta H^{\circ} < 0)$ indicates an exothermic reaction.

(C) The net electric field will be zero at a point outside the charges and closer to the charge which is smaller in magnitude.
Let the point $P$ be at a distance $l$ from the charge $-2q$ at $x = L$. The distance of $P$ from the charge $+8q$ at $x = 0$ is $(L + l)$.
At point $P$,the magnitude of the electric field due to $+8q$ must be equal to the magnitude of the electric field due to $-2q$:
$\frac{k \cdot |8q|}{(L + l)^2} = \frac{k \cdot |2q|}{l^2}$
Simplifying the equation:
$\frac{8}{(L + l)^2} = \frac{2}{l^2}$
$\frac{4}{(L + l)^2} = \frac{1}{l^2}$
Taking the square root on both sides:
$\frac{2}{L + l} = \frac{1}{l}$
$2l = L + l$
$l = L$
The position of point $P$ on the $x$-axis is $x = L + l = L + L = 2L$.

(D) We know that $\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2} = \frac{1}{2} - \frac{1}{2}\cos(2\omega t)$.
This expression represents a simple harmonic motion $(SHM)$ because it is of the form $y = A + B\cos(\omega' t)$,where $A$ and $B$ are constants.
The angular frequency of this motion is $\omega' = 2\omega$.
The time period $T$ is given by $T = \frac{2\pi}{\omega'} = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.
Therefore,the function represents a simple harmonic motion with a period $\pi /\omega$.

(B) Let the initial velocity of the bullet be $u$.
After penetrating $3\, cm$,its velocity becomes $u/2$.
Using the equation of motion $v^2 = u^2 - 2as$,where $a$ is the retardation due to constant resistance:
$(u/2)^2 = u^2 - 2a(3)$
$u^2/4 = u^2 - 6a$
$6a = 3u^2/4$
$a = u^2/8$
Now,let the bullet penetrate an additional distance $x$ before coming to rest $(v=0)$.
Using the equation of motion $v^2 = u^2 - 2as$ for this segment,with initial velocity $u/2$:
$0^2 = (u/2)^2 - 2ax$
$0 = u^2/4 - 2(u^2/8)x$
$u^2/4 = (u^2/4)x$
$x = 1\, cm$.
Therefore,the bullet will penetrate $1\, cm$ further.

(B) Let the centres of the rings be $A$ and $B$. The potential at the centre of a ring of radius $R$ with charge $Q$ is $V = \frac{1}{4\pi \varepsilon_0} \frac{Q}{R}$.
The potential at centre $A$ due to ring $A$ (charge $+q$) is $V_{A1} = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}$.
The potential at centre $A$ due to ring $B$ (charge $-q$) is $V_{A2} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{\sqrt{R^2 + d^2}}$.
So,$V_A = V_{A1} + V_{A2} = \frac{q}{4\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right]$.
The potential at centre $B$ due to ring $B$ (charge $-q$) is $V_{B1} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{R}$.
The potential at centre $B$ due to ring $A$ (charge $+q$) is $V_{B2} = \frac{1}{4\pi \varepsilon_0} \frac{q}{\sqrt{R^2 + d^2}}$.
So,$V_B = V_{B1} + V_{B2} = \frac{q}{4\pi \varepsilon_0} \left[ -\frac{1}{R} + \frac{1}{\sqrt{R^2 + d^2}} \right] = -V_A$.
The potential difference between the centres is $\Delta V = V_A - V_B = V_A - (-V_A) = 2V_A$.
$\Delta V = 2 \times \frac{q}{4\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right] = \frac{q}{2\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right]$.

(D) Number of moles of Helium $(n_1)$ = $\frac{16 \, g}{4 \, g/mol} = 4 \, mol$. Helium is monoatomic,so $f_1 = 3$,$C_{v1} = \frac{3}{2}R$,$C_{p1} = \frac{5}{2}R$.
Number of moles of Oxygen $(n_2)$ = $\frac{16 \, g}{32 \, g/mol} = 0.5 \, mol$. Oxygen is diatomic,so $f_2 = 5$,$C_{v2} = \frac{5}{2}R$,$C_{p2} = \frac{7}{2}R$.
The equivalent $C_v$ of the mixture is $C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{4(\frac{3}{2}R) + 0.5(\frac{5}{2}R)}{4 + 0.5} = \frac{6R + 1.25R}{4.5} = \frac{7.25R}{4.5} = \frac{29R}{18}$.
The equivalent $C_p$ of the mixture is $C_{p,mix} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2} = \frac{4(\frac{5}{2}R) + 0.5(\frac{7}{2}R)}{4 + 0.5} = \frac{10R + 1.75R}{4.5} = \frac{11.75R}{4.5} = \frac{47R}{18}$.
The ratio $\gamma = \frac{C_{p,mix}}{C_{v,mix}} = \frac{47R/18}{29R/18} = \frac{47}{29} \approx 1.62$.

(C) The velocity $v_1$ is given by the derivative of $y_1$ with respect to time $t$: $v_1 = \frac{dy_1}{dt} = 0.1 \times 100\pi \cos(100\pi t + \frac{\pi}{3}) = 10\pi \cos(100\pi t + \frac{\pi}{3})$.
The velocity $v_2$ is given by the derivative of $y_2$ with respect to time $t$: $v_2 = \frac{dy_2}{dt} = 0.1 \times (-100\pi) \sin(100\pi t) = -10\pi \sin(100\pi t)$.
To compare the phase,we express $v_2$ as a cosine function: $v_2 = 10\pi \cos(100\pi t + \frac{\pi}{2})$.
The phase of $v_1$ is $\phi_1 = 100\pi t + \frac{\pi}{3}$ and the phase of $v_2$ is $\phi_2 = 100\pi t + \frac{\pi}{2}$.
The phase difference $\Delta\phi = \phi_1 - \phi_2 = (100\pi t + \frac{\pi}{3}) - (100\pi t + \frac{\pi}{2}) = \frac{\pi}{3} - \frac{\pi}{2} = -\frac{\pi}{6}$.

(A) magnetic needle acts as a magnetic dipole with two poles of equal strength $m$ separated by a distance $2l$.
In a non-uniform magnetic field,the magnetic field strength $B$ varies at different points in space.
Since the force on a magnetic pole is given by $F = mB$,the forces acting on the two poles of the needle will differ in both magnitude and direction because the magnetic field $B$ is not constant.
Because the net force is the vector sum of these unequal forces,the needle experiences a net force.
Because the forces act at different points and are not necessarily collinear,the needle also experiences a net torque.
Therefore,a magnetic needle in a non-uniform magnetic field experiences both a force and a torque.

(C) full circular disc of mass $M_{total}$ and radius $r$ has a moment of inertia about an axis perpendicular to its plane passing through its center given by $I = \frac{1}{2} M_{total} r^2$.
Since a semicircular disc is exactly half of a full circular disc,its mass is $M = \frac{M_{total}}{2}$,which implies $M_{total} = 2M$.
Substituting this into the formula for the moment of inertia,we get:
$I = \frac{1}{2} (2M) r^2 = M r^2$.
Therefore,the moment of inertia of the semicircular disc about the specified axis is $M r^2$.

(C) Let the initial velocity of the $1^{st}$ mass be $\vec{v}_1 = v\hat{i}$. The $2^{nd}$ mass is at rest,so $\vec{v}_2 = 0$.
By the law of conservation of linear momentum: $m\vec{v}_1 + m(0) = m\vec{v}_1' + m\vec{v}_2'$.
Given that the $1^{st}$ mass moves with velocity $\frac{v}{\sqrt{3}}$ in a direction perpendicular to the initial motion (let this be $\hat{j}$),we have $\vec{v}_1' = \frac{v}{\sqrt{3}}\hat{j}$.
Substituting these into the momentum equation: $mv\hat{i} = m(\frac{v}{\sqrt{3}}\hat{j}) + m\vec{v}_2'$.
Dividing by $m$: $\vec{v}_2' = v\hat{i} - \frac{v}{\sqrt{3}}\hat{j}$.
The speed of the $2^{nd}$ mass is the magnitude of $\vec{v}_2'$:
$|\vec{v}_2'| = \sqrt{v^2 + (-\frac{v}{\sqrt{3}})^2} = \sqrt{v^2 + \frac{v^2}{3}} = \sqrt{\frac{4v^2}{3}} = \frac{2}{\sqrt{3}}v$.

(A) In a Young's double slit experiment, the path difference $\Delta x$ at any point $P$ on the screen is given by the difference in distances from the two slits $S_1$ and $S_2$, i.e., $\Delta x = |S_2P - S_1P|$.
For a constant path difference (which corresponds to a specific fringe), the locus of points $P$ such that $|S_2P - S_1P| = \text{constant}$ represents a hyperbola in three-dimensional space.
When this hyperbolic surface intersects a flat screen placed perpendicular to the plane of the slits, the resulting intersection is a hyperbola.
However, for a screen placed at a large distance $D$ from the slits (where $D \gg d$, with $d$ being the slit separation), the hyperbolic fringes appear as nearly straight lines in the central region of the screen.

(D) Step $1$: Calculate the velocity after falling $50\, m$ freely.
Using $v^2 = u^2 + 2as$,where $u = 0$,$a = g = 9.8\, m/s^2$,and $s = 50\, m$:
$v_1^2 = 0 + 2 \times 9.8 \times 50 = 980\, m^2/s^2$.
Step $2$: Calculate the distance covered during deceleration.
Let $h_2$ be the distance covered after the parachute opens. The initial velocity is $v_1$,final velocity $v_f = 3\, m/s$,and deceleration $a = -2\, m/s^2$.
Using $v_f^2 = v_1^2 + 2ah_2$:
$3^2 = 980 + 2(-2)h_2$
$9 = 980 - 4h_2$
$4h_2 = 980 - 9 = 971$
$h_2 = 971 / 4 = 242.75\, m \approx 243\, m$.
Step $3$: Calculate the total height.
Total height $H = h_1 + h_2 = 50 + 243 = 293\, m$.

(D) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 i}{2R}$.
Given,$R = 2\pi \, cm = 2\pi \times 10^{-2} \, m$.
For the first coil with current $i_1 = 3 \, A$:
$B_1 = \frac{\mu_0 \times 3}{2 \times 2\pi \times 10^{-2}} = \frac{4\pi \times 10^{-7} \times 3}{4\pi \times 10^{-2}} = 3 \times 10^{-5} \, T$.
For the second coil with current $i_2 = 4 \, A$:
$B_2 = \frac{\mu_0 \times 4}{2 \times 2\pi \times 10^{-2}} = \frac{4\pi \times 10^{-7} \times 4}{4\pi \times 10^{-2}} = 4 \times 10^{-5} \, T$.
Since the coils are placed at right angles,the resultant magnetic field is $B = \sqrt{B_1^2 + B_2^2}$.
$B = \sqrt{(3 \times 10^{-5})^2 + (4 \times 10^{-5})^2} = \sqrt{9 + 16} \times 10^{-5} = \sqrt{25} \times 10^{-5} = 5 \times 10^{-5} \, Wb/m^2$.

(B) The condition for the resolution of two point objects is given by the Rayleigh criterion: $\theta \geq 1.22 \frac{\lambda}{d}$,where $\theta = \frac{y}{D}$.
Here,$y = 1\, mm = 10^{-3}\, m$ is the separation between the dots,$d = 3\, mm = 3 \times 10^{-3}\, m$ is the diameter of the pupil,and $\lambda = 500\, nm = 5 \times 10^{-7}\, m$ is the wavelength of light.
Substituting these values into the inequality $\frac{y}{D} \geq 1.22 \frac{\lambda}{d}$,we get:
$D \leq \frac{y d}{1.22 \lambda} = \frac{10^{-3} \times 3 \times 10^{-3}}{1.22 \times 5 \times 10^{-7}}$
$D \leq \frac{3 \times 10^{-6}}{6.1 \times 10^{-7}} = \frac{30}{6.1} \approx 4.918\, m$.
Rounding to the nearest integer,the maximum distance is approximately $5\, m$.

(D) The work done to move a particle from the surface of a sphere to infinity is equal to the change in gravitational potential energy,which is the negative of the potential energy at the surface.
$W = U_{\infty} - U_{surface} = 0 - (- \frac{GMm}{R}) = \frac{GMm}{R}$
Given:
$M = 100 \, kg$
$m = 10 \, g = 0.01 \, kg$
$R = 10 \, cm = 0.1 \, m$
$G = 6.67 \times 10^{-11} \, Nm^2/kg^2$
Substituting the values:
$W = \frac{6.67 \times 10^{-11} \times 100 \times 0.01}{0.1}$
$W = \frac{6.67 \times 10^{-11} \times 1}{0.1}$
$W = 6.67 \times 10^{-10} \, J$

(D) Let $L$ be the length of the inclined plane and $\theta = 45^{\circ}$.
For a smooth plane,the acceleration is $a_1 = g \sin \theta = g \sin 45^{\circ} = \frac{g}{\sqrt{2}}$.
The time taken is $t_1 = \sqrt{\frac{2L}{a_1}}$.
For a rough plane with coefficient of kinetic friction $K$,the acceleration is $a_2 = g \sin \theta - Kg \cos \theta = \frac{g}{\sqrt{2}} - \frac{Kg}{\sqrt{2}} = \frac{g}{\sqrt{2}}(1-K)$.
The time taken is $t_2 = \sqrt{\frac{2L}{a_2}}$.
Given $t_2 = n t_1$,so $t_2^2 = n^2 t_1^2$.
Substituting the expressions for $t_1^2$ and $t_2^2$: $\frac{2L}{a_2} = n^2 \frac{2L}{a_1} \implies a_1 = n^2 a_2$.
Substituting the values of $a_1$ and $a_2$: $\frac{g}{\sqrt{2}} = n^2 \left[ \frac{g}{\sqrt{2}}(1-K) \right]$.
$1 = n^2(1-K) \implies 1-K = \frac{1}{n^2}$.
Therefore,$K = 1 - \frac{1}{n^2}$.