AIEEE 2012 Chemistry Question Paper with Answer and Solution

(A) The formula for most probable velocity is $C^* = \sqrt{\frac{2RT}{M}}$.
The formula for average velocity is $\bar{C} = \sqrt{\frac{8RT}{\pi M}}$.
The formula for root mean square velocity is $C = \sqrt{\frac{3RT}{M}}$.
Comparing these values,the ratio is $C^* : \bar{C} : C = \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3}$.

(C) The balanced chemical equation is: $O_{2(g)} + 2SO_{2(g)} \leftrightarrow 2SO_{3(g)}$
Initial moles: $O_2 = 1, SO_2 = 2, SO_3 = 0$
At equilibrium,moles of $SO_3 = 1.6$. Since $2 \, moles$ of $SO_3$ are produced from $1 \, mole$ of $O_2$ and $2 \, moles$ of $SO_2$,the reaction stoichiometry implies:
$SO_3$ produced = $2x = 1.6 \implies x = 0.8$
Equilibrium moles:
$[O_2] = 1 - 0.8 = 0.2 \, mol/L$
$[SO_2] = 2 - 2(0.8) = 2 - 1.6 = 0.4 \, mol/L$
$[SO_3] = 1.6 \, mol/L$
$K_c = \frac{[SO_3]^2}{[O_2][SO_2]^2} = \frac{(1.6)^2}{(0.2)(0.4)^2}$
$K_c = \frac{2.56}{0.2 \times 0.16} = \frac{2.56}{0.032} = 80$

(D) For a single slit diffraction,the condition for the $n^{th}$ minimum is given by $a \sin \theta = n \lambda$.
For the first minimum,$n = 1$.
Given: $\lambda = 5000 \, \mathring{A} = 5000 \times 10^{-10} \, m = 5 \times 10^{-7} \, m$ and $\theta = 30^o$.
Substituting these values into the formula: $a \sin 30^o = 1 \times (5 \times 10^{-7} \, m)$.
Since $\sin 30^o = 0.5$,we have $a \times 0.5 = 5 \times 10^{-7} \, m$.
$a = \frac{5 \times 10^{-7}}{0.5} = 10 \times 10^{-7} \, m = 10^{-6} \, m$.
Converting to centimeters: $a = 10^{-6} \times 10^2 \, cm = 10^{-4} \, cm = 10 \times 10^{-5} \, cm$.

(D) For reaction $(I): N_2 + 2O_2 \rightleftharpoons 2NO_2$,the equilibrium constant is $K_1 = \frac{[NO_2]^2}{[N_2][O_2]^2}$.
For reaction $(II): 2NO_2 \rightleftharpoons N_2 + 2O_2$,the equilibrium constant is $K_2 = \frac{[N_2][O_2]^2}{[NO_2]^2}$.
Clearly,$K_2 = \frac{1}{K_1}$,which implies $K_1 = \frac{1}{K_2}$.
For reaction $(III): NO_2 \rightleftharpoons \frac{1}{2}N_2 + O_2$,the equilibrium constant is $K_3 = \frac{[N_2]^{1/2}[O_2]}{[NO_2]}$.
Comparing $K_3$ with $K_1$,we see that $K_3 = \sqrt{\frac{[N_2][O_2]^2}{[NO_2]^2}} = \sqrt{\frac{1}{K_1}} = \frac{1}{\sqrt{K_1}}$.
Squaring both sides,$K_3^2 = \frac{1}{K_1}$,so $K_1 = \frac{1}{K_3^2}$.
Thus,the correct relation is $K_1 = \frac{1}{K_2} = \frac{1}{K_3^2}$.

(C) The balanced chemical equation is: $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$
Initial moles: $AB_3 = 8, A_2 = 0, B_2 = 0$
Let $x$ be the extent of reaction. At equilibrium,moles are: $AB_3 = 8 - 2x, A_2 = x, B_2 = 3x$
Given that at equilibrium,$A_2 = 2 \ mol$,so $x = 2$.
Equilibrium moles: $AB_3 = 8 - 2(2) = 4 \ mol$,$A_2 = 2 \ mol$,$B_2 = 3(2) = 6 \ mol$.
Since the volume is $1.0 \ dm^3$,the concentrations are: $[AB_3] = 4 \ M, [A_2] = 2 \ M, [B_2] = 6 \ M$.
The equilibrium constant $K_c$ is given by: $K_c = \frac{[A_2][B_2]^3}{[AB_3]^2}$
$K_c = \frac{(2)(6)^3}{(4)^2} = \frac{2 \times 216}{16} = \frac{432}{16} = 27$.

(C) Let the amplitudes be $A_1$ and $A_2$. Given $A_2 = 2A_1$.
Since intensity $I \propto A^2$,we have $I_2 = 4I_1$.
The maximum intensity $I_m$ is given by $I_m = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{I_1} + 2\sqrt{I_1})^2 = (3\sqrt{I_1})^2 = 9I_1$. Thus,$I_1 = \frac{I_m}{9}$.
The resultant intensity $I$ at phase difference $\phi$ is $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
Substituting $I_2 = 4I_1$,we get $I = I_1 + 4I_1 + 2\sqrt{4I_1^2} \cos \phi = 5I_1 + 4I_1 \cos \phi$.
Using the identity $1 + \cos \phi = 2 \cos^2 \frac{\phi}{2}$,we can write $I = I_1 + 4I_1(1 + \cos \phi) = I_1 + 4I_1(2 \cos^2 \frac{\phi}{2}) = I_1(1 + 8 \cos^2 \frac{\phi}{2})$.
Substituting $I_1 = \frac{I_m}{9}$,we get $I = \frac{I_m}{9} (1 + 8 \cos^2 \frac{\phi}{2})$.

(A) thin liquid film has two free surfaces,one on each side of the film.
Therefore,the total upward force due to surface tension is $F = 2T\ell$,where $T$ is the surface tension and $\ell$ is the length of the slider.
Given: Weight $W = mg = 1.5 \times 10^{-2} \, N$,length $\ell = 30 \, cm = 0.3 \, m$.
For equilibrium,the upward force must balance the downward weight:
$2T\ell = mg$
$T = \frac{mg}{2\ell}$
$T = \frac{1.5 \times 10^{-2}}{2 \times 0.3}$
$T = \frac{1.5 \times 10^{-2}}{0.6}$
$T = 0.025 \, N/m$.

(A) Polyamides are polymers containing amide linkages $(-CONH-)$ in their backbone.
$Nylon$ (e.g.,$Nylon-6,6$ or $Nylon-6$) is a well-known example of a polyamide.
$Orlon$ is polyacrylonitrile (an addition polymer).
$Teflon$ is polytetrafluoroethylene (an addition polymer).
$Terylene$ is a polyester (containing ester linkages,$-COO-$).
Therefore,the correct answer is $Nylon$.

(A) Let the output of the first $NAND$ gate be $Y_1 = \overline{A \cdot B}$.
This $Y_1$ is fed as an input to the next two $NAND$ gates.
The outputs of these two gates are $Y_2 = \overline{A \cdot Y_1} = \overline{A \cdot \overline{A \cdot B}}$ and $Y_3 = \overline{B \cdot Y_1} = \overline{B \cdot \overline{A \cdot B}}$.
These are then fed into the final $NAND$ gate,so $Y = \overline{Y_2 \cdot Y_3} = \overline{\overline{A \cdot \overline{A \cdot B}} \cdot \overline{B \cdot \overline{A \cdot B}}}$.
Using De Morgan's theorem,$Y = \overline{\overline{A \cdot \overline{A \cdot B}}} + \overline{\overline{B \cdot \overline{A \cdot B}}} = (A \cdot \overline{A \cdot B}) + (B \cdot \overline{A \cdot B})$.
$Y = (A \cdot (\overline{A} + \overline{B})) + (B \cdot (\overline{A} + \overline{B})) = (A \cdot \overline{A} + A \cdot \overline{B}) + (B \cdot \overline{A} + B \cdot \overline{B})$.
Since $A \cdot \overline{A} = 0$ and $B \cdot \overline{B} = 0$,we get $Y = A \cdot \overline{B} + \overline{A} \cdot B$,which is the $XOR$ gate operation.
The truth table for $XOR$ is:
- If $A=0, B=0$,then $Y=0$.
- If $A=0, B=1$,then $Y=1$.
- If $A=1, B=0$,then $Y=1$.
- If $A=1, B=1$,then $Y=0$.
This matches option $A$.

(D) When the coil oscillates in the magnetic field,the magnetic flux linked with the nearby aluminium plate changes continuously.
According to Faraday's law of electromagnetic induction,this changing magnetic flux induces eddy currents in the aluminium plate.
According to Lenz's law,these eddy currents oppose the motion that produced them.
This effect is known as electromagnetic damping,which brings the oscillating coil to rest quickly.

(C) The resistance of a wire is given by $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.
When the wire is bent at the middle by $180^{\circ}$ and the ends are twisted together,the length of the new wire becomes $l' = \frac{l}{2}$.
Since the two halves are placed side-by-side,the cross-sectional area of the new wire becomes $A' = 2A$.
The new resistance $R'$ is given by $R' = \rho \frac{l'}{A'} = \rho \frac{l/2}{2A} = \frac{1}{4} \left( \rho \frac{l}{A} \right)$.
Therefore,$R' = \frac{R}{4}$.

(A) The work done by a gas during expansion is given by the integral $W = \int_{V_1}^{V_2} P dV$.
Given the process $P = \alpha V$,where $V_1 = V$ and $V_2 = mV$.
Substituting the expression for $P$ into the integral:
$W = \int_{V}^{mV} (\alpha V') dV'$
$W = \alpha \left[ \frac{(V')^2}{2} \right]_{V}^{mV}$
$W = \frac{\alpha}{2} [(mV)^2 - V^2]$
$W = \frac{\alpha}{2} [m^2 V^2 - V^2]$
$W = \frac{\alpha V^2}{2} (m^2 - 1)$