AIEEE 2006 Chemistry Question Paper with Answer and Solution

(C) From Hooke's Law,the elongation $\Delta L$ of a wire of length $L$,cross-sectional area $A$,and Young's modulus $Y$ under a tension $T$ is given by $\Delta L = \frac{TL}{AY}$.
In the first case,a load $W$ is hung from the wire. The tension in the wire is $T_1 = W$. Therefore,the elongation is $\Delta L_1 = \frac{WL}{AY} = l$.
In the second case,the wire passes over a pulley and a weight $W$ is hung at each end. The tension $T_2$ in the wire is equal to the weight $W$ because the system is in equilibrium. Thus,$T_2 = W$.
The elongation in the second case is $\Delta L_2 = \frac{T_2 L}{AY} = \frac{WL}{AY} = l$.
Therefore,the elongation remains $l \, mm$.

(A) The average total energy density $u_{av}$ of an electromagnetic wave is given by the sum of the average electric energy density and average magnetic energy density.
$u_{av} = u_{E} + u_{B} = \frac{1}{2} \epsilon_{0} E_{rms}^{2} + \frac{1}{2} \frac{B_{rms}^{2}}{\mu_{0}}$
Since $E_{rms} = c B_{rms}$,we have $u_{E} = u_{B}$,so $u_{av} = \epsilon_{0} E_{rms}^{2}$.
Given $E_{rms} = 720\, N/C$ and $\epsilon_{0} = 8.854 \times 10^{-12}\, C^2/(N \cdot m^2)$.
$u_{av} = (8.854 \times 10^{-12}) \times (720)^2$
$u_{av} = 8.854 \times 10^{-12} \times 518400$
$u_{av} \approx 4.589 \times 10^{-6}\, J/m^3$.

(A) In the first case,a load $W$ is hung from the wire. The tension in the wire is $T = W$. The elongation is given by $\Delta l = \frac{Tl}{AY} = \frac{Wl}{AY} = l$.
In the second case,the wire passes over a pulley and two weights $W$ each are hung at the two ends. The tension in the wire remains $T = W$ throughout the wire.
Since the tension $T$ is the same in both cases,the elongation $\Delta l$ remains the same.
Therefore,the elongation is $l \, mm$.

(C) The terminal velocity $V_T$ of a sphere of radius $r$ and density $\rho$ falling through a liquid of density $\sigma$ and viscosity $\eta$ is given by $V_T = \frac{2r^2}{9\eta}(\rho - \sigma)g$.
Since $r$,$\eta$,and $\sigma$ are the same for both spheres,$V_T \propto (\rho - \sigma)$.
For gold: $V_{T1} = 0.2 \, m/s$,$\rho_1 = 19.5 \times 10^3 \, kg/m^3$,$\sigma = 1.5 \times 10^3 \, kg/m^3$.
$V_{T1} \propto (19.5 - 1.5) = 18$.
For silver: $V_{T2} = ?$,$\rho_2 = 10.5 \times 10^3 \, kg/m^3$.
$V_{T2} \propto (10.5 - 1.5) = 9$.
Taking the ratio: $\frac{V_{T2}}{V_{T1}} = \frac{9}{18} = 0.5$.
$V_{T2} = 0.5 \times V_{T1} = 0.5 \times 0.2 = 0.1 \, m/s$.

(C) The formula for terminal velocity $V_T$ of a sphere falling through a viscous liquid is given by:
$V_T = \frac{2r^2(\rho - \sigma)g}{9\eta}$
where $\rho$ is the density of the sphere,$\sigma$ is the density of the liquid,$r$ is the radius,and $\eta$ is the coefficient of viscosity.
Since the radius $r$,the liquid density $\sigma$,and the viscosity $\eta$ are the same for both spheres,the terminal velocity is proportional to the difference in densities: $V_T \propto (\rho - \sigma)$.
For gold: $V_{T1} = 0.2 \, m/s$,$\rho_1 = 19.5 \times 10^3 \, kg/m^3$,$\sigma = 1.5 \times 10^3 \, kg/m^3$.
For silver: $V_{T2} = ?$,$\rho_2 = 10.5 \times 10^3 \, kg/m^3$.
Using the ratio: $\frac{V_{T2}}{V_{T1}} = \frac{\rho_2 - \sigma}{\rho_1 - \sigma}$
$\frac{V_{T2}}{0.2} = \frac{10.5 - 1.5}{19.5 - 1.5} = \frac{9}{18} = 0.5$
$V_{T2} = 0.2 \times 0.5 = 0.1 \, m/s$.

(C) For an ideal gas in a rigid box,the heat exchange is given by $Q = n C_v \Delta T$. Since the boxes are in thermal contact and isolated from the surroundings,the heat lost by the hotter gas equals the heat gained by the colder gas.
Nitrogen $(N_2)$ is a diatomic gas,so its molar heat capacity at constant volume is $C_{v1} = \frac{5}{2} R$.
Helium $(He)$ is a monatomic gas,so its molar heat capacity at constant volume is $C_{v2} = \frac{3}{2} R$.
Let $n_1 = 1$ mole $(N_2)$ and $n_2 = 1$ mole $(He)$.
Heat lost by $He$ = Heat gained by $N_2$:
$n_2 C_{v2} (T_{initial, He} - T_f) = n_1 C_{v1} (T_f - T_{initial, N2})$
$1 \cdot \frac{3}{2} R \left( \frac{7}{3} T_0 - T_f \right) = 1 \cdot \frac{5}{2} R (T_f - T_0)$
Multiplying both sides by $\frac{2}{R}$:
$3 \left( \frac{7}{3} T_0 - T_f \right) = 5 (T_f - T_0)$
$7 T_0 - 3 T_f = 5 T_f - 5 T_0$
$12 T_0 = 8 T_f$
$T_f = \frac{12}{8} T_0 = \frac{3}{2} T_0$.

(D) The magnetic flux $\phi$ linked with the coil at any time $t$ is given by $\phi = N B A \cos(\omega t)$.
According to Faraday's law of electromagnetic induction,the induced $emf$ is $e = -\frac{d\phi}{dt}$.
Substituting the expression for $\phi$,we get $e = -\frac{d}{dt}(N B A \cos(\omega t))$.
$e = -N B A \frac{d}{dt}(\cos(\omega t)) = -N B A (-\omega \sin(\omega t))$.
$e = N B A \omega \sin(\omega t)$.
The maximum value of $emf$ $(e_{\max})$ occurs when $\sin(\omega t) = 1$.
Therefore,$e_{\max} = N B A \omega$.

(B) The stability of free radicals is governed by resonance and inductive effects.
$1$. Resonance stabilization: The triphenylmethyl radical $(C_6H_5)_3\dot{C}$ is more stable than the diphenylmethyl radical $(C_6H_5)_2\dot{C}H$ because it has more phenyl groups to delocalize the unpaired electron.
$2$. Inductive effect: Alkyl radicals are stabilized by the $+I$ effect of alkyl groups. The tertiary butyl radical $(CH_3)_3\dot{C}$ is more stable than the isopropyl radical $(CH_3)_2\dot{C}H$.
$3$. Overall order: Resonance-stabilized radicals are significantly more stable than alkyl radicals.
Therefore,the increasing order of stability is: $(CH_3)_2\dot{C}H < (CH_3)_3\dot{C} < (C_6H_5)_2\dot{C}H < (C_6H_5)_3\dot{C}$.

(C) The terminal velocity $V_T$ of a sphere of radius $a$ falling through a viscous liquid is given by the formula:
$V_T = \frac{2 a^2}{9 \eta} (\rho - \sigma) g$
Where $\rho$ is the density of the sphere,$\sigma$ is the density of the liquid,$\eta$ is the coefficient of viscosity,and $g$ is the acceleration due to gravity.
Since the size of the spheres and the liquid are the same,$a$,$\eta$,and $g$ are constant.
Therefore,$V_T \propto (\rho - \sigma)$.
For gold: $V_{T1} = 0.2 \,m/s$,$\rho_1 = 19.5 \,kg/m^3$,$\sigma = 1.5 \,kg/m^3$.
For silver: $V_{T2} = V$,$\rho_2 = 10.5 \,kg/m^3$,$\sigma = 1.5 \,kg/m^3$.
Using the ratio: $\frac{V_{T1}}{V_{T2}} = \frac{\rho_1 - \sigma}{\rho_2 - \sigma}$
$\frac{0.2}{V} = \frac{19.5 - 1.5}{10.5 - 1.5} = \frac{18}{9} = 2$
$V = \frac{0.2}{2} = 0.1 \,m/s$.