AIPMT 2013 Chemistry Question Paper with Answer and Solution

(A) For an adiabatic process, the relationship between pressure $P$ and absolute temperature $T$ is given by $P^{1-\gamma} T^{\gamma} = \text{constant}$, which can be rewritten as $P \propto T^{\frac{\gamma}{\gamma-1}}$.
Given that $P \propto T^3$, we compare the exponents:
$\frac{\gamma}{\gamma-1} = 3$
$\gamma = 3(\gamma - 1)$
$\gamma = 3\gamma - 3$
$2\gamma = 3$
$\gamma = \frac{3}{2} = 1.5$.
Since the ratio ${C_p}/{C_v}$ is equal to the adiabatic index $\gamma$, the value is $1.5$.

(D) Helium is a monoatomic gas,so its degree of freedom $f = 3$.
For a gas at constant volume,the heat energy required is given by $\Delta Q = \mu C_V \Delta T$.
The number of moles $\mu = \frac{\text{mass}}{\text{molar mass}} = \frac{1\, g}{4\, g/mol} = \frac{1}{4}\, mol$.
The molar heat capacity at constant volume is $C_V = \frac{fR}{2} = \frac{3R}{2}$.
Substituting these values,we get $\Delta Q = \frac{1}{4} \times \frac{3R}{2} \times (T_2 - T_1) = \frac{3R}{8}(T_2 - T_1)$.
Since the universal gas constant $R = N_A k_B$,the heat energy is $\Delta Q = \frac{3}{8} N_A k_B (T_2 - T_1)$.

(D) From the ideal gas equation,$PV = nRT$,we can write $V = (\frac{nR}{P})T$.
Comparing this with the equation of a straight line $y = mx$,where $y = V$ and $x = T$,the slope $m$ is given by $m = \frac{nR}{P}$.
Since the slope is inversely proportional to pressure $(m \propto \frac{1}{P})$,a smaller slope corresponds to a higher pressure.
In the $(V-T)$ diagram,the line for $P_1$ makes a smaller angle $\theta_1$ with the $T$-axis compared to the line for $P_2$ (where $\theta_2 > \theta_1$).
Therefore,the slope of the line for $P_1$ is less than the slope of the line for $P_2$.
Since $\text{slope}_1 < \text{slope}_2$,it follows that $\frac{nR}{P_1} < \frac{nR}{P_2}$,which implies $P_1 > P_2$ or $P_2 < P_1$.

(B) The given circuit consists of two $NAND$ gates. The first $NAND$ gate takes inputs $A$ and $B$,producing an output of $\overline{A.B}$.
This output is then fed into the second $NAND$ gate,which has its two input terminals connected together,effectively acting as a $NOT$ gate.
Therefore,the final output $X$ is the inversion of the input to the second gate:
$X = \overline{(\overline{A.B})}$
Using the Boolean algebra property $\overline{\bar{Y}} = Y$,we get:
$X = A.B$
Thus,the correct option is $(b)$.

(D) The equivalent focal length $f_{eq}$ of two thin lenses in contact is given by $\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}$.
For the plano-convex lens (refractive index $\mu_1$): The first surface is plane $(R_1 = \infty)$ and the second surface is convex $(R_2 = -R)$. Using the lens maker's formula: $\frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right) = \frac{\mu_1 - 1}{R}$.
For the plano-concave lens (refractive index $\mu_2$): The first surface is concave $(R_1 = -R)$ and the second surface is plane $(R_2 = \infty)$. Using the lens maker's formula: $\frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{-R} - \frac{1}{\infty} \right) = -\frac{\mu_2 - 1}{R}$.
Adding the two focal powers:
$\frac{1}{f_{eq}} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R} = \frac{\mu_1 - 1 - \mu_2 + 1}{R} = \frac{\mu_1 - \mu_2}{R}$.
Therefore,$f_{eq} = \frac{R}{\mu_1 - \mu_2}$.

(D) Let the total length of the inclined plane be $2s$. The upper half has length $s$ and is smooth,while the lower half has length $s$ and is rough.
According to the work-energy theorem,the total work done by all forces on the block is equal to the change in its kinetic energy.
Since the block starts from rest at the top and comes to rest at the bottom,the change in kinetic energy is $\Delta KE = 0 - 0 = 0$.
The forces doing work are gravity and friction.
Work done by gravity over the total distance $2s$ is $W_g = (mg \sin \theta)(2s)$.
Work done by friction over the lower half of length $s$ is $W_f = -(\mu mg \cos \theta)(s)$.
Setting the total work to zero: $W_g + W_f = 0$.
$(mg \sin \theta)(2s) - (\mu mg \cos \theta)(s) = 0$.
$2mg \sin \theta = \mu mg \cos \theta$.
Dividing both sides by $mg \cos \theta$,we get $\mu = 2 \frac{\sin \theta}{\cos \theta} = 2 \tan \theta$.

(B) Given the formula $P = \frac{a^3 b^2}{cd}$.
To find the relative error in $P$,we use the formula for propagation of errors:
$\frac{\Delta P}{P} = 3 \frac{\Delta a}{a} + 2 \frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}$.
Given percentage errors are $\frac{\Delta a}{a} \times 100 = 1\%$,$\frac{\Delta b}{b} \times 100 = 2\%$,$\frac{\Delta c}{c} \times 100 = 3\%$,and $\frac{\Delta d}{d} \times 100 = 4\%$.
Substituting these values:
$\frac{\Delta P}{P} \times 100 = 3(1\%) + 2(2\%) + 3\% + 4\%$.
$\frac{\Delta P}{P} \times 100 = 3\% + 4\% + 3\% + 4\% = 14\%$.
Thus,the percentage error in $P$ is $14\%$.

(C) The electric potential $V$ in an electric field $\vec{E}$ is related by the relation $dV = -\vec{E} \cdot d\vec{r}$.
This implies that the electric potential decreases as we move in the direction of the electric field.
In the given figure,the electric field $\vec{E}$ is directed towards the right.
Point $B$ is the furthest to the left,followed by point $C$ (which is at the same vertical level as $B$ but slightly to the right),and point $A$ is the furthest to the right.
Since the potential decreases in the direction of the electric field,the point furthest to the left will have the highest potential.
Therefore,the electric potential is maximum at point $B$.

(A) The resistance of a wire is given by $R = \rho \frac{l}{A}$.
Since the volume $V = A \times l$ remains constant when the wire is stretched,we can write $A = \frac{V}{l}$.
Substituting this into the resistance formula,we get $R = \rho \frac{l}{V/l} = \rho \frac{l^2}{V}$.
Since $\rho$ and $V$ are constant,$R \propto l^2$.
If the length is doubled $(l' = 2l)$,the new resistance $R'$ becomes $R' = R \times (2)^2 = 4R$.
Given the initial resistance $R = 4\,\Omega$,the new resistance is $R' = 4 \times 4\,\Omega = 16\,\Omega$.

(A) For an adiabatic process,the relationship between pressure $P$ and temperature $T$ is given by $P^{1-\gamma} T^{\gamma} = \text{constant}$.
Given that $P \propto T^{3}$,we can write this as $P T^{-3} = \text{constant}$.
Comparing the exponents of $P$ and $T$ in the two equations,we have the ratio $\frac{\gamma}{1-\gamma} = -3$.
Solving for $\gamma$: $\gamma = -3(1-\gamma) \Rightarrow \gamma = -3 + 3\gamma \Rightarrow 2\gamma = 3 \Rightarrow \gamma = 1.5$.
Since the ratio $C_P / C_V$ is defined as $\gamma$,the value is $1.5$.

(B) According to the ideal gas equation,$PV = nRT$,which can be rewritten as $V = (nR/P)T$.
For an isobaric process,the pressure $P$ is constant,and $V \propto T$.
Therefore,the $(V-T)$ graph is a straight line passing through the origin,where the slope of the line is given by $m = nR/P$.
Since the slope $m$ is inversely proportional to the pressure $P$ $(m \propto 1/P)$,a larger slope corresponds to a smaller pressure.
From the given diagram,the angle $\theta_2 > \theta_1$,which implies that the slope of the line for $P_2$ is greater than the slope of the line for $P_1$ (i.e.,$\text{slope}_2 > \text{slope}_1$).
Since the slope is inversely proportional to pressure,we conclude that $P_2 < P_1$.

(A) According to Wien's displacement law, the product of the wavelength corresponding to maximum intensity $(\lambda_{m})$ and the absolute temperature $(T)$ of the black body is a constant.
$\lambda_{m} T = b$ (constant)
This implies $\lambda_{m} \propto \frac{1}{T}$.
As the temperature $(T)$ of the iron piece increases, the wavelength $(\lambda_{m})$ of the emitted radiation decreases. Initially, the iron emits longer wavelengths (dull red), and as it gets hotter, it shifts toward shorter wavelengths (yellow and eventually white, which contains all visible colors), confirming the observation.