AIPMT 2015 Chemistry Question Paper with Answer and Solution

(D) According to the de-Broglie hypothesis,the wavelength $\lambda$ associated with a particle of momentum $p$ is given by the relation:
$\lambda = \frac{h}{p}$
where $h$ is Planck's constant.
Rearranging this equation,we get:
$p = \frac{h}{\lambda}$
This implies that the momentum $p$ is inversely proportional to the wavelength $\lambda$ $(p \propto \frac{1}{\lambda})$.
The graph of $p$ versus $\lambda$ for an inverse relationship is a rectangular hyperbola.
Therefore,the correct figure is the one representing a rectangular hyperbola,which corresponds to option $(D)$.

(D) Let us analyze the number of $\sigma$ and $\pi$ bonds in each species:
$A$. $(CN)_2$ (or $N \equiv C-C \equiv N$): It has $3$ $\sigma$ bonds and $4$ $\pi$ bonds.
$B$. $CH_2(CN)_2$ (or $NC-CH_2-CN$): It has $9$ $\sigma$ bonds and $4$ $\pi$ bonds.
$C$. $HCO_3^-$: It has $5$ $\sigma$ bonds and $1$ $\pi$ bond.
$D$. $XeO_4$: The structure consists of a central $Xe$ atom bonded to $4$ oxygen atoms via double bonds. Each double bond consists of $1$ $\sigma$ bond and $1$ $\pi$ bond. Thus,there are $4$ $\sigma$ bonds and $4$ $\pi$ bonds.
Therefore,$XeO_4$ contains an equal number of $\sigma$ and $\pi$ bonds.

(C) The chemical formula of the enolic form of ethyl acetoacetate is $CH_3-C(OH)=CH-COOCH_2CH_3$.
Counting the bonds:
$1$. Sigma $(\sigma)$ bonds: There are $18$ sigma bonds in the molecule.
$2$. Pi $(\pi)$ bonds: There are $2$ pi bonds (one in the $C=C$ double bond and one in the $C=O$ double bond).
Therefore,the correct answer is $18$ sigma bonds and $2$ pi-bonds.

(A) $3,3-$Dimethyl$-1-$butene $(CH_3-C(CH_3)_2-CH=CH_2)$ on treatment with a strong acid undergoes protonation to form a secondary carbocation: $CH_3-C(CH_3)_2-C^+H-CH_3$.
This carbocation undergoes a $1,2-$methyl shift to form a more stable tertiary carbocation: $CH_3-C^+(CH_3)-CH(CH_3)-CH_3$.
Finally,it loses a proton (deprotonation) to form the more substituted alkene,$2,3-$dimethyl$-2-$butene $(CH_3-C(CH_3)=C(CH_3)-CH_3)$,as the major product.

(B) The coefficient of performance $(COP)$ of a refrigerator is given by the formula:
$COP = \frac{T_L}{T_H - T_L}$
where $T_L$ is the temperature inside the freezer and $T_H$ is the temperature of the surroundings.
First,convert the temperature from Celsius to Kelvin:
$T_L = -20 + 273 = 253 \, K$.
Given $COP = 5$,substitute the values into the formula:
$5 = \frac{253}{T_H - 253}$
$5(T_H - 253) = 253$
$5T_H - 1265 = 253$
$5T_H = 1518$
$T_H = 303.6 \, K$.
Now,convert the temperature back to Celsius:
$T_H (^{\circ}C) = 303.6 - 273 = 30.6 \, ^{\circ}C \approx 31 \, ^{\circ}C$.

(A) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function and $\lambda_0$ is the threshold wavelength.
For the first case: $3eV_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$ --- $(1)$
For the second case: $eV_0 = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$ --- $(2)$
Dividing equation $(1)$ by equation $(2)$:
$3 = \frac{\frac{hc}{\lambda} - \frac{hc}{\lambda_0}}{\frac{hc}{2\lambda} - \frac{hc}{\lambda_0}}$
$3 \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right) = \frac{1}{\lambda} - \frac{1}{\lambda_0}$
$\frac{3}{2\lambda} - \frac{3}{\lambda_0} = \frac{1}{\lambda} - \frac{1}{\lambda_0}$
$\frac{3}{2\lambda} - \frac{1}{\lambda} = \frac{3}{\lambda_0} - \frac{1}{\lambda_0}$
$\frac{1}{2\lambda} = \frac{2}{\lambda_0}$
$\lambda_0 = 4\lambda$.

(A) Given: Force $F = 14\,N$,masses $m_A = 4\,kg, m_B = 2\,kg, m_C = 1\,kg$.
Total mass of the system $m = m_A + m_B + m_C = 4 + 2 + 1 = 7\,kg$.
Acceleration of the system $a = \frac{F}{m} = \frac{14}{7} = 2\,m/s^2$.
The contact force between block $A$ and block $B$ is the force required to accelerate blocks $B$ and $C$ together.
Let $F_{AB}$ be the contact force between $A$ and $B$.
$F_{AB} = (m_B + m_C) \times a = (2 + 1) \times 2 = 3 \times 2 = 6\,N$.

(C) When the length of the capillary tube is less than the height $h$ to which water would normally rise,the water will rise to the top of the tube.
At the top,the radius of curvature of the meniscus increases such that the pressure balance is maintained.
The water does not overflow because the pressure at the top surface remains equal to the atmospheric pressure,and the surface tension force adjusts to keep the water in equilibrium.
Therefore,the water rises up to the top of the capillary tube and stays there without overflowing.

(A) Let $f_{o}$ and $f_{e}$ be the focal lengths of the objective and eyepiece respectively.
In normal adjustment,the distance between the objective and the eyepiece is $f_{o} + f_{e}$.
The line of length $L$ on the objective lens acts as an object for the eyepiece.
The distance of this object from the eyepiece is $u = -(f_{o} + f_{e})$.
Using the lens formula for the eyepiece: $1/v - 1/u = 1/f_{e}$.
$1/v = 1/f_{e} + 1/u = 1/f_{e} - 1/(f_{o} + f_{e}) = f_{o} / [f_{e}(f_{o} + f_{e})]$.
So,$v = [f_{e}(f_{o} + f_{e})] / f_{o}$.
The magnification $m$ produced by the eyepiece for this object is $m = v/u = [f_{e}(f_{o} + f_{e}) / f_{o}] / [-(f_{o} + f_{e})] = -f_{e}/f_{o}$.
The ratio of the image length $l$ to the object length $L$ is $|m| = l/L = f_{e}/f_{o}$.
The magnification of the astronomical telescope in normal adjustment is $M = f_{o}/f_{e}$.
Therefore,$M = L/l$.

(A) For a particle in simple harmonic motion with amplitude $A$ and angular frequency $\omega$:
Maximum acceleration is given by $\alpha = A\omega^2$.
Maximum velocity is given by $\beta = A\omega$.
Dividing the two equations: $\frac{\alpha}{\beta} = \frac{A\omega^2}{A\omega} = \omega$.
Therefore,$\omega = \frac{\alpha}{\beta}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$: $T = \frac{2\pi}{(\alpha / \beta)} = \frac{2\pi \beta}{\alpha}$.

(B) According to Bernoulli's principle,the pressure above the roof decreases as the wind speed increases.
The pressure difference $\Delta P$ between the inside and outside of the roof is given by $\Delta P = \frac{1}{2} \rho V^2$.
Given: $\rho = 1.2\,kg/m^3$,$V = 40\,m/s$,and $A = 250\,m^2$.
$\Delta P = \frac{1}{2} \times 1.2 \times (40)^2 = 0.6 \times 1600 = 960\,Pa$.
The force exerted on the roof is $F = \Delta P \times A$.
$F = 960 \times 250 = 240,000\,N = 2.4 \times 10^5\,N$.
Since the pressure outside is lower than the pressure inside,the net force acts in the upward direction.

(A) The number of $\sigma$ and $\pi$ bonds in the given species is calculated as follows:

Molecule	Number of $\sigma$ and $\pi$ bonds
$(CN)_2$ $(N \equiv C-C \equiv N)$	$3 \sigma, 4 \pi$
$CH_2(CN)_2$	$8 \sigma, 4 \pi$
$H_2CO_3$	$6 \sigma, 1 \pi$
$XeO_4$	$4 \sigma, 4 \pi$

In $XeO_4$,there are $4$ $\sigma$ bonds (Xe-$O$ single bonds) and $4$ $\pi$ bonds (Xe=$O$ double bonds). Thus,it contains an equal number of $\sigma$ and $\pi$ bonds.

(D) The initial distance between the centres of the two bodies is $12\,R$. Collision occurs when the distance between their centres is equal to the sum of their radii,which is $R + 2R = 3\,R$.
The total distance covered by both bodies relative to each other until collision is $12\,R - 3\,R = 9\,R$.
Since the bodies move under mutual gravitational attraction only,the position of their centre of mass remains stationary. Let $x$ be the distance covered by the smaller body (mass $M$) and $y$ be the distance covered by the larger body (mass $5M$).
From the property of the centre of mass,$M x = (5M) y$.
We also know that $x + y = 9\,R$,which implies $y = 9\,R - x$.
Substituting $y$ into the first equation: $M x = 5M(9\,R - x)$.
Dividing by $M$: $x = 5(9\,R - x) = 45\,R - 5x$.
$6x = 45\,R$,which gives $x = \frac{45\,R}{6} = 7.5\,R$.

(A) The total mass of the system is $M = m_A + m_B + m_C = 4\, kg + 2\, kg + 1\, kg = 7\, kg$.
The external force applied is $F = 14\, N$.
The acceleration of the system is $a = \frac{F}{M} = \frac{14\, N}{7\, kg} = 2\, m/s^2$.
The contact force between block $A$ and block $B$ is the force required to accelerate blocks $B$ and $C$ together.
Therefore,the contact force $F_{AB} = (m_B + m_C) \times a$.
$F_{AB} = (2\, kg + 1\, kg) \times 2\, m/s^2 = 3\, kg \times 2\, m/s^2 = 6\, N$.

(B) For two particles to collide,the relative velocity vector of one particle with respect to the other must be directed along the line joining their initial positions.
Let the relative position vector be $\vec{r}_{rel} = \vec{r}_1 - \vec{r}_2$.
The relative velocity vector is $\vec{v}_{rel} = \vec{v}_2 - \vec{v}_1$.
For collision,the direction of the relative velocity must be the same as the direction of the relative position vector.
Thus,the unit vectors must be equal:
$\frac{\vec{r}_1 - \vec{r}_2}{|\vec{r}_1 - \vec{r}_2|} = \frac{\vec{v}_2 - \vec{v}_1}{|\vec{v}_2 - \vec{v}_1|}$.

(C) $1$. Coefficient of static friction $(\mu_s)$: When the box just starts to slip,the angle of inclination is the angle of repose. Thus,$\mu_s = \tan \theta = \tan 30^o = \frac{1}{\sqrt{3}} \approx 0.577 \approx 0.6$.
$2$. Coefficient of kinetic friction $(\mu_k)$: The acceleration $a$ of the box sliding down the plank is given by $a = g \sin \theta - \mu_k g \cos \theta$.
$3$. Using the equation of motion $S = ut + \frac{1}{2}at^2$,where $S = 4.0\, m$,$u = 0$,and $t = 4.0\, s$:
$4.0 = 0 + \frac{1}{2} a (4.0)^2 \Rightarrow 4.0 = 8a \Rightarrow a = 0.5\, m/s^2$.
$4$. Substituting $a$ into the force equation (taking $g = 10\, m/s^2$):
$0.5 = 10 \sin 30^o - \mu_k (10) \cos 30^o$
$0.5 = 10(0.5) - \mu_k (10)(\frac{\sqrt{3}}{2})$
$0.5 = 5 - 5\sqrt{3} \mu_k$
$5\sqrt{3} \mu_k = 4.5$
$\mu_k = \frac{4.5}{5 \times 1.732} = \frac{0.9}{1.732} \approx 0.519 \approx 0.5$.
Thus,$\mu_s \approx 0.6$ and $\mu_k \approx 0.5$.

(B) From the first law of thermodynamics,$\Delta U = Q - W$. Since the change in internal energy $\Delta U$ is a state function,it is the same for any path between two states.
For path $ABC$: $\Delta U_{ABC} = \Delta U_{AC} = Q_{ABC} - W_{ABC}$.
Total heat absorbed $Q_{ABC} = Q_{AB} + Q_{BC} = 400 \, J + 100 \, J = 500 \, J$.
The work done $W_{ABC}$ is the area under the path $ABC$ on the $PV$ diagram. Since $AB$ is an isochoric process,$W_{AB} = 0$. For process $BC$,$W_{BC} = P \Delta V = (6 \times 10^4 \, Pa) \times (4 \times 10^{-3} \, m^3 - 2 \times 10^{-3} \, m^3) = 6 \times 10^4 \times 2 \times 10^{-3} = 120 \, J$.
So,$W_{ABC} = 0 + 120 \, J = 120 \, J$.
Thus,$\Delta U_{AC} = 500 \, J - 120 \, J = 380 \, J$.
For the process $AC$,$\Delta U_{AC} = Q_{AC} - W_{AC}$.
The work done $W_{AC}$ is the area under the line $AC$,which is a trapezoid: $W_{AC} = \frac{1}{2} \times (P_A + P_C) \times (V_C - V_A) = \frac{1}{2} \times (2 \times 10^4 + 6 \times 10^4) \times (4 \times 10^{-3} - 2 \times 10^{-3}) = \frac{1}{2} \times 8 \times 10^4 \times 2 \times 10^{-3} = 80 \, J$.
Therefore,$Q_{AC} = \Delta U_{AC} + W_{AC} = 380 \, J + 80 \, J = 460 \, J$.

(D) Since the uranium nucleus is at rest,the initial momentum is $0$. By the law of conservation of linear momentum,the total final momentum must also be $0$.
Therefore,the magnitude of the momentum of the thorium nucleus $(P_{Th})$ must be equal to the magnitude of the momentum of the helium nucleus $(P_{He})$,i.e.,$P_{Th} = P_{He} = P$.
The kinetic energy $(KE)$ of a particle is given by the formula $KE = \frac{P^2}{2m}$,where $m$ is the mass.
Thus,$KE_{Th} = \frac{P^2}{2m_{Th}}$ and $KE_{He} = \frac{P^2}{2m_{He}}$.
Since the mass of the helium nucleus $(m_{He})$ is significantly smaller than the mass of the thorium nucleus $(m_{Th})$,the kinetic energy of the helium nucleus $(KE_{He})$ will be greater than the kinetic energy of the thorium nucleus $(KE_{Th})$.
Hence,option $(D)$ is correct.