AIPMT 2005 Chemistry Question Paper with Answer and Solution

(A) lichen is a symbiotic association between an alga (phycobiont) and a fungus (mycobiont).
In this mutualistic relationship,the alga is responsible for photosynthesis,providing food for both partners.
The fungus provides protection,anchorage,and absorption of water and mineral nutrients from the environment for the alga.
Therefore,the correct function of the fungus is to provide protection,anchorage,and absorption for the alga.

(B) An ectophloic siphonostele is a type of stele in which the phloem is present only on the outer side of the xylem. This arrangement is characteristic of certain Pteridophytes. Specifically,it is found in genera such as $Osmunda$ and $Equisetum$. In these plants,the central pith is surrounded by a ring of xylem,which is in turn surrounded by a ring of phloem.

(C) The pollen tube enters the embryo sac through the filiform apparatus present in the synergids. As the pollen tube approaches the micropylar end,one of the two synergids begins to degenerate. The pollen tube then enters the embryo sac through this degenerated synergid.

(C) The general structure of an amino acid is $R-CH(NH_2)-COOH$.
In this structure,$R$ represents the side chain.
For the simplest amino acid,the side chain $R$ must be the smallest possible group,which is a hydrogen atom $(H)$.
When $R = H$,the amino acid is Glycine $(NH_2-CH_2-COOH)$.
Since it has the smallest side chain,Glycine is considered the simplest amino acid.

(C) The correct answer is $C$. Solar energy is a clean,renewable source of energy that does not contribute to the greenhouse effect. In contrast,the combustion of fossil fuels and biomass releases $CO_2$,which is a greenhouse gas. Nuclear energy produces radioactive waste as a byproduct. Therefore,the pairing of solar energy with the greenhouse effect is incorrect.

(D) The displacement is given by $x = a e^{-\alpha t} + b e^{\beta t}$.
The velocity $v$ is the time derivative of displacement: $v = \frac{dx}{dt}$.
$v = \frac{d}{dt}(a e^{-\alpha t} + b e^{\beta t}) = a(-\alpha) e^{-\alpha t} + b(\beta) e^{\beta t}$.
$v = -a \alpha e^{-\alpha t} + b \beta e^{\beta t}$.
As time $t$ increases,the term $e^{-\alpha t}$ decreases towards zero,while the term $e^{\beta t}$ increases exponentially. Since $b, \beta > 0$,the term $b \beta e^{\beta t}$ will dominate for large $t$,causing the velocity $v$ to increase with time.

(B) The correct order of electron gain enthalpy (with negative sign) is $O < S < F < Cl$.
Generally,electron gain enthalpy becomes more negative (higher magnitude) across a period from left to right due to an increase in effective nuclear charge and a decrease in atomic size.
Down a group,electron gain enthalpy becomes less negative due to an increase in atomic size.
However,$F$ and $O$ have smaller sizes compared to $Cl$ and $S$ respectively,leading to strong inter-electronic repulsions in the $2p$ subshell. Consequently,$Cl$ has a higher electron gain enthalpy than $F$,and $S$ has a higher electron gain enthalpy than $O$.

(C) The energy of an incident photon is given by $E = \frac{hc}{\lambda}$.
Using the approximation $E \approx \frac{12400}{\lambda (\text{in } \mathring{A})} eV$,we get:
$E = \frac{12400}{4100} \approx 3.02\, eV$.
Photoelectric emission occurs if the energy of the incident photon is greater than or equal to the work function $(\phi)$ of the metal.
For metal $A$: $\phi_A = 1.92\, eV$. Since $3.02\, eV > 1.92\, eV$,metal $A$ will emit photoelectrons.
For metal $B$: $\phi_B = 2.0\, eV$. Since $3.02\, eV > 2.0\, eV$,metal $B$ will emit photoelectrons.
For metal $C$: $\phi_C = 5.0\, eV$. Since $3.02\, eV < 5.0\, eV$,metal $C$ will not emit photoelectrons.
Therefore,only metals $A$ and $B$ will emit photoelectrons.

(C) The general equation for a progressive wave is $Y = A \sin(2\pi n t)$,where $n$ is the frequency.
For the first tuning fork,$2\pi n_1 = 500\pi$,so $n_1 = 250 \text{ Hz}$.
For the second tuning fork,$2\pi n_2 = 506\pi$,so $n_2 = 253 \text{ Hz}$.
The number of beats per second is given by $|n_2 - n_1| = |253 - 250| = 3 \text{ beats/second}$.
To find the number of beats per minute,we multiply by $60$ seconds:
$\text{Beats per minute} = 3 \times 60 = 180$.

(C) According to $Fajan's \ rule$,the covalent character of an ionic bond increases with an increase in the polarizing power of the cation.
Polarizing power is directly proportional to the charge density of the cation (charge/size ratio).
Comparing the cations $Na^+$,$Li^+$,and $Be^{2+}$:
$1$. $Na^+$ has a charge of $+1$ and a larger ionic radius.
$2$. $Li^+$ has a charge of $+1$ and a smaller ionic radius than $Na^+$.
$3$. $Be^{2+}$ has a higher charge of $+2$ and the smallest ionic radius among the three.
Thus,the polarizing power increases in the order: $Na^+ < Li^+ < Be^{2+}$.
Consequently,the covalent character increases in the order: $NaCl < LiCl < BeCl_2$.

(B) The angular resolution $\Delta \theta$ of a telescope is given by the formula: $\Delta \theta = \frac{1.22 \lambda}{D}$.
Given:
Diameter of the telescope,$D = 10\,cm = 0.1\,m$.
Wavelength,$\lambda = 5000\,\mathring{A} = 5000 \times 10^{-10}\,m = 5 \times 10^{-7}\,m$.
Substituting the values into the formula:
$\Delta \theta = \frac{1.22 \times 5 \times 10^{-7}}{0.1}$.
$\Delta \theta = 1.22 \times 5 \times 10^{-6}$.
$\Delta \theta = 6.1 \times 10^{-6}\,\text{rad}$.
This value is of the order of $10^{-5}\,\text{rad}$.

(A) In a nuclear fission process,a heavy parent nucleus splits into lighter daughter nuclei (fission products).
According to Einstein's mass-energy equivalence principle,$E = \Delta m c^2$,the total mass of the fission products is slightly less than the mass of the parent nucleus because some mass is converted into energy (binding energy).
Therefore,$\text{mass of fission products} < \text{mass of parent nucleus}$.
Consequently,the ratio $\frac{\text{mass of fission products}}{\text{mass of parent nucleus}} < 1$.

(B) The energy band gap $(E_g)$ is the energy difference between the valence band and the conduction band.
For Carbon (diamond),the band gap is approximately $5.5 \ eV$.
For Silicon,the band gap is approximately $1.1 \ eV$.
For Germanium,the band gap is approximately $0.7 \ eV$.
Comparing these values,we find that $(E_g)_C > (E_g)_{Si} > (E_g)_{Ge}$.
Therefore,the relationship $(E_g)_C > (E_g)_{Si}$ is true.

(B) The energy band gap $(E_g)$ is the energy difference between the conduction band and the valence band.
For Carbon (diamond),Silicon,and Germanium,the energy band gaps are approximately:
$(E_g)_C \approx 5.4 \ eV$
$(E_g)_{Si} \approx 1.1 \ eV$
$(E_g)_{Ge} \approx 0.7 \ eV$
Comparing these values,we observe that $(E_g)_C > (E_g)_{Si} > (E_g)_{Ge}$.
Therefore,the relationship $(E_g)_C > (E_g)_{Si}$ is true.

(D) The angular resolution $(d\theta)$ of a telescope is given by the formula: $d\theta = \frac{1.22 \lambda}{D}$.
Given:
Wavelength $\lambda = 5000\,\mathring{A} = 5000 \times 10^{-10}\,m = 5 \times 10^{-7}\,m$.
Diameter $D = 10\,cm = 0.1\,m$.
Substituting the values into the formula:
$d\theta = \frac{1.22 \times 5 \times 10^{-7}}{0.1} = 1.22 \times 5 \times 10^{-6} = 6.1 \times 10^{-6}\,rad$.
This value is of the order of $10^{-6}\,rad$.

(B) The capacitor $C_4$ is connected directly across the battery of voltage $V$. Therefore,the charge on $C_4$ is $q_4 = C_4 \times V = 4CV$.
The capacitors $C_1$,$C_2$,and $C_3$ are connected in series with each other,and this series combination is connected in parallel with $C_4$ across the battery.
Let the equivalent capacitance of the series combination of $C_1$,$C_2$,and $C_3$ be $C'$.
$\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{C} + \frac{1}{2C} + \frac{1}{3C} = \frac{6+3+2}{6C} = \frac{11}{6C}$.
Thus,$C' = \frac{6C}{11}$.
The charge $q'$ flowing through the series combination of $C_1$,$C_2$,and $C_3$ is given by $q' = C' \times V = \frac{6C}{11} \times V = \frac{6CV}{11}$.
Since $C_1$,$C_2$,and $C_3$ are in series,the charge on each of them is the same,which is $q'$.
Therefore,the charge on $C_2$ is $q_2 = q' = \frac{6CV}{11}$.
The ratio of the charges on $C_2$ and $C_4$ is $\frac{q_2}{q_4} = \frac{6CV/11}{4CV} = \frac{6}{44} = \frac{3}{22}$.

(B) Paramagnetism is observed in species that contain at least one unpaired electron.
$CO_2$,$SO_2$,and $SiO_2$ are all diamagnetic as they have all electrons paired.
$ClO_2$ has a total of $19$ valence electrons $(7 + 6 \times 2 = 19)$,making it an odd-electron species.
Due to the presence of an unpaired electron,$ClO_2$ exhibits paramagnetic behaviour.

(A) The energy band gap $(Eg)$ is the energy difference between the valence band and the conduction band.
For Carbon (diamond),the band gap is approximately $5.5 \, eV$.
For Silicon,the band gap is approximately $1.1 \, eV$.
For Germanium,the band gap is approximately $0.7 \, eV$.
Comparing these values,we find that $(Eg)_C > (Eg)_{Si} > (Eg)_{Ge}$.
Therefore,the relationship $(Eg)_C > (Eg)_{Si}$ is true.

(C) When two cells with $emf$ $E_1$ and $E_2$ and internal resistances $r_1$ and $r_2$ are connected in parallel,the equivalent $emf$ $E_{eq}$ is given by the formula:
$E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$
Given: $E_1 = 18 \, V$,$r_1 = 2 \, \Omega$,$E_2 = 12 \, V$,$r_2 = 1 \, \Omega$.
Substituting the values into the formula:
$E_{eq} = \frac{(18 \times 1) + (12 \times 2)}{2 + 1}$
$E_{eq} = \frac{18 + 24}{3}$
$E_{eq} = \frac{42}{3} = 14 \, V$
The voltmeter connected across the parallel combination will measure the equivalent $emf$ of the circuit.
Therefore,the reading of the voltmeter is $14 \, V$.

(C) The given wave equations are $Y_1 = 4 \sin(500 \pi t)$ and $Y_2 = 2 \sin(506 \pi t)$.
Comparing these with the standard equation $Y = A \sin(\omega t)$,where $\omega = 2 \pi \nu$:
For the first wave: $\omega_1 = 500 \pi \implies 2 \pi \nu_1 = 500 \pi \implies \nu_1 = 250 \text{ Hz}$.
For the second wave: $\omega_2 = 506 \pi \implies 2 \pi \nu_2 = 506 \pi \implies \nu_2 = 253 \text{ Hz}$.
The beat frequency is the difference between the two frequencies: $\nu_{beat} = |\nu_2 - \nu_1| = |253 - 250| = 3 \text{ beats/s}$.
To find the number of beats per minute,multiply the beats per second by $60$: $3 \times 60 = 180 \text{ beats/min}$.

(D) $1$. Diamagnetic materials consist of atoms where the orbital and spin magnetic moments of electrons cancel each other out,resulting in a net atomic magnetic dipole moment of $\mu_d = 0$.
$2$. Paramagnetic materials consist of atoms that possess a permanent magnetic dipole moment due to unpaired electrons,so $\mu_p \neq 0$.
$3$. Ferromagnetic materials consist of atoms that also possess a permanent magnetic dipole moment,so $\mu_f \neq 0$.
$4$. Therefore,the correct relationship is $\mu_d = 0$ and $\mu_p \neq 0$ (and $\mu_f \neq 0$).

(C) The capacitor $C_4$ is connected directly across the battery,so the potential difference across it is $V$. The charge on $C_4$ is $Q_4 = C_4 V = (4C)V = 4CV$.
The capacitors $C_3$,$C_2$,and $C_1$ are in series. Their equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_3} + \frac{1}{C_2} + \frac{1}{C_1} = \frac{1}{3C} + \frac{1}{2C} + \frac{1}{C} = \frac{2+3+6}{6C} = \frac{11}{6C}$.
Thus,$C_{eq} = \frac{6C}{11}$.
The charge on this series combination is $Q_{series} = C_{eq} V = \frac{6CV}{11}$.
Since $C_2$ is part of this series branch,the charge on $C_2$ is $Q_2 = Q_{series} = \frac{6CV}{11}$.
The ratio of the charges is $\frac{Q_2}{Q_4} = \frac{6CV/11}{4CV} = \frac{6}{44} = \frac{3}{22}$.

(D) When a body rolls down an inclined plane without slipping,the force of static friction acts on it.
This frictional force provides the necessary torque to initiate and maintain the rotational motion of the drum.
As the drum moves down,the gravitational potential energy is converted into both translational kinetic energy and rotational kinetic energy.
The frictional force acts to convert a portion of the translational energy into rotational energy,ensuring the condition of pure rolling (no slipping) is maintained.
Therefore,the frictional force converts translational energy into rotational energy.

(C) The energy of an incident photon is given by the formula $E = \frac{hc}{\lambda}$.
Using the approximation $hc \approx 12400\, eV\cdot\mathring{A}$,we get:
$E = \frac{12400\, eV\cdot\mathring{A}}{4100\, \mathring{A}} \approx 3.02\, eV$.
Photoelectric emission occurs if the energy of the incident photon is greater than the work function $(\Phi)$ of the metal.
For metal $A$: $\Phi_A = 1.92\, eV$. Since $3.02\, eV > 1.92\, eV$,metal $A$ will emit photoelectrons.
For metal $B$: $\Phi_B = 2.0\, eV$. Since $3.02\, eV > 2.0\, eV$,metal $B$ will emit photoelectrons.
For metal $C$: $\Phi_C = 5.0\, eV$. Since $3.02\, eV < 5.0\, eV$,metal $C$ will not emit photoelectrons.
Therefore,metals $A$ and $B$ will emit photoelectrons.

(A) The phase angle $\phi$ in an $LCR$ series circuit is given by $\tan \phi = \frac{X_C - X_L}{R}$,where the current leads the voltage when $X_C > X_L$.
Given that the current leads the voltage by $45^{\circ}$,we have $\phi = -45^{\circ}$ (or using the magnitude,$\tan 45^{\circ} = \frac{X_C - X_L}{R}$).
Substituting $X_C = \frac{1}{\omega C}$ and $X_L = \omega L$:
$1 = \frac{\frac{1}{\omega C} - \omega L}{R}$
$R = \frac{1}{\omega C} - \omega L$
$\frac{1}{\omega C} = R + \omega L$
$C = \frac{1}{\omega(R + \omega L)}$
Since $\omega = 2 \pi f$,we get:
$C = \frac{1}{2 \pi f(R + 2 \pi f L)}$

(C) Ex-situ conservation refers to off-site conservation.
It is the process of protecting an endangered species of plant or animal by removing it from an unsafe or threatened habitat and placing it or part of it under human care.
Botanical gardens serve as a primary method for the ex-situ conservation of plant germplasm,ensuring the survival of species outside their natural environment.

(C) The class Mammalia is characterized by having $12$ pairs of cranial nerves. The option stating 'Ten pairs of cranial nerves' is incorrect for mammals,as this is a characteristic feature of amphibians and fishes. Therefore,it is not a typical character of mammals.

(B) The fluid mosaic model describes the cell membrane as a quasi-fluid structure where lipids and proteins can move laterally.
$A$, $C$, and $D$ are consistent with the fluid nature and the structural organization of the membrane.
However, $B$ is incorrect because proteins do not undergo flip-flop movements (transverse diffusion) across the lipid bilayer due to their large size and polar nature, which makes it energetically unfavorable to pass through the hydrophobic core of the membrane.

(D) Retting is a process that uses microorganisms and moisture to dissolve or rot away cellular tissues and the pectin surrounding bast-fiber bundles,facilitating the separation of the fiber from the stem. This process is primarily facilitated by anaerobic butyric acid bacteria,such as species of the genus $Clostridium$ (e.g.,$Clostridium$ $butyricum$).

(B) $G-6-P$ (Glucose$-6-$phosphate) dehydrogenase is an enzyme essential for the proper functioning of red blood cells $(RBCs)$.
It plays a critical role in the pentose phosphate pathway,which generates $NADPH$,a molecule that protects $RBCs$ from oxidative damage.
When this enzyme is deficient,$RBCs$ become susceptible to oxidative stress,leading to their premature destruction,a process known as haemolysis.

(A) Cretinism is a condition characterized by severe physical and mental developmental delays,caused by a deficiency of thyroid hormones during fetal development or early infancy. It is not a genetic or hereditary disease.
Cystic fibrosis,Thalassemia,and Haemophilia are all examples of hereditary diseases caused by mutations in specific genes.

(D) Salivary gland chromosomes,also known as polytene chromosomes,are found in the salivary glands of dipteran larvae.
These chromosomes are formed through a process called endoreduplication,where the $DNA$ replicates multiple times without cell division.
This results in giant chromosomes with distinct banding patterns,which are highly useful for gene mapping and identifying chromosomal aberrations.

(D) Gene flow refers to the transfer of genetic material from one population to another through the movement of individuals or their gametes (e.g.,spores,seeds,pollen).
Gene flow is a fundamental process that maintains the integrity of a species by ensuring that different populations share a common gene pool.
Therefore,it is incorrect to state that gene flow does not occur between populations of a species; in fact,gene flow actively occurs and reduces genetic differences between populations.