AIPMT 2008 Chemistry Question Paper with Answer and Solution

(D) The rod is bent at the midpoint into two equal segments,each of length $l = \frac{L}{2}$ and mass $m = \frac{M}{2}$.
Each segment acts as a rod of length $l$ rotating about one of its ends.
The moment of inertia of a rod of mass $m$ and length $l$ about an axis passing through one end and perpendicular to its length is $I = \frac{1}{3}ml^2$.
Since there are two such segments,the total moment of inertia $I_{\text{total}}$ about the axis passing through the bending point $O$ and perpendicular to the plane is the sum of the moments of inertia of the two segments:
$I_{\text{total}} = I_1 + I_2 = \frac{1}{3} \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2 + \frac{1}{3} \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2$
$I_{\text{total}} = 2 \times \left[ \frac{1}{3} \times \frac{M}{2} \times \frac{L^2}{4} \right]$
$I_{\text{total}} = 2 \times \left[ \frac{ML^2}{24} \right] = \frac{ML^2}{12}$

(B) The initial total resistance in the circuit is $R_{total,1} = R_G + R_1 = 50\,\Omega + 2950\,\Omega = 3000\,\Omega$.
Given the battery voltage $\varepsilon = 3\,V$,the initial current $I_1$ is $I_1 = \frac{\varepsilon}{R_{total,1}} = \frac{3\,V}{3000\,\Omega} = 1 \times 10^{-3}\,A = 1\,mA$.
Since $30$ divisions correspond to $1\,mA$,the current per division is $\frac{1\,mA}{30}$.
To get a deflection of $20$ divisions,the new current $I_2$ must be $I_2 = 20 \times \frac{1\,mA}{30} = \frac{2}{3}\,mA$.
Let the new total resistance be $R_{total,2}$. Using Ohm's law,$I_2 = \frac{\varepsilon}{R_{total,2}}$,so $R_{total,2} = \frac{\varepsilon}{I_2} = \frac{3\,V}{(2/3) \times 10^{-3}\,A} = 4500\,\Omega$.
The new total resistance is $R_{total,2} = R_G + R_{new}$,where $R_{new}$ is the new series resistance.
$4500\,\Omega = 50\,\Omega + R_{new} \Rightarrow R_{new} = 4450\,\Omega$.

(A) The body cavity $(coelom)$ of an earthworm is filled with an alkaline,colourless or milky coelomic fluid containing water,salts,some proteins,and four types of coelomic corpuscles.
During burrowing and locomotion,the contraction of septa (which partition the $coelom$ into a series of coelomic chambers) increases the pressure on the coelomic fluid.
This pressure makes the anterior body segment turgid and elongated,allowing it to function as a hydraulic skeleton.

(C) In albuminous seeds (like $Castor$,$Coconut$,and $Maize$),the endosperm is retained in the mature seed. In exalbuminous or non-albuminous seeds (like $Pea$,$Bean$,and $Groundnut$),the endosperm is completely consumed by the developing embryo during seed development. Therefore,the correct answer is $Pea$.

(D) Intercalary meristem is a type of primary meristem that is found between mature tissues. In monocots like sugarcane,these meristems are responsible for the elongation of internodes. The variation in the length of different internodes is due to the differential activity of the intercalary meristem located at the base of each internode.

(B) According to the Histogen theory proposed by Hanstein,the apical meristem consists of three distinct zones: Dermatogen,Periblem,and Plerome.
$1$. Dermatogen gives rise to the epidermis.
$2$. Periblem gives rise to the cortex.
$3$. Plerome gives rise to the central stele,which includes the vascular tissues (xylem and phloem).
Therefore,vascular tissues in flowering plants develop from the plerome.

(B) The vertical conduction of water from root to aerial parts of a plant is called ascent of sap.
The water molecules remain joined to each other due to a force of attraction called cohesion force.
Attraction between water molecules and the walls of $xylem$ is due to adhesion force.
These factors, along with surface tension, help to ensure the continuity of the water column in $xylem$ and prevent it from breaking or rupturing.

(B) Fermentation is an anaerobic metabolic process in which organic substrates,such as glucose,are incompletely oxidized without the involvement of an external electron acceptor.
In this process,pyruvic acid is converted into products like ethanol and carbon dioxide or lactic acid.
The enzymes $Pyruvic$ $acid$ $decarboxylase$ and $Alcohol$ $dehydrogenase$ catalyze these reactions in yeast.

(C) The chemiosmotic hypothesis,proposed by Peter Mitchell,explains that the synthesis of $ATP$ in oxidative phosphorylation is driven by a proton gradient across the inner mitochondrial membrane.
As electrons pass through the electron transport chain,protons $(H^+)$ are pumped from the mitochondrial matrix into the intermembrane space.
This creates a high concentration of protons in the intermembrane space compared to the matrix,resulting in an electrochemical proton gradient.
The flow of protons back into the matrix through the $F_0-F_1$ $ATP$ synthase complex provides the energy required to phosphorylate $ADP$ into $ATP$.

(B) Senescence is a programmed,active developmental process in plants. In the context of the growth and functioning of a flowering plant,leaf abscission is a clear example of senescence where specific cells undergo programmed death to allow the leaf to detach from the plant body.

(B) The phenomenon of photoperiodism was first discovered by Garner and Allard $(1920-1922)$.
They observed that the Maryland Mammoth variety of tobacco could be made to flower only by reducing the light hours with artificial darkening.

(B) The cornea is the transparent front part of the eye that covers the iris,pupil,and anterior chamber.
It is an avascular structure,meaning it lacks a direct blood supply.
Because the immune system's recognition of foreign tissue relies on the circulation of immune cells through the blood,the lack of blood vessels in the cornea prevents the immune system from easily detecting and attacking the transplanted tissue.
Therefore,cornea transplants are rarely rejected.

(C) The pollen wall consists of two layers: the outer $exine$ and the inner $intine$.
The $exine$ is primarily composed of $sporopollenin$, which is derived from the oxidative polymerization of carotenoids.
$Sporopollenin$ is recognized as one of the most chemically and biologically resistant organic materials known to science.
It is highly resistant to degradation by enzymes, strong acids, and alkalis, which allows pollen grains to be well-preserved in fossil deposits.
Therefore, $Pollen \text{ } exine$ is the correct answer.

(A) Unisexuality of flowers means that a flower is either male (staminate) or female (pistillate).
Because a single flower cannot contain both reproductive organs,it prevents autogamy (self-pollination within the same flower).
However,it does not prevent geitonogamy,which is the transfer of pollen grains from the anther to the stigma of another flower on the same plant.

(B) During spermatogenesis,the process begins with spermatogonia $(2n)$.
These undergo mitotic divisions to form primary spermatocytes $(2n)$.
The primary spermatocytes then undergo the first meiotic division (meiosis-$I$),which is a reductional division,resulting in the formation of two haploid $(n)$ cells known as secondary spermatocytes.
Therefore,at the end of the first meiotic division,the cells formed are secondary spermatocytes.

(A) Statement $(1)$ is correct: $MTP$ is considered relatively safe during the first trimester (up to $12$ weeks of pregnancy).
Statement $(2)$ is incorrect: Lactational amenorrhea is effective only up to $6$ months of exclusive breast-feeding,not two years.
Statement $(3)$ is correct: Intrauterine devices (IUDs) like Copper-$T$ are highly effective and widely used contraceptive methods.
Statement $(4)$ is incorrect: Emergency contraceptive pills are typically effective if taken within $72$ hours ($3$ days) after coitus,not one week.
Therefore,statements $(1)$ and $(3)$ are correct.

(D) The correct answer is $D$. The menstrual fluid consists of blood,endometrial tissue,and secretions from the uterine glands. It does not clot easily because the enzyme plasmin (fibrinolysin) is present in the endometrium,which breaks down fibrin clots. Statement $A$ is correct as menopause leads to a sharp rise in gonadotropins ($FSH$ and $LH$) due to the loss of negative feedback from ovarian steroids. Statement $B$ is correct as the first menstruation at puberty is termed menarche. Statement $C$ is correct as the average blood loss during a normal menstrual cycle is approximately $30-50 \; ml$.

(D) Haploid plants can be produced in large numbers through anther and ovary cultures.
Haploids are highly useful for the isolation of mutants because they contain only a single set of chromosomes.
Consequently,even recessive mutant alleles are expressed in the phenotype in the very first generation after mutagen treatment,whereas in diploids,recessive mutations are masked by the dominant allele.

(D) The primary objective of developing herbicide-resistant genetically modified $(GM)$ crops is to allow farmers to spray herbicides to kill weeds without harming the crop itself. This practice helps in managing weeds efficiently and,when managed correctly,aims to reduce the overall herbicide accumulation in food products,thereby ensuring better health safety.

(C) Biodiversity hotspots are regions characterized by high levels of species richness and endemism. Due to the high density of various species living in close proximity and competing for limited resources,interspecific competition is typically very high in these areas. Therefore,'Lesser interspecific competition' is not observed in biodiversity hotspots.

(D) In $C_4$ plants,the initial fixation of $CO_2$ occurs in the mesophyll cells.
$1$. The enzyme $PEP$ carboxylase $(PEPCase)$ combines $CO_2$ with phosphoenolpyruvate $(PEP)$ to form oxaloacetic acid $(OAA)$.
$2$. $OAA$ is then converted into malic acid (or aspartic acid) in the mesophyll cells.
$3$. This malic acid is then transported to the bundle sheath cells for further processing in the Calvin cycle.

(C) $1$. Calculate the percentage of oxygen: $\% O = 100 - (38.71 + 9.67) = 51.62 \%$.
$2$. Calculate the relative number of moles for each element:
$C = \frac{38.71}{12} = 3.226$
$H = \frac{9.67}{1} = 9.67$
$O = \frac{51.62}{16} = 3.226$
$3$. Determine the simplest molar ratio by dividing by the smallest value $(3.226)$:
$C = \frac{3.226}{3.226} = 1$
$H = \frac{9.67}{3.226} \approx 3$
$O = \frac{3.226}{3.226} = 1$
$4$. The empirical formula is $CH_3O$.