AIPMT 2012 Chemistry Question Paper with Answer and Solution

(D) Cyanobacteria are photosynthetic prokaryotes that contain chlorophyll $a$,similar to green plants. Due to their photosynthetic nature and their characteristic blue-green pigmentation,they are commonly known as blue-green algae. They belong to the Kingdom $Monera$.

(B) Viruses are obligate parasites that require a host cell to replicate.
They exhibit various forms of symmetry,including icosahedral and helical,not just helical symmetry.
While they use the host's machinery to synthesize their nucleic acids and proteins,they do not possess their own independent metabolic machinery for these processes.
Antibiotics target bacterial cell walls or metabolic pathways,which are absent in viruses,making them ineffective against viral infections.
Therefore,the statement that 'All of them have helical symmetry' is incorrect.

(D) Heterotrophic bacteria are the most abundant bacteria in nature.
Many of these are helpful to humans in various ways,such as making curd from milk (e.g.,$Lactobacillus$) and the production of antibiotics (e.g.,$Streptomyces$).
They act as decomposers,help in nitrogen fixation in legumes,and are also involved in the production of various industrial products.
Therefore,the correct category is heterotrophic bacteria.

(D) Multicellular fungi,filamentous algae,and the protonema of mosses all share the ability to reproduce or multiply through the process of fragmentation.
In fragmentation,the organism breaks into smaller pieces (fragments),and each fragment develops into a new individual.
Fungi (like molds),filamentous algae (like Spirogyra),and the protonema stage of mosses (which is a filamentous structure) all exhibit this mode of vegetative propagation.

(A) Let us analyze each statement:
$(a)$ $Equisetum$ is a pteridophyte. In pteridophytes,the female gametophyte is retained on the parent sporophyte for variable periods. This statement is correct.
$(b)$ $Ginkgo$ is a gymnosperm. In gymnosperms,the male gametophyte is highly reduced and is not independent; it is confined within the pollen grain. This statement is correct.
$(c)$ $Riccia$ is a liverwort,while $Polytrichum$ is a moss. The sporophyte of $Riccia$ is the simplest (consisting only of a capsule),whereas the sporophyte of $Polytrichum$ is more complex (differentiated into foot,seta,and capsule). Thus,the sporophyte in $Riccia$ is less developed than in $Polytrichum$. This statement is incorrect.
$(d)$ Sexual reproduction in $Volvox$ is oogamous,not isogamous. This statement is incorrect.
$(e)$ The spores of slime molds possess true cell walls (made of cellulose),which makes them extremely resistant to adverse conditions. This statement is incorrect.
Therefore,only statements $(a)$ and $(b)$ are correct. The total number of correct statements is $2$.

(B) In coconut $(Cocos \, nucifera)$, the coconut water is free-nuclear endosperm, and the white kernel (edible part) is the cellular endosperm. Both represent the endosperm tissue, which provides nutrition to the developing embryo. Therefore, they are equivalent to the endosperm.

(B) In the vascular bundles of monocot stems like $Maize$ ($Zea$ $mays$),the protoxylem elements disintegrate during growth to form a water-containing cavity known as the lysigenous cavity or protoxylem lacuna. This is a characteristic feature of monocot stems.

(A) Companion cells are specialized parenchymatous cells that are closely associated with sieve tube elements in the phloem of angiosperms.
They are connected to the sieve tube elements through pit fields present between their common longitudinal walls.
These cells maintain the pressure gradient in the sieve tube elements and assist in the translocation of food materials.
Therefore,the correct option is $A$.

(B) The common bottle cork is derived from the cork cambium,which is scientifically known as the $Phellogen$.
$Phellogen$ is a meristematic tissue that develops in the cortex of stems and roots.
It produces cork $(Phellem)$ on the outer side and secondary cortex $(Phelloderm)$ on the inner side.
Therefore,the cork is a product of the $Phellogen$.

(D) Hormones are classified based on their chemical nature into lipid-soluble and water-soluble hormones.
Lipid-soluble hormones (such as steroid hormones and thyroid hormones) can easily cross the lipid bilayer of the cell membrane due to their hydrophobic nature.
Once inside the cell,they bind to intracellular receptors,which are often located in the nucleus or cytoplasm,to form a hormone-receptor complex that regulates gene expression.
Cortisol is a steroid hormone,and Testosterone is also a steroid hormone.
Both are lipid-soluble and interact with intracellular receptors.
Insulin,Glucagon,Somatostatin,and Oxytocin are peptide hormones,which are water-soluble and bind to membrane-bound receptors on the cell surface.

(B) The power of the lamp is $P = 200\, W$. Since it is $50\%$ efficient,the power converted into light is $P_{\text{light}} = 0.50 \times 200\, W = 100\, W$.
The energy of a single photon is given by $E = \frac{hc}{\lambda}$,where $h = 6.63 \times 10^{-34}\, J\cdot s$,$c = 3 \times 10^8\, m/s$,and $\lambda = 0.6 \times 10^{-6}\, m$.
Let $n$ be the number of photons emitted per second. The total power is $P_{\text{light}} = nE = n \frac{hc}{\lambda}$.
Rearranging for $n$: $n = \frac{P_{\text{light}} \lambda}{hc} = \frac{100 \times 0.6 \times 10^{-6}}{6.63 \times 10^{-34} \times 3 \times 10^8}$.
Using the approximation $hc \approx 20 \times 10^{-26}\, J\cdot m$ (or $1.989 \times 10^{-25}\, J\cdot m$):
$n = \frac{60 \times 10^{-6}}{19.89 \times 10^{-26}} \approx 3.01 \times 10^{20}$.
Thus,the number of photons emitted per second is $3 \times 10^{20}$.

(A) To convert a millivoltmeter into an ammeter,a shunt resistance $R_s$ is connected in parallel with the meter.
Let $V_m$ be the full-scale voltage of the millivoltmeter and $I$ be the desired full-scale current of the ammeter.
The voltage across the millivoltmeter is $V_m = 25 \times 10^{-3} \, V$.
The current through the shunt is approximately equal to the total current $I = 25 \, A$ because the resistance of the millivoltmeter is very high compared to the shunt.
The voltage across the shunt must equal the voltage across the millivoltmeter:
$V_m = I \times R_s$
$25 \times 10^{-3} = 25 \times R_s$
$R_s = \frac{25 \times 10^{-3}}{25} = 10^{-3} \, \Omega = 0.001 \, \Omega$.

(C) The molecular orbital configuration of $O_2$ ($16$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, (\pi 2p_x^2 = \pi 2p_y^2), (\pi^* 2p_x^1 = \pi^* 2p_y^1)$.
When $O_2$ changes to $O_2^-$ ($17$ electrons),the additional electron enters the $\pi^* 2p$ orbital.

(D) The output voltage $V_{out}$ in a $CE$ transistor amplifier is given by the product of the output current and the load resistance.
$V_{out} = I_c \times R_L$
Given $V_{out} = 2\, V$ and $R_L = 2\, k\Omega = 2000\, \Omega$,the collector current is:
$I_c = \frac{V_{out}}{R_L} = \frac{2}{2000} = 10^{-3}\, A = 1\, mA$.
We know the current amplification factor $\beta = \frac{I_c}{I_b}$. Thus,the base current is:
$I_b = \frac{I_c}{\beta} = \frac{1\, mA}{100} = 0.01\, mA = 10\, \mu A$.
The input signal voltage $V_{in}$ is related to the base current and base resistance $R_i$ by Ohm's law:
$V_{in} = I_b \times R_i$
Given $R_i = 1\, k\Omega = 1000\, \Omega$:
$V_{in} = (10 \times 10^{-6}\, A) \times (1000\, \Omega) = 0.01\, V = 10\, mV$.

(C) The given reaction is the Rosenmund reduction.
In this reaction,acid chlorides $(RCOCl)$ are partially reduced to aldehydes $(RCHO)$ using hydrogen gas $(H_2)$ in the presence of a palladium catalyst supported on barium sulfate $(Pd-BaSO_4)$.
For the reactant benzoyl chloride $(C_6H_5COCl)$,the product $'A'$ is benzaldehyde $(C_6H_5CHO)$.

(B) The position of the particle is given by $x = 8 + 12t - t^3$.
Velocity $v$ is the rate of change of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(8 + 12t - t^3) = 12 - 3t^2$.
Set the velocity to zero to find the time: $12 - 3t^2 = 0 \implies 3t^2 = 12 \implies t^2 = 4 \implies t = 2 \ s$ (since $t > 0$).
Acceleration $a$ is the rate of change of velocity: $a = \frac{dv}{dt} = \frac{d}{dt}(12 - 3t^2) = -6t$.
At $t = 2 \ s$,the acceleration is $a = -6(2) = -12 \ m s^{-2}$.
Retardation is the magnitude of negative acceleration,which is $12 \ m s^{-2}$.

(D) Sucrose is a disaccharide formed by the condensation of two monosaccharides.
These two monosaccharides are $\alpha-D$-glucopyranose and $\beta-D$-fructofuranose.
They are held together by a glycosidic linkage between $C-1$ of $\alpha-D$-glucopyranose and $C-2$ of $\beta-D$-fructofuranose.

(B) The given wave equations are $y_1 = 4 \sin(600\pi t)$ and $y_2 = 5 \sin(608\pi t)$.
Comparing these with the standard form $y = A \sin(2\pi f t)$,we get:
$2\pi f_1 = 600\pi \implies f_1 = 300 \text{ Hz}$.
$2\pi f_2 = 608\pi \implies f_2 = 304 \text{ Hz}$.
The beat frequency is $f_{beat} = |f_2 - f_1| = |304 - 300| = 4 \text{ beats per second}$.
The maximum amplitude is $A_{max} = A_1 + A_2 = 4 + 5 = 9$.
The minimum amplitude is $A_{min} = |A_2 - A_1| = |5 - 4| = 1$.
The intensity ratio is given by $\frac{I_{max}}{I_{min}} = \frac{A_{max}^2}{A_{min}^2} = \frac{9^2}{1^2} = \frac{81}{1}$.

(A) The option $A$ does not correctly represent the order of the property indicated against it.
The correct order of the melting point for the $3d$ series elements is $Ti < V < Cr > Mn$.
Along the $3d$ series,the melting point increases up to $Cr$ and then decreases.
This is because the number of unpaired electrons increases up to $Cr$ $(3d^5 4s^1)$ and then decreases as pairing begins.
$Cr$ has the maximum number of unpaired electrons $(6)$,which leads to stronger metallic bonding and a higher melting point compared to $Mn$ $(3d^5 4s^2)$,which has a stable half-filled $d$-subshell and weaker metallic bonding.

(C) The molecular orbital configuration of $O_2$ ($16$ electrons) is: $\sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2pz}^2, \pi_{2px}^2 = \pi_{2py}^2, \pi_{2px}^{*1} = \pi_{2py}^{*1}$.
When $O_2$ changes to $O_2^-$ ($17$ electrons),the additional electron enters the next available orbital,which is the $\pi^*$ antibonding molecular orbital ($\pi_{2px}^*$ or $\pi_{2py}^*$).

(D) The structure of hypophosphorous acid $(H_3PO_2)$ contains one $P=O$ bond,two $P-H$ bonds,and one $P-OH$ bond.
Since only one hydrogen atom is attached to the oxygen atom,it is a monobasic (monoprotic) acid.
Therefore,the statement that hypophosphorous acid is a diprotic acid is incorrect.

(A) The option $A$ does not correctly represent the order of the property indicated against it.
The correct order of the melting point for $3d$ series elements is $Ti < V < Cr > Mn$.
Along the $3d$ series,the melting point increases up to $Cr$ ($d^5$ configuration) due to the maximum number of unpaired electrons participating in metallic bonding and then decreases.
Therefore,$Mn$ has a lower melting point than $Cr$.

(C) Optical isomerism requires the presence of an asymmetric (chiral) carbon atom,which is a carbon atom bonded to four different groups.
$1$. Lactic acid $(CH_3CH(OH)COOH)$ has a chiral carbon.
$2$. Tartaric acid $(HOOC-CH(OH)-CH(OH)-COOH)$ has chiral carbons.
$3$. $\alpha -$ amino acids $(R-CH(NH_2)-COOH)$ have a chiral carbon (except glycine).
$4$. Maleic acid $(HOOC-CH=CH-COOH)$ is a geometric isomer (cis-isomer) and does not contain any chiral carbon atom.
Therefore,maleic acid does not exhibit optical isomerism.

(A) To determine the nature of the complexes,we analyze their electronic configurations and hybridization:
$1$. $[Ni(NH_3)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. It forms an outer orbital complex ($sp^3d^2$ hybridization) with $2$ unpaired electrons,making it paramagnetic.
$2$. $[Zn(NH_3)_6]^{2+}$: $Zn^{2+}$ is $3d^{10}$. It is diamagnetic as there are no unpaired electrons.
$3$. $[Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ is $3d^3$. It forms an inner orbital complex ($d^2sp^3$ hybridization) and is paramagnetic,but it is not an outer orbital complex.
$4$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. In the presence of strong field ligand $NH_3$,it forms an inner orbital complex ($d^2sp^3$ hybridization) and is diamagnetic.
Therefore,$[Ni(NH_3)_6]^{2+}$ is the correct answer.

(A) $1$. Melting point: The melting points of $3d$ transition elements do not follow a simple increasing trend. $Mn$ has an anomalously low melting point due to its stable $d^5$ configuration,which leads to weak metallic bonding. Thus,the order $Ti < V < Cr < Mn$ is incorrect.
$2$. $2^{nd}$ Ionization Enthalpy: The values generally increase across the period,but $Mn$ has a higher $2^{nd}$ ionization enthalpy than $Cr$ because $Cr^+$ has a $d^5$ configuration,making it relatively stable,while $Mn^+$ has a $d^5s^1$ configuration. The order $Ti < V < Mn < Cr$ is incorrect as $Cr$ should be higher than $Mn$.
$3$. Oxidation states: The number of oxidation states increases from $Ti$ to $Mn$ as the number of unpaired electrons increases. This is correct.
$4$. Magnetic moment: Magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$. For $Ti^{+3} (d^1, n=1)$,$V^{+3} (d^2, n=2)$,$Cr^{+3} (d^3, n=3)$,and $Mn^{+3} (d^4, n=4)$,the magnetic moment increases as $n$ increases. This is correct.
Note: Both $A$ and $B$ are technically incorrect representations. However,in standard competitive chemistry contexts,the anomaly of $Mn$ in melting point is a primary focus.

(A) The circuit consists of a $10 \, V$ voltage source connected in parallel to a resistor $R$ and a $5 \, \Omega$ resistor.
Since the resistors are in parallel,the voltage across each resistor is equal to the source voltage,$V = 10 \, V$.
The power dissipated in the $5 \, \Omega$ resistor is given by $P_1 = \frac{V^2}{R_1} = \frac{10^2}{5} = \frac{100}{5} = 20 \, W$.
The total power dissipated in the circuit is $P_{total} = 30 \, W$.
Therefore,the power dissipated in resistor $R$ is $P_R = P_{total} - P_1 = 30 - 20 = 10 \, W$.
Using the formula for power $P_R = \frac{V^2}{R}$,we have $10 = \frac{10^2}{R}$.
Solving for $R$,we get $R = \frac{100}{10} = 10 \, \Omega$.

(A) The weight of a body at height $h$ is given by $W_h = m g_h$ and at the surface is $W = m g$.
Given that $W_h = \frac{1}{16} W$,it follows that $g_h = \frac{1}{16} g$.
The formula for acceleration due to gravity at height $h$ is $g_h = g \left( \frac{R}{R+h} \right)^2$.
Substituting the given values: $\frac{1}{16} g = g \left( \frac{R}{R+h} \right)^2$.
Taking the square root on both sides: $\frac{1}{4} = \frac{R}{R+h}$.
Cross-multiplying gives $R + h = 4R$.
Therefore,$h = 3R$.

(D) The voltage gain $A_v$ of a $CE$ transistor amplifier is given by the product of current gain $\beta$ and the resistance gain $\frac{R_L}{R_i}$.
$A_v = \beta \times \frac{R_L}{R_i}$
Given: $\beta = 100$,$R_L = 2\,k\Omega$,$R_i = 1\,k\Omega$,and $V_{out} = 2\,V$.
The relationship between input voltage $V_{in}$ and output voltage $V_{out}$ is $V_{out} = A_v \times V_{in}$.
Substituting the values: $2 = (100 \times \frac{2\,k\Omega}{1\,k\Omega}) \times V_{in}$.
$2 = 100 \times 2 \times V_{in}$.
$2 = 200 \times V_{in}$.
$V_{in} = \frac{2}{200} = 0.01\,V$.
Converting to millivolts: $V_{in} = 0.01 \times 1000\,mV = 10\,mV$.

(C) The standard equation for a plane electromagnetic wave is $\vec E = E_0 \cos(kz - \omega t)$.
Comparing this with the given equation $\vec E = \hat i\,40\, \cos(kz - 6 \times 10^8 t)$,we identify the angular frequency $\omega = 6 \times 10^8 \, rad/s$.
In a vacuum,the speed of light $c$ is related to the angular frequency $\omega$ and wave vector $k$ by the relation $c = \frac{\omega}{k}$.
Given $c = 3 \times 10^8 \, m/s$,we can solve for $k$:
$k = \frac{\omega}{c} = \frac{6 \times 10^8}{3 \times 10^8} = 2 \, m^{-1}$.
Thus,the value of the wave vector $k$ is $2 \, m^{-1}$.

(B) The spectral series of a hydrogen-like atom are categorized based on the final energy level $n_f$ of the electron transition.
$1$. The Lyman series $(n_f = 1)$ results in ultraviolet $(U.V.)$ radiation.
$2$. The Balmer series $(n_f = 2)$ results in visible radiation.
$3$. The Paschen series $(n_f = 3)$ results in infrared $(I.R.)$ radiation.
Since the transition $n = 4 \to n = 3$ ends at the third energy level $(n_f = 3)$,it belongs to the Paschen series,which produces infrared radiation.
Therefore,the correct option is $4 \to 3$.

(D) Let $N_0$ be the initial number of nuclei.
At time $t_1$,the amount decayed is $\frac{1}{3}N_0$,so the remaining amount is $N(t_1) = N_0 - \frac{1}{3}N_0 = \frac{2}{3}N_0$.
Using the decay law $N(t) = N_0 (\frac{1}{2})^{t/T_{1/2}}$,we have $\frac{2}{3}N_0 = N_0 (\frac{1}{2})^{t_1/50}$,which implies $(\frac{1}{2})^{t_1/50} = \frac{2}{3}$.
At time $t_2$,the amount decayed is $\frac{2}{3}N_0$,so the remaining amount is $N(t_2) = N_0 - \frac{2}{3}N_0 = \frac{1}{3}N_0$.
Using the decay law,$\frac{1}{3}N_0 = N_0 (\frac{1}{2})^{t_2/50}$,which implies $(\frac{1}{2})^{t_2/50} = \frac{1}{3}$.
Dividing the two equations: $\frac{(\frac{1}{2})^{t_1/50}}{(\frac{1}{2})^{t_2/50}} = \frac{2/3}{1/3} = 2$.
This simplifies to $(\frac{1}{2})^{(t_1-t_2)/50} = 2^1$,or $2^{(t_2-t_1)/50} = 2^1$.
Therefore,$\frac{t_2 - t_1}{50} = 1$,which gives $t_2 - t_1 = 50 \ days$.

(C) The frequency of the echo detected by the driver of the train is calculated using the Doppler effect formula for a moving source and a moving observer.
Here,the train acts as both the source and the observer.
Let $v = 330\, ms^{-1}$ be the speed of sound and $u = 220\, ms^{-1}$ be the speed of the train.
When the sound travels from the train to the stationary object,the object receives a frequency $f_1 = f \left( \frac{v}{v-u} \right)$.
This object then reflects the sound,acting as a stationary source,and the train (moving observer) receives the echo with frequency $f' = f_1 \left( \frac{v+u}{v} \right)$.
Substituting $f_1$ into the equation for $f'$,we get:
$f' = f \left( \frac{v}{v-u} \right) \left( \frac{v+u}{v} \right) = f \left( \frac{v+u}{v-u} \right)$.
Substituting the given values:
$f' = 1000 \left( \frac{330+220}{330-220} \right) = 1000 \left( \frac{550}{110} \right) = 1000 \times 5 = 5000\, Hz$.

(D) Heterotrophic bacteria are the most abundant bacteria in nature.
They play a significant role in human life,such as helping in the production of curd from milk (e.g.,$Lactobacillus$) and the production of antibiotics (e.g.,$Streptomyces$).
Therefore,the correct category is heterotrophic bacteria.

(D) Cartilage is a specialized type of connective tissue. It is flexible and provides structural support to various body parts. In humans,it is found in the external ears (pinna),the tip of the nose,the joints between adjacent bones of the vertebral column,and in the limbs and hands in adults.

(D) In biological reproduction,different organisms use specific vegetative or asexual structures for propagation.
$1$. Ginger reproduces vegetatively through rhizomes,not suckers.
$2$. Chlamydomonas produces zoospores (motile spores) for asexual reproduction,not conidia.
$3$. Yeast reproduces primarily by budding,not zoospores.
$4$. Onion reproduces vegetatively through a bulb,which is a modified underground stem. Therefore,this is the correctly matched pair.

(C) The Test-tube Baby Programme refers to the process of In Vitro Fertilization $(IVF)$,where fertilization occurs outside the body in a laboratory setting.
Following fertilization,the zygote or early embryo (up to $8$ blastomeres) is transferred into the fallopian tube,a technique known as Zygote Intra Fallopian Transfer $(ZIFT)$.
Therefore,the $ZIFT$ technique is specifically employed in the Test-tube Baby Programme to assist in conception.

(C) Bread is produced through the fermentation process by $Saccharomyces$ $cerevisiae$,commonly known as baker's yeast.
Similarly,$Saccharomyces$ $cerevisiae$ is also used in the production of alcoholic beverages like beer and wine,where it is referred to as brewer's yeast.
Therefore,yeast is essential for both bread and beer production.

(C) Golden rice,a variety of $Oryza$ $sativa$,is produced through the genetic engineering of the biosynthesis of beta-carotene,which is a precursor of provitamin-$A$ in the edible parts of the rice grain.
Research leading to the development of golden rice was conducted with the primary goal of helping children in developing countries who suffer from vitamin-$A$ deficiency and associated blindness.
Golden rice has been engineered to be nutritionally enhanced,providing a sustainable solution to combat vitamin-$A$ deficiency in populations dependent on rice as a staple food.

(B) The functional units of an ecosystem include productivity,decomposition,energy flow,and nutrient cycling.
Vertical distribution of different species occupying different levels is called stratification.
For example,in a forest ecosystem,trees occupy the top vertical strata,shrubs occupy the second,and herbs/grasses occupy the bottom layer.
Therefore,stratification is a structural component of an ecosystem,not a functional unit.

(B) The domestic sewage in large cities undergoes secondary treatment in Sewage Treatment Plants $(STPs)$.
In this process,the primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it.
This allows vigorous growth of useful aerobic microbes into flocs,which consume the major part of the organic matter in the effluent,significantly reducing the $BOD$.
Once the $BOD$ of sewage is reduced,the effluent is passed into a settling tank where the bacterial flocs are allowed to sediment.
This sediment is called activated sludge.
$A$ small part of the activated sludge is pumped back into the aeration tank to serve as the inoculum,while the remaining major part is pumped into large tanks called anaerobic sludge digesters.
In these digesters,other kinds of bacteria,which grow anaerobically,digest the bacteria and the fungi in the sludge,producing gases such as methane,hydrogen sulfide,and carbon dioxide.

(B) According to the Stefan-Boltzmann law,the power radiated by a black body is given by $H = \sigma A T^4$,where $A$ is the surface area of the body.
For a star of radius $R$,the surface area is $A = 4 \pi R^2$.
Given that the rate of energy production is $Q$,we have $Q = \sigma (4 \pi R^2) T^4$.
Rearranging the equation to solve for the temperature $T$:
$T^4 = \frac{Q}{4 \pi R^2 \sigma}$.
Taking the fourth root on both sides:
$T = \left( \frac{Q}{4 \pi R^2 \sigma} \right)^{1/4}$.

(D) $K_2Cr_2O_7$ is preferred over $Na_2Cr_2O_7$ in volumetric analysis because $Na_2Cr_2O_7$ is hygroscopic in nature,meaning it absorbs moisture from the atmosphere,which makes it difficult to prepare a standard solution of known concentration. Therefore,the statement that $Na_2Cr_2O_7$ is preferred is incorrect.

(B) The spectral series of a hydrogen-like atom are classified based on the final energy level $n_f$ of the electron transition.
$1$. Lyman series $(n_f = 1)$: Produces ultraviolet $(U.V.)$ radiation.
$2$. Balmer series $(n_f = 2)$: Produces visible radiation.
$3$. Paschen series $(n_f = 3)$: Produces infrared $(I.R.)$ radiation.
Since the transition $n=3$ to $n=1$ results in $U.V.$ radiation (Lyman series),we look for a transition that results in infrared radiation.
The Paschen series corresponds to transitions ending at $n_f = 3$. Therefore,the transition $4 \longrightarrow 3$ belongs to the Paschen series and results in infrared radiation.

(D) $Monascus purpureus$ is a species of yeast that is commercially used in the production of statins.
Statins are bioactive molecules that act as competitive inhibitors of the enzyme responsible for the synthesis of cholesterol.
These are used as blood-cholesterol-lowering agents in patients.
Therefore, the correct option is $D$.