ChemistryQ51–68 of 98 questions
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51
ChemistryMCQAIPMT · 2006
The Earth is assumed to be a sphere of radius $R$. $A$ platform is arranged at a height $R$ from the surface of the Earth. The escape velocity of a body from this platform is $fv$,where $v$ is the escape velocity from the surface of the Earth. The value of $f$ is
A
$\sqrt{2}$
B
$\frac{1}{\sqrt{2}}$
C
$\frac{1}{3}$
D
$\frac{1}{2}$
Solution
(B) The escape velocity $v$ from the surface of the Earth is given by $v = \sqrt{\frac{2GM}{R}}$.
Let the escape velocity from a height $h = R$ be $v'$. According to the principle of conservation of energy,the total energy at the platform must be zero for the body to escape to infinity.
$-\frac{GMm}{R+R} + \frac{1}{2}m(v')^2 = 0$
$-\frac{GMm}{2R} + \frac{1}{2}m(v')^2 = 0$
$(v')^2 = \frac{GM}{R} \Rightarrow v' = \sqrt{\frac{GM}{R}}$.
Given that $v' = fv$,we have $f \sqrt{\frac{2GM}{R}} = \sqrt{\frac{GM}{R}}$.
$f = \frac{\sqrt{GM/R}}{\sqrt{2GM/R}} = \frac{1}{\sqrt{2}}$.
Let the escape velocity from a height $h = R$ be $v'$. According to the principle of conservation of energy,the total energy at the platform must be zero for the body to escape to infinity.
$-\frac{GMm}{R+R} + \frac{1}{2}m(v')^2 = 0$
$-\frac{GMm}{2R} + \frac{1}{2}m(v')^2 = 0$
$(v')^2 = \frac{GM}{R} \Rightarrow v' = \sqrt{\frac{GM}{R}}$.
Given that $v' = fv$,we have $f \sqrt{\frac{2GM}{R}} = \sqrt{\frac{GM}{R}}$.
$f = \frac{\sqrt{GM/R}}{\sqrt{2GM/R}} = \frac{1}{\sqrt{2}}$.
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52
ChemistryMCQAIPMT · 2006
Power dissipated across the $8\,\Omega$ resistor in the circuit shown here is $2\,W$. The power dissipated in watt units across the $3\,\Omega$ resistor is
A
$2$
B
$1$
C
$0.5$
D
$3$
Solution
(D) The $8\,\Omega$ resistor is in parallel with the series combination of $1\,\Omega$ and $3\,\Omega$ resistors.
Let $V$ be the potential difference across the parallel branches.
Given,power dissipated in $8\,\Omega$ resistor is $P_8 = 2\,W$.
Using $P = \frac{V^2}{R}$,we have $2 = \frac{V^2}{8}$,which gives $V^2 = 16$,so $V = 4\,V$.
Since the $1\,\Omega$ and $3\,\Omega$ resistors are in series,the current $I$ flowing through them is $I = \frac{V}{R_{eq}} = \frac{4}{1+3} = \frac{4}{4} = 1\,A$.
The power dissipated in the $3\,\Omega$ resistor is $P_3 = I^2 R = (1)^2 \times 3 = 3\,W$.
Let $V$ be the potential difference across the parallel branches.
Given,power dissipated in $8\,\Omega$ resistor is $P_8 = 2\,W$.
Using $P = \frac{V^2}{R}$,we have $2 = \frac{V^2}{8}$,which gives $V^2 = 16$,so $V = 4\,V$.
Since the $1\,\Omega$ and $3\,\Omega$ resistors are in series,the current $I$ flowing through them is $I = \frac{V}{R_{eq}} = \frac{4}{1+3} = \frac{4}{4} = 1\,A$.
The power dissipated in the $3\,\Omega$ resistor is $P_3 = I^2 R = (1)^2 \times 3 = 3\,W$.
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53
ChemistryMCQAIPMT · 2006
Conifers differ from grasses in the
A
Production of seeds from ovules
B
Lack of xylem tracheids
C
Absence of pollen tubes
D
Formation of endosperm before fertilization
Solution
(D) Conifers belong to the group $Gymnosperms$,while grasses belong to the group $Angiosperms$.
In $Gymnosperms$,the endosperm is formed before fertilization as a result of the development of the female gametophyte (haploid tissue).
In $Angiosperms$ (like grasses),the endosperm is formed after fertilization through the process of double fertilization (specifically,triple fusion involving the fusion of one male gamete with two polar nuclei to form a triploid primary endosperm nucleus).
Therefore,the formation of endosperm before fertilization is a characteristic feature that distinguishes $Gymnosperms$ (conifers) from $Angiosperms$ (grasses).
In $Gymnosperms$,the endosperm is formed before fertilization as a result of the development of the female gametophyte (haploid tissue).
In $Angiosperms$ (like grasses),the endosperm is formed after fertilization through the process of double fertilization (specifically,triple fusion involving the fusion of one male gamete with two polar nuclei to form a triploid primary endosperm nucleus).
Therefore,the formation of endosperm before fertilization is a characteristic feature that distinguishes $Gymnosperms$ (conifers) from $Angiosperms$ (grasses).
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54
ChemistryMCQAIPMT · 2006
Which one of the following is a matching set of phylum and its three examples?
A
Cnidaria - Bonellia,Physalia,Aurelia
B
Platyhelminthes - Planaria,Schistosoma,Enterobius
C
Mollusca - Loligo,Teredo,Octopus
D
Porifera - Spongilla,Euplectella,Pennatula
Solution
(C) The correct matching set is $C$.
$1$. $Loligo$ (Squid),$Teredo$ (Shipworm),and $Octopus$ (Devil fish) are all members of the phylum $Mollusca$.
$2$. In option $A$,$Bonellia$ belongs to $Annelida$,while $Physalia$ and $Aurelia$ belong to $Cnidaria$.
$3$. In option $B$,$Enterobius$ (Pinworm) belongs to $Aschelminthes$,not $Platyhelminthes$.
$4$. In option $D$,$Pennatula$ (Sea pen) belongs to $Cnidaria$,not $Porifera$.
$1$. $Loligo$ (Squid),$Teredo$ (Shipworm),and $Octopus$ (Devil fish) are all members of the phylum $Mollusca$.
$2$. In option $A$,$Bonellia$ belongs to $Annelida$,while $Physalia$ and $Aurelia$ belong to $Cnidaria$.
$3$. In option $B$,$Enterobius$ (Pinworm) belongs to $Aschelminthes$,not $Platyhelminthes$.
$4$. In option $D$,$Pennatula$ (Sea pen) belongs to $Cnidaria$,not $Porifera$.
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55
ChemistryMCQAIPMT · 2006
Sulphur is an important nutrient for optimum growth and productivity in
A
Pulse crops
B
Cereals
C
Fibre crops
D
Oilseed crops
Solution
(D) Sulphur is an essential component of amino acids like cysteine and methionine,which are crucial for protein synthesis. Oilseed crops (such as mustard,groundnut,and sunflower) require high amounts of sulphur because they synthesize sulphur-containing compounds like glucosinolates and oils,which are vital for their growth and productivity. While pulses are rich in protein,oilseeds have a higher physiological requirement for sulphur to maximize their oil content and yield.
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56
ChemistryMCQAIPMT · 2006
Farmers in a particular region were concerned that premature yellowing of leaves of a pulse crop might cause a decrease in the yield. Which treatment could be most beneficial to obtain maximum seed yield?
A
Frequent irrigation of the crop
B
Treatment of the plants with cytokinins along with a small dose of nitrogenous fertilizer
C
Removal of all yellow leaves and spraying the remaining green leaves with $2,4,5$-trichlorophenoxy acetic acid
D
Application of iron and magnesium to promote synthesis of chlorophyll
Solution
(D) Premature yellowing of leaves (chlorosis) in plants is often caused by the deficiency of essential mineral elements like magnesium $(Mg)$ and iron $(Fe)$,which are critical components for chlorophyll synthesis.
Since chlorophyll is essential for photosynthesis,its deficiency leads to yellowing and reduced crop yield.
Therefore,the application of iron and magnesium is the most direct and beneficial treatment to promote chlorophyll synthesis,restore leaf health,and maximize seed yield in pulse crops.
Since chlorophyll is essential for photosynthesis,its deficiency leads to yellowing and reduced crop yield.
Therefore,the application of iron and magnesium is the most direct and beneficial treatment to promote chlorophyll synthesis,restore leaf health,and maximize seed yield in pulse crops.
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57
ChemistryMCQAIPMT · 2006
An enzyme that can stimulate germination of barley seeds is
A
$\alpha$-amylase
B
Lipase
C
Protease
D
Invertase
Solution
(A) Barley seeds contain a significant amount of starch as a stored food reserve. During the germination process,the hormone gibberellin triggers the synthesis of the enzyme $\alpha$-amylase in the aleurone layer. This enzyme then hydrolyzes the stored starch into simpler sugars like glucose and maltose,which provide the necessary energy for the developing embryo. Therefore,$\alpha$-amylase is the key enzyme that stimulates the germination of barley seeds.
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58
ChemistryMCQAIPMT · 2006
How does pruning help in making the hedge dense?
A
It induces the differentiation of new shoots from the rootstock
B
It frees axillary buds from apical dominance
C
The apical shoot grows faster after pruning
D
It releases wound hormone
Solution
(B) Pruning helps in making the hedge dense because it frees the axillary buds from apical dominance.
In plants,the shoot apex contains the highest concentration of the hormone auxin,which suppresses the growth of lateral (axillary) buds while promoting the growth of the apical bud.
When the shoot apex is removed through pruning,the source of apical dominance is eliminated.
As a result,the axillary buds are no longer suppressed and begin to grow,which leads to the development of many lateral branches,making the hedge dense.
In plants,the shoot apex contains the highest concentration of the hormone auxin,which suppresses the growth of lateral (axillary) buds while promoting the growth of the apical bud.
When the shoot apex is removed through pruning,the source of apical dominance is eliminated.
As a result,the axillary buds are no longer suppressed and begin to grow,which leads to the development of many lateral branches,making the hedge dense.
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59
ChemistryMCQAIPMT · 2006
Which one of the following statements is incorrect?
A
The residual air in lungs slightly decreases the efficiency of respiration in mammals
B
The presence of non-respiratory air sacs,increases the efficiency of respiration in birds
C
In insects,circulating body fluids serve to efficiently distribute oxygen to tissues
D
The principle of countercurrent flow facilitates efficient respiration in gills of fishes
Solution
(C) The correct answer is $C$. In insects,the respiratory system consists of a network of tubes called tracheae that open to the outside through spiracles. Oxygen is delivered directly to the tissues via the tracheal system,and the circulating body fluid (hemolymph) does not play a role in oxygen transport. Therefore,the statement that circulating body fluids serve to distribute oxygen in insects is incorrect. Residual air (about $1200\; mL$) remains in the lungs after forceful expiration to prevent lung collapse,but it does not decrease respiratory efficiency. Bird air sacs and fish countercurrent flow are both adaptations that enhance respiratory efficiency.
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60
ChemistryMCQAIPMT · 2006
$A$ steroid hormone which regulates glucose metabolism is
A
Cortisol
B
Corticosterone
C
$11$-deoxycorticosterone
D
Cortisone
Solution
(A) $Cortisol$,also known as hydrocortisone,is the primary glucocorticoid hormone in humans and many other mammals.
It is secreted by the $zona$ $fasciculata$ layer of the adrenal cortex.
This hormone plays a crucial role in regulating glucose metabolism by promoting gluconeogenesis,lipolysis,and proteolysis,especially during periods of fasting or stress,thereby increasing blood glucose levels.
It is secreted by the $zona$ $fasciculata$ layer of the adrenal cortex.
This hormone plays a crucial role in regulating glucose metabolism by promoting gluconeogenesis,lipolysis,and proteolysis,especially during periods of fasting or stress,thereby increasing blood glucose levels.
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61
ChemistryMCQAIPMT · 2006
Which one of the following is not a second messenger in hormone action?
A
Calcium
B
Sodium
C
cAMP
D
cGMP
Solution
(B) Hormones that do not enter the target cell interact with surface receptors to generate second messengers. Common second messengers include $cAMP$,$cGMP$,$IP_3$ (inositol triphosphate),$DAG$ (diacylglycerol),and $Ca^{2+}$ ions. Sodium $(Na^+)$ ions are not involved as second messengers in hormone signaling pathways.
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62
ChemistryMCQAIPMT · 2006
Sertoli cells are regulated by the pituitary hormone known as:
A
$FSH$
B
$GH$
C
$Prolactin$
D
$LH$
Solution
(A) Sertoli cells are regulated by $FSH$ (Follicle Stimulating Hormone).
$FSH$ acts on the Sertoli cells and stimulates the secretion of some factors which help in the process of spermiogenesis.
$FSH$ acts on the Sertoli cells and stimulates the secretion of some factors which help in the process of spermiogenesis.
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63
ChemistryMCQAIPMT · 2006
An important evidence in favour of organic evolution is the occurrence of
A
Homologous and vestigial organs
B
Analogous and vestigial organs
C
Homologous organs only
D
Homologous and analogous organs
Solution
(A) The presence of homologous and vestigial organs serves as significant evidence for organic evolution.
Homologous organs indicate common ancestry and divergent evolution,as they share a similar structural plan despite performing different functions.
Vestigial organs are remnants of structures that were functional in ancestors but have reduced or lost their function in the course of evolution,providing clear evidence of descent with modification.
Homologous organs indicate common ancestry and divergent evolution,as they share a similar structural plan despite performing different functions.
Vestigial organs are remnants of structures that were functional in ancestors but have reduced or lost their function in the course of evolution,providing clear evidence of descent with modification.
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64
ChemistryMCQAIPMT · 2006
In maize,hybrid vigour is exploited by
A
Bombarding the seeds with $DNA$
B
Crossing of two inbred parental lines
C
Harvesting seeds from the most productive plants
D
Inducting mutations
Solution
(B) Hybrid vigour,also known as heterosis,refers to the phenomenon where the offspring of two genetically distinct parents exhibit superior traits compared to both parents. In maize,this is commercially exploited by crossing two genetically different inbred parental lines to produce high-yielding hybrid seeds.
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65
ChemistryMCQAIPMT · 2006
Niche overlap indicates
A
Active cooperation between two species
B
Two different parasites on the same host
C
Sharing of one or more resources between the two species
D
Mutualism between two species
Solution
(C) Niche overlap refers to the situation where two or more species utilize the same resources or occupy similar habitats within an ecosystem.
It indicates that the species have similar ecological requirements.
When resources such as food or space are limiting,niche overlap often leads to interspecific competition.
For example,two different species of parasites competing for the same host represent a form of niche overlap.
It indicates that the species have similar ecological requirements.
When resources such as food or space are limiting,niche overlap often leads to interspecific competition.
For example,two different species of parasites competing for the same host represent a form of niche overlap.
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66
ChemistryMCQAIPMT · 2006
Which one of the following is not used for the construction of ecological pyramids?
A
Dry weight
B
Number of individuals
C
Rate of energy flow
D
Fresh weight
Solution
(D) Ecological pyramids are graphical representations of the trophic structure and function at successive trophic levels.
Ecological pyramids are generally of three types:
$(i)$ Pyramid of number: Represents the number of organisms at each trophic level.
$(ii)$ Pyramid of biomass: Represents the total dry weight of living organisms at each trophic level.
$(iii)$ Pyramid of energy: Represents the rate of energy flow or productivity at successive trophic levels.
Fresh weight is not used because it includes water content,which varies significantly and does not provide a reliable measure of the organic matter or energy available at a trophic level. Therefore,fresh weight is not used for the construction of ecological pyramids.
Ecological pyramids are generally of three types:
$(i)$ Pyramid of number: Represents the number of organisms at each trophic level.
$(ii)$ Pyramid of biomass: Represents the total dry weight of living organisms at each trophic level.
$(iii)$ Pyramid of energy: Represents the rate of energy flow or productivity at successive trophic levels.
Fresh weight is not used because it includes water content,which varies significantly and does not provide a reliable measure of the organic matter or energy available at a trophic level. Therefore,fresh weight is not used for the construction of ecological pyramids.
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67
ChemistryMCQAIPMT · 2006
Which of the following pairs of an animal and a plant represents endangered organisms in India?
A
Bentinckia nicobarica and red panda
B
Tamarind and rhesus monkey
C
Cinchona and leopard
D
Banyan and black buck
Solution
(A) The plant $Bentinckia$ $\text{nicobarica}$ (a member of the family $Arecaceae$) and the animal, red panda $(Ailurus$ $\text{fulgens})$, are both classified as endangered species in India.
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68
ChemistryMCQAIPMT · 2006
Photochemical smog pollution does not contain
A
Ozone
B
Nitrogen dioxide
C
Carbon dioxide
D
$PAN$
Solution
(C) Photochemical smog is primarily composed of nitrogen oxides $(NO_x)$,ozone $(O_3)$,and peroxyacetyl nitrate $(PAN)$.
These components are formed through complex photochemical reactions involving sunlight and precursors like hydrocarbons and nitrogen oxides.
$CO_2$ is a greenhouse gas and a product of combustion,but it is not a constituent of photochemical smog.
Therefore,photochemical smog pollution does not contain $CO_2$.
These components are formed through complex photochemical reactions involving sunlight and precursors like hydrocarbons and nitrogen oxides.
$CO_2$ is a greenhouse gas and a product of combustion,but it is not a constituent of photochemical smog.
Therefore,photochemical smog pollution does not contain $CO_2$.
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