datavorous/entrance-exam-dataset huggingface dataset:
A solution of CuSO4is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?

Options:
A 0.3 grams
B 0.6 grams
C 0.9 grams
D 1.2 grams

The correct answer is:
0.3 grams

Solution:
Correct Option is : (A)
0.3 grams

Topic: Chemistry, Electrochemistry, NEET
A solution of glucose in water is labelled as 10 percentWW . If the density of the solution is 1.2 g mL-1 , then what shall be the molarity of the solution?

Options:
A 0.17 M
B 0.67 M
C 0.6 M
D 0.76 M

The correct answer is:
0.67 M

Solution:
mol of glucose =180
10% WW solution of glucose means 100 g solution contains 10 g glucose.
Molarity=moles of solutevolume of solution(mL)Ã1000=101801001.2Ã1000 M
=10Ã1.2Ã1000180Ã100=0.67 M

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A solution of glucose in water is labelled as 10% w/w, what would be the molality?

Options:
A 0.62 m
B 1.42 m
C 2.34 m
D 3.67 m

The correct answer is:
0.62 m

Solution:
Correct Option is : (A)
0.62 m

Topic: Chemistry, Solutions, NEET
A solution of glucose in water is labelled as 10% w/w, what would be the mole fraction of each component in the solution?

Options:
A Mole fraction of Glucose=0.28 and Mole fraction of Water=0.72
B Mole fraction of Glucose=0.99 and Mole fraction of Water=0.01
C Mole fraction of Glucose=0.01 and Mole fraction of Water=0.99
D Mole fraction of Glucose=0.72 and Mole fraction of Water=0.28

The correct answer is:
Mole fraction of Glucose=0.01 and Mole fraction of Water=0.99

Solution:
Correct Option is : (C)
Mole fraction of Glucose=0.01 and Mole fraction of Water=0.99

Topic: Chemistry, Solutions, NEET
A solution of glucose in water is labelled as 10% w/w,If the density of solution is 1.2 g mL–1, then what shall be the molarity of the solution?

Options:
A 0.67M
B 0.97M
C 1.34M
D 0.23M

The correct answer is:
0.67M

Solution:
Correct Option is : (A)
0.67M

Topic: Chemistry, Solutions, NEET
A solution of KMnO4on reduction yields either a colorless solution or a brown precipitate or a green solution depending on the pH of the solution. What different stages of the reduction do these represent and how are they carried out?

Options:
A The colorless solution represents the complete reduction to Mn(II) ions, while the brown precipitate represents partial reduction to Mn(IV) oxide.
B The colorless solution represents partial reduction to Mn(IV) oxide, while the brown precipitate represents complete reduction to Mn(II) ions.
C The colorless solution represents the partial reduction to Mn(IV) oxide, while the green solution represents complete reduction to Mn(II) ions.
D The brown precipitate represents the complete reduction to Mn(II) ions, while the green solution represents partial reduction to Mn(IV) oxide.

The correct answer is:
The colorless solution represents the complete reduction to Mn(II) ions, while the brown precipitate represents partial reduction to Mn(IV) oxide.

Solution:
Correct Option is : (A)
The colorless solution represents the complete reduction to Mn(II) ions, while the brown precipitate represents partial reduction to Mn(IV) oxide.

Topic: Chemistry, d and f Block Elements, NEET
A solution of Ni(NO₃)₂ is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Options:
A 2.87 g
B 1.84 g
C 0.83 g
D 3.42 g

The correct answer is:
1.84 g

Solution:
Correct Option is : (B)
1.84 g

Topic: Chemistry, Electrochemistry, NEET
A solution of sucrose (molar mass =342 gmol-1 ) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be
( Kf for water =1.86 Kkgmol-1 )

Options:
A -0.570 C
B -0.372 C
C -0.520 C
D +0.372 C

The correct answer is:
-0.372 C

Solution:
ΔTf=Kf.wmol.wt.×Wsol.×1000
ΔTf=68.5×1000342×1000×1.86
ΔTf=0.372
F.pt. of solution =0oC-0.372=-0.372C

Topic: Chemistry, Solutions, NEET
A solution of sucrose (molar mass $=342 \mathrm{~g} \mathrm{~mol}^{-1}$ ) has been prepared by dissolving $68.5 \mathrm{~g}$ of sucrose in $1000 \mathrm{~g}$ of water. The freezing point of the solution obtained will be $\left(\mathrm{k}_{\mathrm{f}}\right.$ for water $=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )

Options:
A $-0.372^{\circ} \mathrm{C}$
B $-0.520^{\circ} \mathrm{C}$
C $+0.372^{\circ} \mathrm{C}$
D $-0.570^{\circ} \mathrm{C}$

The correct answer is:
$-0.372^{\circ} \mathrm{C}$

Solution:
Depression in freezing point,
$$
\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{k}_{\mathrm{f}} \times \mathrm{m} \\
& \text { where, } \begin{aligned}
& \mathrm{m}=\text { molality }=\frac{\mathrm{w}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{B}} \cdot \mathrm{W}_{\mathrm{A}}} \\
&=\frac{68.5 \times 1000}{342 \times 1000}=\frac{68.5}{342} \\
& \Delta \mathrm{T}_{\mathrm{f}}=1.86 \times \frac{68.5}{342} \\
&= 0.372^{\circ} \mathrm{C} \\
& \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{T}^0-\mathrm{T}_s \\
&= 0-0.372^{\circ} \mathrm{C} \\
&=-0.372^{\circ} \mathrm{C}
\end{aligned}
\end{aligned}
$$

Topic: Chemistry, Solutions, NEET, NEET 2010 (Screening)
A solution of sucrose (Molar mass = 342 g/mol) is prepared by dissolving 68.4 g of it per litre of solution, what is its osmotic pressure (R=0.082 L atmK-1 mol-1) at 273 K?

Options:
A 4.48 atm
B 2 atm
C 1 atm
D 5 atm

The correct answer is:
4.48 atm

Solution:
Sucrose is non ionisable so i = 1
π=i CRT
=1× 68.4 342 1 0.082 ×273
=4.477atm
≃4.48atm

Topic: Chemistry, Solutions, NEET
A solution of sucrose (Molar mass = 342) has been prepared by dissolving 68.4g of sucrose in one kg of water. Kf for water is 1.86 K kg mol-1 .

The freezing point of the solution will be

Options:
A -0.6840C
B -0.3420C
C -0.3720C
D -0.1860C

The correct answer is:
-0.3720C

Solution:
ΔTf=Kf.m
ΔTf=i×kf×m
Tf0-TfS=i×kf×m
0-Tfs=11.86kkgmol-168.43421mol/kg
=0.3720k=0.3720C
TfS=-0.3720C

Topic: Chemistry, Solutions, NEET
A solution of urea (mol. mass $\left.56 \mathrm{~g} \mathrm{~mol}^{-1}\right)$ boils at $100.18^{\circ} \mathrm{C}$ at the atmospheric pressure. If $\mathrm{K}_f$ and $\mathrm{K}_b$ for water for 1.86 and $0.512 \mathrm{Kkmol}^{-1}$ respectively, the above solution will freeze at:

Options:
A $0.654^{\circ} \mathrm{C}$
B $-0.654^{\circ} \mathrm{C}$
C $6.54^{\circ} \mathrm{C}$
D $-6.54^{\circ} \mathrm{C}$

The correct answer is:
$-0.654^{\circ} \mathrm{C}$

Solution:
$$
\begin{aligned}
\Delta \mathrm{T}_f & =\mathrm{K}_f m \ldots(1) \\
\Delta \mathrm{T}_b & =\mathrm{K}_b m \ldots(2) \\
\Rightarrow \quad \frac{\Delta \mathrm{T}_f}{\Delta \mathrm{T}_b} & =\frac{\mathrm{K}_f}{\mathrm{~K}_b} \ldots(3)
\end{aligned}
$$
$$
\begin{aligned}
& \text { b.p. of water }=100^{\circ} \mathrm{C} ; \mathrm{K}_f=1.86 \mathrm{~kg} \\
& \mathrm{~mol}^{-1} \\
& \text { b.p. of urea in water }=100.18^{\circ} \mathrm{C} ; \mathrm{K}_b \\
& =0.512 \mathrm{~kg} \mathrm{~mol}^{-} \\
& \Rightarrow \quad \Delta \mathrm{T}_b=0.18 \\
& \text { f.p. of water }=0^{\circ} \mathrm{C} \\
& f . p . \text { of urea in water }=-\mathrm{T}^{\circ} \mathrm{C} \\
& \Rightarrow \quad \Delta \mathrm{T}_f=\mathrm{T} \\
& \Rightarrow \text { from eq. }(3) \\
& \qquad \frac{\mathrm{T}}{0.18}=\frac{1.86}{0.512} \\
& \Rightarrow \quad \mathrm{T}=0.6539 \\
& \Rightarrow f . p . \text { of urea in water }=-0.654^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Chemistry, Solutions, NEET, NEET 2005
A solution which is 10-3 M each in Mn2+,Fe2+,Zn2+ and Hg2+ is treated with 10-16 M sulphide ion. If Ksp of MnS , FeS, ZnS and HgS are 10-13,10-18,10-24 and 10-53 respectively. Which one will precipitate first?

Options:
A FeS
B MgS
C HgS
D ZnS

The correct answer is:
HgS

Solution:
[S2-] needed for precipitation of
FeS=10-1810-3=10-15M
MnS=10-1310-3=10-10M
HgS=10-5310-3=10-50M
ZnS=10-2410-3=10-21M
Thus, minimum [S2-] is for HgS and so it will be precipitated first.

Topic: Chemistry, Ionic Equilibrium, NEET
A solution which is 10–3 M each in Mn2 + , Fe2 + , Zn2 + and Hg2 + is treated with 10 - 16 M sulphide ion. If Ksp of MnS, FeS, ZnS and HgS are 10 - 13 , 10–18 , 10–24 and 10–53 respectively, which one will precipitate first ?

Options:
A FeS
B MgS
C HgS
D ZnS

The correct answer is:
HgS

Solution:
[S2 - ] need for precipitation of
FeS = 1 0 - 1 8 1 0 - 3 = 1 0 - 1 5 M
MnS = 1 0 - 1 3 1 0 - 3 = 1 0 - 1 0 M
HgS = 1 0 - 5 3 1 0 - 3 = 1 0 - 5 0 M
ZnS = 1 0 - 2 4 1 0 - 3 = 1 0 - 2 1 M
Thus minimum [S2 - ] for precipitation is for HgS it will be precipitated first.

Topic: Chemistry, Ionic Equilibrium, NEET
A somaclonal variation appears in plants

Options:
A Growing in polluted soil or water
B Exposed to gamma rays
C Raised in tissue culture
D Transformed by recombinant DNA technology

The correct answer is:
Raised in tissue culture

Solution:
Somaclonal variation is the variation seen in plants that have been produced by plant tissue culture. Somaclonal variation is not restricted to but is particularly common in, plants regenerated from callus. The variations can be genotypic or phenotypic, which in the latter case can be either genetic or epigenetic in origin.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
A somatic cell in a human male contains ___________.

Options:
A no gene on the sex chromosome
B genes on only on sex chromosomes
C two genes for every sex-linked character
D only one sex-linked gene for each character

The correct answer is:
only one sex-linked gene for each character

Solution:
Human male contains 44 autosomes and a pair of sex chromosome having one X and one Y chromosome. Genes located on the X and Y chromosomes are called a sex-linked gene. Genes present on X chromosome are called X linked gene and gene present on Y chromosome are called a Y linked gene. Male is hemizygous for both X linked and Y linked genes.

Topic: Biology, Reproduction in Organisms, NEET
A somatic cell that has just completed the S phase of its cell cycle, as compared to gamete of the same species, has:

Options:
A Twice the number of chromosomes and twice the amount of DNA
B Same number of chromosomes but twice the amount of DNA
C Twice the number of chromosomes and four times the amount of DNA
D Four times the number of chromosomes and twice the amount of DNA

The correct answer is:
Twice the number of chromosomes and four times the amount of DNA

Solution:
S-phase (synthesis phase) is the part of the cell cycle in which DNA is replicated, but a number of chromosome does not change.
If a cell is diploid then after S phase DNA content in the cell will be 4 C and Chromosome in the cell will be 2 N.
While in a gamete,
DNA content in gamete = C
Chromosome in gamete = N
Hence, the Number of chromosome in a somatic cell will be twice than gamete while DNA content will be four times.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2015 (Phase 1)
A somatic cell that has just completed the S phase of its cell cycle, as compared to gamete of the same species, has:

Options:
A Twice the number of chromosomes and twice the amount of DNA
B Same number of chromosomes but twice the amount of DNA
C Twice the number of chromosomes and four times the amount of DNA
D Four times the number of chromosomes and twice the amount of DNA

The correct answer is:
Twice the number of chromosomes and four times the amount of DNA

Solution:
S-phase (synthesis phase) is the part of the cell cycle in which DNA is replicated, but a number of chromosome does not change.
If a cell is diploid then after S phase DNA content in the cell will be 4 C and Chromosome in the cell will be 2 N.
While in a gamete,
DNA content in gamete = C
Chromosome in gamete = N
Hence, the Number of chromosome in a somatic cell will be twice than gamete while DNA content will be four times.

Topic: Biology, Cell Cycle and Cell Division, NEET
A sound of ____dB is generated by take-off of a jet plane.

Options:
A 80
B 50
C 100
D 150

The correct answer is:
150

Solution:
Eardrums may get damaged by brief exposure to an extremely high sound level like 150 dB or more leading to permanently impairing hearing. Even chronic exposure to a low-level noise (80-100 dB) as found in many industries and metropolitan cities near the roads may permanently damage hearing ability of humans. Jet plane and rockets generate about 150 dB of sound during take off.

Topic: Biology, Environmental Issues, NEET
A sound wave of frequency f propagating through the air with a velocity c , is reflected from a wall which is moving away from the stationary source with a constant speed v . The frequency of the reflected wave, measured by the observer at the position of the source is

Options:
A f(c-v)c+v
B f(c+v)c-v
C f(c+2v)c+v
D f(c-v)c-2v

The correct answer is:
f(c-v)c+v

Solution:
In first case observer (surface) is moving away. perceiver frequency
f ′ =f( c−v c )
In second case source (surface) is moving away. Perceived frequency
f ″ = f ′ ( c c+v )=f( c−v c+v )

Topic: Physics, Waves and Sound, NEET
A sounding source of frequency 500 Hz moves towards a stationary observer with a velocity of 30 ms-1. If the velocity of sound in air is 330 ms-1, find frequency heard by the observer.

Options:
A 500 Hz
B 550 Hz
C 355 Hz
D 55.5 Hz

The correct answer is:
550 Hz

Solution:
f=500 Hz vo=0
vs=30 m s-1, v=330 m s-1
From ,
f'=fv-vov-vs=500300330-30=550 Hz

Topic: Physics, Waves and Sound, NEET
A source emits a sound of a frequency of 400 Hz, but the listener hears it 390 Hz. Then

Options:
A The listener is moving towards the source
B The source is moving towards the listener
C The listener is moving away from the source
D The listener has a defective ear

The correct answer is:
The listener is moving away from the source

Solution:
Real frequency ν=400 Hz
Apparent frequency ν'=390 Hz
ν' <ν
So, the distance between the source and listener increases or the listener is going away from the source.

Topic: Physics, Waves and Sound, NEET
A source is moving on a circle of radius 3 m with constant angular velocity Ï=5 rad s-1. If the observer is at a distance 5 m from the centre of the circle, the time interval between the maximum and minimum frequency received by the observer is

Options:
A π5
B 25cos-1⁡34
C 25cos-1⁡35
D 25sin-1⁡35

The correct answer is:
25cos-1⁡35

Solution:
When the source is at point P, the frequency received is minimum and when the source is at point Q, the frequency received is maximum.
2θ=ωt
t=25cos-1⁡35

Topic: Physics, Waves and Sound, NEET
A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 m s-1 at an angle of 60o with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer corresponding to the wave emitted by the source at the instant shown is (velocity of sound in air 330 m s-1 )
â

Options:
A 103âHz
B 106 Hz
C 97 Hz
D 100 Hz

The correct answer is:
103 Hz

Solution:
f=f0v-v0v-vs
f=100v-0v-+9.7
f=100vv1-9.7v
f=1001+9.7330=103 Hz

Topic: Physics, Waves and Sound, NEET, NEET 2015 (Phase 2)
A source of sound S is moving with a velocity of 50 m s-1 towards a stationary observer. The observer measures the frequency of the source as 1000 Hz. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? The velocity of the sound in the medium is 350 m s-1.

Options:
A 750 Hz

B 857 Hz
C 1143 Hz
D 1333 Hz

The correct answer is:
750 Hz

Solution:
Here, us=50 m s-1, vL=0, u=350 m s-1
When the source is moving towards the observer,
ν′=1000
ν′=u×νu-us
ν=u-usν′u
=(350-50)1000350=60007 Hz
When the source is moving away from the observer,
ν′=u×νu+vs
=350(350+50)×60007
=750 Hz

Topic: Physics, Waves and Sound, NEET
A source of the sound of frequency 600âHz is placed inside water. The speed of sound in water is 1500âm s-1 and in the air, it is 300âm s-1 The frequency of sound recorded by an observer who is standing in the air is

Options:
A 200âHz
B 3000 Hz
C 120 Hz
D 600 Hz

The correct answer is:
600 Hz

Solution:
The frequency is a characteristic of source. It is independent of the medium. Hence, the correct option is 600 Hz .

Topic: Physics, Waves and Sound, NEET
A source of unknown frequency gives 4 beats/s when sounded with a source of known frequency $250 \mathrm{~Hz}$. The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency $513 \mathrm{~Hz}$. The unknown frequency is

Options:
A $254 \mathrm{~Hz}$
B $246 \mathrm{~Hz}$
C $240 \mathrm{~Hz}$
D $260 \mathrm{~Hz}$

The correct answer is:
$254 \mathrm{~Hz}$

Solution:
Here,
We know beat frequency is the difference of the frequencies of the sources. When unknown source is sounded with known source of frequency 250 Hz, it gives 4 beats/s. It means the frequency of unknown source may be 254 Hz or 246 Hz. Now second harmonic of the source of unknown frequency gives five beats per second, when sounded with a source of frequency 513 Hz. It means the frequency of unknown source may be 518 Hz or 508 Hz.
\(\text {As } \frac{518}{2}=259 \mathrm{~Hz} \text { and } \frac{508}{2}=254 \mathrm{~Hz}\)
Hence unknown frequency is 254 Hz.

Topic: Physics, Waves and Sound, NEET, NEET 2013 (All India)
A source $\mathrm{S}_1$ is producing, $10^{15}$ photons/s of wavelength $5000 Å$. Another source $S_2$ is producing $1.02 \times 10^{15}$ photons per second of wavelength $5100 Å$. Then, (power of $\left.S_2\right) /\left(\right.$ power of $S_1$ ) is equal to

Options:
A 1.00
B 1.02
C 1.04
D 0.98

The correct answer is:
1.00

Solution:
Number of photons emitted per second
$$
\begin{array}{cc}
\quad \mathrm{n}=\frac{\mathrm{P}}{\left(\frac{\mathrm{hc}}{\lambda}\right)} \\
\therefore \quad \mathrm{P}=\frac{\mathrm{nhc}}{\lambda} \\
\Rightarrow \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{\mathrm{n}_2 \lambda_1}{\mathrm{n}_1 \lambda_2}=\frac{1.02 \times 10^{15} \times 5000}{10^{15} \times 5100}=1
\end{array}
$$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2010 (Screening)
A space station is at a height equal to the radius of the Earth. If VE is the escape velocity on the surface of the Earth, the same on the space station is......times VE.

Options:
A 12
B 14
C 12
D 13

The correct answer is:
12

Solution:
Escape velocity at a height h from the surface of Earth vEâ²=2GMR+h vE=2GMR
Given, h=R
â vEâ²=2GMR+R=122GMR=vE2

Topic: Physics, Gravitation, NEET
A space-ship travelling in the interstellar medium is picking up dust at a rate dMdt=Î±v, where Î± is a positive constant, v is the instantaneous velocity and M is the total mass of the space-ship at any instant. The instantaneous acceleration of the space-ship is

Options:
A â2Î± v 2 M
B âÎ± v 2 M
C −α v 2 2M
D −α v 2

The correct answer is:
−α v 2 M

Solution:
dMdt=Î±v

Fthrust=-vdMdt=Ma

â-vÎ±v=Ma

âa=âÎ±v2M

Topic: Physics, Center of Mass Momentum and Collision, NEET
A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution (Qsp) becomes greater than its solubility product. If the solubility of BaSO4in water is 8 × 10–4mol dm–3. Calculate its solubility in 0.01 mol dm–3of H2SO4.

Options:
A 3 × 10–4mol dm–3
B 12 × 10–4mol dm–3
C 6 × 10–4mol dm–3
D 2 × 10–4mol dm–3

The correct answer is:
6 × 10–4mol dm–3

Solution:
Correct Option is : (C)
6 × 10–4mol dm–3

Topic: Chemistry, Ionic Equilibrium, NEET
A special type of diffusion when water is absorbed by solids, colloids causing increase in volume is called

Options:
A reverse osmosis
B imbibition
C osmosis
D Both A & B

The correct answer is:
imbibition

Solution:
A special type of diffusion where water is absorbed by solids and colloids is called imbibitions. The volume of the substance absorbing water through imbibitions increases to a great extent. An example of this type of diffusion is absorption of water by dry wood and seeds. Imbibition is an important process as without it the seedlings cannot come out of the soil into the open. In imbibitionstoo the water transport occurs along the concentration gradient.

Topic: Biology, Transport in Plants, NEET
A specialised centre called respiratory rhythm centre is primarily responsible for the regulation of respiration is located in:

Options:
A Medulla oblongata
B Carotid bodies
C Amygdala
D Hypothalamus

The correct answer is:
Medulla oblongata

Solution:
Medulla oblongata: The lower half of the brainstem that contains the cardiac, respiratory, vomiting, and vasomotor centers and regulates autonomic, involuntary functions such as breathing, heart rate, and blood pressure.

Topic: Biology, Neural Control and Coordination, NEET
A specialised nodal tissue embedded in the lower corner of the right atrium, close to atrio-ventricular septum, delays the spreading of impulses to heart apex for about $0.1 \mathrm{sec}$. The delay allows

Options:
A blood to enter aorta
B the ventricles to empty completely
C blood to enter pulmonary arteries
D the atria to empty completely

The correct answer is:
the atria to empty completely

Solution:
Atrio-Ventricular Node (AVN) present in the lower corner of the right atrium, delays the spreading of impulses to heart ventricles for about 0.1 second. This pause allows the atria to empty completely into the ventricles before the ventricles pump out the blood.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2019 (Odisha)
A specialized center called respiratory rhythm center is primarily responsible for the regulation of respiration. It is located in

Options:
A cerebellum
B the lowest part of the brain
C the top portion of the brainstem
D none of these

The correct answer is:
the lowest part of the brain

Solution:
Medulla oblongata, the lower half of the brain, or brainstem contains the cardiac, respiratory, vomiting and vasomotor centres and regulates autonomic, involuntary functions such as breathing, heart rate and blood pressure.

Topic: Biology, Breathing and Exchange of Gases, NEET
A specialized centre present in the medulla region of the brain called _____A______ is primarily responsible for this regulation. Another centre present in the pons region of the brain called _____B______ can moderate the functions of the _____A______. Neural signal from this centre can _____C______the duration of inspiration and thereby alter the respiratory rate. _____D______is situated adjacent to the rhythm centre which is highly sensitive to _____E______.

Options:
A

A	B	C	D	E

Chemosensitive area	Pneumotaxic centre	Increase	Respiratory rhythm centre	O2

B

A	B	C	D	E

Pneumotaxic centre	Respiratory rhythm centre	Reduce	Chemosensitive area	O2 and hydrogen ions

C

A	B	C	D	E

Respiratory rhythm centre	Pneumotaxic centre	Increase	Chemosensitive area	CO2 and hydrogen ions

D

A	B	C	D	E

Respiratory rhythm centre	Pneumotaxic centre	Reduce	Chemosensitive area	CO2 and hydrogen ions

The correct answer is:

A	B	C	D	E

Respiratory rhythm centre	Pneumotaxic centre	Reduce	Chemosensitive area	CO2 and hydrogen ions

Solution:
A specialised centre present in the medulla region of the brain called respiratory rhythm centre is primarily responsible for this regulation. Another centre present in the pons region of the brain called pneumotaxic centre can moderate the functions of the respiratory rhythm centre. Neural signal from this centre can reduce the duration of inspiration and thereby alter the respiratory rate. A chemosensitive area is situated adjacent to the rhythm centre which is highly sensitive to CO2 and hydrogen ions. Increase in these substances can activate this centre, which in turn can signal the rhythm centre to make necessary adjustments in the respiratory process by which these substances can be eliminated. Receptors associated with aortic arch and carotid artery also can recognise changes in CO2 and H+ concentration and send necessary signals to the rhythm centre for remedial actions. The role of oxygen in the regulation of respiratory rhythm is quite insignificant.

Topic: Biology, Breathing and Exchange of Gases, NEET