datavorous/entrance-exam-dataset huggingface dataset:
Drift speed of electrons, when 1.5 A current flows in a copper wire of cross section 5 mm2 is vd. If the electron density in copper is 9×1028 m-3 the value of vd in mm s-1 is close to (Take charge of an electron to be =1.6×10-19 C)

Options:
A 0.2
B 3
C 2
D 0.02

The correct answer is:
0.02

Solution:
Given,

Current, i=1.5 A,

Cross-section of the wire, A=5×10-6 m2 and

Electron density, n=9×1028 m-3.

As current, i=neAvd
⇒1.5=9×1028×1.6×10-19×5×10-6×vd
⇒vd=0.02 mm s-1.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Drift velocity $v_d$ varies with the intensity of electric field as per the relation

Options:
A $v_d \propto E$
B $v_d \propto \frac{1}{E}$
C $v_d=$ constant
D $v_d \propto E^2$

The correct answer is:
$v_d \propto E$

Solution:
$v_d=\frac{e}{m} \times \frac{V}{l} \tau$ or $v_d=\frac{e}{m} \cdot \frac{E l}{l} \tau$ (Since $V=E I$ )
$\therefore v_d \propto E$

Topic: Physics, Current Electricity, JEE Main
Drug which helps to reduce anxiety and brings about calmness is called

Options:
A tranquilizer
B diuretic
C analgesic
D antihistamine

The correct answer is:
tranquilizer

Solution:
A substance which is used for the purpose of diagnosis, prevention, cure or relief of a disease is called drug. Drugs are chemical of low molecular masses, these interact with macromolecular targets and produce a biological response. When the biological response is therapeutic and useful, these chemicals are called medicines and are used in diagnosis, prevention and treatment of diseases.

Tranquilizers are a class of chemical compounds used for the treatment of stress, and mild or severe mental diseases. These relieve anxiety, stress, excitement or irritability by inducing a sense of well-being.

Diuretics are a class of drugs that increase the flow of urine.

Analgesics are the drugs used to reduce or abolish pain without causing impairment of consciousness, mental confusion or paralysis or some other nervous system disturbances.

Antihistamines are used to relieve or prevent the symptoms of hay fever and other types of allergy.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Drug which reduce fever is known as

Options:
A Barbiturates
B Antiseptic
C Antipyretic
D Antibiotic

The correct answer is:
Antipyretic

Solution:
The correct answer is Antipyretic.
- Antipyretic is a drug that reduces fever.
- Antipyretics cause the hypothalamus to override a prostaglandin induced increase in temperature.
- The body then works to lower the temperature, which results in a reduction in fever.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Drugs used to bind to receptors, inhibiting its natural function and blocking a message are called

Options:
A antagonists.
B agonists.
C allosteric.
D lead compounds.

The correct answer is:
antagonists.

Solution:
Drugs that bind to receptor site and inhibit the natural functions are known as antagonists. Agonist are type of drugs that activate the natural messenger by switching on the receptor, useful when lack of chemical messenger. If the drug binds to a site other than active site in an enzyme, it is known as allosteric site.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2022 (25 Jul Shift 1)
Dry distillation of calcium acetate and calcium formate forms:

Options:
A methanol
B ethanal
C ethanol
D acetone

The correct answer is:
ethanal

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
→ Dry Ether Na

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Dry heating of calcium acetate gives

Options:
A Acetaldehyde
B Ethane
C Acetic acid
D Acetone

The correct answer is:
Acetone

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Dry ice is

Options:
A
solid $\mathrm{NH}_3$

B
solid $\mathrm{SO}_2$

C
solid $\mathrm{CO}_2$

D
solid $\mathrm{N}_2$

The correct answer is:
solid $\mathrm{CO}_2$

Solution:
Carbon dioxide can be obtained as a solid in the form of dry ice, allowing the liquified $\mathrm{CO}_2$ to expand rapidly.
It is used to maintain the low temperature in laboratory.
Dry ice is solid $\mathrm{CO}_2$

Topic: Chemistry, p Block Elements (Group 13 & 14)
Dry indehiscent single-seeded fruit formed bicarpellary syncarpous inferior ovary is

Options:
A Cremocarp
B Caryopsis
C Cypsela
D Berry

The correct answer is:
Cypsela

Solution:
Cypsela - Asteraceae.
Cremocarp - Shizocarpic fruit, as in umbelliferae.
Caryopsis - Indehiscent fruit formed from monocarpellary and superior ovary with basal placentation.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2008 (Mains)
Dry indehiscent single-seeded fruit formed from bicarpellary syncarpous inferior ovary is

Options:
A caryopsis
B cypsela
C berry
D cremocarp

The correct answer is:
cypsela

Solution:
Cypsela is dry, indehiscent, single seeded fruit develops from an unilocular, single ovulate inferior ovary of bicarpellary, syncarpous, gynoecium possessing basal placentation. The fruit wall develops from pericarp and thalamus and is thin and remains attached to the seed at one point, eg, Heltanthus.
Caryopsis are dry, indehiscent, small, single seeded fruit develop from unilocular, single ovuled, superior ovary of multicarpellary gynoecium. It differs from typical achenes as their pericarp is completely fused with the seed coat or testa, eg, Poaceae.
Cremocarp are bilocular and two seeded schizocarpic fruits (small, dry) developed from inferior ovary of bicarpellary, syncarpous gynoecium possessing persistant stylopodium, $e g$, Apiaceae.
Berry is a fleshy, indehiscent, few to multiseeded fruit derived from multicarpellary syncarpous gynoecium. The fleshy pericarp of berry consists of epicarp, mesocarp and endocarp.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2008 (Screening)
Due to an explosion underneath water, a bubble started oscillating. If this oscillation has time period $T$, which is proportional to $p^\alpha S^\beta E^\gamma$, where $p$ is static pressure, $S$ is density of water and $E$ is total energy of explosion. Determine $\alpha, \beta$ and $\gamma$.

Options:
A $\alpha=-\frac{3}{2}, \beta=\frac{1}{3}, \gamma=-\frac{5}{6}$
B $\alpha=-\frac{5}{6}, \beta=\frac{1}{2}, \gamma=\frac{1}{3}$
C $\alpha=\frac{1}{2}, \beta=-\frac{5}{6}, \gamma=\frac{7}{4}$
D $\alpha=\frac{1}{3}, \beta=\frac{3}{2}, \gamma=\frac{4}{3}$

The correct answer is:
$\alpha=-\frac{5}{6}, \beta=\frac{1}{2}, \gamma=\frac{1}{3}$

Solution:
Given, time-period of oscillation $T$ is
$T \propto p^\alpha S^\beta E^\gamma \text { or } T=k p^\alpha S^\beta E^\gamma$
Now, subtituting dimensions of $T, p, S$ and $E$, we have
$\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{\mathrm{l}}\right]=k\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]^\alpha\left[\mathrm{ML}^{-3}\right]^\beta\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]^\gamma$
Equating powers of similar terms, we have
Solving, we get $\alpha=\frac{-5}{6}, \beta=\frac{1}{2}, \gamma=\frac{1}{3}$

Topic: Physics, Units and Dimensions, JEE Main
Due to cold weather, a 1 m water pipe of cross-sectional area 1 cm2 is filled with ice at -10 °C. Resistive heating is used to melt the ice. Current of 0.5 A is passed through 4 kΩ resistance. Assuming that all the heat produced is used for melting, what is the minimum time required?

(Given latent heat of fusion for water/ice=3.33×105 J kg-1, specific heat of ice =2×103 J kg-1 °C-1 and density of ice =103 kg m-3 )

Options:
A 3.53 s
B 0.353 s
C 35.3 s
D 70.6 s

The correct answer is:
35.3 s

Solution:
The mass of ice, m=ρAL=103×10-4×1=10-1 kg,

Energy required to melt the ice, =mSΔT+mL,

⇒i2Rt=mSΔT+mL
⇒(1)2×103×t=103×1×10-4×2000×10+103×1×10-4×330×103⇒(1)2×103×t=3.53×104⇒t=35.3 s

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Due to Frenkel defect, the density of ionic solids

Options:
A Increases
B Decreases
C Does not change
D Changes

The correct answer is:
Does not change

Solution:
Since no ions are missing from the crystal as a whole, there is no effect on density.

Topic: Chemistry, Solid State, JEE Main
Due to the presence of an unpaired electron, free radicals are:

Options:
A
Chemically reactive

B
Chemically inactive

C
Anions

D
Cations

The correct answer is:
Chemically reactive

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2005
Due to the presence of the current $\mathrm{l}_1$ at the origin

Options:
A
The forces on $A B$ and $D C$ are zero

B
The forces on $A D$ and $B C$ are zero

C
The magnitude of the net force on the loop is given by $\frac{\mu_0 I_1}{4 \pi}\left[2(b-a)+\frac{\pi}{3}(a+b)\right]$

D
The magnitude of the net force on the loop is given by $\frac{\mu_0 \|_1}{24 a b}(b-a)$

The correct answer is:
The forces on $A D$ and $B C$ are zero

Solution:
The forces on $A D$ and $B C$ are zero because magnetic field due to a straight wire on $A D$ and $B C$ is parallel to elementary length of the loop.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2009
Due to which of the following reasons can potassium permanganate not be prepared as a primary standard?

Options:
A It has a relatively low molar mass.
B It only reacts in an acidic solution.
C Its solutions are volatile.
D It is not highly soluble in water.

The correct answer is:
Its solutions are volatile.

Solution:
Potassium permanganate has the chemical formula $\mathrm{KMnO}_4$.
It is a flexible and potent oxidant that may be used to classify a variety of substances by direct or indirect titration.
It is not used as a major standard since obtaining the pure form of $\mathrm{KMnO}_4$ is challenging because it is not free of $\mathrm{MnO}_2$.
Furthermore, the color of Potassium permanganate is so vibrant that it serves as its own indicator.
Hence, $\mathrm{KMnO}_4$ is not used as a primary standard, because it is difficult to obtain the pure state of $\mathrm{KMnO} 4$ as it is not free from $\mathrm{MnO} 2$. Also, the colour is so intense that it acts as its own indicator.

Topic: Chemistry, Redox Reactions, JEE Main
Duplication of centrioles is observed during

Options:
A S-Phase
B G2 Phase
C G1 Phase
D both B and C

The correct answer is:
S-Phase

Solution:
The cell cycle is divided into two phases - Interphase and Mitosis phase or the M phase. Interphase is further divide into 3 phases - the Gap 1 or G1 phase, the Synthesis phase or S phase and the Gap 2 or G2 phase. G1 phase is the interval between mitosis and initiation of DNA replication during which the cell is metabolically active and grows but does not replicate. During the S or synthesis phase DNA synthesis and replication occurs. The amount of DNA in the cell doubles during this phase but the number of chromosomes remain same. The centrioles also duplicate in the cytoplasm during S phase. In the G2 phase, proteins are synthesised in preparation for mitosis while cell growth continues.

Topic: Biology, Cell Cycle and Cell Division, NEET
During an adiabatic compression, $830 \mathrm{~J}$ of work is done on 2 moles of a diatomic ideal gas to reduce its volume by $50 \%$. The change in its temperature is nearly: $\left(\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$

Options:
A
$40 \mathrm{~K}$

B
$33 \mathrm{~K}$

C
$20 \mathrm{~K}$

D
$14 \mathrm{~K}$

The correct answer is:
$20 \mathrm{~K}$

Solution:
Given : work done, $W=830 \mathrm{~J}$
No. of moles of gas, $\mu=2$
For diatomic gas $\gamma=1.4$
Work done during an adiabatic change
$$
\begin{aligned}
&W=\frac{\mu R\left(T_1-T_2\right)}{\gamma-1} \\
&\Rightarrow 830=\frac{2 \times 8.3(\Delta T)}{1.4-1}=\frac{2 \times 8.3(\Delta T)}{0.4} \\
&\Rightarrow \Delta T=\frac{830 \times 0.4}{2 \times 8.3}=20 \mathrm{~K}
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main
During an adiabatic compression, $830 \mathrm{~J}$ of work is done on 2 moles of a diatomic ideal gas to reduce its volume by $50 \%$. The change in its temperature is nearly: $\left(\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$

Options:
A
$40 \mathrm{~K}$

B
$33 \mathrm{~K}$

C
$20 \mathrm{~K}$

D
$14 \mathrm{~K}$

The correct answer is:
$20 \mathrm{~K}$

Solution:
Given : work done, $W=830 \mathrm{~J}$
No. of moles of gas, $\mu=2$
For diatomic gas $\gamma=1.4$
Work done during an adiabatic change
$$
\begin{aligned}
&W=\frac{\mu R\left(T_1-T_2\right)}{\gamma-1} \\
&\Rightarrow 830=\frac{2 \times 8.3(\Delta T)}{1.4-1}=\frac{2 \times 8.3(\Delta T)}{0.4} \\
&\Rightarrow \Delta T=\frac{830 \times 0.4}{2 \times 8.3}=20 \mathrm{~K}
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2014 (11 Apr Online)
During an adiabatic process, if the internal energy of the system increases, what happens to the work done by the system?

Options:
A It increases
B It decreases
C It remains constant
D It becomes zero

The correct answer is:
It increases

Solution:
When the internal energy of a system increases in an adiabatic process, the system must do work on its surroundings, leading to an increase in the work done by the system.

Topic: Chemistry, Thermodynamics (C), NEET
During an adiabatic process, if the pressure of a gas is found to be proportional to the cube of its absolute temperature, then the ratio of $\frac{C_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}$ for the gas is :

Options:
A $\frac{5}{3}$
B $\frac{9}{7}$
C $\frac{3}{2}$
D $\frac{7}{5}$

The correct answer is:
$\frac{3}{2}$

Solution:
$\begin{aligned}
& \mathrm{P} \propto \mathrm{T}^3 \\
& \mathrm{PT}^{-3}=\text { constant } \\
& \because \frac{\mathrm{PV}}{\mathrm{T}}=\mathrm{nR}=\text { constant from ideal gas equation } \\
& (\mathrm{P})(\mathrm{PV})^{-3}=\text { constant } \\
& \mathrm{P}^{-2} \mathrm{~V}^{-3}=\text { cosntant }...(i)
\end{aligned}$
$\because$ Process equation for adiabatic process is
$\mathrm{PV}^{\mathrm{y}}=\text { constant }...(ii)$
Comparing equation (1) and (2)
$\frac{C_P}{C_V}=y=\frac{3}{2}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (05 Apr Shift 2)
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio $C_\rho / C_v$ for the gas is

Options:
A $\frac{3}{2}$
B $\frac{4}{3}$
C $2$
D $\frac{5}{3}$

The correct answer is:
$\frac{3}{2}$

Solution:
Given $P \propto T^3$, but we know for an adiabatic process, the pressure $P \propto T^{\gamma / y-1}$
So $\frac{\gamma}{\gamma-1}=3 \Rightarrow \gamma=\frac{3}{2} \Rightarrow \frac{C_p}{C_v}=\frac{3}{2}$

Topic: Physics, Thermodynamics, JEE Main
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio CPCV for the gas is

Options:
A 43
B 2
C 53
D 32

The correct answer is:
32

Solution:
In an adiabatic process, T Î³ = constant P Î³ - 1
or T Î³ / Î³ - 1 = constant P
Given T 3 = constant P
â´ Î³ Î³ - 1 = 3 â 3 Î³ - 3 = Î³
or 2 Î³ = 3 â Î³ = 3 / 2
For monoatomic gas, Î³ = 5 3 = 1.67
For diatomic gas, Î³ = 7 5 = 1.4
when Î³ = 1.5, the gas must be a suitable mixture of monoatomic and diatomic gases
â´ Î³ = 3/2.

Topic: Physics, Thermodynamics, JEE Main
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio $\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}$ for the gas is

Options:
A
$\frac{4}{3}$

B
2

C
$\frac{5}{3}$

D
$\frac{3}{2}$

The correct answer is:
$\frac{3}{2}$

Solution:
$\mathrm{P} \propto \mathrm{T}^3 \quad\left(\frac{\mathrm{P}_1}{\mathrm{P}_2}\right)=\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^3$
Comparing it with standard eq. $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma=\frac{3}{2}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2003
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The value of ratio CPCV for the gas is

Options:

The correct answer is:
1.5

Solution:
In an adiabatic process, T γ = constant P γ - 1
or T γ / γ - 1 = constant P
Given T 3 = constant P
∴ γ γ - 1 = 3 ⇒ 3 γ - 3 = γ
or 2 γ = 3 ⇒ γ = 3 / 2

⇒ CpCv=1.5

Topic: Physics, Thermodynamics, JEE Main
During an adiabatic process, the pressure of an ideal gas is found to be proportional to the cube of its absolute temperature. The ratio CpCv for the gas is

Options:
A 3 2
B 4 3
C 2
D 5 3

The correct answer is:
3 2

Solution:
Given P ∝ T 3 . But for adiabatic process P ∝ T γ / γ - 1 . So,
γ γ - 1 = 3 ⇒ γ = 3 2 ⇒ C p C v = 3 2

Topic: Physics, Thermodynamics, JEE Main
During an atomic explosion, the energy released is due to

Options:
A Conversion of chemical energr into nuclear energy
B Conversion of protons to neutrons
C Conversion of mechanical energy into nuclear energy
D Conversion of mass into energy

The correct answer is:
Conversion of mass into energy

Solution:
Atomic explosion is carried out by via nuclear reaction in which there is net change in mass of the product and reactant and this change in mass gets converted into energy due to which a huge amount of energy in the form of heat is released.

Topic: Physics, Nuclear Physics, JEE Main
During an electrolysis of conc H2SO4, peroxydisulphuric acid H2S2O8 and O2 form in an equimolar amount. The moles of H2 that will be formed simultaneously will be

Options:
A Thrice that of O2
B Twice that of O2
C Equal to that of O2
D Half of the of O2

The correct answer is:
Thrice that of O2

Solution:
2H2O→O2+4H++4e-
2SO42-→S2O82-+2e-
For 1 mole each of O2 and S2O82- , no of mole of e- = 6 mole e-
So 6 mole e-
6H++6e-→3H2
Hence it produces 3 moles H 2

Topic: Chemistry, Electrochemistry, JEE Main
During anaphase-I of meiosis

Options:
A Homologous chromosomes separate
B Non-homologous chromosomes separate
C Sister chromatids chromosomes separate
D Non Sister chromatids chromosomes separate

The correct answer is:
Homologous chromosomes separate

Solution:
The homologous chromosomes separate, while sister chromatids remain associated at their centromeres. Separation of homologous chromosomes at anaphase is called disjunction.

Topic: Biology, Cell Cycle and Cell Division, NEET
During Anaphase-I of meiosis

Options:
A
homologous chromosomes separate

B
non-homologous autosomes separate

C
sister chromatids separate

D
non-sister chromatids separate

The correct answer is:
homologous chromosomes separate

Solution:
During Anaphase-I, homologous chromosomes separate, while sister chromatids remain associated at their centromeres.

Topic: Biology, Cell Cycle and Cell Division
During cell growth, DNA synthesis takes place in

Options:
A S phase
B G1 phase
C G2 phase
D M phase

The correct answer is:
S phase

Solution:
DNA replication occurs in the S phase of the cell cycle. S phase, or synthesis, is the phase of the cell cycle when DNA packaged into chromosomes is replicated. The amount of nucleic acid, i.e., DNA doubles but the number of chromosomes remains the same.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2016 (Phase 2)
During change of $\mathrm{O}_2$ to $\mathrm{O}_2^{-}$ion, the electron adds on which one of the following orbitals?

Options:
A $\pi^*$ orbital
B $\pi$ orbital
C $\widetilde{\circ}$ orbital
D $\sigma$ orbital

The correct answer is:
$\pi^*$ orbital

Solution:
Electronic configuration of \(\mathrm{O}_2\)
\(\begin{aligned}
& \sigma(1 s)^2 \sigma^*(1 s)^2 \sigma(2 s)^2 \sigma^*(2 s)^2 \sigma\left(2 p_z\right)^2 \pi\left(2 p_x\right)^2=\pi\left(2 p_y\right)^2 \\
& \pi^*\left(2 p_x\right)^1=\pi^*\left(2 p_y\right)^1
\end{aligned}\)
Thus, the incoming electron will enter in \(\pi^* 2 p_x\) to form \(\mathrm{O}_2\).

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2012 (Mains)
During chloride shift or Hamburger phenomenon, when the whole blood is saturated with $\mathrm{CO}_2$, following changes occurs. Which one of them is not correct?

Options:
A Bicarbonate content of plasma and corpuscles increase
B Chloride content of plasma diminished and that of the cell is increased
C Total base of blood remain unchanged
D Water content and volume of corpuscles decrease

The correct answer is:
Water content and volume of corpuscles decrease

Solution:
According to Hamburger phenomenon, when the whole blood is saturated with carbon dioxide, the following changes are seen
(i) The bicarbonate contents of plasma and corpuscles increase.
(ii) The chloride content of plasma is diminished and that of the cell is increased.
(iii) The total base (cations) of both plasma and corpuscles remain unchanged.
(iv) The water content and the volume of corpuscles increase.
(v) When the carbon dioxide is removed from the blood reverse changes takes place.
(vi) When $\mathrm{CO}_2$ enters blood, $\mathrm{Cl}^{-}$from plasma enters RBCs leaving $\mathrm{Na}^{+}$behind. When $\mathrm{CO}_2$ escapes from blood. $\mathrm{Cl}^{-}$leaves the cell and combines with $\mathrm{Na}^{+}$again. Due to this alternate movement of $\mathrm{Cl}^{-}$ ions, this phenomenon is called chloride shift.

Topic: Biology, Breathing and Exchange of Gases, JIPMER, JIPMER 2016
During contraction of a skeletal muscle:

Options:
A A - band shortens
B I - band remain same
C I - band shortens
D Z - line is stretched

The correct answer is:
I - band shortens

Solution:
The change in the tension in the skeletal muscle is called muscle contraction. The muscle contraction is achieved by the sliding of actin filaments over myosin filaments. During muscle contraction, two Z lines come close, the Light band (I band) shortens and the length of dark band (A band) remains the same. H zone or Henson's zone is the area in the middle of the dark band where actin filaments don't overlap myosin filaments in a relaxed muscle fibre. During contraction, the actin filaments completely overlap the myosin filaments and thus there is no H zone.

Topic: Biology, Locomotion and Movement, NEET
During debromination of meso-dibromobutane, the major compound formed is

Options:
A n-butane
B 1-butane
C cis-2-butene
D trans-2-butene

The correct answer is:
trans-2-butene

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
During dehydration of alcohols to alkenes by heating with conc. $\mathrm{H}_2 \mathrm{SO}_4$ the initiation step is

Options:
A
formation of carbocation

B
elimination of water

C
formation of an ester

D
protonation of alcohol molecule

The correct answer is:
protonation of alcohol molecule

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
During dehydration of alcohols to alkenes by heating with conc. $\mathrm{H}_2 \mathrm{SO}_4$ the initiation step is

Options:
A
formation of carbocation

B
elimination of water

C
formation of an ester

D
protonation of alcohol molecule

The correct answer is:
protonation of alcohol molecule

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2003
During ecological succession:

Options:
A The establishment of a new biotic community is very fast in its primary phase.
B The numbers and types of animals remain constant.
C The changes lead to a community that is in near equilibrium with the environment and is called pioneer community.
D The gradual and predictable change in species composition occurs in a given area.

The correct answer is:
The gradual and predictable change in species composition occurs in a given area.

Solution:
Ecological succession involves a gradual and fairly predictable change in the species composition of a given area, i.e., the numbers and types of animals. The establishment of a new biotic community is very slow in its primary phase.

The first set of species or community that develops in a bare area in ecological succession is the pioneer community. Pioneer species facilitate succession. The final steady stable self-sustaining community in an ecological succession that lead to a community that is in near equilibrium with the environment and is called climax community.

Topic: Biology, Ecosystem, NEET, NEET 2015 (Phase 2)