datavorous/entrance-exam-dataset huggingface dataset:
When an aqueous solution of $\mathrm{CuCl}_2$ is electrolysed using Pt inert electrodes, the reaction at cathode and anode respectively are

Options:
A $4 \mathrm{H}_2 \mathrm{O}(f) \stackrel{+4 \mathrm{a}^{-}}{\longrightarrow} 2 \mathrm{H}_2(g)+4 \mathrm{OH}_2(\mathrm{aq})$;
$$
2 \mathrm{H}_2 \mathrm{O}(l) \stackrel{-4 \mathrm{a}^{-}}{\longrightarrow} \mathrm{O}_2(g)+4 \mathrm{H}^{+}(\mathrm{aq})
$$
B $2 \mathrm{Cu}^{2 *}(\mathrm{aq}) \stackrel{+\mathrm{se}^{-}}{\longrightarrow} 2 \mathrm{Cu}(\mathrm{s})$;
$$
2 \mathrm{H}_2 \mathrm{O}(l) \stackrel{-4 \mathrm{~s}^{-}}{\longrightarrow} \mathrm{O}_2(g)+4 \mathrm{H}^*(a q)
$$
C $\mathrm{Cu}^{2 *}(\mathrm{aq}) \stackrel{+2 \mathrm{a}^{-}}{\longrightarrow}, \mathrm{Cu}(\mathrm{s}) ; 2 \mathrm{Cl}^{-}(\mathrm{aq}) \stackrel{-2 \mathrm{a}^{-}}{\longrightarrow} \mathrm{Cl}_2(\mathrm{~g})$
D $2 \mathrm{H}_2 \mathrm{O}(f) \stackrel{+2 \mathrm{a}^{-}}{\longrightarrow} \mathrm{H}_2(g)+2 \overline{\mathrm{O}} \mathrm{H}(\mathrm{aq})$;
$$
2 \mathrm{Cl}^{-}(a q) \stackrel{-2 \mathrm{an}^{-}}{\longrightarrow} \mathrm{Cl}_2(g)
$$

The correct answer is:
$\mathrm{Cu}^{2 *}(\mathrm{aq}) \stackrel{+2 \mathrm{a}^{-}}{\longrightarrow}, \mathrm{Cu}(\mathrm{s}) ; 2 \mathrm{Cl}^{-}(\mathrm{aq}) \stackrel{-2 \mathrm{a}^{-}}{\longrightarrow} \mathrm{Cl}_2(\mathrm{~g})$

Solution:
$$
\text { } \mathrm{CuCl}_2 \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)
$$
$\mathrm{Cl}^{-}$show oxidation at anode and form $\mathrm{Cl}_2$ gas.
$$
2 \mathrm{Cl}^{-} \stackrel{-2}{2} \stackrel{e^{-}}{\longrightarrow} \mathrm{Cl}_2(g)
$$
Due to less reactivity, $\mathrm{Cu}^{2+}$ gain electrons and deposited at cathode electrode.
$$
\mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu}
$$

Topic: Chemistry, Electrochemistry, JEE Main
When an electric current is passed through acidified water, 112 mL of hydrogen gas at N.T.P was collected at the cathode in 965 seconds. The current passed, in ampere, is:

Options:
A 2.0
B 0.1
C 0.5
D 1.0

The correct answer is:
1.0

Solution:
Cathode
2e-+ 2H2O⟶H2+ 2OH- v.f.H2=2
Mole =i×tv.f.×96500
11222400=i×9652×96500
12= i2
i=1 amp

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2018 (15 Apr)
When an electric current is passed through acidulated, water, 112 mL of hydrogen gas at NTP collects at the cathode in 965 seconds. The current passed, in amperes, is

Options:

The correct answer is:
1

Solution:
22,400 mL of hydrogen at STP (or NTP) = 2g
So 112 mL of hydrogen at STP =2g×112mL22,400mL=10-2g
2H++2e-2F→H21 mol
=2×96500 C=2g
2 g hydrogen is deposited by =2×96500C
So 10-2 g hydrogen will be deposited by
=2×96500×10-2g2g=965 C
Charge Q=it
So i=Qt=965 C965 s=1 Cs-1=1 A

Topic: Chemistry, Electrochemistry, JEE Main
When an electric current is passes through acidified water, $112 \mathrm{~mL}$ of hydrogen gas at N.T.P. was collected at the cathode in $965$ seconds. The current passed, in ampere, is :

Options:
A
$2.0$

B
$0.1$

C
$0.5$

D
$1.0$

The correct answer is:
$1.0$

Solution:
Reduction at cathode:
$2 \mathrm{e}^{-}+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2+2 \mathrm{OH}^{-}$
(valence factor) $\mathrm{H}_2=2$
At NTP $22400 \mathrm{~mL}$ of $\mathrm{H}_2=1$ mole of $\mathrm{H}_2 ~112 \mathrm{~mL}$ of $\mathrm{H}_2=\frac{1}{22400} \times 112=0.005 \text { mole of } \mathrm{H}_2$
Moles of $\mathrm{H}_2$ produced
$\begin{aligned}
&=\frac{\mathrm{I} \times \mathrm{t}}{96500} \times \text { mole ratio } \\
&0.005=\frac{\mathrm{I} \times 965}{96500} \times \frac{1 \text { mole of } \mathrm{H}_2}{2 \text { mole of }e^{-}} \\
&\mathrm{I}=1.0 \mathrm{~A}
\end{aligned}$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
When an electric current passes through a solenoid, it acts as a/an _____

Options:
A electric bell
B bar magnet
C resistor
D insulator

The correct answer is:
bar magnet

Solution:
- The solenoid is a long coil containing a large number of close turns of insulated copper wire.
- When electricity is passed through the solenoid, it acts as an electro magnet.

Topic: Physics, Magnetic Properties of Matter, JEE Main
When an electron in a hydrogen atom jumps from the third orbit to the second
orbit, it emits a photon of wavelength ${\lambda} \lambda^{\prime}$. When it jumps from the fourth orbit to
third orbit, the wavelength emitted by the photon will be

Options:
A $\frac{20}{13} \lambda$
B $\frac{16}{25} \lambda$
C $\frac{9}{16} \lambda$
D $\frac{20}{7} \lambda$

The correct answer is:
$\frac{20}{7} \lambda$

Solution:
$\begin{aligned} \frac{1}{\lambda} &=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{R \times 5}{36} \\ \frac{1}{\lambda^{\prime}} &=R\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)=R\left(\frac{1}{9}-\frac{1}{16}\right)=R \times \frac{7}{144} \\ \frac{\lambda^{\prime}}{\lambda} &=\frac{5}{36} \times \frac{144}{7}=\frac{20}{7} \\ \therefore \lambda^{\prime} &=\frac{20}{7} \lambda \end{aligned}$

Topic: Physics, Atomic Physics, JEE Main
When an electron in Hydrogen atom revolves in stationary orbit, it

Options:
A does not radiate light though its velocity changes.
B does not radiate light and velocity remains unchanged.
C radiates light but its velocity is unchanged.
D radiates light with the change of energy.

The correct answer is:
does not radiate light though its velocity changes.

Solution:
For an electron revolving in stationary orbits it doesn't radiate light through its velocity change.

Topic: Physics, Atomic Physics, MHT CET, MHT CET 2016
When an electron in Hydrogen atom revolves in stationary orbit, it

Options:
A does not radiate light though its velocity changes.
B does not radiate light and velocity remains unchanged.
C radiates light but its velocity is unchanged.
D radiates light with the change of energy.

The correct answer is:
does not radiate light though its velocity changes.

Solution:
For an electron revolving in stationary orbits it doesn't radiate light through its velocity change.

Topic: Physics, Atomic Physics, NEET
When an electron jumps from a level n=4 to n=1, momentum of the recoiled hydrogen atom will be

Options:
A 6.8Ã10-27kg ms-1
B 12.75Ã10-19kg ms-1
C 13.6Ã10-19kg ms-1
D zero

The correct answer is:
6.8Ã10-27kg ms-1

Solution:
E=E4-E1=-13.642â-13.612
=-0.85+13.6=12.75 eV
=12.75Ã1.6Ã10-19 J
P=Ec=12.75Ã1.6Ã10-193Ã108
=6.8Ã10-27kg ms-1
This must be the momentum of recoiled hydrogen atom (in opposite direction)

Topic: Physics, Atomic Physics, JEE Main
When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas, is

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

The correct answer is:
$\frac{5}{7}$

Solution:
Fraction of supplied energy which in creases the internal energy is given by
$f=\frac{\Delta U}{(\Delta Q)_P}=\frac{(\Delta Q)_V}{(\Delta Q)_P}=\frac{\mu C_V \Delta T}{\mu C_P \Delta T}=\frac{1}{\gamma}$
For diatomic gas ${ }^{\gamma=\frac{7}{5}} \Rightarrow \quad f=\frac{5}{7}$

Topic: Physics, Thermodynamics, JEE Main
When an ideal gas is compressed by an external pressure greater than its internal pressure, what happens to the volume of the gas?

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

The correct answer is:
The volume decreases

Solution:
When an ideal gas is compressed by an external pressure greater than its internal pressure, the volume of the gas decreases.

Topic: Chemistry, Thermodynamics (C), NEET
When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules

Options:
A Are above the inversion temperature
B Exert no attractive force on each other
C Do work equal to loss in kinetic energy
D Collide without loss of energy

The correct answer is:
Exert no attractive force on each other

Solution:
Ideal gas has no attractive force between the particles

Topic: Chemistry, States of Matter, JEE Main
When an inorganic compound X having 3câ2eâ as well as 2 c - 2 e - bonds react with NH3 gas at a certain temperature, gives a compound Y, isostructural with benzene. Compound X with ammonia at a high temperature produces a substance Z, then

Options:
A X is BH3, Y is B2N2H3, Z is inorganic Benzene
B X is B2H6, Y is B3N3H6, Z is Boron nitride
C X is borax, Y is B2O3, Z is inorganic Benzene
D Reactions insufficient to predict

The correct answer is:
X is B2H6, Y is B3N3H6, Z is Boron nitride

Solution:
This problem is based on preparation of inorganic benzene and structural charactristics of diborane. Student must be familiar with concept of bonding in diborane to solve this problem. Identify the compounds and complete the chemical reaction using information supplied in the question.
B 2 H 6 X + 2NH 3 excess → Δ Low temperature B 2 H 6 ⋅2 NH 3 → Δ 473K B 3 N 3 H 6 borozole  Y
Y has structure similar to benzene (inorganic benzene)
B 2 H 6 + NH 3 excess → Δ High temperature BN X boron nitride   Z

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
When an inorganic compound (X) having 3c-2e as well as 2c-2e bonds react with ammonia gas at a certain temperature, gives a compound (Y) isostructural with benzene. Compound (X) with ammonia at a high temperature produces a substance (Z). Which option is not correct?

Options:
A (X) is B2H6
B (Z) is known as inorganic graphite
C (Y) is B3N3H6
D (Z) is soft like graphite

The correct answer is:
(Z) is soft like graphite

Solution:
Diborane, B2H6 , is a compound having 3c-2e and 2c-2e bonds.
B 2 H 6 â+â 2NH 3 â lowâtemp. B 2 H 6 â 2NH 3 â 200 â Î¿ C B 3 N 3 H 6 Y
(Y) has structure similar to benzene. It is called inorganic benzene.
B 2 H 6  +  NH 3 → high temp. (BN) x Z .
(Z) is called inorganic graphite and is a hard substance.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
When an object is kept at a distance of 30 cm from a concave mirror, the image is formed at a distance of 10 cm from the mirror. If the object is moved with a speed of 9 cm s-1 , the speed (in cm s-1) with which image moves at that instant is

Options:

The correct answer is:
1

Solution:
Velocity of image

vi=-v2u2×v0

=-1023029

=-1 cm sec-1

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (03 Sep Shift 2)
When an object is placed at a distance of \( 25 \mathrm{~cm} \) from a mirror, the magnification is \( m_{1} . \) But the magnification becomes \( m_{2} \), when the object is moved \( 15 \mathrm{~cm} \) farther away with respect to the earlier position and \( \frac{m_{1}}{m_{2}}=4 \), then find the focal length of the mirror and what type of mirror it is?

Options:
A \( 20 \mathrm{~cm} \), convex
B \( 20 \mathrm{~cm} \), concave
C \( 10 \mathrm{~cm} \), convex
D \( 10 \mathrm{~cm} \), concave

The correct answer is:
\( 20 \mathrm{~cm} \), concave

Solution:
m=-vu=ff-u
so,m1=ff--25=ff+25
Similarly,m2=ff+40
andm1m2=4
∴f+40f+25=4
f=-20 cm

Negative sign shows that the mirror is concave.

Topic: Physics, Ray Optics, JEE Main
When an object is viewed with a light of wavelength 6000 â« under a microscope, its resolving power is 104. The resolving power of the microscope when the same object is viewed with a light of wavelength 4000 â«, is nÃ103. The value of n is

Options:

The correct answer is:
15

Solution:
The resolving power of a microscope is given by the following formula
R=2 μsin⁡θ1.22 λ
Where, μ= refractive index
λ= the wavelength of the light
∴ R∝1λ
R2R1=λ1λ2 =6000Å4000Å
R2=32×R1=32×104
=15×103

Topic: Physics, Wave Optics, JEE Main
When an object is viewed with a light of wavelength 6000 â« under a microscope, its resolving power is 104. The resolving power of the microscope when the same object is viewed with a light of wavelength 4000 â«, is nÃ103. The value of n is

Options:

The correct answer is:
15

Solution:
The resolving power of a microscope is given by the following formula
R=2 μsin⁡θ1.22 λ
Where, μ= refractive index
λ= the wavelength of the light
∴ R∝1λ
R2R1=λ1λ2 =6000 Å4000 Å
R2=32×R1=32×104
=15×103

Topic: Physics, Wave Optics, JEE Main
When an open organ is dipped in water upto half of its height, then its frequency will become

Options:
A half
B double
C remain same
D four time

The correct answer is:
remain same

Solution:
Frequency of open organ pipe is double than that of closed pipe.
If frequency of open pipe $=n$
$\therefore$ Frequency of closed pipe $=\frac{n}{2}$
On inserting an open pipe into water upto half of its height, it will behave like a closed pipe of half length and frequency is inversely proportional to the length of air column.
Hence, frequency of closed pipe will become double $=2 \times \frac{n}{2}=n$

Topic: Physics, Waves and Sound, JIPMER, JIPMER 2014
When an optically active amine 'A' having molecular formula C4H11N is subjected to Hoffmann's exhaustive methylation followed by hydrolysis, an alkene 'B' is produced which upon ozonolysis and subsequent hydrolysis yields formaldehyde and propanal. The amine 'A' is

Options:
A
B
C
D CH3CH2CH2CH2NH2

The correct answer is:

Solution:
The only optically active molecule is
ââAgOH, ÎâCH2=CHB-CH2CH3ââi O3ii Zn, H2OâHCHOFormaldehyde+CH3CH2CHOPropanal

Topic: Chemistry, Amines, JEE Main
When an ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum kinetic energy of the photoelectrons is Kmax . When X-rays are incident on the same cell, then

Options:
A V0 and Kmax both increase.
B V0 and Kmax both decrease.
C V0 increases but Kmax remains the same.
D Kmax increases by V0 remains the same.

The correct answer is:
V0 and Kmax both increase.

Solution:
The frequency of X- ray is higher than that of ultraviolet light.
Now, Kmax=hv-v0 .
Hence Kmax increases as v is increased.
Also Kmax=eV0 , where V0 is the stopping potential.

Topic: Physics, Dual Nature of Matter, NEET
When an unpolarised light falls on a polaroid:

Options:
A The electric vectors oscillating perpendicular to the pass-axis of polaroid pass.
B The electric vectors oscillating parallel to the pass-axis of polaroid pass.
C Depending on the direction sometimes all the electric vectors are passed.
D Depending on the direction none of the electric vector passes.

The correct answer is:
The electric vectors oscillating parallel to the pass-axis of polaroid pass.

Solution:
An unpolarised light consists of electric vectors in all possible directions. When it falls on a polaroid, the electric vectors of a light wave which are parallel to the direction of aligned molecules of polaroid, get absorbed. The electric vectors which are parallel to the pass-axis of polaroid are passed. Neither all the electric vectors are passed nor all the vectors are absorbed in any condition.

Topic: Physics, Wave Optics, JEE Main
When an unpolarized light of intensity $I_0$ is incident on a polarizing sheet, the intensity of the light which does not get transmitted is

Options:
A Zero
B $I_0$
C $\frac{1}{2} I_0$
D $\frac{1}{4} I_0$

The correct answer is:
$\frac{1}{2} I_0$

Solution:
If an unpolarised light is converted into plane polarised light by passing through a polaroid, it's intensity becomes half.

Topic: Physics, Wave Optics, JEE Main
When an unpolarized light of intensity $I_0$ is incident on a polarizing sheet, the intensity of the light which does not get transmitted is

Options:
A
$\frac{1}{2} \mathrm{I}_0$

B
$\frac{1}{4} \mathrm{I}_0$

C
zero

D
$\mathrm{I}_0$

The correct answer is:
$\frac{1}{2} \mathrm{I}_0$

Solution:
When unpolarised light of intensity $\mathrm{I}_0$ is incident on a polarizing sheet, only $\mathrm{I}_0 / 2$ is transmitted.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2005
When and where does reduction division take place in the life cycle of a liverwort, a moss, a fern, a gymnosperm and an angiosperm?

Options:

None

Solution:
1. Liverwort-The main part of the body of liverwort is thalloid. Haploid gametes are produced from the male and female sex organs gamets fuses to form zygote. Zygote develops in the form of sporophytes. These sporophytes are further differentiated into foot, seta and capsule. As a result of reduction division many haploid spores are produced in capsule.
II. Moss - In the first stage in moss primary protonema develops into secondary protonema. Both these stages are haploid. Zygote formed by the fusion of gametes further produce sporophytes.
III. Fern-Leaves of sporophyte bear sporangia in which spores are produced by reduction division in meiosis.
IV. Gymnosperm - In microsporophylls and megasporophylls that bear microsporangia and megasporangia respectively, reduction division occurs to produce microspores (pollen grains) and megaspore.
V. Angiosperm - Main part of the body is sporophytic and bears flowers. Reduction division takes place in anthers of stamen i.e. haploid pollen grains and in the ovary of pistil producing eggs.

Topic: Biology, Plant Kingdom
When aniline is nitrated with nitrating mixture in ice cold condition, the major product obtained is

Options:
A $p$ -nitroaniline
B 2,4 dinitroaniline
C $o$ -nitroaniline
D $m$ nitroaniline

The correct answer is:
$p$ -nitroaniline

Solution:
Direct nitration of aniline with nitric acid gives a complex mixture of mono, di- and tri-nitro compounds and oxidation products. If $-\mathrm{NH}_{2}$ group is protected by acetylation and then nitrated with nitrating mixture, $p$ -isomer is the main product.

Topic: Chemistry, Amines, WBJEE, WBJEE 2013
When aniline reacts with $\mathrm{NaNO}_2$ and dil. $\mathrm{HCl}$ at 0-5 degree celcius the product formed is

Options:
A Nitroaniline
B Benzene diazonium chloride
C Benzene
D Trinitroaniline

The correct answer is:
Benzene diazonium chloride

Solution:

Topic: Chemistry, Amines, JEE Main
When are substances said to exhibit ferrimagnetism?

Options:
A When all domains have their magnetic moments aligned.
B When domains have their magnetic moments aligned in parallel directions.
C When domains have their magnetic moments aligned in anti-parallel directions.
D When domains have their magnetic moments aligned in unequal numbers in both parallel and anti-parallel directions.

The correct answer is:
When domains have their magnetic moments aligned in unequal numbers in both parallel and anti-parallel directions.

Solution:
Substances are said to exhibit ferrimagnetism when the magnetic moments of domains are aligned in unequal numbers in both parallel and anti-parallel directions.

Topic: Chemistry, Solid State, NEET
When atoms are bombarded with alpha particles, only a few in million suffer deflection, others pass out undeflected. This is because

Options:
A The force of repulsion on the moving alpha particle is small
B The force of attraction on the alpha particle to the oppositely charged electrons is very small
C There is only one nucleus and large number of electrons
D The nucleus occupies much smaller volume compared to the volume of the atom

The correct answer is:
The nucleus occupies much smaller volume compared to the volume of the atom

Solution:
The nucleus occupies much smaller volume compared to the volume of the atom.

Topic: Chemistry, Structure of Atom, JEE Main
When attraction between \( A-B \) is more than that of \( A-A \) and \( B-B \), the solution will show

Options:
A Positive deviation from Raoult's law
B Negative deviation from Raoult's law
C No deviation from Raoult's law
D Cannot be predicted

The correct answer is:
Negative deviation from Raoult's law

Solution:
When A-B attractions are greater than A-A and B-B, then less vapours are formed and hence, solution shows negative deviation.

Topic: Chemistry, Solutions, JEE Main
When benzene or its derivative is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride, it gives

Options:
A Benzaldehyde
B Benzophenone
C Benzyl alcohol
D Benzal chloride

The correct answer is:
Benzaldehyde

Solution:
By Gatterman-Koch reaction- When benzene or its derivative is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride, it gives benzaldehyde.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
When benzoic acid is treated with \( \mathrm{LiAlH}_{4} \), it forms:

Options:
A benzaldehyde.
B benzyl alcohol.
C benzene.
D toluene.

The correct answer is:
benzyl alcohol.

Solution:
LiAlH4 reduces carboxylic acid group to alcohol. Hence, benzyl alcohol will be formed.

Ph-COOHLiAlH4 Reduction Ph-CH2OH

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
When benzoic acid is treated with $\mathrm{PCl}_5$ at $100^{\circ} \mathrm{C}$, it gives

Options:
A Benzoyl chloride
B o-chlorobenzoic acid
C p-chlorobenzoic acid
D Benzyl chloride

The correct answer is:
Benzoyl chloride

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
When benzyl alcohol is oxidised with $\mathrm{KMnO}_4$, the product obtained is

Options:
A Benzaldehyde
B Benzoic acid
C $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$
D None of these

The correct answer is:
Benzoic acid

Solution:
$\mathrm{C}_6 \mathrm{H}_3 \mathrm{CH}_2 \mathrm{OH}+[\mathrm{O}] \xrightarrow{\mathrm{KNanO}_4} \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
When bob A of a pendulum, released from 30 degrees to the vertical, hits another bob B of the same mass at rest on a table in an elastic collision, how high does bob A rise after the collision?

Options:
A Bob A rises to the same height as its initial release point.
B Bob A does not rise; it remains at the same level.
C Bob A rises to a height less than its initial release point.
D Bob A rises to a height greater than its initial release point.

The correct answer is:
Bob A does not rise; it remains at the same level.

Solution:
Correct Option is : (B)
Bob A does not rise; it remains at the same level.

Topic: Chemistry, Work Power Energy, NEET
When both ovaries of rat are removed then which hormone is decreased in blood?

Options:
A Oxytocin
B Prolactin
C Estrogen
D Gonadotrophic releasing factor

The correct answer is:
Estrogen

Solution:
Estrogens are produced primarily by the ovaries. They are released by the follicles on the ovaries and are also secreted by the corpus luteum after the egg has been released from the follicle and from the placenta. Thus, if the ovaries of rat are removed, then the level of estrogen is decreased in blood.
Related Theory:-
Secretion of progesterone from corpus luteum maintains the endometrium which is necessary for implantation of fertilized ovum and other events of pregnancy.

Topic: Biology, Human Reproduction, NEET, NEET 2002
When breast feeding is replaced by less nutritive food low in proteins and calories, the infants below the age of one year are likely to suffer from :

Options:
A Pellagra
B Marasmus
C Rickets
D Kwashiorkor

The correct answer is:
Marasmus

Solution:
No Solution Available

Topic: Biology, Digestion and Absorption, NEET, NEET 2009 (Mains)
When but-3-en-2-ol reacts with aq HBr, the product formed is

Options:
A 3-bromobut-1-ene
B 1-bromobut-2-ene
C a mixture of both A and B
D 2-bromobut-2-ene

The correct answer is:
a mixture of both A and B

Solution:
The reactants form allyl carbocation, which is stability by delocalization of electrons.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
When CuSO4 solution in water is treated with concentrated HCl it turns

Options:
A Violet
B Yellow
C Purple
D Green

The correct answer is:
Yellow

Solution:
CuSO4aq.+2HClaq.→CuCl2Yellowaq.+H2SO4aq.

Topic: Chemistry, Redox Reactions, JEE Main
When $\mathrm{CH}_2=\mathrm{CH}-\mathrm{COOH}$ is reduced with $\mathrm{LiAlH}_4$, the compound obtained will be

Options:
A
$\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2 \mathrm{OH}$

B
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{OH}$

C
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CHO}$

D
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{COOH}$

The correct answer is:
$\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2 \mathrm{OH}$

Solution:
$\mathrm{LiAl}^{\mathrm{H}}{ }_4$ can reduce $\mathrm{COOH}$ group and not the double bond
$\mathrm{CH}_2=\mathrm{CH}-\mathrm{COOH} \underset{\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2 \mathrm{OH}}{\stackrel{\mathrm{LAH}}{\longrightarrow}}$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2003