datavorous/entrance-exam-dataset huggingface dataset:
When $\mathrm{H}_2 \mathrm{~S}$ is passed through $\mathrm{Hg}_2 \mathrm{~S}$ we get

Options:
A $\mathrm{HgS}$
B $\mathrm{HgS}+\mathrm{Hg}_2 \mathrm{~S}$
C $\mathrm{Hg}_2 \mathrm{~S}+\mathrm{Hg}$
D $\mathrm{Hg}_2 \mathrm{~S}$

The correct answer is:
$\mathrm{Hg}_2 \mathrm{~S}+\mathrm{Hg}$

Solution:
When $\mathrm{H}_2 \mathrm{~S}$ is passed through $\mathrm{Hg}_2 \mathrm{~S}$, it breaks down the $\mathrm{Hg}_2 \mathrm{~S}$ into $\mathrm{Hg}$ and $\mathrm{HgS}$. This is a disproportionation reaction, which is a type of reaction in which a single reactant breaks down into two different products, one of which has a higher oxidation state than the reactant and the other has a lower oxidation state. In the case of $\mathrm{Hg}_2 \mathrm{~S}$, the mercury atoms have an oxidation state of +1 . When $\mathrm{Hg}_2 \mathrm{~S}$ breaks down, one mercury atom goes to +2 (in $\mathrm{HgS}$) and the other goes to 0 (in $\mathrm{Hg})$.
The chemical equation for the reaction is as follows:
$\mathrm{Hg}_2 \mathrm{~S}+\mathrm{H}_2 \mathrm{~S} \rightarrow \gt \mathrm{HgS}+\mathrm{Hg}+\mathrm{H}_2$
This reaction is used in the production of mercury. HgS is the main ore of mercury, and when it is heated with a reducing agent such as $\mathrm{H}_2 \mathrm{~S}$, it decomposes to produce mercury metal. Hg metal is then condensed and collected.

Topic: Chemistry, Practical Chemistry, JEE Main
When $\mathrm{H}_2 \mathrm{~S}$ is passed through $\mathrm{Hg}_2 \mathrm{~S}$ we get

Options:
A
$\mathrm{HgS}$

B
$\mathrm{HgS}+\mathrm{Hg}_2 \mathrm{~S}$

C
$\mathrm{Hg}_2 \mathrm{~S}$

D
$\mathrm{Hg}_2 \mathrm{~S}_2$

The correct answer is:
$\mathrm{Hg}_2 \mathrm{~S}$

Solution:
No Solution Available

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2002
When $\mathrm{H}_2 \mathrm{~S}$ is passed through in II group sometimes solution becomes milky. It indicates the presence of

Options:
A Acidic salt
B An oxidising agent
C Thiosulphate
D A reducing agent

The correct answer is:
An oxidising agent

Solution:
For the solution to become milky $\mathrm{H}_2 \mathrm{~S}$ must be oxidised to $\mathrm{SO}_2$

Topic: Chemistry, Practical Chemistry, JEE Main
When huge amount of sewage is dumped into a river, its BOD will

Options:
A Increase
B Decrease
C Sharply decrease
D Remain unchanged

The correct answer is:
Increase

Solution:
By dumping of a huge amount of sewage, the oxygen levels are depleted, which are reflected in terms of BOD values of water. The number of microbes also increases tremendously and these also consume most of the oxygen. Thus, BOD of the river water will increase.

Topic: Biology, Environmental Issues, NEET
When initial concentration of the reactant is doubled, the half-life period of a zero-order reaction

Options:
A Is tripled
B Is doubled
C Is halved
D Remains unchanged

The correct answer is:
Is doubled

Solution:
Half-life of zero order
t1/2=[A0]2k i.e., t1/2∝ a or (A0)
t1/2 will be doubled on doubling the initial concentration.

Topic: Chemistry, Chemical Kinetics, NEET
When initial concentration of the reactant is doubled, what happens to the half-life period of a zero order reaction?

Options:
A It is tripled.
B It is doubled.
C It is halved.
D It remains unchanged.

The correct answer is:
It is doubled.

Solution:
Order of reaction =0

Half lifer
t12 = A02Kt12∝A0
∴ Initial concentration doubled then half-life i.e t 1 2 becomes double

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2018
When $\mathrm{KBr}$ is dissolved in water, $\mathrm{K}^{+}$ions are

Options:
A oxidised
B reduced
C hydrolysed
D hydrated

The correct answer is:
hydrated

Solution:
When $\mathrm{KBr}$ is dissolved in water, $\mathrm{K}^{+}$ions are hydrated due to ionic nature. $\mathrm{KBr}$ is soluble in wuter due to hydration of $\mathrm{K}^{+}$ion $\mathrm{by} \mathrm{H}_{2} \mathrm{O}$ molecule.

Topic: Chemistry, Redox Reactions, JEE Main
When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be:

Options:
A $6 \%$
B $600 \%$
C $60 \%$
D $500 \%$

The correct answer is:
$500 \%$

Solution:
Kinetic energy $(K)=\frac{P^2}{2 m}$
$\Rightarrow \mathrm{P}=\sqrt{2 \mathrm{mK}}$
If $K_f=36 K_i$
So, $\mathrm{P}_{\mathrm{f}}=6 \mathrm{P}_{\mathrm{i}}$
$\begin{aligned}
\% \text { increase in momentum } & =\frac{\mathrm{P}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{i}}} \times 100 \% \\
& =\frac{6 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{i}}} \times 100 \% \\
& =500 \%
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2024 (06 Apr Shift 2)
When $\mathrm{KMnO}_4$ acts as an oxidising agent and ultimately forms $\left[\mathrm{MnO}_4\right]^{-1}, \mathrm{MnO}_2, \mathrm{Mn}_2 \mathrm{O}_3, \mathrm{Mn}^{+2}$ then the number of electrons transferred in each case respectively is

Options:
A
4, 3, 1, 5

B
1, 5, 3, 7

C
1, 3, 4, 5

D
3, 5, 7, 1

The correct answer is:
1, 3, 4, 5

Solution:

Topic: Chemistry, d and f Block Elements, JEE Main
When $\mathrm{KMnO}_4$ acts as an oxidising agent and ultimately forms $\left[\mathrm{MnO}_4\right]^{-1}, \mathrm{MnO}_2, \mathrm{Mn}_2 \mathrm{O}_3, \mathrm{Mn}^{+2}$ then the number of electrons transferred in each case respectively is

Options:
A
4, 3, 1, 5

B
1, 5, 3, 7

C
1, 3, 4, 5

D
3, 5, 7, 1

The correct answer is:
1, 3, 4, 5

Solution:

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2002
When $\mathrm{KMnO}_4$ acts as an oxidizing agent and ultimately forms $\left[\mathrm{MnO}_4\right]^{-2}, \mathrm{MnO}_2, \mathrm{Mn}_2 \mathrm{O}_3, \mathrm{Mn}^{+2}$ then the number of electrons transferred in each case respectively is

Options:
A $4,3,1,5$
B $1,5,3,7$
C $1,3,4,5$
D $3,5,7,1$

The correct answer is:
$1,3,4,5$

Solution:
In \(\mathrm{KMnO}_4\) the \(\mathrm{O} . \mathrm{N}\). of \(\mathrm{Mn}\) is +7, in \(\mathrm{MnO}_4^{-2}+6\), in \(\mathrm{MnO}_2+4\) in \(M n_2 \mathrm{O}_3+3\) and in \(\mathrm{Mn}^{2+}\) is +2. The difference being \(1,3,4\)

Topic: Chemistry, Redox Reactions, JEE Main
When lead storage battery is charged

Options:
A $\mathrm{PbO}_2$ is dissolved
B $\mathrm{H}_2 \mathrm{SO}_4$ is regenerated
C $\mathrm{PbSO}_4$ is deposited on lead electrode
D Lead is deposited on lead electrode

The correct answer is:
$\mathrm{PbSO}_4$ is deposited on lead electrode

Solution:
When a lead storage battery is charged, the chemical reactions are reversed from when it is discharged. During charging, lead sulfate $\left(\mathrm{PbSO}_4\right)$ is deposited on the lead electrode. Sulfuric acid $\left(\mathrm{H}_2 \mathrm{SO}_4\right)$ is regenerated, not decreased.
Here's a more detailed explanation of the chemical reactions that take place when a lead storage battery is charged:
- Positive electrode (anode): $\mathrm{PbSO}_4$ (lead sulfate) $+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}$ (lead) $+\mathrm{SO}_4{ }^{2-}$ (sulfate ion)
- Negative electrode (cathode): $\mathrm{PbSO}_4$ (lead sulfate) $+4 \mathrm{H}^{+}$(hydrogen ion) $+2 \mathrm{e}^{-}$ $\rightarrow \mathrm{PbO}_2$ (lead dioxide) $+2 \mathrm{H}_2 \mathrm{O}$ (water)
As you can see, the lead sulfate on the electrodes is converted back into lead (at the anode) and lead dioxide (at the cathode) during charging. The sulfuric acid concentration also increases because sulfate ions $\left(\mathrm{SO}_4{ }^2\right)$ are not used up in the reaction, but instead recombine with hydrogen ions $\left(\mathrm{H}^{+}\right)$to form sulfuric acid $\left(\mathrm{H}_2 \mathrm{SO}_4\right)$.

Topic: Chemistry, Electrochemistry, JEE Main
When light enters glass from vacuum then the wavelength of light

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

The correct answer is:
decreases.

Solution:
When a light enters glass form vacuum, then the wavelnght of light decrease.
This is because the speed of light in glass in less than that in the vacuum.
From Snell's law the refractive index in a medium is given by.
$\mu=\frac{c}{v}$
The glass is of higher refractive index than vaccum, so speed of light decrease and as teh speed of light is drectly proportional to the wavelenght ( $c \equiv v \lambda$ ), So the wavelenght also be decrease.

Topic: Physics, Ray Optics, MHT CET, MHT CET 2020 (14 Oct Shift 1)
When light enters glass from vacuum, then the wavelength of light

Options:
A Decrease
B Become zero
C Remains same
D Increases

The correct answer is:
Decrease

Solution:
When a light enters glass from vacuum, then the wavelength of light decreases.
This is because the speed of light in glass is less than that in the vacuum.
From Snell’s law, the refractive index in a medium is given by
μ=cv
The glass is of higher refractive index than vacuum, so speed of light decreases and as the speed of light is directly proportional to the wavelength c=vλ , so, the wavelength also be decreases.

Topic: Physics, Ray Optics, MHT CET, MHT CET 2019 (Shift 2)
When light is refracted from a surface, which of its following physical parameters does not change?

Options:
A Velocity
B Amplitude
C Frequency
D Wavelength

The correct answer is:
Frequency

Solution:
When light is refracted its frequency will remain unchanged.

Topic: Physics, Ray Optics, WBJEE, WBJEE 2015
When light of an unknown polarisation is examined with a polaroid, it is found to exhibit maximum intensity $I_0$ along $y$-axis and minimum intensity $\frac{2 Y_0}{3}$ along $x$-axis. The intensity transmited through a polaroid with pass axis at $45^{\circ}$ to $y$-axis (in $x$ - $y$ plane) is

Options:
A $\left.\frac{5}{8}\right\}_0$
B $\frac{f_0}{2}$
C $\frac{5}{6} f_0$
D $\frac{f_0}{4}$

The correct answer is:
$\frac{5}{6} f_0$

Solution:
Given, maximum intensity along y axis,
$$
Y_Y=Y_0
$$
and minimum intensity along $x$ axis,
$$
Y_X=\frac{2 Y_0}{3}
$$
The intensity transmitted, through a polaroid with pass axis $45^{\circ}$ to $y$ axis (ie., $45^{\circ}$ to $x$ axis also) is given by Brewster's law,
$$
\begin{aligned}
Y & =Y_x \cos ^2 45^{\circ}+Y_y \cos ^2 45^{\circ} \\
& =Y_0 \cos ^2 45^{\circ}+\frac{2 Y_0}{3} \cos ^2 45^{\circ} \\
& =Y_0 \cdot \frac{1}{2}+\frac{2 Y_0}{3} \cdot \frac{1}{2}=\frac{Y_0}{2}+\frac{Y_0}{3}=\frac{5 Y_0}{6}
\end{aligned}
$$

Topic: Physics, Wave Optics, JEE Main
When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v1. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v2. If v2=xv1, the value of x will be _____ .

Options:

The correct answer is:
2

Solution:
Maximum kinetic energy is expressed as Kmax=hf-ϕ, here, ϕ is work function of metal.

For first case, 12mv12=h2f0−f0

For second case, 12mv22=h5f0−f0

⇒v22v12=41⇒v2v1=2

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2022 (24 Jun Shift 1)
When light of wavelength Î» is incident on photosensitive surface, the stopping potential is V. When light of wavelength 3Î» is incident on same surface, the stopping potential is V6. Threshold wavelength for the surface is

Options:
A 2λ
B 3λ
C 4λ
D 5λ

The correct answer is:
5λ

Solution:
Using hcÎ»-Ï=eV
When, wavelength Î» is incident on metallic surface, the sopping potential required to stop most energetic electron is V and when wavelength increases to 3Î» , stopping potential required as V6
â´ hcÎ»=eV+Ï ..... (i)
And hc3Î»=eV6+Ï .... (ii)
Thus equation (ii) can be rewritten as 2hcÎ»=eV+6Ï ..... (iii)
Subtracting equation (i) from (iii) we get
hcÎ»=5Ï and Ï=hcÎ»0
Thus Î»0=5Î»

Topic: Physics, Dual Nature of Matter, JEE Main
When light rays are incident on a prism at an angle of $45^{\circ}$, the minimum deviation is obtained. If refractive index of the material of prism is $\sqrt{2}$, then the angle of prism will be

Options:
A $30^{\circ}$
B $40^{\circ}$
C $50^{\circ}$
D $60^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
$\frac{\sin \frac{A+\delta m}{2}}{\sin \frac{A}{2}}=\mu$ But $\frac{A+\delta_m}{2}=i=45^{\circ}$
So $\frac{\sin 45^{\circ}}{\sin (A / 2)}=\sqrt{2} \Rightarrow \frac{1}{2}=\sin \frac{A}{2} \Rightarrow A=60^{\circ}$

Topic: Physics, Ray Optics, JEE Main
When light rays from the sun fall on a convex lens along a direction parallel to its axis

Options:
A Focal length for all colours is the same
B Focal length for violet colour is the shortest
C Focal length for yellow colour is the longest
D Focal length for red colour is the shortest

The correct answer is:
Focal length for violet colour is the shortest

Solution:
Focal length for voilet colours is minimum

Topic: Physics, Ray Optics, JEE Main
When light travels from one medium to the other of which the refractive index is different, then which of the following will change

Options:
A Frequency, wavelength and velocity
B Frequency and wavelength
C Frequency and velocity
D Wavelength and velocity

The correct answer is:
Wavelength and velocity

Solution:
Velocity and wavelength change but frequency remains same.

Topic: Physics, Ray Optics, JEE Main
When lime stone is heated strongly, it gives off. In metallurgy this process is known as

Options:
A Calcination
B Roasting
C Smelting
D Ore dressing

The correct answer is:
Calcination

Solution:
$\mathrm{CaCO}_3 \rightarrow \mathrm{CaO}+\mathrm{CO}_2$
This process is known as Caldnation.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
When M1 gram of ice at -10oC (specific heat =0.5 cal g-1 ℃-1 ) is added to M2 gram of water at 50 oC, finally no ice is left and the water is at 0 oC . The value of latent heat of ice, in cal g-1 is:

Options:
A 50M2M1
B 5M1M2-50
C 5M2M1-5
D 50M2M1-5

The correct answer is:
50M2M1-5

Solution:
Principle of calorimetry
M1Sice0--10+M1Lf=M2Swater50-0
M10.510+M1Lf=M250
5M1+M1Lf=M250
Lf=50M2M1-5

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2019 (12 Apr Shift 1)
When m-chlorobenzaldehyde is treated with $50 \% \mathrm{KOH}$ solution, the products (s) obtained is (are):

Options:
A

B

C

D

The correct answer is:

Solution:
When m-chlorobenzaldehyde is treated with $50 \% \mathrm{KOH}$ solution, the product(s) obtained are potassium metachloro benzoate and metachloro benzyl alcohol. It undergoes Cannizzaro reaction in which one molecule is oxidized to metachloro benzoic acid and other molecule is reduced to metachloro benzyl alcohol.
Related Theory
The Cannizzaro reaction is not given by Acetaldehyde. Aldehydes that contains alpha hydrogen atoms are not involved in Cannizzaro reaction and aldehyde like acetaldehyde contains alpha hydrogen in it.
Caution
m-chlorobenzaldehyde has no $\alpha$-hydrogen and hence gives Cannizzaro's reaction.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
When metal M is treated with NaOH, a white gelatinous precipitate X is obtained, which is soluble in excess of NaOH. Compound X when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal M is

Options:
A Fe
B Zn
C Ca
D Al

The correct answer is:
Al

Solution:
Alumina Al2O3 is used in chromatography

Al+NaOH→Al(OH)3(gelatinous)→NaOH(excess)[Al(OH)4]−(soluble)→ΔAl2O3

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2018 (08 Apr)
When metal M is treated with NaOH, a white gelatinous precipitate X is obtained, which is soluble in excess of NaOH. Compound X when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal M is

Options:
A Fe
B Zn
C Ca
D Al

The correct answer is:
Al

Solution:
Alumina Al2O3 is used in chromatography

Al+NaOH→Al(OH)3(gelatinous)→NaOH(excess)[Al(OH)4]−(soluble)→ΔAl2O3

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
When metal 'M' is treated with NaOH, a white gelatinous precipitate 'X' is obtained, which is soluble in excess of NaOH. Compound 'X' when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal 'M' is

Options:
A Fe
B Zn
C Ca
D Al

The correct answer is:
Al

Solution:
2Al+6H2O→2AlOH3↓+3H2
AlOH3+NaOH→NaOH4 soluble

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
When methyl iodide is heated with ammonia, the product obtained is

Options:
A Methylamine
B Dimethylamine
C Trimethylamine
D A mixture of the above three amines

The correct answer is:
A mixture of the above three amines

Solution:
$\mathrm{CH}_3 \mathrm{I} \xrightarrow[\Delta]{\mathrm{NH}_3} \underset{\text { Methyl mmine }}{\mathrm{CH}_3 \mathrm{NH}_2} \xrightarrow{\mathrm{CH}_3 \mathrm{I}} \underset{\text { Dimethyl mmine }}{\left(\mathrm{CH}_3\right)_2 \mathrm{NH}} \xrightarrow{\mathrm{CH}_3 \mathrm{I}} \underset{\text { Trimetyhumine }}{\left(\mathrm{CH}_3\right)_3 \mathrm{~N}}$

Topic: Chemistry, Amines, JEE Main
When MnO 2 is fused with KOH and KNO 3 , a coloured compound is formed. Choose the right compound with the appropriate colour.

Options:
A K 2 MnO 4 , green
B KMnO 4 , purple
C Mn 2 O 3 , brown
D Mn 3 O 4 , black
E MnO 2 , black

The correct answer is:
K 2 MnO 4 , green

Solution:
When MnO 2 reacts with fused mixture of KOH and KNO 3 , it forms K 2 MnO 4 (Green compound) as follows:
MnO 2 + KNO 3 +2 KOH → Δ K 2 MnO 4 green + KNO 2 + H 2 O

Topic: Chemistry, d and f Block Elements, JEE Main
When monochromatic radiation of intensity $I$ falls on a metal surface, the number of photoelectron and their maximum kinetic energy are $N$ and $T$ respectively. If the intensity of radiation is 2I, the number of emitted electrons and their maximum kinetic energy are respectively

Options:
A $\mathrm{N}$ and $2 \mathrm{~T}$
B $2 \mathrm{~N}$ and $\mathrm{T}$
C $2 \mathrm{~N}$ and $2 \mathrm{~T}$
D $\mathrm{N}$ and $\mathrm{T}$

The correct answer is:
$2 \mathrm{~N}$ and $\mathrm{T}$

Solution:
(i) Kinetic energy of photoelectrons depends on the frequency of incident radiations and is independent of the intensity of illumination.
(ii) Number of photoelectrons depends upon the intensity.
So, the number of emitted electrons and their maximum kinetic energy are $2 N$ and $T$ respectively.

Topic: Physics, Dual Nature of Matter, NEET, NEET 2010 (Mains)
When $\mathrm{NaOH}$ crystals are left in open air, they acquire a fluid layer around each crystal as

Options:
A They start melting
B They absorb moisture from air
C They react with air to form a liquid compound
D They absorb $\mathrm{CO}_2$ from air

The correct answer is:
They absorb moisture from air

Solution:
$\mathrm{NaOH}$ is a deliquescent white crystalline solid. It absorbs moisture from the atmosphere.

Topic: Chemistry, s Block Elements, JEE Main
When neopentyl alcohol is heated with an acid, it slowly converted into an 85 : 15 mixture of alkenes A and B, respectively. What are these alkenes?

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2020 (04 Sep Shift 1)
When $\mathrm{NH}_{3}$ is passed over heated metal $\mathrm{A}$, its amide is formed. The metal is

Options:
A $\mathrm{Mg}$
B $\mathrm{K}$
C $\mathrm{Al}$
D $\mathrm{Pb}$

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

Solution:
When Potassium is treated with ammonia, then potassium amide is obtained

$\mathrm{K}+\mathrm{NH}_{3} \longrightarrow \underset{\text { Pot. amide }}{\mathrm{KNH}_{2}}+\frac{1}{2} \mathrm{H}_{2}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
When npn transistor is used as amplifier

Options:
A
electrons move from base to collector

B
holes move from emitter to base

C
electrons move from collector to base

D
holes move from base to emitter.

The correct answer is:
electrons move from base to collector

Solution:
When $n p n$ transistor is used, majority charge carrier electrons of $\mathrm{n}$ type emitter move from emitter to base and then base to collector.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2004
When NPN transistor is used as an amplifier

Options:
A electrons move from base to collector
B holes move from emitter to base
C electrons move from collector to base
D holes move from base to collector

The correct answer is:
electrons move from base to collector

Solution:
Elelctrons of n-type emitter moves from emitter to base and then base to the collector when NPN transistor is used as an amplifier.

Topic: Physics, Semiconductors, JEE Main
When one light ray is reflected from a plane mirror with 30° angle of reflection, the angle of deviation of the ray after reflection is:

Options:
A 120°
B 110°
C 140°
D 130°

The correct answer is:
120°

Solution:
It is given that i=30°.

The formula for angle of deviation is
δ=180°-2i

Substituting the values,
δ=180°-60°
=120°

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (11 Apr Shift 2)
When one mole of zinc dust and one mole of sulphuric acid react in a bomb calorimeter, then ΔU and w correspond to

Options:
A ΔU>0, w=0
B ΔU<0, w=0
C ΔU=0, w=0
D ΔU<0, w>0

The correct answer is:
ΔU>0, w=0

Solution:
Bomb calorimeter is commonly used to find the heat of combustion of organic substances which consists of a sealed combustion chamber, called a bomb. If a process is run in a sealed container then no expansion or compression is allowed, so w= 0 and ÎU=q=-ve (exothermic reaction) .
ÎU <0,w= 0

Topic: Chemistry, Thermodynamics (C), JEE Main
When orthoboric acid $\left(\mathrm{H}_3 \mathrm{BO}_3\right)$ is heated, the residue left is

Options:
A Metaboric acid
B Boron
C Boric anhydride
D Borax

The correct answer is:
Boric anhydride

Solution:
$2 \mathrm{H}_3 \mathrm{BO}_3 \rightarrow \mathrm{B}_2 \mathrm{O}_3+3 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
When p-n junction diode is forward biased, then

Options:
A the depletion region is reduced and barrier height is increased
B the depletion region is widened and barrier height is reduced
C both the depletion region and barrier height are reduced
D both the depletion region and barrier height are increased

The correct answer is:
both the depletion region and barrier height are reduced

Solution:
When p-n junction diode is forward biased, both the depletion region and barrier height are reduced.

Topic: Physics, Semiconductors, JEE Main
When $\mathrm{p}-\mathrm{n}$ junction diode is reverse biased, then the width of the barrier potential will

Options:
A increase and it will offer more resistance.
B decrease and it will offer zero resistance.
C remain constant and it will not offer resistance.
D decrease and it will offer more resistance.

The correct answer is:
increase and it will offer more resistance.

Solution:
No Solution.

Topic: Physics, Semiconductors, MHT CET, MHT CET 2020 (20 Oct Shift 2)
When $\alpha$-particles are sent through a thin metal foil, most of them go straight through the foil because (one or more are correct)

Options:
A Alpha particles are much heavier than electrons
B Alpha particles are positively charged
C Most part of the atom is empty space
D Alpha particles move with high velocity

The correct answer is:
Most part of the atom is empty space

Solution:
$\alpha$-particles pass through because most part of the atom is empty

Topic: Chemistry, Structure of Atom, JEE Main
When $\beta$-particles are sent through a tin metal foil, most of them go straight through the foil as

Options:
A $\beta$-particles are much heavier than electrons
B $\beta$-particles are positively charged
C Most part of the atom is empty space
D $\beta$-particles move with high velocity

The correct answer is:
Most part of the atom is empty space

Solution:
$\alpha$-particles pass through because most part of the atom is empty.

Topic: Chemistry, Structure of Atom, JEE Main
When phenol is allowed to react with $\mathrm{Br}_2$ in (i) $\mathrm{CS}_2$ solution and (ii) in aqueous solution, the resulting compounds are

Options:
A (i) 2, 4, 6-tribromophenol and
(ii) $o$-and $p$-bromophenol
B (i) $m$-bromophenol and
(ii) 2, 3, 4-tribromophenol
C (i) o-and p-bromophenol and
(ii) 2, 4, 6-tribromophenol
D (i) $o$ - and $m$-bromophenol and
(ii) 2, 3, 4-tribromophenol

The correct answer is:
(i) o-and p-bromophenol and
(ii) 2, 4, 6-tribromophenol

Solution:
In aq. Solution phenol ionizes to give phenoxide in which highly activates benzene ring and give trisubstituted product while in presence of \(\mathrm{CS}_2\) an intert solvant phenol is unable to ionize due to which benzene ring in slightly activated. Hence, monosubstituted product is obtained.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main