datavorous/entrance-exam-dataset huggingface dataset:
The function of time representing a simple harmonic motion with a period of πω is :

Options:
A sin(ωt)+cos(ωt)
B cos(ωt)+cos(2ωt)+cos(3ωt)
C sin2(ωt)
D 3cosπ4-2ωt

The correct answer is:
3cosπ4-2ωt

Solution:
Time period T=2πω'

πω=2πω'

ω'=2ω→ Angular frequency of SHM

Option (3)

sin2ωt=122sin2ωt=121-cos2ωt

Angular frequency of 12-12cos2ωt is 2ω

Option (4)

Angular frequency of SHM

3cosπ4-2ωt is 2ω.

So option (3) & (4) both have angular frequency 2ω but option (4) is direct answer.

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (18 Mar Shift 2)
The function $\sin ^2(\omega t)$ represents

Options:
A
a periodic, but not simple harmonic motion with a period $2 \pi / \omega$

B
a periodic, but not simple harmonic motion with a period $\pi / \omega$

C
a simple harmonic motion with a period $2 \pi / \omega$

D
a simple harmonic motion with a period $\pi / \omega$

The correct answer is:
a simple harmonic motion with a period $\pi / \omega$

Solution:
$y=\frac{(1-\cos 2 \omega t)}{2}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2005
The function $\frac{x-2}{x+1},(x \neq-1)$ is increasing on the interval

Options:
A $(-\infty, 0]$
B $[0, \infty)$
C $R$
D None of these

The correct answer is:
$R$

Solution:
$\begin{array}{ll}\text { Since } & f^{\prime}(x)=\frac{3}{(x+1)^2} \text { is greater than '0' in interval }(-\infty, \infty) \text {, therefore } f(x)=\frac{x-2}{x+1} \text { is increasing in interval } \\ \text { is }\end{array}$ $(-\infty, \infty)$ or $R$.

Topic: Mathematics, Application of Derivatives, JEE Main
The function $x^4-4 x$ is decreasing in the interval

Options:
A $[-1,1]$
B $(-\infty, 1)$
C $[1,+\infty)$
D None of these

The correct answer is:
$(-\infty, 1)$

Solution:
Let $f(x)=x^4-4 x \Rightarrow f^{\prime}(x)=4 x^3-4$
So, $4 x^3-4 \lt 0$ or $x^3 \lt 1$
Hence function of decreasing in $(-\infty, 1)$

Topic: Mathematics, Application of Derivatives, JEE Main
The function x5-5x4+5x3-1 is

Options:
A maximum at x=3 and minimum at x=1
B minimum at x=1
C neither maximum nor minimum at x=0
D maximum at x=0

The correct answer is:
neither maximum nor minimum at x=0

Solution:
Let, fx=x5-5x4+5x3-1
â f â² x =5 x 4 â20 x 3 +15 x 2 =0
â´ x 2 x-3 x-1=0 or x=3,1,0
Now, f â³ x =20 x 3 â60 x 2 +30x
Put x=3,1,0, we get f â³ 3 >0,â f â³ 1 <0,â f â³ 0 =0
Hence fx is minimum at x=3, maximum at x=1 , neither maximum nor minimum at x=0.

Topic: Mathematics, Application of Derivatives, JEE Main
The function $y=e^{-|x|}$ is

Options:
A Continuous and differentiable at $x=0$
B Neither continuous nor differentiable at $x=0$
C Continuous but not differentiable at $x=0$
D Not continuous but differentiable at $x=0$

The correct answer is:
Continuous but not differentiable at $x=0$

Solution:
We have,
$f(x)= \begin{cases}e^{-x}, & x \geq 0 \\ e^x, & x \lt 0\end{cases}$
Clearly, $f(x)$ is continuous and differentiable for all non zero $x$.
Now $\lim _{x \rightarrow 0-} f(x)=\lim _{x \rightarrow 0} e^x=1 \quad \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) e^{-x}=1$
Also, $f(0)=e^0=1$. So, $f(x)$ is continuous for all $x$.
$($ LHD at $x=0)=\left(\frac{d}{d x}\left(e^x\right)\right)_{x=0}=1$
(RHD at $x=0)=\left(\frac{d}{d x}\left(e^{-x}\right)\right)_{x=0}=-1$
So, $f(x)$ is not differentiable at $x=0$.
Hence $f(x)=e^{-|x|}$ is everywhere continuous but not differentiable at $x=0$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function $y=|\sin x|$ is continuous for any $x$ but it is not differentiable at

Options:
A $x=0$ only
B $x=\pi$ only
C $x=k \pi(k$ is an integer) only
D $x=0$ and $x=k \pi(k$ is an integer)

The correct answer is:
$x=0$ and $x=k \pi(k$ is an integer)

Solution:
It can be easily seen from the graph of $f(x)=|\sin x|$ that it is every where continuous but not differentiable at integer multiples of $\pi$ and at $x=0$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function $y=\tan ^{-1} x-x \quad$

Options:
A is always decreasing
B is always increasing
C first increases and then decreases
D first decreases and then increases

The correct answer is:
is always decreasing

Solution:
Let $y=\tan ^{-1} x-x$
On differentiating w.r.t $x$, we get $\frac{d y}{d x}=\frac{1}{1+x^{2}}-1=\frac{1-1-x^{2}}{1+x^{2}}=\frac{-x^{2}}{1+x^{2}}$
$\Rightarrow \frac{d y}{d x} < 0, \forall x \in R$
Hence, function is always decreasing.

Topic: Mathematics, Application of Derivatives, JEE Main
The function y=x4-8x3+22x2-24x+10 attains local maximum or minimum at x=a, x=b and x=c a<b<c. Then a, b and c are in

Options:
A Geometric progression
B Harmonic progression
C Arithmetic progression
D None of these

The correct answer is:
Arithmetic progression

Solution:
dydx=4x3-24x2+44x-24
=4x3-6x2+11x-6
=4x-1x-2x-3
So, dydx=0 at x=1, 2, 3
which are the points of local extremum.
i.e., a=1
b=2
c=3

Hence, a, b, c are in AP

Topic: Mathematics, Application of Derivatives, JEE Main
The function \(y=\sin ^{-1}(\cos x)\) is not differentiable at ____

Options:
A \(\pi\) only
B \(2 \pi\) only
C \(-2 \pi\) only
D All options are correct

The correct answer is:
All options are correct

Solution:
Domain for \(\sin ^{-1} x\)
is, \(x \in(-1,1)\) both values -1 and 1 are excluded
Now, \(\cos \pi=-1\)
and \(\cos 2 \pi=1\)
Also, \(\cos -2 \pi=1\)
Hence, \(\sin ^{-1}(\cos x)\) is non differentiable at \(\pi, 2 \pi\) and \(-2 \pi\).

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function \(y=\sin ^{-1}(\cos x)\) is not differentiable at ____

Options:
A \(\pi\) only
B \(2 \pi\) only
C \(-2 \pi\) only
D All options are correct

The correct answer is:
All options are correct

Solution:
Domain for \(\sin ^{-1} x\)
is, \(x \in(-1,1)\) both values -1 and 1 are excluded
Now, \(\cos \pi=-1\)
and \(\cos 2 \pi=1\)
Also, \(\cos -2 \pi=1\)
Hence, \(\sin ^{-1}(\cos x)\) is non differentiable at \(\pi, 2 \pi\) and \(-2 \pi\).

Topic: Mathematics, Continuity and Differentiability, AP EAMCET, AP EAMCET 2020 (17 Sep Shift 2)
The functional group that shows negative resonance effect is:

Options:
A -NH2
B -OH
C -COOH
D -OR

The correct answer is:
-COOH

Solution:
The resonance effect definition can be given as a chemical phenomenon, observed in the characteristic compounds containing double bonds in the organic compounds. The organic compounds have these double bonds in the structures and have the overlapping of the p-orbitals, usually on the two adjacent sides of carbon atoms.

Negative Resonance Effect: The Negative resonance effect happens when the groups withdraw the electrons from the other molecules by the delocalisation process. Usually, the groups are denoted either by -R or -M. The molecular electron density is said to decrease in this process. The negative resonance effect examples are, C=O, -COOH, -C≡N and -NO2.

-COOH shows –R effect, while rest 3 groups shows +R effect via lone pair.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (01 Feb Shift 2)
The functional group which is formed when Phenol is made to react with Chloroform in the presence of dilute Sodium hydroxide

Options:
A \( -\mathrm{CH}_{2} \mathrm{Cl} \)
B \( -\mathrm{COOH} \)
C \( -\mathrm{CHCl}_{2} \)
D \( -\mathrm{CHO} \)

The correct answer is:
\( -\mathrm{CHO} \)

Solution:
The reaction of phenol with chloroform in presence of dilute sodium hydroxide finally introduces aldehyde -CHO functional group and hydroxy group at -ortho position. The reaction is known as the Reimer Tiemann reaction.

Reimer Tiemann reaction:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The functional group, which is found in amino acid is

Options:
A
$\mathrm{COOH}$ group

B
$\mathrm{NH}_2$ group

C
$\mathrm{CH}_3$ group

D
both (a) and (b)

The correct answer is:
both (a) and (b)

Solution:
No Solution Available

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2002
The functional groups present in 'salol' are

Options:
A $-\mathrm{NH}_2$ and $-\mathrm{OR}$
B $\mathrm{OH}$ and $-\mathrm{COR}$
C $-\mathrm{NH}_2$ and $-\mathrm{COOH}$
D $-\mathrm{OH}$ and $-\mathrm{COOR}$

The correct answer is:
$-\mathrm{OH}$ and $-\mathrm{COOR}$

Solution:
The functional group present in salol are $-\mathrm{OH}$ and $-\mathrm{COOR}$. Salol is phenyl salicylate. It's structure is as :

Topic: Chemistry, General Organic Chemistry, JEE Main
The functional megaspore referred to as:

Options:
A The megaspore that degenerates after formation.
B The megaspore that only develops in female gametophyte.
C The megaspore that undergoes reduction division.
D The megaspore that is functionally inactive.

The correct answer is:
The megaspore that only develops in female gametophyte.

Solution:
The integuments of the ovule encloses a tissue called nucellus except at an opening, micropyle. Within the nucellus a cell, namely, megaspore mother cell forms four haploid megaspores via meiosis forming a linear tetrad. Only one of the four remains functional which later develops into embryo sac or female gametophyte. This, daughter cell of the tetrad is referred to as functional megaspore. Functional megaspore can either be atmicropylar end or chalazal end depending upon the species of the plant.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The functions of antihistamine are :

Options:
A Antiallergic and Analgesic
B Antacid and Antiallergic
C Analgesic and antacid
D Antiallergic and antidepressant

The correct answer is:
Antacid and Antiallergic

Solution:
Antihistamines are used to relieve or prevent the symptoms of hay fever and other types of allergy. They work by preventing the effects of a substance called histamine, which is produced by the body. Histamines can cause itching, sneezing, runny nose, and watery eyes. Hence, antihistamines are used as antacid and Antiallergic.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2021 (16 Mar Shift 1)
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60 cm, the length of the closed pipe will be :

Options:
A 60 cm
B 45 cm
C 30 cm
D 15 cm

The correct answer is:
15 cm

Solution:
λ4=L1 2λ2=L2

As, v=fλ ⇒f2=2v2L2

⇒v=f14L1 ⇒f2=vL2

⇒f1=v4L1

As per question, f1=f2

v4L1=vL2

⇒L2=4L1

⇒60=4×L1

⇒L1=15 cm

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of the organ pipe open at both the ends is

Options:
A 80 cm
B 100 cm
C 120 cm
D 140 cm

The correct answer is:
120 cm

Solution:
For closed organ pipe,
\(n_c=\frac{V}{4 l}\)
For open organ pipe,
\(n_0=\frac{V}{2 l^{\prime}}\)
For the second overtone of an open organ pipe,
\(\begin{aligned}
& n^{\prime}=3 n_0=\frac{3 V}{2 l^{\prime \prime}} \\
& n_c=n^{\prime} \\
& \frac{V}{4 l}=\frac{3 V}{2 l^{\prime}} \\
& l^{\prime}=6 l=6 \times 20 \\
& l^{\prime}=120 \mathrm{~cm}
\end{aligned}\)

Topic: Physics, Waves and Sound, JEE Main
The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of the organ pipe open at both the ends is

Options:
A 80 cm
B 100 cm
C 120 cm
D 140 cm

The correct answer is:
120 cm

Solution:
For closed organ pipe,
nc=V4l
For open organ pipe,
n0=V2l′
For the second overtone of an open organ pipe,
n′=3n0=3V2l′
nc=n′
V4l=3V2l′
l′=6l=6×20
l′=120 cm

Topic: Physics, Waves and Sound, JEE Main
The fundamental frequency of a closed pipe is $400 \mathrm{~Hz}$. If $1 / 3^{\text {rd }}$ pipe is filled with
water, the frequency of $2^{\text {nd }}$ harmonic of the pipe will be (Neglect end correction)

Options:
A $600 \mathrm{~Hz}$
B $1800 \mathrm{~Hz}$
C $1200 \mathrm{~Hz}$
D $300 \mathrm{~Hz}$

The correct answer is:
$1800 \mathrm{~Hz}$

Solution:
Fundamental frequency of closed pipe,
$\begin{array}{l}
n=\frac{V}{4 L}=400 \\
V=400 \times 4 L
\end{array}$
If $\frac{1}{3}$ rd of pipe is filled with water, then remaining length of air column is $L-\frac{L}{3}=\frac{2 L}{3}$. Now, fundamental frequency $=\frac{v}{4\left(\frac{2 L}{3}\right)}=\frac{3 v}{8 L}$
The second harmonic of the pipe $=3 \times$ fundamental frequency
$\begin{array}{l}
=3 \times\left(\frac{3 \mathrm{v}}{8 \mathrm{~L}}\right) \\
=3 \times \frac{3 \times 400 \times 4 \mathrm{~L}}{8 \mathrm{~L}} \\
=1800 \mathrm{~Hz}
\end{array}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2020 (19 Oct Shift 2)
The fundamental frequency of a closed pipe is equal to the frequency of the second harmonic of an open pipe. The ratio of their lengths is

Options:
A $1: 2$
B $1: 4$
C $1: 8$
D $1: 16$

The correct answer is:
$1: 4$

Solution:
Let the length of closed pipe is $L_{1}$ and that of pipe is $L_{2}$. Fundamental frequency of closed pipe, $v_{1}=\frac{v}{4 L_{1}}$
and frequency of second harmonic of open pipe.
$$
v_{2}=2 \times \frac{v}{2 L_{2}}
$$
Given,
$$
v_{1}=v_{2}
$$
$i, \theta$
$$
\frac{v}{4 L_{1}}=\frac{2 v}{2 L_{2}} \Rightarrow \frac{L_{1}}{L_{2}}=\frac{1}{4}
$$

Topic: Physics, Waves and Sound, JEE Main
The fundamental frequency of an open organ pipe is $300 \mathrm{~Hz}$. The first overtone of this pipe has same frequency as first overtone of a closed organ pipe. If speed of sound is $330 \mathrm{~m} / \mathrm{s}$, then the length of closed organ pipe is

Options:
A $41 \mathrm{~cm}$
B $37 \mathrm{~cm}$
C $31 \mathrm{~cm}$
D $80 \mathrm{~cm}$

The correct answer is:
$41 \mathrm{~cm}$

Solution:
For open pipe, $\mathrm{n}=\frac{\mathrm{v}}{2 \ell},$ where $\mathrm{n}_{0}$ is the
fundamental frequency of open pipe. $\therefore \ell=\frac{\mathrm{v}}{2 \mathrm{n}}=\frac{330}{2 \times 300}=\frac{11}{20}$
As freq. of 1 st overtone of open pipe $=$ freq. of 1 st overtone of closed pipe
$\begin{aligned} \therefore \quad 2 \frac{\mathrm{v}}{2 \ell}=3 \frac{\mathrm{v}}{4 \ell^{\prime}} \Rightarrow \ell^{\prime} &=\frac{3 \ell}{4}=\frac{3}{4} \times \frac{11}{20} \\ &=41.25 \mathrm{~cm} \end{aligned}$

Topic: Physics, Waves and Sound, JEE Main
The fundamental frequency of vibration of a string between two rigid support is 50 Hz. The mass of the string is 18 g and its linear mass density is 20 g m-1. The speed of the transverse waves so produced in the string is _______ m s–1.

Options:

The correct answer is:
90

Solution:
The fundamental frequency is f=50 Hz.

The length of the string is l=λ2

Linear mass density is μ=ml=20 g m-1

Using v=fλ

v2l=50

⇒v=100×l=100×1820=90 m s-1

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The fundamental note produced by a closed organ pipe is of frequency $f$. The fundamental note produced by an open organ pipe of same length will be of frequency

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

The correct answer is:
$2 f$

Solution:
$\begin{aligned} & n_{\text {closed }}=\frac{v}{4 l}, n_{\text {open }}=\frac{v}{2 l} \\ & \Rightarrow n_{\text {open }}=2 n_{\text {closed }}=2 f\end{aligned}$

Topic: Physics, Waves and Sound, JEE Main
The fusion of chromite ore with sodium carbonate in the presence of air leads to the formation of products \(\mathrm{A}\) and \(\mathrm{B}\) along with the evolution of \(\mathrm{CO}_2\). The sum of spin-only magnetic moment values of \(A\) and \(B\) is ______ B.M. (Nearest integer) [Given atomic number : \(\mathrm{C}: 6, \mathrm{Na}: 11, \mathrm{O}: 8, \mathrm{Fe}: 26, \mathrm{Cr}: 24\)]

Options:

The correct answer is:
6

Solution:
Spin only magnetic moment
For $\mathrm{Na}_2 \mathrm{CrO}_4 \quad \mu_{\mathrm{B}}=0$
For $\mathrm{Fe}_2 \mathrm{O}_3 \quad \mu_{\mathrm{B}}=5.9$ sum $=5.9$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The galvanometer deflection, when key K1 is closed but K2 is open, equals θ0 (see figure). On closing K2 also and adjusting R2 to 5Ω, the deflection in galvanometer becomes θ05. The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]:

Options:
A 12 Ω
B 22 Ω
C 5 Ω
D 25 Ω

The correct answer is:
22 Ω

Solution:
θ​0∝iG⇒θ​0C=iG
I-case Cθ0=V220+R ......(i)
II-case Cθ05=V220+5R5+R×55+R ....(ii)
From (i) and (ii) R=22Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The gametophyte is not an independent, free living generation in

Options:
A Adiantum
B Marchantia
C Pinus
D Polytrichum

The correct answer is:
Pinus

Solution:
Unlike bryophytes and pteridophytes, in gymnosperms (eg, Pinus, Cycas, etc.), the male and female gametophytes do not have an independent free-living existence. They remain within the sporangia retained on the sporophytes.

Topic: Biology, Plant Kingdom, NEET, NEET 2011 (Screening)
The gametophyte is not an independent, free-living generation in:

Options:
A Marchantia
B Polytrichum
C Adiantum
D Pinus

The correct answer is:
Pinus

Solution:
Pinus is member of gymnosperms. In gymnosperms, the gametophytes do not have an independent free-living existence. They remain within the sporangia retained of the sporophytes.

Topic: Biology, Plant Kingdom, NEET
The gap between the plates of a parallel plate capacitor of area A and distance between plates d, is filled with a dielectric whose relative permittivity varies linearly from ϵ1 at one plate to ϵ2 at the other. The capacitance of the capacitor is

Options:
A ϵ0∈2-∈1Adln∈2/∈1
B ϵ0∈2+∈1A2d
C ϵ0∈1+∈2Ad
D ϵ0Adln∈2/∈1

The correct answer is:
ϵ0∈2-∈1Adln∈2/∈1

Solution:
1dC=dxϵ0ϵA

=dxϵ0ϵ1+ϵ2-ϵ1dxA

1C=∫1dC=1∈0A∫0xdx∈1+∈2-∈1d

=ln∈2/∈1d∈0A∈2-∈1

C=∈0A∈2-∈1ln∈2/∈1·d

Topic: Physics, Capacitance, JEE Main
The gap between the plates of a parallel plate capacitor of area A and distance between plates d, is filled with a dielectric whose relative permittivity varies linearly from ϵ1 at one plate to ϵ2 at the other. The capacitance of the capacitor is

Options:
A ϵ0∈2-∈1Adln∈2/∈1
B ϵ0∈2+∈1A2d
C ϵ0∈1+∈2Ad
D ϵ0Adln∈2/∈1

The correct answer is:
ϵ0∈2-∈1Adln∈2/∈1

Solution:
1dC=dxϵ0ϵA

=dxϵ0ϵ1+ϵ2-ϵ1dxA

1C=∫1dC=1∈0A∫0xdx∈1+∈2-∈1d

=ln∈2/∈1d∈0A∈2-∈1

C=∈0A∈2-∈1ln∈2/∈1·d

Topic: Physics, Capacitance, JEE Main, JEE Main 2014 (19 Apr Online)
The gas A2 in the left flask allowed to react with gas B2 present in right flask as, A 2 g + B 2 g â 2 AB g ; Kc = 4 at 27oC. What is the concentration of AB when equilibrium is established?

Options:
A 1.33 M
B 2.66 M
C 0.66 M
D 0.33 M

The correct answer is:
0.66 M

Solution:
Kc=AB2A2B2
Kc=4x22-x4-x
â x = 3 2 2 4 = 1.33 mole
AB g = 2Ã1.33 4 =0.66 M

Topic: Chemistry, Chemical Equilibrium, JEE Main
The gas evolved on heating CH3MgBr in methanol is:

Options:
A Methane
B Ethane
C Propane
D HBr

The correct answer is:
Methane

Solution:
CH3OH having acidic H so in CH3MgBr the CH3- will react as a base and form CH4 by acid-base reaction.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2016 (09 Apr Online)
The gas evolved on heating CaF2 and SiO2 with concentrated H2SO4, on hydrolysis gives a white gelatinous precipitate. The precipitate is:

Options:
A Silicic acid
B Calcium fluorosilicate
C Hydrofluorosilicic acid
D Silica gel

The correct answer is:
Hydrofluorosilicic acid

Solution:
Reactions involved are
CaF2+H2SO4Conc.→ΔCaSO4+2HFgSiO2+4HFg→ΔSiF4g+2H2O3SiF4+2H2O→2H2SiF6+SiO2

The mixture after hydrolysis apperas as white gelatinous precipitate.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2014 (09 Apr Online)
The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was :

Options:
A Ammonia
B Phosgene
C Methyl isocyanate
D Methylamine

The correct answer is:
Methyl isocyanate

Solution:
Methyl isocyanate (M.I.C.)
CH 3 - N = C = O

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2013 (07 Apr)
The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was :

Options:
A Ammonia
B Phosgene
C Methyl isocyanate
D Methylamine

The correct answer is:
Methyl isocyanate

Solution:
Methyl isocyanate (M.I.C.)
CH 3 - N = C = O

Topic: Chemistry, Amines, JEE Main
The gas liberated by the electrolysis of Dipotassium succinate solution is:

Options:
A
Ethane

B
Ethyne

C
Ethene

D
Propene

The correct answer is:
Ethene

Solution:
Ethene is obtained by electrolysis of dipotassium succinate as follows

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2014 (11 Apr Online)
The gas obtained by roasting of sulphide ore is reacted with acidified potassium dichromate. A green coloured compound ' X ' is formed. The compound X can be:

Options:
A SO2
B Cr2SO43
C Cl2
D CrO2Cl2

The correct answer is:
Cr2SO43

Solution:
Sulphide ore  → roasting SO 2
SO 2 → K 2 Cr 2 O 7 +H 2 SO 4 Cr 2 (SO 4 ) 3 Green Colour

Topic: Chemistry, d and f Block Elements, JEE Main
The gas phase decomposition of dimethylether follows first order kinetics
CH3-O-CH3g→CH4g+H2g+COg
The reaction is carried out in constant volume container at 500oC and has a half-life of 14.5 min . Initially only dimethylether is present at a pressure of 0.40 atm . The total pressure of the system after 12 min is x 100  atm . The value of x is
[Given 10 0.25  = 1.778 ]

Options:

The correct answer is:
75

Solution:
Now: 12=2.3030.693Ã14.5logâ¡0.40.4-x
0.25=logâ¡0.40.4-x
100.25=0.40.4-x
0.40.4-x=1.778
âx=0.175
Total pressure =0.4-0.75+0.175+0.175+0.175=0.75 atm
now: x100=0.75
x=75

Topic: Chemistry, Chemical Kinetics, JEE Main
The gas phase reaction

2 Ag⇌A2g

at 400 K has ΔGo=+25.2 kJ mol-1.

The equilibrium constant KC for this reaction is ___ ×10-2. (Round off to the Nearest integer)

Use : R=8.3 J mol-1 K-1,ln10=2.3 log102=0.30, 1 atm=1 bar

antilog (-0.3)=0.501

Options:

The correct answer is:
2

Solution:
Using formula

ΔrG0=-RTlnKp

25200=-2.3×8.3×400logKp

Kp=10-3.3=10-3×0.501

=5.01×10-4Bar-1

=5.01×10-9 Pa-1

=KC8.3×400

KC=1.66×10-5 m3/mole

=1.66×10-2 L/mol

=2

Gibbs free energy can be visualized as the amount of useful energy present in a thermodynamic system that can be utilized to perform some useful work.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (18 Mar Shift 2)
The gas phase reaction 2NO2g → N2O4 g is an exothermic reaction. The decomposition of N2O4 , in equilibrium mixture of NO2 g and N2O4 (g) can be increased by:

Options:
A Addition of an inert gas at constant pressure
B Lowering the temperature
C Increasing the pressure
D Addition of an inert gas at constant volume

The correct answer is:
Addition of an inert gas at constant pressure

Solution:
2 NO2g→ N2O4 g ∆H=(-)ve
By addition of inert gas at constant pressure, volume increases and thus, the pressure decreases. The reaction now moves in backward direction as this would increase pressure and hence, decomposition of N2O4 increases.

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2018 (16 Apr Online)
The gas produced by the passage of air over hot coke is

Options:
A carbon monoxide
B carbon dioxide
C producer gas
D water gas

The correct answer is:
producer gas

Solution:
The gas produced by passage of air over hot coke is called producer gas. The reaction proceeds as:
$$
2 \mathrm{C}+\underbrace{\mathrm{O}_2+\mathrm{N}_2}_{\text {Coke }} \stackrel{\Delta}{\longrightarrow} \underbrace{2 \mathrm{CO}+\mathrm{N}_2}_{\text {Producer gas }}
$$
Other carbonaceous (carbon containing) fuels also produce producer gas of different compositions.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The gas produced by the passage of air over hot coke is

Options:
A Carbon monoxide
B Carbon dioxide
C Producer gas
D Water gas

The correct answer is:
Producer gas

Solution:
2 C s + O 2 g + N 2 g → 1263 K 2 CO(g)+4  N 2 (g) ︸ producer gas

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The gas produced by treating an aqueous solution of ammonium chloride with sodium nitrite is

Options:
A NH3
B N2
C N2O
D Cl2

The correct answer is:
N2

Solution:
In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite.

NH4Cl(aq)+NaNO2(aq)→N2(g)+2H2O(l)+NaCl(aq)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (27 Jun Shift 2)