datavorous/entrance-exam-dataset huggingface dataset:
RNA is different from DNA because RNA contains

Options:
A
ribose sugar and thymine

B
ribose sugar and uracil

C
deoxyribose sugar and thymine

D
deoxyribose sugar and uracil

The correct answer is:
ribose sugar and uracil

Solution:
No Solution Available

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2002
Roasted copper pyrite on smelting with sand produces

Options:
A $\mathrm{FeSiO}_{3}$ as fusible slag and $\mathrm{Cu}_{2} \mathrm{S}$ as matte
B $\mathrm{CaSiO}_{3}$ as infusible slag and $\mathrm{Cu}_{2} \mathrm{O}$ as matte
C $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$ as fusible slag and $\mathrm{Cu}_{2} \mathrm{S}$ as matte
D $\mathrm{Fe}_{3}\left(\mathrm{PO}_{4}\right)_{2}$ as infusible slag and $\mathrm{Cu}_{2} \mathrm{S}$ as matte

The correct answer is:
$\mathrm{FeSiO}_{3}$ as fusible slag and $\mathrm{Cu}_{2} \mathrm{S}$ as matte

Solution:
Roasted copper pyrite on smelting with sand produces $\mathrm{FeSiO}_{3}$ as fusible slag and $\mathrm{Cu}_{2}$ S as matte
For removing the gangue, FeS, silica present in the lining of the Bessemer converter, acts as a flux and forms slag (iron silicate) on reaction with FeO.
$\mathrm{FeS}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{FeO}+2 \mathrm{SO}_{2}$
$\mathrm{FeO}+\mathrm{SiO}_{2} \longrightarrow \mathrm{FeSiO}_{3}$
Copper matte mainly contains $\mathrm{Cu}_{2} \mathrm{S}$ and FeS.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, WBJEE, WBJEE 2015
Rohu and mackerel are variety of

Options:
A Marine fish
B Freshwater fish
C Rohu is marine but Mackerel is freshwater fish
D Rohu is freshwater but mackerel is marine fish

The correct answer is:
Rohu is freshwater but mackerel is marine fish

Solution:
A large section of the Indian population uses fish as food. Fish is an easily available source of protein. It is highly nutritious and easily digestible. In India, edible fish are abundantly available from sea, rivers, lakes, ponds and marshes. The marine fishery is divided into coastal fisheries (about 3 miles from the coast line), and deep sea fisheries, which have different fauna of fish, in India, about 75% of the marine fish catch is from the western coast. The main marine fisheries are Sardines (26%), Mackerel (9.7%) and the Bombay duck (11%). Fresh water fishes are Rohu (Labeo rohita), Calbasu (Labeo calbasu), Catla (Catla catla), Singhara (Mystus singhala), Magur (Clarius batrachus), Singhi (Heteropneustes).

Topic: Biology, Strategies for Enhancement in Food Production, NEET
Role of ethylene glycol in the radiator of vehicles is to

Options:
A Decrease the freezing point of water in the winter and increase the boiling point of water in the summer
B Only decrease the freezing point of water
C Only increase the boiling point of water
D Prevent corrosion of automobile parts

The correct answer is:
Decrease the freezing point of water in the winter and increase the boiling point of water in the summer

Solution:
When ethylene glycol is added to H2O as antifreeze, it decreases the freezing point of H2O in winter and increase the boiling point of water in the summer.

Topic: Chemistry, Solutions, JEE Main
Rolle's theorem is not applicable to the function $f(x) \neq x$ defined on $[-1,1]$ because

Options:
A $f$ is not continuous on $[-1,1]$
B $f$ is not differentiable on $(-1,1)$
C $f(-1) \neq f(1)$
D $f(-1)=f(1) \neq 0$

The correct answer is:
$f$ is not differentiable on $(-1,1)$

Solution:
$f(x)=\left\{\begin{array}{c}
-x, \text { when }-1 \leq x \lt 0 \\
x, \text { when } 0 \leq x \leq 1\end{array}\right.$
Clearly $f(-1)=|-1|=1=f(1)$
$\begin{aligned}& \text { But } R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{|h|}{h}=\lim _{h \rightarrow 0} \frac{h}{h}=1 \\
& L f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{|-h|}{-h}=\lim _{h \rightarrow 0} \frac{h}{-h}=-1 \\
& \therefore R f^{\prime}(0) \neq L f^{\prime}(0)\end{aligned}$
But
Hence it is not differentiable on $(-1,1)$.

Topic: Mathematics, Application of Derivatives, JEE Main
Root development is promoted by

Options:
A abscisic acid
B auxin
C gibberellin
D ethylene

The correct answer is:
auxin

Solution:
The higher concentration of auxin causes inhibition of root elongation, but number of lateral branches in the root is considerably increased. In plant tissue cultures, relative amounts of auxin and cytokinin induce the development of root, shoot or undifferentiated mass of tissues. High auxin to cytokinin ratio favours the root development high cytokinin to auxin ratio favours shoot development, while intermediate concentration results in the formation of undifferentiated cell mass.

Topic: Biology, Plant - Growth and Development, NEET, NEET 2010 (Mains)
Round the number 234555359 to 3 significant figures.

Options:
A 234000000
B 234000
C 235000000
D 234500000

The correct answer is:
234000000

Solution:
Upto 3 significant figures : 234000000 . Since the 3 rd digit is an even number and the 4 th digit is 5 , it remains same as 4 .

Topic: Chemistry, Some Basic Concepts of Chemistry, TS EAMCET, TS EAMCET 2022 (20 Jul Shift 2)
' \( R^{\prime} \) is

Options:
A sulphanilamide
B p-bromo sulphanilamide
C o-bromo sulphanilic acid
D sulphanilic acid

The correct answer is:
sulphanilic acid

Solution:
The reaction involved is,
(R) Sulphanilic acid

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Rubisco enzyme can act both as carboxylase and as oxygenase. In C4 cycle it functions as

Options:
A Oxygenase only
B Carboxylase only
C Mainly carboxylase activity minimizing oxygenase.
D Oxygenase but sometimes as carboxylase

The correct answer is:
Mainly carboxylase activity minimizing oxygenase.

Solution:
In C4 plants photorespiration does not occur. This is because they have a mechanism that increases the concentration of CO2 at the enzyme site. This takes place when the C4 acid from the mesophyll is broken down in the bundle sheath cells to release CO2 – this results in increasing the intracellular concentration of CO2. In turn, this ensures that the RuBisCO functions as a carboxylase minimizing the oxygenase activity.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Rutherford scattering experiment is based on:

Options:
A alpha particle scattering from gold foil
B beta particle scattering from gold foil
C beta particle scattering from $\mathrm{Zn}$ foil
D ganmma rays from aluminium foil

The correct answer is:
alpha particle scattering from gold foil

Solution:
- Rutherford conducted an experiment by bombarding a thin sheet of gold with \(\alpha\) particles.
- And then studied the trajectory of these particles after their interaction with the gold foil.

- The observations made by Rutherford led him to conclude that

Topic: Physics, Atomic Physics, JEE Main
Rutherford's alpha particle scattering experiment eventually led to the conclusion that

Options:
A Mass and energy are related
B Electrons occupy space around the nucleus
C Neutrons are buried deep in the nucleus
D The point of impact with matter can be precisely determined

The correct answer is:
Electrons occupy space around the nucleus

Solution:
Electrons in an atom occupy the extra nuclear region.

Topic: Chemistry, Structure of Atom, JEE Main
Rutherford’s alpha particle scattering experiment eventually led to the conclusion that

Options:
A mass and energy are related

B electrons occupy space around the nucleus
C neutrons are burned deep in the nucleus

D the point of impact with matter can be precisely determined

The correct answer is:
electrons occupy space around the nucleus

Solution:
According to Rutherford’s model, there is a heavily positively charged nucleus and negatively charged electrons occupies space around it in order to maintain electro - neutrality of atom.

Topic: Chemistry, Structure of Atom, NEET
Rydberg constant \( R \) is equal to

Options:
A \( \frac{m e^{2}}{8 \epsilon_{0}^{2} c h^{3}} \)
B \( \frac{m e^{4}}{8 \epsilon_{0}^{2} c h^{3}} \)
C \( \frac{m^{2} e^{4}}{8 \epsilon_{0}^{2} c h^{3}} \)
D \( \frac{m^{4} e^{4}}{8 \epsilon_{0}^{2} c h^{3}} \)

The correct answer is:
\( \frac{m e^{4}}{8 \epsilon_{0}^{2} c h^{3}} \)

Solution:
Rydberg constant R represents the wavelengths or light frequencies of different sets of similar spectral lines, more specifically those of the Balmer series emitted by hydrogen atoms. The magnitude of this constant is built on the hypothesis that, relative to a single orbiting electron, the nucleus of the atom producing light is enormously huge.

It is given by R=me48ϵ02ch3, here m represents rest mass of the electron, e represents the charge of a single electron, ϵ0 represents vacuum permittivity, h is the Planck constant and c the speed of light in vacuum.

Topic: Physics, Atomic Physics, JEE Main
$S_1, S_2, \ldots \ldots, S_{10}$ are the speakers in a conference. If $S_1$ addresses only after $S_2$, then the number of ways the speakers address is

Options:
A $10 !$
B $9 !$
C $10 \times 8 !$
D $\frac{(10 !)}{2}$

The correct answer is:
$\frac{(10 !)}{2}$

Solution:
Since, $S_1$ speak after $S_2$, therefore two places can be chosen out of 10 place is ${ }^{10} C_2$ ways and rest of the 8 speakers can speak in 8 ! ways.
$\therefore$ Required number of ways $={ }^{10} C_2 \cdot 8$ !
$$
\begin{aligned}
& =\frac{10 !}{2 ! 8 !} \cdot 8 ! \\
& =\frac{10 !}{2 !} .
\end{aligned}
$$

Topic: Mathematics, Permutation Combination, JEE Main
SbF3 reacts with XeF4 and XeF6 to form ionic compounds XeF3+SbF6- and XeF5+SbF6- then molecular shape of XeF3+ ion and XeF5+ ion respectively

Options:
A Square pyramidal, T-shaped
B Bent-T-shape, square pyramidal
C See-saw, square pyramidal
D Square pyramidal, See-saw

The correct answer is:
Bent-T-shape, square pyramidal

Solution:
The shapes of these ions are as follows
[XeF3+]: Bent-T shape ( sp3d -hybridisation)
[XeF5+]: Square pyramidal ( sp3d2 -hybridisation)

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
SbF5 reacts with XeF4 to form an adduct. The shapes of cation and anion in the adduct are respectively

Options:
A Square planar, trigonal bipyramidal
B T-shaped, octahedral
C Square pyramidal, octahedral
D Square planar, octahedral

The correct answer is:
T-shaped, octahedral

Solution:
XeF 4  +  SbF 5 → [XeF 3 ] +  [SbF 6 ] − → [XeF 3 ] + sp 3 d T-Shaped  +  [SbF 6 ] − sp 3 d 2 octahedral
As [XeF3]+ has 3 bond pair and 2 lone pair of electron so it has sp3d hybridisation and T-shape.
[SbF6]- has 6 bond pair electrons so it is sp3d2 hybridise and octahedral in shape.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
∫sec2x·cot43xdx is equal to

Options:
A 3tan-13x+C
B -34tan-43x+C
C -3tan-13x+C
D -3cot-13x+C

The correct answer is:
-3tan-13x+C

Solution:
Given integral can be written as

I=∫sec2xtanx43 dx

Let tanx=t

⇒sec2xdx=dt

⇒I=∫t-43 dt

Using ∫xndx=xn+1n+1+C, we get

I=t-13-13+ C

⇒I=-3t13+ C

⇒I=-3tan-13x+C.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2019 (09 Apr Shift 1)
SiCl4→H2OA→HeatB→HeatNa2CO3C,
The compound (C) is

Options:
A SiO2
B Si
C SiC
D Na2SiO3

The correct answer is:
Na2SiO3

Solution:
SiCl 4  +  H 2 O→ Si OH 4 A → Heat SiO 2 B → Heat Na 2 CO 3 Na 2 SiO 3 C

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
∫sin5x2sinx2dx, is equal to

Options:
A x+2sinx+sin2x+c
B 2x+sinx+sin2x+c
C x+2sinx+2sin2x+c
D 2x+sinx+2sin2x+c

The correct answer is:
x+2sinx+sin2x+c

Solution:
We have, I=∫sin5x2sinx2dx

=∫sin2x+x2sinx2dx

Using sinA+B=sinAcosB+cosAsinB

=∫sin2xcosx2+cos2xsinx2sinx2dx

=∫2 sinx cosxcosx2sinx2+cos 2xdx

=∫22sinx2cosx2cosx cosx2sinx2+cos 2xdx

=∫4cos2x2cosx+cos2xdx

Using cos2x=2cos2x-1⇒2cos2x=1+cos2x

=∫21+cosxcosx+cos2xdx

=∫2cosx+2cos2x+cos2xdx

=∫2cosx+1+cos2x+cos2xdx

=∫2cosx+2cos2x+1dx

=2sinx+sin2x+x+c, where c is the constant of integration.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2019 (08 Apr Shift 1)
$\frac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)}=$

Options:
A $\frac{\cos B+\sin B}{\cos B-\sin B}$
B $\frac{\cos A+\sin A}{\cos A-\sin A}$
C $\frac{\cos A-\sin A}{\cos A+\sin A}$
D None of these

The correct answer is:
$\frac{\cos A+\sin A}{\cos A-\sin A}$

Solution:
$\begin{aligned}
& \frac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)} \\
&=\frac{\sin (B+A)+\sin \left(90^{\circ}-B-A\right)}{\sin (B-A)+\sin \left(90^{\circ}-A+B\right)} \\
&=\frac{2 \sin \left(A+45^{\circ}\right) \cos \left(45^{\circ}-B\right)}{2 \sin \left(45^{\circ}-A\right) \cos \left(45^{\circ}-B\right)}
\end{aligned}$
$=\frac{\sin \left(A+45^{\circ}\right)}{\sin \left(45^{\circ}-A\right)}=\frac{\cos A+\sin A}{\cos A-\sin A}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\frac{(\cos \theta+i \sin \theta)^4}{(\sin \theta+i \cos \theta)^5}$ is equal to

Options:
A $\cos \theta-i \sin \theta$
B $\cos 9 \theta-i \sin 9 \theta$
C $\sin \theta-i \cos \theta$
D $\sin 9 \theta-i \cos 9 \theta$

The correct answer is:
$\sin 9 \theta-i \cos 9 \theta$

Solution:
$\begin{aligned}& \frac{(\cos \theta+i \sin \theta)^4}{(\sin \theta+i \cos \theta)^5}=\frac{(\cos +i \sin \theta)^4}{i^5\left(\frac{1}{i} \sin \theta+\cos \theta\right)^5} \\& =\frac{(\cos \theta+i \sin \theta)^4}{i(\cos \theta-i \sin \theta)^5}=\frac{(\cos \theta+i \sin \theta)^4}{i(\cos \theta+i \sin \theta)^{-5}} (By property) \\& =\frac{1}{i}(\cos \theta+i \sin \theta)^9=\sin 9 \theta-i \cos 9 \theta .\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
$\int \frac{\sqrt{\cos 2 x}}{\sin x} d x=$

Options:
A $\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2}+\sqrt{1-\tan ^2 x}}{\sqrt{2}-\sqrt{1-\tan ^2 x}}\right|$
$$
-\frac{1}{2} \log \left|\frac{1-\sqrt{1-\tan ^2 x}}{1+\sqrt{1-\tan ^2 x}}\right|+c
$$
B $\frac{1}{\sqrt{2}} \log \left|\frac{\sqrt{2}+\sqrt{1-\tan ^2 x}}{\sqrt{2}-\sqrt{1-\tan ^2 x}}\right|$
$$
-\frac{1}{2} \log \left|\frac{1+\sqrt{1-\tan ^2 x}}{1-\sqrt{1-\tan ^2 x}}\right|+c
$$
C $\frac{1}{4 \sqrt{2}} \log \left|\frac{\sqrt{2}-\sqrt{1-\tan ^2 x}}{\sqrt{2}+\sqrt{1-\tan ^2 x}}\right|$
$$
+\frac{1}{2} \log \left|\frac{1-\sqrt{1-\tan ^2 x}}{1+\sqrt{1-\tan ^2 x}}\right|+c
$$
D $\frac{1}{4 \sqrt{2}} \log \left|\frac{2-\sqrt{1-\tan ^2 x}}{2+\sqrt{1-\tan ^2 x}}\right|$
$$
+\frac{1}{2 \sqrt{2}} \log \left|\frac{1-\sqrt{1-\tan ^2 x}}{1+\sqrt{1-\tan ^2 x}}\right|+c
$$

The correct answer is:
$\frac{1}{\sqrt{2}} \log \left|\frac{\sqrt{2}+\sqrt{1-\tan ^2 x}}{\sqrt{2}-\sqrt{1-\tan ^2 x}}\right|$
$$
-\frac{1}{2} \log \left|\frac{1+\sqrt{1-\tan ^2 x}}{1-\sqrt{1-\tan ^2 x}}\right|+c
$$

Solution:
We have, $I=\int \frac{\sqrt{\cos 2 x}}{\sin x} d x$
Put $\cos 2 x=\frac{1-\tan ^2 x}{1+\tan ^2 x}$
$\because \quad I=\int \frac{\sqrt{1-\tan ^2 x}}{\sec x \sin x} d x$
Put $\quad 1-\tan ^2 x=t^2$
$$
\begin{aligned}
& \Rightarrow \quad-2 \tan x \sec ^2 x d x=2 t d t \\
& I=-\int \frac{t^2}{\left(1-t^2\right)\left(2-t^2\right)} d t=-\int \frac{t^2}{\left(t^2-1\right)\left(t^2-2\right.} d t \\
& I=-\int\left(\frac{2}{t^2-2}-\frac{1}{t^2-1}\right) d t=2 \int \frac{-1}{t^2-2} d t+\int \frac{d t}{t^2-1} \\
& I=\frac{-2}{2 \sqrt{2}} \log \left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right|+\frac{1}{2} \log \left|\frac{t-1}{t+1}\right|+c \\
& I=\frac{-1}{\sqrt{2}} \log \left|\frac{\sqrt{1-\tan ^2 x}-\sqrt{2}}{\sqrt{1-\tan ^2 x}+\sqrt{2}}\right|+\frac{1}{2} \log \left|\frac{\sqrt{1-\tan ^2 x}-1}{\sqrt{1-\tan ^2 x}+1}\right|+c \\
& I=\frac{1}{\sqrt{2}} \log \left|\frac{\sqrt{2}+\sqrt{1-\tan ^2 x}}{\sqrt{2}-\sqrt{1-\tan ^2 x}}\right|-\frac{1}{2} \log \left|\frac{1+\sqrt{1-\tan ^2 x}}{1-\sqrt{1-\tan ^2 x}}\right|+c
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main
SO2 turns into which acid when mixed with water

Options:
A H2SO2
B H2SO3
C H2SO4
D None of the above

The correct answer is:
H2SO3

Solution:
The reaction involved in this case is
SO2+ H2O → H2SO3

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
SOCl2+HOH→X+[Y]

Which of the following is/are incorrect statements(s)?

(I) One of the products is a gas having sp3d hybridization.

(II) Both the products are strong acids.

(III) One of the product has one pπ-dπ bond.

(IV) One of the product when react with NH3 gives white fumes.

Options:
A II, IV
B I, II
C I, II, III
D II, III

The correct answer is:
I, II

Solution:
SOCl2+HOH→SO2+HCl

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
$S=\tan ^{-1}\left(\frac{1}{n^2+n+1}\right)+\tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)+\ldots$ $+\tan ^{-1}\left(\frac{1}{1+(n+19)(n+20)}\right)$, then $\tan S$ is equal to :

Options:
A
$\frac{20}{401+20 n}$

B
$\frac{n}{n^2+20 n+1}$

C
$\frac{20}{n^2+20 n+1}$

D
$\frac{n}{401+20 n}$

The correct answer is:
$\frac{20}{n^2+20 n+1}$

Solution:
We know that,
$$
\begin{array}{r}
\tan ^{-1} \frac{1}{1+2}+\tan ^{-1} \frac{1}{1+2 \times 3}+\tan ^{-1} \frac{1}{1+3 \times 4}+\ldots \ldots+ \\
\tan ^{-1} \frac{1}{1+(n-1) n}+\tan ^{-1} \frac{1}{1+n(n+1)}+\ldots \ldots+ \\
\tan ^{-1} \frac{1}{1+(n+19)(n+20)}=\tan ^{-1} \frac{n+19}{n+21} \\
\Rightarrow \tan ^{-1} \frac{n-1}{n+1}+\tan ^{-1} \frac{1}{1+n(n+1)}+\tan ^{-1} \frac{1}{1+(n+1)(n+2)} \\
+\ldots \ldots+\frac{1}{1+(n+19)(n+20)}=\tan ^{-1} \frac{n+19}{n+21} \\
\tan ^{-1} \frac{1}{1+n(n+1)}+\tan ^{-1} \frac{1}{1+(n+1)(n+2)}+\ldots \ldots+ \\
\frac{1}{1+(n+19)(n+20)}=\tan ^{-1} \frac{n+19}{n+21}-\tan ^{-1} \frac{n-1}{n+1}
\end{array}
$$
$$
\begin{aligned}
& \tan ^{-1}\left(\frac{1}{n^2+n+1}\right)+\tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)+\ldots \ldots+ \\
& \quad=\tan ^{-1}\left(\frac{\frac{n+19}{n+21}-\frac{n-1}{n+1}}{1+\frac{n+19}{n+21} \times \frac{n-1}{n+1}}\right) \\
& \quad=\tan ^{-1} \frac{1}{1+(n+19)(n+20)} \frac{20}{n^2+20 n+1}=\mathrm{S} \\
& \therefore \quad \tan ^{-1} \mathrm{~S}=\frac{20}{n^2+20 n+1}
\end{aligned}
$$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2013 (23 Apr Online)
Sacred grooves are specially useful in

Options:
A generating environmental awareness
B preventing soil erosion
C year-round flow of water in rivers
D conserving rare and threatened species

The correct answer is:
conserving rare and threatened species

Solution:
Biodiversity is being threatened by reduction in space, smaller and fragmented habitat, over exploitation by human, climatic change, etc. Sacred grooves are especially useful in conserving rare and threatened species.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2012 (Mains)
Salicylaldehyde is synthesized from phenol, when reacted with

Options:
A
B CO2 , NaOH
C CCl4 , NaOH
D HCCl3 , NaOH

The correct answer is:
HCCl3 , NaOH

Solution:
Salicylaldehyde is prepared by heating phenol and chloroform together in the presence of potassium hydroxide or sodium hydroxide. This reaction is called the Reimer-Tiemann reaction.

The Reimer Tiemann reaction is an organic chemical reaction where phenol is converted into an ortho hydroxy benzaldehyde using chloroform, a base, and an acid workup. This reaction can also be described as the chemical reaction used for the ortho-formylation of phenols.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Salicylic acid is prepared from phenol by

Options:
A Reimer Tiemann reaction
B Kolbe's reaction
C Kolbe-electrolysis reaction
D None of these

The correct answer is:
Reimer Tiemann reaction

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Salicylic acid is produced when phenol in alcoholic KOH is treated with

Options:
A CH3Cl
B CHCl3
C CH2Cl2
D CCl4

The correct answer is:
CCl4

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Saline hydrides react explosively with water, such fires can be extinguished by

Options:
A Water
B Carbon dioxide
C Sand
D None of these

The correct answer is:
Sand

Solution:
Fire due to action of water on saline hydrides cannot be extinguished with water or $\mathrm{CO}_2$. These hydrides can reduce $\mathrm{CO}_2$ at high temperature to produce $\mathrm{O}_2$.

Topic: Chemistry, Hydrogen, JEE Main
Salts of A (atomic mass 15), B (atomic mass 27) and C (atomic mass 48) were electrolysed using same amount of charge. It was found that when 4.5 g of A was deposited, the masses of B and C deposited were 2.7 g and 9.6 g. The valencies of A, B and C were respectively

Options:
A 1, 3 and 2
B 3, 1 and 3
C 3, 1 and 2
D 2, 3 and 2

The correct answer is:
3, 1 and 2

Solution:
According to Faraday's second law
Mass of A Equivalent mass of A = Mass of B Equivalent mass of B = Mass of C Equivalent mass of C
4.515/n1=2.727/n2=9.648/n3
n1:n2:n3
0.3:0.1:0.2
3:1:2 .

Topic: Chemistry, Electrochemistry, JEE Main
Salts of A (atomic weight 8), B (atomic weight 18) and C (atomic weight 50) were electrolysed under identical conditions using the same quantity of electricity. It was found that 2.4 g of A was deposited, the weight of B and C deposited are 1.8 g and 7.5g respectively. The valences of A, B and C are, respectively,

Options:
A 3, 1 and 2
B 1, 2 and 3
C 1, 3 and 2
D 3, 2 and 1

The correct answer is:
1, 3 and 2

Solution:
According to the Faraday's law of electrolysis,

W=ZQ=E96500Q

For the compound A,

Equivalent mass of A = Atomic massValency=8x

x is the valency factor for compound A.

So, W=E96500Q=8x96500Qx=72.4×96500Q

For compound B,

Equivalent mass of B=Atomic massValency=18y

y is the valency factor for compound B.

So, W=E96500Q=18y96500Qy=271.8×96500Q

For compound C,

Equivalent mass of C = Atomic massValency=50z

z is the valency factor for compound C.

So, W=E96500Q=50z96500Qz=507.5×96500Q

Hence, x:y:z=82.4 : 181.8 : 507.5=3.33 : 10 : 6.66x:y:z = 1:3:2

Hence, the valencies of A,B and C respectively are 1,3,2.

Topic: Chemistry, Electrochemistry, JEE Main
Same current is flowing in two alternating circuits. The first circuit contains only inductance and the other contains only a capacitor. If the frequency of the e.m.f. of ac is increased, the effect on the value of the current will be

Options:
A Increases in the first circuit and decreases in the other
B Increases in both the circuits
C Decreases in both the circuits
D Decreases in the first circuit and increases in the other

The correct answer is:
Decreases in the first circuit and increases in the other

Solution:
For the first circuit $i=\frac{V}{Z}=\frac{V}{\sqrt{R^2+\omega^2 L^2}}$
$\therefore$ Increase in $\omega$ will cause a decrease in $i$.
For the second circuit $i=\frac{V}{\sqrt{R^2+\frac{1}{\omega^2 C^2}}}$
$\therefore$ Increase in $\omega$ will cause an increase in $i$.

Topic: Physics, Alternating Current, JEE Main
Sand is being dropped on a conveyer belt at the rate of $2 \mathrm{~kg}$ per second. The force necessary to keep the belt moving with a constant speed of $3 \mathrm{~ms}^{-1}$ will be

Options:
A
$12 \mathrm{~N}$

B
$6 \mathrm{~N}$

C
zero

D
$18 \mathrm{~N}$

The correct answer is:
$6 \mathrm{~N}$

Solution:
Here, rate of fall of sand on conveyer belt,
$$
\frac{d m}{d t}=2 \mathrm{~kg} \mathrm{~s}^{-1}
$$
Conveyer belt speed $v=3 \mathrm{~ms}^{-1}$ Force $F=$ ?
$$
F=v \frac{d m}{d t}=3 \times 2=6 \mathrm{~N}
$$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2012 (19 May Online)
Sand is being dropped on a conveyor belt at the rate of $\mathrm{M} \mathrm{kg} / \mathrm{s}$. The force necessary to keep the belt moving with a constant velocity of $v \mathrm{~m} / \mathrm{s}$ will be

Options:
A Zero
B $\mathrm{M} v$ newton
C $2 \mathrm{M} v$ newton
D $\frac{M v}{2}$ newton

The correct answer is:
$\mathrm{M} v$ newton

Solution:
Force $=v \frac{d u}{d t}=M v$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2008 (Mains)