datavorous/entrance-exam-dataset huggingface dataset:
Moment of inertia of a disc of mass M and radius 'R' about any of its diameter is MR24. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, x2MR2. The value of x is ______.

Options:

The correct answer is:
3

Solution:
Given here,

Moment of inertia of a disc of mass M and radius 'R' about any of its diameter is MR24.

Moment of inertia of disc about centre is Icm=MR24+MR24=MR22

Using parallel axis theorem, we can write

I=Icm+Md2

=MR22+MR2

=32MR2

Hence, the value of x=3.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Moment of inertia of a hollow cylinder of mass $M$ and radius $r$ about its own axis is

Options:
A $\frac{2}{3} M r^2$
B $\frac{2}{5} M r^2$
C $\frac{1}{3} M r^2$
D $M \boldsymbol{r}^2$

The correct answer is:
$M \boldsymbol{r}^2$

Solution:
Moment of inertia of a hollow cylinder of mass $\mathrm{M}$ and radius $\mathrm{r}$ about its own axis is $M r^2$.

Topic: Physics, Rotational Motion, JEE Main
Moment of inertia of a square plate of side l about the axis passing through one of the corner and perpendicular to the plane of square plate is given by:

Options:
A Ml26
B 23Ml2
C Ml2
D Ml212

The correct answer is:
23Ml2

Solution:
Use parallel axis theorem to get the magnitude of the moment of inertia of a square plate of side l about the axis passing through one of the corners and perpendicular to the plane of a square plate,

Iz=Icm+Ml22

⇒Iz=Ml26+Ml22=4Ml26=2Ml23.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (27 Aug Shift 1)
Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane is I0 as shown in the figure. A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is:

Options:
A 7 8 I 0
B 15 16 I 0
C 3 4 I 0
D 31I032

The correct answer is:
15 16 I 0

Solution:
Given,

AB=BC=AC=l
Moment of inertia of a triangular lamina ABC
I0=kml2
DE=EF=DF=12,

AB=l2
∴ Moment of inertia of ∆DEF
IDEF=km4l22
IDEF=k16 ml2.
IDEF=I016
Moment of inertia of the remaining part
Iremain=I0-I016
=15I016

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2017 (08 Apr Online)
Moment of inertia of big drop is $I$. If 8 droplets are formed from big drop, then moment of inertia of small droplet is

Options:
A $\frac{I}{32}$
B $\frac{I}{16}$
C $\frac{I}{8}$
D $\frac{I}{4}$

The correct answer is:
$\frac{I}{32}$

Solution:
Moment of inertia of big drop is $I=\frac{2}{5} M R^{2}$ When small droplets are formed from big drop volume of liquid remain same
$$
\begin{aligned}
n \frac{4}{3} \pi r^{3} &=\frac{4}{3} \pi R^{3} \\
n^{1 / 3} r &=R \text { as } n=8 \Rightarrow r=\frac{R}{2}
\end{aligned}
$$
Mass of each small droplet $=\frac{M}{8}$
$\therefore$ Moment of inertia of each small droplet
$$
\begin{array}{l}
=\frac{2}{5}\left[\frac{M}{8}\right]\left[\frac{R}{2}\right]^{2} \\
=\frac{1}{32}\left[\frac{2}{5} M R^{2}\right]=\frac{I}{32}
\end{array}
$$

Topic: Physics, Rotational Motion, MHT CET, MHT CET 2009
Mond's process is used for preparing

Options:
A $\mathrm{Ni}$
B $\mathrm{H}_2 \mathrm{SO}_4$
C $\mathrm{NH}_3$
D $\mathrm{HNO}_3$

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

Solution:
Correct Option is 1 $\mathrm{Ni}$
Monds Process;

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Monocarboxylic acids are functional isomers of:

Options:
A
Ethers

B
Amines

C
Esters

D
Alcohols

The correct answer is:
Esters

Solution:
Mono-carboxylic acids are functional isomers of esters. e.g.,
$$
\mathrm{CH}_3 \mathrm{COOH} \quad \mathrm{HCOOCH}_3
$$
Acetic acid Methyl formate

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2013 (23 Apr Online)
Monocarpellary ovary, diadelphous androecium and marginal placentation is found in

Options:
A Cruciferae.
B Solanaceae.
C Liliaceae.
D Papilionaceae.

The correct answer is:
Papilionaceae.

Solution:
In Papilionaceae, only one carpel is present so a monocarpellary ovary is found. When inside of the stamens' filaments get fused and anthers remain free then that condition is called adelphous. When filaments form two bundles then that is known as diadelphous(9+1) condition. It is the subfamily of Leguminosae. In the unilocular ovary, marginal placentation is present. Ovules are present on the margin of the locule.
In Cruciferae and Compositae, the gynoecium is bi-carpellary. In Liliaceae, the tricarpellary ovary is found.

Topic: Biology, Morphology of Flowering Plants, NEET
Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is μ , a ray, incident at an angle θ , on the face AB would get transmitted through the face AC of the prism provided:

Options:
A θ < cos - 1 μ sin A + sin - 1 1 μ
B Î¸>sin-1â¡Î¼ sinâ¡A-sin-1â¡1Î¼
C θ < sin - 1 μ sin A - sin - 1 1 μ
D θ>cos-1â¡μ sinâ¡A+sin-1â¡1μ

The correct answer is:
Î¸>sin-1â¡Î¼ sinâ¡A-sin-1â¡1Î¼

Solution:
Central Idea: The ray will get transmitted through face AC if iAC<iC

Consider the ray diagram is shown below.

A ray of light incident on face AB at an angle θ

r1= Angle of refraction on face AB

r2= Angle of incidence at face AC

For transmission of light through face AC

iAC<iC

or A-r1<iC

or sinA-r1<siniC

or sinA-r1<1μ

A-r1<sin-11μ

or sinr1>sinA-sin-11μ

Now, applying Snell's law at the face AB

1×sinθ=μsinr1

or sinr1=sinθμ

⇒sinθμ>sinA-sin-11μ

or θ>sin-1μsinA-sin-11μ

Topic: Physics, Ray Optics, JEE Main, JEE Main 2015 (04 Apr)
Monochromatic light of frequency 6×1014 Hz is produced by a laser. The power emitted is 2×10-3 W. How many photons per second on an average, are emitted by the source?

(Given h=6.63×10-34 J s)

Options:
A 9×1018
B 6×1015
C 5×1015
D 7×1016

The correct answer is:
5×1015

Solution:
Energy of a single photon is given by, hν. Therefore, power can be written as

P=nhv

⇒n=Phν=2×10-36.63×10-34×6×1014

=5×1015

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Monochromatic light of frequency 6.0Ã1014 Hz is produced by a laser. The power emitted is 2Ã10-3 W. The number of photons emitted, on an average, by the source per second is nÃ1015 where n is an integer. Find the value of n?

Options:

The correct answer is:
5

Solution:
P=2Ã10-3 W
Energy of photon,
E=hv
= 6.6Ã10-34Ã6Ã1014 J
Where h being Planck's constant.
Number of photons emitted per second
n=p E
= 2Ã10-36.6Ã10-34Ã6Ã1014=5Ã1015

Topic: Physics, Dual Nature of Matter, JEE Main
Monochromatic light of wavelength $500 \mathrm{~nm}$ is used in Young's double slit experiment. An interference pattern is obtained on a screen. When one of the slits is covered with a very thin glass plate (refractive index $=1.5$ ), the central maximum is shifted to a position previously occupied by the $4^{\text {th }}$ bright fringe. The thickness of the glass-plate is ______ $\mu \mathrm{m}$.

Options:

The correct answer is:
4

Solution:
$\begin{aligned} & (\mu-1) \mathrm{t}=\mathrm{n} \lambda \\ & (1.5-1) \mathrm{t}=4 \times 500 \times 10^{-9} \mathrm{~m} \\ & \mathrm{t}=4000 \times 10^{-9} \mathrm{~m} \\ & \mathrm{t}=4 \mu \mathrm{m}\end{aligned}$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Monochromatic light of wavelength $589 \mathrm{~nm}$ is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is $1.33$.

Options:

None

Solution:
(a) For reflected light, wavelength, frequency and speed remains same as incident light. $\lambda=589 \mathrm{~nm}, \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}$. $v=\frac{\mathrm{c}}{\lambda}=\frac{3 \times 10^8}{589 \times 10^{-9}}=5.09 \times 10^{14} \mathrm{~Hz}$
(b) For refracted light the frequency of the refracted light is the same as the incident frequency. The speed changes due to change in wavelength.
$\because \mathrm{n}=1.33$
$\therefore v=\frac{\mathrm{c}}{\lambda}=\frac{3 \times 10^8}{1.33}=2.26 \times 10^8 \mathrm{~ms}^{-1}$.
Wavelength, $\lambda=\frac{\mathrm{v}}{\mathrm{v}}=\frac{2.26 \times 10^8}{5.09 \times 1014}$ $=0.444 \times 10^{-6}=444 \times 10^{-9} \mathrm{~m}

Topic: Physics, Wave Optics
Monochromatic light of wavelength $667 \mathrm{~nm}$ is produced by a helium neon laser. The power emitted is $9 \mathrm{~mW}$. The number of photons arriving per sec on the average at a target irradiated by this beam is:

Options:
A $3 \times 10^{16}$
B $9 \times 10^{15}$
C $3 \times 10^{19}$
D $9 \times 10^{17}$

The correct answer is:
$3 \times 10^{16}$

Solution:
$\lambda=667 \times 10^{-9} m, \mathrm{P}=9 \times 10^{-3} \mathrm{~W}$

$\mathrm{P}=\frac{\text { Nhc }}{\lambda}, \mathrm{N}=$ No. of photons emitted/sec.

$\mathrm{N}=\frac{9 \times 10^{-3} \times 667 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^{8}}=3 \times 10^{16} / \mathrm{sec}$

Topic: Physics, Dual Nature of Matter, JEE Main
Monochromatic light of wavelength $667 \mathrm{~nm}$ is produced by a helium neon laser. The power emitted is $9 \mathrm{~mW}$. The number of photons arriving per second on the average at a target irradiated by this beam is :

Options:
A $3 \times 10^{19}$
B $9 \times 10^{17}$
C $3 \times 10^{16}$
D $9 \times 10^{15}$

The correct answer is:
$3 \times 10^{16}$

Solution:
$$
\begin{array}{r}
\text { No. of photons }=\frac{E}{(\mathrm{hc} / \lambda)} \\
=\frac{9 \times 10^{-3} \times 6.67 \times 10^{-7}}{6.6 \times 10^{-34} \times 3 \times 10^8} \\
=3 \times 10^{16}
\end{array}
$$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2009 (Mains)
Monochromatic light passes through a prism. Compares to that in air, inside the prism the light's

Options:
A speed and wavelength are different but frequency remains same.
B speed and frequency are different but wavelength remains same.
C frequency and wavelength are different but speed remains same.
D speed, wavelength and frequency are all different.

The correct answer is:
speed and wavelength are different but frequency remains same.

Solution:
No Solution

Topic: Physics, Wave Optics, KVPY, KVPY 2014 (SB/SX)
Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be $3.57 \mathrm{~V}$. The threshold frequency of the material is

Options:
A $4 \times 10^{15} \mathrm{~Hz}$
B $5 \times 10^{15} \mathrm{~Hz}$
C $1.6 \times 10^{15} \mathrm{~Hz}$
D $2.5 \times 10^{15} \mathrm{~Hz}$

The correct answer is:
$1.6 \times 10^{15} \mathrm{~Hz}$

Solution:
Energy released from emission of electron
$\begin{aligned}
E & =(-3.4)-(-13.6) \\
& =10.2 \mathrm{eV}
\end{aligned}$
From photo electric equation.
Work function
$\begin{aligned}
\phi & =E-e V=h v \\
v & =\frac{E-e V}{h} \\
& =\frac{(10.2-3.57) e}{6.67 \times 10^{-34}} \\
v & =\frac{6.63 \times 1.6 \times 10^{-19}}{6.67 \times 10^{-34}} \\
& =1.6 \times 10^{15} \mathrm{~Hz}
\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main
Monochromatic radiation of wavelength λ is incident on a hydrogen sample in ground state. Hydrogen atom absorbs a fraction of light and subsequently emits radiations of six different wavelengths. The wavelength λ is

Options:
A 97.5 nm
B 121.6 nm
C 110.3 nm
D 45.2 nm

The correct answer is:
97.5 nm

Solution:
As H-atom emits 6 spectral lines nn-12=6

∴ n=4

ΔE=E4-E1=13.6-0.85

=12.75 eV

λ=hc∆E=1242 nm12.75

∴λ=97.5 nm

Topic: Physics, Atomic Physics, JEE Main
Monochromatic radiation of wavelength λ is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. Find the value of λ.

Options:
A 102 nm
B 100 nm
C 97.5 nm
D 94.5 nm

The correct answer is:
97.5 nm

Solution:
As the hydrogen atoms emit radiation of six different wavelengths, some of them must have been excited to n=4 , The energy in n=4 state is
E 4 = E 1 4 2 = - 1 3 Â· 6 e V 1 6 = - 0 Â· 8 5 eV.
The energy needed to take a hydrogen atom from its ground state to n=4 is
13.6 eV - 0.85 eV = 12.75 eV .
The photons of the incident radiation should have 12.75ââeV of energy. So
hc Î» = 1 2 Â· 7 5 eV
or , Î» = hc 1 2 Â· 7 5 eV
= 1 2 4 2 eV nm 1 2 Â· 7 5 eV
= 97.5 nm .

Topic: Physics, Atomic Physics, JEE Main
Monoclinic crystal has dimension

Options:
A $a \neq b \neq c, \alpha=\gamma=90^{\circ}, \beta \neq 90^{\circ}$
B $a=b=c, \alpha=\beta=\gamma=90^{\circ}$
C $a=b \neq c, \alpha=\beta=\gamma=90^{\circ}$
D $a \neq b \neq c, \alpha \neq \beta \neq \gamma \neq 90^{\circ}$

The correct answer is:
$a \neq b \neq c, \alpha=\gamma=90^{\circ}, \beta \neq 90^{\circ}$

Solution:
The conditions for monoclinic crystal system.

Topic: Chemistry, Solid State, JEE Main
Monoclinic sulphur is an example of monoclinic crystal system. What are the characteristics of the crystal system?

Options:
A a≠b≠c,α=β=γ=90°
B a≠b≠c,α≠β≠γ≠90°
C a=b≠c,α=β=γ=90°
D a≠b≠c,α=γ=90°,β≠90°

The correct answer is:
a≠b≠c,α=γ=90°,β≠90°

Solution:
For monoclinic crystal system,
a≠b≠c,α=γ=90°,β≠90°

Topic: Chemistry, Solid State, JEE Main
Monomer of Novolac is :

Options:
A 3-Hydroxybutanoic acid.
B 1,3 -Butadiene and styrene.
C phenol and melamine.
D o-Hydroxymethylphenol.

The correct answer is:
o-Hydroxymethylphenol.

Solution:
Novalac : It is linear polymer of o-Hydroxymethylphenol.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2021 (31 Aug Shift 1)
Monomer units of Dacron polymer are :

Options:
A glycerol and phthalic acid
B ethylene glycol and phthalic acid
C ethylene glycol and terephthalic acid
D glycerol and terephthalic acid

The correct answer is:
ethylene glycol and terephthalic acid

Solution:

Topic: Chemistry, Polymers, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Montreal protocol aims at :

Options:
A Control of $\mathrm{CO}_2$ emission
B Reduction of ozone depleting substances
C Biodiversity conservation
D Control of water pollution

The correct answer is:
Reduction of ozone depleting substances

Solution:
No Solution Available

Topic: Biology, Environmental Issues, NEET, NEET 2009 (Mains)
Montreal protocol was signed in

Options:
A 1987 and became effective in 1989 to control the emission of ozone depletion substances.
B 1989 to reduce emission of green house of gases.
C 1997 to mitigate global climate changes.
D 1992 to conserve biodiversity.

The correct answer is:
1987 and became effective in 1989 to control the emission of ozone depletion substances.

Solution:
In 1987 in Montreal, Canada. Twenty-seven industrialized countries signed an international treaty called Montreal protocol.
​Its main objective was to protect the stratospheric ozone by limiting the production and use of ozone-depleting substances, developing alternatives to chlorofluorocarbons, phasing out of ozone depleting substances.

Topic: Biology, Environmental Issues, NEET
More number of oxidation state are exhibited by the actinoids, than by lanthanoids. The main reason for this is

Options:
A greater metallic character of lanthanoids than that of their corresponding actinoids
B more active nature of actinoids
C more energy difference between $5 f$ and
$6 d$-orbitals, than that between $4 f$ and $5 d$-orbitals
D lesser energy difference between $5 f$ and $6 d$-orbitals, than that between $4 f$ and $5 d$-orbitals

The correct answer is:
lesser energy difference between $5 f$ and $6 d$-orbitals, than that between $4 f$ and $5 d$-orbitals

Solution:
More number of oxidation state are exhibited by the actinoids, than by lanthanoids. The main reason for this is lesser energy difference between $5 f$ and $6 d$-orbitals, than that between $4 f$ and $5 d$-orbitals.

Topic: Chemistry, d and f Block Elements, COMEDK, COMEDK 2015
More number of oxidation states are exhibited by the actinides than those by the lanthanides, the main reason being

Options:
A More reactive nature of the actinides than the lanthanides
B 4f orbitals more diffused than the 5f orbitals
C More energy difference between 5f and 6d than between 4f and 5d orbitals
D Lesser energy difference between 5f and 6d than between 4f and 5d orbitals

The correct answer is:
Lesser energy difference between 5f and 6d than between 4f and 5d orbitals

Solution:
Actinides show different oxidation states such as +2, +3, +4, +5, +6 and +7 . However +3 oxidation state is most common among all the lanthanides.The wide range of oxidation states of actinides is attributed to the fact that the 5f ,6d and 7s energy levels are of comparable energies. Therefore all these three subshells can participate.

Topic: Chemistry, d and f Block Elements, JEE Main
More number of oxidation states are exhibited by the actinoids than by the lanthanoids. The main reason for this is:

Options:
A more active nature of the actinoids
B more energy difference between $5 f$ and $6 d$ orbitals than that between $4 f$ and $5 d$ orbitals
C lesser energy difference between $5 f$ and $6 d$ orbitals than that between $4 f$ and $5 d$ orbitals
D greater metallic character of the lanthanoids than that of the corresponding actinoids

The correct answer is:
lesser energy difference between $5 f$ and $6 d$ orbitals than that between $4 f$ and $5 d$ orbitals

Solution:
Actinides exhibit more number of oxidation states than lanthanides because of lesser energy difference between the $5 f$ and $6 d$ orbitals than that between the $4 f$ and $5 d$ orbitals.
Related Theory
Actinides have a very small energy gap between $5 f$ and $6 d$ orbitals, so they can easily form a variety of configurations and hence they are found to be stable in various oxidation states. Lanthanides are comparatively less electropositive than actinides and due to that reason, they are less reactive than actinides.

Topic: Chemistry, d and f Block Elements, NEET, NEET 2006
More than $70 \%$ of world's freshwater is contained in:

Options:
A polar ice
B glaciers and mountains
C Antarctica
D Greenland.

The correct answer is:
polar ice

Solution:
The distribution of water on the Earth's surface is extremely uneven. Only $3 \%$ of water on the surface is fresh; the remaining $97 \%$ resides in the ocean. Of freshwater, $69 \%$ resides in glaciers, $30 \%$ underground, and less than $1 \%$ is located in lakes, rivers, and swamps. Freshwater is water that contains little or no dissolved salt. Most freshwater is frozen in ice caps and glaciers.

Topic: Biology, Ecosystem, NEET, NEET 2005
Mosley's name is most closely associated with the discovery of

Options:
A Positron
B Deutrons
C Atomic number
D Atomic weight

The correct answer is:
Atomic number

Solution:
Henry Moseley is best known for his discovery of atomic numbers in 1913.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Most algal genera show haplontic life style. Name an alga which is
(a) Haplo diplontic
(b) Diplontic

Options:

None

Solution:
Haplo diplontic type of life cycle is exhibited by Ectocarpus, Polysiphonia and Kelps. The main plant body is saprophytic in Fucus and it shows diplontic type of life cycle.

Topic: Biology, Plant Kingdom
Most animals that live in deep oceanic waters are:

Options:
A Secondary consumers
B Tertiary consumers
C Detritivores
D Primary consumers

The correct answer is:
Detritivores

Solution:
Detritivores are heterotrophic organisms that obtain nutrients by consuming detritus that is the decomposing plant and animal remains.

Benthos is the community of organisms that live on, in, or near the sea bed, also known as the benthic zone. As no light is available at this zone of sea, the energy source for deep benthic ecosystems is often organic matter from higher up in the water column which drifts down to the depths. This dead and decaying matter sustains the benthic organisms, and therefore, most organisms in benthic zone i.e., in deep oceanic waters, are scavengers or detritivores.

Topic: Biology, Ecosystem, NEET, NEET 2015 (Phase 2)
Most basic hydroxide among the following is

Options:
A $\mathrm{Lu}(\mathrm{OH})_{3}$
B $\mathrm{Eu}(\mathrm{OH})_{3}$
C $\mathrm{Yb}(\mathrm{OH})_{3}$
D $\mathrm{Ce}(\mathrm{OH})_{3}$

The correct answer is:
$\mathrm{Ce}(\mathrm{OH})_{3}$

Solution:
Due to lanthanide contraction, the size of $M^{3+}$ ions [ie, $\mathrm{Lu}^{3+}, \mathrm{Eu}^{3+}, \mathrm{Yb}^{3+}$ and $\mathrm{Ce}^{3+}$ ] decreases
and thus, the basic strength of their hydroxides decreases.
The order of size of given $M^{3+}$ ions is $\mathrm{Ce}^{3+}>\mathrm{Eu}^{3+}>\mathrm{Yb}^{3+}>\mathrm{Lu}^{3+}$
$\therefore$ The order of basic strength of hydroxides is $\mathrm{Ce}(\mathrm{OH})_{3}>\mathrm{Eu}(\mathrm{OH})_{3}>\mathrm{Yb}(\mathrm{OH})_{3}>\mathrm{Lu}(\mathrm{OH})_{3}$
(most basic)

Topic: Chemistry, d and f Block Elements, JEE Main
Most common oxidation states of Ce (cerium) are

Options:
A
+2, +3

B
+2, +4

C
+3, +4

D
+3, +5

The correct answer is:
+3, +4

Solution:
No Solution Available

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2002
Most common oxidation states shown by cerium are

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

The correct answer is:
+3,+4

Solution:
The electronic configuration of Ce is:
Ce :Xe 4f1 5d1 6s2
The most common oxidation states shown by cerium are +3 and +4 .
As energy difference between 4f and 5d-orbitals is very less, so electrons of f-orbitals also take part in bond formation

Topic: Chemistry, d and f Block Elements, JEE Main
Most of the indicators have a useful colour change over a \( \mathrm{pH} \) range of
units,

Options:

The correct answer is:
2

Solution:
Most of the indicators have a useful colour change over a pH range of 2 units. For example, the most common indicator, phenolphthalein, changes from colourless to pink in the pH range from 8.3 to 10.0. This solves the problem for the fact that we may not have an indicator for which pKrn is exactly equal to the pH at the equivalence point.

Topic: Chemistry, Ionic Equilibrium, JEE Main
Most of the transition metals exhibit
(i) paramagnetic behaviour
(ii) diamagnetic behaviour
(iii) variable oxidation states
(iv) formation of coloured ions

Options:
A (ii), (iii) and (iv)
B (i), (iii) and (iv)
C (i), (ii) and (iii)
D (i), (ii) and (iv)

The correct answer is:
(i), (iii) and (iv)

Solution:
Due to presence of unpaired electrons in n-1d orbitals, the most of the transition metal ions and their compounds are paramagnetic. They form coloured ions and show variable oxidation states also due to presence of vacant d-orbitals.

Topic: Chemistry, d and f Block Elements, JEE Main
Most suitable reason why SA node act as pacemaker for heart.

Options:
A It is located in the right atrium.
B It is made up of neural tissue.
C It is first to generate the nerve impulse.
D It can generate the maximum number of action potentials per unit time.

The correct answer is:
It can generate the maximum number of action potentials per unit time.

Solution:
Sino-atrial node (SAN) - SAN is a patch of nodal tissue present in the right upper corner of the right atrium.
It can generate the maximum number of action potentials : 70 - 75 / min. It is also termed as the pacemaker because it is responsible for initiating and maintaining the rhythmic contractile activity of the heart. Normal heart rate = 70-75 times in a minute (72 beats/min).

Topic: Biology, Body Fluids and Circulation, NEET
Most suitable salt which can be used for efficient clotting of blood will be?

Options:
A FeCl3
B MgHCO32
C FeSO4
D NaHCO3

The correct answer is:
FeCl3

Solution:
Blood is a negatively charged sol.

Blood being a colloidal solution, its coagulation by Hardy-Shuze's law states " Higher the charge on cation higher will be its efficiency to coagulate the colloidal solution"

According to Hardy-Schulz's rule, the cation with high charge has high coagulation power.

FeCl3 has Fe+3 ions.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2021 (24 Feb Shift 2)
Motion of a particle in x-y plane is described by a set of following equations x=4sinπ2-ωt m and y=4sinωt m. The path of the particle will be

Options:
A circular
B helical
C parabolic
D eliptical

The correct answer is:
circular

Solution:
x=4sinπ2-ωt=4cosωt,

y=4sinωt

Clearly, x2+y2=4cosωt2+4sinωt2⇒x2+y2=16

Hence, the path covered by particle is a circle.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (28 Jun Shift 1)