datavorous/entrance-exam-dataset huggingface dataset:
Some ideal monoatomic gas A in an enclosure has a pressure P and the temperature T. Another ideal monoatomic gas B enclosed in a container of the same volume has a pressure 2P and the temperature T 2. The ratio of the average kinetic energy per molecule of gas A to gas B is

Options:
A 4 :1
B 2 :1
C 1 :2
D 1 :4

The correct answer is:
2 :1

Solution:
The average kinetic energy, K=32kT
∴K∝T
∴K1K2=T1T2=TT2=21
∴K1 :K2=2 :1

Topic: Physics, Thermodynamics, JEE Main
Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as: (where λ is the decay constant)

Options:
A 12ln2λ
B ln2λ
C 2ln2λ
D ln32λ

The correct answer is:
ln32λ

Solution:
3N04=N0e-λt1

N02=N0e-λ2

ln3/4=-λt1 … (1)

ln1/2=-λt2 … (2)

ln3/4-ln1/2=λt2-t1 … (3)

Δt=ln3/2λ

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Some of the properties of the two species, $\mathrm{NO}_3^{-}$ and $\mathrm{H}_3 \mathrm{O}^{+}$are described below. Which one of them is correct?

Options:
A Dissimilar in hybridisation for the central atom with different structures
B Isostructural with same hybridisation for the central atom
C Isostructural with different hybridisation for the central atom
D Similar in hybridisation for the central atom with different structures

The correct answer is:
Dissimilar in hybridisation for the central atom with different structures

Solution:
$$
\begin{aligned}
& \text { In } \mathrm{NO}_3^{-}, \\
& \mathrm{H}=\frac{1}{2}[5+0-0+1]=3
\end{aligned}
$$
Thus, in $\mathrm{NO}_3^{-}$, the central atom is $\mathrm{sp}^2$ hybridised and it has trigonal planar geometry.
In $\mathrm{H}_3 \mathrm{O}^{+}$,
$$
\mathrm{H}=\frac{1}{2}[6+3-1+0]=4
$$
Thus, $\mathrm{O}$ is sp ${ }^3$ hybridised in $\mathrm{H}_3 \mathrm{O}^{+}$and $\mathrm{H}_3 \mathrm{O}^{+}$ has pyramidal geometry due to the presence of one lone pair of electrons.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2010 (Mains)
Some plants show slender lateral branches which arise from the base of main axis and arch downwards after its initial aerial growth to touch the ground. Examples of such plants include:

Options:
A Pistia and Eichhornia
B Mint and Jasmine
C Banana, Pineapple and Chrysanthemum
D Opuntia and Euphorbia

The correct answer is:
Mint and Jasmine

Solution:
In the given question, description of stolon is represented which are seen in mint and jasmine. Pistia and Eichhornia are examples of offset. Banana, pineapple and Chrysanthemum are examples of suckers. Opuntia is an example of phylloclade while Euphorbia is an example of cladode.

Topic: Biology, Reproduction in Organisms, NEET
Some properties of system are given below out of which \( x \) are number of extensive property and \( y \) are the number of intensive property. find \( (\mathrm{x}-\mathrm{y}) \). Boiling point, surface tension, heat capacity, gibbs free energy, vapour pressure, refractive index, density, temperature, entropy, enthalpy, mass, length, volume, area, internal energy.

Options:

The correct answer is:
03

Solution:
Extensive properties → Heat capacity, gibbs free energy, entropy, enthalpy, mass, length, volume, area, internal energy
x = 9
Intensive properties → Surface tension, boiling point, vapour pressures, refrective index, density, temperature
y = 6,
x – y = 03

Topic: Chemistry, Thermodynamics (C), JEE Main
Some reactions of NO2 relevant to photochemical smog formation are

Identify A,B,X and Y

Options:
A X=[O],Y=NO,A=O2, B=O3
B X=N2O,Y=[O],A=O3, B=NO
C X=12O2,Y=NO2, A=O3, B=O2
D X=NO,Y=[O],A=O2, B=N2O3

The correct answer is:
X=[O],Y=NO,A=O2, B=O3

Solution:
When fossil fuels are burnt, a variety of pollutants are emitted into the earth’s troposphere. Two of the pollutants that are emitted are hydrocarbons (unburnt fuels) and nitric oxide (NO). When these pollutants build up to sufficiently high levels, a chain reaction occurs from their interaction with sunlight in which NO is converted into nitrogen dioxide NO2. This NO2 in turn absorbs energy from sunlight and breaks up into nitric oxide and a free oxygen atom.

NO2 (g) →hν NO (g) + O(g)

Oxygen atoms are very reactive and combine with the O2 in air to produce ozone.

O (g) + O2 (g) ⇌ O3 (g)

Hence, X = O, Y = NO, A = O2, B = O3

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Some statements about heavy water are given below
(A) Heavy water is used as moderator in nuclear reactors.
(B) Heavy water is more associated than ordinary water.
(C) Heavy water is more effective solvent than ordinary water.
Which of the above statements are correct?

Options:
A $\mathrm{A}$ and $\mathrm{B}$
B A, B and C
C B and C
D A and C

The correct answer is:
$\mathrm{A}$ and $\mathrm{B}$

Solution:
Heavy water is used as a moderator in nuclear reactors. Its boiling point is higher as compared to the ordinary water. Thus, it is more associated as compared to ordinary water. The dielectric constant is however higher for $\mathrm{H}_2 \mathrm{O}$, thus, $\mathrm{H}_2 \mathrm{O}$ is a more effective solvent than heavy water $\left(\mathrm{D}_2 \mathrm{O}\right)$.

Topic: Chemistry, Hydrogen, NEET, NEET 2010 (Mains)
Some varieties of wheat are known as spring wheat while others are called winter wheat. Former variety is sown and planted in spring and is harvested by the end of the same season. However, winter varieties, if planted in spring, fail to flower or produce mature grains within a span of a flowering season. Explain, why?

Options:

None

Solution:
Some annual plants such as wheat do not flower, unless theyexperience a low temperature during spring they remain vegetative but after receiving low temperature (in winter) they grow further to bear flowers and fruits.
During winter the low temperature prevents precocious reproductive development in autumn, thus enabling the plant to reach vegetative maturity before reproductive phase. Thus, in spring when spring varieties are planted they flower and bear fruits prior to end of growing season. But, if the winter varieties are planted in spring, they fail to flower and produce mature grains before the end of growing season, as they could not perceive low temperature of winters.

Topic: Biology, Plant - Growth and Development
Some vascular bundles are described as open because these

Options:
A are surrounded by pericycle but to endodermis
B are capable of producing secondary xylem and phloem
C possess conjunctive tissue between xylem and phloem
D are not surrounded by pericycle

The correct answer is:
are capable of producing secondary xylem and phloem

Solution:
In open vascular bundle, cambium is present between xylem and phloem of the same vascular bundle, $e g$, dicot stems. Due to the presence of cambium, secondary growth in stem occurs and secondary permanent tissues (secondary xylem, secondary phloem and vascular rays) are formed.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2011 (Mains)
Sometimes due to biotic/abiotic factor the climax remain in a particular seral stage (pre climax) without reaching climax. Do you agree with this statement. If yes give a suitable example.

Options:

None

Solution:
Climax sometimes remains in a particular seral stage without reaching to the climax because any change in abiotic and biotic component during ecological succession may affect the particular seral stage, leading to preclimax stage before the climax is achieved.
This type of condition occurs in presence of seeds and other propagules. This secondarily based area may be invaded by moss or exotic weeds thus seriously exhibiting succession without the regeneration of the climax community. In the case of natural calamities like-fire landslide, floods, change in soil texture.

Topic: Biology, Ecosystem
Sometimes it is possible to separate two sulphide ores by adjusting the proportion of oil to water or by using depressants in froth flotation. When a depressant NaCN is added to an ore containing ZnS and PbS, what is the correct observation?

Options:
A NaCN prevents PbS from coming to the froth but allows ZnS to come with froth
B NaCN prevents ZnS from coming to the froth but allows PbS to come with froth
C NaCN prevents frothing of both ZnS and PbS, hence no froth is formed
D NaCN does not act as depressant hence a mixture of PbS and ZnS is found in froth

The correct answer is:
NaCN prevents ZnS from coming to the froth but allows PbS to come with froth

Solution:
NaCN acts as a depressant, it selectivley prevents ZnS from coming to froth by forming a complex Na2[Zn(CN)4] but allows PbS to come with froth.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Sound becomes hazardous noise pollution at level

Options:
A Above 30 dB
B Above 80 dB
C Above 100 dB
D Above 120 dB

The correct answer is:
Above 80 dB

Solution:
Above 80 dB sound becomes hazardous. Noise pollution, also known as environmental noise or sound pollution, is the propagation of noise with harmful impact on the activity of human or animal life. The source of outdoor noise worldwide is mainly caused by machines, transport and transportation systems.

Topic: Biology, Environmental Issues, NEET
Sound travels in a mixture of two moles of helium and n moles of hydrogen. If rms speed of gas molecules in the mixture is 2 times the speed of sound, then the value of n will be

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Speed of sound in a gas is given by, vs=γRTM

The formula for RMS speed is, vrms=3RTM

It is given that vrms=2vs

⇒vsvrms=γ3=12⇒γ3=12⇒γ=32

For γ of mixture,

γmix=CP,mixCV,mix=2CP,H+nCP,He2+n2CV,H+nCV,He2+n

Since Hydrogen is a diatomic gas, CV,H=52R, CP,H=72R and Helium is monatomic gas, CV,He=32R, CP,He=52R. Therefore,

γmix=2CP,H+nCP,He2+n2CV,H+nCV,He2+n⇒32=2×72R+n×52R2×52R+n×32R

⇒n=2

Topic: Physics, Waves and Sound, JEE Main
Sound travels in a mixture of two moles of helium and n moles of hydrogen. If rms speed of gas molecules in the mixture is 2 times the speed of sound, then the value of n will be

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Speed of sound in a gas is given by, vs=γRTM

The formula for RMS speed is, vrms=3RTM

It is given that vrms=2vs

⇒vsvrms=γ3=12⇒γ3=12⇒γ=32

For γ of mixture,

γmix=CP,mixCV,mix=2CP,H+nCP,He2+n2CV,H+nCV,He2+n

Since Hydrogen is a diatomic gas, CV,H=52R, CP,H=72R and Helium is monatomic gas, CV,He=32R, CP,He=52R. Therefore,

γmix=2CP,H+nCP,He2+n2CV,H+nCV,He2+n⇒32=2×72R+n×52R2×52R+n×32R

⇒n=2

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Sound travels in rocks in the form of

Options:
A Longitudinal elastic waves only
B Transverse elastic waves only
C Both longitudinal and transverse elastic waves
D Non-elastic waves

The correct answer is:
Both longitudinal and transverse elastic waves

Solution:
1. In rock sound travels in the form of Iongitudinal, transverse and elastic waves,
2. In longitudinal wave, the vibration of particles along the direction of propagation of wave.
3. Transverse waves cause the medium to move perpendicular to the direction of the wave.
4. When particles are displaced in a medium, an elastic wave occurs in which a force proportionate to the displacement acts on the particles to return them to their original position. Sound waves are elastic waves.
5. Sound travel in the air only in the form of Iongitudinal wave.

Topic: Physics, Waves and Sound, JEE Main
Sound velocity is maximum in

Options:
A $\mathrm{H}_2$
B $\mathrm{N}_2$
C $\mathrm{He}$
D $\mathrm{O}_2$

The correct answer is:
$\mathrm{H}_2$

Solution:
$v=\sqrt{\frac{\gamma R T}{M}} \Rightarrow v \propto \frac{1}{\sqrt{M}}$
Since $M$ is minimum for $\mathrm{H}_2$ so sound velocity is maximum in $\mathrm{H}_2$

Topic: Physics, Waves and Sound, JEE Main
Sound waves in the air are always longitudinal because

Options:
A Of the inherent characteristics of sound waves in air
B Air does not have a modulus of rigidity
C Air is a mixture of several gases
D Density of air is very small

The correct answer is:
Air does not have a modulus of rigidity

Solution:
In the transmission of sound through air, there is no actual movement of air from the sound producing body to our ear. The air layers only vibrate back and forth, and transfer the sound energy from one layer to the next layer till it reaches our ear. This back and forth motion causes the compression and rarefaction in a sound wave. This motion is along the direction of propagation of sound and hence, the sound waves are longitudinal. Note that the layers of air consist of molecules of gases. Note that the layers of air consist of molecules of gases. So, when the air layers vibrate back and forth, we actually mean that the molecules in air layers vibrate back and forth by a small distance.
Therefore, it simply means that air does not have modulus of rigidity.

Topic: Physics, Waves and Sound, JEE Main
Sound waves of Î½=600 Hz fall normally on a perfectly reflecting wall. The shortest distance from the wall at which all particles will have a maximum amplitude of vibration will be (speed of sound=300 m s-1)

Options:
A 78 m
B 38 m
C 18 m
D 14 m

The correct answer is:
18 m

Solution:
The wall acts like a rigid boundary and reflects this wave and sends it back towards the open end. At the open an antinode is formed and a node is formed at the wall. The distance between antinode and node is
λ4
Therefore, if v be the frequency of note emitted then
λ=vv
⟹ λ=300600=12 m
Maximum amplitude is obtained at a distance
=λ4=12×14=18 m

Topic: Physics, Waves and Sound, JEE Main
Sound waves of wavelength greater than that of audible sound are called

Options:
A Seismic waves
B Sonic waves
C Ultrasonic waves
D Infrasonic waves

The correct answer is:
Infrasonic waves

Solution:
Infrasonic waves have frequency less than $(20 \mathrm{~Hz})$ audible sound and wavelength more than audible sound

Topic: Physics, Waves and Sound, JEE Main
sp3d2 hybridization is not displayed by

Options:
A SF6
B BrF5
C PF5
D CrF63-

The correct answer is:
PF5

Solution:
SF6=sp3d2 BrF5=sp3d2
CrF63-=sp3d2 PF5=sp3d

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2017 (08 Apr Online)
Space between two concentric conducting spheres of radii a and b (b>a) is filled with a medium of resistivity ρ . The resistance between the two spheres will be:

Options:
A ρ4π1a+1b
B ρ2π1a+1b
C ρ4π1a-1b
D ρ2π1a-1b

The correct answer is:
ρ4π1a-1b

Solution:
Consider small volume element between r and r+dr
Resistance of this volume along rational directions,
dR=ρdr4πr2
As all element resistance are in series,
R=∫r=ar=bρdr4πr2=ρ4π-1rr=ar=b=ρ4π1a-1b

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Specialized cells for fixing atmospheric nitrogen in Nostoc are:

Options:
A heterocysts
B hormogonia
C nodules
D akinetes

The correct answer is:
heterocysts

Solution:
Heterocyst are found in many species of filamentous blue-green algae. They are cells of slightly larger size and with a more thickened wall than the vegetative cells. heterocyst are the site of nitrogen fixation in blue-green algae. The role includes: (i) fixation of nitrogen by all heterocystous algae, (ii) inhibition of heterocyst formation by combined nitrogen and (iii) direct observations on acetylene reduction by isolated heterocyst.
Related Theory
Heterocysts are specialized, paleyellow, thick-walled cells with the function of nitrogen-fixing formed during nitrogen starvation by some filamentous cyanobacteria, such as Nostoc and Anabaena. They fix nitrogen from dinitrogen $\left(N_2\right)$ in the air using the enzyme nitrogenase, in order to provide the cells in the filament with nitrogen for biosynthesis. Nitrogenase is inactivated by oxygen, so the heterocyst must create a micro anaerobic environment. Hence, their main function is nitrogen fixation.

Topic: Biology, Mineral Nutrition, NEET, NEET 2013 (Karnataka)
Species acting as both Bronsted acid and base is

Options:
A
$\left(\mathrm{HSO}_4\right)^{-1}$

B
$\mathrm{Na}_2 \mathrm{CO}_3$

C
$\mathrm{NH}_3$

D
$\mathrm{OH}^{-1}$

The correct answer is:
$\left(\mathrm{HSO}_4\right)^{-1}$

Solution:
$\left(\mathrm{HSO}_4\right)^{-}$ can accept and donate a proton
$\left(\mathrm{HSO}_4\right)^{-}+\mathrm{H}^{+} \rightarrow \mathrm{H}_2 \mathrm{SO}_4$
$\left(\mathrm{HSO}_4\right)^{-}-\mathrm{H}^{+} \rightarrow \mathrm{SO}_4^{2-}$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2002
Species diversity _______ as we move away from the _______ towards ______.

Options:
A decreases, equator, poles
B increases, equator, poles
C decreases, low altitude, high altitude
D More than one option is correct

The correct answer is:
More than one option is correct

Solution:
The region near the middle of the Earth between the Tropic of Cancer and Tropic of Capricon is considered as the Tropic region. This region receives the most amount of solar energy which leads to better productivity and higher species diversity. Also, the seasonal variations in such regions are low and environment mainly remains the same supporting high species richness and promoting niche specialization. Species-area relationship show a rectangular hyperbolic function. Thus, species diversity decreases as we move away from the equator towards poles.
Also as we move from low altitude to higher altidue, the species diversity decreases.

Topic: Biology, Biodiversity and its Conservation, NEET
Species like SbCl6-, SnCl62â, XeF5+ and IO65- has hybridization as sp3dx-1. The value of âxâ is

Options:

The correct answer is:
3.00

Solution:
Finding the number of hybridized orbitals (by using the formula) in the central atom to find the hybridization:
H = 1 2 V + M - C + A
SbCl6- â H = 12(5+6+1)=6, sp3d2
SnCl62- â H = 12(4+6+2)=6, sp3d2
XeF5+ â H = 12(8+5-1)=6, sp3d2
IO65- â H = 12(7+0+5)=6, sp3d2

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Specific resistance of a conductor increase with:

Options:
A increase in the temperature
B increase in the cross-section area
C increase in the cross-section and decrease in the length
D increase in the cross section area

The correct answer is:
increase in the temperature

Solution:
Specific resistance of a conduction increases with increases in the temperature.

Topic: Physics, Current Electricity, JEE Main
Specify the co-ordination geometry around and hybridisation of N and B atoms in a 1:1 complex of BF3 and NH3

Options:
A N: tetrahedral, sp3 ; B : tetrahedral, sp3
B N: pyramidal, sp3 ; B : tetrahedral, sp3
C N: pyramidal, sp3 ; B : planar, sp2
D N: pyramidal, sp3 ; B : pyramidal, sp3

The correct answer is:
N: tetrahedral, sp3 ; B : tetrahedral, sp3

Solution:
In the 1 : 1 complex H3N:→BF3 , both N and B have sp3 hybridised and tetrahedral geometry as both N & B atoms have 4 bp electrons .

Topic: Chemistry, Coordination Compounds, JEE Main
Speed of an electron in Bohr’s 7th orbit for Hydrogen atom is 3.6 × 106 m s-1. The corresponding speed of the electron in 3rd orbit, in m s-1 is :

Options:
A 1.8×106
B 7.5×106
C 3.6×106
D 8.4×106

The correct answer is:
8.4×106

Solution:
For hydrogen like atom, the velocity of electron in nth orbit is given by

v=z2e22ε0hn

∴v∝1n

⇒v1v2=n2n1

⇒3.6×106v2=37

v2=8.4×106 m s-1

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Speed of sound at constant temperature depends on

Options:
A Pressure
B Density of gas
C Above both
D None of the above

The correct answer is:
None of the above

Solution:
Speed do sound, doesn't depend up on pressure and density medium.

Topic: Physics, Waves and Sound, JEE Main
Speed of two identical cars are $u$ and $4 u$ at a specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is

Options:
A 1;1
B 1:4
C 1:8
D 1:16

The correct answer is:
1:16

Solution:
$S \propto u^2 \Rightarrow \frac{S_1}{S_2}=\left(\frac{1}{4}\right)^2=\frac{1}{16}$

Topic: Physics, Motion In One Dimension, JEE Main
Speeds of two identical cars are $u$ and $4 u$ at the specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is

Options:
A
1 : 1

B
1 : 4

C
1 : 8

D
1 : 16

The correct answer is:
1 : 16

Solution:
Use $u^2=2$ as. a is same for both cases
$$
\mathrm{s}_1=\mathrm{u}^2 / 2 \mathrm{a} ; \mathrm{s}_2=16 \mathrm{u}^2 / 2 \mathrm{a}=16 \mathrm{~s}_1 \Rightarrow \mathrm{s}_1: \mathrm{s}_2=1: 16
$$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2002
Spermatogenesis is under the regulatory influence of

Options:
A ADH
B FSH
C LH
D STH

The correct answer is:
FSH

Solution:
Follicle Stimulating Hormone (FSH) is produced by basophilic cells of adenohypophysis of pituitary gland. In females, this hormone is responsible for the growth of ovarian follicles up to ovulation, while in males, its functions are development of seminiferous tubules and maintenance of spermatogenesis.

Topic: Biology, Human Reproduction, JIPMER, JIPMER 2012
Spherical balls of radius ' $r$ ' are falling in a viscous fluid of viscosity ' $\eta$ ' with a velocity ' $v$ '. The retarding viscous force acting on the spherical ball is

Options:
A Inversely proportional to ' $r$ ' but directly proportional to velocity ' $v$
B Directly proportional to both radius ' $r$ ' and velocity ' $v$
C Inversely proportional to both radius ' $r$ ' and velocity ' $v$
D Directly proportional to ' $r$ ' but inversely proportional to ' $v$

The correct answer is:
Directly proportional to both radius ' $r$ ' and velocity ' $v$

Solution:
Retarding force acting on a ball falling into a viscous fluid
$\mathrm{F}=6 \pi \eta \mathrm{Rv}$
Where $\mathrm{R}=$ Radius of ball
$\mathrm{v}=$ velocity of ball
$\eta=$ coefficient of viscosity
$\therefore \mathrm{F} \propto \mathrm{R}$ and $\mathrm{F} \propto \mathrm{v}$ or in words,
Retarding force is directly proportional to both $\mathrm{R}$ and $\mathrm{v}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Spherical balls of radius $R$ are falling in a viscous fluid of viscosity $\eta$ with a velocity $v$. The retarding viscous force acting on the spherical ball is

Options:
A
directly proportional to $\mathrm{R}$ but inversely proportional to $\mathrm{v}$.

B
directly proportional to both radius $R$ and velocity $v$.

C
inversely proportional to both radius $\mathrm{R}$ and velocity $\mathrm{v}$.

D
inversely proportional to $\mathrm{R}$ but directly proportional to velocity $\mathrm{v}$.

The correct answer is:
directly proportional to both radius $R$ and velocity $v$.

Solution:
Retarding viscous force $=6 \pi \eta \mathrm{R} v$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2004
Spherical insulating ball and a spherical metallic ball of same size and mass are dropped from the same height. Choose the correct statement out of the following {Assume negligible air friction}

Options:
A Time taken by them to reach the earth's surface will be independent of the properties of their materials
B Insulating ball will reach the earth's surface earlier than the metal ball
C Both will reach the earth's surface simultaneously
D Metal ball will reach the earth's surface earlier than the insulating ball

The correct answer is:
Insulating ball will reach the earth's surface earlier than the metal ball

Solution:
When a spherical insulating ball and a spherical metallic ball of same size and mass are dropped from the same height, then magnetic field will act on metallic ball. If the earth's magnetic field is taken into account, then an induced eddy current will form inside the spherical metallic ball, thus retarding its motion. Hence, it will take more time than insulating ball.

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (31 Jan Shift 1)