datavorous/entrance-exam-dataset huggingface dataset:
Low $\mathrm{Ca}^{2+}$ in the body fluid may be the cause of

Options:
A tetany
B anaemia
C angina pectoris
D gout

The correct answer is:
tetany

Solution:
The usual cause of tetany is lack of calcium. But excess of phosphate can also trigger the spasms. Hypoparathyroidism (hyposecretion of parathormone) can lead to tetany. In this, some skeletal muscles, usually of hands and feet fail to relax after a contraction and remain in sustained condition. Tetany of laryngeal, thoracic and phrenic muscles which help in breathing causes death, because the patient fails to breath (asphyxia).

Topic: Biology, Chemical Coordination and Integration, NEET, NEET 2010 (Screening)
Low oxidation state of metals in their complexes are common when ligands

Options:
A have good π-accepting character
B have good σ-donor character
C are havind good π-donating ability
D are havind poor σ-donating ability

The correct answer is:
have good π-accepting character

Solution:
When metal has low oxidation state, it has more electron density in d orbital. So it has more tendency to back donate electrons. Thus compound must have good π acceptor ligand.

π backbonding, also called π backdonation, is when electrons move from an atomic orbital on one atom to an appropriate symmetry antibonding orbital on a π-acceptor ligand.It is especially common in the organometallic chemistry of transition metals with multi-atomic ligands such as carbon monoxide, ethylene or the nitrosonium cation. Electrons from the metal are used to bond to the ligand, in the process relieving the metal of excess negative charge. Compounds where π backbonding occurs include Ni(CO)4 and Zeise's salt.

The metal to ligand bonding creates a synergic effect which strengthens the bond between ligand and the metal

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2022 (27 Jul Shift 2)
Low spin complex of $d^6$-cation in an octahedral field will have the following energy
$\left(\Delta_o=\right.$ crystal field splitting energy in an octahedral field, $P=$ electron pairing energy)

Options:
A $\frac{-12}{5} \Delta_o+P$
B $\frac{-12}{5} \Delta_o+3 P$
C $\frac{-2}{5} \Delta_o+2 P$
D $\frac{-2}{5} \Delta_o+P$

The correct answer is:
$\frac{-12}{5} \Delta_o+3 P$

Solution:
\(\text {C.F.S.E. }=(-0.4 x+0.6 y) \Delta_o+z P\)
where \(\mathrm{x}=\) number of electrons occupying \(\mathrm{t}_{2 g}\) orbital
\(y=\) number of electrons occupying \(e_g\) orbital
\(z=\) number of pairs of electrons
For low spin \(\mathrm{d}^6\) complex electronic configuration
\(\begin{aligned}
& =t_{2 g}^6 e_g^0 \text { or } t_{2 g}^{2,2,2} e_g^0 \\
& \therefore \quad x=6, y=0, z=3 \\
& \text { C.F.S.E. }=(-0.4 \times 6+0 \times 0.6) \Delta_o+3 P \\
& =\frac{-12}{5} \Delta_o+3 P
\end{aligned}\)

Topic: Chemistry, Coordination Compounds, NEET, NEET 2012 (Mains)
Lowest Oxidation number of an atom in a compound A2B is -2. The number of an electron in its valence shell is

Options:

The correct answer is:
6

Solution:
A2 B→2 A++B-2, B-2 has complete octet in its dianionic form, thus in its atomic state it has 6 electrons in its valence shell. As it has negative charge, it has acquired two electrons to complete its octet. If the lowest oxidation state of an atom is -2, this typically implies that the atom can gain two electrons to reach a stable configuration. Therefore, the number of valence electrons in that atom would be six, because gaining two electrons would fill its outer shell.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2024 (01 Feb Shift 1)
LPG mainly consist of the following

Options:
A methane
B hydrogen
C acetylene
D butane

The correct answer is:
butane

Solution:
Liquified petroleum gas (LPG) mainly consist butane. Besides this, it also consists propane, methane, nitrogen and ethane in little amount.

Topic: Chemistry, General Organic Chemistry, MHT CET, MHT CET 2012
L-shaped chromosomes are also called

Options:
A acrocentric
B telocentric
C sub-metacentric
D None of these

The correct answer is:
sub-metacentric

Solution:
The shape of the chromosome is determined by the position of centromere on the chromosome. Chromosomes may be of following five types on the basis of their shape.
(i) Acentric- Centromere remains absent.
(ii)Telocentric- Centromere present at one end of chromosome.
(iii)Acrocentric- When chromosome is divided into a very small segment at one end and a very large segment on the other end.
(iv)Submetacentric- When chromosome have two unequal segments, one of which is slightly larger than other and forms L-shaped chromosome.
(v)Metacentric- When two segments are equal and forming $V$-shaped structure.

Topic: Biology, Cell - The Unit of Life, JIPMER, JIPMER 2009
Lucas reagent is

Options:
A Concentrated $\mathrm{HCl}+$ anhydrous $\mathrm{ZnCl}_2$
B Dilute $\mathrm{HCl}+$ hydrated $\mathrm{ZnCl}_2$
C Concentrated $\mathrm{HNO}_3+$ anhydrous $\mathrm{ZnCl}_2$
D Concentrated $\mathrm{HCl}+$ anhydrous $\mathrm{MgCl}_2$

The correct answer is:
Concentrated $\mathrm{HCl}+$ anhydrous $\mathrm{ZnCl}_2$

Solution:
Luca’s reagent is a solution of anhydrous zinc chloride in concentrated hydrochloric acid.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Lucas test is used to make distinguation between \( 1^{\circ}, 2^{\circ} \) and \( 3^{\circ} \) alcohols.
This shown that -

Options:
A \( \mathrm{ROH} \) behaves as a base
B greater the value of \( \mathrm{pK}_{\mathrm{a}} \) (alcohols), greater the reactivity with conc. \( \mathrm{HCl} \) and thus sooner the formation of white turbidity
C both of the above are correct
D none of the above is correct

The correct answer is:
both of the above are correct

Solution:
ROH behaves as a base and gets protonated. Greater the value of pKa (teritary alcohols), greater the reactivity with conc. HCl and thus sooner the formation of turbidity

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Lungs does not collapse between breaths and some amount of air will always remain in the lungs which can never be expelled because,

Options:
A There is a negative pressure in the lungs
B There is a negative intrapleural pressure pulling at the lung walls
C There is a positive intrapleural pressure
D Pressure in the lungs is higher than the atmospheric pressure

The correct answer is:
There is a negative intrapleural pressure pulling at the lung walls

Solution:
The pressure within the pleural cavity is slightly less than that of atmospheric pressure and this is referred to as negative pressure. Hence, the lungs do not collapse between breaths and some air always remains in it. Even after the most forceful expiration, there will be sufficient volume of air in the lungs. This is called residual volume. It is about 1.2 litre in males and 1.1 litre in females.

Topic: Biology, Breathing and Exchange of Gases, NEET, NEET 2016 (Phase 2)
Lupinus arcticus is an example of

Options:
A 2000 years old viable seed.
B plants that produce flowers once in its lifetime.
C plants that produces smallest pollen.
D 10000 years old viable seed.

The correct answer is:
10000 years old viable seed.

Solution:
Seeds of a large number of species live for several years.
Some seeds can remain alive for hundreds of years. There are several records of very old yet viable seeds. The oldest is that of a lupine, Lupinus arcticus excavated from Arctic Tundra. The seed germinated and flowered after an estimated record of 10,000 years of dormancy.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Lying in the plane x+y+z=6 is a line L passing through 1,2,3 and perpendicular to the line of intersection of planes x+y+z=6 and 2x-y+z=4, then the equation of L is

Options:
A x-14=y-2-7=z-33
B x-12=y-21=z-3-3
C x-14=y-2-5=z-31
D x-13=y-21=z-3-4

The correct answer is:
x-14=y-2-5=z-31

Solution:
Required line lies in the plane x+y+z=6 so it is perpendicular to i^+j^+k^ and it is also perpendicular to line of intersection of the two given planes.
Let, nâ1=i^+j^+k^,nâ2=2i^-j^+k^ & Î½â be a vector along the required line
So, Î½â=nâ1Ãnâ1Ãnâ2
=2i^+j^+k^-32i^-j^+k^
=-4i^+5j^-k^
Hence, equation of the line is x-14=y-2-5=z-31

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Lysigenous cavity in monocot stems vascular bundles developed by the dissolution of _________.

Options:
A protoxylem
B metaxylem
C phloem
D ground tissue

The correct answer is:
protoxylem

Solution:
There are four tissue systems present in monocot stems i.e. dermal tissue system, hypodermal tissue system, ground tissue system, and vascular bundle system. There are barrel-shaped parenchymatous cells found in the epidermis. Trichomes are absent in the monocot stem and below the epidermis, there is a layer of the hypodermal cell present. The ground tissue is undifferentiated in monocot stem which is made up of loosely arranged parenchyma cells. There is a vascular bundle which is found scattered in the ground tissue. Xylem is found on the inner side whereas phloem on the outer side and cambium is absent in monocot stem. So, the inner protoxylem is non-functional and remains water-filled. This water-filled cavity is called the lysigenous cavity or protoxylem cavity.

Topic: Biology, Anatomy of Flowering Plants, NEET
$\sum_{m=1}^n \tan ^{-1}\left(\frac{2 m}{m^4+m^2+2}\right)$ is equal to

Options:
A $\tan ^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)$
B $\tan ^{-1}\left(\frac{n^2-n}{n^2-n+2}\right)$
C $\tan ^{-1}\left(\frac{n^2+n+2}{n^2+n}\right)$
D None of these

The correct answer is:
$\tan ^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)$

Solution:
We have $\sum_{m=1}^n \tan ^{-1}\left(\frac{2 m}{m^4+m^2+2}\right)$
$\begin{aligned}
& =\sum_{m=1}^n \tan ^{-1}\left(\frac{2 m}{1+\left(m^2+m+1\right)\left(m^2-m+1\right)}\right) \\
& =\sum_{m=1}^n \tan ^{-1}\left(\frac{\left(m^2+m+1\right)-\left(m^2-m+1\right)}{1+\left(m^2+m+1\right)\left(m^2-m+1\right)}\right) \\
& =\sum_{m=1}^n\left[\tan ^{-1}\left(m^2+m+1\right)-\tan ^{-1}\left(m^2-m+1\right)\right] \\
& =\left(\tan ^{-1} 3-\tan ^{-1} 1\right)+\left(\tan ^{-1} 7-\tan ^{-1} 3\right)+ \\
& \left(\tan ^{-1} 13-\tan ^{-1} 7\right)+\ldots \ldots+\left[\tan ^{-1}\left(n^2+n+1\right)-\tan ^{-1}\left(n^2-n+1\right)\right] \\
& =\tan ^{-1}\left(n^2+n+1\right)-\tan ^{-1} 1=\tan ^{-1}\left(\frac{n^2+n}{2+n^2+n}\right) .
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
MeC≡C-COCl→H2+Lindlar'scatalystA
The product 'A' is

Options:
A Me-C≡C-CHO
B
C
D

The correct answer is:

Solution:
Lindlar's catalyst reduces C≡C to C=C in syn (cis) addition and simultaneously reduces -COCl group to -CHO group.

Topic: Chemistry, Hydrocarbons, JEE Main
Mg(s)|Mg2+(aq)||Zn2+(aq)|Zn(s);E°=+3.13V
The correct plot of Ecell versus log⁡[Mg2+][Zn2+] will be represented as

Options:
A
B
C
D

The correct answer is:

Solution:
For the given cell,
E =  E o − 0.059 2 log 10 Mg 2+ Zn 2+
This equation is of the form y = mx + c, its slope is negative and intercept equal to E o
Slope=-0.0592
Intercept =  E o .

Topic: Chemistry, Electrochemistry, JEE Main
M grams of steam at 100oC is mixed with 200g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40oC [heat of vaporization of water is 540cal/g and heat of fusion of ice is 80cal/g ], the value of M is ________

Options:

The correct answer is:
40

Solution:
MiceLf+Mice40-0Cw=MsteamLv+Msteam100-40Cw
⇒20080+401=M540+601
⇒200120=M600
M=40 gm

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Mn2CO10 is an organometallic compound due to the presence of:

Options:
A Mn-O bond
B Mn-Mn bond
C C-O bond
D Mn-C bond

The correct answer is:
Mn-C bond

Solution:
The bond between metal and carbon has covalent nature. For this reason, the compound is considered as an organometallic compound.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Mn2+ can be converted into Mn7+ by reacting with

Options:
A SO2
B Cl2
C PbO2
D SnCl2

The correct answer is:
PbO2

Solution:
Strong oxidising agents such as PbO2 or sodium bismuthate (NaBiO3) oxidise Mn2+ to MnO4- or Mn7+.
2 Mn 2 + + 5 PbO 2 + 4 H + â 2 MnO 4 - + 5 Pb 2 + + 2 H 2 O

Topic: Chemistry, d and f Block Elements, JEE Main
MnO4- is a good oxidising agent in different media where it changes to
→Mn2+
→MnO42-
→MnO2
→Mn2O3
The oxidation number of Mn correspondingly reduces by

Options:
A -1,-3,-4,-5
B -5,-4,-3,-2
C -5,-1,-3,-4
D -2,-6,-4,-3

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

Solution:

Topic: Chemistry, d and f Block Elements, JEE Main
(M) number of chiral carbon in (M) = x

(Number of Geometrical isomers in (N) = y ).
The value of yx is

Options:

The correct answer is:
1

Solution:
(i) Trans, Trans, Trans
(ii) Trans, Trans, Cis
(iii) Cis, Trans, Cis
(iv) Cis, Cis, Cis
(v) Cis, Cis, Trans
(vi) Trans, Cis, Trans
x=6, y=6, yx=1

Topic: Chemistry, General Organic Chemistry, JEE Main
∧mo for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2mol-1 respectively. If the conductivity of 0.001 M HA is 5×10-5S cm-1, degree of dissociation of HA is

Options:
A 0.125
B 0.75
C 0.25
D 0.50

The correct answer is:
0.125

Solution:
Kohlrausch's law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations.

λmoHA=λmoHCl+λmoNaA-λoNaCl
=425.9+100.5-126.4
=400
λmo=K×1000M=5×10-5×10310-3=50
α=50400=0.125

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2019 (12 Jan Shift 2)
Mpdenotes the mass of a proton and Mn that of a neutron. A given nucleus, of binding energy B, contains Z protons and N neutrons. The mass M(N, Z) of the nucleus is given by (c is the velocity of light)

Options:
A M(N,Â Z)=NMn+ZMp-Bc2
B MN,Â Z=NMn+ZMp+Bc2
C M(N, Z)=NMn+ZMp-B/c2
D MN,Â Z=NMn+ZMp+B/c2

The correct answer is:
M(N, Z)=NMn+ZMp-B/c2

Solution:
Mass defect = mass of nucleon - mass of nucleus
B= ZMp+NMn-MN,Zc2
âM(M, Z)=ZMp+NMn-B/c2

Topic: Physics, Nuclear Physics, JEE Main
M→ possible alkynes
Write the sum of value of M + N.

Options:

The correct answer is:
9

Solution:
3+6=9

Topic: Chemistry, General Organic Chemistry, JEE Main
$m v^2$
A body of mass $m$ is moving in a circle of radius $r$ with a constant speed $v$. The force on the body is $\frac{m v^2}{r}$ and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle

Options:
A $\frac{m v^2}{\pi r^2}$
B Zero
C $\frac{m v^2}{r^2}$
D $\frac{\pi r^2}{m v^2}$

The correct answer is:
Zero

Solution:
Work done by centripetal force is always zero, because force and instantaneous displacement are always perpendicular.
$W=\vec{F} \cdot \vec{s}=F s \cos \theta=F s \cos \left(90^{\circ}\right)=0$

Topic: Physics, Work Power Energy, JEE Main
Macronutrients are present in plant tissue in an amount:

Options:
A >10 mol/kg of dry matter
B >10 mmol/kg of dry matter
C <10 nmol/kg of dry matter
D <10 mmol/kg of dry matter

The correct answer is:
>10 mmol/kg of dry matter

Solution:
Macronutrients are generally present in plant tissues in large amounts (in excess of 10 mmole kg–1 of dry matter). The macronutrients include carbon, hydrogen, oxygen, nitrogen, phosphorous, sulphur, potassium, calcium and magnesium. Of these, carbon, hydrogen and oxygen are mainly obtained from CO2 and H2O, while the others are absorbed from the soil as mineral nutrition.

Topic: Biology, Mineral Nutrition, NEET
Magnesium powder burns in air to give

Options:
A MgO and Mg3N2.
B MgO only.
C MgNO32 and Mg3N2.
D MgO and MgNO32.

The correct answer is:
MgO and Mg3N2.

Solution:
N2 and O2 are present in the air as the main constituent, so the following reaction takes place:

Mg+O2→MgO, Mg+N2→Mg3N2 will form.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2019 (09 Apr Shift 1)
Magnetic dipole moment of a rectangular loop is

Options:
A Inversely proportional to current in loop
B Inversely proportional to area of loop
C Parallel to plane of loop and proportional to area of loop
D Perpendicular to plane of loop and proportional to area of loop

The correct answer is:
Perpendicular to plane of loop and proportional to area of loop

Solution:
Magnetic dipole is given by, \(M=N i A\)
So, \(M \propto A\)
So, Magnetic moment is perpendicular to plane of loop and proportional to area of loop.

Topic: Physics, Magnetic Effects of Current, JEE Main
Magnetic field at point \( P \) due to given current distribution is

Options:
A \( \frac{\mu_{0} I}{4 \pi a} \odot \)
B \( \frac{\mu_{0} I}{2 \pi a} \odot \)
C \( \frac{\mu_{0} I}{\pi a} \otimes \)
D zero

The correct answer is:
\( \frac{\mu_{0} I}{2 \pi a} \odot \)

Solution:
There will be no magnetic field due to segments 3 and 4 as the point P lies along them.

The magnetic field due to segment 1 is,

B1=μ0I4πasin90°+sin0°=μ0I4πa outwards.

The magnetic field due to segment 2 is

B2=μ0I4πasin90°+sin0°=μ0I4πa outwards.

Bnet=B1+B2=2×μ0i4πa=μ0i2πa outwards.

Topic: Physics, Magnetic Effects of Current, JEE Main
Magnetic field due to $0.1 \mathrm{~A}$ current flowing through a circular coil of radius $0.1 \mathrm{~m}$ and 1000 turns at the centre of the coil is

Options:
A $2 \times 10^{-1} \mathrm{~T}$
B $4.31 \times 10^{-2} \mathrm{~T}$
C $6.28 \times 10^{-4} \mathrm{~T}$
D $9.81 \times 10^{-4} \mathrm{~T}$

The correct answer is:
$6.28 \times 10^{-4} \mathrm{~T}$

Solution:
$B=\frac{\mu_0 N i}{2 r}=\frac{4 \pi \times 10^{-7} \times 1000 \times 0.1}{2 \times 0.1}=6.28 \times 10^{-4} \mathrm{~T}$

Topic: Physics, Magnetic Effects of Current, JEE Main
Magnetic field in a plane electromagnetic wave is given by,
B→=B0sinkx+ωtj^ T. Expression for corresponding electric field will be:
(Where c is speed of light)

Options:
A E→=-B0csinkx+ωtk^ V m-1
B E→=B0csinkx-ωtk^ V m-1
C E→=B0csinkx+ωtk^ V m-1
D E→=B0csinkx+ωtk^ V m-1

The correct answer is:
E→=B0csinkx+ωtk^ V m-1

Solution:
For electromagnetic wave speed of light is given by
c=E0B0
Electric field is expressed as E0=cB0 , E→ is ⊥ to direction x and direction of B→
The expression of electric field is E→=cB0 sin⁡(kx+wt)k^ V m-1

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2017 (08 Apr Online)
Magnetic field induction at the centre of a circular coil of radius $5 \mathrm{~cm}$ and carrying a current $0.9 \mathrm{~A}$ is (in SI units) $\left(\varepsilon_0=\right.$ absolute permittivity of air in SI units, velocity of light $=3 \times 10^8 \mathrm{~ms}^{-1}$ )

Options:
A $\frac{1}{\varepsilon_0 10^{16}}$
B $\frac{10^{16}}{\varepsilon_0}$
C $\frac{\varepsilon_0}{10^{16}}$
D $10^{16} \varepsilon_0$

The correct answer is:
$\frac{1}{\varepsilon_0 10^{16}}$

Solution:
Magnetic field induction at the centre of circular coil,
$\begin{aligned}
& c^2=\frac{1}{\mu_0 \varepsilon_0} \\
& \therefore \quad \mu_0=\frac{1}{\varepsilon_0 c^2} \\
&
\end{aligned}$
So, from Eq. (i), we get
$\begin{aligned}
B & =\frac{1}{\varepsilon_0 c^2} \frac{I}{2 r} \\
& =\frac{1}{\varepsilon_0 \times\left(3 \times 10^8\right)^2} \times \frac{0.9}{2 \times 5 \times 10^{-2}} \\
& =\frac{1 \times 9}{\varepsilon_0 \times 9 \times 10^{16}}=\frac{1}{\varepsilon_0 \times 10^{16}}
\end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main
Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 0.2 m from the centre are in the ratio 8:1. The radius of coil is _______ .

Options:
A 0.15 m
B 0.2 m
C 0.1 m
D 1.0 m

The correct answer is:
0.1 m

Solution:
We know, the magnetic field on the axis of a current carrying circular ring is given by

B=μ04π2NIAR2+x23/2

∴ B1 B2=81=R2+0.22R2+0.0523/2

4R2+0.052=R2+0.22

4R2-R2=0.22-4×0.052

4R2-R2=0.22-0.12

3R2=0.3×0.1

R2=0.12⇒R=0.1

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (25 Feb Shift 1)
Magnetic flux (in weber) in a closed circuit of resistance 20 Ω varies with time ts as ϕ=8t2-9t+5. The magnitude of the induced current at t=0.25 s will be _____ mA.

Options:

The correct answer is:
250

Solution:
Given that magnetic flux ϕ=8t2-9t+5

According to Faraday's law of electromagnetic induction

emf=-dϕdt=-16t-9

At t=0.25 s

emf=-16×0.25-9=5 V

Current =emf Resistance =5 V20 Ω

=14 A=10004 mA=250 mA

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Magnetic flux $\varphi$ (in weber) linked with a closed circuit of resistance 10 ohm varies with time $t$ (in seconds) as
$\phi=5 t^2-4 t+1$ The induced electromotive force in the circuit at $t=0.2 \mathrm{sec}$. is

Options:
A $0.4$ volts
B $-0.4$ volts
C $-2.0$ volts
D $2.0$ volts

The correct answer is:
$2.0$ volts

Solution:
$e=-\frac{d \phi}{d t}=-(10 t-4)$ $\Rightarrow(e)_{t=2}=-(10 \times0.2-4)=2 \mathrm{volt}$

Topic: Physics, Electromagnetic Induction, JEE Main
Magnetic flux Ï linked with a stationary loop of resistance R varies with time t as Ï=at(Tât). The amount of heat generated in the loop, during the time interval T, is

Options:
A aT3R
B a2T23R
C a2T2R
D a2T33R

The correct answer is:
a2T33R

Solution:
|E|=dϕdt=(aT–2at)
H=∫E2Rdt=a2R∫0TT-2t2dt
=a2R∫0TT2+4t2-4Ttdt
=a2RT2t0T+43t30T-2Tt20T=a2T33R

Topic: Physics, Electromagnetic Induction, JEE Main
Magnetic flux through a coil of resistance $10 \Omega$ is changed by $\Delta \phi$ in $0.1 \mathrm{~s}$. The resulting current in the coil varies with time as shown in the figure. Then $|\Delta \phi|$ is equal to (in weber)

Options:
A
6

B
4

C
2

D
8

The correct answer is:
2

Solution:
$$
\text { As } \begin{aligned}
e=\frac{\Delta \phi}{\Delta t} & \text { or } R i=\frac{\Delta \phi}{\Delta t} \quad(\because e=R i) \\
\Rightarrow \quad \Delta \phi & =R(i . \Delta t) \\
& =R \times \text { area under } i-t \text { graph } \\
= & 10 \times \frac{1}{2} \times 4 \times 0.1=2 \text { weber }
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2012 (12 May Online)
Magnetic induction at the centre of a circular loop of area $\pi \mathrm{m}^2$ is 0.1 tesla. The magnetic moment of the loop is ( $\mu_0=$ permeability of air $)$

Options:
A $\frac{0.1 \pi}{\mu_0}$
B $\frac{0.2 \pi}{\mu_0}$
C $\frac{0.3 \pi}{\mu_0}$
D $\frac{0.4 \pi}{\mu_0}$

The correct answer is:
$\frac{0.2 \pi}{\mu_0}$

Solution:
Magnetic induction at the centre of circular loop
$$
B=\frac{\mu_0}{2} \frac{n i}{r}
$$
Magnetic moment of the loop
$$
\begin{array}{rlr}
M & =n i A=\frac{2 B r A}{\mu_0} & \\
& =\frac{2 \times 0.1 \times 1 \times \pi \times(1)^2}{\mu_0} & {[\because r=1]} \\
& =\frac{0.2 \pi}{\mu_0} &
\end{array}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
Magnetic materials used for making permanent magnets P and magnets in a transformer T have different properties of the following, which property best matches for the type of magnet required?

Options:
A T : Large retentivity, small coercivity
B P : small retentivity, large coercivity
C T : Large retentivity, large coercivity
D P: large retentivity, large coercivity

The correct answer is:
P: large retentivity, large coercivity

Solution:
As per the properties of Magnetic substances, materials favorable for permanent magnets should have high retentivity, large coercivity and high permeability.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2020 (02 Sep Shift 1)
Magnetic moment of $\mathrm{Gd}^{3+}$ ion $(\mathrm{Z}=64)$ is

Options:
A
3.62 BM

B
$9.72 \mathrm{BM}$

C
7.9 BM

D
$10.60 \mathrm{BM}$

The correct answer is:
7.9 BM

Solution:
$\mathrm{Gd}(64)=[\mathrm{Xe}] 4 f^7 5 d^1 6 s^2$
$\mathrm{Gd}^{3+}=[\mathrm{Xe}] 4 f^7 5 d^0 6 s^0$
i.e. no. of unpaired electrons $=7$
$$
\begin{aligned}
\mu & =\sqrt{n(n+2)}=\sqrt{7(7+2)} \\
& =\sqrt{63}=7.93 \mathrm{BM}
\end{aligned}
$$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2012 (12 May Online)
Magnetic susceptibility of a paramagnetic substance is

Options:
A Large and positive
B Small and positive
C Small and negative
D Large and negative

The correct answer is:
Small and positive

Solution:
Key Idea For paramagnetic substance, the magnetic susceptibility is small and positive, because they get feebly magnetised when placed in a magnetic field.
The magnetic susceptibility of a substance shows how easily a substance can be magnetised and given by χm=lH
Where, l = intensity of magnetisation,
and H = magnetic intensity of the field

Topic: Physics, Magnetic Properties of Matter, JEE Main