datavorous/entrance-exam-dataset huggingface dataset:
What is the strength of transverse magnetic field required to bend all the photoelectrons within a circle of a radius 50 cm when light of wavelength 3800 Aâ is incident on a barium emitter? (Given that work function of barium is 2.5 eV; h=6.63Ã10-34 J s; e=1.6Ã10-19 C; m=9.1Ã10-31 kg)

Options:
A 4.88×10-4 T
B 32×10-5 T
C 6.32×10-6 T
D 7.6×10-8 T

The correct answer is:
6.32×10-6 T

Solution:
12mvmax2=hcÎ»-Ï0 or vmax2=2mhcÎ»-Ï0
=29.1Ã10-31Ã6.63Ã10-34Ã3Ã1083.8Ã10-7-2.5Ã1.6Ã10-19
=27.12Ã1010 m2 s-2
Now, B=mvmaxeRmax=9.1Ã10-31Ã5.21Ã1051.6Ã10-19Ã0.5
=6.32Ã10-6 T

Topic: Physics, Dual Nature of Matter, JEE Main
What is the structure in the ovary that secretes mainly progesterone?

Options:
A Ovarian follicles
B Stromal tissues
C Corpus luteum
D Alveoli

The correct answer is:
Corpus luteum

Solution:
After ovulation, the ruptured follicle is converted to a structure called corpus luteum, which secretes mainly progesterone.

Topic: Biology, Chemical Coordination and Integration, NEET
What is the structure of C

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2024 (09 Apr Shift 1)
What is the structure of the major product when phenol is treated with bromine water?

Options:
A

B

C

D

The correct answer is:

Solution:
Phenol has activating (electron releasing) $-\mathrm{OH}$ group and bromine water supplies $\mathrm{Br}^{+}$ ion easily, hence under such conditions reaction does not stop at monobromo or dibromo stage but a fully brominated $(2,4,6$,-tribromophenol) compound is the final product.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
What is the structure of the major product when phenol is treated with bromine water?

Options:
A

B

C

D

The correct answer is:

Solution:
Phenol has activating (electron releasing) $-\mathrm{OH}$ group and bromine water supplies $\mathrm{Br}^{+}$ ion easily, hence under such conditions reaction does not stop at monobromo or dibromo stage but a fully brominated $(2,4,6$,-tribromophenol) compound is the final product.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2013 (22 Apr Online)
What is the sum of all natural numbers n such that the product of the digits of n (in base 10 ) is equal to n2-10n-36?

Options:
A 12
B 13
C 124
D 2612

The correct answer is:
13

Solution:
Given, n2-10n-36

n is a natural number.
∴ Product of its digits is ≥0

∴n2-10n-36≥0

n=10±100+1442

n=5±61

∴ n∈(-∞,5-61)∪(5+61,∞)

But n is positive integer.

∴ n≥13

When n is two digits numbers, then maximum product =9×9=81

∴ n2-10n-36≤81

n2-10n-117≤0

∴ n∈[5-142,5+142]

n is taken two digit number.

∴ n∈[13,17)=13,14,15,16

∴ Product of digits =3,4,5,6

When put n=13

132-10×13-36=169-166=3

n=13 satisfies

Topic: Mathematics, Quadratic Equation, JEE Main
What is the sum of all two-digit numbers which when divided by 3 leave 2 as the remainder?

Options:
A 1565
B 1585
C 1635
D 1655

The correct answer is:
1635

Solution:
The numbers which divided by 3, leaving remainder 2 will be of the form $3 \mathrm{x}+2$ Given, $3 \mathrm{x}+2$ is 2-digit number So, $x$ can be from 3 to 32
Sum of numbers $=\sum_{\mathrm{x}=3}^{32}(3 \mathrm{x}+2)$
$=3(3+4+5+\ldots .+32)+(2+2+\ldots .)$
$3,4,5 \ldots .32$ is an $\mathrm{A}$. $\mathrm{P}$ with $\mathrm{a}=3, \mathrm{~d}=1, \mathrm{~T}_{\mathrm{n}}=32$
$\therefore \mathrm{T}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) 1$
$32=3+(\mathrm{n}-1) 1$
$\Rightarrow \mathrm{n}-1=29 \Rightarrow \mathrm{n}=30$ terms
$\therefore(1) \Rightarrow 3(3+4+5+\ldots .+32)+(2+2+\ldots .30$ times $)$
$=3\left(\frac{30}{2}(3+32)\right)+(2 \times 30)\left(\because \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+\ell)\right)$
$=\frac{90}{2}(35)+60$
$=(45 \times 35)+60$
$=1575+60=1635$

Topic: Mathematics, Sequences and Series, NDA, NDA 2018 (Phase 1)
What is the sum of focal radii of any point on an ellipse equal to?

Options:
A Length oflatusrectum
B Length of major-axis
C Length of minor-axis
D Length of semi-latusrectum

The correct answer is:
Length of major-axis

Solution:
We know sum of the focal distances of a point on the ellipse is constant and is equal to the length of the major axis. Thus, we know that, the sum of focal radii of any point on ellipse is equal to length of major axis

Topic: Mathematics, Ellipse, JEE Main
What is the sum of the roots of the equation $x^3-6 x^2+11 x-6=0$ ?

Options:
A $6$
B $11$
C $-6$
D $1$

The correct answer is:
$6$

Solution:
To find the sum of the roots of a polynomial equation, we can use Vieta's formulas. For a cubic equation $a x^3+b x^2+c x+d=0$, the sum of the roots is given by $-\frac{b}{a}$.
In this equation $x^3-6 x^2+11 x-6=0$, we can see that $a=1$ (coefficient of $x^3$ ) and $b=-6$ (coefficient of $x^2$ ).
So, the sum of the roots $=-\frac{b}{a}=-\frac{-6}{1}=6$.
Therefore, the correct answer is option 1) 6 .

Topic: Mathematics, Quadratic Equation, JEE Main
What is the symbol of element if it's atomic number is 116?

Options:
A UuS
B Uut
C Uuh
D Uun

The correct answer is:
Uuh

Solution:
\begin{array}{|c|l|c|c|c|}
\hline Atomic number & Temp. Name & Temp. Symbol & Name of the element & Symbol \\
\hline 109 & Unnilennium & Une & Meitnerium & \mathrm{Mt} \\
\hline 110 & Ununnilium & Uun & Darmstadtium & \mathrm{Ds} \\
\hline 111 & Unununium & Uuu & Roentgenium & \mathrm{Rg} \\
\hline 112 & Ununbium & Uub & Copernicium & \mathrm{Cn} \\
\hline 113 & Ununtrium & Uut & Nihonium & \mathrm{Nh} \\
\hline 114 & Ununquadium & Uuq & Flerovium & \mathrm{Fl} \\
\hline 115 & Ununpentium & Uup & Moscovium & \mathrm{Mc} \\
\hline 116 & Ununhexium & Uuh & Livermorium & \mathrm{Lv} \\
\hline 117 & Ununseptium & Uus & Tennessine & \mathrm{Ts} \\
\hline 118 & Ununoctium & Uuo & Oganesson & \mathrm{Og} \\
\hline
\end{array}

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
What is the temperature of the junction of an iron bar (L1 = 0.1 m, A1 = 0.02 m2, K1 = 79 W ${{m}^{-1}}$ ${{K}^{-1}}$) and a brass bar (L2 = 0.1 m, A2 = 0.02 m2, K2 = 109 W ${{m}^{-1}}$ ${{K}^{-1}}$) when their free ends are maintained at 373 K and 273 K respectively?

Options:
A 273 K
B 315 K
C 328 K
D 350 K

The correct answer is:
315 K

Solution:
Correct Option is : (B)
315 K

Topic: Chemistry, Thermal Properties of Matter, NEET
What is the tendency of amorphous solids similar to?

Options:
A They have a tendency to become transparent.
B They have a tendency to remain unchanged.
C They have a tendency to flow, though very slowly.
D They have a tendency to become extremely rigid.

The correct answer is:
They have a tendency to flow, though very slowly.

Solution:
Like liquids, amorphous solids have a tendency to flow, though very slowly, as mentioned in the paragraph.

Topic: Chemistry, Solid State, NEET
What is the term for the phenomenon where oppositely oriented domains cancel out each other's magnetic moment?

Options:
A Ferromagnetism
B Paramagnetism
C Antiferromagnetism
D Diamagnetism

The correct answer is:
Antiferromagnetism

Solution:
Antiferromagnetism is the phenomenon where oppositely oriented domains cancel out each other's magnetic moment.

Topic: Chemistry, Solid State, NEET
What is the term used for the remaining material after the tissue is dried?

Options:
A Ash
B Elemental analysis
C Biomolecules
D Dry weight

The correct answer is:
Dry weight

Solution:
The term used for the remaining material after the tissue is dried is dry weight. This material is obtained by weighing a small amount of a living tissue (wet weight) and drying it until all the water has evaporated.

Topic: Biology, Biomolecules (B), NEET
What is the term used to describe the process of maintaining the Earth's temperature and making it suitable for life?

Options:
A Carbon footprint
B Greenhouse effect
C Photosynthesis
D Ozone layer

The correct answer is:
Greenhouse effect

Solution:
The greenhouse effect is the process that maintains the Earth's temperature and makes it suitable for life.

Topic: Chemistry, Environmental Chemistry, NEET
What is the time period of the seconds pendulum, on a planet similar to earth, but mass and radius three times as that of the earth?

Options:
A 2 s
B 6 s
C 23 s
D 23 s

The correct answer is:
23 s

Solution:
g=GMR2
gpge=MpMeÃReRp2=3Ã19=13
T=2Ïlg
TpTe=gegp=3
Tp=3Te
Time period at earth for seconds pendulum =2 s
∴Tp=23 s

Topic: Physics, Oscillations, JEE Main
What is the total energy of two molecular orbitals compared to two original atomic orbitals?

Options:
A It is higher.
B It is lower.
C It remains the same.
D It is zero.

The correct answer is:
It remains the same.

Solution:
The total energy of two molecular orbitals remains the same as that of two original atomic orbitals.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
What is the trend in atomic radius as you move down Group 17 (halogens)?

Options:
A Increases
B Decreases
C Remains constant
D Shows irregular variation

The correct answer is:
Increases

Solution:
The trend in atomic radius as you move down Group 17 (halogens)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the trend in boiling points as alkenes increase in size by adding a -CH2 group?

Options:
A Boiling points decrease
B Boiling points remain constant
C Boiling points increase
D Boiling points fluctuate

The correct answer is:
Boiling points increase

Solution:
Boiling points of alkenes increase with the addition of each -CH2 group, typically by 20–30 K.

Topic: Chemistry, Hydrocarbons, NEET
What is the trend in boiling points of alkanes as molecular size or surface area increases?

Options:
A Boiling points decrease
B Boiling points remain constant
C Boiling points increase
D Boiling points follow an unpredictable pattern

The correct answer is:
Boiling points increase

Solution:
The boiling points of alkanes increase with an increase in molecular size or surface area due to the stronger intermolecular van der Waals forces.

Topic: Chemistry, Hydrocarbons, NEET
What is the trend in boiling points of isomeric haloalkanes as branching increases?

Options:
A They remain constant.
B They increase.
C They decrease.
D They become unpredictable.

The correct answer is:
They decrease.

Solution:
The boiling points of isomeric haloalkanes decrease with an increase in branching.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
What is the trend in electron gain enthalpy as you move down Group 17 (halogens)?

Options:
A Increases
B Decreases
C Remains constant
D Varies unpredictably

The correct answer is:
Decreases

Solution:
Electron gain enthalpy becomes less negative as you move down Group 17 (halogens).**

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the trend in hydration enthalpy of alkali metal ions as you move down the group?

Options:
A Increases
B Decreases
C Remains constant
D Fluctuates unpredictably

The correct answer is:
Decreases

Solution:
The trend in hydration enthalpy of alkali metal ions is that it decreases as you move down the group. This is because larger ions have a lower charge density and are less effective at attracting and binding water molecules.

Topic: Chemistry, s Block Elements, NEET
What is the trend in ionic radii for the elements in Group 15 in a particular state?

Options:
A They decrease.
B They remain constant.
C They increase significantly.
D They increase slightly.

The correct answer is:
They increase significantly.

Solution:
The paragraph doesn't explicitly mention the trend in ionic radii but does mention the increase in covalent radius.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the trend in ionisation enthalpy as you move down Group 17 (halogens)?

Options:
A Increases
B Decreases
C Remains constant
D Varies unpredictably

The correct answer is:
Decreases

Solution:
Ionisation enthalpy decreases as you move down Group 17 (halogens) due to the increase in atomic size and the addition of electron shells.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the trend in ionization enthalpy values expected as you move down the group?

Options:
A They decrease smoothly.
B They increase steadily.
C They remain constant.
D They show discontinuity.

The correct answer is:
They decrease smoothly.

Solution:
The expected trend in ionization enthalpy values as you move down the group is that they decrease smoothly.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
What is the trend in melting points among alkali metals as you move down the periodic table?

Options:
A Melting points decrease
B Melting points increase
C Melting points remain constant
D Melting points are unpredictable

The correct answer is:
Melting points decrease

Solution:
The melting points of alkali metals generally decrease as you move down the periodic table, which is a consequence of weaker metallic bonding due to the increased atomic size and greater distance between the valence electrons and the nucleus.

Topic: Chemistry, s Block Elements, NEET
What is the trend in melting points from top to bottom in Group 15?

Options:
A Decreasing
B Increasing
C Remaining constant
D Becoming erratic

The correct answer is:
Increasing

Solution:
The paragraph indicates that the melting points increase from top to bottom in the group.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the trend in oxidation states from Carbon (C) to Lead (Pb) in Group 14?

Options:
A The common oxidation state decreases as you move from Carbon to Lead.
B The common oxidation state remains the same for all elements in Group 14.
C The common oxidation state increases as you move from Carbon to Lead.
D Carbon and Lead have the same common oxidation state.

The correct answer is:
The common oxidation state increases as you move from Carbon to Lead.

Solution:
Correct Option is : (C)
The common oxidation state increases as you move from Carbon to Lead.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the trend in the acidic character of the oxides as you move down the Group 15 elements?

Options:
A It increases
B It decreases
C It remains the same
D It depends on the oxidation state

The correct answer is:
It decreases

Solution:
The paragraph indicates that the acidic character decreases as you move down the group.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the trend in the ionic character of metal halides in the order of F > Cl > Br > I?

Options:
A Increases
B Decreases
C Remains constant
D Depends on the metal

The correct answer is:
Decreases

Solution:
The ionic character of the halides decreases in the order F > Cl > Br > I.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the trend in the relative stability of the +1 oxidation state as you move down the group?

Options:
A It decreases
B It increases
C It remains constant
D It cannot be determined from the paragraph

The correct answer is:
It increases

Solution:
The relative stability of +1 oxidation state progressively increases for heavier elements: Al

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Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
What is the trend in the solubility and nature of oxides of Group 2 elements?

Options:
A Increases down the group, becoming more basic
B Decreases down the group, becoming more acidic
C Remains constant across the group
D Increases down the group, becoming more acidic

The correct answer is:
Increases down the group, becoming more basic

Solution:
Correct Option is : (A)
Increases down the group, becoming more basic

Topic: Chemistry, s Block Elements, NEET
What is the trend in the stability of the -2 oxidation state as you move down Group 16?

Options:
A It increases
B It remains constant
C It varies unpredictably
D It decreases

The correct answer is:
It decreases

Solution:
The paragraph mentions that the stability of the -2 oxidation state decreases down the group.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the trend in the stability of the +6 oxidation state as you move down Group 16?

Options:
A It increases
B It remains constant
C It varies unpredictably
D It decreases

The correct answer is:
It decreases

Solution:
The paragraph explains that the stability of the +6 oxidation state decreases down the group.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the trend of electronegativity across a period in the periodic table?

Options:
A It increases from left to right.
B It decreases from left to right.
C It remains constant.
D It depends on the specific period.

The correct answer is:
It increases from left to right.

Solution:
Electronegativity generally increases across a period from left to right in the periodic table.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
What is the trend of electronegativity values as you move down Group 15?

Options:
A They increase.
B They remain constant.
C They decrease.
D They fluctuate.

The correct answer is:
They decrease.

Solution:
The paragraph states that electronegativity values generally decrease down the group with increasing atomic size.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the true regarding complete combustion of gaseous isobutane‐

Options:
A ΔH=ΔE
B ΔH>ΔE
C ΔH=ΔE=0
D ΔH<ΔE

The correct answer is:
ΔH<ΔE

Solution:
C4H10g+6.5O2gâ4CO2g+5H2O(l)
În=4-7.5=-3.5
ÎH=ÎE+ÎngRT
â´ ÎH<ÎE

Topic: Chemistry, Thermodynamics (C), JEE Main
What is the type of movable joint present between the atlas and axis?

Options:
A Pivot
B Saddle
C Hinge
D Gliding

The correct answer is:
Pivot

Solution:
Pivot joint allows only a rotatory movement of one bone on the other, which remains stationary. Example: joint between atlas and axis vertebrae which enables the head to turn from side to side.

Topic: Biology, Locomotion and Movement, NEET
What is the typical behavior of domains in antiferromagnetic substances when an external magnetic field is applied?

Options:
A The domains align in the same direction.
B The domains align randomly.
C The domains repel each other.
D The domains remain oppositely oriented.

The correct answer is:
The domains align in the same direction.

Solution:
When an external magnetic field is applied to antiferromagnetic substances, the domains align in the same direction, overcoming their oppositely oriented arrangement.

Topic: Chemistry, Solid State, NEET
What is the ultimate product formed from the radioactive disintegration of uranium?

Options:
A Iron
B Radium
C Thorium
D Lead

The correct answer is:
Radium

Solution:
The ultimate product formed from the radioactive disintegration of uranium is Radium.

Topic: Physics, Nuclear Physics, JEE Main
What is the value of $\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ} ?$

Options:
A 4
B 2
C 1
D 0

The correct answer is:
4

Solution:
$\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$
$=\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}=\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$
$=\frac{4 \cdot \frac{1}{2}\left(\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}\right)}{2 \sin 20^{\circ} \cos 20^{\circ}}$
$=\left(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right)\left(\frac{4}{2 \sin 20^{\circ} \cos 20^{\circ}}\right)$
$=\left(\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ}\right)\left(\frac{4}{\sin 40^{\circ}}\right)$
$=\sin 40^{\circ} \frac{4}{\sin 40^{\circ}}=4$
$(\because \sin (A-B)=\sin A \cos B-\cos A \sin B)$

Topic: Mathematics, Trigonometric Equations, JEE Main
What is the value of electron gain enthalpy of $\mathrm{Na}^{+}$if $\mathrm{IE}_1$ of $\mathrm{Na}=5.1 \mathrm{eV}$ ?

Options:
A $-5.1 \mathrm{eV}$
B $-10.2 \mathrm{eV}$
C $+2.55 \mathrm{eV}$
D $+10.2 \mathrm{eV}$

The correct answer is:
$-5.1 \mathrm{eV}$

Solution:
$\mathrm{IE}_1$ of $\mathrm{Na}=-$ Electron gain enthalpy of $\mathrm{Na}^{+}$ion $=-5.1 \mathrm{eV}$.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET, NEET 2011 (Mains)
What is the value of frequency at which electromagnetic wave must be propagated for the D-region of atmosphere to have a refractive index of 0.5? Electron density for D-region is 400 electrons/cc

Options:
A 200 kHz

B 104.2 kHz
C 208.4 kHz
D 312.6 kHz

The correct answer is:
208.4 kHz

Solution:
Here, N=400 electrons c/cc=400Ã106 electrons m-3
Î¼=0.5=1-81.56 Nv2
=1-81.56Ã400Ã106v2
On solving, v=208.4 kHz.

Topic: Physics, Communication System, JEE Main
What is the value of Î G kJ / mol at 298 K at some non-equilibrium condition? Given the concentrations of [NH3] is 0.05 M and [NH4+] = [OH-] = 0.002 M in the presence of excess water. Also Î G Reaction â = + 2 6 Â· 8 1 KJ mol .

NH 3 (aq)â+â H 2 O(l)â NH 4 + (aq)â+â OH â (aq)

Options:
A + 3.437
B - 9.433
C + 50.18
D - 50.18

The correct answer is:
+ 3.437

Solution:
Î G = Î G â + 2 . 3 0 3 RT log Q
= 2 6 . 8 1 kJ + 2 . 3 0 3 Ã 8 . 3 1 4 Ã 2 9 8 log 2 Ã 1 0 - 3 Ã 2 Ã 1 0 - 3 5 Ã 1 0 - 2
= 2 6 . 8 1 kJ + 5 7 0 6 J log 4 5 Ã 1 0 - 4
= 2 6 . 8 1 + 5 . 7 0 6 - 4 + 0 . 6 0 - 0 . 7
= 2 6 . 8 1 + 5 . 7 0 6 - 4 . 1
= 2 6 . 8 1 - 2 3 . 3 7
=3.44 kJ mol −1

Topic: Chemistry, Thermodynamics (C), JEE Main
What is the value of inductance $L$ for which the current is a maximum in a series $L C R$ circuit with $C=10 \mu \mathrm{F}$ and $\omega=1000 \mathrm{~s}^{-1}$ ?

Options:
A $100 \mathrm{mH}$
B $1 \mathrm{mH}$
C Cannot be calculated unless $R$ is known
D $10 \mathrm{mH}$

The correct answer is:
$100 \mathrm{mH}$

Solution:
Current in $L C R$ series circuit,
$i=\frac{V}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}$
where $V$ is rms value of current, $R$ is resistance, $X_{L}$ is inductive reactance and $X_{C}$ is capacitive reactance.
For current to be maximum, denominator should be minimum which can be done, if
$$
X_{L}=X_{C}
$$
This happens in resonance state of the circuit ie,
$$
\omega L=\frac{1}{\omega C}
$$
or $\quad L=\frac{1}{\omega^{2} C}$ ...(i)
$\begin{aligned} \text { Given, } \omega &=1000 \mathrm{~s}^{-1}, C=10 \mu \mathrm{F} \\ &=10 \times 10^{-6} \mathrm{~F} \\ \text { Hence, } L &=\frac{1}{(1000)^{2} \times 10 \times 10^{-6}} \\ &=0.1 \mathrm{H} \\ &=100 \mathrm{mH} \end{aligned}$

Topic: Physics, Alternating Current, JEE Main
What is the value of $k$ if the area bounded by the curve $y=\sin k x, y=0, x=\pi / k, x=\pi /(3 k)$ is 3 sq unit?

Options:
A $1 / 2$
B 1
C $3 / 2$
D 2

The correct answer is:
$1 / 2$

Solution:
Given $x=\pi / k, x=\frac{\pi}{3 k}$ and $y=\sin k x$
Let the required area be $A$ So, $\mathrm{A}=3$ (Given)
ThereforeArea, $A=\int_{\pi / 3 k}^{\pi / k} \sin k x d x$
$\Rightarrow 3=-\left[\frac{\cos k x}{k}\right]_{\pi / 3 k}^{\pi / k}$
$\Rightarrow 3=-\frac{1}{k}\left[\cos \pi-\cos \frac{\pi}{3}\right]$
$\Rightarrow 3=-\frac{1}{k}\left[-1-\frac{1}{2}\right]$
$(\because \cos \pi=\cos (\pi / 2+\pi / 2)$
$\quad=-\sin \pi / 2$ and $\left.\cos 60^{\circ}=\frac{1}{2}\right)$
$\Rightarrow 3=\frac{3}{2 k} \Rightarrow k=\frac{1}{2}$

Topic: Mathematics, Area Under Curves, JEE Main
What is the value of linear velocity, if vector $r=3 i-4 j+k$ and $\omega=5 i-6 j+6 k$

Options:
A $6 \hat{i}+2 \hat{j}-3 \hat{k}$
B $-18 \hat{i}-13 \hat{j}+2 \hat{k}$
C $4 \hat{i}-13 \hat{j}+6 \hat{k}$
D $6 \hat{i}-2 \hat{j}+8 \hat{k}$

The correct answer is:
$-18 \hat{i}-13 \hat{j}+2 \hat{k}$

Solution:
$\vec{v}=\vec{\omega} \times \vec{r}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 1 \\ 5 & -6 & 6\end{array}\right|=-18 \hat{i}-13 \hat{j}+2 \hat{k}$

Topic: Physics, Motion In Two Dimensions, JEE Main
What is the value of n in the following half equation,

Cr(OH) 4 −  +  OH − → CrO 4 2−  +  H 2 O +  ne −  ?

Options:
A 3
B 6
C 5
D 2

The correct answer is:
3

Solution:
Cr(OH)4-+OH-→CrO42-+H2O
To balance H and O atoms, add 3H2O to R.H.S. and 30H⊖ to the L.H.S. we have,
Cr(OH)4-+4OH-→CrO42-+4H2O
To balance charge, add 3e- to the R.H.S. we have
Cr(OH)4-+4OH-→CrO42-+4H2O+3e-

Topic: Chemistry, Redox Reactions, JEE Main