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25 Jan 2023 Shift 2

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Physics30 questions

Q1JEE Main 2023MCQ4MThermodynamics

Match List I with List II List List II A. Young's Modulus (Y) I. $\;\; [ {\text{M}} {\text{L}}^{-1} {\text{T}}^{-1}]$ B. Co-efficient of Viscosity $(η)$ II. $\;\; [ {\text{M}} {\text{L}}^2 {\text{T}}^{-1}]$ C. Planck's Constant (h) III. $\;\; [ {\text{M}} {\text{L}}^{-1} {\text{T}}^{-2}]$ D. Work Function $(\varphi )$ IV. $\;\; [ {\text{M}} {\text{L}}^2 {\text{T}}^{-2}]$ Choose the correct answer from the options given below: [25-Jan-2023 Shift 2]

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📖 Explanation

$\; {\text{Y}}=\;\frac{\;\text{ Stress }\;}{\;\text{ Strain }\;}=\;{ {\text{F}} / {\text{A}}}/{\Delta ℓ / ℓ}=\;{[ {\text{MLT}}^{-2}]\$ }/{[ {\text{L}}^2]$ }= [ {\text{ML}}^{-1} {\text{T}}^{-2}]; {\text{F}}=6 \pi η {\text{rv}} \Rightarrow η=;{ {\text{F}}}/{6 \pi {\text{rv}}};{[η]=;{[ {\text{MLT}}^{-2}]$ }/{[ {\text{L}}] $[ {\text{LT}}^{-1}]$ }= $[ {\text{ML}}^{-1} {\text{T}}^{-1}]$ }; {\text{E}}= {\text{h}} v \Rightarrow {\text{h}}=;{ {\text{E}}}/{v}=;{[ {\text{ML}}^2 {\text{T}}^{-2}]$ }/{[ {\text{T}}^{-1}]$ }= [ {\text{ML}}^2 {\text{T}}^{-1}] $Work function has same dimension as that of energy, so$ [φ]= [ {\text{ML}}^2 {\text{T}}^{-2}]$

Q2JEE Main 2023MCQ4MThermodynamics

According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is :- [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Diatomic gas molecules have three translational degree of freedom, two rotational degree of freedom & it is given that it has one vibrational mode so there are two additional degree of freedom corresponding to one vibrational mode, so total degree of freedom $=7$ ${\text{C}}_{ {\text{v}}}=\;{ {\text{fR}}}/{2}=\;{7 {\text{R}}}/{2}$

Q3JEE Main 2023MCQ4MRay Optics and Optical Instruments

The light rays from an object have been reflected towards an observer from a standard flat mirror, the image observed by the observer are :-A. RealB. ErectC. Smaller in size then objectD. Laterally invertedChoose the most appropriate answer from the options given below : [25-Jan-2023 Shift 2]

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Q4JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

For a moving coil galvanometer, the deflection in the coil is $0.05 {\text{rad}}$ when a current of $10 {\text{mA}}$ is passed through it. If the torsional constant of suspension wire is $4.0 \times 10^{-5} {\text{Nm}} {\text{rad}}^{-1}$ , the magnetic field is $0.01 {\text{T}}$ and the number of turns in the coil is 200 , the area of each turn (in ${\text{cm}}^2$ ) is : [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\tau = {\text{K}} \theta$ ${\text{NiAB}}= {\text{K}} \theta$ ${\text{A}}=\;{ {\text{K}} \theta}/{ {\text{NiB}}}=\;\frac{4 \times 10^{-5} \times 0.05}{200 \times 10 \times 10^{-3} \times 0.01}$ On solving $A=10^{-4} {\text{m}}^2=1 {\text{cm}}^2$

Q5JEE Main 2023MCQ4MThermodynamics

The graph between two temperature scales $P$ and $Q$ is shown in the figure. Between upper fixed point and lower fixed point there are 150 equal divisions of scale ${\text{P}}$ and 100 divisions on scale ${\text{Q}}$ . The relationship for conversion between the two scales is given by : [25-Jan-2023 Shift 2]

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📖 Explanation

$\;\;\frac{\;\text{ reading on scale }\;-\;\text{ Lower fixed point }\;}{\;\text{ upper fixed point-lower fixed point }\;}=\;\text{ constant }\;$ $\;\;\frac{t_P-30}{180-30}=\;\frac{t_Q-0}{100-0}$ $\;\;\frac{t_P-30}{150}=\;\frac{t_Q}{100}$

Q6JEE Main 2023MCQ4MElectromagnetic Waves

Match List I with List II : A. Gauss'sLaw inElectrostat ics I. $∮ \vec{E} \cdot d \vec{l}=-\;\frac{d \varphi _B}{d t}$ B. Faraday'sLaw II. $∮ { \vec{\text{B}} \cdot {\text{d}} { \vec{\text{A}}=0$ C. Gauss'sLaw inMagnetism III. $∮ \vec{B} . d \vec{l}=\mu _0 i_c+\mu _0 \in _0 \;\frac{d \varphi _E}{d t}$ D. Ampere-MaxwellLaw IV. $∮ { \vec{\text{E}} \cdot {\text{d}} { \vec{\ s}=\;{ {\text{q}}}/{E_0}$ Choose the correct answer from the options given below : [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Gauss's Law of electrostatic $\varphi =∮ { \vec{\text{E}} \cdot {\text{d}} { \vec{\ s}=\;{ {\text{q}}}/{E_0}$ Faraday's law $∮ \vec{E} \cdot d \vec{l}=\;\frac{-d \varphi _B}{d t}$ Gauss's law of magnetism $∮ { \vec{\text{B}} \cdot {\text{d}} { \vec{\text{A}}=0$ Ampere's Maxwell law $∮ \vec{B} \cdot d \vec{l}=\mu _0 i_C+\mu _0 \in _0 \;\frac{d \varphi _E}{d t}$ Where $i_C$ : Conduction current $E_0 \;{ {\text{d}} \varphi _{ {\text{E}}}}/{ {\text{dt}}}:$ Displacement current

Q7JEE Main 2023MCQ4MSemiconductor Electronics

Statement I : When a ${\text{Si}}$ sample is doped with Boron, it becomes $P$ type and when doped by Arsenic it becomes ${\text{N}}$ -type semi conductor such that P-type has excess holes and ${\text{N}}$ -type has excess electrons.Statement II : When such P-type and N-type semi-conductors, are fused to make a junction, a current will automatically flow which can be detected with an externally connected ammeter.In the light of above statements, choose the most appropriate answer from the options given below. Options: [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Statement - I is correctWhen P-N junction is formed an electric field is generated form ${\text{N}}$ -side to ${\text{P}}$ -side due to which barrier potential arises & majority charge carrier can not flow through the junction due to barrier potential so current is zero unless we apply forward bias voltage.

Q8JEE Main 2023MCQ4MLaws of Motion

Consider a block kept on an inclined plane (inclined at $45^{\circ}$ ) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane $(\mu )$ is equal to : [25-Jan-2023 Shift 2]

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📖 Explanation

$\; {\text{F}}_1= {\text{mg}} sin \;45^{\circ}+ {\text{f}}= {\text{mg}} sin \;45^{\circ}+\mu {\text{N}}$ $\; {\text{F}}_1=\;{ {\text{mg}}}/{{\sqrt{2}}}+\mu {\text{mg}} \cos 45^{\circ}$ $\; {\text{F}}_1=\;{ {\text{mg}}}/{{\sqrt{2}}}(1+\mu )$ $\;F_2=m g sin \;45^{\circ}- {\text{f}}= {\text{mg}} sin \;45^{\circ}-\mu {\text{N}}$ $\;=\;{ {\text{mg}}}/{{\sqrt{2}}}(1-\mu )$ $\; {\text{F}}_1=2 {\text{F}}_2$ $\;\;{ {\text{mg}}}/{{\sqrt{2}}}(1+\mu )=2 \;{ {\text{mg}}}/{{\sqrt{2}}}(1-\mu )$ $\;1+\mu =2-2 \mu$ $\;\mu =1 / 3=0.33$

Q9JEE Main 2023MCQ4MElectrostatics

A point charge of $10 \mu {\text{C}}$ is placed at the origin. At what location on the ${\text{X}}$ -axis should a point charge of $40 \mu {\text{C}}$ be placed so that the net electric field is zero at ${\text{x}}=2 {\text{cm}}$ on the ${\text{X}}$ -axis ? [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$E_p=\;\frac{K \times 10}{2^2}-\;\frac{K \times 40}{ (x_0-2) ^2}=0$ $\;\frac{1}{2}=\;\frac{2}{x_0-2}$ $x_0-2=4$ ${\text{x}}_0=6 {\text{cm}}$

Q10JEE Main 2023MCQ4MAtoms and Nuclei

The energy levels of an atom is shown is figure. Which one of these transitions will result in the emission of a photon of wavelength $124.1 {\text{nm}}$ ? Given $( {\text{h}}=6.62 \times 10^{-34} {\text{Js}})$ [25-Jan-2023 Shift 2]

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📖 Explanation

$\;\lambda =\;{ {\text{hc}}}/{\Delta {\text{E}}}$ $\;\Delta {\text{E}}_{ {\text{A}}}=2.2 {\text{eV}}$ $\;\Delta {\text{E}}_{ {\text{B}}}=5.2 {\text{eV}}$ $\;\Delta {\text{E}}_{ {\text{C}}}=3 {\text{eV}}$ $\;\Delta {\text{E}}_{ {\text{D}}}=10 {\text{eV}}$ $\;\lambda _{ {\text{A}}}=\;\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{2.2 \times 1.6 \times 10^{-19}}$ $\;=\;\frac{12.41 \times 10^{-7}}{2.2} {\text{m}}$ $\;=\;\frac{1241}{2.2} {\text{nm}}=564 {\text{nm}}$ $\;\lambda _{ {\text{B}}}=\;\frac{1241}{5.2} {\text{nm}}=238.65 {\text{nm}}$ $\;\lambda _{ {\text{C}}}=\;\frac{1241}{3} {\text{nm}}=413.66 {\text{nm}}$ $\;\lambda _{ {\text{D}}}=\;\frac{1241}{10}=124.1 {\text{nm}}$

Q11JEE Main 2023MCQ4MOscillations

A particle executes simple harmonic motion between ${\text{x}}=- {\text{A}}$ and ${\text{x}}=+ {\text{A}}$ . If time taken by particle to go from ${\text{x}}=0$ to $\;{ {\text{A}}}/{2}$ is $2 {\ s}$ ; then time taken by particle in going from ${\text{x}}=\;{ {\text{A}}}/{2}$ to ${\text{A}}$ is : [25-Jan-2023 Shift 2]

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📖 Explanation

Let time from 0 to ${\text{A}} / 2$ is ${\text{t}}_1$ $\&$ from ${\text{A}} / 2$ to ${\text{A}}$ is ${\text{t}}_2$ then $\omega {\text{t}}_1=\pi / 6$ $\omega t_2=\pi / 3$ $\;\frac{t_1}{t_2}=\;\frac{1}{2}$ $t_2=2 t_1=2 \times 2=4 {\ sec}$

Q12JEE Main 2023MCQ4MThermodynamics

Match List I with List II : List I List II A. IsothermalProcess I. Work done by the gas decreases internal energy B. Adiabatic Process II. No change in internal energy C. Isochoric Process III. The heat absorbed goes partly to increase internal energy and partly to do work D. Isobaric Process IV. No work is done on or by the gasChoose the correct answer from the options given below : [25-Jan-2023 Shift 2]

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📖 Explanation

$\Delta {\text{U}}= {\text{nC}}_{ {\text{v}}} \Delta {\text{T}}$ For isothermal process ${\text{T}}$ is constantSo $\Delta {\text{U}}=0$ ${\text{A}} \longrightarrow {\text{II}}$ Adiabatic process $\;\Delta {\text{Q}}= 0$ $\;\Delta {\text{Q}}= \Delta {\text{U}}+\Delta {\text{W}}$ $\;\Delta {\text{U}}=-\Delta {\text{W}}$ Work done by gas is positiveSo $\Delta {\text{U}}$ is negative $\;\text{ B }\; \longrightarrow \;\text{ I }\;$ For Isochoric process $\Delta {\text{W}}=0$ ${\text{C}} \longrightarrow {\text{IV}}$ For Isobaric process $\;\Delta {\text{W}}= {\text{P}} \Delta {\text{V}} \ne 0$ $\;\Delta {\text{U}}= {\text{nC}}_{ {\text{V}}} \Delta {\text{T}} \ne 0$ Heat absorbed goes partly to increase internal energy and partly do work.

Q13JEE Main 2023MCQ4MThermodynamics

Match List I with List II List I List II A. Troposphere I. Approximate 65-75 km over Earth's surface B. E-Part ofStratosphere II. Approximate 300 km over Earth's surface C. ${\text{F}}_2- {\text{Part}}$ ofThermosphere III. Approximate $10 {\text{km}}$ over Earth's surface D. D-Part ofStratosphere IV. Approximate $100 {\text{km}}$ over Earth's surfaceChoose the correct answer from the options givenbelow: [25-Jan-2023 Shift 2]

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Q14JEE Main 2023MCQ4MGravitation

A body of mass is taken from earth surface to the height $h$ equal to twice the radius of earth $( {\text{R}}_{ {\text{e}}})$ , the increase in potential energy will be : ( $g=$ acceleration due to gravity on the surface of Earth) [25-Jan-2023 Shift 2]

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📖 Explanation

${\text{U}}=\;{- {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ {\text{r}}}$ ${\text{U}}_{ {\text{i}}}=\;{- {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ {\text{R}}_{ {\text{e}}}}$ ${\text{U}}_{ {\text{f}}}=\;{- {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ ( {\text{R}}_{ {\text{e}}}+ {\text{h}}) }=\;{- {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ {\text{R}}_{ {\text{e}}}+2 {\text{R}}_{ {\text{e}}}}$ $\;{- {\text{GM}}_e {\text{m}}}/{3 {\text{R}}_{ {\text{e}}}}$ Increase in internal energy $\Delta {\text{U}}= {\text{U}}_{ {\text{f}}}- {\text{U}}_{ {\text{i}}}$ $\;=\;\frac{2}{3} \;{ {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ {\text{R}}_{ {\text{e}}}}$ $\;\;\frac{2}{3} \;{ {\text{GM}}_{ {\text{e}}}}/{ {\text{R}}_{ {\text{e}}}^2} {\text{mR}}_{ {\text{e}}}$ $\;=\;\frac{2}{3} {\text{mgR}}_{ {\text{e}}}$

Q15JEE Main 2023MCQ4MElectromagnetic Induction

A wire of length $1 {\text{m}}$ moving with velocity $8 {\text{m}} / {\ s}$ at right angles to a magnetic field of $2 {\text{T}}$ . The magnitude of induced emf, between the ends of wire will be ________. [25-Jan-2023 Shift 2]

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📖 Explanation

Induced emf across the ends $= {\text{Bv}} ℓ$ $=2 \times 8 \times 1=16 {\text{V}}$

Q16JEE Main 2023MCQ4MMotion in a Straight Line

The distance travelled by a particle is related to time $t$ as $x=4 t^2$ . The velocity of the particle at $t=$ $5 {\ s}$ is. [25-Jan-2023 Shift 2]

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📖 Explanation

$\; {\text{x}}=4 {\text{t}}^2$ $\; {\text{v}}=\;{ {\text{dx}}}/{ {\text{dt}}}=8 {\text{t}}$ At ${\text{t}}=5 {\ sec}$ ${\text{v}}=8 \times 5=40 {\text{m}} / {\ s}$

Q17JEE Main 2023MCQ4MMotion in a Plane

Two objects are projected with same velocity 'u' however at different angles $\alpha$ and $\beta$ with the horizontal. If $\alpha+\beta=90^{\circ}$ , the ratio of horizontal range of the first object to the $2^{\;\text{nd }\;}$ object will be : [25-Jan-2023 Shift 2]

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📖 Explanation

$\;\text{ Range }\;=\;{ {\text{u}}^2 sin \;2 \theta}/{ {\text{g}}}$ Range for projection angle " $\alpha$ " ${\text{R}}_{ {\text{l}}}=\;{ {\text{u}}^2 sin \;2 \alpha}/{ {\text{g}}}$ Range for projection angle " $\beta$ " $\; {\text{R}}_2=\;{ {\text{u}}^2 sin \;2 \beta}/{ {\text{g}}}$ $\;\alpha+\beta=90^{\circ} \;\text{ (Given) }\;$ $\;\Rightarrow \beta=90^{\circ}-\alpha$ $\; {\text{R}}_2=\;{ {\text{u}}^2 sin \;2 (90^{\circ}-\alpha) }/{ {\text{g}}}$ $\; {\text{R}}_2=\;{ {\text{u}}^2 sin \; (180^{\circ}-2 \alpha) }/{ {\text{g}}}$ $\; {\text{R}}_2=\;{ {\text{u}}^2 sin \;2 \alpha}/{ {\text{g}}}$ $\;\Rightarrow \;{ {\text{R}}_1}/{ {\text{R}}_2}=\;{ ({ {\text{u}}^2 sin \;2 \alpha}/{ {\text{g}}}) }/{ (\;{ {\text{u}}^2 sin \;2 \alpha}/{ {\text{g}}}) }=\;\frac{1}{1}$ $\;$

Q18JEE Main 2023MCQ4MCurrent Electricity

The resistance of a wire is $5 Ω$ . It's new resistance in ohm if stretched to 5 times of it's original length will be : [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{R}}_{\;\text{initial }\;}=\;{\rho ℓ}/{ {\text{A}}}=5 Ω$ $\because$ Volume of wire is constant in stretching $\; {\text{V}}_{ {\text{i}}}= {\text{V}}_{ {\text{f}}}$ $\; {\text{A}}_{ {\text{i}}} ℓ_{ {\text{i}}}= {\text{A}}_{ {\text{f}}} ℓ_{ {\text{f}}}$ $\; {\text{A}} ℓ= {\text{A}}^{'}(5 ℓ)$ $\; {\text{A}}^{'}=\;{ {\text{A}}}/{5}$ $\; {\text{R}}_{ {\text{f}}}=\;{\rho ℓ_{ {\text{f}}}}/{ {\text{A}}_{ {\text{f}}}}=\;{\rho (5 ℓ)}/{ (\;{ {\text{A}}}/{5}) }$ $\;=25 (\;{\rho ℓ}/{ {\text{A}}})$ $\;=25 \times 5=125 Ω$

Q19JEE Main 2023MCQ4MDual Nature of Matter and Radiation

Given below are two statements :Statement I : Stopping potential in photoelectric effect does not depend on the power of the light source.Statement II : For a given metal, the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light.In the light of above statements, choose the most appropriate answer from the options given below.Options : [25-Jan-2023 Shift 2]

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📖 Explanation

Stopping potential ${\text{V}}_{ {\text{S}}}=\;{ {\text{KE}}_{\max }}/{ {\text{e}}}$ ${\text{V}}_{ {\text{S}}}=\;{\;{ {\text{hC}}}/{\lambda }-\varphi }/{ {\text{e}}}$ Stopping potential does not depend on intensity or power of light used, it only depends on frequency or wavelength of incident light.So both statements I and II are correct

Q20JEE Main 2023MCQ4MGravitation

Every planet revolves around the sun in an elliptical orbit :A. The force acting on a planet is inversely proportional to square of distance from sun.B. Force acting on planet is inversely proportional to product of the masses of the planet and the sunC. The centripetal force acting on the planet is directed away from the sun.D. The square of time period of revolution of planet around sun is directly proportional to cube of semi-major axis of elliptical orbit.Choose the correct answer from the options given below :Options : [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{F}}=\;{ {\text{Gm}}_1 {\text{m}}_2}/{ {\text{r}}^2}$ $\;\Rightarrow {\text{F}} \propto \;{1}/{ {\text{r}}^2}$ $\;\Rightarrow {\text{F}} \propto {\text{m}}_1 {\text{m}}_2$ $\;\Rightarrow \;\text{ This force provides centripetal force and acts }\;$ $\;\;\text{ towards sun }\;$ $\;\Rightarrow {\text{T}}^2 \propto {\text{a}}^3 \;\text{ (Kepler's third law) }\;$

Q21JEE Main 2023NAT4MElectrostatic Potential and Capacitance

A capacitor has capacitance $5 \mu {\text{F}}$ when it's parallel plates are separated by air medium of thickness $d$ . A slab of material of dielectric constant $1.5$ having area equal to that of plates but thickness $\;{ {\text{d}}}/{2}$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be _______ $\mu {\text{F}}.$ [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{C}}_{\;\text{new }\;}=\;{E_0 {\text{A}}}/{{ (\;{ {\text{d}}}/{2}) }/{1.5}+\;{ ({ {\text{d}}}/{2}) }/{1}}$ $\;=\;{E_0 A}/{ (\;{ {\text{d}}}/{3}+\;{ {\text{d}}}/{2}) }=\;{6 E_0 {\text{A}}}/{5 {\text{d}}}$ $\;=\;\frac{6}{5} \times 5 \mu {\text{F}}=6 \mu {\text{F}}$ $\;$

Q22JEE Main 2023NAT4MWaves

A train blowing a whistle of frequency $320 {\text{Hz}}$ approaches an observer standing on the platform at a speed of $66 {\text{m}} / {\ s}$ . The frequency observed by the observer will be (given speed of sound $=330 {\text{ms}}^{-1}$ ) ________ ${\text{Hz}}$ . [25-Jan-2023 Shift 2]

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📖 Explanation

$\;f_{a p p}=f (\;\frac{v}{v-v_S})$ $\;=320 (\;\frac{330}{330-66})$ $\;=400 {\text{Hz}}$

Q23JEE Main 2023NAT4MRay Optics and Optical Instruments

An object is placed on the principal axis of convex lens of focal length $10 {\text{cm}}$ as shown. A plane mirror is placed on the other side of lens at a distance of $20 {\text{cm}}$ . The image produced by the plane mirror is $5 {\text{cm}}$ inside the mirror. The distance of the object from the lens is _____ ${\text{cm}}$ . [25-Jan-2023 Shift 2]

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📖 Explanation

$\; {\text{f}}=10 {\text{cm}}$ $\;\;{1}/{ {\text{v}}}-\;{1}/{ {\text{u}}}=\;{1}/{ {\text{f}}}$ $\;\;\frac{1}{15}-\;{1}/{- {\text{u}}}=\;\frac{1}{10}$ $\;\Rightarrow \;{1}/{ {\text{u}}}=\;\frac{1}{10}-\;\frac{1}{15}$ On solving we get value of ${\text{u}}$ as $30 {\text{cm}}$ .

Q24JEE Main 2023NAT4MMagnetic Effects of Current and Magnetism

Two long parallel wires carrying currents $8 {\text{A}}$ and $15 {\text{A}}$ in opposite directions are placed at a distance of $7 {\text{cm}}$ from each other. A point $P$ is at equidistant from both the wires such that the lines joining the point $P$ to the wires are perpendicular to each other. The magnitude of magnetic field at $P$ is ______ $\times 10^{-6} {\text{T}}$ . (Given : ${\sqrt{2}}=1.4)$ [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Magnetic fields due to both wires will be perpendicular to each other. $\; {\text{B}}_1=\;{\mu _0 {\text{i}}_1}/{2 \pi {\text{d}}} \;\; {\text{B}}_2=\;{\mu _0 {\text{i}}_2}/{2 \pi {\text{d}}}$ $\; {\text{B}}_{\;\text{net }\;}=\sqrt{ {\text{B}}_1^{2+} {\text{B}}_2^2} \Rightarrow \;{\mu _0}/{2 \pi {\text{d}}} \sqrt{ {\text{i}}_1^{2+} {\text{i}}_2^2}$ $\;\Rightarrow \;{4 \pi \times 10^{-7}}/{2 \pi \times (7 / {\sqrt{2}}) \times 10^{-2}} \times {\sqrt{8^{2+}15^2}} ( {\text{d}}=\;{7}/{{\sqrt{2}}} {\text{cm}})$ $\;\Rightarrow 68 \times 10^{-6} {\text{T}}$

Q25JEE Main 2023NAT4MProperties of Matter

A spherical drop of liquid splits into 1000 identical spherical drops. If ${\text{u}}_{ {\text{i}}}$ is the surface energy of the original drop and ${\text{u}}_{ {\text{f}}}$ is the total surface energy of the resulting drops, the (ignoring evaporation). $\;{ {\text{u}}_{ {\text{f}}}}/{ {\text{u}}_{ {\text{i}}}}= (\;{10}/{ {\text{x}}})$ . Then value of ${\text{x}}$ is _______: [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Surface Tension $=T$ $R$ : Radius of bigger drop $r$ : Radius of smaller dropVolume will remain same $\;\;\frac{4}{3} \pi {\text{R}}^3=1000 \times \;\frac{4}{3} \pi {\text{r}}^3$ $\; {\text{R}}=10 {\text{r}}$ $\; {\text{u}}_{ {\text{i}}}= {\text{T}} \cdot 4 \pi {\text{R}}^2$ $\; {\text{u}}_{ {\text{f}}}= {\text{T}} \cdot 4 \pi {\text{r}}^2 \times 1000$ $\;\;{ {\text{u}}_{ {\text{f}}}}/{ {\text{u}}_{ {\text{i}}}}=\;{1000 {\text{r}}^2}/{ {\text{R}}^2}$ $\;\;{ {\text{u}}_{ {\text{f}}}}/{ {\text{u}}_{ {\text{i}}}}=\;\frac{10}{1}$

Q26JEE Main 2023NAT4MCentre of Mass and Collision

A body of mass $1 {\text{kg}}$ collides head on elastically with a stationary body of mass $3 {\text{kg}}$ . After collision, the smaller body reverses its direction of motion and moves with a speed of $2 {\text{m}} / {\ s}$ . The initial speed of the smaller body before collision is _______ ${\text{ms}}^{-1}$ . [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$1 \times {\text{u}}_1=-2+3 {\text{v}}\;\Rightarrow {\text{u}}_1=-2+3 {\text{v}}$ . . . (1) $1=\;{ {\text{v}}+2}/{ {\text{u}}_1}\;\Rightarrow {\text{v}}+2= {\text{u}}_1$ . . . (2)Solving (1) and (2) ${\text{u}}_1=4 {\text{m}} / {\ s}$

Q27JEE Main 2023NAT4MAtoms and Nuclei

A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ . The ratio of their nuclear sizes will be $(\;{ {\text{x}}}/{3}) ^{\;\frac{1}{3}}$ . The value of ' ${\text{x}}$ ' is: [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\;{ {\text{v}}_1}/{ {\text{v}}_2}=\;\frac{3}{2}$ $\; {\text{m}}_1 {\text{v}}_1= {\text{m}}_2 {\text{v}}_2 \Rightarrow \;{ {\text{m}}_1}/{ {\text{m}}_2}=\;\frac{2}{3}$ Since, Nuclear mass density is constant $\;\;{ {\text{m}}_1}/{\frac{4}{3} \pi r_1^3}=\;{ {\text{m}}_2}/{\frac{4}{3} \pi {\text{r}}_2^3}$ $\; (\;\frac{r_1}{r_2}) ^3=\;\frac{m_1}{m_2}$ $\;\;{ {\text{r}}_1}/{ {\text{r}}_2}= (\;\frac{2}{3}) ^{\;\frac{1}{3}}$ $\;\;\text{ So, }\; x=2$ $\;$

Q28JEE Main 2023NAT4MCurrent Electricity

Two cells are connected between points $A$ and $B$ as shown. Cell 1 has emf of $12 {\text{V}}$ and internal resistance of $3 Ω$ . Cell 2 has emf of $6 {\text{V}}$ and internal resistance of $6 Ω$ . An external resistor ${\text{R}}$ of $4 Ω$ is connected across ${\text{A}}$ and ${\text{B}}$ . The current flowing through ${\text{R}}$ will be _______ A. [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{E}}_{ {\text{eq}}}=\;\frac{\;\frac{12}{3}-\frac{6}{6}}{\frac{1}{3}+\;\frac{1}{6}}$ $\; {\text{E}}_{ {\text{eq}}}=6 {\text{V}}$ $\; {\text{r}}_{ {\text{eq}}}=2 Ω$ $\; {\text{R}}=4 Ω$ So, $i=\;\frac{6}{2+4}=1 {\text{A}}$

Q29JEE Main 2023NAT4MAlternating Current

A series LCR circuit is connected to an ${\text{AC}}$ source of $220 {\text{V}}, 50 {\text{Hz}}$ . The circuit contains a resistance ${\text{R}}=80 Ω$ , an inductor of inductive reactance ${\text{X}}_{ {\text{L}}}=70 Ω$ , and a capacitor of capacitive reactance ${\text{X}}_{ {\text{C}}}=130 Ω$ . The power factor of circuit is $\;{ {\text{X}}}/{10}$ . The value of $x$ is : [25-Jan-2023 Shift 2]

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📖 Explanation

$\;\cos \varphi =\;\frac{R}{Z}=\;{ {\text{R}}}/{\sqrt{ {\text{R}}^{2+} ( {\text{X}}_{ {\text{C}}}- {\text{X}}_{ {\text{L}}}) ^2}}$ $\;\cos \varphi =\;{80}/{{\sqrt{(80)^{2+}(60)^2}}}$ $\;\cos \varphi =\;\frac{80}{100} \Rightarrow \;\frac{8}{10}$ So, $x=8$

Q30JEE Main 2023NAT4MRotational Motion

If a solid sphere of mass $5 {\text{kg}}$ and a disc of mass 4 ${\text{kg}}$ have the same radius. Then the ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be $\;\frac{x}{7}$ . The value of $x$ is ______. [25-Jan-2023 Shift 2]

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📖 Explanation

$\;I_1=\;\frac{2}{5} m_1 R^{2+}m_1 R^2$ $\;I_1=m_1 R^2 (\;\frac{7}{5})$ $\;I_1=7 R^2$ $\; {\text{I}}_2=\;{ {\text{m}}_2 {\text{R}}^2}/{4}+ {\text{m}}_2 {\text{R}}^2$ $\; {\text{I}}_2=\;\frac{5}{4} {\text{m}}_2 {\text{R}}^2$ $\; {\text{I}}_2=5 {\text{R}}^2$ $\;\;{ {\text{I}}_2}/{ {\text{I}}_1}=\;\frac{5}{7}$ $\; {\text{x}}=5$

Chemistry30 questions

Q31JEE Main 2023MCQ4Md and f Block Elements

Match List I with List II List I List II A. Cobalt catalyst I. $( {\text{H}}_2+ {\text{Cl}}_2)$ production B. Syngas II. Water gas production C. Nickel catalyst III. Coalgasification D. Brine solution IV. MethanolproductionChoose the correct answer from the options given below :- [25-Jan-2023 Shift 2]

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📖 Explanation

Cobalt catalyst $\to$ Methanol productionSyn gas $\to$ Coal gasification $( {\text{C}}_{\;\text{(Redhot coke) }\;}+ {\text{H}}_2 {\text{O}}( {\text{g}}) \to {\text{CO}}+ {\text{H}}_2)$ Nickel catalyst $\to$ Water gas productionBrine solution $\to$ Production $($ aq. ${\text{NaCl}}) \;\; {(\text{table} {\text{H}}_2 \to \;\text{ Cathode }\; ; {\text{Cl}}_2 \to \;\text{ anode }\;)}$

Q32JEE Main 2023MCQ4Md and f Block Elements

Given below are two statements :-Statement I :- In froth floatation method a rotating paddle agitates the mixture to drive air out of it.Statement II :- Iron pyrites are generally avoided for extraction of iron due to environmental reasons.In the light of the above statements, choose the correct answer from the options given below :- [25-Jan-2023 Shift 2]

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Q33JEE Main 2023MCQ4MClassification of Elements

Which of the following represents the correct order of metallic character of the given elements? [25-Jan-2023 Shift 2]

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Q34JEE Main 2023MCQ4MSalt Analysis

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason RAssertion A :- The alkali metals and their salts impart characteristic colour to reducing flame.Reason R :- Alkali metals can be detected using flame tests.In the light of the above statements, choose the most appropriate answer form the options given below [25-Jan-2023 Shift 2]

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Q35JEE Main 2023MCQ4MSolutions

What is the mass ratio of ethylene glycol $( {\text{C}}_2 {\text{H}}_6 {\text{O}}_2.$ , molar mass $=62 {\text{g}} / {\text{mol}}$ ) required for making $500 {\text{g}}$ of $0.25$ molal aqueous solution and $250 {\text{mL}}$ of $0.25$ molar aqueous solution ? [25-Jan-2023 Shift 2]

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📖 Explanation

Assume : Mass of solvent $\approx$ Mass of solutionCase I :- $0.25=\;{ {\text{W}}_1}/{62} \times \;\frac{1000}{500}$ Case II :- $\;0.25=\;{ {\text{W}}_2}/{62} \times \;\frac{1000}{250}$ $\;\;{ {\text{W}}_1}/{ {\text{W}}_2}=\;\frac{2}{1}$

Q36JEE Main 2023MCQ4MChemical Bonding and Molecular Structure

Statement I :- Dipole moment is a vector quantity and by convention it is depicted by a small arrow with tail on the negative centre and head pointing towards the positive centre.Statement II :- The crossed arrow of the dipole moment symbolizes the direction of the shift of charges in the molecules.In the light of the above statements, choose the most appropriate answer from the options given below :- [25-Jan-2023 Shift 2]

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Q37JEE Main 2023MCQ4MAlcohols, Phenols and Ethers

Given below are two statements, one is labelled as Assertion $A$ and the other is labelled as Reason $R$ Assertion A :- Butylated hydroxyl anisole when added to butter increases its shelf life.Reason R :- Butylated hydroxyl anisole is more reactive towards oxygen than food.In the light of the above statements, choose the most appropriate answer from the options given below :- [25-Jan-2023 Shift 2]

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Q38JEE Main 2023MCQ4MHaloalkanes and Haloarenes

A. Ammonium salts produce haze in atmosphere.B. Ozone gets produced when atmospheric oxygen reacts with chlorine radicals.C. Polychlorinated biphenyls act as cleansing solvents.D. 'Blue baby' syndrome occurs due to the presence of excess of sulphate ions in water.Choose the correct answer from the options given below :- [25-Jan-2023 Shift 2]

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📖 Explanation

B. ${\text{Cl}}^{\bullet }+ {\text{O}}_3 \longrightarrow {\text{O}}_2+ {\text{ClO}}^{\bullet }$ D. 'Blue baby' syndrome occurs due to the presence of excess of nitrate ions in water.

Q39JEE Main 2023MCQ4MAmines

Match List I with List II List I (Amines) List II $( {\text{pK}}_{ {\text{b}}})$ A. Aniline I. $3.25$ B. Ethanamine II. 3.00 C. N-Ethylethanamine III. 9.38 D. N, N-Diethylethanamine IV. 3.29Choose the correct answer from the options given below :- [25-Jan-2023 Shift 2]

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📖 Explanation

Basic Strength $\alpha \;{1}/{ {\text{pK}}_{ {\text{b}}}}$ Order for ${\text{pK}}_{ {\text{b}}}: {\text{A}} > {\text{B}} > {\text{D}} > {\text{C}}$

Q40JEE Main 2023MCQ4MElectrochemistry

Which one among the following metals is the weakest reducing agent ? [25-Jan-2023 Shift 2]

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Q41JEE Main 2023MCQ4MCoordination Compounds

Match List I with List II. List I Isomeric pairs List II Type of isomers A. Propanamine and N-Methylethanamine I. Metamers B. Hexan-2-one andHexan-3-one II. Positional isomers C. Ethanamide andHydroxyethanimine III. Functional isomers D. o-nitrophenol and pnitrophenol IV. TautomersChoose the correct answer from the options given below :- [25-Jan-2023 Shift 2]

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Q42JEE Main 2023MCQ4MCoordination Compounds

Match List I with List II List I (Amines) List II $( {\text{pK}}_{ {\text{b}}})$ A. Glyptal I. Flexible pipes B. Neoprene II. Synthetic wool C. Acrilan III. Paints and Lacquers D. LDP IV. GasketsChoose the correct answer from the options given below :- [25-Jan-2023 Shift 2]

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Q43JEE Main 2023MCQ4Mp Block Elements

Given below are two statements, one is labelled as Assertion $A$ and the other is labelled as Reason $R$ Assertion A :- Carbon forms two important oxides $- {\text{CO}}$ and ${\text{CO}}_2 . {\text{CO}}$ is neutral whereas ${\text{CO}}_2$ is acidic in nature.Reason $R$ :- ${\text{CO}}_2$ can combine with water in a limited way to form carbonic acid, while ${\text{CO}}$ is sparingly soluble in water.In the light of the above statements, choose the most appropriate answer from the options given below :- [25-Jan-2023 Shift 2]

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Q44JEE Main 2023MCQ4MRedox Reactions

Potassium dichromate acts as a strong oxidizing agent in acidic solution. During this process, the oxidation state changes from [25-Jan-2023 Shift 2]

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Q45JEE Main 2023MCQ4MIonic Equilibrium

When the hydrogen ion concentration $[ {\text{H}}^{+}]$ changes by a factor of 1000 , the value of ${\text{pH}}$ of the solution _______. [25-Jan-2023 Shift 2]

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📖 Explanation

$\;\Delta \$ [ {\text{H}}^{+}]$ =1000; ;; \Delta {\text{pH}}=-\log \Delta $[ {\text{H}}^{+}]$ =-\log 10^3;=-3$

Q46JEE Main 2023MCQ4MCoordination Compounds

Match List I with List II List I Coordination entity List II Wavelength of light absorbed in ${\text{nm}}$ A. $[ {\text{CoCl}} ( {\text{NH}}_3) _5] ^{2+}$ I. 310 B. $[ {\text{Co}} ( {\text{NH}}_3) _6] ^{3+}$ II. 475 C. $[ {\text{Co}}( {\text{CN}})_6] ^{3-}$ III. 535 D. $[ {\text{Cu}} ( {\text{H}}_2 {\text{O}}) _4] ^{2+}$ IV. 600 Choose the correct answer from the options given below :- [25-Jan-2023 Shift 2]

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📖 Explanation

List I Coordination entity List II Wavelength of light absorbed in ${\text{nm}}$ A. $[ {\text{CoCl}} ( {\text{NH}}_3) _5] ^{2+}$ I. 535 B. $[ {\text{Co}} ( {\text{NH}}_3) _6] ^{3+}$ II. 475 C. $[ {\text{Co}}( {\text{CN}})_6] ^{3-}$ III. 310 D. $[ {\text{Cu}} ( {\text{H}}_2 {\text{O}}) _4] ^{2+}$ IV. 600 $\;E=\;\frac{h c}{\lambda } \Rightarrow E \propto \;\frac{1}{\lambda }$ $\;\Rightarrow \Delta (\;\text{ CFSE }\;) \propto \;{1}/{\lambda _{\;\text{absorb }\;}} \propto \;\text{ strength of ligand. }\;$

Q47JEE Main 2023MCQ4MAlcohols, Phenols and Ethers

Find out the major product from the following reaction. [25-Jan-2023 Shift 2]

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Q48JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

' $A$ ' in the given reaction is [25-Jan-2023 Shift 2]

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Q49JEE Main 2023MCQ4MHaloalkanes and Haloarenes

The isomeric deuterated bromide with molecular formula ${\text{C}}_4 {\text{H}}_8 {\text{DBr}}$ having two chiral carbon atoms is [25-Jan-2023 Shift 2]

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Q50JEE Main 2023MCQ4MSalt Analysis

A chloride salt solution acidified with dil. ${\text{HNO}}_3$ gives a curdy white precipitate, $[ {\text{A}}]$ , on addition of ${\text{AgNO}}_3$ . [A] on treatment with ${\text{NH}}_4 {\text{OH}}$ gives a clear solution, B. [25-Jan-2023 Shift 2]

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Q51JEE Main 2023NAT4MStructure of Atom

The number of given orbitals which have electron density along the axis is ________ $p_x, p_y, p_z, d_{x y}, d_{y z}, d_{x z}, d_z, d_{x^{2-}y^2}$ [25-Jan-2023 Shift 2]

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Q52JEE Main 2023NAT4MPractical Organic Chemistry

Number of compounds giving (i) red colouration with ceric ammonium nitrate and also (ii) positive iodoform test from the following is [25-Jan-2023 Shift 2]

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Q53JEE Main 2023NAT4MSolutions

The number of pairs of the solution having the same value of the osmotic pressure from the following is ________.(Assume 100% ionization)A. $0.500 {\text{M}} {\text{C}}_2 {\text{H}}_5 {\text{OH}}( {\text{aq}})$ and $0.25 {\text{M}} {\text{KBr}}( {\text{aq}})$ B. $0.100 {\text{M}} {\text{K}}_4 [ {\text{Fe}}( {\text{CN}})_6]$ (aq) and $0.100 {\text{M}}$ ${\text{FeSO}}_4 ( {\text{NH}}_4) _2 {\text{SO}}_4( {\text{aq}})$ C. $0.05 {\text{M}} {\text{K}}_4 [ {\text{Fe}}( {\text{CN}})_6] ( {\text{aq}})$ and $0.25 {\text{M}} {\text{NaCl}}$ (aq)D. $0.15 {\text{M}} {\text{NaCl}}( {\text{aq}})$ and $0.1 {\text{M}} {\text{BaCl}}_2$ (aq)E. $0.02 {\text{M}} {\text{KCl}} . {\text{MgCl}}_2 .6 {\text{H}}_2 {\text{O}}( {\text{aq}})$ and $0.05 {\text{M}}$ ${\text{KCl}}( {\text{aq}})$ [25-Jan-2023 Shift 2]

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📖 Explanation

$\;\pi= {\text{iCRT}}$ $\;\pi \propto {\text{iC}}$ A, B, D and E have same value of osmatic pressure.

Q54JEE Main 2023NAT4MChemical Thermodynamics

$28.0 {\text{L}}$ of ${\text{CO}}_2$ is produced on complete combustion of $16.8 {\text{L}}$ gaseous mixture of ethene and methane at $25^{\circ} {\text{C}}$ and 1 atm. Heat evolved during the combustion process is _________ ${\text{kJ}}$ .Given : $\Delta {\text{H}}_{ {\text{C}}} ( {\text{CH}}_4) =-900 {\text{kJ}} {\text{mol}}^{-1}$ $\Delta {\text{H}}_{ {\text{C}}} ( {\text{C}}_2 {\text{H}}_4) =-1400 {\text{kJ}} {\text{mol}}{ }^{-1}$ [25-Jan-2023 Shift 2]

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📖 Explanation

Let, Volume of ${\text{C}}_2 {\text{H}}_4$ is ${\text{x}}$ litre Total volume of ${\text{CO}}_2=2 {\text{x}}+16.8- {\text{x}}$ $\;\Rightarrow \;\; 28=16.8+ {\text{x}}$ $\; \;\; {\text{x}}=11.2 {\text{L}}$ $\; {\text{n}}_{ {\text{CH}}_4}=\;{ {\text{PV}}}/{ {\text{RT}}}=\;\frac{1 \times 5.6}{0.082 \times 298}=0.229 \;\text{ mole }\;$ $\; {\text{n}}_{ {\text{C}}_2 {\text{H}}_2}=\;\frac{11.2}{0.082 \times 298}=0.458 \;\text{ mole }\;$ $\; \therefore \;\; \;\text{ Heat evolved }\;=0.229 \times 900+0.458 \times 1400$ $\;=206.1+641.2$ $\;=847.3 {\text{kJ}}$

Q55JEE Main 2023NAT4MCoordination Compounds

Total number of moles of ${\text{AgCl}}$ precipitated on addition of excess of ${\text{AgNO}}_3$ to one mole each of the following complexes $[ {\text{Co}} ( {\text{NH}}_3) _4 {\text{Cl}}_2] {\text{Cl}}$ , $[ {\text{Ni}} ( {\text{H}}_2 {\text{O}}) _6] {\text{Cl}}_2, [ {\text{Pt}} ( {\text{NH}}_3) _2 {\text{Cl}}_2]$ and $[ {\text{Pd}} ( {\text{NH}}_3) _4] {\text{Cl}}_2$ is [25-Jan-2023 Shift 2]

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📖 Explanation

$[ {\text{Co}} ( {\text{NH}}_3) _4 {\text{Cl}}_2] {\text{Cl}} \Rightarrow$ Gives 1 mole ${\text{AgCl}}$ $[ {\text{Ni}} ( {\text{H}}_2 {\text{O}}) _6] {\text{Cl}}_2 \Rightarrow$ Gives 2 moles ${\text{AgCl}}$ $[ {\text{Pt}} ( {\text{NH}}_3) _2 {\text{Cl}}_2] \Rightarrow$ Gives No ${\text{AgCl}}$ $[ {\text{Pd}} ( {\text{NH}}_3) _4] {\text{Cl}}_2 \Rightarrow$ Gives 2 moles ${\text{AgCl}}$ Total number of moles of ${\text{AgCl}}=5$ mole.

Q56JEE Main 2023NAT4MOrganic Chemistry - Some Basic Principles

Number of hydrogen atoms per molecule of a hydrocarbon A having $85.8 \%$ carbon is _______(Given : Molar mass of ${\text{A}}=84 {\text{g}} {\text{mol}}^{-1}$ ) [25-Jan-2023 Shift 2]

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📖 Explanation

Element Percentage Mole Mole ratio C 85.8 $\;\frac{85.8}{12}=7.15$ 1 H 14.2 $\;\frac{14.2}{1}=14.2$ 2Empirical formula $( {\text{CH}}_2)$ $14 \times n\;=84$ $n\;=6$ $\therefore$ Molecular formula ${\text{C}}_6 {\text{H}}_{12}$

Q57JEE Main 2023NAT4MElectrochemistry

${\text{Pt}}( {\ s}) {\text{H}}_2( {\text{g}})(1 {\bar}) | {\text{H}}^{+}( {\text{aq}})(1 {\text{M}})| | {\text{M}}^{3+}( {\text{aq}}), {\text{M}}^{+}( {\text{aq}})| {\text{Pt}}( {\ s})$ The ${\text{E}}_{\;\text{cell }\;}$ for the given cell is $0.1115 {\text{V}}$ at $298 {\text{K}}$ when $\;{[ {\text{M}}^{+}( {\text{aq}})]\$ }/{[ {\text{M}}^{3+}( {\text{aq}})]$ }=10^{ {\text{a}}} $The value of a is _________Given :$ {\text{E}}^\theta {\text{M}}^{3+} / {\text{M}}^{+}=0.2 {\text{V}};{2.303 {\text{RT}}}/{ {\text{F}}}=0.059 {\text{V}}$ [25-Jan-2023 Shift 2]

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📖 Explanation

Overall reaction :- $\; {\text{H}}_{2( {\text{g}})}+ {\text{M}}_{\;\text{(aq) }\;}^{3+} \longrightarrow {\text{M}}_{\;\text{(aq) }\;}^{+}+2 {\text{H}}_{\;\text{(aq) }\;}^{+}$ $\; {\text{E}}_{\;\text{Cell }\;}= {\text{E}}_{\;\text{Cathode }\;}^{0-} {\text{E}}_{\;\text{anode }\;}^{0-}\;\frac{0.059}{2} \log \;{[ {\text{M}}^{+}]\$ × 1^2}/{[ {\text{M}}^{+3}]$ 1};0.1115=0.2-;\frac{0.059}{2} \log ;{[ {\text{M}}^{+}]$ }/{[ {\text{M}}^{+3}]$ }; ;; 3=\log ;{[ {\text{M}}^{+}]$ }/{[ {\text{M}}^{+3}]$ }; \therefore {\text{a}}=3$

Q58JEE Main 2023NAT4MStates of Matter

Based on the given figure, the number of correct statement $/ {\ s}$ is/are _______A. Surface tension is the outcome of equal attractive and repulsion forces acting on the liquid molecule in bulk.B. Surface tension is due to uneven forces acting on the molecules present on the surface.C. The molecule in the bulk can never come to the liquid surface.D. The molecules on the surface are responsible for vapour pressure if the system is a closed system. [25-Jan-2023 Shift 2]

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Q59JEE Main 2023NAT4MChemical Kinetics

A first order reaction has the rate constant, ${\text{k}}=4.6$ $\times 10^{-3} {\ s}^{-1}$ . The number of correct statement $/ {\ s}$ from the following is/are ________.Given : $\log 3=0.48$ A. Reaction completes in $1000 {\ s}$ .B. The reaction has a half-life of $500 {\ s}$ .C. The time required for $10 \%$ completion is 25 times the time required for $90 \%$ completion.D. The degree of dissociation is equal to $(1-e^{-k t})$ .E. The rate and the rate constant have the same unit. [25-Jan-2023 Shift 2]

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📖 Explanation

$\; {\text{t}}_{10 \%}=\;{1}/{ {\text{K}}} \ln (\;{ {\text{a}}}/{ {\text{a}}- {\text{x}}}) =\;{1}/{ {\text{K}}} \ln (\;\frac{100}{90})$ $\; {\text{t}}_{10 \%}=\;{2.303}/{ {\text{K}}}(\log 10-\log 9)$ $\; {\text{t}}_{10 \%}=\;{2.093}/{ {\text{K}}} \times (0.04)$ $\;\;\text{ Similarly }\;$ $\; {\text{t}}_{90 \%}=\;{1}/{ {\text{K}}} \ln (\;\frac{100}{10})$ $\; {\text{t}}_{90 \%}=\;{2.303}/{ {\text{K}}}$ $\;\;\frac{t_{90 \%}}{t_{10 \%}}=\;\frac{1}{0.04}=25$ $\;e^{k t}=\;\frac{a}{a-x}$ $\;\;\frac{a-x}{a}=e^{-k t}$ $\;1-\;\frac{x}{a}=e^{-k t}$ $\;x=a (1-e^{-k t})$ $\;\alpha=\;\frac{x}{a}= (1-e^{-k t})$

Q60JEE Main 2023NAT4MChemical Thermodynamics

The number of incorrect statement/s from the following is/are ________A. Water vapours are adsorbed by anhydrous calcium chloride.B. There is a decrease in surface energy during adsorption.C. As the adsorption proceeds, $\Delta {\text{H}}$ becomes more and more negative.D. Adsorption is accompanied by decrease in entropy of the system. [25-Jan-2023 Shift 2]

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📖 Explanation

'A' water vapours are absorbed by calcium chloride. C. As the adsorption proceeds, $\Delta {\text{H}}$ becomes less and less negative.

Mathematics30 questions

Q61JEE Main 2023MCQ4MApplication of Derivatives

Let the function $f(x)=2 x^{3+}(2 p-7) x^{2+}3(2 p-9) x-6$ have a maxima for some value of ${\text{x}}< 0$ and a minima for some value of $x>0$ . Then, the set of all values of ${\text{p}}$ is [25-Jan-2023 Shift 2]

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📖 Explanation

$\;f(x)=2 x^{3+}(2 p-7) x^{2+}3(2 p-9) x-6$ $\;f^{'}(x)=6 x^{2+}2(2 p-7) x+3(2 p-9)$ $\;f^{'}(0) < 0$ $\; \therefore 3(2 p-9) < 0$ $\; \;\; p < \;\frac{9}{2}$ $\; \;\; p \in (-\infty , \;\frac{9}{2})$

Q62JEE Main 2023MCQ4MComplex Numbers

Let $z$ be a complex number such that $|\;\frac{z-2 i}{z+i}|=2, z \ne -i$ . Then $z$ lies on the circle of radius 2 and centre [25-Jan-2023 Shift 2]

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📖 Explanation

$\;(z-2 i)(\overline{z}+2 i)=4(z+i)(\overline{z}-i)$ $\;z \overline{z}+4+2 i(z-\overline{z})=4(z \overline{z}+1+i(\overline{z}-z))$ $\;3 z \overline{z}-6 i(z-\overline{z})=0$ $\;x^{2+}y^{2-}2 i(2 i y)=0$ $\;x^{2+}y^{2+}4 y=0$

Q63JEE Main 2023MCQ4MLimits, Continuity and Differentiability

If the function $f(x)={\begin{cases}(1+{|\cos x|}) \\ \\ \frac{\lambda }{{|\cos x|}} & {, \\ \\ } 0 < x < \\ \\ \frac{\pi}{2} \\ \mu & {, \\ \\ } x=\\ \\ \frac{\pi}{2} \\ e^{\\ \\ \frac{\cot 6 x}{\cot 4 x}} ,{ & \\ \\ } \\ \\ \frac{\pi}{2} < x < \pi }\end{cases}$ is continuous at $x=\;\frac{\pi}{2}$ , then $9 \lambda +6 \log _e \mu +\mu ^{6-}e^{6 \lambda }$ is equal to [25-Jan-2023 Shift 2]

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📖 Explanation

$\;\Rightarrow \lim _{x \to \;\frac{\pi^{+}}{2}} e^{\;\frac{\cot 6 x}{\cot 4 x}}=\lim _{x \to \;\frac{\pi^{+}}{2}} e^{\;\frac{ sin \;4 x \cdot \cos 6 x}{ sin \;6 x \cdot \cos 4 x}}=e^{2 / 3}$ $\;\Rightarrow \lim _{x \to \;\frac{\pi^{-}}{2}} (1+|\cos x|)^{\;\frac{\lambda }{|\cos x|} }=e^\lambda .$ $\;\Rightarrow f(\pi / 2)=\mu$ For continuous function $\Rightarrow {\text{e}}^{2 / 3}= {\text{e}}^\lambda =\mu$ $\lambda =\;\frac{2}{3}, \mu = {\text{e}}^{2 / 3}$ Now, $9 \lambda +6 \log _e \mu +\mu ^{6-}e^{6 \lambda }=10$

Q64JEE Main 2023MCQ4MFunctions

Let ${\text{f}}( {\text{x}})=2 {\text{x}}^{ {\text{n}}}+\lambda , \lambda \in ℝ, {\text{n}} \in ℕ$ , and ${\text{f}}(4)=133$ , $f(5)=255$ . Then the sum of all the positive integer divisors of $( {\text{f}}(3)- {\text{f}}(2))$ is [25-Jan-2023 Shift 2]

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📖 Explanation

$\;f( {\text{x}})=2 {\text{x}}^{ {\text{n}}}+\lambda$ $\;f(4)=133$ $\;f(5)=255$ $\;133=2 \times 4^{ {\text{n}}}+\lambda$ . . . (1) $\;255=2 \times 5^{ {\text{n}}}+\lambda$ . . . (2) $\;\;\text{ (2) }\;-(1)$ $\;122=2 (5^{ {\text{n}}}-4^{ {\text{n}}})$ $\;\Rightarrow 5^{ {\text{n}}}-4^{ {\text{n}}}=61$ $\; \therefore {\text{n}}=3 & \lambda =5$ Now, $f(3)-f(2)=2 (3^{3-}2^3) =38$ Number of Divisors is $1,2,19,38 ; \&$ their sum is 60

Q65JEE Main 2023MCQ4MVector Algebra

If the four points, whose position vectors are $3 { {\text{i}}^{\^}-4 { {\text{j}}^{\^}+2 { {\text{k}}^{\^}, { {\text{i}}^{\^}+2 { {\text{j}}^{\^}-{ {\text{k}}^{\^},-2 { {\text{i}}^{\^}-{ {\text{j}}^{\^}+3 { {\text{k}}^{\^} \;\;$ and $5 { {\text{i}}^{\^}-2 \alpha { {\text{j}}^{\^}+4 { {\text{k}}^{\^}$ are coplanar, then $\alpha$ is equal to [25-Jan-2023 Shift 2]

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📖 Explanation

Let $A:(3,-4,2)$ ${\text{C}}:(-2,-1,3)$ $\;\text{ B : }\;(1,2,-1) \;\; \;\text{ D : }\;(5,-2 \alpha, 4)$ A, B, C, D are coplanar points, then $\;\Rightarrow {\begin{vmatrix}1-3 & 2+4 & -1-2 \\ -2-3 & -1+4 & 3-2 \\ 5-3 & -2 \alpha+4 & 4-2\end{vmatrix}}=0$ $\;\Rightarrow \alpha=\;\frac{73}{17}$

Q66JEE Main 2023MCQ4MMatrices and Determinants

Let $A={\begin{bmatrix}\\ \\ {1}/{{\sqrt{10}}} & \\ \\ {3}/{{\sqrt{10}}} \\ \\ \\ {-3}/{{\sqrt{10}}} & \\ \\ {1}/{{\sqrt{10}}}\end{bmatrix}}$ and $B={\begin{bmatrix}1 & -i \\ 0 & 1\end{bmatrix}}$ , where ${\text{i}}={\sqrt{-1}}$ . If ${\text{M}}= {\text{A}}^{ {\text{T}}} {\text{BA}}$ , then the inverse of the matrix $AM^{2023} {\text{A}}^{ {\text{T}}}$ is [25-Jan-2023 Shift 2]

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📖 Explanation

$\; {\text{AA}}^{ {\text{T}}}={\begin{bmatrix}\\ \\ {1}/{{\sqrt{10}}} & \\ \\ {3}/{{\sqrt{10}}} \\ \\ \\ {-3}/{{\sqrt{10}}} & \\ \\ {1}/{{\sqrt{10}}}{cc}\\ \\ {1}/{{\sqrt{10}}} & \\ \\ {-3}/{{\sqrt{10}}} \\ \\ \\ {3}/{{\sqrt{10}}} & \\ \\ {1}/{{\sqrt{10}}}\end{bmatrix}}={\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}$ $\; {\text{B}}^2={\begin{bmatrix}1 & - {\text{i}} \\ 0 & 1\end{bmatrix}}{\begin{bmatrix}1 & - {\text{i}} \\ 0 & 1\end{bmatrix}}={\begin{bmatrix}1 & -2 {\text{i}} \\ 0 & 1\end{bmatrix}}$ $\; {\text{B}}^3={\begin{bmatrix}1 & -3 {\text{i}} \\ 0 & 1\end{bmatrix}}$ . . . $\; {\text{B}}^{2023}={\begin{bmatrix}1 & -2023 {\text{i}} \\ 0 & 1\end{bmatrix}}$ $\; {\text{M}}= {\text{A}}^{ {\text{T}}} {\text{BA}}$ $\; {\text{M}}^2= {\text{M}} \cdot {\text{M}}= {\text{A}}^{ {\text{T}}} {\text{BA}} {\text{A}}^{ {\text{T}}} {\text{BA}}= {\text{A}}^{ {\text{T}}} {\text{B}}^2 {\text{A}}$ $\; {\text{M}}^3= {\text{M}}^2 \cdot {\text{M}}= {\text{A}}^{ {\text{T}}} {\text{B}}^2 {\text{AA}}^{ {\text{T}}} {\text{BA}}= {\text{A}}^{ {\text{T}}} {\text{B}}^3 {\text{A}}$ . . . $\; {\text{M}}^{2023}=. . . ⋯ ⋯ ⋯ ⋯ \cdot {\text{A}}^{ {\text{T}}} {\text{B}}^{2023} {\text{A}}$ $\; {\text{AM}}^{2023} {\text{A}}^{ {\text{T}}}= {\text{AA}}^{ {\text{T}}} {\text{B}}^{2023} {\text{AA}}^{ {\text{T}}}= {\text{B}}^{2023}$ $\;={\begin{bmatrix}1 & -2023 {\text{i}} \\ 0 & 1\end{bmatrix}}$ Inverse of $( {\text{AM}}^{2023} {\text{A}}^{ {\text{T}}})$ is ${\begin{bmatrix}1 & 2023 i \\ 0 & 1\end{bmatrix}}$

Q67JEE Main 2023MCQ4MSets and Relations

Let $\Delta , ∇ \in \{∧, ∨\}$ be such that $(p \to q) \Delta (p ∇ q)$ is a tautology. Then [25-Jan-2023 Shift 2]

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📖 Explanation

${\text{Given}}( {\text{p}} \to {\text{q}}) \Delta ( {\text{p}} ∇ {\text{q}})$ Option I $\Delta =∧, ∇=∨$ Option $2 \;\; \Delta =∨, ∇=∧$ Option $3 \;\; \Delta =∨, ∇=∨$ Hence, it is tautology. Option $4 \;\; \Delta =∧, ∇=∧$

Q68JEE Main 2023MCQ4MPermutations and Combinations

The number of numbers, strictly between 5000 and 10000 can be formed using the digits $1,3,5,7,9$ without repetition, is [25-Jan-2023 Shift 2]

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📖 Explanation

Numbers between $5000 & 10000$ Using digits $1,3,5,7,9$ Total Numbers $=3 \times 4 \times 3 \times 2=72$

Q69JEE Main 2023MCQ4MFunctions

The number of functions ${\text{f}}:\{1,2,3,4\} \to \{ {\text{a}} \in ℤ:| {\text{a}}| \le 8\}$ satisfying ${\text{f}}( {\text{n}})+$ $\;{1}/{ {\text{n}}} {\text{f}}( {\text{n}}+1)=1, ∀ {\text{n}} \in \{1,2,3\}$ is [25-Jan-2023 Shift 2]

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📖 Explanation

$\;f:\{1,2,3,4\} \to \{ {\text{a}} \in ℤ:| {\text{a}}| \le 8\}$ $\;f( {\text{n}})+\;{1}/{ {\text{n}}} {\text{f}}( {\text{n}}+1)=1, ∀ {\text{n}} \in \{1,2,3\}$ $f( {\text{n}}+1)$ must be divisible by ${\text{n}}$ $\;f(4) \Rightarrow -6,-3,0,3,6$ $\;f(3) \Rightarrow -8,-6,-4,-2,0,2,4,6,8$ $\;f(2) \Rightarrow -8, . . . . . . . . . . . . . . ., 8$ $\;f(1) \Rightarrow -8, . . . . . . . . . . . . . . ., 8$ $\;\frac{f(4)}{3}$ must be odd since $f(3)$ should be even therefore 2 solution possible.

Q70JEE Main 2023MCQ4MParabola

The equations of two sides of a variable triangle are ${\text{x}}=0$ and ${\text{y}}=3$ , and its third side is a tangent to the parabola $y^2=6 x$ . The locus of its circumcentre is: [25-Jan-2023 Shift 2]

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📖 Explanation

$\;y^2=6 x & y^2=4 a x$ $\;\Rightarrow 4 a=6 \Rightarrow a=\;\frac{3}{2}$ $y=m x+\;\frac{3}{2 m} ;( {\text{m}} \ne 0)$ ${\text{h}}=\;{6 {\text{m}}-3}/{4 {\text{m}}^2}, {\text{k}}=\;{6 {\text{m}}+3}/{4 {\text{m}}}$ , Now eliminating ${\text{m}}$ and we get $\;\Rightarrow 3 {\text{h}}=2 (-2 {\text{k}}^{2+}9 {\text{k}}-9)$ $\;\Rightarrow 4 {\text{y}}^{2-}18 {\text{y}}+3 {\text{x}}+18=0$

Q71JEE Main 2023MCQ4MLogarithms

Let $f: ℝ \to ℝ$ be a function defined by $f(x)=$ $\log _{{\sqrt{m}}}\{{\sqrt{2}}( sin \;x-\cos x)+m-2\}$ , for some $m$ , such that the range of ${\text{f}}$ is $[0,2]$ . Then the value of ${\text{m}}$ is _______ [25-Jan-2023 Shift 2]

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📖 Explanation

Since, $-{\sqrt{2}} \le sin \;x-\cos x \le {\sqrt{2}}$ $\therefore \;\; -2 \le {\sqrt{2}}( sin \;x-\cos x) \le 2$ $\;\;\text{ (Assume }\; {\sqrt{2}}( sin \;x-\cos x)=k)$ $\;-2 \le {\text{k}} \le 2 \;\; . . . \;\text{ (i) }\;$ $\;f( {\text{x}})=\log _{{\sqrt{m}}}(k+ {\text{k}}-2)$ Given, $\;0 \le f( {\text{x}}) \le 2$ $\;0 \le \log _{{\sqrt{\;\text{ min }\;}}}( {\text{k}}+ {\text{m}}-2) \le 2$ $\;1 \le {\text{k}}+ {\text{m}}-2 \le {\text{m}}$ $\;- {\text{m}}+3 \le {\text{k}} \le 2 . . . \;\text{..ii) }\;$ From eq. (i) & (ii), we get $- {\text{m}}+3=-2$ $\Rightarrow {\text{m}}=5$

Q72JEE Main 2023MCQ4MMatrices and Determinants

Let $A, B, C$ be $3 \times 3$ matrices such that $A$ is symmetric and B and ${\text{C}}$ are skew-symmetric.Consider the statements $( {\text{S}} 1) {\text{A}}^{13} {\text{B}}^{26}- {\text{B}}^{26} {\text{A}}^{13}$ is symmetric(S2) ${\text{A}}^{26} {\text{C}}^{13}- {\text{C}}^{13} {\text{A}}^{26}$ is symmetricThen. [25-Jan-2023 Shift 2]

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📖 Explanation

Given, $A^{ {\text{T}}}=A, B^{ {\text{T}}}=- {\text{B}}, {\text{C}}^{ {\text{T}}}=- {\text{C}}$ Let $M=A^{13} B^{26}-B^{26} A^{13}$ Then, $M^T= (A^{13} B^{26}-B^{26} A^{13}) ^T$ $\;= ( {\text{A}}^{13} {\text{B}}^{26}) ^{ {\text{T}}}- ( {\text{B}}^{26} {\text{A}}^{13}) ^{ {\text{T}}}$ $\;= ( {\text{B}}^{ {\text{T}}}) ^{26} ( {\text{A}}^{ {\text{T}}}) ^{13}- ( {\text{A}}^{ {\text{T}}}) ^{13} ( {\text{B}}^{ {\text{T}}}) ^{26}$ $\;= {\text{B}}^{26} {\text{A}}^{13}- {\text{A}}^{13} {\text{B}}^{26}=- {\text{M}}$ Hence, $M$ is skew symmetricLet, ${\text{N}}= {\text{A}}^{26} {\text{C}}^{13}- {\text{C}}^{13} {\text{A}}^{26}$ then, $N^{ {\text{T}}}= ( {\text{A}}^{26} {\text{C}}^{13}) ^{ {\text{T}}}- ( {\text{C}}^{13} {\text{A}}^{26}) ^{ {\text{T}}}$ $=-( {\text{C}})^{13}( {\text{A}})^{26}+ {\text{A}}^{26} {\text{C}}^{13}= {\text{N}}$ Hence, ${\text{N}}$ is symmetric. $\therefore$ Only S2 is true.

Q73JEE Main 2023MCQ4MDifferential Equations

Let $y=y(t)$ be a solution of the differential equation $\;\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}$ Where, $\alpha > 0, \beta > 0$ and $\gamma > 0$ . Then ${\text{Lim}}_{t \to \infty } y(t)$ [25-Jan-2023 Shift 2]

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📖 Explanation

$\;\;\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}$ $\;\;\text{ I.F. }\;=e^{\int \alpha \text{dt}}=e^{\alpha t}$ $\;\;\text{ Solution }\; \Rightarrow y \cdot e^{\alpha t}=\int \gamma e^{-\beta T} \cdot e^{\alpha t} \text{dt}$ $\;\Rightarrow y e^{\alpha t}=\gamma \;\frac{e^{(\alpha-\beta) t}}{(\alpha-\beta)}+c$ $\;\Rightarrow y=\;\frac{\gamma }{e^{\beta t}(\alpha-\beta)}+\;\frac{c}{e^{\alpha t}}$ So, $\lim _{t \to \infty } y(t)=\;\frac{\gamma }{\infty }+\;\frac{c}{\infty }=0$

Q74JEE Main 2023MCQ4MPermutations and Combinations

$sum_{{k=0}}^6 \;\;{}^{51- {\text{k}}} {\text{C}}_3$ is equal to [25-Jan-2023 Shift 2]

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📖 Explanation

$sum_{\text{k}=0}^6{}^{51- {\text{k}}} {\text{C}}_3$ $={ }^{51} {\text{C}}_3+{ }^{50} {\text{C}}_3+{ }^{49} {\text{C}}_3+. . .+{ }^{45} {\text{C}}_3$ $=\;{ }^{45} {\text{C}}_3+{ }^{46} {\text{C}}_3+. . . .+{ }^{51} {\text{C}}_3$ $=\;{ }^{45} {\text{C}}_4+{ }^{45} {\text{C}}_3+{ }^{46} {\text{C}}_3+. . . .+{ }^{51} {\text{C}}_3-{ }^{45} {\text{C}}_4$ $\; ({ }^{ {\text{n}}} {\ C}_{ {\text{r}}}+{ }^{ {\text{n}}} {\text{C}}_{ {\text{r}}-1}={ }^{ {\text{n}}+1} {\text{C}}_{ {\text{r}}})$ $=\;{ }^{52} {\text{C}}_4-{ }^{45} {\text{C}}_4$

Q75JEE Main 2023MCQ4MThree Dimensional Geometry

The shortest distance between the lines $x+1=2 y=- 12 z$ and $x=y+2=6 z-6$ is [25-Jan-2023 Shift 2]

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📖 Explanation

$\;\;\frac{x+1}{1}=\;\frac{y}{\frac{1}{2}}=\;\frac{z}{\frac{-1}{12}} \;\text{ and }\; \;\frac{x}{1}=\;\frac{y+2}{1}=\;\frac{z-1}{\frac{1}{6}}$ $\;\Rightarrow \;\text{ Shortest distance }\;={(\vec{\text{b}}-\vec{\text{a}}) \cdot ( \vec{\text{p}} \times \vec{\text{q}})}/{|{ \vec{\text{p}} \times { \vec{\text{q}}|}$ S.D. $=(-\;{ {\text{i}}}^{\^}+2 \;{ {\text{j}}}^{\^}-\;{ {\text{k}}}^{\^}) \cdot \;{( \vec{\text{p}} \times \vec{\text{q}})}/{|{ \vec{\text{p}} \times { \vec{\text{q}}|}$ $\; \{{ \vec{\text{p}} \times { \vec{\text{q}} \equiv {\begin{vmatrix}{ {\text{i}}^{\^} , { {\text{j}}^{\^} , { {\text{k}}^{\^} \\ 1 & \\ \\ \frac{1}{2} & \\ \\ \frac{-1}{12} \\ 1 & 1 & \\ \\ \frac{1}{6}\end{vmatrix}}=\;\frac{1}{6} { {\text{i}}^{\^}-\;\frac{1}{4} { {\text{j}}^{\^}+\;\frac{1}{2} { {\text{k}}^{\^} \;\text{ or }\; 2 { {\text{i}}^{\^}-3 { {\text{j}}^{\^}+6 { {\text{k}}^{\^}\}$ $\;\;\text{ S.D. }\;=\;{(- {\text{i}}^{\^}+2 {\text{j}}^{\^}- {\text{k}}^{\^}) \cdot (2 {\text{i}}^{\^}-3 {\text{j}}^{\^}+6 {\text{k}}^{\^})}/{{\sqrt{2^{2+}3^{2+}6^2}}}= {|\;\frac{-14}{7}|}=2$ $\;$

Q76JEE Main 2023MCQ4MProbability

Let $N$ be the sum of the numbers appeared when two fair dice are rolled and let the probability that ${\text{N}}-2, \sqrt{3 {\text{N}}}, {\text{N}}+2$ are in geometric progression be $\;{ {\text{k}}}/{48}$ . Then the value of ${\text{k}}$ is [25-Jan-2023 Shift 2]

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📖 Explanation

${\text{n}}( {\ s})=36$ Given : ${\text{N}}-2, \sqrt{3 {\text{N}}}, {\text{N}}+2$ are in G.P. $\;3 {\text{N}}=( {\text{N}}-2)( {\text{N}}+2)$ $\;3 {\text{N}}= {\text{N}}^{2-}4$ $\;\Rightarrow {\text{N}}^{2-}3 {\text{N}}-4=0$ $\;( {\text{N}}-4)( {\text{N}}+1)=0 \Rightarrow {\text{N}}=4 \;\text{ or }\; {\text{N}}=-1 \;\text{ rejected }\;$ $\; \;\; ( {\text{Sum}}=4) \equiv \{(1,3),(3,1),(2,2)\}$ $\; {\text{n}}( {\text{A}})=3$ $\; {\text{P}}( {\text{A}})=\;\frac{3}{36}=\;\frac{1}{12}=\;\frac{4}{48} \Rightarrow {\text{k}}=4$

Q77JEE Main 2023MCQ4MDefinite Integrals

The integral $16 int_{1}^{2} \;\frac{d x}{x^3 (x^{2+}2) ^2}$ is equal to [25-Jan-2023 Shift 2]

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📖 Explanation

$\;I=16 int_{1}^{2} \;\frac{d x}{x^3 (x^{2+}2) ^2}$ $\;=16 int_{1}^{2} \;\frac{d x}{x^3 x^4 (1+\;\frac{2}{x^2}) ^2}$ Let, $1+\;\frac{2}{x^2}=t \Rightarrow \;\frac{-4}{x^3} \text{dx}=d t$ $I=-4 int_{3}^{\;\frac{3}{2}} \;\frac{d t}{ (\;\frac{2}{t-1}) ^2 t^2}$ $I=-4 int_{3}^{\;\frac{3}{2}} (\;\frac{t-1}{2})^2 \;\frac{d t}{t^2}$ $I=-\;\frac{4}{4} int_{3}^{\;\frac{3}{2}} (1-\;\frac{2}{t}+\;\frac{1}{t^2}) \text{dt}$ $\;I=-1 [t-2 \ln |t|-\;\frac{1}{t}] _3^{\;\frac{3}{2}}$ $\;I=-1 [ (\;\frac{3}{2}-2 ℓ \ln \;\frac{3}{2}-\;\frac{2}{3}) - (3-2 ℓ \ln 3-\;\frac{1}{3}) ]$ $\;I=-1 [2 ℓ {\text{n}} 2-\;\frac{11}{6}]$ $\;I=\;\frac{11}{6}-ℓ {\text{n}} 4$

Q78JEE Main 2023MCQ4MArea Under The Curves

Let ${\text{T}}$ and ${\text{C}}$ respectively be the transverse and conjugate axes of the hyperbola $16 {\text{x}}^{2-}$ $y^{2+}64 x+4 y+44=0$ . Then the area of the region above the parabola $x^2=y+4$ , below the transverse axis ${\text{T}}$ and on the right of the conjugate axis ${\text{C}}$ is: [25-Jan-2023 Shift 2]

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📖 Explanation

$\;16 (x^{2+}4 x) - (y^{2-}4 y) +44=0$ $\;16(x+2)^{2-}64-(y-2)^{2+}4+44=0$ $\;16(x+2)^{2-}(y-2)^2=16$ $\;\;\frac{(x+2)^2}{1}-\;\frac{(y-2)^2}{16}=1$ $\;A=int_{-2}^{{\sqrt{6}}} (2- (x^{2-}4) ) \text{dx}$ $\;A=int_{-2}^{{\sqrt{6}}} (6-x^2) \text{dx}= (6 x-\;\frac{x^3}{3}) _{-2}^{{\sqrt{6}}}$ $\;A= (6 {\sqrt{6}}-\;{6 {\sqrt{6}}}/{3}) - (-12+\;\frac{8}{3})$ $\;A=\;{12 {\sqrt{6}}}/{3}+\;\frac{28}{3}$ $\;A=4 {\sqrt{6}}+\;\frac{28}{3}$

Q79JEE Main 2023MCQ4MVector Algebra

Let $\vec{a}=-\hat{i}-\hat{j}+\hat{k}, \vec{a} \cdot \vec{b}=1$ and $\vec{a} \times \vec{b}=\hat{i}-\hat{j}$ . Then $\vec{a}-6 \vec{b}$ is equal to [25-Jan-2023 Shift 2]

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📖 Explanation

${ \vec{\text{a}} \times { \vec{\text{b}}=({ {\text{i}}^{\^}-{ {\text{j}}^{\^})$ Taking cross product with $\vec{a}$ $\;\Rightarrow \;\; \vec{a} \times (\vec{a} \times \vec{b})=\vec{a} \times (\hat{i}-\hat{j})$ $\;\Rightarrow \;\; (\vec{a} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{a}) \vec{b}=\hat{i}+\hat{j}+2 \hat{k}$ $\;\Rightarrow \;\; \vec{a}-3 \vec{b}=\hat{i}+\hat{j}+2 \hat{k}$ $\;\Rightarrow \;\; 2 \vec{a}-6 \vec{b}=2 \hat{i}+2 \hat{j}+4 \hat{k}$ $\;\Rightarrow \;\; \;\; \vec{a}-6 \vec{b}=3 \hat{i}+3 \hat{j}+3 \hat{k}$

Q80JEE Main 2023MCQ4MThree Dimensional Geometry

The foot of perpendicular of the point $(2,0,5)$ on the line $\;\frac{x+1}{2}=\;\frac{y-1}{5}=\;\frac{z+1}{-1}$ is $(\alpha, \beta, \gamma )$ . Then. Which of the following is NOT correct? [25-Jan-2023 Shift 2]

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📖 Explanation

${\text{L}}: \;{ {\text{x}}+1}/{2}=\;{ {\text{y}}-1}/{5}=\;{ {\text{z}}+1}/{-1}=\lambda$ (let)Let foot of perpendicular is $\; {\text{P}}(2 \lambda -1,5 \lambda +1,-\lambda -1)$ $\;{ \vec{\text{PA}}=(3-2 \lambda ) { {\text{i}}^{\^}-(5 \lambda +1) { {\text{j}}^{\^}+(6+\lambda ) { {\text{k}}^{\^}$ Direction ratio of line $\Rightarrow { \vec{\text{b}}=2 { {\text{i}}^{\^}+5 { {\text{j}}^{\^}-{ {\text{k}}^{\^}$ $\;\;\text{ Now, }\; \Rightarrow \vec{P A} \cdot \vec{b}=0$ $\;\Rightarrow 2(3-2 \lambda )-5(5 \lambda +1)-(6+\lambda )=0$ $\;\Rightarrow \lambda =\;\frac{-1}{6}$ $\;P(2 \lambda -1,5 \lambda +1,-\lambda -1) \equiv P(\alpha, \beta, \gamma )$ $\;\Rightarrow \alpha=2 (-\;\frac{1}{6}) -1=-\;\frac{4}{3} \Rightarrow \alpha=-\;\frac{4}{3}$ $\;\Rightarrow \beta=5 (-\;\frac{1}{6}) +1=\;\frac{1}{6} \Rightarrow \beta=\;\frac{1}{6}$ $\;\Rightarrow \gamma =-\lambda -1=\;\frac{1}{6}-1 \Rightarrow \gamma =-\;\frac{5}{6}$ $\therefore$ Check options

Q81JEE Main 2023NAT4MSequences and Series

For the two positive numbers ${\text{a}}, {\text{b}}$ , if ${\text{a}}, {\text{b}}$ and $\;\frac{1}{18}$ are in a geometric progression, while $\;{1}/{ {\text{a}}}, 10$ and $\;{1}/{ {\text{b}}}$ are in an arithmetic progression, then, $16 {\text{a}}+12 {\text{b}}$ is equal to ________. [25-Jan-2023 Shift 2]

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📖 Explanation

$\; {\text{a}}, {\text{b}}, \;\frac{1}{18} \to {\text{GP}}$ $\;\;{ {\text{a}}}/{18}= {\text{b}}^2$ $\;\;{1}/{ {\text{a}}}, 10, \;{1}/{ {\text{b}}} \to {\text{AP}}$ $\;\;{1}/{ {\text{a}}}+\;{1}/{ {\text{b}}}=20$ $\;\Rightarrow {\text{a}}+ {\text{b}}=20 {\text{ab}}, \;\text{ from eq. (i) }\; ; \;\text{ we get }\;$ $\;\Rightarrow 18 {\text{b}}^{2+} {\text{b}}=360 {\text{b}}^3$ $\;\Rightarrow 360 {\text{b}}^{2-}18 {\text{b}}-1=0 \;\; \{ \because {\text{b}} \ne 0\}$ $\;\Rightarrow {\text{b}}=\;{18 \pm {\sqrt{324+1440}}}/{720}$ $\;\Rightarrow {\text{b}}=\;{18+{\sqrt{1764}}}/{720} \;\; \{ \because {\text{b}}>0\}$ $\;\Rightarrow {\text{b}}=\;\frac{1}{12}$ $\;\Rightarrow =18 \times \;\frac{1}{144}=\;\frac{1}{8}$ Now, $16 a+12 b=16 \times \;\frac{1}{8}+12 \times \;\frac{1}{12}=3$

Q82JEE Main 2023NAT4MCircles

Points ${\text{P}}(-3,2), {\text{Q}}(9,10)$ and ${\text{R}}(\alpha, 4)$ lie on a circle ${\text{C}}$ with ${\text{PR}}$ as its diameter. The tangents to ${\text{C}}$ at the points ${\text{Q}}$ and ${\text{R}}$ intersect at the point ${\text{S}}$ . If ${\text{S}}$ lies on the line $2 {\text{x}}- {\text{ky}}=1$ , then ${\text{k}}$ is equal to _______. [25-Jan-2023 Shift 2]

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📖 Explanation

$\; {\text{m}}_{ {\text{PQ}}} \cdot {\text{m}}_{ {\text{QR}}}=-1$ $\;\Rightarrow \;\frac{10-2}{9+3} \times \;\frac{10-4}{9-\alpha}=-1 \Rightarrow \alpha=13$ $\; {\text{m}}_{ {\text{OP}}} \cdot {\text{m}}_{ {\text{QS}}}=-1 \Rightarrow {\text{m}}_{ {\text{QS}}}=-\;\frac{4}{7}$ Equation of QS $\; {\text{y}}-10=-\;\frac{4}{7}( {\text{x}}-9)$ $\;\Rightarrow 4 {\text{x}}+7 {\text{y}}=106 . . .(1)$ $\; {\text{m}}_{0 {\text{R}}} \cdot {\text{m}}_{ {\text{RS}}}=-1 \Rightarrow {\text{m}}_{ {\text{RS}}}=-8$ Equation of RS $\;y-4=-8(x-13)$ $\;\Rightarrow 8 x+y=108$ Solving eq. (1) & (2) ${\text{x}}_1=\;\frac{25}{2} {\text{y}}_1=8$ ${\text{S}} ( {\text{x}}_1, {\text{y}}_1)$ lies on $2 {\text{x}}- {\text{ky}}=1$ $\;25-8 {\text{k}}=1$ $\;\Rightarrow 8 {\text{k}}=24$ $\;\Rightarrow {\text{k}}=3$

Q83JEE Main 2023NAT4MQuadratic Equation and Inequalities

Let $a \in {\text{R}}$ and let $\alpha, \beta$ be the roots of the equation $x^{2+}60^{\;\frac{1}{4}} x+a=0$ . If $\alpha^{4+}\beta^4=-30$ , then the product of all possible values of a is ________. [25-Jan-2023 Shift 2]

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📖 Explanation

$\alpha+\beta=-60^{\;\frac{1}{4}} \;\; & \;\; \alpha \beta=a$ Given $\alpha^{4+}\beta^4=-30$ $\;\Rightarrow (\alpha^{2+}\beta^2) ^{2-}2 \alpha^2 \beta^2=-30$ $\;\Rightarrow \{(\alpha+\beta)^{2-}2 \alpha \beta\} ^{2-}2 a^2=-30$ $\;\Rightarrow \{60^{\;\frac{1}{2}}-2 a\} ^{2-}2 a^2=-30$ $\;\Rightarrow 60+4 a^{2-}4 a \times 60^{\;\frac{1}{2}}-2 a^2=-30$ $\;\Rightarrow 2 a^{2-}4.60^{\;\frac{1}{2}} a+90=0$ $\;\;\text{ Product }\;=\;\frac{90}{2}=45$

Q84JEE Main 2023NAT4MPermutations and Combinations

Suppose Anil's mother wants to give 5 whole fruits to Anil from a basket of 7 red apples, 5 white apples and 8 oranges. If in the selected 5 fruits, at least 2 orange, at least one red apple and at least one white apple must be given, then the number of ways, Anil's mother can offer 5 fruits to Anil is _______ [25-Jan-2023 Shift 2]

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📖 Explanation

7 Red apple(RA), 5 white apple(WA), 8 oranges $(O)$ 5 fruits to be selected (Note:- fruits taken different)Possible selections :- $(2 {\text{O}}, 1 {\text{RA}}, 2 {\text{WA}})$ or $(2 {\text{O}}$ ,2RA, 1WA) or (3O, 1RA, 1WA) $\;\Rightarrow { }^8 {\text{C}}_2{ }^7 {\text{C}}_1{ }^5 {\text{C}}_2+{ }^8 {\text{C}}_2{ }^7 {\text{C}}_2{ }^5 {\text{C}}_1+{ }^8 {\text{C}}_3{ }^7 {\text{C}}_1{ }^5 {\text{C}}_1$ $\;\Rightarrow 1960+2940+1960$ $\;\Rightarrow 6860$

Q85JEE Main 2023NAT4MTrigonometric Ratios and Identities

If ${\text{m}}$ and ${\text{n}}$ respectively are the numbers of positive and negative value of $\theta$ in the interval $[-\pi, \pi]$ that satisfy the equation $\cos 2 \theta \cos \;\frac{\theta}{2}=\cos 3 \theta \cos \;\frac{9 \theta}{2}$ , then ${mn}$ is equal to ________. [25-Jan-2023 Shift 2]

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📖 Explanation

$\;\cos 2 \theta \cdot \cos \;\frac{\theta}{2}=\cos 3 \theta \cdot \cos \;\frac{9 \theta}{2}$ $\;\Rightarrow 2 \cos 2 \theta \cdot \cos \;\frac{\theta}{2}=2 \cos \;\frac{9 \theta}{2} \cdot \cos 3 \theta$ $\;\Rightarrow \cos \;\frac{5 \theta}{2}+\cos \;\frac{3 \theta}{2}=\cos \;\frac{15 \theta}{2}+\cos \;\frac{3 \theta}{2}$ $\;\Rightarrow \cos \;\frac{15 \theta}{2}=\cos \;\frac{5 \theta}{2}$ $\;\Rightarrow \;\frac{15 \theta}{2}=2 {\text{k}} \pi \pm \;\frac{5 \theta}{2}$ $\;5 \theta=2 {\text{k}} \pi \;\text{ or }\; 10 \theta=2 {\text{k}} \pi$ $\;\theta=\;{2 {\text{k}} \pi}/{5}$ $\; \therefore \theta= \{ -\pi, \;\frac{-4 \pi}{5}, \;\frac{-3 \pi}{5}, \;\frac{-2 \pi}{5}, \;\frac{-\pi}{5}, 0, \;\frac{\pi}{5}, \;\frac{2 \pi}{5}, \;\frac{3 \pi}{5}, \;\frac{4 \pi}{5}, \pi \}$ $\; \therefore {\text{m}} \cdot {\text{n}}=25$

Q86JEE Main 2023NAT4MDefinite Integrals

If $int_{{\;\frac{1}{3}}}^{3} |\log _e x| \text{dx}=\;{ {\text{m}}}/{ {\text{n}}} \log _ {\text{e}} (\;{ {\text{n}}^2}/{ {\text{e}}})$ , where ${\text{m}}$ and ${\text{n}}$ are coprime natural numbers, then ${\text{m}}^{2+} {\text{n}}^{2-}5$ is equal to ________. [25-Jan-2023 Shift 2]

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📖 Explanation

$\;int_{{\;\frac{1}{3}}}^{3} {|ℓ\text{nx}}| {\text{dx}}=int_{{\;\frac{1}{3}}}^{1}(-ℓ {\text{nx}}) {\text{dx}}+int_{1}^{3}(ℓ {\text{nx}}) {\text{dx}}$ $\;=-[ {\text{x}} ℓ {\text{nx}}- {\text{x}}]_{1 / 3}^{1+}[ {\text{x}} ℓ {\text{nx}}- {\text{x}}]_1^3$ $\;=- [-1- (\;\frac{1}{3} ℓ {\text{n}} \;\frac{1}{3}-\;\frac{1}{3}) ] +[3 ℓ {\text{n}} 3-3-(-1)]$ $\;= [-\;\frac{2}{3}-\;\frac{1}{3} ℓ {\text{n}} \;\frac{1}{3}] +[3 ℓ \ln 3-2]$ $\;=-\;\frac{4}{3}+\;\frac{8}{3} ℓ {\text{nn}} 3$ $\;=\;\frac{4}{3}(2 ℓ {\text{n}} 3-1)$ $\;=\;\frac{4}{3} (ℓ {\text{n}} \;{9}/{ {\text{e}}})$ $\; \therefore {\text{m}}=4, {\text{n}}=3$ Now, ${\text{m}}^{2+} {\text{n}}^{2-}5=16+9-5=20$

Q87JEE Main 2023NAT4MBinomial Theorem

The remainder when $(2023)^{2023}$ is divided by 35 is ________. [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\;(2023)^{2023}$ $\;=(2030-7)^{2023}$ $\;=(35 {\text{K}}-7)^{2023}$ $\;={ }^{2023} {\text{C}}_0(35 {\text{K}})^{2023}(-7)^{0+}{ }^{2023} {\text{C}}_1(35 {\text{K}})^{2022}(-7)+$ $\;. . . . .+. . . . . . .+{ }^{2023} {\text{C}}_{2023}(-7)^{2023}$ $\;=35 {\text{N}}-7^{2023} .$ $\;\;\text{ Now, }\;-7^{2023}=-7 \times 7^{2022}=-7 (7^2) ^{1011}$ $\;=-7(50-1)^{1011}$ $\;=-7 ({ }^{1011} {\text{C}}_0 50^{1011}-{ }^{1011} {\text{C}}_1(50)^{1010}+. . . . . .{ }^{1011} {\text{C}}_{1011})$ $\;=-7(5 \lambda -1)$ $\;=-35 \lambda +7$ $\therefore$ when $(2023)^{2023}$ is divided by 35 remainder is 7

Q88JEE Main 2023NAT4MThree Dimensional Geometry

If the shortest distance between the line joining the points $(1,2,3)$ and $(2,3,4)$ , and the line $\;\frac{x-1}{2}=\;\frac{y+1}{-1}=\;\frac{z-2}{0}$ is $\alpha$ , then $28 \alpha^2$ is equal to _______. [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\;{ \vec{\text{r}}=({ {\text{i}}^{\^}+2 { {\text{j}}^{\^}+3 { {\text{k}}^{\^})+\lambda ({ {\text{i}}^{\^}+{ {\text{j}}^{\^}+{ {\text{k}}^{\^}) \;\; { \vec{\text{r}}={ \vec{\text{a}}+\lambda { \vec{\text{p}}$ $\;{ \vec{\text{r}}=(+{ {\text{i}}^{\^}-{ {\text{j}}^{\^}+2 { {\text{k}}^{\^})+\mu (2 { {\text{i}}^{\^}-{ {\text{j}}^{\^}) \;\; { \vec{\text{r}}={ \vec{\text{b}}+\mu { \vec{\text{q}}$ ${ {\text{p}}}^{\to } \times { {\text{q}}}^{\to }={\begin{vmatrix}{ {\text{i}}^{\^}} & { {\text{j}}^{\^}} & { {\text{k}}^{\^}} \\ 1 & 1 & 1 \\ 2 & -1 & 0\end{vmatrix}}$ $=\hat{i}+2 \hat{j}-3 \hat{k}$ ${\text{d}}= {\begin{vmatrix}{( \vec{\text{b}}- \vec{\text{a}}) \cdot ( \vec{\text{p}} \times \vec{\text{q}})}/{\end{vmatrix} \vec{\text{p}} \times \vec{\text{q}}|}|}$ $\; {\text{d}}= {\begin{vmatrix}{(-3 {\text{j}}^{\^}- {\text{k}}^{\^}) \cdot ( {\text{i}}^{\^}+2 {\text{j}}^{\^}-3 {\text{k}}^{\^}) }/ {\sqrt{14}}\end{vmatrix}}$ $\;={ |\;{-6+3}/{{\sqrt{14}}}|}=\;{3}/{{\sqrt{14}}}$ $\;\alpha=\;{3}/{{\sqrt{14}}}$ Now, $28 \alpha^2=2 / 2 \times \;\frac{9}{14}=18$

Q89JEE Main 2023NAT4MProbability

$25 \%$ of the population are smokers. A smoker has 27 times more chances to develop lung cancer then a non-smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is $\;{ {\text{k}}}/{10}$ . Then the value of ${\text{k}}$ is ________. [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{E}}_1: \;\text{ Smokers }\;$ $\; {\text{P}} ( {\text{E}}_1) =\;\frac{1}{4}$ $\; {\text{E}}_2: \;\text{ non-smokers }\;$ $\; {\text{P}} ( {\text{E}}_2) =\;\frac{3}{4}$ ${\text{E}}$ : diagnosed with lung cancer $\; {\text{P}} ( {\text{E}} / {\text{E}}_1) =\;\frac{27}{28}$ $\; {\text{P}} ( {\text{E}} / {\text{E}}_2) =\;\frac{1}{28}$ $\; {\text{P}} ( {\text{E}}_1 / {\text{E}}) =\;{ {\text{P}} ( {\text{E}}_1) {\text{P}} ( {\text{E}} / {\text{E}}_1) }/{ {\text{P}}( {\text{E}})}$ $\;=\;\frac{\;\frac{1}{4} \times \frac{27}{28}}{\frac{1}{4} \times \;\frac{27}{28}+\;\frac{3}{4} \times \;\frac{1}{28}}=\;\frac{27^9}{30_{10}}=\;\frac{9}{10}$ $\; {\text{K}}=9$

Q90JEE Main 2023NAT4MPermutations and Combinations

A triangle is formed by ${\text{X}}$ - axis, ${\text{Y}}$ - axis and the line $3 x+4 y=60$ . Then the number of points $P(a$ , b)which lie strictly inside the triangle, where a is an integer and ${\text{b}}$ is a multiple of ${\text{a}}$ , is ________. [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

If ${\text{x}}=1, {\text{y}}=\;\frac{57}{4}=14.25$ $(1,1)(1,2)-(1,14) \;\; \Rightarrow 14 \;\text{ pts. }\;$ If $x=2, y=\;\frac{27}{2}=13.5$ $(2,2)(2,4) . . .(2,12) \;\; \Rightarrow 6$ pts.If $x=3, y=\;\frac{51}{4}=12.75$ $(3,3)(3,6)-(3,12) \;\; \Rightarrow 4$ pts.If $x=4, y=12$ $(4,4)(4,8) \;\; \Rightarrow 2$ pts.If $x=5 . y=\;\frac{45}{4}=11.25$ $(5,5),(5,10) \;\; \Rightarrow 2$ pts.If $x=6, y=\;\frac{21}{2}=10.5$ $(6,6) \;\; \Rightarrow 1 {\text{pt}}$ If $x=7, y=\;\frac{39}{4}=9.75$ $(7,7) \;\; \Rightarrow 1 {\text{pt}}$ .If $x=8, y=9$ $(8,8) \;\; \Rightarrow 1 {\text{pt}}$ .If ${\text{x}}=9 {\text{y}}=\;\frac{33}{4}=8.25 \Rightarrow$ no pt.Total $=31 {\text{pts}}$ .

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