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11 Apr 2023 Shift 2

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Physics30 questions

Q1JEE Main 2023MCQ4MProperties of Matter

Eight equal drops of water are falling through air with a steady speed of $10 {\text{cm}} / {\ s}$ . If the drops collapse, the new velocity is :- [11-Apr-2023 shift 2]

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📖 Explanation

$\;8 \times \;\frac{4}{3} \pi r^3=\;\frac{4}{3} \pi R^3$ $\;R=2 r$ $\;V=\;\frac{2 r^2}{9 η} (\rho _b-\rho _{a i r})$ $\;V \propto r^2$ $\;\;\frac{V_1}{V_2}= (\;\frac{r}{R}) ^2$ ${\text{V}}_2= {\text{V}}_1 \times 4=10 \times 4=40 {\text{km}} / {\ s}^{-1}$

Q2JEE Main 2023MCQ4MElectromagnetic Induction

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason RAssertion A : A bar magnet dropped through a metallic cylindrical pipe takes more time to come down compared to a non-magnetic bar with same geometry and mass.Reason R : For the magnetic bar, Eddy currents are produced in the metallic pipe which oppose the motion of the magnetic bar.In the light of the above statements, choose the correct answer from the options given below [11-Apr-2023 shift 2]

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Q3JEE Main 2023MCQ4MGravitation

A space ship of mass $2 \times 10^4 {\text{kg}}$ is launched into a circular orbit close to the earth surface. The additional velocity to be imparted to the space ship in the orbit to overcome the gravitational pull will be (if $g=10 {\text{m}} / {\ s}^2$ and radius of earth $=6400 {\text{km}}$ ): [11-Apr-2023 shift 2]

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📖 Explanation

$\;\Delta {\text{V}}=\sqrt{\;{ {\text{GM}}}/{ {\text{R}}}}({\sqrt{2}}-1)$ $\;=\sqrt{\;{ {\text{GM}}}/{ {\text{R}}^2} \times {\text{R}}}({\sqrt{2}}-1)$ $\;=\sqrt{ {\text{gR}}}({\sqrt{2}}-1)=8000({\sqrt{2}}-1) {\text{ms}}^{-1}$ $\;=8({\sqrt{2}}-1) {\text{kms}}^{-1}$ $\;$

Q4JEE Main 2023MCQ4MMotion in a Plane

A projectile is projected at $30^{\circ}$ from horizontal with initial velocity $40 {\text{ms}}^{-1}$ . The velocity of the projectile at ${\text{t}}=$ $2 {\ s}$ from the start will be :(Given ${\text{g}}=10 {\text{m}} / {\ s}^2$ ) [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{U}}_{ {\text{x}}}=40 \cos 30=20 {\sqrt{3}}$ $\; {\text{U}}_{ {\text{y}}}=40 sin \;30=20$ $\; {\text{Vx}}=20 {\sqrt{3}}$ $\; {\text{V}}_{ {\text{y}}}= {\text{u}}_{ {\text{y}}}- {\text{gt}}=20-10 \times 2=0$ $\; {\text{V}}=\sqrt{ {\text{v}}_{ {\text{x}}}{ }^{2+} {\text{v}}_{ {\text{y}}}{ }^2}=20 {\sqrt{3}}= {\text{ms}}^{-1}$

Q5JEE Main 2023MCQ4MElectromagnetic Waves

A plane electromagnetic wave of frequency $20 {\text{MHz}}$ propagates in free space along ${\text{x}}$ -direction. At a particular space and time, ${ {\text{E}}}^{\to }=6.6 { {\text{j}}^{\^}} {\text{v}} / {\text{m}}$ . What is ${ {\text{B}}}^{\to }$ at this point? [11-Apr-2023 shift 2]

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📖 Explanation

$|B|=\;{| {\text{E}}|}/{ {\text{C}}}=\;\frac{6.6}{3 \times 10^8}=2.2 \times 10^{-8}$ For direction of ${ {\text{B}}}^{\to }$ $\;={ {\text{E}}}^{\to } \times { {\text{B}}}^{\to }={ {\text{C}}}^{\to }$ $\;={ {\text{j}}^{\^}} \times { {\text{B}}}^{\to }={ {\text{i}}^{\^}}$ $\;{ {\text{B}}}^{\to }= (2.2 \times 10^{-8}) { {\text{k}}^{\^}} {\text{T}}$

Q6JEE Main 2023MCQ4MWaves

A car P travelling at $20 {\text{ms}}^{-1}$ sounds its horn at a frequency of $400 {\text{Hz}}$ . Another car ${\text{Q}}$ is travelling behind the first car in the same direction with a velocity $40 {\text{ms}}^{-1}$ . The frequency heard by the passenger of the car ${\text{Q}}$ is approximately $[Take, velocity of sound $=360 {\text{ms}}^{-1}$ ]$ [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{V}}_{ {\text{c}}}=20 {\text{ms}}^{-1}$ $\; {\text{f}}=400 {\text{Hz}}$ $\;f_{a p p}= [\;\frac{V s- (-V_Q) }{V_S- (-V_p) }] f$ $\;= (\;\frac{360+40}{360+20}) \times 400=\;\frac{400}{380} \times 400=421 {\text{Hz}}$ Ans. (3)

Q7JEE Main 2023MCQ4MLaws of Motion

A body of mass $500 {\text{g}}$ moves along ${\text{x}}$ -axis such that it's velocity varies with displacement ${\text{x}}$ according to the relation $v=10 \sqrt{ {\text{x}}} {\text{m}} / {\ s}$ the force acting on the body is:- [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{v}}=10 \sqrt{ {\text{x}}} \Rightarrow \;{ {\text{dv}}}/{ {\text{dx}}}=\;{10}/{2 \sqrt{ {\text{x}}}}=\;{5}/{\sqrt{ {\text{x}}}}$ $\; {\text{a}}= {\text{v}} \;{ {\text{dv}}}/{ {\text{dx}}}$ $\; {\text{a}}= {\text{v}} \times \;{5}/{\sqrt{ {\text{x}}}}=10 \sqrt{ {\text{x}}} \times \;{5}/{\sqrt{ {\text{x}}}}=50 {\text{ms}}^{-2}$ $\; {\text{F}}= {\text{ma}}=\;\frac{500}{1000} \times 50=25 {\text{N}}$

Q8JEE Main 2023MCQ4MDual Nature of Matter and Radiation

The ratio of the de-Broglie wavelengths of proton and electron having same Kinetic energy: (Assume ${\text{m}}_{ {\text{p}}}= {\text{m}}_{ {\text{e}}} \times 1849$ ) [11-Apr-2023 shift 2]

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📖 Explanation

$\;\lambda =\;{ {\text{h}}}/{ {\text{P}}}=\;{ {\text{h}}}/{\sqrt{2 {\text{mK}}}}$ $\;\;{\lambda _{ {\text{P}}}}/{\lambda _{ {\text{e}}}}=\sqrt{\;{ {\text{m}}_{ {\text{e}}}}/{ {\text{m}}_{ {\text{p}}}}}=\sqrt{\;{ {\text{m}}_{ {\text{e}}}}/{1840 {\text{me}}}}=\;{1}/{{\sqrt{1840}}}$ $\;\;{\lambda _{ {\text{P}}}}/{\lambda _{ {\text{e}}}}=\;\frac{1}{43} \;\; \;\text{ Ans. (3) }\;$ Ans. (3)

Q9JEE Main 2023MCQ4MUnits and Measurements

If force (F), velocity (V) and time (T) are considered as fundamental physical quantity, then dimensional formula of density will be: [11-Apr-2023 shift 2]

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📖 Explanation

$\rho =F^x V^y T^z$ ${[M L^{-3}]\$ = $[M L T^{-2}]$ ^x $[{L T^{-1}}^y]$ ^y[T]^z}1=x;;;;;;;x=1-3=x+y;;;;;;;y=-40=-2 x-y+z;;;;;;z=2 x+4=2-4=-2ρ=F V^{-4} T^{-2};;\text{ Ans. (3) };$

Q10JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

An electron is allowed to move with constant velocity along the axis of current carrying straight solenoid.A. The electron will experience magnetic force along the axis of the solenoid.B. The electron will not experience magnetic force.C. The electron will continue to move along the axis of the solenoid.D. The electron will be accelerated along the axis of the solenoid.E. The electron will follow parabolic path-inside the solenoid.Choose the correct answer from the options given below: [11-Apr-2023 shift 2]

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📖 Explanation

We know thatIn the solenoid magnetic field along the axis of solenoid.When charge particle moving inside solenoid along the axis ${\text{F}}=0$ $\;{ {\text{F}}_{ {\text{m}}}^{\to }}= {\text{q}}({ {\text{V}}}^{\to } \times { {\text{B}}}^{\to })$ $\;\;\text{ So, }\; { {\text{F}}_{ {\text{m}}}^{\to }}=0$ And it moves with constant velocity.

Q11JEE Main 2023MCQ4MThermodynamics

The Thermodynamic process, in which internal energy of the system remains constant is [11-Apr-2023 shift 2]

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📖 Explanation

$\Delta {\text{U}}= {\text{nC}}_{ {\text{V}}} \Delta {\text{T}}$ , for all processFor isothermal process, $\Delta {\text{T}}=0$ So, $\Delta {\text{U}}=0$ That means internal energy of system remains constant.

Q12JEE Main 2023MCQ4MElectromagnetic Waves

In satellite communication, the uplink frequency band used is: [11-Apr-2023 shift 2]

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Q13JEE Main 2023MCQ4MSemiconductor Electronics

The logic operations performed by the given digital circuit is equivalent to: [11-Apr-2023 shift 2]

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📖 Explanation

${\text{Y}}\;=\overline{\overline{( {\text{A}}+ {\text{B}}) \cdot ( {\text{AB}})}}$ ${\text{Y}}\;=( {\text{A}}+ {\text{B}}) \cdot ( {\text{AB}})$ ${\text{Y}}\;= {\text{AB}}+ {\text{AB}}$ ${\text{Y}}\;=( {\text{A}} \cdot {\text{B}})$ ${\text{Y}}\;= {\text{AND}} \;\text{ Gate }\;$

Q14JEE Main 2023MCQ4MCurrent Electricity

The current flowing through ${\text{R}}_2$ is: [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{R}}_{ {\text{eq}}}=4 Ω$ $\; {\text{i}}=\;\frac{8}{4}=2 {\text{A}}$ $\; {\text{i}}_1=\;\frac{2 \times 3}{3+6}=\;\frac{2}{3} {\text{A}}$ $\; {\text{i}}_2=\;\frac{2 / 3}{2}=\;\frac{1}{3} {\text{A}}$

Q15JEE Main 2023MCQ4MMotion in a Plane

When vector $\vec{A}=2 \hat{i}+3 \hat{j}+2 \hat{k}$ is subtracted from vector $\vec{B}$ , it gives a vector equal to $2 \hat{j}$ . Then the magnitude of vector $\vec{B}$ will be: [11-Apr-2023 shift 2]

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📖 Explanation

$\;{ {\text{A}}}^{\to }=2 { {\text{i}}^{\^}}+3 { {\text{j}}^{\^}}+2 { {\text{k}}^{\^}}$ $\;{ {\text{B}}}^{\to }-{ {\text{A}}}^{\to }=2 { {\text{j}}^{\^}}$ $\;{ {\text{B}}}^{\to }=2 { {\text{j}}^{\^}}+{ {\text{A}}}^{\to }=2 { {\text{i}}^{\^}}+5 { {\text{j}}^{\^}}+2 { {\text{k}}^{\^}}$ $\;|{ {\text{B}}}^{\to }|={\sqrt{2^{2+}5^{2+}2^2}}={\sqrt{4+25+4}}={\sqrt{8+25}}={\sqrt{33}}$

Q16JEE Main 2023MCQ4MGravitation

If ${\text{V}}$ is the gravitational potential due to sphere of uniform density on it's surface, then it's value at the center of sphere will be:- [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{V}}=-\;{ {\text{GM}}}/{ {\text{R}}^3} (1.5 {\text{R}}^{2-}0.5 {\text{r}}^2)$ $\; {\text{V}}=-\;{ {\text{GM}}}/{ {\text{R}}}[\;\text{ At the surface] }\;$ $\; {\text{V}}_{\;\text{centre }\;}=-\;{3 {\text{GM}}}/{2 {\text{R}}}=\;\frac{3}{2} {\text{V}}$

Q17JEE Main 2023MCQ4MThermodynamics

The root mean square speed of molecules of nitrogen gas at $27^{\circ} {\text{C}}$ is approximately : (Given mass of a nitrogen molecule $=4.6 \times 10^{-26} {\text{kg}}$ and take Boltzmann constant ${\text{k}}_{ {\text{B}}}=1.4 \times 10^{-23} {\text{JK}}^{-1}$ ) [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{V}}_{\;\text{rms }\;}=\sqrt{\;{3 {\text{RT}}}/{ {\text{M}}_{ {\text{w}}}}}=\sqrt{\;{3 {\text{RT}}}/{ {\text{mN}}_{ {\text{A}}}}}=\sqrt{\;{3 {\text{KT}}}/{ {\text{m}}}}$ $\;=\sqrt{\;\frac{3 \times 1.4 \times 10^{-23} \times 300}{4.6 \times 10^{-26}}}$ $\;=\sqrt{\;\frac{9 \times 1.4 \times 10^5}{4.6}}$ $\;={\sqrt{2.73 \times 10^5}}$ $\;={\sqrt{27.3 \times 10^4}}$ $\;=522.4 {\text{ms}}^{-1}$ $\;=523 {\text{ms}}^{-1}$

Q18JEE Main 2023MCQ4MAtoms and Nuclei

The energy of ${\text{He}}^{+}$ ion in its first excited state is, (The ground state energy for the Hydrogen atom is $-13.6 {\text{eV}}$ ): [11-Apr-2023 shift 2]

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📖 Explanation

${\text{E}}=- (\;{13.6 {\text{z}}^2}/{ {\text{n}}^2}) {\text{eV}}$ First excited state $\; {\text{n}}=2$ $\; {\text{E}}=\;{-13.6 \times {\text{z}}^2}/{2^2}=-13.6 {\text{eV}}$

Q19JEE Main 2023MCQ4MRay Optics and Optical Instruments

When one light ray is reflected from a plane mirror with $30^{\circ}$ angle of reflection, the angle of deviation of the ray after reflection is: [11-Apr-2023 shift 2]

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📖 Explanation

$\;\delta =Π-2 {\text{i}}=Π-2 \times 30$ $\;=180-60=120^{\circ}$

Q20JEE Main 2023MCQ4MElectrostatic Potential and Capacitance

A capacitor of capacitance ${\text{C}}$ is charge to a potential V. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is : [11-Apr-2023 shift 2]

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📖 Explanation

$\;\varphi =\;{ {\text{q}}}/{\epsilon_0}$ $\;\varphi =\;{ {\text{CV}}}/{\epsilon_0}$

Q21JEE Main 2023NAT4MElectromagnetic Induction

A coil has an inductance of $2 {\text{H}}$ and resistance of $4 Ω$ . A $10 {\text{V}}$ is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be ________ $\times 10^{-2} {\text{J}}$ . [11-Apr-2023 shift 2]

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📖 Explanation

At steady state, inductor will act as a short circuit. $\; {\text{I}}=\;{ {\text{V}}}/{ {\text{R}}}=\;\frac{10}{4}=\;\frac{5}{2} {\text{A}}$ $\; {\text{E}}=\;\frac{1}{2} {\text{LI}}^2=\;\frac{1}{2} \times 2 \times (\;\frac{5}{2}) ^2=\;\frac{25}{4}=6.25=625 \times 10^{-2} {\text{J}}$ .

Q22JEE Main 2023NAT4MElectromagnetic Induction

A metallic cube of side $15 {\text{cm}}$ moving along y-axis at a uniform velocity of $2 {\text{ms}}^{-1}$ . In a region of uniform magnetic field of magnitude $0.5 {\text{T}}$ directed along z-axis. In equilibrium the potential difference between the faces of higher and lower potential developed because of the motion through the field will be ________ ${\text{mV}}$ . [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{qVB}}= {\text{qE}}$ $\; {\text{E}}= {\text{VB}}$ $\;\Delta {\text{V}}= {\text{EL}}= {\text{VBL}}$ $\;\Delta {\text{V}}=2 \times 0.5 \times \;\frac{15}{100}=\;\frac{15}{100} {\text{volt}}$ $\;=15 \times 10^{-2} {\text{volt}}$ $\;=150 \times 10^{-3} {\text{v}}$

Q23JEE Main 2023NAT4MElectrostatic Potential and Capacitance

In the given circuit, ${\text{C}}_1=2 \mu {\text{F}}, {\text{C}}_2=0.2 \mu {\text{F}}, {\text{C}}_3=2 \mu {\text{F}}, {\text{C}}_4=4 \mu {\text{F}}, {\text{C}}_5=2 \mu {\text{F}}, {\text{C}}_6=2 \mu {\text{f}}$ , The charge stored on capacitor ${\text{C}}_4$ is ________ $\mu {\text{C}}$ . [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{C}}_{ {\text{eq}}}=0.5 \mu {\text{F}}$ $\; {\text{Q}}=0.5 \times 10=5 \mu {\text{C}}$ $\; {\text{Q}}^{'}=\;{5 \mu {\text{C}} \times 0.8}/{0.8+0.2}=4 \mu {\text{C}}$

Q24JEE Main 2023NAT4MAtoms and Nuclei

A nucleus disintegrates into two nuclear parts, in such a way that ratio of their nuclear sizes is $1: 2^{1 / 3}$ . Their respective speed have a ratio of $n: 1$ . The value of $n$ is _______ [11-Apr-2023 shift 2]

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📖 Explanation

From LCM : $\; {\text{m}}_1 {\text{V}}_1= {\text{m}}_2 {\text{V}}_2$ $\;\;{ {\text{V}}_1}/{ {\text{V}}_2}=\;{ {\text{m}}_2}/{ {\text{m}}_1}=\;{ {\text{r}}_2^3}/{ {\text{r}}_1{ }^3}= (\;{ {\text{r}}_2}/{ {\text{r}}_1}) ^3$ $\;\;{ {\text{V}}_1}/{ {\text{V}}_2}= (\;{2^{\frac{1}{3}}}/{1}) ^3=\;\frac{2}{1}$ $\; {\text{n}}=2$

Q25JEE Main 2023NAT4MProperties of Matter

The surface tension of soap solution is $3.5 \times 10^{-2} {\text{Nm}}^{-1}$ . The amount of work done required to increase the radius of soap bubble from $10 {\text{cm}}$ to $20 {\text{cm}}$ is ________ $\times 10^{-4} {\text{J}}$ . (take $\pi=22 / 7$ ) [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{W}}=\Delta {\text{U}}$ $\; {\text{W}}=25 \times ( {\text{A}}_{ {\text{f}}}- {\text{A}}_{ {\text{i}}})$ $\;=2 \times 5 \times 4 \pi ( {\text{r}}_{ {\text{f}}}^{2-} {\text{r}}_{ {\text{i}}} 2)$ $\;=2 \times 3.5 \times 10^{-2} \times 4 \times \;\frac{22}{7} \times 10^{-4}(300)$ $\;=264 \times 10^{-4} {\text{J}}$

Q26JEE Main 2023NAT4MRotational Motion

A circular plate is rotating horizontal plane, about an axis passing through its center perpendicular to the plate, with an angular velocity $\omega$ . A person sits at the center having two dumbbells in his hands. When he stretches out his hands, the moment of inertia of the system becomes triple. If ${\text{E}}$ be the initial Kinetic energy of the system, then final Kinetic energy will be $\;\frac{E}{x}$ . The value of $x$ is ______ [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{I}}_1 \omega _1= {\text{I}}_2 \omega _2$ $\; {\text{I}} \omega =3 {\text{I}} \omega _2$ $\;\omega _2=\;\frac{\omega }{3}$ $\; {\text{E}}=\;\frac{1}{2} {\text{I}} \omega ^2$ $\; {\text{E}}_{ {\text{f}}}=\;\frac{1}{2} \times 3 {\text{I}} \times (\;\frac{\omega }{3}) ^2=\;{\;\frac{1}{2} {\text{I}} \omega ^2}/{3}$ $\; {\text{x}}=3$

Q27JEE Main 2023NAT4MWork, Power and Energy

A block of mass $5 {\text{kg}}$ starting from rest pulled up on a smooth incline plane making an angle of $30^{\circ}$ with horizontal with an affective acceleration of $1 {\text{ms}}^{-2}$ . The power delivered by the pulling force at ${\text{t}}=10 {\ s}$ from the starts is ________ W.$[use ${\text{g}}=10 {\text{ms}}^{-2}$ ]$ (Calculate the nearest integer value) [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{F}}- {\text{mg}} sin \;30= {\text{ma}}$ $\; {\text{F}}=5 \times 1+25=30 {\text{N}}$ $\; {\text{V}}= {\text{u}}+ {\text{at}}=0+1 \times 10=10$ $\; {\text{P}}= {\text{FV}}=30 \times 10=300 {\text{watt}}$

Q28JEE Main 2023NAT4MRay Optics and Optical Instruments

As shown in the figure, a plane mirror is fixed at a height of $50 {\text{cm}}$ from the bottom of tank containing water $(\mu =\;\frac{4}{3})$ . The height of water in the tank is $8 {\text{cm}}$ . A small bulb is placed at the bottom of the water tank. The distance of image of the bulb formed by mirror from the bottom of the tank is ________ ${\text{cm}}$ . [11-Apr-2023 shift 2]

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📖 Explanation

Apparent depth of ${\text{O}}=\;{ {\text{d}}}/{\mu }=6$ Distance between ${\text{O}}$ and ${\text{I}}_2=48+50=98 {\text{cm}}$

Q29JEE Main 2023NAT4MCurrent Electricity

Two identical cells each of emf $1.5 {\text{V}}$ are connected in series across a $10 Ω$ resistance. An ideal voltmeter connected across $10 Ω$ resistance reads $1.5 {\text{V}}$ . The internal resistance of each cell is ________ $Ω$ . [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{I}}=\;{2 \epsilon}/{10+2 {\text{r}}}=\;\frac{1.5}{10}$ $\;20 \epsilon=15+3 {\text{r}}$ $\;\Rightarrow 20 \times 1.5=15+3 {\text{r}}$ $\;\Rightarrow 30=15+3 {\text{r}}$ $\; {\text{r}}=5 Ω$

Q30JEE Main 2023NAT4MWaves

A wire of density $8 \times 10^3 {\text{kg}} / {\text{m}}^3$ is stretched between two clamps $0.5 {\text{m}}$ apart. The extension developed in the wire is $3.2 \times 10^{-4} {\text{m}}$ . If ${\text{Y}}=8 \times 10^{10} {\text{N}} / {\text{m}}^2$ , the fundamental frequency of vibration in the wire will be ________ Hz. [11-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{f}}=\;\frac{1}{2 ℓ} \sqrt{\;{ {\text{T}}}/{\mu }}$ $\; {\text{f}}=\;\frac{1}{2 ℓ} \sqrt{\;{ {\text{YA}} \Delta ℓ}/{ℓ \mu }}$ $\;f=\;\frac{1}{2 \times 0.5} \sqrt{\;\frac{8 \times 10^{10} \times 3.2 \times 10^{-4}}{8 \times 10^3 \times 0.5}}$ $\; {\text{f}}=\;\frac{1}{1} {\sqrt{6400}}$ $\; {\text{f}}=80 {\text{Hz}}$ $\;$

Chemistry30 questions

Q31JEE Main 2023MCQ4MCoordination Compounds

The magnetic moment is measured in Bohr Magneton (BM). Spin only magnetic moment of ${\text{Fe}}$ in $[ {\text{Fe}} ( {\text{H}}_2 {\text{O}}) _6] ^{3+}$ and $[ {\text{Fe}}( {\text{CN}})_6] ^{3-}$ complexes respectively is: [11-Apr-2023 shift 2]

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📖 Explanation

$\;{ [ {\text{Fe}} ( {\text{H}}_2 {\text{O}}) _6] ^{3+}}$ $\; {\text{Fe}}^{3+} \Rightarrow [ {\text{Ar}}] 3 {\text{d}}^5 4 {\ s}^0$ No pairing ↿ ↿ ↿ ↿ ↿ $\; \therefore \;\text{ Unpaired }\; {\text{e}}^{-}=5$ $\;\mu =\sqrt{ {\text{n}}( {\text{n}}+2)}$ $\;={\sqrt{5(5+2)}}$ $\;\mu ={\sqrt{35}}=5.92 \;\text{ B.M. }\;$ $\;{ [ {\text{Fe}}( {\text{CN}})_6] ^{-3}}$ ↿⇂ ↿⇂ ↿ ↿ ↿ $\; \therefore \;\text{ Unpaired }\; {\text{e}}^{-}=1$ $\;\mu =\sqrt{ {\text{n}}( {\text{n}}+2)}$ $\;={\sqrt{1(1+2)}}={\sqrt{3}}=1.732 \;\text{ B.M. }\;$

Q32JEE Main 2023MCQ4MSolid State

Which one of the following pairs is an example of polar molecular solids? [11-Apr-2023 shift 2]

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📖 Explanation

${\text{SO}}_2$ and ${\text{NH}}_3$ are polar molecules. They are constituent particles of polar molecular solids.

Q33JEE Main 2023MCQ4MCoordination Compounds

Match List I with List II List I Complex List I Colour A. $\;\; {\text{Mg}} ( {\text{NH}}_4) {\text{PO}}_4$ I. Brown B. $\;\; {\text{K}}_3 [ {\text{Co}} ( {\text{NO}}_2) _6]$ II. White C. ${\text{MnO}}( {\text{OH}})_2$ III. Yellow D. ${\text{Fe}}_4 [ {\text{Fe}}( {\text{CN}})_6] _3$ IV. blueChoose the correct answer from the options given below: [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{Mg}} ( {\text{NH}}_4) {\text{PO}}_4 \Rightarrow \;\text{ White }\;$ $\; {\text{K}}_3 [ {\text{Co}} ( {\text{NO}}_2) _6] \Rightarrow \;\text{ Yellow }\;$ $\; {\text{MnO}}( {\text{OH}})_2 \Rightarrow \;\text{ Brown }\;$ $\; {\text{Fe}}_4 [ {\text{Fe}}( {\text{CN}})_6] _3 \Rightarrow \;\text{ Blue }\;$

Q34JEE Main 2023MCQ4MSolutions

A solution is prepared by adding $2 {\text{g}}$ of " ${\text{X}}$ " to 1 mole of water. Mass percent of " ${\text{X}}$ " in the solution is [11-Apr-2023 shift 2]

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📖 Explanation

Solute $(X)=2 g$ Solvent $( {\text{H}}_2 {\text{O}}) =1$ mole $=18 {\text{g}}$ Total mass $=2+18=20 {\text{g}}$ $\%$ mass of $X=\;\frac{2}{20} \times 100=10 \%$

Q35JEE Main 2023MCQ4MCoordination Compounds

If ${\text{Ni}}^{2+}$ is replaced by ${\text{Pt}}^{2+}$ in the complex $[ {\text{NiCl}}_2 {\text{Br}}_2] 2^{-}$ , which of the following properties are expected to get changed?A. GeometryB. Geometrical isomerismC. Optical isomerismD. Magnetic properties [11-Apr-2023 shift 2]

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📖 Explanation

$[ {\text{NiBr}}_2 {\text{Cl}}_2] ^{2-} \to$ This complex species is tetrahedral as ${\text{Br}}^{Θ} & {\text{Cl}}^{Θ}$ are weak field ligands. $[ {\text{PtBr}}_2 {\text{Cl}}_2] ^{2-} \to$ As Pt belongs to $5 {\text{d}}$ series. This complex species is square planar.Both the complex species are optically inactive. $[ {\text{NiBr}}_2 {\text{Cl}}_2] ^{2-}$ , being tetrahedral does not show Geometrical Isomerism. $[ {\text{PtBr}}_2 {\text{Cl}}_2] ^{2-}$ shows two Geometrical Isomers.

Q36JEE Main 2023MCQ4Md and f Block Elements

Given below are two statements :Statement I : In the metallurgy process, sulphide ore is converted to oxide before reduction.Statement II : Oxide ores in general are easier to reduce.In the light of the above statements, choose the most appropriate answer from the options given below: [11-Apr-2023 shift 2]

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📖 Explanation

$2 {\text{ZnS}}+3 {\text{O}}_2 \to 2 {\text{ZnO}}+2 {\text{SO}}_2$ Oxides on carbon reduction forms ${\text{CO}}_2$ while sulphide on carbon reduction gives ${\text{CS}}_2$ . ${\text{CO}}_2$ is more volatile compared to ${\text{CS}}_2$ therefore oxides are easy to reduce.

Q37JEE Main 2023MCQ4MHaloalkanes and Haloarenes

${\text{H}}_3 {\text{C}}- {\text{CH}}_2- {\text{CH}}_{|}_{\text{OH}}- {\text{CH}}_3 { {\to }^{\;\text{ (ii) }\; {\text{Mg}} \;\text{, Dry ether }\;}}^{(\text{i}) {\text{NaI}}, {\text{H}}_3 {\text{PO}}_4}_{(\text{iii}) \text{D}_2\text{O}}{[ {\text{X}}]}_{\text{Product}}$ Product $[ {\text{X}}]$ formed in the above reaction is: [11-Apr-2023 shift 2]

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Q38JEE Main 2023MCQ4MBiomolecules

Given below are two statements :Statement I : Ethene at 333 to $343 {\text{K}}$ and 6-7 atm pressure in the presence of ${\text{AlEt}}_3$ and ${\text{TiCl}}_4$ undergoes addition polymerization to give LDP.Statement II : Caprolactam at 533-543 K in ${\text{H}}_2 {\text{O}}$ through step growth polymerizes to give Nylon 6.In the light of the above statements, choose the correct answer from the options given below: [11-Apr-2023 shift 2]

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📖 Explanation

(Fact-Based)Statement-I : Ethane at 333 to $343 {\text{K}}$ and 6-7 atm pressure of ${\text{AlEt}}_3$ and ${\text{TiCl}}_4$ undergo addition polymerization to give HDPE not LDPEStatement-II :

Q39JEE Main 2023MCQ4MChemical Kinetics

For a chemical reaction ${\text{A}}+ {\text{B}} \to$ Product, the order is 1 with respect to ${\text{A}}$ and ${\text{B}}$ . $\text{Rate}$ ${\text{molL}}^{-1} {\text{S}}^{-1}$ [A] ${\text{molL}}^{-1}$ [B] ${\text{molL}}^{-1}$ 0.10 20 0.5 0.40 x 0.5 0.80 40 YWhat is the value of $x$ and $y$ ? [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{r}}= {\text{K}}[ {\text{A}}]^1[ {\text{B}}]^1$ $\;0.1= {\text{K}}(20)^1(0.5)^1$ . . . (i) $\;0.40= {\text{K}}( {\text{x}})^1(0.5)^1$ . . . (ii) $\;0.80= {\text{K}}(40)^1( {\text{y}})^1$ . . . (iii)From (i) and (ii) $x=80$ From (i) and (iii) $y=2$

Q40JEE Main 2023MCQ4MHaloalkanes and Haloarenes

Which of the following compounds is an example of Freon? [11-Apr-2023 shift 2]

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Q41JEE Main 2023MCQ4MAmines

Compound ' ${\text{B}}$ ' is [11-Apr-2023 shift 2]

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📖 Explanation

The starting compound is 4-methyl-2-isopropylphenol (thymol).

Formation of Compound A: Reaction with $\text{NaNO}_2/\text{HCl}$ (nitrous acid) at low temperature performs nitrosation. The phenolic -OH group is strongly activating and ortho/para directing. The para position is occupied by a methyl group, and one ortho position is occupied by an isopropyl group. Therefore, the nitrosyl group (\text{NO}^+$) attacks the remaining available ortho position to the -OH group, which is position 6.
Compound A is 4-methyl-2-isopropyl-6-nitrosophenol.
Formation of Compound B: Compound A is treated with $\text{NH}_4\text{SH}$ (ammonium hydrosulfide) in water. $\text{NH}_4\text{SH}$ is a selective reducing agent that reduces a nitroso group (-NO) to an amino group (- $\text{NH}_2$ ).
Therefore, the -NO group at position 6 in Compound A is reduced to an - $\text{NH}_2$ group.
Compound B is 4-methyl-2-isopropyl-6-aminophenol. Its structure is a benzene ring with -OH, -CH(CH $_3$ ) $_2$ , -CH $_3$ , and -NH $_2$ groups at positions 1, 2, 4, and 6 respectively.

The structure shown in the target option 'A' depicts 2-isopropyl-4-methyl-6-aminophenol.

Q42JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.Assertion Acan be subjected to Wolff-Kishner reduction to giveReason R:Wolff-Kishner reduction is used to convertIn the light of the above statements, choose the correct answer from the options given below: [11-Apr-2023 shift 2]

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📖 Explanation

Analysis of Assertion A:
The substrate is 3-chloro-4-methylpentan-2-one ( $CH_3-CH(CH_3)-CH(Cl)-CO-CH_3$ ). The reaction conditions are those for Wolff-Kishner reduction ( $NH_2-NH_2$ , $OH^-$ ). The asserted product is 4-methylpent-2-ene ( $CH_3-CH(CH_3)-CH=CH-CH_3$ ).
Wolff-Kishner reduction typically reduces a ketone's carbonyl group (C=O) to a methylene group ( $CH_2$ ). However, the substrate is an $\alpha$ -halo ketone. Under the strong basic conditions of Wolff-Kishner, $\alpha$ -halo ketones are highly susceptible to Favorskii rearrangement if an $\alpha'$ -hydrogen is present (which it is, on the terminal methyl group adjacent to the carbonyl). Favorskii rearrangement would lead to the formation of a carboxylate or carboxylic acid derivative, not an alkene. While it's conceivable that reduction of the carbonyl and subsequent E2 elimination of HCl could occur, forming an alkene, the Favorskii rearrangement is a significant competing pathway for $\alpha$ -halo ketones under these conditions. Therefore, stating that 4-methylpent-2-ene is the major product from Wolff-Kishner reduction of this specific $\alpha$ -halo ketone is generally considered false due to competitive side reactions like Favorskii rearrangement.

Analysis of Reason R:
Reason R states that Wolff-Kishner reduction is used to convert acetone ( $CH_3-CO-CH_3$ ) to propane ( $CH_3-CH_2-CH_3$ ). This is the standard and classic application of the Wolff-Kishner reduction, where a ketone is deoxygenated to an alkane. This statement is entirely true.

Conclusion:
Assertion A is false because the specific outcome (major product as an alkene) for an $\alpha$ -halo ketone under Wolff-Kishner conditions is questionable due to competing reactions like Favorskii rearrangement. Reason R is true as it correctly describes the fundamental Wolff-Kishner reaction. Therefore, A is false but R is true.

The final answer is $\boxed{D}$

Q43JEE Main 2023MCQ4MCoordination Compounds

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.Assertion A : $[ {\text{CoCl}} ( {\text{NH}}_3) _5] ^{2+}$ absorbs at lower wavelength of light with respect to $[ {\text{CoCl}} ( {\text{NH}}_3) _5 ( {\text{H}}_2 {\text{O}}) ] ^{3+}$ Reason R : It is because the wavelength of the light absorbed depends on the oxidation state of the metal ion.In the light of the above statements, choose the correct answer from the options given below: [11-Apr-2023 shift 2]

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📖 Explanation

Since ${\text{H}}_2 {\text{O}}$ is strong field ligand compared to chloride and ${\text{Co}}^{3+}$ ion is present. $\therefore$ CFSE is higher for $[ {\text{Co}} ( {\text{NH}}_3) 5 {\text{H}}_2 {\text{O}}] ^{+3}$ , hence it will absorb at lower wavelength.

Q44JEE Main 2023MCQ4MHaloalkanes and Haloarenes

Compound from the following that will not produce precipitate on reaction with ${\text{AgNO}}_3$ is : [11-Apr-2023 shift 2]

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📖 Explanation

The reaction of an organic halide with $\text{AgNO}_3$ proceeds via an $\text{SN}1$ mechanism, forming a carbocation intermediate and $\text{AgBr}$ precipitate. The likelihood of precipitate formation depends on the stability of the carbocation.

A. 1-bromocyclopropene forms a cyclopropenyl cation, which is aromatic (2 $\pi$ electrons) and highly stable. Thus, it will form a precipitate.
B. 1-bromocyclohexene is a vinyl halide. The bromine is directly attached to an $\text{sp}^2$ hybridized carbon of an alkene. The resulting vinyl carbocation is highly unstable due to the positive charge on an $\text{sp}^2$ carbon and angle strain. Therefore, it will not readily react to form a precipitate.
C. Benzyl bromide forms a benzylic carbocation, which is highly stabilized by resonance with the benzene ring. Thus, it will form a precipitate.
D. Cinnamyl bromide forms a resonance-stabilized allylic carbocation, further stabilized by the phenyl group. Thus, it will form a precipitate.

Only the compound in option B forms a highly unstable carbocation, preventing the formation of $\text{AgBr}$ precipitate.

Q45JEE Main 2023MCQ4MBiomolecules

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.Assertion A : A solution of the product obtained by heating a mole of glycine with a mole of chlorine in presence of red phosphorous generates chiral carbon atom.Reason R : A molecule with 2 chiral carbons is always optically active.In the light of the above statements, choose the correct answer from the options given below: [11-Apr-2023 shift 2]

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Q46JEE Main 2023MCQ4MClassification of Elements

Alkali metal from the following with least melting point is: [11-Apr-2023 shift 2]

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Q47JEE Main 2023MCQ4MChemical Bonding and Molecular Structure

Which hydride among the following is less stable? [11-Apr-2023 shift 2]

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📖 Explanation

${\text{BeH}}_2$ is hypovalent

Q48JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

The major product formed in the following reaction is ${\text{C}}_6 {\text{H}}_5- {\text{CH}}( {\text{OH}})- {\text{CH}}_{|}_{\text{CH}_3}- {\text{CH}}_2 {\text{CHO}} {\to }_{\Delta}^{ {\text{Zn}}( {\text{Hg}}) / {\text{HCl}}} \;\text{ Major products }\;$ Choose the correct answer from the options given below: [11-Apr-2023 shift 2]

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Q49JEE Main 2023MCQ4Mp Block Elements

One mole of ${\text{P}}_4$ reacts with 8 moles ${\text{SOCl}}_2$ to give 4 moles of ${\text{A}}, {\text{x}}$ mole of ${\text{SO}}_2$ and 2 moles of ${\text{B}} . {\text{A}}, {\text{B}}$ and ${\text{x}}$ respectively are [11-Apr-2023 shift 2]

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📖 Explanation

${\text{P}}_4+8 {\text{SOCl}}_2 \to 4 {\text{PCl}}_3+2 {\text{S}}_2 {\text{Cl}}_2+4 {\text{SO}}_2$

Q50JEE Main 2023MCQ4MSolutions

What weight of glucose must be dissolved in $100 {\text{g}}$ of water to lower the vapour pressure by $0.20 {\text{mmHg}}$ ?(Assume dilute solution is being formed)Given : Vapour pressure of pure water is $54.2 {\text{mmHg}}$ at room temperature. Molar mass of glucose is $180 {\text{g}} {\text{mol}}^{-1}$ [11-Apr-2023 shift 2]

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📖 Explanation

$\;{ {\text{P}}^{0-} {\text{P}}_{ {\ s}}}/{ {\text{P}}^0}=\;{ {\text{n}}}/{ {\text{N}}}$ (for dilute solution) $\;\frac{0.2}{54.2}=\;{ {\text{n}} \times 18}/{100}$ $\; {\text{n}}=\;\frac{100}{271 \times 18}$ $\; {\text{w}}=\;\frac{100 \times 180}{271 \times 18} ; {\text{w}}=3.69 {\text{g}}$

Q51JEE Main 2023NAT4MChemical Thermodynamics

The total number of intensive properties from the following is ______ new line volume, Molar heat capacity, Molarity, ${\text{E}}^\theta$ cell, Gibbs free energy change, Molar mass, Mole [11-Apr-2023 shift 2]

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📖 Explanation

Extensive $\Rightarrow$ Mole, Volume, Gibbs free energy.Intensive $\Rightarrow$ Molar mass, Molar heat capacity, Molarity, ${\text{E}}^\theta$ cell.

Q52JEE Main 2023NAT4MSome Basic Concepts of Chemistry

The volume of hydrogen liberated at STP by treating $2.4 {\text{g}}$ of magnesium with excess of hydrochloric acid is ________ $\times 10^{-2} {\text{L}}$ .Given: Molar volume of gas is $22.4 {\text{L}}$ at STP. Molar mass of magnesium is $24 {\text{g}} {\text{mol}}^{-1}$ [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{Mg}}+2 {\text{HCl}} \to {\text{MgCl}}_2+ {\text{H}}_2 ↑$ $\; {\text{w}}=2.4 {\text{g}}$ $\; {\text{N}}=\;\frac{2.4}{24}=0.1 {\text{mole}}$ $\;1 \;\text{ mole of gas at STP }\; \Rightarrow 22.4 \;\text{ lit. }\;$ $\; \therefore 0.1 \;\text{ mole of gas }\;=0.1 \times 22.4$ $\;=2.24 \;\text{ lit. }\;=224 \times 10^{-2} \;\text{ litre }\;$

Q53JEE Main 2023NAT4MChemical Kinetics

The number of correct statements about modern adsorption theory of heterogeneous catalysis from the following isA. The catalyst is diffused over the surface of reactants.B. Reactants are adsorbed on the surface of the catalyst.C. Occurrence of chemical reaction on the catalyst's surface through formation of an intermediate.D. It is a combination of intermediate compound formation theory and the old adsorption theory.E. It explains the action of the catalyst as well as those of catalytic promoters and poisons. [11-Apr-2023 shift 2]

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Q54JEE Main 2023NAT4MStructure of Atom

The number of correct statements from the following isA. For $1 {\ s}$ orbital, the probability density is maximum at the nucleusB. For $2 {\ s}$ orbital, the probability density first increases to maximum and then decreases sharply to zero.C. Boundary surface diagrams of the orbitals encloses a region of $100 \%$ probability of finding the electron.D. p and d-orbitals have 1 and 2 angular nodes respectivelyE. probability density of p-orbital is zero at the nucleus [11-Apr-2023 shift 2]

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📖 Explanation

${\text{A}}, {\text{D}}$ and ${\text{E}}$ statements are correct.For $2 {\ s}$ orbital, the probability density first decreases and then increases.At any distance from nucleus the probability density of finding electron is never zero and it always have some finite value.

Q55JEE Main 2023NAT4MElectrochemistry

The number of correct statements from the following is ________A. ${\text{E}}_{\;\text{cell }\;}$ is an intensive parameterB. A negative ${\text{E}} ⊖$ means that the redox couple is a stronger reducing agent than the ${\text{H}}^{+} / {\text{H}}_2$ couple.C. The amount of electricity required for oxidation or reduction depends on the stoichiometry of the electrode reaction.D. The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte. [11-Apr-2023 shift 2]

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Q56JEE Main 2023NAT4MClassification of Elements

${\text{Mg}} ( {\text{NO}}_3) _2 {\text{XH}}_2 {\text{O}}$ and ${\text{Ba}} ( {\text{NO}}_3) _2 {\text{YH}}_2 {\text{O}}$ , represent formula of the crystalline forms of nitrate salts. Sum of ${\text{X}}$ and ${\text{Y}}$ is _______ [11-Apr-2023 shift 2]

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📖 Explanation

${\text{Mg}} ( {\text{NO}}_3) _2 \cdot 6 {\text{H}}_2 {\text{O}}$ is a hydrated salt whereas ${\text{Ba}} ( {\text{NO}}_3) _2$ is a anhydrous salt. $\therefore {\text{x}}+ {\text{y}}=6$

Q57JEE Main 2023NAT4MHydrocarbons

The number of possible isomeric products formed when 3-chloro-1-butene reacts with ${\text{HCl}}$ through carbocation formation is _________ [11-Apr-2023 shift 2]

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📖 Explanation

Total Possible Isomeric product $=1+3=4$

Q58JEE Main 2023NAT4MChemical Equilibrium

4.5 moles each of hydrogen and iodine is heated in a sealed ten litre vessel. At equilibrium, 3 moles of ${\text{HI}}$ were found. The equilibrium constant for ${\text{H}}_2( {\text{g}})+ {\text{I}}_2( {\text{g}}) ⇌ 2 {\text{HI}}( {\text{g}})$ is ________ [11-Apr-2023 shift 2]

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📖 Explanation

${\text{K}}_{ {\text{c}}}=\;{[ {\text{HI}}]^2}/{ [ {\text{H}}_2] [ {\text{I}}_2] }=\;\frac{(3)^2}{3 \times 3}=\;\frac{9}{9}=1$

Q59JEE Main 2023NAT4MBiomolecules

Number of compounds from the following which will not produce orange red precipitate with Benedict solution is ________ Glucose, maltose, sucrose, ribose, 2-deoxyribose, amylose, lactose [11-Apr-2023 shift 2]

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📖 Explanation

$\;\text{ Benedict test : }\;\;$ $\;\text{ Glucose }\;\;-\;✔$ $\;\text{ maltose }\;\;-\;✔$ $\;\text{ sucrose }\;\;-\;\times$ $\;\text{ ribose }\;\;-\;✔$ $\;\text{ 2-deoxyribose }\;\;-\;✔$ $\;\text{ amylose }\;\;-\;\times$ $\;\text{ lactose }\;\;-\;✔$

Q60JEE Main 2023NAT4MChemical Bonding and Molecular Structure

The maximum number of lone pairs of electrons on the central atom from the following species is ________ ${\text{ClO}}_3^{-}, {\text{XeF}}_4, {\text{SF}}_4$ and ${\text{I}}_3^{-}$ [11-Apr-2023 shift 2]

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📖 Explanation

To determine the number of lone pairs on the central atom for each species, we use the formula:
$\text{Lone pairs} = \frac{1}{2} (\text{Valence electrons of central atom} + \text{Number of monovalent atoms} \pm \text{Charge})$

$\text{ClO}_3^{-}$ :
- Central atom: Cl (7 valence electrons).
- Oxygen is not monovalent.
- Total valence electrons: $7 (\text{Cl}) + 3 \times 6 (\text{O}) + 1 (\text{charge}) = 26$ .
- Number of bonds = 3 (to O atoms). $3 \times 2 = 6$ bonding electrons.
- Remaining electrons = $26 - 6 = 20$ .
- Each O needs $8 - 2 = 6$ electrons (3 lone pairs). $3 \times 6 = 18$ electrons for O lone pairs.
- Remaining electrons for central atom = $20 - 18 = 2$ .
- Number of lone pairs on Cl = $2/2 = \mathbf{1}$ .
$\text{XeF}_4$ :
- Central atom: Xe (8 valence electrons).
- Monovalent atoms: 4 F.
- Lone pairs on Xe = $\frac{1}{2} (8 - 4) = \frac{1}{2} (4) = \mathbf{2}$ .
$\text{SF}_4$ :
- Central atom: S (6 valence electrons).
- Monovalent atoms: 4 F.
- Lone pairs on S = $\frac{1}{2} (6 - 4) = \frac{1}{2} (2) = \mathbf{1}$ .
$\text{I}_3^{-}$ :
- Central atom: I (7 valence electrons).
- Monovalent atoms bonded to central I = 2 (terminal I atoms).
- Charge = -1 (add 1 electron).
- Lone pairs on central I = $\frac{1}{2} (7 - 2 + 1) = \frac{1}{2} (6) = \mathbf{3}$ .

Comparing the lone pairs: $\text{ClO}_3^{-}$ (1), $\text{XeF}_4$ (2), $\text{SF}_4$ (1), $\text{I}_3^{-}$ (3).
The maximum number of lone pairs on the central atom is 3, found in $\text{I}_3^{-}$ .

[CORRECT_OPTION: Not provided, but the answer is 3]

Mathematics30 questions

Q61JEE Main 2023MCQ4MTrigonometric Ratios and Identities

The angle of elevation of the top $P$ of a tower from the feet of one person standing due South of the tower is $45^{\circ}$ and from the feet of another person standing due west of the tower is $30^{\circ}$ . If the height of the tower is 5 meters, then the distance ( in meters) between the two persons is equal to [11-Apr-2023 shift 2]

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📖 Explanation

Tower ${\text{AB}}=5 {\text{m}}$ $\;∠ {\text{APB}}=45^{\circ}$ $\;∠ {\text{PAB}}=90^{\circ}$ $\;\tan 45^{\circ}=\;{ {\text{AB}}}/{ {\text{AP}}}$ $\;1=\;{ {\text{AB}}}/{ {\text{AP}}}$ $\; {\text{AP}}=5 {\text{m}}$ $\tan 30^{\circ}=\;{ {\text{AB}}}/{ {\text{AQ}}}$ $\;{1}/{1 {\sqrt{3}}}=\;{5}/{ {\text{AQ}}}$ ${\text{AQ}}=5 {\sqrt{3}}$ $\; {\text{AP}}^{2+} {\text{AQ}}^2= {\text{PQ}}^2$ $\; {\text{PQ}}^2=5^{2+}(5 {\sqrt{3}})^2$ $\; {\text{PQ}}^2=25+75=100$ $\; {\text{PQ}}=10 {\text{cm}}$ Option (A) $10 {\text{cm}}$ correct.

Q62JEE Main 2023MCQ4MFunctions

Let ${\text{a}}, {\text{b}}, {\text{c}}$ and ${\text{d}}$ be positive real numbers such that ${\text{a}}+ {\text{b}}+ {\text{c}}+ {\text{d}}=11$ . If the maximum value of ${\text{a}}^5 {\text{b}}^3 {\text{c}}^2 {\text{d}}$ is $3750 \beta$ , then the value of $\beta$ is [11-Apr-2023 shift 2]

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📖 Explanation

Given ${\text{a}}+ {\text{b}}+ {\text{c}}+ {\text{d}}=11$ $( {\text{a}}, {\text{b}}, {\text{c}}, {\text{d}} > 0\}$ $(a^5 b^3 c^2 d)$ max. $=$ ?Let assume Numbers - $\;{ {\text{a}}}/{5}, \;{ {\text{a}}}/{5}, \;{ {\text{a}}}/{5}, \;{ {\text{a}}}/{5}, \;{ {\text{a}}}/{5}, \;{ {\text{b}}}/{3}, \;{ {\text{b}}}/{3}, \;{ {\text{b}}}/{3}, \;{ {\text{c}}}/{2}, \;{ {\text{c}}}/{2}$ ,We know A.M. $\ge$ G.M. $\;\;{\;{ {\text{a}}}/{5}+{ {\text{a}}}/{5}+{ {\text{a}}}/{5}+{ {\text{a}}}/{5}+{ {\text{a}}}/{5}+{ {\text{b}}}/{3}+{ {\text{b}}}/{3}+{ {\text{b}}}/{3}+{ {\text{c}}}/{2}+{ {\text{c}}}/{2}+ {\text{d}}}/{11} \ge (\;{ {\text{a}}^5 {\text{b}}^3 {\text{c}}^2 {\text{d}}}/{5^5 \cdot 3^3 \cdot 2^2 \cdot 1}) ^{\;\frac{1}{11}}$ $\;\;\frac{11}{11} \ge (\;{ {\text{a}}^5 {\text{b}}^3 {\text{c}}^2 {\text{d}}}/{5^5 \cdot 3^3 \cdot 2^2 \cdot 1}) ^{\;\frac{1}{11}}$ $\; {\text{a}}^5 \cdot {\text{b}}^3 \cdot {\text{c}}^2 \cdot {\text{d}} \le 5^5 \cdot 3^3 \cdot 2^2$ $\;\max ( {\text{a}}^5 {\text{b}}^3 {\text{c}}^2 {\text{d}}) =5^5 \cdot 3^3 \cdot 2^2=337500$ $\;=90 \times 3750=\beta \times 3750$ $\;\beta=90$ Option (C) 90 correct

Q63JEE Main 2023MCQ4MDefinite Integrals

If $f: R \to R$ be a continuous function satisfying $int_{0}^{\;\frac{\pi}{2}} f( sin \;2 x) sin \;x \text{dx}+\alpha int_{0}^{\;\frac{\pi}{4}} f(\cos 2 x) \cos x \text{dx}=0$ , then the value of $\alpha$ is [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{F}}: {\text{R}} \to {\text{R}}$ $\;\Rightarrow int_{0}^{\;\frac{\pi}{2}} F( sin \;2 x) sin \;\text{dx}+\alpha int_{0}^{\;\frac{\pi}{4}} F(\cos 2 x) \cdot \cos x \text{dx}=0$ $\;\Rightarrow int_{0}^{\;\frac{\pi}{4}} {\text{F}}( sin \;2 x) sin \;x \text{dx}+int_{\;\frac{\pi}{4}}^{\;\frac{\pi}{2}} F( sin \;2 x) \cdot sin \;x \text{dx}+\alpha int_{0}^{\;\frac{\pi}{4}} F(\cos 2 x) \cdot \cos x \text{dx}=0$ $\;int_{0}^{a} F(x) \text{dx}=int_{0}^{a} F(a-x) \text{dx}$ $\;\;\text{ Let }\; {\text{x}}= {\text{t}}+\;\frac{\pi}{4}$ $\;int_{0}^{\;\frac{\pi}{4}} {\text{F}}(\cos 2 {\text{x}}) sin \; (\;\frac{\pi}{4}- {\text{x}}) {\text{dx}}+int_{0}^{\;\frac{\pi}{4}} {\text{F}}(\cos 2 {\text{t}}) sin \; ( {\text{t}}+\;\frac{\pi}{4}) +\alpha int_{0}^{\;\frac{\pi}{4}} {\text{F}}(\cos 2 {\text{x}}) \cos {\text{x}} \text{dx}=0$ $\;int_{0}^{\;\frac{\pi}{4}} F(\cos 2 x) \{ sin \; (\;\frac{\pi}{4}-x) + sin \; (x+\;\frac{\pi}{4}) +\alpha \cos x=0.$ $\;int_{0}^{\;\frac{\pi}{4}} {\text{F}}(\cos 2 x)\{({\sqrt{2}}+\alpha) \cos x\} \text{dx}=0$ $\;({\sqrt{2}}+\alpha) int_{0}^{\;\frac{\pi}{4}} {\text{F}}(\cos 2 x) \cos x \text{dx}=0$ $\;$ $\because$ in interval $(0, \;\frac{\pi}{4}) \Rightarrow {\text{F}}(\cos 2 {\text{x}}) & \cos {\text{x}}$ is NOT Zero. $\; \therefore {\sqrt{2}}+\alpha=0$ $\;\alpha=-{\sqrt{2}}$

Q64JEE Main 2023MCQ4MLimits, Continuity and Differentiability

Let $f$ and $g$ be two functions defined by $f(x)={\begin{cases}x+1 & x < 0 \\ {|x-1|,} & x \ge 0}\end{cases}$ and $g(x)={\begin{cases}x+1 & x < 0 \\ 1 & x \ge 0}\end{cases}$ Then (gof) (x) is [11-Apr-2023 shift 2]

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📖 Explanation

$\;f(x)={\begin{cases}x+1 & x< 0 \\ 1-x & 0 \le x< 1 \\ x-1 & 1 \le x}\end{cases}$ $\;g(x)={\begin{cases}x+1 & x< 0 \\ 1 & x \ge 0}\end{cases}$ $g(f(x))={\begin{cases}x+2 & x< -1 \\ 1 & x \ge -1 }\end{cases}$ $\therefore {\text{g}}( {\text{f}}( {\text{x}}))$ is continuous everywhere ${\text{g}}( {\text{f}}( {\text{x}}))$ is not differentiable at ${\text{x}}=-1$ Differentiable everywhere else

Q65JEE Main 2023MCQ4MEllipse

If the radius of the largest circle with centre $(2,0)$ inscribed in the ellipse $x^{2+}4 y^2=36$ is $r$ , then $12 r^2$ is equal to [11-Apr-2023 shift 2]

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📖 Explanation

${\text{C}}(2,0)$ Ellipse $x^{2+}4 y^2=36$ $\;\frac{x^2}{36}+\;\frac{y^2}{9}=1$ Equation of Normal at ${\text{P}}(6 \cos \theta, 3 sin \;\theta)$ is $(6 \ sec \theta) {\text{x}}-(3 {\text{cosec}} \theta) {\text{y}}=27$ It passes through $(2,0)$ $\Rightarrow \ sec \theta=\;\frac{27}{12}=\;\frac{9}{4}$ $\cos \theta \;\frac{4}{9}, sin \;\theta=\;{{\sqrt{65}}}/{9}$ ${\text{P}} (\;\frac{8}{3}, \;{{\sqrt{65}}}/{3})$ $\;{\gamma }/{ {\text{P}} (\;\frac{8}{3}, \;{{\sqrt{65}}}/{3}) {\text{c}}(2,0)}$ $\gamma =\sqrt{ (\;\frac{8}{3}-2) ^{2+} (\;{{\sqrt{65}}}/{3}) ^2}=\;{{\sqrt{69}}}/{3}$ Value of $12 \gamma ^2= (\;{{\sqrt{69}}}/{3}) ^2 \times 12$ $\Rightarrow \;\frac{12 \times 69}{9}=92$

Q66JEE Main 2023MCQ4MStatistics

Let the mean of 6 observations $1,2,4,5, {\text{x}}$ and ${\text{y}} \text{be} 5$ and their variance be 10 . Then their mean deviation about the mean is equal to [11-Apr-2023 shift 2]

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📖 Explanation

Mean of $1,2,4,5, x, y$ is 5 and variance is 10 $\Rightarrow \;\text{ mean }\; \Rightarrow \;{12+ {\text{x}}+ {\text{y}}}/{6}=5$ $\;12+x+y=30$ $\;x+y=18$ and by variance $\;\frac{x^{2+}y^{2+}46}{6}-5^2=10$ $\;x^{2+}y^2=164$ $\;x=8 \;\; y=10$ $\;\;\text{ mean daviation }\;=\;\frac{|x-\overline{x}|}{6}$ $\;\Rightarrow \;\frac{4+3+1+0+3+5}{6}=\;\frac{16}{6}=\;\frac{8}{3}$

Q67JEE Main 2023MCQ4MSets and Relations

Let ${\text{A}}=\{1,3,4,6,9\}$ and ${\text{B}}=\{2,4,5,8,10\}$ . Let ${\text{R}}$ be a relation defined on ${\text{A}} \times {\text{B}}$ such that ${\text{R}}= \{ ( ( {\text{a}}_1, {\text{b}}_1) , ( {\text{a}}_2, {\text{b}}_2...$ )): $a_1 \le b_2$ and $b_1 \le a_2\}$ . Then the number of elements in the set $R$ is [11-Apr-2023 shift 2]

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📖 Explanation

Let ${\text{a}}_1=1 \Rightarrow 5$ choices of ${\text{b}}_2$ $a_1=3 \Rightarrow 4$ choices of $b_2$ $a_1=4 \Rightarrow 4$ choices of $b_2$ $a_1=6 \Rightarrow 2$ choices of $b_2$ $a_1=9 \Rightarrow 1$ choices of $b_2$ For $(a_1, b_2) 16$ ways .Similarly, ${\text{b}}_1=2 \Rightarrow 4$ choices of ${\text{a}}_2$ $b_1=4 \Rightarrow 3$ choices of $a_2$ ${\text{b}}_1=5 \Rightarrow 2$ choices of ${\text{a}}_2$ ${\text{b}}_1=8 \Rightarrow 1$ choices of ${\text{a}}_2$ Required elements in ${\text{R}}=160$

Q68JEE Main 2023MCQ4MThree Dimensional Geometry

Let $P$ be the plane passing through the points $(5,3,0),(13,3,-2)$ and $(1,6,2)$ . For $\alpha \in {\text{N}}$ , if the distances of the points ${\text{A}}(3,4, \alpha)$ and ${\text{B}}(2, \alpha, a)$ from the plane ${\text{P}}$ are 2 and 3 respectively, then the positive value of a is [11-Apr-2023 shift 2]

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📖 Explanation

${\begin{vmatrix}{ {\text{i}}^{\^}} & { {\text{j}}^{\^}} & { {\text{k}}^{\^}} \\ 8 & 0 & -2 \\ 4 & -3 & -2\end{vmatrix}}={ {\text{i}}^{\^}}(-6)+8 { {\text{j}}^{\^}}-24 { {\text{k}}^{\^}}$ Normal of the plane $=3 \hat{i}-4 \hat{j}+12 \hat{k}$ Plane : $3 x-4 y+12 z=3$ Distance from ${\text{A}}(3,4, \alpha)$ $\;{| \;\frac{9-16+12 \alpha-3}{13} |}=2$ $\;\alpha=3$ $\;\alpha=-8 \;\text{ (rejected) }\;$ Distance from ${\text{B}}(2,3, {\text{a}})$ $\;{| \;\frac{6-12+12 a-3}{13} |}=3$ $\; {\text{a}}=4$

Q69JEE Main 2023MCQ4MPermutations and Combinations

If the letters of the word MATHS are permuted and all possible words so formed are arranged as in a dictionary with serial number, then the serial number of the word THAMS is [11-Apr-2023 shift 2]

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📖 Explanation

$\;\Rightarrow 4 \times 4 !+3 ! \times 1+0+0+0$ $\;\Rightarrow 96+6=102$ $\;\;\text{ Ran k THAMS }\;=102+1=103$

Q70JEE Main 2023MCQ4MVector Algebra

If four distinct points with position vectors ${ {\text{a}}}^{\to }, { {\text{b}}}^{\to }, { {\text{c}}}^{\to }$ and ${ {\text{d}}}^{\to }$ are coplanar, then $[{ {\text{a}}}^{\to } { {\text{b}}}^{\to } { {\text{c}}}^{\to }]$ is equal to [11-Apr-2023 shift 2]

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📖 Explanation

$\;{ {\text{a}}}^{\to }, { {\text{b}}}^{\to }, { {\text{c}}}^{\to }, { {\text{d}}}^{\to } \to \;\text{ coplanar }\;$ $\;{[{ {\text{a}}}^{\to } { {\text{b}}}^{\to } { {\text{c}}}^{\to }]\$ =?};{ {\text{b}}}^{→ }-{ {\text{a}}}^{→ }, { {\text{c}}}^{→ }-{ {\text{b}}}^{→ }, { {\text{d}}}^{→ }-{ {\text{c}}}^{→ } → ;\text{ coplanar };;{[{ {\text{b}}}^{→ }-{ {\text{a}}}^{→ } { {\text{c}}}^{→ }-{ {\text{b}}}^{→ }, { {\text{d}}}^{→ }-{ {\text{c}}}^{→ }]$=0};\Rightarrow ({ {\text{b}}}^{\to }-{ {\text{a}}}^{\to }) \cdot (({ {\text{c}}}^{\to }-{ {\text{b}}}^{\to }) \times ({ {\text{d}}}^{\to }-{ {\text{c}}}^{\to }))=0;({ {\text{b}}}^{\to }-{ {\text{a}}}^{\to }) \cdot ({ {\text{c}}}^{\to } \times { {\text{b}}}^{\to }-{ {\text{c}}}^{\to } \times { {\text{a}}}^{\to }-{ {\text{a}}}^{\to } \times { {\text{d}}}^{\to })=0;{[ {\text{bcd}}]-[ {\text{bca}}]-[ {\text{bad}}]-[ {\text{acd}}]=0};{[{ {\text{a}}}^{\to } { {\text{b}}}^{\to } { {\text{c}}}^{\to }]$=$[{ {\text{d}}}^{\to } { {\text{c}}}^{\to } {\vec{a }]$+$[{ {\text{b}}}^{\to } { {\text{d}}}^{\to } { {\text{a}}}^{\to }]$+$[{ {\text{c}}}^{\to } { {\text{d}}}^{\to } { {\text{b}}}^{\to }]$}$

Q71JEE Main 2023MCQ4MBinomial Theorem

The sum of the coefficients of three consecutive terms in the binomial expansion of $(1+x)^{n+2}$ , which are in the ratio $1: 3: 5$ , is equal to [11-Apr-2023 shift 2]

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📖 Explanation

$\;{ }^{ {\text{n}}+2} {\text{c}}_{ {\text{r}}-1}:{ }^{ {\text{n}}+2} {\text{c}}_{ {\text{r}}}:{ }^{ {\text{n}}+2} {\text{c}}_{ {\text{r}}+1}:: 1: 3: 5$ $\;\;\frac{(n+2) !}{(r-1) !(n-r+3) !} \times \;\frac{r !(n+2-r) !}{(n+2) !}=\;\frac{1}{3}$ $\;\;{ {\text{r}}}/{( {\text{n}}- {\text{r}}+3)}=\;\frac{1}{3} \Rightarrow {\text{n}}- {\text{r}}+3=3 {\text{r}}$ $\; {\text{n}}=4 {\text{r}}-3-0$ $\;\;\frac{(n+1) !}{r !(n+2-r) !} \times \;\frac{(r+1) !(n-r+1) !}{(n+2) !}=\;\frac{3}{5}$ $\;\;{ {\text{r}}+1}/{ {\text{n}}+2- {\text{r}}}=\;\frac{3}{5}$ $\;8 r-1=3 n$ . . . (2)By equation 1 and 2 $\;{8 {\text{r}}-1}/{3}=4 {\text{r}}-3\; {\text{n}}=4 {\text{r}}-3$ ${\text{r}}=2\; {\text{n}}=4 \times 2-3$ ${\text{n}}=5$ Sum: ${ }^7 {\text{C}}_1+{ }^7 {\text{C}}_2+{ }^7 {\text{C}}_3=7+21+35=63$

Q72JEE Main 2023MCQ4MDifferential Equations

Let ${\text{y}}= {\text{y}}( {\text{x}})$ be the solution of the differential equation $\;{ {\text{dy}}}/{ {\text{dx}}}+\;{5}/{ {\text{x}} ( {\text{x}}^{5+}1) } {\text{y}}=\;{ ( {\text{x}}^{5+}1) ^2}/{ {\text{x}}^7}, {\text{x}} > 0$ . If ${\text{y}}(1)=2$ , then $y(2)$ is equal to [11-Apr-2023 shift 2]

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📖 Explanation

$\;\text{ I.F }\;==e^{\int \;\frac{5 \text{dx}}{x (x^{5+}1) }}=e^{e^{\;\frac{5 x^{-6}-6 x}{ (x^{-5}+1) }}}$ Put, $1+ {\text{x}}^{-5}= {\text{t}} \Rightarrow -5 {\text{x}}^{-6} {\text{dx}}= {\text{dt}}$ $\;\Rightarrow {\text{e}}^{\int \;{- {\text{dt}}}/{1}}=\;{1}/{ {\text{t}}}=\;{ {\text{x}}^5}/{1+ {\text{x}}^5}$ $\; {\text{y}} \cdot \;{ {\text{x}}^5}/{1+ {\text{x}}^5}=\int \;{ {\text{x}}^5}/{ (1+ {\text{x}}^5) } \times \;{ (1+ {\text{x}}^5) ^2}/{ {\text{x}}^7} {\text{dx}}$ $\;=\int {\text{x}}^3 {\text{dx}}+\int {\text{x}}^{-2} {\text{dx}}$ $\; {\text{y}} \cdot \;{ {\text{x}}^5}/{1+ {\text{x}}^5}=\;{ {\text{x}}^4}/{4}-\;{1}/{ {\text{x}}}+ {\text{c}}$ Given than: $x=1 \Rightarrow y=2$ $\;2 \cdot \;\frac{1}{2}=\;\frac{1}{4}-1+ {\text{c}}$ $\; {\text{c}}=\;\frac{7}{4}$ $y \cdot \;\frac{x^5}{1+x^5}=\;\frac{x^4}{4}-\;\frac{1}{x}+\;\frac{7}{4}$ Now put, $x=2$ $\;y \cdot (\;\frac{32}{33}) =\;\frac{21}{4}$ $\;y=\;\frac{693}{128}$

Q73JEE Main 2023MCQ4MSets and Relations

The converse of $((∼ p) ∧ q) \Rightarrow r$ is [11-Apr-2023 shift 2]

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📖 Explanation

$((- {\text{P}}) ∧ 2) \Rightarrow {\text{r}}$ Converse .... $\;∼((∼ {\text{P}}) ∧ {\text{q}}) \Rightarrow (∼ {\text{r}})$ $\;( {\text{P}} ∨(∼ {\text{q}})) \Rightarrow (∼ {\text{r}})$

Q74JEE Main 2023MCQ4MBinomial Theorem

If the $1011^{\;\text{th }\;}$ term from the end in the binominal expansion of $(\;\frac{4 x}{5}-\;\frac{5}{2 x}) ^{2022}$ is 1024 times $1011^{\;\text{th }\;}$ term from the beginning, the $|x|$ is equal to [11-Apr-2023 shift 2]

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📖 Explanation

${\text{T}}_{1011}$ from beginning $= {\text{T}}_{1010+1}$ $={ }^{2022} {\text{C}}_{1010} (\;{4 {\text{x}}}/{5}) ^{1012} (\;{-5}/{2 {\text{x}}}) ^{1010}$ ${\text{T}}_{1011}$ from end $\;={ }^{2022} C_{1010} (\;\frac{-5}{2 x}) ^{1012} (\;\frac{4 x}{5}) ^{1010}$ $\;\;\text{ Given: }\;={ }^{2022} C_{1010} (\;\frac{-5}{2 x}) ^{1012} (\;\frac{4 x}{5}) ^{1010}$ $\;=2^{10} \cdot { }^{2022} C_{1010} (\;\frac{-5}{2 x}) ^{1010} (\;\frac{4 x}{5}) ^{1012}$ $\; (\;\frac{-5}{2 x}) ^2=2^{10} (\;\frac{4 x}{5}) ^2$ $\;x^4=\;\frac{5^4}{2^{16}}$ $\;|x|=\;\frac{5}{16}$

Q75JEE Main 2023MCQ4MMatrices and Determinants

If the system of linear equations $\;7 x+11 y+\alpha z=13$ $\;5 x+4 y+7 z=\beta$ $\;175 x+194 y+57 z=361$ has infinitely many solutions, then $\alpha+ {\text{B}}+2$ is equal to: [11-Apr-2023 shift 2]

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📖 Explanation

$\;7 x+11 y+\alpha z=13$ $\;5 x+4 y+7 z=\beta$ $\;175 x+194 y+57 z=361$ $4 {\ sc}$ condition of Infinite Many solution $\Delta =0 & \Delta {\text{x}}, \Delta {\text{y}}, \Delta {\text{z}}=0$ check.After solving we get $\alpha+13+2=4$

Q76JEE Main 2023MCQ4MThree Dimensional Geometry

Let the line passing through the point $P(2,-1,2)$ and $Q(5,3,4)$ meet the plane $x-y+z=4$ at the point $T$ . Then the distance of the point $R$ from the plane $x+2 y+3 z+2=0$ measured parallel to the line $\;\frac{x-7}{2}=\;\frac{y+3}{2}=\;\frac{z-2}{1}$ is equal to [11-Apr-2023 shift 2]

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📖 Explanation

Line: $\;\frac{x-5}{3}=\;\frac{y-3}{4}=\;\frac{z-4}{2}=\lambda$ $\; {\text{R}}(3 \lambda +5,4 \lambda +3,2 \lambda +4)$ $\; \therefore 3 \lambda +5-4 \lambda -3+2 \lambda +4=4$ $\;\lambda +6=4 \;\; \therefore \lambda =-2$ $\; \therefore {\text{R}} \equiv (-1,-5,0)$ Line: $\;\frac{x+1}{2}=\;\frac{y+5}{2}=\;\frac{z-0}{1}=\mu$ Point ${\text{T}}=(2 \mu -1,2 \mu -5, \mu )$ It lies on plane $\;2 \mu -1+2(2 \mu -5)+3 \mu +2=0$ $\;\mu =1$ $\; \therefore {\text{T}}=(1,-3,1)$ $\; \therefore {\text{RT}}=3$

Q77JEE Main 2023MCQ4MDefinite Integrals

Let the function ${\text{f}}:[0,2] \to R$ be defined as $f(x)={\begin{cases}e^{\min \{x^2, x-[x]\} } & x \in [0 1) \\ e^{ [x-\log _c x] } & x \in [1 2) }\end{cases}$ where $[t]$ denotes the greatest integer less than or equal to $t$ . Then the value of the integral $int_{0}^{2} x f(x) \text{dx}$ is [11-Apr-2023 shift 2]

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📖 Explanation

$\;F[0,2] \to R$ $\;F(x)={\begin{cases}\min \{x^2,\{x\}\} { \\ \\ \\ } x \in [0,1) \\ { [x-\log _e x] =1{ \\ \\ \\ } x \in [1,2)}}\end{cases}$ ${\text{F}}( {\text{x}})=$ ${\begin{cases}{\text{e}}^{ {\text{x}}^2}: {\text{x}} \in [0 & 1) \\ {\text{e}} \\ \\ \\ \\ {\text{x}} \in [1 & 2)}\end{cases}$ $\;int_{0}^{2} x f(x) \text{dx}=int_{0}^{1} x \cdot e^{x^2} \text{dx}+int_{1}^{2} x \cdot e \text{dx}$ $\;=\;\frac{1}{2}(e-1)+\;\frac{1}{2}(4-1) e$ $\;\Rightarrow 2 e-\;\frac{1}{2}$

Q78JEE Main 2023MCQ4MComplex Numbers

For $a \in C$ , let $A=\{z \in C: {\text{Re}}(a+\overline{z}) > {\text{Im}}(\overline{a}+z)\}$ and $B=\{z \in C: {\text{Re}}(a+\overline{z}) < {\text{Im}}(\overline{a}+z)\}$ . The among the two statements:(S1) : If ${\text{Re}}(a), {\text{Im}}(a) > 0$ , then the set A contains all the real numbers(S2) : If ${\text{Re}}( {\text{a}}), {\text{Im}}( {\text{a}}) < 0$ , then the set $B$ contains all the real numbers, [11-Apr-2023 shift 2]

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📖 Explanation

Let $a=x_1+i y_1 z=x+i y$ $\;{\text{Now}} {\text{Re}}( {\text{a}}+\overline{ {\text{z}}}) > {\text{Im}}(\overline{ {\text{a}}}+ {\text{z}})$ $\; \therefore {\text{x}}_1+ {\text{x}}>- {\text{y}}_1+ {\text{y}}$ $\; {\text{x}}_1=2, {\text{y}}_1=10, {\text{x}}=-12, {\text{y}}=0$ Given inequality is not valid for these values.S1 is false. $\;\;\text{ Now }\; {\text{Re}}( {\text{a}}+\overline{ {\text{z}}}) < {\text{Im}}(\overline{ {\text{a}}}+ {\text{z}})$ $\; {\text{x}}_1+ {\text{x}} < - {\text{y}}_1+ {\text{y}}$ $\; {\text{x}}_1=-2, {\text{y}}_1=-10, {\text{x}}=12, {\text{y}}=0$ Given inequality is not valid for these values. ${\text{S}} 2$ is false.

Q79JEE Main 2023MCQ4MMatrices and Determinants

If ${\begin{vmatrix}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda ^2\end{vmatrix}}=\;\frac{9}{8}(103 x+81)$ , then $\lambda , \;\frac{\lambda }{3}$ are the roots of the equation [11-Apr-2023 shift 2]

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📖 Explanation

${\begin{vmatrix}x+1 & x & x \\ x & x+d & x \\ x & x & x+d^2\end{vmatrix}}=\;\frac{9}{8}(103 x+81)$ Put ${\text{x}}=0$ $\;{\begin{vmatrix}1 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda ^2\end{vmatrix}}=\;\frac{9}{8} \times 81$ $\;\lambda ^3=\;\frac{9^3}{8}$ $\;\lambda =\;\frac{9}{2}$ $\;\;\frac{\lambda }{3}=\;\frac{9}{2 \times 3} \Rightarrow \;\frac{3}{2}$ $\;\;\frac{\lambda }{3}=\;\frac{3}{2}$ Option (C) $4 {\text{x}}^{2-}24 {\text{x}}+27=0$ has Root $\;\frac{3}{2}, \;\frac{9}{2}$

Q80JEE Main 2023MCQ4MFunctions

The domain of the function $f(x)=\;{1}/{{\sqrt{[x]^{2-}3[x]-10}}}$ is (where $[x]$ denotes the greatest integer less than or equal to ${\text{x}}$ ) [11-Apr-2023 shift 2]

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📖 Explanation

$\;F(x)=\;{1}/{{\sqrt{[x]^{2-}3[x]-10}}}$ $\;{[x]^{2-}3[x]-10 > 0}$ $\;([x]+2)([x]-5) > 0$ $\;{[x] < -2 \;\text{ or }\;[x]>5}$ $\;{[x] \le -3 \;\text{ or }\;[x] \ge 6}$ $\;x < -2 \;\text{ or }\; x \ge 6$ $\;x \in (-\infty ,-2) ∪[6, \infty )$

Q81JEE Main 2023NAT4MArea Under The Curves

If ${\text{A}}$ is the area in the first quadrant enclosed by the curve ${\text{C}}: 2 {\text{x}}^{2-} {\text{y}}+1=0$ , the tangent to ${\text{C}}$ at the point $(1,3)$ and the line $x+y=1$ , then the value of $60 {\text{A}}$ is ________. [11-Apr-2023 shift 2]

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📖 Explanation

$y=2 x^{2+}1$ Tangenet at $(1,3)$ $\; {\text{y}}=4 {\text{x}}-1$ $\; {\text{A}}=int_{0}^{1} (2 {\text{x}}^{2+}1) {\text{dx}}-\;\text{ area of }\;(\triangle {\text{QOT}})-\;\text{ area of }\;$ $\;(\triangle {\text{PQR}})+\;\text{ area of }\;(\Delta {\text{QRS}})$ $\; {\text{A}}= (\;\frac{2}{3}+1) -\;\frac{1}{2}-\;\frac{9}{8}+\;\frac{9}{40}=\;\frac{16}{60}$

Q82JEE Main 2023NAT4MPermutations and Combinations

Let $A=\{1,2,3,4,5\}$ and $B=\{1,2,3,4,5,6\}$ . Then the number of functions $f: A \to B$ satisfying $f(1)+f(2)=$ ${\text{f}}(4)-1$ is equal to ________. [11-Apr-2023 shift 2]

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📖 Explanation

$\;f(1)+f(2)+1=f(4) \le 6$ $\;f(1)+f(2) \le 5$ Case (i) ${\text{f}}(1)=1 \Rightarrow {\text{f}}(2)=1,2,3,4 \Rightarrow 4$ mappingsCase (ii) ${\text{f}}(1)=2 \Rightarrow {\text{f}}(2)=1,2,3 \Rightarrow 3$ mappingsCase (iii) ${\text{f}}(1)=3 \Rightarrow {\text{f}}(2)=1,2 \Rightarrow 2$ mappingsCase (iv) ${\text{f}}(1) 4 \Rightarrow {\text{f}}(2)=1 \Rightarrow 1$ mapping $f(5) & {\text{f}}(6)$ both have 6 mappings eachNumber of functions $=(4+3+2+1) \times 6 \times 6=360$

Q83JEE Main 2023NAT4MParabola

Let the tangent to the parabola $y^2=12 x$ at the point $(3, \alpha)$ be perpendicular to the line $2 x+2 y=3$ . Then the square of distance of the point $(6,-4)$ from the normal to the hyperbola $\alpha^2 x^{2-}9 y^2=9 \alpha^2$ at its point $(\alpha-1, \alpha+$ 2 ) is equal to _______. [11-Apr-2023 shift 2]

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📖 Explanation

$\because {\text{P}}(3, \alpha)$ lies on ${\text{y}}^2=12 {\text{x}}$ $\Rightarrow \alpha= \pm 6$ But, $\;{ {\text{dy}}}/{ {\text{dx}}} { |}_{(3, \alpha)}=\;\frac{6}{\alpha}=1 \Rightarrow \alpha=6(\alpha=-6$ reject $)$ Now, hyperbola $\;\frac{x^2}{9}-\;\frac{y^2}{36}=1$ , normal at ${\text{Q}}(\alpha-1, \alpha+2) \;\text{ is }\; \;{9 {\text{x}}}/{5}+\;{36 {\text{y}}}/{8}=45$ $\Rightarrow 2 {\text{x}}+5 {\text{y}}-50=0$ Now, distance of $(6,-4)$ from $2 x+5 y-50=0$ is equal to $\;{| \;{2(6)-5(4)-50}/{{\sqrt{2^{2+}5^2}}} |}=\;{58}/{{\sqrt{29}}}$ $\;\Rightarrow \;\text{ Square of distance }\;=116$

Q84JEE Main 2023NAT4MSequences and Series

For ${\text{k}} \in {\text{N}}$ , if the sum of the series $1+\;{4}/{ {\text{k}}}+\;{8}/{ {\text{k}}^2}+\;{13}/{ {\text{k}}^3}+\;{19}/{ {\text{k}}^4}+. . . . . .$ is 10 , then the value of ${\text{k}}$ is ______ [11-Apr-2023 shift 2]

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📖 Explanation

$\;10=1+\;{4}/{ {\text{k}}}+\;{8}/{ {\text{k}}^2}+\;{13}/{ {\text{k}}^3}+\;{19}/{ {\text{k}}^4}+. . . . . . \;\text{. upto }\; \infty$ $\;9=\;{4}/{ {\text{k}}}+\;{8}/{ {\text{k}}^2}+\;{13}/{ {\text{k}}^3}+\;{19}/{ {\text{k}}^4}+. . . . . . . \;\text{ upto }\; \infty$ $\;\;{9}/{ {\text{k}}}=\;{4}/{ {\text{k}}^2}+\;{8}/{ {\text{k}}^3}+\;{13}/{ {\text{k}}^4}+. . . . . \;\text{.upto }\; \infty$ _________________________________ $\; {\text{S}}=9 (1-\;{1}/{ {\text{k}}}) =\;{4}/{ {\text{k}}}+\;{4}/{ {\text{k}}^2}+\;{5}/{ {\text{k}}^3}+\;{6}/{ {\text{k}}^4} . . . . . . \;\text{ upto }\; \infty$ $\;\;{ {\text{S}}}/{ {\text{k}}}=\;{4}/{ {\text{k}}^2}+\;{4}/{ {\text{k}}^3}+\;{5}/{ {\text{k}}^4}+. . . . . \;\text{.upto }\; \infty$ __________________________________ $\; (1-\;{1}/{ {\text{k}}}) {\text{S}}=\;{4}/{ {\text{k}}}+\;{1}/{ {\text{k}}^3}+\;{1}/{ {\text{k}}^4}+\;{1}/{ {\text{k}}^5}+. . . . . \infty$ $\;9 (1-\;{1}/{ {\text{k}}}) ^2=\;{4}/{ {\text{k}}}+\;{\;{1}/{ {\text{k}}^3}}/{ (1-\;{1}/{ {\text{k}}}) }$ $\;9( {\text{k}}-1)^3=4 {\text{k}}( {\text{k}}-1)+1$ $\;k=2$ $\;$

Q85JEE Main 2023NAT4MThree Dimensional Geometry

Let the line $ℓ: x=\;\frac{1-y}{-2}=\;\frac{z-3}{\lambda }, \lambda \in R$ meet the plane $P: x+2 y+3 z=4$ at the point $(\alpha, \beta, \gamma )$ . If the angle between the line $ℓ$ and the plane ${\text{P}}$ is $\cos ^{-1} (\sqrt{\;\frac{5}{14}})$ , then $\alpha+2 \beta+6 \gamma$ is equal to _______. [11-Apr-2023 shift 2]

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📖 Explanation

$ℓ: x=\;\frac{y-1}{2}=\;\frac{z-3}{\lambda }, \lambda \in ℝ$ Dr's of line $ℓ(1,2, \lambda )$ Dr's of normal vector of plane ${\text{P}}: {\text{x}}+2 {\text{y}}+3 {\text{z}}=4$ are $(1,2,3)$ Now, angle between line $ℓ$ and plane ${\text{P}}$ is given by $\; sin \;\theta={| \;{1+4+3 \lambda }/{{\sqrt{5+\lambda ^2}} \cdot {\sqrt{14}}} |}=\;{3}/{{\sqrt{14}}} (\;\text{ given }\; \cos \theta=\sqrt{\;\frac{5}{14}})$ $\;\Rightarrow \lambda =\;\frac{2}{3}$ Let variable point on line $ℓ$ is $(t, 2 t+1, \;\frac{2}{3} t+3)$ line of plane $P$ . $\;\Rightarrow {\text{t}}=-1$ $\;\Rightarrow (-1,-1, \;\frac{7}{3}) \equiv (\alpha, \beta, \gamma )$ $\;\Rightarrow \alpha+2 \beta+6 \gamma =11$

Q86JEE Main 2023NAT4MFunctions

The number of points where the curve $f(x)=e^{8 x}-e^{6 x}-3 e^{4 x}-e^{2 x}+1, x \in R$ cuts $x$ -axis, is equal to _______ [11-Apr-2023 shift 2]

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📖 Explanation

Let ${\text{e}}^{2 {\text{x}}}= {\text{t}}$ $\;\Rightarrow {\text{t}}^{4-} {\text{t}}^{3-}3 {\text{t}}^{2-} {\text{t}}+1=0$ $\;\Rightarrow {\text{t}}_2+\;{1}/{ {\text{t}}_2}- ( {\text{t}}+\;{1}/{ {\text{t}}}) -3=0$ $\;\Rightarrow ( {\text{t}}+\;{1}/{ {\text{t}}}) ^{2-} ( {\text{t}}+\;{1}/{ {\text{t}}}) -5=0$ $\;\Rightarrow {\text{t}}+\;{1}/{ {\text{t}}}=\;{1+{\sqrt{21}}}/{2}$ Two real values of $t$ .

Q87JEE Main 2023NAT4MStraight Lines and Pair of Straight Lines

If the line $l_1: 3 {\text{y}}-2 {\text{x}}=3$ is the angular bisector of the line $l_2: {\text{x}}- {\text{y}}+1=0$ and $l_3: {\text{ax}}+\beta {\text{y}}+17$ , then $\alpha^{2+}\beta^{2-}$ $\alpha-\beta$ is equal to _______. [11-Apr-2023 shift 2]

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📖 Explanation

Point of intersection of $ℓ_1: 3 y-2 x=3$ $ℓ_2: {\text{x}}- {\text{y}}+1=0$ is ${\text{P}} \equiv (0,1)$ Which lies on $ℓ_3: \alpha x-\beta y+17=0$ , $\Rightarrow \beta=-17$ Consider a random point ${\text{Q}} \equiv (-1,0)$ on $ℓ_2: {\text{x}}- {\text{y}}+1=0$ , image of ${\text{Q}}$ about $ℓ_2: {\text{x}}- {\text{y}}+1=0$ , is ${\text{Q}}^{'} \equiv (\;\frac{-17}{13}, \;\frac{6}{13})$ which is calculated by formulae $\;{ {\text{x}}-(-1)}/{2}=\;{ {\text{y}}-0}/{-3}=2 (\;\frac{-2+3}{13})$ Now, $Q^{'}$ lies in $ℓ_3: \alpha x+\beta y+17=0$ $\Rightarrow \alpha=7$ Now, $\alpha^{2+}\beta^{2-}\alpha-\beta=348$

Q88JEE Main 2023NAT4MProbability

Let the probability of getting head for a biased coin be $\;\frac{1}{4}$ . It is tossed repeatedly until a head appears. Let ${\text{N}}$ be the number of tosses required. If the probability that the equation $64 x^{2+}5 N x+1=0$ has no real root is $\;\frac{p}{q}$ , where ${\text{p}}$ and ${\text{q}}$ are co-prime, then ${\text{q}}- {\text{p}}$ is equal to _______. [11-Apr-2023 shift 2]

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📖 Explanation

$\;64 {\text{x}}^{2+}5 {\text{Nx}}+1=0$ $\; {\text{D}}=25 {\text{N}}^{2-}256 < 0$ $\;\Rightarrow {\text{N}}^2 < \;\frac{256}{25} \Rightarrow {\text{N}} < \;\frac{16}{5}$ $\; \therefore {\text{N}}=1,2,3$ $\; \therefore \;\text{ Probability }\;=\;\frac{1}{4}+\;\frac{3}{4} \times \;\frac{1}{4}+\;\frac{3}{4} \times \;\frac{3}{4} \times \;\frac{1}{4}=\;\frac{37}{64}$ $\; \therefore {\text{q}}- {\text{p}}=27$

Q89JEE Main 2023NAT4MVector Algebra

Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$ and $\vec{b}=\hat{i}+\hat{j}-\hat{k}$ . If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c}=11, \vec{b} \cdot (\vec{a} \times \vec{c})=27$ and $\vec{b} \cdot \vec{c}=-{\sqrt{3}}|\vec{b}|$ , then $|{\text{bo} {a}}^{\to } \times {\text{bo} {c}}^{\to }|^2$ is equal to _______. [11-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\;{ {\text{a}}}^{\to }={ {\text{i}}^{\^}}+2 { {\text{j}}^{\^}}+3 { {\text{k}}^{\^}}, { {\text{b}}}^{\to }={ {\text{i}}^{\^}}+{ {\text{j}}^{\^}}-{ {\text{k}}^{\^}}$ $\;{ {\text{b}}}^{\to } \cdot ({ {\text{a}}}^{\to } \times { {\text{c}}}^{\to })=27, { {\text{a}}}^{\to } \cdot { {\text{b}}}^{\to }=0$ $\;{ {\text{b}}}^{\to } \times ({ {\text{a}}}^{\to } \times { {\text{c}}}^{\to })=-3 { {\text{a}}}^{\to }$ Let $\theta$ be angle between $\vec{b}, \vec{a} \times \vec{c}$ Then $|{ {\text{b}}}^{\to }| \cdot |{ {\text{a}}}^{\to } \times { {\text{c}}}^{\to }| sin \;\theta=3 {\sqrt{14}}$ $|{ {\text{b}}}^{\to }| \cdot |{ {\text{a}}}^{\to } \times { {\text{c}}}^{\to }| \cos \theta=27$ $\;\Rightarrow sin \;\theta=\;{{\sqrt{14}}}/{{\sqrt{95}}}$ $\; \therefore |{ {\text{b}}}^{\to }| \times |{ {\text{a}}}^{\to } \times { {\text{c}}}^{\to }|=3 {\sqrt{95}}$ $\;\Rightarrow |{ {\text{a}}}^{\to } \times { {\text{c}}}^{\to }|={\sqrt{3}} \times {\sqrt{95}}$

Q90JEE Main 2023NAT4MComplex Numbers

Let $S= \{z \in C-\{i, 2 i\}: \;\frac{z^{2+}8 i z-15}{z^{2-}3 i z-2} \in R\}$ . If $\alpha-\;\frac{13}{11} i \in S, a \in R-\{0\}$ , then $242 \alpha^2$ is equal to _______. [11-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; (\;\frac{z^{2+}8 i z-15}{z^{2-}3 i z-2}) \in R$ $\;\Rightarrow 1+\;\frac{(11 i z-13)}{ (z^{2-}3 i z-2) } \in R$ Put $Z=\alpha-\;\frac{13}{11} {\text{i}}$ $\Rightarrow ( {\text{z}}^{2-}3 {\text{iz}}-2)$ is imaginaryPut $z=x+$ iy $\;\Rightarrow (x^{2-}y^{2+}2 x y i-3 i x+3 y-2) \in \;\text{ Imaginary }\;$ $\;\Rightarrow {\text{Re}} (x^{2-}y^{2+}3 y-2+(2 x y-3 x) i) =0$ $\;\Rightarrow x^{2-}y^{2+}3 y-2=0$ $\;x^2=y^{2-}3 y+2$ $\;x^2=(y-1)(y-2) \therefore z=\alpha-\;\frac{13}{11} i$ Put $x=\alpha, y=\;\frac{-13}{11}$ $\alpha^2= (\;\frac{-13}{11}-11) (\;\frac{-13}{11}-2)$ $\;\alpha^2=\;\frac{(24 \times 35)}{121}$ $\;242 \alpha^2=48 \times 35=1680$

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