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13 Apr 2023 Shift 2

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Physics30 questions

Q1JEE Main 2023MCQ4MAlternating Current

Given below are two statements :Statements I : An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor.Statement II : An AC circuit containing a pure capacitor or a pure inductor consumes high power due to its non-zero power factor.In the light of above statements, choose the correct answer form the options given below : [13-Apr-2023 shift 2]

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📖 Explanation

Statement-I: An AC circuit for resonance inductor and capacitor both should required.Statement-II: An AC circuit containing a pure capacitor and pure inductor have no power lossFor resonance, $\varphi =0$ means both capacitor and inductor must be present.

Q2JEE Main 2023MCQ4MMotion in a Straight Line

A passenger sitting in a train A moving at $90 {\text{km}} / {\text{h}}$ observes another train B moving in the opposite direction for $8 {\ s}$ . if the velocity of the train B is $54 {\text{km}} / {\text{h}}$ , then length of train B is : [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{V}}_{ {\text{A}}}=\;{90 {\text{km}}}/{ {\text{hr}}}=25 {\text{ms}}^{-1}$ $\; {\text{V}}_{ {\text{B}}}=\;{54 {\text{km}}}/{ {\text{hr}}}=15 {\text{ms}}^{-1}$ $\;{ {\text{V}}_{ {\text{BA}}}^{\to }}={ {\text{V}}_{ {\text{B}}}^{\to }}-{ {\text{VA}}}^{\to }=40 {\text{ms}}^{-1}$ $\;\;\text{ Time of crossing }\;=\;\frac{\;\text{ Length of train }\;}{\;\text{ Re lative velocity }\;}$ $\;8=\;\frac{ℓ}{40}$ $\;ℓ=8 \times 40=320 \;\text{ meter }\;$

Q3JEE Main 2023MCQ4MSemiconductor Electronics

The output from a NAND gate having inputs A and B given below will be, [13-Apr-2023 shift 2]

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Q4JEE Main 2023MCQ4MMotion in a Straight Line

The distance travelled by an object in time $t$ is given by ${\ s}=(2.5) {\text{t}}^2$ . The instantaneous speed of the object at ${\text{t}}=$ $5 {\ s}$ will be : [13-Apr-2023 shift 2]

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📖 Explanation

${\text{S}}=2.5 {\text{t}}^2$ At, ${\text{t}}=5 {\sec}$ . ${\text{v}}=5 \times 5=25 {\text{ms}}^{-1}$

Q5JEE Main 2023MCQ4MWave Optics

In a Young's double slits experiment, the ratio of amplitude of light coming from slits is $2: 1$ . The ratio of the maximum to minimum intensity in the interference pattern is : [13-Apr-2023 shift 2]

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📖 Explanation

Given that $\;\;\frac{A_1}{A_2}=\;\frac{2}{1}$ $\;\;\frac{I_{\max }}{I_{\min }}=\;\frac{ (A_1+A_2) ^2}{ (A_1-A_2) ^2}=\;\frac{9}{1}$ $\;=9: 1$

Q6JEE Main 2023MCQ4MAtoms and Nuclei

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason RAssertion A : The binding energy per nucleon is practically independent of the atomic number for nuclei of mass number in the range 30 to 170 .Reason R : Nuclear force is short ranged.In the light of the above statements, choose the correct answer from the options given below [13-Apr-2023 shift 2]

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📖 Explanation

Assertion A is true: the binding energy per nucleon curve shows a relatively constant value for nuclei with mass numbers between approximately $30$ and $170$ .
Reason R is true: the nuclear force is a strong, attractive force but acts only over very short distances (on the order of a few femtometers).
The short-range nature of the nuclear force leads to its saturation property, meaning each nucleon interacts only with its nearest neighbors. This "saturation" is why the binding energy per nucleon becomes nearly constant for intermediate-sized nuclei, as adding more nucleons does not significantly increase the total number of bonds per nucleon. Thus, R correctly explains A.

Q7JEE Main 2023MCQ4MGravitation

Two planets ${\text{A}}$ and ${\text{B}}$ of radii ${\text{R}}$ and $1.5 {\text{R}}$ have densities $\rho$ and $\rho / 2$ respectively. The ratio of acceleration due to gravity at the surface of ${\text{B}}$ to ${\text{A}}$ is : [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{g}}=\;{ {\text{GM}}}/{ {\text{R}}^2}=\;{ {\text{G}} \times \frac{4}{3} \pi {\text{R}}^3 \times \rho }/{ {\text{R}}^2}=\;\frac{4}{3} {\text{G}} \pi {\text{PR}}$ $\;\;{ {\text{g}}_2}/{ {\text{g}}_1}=\;{ {\text{R}}_2}/{ {\text{R}}_1} \times \;\frac{\rho _2}{\rho _1}=1.5 \times \;\frac{1}{2}=\;\frac{3}{4}$

Q8JEE Main 2023MCQ4MThermodynamics

The mean free path of molecules of a certain gas at STP is $1500 {\text{d}}$ , where ${\text{d}}$ is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at $373 {\text{K}}$ is approximately : [13-Apr-2023 shift 2]

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📖 Explanation

mean free path $\lambda$ $\;\lambda =\;{ {\text{RT}}}/{{\sqrt{2}} {\text{Cd}}^2 {\text{N}}_{ {\text{A}}} {\text{P}}}$ $\;\lambda \propto {\text{T}}$ $\;\;{1500 {\text{d}}}/{\lambda }=\;\frac{273}{373}$ $\;\lambda =2049 {\text{d}}$

Q9JEE Main 2023MCQ4MElectromagnetic Waves

To radiate EM signal of wavelength $\lambda$ with high efficiency, the antennas should have a minimum size equal to: [13-Apr-2023 shift 2]

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📖 Explanation

Minimum length of antennaShould be $\;\frac{\lambda }{4}$

Q10JEE Main 2023MCQ4MOscillations

A particle executes SHM of amplitude A. The distance from the mean position when its's kinetic energy becomes equal to its potential energy is : [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{K}} . {\text{E}}= {\text{P}} . {\text{E}}$ $\;\;\frac{1}{2} M \omega ^2 (A^{2-}x^2) =\;\frac{1}{2} M \omega ^2 x^2$ $\; (A^{2-}x^2) =x^2$ $\;2 x^2=A^2$ $\;x= \pm \;{A}/{{\sqrt{2}}}$

Q11JEE Main 2023MCQ4MElectromagnetic Waves

In an electromagnetic wave, at an instant and at a particular position, the electric field is along the negative ${\text{z}}$ axis and magnetic field is along the positive ${\text{x}}$ -axis. Then the direction of propagation of electromagnetic wave is : [13-Apr-2023 shift 2]

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📖 Explanation

Direction of propagation of EM wave will be in the direction of $(\vec{E} \times \vec{B})$

Q12JEE Main 2023MCQ4MDual Nature of Matter and Radiation

Given below are two statements :Statement I : Out of microwaves, infrared rays and ultraviolet rays, ultraviolet rays are the most effective for the emission of electrons from a metallic surface.Statement II : Above the threshold frequency, the maximum kinetic energy of photoelectrons is inversely proportional to the frequency of the incident light.In the light of above statements, choose the correct answer form the options given below [13-Apr-2023 shift 2]

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📖 Explanation

UV rays have maximum frequency hence are most effective for emission of electrons from the metallic surface. $K E_{\max }=h f-h f_0$

Q13JEE Main 2023MCQ4MGravitation

Given below are two statements :Statement I : For a planet, if the ratio of mass of the planet to its radius increases, the escape velocity from the planet also increases.Statement II : Escape velocity is independent of the radius of the planet.In the light of above statements, choose the most appropriate answer form the options given below [13-Apr-2023 shift 2]

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📖 Explanation

${\text{V}}_{ {\text{e}}}=\sqrt{\;{2 {\text{GM}}}/{ {\text{R}}}}$ As, $\;\frac{M}{R}$ increases $\Rightarrow V_e$ increases ${\text{V}}_{ {\text{e}}} \propto \;{1}/{\sqrt{ {\text{R}}}}$ As, $V_e$ depends on $R$

Q14JEE Main 2023MCQ4MCircular Motion

A vehicle of mass $200 {\text{kg}}$ is moving along a levelled curved road of radius $70 {\text{m}}$ with angular velocity of 0.2 ${\text{rad}} / {\ s}$ . The centripetal force acting on the vehicle is: [13-Apr-2023 shift 2]

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📖 Explanation

$\;F_c=m \omega ^2 r=200 \times (0.2)^2 \times 70$ $\;=560 N$

Q15JEE Main 2023MCQ4MElectrostatics

A $10 \mu {\text{C}}$ charge is divided into two parts and placed at $1 {\text{cm}}$ distance so that the repulsive force between them is maximum. The charges of the two parts are: [13-Apr-2023 shift 2]

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📖 Explanation

Divide $q=10 \mu$ cinto parts $(x)$ and $(q-x)$ ${\text{F}}=\;{( {\text{K}})( {\text{x}})( {\text{q}}- {\text{x}})}/{ {\text{r}}^2}$ For ${\text{F}}$ to be maximum $\;\;\frac{d F}{d x}=0$ $\;x=\;\frac{q}{2}=\;\frac{10 \mu C}{2}=5 \mu C$ $\;q-x=10 \mu C-5 \mu C=5 \mu C$

Q16JEE Main 2023MCQ4MUnits and Measurements

In the equation $[ {\text{x}}+\;{ {\text{a}}}/{ {\text{y}}^2}] [ {\text{Y}}- {\text{b}}]= {\text{RT}}, {\text{X}}$ is pressure, ${\text{Y}}$ is volume, ${\text{R}}$ is universal gas constant ant ${\text{T}}$ is temperature. The physical quantity equivalent to the ratio $\;{ {\text{a}}}/{ {\text{b}}}$ is : [13-Apr-2023 shift 2]

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📖 Explanation

$x$ and $\;\frac{a}{y^2}$ have same dimensions $y$ and $b$ have same dimensions $\;{[ {\text{a}}]= \$ [ {\text{ML}}^5 {\text{T}}^{-2}]$ };{[ {\text{b}}]= $[ {\text{L}}^3]$ };;{[ {\text{a}}]}/{[ {\text{b}}]}= {\text{ML}}^2 {\text{T}}^{-2} ;\text{ has dimension of energy };$

Q17JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

An electron is moving along the positive ${\text{x}}$ -axis. If the uniform magnetic field is applied parallel to the negative z-axis, thenA. The electron will experience magnetic force along positive y-axisB. The electron will experience magnetic force along negative y-axisC. The electron will not experience any force in magnetic fieldD. The electron will continue to move along the positive ${\text{x}}$ -axisE. The electron will move along circular path in magnetic field Choose the correct answer from the options given below : [13-Apr-2023 shift 2]

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📖 Explanation

$\;{ {\text{F}}}^{\to }= {\text{q}}({ {\text{V}}}^{\to } \times { {\text{B}}}^{\to })$ $\;{ {\text{F}}}^{\to }=- {\text{e}}({ {\text{V}}}^{\to } \times { {\text{B}}}^{\to })$ Force will be along -ve yaxis As magnetic force is $⟂_r$ to velocity, path of electron must be circle.

Q18JEE Main 2023MCQ4MProperties of Matter

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason RAssertion A : A spherical body of radius $(5 \pm 0.1) {\text{mm}}$ having a particular density is falling through a liquid of constant density. The percentage error in the calculation of its terminal velocity is $4 \%$ Reason $\text{bo} {R}$ : The terminal velocity of the spherical body falling through the liquid is inversely proportional to its radius.In the light of the above statements, choose the correct answer from the options given below [13-Apr-2023 shift 2]

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📖 Explanation

Terminal velocity of a spherical body in liquid $\;V_t \propto r^2$ $\;\;\frac{\Delta V_t}{V_t}=2 \;\frac{\Delta r}{r}$ $\;\;\frac{\Delta V_t}{V_t} \times 100 \%=2 \times \;\frac{0.1}{5} \times 100=4 \%$ $\;\;\text{ Also, }\; V_t \propto r^2$ $\;\;\text{ Reason (R) is false }\;$

Q19JEE Main 2023MCQ4MThermodynamics

The initial pressure and volume of an ideal gas are $P_o$ and $V_0$ . The final pressure of the gas when the gas is suddenly compressed to volume $\;\frac{V_o}{4}$ will be :(Given $\gamma =$ ratio of specific heats at constant pressure and at constant volume) [13-Apr-2023 shift 2]

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📖 Explanation

As, gas in suddenly compressed, the process is adiabatic So equation of gas for adiabatic process is: $\; {\text{PV}}^\gamma =\;\text{ constant }\;$ $\; {\text{P}}_0 {\text{V}}_0^\gamma = {\text{P}}_2 (\;{ {\text{V}}_0}/{4}) ^\gamma$ $\; {\text{P}}_2= {\text{P}}_0(4)^\gamma$

Q20JEE Main 2023MCQ4MElectrostatic Potential and Capacitance

In the network shown below, the charge accumulated in the capacitor in steady state will be : [13-Apr-2023 shift 2]

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📖 Explanation

In steady state, no current will pass through capacitor hence capacitor will act as open circuit. $\; {\text{i}}_2=0$ $\; {\text{i}}_1=\;\frac{3}{6+4}=\;\frac{3}{10} {\text{A}}$ Potential difference on $6 Ω$ resistor $=6 \times \;\frac{3}{10}=1.8$ volt capacitor will have same potential so charge $= {\text{cv}}=4 \times 1.8=7.2 \mu {\text{c}}$

Q21JEE Main 2023NAT4MWaves

In an experiment with sonometer when a mass of $180 {\text{g}}$ is attached to the string, it vibrates with fundamental frequency of $30 {\text{Hz}}$ . When a mass ${\text{m}}$ is attached, the string vibrates with fundamental frequency of $50 {\text{Hz}}$ . The value of ${\text{m}}$ is ________ g. [13-Apr-2023 shift 2]

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📖 Explanation

$\;f=\;\frac{1}{2 ℓ} \sqrt{\;\frac{T}{\mu }}$ $\;\;\frac{f_2}{f_1}=\sqrt{\;\frac{T_2}{T_1}}$ $\; (\;\frac{50}{30}) ^2=\;{ {\text{mg}}}/{180 {\text{g}}}$ $\; {\text{m}}=\;\frac{25}{9} \times 180=500 \;\text{ gram }\;$

Q22JEE Main 2023NAT4MThermodynamics

Two plates A and B have thermal conductivities $84 {\text{Wm}}^{-1} {\text{K}}^{-1}$ and $126 {\text{Wm}}^{-1} {\text{K}}^{-1}$ respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of ${\text{A}}$ and ${\text{B}}$ are kept at $100^{\circ} {\text{C}}$ and $0^{\circ} {\text{C}}$ respectively, then the temperature of the surface of contact in steady state is ______ ${ }^{\circ} {\text{C}}$ . [13-Apr-2023 shift 2]

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📖 Explanation

Let the temperature of contact surface is ${\text{T}}$ then, $\;\;{ {\text{K}}_A {\text{A}} ( {\text{T}}_{ {\text{A}}}-T) }/{ {\text{L}}}=\;{ {\text{K}}_{ {\text{B}}} {\text{A}} ( {\text{T}}- {\text{T}}_{ {\text{B}}}) }/{ {\text{L}}}$ $\;84(100-T)=126( {\text{T}}-0)$ $\;T=40^{\circ} {\text{C}}$

Q23JEE Main 2023NAT4MElectrostatic Potential and Capacitance

In the circuit shown, the energy stored in the capacitor is $n \mu J$ . The value of $n$ is [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{I}}_1=\;\frac{12}{3+9}=1 {\text{A}}$ $\; {\text{I}}_2=\;\frac{12}{4+2}=2 {\text{A}}$ $\; {\text{V}}_{ {\text{A}}}- {\text{V}}_{ {\text{C}}}=3 {\text{I}}_1=3 {\text{V}}$ $\; {\text{V}}_{ {\text{A}}}- {\text{V}}_{ {\text{D}}}=2 \times 4=8 {\text{V}}$ $\;\;\text{ So, }\; {\text{V}}_{ {\text{A}}}- {\text{V}}_{ {\text{D}}}=5 {\text{V}}$ $\; {\text{U}}=\;\frac{1}{2} {\text{CV}}^2=\;\frac{1}{2} \times 6 \times 5^2=75 \mu {\text{J}}$

Q24JEE Main 2023NAT4MRotational Motion

A light rope is wound around a hollow cylinder of mass $5 {\text{kg}}$ and radius $70 {\text{cm}}$ . The rope is pulled with a force of $52.5 {\text{N}}$ . The angular acceleration of the cylinder will be _______ ${\text{rads}}^{-2}$ . [13-Apr-2023 shift 2]

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📖 Explanation

$\;\tau = {\text{l}} \alpha$ $\; {\text{FR}}= {\text{mR}}^2 \alpha$ $\;\propto =\;{ {\text{F}}}/{ {\text{mR}}}=\;\frac{52.5}{5 \times 0.7}=15 {\text{rads}}^{-2}$

Q25JEE Main 2023NAT4MMagnetic Effects of Current and Magnetism

A straight wire $A B$ of mass $40 {\text{g}}$ and length $50 {\text{cm}}$ is suspended by a pair of flexible leads in uniform magnetic field of magnitude $0.40 {\text{T}}$ as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is _______ A. $(.$ Take $g=10 {\text{ms}}^{-2})$ [13-Apr-2023 shift 2]

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📖 Explanation

For equilibrium : $\; {\text{Mg}}= {\text{IlB}}$ $\; {\text{I}}=\;{ {\text{Mg}}}/{ {\text{IB}}}=\;\frac{40 \times 10^{-3} \times 10}{50 \times 10^{-2} \times 0.4}=2 {\text{A}}$

Q26JEE Main 2023NAT4MElectromagnetic Induction

An insulated copper wire of 100 turns is wrapped around a wooden cylindrical core of the cross-sectional area $24 {\text{cm}}^2$ . The two ends of the wire are connected to a resistor. The total resistance in the circuit is $12 Ω$ . If an externally applied uniform magnetic field in the core along its axis changes from $1.5 {\text{T}}$ in one direction to $1.5 {\text{T}}$ in the opposite direction, the charge flowing through a point in the circuit during the change of magnetic field will be _________ ${\text{mC}}$ . [13-Apr-2023 shift 2]

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📖 Explanation

$\;|\Delta {\text{Q}}|=\;{\Delta \varphi }/{ {\text{R}}}=\;{2 {\text{NBA}}}/{ {\text{R}}}$ $\;=\;\frac{2 \times 100 \times 1.5 \times 24 \times 10^{-4}}{12}$ $\;=6 \times 10^{-2} {\text{c}}$ $\;=60 {\text{mc}}$

Q27JEE Main 2023NAT4MRay Optics and Optical Instruments

A bi convex lens of focal length $10 {\text{cm}}$ is cut in two identical parts along a plane perpendicular to the principal axis. The power of each lens after cut is ________ D. [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{P}}_1+ {\text{P}}_1= {\text{P}}=\;{1}/{ {\text{f}}}$ $\;2 {\text{P}}_1=\;\frac{1}{0.1}$ $\; {\text{P}}_1=5 {\text{D}}$

Q28JEE Main 2023NAT4MElectrostatics

Three point charges $q,-2 q$ and $2 q$ are placed on $x$ -axis at a distance $x=0, x=\;\frac{3}{4} R$ and $x=R$ respectively from origin as shown. If ${\text{q}}=2 \times 10^{-6} {\text{C}}$ and ${\text{R}}=2 {\text{cm}}$ , the magnitude of net force experienced by the charge $-2 {\text{q}}$ is _______ N. [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{F}}_{ {\text{BA}}}=\;{32 {\text{kq}}^2}/{9 {\text{q}}^2} \;\; {\text{F}}_{ {\text{BC}}}=\;{64 {\text{kq}}^2}/{ {\text{R}}^2}$ $\; {\text{F}}_{ {\text{B}}}= {\text{F}}_{ {\text{BC}}}- {\text{F}}_{ {\text{BA}}}=\;{544 {\text{kq}}^2}/{9 {\text{R}}^2}$ $\;=5440 {\text{N}}$

Q29JEE Main 2023NAT4MAtoms and Nuclei

An atom absorbs a photon of wavelength $500 {\text{nm}}$ and emits another photon of wavelength $600 {\text{nm}}$ . The net energy absorbed by the atom in this process is ${\text{n}} \times 10^{-4} {\text{eV}}$ . The value of ${\text{n}}$ is [Assume the atom to be stationary during the absorption and emission process] $(.$ Take ${\text{h}}=6.6 \times 10^{-34} {\text{Js}}$ and ${\text{c}}=3 \times 10^8 {\text{m}} / {\ s})$ [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{E}}= {\text{E}}_1- {\text{E}}_2= {\text{hc}} (\;\frac{1}{\lambda _1}-\;\frac{1}{\lambda _2})$ $\; {\text{E}}=6.6 \times 10^{-20} {\text{J}}$ $\; {\text{E}}=4.125 \times 10^{-1} {\text{eV}}$ $\; {\text{E}}=4125 \times 10^{-4} {\text{eV}}$

Q30JEE Main 2023NAT4MWork, Power and Energy

A car accelerates from rest to $u {\text{m}} / {\ s}$ . The energy spent in this process is ${\text{E}} {\text{J}}$ . The energy required to accelerate the car from ${\text{u}} {\text{m}} / {\ s}$ to $2 {\text{u}} {\text{m}} / {\ s}$ is ${\text{nE}} {\text{J}}$ . The value of ${\text{n}}$ is _______. [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{E}}_1=\;\frac{1}{2} {\mu}^{2-}0=\;\frac{1}{2} {\mu}^2= {\text{E}}$ $\; {\text{E}}_2=\;\frac{1}{2} {\text{m}}(24)^{2-}\;\frac{1}{2} {\mu}^2$ $\;=\;\frac{3}{2} {\mu}^2=3 {\text{E}}$

Chemistry30 questions

Q31JEE Main 2023MCQ4MSome Basic Concepts of Chemistry

Which of the following are the Green house gases?(A) Water vapour(B) Ozone(C) ${\text{I}}_2$ (D) Molecular hydrogenChoose the most appropriate answer from the options given below : [13-Apr-2023 shift 2]

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📖 Explanation

Green house gases are ${\text{CO}}_2, {\text{CH}}_4$ , water vapour, nitrous oxide, ${\text{CFC}}_{ {\ s}}$ and ozone.

Q32JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

The major product for the following reaction is : [13-Apr-2023 shift 2]

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Q33JEE Main 2023MCQ4MSalt Analysis

In the wet tests for detection of various cations by precipitation, ${\text{Ba}}^{2+}$ cations are detected by obtaining precipitate of : [13-Apr-2023 shift 2]

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📖 Explanation

In wet testing, $( {\text{NH}}_4) _2 {\text{CO}}_3$ is used as group reagent for $5^{\;\text{th }\;}$ group cations $( {\text{Ba}}^{2+}, {\text{Ca}}^{2+}, {\text{Sr}}^{2+})$ ${\text{Ba}}^{+2}+ ( {\text{NH}}_4) _2 {\text{CO}}_3 \to { {\text{BaCO}}_3 ↓}_{\;\text{ (white precipitate) }\;}+ {{\text{NH}}^{⊕}_4}$

Q34JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

Compound A from the following reaction sequence is : [13-Apr-2023 shift 2]

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Q35JEE Main 2023MCQ4MStructure of Atom

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.Assertion A : Isotopes of hydrogen have almost same chemical properties, but difference in their rates of reaction.Reason R : Isotopes of hydrogen have different enthalpy of bond dissociation.In the light of the above statements, choose the most appropriate answer from the options given below : [13-Apr-2023 shift 2]

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Q36JEE Main 2023MCQ4MChemical Thermodynamics

Given below are statements related to Ellingham diagram :Statement I : Ellingham diagram can be constructed for oxides, sulfides and halides of metals.Statement II : It consists of plosts of $\Delta _{ {\text{f}}} {\text{H}}^0$ vs for formation of oxides of Clements.In the light of the above statements, choose the most appropriate answer from the options given below : [13-Apr-2023 shift 2]

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📖 Explanation

Statement I is correct, Ellingham diagram can be constructed for formation of oxides, sulphides and halides of metals. (Ref: NCERT) Statement II is incorrect because Ellingham diagram consists of $\Delta _{ {\text{f}}} {\text{G}}^0$ vs T for formation of oxides of elements.

Q37JEE Main 2023MCQ4MClassification of Elements

Better method for preparation of ${\text{BeF}}_2$ , among the following is [13-Apr-2023 shift 2]

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📖 Explanation

As per NCERT (s block), the better method of preparation of ${\text{BeF}}_2$ is heating $( {\text{NH}}_4) _2 {\text{BeF}}_4$ $( {\text{NH}}_4) _2 {\text{BeF}}_4 {\to }^{\Delta } {\text{BeF}}_2+ {\text{NH}}_4 {\text{F}}$

Q38JEE Main 2023MCQ4MChemical Thermodynamics

Identify the correct order of standard enthalpy of formation of sodium halides. [13-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

For a given metal $\Delta _{ {\text{f}}} {\text{H}}^0$ always becomes less negative from fluoride to iodide.

Q39JEE Main 2023MCQ4MCoordination Compounds

Which of the following complexes will exhibit maximum attraction to an applied magnetic field? [13-Apr-2023 shift 2]

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📖 Explanation

Complex with maximum number of unpaired electron will exhibit maximum attraction to an applied magnetic field $\;{ [ {\text{Zn}} ( {\text{H}}_2 {\text{O}}) _6] ^{2+} \to d^{10} \;\text{ system }\; \to t_{2 {\text{g}}}^6 \;\text{ eg }\;^4, 0 \;\text{ unpaired }\; {\text{e}}^{-}}$ $\;{ [ {\text{Co}} ( {\text{H}}_2 {\text{O}}) _6] ^{2+} \to {\text{d}}^7 \;\text{ system }\; \to {\text{t}}_{2 {\text{g}}}^5 \;\text{ eg }\;^2, 3 \;\text{ unpaired }\; {\text{e}}^{-}}$ $\;{ [ {\text{Co}}( {\text{en}})_3] ^{3+} \to {\text{d}}^6 \;\text{ system }\; \to {\text{t}}_{2 {\text{g}}}^6 {\text{eg}}^0, 0 \;\text{ unpaired }\; {\text{e}}^{-}}$ $\;{ [ {\text{Ni}} ( {\text{H}}_2 {\text{O}}) _6] ^{2+} \to {\text{d}}^8 \;\text{ system }\; \to {\text{t}}_{2 {\text{g}}}^6 \;\text{ eg }\;^2, 2 \;\text{ unpaired }\; {\text{e}}^{-}}$

Q40JEE Main 2023MCQ4MRedox Reactions

The correct group of halide ions which can be oxidized by oxygen in acidic medium is [13-Apr-2023 shift 2]

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📖 Explanation

Only ${\text{I}}^{-}$ among halides can be oxidised to Iodine by oxygen in acidic medium $4 {\text{I}}^{-}( {\text{aq}})+4 {\text{H}}^{+}( {\text{aq}})+ {\text{O}}_2( {\text{g}}) \to 2 {\text{I}}_2( {\ s})+2 {\text{H}} 2 {\text{O}}( {\text{l}})$

Q41JEE Main 2023MCQ4MCoordination Compounds

The total number of stereoisomers for the complex $[ {\text{Cr}}( {\text{ox}})_2 {\text{ClBr}}] ^{3-}($ where ox $=$ oxalate $)$ is : [13-Apr-2023 shift 2]

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📖 Explanation

$[ {\text{Cr}}( {\text{Ox}})_2 {\text{ClBr}}] ^{-3}$ - No. of isomers-- This structure has plane of symmetry, So no optical isomerism will be shown.- This structure does not contain plane of symmetry, So two forms ${\text{d}}$ as well as 1 will be shown.

Q42JEE Main 2023MCQ4MHaloalkanes and Haloarenes

Match List I with List II I - Bromopropane is reacted with reagents in List I to give product in List II LIST I-Reagent LIST II - Product A ${\text{KOH}}( {\text{alc}})$ I. Nitrile B. ${\text{KCN}}$ (alc) II. Ester C. ${\text{AgNO}}_2$ III. Alkene D. ${\text{H}}_3 {\text{CCOOAg}}$ IV. NitroalkaneChoose the correct answer from the options given below : [13-Apr-2023 shift 2]

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📖 Explanation

${\text{CH}}_3- {\text{CH}}_2- {\text{CH}}_2- {\text{Br}}+ {\text{KOH}}( {\text{Alc}}) \to {\text{CH}}_3-\;{ {\text{CH}}}_{\;\text{ (Allene) }\;}= {\text{CH}}_2$ ${\text{CH}}_3- {\text{CH}}_2- {\text{CH}}_2- {\text{Br}}+ {\text{KCN}}( {\text{Alc}}) \to {\text{CH}}_3- {{\text{CH}}_2- {\text{CH}}_2}_{(\text{Nitrle})}- {\text{CN}}$ $\; {\text{CH}}_3- {\text{CH}}_2- {\text{CH}}_2- {\text{Br}}+ {\text{AgNO}}+ {\text{CH}}_3-\;{ {\text{CH}}_2}_{\;\text{ (Niroallkane) }\;}- {\text{CH}}_2- {\text{NO}}_2+ {\text{AgBr}} ↓$ $\; {\text{CH}}_3- {\text{CH}}_2- {\text{CH}}_2- {\text{Br}}+ {\text{CH}}_3- {\text{COOAg}} \to {\text{CH}}_3- {\text{COO}}- {\text{CH}}_2- {{\text{CH}}_2}_{(\text{Ester})}- {\text{CH}}_3+ {\text{AgBr}} ↓$

Q43JEE Main 2023MCQ4Mp Block Elements

The covalency and oxidation state respectively of boron in $[ {\text{BF}}_4] ^{-}$ , are : [13-Apr-2023 shift 2]

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📖 Explanation

Number of covalent bond formed by Boron is 4 Oxidation number of fluorine is -1 , Oxidation number of ${\text{B}}+4 \times (-1)=-1$ , Thus, Oxidation number of ${\text{B}}=+3$

Q44JEE Main 2023MCQ4MChemical Thermodynamics

What happens when methance undergoes combustion in systems A and B respectively? [13-Apr-2023 shift 2]

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Q45JEE Main 2023MCQ4MBiomolecules

The naturally occurring amino acid that contains only one basic functional group in its chemical structure is : [13-Apr-2023 shift 2]

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Q46JEE Main 2023MCQ4MChemical Bonding and Molecular Structure

Given below are two statements :Statement I: ${\text{SO}}_2$ and ${\text{H}}_2 {\text{O}}$ both possess ${\text{V}}$ -shaped structure.Statement II : The bond angle of ${\text{SO}}_2$ less than that of ${\text{H}}_2 {\text{O}}$ In the light of the above statements, choose the most appropriate answer from the options given below : [13-Apr-2023 shift 2]

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📖 Explanation

Both are bent in shape.Bond angle of ${\text{SO}}_2 ( {\ sp}^2)$ is greater than that of ${\text{H}}_2 {\text{O}} ( {\ sp}^3)$ due to higher repulsion of multiple bonds.

Q47JEE Main 2023MCQ4MSolutions

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.Assertion A : The diameter of colloidal particles in solution should not be much smaller than wavelength of light to show Tyndall effect.Reason R: The light scatters in all direction when the size of particles is large enough.In the light of the above statements, choose the correct answer from the options given below : [13-Apr-2023 shift 2]

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Q48JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

Match List I with List II LIST I LIST II A. Weak intermolecular farces of attraction I. Hexamethylenendiamine + adipic B. Hydrogen bonding II. $\;\; {\text{AlEt}}_3+ {\text{TiCl}}_4$ C. Heavily branched polymer III. 2-chloro-1,3-butadiene D. High density polymer IV. Phenol + formaldehydeChoose the correct answer from the options given below : [13-Apr-2023 shift 2]

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📖 Explanation

Hexamethylenediamine on reaction with adipic acid forms Nylon 6, 6 which shows H-bonding due to presence of amide group.- ${\text{AlEt}}_3+ {\text{TiCl}}_4$ is Ziegler-Natta catalyst used to prepare high density polyethylene.- 2-chloro-1, 3-butadiene (chloroprene) is monomer of neoprene which is a rubber (an elastomer)- Phenol - formaldehyde forms Bakelite which is heavily branched (cross-linked) polymer

Q49JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

Given below are two statements :Statement I : Tropolone is an aromatic compound and has $8 \pi$ electrons.Statement II : $\pi$ electrons of $> {\text{C}}= {\text{O}}$ group in tropolone is incolved in aromaticityIn the light of the above statements, choose the correct answer from the options given below : [13-Apr-2023 shift 2]

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📖 Explanation

Tropolone is an aromatic compound and has $8 \pi$ electrons ( $6 \pi {\text{e}}^{-}$ are endocyclic and $2 \pi {\text{e}}^{-}$ are exocyclic) and $\pi$ electrons of ${\text{C}}= {\text{O}}$ group in tropolone is not involved in aromaticity.

Q50JEE Main 2023MCQ4MAlcohols, Phenols and Ethers

Given below are two statements, one is labelled as Assertion $\text{bo} {A}$ and the other is labelled as Reason $\text{bo} {R}$ . Assertion A : Order of acidic nature of the following compounds is A $> {\text{B}} > {\text{C}}$ .Reason R : Fluoro is a stronger electron withdrawing group than Chloro group.In the light of the above statements, choose the correct answer from the options given below : [13-Apr-2023 shift 2]

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📖 Explanation

Acidic strength $\alpha- {\text{I}}$ effect $\alpha \;{ {\text{I}}}/{+ {\text{I}}} \;\text{ effect }\;$ ${\text{F}}, {\text{Cl}}$ exerts -1 effect, Methyl exerts $+ {\text{I}}$ effect, ${\text{C}}$ is least acidic.Among A and B; since inductive effect is distance dependent, Extent of $- {\text{I}}$ effect is higher in A followed by B even though ${\text{F}}$ is stronger electron withdrawing group than ${\text{Cl}}$ . Thus, ${\text{A}}$ is more acidic than ${\text{B}}$ .

Q51JEE Main 2023NAT4Mp Block Elements

If the formula of Borax is ${\text{Na}}_2 {\text{B}}_4 {\text{O}}_{ {\text{x}}}( {\text{OH}})_{ {\text{y}}}$ . ${\text{zH}}_2 {\text{O}}$ , then ${\text{x}}+ {\text{y}}+ {\text{z}}=$ _______. [13-Apr-2023 shift 2]

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📖 Explanation

Formula of borax is ${\text{Na}}_2 {\text{B}}_4 {\text{O}}_5( {\text{OH}})_4 \cdot 8 {\text{H}}_2 {\text{O}}$

Q52JEE Main 2023NAT4MSolutions

Sea water contains $29.25 \% {\text{NaCl}}$ and $19 \% {\text{MgCl}}_2$ by weight of solution. The normal boiling point of the sea water is _______ ${ }^{\circ} {\text{C}}$ (Nearest integer)Assume ${100 \%}$ ionization for both ${\text{NaCl}}$ and ${\text{MgCl}}_2$ Given : ${\text{K}}_{ {\text{b}}} ( {\text{H}}_2 {\text{O}}) =0.52 {\text{K}} {\text{kg}} {\text{mol}}^{-1}$ Molar mass of ${\text{NaCl}}$ and ${\text{MgCl}}_2$ is 58.5 and $95 {\text{g}} {\text{mol}}^{-1}$ respectivley. [13-Apr-2023 shift 2]

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📖 Explanation

$\;\;\text{ Amount of solvent }\;=100-(29.25+19)=51.75 {\text{g}}$ $\Delta {\text{T}}_{ {\text{b}}}\;= [\;\frac{2 \times 29.25 \times 1000}{58.5 \times 51.75}+\;\frac{3 \times 19 \times 1000}{95 \times 51.75}] \times 0.52$ $\Delta {\text{T}}_{ {\text{b}}}\;=16.075$ $\Delta {\text{T}}_{ {\text{b}}}\;= ( {\text{T}}_{ {\text{b}}}) _{\;\text{solution }\;}- ( {\text{T}}_{ {\text{b}}}) _{\;\text{solvent }\;}$ ${(\text{table} {\text{t}})}_{\;\text{solution }\;}\;=100+16.07$ $\;=116.07^{\circ} {\text{C}}$

Q53JEE Main 2023NAT4MIonic Equilibrium

$20 {\text{mL}}$ of $0.1 {\text{M}} {\text{NaOH}}$ is added to $50 {\text{mL}}$ of $0.1 {\text{M}}$ acetic acid solution. The ${\text{pH}}$ of the resulting solution is _______ $\times 10^{-2}$ (Nearest integer) $\;\;\text{ Given : }\; {\text{pKa}} ( {\text{CH}}_3 {\text{COOH}}) =4.76$ $\;\log 2=0.30$ $\;\log 3=0.48$ [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{pH}}= {\text{pKa}}+\log _{10} \;\frac{[\;\text{ salt }\;]}{[\;\text{ acid }\;]}$ $\; {\text{pH}}=4.76+\log _{10} \;\frac{2}{3}$ $\; {\text{pH}}=4.58=458 \times 10^{-2}$

Q54JEE Main 2023NAT4MElectrochemistry

At $298 {\text{K}}$ , the standard reduction potential for ${\text{Cu}}^{2+} / {\text{Cu}}$ electrode is $0.034 {\text{V}}$ .Given : ${\text{K}}_{ {\ sp}} {\text{Cu}}( {\text{OH}})_2=1 \times 10^{-20}$ Take $\;\frac{2.303 R T}{F}=0.059 {\text{V}}$ The reduction potential at ${\text{pH}}=14$ for the above couple is $(-) {\text{x}} \times 10^{-2} {\text{V}}$ . The value of $x$ is ________. [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{Cu}}( {\text{OH}})_2( {\ s}) \;\; {\text{Cu}}^{2+}( {\text{aq}})+2 {\text{OH}}^{-}( {\text{aq}})$ $\; {\text{Ksp}}= \$ [ {\text{Cu}}^{2+}]$ $[ {\text{OH}}^{-}]$ ^2; {\text{pH}}=14 ; {\text{pOH}}=0 ; $[ {\text{OH}}^{-}]$ =1 {\text{M}}; \therefore $[ {\text{Cu}}^{2+}]$ =;{ {\text{Ksp}}}/{[1]^2}=10^{-20} {\text{M}}; {\text{Cu}}^{2+}( {\text{aq}})+2 {\text{e}}^{-} \to {\text{Cu}}( {\ s}); {\text{E}}= {\text{E}}^{\circ}-;\frac{0.059}{2} \log _{10} ;{1}/{[ {\text{Cu}}^{2+}]$ };=0.34-;\frac{0.059}{2} \log _{10} ;\frac{1}{10^{-20}};=-0.25=-25 × 10^{-2}$

Q55JEE Main 2023NAT4MSolid State

Sodium metal crystallizes in a body centred cubic lattice with unit cell edge length of $4 Å$ . The radius of sodium atom is ______ $\times 10^{-1} Å$ (Nearest integer) [13-Apr-2023 shift 2]

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📖 Explanation

$\;{\sqrt{3}} {\text{a}}=4 {\text{r}}$ $\;{\sqrt{3}} \times 4=4 {\text{r}}$ $\; {\text{r}}=1.732 Å$ $\;=17.32 \times 10^{-1}$

Q56JEE Main 2023NAT4MChemical Kinetics

$A(g) \to 2 B(g)+C(g)$ is first order reaction. The initial pressure of the system was found to be $800 {\text{mm}} {\text{Hg}}$ which increased to $1600 {\text{mm}} {\text{Hg}}$ after $10 {\min}$ . The total pressure of the system after $30 {\min}$ will be _____ ${\text{mm}}$ Hg. (Nearest integer) [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{t}}_{\;\frac{1}{2}}=10 \;\text{ minutes }\;$ $\; ( {\text{P}}_{ {\text{A}}}) _{30 \min }= ( {\text{P}}_{ {\text{A}}}) _0 (\;\frac{1}{2}) ^{30 / 10}$ $\; ( {\text{P}}_{ {\text{A}}}) _{30 \min }=100 {\text{mm}} {\text{Hg}}$

Q57JEE Main 2023NAT4MSome Basic Concepts of Chemistry

$1 {\text{g}}$ of a carbonate $( {\text{M}}_2 {\text{CO}}_3)$ on treatment with excess ${\text{HCl}}$ produces $0.01 {\text{mol}}$ of ${\text{CO}}_2$ . The molar mass of ${\text{M}}_2 {\text{CO}}_3$ is _______ ${\text{g}} {\text{mol}}^{-1}$ . (Nearest integer) [13-Apr-2023 shift 2]

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📖 Explanation

$\;{{ {\text{M}}_2 {\text{CO}}}_{{1 {\text{gm}}}}_3}+\;{2 {\text{HCl}}}_{\;\text{ Excess }\;} \to \;{2 {\text{MCl}}}_{0.02 \;\text{ mole }\;}+ {\text{H}}_2 {\text{O}}+\;{ {\text{CO}}_2}_{\;\text{ 0.01mole }\;}$ From principle of atomic conservation of carbon atom, Mole of ${\text{M}}_2 {\text{CO}}_3 \times 1=$ Mole of ${\text{CO}}_2 \times 1$ $\;\;{1 {\text{gm}}}/{\;\text{ molar mass of }\; {\text{M}}_2 {\text{CO}}_3}=0.01 \times 1$ $\; \therefore \;\text{ Molar mass of }\; {\text{M}}_2 {\text{CO}}_3=100 {\text{gm}} / {\text{mole}}$

Q58JEE Main 2023NAT4MPractical Organic Chemistry

$0.400 {\text{g}}$ of an organic compound $( {\text{X}})$ gave $0.376 {\text{g}}$ of ${\text{AgBr}}$ in Carius method for estimation of bromine. % of bromine in the compound $( {\text{X}})$ is ______ . (Given: Molar mass ${\text{AgBr}}=188 {\text{g}} {\text{mol}}^{-1}, {\text{Br}}=80 {\text{g}} {\text{mol}}^{-1}$ ) [13-Apr-2023 shift 2]

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📖 Explanation

$\;\;\text{ mole of }\; {\text{AgBr}}=\;\frac{0.376}{188}$ $\;\;\text{ mole of }\; {\text{Br}}^{-}=\;\text{mole of }\; {\text{AgBr}}=\;\frac{0.376}{188}$ $\;\;\text{ mass of }\; {\text{Br}}^{-}=\;\frac{0.376}{188} \times 80$ $\;\% \;\text{ of }\; {\text{Br}}^{-}=\;\frac{0.376 \times 80}{188 \times 0.4} \times 100=40 \%$

Q59JEE Main 2023NAT4MStructure of Atom

The orbital angular momentum of an electron in $3 {\ s}$ orbital is $\;{ {\text{xh}}}/{2 \pi}$ . The value of ${\text{x}}$ is _______ (nearest integer) [13-Apr-2023 shift 2]

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📖 Explanation

Orbital angular momentum $={\sqrt{1(1+1)}} \;{ {\text{h}}}/{2 \pi}$ Value of 1 for $s=0$

Q60JEE Main 2023NAT4MRedox Reactions

See the following chemical reaction: ${\text{Cr}}_2 {\text{O}}_7^{2-}+ {\text{XH}}^{+}+6 {\text{Fe}}^{2+} \to {\text{YCr}}^{3+}+6 {\text{Fe}}^{3+}+ {\text{ZH}}_2 {\text{O}}$ The sum of ${\text{X}}, {\text{Y}}$ and ${\text{Z}}$ is _______ [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{Cr}}_2 {\text{O}}_7^{2-}+14 {\text{H}}^{+}+6 {\text{Fe}}^{2+} \to 6 {\text{Fe}}^{3+}+2 {\text{Cr}}^{3+}+7 {\text{H}}_2 {\text{O}}$ $\; {\text{x}}=14$ $\; {\text{y}}=2$ $\; {\text{z}}=7$ $\;\;\text{ Hence }\;(x+y+z)=14+2+7=23$

Mathematics30 questions

Q61JEE Main 2023MCQ4MArea Under The Curves

The area of the region $\{(x \cdot y): x^2 \le y \le |x^{2-}4 |, y \ge 1\}$ is: [13-Apr-2023 shift 2]

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Q62JEE Main 2023MCQ4MLimits, Continuity and Differentiability

If $\lim _{x \to 0} \;{e^{a x}-\cos (b x)-\frac{c x e^{-c x}}{2}}/{1-\cos (2 x)}=17$ , then $5 a^{2+}b^2$ is equal to [13-Apr-2023 shift 2]

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📖 Explanation

$\lim _{x \to 0} \;{e^{a x}-\cos b x-\frac{c x e^{-c x}}{2}}/{1-\cos 2 x}=17$ On expansion $\lim _{x \to 0} \;\frac{ (1+a x+\frac{(a x)^2}{2 !}+. . .) - (1-\frac{(b x)^2}{2 !}+. . .) -\frac{c x}{2} (1-c x+\frac{(c x)^2}{2 !}) }{ (\;\frac{1-\cos 2 x}{(2 x)^2}) \times (2 x)^2}=17$ $\lim _{x \to 0} \;\frac{x (a-\frac{c}{2}) +x^2 (\frac{a^2}{2}+\frac{b^2}{2}+\frac{c^2}{2}) }{\frac{1}{2} (4 x^2) }=17$ For limit to be exist ${\text{a}}-\;{ {\text{c}}}/{2}=0 \Rightarrow {\text{c}}=2 {\text{a}}$ $\;\Rightarrow \;{\;{ {\text{a}}^2}/{2}+{ {\text{b}}^2}/{2}+{ {\text{c}}^2}/{2}}/{2}=17$ $\;\Rightarrow \;{ {\text{a}}^2}/{2}+\;{ {\text{b}}^2}/{2}+\;{4 {\text{a}}^2}/{2}=34$ $\;\Rightarrow 5 {\text{a}}^{2+} {\text{b}}^2=68$

Q63JEE Main 2023MCQ4MThree Dimensional Geometry

The line, that is coplanar to the line $\;{ {\text{x}}+3}/{-3}=\;{ {\text{y}}-1}/{1}=\;{ {\text{z}}-5}/{5}$ , is [13-Apr-2023 shift 2]

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📖 Explanation

Condition of co-planarity ${\begin{vmatrix}{\text{x}}_2- {\text{x}}_1 & {\text{a}}_1 & {\text{a}}_2 \\ {\text{y}}_2- {\text{y}}_1 & {\text{b}}_1 & {\text{b}}_2 \\ {\text{z}}_2- {\text{z}}_1 & {\text{c}}_1 & {\text{c}}_2\end{vmatrix}}=0$ Where a1, b1, c1 are direction cosine of $1^{\;\text{st }\;}$ line and a2, b2, c2 are direction cosine of $2^{\;\text{nd }\;}$ line. Now. Solving optionsPoint $(-3,1,5) &$ point $(-1,2,5)$ $\;(1){\begin{vmatrix}-3 & 1 & 5 \\ 1 & 2 & 5 \\ -2 & -1 & 0\end{vmatrix}}$ $\;=-3(5)-(10)+5(-1+4)$ $\;=-15-10+15=-10$ (2) point $(-1,2,5)$ $\;{\begin{vmatrix}-3 & 1 & 5 \\ -1 & 2 & 5 \\ -2 & -1 & 0\end{vmatrix}}$ $\;=3(5)-(10)+5(1+4)$ $\;-25+5=0$ (3) point $(-1,2,5)$ $\;{\begin{vmatrix}-3 & 1 & 5 \\ -1 & 2 & 4 \\ -2 & -1 & 0\end{vmatrix}}$ $\;-3(4)-(8)+5(1+4)$ $\;-12-8+25=5$ (4) point $(-1,2,5)$ $\;{\begin{vmatrix}-3 & 1 & 5 \\ -1 & 2 & 5 \\ 4 & 1 & 0\end{vmatrix}}$ $\;-3(-5)-(-20)+5(-1-8)$ $\;15+20-45=-10$

Q64JEE Main 2023MCQ4MThree Dimensional Geometry

The plane, passing through the points $(0,-1,2)$ and $(-1,2,1)$ and parallel to the line passing through $(5,1,-7)$ and $(1,-1,-1)$ , also passes through the point [13-Apr-2023 shift 2]

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📖 Explanation

Plane passing through $(0,-1,0)$ and $(-1,2,1)$ Then vector in plane $⟨-1,3,-1⟩$ vector parallel to plane is $⟨ 4,2,-6⟩$ Normal vector to plane $({ {\text{n}}}^{\to })={\begin{vmatrix}{ {\text{i}}^{\^}} & { {\text{j}}^{\^}} & { {\text{k}}^{\^}} \\ -1 & 3 & -1 \\ 4 & 2 & -6\end{vmatrix}}$ $\;={ {\text{i}}^{\^}}(16)-{ {\text{j}}^{\^}}(10)+{ {\text{k}}^{\^}}(-14)$ $\;{ {\text{n}}}^{\to }=⟨ 8,5,7⟩$ Equation of plane $\;8( {\text{x}}-0)+5( {\text{y}}+1)+7( {\text{z}}-2)=0$ $\;\Rightarrow 8 {\text{x}}+5 {\text{y}}+7 {\text{z}}=9$ From given options point $(-2,5,0)$ lies on plane.

Q65JEE Main 2023MCQ4MVector Algebra

Let for a triangle ${\text{ABC}}$ , $\;{ {\text{AB}}}^{\to }=-2 \hat{i}+\hat{j}+3 \hat{k}$ $\;{ {\text{CB}}}^{\to }=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ $\;{ {\text{CA}}}^{\to }=4 \hat{i}+3 \hat{j}+\delta \hat{k}$ If $\delta >0$ and the area of the triangle ${\text{ABC}}$ is $5 {\sqrt{6}}$ , then ${ {\text{CB}}}^{\to } \cdot { {\text{CA}}}^{\to }$ is equal to [13-Apr-2023 shift 2]

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📖 Explanation

$\;{ {\text{CA}}}^{\to }+{ {\text{AB}}}^{\to }={ {\text{CB}}}^{\to }$ $\;⟨ 4,3, \delta ⟩ \cdot +⟨-2,1,3⟩={ {\text{CB}}}^{\to }$ $\;\Rightarrow { {\text{CB}}}^{\to }=⟨ 2,4,3+\delta ⟩$ $\;{ {\text{AB}}}^{\to } \times { {\text{AC}}}^{\to }={\begin{vmatrix}{ {\text{i}}^{\^}} & { {\text{j}}^{\^}} & { {\text{k}}^{\^}} \\ -2 & 1 & 3 \\ -4 & -3 & -\delta \end{vmatrix}}$ $\;={ℓ}^{\^}(9-\delta )-{ {\text{j}}^{\^}}(2 \delta +12)+ {\text{k}}(10)$ $\;| {\text{AB}} \times {\text{AC}}|^2=(9-\delta )^{2+}(2 \delta +12)^{2+}(10)^2$ $\;=5 \delta ^{2+}30 \delta +325$ $\;\;\text{ Area of }\; \Delta {\text{ABC}}=5 {\sqrt{6}}$ $\;\Rightarrow \;\frac{1}{2}|{ {\text{AB}}}^{\to } \times { {\text{AC}}}^{\to }|=5 {\sqrt{6}}$ $\;\Rightarrow |{ {\text{AB}}}^{\to } \times { {\text{AC}}}^{\to }|^2=600$ $\;\Rightarrow 5 \delta ^{2+}30 \delta -275=0$ $\;\Rightarrow {\text{S}}^{2+}6 \delta -55=0$ $\;\Rightarrow (\delta +11)(\delta -5)=0$ $\;\delta =5$ $\;{ {\text{CB}}}^{\to }=< 2,3,8>$ $\;{ {\text{CB}}}^{\to } \cdot { {\text{CA}}}^{\to }=< 2,4,8>.< 4,3,5>$ $\;=8+12+40=60$

Q66JEE Main 2023MCQ4MMatrices and Determinants

Let for $A={\begin{bmatrix}1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2\end{bmatrix}},|A|=2$ . If $|2 {\text{adj}}(2 {\text{adj}}(2 A))|=32^n$ , then $3 n+\alpha$ is equal to [13-Apr-2023 shift 2]

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📖 Explanation

$\;|A|=2$ $\;{\text{adj}}( {\text{kA}})= {\text{k}}^{ {\text{m}}-1} {\text{adjA}} \;\; \{ {\text{m}}=\;\text{ order of matrix }\;\}$ $\;{\text{adj}}(2 {\text{A}})=2^2 {\text{adj}} {\text{A}}=4 {\text{adj}}( {\text{A}})$ $\;{\text{adj}}(2 {\text{adj}}(2 A))={\text{adj}}(8 {\text{adj}} A)$ $\;=8^2 {\text{adj}} {\text{adj}}( {\text{A}})$ $\;|2 {\text{adj}} 2 {\text{adj}}(2 {\text{A}})|={| 2^7 {\text{adj}} {\text{adj}}( {\text{A}}) |}$ $\;= (2^7) ^3| {\text{A}}|^{2^2}$ $\;=2^{21}| {\text{A}}|^4$ $\;=2^{21} \cdot 2^4$ $\;\Rightarrow 2^{25}=(32)^{ {\text{n}}}$ $\;\Rightarrow 2^{25}=2^{5 {\text{n}}}$ $\;\Rightarrow {\text{n}}=5$ $\;| {\text{A}}|=2$ $\;(6-1)-2(2 \alpha-1)+3(\alpha-3)=2$ $\;\Rightarrow 5-4 \alpha+2+3 \alpha-9$ $\;\Rightarrow \alpha=-4$ $\;3 n+\alpha=11$ $\;$

Q67JEE Main 2023MCQ4MInverse Trigonometric Functions

The range of $f(x)=4 sin \;^{-1} (\;\frac{x^2}{x^{2+}1})$ is [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{f}}( {\text{x}})=4 sin \;^{-1} (\;{ {\text{x}}^2}/{1+ {\text{x}}^2})$ $\;0 \le \;{ {\text{x}}^2}/{1+ {\text{x}}^2} < 1$ $\;\Rightarrow 0 \le sin \;^{-1} (\;{ {\text{x}}^2}/{1+ {\text{x}}^2}) < \;\frac{\pi}{2}$ $\;\Rightarrow 0 \le 4 sin \;^{-1} (\;{ {\text{x}}^2}/{1+ {\text{x}}^2}) < 2 \pi$ Range : $[0,2 \pi)$

Q68JEE Main 2023MCQ4MSequences and Series

Let $a_1, a_2, a_3, . . .$ . be a G. P. of increasing positive numbers. Let the sum of its $6^{\;\text{th }\;}$ and $8^{\;\text{th }\;}$ terms be 2 and the product of its $3 r^d$ and $5^{\;\text{th }\;}$ terms be $\;\frac{1}{9}$ . Then $6 (a_2+a_4) (a_4+a_6)$ is equal to [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{a}}_3 \cdot {\text{a}}_5=\;\frac{1}{9}$ $\;\Rightarrow {\text{ar}}^2 \cdot {\text{ar}}^4=\;\frac{1}{9}$ $\;\Rightarrow ( {\text{ar}}^3) ^2=\;\frac{1}{9}$ $\;\Rightarrow {\text{ar}}^3=\;\frac{1}{3}$ . . . (i) $\; {\text{a}}_6+ {\text{a}}_8=2$ $\;\Rightarrow {\text{ar}}^{5+} {\text{ar}}^7=2$ $\;\Rightarrow {\text{ar}}^3 ( {\text{r}}^{2+} {\text{r}}^4) =2$ $\;\Rightarrow \;\frac{1}{3} r^2 (1+r^2) =2$ $\;\Rightarrow {\text{r}}^2 (1+ {\text{r}}^2) =2 \times 3$ $\;\Rightarrow r^2=2 \Rightarrow r={\sqrt{2}}$ $\; {\text{a}}=\;\frac{1}{3} \times \;{1}/{ {\text{r}}^3}$ $\;=\;\frac{1}{3} \times \;{1}/{2 {\sqrt{2}}}=\;{1}/{6 {\sqrt{2}}}$ $\;6 ( {\text{a}}_2+ {\text{a}}_4) ( {\text{a}}_4+ {\text{a}}_6)$ $\;\Rightarrow 6 ( {\text{ar}}+ {\text{ar}}^3) ( {\text{ar}}^{3+} {\text{ar}}^5)$ $\;\Rightarrow 6 (\;{ {\text{ar}}^3}/{ {\text{r}}^2}+\;\frac{1}{3}) (\;\frac{1}{3}+\;\frac{1}{3} {\text{r}}^2) =3$ $\;$

Q69JEE Main 2023MCQ4MMatrices and Determinants

If the system of equations $\;2 x+y-z=5$ $\;2 x-5 y+\lambda z=\mu$ $\;x+2 y-5 z=7$ has infinitely many solutions, then $(\lambda +\mu )^{2+}(\lambda -\mu )^2$ is equal to [13-Apr-2023 shift 2]

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📖 Explanation

$\;\Delta =0$ $\;\Rightarrow {\begin{vmatrix}2 & 1 & -1 \\ 2 & -5 & \lambda \\ 1 & 2 & -5\end{vmatrix}}=0$ $\;\Rightarrow 2(25-2 \lambda )-1(-10-\lambda )-1(4+5)=0$ $\;\Rightarrow 51-3 x=0$ $\;\Rightarrow \lambda =17$ $\;\Delta _x=0$ $\;{\begin{vmatrix}5 & 1 & -1 \\ \mu & -5 & 17 \\ 7 & 2 & -5\end{vmatrix}}=0$ $\;\Rightarrow 5(25-34)-1(-5 \mu -119)-1(2 \mu +35)=0$ $\;\Rightarrow -45+5 \mu +119-2 \mu -35=0$ $\;\Rightarrow 39+3 \mu =0 \Rightarrow \mu =-13$ $\;(\lambda +\mu )^{2+}(\lambda -\mu )^2=4^{2+}(30)^2$ $\;=916$

Q70JEE Main 2023MCQ4MSets and Relations

The statement $( {\text{p}} ∧(∼ {\text{q}})) ∨((∼ {\text{p}}) ∧ {\text{q}}) ∨((∼ {\text{p}}) ∧(∼ {\text{q}}))$ is equivalent to [13-Apr-2023 shift 2]

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📖 Explanation

$( {\text{p}} ∧ ∼ {\text{q}}) ∨(∼ {\text{p}} ∧ {\text{q}}) ∨(∼ {\text{p}} ∧ ∼ {\text{q}})$ ${\text{p}} ∧ ∼ {\text{q}} \Rightarrow$ $∼ {\text{p}} ∧ {\text{q}}=$ $∼ {\text{p}} ∧ ∼ {\text{q}}=$ $\;( {\text{p}} ∧ ∼ {\text{q}}) ∨(∼ {\text{p}} ∧ {\text{q}})(∼ {\text{p}} ∧ ∼ {\text{q}})$ $\;(\alpha, \beta)$ $\;(∼ {\text{p}}) ∨(∼ {\text{q}})$ Plane passing through $(0,-1,2)$ and $(-1,2,1)$ then vector in plane $< -1,3,-1 >$ vector parallel to plane is $⟨ 4,2,-6⟩$ normal vector to plane $L_2$ $\;({ {\text{n}}}^{\to })={\begin{vmatrix}{ {\text{i}}^{\^}} & { {\text{j}}^{\^}} & { {\text{k}}^{\^}} \\ -1 & 3 & -1 \\ 4 & 2 & -6\end{vmatrix}}$ $\;= {\text{i}}(-16)-{ {\text{j}}^{\^}}(10)+{ {\text{k}}^{\^}}(-14)$ $\;{ {\text{n}}}^{\to }=< 8,5,7>$ Equation of plane $\;8( {\text{x}}-0)+5( {\text{y}}+1)+7( {\text{z}}-2)=0$ $\;\Rightarrow 8 {\text{x}}+5 {\text{y}}+7 {\text{z}}=9$ From given options point $(-2,5,0)$ lies on plane.

Q71JEE Main 2023MCQ4MComplex Numbers

Let $S= \{z \in C: \overline{z}=i (z^{2+}{\text{Re}}(\overline{z})) \}$ . Then $sum_{\text{z}\in \text{S}} {|\text{z}|}^{2}$ is equal to [13-Apr-2023 shift 2]

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📖 Explanation

Let $z=x+i y$ $\;\overline{z}=i (z^{2+}{\text{Re}}(\overline{z}))$ $\Rightarrow x-i y=i (x^{2-}y^{2+}2 i x y+x)$ $\Rightarrow x-i y=-2 x y+i (x^{2-}y^{2+}x)$ $\;x+2 x y=0 \;\text{ and }\; x^{2-}y^{2+}x+y=0$ $\;x(1+2 y)=0 \;\text{ and }\; x^{2-}y^{2+}x+y=0$ If $x=0$ then $-y^{2+}y=0$ $\Rightarrow {\text{y}}=1,0$ If $y=\;\frac{-1}{2}$ then $x^{2-}\;\frac{1}{4}+x-\;\frac{1}{2}=0$ $\;\Rightarrow {\text{x}}=-\;\frac{3}{2}, \;\frac{1}{2}$ $\;= \{0+ {\text{i}} 0,0+ {\text{i}},-\;\frac{3}{2}-\;\frac{1}{2} {\text{i}}, \;\frac{1}{2}-\;\frac{1}{2} {\text{i}}\}$ $\;sum_{{ {\text{Z}} \in {\text{S}}}| {\text{Z}}|}^{2=0+1+\;}\frac{9}{4}+\;\frac{1}{4}+\;\frac{1}{4}+\;\frac{1}{4}=4$

Q72JEE Main 2023MCQ4MQuadratic Equation and Inequalities

Let $\alpha, \beta$ be the roots of the equation $x^{2-}{\sqrt{2}} x+2=0$ , Then $\alpha^{14}+\beta^{14}$ is equal to [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{x}}^{2-}{\sqrt{2}} {\text{x}}+2=0$ $\; {\text{x}}=\;{{\sqrt{2}} \pm {\sqrt{-6}}}/{2}$ $\;={\sqrt{2}} (\;{1 \pm {\text{i}} {\sqrt{3}}}/{2})$ $\;=-{\sqrt{2}} \omega ,-{\sqrt{2}} \omega ^2$ $\;\Rightarrow \alpha-{\sqrt{2}} \omega , \beta=-{\sqrt{2}} \omega ^2$ $\; \;\; \alpha^{14}+\beta^{14}=2^7 (\omega ^{14}+\omega ^{28}) =2^7 (\omega ^{2+}\omega ) =-128$

Q73JEE Main 2023MCQ4MVector Algebra

Let $|\vec{a}|=2,|\vec{b}|=3$ and the angle between the vectors $\vec{a}$ and $\vec{b}$ be $\;\frac{\pi}{4}$ . Then $|(\vec{a}+2 \vec{b}) \times (2 \vec{a}-3 \vec{b})|^2$ is equal to [13-Apr-2023 shift 2]

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📖 Explanation

$\;\cos (\;\frac{\pi}{4}) =\;\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$ $\;\Rightarrow \;{1}/{{\sqrt{2}}}=\;\frac{\vec{a} \cdot \vec{b}}{(2)(3)} \Rightarrow \vec{a} \cdot \vec{b}=3 {\sqrt{2}}$ Let $\vec{p}=\vec{a}+2 \vec{b}$ $\;{ {\text{q}}}^{\to }=2 { {\text{a}}}^{\to }-3 { {\text{b}}}^{\to }$ $\;|{ {\text{p}}}^{\to }|^2=|{ {\text{a}}}^{\to }|^{2+}4|{ {\text{b}}}^{\to }|^{2+}4({ {\text{a}}}^{\to } \cdot { {\text{b}}}^{\to })$ $\;=4+36+12 {\sqrt{2}}$ $\;=40+12 {\sqrt{2}}$ $\;|{ {\text{q}}}^{\to }|^2=4|{ {\text{a}}}^{\to }|^{2+}9|{ {\text{b}}}^{\to }|^{2-}12({ {\text{a}}}^{\to }-{ {\text{b}}}^{\to })$ $\;=16+81-36 {\sqrt{2}}$ $\;=97-36 {\sqrt{2}}$ $\;{ {\text{p}}}^{\to } \cdot { {\text{q}}}^{\to }=2|{ {\text{a}}}^{\to }|^{2-}6|{ {\text{b}}}^{\to }|^{2+}{ {\text{a}}}^{\to } \cdot { {\text{b}}}^{\to }$ $\;=8-54+3 {\sqrt{2}}$ $\;=-46+3 {\sqrt{2}}$ $\;|{ {\text{p}}}^{\to } \times { {\text{q}}}^{\to }|=(|{ {\text{p}}}^{\to }||{ {\text{q}}}^{\to }|)^{2-}({ {\text{p}}}^{\to } \cdot { {\text{q}}}^{\to })^2$ $\;=(40+12 {\sqrt{2}}(97-36 {\sqrt{2}}))-(3 {\sqrt{2}}-46)^2$ $\;=(3016-276 {\sqrt{2}})-(2134-276 {\sqrt{2}})$ $\;=882$

Q74JEE Main 2023MCQ4MDefinite Integrals

The value of $\;{e^{-\frac{\pi}{4}}+int_{0}^{\frac{\pi}{4}} e^{-x} \tan ^{50} x \text{dx}}/{int_{0}^{\;\frac{\pi}{4}} e^{-x} (\tan ^{49} x+\tan ^{51} x) \text{dx}}$ is [13-Apr-2023 shift 2]

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📖 Explanation

$\;\;\text{ let }\; I_1=e^{-\pi / 4}+int_{0}^{\pi / 4} e^{-x} \tan ^{50} x \text{dx}$ $\;I_2=int_{0}^{\pi / 4} e^{-x} (\tan ^{49} x+\tan ^{51} x) \text{dx}$ $\;=int_{0}^{\pi / 4} {\text{e}}^{- {\text{x}}} \tan ^{49} {\text{x}} (\ sec ^2 {\text{x}}) {\text{dx}}$ $\;={| {\text{e}}^{- {\text{x}}} \;{\tan ^{50} {\text{x}}}/{50} |}_0^{\pi / 4}+\;\frac{1}{50} int_{0}^{\pi / 4} {\text{e}}^{- {\text{x}}} \tan ^{50} {\text{xdx}}$ $\;=\;{ {\text{e}}^{-\pi / 4}}/{50}+\;\frac{1}{50} int_{0}^{\pi / 4} {\text{e}}^{- {\text{x}}} \tan ^{50} {\text{xdx}}=\;{ {\text{I}}_1}/{50}$ $\;\;\text{ then }\; \;{ {\text{I}}_1}/{ {\text{I}}_2}=50$

Q75JEE Main 2023MCQ4MBinomial Theorem

The coefficient of $x^5$ in the expansion of $(2 x^{3-}\;\frac{1}{3 x^2}) ^5$ is [13-Apr-2023 shift 2]

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📖 Explanation

general term for $(2 x^{3-}\;\frac{1}{3 x^2}) ^5$ $\; {\text{T}}_{ {\text{r}}+1}={ }^5 {\text{C}}_{ {\text{r}}} (-\;{1}/{3 {\text{x}}^2}) ^{ {\text{r}}} (2 {\text{x}}^3) ^{5- {\text{r}}}$ $\; \;\; { }^5 {\text{C}}_{ {\text{r}}}(-1)^{ {\text{r}}} 3^{- {\text{r}}} 2^{5- {\text{r}}} \cdot {\text{x}}^{15-5 {\text{r}}}$ $\;15-5 {\text{r}}=5 \Rightarrow {\text{r}}=2$ Coeff. Of $x^5={ }^5 {\text{C}}_2(-1)^2 3^{-2} 2^3$ $=10 \times \;\frac{1}{9} \times 8$ $=\;\frac{80}{9}$

Q76JEE Main 2023MCQ4MProbability

The random variable ${\text{X}}$ follows binomial distribution ${\text{B}}( {\text{n}}, {\text{p}})$ , for which the difference of the mean and the variance is 1 . If $2 {\text{P}}(x=2)=3 {\text{P}}( {\text{x}}=1)$ , then ${\text{n}}^2 {\text{P}}( {\text{X}}>1)$ is equal to [13-Apr-2023 shift 2]

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📖 Explanation

$\;2 {\text{P}}( {\text{x}}=2)=3 {\text{P}}( {\text{x}}=1)$ $\;2 \times { }^{ {\text{n}}} {\text{c}}_2 {\text{P}}^2(1- {\text{P}})^{ {\text{n}}-2}=3^{ {\text{n}}} {\text{c}}_1 {\text{P}}^1(1- {\text{P}})^{ {\text{n}}-1}$ $\;\Rightarrow 2 \;{ {\text{n}}( {\text{n}}-1)}/{2} \cdot {\text{P}}=3 {\text{n}}(1- {\text{P}})$ $\;\Rightarrow ( {\text{n}}-1) {\text{P}}=3(1- {\text{P}}) ⋯( {\text{i}})$ $\; {\text{nP}}- {\text{nPq}}=1$ $\;\Rightarrow {\text{nP}}- {\text{nP}}(1- {\text{p}})=1$ $\;\Rightarrow {\text{nP}}=1 \Rightarrow {\text{n}}=\;{1}/{ {\text{p}}^2}$ $\;\Rightarrow {\text{put}} \;\text{ \in equ (i) }\;$ $\; (\;{1}/{ {\text{p}}^2}-1) {\text{P}}=3(1- {\text{p}})$ $\;\;{1}/{ {\text{p}}}- {\text{P}}=3-3 {\text{P}}$ $\;\Rightarrow 1- {\text{P}}^2=3 {\text{P}}-3 {\text{p}}^2$ $\;\Rightarrow 2 {\text{P}}^{2-}3 {\text{P}}+1=0$ $\;\Rightarrow 2 {\text{P}}^{2-}2 {\text{P}}- {\text{P}}+1=0$ $\;\Rightarrow 2 {\text{P}}( {\text{P}}-1)-1( {\text{P}}-1)=0$ $\;\Rightarrow {\text{P}}=\;\frac{1}{2}, {\text{P}}=1\{{\text{Re}} j e c t e d\}$ $\; {\text{n}}=\;\frac{1}{(1 / 2)^2}=4$ $\; {\text{n}}^2 {\text{P}}( {\text{x}} > 1)= {\text{n}}^2 {\text{p}}(1- {\text{P}}( {\text{x}}=0)- {\text{P}}( {\text{x}}=1))$ $\;\Rightarrow {\text{n}}^2 {\text{P}} (1-(1- {\text{P}})^{ {\text{n}}}- {\text{nP}}(1- {\text{P}})^{ {\text{n}}-1})$ $\;\Rightarrow (4)^2 (1- (\;\frac{1}{2}) ^{4-}4 (\;\frac{2}{2}) ^4)$ $\;\Rightarrow 16-1-4=11$

Q77JEE Main 2023MCQ4MCircles

Let the centre of a circle ${\text{C}}$ be $(\alpha, \beta)$ and its radius ${\text{r}}< 8$ . Let $3 {\text{x}}+4 {\text{y}}=24$ and $3 {\text{x}}-4 {\text{y}}=32$ be two tangents and $4 x+3 y=1$ be a normal to $C$ . Then $(\alpha-\beta+r)$ is equal to [13-Apr-2023 shift 2]

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📖 Explanation

Centre lies on normal $\Rightarrow 4 \alpha+3 \beta=1$ Distance of $(\alpha, \beta)$ from $L_1$ and $L_2$ are equal ${| \;\frac{3 \alpha+4 \beta-24}{5} |}={| \;\frac{3 \alpha-4 \beta-32}{5} |}$ $\;3 \alpha+4 \beta-24=-3 \alpha+4 \beta+32$ $\;\Rightarrow 6 \alpha=56$ $\;\Rightarrow \alpha=\;\frac{28}{3}, \beta=\;\frac{-109}{3}$ $\;r=\sqrt{ (\;\frac{28}{3}-4) ^{2+} (\;\frac{-109}{3}+5) ^2} > 8$ $\;\;\text{ (reject) }\;$ $\;3 \alpha+4 \beta-24=-3 \alpha-4 \beta-32$ $\;8 \beta=-8$ $\;\beta=-1, \alpha=1$ $\;\gamma ={\sqrt{(4-1)^{2+}(-5+1)^2}}=5$ $\;\alpha-\beta+\gamma =7$

Q78JEE Main 2023MCQ4MThree Dimensional Geometry

Let ${\text{N}}$ be the foot of perpendicular from the point ${\text{P}}(1,-2,3)$ on the line passing through the points $(4,5,8)$ and $(1,-7.5)$ . Then the distance of ${\text{N}}$ from the plane $2 {\text{x}}-2 {\text{y}}+ {\text{z}}+5=0$ is [13-Apr-2023 shift 2]

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📖 Explanation

$\;{ {\text{PN}}}^{\to }=⟨\lambda , 4 \lambda -5, \lambda +2⟩$ $\;{ {\text{PN}}}^{\to } .< 1,4,1>=0$ $\;\Rightarrow \lambda +16 \lambda -20+\lambda +2=0$ $\;\Rightarrow \lambda =1$ $\; {\text{N}}(2,-3,6)$ Distance of ${\text{N}}$ from $2 {\text{x}}-2 {\text{y}}+ {\text{z}}+5=0$ is $\; {\text{d}}={| \;{2(2)-2(-3)+6+5}/{{\sqrt{2^{2+}(-2)^{2+}(1)^2}}} |}$ $\;={| \;\frac{21}{3} |}=7$

Q79JEE Main 2023MCQ4MPermutations and Combinations

All words, with or without meaning, are made using all the letters of the word MONDAY. These words are written as in a dictionary with serial numbers. The serial number of the word MONDAY is [13-Apr-2023 shift 2]

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📖 Explanation

$\;\;\text{ Rank }\;=120+120+24+24+24+6+6+2+1$ $\; \;\; =327$

Q80JEE Main 2023MCQ4MStraight Lines and Pair of Straight Lines

Let $(\alpha, \beta)$ be the centroid of the triangle formed by the lines $15 x-y=82,6 x-5 y=-4$ and $9 x+4 y=17$ . Then $\alpha+2 \beta$ and $2 \alpha-\beta$ are the roots of the equation [13-Apr-2023 shift 2]

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📖 Explanation

$\;\;\text{ Centroid }\;(\alpha, \beta)= (\;\frac{6+1+5}{3}, \;\frac{8-7+2}{3}) =(4,1)$ $\;\alpha+2 \beta=4+2=6$ $\;2 \alpha-\beta=8-1=7$ Quadratic equation $\;x^{2-}(6+7) x+(6 \times 7)=0$ $\;\Rightarrow x^{2-}13 x+42=0$

Q81JEE Main 2023NAT4MSets and Relations

Let ${\text{A}}=\{-4,-3,-2,0,1,3,4\}$ and ${\text{R}}= \{( {\text{a}}, {\text{b}}) \in {\text{A}} \times {\text{A}}: {\text{b}}=| {\text{a}}|.$ or $^2= {\text{a}}+1\}$ be a relation on ${\text{A}}$ . Then the minimum number of elements, that must be added to the relation ${\text{R}}$ so that it becomes reflexive and symmetric, is ________ [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{R}}=[(-4,4),(-3,3),(3,-2),(0,1),(0,0),(1,1),$ $\;(4,4),(3,3)\}$ For reflexive, add $\Rightarrow (-2,-2),(-4,-4),(-3,-3)$ For symmetric, add $\Rightarrow (4,-4),(3,-3),(-2,3),(1,0)$

Q82JEE Main 2023NAT4MDefinite Integrals

Let $f_n=int_{0}^{\;\frac{\pi}{2}} (sum_{k=1}^{n} sin \;^{k-1} x) (sum_{k=1}^{n}(2 k-1) sin \;^{k-1} x) \cos x \text{dx}, n \in N$ . Then $f_{21}-f_{20}$ is equal to _________ [13-Apr-2023 shift 2]

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📖 Explanation

$f_n=int_{0}^{\pi / 2} (sum_{k=1}^{n} sin \;^{k-1} x) (sum_{k=1}^{n}(2 k-1) sin \;^{k-1} x) \cos x \text{dx}$ $\; sin \;x=t$ $\;\cos x \text{dx}=d t$ $f_n int_{0}^{1} (sum_{{k=1}^n t}^{k-1}) (sum_{{k=1}^n(2 k-1) t}^{k-1}) \text{dt}$ $\;=int_{0}^{1} (1+t+t^2 . . . t^n) (1+3 t+5 t^{2+}. . . .+(2 n+1) t^n) \text{dt}$ $\;=int_{0}^{1} (1+t+t^{2+}. . . t^{n-1}) (1+3 t+5 t^{2+}. . .+(2 n-1) t^{n-1}) \text{dt}$ $\;+int_{0}^{1} (1+3 t+5 t^{2+}. . .(2 n+1)) t^n \text{dt}$ $\;+int_{0}^{1} (1+t+t^{2+}. . .+t^{n-1}) (2 n+1) t^n \text{dt}$ $\;f_{n+1}-f_n=int_{0}^{1} (1+3 t+5 t^{2+}. . .+(2 n+1) t^n) t^n \text{dt}$ $\;+int_{0}^{1} (1+t+t^{2+}. . . t^{n+1}) ((2 n+1) t^n) \text{dt}$ put $n=20$ $\;f_{21}-f_{20}=int_{0}^{1} (1+3 t+5 t^2 . . . 41 \cdot t^{20}) t^{20} \text{dt}+int_{0}^{1} (1+t+t^2 . . . t^{19}) (41 \cdot t^{20}) \text{dt}$ $\;= (\;\frac{1}{21}+\;\frac{3}{22}+\;\frac{5}{23}+. . .+\;\frac{39}{40}+\;\frac{41}{41}) + (\;\frac{41}{21}+\;\frac{41}{22}+\;\frac{41}{40})$ $\;=\;\frac{1+41}{21}+\;\frac{3+41}{22}+. . .+\;\frac{39+41}{40}+1=40+1=41$

Q83JEE Main 2023NAT4MDifferential Equations

If $y=y(x)$ is the solution of the differential equation $\;\frac{d y}{d x}+\;\frac{4 x}{ (x^{2-}1) } y=\;{x+2}/{ (x^{2-}1) ^{\;\frac{5}{2}}}, x > 1$ such that ${\text{y}}(2)=\;\frac{2}{9} \log _{ {\text{e}}}(2+{\sqrt{3}})$ and ${\text{y}}({\sqrt{2}})=\alpha \log _{ {\text{e}}}({\sqrt{\alpha}}+\beta)+\beta-{\sqrt{\gamma }}, \alpha, \beta, \gamma , \in {\text{N}}$ , then $\alpha \beta \gamma$ is equal to _______ [13-Apr-2023 shift 2]

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📖 Explanation

given differential equation $\;\frac{d y}{d x}+\;\frac{4 x}{ (x^{2-}1) } y=\;\frac{x+2}{ (x^{2-}1) ^{5 / 2}}$ is linear D.E. $\;\;\text{ I.F. }\;=int_{e \;\frac{4 x}{x}^{2-1}} \text{dx}=e_e 2 \ln (x^{2-}1) =e_e \ln (x^{2-}1) ^2= (x^{2-}1) ^2$ $\;y (x^{2-}1) ^2=\int \;\frac{x+2}{ (x^{2-}1) ^{5 / 2}} (x^{2-}1) ^2 \text{dx}$ $\;=\int \;{x}/{{\sqrt{x^{2-}1}}} \text{dx}+\int \;{2 \text{dx}}/{{\sqrt{x^{2-}1}}}$ $\;={\sqrt{x^{2-}1}}+2 \ln \$ [x+{\sqrt{x^{2-}1}}]$ +C;;\text{ put }; y(2)=;\frac{2}{9} \ln (2+{\sqrt{3}});;\frac{2}{9} \ln (2+{\sqrt{3}})(9)={\sqrt{3}}+2 \ln [2+{\sqrt{3}}]+C;=C=-{\sqrt{3}};;\text{ put }; x={\sqrt{2}};y=1+2 \ln [{\sqrt{2}}+1]-{\sqrt{3}};\alpha=2, \beta=1=\gamma =3;\alpha \beta \gamma =2(1)(3)=6$

Q84JEE Main 2023NAT4MPermutations and Combinations

Total numbers of 3-digit numbers that are divisible by 6 and can be formed by using the digits 1, 2, 3, 4, 5 with repetition, is _______. [13-Apr-2023 shift 2]

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📖 Explanation

a b 2 $\;( {\text{a}}, {\text{b}})=(1,3),(3,1),(2,2),(2,5),(5,2),(3,4),(4,3),(5,5)$ $\;=8 \;\text{ numbers }\;$ a b 4 $\;( {\text{a}}, {\text{b}})=(1,1),(1,4),(4,1),(2,3),(3,2)$ $\;(4,4),(3,5),(5,3)=8 \;\text{ numbers }\;$ $\;\;\text{ total }\; 8+8=16$

Q85JEE Main 2023NAT4MSequences and Series

The remainder, when $7^{103}$ is divided by 17 , is ______. [13-Apr-2023 shift 2]

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📖 Explanation

$\;7^{103}=7.7^{102}$ $\;=7 (7^2) ^{51}$ $\;=7(51-2)^{51} \to \;\text{ remainder }\;=7(-2)^{51}$ $\;-7 (2^3) (16)^{12}=-56(17-1)^{12} \to \;\text{ remainder }\;=-56(-1)^{12}$ $\;\;\text{ Remainder }\;=-56+17 {\text{k}}$ $\;=-56+68$ $\;=12$

Q86JEE Main 2023NAT4MDifferentiation

Let $f(x)=sum_{{k=1}}^{{10}} k x^k, x \in R$ If $2 f'(2)-f^{'}(2)=119(2)^n+1$ then $n$ is equal to _______ [13-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{f}}( {\text{x}})=sum_{{ {\text{k}}=1}}^{{10}} {\text{kx}} {\text{x}}^{ {\text{k}}}$ $\; \;\; \Rightarrow {\text{f}}( {\text{x}})= {\text{x}}+2 {\text{x}}^{2+}3 {\text{x}}^{3+}⋯+9 {\text{x}}^{9+}10 {\text{x}}^{10}-( {\text{i}})$ $\; {\text{xf}}( {\text{x}})= {\text{x}}^{2+}2 {\text{x}}^{3+}. . .+9 {\text{x}}^{10}+10 {\text{x}}^{11} . . . \;\text{ (ii) }\;$ $\; \;\; "( {\text{i}})-( {\text{ii}})^1$ $\; {\text{f}}( {\text{x}})(1- {\text{x}})= {\text{x}}+ {\text{x}}^{2+} {\text{x}}^{3+}⋯+ {\text{x}}^{10}-10 {\text{x}}^{11}$ $\; {\text{f}}( {\text{x}})(1- {\text{x}})=\;{ {\text{x}} (1- {\text{x}}^{10}) }/{1- {\text{x}}}-10 {\text{x}}^{11}$ $\; {\text{f}}( {\text{x}})=\;{ {\text{x}} (1- {\text{x}}^{10}) }/{(1- {\text{x}})^2}-\;{10 {\text{x}}^{11}}/{(1- {\text{x}})}$ $\; {\text{f}}(2)=2+ {\text{g}}(2)^{11}$ $\;(1- {\text{x}})^2 {\text{f}}( {\text{x}})= {\text{x}} (1- {\text{x}}^{10}) -10 {\text{x}}^{11}(1- {\text{x}})$ $\;\;\text{ diff. w.r.t. }\; {\text{x}}$ $\;(1- {\text{x}})^2 {\text{f}}^{'}(2)+ {\text{f}}(2) 2(1- {\text{x}})(-1)$ $\;= {\text{x}} (-10 {\text{x}}^9) + (1- {\text{x}}^{10}) -10 {\text{x}}^{11}(-1)-(1- {\text{x}})(110) {\text{x}}^{10}$ $\;\;\text{ put }\; x=2$ $\;f^{'}(2)+f(2)(2)=-10(2)^{10}+1-2^{10}+10(2)^{11}-110(2)^{10}+110(2)^{11}$ $\;=(-121) 2^{10}+(120) 2^{11}+1$ $\;=2^{10}(240-121)+1$ $\;=119(2)^{10}+1$ $\; {\text{n}}=10$

Q87JEE Main 2023NAT4MInverse Trigonometric Functions

For $x \in (-1,1]$ , the number of solutions of the equation $sin \;^{-1} x=2 \tan ^{-1} x$ is equal to ________. [13-Apr-2023 shift 2]

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📖 Explanation

$\; sin \;^{-1} x=2 \tan ^{-1} x$ $\; sin \;^{-1} x= sin \;^{-1} (\;\frac{2 x}{1+x^2})$ $\;\Rightarrow x=\;\frac{2 x}{1+x^2}$ $\;\Rightarrow x (\;\frac{2}{1+x^2}-1) =0$ $\;\Rightarrow x=0,1,-1 \;\text{ but }\;-1 \;\text{ is not included. }\;$ Answer 2 solutions

Q88JEE Main 2023NAT4MStatistics

The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively, Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is _______. [13-Apr-2023 shift 2]

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📖 Explanation

$\;\;\text{ Mean }\;=\;\frac{Σ x_i}{10}$ $\;\Rightarrow 50=\;\frac{Σ x_i}{10}$ $\;\Rightarrow Σ x_i=500$ $\;\;\text{ correct }\; \sum x_i=500-45-50+20+25=450$ $\;\sigma^2=\;\frac{Σ x_i^2}{10}=(\overline{x})^2$ $\;\Rightarrow 144=\;\frac{Σ x_i^2}{10}-2500$ $\;\Rightarrow Σ x_i^2=26440$ $\;\;\text{ correct }\; \sum x_i^2=26440-(45)^{2-}(50)^{2+}(20)^{2+}(25)^2$ $\;=26440-2025-2500+400+625$ $\;=22940$ $\;\sigma^2=\;\frac{\;\text{ correct }\; Σ x_i^2}{10}- (\;\frac{\;\text{ correct }\; Σ x_i}{10}) ^2$ $\;=\;\frac{22940}{10}- (\;\frac{450}{10}) ^2=2294-2025$ $\;=269$

Q89JEE Main 2023NAT4MHyperbola

The foci of a hyperbola are $( \pm 2,0)$ and its eccentricity is $\;\frac{3}{2}$ . A tangent, perpendicular to the line $2 x+3 y=6$ , is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the ${\text{x}}$ and ${\text{y}}$ axes are a and b respectively, then $|6 a|+|5 b|$ is equal to _______. [13-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\;2 a e=4$ $\;2 a (\;\frac{3}{2}) =4$ $\;\Rightarrow a=\;\frac{4}{3}$ $\;e^2=1+\;\frac{b^2}{a^2}$ $\;\Rightarrow \;\frac{9}{4}=1+b^2 (\;\frac{9}{16})$ $\;\Rightarrow b^2= (\;\frac{5}{4}) (\;\frac{16}{9}) =\;\frac{20}{9}$ slope of tangent $m=\;\frac{3}{2}$ equation of tangent is $\;y=m x \pm {\sqrt{a^2 m^{2-}b^2}}$ $\;y=\;\frac{3}{2} x \pm \sqrt{\;\frac{16}{9} (\;\frac{9}{4}) -\;\frac{20}{9}}$ $\;\Rightarrow y=\;\frac{3 x}{2} \pm \;\frac{4}{3}$ $\;y=0 \Rightarrow a= \pm \;\frac{8}{9}$ $\;x=0 \Rightarrow b= \pm \;\frac{4}{3}$ $\;|6 a|+|5 b|=\;\frac{16}{3}+\;\frac{20}{3}=12$

Q90JEE Main 2023NAT4MSequences and Series

Let $[\alpha]$ denote the greatest integer $\le \alpha$ . Then $[{\sqrt{1}}]+[{\sqrt{2}}]+[{\sqrt{3}}]+. . .+[{\sqrt{120}}]$ is equal to _______. [13-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{S}}=[{\sqrt{1}}]+[{\sqrt{2}}]+[{\sqrt{3}}]+. . .+[{\sqrt{120}}]$ $\;{[{\sqrt{1}}] \to [{\sqrt{3}}]=1 \times 3}$ $\;{[{\sqrt{4}}] \to [{\sqrt{8}}]=2 \times 5}$ $\;{[{\sqrt{9}}] \to [{\sqrt{15}}]=3 \times 7}$ $\;⋮$ $\;{[{\sqrt{100}}] \to [{\sqrt{120}}]=10 \times 21}$ $\; {\text{S}}=1 \times 3+2 \times 5+3 \times 7+. . .+10 \times 21$ $\;=sum_{{ {\text{r}}=1}}^{{10}} {\text{r}}(2 {\text{r}}+1)$ $\;=2 sum_{{ {\text{r}}=1}}^{{10}} {\text{r}}^{2+}sum_{{ {\text{r}}=1}}^{{10}} {\text{r}}$ $\;=\;\frac{2 \times 10 \times 11 \times 21}{6}+\;\frac{10 \times 11}{2}$ $\;=770+55$ $\;=825$

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