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31 Jan 2023 Shift 2

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Physics30 questions

Q1JEE Main 2023MCQ4MCurrent Electricity

The ${\text{H}}$ amount of thermal energy is developed by a resistor in $10 {\ s}$ when a current of $4 {\text{A}}$ is passed through it. If the current is increased to $16 {\text{A}}$ , the thermal energy developed by the resistor in $10 {\ s}$ will be: [31-Jan-2023 Shift 2]

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📖 Explanation

$H \propto i^2$ for $t=$ constant $\Rightarrow \;\frac{H}{H^{'}}= (\;\frac{4}{16}) ^2$ $\Rightarrow H^{'}=16 H$

Q2JEE Main 2023MCQ4MCircular Motion

A body is moving with constant speed, in a circle of radius $10 {\text{m}}$ . The body completes one revolution in $4 {\ s}$ . At the end of $3 {\text{rd}}$ second, the displacement of body (in ${\text{m}}$ ) from its starting point is: [31-Jan-2023 Shift 2]

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📖 Explanation

$\omega =\;{2 \pi}/{ {\text{T}}}=\;\frac{2 \pi}{4}=\;\frac{\pi}{2} {\text{rad}} / {\ s}$ $\theta=\omega {\text{t}}$ $\theta=\;\frac{\pi}{2} \times 3$ $\theta=\;\frac{3 \pi}{2} {\text{rad}}$ $r=10 {\text{m}}$ $T=4 {\ sec}$ $d={\sqrt{2}}(10) {\text{m}}$

Q3JEE Main 2023MCQ4MRay Optics and Optical Instruments

A microscope is focused on an object at the bottom of a bucket. If liquid with refractive index - is poured inside the bucket, then microscope have to be raised by $30 {\text{cm}}$ to focus the object again. The height of the liquid in the bucket is : [31-Jan-2023 Shift 2]

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📖 Explanation

$\;\text{ Shift }\;= (d-\;\frac{d}{\mu }) =30 {\text{cm}}$ $\Rightarrow d [1-\;\frac{1}{\frac{5}{3}}] =30$ $\Rightarrow d=\;\frac{30 \times 5}{2}=75 {\text{cm}}$

Q4JEE Main 2023MCQ4MCircular Motion

A stone of mass $1 {\text{kg}}$ is tied to end of a massless string of length $1 {\text{m}}$ . If the breaking tension of the string is $400 {\text{N}}$ , then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is: [31-Jan-2023 Shift 2]

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📖 Explanation

$T sin \;\theta=\;\frac{m v^2}{l sin \;\theta}$ $\cos \theta=\;\frac{m g}{T} . . . . . . . . .(i)$ $sin \;^2 \theta=\;\frac{m v^2}{T l} . . . . . . . . . (ii)$ From (i) and (ii), $1= (\;\frac{m g}{T}) ^{2+}\;\frac{m v^2}{T l}$ $\Rightarrow 1= (\;\frac{10}{400}) ^{2+}\;\frac{v^2}{400}$ $\Rightarrow v^2=399.78$ $\Rightarrow v=20 {\text{m}} / {\ s}$

Q5JEE Main 2023MCQ4MWaves

For a solid rod, the Young's modulus of elasticity is $3.2 \times 10^{11} {\text{Nm}}^{-2}$ and density is $8 \times 10^3 {\text{kg}} {\text{m}}^{-3}$ . The velocity of longitudinal wave in the rod will be. [31-Jan-2023 Shift 2]

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📖 Explanation

$v\;=\sqrt{\;\frac{Y}{\rho }}=\sqrt{\;\frac{3.2 \times 10^{11}}{8 \times 10^3}}$ $\;={\sqrt{0.4 \times 10^8}}$ $\;={\sqrt{40 \times 10^6}}$ $\;=6.32 \times 10^3 {\text{m}} / {\ s}$

Q6JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

A long conducting wire having a current I flowing through it, is bent into a circular coil of ${\text{N}}$ turns. Then it is bent into a circular coil of ${\text{n}}$ turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is : [31-Jan-2023 Shift 2]

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📖 Explanation

$I\;=(2 \pi r) n$ $r\;\propto (\;\frac{I}{n})$ $B\;=n (\;\frac{\mu _0 i}{2 r}) \propto (\;\frac{\mu _0 i}{2 L}) n^2$ $\;\frac{B_1}{B_2}\;= (\;\frac{N^2}{n^2})$

Q7JEE Main 2023MCQ4MThermodynamics

Heat energy of $735 {\text{J}}$ is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but do not oscillate. The increase in the internal energy of the gas will be: [31-Jan-2023 Shift 2]

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📖 Explanation

$\Delta Q=n C_P \Delta T=735 {\text{J}}$ $\Rightarrow \;\frac{5 n R \Delta T}{2}=735 {\text{J}}$ $\Delta U=n C_V \Delta T=\;\frac{3}{2}(n R \Delta T)=\;\frac{3}{2} \times \;\frac{2}{5} \times 735$ $=441 {\text{J}}$

Q8JEE Main 2023MCQ4MElectromagnetic Waves

Given below are two statements :Statement I: For transmitting a signal, size of antenna $(l)$ should be comparable to wavelength of signal (at least $l=\;\frac{\lambda }{4}$ in dimension)Statement II: In amplitude modulation, amplitude of carrier wave remains constant (unchanged).In the light of the above statements, choose the most appropriate answer from the options given below

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📖 Explanation

In amplitude modulation frequency of carrier wave remains unchanged.Minimum size of antenna should be $\;\frac{1}{4}$ th of wavelength.

Q9JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by $50 \%$ . The percentage change in voltage sensitivity of the galvanometer will be: [31-Jan-2023 Shift 2]

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📖 Explanation

Current sensitivity $=$ Voltage sensitivity $\times R$ Current sensitivity is made $1.5$ times. $R$ also increase $1.5$ times.Hence voltage sensitivity $=\;\frac{1.5 \times \;\text{ current sensitivity }\;}{1.5 \times R}$ $=$ no change

Q10JEE Main 2023MCQ4MDual Nature of Matter and Radiation

If the two metals ${\text{A}}$ and ${\text{B}}$ are exposed to radiation of wavelength $350 {\text{nm}}$ . The work functions of metals ${\text{A}}$ and ${\text{B}}$ are $4.8 {\text{eV}}$ and $2.2 {\text{eV}}$ . Then choose the correct option [31-Jan-2023 Shift 2]

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📖 Explanation

$\varphi =\;\frac{h c}{\lambda }=\;\frac{1240}{350} {\text{eV}}=3.54 {\text{eV}}$ $\therefore$ Only metal B will emit photoelectron.

Q11JEE Main 2023MCQ4MGravitation

A body weight $W$ , is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be: [31-Jan-2023 Shift 2]

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📖 Explanation

$g_h\;=\;{g}/{ (1+\;{ {\text{h}}}/{ {\text{R}}}) ^2}$ ${\text{h}}\;=9 {\text{R}}$ ${\text{g}}_{ {\text{h}}}\;=\;{ {\text{g}}}/{(1+9)^2}=\;{ {\text{g}}}/{100}$ ${\text{w}}_{ {\text{h}}}\;=\;{ {\text{mg}}}/{100}=\;{ {\text{w}}}/{100}$

Q12JEE Main 2023MCQ4MUnits and Measurements

Match List-I with List-II. List-I List-II A. Angular momentum I. $[ {\text{ML}}^2 {\text{T}}^{-2}].$ B. Torque II. $[ {\text{ML}}^{-2} {\text{T}}^{-2}]$ C. Stress III. $[ {\text{ML}}^2 {\text{T}}^{-1}]$ D. Pressure gradient IV. $[ {\text{ML}}^{-1} {\text{T}}^{-2}].$ Choose the correct answer from the options given below: [31-Jan-2023 Shift 2]

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📖 Explanation

$\vec{L}=\vec{r} \times \vec{p} \Rightarrow [ {\text{L}}]\;= \$ [ {\text{M}}^0 {\text{L}}^1 {\text{T}}^0]$ [ {\text{M}}^1 {\text{L}}^1 {\text{T}}^{-1}]=;{[ {\text{M}}^1 {\text{L}}^2 {\text{T}}^{-1}]$ }\vec{τ}=\vec{r} × \vec{F} \Rightarrow [τ];= $[ {\text{L}}^1]$ [ {\text{MLT}}^{-2}];= [ {\text{ML}}^2 {\text{T}}^{-2}];\text{ Stress }; ≡ ;\text{ Pressure };=;\frac{F}{A} \Rightarrow [;\text{ Stress };]= [ {\text{ML}}^{-1} {\text{T}}^{-2}] $Pressure Gradient$ =;\frac{d P}{d x} \Rightarrow [ $Pressure Gradient$ ]= [ {\text{ML}}^{-2} {\text{T}}^{-2}]$

Q13JEE Main 2023MCQ4MAlternating Current

An alternating voltage source ${\text{V}}=260 sin \;(628 {\text{t}})$ is connected across a pure inductor of $5 {\text{mH}}$ . Inductive reactance in the circuit is: [31-Jan-2023 Shift 2]

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📖 Explanation

$\omega =628 {\text{rad}} / {\ s}$ $X_L=L \omega$ $=5 {\text{mH}} \times 628$ $=3.14 Ω$

Q14JEE Main 2023MCQ4MElectromagnetic Waves

Match List-I with List-II. List-I List-II A. Microwaves I. Physiotherapy B. UV rays II. Treatment of cancer C. Infra-red rays III.Lasik eye surgery D. X-rays IV. Aircraft navigationChoose the correct answer from the option given below: [31-Jan-2023 Shift 2]

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Q15JEE Main 2023MCQ4MAtoms and Nuclei

The radius of electron's second stationary orbit in Bohr's atom is $R$ . The radius of 3 rd orbit will be [31-Jan-2023 Shift 2]

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📖 Explanation

$r \propto \;\frac{n^2}{Z}$ $\;{r_{2 n d}}/{r_{3 {\text{rd}}}}= (\;\frac{n_2}{n_3}) ^2$ $\Rightarrow \;\frac{R}{r_{3 r d}}= (\;\frac{2}{3}) ^2$ $\Rightarrow r_{3 {\text{rd}}}=\;\frac{9}{4} R$ $=2.25 R$

Q16JEE Main 2023MCQ4MProperties of Matter

Under the same load, wire A having length $5.0 {\text{m}}$ and cross section $2.5 \times 10^{-5} {\text{m}}^2$ stretches uniformly by the same amount as another wire B of length $6.0 {\text{m}}$ and a cross section of $3.0 \times 10^{-5} {\text{m}}^2$ stretches. The ratio of the Young's modulus of wire A to that of wire B will be: [31-Jan-2023 Shift 2]

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📖 Explanation

$\Delta ℓ=\;\frac{F ℓ}{S Y}$ $F$ is same for both wire and $\Delta ℓ$ is also same $\;\frac{\Delta ℓ}{F}=\;\frac{ℓ}{S Y}$ $\Rightarrow \;\frac{ℓ_A}{S_A Y_A}=\;\frac{ℓ_B}{S_B Y_B}$ $\Rightarrow \;\frac{5}{2.5 \times Y_A}=\;\frac{6}{3 \times Y_B}$ $\Rightarrow \;\frac{Y_A}{Y_B}=1$

Q17JEE Main 2023MCQ4MElectrostatics

Considering a group of positive charges, which of the following statements is correct? [31-Jan-2023 Shift 2]

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📖 Explanation

$V=\;\frac{Σ K Q_i}{r_i}$ Here, $Q_i$ and $r_i$ are positive. $\therefore V>0$ The correct statement is:(A) Net potential of the system cannot be zero at a point but net electric field can be zero at that point.Explanation:In a group of positive charges, the net potential at a point is the sum of the potentials due to each individual charge. The potential due to a point charge is given by the Coulomb's law, which is non-zero except at the location of the charge itself. Therefore, the net potential due to a group of positive charges can never be zero at a point.On the other hand, the net electric field at a point is the vector sum of the electric fields due to each individual charge. If the charges are arranged in such a way that their electric fields cancel out at a particular point, then the net electric field at that point can be zero, even though the charges are present. This can happen, for example, in a symmetrical arrangement of charges.

Q18JEE Main 2023MCQ4MLaws of Motion

A body of mass $10 {\text{kg}}$ is moving with an initial speed of $20 {\text{m}} / {\ s}$ . The body stops after $5 {\ s}$ due to friction between body and the floor. The value of the coefficient of friction is: (Take acceleration due to gravity ${\text{g}}=10 {\text{ms}}^{-2}$ ) [31-Jan-2023 Shift 2]

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📖 Explanation

$a=-\mu g$ $\because v=u+a t$ $0=20+(-\mu \times 10) \times 5$ $50 \mu =20$ $\mu =\;\frac{2}{5}=0.4$

Q19JEE Main 2023MCQ4MThermodynamics

A hypothetical gas expands adiabatically such that its volume changes from 08 litres to 27 litres. If the ratio of final pressure of the gas to initial pressure of the gas is $\;\frac{16}{81}$ . Then the ratio of $\;{ {\text{Cp}}}/{ {\text{Cv}}}$ will be.

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📖 Explanation

Let $\gamma$ be the ratio of $\;\frac{C_p}{C_v}$ Then for adiabatic process $P V^\gamma =\;\text{ Constant }\;$ $\Rightarrow \;\frac{P_i}{P_f}= (\;\frac{V_f}{V_i}) ^\gamma$ $\Rightarrow \;\frac{81}{16}= (\;\frac{27}{8}) ^\gamma$ $\Rightarrow \gamma =\;\frac{4}{3}$

Q20JEE Main 2023MCQ4MSemiconductor Electronics

Given below are two statements:Statement I: In a typical transistor, all three regions emitter, base and collector have same doping level.Statement II: In a transistor, collector is the thickest and base is the thinnest segment.In the light of the above statements, choose the most appropriate answer from the options given below. [31-Jan-2023 Shift 2]

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Q21JEE Main 2023NAT4MAlternating Current

A series LCR circuit consists of $R=80 Ω {\text{X}}_{ {\text{L}}}=100 Ω$ , and ${\text{X}}_{ {\text{C}}}=40 Ω$ The input voltage is $2500 \cos (100 \pi {\text{t}})$ V. The amplitude of current, in the circuit, is ________ A. [31-Jan-2023 Shift 2]

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📖 Explanation

$\omega =100 \pi$ $\;\text{ So }\; Z={\sqrt{R^{2+} (X_L-X_C) ^2}}$ $={\sqrt{80^{2+}(100-40)^2}}$ $=100 Ω$ $i_0=\;\frac{V_0}{Z}=\;\frac{2500}{100} {\text{A}}=25 {\text{A}}$

Q22JEE Main 2023NAT4MWave Optics

Two light waves of wavelengths 800 and $600 {\text{nm}}$ are used in Young's double slit experiment to obtain interference fringes on a screen placed $7 {\text{m}}$ away from plane of slits. If the two slits are separated by $0.35 {\text{mm}}$ , then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be _______ ${\text{mm}}$ . [31-Jan-2023 Shift 2]

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📖 Explanation

$\omega _1=\;\frac{\lambda _1 D}{d}$ and $\omega _2=\;\frac{\lambda _2 D}{d}$ $\omega _1=16 {\text{mm}}$ and $\omega _2=12 {\text{mm}}$ So ${\text{LCM}} (\omega _1, \omega _2) =48 {\text{mm}}$ So at $48 {\text{mm}}$ distance both bright fringes will be found.

Q23JEE Main 2000NAT4MThermodynamics

A water heater of power $2000 {\text{W}}$ is used to heat water. The specific heat capacity of water is $4200 {\text{J}}$ ${\text{kg}}^{-1} {\text{K}}^{-1}$ . The efficiency of heater is $70 \%$ . Time required to heat $2 {\text{kg}}$ of water from $10^{\circ} {\text{C}}$ to $60^{\circ} {\text{C}}$ is _______ s.(Assume that the specific heat capacity of water remains constant over the temperature range of the water). [31-Jan-2023 Shift 2]

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📖 Explanation

The amount of heat energy required to raise the temperature of a substance can be calculated as: ${\text{Q}}= {\text{m}} \times {\text{c}} \times \Delta {\text{T}}$ where $Q$ is the heat energy required, ${\text{m}}$ is the mass of the substance, ${\text{c}}$ is its specific heat capacity, and $\Delta T$ is the change in temperature.The time required to heat a substance can be calculated as : ${\text{t}}=\;\frac{Q}{P}$ where $t$ is the time required, and $P$ is the power of the heating device.The actual power output of the heating device can be calculated as: $P_{\;\text{actual }\;}=P_{\;\text{input }\;} \times \;\text{ efficiency }\;$ where $P_{-}$ input is the input power to the device and efficiency is the fraction of input power that is actually converted to useful power output.Substituting the given values: ${\text{Q}}=2 {\text{kg}} \times 4200 {\text{J}} / {\text{kg}} / {\text{K}} \times (60-10)=2 {\text{kg}} \times 4200 {\text{J}} / {\text{kg}} / {\text{K}} \times 50 {\text{K}}=4200 \times 50 \times 2 {\text{J}}=420,000 {\text{J}}$ $P_{\;\text{input }\;}=2000 {\text{W}}=2000 {\text{J}} / {\ s}$ ${\text{P}}_{\;\text{actual }\;}=2000 \times 0.7=1400 {\text{J}} / {\ s}$ ${\text{t}}=\;{Q}/{P_{\;\text{actual }\;}}=\;\frac{420,000}{1400} {\text{J}} / {\ s}=300 {\ s}$ So, the time required to heat $2 {\text{kg}}$ of water from $10^{\circ} {\text{C}}$ to $60^{\circ} {\text{C}}$ is approximately $300 {\ s}$ .

Q24JEE Main 2023NAT4MCentre of Mass and Collision

A ball is dropped from a height of $20 {\text{m}}$ . If the coefficient of restitution for the collision between ball and floor is $0.5$ , after hitting the floor, the ball rebounds to a height of ________ ${\text{m}}$ . [31-Jan-2023 Shift 2]

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📖 Explanation

We know, $h^{'}=e^2 h$ $h^{'}=(0.5)^2 \times 20 m=5 m$

Q25JEE Main 2023NAT4MRotational Motion

Two discs of same mass and different radii are made of different materials such that their thicknesses are $1 {\text{cm}}$ and $0.5 {\text{cm}}$ respectively. The densities of materials are in the ratio 3:5. The moment of inertia of these discs respectively about their diameters will be in the ratio of $-$ . The value of $x$ is _______. [31-Jan-2023 Shift 2]

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📖 Explanation

$m=\rho \pi R^2 t$ $\;\text{ so }\; R^2=\;\frac{m}{\rho \pi t}$ $I=\;\frac{m R^2}{4}=\;\frac{m^2}{4 \rho \pi t}$ So $\;\frac{I_1}{I_2}=\;\frac{\rho _2 t_2}{\rho _1 t_1}=\;\frac{5}{3} \times \;\frac{0.5}{1}=\;\frac{5}{6}$ So $x=5$

Q26JEE Main 2023NAT4MAtoms and Nuclei

If the binding energy of ground state electron in a hydrogen atom is $13.6 {\text{eV}}$ , then, the energy required to remove the electron from the second excited state of ${\text{Li}}^{2+}$ will be: $x \times 10^{-1} {\text{eV}}$ . The value of $x$ is _______ . [31-Jan-2023 Shift 2]

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📖 Explanation

${\text{E}}_H=13.6$ ${\text{E}}_{ {\text{Li}}^{2+}}=13.6 \;\frac{Z^2}{n^2}=13.6 \times \;\frac{9}{9}=13.6 {\text{eV}}$ $=136 \times 10^{-1} {\text{eV}}$

Q27JEE Main 2023NAT4MCurrent Electricity

For the given circuit, in the steady state, $| {\text{V}}_{ {\text{B}}}- {\text{V}}_{ {\text{D}}}|=$ [31-Jan-2023 Shift 2]

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📖 Explanation

In steady state, capacitor behaves as an open circuit. Circuit is : $\Rightarrow i_{A B}=\;\frac{6}{3}=2 {\text{A}}$ $i_{A D}=\;\frac{6}{12}=0.5 {\text{A}}$ $\Rightarrow V_B+2 \times 2-10 \times 0.5=V_D$ $\Rightarrow V_B-V_D=1 \;\text{ volt }\;$

Q28JEE Main 2023NAT4MElectrostatic Potential and Capacitance

Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu {\text{F}}$ are individually charged by a $100 {\text{V}}$ D.C. source. Capacitor ${\text{C}}_1$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor ${\text{C}}_2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor ${\text{C}}_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be _______ V. (Assuming Dielectric constant $=10$ ) [31-Jan-2023 Shift 2]

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📖 Explanation

Charge on ${\text{C}}_1= {\text{KCV}}$ And charge on ${\text{C}}_2= {\text{CV}}$ When they are connected in parallel charge will be equally divided so charge on one capacitor is $q=\;\frac{K+1}{2} {\text{CV}}$ So, $V=\;\frac{q}{K C}=\;\frac{K+1}{2 K}=55 {\text{V}}$

Q29JEE Main 2023NAT4MWaves

The displacement equations of two interfering waves are given by $y_1=10 sin \; (\omega t+\;\frac{\pi}{3}) {\text{cm}},$ $y_2=5[ sin \;\omega t+{\sqrt{3}} \cos \omega t] {\text{cm}} \;\text{ respectively. }\;$ The amplitude of the resultant wave is ____ ${\text{cm}}$ . [31-Jan-2023 Shift 2]

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📖 Explanation

Given, $y_1=10 sin \; (\omega t+\;\frac{\pi}{3}) {\text{cm}}$ and $y_2=5( sin \;\omega t+{\sqrt{3}} \cos \omega t)$ $=10 sin \; (\omega t+\;\frac{\pi}{3})$ Thus the phase difference between the waves is 0 . $\;\text{ so }\; A=A_1+A_2=20 {\text{cm}}$

Q30JEE Main 2023NAT4MMotion in a Plane

Two bodies are projected from ground with same speeds $40 {\text{ms}}^{-1}$ at two different angles with respect to horizontal. The bodies were found to have same range. If one of the body was projected at an angle of $60^{\circ}$ , with horizontal then sum of the maximum heights, attained by the two projectiles, is ________ ${\text{m}}$ . (Given ${\text{g}}=10 {\text{ms}}^{-2}$ ) [31-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Since range is same. $\Rightarrow \theta_1+\theta_2=90^{\circ}$ $\Rightarrow \theta_2=30^{\circ}$ $\Rightarrow (H_{\max }) _1+ (H_{\max }) _2=\;\frac{u^2 sin \;^2 \theta_1}{2 g}+\;\frac{u^2 sin \;^2 \theta_2}{2 g}$ $=\;\frac{40^2}{20} (\;\frac{1}{4}+\;\frac{3}{4}) =80 {\text{m}}$

Chemistry30 questions

Q31JEE Main 2023MCQ4MHaloalkanes and Haloarenes

In the following halogenated organic compounds the one with maximum number of chlorine atoms in its structure is : [31-Jan-2023 Shift 2]

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Q32JEE Main 2023MCQ4MIonic Equilibrium

Incorrect statement for the use of indicators in acid-base titration is : [31-Jan-2023 Shift 2]

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Q33JEE Main 2023MCQ4MAlcohols, Phenols and Ethers

Which of the following compounds are not used as disinfectants?(A) Chloroxylenol(B) Bithional(C) Veronal(D) Prontosil(E) TerpineolChoose the correct answer from the options given below: [31-Jan-2023 Shift 2]

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Q34JEE Main 2023MCQ4MHydrocarbons

A hydrocarbon ' ${\text{X}}$ ' with formula ${\text{C}}_6 {\text{H}}_8$ uses two moles of ${\text{H}}_2$ on catalytic hydrogenation of its one mole. On ozonolysis, ' ${\text{X}}$ ' yields two moles of methane dicarbaldehyde. The hydrocarbon ' ${\text{X}}$ ' is: [31-Jan-2023 Shift 2]

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Q35JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

Cyclohexylamine when treated with nitrous acid yields (P). On treating (P) with PCC results in (Q). When (Q) is heated with dil. ${\text{NaOH}}$ we get (R) The final product (R) is: [31-Jan-2023 Shift 2]

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Q36JEE Main 2023MCQ4MSalt Analysis

Given below are two statements:Statement I : Upon heating a borax bead dipped in cupric sulphate in a luminous flame, the colour of the bead becomes green.Statement II : The green colour observed is due to the formation of copper(I) metaborate.In the light of the above statements, choose the most appropriate answer from the options given below : [31-Jan-2023 Shift 2]

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📖 Explanation

(Borax Bead Test)On treatment with metal salt, boric anhydride forms metaborate of the metal which gives different colours in oxidising and reducing flame. For example, in the case of copper sulphate, following reactions occur. ${\text{CuSO}}_4+ {\text{B}}_2 {\text{O}}_3 \;{\to }^{\;\text{ Non -luminous flame }\;}_{(\;\text{ Oxidising }\;)} {\text{Cu}} ( {\text{BO}}_2) _2+ {\text{SO}}_3$ Two reactions may take place in reducing flame (Luminous flame)(i) The blue-green ${\text{Cu}} ( {\text{BO}}_2) _2$ is reduced to colourless cuprous metaborate as: $\;2 {\text{Cu}} ( {\text{BO}}_2) _2+2 {\text{NaBO}}_2+ {\text{C}} \;{ {\to }^{\;\text{ Luminous }\;}_{\;\text{ flame }\;}}$ $\;2 {\text{CuBO}}_2+ {\text{Na}}_2 {\text{B}}_4 {\text{O}}_7+ {\text{CO}}$ (ii) Cupric metaborate may be reduced to metallic copper and bead appears red opaque. $\;2 {\text{Cu}} ( {\text{BO}}_2) _2+4 {\text{NaBO}}_2+2 {\text{C}} \;{ {\to }^{\;\text{ Luminous }\;}_{\;\text{ flame }\;}}$ $\;2 {\text{Cu}}+2 {\text{Na}}_2 {\text{B}}_4 {\text{O}}_7+2 {\text{CO}}$

Q37JEE Main 2023MCQ4MSolutions

Evaluate the following statements for their correctness.(A) The elevation in boiling point temperature of water will be same for $0.1 {\text{M}} {\text{NaCl}}$ and $0.1 {\text{M}}$ urea.(B) Azeotropic mixtures boil without change in their composition(C) Osmosis always takes place from hypertonic to hypotonic solution(D) The density of $32 \% \;\; {\text{H}}_2 {\text{SO}}_4$ solution having molarity $4.09 {\text{M}}$ is approximately $1.26 {\text{g}} {\text{mL}}^{-1}$ (E) A negatively charged sol is obtained when ${\text{KI}}$ solution is added to silver nitrate solution.Choose the correct answer from the options given below : [31-Jan-2023 Shift 2]

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📖 Explanation

(A) $\Delta {\text{T}}_{ {\text{b}}} \propto {\text{i}} \times {\text{c}}$ (B) Azeotropic mixtures have same composition in both liquid and vapour phase.(C) Osmosis always takes place from hypotonic to hypertonic solution.(D) ${\text{M}}=\;\frac{30 \times 10 \times 1.26}{98} \approx 4.09 {\text{M}}$ (E) When ${\text{KI}}$ solutions is added to ${\text{AgNO}}_3$ solution, positively charged solution results due to adsorption of ${\text{Ag}}^{\;\text{ions }\;}$ iom dispersion medium ${\text{AgI}} / {\text{Ag}}^{+}$ Positively charged

Q38JEE Main 2023MCQ4MBiomolecules

Compound ${\text{A}}, {\text{C}}_5 {\text{H}}_{10} {\text{O}}_5$ , given a tetraacetate with ${\text{Ac}}_2 {\text{O}}$ and oxidation of ${\text{A}}$ with ${\text{Br}}_2- {\text{H}}_2 {\text{O}}$ gives an acid, ${\text{C}}_5 {\text{H}}_{10} {\text{O}}_6$ . Reduction of ${\text{A}}$ with ${\text{HI}}$ gives isopentane. The possible structure of ${\text{A}}$ is : [31-Jan-2023 Shift 2]

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📖 Explanation

(i) Formation of tetraacetete with ${\text{Ac}}_2 {\text{O}}$ means compound ${\text{A}}$ has four $- {\text{OH}}$ linkage.Reduction of ${\text{A}}$ with ${\text{HI}}$ gives Isopentane i.e. molecule contains five carbon atom.

Q39JEE Main 2023MCQ4MStructure of Atom

Arrange the following orbitals in decreasing order of energy ?(A) ${\text{n}}=3,1=0, {\text{m}}=0$ (B) ${\text{n}}=4,1=0, {\text{m}}=0$ (C) ${\text{n}}=3,1=1, {\text{m}}=0$ (D) ${\text{n}}=3, {\text{l}}=2, {\text{m}}=1$ The correct option for the order is : [31-Jan-2023 Shift 2]

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📖 Explanation

(A) ${\text{n}}=3 ; {\text{l}}=0 ; {\text{m}}=0 ; 3 {\ s}$ orbital(B) ${\text{n}}=4 ; 1=0 ; {\text{m}}=0 ; 4 {\ s}$ orbital(C) ${\text{n}}=3 ; 1=1 ; {\text{m}}=0 ; 3 {\text{p}}$ orbital(D) ${\text{n}}=3 ; 1=2 ; {\text{m}}=0 ; 3 {\text{d}}$ orbitalAs per Hund's rule energy is given by $( {\text{n}}+l)$ value.If value of $( {\text{n}}+l)$ remains same then energy is given by ${\text{n}}$ only.

Q40JEE Main 2023MCQ4Mp Block Elements

The Lewis acid character of boron tri halides follows the order: [31-Jan-2023 Shift 2]

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📖 Explanation

Extent of back bonding, reduces down the group leading to more Lewis acidic strength ${\text{BF}}_3 > {\text{BCl}}_3 > {\text{BBr}}_3 > {\text{BI}}_3$ (extent of back bonding) $(2 p-2 p)(2 p-3 p)(2 p-4 p)(2 p-5 p)$ ${\text{BF}}_3 < {\text{BCl}}_3 < {\text{BBr}}_3 < {\text{BI}}_3$ (lewis acidic nature $)$

Q41JEE Main 2023MCQ4MChemical Kinetics

Match List-I with List-II List-I List-II (A) Physisorption I Singlelayer adsorption (B) Chemisorption II $20-40 {\text{kJ}} {\text{mol}}^{-1}$ (C) ${\text{N}}_2( {\text{g}})+3 {\text{H}}_2( {\text{g}}) {\to }^{ {\text{Fe}}( {\ s})} 2 {\text{NH}}_3( {\text{g}})$ III Chromatography (D) Analytical Application or Adsorption IV Heterogeneous catalysisChoose the correct answer from the options given below : [31-Jan-2023 Shift 2]

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📖 Explanation

(A) Physisorption $=20-40 {\text{kJ}} / {\text{mol}}$ and Chemisorption $=80-240 {\text{kJ}} / {\text{mol}}$ (B) Physisorption is multi-layered and chemisorption is unimolecular layered.(C) In heterogeneous catalysis, medium and catalyst are in different phases.(D) Chromatography uses adsorption to purify/separate mixtures.

Q42JEE Main 2023MCQ4MClassification of Elements

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R)Assertion (A) : The first ionization enthalpy of $3 {\text{d}}$ series elements is more than that of group 2 metalsReason (R): In 3d series of elements successive filling of d-orbitals takes place.In the light of the above statements, choose the correct answer from the options given below : [31-Jan-2023 Shift 2]

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📖 Explanation

From Sc to Mn ionization energy is less than that of ${\text{Mg}}$ For 3 d series: Sc Ti V Cr Mn IE $( {\text{KJ}} / {\text{mol}})$ 631 656 650 653 717 Fe Co Ni Cu Zn IE $( {\text{KJ}} / {\text{mol}})$ 762 758 736 745 906For $2^{\;\text{ad }\;}$ Group Be Mg Ca Sr Ba Ra IE $( {\text{KJ}} / {\text{mol}})$ 631 656 650 653 717 762

Q43JEE Main 2023MCQ4MBiomolecules

The element playing significant role in neuromuscular function and interneuronal transmission is: [31-Jan-2023 Shift 2]

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Q44JEE Main 2023MCQ4MRedox Reactions

Given below are two statements:Statement I : ${\text{H}}_2 {\text{O}}_2$ is used in the synthesis of CephalosporinStatement II : ${\text{H}}_2 {\text{O}}_2$ is used for the restoration of aerobic conditions to sewage wastes.In the light of the above statements, choose the most appropriate answer from the options given below : [31-Jan-2023 Shift 2]

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Q45JEE Main 2023MCQ4MIonic Equilibrium

The normal rain water is slightly acidic and its ${\text{pH}}$ value is $5.6$ because of which one of the following ? [31-Jan-2023 Shift 2]

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📖 Explanation

We are aware that normally rain water has a pH of $5.6$ due to the presence of ${\text{H}}^{+}$ ions formed by the reactions of rain water with carbon dioxide present in the atmosphere. $\; {\text{H}}_2 {\text{O}}( {\text{l}})+ {\text{CO}}_2( {\text{g}}) ⇋ {\text{H}}_2 {\text{CO}}_3( {\text{aq}})$ $\; {\text{H}}_2 {\text{CO}}_3( {\text{aq}}) ⇋ {\text{H}}^{+}( {\text{aq}})+ {\text{HCO}}_3^{-}( {\text{aq}})$ $\;$

Q46JEE Main 2023MCQ4MHydrocarbons

When a hydrocarbon A undergoes complete combustion it requires 11 equivalents of oxygen and produces 4 equivalents of water. What is the molecular formula of A ? [31-Jan-2023 Shift 2]

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📖 Explanation

$\; {\text{C}}_{ {\text{x}}} {\text{H}}_{ {\text{y}}}+ ( {\text{x}}+\;{ {\text{y}}}/{4}) {\text{O}}_2 \to {\text{xCO}}_2+\;{ {\text{y}}}/{2} {\text{H}}_2 {\text{O}}$ $\;\;{ {\text{y}}}/{2}=4 \therefore {\text{y}}=8$ $\; {\text{x}}+\;\frac{8}{4}=11$ $\; \therefore {\text{x}}=9$ $\; \therefore \;\text{ Hydrocarbon will be }\;= {\text{C}}_9 {\text{H}}_8$

Q47JEE Main 2023MCQ4MAmines

An organic compound $[ {\text{A}}] ( {\text{C}}_4 {\text{H}}_{11} {\text{N}})$ , shows optical activity and gives ${\text{N}}_2$ gas on treatment with ${\text{HNO}}_2$ . The compound $[ {\text{A}}]$ reacts with ${\text{PhSO}}_2 {\text{Cl}}$ producing a compound which is soluble in ${\text{KOH}}$ . The structure of ${\text{A}}$ is: [31-Jan-2023 Shift 2]

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📖 Explanation

${\text{C}}_4 {\text{H}}_{11} {\text{N}}$ releases ${\text{N}}_2$ with ${\text{HNO}}_2$ i.e. it is primary amine.After reacting with Hinsberg reagent it forms a compound which is soluble in ${\text{KOH}}$ ,Hence, the amine is primary.

Q48JEE Main 2023MCQ4Md and f Block Elements

Which one of the following statements is incorrect? [31-Jan-2023 Shift 2]

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📖 Explanation

Van - Arkel process is used for purification of ${\text{Ti}}$ , ${\text{Zr}}, {\text{Hf}}$ and ${\text{B}}$ .

Q49JEE Main 2023MCQ4Md and f Block Elements

Which of the following elements have half-filled f-orbitals in their ground state?(Given : atomic number ${\text{Sm}}=62 ; {\text{Eu}}=63 ; {\text{Tb}}=65 ; {\text{Gd}}=64, {\text{Pm}}=61 \;\text{ ) }\;$ A. ${\text{Sm}}$ B. ${\text{Eu}}$ C. ${\text{Tb}}$ D. GdE. ${\text{Pm}}$ Choose the correct answer from the options given below: [31-Jan-2023 Shift 2]

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📖 Explanation

$62 {\text{Sm}}: 4 {\text{f}}^6 6 {\ s}^2$ 2. ${ }_{64} {\text{Gd}}: 4 {\text{f}}^7 5 {\text{d}}^1 6 {\ s}^2$ 3. ${ }_{63} {\text{Eu}}: 4 {\text{f}}^7 6 {\ s}^2$ 4. ${ }_{65} {\text{Tb}}: 4 {\text{f}}^9 6 {\ s}^2$ 5. ${ }_{61} {\text{Pm}}: 4 f^5 6 {\ s}^2$

Q50JEE Main 2023MCQ4MPractical Organic Chemistry

In Dumas method for the estimation of ${\text{N}}_2$ , the sample is heated with copper oxide and the gas evolved is passed over: [31-Jan-2023 Shift 2]

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📖 Explanation

Duma's method.The nitrogen containing organic compound, when heated with ${\text{CuO}}$ in a atmosphere of ${\text{CO}}_2$ , yields free ${\text{N}}_2$ in addition to ${\text{CO}}_2$ and ${\text{H}}_2 {\text{O}}$ . $\; {\text{C}}_{ {\text{x}}} {\text{H}}_{ {\text{y}}} {\text{N}}_{ {\text{z}}}+ (2 {\text{x}}+\;{ {\text{y}}}/{2}) {\text{CuO}} \to$ $\; \;\; {\text{x}} {\text{CO}}_2+\;{ {\text{y}}}/{2} {\text{H}}_2 {\text{O}}+\;{ {\text{z}}}/{2} {\text{N}}_2+ (2 {\text{x}}+\;{ {\text{y}}}/{2}) {\text{Cu}}$ Traces of nitrogen oxides formed, if any, are reduced to nitrogen by passing the gaseous mixture over heated copper gauze.

Q51JEE Main 2023NAT4MCoordination Compounds

If the CFSE of $[ {\text{Ti}} ( {\text{H}}_2 {\text{O}}) _6] ^{3+}$ is $-96.0 {\text{kJ}} / {\text{mol}}$ , this complex will absorb maximum at wavelength ____ ${\text{nm}}$ . (nearest integer)Assume Planck's constant $( {\text{h}})=6.4 \times 10^{-34} {\text{Js}}$ Speed of light $( {\text{c}})=3.0 \times 10^8 {\text{m}} / {\ s}$ and Avogadro's constant $( {\text{N}}_{ {\text{A}}}) =6 \times 10^{23} / {\text{mol}}$ [31-Jan-2023 Shift 2]

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📖 Explanation

$\; ( {\text{Ti}}^{+3} ( {\text{H}}_2 {\text{O}}) _6) ^{3+}$ $\; {\text{Ti}}^{+3}: 3 {\text{d}}^1$ $\;\;\text{ C.F.S.E. }\;=-0.4 \times \Delta _0$ $\;=-\;{96 \times 10^3}/{ {\text{N}}_0} {\text{J}}$ $\;\Delta _0=\;\frac{96 \times 10^3}{0.4 \times 6 \times 10^{23}}$ $\;\Rightarrow \;\; \;{ {\text{hc}}}/{\lambda }=\;\frac{96 \times 10^3}{0.4 \times 6 \times 10^{23}}$ $\;\lambda =\;\frac{0.4 \times 6 \times 10^{23} \times 6.4 \times 10^{-34} \times 3 \times 10^8}{96 \times 10^3}$ $\;=0.48 \times 10^{-6} {\text{m}}$ $\;=480 \times 10^{-9} {\text{m}}$ $\;=480 {\text{nm}}$ $\;$

Q52JEE Main 2023NAT4MChemical Bonding and Molecular Structure

Amongst the following, the number of species having the linear shape is _______. ${\text{XeF}}_2, {\text{I}}_3^{+}, {\text{C}}_3 {\text{O}}_2, {\text{I}}_3^{-}, {\text{CO}}_2, {\text{SO}}_2, {\text{BeCl}} \;\text{ and }\; {\text{BCl}}_2^{Θ}$ [31-Jan-2023 Shift 2]

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Q53JEE Main 2023NAT4MElectrochemistry

The resistivity of a $0.8 {\text{M}}$ solution of an electrolyte is _______ $5 \times 10^{-3} Ω {\text{cm}}$ . Its molar conductivity is $\times 10^4 Ω^{-1} {\text{cm}}^2 {\text{mol}}^{-1}$ . (Nearest integer) [31-Jan-2023 Shift 2]

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📖 Explanation

$Lambda_{ {\text{m}}}=\;\;{κ \times 1000}/{ {\text{M}}}$ $Lambda_{ {\text{m}}}=\;\;\frac{1}{\rho } \times \;{1000}/{ {\text{M}}}$ $\;\;\frac{1}{5 \times 10^{-3}} \times \;\frac{1000}{0.8}$ Ans. $25 \times 10^4 Ω^{-1} {\text{cm}}^{-2} {\text{mol}}^{-1}$

Q54JEE Main 2023NAT4MIonic Equilibrium

At $298 {\text{K}}$ , the solubility of silver chloride in water is $1.434 \times 10^{-3} {\text{g}} {\text{L}}^{-1}$ . The value of $-\log {\text{K}}_{ {\ sp}}$ for silver chloride is(Given mass of ${\text{Ag}}$ is $107.9 {\text{g}} {\text{mol}}^{-1}$ and mass of ${\text{Cl}}$ is _______ $35.5 {\text{g}} {\text{mol}}^{-1})$ [31-Jan-2023 Shift 2]

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📖 Explanation

$\; {\text{AgCl}}( {\ s}) \to \;{{ {\text{Ag}}^{+}}}_{{ {\text{S}}}} \;\text{(aq. }\;) +\;{{ {\text{Cl}}^{-}}}_{{ {\text{Cl}}}} \;\text{(aq.) }\;$ $\; {\text{K}}_{ {\ sp}}= {\text{S}}^2= (\;\frac{1.43}{143.4} \times 10^{-3}) ^2=10^{-10}$ $\;-\log {\text{K}}_{ {\ sp}}=10$

Q55JEE Main 2023NAT4MSolid State

A sample of a metal oxide has formula ${\text{M}}_{0.83} {\text{O}}_{1.00}$ . The metal $M$ can exist in two oxidation states $+2$ and $+3$ . In the sample of ${\text{M}}_{0.83} {\text{O}}_{1.00}$ , the percentage of metal ions existing in $+2$ oxidation state is _______ $\%$ (nearest integer) [31-Jan-2023 Shift 2]

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📖 Explanation

$\; {\text{M}} \;{\to +x}_{x+3}(0.83-x)$ $\;2 x+3(0.83-x)=2$ $\;x=0.49$ $\;\% M^{2+}=\;\frac{0.49}{0.83} \times 100$ $\;=59 \%$ $\;$

Q56JEE Main 2023NAT4MSome Basic Concepts of Chemistry

Assume carbon burns according to following equation : $2 {\text{C}}_{( {\ s})}+ {\text{O}}_{2( {\text{g}})} \to 2 {\text{CO}}( {\text{g}})$ When $12 {\text{g}}$ carbon is burnt in $48 {\text{g}}$ of oxygen, the volume of carbon monoxide produced is _______ $\times 10^{-1} {\text{L}}$ at STP [nearest integer]$[Given : Assume ${\text{CO}}$ as ideal gas, Mass of ${\text{C}}$ is $12 {\text{g}} {\text{mol}}^{-1}$ , Mass of ${\text{O}}$ is $16 {\text{g}} {\text{mol}}^{-1}$ and molar volume of an ideal gas at STP is $22.7 {\text{L}} {\text{mol}}^{-1}$ ]$ [31-Jan-2023 Shift 2]

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📖 Explanation

$\;2 {\text{C}}( {\ s})+ {\text{O}}_2( {\text{g}}) \to 2 {\text{CO}}( {\text{g}})$ $\;1 {\text{mol}} 1.5 {\text{mol}}$ Limiting reagent is carbon. One mole carbon produces one mole ${\text{CO}}$ . Hence, volume at STP is $227 \times 10^{-1}$ litre

Q57JEE Main 2023NAT4MClassification of Elements

The number of alkali metal(s), from ${\text{Li}}, {\text{K}}, {\text{Cs}}, {\text{Rb}}$ having ionization enthalpy greater than $400 {\text{kJ}} {\text{mol}}^{-1}$ and forming stable super oxide is ________. [31-Jan-2023 Shift 2]

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📖 Explanation

${\text{K}}, {\text{Rb}}$ and ${\text{Cs}}$ form stable super oxides but Cs has ionisation enthalpy less than $400 {\text{kJ}}$ .

Q58JEE Main 2023NAT4MChemical Thermodynamics

Enthalpies of formation of ${\text{CCl}}_4( {\text{g}}), {\text{H}}_2 {\text{O}}( {\text{g}}), {\text{CO}}_2( {\text{g}})$ and ${\text{HCl}}( {\text{g}})$ are $-105,-242,-394$ and $-92 {\text{kJ}} {\text{mol}}^{-1}$ respectively. The magnitude of enthalpy of the reaction given below is ________ ${\text{kJ}} {\text{mol}}{ }^{-1}$ (nearest integer) ${\text{CCl}}_4( {\text{g}})+2 {\text{H}}_2 {\text{O}}( {\text{g}}) \to {\text{CO}}_2( {\text{g}})+4 {\text{HCl}}( {\text{g}})$ [31-Jan-2023 Shift 2]

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📖 Explanation

$\;\Delta _{ {\text{r}}} {\text{H}}=\sum {\text{H}}_{ {\text{p}}}-\sum {\text{H}}_{ {\text{R}}}$ $\; \;\; =(-394+4 \times -92)-(-105+(2 \times -242))$ $\; \;\; =-173 {\text{kJ}} / {\text{mol}}$

Q59JEE Main 2023NAT4MPractical Organic Chemistry

The number of molecules which gives haloform test among the following molecules is [31-Jan-2023 Shift 2]

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Q60JEE Main 2023NAT4MChemical Kinetics

The rate constant for a first order reaction is $20 {\min}^{-1}$ . The time required for the initial concentration of the reactant to reduce to its $\;\frac{1}{32}$ level is _______ $\times 10^{-2} {\min}$ . (Nearest integer)(Given : $\;\ln 10=2.303$ $\;\log 2=0.3010)$ [31-Jan-2023 Shift 2]

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📖 Explanation

${\text{C}}=\;{ {\text{C}}_o}/{2^{ {\text{n}}}}\;=\;{ {\text{C}}_{ {\text{o}}}}/{32}$ ${\text{n}}\;=5$ ${\text{t}}\;=5 {\text{t}}_{1 / 2}$ $\;=\;\frac{5 \times 0.693}{20}=\;\frac{0.693}{4}$ $\;=0.17325 {\min}=17.325 \times 10^{-2} {\min} .$

Mathematics30 questions

Q61JEE Main 2023MCQ4MDifferential Equations

If $\varphi (x)=\;{1}/{{\sqrt{x}} \;\frac{\pi}{4}} \int^x (4 {\sqrt{2}} sin \;t-3 \varphi ^{'}(t)) \text{dt}, x>0$ then $\varphi ^{'} (\;\frac{\pi}{4})$ is equal to : [31-Jan-2023 Shift 2]

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📖 Explanation

$\;\varphi ^{'}( {\text{x}})=\;{1}/{\sqrt{ {\text{x}}}} \$ [ (4 {\sqrt{2}} sin ; {\text{x}}-3 φ^{'}( {\text{x}})) ⋅ 1-0]$ -;\frac{1}{2} {\text{x}}^{-3 / 2};int_{{;\frac{\pi}{4}}}^{ {\text{x}}} (4 {\sqrt{2}} sin ; {\text{t}}-3 \varphi ^{'}( {\text{t}})) {\text{dt}};\varphi ^{'} (;\frac{\pi}{4}) =;{2}/{{\sqrt{\pi}}} $[4-3 φ^{'} (;\frac{pi}{4}) ]$ +0; (1+;{6}/{{\sqrt{\pi}}}) \varphi ^{'} (;\frac{\pi}{4}) =;{8}/{{\sqrt{\pi}}};\varphi ^{'} (;\frac{\pi}{4}) =;{8}/{{\sqrt{\pi}}+6}$

Q62JEE Main 2023MCQ4MMatrices and Determinants

If a point ${\text{P}}(\alpha, \beta, \gamma ) \;\;$ satisfying $(\alpha \beta \gamma ){(\text{table} 2 , 10 , 8 ; 9 , 3 , 8 ; 8 , 4 , 8)}={(\text{table} 0 , 0 , 0)}$ lies on the plane $2 x+4 y+3 z=5$ , then $6 \alpha+9 \beta+7 \gamma$ is equal to: [31-Jan-2023 Shift 2]

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📖 Explanation

$\;2 \alpha+4 \beta+3 \gamma =5$ . . . (1) $\;2 \alpha+9 \beta+8 \gamma =0$ . . . (2) $\;10 \alpha+3 \beta+4 \gamma =0$ . . . (3) $\;8 \alpha+8 \beta+8 \gamma =0$ . . . (4)Subtract (4) from (2) $\;-6 \alpha+\beta=0$ $\;\beta=6 \alpha$ . . . (5)From equation (4) $\;8 \alpha+48 \alpha+8 \gamma =0$ $\;\gamma =-7 \alpha$ . . . (6)From equation (1) $\;2 \alpha+24 \alpha-21 \alpha=5$ $\;5 \alpha=5$ $\;\alpha=1$ $\;\beta=+6, \;\; \gamma =-7$ $\; \therefore 6 \alpha+9 \beta+7 \gamma$ $\;=6+54-49$ $\;=11$

Q63JEE Main 2023MCQ4MSequences and Series

Let $a_1, a_2, a_3, . . .$ . be an A.P. If $a_7=3$ , the product $a_1 a_4$ is minimum and the sum of its first ${\text{n}}$ terms is zero, then ${\text{n}} !-4 {\text{a}}_{ {\text{n}}( {\text{n}}+2)}$ is equal to : [31-Jan-2023 Shift 2]

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📖 Explanation

$\;a+6 d=3$ . . . (1) $\;Z=a(a+3 d)$ $\;=(3-6 d)(3-3 d)$ $\;=18 d^{2-}27 d+9$ Differentiating with respect to ${\text{d}}$ $\Rightarrow 36 {\text{d}}-27=0$ $\Rightarrow {\text{d}}=\;\frac{3}{4}$ , from (1) ${\text{a}}=\;\frac{-3}{2},( {\text{Z}}=$ minimum $)$ Now, ${\text{S}}_{ {\text{n}}}=\;{ {\text{n}}}/{2} (-3+( {\text{n}}-1) \;\frac{3}{4}) =0$ $\Rightarrow n=5$ Now, $\; {\text{n}} !-4 a_{n(n+2)}=120-4 (a_{35})$ $\;=120-4(a+(35-1) d)$ $\;=120-4 (\;\frac{-3}{2}+34 \cdot (\;\frac{3}{4}) )$ $\;=120-4 (\;\frac{-6+102}{4})$ $\;=120-96=24$

Q64JEE Main 2023MCQ4MInverse Trigonometric Functions

Let $( {\text{a}}, {\text{b}}) ⊂(0,2 \pi)$ be the largest interval for which $sin \;^{-1}( sin \;\theta)-\cos ^{-1}( sin \;\theta)>0, \theta \in (0,2 \pi)$ holds. If $\;\alpha x^{2+}\beta x+ sin \;^{-1} (x^{2-}6 x+10) +\cos ^{-1}$ $\; (x^{2-}6 x+10) =0$ and $\alpha-\beta=b-a$ , then $\alpha$ is equal to : [31-Jan-2023 Shift 2]

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📖 Explanation

$\; sin \;^{-1} sin \;\theta- (\;\frac{\pi}{2}- sin \;^{-1} sin \;\theta) > 0$ $\;\Rightarrow sin \;^{-1} sin \;\theta > \;\frac{\pi}{4}$ $\;\Rightarrow sin \;\theta>\;{1}/{{\sqrt{2}}}$ $\;\;\text{ So, }\; \theta \in (\;\frac{\pi}{4}, \;\frac{3 \pi}{4})$ $\; \;\;\;\; \theta \in (\;\frac{\pi}{4}, \;\frac{3 \pi}{4}) =(a, b)$ $\;\Rightarrow \alpha-a=\;\frac{\pi}{2}=\alpha-\beta$ $\;\Rightarrow \beta=\alpha-\;\frac{\pi}{2}$ $\; \;\;\;\; x+\beta x+ sin \;^2 \$ [( {\text{x}}-3)^{2+}1]$ +\cos ^{-1} $[( {\text{x}}-3)^{2+}1]$ =0; ;; {\text{x}}=3,9 \alpha+3 \beta+;\frac{\pi}{2}+0=0;\Rightarrow 9 \alpha+3 (\alpha-;\frac{\pi}{2}) +;\frac{\pi}{2}=0;\Rightarrow 12 \alpha-\pi=0;\alpha=;\frac{\pi}{12}$

Q65JEE Main 2023MCQ4MDifferential Equations

Let $y=y(x)$ be the solution of the differential equation $(3 y^{2-}5 x^2) y \text{dx}+2 x (x^{2-}y^2) d y=0$ such that $y(1)=1$ . then $|(y(2))^{3-}12 y(2)|$ is equal to: [31-Jan-2023 Shift 2]

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📖 Explanation

$\; (3 y^{2-}5 x^2) y \cdot \text{dx}+2 x (x^{2-}y^2) d y=0$ $\;\Rightarrow \;\frac{d y}{d x}=\;\frac{y (5 x^{2-}3 y^2) }{2 x (x^{2-}y^2) }$ Put ${\text{y}}= {\text{mx}}$ $\Rightarrow {\text{m}}+ {\text{x}} \cdot \;{ {\text{dm}}}/{ {\text{dx}}}=\;{ {\text{m}} (5-3 {\text{m}}^2) }/{2 (1- {\text{m}}^2) }$ $\; {\text{x}} \cdot \;{d {\text{m}}}/{ {\text{dx}}}=\;{ (5-3 {\text{m}}^2) {\text{m}}-2 {\text{m}} (1- {\text{m}}^2) }/{2 (1- {\text{m}}^2) }$ $\Rightarrow \;{ {\text{dx}}}/{ {\text{x}}}=\;{2 ( {\text{m}}^{2-}1) }/{ {\text{m}} ( {\text{m}}^{2-}3) } {\text{dm}}$ $\Rightarrow \;{ {\text{dx}}}/{ {\text{x}}}= (\;{2}/{ {\text{m}}}-\;{\;\frac{4}{3}}/{ {\text{m}}}+\;{\;{4 {\text{m}}}/{3}}/{ {\text{m}}^{2-}3}) {\text{dm}}$ $\;\Rightarrow \int \;{ {\text{dx}}}/{ {\text{x}}}=\int \;{ (\frac{2}{3}) }/{ {\text{m}}}+\int \;\frac{2}{3} (\;{2 {\text{m}}}/{ {\text{m}}^{2-}3}) {\text{dm}}$ $\;\Rightarrow \ln | {\text{x}}|=\;\frac{2}{3} \ln | {\text{m}}|+\;\frac{2}{3} \ln | {\text{m}}^{2-}3|+ {\text{C}}$ $\;\;\text{ Or, }\; \ln | {\text{x}}|=\;\frac{2}{3} \ln |\;{ {\text{y}}}/{ {\text{x}}}|+\;\frac{2}{3} \ln || \;{ {\text{y}}}/{ {\text{x}}}) ^{2-}3 \mid + {\text{C}}$ $\;\;\text{ Put }\;( {\text{x}}=1, {\text{y}}=1): \;\text{ we get }\; {\text{c}}=-\;\frac{2}{3} \ln (2)$ $\;\Rightarrow \ln | {\text{x}}|=\;\frac{2}{3} \ln |\;{ {\text{y}}}/{ {\text{x}}}|+\;\frac{2}{3} \ln | (\;{ {\text{y}}}/{ {\text{x}}}) ^{2-}3|-\;\frac{2}{3} \ln (2)$ $\;\Rightarrow (\;{ {\text{y}}}/{ {\text{x}}}) \$ [ (;{ {\text{y}}}/{ {\text{x}}}) ^{2-}3]$ =2 . ( {\text{x}}^{3 / 2});;\text{ Put }; {\text{x}}=2 ;\text{ to get }; {\text{y}}(2);\Rightarrow {\text{y}} ( {\text{y}}^{2-}12) =4 \times 2 \times 2 \times 2 {\sqrt{2}};\Rightarrow {\text{y}}^{3-}12 {\text{y}}=32 {\sqrt{2}};\Rightarrow | {\text{y}}^3(2)-12 {\text{y}}(2)|=32 {\sqrt{2}}$

Q66JEE Main 2023MCQ4MCircles

The set of all values of ${\text{a}}^2$ for which the line ${\text{x}}+ {\text{y}}=0$ bisects two distinct chords drawn from ${\text{a}}$ point ${\text{P}} (\;{1+ {\text{a}}}/{2}, \;{1- {\text{a}}}/{2})$ on the circle $2 x^{2+}2 y^{2-}(1+a) x-(1-a) y=0$ is equal to: [31-Jan-2023 Shift 2]

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📖 Explanation

$\; {\text{x}}^{2+} {\text{y}}^{2-}\;{(1+ {\text{a}}) {\text{x}}}/{2}-\;{(1- {\text{a}}) {\text{y}}}/{2}=0$ $\;\;\text{ Centre }\; (\;{1+ {\text{a}}}/{4}, \;{1- {\text{a}}}/{4}) \Rightarrow ( {\text{h}}, {\text{k}})$ $\; {\text{P}} (\;{1+ {\text{a}}}/{2}, \;{1- {\text{a}}}/{2}) \Rightarrow (2 {\text{h}}, 2 {\text{k}})$ Equation of chord $\Rightarrow {\text{T}}= {\text{S}}_1$ $\;\Rightarrow ( {\text{x}}- {\text{y}}) \lambda -\;{2 {\text{h}}( {\text{x}}+\lambda )}/{2}-\;{(2 {\text{k}})( {\text{y}}-\lambda )}/{2}$ $\;=2 \lambda ^{2-}2 {\text{h}}(\lambda )+2 {\text{k}} \lambda$ Now, $\lambda (2 {\text{h}}, 2 {\text{k}})$ satisfies the chord $\; \therefore (2 {\text{h}}-2 {\text{k}}) \lambda - {\text{h}}( {\text{x}}+\lambda )- {\text{k}}( {\text{y}}-\lambda )$ $\;\Rightarrow 2 \lambda ^{2+}4 {\text{k}} \lambda -4 {\text{h}} \lambda + {\text{h}} \lambda - {\text{k}} \lambda + {\text{hx}}+ {\text{ky}}=0$ $\;\Rightarrow 2 \lambda ^{2+}\lambda (3 {\text{k}}-3 {\text{h}})+ {\text{ky}}+ {\text{hx}}=0$ $\;\Rightarrow {\text{D}}>0$ $\;\Rightarrow 9( {\text{k}}- {\text{h}})^{2-}8( {\text{ky}}+ {\text{hx}}) > 0$ $\;\Rightarrow 9( {\text{k}}- {\text{h}})^{2-}8 (2 {\text{k}}^{2+}2 {\text{h}}^2) > 0$ $\;\Rightarrow -7 {\text{k}}^{2-}7 {\text{h}}^{2-}18 {\text{kh}} > 0$ $\;\Rightarrow 7 {\text{k}}^{2+}7 {\text{h}}^{2+}18 {\text{kh}} < 0$ $\;\Rightarrow 7 (\;{1- {\text{a}}}/{4}) ^{2+}7 (\;{1+ {\text{a}}}/{4}) ^{2+}18 (\;{1- {\text{a}}^2}/{16}) < 0$ $\;\Rightarrow 7 \$ [;{2 (1+ {\text{a}}^2) }/{16}]$ +;{18 (1- {\text{a}}^2) }/{16} < 0, ;; {\text{a}}^2= {\text{t}};\Rightarrow ;\frac{7}{8}(1+ {\text{t}})+;{18(1- {\text{t}})}/{16} < 0;\Rightarrow ;{14+14 {\text{t}}+18-18 {\text{t}}}/{16} < 0;\Rightarrow 4 {\text{t}} > 32; {\text{t}}>8 ;; {\text{a}}^2 > 8$

Q67JEE Main 2023MCQ4MSets and Relations

Among the relations ${\text{S}}= \{( {\text{a}}, {\text{b}}): {\text{a}}, {\text{b}} \in ℝ-\{0\}, 2+\;{ {\text{a}}}/{ {\text{b}}} > 0\}$ And $T= \{(a, b): a, b \in ℝ, a^{2-}b^2 \in Z\}$ , [31-Jan-2023 Shift 2]

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📖 Explanation

For relation ${\text{T}}= {\text{a}}^{2-} {\text{b}}^2=- {\text{I}}$ Then, $(b, a)$ on relation ${\text{R}}$ $\Rightarrow {\text{b}}^{2-} {\text{a}}^2=- {\text{I}}$ $\therefore {\text{T}}$ is symmetric $\; {\text{S}}= \{( {\text{a}}, {\text{b}}): {\text{a}}, {\text{b}} \in {\text{R}}-\{0\}, 2+\;{ {\text{a}}}/{ {\text{b}}}>0\}$ $\;2+\;{ {\text{a}}}/{ {\text{b}}} > 0 \Rightarrow \;{ {\text{a}}}/{ {\text{b}}} > -2, \Rightarrow \;{ {\text{b}}}/{ {\text{a}}} < \;\frac{-1}{2}$ If $(b, a) \in S$ then $2+\;\frac{b}{a}$ not necessarily positive $\therefore {\text{S}}$ is not symmetric

Q68JEE Main 2023MCQ4MQuadratic Equation and Inequalities

The equation $e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^x+1=0, x \in R$ has: [31-Jan-2023 Shift 2]

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📖 Explanation

$e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^x+1=0$ Let $e^x=t$ Now, ${\text{t}}^{4+}8 {\text{t}}^{3+}13 {\text{t}}^{2-}8 {\text{t}}+1=0$ Dividing equation by $t^2$ , $\;t^{2+}8 t+13-\;\frac{8}{t}+\;\frac{1}{t^2}=0$ $\;t^{2+}\;\frac{1}{t^2}+8 (t-\;\frac{1}{t}) +13=0$ $\; (t-\;\frac{1}{t}) ^{2+}2+8 (t-\;\frac{1}{t}) +13=0$ Let ${\text{t}}-\;{1}/{ {\text{t}}}= {\text{z}}$ $\;z^{2+}8 z+15=0$ $\;(z+3)(z+5)=0$ $\;z=-3 \;\text{ or }\; z=-5$ So, $t-\;{1}/{ {\text{t}}}=-3$ or ${\text{t}}-\;{1}/{ {\text{t}}}=-5$ $\; {\text{t}}^{2+}3 {\text{t}}-1=0 \;\text{ or }\; {\text{t}}^{2+}5 {\text{t}}-1=0$ $\; {\text{t}}=\;{-3 \pm {\sqrt{13}}}/{2} \;\text{ or }\; {\text{t}}=\;{-5 \pm {\sqrt{29}}}/{2}$ as ${\text{t}}= {\text{e}}^{ {\text{x}}}$ so ${\text{t}}$ must be positive, $t=\;{{\sqrt{13}}-3}/{2} \;\text{ or }\; \;{{\sqrt{29}}-5}/{2}$ So, ${\text{x}}=\ln (\;{{\sqrt{13}}-3}/{2})$ or ${\text{x}}=\ln (\;{{\sqrt{29}}-5}/{2})$ Hence two solution and both are negative.

Q69JEE Main 2023MCQ4MSets and Relations

The number of values of $r \in \{p, q, ∼ p, ∼ q\}$ for which $((p ∧ q) \Rightarrow (r ∨ q)) ∧((p ∧ r) \Rightarrow q)$ is a tautology, is: [31-Jan-2023 Shift 2]

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📖 Explanation

$((p ∧ q) \Rightarrow (r ∨ q)) ∧((p ∧ r) \Rightarrow q)$ We know, ${\text{p}} \Rightarrow {\text{q}}$ is equivalent to $\;∼ {\text{p}} ∨ {\text{q}}$ $\;(∼( {\text{p}} ∧ {\text{q}}) ∨( {\text{r}} ∨ {\text{q}})) ∧(∼( {\text{p}} ∧ {\text{r}})) ∨ {\text{q}}))$ $\;\Rightarrow (∼ {\text{p}} ∨ ∼ {\text{q}} ∨ {\text{r}} ∨ {\text{q}}) ∧(∼ {\text{p}} ∨ ∼ {\text{r}} ∨ {\text{q}})$ $\;\Rightarrow (∼ {\text{p}} ∨ {\text{r}} ∨ {\text{t}}) ∧(∼ {\text{p}} ∨ ∼ {\text{r}} ∨ {\text{q}})$ $\;\Rightarrow ( {\text{t}}) ∧(∼ {\text{p}} ∨ ∼ {\text{r}} ∨ {\text{q}})$ For this to be tautology, $(∼ {\text{p}} ∨ ∼ {\text{r}} ∨ {\text{q}})$ must be always true which follows for $r=∼$ p or $r=q$ .

Q70JEE Main 2023MCQ4MFunctions

Let $f: ℝ-\{2,6\} \to ℝ$ be real valued function defined as $f(x)=\;\frac{x^{2+}2 x+1}{x^{2-}8 x+12}$ . Then range of $f$ is [31-Jan-2023 Shift 2]

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📖 Explanation

Let $y=\;\frac{x^{2+}2 x+1}{x^{2-}8 x+12}$ By cross multiplying $\; {\text{yx}}^{2-}8 {\text{xy}}+12 {\text{y}}- {\text{x}}^{2-}2 {\text{x}}-1=0$ $\; {\text{x}}^2( {\text{y}}-1)- {\text{x}}(8 {\text{y}}+2)+(12 {\text{y}}-1)=0$ Case $1, y \ne 1$ $\;D \ge 0$ $\;\Rightarrow (8 y+2)^{2-}4(y-1)(12 y-1) \ge 0$ $\;\Rightarrow y(4 y+21) \ge 0$ ${\text{y}} \in (-\infty , \;\frac{-21}{4}] ∪\$ [0, ∞ )-{1} $Case 2,$ {\text{y}}=1; {\text{x}}^{2+}2 {\text{x}}+1= {\text{x}}^{2-}8 {\text{x}}+12;10 {\text{x}}=11; {\text{x}}=;\frac{11}{10} ;; ;\text{ So, }; {\text{y}} ;\text{ can be }; 1 $Hence$ {\text{y}} ∈ (-∞ , ;\frac{-21}{4}]$ ∪[0, \infty )$

Q71JEE Main 2023MCQ4MLimits, Continuity and Differentiability

$\lim _{x \to \infty } \;{({\sqrt{3 x+1}}+{\sqrt{3 x-1}})^{6+}({\sqrt{3 x+1}}-{\sqrt{3 x-1}})^6}/{ (x+{\sqrt{x^{2-}1}}) ^{6+} (x-{\sqrt{x^{2-}1}}) ^6} x^3$ [31-Jan-2023 Shift 2]

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📖 Explanation

$\;\lim _{x \to \infty } \;{({\sqrt{3 x+1}}+{\sqrt{3 x-1}})^{6+}({\sqrt{3 x+1}}-{\sqrt{3 x-1}})^6}/{ (x+{\sqrt{x^{2-}1}}) ^{6+} (x-{\sqrt{x^{2-}1}}) ^6} x^3$ $\;\lim _{x \to \infty } x^3 \times \{\;{x^3 \{ (\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}) ^{6+} (\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}) ^6\} }/{x^6 \{ (1+\sqrt{1-\;\frac{1}{x^2}}) ^{6+} (1-\sqrt{1-\;\frac{1}{x^2}}) ^6\} }\}$ $\;=\;{(2 {\sqrt{3}})^{6+}0}/{2^{6+}0}=3^3=(27)$

Q72JEE Main 2023MCQ4MThree Dimensional Geometry

Let $P$ be the plane, passing through the point $(1,-1,-5)$ and perpendicular to the line joining the points $(4,1,-3)$ and $(2,4,3)$ . Then the distance of $P$ from the point $(3,-2,2)$ is [31-Jan-2023 Shift 2]

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📖 Explanation

Equation of Plane : $\;2(x-1)-3(y+1)-6(z+5)=0$ $\;\;\text{ Or }\; 2 x-3 y-6 z=35$ $\;\Rightarrow \;\text{ Re quired distance }\;=$ $\;\;{|2(3)-3(-2)-6(2)-35|}/{{\sqrt{4+9+36}}}$ $\;=5$

Q73JEE Main 2023MCQ4MApplication of Derivatives

The absolute minimum value, of the function $f(x)=x^{2-}x+1|+ [x^{2-}x+1]$ , where $[t]$ denotes the greatest integer function, in the interval $[-1,2]$ , is : [31-Jan-2023 Shift 2]

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📖 Explanation

$f(x)= |x^{2-}x+1|+ \$ [x^{2-}x+1]$ ; x \in [-1,2] $Let$ g( {\text{x}})= {\text{x}}^{2-} {\text{x}}+1= ( {\text{x}}-;\frac{1}{2}) ^{2+};\frac{3}{4}\because | {\text{x}}^{2-} {\text{x}}+1| ;\text{ and }; [ {\text{x}}^{2-} {\text{x}}+2] $Both have minimum value at$ {\text{x}}=1 / 2;\Rightarrow ;\text{ Minimum }; {\text{f}}( {\text{x}})=;\frac{3}{4}+0;=;\frac{3}{4}$

Q74JEE Main 2023MCQ4MThree Dimensional Geometry

Let the plane $P: 8 x+\alpha_1 y+\alpha_2 z+12=0$ be parallel to the line $L: \;\frac{x+2}{2}=\;\frac{y-3}{3}=\;\frac{z+4}{5}$ . If the intercept of $P$ on the $y$ -axis is 1 , then the distance between ${\text{P}}$ and ${\text{L}}$ is : [31-Jan-2023 Shift 2]

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📖 Explanation

P: $8 {\text{x}}+\alpha_1 {\text{y}}+\alpha_2 {\text{z}}+12=0$ L: $\;{ {\text{x}}+2}/{2}=\;{ {\text{y}}-3}/{3}=\;{ {\text{z}}+4}/{5}$ $\because {\text{P}}$ is parallel to ${\text{L}}$ $\;\Rightarrow 8(2)+\alpha_1(3)+5 (\alpha_2) =0$ $\;\Rightarrow 3 \alpha_1+5 (\alpha_2) =-16$ Also ${\text{y}}$ -intercept of plane ${\text{P}}$ is 1 $\Rightarrow \alpha_1=-12$ And $\alpha_2=4$ $\Rightarrow$ Equation of plane $P$ is $2 {\text{x}}-3 {\text{y}}+ {\text{z}}+3=0$ $\Rightarrow$ Distance of line $L$ from Plane $P$ is $\;= |\;{0-3(6)+1+3}/{{\sqrt{4+9+1}}}|$ $\;={\sqrt{14}}$

Q75JEE Main 2023MCQ4MThree Dimensional Geometry

The foot of perpendicular from the origin ${\text{O}}$ to a plane ${\text{P}}$ which meets the co-ordinate axes at the points $A, B, C$ is $(2, a, 4), a \in N$ . If the volume of the tetrahedron ${\text{OABC}}$ is $144 {\text{unit}}^3$ , then which of the following points is NOT on P? [31-Jan-2023 Shift 2]

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📖 Explanation

Equation of Plane: $\;(2 \;{ {\text{i}}}^{\^}+ {\text{a}} \;{ {\text{j}}}^{\^}+4 \;{ {\text{k}}}^{\^}) \cdot \$ [( {\text{x}}-2) ;{ {\text{i}}}^{^}+( {\text{y}}- {\text{a}}) ;{ {\text{j}}}^{^}+( {\text{z}}-4) ;{ {\text{k}}}^{^}]$=0;\Rightarrow 2 {\text{x}}+ {\text{ay}}+4 {\text{z}}=20+ {\text{a}}^2;\Rightarrow {\text{A}} \equiv (;{20+ {\text{a}}^2}/{2}, 0,0); {\text{B}} \equiv (0, ;{20+ {\text{a}}^2}/{ {\text{a}}}, 0); {\text{C}} \equiv (0,0, ;{20+ {\text{a}}^2}/{4});\Rightarrow ;\text{ Volume of tetrahedron };;=;\frac{1}{6}[{ \vec{\text{a}} { \vec{\text{b}} { \vec{\text{c}}];=;\frac{1}{6} { \vec{\text{a}} .({ \vec{\text{b}} \times { \vec{\text{c}});\Rightarrow ;\frac{1}{6} (;{20+ {\text{a}}^2}/{2}) \cdot (;{20+ {\text{a}}^2}/{ {\text{a}}}) \cdot (;{20+ {\text{a}}^2}/{4}) =144;\Rightarrow (20+ {\text{a}}^2) ^3=144 \times 48 \times {\text{a}};\Rightarrow {\text{a}}=2\Rightarrow $Equation of plane is$ 2 x+2 y+4 z=24 $Or$ {\text{x}}+ {\text{y}}+2 {\text{z}}=12\Rightarrow (3,0,4) $Not lies on the Plane$ {\text{x}}+ {\text{y}}+2 {\text{z}}=12$

Q76JEE Main 2023MCQ4MStatistics

Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and $\alpha( > 0)$ , and the mean and standard deviation of marks of class B of ${\text{n}}$ students be respectively 55 and $30-\alpha$ . If the mean and variance of the marks of the combined class of $100+ {\text{n}}$ students are respectively 50 and 350 , then the sum of variances of classes A and B is: [31-Jan-2023 Shift 2]

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📖 Explanation

$\;\overline{x}=\;\frac{100 \times 40+55 n}{100+n}$ $\;5000+50 n=4000+55 n$ $\;1000=5 n$ $\;n=200$ $\;\sigma_1^2=\;{\sum {\text{x}}_{ {\text{i}}}^2}/{100}-40^2$ $\;\sigma_2^2=\;{\sum {\text{x}}_{ {\text{j}}}^2}/{100}-55^2$ $\;350=\sigma^2=\;{\sum {\text{x}}_{ {\text{i}}}^{2+}\sum {\text{x}}_{ {\text{j}}}^2}/{300}-(\overline{ {\text{x}}})^2$ $\;2850=\;\frac{ (1600+\alpha^2) \times 100+ \$ [(30-alpha)^{2+}3025]$ \times 200}{300}-(50)^2;8550=\alpha^{2+}2(30-\alpha)^{2+}7650;\alpha^{2+}2(30-\alpha)^2=900;\alpha^{2-}40 \alpha+300=0;\alpha=10,30;\sigma_1^{2+}\sigma_2^2=10^{2+}20^2=500$

Q77JEE Main 2023MCQ4MVector Algebra

Let : $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and ${ \vec{\text{c}}=5 \;{ {\text{i}}}^{\^}-3 \;{ {\text{j}}}^{\^}+3 \;{ {\text{k}}}^{\^}$ be there vectors. If ${ \vec{\text{r}}$ is a vector such that, ${ \vec{\text{r}} \times { \vec{\text{b}}={ \vec{\text{c}} \times { \vec{\text{b}}$ and ${ \vec{\text{r}} \cdot { \vec{\text{a}}=0$ . Then $25|\vec{r}|^2$ is equal to [31-Jan-2023 Shift 2]

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📖 Explanation

$\;\;\text{ Sol. }\; \vec{a}=\;{i}+2 \;{j}+3 \;{k}$ $\;\vec{b}=\;{i}-\;{j}+2 \;{k}$ $\;{ \vec{\text{c}}=\;{5 {\text{i}}}^{\^}-3 \;{ {\text{j}}}^{\^}+3 \;{ {\text{k}}}^{\^}$ $\;({ \vec{\text{r}}-{ \vec{\text{c}}) \times { \vec{\text{b}}=0, { \vec{\text{r}} \cdot { \vec{\text{a}}=0$ $\;\Rightarrow { \vec{\text{r}}-{ \vec{\text{c}}=\lambda { \vec{\text{b}}$ $\;\;\text{ Also, }\;({ \vec{\text{c}}+\lambda { \vec{\text{b}}) \cdot { \vec{\text{a}}=0$ $\;\Rightarrow \vec{a} \cdot \vec{c}+\lambda (\vec{a} \cdot \vec{b})=0$ $\; \therefore \lambda =\;\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}=\;\frac{-8}{5}$ $\;{ \vec{\text{r}}=\;{5(5 \;{ {\text{i}}}^{\^}-3 \;{ {\text{i}}}^{\^}+3 \;{ {\text{k}}}^{\^})-8(\;{ {\text{i}}}^{\^}-\;{ {\text{j}}}^{\^}+2 \;{ {\text{k}}}^{\^})}/{5}$ $\;{ \vec{\text{r}}=\;{17 \;{ {\text{i}}}^{\^}-7 \;{ {\text{j}}}^{\^}+\;{ {\text{k}}}^{\^}}/{5}$ $\;|\vec{r}|^2=\;\frac{1}{25}(289+50)$ $\;25|{ \vec{\text{r}}|^2=339$ $\;$

Q78JEE Main 2023MCQ4MHyperbola

Let ${\text{H}}$ be the hyperbola, whose foci are $(1 \pm {\sqrt{2}}, 0)$ and eccentricity is ${\sqrt{2}}$ . Then the length of its lat us rectum is [31-Jan-2023 Shift 2]

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📖 Explanation

$\;2 {\text{ae}}=|(1+{\sqrt{2}})-(1+{\sqrt{2}})|=2 {\sqrt{2}}$ $\; {\text{ae}}={\sqrt{2}}$ $\; {\text{a}}=1$ $\;\Rightarrow {\text{b}}=1 \;\; \because {\text{e}}={\sqrt{2}} \Rightarrow \;\text{ Hyperbola is rectangular }\;$ $\;\Rightarrow {\text{L}} \cdot {\text{R}}=\;{2 {\text{b}}^2}/{ {\text{a}}}=2$

Q79JEE Main 2023MCQ4MDefinite Integrals

Let $\alpha > 0$ . If $int_{0}^{\alpha} \;{x}/{{\sqrt{x+\alpha}}-\sqrt{ {\text{x}}}} \text{dx}=\;{16+20 {\sqrt{2}}}/{15}$ , then $\alpha$ is equal to : [31-Jan-2023 Shift 2]

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📖 Explanation

After rationalising $int_{0}^{\alpha} \;{ {\text{x}}}/{\alpha}(\sqrt{ {\text{x}}+\alpha}+\sqrt{ {\text{x}}})$ $\;\;\frac{1}{\alpha} \$ [;\frac{2}{5}( {\text{x}}+alpha)^{5 / 2}-alpha ;\frac{2}{3}( {\text{x}}+alpha)^{3 / 2}+;\frac{2}{5} {\text{x}}^{5 / 2}]$ |_0 ^\alpha;=;\frac{1}{\alpha} (;\frac{5}{2}(2 \alpha)^{5 / 2}-;\frac{2 \alpha}{3}(2 \alpha)^{3 / 2}+;\frac{2}{5} \alpha^{5 / 2}-;\frac{2}{5} \alpha^{5 / 2}+;\frac{2}{3} \alpha^{5 / 2});=;\frac{1}{\alpha} (;\frac{2^{7 / 2} \alpha^{5 / 2}}{5} ;\frac{2^{5 / 2} \alpha^{5 / 2}}{3}+;\frac{2}{3} \alpha^{5 / 2});=\alpha^{3 / 2} (;\frac{2^{7 / 2}}{5}-;\frac{2^{5 / 2}}{3}+;\frac{2}{3});=;\frac{\alpha^{3 / 2}}{15}(24 {\sqrt{2}}-20 {\sqrt{2}}+10)=;\frac{\alpha^{3 / 2}}{15}(4 {\sqrt{2}}+10) $Now,$ ;;\frac{\alpha^{3 / 2}}{15}(4 {\sqrt{2}}+10)=;{16+20 {\sqrt{2}}}/{15};\Rightarrow \alpha=2$

Q80JEE Main 2023MCQ4MComplex Numbers

The complex number ${\text{z}}=\;{ {\text{i}}-1}/{\cos \;\frac{\pi}{3}+ {\text{i}} sin \;\;\frac{\pi}{3}}$ is equal to: [31-Jan-2023 Shift 2]

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📖 Explanation

$\; {\text{Z}}=\;{ {\text{i}}-1}/{\cos \;\frac{\pi}{3}+ {\text{i}} sin \;\;\frac{\pi}{3}}=\;{i-1}/{\frac{1}{2}+\;{{\sqrt{3}}}/{2} {\text{i}}}$ $\;=\;{ {\text{i}}-1}/{\frac{1}{2}+\;{{\sqrt{3}}}/{2} {\text{i}}} \times \;{\;\frac{1}{2}-\sqrt{\frac{3}{2} {\text{i}}}}/{\frac{1}{2}-{\sqrt{3 / 2}} {\text{i}}}=\;{{\sqrt{3}}-1}/{2}+\;{{\sqrt{3}}+1}/{2} {\text{i}}$ Apply polar form, $\;r \cos \theta=\;{{\sqrt{3}}-1}/{2}$ $\;r sin \;\theta=\;{{\sqrt{3}}+1}/{2}$ Now, $\tan \theta=\;{{\sqrt{3}}+1}/{{\sqrt{3}}-1}$ So, $\;\; \theta=\;\frac{5 \pi}{12}$

Q81JEE Main 2023NAT4MBinomial Theorem

The Coefficient of ${\text{x}}^{-6}$ , in the expansion of $(\;\frac{4 x}{5}+\;\frac{5}{2 x^2}) ^9$ , is _______ [31-Jan-2023 Shift 2]

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📖 Explanation

$(\;{4 {\text{x}}}/{5}+\;{5}/{2 {\text{x}}^2}) ^9$ Now, ${\text{T}}_{ {\text{r}}+1}={ }^9 {\text{C}}_{ {\text{r}}} \cdot (\;{4 {\text{x}}}/{5}) ^{9- {\text{r}}} (\;{5}/{2 {\text{x}}^2}) ^{ {\text{r}}}$ $={ }^9 {\text{C}}_{ {\text{r}}} \cdot (\;\frac{4}{5}) ^{9- {\text{r}}} (\;\frac{5}{2}) ^{ {\text{r}}} \cdot {\text{x}}^{9-3 {\text{r}}}$ Coefficient of $x^{-6}$ i.e. $9-3 r=-6 \Rightarrow r=5$

Q82JEE Main 2023NAT4MArea Under The Curves

Let the area of the region $\{( {\text{x}}, {\text{y}}): 2 {\text{x}}-1| \le {\text{y}} \le | {\text{x}}^{2-} {\text{x}}|, 0 \le {\text{x}} \le 1\}$ be A. Then $(6 {\text{A}}+11)^2$ is equal to ______. [31-Jan-2023 Shift 2]

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📖 Explanation

${\text{y}} \ge |2 {\text{x}}-1|, \;\; {\text{y}} \le | {\text{x}}^{2-} {\text{x}}|$ Both curve are symmetric about ${\text{x}}=\;\frac{1}{2}$ Hence $A=2 int_{{\;{3-{\sqrt{5}}}/{2}}}^{\;\frac{1}{2}} ( (x-x^2) -(1-2 x)) \text{dx}$ On solving $6 {\text{A}}+11=5 {\sqrt{5}}$ $(6 {\text{A}}+11)^2=125$

Q83JEE Main 2023NAT4MPermutations and Combinations

If ${ }^{2 n+1} P_{n-1}:{ }^{2 n-1} P_n=11: 21$ , then $n^{2+}n+15$ is equal to: [31-Jan-2023 Shift 2]

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📖 Explanation

$\;\;{(2 {\text{n}}+1) !( {\text{n}}-1) !}/{( {\text{n}}+2) !(2 {\text{n}}-1) !}=\;\frac{11}{21}$ $\;\Rightarrow \;{(2 {\text{n}}+1)(2 {\text{n}})}/{( {\text{n}}+2)( {\text{n}}+1) {\text{n}}}=\;\frac{11}{21}$ $\;\Rightarrow \;{2 {\text{n}}+1}/{( {\text{n}}+1)( {\text{n}}+2)}=\;\frac{11}{42}$ $\;\Rightarrow {\text{n}}=5$ $\;\Rightarrow {\text{n}}^{2+} {\text{n}}+15=25+5+15=45$

Q84JEE Main 2023NAT4MBinomial Theorem

If the constant term in the binomial expansion of $(\;{x^{\frac{5}{2}}}/{2}-\;\frac{4}{x^{ℓ}}) ^9$ is $-84$ and the Coefficient of ${\text{x}}^{-3 ℓ}$ is $2^\alpha \beta$ , where $\beta < 0$ is an odd number, Then $|\alpha ℓ-\beta|$ is equal to _________ [31-Jan-2023 Shift 2]

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📖 Explanation

$\;\;\text{ In, }\; (\;{x^{\frac{5}{2}}}/{2}-\;\frac{4}{x^{ℓ}}) ^9$ $\;T_{r+1}={ }^9 C_r \;\frac{ (x^{5 / 2}) ^{9-r}}{2^{9-r}} (\;\frac{-4}{x^{ℓ}}) ^r$ $\;=(-1)^r \;\frac{{ }^9 C_r}{2^{9-r}} 4^r x^{\;\frac{45}{2}-\;\frac{5 r}{2}- {\text{r}}}$ $\;=45-5 r-21 r=0$ $\;r=\;\frac{45}{5+21}$ . . . (1) Now, according to the question, $(-1)^{ {\text{r}}} \;{{ }^9 {\text{C}}_{ {\text{r}}}}/{2^{9- {\text{r}}}} 4^{ {\text{r}}}=-84$ $=(-1)^{ {\text{r}}}{ }^9 {\text{C}}_{ {\text{r}}} 2^{3 {\text{r}}-9}=21 \times 4$ Only natural value of $r$ possible if $3 r-9=0$ ${\text{r}}=3$ and ${ }^9 {\text{C}}_3=84$ $\therefore 1=5$ from equation (1)Now, coefficient of ${\text{x}}^{-31}= {\text{x}}^{\;\frac{45}{2}-\;{5 {\text{r}}}/{2}- {\text{lr}}}$ at $1=5$ , gives $\;r=5$ $\; \therefore { }^9 c_5(-1) \;\frac{4^5}{2^4}=2^\alpha \times \beta$ $\;=-63 \times 2^7$ $\;\Rightarrow \alpha=7, \beta=-63$ $\; \therefore \;\; \;\text{ value of }\;|\alpha ℓ-\beta|=98$

Q85JEE Main 2023NAT4MVector Algebra

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $|{ \vec{\text{a}}|={\sqrt{31}}, 4|{ \vec{\text{b}}|=|{ \vec{\text{c}}|=2$ and $2({ \vec{\text{a}} \times { \vec{\text{b}})=3({ \vec{\text{c}} \times { \vec{\text{a}})$ .If the angle between $\vec{b}$ and $\vec{c}$ is $\;\frac{2 \pi}{3}$ , then $(\;\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}}) ^2$ is equal to _________. [31-Jan-2023 Shift 2]

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📖 Explanation

$\;\;\text{ Sol. }\; 2({ \vec{\text{a}} \times { \vec{\text{b}})=3({ \vec{\text{c}} \times { \vec{\text{a}})$ $\;{ \vec{\text{a}} \times (2 { \vec{\text{b}}+3 { \vec{\text{c}})=0$ $\;{ \vec{\text{a}}=\lambda (2 { \vec{\text{b}}+3 { \vec{\text{c}})$ $\;|{ \vec{\text{a}}|^2=\lambda ^2|2 { \vec{\text{b}}+3 { \vec{\text{c}}|^2$ $\;|{ \vec{\text{a}}|^2=\lambda ^2 (4|{ \vec{\text{b}}|^{2+}9|{ \vec{\text{c}}|^{2+}12 { \vec{\text{b}} \cdot { \vec{\text{c}})$ $\;31=31 \lambda ^2 \Rightarrow \lambda =\pm 1$ $\;{ \vec{\text{a}}=\pm (2 { \vec{\text{b}}+3 { \vec{\text{c}})$ $|{ \vec{\text{b}} \times { \vec{\text{c}}|^2=|{ \vec{\text{b}}|^2|{ \vec{\text{c}}|^{2-}({ \vec{\text{b}} \cdot { \vec{\text{c}})^2=\;\frac{3}{4}$ $(\;\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}}) ^2=3$

Q86JEE Main 2023NAT4MParabola

Let ${\text{S}}$ be the set of all $a \in N$ such that the area of the triangle formed by the tangent at the point $P(b, c), b, c \in N$ , on the parabola $y^2=2 a x$ and the lines ${\text{x}}= {\text{b}}, {\text{y}}=0$ is 16 unit $^2$ , then $sum_{ {\text{a}} \in {\text{S}}} {\text{a}}$ is equal to __________. [31-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

As $P(b, c)$ lies on parabola so ${\text{c}}^2=2 {\text{ab}}$ . . . (1)Now equation of tangent to parabola $y^2=2 {\text{ax}}$ in point form is ${\text{yy}}_1=2 {\text{a}} \;{ ( {\text{x}}+ {\text{x}}_1) }/{2}, ( {\text{x}}_1, {\text{y}}_1) =( {\text{b}}, {\text{c}})$ $\Rightarrow {\text{yc}}= {\text{a}}( {\text{x}}+ {\text{b}})$ For point ${\text{B}}$ , put ${\text{y}}=0$ , now ${\text{x}}=- {\text{b}}$ $\;\;\text{ So, area of }\; \triangle {\text{PBA}}, \;\frac{1}{2} \times {\text{AB}} \times {\text{AP}}=16$ $\;\Rightarrow \;\frac{1}{2} \times 2 {\text{b}} \times {\text{c}}=16$ $\;\Rightarrow {\text{bc}}=16$ As $b$ and $c$ are natural number so possible values of $(b, c)$ are $(1,16),(2,8),(4,4),(8,2)$ and $(16,1)$ Now from equation (1) ${\text{a}}=\;{ {\text{c}}^2}/{2 {\text{b}}}$ and $a \in N$ , so values of $( {\text{b}}, {\text{c}})$ are $(1,16),(2,8)$ and $(4,4)$ now values of are 128,16 and 2 .Hence sum of values of a is 146.

Q87JEE Main 2023NAT4MSequences and Series

The sum $1^{2-}2.3^{2+}3.5^{2-}4.7^{2+}5.9^{2-}. . . . . . . .+15.29^2$ is ________. [31-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Separating odd placed and even placed terms we get $\; {\text{S}}= (1.1^{2+}3.5^{2+}. . . .15 \cdot (29)^2) - (2.3^{2+}4.7^2.$ $\;+. . .+14 .(27)^2$ $\; {\text{S}}=sum_{{ {\text{n}}=1}}^{8}(2 {\text{n}}-1)(4 {\text{n}}-3)^{2-}sum_{{ {\text{n}}=1}}^{7}(2 {\text{n}})(4 {\text{n}}-1)^2$ Applying summation formula we get $=29856-22904=6952$

Q88JEE Main 2023NAT4MProbability

Let ${\text{A}}$ be the event that the absolute difference between two randomly choosen real numbers in the sample space $[0,60]$ is less than or equal to a If ${\text{P}}( {\text{A}})=\;\frac{11}{36}$ , then ${\text{a}}$ is equal to _______. [31-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$| {\text{x}}- {\text{y}}| < {\text{a}} \Rightarrow - {\text{a}} < {\text{x}}- {\text{y}} < {\text{a}}$ $\Rightarrow {\text{x}}- {\text{y}} < {\text{a}} \;\text{ and }\; {\text{x}}- {\text{y}} > - {\text{a}}$ $\; {\text{P}}( {\text{A}})=\;{{\text{ar}}( {\text{OACDEG}})}/{( {\text{OBDF}})}$ $\;=\;{{\text{ar}}( {\text{OBDF}})-{\text{ar}}( {\text{ABC}})-{\text{ar}}( {\text{EFG}})}/{{\text{ar}}( {\text{OBDF}})}$ $\;\Rightarrow \;\frac{11}{36}=\;{(60)^{2-}\frac{1}{2}(60- {\text{a}})^{2-}\frac{1}{2}(60- {\text{a}})^2}/{3600}$ $\;\Rightarrow \;\; 1100=3600-(60- {\text{a}})^2$ $\;\Rightarrow \;\; (60- {\text{a}})^2=2500 \Rightarrow 60- {\text{a}}=50$ $\;\Rightarrow \;\; {\text{a}}=10$

Q89JEE Main 2023NAT4MMatrices and Determinants

Let ${\text{A}}= [ {\text{a}}_{ {\text{ij}}}] , {\text{a}}_{ {\text{ij}}} \in {\text{Z}} ∩[0,4], 1 \le {\text{i}}, {\text{j}} \le 2$ . The number of matrices A such that the sum of all entries is a prime number ${\text{p}} \in (2,13)$ is _______. [31-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

As given ${\text{a}}+ {\text{b}}+ {\text{c}}+ {\text{d}}=3$ or 5 or 7 or 11if sum $=3$ $\; (1+x+x^{2+}. . .+x^4) ^4 \to x^3$ $\; (1-x^5) ^4(1-x)^{-4} \to x^3$ $\; \therefore { }^{4+3-1} C_3={ }^6 C_3=20$ If ${\ \sum}=5$ $\; (1-4 {\text{x}}^5) (1- {\text{x}})^{-4} \to {\text{x}}^5$ $\;\Rightarrow { }^{4+5-1} {\text{C}}_5-4 {\text{x}}^{4.4+0-1} {\text{C}}_0={ }^8 {\text{C}}_5-4=52$ If sum $=7$ $\; (1-4 {\text{x}}^5) (1- {\text{x}})^{-4} \to {\text{x}}^7$ $\;\Rightarrow { }^{4+5-1} {\text{C}}_4-{ }^{4.4+0-1} {\text{C}}_0={ }^8 {\text{C}}_5-4=52$ $\;\;\text{ If \sum }\;=11$ $\; \;\;\;\; (1-4 {\text{x}}^{5+}6 {\text{x}}^{10}) (1- {\text{x}})^{-4} \to {\text{x}}^{11}$ $\;\Rightarrow { }^{4+11-1} {\text{C}}_{11}-4 \cdot { }^{4+6-4} {\text{C}}_6+6 \cdot { }^{4+1-1} {\text{C}}_1$ $\;={ }^{14} {\text{C}}_{11}-4 \cdot { }^9 {\text{C}}_6+6.4=364-336+24=52$ $\; \therefore \;\text{ Total matrices }\;=20+52+80+52=204$

Q90JEE Main 2023NAT4MMatrices and Determinants

Let $A$ be a $n \times n$ matrix such that $|A|=2$ . If the determinant of the matrix ${\text{Adj}} (2 . {\text{Adj}} (2 {\text{A}}^{-1}) )$ . is $2^{84}$ , then ${\text{n}}$ is equal to ______. [31-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; |{\text{Adj}} (2 {\text{Adj}} (2 A^{-1}) ) |$ $\;= |2 {\text{Adj}} ({\text{Adj}} (2 A^{-1}) ) |^{n-1}$ $\;=2^{n(n-1)} |{\text{Adj}} (2 A^{-1}) |^{n-1}$ $\;=2^{n(n-1)} | (2 A^{-1}) |^{(n-1)(n-1)}$ $\;=2^{n(n-1)} 2^{n(n-1)(n-1)} |A^{-1}|^{(n-1)(n-1)}$ $\;=2^{n(n-1)+n(n-1)(n-1)} \;\frac{1}{|A|^{(n-1)^2}}$ $\;=\;\frac{2^{n(n-1)+n(n-1)(n-1)}}{2^{(n-1)^2}}$ $\;=2^{n(n-1)+n(n+1)^{2-}(n-1)^2}$ $\;=2^{(n-1) (n^{2-}n+1) }$ Now, $2^{( {\text{n}}-1) ( {\text{n}}^{2-} {\text{n}}+1) }$ $2^{( {\text{n}}-1) ( {\text{n}}^{2-} {\text{n}}+1) }=2^{84}$ So, ${\text{n}}=5$

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