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31 Jan 2023 Shift 1

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Physics30 questions

Q1JEE Main 2023MCQ4MMagnetism and Matter

A bar magnet with a magnetic moment $5.0 {\text{Am}}^2$ is placed in parallel position relative to a magnetic field of $0.4 {\text{T}}$ . The amount of required work done in turning the magnet form parallel to antiparallel position relative to the field direction is ________. [31-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{u}}=- {\text{MB}} \cos \theta$ $\; {\text{W}}=\Delta {\text{u}}$ $\; {\text{W}}=- {\text{MB}} \cos 180^{\circ} (- {\text{mB}} \cos 0^{\circ})$ $\; {\text{W}}=2 {\text{MB}}=2 \times 5 \times 0.4=4 {\text{J}}$ Option 1

Q2JEE Main 2023MCQ4MDual Nature of Matter and Radiation

If a source of electromagnetic radiation having power $15 {\text{kW}}$ produces $10^{16}$ photons per second, the radiation belongs to a part of spectrum is.(Take Planck constant ${\text{h}}=6 \times 10^{-34} {\text{Js}}$ ) [31-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\text{ Energy of one photon }\;=\;\frac{\;\text{ Power }\;}{\;\text{ Photon frequency }\;}$ $\;E=h v=\;\frac{15 \times 10^3}{10^{16}}$ $\;v=\;\frac{15 \times 10^{-13}}{6 \times 10^{-34}}=2.5 \times 10^{21}$ So gamma Rays. Option 3

Q3JEE Main 2023MCQ4MElectromagnetic Waves

The amplitude of $15 sin \;(1000 \pi {\text{t}})$ is modulated by $10 sin \;(4 \pi {\text{t}})$ signal. The amplitude modulated signal contains frequency(ies) of(A) $500 {\text{Hz}}$ (B) $2 {\text{Hz}}$ (C) $250 {\text{Hz}}$ (D) $498 {\text{Hz}}$ (E) $502 {\text{Hz}}$ Choose the correct answer from the options given below: [31-Jan-2023 Shift 1]

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📖 Explanation

Carrier wave frequency ${\text{V}}_{ {\text{C}}}=\;\frac{100 \pi}{2 \pi}=500 {\text{Hz}}$ Modulating wave frequency $\; {\text{V}}_{ {\text{m}}}=\;\frac{4 \pi}{2 \pi}=2 {\text{Hz}}$ $\; \therefore \;\; {\text{V}}_{ {\text{C}}}- {\text{V}}_{ {\text{m}}}, {\text{V}}_{ {\text{C}}}, {\text{V}}_{ {\text{C}}}+ {\text{V}}_{ {\text{m}}}$ $\; \;\; =498 {\text{Hz}} .500 {\text{Hz}}, 502 {\text{Hz}}$

Q4JEE Main 2023MCQ4MLaws of Motion

As shown in figure, a $70 {\text{kg}}$ garden roller is pushed with a force of ${ \vec{\text{F}}=200 {\text{N}}$ at an angle of $30^{\circ}$ with horizontal. The normal reaction on the roller is (Given ${\text{g}}=10 {\text{m}} {\ s}^{-2}$ ) [31-Jan-2023 Shift 1]

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📖 Explanation

${\text{N}}\;= {\text{mg}}+ {\text{F}} sin \;30^{\circ}$ $\;=700+200 \times \;\frac{1}{2}=800 \;\text{ newton. }\;$

Q5JEE Main 2023MCQ4MMotion in a Plane

The initial speed of a projectile fired from ground is ${\text{u}}$ . At the highest point during its motion, the speed of projectile is $\;{{\sqrt{3}}}/{2} {\text{u}}$ . The time of flight of the projectile is: [31-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{u}} \cos \theta=\;{{\sqrt{3}} {\text{u}}}/{2} \Rightarrow \cos \theta=\;{{\sqrt{3}}}/{2}$ $\;\Rightarrow \theta=30^{\circ}$ $\; {\text{T}}=\;{2 {\text{u}} sin \;30^{\circ}}/{ {\text{g}}}=\;{ {\text{u}}}/{ {\text{g}}}$ Option 2.

Q6JEE Main 2023MCQ4MMotion in a Straight Line

Spherical insulating ball and a spherical metallic ball of same size and mass are dropped from the same height. Choose the correct statement out of the following {Assume negligible air friction} [31-Jan-2023 Shift 1]

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Q7JEE Main 2023MCQ4MAtoms and Nuclei

A free neutron decays into a proton but a free proton does not decay into neutron. This is because [31-Jan-2023 Shift 1]

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Q8JEE Main 2023MCQ4MSemiconductor Electronics

The effect of increase in temperature on the number of electrons in conduction band $(n_e)$ and resistance of a semiconductor will be as: [31-Jan-2023 Shift 1]

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Q9JEE Main 2023MCQ4MOscillations

The maximum potential energy of a block executing simple harmonic motion is 25 J. A is amplitude of oscillation. At A $/ 2$ , the kinetic energy of the block is [31-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{u}}_{\max }=\;\frac{1}{2} {\text{m}} \omega ^2 {\text{A}}^2=25 {\text{J}}$ $\; {\text{KE}} \;\text{ at }\; \;{ {\text{A}}}/{2}=\;\frac{1}{2} {\text{m}} {\text{v}}_1^2=\;\frac{1}{2} {\text{m}} \omega ^2 ( {\text{A}}^{2-}\;{ {\text{A}}^2}/{4})$ $\; {\text{KE}}=\;\frac{1}{2} {\text{m}} \omega ^2 \;{3 {\text{A}}^2}/{4}=\;\frac{3}{4} (\;\frac{1}{2} {\text{m}} \omega ^2 {\text{A}}^2)$ $\; {\text{KE}}=\;\frac{3}{4} \times 25=18.75 {\text{J}}$

Q10JEE Main 2023MCQ4MThermodynamics

The pressure of a gas changes linearly with volume from A to B as shown in figure. If no heat is supplied to or extracted from the gas then change in the internal energy of the gas will be [31-Jan-2023 Shift 1]

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📖 Explanation

As $\Delta {\text{q}}=0$ $\Delta {\text{u}}\;=- {\text{W}}$ ${\text{W}}\;=\int {\text{PdV}}$ $\Delta {\text{u}}\;=- {\text{W}}=30 \times 10^3 \times 150 \times 10^{-6}$ $\;=4500 \times 10^{-3}$ $\;=4.5 {\text{J}}$

Q11JEE Main 2023MCQ4MElectrostatic Potential and Capacitance

Which of the following correctly represents the variation of electric potential (V) of a charged spherical conductor of radius (R) with radial distance (r) from the centre? [31-Jan-2023 Shift 1]

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Q12JEE Main 2023MCQ4MDual Nature of Matter and Radiation

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: The beam of electrons shows wave nature and exhibit interference and diffraction. Reason R : Davisson Germer Experimentally verified the wave nature of electrons.In the light of the above statements. Choose the most appropriate answer from the options given below : [31-Jan-2023 Shift 1]

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Q13JEE Main 2023MCQ4MCurrent Electricity

The drift velocity of electrons for a conductor connected in an electrical circuit is ${\text{V}}_{ {\text{d}}}$ . The conductor in now replaced by another conductor with same material and same length but double the area of cross section. The applied voltage remains same. The new drift velocity of electrons will be [31-Jan-2023 Shift 1]

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📖 Explanation

${\text{V}}_{ {\text{d}}}=\;{ {\text{eE}}}/{ {\text{m}}} \tau$ that is independent of area

Q14JEE Main 2023MCQ4MGravitation

At a certain depth "d" below surface of earth. value of acceleration due to gravity becomes four times that of its value at a height $3 R$ above earth surface. Where ${\text{R}}$ is Radius of earth (Take ${\text{R}}=6400 {\text{km}}$ ). The depth ${\text{d}}$ is equal to [31-Jan-2023 Shift 1]

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📖 Explanation

$\;\;{ {\text{GM}}}/{ {\text{R}}^2} [1-\;{ {\text{d}}}/{ {\text{R}}}] =\;{4 \times {\text{GM}}}/{(4 {\text{R}})^2}$ $\;1-\;{ {\text{d}}}/{ {\text{R}}}=\;\frac{1}{4} \Rightarrow \;{ {\text{d}}}/{ {\text{R}}}=\;\frac{3}{4} \Rightarrow {\text{d}} \;\frac{3}{4} {\text{R}}$ $\; {\text{d}}=4800 {\text{km}}$

Q15JEE Main 2023MCQ4MProperties of Matter

If 1000 droplets of water of surface tension $0.07 {\text{N}} / {\text{m}}$ . having same radius $1 {\text{mm}}$ each, combine to from a single drop. In the process the released surface energy is- $(.$ Take $\pi=\;\frac{22}{7})$ [31-Jan-2023 Shift 1]

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📖 Explanation

$\;1000 \times \;\frac{4 \pi}{3}(1)^3=\;\frac{4 \pi}{3} {\text{R}}^3$ $\; {\text{R}}=10 {\text{mm}}$ $\; {\text{T}} \times 1000 \times 4 \pi (10^{-3}) ^{2-} {\text{T}} \times 4 \pi (10 \times 10^{-3}) ^2=\Delta {\text{E}}$ $\;\Delta {\text{E}}=4 \times \pi \times 7 \times 10^{-2}[1000-100] \times 10^{-6}$ $\;\Delta {\text{E}}=7.92 \times 10^{-4} {\text{J}}$ Option 2.

Q16JEE Main 2023MCQ4MMagnetism and Matter

A rod with circular cross-section area $2 {\text{cm}}^2$ and length $40 {\text{cm}}$ is wound uniformly with 400 turns of an insulated wire. If a current of $0.4 {\text{A}}$ flows in the wire windings, the total magnetic flux produced inside windings is $4 \pi \times 10^{-6} {\text{Wb}}$ . The relative permeability of the rod is(Given : Permeability of vacuum $\mu _0=4 \pi \times 10^{-7} {\text{NA}}^{-2} \;\text{ ) }\;$ [31-Jan-2023 Shift 1]

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📖 Explanation

$\;\varphi =\mu _{ {\text{r}}} \mu _0 \;{ {\text{N}}}/{ℓ} {\text{I}} {\text{xA}}$ $\;\mu _{ {\text{r}}}=125$ Option 3.

Q17JEE Main 2023MCQ4MThermodynamics

The correct relation between $\gamma =\;{ {\text{C}}_{ {\text{p}}}}/{ {\text{c}}_{ {\text{v}}}}$ and temperature $T$ is: [31-Jan-2023 Shift 1]

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📖 Explanation

$\gamma$ is independent of temperatureOption 2

Q18JEE Main 2023MCQ4MWave Optics

Two polaroide ${\text{A}}$ and ${\text{B}}$ are placed in such a way that the pass-axis of polaroids are perpendicular to each other. Now, another polaroid ${\text{C}}$ is placed between ${\text{A}}$ and ${\text{B}}$ bisecting angle between them. If intensity of unpolarised light is ${\text{I}}_0$ then intensity of transmitted light after passing through polaroid B will be : [31-Jan-2023 Shift 1]

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📖 Explanation

$\;I_A=\;\frac{I_o}{2}$ $\;I C=\;\frac{I_{\circ }}{2} \cos ^2 45=\;\frac{I_o}{4}$ $\;I_B=I_C \cos ^2 45=\;\frac{I_o}{8}$ Option 3.

Q19JEE Main 2023MCQ4MAlternating Current

If ${\text{R}}, {\text{X}}_{ {\text{L}}}$ and ${\text{X}}_{ {\text{C}}}$ represent resistance, inductive reactance and capacitive reactance. Then which of the following is dimensionless: [31-Jan-2023 Shift 1]

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📖 Explanation

All three have same dimension therefore $\;{ {\text{R}}}/{\sqrt{ {\text{X}}_{ {\text{L}}} {\text{X}}_{ {\text{C}}}}}$ is dimensionless.Option 2

Q20JEE Main 2023MCQ4MCentre of Mass and Collision

100 balls each of mass ${\text{m}}$ moving with speed ${\text{v}}$ simultaneously strike a wall normally and reflected back with same speed, in time $t {\ s}$ . The total force exerted by the balls on the wall is [31-Jan-2023 Shift 1]

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📖 Explanation

${\text{P}}_{ {\text{i}}}= {\text{Nm}} ∪ \;{ {\text{i}}}^{\^} \;\; { \vec{\text{P}}_{ {\text{f}}}=- {\text{Nm}} v \;{ {\text{i}}}^{\^}$ ${\text{N}}$ is Number of balls strikes with wall $\; {\text{N}}=100$ $\;\Delta { \vec{\text{p}}={ \vec{\text{P}}_{ {\text{f}}}-{ \vec{\text{P}}_{ {\text{i}}}=-2 {\text{Nm}} v \;{ {\text{i}}}^{\^}$ $\;=-200 {\text{Nm}} \;{\;\hat{i}^{\^}$ $\;{ \vec{\text{F}}_{\;\text{Total }\;}=\;{\Delta { \vec{\text{P}}}/{\Delta {\text{t}}}=-\;{200 {\ mvt}}/{ {\text{t}}}$ $\;|{ \vec{\text{F}}|=\;{200 {\ mv}}/{ {\text{t}}}$ $\;$

Q21JEE Main 2023NAT4MProperties of Matter

A thin rod having a length of $1 {\text{m}}$ and area of cross-section $3 \times 10^{-6} {\text{m}}^2$ is suspended vertically from one end. The rod is cooled from $210^{\circ} {\text{C}}$ to $160^{\circ} {\text{C}}$ . After cooling, a mass ${\text{M}}$ is attached at the lower end of the rod such that the length of rod again becomes $1 {\text{m}}$ . Young's modulus and coefficient of linear expansion of the rod are $2 \times 10^{11} {\text{Nm}}^{-2}$ and $2 \times 10^{-5} {\text{K}}^{-1}$ , respectively. The value of $M$ is _______ ${\text{kg}}$ . (Take ${\text{g}}=10 {\text{m}} {\ s}^{-2}$ ) [31-Jan-2023 Shift 1]

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📖 Explanation

If $\Delta ℓ$ is decease in length of rod due to decease in temperature $\;\Delta ℓ=ℓ \alpha \Delta {\text{T}}$ $\;\alpha=2 \times 10^{-5} {\text{K}}^{-1}, \Delta {\text{T}}=(210-160)$ $\;=50 {\text{K}}$ $\;\Delta ℓ=1 \times 2 \times 10^{-5} \times 50=10^{-3} {\text{m}}$ $\;\;\text{ Young Modulus }\;= {\text{Y}}=\;{ {\text{F}} / {\text{A}}}/{\Delta ℓ / ℓ} \;\; {\text{A}}=3 \times 10^{-6} {\text{m}}^2$ $\;2 \times 10^{11}=\;{ {\text{Mg}} / 3 \times 10^{-6}}/{10^{-3} / 1}$ $\; {\text{Mg}}=2 \times 10^{11} \times 3 \times 10^{-9}=6 \times 10^{-2}$ $\;M=60 {\text{kg}}$ $\;$ Ans is 60 .

Q22JEE Main 2023NAT4MMotion in a Plane

The speed of a swimmer is $4 {\text{km}} {\text{h}}^{-1}$ in still water. If the swimmer makes his strokes normal to the flow of river of width $1 {\text{km}}$ , he reaches a point $750 {\text{m}}$ down the stream on the opposite bank.The speed of the river water is _______ ${\text{km}} {\text{h}}^{-1}$ . [31-Jan-2023 Shift 1]

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📖 Explanation

time to cross the River width $\omega =1000 {\text{m}}$ $\;\text{ is }\;=\;{1 {\text{km}}}/{4 {\text{km}} / {\text{h}}}$ Drift ${\text{x}}= {\text{Vm}} / {\text{g}} \times {\text{t}}$ Where ${\text{Vm}} / {\text{g}}$ is velocity of River w.r. to ground. $\; {\text{x}}= {\text{Vm}} / {\text{g}} \times \;\frac{1}{4}=750 {\text{m}}=\;\frac{3}{4} {\text{km}}$ $\; {\text{Vm}} / {\text{g}}=3 {\text{km}} / {\text{hr}}$ Ans is $3 {\text{km}} / {\text{hr}}$ .

Q23JEE Main 2023NAT4MOscillations

In the figure given below. a block of mass ${\text{M}}=490 {\text{g}}$ placed on a frictionless table is connected with two springs having same spring constant $( {\text{K}}=2 {\text{N}} {\text{m}}^{-1})$ . If the block is horizontally displaced through ' ${\text{X}}$ 'm then the number of complete oscillations it will make in $14 \pi$ seconds will be _______ [31-Jan-2023 Shift 1]

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📖 Explanation

$\text{Keff}=\text{K}+\text{K}$ as both springs are in use in parallel $=2 {\text{k}}\;$ $=2 \times 2=4 {\text{N}} / {\text{m}} \;\; {\text{m}}\;=490 {\text{gm}}$ $\;=0.49 {\text{kg}}$ $T\;=2 \pi \sqrt{\;{ {\text{m}}}/{ {\text{Keff}}}}=2 \pi \sqrt{\;{0.49 {\text{kg}}}/{4}}$ $\;=2 \pi \sqrt{\;\frac{49}{400}}=2 \pi \;\frac{7}{20}=\;\frac{7 \pi}{10}$ No. of oscillation in the $14 \pi$ is ${\text{N}}=\;{\;\text{ time }\;}/{ {\text{T}}}=\;\frac{14 \pi}{7 \pi / 10}=20$ Ans in 20 .

Q24JEE Main 2023NAT4MElectromagnetic Waves

In a medium the speed of light wave decreases to $0.2$ times to its speed in free space The ratio of relative permittivity to the refractive index of the medium is $x: 1$ . The value of $x$ is _______.(Given speed of light in free space $=3 \times 10^8 {\text{m}} {\ s}^{-1}$ and for the given medium $\mu _{ {\text{r}}}=1$ ) [31-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{V}}=\;{ {\text{C}}}/{\mu } \Rightarrow \mu =\;{ {\text{C}}}/{ {\text{V}}}=\;{ {\text{C}}}/{0.2 {\text{C}}}$ $\;\mu =5$ $\;\mu =\sqrt{E_{ {\text{r}}} \mu _{ {\text{r}}}}$ $\;\Rightarrow E_{ {\text{r}}}=\;{\mu ^2}/{\mu _{ {\text{r}}}}$ $\; \therefore \;{E_{ {\text{r}}}}/{\mu }=\;{\mu }/{\mu _{ {\text{r}}}}=5$

Q25JEE Main 2023NAT4MRotational Motion

A solid sphere of mass $1 {\text{kg}}$ rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{-3} {\text{J}}$ . The speed of the centre of mass of the sphere is _______ ${\text{cm}} {\ s}^{-1}$ . [31-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\frac{1}{2} {\ mv}^{2+}\;\frac{1}{2} {\text{I}}^2=7 \times 10^{-3}$ $\;\;\frac{1}{2} {\ mv}^{2+}\;\frac{1}{2} (\;\frac{2}{5} {\text{MR}}^2) (\;{ {\text{V}}}/{ {\text{R}}}) ^2=7 \times 10^{-3}$ $\;\;\frac{1}{2} {\text{MV}}^2 [1+\;\frac{2}{5}] =7 \times 10^{-3}$ $\;\;\frac{1}{2}(1) ( {\text{V}}^2) (\;\frac{7}{5}) =7 \times 10^{-3}$ $\; {\text{V}}^2=10^{-2}$ $\; {\text{V}}=10^{-1}=0.1 {\text{m}} / {\ s}=10 {\text{cm}} / {\ s}$ $\;\;\text{ Ans: }\; 10$

Q26JEE Main 2023NAT4MAlternating Current

An inductor of $0.5 {\text{mH}}$ , a capacitor of $20 \mu {\text{F}}$ and resistance of $20 Ω$ are connected in series with a $220 {\text{V}}$ ac source. If the current is in phase with the emf, the amplitude of current of the circuit is $\sqrt{ {\text{x}}} {\text{A}}$ . The value of ${\text{x}}$ is - [31-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{X}}_{ {\text{L}}}= {\text{X}}_{ {\text{C}}}$ $\;\;\text{ So, }\; Z=R=20 Ω$ $\; {\text{i}}_{ {\text{ms}}}=\;\frac{220}{20}=11$ $\; {\text{i}}_{\max }=11 {\sqrt{2}}={\sqrt{242}}$ Ans: 242

Q27JEE Main 2023NAT4MElectrostatics

Expression for an electric field is given by ${ \vec{\text{E}}=4000 x^2 \;{ {\text{i}}}^{\^} \;{ {\text{V}}}/{ {\text{m}}}$ . The electric flux through the cube of side $20 {\text{cm}}$ when placed in electric field (as shown in the figure) is ________ ${\text{V}} {\text{cm}}$ . [31-Jan-2023 Shift 1]

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📖 Explanation

$\;\text{ Flux }\;\;={ \vec{\text{E}} \cdot { \vec{\text{A}}$ $\;=4000(0 \cdot 2)^2 \;{ {\text{V}}}/{ {\text{m}}} \cdot (0 \cdot 2)^2 {\text{m}}^2$ $\;=4000 \times 16 \times 10^{-4} {\text{Vm}}$ $\;=640 {\text{Vcm}}$ Ans. 640

Q28JEE Main 2023NAT4MWork, Power and Energy

A lift of mass ${\text{M}}=500 {\text{kg}}$ is descending with speed of $2 {\text{ms}}^{-1}$ . Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of $2 {\text{ms}}^{-2}$ . The kinetic energy of the lift at the end of fall through to a distance of $6 {\text{m}}$ will be _______ ${\text{kJ}}$ . [31-Jan-2023 Shift 1]

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📖 Explanation

${\text{v}}^2\;= {\text{u}}^{2+}2 {\text{as}}$ $\;=2^{2+}2(2)(6)$ $\;=4+24=28$ ${\text{KE}}\;=\;\frac{1}{2} {\ mv}^2$ $\;=\;\frac{1}{2}(500) 28$ $\;=7000 {\text{J}}$ $\;=7 {\text{kJ}}$ Ans. 7

Q29JEE Main 2023NAT4MAtoms and Nuclei

For hydrogen atom, $\lambda _1$ and $\lambda _2$ are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of $\lambda _1$ and $\lambda _2$ is $\;\frac{x}{32}$ . The value of $x$ is _______. [31-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\frac{1}{\lambda }= {\text{Rz}}^2 [\;{1}/{ {\text{n}}_1^2}-\;{1}/{ {\text{n}}_2^2}]$ $\;\;\frac{1}{\lambda _1}= {\text{Rz}}^2 \$ [;\frac{1}{1^2}-;\frac{1}{3^2}]$ =;\frac{8}{9} {\text{Rz}}^2 $. . . (1)$ ;;\frac{1}{\lambda _2}= {\text{Rz}}^2 $[;\frac{1}{1^2}-;\frac{1}{2^2}]$ =;\frac{3}{4} {\text{Rz}}^2 $. . . (2)$ 1 / 2 \Rightarrow ;\frac{\lambda _2}{\lambda _1}=;\frac{8}{9} \times ;\frac{4}{3}=;\frac{32}{27};\frac{\lambda _1}{\lambda _2}=;\frac{27}{32}$Ans. 27

Q30JEE Main 2023NAT4MCurrent Electricity

Two identical cells, when connected either in parallel or in series gives same current in an external resistance $5 Ω$ . The internal resistance of each cell will be ________ $Ω$ . [31-Jan-2023 Shift 1]

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📖 Explanation

$i=\;\frac{2 \epsilon}{5+2 r}$ (1) $\;\;$ i $=\;{\epsilon}/{{ {\text{r}}}/{2}+5}$ . . . (2)Equating (1) and (2) $\;\;{2 \epsilon}/{5+2 {\text{r}}}=\;{\epsilon}/{{ {\text{r}}}/{2}+5} \Rightarrow {\text{r}}+10=5+2 {\text{r}}$ $\; {\text{r}}=5 \;\text{ Ans. }\; 5$

Chemistry30 questions

Q31JEE Main 2023MCQ4Md and f Block Elements

${\text{Nd}}^{2+}=$ _______ [31-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{Nd}}(60)=[ {\text{Xe}}] 4 {\text{f}}^4 5 {\text{d}}^0 6 {\ s}^2$ $\; {\text{Nd}}^{2+}=[ {\text{Xe}}] 4 {\text{f}}^4 5 {\text{d}}^0 5 {\ s}^0$

Q32JEE Main 2023MCQ4Md and f Block Elements

The methods NOT involved in concentration of ore are(A) Liquation(B) Leaching(C) Electrolysis(D) Hydraulic washing(E) Froth floatationChoose the correct answer from the options given below : [31-Jan-2023 Shift 1]

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Q33JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

Consider the following reactionPropanal $+$ Methanal $= {\to }^{\;\text{ (i)dil. }\; {\text{NaOH}}}_{(\text{ii})\Delta }_{(\text{iii})\text{NaCN}}_{(\text{iv})\text{H}_3\text{O}^+} {\text{Product} \text{B}}_{(\text{C}_5\text{H}_8\text{O}_3)}$ The correct statement for product ${\text{B}}$ is. It is [31-Jan-2023 Shift 1]

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📖 Explanation

${\text{CH}}_3- {\text{CH}}_2- {\text{CHO}}+ {\text{HCHO}} \; {\to }^{\text{OH}^{-}}_{\Delta }$ Carboxylic acid will give ${\text{CO}}_2$ gas, with ${\text{NaHCO}}_3$ solution

Q34JEE Main 2023MCQ4Md and f Block Elements

The correct order of basicity of oxides of vanadium is [31-Jan-2023 Shift 1]

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📖 Explanation

With increase in $\%$ of oxygen acidic nature of oxide of an element increase and basic nature decreases

Q35JEE Main 2023MCQ4Md and f Block Elements

When ${\text{Cu}}^{2+}$ ion is treated with ${\text{KI}}$ , a white precipitate, ${\text{X}}$ appears in solution. The solution is titrated with sodium thiosulphate, the compound ${\text{Y}}$ is formed. ${\text{X}}$ and ${\text{Y}}$ respectively are [31-Jan-2023 Shift 1]

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📖 Explanation

${\text{Cu}}^{2+}+2 {\text{KI}} \to \;{ {\text{CuI}}_2}_{\;\text{ Unstable }\;} ↓+2 {\text{K}}^{+}$ $Γ^{-}$ is strong R.A it reduces ${\text{Cu}}^{2+}$ to ${\text{Cu}}^{+}$ $2 {\text{CuI}}_2 \to \ { {\text{Cu}}_2 {\text{I}}_2 ↓}_{(\text{White}) '\text{X}'}+ {\text{I}}_2$ ${\text{KI}}+ {\text{I}}_2 \to {\text{K}}^{+} {\text{I}}_3^{-} \;\text{(Brown solution) }\;$ ${\text{I}}_3^{-} ⇌ {\text{I}}_2+ {\text{I}}^{-}$ ${\text{KI}}_3+ {\text{Na}}_2 {\text{S}}_2 {\text{O}}_3 \to {\text{KI}}+{ {\text{Na}}_2 {\text{S}}_4 {\text{O}}_6}_{(\text{Y})}$

Q36JEE Main 2023MCQ4MAmines

${\to }^ {{\text{H}}_2 / {\text{Pd}}}_{\text{C}_2\text{H}_5\text{OH}}[\text{A}] {\to }^ {( {\text{CH}}_3 {\text{CO}}) _2 {\text{O}} }_{\text{Pyridine}}\to [ {\text{B}}]$ [31-Jan-2023 Shift 1]

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Q37JEE Main 2023MCQ4MCoordination Compounds

Cobalt chloride when dissolved in water forms pink colored complex ${\text{X}}$ which has octahedral geometry. This solution on treating with cone ${\text{HCl}}$ forms deep blue complex, $Y$ which has a $Z$ geometry. ${\text{X}}, {\text{Y}}$ and ${\text{Z}}$ , respectively, are [31-Jan-2023 Shift 1]

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Q38JEE Main 2023MCQ4Mp Block Elements

Identify ${\text{X}}, {\text{Y}}$ and ${\text{Z}}$ in the following reaction. (Equation not balanced) ${\text{Cl}} ˙{ {\text{O}}}+ {\text{NO}}_2 \to x{ {\text{X}}} {\to }^{ {\text{H}}_2 {\text{O}}} x{ {\text{Y}}}+x{ {\text{Z}}}$ [31-Jan-2023 Shift 1]

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📖 Explanation

${\text{Cl}} ˙{ {\text{O}}}+ {\text{NO}}_2 \to {{\text{ClONO}}_2 }_{(\text{X})} {\to }^{+ {\text{H}}, {\text{O}}} {\text{HOCl}}_{(\text{Y})}+ {{\text{HNO}}_3}_{(\text{Z})}$

Q39JEE Main 2023MCQ4MHaloalkanes and Haloarenes

The correct order of melting point of dichlorobenzenes is [31-Jan-2023 Shift 1]

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Q40JEE Main 2023MCQ4MBiomolecules

A protein ' ${\text{X}}$ ' with molecular weight of $70,000 {\text{u}}$ , on hydrolysis gives amino acids. One of these amino acid is [31-Jan-2023 Shift 1]

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📖 Explanation

Only in option (2) $\alpha$ -Amino acid is given all the other options are not $\alpha$ -Amino acids.

Q41JEE Main 2023MCQ4MStructure of Atom

Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from ${\text{n}}=4$ to ${\text{n}}=2$ of ${\text{He}}^{+}$ spectrum [31-Jan-2023 Shift 1]

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📖 Explanation

${\text{He}}^{+}$ ion: $\;\;{1}/{\lambda ( {\text{H}})}= {\text{R}}(1)^2 [\;{1}/{ {\text{n}}_1^2}-\;{1}/{ {\text{n}}_2^2}]$ $\;\;{1}/{\lambda ( {\text{He}}^{+}) }= {\text{R}}(2)^2 [\;\frac{1}{2^2}-\;\frac{1}{4^2}]$ Given $\lambda ( {\text{H}})=\lambda ( {\text{He}}^{+})$ $\; {\text{R}}(1)^2 \$ [;{1}/{ {\text{n}}_1^2}-;{1}/{ {\text{n}}_2^2}]$ = {\text{R}}(4) [;\frac{1}{2^2}-;\frac{1}{4^2}];;{1}/{ {\text{n}}_1^2}-;{1}/{ {\text{n}}_2^2}=;\frac{1}{1^2}-;\frac{1}{2^2} $On comparing$ {\text{n}}_1=1 & {\text{n}}_2=2$Ans. 1

Q42JEE Main 2023MCQ4MPractical Organic Chemistry

Match items of column I and II Column I (Mixture of compounds) Column II (Separation Technique) A. ${\text{H}}_2 {\text{O}} / {\text{CH}}_2 {\text{Cl}}_2$ i. Crystallization ii. Differential solvent extraction C. Kerosene/Naphthalene iii. Column chromatography D. ${\text{C}}_6 {\text{H}}_{12} {\text{O}}_6 / {\text{NaCl}}$ iv. Fractional DistillationCorrect match is: [31-Jan-2023 Shift 1]

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📖 Explanation

A. ${\text{H}}_2 {\text{O}} / {\text{CH}}_2 {\text{Cl}}_2 \to$ ii, ${\text{CH}}_2 {\text{Cl}}_2 > {\text{H}}_2 {\text{O}}$ (density) so they can be separated by differential solvent extraction.B.iii. columnchromatography Due to ${\text{H}}$ -bonding in it can be separated fromby column chromatography. C. Kerosene / Naphthalene $\to$ iv. Fractional distillation.Due to different B.P. of kerosene and Naphthalene it can be separated by fractional distillation.D. ${\text{C}}_6 {\text{H}}_{12} {\text{O}}_6 / {\text{NaCl}} \to$ i. Crystallization. ${\text{NaCl}}$ (ionic compound) can be crystallized.

Q43JEE Main 2023MCQ4MClassification of Elements

The correct increasing order of the ionic radii is [31-Jan-2023 Shift 1]

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📖 Explanation

In isoelectronic species size $\propto \;\frac{1}{Z}$ $_{}_{\text{Z}:} { {\text{Ca}}^{2+}}_{}_{20} { < {\text{K}}^{+} }_{}_{19}{ < {\text{Cl}}^{-}}_{}_{17}{ < {\text{S}}^{2-}}_{}_{18}: \;\text{ Size }\;$ $Z: 20$ 191718

Q44JEE Main 2023MCQ4MRedox Reactions

${\text{H}}_2 {\text{O}}_2$ acts as a reducing agent in [31-Jan-2023 Shift 1]

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Q45JEE Main 2023MCQ4MBiomolecules

Which of the following artificial sweeteners has the highest sweetness value in comparison to cane sugar? [31-Jan-2023 Shift 1]

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📖 Explanation

Sweetness value order wrt cane sugar Alitame $>$ Sucralose $>$ Saccharin $>$ Aspartame

Q46JEE Main 2023MCQ4MChemical Bonding and Molecular Structure

Match List I with List II List I List II A. ${\text{XeF}}_4$ I. See - saw B. ${\text{SF}}_4$ II. Square planar C. ${\text{NH}}_4^{+}$ III. Bent ${\text{T}}-$ shaped D. ${\text{BrF}}_3$ IV. TetrahedralChoose the correct answer from the options given below : [31-Jan-2023 Shift 1]

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Q47JEE Main 2023MCQ4MHydrocarbons

Choose the correct set of reagents for the following conversion trans $( {\text{Ph}}- {\text{CH}}= {\text{CH}}- {\text{CH}}_3) \to {\text{cis}} ( {\text{Ph}}- {\text{CH}}= {\text{CH}}- {\text{CH}}_3)$ [31-Jan-2023 Shift 1]

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Q48JEE Main 2023MCQ4MSolutions

Adding surfactants in non polar solvent, the micelles structure will look [31-Jan-2023 Shift 1]

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Q49JEE Main 2023MCQ4MAlcohols, Phenols and Ethers

An organic compound ' $A$ ' with empirical formula ${\text{C}}_6 {\text{H}}_6 {\text{O}}$ gives sooty flame on burning. Its reaction with bromine solution in low polarity solvent results in high yield of $B . B$ is [31-Jan-2023 Shift 1]

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Q50JEE Main 2023MCQ4MElectrochemistry

Which one of the following statements is correct for electrolysis of brine solution? [31-Jan-2023 Shift 1]

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📖 Explanation

Electrolysis of brine solution ${\text{NaCl}} \;\text{ (aq.) }\; \to {\text{Na}}_{\;\text{(aq) }\;}^{+}+ {\text{Cl}}_{\;\text{(aq) }\;}^{+}$ At anode : $2 {\text{Cl}}_{( {\text{aq}} .)}^{+} \to {{\text{Cl}}_2( {\text{g}})}_{\text{Major}}+2 {\text{e}}^{-}$ $2 {\text{H}}_2 {\text{O}}_{(ℓ)} \to {{\text{O}}_{2( {\text{g}})}}_{\text{Minor}}+4 {\text{H}}_{( {\text{aq}})}^{+}+4 {\text{e}}^{-}$ At Cathode : $2 {\text{H}}_2 {\text{O}}_{(ℓ)}+2 {\text{e}}^{-} \to {\text{H}}_{2( {\text{g}})} ↑+2 {\text{OH}}_{\;\text{(aq) }\;}^{-}$ $2 {\text{Na}}^{+}+2 {\text{OH}}^{-} \to 2 {\text{NaOH}}$

Q51JEE Main 2023NAT4MElectrochemistry

The logarithm of equilibrium constant for the reaction ${\text{Pd}}^{2+}+4 {\text{Cl}}^{-} ⇌ {\text{PdCl}}_4^{2-}$ is _______ (Nearest integer) $\;\;\text{ Given }\;: \;{2.303 {\text{RT}}}/{ {\text{F}}}=0.06 {\text{V}}$ $\; {\text{Pd}}_{( {\text{aq}})}^{2+}+2 {\text{e}}^{-} ⇌ {\text{Pd}}( {\ s}) \;\; {\text{E}}^{\circ}=0.83 {\text{V}}$ $\; {\text{PdCl}}_4^{2-}( {\text{aq}})+2 {\text{e}}^{-} ⇌ {\text{Pd}}( {\ s})+4 {\text{Cl}}^{-}( {\text{aq}})$ $\; {\text{E}}^{\circ}=0.65 {\text{V}}$ [31-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\text{ Sol. }\; \;\; \Delta {\text{G}}^{\circ}=- {\text{RT}} ℓ {\text{n}} K$ $\;- {\text{nFE}}_{\;\text{cell }\;}^0=- {\text{RT}} \times 2.303 (\log _{10} {\text{K}})$ $\;\;{ {\text{E}}_{\;\text{Cell }\;}^0}/{0.06} \times {\text{n}}=\log {\text{K}}$ . . . (1) $\; {\text{Pd}}^{+2} \;\text{ (aq.) }\;+\text{not} \mathcal{L e}^{-} ⇌ {\text{Pd}}( {\ s}), {\text{E}}_{\;\text{cat,red }\;}^{\circ}=0.83$ $\; {\text{Pd}}( {\ s})+4 {\text{Cl}}^{-} \;\text{(aq.) }\; ⇌ {\text{PdCl}}_4^{2-},( {\text{aq}})+2 {\text{e}}^{-}, {\text{E}}_{\;\text{mat.ouir }\;}^0=0.65$ $\;\;\text{ Net Reaction }\; \to {\text{Pd}}^{2+} \;\text{ (aq.) }\;+4 {\text{Cl}}^{-} \;\text{(aq.) }\; ⇌ {\text{PdCl}}_4{ }^{2-} \;\text{ (aq.) }\;$ $\; {\text{E}}_{\;\text{cell }\;}^{\circ}= {\text{E}}_{\;\text{cat,red }\;}^{\circ}- {\text{E}}_{\;\text{Alode }\;, 0 {\text{c}} {\text{d}}^{ {\text{a}}}}^{\circ}$ $\; {\text{E}}_{\;\text{cell }\;}^{\circ}=0.83-0.65$ $\; {\text{E}}_{\;\text{cell }\;}^{\circ}=0.18$ . . . (2) $\;\;\text{ Also }\; {\text{n}}=2$ . . . (3) $\;\;\text{ Using equation (1), (2) & (3) }\;$ $\;\log K=6$ $\;$

Q52JEE Main 2023NAT4MChemical Kinetics

${\text{A}} \to {\text{B}}$ The rate constants of the above reaction at $200 {\text{K}}$ and $300 {\text{K}}$ are $0.03 {\min}^{-1}$ and $0.05 {\min}^{-1}$ respectively. The activation energy for the reaction is ________ ${\text{J}}$ (Nearest integer)(Given : In $10=2.3$ $\; {\text{R}}=8.3 {\text{J}} {\text{K}}^{-1} {\text{mol}}^{-1}$ $\;\log 5=0.70$ $\;\log 3=0.48$ $\;\log 2=0.30$ [31-Jan-2023 Shift 1]

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📖 Explanation

$\;\log \;{ {\text{K}}_{300}}/{ {\text{K}}_{200}}=\;{ {\text{E}}_{ {\text{a}}}}/{2.3 \times 8.314} (\;{1}/{ {\text{T}}_1}-\;{1}/{ {\text{T}}_2})$ $\;\log \;\frac{0.05}{0.03}=\;{ {\text{Ea}}}/{2.305 \times 8.314} \times [\;\frac{1}{200}-\;\frac{1}{300}]$ $\; {\text{E}}_{ {\text{a}}}=2519.88 {\text{J}} \Rightarrow {\text{E}}_{ {\text{a}}}=2520 {\text{J}}$

Q53JEE Main 2023NAT4MChemical Thermodynamics

The enthalpy change for the conversion of $\;\;\frac{1}{2} {\text{Cl}}_2( {\text{g}}) \;\text{ to }\; {\text{Cl}}^{-}( {\text{aq}}) \;\text{ is }\;(-)$ ________ $\; {\text{kJ}} {\text{mol}}^{-1} ( {\text{Nearest}}^{-}. \;\text{integer) }\;$ $\;\;\text{ Given : }\; \Delta _{ {\text{dis}}} {\text{H}}_{ {\text{Cl}}_{2( {\text{g}})}^{\circ}}^{\circ}=240 {\text{kJmol}}^{-1}$ $\;\Delta _{ {\text{eg}}} {\text{H}}_{ {\text{Cl}}_{( {\text{g}})}}^{\circ}=-350 {\text{kJmol}}^{-1}$ $\;\Delta _{ {\text{hyd}}} {\text{H}}_{ {\text{Cl}}_{( {\text{g}})}^{\circ}}^{\circ}=-380 {\text{kJmol}}^{-1}$ [31-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\frac{1}{2} {\text{Cl}}_{2( {\text{g}})} \to {\text{Cl}}_{( {\text{g}})} \to {\text{Cl}}_{( {\text{g}})}^{-} \to {\text{Cl}}_{( {\text{aq}} .)}^{-}$ $\;\Delta {\text{H}}^{\circ}=\;\frac{1}{2} \times 240+(-350)+(-380)$ $\;=-610 \;\text{ ans. }\;$

Q54JEE Main 2023NAT4MOrganic Chemistry - Some Basic Principles

On complete combustion, $0.492 {\text{g}}$ of an organic compound gave $0.792 {\text{g}}$ of ${\text{CO}}_2$ .The $\%$ of carbon in the organic compound is _______ (Nearest integer) [31-Jan-2023 Shift 1]

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📖 Explanation

weight of ${\text{C}}$ in $0.792 {\text{gm}} {\text{CO}}_2$ $\;=\;\frac{12}{44} \times 0.792=0.216$ $\;\% \;\text{ of }\; {\text{C}} \;\text{ \in compound }\;=\;\frac{0.216}{0.492} \times 100$ $\;=43.90 \%$ Ans: 44

Q55JEE Main 2023NAT4MSolutions

At $27^{\circ} {\text{C}}$ , a solution containing $2.5 {\text{g}}$ of solute in $250.0 {\text{mL}}$ of solution exerts an osmotic pressure of $400 {\text{Pa}}$ . The molar mass of the solute is _______ ${\text{g}}$ ${\text{mol}}^{-1}$ (Nearest integer)(Given : ${\text{R}}=0.083 {\text{L}} {\bar} {\text{K}}^{-1} {\text{mol}}^{-1})$ [31-Jan-2023 Shift 1]

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📖 Explanation

$\pi=$ CRT $\;{400 {\text{Pa}}}/{10^5}=\;{\;{2.5 {\text{g}}}/{ {\text{M}}_{\circ }}}/{250 / 1000 {\text{L}}} \times 0.83 \;{ {\text{L}}- {\bar}}/{ {\text{K}} \cdot {\text{mol}}} \times 300 {\text{K}}$ $M_0=62250$

Q56JEE Main 2023NAT4MSome Basic Concepts of Chemistry

Zinc reacts with hydrochloric acid to give hydrogen and zinc chloride. The volume of hydrogen gas produced at STP from the reaction of $11.5 {\text{g}}$ of zinc with excess ${\text{HCl}}$ is _______ ${\text{L}}$ (Nearest integer)(Given : Molar mass of ${\text{Zn}}$ is $65.4 {\text{g}} {\text{mol}}^{-1}$ and Molar volume of ${\text{H}}_2$ at ${\text{STP}}=22.7 {\text{L}}$ ) [31-Jan-2023 Shift 1]

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📖 Explanation

${\text{Zn}}+2 {\text{HCl}} \to {\text{ZnCl}}_2+ {\text{H}}_2 ↑$ Moles of ${\text{Zn}}$ used $=\;\frac{11.5}{65.4}=$ Moles of ${\text{H}}_2$ evolvedVolume of ${\text{H}}_2=\;\frac{11.5}{65.4} \times 22.7 {\text{L}}=3.99 {\text{L}}$

Q57JEE Main 2023MCQ4MAmines

How many of the transformation given below would result in aromatic amines? [31-Jan-2023 Shift 1]

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📖 Explanation

Product in the given reactions are as follow-2. No reactions will be observed as in Gabriel phthalimide synthesisis poor substrate for ${\text{SN}}^2$ 3. 4. $+ {\text{CH}}_3 {\text{COOH}}$ Aromatic amines will he formed in 1,3 & 4 Ans: 3

Q58JEE Main 2023NAT4MChemical Equilibrium

For reaction: ${\text{SO}}_2( {\text{g}})+\;\frac{1}{2} {\text{O}}_2( {\text{g}}) ⇌ {\text{SO}}_3( {\text{g}})$ $K_P=2 \times 10^{12}$ at $27^{\circ} {\text{C}}$ and 1 atm pressure. The ${\text{K}}_{ {\text{c}}}$ for the same reaction is _______ $\times 10^{13}$ . (Nearest integer)(Given ${\text{R}}=0.082 {\text{L}} {\text{atm}} {\text{K}}^{-1} {\text{mol}}^{-1}$ ) [31-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{SO}}_{2( {\text{g}})}+\;\frac{1}{2} {\text{O}}_{2( {\text{g}})} ⇌ {\text{SO}}_{3( {\text{g}})}$ $\; {\text{K}}_{ {\text{p}}}=2 \times 10^{12} \;\text{ at }\; 300 {\text{K}}$ $\; {\text{K}}_{ {\text{P}}}= {\text{K}}_{ {\text{C}}} \times ( {\text{RT}})^{\Delta n_{ {\text{g}}}}$ $\;2 \times 10^{12}= {\text{K}}_{ {\text{c}}} \times (0.082 \times 300)^{-1 / 2}$ $\; {\text{K}}_{ {\text{C}}}=9.92 \times 10^{12}$ $\; {\text{K}}_{ {\text{C}}}=0.992 \times 10^{13}$ Ans. 1

Q59JEE Main 2023NAT4Mp Block Elements

The oxidation sate of phosphorus in hypophosphoric acid is + ________.. [31-Jan-2023 Shift 1]

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📖 Explanation

${\text{H}}_4 {\text{P}}_2 {\text{O}}_6$ ${\text{HO}}- {\text{P}}_{|}_{\text{O}}_{\text{H}}^{||}^{\text{O}}- {\text{P}}_{|}_{\text{O}}_{\text{H}}^{||}^{\text{O}}- {\text{OH}}$ O.S. of $P$ is $+4$

Q60JEE Main 2023NAT4MStates of Matter

The total pressure of a mixture of non-reacting gases $X(0.6 {\text{g}})$ and $Y(0.45 {\text{g}})$ in a vessel is 740 ${\text{mm}}$ of ${\text{Hg}}$ . The partial pressure of the gas $X$ is ${\text{mm}}$ of ${\text{Hg}}$ . (Nearest Integer)(Given : molar mass ${\text{X}}=20$ and ${\text{Y}}=45 {\text{g}} {\text{mol}}^{-1}$ ) [31-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{P}}_{ {\text{X}}}=χ_{ {\text{X}}}{ }^{P_T}$ $\;=\;\frac{\;\frac{0.6}{20}}{\frac{0.6}{20}+\;\frac{0.45}{45}} \times 740$ $\; {\text{P}}_{ {\text{X}}}=555 {\text{mm}} {\text{Hg}}$

Mathematics30 questions

Q61JEE Main 2023MCQ4MEllipse

If the maximum distance of normal to the ellipse $\;{ {\text{x}}^2}/{4}+\;{ {\text{y}}^2}/{ {\text{b}}^2}=1, {\text{b}}< 2$ , from the origin is 1 , then the eccentricity of the ellipse is: [31-Jan-2023 Shift 1]

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📖 Explanation

Equation of normal is $2 x \ sec \theta-b y {\text{cosec}} \theta=4-b^2$ Distance from $(0,0)=\;{4- {\text{b}}^2}/{\sqrt{4 \ sec ^2 \theta+ {\text{b}}^2 {\text{cosec}}^2 \theta}}$ Distance is maximum if $4 \ sec ^2 \theta+b^2 {\text{cosec}}^2 \theta$ is minimum $\Rightarrow \tan ^2 \theta=\;{ {\text{b}}}/{2}$ $\Rightarrow \;{4-b^2}/{\sqrt{4 \cdot \;\frac{b+2}{2}+b^2 \cdot \;\frac{b+2}{b}}}=1$ $\Rightarrow 4-b^2=b+2 \Rightarrow b=1 \Rightarrow {\text{e}}=\;{{\sqrt{3}}}/{2}$

Q62JEE Main 2023MCQ4MComplex Numbers

For all ${\text{z}} \in {\text{C}}$ on the curve $C_1: |{\text{z}}|=4$ , let the locus of the point $z+\;\frac{1}{z}$ be the curve $C_2$ . Then [31-Jan-2023 Shift 1]

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📖 Explanation

Let $w=z+\;\frac{1}{z}=4 e^{i \theta}+\;\frac{1}{4} {\text{e}}^{- {\text{i}} \theta}$ $\Rightarrow {\text{w}}=\;\frac{17}{4} \cos \theta+ {\text{i}} \;\frac{15}{4} sin \;\theta$ So locus of $w$ is ellipse $\;\frac{x^2}{ (\;\frac{17}{4}) ^2}+\;\frac{y^2}{ (\;\frac{15}{4}) ^2}=1$ Locus of ${\text{z}}$ is circle ${\text{x}}^{2+} {\text{y}}^2=16$ So intersect at 4 points

Q63JEE Main 2023MCQ4MApplication of Derivatives

A wire of length $20 {\text{m}}$ is to be cut into two pieces.A piece of length $ℓ_1$ is bent to make a square of area ${\text{A}}_1$ and the other piece of length $ℓ_2$ is made into a circle of area ${\text{A}}_2$ . If $2 {\text{A}}_1+3 {\text{A}}_2$ is minimum then $(\pi ℓ_1) : ℓ_2$ is equal to: [31-Jan-2023 Shift 1]

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📖 Explanation

$\;ℓ_1+ℓ_2=20 \Rightarrow \;{ {\text{d}} ℓ_2}/{ {\text{d}} ℓ_1}=-1$ $\; {\text{A}}_1= (\;\frac{ℓ_1}{4}) ^2 \;\text{ and }\; {\text{A}}_2=\pi (\;\frac{ℓ_2}{2 \pi}) ^2$ Let ${\text{S}}=2 {\text{A}}_1+3 {\text{A}}_2=\;\frac{ℓ_1^2}{8}+\;\frac{3 ℓ_2^2}{4 \pi}$ $\;\Rightarrow \;\frac{ℓ_1}{4}=\;\frac{6 ℓ_2}{4 \pi} \Rightarrow \;\frac{\pi ℓ_1}{ℓ_2}=6$

Q64JEE Main 2023MCQ4MMatrices and Determinants

For the system of linear equations $\;x+y+z=6$ $\;\alpha x+\beta y+7 z=3$ $\;x+2 y+3 z=14$ which of the following is NOT true? [31-Jan-2023 Shift 1]

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📖 Explanation

By equation 1 and 3 $\;y+2 z=8$ $\;y=8-2 z$ $\;\text{ And }\; \;\; {\text{x}}=-2+ {\text{z}}$ Now putting in equation 2 $\;\alpha(z-2)+\beta(-2 z+8)+7 z=3$ $\;\Rightarrow (\alpha-2 \beta+7) z=2 \alpha-8 \beta+3$ So equations have unique solution if $\alpha-2 \beta+7 \ne 0$ And equations have no solution if $\alpha-2 \beta+7=0$ and $2 \alpha-8 \beta+3 \ne 0$ And equations have infinite solution if $\alpha-2 \beta+7=0$ and $2 \alpha-8 \beta+3=0$

Q65JEE Main 2023MCQ4MThree Dimensional Geometry

Let the shortest distance between the lines ${\text{L}}: \;{ {\text{x}}-5}/{-2}=\;{ {\text{y}}-\lambda }/{0}=\;{ {\text{z}}+\lambda }/{1}, \lambda \ge 0$ and ${\text{L}}_1: {\text{x}}+1= {\text{y}}-$ $1=4-z$ be $2 {\sqrt{6}}$ . If $(\alpha, \beta, \gamma )$ lies on $L$ , then which of the following is NOT possible? [31-Jan-2023 Shift 1]

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📖 Explanation

$\;{ {\text{b}}_1}^{\to } \times { {\text{b}}_2}^{\to }={\begin{vmatrix}{\text{bo} \hat{i}} & {\text{bo} \hat{j}} & {\text{bo} \hat{k}} \\ -2 & 0 & 1 \\ 1 & 1 & -1\end{vmatrix}}=-\;{ {\text{i}}}^{\^}-\;{ {\text{j}}}^{\^}-2 \;{ {\text{k}}}^{\^}$ $\;{ {\text{a}}_2}-{ {\text{a}}_1}=6 \;{ {\text{i}}^{\^}+(\lambda -1) \;{ {\text{j}}}^{\^}+(-\lambda -4) \;{ {\text{k}}}^{\^}$ $\;2 {\sqrt{6}}= |\;{-6-\lambda +1+2 \lambda +8}/{{\sqrt{1+1+4}}}|$ $\;|\lambda +3|=12 \Rightarrow \lambda =9,-15$ $\;\alpha=-2 {\text{k}}+5, \gamma = {\text{k}}-\lambda \;\text{ where }\; {\text{k}} \in {\text{R}}$ $\;\Rightarrow \alpha+2 \gamma =5-2 \lambda =-13,35$

Q66JEE Main 2023MCQ4MInverse Trigonometric Functions

Let $y=f(x)$ represent a parabola with focus $(-\;\frac{1}{2}, 0)$ and directrix ${\text{y}}=-\;\frac{1}{2}$ .Then $S= \{x \in R: \tan ^{-1} ({\sqrt{f(x)}}+ sin \;^{-1}({\sqrt{f(x)+1}})) =\;\frac{\pi}{2}\} :$ [31-Jan-2023 Shift 1]

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📖 Explanation

$\; (x+\;\frac{1}{2}) ^2= (y+\;\frac{1}{4})$ $\;y= (x^{2+}x)$ $\;\tan ^{-1} {\sqrt{x(x+1)}}+ sin \;^{-1} {\sqrt{x^{2+}x+1}}=\pi / 2$ $\;0 \le x^{2+}x+1 \le 1$ $\;x^{2+}x \le 0$ . . . (1) $\;\;\text{ Also }\; x^{2+}x \ge 0$ . . . (2) $\; \therefore x^{2+}x=0 \Rightarrow x=0,-1$ ${\text{S}}$ contains 2 element.

Q67JEE Main 2023MCQ4MMatrices and Determinants

Let $A={(\text{table} 1 , 0 , 0 ; 0 , 4 , -1 ; 0 , 12 , -3)}$ . Then the sum of the diagonal elements of the matrix $( {\text{A}}+ {\text{I}})^{11}$ is equal to: [31-Jan-2023 Shift 1]

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📖 Explanation

$\;A^2={\begin{bmatrix}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{bmatrix}}{\begin{bmatrix}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{bmatrix}}$ $\;\; ={\begin{bmatrix}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{bmatrix}}= {\text{A}}$ $\;\Rightarrow {\text{A}}^3= {\text{A}}^4=. . . . . .= {\text{A}}$ $\;( {\text{A}}+ {\text{I}})^{11}={ }^{11} {\text{C}}_0 {\text{A}}^{11}+{ }^{11} {\text{C}}_1 {\text{A}}^{10}+. . . .{ }^{11} {\text{C}}_{10} {\text{A}}+{ }^{11} {\text{C}}_{11} {\text{I}}$ $\;= ({ }^{11} {\text{C}}_0+{ }^{11} {\text{C}}_1+. . . .{ }^{11} {\text{C}}_{10}) {\text{A}}+ {\text{I}}$ $\;= (2^{11}-1) {\text{A}}+ {\text{I}}=2047 {\text{A}}+ {\text{I}}$ $\; \therefore \;\text{ Sum of diagonal elements }\;=2047(1+4-3)+3$ $\;=4094+3=4097$

Q68JEE Main 2023MCQ4MSets and Relations

Let $R$ be a relation on ${\text{N}} \times {\text{N}}$ defined by $( {\text{a}}, {\text{b}}) {\text{R}}$ $(c, d)$ if and only if ad $(b-c)=b c(a-d)$ . Then $R$ is [31-Jan-2023 Shift 1]

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📖 Explanation

$(a, b) R(c, d) \Rightarrow a d(b-c)=b c(a-d)$ Symmetric: $(c, d) R(a, b) \Rightarrow {\text{cb}}( {\text{d}}- {\text{a}})= {\text{da}}( {\text{c}}- {\text{b}}) \Rightarrow$ SymmetricReflexive:(a, b) $R(a, b) \Rightarrow a b(b-a) \ne b a(a-b) \Rightarrow$ Not reflexiveTransitive: $(2,3) {\text{R}}(3,2)$ and $(3,2) {\text{R}}(5,30)$ but $((2,3),(5,30)) ∉ R \Rightarrow$ Not transitive

Q69JEE Main 2023MCQ4MDifferentiation

Let $y=f(x)= sin \;^3 (\;\frac{\pi}{3} (\cos (\;{\pi}/{3 {\sqrt{2}}} (-4 x^{3+}5 x^{2+}1) ^{\;\frac{3}{2}}) ) )$ Then, at ${\text{x}}=1$ , [31-Jan-2023 Shift 1]

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📖 Explanation

$\;y= sin \;^3(\pi / 3 \cos g(x))$ $\;g(x)=\;{\pi}/{3 {\sqrt{2}}} (-4 x^{3+}5 x^{2+}1) ^{3 / 2}$ $\;g(1)=2 \pi / 3$ $\;y^{'}=3 sin \;^2 (\;\frac{\pi}{3} \cos g(x)) \times \cos (\;\frac{\pi}{3} \cos g(x))$ $\;y^{'}(1)=3 sin \;^2 (-\;\frac{\pi}{6}) \cdot \cos (\;\frac{\pi}{6}) \cdot \;\frac{\pi}{3} (- sin \;\;\frac{2 \pi}{3}) g^{'}(1)$ $\;g^{'}(x)=\;{\pi}/{3 {\sqrt{2}}} (-4 x^{3+}5 x^{2+}1) ^{1 / 2} (-12 x^{2+}10 x)$ $\;g^{'}(1)=\;{\pi}/{2 {\sqrt{2}}}({\sqrt{2}})(-2)=-\pi$ $\;y^{'}(1)=\;\frac{\text{not} 6}{4} \cdot \;{{\sqrt{3}}}/{2} \cdot \;\frac{\pi}{\text{not}} (\;{-{\sqrt{3}}}/{2}) (-\pi)=\;\frac{3 \pi^2}{16}$ $\;y^{'}(1)= sin \;^3(\pi / 3 \cos 2 \pi / 3)=-\;\frac{1}{8}$ $\;2 y^{'}(1)+3 \pi^2 y(1)=0$

Q70JEE Main 2023MCQ4MSequences and Series

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is [31-Jan-2023 Shift 1]

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📖 Explanation

$\;a, a r, a r^2, a r^3(a, r > 0)$ $\;a^4 r^6=1296$ $\;a^2 r^3=36$ $\;a=\;\frac{6}{r^{3 / 2}}$ $\;a+a r+a r^{2+}a r^3=126$ $\;\;\frac{1}{r^{3 / 2}}+\;\frac{r}{r^{3 / 2}}+\;\frac{r^2}{r^{3 / 2}}+\;\frac{r^3}{r^{3 / 2}}=\;\frac{126}{6}=21$ $\; (r^{-3 / 2}+r^{3 / 2}) + (r^{1 / 2}+r^{-1 / 2}) =21$ $\;r^{1 / 2}+r^{-1 / 2}=A$ $\;r^{-3 / 2}+r^{3 / 2}+3 A=A^3$ $\;A^{3-}3 A+A=21$ $\;A^{3-}2 A=21$ $\;A=3$ $\;{\sqrt{r}}+\;{1}/{{\sqrt{r}}}=3$ $\;r_r+1=3 {\sqrt{r}}$ $\;r^{2+}2 r+1=9 r$ $\;r^{2-}7 r+1=0$

Q71JEE Main 2023MCQ4MQuadratic Equation and Inequalities

The number of real roots of the equation ${\sqrt{x^{2-}4 x+3}}+{\sqrt{x^{2-}9}}={\sqrt{4 x^{2-}14 x+6}}$ , is: [31-Jan-2023 Shift 1]

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📖 Explanation

$\;{\sqrt{(x-1)(x-3)}}+{\sqrt{(x-3)(x+3)}}$ $\;=\sqrt{4 (x-\;\frac{12}{4}) (x-\;\frac{2}{4}) }$ $\;\Rightarrow {\sqrt{x-3}}=0 \Rightarrow x=3 \;\text{ which is \in domain }\;$ $\;{\sqrt{x-1}}+{\sqrt{x+3}}={\sqrt{4 x-2}}$ $\;2 {\sqrt{(x-1)(x+3)}}=2 x-4$ $\;x^{2+}2 x-3=x^{2-}4 x+4$ $\;6 x=7$ $\;x=7 / 6 \;\text{ or }\;$

Q72JEE Main 2023MCQ4MDifferential Equations

Let a differentiable function $f$ satisfy $f(x)+int_{3}^{x} \;\frac{f(t)}{t} \text{dt}={\sqrt{x+1}}, x \ge 3$ . Then $12 f(8)$ is equal to: [31-Jan-2023 Shift 1]

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📖 Explanation

Differentiate w.r.t. x $\;f^{'}(x)+\;\frac{f(x)}{x}=\;{1}/{2 {\sqrt{x+1}}}$ $\;\;\text{ I.F. }\;=e^{\int \;\frac{1}{x} \text{dx}}=e^{\ln x}=x$ $\;x f(x)=\int \;{x}/{2 {\sqrt{x+1}}} \text{dx}$ $\;x+1=t^2$ $\;=\int \;\frac{t^{2-}1}{2 t} 2 t \text{dt}$ $\;x f(x)=\;\frac{t^3}{3}-t+c$ $\;x f(x)=\;\frac{(x+1)^{3 / 2}}{3}-{\sqrt{x+1}}+c$ Also putting ${\text{x}}=3$ in given equation ${\text{f}}(3)+0={\sqrt{4}}$ $f(3)=2$ $\;\Rightarrow C=8-\;\frac{8}{3}=\;\frac{16}{3}$ $\;f(x)=\;{\;\frac{(x+1)^{3 / 2}}{3}-{\sqrt{x+1}}+\frac{16}{3}}/{x}$ $\;f(8)=\;\frac{9-3+\frac{16}{3}}{8}=\;\frac{34}{24}$ $\;\Rightarrow 12 f(8)=17$

Q73JEE Main 2023MCQ4MFunctions

If the domain of the function $f(x)=\;\frac{[x]}{1+x^2}$ , where $[x]$ is greatest integer $\le x$ , is $[2,6)$ , then its range is [31-Jan-2023 Shift 1]

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📖 Explanation

$f(x)=\;\frac{2}{1+x^2}\;x \in [2,3)$ $f(x)=\;\frac{3}{1+x^2}\;x \in [3,4)$ $f(x)=\;\frac{4}{1+x^2}\;x \in [4,5)$ $f(x)=\;\frac{5}{1+x^2}\;x \in [5,6)$ $(\;\frac{5}{37}, \;\frac{2}{5}]$

Q74JEE Main 2023MCQ4MVector Algebra

Let ${ \vec{\text{a}}=2 \;{ {\text{i}}}^{\^}+\;{ {\text{j}}}^{\^}+\;{ {\text{k}}}^{\^}$ , and ${ \vec{\text{b}}$ and ${ \vec{\text{c}}$ be two nonzero vectors such that $|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$ and $\vec{b} . \vec{c}=0$ . Consider the following two statement:(A) $|{ \vec{\text{a}}+\lambda { \vec{\text{c}}| \ge |{ \vec{\text{a}}|$ for all $\lambda \in ℝ$ .(B) $\vec{a}$ and $\vec{c}$ are always parallel [31-Jan-2023 Shift 1]

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📖 Explanation

$\;|\vec{a}+\vec{b}+\vec{c}|^2=|\vec{a}+\vec{b}-\vec{c}|^2$ $\;2 \vec{a} \cdot \vec{b}+2 \vec{b} \cdot \vec{c}+2 \vec{c} \cdot \vec{a}=2 \vec{a} \cdot \vec{b}-2 \vec{b} \cdot \vec{c}-2 \vec{c} \cdot \vec{a}$ $\;4 \vec{a} \cdot \vec{c}=0$ B is incorrect $\;|{ \vec{\text{a}}+\lambda { \vec{\text{c}}|^2 \ge |{ \vec{\text{a}}|^2$ $\;\lambda ^2 {\text{c}}^2 \ge 0$ True $∀ \lambda \in {\text{R}}$ (A) is correct.

Q75JEE Main 2023MCQ4MDefinite Integrals

Let $\alpha \in (0,1)$ and $\beta=\log _e(1-\alpha)$ . Let $P_n(x)=x+\;\frac{x^2}{2}+\;\frac{x^3}{3}+. . . . .+\;\frac{x^n}{n}, x \in (0,1)$ .Then the integral $int_{0}^{\alpha} \;{t^{50}}/{1- {\text{t}}} {\text{dt}}$ is equal to [31-Jan-2023 Shift 1]

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📖 Explanation

$\;int_{0}^{\alpha} \;{t^{50}-1+1}/{1- {\text{t}}}=-int_{0}^{\alpha} (1+ {\text{t}}+. . . . .+ {\text{t}}^{49}) +int_{0}^{\alpha} \;{1}/{1- {\text{t}}} {\text{dt}}$ $\;=- (\;\frac{\alpha^{50}}{50}+\;\frac{\alpha^{49}}{49}+. . . . .+\;\frac{\alpha^1}{1}) + (\;{\ln (1- {\text{f}})}/{-1}) _0^\alpha$ $\;=- {\text{P}}_{50}(\alpha)-\ln (1-\alpha)$ $\;=- {\text{P}}_{50}(\alpha)-\beta$

Q76JEE Main 2023MCQ4MInverse Trigonometric Functions

If $sin \;^{-1} \;\frac{\alpha}{17}+\cos ^{-1} \;\frac{4}{5}-\tan ^{-1} \;\frac{77}{36}=0,0 < \alpha < 13$ , then $sin \;^{-1}( sin \;\alpha)+\cos ^{-1}(\cos \alpha)$ is equal to [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\cos ^{-1} \;\frac{4}{5}=\tan ^{-1} \;\frac{3}{4}$ $\; \therefore sin \;^{-1} \;\frac{\alpha}{17}=\tan ^{-1} \;\frac{77}{36}-\tan ^{-1} \;\frac{3}{4}=\tan ^{-1} (\;\frac{\;\frac{77}{36}-\frac{3}{4}}{1+\;\frac{77}{36} \cdot \;\frac{3}{4}})$ $\; sin \;-1 \;\frac{\alpha}{17}=\tan ^{-1} \;\frac{8}{15}= sin \;^{-1} \;\frac{8}{17}$ $\;\Rightarrow \;\frac{\alpha}{17}=\;\frac{8}{17} \Rightarrow \alpha=8$ $\; \therefore sin \;^{-1}( sin \;8)+\cos ^{-1}(\cos 8)$ $\;=3 \pi-8+8-2 \pi$ $\;=\pi$

Q77JEE Main 2023MCQ4MCircles

Let a circle $C_1$ be obtained on rolling the circle $x^{2+}y^{2-}4 x-6 y+11=0$ upwards 4 units on the tangent ${\text{T}}$ to it at the point $(3,2)$ . Let ${\text{C}}_2$ be the image of $C_1$ in $T$ . Let $A$ and $B$ be the centers of circles ${\text{C}}_1$ and ${\text{C}}_2$ respectively, and ${\text{M}}$ and ${\text{N}}$ be respectively the feet of perpendiculars drawn from ${\text{A}}$ and ${\text{B}}$ on the ${\text{x}}$ -axis. Then the area of the trapezium AMNB is : [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{C}}=(2,3), {\text{r}}={\sqrt{2}}$ $\;\;\text{ Centre of }\; {\text{G}}= {\text{A}}=2+4 \;{1}/{{\sqrt{2}}},$ $\;3+\;{4}/{{\sqrt{2}}}=(2+2 {\sqrt{2}}, 3+2 {\sqrt{2}})$ $\; {\text{A}}(2+2 {\sqrt{2}}, 3+2 {\sqrt{2}})$ $\; {\text{B}}(4+2 {\sqrt{2}}, 1+2 {\sqrt{2}})$ $\;\;{ {\text{x}}-(2+2 {\sqrt{2}})}/{1}=\;{ {\text{y}}-(3+2 {\sqrt{2}})}/{-1}=2$ $\; \therefore \;\text{ area of trapezium: }\;$ $\;\;\frac{1}{2}(4+4 {\sqrt{2}}) 2=4(1+{\sqrt{2}})$

Q78JEE Main 2023MCQ4MSets and Relations

$( {\text{S}} 1)( {\text{p}} \Rightarrow {\text{q}}) ∨( {\text{p}} ∧(∼ {\text{q}}))$ is a tautology $( {\text{S}} 2)((∼ {\text{p}}) \Rightarrow (∼ {\text{q}})) ∧((∼ {\text{p}}) ∨ {\text{q}})$ is aContradiction. Then [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

p q ${\text{p}} \Rightarrow {\text{q}}$ $∼ q$ $p ∧ ∼ q$ $( {\text{p}}= {\text{q}}) ∨( {\text{p}} ∧ ∼ {\text{q}})$ T T T F F T T F F T T T F T T F F T F F T T F T $∼ p$ $∼ q$ $∼ {\text{p}} \Rightarrow ∼ {\text{q}}$ $∼ {\text{p}} ∨ {\text{q}}$ $((∼ p) \Rightarrow (∼ q)) ∧(∼ p) ∨ q)$ F F T T T F T T F F T F F T F T T T T T

Q79JEE Main 2023MCQ4MDefinite Integrals

The value of $int_{{\;\frac{\pi}{3}}}^{\;\frac{\pi}{2}} \;\frac{(2+3 sin \;x)}{ sin \;x(1+\cos x)} \text{dx}$ is equal to [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;int_{\pi / 3}^{\pi / 2} (\;\frac{2+3 sin \;x}{ sin \;x(1+\cos x)}) \text{dx}=2 int_{\pi / 3}^{\pi / 2} \;\frac{d x}{ sin \;x+ sin \;x \cos x}+3$ $\;3 int_{\pi / 3}^{\pi / 2} \;\frac{d x}{1+\cos x}$ $\;int_{\pi / 3}^{\pi / 2} \;\frac{d x}{1+\cos x}=int_{\pi / 3}^{\pi / 2} \;\frac{1-\cos x}{ sin \;^2 x} \text{dx}$ $\;=int_{\pi / 3}^{\pi / 2} ({\text{cosec}}^2 x-\cot x {\text{cosec}} x) \text{dx}$ $\;=({\text{cosecx}}-\cot x) int_{\pi / 3}^{\pi / 2}=(1)- (\;{2}/{{\sqrt{3}}}-\;{1}/{{\sqrt{3}}}) =1-\;{1}/{{\sqrt{3}}}$ $\;int_{\pi / 3}^{\pi / 2} \;\frac{d x}{ sin \;x(1+\cos x)}=$ $\;\int \;{d x}/{(2 \tan {\text{x}} / 2) (1+1-\tan ^2 {\text{x}} / 2) }$ $\;=\int \;{ (1+\tan ^2 {\text{x}} / 2) \ sec ^2 {\text{x}} / 2 {\text{dx}}}/{2 \tan {\text{x}} / 22}$ $\;\tan {\text{x}} / 2= {\text{t}}$ $\;\;\frac{1}{2} \int (\;{1+ {\text{t}}^2}/{ {\text{t}}}) {\text{dt}}=\;\frac{1}{2} \$ [ℓ {\text{nt}}+;{ {\text{t}}^2}/{2}]$ _{;{1}/{{\sqrt{3}}}}^1;=;\frac{1}{2} [ (0+;\frac{1}{2}) - (ℓ {\text{n}} ;{1}/{{\sqrt{3}}}+;\frac{1}{6}) ] = (;\frac{1}{3}+ℓ {\text{n}} {\sqrt{3}}) ;\frac{1}{2};= (;\frac{1}{6}+;\frac{1}{2} ℓ {\text{n}} {\sqrt{3}}) {\text{dt}};2 (;\frac{1}{6}+;\frac{1}{2} ℓ {\text{n}} {\sqrt{3}}) +3 (1-;{1}/{{\sqrt{3}}});=;\frac{1}{3}+ℓ {\text{n}} {\sqrt{3}}+3-{\sqrt{3}}=;\frac{10}{3}+ℓ {\text{n}} {\sqrt{3}}-{\sqrt{3}}$

Q80JEE Main 2023MCQ4MProbability

A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\;{{ }^5 {\text{C}}_2+{ }^6 {\text{C}}_2}/{{ }^2 {\text{C}}_2+{ }^3 {\text{C}}_2+{ }^4 {\text{C}}_2+{ }^5 {\text{C}}_2+{ }^8 {\text{C}}_2}=\;\frac{10+15}{1+3+6+10+15}$ $\;=\;\frac{25}{35}=\;\frac{5}{7}$

Q81JEE Main 2023NAT4MPermutations and Combinations

Let 5 digit numbers be constructed using the digits $0,2,3,4,7,9$ with repetition allowed, and are arranged in ascending order with serial numbers. Then the serial number of the number 42923 is _______. [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;42920=1$ $\;42922=1$ $\;42923=1$ $\;=2997$

Q82JEE Main 2023NAT4MSequences and Series

Let $a_1, a_2, . . . . . ., a_n$ be in A.P. If $a_5=2 a_7$ and $a_{11}=18$ , then $12 (\;{1}/{\sqrt{a_{10}}+\sqrt{a_{11}}}+\;{1}/{\sqrt{a_{11}}+\sqrt{a_{12}}}+. . . . \;{1}/{\sqrt{a_{17}}+\sqrt{a_{18}}})$ is equal to _______. [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;2 a_7=a_5 \;\text{ (given) }\;$ $\;2 (a_1+6 d) =a_1+4 d$ $\;a_1+8 d=0$ . . . (1) $\;a_1+10 d=18$ . . . (2) $\;\;\text{ By }\;(1) \;\text{ and }\;(2) \;\text{ we get }\; a_1=-72, d=9$ $\;a_{18}=a_1+17 d=-72+153=81$ $\;a_{10}=a_1+9 d=9$ $\;12 (\;{\sqrt{a_{11}}-\sqrt{a_{10}}}/{d}+\;{\sqrt{a_{12}}-\sqrt{a_{11}}}/{d}+. . . . . . \;{\sqrt{a_{18}}-\sqrt{a_{17}}}/{d})$ $\;12 (\;{\sqrt{a_{18}}-\sqrt{a_{10}}}/{d}) =\;\frac{12(9-3)}{9}=\;\frac{12 \times 6}{6}=8$

Q83JEE Main 2023NAT4MThree Dimensional Geometry

Let $\theta$ be the angle between the planes $P_1=\vec{r} \cdot ({i}^{ \wedge }+{i}^{ \wedge }+2{k}^{ \wedge })=9$ and $P_2=\vec{r} \cdot (2 {i}^{ \wedge }-{i}^{ \wedge }+{k}^{ \wedge })=15$ .Let ${\text{L}}$ be the line that meets $P_2$ at the point $(4,-2,5)$ and makes an angle $\theta$ with the normal of $P_2$ . If $\alpha$ is the angle between $L$ and $P_2$ then $(\tan ^2 \theta) (\cot ^2 \alpha)$ is equal to _______. [31-Jan-2023 Shift 1]

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📖 Explanation

$\;\cos \theta=\;{(\;{ {\text{i}}}^{\^}+\;{ {\text{j}}}^{\^}+2 \;{ {\text{k}}}^{\^}) \cdot (2 \;{ {\text{i}}}^{\^}-\;{ {\text{j}}}^{\^}+\;{ {\text{k}}}^{\^})}/{6}=\;\frac{2-1+2}{6}=\;\frac{1}{2}$ $\;\theta=\pi / 3 \;\; \alpha=\pi / 6$ $\; (\tan ^2 \theta) (\cot ^2 \alpha)$ (3) (3)=9

Q84JEE Main 2023NAT4MBinomial Theorem

Let $\alpha>0$ , be the smallest number such that the expansion of $(x^{\;\frac{2}{3}}+\;\frac{2}{x^3}) ^{30}$ has a term $\beta x^{-\alpha}, \beta \in N$ .Then $\alpha$ is equal to _______. [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{T}}_{ {\text{r}}+1}={ }^{30} {\text{C}}_{ {\text{r}}} ( {\text{x}}^{2 / 3}) ^{30- {\text{r}}} (\;{2}/{ {\text{x}}^3}) ^{ {\text{r}}}$ $\;={ }^{30} {\text{C}}_{ {\text{r}}} \cdot 2 \cdot {\text{x}}^{\;{60-11 {\text{r}}}/{3}}$ $\;\;{60-11 {\text{r}}}/{3} < 0 \Rightarrow 11 {\text{r}} > 60 \Rightarrow {\text{r}} > \;\frac{60}{11} \Rightarrow {\text{r}}=6$ $\; {\text{T}}_7={ }^{30} {\text{C}}_6 \cdot 2^6 {\text{x}}^{-2}$ We have also observed $\beta={ }^{30} {\text{C}}_6(2)^6$ is a natural number. $\therefore \alpha=2$

Q85JEE Main 2023NAT4MVector Algebra

Let ${ \vec{\text{a}}$ and ${ \vec{\text{b}}$ be two vector such that $|\vec{a}|={\sqrt{14}}$ , $|\vec{b}|={\sqrt{6}}$ and $|\vec{a} \times \vec{b}|={\sqrt{48}}$ . Then $(\vec{a} \cdot \vec{b})^2$ is equal to _______. [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;|{ \vec{\text{a}}|={\sqrt{14}},|{ \vec{\text{b}}|={\sqrt{6}} \;\; {| \vec{\text{a}} \times \vec{\text{b}}|}={\sqrt{48}}$ $\;{| \vec{\text{a}} \times \vec{\text{b}}|}^{2+}{| \vec{\text{a}} \cdot \vec{\text{b}}|}^2={| \vec{\text{a}}|}^2 \times {| \vec{\text{b}}|}^2$ $\;\Rightarrow ( \vec{\text{a}} \cdot \vec{\text{b}})^2=84-48=36$

Q86JEE Main 2023NAT4MThree Dimensional Geometry

Let the line ${\text{L}}: \;{ {\text{x}}-1}/{2}=\;{ {\text{y}}+1}/{-1}=\;{ {\text{z}}-3}/{1}$ intersect the plane $2 x+y+3 z=16$ at the point P. Let the point $Q$ be the foot of perpendicular from the point ${\text{R}}(1,-1,-3)$ on the line ${\text{L}}$ . If $\alpha$ is the area of triangle $P Q R$ . then $\alpha^2$ is equal to ________. [31-Jan-2023 Shift 1]

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📖 Explanation

Any point on ${\text{L}}((2 \lambda +1),(-\lambda -1),(\lambda +3))$ $\;2(2 \lambda +1)+(-\lambda -1)+3(\lambda +3)=16$ $\;6 \lambda +10=16 \Rightarrow \lambda =1$ $\; \therefore P=(3,-2,4)$ ${\text{DR}}$ of ${\text{QR}}=⟨ 2 \lambda ,-\lambda , \lambda +6⟩$ ${\text{DR}}$ of ${\text{L}}=⟨ 2,-1,1⟩$ $4 \lambda +\lambda +\lambda +6=0 \;\; 6 \lambda +6=0 \Rightarrow \lambda =-1$ ${\text{Q}}=(-1,0,2)$ ${ \vec{\text{QR}}=2 \;{ {\text{i}}}^{\^}-\;{ {\text{j}}}^{\^}-5 \;{ {\text{k}}}^{\^} \;\; { \vec{\text{QP}}=4 \;{ {\text{i}}}^{\^}-2 \;{ {\text{j}}}^{\^}+2 \;{ {\text{k}}}^{\^}$ $\;{ \vec{\text{QR}} \times { \vec{\text{QP}}={\begin{vmatrix}\\ \\ { {\text{i}}}^{\^} & \\ \\ { {\text{j}}}^{\^} & \\ \\ { {\text{k}}}^{\^} \\ 2 & -1 & -5 \\ 4 & -2 & 2\end{vmatrix}}=-12 \;{ {\text{i}}}^{\^}-24 \;{ {\text{j}}}^{\^}$ $\;\alpha=\;\frac{1}{2} \times {\sqrt{144+576}} \Rightarrow \alpha^2=\;\frac{720}{4}=180$

Q87JEE Main 2023NAT4MBinomial Theorem

The remainder on dividing $5^{99}$ by 11 is ________. [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;5^{99}=5^4 \cdot 5^{95}$ $\;=625 \$ [5^5]$ ^{19};=625[3125]^{19};=625[3124+1]^{19};=625[11 {\text{k}} \times 19+1];=625 \times 11 {\text{k}} \times 19+625;=11 {\text{k}}_1+616+9;=11 ( {\text{k}}_2) +9;;\text{ Remainder };=9$

Q88JEE Main 2023NAT4MStatistics

If the variance of the frequency distribution $\text{bo} {X}_{\text{bo} {i}}$ 2 3 4 5 6 7 8 Frequency $f_i$ 3 6 16 alpha 9 5 6 [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\sigma_{ {\text{x}}}^2=\sigma_{ {\text{d}}}^2=\;{\sum {\text{f}}_{ {\text{i}}} {\text{d}}_{ {\text{i}}}^2}/{\sum {\text{f}}_{ {\text{i}}}}- (\;{\sum {\text{f}}_{ {\text{i}}} {\text{d}}_{ {\text{i}}}}/{\sum {\text{f}}_{ {\text{i}}}}) ^2$ $\;=\;\frac{150}{45+\alpha}-0=3$ $\;\Rightarrow 150=135+3 \alpha$ $\;\Rightarrow 3 \alpha=15 \Rightarrow \alpha=5$

Q89JEE Main 2023NAT4MArea Under The Curves

Let for ${\text{x}} \in {\text{R}}$ $f(x)=\;\frac{x+|x|}{2}$ and $g(x)={\begin{cases}x & {,} x < 0 \\ x^2 {,} & x \ge 0}$ Then area bounded by the curve $y=(f o g)(x)$ and the lines ${\text{y}}=0,2 {\text{y}}- {\text{x}}=15$ is equal to ________\end{cases} [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;f(x)=\;\frac{x+|x|}{2}={\begin{bmatrix}x & x \ge 0 \\ 0 & x< 0}.$ $\ \\ g(x)={[\text{table} x^2 , x \ge 0 \\ x & x < 0}.$ $\ \\ f o g(x)=f[g(x)\end{bmatrix}={[\text{table} g(x) , g(x) \ge 0 ; 0 , g(x) < 0}.$ $\;f o g(x)={[\text{table} x^2 , x \ge 0 ; 0 , x < 0}.$ $\;2 y-x=15$ $\;\;\frac{3}{0} (\;\frac{x+15}{2}-x^2) \text{dx}+\;\frac{1}{2} \times \;\frac{15}{2} \times 15$ $\;\;\frac{x^2}{4}+\;\frac{15 x}{2}-\;\frac{x^3}{3}|_0 ^{3+}\;\frac{225}{4}$ $\;=\;\frac{9}{4}+\;\frac{45}{2}-9+\;\frac{225}{4}=\;\frac{99-36+225}{4}$ $\;=\;\frac{288}{4}=72$

Q90JEE Main 2023NAT4MPermutations and Combinations

Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , is equal to ________. [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$1000-2799$ Divisible by 3 $1002+(n-1) 3=2799$ $n=600$ Divisible by 11 $\;1-2799 \to [\;\frac{2799}{11}] =[254]=254$ $\;1-999= [\;\frac{999}{11}] =90$ $\;1000-2799=254-90=164$ Divisible by 33 $\;1-2799 \to [\;\frac{2799}{33}] =84$ $\;1-999 \to [\;\frac{999}{33}] =30$ $\;1000-2799 \to 54$ $\; \therefore {\text{n}}(3)+ {\text{n}}(11)- {\text{n}}(33)$ $600+164-54=710$

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