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13 Apr 2023 Shift 1

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Physics30 questions

Q1JEE Main 2023MCQ4MCurrent Electricity

Different combination of 3 resistors of equal resistance $R$ are shown in the figures. The increasing order for power dissipation is: [13-Apr-2023 shift 1]

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📖 Explanation

Power dissipation, ${\text{P}}= {\text{I}}^2 {\text{R}}$ (A) ${\text{R}}_{ {\text{eq}}}=\;{ {\text{R}}}/{2}+ {\text{R}}=\;{3 {\text{R}}}/{2}$ (B) $R_{\;\text{eq }\;}=\;\frac{(2 R)(R)}{2 R+R}=\;\frac{2 R}{3}$ (C) ${\text{R}}_{ {\text{eq}}}=\;{ {\text{R}}}/{3}$ (D) $\;R_{\;\text{eq }\;}=3 R$ $\;R_D > R_A > R_B > R_C$ Since, $P \propto R_{\;\text{eq }\;}$ ${\text{P}}_{ {\text{D}}} > {\text{P}}_{ {\text{A}}} > {\text{P}}_{ {\text{B}}} > {\text{P}}_{ {\text{C}}}$

Q2JEE Main 2023MCQ4MSemiconductor Electronics

For the following circuit and given inputs ${\text{A}}$ and ${\text{B}}$ , chose the correct option for output ' ${\text{Y}}$ ' [13-Apr-2023 shift 1]

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📖 Explanation

Output, $\;\; {\text{y}}=\overline{\overline{ {\text{A}}} \cdot {\text{B}}}=\overline{\overline{ {\text{A}}}}+ {\text{B}}$ $\; {\text{y}}= {\text{A}}+\overline{ {\text{B}}}$ ${\text{t}}_1 \;\text{ to }\; {\text{t}}_2,\; {\text{A}}=0, {\text{B}}=1, {\text{Y}}=0$ ${\text{t}}_2 \;\text{ to }\; {\text{t}}_3\; {\text{A}}=1, {\text{B}}=1, {\text{Y}}=1$ ${\text{t}}_3 \;\text{ to }\; {\text{t}}_4\; {\text{A}}=0, {\text{B}}=0, {\text{Y}}=1$ ${\text{t}}_4 \;\text{ to }\; {\text{t}}_5,\; {\text{A}}=1, {\text{B}}=1, {\text{Y}}=1$ ${\text{t}}_5 \;\text{ to }\; {\text{t}}_6,\; {\text{A}}=1, {\text{B}}=0, {\text{Y}}=1$ $\;\text{ After }\;_6,\; {\text{A}}=0, {\text{B}}=0, {\text{Y}}=1$

Q3JEE Main 2023MCQ4MLaws of Motion

A bullet of $10 {\text{g}}$ leaves the barrel of gun with a velocity of $600 {\text{m}} / {\ s}$ . If the barrel of gun is $50 {\text{cm}}$ long and mass of gun is $3 {\text{kg}}$ , then value of impulse supplied to the gun will be: [13-Apr-2023 shift 1]

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📖 Explanation

Impulse, $|{ {\text{I}}}^{\to }|\;=|\Delta { {\text{p}}}^{\to }|$ $\;= {\text{mV}}-0$ $\;= (10 \times 10^{-3} {\text{kg}}) (600 {\text{m}} / {\ s})$ ${\text{I}}\;=6 {\text{N}}- {\text{S}}$

Q4JEE Main 2023MCQ4MElectromagnetic Induction

Which of the following Maxwell's equation is valid for time varying conditions but not valid for static conditions: [13-Apr-2023 shift 1]

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📖 Explanation

For static conditions $∮ { {\text{E}}}^{\to } \cdot {\text{d}} { {\text{l}}}^{\to }=0$ For time varying condition, $∮ { {\text{E}}}^{\to } \cdot {\text{d}} { {\text{l}}}^{\to }=-\;{∂ \varphi _{ {\text{B}}}}/{∂ {\text{t}}}$

Q5JEE Main 2023MCQ4MElectromagnetic Waves

Match List - I with List - II List - I (Layer of atmosphere) List -II (Approximate height over earth's surface) (A) F1 - Layer (I) $10 {\text{km}}$ (B) D - Layer (II) $170-190 {\text{km}}$ (C) Troposphere (III) $100 {\text{km}}$ (D) E-layer (IV) $65-75 {\text{km}}$ Choose the correct answer from the options given below: [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{F}}_1 \to \;\text{ Lower part of }\; {\text{F}} \;\text{ layer of ionosphere }\;(170-190 {\text{Km}})$ $\; {\text{D}} \to \;\text{ Lowest layer of ionosphere }\;(65-75 {\text{Km}})$ $\; \;\; \;\text{ Troposphere }\; \to \;\text{ Lowest layer of atmosphere }\;(10 {\text{Km}})$ $\; {\text{E}} \to \;\text{ Middle part of ionosphere }\;(100 {\text{Km}})$

Q6JEE Main 2023MCQ4MThermodynamics

The rms speed of oxygen molecule in a vessel at particular temperature is $(1+\;\frac{5}{x}) ^{\;\frac{1}{2}} v$ , where $v$ is the average speed of the molecule. The value of $x$ will be: (Take $\pi=\;\frac{22}{7}$ ) [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{V}}_{ {\text{mms}}}=\sqrt{\;{3 {\text{RT}}}/{ {\text{M}}}}$ $\; {\text{V}}_{ {\text{avg}}}=v=\sqrt{\;{8 {\text{RT}}}/{\pi {\text{M}}}}$ $\; {\text{V}}_{ {\text{mms}}}=\sqrt{\;\frac{3 \pi}{8} v}$ $\; {\text{V}}_{ {\text{mms}}}=\sqrt{\;\frac{3}{8} \times \;\frac{22}{7}} {\text{v}}= (\;\frac{33}{28}) ^{1 / 2} v$ $\; {\text{V}}_{ {\text{mms}}}= (1+\;\frac{5}{28}) ^{1 / 2} v$ $\; {\text{x}}=28$

Q7JEE Main 2023MCQ4MWork, Power and Energy

The ratio of powers of two motors is $\;{3 \sqrt{ {\text{x}}}}/{\sqrt{ {\text{x}}}+1}$ , that are capable of raising $300 {\text{kg}}$ water in 5 minutes and $50 {\text{kg}}$ water in 2 minutes respectively from a well of $100 {\text{m}}$ deep. The value of ${\text{x}}$ will be [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{P}}=\;\frac{\;\text{ Work }\;}{\;\text{ Time }\;}$ $\; {\text{P}}_1=\;{ {\text{mgh}}}/{ {\text{t}}_1}=\;{(300) {\text{g}}(100)}/{5}$ $\; {\text{P}}_2=\;{(50) {\text{g}}(100)}/{2}$ $\;\;{ {\text{P}}_1}/{ {\text{P}}_2}=\;\frac{600}{250}=\;\frac{12}{5}=\;\frac{3 \times 4}{4+1}$ $\;\;{ {\text{P}}_1}/{ {\text{P}}_2}=\;{3 {\sqrt{16}}}/{{\sqrt{16}}+1}$ $\; {\text{x}}=16$

Q8JEE Main 2023MCQ4MMotion in a Straight Line

Two trains 'A' and 'B' of length ' $l$ ' and ' $4 l$ ' are travelling into a tunnel of length ' ${\text{L}}$ ' in parallel tracks from opposite directions with velocities $108 {\text{km}} / {\text{h}}$ and $72 {\text{km}} / {\text{h}}$ , respectively. If train 'A' takes $35 {\ s}$ less time than train 'B' to cross the tunnel then, length ' $L$ ' of tunnel is:(Given ${\text{L}}=60 {\text{l}}$ ) [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{V}}_{ {\text{A}}}=108 \times \;\frac{5}{18}=30 {\text{m}} / {\ s}$ $\; {\text{V}}_{ {\text{B}}}=72 \times \;\frac{5}{18}=20 {\text{m}} / {\ s}$ $\; {\text{T}}_{ {\text{A}}}=\;{ℓ+ {\text{L}}}/{30}, {\text{T}}_{ {\text{B}}}=\;{4 ℓ+ {\text{L}}}/{20}$ $\; {\text{T}}_{ {\text{A}}}= {\text{T}}_{ {\text{B}}}-35$ $\;\;{ℓ+ {\text{L}}}/{30}=\;{4 ℓ+ {\text{L}}}/{20}-35$ $\; {\text{Given}}, \;\; {\text{L}}=60 ℓ$ $\;\;\frac{61 ℓ}{30}=\;\frac{64 ℓ}{20}-35$ $\;\;\frac{192 ℓ-122 ℓ}{60}=35$ $\;70 ℓ=60 \times 35$ $\;ℓ=30 {\text{m}}$ $\; {\text{L}}=60 ℓ=1800 {\text{m}}$

Q9JEE Main 2023MCQ4MLaws of Motion

Two bodies are having kinetic energies in the ratio $16: 9$ . If they have same linear momentum, the ratio of their masses respectively is: [13-Apr-2023 shift 1]

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📖 Explanation

Kinetic energy, $K E=\;\frac{P^2}{2 m}$ $\;\;{ {\text{k}}_1}/{ {\text{k}}_2}=\;{ {\text{m}}_2}/{ {\text{m}}_1}$ $\;\;\frac{16}{9}=\;{ {\text{m}}_2}/{ {\text{m}}_1}$ $\;\;{ {\text{m}}_1}/{ {\text{m}}_2}=\;\frac{9}{16}$

Q10JEE Main 2023MCQ4MProperties of Matter

The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross - section. Cross sectional areas at $A$ is $1.5 {\text{cm}}^2$ , and $B$ is $25 {\text{mm}}^2$ , if the speed of liquid at $B$ is $60 {\text{cm}} / {\ s}$ then $(P_A-P_B)$ is: (Given ${\text{P}}_{ {\text{A}}}$ and ${\text{P}}_{ {\text{B}}}$ are liquid pressures at ${\text{A}}$ and ${\text{B}}$ points)Density $\rho =1000 {\text{kg}} {\text{m}}^{-3}$ $A$ and $B$ are on the axis of tube [13-Apr-2023 shift 1]

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📖 Explanation

By equation of continuity, $\; {\text{A}}_1 {\text{V}}_1= {\text{A}}_2 {\text{V}}_2$ $\; (1.5 \times 10^{-4}) {\text{V}}_{ {\text{A}}}= (25 \times 10^{-6}) 60 {\text{cm}} / {\ s}$ $\; {\text{V}}_{ {\text{A}}}=10 {\text{cm}} / {\ s}$ By Bernoulli's theorem. $\; {\text{P}}_{ {\text{A}}}+\;\frac{1}{2} \rho {\text{V}}_{ {\text{A}}}^2= {\text{P}}_{ {\text{B}}}+\;\frac{1}{2} \rho {\text{V}}_{ {\text{B}}}^2$ $\; {\text{P}}_{ {\text{A}}}- {\text{P}}_{ {\text{B}}}=\;\frac{\rho }{2} ( {\text{V}}_{ {\text{B}}}^{2-} {\text{V}}_{ {\text{A}}}^2)$ $\; {\text{P}}_{ {\text{A}}}- {\text{P}}_{ {\text{B}}}=\;\frac{1000}{2} (60^{2-}10^2) \times 10^{-4}$ $\; {\text{P}}_{ {\text{A}}}- {\text{P}}_{ {\text{B}}}=175 {\text{Pa}}$

Q11JEE Main 2023MCQ4MAtoms and Nuclei

${ }_{92}^{238} {\text{A}} \to { }_{90}^{234} {\text{B}}+{ }_2^4 {\text{D}}+ {\text{Q}}$ In the given nuclear reaction, the approximate amount of energy released will be: [Given, mass of ${ }_{92}^{238} {\text{A}}=238.05079 \times 931.5 {\text{MeV}} / {\text{c}}^2$ , $\;\;\text{ mass of }\;{ }_{90}^{234} {\text{B}}=234.04363 \times 931.5 {\text{MeV}} / {\text{c}}^2 \;\text{, }\;$ $\;\;\text{ mass of }\; { }_2^4 {\text{D}}=4.00260 \times 931.5 {\text{MeV}} / {\text{c}}^2]$ [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{Q}}=\Delta {\text{mC}}^2$ $\; {\text{Q}}=(238.05079-234.04363-4.00260) \times 931.5 {\text{MeV}}$ $\; {\text{Q}}=0.00456 \times 931.5 {\text{MeV}}$ $\; {\text{Q}}=4.25 {\text{MeV}}$

Q12JEE Main 2023MCQ4MRotational Motion

A disc is rolling without slipping on a surface. The radius of the disc is ${\text{R}}$ . At ${\text{t}}=0$ , the top most point on the disc is ${\text{A}}$ as shown in figure. When the disc completes half of its rotation, the displacement of point ${\text{A}}$ from its initial position is [13-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Displacement }\;= {\text{A}}^{'} {\text{A}}$ $\; {\text{A}}^{'} {\text{A}}=\sqrt{(\pi {\text{R}})^{2+}(2 {\text{R}})^2}$ $\; {\text{A}}^{'} {\text{A}}= {\text{R}} {\sqrt{\pi^{2+}4}}$

Q13JEE Main 2023MCQ4MOscillations

Which graph represents the difference between total energy and potential energy of a particle executing SHM vs it's distance from mean position? [13-Apr-2023 shift 1]

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📖 Explanation

Total energy in ${\text{SHM}}= {\text{E}}$ $\;E=K+U$ $\;E-U=K$ $\;E-U=\;\frac{1}{2} m \omega ^2 (A^{2-}x^2)$

Q14JEE Main 2023MCQ4MElectrostatics

Two charges each of magnitude $0.01 {\text{C}}$ and separated by a distance of $0.4 {\text{mm}}$ constitute an electric dipole. If the dipole is placed in an uniform electric field ' ${ {\text{E}}}^{\to }$ ' of 10 dyne/C making $30^{\circ}$ angle with ${ {\text{E}}}^{\to }$ , the magnitude of torque acting on dipole is [13-Apr-2023 shift 1]

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📖 Explanation

Dipole moment, ${\text{P}}= {\text{qd}}$ $\; {\text{P}}=0.01 \times 0.4 \times 10^{-3}$ $\; {\text{P}}=4 \times 10^{-6} {\text{C}}- {\text{m}}$ Torque, $\tau \;=p E sin \;\theta$ $\tau \;=4 \times 10^{-6} \times (10 \times 10^{-5}) \times sin \;30^{\circ}$ $\tau \;=4 \times 10^{-10} {\text{N}}- {\text{m}}$

Q15JEE Main 2023MCQ4MProperties of Matter

Under isothermal condition, the pressure of a gas is given by ${\text{P}}= {\text{aV}}^{-3}$ , where ${\text{a}}$ is a constant and ${\text{V}}$ is the volume of the gas. The bulk modulus at constant temperature is equal to [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{P}}= {\text{aV}}^{-3}$ $\;\;{ {\text{dP}}}/{ {\text{dV}}}=-3 {\text{aV}}^{-4}$ Bulk modulus, $B=-V \;\frac{d P}{d V}$ $\; {\text{B}}=- {\text{V}} (\;{-3 {\text{a}}}/{ {\text{V}}^4})$ $\; {\text{B}}=3 \;{ {\text{a}}}/{ {\text{V}}^3}=3 {\text{P}}$

Q16JEE Main 2023MCQ4MGravitation

A planet having mass $9 {\text{Me}}$ and radius $4 {\text{Re}}$ , where ${\text{Me}}$ and ${\text{Re}}$ are mass and radius of earth respectively, has escape velocity in ${\text{km}} / {\ s}$ given by:(Given escape velocity on earth ${\text{V}}_{ {\text{e}}}=11.2 \times 10^3 {\text{m}} / {\ s}$ ) [13-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Escape velocity, }\; {\text{v}}_{ {\text{e}}}=\sqrt{\;{2 {\text{GM}}}/{ {\text{R}}}}$ $\; {\text{V}}_{ {\text{p}}}=\sqrt{\;{2 {\text{G}} (9 {\text{m}}_{ {\text{e}}}) }/{4 {\text{R}}_{ {\text{e}}}}}=\;\frac{3}{2} ( {\text{V}}_{ {\text{e}}}) _{\;\text{earth }\;}$ $\; {\text{v}}_{ {\text{p}}}=\;\frac{3}{2} \times 11.2 {\text{km}} / {\ s}$ $\; {\text{v}}_{ {\text{p}}}=16.8 {\text{km}} / {\ s}$

Q17JEE Main 2023MCQ4MUnits and Measurements

A body of mass $(5 \pm 0.5) {\text{kg}}$ is moving with a velocity of $(20 \pm 0.4) {\text{m}} / {\ s}$ . Its kinetic energy will be [13-Apr-2023 shift 1]

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📖 Explanation

Kinetic energy, ${\text{KE}}\;=\;\frac{1}{2} {\ mv}^2$ ${\text{KE}}\;=\;\frac{1}{2} \times 5 \times 20^2$ ${\text{KE}}=1000 {\text{J}}$ $\;\;{\Delta {\text{K}}}/{ {\text{K}}}=\;{\Delta {\text{m}}}/{ {\text{m}}}+\;{2 \Delta {\text{v}}}/{ {\text{v}}}$ $\;\;{\Delta {\text{K}}}/{1000}=\;\frac{0.5}{5}+2 \times \;\frac{0.4}{20}$ $\;\Delta {\text{K}}=1000(0.1+0.04)$ $\;\Delta {\text{K}}=1000 \times 0.14$ $\;\Delta {\text{K}}=140 {\text{J}}$ $\; {\text{KE}}=(1000 \pm 140) {\text{J}}$

Q18JEE Main 2023MCQ4MDual Nature of Matter and Radiation

The difference between threshold wavelengths for two metal surfaces A and B having work function $\varphi _{ {\text{A}}}=9 {\text{eV}}$ and $\varphi _{ {\text{B}}}=4.5 {\text{eV}}$ in ${\text{nm}}$ is: $\{$ Given, hc $=1242 {\text{eV}} {\text{nm}}\}$ [13-Apr-2023 shift 1]

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📖 Explanation

$\;\varphi =\;{ {\text{hc}}}/{\lambda }$ $\;\lambda _{ {\text{A}}}=\;\frac{1242}{9}=138 {\text{nm}}$ $\;\lambda _{ {\text{B}}}=\;\frac{1242}{4.5}=276 {\text{nm}}$ $\;\lambda _{ {\text{B}}}-\lambda _{ {\text{A}}}=276-138=138 {\text{nm}}$

Q19JEE Main 2023MCQ4MElectromagnetic Waves

The source of time varying magnetic field may be(A) A permanent magnet(B) An electric field changing linearly with time(C) Direct current(D) A decelerating charge particle(E) An antenna fed with a digital signalChoose the correct answer from the options given below: [13-Apr-2023 shift 1]

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📖 Explanation

Accelerated charge particle produces EMW which has time varying ${\text{E}}$ and ${\text{B}}$ . If ${\text{E}}$ is linear function of time then ${\text{B}}$ will be constant.

Q20JEE Main 2023MCQ4MRay Optics and Optical Instruments

A vessel of depth ' $d$ ' is half filled with oil of refractive index ${\text{n}}_1$ and the other half is filled with water of refractive index ${\text{n}}_2$ . The apparent depth of this vessel when viewed from above will be - [13-Apr-2023 shift 1]

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📖 Explanation

$\; (d_{\;\text{spp }\;}) _1=\;{d}/{2 (\;{ {\text{n}}_1}/{ {\text{n}}_2}) }=\;{ {\text{n}}_2 {\text{d}}}/{2 {\text{n}}_1}$ $\; ( {\text{d}}_{ {\text{app}}}) _2=\;{ ( {\text{d}}_{ {\ spp}}) _1+{ {\text{d}}}/{2}}/{ {\text{n}}_2}$ $\;=\;{ ({ {\text{n}}_2}/{ {\text{n}}_1}+1) { {\text{d}}}/{2}}/{ {\text{n}}_2}$ $\; ( {\text{d}}_{ {\ spp}}) _2=\;{ ( {\text{n}}_1+ {\text{n}}_2) {\text{d}}}/{2 {\text{n}}_1 {\text{n}}_2}$

Q21JEE Main 2023NAT4MCurrent Electricity

When a resistance of $5 Ω$ is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of $250 {\text{mA}}$ , however when $1050 Ω$ resistance is connected with it in series, it gives full scale deflection for 25 volt. The resistance of galvanometer is ________ $Ω$ . [13-Apr-2023 shift 1]

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📖 Explanation

$\;I_g(G)= (250-I_g) 5$ $\;I_g=\;\frac{1250}{5+G} m A$ For voltmeter, $\; {\text{V}}= {\text{I}}_{ {\text{g}}} {\text{R}}$ $\;25= {\text{I}}_{ {\text{g}}}( {\text{G}}+1050)$ From equation (1), $\;25=\;{1250 \times 10^{-3}}/{ {\text{G}}+5}( {\text{G}}+1050)$ $\;20( {\text{G}}+5)= {\text{G}}+1050$ $\;19 {\text{G}}=1050-100$ $\; {\text{G}}=\;\frac{950}{19}=50 Ω$

Q22JEE Main 2023NAT4MAtoms and Nuclei

The radius of $2^{\;\text{nd }\;}$ orbit of ${\text{He}}^{+}$ of Bohr's model is $r_1$ and that of fourth orbit of ${\text{Be}}^{3+}$ is represented as $r_2$ . Now the ratio $\;\frac{r_2}{r_1}$ is $x: 1$ . The value of $x$ is __________ [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{r}} \propto \;{ {\text{n}}^2}/{ {\text{Z}}}$ $\;\;{ {\text{r}}_2}/{ {\text{r}}_1}= (\;{ {\text{n}}_2}/{ {\text{n}}_1}) ^2 \times \;{ {\text{z}}_1}/{ {\text{z}}_2}$ $\;\;{ {\text{r}}_2}/{ {\text{r}}_1}= (\;\frac{4}{2}) ^2 \times \;\frac{2}{4}$ $\;\;{ {\text{r}}_2}/{ {\text{r}}_1}=2$ $\; {\text{x}}=2$

Q23JEE Main 2023NAT4MRotational Motion

A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is $\pi: 22$ the, the value of its angular speed will be ${\text{rad}} / {\ s}$ . [13-Apr-2023 shift 1]

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📖 Explanation

Angular momentum, $\; {\text{L}}=I \omega$ $\; {\text{L}}=\;\frac{2}{5} {\text{MR}}^2 \omega$ $\;\;\text{ Energy }\;=\;\frac{1}{2} {\text{MV}}^{2+}\;\frac{1}{2} {\text{I}} \omega ^2$ $\; {\text{E}}=\;\frac{1}{2} {\text{M}}(\omega {\text{R}})^{2+}\;\frac{1}{2} (\;\frac{2}{5} {\text{MR}}^2) \omega ^2$ $\;=\;\frac{7}{10} {\text{M}} \omega ^2 {\text{R}}^2$ $\;\;{ {\text{L}}}/{ {\text{E}}}=\;\frac{4}{7 \omega }=\;\frac{\pi}{22}$ $\;\omega =\;\frac{88}{7 \pi}=\;\frac{88}{7 \times \;\frac{22}{7}}=4 {\text{rad}} / {\ s}$

Q24JEE Main 2023NAT4MRay Optics and Optical Instruments

A fish rising vertically upward with a uniform velocity of $8 {\text{ms}}^{-1}$ , observes that a bird is diving vertically downward towards the fish with the velocity of $12 {\text{ms}}^{-1}$ . If the refractive index of water is $\;\frac{4}{3}$ , then the actual velocity of the diving bird to pick the fish, will be _________ ${\text{ms}}^{-1}$ . [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{d}}_{\;\text{app }\;}= {\text{d}}_1+\mu {\text{d}}$ $\; {\text{v}}_{\;\text{app }\;}= {\text{v}}_1+\mu {\text{v}}$ $\;12=8+\;\frac{4}{3} {\text{v}}$ $\;4=\;\frac{4}{3} {\text{v}}$ $\; {\text{v}}=3 {\text{m}} / {\ s}$

Q25JEE Main 2023NAT4MProperties of Matter

The elastic potential energy stored in a steel wire of length $20 {\text{m}}$ stretched through $2 {\text{cm}}$ is $80 {\text{J}}$ . The cross sectional area of the wire is ________ $(.$ Given, $y=2.0 \times 10^{11} {\text{Nm}}^{-2}$ ) ${\text{mm}}^2$ . [13-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Energy, }\; {\text{U}}=\;\frac{1}{2} {\text{kx}}^2$ $\;80=\;\frac{1}{2} {\text{k}} (2 \times 10^{-2}) ^2$ $\; {\text{k}}=\;\frac{160}{4 \times 10^{-4}}$ $\; {\text{k}}=4 \times 10^5 {\text{N}} / {\text{m}}$ $\;\;{ {\text{yA}}}/{ℓ}=4 \times 10^5$ $\; {\text{A}}=\;\frac{4 \times 10^5 \times 20}{2 \times 10^{11}}$ $\; {\text{A}}=40 \times 10^{-6} {\text{m}}^2$ $\; {\text{A}}=40 {\text{mm}}^2$

Q26JEE Main 2023NAT4MSemiconductor Electronics

From the given transfer characteristic of a transistor in CE configuration, the value of power gain of this configuration is $10^{ {\text{x}}}$ , for ${\text{R}}_{ {\text{B}}}=10 {\text{k}} Ω, {\text{R}}_{ {\text{C}}}=1 {\text{k}} Ω$ . The value of ${\text{x}}$ is ________ [13-Apr-2023 shift 1]

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📖 Explanation

Current gain, $\beta=\;{\Delta {\text{I}}_{ {\text{C}}}}/{\Delta {\text{I}}_{ {\text{B}}}}$ $\;\beta=\;{10 {\text{mA}}}/{100 \mu {\text{A}}}$ $\;\beta=100$ Power gain $\beta^2 \;\frac{R_C}{R_B}$ $\;=10^4 \times \;\frac{1}{10}$ $\;=10^3$ So, ${\text{x}}=3$

Q27JEE Main 2023NAT4MElectromagnetic Induction

In the given figure, an inductor and a resistor are connected in series with a batter of emf $E$ volt. $\;\frac{E^a}{2 b} J / s$ represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of $\;\frac{b}{a}$ will be _________ [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{U}}=\;\frac{1}{2} {\text{LI}}^2$ $\; {\text{I}}= {\text{I}}_0 (1- {\text{e}}^{- {\text{t}} / {\text{t}}})$ Rate of energy, $P=\;{ {\text{dU}}}/{ {\text{dt}}}$ $\; {\text{P}}= {\text{LI}} \;{ {\text{dI}}}/{ {\text{dt}}}$ $\;\;{ {\text{dP}}}/{ {\text{dt}}}= {\text{L}} ( {\text{I}} \;{ {\text{d}}^2 {\text{I}}}/{ {\text{dt}}^2}+ (\;{ {\text{dI}}}/{ {\text{dt}}}) ^2)$ For maximum rate, $\;{ {\text{dP}}}/{ {\text{dt}}}=0$ $\; {\text{I}} \;{ {\text{d}}^2 {\text{I}}}/{ {\text{dt}}^2}=- (\;{ {\text{dI}}}/{ {\text{dt}}}) ^2$ $\; {\text{I}}= {\text{I}}_0 (1- {\text{e}}^{ {\text{t}} / {\tau }})$ $\;\;{ {\text{dI}}}/{ {\text{dt}}}=\;{ {\text{I}}_0}/{\tau } {\text{e}}^{- {\text{t}} / {\tau }}$ $\;\;{ {\text{d}}^2 {\text{I}}}/{ {\text{dt}}^2}=-\;{ {\text{I}}_0}/{\tau ^2} {\text{e}}^{- {\text{t}} / \tau }$ By equation (1), $I_0 (1-e^{-t / \tau }) \times \;\frac{I_0}{\tau ^2} e^{-t / \tau }=\;\frac{-I_0^2}{\tau ^2} e^{-2 t / \tau }$ Let ${\text{e}}^{- {\text{t}} / {\text{t}}}= {\text{x}}$ $\;x-x^2=x^2$ $\;x=\;\frac{1}{2}$ Maximum power, $\; {\text{P}}= {\text{LI}} \;{ {\text{dI}}}/{ {\text{dt}}}$ $\; {\text{P}}= {\text{LI}}_0 (1-\;\frac{1}{2}) (\;{ {\text{I}}_0}/{\tau } \times \;\frac{1}{2})$ $\; {\text{P}}=\;\frac{L I_0^2}{4 \times \;\frac{L}{R}}=\;\frac{I_0^2 R}{4}$ $\;P=\;\frac{E^2}{4 R}$ $\;a=2,2 b=4 R$ $\;b=2 R=50$ $\;\;\frac{b}{a}=25$

Q28JEE Main 2023NAT4MCurrent Electricity

A potential ${\text{V}}_0$ is applied across a uniform wire of resistance ${\text{R}}$ . The power dissipation is ${\text{P}}_1$ . The wire is then cut into two equal halves and a potential of ${\text{V}}_0$ is applied across the length of each half. The total power dissipation across two wires is $P_2$ . The ratio $P_2: P_1$ is ${\sqrt{x}}: 1$ . The value of $x$ is ________ [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{P}}_2=\;{ {\text{v}}_0^2}/{ (\;{ {\text{R}}}/{2}) }+\;{ {\text{v}}_0^2}/{ (\;{ {\text{R}}}/{2}) }$ $\; {\text{P}}_2=4 {\text{P}}_1$ $\;\;{ {\text{P}}_2}/{ {\text{P}}_1}=\;\frac{4}{1}=\;{\sqrt{ {\text{x}}}}/{1}$ $\; {\text{x}}=16$

Q29JEE Main 2023NAT4MOscillations

At a given point of time the value of displacement of a simple harmonic oscillator is given as $y=A \cos (30^{\circ})$ . If amplitude is $40 {\text{cm}}$ and kinetic energy at that time is $200 {\text{J}}$ , the value of force constant is $1.0 \times 10^{ {\text{x}}} {\text{Nm}}^{-1}$ . The value of $x$ is _______ [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{v}}=\omega \sqrt{ {\text{A}}^{2-} {\text{x}}^2}$ $\; {\text{y}}= {\text{A}} \times \;{{\sqrt{3}}}/{2}$ $\; {\text{v}}=\omega \sqrt{ {\text{A}}^{2-}\;{3 {\text{A}}^2}/{4}}=\;{\omega {\text{A}}}/{2}$ $\;\;\text{ Given, }\; {\text{KE}}=200 {\text{J}}$ $\;\;\frac{1}{2} {\text{m}} \;{\omega ^2 {\text{A}}^2}/{4}=200$ $\; {\text{KA}}{ }^2=1600 \;\; ( {\text{K}}= {\text{m}} \omega ^2)$ $\; {\text{K}}=\;\frac{1600}{ (40 \times 10^{-2}) ^2}$ $\; {\text{K}}=10^4 {\text{N}} / {\text{m}}$ $\; {\text{x}}=4$

Q30JEE Main 2023NAT4MElectrostatics

A thin infinite sheet charge and an infinite line charge of respective charge densities $+\sigma$ and $+\lambda$ are placed parallel at $5 {\text{m}}$ distance from each other. Points ' ${\text{P}}$ ' and ' ${\text{Q}}$ ' are at $\;\frac{3}{\pi} {\text{m}}$ and $\;\frac{4}{\pi} {\text{m}}$ perpendicular distances from line charge towards sheet charge, respectively. ' $E_{ {\text{P}}}$ ' and ' ${\text{E}}_{ {\text{Q}}}$ ' are the magnitudes of resultant electric field intensities at point 'P' and 'Q' respectively. If $\;\frac{E_P}{E_Q}=\;\frac{4}{a}$ for $2|\sigma|=|\lambda |$ , then the value of a is ________ [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{E}}_{ {\text{p}}}=\;{2 {\text{K}} \lambda }/{ {\text{r}}}-\;\frac{\sigma}{2 \epsilon_0}$ $\; {\text{E}}_{ {\text{p}}}=\;\frac{\sigma}{2 \epsilon_0}-\;\frac{\lambda }{2 \pi \epsilon_0 (\;\frac{3}{\pi}) }$ $\; {\text{E}}_{ {\text{p}}}=\;\frac{2 \sigma}{2 \epsilon_0}-\;\frac{2 \sigma}{6 \epsilon_0}=\;\frac{\sigma}{6 \epsilon_0}$ $\;\;\text{ Similarly, }\; {\text{E}}_{ {\text{Q}}}=\;\frac{\sigma}{2 \epsilon_0}-\;\frac{2 \sigma}{2 \pi \epsilon_0 (\;\frac{4}{\pi}) }=\;\frac{\sigma}{4 \epsilon_0}$ $\;\;{ {\text{E}}_{ {\text{p}}}}/{ {\text{E}}_{ {\text{Q}}}}=\;\frac{4}{6}$ $\; {\text{a}}=6$

Chemistry30 questions

Q31JEE Main 2023MCQ4MSome Basic Concepts of Chemistry

Given below are two statements:Statement I: Permutit process is more efficient compared to the synthetic resin method for the softening of water.Statement II: Synthetic resin method results in the formation of soluble sodium salts.In the light of the above statements, choose the most appropriate answer from the options given below: [13-Apr-2023 shift 1]

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Q32JEE Main 2023MCQ4Md and f Block Elements

Which one of the following is most likely a mismatch? [13-Apr-2023 shift 1]

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Q33JEE Main 2023MCQ4MStructure of Atom

The energy of an electron in the first Bohr orbit of hydrogen atom is $-2.18 \times 10^{-18} {\text{J}}$ . Its energy in the third Bohr orbit is _______. [13-Apr-2023 shift 1]

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📖 Explanation

$\;E_{1,1}=-2.18 \times 10^{-18} J$ $\;E_{3,1}=E_{1,1} \times \;\frac{1^2}{3^2}$ $\;E_{3,1}=\;\frac{1}{9} \times E_{1,1}$

Q34JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

In the above reaction, left hand side and right hand side rings are named as ' A ' and ' ${\text{B}}$ ' respectively. They undergo ring expansion. The correct statement for this process is: [13-Apr-2023 shift 1]

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Q35JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

Match The following Column-A Column-B a) Nylon 6 I. Natural Rubber b) Vulcanized Rubber II. Cross Linked c) cis-1, 4-polyisoprene III. Caprolactam d) Polychloroprene IV. NeopreneChoose the correct answer from options given below: [13-Apr-2023 shift 1]

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Q36JEE Main 2023MCQ4MSolutions

What happens when a lyophilic sol is added to a lyophobic sol? [13-Apr-2023 shift 1]

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Q37JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

In the reaction given below [13-Apr-2023 shift 1]

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Q38JEE Main 2023MCQ4MHydrocarbons

In the following reaction ' ${\text{X}}$ ' is ${\text{CH}}_3 ( {\text{CH}}_2) _4 {\text{CH}}_3 {\to }_{\text{HCl}}_{\Delta}^{\;\text{ Anhy. }\; {\text{AlCl}}_3} ' {\text{X}}_{\text{major} \text{product}}$ ' [13-Apr-2023 shift 1]

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📖 Explanation

n-alkanes on heating in this presence of anhydrous ${\text{AlCl}}_3$ and hydrogen chloride gas isomerise to branched chain alkanes. The major product has one methyl side chain. ${\text{CH}}_3 ( {\text{CH}}_2) _4 {\text{CH}}_3 {\to }_{\text{HCl}}_{\Delta}^{\;\text{ Anhy. }\; {\text{AlCl}}_3} {\text{CH}}_3- {\text{CH}}_{|}_{\text{CH}_3}_{2- \text{methylpentane}(\text{major})}- ( {\text{CH}}_2) _2- {\text{CH}}_3$

Q39JEE Main 2023MCQ4MHaloalkanes and Haloarenes

2-Methyl propyl bromide reacts with ${\text{C}}_2 {\text{H}}_5 {\text{O}}^{-}$ and gives ' ${\text{A}}$ ' whereas on reaction with ${\text{C}}_2 {\text{H}}_5 {\text{OH}}$ it gives ' ${\text{B}}$ '. The mechanism followed in these reactions and the products ' ${\text{A}}$ ' and ' ${\text{B}}$ ' respectively are : [13-Apr-2023 shift 1]

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📖 Explanation

${\text{C}}_2 {\text{H}}_5 {\text{OH}}$ is weak nucleophile.

Q40JEE Main 2023MCQ4MAlcohols, Phenols and Ethers

In the reaction given below [13-Apr-2023 shift 1]

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Q41JEE Main 2023MCQ4MBiomolecules

$\;\text{ D-(+) Glyceraldehyde }\; \;{\;\text{ i) }\; {\text{HCN}}}/{\;{\;\text{iii) }\; {\text{HNO}}}_{\;\text{ iii) }\; {\text{H}}_2 {\text{O}} / {\text{H}}^{+}_3}}$ The products formed in the above reaction are [13-Apr-2023 shift 1]

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Q42JEE Main 2023MCQ4Mp Block Elements

${\text{CIF}}_5$ at room temperature is a: [13-Apr-2023 shift 1]

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📖 Explanation

${\text{ClF}}_5$ is colourless liquid.

Q43JEE Main 2023MCQ4Md and f Block Elements

The pair of lanthanides in which both elements have high third - ionization energy is: [13-Apr-2023 shift 1]

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📖 Explanation

${\text{Eu}}^{+2}:[ {\text{Xe}}] 4 {\text{f}}^7$ ${\text{Yb}}^{+2}:[ {\text{Xe}}] 4 {\text{f}}^{14}$ High IE due to half filled & fully filled configurations

Q44JEE Main 2023MCQ4MCoordination Compounds

The mismatched combinations areA. Chlorophyll - CoB. Water hardness - EDTAC. Photography $- [ {\text{Ag}}( {\text{CN}})_2] ^{-}$ D. Wilkinson catalyst - $[ ( {\text{Ph}}_3 {\text{P}}) _3 {\text{RhCl}}]$ E. Chelating ligand - D-PenicillamineChoose the correct answer from the options given below : [13-Apr-2023 shift 1]

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📖 Explanation

${\text{Mg}}$ is present in chlorophyll and in black and white photography the developed film is fixed by washing with hypo solution which dissolves the undecomposed ${\text{AgBr}}$ to form a complex ion $[ {\text{Ag}} ( {\text{S}}_2 {\text{O}}_3) _2] ^{3-}$

Q45JEE Main 2023MCQ4MClassification of Elements

Which of the following statements are not correct?A. The electron gain enthalpy of ${\text{F}}$ is more negative than that of ${\text{Cl}}$ .B. Ionization enthalpy decreases in a group of periodic table.C. The electronegativity of an atom depends upon the atoms bonded to it.D. ${\text{Al}}_2 {\text{O}}_3$ and ${\text{NO}}$ are examples of amphoteric oxides.Choose the most appropriate answer from the options given below : [13-Apr-2023 shift 1]

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📖 Explanation

Electronegativity of an element depends on the atom with which it is attached. ${\text{NO}}=$ neutral oxide ${\text{Al}}_2 {\text{O}}_3=$ amphoteric oxide

Q46JEE Main 2023MCQ4MSome Basic Concepts of Chemistry

The radical which mainly causes ozone depletion in the presence of UV radiations is : [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{O}}_2( {\text{g}}) {\to }^{ {\text{uv}}} {\text{O}}( {\text{g}})+ {\text{O}}( {\text{g}})$ $\; {\text{O}}_2( {\text{g}})+ {\text{O}}( {\text{g}}) \to {\text{O}}_3( {\text{g}})$ $\; {\text{CF}}_2 {\text{Cl}}_2( {\text{g}}) {\to }^{ {\text{uv}}} ˙{ {\text{Cl}}}( {\text{g}})+˙{ {\text{C}}}_2 {\text{Cl}}( {\text{g}})$ $\;˙{ {\text{Cl}}}( {\text{g}})+ {\text{O}}_3( {\text{g}}) \to {\text{ClO}}( {\text{g}})+ {\text{O}}_2( {\text{g}})$ $\;˙{ {\text{ClO}}}( {\text{g}})+ {\text{O}}( {\text{g}}) \to ˙{ {\text{Cl}}}( {\text{g}})+ {\text{O}}_2( {\text{g}})$

Q47JEE Main 2023MCQ4MChemical Bonding and Molecular Structure

In which of the following processes, the bond order increases and paramagnetic character changes to diamagnetic one? [13-Apr-2023 shift 1]

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📖 Explanation

${\text{NO}}$ is paramagnetic with ${\text{BO}}=2.5, {\text{NO}}^{+}$ is diamagnetic with ${\text{BO}}=3$

Q48JEE Main 2023MCQ4Mp Block Elements

The incorrect statement from the following for borazine is: [13-Apr-2023 shift 1]

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📖 Explanation

Borazine is ${\text{B}}_3 {\text{N}}_3 {\text{H}}_6$ ${\text{B}}_3 {\text{N}}_3 {\text{H}}_6+9 {\text{H}}_2 {\text{O}} \to 3 {\text{NH}}_3+3 {\text{H}}_3 {\text{BO}}_3+3 {\text{H}}_2$

Q49JEE Main 2023MCQ4MChemical Bonding and Molecular Structure

Among the following compounds, the one which shows highest dipole moment is [13-Apr-2023 shift 1]

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Q50JEE Main 2023MCQ4MClassification of Elements

${\text{Be}}( {\text{OH}})_2$ reacts with ${\text{Sr}}( {\text{OH}})_2$ to yield an ionic salt. Choose the incorrect option related to this reaction from the following : [13-Apr-2023 shift 1]

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📖 Explanation

${\text{Be}}( {\text{OH}})_2$ is amphoteric in nature. ${\text{Sr}}( {\text{OH}})_2$ is basic in nature.These two undergo acid - base reaction to form a salt. ${\text{Be}}( {\text{OH}})_2+ {\text{Sr}}( {\text{OH}})_2 \to \;{ {\text{Sr}} [ {\text{Be}}_{\;\text{ (salt) }\;}( {\text{OH}})_4] }$

Q51JEE Main 2023NAT4MSolutions

Solution of $12 {\text{g}}$ of non-electrolyte (A) prepared by dissolving it in $1000 {\text{mL}}$ of water exerts the same osmotic pressure as that of $0.05 {\text{M}}$ glucose solution at the same temperature. The empirical formula of ${\text{A}}$ is ${\text{CH}}_2 {\text{O}}$ . The molecular mass of $A$ is _________ g. (Nearest integer) [13-Apr-2023 shift 1]

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📖 Explanation

$\;\pi_{ {\text{A}}}=\pi_{\;\text{glucose }\;}$ $\; {\text{C}}_{ {\text{A}}} {\text{RT}}= {\text{CRT}}$ $\;\;{12 / {\text{M}}_{ {\text{A}}}}/{1}=0.05$ $\; {\text{M}}_{ {\text{A}}}(\;\text{ Molar mass of }\; {\text{A}})=\;\frac{12}{0.05}=\;\frac{1200}{5}=240 {\text{gm}}$

Q52JEE Main 2023NAT4MRedox Reactions

${\text{KMnO}}_4$ is titrated with ferrous ammonium sulphate hexahydrate in presence of dilute ${\text{H}}_2 {\text{SO}}_4$ . Number of water molecules produced for 2 molecules of ${\text{KMnO}}_4$ is ________. [13-Apr-2023 shift 1]

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📖 Explanation

$2 {\text{KMnO}}_4+8 {\text{H}}_2 {\text{SO}}_4+10 {\text{FeSO}} {\text{F}}_4 \cdot ( {\text{NH}}_4) _2 {\text{SO}}_4 \cdot 6 {\text{H}}_2 {\text{O}} \to {\text{K}}_2 {\text{SO}}_4+2 {\text{MnSO}}_4+5 {\text{Fe}}_2 ( {\text{SO}}_4) _3+10 ( {\text{NH}}_4) _2 {\text{SO}}_4+68 {\text{H}}_2 {\text{O}}$ On the basis of above equation, 68 molecules of water will be produced from 2 molecules of ${\text{KMnO}}_4$ .

Q53JEE Main 2023NAT4MIonic Equilibrium

$20 {\text{mL}}$ of calcium hydroxide was consumed when it was reacted with $10 {\text{mL}}$ of unknown solution of ${\text{H}}_2 {\text{SO}}_4$ . Also $20 {\text{mL}}$ standard solution of $0.5 {\text{MHCl}}$ containing 2 drops of phenolphthalein was titrated with calcium hydroxide, the mixture showed pink colour when burette displayed the value of $35.5 {\text{mL}}$ whereas the burette showed $25.5 {\text{mL}}$ initially. The concentration of ${\text{H}}_2 {\text{SO}}_4$ is _______ M.(Nearest integer) [13-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ miliequivalent of }\; {\text{Ca}}( {\text{OH}})_2=\;\text{ miliequivalent of }\; {\text{H}}_2 {\text{SO}}_4$ $\; {\text{M}}_1 \times 2 \times 20= {\text{M}}_2 \times 2 \times 10$ $\;2 {\text{M}}_1= {\text{M}}_2$ $\;\;\text{ miliequivalent of }\; {\text{HCl}}=\;\text{ miliequivalent of }\; {\text{Ca}}( {\text{OH}})_2$ $\;20 \times 0.5=10 \times {\text{M}}_1 \times 2$ $\; {\text{M}}_1=0.5 {\text{M}}$ $\;\;\text{ Concentration of }\; {\text{H}}_2 {\text{SO}}_4= {\text{M}}_2=2 {\text{M}}_1$ $\;=2 \times 0.5$ $\;=1 {\text{M}}$ $\;$

Q54JEE Main 2023NAT4MChemical Kinetics

$t_{87.5}$ is the time required for the reaction to undergo $87.5 \%$ completion and $t_{50}$ is the time required for the reaction to undergo $50 \%$ completion. The relation between $t_{87.5}$ and $t_{50}$ for a first order reaction is _________ $t_{87.5}$ $=x \times t_{50}$ . The value of $x$ is _________ . (Nearest integer) [13-Apr-2023 shift 1]

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Q55JEE Main 2023NAT4MStates of Matter

A certain quantity of real gas occupies a volume of $0.15 {\text{dm}}^3$ at $100 {\text{atm}}$ and $500 {\text{K}}$ when its compressibility factor is 1.07 . Its volume at $300 {\text{atm}}$ and $300 {\text{K}}$ (When its compressibility factor is 1.4 ) is ____ $\times 10^{-4} {\text{dm}}^3$ . (Nearest integer) [13-Apr-2023 shift 1]

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📖 Explanation

${\text{Z}}\;=\;{ {\text{PV}}}/{ {\text{nRT}}}$ $\;{ {\text{Z}}_1}/{ {\text{Z}}_2}\;= (\;{ {\text{P}}_1 {\text{V}}_1}/{ {\text{nRT}}_1}) \times (\;{ {\text{nRT}}_2}/{ {\text{P}}_2 {\text{V}}_2})$ $\;\frac{1.07}{1.4}\;= (\;\frac{100 \times 0.15}{500}) (\;{300}/{300 \times {\text{V}}_2})$ ${\text{V}}_2\;=\;\frac{0.03 \times 1.4}{1.07}=0.03925$ $\;=392 \times 10^{-4} {\text{dm}}^3$

Q56JEE Main 2023NAT4MElectrochemistry

A metal surface of $100 {\text{cm}}^2$ area has to be coated with nickel layer of thickness $0.001 {\text{mm}}$ . A current of $2 {\text{A}}$ was passed through a solution of ${\text{Ni}} ( {\text{NO}}_3) _2$ for ' ${\text{x}}$ ' seconds to coat the desired layer. The value of ${\text{x}}$ is (Nearest integer) $(\rho _{ {\text{Ni}}} (..$ density of Nickel) is $10 {\text{g}} {\text{mL}}^{-1}$ , Molar mass of Nickel is ________ $60 {\text{g}} {\text{mol}}^{-1} {\text{F}}=96500 {\text{C}} {\text{mol}}^{-1}$ ) [13-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Volume of nickel required }\;=100 \times 0.001 \times 10^{-3} \times 100$ $\;=0.01 {\text{cm}}^3$ $\;\;\text{ Mass of Nickel required }\;=0.01 \times 10$ $\;=0.1 {\text{gm}}$ $\;\;\text{ Moles }\;=\;\frac{0.1}{60}=\;\frac{1}{600} {\text{mol}}$ $\; {\text{Ni}}^{2+}+2 {\text{e}}^{-} \to {\text{Ni}}( {\ s})$ $\;\;\text{ for coating of }\; 1 {\text{mol}} {\text{Ni}} \;\text{, charge required }\;=2 \times 96500 {\text{C}}$ $\;\;\text{ for coating of }\; \;\frac{1}{600} \;\text{ mol, charge required }\;=2 \times 96500 \times \;\frac{1}{600} {\text{C}}$ $\;=\;\frac{965}{3} {\text{C}}$ $\;I=\;{q}/{ {\text{t}}}$ $\; {\text{t}}=\;\frac{965 / 3}{2}=160.83 {\ sec} \approx 161$ $\;$

Q57JEE Main 2023NAT4MIonic Equilibrium

$25.0 {\text{mL}}$ of $0.050 {\text{MBa}} ( {\text{NO}}_3) _2$ is mixed with $25.0 {\text{mL}}$ of $0.020 {\text{M}} {\text{NaF}} . {\text{K}}_{ {\ sp}}$ of ${\text{BaF}}_2$ is $0.5 \times 10^{-6}$ at $298 {\text{K}}$ . The ratio of $[ {\text{Ba}}^{2+}] \$ [ {\text{F}}^{-}]$ ^2 $and$ {\text{K}}_{ {\ sp}}$ is _________ . (Nearest integer) [13-Apr-2023 shift 1]

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📖 Explanation

$\;{[ {\text{Ba}}^{+2}]\$ =;\frac{25 × 0.05}{50}=0.025 {\text{M}}};{[ {\text{F}}^{-}]$ =;\frac{25 \times 0.02}{50}=0.01 {\text{M}}};{[ {\text{Ba}}^{+2}]$ $[ {\text{F}}^{-}]$ ^2=25 × 10^{-7}}; {\text{K}}{;\text{sp };}=5 × 10^{-7} ;\text{ (given) };;;\text{ Ratio };=;{[ {\text{Ba}}^{+2}]$ $[ {\text{F}}^{-}]$ ^2}/{ {\text{K}}{ {\ sp}}}=5$

Q58JEE Main 2023NAT4MChemical Thermodynamics

${\text{A}}_2+ {\text{B}}_2 \to 2 {\text{AB}} \cdot \Delta {\text{H}}_{ {\text{f}}}^0=-200 {\text{kJ}} {\text{mol}}^{-1}$ new line ${\text{AB}}, {\text{A}}_2$ and ${\text{B}}_2$ are diatomic molecules. If the bond enthalpies of ${\text{A}}_2, {\text{B}}_2$ and ${\text{AB}}$ are in the ratio $1: 0.5: 1$ , then the bond enthalpy of ${\text{A}}_2$ is ________ ${\text{kJ}} {\text{mol}}{ }^{-1}$ . (Nearest integer) [13-Apr-2023 shift 1]

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📖 Explanation

${\text{A}}_2+ {\text{B}}_2 \to 2 {\text{AB}} \;\; \Delta {\text{H}}_{ {\text{f}}}^{\circ}=-200 {\text{kJ}} / {\text{mol}}$ Bond enthalpy of ${\text{A}}_2= {\text{x}}$ Bond enthalpy of ${\text{B}}_2=0.5 {\text{x}}$ Bond enthalpy of $A B=x$ $\;\Delta {\text{H}}_{ {\text{f}}}^{\circ}= {\text{x}}+0.5 {\text{x}}-2 {\text{x}}=-2(200)$ $\;-0.5 {\text{x}}=-400$ $\; {\text{x}}=\;\frac{400}{0.5}=800 {\text{kJ}} / {\text{mol}}$ Bond enthalpy of ${\text{A}}_2= {\text{x}}=800 {\text{kJ}} / {\text{mol}}$

Q59JEE Main 2023NAT4MOrganic Chemistry - Some Basic Principles

An organic compound gives $0.220 {\text{g}}$ of ${\text{CO}}_2$ and $0.126 {\text{g}}$ of ${\text{H}}_2 {\text{O}}$ on complete combustion. If the $\%$ of carbon is 24 then the $\%$ of hydrogen is ________ $\times 10^{-1}$ . (Nearest integer) [13-Apr-2023 shift 1]

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📖 Explanation

$\% \;\text{ of carbon }\;=\;\frac{\;\frac{0.220}{44} \times 12}{x} \times 100$ ({\text{x}}= $mass of organic compound$ );24=;\frac{6}{x};x=0.25 {\text{gm}};% ;\text{ of }; {\text{H}}=;\frac{;\frac{0.126}{18} × 2 × 1}{0.25} × 100;=5.6=56 × 10^{-1}$

Q60JEE Main 2023NAT4MOrganic Chemistry - Some Basic Principles

For the given reactionThe total number of possible products formed by tertiary carbocation of ${\text{A}}$ is _________. [13-Apr-2023 shift 1]

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Mathematics30 questions

Q61JEE Main 2023MCQ4MDefinite Integrals

$int_{0}^{\infty } \;\frac{6}{e^{3 x}+6 e^{2 x}+11 e^x+6} \text{dx}$ = [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{I}}=int_{0}^{\infty } \;{6}/{ (e^x+1) ( {\text{e}}^x+2) ( {\text{e}}^x+3) } {\text{dx}}$ $\;\; =6 int_{0}^{\infty } (\;{\;\frac{1}{2}}/{ {\text{e}}^{ {\text{x}}}+1}+\;{-1}/{ {\text{e}}^x+2}+\;{\;\frac{1}{2}}/{ {\text{e}}^x+3}) \text{dx}$ $\;\; =3 int_{0}^{\infty } \;{ {\text{e}}^{-x}}/{1+ {\text{e}}^{- {\text{x}}}} {\text{dx}}-6 int_{0}^{\infty } \;{ {\text{e}}^{-x} {\text{dx}}}/{1+2 {\text{e}}^{- {\text{x}}}}+3 int_{0}^{\infty } \;{ {\text{e}}^{-x}}/{1+3 {\text{e}}^{- {\text{x}}}} {\text{dx}}$ $\;\; =3 \$ [-\ln (1+ {\text{e}}^{- {\text{x}}}) ]$ _0^{\infty }+6 ;\frac{1}{2} $[\ln (1+2 {\text{e}}^{- {\text{x}}}) ]$ _0^{\infty };; -;\frac{3}{3} $[\ln (1+3 {\text{e}}^{- {\text{x}}}) ]$ _0^{\infty };; =3 \ln 2-3 \ln 3+\ln 4;; =3 \ln ;\frac{2}{3}+\ln 4;; =\ln ;\frac{32}{27}$

Q62JEE Main 2023MCQ4MLimits, Continuity and Differentiability

Among(S1): $\lim _{n \to \infty } \;\frac{1}{n^2}(2+4+6+. . . . .+2 n)=1$ (S2): $\lim _{n \to \infty } \;\frac{1}{n^{16}} (1^{15}+2^{15}+3^{15}+. . . . .+n^{15}) =\;\frac{1}{16}$ [13-Apr-2023 shift 1]

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📖 Explanation

$\;S_1: \lim _{n \to \infty } \;\frac{n(n+1)}{n^2}=1 \Rightarrow \;\text{ True }\;$ $\;S_2: \lim _{n \to \infty } \;\frac{1}{n^{16}} (\sum r^{15}) =\lim _{n \to \infty } \;\frac{1}{n} \sum (\;\frac{r}{n}) ^{15}$ $\;=int_{0}^{1} x^{15} \text{dx}=\;\frac{1}{16} \Rightarrow \;\text{ True }\;$

Q63JEE Main 2023MCQ4MMatrices and Determinants

The number of symmetric matrices of order 3 , with all the entries from the set $\{0,1,2,3,4,5,6,7,8,9\}$ , is [13-Apr-2023 shift 1]

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📖 Explanation

$A={\begin{bmatrix}a & b & c \\ b & d & e \\ c & e & f\end{bmatrix}}, a, b, c, d, e, f \in \{0,1,2, . . . .9\}$ , Number of matrices $=10^6$

Q64JEE Main 2023MCQ4MVector Algebra

Let $\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$ and $\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$ . If a vector $\vec{d}$ satisfies $\vec{d} \times \vec{b}=\vec{c} \times \vec{b}$ and $\vec{d} \cdot \vec{a}=24$ , then $|{ {\text{d}}}^{\to }|^2$ is equal to - [13-Apr-2023 shift 1]

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📖 Explanation

$\;{ {\text{d}}}^{\to } \times { {\text{b}}}^{\to }={ {\text{c}}}^{\to } \times { {\text{b}}}^{\to }$ $\;\Rightarrow ({ {\text{d}}}^{\to }-{ {\text{c}}}^{\to }) \times { {\text{b}}}^{\to }=0$ $\;\Rightarrow { {\text{d}}}^{\to }={ {\text{c}}}^{\to }+\lambda { {\text{b}}}^{\to }$ Also ${\;\text{ dä }\;}^{\to }=24$ $\;\Rightarrow ({ {\text{c}}}^{\to }+\lambda { {\text{b}}}^{\to }) \cdot { {\text{a}}}^{\to }=24$ $\;\;\lambda = \;{24-{ {\text{a}}}^{\to } \cdot { {\text{c}}}^{\to }}/{{ {\text{b}}}^{\to } \cdot { {\text{a}}}^{\to }}=\;\frac{24-6}{9}=2$ $\;\Rightarrow { {\text{d}}}^{\to }={ {\text{c}}}^{\to }+2({ {\text{b}}}^{\to })$ $\; \;\; =8 { {\text{i}}^{\^}}-5 { {\text{j}}^{\^}}+18 { {\text{k}}^{\^}}$ $\;\Rightarrow |{ {\text{d}}}^{\to }|^2=64+25+324=413$

Q65JEE Main 2023MCQ4MProbability

A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If ${\text{X}}$ denotes the number of tosses of the coin, then the mean of ${\text{X}}$ is- [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{P}}( {\text{H}})=\;\frac{3}{4}$ $\; {\text{P}}( {\text{T}})=\;\frac{1}{4}$ $\;\;\text{ Mean }\; \overline{X}=\;\frac{3}{4}+\;\frac{3}{8}+3 (\;\frac{1}{64}+\;\frac{3}{64})$ $\;=\;\frac{3}{4}+\;\frac{3}{8}+\;\frac{3}{16}$ $\;=3 (\;\frac{7}{16}) =\;\frac{21}{16}$

Q66JEE Main 2023MCQ4MApplication of Derivatives

$\max _{0 \le x \le \pi} \{x-2 sin \;x \cos x+\;\frac{1}{3} sin \;3 x\} =$ [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{f}}( {\text{x}})= {\text{x}}- sin \;2 {\text{x}}+\;\frac{1}{3} sin \;3 {\text{x}}$ $\; {\text{f}}^{'}( {\text{x}})=1-2 \cos 2 {\text{x}}+\cos 3 {\text{x}}=0$ $\; {\text{x}}=\;\frac{5 \pi}{6}, \;\frac{\pi}{6}$ $\; \therefore {\text{f}}^{' '}( {\text{x}})=4 sin \;2 {\text{x}}-3 sin \;3 {\text{x}}$ $\; {\text{f}}^{' '} (\;\frac{5 \pi}{6}) < 0$ $\;\Rightarrow (\;\frac{5 \pi}{6}) \;\text{ is point of maxima }\;$ $\; {\text{f}} (\;\frac{5 \pi}{6}) =\;\frac{5 \pi}{6}+\;{{\sqrt{3}}}/{2}+\;\frac{1}{3}$

Q67JEE Main 2023MCQ4MFunctions

The set of all ${\text{a}} \in {\text{R}}$ for which the equation ${\text{x}}| {\text{x}}-1|+| {\text{x}}+2|+ {\text{a}}=0$ has exactly one real root, is [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{f}}( {\text{x}})= {\text{x}}| {\text{x}}-1|+| {\text{x}}+2|$ $\; {\text{x}}| {\text{x}}-1|+| {\text{x}}+2|+ {\text{a}}=0$ $\; {\text{x}}| {\text{x}}-1|+| {\text{x}}+2|=- {\text{a}}$ All values are increasing.

Q68JEE Main 2023MCQ4MParabola

Let $P Q$ be a focal chord of the parabola $y^2=36 x$ of length 100 , making an acute angle with the positive $x$ -axis. Let the ordinate of $P$ be positive and $M$ be the point on the line segment $P Q$ such that $P M: M Q=3: 1$ . Then which of the following points does NOT lie on the line passing through ${\text{M}}$ and perpendicular to the line ${\text{PQ}}$ ? [13-Apr-2023 shift 1]

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📖 Explanation

$\;9 ( {\text{t}}+\;{1}/{ {\text{t}}}) ^2=100$ $\; {\text{t}}=3$ $\;\Rightarrow {\text{P}}(81,54) & {\text{Q}}(1,-6)$ $\; {\text{M}}(21,9)$ $\Rightarrow L$ is $(y-9)=\;\frac{-4}{3}(x-21)$ $\;3 y-27=-4 x+84$ $\;4 x+3 y=111$

Q69JEE Main 2023MCQ4MMatrices and Determinants

For the system of linear equations $\;2 x+4 y+2 a z=b$ $\;x+2 y+3 z=4$ $\;2 x-5 y+2 z=8$ which of the following is NOT correct? [13-Apr-2023 shift 1]

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📖 Explanation

$\Delta \;={\begin{vmatrix}2 & 4 & 2 {\text{a}} \\ 1 & 2 & 3 \\ 2 & -5 & 2\end{vmatrix}}=18(3- {\text{a}})$ $\Delta _{ {\text{x}}}\;={\begin{vmatrix}{\text{b}} & 4 & 2 {\text{a}} \\ 4 & 2 & 3 \\ 8 & -5 & 2\end{vmatrix}}=(64+19 {\text{b}}-72 {\text{a}})$ For unique solution $\Delta =0$ $\Rightarrow {\text{a}} \ne 3 \;\text{ and }\; {\text{b}} \in {\text{R}}$ For infinitely many solution : $\;\Delta =\Delta _x=\Delta _y=\Delta _z=0$ $\;\Rightarrow a=3 \;\; \because \Delta =0$ $\;\;\text{ and }\; {\text{b}}=8 \;\; \because \Delta _{ {\text{x}}}=0$

Q70JEE Main 2023MCQ4MSequences and Series

Let ${\ s}_1, {\ s}_2, {\ s}_3, . . . . . . ., {\ s}_{10}$ respectively be the sum to 12 terms of 10 A.P. ${\ s}_{ {\text{m}}}$ whose first terms are $1,2,3, . . ., 10$ and the common differences are $1,3,5, . . . . . . . . . ., 19$ respectively. Then $sum_{{ {\text{i}}=1}}^{{10}} {\ s}_{ {\text{i}}}$ is equal to [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{S}}_{ {\text{k}}}=6(2 {\text{k}}+(11)(2 {\text{k}}-1))$ $\; {\text{S}}_{ {\text{k}}}=6(2 {\text{k}}+22 {\text{k}}-11)$ $\; {\text{S}}_{ {\text{k}}}=144 {\text{k}}-66$ $\;sum_{1}^{10} {\text{S}}_{ {\text{k}}}=144 sum_{{ {\text{k}}=1}}^{{10}} {\text{k}}-66 \times 10$ $\;=144 \times \;\frac{10 \times 11}{2}-660$ $\;=7920-660$ $\;=7260$

Q71JEE Main 2023MCQ4MDifferentiation

For the differentiable function $f: R-\{0\} \to R$ , let $3 f(x)+2 f (\;\frac{1}{x}) =\;\frac{1}{x}-10$ , then $|f(3)+f^{'} (\;\frac{1}{4}) |$ is equal to [13-Apr-2023 shift 1]

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📖 Explanation

$\;{ [3 {\text{f}}( {\text{x}})+2 {\text{f}} (\;{1}/{ {\text{x}}}) =\;{1}/{ {\text{x}}}-10] \times 3}$ $\;{ [2 {\text{f}}( {\text{x}})+3 {\text{f}} (\;{1}/{ {\text{x}}}) = {\text{x}}-10] \times 2}$ $\;5 {\text{f}}( {\text{x}})=\;{3}/{ {\text{x}}}-2 {\text{x}}-10$ $\; {\text{f}}( {\text{x}})=\;\frac{1}{5} (\;{3}/{ {\text{x}}}-2 {\text{x}}-10)$ $\; {\text{f}}^{'}( {\text{x}})=\;\frac{1}{5} (-\;{3}/{ {\text{x}}^2}-2)$ $\;{| {\text{f}}(3)+ {\text{f}}^{'} (\;\frac{1}{4}) |}={| \;\frac{1}{5}(1-6-10)+\;\frac{1}{5}(-48-2) |}$ $\;=|-3-10|=13$

Q72JEE Main 2023MCQ4MSets and Relations

The negation of the statement $((A ∧(B ∨ C)) \Rightarrow (A ∨ B)) \Rightarrow A$ is [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{p}}:(( {\text{A}} ∧( {\text{B}} ∨ {\text{C}})) \Rightarrow ( {\text{A}} ∨ {\text{B}})) \Rightarrow {\text{A}}$ $\;{[∼( {\text{A}} ∧( {\text{B}} ∨ {\text{C}})) ∨( {\text{A}} ∨ {\text{B}})] \Rightarrow {\text{A}}}$ $\;{[( {\text{A}} ∧( {\text{B}} ∨ {\text{C}})) ∧ ∼( {\text{A}} ∨ {\text{B}})] ∨ {\text{A}}}$ $\;( {\text{f}} ∨ {\text{A}})= {\text{A}}$ $\;∼ {\text{p}} \equiv ∼ {\text{A}}$

Q73JEE Main 2023MCQ4MEllipse

Let the tangent and normal at the point $(3 {\sqrt{3}}, 1)$ on the ellipse $\;\frac{x^2}{36}+\;\frac{y^2}{4}=1$ meet the $y$ -axis at the points $A$ and ${\text{B}}$ respectively. Let the circle ${\text{C}}$ be drawn taking ${\text{AB}}$ as a diameter and the line ${\text{x}}=2 {\sqrt{5}}$ intersect ${\text{C}}$ at the points $P$ and $Q$ . If the tangents at the points $P$ and $Q$ on the circle intersect at the point $(\alpha, \beta)$ , then $\alpha^{2-}\beta^2$ is equal to [13-Apr-2023 shift 1]

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📖 Explanation

Given ellipse $\;\frac{x^2}{36}+\;\frac{y^2}{4}=1$ $\;\;{x}/{4 {\sqrt{3}}}+\;\frac{y}{4}=1$ $\;y=4$ $\;\;{ {\text{x}}}/{4}-\;{4}/{4 {\sqrt{3}}}=\;{2}/{{\sqrt{3}}}$ $\; {\text{y}}=-8$ $\; {\text{x}}^{2+} {\text{y}}^{2+}4 {\text{y}}-32=0$ $\; {\text{hx}}+ {\text{ky}}+2( {\text{y}}+ {\text{k}})-32=0$ $\; {\text{k}}=-2$ $\; {\text{hx}}+2 {\text{k}}-32=0$ $\; {\text{hx}}=36$ $\;\alpha= {\text{h}}=\;{36}/{2 {\sqrt{5}}}$ $\;\beta= {\text{k}}=-2$ $\;\alpha^{2-}\beta^2=\;\frac{304}{5}$

Q74JEE Main 2023MCQ4MThree Dimensional Geometry

The distance of the point $(-1,2,3)$ from the plane $\vec{r} \cdot (\hat{i}-2 \hat{j}+3 \hat{k})=10$ parallel to the line of the shortest distance between the lines ${ {\text{r}}}^{\to }=({ {\text{i}}^{\^}}-{ {\text{j}}^{\^}})+\lambda (2 { {\text{i}}^{\^}}+{ {\text{k}}^{\^}})$ and ${ {\text{r}}}^{\to }=(2 { {\text{i}}^{\^}}-{ {\text{j}}^{\^}})+\mu ({ {\text{i}}^{\^}}-{ {\text{j}}^{\^}}+{ {\text{k}}^{\^}})$ is [13-Apr-2023 shift 1]

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📖 Explanation

Let ${\text{L}}_1: { {\text{r}}}^{\to }=({ {\text{i}}^{\^}}-{ {\text{j}}^{\^}})+\lambda (2 { {\text{i}}^{\^}}+{ {\text{k}}^{\^}})$ $\; {\text{L}}_2: { {\text{r}}}^{\to }=(2 { {\text{i}}^{\^}}-{ {\text{j}}^{\^}})+\mu ({ {\text{i}}^{\^}}-{ {\text{j}}^{\^}}+{ {\text{k}}^{\^}})$ $\;{ {\text{n}}}^{\to }={\begin{vmatrix}{ {\text{i}}^{\^}} & { {\text{j}}^{\^}} & { {\text{k}}^{\^}} \\ 2 & 0 & 1 \\ 1 & -1 & 1\end{vmatrix}}$ $\;{ {\text{n}}}^{\to }={ {\text{i}}^{\^}}-{ {\text{j}}^{\^}}-2 { {\text{k}}^{\^}}$ Equation of line along shortest distance of ${\text{L}}_1$ and ${\text{L}}_2$ $\;\;\frac{x+1}{1}=\;\frac{y-2}{-1}=\;\frac{z-3}{-2}=r$ $\;\Rightarrow (x, y, z) \equiv (r-1,2-r, 3-2 r)$ $\;\Rightarrow (r-1)-2(2-r)+3(3-2 r)=10$ $\;\Rightarrow r=-2$ $\;\Rightarrow Q(x, y, z) \equiv (-3,4,7)$ $\;\Rightarrow P Q={\sqrt{4+4+16}}=2 {\sqrt{6}}$

Q75JEE Main 2023MCQ4MMatrices and Determinants

Let $B={\begin{bmatrix}1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4\end{bmatrix}}, \alpha>2$ be the adjoint of a matrix $A$ and $|A|=2$ . then ${\begin{bmatrix}\alpha & -2 \alpha & \alpha\end{bmatrix}} {\text{B}}{\begin{bmatrix}\alpha \\ -2 \alpha \\ \alpha\end{bmatrix}}$ is equal to [13-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Given, }\; B={\begin{bmatrix}1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4\end{bmatrix}}$ $\;|B|=4$ $\;1(8-3 \alpha)-3(4-3 \alpha)+\alpha(\alpha-2 \alpha)=4$ $\;-\alpha^{2+}6 \alpha-8=0$ $\;\alpha=2,4$ Given $\alpha > 2$ So, $\alpha=2$ is rejected ${\begin{bmatrix}4 & -8 & 4\end{bmatrix}}{\begin{bmatrix}1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4\end{bmatrix}}{\begin{bmatrix}4 \\ -8 \\ 4\end{bmatrix}}=[-16]_{1 \times 1}$

Q76JEE Main 2023MCQ4MFunctions

For ${\text{x}} \in {\text{R}}$ , two real valued functions ${\text{f}}( {\text{x}})$ and ${\text{g}}( {\text{x}})$ are such that, ${\text{g}}( {\text{x}})=\sqrt{ {\text{x}}}+1$ and ${\text{fog}}( {\text{x}})= {\text{x}}+3-\sqrt{ {\text{x}}}$ . Then ${\text{f}}(0)$ is equal to [13-Apr-2023 shift 1]

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📖 Explanation

$\;g(x)={\sqrt{x}}+1$ $\;\;\text{ fog }\;(x)=x+3-{\sqrt{x}}$ $\;=({\sqrt{x}}+1)^{2-}3({\sqrt{x}}+1)+5$ $\;=g^2(x)-3 g(x)+5$ $\;\Rightarrow f(x)=x^{2-}3 x+5$ $\; \therefore f(0)=5$ But, if we consider the domain of the composite function fog $(x)$ then in that case $f(0)$ will be not defined as $g(x)$ cannot be equal to zero.

Q77JEE Main 2023MCQ4MThree Dimensional Geometry

Let the equation of plane passing through the line of intersection of the planes $x+2 y+a z=2$ and $x-y+z=3$ be $5 x$ $-11 y+b z=6 a-1$ . For $c \in Z$ , if the distance of this plane from the point $(a,-c, c)$ is $\;{2}/{{\sqrt{a}}}$ , then $\;\frac{a+b}{c}$ is equal to [13-Apr-2023 shift 1]

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📖 Explanation

$\;(x+2 y+a z-2)+\lambda (x-y+z-3)=0$ $\;\;\frac{1+\lambda }{5}=\;\frac{2-\lambda }{-11}=\;\frac{a+\lambda }{b}=\;\frac{2+3 \lambda }{6 a-1}$ $\;\lambda =-\;\frac{7}{2}, a=3, b=1$ $\;\;{2}/{{\sqrt{a}}}={| \;{5 a+11 c+b c-6 a+1}/{{\sqrt{25+121+1}}} |}$ $\;c=-1$ $\; \therefore \;\frac{a+b}{c}=\;\frac{3+1}{-1}=-4$

Q78JEE Main 2022MCQ4MBinomial Theorem

Fractional part of the number is $\;\frac{4^{2022}}{15}$ equal to [13-Apr-2023 shift 1]

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📖 Explanation

$\{\;\frac{4^{2022}}{15}\} = \{\;\frac{2^{4044}}{15}\} = \{\;\frac{(1+15)^{1011}}{15}\} =\;\frac{1}{15}$

Q79JEE Main 2023MCQ4MDifferential Equations

Let $y=y_1(x)$ and $y=y_2(x)$ be the solution curves of the differential equation $\;\frac{d y}{d x}=y+7$ with initial conditions $y_1(0)=0$ and $y_2(0)=1$ respectively. Then the curves $y=y_1(x)$ and $y=y_2(x)$ intersect at [13-Apr-2023 shift 1]

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📖 Explanation

$\;\;\frac{d y}{d x}=y+7 \Rightarrow \;\frac{d y}{d x}-y=7$ $\;\;\text{ I.F. }\;=e^{-x}$ $\;y e^{-x}=\int 7 e^{-x} \text{dx}$ $\;\Rightarrow y e^{-x}=-7 e^{-x}+c$ $\;\Rightarrow y=-7+c e^x$ $\;-7+7 e^x=-7+8 e^x \Rightarrow e^x=0$ No solution

Q80JEE Main 2023MCQ4MArea Under The Curves

The area of the region enclosed by the curve ${\text{f}}( {\text{x}})=\max \{ sin \; {\text{x}}, \cos {\text{x}}\},-\pi \le {\text{x}} \le \pi$ and the ${\text{x}}$ -axis is [13-Apr-2023 shift 1]

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📖 Explanation

Area $=$ ${| int_{-\pi}^{\;\frac{-3 \pi}{4}} sin \;x \text{dx} |}+{| int_{\;\frac{-3 \pi}{4}}^{\;\frac{-\pi}{2}} \cos x \text{dx} |}+int_{\;\frac{-\pi}{2}}^{\;\frac{\pi}{4}} \cos x \text{dx}+int_{{\;\frac{\pi}{4}}}^{\pi} sin \;x \text{dx}=4$

Q81JEE Main 2023NAT4MSequences and Series

The sum to 20 terms of the series $2.2^{2-}3^{2+}2.4^{2-}5^{2+}2.6^{2-}. . . . . . .$ . is equal to _______. [13-Apr-2023 shift 1]

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📖 Explanation

$\; (2^{2-}3^{2+}4^{2-}5^{2+}20 \;\text{ terms }\;) + (2^{2+}4^{2+}. . . .+10 \;\text{ terms }\;)$ $\;-(2+3+4+5+. . . .+11)+4 [1+2^{2+}. . . . .10^2]$ $\;- [\;\frac{21 \times 22}{2}-1] +4 \times \;\frac{10 \times 11 \times 21}{6}$ $\;=1-231+14 \times 11 \times 10$ $\;=1540+1-231$ $\;=1310$

Q82JEE Main 2023NAT4MStatistics

Let the mean of the data $x$ 1 3 5 7 9 Frequency (f) 4 24 28 $a$ 8be 5 . If ${\text{m}}$ and $\sigma^2$ are respectively the mean deviation about the mean and the variance of the data, then $\;{3 \alpha}/{ {\text{m}}+\sigma^2}$ is equal to _____. [13-Apr-2023 shift 1]

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📖 Explanation

$\;5=\overline{x}=\;\frac{Σ x_i f_i}{Σ f_i}=\;\frac{4+72+140+7 \alpha+72}{64+\alpha}$ $\;\Rightarrow 320+5 \alpha=288+7 \alpha \Rightarrow 2 \alpha=32 \Rightarrow \alpha=16$ $\;\;\text{ M.D. }\;(\overline{ {\text{x}}})=\;{Σ {\text{f}}_{ {\text{i}}}/{| {\text{x}}_{ {\text{i}}}-\overline{ {\text{x}}} |}}{\sum {\text{f}}_{ {\text{i}}}} \;\text{ where }\; \sum {\text{f}}_{ {\text{i}}}=64+16=80$ $\;\;\text{ M.D. }\;(\overline{ {\text{x}}})=\;\frac{4 \times 4+24 \times 2+28 \times 0+16 \times 2+8 \times 4}{80}=\;\frac{8}{5}$ $\;\;\text{ Variance }\;=\;{Σ {\text{f}}_{ {\text{i}}} ( {\text{x}}_{ {\text{i}}}-\overline{ {\text{x}}}) ^2}/{Σ {\text{f}}_{ {\text{i}}}}$ $\;=\;\frac{4 \times 16+24 \times 4+0+16 \times 4+8 \times 16}{80}=\;\frac{352}{80}$ $\; \therefore \;\frac{3 \alpha}{m+\sigma^2}=\;\frac{3 \times 16}{\frac{128}{80}+\;\frac{352}{80}}=8$

Q83JEE Main 2023NAT4MBinomial Theorem

Let $\alpha$ be the constant term in the binomial expansion of $(\sqrt{ {\text{x}}}-\;{6}/{ {\text{x}}^{\;\frac{3}{2}}}) ^{ {\text{n}}}, {\text{n}} \le 15$ . If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of ${\text{x}}^{- {\text{n}}}$ is $\lambda \alpha$ , then $\lambda$ is equal to _______. [13-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{T}}_{ {\text{k}}+1}={ }^{ {\text{n}}} {\text{C}}_{ {\text{k}}}( {\text{x}})^{\;{ {\text{n}}- {\text{k}}}/{2}}(-6)^{ {\text{k}}}( {\text{x}})^{\;\frac{-3}{2} {\text{k}}}$ $\;\;{ {\text{n}}- {\text{k}}}/{2}-\;\frac{3}{2} {\text{k}}=0$ $\; {\text{n}}-4 k=0$ $\;(-5)^{ {\text{n}}}- ({ }_n C_{\;{ {\text{n}}}/{}}(-6)^{\;{ {\text{n}}}/{4}}) =649$ By observation $(625+24=649)$ , we get ${\text{n}}=4$ $\because {\text{n}}=4 & {\text{k}}=1$ Required is coefficient of $\;x^{-4} \;\text{ is }\; ({\sqrt{4}}-\;{6}/{x^{\;\frac{3}{2}}}) ^4$ $\;{ }^4 C_1(-6)^3$ By calculating we will get $\lambda =36$

Q84JEE Main 2023NAT4MComplex Numbers

Let $\omega =z \overline{z}+k_1 z+k_2 i z+\lambda (1+i), k_1, k_2 \in R$ . Let ${\text{Re}}(\omega )=0$ be the circle $C$ of radius 1 in the first quadrant touching the line ${\text{y}}=1$ and the ${\text{y}}$ -axis. If the curve ${\text{Im}}(\omega )=0$ intersects ${\text{C}}$ at ${\text{A}}$ and ${\text{B}}$ , then $30( {\text{AB}})^2$ is equal to _______. [13-Apr-2023 shift 1]

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📖 Explanation

$\;\omega =z \overline{z}+ {\text{k}}_1 z+ {\text{k}}_2 {\text{i}} z+\lambda (1+ {\text{i}})$ $\;{\text{Re}}(\omega )= {\text{x}}^{2+} {\text{y}}^{2+} {\text{k}}_1 {\text{x}}- {\text{k}}_2 {\text{y}}+\lambda =0$ $\;\;\text{ Centre }\; \equiv (\;{- {\text{k}}_1}/{2}, \;{ {\text{k}}_2}/{2}) \equiv (1,2)$ $\;\Rightarrow {\text{k}}_1=-2, {\text{k}}_2=4$ $\;\;\text{ radius }\;=1 \Rightarrow \lambda =4$ $\;{\text{Im}}= {\text{k}}_1 {\text{y}}+ {\text{k}}_2 {\text{x}}+\lambda =0$ $\; \therefore 2 {\text{x}}- {\text{y}}+2=0$ $\; {\text{d}}=\;{2}/{{\sqrt{5}}}$ $\;\;\frac{1^2}{4}=1-\;\frac{4}{5}=\;\frac{1}{5}$ $\; \therefore 301^2=24$

Q85JEE Main 2023NAT4MVector Algebra

Let $\vec{a}=3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+3 \hat{k}$ . If $\vec{b}$ is a vector such that $\vec{a}=\vec{b} \times \vec{c}$ and $|\vec{b}|^2=50$ , then $|72-| \vec{b}+\vec{c} |^2 \mid$ is equal to _______. [13-Apr-2023 shift 1]

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📖 Explanation

$\;|{ {\text{a}}}^{\to }|={\sqrt{11}},|{ {\text{c}}}^{\to }|={\sqrt{22}}$ $\;|{ {\text{a}}}^{\to }|=|{ {\text{b}}}^{\to } \times { {\text{c}}}^{\to }|=|{ {\text{b}}}^{\to }||{ {\text{c}}}^{\to }| sin \;\theta$ $\;{\sqrt{11}}={\sqrt{50}} {\sqrt{22}} sin \;\theta$ $\;\Rightarrow sin \;\theta=\;\frac{1}{10}$ $\;|{ {\text{b}}}^{\to }+{ {\text{c}}}^{\to }|^2=|\vec{b}|^{2+}|{ {\text{c}}}^{\to }|^{2+}2 { {\text{b}}}^{\to } \cdot { {\text{c}}}^{\to }$ $\;=|{ {\text{b}}}^{\to }|^{2+}|{ {\text{c}}}^{\to }|^{2+}2|{ {\text{b}}}^{\to }||\vec{c}| \cos \theta$ $\;=50+22+2 \times {\sqrt{50}} \times {\sqrt{22}} \times \;{{\sqrt{99}}}/{10}$ $\;=72+66$ $|72-|\vec{b}+\vec{c} |^2 \mid =66$

Q86JEE Main 2023NAT4MHyperbola

Let ${\text{m}}_1$ , and ${\text{m}}_2$ be the slopes of the tangents drawn from the point ${\text{P}}(4,1)$ to the hyperbola ${\text{H}}: \;{ {\text{y}}^2}/{25}-\;{ {\text{x}}^2}/{16}=1$ . If ${\text{Q}}$ is the point from which the tangents drawn to $H$ have slopes ${| {\text{m}}_1 |}$ and ${| {\text{m}}_2 |}$ and they make positive intercepts $\alpha$ and $\beta$ on the $x$ -axis, then $\;\frac{(P Q)^2}{\alpha \beta}$ is equal to _______. [13-Apr-2023 shift 1]

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📖 Explanation

Equation of tangent to the hyperbola $\;\frac{y^2}{a^2}-\;\frac{x^2}{b^2}=1$ $y=m x \pm {\sqrt{a^{2-}b^2 m^2}}$ passing through $(4,1)$ $\;1=4 {\text{m}} \pm \sqrt{25-16 {\text{m}}^2} \Rightarrow 4 {\text{m}}^{2-} {\text{m}}-3=0$ $\;\Rightarrow {\text{m}}=1, \;\frac{-3}{4}$ Equation of tangent with positive slopes $1 & \;\frac{3}{4}$ $4 y=3 x-16$ $y=x-3\}$ with positive intercept on $x$ -axis. $\alpha=\;\frac{16}{3}, \beta=3$ Intersection points: $\;Q:(-4,-7)$ $\;P:(4,1)$ $\;P Q^2=128$ $\;\;\frac{P^2}{\alpha \beta}=\;\frac{128}{16}=8$

Q87JEE Main 2023NAT4MThree Dimensional Geometry

Let the image of the point $(\;\frac{5}{3}, \;\frac{5}{3}, \;\frac{8}{3})$ in the plane $x-2 y+z-2=0$ be $P$ . If the distance of the point $Q(6,-2$ , $\alpha), \alpha > 0$ , from $P$ is 13 , then $\alpha$ is equal to ________. [13-Apr-2023 shift 1]

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📖 Explanation

Image of point $(\;\frac{5}{3}, \;\frac{5}{3}, \;\frac{8}{3})$ $\;\frac{x-\frac{5}{3}}{1}=\;\frac{y-\frac{5}{3}}{-2}=\;\frac{z-\frac{8}{3}}{1}=\;\frac{-2 (1 \times \frac{5}{3}+(-2) \times \frac{8}{3}+1 \times \frac{8}{3}-2) }{1^{2+}2^{2+}1^2}$ $=\;\frac{1}{3}$ $\therefore x=2, y=1, z=3$ $13^2=(6-2)^{2+}(-2-1)^{2+}(\alpha-3)^2$ $\Rightarrow (\alpha-3)^2=144 \Rightarrow \alpha=15( \because \alpha > 0)$

Q88JEE Main 2023NAT4MDefinite Integrals

Let for ${\text{x}} \in {\text{R}}, {\text{S}}_0( {\text{x}})= {\text{x}}, {\text{S}}_{ {\text{k}}}( {\text{x}})= {\text{C}}_{ {\text{k}}} {\text{x}}+ {\text{k}} int_{0}^{ {\text{x}}} {\text{S}}_{ {\text{k}}-1}( {\text{t}}) {\text{dt}}$ where ${\text{C}}_0=1, {\text{C}}_{ {\text{k}}}=1-int_{0}^{1} {\text{S}}_{ {\text{k}}-1}( {\text{x}}) {\text{dx}}, {\text{k}}=1,2,3, . . .$ Then ${\text{S}}_2(3)+6 {\text{C}}_3$ is equal to _______. [13-Apr-2023 shift 1]

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📖 Explanation

Given, ${\text{S}}_{ {\text{k}}}( {\text{x}})= {\text{C}}_{ {\text{k}}} {\text{x}}+ {\text{k}} int_{0}^{ {\text{x}}} {\text{S}}_{ {\text{k}}-1}( {\text{t}}) {\text{dt}}$ put ${\text{k}}=2$ and ${\text{x}}=3$ ${\text{S}}_2(3)= {\text{C}}_2(3)+2 int_{0}^{3} {\text{S}}_1( {\text{t}}) {\text{dt}}$ Also, $S_1(x)=C_1(x)+int_{0}^{x} S_0(t) \text{dt}$ . . . . . . (1) $=C_1 x+\;\frac{x^2}{2}$ $S_2(3)=3 C_2+2 int_{0}^{3} (C_1 t+\;\frac{t^2}{2}) \text{dt}$ $=3 C_2+9 C_1+9$ Also, $C_1=1-int_{0}^{1} {\text{S}}_0(x) \text{dx}=\;\frac{1}{2}$ $\;C_2=1-int_{0}^{1} {\text{S}}_1(x) \text{dx}=0$ $\;C_3=1-int_{0}^{1} S_2(x) \text{dx}$ $\;=1-int_{0}^{1} (C_2 x+C_1 x^{2+}\;\frac{x^3}{3}) \text{dx}=\;\frac{3}{4}$ $\;S_2(x)=C_2 x+2 int_{0}^{x} S_1(t) \text{dt}$ $\;=C_2 x+C_1 x^{2+}\;\frac{x^3}{3}$ $\;\Rightarrow {\text{S}}_2(3)+6 {\text{C}}_3=6 {\text{C}}_3+3 {\text{C}}_2+9 {\text{C}}_1+9=18$

Q89JEE Main 2023NAT4MInverse Trigonometric Functions

If $S= \{x \in R: sin \;^{-1} (\;{x+1}/{{\sqrt{x^{2+}2 x+2}}}) - sin \;^{-1} (\;{x}/{{\sqrt{x^{2+}1}}}) =\;\frac{\pi}{4}\}$ , then $sum_{x \in s} ( sin \; ( (x^{2+}x+5) \;\frac{\pi}{2}) -\cos ( (x^{2+}x+5) \pi) )$ is equal to _______. [13-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\; sin \;^{-1} (\;{(x+1)}/{{\sqrt{(x+1)^{2+}1}}}) - sin \;^{-1} (\;{x}/{{\sqrt{x^{2+}1}}}) =\;\frac{\pi}{4}$ $\; \because \;{t}/{{\sqrt{t^{2+}1}}} \in (-1,1)$ $\; sin \;^{-1} (\;{( {\text{x}}+1)}/{\sqrt{( {\text{x}}+1)^{2+}1}}) = sin \;^{-1} (\;{ {\text{x}}}/{\sqrt{ {\text{x}}^{2+}1}}) +\;\frac{\pi}{4}$ $\;\;{(x+1)}/{{\sqrt{(x+1)^{2+}1}}}= (\;{1}/{{\sqrt{2}}}) \cos ( sin \;^{-1} (\;{x}/{{\sqrt{x^{2+}1}}}) ) +\;{1}/{{\sqrt{2}}} (\;{x}/{{\sqrt{x^{2+}1}}})$ $\;=\;{1}/{{\sqrt{2}}} (\;{1}/{{\sqrt{x^{2+}1}}}+\;{x}/{{\sqrt{x^{2+}1}}})$ $\;\;{(x+1)}/{{\sqrt{(x+1)^{2+}1}}}=\;{1}/{{\sqrt{2}}} (\;{1+x}/{{\sqrt{x^{2+}1}}})$ After solving this equation, we get $\; {\text{x}}=-1 \;\text{ or }\; {\text{x}}=0$ $\;S=\{-1,0\}$ $\;sum_{x \in R} ( sin \; ( (x^{2+}x+5) \;\frac{\pi}{2}) -\cos ( (x^{2+}x+5) \pi) )$ $\;= [ sin \; (\;\frac{5 \pi}{2}) -\cos (5 \pi)] + [ sin \; (\;\frac{5 \pi}{2}) -\cos (5 \pi)] =4$

Q90JEE Main 2023NAT4MPermutations and Combinations

The number of seven digit positive integers formed using the digits 1, 2, 3 and 4 only and sum of the digits equal to 12 is _______. [13-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$x_1+x_2+x_3+x_4+x_5+x_6+x_7=12, x_i \in \{1,2,3,4\}$ No. of solutions $={ }^{5+7-1} C_{7-1}-\;\frac{7 !}{6 !}-\;\frac{7 !}{5 !}=413$

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