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12 Apr 2023 Shift 1

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Physics30 questions

Q1JEE Main 2023MCQ4MGravitation

The ratio of escape velocity of a planet to the escape velocity of earth will be :-Given : Mass of the planet is 16 times mass of earth and radius of the planet is 4 times the radius of earth. [12-Apr-2023 shift 1]

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📖 Explanation

Given mass of planet $=16$ mass of the earth $( {\text{m}}_{ {\text{p}}}=16 {\text{m}}_{ {\text{e}}})$ $r_p=4 r_{\;\text{o }\;}$ We know $V_{\;\text{escape }\;}=\sqrt{\;\frac{2 G m}{R}}, \;{V_{\;\text{esplanet }\;}}/{V_{\;\text{esearth }\;}}=$ ? $\;{V_{\;\text{es plavet }\;}}/{V_{\;\text{es earth }\;}}\;=\sqrt{\;{2 {\text{GM}}_{ {\text{p}}}}/{ {\text{R}}_{ {\text{p}}}} \times \;{ {\text{R}}_{ {\text{e}}}}/{2 {\text{GM}}_{ {\text{e}}}}}$ $\;=\sqrt{\;{ {\text{M}}_{ {\text{p}}}}/{ {\text{M}}_{ {\text{e}}}} \times \;{ {\text{R}}_{ {\text{e}}}}/{ {\text{R}}_{ {\text{p}}}}}$ $\;=\sqrt{\;{16 {\text{m}}_{ {\text{e}}}}/{ {\text{m}}_{ {\text{e}}}} \times \;{ {\text{R}}_{ {\text{e}}}}/{4 {\text{R}}_{ {\text{e}}}}}$ $\;=\sqrt{\;\frac{16}{4}}=2: 1$ $\;{ {\text{V}}_{\;\text{esplanet }\;}}/{ {\text{V}}_{\;\text{eseath }\;}}\;=\;\frac{2}{1}$

Q2JEE Main 2023MCQ4MThermodynamics

An engine operating between the boiling and freezing points of water will haveA. efficiency more than $27 \%$ B. efficiency less than the efficiency of a Carnot engine operating between the same two temperatures.C. efficiency equal to $27 \%$ D. efficiency less than $27 \%$ Choose the correct answer from the options given below : [12-Apr-2023 shift 1]

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📖 Explanation

Engine is operating between FP and BP of water.We asked to find efficiency : $\; {\text{T}}_1=100^{\circ} {\text{C}} \;\; \Rightarrow 100+273=373 {\text{K}}$ $\; {\text{T}}_2=0^{\circ} {\text{C}} \;\; \Rightarrow 0+273=273 {\text{K}}$ We know $\;η=1-\;\frac{T_2}{T_1}$ $\;η=1-\;\frac{273}{373}$ $\;η=\;\frac{373-273}{373}=\;\frac{100}{373}$ $\;η=0.268$ and $η \%\;=0.268 \times 100$ $η \%\;=26.8 \%$

Q3JEE Main 2023MCQ4MCurrent Electricity

A wire of resistance $160 Ω$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be [12-Apr-2023 shift 1]

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📖 Explanation

Voulme remain same $\; {\text{A}} ℓ= {\text{A}}^{'} ℓ^{'}$ $\; {\text{A}}^{'}=\;{ {\text{A}} ℓ}/{ℓ^{'}}=\;{ {\text{A}} ℓ}/{ℓ / 4}$ $\; {\text{A}}^{'}=4 {\text{A}}$ $\; {\text{L}}^{'}=\;{ {\text{L}}}/{4}$ $\;\;{ {\text{R}}^{'}}/{ {\text{R}}^{'}}=\;{\rho { {\text{L}}^{'}}/{ {\text{A}}^{'}}}/{{ {\text{L}}}/{ {\text{A}}}}$ $\; {\text{R}}^{'}= {\text{R}} [\;{ {\text{L}}^{'}}/{ {\text{A}}} \times \;{ {\text{A}}}/{ {\text{L}}}]$ $\; {\text{R}}^{'}=160 [\;{ {\text{L}}}/{4 {\text{L}}} \times \;{ {\text{A}}}/{4 {\text{A}}}]$ $\; {\text{R}}^{'}=10 Ω$

Q4JEE Main 2023MCQ4MLaws of Motion

Three forces $F_1=10 {\text{N}}, F_2=8 {\text{N}}, F_3=6 {\text{N}}$ are acting on a particle of mass $5 {\text{kg}}$ . The forces $F_2$ and $F_3$ are applied perpendicularly so that particle remains at rest. If the force ${\text{F}}_1$ is removed, then the acceleration of the particle is : [12-Apr-2023 shift 1]

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📖 Explanation

$F_1\;=10 {\text{N}}$ $F_2\;=8 {\text{N}}$ $F_3\;=6 {\text{N}}$ ${ {\text{a}}}^{\to }\;=\;{ {\text{F}}_2 { {\text{i}}^{\^}}+ {\text{F}}_3 { {\text{j}}^{\^}}}/{5}$ ${ {\text{a}}}^{\to }\;=\;{8 { {\text{i}}^{\^}}+6 { {\text{j}}^{\^}}}/{5}$ $|{ {\text{a}}}^{\to }|\;=\;{{\sqrt{64+36}}}/{5}$ $\;=\;\frac{10}{5}=2 {\text{m}} / {\ sec}^2$ OR ${\text{F}}_1^2= {\text{F}}_2^{2+} {\text{F}}_3^2$ Now ${\text{F}} 1$ is removed then net force $\; {\text{F}}_{\;\text{net }\;}=\sqrt{ {\text{F}}_2^{2+} {\text{F}}_3^2}$ $\; \;\; ={\sqrt{64+36}}$

Q5JEE Main 2023MCQ4MElectromagnetic Waves

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A : EM waves used for optical communication have longer wavelength than that of microwave, employed in Radar technology.Reason R : Infrared EM waves are more energetic than microwaves, (used in Radar) In the light of given statements, choose the correct answer from the options given below. [12-Apr-2023 shift 1]

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📖 Explanation

Assertion A is false because electromagnetic waves used for optical communication, such as infrared waves, have shorter wavelengths than microwaves used in radar. According to the energy-wavelength relationship, $E = hc/\lambda$ , energy is inversely proportional to wavelength. Since infrared waves have a shorter wavelength, they are more energetic than microwaves. Therefore, Reason R is true.

Q6JEE Main 2023MCQ4MThermodynamics

If the r. ${\text{m}}$ . s speed of chlorine molecule is $490 {\text{m}} / {\ s}$ at $27^{\circ} {\text{C}}$ , the ${\text{r}} . {\text{m}} . {\ s}$ speed of argon molecules at the same temperature will be (Atomic mass of argon $=39.9 {\text{u}}$ , molecular mass of chlorine $=70.9 {\text{u}}$ ) [12-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Molar mass of chlorine }\;=70.9 {\text{u}}$ $\;\;\text{ Atomic mass of argon }\;=39.9 {\text{u}}$ $\; {\text{V}}_{\;\text{rms }\;}=\sqrt{\;{3 {\text{RT}}}/{ {\text{m}}}}$ $\; {\text{V}}_{\;\text{rms }\;} \propto \sqrt{\;{1}/{ {\text{m}}}}$ $\;\;{ {\text{V}}_{\;\text{rmschlorine }\;}}/{ {\text{V}}_{\;\text{rmsarg on }\;}}=\sqrt{\;{ {\text{m}}_{\;\text{arg on }\;}}/{ {\text{m}}_{\;\text{chlorine }\;}}}$ $\;\;{490}/{ {\text{V}}_{\;\text{rmsargon }\;}}=\sqrt{\;\frac{39.9}{70.9}}$ $\; {\text{V}}_{\;\text{rmsargon }\;}=490 \times 1.33$ $\; {\text{V}}_{\;\text{msargon }\;}=651.7 {\text{m}} / {\ sec}$

Q7JEE Main 2023MCQ4MGravitation

Two satellites A and B move round the earth in the same orbit. The mass of A is twice the mass of B. The quantity which is same for the two satellites will be [12-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Speed }\; V=\sqrt{\;\frac{G M_e}{r}}$ $\; {\text{m}}_e=\;\text{ mass of earth }\;$ $\;r=\;\text{ radius of orbit }\;$ Independent on mass of satellite so speed is same Other three quantities are mass dependent $\;\to \;\; {\text{KE}}=\;{ {\text{Gm}}_{ {\text{e}}} {\text{m}}}/{2 {\text{r}}}$ $\;\to \;\; P E=\;{- {\text{Gm}}_{ {\text{e}}} {\text{m}}}/{ {\text{r}}}$ $\;\to \;\; {\text{TE}}=\;{- {\text{Gm}}_{ {\text{e}}} {\text{m}}}/{2 {\text{r}}}$ $\;$

Q8JEE Main 2023MCQ4MAlternating Current

Given below are two statements:Statements I : When the frequency of an a.c source in a series LCR circuit increases, the current in the circuit first increases, attains a maximum value and then decreasesStatements II : In a series LCR circuit, the value of power factor at resonance is one.In the light of given statements, choose the most appropriate answer from the options given below. [12-Apr-2023 shift 1]

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📖 Explanation

(i) At resonance frequency power factor of RLC circuit is one(ii) Graph of RLC circuit is: $\;\text{ At }\; \;\omega =\omega _{ {\text{R}}}$ $\; {\text{Z}}= {\text{R}}$ $\;\cos \varphi =\;{ {\text{R}}}/{ {\text{Z}}}=1$

Q9JEE Main 2023MCQ4MOscillations

A particle is executing simple harmonic motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be [12-Apr-2023 shift 1]

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📖 Explanation

Ratio of ${\text{PE}}$ and ${\text{KE}}=$ ?When $\;{ {\text{x}}}_{{{ {\text{x}}} \\ \;\text{ disp. }\;}}=\;{ {\text{A}}}/{2}$ ${\text{KE}}\;=\;\frac{1}{2} {\text{m}} \omega ^2 [ {\text{A}}^{2-} {\text{x}}^2]$ $\;=\;\frac{1}{2} {\text{m}} \omega ^2 [ {\text{A}}^{2-}\;{ {\text{A}}^2}/{4}]$ $\; {\text{KE}}==\;\frac{1}{2} {\text{m}} \omega ^2 [\;{3 {\text{A}}^2}/{4}]$ ... (1) $\;\;\text{ And }\; {\text{PE}}=\;\frac{1}{2} {\text{m}} \omega ^2 {\text{x}}^2$ $\;=\;\frac{1}{2} {\text{m}} \omega ^2 \;{ {\text{A}}^2}/{4}$ $\;\;{ {\text{PE}}}/{ {\text{KE}}}=\;{ {\text{m}} \omega ^2 {\text{A}}^2 \times 8}/{8 {\text{m}} \omega ^2 {\text{A}}^2 \times 3}=\;\frac{1}{3}$ $\;\Rightarrow 1: 3$ $\;$

Q10JEE Main 2023MCQ4MElectrostatics

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R Assertion A : If an electric dipole of dipole moment $30 \times 10^{-5} {\text{C}} {\text{m}}$ is enclosed by a closed surface, the net flux coming out of the surface will be zero.Reason R : Electric dipole consists of two equal and opposite charges.In the light of above, statements, choose the correct answer from the options given below. [12-Apr-2023 shift 1]

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Q11JEE Main 2023MCQ4MMagnetism and Matter

Given below are two statements:Statements I : The diamagnetic property depends on temperature.Statements II : The induced magnetic dipole moment in a diamagnetic sample is always opposite to the magnetizing field.In the light of given statements, choose the correct answer from the options given below. [12-Apr-2023 shift 1]

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Q12JEE Main 2023MCQ4MUnits and Measurements

LIST I LIST II A. Spring constant I. $[ {\text{T}}^{-1}]$ B. Angular speed II. $[ {\text{MT}}^{-2}]$ C. Angular momentum III. $[ {\text{ML}}^2]$ D. Moment of Inertia IV. $[ {\text{ML}}^2 {\text{T}}^{-1}]$ Choose the correct answer from the options given below : [12-Apr-2023 shift 1]

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📖 Explanation

(i) Spring constant $\; {\text{F}}= {\text{Kx}}$ $\; {\text{K}}=\;{ {\text{F}}}/{ {\text{x}}}=\;{ {\text{kgm}} {\ sec}}/{ {\text{m}}}= {\text{kg}} \cdot {\ sec}^{-2}= [ {\text{ML}}^0 {\text{T}}^{-2}]$ (ii) Angular speed : (ω) $\omega \;=\;{2 \pi}/{ {\text{T}}}$ $\omega \;= [ {\text{M}}^0 {\text{L}}^0 {\text{T}}^{-1}]$ (iii) Angular momentum $\; {\text{L}}= {\text{MVR}} \Rightarrow {\text{Kg}} \cdot {\text{m}} {\ sec}{ }^{-1} {\text{m}}= {\text{kgm}}^2 {\ sec}^{-1}$ $\;= [ {\text{ML}}^2 {\text{T}}^{-1}]$ $\;$ (iv) Moment of Inertia ${\text{I}}\;= {\text{MR}}^2= {\text{Kg}} \cdot {\text{m}}^2$ $\;= [ {\text{ML}}^2 {\text{T}}^0]$

Q13JEE Main 2023MCQ4MRay Optics and Optical Instruments

An ice cube has a bubble inside. When viewed from one side the apparent distance of the bubble is $12 {\text{cm}}$ . When viewed from the opposite side, the apparent distance of the bubble is observed as $4 {\text{cm}}$ . If the side of the ice cube is $24 {\text{cm}}$ , the refractive index of the ice cube is [12-Apr-2023 shift 1]

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📖 Explanation

Question says small bubble is trapped inside a cube of length $24 {\text{cm}}$ . If apparent distance of bubble is $12 {\text{cm}}$ and $4 {\text{cm}}$ from one and other side and find ${\text{x}}=$ ? $\;\;\text{ Using }\; \;\frac{\;\text{ Apparent depth }\;}{\;\text{ Realdepth }\;}=\;\frac{\mu _2}{\mu _1}$ $\;\mu _2=1 \;\; \mu _1 \;\text{ is glass }\;$ $\;\mu _1=\mu \;\; \mu _2 \;\text{ is air }\;$ $\;\;{12}/{ {\text{d}}_1}=\;\frac{1}{\mu }$ $\;\;{12 \mu = {\text{d}}_1}/{{4}/{ {\text{d}}_2}}=\;\frac{1}{\mu } \;\; \Rightarrow (1)$ $\; {\text{d}}_1+ {\text{d}}_2=24 {\text{m}} \Rightarrow \;\; 12 \mu +4 \mu =24 \Rightarrow {\text{d}}_2$

Q14JEE Main 2023MCQ4MWaves

The amplitude of $15 sin \;(100 \pi {\text{t}})$ is modulated by $10 sin \;(4 \pi {\text{t}})$ signal. The amplitude modulated signal contains frequencies ofA. $500 {\text{Hz}}$ 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 : [12-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Signal }\;=15 sin \;(1000 \pi {\text{t}}) \;\text{ is modulated by }\; 10 sin \;(4 \pi {\text{t}}) \;$ $\omega _{ {\text{m}}}=4 \pi$ $2 \pi {\text{f}}_{ {\text{m}}}=4 \pi$ ${\text{f}}_{ {\text{m}}}=2 {\text{H}}$

Q15JEE Main 2023MCQ4MAtoms and Nuclei

A $12.5 {\text{eV}}$ electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be : [12-Apr-2023 shift 1]

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📖 Explanation

$\;12.5=-13.6 \$ [;{1}/{ {\text{n}}{ {\text{f}}}^2}-;{1}/{ {\text{n}}{ {\text{i}}}^2}]$ ;; {\ni}=1;12.5=-13.6 [;{1}/{ {\text{n}}{ {\text{f}}}^2}-1];12.5=13.6-;{13.6}/{ {\text{n}}{ {\text{f}}}^2};-1.1=-;{13.6}/{ {\text{n}}{ {\text{f}}}^2}; {\text{n}}{ {\text{f}}}=3.5n_{ {\text{f}}}=3;;\text{ so transition is }; 3 ;\text{ to }; 1 $No. of lines$ ={ }^3 {\text{C}}2 ;; ({ }^n {\text{C}}{ {\text{r}}}=;{x{ {\text{n}}}}/{x{\mid {\text{n}}- {\text{r}}) \mid x{r}}});=;\frac{{3}}{(3-2)(2]};;\frac{3 \times 2 \times 1}{1 \times 2}=3 ;\text{ lines };;$

Q16JEE Main 2023MCQ4MMotion in a Straight Line

A ball is thrown vertically upward with an initial velocity of $150 {\text{m}} / {\ s}$ . The ratio of velocity after $3 {\ s}$ and $5 {\ s}$ is $\;\frac{x+1}{x}$ . The value of $x$ is _______. {take, $g=10 {\text{m}} / {\ s}^2$ } [12-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ ToF }\;=\;\frac{24}{g}=\;\frac{2 \times 150}{10}=30 {\ sec}$ $\;\;\text{ ToA }\;=15 {\ sec} .$ $\;\;{ {\text{V}}_1}/{ {\text{V}}_2}=\;{ {\text{u}}+ {\text{at}}_1}/{ {\text{u}}+ {\text{at}}_2}$ $\;\;{ {\text{V}}_1}/{ {\text{V}}_2}=\;\frac{150-10 \times 3}{150-10 \times 5}=\;\frac{120}{100}$ $\;\;{ {\text{V}}_1}/{ {\text{V}}_2}=\;\frac{6}{5}$ $\; {\text{x}}+1=6$ $\; {\text{x}}=5$

Q17JEE Main 2023MCQ4MSemiconductor Electronics

In an n-p-n common emitter (CE) transistor the collector current changes from $5 {\text{mA}}$ to $16 {\text{mA}}$ for the change in base current from $100 \mu {\text{A}}$ and $200 \mu {\text{A}}$ , respectively. The current gain of transistor is _______. [12-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ (1) }\; \beta=\;{\Delta {\text{I}}_{ {\text{C}}}}/{\Delta {\text{I}}_{ {\text{B}}}} \;\; \;\;\;\Delta {\text{I}}_{ {\text{C}}}=16-5=11 {\text{mA}}, \Delta {\text{I}}_{ {\text{B}}}=200-100=100 \mu {\text{A}}$ $\;\beta=\;\frac{11 \times 10^{-3}}{100 \times 10^{-6}}$ $\;\beta=\;\frac{11000}{100}$ $\;\beta=110$

Q18JEE Main 2023MCQ4MThermodynamics

A body cools from $80^{\circ} {\text{C}}$ to $60^{\circ} {\text{C}}$ in 5 minutes. The temperature of the surrounding is $20^{\circ} {\text{C}}$ . The time it takes to cool from $60^{\circ} {\text{C}}$ to $40^{\circ} {\text{C}}$ is: [12-Apr-2023 shift 1]

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📖 Explanation

Body cools from $80^{\circ} {\text{C}}$ to $60^{\circ} {\text{C}}$ in $5 {\min}$ ${\text{T}}_{\;\text{suroumding }\;}=20^{\circ} {\text{C}}$ Time taken to cool from $60^{\circ} {\text{C}}$ to $40^{\circ} {\text{C}}$ ? we know newton's law of cooling is $-\;{ {\text{d}} \theta}/{ {\text{dt}}}= {\text{b}} (\theta-\theta_0)$ surrounding temp. We take approx value of ${\text{v}}$ as $\;\frac{80+60}{2}=70^{\circ}$ $\;\;\frac{20}{5}= {\text{b}}[70-20]$ . . . (1) $\;\;{20}/{ {\text{t}}}= {\text{b}}[50-20]$ . . . (2)After solving ${\text{t}}=\;\frac{25}{3} {\min}$ In second ${\text{t}}=\;\frac{25}{3} \times 60$ $=500 {\ sec} \;\text{. }\;$

Q19JEE Main 2023MCQ4MDual Nature of Matter and Radiation

A proton and an $\alpha$ -particle are accelerated from rest by $2 {\text{V}}$ and $4 {\text{V}}$ potentials, respectively. The ratio of their de-Broglie wavelength is : [12-Apr-2023 shift 1]

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📖 Explanation

De Broglie wavelength $\lambda =\;{ {\text{h}}}/{\sqrt{2 {\text{mqV}}}}$ $\; {\text{m}}_alpha=4 {\text{m}} \to 4 {\text{V}}$ $\; {\text{m}}_{ {\text{p}}}= {\text{m}} \to 2 {\text{V}}$ $\;\lambda =\;{ {\text{h}}}/{\sqrt{2 {\text{mKE}}}}$ $\lambda _{ {\text{p}}}=\;{ {\text{h}}}/{\sqrt{2 {\text{mq}}(2 {\text{V}})}}$ . . . . (1), $\;\;\;\lambda _alpha=\;{ {\text{h}}}/{\sqrt{4 {\text{mq}}(4 {\text{V}})}}$ $\;\;{\lambda _{ {\text{p}}}}/{\lambda _alpha}=\;{ {\text{h}}}/{\sqrt{2 {\text{mq}}(2 {\text{V}})}} \times \;{\sqrt{4 {\text{mq}}(4 {\text{V}})}}/{ {\text{h}}}$ $\;\;{\lambda _{ {\text{p}}}}/{\lambda _alpha}=4 \;\; \Rightarrow \;\; \;{\lambda _{ {\text{p}}}}/{\lambda _alpha}=4: 1$

Q20JEE Main 2023MCQ4MLaws of Motion

Given below are two statements:Statements I : A truck and a car moving with same kinetic energy are brought to rest by applying breaks which provide equal retarding forces. Both come to rest in equal distance.Statements II : A car moving towards east takes a turn and moves towards north, the speed remains unchanged. The acceleration of the car is zero.In the light of given statements, choose the most appropriate answer from the options given below [12-Apr-2023 shift 1]

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Q21JEE Main 2023NAT4MWork, Power and Energy

To maintain a speed of $80 {\text{km}} / {\text{h}}$ by a bus of mass $500 {\text{kg}}$ on a plane rough road for $4 {\text{km}}$ distance, the work done by the engine of the bus will be _______ KJ. [The coefficient of fiction between tyre of bus and road is 0.04 .] [12-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{F}}=0.04 \times 500 \times 9.8$ $\;=20 \times 9.8=196$ $\; {\text{WD}}= {\text{F}} \times {\text{disc}}$ $\;=196 \times 4000$ $\;=784 {\text{kJ}}$

Q22JEE Main 2023NAT4MElectromagnetic Induction

A conducting circular loop is placed in a uniform magnetic field of $0.4 {\text{T}}$ with its plane perpendicular to the field. Somehow, the radius of the loop starts expending at a constant rate of $1 {\text{mm}} / {\ s}$ . The magnitude of induced emf in the loop at an instant when the radius of the lop is $2 {\text{cm}}$ will be ________ $\mu {\text{V}}$ . [12-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{B}}=0.4 {\text{T}}$ $\;\;\text{ Rate }\; \;{ {\text{dr}}}/{ {\text{dt}}}=1 {\text{mm}} / {\ sec} \;\; \;\text{ Einduced }\;=?$ $\; {\text{R}}=2 {\text{cm}}$ $\;\varphi = {\text{B}} \pi {\text{r}}^2 \;\; ((\varphi = {\text{B}} . {\text{A}})$ $\;\epsilon=\;{ {\text{d}} \varphi }/{ {\text{dt}}}= {\text{B}} \pi(2 {\text{r}}) \;{ {\text{dr}}}/{ {\text{dt}}}$ $\;\;\text{ Ein }\;=0.4 \times \pi \times 4 \times 10^{-2} \times 10^{-3}$ $\;\;\text{ Ein }\;=16 \pi \times 10^{-6}$ $\;\;\text{ Ein }\;=50 \mu {\text{V}}$

Q23JEE Main 2023NAT4MAtoms and Nuclei

A common example of alpha decay is $\;{ }_{92}^{238} {\text{U}} \to { }_{90}^{234} {\text{Th}}+{ }_2 {\text{He}}^{4+} {\text{Q}}$ $\;\;\text{ Given : }\;$ $\;{ }_{92}^{238} {\text{U}}=238.05060 {\text{u}},$ $\;{ }^{234} {\text{Th}}=234.04360 {\text{u}},$ $\;90$ $\;{ }_2^4 {\text{He}}=4.00260 {\text{u}} \;\text{ and }\;$ $\; {\text{lu}}=931.5 \;{ {\text{MeV}}}/{ {\text{c}}^2}$ The energy released (Q) during the alpha decay of ${ }^{238} {\text{U}}$ is _______ ${\text{MeV}}$ [12-Apr-2023 shift 1]

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📖 Explanation

$\;\alpha= ( {\text{m}}_{ {\text{u}}}- {\text{M}}_{\;\text{th }\;}- {\text{M}}_{ {\text{Ho}}}) {\text{C}}^2$ $\;\alpha=0.00456 \times 931.5$ $\;\alpha=4.25 {\text{MeV}}$

Q24JEE Main 2023NAT4MElectrostatic Potential and Capacitance

64 identical drops each charged upto potential of $10 {\text{mV}}$ are combined to form a bigger drop. The potential of the bigger drop will be _______ ${\text{mV}}$ . [12-Apr-2023 shift 1]

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📖 Explanation

We know $V=\;\frac{k q}{r}, \;\; q^{'}=64 q$ Voulme remain const so $\;\;\frac{4}{3} \pi r^3 \times 64=\;\frac{4}{3} \pi R^3$ $\;R=4 r$ Now new potential ${\text{V}}^{'}=\;{ {\text{K}} 64 {\text{q}}}/{4 {\text{r}}}=\;{16 {\text{kq}}}/{ {\text{r}}}$ $\; {\text{V}}^{'}=16 . {\text{V}} \;\text{ and }\; {\text{V}}=\;{ {\text{Kq}}}/{ {\text{r}}}=10 {\text{mV}}$ $\; {\text{V}}^{'}=16 \times 10 \times 10^{-3}$ $\; {\text{V}}^{'}=0.16 {\text{V}}$ OR $V^{'}=160 {\text{mV}}$

Q25JEE Main 2023NAT4MRotational Motion

For a rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is $\;\frac{x}{5}$ . The value of $x$ is ______. [12-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{KE}}_{ {\text{T}}} \;\frac{1}{2} {\text{mV}}^2$ $\; {\text{KE}}_{ {\text{R}}}=\;\frac{1}{2} (\;\frac{2}{5} {\text{mR}}^2) \omega ^2=\;\frac{1}{3} {\text{mR}}^2 \cdot \;{ {\text{V}}^2}/{ {\text{R}}^2}=\;\frac{1}{3} {\text{mV}}^2$ $\; {\text{KE}}_{ {\text{T}}}=\;\frac{1}{2} {\text{mV}}^{2+}\;{ {\text{mV}}^2}/{3}$ $\; {\text{KE}}_{ {\text{T}}}=\;\frac{5}{6} {\text{mV}}^2$ $\;\;{ {\text{K}}_{ {\text{R}}}}/{ {\text{K}}_{ {\text{T}}}}=\;{1 / 3 {\text{mV}}^2}/{5 / 6 {\text{mV}}^2}=\;\frac{2}{5}$ $\; {\text{x}}=2$

Q26JEE Main 2023NAT4MRay Optics and Optical Instruments

Two convex lenses of focal length $20 {\text{cm}}$ each are placed coaxially with a separation of $60 {\text{cm}}$ between them. The image of the distant object formed by the combination is at _______ ${\text{cm}}$ from the first lens. [12-Apr-2023 shift 1]

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📖 Explanation

For ${\text{I}}^{\;\text{st }\;}$ lens $\;{1}/{ {\text{V}}_1}-\;\frac{1}{\infty }=\;\frac{1}{20}$ ${\text{V}}_1=20 {\text{cm}}$ For $2^{\;\text{nd }\;}$ lens $\;\;{1}/{ {\text{V}}_2}-\;\frac{1}{-40}=\;\frac{1}{20}$ $\; {\text{V}}_2=40 {\text{cm}}$ So dist $=40+60=100 {\text{cm}}$

Q27JEE Main 2023NAT4MMagnetic Effects of Current and Magnetism

A compass needle oscillates 20 times per minute at a place where the dip is $30^{\circ}$ and 30 times per minute where the dip is $60^{\circ}$ . The ratio of total magnetic field due to the earth at two places respectively is $\;{4}/{\sqrt{ {\text{x}}}}$ . The value of $x$ is [12-Apr-2023 shift 1]

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📖 Explanation

Period of oscillation $\alpha \;{1}/{{\sqrt{B_H}}}$ $\;T \alpha \;{1}/{{\sqrt{B \cos \theta}}} \Rightarrow \;\frac{T_1}{T_2}=\sqrt{\;\frac{B_2 \cos \theta_2}{B_1 \cos \theta_1}}$ $\;\Rightarrow \;\frac{60 / 20}{60 / 30}=\sqrt{\;\frac{B_2}{B_1} \;\frac{\cos 60^{\circ}}{\cos 30^{\circ}}} \Rightarrow \;\frac{3}{2}=\sqrt{\;{B_2}/{{\sqrt{3}} B_1}}$ $\;\Rightarrow \;\frac{9}{4}=\;{B_2}/{{\sqrt{3}} B_1} \Rightarrow \;\frac{B_1}{B_2}=\;{4}/{9 {\sqrt{3}}}=\;{4}/{{\sqrt{243}}}$

Q28JEE Main 2023NAT4MWaves

For a certain organ pipe, the first three resonance frequencies are in the ratio of $1: 3: 5$ respectively. If the frequency of fifth harmonic is $405 {\text{Hz}}$ and the speed of sound in air is $324 {\text{ms}}^{-1}$ the length of the organ pipe is ________ ${\text{m}}$ [12-Apr-2023 shift 1]

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📖 Explanation

Resonance frequency in closed organ pipe ${\text{f}}=(2 {\text{n}}+1) \;{ {\text{V}}}/{4 ℓ}$ Given is $5^{\;\text{th }\;}$ harmonicSo $5 \;{ {\text{V}}}/{4 ℓ}=405$ $\;\;\frac{5 \times 324}{4 \times 405}=ℓ$ $\;ℓ=1 {\text{m}}$

Q29JEE Main 2023NAT4MProperties of Matter

Glycerin of density $1.25 \times 10^3 {\text{kg}} {\text{m}}^{-3}$ is flowing through the conical section of pipe. The area of cross-section of the pipe at its ends are $10 {\text{cm}}^2$ and $5 {\text{cm}}^2$ and pressure drop across its length is $3 {\text{Nm}}^{-2}$ . The rate of flow of glycerin through the pipe is ${\text{x}} \times 10^{-5} {\text{m}}^3 {\ s}^{-1}$ . The value of ${\text{x}}$ is _______. [12-Apr-2023 shift 1]

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📖 Explanation

${\text{A}}_1=10 {\text{cm}}^2$ $\Delta {\text{P}}= {\text{P}}_1- {\text{P}}_2=3 {\text{N}} / {\text{m}}^2 \;\text{ (given) }\;$ By continuity equation $\; {\text{A}}_1 {\text{v}}_1= {\text{A}}_2 {\text{v}}_2$ $\; \therefore {\text{v}}_1=\;{ {\text{A}}_2}/{ {\text{A}}_1} {\text{v}}_2$ . . . . . (1) By bernoulli's equation $\;P_1+\;\frac{1}{2} \rho v_1{ }^2=P_2+\;\frac{1}{2} \rho v_2{ }^2$ $\;P_1-P_2=\;\frac{1}{2} \rho (v_2{ }^{2-}v_1{ }^2)$ $\;\Delta P=\;\frac{1}{2} \rho (v_2{ }^{2-}\;\frac{A_2^2}{A_1^2} v_2{ }^2)$ $\;\Delta P=\;\frac{1}{2} \rho \$ [1- (;\frac{A_2}{A_1}) ^2]$ v_2^2;3=;\frac{1}{2} \times 1.25 \times 10^3 (1- (;\frac{5}{10}) ^2) v_2^2;3=;\frac{1}{2} \times 1.25 \times 10^3 [1-;\frac{1}{4}] v_2^2;3=;\frac{1}{2} \times 1.25 \times 10^3 \times ;\frac{3}{4} v_2^2\therefore {\text{v}}_2=8 \times 10^{-2} {\text{m}} / {\ s} $So discharge rate$ = {\text{A}}_2 {\text{V}}_2;=5 \times 10^{-4} \times 8 \times 10^{-2};=4 \times 10^{-5} {\text{m}}^3 / {\ s}$

Q30JEE Main 2023NAT4MCurrent Electricity

The current flowing through a conductor connected across a source is $2 {\text{A}}$ and $1.2 {\text{A}}$ at $0^{\circ} {\text{C}}$ and $100^{\circ} {\text{C}}$ respectively. The current flowing through the conductor at $50^{\circ} {\text{C}}$ will be ______ $\times 10^2 {\text{mA}}$ [12-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{i}}_0 {\text{R}}_0= {\text{i}}_{100} {\text{R}}_{100} \;\; \;\text{ (For same source) }\;$ $\;\Rightarrow 2 {\text{R}}_0=1.2 {\text{R}}(1+100 \alpha)$ $\;\Rightarrow 1+100 \alpha=\;\frac{5}{3} \Rightarrow 100 \alpha=\;\frac{2}{3}$ $\;\Rightarrow 50 \alpha=\;\frac{1}{3}$ $\; \therefore {\text{i}}_{50} {\text{R}}_{50}= {\text{i}}_0 {\text{R}}_0$ $\;\Rightarrow {\text{i}}_{50}=\;{ {\text{i}}_0 {\text{R}}_0}/{ {\text{R}}_{50}}=\;{2 \times {\text{R}}_0}/{ {\text{R}}_0(1+50 \alpha)}=\;\frac{2}{1+\;\frac{1}{3}}=1.5 {\text{A}}$ $\;=15 \times 10^2 {\text{mA}}$

Chemistry30 questions

Q31JEE Main 2023MCQ4MSome Basic Concepts of Chemistry

Match list I with list II LIST I LIST II A. Nitrogen oxides in air I. Eutrophication B. Methane in air II. ${\text{pH}}$ of rain water becomes 5.6 C. Carbon dioxide III. Global warming D. phosphate fertilisers in water IV. Acid rainChoose the correct answer from the options given below: [12-Apr-2023 shift 1]

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Q32JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

A in the above reaction is: [12-Apr-2023 shift 1]

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Q33JEE Main 2023MCQ4Mp Block Elements

In the given reaction cycle ${\text{X}}, {\text{Y}}$ and ${\text{Z}}$ respectively are [12-Apr-2023 shift 1]

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📖 Explanation

The first reaction, ${\text{CaCl}}_2 + {\text{Na}}_2{\text{CO}}_3 \to {\text{X}} + {\text{Y}}$ , is a double displacement reaction. It produces a precipitate of calcium carbonate, ${\text{CaCO}}_3$ (X), and aqueous sodium chloride, ${\text{NaCl}}$ (Y). The reaction cycle shows that X reacts with Z. Calcium carbonate ({\text{CaCO}}_3 $) is a base that reacts with an acid. Given the options, Z must be hydrochloric acid ({\text{HCl}}$ ), which reacts with ${\text{CaCO}}_3$ to form ${\text{CaCl}}_2$ , water, and carbon dioxide, completing the cycle.

Q34JEE Main 2023MCQ4Mp Block Elements

Given below are two statements :Statement I : ${\text{SbCl}}_5$ is more covalent than ${\text{SbCl}}_3$ Statement II : The higher oxides of halogens also tend to be more stable than lower ones.In the light of the above statements, choose the most appropriate answer from the options given below [12-Apr-2023 shift 1]

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📖 Explanation

${\text{I}} \to {\text{SbCl}}^{+5}$ is more covalent due to sb in higher 0.5 more covalent due to more charge.II $\to$ Higher oxides of halogen tend to be more stable because higher oxidation states are less reactive and also the size of the atoms are more higher so they are less reactive.

Q35JEE Main 2023MCQ4MSome Basic Concepts of Chemistry

A metal chloride contains $55.0 \%$ of chlorine by weight . $100 {\text{mL}}$ vapours of the metal chloride at STP weigh $0.57 {\text{g}}$ . The molecular formula of the metal chloride is (Given: Atomic mass of chlorine is $35.5 {\text{u}}$ ) [12-Apr-2023 shift 1]

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📖 Explanation

Molecular. weight of metal chloride $\;=\;\frac{0.57}{100} \times 22700$ $\;=129.39$ $\;\;\text{ weight of }\; {\text{Cl}}=129.39 \times 0.55$ $\;=71.1645$ $\; \therefore \;\text{ Mole of }\; {\text{Cl}} {\text{Cl}}=\;\frac{71.1645}{35.5} ≅ 2$ $\;\;\text{ Hence }\; {\text{MCl}}_2$

Q36JEE Main 2023MCQ4MStates of Matter

Four gases A, B, C and D have critical temperature 5.3, 33.2, 126.0 and $154.3 {\text{K}}$ respectively. For their adsorption on a fixed amount of charcoal, the correct order is: [12-Apr-2023 shift 1]

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📖 Explanation

Extent of adsorption $\alpha$ critical temp.

Q37JEE Main 2023MCQ4MChemical Bonding and Molecular Structure

The bond order and magnetic property of acetylide ion are same as that of [12-Apr-2023 shift 1]

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📖 Explanation

${\text{CH}} \equiv {\text{C}}^{-}\;\;\;\; {\text{NO}}^{+} \to 14$ $\;\text{ B.O }\;=3\;\;\;\;\;1 {\text{B}} \cdot {\text{O}}=3$ Both are diamagnetic because both have absence of unpaired electron.

Q38JEE Main 2023MCQ4MElectrochemistry

For lead storage battery pick the correct statementsA. During charging of battery, ${\text{PbSO}}_4$ on anode is converted into ${\text{PbO}}_2$ B. During charging of battery, ${\text{PbSO}}_4$ on cathode is converted into ${\text{PbO}}_2$ C. Lead storage battery consists of grid of lead packed with ${\text{PbO}}_2$ as anodeD. Lead storage battery has $∼ 38 \%$ solution of sulphuric acid as an electrolyte Choose the correct answer from the options given below: [12-Apr-2023 shift 1]

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📖 Explanation

Lead storage battery consists of lead anode and a grid of lead packed with lead oxide $( {\text{PbO}}_2)$ as cathode, a $38 \%$ solution of ${\text{H}}_2 {\text{SO}}_4$ is used as an electrolyte.On charging the battery the reaction is reversed and ${\text{PbSO}}_4( {\ s})$ on anode and cathode is converted into ${\text{Pb}}$ and ${\text{PbO}}_2$ respectively.

Q39JEE Main 2023MCQ4MCoordination Compounds

Match List I with List II LIST I Complex LIST II CFSE $(\Delta(\Delta_\text{o})$ A. $[ {\text{Cu}} ( {\text{NH}}_3) _6] ^{2+}$ I. $-0.6$ B. $[ {\text{Ti}} ( {\text{H}}_2 {\text{O}}) _6] ^{3+}$ II. -2.0 C. $[ {\text{Fe}}( {\text{CN}})_6] ^{3-}$ III. -1.2 D. $[ {\text{NiF}}_6] ^{4-}$ IV. -0.4Choose the correct answer from the options given below: [12-Apr-2023 shift 1]

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📖 Explanation

The Crystal Field Stabilization Energy (CFSE) is calculated from the d-electron configuration using the formula ${\text{CFSE}} = (-0.4n_{t_{2g}} + 0.6n_{e_g})\Delta_o$ .
A. $[{\text{Cu}}({\text{NH}}_3)_6]^{2+}$ is a d⁹ system ( $t_{2g}^6 e_g^3$ ), so CFSE = $-0.6\Delta_o$ .
B. $[{\text{Ti}}({\text{H}}_2{\text{O}})_6]^{3+}$ is a d¹ system ( $t_{2g}^1 e_g^0$ ), so CFSE = $-0.4\Delta_o$ .
C. $[{\text{Fe}}({\text{CN}})_6]^{3-}$ is a d⁵ low-spin system ( $t_{2g}^5 e_g^0$ ), so CFSE = $-2.0\Delta_o$ .
D. $[{\text{NiF}}_6]^{4-}$ is a d⁸ system ( $t_{2g}^6 e_g^2$ ), so CFSE = $-1.2\Delta_o$ .
The correct matching is A-I, B-IV, C-II, D-III.

Q40JEE Main 2023MCQ4MClassification of Elements

The density of alkali metals metals is in the order [12-Apr-2023 shift 1]

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📖 Explanation

${\text{K}} < {\text{Na}} < {\text{Rb}} < {\text{Cs}}$ ${\text{D}}=\;\frac{\;\text{ Mass }\;}{\;\text{ volume }\;} \to$ generally do min antPotasium have extra volume due to presence at vaccant $3 {\text{d}}$ .

Q41JEE Main 2023MCQ4Mp Block Elements

Match List I with List II LIST I Type of Hydride LIST II Example A. Electron deficient hydride I. ${\text{MgH}}_2$ B. Electron rich hydride II. HF C. Electron precise hydride III. ${\text{B}}_2 {\text{H}}_6$ D. Saline hydride IV. ${\text{CH}}_4$ Choose the correct answer from the options given below: [12-Apr-2023 shift 1]

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📖 Explanation

$( {\text{B}}_2 {\text{H}}_6) -$ Electron dificent hydride(Hf) - electron rich ${\text{CH}}_4$ - electron precise hydride ${\text{MgH}}_2$ - saline hydride

Q42JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

Match List I with List II LIST I (Examples) LIST II (Tуре) A. 2-Chloro-1, 3-butadiene I. Biodegradable polymer B. Nylon 2-nylone 6 II. Synthetic Rubber C. Polyacrylonitrile III. Polyester D. Dacron IV. Addition Polymer Choose the correct answer from the options given below: [12-Apr-2023 shift 1]

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Q43JEE Main 2023MCQ4MChemical Thermodynamics

Given below are two statements: one is labelled as Assertion $\text{bo} {A}$ and the other is labelled as Reason RAssertion A : In the Ellingham diagram, a sharp change in slope of the line is observed for ${\text{Mg}} \to {\text{MgO}}$ at $∼ 1120^{\circ} {\text{C}}$ Reason R: There is a large change of entropy associated with the change of stateIn the light of the above statements, choose the correct answer from the options given below [12-Apr-2023 shift 1]

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📖 Explanation

(A) shape change slope of $1120^{\circ} {\text{C}}$ (R) During slate change there is large change in entropy.

Q44JEE Main 2023MCQ4MAmines

The incorrect statement regarding the reaction given below is [12-Apr-2023 shift 1]

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📖 Explanation

Here electrophile ${\text{NO}}^{+}$ attack on para position at low temperature

Q45JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

The major product ' ${\text{P}}$ ' formed in the following sequence of reactions is [12-Apr-2023 shift 1]

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📖 Explanation

The reaction sequence proceeds as follows:
(i) The carboxylic acid is converted to an acyl chloride using ${\text{SOCl}}_2$ .
(ii) The acyl chloride reacts with the primary amine ${\text{R-NH}}_2$ to form an N-substituted amide.
(iii) Lithium aluminium hydride ({\text{LiAlH}}_4 $) is a strong reducing agent that reduces the amide's carbonyl group (C=O) to a methylene group (CH₂), forming a secondary amine. The double bond is not affected. (iv)$ {\text{H}}_3{\text{O}}^+$ is for the work-up step.
The final product 'P' is a secondary amine.

Q46JEE Main 2023MCQ4Md and f Block Elements

Given below are two statement: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: $5 {\text{f}}$ electrons can participate in bonding to a far greater extent than $4 {\text{f}}$ electrons Reason R: $5 {\text{f}}$ orbitals are not as buried as $4 {\text{f}}$ orbitals In the light of the above statements, choose the correct answer from the options given below [12-Apr-2023 shift 1]

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Q47JEE Main 2023MCQ4MAlcohols, Phenols and Ethers

In the following reaction [12-Apr-2023 shift 1]

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📖 Explanation

First, we determine the molar mass (M) of the metal chloride vapor. At STP, 1 mole of any gas occupies 22.7 L (or 22700 mL). Given that 100 mL of the vapor weighs 0.57 g:
$\text{Moles} = \frac{0.100 \, \text{L}}{22.7 \, \text{L/mol}} \approx 0.0044 \, \text{mol}$
$M = \frac{\text{mass}}{\text{moles}} = \frac{0.57 \, \text{g}}{0.0044 \, \text{mol}} \approx 129.4 \, \text{g/mol}$
The compound contains 55.0% chlorine by weight. The mass of chlorine in one mole is:
$\text{Mass of Cl} = 129.4 \, \text{g/mol} \times 0.550 \approx 71.17 \, \text{g}$
The number of chlorine atoms is this mass divided by the atomic mass of chlorine (35.5 u):
$\text{Number of Cl atoms} = \frac{71.17 \, \text{g}}{35.5 \, \text{g/mol}} \approx 2$
Thus, the molecular formula is MCl₂.

Q48JEE Main 2023MCQ4Mp Block Elements

Given below are two statements:Statement I : Boron is extremely hard indicating its high lattice energyStatement II : Boron has highest melting and boiling point compared to its other group members.In the light of the above statements, choose the most appropriate answer from the options given below [12-Apr-2023 shift 1]

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Q49JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

Correct statements for the given reaction are :A. Compound ' B' is aromaticB. The completion of above reaction is very slowC. 'A' shows tautomerismD. The bond lengths of ${\text{C}}- {\text{C}}$ in compound ${\text{B}}$ are found to be same Choose the correct answer from the options given below: [12-Apr-2023 shift 1]

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📖 Explanation

(i) B is Aromatic(ii) Completion of reaction is very fast due to formation of aromatic compound(iii) A show keto-enol tautomerism(iv) ${\text{B}}$ is aromatic so ${\text{C}}- {\text{C}}$ bond length are same.

Q50JEE Main 2023MCQ4MHydrocarbons

2-hexene $\;{\to }^{(\text{i}) \text{O}_3}_{\;\text{ (ii) }\; {\text{H}}_2 {\text{O}}}$ Products The two products formed in above reaction are - [12-Apr-2023 shift 1]

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📖 Explanation

${\text{C}}- {\text{C}}- {\text{C}}- {\text{C}}= {\text{C}}- {\text{C}} {\to }_{(\text{ii}) \text{H}_2\text{O}}^{\;\text{ (i) }\; {\text{O}}_3} \to { {\text{C}}- {\text{C}}- {\text{C}}- {\text{CHO}}}_{(\text{Butanal})}+{ {\text{OHC}}- {\text{CH}}_3}_{\text{acetaldehyde}}$

Q51JEE Main 2023NAT4MOrganic Chemistry - Some Basic Principles

The value of $x$ in compound ' $D$ ' is _______. [12-Apr-2023 shift 1]

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Q52JEE Main 2023NAT4MChemical Kinetics

The reaction $2 {\text{NO}}+ {\text{Br}}_2 \to 2 {\text{NOBr}}$ takes places through the mechanism given below: $\; {\text{NO}}+ {\text{Br}}_2 ⇔ {\text{NOBr}}_2 \;\text{ (fast) }\;$ $\; {\text{NOBr}}_2+ {\text{NO}} \to 2 {\text{NOBr}} \;\text{ (slow) }\;$ The overall order of the reaction is _______. [12-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{RDS}}: {\text{NOBr}}_2+ {\text{NO}} \to 2 {\text{NOBr}}$ $\; {\text{r}}= {\text{K}} [ {\text{NOBr}}_2] [ {\text{NO}}]$ . . . . . (i) $\; {\text{Keq}}=\;{ [ {\text{NOBr}}_2] }/{ [ {\text{NO}}_2 [ {\text{B}}_2] .}$ . . . . . (ii) $\;\;\text{ From (i) & (ii) }\;$ $\; {\text{R}}= {\text{K}} . {\text{Keq}} \cdot [ {\text{NO}}] [ {\text{Br}}_2] [ {\text{NO}}]$ $\; {\text{R}}= {\text{K}}^{'}[ {\text{NO}}]^2 [ {\text{Br}}_2]$ $\;\;\text{ Overall order }\;=3$ $\;$

Q53JEE Main 2023NAT4MStates of Matter

At $600 {\text{K}}$ , the root mean square (rms) speed of gas $X$ (molar mass $=40$ ) is equal to the most probable speed of gas ${\text{Y}}$ at $90 {\text{K}}$ . The molar mass of the gas ${\text{Y}}$ is ______ ${\text{gmol}}^{-1} \cdot ($ Nearest integer $)$ [12-Apr-2023 shift 1]

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📖 Explanation

${\text{U}}_{\;\text{rms }\; {\text{x}}, 600}= {\text{U}}_{ {\mp} {\text{Y}}, 90}$ $\;\sqrt{\;{3 \times {\text{R}} \times 600}/{400}}=\sqrt{\;{2 \times {\text{R}} \times 20}/{ {\text{M}}}}$ $\; {\text{M}}=4$ $\;$

Q54JEE Main 2023NAT4MIonic Equilibrium

An analyst wants to convert $1 {\text{L}} {\text{HCl}}$ of ${\text{pH}}=1$ to a solution of ${\text{HCl}}$ of ${\text{pH}} 2$ . The volume of water needed to do this dilution is _______ mL. (Nearest integer) [12-Apr-2023 shift 1]

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📖 Explanation

$( {\text{M}}_1 \times {\text{V}}_1) \; ( {\text{M}}_2 \times {\text{V}}_2)$ $-1 \;\; =\;-2$ $10 \times 1\;10 \times {\text{V}}_2$ ${\text{V}}_2=10 {\text{L}}$ $\;\text{ Water added }\;=10-1$ $=9 {\text{Litre}}$ $=9000 {\text{mL}}$

Q55JEE Main 2023NAT4MBiomolecules

In an oligopeptide named Alanylglycylphenyl alanyl isoleucine, the number of ${\ sp}^2$ hybridised carbons is [12-Apr-2023 shift 1]

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Q56JEE Main 2023NAT4MSolutions

80 mole percent of ${\text{MgCl}}_{-} 2$ is dissociated in aqueous solution. The vapour pressure of 1.0 molal aqueous solution of ${\text{MgCl}}_2$ at $38^{\circ} {\text{C}}$ is _______ ${\text{mmHg}}$ . (Nearest integer)Given : Vapour pressure of water at $38^{\circ} {\text{C}}$ is $50 {\text{mm}} {\text{Hg}}$ . [12-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{MgCl}}_2 \to {\text{Mg}}^{+2}+2 {\text{Cl}}^{-}$ $\;1-\alpha \;\; \alpha \;\; 2 \alpha$ $\; {\text{i}}=1+2 \alpha(\alpha=0.8)$ $\; {\text{i}}=2.6$ $\;\;{\Delta {\text{p}}}/{ {\text{p}}^{\circ}}=\;{ {\text{i}} \times {\text{n}}_2}/{ {\text{n}}_1}$ $\;\Delta {\text{p}}=2.34$ $\; {\text{p}}_{ {\ s}}=47.66$ $\; {\text{p}}_{ {\ s}} ≅ 48$

Q57JEE Main 2023NAT4MPractical Organic Chemistry

Three organic compounds A, B and C were allowed to run in thin layer chromatography using hexane and gave the following result (see figure). The $R_f$ value of the most polar compound is ______ $\times 10^{-2}$ . [12-Apr-2023 shift 1]

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📖 Explanation

Most polar compound - ${\text{C}}$ - because of lowest moulmerls in upper direction ${\text{R}}_{ {\text{f}}}=\;\frac{\;\text{ Dis tan ce covered by compound }\;}{\;\text{ Dis tance covered by solvent }\;}=\;\frac{2}{8}=0.25=25 \times 10^{-2}$

Q58JEE Main 2023NAT4MChemical Thermodynamics

One mole of an ideal gas at $350 {\text{K}}$ is in a $2.0 {\text{L}}$ vessel of thermally conducting walls, which are in contact with the surroundings. It undergoes isothermal reversible expansion from $2.0 {\text{L}}$ to $3.0 {\text{L}}$ against a constant pressure of $4 {\text{atm}}$ . The change in entropy of the surroundings $(\Delta S)$ is _______ ${\text{JK}}^{-1}$ (Nearest integer) Given: ${\text{R}}=8.314 {\text{J}} {\text{K}}^{-1} {\text{mol}}^{-1}$ [12-Apr-2023 shift 1]

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📖 Explanation

$\;\Delta {\text{S}}_{\;\text{System }\;}=n R ℓ (\;{ {\text{V}}_2}/{ {\text{V}}_1}) =1 \times 8.314 \ln (\;\frac{3}{2})$ $\;\Delta {\text{S}}_{\;\text{System }\;}=3.37$ $\;\Delta {\text{S}}_{\;\text{Sur. }\;}=3.37$

Q59JEE Main 2023NAT4MStructure of Atom

Values of work function ({\text{W}}_0) $for a few metals are gives below Metal Li Na K Mg Cu Ag$ {\text{W}}_0 / {\text{eV}} $2.42 2.3 2.25 3.7 4.8 4.3The number of metals which will show photoelectric effect when light of wavelength$ 400 {\text{nm}} $falls on it isGiven:$ {\text{h}}=6.6 × 10^{-34} {\text{J}} {\ s}; {\text{c}}=3 × 10^8 {\text{ms}}^{-1}; {\text{e}}=1.6 × 10^{-19} {\text{C}}$ [12-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{E}}( {\text{ev}})=\;\frac{1240}{400}=3.1 {\text{ev}}$ $\; {\text{Mg}}, {\text{Cu}}, {\text{Ag}}$

Q60JEE Main 2023NAT4MAldehydes, Ketones and Carboxylic Acids

The mass of ${\text{NH}}_3$ produced when $131.8 {\text{kg}}$ of cyclohexanecarbaldehyde undergoes Tollen's test is ${\text{kg}}$ . (Nearest Integer)Molar Mass of ${\text{C}}=12 {\text{g}} / {\text{mol}}$ $\; {\text{N}}=14 {\text{g}} / {\text{mol}}$ $\; {\text{O}}=16 {\text{g}} / {\text{mol}}$ [12-Apr-2023 shift 1]

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📖 Explanation

The reaction is the Tollen's test for an aldehyde, where cyclohexanecarbaldehyde is oxidized. The balanced chemical equation used in the provided solution is:
$\text{C}_6\text{H}_{11}\text{CHO} + 2[\text{Ag}(\text{NH}_3)_2]\text{OH} \rightarrow \text{C}_6\text{H}_{11}\text{COONH}_4 + 2\text{Ag} + 3\text{NH}_3 + \text{H}_2\text{O}$
This shows that 1 mole of aldehyde produces 3 moles of ammonia (NH₃).
First, calculate the moles of cyclohexanecarbaldehyde (Molar Mass = 112 g/mol):
$\text{Moles of aldehyde} = \frac{131.8 \, \text{kg} \times 1000 \, \text{g/kg}}{112 \, \text{g/mol}} \approx 1176.8 \, \text{mol}$
Next, calculate the moles of NH₃ produced using the 1:3 stoichiometry:
$\text{Moles of NH}_3 = 1176.8 \, \text{mol} \times 3 \approx 3530.4 \, \text{mol}$
Finally, convert moles of NH₃ to mass (Molar Mass = 17 g/mol):
$\text{Mass of NH}_3 = 3530.4 \, \text{mol} \times 17 \, \text{g/mol} \approx 60017 \, \text{g} \approx 60 \, \text{kg}$
Note: There is a discrepancy between this calculated value and the provided answer key. The explanation follows the method shown in the solution PDF.

Mathematics30 questions

Q61JEE Main 2023MCQ4MDifferential Equations

Let $y=y(x), y > 0$ , be a solution curve of the differential equation $(1+x^2) d y=y(x-y) \text{dx}$ . If $y(0)=1$ and ${\text{y}}(2 {\sqrt{2}})=\beta$ , then [12-Apr-2023 shift 1]

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📖 Explanation

$\; (1+x^2) d y=y(x-y) \text{dx}$ $\;Y(0)=1 \cdot y(2 {\sqrt{2}})=\beta$ $\;\;\frac{d y}{d x}=\;\frac{y x-y^2}{1+x^2}$ $\;\;\frac{d y}{d x}+y (\;\frac{-x}{1+x^2}) = (\;\frac{-1}{1+x^2}) y^2$ $\;\;\frac{1}{y^2} \;\frac{d y}{d x}+\;\frac{1}{y} (\;\frac{-x}{1+x^2}) =\;\frac{-1}{1+x^2}$ $\;\;\text{ Put }\; \;\frac{1}{y}=t \;\text{ then }\; \;\frac{-1}{y^2} \;\frac{d y}{d x}=\;\frac{d t}{d x}$ $\;\;\frac{d t}{d x}+t \;\frac{x}{1+x^2}=\;\frac{1}{1+x^2}$ $\;\;\text{ If }\;=e^{\int \;\frac{x}{1+x^2} \text{dx}}=e^{\;\frac{1}{2} \ln (1+x^2) } {\sqrt{1+x^2}}$ $\;t {\sqrt{1+x^2}}=\int \;{1}/{{\sqrt{1+x^2}}} \text{dx}$ $\;\;{{\sqrt{1+x^2}}}/{y}=\ln (x+{\sqrt{x^{2+}1}}) +c$ $\;y(0)=1 \Rightarrow c=1$ $\;\Rightarrow {\sqrt{1+x^2}}=y \ln (e (x+{\sqrt{x^{2+}1}}) )$ $\;\beta=\;{3}/{\ln (e(3+2 {\sqrt{2}}))} \Rightarrow \;\frac{3}{\beta}=\ln (e(3+2 {\sqrt{2}}))$ $\;e^{\;\frac{3}{\beta}}=e(3+2 {\sqrt{2}})$

Q62JEE Main 2023MCQ4MFunctions

Let $D$ be the domain of the function $f(x)= sin \;^{-1} (\log _{3 x} (\;\frac{6+2 \log _3 x}{-5 x}) )$ . If the range of the function $g: D \to R$ defined by $g(x)=x-[x],([x]$ is the greatest integer function $)$ , is $(\alpha, \beta)$ , then $\alpha^{2+}\;\frac{5}{\beta}$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

$\;\;\frac{6+2 \log _3 x}{-5 x} > 0 & x > 0 & x \ne \;\frac{1}{3}$ $\;\;\text{ this gives }\; x \in (0, \;\frac{1}{27}) . . .(1)$ $\;-1 \le \log _{3 x} (\;\frac{6+2 \log _3 x}{-5 x}) \le 1$ $\;3 x \le \;\frac{6+2 \log _3 x}{-5 x} \le \;\frac{1}{3 x}$ $\;15 x^{2+}6+2 \log _3 x \ge 0 \;\; 6+2 \log _3 x+\;\frac{5}{3} \ge 0$ $\;x \in (0, \;\frac{1}{27}) . . .(2) \;\; x \ge 3^{-\;\frac{23}{6}} . . . .(3)$ $\;\;\text{ form (1), (2) & (3) }\;$ $\;x \in [3^{-\;\frac{23}{6}}, \;\frac{1}{27})$ $\therefore \alpha$ is small positive quantity $\& \beta=\;\frac{1}{27}$ $\therefore \alpha^{2+}\;\frac{5}{\beta}$ is just greater than 135

Q63JEE Main 2023MCQ4MComplex Numbers

Let ${\text{C}}$ be the circle in the complex plane with centre ${\text{z}}_0=\;\frac{1}{2}(1+3 i)$ and radius $r=1$ . Let $z_1=1+i$ and the complex number $z_2$ be outside the circle $C$ such that ${|z_1-z_0 |}{z_2-z_0 |}=1$ . If $z_0 \cdot z_1$ and $z_2$ are collinear, then the smaller value of ${|z_2 |}^2$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

$\;{| z_1-z_0 |}={| \;\frac{1-i}{2} |}=\;\frac{1}{2}$ $\;\Rightarrow {| z_2-z_0 |}={\sqrt{2}}: \;\text{ centre }\; (\;\frac{1}{2}, \;\frac{3}{2})$ $z_0 (\;\frac{1}{2}, \;\frac{3}{2})$ and $z_1(1,1)$ $\;\tan \theta=-1 \Rightarrow \theta=135^{\circ}$ $\;z_2 (\;\frac{1}{2}+{\sqrt{2}} \cos 135^{\circ}, \;\frac{3}{2}+{\sqrt{2}} sin \;135^{\circ})$ or $\; (\;\frac{1}{2}-{\sqrt{2}} \cos 135^{\circ}, \;\frac{3}{2}-{\sqrt{2}} sin \;135^{\circ})$ $\;\Rightarrow {\text{z}}_2 (-\;\frac{1}{2}, \;\frac{5}{2}) \;\text{ or }\; {\text{z}}_2 (\;\frac{3}{2}, \;\frac{1}{2})$ $\;\Rightarrow {| {\text{z}}_3 |}^2=\;\frac{26}{4}, \;\frac{5}{2}$ $\;\Rightarrow {| {\text{z}}_2 |}_{\min }^2=\;\frac{5}{2}$

Q64JEE Main 2023MCQ4MSequences and Series

Let $< a_n >$ be a sequence such that $a_1+a_2+. .+a_n=\;\frac{n^{2+}3 n}{(n+1)(n+2)}$ . If $28 sum_{{ {\text{k}}=1}}^{{10}} \;\frac{1}{a_k}=p_1 p_2 p_3 . . . p_m$ , where $p_1, p_2 . . . . . .$ ${\text{p}}_{ {\text{m}}}$ are the first ${\text{m}}$ prime numbers, then ${\text{m}}$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{a}}_{ {\text{n}}}= {\text{S}}_{ {\text{n}}}- {\text{S}}_{ {\text{n}}-1}=\;{ {\text{n}}^{2+}3 {\text{n}}}/{( {\text{n}}+1)(1+2)}-\;{( {\text{n}}-1)( {\text{n}}+2)}/{ {\text{n}}( {\text{n}}+1)}$ $\;\Rightarrow {\text{a}}_{ {\text{n}}}=\;{4}/{ {\text{n}}( {\text{n}}+1)(1+2)}$ $\;\Rightarrow 28 sum_{{ {\text{k}}-1}}^{{10}} \;{1}/{ {\text{a}}_{ {\text{k}}}}=28 sum_{{ {\text{k}}-1}}^{{10}} \;{ {\text{k}}( {\text{k}}+1)( {\text{k}}+2)}/{4}$ $\;=\;\frac{7}{4} sum_{{ {\text{k}}-1}}^{{10}}( {\text{k}}( {\text{k}}+1)( {\text{k}}+2)( {\text{k}}+3)-( {\text{k}}-1) {\text{k}}( {\text{k}}+1)( {\text{k}}+2))$ $\;=\;\frac{7}{4} \cdot 10 \cdot 11 \cdot 12 \cdot 13=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ So ${\text{m}}=6$

Q65JEE Main 2023MCQ4MThree Dimensional Geometry

Let the plane P: $4 x-y+z=10$ be rotated by an angle $\;\frac{\pi}{2}$ about its line of intersection with the plane $x+y-z$ $=4$ . If $\alpha$ is the distance of the point $(2,3,-4)$ from the new position of the plane $P$ , then $35 \alpha$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

Let equation in new position is $\;(4 x-y+z-10)+\lambda (x+y-z-4)=0$ $\;4(4+\lambda )-1 \cdot (-1+\lambda )+1 \cdot (1-\lambda )=0$ $\;\Rightarrow \lambda =-9$ So equation in new position is $\;-5 x-10 y+10 z+26=0$ $\;\Rightarrow \alpha=\;\frac{54}{15}$

Q66JEE Main 2023MCQ4MSets and Relations

Among the two statements(S1): $(p \Rightarrow q) ∧(p ∧(∼ q))$ is a contradiction and $(S 2):(p ∧ q) ∨((∼ p) ∧ q) ∨(p ∧(∼ q)) ∨((∼ p) ∧(∼ q))$ is a tautology [12-Apr-2023 shift 1]

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📖 Explanation

${\text{S}}_1:( {\text{p}} \to {\text{q}}) ∧( {\text{p}} ∧(∼ {\text{q}}))$ P Q p → q $\text{p} ∧(∼ q)$ S1 T T T F F T F F T F F T T F F F F T F F $\Rightarrow {\text{S}}_1$ is Contradiction ${\text{S}}_2$ p q ${\text{p}} ∧ {\text{q}}$ $(p ∧ ∼ q)$ $(p ∧ ∼ q)$ $(∼ {\text{p}}) ∧(∼ {\text{p}})$ ${\text{S}}_2$ T T T F F F T T F F F T F T F T F T F F T F F F F F T T ${\text{S}}_2$ is tautology

Q67JEE Main 2023MCQ4MStatistics

Two dice $A$ and $B$ are rolled. Let numbers obtained on $A$ and $B$ be $\alpha$ and $\beta$ respectively. If the variance of $\alpha-\beta$ is $\;\frac{p}{q}$ , where $p$ and $q$ are co-prime, then the sum of the positive divisior of $p$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

$\alpha-\beta$ Case P 5 (6,1) 1/36 4 (6,2) (5,1) 2/36 3 (6,3) (5,2) (4,1) 3/36 2 (6,4) (5,3) (4,3) (3,1) 4/36 1 (6,5) (5,4) (4,3) (3,2) (2,1) 5/36 0 (6,6) (5,5) . . . . . . (1,1) 6/36 -1 - - - - - 5/36 -2 - - - - - 4/36 -3 - - - - - 3/36 -4 (2,6) (1,5) 2/36 -5 (1,6) 1/36 $\;\sum (x^2) =\sum x^2 P(x)=2 [\;\frac{25}{36}+\;\frac{32}{36}+\;\frac{27}{36}+\;\frac{16}{36}+\;\frac{5}{36}]$ $\;=\;\frac{105}{18}=\;\frac{35}{6}$ $\mu =\sum( {\text{x}})=0$ as data is symmetric $\sigma^2=\sum ( {\text{x}}^2) =\sum {\text{x}}^2 {\text{P}}( {\text{x}})=\;\frac{35}{6} {\text{P}}=35=5 \times 7$ Sum of divisors $= (5^{\circ}+5^1) (7^{0+}7^1) =6 \times 8=48$

Q68JEE Main 2023MCQ4MPermutations and Combinations

The number of five digit numbers, greater then 40000 and divisible by 5 , which can be formed using the digits $0,1,3,5,7$ and 9 without repetition, is equal to [12-Apr-2023 shift 1]

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📖 Explanation

So Required numbers $=5 \times { }^4 {\text{P}}_3=120$

Q69JEE Main 2023MCQ4MQuadratic Equation and Inequalities

Let $\alpha, \beta$ be the roots of the quadratic equation $x^{2+}{\sqrt{6}} x+3=0$ . Then $\;\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

$\alpha, \beta=\;{-{\sqrt{6}} \pm {\sqrt{6-12}}}/{2}=\;{-{\sqrt{6}} \pm {\sqrt{6}} i}/{2}$ $=\sqrt{3 e^{ \pm \;\frac{3 \pi i}{4}}}$ Required expression $\;\;{({\sqrt{3}})^{23} (2 \cos \frac{69 \pi}{4}) +({\sqrt{3}})^{14} (2 \cos \frac{42 \pi}{4}) }/{({\sqrt{3}})^{15} (2 \cos \;\frac{45 \pi}{4}) +({\sqrt{3}})^{10} (2 \cos \;\frac{30 \pi}{4}) }$ $\;({\sqrt{3}})^8=81$

Q70JEE Main 2023MCQ4MApplication of Derivatives

If the local maximum value of the function $f(x)= (\;{{\sqrt{3 e}}}/{2 sin \;x}) ^{ sin \;^2 x}, x \in (0, \;\frac{\pi}{2})$ , is $\;\frac{k}{e}$ , then $(\;\frac{k}{e}) ^{8+}\;\frac{k^8}{e^5}+k^8$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Let }\; y= (\;{{\sqrt{3}} {\text{e}}}/{2 sin \;x}) ^{ sin \;^2 {\text{x}}}$ $\;\ln {\text{y}}= sin \;^2 {\text{x}} \cdot \ln (\;{{\sqrt{3}} {\text{e}}}/{2 sin \; {\text{x}}})$ $\;\;{1}/{ {\text{y}}} {\text{y}}^{'}=\ln (\;{{\sqrt{3}} {\text{e}}}/{2 sin \; {\text{x}}}) 2 sin \;x \cos {\text{x}}+ sin \;^2 {\text{x}} \;{2 sin \; {\text{x}}}/{{\sqrt{3}} {\text{e}}} \;{{\sqrt{3}} {\text{e}}}/{2}(-{\text{cosecx}} \cot {\text{x}})$ $\;\Rightarrow sin \;x \cos [2 \ln (\;{{\sqrt{3}} {\text{e}}}/{2 sin \; {\text{x}}}) -1] =0$ $\;\Rightarrow \ln (\;{3 {\text{e}}}/{4 sin \;^2 {\text{x}}}) =1 \Rightarrow \;{3 {\text{e}}}/{4 sin \;^2 {\text{x}}}= {\text{e}} \Rightarrow sin \;^2 {\text{x}}=\;\frac{3}{4}$ $\;\Rightarrow sin \;x=\;{{\sqrt{3}}}/{2} (\;\text{ as }\; x \in (0, \;\frac{\pi}{2}) )$ $\;\Rightarrow {\text{local}} \max \;\text{ value }\;= (\;{{\sqrt{3 e}}}/{{\sqrt{3}}}) ^{3 / 4}= {\text{e}}^{3 / 8}=\;{ {\text{k}}}/{ {\text{e}}}$ $\;\Rightarrow {\text{k}}^8= {\text{e}}^{11}$ $\;\Rightarrow (\;{ {\text{k}}}/{ {\text{e}}}) ^{8+}\;{ {\text{k}}^8}/{ {\text{e}}^5}+ {\text{k}}^8= {\text{e}}^{3+} {\text{e}}^{6+} {\text{e}}^{11}$

Q71JEE Main 2023MCQ4MVector Algebra

Let $\lambda \in Z, \vec{a}=\lambda \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$ . Let $\vec{c}$ be a vector such that $(\vec{a}+\vec{b}+\vec{c}) \times \vec{c}=\vec{0}, \vec{a} \cdot \vec{c}=-17$ and $\vec{b} \cdot \vec{c}$ $=-20$ . Then $|{ {\text{c}}}^{\to } \times (\lambda { {\text{i}}^{\^}}+{ {\text{j}}^{\^}}+{ {\text{k}}^{\^}})|^2$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

$\;(\vec{a}+\vec{b}+\vec{c})_{\times } \vec{c}=0$ $\;(\vec{a}+\vec{b}) \times \vec{c}=0$ $\;{ {\text{c}}}^{\to }=\alpha({ {\text{a}}}^{\to }+{ {\text{b}}}^{\to })=\alpha(\lambda +3) { {\text{i}}^{\^}}+\alpha { {\text{k}}^{\^}}$ $\;{ {\text{b}}}^{\to } \cdot { {\text{c}}}^{\to }=-20 \Rightarrow 3 \alpha(\lambda +3)+2 \alpha=-20$ $\;\vec{a} \cdot \vec{c}=-17 \Rightarrow \alpha \lambda (\lambda +3)-\alpha=-17$ $\;\Rightarrow \alpha(3 \lambda +9+2)=-20$ $\;\alpha (\lambda ^{2+}3 \lambda -1) =-17$ $\;17(3 \lambda +11)=20 (\lambda ^{2+}3 \lambda -1)$ $\;20 \lambda ^{2+}9 \lambda -207=0$ $\;\lambda =3 \;\; (\lambda \in Z)$ $\;\Rightarrow \alpha=-1 \Rightarrow { {\text{c}}}^{\to }=-(6 { {\text{i}}^{\^}}+{ {\text{k}}^{\^}})$ $\;{ {\text{v}}}^{\to }={ {\text{c}}}^{\to } \times (3 { {\text{i}}^{\^}}+{ {\text{j}}^{\^}}+{ {\text{k}}^{\^}})$ $\;={\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ -6 & 0 & -1 \\ 3 & 1 & 1\end{vmatrix}}=\hat{i}+3 \hat{j}-6 \hat{k}$ $\;|{ {\text{v}}}^{\to }|^2=(-1)^{2+}3^{2+}6^2=46$ $\;$

Q72JEE Main 2023MCQ4MTrigonometric Ratios and Identities

In a triangle $A B C$ , if $\cos A+2 \cos B+\cos C=2$ and the lengths of the sides opposite to the angles $A$ and $C$ are 3 and 7 respectively, then $\cos A-\cos C$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

$\;\cos A+\cos C=2(1-\cos B)$ $\;2 \cos \;\frac{A+C}{2} \cos \;\frac{A-C}{2}=4 sin \;^2 B / 2$ $\;\;\text{ as }\; \cos (\;\frac{A+C}{2}) = sin \;\;\frac{B}{2}$ $\;\;\text{ so }\; \cos \;\frac{A-C}{2}=2 sin \;\;\frac{B}{2}$ $\;2 \cos B / 2 \cos \;\frac{A-C}{2}=4 sin \;B / 2 \cos B / 2$ $\;2 sin \; (\;\frac{A+C}{2}) \cos (\;\frac{A-C}{2}) =4 sin \;B / 2 \cos B / 2$ $\; sin \;A+ sin \;C=2 sin \;B$ $\;a+c=2 b \Rightarrow a=3, c=7, b=5$ $\;\cos A-\cos C=\;\frac{b^{2+}c^{2-}a^2}{2 b c}-\;\frac{a^{2+}b^{2-}c^2}{2 a b}$ $\;=\;\frac{25+49-9}{70}-\;\frac{9+25-49}{30}$ $\;\;\frac{65}{70}+\;\frac{1}{2}=\;\frac{20}{14}=\;\frac{10}{7}$

Q73JEE Main 2023MCQ4MStraight Lines and Pair of Straight Lines

If the point $(\alpha, \;{7 {\sqrt{3}}}/{3})$ lies on the curve traced by the mid-points of the line segments of the lines ${\text{x}} \cos \theta+ {\text{y}}$ $sin \;\theta=7, \theta \in (0, \;\frac{\pi}{2})$ between the co-ordinates axes, then $\alpha$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

${\text{pt}} (\alpha, \;{7 {\sqrt{3}}}/{3})$ $\; {\text{x}} \cos \theta+ {\text{y}} sin \;\theta=7$ $\; {\text{x}} \;\text{ - intercept }\;=\;\frac{7}{\cos \theta}$ $\; {\text{y}}-\;\text{ intercept }\;=\;\frac{7}{ sin \;\theta}$ $\; {\text{A}}: (\;\frac{7}{\cos \theta}, 0) {\text{B}}: (0, \;\frac{7}{ sin \;\theta})$ Locus of mid pt M : (h, k) $\; {\text{h}}=\;\frac{7}{2 \cos \theta}, {\text{k}}=\;\frac{7}{2 sin \;\theta}$ $\;\;\frac{7}{2 sin \;\theta}=\;{7 {\sqrt{3}}}/{3} \Rightarrow sin \;\theta=\;{{\sqrt{3}}}/{2} \Rightarrow \theta=\;\frac{\pi}{3}$ $\;\alpha=\;\frac{7}{2 \cos \theta}=7$

Q74JEE Main 2023MCQ4MEllipse

Let $P (\;{2 {\sqrt{3}}}/{{\sqrt{7}}}, \;{6}/{{\sqrt{7}}}) , {\text{Q}}, {\text{R}}$ , and ${\text{S}}$ be four points on the ellipse $9 {\text{x}}^{2+}4 {\text{y}}^2=36$ . Let $P Q$ and $R S$ be mutually perpendicular and pass through the origin. If $\;{1}/{( {\text{PQ}})^2}+\;{1}/{( {\text{RS}})^2}=\;{ {\text{p}}}/{ {\text{q}}}$ , where ${\text{p}}$ and ${\text{q}}$ are coprime, then ${\text{p}}+ {\text{q}}$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

Let ${\text{R}}(2 \cos \theta, 3 sin \;\theta)$ As ${\text{OP}} ⟂ {\text{OR}}$ $\;\;\text{ so }\; \;\frac{3 sin \;\theta}{2 \cos \theta} \times \;{\;{6}/{{\sqrt{7}}}}/{{2 {\sqrt{3}}}/{{\sqrt{7}}}}=-1$ $\;\Rightarrow \tan \theta=\;{-2}/{3 {\sqrt{3}}}$ $\;\Rightarrow {\text{R}} (\;{-6 {\sqrt{3}}}/{{\sqrt{31}}}, \;{6}/{{\sqrt{31}}}) \;\text{ or }\; {\text{R}} (\;{6 {\sqrt{3}}}/{{\sqrt{31}}}, \;{-6}/{{\sqrt{31}}})$ $\;\;\text{ Now }\;=\;{1}/{( {\text{PQ}})^2}+\;{1}/{( {\text{RS}})^2}=\;\frac{1}{4} (\;{1}/{( {\text{OP}})^2}+\;{1}/{( {\text{OR}})^2})$ $\;=\;\frac{1}{4} (\;\frac{1}{\frac{48}{7}}+\;\frac{1}{144}) =\;\frac{1}{4} (\;\frac{7}{48}+\;\frac{31}{144})$ $\;=\;\frac{13}{144}$ $\;\Rightarrow {\text{p}}+ {\text{q}}=157$

Q75JEE Main 2023MCQ4MThree Dimensional Geometry

Let the lines $1_1: \;\frac{x+5}{3}=\;\frac{y+4}{1}=\;\frac{z-\alpha}{-2}$ and $l_2: 3 x+2 y+z-2=0=x-3 y+2 z-13$ be coplanar. If the point $P(a, b, c)$ on $l_1$ is nearest to the point $Q(-4,-3,2)$ , then $|a|+|b|+|c|$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

$\;(3 x+2 y+z-2)+\mu (x-3 y+2 z-13)=0$ $\;3(3+\mu )+1 \cdot (2-3 \mu )-2(1+2 \mu )=0$ $\;9-4 \mu =0$ $\;\mu =\;\frac{9}{4}$ $\;4(-15-8+\alpha-2)+9(-5+12+2 \alpha-13)=0$ $\;-100+4 \alpha-54+18 \alpha=0$ $\;\Rightarrow \alpha=7$ $\;\;\text{ Let }\; P \equiv (3 \lambda -5, \lambda -4,-2 \lambda +7)$ Direction ratio of PQ $(3 \lambda -1, \lambda -1,-2 \lambda +5)$ But ${\text{PQ}} ⟂ ℓ_1$ $\;\Rightarrow 3(3 \lambda -1)+1 \cdot (\lambda -1)-2(-2 \lambda +5)=0$ $\;\Rightarrow \lambda =1$ $\;P(-2,-3,5) \Rightarrow |a|+|b|+|c|=10$

Q76JEE Main 2023MCQ4MBinomial Theorem

If $\;{1}/{ {\text{n}}+1}{ }^{ {\text{n}}} {\text{C}}_{ {\text{n}}}+\;{1}/{ {\text{n}}}{ }^{ {\text{n}}} {\text{C}}_{ {\text{n}}-1}+. . .+\;\frac{1}{2}{ }^{ {\text{n}}} {\text{C}}_1+{ }^{ {\text{n}}} {\text{C}}_0=\;\frac{1023}{10}$ then ${\text{n}}$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

$sum_{{ {\text{T}}=0}}^{{ {\text{n}}}}$ $\;\frac{{ }^n C_r}{r+1}=$ $\;\frac{1}{n+1} sum_{{r=0}^n{ }}^{{n+1}} C_{r+1}$ $\;=\;{1}/{ {\text{n}}+1} (2^{ {\text{n}}+1}-1) =\;\frac{1023}{10}$ $\; {\text{n}}+1=10 \Rightarrow {\text{n}}=9$

Q77JEE Main 2023MCQ4MBinomial Theorem

The sum, of the coefficients of the first 50 terms in the binomial expansion of $(1-x)^{100}$ , is equal to [12-Apr-2023 shift 1]

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📖 Explanation

$\;(1-x)^{100}=C_0-C_1 x+C_2 x^{2-}$ $\;C_3 x^{3+}. . . . . . C_{99} x^{99}+C_{100} x^{100}$ $\;\Rightarrow C_0-C_1+C_2-C_3+. . .-C_{99}+C_{100}=0$ $\;C_0-C_1+C_2+. . . C_{99}=-\;\frac{1}{2}{ }^{100} C_{50}$ $\;-\;\frac{1}{2} \;\frac{100 !}{50 ! 50 !}=-\;\frac{1}{2} \times \;\frac{100 \times 99 !}{50 ! 50 !}=-{ }^{99} C_{49}$

Q78JEE Main 2023MCQ4MVector Algebra

Let $a, b, c$ be three distinct real numbers, none equal to one. If the vectors $a \hat{i}+\hat{j}+\hat{k}, \hat{i}+b \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+c \hat{k}$ are coplanar, then $\;\frac{1}{1-a}+\;\frac{1}{1-b}+\;\frac{1}{1-c}$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

$\;{\begin{vmatrix}{\text{a}} & 1 & 1 \\ 1 & {\text{b}} & 1 \\ 1 & 1 & {\text{c}}\end{vmatrix}}=0$ $\; {\text{C}}_2 \to {\text{C}}_2- {\text{C}}_1, {\text{C}}_3 \to {\text{C}}_3- {\text{C}}_1$ ${\begin{vmatrix}a & 1-a & 1-a \\ 1 & b-1 & 0 \\ 1 & 0 & c-1\end{vmatrix}}=0$ $\;a(b-1)(c-1)-(1-a)(c-1)+(1-a)(1-b)=0$ $\;a(1-b)(1-c)+(1-a)(1-c)+(1-a)(1-b)=0$ $\;\;\frac{a}{1-a}+\;\frac{a}{1-b}+\;\frac{a}{1-c}=0$ $\;\Rightarrow -1+\;\frac{1}{1-a}+\;\frac{1}{1-b}+\;\frac{1}{1-c}=0$ $\;\Rightarrow \;\frac{1}{1-a}+\;\frac{1}{1-b}+\;\frac{1}{1-c}=1$

Q79JEE Main 2023MCQ4MMatrices and Determinants

Let $A={\begin{bmatrix}1 & \\ \\ \frac{1}{51} \\ 0 & 1\end{bmatrix}}$ . If $B={\begin{bmatrix}1 & 2 \\ -1 & -1\end{bmatrix}} A{\begin{bmatrix}-1 & -2 \\ 1 & 1\end{bmatrix}}$ , then the sum of all the elements of the matrix $sum_{{n=1}}^{{50}} B^n$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Let }\; {\text{C}}={\begin{bmatrix}1 & 2 \\ -1 & -1\end{bmatrix}}, {\text{D}}={\begin{bmatrix}-1 & -2 \\ 1 & 1\end{bmatrix}}$ $\; {\text{DC}}={\begin{bmatrix}1 & 2 \\ -1 & -1\end{bmatrix}}{\begin{bmatrix}-1 & -2 \\ 1 & 1\end{bmatrix}}={\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}= {\text{I}}$ $\; {\text{B}}= {\text{CAD}}$ $\;B^n=⏟{( {\text{CAD}})( {\text{CAD}})( {\text{CAD}}) . . . . . .( {\text{CAD}})}_{\;\text{n-\times }\;}$ $\;\Rightarrow {\text{B}}^{ {\text{n}}}= {\text{CA}}^{ {\text{n}}} {\text{D}} \;\; . . .(1)$ $A^3={\begin{bmatrix}1 & \\ \\ \frac{3}{51} \\ 0 & 1\end{bmatrix}}$ Similarly ${\text{A}}^{ {\text{n}}}=$ $A^3={\begin{bmatrix}1 & \\ \\ \frac{n}{51} \\ 0 & 1\end{bmatrix}}$ Bn ${\begin{bmatrix}1 & 2 \\ -1 & -1\end{bmatrix}}{\begin{bmatrix}1 & \\ \\ { {\text{n}}}/{51} \\ 0 & 1\end{bmatrix}}{\begin{bmatrix}-1 & -2 \\ 1 & 1\end{bmatrix}}$ $\;={\begin{bmatrix}1 & \\ \\ { {\text{n}}}/{51}+2 \\ -1 & -\\ \\ { {\text{n}}}/{51}-1\end{bmatrix}$ ${\begin{bmatrix}-1 & -2 \\ 1 & 1\end{bmatrix}}$ $={\begin{bmatrix}\\ \\ { {\text{n}}}/{51}+1 & \\ \\ { {\text{n}}}/{51} \\ -\\ \\ { {\text{n}}}/{51} & 1-\\ \\ { {\text{n}}}/{51}\end{bmatrix}}$ $sum_{{n=1}}^{{50}} B^n={\begin{bmatrix}25-50 & 25 \\ -25 & -25-50\end{bmatrix}}={\begin{bmatrix}75 & 25 \\ -25 & 25\end{bmatrix}}$ Sum of the elements $=100$

Q80JEE Main 2023MCQ4MArea Under The Curves

The area of the region enclosed by the curve $y=x^3$ and its tangent at the point $(-1,-1)$ is [12-Apr-2023 shift 1]

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📖 Explanation

equation of tangent: $y+1=3(x+1)$ i.e. $y=3 x+2$ Point of intersection with curve $(2,8)$ So Area $=int_{-1}^{2} ((3 x+2)- {\text{x}}^3) {\text{dx}}=\;\frac{27}{4}$

Q81JEE Main 2023NAT4MStatistics

Let the positive numbers ${\text{a}}_1, {\text{a}}_2, {\text{a}}_3, {\text{a}}_4$ and ${\text{a}}_5 {\text{be}}$ in a G.P. Let their mean and variance be $\;\frac{31}{10}$ and $\;{ {\text{m}}}/{ {\text{n}}}$ respectively, where ${\text{m}}$ and ${\text{n}}$ are co-prime. If the mean of their reciprocal is $\;\frac{31}{40}$ and ${\text{a}}_3+ {\text{a}}_4+ {\text{a}}_5=14$ , then ${\text{m}}+ {\text{n}}$ is equal to ______. [12-Apr-2023 shift 1]

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📖 Explanation

Let $\;{ {\text{a}}}/{ {\text{r}}^2}, \;{ {\text{a}}}/{ {\text{r}}}, {\text{a}}, {\text{ar}}, {\text{ar}}^2$ Given $\;{ {\text{a}}}/{ {\text{r}}^2}+\;{ {\text{a}}}/{ {\text{r}}}+ {\text{a}}+ {\text{ar}}+ {\text{ar}}^2=5 \times \;\frac{31}{10} . . . (1)$ And $\;{ {\text{r}}^2}/{ {\text{a}}}+\;{ {\text{r}}}/{ {\text{a}}}+\;{1}/{ {\text{a}}}+\;{1}/{ {\text{ar}}}+\;{1}/{ {\text{ar}}^2}=5 \times \;\frac{31}{40} . . . (2)$ $\;(1) \div (2) a^2=4 \Rightarrow a=2$ $\; \therefore r+\;\frac{1}{r}=5 / 2 \;\; (a \ne -2)$ $\;\Rightarrow r=2$ $\; \therefore \;\text{ Now }\; \;\frac{1}{2}, 1,2.4,8$ $\therefore$ Now $\;\frac{1}{2}, 1,2.4,8$ $\; \therefore \sigma^2=\;{Σ {\text{x}}^2}/{ {\text{N}}}- (\;{Σ {\text{x}}}/{ {\text{N}}}) ^2$ $\;=\;\frac{186}{25}=\;{ {\text{M}}}/{ {\text{N}}} \Rightarrow 211= {\text{m}}+ {\text{n}}$

Q82JEE Main 2023NAT4MMatrices and Determinants

Let $D_k={\begin{vmatrix}1 & 2 k & 2 k-1 \\ n & n^{2+}n+2 & n^2 \\ n & n^{2+}n & n^{2+}n+2\end{vmatrix}}$ . If $sum_{k=1}^{n} D_k=96$ , then $n$ is equal to [12-Apr-2023 shift 1]

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📖 Explanation

$D_k={\begin{vmatrix}1 & 2 k & 2 k-1 \\ n & n^{2+}n+2 & n^2 \\ n & n^{2+}n & n^{2+}n+2\end{vmatrix}}$ $sum_{k=1}^{n} D_k=96 \Rightarrow$ ${\begin{vmatrix}sum_{{ {\text{k}}=1}}^{{ {\text{n}}}} 1 & \sum 2 {\text{k}} & \sum(2 {\text{k}}-1) \\ {\text{n}} & {\text{n}}^{2+} {\text{n}}+2 & {\text{n}}^2 \\ {\text{n}} & {\text{n}}^{2+} {\text{n}} & {\text{n}}^{2+} {\text{n}}+2\end{vmatrix}}=96$ $\;\Rightarrow {\begin{vmatrix}n & n^{2+}n & n^2 \\ n & n^{2+}n+2 & n^2 \\ n & n^{2+}n & n^{2+}n+2\end{vmatrix}}=96$ $\; {\text{R}}_2 \to {\text{R}}_2- {\text{R}}_1 \;\text{ and }\; {\text{R}}_3 \to {\text{R}}_3- {\text{R}}_1$ $\;{\begin{vmatrix}{\text{n}} & {\text{n}}^{2+} {\text{n}} & {\text{n}}^2 \\ 0 & 2 & 0 \\ 0 & 0 & {\text{n}}+2\end{vmatrix}}=96$ $\;\Rightarrow {\text{n}}(2 {\text{n}}+4)=96 \Rightarrow {\text{n}}( {\text{n}}+2)=48 \Rightarrow {\text{n}}=6$ $\;$

Q83JEE Main 2023NAT4MDefinite Integrals

If ${int_{-0.15}^{0.15} {|100 {\text{x}}^{2-}1 |} {\text{dx}}=\;{ {\text{k}}}/{3000}$ , then ${\text{k}}$ is equal to ______. [12-Apr-2023 shift 1]

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📖 Explanation

$int_{-0.15}^{0.15}{| 100 x^{2-}1 |} \text{dx}=2 int_{0}^{0.15}{| 100 x^{2-}1 |} \text{dx}$ Now $100 {\text{x}}^{2-}1=0 \Rightarrow {\text{x}}^2=\;\frac{1}{100} \Rightarrow {\text{x}}=0.1$ $I=2 [int_{0}^{0.1} (1-100 x^2) \text{dx}+int_{0.1}^{0.15} (100 x^{2-}1) \text{dx}]$ $\;I=2 \$ [x-;\frac{100}{3} x^3]$ _0^{0.1}+2 $[;\frac{100 x^3}{3}-x]$ _{0.1}^{0.15};=2 [0.1-;\frac{0.1}{3}] +2 [;\frac{0.3375}{3}-0.15-;\frac{0.1}{3}+0.1];=2 [0.2-;\frac{0.2}{3}+0.1125-0.15];=2 [;\frac{5}{100}-;\frac{2}{30}+;\frac{1125}{10000}] =2 (;\frac{1500-2000+3375}{30000});=;\frac{575}{3000} \Rightarrow {\text{k}}=575$

Q84JEE Main 2023NAT4MCircles

Two circles in the first quadrant of radii $r_1$ and $r_2$ touch the coordinate axes. Each of them cuts off an intercept of 2 units with the line $x+y=2$ . Then $r_1^{2+}r_2^{2-}r_1 r_2$ is equal to _______. [12-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Circle }\;(x-a)^{2+}(y-a)^2=a^2$ $\;x^{2+}y^{2-}2 a x-2 a y+a^2=0$ $\;\;\text{ Intercept }\;=2$ $\;\Rightarrow 2 {\sqrt{a^{2-}d^2}}=2$ Where $d=$ perpendicular distance of centre from line $x+y=2$ $\;\Rightarrow 2 \sqrt{a^{2-} (\;{a+a-2}/{{\sqrt{2}}}) ^2}=2$ $\;\Rightarrow a^{2-}\;\frac{(2 a-2)^2}{2}=1 \Rightarrow 2 a^{2-}4 a^{2+}8 a-4=2$ $\;\Rightarrow 2 a^{2-}8 a+6=0 \Rightarrow a^{2-}4 a+3=0$ $\; \therefore r_1+r_2=4 \;\text{ and }\; r_1 r_2=3$ $\; \therefore r_1^{2+}r_2^{2-}r_1 r_2= (r_1+r_2) ^{2-}3 r_1 r_2$ $\;=16-9=7$

Q85JEE Main 2023NAT4MProbability

A fair $n( n > 1)$ faces die is rolled repeatedly until a number less than $n$ appears. If the mean of the number of tosses required is $\;{ {\text{n}}}/{9}$ , then ${\text{n}}$ is equal to _______. [12-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Mean }\;=1 \cdot \;{ {\text{n}}-1}/{ {\text{n}}}+2 \;{1}/{ {\text{n}}} (\;{ {\text{n}}-1}/{ {\text{n}}}) +3 (\;{1}/{ {\text{n}}}) ^2 (\;{ {\text{n}}-1}/{ {\text{n}}}) . . .$ $\;\;{ {\text{n}}}/{9}= (\;{ {\text{n}}-1}/{ {\text{n}}}) (1+2 (\;{1}/{ {\text{n}}}) +3 (\;{1}/{ {\text{n}}}) ^2 . . . . . . .)$ $\;\;{ {\text{n}}}/{9}= (\;{ {\text{n}}-1}/{ {\text{n}}}) (1-\;{1}/{ {\text{n}}}) ^{-2}= (\;{ {\text{n}}-1}/{ {\text{n}}}) \cdot \;{ {\text{n}}^2}/{( {\text{n}}-1)^2}$ $\;\;{ {\text{n}}}/{9}=\;{ {\text{n}}}/{ {\text{n}}-1} \Rightarrow {\text{n}}=10$

Q86JEE Main 2023NAT4MPermutations and Combinations

Let the digits a, b, c be in A.P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed? [12-Apr-2023 shift 1]

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📖 Explanation

${\text{abc}}$ or cbaabc - - - - - cba $\;{{ }^7 {\text{C}}_1 \times 2 \times 6 !}/{2 ! 2 ! 2 !}=1260$

Q87JEE Main 2023NAT4MThree Dimensional Geometry

Let the plane $x+3 y-2 z+6=0$ meet the co-ordinate axes at the points A, B, C. If the orthocenter of the triangle ${\text{ABC}}$ is $(\alpha, \beta, \;\frac{6}{7})$ , then $98(\alpha+\beta)^2$ is equal to _______. [12-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{A}}(-6,0,0) {\text{B}}(0,-2,0) {\text{C}}=(0,0,3)$ $\;{ {\text{AB}}}^{\to }=6 { {\text{i}}^{\^}}-2 { {\text{j}}^{\^}}, { {\text{BC}}}^{\to }=2 { {\text{j}}^{\^}}+3 { {\text{k}}^{\^}}$ $\;{ {\text{AC}}}^{\to }=6 { {\text{i}}^{\^}}+3 { {\text{k}}^{\^}}$ ${ {\text{AH}}}^{\to } \cdot { {\text{BC}}}^{\to }=0$ $\; (\alpha+6, \beta, \;\frac{6}{7}) \cdot (0,2,3)=0$ $\;\beta=\;\frac{-9}{7}$ $\;{ {\text{CH}}}^{\to } \cdot { {\text{AB}}}^{\to }=0$ $\; (\alpha, \beta, \;\frac{-15}{7}) \cdot (6,-2,0)=0$ $\;6 \alpha-2 \beta=0$ $\;\alpha=\;\frac{-3}{7}$ $\;98(\alpha+\beta)^2=(98) \;\frac{(144)}{49}=288$

Q88JEE Main 2023NAT4MSets and Relations

The number of the relations, on the set $\{1,2,3\}$ containing $(1,2)$ and $(2,3)$ , which are reflexive and transitive but not symmetric, is ________. [12-Apr-2023 shift 1]

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📖 Explanation

${\text{A}}=\{1,2,3\}$ For Reflexive $(1,1)(2,2),(3,3) \in R$ For transitive : $(1,2)$ and $(2,3) \in R \Rightarrow (1,3) \in R$ Not symmetric : $(2,1)$ and $(3,2) ∉ R$ $\; {\text{R}}_1=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$ $\; {\text{R}}_2=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)(2,1)\}$ $\; {\text{R}}_3=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)(2,1)\}$

Q89JEE Main 2023NAT4MLimits, Continuity and Differentiability

Let $[ {\text{x}}]$ be the greatest integer $\le {\text{x}}$ . Then the number of points in the interval $(-2,1)$ , where the function ${\text{f}}( {\text{x}})=\mid [{\text{x}}]\mid +\sqrt{ {\text{x}}-[ {\text{x}}]}$ is discontinuous, is _______. [12-Apr-2023 shift 1]

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📖 Explanation

Need to check at doubtful pointsdiscont at ${\text{x}}$ I only $\;\;\text{ at }\; {\text{x}}=-1 \Rightarrow {\text{f}} (-1^{+}) =1+0=1$ $\;\Rightarrow {\text{f}} (-1^{-}) =2+1=3$ $\;\;\text{ at }\; {\text{x}}=0 \Rightarrow {\text{f}} (0^{+}) =0+0=0$ $\;\Rightarrow {\text{f}} (0^{-}) =1+1=2$ $\;\;\text{ at }\; {\text{x}}=1 \Rightarrow {\text{f}} (1^{+}) =1+0=1$ $\;\Rightarrow {\text{f}} (1^{-}) =0+1=1$ discont. at two points

Q90JEE Main 2023NAT4MIndefinite Integrals

Let $I(x)=\int \sqrt{\;\frac{x+7}{x}} \text{dx}$ and $I(9)=12+7 \log _e 7$ . If $I(1)=\alpha+7 \log _e(1+2 {\sqrt{2}})$ , then $\alpha^4$ is equal to ________. [12-Apr-2023 shift 1]

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📖 Explanation

$\;\int \sqrt{\;\frac{x+7}{x}} \text{dx}$ $\;\;\text{ Put }\; x=t^2$ $\; {\text{dx}}=2 {\text{tdt}}$ $\;\int 2 \sqrt{ {\text{t}}^{2+}7} {\text{dt}}=2 \int \sqrt{ {\text{t}}^{2+}{\sqrt{7^2}}} {\text{dt}}$ $\; {\text{I}}( {\text{t}})=2 \$ [;{ {\text{t}}}/{2} \sqrt{ {\text{t}}^{2+}7}+;\frac{7}{2} \ln {| \sqrt{ {\text{t}}^{2+}7} |}]$ + {\text{C}}; {\text{I}}( {\text{x}})=\sqrt{ {\text{x}}} \sqrt{ {\text{x}}+7}+7 \ln |\sqrt{ {\text{x}}}+\sqrt{ {\text{x}}+7}|+ {\text{C}}; {\text{I}}(9)=12+7 \ln 7=12+7(\ln (3+4))+ {\text{C}};\Rightarrow {\text{C}}=0; {\text{I}}( {\text{x}})=\sqrt{ {\text{x}}} \sqrt{ {\text{x}}+7}+7 \ln (\sqrt{ {\text{x}}}+\sqrt{ {\text{x}}+7}); {\text{I}}(1)=1 {\sqrt{8}}+7 \ln (1+{\sqrt{8}}); {\text{I}}(1)={\sqrt{8}}+7 \ln (1+2 {\sqrt{2}});\alpha={\sqrt{8}};\alpha^4= (8^{1 / 2}) ^4;\alpha^4=8^2=64$

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