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25 Jul 2022 Shift 1

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Physics30 questions

Q1JEE Main 2022MCQ4MUnits and Measurements

If momentum $[ {P}]$ , area $[ {A}]$ and time $[ {T}]$ are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is: [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$[η]=[M L^{-1} T^{-1}]$ Now if $[η]=[P]^{a}[A]^{b}[T]^{c}$ $\Rightarrow \$ [M L^{-1} T^{-1}]$=$[M L^{1} T^{-1}]^{a}$[L^{2}]^{b}[T]^{c}\Rightarrow a=1, a+2 b=-1,-a+c=-1\Rightarrow a=1, b=-1, c=0\Rightarrow [η]=[P][A]^{-1}[T]^{0}=[P A^{-1} T^{0}]$

Q2JEE Main 2022MCQ4MUnits and Measurements

Which of the following physical quantities have the same dimensions? [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Electric displacement $(\vec{D})=\epsilon_{0} \vec{E}$ $\Rightarrow [\overline{D}]=[\epsilon_{0}][\overline{E}]$ $= M^{-1} L^{-3} T^{4} A^{2}][M^{1} L^{1} A^{-1} T^{-3}$ ${[\overline{D}]=\$ [L^{-2} T^{1} A^{1}]$} $[Surface charge density]$ =;\frac{[Q]}{[A]}{[\sigma]=$[A T L^{-2}]$}\Rightarrow \vec{D} ;\text{ and };[\sigma] ;\text{ have same dimensions };$

Q3JEE Main 2022MCQ4MCircular Motion

A person moved from $A$ to $B$ on a circular path as shown in figure. If the distance travelled by him is $60 {m}$ , then the magnitude of displacement would be:(Given $\cos 185^{\circ}--0.7$ ) [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Distance travelled $=60 {\text{m}}$ $\Rightarrow$ Angle covered $=135^{\circ}$ Displacement $=2 R \sin (\;\frac{135^{\circ}}{2})$ $=2(\;\frac{60}{135} \times \;\frac{180}{\pi})\$ [;\frac{1-\cos (135^{\circ})}{2}]^{1 / 2}=2(;\frac{80}{pi})(0.85)^{1 / 2}≈ 47 {\text{m}}$

Q4JEE Main 2022MCQ4MWork, Power and Energy

A body of mass $0.5 {kg}$ travels on straight line path with velocity $v=(3 x^{2}+4) {m} / {s}$ . The net workdone by the force during its displacement from $x=0$ to $x=2 {m}$ is : [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$v=3 x^{2}+4$ at $x=0, v_{1}=4 {m} / {s}$ $x=2, v_{n}-16 {m} / {s}$ $\Rightarrow$ Work done $=\Delta$ kinetic energy $-\;\frac{1}{2} \times m(v_{2}^{2}-v_{1}^{2})$ $=\;\frac{1}{4}(256-16)$ $=60 {J}$

Q5JEE Main 2022MCQ4MRotational Motion

A solid cylinder and a solid sphere, having same mass $M$ and radius $R_{v}$ roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be : [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$a=\;{g \sin \theta}/{1+\;\frac{K^{2}}{R^{2}}}$ $v=\sqrt{\;{2 S g \sin \theta}/{1+\;\frac{K^{2}}{R^{2}}}}$ $\Rightarrow \;\frac{v_{c}}{v_{s s}} \sqrt{\;{1+\frac{K_{s s}^{2}}{R^{2}}}/{1+\;\frac{K_{c}^{2}}{R^{2}}}}=\sqrt{\;\frac{1+\frac{2}{5}}{1+\;\frac{1}{2}}}$ $\Rightarrow \sqrt{\;\frac{\;\frac{7}{5}}{\frac{3}{2}}}=\sqrt{\;\frac{14}{15}}$

Q6JEE Main 2022MCQ4MGravitation

Three identical particles ${A}, {B}$ and ${C}$ of mass $100 {kg}$ each are placed in a straight line with ${AB}= {BC}=13 {m}$ . The gravitational force on a fourth particle ${P}$ of the same mass is ${F}$ , when placed at a distance $13 {m}$ from the particle ${B}$ on the perpendicular bisector of the line ${AC}$ . The value of ${F}$ will be approximately: [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${m}=100 {kg}$ $F_{A P}=\;\frac{G m^{2}}{(13 \sqrt{2})^{2}}$ $F_{B P}=\;\frac{G m^{2}}{13^{2}}$ $F_{C P}=\;\frac{G m^{2}}{(13 \sqrt{2})^{2}}$ $F_{n e t}=F_{B P}+F_{A P} \cos 45^{\circ}+F_{C P} \cos 45^{\circ}$ $=\;\frac{G m^{2}}{13^{2}}(1+\;\frac{1}{\sqrt{2}})$ $=\;\frac{G 100^{2}}{169}(1+0.707)$ $≃ 100 G$

Q7JEE Main 2022MCQ4MThermodynamics

A certain amount of gas of volume ${V}$ at $27^{\circ} {C}$ temperature and pressure $2 \times 10^{7} {Nm}^{-2}$ expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (Use $\gamma =1.5)$ : [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Let $A B$ is isothermal process and $B C$ is adiabatic process then for $A B$ process ${P}_{ {A}} {V}_{ {A}}= {P}_{ {B}} {V}_{ {B}}$ $\Rightarrow {P}_{ {B}}=10^{7} {Nm}^{-2}$ For process BC ${P}_{ {B}} {V}_{B}^{r}= {P}_{ {C}} {V}_{C}^{r}$ $P_{C}=3.536 \times \times 10^{6} {Pa}$

Q8JEE Main 2022MCQ4MThermodynamics

Following statements are given :(A) The average kinetic energy of a gas molecule decreases when the temperature is reduced.(B) The average kinetic energy of a gas molecule increases with increase in pressure at constant temperature.(C) The average kinetic energy of a gas molecule decreases with increase in volume.(D) Pressure of a gas increases with increase in temperature at constant pressure.(E) The volume of gas decreases with increase in temperature.Choose the correct answer from the options given below : [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Because ${KE} \propto {T}$ so ${A}$ is correct, ${B}$ is incorrect, statement ${C}$ cannot be said, statement ${D}$ is contradicting itself, statement ${E}$ is incorrect (Isothermal process)So no answer correct (Bonus)If the statement of $D$ would have been."Pressure of gas increases with increase in temperature at constant volume, "then statement ${D}$ would have been correct, so in that case answer would have been 'A'.

Q9JEE Main 2022MCQ4MOscillations

In figure (A), mass '2 m' is fixed on mass ' $m$ ' which is attached to two springs of spring constant $k$ . In figure (B), mass ' $m$ ' is attached to two springs of spring constant ' $k$ ' and ' $2 k^{'}$ . If mass ' $m$ ' in (A) and in (B) are displaced by distance ' $x$ ' horizontally and then released, then time period ${T}_{1}$ and ${T}_{2}$ corresponding to $( {A})$ and $( {B})$ respectively follow the relation. [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Both the springs are in parallel combination in both the diagrams so $T_{1}=2 \pi \sqrt{\;\frac{3 m}{2 k}}$ $\;\text{ and }\; T_{2}=2 \pi \sqrt{\;\frac{m}{3 k}}$ So, $\;\frac{T_{1}}{T_{2}}=\;\frac{3}{\sqrt{2}}$

Q10JEE Main 2022MCQ4MElectrostatic Potential and Capacitance

A condenser of $2 \mu {F}$ capacitance is charged steadily from 0 to $5 {C}$ . Which of the following graph represents correctly the variation of potential difference $( {V})$ across it's plates with respect to thecharge $(Q)$ on the condenser? [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$Q=C V$ As capacitance is constant ${Q} \propto {V}$ and $V_{f}=\;\frac{Q_{f}}{C}=\;\frac{5}{2 \times 10^{-6}}=2.5 \times 10^{6} {V}$ So correct graph will be A.

Q11JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of radii of their circular paths is $6: 5$ and their respective masses ratio is $9: 4$ . Then, the ratio of their charges will be : [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

We know that $R=\;\frac{m v}{B q}=\sqrt{\;\frac{2 m K}{B q}}$ $\Rightarrow$ Ratio of radii $=\;\frac{R_{1}}{R_{2}}=\sqrt{\;\frac{m_{1}}{m_{2}}} \;\frac{q_{2}}{q_{1}}$ $\Rightarrow \;\frac{6}{5}=\sqrt{\;\frac{9}{4}} \;\frac{q_{2}}{q_{1}}$ $\Rightarrow \;\frac{q_{1}}{q_{2}}=\;\frac{3}{2} \times \;\frac{5}{6}=\;\frac{5}{4}$

Q12JEE Main 2022MCQ4MAlternating Current

To increase the resonant frequency in series LCR circuit, [25-Jul-2022-Shift-1]

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📖 Explanation

Resonant frequency $=\;\frac{1}{\sqrt{L C}}=\omega _{0}$ $\Rightarrow$ If we decrease $C, \omega _{0}$ would increase $\Rightarrow$ Another capacitor should be added in series.

Q13JEE Main 2022MCQ4MElectromagnetic Induction

A small square loop of wire of side $l$ is placed inside a large square loop of wire ${L}( {L}>>l)$ . Both loops are coplanar and their centres coincide at point ${O}$ as shown in figure. The mutual inductance of the system is : [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

We know $\varphi =M i$ Let i current be flowing in the larger loop $\Rightarrow \varphi ≃[4 \times \;\frac{\mu _{0} i}{4 \pi(L_{L} / 2)}\$ [\sin 45^{\circ}+\sin 45^{\circ}]$] \times ;\text{ Area };=;\frac{2 \sqrt{2} \mu _{0} i}{\pi L} \times I^{2}\Rightarrow M=;\frac{\varphi }{i}=;\frac{2 \sqrt{2} \mu _{0} I^{2}}{\pi L}$

Q14JEE Main 2022MCQ4MAlternating Current

The rms value of conduction current in a parallel plate capacitor is $6.9 \mu {A}$ . The capacity of this capacitor, if it is connected to $230 {V}$ ac supply with an angular frequency of $600 {rad} / {s}$ , will be: [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$Z_{C}=\;\frac{V}{I}$ $\Rightarrow \;\frac{1}{\omega C}=\;\frac{230}{6.9} M Ω$ $\Rightarrow C=\;\frac{6.9}{230 \omega } \mu F$ $=\;\frac{6.9}{230 \times 600} \mu F$ $C=50 p F$

Q15JEE Main 2022MCQ4MRay Optics and Optical Instruments

Which of the following statement is correct? [25-Jul-2022-Shift-1]

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Q16JEE Main 2022MCQ4MRay Optics and Optical Instruments

Time taken by light to travel in two different materials $A$ and $B$ of refractive indices $\mu _{A}$ and $\mu _{B}$ of same thickness is $t_{1}$ and $t_{2}$ respectively. If $t_{2}-t_{1}=5 \times 10^{-10}$ s and the ratio of $\mu _{A}$ to $\mu _{B}$ is $1: 2$ . Then, the thickness of material, in meter is: (Given $v_{ {A}}$ and $v_{ {B}}$ are velocities of light in $A$ and $B$ materials respectively.) [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$t_{2}-t_{1}=5 \times 10^{-10}$ $\Rightarrow \;\frac{d}{v_{B}}-\;\frac{d}{v_{A}}=5 \times 10^{-10}$ and, $\;\frac{v_{B}}{v_{A}}=\;\frac{\mu _{A}}{\mu _{B}}=\;\frac{1}{2}$ $\Rightarrow d(1-\;\frac{v_{B}}{v_{A}})=5 \times 10^{-10} \times v_{B}$ $\Rightarrow d(1-\;\frac{1}{2})=5 \times 10^{-10} \times v_{B}$ $\Rightarrow d=10 \times 10^{-10} \times v_{B} m$ $\Rightarrow d=5 \times 10^{-10} \times v_{A} m$

Q17JEE Main 2022MCQ4MDual Nature of Matter and Radiation

A metal exposed to light of wavelength $800 {nm}$ and and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength $500 {nm}$ is used. The workfunction of the metal is : (Take hc $=1230 {eV}- {nm}$ ). [25-Jul-2022-Shift-1]

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📖 Explanation

$\because K_{m}=\;\frac{h c}{\lambda }-\varphi$ $\Rightarrow K=\;\frac{1230}{800}-\varphi$ and, $2 K=\;\frac{1230}{500}-\varphi$ $\Rightarrow 2 \times \;\frac{1230}{800}-2 \varphi =\;\frac{1230}{500}-\varphi$ $\Rightarrow \varphi =0.615 {eV}$

Q18JEE Main 2022MCQ4MAtoms and Nuclei

The momentum of an electron revolving in ${n}^{\;\text{th }\;}$ orbit is given by:(Symbols have their usual meanings) [25-Jul-2022-Shift-1]

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📖 Explanation

$\because m v r=\;\frac{n h}{2 \pi}$ $\Rightarrow m v=\;\frac{n h}{2 \pi r}$

Q19JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

The magnetic moment of an electron (e) revolving in an orbit around nucleus with an orbital angular momentum is given by : [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\because \vec{\mu }=\;\frac{\vec{q}}{2 m}$ $\Rightarrow \vec{\mu }=\;\frac{-e \vec{L}}{2 m}$

Q20JEE Main 2022MCQ4MSemiconductor Electronics

In the circuit, the logical value of $A=1$ or $B=1$ when potential at $A$ or $B$ is $5 {\text{V}}$ and the logical value of $A=0$ or $B=0$ when potential at $A$ or $B$ is $0 {\text{V}}$ . [25-Jul-2022-Shift-1] The truth table of the given circuit will be :

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📖 Explanation

Given circuit is equivalent to an AND gate. $A\;B\;Y$ $0\;0\;0$ $0\;1\;0$ $1\;0\;0$ $1\;1\;1$

Q21JEE Main 2022NAT4MMotion in a Straight Line

A car is moving with speed of $150 {km} / {h}$ and after applying the break it will move $27 {m}$ before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling m distance. [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$F_{R} d=\;\frac{1}{2} m v^{2}$ $\;\frac{d_{2}}{d_{1}}=(\;\frac{v_{2}}{v_{1}})^{2}=(\;\frac{1}{3})^{2}$ $d_{2}=d_{1} \times \;\frac{1}{9}=3 m$

Q22JEE Main 2022NAT4MLaws of Motion

Four forces are acting at a point ${P}$ in equilibrium as shown in figure. The ratio of force ${F}_{1}$ to ${F}_{2}$ is $1: x$ where $x=$ [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$F_{1}=+2 \times \;\frac{1}{\sqrt{2}}-\;\frac{1}{\sqrt{2}}=\;\frac{1}{\sqrt{2}}$ $F_{2}=2 \times \;\frac{1}{\sqrt{2}}+\;\frac{1}{\sqrt{2}}=\;\frac{3}{\sqrt{2}}$ $\;\frac{F_{1}}{F_{2}}=\;\frac{1}{3}=\;\frac{1}{x} \Rightarrow x=3$

Q23JEE Main 2022NAT4MProperties of Matter

A wire of length ${L}$ and radius ${r}$ is clamped rigidly at one end. When the other end of the wire is pulled by a force ${F}$ , its length increases by $5 {cm}$ . Another wire of the same material of length $4 {L}$ and radius $4 {r}$ is pulled by a force $4 {F}$ under same conditions. The increase in length of this wire is ${cm}$ [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{F / A}{\Delta L / L}=Y$ $\Rightarrow \Delta L=\;\frac{F L}{A Y}$ $\;\frac{\Delta L_{2}}{\Delta L_{1}}=(\;\frac{F_{2}}{F_{1}}) \times (\;\frac{L_{2}}{L_{1}}) \times (\;\frac{A_{1}}{A_{2}})$ $=4 \times 4 \times \;\frac{1}{16}=1$ $\Delta L_{2}=\Delta L_{1}=5 {cm} .$

Q24JEE Main 2022NAT4MProperties of Matter

A unit scale is to be prepared whose length does not change with temperature and remains $20 {cm}$ , using a bimetallic strip made of brass and iron each of different length. The length of both components would change in such a way that difference between their lengths remains constant. If length of brass is $40 {cm}$ and length of iron will be ${cm} .(\alpha_{\;\text{iron }\;}=1.2 \times 10^{-5} {K}^{-1}$ and $\alpha_{\;\text{brass }\;}=1.8 \times 10^{-5} {K}^{-1})$ [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\Delta L_{1}=\alpha_{1} L_{1} \Delta T$ $\Delta L_{2}=\alpha_{2} L_{2} \Delta T$ $\alpha_{1} L_{1}=\alpha_{2} L_{2}$ $1.2 \times 10^{-5} \times L_{1}=1.8 \times 10^{-5} \times L_{2}$ $L_{1}=\;\frac{1.8}{1.2} \times 40=60 {cm}$

Q25JEE Main 2022NAT4MWaves

An observer is riding on a bicycle and moving towards a hill at $18 {kmh}^{-1}$ . He hears a sound from a source at some distance behind him directly as well as after its reflection from the hill. If the original frequency of the sound as emitted by source is $640 {Hz}$ and velocity of the sound in air is $320 {m} / {s}$ , the beat frequency between the two sounds heard by observer will be ${Hz}$ . [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$f_{1}=f_{0}(\;\frac{320-5}{320})=640(\;\frac{315}{320})$ $=630 {Hz}$ $f_{3}=f_{0}[$ No relative motion] $f_{2}=f_{0}[\;\frac{320+5}{320}]=640(\;\frac{325}{320})$ $=650$ Beat frequency $=f_{2}-f_{1}$ $=650-630=20 {Hz}$

Q26JEE Main 2022NAT4MElectrostatics

The volume charge density of a sphere of radius $6 {m}$ is $2 \mu {C} {cm}^{-3}$ . The number of lines of force per unit surface area coming out from the surface of the sphere is $\times 10^{10} {NC}^{-1}$ .[Given : Permittivity of vacuum $E_{0}=8.85 \times 10^{-12} {C}^{2} {N}^{-1}- {m}^{-2}$ ) [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\rho =2 \mu {c} / {cm}^{3}$ ${R}=6 {m}$ Number of lines of force per unit area = Electric field at surface. $=\;\frac{K Q}{R^{2}}$ $=\;\frac{1}{4 \pi \epsilon_{0}} \;\frac{\rho \frac{4}{3} \pi R^{3}}{R^{2}}$ $=\;\frac{\rho R}{3 E_{0}}$ $=\;\frac{2 \times 10^{-6} \times 10^{6} \times 6}{3 \times 8.85 \times 10^{-12}}$ $=0.45197 \times 10^{12}$ $=45.19 \times 10^{10} {N} / {C}$ $∼ 45 \times 10^{10}$

Q27JEE Main 2022NAT4MCurrent Electricity

In the given figure, the value of ${V}_{0}$ will be V. [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Using Kirchhoff's junction rule. $\;\frac{2-V_{0}}{1}+\;\frac{4-V_{0}}{1}+\;\frac{6-V_{0}}{1}=0$ $12-3 V_{0}=0$ $V_{0}=4 V$

Q28JEE Main 2022NAT4MCurrent Electricity

Eight copper wire of length $l$ and diameter $d$ are joined in parallel to form a single composite conductor of resistance $R$ . If a single copper wire of length $2 l$ have the same resistance $(R)$ then its diameter will be d. [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$R A B=R$ $R=\;\frac{1}{8}$ (Resistance of one wire) $=\;\frac{1}{8} \rho \;{l}/{\pi \;\frac{d^{2}}{4}}=\;\frac{\rho l}{2 \pi d^{2}}$ Resistance of copper wire of length $2 l$ and diameter $x=R$ $\rho \;{2 l}/{\pi \;\frac{x^{2}}{4}}=R$ $\;\frac{8 \rho l}{\pi x^{2}}=\;\frac{\rho l}{2 \pi d^{2}}$ $16 d^{2}=x^{2}$ $x=4 d$

Q29JEE Main 2022NAT4MSemiconductor Electronics

The energy band gap of semiconducting material to produce violet (wavelength $=4000 {A}^{o}$ ) LED is ${eV}$ . (Round off to the nearest integer). [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Energy corresponding to wavelength $4000 {A}^{\circ}$ $E=\;\frac{h c}{\pi}$ $=\;\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{4000 \times 10^{-10} \times 1.6 \times 10^{-19}} {eV}$ $=\;\frac{12400}{4000}$ $=3.1 {eV}$ $\approx 3 {eV}$

Q30JEE Main 2022NAT4MElectromagnetic Waves

The required height of a TV tower which can cover the population of $6.03$ lakh is $h$ . If the average population density is 100 per square ${km}$ and the radius of earth is $6400 {km}$ , then the value of $h$ will be [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$r=\sqrt{(h+R)^{2}-R^{2}} ≅ \sqrt{2 h R}$ $A=\;\frac{6.03 \times 10^{5}}{100}$ $\pi r^{2}=6.03 \times 10^{3}$ $\pi 2 R h=6.03 \times 10^{3}$ $h=\;\frac{6.03 \times 10^{3}}{2 \times \pi \times R}=0.015 \times 10 \times 10^{3} {m}$ $=150 {m}$

Chemistry30 questions

Q31JEE Main 2022MCQ4MSome Basic Concepts of Chemistry

${SO}_{2} {Cl}_{2}$ on reaction with excess of water results into acidic mixture ${SO}_{2} {Cl}_{2}+2 {H}_{2} {O} \to {H}_{2} {SO}_{4}+2 {HCl}$ 16 moles of ${NaOH}$ is required for the complete neutralisation of the resultant acidic mixture. The number of moles of ${SO}_{2} {Cl}_{2}$ used is: [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{SO}}_2 {\text{Cl}}_2+2 {\text{H}}_2 {\text{O}} \to {\text{H}}_2 {\text{SO}}_4+2 {\text{HCl}}$ Moles of ${\text{NaOH}}$ required for complete neutralization of resultant acidic mixture $=16$ molesAnd 1 mole of ${\text{SO}}_2 {\text{Cl}}_2$ produced 4 moles of ${\text{H}}^{+}$ . $\therefore$ Moles of ${\text{SO}}_2 {\text{Cl}}_2$ used will be $=\;\frac{16}{4}=4$ moles

Q32JEE Main 2022MCQ4MStructure of Atom

Which of the following sets of quantum numbers is not allowed? [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

If $n=3$ , then possible values of $l=0,1,2$ But in option (C), the value of 1 is given ' 3 ', this is not possible.

Q33JEE Main 2022MCQ4MSolutions

The depression in freezing point observed for a formic acid solution of concentration $0.5 {mL} {L}^{-1}$ is $0.0405^{\circ} {C}$ . Density of formic acid is $1.05 {g} {mL}^{-1}$ . The Van't Hoff factor of the formic acid solution is nearly : (Given for water ${k}_{ {f}}=1.86 {k} {kg} {mol}^{-1}$ ) [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\Delta {\text{T}}_{ {\text{f}}}$ of formic acid $=0.0405^{\circ} {\text{C}}$ Concentration $=0.5 {\text{mL}} / {\text{L}}$ and density $=1.05 {\text{g}} / {\text{mL}}$ $\therefore$ Mass of formic acid in solution $=1.05 \times 0.5 {\text{g}}$ $=0.525 {\text{g}}$ $\therefore$ According to Van't Hoff equation, $\Delta {\text{T}}_{ {\text{f}}}= {\text{ik}}_{ {\text{f}}} \cdot {\text{m}}$ $0.0405= {\text{i}} \times 1.86 \times \;\frac{0.525}{46 \times 1}$ (Assuming mass of $1 {\text{L}}$ water $= {\text{kg}}$ ) ${\text{i}}=\;\frac{0.0405 \times 46}{1.86 \times 0.525}=1.89 \approx 1.9$

Q34JEE Main 2022MCQ4MIonic Equilibrium

$20 {\text{mL}}$ of $0.1$ ${\text{M}} {\text{NH}}_4 {\text{OH}}$ is mixed with $40 {\text{mL}}$ of $0.05 {\text{MHCl}}$ . The ${\text{pH}}$ of the mixture is nearest to:(Given: $K_b ( {\text{NH}}_4 {\text{OH}}) =1 \times 10^{-5}, \log 2=0.30, \log 3=0.48, \log 5=0.69, \log 7=0.84, \log 11=1.04$ ) [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{NH}}_4 {\text{OH}}+ {\text{HCl}} \to {\text{NH}}_4 {\text{Cl}}+ {\text{H}}_2 {\text{O}}$ mmole 22 $[ {\text{NH}}_4^{+}]=\;{2 {\text{mmole}}}/{60 {\text{ml}}}=\;\frac{1}{30} {\text{M}}$ ${\text{pH}}=\;{ {\text{pK}}_{ {\text{w}}}- {\text{pK}}_{ {\text{b}}}-\log {\text{C}}}/{2}=\;\frac{14-5+1.48}{2}=5.24$

Q35JEE Main 2022MCQ4Md and f Block Elements

List I List II (A) ${N}_{2}( {g})+3 {H}_{2}( {g}) \to 2 {NH}_{3}( {g})$ (I) $Cu$ (B) ${CO}( {g})+3 {H}_{3}( {g}) \to {CH}_{4}( {g})+ {H}_{2} {O}( {g})$ (II) ${Cu} / {ZnO}- {Cr}_{2} {O}_{3}$ (C) ${CO}( {g})+ {H}_{2}( {g}) \to {HCHO}( {g})$ (III) ${Fe}_{x} {O}_{y}+ {K}_{2} {O}+ {Al}_{2} {O}_{3}$ (D) ${CO}( {g})+2 {H}_{2}( {g}) \to {CH}_{3} {OH}( {g})$ (IV) $\text{Ni}$ Choose the correct answer from the options given below : [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Here, we have to match the reactions with their correct catalyst :(A) ${\text{N}}_2( {\text{g}})+3 {\text{H}}_2( {\text{g}})^{ {\text{Fe}}_{ {\text{x}}} {\text{O}}_{ {\text{y}}}+ {\text{K}}_2 {\text{O}}+ {\text{Al}}_2 {\text{O}}_3} 2 {\text{NH}}_3( {\text{g}})$ (B) ${\text{CO}}( {\text{g}})+3 {\text{H}}_2( {\text{g}}) {\longrightarrow }^{ {\text{Ni}}} {\text{CH}}_4( {\text{g}})+ {\text{H}}_2 {\text{O}}( {\text{g}})$ (C) ${\text{CO}}( {\text{g}})+ {\text{H}}_2( {\text{g}}) {\longrightarrow }^{ {\text{Cu}}} {\text{HCHO}}( {\text{g}})$ (D) ${\text{CO}}( {\text{g}})+2 {\text{H}}_2 \$ (g) {⟶}^{ {\text{Cu}} / {\text{ZnO}}^{- {\text{Zr}}_2 {\text{O}}_3}} {\text{CH}}_3- {\text{OH}}( {\text{g}})$

Q36JEE Main 2022MCQ4MClassification of Elements

The IUPAC nomenclature of an element with electronic configuration $[ {Rn}] 5 {f}^{14} 6 {d}^{1} 7 {s}^{2}$ is : [25-Jul-2022-Shift-1]

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📖 Explanation

The element with electronic configuration $[R {\text{n}}] 5 {\text{f}}^{14} 6 {\text{d}}^1 7 {\text{s}}^2$ has atomic number $\to 103$ $\therefore$ Its IUPAC name is: Unniltrium

Q37JEE Main 2022MCQ4Md and f Block Elements

The compound(s) that is (are) removed as slag during the extraction of copper is:(A) ${CaO}$ (B) FeO(C) ${Al}_{2} {O}_{3}$ (D) ${ZnO}$ (E) NiO Choose the correct answer from the options given below: [25-Jul-2022-Shift-1]

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📖 Explanation

The compound(s) that are removed as a slag during the extraction of copper is : ${\text{FeS}} {\to }^{ {\text{O}}_2 / {\text{SiO}}_2} {\text{FeSiO}}_3+ {\text{SO}}_2$ $\therefore$ Only iron oxide (FeO) formed slag during extraction of copper.

Q38JEE Main 2022MCQ4MRedox Reactions

The reaction of ${H}_{2} {O}_{2}$ with potassium permanganate in acidic medium leads to the formation of mainly : [25-Jul-2022-Shift-1]

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📖 Explanation

${\text{H}}_2 {\text{O}}_2+ {\text{MnO}}_4^{-} \to {\text{Mn}}^{2+}+ {\text{O}}_2$ (unbalanced)

Q39JEE Main 2022MCQ4MClassification of Elements

Choose the correct order of density of the alkali meals: [25-Jul-2022-Shift-1]

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📖 Explanation

Density ( $\rho$ ) is defined as mass per unit volume ( $\rho = \frac{m}{V}$ ).
For alkali metals, as we move down the group from Lithium (Li) to Cesium (Cs), the atomic mass ( $m$ ) increases significantly.
While the atomic volume ( $V$ ) also increases, the increase in atomic mass is generally more pronounced than the increase in atomic volume.
Therefore, the density of alkali metals generally increases as we go down the group.
Following this general trend, the order of increasing density is ${Li}< {Na}< {K}< {Rb}< {Cs}$ .

Q40JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

The geometry around boron in the product ' $B$ ' formed from the following reaction is ${BF}_{3}+ {NaH} {\to }^{450 {K}} {A}+ {NaF}$ ${A}+ {NMe}_{3} \to {B}$ [25-Jul-2022-Shift-1]

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📖 Explanation

${\text{BF}}_3+ {\text{NaH}} {\longrightarrow }^{450 {\text{K}}} \;{{\text{B}}_2 {\text{H}}_6}_{\;\text{ (diborane) }}+ {\text{NaF}}$ ${\text{B}}_2 {\text{H}}_6+ {\text{NMe}}_3 \longrightarrow 2[ {\text{BH}}_3 ← {\text{NMe}}_3]_{\text{symmetrical} \text{cleavage}}$

Q41JEE Main 2022MCQ4Mp Block Elements

The interhalogen compound formed from the reaction of bromine with excess of fluorine is a: [25-Jul-2022-Shift-1]

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📖 Explanation

${\text{Br}}_2+\;{5 {\text{F}}}_{(\;\text{ Excess }\;)_2} \longrightarrow 2 {\text{BrF}}_5$ If ${\text{BrF}}_5$ undergoes hydrolysis it will produce halide.

Q42JEE Main 2022MCQ4Mp Block Elements

The photochemical smog does not generally contain: [25-Jul-2022-Shift-1]

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📖 Explanation

Photochemical smog contain:Ozone, nitric oxide, organic compounds, nitrogen dioxide, formaldehyde. $\therefore {\text{SO}}_2$ is not the part of photochemical smog.

Q43JEE Main 2022MCQ4MHydrocarbons

A compound $'A '$ on reaction with $' {X}{'}$ and $' {Y}{'}$ produces the same major product but different by product ' $a$ ' and ' $b$ '. Oxidation of ' $a$ ' gives a substance produced by ants. ' $X$ ' and 'Y' respectively are [25-Jul-2022-Shift-1]

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Q44JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

Most stable product of the following reaction is: [25-Jul-2022-Shift-1]

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Q45JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Which one of the following reactions does not represent correct combination of substrate and product under the given conditions? [25-Jul-2022-Shift-1]

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Q46JEE Main 2022MCQ4MAmines

An organic compound 'A' on reaction with ${NH}_{3}$ followed by heating gives compound ${B}$ . Which on further strong heating gives compound ${C}( {C}_{8} {H}_{5} {NO}_{2})$ . Compound ${C}$ on sequential reaction withethanolic ${KOH}$ , alkyl chloride and hydrolysis with alkali gives a primary amine. The compound ${A}$ is : [25-Jul-2022-Shift-1]

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📖 Explanation

Diamagnetic species are: ${\text{N}}_2, {\text{O}}_2{ }^{2-}$

Q47JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Melamine polymer is formed by the condensation of: [25-Jul-2022-Shift-1]

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📖 Explanation

Melamine :Formaldehyde ${\text{HCHO}}$ Melamine formaldehyde Resin is melamine polymer

Q48JEE Main 2022MCQ4MBiomolecules

During the denaturation of proteins, which of these structures will remain intact? [25-Jul-2022-Shift-1]

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Q49JEE Main 2022MCQ4MBiomolecules

Drugs used to bind to receptors, inhibiting is natural function and blocking a message are called: [25-Jul-2022-Shift-1]

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Q50JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

Given below are two statements :Statement I: On heating with ${KHSO}_{4}$ , glycerol is dehydrated and acrolein is formed.Statement II : Acrolein has fruity odour and can be used to test glycerol's presence.Choose the correct option. [25-Jul-2022-Shift-1]

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Q51JEE Main 2022NAT4MChemical Bonding and Molecular Structure

Among the following species ${N}_{2}, {N}_{2}+, {N}_{2}^{-}, {N}_{2}^{2-}, {O}_{2}, {O}_{2}^{+}, {O}_{2}^{-}, {O}_{2}^{2-}$ the number of species showing diamagnesim is [25-Jul-2022-Shift-1]

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📖 Explanation

Those species which have unpaired electrons are called paramagnetic species.And those species which have no unpaired electrons are called diamagnetic species. (1) $N_2$ has 14 electrons.Moleculer orbital configuration of $N_2$ $=\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \pi_{2 p_x^2}=\pi_{2 p_y^2} \sigma_{2 p_z^2}$ Here no unpaired electron present, so it is diamagnetic.(2) Moleculer orbital configuration of $N_2^{+}$ (13 electrons) $=\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \pi_{2 p_x^2}=\pi_{2 p_y^2} \sigma_{2 p_z^1}$ Here in $N_2^{+}, 1$ unpaired electron present, so it is paramagnetic.(3) ${\text{N}}_2^{2-}$ has 16 electrons. Moleculer orbital configuration of ${\text{N}}_2^{2-}$ is $\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \sigma_{2 p_z^2} \pi_{2 p_x^2}=\pi_{2 p_y^2} \pi_{2 p_x^1}^*=\pi_{2 p_y^1}^*$ Here 2 unpaired electron present, so it is paramagnetic.(4) ${\text{N}}_2^{-}$ has 15 electrons.Moleculer orbital configuration of ${\text{N}}_2^{-}$ is $\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \sigma_{2 p_z^2} \pi_{2 p_x^2}=\pi_{2 p_y^2} \pi_{2 p_x^1}^*=\pi_{2 p_y^0}^*$ Here 1 unpaired electron present, so it is paramagnetic.(a) $O_2^{2-}$ has 18 electrons. Moleculer orbital configuration of $O_2^{2-}$ is $\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \sigma_{2 p_z^2} \pi_{2 p_x^2}=\pi_{2 p_y^2} \pi_{2 p_x^2}^*=\pi_{2 p_y^2}^*$ Here is no unpaired electron so it is diamagnetic.(b) ${\text{O}}_2^{-}$ has 17 electrons.Moleculer orbital configuration of $O_2^{2-}$ is $\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \sigma_{2 p_z^2} \pi_{2 p_x^2}=\pi_{2 p_y^2} \pi_{2 p_x^2}^*=\pi_{2 p_y^1}^*$ Here 1 unpaired electron present, so it is paramagnetic.(c) ${\text{O}}_2$ has 16 electrons. Moleculer orbital configuration of $O_2$ is $\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \sigma_{2 p_z^2} \pi_{2 p_x^2}=\pi_{2 p_y^2} \pi_{2 p_x^1}^*=\pi_{2 p_y^1}^*$ Here 2 unpaired electron present, so it is paramagnetic.(d) ${\text{O}}_2^{+}$ has 15 electrons.Moleculer orbital configuration of $O_2^{+}$ is $\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \sigma_{2 p_z^2} \pi_{2 p_x^2}=\pi_{2 p_y^2} \pi_{2 p_x^1}^*=\pi_{2 p_y^0}^*$ Here 1 unpaired electron present, so it is paramagnetic.

Q52JEE Main 2022NAT4MChemical Thermodynamics

The enthalpy of combustion of propane, graphite and dihydrogen at $298 {K}$ are $-2220.0 {kJ} {mol}^{-1},-393.5 {kJ} {mol}^{-1}$ and $-285.8 {kJ} {mol}^{-1}$ respectively. The magnitude ofenthalpy of formation of propane $( {C}_{3} {H}_{8})$ is ${kJ} {mol}^{-1} .$ (Nearest integer) [25-Jul-2022-Shift-1]

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📖 Explanation

Enthalpy of combustion of propane, graphite and ${\text{H}}_2$ at $298 {\text{K}}$ are ${\text{C}}_3 {\text{H}}_8( {\text{g}})+5 {\text{O}}_2( {\text{g}}) \to 3 {\text{CO}}_2( {\text{g}})+4 {\text{H}}_2 {\text{O}}( {\text{I}}), \Delta {\text{H}}_1=-2220 {\text{kJ}} {\text{mol}}^{-1}$ ${\text{C}}(\;\text{ graphite }\;)+ {\text{O}}_2( {\text{g}}) \to {\text{CO}}_2( {\text{g}}), \;\; \Delta {\text{H}}_2=-393.5 {\text{kJ}} {\text{mol}}^{-1}$ ${\text{H}}_2( {\text{g}})+\;\frac{1}{2} {\text{O}}_2( {\text{g}}) \to {\text{H}}_2 {\text{O}}( {\text{I}}), \;\; \Delta {\text{H}}_3=-285.8 {\text{kJ}} {\text{mol}}^{-1}$ The desired reaction is $3 {\text{C}} \;\text{ (graphite) }\;+4 {\text{H}}_2( {\text{g}}) \to {\text{C}}_3 {\text{H}}_8( {\text{g}})$ $\Delta {\text{H}}_{ {\text{f}}}=3 \Delta {\text{H}}_2+4 \Delta {\text{H}}_3-\Delta {\text{H}}_1$ $=3(-393.5)+4(-285.8)-(-2220)$ $=-103.7 {\text{kJ}} {\text{mol}}^{-1}$ $|\Delta {\text{H}}_{ {\text{f}}}| ≃ 104 {\text{kJ}} {\text{mol}}^{-1}$

Q53JEE Main 2022NAT4MStates of Matter

The pressure of a moist gas at $27^{\circ} {C}$ is $4 {a} \cdot {m}$ . The volume of the container is doubled at the same temperature. The new pressure of the moist gas is $\times 10^{-1} {atm} .$ (Nearest integer)(Given : The vapour pressure of water at $27^{\circ} {C}$ is $0.4 {atm} .$ ) [25-Jul-2022-Shift-1]

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📖 Explanation

From ideal gas equation, $P \propto \;\frac{1}{V}$ ${\text{P}}_1 {\text{V}}_1= {\text{P}}_2 {\text{V}}_2$ Pressure of the gas $=4-0.4=3.6 {\text{atm}}$ $3.6 {\text{V}}_1= {\text{P}}_2(2 {\text{V}}_1)$ ${\text{P}}_2=1.8 {\text{atm}}$ Hence, new pressure of moist gas is $1.8+0.4=2.2 {\text{atm}}=22 \times 10^{-1} {\text{atm}}$

Q54JEE Main 2022NAT4MElectrochemistry

The cell potential for ${Zn}| {Zn}^{2+}( {aq})|| {Sn}^{x+}| {Sn}$ is $0.801 {V}$ at $298 {K}$ . The reaction quotient for the above reaction is $10^{-2}$ . The number of electrons involved in the given electrochemical cell reaction is( Given : ${E}_{ {Zn}^{2+} \mid {Zn}}^{ {o}}=-0.763 {V}, {E}_{ {Sn}^{x+} \mid {Sn}}^{ {o}}=+0.008 {V}$ and $\;\frac{2.303 {RT}}{ {F}}=0.06 {V}$ ) [25-Jul-2022-Shift-1]

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📖 Explanation

${\text{A}}: {\text{Zn}} \to {\text{Zn}}^{2+}+2 e^{-}$ ${\text{C}}: {\text{Sn}}^{+ {\text{x}}}+ {\text{xe}}^{-} \to {\text{Sn}}$ ${\text{E}}_{ {\text{Cell}}}^{\circ}= {\text{E}}_{ {\text{Zn}} \mid {\text{Zn}}^{2+}}^{\circ}+ {\text{E}}_{ {\text{Sn}}^{+x} \mid {\text{Sn}}}^{\circ}$ $\Rightarrow 0.763+0.008=0.771 {\text{V}}$ From the Nernst equation, ${\text{E}}_{\;\text{Cell }\;}= {\text{E}}_{\;\text{Cell }\;}^{\circ} \;{-2.303 {\text{RT}}}/{ {\text{nF}}} \log {\text{Q}}$ $0.801=0.771-\;{0.06}/{ {\text{n}}} \log 10^{-2}$ $0.03=\;\frac{0.06}{n} \times 2$ ${\text{n}}=4$

Q55JEE Main 2022NAT4MChemical Kinetics

The half life for the decomposition of gaseous compound ${A}$ is $240 {s}$ when the gaseous pressure was 500 Torr initially. When the pressure was 250 Torr, the half life was found to be $4.0 {min}$ . The order ofthe reaction is (Nearest integer) [25-Jul-2022-Shift-1]

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📖 Explanation

$( {\text{t}}_{1 / 2})_{ {\text{A}}}=240 {\text{s}}$ when ${\text{P}}=500$ torr $(t_{1 / 2})_A=4 {\min}=4 \times 60=240 {\sec}$ when $P=250$ torrIf means half-life is independent of concentration of reactant present. $\therefore$ Order of reaction $=1$

Q56JEE Main 2022NAT4MCoordination Compounds

Consider the following metal complexes: ${[ {\text{Co}}( {\text{NH}}_{3})_{6}]^{3+}}$ ${[ {\text{CoCl}}( {\text{NH}}_{3})_{5}]^{2+}}$ ${[ {\text{Co}}( {\text{CN}})_{6}]^{3-}}$ ${[ {\text{Co}}( {\text{NH}}_{3})_{5}( {\text{H}}_{2} {\text{O}})]^{3+}}$ The spin-only magnetic moment value of the complexthat absorbes light with shortest wavelength is B. M. (Nearest integer) [25-Jul-2022-Shift-1]

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📖 Explanation

$\Delta _0 \propto \;\frac{1}{\lambda }$ Here, ${\text{CN}}^{-}$ being SFL will have maximum CFSESo, $[ {\text{Co}}( {\text{CN}})_6]^{3-}$ will be ${\text{d}}^2 {\ sp}^3, \mu =0$

Q57JEE Main 2022NAT4Md and f Block Elements

Among ${Co}^{3+}, {Ti}^{2+}, {V}^{2+}$ and ${Cr}^{2+}$ ions, one if used as a reagent cannot liberate ${H}_{2}$ from dilute mineral acid solution, its spin-only magnetic moment in gaseous state is B.M. (Nearest integer) [25-Jul-2022-Shift-1]

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📖 Explanation

${\text{Co}}^{3+}$ will not liberate ${\text{H}}_2$ gas an reaction with dilute acid ${\text{E}}_{ {\text{Co}}^{3+} / {\text{Co}}^{2+}}^{ {\text{O}}}=+1.97$ And ${\text{Co}}^{3+}$ has electronic configuration $=[ {\text{Ar}}] 3 d^6$ $\therefore 4$ unpaired ${\text{e}}^{-}$ are present in it $\therefore$ Spin-only magnetic moment $=\sqrt{4(4+2)}=4.92 \approx 5$

Q58JEE Main 2022NAT4MPractical Organic Chemistry

While estimating the nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from $0.25 {g}$ of the compound neutralized $2.5 {mL}$ of $2 {M} {H}_{2} {SO}_{4}$ . The percentage of nitrogen present in organic compound is [25-Jul-2022-Shift-1]

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📖 Explanation

${\text{NH}}_3$ gas is neutralized by $2.5 {\text{mL}}$ of $2 {\text{M}} {\text{H}}{ }_2 {\text{SO}}_4$ $\therefore$ Moles of ${\text{NH}}_3$ neutralized $=2.5 \times 2 \times 2$ millimole $=10 \times 10^{-3}$ moles $\therefore$ Weight of ${\text{N}}$ present in the compound will be $=10 \times 10^{-3} \times 14$ $=0.14 {\text{g}}$ $\therefore \%$ of ' ${\text{N}}^{'}$ in compound $=\;\frac{0.14}{0.25} \times 100$ $=56 \%$

Q59JEE Main 2022NAT4MOrganic Chemistry - Some Basic Principles

The number of ${sp}^{3}$ hybridised carbons in an acyclic neutral compound with molecular formula ${C}_{4} {H}_{5} {N}$ is [25-Jul-2022-Shift-1]

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📖 Explanation

${\text{DU}}=4+1-(\;\frac{5-1}{2})=3$ ${\text{H}}_3 {\text{C}}_{↑}_{{\ sp}^3}- {\text{CH}}= {\text{CH}}- {\text{C}} \equiv {\text{N}}$ ${\text{CH}}_2= {\text{C}}= {\text{CH}}_{\text{Zero} {\ sp}^3 \text{carbon}}= {\text{CH}}= {\text{NH}}$

Q60JEE Main 2022NAT4MAldehydes, Ketones and Carboxylic Acids

In the given reactionThe number of chiral carbon/s in product $A$ is___ [25-Jul-2022-Shift-1]

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Mathematics30 questions

Q61JEE Main 2022MCQ4MPermutations and Combinations

The total number of functions, $f:\{1,2,3,4\} \to \{1,2,3,4,5,6\}$ such that $f(1)+f(2)=f(3)$ , is equal to : [25-Jul-2022-Shift-1]

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📖 Explanation

Given, $f(1)+f(2)=f(3)$ It means $f(1), f(2)$ and $f(3)$ are dependent on each other. But there is no condition on $f(4)$ , so $f(4)$ can be $f(4)=1,2,3,4,5,6$ .For $f(1), f(2)$ and we have to find how many functions possible which will satisfy the condition $f(1)+f(2)=f(3)$ Case 1:When $f(3)=2$ then possible values of $f(1)$ and $f(2)$ which satisfy $f(1)+f(2)=f(3)$ is $f(1)=1$ and $f(2)=1$ .And $f(4)$ can be $=1,2,3,4,5,6$ $\therefore$ Total possible functions $=1 \times 6=6$ Case 2 : When $f(3)=3$ then possible values(1) $f(1)=1$ and $f(2)=2$ (2) $f(1)=2$ and $f(2)=1$ And $f(4)$ can be $=1,2,3,4,5,6$ . $\therefore$ Total functions $=2 \times 6=12$ Case 3 :When $f(3)=4$ then(1) $f(1)=1$ and $f(2)=3$ (2) $f(1)=2$ and $f(2)=2$ (3) $f(1)=3$ and $f(2)=1$ And $f(4)$ can be $=1,2,3,4,5,6$ $\therefore$ Total functions $=3 \times 6=18$ Case 4 :When $f(3)=5$ then(1) $f(1)=1$ and $f(4)=4$ (2) $f(1)=2$ and $f(4)=3$ (3) $f(1)=3$ and $f(4)=2$ (4) $f(1)=4$ and $f(4)=1$ And $f(4)$ can be $=1,2,3,4,5$ and 6 $\therefore$ Total functions $=4 \times 6=24$ Case 5 :When $f(3)=6$ then (1) $f(1)=1$ and $f(2)=5$ (2) $f(1)=2$ and $f(2)=4$ (3) $f(1)=3$ and $f(2)=3$ (4) $f(1)=4$ and $f(2)=2$ (5) $f(1)=5$ and $f(2)=1$ And $f(4)$ can be $=1,2,3,4,5$ and 6 $\therefore$ Total possible functions $=5 \times 6=30$ $\therefore$ Total functions from those 5 cases we get $=6+12+18+24+30$ $=90$

Q62JEE Main 2022MCQ4MComplex Numbers

If $\alpha, \beta, \gamma , \delta$ are the roots of the equation $x^{4}+x^{3}+x^{2}+x+1=0$ , then $\alpha^{2021}+\beta^{2021}+\gamma ^{2021}+\delta ^{2021}$ is equal to : [25-Jul-2022-Shift-1]

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📖 Explanation

When, $x^5=1$ then $x^{5-}1=0$ $\Rightarrow (x-1)(x^{4+}x^{3+}x^{2+}x+1)=0$ Given, $x^{4+}x^{3+}x^{2+}x+1=0$ has roots $\alpha, \beta, \gamma$ and 8 . $\therefore$ Roots of $x^{5-}1=0$ are $1, \alpha, \beta, \gamma$ and 8We know, Sum of $p^{\;\text{th }\;}$ power of $n^{\;\text{th }\;}$ roots of unity $=0$ . (If $p$ is not multiple of $n$ ) or $n$ (If $p$ is multiple of $n$ ) $\therefore$ Here, Sum of ${\text{p}}^{\;\text{th }\;}$ power of ${\text{n}}^{\;\text{th }\;}$ roots of unityHere, $p=2021$ , which is not multiple of 5 . $\therefore 1^{2021}+\alpha^{2021}+\beta^{2021}+\gamma ^{2021}+8^{2021}=0$ $\Rightarrow \alpha^{2021}+\beta^{2021}+\gamma ^{2021}+8^{2021}=-1$

Q63JEE Main 2022MCQ4MComplex Numbers

For ${n} \in N$ , let ${S}_{ {n}}=\{z \in C:| z-3+2 i|=\;\frac{ {n}}{4}\}$ and ${T}_{ {n}}=\{z \in C:| z-2+3 i|=\;\frac{1}{ {n}}\}$ . Then the number of elements in the set $\{n \in N: S_{n} ∩ T_{n}=\varphi \}$ is : [25-Jul-2022-Shift-1]

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📖 Explanation

$S_n=\{z \in C:|z-3+2 i|=\;\frac{n}{4}\}$ represents a circle with centre $C_1(3,-2)$ and radius $r_1=\;\frac{n}{4}$ Similarly $T_n$ represents circle with centre $C_2(2,-3)$ and radius $r_2=\;\frac{1}{n}$ As $S_n ∩ T_n=\varphi$ $C_1 C_2>r_1+r_2 \;\;$ OR $\;\; C_1 C_2< |r_1-r_2|$ $\sqrt{2}>\;\frac{n}{4}+\;\frac{1}{n} \;\;$ OR $\;\; \sqrt{2}< |\;\frac{n}{4}-\;\frac{1}{n}|$ $n=1,2,3,4 \;\; n$ may take infinite values

Q64JEE Main 2022MCQ4MMatrices and Determinants

The number of $\theta \in (0,4 \pi)$ for which the system of linear equations $3(\sin 3 \theta) x-y+z=2$ $3(\cos 2 \theta) x+4 y+3 z=3$ $6 x+7 y+7 z=9$ has no solution, is : [25-Jul-2022-Shift-1]

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📖 Explanation

Given, $3(\sin 3 \theta) x-y+z=2$ $3(\cos 2 \theta) x+4 y+3 z=3$ $6 x+7 y+7 z=9$ For no solutions determinant of coefficient will be $=0$ $\therefore D=\begin{vmatrix}3 \sin 3 \theta & -1 & 1 \\ 3 \cos 2 \theta & 4 & 3 \\ 6 & 7 & 7\end{vmatrix}=0$ $\Rightarrow 3 \sin 3 \theta(28-21)+1(21 \cos 2 \theta-18)+1(21 \cos 2 \theta-24)=0$ $\Rightarrow 21 \sin 3 \theta+42 \cos 2 \theta-42=0$ $\Rightarrow \sin 3 \theta+2 \cos 2 \theta-2=0$ $\Rightarrow 3 \sin \theta-4 \sin ^3 \theta+2(1-2 \sin ^2 \theta)-2=0$ $\Rightarrow 3 \sin \theta-4 \sin ^3 \theta-4 \sin ^2 \theta=0$ $\Rightarrow 4 \sin ^3 \theta+4 \sin ^2 \theta-3 \sin \theta=0$ $\Rightarrow \sin \theta(4 \sin ^2 \theta+4 \sin \theta-3)=0$ $\therefore \sin \theta=0$ $\Rightarrow \theta=\pi, 2 \pi, 3 \pi$ when $\theta \in (0,4 \pi)$ or, $4 \sin { }^2 \theta+4 \sin \theta-3=0$ $\Rightarrow 4 \sin 2 \theta+6 \sin \theta-2 \sin \theta-3=0$ $\Rightarrow 2 \sin \theta(2 \sin \theta+3)-1(2 \sin \theta+3)=0$ $\Rightarrow (2 \sin \theta-1)(2 \sin \theta+3)=0$ $\therefore \sin \theta=\;\frac{1}{2}$ or, $\sin \theta=-\;\frac{3}{2}$ [not possible as $\sin \in [-1,1]$ ] $\therefore \sin \theta=\;\frac{1}{2}$ $\Rightarrow \theta=\;\frac{\pi}{6}, \;\frac{5 \pi}{6}, \;\frac{13 \pi}{6}, \;\frac{17 \pi}{6}$ $\therefore$ Possible values of $\theta=\pi, 2 \pi, 3 \pi, \;\frac{\pi}{6}, \;\frac{5 \pi}{6}, \;\frac{13 \pi}{6}, \;\frac{17 \pi}{6}$ $\therefore$ Total 7 values of $\theta$ possible.

Q65JEE Main 2022MCQ4MLimits, Continuity and Differentiability

If $\lim _{n \to \infty }(\sqrt{n^{2}-n-1}+n \alpha+\beta)=0$ , then $8(\alpha+\beta)$ is equal to: [25-Jul-2022-Shift-1]

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📖 Explanation

$\lim _{n \to \alpha}(\sqrt{n^{2-}n-1}+n \alpha+\beta)=0$ [ This limit will be zero when $\alpha< 0$ as when $\alpha>0$ then overall limit will be $\infty$ .] $\Rightarrow \lim _{n \to \alpha} \;\frac{(\sqrt{n^{2-}n-1}+n \alpha+\beta)(\sqrt{n^{2-}n-1}-(n \alpha+\beta))}{\sqrt{n^{2-}n-1}-(n \alpha+\beta)}=0$ $\Rightarrow \lim _{n \to \alpha} \;\frac{(n^{2-}n-1)-(n \alpha+\beta)^2}{\sqrt{n^{2-}n-1}-(n \alpha+\beta)}=0$ $\Rightarrow \lim _{n \to \alpha} \;\frac{n^{2-}n-1-n^2 \alpha^{2-}2 n \alpha \beta-\beta^2}{\sqrt{n^{2-}n-1}-(n \alpha+\beta)}=0$ $\Rightarrow \lim _{n \to \alpha} \;\frac{n^2(1-\alpha^2)-n(1+2 \alpha \beta)-(1+\beta^2)}{\sqrt{n^{2-}n-1}-(n \alpha+\beta)}$ Here power of " $n$ " in the numerator is 2 and power of " $n$ " in the denominator is 1.To get the value of limit equal to zero power of " $n$ " should be equal in both numerator and denominator, otherwise value of limit will be infinite $(\infty )$ . $\therefore$ Coefficient of $n^2$ should be 0 in this case. $\therefore 1-\alpha^2=0$ $\Rightarrow \alpha=\pm 1$ But $\alpha$ should be $< 0$ $\therefore \alpha=+1 \;\text{ not possible }\;$ $\therefore \alpha=-1$ $\Rightarrow \lim _{n \to \alpha} \;{0-n(1+2 \alpha \beta)-(1+\beta)}/{n\$ [\sqrt{1-;\frac{1}{n}-;\frac{1}{n^2}}-alpha-;\frac{beta}{n}]$}=0 $Divide numerator and denominator by$ {\text{n}} $then we get,$ \Rightarrow \lim _{n \to \alpha} ;{-(1+2 \alpha \beta)-\frac{(1+\beta)}{n}}/{\sqrt{1-;\frac{1}{n}-;\frac{1}{n^2}}-\alpha-;\frac{\beta}{n}}=0\Rightarrow ;\frac{-(1+2 \alpha \beta)-0}{\sqrt{1-0-0}-\alpha-0}=0\Rightarrow ;\frac{-(1+2 \alpha \beta)}{1-\alpha}=0\Rightarrow -(1+2 \alpha \beta)=0\Rightarrow 1+2 \alpha \beta=0\Rightarrow 2 \alpha \beta=-1\Rightarrow \beta=-;\frac{1}{2 \alpha}=-;\frac{1}{2(-1)}=;\frac{1}{2}\therefore 8(\alpha+\beta)=8(-1+;\frac{1}{2})=8 \times -;\frac{1}{2}=-4$

Q66JEE Main 2022MCQ4MApplication of Derivatives

If the absolute maximum value of the function $f(x)=(x^{2}-2 x+7) {e}^{(4 x^{3}-12 x^{2}-180 x+31)}$ in the interval $[-3,0]$ is $f(\alpha)$ , then : [25-Jul-2022-Shift-1]

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📖 Explanation

Given, $f(x)={(x^{2-}2 x+7)}_{}_{╰───🡫───╯} _{f_1(x)} {e^{(4 x^{3-}12 x^{2-}180 x+31)}}_{}_{╰─────🡫──────╯} _{f_2(x)}$ $f_1(x)=x^{2-}2 x+7$ $f_1^{'}(x)=2 x-2$ , so $f(x)$ is decreasing in $[-3,0]$ and positive also $f_2(x)=e^{4 x^{3-}12 x^{2-}180 x+31}$ $f_2^{'}(x)=e^{4 x^{3-}12 x^{2-}180 x+31} \cdot 12 x^{2-}24 x-180$ $=12(x-5)(x+3) e^{4 x^{3-}12 x^{2-}180 x+31}$ So, $f_2(x)$ is also decreasing and positive in $\{-3,0\}$ $\therefore$ absolute maximum value of $f(x)$ occurs at $x=-3$

Q67JEE Main 2022MCQ4MApplication of Derivatives

The curve $y(x)=a x^{3}+b x^{2}+c x+5$ touches the $x$ -axis at the point ${P}(-2,0)$ and cuts the $y$ axis at the point $Q$ , where $y^{'}$ is equal to 3 . Then the local maximum value of $y(x)$ is: [25-Jul-2022-Shift-1]

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📖 Explanation

$f(x)=y=a x^{3+}b x^{2+}c x+5 \;\; . . . \;\text{ (i) }\;$ $\;\frac{d y}{d x}=3 a x^{2+}2 b x+c \;\; . . .(i i)$ Touches $x$ -axis at $P(-2,0)$ $.\Rightarrow y|_{x=-2}=0 \Rightarrow -8 a+4 b-2 c+5=0 \;\; . . . . . . \;\text{ (iii) }\;$ Touches $x$ -axis at $P(-2,0)$ also implies $.\;\frac{d y}{d x}|_{x=-2}=0 \Rightarrow 12 a-4 b+c=0 \;\; . . . . . .(i v)$ $y=f(x)$ cuts $y$ -axis at $(0,5)$ Given, $.\;\frac{d y}{d x}|_{x=0}=c=3$ From (iii), (iv) and (v) $a=-\;\frac{1}{2}, b=-\;\frac{3}{4}, c=3$ $\Rightarrow f(x)=\;\frac{-x^2}{2}-\;\frac{3}{4} x^{2+}3 x+5$ $f^{'}(x)=\;\frac{-3}{2} x^{2-}\;\frac{3}{2} x+3$ $=\;\frac{-3}{2}(x+2)(x-1)$ $f^{'}(x)=0$ at $x=-2$ and $x=1$ By first derivative test $x=1$ in point of local maximum Hence local maximum value of $f(x)$ is $f(1)$ i.e., $\;\frac{27}{4}$

Q68JEE Main 2022MCQ4MArea Under The Curves

The area of the region given by $A=\{(x, y): x^{2} \le y \le \min \{x+2,4-3 x\}\}$ is : [25-Jul-2022-Shift-1]

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📖 Explanation

$A=\{(x, y): x^2 \le y \le \min \{x+2,4-3 x\}.$ So area of required region $A=int_{-1}^{\;\frac{1}{2}}(x+2-x^2) d x+int_{\;\frac{1}{2}}^1(4-3 x-x^2) d x$ $=\$ [;\frac{x^2}{2}+2 x-;\frac{x^3}{3}]{-1}^{;\frac{1}{2}}+$[4 x-;\frac{3 x^2}{2}-;\frac{x^3}{3}]{;\frac{1}{2}}^1=(;\frac{1}{8}+1-;\frac{1}{24})-(;\frac{1}{2}-2+;\frac{1}{3})+(4-;\frac{3}{2}-;\frac{1}{3})-(2-;\frac{3}{8}-;\frac{1}{24})=;\frac{17}{6}$

Q69JEE Main 2022MCQ4MDefinite Integrals

For any real number $x$ , let $[x]$ denote the largest integer less than equal to $x$ . Let $f$ be a real valued function defined on the interval $[-10,10]$ by $f(x)={\begin{cases}x-[x]{,} & \\ \\ \text{ if }\\ \\ [x] \\ \\ \text{ is odd }\\ \\ \\ 1+[x]-x{,} & \\ \\ \text{ if }\\ \\ [x]\\ \\ \text{ is even }\\ \\ }\end{cases}$ Then the value of $\;\frac{\pi^2}{10} int_{-10}^{10} f(x) \cos \pi x \text{dt}$ is : [25-Jul-2022-Shift-1]

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📖 Explanation

Case 1:Let $0 \le x< 1$ then $[x]=0$ , which is even $\therefore f(x)=1+[x]-x$ $=1+0-x$ $=1-x$ Case 2 : Let $1 \le x < 2$ then $[x]=1$ , which is odd $\therefore f(x)=x-[x]$ $=x-1$ Case 3 : Let $2 \le x < 3$ then $[x]=2$ , which is even $\therefore f(x)=1+[x]-x$ $=1+2-x$ $=3-x$ Case 4 : Let $3 \le x < 4$ then $[x]=3$ , which is odd $\therefore f(x)=x-[x]$ $=x-3$ $\therefore f(x)=\begin{cases}1-x & 0 \le x < 1 \\ x-1 & 1 \le x< 2 \\ 3-x & 2 \le x< 3 \\ x-3 & 3 \le x< 4\end{cases}$ $\therefore f(x)$ is periodic and period of $f(x)=2$ And period of $\cos \pi x=\;\frac{2 \pi}{\pi}=2$ $\therefore$ Period of $f(x) \cos \pi x=2$ Now, $I=\;\frac{\pi^2}{10} int_{-10}^{10} f(x) \cos \pi x d x$ $=\;\frac{\pi^2}{10} int_{-10}^{-10+10 \times 2} f(x) \cos \pi x d x$ $=\;\frac{\pi^2}{10} int_0^{10 \times 2} f(x) \cos \pi x d x$ $=\;\frac{\pi^2}{10} \times 10 int_0^2 f(x) \cos \pi x d x$ $=\pi^2 int_0^2 f(x) \cos \pi x d x$ $\therefore I=\pi^2[int_0^1 f(x) \cos \pi x d x+int_1^2 f(x) \cos \pi x d x]$ $=\pi^2[int_0^1(1-x) \cos \pi x d x+int_1^2(x-1) \cos \pi x d x]$ $=\pi^2[int_0^1 \cos \pi x d x-int_0^1 x \cos \pi x d x+int_1^2 x \cos \pi x d x-int_1^2 \cos \pi x d x]$ $=\pi^2[\;\frac{1}{\pi}[\sin \pi x]_0^{1-}int_0^1 x \cos \pi x d x+int_1^2 x \cos \pi x d x-\;\frac{1}{\pi}[\sin \pi x]_1^2]$ $=\pi^2[0-int_0^1 x \cos \pi x d x+int_1^2 x \cos \pi x d x-0]$ $=\pi^2[-\$ [x ;\frac{\sin pi x}{pi}+;\frac{1}{pi^2} \cos pi x]_0^{1+}$[x ;\frac{\sin \pi x}{\pi}+;\frac{1}{\pi^2} \cos \pi x]_1^2] $[As$ .\int x \cos \pi x d x=x \cdot \int \cos \pi x-\int(1 . ;\frac{\sin \pi x}{\pi}) d x=x \cdot ;\frac{\sin \pi x}{\pi}+;\frac{1}{\pi^2} \cos \pi x+c]=\pi^2[-$[(1 ⋅ ;\frac{\sin pi}{pi}+;\frac{1}{pi^2} ⋅ \cos pi)-(0+;\frac{1}{pi^2} ⋅ \cos 0)]$+$[(2 ⋅ ;\frac{\sin 2 pi}{pi}+;\frac{1}{pi^2} \cos 2 pi)-(1.;\frac{\sin pi}{pi}+;\frac{1}{pi^2} \cos pi)]$]=\pi^2$[-{(-;\frac{1}{pi^2})-(;\frac{1}{pi^2})}+((+;\frac{1}{pi^2})-(-;\frac{1}{pi^2})]$.=\pi^2[-(-;\frac{2}{\pi^2})+;\frac{2}{\pi^2}]=\pi^2[;\frac{2}{\pi^2}+;\frac{2}{\pi^2}]=\pi^2 \times ;\frac{4}{\pi^2}=4$

Q70JEE Main 2022MCQ4MDifferential Equations

The slope of the tangent to a curve $C: y=y(x)$ at any point $(x, y)$ on it is $\;\frac{2 {e}^{2 x}-6 {e}^{-x}+9}{2+9 {e}^{-2 x}}$ . If $C$ passes through the points $(0, \;\frac{1}{2}+\;\frac{\pi}{2 \sqrt{2}})$ and $(\alpha, \;\frac{1}{2} {e}^{2 \alpha})$ , then ${e}^{\alpha}$ is equal to : [25-Jul-2022-Shift-1]

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📖 Explanation

$\;\frac{d y}{d x}=\;\frac{2 e^{2 x}-6 e^{-x}+9}{2+9 e^{-2 x}}=e^{2 x}-\;\frac{6 e^{-x}}{2+9 e^{-2 x}}$ $\int d y=\int e^{2 x} d x-3 \int {1+(\;\frac{3 e^{-x}}{\sqrt{2}})^2} d x ({\;\text{put }\; e^{-x}=t})$ $=\;\frac{e^{2 x}}{2}+3 \int \;{d t}/{1+(\;\frac{3 t}{\sqrt{2}})^2}$ $=\;\frac{e^{2 x}}{2}+\sqrt{2} \tan ^{-1} \;\frac{3 t}{\sqrt{2}}+C$ $y=\;\frac{e^{2 x}}{2}+\sqrt{2} \tan ^{-1}(\;\frac{3 e^{-x}}{\sqrt{2}})+C$ It is given that the curve passes through $(0, \;\frac{1}{2}+\;\frac{\pi}{2 \sqrt{2}})$ $\;\frac{1}{2}+\;\frac{\pi}{2 \sqrt{2}}=\;\frac{1}{2}+\sqrt{2} \tan ^{-1}(\;\frac{3}{\sqrt{2}})+C$ $\Rightarrow \;\; C=\;\frac{\pi}{2 \sqrt{2}}-\sqrt{2} \tan ^{-1}(\;\frac{3}{\sqrt{2}})$ Now if $(\alpha, \;\frac{1}{2} e^{2 \alpha})$ satisfies the curve, then $\;\frac{1}{2} e^{2 \alpha}=\;\frac{e^{2 \alpha}}{2}+\sqrt{2} \tan ^{-1}(\;\frac{3 e^{-\alpha}}{\sqrt{2}})+\;\frac{\pi}{2 \sqrt{2}}-\sqrt{2} \tan ^{-1}(\;\frac{3}{\sqrt{2}})$ $\tan ^{-1}(\;\frac{3}{\sqrt{2}})-\tan ^{-1}(\;\frac{3 e^{-\alpha}}{\sqrt{2}})=\;\frac{\pi}{2 \sqrt{2}} \times \;\frac{1}{\sqrt{2}}=\;\frac{\pi}{4}$ $\;{\;\frac{3}{\sqrt{2}}-\frac{3 e^{-\alpha}}{\sqrt{2}}}/{1+\;\frac{9}{2} e^{-\alpha}}=1$ $\;\frac{3}{\sqrt{2}} e^\alpha-\;\frac{3}{\sqrt{2}}=e^\alpha+\;\frac{9}{2}$ $e^\alpha=\;{\;\frac{9}{2}+\frac{3}{\sqrt{2}}}/{\frac{3}{\sqrt{2}}-1}=\;\frac{3}{\sqrt{2}}(\;\frac{3+\sqrt{2}}{3-\sqrt{2}})$

Q71JEE Main 2022MCQ4MDifferential Equations

The general solution of the differential equation $(x-y^{2}) {d} x+y(5 x+y^{2}) {d} y=0$ is: [25-Jul-2022-Shift-1]

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📖 Explanation

$(x-y^2) d x+y(5 x+y^2) d y=0$ $y \;\frac{d y}{d x}=\;\frac{y^{2-}x}{5 x+y^2}$ Let $y^2=t \;\frac{1}{2} \cdot \;\frac{d t}{d x}=\;\frac{t-x}{5 x+t}$ Now substitute, $t=v x$ $\;\frac{d t}{d x}=v+x \;\frac{d v}{d x}$ $\;\frac{1}{2}\{v+x \;\frac{d v}{d x}\}=\;\frac{v-1}{5+v}$ $x \;\frac{d v}{d x}=\;\frac{2 v-2}{5+v}-v=\;\frac{-3 v-v^{2-}2}{5+v}$ $\int \;\frac{5+v}{v^{2+}3 v+2} d v=\int-\;\frac{d x}{x}$ $\int \;\frac{4}{v+1} d v-\int \;\frac{3}{v+2} d v=-\int \;\frac{d x}{x}$ $4 \ln |v+1|-3 \ln |v+2|=-\ln x+\ln C$ $|\;\frac{(v+1)^4}{(v+2)^3}|=\;\frac{c}{x}$ $|{(\frac{y^2}{x}+1) ^4} / { (\;\frac{y^2}{x}+2)^3 }=\;\frac{c}{x}|$ $|(y^{2+}x)^4|=C|(y^{2+}2 x)^3|$

Q72JEE Main 2022MCQ4MStraight Lines and Pair of Straight Lines

A line, with the slope greater than one, passes through the point $A(4,3)$ and intersects the line $x-y-2=0$ at the point ${B}$ . If the length of the line segment $A B$ is $\;\frac{\sqrt{29}}{3}$ , then $B$ also lies on the line: [25-Jul-2022-Shift-1]

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📖 Explanation

Let inclination of required line is $\theta$ ,So the coordinates of point $B$ can be assumed as $(4-\;\frac{\sqrt{29}}{3} \cos \theta, 3-\;\frac{\sqrt{29}}{3} \sin \theta)$ Which satisfices $x-y-2=0$ $4-\;\frac{\sqrt{29}}{3} \cos \theta-3+\;\frac{\sqrt{29}}{3} \sin \theta-2=0$ $\sin \theta-\cos \theta=\;\frac{3}{\sqrt{29}}$ By squaring $\sin 2 \theta=\;\frac{20}{29}=\;\frac{2 \tan \theta}{1+\tan ^2 \theta}$ $\tan \theta=\;\frac{5}{2}$ only (because slope is greater than 1 ) $\sin \theta=\;\frac{5}{\sqrt{29}}, \cos \theta=\;\frac{2}{\sqrt{29}}$ Point $B:(\;\frac{10}{3}, \;\frac{4}{3})$ Which also satisfies $x+2 y=6$

Q73JEE Main 2022MCQ4MParabola

Let the locus of the centre $(\alpha, \beta), \beta>0$ , of the circle which touches the circle $x^{2}+(y-1)^{2}=1$ externally and also touches the $x$ -axis be ${L}$ . Then the area bounded by ${L}$ and the line $y=4$ is: [25-Jul-2022-Shift-1]

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📖 Explanation

Radius of circle $S$ touching $x$ -axis and centre $(\alpha, \beta)$ is $|\beta|$ . According to given conditions $\alpha^{2+}(\beta-1)^2=(|\beta|+1)^2$ $\alpha^{2+}\beta^{2-}2 \beta+1=\beta^{2+}1+2|\beta|$ $\alpha^2=4 \beta \;\text{ as }\; \beta>0$ $\therefore \;\;$ Required louse is $L: x^2=4 y$ The area of shaded region $=2 int_0^4 2 \sqrt{y} d y$ $=4 .\$ [;{y^{\frac{3}{2}}}/{\frac{3}{2}}]_0^4=;\frac{64}{3}$ square units.

Q74JEE Main 2022MCQ4MThree Dimensional Geometry

Let ${P}$ be the plane containing the straight line $\;\frac{x-3}{9}=\;\frac{y+4}{-1}=\;\frac{z-7}{-5}$ and perpendicular to the plane containing the straight lines $\;\frac{x}{2}=\;\frac{y}{3}=\;\frac{z}{5}$ and $\;\frac{x}{3}=\;\frac{y}{7}=\;\frac{z}{8}$ . If ${d}$ is the distance of ${P}$ from the point $(2,-5,11)$ , then ${d}^{2}$ is equal to : [25-Jul-2022-Shift-1]

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📖 Explanation

Let $⟨ a, b, c⟩$ be direction ratios of plane containing ${\text{lines}} \;\frac{x}{2}=\;\frac{y}{3}=\;\frac{z}{5}$ and $\;\frac{x}{3}=\;\frac{y}{7}=\;\frac{z}{8}$ . $\therefore \;\; 2 a+3 b+5 c=0 \;\; . . . \;\text{ (i) }\;$ and $3 a+7 b+8 c=0 \;\; . . .$ (ii)from eq. (i) and (ii) : $\;\frac{a}{24-35}=\;\frac{b}{15-16}=\;\frac{c}{14-9}$ $\therefore \;\;$ D. $R^S$ . of plane are $< 11,1,-5>$ Let $D \cdot R^S$ of plane $P$ be $< a_1, b_1, c_1>$ then. $11 a_1+b_1-5 c_1=0$ and $9 a_1-b_1-5 c_1=0$ From eq. (iii) and (iv) : $\;\frac{a_1}{-5-5}=\;\frac{b_1}{-45+55}=\;\frac{c_1}{-11-9}$ $\therefore \;\;$ D. A ${ }^5$ . of plane $P$ are $< 1,-1,2>$ Equation plane $P$ is : $1(x-3)-1(y+4)+2(z-7)=0$ $\Rightarrow x-y+2 z-21=0$ Distance from point $(2,-5,11)$ is $d=\;\frac{|2+5+22-2|}{\sqrt{6}}$ $\therefore d^2=\;\frac{32}{3}$

Q75JEE Main 2022MCQ4MVector Algebra

Let ${ABC}$ be a triangle such that ${ \vec{BC}={ \vec{a}, { \vec{CA}={ \vec{b}, { \vec{AB}={ \vec{c}, { {a} \mid }=6 \sqrt{2}, \overrightarrow{\mid \vec{b} \mid =2 \sqrt{3}$ and $\vec{b} \cdot \vec{c}=12$ . Consider the statements : $( {S} 1): \mid ({ \vec{a} \times { \vec{b})+{( \vec{c} \times { \vec{b})|-| \vec{c} \mid =6(2 \sqrt{2}-1)$ $( {S} 2): ∠ {ACB}=\cos ^{-1}(\sqrt{\;\frac{2}{3}})$ Then [25-Jul-2022-Shift-1]

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📖 Explanation

$\because \vec{a}+\vec{b}+\vec{c}=0$ then $\overline{a}+\vec{c}=-\vec{b}$ then $(\vec{a}+\vec{c}) \times \vec{b}=-\vec{b} \times \overline{b}$ $\therefore \;\; \vec{a} \times \vec{b}+\vec{c} \times \vec{b}=\overline{0} \;\; . . .(i)$ For $(S 1):|\vec{a} \times \vec{b}+\vec{c} \times \vec{b}|-|\vec{c}|=6(2 \sqrt{2}-1)$ $|(\vec{a}+\vec{c}) \times \vec{b}|-|\vec{c}|=6(2 \sqrt{2}-1)$ $|\vec{c}|=6-12 \sqrt{2}($ not possible $)$ For (S2) : from (i) $\vec{b}+\vec{c}=-\vec{a}$ $\Rightarrow \vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{b}=-\vec{a} \cdot \vec{b}$ $\Rightarrow \;\; 12+12=-6 \sqrt{2} \cdot 2 \sqrt{3} \cos (\pi-∠ A C B)$ $\therefore \;\; \cos (∠ A C B)=\sqrt{\;\frac{2}{3}}$ $\therefore \;\; ∠ A C B=\cos ^{-1} \sqrt{\;\frac{2}{3}}$ $\therefore \;\; S(2)$ is correct.

Q76JEE Main 2022MCQ4MProbability

If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is: [25-Jul-2022-Shift-1]

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📖 Explanation

If $n$ is number of trails, $p$ is probability of success and $q$ is probability of unsuccess then,Mean $=n p$ and variance $=n p q$ .Here $n p+n p q=24 \;\; . . .(i)$ $n p . n p q=128 \;\; . . .$ (ii)and $q=1-p \;\; . . .(i i i)$ from eq. (i), (ii) and (iii) : $p=q=\;\frac{1}{2}$ and $n=32$ . $\therefore$ Required probability $=p(X=1)+p(X=2)$ $={ }^{32} C_1 \cdot (\;\frac{1}{2})^{32}+{ }^{32} C_2 \cdot (\;\frac{1}{2})^{32}$ $=(32+\;\frac{32 \times 31}{2}) \cdot \;\frac{1}{2^{32}}$ $=\;\frac{33}{2^{28}}$

Q77JEE Main 2022MCQ4MProbability

If the numbers appeared on the two throws of a fair six faced die are $\alpha$ and $\beta$ , then the probability that $x^{2}+\alpha x+\beta>0$ , for all $x \in R$ , is : [25-Jul-2022-Shift-1]

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📖 Explanation

For $x^{2+}\alpha x+\beta>0 ∀ x \in R$ to hold, we should have $\alpha^{2-}4 \beta< 0$ If $\alpha=1, \beta$ can be $1,2,3,4,5$ , 6 i.e., 6 choicesIf $\alpha=2, \beta$ can be $2,3,4,5,6$ i.e., 5 choicesIf $\alpha=3, \beta$ can be $3,4,5,6$ i.e., 4 choicesIf $\alpha=4, \beta$ can be 5 or 6 i.e., 2 choicesIf $\alpha=6$ , No possible value for $\beta$ i.e., 0 choicesHence total favourable outcomes $=6+5+4+2+0+0$ $=17$ Total possible choices for $\alpha$ and $\beta=6 \times 6=36$ Required probability $=\;\frac{17}{36}$

Q78JEE Main 2022MCQ4MTrigonometric Ratios and Identities

The number of solutions of $|\cos x|=\sin x$ , such that $-4 \pi \le x \le 4 \pi$ is : [25-Jul-2022-Shift-1]

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📖 Explanation

Period of $|\cos x|=\pi$ And period of $\sin x=2 \pi$ Graph of $\sin x$ and $|\cos x|$ cuts each other at two points ${\text{A}}$ and ${\text{B}}$ in $[0,2 \pi]$ So, in $[-4 \pi, 4 \pi]$ , total 4 similar graph will be present and graph of $\sin x$ and $|\cos x|$ will cut $4 \times 2=8$ times. $\therefore$ Total possible solutions $=8$

Q79JEE Main 2022MCQ4MTrigonometric Ratios and Identities

A tower PQ stands on a horizontal ground with base $Q$ on the ground. The point $R$ divides the tower in two parts such that $Q R=15 {m}$ . If from a point $A$ on the ground the angle of elevation of $R$ is $60^{\circ}$ and the part $P R$ of the tower subtends an angle of $15^{\circ}$ at $A$ , then the height of the tower is : [25-Jul-2022-Shift-1]

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📖 Explanation

For $\triangle {\text{AQR}}$ , $\tan 60^{\circ}=\;\frac{15}{x} . . . . . .(1$ From $\triangle {\text{AQP}}$ , $\tan 75^{\circ}=\;\frac{h}{x}$ $\Rightarrow (2+\sqrt{3})=\;\frac{h}{x}[ \because \tan 75^{\circ}=2+\sqrt{3}]$ $\Rightarrow h=(2+\sqrt{3}) x$ $=(2+\sqrt{3}) \;\frac{15}{\sqrt{3}}[\;\text{ From }\;(1)]$ $=(2+\sqrt{3}) \times \;\frac{15 \sqrt{3}}{3}$ $=(2+\sqrt{3}) \times 5 \sqrt{3}$ $=5(2 \sqrt{3}+3) {\text{m}}$

Q80JEE Main 2022MCQ4MSets and Relations

Which of the following statements is a tautology? [25-Jul-2022-Shift-1]

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📖 Explanation

A statement is a tautology if it is always true.
Consider option D: $q \Rightarrow ((∼ p) ∨ q)$ .
Using the logical equivalence $A \Rightarrow B \equiv ∼A ∨ B$ , we can rewrite the statement:
$q \Rightarrow ((∼ p) ∨ q) \equiv ∼q ∨ ((∼ p) ∨ q)$
Apply the associative property of disjunction:
$∼q ∨ ((∼ p) ∨ q) \equiv (∼q ∨ q) ∨ (∼ p)$
We know that $∼q ∨ q$ is always true (T):
$(∼q ∨ q) ∨ (∼ p) \equiv T ∨ (∼ p)$
Since $T ∨ X$ is always true for any proposition $X$ :
$T ∨ (∼ p) \equiv T$
Thus, $q \Rightarrow ((∼ p) ∨ q)$ is a tautology.

Q81JEE Main 2022NAT4MMatrices and Determinants

Let $A=(\text{table} 2 , -1 , -1 ; 1 , 0 , -1 ; 1 , -1 , 0)$ and $B=A-I$ . If $\omega =\;\frac{\sqrt{3} i-1}{2}$ , then the number of elements in the set $\{n \in \{1,2, . . ., 100\}: A^{n}+(\omega B)^{n}=A+B\}$ is equal to [25-Jul-2022-Shift-1]

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📖 Explanation

Here $A=(\text{table} 2 , -1 , -1 ; 1 , 0 , -1 ; 1 , -1 , 0)$ We get $A^2=A$ and similarly for $B=A-I=\begin{bmatrix}1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1\end{bmatrix}$ We get $B^2=-B \Rightarrow B^3=B$ $\therefore A^n+(\omega B)^n=A+(\omega B)^n \;\; \;\text{ for }\; n \in {\text{N}}$ For $\omega ^n$ to be unity $n$ shall be multiple of 3 and for $B^n$ to be $B$ . $n$ shell be $3,5,7, . . . 99$ $\therefore n=\{3,9,15, . . . . .99\}$ Number of elements $=17$

Q82JEE Main 2022NAT4MPermutations and Combinations

The letters of the word 'MANKIND' are written in all possible orders and arranged in serial order as in an English dictionary. Then the serial number of the word 'MANKIND' is [25-Jul-2022-Shift-1]

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📖 Explanation

MANKIND $(\;\frac{4 \times 6 !}{2 !})+(5 ! \times 0)+(\;\frac{4 ! \times 3}{2 !})+(3 ! \times 2)+(2 ! \times 1)+(1 ! \times 1)+(0 ! \times 0)+1=1492$

Q83JEE Main 2022NAT4MBinomial Theorem

If the maximum value of the term independent of $t$ in the expansion of $( {t}^{2} x^{\;\frac{1}{5}}+\;{(1-x)^{\frac{1}{10}}}/{ {t}})^{15}, x \ge slant 0$ , is ${K}$ , then $8 {K}$ is equal to [25-Jul-2022-Shift-1]

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📖 Explanation

General term of $(t^2 x^{\;\frac{1}{5}}+\;{(1-x)^{\frac{1}{10}}}/{t})^{15}$ is $T_{r+1}={ }^{15} C_r \cdot (t^2 x^{\;\frac{1}{5}})^{15-r} \cdot (\;{(1-x)^{\frac{1}{10}}}/{t})^r$ $={ }^{15} C_r \cdot t^{30-2 r} \cdot x^{\;\frac{15-r}{5}} \cdot (1-x)^{\;\frac{r}{10}} \cdot t^{-r}$ $={ }^{15} C_r \cdot t^{30-3 r} \cdot x^{\;\frac{15-r}{5}} \cdot (1-x)^{\;\frac{r}{10}}$ Term will be independent of t when $30-3 r=0 \Rightarrow r=10$ $\therefore T_{10+1}=T_{11}$ will be independent of ${\text{t}}$ $\therefore T_{11}={ }^{15} C_{10} \cdot x^{\;\frac{15-10}{5}} \cdot (1-x)^{\;\frac{10}{10}}$ $={ }^{15} C_{10} \cdot x^1 \cdot (1-x)^1$ ${\text{T}}_{11}$ will be maximum when $x(1-x)$ is maximum.Let $f(x)=x(1-x)=x-x^2$ $f(x)$ is maximum or minimum when $f^{'}(x)=0$ $\therefore f^{'}(x)=1-2 x$ For maximum $/$ minimum $f^{'}(x)=0$ $\therefore 1-2 x=0$ $\Rightarrow x=\;\frac{1}{2}$ Now, $f^{' '}(x)=-2< 0$ $\therefore$ At $x=\;\frac{1}{2}, f(x)$ maximum $\therefore$ Maximum value of $T_{11}$ is $={ }^{15} C_{10} \cdot \;\frac{1}{2}(1-\;\frac{1}{2})$ $={ }^{15} C_{10} \cdot \;\frac{1}{4}$ Given $K={ }^{15} C_{10} \cdot \;\frac{1}{4}$ Now, $8 K=2({ }^{15} C_{10})$ $=6006$

Q84JEE Main 2022NAT4MQuadratic Equation and Inequalities

Let $a, b$ be two non-zero real numbers. If $p$ and $r$ are the roots of the equation $x^{2}-8 {a} x+2 {a}=0$ and $q$ and $s$ are the roots of the equation $x^{2}+12 {b} x+6 {b}=0$ , such that $\;\frac{1}{ {p}}, \;\frac{1}{ {q}}, \;\frac{1}{ {r}}, \;\frac{1}{ {s}}$ are in A.P., then ${a}^{-1}- {b}^{-1}$ is equal to [25-Jul-2022-Shift-1]

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📖 Explanation

$\because$ Roots of $2 a x^{2-}8 a x+1=0$ are $\;\frac{1}{p}$ and $\;\frac{1}{r}$ and roots of $6 b x^{2+}12 b x+1=0$ are $\;\frac{1}{q}$ and $\;\frac{1}{s}$ Let $\;\frac{1}{p}, \;\frac{1}{q}, \;\frac{1}{r}, \;\frac{1}{s}$ as $\alpha-3 \beta, \alpha-\beta, \alpha+\beta, \alpha+3 \beta$ So sum of roots $2 \alpha-2 \beta=4$ and $2 \alpha+2 \beta=-2$ Clearly $\alpha=\;\frac{1}{2}$ and $\beta=-\;\frac{3}{2}$ Now product of roots, $\;\frac{1}{p} \cdot \;\frac{1}{r}=\;\frac{1}{2 a}=-5 \Rightarrow \;\frac{1}{a}=-10$ and $\;\frac{1}{q} \cdot \;\frac{1}{x}=\;\frac{1}{6 b}=-8 \Rightarrow \;\frac{1}{b}=-48$ So, $\;\frac{1}{a}-\;\frac{1}{b}=38$

Q85JEE Main 2022NAT4MSequences and Series

Let $a_{1}=b_{1}=1, a_{n}=a_{n-1}+2$ and $b_{n}=a_{n}+b_{n-1}$ for every natural number $n \ge slant 2$ . Then $sum_{n=1}^{15} a_{n} \cdot b_{n}$ is equal to [25-Jul-2022-Shift-1]

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📖 Explanation

Given, $a_n=a_{n-1}+2$ $\Rightarrow a_n-a_{n-1}=2$ $\therefore$ In this series between any two consecutives terms difference is 2 . So this is an A.P. with common difference 2.Also given $a_1=1$ $\therefore$ Series is $=1,3,5,7 . . . . . .$ $\therefore a_n=1+(n-1) 2=2 n-1$ Also $b_n=a_n+b_{n-1}$ When $n=2$ then $b_2-b_1=a_2=3$ $.\Rightarrow b_2-1=3 \;\text{ [Given }\; b_1=1]$ $\Rightarrow b_2=4$ When $n=3$ then $b_3-b_2=a_3$ $\Rightarrow b_3-4=5$ $\Rightarrow b_3=9$ $\therefore$ Series is $=1,4,9 . . . . . .$ $=1^2, 2^2, 3^2 . . . . . . n^2$ $\therefore b_n=n^2$ Now, $sum_{n=1}^{15}(a_n . b_n)$ $=sum_{n=1}^{15}[(2 n-1) n^2]$ $=sum_{n=1}^{15} 2 n^{3-}sum_{n=1}^{15} n^2$ $=2(1^{3+}2^{3+}. . . 15^3)-(1^{2+}2^{2+}. . . 15^2)$ $=2 \times (\;\frac{15 \times 16}{2})^{2-}(\;\frac{15(16) \times 31}{6})$ $=27560$

Q86JEE Main 2022NAT4MLimits, Continuity and Differentiability

Let $f(x)={\begin{cases}{|4 x^{2-}8 x+5 \mid }{,}\text{if} \ \\ 8 x^{2-}6 x+1 \ge 0 \\ {[4 x^{2-}8 x+5]\$ , \text{if} \ \ 8 x^{2-}6 x+1< 0}} $, where$ [alpha] $denotes the greatest integer less than or equal to$ alpha $\end{cases} Then the number of points \in$ R $where$ f$ is not differentiable is [25-Jul-2022-Shift-1]

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📖 Explanation

$f(x)=\begin{cases}\text{|} 4 x^{2-}8 x+5\text{|} & \ \\ \text{ if }\ \\ 8 x^{2-}6 x+1 \ge 0 \\ {[4 x^{2-}8 x+5]\$ ,} & \ \ \text{ if }\ \ 8 x^{2-}6 x+1 < 0\end{cases}=\begin{cases}4 x^{2-}8 x+5 & \ \ \text{ if }\ \ x ∈[-∞ & \ \ \frac{1}{4}] ∪[\ \ \frac{1}{2} & ∞ ) \ {[4 x^{2-}8 x+5]$} & \ \ \text{ if }\ \ x \in (\ \ \frac{1}{4} & \ \ \frac{1}{2})\end{cases}f(x)=\begin{cases}4 x^{2-}8 x+5 & \ \ \text{ if }\ \ x \in (-\infty & \ \ \frac{1}{4}] ∪[\ \ \frac{1}{2} & \infty ) \ 3 & x \in (\ \ \frac{1}{4} & \ \ \frac{2-\sqrt{2}}{2}) \ 2 & x \in [\ \ \frac{2-\sqrt{2}}{2} & \ \ \frac{1}{2})\end{cases}\therefore ;; $Non-diff at$ x=;\frac{1}{4}, ;\frac{2-\sqrt{2}}{2}, ;\frac{1}{2}$

Q87JEE Main 2022NAT4MDefinite Integrals

If $\lim _{n \to \infty } \;\frac{(n+1)^{k-1}}{n^{k+1}}[(n k+1)+(n k+2)+. . .+(n k+n)]$ , then the integral value of ${k}$ is $=33 \cdot \lim _{n \to \infty } \;\frac{1}{n^{k+1}} \cdot [1^{k}+2^{k}+3^{k}+. . .+n^{k}]$ equal to [25-Jul-2022-Shift-1]

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📖 Explanation

$\lim _{n \to \infty }(\;\frac{n+1}{n})^{k-1} \;\frac{1}{n} sum_{r=1}^n(k+\;\frac{r}{n})=33 \lim _{n \to \infty } \;\frac{1}{n} sum_{k=1}^n(\;\frac{r}{n})^k$ $\Rightarrow int_0^1(k+x) d x=33 int_0^1 x^k d x$ $\Rightarrow \;\; \;\frac{2 k+1}{2}=\;\frac{33}{k+1}$ $\Rightarrow \;\; k=5$

Q88JEE Main 2022NAT4MHyperbola

Let the equation of two diameters of a circle $x^{2}+y^{2}-2 x+2 f y+1=0$ be $2 p x-y=1$ and $2 x+p y=4 p$ . Then the slope ${m} \in (0, \infty )$ of the tangent to the hyperbola $3 x^{2}-y^{2}=3$ passing through the centre of the circle is equal to [25-Jul-2022-Shift-1]

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📖 Explanation

$2 p+f-1=0 \;\; . . .(1)$ $2-p f-4 p=0 \;\; . . .(2)$ $2=p(f+4)$ $p=\;\frac{2}{f+4}$ $2 p=1-f$ $\;\frac{4}{f+4}=1-f$ $f^{2+}3 f=0$ $f=0 \;\text{ or }\;-3$ Hyperbola $3 x^{2-}y^2=3, x^{2-}\;\frac{y^2}{3}=1$ ${\text{y}}= {\text{mx}} \pm \sqrt{ {\text{m}}^{2-}3}$ It passes $(1,0)$ $o=m \pm \sqrt{m^{2-}3}$ $m$ tends $\infty$ It passes $(1,3)$ $3=m \pm \sqrt{m^{2-}3}$ $(3-m)^2=m^{2-}3$ $m=2$

Q89JEE Main 2022NAT4MCircles

The sum of diameters of the circles that touch (i) the parabola $75 x^{2}=64(5 y-3)$ at the point $(\;\frac{8}{5}, \;\frac{6}{5})$ and (ii) the $y$ -axis, is equal to [25-Jul-2022-Shift-1]

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📖 Explanation

Equation of tangent to the parabola at $P(\;\frac{8}{5}, \;\frac{6}{5})$ $75 x \cdot \;\frac{8}{5}=160(y+\;\frac{6}{5})-192$ $\Rightarrow 120 x=160 y$ $\Rightarrow 3 x=4 y$ Equation of circle touching the given parabola at ${\text{P}}$ can be taken as $(x-\;\frac{8}{5})^{2+}(y-\;\frac{6}{5})^{2+}\lambda (3 x-4 y)=0$ If this circle touches $y$ -axis then $\;\frac{64}{25}+(y-\;\frac{6}{5})^{2+}\lambda (-4 y)=0$ $\Rightarrow y^{2-}2 y(2 \lambda +\;\frac{6}{5})+4=0$ $\Rightarrow D=0$ $\Rightarrow (2 \lambda +\;\frac{6}{6})^2=4$ $\Rightarrow \lambda =\;\frac{2}{5} \;\text{ or }\;-\;\frac{8}{5}$ Radius $=1$ or 4Sum of diameter $=10$

Q90JEE Main 2022NAT4MThree Dimensional Geometry

The line of shortest distance between the lines $\;\frac{x-2}{0}=\;\frac{y-1}{1}=\;\frac{z}{1}$ and $\;\frac{x-3}{2}=\;\frac{y-5}{2}=\;\frac{z-1}{1}$ makes an angle of $\cos ^{-1}(\sqrt{\;\frac{2}{27}})$ with the plane ${P}: {a} x-y-z=0,(a>0)$ . If the image of the point $(1,1,-5)$ in the plane $P$ is $(\alpha, \beta, \gamma )$ , then $\alpha+\beta-\gamma$ is equal to [25-Jul-2022-Shift-1]

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📖 Explanation

DR's of line of shortest distance $\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 2 & 2 & 1\end{vmatrix}=-\hat{i}+2 \hat{j}-2 \hat{k}$ angle between line and plane is $\cos ^{-1} \sqrt{\;\frac{2}{27}}=\alpha$ $\cos \alpha=\sqrt{\;\frac{2}{27}}, \sin \alpha=\;\frac{5}{3 \sqrt{3}}$ DR's normal to plane $(1,-1,-1)$ $\sin \alpha=|\;\frac{-a-2+2}{\sqrt{4+4+1} \sqrt{a^{2+}1+1}}|=\;\frac{5}{3 \sqrt{3}}$ DR's normal to plane $(1,-1,-1)$ $\sin \alpha=|\;\frac{-a-2+2}{\sqrt{4+4+1} \sqrt{a^{2+}1+1}}|$ $=\;\frac{5}{3 \sqrt{3}}$ $\sqrt{3}|a|$ $=5 \sqrt{a^{2+}2}$ $3 a^2=25 a^{2+}50$ No value of (a) (No such plane exists)

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