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28 Jul 2022 Shift 1

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Physics30 questions

Q1JEE Main 2022MCQ4MUnits and Measurements

The dimensions of $(\;{ {\text{B}}^2}/{\mu _0})$ will be : (if $\mu _0$ : permeability of free space and $B$ : magnetic field) [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${[\;\frac{B^2}{\mu _0}]\$ =} $[Energy density]$ =;\frac{M L^2 T^{-2}}{L^3}=M L^{-1} T^{-2}$

Q2JEE Main 2022MCQ4MMotion in a Plane

A NCC parade is going at a uniform speed of $9 {\text{km}} / {\text{h}}$ under a mango tree on which a monkey is sitting at a height of $19.6 {\text{m}}$ . At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is: (Given $g-9.8 {\text{m}} / {\text{s}}^2$ ) [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$H=\;\frac{1}{2} g t^2$ $19.6=4.9 t^2$ $t=2 {\ sec}$ $D=9 \times \;\frac{5}{18} \times 2=5 {\text{m}}$

Q3JEE Main 2022MCQ4MCentre of Mass and Collision

In two different experiments, an object of mass $5 {\text{kg}}$ moving with a speed of $25 {\text{ms}}^{-1}$ hits two different walls and comes to rest within (i) 3 second, (ii) 5 seconds, respectively. Choose the correct option out of the following : [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Impulse $=$ change in momentum ${\text{I}}=\Delta {\text{P}}$ ${\text{F}}_{\;\text{aug }\;}=\;{\Delta {\text{P}}}/{\Delta {\text{t}}}$ $\Delta {\text{t}}_1=3 \;\; \Delta {\text{t}}_2=5$ $\Delta {\text{P}}_1=\Delta {\text{P}}_2$ ${\text{I}}_1= {\text{I}}_2$ ${\text{F}}_{\;\text{avg }\;}$ in case (i) is more than (ii)

Q4JEE Main 2022MCQ4MLaws of Motion

A balloon has mass of $10 {\text{g}}$ in air. The air escapes from the balloon at a uniform rate with velocity $4.5 {\text{cm}} / {\text{s}}$ . If the balloon shrinks in $5 {\text{s}}$ completely. Then, the average force acting on that balloon will be (in dyne). [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{F}}=\;{ {\text{dm}}}/{ {\text{dt}}} {\text{v}}$ $=\;{10 {\text{g}}}/{5 {\ s}}(4.5 \;{ {\text{cm}}}/{ {\ s}})=9 \;{ {\text{gcm}}}/{ {\ s}^2}=9$ dyne

Q5JEE Main 2022MCQ4MGravitation

If the radius of earth shrinks by $2 \%$ while its mass remains same. The acceleration due to gravity on the earth's surface will approximately : [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$g=\;{ {\text{GM}}}/{ {\text{R}}^2}$ ${\text{M}}=\;\text{ constant }\; {\text{g}}< \;{1}/{ {\text{R}}^2}$ $100 \;{\Delta {\text{g}}}/{ {\text{g}}}=-2 \;{\Delta {\text{R}}}/{ {\text{R}}} 100$ $\%$ change $=-2(-2)$ $\%$ change in ${\text{g}}=4 \%$ increase by $4 \%$

Q6JEE Main 2022MCQ4MProperties of Matter

The force required to stretch a wire of cross-section $1 {\text{cm}}^2$ to double its length will be : (Given Yong's modulus of the wire $-2 \times 10^{11} {\text{N}} / {\text{m}}^2$ ) [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{F}}=\gamma {\text{A}} \;\frac{\Delta ℓ}{ℓ}$ $=2 \times 10^{11} \times 10^{-4}(\;\frac{2 ℓ-ℓ}{ℓ})$ $=2 \times 10^7 {\text{N}}$

Q7JEE Main 2022MCQ4MThermodynamics

A Carnot engine has efficiency of $50 \%$ . If the temperature of sink is reduced by $400^{\circ} {\text{C}}$ , its efficiency increase by $30 \%$ . The temperature of the source will be: [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$η=1-\;{ {\text{T}}_{ {\text{L}}}}/{ {\text{T}}_{ {\text{H}}}}$ $\;\frac{1}{2}=1-\;{ {\text{T}}_{ {\text{L}}}}/{ {\text{T}}_{ {\text{H}}}}$ $\;\frac{1}{2}(1 \cdot 3)=1-(\;{ {\text{T}}_{ {\text{L}}}-40}/{ {\text{T}}_{ {\text{H}}}})$ $\;\frac{1}{2}(1 \cdot 3)=\;\frac{1}{2}+\;{40}/{ {\text{T}}_{ {\text{H}}}} \;\; {\text{T}}_{ {\text{H}}}=266.7 {\text{K}}$

Q8JEE Main 2022MCQ4MThermodynamics

Statement I : The average momentum of a molecule in a sample of an ideal gas depends on temperature. Statement II : The rms speed of oxygen molecules in a gas is $v$ . If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become $2 v$ . In the light of the above statements, choose the correct answer from the options given below: [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${[ {\text{P}}_{\;\text{avg }\;}=0]}$ (due to random motion) ${\text{v}}_{\;\text{rms }\;}=\sqrt{\;{3 {\text{RT}}}/{ {\text{M}}}}$ ${\text{T}}_{\;\text{new }\;}=2 {\text{T}}$ ${\text{M}}_{\;\text{new }\;}=\;{ {\text{M}}}/{2}$ $\;{ {\text{v}}_{\;\text{new }\;}}/{ {\text{v}}}=\;{\sqrt{{2 {\text{T}}}/{ {\text{M}} / 2}}}/{\sqrt{\;{ {\text{T}}}/{ {\text{M}}}}}$ ${\text{v}}_{\;\text{new }\;}=2 {\text{v}}$

Q9JEE Main 2022MCQ4MWaves

In the wave equation $y-0.5 \sin \;\frac{2 \pi}{\lambda }(400 {\text{t}}-x) {\text{m}}$ the velocity of the wave will be: [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{y}}=0.5 \ sin (\;\frac{2 \pi}{\lambda } 400 {\text{t}}-\;\frac{2 \pi}{\lambda } {\text{x}})$ $\omega =\;\frac{2 \pi}{\lambda } 400$ ${\text{K}}=\;\frac{2 \pi}{\lambda }$ ${\text{v}}=\;{\omega }/{ {\text{k}}} \;\; [ {\text{v}}=400 {\text{m}} / {\ s}]$

Q10JEE Main 2022MCQ4MElectrostatic Potential and Capacitance

Two capacitors, each having capacitance $40 \mu {\text{F}}$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant ${\text{K}}$ such that the equivalence capacitance of the system became $24 \mu {\text{F}}$ . The value of ${\text{K}}$ will be : [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{C}}_{\;\text{eq }\;}=\;{ {\text{C}}( {\text{KC}})}/{ {\text{C}}+ {\text{KC}}}=\;{ {\text{KC}}}/{ {\text{K}}+1}$ $24=\;{ {\text{K}} 40}/{ {\text{K}}+1}$ ${[ {\text{K}}=1 \cdot 5] }$

Q11JEE Main 2022MCQ4MCurrent Electricity

A wire of resistance ${\text{R}}_1$ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is: [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{R}}_1=\rho \;{ {\text{L}}_1}/{ {\text{A}}_1}$ ${\text{R}}_2=\rho (\;{3 {\text{L}}_1}/{ {\text{A}}_1 / 3})=9 \rho \;{ {\text{L}}_1}/{ {\text{A}}_1}$ $\therefore \;{ {\text{R}}_2}/{ {\text{R}}_1}=9$

Q12JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

The current sensitivity of a galvanometer can be increased by : (A) decreasing the number of turns (B) increasing the magnetic freld (C) decreasing the area of the coil (D) decreasing the torsional constant of the spring Choose the most appropriate answer from the options given below : [28-Jul-2022-Shift-1]

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📖 Explanation

$N i A B=k \theta$ $\Rightarrow \;\frac{\theta}{i}=\;\frac{N A B}{k}$ $\Rightarrow$ Sensitivity increases if $B ↑$ and $k ↓$

Q13JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

As shown in the figure, a metallic rod of linear density $0.45 {\text{kg}} {\text{m}}^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of $45^{\circ}$ with the horizontal. The minimum current flowing in the rod required to keep it stationary, when $0.15 {\text{T}}$ magnetic field is acting on it in the vertical upward direction, will be : $\{.$ Use $.g=10 {\text{m}} / {\text{s}}^2\}$ [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$m g \times \;\frac{1}{\sqrt{2}}=\;\frac{i l B}{\sqrt{2}}$ $\Rightarrow i=\;\frac{m g}{B l}$ $=\;\frac{0.45 \times 10}{0.15}=30 {\text{A}}$

Q14JEE Main 2022MCQ4MAlternating Current

The equation of current in a purely inductive circuit is $5 \sin (49 \pi t-30^{\circ})$ . If the inductance is $30 {\text{mH}}$ then the equation for the voltage across the inductor, will be : $\{.$ Let $.\pi=\;\frac{22}{7}\}$ [28-Jul-2022-Shift-1]

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📖 Explanation

$V(t)=I \omega L \ sin (49 \pi t-30^{\circ}+90^{\circ})$ $=5 \times 49 \pi \times \;\frac{30}{1000} \ sin (49 \pi t+60^{\circ})$ $=23.1 \ sin (49 \pi t+60^{\circ})$

Q15JEE Main 2022MCQ4MElectromagnetic Waves

As shown in the figure, after passing through the medium 1 . The speed of light $v_2$ in medium 2 will be: $(.$ Given $. {\text{c}}=3 \times 10^8 {\text{ms}}^{-1})$ [28-Jul-2022-Shift-1]

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📖 Explanation

$V=\;\frac{1}{\sqrt{\mu \epsilon}}=\;\frac{1}{\sqrt{\mu _r \epsilon r \mu _0 \epsilon_0}}$ $\Rightarrow V_2=\;\frac{c}{\sqrt{9}}=10^8 {\text{m}} / {\ s}$

Q16JEE Main 2022MCQ4MRay Optics and Optical Instruments

In normal adujstment, for a refracting telescope, the distance between objective and eye piece is $30 {\text{cm}}$ . The focal length of the objective, when the angular magnification of the telescope is 2 , will be [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\because m=\;\frac{f_o}{f_e}$ $\Rightarrow 2=\;\frac{f_o}{f_e} . . . . . .$ (i)and, $l=f_o+f_e$ $\Rightarrow 30=f_o+f_e . . . .$ . (ii) $\Rightarrow 30=f_o+\;\frac{f_o}{2}$ $\Rightarrow 30 \times \;\frac{2}{3}=f_o$ $\Rightarrow f_o=20 {\text{cm}}$

Q17JEE Main 2022MCQ4MDual Nature of Matter and Radiation

The equation $\lambda =\;\frac{1.227}{x} {\text{nm}}$ can be used to find the de-Brogli wavelength of an electron. In this equation $x$ stands for: Where ${\text{m}}=$ mass of electron ${\text{P}}-$ momentum of electron ${\text{K}}=$ Kinetic energy of electron ${\text{V}}=$ Accelerating potential in volts for electron [28-Jul-2022-Shift-1]

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📖 Explanation

$\because \lambda =\;\frac{1.227}{\sqrt{V}} {\text{nm}}$ $\Rightarrow x=\sqrt{V}$

Q18JEE Main 2022MCQ4MAtoms and Nuclei

The half life period of a radioactive substance is 60 days. The time taken for $\;\frac{7}{8}$ th of its original mass to disintegrate will be : [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\because N=\;{N_0}/{2^{\;\frac{t}{T_{1 / 2}}}}$ $\Rightarrow 2^{\;\frac{t}{T_{1 / 2}}}=\;\frac{N_0}{N}=\;\frac{N_0}{(\;\frac{N_0}{8})}$ $\Rightarrow 2^{\;\frac{t}{T_{1 / 2}}}=8=2^3$ $\Rightarrow t=3 \times T_{1 / 2}=3 \times 60$ $=180$ days

Q19JEE Main 2022MCQ4MSemiconductor Electronics

Identify the solar cell characteristics from the following options : [28-Jul-2022-Shift-1]

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Q20JEE Main 2022MCQ4MElectromagnetic Waves

In the case of amplitude modulation to avoid distortion the modulation index $(\mu )$ should be: [28-Jul-2022-Shift-1]

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📖 Explanation

$\mu =\;\frac{A_m}{A_c}$ $\mu \le 1$ to avoid distortionbecause $\mu >1$ will result in interference between career frequency & message frequency.

Q21JEE Main 2022NAT4MWork, Power and Energy

If the projection of $2 \hat{i}+4 \hat{j}-2 \hat{k}$ on $\hat{i}+2 \hat{j}+\alpha \hat{k}$ is zero. Then, the value of $\alpha$ will be _______. [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${ \vec{\text{a}} \cdot { \vec{\text{b}}=0$ $\therefore { \vec{\text{a}} \cdot { \vec{\text{b}}=0$ $\therefore 2 \times 1+4 \times 2-2 \times \alpha=0$ $\therefore \alpha=5$

Q22JEE Main 2022NAT4MAtoms and Nuclei

A freshly prepared radioactive source of half life 2 hours 30 minutes emits radiation which is 64 times the permissible safe level. The minimum time, after which it would be possible to work safely with source, will be _______ hours. [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$A=A_0 \times 2^{-t / T}$ $\;\frac{A_0}{64}=A_0 \times 2^{-t / T}$ $\therefore \;\; t=6 T=6 \times 2 \cdot 5=15$ hours

Q23JEE Main 2022NAT4MWave Optics

In a Young's double slit experiment, a laser light of $560 {\text{nm}}$ produces an interference pattern with consecutive bright fringes' separation of $7.2 {\text{mm}}$ . Now another light is used to produce an interference pattern with consecutive bright fringes' separation of $8.1 {\text{mm}}$ . The wavelength of second light is ________ ${\text{nm}}$ . [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\beta \propto \lambda$ $\lambda _2=\;\frac{9}{8} \lambda _1$ $\therefore \beta_2=\;\frac{9}{8} \beta_1=\;\frac{9}{8} \times 560=630 {\text{nm}}$ .

Q24JEE Main 2022NAT4MAlternating Current

The frequencies at which the current amplitude in an LCR series circuit becomes $\;\frac{1}{\sqrt{2}}$ times its maximum value, are $212 {\text{rad}} {\text{s}}^{-1}$ and $232 {\text{rad}} {\text{s}}^{-1}$ . The value of resistance in the circuit is $R=5 Ω$ . The self inductance in the circuit is _______ ${\text{mH}}$ . [28-Jul-2022-Shift-1]

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📖 Explanation

Band width $=232-212=\;{ {\text{R}}}/{ {\text{L}}}$ $\therefore {\text{L}}=\;\frac{5}{20}=250 {\text{mH}}$

Q25JEE Main 2022NAT4MCurrent Electricity

As shown in the figure, a potentiometer wire of resistance $20 Ω$ and length $300 {\text{cm}}$ is connected with resistance box (R.B.) and a standard cell of emf 4 V. For a resistance ' $R$ ' of resistance box introduced into the circuit, the null point for a cell of $20 {\text{mV}}$ is found to be $60 {\text{cm}}$ . The value of ' $R$ ' _______ is $Ω$ . [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{E}}=\;{ {\text{AC}}}/{ {\text{AB}}}( {\text{V}}_{ {\text{A}}}- {\text{V}}_{ {\text{B}}})$ $\therefore 20 \times 10^{-3}=\;\frac{60}{300} \times \;{4 \times 20}/{ {\text{R}}+20}$ $\therefore {\text{R}}=780$

Q26JEE Main 2022NAT4MElectrostatics

Two electric dipoles of dipole moments $1.2 \times 10^{-30} {\text{Cm}}$ and $2.4 \times 10^{-30} {\text{Cm}}$ are placed in two different uniform electric fields of strengths $5 \times 10^4 {\text{NC}}^{-1}$ and $15 \times 10^4 {\text{NC}}^{-1}$ respectively. The ratio of maximum torque experienced by the electric dipoles will be $\;\frac{1}{x}$ . The value of $x$ is _______. [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$|\tau |_{\max }= {\text{PE}}$ $\;\frac{\tau _1}{\tau _2}=\;{ {\text{P}}_1 {\text{E}}_1}/{ {\text{P}}_2 {\text{E}}_2}=\;\frac{1 \cdot 2 \times 10^{-30} \times 5 \times 10^4}{2 \cdot 4 \times 10^{-30} \times 15 \times 10^4}=\;\frac{1}{6}$ Hence $x=6$

Q27JEE Main 2022NAT4MWaves

The frequency of echo will be _______ ${\text{Hz}}$ if the train blowing a whistle of frequency $320 {\text{Hz}}$ is moving with a velocity of $36 {\text{km}} / {\text{h}}$ towards a hill from which an echo is heard by the train driver. Velocity of sound in air is $330 {\text{m}} / {\text{s}}$ . [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

The hill will be a secondary source ${\text{f}}_1=$ frequency of the car w.r.t. the hill ${\text{f}}_1=(\;{ {\text{v}}}/{ {\text{v}}- {\text{v}}_{ {\ s}}}) {\text{f}}=(\;\frac{330}{320}) \times 320=330 {\text{Hz}}$ ${\text{f}}_2=$ Frequency of the sound reflected by hill w.r.t. the car (echo) ${\text{f}}_2=(\;{ {\text{v}}+ {\text{v}}_0}/{ {\text{v}}}) {\text{f}}_1=\;\frac{(330+10)}{330} \times 330=340 {\text{Hz}}$

Q28JEE Main 2022NAT4MProperties of Matter

The diameter of an air bubble which was initially $2 {\text{mm}}$ , rises steadily through a solution of density $1750 {\text{kg}} {\text{m}}^{-3}$ at the rate of $0.35 {\text{cms}}^{-1}$ . The coefficient of viscosity of the solution is _______ poise (in nearest integer). (the density of air is negligible). [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

As the bubble is rising steadily the net force acting on it will be zero (Because of density of air the value of ${\text{mg}}$ can be neglected)So $B=F \Rightarrow \;\frac{4 \pi}{3} R^3 \rho g=6 \pi η R v$ Putting $\;\; {\text{R}}=1 {\text{mm}}=10^{-3} {\text{m}}$ ${c} \rho =1.75 \times 10^3 {\text{kg}} / {\text{m}}^3$ ${\text{g}}=10 {\text{m}} / {\ s}^2$ ${\text{v}}=0.35 \times 10^{-2} {\text{m}} / {\ s}$ $η=\;\frac{10}{9} ≃ 1 \cdot 11 \;\text{ SI unit }\;=11 \;\text{ poise }\;( {\text{CGS}})$

Q29JEE Main 2022NAT4MWork, Power and Energy

A block of mass 'Im' (as shown in figure) moving with kinetic energy E compresses a spring through a distance $25 {\text{cm}}$ when, its speed is halved. The value of spring constant of used spring will be ${\text{nE}} {\text{Nm}}{ }^{-1}$ for ${\text{n}}=$ _______. [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Using work - energy theorem ${\text{W}}_{ {\text{net}}}=( {\text{K}}_{ {\text{f}}}- {\text{K}}_{ {\text{i}}})$ $\Rightarrow -\;\frac{1}{2} {\text{Kx}}^2=\;\frac{1}{2} {\text{m}}(\;{ {\text{v}}}/{2})^{2-}\;\frac{1}{2} {\ mv}^2=\;{ {\text{E}}}/{4}- {\text{E}}$ $\Rightarrow \;\frac{1}{2} {\text{Kx}}^2=\;{3 {\text{E}}}/{4} \Rightarrow {\text{K}}=\;{3 {\text{E}}}/{2 {\text{x}}^2}$ $\Rightarrow {\text{K}}=\;{3 {\text{E}}}/{2 \times (\;\frac{1}{4})^2}=24 {\text{E}}$ ${\text{n}}=24$

Q30JEE Main 2022NAT4MRotational Motion

Four identical discs each of mass ' $M$ ' and diameter 'a' are arranged in a small plane as shown in figure. If the moment of inertia of the system about ${\text{OO}}^{'}$ is $\;\frac{x}{4} {\text{Ma}}^2$ . Then, the value of $x$ will be ________. [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{I}}_1= {\text{I}}_3=\;{ {\text{MR}}^2}/{4}$ ${\text{I}}_2=\;{ {\text{MR}}^2}/{4}+ {\text{MR}}^2=\;\frac{5}{4} {\text{MR}}^2= {\text{I}}_4$ So ${\text{I}}= {\text{I}}_1+ {\text{I}}_2+ {\text{I}}_3+ {\text{I}}_4$ $=\;{ {\text{MR}}^2}/{2}+\;\frac{5}{2} {\text{MR}}^2$ $=3 {\text{MR}}^2, \;\text{ Putting }\; {\text{R}}=\;{ {\text{a}}}/{2}$ ${\text{I}}=\;{3 {\text{Ma}}^2}/{4}, \;\text{ So }\; {\text{x}}=3$

Chemistry30 questions

Q31JEE Main 2022MCQ4MStructure of Atom

Identify the incorrect statement from the following. [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Atomic orbital is characterised by the quantum numbers ${\text{n}}, l$ and ${\text{m}}$ .Hence option D is incorrect.

Q32JEE Main 2022MCQ4MChemical Thermodynamics

Which of the following relation is not correct? [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

If ${\text{U}}+ {\text{Pv}}$ (By definition) $\Delta 14=\Delta {\text{U}}+\Delta ({\Pr})$ at constant pressure $\Delta {\text{H}}=\Delta {\text{U}}+ {\text{P}} \Delta {\text{V}}$

Q33JEE Main 2022MCQ4MElectrochemistry

Match List - I with List - II. List - I List - II (A) ${\text{Cd}}(s)+2 {\text{Ni}}( {\text{OH}})_3(s) \to {\text{CdO}}(s)+2 {\text{Ni}}( {\text{OH}})_2(s)+ {\text{H}}_2 {\text{O}}( {\text{l}})$ (l) Primary battery (B) $Z n(H g)+H g O(s) \to Z n O(s)+H g(l)$ (II) Discharging of secondary battery (C) $2 {\text{PbSO}}_4(s)+2 {\text{H}}_2 {\text{O}}(l) \to {\text{Pb}}(s)+ {\text{PbO}}_2(s)+2 {\text{H}}_2 {\text{SO}}_4(a q)$ (III) Fuel cell (D) $2 {\text{H}}_2(g)+ {\text{O}}_2(g) \to 2 {\text{H}}_2 {\text{O}}(l)$ (IV) Charging of secondary batteryChoose the correct answer from the options given below: [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

(a) ${\text{Cd}}( {\ s})+2 {\text{Ni}}( {\text{OH}})_3( {\ s}) \to {\text{CdO}}( {\ s})+2 {\text{Ni}}( {\text{OH}})_2( {\ s})+ {\text{H}}_2 {\text{O}}(l)$ Discharge of secondary Battery(b) ${\text{Zn}}( {\text{Hg}})+ {\text{HgO}}( {\ s}) \to {\text{ZnO}}( {\ s})+ {\text{Hg}}(l)$ (Primary Battery Mercury cell)(c) $2 {\text{PbSO}}_4( {\ s})+2 {\text{H}}_2 {\text{O}}(l) \to {\text{Pb}}( {\ s})+ {\text{PbO}}_2( {\ s})+2 {\text{H}}_2 {\text{SO}}_4( {\text{aq}})$ (Charging of secondary Battery)(d) $2 {\text{H}}_2( {\text{g}})+ {\text{O}}_2( {\text{g}}) \to 2 {\text{H}}_2 {\text{O}}(l)$ (Fuel cell)

Q34JEE Main 2022MCQ4MChemical Kinetics

Choose the correct answer from the options given below: List - I Reaction (A) List - II Catalyst (A) $4 {\text{NH}}_3(g)+5 {\text{O}}_2(g) \to 4 {\text{NO}}(g)+6 {\text{H}}_2 {\text{O}}(g)$ (I) $N O(g)$ (B) ${\text{N}}_2(g)+3 H_2(g) \to 2 N_3(g)$ (II) ${\text{H}}_2 {\text{SO}}_4(l)$ (C) ${\text{C}}_{12} {\text{H}}_{22} {\text{O}}_{11}(a q)+ {\text{H}}_2 {\text{O}}(l) \to {{ {\text{C}}}\;_6 {\text{H}}_{12} {\text{O}}_6}_{\text{Glucose}}+\;{ {\text{C}}}{_6 {\text{H}}_{12} {\text{O}}_6}_{\text{Fructose}}$ (III) ${\text{Pt}}(s)$ (D) $2 {\text{SO}}_2(g)+ {\text{O}}_2(g) \to 2 {\text{SO}}_3(g)$ (IV) $F e(s)$ [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

(A) $4 {\text{NH}}_3( {\text{g}})+5 {\text{O}}_2( {\text{g}}) {\longrightarrow }^{ {\text{Pt}}( {\ s})} 4 {\text{NO}}( {\text{g}})+6 {\text{H}}_2 {\text{O}}( {\text{g}})$ (B) ${\text{N}}_2( {\text{g}})+3 {\text{H}}_2( {\text{g}}) {\longrightarrow }^{ {\text{Fe}}( {\ s})} 2 {\text{NH}}_3( {\text{g}})$ (C) ${\text{C}}_{12} {\text{H}}_{22} {\text{O}}_{11}( {\text{aq}})+ {\text{H}}_2 {\text{O}}( {\text{l}}) {\longrightarrow }^{ {\text{H}}_2 {\text{SO}}_4(1)} \;{ {\text{C}}_6 {\text{H}}_{12} {\text{O}}_6}_{\;\text{ Glucose }\;}+\;{ {\text{C}}_6 {\text{H}}_{12} {\text{O}}_6}_{\;\text{ Fructose }\;}$ (D) $2 {\text{SO}}_2( {\text{g}})+ {\text{O}}_2( {\text{g}}) {\longrightarrow }^{ {\text{NO}}( {\text{g}})} 2 {\text{SO}}_3$

Q35JEE Main 2022MCQ4MClassification of Elements

In which of the following pairs, electron gain enthalpies of constituent elements are nearly the same or identical? (A) ${\text{Rb}}$ and ${\text{Cs}}$ (B) Na and K (C) Ar and ${\text{Kr}}$ (D) I and At Choose the correct answer from the options given below: [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{Rb}} \& {\text{Cs}}$ have nearly same electron gain enthalpy electron gain enthalpy $=-46 {\text{kj}} / {\text{ml}}$ Ar & Kr have same $\Delta {\text{H}}_{ {\text{eq}}}$ . Value is $+96 {\text{kj}} / {\text{ml}}$

Q36JEE Main 2022MCQ4Mp Block Elements

Which of the reaction is suitable for concentrating ore by leaching process? [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Leaching involves the treatment of ore with a suitable reagent so as it make it soluble while impurities remain insoluble. ${\text{Al}}_2 {\text{O}}_3+2 {\text{NaOH}}+3 {\text{H}}_2 {\text{O}} \to 2 {\text{Na}}[ {\text{Al}}( {\text{OH}})_4]$ Soluble complex

Q37JEE Main 2022MCQ4MIonic Equilibrium

The metal salts formed during softening of hardwater using Clark's method are : [28-Jul-2022-Shift-1]

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📖 Explanation

In Clark's method, calculated amount of lime is added to hard water. It precipitates out calcium carbonate and magnesium hydroxide which can filtered off. ${\text{Ca}}( {\text{HCO}}_3)_2+ {\text{Ca}}( {\text{OH}})_2 \to 2 {\text{CaCO}}_3 ↓+2 {\text{H}}_2 {\text{O}}$ ${\text{Mg}}( {\text{HCO}}_3)_2+2 {\text{Ca}}( {\text{OH}})_2 \to 2 {\text{CaCO}}_3 ↓+ {\text{Mg}}( {\text{OH}})_2+2 {\text{H}}_2 {\text{O}}$

Q38JEE Main 2022MCQ4MClassification of Elements

Which of the following statement is incorrect? [28-Jul-2022-Shift-1]

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📖 Explanation

Low solubility of LiF in water is due to the fact that though ${\text{Li}}^{+}$ is having high hydration enthalpy but it has higher lattice enthalpy when present in LiF. Due to higher lattice enthalpy its solubility is less.

Q39JEE Main 2022MCQ4Mp Block Elements

Match List - I with List - II, match the gas evolved during each reaction. List - I List - II (A) $( {\text{NH}}_4)_2 {\text{Cr}}_2 {\text{O}}_7 {\longrightarrow }^{\Delta }$ (I) ${\text{H}}_2$ (B) ${\text{KMnO}}_4+ {\text{HCl}} \to$ (II) ${\text{N}}_2$ (C) ${\text{Al}}+ {\text{NaOH}}+ {\text{H}}_2 {\text{O}} \to$ (III) ${\text{O}}_2$ (D) ${\text{NaNO}}_3 {\longrightarrow }^{\Delta }$ (IV) ${\text{Cl}}_2$ Choose the correct answer from the options given below: [28-Jul-2022-Shift-1]

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📖 Explanation

$( {\text{NH}}_4)_2 {\text{Cr}}_2 {\text{O}}_7 {\longrightarrow }^{\Delta } {\text{N}}_2+4 {\text{H}}_2 {\text{O}}+ {\text{Cr}}_2 {\text{O}}_3$ ${\text{KMnO}}_4+ {\text{HCl}} \longrightarrow {\text{KCl}}+ {\text{MnCl}}_2+ {\text{Cl}}_2+ {\text{H}}_2 {\text{O}}$ ${\text{Al}}+ {\text{NaOH}}+ {\text{H}}_2 {\text{O}} \longrightarrow {\text{Na}}( {\text{Al}}( {\text{OH}})_4)+ {\text{H}}_2$ $2 {\text{NaNO}}_3( {\ s}) {\longrightarrow }^{\Delta } 2 {\text{NaNO}}_2( {\ s})+ {\text{O}}_2$

Q40JEE Main 2022MCQ4MElectrochemistry

Which of the following has least tendency to liberate ${\text{H}}_2$ from mineral acids? [28-Jul-2022-Shift-1]

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📖 Explanation

The metal atom whose oxidation potential is less than that of hydrogen can release ${\text{H}}_2$ from mineral acids. ${\text{E}}_{ {\text{Zn}} / {\text{Zn}}^{+2}}=0.76$ ${\text{E}}_{ {\text{Ni}} / {\text{Ni}}^{\circ 2}}=0.25$ $E_{M n / M n^{+2}}=1.18$ ${\text{E}}_{ {\text{Cu}}^{\circ} / {\text{Cu}}^{+2}}=-0.34$

Q41JEE Main 2022MCQ4MBiomolecules

Given below are two statements: Statement I : In polluted water values of both dissolved oxygen and ${\text{BOD}}$ are very low. Statement II : Eutrophication results in decrease in the amount of dissolved oxygen. In the light of the above statements, choose the most appropriate answer from the options given below: [28-Jul-2022-Shift-1]

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📖 Explanation

Since eutrophication is result of excessive growth of weed in water bodies, which consume dissolved oxygen of water bodies. $\therefore$ Eutrophication decreases amount of dissolved oxygen in water bodies.Polluted water has low value of dissolved oxygen, but high valueof BOD (Biological oxygen demand), since chemical and organic matter requires dissolved oxygen to get decompose.

Q42JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Match List - I with List - II. List - I List - II (l)Spiro compound (II)Aromatic compound (III)Non-planar Heterocyclic compound (IV)Bicyclo compoundChoose the correct answer from the options given below: [28-Jul-2022-Shift-1]

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Q43JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

Choose the correct option for the following reactions. [28-Jul-2022-Shift-1]

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Q44JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Among the following marked proton of which compound shows lowest ${\text{pK}}_{ {\text{a}}}$ value? [28-Jul-2022-Shift-1]

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📖 Explanation

The conjugate base of compound (C) is stabilized by extended conjugation. Hence the indicated proton of compound $C$ is most acidic i.e. will have lowest ${\text{pK}}_{ {\text{a}}}$ .

Q45JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

Identify the major products ${\text{A}}$ and ${\text{B}}$ for the below given reaction sequence. [28-Jul-2022-Shift-1]

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Q46JEE Main 2022MCQ4MAmines

Identify the correct statement for the below given transformation. [28-Jul-2022-Shift-1]

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Q47JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Terylene polymer is obtained by condensation of : [28-Jul-2022-Shift-1]

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Q48JEE Main 2022MCQ4MBiomolecules

For the below given cyclic hemiacetal $( {\text{X}})$ , the correct pyranose structure is : [28-Jul-2022-Shift-1]

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Q49JEE Main 2022MCQ4MBiomolecules

Statements about Enzyme Inhibitor Drugs are given below: (A) There are Competitive and Non-competitive inhibitor drugs. (B) These can bind at the active sites and allosteric sites. (C) Competitive Drugs are allosteric site blocking drugs. (D) Non-competitive Drugs are active site blocking drugs. Choose the correct answer from the options given below: [28-Jul-2022-Shift-1]

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Q50JEE Main 2022MCQ4MChemical Kinetics

For kinetic study of the reaction of iodide ion with ${\text{H}}_2 {\text{O}}_2$ at room temperature : (A) Always use freshly prepared starch solution. (B) Always keep the concentration of sodium thiosulphate solution less than that of KI solution. (C) Record the time immediately after the appearance of blue colour. (D) Record the time immediately before the appearance of blue colour. (E) Always keep the concentration of sodium thiosulphate solution more than that of KI solution. Choose the correct answer from the options given below: [28-Jul-2022-Shift-1]

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📖 Explanation

To minimize contamination, use freshly prepared starch solution to determine end point. As KI is used in excess to consume all the ${\text{H}}_2 {\text{O}}_2$ the concentration of sodium thiosulphate solution is less than KI solution. After appearance of blue colour record the time immediately.

Q51JEE Main 2022NAT4MSome Basic Concepts of Chemistry

In the given reaction, $X+Y+3 Z ⇆ X YZ_3$ if one mole of each of $X$ and $Y$ with $0.05 {\text{mol}}$ of $Z$ gives compound $X Y Z_3$ - (Given : Atomic masses of $X, Y$ and $Z$ are 10, 20 and $30 {\text{amu}}$ , respectively.) The yield of $X Y Z_3$ is _______g. (Nearest integer) [28-Jul-2022-Shift-1]

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📖 Explanation

$\;{X}_{n_{\;\text{moles }\;}=1}+\;{Y}_{1}+\;{3 Z}_{0.05} ⇌ x y z_3$ Limiting reagent is $Z=\;\frac{0.05}{3}=.016$ 3 moles of $Z \to 1$ mole of $X Y Z_3$ $0.05$ mole of $Z \to \;\frac{1}{3} \times 0.05$ mole of $X Y Z_3$ M.wt. of ${\text{XYZ}}_3=10+20+90$ $=120 {\text{amu}}$ Wt. of $X Y Z_3=\;\frac{.05}{3} \times 120$ $=2 {\text{g}}$

Q52JEE Main 2022NAT4MSolid State

An element ${\text{M}}$ crystallises in a body centred cubic unit cell with a cell edge of $300 {\pm}$ . The density of the element is $6.0 {\text{g}} {\text{cm}}^{-3}$ . The number of atoms present in $180 {\text{g}}$ of the element is _________ $\times 10^{23}$ (Nearest integer) [28-Jul-2022-Shift-1]

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📖 Explanation

$M$ is body certred cubic, $\therefore Z=2$ Let mass of 1 atom of $M$ is $A$ Edge length $=300 {\pm}$ Density $=6 {\text{g}} / {\text{cm}}^3$ $\therefore 6 {\text{g}} / {\text{cm}}^3=\;{ {\text{Z}} \times {\text{A}}}/{(300 \times 10^{-10})^3}=\;{2 \times {\text{A}}}/{27 \times 10^{-24}}$ $A=81 \times 10^{-24} {\text{g}}$ $\therefore$ Atomic mass $=48.6 {\text{g}}$ $\therefore$ Mole in $180 {\text{g}}=\;\frac{180}{48.6}=3.7$ molesAtoms of ${\text{M}}=3.7 \times 6 \times 10^{23}=22.22 \times 10^{23}$ atoms

Q53JEE Main 2022NAT4MChemical Bonding and Molecular Structure

The number of paramagnetic species among the following is _______. ${\text{B}}_2, {\text{Li}}_2, {\text{C}}_2, {\text{C}}_2^{-}, {\text{O}}_2^{2-}, {\text{O}}_2^{+} \;\text{and }\; {\text{He}}_2^{+}$ [28-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. $B_2$ has 10 electrons.Molecular orbital configuration of $B_2$ is $\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \pi_{2 p_x^1}=\pi_{2 p_y^1}$ Here two unpaired electrons present. So it is paramagnetic. $O_2^{2-}$ 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. $O_2^{+}$ 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. $C_2$ has 12 electrons.Moleculer orbital configuration of $C_2$ $=\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \pi_{2 p_x^2}=\pi_{2 p_y^2}$ Here no unpaired electron present, so it is diamagnetic. $C_2^{-}$ has 13 electrons. Moleculer orbital configuration of $C_2^{-}$ is $\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \pi_{2 p_x^2}=\pi_{2 p_y^2} \sigma_{2 p_z^1}$ Here 1 unpaired electron present, so it is paramagnetic. ${\text{Li}}_2$ has 6 electrons. ${\text{Li}}_2=\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2}$ Here no unpaired electron present, so it is diamagnetic.Configuration of ${\text{He}}_2^{+}$ (3 electrons) is $=\sigma_{1 s^2} \sigma_{1 s^1}^*$ Here 1 unpaired electron present, so it is paramagnetic.

Q54JEE Main 2022NAT4MSolutions

$150 {\text{g}}$ of acetic acid was contaminated with $10.2 {\text{g}}$ ascorbic acid $( {\text{C}}_6 {\text{H}}_8 {\text{O}}_6)$ to lower down its freezing point by $(x \times 10^{-1})^{\circ} {\text{C}}$ . The value of $x$ is _______. (Nearest integer) [Given ${\text{K}}_f=3.9 {\text{K}} {\text{kg}} {\text{mol}}^{-1}$ ; molar mass of ascorbic acid $.=176 {\text{g}} {\text{mol}}^{-1}]$ [28-Jul-2022-Shift-1]

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📖 Explanation

M.wt. of Acetic acid $=60 {\text{g}}$ M.wt. of Ascorbic acid $=176 {\text{g}}$ $\Delta T_f \;\; =K_f m$ $\Delta T_f \;\; =\;\frac{3.9 \times 10.2 \times 1000}{176 \times 150}$ $\Delta T_f \;\; =1.506$ $\;\; =15.06 \times 10^{-1}$ $\;\; =15$

Q55JEE Main 2022NAT4MIonic Equilibrium

${\text{K}}_{ {\text{a}}}$ for butyric acid $( {\text{C}}_3 {\text{H}}_7 {\text{COOH}})$ is $2 \times 10^{-5}$ . The ${\text{pH}}$ of $0.2 {\text{M}}$ solution of butyric acid is _______ $\times 10^{-1}$ . (Nearest integer) [Given $\log 2=0.30]$ [28-Jul-2022-Shift-1]

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📖 Explanation

${\text{K}}_{ {\text{a}}}$ of Butyric acid $\Rightarrow 2 \times 10^{-5} {\text{PKa}}=4.7$ ${\text{pH}}$ of $0.2 {\text{M}}$ solution ${\text{pH}}=\;\frac{1}{2} {\text{pK}}_{ {\text{a}}}-\;\frac{1}{2} \log {\text{C}}$ $=\;\frac{1}{2}(4 \cdot 7) \;\frac{1}{2} \log (0.2)$ $=2.35+0.35=2.7$ ${\text{pH}}=27 \times 10^{-1}$

Q56JEE Main 2022NAT4MChemical Kinetics

For the given first order reaction ${\text{A}} \to {\text{B}}$ the half life of the reaction is $0.3010 {\min}$ . The ratio of the initial concentration of reactant to the concentration of reactant at time $2.0 {\min}$ will be equal to ________. (Nearest integer) [28-Jul-2022-Shift-1]

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📖 Explanation

${\text{t}}_{1 / 2}=\;{0.693}/{ {\text{K}}} \;\; {\text{t}}_{1 / 2} \;\text{ given }\;=0.3010$ $K=\;\frac{0.693}{0.3010}$ $K=2.30$ ${\text{K}}=\;{2.303}/{ {\text{t}}} \log \;{( {\text{A}}_0)}/{( {\text{A}}_{ {\text{t}}})}$ $A_0 \to$ initial concentration of reactant $A_t \to$ concentration of reactant at time $t$ $2.303=\;\frac{2.303}{2} \log \;\frac{(A_0)}{(A_t)}$ $2=\log \;\frac{(A_0)}{(A_t)}$ $100=\;\frac{A_0}{A_t}$

Q57JEE Main 2022NAT4MChemical Bonding and Molecular Structure

The number of interhalogens from the following having square pyramidal structure is : ${\text{ClF}}_3, {\text{TF}}_7, {\text{BrF}}_5, {\text{BrF}}_3, {\text{I}}_2 {\text{Cl}}_6, {\text{IF}}_5, {\text{ClF}}, {\text{ClF}}_5$ [28-Jul-2022-Shift-1]

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📖 Explanation

${\text{CIF}}_3 \to 3 \sigma$ bond $+2$ lone pair ${\text{IF}}_7 \to 7 \sigma$ bond $+0$ lone pair ${\text{BrF}}_5 \to 5 \sigma$ bond $+1$ lone pair $\to$ Square pyramidal ${\text{BrF}}_3 \to 3 \sigma$ bond $+2$ lone pair ${\text{I}}_2 {\text{Cl}}_6 \to 4 \sigma$ bond $+2$ lone pair ${\text{IF}}_5 \to 5 \sigma$ bond $+1$ lone pair $\to$ Square pyramidal ${\text{CIF}} \to 1 \sigma$ bond $+3$ lone pair ${\text{CIF}}_5 \to 5 \sigma$ bond $+1$ lone pair $\to$ Square pyramidal

Q58JEE Main 2022NAT4Md and f Block Elements

The disproportionation of ${\text{MnO}}_4^{2-}$ in acidic medium resulted in the formation of two manganese compounds $A$ and $B$ . If the oxidation state of $M n$ in $B$ is smaller than that of $A$ , then the spin-only magnetic moment $(\mu )$ value of $B$ in $B M$ is _______.(Nearest integer) [28-Jul-2022-Shift-1]

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📖 Explanation

$3 {\text{MnO}}_4^{-2}+4 {\text{H}}^{+} \to {\text{MnO}}_2+ {\text{MnO}}_4^{-}$ ${\text{Mn}} \to 4 s^2 3 d^5$ ${\text{Mn}}^{+4} \to 3 d^3$ ${\text{n}}=3$ $\mu =\sqrt{n(n+2)}$ $\;\; =\sqrt{3(5)}$ $=\sqrt{15}$ $=3.87 \approx 4$ B.M.

Q59JEE Main 2022NAT4MCoordination Compounds

Total number of relatively more stable isomer(s) possible for octahedral complex $[ {\text{Cu}}( {\text{en}})_2( {\text{SCN}})_2]$ will be _______. [28-Jul-2022-Shift-1]

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📖 Explanation

$[ {\text{Cu}}( {\text{en}})_2( {\text{SCN}})_2]$

Q60JEE Main 2022NAT4MOrganic Chemistry - Some Basic Principles

On complete combustion of $0.492 {\text{g}}$ of an organic compound containing ${\text{C}}, {\text{H}}$ and ${\text{O}}, 0.7938 {\text{g}}$ of ${\text{CO}}_2$ and $0.4428 {\text{g}}$ of ${\text{H}}_2 {\text{O}}$ was produced. The % composition of oxygen in the compound is _______. [28-Jul-2022-Shift-1]

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📖 Explanation

$\%$ of ${\text{H}}=\;\frac{2}{18} \times \;{\;\text{ wt. of }\; {\text{H}}_2 {\text{O}}}/{\;\text{ wt. of organic compound }\;} \times 100$ $=\;\frac{2}{18} \times \;\frac{0.4428}{0.492} \times 100$ $=0.11 \times 0.9 \times 100$ $=0.099 \times 100=9.9$ $\%$ of $C=\;\frac{12}{44} \times \;\frac{0.7938}{0.492} \times 100$ $=0.27 \times 1.61 \times 100$ $=43.47$ $\%$ Oxygen $=100-(43.47+9.9)$ $=100-53.37 ≃ 46$

Mathematics30 questions

Q61JEE Main 2022MCQ4MDifferential Equations

Let the solution curve of the differential equation $x {\text{d}} y=(\sqrt{x^{2+}y^2}+y) {\text{d}} x, x > 0$ , intersect the line $x=1$ at $y=0$ and the line $x=2$ at $y=\alpha$ . Then the value of $\alpha$ is : [28-Jul-2022-Shift-1]

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📖 Explanation

$\;\frac{x d y-y d x}{\sqrt{x^{2+}y^2}}=d x$ $\Rightarrow \;\frac{d y}{d x}=\;\frac{\sqrt{x^{2+}y^2}}{x}+\;\frac{y}{x}$ $\Rightarrow \;\frac{d y}{d x}=\sqrt{1+\;\frac{y^2}{x^2}}+\;\frac{y}{x}$ Let $\;\frac{y}{x}=v$ $\Rightarrow v+x \;\frac{d v}{d x}=\sqrt{1+v^2}+v$ $\Rightarrow \;\frac{d v}{\sqrt{1+v^2}}=\;\frac{d x}{x}$ OR $\ln (v+\sqrt{1+v^2})=\ln x+C$ at $x=1, y=0$ $\Rightarrow C=0$ $\;\frac{y}{x}+\sqrt{1+\;\frac{y^2}{x^2}}=x$ At $x=2$ $\;\frac{y}{2}+\sqrt{1+\;\frac{y^2}{4}}=2$ $\Rightarrow 1+\;\frac{y^2}{4}=4+\;\frac{y^2}{4}-2 y$ OR $y=\;\frac{3}{2}$

Q62JEE Main 2022MCQ4MInverse Trigonometric Functions

Considering only the principal values of the inverse trigonometric functions, the domain of the function $f(x)=\cos ^{-1}(\;\frac{x^{2-}4 x+2}{x^{2+}3})$ is : [28-Jul-2022-Shift-1]

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📖 Explanation

$-1 \le \;\frac{x^{2-}4 x+2}{x^{2+}3} \le 1$ $\Rightarrow -x^{2-}3 \le x^{2-}4 x+2 \le x^{2+}3$ $\Rightarrow 2 x^{2-}4 x+5 \ge 0 \ \;\; -4 x \le 1$ $x \in R \& x \ge -\;\frac{1}{4}$ So domain is $[-\;\frac{1}{4}, \infty )$

Q63JEE Main 2022MCQ4MVector Algebra

Let the vectors $\vec{a}=(1+t)^{} \hat{i}+(1-t)^{} \hat{j}+\hat{k}, \vec{b}=(1-t)^{} \hat{i}+(1+t) \hat{j}+2 \hat{k}$ and $\vec{c}-t \hat{i}-t \hat{j}+\hat{k}, t \in R$ be such that for $\alpha, \beta, \gamma \in R, \alpha^{} \vec{a}+\beta^{} \vec{b}+\gamma ^{} \vec{c}=\vec{0} \Rightarrow \alpha=\beta=\gamma =0$ . Then, the setof all values of $t$ is : [28-Jul-2022-Shift-1]

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📖 Explanation

Clearly $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar $\begin{vmatrix}1+t & 1-t & 1 \\ 1-t & 1+t & 2 \\ t & -t & 1\end{vmatrix} \ne 0$ $\Rightarrow (1+t)(1+t+2 t)-(1-t)(1-t-2 t)+1(t^{2-}t-t-t^2) \ne 0$ $\Rightarrow (3 t^{2+}4 t+1)-(1-t)(1-3 t)-2 t \ne 0$ $\Rightarrow (3 t^{2+}4 t+1)-(3 t^{2-}4 t+1)-2 t \ne 0$ $\Rightarrow t \ne 0$

Q64JEE Main 2022MCQ4MInverse Trigonometric Functions

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $\cos ^{-1}(x)-2 \sin ^{-1}(x)=\cos ^{-1}(2 x)$ is equal to : [28-Jul-2022-Shift-1]

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📖 Explanation

$\cos ^{-1} x-2 \ sin ^{-1} x=\cos ^{-1} 2 x$ For Domain : $x \in [\;\frac{-1}{2}, \;\frac{1}{2}]$ $\cos ^{-1} x-2(\;\frac{\pi}{2}-\cos ^{-1} x)=\cos ^{-1}(2 x)$ $\Rightarrow \cos ^{-1} x+2 \cos ^{-1} x=\pi+\cos ^{-1} 2 x$ $\Rightarrow \cos (3 \cos ^{-1} x)=-\cos (\cos ^{-1} 2 x)$ $\Rightarrow 4 x^3=x$ $\Rightarrow x=3, \pm \;\frac{1}{2}$

Q65JEE Main 2022MCQ4MSets and Relations

Let the operations $*, ⊙ \in \{∧, ∨\}$ . If $( {\text{p}} * {\text{q}}) ⊙( {\text{p}} ⊙ ∼ {\text{q}})$ is a tautology, then the ordered pair $(*, ⊙)$ is: [28-Jul-2022-Shift-1]

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📖 Explanation

$*, ⊙ \in \{∧, ∨\}$ Now for $(p * q) ⊙(p ⊙ ∼ q)$ is tautology(A) $(∨, ∧):(p ∨ q) ∧(p ∧ ∼ q)$ not a tautology(B) $(∨, ∨):(p ∨ q) ∨(p ∨ ∼ q)$ $=P ∨ T$ is tautology(C) $(∧, ∧):(p ∧ q) ∧(p ∧ ∼ q)$ $=(p ∧ p) ∧(q ∧ ∼ q)=p ∧ F$ not a tautology (Fallasy)(D) $(∧, ∨):(p ∧ q) ∨(p ∨ ∼ q)$ not a tautology

Q66JEE Main 2022MCQ4MVector Algebra

Let a vector $\vec{a}$ , has magnitude 9. Let a vector $\vec{b}$ be such that for every $(x, y) \in R \times R-\{(0,0)\}$ , the vector $(x \vec{a}+y \vec{b})$ is perpendicular to the vector $(6 y \vec{a}-18 x \vec{b})$ . Then the value of $|\vec{a} \times \vec{b}|$ is equal to: [28-Jul-2022-Shift-1]

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📖 Explanation

$(\vec{x a}+\vec{y b}) \cdot (6 \vec{y a}-18 x \vec{b})=0$ $\Rightarrow (6 x y|\vec{a}|^{2-}18 x y|\vec{b}|^2)+(6 y^{2-}18 x^2) \vec{a} \cdot \vec{b}=0$ As given equation is identityCoefficient of $x^2=$ coefficient of $y^2=$ coefficient of $x y=0$ $\Rightarrow |\vec{a}|^2=.3 \vec{\mid b}|^2 \Rightarrow \vec{b \mid }=3 \sqrt{3}$ and ${a . \vec\vec{b}=0$ $|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \ sin \theta$ $=9.3 \sqrt{3} .1=27 \sqrt{3}$

Q67JEE Main 2022MCQ4MTrigonometric Ratios and Identities

For ${\text{t}} \in (0,2 \pi)$ , if ${\text{ABC}}$ is an equilateral triangle with vertices ${\text{A}}(\sin t,-\cos {\text{t}}), {\text{B}}({\text{cost}}, \sin t)$ and $C(a, b)$ such that its orthocentre lies on a circle with centre $(1, \;\frac{1}{3})$ , then $(a^{2-}b^2)$ is equal to: [28-Jul-2022-Shift-1]

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📖 Explanation

Let $P(h, k)$ be the orthocentre of $\triangle {\text{ABC}}$ Then $h=\;\frac{\ sin t+\cos t+a}{3}, k=\;\frac{-\cos t+\ sin t+b}{3}$ (orthocentre coincide with centroid) $\therefore (3 h-a)^{2+}(3 k-b)^2=2$ $\therefore (h-\;\frac{a}{3})^{2+}(k-\;\frac{b}{3})^2=\;\frac{2}{9}$ $\because$ orthocentre lies on circle with centre $(1, \;\frac{1}{3})$ $\therefore a=3, b=1$ $\therefore a^{2-}b^2=8$

Q68JEE Main 2022MCQ4MSets and Relations

For $\alpha \in N$ , consider a relation ${\text{R}}$ on $N$ given by ${\text{R}} .=\{(x, y): 3 x+\alpha y$ is a multiple of 7 $\}$ . The relation $R$ is an equivalence relation if and only if : [28-Jul-2022-Shift-1]

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📖 Explanation

$R=\{(x, y): 3 x+\alpha y$ is multiple of 7 $\}$ , now ${\text{R}}$ to be an equivalence relation(1) R should be reflexive : $(a, a) \in R ∀ a \in N$ $\therefore 3 a+a \alpha=7 k$ $\therefore (3+\alpha) a=7 k$ $\therefore 3+\alpha=7 k_1 \Rightarrow \alpha=7 k_1-3$ $=7 k_1+4$ (2) R should be symmetric : $a R b ⇔ b R a$ $a R b: 3 a+(7 k-3) b=7 m$ $\Rightarrow 3(a-b)+7 k b=7 m$ $\Rightarrow 3(b-a)+7 k a=7 m$ So, $a R b \Rightarrow b R a$ $\therefore$ R will be symmetric for $a=7 k_1-3$ (3) Transitive : Let $(a, b) \in R,(b, c) \in R$ $\Rightarrow 3 a+(7 k-3) b=7 k_1$ and $3 b+(7 k_2-3) c=7 k_3$ Adding $3 a+7 k b+(7 k_2-3) c=7(k_1+k_3)$ $3 a+(7 k_2-3) c=7 m$ $\therefore (a, c) \in R$ $\therefore {\text{R}}$ is transitive $\therefore \alpha=7 k-3=7 k+4$

Q69JEE Main 2022MCQ4MProbability

Out of $60 \%$ female and $40 \%$ male candidates appearing in an exam, $60 \%$ candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is : [28-Jul-2022-Shift-1]

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📖 Explanation

$P($ Female $)=\;\frac{60}{100}=\;\frac{3}{5}$ $P($ Male $)=\;\frac{2}{5}$ $P($ Female/Qualified $)=\;\frac{40}{60}=\;\frac{2}{3}$ $P($ Male/qualified $)=\;\frac{20}{60}=\;\frac{1}{3}$

Q70JEE Main 2022MCQ4MDifferential Equations

If $y=y(x), x \in (0, \pi / 2)$ be the solution curve of the differential equation $(\sin ^2 2 x) \;\frac{d y}{d x}+(8 \sin ^2 2 x+2 \sin 4 x) y-2 {\text{e}}^{-4 x}(2 \sin 2 x+\cos 2 x),$ with $y(\pi / 4)= {\text{e}}^{-\pi}$ , then $y(\pi / 6)$ is equal to: [28-Jul-2022-Shift-1]

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📖 Explanation

$(\ sin ^2 2 x) \;\frac{d y}{d x}+(8 \ sin ^2 2 x+2 \ sin 4 x) y$ $=2 e^{-4 x}(2 \ sin 2 x+\cos 2 x)$ $\;\frac{d y}{d x}+(8+4 \cot 2 x) y=2 e^{-4 x}(\;\frac{2 \ sin 2 x+\cos 2 x}{\ sin ^2 2 x})$ Integrating factor $(I . F .)=e^{\int(8+4 \cot 2 x) d x}$ $=e^{8 x+2 \ln \ sin 2 x}$ Solution of differential equationy. $e^{8 x+2 \ln \ sin 2 x}$ $=\int 2 e^{(4 x+2 \ln \ sin 2 x)} \;\frac{(2 \ sin 2 x+\cos 2 x)}{\ sin ^2 2 x} d x$ $=2 \int e^{4 x}(2 \ sin 2 x+\cos 2 x) d x$ y. $e^{8 x+2 \ln \ sin 2 x}=e^{4 x} \ sin 2 x+c$ $y(\;\frac{\pi}{4})=e^{-\pi}$ $e^{-\pi} \cdot e^{2 \pi}=e^\pi+c \Rightarrow c=0$ $y(\;\frac{\pi}{6})=\;{e^{\frac{2 \pi}{3} \frac{\sqrt{3}}{2}}}/{e^{(\;\frac{4 \pi}{3}+2 \ln \;\frac{\sqrt{3}}{2})}}$ $=e^{\;\frac{-2 \pi}{3}} \cdot \;\frac{2}{\sqrt{3}}$

Q71JEE Main 2022MCQ4MParabola

If the tangents drawn at the points ${\text{P}}$ and ${\text{Q}}$ on the parabola $y^2=2 x-3$ intersect at the point $R(0,1)$ , then the orthocentre of the triangle $P Q R$ is: [28-Jul-2022-Shift-1]

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📖 Explanation

Equation of chord $P Q$ $\Rightarrow y \times 1=x-3$ $\Rightarrow x-y=3$ For point $P$ & $Q$ Intersection of ${\text{PQ}}$ with parabola $P:(6,3) Q:(2,-1)$ Slope of $R Q=-1$ & slope of $P Q=1$ Therefore $∠ P Q R=90^{\circ} \Rightarrow$ Orthocentre is at $Q:(2,-1)$

Q72JEE Main 2022MCQ4MCircles

Let $C$ be the centre of the circle $x^{2+}y^{2-}x+2 y=\;\frac{11}{4}$ and $P$ be a point on the circle. A line passes through the point $C$ , makes an angle of $\;\frac{\pi}{4}$ with the line $C P$ and intersects the circle at the points $Q$ and $R$ .. Then the area of the triangle $P Q R$ (in unit ${ }^2$ ) is : [28-Jul-2022-Shift-1]

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📖 Explanation

$Q R=2 r=4$ $P=(\;\frac{1}{2}+2 \cos \;\frac{\pi}{4},-1+2 \ sin \;\frac{\pi}{4})$ $=(\;\frac{1}{2}+\sqrt{2},-1+\sqrt{2})$ Area of $\triangle P Q R=\;\frac{1}{2} \times 4 \times \sqrt{2}$ $=2 \sqrt{2}$ sq. units

Q73JEE Main 2022MCQ4MBinomial Theorem

The remainder when $7^{2022}+3^{20222}$ is divided by 5 is : [28-Jul-2022-Shift-1]

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📖 Explanation

$7^{2022}+3^{2022}$ $=(49)^{1011}+(9)^{1011}$ $=(50-1)^{1011}+(10-1)^{1011}$ $=5 \lambda -1+5 k-1$ $=5 m-2$ $\;\text{ Remainder }\;=5-2=3$

Q74JEE Main 2022MCQ4MMatrices and Determinants

Let the matrix $A=\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{bmatrix}$ and the matrix $B_0=A^{49}+2 A^{98}$ . If $B_n=A d j(B_{n-1})$ for all $n>1$ , then ${\det}(B_4)$ is equal to : [28-Jul-2022-Shift-1]

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📖 Explanation

$A=\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{bmatrix}$ $\Rightarrow A^2=\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{bmatrix} \times \begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{bmatrix}=\begin{bmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}$ $\Rightarrow A^3=\begin{bmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=l$ Now $B_0=A^{49}+2 A^{98}=(A^3)^{16} \cdot A+2(A^3)^{32} \cdot A^2$ $B_0=A+2 A^2=\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{bmatrix}+\begin{bmatrix}0 & 0 & 2 \\ 2 & 0 & 0 \\ 0 & 2 & 0\end{bmatrix}=\begin{bmatrix}0 & 1 & 2 \\ 2 & 0 & 1 \\ 1 & 2 & 0\end{bmatrix}$ $|B_0|=9$ Since, $B_n=A d j|B_{n-1}| \Rightarrow |B_n|=|B_{n-1}|^2$ Hence $|B_4|=|B_3|^2=|B_2|^4=|B_1|^8=|B_0|^{16}$ $=|3^2|^{16}=3^{32}$

Q75JEE Main 2022MCQ4MComplex Numbers

Let $S_1=\{z_1 \in C: z_1-3|=\;\frac{1}{2}\}$ and $S_2=\{z_2 \in C: z_2-| z_2+1|| = z_2+| z_2-1|| \}$ . Then, for $z_1 \in S_1$ and $z_2 \in S_2$ , the least value of $z_2-z_1|$ is : [28-Jul-2022-Shift-1]

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📖 Explanation

$\because |Z_2+| Z_2-1||^2=|Z_2-| Z_2+1||^2$ $\Rightarrow (Z_2+|Z_2-1|)(\overline{Z}_2+|Z_2-1|)=(Z_2-|Z_2+1|)(\overline{Z}_2-|Z_2+1|)$ $\Rightarrow Z_2(|Z_2-1|+|Z_2+1|)+\overline{Z}_2(|Z_2-1|+|Z_2+1|)=|Z_2+1|^{2-}|Z_2-1|^2$ $\Rightarrow (Z_2+\overline{Z}_2)(|Z_2+1|+|Z_2-1|)=2(Z_2+\overline{Z}_2)$ $\Rightarrow \;\text{ Either }\; Z_2+\overline{Z}_2=0 \;\text{ or }\;|Z_2+1|+|Z_2-1|=2$ So, $Z_2$ lies on imaginary axis or on real axis within $[-1,1]$ Also $|Z_1-3|=\;\frac{1}{2} \Rightarrow Z_1$ lies on the circle having center 3 and radius $\;\frac{1}{2}$ .Clearly $|Z_1-Z_2|_{\min }=\;\frac{3}{2}$

Q76JEE Main 2022MCQ4MThree Dimensional Geometry

The foot of the perpendicular from a point on the circle $x^{2+}y^{2-}1, z-0$ to the plane $2 x+3 y+z=6$ lies on which one of the following curves? [28-Jul-2022-Shift-1]

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📖 Explanation

Any point on $x^{2+}y^2=1, z=0$ is $p(\cos \theta, \ sin \theta, 0)$ If foot of perpendicular of ${\text{p}}$ on the plane $2 x+3 y+z=6$ is $(h, k, l)$ then $\;\frac{h-\cos \theta}{2}=\;\frac{k-\ sin \theta}{3}=\;\frac{l-0}{1}$ $=-(\;\frac{2 \cos \theta+3 \ sin \theta+0-6}{2^{2+}3^{2+}1^2})=r(\;\text{ let }\;)$ $h=2 r+\cos \theta, k=3 r+\ sin \theta, l=r$ Hence, $h-2 l=\cos \theta$ and $k-3 l=\ sin \theta$ Hence, $(h-2 l)^{2+}(k-3 l)^2=1$ When $l=6-2 h-3 k$ Hence required locus is $(x-2(6-2 x-3 y))^{2+}(y-3(6-2 x-3 y))^2=1$ $\Rightarrow (5 x+6 y-12)^{2+}4(3 x+5 y-9)^2=1, z=6-2 x-3 y$

Q77JEE Main 2022MCQ4MApplication of Derivatives

If the minimum value of $f(x)=\;\frac{5 x^2}{2}+\;\frac{\alpha}{x^5}, x > 0$ , is 14 , then the value of $\alpha$ is equal to: [28-Jul-2022-Shift-1]

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📖 Explanation

$f(x)=\;\frac{5 x^2}{2}+\;\frac{\alpha}{x^5}\{x>0\}$ $f^{'}(x)=5 x-\;\frac{5 \alpha}{x^6}=0$ $\Rightarrow x=(\alpha)^{\;\frac{1}{7}}$ $f(x)_{\min }=\;{5(\alpha)^{\frac{2}{7}}}/{2}+\;{\alpha}/{\alpha^{\;\frac{5}{7}}}=14$ $\;\frac{5}{2} \alpha^{\;\frac{2}{7}}+\alpha^{\;\frac{2}{7}}=14$ $\;\frac{7}{2} \alpha^{\;\frac{2}{7}}=14$ $\alpha=128$

Q78JEE Main 2022MCQ4MFunctions

Let $\alpha, \beta$ and $\gamma$ be three positive real numbers. Let $f(x)=\alpha x^{5+}\beta x^{3+}\gamma x, x \in R$ and $g: R \to R$ . be such that $g(f(x))=x$ for all $x \in R$ . If ${\text{a}}_1, {\text{a}}_2, {\text{a}}_3, . . ., {\text{a}}_{ {\text{n}}}$ be in arithmetic progression with mean zero, then the value of $f(g(\;{1}/{ {\text{n}}} sum_{i=1}^{ {\text{n}}} f( {\text{a}}_i)))$ is equal to: [28-Jul-2022-Shift-1]

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📖 Explanation

$f(g(\;\frac{1}{n} sum_{i=1}^{n} f(a_i)))$ $\;\frac{a_1+a_2+a_3+. . . . . .+a_n}{n}=0$ $\therefore$ First and last term, second and second last and so on are equal in magnitude but opposite in sign. $f(x)=\alpha x^{5+}\beta x^{3+}\gamma x$ $sum_{i=1}^{n} f(a_i)=\alpha(a_1^{5+}a_2^{5+}a_3^{5+}. . .+a_n^5)+\beta(a_1^{3+}a_2^{3+}. . .+a_n^3)+\gamma (a_1+a_2+. . .+a_n)$ $=0 \alpha+0 \beta+0 \gamma$ $=0$ $\therefore f(g(\;\frac{1}{n} sum_{i=1}^{n} f(a_i)))=\;\frac{1}{n} sum_{i=1}^{n} f(a_i)=0$

Q79JEE Main 2022MCQ4MSequences and Series

Consider the sequence $a_1, a_2, a_3, . . .$ such that $a_1-1, a_2-2$ and $a_{n+2}=\;\frac{2}{a_{n+1}}+a_{-n}$ for ${\text{n}}-1,2,3, . . .$ If $(\;\frac{a_1+\frac{1}{a_2}}{a_3}) \cdot (\;\frac{a_2+\frac{1}{a_3}}{a_4}) \cdot (\;\frac{a_3+\frac{1}{a_4}}{a_5}) . . .(\;{a_{30}+\frac{1}{a_{31}}}/{a_{32}})=2^\alpha({ }^{61} C_{31})$ , then $\alpha$ is equal to : [28-Jul-2022-Shift-1]

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📖 Explanation

$a_{n+2}=\;\frac{2}{a_{n+1}}+a_n$ $\Rightarrow a_n a_{n+1}+1=a_{n+1} a_{n+2}-1$ $\Rightarrow a_{n+2} a_{n+1}-a_n \cdot a_{n+1}=2$ For $n=1\;a_3 a_2-a_1 a_2=2$ $n=2\;a_4 a_3-a_3 a_2=2$ $n=3\;a_5 a_4-a_4 a_3=2$ . . . $n=n \;\; \;\frac{a_{n+2} a_{n+1}-a_n a_{n+1}=2}{a_{n+2} a_{n+1}=2 n+a_1 a_2}$ Now, $\;\frac{(a_1 a_2+1)}{a_2 a_3} \cdot \;\frac{(a_2 a_3+1)}{a_3 a_4} \cdot \;\frac{(a_3 a_4+1)}{a_4 a_5} . . . . . . \cdot \;\frac{(a_{30} a_{31}+1)}{a_{31} a_{32}}$ $=\;\frac{3}{4} \cdot \;\frac{5}{6} \cdot \;\frac{7}{8} . . . . . . \cdot \;\frac{61}{62}$ $=2^{-60}({ }^{61} C_{31})$

Q80JEE Main 2022MCQ4MDifferential Equations

The minimum value of the twice differentiable function $f(x)=int_0^x e^{x-t} f^{'}( {\text{t}}) {\text{dt}}-(x^{2-}x+1) {\text{e}}^x$ , $x \in R_{\;\text{, is }\;}$ : [28-Jul-2022-Shift-1]

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📖 Explanation

$f(x)=int_{0^x e}^{x-t} f^{'}(t) d t-(x^{2-}x+1) e^x$ $f(x)=e^x int_{0^x e}^{-t} f^{'}(t) d t-(x^{2-}x+1) e^x$ $e^{-x} f(x)=int_{0^x e}^{-t} f^{'}(t) d t-(x^{2-}x+1)$ Differentiate on both side $e^{-x} f^{'}(x)+(-f(x) e^{-x})=e^{-x} f^{'}(x)-2 x+1$ $f(x)=e^x(2 x-1)$ $f^{'}(x)=e^x(2)+e^x(2 x-1)$ $=e^x(2 x+1)$ $x=-\;\frac{1}{2}$ $f^{' '}(x)=e^x(2)+(2 x+1) e^x$ $=e^x(2 x+3)$ $\;\text{ For }\; x=-\;\frac{1}{2} f^{' '}(x)>0$ $\Rightarrow$ Maxima $\therefore$ Max. $=e^{-\;\frac{1}{2}}(-1-1)$ $\therefore -\;\frac{2}{\sqrt{e}}$

Q81JEE Main 2022NAT4MPermutations and Combinations

Let $S$ be the set of all passwords which are six to eight characters long, where each character is either an alphabet from $\{A, B, C, D, E\}$ or a number from $\{1,2,3,4,5\}$ with the repetition of characters allowed. If the number of passwords in $S$ whose at least one character is a number from $\{1,2,3,4,5\}$ is $\alpha \times 5^6$ , then $\alpha$ is equal to _______. [28-Jul-2022-Shift-1]

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📖 Explanation

If password is 6 character long, thenTotal number of ways having atleast one number $=10^{6-}5^6$ Similarly, if 7 character long $=10^{7-}5^7$ and if 8-character long $=10^{8-}5^8$ Number of password $=(10^{6+}10^{7+}10^8)-(5^{6+}5^{7+}5^8)$ $=5^6(2^{6+}5.2^{7+}25.2^{8-}1-5-25)$ $=5^6(64+640+6400-31)$ $=7073 \times 5^6$ $\therefore \alpha=7073$

Q82JEE Main 2022NAT4MThree Dimensional Geometry

Let ${\text{P}}(-2,-1,1)$ and ${\text{Q}}(\;\frac{56}{17}, \;\frac{43}{17}, \;\frac{111}{17})$ be the vertices of the rhombus ${\text{PRQS}}$ . If the direction ratios of the diagonal ${\text{RS}}$ are $\alpha,-1, \beta$ , where both $\alpha$ and $\beta$ are integers of minimum absolute values, then $\alpha^{2+}\beta^2$ is equal to _______. [28-Jul-2022-Shift-1]

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📖 Explanation

d.r's of $R S=< \alpha,-1, \beta>$ d.r's of $P Q=< \;\frac{90}{17}, \;\frac{60}{17}, \;\frac{94}{17}>=< 45,30,47>$ as ${\text{PQ}}$ and ${\text{RS}}$ are diagonals of rhombus $\alpha(45)+30(-1)+47(\beta)=0$ $\Rightarrow 45 \alpha+47 \beta=30$ i.e., $\alpha=\;\frac{30-47 \beta}{45}$ for minimum integral value $\alpha=-15$ and $\beta=15$ $\Rightarrow \alpha^{2+}\beta^2=450$

Q83JEE Main 2022NAT4MApplication of Derivatives

Let $f:[0,1] \to R$ be a twice differentiable function in $(0,1)$ such that $f(0)-3$ and $f(1)-5$ . If the line $y-2 x+3$ intersects the graph of $f$ at only two distinct points in $(0,1)$ , then the least number of points $x \in (0,1)$ , at which $f^{' '}(x)-0$ , is _______. [28-Jul-2022-Shift-1]

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📖 Explanation

If a graph cuts $y=2 x+5$ in $(0,1)$ twice then its concavity changes twice. $\therefore f^{'}(x)=0$ at at least two points.

Q84JEE Main 2022NAT4MDefinite Integrals

If $int_0^{\sqrt{3}} \;{15 x^3}/{\sqrt{1+x^{2+}\sqrt{(1+x^2)^3}}} {\text{d}} x-\alpha \sqrt{2}+\beta \sqrt{3}$ , where $\alpha, \beta$ are integers, then $\alpha+\beta$ is equal to ________. [28-Jul-2022-Shift-1]

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📖 Explanation

Put $x=\tan \theta \Rightarrow d x=\ sec ^2 \theta d \theta$ $\Rightarrow I=int_{0}^{\;\frac{\pi}{3}} \;{15 \tan ^3 \theta \cdot \ sec ^2 \theta d \theta}/{\sqrt{1+\tan ^2 \theta+\sqrt{\ sec ^6 \theta}}}$ $\Rightarrow I=int_{0}^{\;\frac{\pi}{3}} \;\frac{15 \tan ^2 \theta \ sec ^2 \theta d \theta}{\ sec \theta \sqrt{1+\ sec \theta}}$ $\Rightarrow I=int_{0}^{\;\frac{\pi}{3}} \;\frac{15(\ sec ^2 \theta-1) \ sec \theta \tan \theta d \theta}{(\sqrt{1+\ sec \theta})}$ Now put $1+\ sec \theta=t^2$ $\Rightarrow \ sec \theta \tan \theta d \theta=2 t d t$ $\Rightarrow I=int_{\sqrt{2}}^{\sqrt{3}} \;\frac{15((t^{2-}1)^{2-}1) 2 t d t}{t}$ $\Rightarrow I=30 int_{\sqrt{2}}^{\sqrt{3}}(t^{4-}2 t^{2+}1-1) d t$ $\Rightarrow I=30 int_{\sqrt{2}}^{\sqrt{3}}(t^{4-}2 t^2) d t$ $\Rightarrow I=30 int_{\sqrt{2}}^{\sqrt{3}}(t^{4-}2 t^2) d t$ $\Rightarrow I=.30(\;\frac{t^5}{5}-\;\frac{2 t^3}{3})|_{\sqrt{2}} ^{\sqrt{3}}$ $=30[(\;\frac{9}{5} \sqrt{3}-2 \sqrt{3})-(\;\frac{4 \sqrt{2}}{5}-\;\frac{4 \sqrt{2}}{3})]$ $=(54 \sqrt{3}-60 \sqrt{3})-(24 \sqrt{2}-40 \sqrt{2})$ $=16 \sqrt{2}-6 \sqrt{3}$ $\therefore \alpha=16$ and $\beta=-6$ $\alpha+\beta=10$

Q85JEE Main 2022NAT4MMatrices and Determinants

Let $A=\begin{bmatrix}1 & -1 \\ 2 & \alpha\end{bmatrix}$ and $B=\begin{bmatrix}\beta & 1 \\ 1 & 0\end{bmatrix}, \alpha, \beta \in R$ . Let $\alpha_1$ be the value of $\alpha$ which satisfies $( {\text{A}}+ {\text{B}})^{2-} {\text{A}}^{2+}\begin{bmatrix}2 & 2 \\ 2 & 2\end{bmatrix}$ and $\alpha_2$ be the value of $\alpha$ which satisfies $( {\text{A}}+ {\text{B}})^{2-} {\text{B}}^2$ . Then $|\alpha_1-\alpha_2|$ is equal to _______. [28-Jul-2022-Shift-1]

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📖 Explanation

$(A+B)^2=A^{2+}B^{2+}A B+B A$ $=A^{2+}\begin{bmatrix}2 & 2 \\ 2 & 2\end{bmatrix}$ $\therefore B^{2+}A B+B A=\begin{bmatrix}2 & 2 \\ 2 & 2\end{bmatrix} . . . . . . .(1)$ $A B=\begin{bmatrix}1 & -1 \\ 2 & \alpha\end{bmatrix}\begin{bmatrix}\beta & 1 \\ 1 & 0\end{bmatrix}=\begin{bmatrix}\beta-1 & 1 \\ \alpha+2 \beta & 2\end{bmatrix}$ $B A=\begin{bmatrix}\beta & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}1 & -1 \\ 2 & \alpha\end{bmatrix}=\begin{bmatrix}\beta+2 & \alpha-\beta \\ 1 & -1\end{bmatrix}$ $B^2=\begin{bmatrix}\beta & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}\beta & 1 \\ 1 & 0\end{bmatrix}=\begin{bmatrix}\beta^{2+}1 & \beta \\ \beta & 1\end{bmatrix}$ By (1) we get ${\begin{bmatrix}\beta^{2+}2 \beta & \alpha+1 \\ \alpha+3 \beta+1 & 2\end{bmatrix}\$ =\begin{bmatrix}2 & 2 \ 2 & 2\end{bmatrix}}∴ alpha=1 beta=0 \Rightarrow alpha_1=1 $Similarly if$ A^{2+}A B+B A=0 $then$ (A^2=\begin{bmatrix}1 & -1 \ 2 & alpha\end{bmatrix}\begin{bmatrix}1 & -1 \ 2 & alpha\end{bmatrix}=$\begin{bmatrix}-1 & -1-\alpha \ 2+2 \alpha & \alpha^{2-}2\end{bmatrix}$){\begin{bmatrix}2 beta & alpha-beta+1-1-alpha \ alpha+2 beta+1+2+2 alpha & alpha^{2-}2+1\end{bmatrix}$=\begin{bmatrix}0 & 0 \ 0 & 0\end{bmatrix}}\Rightarrow \beta=0 ;\text{ and }; \alpha=-1 \Rightarrow \alpha_2=-1\therefore |\alpha_1-\alpha_2|=|2|=2$

Q86JEE Main 2022NAT4MQuadratic Equation and Inequalities

For ${\text{p}}, {\text{q}} \in R_{\;\text{s }\;}$ consider the real valued function $f(x)=(x- {\text{p}})^{2-} {\text{q}}, x \in R$ and ${\text{q}}>0$ . Let ${\text{a}}_1, {\text{a}}_{2^{'}}$ ${\text{a}}_3$ and ${\text{a}}_4$ be in an arithmetic progression with mean ${\text{p}}$ and positive common difference. If $f( {\text{a}}_i)|=500$ for all $i=1,2,3,4$ , then the absolute difference between the roots of $f(x)=0$ is _______. [28-Jul-2022-Shift-1]

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📖 Explanation

$\because a_1, a_2, a_3, a_4$ $\therefore a_2=p-3 d, a_2=p-d, a_3=p+d$ and $a_4=p+3 d$ Where $d>0$ $\because |f(a_i)|=500$ $\Rightarrow |9 d^{2-}q|=500$ and $|d^{2-}q|=500$ either $9 d^{2-}q=d^{2-}q$ $\Rightarrow d=0$ not acceptable $\therefore 9 d^{2-}q=q-d^2$ $\therefore 5 d^{2-}q=0$ Roots of $f(x)=0$ are $p+\sqrt{q}$ and $p-\sqrt{q}$ $\therefore$ absolute difference between roots $=|2 \sqrt{q}|=50$

Q87JEE Main 2022NAT4MEllipse

For the hyperbola ${\text{H}}: x^{2-}y^2=1$ and the ellipse ${\text{E}}: \;{x^2}/{ {\text{a}}^2}+\;{y^2}/{ {\text{b}}^2}=1, {\text{a}} > {\text{b}} > 0$ , let the (1) eccentricity of ${\text{E}}$ be reciprocal of the eccentricity of ${\text{H}}$ , and (2) the line $y=\sqrt{\;\frac{5}{2}} x+ {\text{K}}$ be a common tangent of ${\text{F}}$ and ${\text{H}}$ . Then $4( {\text{a}}^{2+} {\text{b}}^2)$ is equal to ________. [28-Jul-2022-Shift-1]

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📖 Explanation

The equation of tangent to hyperbola $x^{2-}y^2=1$ within slope $m$ is equal to $y=m x \pm \sqrt{m^{2-}1}$ .And for same slope $m$ , equation of tangent to ellipse $\;\frac{x^2}{a^2}+\;\frac{y^2}{b^2}=1$ is $y=m x \pm \sqrt{a^2 m^{2+}b^2} . . . . . .$ . (ii) $\because$ Equation (i) and (ii) are identical $\therefore a^2 m^{2+}b^2=m^{2-}1$ $\therefore m^2=\;\frac{1+b^2}{1-a^2}$ But equation of common tangent is $y=\sqrt{\;\frac{5}{2}} x+k$ $\therefore m=\sqrt{\;\frac{5}{2}} \Rightarrow \;\frac{5}{2}=\;\frac{1+b^2}{1-a^2}$ $\therefore 5 a^{2+}2 b^2=3 . . . . . . \;\text{ (i) }\;$ $\;\text{ eccentricity of ellipse }\;=\;\frac{1}{\sqrt{2}}$ $\therefore 1-\;\frac{b^2}{a^2}=\;\frac{1}{2}$ $\Rightarrow a^2=2 b^2 . . . . . . . . \;\text{ (ii) }\;$ From equation (i) and (ii) : $a^2=\;\frac{1}{2}, b^2=\;\frac{1}{4}$ $\therefore 4(a^{2+}b^2)=3$

Q88JEE Main 2022NAT4MSequences and Series

Let $x_1, x_2, x_3, . . ., x_{20}$ be in geometric progression with $x_1=3$ and the common ratio $\;\frac{1}{2}$ . A new data is constructed replacing each $x_i$ by $(x_i-i)^2$ . If $\overline{x}$ is the mean of new data, then the greatest integer less than or equal to $\overline{x}$ is ________. [28-Jul-2022-Shift-1]

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📖 Explanation

$x_1, x_2, x_3, . . . ., x_{20}$ are in G.P. $x_1=3, r=\;\frac{1}{2}$ $\overline{x}=\;\frac{\sum x_i^{2-}2 x_i i+i^2}{20}$ $=\;\frac{1}{20}[12(1-\;\frac{1}{2^{40}})-6(4-\;\frac{11}{2^{18}})+70 \times 41]$ $\begin{cases}S=1+2 \cdot \\ \\ \frac{1}{2}+3 \cdot \\ \\ \frac{1}{2^2}+\end{cases} . . ; \;\; \;\frac{S}{2}=\;\frac{1}{2}+\;\frac{2}{2^2}+. . . .$ $.\;\frac{S}{2}=2(1-\;\frac{1}{2^{20}})-\;\frac{20}{2^{20}}=4-\;\frac{11}{2^{18}}\}$ $\therefore [\overline{x}]=[\;\frac{2858}{20}-(\;\frac{12}{240}-\;\frac{66}{2^{18}}) \cdot \;\frac{1}{20}]$ $=142$

Q89JEE Main 2022NAT4MLimits, Continuity and Differentiability

$\lim _{x \to 0}(\;\frac{(x+2 \cos x)^{3+}2(x+2 \cos x)^{2+}3 \sin (x+2 \cos x)}{(x+2)^{3+}2(x+2)^{2+}3 \sin (x+2)})^{\;\frac{100}{x}}$ is equal to _______. [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Let $x+2 \cos x=a$ $x+2=b$ as $x \to 0, a \to 2$ and $b \to 2$ $\lim _{x \to 0}(\;\frac{a^{3+}2 a^{2+}3 \ sin a}{b^{3+}2 b^{2+}3 \ sin b})^{\;\frac{100}{x}}$ $=e^{\lim _{x \to 0} \cdot \;\frac{100}{x} \cdot \;\frac{(a^{3-}b^3)+2(a^{2-}b^2)+3(\ sin a-\ sin b)}{b^{3+}2 b^{2+}3 \ sin b}}$ $\because \lim _{x \to 0} \;\frac{a-b}{x}=\lim _{x \to 0} \;\frac{2(\cos x-1)}{x}=0$ $=e^0$ $=1$

Q90JEE Main 2022NAT4MQuadratic Equation and Inequalities

The sum of all real values of $x$ for which $\;\frac{3 x^{2-}9 x+17}{x^{2+}3 x+10}-\;\frac{5 x^{2-}7 x+19}{3 x^{2+}5 x+12}$ is equal to ________. [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{3 x^{2-}9 x+17}{x^{2+}3 x+10}=\;\frac{5 x^{2-}7 x+19}{3 x^{2+}5 x+12}$ $\Rightarrow \;\frac{3 x^{2-}9 x+17}{5 x^{2-}7 x+19}=\;\frac{x^{2+}3 x+10}{3 x^{2+}5 x+12}$ $\;\frac{-2 x^{2-}2 x-2}{5 x^{2-}7 x+19}=\;\frac{-2 x^{2-}2 x-2}{3 x^{2+}5 x+12}$ Either $x^{2+}x+1=0$ or No real roots $\Rightarrow 5 x^{2-}7 x+19=3 x^{2+}5 x+12$ $2 x^{2-}12 x+7=0$ sum of roots $=6$

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