Previous Year Paper

25 Jul 2022 Shift 2

90 Questions360 Marks180 MinTake as Timed Mock →

🎯 Practice smarter, not harder

Practice 25 Jul 2022 Shift 2 questions with full analytics — free

✓ Attempt tracking & analytics
✓ Timer & exam-mode simulation
✓ Adaptive difficulty (Easy → Hard)
✓ Bookmarks, notes & mistake review

Just sign in to unlock everything. Free for all students.

🚀 Start Practice Mode Sign in free →

Physics30 questions

Q1JEE Main 2022MCQ4MSemiconductor Electronics

In AM modulation, a signal is modulated on a carrier wave such that maximum and minimum amplitudes are found to be $6 {V}$ and $2 {V}$ respectively. The modulation index is: [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${A}_{\max }=6 {V}$ ${A}_{\min }=2 {V}$ $\mu =\;\frac{A_{\max }-A_{\min }}{A_{\max }+A_{\min }}=\;\frac{6-2}{6+2}=0.5$ $\mu =50 \%$

Q2JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

The electric current in a circular coil of 2 turns produces a magnetic induction ${B}_{1}$ at its centre. The coil is unwound and in rewound into a circular coil of 5 tuns and the same current produces a magnetic induction $B_{2}$ at its centre. The ratio of $\;\frac{B_{2}}{B_{1}}$ is [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$B=\;\frac{n \mu _{0} I}{2 R}$ $B_{1}=\;\frac{2 \mu _{0} I}{2 R_{1}}$ $B_{2}=\;\frac{5 \mu _{0} I}{2 R_{2}}$ $R_{2}=\;\frac{2 R_{1}}{5}$ $\Rightarrow \;\frac{B_{2}}{B_{1}}=\;\frac{5}{2} \times \;\frac{R_{1}}{R_{2}}=\;\frac{25}{4}$

Q3JEE Main 2022MCQ4MProperties of Matter

A drop of liquid of density $\rho$ is floating half immersed in a liquid of density $\boldsymbol{\sigma}$ and surface tension $7.5 \times 10^{-4} {Ncm}^{-1}$ . The radius of drop in ${cm}$ will be : $(g=10 m s^{-2})$ [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Balancing the forces on drop $2 \pi R T+\;\frac{4}{3} \pi R^{3} \rho g=\;\frac{2}{3} \pi R^{3} \sigma g$ $\Rightarrow 2 T=\;\frac{2 R^{2}}{3}(\sigma-2 \rho ) \times 10$ $\Rightarrow \;\frac{15 \times 10^{-2} \times 3}{10(\sigma-2 \rho ) 2}=R^{2}$ $R=\;\frac{3}{2 \times 10} \sqrt{\;\frac{1}{(\sigma-2 \rho )}}$ $=\;\frac{3}{20} \sqrt{\;\frac{1}{\sigma-2 \rho }}(\;\text{ \in m }\;)$ $( {R})$ in ${cm}=\;\frac{3 \times 100}{20} \sqrt{\;\frac{1}{\sigma-2 \rho }}=15 \times \sqrt{\;\frac{1}{\sigma-2 \rho }}$ Now if $2 \rho >\sigma(R_{\;\text{\in }\; c m})=\;\frac{15}{\sqrt{2 \rho -\sigma}}$

Q4JEE Main 2022MCQ4MCentre of Mass and Collision

Two billiard balls of mass $0.05 {kg}$ each moving in opposite directions with $10 {ms}^{-1}$ collide and rebound with the same speed. If the time duration of contact is $t=0.005 {s}$ , then what is the force exerted on the ball due to each other? [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Change in momentum of one ball $=2 \times (0.05)(10) {kg} {m} / {s}$ $=1 {kg} {m} / {s}$ $\Rightarrow F_{a v g}=\;\frac{1}{\Delta t}=\;\frac{1}{0.005} {N}$ $=200 {N}$

Q5JEE Main 2022MCQ4MLaws of Motion

For a free body diagram shown in the figure, the four forces are applied in the ' $x$ ' and ' $y$ ' directions. What additional force must be applied and at what angle with positive $x$ -axis so that the net acceleration of body is zero? [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Resultant of already applied forces $=-\hat{i}-\hat{j}$ $\Rightarrow$ Force required to balance $=\hat{i}+\hat{j}$ $\Rightarrow$ Force required $=\sqrt{2} {N}$ in magnitude at angle $45^{\circ}$ with +ve $x$ -axis

Q6JEE Main 2022MCQ4MElectrostatic Potential and Capacitance

Capacitance of an isolated conducting sphere of radius ${R}_{1}$ becomes ${n}$ times when it is enclosed by a concentric conducting sphere of radius $R_{2}$ connected to earth. The ratio of their radii $(\;\frac{R_{2}}{R_{1}})$ is : [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Initially $=C_{0}=4 \pi \epsilon_{0} R_{1}$ Finally $\;\frac{4 \pi \epsilon_{0} R_{1} R_{2}}{R_{2}-R_{1}}=n C_{0}=4 \pi \epsilon_{0} n R_{1}$ $\;\frac{R_{2}}{R_{2}-R_{1}}=n$ $1-\;\frac{R_{1}}{R_{2}}=\;\frac{1}{n}$ $\;\frac{R_{1}}{R_{2}}=\;\frac{n-1}{n}$ $\;\frac{R_{2}}{R_{1}}=\;\frac{n}{n-1}$

Q7JEE Main 2022MCQ4MDual Nature of Matter and Radiation

The ratio of wavelengths of proton and deuteron accelerated by potential $V_{p}$ and $V_{d}$ is $1: \sqrt{2}$ . Then the ratio of $V_{p}$ to $V_{d}$ will be : [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\lambda =\;\frac{h}{m v}=\;\frac{h}{\sqrt{2 m e V}}$ so $\;\frac{\lambda _{p}}{\lambda _{d}}=\;{\sqrt{m_{d} V_{d}}}/{\sqrt{m_{p} V_{p}}}=\;\frac{1}{\sqrt{2}}$ $\;\frac{2 V_{d}}{V_{p}}=\;\frac{1}{2}$ $\;\frac{V_{p}}{V_{d}}=\;\frac{4}{1}$

Q8JEE Main 2022MCQ4MRay Optics and Optical Instruments

For an object placed at a distance $2.4 {m}$ from a lens, a sharp focused image is observed on a screen placed at a distance $12 {cm}$ from the lens. A glass plate of refractive index $1.5$ and thickness $1 {cm}$ is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen? [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

The shift produced by the glass plate is $d=t(1-\;\frac{1}{\mu })=1 \times (1-\;\frac{1}{1.5})=\;\frac{1}{3} {cm}$ So final image must be produced at $(12-\;\frac{1}{3}) {cm}=\;\frac{35}{3} {cm}$ from lens so that glass plate must shift it to produce image at screen. So $\;\frac{1}{12}-\;\frac{1}{-240}=\;\frac{1}{f}=\;\frac{1}{35 / 3}-\;\frac{1}{u}$ $\;\frac{1}{u}=\;\frac{3}{35}-\;\frac{1}{12}-\;\frac{1}{240}$ or $u=-560 {cm}$ $\;\text{ so }\; \;\text{ shift }\;=5.6-2.4=3.2 {m}$

Q9JEE Main 2022MCQ4MElectromagnetic Waves

Light wave travelling in air along x-direction is given by $E_{y}=540 \sin \pi \times 10^{4}(x-c t) {Vm}^{-1}$ . Then, the peak value of magnetic field of wave will be (Given ${c}=3 \times 10^{8} {ms}^{-1}$ ) [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$c=3 \times 10^{8} {m} / {sec}$ $B=\;\frac{E}{c}=\;\frac{540}{3 \times 10^{8}}=18 \times 10^{-7} T$

Q10JEE Main 2022MCQ4MElectromagnetic Induction

When you walk through a metal detector carrying a metal object in your pocket, it raises an alarm. This phenomenon works on : [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q11JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

An electron with energy $0.1 {keV}$ moves at right angle to the earth's magnetic field of $1 \times 10^{-4} {Wbm}^{-}$ 2. The frequency of revolution of the electron will be(Take mass of electron $=9.0 \times 10^{-31} {kg}$ ) [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$T=\;\frac{2 \pi m}{B q}$ $\Rightarrow$ Frequency $f=\;\frac{B q}{2 \pi m}$ $=\;\frac{10^{-4} \times 1.6 \times 10^{-19}}{2 \pi \times 9 \times 10^{-31}}$ $≃ 2.8 \times 10^{6} {Hz}$

Q12JEE Main 2022MCQ4MCurrent Electricity

A current of $15 {mA}$ flows in the circuit as shown in figure. The value of potential difference between the points $A$ and $B$ will be: [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Effective $R=[5+\;\frac{5 \times 10}{5+10}+10] k Ω$ $=\;\frac{275}{15} k Ω$ $\Rightarrow \Delta V_{A B}=15 m A \times \;\frac{275}{15} k Ω$ $=275 {V}$

Q13JEE Main 2022MCQ4MOscillations

The length of a seconds pendulum at a height $h=2 R$ from earth surface will be:(Given ${R}=$ Radius of earth and acceleration due to gravity at the surface of earth, ${g}=\pi^{2} {ms}-2$ ) [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$g=\;\frac{G M}{(R+h)^{2}}=\;\frac{G M}{9 R^{2}}=\;\frac{g_{0}}{9}$ $\Rightarrow T=2 \pi \sqrt{\;\frac{l}{g}}=2 \pi \sqrt{\;{l}/{\frac{g_{0}}{9}}}$ $\Rightarrow 2=2 \pi \sqrt{\;\frac{9 l}{g_{0}}}$ $\Rightarrow l=\;\frac{g_{0}}{9 \pi^{2}}=\;\frac{1}{9} m$

Q14JEE Main 2022MCQ4MThermodynamics

Sound travels in a mixture of two moles of helium and ${n}$ moles of hydrogen. If rms speed of gas molecules in the mixture is $\sqrt{2}$ times the speed of sound, then the value of ${n}$ will be : [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Molar mass $M-\;\frac{2 \times 4+n \times 1}{2+n} . . . . .$ (i)Also, $\gamma =\;{n_{1} C_{P_{1}}+n_{2} C_{P_{2}}}/{n_{1} C_{V_{1}}+n_{2} C_{V_{2}}}=\;\frac{2 \times 5 R+n \times 7 R}{2 \times 3 R+n \times 5 R}$ $\Rightarrow \gamma =\;\frac{10+7 n}{6+5 n} . . . . . . \;\text{ (ii) }\;$ Given that $V_{r m s}=\sqrt{2} V_{\;\text{sound }\;}$ $\Rightarrow \sqrt{\;\frac{3 R T}{M}}=\sqrt{2} \sqrt{\;\frac{\gamma R T}{M}}$ $\Rightarrow \gamma =\;\frac{3}{2}$ $\Rightarrow n=2$

Q15JEE Main 2022MCQ4MThermodynamics

Let $η_{1}$ is the efficiency of an engine at $T_{1}=447^{\circ} {C}$ and ${T}_{2}=147^{\circ} {C}$ while $η_{2}$ is the efficiency at ${T}_{1}=947^{\circ} {C}$ and ${T}_{2}=47^{\circ} {C}$ The ratio $\;\frac{η_{1}}{η_{2}}$ will be: [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$η_{1}=1-\;\frac{420}{720}=\;\frac{300}{720}$ And $η_{2}=1-\;\frac{320}{1220}=\;\frac{900}{1220}$ $\Rightarrow \;\frac{η_{1}}{η_{2}}=\;\frac{300}{720} \times \;\frac{1220}{900}$ $≃ 0.56$

Q16JEE Main 2022MCQ4MGravitation

An object is taken to a height above the surface of earth at a distance $\;\frac{5}{4} {R}$ from the centre of the earth. Where radius of earth, $R=6400 {km}$ . The percentage decrease in the weight of the object will be [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$w=m g$ $w^{'}=\;\frac{m g}{(1+\;\frac{h}{R})^{2}}=\;\frac{m g}{(\;\frac{5}{4})^{2}}=\;\frac{16}{25} m g$ $\therefore \%$ decrease in weight $=(1-\;\frac{16}{25}) \times 100 \%$ $=36 \%$

Q17JEE Main 2022MCQ4MCentre of Mass and Collision

A bag of sand of mass $9.8 {kg}$ is suspended by a rope. A bullet of $200 {g}$ travelling with speed $10 {ms}^{-1}$ gets embedded in it, then loss of kinetic energy will be : [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Loss in $K E=\;\frac{1}{2} \times \;\frac{m_{1} m_{2}}{m_{1}+m_{2}} \times v^{2}$ $=\;\frac{1}{2} \times \;\frac{9.8 \times 0.2}{10} \times (10)^{2}$ $=9.8 {J}$

Q18JEE Main 2022MCQ4MMotion in a Plane

A ball is projected from the ground with a speed $15 {ms}^{-1}$ at an angle $\theta$ with horizontal so that its range and maximum height are equal, then 'tan $\theta^{'}$ will be equal to: [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\because \tan \theta=\;\frac{4 H}{R}$ $\Rightarrow \tan \theta=4 \times 1$ $\Rightarrow \tan \theta=4$

Q19JEE Main 2022MCQ4MCurrent Electricity

The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are $1 \%, 2 \%$ and $3 \%$ respectively. The maximum percentage error in the detection of the dissipated heat will be : [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\because H=i^{2} R t$ $\because \%$ error in $H=2 \times 2 \%+1 \%+3 \%$ $=8 \%$

Q20JEE Main 2022MCQ4MAtoms and Nuclei

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength $\lambda$ . The value of principal quantum number ' $n$ ' of the excited state will be : (R : Rydberg constant) [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\because \;\frac{1}{\lambda }=R(\;\frac{1}{1^{2}}-\;\frac{1}{n^{2}})$ $\Rightarrow \;\frac{1}{\lambda R}=1-\;\frac{1}{n^{2}}$ $\Rightarrow \;\frac{1}{n^{2}}=1-\;\frac{1}{\lambda R}=\;\frac{\lambda R-1}{\lambda R}$ $\Rightarrow n=\sqrt{\;\frac{\lambda R}{\lambda R-1}}$

Q21JEE Main 2022NAT4MMotion in a Straight Line

A particle is moving in a straight line such that its velocity is increasing at $5 {ms}^{-1}$ per meter. The acceleration of the particle is _____ ${ms}^{-2}$ at a point where its velocity is $20 {ms}^{-1}$ . [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{d v}{d x}=5 {ms}^{-1} / {m}$ Acceleration of particle when $v=20 {m} / {s}$ $a=v \;\frac{d v}{d x}=20(5) {m} / {s}^{2}=100 {m} / {s}^{2}$

Q22JEE Main 2022NAT4MCentre of Mass and Collision

Three identical spheres each of mass $M$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $3 {m}$ each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be $\sqrt{x} {m}$ . The value of $x$ is___ [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$d_{c m}=3 \sin 45^{\circ}=\;\frac{3}{\sqrt{2}}$ $d_{c m}=\;\frac{2}{3} \times \;\frac{3}{\sqrt{2}}=\sqrt{2}=\sqrt{x}$ $x=2$

Q23JEE Main 2022NAT4MThermodynamics

A block of ice of mass $120 {g}$ at temperature $0^{\circ} {C}$ is put in $300 {g}$ of water at $25^{\circ} {C}$ . The $\times {g}$ of ice melts as the temperature of the water reaches $0^{\circ} {C}$ . The value of $x$ is____$[Use specific heat capacity of water $=4200 {Jkg}^{-1} {K}^{-1}$ , Latent heat of ice $=3.5 × 10^{5} {Jkg}^{-1}$ ]$ [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Heat lost by water $=$ Heat gained by ice $0.3 \times 4200 \times 25=x \times 3.5 \times 10^{5}$ $x=\;\frac{0.3 \times 4200 \times 25}{3.5 \times 10^{5}}$ $=90 \times 100 \times 10^{5} \times 10^{3} {gram}=90 {gm}$

Q24JEE Main 2022NAT4MAtoms and Nuclei

$\;\frac{x}{x+4}$ is the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its(i) third permitted energy level to the second level and(ii) the highest permitted energy level to the second permitted level.The value of $x$ will be [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$E_{n}=-\;\frac{13.6}{n^{2}} e V$ $\;{\;\frac{1}{2^{2}}-\frac{1}{3^{2}}}/{\frac{1}{2^{2}}}=\;\frac{x}{x+4}$ $\Rightarrow \;\frac{9-4}{9 \times 4 \times \;\frac{1}{4}}=\;\frac{x}{x+4}=\;\frac{5}{9}$ $x=5$

Q25JEE Main 2022NAT4MCurrent Electricity

In a potentiometer arrangement, a cell of emf $1.20 {V}$ gives a balance point at $36 {cm}$ length of wire. This cell is now replaced by another cell of emf $1.80 {V}$ . The difference in balancing length of potentiometer wire in above conditions will be ____ cm. [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$E \propto I$ $\;\frac{1.2}{1.8}=\;\frac{36}{I^{'}}$ $I^{'}=\;\frac{3}{2} \times 36=54 {cm}$ $\Delta I=I^{'}-I=54-36=18 {cm}$

Q26JEE Main 2022NAT4MSemiconductor Electronics

Two ideal diodes are connected in the network as shown in figure. The equivalent resistance between $A$ and $B$ is ____ $Ω$ . [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$R=\;\frac{20 \times 20}{40}+15=25 Ω$

Q27JEE Main 2022NAT4MWaves

Two waves executing simple harmonic motions travelling in the same direction with same amplitude and frequency are superimposed. The resultant amplitude is equal to the $\sqrt{3}$ times of amplitude of individual motions. The phase difference between the two motions is ____(degree). [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$A_{n e t}=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \varphi }$ $\sqrt{3} A=\sqrt{A^{2}+A^{2}+2 A^{2} \cos \varphi }$ $3 A^{2}=2 A^{2}+2 A^{2} \cos \varphi$ $\cos \varphi =\;\frac{1}{2}$ $\varphi =60^{\circ}$

Q28JEE Main 2022NAT4MElectrostatic Potential and Capacitance

Two parallel plate capacitors of capacity ${C}$ and $3 {C}$ are connected in parallel combination and charged to a potential difference $18 {V}$ . The battery is then disconnected and the space between the plates of the capacitor of capacity ${C}$ is completely filled with a material of dielectric constant 9 . The final potential difference across the combination of capacitors will be ____V. [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$V_{\;\text{common }\;}=\;\frac{18 C V+54 C {CV}}{3 C+9 {C}}=6 {V}$

Q29JEE Main 2022NAT4MRay Optics and Optical Instruments

A convex lens of focal length $20 {cm}$ is placed in front of a convex mirror with principal axis coinciding each other. The distance between the lens and mirror is $10 {cm}$ . A point object is placed on principal axis at a distance of $60 {cm}$ from the convex lens. The image formed by combination coincides the object itself. The focal length of the convex mirror is ___ ${cm}$ . [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{1}{v}+\;\frac{1}{u}=\;\frac{1}{f}$ $\;\frac{1}{v}-\;\frac{1}{-60}=\;\frac{1}{20}$ $\;\frac{1}{v}=-\;\frac{1}{60}+\;\frac{1}{20}=\;\frac{-1+3}{60}=\;\frac{2}{60}$ $\Rightarrow v=+30 {cm}$ $\therefore$ Radius of curvature of mirror $=30-10=20 {cm}$ $\Rightarrow f_{\;\text{mirror }\;}=\;\frac{20}{2}=10 {cm}$

Q30JEE Main 2022NAT4MElectromagnetic Induction

Magnetic flux (in weber) in a closed circuit of resistance $20 Ω$ varies with time $t(s)$ at $\varphi =8 t^{2}-9 t+5$ . The magnitude of the induced current at $t=0.25 {s}$ will be ____ ${mA}$ . [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$R=20 Ω$ $\varphi =8 t^{2}-9 t+5$ $\epsilon=|-\;\frac{d \varphi }{d t}|=|16 t-9|=|16(0.25)-9|=5$ $i=\;\frac{\epsilon}{R}=\;\frac{5}{20}=0.25 A=\;\frac{0.25}{10^{3}} \times 10^{3} A=250 {mA}$

Chemistry30 questions

Q31JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

List I (molecule) List II (hybridization ; shape) (A) $XeO_3$ (I) $sp^3d$ ; linear (B) $XeF_2$ (II) $sp^3$ ; pyramidal (C) $XeOF_4$ (III) $sp^3d^3$ ; distorted octahedral (D) $XeF_6$ (IV) $sp^3d^2$ ; square pyramidal Choose the correct answer from the options given below: [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{XeO}}_3-s p^3$ , Pyramidal ${\text{XeF}}_2-s p^3 d$ , linear ${\text{XeOF}}_4-s p^3 d^2$ , Square Pyramidal ${\text{XeF}}_6-s p^3 d^3$ , distorted octahedral

Q32JEE Main 2022MCQ4MSolutions

Two solutions $A$ and $B$ are prepared by dissolving $1 {g}$ of non-volatile solutes $X$ and $Y$ , respectively in 1 ${kg}$ of water. The ratio of depression in freezing points for $A$ and $B$ is found to be $1: 4$ . The ratio of molar masses of ${X}$ and ${Y}$ is [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\Delta T_f=i k_f \times m$ $\;{\Delta T_{ {\text{f}}( {\text{A}})}}/{\Delta {\text{T}}_{ {\text{f}}( {\text{B}})}}=\;\frac{1}{4}$ $\;{ {\text{i}} \times {\text{K}}_{ {\text{f}}} \times {1}/{ {\text{M}}_{ {\text{A}}}} \times 1}/{ {\text{i}} \times {\text{K}}_{ {\text{f}}} \times \;{1}/{ {\text{M}}_{ {\text{B}}}} \times 1}=\;\frac{1}{4}$ $\;\frac{M_B}{M_A}=\;\frac{1}{4}$ $M_A: M_B=4: 1$

Q33JEE Main 2022MCQ4MIonic Equilibrium

$K_{a_{1}}, K_{a_{2}}$ and $K_{a_{3}}$ are the respective ionization constants for the following reactions (a), (b) and (c).(a) ${H}_{2} {C}_{2} {O}_{4} ⇄ {H}^{+}+ {HC}_{2} {O}_{4}^{-}$ (b) ${HC}_{2} {O}_{4}^{-} ⇄ {H}^{+}+ {HC}_{2} {O}_{4}^{2-}$ (c) ${H}_{2} {C}_{2} {O}_{4} ⇄ 2 {H}^{+}+ {C}_{2} {O}_{4}^{2-}$ The relationship between $K_{a_{1}}, K_{a_{2}}$ and $K_{a_{3}}$ is given as [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{H}}_2 {\text{C}}_2 {\text{O}}_4 ⇌ 2 {\text{H}}^{+}+ {\text{C}}_2 {\text{O}}_4^{2-} \;\;{\text{K}}_{ {\text{a}}3}$ ${\text{H}}_2 {\text{C}}_2 {\text{O}}_4 ⇌ {\text{H}}^{+}+ {\text{HC}}_2 {\text{O}}_4^{-} \;\; {\text{K}}_{ {\text{a}}1}$ ${\text{HC}}_2 {\text{O}}_4^{-} ⇌ {\text{H}}^{+}+ {\text{C}}_2 {\text{O}}_4^{2-} \;\;{\text{K}}_{ {\text{a}}2}$ ${\text{K}}_{ {\text{a}}3}=\;{[ {\text{H}}^{+}]^2\$ [ {\text{C}}_2 {\text{O}}_4^{2-}]$}/{[ {\text{H}}_2 {\text{C}}_2 {\text{O}}4]}{\text{K}}{ {\text{a}}1}=;{[ {\text{H}}^{+}]$$[ {\text{HC}}_2 {\text{O}}_4^{-}]$}/{[ {\text{H}}_2 {\text{C}}_2 {\text{O}}4]}, {\text{K}}{ {\text{a}}2}=;{[ {\text{H}}^{+}]$$[ {\text{C}}_2 {\text{O}}4^{-}]$}/{[ {\text{HC}}2 {\text{O}}4^{-}]$}{\text{K}}{ {\text{a}}3}= {\text{K}}{ {\text{a}}1} × {\text{K}}{ {\text{a}}2}$

Q34JEE Main 2022MCQ4MElectrochemistry

The molar conductivity of a conductivity cell filled with 10 moles of $20 {mL} {NaCl}$ solution is $Lambda_{m 1}$ and that of 20 moles another identical cell heaving $80 {mL} {NaCl}$ solution is $Lambda_{m 2}$ . The conductivities exhibited by these two cells are same. The relationship between $Lambda_{m 2}$ and $Lambda_{m 1}$ is [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$Lambda_{ {\text{m}}_1}=\;{ {\text{k}}_1 \times 1000}/{ {\text{M}}_1}=\;{ {\text{k}} \times 1000}/{\frac{10}{0.02}}$ $Lambda_{ {\text{m}}_2}=\;{ {\text{k}}_2 \times 1000}/{\frac{20}{0.08}}$ It is given that ${\text{k}}_1= {\text{k}}_2$ ${\text{k}}_1=\;{Lambda_{ {\text{m}}_1}}/{2} \;\; {\text{k}}_2=\;{Lambda_{ {\text{m}}_2}}/{4}$ Applying the given condition on conductivity. $\;{Lambda_{ {\text{m}}_1}}/{2}=\;{Lambda_{ {\text{m}}_2}}/{4}$ $Lambda_{ {\text{m}}_2}=2 Lambda_{ {\text{m}}_1}$

Q35JEE Main 2022MCQ4MChemical Thermodynamics

For micelle formation, which of the following statements are correct?A. Micelle formation is an exothermic process.B. Micelle formation is an endothermic process.C. The entropy change is positive.D. The entropy change is negative. [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Micelle formation is an endothermic process with positive entropy change. For micelle formation, DS $>0$ (hydrophobic effect) This is possible because, the decrease in entropy due to clustering is offset by increase in entropy due to desolvation of the surfactant, Also ${\text{DH}}>06$ .

Q36JEE Main 2022MCQ4MClassification of Elements

The first ionization enthalpies of ${Be}, {B}, {N}$ and ${O}$ follow the order [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

The first ionization energy increase from left to right along $2^{\;\text{nd }\;}$ period with the following exceptions ${\text{IE}}_1: {\text{Be}}> {\text{B}}$ and ${\text{N}}> {\text{O}}$ This is due to stable configuration of ${\text{Be}}$ in comparison to ${\text{B}}$ and that of ${\text{N}}$ in comparison to ${\text{O}}$ .Hence the correct order is ${\text{N}}> {\text{O}}> {\text{Be}}> {\text{B}}$

Q37JEE Main 2022MCQ4Md and f Block Elements

Given below are two statements.Statement I : Pig iron is obtained by heating cast iron with scrap iron.Statement II : Pig iron has a relatively lower carbon content than that of cast iron.In the light of the above statements, choose the correct answer from the options given below. [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q38JEE Main 2022MCQ4MElectrochemistry

High purity (> 99.95%) dihydrogen is obtained by [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

High purity (>99.95 %) ${\text{H}}_2$ is obtained by electrolysing warm aqueous ${\text{Ba}}( {\text{OH}})_2$ solution between nickel electrodes.

Q39JEE Main 2022MCQ4MClassification of Elements

The correct order of density is [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Density of ${\text{Sr}}=2.63 {\text{g}} / {\text{cm}}^3$ Density of ${\text{Be}}=1.84 {\text{g}} / {\text{cm}}^3$ Density of ${\text{Mg}}=1.74 {\text{g}} / {\text{cm}}^3$ Density of ${\text{Ca}}=1.55 {\text{g}} / {\text{cm}}^3$

Q40JEE Main 2022MCQ4Mp Block Elements

The total number of acidic oxides from the following list is $\;\text{ NO, }\; {N}_{2} {O}, {B}_{2} {O}_{3}, {N}_{2} {O}_{5}, {CO}, {SO}_{3}, {P}_{4} {O}_{10}$ [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{NO}}, {\text{N}}_2 {\text{O}}, {\text{CO}}$ - neutral oxides ${\text{B}}_2 {\text{O}}_3, {\text{N}}_2 {\text{O}}_5, {\text{SO}}_3, {\text{P}}_4 {\text{O}}_{10}-$ acidic oxides

Q41JEE Main 2022MCQ4MCoordination Compounds

The correct order of energy of absorption for the following metal complexes is ${A}:[ {\text{Ni}}( {\text{en}})_{3}]^{2+}, {B}:[ {\text{Ni}}( {\text{NH}}_{3})_{6}]^{2+}, {C}:[ {\text{Ni}}( {\text{H}}_{2} {O})_{6}]^{2+}$ [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Stronger is ligand attached to metal ion, greater will be the splitting between $t_2 g$ and eg (hence greater will be $\Delta {\text{U}}$ ) $\therefore$ greater will be absorption of energy.Hence correct order $[ {\text{Ni}}( {\text{en}})_3]^{2+}>[ {\text{Ni}}( {\text{NH}}_3)_6]^{2+}>[ {\text{Ni}}( {\text{H}}_2 {\text{O}})_6]^{2+}$

Q42JEE Main 2022MCQ4Mp Block Elements

List I List II (A) Sulphate (I Pesticide (B) Fluoride (II) Bending of bones (C) Nicotine (III) Laxative effect (D) Sodium arsinite (IV) Herbicide Choose the correct answer from the options given below: [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q43JEE Main 2022MCQ4MHydrocarbons

Major product of the following reaction is : [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q44JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

What is the major product of the following reaction? [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q45JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

Arrange the following in decreasing acidic strength: [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q46JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

${CH}_{3}- {CH}_{2}-C N \;{{\to }^{C H_{3} M g B r}_{\;\text{ Ether } }} A {\to }^{H_{3} O^{+}} B \;{{\to }^{ {Zn-Hg}}_{ {HCl}}} C$ The correct structure of [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q47JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

List I List II (A) Nylon 6,6 (I) Buckets (B) Low density polythene (II) Non-stick utensils (C) High density polythene (III) Bristles of brushes (D) Teflon (IV) Toys Choose the correct answer from the options given below : [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Nylon $6,6 \to$ used in making bristles of brushesLow density polythene $\to$ used in making ToysHigh density polythene $\to$ used in making BucketsTeflon $\to$ used in making non-stick utensils

Q48JEE Main 2022MCQ4MBiomolecules

Glycosidic linkage between C1 of $\alpha$ -glucose and C2 of $\beta$ -fructose is found in [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Sucrose is a disaccharide composed of one unit of $\alpha$ -D-glucose and one unit of $\beta$ -D-fructose. The glycosidic linkage in sucrose is formed between the anomeric carbon (C1) of $\alpha$ -D-glucose and the anomeric carbon (C2) of $\beta$ -D-fructose, specifically an $\alpha$ -1,2-glycosidic bond. Other options do not fit this description.

Q49JEE Main 2022MCQ4MBiomolecules

Some drugs bind to a site other than the active site of an enzyme. This site is known as [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q50JEE Main 2022MCQ4MIonic Equilibrium

In base vs. acid titration, at the end point methyl orange is present as [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Methyl orange is an acid-base indicator that changes color based on pH. Its color transition range is pH 3.1 (red) to pH 4.4 (yellow). The structural change responsible for this color variation is as follows:

In acidic solutions (below pH 3.1), methyl orange exists predominantly in its protonated form, which is red. This red color is attributed to the formation of a highly conjugated quinonoid structure.
In alkaline solutions (above pH 4.4), methyl orange exists predominantly in its deprotonated form, which is yellow. This yellow color is due to a less conjugated benzenoid structure.

In a "base vs. acid titration" (meaning acid is added to a base), as acid is added, the pH decreases. When the solution reaches the endpoint (where the indicator changes color from yellow to red), the methyl orange molecules are protonated and convert to their red, quinonoid form.

Thus, at the endpoint where methyl orange signals the completion of the titration (by turning red), it is present in its quinonoid form.

The final answer is $\boxed{A}$

Q51JEE Main 2022NAT4MSome Basic Concepts of Chemistry

56.0 ${L}$ of nitrogen gas is mixed with excess hydrogen gas and it is found that $20 {L}$ of ammonia gas is produced. The volume of unused nitrogen gas is found to be______ L. [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{N}}_2( {\text{g}})+3 {\text{H}}_2( {\text{g}}) \to 2 {\text{NH}}_3( {\text{g}})$ Since ${\text{H}}_2$ is in excess and $20 {\text{L}}$ of ammonia gas is produced.Hence, 2 moles ${\text{NH}}_3 \equiv 1$ mole ${\text{N}}_2 \;\; (v \propto {\text{n}})$ $20 {\text{LNH}}_3 \equiv 10 {\text{L}} {\text{N}}_2$ Volume of ${\text{N}}_2$ left $=56-10$ $=46 {\text{L}}$

Q52JEE Main 2022NAT4MStates of Matter

A sealed flask with a capacity of $2 {dm}^{3}$ contains $11 {g}$ of propane gas. The flask is so weak that it will burst if the pressure becomes $2 {MPa}$ . The minimum temperature at which the flask will burst is__________ ${ }^{\circ} {C}$ . [Nearest integer](Given : ${R}=8.3 {J} {K}^{-1} {mol}^{-1}$ , Atomic masses of ${C}$ and ${H}$ are $12 {u}$ and $1 {u}$ , respectively.) (Assume that propane behaves as an ideal gas.) [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

From ideal gas equation, ${\text{PV}}= {\text{nRT}}$ $P=2 \times 10^6 {\text{Pa}}$ $V=2 {\text{dm}}^3=2 \times 10^{-3} {\text{m}}^3$ ${\text{R}}=8.3 {\text{JK}}^{-1} {\text{mol}}^{-1}$ ${\text{n}}=\;\frac{11}{44} {\text{mol}}$ $2 \times 10^6 \times 2 \times 10^{-3}=\;\frac{11}{44} \times 8.3 \times {\text{T}}$ ${\text{T}}=1927.7 {\text{K}}$ ${\text{T}}(\;\text{ \in }\;{ }^{\circ} {\text{C}})=1927.7-273 ≃ 1655^{\circ} {\text{C}}$

Q53JEE Main 2022NAT4MStructure of Atom

When the excited electron of a ${H}$ atom from $n=5$ drops to the ground state, the maximum number of emission lines observed are______ [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Maximum number of emission lines $=\;\frac{(n_2-n_1)(n_2-n_1+1)}{2}$ ${\text{n}}_2=5$ ${\text{n}}_1=1$ $\Rightarrow \;\frac{(5-1)(5-1+1)}{2}=10$ Hence maximum number of emission lines observed are 10 .

Q54JEE Main 2022NAT4MChemical Thermodynamics

While performing a thermodynamics experiment, a student made the following observations. ${HCl}+ {NaOH} \to {NaCl}+ {H}_{2} {O} \Delta {H}=-57.3 {kJ} {mol}^{-1}$ ${CH}_{3} {COOH}+ {NaOH} \to {CH}_{3} {COONa}+ {H}_{2} {O} \Delta {H}=-55.3 {kJ} {mol}^{-1}$ The enthalpy of ionization of ${CH}_{3} {COOH}$ as calculated by the student is _____ ${kJ} {mol}^{-1}$ . (nearest integer) [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

(I) ${\text{HCl}}+ {\text{NaOH}} \to {\text{NaCl}}+ {\text{H}}_2 {\text{O}}$ $\Delta {\text{H}}_1=-57.3 {\text{KJmol}}^{-1}$ (II) ${\text{CH}}_3 {\text{COOH}}+ {\text{NaOH}} \to {\text{CH}}_3 {\text{COONa}}+ {\text{H}}_2 {\text{O}}$ $\Delta {\text{H}}_2=-55.3 {\text{KJmol}}^{-1}$ Reaction (I) can be written as(III) ${\text{NaCl}}+ {\text{H}}_2 {\text{O}} \to {\text{HCl}}+ {\text{NaOH}}$ $\Delta {\text{H}}_3=57.3 {\text{KJmol}}^{-1}$ By adding (II) and (III) ${\text{CH}}_3 {\text{COOH}}+ {\text{NaCl}}\;\to {\text{CH}}_3 {\text{COONa}}+ {\text{HCl}} \;\; \Delta {\text{H}}_{ {\text{r}}}$ $\Delta {\text{H}}_{ {\text{r}}}=\Delta {\text{H}}_3+\Delta {\text{H}}_2 \;\; =57.3-55.3$ $\;\; =2 {\text{kJ}} {\text{mol}}^{-1}$

Q55JEE Main 2022NAT4MChemical Kinetics

For the decomposition of azomethane. ${CH}_{3} {N}_{2} {CH}_{3}( {g}) \to {CH}_{3} {CH}_{3}( {g})+ {N}_{2}( {g})$ , a first order reaction, the variation in partial pressure with time at $600 {K}$ is given as The half life of the reaction is_____ $\times 10^{-5} {s}$ . [Nearest integer] [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

For first order reaction, $\ln A=\ln A_0-k t$ $\;\text{ Hence Slope }\;=-k$ $-k=-3.465 \times 10^4$ $k=\;\frac{0.693}{t_{1 / 2}}$ $3.465 \times 10^4=\;\frac{0.693}{t_{1 / 2}}$ $t_{1 / 2}=2 \times 10^{-5} {\text{s}}$

Q56JEE Main 2022NAT4MChemical Bonding and Molecular Structure

The sum of number of lone pairs of electrons present on the central atoms of ${XeO}_{3}, {XeOF}_{4}$ and ${XeF}_{6}$ , is [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q57JEE Main 2022NAT4Md and f Block Elements

The spin-only magnetic moment value of ${M}^{3+}$ ion (in gaseous state) from the pairs ${Cr}^{3+} / {Cr}^{2+}, {Mn}^{3+} /$ ${Mn}^{2+}, {Fe}^{3+} / {Fe}^{2+}$ and ${Co}^{3+} / {Co}^{2+}$ that has negative standard electrode potential, is _____ B.M. [Nearest integer] [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Among the pairs given, ${\text{Cr}}^{3+} / {\text{Cr}}^{2+}$ has negative reduction potential which is $-0.41 {\text{V}}$ . ${\text{Cr}}( {\text{III}}) \Rightarrow d^3$ Number of unpaired electrons $=3$ $\mu =\sqrt{3(3+2)}=\sqrt{15} ≃ 4 \;\text{ B.M. }\;$

Q58JEE Main 2022NAT4MAlcohols, Phenols and Ethers

A sample of $4.5 {mg}$ of an unknown monohydric alcohol, R-OH was added to methylmagnesium iodide. A gas is evolved and is collected and its volume measured to be $3.1 {mL}$ . The molecular weight of the unknown alcohol is _____ ${g} / {mol}$ . [Nearest integer] [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{R}}- {\text{OH}}+ {\text{CH}}_3 {\text{Mgl}} \Rightarrow {\text{R}}- {\text{OMgl}}+ {\text{CH}}_4$ moles of alcohol $( {\text{ROH}}) \equiv$ moles of ${\text{CH}}_4$ At STP, [Assuming STP]1 mole corresponds to $22.7 {\text{L}}$ Hence, $3.1 {\text{mL}} \equiv \;\frac{3.1}{22700} {\text{mol}}$ So, moles of alcohol $=\;\frac{3.1}{22700}$ $\Rightarrow \;\frac{3.1}{22700}=\;{4.5 \times 10^{-3}}/{ {\text{M}}}$ $M ≃ 33 {\text{g}} / {\text{mol}}$

Q59JEE Main 2022NAT4MPractical Organic Chemistry

The separation of two coloured substances was done by paper chromatography. The distances travelled by solvent front, substance A and substance $B$ from the base line are $3.25 {cm}, 2.08 {cm}$ and $1.05 {cm}$ , respectively. The ratio of $R_{f}$ values of $A$ to $B$ is_____ [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{R}}_{ {\text{f}}}=\;\frac{\;\text{ Distance travelled by the substance }\;}{\;\text{ Distance travelled by the solvent front }\;}$ $(R_f)_A=\;\frac{2.08}{3.25}$ $(R_f)_B=\;\frac{1.05}{3.25}$ $\;\frac{(R_f)_A}{(R_f)_B} ≃ 2$

Q60JEE Main 2022NAT4MHaloalkanes and Haloarenes

The total number of monobromo derivatives formed by he alkanes with molecular formula ${C}_{5} {H}_{12}$ is (excluding stereo isomers) [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

It is sum of distance of $z$ from $(3 \sqrt{2}, 0)$ and $(0, p \sqrt{2})$ For minimising, ${\text{z}}$ should lie on ${\text{AB}}$ and $A B=5 \sqrt{2}$ $(A B)^2=18+2 p^2$ $p=\pm 4$

Mathematics30 questions

Q61JEE Main 2022MCQ4MComplex Numbers

For $z \in C$ if the minimum value of $(|z-3 \sqrt{2}|+|z-p \sqrt{2} i|)$ is $5 \sqrt{2}$ , then a value Question: of $p$ is [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

It is sum of distance of ${\text{z}}$ from $(3 \sqrt{2}, 0)$ and $(0, p \sqrt{2})$ For minimising, ${\text{z}}$ should lie on ${\text{AB}}$ and $A B=5 \sqrt{2}$ $(A B)^2=18+2 p^2$ $p=\pm 4$

Q62JEE Main 2022MCQ4MMatrices and Determinants

The number of real values of $\lambda$ , such that the system of linear equations $2 x-3 y+5 z=9$ $x+3 y-z=-18$ $3 x-y+(\lambda ^{2}-|\lambda |) z=16$ has no solutions, is [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\Delta ={\begin{vmatrix}2 & -3 & 5 \\ 1 & 3 & -1 \\ 3 & -1 & \lambda ^{2-}\text{\end{vmatrix}\lambda |} |}$ $=2(3 \lambda ^{2-}3|\lambda |-1)+3(\lambda ^{2-}|\lambda |+3)+5(-1-9)$ $=9 \lambda ^{2-}9|\lambda |-43$ $=9|\lambda |^{2-}9|\lambda |-43$ $\Delta =0$ for 2 values of $|\lambda |$ out of which one is $-$ ve and other is $+$ veSo, 2 values of $\lambda$ satisfy the system of equations to obtain no solution.

Q63JEE Main 2022MCQ4MPermutations and Combinations

The number of bijective functions $f:\{1,3,5,7, . . ., 99\} \to \{2,4,6,8, . . . 100\}$ , such that $f(3) \ge f(9) \ge f(15) \ge f(21) \ge . . . . f(99)$ , is [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction $f(3)>f(9)>f(15) . . . . . .>f(99)$ So number of ways $={ }^{50} C_{17} \cdot 1.33$ ! $={ }^{50} P_{33}$

Q64JEE Main 2022MCQ4MBinomial Theorem

The remainder when $(11)^{1011}+(1011)^{11}$ is divided by 9 is [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{Re}}(\;\frac{(11)^{1011}+(1011)^{11}}{9})={\text{Re}}(\;\frac{2^{1011}+3^{11}}{9})$ For ${\text{Re}}(\;\frac{2^{1011}}{9})$ $2^{1011}=(9-1)^{337}={ }^{337} C_0 9^{337}(-1)^{0+}{ }^{337} C_1 9^{336}(-1)^{1+}{ }^{337} C_2 9^{335}(-1)^{2+}. . . . . .+{ }^{337} C_{337} 9^0(-1)^{337}$ So, remainder is 8and ${\text{Re}}(\;\frac{3^{11}}{9})=0$ So, remainder is 8

Q65JEE Main 2022MCQ4MSequences and Series

The sum $sum_{n=1}^{21} \;\frac{3}{(4 n-1)(4 n+3)}$ is equal to [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$sum_{n=1}^{21} \;\frac{3}{(4 n-1)(4 n+3)}=\;\frac{3}{4} sum_{n=1}^{21} \;\frac{1}{4 n-1}-\;\frac{1}{4 n+3}$ $=\;\frac{3}{4}[(\;\frac{1}{3}-\;\frac{1}{7})+(\;\frac{1}{7}-\;\frac{1}{11})+. . .+(\;\frac{1}{83}-\;\frac{1}{87})]$ $=\;\frac{3}{4}[\;\frac{1}{3}-\;\frac{1}{87}]=\;\frac{3}{4} \;\frac{84}{3.87}=\;\frac{7}{29}$

Q66JEE Main 2022MCQ4MLimits, Continuity and Differentiability

$\lim _{x \to \;\frac{\pi}{4}} \;\frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}$ is equal to [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\lim _{x \to \;\frac{\pi}{4}} \;\frac{8 \sqrt{2}-(\cos x+\sin x)^7}{\sqrt{2}-\sqrt{2} \sin 2 x} \;\; (\;\frac{0}{0}.$ form $)$ $=\lim _{x \to \;\frac{\pi}{4}} \;\frac{-7(\cos x+\sin x)^6(-\sin x+\cos x)}{-2 \sqrt{2} \cos 2 x}$ using L $-$ H Rule $=\lim _{x \to \;\frac{\pi}{4}} \;\frac{56(\cos x-\sin x)}{2 \sqrt{2} \cos 2 x}(\;\frac{0}{0})$ $=\lim _{x \to \;\frac{\pi}{4}} \;\frac{-56(\sin x+\cos x)}{-4 \sqrt{2} \sin 2 x}$ using L - H Rule $=7 \sqrt{2} \cdot \sqrt{2}=14$

Q67JEE Main 2022MCQ4MDefinite Integrals

$\lim _{n \to \infty } \;\frac{1}{2^{n}}(\;{1}/{\sqrt{1-\;\frac{1}{2^{n}}}}+\;{1}/{\sqrt{1-\;\frac{2}{2^{n}}}}+\;{1}/{\sqrt{1-\;\frac{3}{2^{n}}}}+. . .+\;{1}/{\sqrt{1-\;\frac{2^{n}-1}{2^{n}}}})$ is equal to [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$I=\lim _{n \to \infty } \;\frac{1}{2^n}(\;{1}/{\sqrt{1-\;\frac{1}{2^n}}}+\;{1}/{\sqrt{1-\;\frac{2}{2^n}}}+\;{1}/{\sqrt{1-\;\frac{3}{2^n}}}+. . . . . .+\;{1}/{\sqrt{1-\;\frac{2^n-1}{2^n}}})$ Let $2^n=t$ and if $n \to \infty$ then $t \to \infty$ $I=\lim _{n \to \infty } \;\frac{1}{t}(sum_{r=1}^{t-1} \;{1}/{\sqrt{1-\;\frac{r}{t}}})$ $I=int_0^1 \;\frac{d x}{\sqrt{1-x}}=int_0^1 \;\frac{d x}{\sqrt{x}} \;\; (int_0^a f(x) d x=int_0^a f(a-x) d x.$ $=\$ [2 x^{;\frac{1}{2}}]_0^1=2$

Q68JEE Main 2022MCQ4MProbability

If $A$ and $B$ are two events such that $P(A)=\;\frac{1}{3}, P(B)=\;\frac{1}{5}$ and $P(A ∪ B)=\;\frac{1}{2}$ , then $P(A \mid B^{'})+P(B \mid A^{'})$ is equal to [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$P(A)=\;\frac{1}{3}, P(B)=\;\frac{1}{5}$ and $P(A ∪ B)=\;\frac{1}{2}$ $\therefore P(A ∩ B)=\;\frac{1}{3}+\;\frac{1}{5}-\;\frac{1}{2}=\;\frac{1}{30}$ Now, $P(A \mid B^{'})+P(B \mid A^{'})=\;\frac{P(A ∩ B^{'})}{P(B^{'})}+\;\frac{P(B ∩ A^{'})}{P(A^{'})}$ $=\;\frac{\;\frac{9}{30}}{\frac{4}{5}}+\;\frac{\;\frac{5}{30}}{\frac{2}{3}}=\;\frac{5}{8}$

Q69JEE Main 2022MCQ4MDefinite Integrals

Let $|t|$ denote the greatest integer less than or equal to $t$ . Then the value of the integral $int_{-3}^{101}([\sin (\pi x)]+e^{[\cos (2 \pi x)]}) d x$ is equal to [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$I=int_{-3}^{101}([\sin (\pi x)]+e^{[\cos (2 \pi x)]}) d x$ $[\sin \pi x]$ is periodic with period 2 and $e^{[\cos (2 \pi x)]}$ is periodic with period 1 .So, $I=52 int_0^2([\sin \pi x]+e^{[\cos 2 \pi x]}) d x$ $=52\{int_1^{2-}1 d x+int_{\;\frac{1}{4}}^{\;\frac{3}{4}} e^{-1} d x+int_{\;\frac{5}{4}}^{\;\frac{7}{4}} e^{-1} d x+int_0^{\;\frac{1}{4}} e^0 d x+int_{\;\frac{3}{4}}^{\;\frac{5}{4}} e^0 d x+int_{\;\frac{7}{4}}^2 e^0 d x\}$ $=\;\frac{52}{e}$

Q70JEE Main 2022MCQ4MStraight Lines and Pair of Straight Lines

Let the point $P(\alpha, \beta)$ be at a unit distance from each of the two lines $L_{1}: 3 x-4 y+12=0$ , and $L_{2}: 8 x+6 y+11=0$ . If $P$ lies below $L_{1}$ and above $L_{2}$ , then $100(\alpha+\beta)$ is equal to [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$L_1: 3 x-4 y+12=0$ $L_2: 8 x+6 y+11=0$ Equation of angle bisector of $L_1$ and $L_2$ of angle containing origin $2(3 x-4 y+12)=8 x+6 y+11$ $2 x+14 y-13=0 . . . . . . . \;\text{ (i) }\;$ $\;\frac{3 \alpha-4 \beta+12}{5}=1$ $\Rightarrow 3 \alpha-4 \beta+7=0 \;\text{..... (ii) }\;$ Solution of $2 x+14 y-13=0$ and $3 x-4 y+7=0$ gives the required point $P(\alpha, \beta), \alpha=\;\frac{-23}{25}, \beta=\;\frac{53}{50}$ $100(\alpha+\beta)=14$

Q71JEE Main 2022MCQ4MDifferential Equations

Let a smooth curve $y=f(x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $(\;\frac{-y}{x})$ . If the curve passes through the points $(1,2)$ and $(8,1)$ , then $|y(\;\frac{1}{8})|$ is equal to [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{d y}{d x} \propto \;\frac{-y}{x}$ $\;\frac{d y}{d x}=\;\frac{-k y}{x} \Rightarrow \int \;\frac{d y}{y}=-K \int \;\frac{d x}{x}$ $\ln |y|=-K \ln |x|+C$ If the above equation satisfy $(1,2)$ and $(8,1)$ $\ln 2=-K \times 0+C \Rightarrow C=\ln 2$ $\ln 1=-K \ln 8+\ln 2 \Rightarrow K=\;\frac{1}{3}$ So, at $x=\;\frac{1}{8}$ $\ln |y|=-\;\frac{1}{3} \ln (\;\frac{1}{8})+\ln 2=2 \ln 2$ $|y|=4$

Q72JEE Main 2022MCQ4MEllipse

If the ellipse $\;\frac{x^{2}}{a^{2}}+\;\frac{y^{2}}{b^{2}}=1$ meets the line $\;\frac{x}{7}+\;\frac{y}{2 \sqrt{6}}=1$ on the $x$ -axis and the line $\;\frac{x}{7}-\;\frac{y}{2 \sqrt{6}}=1$ on the $y$ -axis, then the eccentricity of the ellipse is [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{x^2}{a^2}+\;\frac{y^2}{b^2}=1$ meets the line $\;\frac{x}{7}+\;\frac{y}{2 \sqrt{6}}=1$ on the ${\text{x}}$ -axis So, $a=7$ and $\;\frac{x^2}{a^2}+\;\frac{y^2}{b^2}=1$ meets the line $\;\frac{x}{7}-\;\frac{y}{2 \sqrt{6}}=1$ on the $y$ -axis So, $b=2 \sqrt{6}$ Therefore, $e^2=1-\;\frac{b^2}{a^2}=1-\;\frac{24}{49}$ $e=\;\frac{5}{7}$

Q73JEE Main 2022MCQ4MParabola

The tangents at the points $A(1,3)$ and $B(1,-1)$ on the parabola $y^{2}-2 x-2 y=1$ meet at the point $P$ . Then the area (in unit ${ }^{2}$ ) of the triangle $P A B$ is: [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Given curve : $y^{2-}2 x-2 y=1$ .Can be written as $(y-1)^2=2(x+1)$ And, the given information can be plotted as shown in figureTangent at $A: 2 y-x-5=0$ using $T=0\}$ Intersection with $y=1$ is $x=-3$ Hence, point $P$ is $(-3,1)$ Taking advantage of symmetryArea of $\triangle P A B=2 \times \;\frac{1}{2} \times (1-(-3)) \times (3-1)$ $=8$ sq. units

Q74JEE Main 2022MCQ4MHyperbola

Let the foci of the ellipse $\;\frac{x^{2}}{16}+\;\frac{y^{2}}{7}=1$ and the hyperbola $\;\frac{x^{2}}{144}-\;\frac{y^{2}}{\alpha}=\;\frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is: [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Ellipse: $\;\frac{x^2}{16}+\;\frac{y^2}{7}=1$ Eccentricity $=\sqrt{1-\;\frac{7}{16}}=\;\frac{3}{4}$ ${\text{Foci}} \equiv (\pm a e, 0) \equiv (\pm 3,0)$ Hyperbola : $\;\frac{x^2}{(\;\frac{144}{25})}-\;\frac{y^2}{(\;\frac{\alpha}{25})}=1$ Eccentricity $=\sqrt{1+\;\frac{\alpha}{144}}=\;\frac{1}{12} \sqrt{144+\alpha}$ Foci $\equiv (\pm a e, 0) \equiv (\pm \;\frac{12}{5} \cdot \;\frac{1}{12} \sqrt{144+\alpha}, 0)$ If foci coincide then $3=\;\frac{1}{5} \sqrt{144+\alpha} \Rightarrow \alpha=81$ Hence, hyperbola is $\;\frac{x^2}{(\;\frac{12}{5})^2}-\;\frac{y^2}{(\;\frac{9}{5})^2}=1$ Length of latus rectum $=2 \cdot \;\frac{\;\frac{81}{25}}{\frac{12}{5}}=\;\frac{27}{10}$

Q75JEE Main 2022MCQ4MThree Dimensional Geometry

A plane $E$ is perpendicular to the two planes $2 x-2 y+z=0$ and $x-y+2 z=4$ , and passes through the point $P(1,-1,1)$ . If the distance of the plane $E$ from the point $Q(a, a, 2)$ is $3 \sqrt{2}$ , then $(P Q)^{2}$ is equal to [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

First plane, $P_1=2 x-2 y+z=0$ , normal vector $\equiv \overline{n}_1=(2,-2,1)$ Second plane, $P_2 \equiv x-y+2 z=4$ , normal vector $\equiv \overline{n}_2=(1,-1,2)$ Plane perpendicular to ${\text{P}}_1$ and ${\text{P}}_2$ will have normal vector $\overline{n}_3$ Where $\overline{n}_3=(\overline{n}_1 \times \overline{n}_2)$ Hence, $\overline{n}_3=(-3,-3,0)$ Equation of plane E through $P(1,-1,1)$ and $\overline{n}_3$ as normal vector $(x-1, y+1, z-1) \cdot (-3,-3,0)=0$ $\Rightarrow x+y=0 \equiv E$ Distance of $P Q(a, a, 2)$ from $E=|\;\frac{2 a}{\sqrt{2}}|$ as given, $|\;\frac{2 a}{\sqrt{2}}|=3 \sqrt{2} \Rightarrow a=\pm 3$ Hence, $Q \equiv (\pm 3, \pm 3,2)$ Distance $7 Q=\sqrt{21} \Rightarrow (P Q)^2=21$

Q76JEE Main 2022MCQ4MVector Algebra

The shortest distance between the lines $\;\frac{x+7}{-6}=\;\frac{y-6}{7}=z$ and $\;\frac{7-x}{2}=y-2=z-6$ is [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$L_1: \;\frac{x+7}{6}=\;\frac{y-6}{7}=\;\frac{z-0}{1}$ Any point on it $\vec{a}_1(-7,6,0)$ and $L_1$ is parallel to $\vec{b}_1(-6,7,1)$ $L_2: \;\frac{x-7}{-2}=\;\frac{y-2}{1}=\;\frac{z-6}{1}$ Any point on it $\vec{a}_2(7,2,6)$ and $L_2$ is parallel to $\vec{b}_2(-2,1,1)$ Shortest distance between ${\text{L}}_1$ and ${\text{L}}_2$ $=|\;{(\vec{a}_2-\vec{a}_1) \cdot \vec{(b_1} \times \vec{b}_2)}/{|{b_\vec{1} \times \vec{b}_2}|}|=|\;\frac{(-14,4,-6) \cdot (3,2,4)}{\sqrt{9+4+16} }|$ $=2 \sqrt{29}$

Q77JEE Main 2022MCQ4MVector Algebra

Let $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$ and let $\vec{b}$ be a vector such that $\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{a} \cdot \vec{b}=3$ . Then the projection of $\vec{b}$ on the vector $\vec{a}-\vec{b}$ is: [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$ $\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$ $\vec{a} \cdot \vec{b}=3$ $|\vec{a} \times \vec{b}|^{2+}|\vec{a} \cdot \vec{b}|^2=|\vec{a}|^2 \cdot \vec{|b|^2}$ $\Rightarrow 5+9=6|\vec{b}|^2$ $\Rightarrow |b|^2=\;\frac{7}{3}$ $|\vec{a}-\vec{b}|=\sqrt{|\vec{a}|^{2+}.\vec{|b}|^{2-}2 \vec{a} \cdot \vec{b}}=\sqrt{\;\frac{7}{3}}$ projection of $\vec{b}$ on $\vec{a}-\vec{b}=\;\frac{\vec{b} \cdot (\vec{a}-\vec{b})}{|\vec{a}-\vec{b}|}$ $=\;\frac{\vec{b} \cdot \vec{a}-|\vec{b}|^2}{\vec{a}-\vec{b} \mid }=\;{3-\frac{7}{3}}/{\sqrt{\;\frac{7}{3}}}$ $=\;\frac{2}{\sqrt{21}}$

Q78JEE Main 2022MCQ4MStatistics

If the mean deviation about median for the numbers $3,5,7,2 {k}, 12,16,21,24$ , arranged in the ascending order, is 6 then the median is [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Median $=\;\frac{2 k+12}{2}=k+6$ Mean deviation $=\sum \;\frac{|x_i-M|}{n}=6$ $\Rightarrow \;\frac{(k+3)+(k+1)+(k-1)+(6-k)+(6-k)+(10-k)+(15-k)+(18-k)}{8}$ $\therefore \;\frac{58-2 k}{8}=6$ $k=5$ Median $=\;\frac{2 \times 5+12}{2}=11$

Q79JEE Main 2022MCQ4MTrigonometric Ratios and Identities

$2 \sin (\;\frac{\pi}{22}) \sin (\;\frac{3 \pi}{22}) \sin (\;\frac{5 \pi}{22}) \sin (\;\frac{7 \pi}{22}) \sin (\;\frac{9 \pi}{22})$ is equal to : [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$2 \sin \;\frac{\pi}{22} \sin \;\frac{3 \pi}{22} \sin \;\frac{5 \pi}{22} \sin \;\frac{7 \pi}{22} \sin \;\frac{9 \pi}{22}$ $=2 \sin (\;\frac{11 \pi-10 \pi}{22}) \sin (\;\frac{11 \pi-8 \pi}{22}) \sin (\;\frac{11 \pi-6 \pi}{22}) \sin (\;\frac{11 \pi-4 \pi}{22}) \sin (\;\frac{11 \pi-2 \pi}{22})$ $=2 \cos \;\frac{\pi}{11} \cos \;\frac{2 \pi}{11} \cos \;\frac{3 \pi}{11} \cos \;\frac{4 \pi}{11} \cos \;\frac{5 \pi}{11}$ $=\;\frac{2 \sin \frac{32 \pi}{11}}{2^5 \sin \;\frac{\pi}{11}}$ $=\;\frac{1}{16}$

Q80JEE Main 2022MCQ4MSets and Relations

Consider the following statements: $P$ : Ramu is intelligent.Q : Ramu is rich.R : Ramu is not honest.The negation of the statement "Ramu is intelligent and honest if and only if Ramu is not rich" can be expressed as: [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$P:$ Ramu is intelligentQ: Ramu is rich ${\text{R}}$ : Ramu is not honestGiven statement, "Ramu is intelligent and honest if and only if Ramu is not rich" $=(P ∧ ∼ R) ⇔ ∼ Q$ So, negation of the statement is $∼[(P ∧ ∼ R) ⇔ ∼ Q]$ $=∼[\{∼(P ∧ ∼ R) ∨ ∼ Q\} ∧\{Q ∨(P ∧ ∼ R)\}]$ $=((P ∧ ∼ R) ∧ Q) ∨(∼ Q ∧(∼ P ∨ R))$

Q81JEE Main 2022NAT4MPermutations and Combinations

Let $A=\{1,2,3,4,5,6,7\}$ . Define $B=\{T ⊆ A$ : either $1 ∉ T$ or $2 \in T\}$ and $C=\{T ⊆ A: T$ the sum of all the elements of $T$ is a prime number $\}$ . Then the number of elements in the set $B ∪ C$ is____ [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\because (B ∪ C)^{'}=B^{'} ∩ C^{'}$ $B^{'}$ is a set containing sub sets of $A$ containing element 1 and not containing 2.And $C^{'}$ is a set containing subsets of $A$ whose sum of elements is not prime.So, we need to calculate number of subsets of $\{3,4,5,6,7\}$ whose sum of elements plus 1 is composite.Number of such 5 elements subset $=1$ Number of such 4 elements subset $=3$ (except selecting 3 or 7 )Number of such 3 elements subset $=6$ (except selecting $\{3,4,5\},\{3,6,7\},\{4,5,7\}$ or $\{5,6,7\}$ )Number of such 2 elements subset $=7$ (except selecting $\{3,7\},\{4,6\},\{5,7\}$ )Number of such 1 elements subset $=3$ (except selecting $\{4\}$ or $\{6\}$ )Number of such 0 elements subset $=1$ $n(B^{'} ∩ C^{'})=21 \Rightarrow n(B ∪ C)=2^{7-}21=107$

Q82JEE Main 2022NAT4MQuadratic Equation and Inequalities

Let $f(x)$ be a quadratic polynomial with leading coefficient 1 such that $f(0)=p, p \ne 0$ , and $f(1)=\;\frac{1}{3}$ . If the equations $f(x)=0$ and $f \circ f \circ f \circ f(x)=0$ have a common real root, then $f(-3)$ is equal to___ [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Let $f(x)=(x-\alpha)(x-\beta)$ It is given that $f(0)=p \Rightarrow \alpha \beta=p$ and $f(1)=\;\frac{1}{3} \Rightarrow (1-\alpha)(1-\beta)=\;\frac{1}{3}$ Now, let us assume that, $\alpha$ is the common root of $f(x)=0$ and fofofof $(x)=0$ fofofof $(x)=0$ $\Rightarrow \;\text{ fofof }\;(0)=0$ $\Rightarrow f o f(p)=0$ So, $f(p)$ is either $\alpha$ or $\beta$ . $(p-\alpha)(p-\beta)=\alpha$ $(\alpha \beta-\alpha)(\alpha \beta-\beta)=\alpha \Rightarrow (\beta-1)(\alpha-1) \beta=1( \because \alpha \ne 0)$ So, $\beta=3$ $(1-\alpha)(1-3)=\;\frac{1}{3}$ $\alpha=\;\frac{7}{6}$ $f(x)=(x-\;\frac{7}{6})(x-3)$ $f(-3)=(-3-\;\frac{7}{6})(3-3)=25$

Q83JEE Main 2022NAT4MMatrices and Determinants

Let $A=\begin{bmatrix}1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1\end{bmatrix}, a, b \in C$ . If for some $n \in ℕ, A^{n}=\begin{bmatrix}1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1\end{bmatrix}$ then $n+a+b$ is equal to [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$A=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}+\begin{bmatrix}0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0\end{bmatrix}=I+B$ $B^2=\begin{bmatrix}0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0\end{bmatrix}+\begin{bmatrix}0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0\end{bmatrix}=\begin{bmatrix}0 & 0 & a b \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$ $B^3=0$ $\therefore A^n=(1+B)^n={ }^n C_0 I+{ }^n C_1 B+{ }^n C_2 B^{2+}{ }^n C_3 B^{3+}. . .$ $=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}+\begin{bmatrix}0 & n a & n a \\ 0 & 0 & n b \\ 0 & 0 & 0\end{bmatrix}+\begin{bmatrix}0 & 0 & \\ \\ \frac{n(n-1) a b}{2} \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$ $=\begin{bmatrix}1 & n a & n a+\\ \\ \frac{n(n-1)}{2} a b \\ 0 & 1 & n b \\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}1 & 48 & 2160 \\ 0 & 1 & 48 \\ 0 & 0 & 1\end{bmatrix}$ On comparing we get $n a=48, n b=96$ and $n a+\;\frac{n(n-1)}{2} a b=2160$ $\Rightarrow a=4, n=12 \;\text{ and }\; b=8$ $n+a+b=24$

Q84JEE Main 2022NAT4MApplication of Derivatives

The sum of the maximum and minimum values of the function $f(x)=|5 x-7|+[x^{2}+2 x]$ in the interval $[\;\frac{5}{4}, 2]$ , where $[t]$ is the greatest integer $< t$ , is__ [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$f(x)=|5 x-7|+[x^{2+}2 x]$ $=|5 x-7|+\$ [(x+1)^2]$-1 $Critical points of$ f(x)=;\frac{7}{5}, \sqrt{5}-1, \sqrt{6}-1, \sqrt{7}-1, \sqrt{8}-1,2\therefore $Maximum or minimum value of$ f(x) $occur at critical points or boundary points$ \therefore f(;\frac{5}{4})=;\frac{3}{4}+4=;\frac{19}{4}f(;\frac{7}{5})=0+4=4 $as both$ |5 x-7| $and$ x^{2+}2 x $are increasing in nature after$ x=;\frac{7}{5}\therefore f(2)=3+8=11\therefore f(;\frac{7}{5}){\min }=4 $and$ f(2){\max }=11 $Sum is$ 4+11=15$

Q85JEE Main 2022NAT4MDifferential Equations

Let $y=y(x)$ be the solution of the differential equation $\;\frac{d y}{d x}=\;\frac{4 y^{3}+2 y x^{2}}{3 x y^{2}+x^{3}}, y(1)=1 .$ If for some $n \in ℕ, y(2) \in [n-1, n)$ , then $n$ is equal to __ [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{d y}{d x}=\;\frac{y}{x} \;\frac{(4 y^{2+}2 x^2)}{(3 y^{2+}x^2)}$ Put $y=v x$ $\Rightarrow \;\frac{d y}{d x}=v+x \;\frac{d v}{d x}$ $\Rightarrow v+x \;\frac{d v}{d x}=\;\frac{v(4 v^{2+}2)}{(3 v^{2+}1)}$ $\Rightarrow x \;\frac{d v}{d x}=v(\;\frac{(4 v^{2+}2-3 v^{2-}1)}{3 v^{2+}1})$ $\Rightarrow \int(3 v^{2+}1) \;\frac{d v}{v^{3+}v}=\int \;\frac{d x}{x}$ $\Rightarrow \ln |v^{3+}v|=\ln x+c$ $\Rightarrow \ln |(\;\frac{y}{x})^{3+}(\;\frac{y}{x})|=\ln x+C$ $↓ y(1)=1$ $\Rightarrow C=\ln 2$ $\therefore$ for $y(2)$ $\ln (\;\frac{y^3}{8}+\;\frac{y}{2})=2 \ln 2 \Rightarrow \;\frac{y^3}{8}+\;\frac{y}{2}=4$ $\Rightarrow [y(2)]=2$ $\Rightarrow n=3$

Q86JEE Main 2022NAT4MDifferential Equations

Let $f$ be a twice differentiable function on $R$ . If $f^{'}(0)=4$ and $f(x)+int_{0}^{x}(x-t) f^{'}(t) d t=(e^{2 x}+e^{-2 x}) \cos 2 x+\;\frac{2}{a} x$ , then $(2 a+1)^{5} a^{2}$ is equal to ___ [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Let the given equation be
$f(x) + \int_{0}^{x}(x-t) f'(t) dt = (e^{2 x}+e^{-2 x}) \cos 2 x+\;\frac{2}{a} x$
First, evaluate the equation at $x=0$ :
$f(0) + \int_{0}^{0}(0-t) f'(t) dt = (e^{0}+e^{0}) \cos 0+\;\frac{2}{a} \cdot 0$
$f(0) + 0 = (1+1) \cdot 1 + 0 \implies f(0) = 2$ .

Next, differentiate the integral term with respect to $x$ . Let $g(x) = \int_{0}^{x}(x-t) f'(t) dt$ .
We can write $g(x) = x \int_{0}^{x} f'(t) dt - \int_{0}^{x} t f'(t) dt$ .
Differentiating using the Leibniz rule:
$g'(x) = \left(1 \cdot \int_{0}^{x} f'(t) dt + x \cdot f'(x)\right) - x f'(x)$
$g'(x) = \int_{0}^{x} f'(t) dt = f(x) - f(0)$ .

Now, differentiate the entire given equation with respect to $x$ :
$f'(x) + g'(x) = \frac{d}{dx} \left( (e^{2x} + e^{-2x}) \cos 2x + \frac{2}{a} x \right)$
$f'(x) + (f(x) - f(0)) = (2e^{2x} - 2e^{-2x}) \cos 2x - 2(e^{2x} + e^{-2x}) \sin 2x + \frac{2}{a}$ .

Substitute $x=0$ into this differentiated equation:
$f'(0) + (f(0) - f(0)) = (2e^0 - 2e^0) \cos 0 - 2(e^0 + e^0) \sin 0 + \frac{2}{a}$
$f'(0) + 0 = (2-2) \cdot 1 - 2(1+1) \cdot 0 + \frac{2}{a}$
$f'(0) = 0 - 0 + \frac{2}{a}$
$f'(0) = \frac{2}{a}$ .

Given $f'(0)=4$ , we have:
$4 = \frac{2}{a} \implies a = \frac{2}{4} = \frac{1}{2}$ .

Finally, calculate $(2a+1)^5 a^2$ :
$(2(\frac{1}{2})+1)^5 (\frac{1}{2})^2 = (1+1)^5 \left(\frac{1}{4}\right) = 2^5 \cdot \frac{1}{4} = 32 \cdot \frac{1}{4} = 8$ .

The final answer is $\boxed{8}$ .
[CORRECT_OPTION: X]

Q87JEE Main 2022NAT4MDefinite Integrals

Let $a_{n}=int_{-1}^{n}(1+\;\frac{x}{2}+\;\frac{x^{2}}{3}+. . . . .+\;\frac{x^{n-1}}{n}) d x$ for every ${n} \in {N}$ . Then the sum of all the elements of the set $\{ {n} \in {N}: {a}_{ {n}} \in (2,30)\}$ is___ [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\because a_n=int_{-1}^n(1+\;\frac{x}{2}+\;\frac{x^2}{3}+. . .+\;\frac{x^{n-1}}{n}) d x$ $=\$ [x+;\frac{x^2}{2^2}+;\frac{x^3}{3^2}+. . . . . .+;\frac{x^n}{n^2}]_{-1}^na_n=;\frac{n+1}{1^2}+;\frac{n^{2-}1}{2^2}+;\frac{n^{3+}1}{3^2}+;\frac{n^{4-}1}{4^2}+. . .+;\frac{n^n+(-1)^{n+1}}{n^2} $Here,$ a_1=2, a_2=;\frac{2+1}{1}+;\frac{2^{2-}1}{2}=3+;\frac{3}{2}=;\frac{9}{2}a_3=4+2+;\frac{28}{9}=;\frac{100}{9}a_4=5+;\frac{15}{4}+;\frac{65}{9}+;\frac{255}{16}>31∴ $The required set is$ {2,3} . \because a_n ∈(2,30)∴ $Sum of elements$ =5$.

Q88JEE Main 2022NAT4MCircles

If the circles $x^{2}+y^{2}+6 x+8 y+16=0$ and $x^{2}+y^{2}+2(3-\sqrt{3}) x+2(4-\sqrt{6}) y=k+6 \sqrt{3}+8 \sqrt{6}, k>0$ , touch internally at the point $P(\alpha, \beta)$ , then $(\alpha+\sqrt{3})^{2}+(\beta+\sqrt{6})^{2}$ is equal to [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

The circle $x^{2+}y^{2+}6 x+8 y+16=0$ has centre $(-3,-4)$ and radius 3 units,The circle $x^{2+}y^{2+}2(3-\sqrt{3}) x+2(4-\sqrt{6}) y=k+6 \sqrt{3}+8 \sqrt{6}, k>0$ has centre $(\sqrt{3}-3, \sqrt{6}-4)$ and radius $\sqrt{k+34}$ $\because$ These two circles touch internally hence $\sqrt{3+6}=|\sqrt{k+34}-3|$ Here, $k=2$ is only possible $( \because k>0)$ Equation of common tangent to two circles is $2 \sqrt{3} x+2 \sqrt{6} y+16+6 \sqrt{3}+8 \sqrt{6}+k=0$ $\because k=2$ then equation is $x+\sqrt{2} y+3+4 \sqrt{2}+3 \sqrt{3}=0 . . . . . . \;\text{ (i) }\;$ $\because (\alpha, \beta)$ are foot of perpendicular from $(-3,-4)$ To line (i) then $\;\frac{\alpha+3}{1}=\;\frac{\beta+4}{\sqrt{2}}=\;\frac{-(-3-4 \sqrt{2}+3+4 \sqrt{2}+3 \sqrt{3})}{1+2}$ $\therefore \alpha+3=\;\frac{\beta+4}{\sqrt{2}}=-\sqrt{3}$ $\Rightarrow (\alpha+\sqrt{3})^2=9$ and $(\beta+\sqrt{6})^2=16$ $\therefore (\alpha+\sqrt{3})^{2+}(\beta+\sqrt{6})^2=25$

Q89JEE Main 2022NAT4MArea Under The Curves

Let the area enclosed by the $x$ -axis, and the tangent and normal drawn to the curve $4 x^{3}-3 x y^{2}+6 x^{2}-5 x y-8 y^{2}+9 x+14=0$ at the point $(-2,3)$ be ${A}$ . Then $8 {A}$ is equal to [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

First, we find the slope of the tangent to the curve $F(x,y) = 4 x^{3}-3 x y^{2}+6 x^{2}-5 x y-8 y^{2}+9 x+14=0$ at $P(-2,3)$ using implicit differentiation.
$\frac{\partial F}{\partial x} = 12x^2 - 3y^2 + 12x - 5y + 9$ and $\frac{\partial F}{\partial y} = -6xy - 5x - 16y$ .
At $(-2,3)$ , $\frac{\partial F}{\partial x} \Big|_{(-2,3)} = 12(4) - 3(9) + 12(-2) - 5(3) + 9 = 48 - 27 - 24 - 15 + 9 = -9$ .
At $(-2,3)$ , $\frac{\partial F}{\partial y} \Big|_{(-2,3)} = -6(-2)(3) - 5(-2) - 16(3) = 36 + 10 - 48 = -2$ .
The slope of the tangent is $m_T = -\frac{\partial F / \partial x}{\partial F / \partial y} = -\frac{-9}{-2} = -\frac{9}{2}$ .
The equation of the tangent at $(-2,3)$ is $y - 3 = -\frac{9}{2}(x + 2)$ , which simplifies to $9x + 2y + 12 = 0$ .
The x-intercept of the tangent is found by setting $y=0$ : $9x + 12 = 0 \Rightarrow x_T = -\frac{12}{9} = -\frac{4}{3}$ .

The slope of the normal is $m_N = -\frac{1}{m_T} = \frac{2}{9}$ .
The equation of the normal at $(-2,3)$ is $y - 3 = \frac{2}{9}(x + 2)$ , which simplifies to $2x - 9y + 31 = 0$ .
The x-intercept of the normal is found by setting $y=0$ : $2x + 31 = 0 \Rightarrow x_N = -\frac{31}{2}$ .

The area $A$ is that of a triangle with vertices $P(-2,3)$ , $X_T(-\frac{4}{3}, 0)$ , and $X_N(-\frac{31}{2}, 0)$ .
The base of the triangle on the x-axis is $b = |x_N - x_T| = |-\frac{31}{2} - (-\frac{4}{3})| = |-\frac{93}{6} + \frac{8}{6}| = |-\frac{85}{6}| = \frac{85}{6}$ .
The height of the triangle is the absolute value of the y-coordinate of $P$ , which is $h = |3| = 3$ .
The area $A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{85}{6} \times 3 = \frac{85}{4}$ .
Finally, $8A = 8 \times \frac{85}{4} = 2 \times 85 = 170$ .

The final answer is $\boxed{170}$ .

Q90JEE Main 2022NAT4MInverse Trigonometric Functions

Let $x=\sin (2 \tan ^{-1} \alpha)$ and $y=\sin (\;\frac{1}{2} \tan ^{-1} \;\frac{4}{3})$ . If $S=\{a \in R: y^{2}=1-x\}$ , then $sum_{\alpha \in S} 16 \alpha^{3}$ is equal to [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\because x=\sin (2 \tan ^{-1} \alpha)=\;\frac{2 \alpha}{1+\alpha^2} . . . . . . \;\text{ (i) }\;$ and $y=\sin (\;\frac{1}{2} \tan ^{-1} \;\frac{4}{3})=\sin (\sin ^{-1} \;\frac{1}{\sqrt{5}})=\;\frac{1}{\sqrt{5}}$ Now, $y^2=1-x$ $\;\frac{1}{5}=1-\;\frac{2 \alpha}{1+\alpha^2}$ $\Rightarrow 1+\alpha^2=5+5 \alpha^{2-}10 \alpha$ $\Rightarrow 2 \alpha^{2-}5 \alpha+2=0$ $\therefore \alpha=2, \;\frac{1}{2}$ $\therefore sum_{\alpha \in S} 16 \alpha^3=16 \times 2^{3+}16 \times \;\frac{1}{2^3}=130$

← All JEE Main 2022 papers Take as Timed Mock →