Previous Year Paper

29 Jun 2022 Shift 2

90 Questions360 Marks180 MinTake as Timed Mock →

🎯 Practice smarter, not harder

Practice 29 Jun 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 2022MCQ4MMotion in a Straight Line

A small toy starts moving from the position of rest under a constant acceleration. If it travels a distance of $10 {\text{m}}$ in ${\text{t}} {\text{s}}$ , the distance travelled by the toy in the next ${\text{t}} {\text{s}}$ will be : [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{u}}=0, \;\;$ Say acceleration is aFort s $\;\; 10=\;\frac{1}{2} {\text{at}}^2$ For $2 {\text{t}} {\ s}$ $10+x=\;\frac{1}{2} a(2 t)^2$ $\;\frac{10+x}{10}=\;\frac{4}{1}$ $x=30 m$

Q2JEE Main 2022MCQ4MProperties of Matter

At what temperature a gold ring of diameter $6.230 {\text{cm}}$ be heated so that it can be fitted on a wooden bangle of diameter $6.241 {\text{cm}}$ ? Both the diameters have been measured at room temperature $(27^{\circ} {\text{C}})$ .(Given : coefficient of linear thermal expansion of gold $\alpha_{ {\text{L}}}=1.4 \times 10^{-5} {\text{K}}^{-1}$ ) [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\Delta D=D \alpha \Delta T$ $\Delta T=\;\frac{0.011}{6.230 \times 1.4 \times 10^{-5}}$ $=126.11^{\circ} {\text{C}}$ $\Rightarrow T_f= {\text{T}}+\Delta T$ $=(27+126.11)^{\circ} {\text{C}}$ $=153.11^{\circ} {\text{C}}$

Q3JEE Main 2022MCQ4MElectrostatics

Two point charges $Q$ each are placed at a distance $d$ apart. A third point charge q is placed at a distance $x$ from mid-point on the perpendicular bisector. The value of $x$ at which charge q will experience the maximum Coulomb's force is : [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Force experienced by the charge $q$ $F=\;{k Q q x}/{[ (\;\frac{d}{2}) ^{2+}x^2]\$ ^{;\frac{3}{2}}} $For maximum Coulomb's force for$ {\text{x}};\frac{d F}{d x}=0 $On solving$ x=;{d}/{2 {\sqrt{2}}}$

Q4JEE Main 2022MCQ4MRay Optics and Optical Instruments

The speed of light in media 'A' and 'B' are $2.0 \times 10^{10} {\text{cm}} / {\text{s}}$ and $1.5 \times 10^{10} {\text{cm}} / {\text{s}}$ respectively. A ray of light enters from the medium $B$ to $A$ at an incident angle ' $\theta$ '. If the ray suffers total internal reflection, then [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\mu _A=\;\frac{3 \times 10^8}{2 \times 10^8}=1.5$ $\mu _B=\;\frac{3 \times 10^8}{1.5 \times 10^8}=2$ For TIR $\theta>i_c$ $\theta> sin \;^{-1} (\;\frac{1.5}{2})$ $\theta> sin \;^{-1} (\;\frac{3}{4})$

Q5JEE Main 2022MCQ4MAtoms and Nuclei

In the following nuclear reaction, $D {\longrightarrow }^{\alpha} D_1 {\longrightarrow }^{\beta^{-}} D_2 {\longrightarrow }^{\alpha} D_3 {\longrightarrow }^{\gamma } D_4$ Mass number of $D$ is 182 and atomic number is 74 . Mass number and atomic number of $D_4$ respectively will be____ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Equivalent reaction can be written as $D \longrightarrow D_4+2 \alpha+\beta^{-}+\gamma$ $\Rightarrow \;\text{ Mass number of }\; D_4=\;\text{ Mass number of }\; D-2 \times 4$ $=182-8=174$ $\Rightarrow \;\text{ Atomic number of }\; D_4$ $=\;\text{ Atomic number of }\; D-2 \times 2+1$ $=74-4+1=71$

Q6JEE Main 2022MCQ4MDual Nature of Matter and Radiation

The electric field at a point associated with a light wave is given by $E=200\$ [\sin (6 × 10^{15}) t+\sin (9 × 10^{15}) t]$ V^{-1} $Given :$ h=4.14 \times 10^{-15} {\text{eVs}} $If this light falls on a metal surface having a work function of$ 2.50 {\text{eV}}$, the maximum kinetic energy of the photoelectrons will be [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Frequency of EM waves $=\;\frac{6}{2 \pi} \times 10^{15}$ and $\;\frac{9}{2 \pi} \times 10^{15}$ Energy of one photon of these waves $= (4.14 \times 10^{-15} \times \;\frac{6}{2 \pi} \times 10^{15}) {\text{eV}}$ and $(4.14 \times 10^{-15} \times \;\frac{9}{2 \pi} \times 10^{15}) {\text{eV}}$ $=3.95 {\text{eV}}$ and $5.93 {\text{eV}}$ $\Rightarrow$ Energy of maximum energetic electrons $=5.93-2.50=3.43 {\text{eV}}$

Q7JEE Main 2022MCQ4MCurrent Electricity

A capacitor is discharging through a resistor $R$ . Consider in time $t_1$ , the energy stored in the capacitor reduces to half of its initial value and in time $t_2$ , the charge stored reduces to one eighth of its initial value. The ratio ${\text{t}}_1 / {\text{t}}_2$ will be [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

For a discharging capacitor when energy reduces to half the charge would become $\;{1}/{{\sqrt{2}}}$ times the initial value. $\Rightarrow (\;\frac{1}{2}) ^{1 / 2}=e^{-t_1 / \tau }$ Similarly, $(\;\frac{1}{2}) ^3=e^{-t_2 / \tau }$ $\Rightarrow \;\frac{t_1}{t_2}=\;\frac{1}{6}$

Q8JEE Main 2022MCQ4MThermodynamics

Starting with the same initial conditions, an ideal gas expands from volume $V_1$ to $V_2$ in three different ways. The work done by the gas is ${\text{W}}_1$ if the process is purely isothermal, ${\text{W}}_2$ , if the process is purely adiabatic and ${\text{W}}_3$ if the process is purely isobaric. Then, choose the correct option [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Comparing the area under the PV graph $A_3>A_1>A_2$ $\Rightarrow W_3>W_1>W_2$

Explanation Figure 1

Q9JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

Two long current carrying conductors are placed parallel to each other at a distance of $8 {\text{cm}}$ between them. The magnitude of magnetic field produced at mid-point between the two conductors due to current flowing in them is $300 \mu T$ . The equal current flowing in the two conductors is: [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

As $B_{\;\text{net }\;} \ne 0$ that is the wires are carrying current in opposite direction. $\;\frac{\mu _0 I \times 2}{2 \pi (4 \times 10^{-2}) }=30 \times 10^{-6} {\text{T}}$ $\Rightarrow I=\;\frac{30 \times 10^{-6}}{10^{-6}} {\text{A}}=30 {\text{A}} \;\text{ \in opposite direction. }\;$

Q10JEE Main 2022MCQ4MGravitation

The time period of a satellite revolving around earth in a given orbit is 7 hours. If the radius of orbit is increased to three times its previous value, then approximate new time period of the satellite will be [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$T_2^2= (\;\frac{R_2}{R_1}) ^3 T_1^2$ $\Rightarrow T_2=(3)^{3 / 2} \times 7 \approx 5.2 \times 7$ $T_2 ≅ 36 {\text{hrs}}$

Q11JEE Main 2022MCQ4MElectromagnetic Waves

The TV transmission tower at a particular station has a height of $125 {\text{m}}$ . For doubling the coverage of its range, the height of the tower should be increased by [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Range $R=\sqrt{2 h {\text{Re}}}$ Let the height be $h^{'}$ to double the range so $2 R=\sqrt{2 h^{'} {\text{Re}}}$ On solving $h^{'}=4 h$ ${\text{h}}^{'}=500 {\text{m}}$ So $\Delta {\text{h}}=375 {\text{m}}$

Q12JEE Main 2022MCQ4MOscillations

The motion of a simple pendulum executing S.H.M. is represented by the following equation. $y=A \sin (\pi t+\varphi )$ , where time is measured in second. The length of pendulum is [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\omega =\pi=\sqrt{\;\frac{g}{l}}$ So, $l=\;\frac{g}{\pi^2}$ $≃ 99.4 {\text{cm}}$ (Nearest value)

Q13JEE Main 2022MCQ4MThermodynamics

A vessel contains $16 {\text{g}}$ of hydrogen and $128 {\text{g}}$ of oxygen at standard temperature and pressure. The volume of the vessel in ${\text{cm}}^3$ is : [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Total number of moles are $n=n_{H_2}+n_{O_2}$ $=\;\frac{16}{2}+\;\frac{128}{32}$ $=12 \;\text{ moles }\;$ $\;\text{ Using }\; P V=n R T$ $V=\;\frac{n R T}{P}$ $=\;\frac{12 \times 8.31 \times 273.15}{10^5} {\text{m}}^3$ $=0.27 {\text{m}}^3=27 \times 10^4 {\text{cm}}^3$

Q14JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

Given below are two statements :Statement I : The electric force changes the speed of the charged particle and hence changes its kinetic energy; whereas the magnetic force does not change the kinetic energy of the charged particle.Statement II : The electric force accelerates the positively charged particle perpendicular to the direction of electric field. The magnetic force accelerates the moving charged particle along the direction of magnetic field.In the light of the above statements, choose the most appropriate answer from the options given below : [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Electric field accelerates the particle in the direction of field $(\vec{F}=q \vec{E}=m \vec{a})$ and magnetic field accelerates the particle perpendicular to the field $(\vec{F}=\vec{q v} \times \vec{B}=m \vec{a})$ .

Q15JEE Main 2022MCQ4MLaws of Motion

A block of mass $40 {\text{kg}}$ slides over a surface, when a mass of $4 {\text{kg}}$ is suspended through an inextensible massless string passing over frictionless pulley as shown below.The coefficient of kinetic friction between the surface and block is $0.02$ . The acceleration of block is. (Given ${\text{g}}=10 {\text{ms}}^{-2}$ .) [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$f_{r_{\max }}=\mu N$ $=0.02 \times 400$ $=8 {\text{N}}$ Let the acceleration is a as shown then. $40-T=4 a$ $T-8=40 a$ $\Rightarrow a=\;\frac{32}{44}=\;\frac{8}{11} {\text{m}} / {\ s}^2$

Explanation Figure 2

Q16JEE Main 2022MCQ4MWork, Power and Energy

In the given figure, the block of mass $m$ is dropped from the point ' $A$ '. The expression for kinetic energy of block when it reaches point ' $B$ ' is [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Loss in potential energy = gain in kinetic energy $- (m g (y-y_0) -m g y) =K E-0$ $\Rightarrow {\text{KE}}= {\text{mgy}}_0$

Q17JEE Main 2022MCQ4MLaws of Motion

A block of mass M placed inside a box descends vertically with acceleration 'a'. The block exerts a force equal to one-fourth of its weight on the floor of the box. The value of 'a' will be [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Using Newton's second law $m g-\;\frac{m g}{4}=m a$ $\Rightarrow a=\;\frac{3 g}{4}$

Explanation Figure 1

Q18JEE Main 2022MCQ4MElectrostatic Potential and Capacitance

If the electric potential at any point $(x, y, z) m$ in space is given by $V=3 x^2$ volt. The electric field at the point $(1,0,3) {\text{m}}$ will be : [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\vec{E}=-\;\frac{d V}{d x} \hat{i}$ $\vec{E}=-6 x \hat{i}$ So, $\vec{E}$ at $(1,0,3)$ is $\vec{E}=-6 \hat{i} {\text{V}} / {\text{m}}$

Q19JEE Main 2022MCQ4MCurrent Electricity

The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of $2 Ω$ . The value of internal resistance of each cell is [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

From diagram $i_p=\;\frac{E}{2+\;\frac{r}{2}}$ and $i_s=\;\frac{2 E}{2+2 r}$ given $i_p=i_s$ $\;\frac{1}{2+\;\frac{r}{2}}=\;\frac{1}{1+r}$ $1+r=2+\;\frac{r}{2}$ $r=2 Ω$

Explanation Figure 1

Q20JEE Main 2022MCQ4MMotion in a Plane

A person can throw a ball upto a maximum range of $100 {\text{m}}$ . How high above the ground he can throw the same ball? [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$R=\;{ {\text{u}}^2 sin \;2 \theta}/{ {\text{g}}} {\text{R}}_{\max }=\;{ {\text{u}}^2}/{ {\text{g}}}=100$ ${\text{H}}_{\max }=\;{ {\text{u}}^2}/{2 {\text{g}}}=\;\frac{100}{2}=50 {\text{m}}$

Q21JEE Main 2022NAT4MUnits and Measurements

The Vernier constant of Vernier callipers is $0.1 {\text{mm}}$ and it has zero error of $(-0.05) {\text{cm}}$ . While measuring diameter of a sphere, the main scale reading is $1.7 {\text{cm}}$ and coinciding vernier division is 5 . The corrected diameter will be ____ $\times 10^{-2} {\text{cm}}$ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Since zero error is negative, we will add $0.05 {\text{cm}}$ . $\Rightarrow \;\text{ Corrected reading }\;=1.7 {\text{cm}}+5 \times 0.1 {\text{mm}}+0.05 {\text{cm}}$ $=180 \times 10^{-2} {\text{cm}}$

Q22JEE Main 2022NAT4MProperties of Matter

A small spherical ball of radius $0.1 {\text{mm}}$ and density $10^4 {\text{kg}} {\text{m}}^{-3}$ falls freely under gravity through a distance h before entering a tank of water. If, after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of h will be _____m.(Given ${\text{g}}=10 {\text{ms}}^{-2}$ , viscosity of water $=1.0 \times 10^{-5} {\text{N}}- {\text{sm}}^{-2}$ ). [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\sqrt{2 g h}}=\;\text{ terminal speed }\;$ $\Rightarrow {\sqrt{2 g h}}=\;\frac{2}{9} \;\frac{r^2 g (\rho -\rho ^{'}) }{η}$ $=\;\frac{2}{9} \times \;\frac{10^{-8} \times 10 \times 9000}{10^{-5}}$ $\Rightarrow h=\;\frac{400}{2 g}$ $\Rightarrow h=20 {\text{m}}$

Q23JEE Main 2022NAT4MWaves

In an experiment to determine the velocity of sound in air at room temperature using a resonance tube, the first resonance is observed when the air column has a length of $20.0 {\text{cm}}$ for a tuning fork of frequency $400 {\text{Hz}}$ is used. The velocity of the sound at room temperature is $336 {\text{ms}}^{-1}$ . The third resonance is observed when the air column has a length of ___ ${\text{cm}}$ . [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$400=\;\frac{v}{4 (L_1+e) } . . . .$ . (i) $400=\;\frac{5 v}{4 (L_2+e) } . . . .$ . (ii) $\Rightarrow L_1+e=\;\frac{\lambda }{4}=21 {\text{cm}}$ $L_2+e=\;\frac{5 \lambda }{4}=105 {\text{cm}}$ $\Rightarrow {\text{e}}=1 {\text{cm}} & {\text{L}}_2=104 {\text{cm}}$

Q24JEE Main 2022NAT4MCurrent Electricity

Two resistors are connected in series across a battery as shown in figure. If a voltmeter of resistance $2000 Ω$ is used to measure the potential difference across $500 Ω$ resistor, the reading of the voltmeter will be _____V. [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

New $R_{\;\text{eff }\;}=\;\frac{2000 \times 500}{2500}+600 Ω=1000 Ω$ $\Rightarrow$ Reading of voltmeter $=\;\frac{400}{1000} \times 20=8$ volts

Q25JEE Main 2022NAT4MSemiconductor Electronics

A potential barrier of $0.4 {\text{V}}$ exists across a p-n junction. An electron enters the junction from the ${\text{n}}$ -side with a speed of $6.0 \times 10^5 {\text{ms}}^{-1}$ . The speed with which electron enters the ${\text{p}}$ side will be $\;\frac{x}{3} \times 10^5 {\text{ms}}^{-1}$ the value of ${\text{x}}$ is____(Given mass of electron $=9 \times 10^{-31} {\text{kg}}$ , charge on electron $=1.6 \times 10^{-19} {\text{C}}$ .) [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Conserving energy, $\;\frac{1}{2} m v^2=\;\frac{1}{2} m (6 \times 10^5) ^{2-}0.4 {\text{eV}}$ $\Rightarrow v=\sqrt{ (6 \times 10^5) ^{2-}\;\frac{2 \times 1.6 \times 10^{-19} \times 0.4}{9 \times 10^{-31}}}$ $=\sqrt{36 \times 10^{10}-\;\frac{1.28}{9} \times 10^{12}}$ $\Rightarrow v=\;\frac{14}{3} \times 10^5 {\text{m}} / {\ s}$ $\Rightarrow x=14$

Q26JEE Main 2022NAT4MElectrostatic Potential and Capacitance

The displacement current of $4.425 \mu {\text{A}}$ is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of $10^6 {\text{Vs}}^{-1}$ . The area of each plate of the capacitor is $40 {\text{cm}}^2$ . The distance between each plate of the capacitor is $x \times 10^{-3} {\text{m}}$ . The value of $x$ is_____(Permittivity of free space, ${\text{E}}_0=8.85 \times 10^{-12} {\text{C}}^2 {\text{N}}^{-1} {\text{m}}^{-2}$ ). [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$4.425 \mu A=\;\frac{E_0 A}{d} \times \;\frac{d V}{d t}$ $\Rightarrow d=\;\frac{8.85 \times 10^{-12} \times 40 \times 10^{-4}}{4.425 \times 10^{-6}} \times 10^6$ $\Rightarrow d=8 \times 10^{-3} {\text{m}}$ $\Rightarrow x=8$

Q27JEE Main 2022NAT4MRotational Motion

The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is $I_1$ . The same rod is bent into a ring and its moment of inertia about a diameter is $I_2$ . If $\;\frac{I_1}{I_2}$ is $\;\frac{x \pi^2}{3}$ , then the value of $x$ will be___ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$I_1=\;\frac{M L^2}{3} . . . . .(1)$ For ring: $I_2=\;\frac{M R^2}{2}$ and $2 \pi R=L$ $\Rightarrow I_2=\;\frac{M}{2} (\;\frac{L^2}{4 \pi^2}) . . . . .$ $\Rightarrow \;\frac{I_1}{I_2}=\;\frac{8 \pi^2}{3}$ $\Rightarrow x=8$

Q28JEE Main 2022NAT4MAtoms and Nuclei

The half life of a radioactive substance is 5 years. After $x$ years a given sample of the radioactive substance gets reduced to $6.25 \%$ of its initial value. The value of ${\text{x}}$ is____ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$N=N_0 e^{-\lambda t}$ $\Rightarrow \;\frac{6.25}{100}=e^{-\lambda t}$ $\Rightarrow e^{-\lambda t}=\;\frac{1}{16}= (\;\frac{1}{2}) ^4$ $\Rightarrow t=4 t_{1 / 2}$ $\Rightarrow t=20 \;\text{ years }\;$

Q29JEE Main 2022NAT4MWave Optics

In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by $5 \times 10^{-2} {\text{m}}$ towards the slits, the change in fringe width is $3 \times 10^{-3} {\text{cm}}$ . If the distance between the slits is $1 {\text{mm}}$ , then the wavelength of the light will be____ ${\text{nm}}$ . [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Fringe width $\beta=\;\frac{\lambda D}{d}$ $\Rightarrow |d \beta|=\;\frac{\lambda }{d}|d(D)|$ $\Rightarrow 3 \times 10^{-3} {\text{cm}}=\;\frac{\lambda }{1 m m} (5 \times 10^{-2} {\text{m}})$ $\Rightarrow \lambda =\;\frac{3 \times 10^{-8}}{5 \times 10^{-2}} {\text{m}}$ $\Rightarrow \lambda =600 {\text{nm}}$

Q30JEE Main 2022NAT4MAlternating Current

An inductor of $0.5 {\text{mH}}$ , a capacitor of $200 \mu {\text{F}}$ and a resistor of $2 Ω$ are connected in series with a $220 {\text{V}}$ ac source. If the current is in phase with the emf, the frequency of ac source will be ______ $\times 10^2 {\text{Hz}}$ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Current will be in phase with emf when $\omega L=\;\frac{1}{\omega C}$ $\Rightarrow \omega =\;{1}/{{\sqrt{L C}}}=\;{1}/{\sqrt{5 \times 10^{-4} \times 2 \times 10^{-4}}}$ $\Rightarrow \omega =\;{10^4}/{{\sqrt{10}}} {\text{rad}} / {\ s}$ $\Rightarrow f=\;\frac{1}{2 \pi} \times \;{10^4}/{{\sqrt{10}}} {\text{Hz}}$ $\Rightarrow {\text{f}} ≃ 500 {\text{Hz}}$

Chemistry30 questions

Q31JEE Main 2022MCQ4MSome Basic Concepts of Chemistry

Using the rules for significant figures, the correct answer for the expression $\;\frac{0.02858 \times 0.112}{0.5702}$ will be [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{0.02858 \times 0.112}{0.5702}=0.00561$ Reported answer should not be more precise than least precise term $(0.112$ is the least precise term with three significant figures) in calculations, so there should be three significant figures in reported answer.

Q32JEE Main 2022MCQ4MStructure of Atom

Which of the following is the correct plot for the probability density $ψ^2$ ( $r$ ) as a function of distance ' $r$ ' of the electron from the nucleus for $2 s$ orbital? [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Formula for number of radial nodes in ${\text{n}}^{\;\text{th }\;}$ orbital $= {\text{n}}-l-1$ For $2 {\ s}$ , number of radial nodes $=2-0-1=1$ and value of $ψ^2$ is always positive.

Q33JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

Consider the species ${\text{CH}}_4, {\text{NH}}_4^{+}$ and ${\text{BH}}_4^{-}$ . Choose the correct option with respect to the these species. [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

Q34JEE Main 2022MCQ4MChemical Equilibrium

$4.0$ moles of argon and $5.0$ moles of ${\text{PCl}}_5$ are introduced into an evacuated flask of 100 litre capacity at $610 {\text{K}}$ . The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be $6.0 {\text{atm}}$ . The ${\text{K}}_{ {\text{p}}}$ for the reaction is[Given : $. {\text{R}}=0.082 {\text{L}} {\text{atm}} {\text{K}}^{-1} {\text{mol}}^{-1}]$ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\text{table} ,,,,\text{Total Moles:};,{\text{PCl}}_5( {\text{g}}) ⇌ ,{\text{PCl}}_3( {\text{g}})+ ,{\text{Cl}}_2( {\text{g}}),;\text{At} \; t=0:, 5,0,0, 5+4(\text{For} \text{Ne} \text{gas})=9 ;\text{At}\; t=t_{eq}: , 5-x, x, x, 5-x+x+x+4(\text{For} \text{Ne} \text{gas})=9 +x ;\text{Mole Fraction:} , \frac{5-x}{9+x}, x/{9+x}, x/{9+x};\text{Partial Pressure:}, (\frac{5-x}{9+x})P, (x/{9+x})^p, (x/{9+x})^P$ Here 4 moles of inert gas argon also present. $\therefore$ Total moles of mixture present at equilibrium, $n_T=5+x+4$ $=9+x$ At equilibrium, total pressure $( {\text{p}}_{ {\text{T}}}) =6 {\text{atm}}$ Volume $(v)=100 {\text{L}}$ Temperature $=610 {\text{K}}$ $\therefore$ using ideal gas equation, $P_T V=n_T R T$ $\Rightarrow 6 \times 100=(9+x) \times 0.082 \times 610$ $\Rightarrow x=3$ Now, $K_P=\;{P_{P {\text{Cl}}_3} \times P_{ {\text{Cl}}_2}}/{P_{P C l_5}}$ $=\;\frac{ [\frac{3}{9+3} \times 6] \times [\frac{3}{9+3} \times 6] }{ [\;\frac{5}{9+3} \times 6] }$ $=\;\frac{27}{12}$ $=\;\frac{9}{4}$ $=2.25 {\text{atm}}$ Note :Inert gas always contribute to total mole and pressure calculation.

Q35JEE Main 2022MCQ4MSolutions

A $42.12 \%( {\text{w}} / {\text{v}})$ solution of ${\text{NaCl}}$ causes precipitation of a certain sol in 10 hours. The coagulating value of ${\text{NaCl}}$ for the sol in 2 hours is :$[Given : Molar mass : ${\text{Na}}=23.0 {\text{g}} {\text{mol}}^{-1} ; {\text{Cl}}=35.5 {\text{g}} {\text{mol}}^{-1}$ ]$ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

Q36JEE Main 2022MCQ4MClassification of Elements

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.Assertion A : The first ionization enthalpy for oxygen is lower than that of nitrogen.Reason ${\text{R}}$ : The four electrons in $2 {\text{p}}$ orbitals of oxygen experience more electron-electron repulsion.In the light of the above statements, choose the correct answer from the options given below. [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Nitrogen has half-filled p-orbitals which is stable. Due to this, its $1^{\;\text{st }\;}$ ionization energy is more than oxygen.

Q37JEE Main 2022MCQ4Md and f Block Elements

Match List I with List II. List-l List-II (A) Siderite (I) $FeCO_3$ (B) Malachite (II) $CuCO_3 \;.Cu(OH)_2$ (C) Sphalerite (III) $ZnS$ (D) Calamine (IV) $ZnCO_3$ Choose the correct answer from the options given below : [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Siderite $\to {\text{FeCO}}_3$ Malachite $\to {\text{CuCO}}_3 . {\text{Cu}}( {\text{OH}})_2$ Sphalerite $\to {\text{ZnS}}$ Calamine $\to {\text{ZnCO}}_3$

Q38JEE Main 2022MCQ4MCoordination Compounds

Given below are two statements.- Statement I : In ${\text{CuSO}}_4 \cdot 5 {\text{H}}_2 {\text{O}}, {\text{Cu}}- {\text{O}}$ bonds are present.- Statement II : In ${\text{CuSO}}_4 \cdot 5 {\text{H}}_2 {\text{O}}$ , ligands coordinating with ${\text{Cu}}( {\text{II}})$ ion are O-and S-based ligands.In the light of the above statements, choose the correct answer from the options given below. [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Statement I is true but statement II is false. Only oxygen atom forms a Co-ordinate bond with ${\text{Cu}}^{+2}$ in ${\text{CuSO}}_4 \cdot 5 {\text{H}}_2 {\text{O}}$

Explanation Figure 1

Q39JEE Main 2022MCQ4MSalt Analysis

Amongst baking soda, caustic soda and washing soda, carbonate anion is present in [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Baking soda $\to {\text{NaHCO}}_3$ Washing soda $\to {\text{Na}}_2 {\text{CO}}_3 \cdot 10 {\text{H}}_2 {\text{O}}$ Caustic soda $\to {\text{NaOH}}$ ${\text{CO}}_3^{-2}$ ion is present only in washing soda.

Q40JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

Number of lone pair(s) of electrons on central atom and the shape ${\text{BrF}}_3$ molecule respectively, are [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Steric no. $=5 ( {\ sp}^3 {\text{d}})$ , lone pair $=2$ Bent T shape.

Explanation Figure 1

Q41JEE Main 2022MCQ4Mp Block Elements

Aqueous solution of which of the following boron compounds will be strongly basic in nature? [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{Na}}_2 {\text{B}}_4 {\text{O}}_7+7 {\text{H}}_2 {\text{O}} \longrightarrow 2 {\text{H}}_3 {\text{BO}}_3+2 {\text{Na}} [ {\text{B}}( {\text{OH}})_4]$ Aqueous solution of borax is buffer whose ${\text{pH}} ≃ 9$ Other compounds are less basic than this.

Q42JEE Main 2022MCQ4Mp Block Elements

Sulphur dioxide is one of the components of polluted air. ${\text{SO}}_2$ is also a major contributor to acid rain. The correct and complete reaction to represent acid rain caused by ${\text{SO}}_2$ is [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$2 {\text{SO}}_2+ {\text{O}}_2+2 {\text{H}}_2 {\text{O}} \longrightarrow 2 {\text{H}}_2 {\text{SO}}_4$ Acid rain occurs due to increased concentration of oxides of sulphur and Nitrogen.

Q43JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Which of the following carbocations is most stable? [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

Q44JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

The stable carbocation formed in the above reaction is [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Initially ${\text{CH}}_3- {\text{CH}}_2- {\text{CH}}_2^{+}$ is formed. On rearrangement ${\text{CH}}_3- {\text{CH}}- {\text{CH}}_3$ stable carbocation is formed.

Q45JEE Main 2022MCQ4MHaloalkanes and Haloarenes

Two isomers $( {\text{A}})$ and $( {\text{B}})$ with Molar mass $184 {\text{g}} / {\text{mol}}$ and elemental composition ${\text{C}}, 52.2 \% ; {\text{H}}, 4.9 \%$ and ${\text{Br}} 42.9 \%$ gave benzoic acid and p-bromobenzoic acid, respectively on oxidation with ${\text{KMnO}}_4$ . Isomer 'A' is optically active and gives a pale yellow precipitate when warmed with alcoholic ${\text{AgNO}}_3$ . Isomer ' $A$ ' and ' ${\text{B}}$ ' are, respectively [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

Q46JEE Main 2022MCQ4MAmines

In Friedel-Crafts alkylation of aniline, one gets [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

Q47JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R.Assertion A : Dacron is an example of polyester polymer.Reason $R$ : Dacron is made up of ethylene glycol and terephthalic acid monomers.In the light of the above statements, choose the most appropriate answer from the options given below. [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

Q48JEE Main 2022MCQ4MBiomolecules

The structure of protein that is unaffected by heating is [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

Q49JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

The mixture of chloroxylenol and terpineol is an example of [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

Q50JEE Main 2022MCQ4MSalt Analysis

A white precipitate was formed when ${\text{BaCl}}_2$ was added to water extract of an inorganic salt. Further, a gas ' $X$ ' with characteristic odour was released when the formed white precipitate was dissolved in dilute ${\text{HCl}}$ . The anion present in the inorganic salt is [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Anion is ${\text{SO}}_3^{-2}$ ${\text{BaSO}}_3 {\longrightarrow }^{\;\text{ dil }\; {\text{HCl}}} {\text{SO}}_2_{x(gas)} ↑$ Gas is released with smell of burning sulphur.

Q51JEE Main 2022NAT4MStates of Matter

A box contains $0.90 {\text{g}}$ of liquid water in equilibrium with water vapour at $27^{\circ} {\text{C}}$ . The equilibrium vapour pressure of water at 27 ${ }^{\circ} {\text{C}}$ is $32.0$ Torr. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates, then the volume of the box must be ____litre. [nearest integer](Given : ${\text{R}}=0.082 {\text{L}} {\text{atm}} {\text{K}}^{-1} {\text{mol}}^{-1}$ )(Ignore the volume of the liquid water and assume water vapours behave as an ideal gas.) [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

We know, 760 Torr $=1 {\text{atm}}$ $\therefore 32$ Torr $=\;\frac{32}{760} {\text{atm}}$ As all the liquid water evaporates so entire water is in gaseous state. $\therefore$ Weight of water vapour $=0.9 {\text{g}}$ $\therefore$ Moles of water vapour $(n)=\;\frac{0.9}{18}$ Pressure $(P)=\;\frac{32}{760} {\text{atm}}$ Temperature $(T)=(27+273) K=300 {\text{K}}$ ${\text{R}}=0.082 {\text{L}}$ atm ${\text{K}}^{-1} {\text{mol}}^{-1}$ Given water vapour act as an ideal gas, so we can apply ideal gas equation.From ideal gas equation, ${\text{PV}}= {\text{nRT}}$ $\Rightarrow \;\frac{32}{760} \times v=\;\frac{0.9}{18} \times 0.082 \times 300$ $\Rightarrow v=29 {\text{L}}$

Q52JEE Main 2022NAT4MChemical Thermodynamics

$2.2 {\text{g}}$ of nitrous oxide $( {\text{N}}_2 {\text{O}})$ gas is cooled at a constant pressure of $1 {\text{atm}}$ from $310 {\text{K}}$ to $270 {\text{K}}$ causing the compression of the gas from $217.1 {\text{mL}}$ to $167.75 {\text{mL}}$ . The change in internal energy of the process, $\Delta U$ is ' $-x^{'} J$ . The value of ' $x$ ' is ______[nearest integer](Given : atomic mass of ${\text{N}}=14 {\text{g}} {\text{mol}}^{-1}$ and of ${\text{O}}=16 {\text{g}} {\text{mol}}^{-1}$ . Molar heat capacity of ${\text{N}}_2 {\text{O}}$ is $100 {\text{J}} {\text{K}}^{-1} {\text{mol}}^{-1}$ ) [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\Delta T\;=-40 {\text{K}}$ $\Delta U\;=q+w$ $\;=\;\frac{100 \times 2.2}{44}(-40)-(-49.39) \times 10^{-3} \times 101.325$ $\;=-200+5$ $\;=-195 {\text{J}}$ ${\text{x}}=\;195$

Q53JEE Main 2022NAT4MSolutions

Elevation in boiling point for $1.5$ molal solution of glucose in water is $4 {\text{K}}$ . The depression in freezing point for $4.5$ molal solution of glucose in water is $4 {\text{K}}$ . The ratio of molal elevation constant to molal depression constant $( {\text{K}}_{ {\text{b}}} / {\text{K}}_{ {\text{f}}})$ is____ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\Delta {\text{T}}_{ {\text{b}}}= {\text{iK}}_{ {\text{b}}} {\text{m}}$ $\Delta {\text{T}}_{ {\text{f}}}= {\text{iK}}_{ {\text{f}}} {\text{m}}$ $\;\frac{4}{4}=\;{ {\text{K}}_{ {\text{b}}} 1.5}/{ {\text{K}}_{ {\text{f}}} 4.5}$ $\;{ {\text{K}}_{ {\text{b}}}}/{ {\text{K}}_{ {\text{f}}}}=3$

Q54JEE Main 2022NAT4MElectrochemistry

The cell potential for the given cell at $298 {\text{K}}$ ${\text{Pt}}| {\text{H}}_2( {\text{g}}, 1 {\bar})| {\text{H}}^{+}( {\text{aq}}) \| {\text{Cu}}^{2+}( {\text{aq}}) \mid {\text{Cu}}( {\text{s}})$ is $0.31 {\text{V}}$ . The ${\text{pH}}$ of the acidic solution is found to be 3 , whereas the concentration of ${\text{Cu}}^{2+}$ is $10^{-x} {\text{M}}$ . The value of ${\text{x}}$ is_____(Given : $E_{C u^{2+} / C u}^{Θ}=0.34 {\text{V}}$ and $\;\frac{2.303 R T}{F}=0.06 {\text{V}}$ ) [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{Q}}=\;{[ {\text{H}}^{+}]\$ ^2}/{[ {\text{Cu}}^2]$ {\text{pH}}_2}=;{10^{-6}}/{ {\text{C}}} ;; {\text{pH}}2=1E=E{;\text{cell };}^{\circ}-;\frac{0.06}{n} \log Q0.31=0.34-;\frac{0.06}{2} \log ;\frac{10^{-6}}{C}\log ;\frac{10^{-6}}{C}=1C=10^{-7} Mx=7$

Q55JEE Main 2022NAT4MChemical Kinetics

The equation $k=(6.5 \times 10^{12} {\text{S}}^{-1}) {\text{e}}^{-26000 {\text{K}} / {\text{T}}}$ is followed for the decomposition of compound ${\text{A}}$ . The activation energy for the reaction is ${\text{kJ}} {\text{mol}}^{-1}$ . [nearest integer](Given : ${\text{R}}=8.314 {\text{JK}}^{-1} {\text{mol}}^{-1}$ ) [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$k=A e^{\;\frac{-E_a}{R T}}$ $\;\frac{E_a}{R T}\;=\;\frac{26000}{T}$ $E_a\;=26000 \times 8.314$ $\;=216164 {\text{J}}$ $\;=216 {\text{kJ}}$

Q56JEE Main 2022NAT4MCoordination Compounds

Spin only magnetic moment of $[ {\text{MnBr}}_6]^{4-}$ is _____ B.M. (round off to the closest integer) [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{Mn}}^{2+} \to {\text{t}}_{2 {\text{g}}^{111} e_{ {\text{g}}^{11}}}$ $\mu _{ {\ s}}={\sqrt{35}}$ $=5.91$ $=6$

Q57JEE Main 2022NAT4MCoordination Compounds

For the reaction given below: ${\text{CoCl}}_3 \cdot {\text{xNH}}_3+ {\text{AgNO}}_3( {\text{aq}}) \to$ If two equivalents of ${\text{AgCl}}$ precipitate out, then the value of ${\text{x}}$ will be_____ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{CoCl}}_3 \cdot {\text{xNH}}_3+ {\text{AgNO}}_3 \to \;{{ {\text{AgCl}}}}_{{2 {\text{mol}}}} ↓$ $[ {\text{Co}} ( {\text{NH}}_3) _5 {\text{Cl}}] {\text{Cl}}_2+ {\text{AgNO}}_3 \to \;{{ {\text{AgCl}}}}_{{2 {\text{mol}}}} ↓$ $x=5$

Q58JEE Main 2022NAT4MAlcohols, Phenols and Ethers

The number of chiral alcohol(s) with molecular formula ${\text{C}}_4 {\text{H}}_{10} {\text{O}}$ is_____ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

Q59JEE Main 2022NAT4MAldehydes, Ketones and Carboxylic Acids

In the given reaction, the number of ${sp}^2$ hybridised carbon(s) in compound ' ${\text{X}}$ ' is __ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

Q60JEE Main 2022NAT4MAlcohols, Phenols and Ethers

In the given reaction, The number of $\pi$ electrons present in the product ' ${\text{P}}$ ' is [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

Mathematics30 questions

Q61JEE Main 2022MCQ4MComplex Numbers

Let $\alpha$ be a root of the equation $1+x^{2+}x^4=0$ . Then, the value of $\alpha^{1011}+\alpha^{2022}-\alpha^{3033}$ is equal to : [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Given, $\alpha$ is a root of the equation $1+x^{2+}x^4=0$ $\therefore \alpha$ will satisfy the equation. $\therefore 1+\alpha^{2+}\alpha^4=0$ $\alpha^2=\;{-1 \pm {\sqrt{1-4}}}/{2}$ $=\;{-1 \pm {\sqrt{3}} i}/{2}$ $\therefore \alpha^2=\omega a r \omega ^2$ Now, $\alpha^{1011}+\alpha^{2022}-\alpha^{3033}$ $=\alpha \cdot (\alpha^2) ^{505}+ (\alpha^2) ^{1011}-\alpha \cdot (\alpha^2) ^{1516}$ $=\alpha(\omega )^{505}+(\omega )^{1011}-\alpha \cdot (\omega )^{1516}$ $=\alpha \cdot (\omega ^3) ^{168} \cdot \omega + (\omega ^3) ^{337}-\alpha \cdot (\omega ^3) ^{505} \cdot \omega$ $=\alpha \omega +1-\alpha \omega$ $=1$

Q62JEE Main 2022MCQ4MComplex Numbers

Let $\arg (z)$ represent the principal argument of the complex number $z$ . Then, $|z|=3$ and $\arg (z-1)-\arg (z+1)=\;\frac{\pi}{4}$ intersect [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Let $z=x+i y$ $\therefore |z|={\sqrt{x^{2+}y^2}}$ Given, $|z|=3$ $\therefore {\sqrt{x^{2+}y^2}}=3$ $\Rightarrow x^{2+}y^2=9=3^2$ This represent a circle with center at $(0,0)$ and radius $=3$ Now, given $\arg (z-1)-\arg (z+1)=\;\frac{\pi}{4}$ $\Rightarrow \arg (x+i y-1)-\arg (x+i y+1)=\;\frac{\pi}{4}$ $\Rightarrow \arg (x-1+i y)-\arg (x+1+i y)=\;\frac{\pi}{4}$ $\Rightarrow \tan ^{-1} (\;\frac{y}{x-1}) -\tan ^{-1} (\;\frac{y}{x+1}) =\;\frac{\pi}{4}$ $\Rightarrow \tan ^{-1} (\;\frac{\;\frac{y}{x-1}-\frac{y}{x+1}}{1+\;\frac{y}{x-1} \times \;\frac{y}{x+1}}) =\;\frac{\pi}{4}$ $\Rightarrow \tan ^{-1} (\;\frac{\;\frac{x y+y-x y+y}{x^{2-}1}}{\frac{x^{2-}1+y^2}{x^{2-}1}}) =\;\frac{\pi}{4}$ $\Rightarrow \tan ^{-1} (\;\frac{x y+y-x y+y}{x^{2-}1+y^2}) =\;\frac{\pi}{4}$ $\Rightarrow \;\frac{2 y}{x^{2-}1+y^2}=\tan (\;\frac{\pi}{4})$ $\Rightarrow 2 y=x^{2+}y^{2-}1$ $\Rightarrow x^{2+}y^{2-}2 y-1=0$ $\Rightarrow x^{2+}(y-1)^2=({\sqrt{2}})^2$ This represent a circle with center at $(0,1)$ and radius ${\sqrt{2}}$ . From diagram you can see both the circles do not cut anywhere.

Explanation Figure 1

Q63JEE Main 2022MCQ4MBinomial Theorem

Let $A=(\text{table} 2 , -1 ; 0 , 2)$ . If $B=I-{ }^5 C_1(\text{adj} \;A)+{ }^5 C_2(\text{adj} \;A)^{2-}. . . . .-{ }^5 C_5(\text{adj} \;A)^5$ , then the sum of all elements of the matrix $B$ is [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Given $A={\begin{bmatrix}2 & -1 \\ 0 & 2\end{bmatrix}}$ and $B=I-5_{C_1}({\text{adj}} A)+5_{C_2}({\text{adj}} A)^{2-}5_{C_3}({\text{adj}} A)^{3+}5_{C_4}({\text{adj}} A)^{4-}5_{C_5}({\text{adj}} A)^5$ $=(I-({\text{adj}} A))^5$ $\;\text{ Cofactor of }\; A={\begin{bmatrix}(-1)^{1+1} \cdot 2 & (-1)^{1 + 2} \cdot 0 \\ (-1)^{2+1} \cdot (-1) & (-1)^{2+2} \cdot 2\end{bmatrix}\$ }={\begin{bmatrix}2 & 0 \ 1 & 2\end{bmatrix}} $Transpose of cofactor of$ A={\begin{bmatrix}2 & 1 \ 0 & 2\end{bmatrix}}∴ {\text{adj}} A={\begin{bmatrix}2 & 1 \ 0 & 2\end{bmatrix}} $Now,$ I-a d j A={\begin{bmatrix}1 & 0 \ 0 & 1\end{bmatrix}}-{\begin{bmatrix}2 & 1 \ 0 & 2\end{bmatrix}}={\begin{bmatrix}-1 & -1 \ 0 & -1\end{bmatrix}} $Now let,$ P=I-{\text{adj}} A={\begin{bmatrix}-1 & -1 \ 0 & -1\end{bmatrix}}∴ P^2={\begin{bmatrix}-1 & -1 \ 0 & -1\end{bmatrix}}{\begin{bmatrix}-1 & -1 \ 0 & -1\end{bmatrix}}={\begin{bmatrix}1 & 2 \ 0 & 1\end{bmatrix}}P^4=P^2 ⋅ P^2={\begin{bmatrix}1 & 2 \ 0 & 1\end{bmatrix}}{\begin{bmatrix}1 & 2 \ 0 & 1\end{bmatrix}}={\begin{bmatrix}1 & 4 \ 0 & 1\end{bmatrix}}P^5=P^4 ⋅ P={\begin{bmatrix}1 & 4 \ 0 & 1\end{bmatrix}}{\begin{bmatrix}-1 & -1 \ 0 & -1\end{bmatrix}}={\begin{bmatrix}-1 & -5 \ 0 & -1\end{bmatrix}}∴ B={\begin{bmatrix}-1 & -5 \ 0 & -1\end{bmatrix}} $Now \sum of elements$ =-1-5-1+0=-7$

Q64JEE Main 2022MCQ4MSequences and Series

The sum of the infinite series $1+\;\frac{5}{6}+\;\frac{12}{6^2}+\;\frac{22}{6^3}+\;\frac{35}{6^4}+\;\frac{51}{6^5}+\;\frac{70}{6^6}+. . . . . .$ is equal to: [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Let the sum of the infinite series be $S$ .
$S = 1+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+\frac{35}{6^4}+\dots$
This is an arithmetico-geometric series. We apply the method of differences. Multiply the series by the common ratio of the geometric part, which is $\frac{1}{6}$ .
$\frac{1}{6}S = \frac{1}{6}+\frac{5}{6^2}+\frac{12}{6^3}+\frac{22}{6^4}+\dots$
Subtracting this from the original series:
$S - \frac{1}{6}S = (1-0) + (\frac{5}{6}-\frac{1}{6}) + (\frac{12}{6^2}-\frac{5}{6^2}) + (\frac{22}{6^3}-\frac{12}{6^3}) + \dots$
$\frac{5}{6}S = 1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+\dots$ . Let this be equation (1).
The numerators of the terms on the right (after the first term) form an arithmetic progression $4, 7, 10, \dots$ . Let's apply the method again. Multiply (1) by $\frac{1}{6}$ :
$\frac{1}{6}\left(\frac{5}{6}S\right) = \frac{1}{6}+\frac{4}{6^2}+\frac{7}{6^3}+\frac{10}{6^4}+\dots$ . Let this be equation (2).
Subtracting (2) from (1):
$\frac{5}{6}S - \frac{5}{36}S = (1-0) + (\frac{4}{6}-\frac{1}{6}) + (\frac{7}{6^2}-\frac{4}{6^2}) + \dots$
$\frac{25}{36}S = 1 + \frac{3}{6} + \frac{3}{6^2} + \frac{3}{6^3} + \dots = 1 + \frac{3}{6}\left(1+\frac{1}{6}+\frac{1}{6^2}+\dots\right)$
This contains a simple geometric series with sum $\frac{1}{1-1/6} = \frac{6}{5}$ .
$\frac{25}{36}S = 1 + \frac{1}{2}\left(\frac{6}{5}\right) = 1 + \frac{3}{5} = \frac{8}{5}$ .
Therefore, $S = \frac{8}{5} \times \frac{36}{25} = \frac{288}{125}$ .

Q65JEE Main 2022MCQ4MLimits, Continuity and Differentiability

The value of $\lim _{x \to 1} \;\frac{(x^{2-}1) \sin ^2(\pi x)}{x^{4-}2 x^{3+}2 x-1}$ is equal to: [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$S=1+\;\frac{5}{6}+\;\frac{12}{6^2}+\;\frac{22}{6^3}+. . . . . .(1)$ $\;\frac{1}{6} S=\;\frac{1}{6}+\;\frac{5}{6^2}+\;\frac{12}{6^3}+. . . . . .(2)$ $S-\;\frac{1}{6} S=1+\;\frac{4}{6}+\;\frac{7}{6^2}+\;\frac{10}{6^3}+. . . . . . . . .$ $\Rightarrow \;\frac{5 S}{6}=1+\;\frac{4}{6}+\;\frac{7}{6^2}+\;\frac{10}{6^3}+. . . . . . \;\text{ (3) }\;$ Now, multiplying both sides by $\;\frac{1}{6}$ , we get $\Rightarrow \;\frac{5 S}{36}=\;\frac{1}{6}+\;\frac{4}{6^2}+\;\frac{7}{6^3}+\;\frac{10}{6^4}+.$ Subtract equation (4) from equation (3), we get $\;\frac{25}{36} S=1+\;\frac{3}{6}+\;\frac{3}{6^2}+\;\frac{3}{6^3}+. . . . . .$ $\Rightarrow \;\frac{25 S}{36}=1+\;\frac{\;\frac{3}{5}}{1-\;\frac{1}{6}}$ $=1+\;\frac{3}{6} \times \;\frac{6}{5}$ $=1+\;\frac{3}{5}=\;\frac{8}{5}$ $\Rightarrow S=\;\frac{8}{5} \times \;\frac{36}{25}=\;\frac{288}{125}$

Q66JEE Main 2022MCQ4MApplication of Derivatives

Let $f: R \to R$ be a function defined by $f(x)=(x-3)^{n_1}(x-5)^{n_2}, n_1, n_2 \in N$ . Then, which of the following is NOT true? [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\lim _{x \to 1} \;\frac{ (x^{2-}1) sin \;^2(\pi x)}{x^{4-}2 x^{2+}2 x-1}$ $=\lim _{x \to 1} \;\frac{ (x^{2-}1) sin \;^2(\pi x)}{ (x^{2-}1) (x-1)^2}$ $=\lim _{x \to 1} \;\frac{ sin \;^2(\pi x)}{(x-1)^2}$ Let $x=1+h$ $\therefore$ when $x \to 1$ then $h \to 0$ $=\lim _{h \to 0} \;\frac{ sin \;^2(\pi(1+h))}{(1+h-1)^2}$ $=\lim _{h \to 0} \;\frac{ sin \;^2(\pi h)}{h^2}$ $=\lim _{h \to 0} \pi^2 \times \;\frac{ sin \;^2(\pi h)}{(\pi h)^2}$ $=\pi^2 \times 1$ $=\pi^2$

Q67JEE Main 2022MCQ4MDefinite Integrals

Let ${\text{f}}$ be a real valued continuous function on $[0,1]$ and $f(x)=x+int_0^{1}(x-t) f(t) d t$ Then, which of the following points $(x, y)$ lies on the curve $y=f(x)$ ? [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{f}}( {\text{x}})= (1+int_{0}^{1} {\text{f}}( {\text{t}}) {\text{dt}}) {\text{x}}-int_{0}^{1} {\text{tf}}( {\text{t}}) {\text{dt}}$ ${\text{f}}( {\text{x}})= {\text{Ax}}- {\text{B}}$ ${\text{A}}=1+int_{0}^{1} {\text{f}}( {\text{t}}) {\text{dt}}=1+int_{0}^{1}( {\text{At}}- {\text{B}}) {\text{dt}}$ $\Rightarrow {\text{A}}=2(1- {\text{B}})$ Also ${\text{B}}=int_{0}^{1} {\text{tf}}( {\text{t}}) {\text{dt}}=int_{0}^{1} ( {\text{At}} {\text{t}}^{2-} {\text{Bt}}) {\text{dt}}$ ${\text{A}}=\;\frac{9}{2} {\text{B}}$ From $(2)$ , (3) ${\text{A}}=\;\frac{18}{13}, {\text{B}}=\;\frac{4}{13}$ so ${\text{f}}(6)=8$

Q68JEE Main 2022MCQ4MDefinite Integrals

If $int_0^2(\sqrt{2 x}-\sqrt{2 x-x^2}) d x=int_0^1(1-\sqrt{1-y^2}-\;\frac{y^2}{2}) d y+int_1^2(2-\;\frac{y^2}{2}) d y+I$ , then I equals [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

LHS $=int_{0}^{2} (\sqrt{2 {\text{x}}}-\sqrt{2 {\text{x}}- {\text{x}}^2}) {\text{dx}}=\;\frac{8}{3}-\;\frac{\pi}{2}$ RHS $=int_{0}^{1} (1-\sqrt{1- {\text{y}}^2}-\;{ {\text{y}}^2}/{2}) {\text{dy}}+int_{1}^{2} (2-\;{ {\text{y}}^2}/{2}) {\text{dy}}+ {\text{I}}$ ${\text{I}}+\;\frac{5}{3}-\;\frac{\pi}{4}$ So, ${\text{I}}=1-\;\frac{\pi}{4}=int_{0}^{1} (1-\sqrt{1- {\text{y}}^2}) {\text{dy}}$

Q69JEE Main 2022MCQ4MDifferential Equations

If ${\text{y}}= {\text{y}}( {\text{x}})$ is the solution of the differential equation $(1+e^{2 x}) \;\frac{d y}{d x}+2(1+y^2) e^x=0$ and ${\text{y}}(0)=0$ then $6(y^{'}(0)+(y(\log _e \sqrt{3}))^2)$ is equal to [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;{ {\text{dy}}}/{1+ {\text{y}}^2}+\;{2 {\text{e}}^{ {\text{x}}}}/{1+ {\text{e}}^{2 {\text{x}}}} {\text{dx}}=0$ on integration $\tan ^{-1} {\text{y}}+2 \tan ^{-1} {\text{e}}^{ {\text{x}}}= {\text{c}}$ $\because {\text{y}}(0)=0$ so, $C=\;\frac{\pi}{2} \Rightarrow \tan ^{-1} y+2 \tan ^{-1} e^x=\;\frac{\pi}{4}$ from eq.(i), $(\;{ {\text{dy}}}/{ {\text{dx}}}) _{ {\text{x}}=0}=-1$ $\arg y(\ln {\sqrt{3}})=-\;{1}/{{\sqrt{3}}}$ $6 \$ [y^{'}(0)+ (y(\ln {\sqrt{3}})^2]$ =6 [-1+;\frac{1}{3}] =-4.$

Q70JEE Main 2022MCQ4MParabola

Let $P: y^2=4 a x, a>0$ be a parabola with focus $S$ . Let the tangents to the parabola $P$ make an angle of $\;\frac{\pi}{4}$ with the line $y=3 x+5$ touch the parabola $P$ at $A$ and $B$ . Then the value of a for which $A, B$ and $S$ are collinear is [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\tan \;\frac{\pi}{4}= |\;\frac{m-3}{1+3 m}{{}/{}|}$ $\Rightarrow \;\frac{m-3}{1+3 m}=\pm 1$ $\Rightarrow \;\frac{m-3}{1+3 m}=+1 \;\text{ and }\; \;\frac{m-3}{1+3 m}=-1$ $\Rightarrow m-3=1+3 m \;\text{ and }\; m-3=-1-3 m$ $\Rightarrow 2 m=-4 \;\text{ and }\; 4 m=2$ $m=-2 \;\text{ and }\; m=\;\frac{1}{2}$ We know, Equation of tangent to the parabola $y^2=4 m$ is $y=m x+\;\frac{a}{m}$ and point of contact is $(\;\frac{a}{m^2}, \;\frac{2 a}{m})$ $\therefore$ Equation of tangent $y=-2 x-\;\frac{a}{2}$ and $y=\;\frac{x}{2}+2 a$ $\therefore$ Point of contact A and B are $A (\;\frac{a}{(-2)^2}, \;\frac{2 a}{-2}) =A (\;\frac{a}{4},-a)$ $B (\;\frac{a}{ (\;\frac{1}{2}) ^2}, \;\frac{2 a}{ (\;\frac{1}{2}) }) =B(4 a, 4 a)$ As points A, B and S are colinear so area of triangle formed by those 3 points are zero. $\;\text{ Area of }\; \triangle {\text{ABS}}=\;\frac{1}{2}{\begin{vmatrix}\\ \\ \frac{a}{4} & -a & 1 \\ 4 a & 4 a & 1 \\ a & 0 & 1\end{vmatrix}}$ $=\;\frac{a}{4}(4 a-0)+a(4 a-a)+1 (0-4 a^2)$ $=a^{2+}3 a^{2-}4 a^2=0$ $\therefore$ Area of triangle is independent of value of a.So, for all value of $a>0$ (already given a must be greater than 0 ) point $A, B$ and $S$ will be collinear.

Explanation Figure 1

Q71JEE Main 2022MCQ4MCircles

Let a triangle ${\text{ABC}}$ be inscribed in the circle $x^{2-}\sqrt{2}(x+y)+y^2=0$ such that $∠ B A C=\;\frac{\pi}{2}$ . If the length of side $A B$ is $\sqrt{2}$ , then the area of the $\triangle A B C$ is equal to : [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

For equation of circle $x^{2+}y^{2+}2 g x+2 f y+c=0$ , center is $(-g,-f)$ and radius $r={\sqrt{g^{2+}f^{2-}c}}$ Given,equation of circle is $x^{2-}{\sqrt{2}}(x+y)+y^2=0$ $\Rightarrow x^{2+}y^{2-}{\sqrt{2}} x-{\sqrt{2}} y=0$ $\Rightarrow x^{2+}y^{2+}2 (-\;{1}/{{\sqrt{2}}}) x+2 (-\;{1}/{{\sqrt{2}}}) =0$ $\therefore g=-\;{1}/{{\sqrt{2}}} \;\text{ and }\; f=-\;{1}/{{\sqrt{2}}}$ $\therefore \;\text{ Center }\;=(-g,-f)= (\;{1}/{{\sqrt{2}}}, \;{1}/{{\sqrt{2}}})$ $\;\text{ And Radius }\;=r=\sqrt{ (-\;{1}/{{\sqrt{2}}}) ^{2+} (-\;{1}/{{\sqrt{2}}}) ^{2-}0}$ $=\sqrt{\;\frac{1}{2}+\;\frac{1}{2}}={\sqrt{1}}=1$ As $A B$ and $A C$ makes an angle $90^{\circ}$ then line $B C$ passes through the center of circle and $B C$ is the diameter of the circle. $\therefore$ Length of $B C=2 r=2 \times 1=2$ $\therefore A C^2=B C^{2-}A B^2$ $=2^{2-}({\sqrt{2}})^2$ $=2$ $\Rightarrow A C={\sqrt{2}}$ $\therefore$ Area of right angle triangle $A B C$ $=\;\frac{1}{2} \times A C \times A B$ $=\;\frac{1}{2} \times {\sqrt{2}} {\sqrt{2}}$ $=1 \;\text{ square unit. }\;$

Explanation Figure 1

Q72JEE Main 2022MCQ4MThree Dimensional Geometry

Let $\;\frac{x-2}{3}=\;\frac{y+1}{-2}=\;\frac{z+3}{-1}$ lie on the plane $p x-q y+z=5$ , for some ${\text{p}}, {\text{q}} \in {\text{R}}$ . The shortest distance of the plane from the origin is : [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$(2,-1,-3)$ satisfy the given plane.So $2 p+q=8$ Also given line is perpendicular to normal plane so $3 p+2 q-1=0$ $\Rightarrow p=15, q=-22$ Eq. of plane $15 x-22 y+z-5=0$ its distance from origin $=\;{6}/{{\sqrt{710}}}=\sqrt{\;\frac{5}{142}}$

Q73JEE Main 2022MCQ4MStraight Lines and Pair of Straight Lines

The distance of the origin from the centroid of the triangle whose two sides have the equations $x-2 y+1=0$ and $2 x-y-1=0$ and whose orthocenter is $(\;\frac{7}{3}, \;\frac{7}{3})$ is : [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

For point A, $2 x-y-1=0$ $x-2 y+1=0$ Solving (1) and (2), we get $x=1, y=1$ $\therefore$ Point $A=(1,1)$ Altitude from B to line $A C$ is perpendicular to line $A C$ . $\therefore$ Equator of altitude ${\text{BH}}$ is $2 x+y+\lambda =0$ It passes through point $H (\;\frac{7}{3}, \;\frac{7}{3})$ so it satisfy the equation (3). $\;\frac{14}{3}+\;\frac{7}{3}+\lambda =0$ $\Rightarrow \alpha=-7$ $\therefore$ Altitude ${\text{BH}}=2 {\text{x}}+ {\text{y}}-7=0$ Solving equation (1) and (4), we get $x=2, y=3$ $\therefore$ Point $B=(2,3)$ Altitude from $C$ to line $A B$ is perpendicular to line $A B$ . $\therefore$ Equation of altitude ${\text{CH}}$ is $x+2 y+\lambda =0$ It passes through point $H (\;\frac{7}{3}, \;\frac{7}{3})$ so it satisfy equation (5). $\;\frac{7}{3}+\;\frac{14}{3}+\lambda =0$ $\Rightarrow \lambda =-7$ $\therefore$ Altitude ${\text{CH}}= {\text{x}}+2 {\text{y}}-7=0$ Solving equation (2) and (6), we get $x=3, y=2$ $\therefore$ Point $C=(3,2)$ Centroid ${\text{G}}( {\text{x}}, {\text{y}})$ of triangle ${\text{A}}(1,1), {\text{B}}(2,3)$ and ${\text{C}}(3,2)$ is $x=\;\frac{1+2+3}{3}=2, y=\;\frac{1+2+3}{3}=2$ Now dDistance of point ${\text{G}}(2,2)$ from center ${\text{O}}(0,0)$ is $O G={\sqrt{2^{2+}2^2}}=2 {\sqrt{2}}$

Explanation Figure 1

Q74JEE Main 2022MCQ4MThree Dimensional Geometry

Let $Q$ be the mirror image of the point $P(1,2,1)$ with respect to the plane $x+2 y+2 z=16$ . Let $T$ be a plane passing through the point ${\text{Q}}$ and contains the line $\vec{r}=-\hat{k}+\lambda (\hat{i}+\hat{j}+2 \hat{k}), \lambda \in R$ . Then, which of the following points lies on ${\text{T}}$ ? [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Image of $P(1,2,1)$ in $x+2 y+2 z-16=0$ is given by ${\text{Q}}(4,8,7)$ Eq. of plane ${\text{T}}={\begin{vmatrix}{\text{x}} & {\text{y}} & {\text{z}}+1 \\ 4 & 8 & 6 \\ 1 & 1 & 2\end{vmatrix}}=0$ $\Rightarrow 2 {\text{x}}- {\text{z}}=1$ so ${\text{B}}(1,2,1)$ lies on it.

Q75JEE Main 2022MCQ4MVector Algebra

Let $A, B, C$ be three points whose position vectors respectively are $\vec{a}=\hat{i}+4 \hat{j}+3 \hat{k}$ $\vec{b}=2 \hat{i}+\alpha \hat{j}+4 \hat{k}, \alpha \in R$ $\vec{c}=3 \hat{i}-2 \hat{j}+5 \hat{k}$ If $\alpha$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are noncollinear, then the length of the median, in $\Delta {\text{ABC}}$ , through ${\text{A}}$ is : [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${ \vec{\text{AB}} \| { \vec{\text{AC}}$ if $\;\frac{1}{2}=\;\frac{\alpha-4}{-6}=\;\frac{1}{2} \Rightarrow \alpha=1$ ${ \vec{\text{a}}, { \vec{\text{b}}, { \vec{\text{c}}$ are non-collinear for $\alpha=2$ (smallest positive integer)Mid-point of ${\text{BC}}= {\text{M}} (\;\frac{5}{2}, 0, \;\frac{9}{2})$ ${\text{AM}}=\sqrt{\;\frac{9}{4}+16+\;\frac{9}{4}}=\;{{\sqrt{82}}}/{2}$

Q76JEE Main 2022MCQ4MProbability

The probability that a relation ${\text{R}}$ from $\{ {\text{x}}, {\text{y}}\}$ to $\{ {\text{x}}, {\text{y}}\}$ is both symmetric and transitive, is equal to [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Total no. of relations $=2^{2 \times 2}=16$ Fav. relation $=\varphi ,\{( {\text{x}}, {\text{x}})\},\{( {\text{y}}, {\text{y}})\},\{( {\text{x}}, {\text{x}})( {\text{y}}, {\text{y}})\}$ $\{(x, x),(y, y),(x, y)(y, x)\}$ $\;\text{ Prob. }\;=\;\frac{5}{16}$

Q77JEE Main 2022MCQ4MStatistics

The number of values of $a \in N$ such that the variance of $3,7,12, a, 43-a$ is a natural number is : [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Mean $=13$ Variance $=\;{9+49+144+ {\text{a}}^{2+}(43- {\text{a}})^2}/{5}-13^2 \in {\text{N}}$ $\Rightarrow \;\frac{2 a^{2-}a+1}{5} \in {\text{N}}$ $\Rightarrow 2 a^{2-}a+1-5 n=0$ must have solution asnatural numbersits $D=40 n-7$ always has 3 at unit place $\Rightarrow {\text{D}}$ can't be perfect squareSo, a can't be integer.

Q78JEE Main 2022MCQ4MTrigonometric Ratios and Identities

From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is $60^{\circ}$ . The pole subtends an angle $30^{\circ}$ at the top of the tower. Then the height of the tower is [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Here $A B$ is a tower and $C D$ is a pole.In triangle ${\text{ABC}}$ , $\tan 60^{\circ}=\;\frac{A B}{A C}=\;\frac{20+h}{x} . . . . . . .(1)$ In triangle ${\text{BED}}$ , $\tan 30^{\circ}=\;\frac{h}{x} . . . . . .(2)$ Divide equation (1) by equation (2), we get $\;\frac{\tan 60^{\circ}}{\tan 30^{\circ}}=\;\frac{20+h}{x} \times \;\frac{x}{h}$ $\Rightarrow \;{{\sqrt{3}}}/{{1}/{{\sqrt{3}}}}=\;\frac{20+h}{h}$ $\Rightarrow 3=\;\frac{20+h}{h}$ $\Rightarrow 3 h=20+h$ $\Rightarrow h=10 {\text{m}}$ $\therefore \;\text{ Height of tower }\;=20+10=30 {\text{m}}$

Explanation Figure 1

Q79JEE Main 2022MCQ4MSets and Relations

Negation of the Boolean statement $(p ∨ q) \Rightarrow ((∼ r) ∨ p)$ is equivalent to [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Given, $(p ∨ q) \Rightarrow ((∼ r) ∨ p)$ Negation is $∼((p ∨ q) \Rightarrow (∼ r) ∨ p))$ $=(p ∨ q) ∧ ∼((∼ r) ∨ p)$ $=(p ∨ q) ∧(r ∧ ∼ p)$ ${[(p ∧ ∼ p) ∨(q ∧ ∼ p)] ∧ r}$ $=q ∧ ∼ p ∧ r$

Q80JEE Main 2022MCQ4MBinomial Theorem

Let $n \ge 5$ be an integer. If $9^n-8 n-1=64 \alpha$ and $6^n-5 n-1=25 \beta$ , then $\alpha-\beta$ is equal to [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Given, $9^n-8 n-1=64 \alpha$ $\Rightarrow \alpha=\;\frac{(1+8)^n-8 n-1}{64}$ $=\;\frac{ ({ }^n C_0 \cdot 1+{ }^n C_1 \cdot 8^{1+}{ }^n C_2 \cdot 8^{2+}. . . . . .+{ }^n C_n \cdot 8^n) -8 n-1}{8^2}$ $=\;\frac{1+8 n+{ }^n C_2 \cdot 8^{2+}. . . .+{ }^n C_n \cdot 8^n-8 n-1}{8^2}$ $=\;\frac{{ }^n C_2 \cdot 8^{2+}{ }^n C_3 \cdot 8^{3+}. . . .+{ }^n C_n \cdot 8^n}{8^2}$ $={ }^n C_2+{ }^n C_3 \cdot 8+{ }^n C_4 \cdot 8^{2+}. . .{ }^n C_n \cdot 8^{n-2}$ Also given, $6^n-5 n-1=25 \beta$ $\Rightarrow \beta=\;\frac{(1+5)^n-5 n-1}{25}$ $=\;\frac{{ }^n C_0 \cdot 1+{ }^n C_1 \cdot 5+{ }^n C_2 \cdot 5^{2+}. . . . .+{ }^n C_n \cdot 5^n-5 n-1}{5^2}$ $=\;\frac{1+5 n+{ }^n C_2 \cdot 5^{2+}{ }^n C_3 \cdot 5^{3+}. . . .+{ }^n C_n \cdot 5^{2-}5 n-1}{5^2}$ $=\;\frac{{ }^n C_2 \cdot 5^{2+}{ }^n C_3 \cdot 5^{3+}{ }^n C_4 \cdot 5^{4+}. . . .+{ }^n C_n \cdot 5^n}{5^2}$ $={ }^n C_2+{ }^n C_3 \cdot 5+{ }^n C_4 \cdot 5^{2+}. . . . . . .+{ }^n C_n \cdot 5^{n-2}$ $\therefore \alpha-\beta$ $= ({ }^n C_2+{ }^n C_3 \cdot 8+{ }^n C_4 \cdot 8^{2+}. . . .+{ }^n C_n \cdot 8^{n-2}) - ({ }^n C_2+{ }^n C_3 \cdot 5+{ }^n C_4 \cdot 5^{2+}. . .+{ }^n C_n \cdot 5^{n-2}.$ $={ }^n C_3 \cdot (8-5)+{ }^n C_4 \cdot (8^{2-}5^2) +. . .+{ }^n C_n (8^{n-2}-5^{n-2})$

Q81JEE Main 2022NAT4MVector Algebra

Let $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}, \vec{b}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a}+(\vec{b} \times \vec{c})=\vec{0}$ and $\vec{b} \cdot \vec{c}=5$ . Then the value of $3(\vec{c} \cdot \vec{a})$ is equal to____ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\vec{a}+\vec{b} \times \vec{c}=0$ $\vec{a} \times \vec{b}+|\vec{b}|^2 \vec{c}-5 \vec{b}=0$ It gives $\vec{c}=\;\frac{1}{3}(10 \hat{i}+3 \hat{j}+2 \hat{k})$ so $3 \vec{a} \cdot \vec{c}=10$ But it does not satisfy $\vec{a}+\vec{b} \times \vec{c}=0$ .This question has data error.Alternate (Explanation) :According to given $\vec{a} & \vec{b}$ $\vec{a} \cdot \vec{b}=1-2+3=2$ ... (i)but given equation ${ \vec{\text{a}}=-({ \vec{\text{b}} \times { \vec{\text{c}})$ $\Rightarrow \vec{a} ⟂ \vec{b} \Rightarrow \vec{a} \cdot \vec{b}=0$ which contradicts.

Q82JEE Main 2022NAT4MDifferential Equations

Let $y=y(x), x>1$ , be the solution of the differential equation $(x-1) \;\frac{d y}{d x}+2 x y=\;\frac{1}{x-1}$ , with $y(2)=\;\frac{1+e^4}{2 e^4}$ . If $y(3)=\;\frac{e^\alpha+1}{\beta e^\alpha}$ , then the value of $\alpha+\beta$ is equal to____ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{d y}{d x}+\;\frac{2 x}{x-1} \cdot y=\;\frac{1}{(x-1)^2}$ $y=\;\frac{1}{(x-1)^2} [\;\frac{e^{2 x}+1}{2 e^{2 x}}]$ $y(3)=\;\frac{e^{6+}1}{8 e^6}$ $\alpha+\beta=14$

Q83JEE Main 2022NAT4MSequences and Series

Let $3,6,9,12$ , upto 78 terms and $5,9,13,17$ , upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to_____ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

1st AP : $3,6,9,12, . . . . . .$ . upto 78 terms $t_{78}=3+(78-1) 3$ $=3+77 \times 3$ $=234$ 2nd AP: $5,9,13,17, . . . . . .$ upto 59 terms $t_{59}=5+(59-1) 4$ $=5+58 \times 4$ $=237$ Common term's AP :First term $=9$ Common difference of first ${\text{AP}}=3$ And common difference of second ${\text{AP}}=4$ $\therefore$ Common difference of common terms ${\text{AP}}={\text{LCM}}(3,4)=12$ $\therefore \;\text{ New AP }\;=9,21,33, . . . . . .$ ${\text{t}}_{ {\text{n}}}=9+( {\text{n}}-1) 12 \le 234$ $\Rightarrow n \le \;\frac{237}{12}$ $\Rightarrow n=19$ $\therefore S_{19}=\;\frac{19}{2}[2.9+(19-1) 12]$ $=19(9+108)$ $=2223$

Q84JEE Main 2022NAT4MTrigonometric Ratios and Identities

The number of solutions of the equation $\sin x=\cos ^2 x$ in the interval $(0,10)$ is____ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$sin \;x=\cos ^2 x, x \in (0,10)$ $\Rightarrow sin \;x=1- sin \;^2 x$ $\Rightarrow sin \;^2 x+ sin \;x-1=0$ $\therefore sin \;x=\;{-1 \pm {\sqrt{1+4}}}/{2}$ $\Rightarrow sin \;x=\;{-1 \pm {\sqrt{5}}}/{2}$ We know $sin \;\in (-1,1)$ $\therefore \;{-1-{\sqrt{5}}}/{2}$ can't be a value of $sin \;x$ $\therefore sin \;x=\;{{\sqrt{5}}-1}/{2}$ $3 \pi=3 \times 3.14=9.42< 10$ $\;\frac{7 \pi}{2}=\;\frac{7}{2} \times 3.14=10.99>10$ $\therefore 10$ will be in between $3 \pi$ and $\;\frac{7 \pi}{2}$ .There are 4 intersection at $A, B, C$ and $D$ between $sin \;x$ graph and $y=\;{{\sqrt{5}}-1}/{2}$ graph. $\therefore$ possible solution $=4$

Explanation Figure 1

Q85JEE Main 2022NAT4MArea Under The Curves

For real numbers $a, b(a>b>0)$ , letArea $\{(x, y): x^{2+}y^2 \le a^2.$ and $.\;\frac{x^2}{a^2}+\;\frac{y^2}{b^2} \ge 1\}=30 \pi$ andArea $\{(x, y): x^{2+}y^2 \le b^2.$ and $.\;\frac{x^2}{a^2}+\;\frac{y^2}{b^2} \le 1\}=18 \pi$ Then, the value of $(a-b)^2$ is equal to____ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

given $\pi {\text{a}}^{2-}\pi {\text{ab}}=30 \pi$ and $\pi {\text{ab}}-\pi {\text{b}}^2=18 \pi$ on subtracting, we get $(a-b)^2=a^{2-}2 a b+b^2=12$

Q86JEE Main 2022NAT4MApplication of Derivatives

Let $f$ and $g$ be twice differentiable even functions on $(-2,2)$ such that $f(\;\frac{1}{4})=0, f(\;\frac{1}{2})=0$ , $f(1)=1$ and $g(\;\frac{3}{4})=0, g(1)=2$ . Then, the minimum number of solutions of $f(x) g^{' '}(x)+f^{'}(x) g^{'}(x)=0$ in $(-2,2)$ is equal to____ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Let $h(x)=f(x) g^{'}(x) \to 5$ roots $\because {\text{f}}( {\text{x}})$ is even $\Rightarrow$ ${\text{f}} (\;\frac{1}{4}) = {\text{f}} (\;\frac{1}{2}) = {\text{f}} (-\;\frac{1}{2}) = {\text{f}} (\;\frac{1}{4}) =0$ ${\text{g}}( {\text{x}})$ is even $\Rightarrow {\text{g}} (\;\frac{3}{4}) = {\text{g}} (-\;\frac{3}{4}) =0$ ${\text{g}}^{'}( {\text{x}})=0$ has minimum one root $h^{'}( {\text{x}})$ has at last 4 roots

Q87JEE Main 2022NAT4MBinomial Theorem

Let the coefficients of ${\text{x}}^{-1}$ and ${\text{x}}^{-3}$ in the expansion of $(2 x^{\;\frac{1}{5}}-\;{1}/{x^{\;\frac{1}{5}}})^{15}, x>0$ , be ${\text{m}}$ and ${\text{n}}$ respectively. If ${\text{r}}$ is a positive integer such that $m n^2={ }^{15} C_r \cdot 2^r$ , then the value of ${\text{r}}$ is equal to___ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{T}}_{ {\text{r}}+1}=(-1)^{ {\text{r}}} \cdot { }^{15} {\text{C}}_{ {\text{r}}} \cdot 2^{15- {\text{r}}} \times { }^{\;{15-2 {\text{r}}}/{5}}$ ${\text{m}}={ }^{15} {\text{C}}_{10} 2^5$ ${\text{n}}=-1$ so ${mn}^2={ }^{15} {\text{C}}_5 2^5$

Q88JEE Main 2022NAT4MPermutations and Combinations

The total number of four digit numbers such that each of first three digits is divisible by the last digit, is equal to____ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Let the number is abcd, where a,b,c are divisible by d. $\text{ No. of such numbers }\;$ ${\text{d}}=1,\;9 \times 10 \times 10=900$ ${\text{d}}=2,\;4 \times 5 \times 5=100$ ${\text{d}}=3,\;3 \times 4 \times 4=48$ ${\text{d}}=4,\;2 \times 3 \times 3=18$ ${\text{d}}=5,\;1 \times 2 \times 2=4$ ${\text{d}}=6,7,8,9\;\;\;\;\;4 \times 4=16$ Total =1086

Q89JEE Main 2022NAT4MMatrices and Determinants

Let $M=\begin{bmatrix}0 & -\alpha \\ \alpha & 0\end{bmatrix}$ , where $\alpha$ is a non-zero real number an $N=sum_{k=1}^{49} M^{2 k}$ . If $(I-M^2) N=-2 I$ , then the positive integral value of $\alpha$ is____ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{M}}={\begin{bmatrix}0 & -\alpha \\ \alpha & 0\end{bmatrix}} ; {\text{M}}^2={\begin{bmatrix}-\alpha^2 & 0 \\ 0 & -\alpha^2\end{bmatrix}\$ }=-alpha^2 {\text{I}}{\text{N}}= {\text{M}}^{2+} {\text{M}}^{4+}. . . . . .+ {\text{M}}^{98}= $[-\alpha^{2+}\alpha^{4-}\alpha^{6+}. . .]$ {\text{I}}=-alpha^2 ;\frac{ (1- (-alpha^2) ^{49}) }{1+alpha^2} ⋅ {\text{I}}{\text{I}}- {\text{M}}^2= (1+alpha^2) {\text{I}}( {\text{I}}- {\text{M}}^2) {\text{N}}=-alpha^2 (alpha^{98}+1) =-2alpha=1$

Q90JEE Main 2022NAT4MFunctions

Let $f(x)$ and $g(x)$ be two real polynomials of degree 2 and 1 respectively. If $f(g(x))=8 x^{2-}2 x$ and $g(f(x))=4 x^{2+}6 x+1$ , then the value of $f(2)+g(2)$ is_____ [29-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$f (g(x)=8 x^{2-}2 x.$ $g (f(x)=4 x^{2+}6 x+1.$ $\;\text{ So, }\; g(x)=2 x-1$ $& f(x)=2 x^{2+}3 x+1$ $f(2)=8+6+1=15$ Ans. 18

← All JEE Main 2022 papers Take as Timed Mock →