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27 Jun 2022 Shift 2

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Physics30 questions

Q1JEE Main 2022MCQ4MUnits and Measurements

The SI unit of a physical quantity is pascal-second. The dimensional formula of this quantity will be : [27-Jun-2022-Shift-2]

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📖 Explanation

$\;\text{ [pascal-second] }\;=\;\frac{M L T^{-2}}{L^2} \times T$ $=M L^{-1} T^{-1}$

Q2JEE Main 2022MCQ4MUnits and Measurements

The distance of the Sun from earth is $1.5 \times 10^{11} {\text{m}}$ and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be : [27-Jun-2022-Shift-2]

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📖 Explanation

Diameter $= {\text{r}} \times \delta$ $=1.5 \times 10^{11} \times (2000) \times (\;\frac{1}{3600}) \times (\;\frac{\pi}{180})$ $=1.45 \times 10^9 {\text{m}}$

Q3JEE Main 2022MCQ4MMotion in a Straight Line

When a ball is dropped into a lake from a height $4.9 {\text{m}}$ above the water level, it hits the water with a velocity $v$ and then sinks to the bottom with the constant velocity v. It reaches the bottom of the lake $4.0 {\text{s}}$ after it is dropped. The approximate depth of the lake is : [27-Jun-2022-Shift-2]

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📖 Explanation

$t_1=\sqrt{\;\frac{2 h}{g}}$ $=\sqrt{\;\frac{2 \times 4.9}{9.8}}=1 {\ s}$ $\Delta t=4-1=3 {\ s}$ , $v={\sqrt{2 g h}}={\sqrt{2 \times 9.8 \times 4.9}}=9.8 {\text{m}} / {\ s}$ $\therefore$ depth $=9.8 \times 3=29.4 {\text{m}}$

Q4JEE Main 2022MCQ4MCircular Motion

One end of a massless spring of spring constant ${\text{k}}$ and natural length ${\text{I}}_0$ is fixed while the other end is connected to a small object of mass ${\text{m}}$ lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $\omega$ about an axis passing through fixed end, then the elongation of the spring will be : [27-Jun-2022-Shift-2]

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📖 Explanation

$m \omega ^2 (l_0+x) =k x$ $\Rightarrow m \omega ^2 l_0= (k-m \omega ^2) \times x$ $\Rightarrow x=\;\frac{m \omega ^2 l_0}{ (k-m \omega ^2) }$

Q5JEE Main 2022MCQ4MCircular Motion

A stone tide to a spring of length $L$ is whirled in a vertical circle with the other end of the spring at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is $\sqrt{x(u^{2-}g L)}$ . The value of ${\text{x}}$ is - [27-Jun-2022-Shift-2]

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📖 Explanation

$\vec{v}={\sqrt{u^{2-}2 g L}} \hat{j}$ $\vec{u}=u \hat{i}$ $\therefore |\vec{v}-\vec{u}|={\sqrt{ (u^{2-}2 g L) +u^2}}$ $={\sqrt{2 u-2 g L}}$ $\therefore x=2$

Q6JEE Main 2022MCQ4MGravitation

Four spheres each of mass m from a square of side d (as shown in figure). A fifth sphere of mass M is situated at the centre of square. The total gravitational potential energy of the system is : [27-Jun-2022-Shift-2]

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📖 Explanation

Total gravitational potential energy $=- \{\;{4 G M m}/{d / {\sqrt{2}}}+\;\frac{4 G m^2}{d}+\;{2 G m^2}/{{\sqrt{2}} d}\}$ $=-\;\frac{G m}{d}\{M 4 {\sqrt{2}}+(4+{\sqrt{2}}) m\}$ $=-\;\frac{G m}{d}\{4 {\sqrt{2}} M+(4+{\sqrt{2}}) m\}$

Q7JEE Main 2022MCQ4MThermodynamics

For a perfect gas, two pressures $P_1$ and $P_2$ are shown in figure. The graph shows: [27-Jun-2022-Shift-2]

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📖 Explanation

As per ideal gas equation, $V=\;\frac{n R}{P} T$ $\Rightarrow$ Slope of V-T graph is inversely proportional to ${\text{P}}$ .As $m_2>m_1 \Rightarrow P_1>P_2$

Q8JEE Main 2022MCQ4MThermodynamics

According to kinetic theory of gases,A. The motion of the gas molecules freezes at $0^{\circ} {\text{C}}$ .B. The mean free path of gas molecules decreases if the density of molecules is increased.C. The mean free path of gas molecules increases if temperature is increased keeping pressure constant.D. Average kinetic energy per molecule per degree of freedom is $\;\frac{3}{2} k_B T$ (for monoatomic gases).Choose the most appropriate answer from the options given below : [27-Jun-2022-Shift-2]

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📖 Explanation

According to kinetic theory of gases,A. The motion of the gas molecules freezes at $0 {\text{K}}$ .B. The mean free path decreases on increasing the number density of the molecules as $\mu =\;{1}/{{\sqrt{2}} \pi n d^2} \Rightarrow \mu \propto \;\frac{1}{n} \;\text{. }\;$ C. The mean free path increases on increasing the volume. Now if temperature is increased by keeping the pressure constant the volume should increase that is mean free path increases.D. K.E.avg per molecule per degree of freedom is $\;\frac{1}{2} k_B T$ .

Q9JEE Main 2022MCQ4MThermodynamics

A lead bullet penetrates into a solid object and melts. Assuming that $40 \%$ of its kinetic energy is used to heat it, the initial speed of bullet is :(Given : initial temperature of the bullet $=127^{\circ} {\text{C}}$ , Melting point of the bullet $=327^{\circ} {\text{C}}$ , Latent heat of fusion of lead $=2.5 \times 10^4 {\text{J}}$ ${\text{kg}}^{-1}$ , Specific heat capacity of lead $.=125 {\text{J}} / {\text{kg}} {\text{K}})$ [27-Jun-2022-Shift-2]

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📖 Explanation

$\;\frac{2}{5} \times \;\frac{1}{2} m v^2=m L+m s \Delta T$ $\Rightarrow \;\frac{v^2}{5}=2.5 \times 10^{4+}125+200$ $\Rightarrow \;\frac{v^2}{5}=5 \times 10^4$ $\Rightarrow v=500 {\text{m}} / {\ s}$

Q10JEE Main 2022MCQ4MOscillations

The equation of a particle executing simple harmonic motion is given by $x=\sin \pi(t+\;\frac{1}{3}) m$ . At $t=1 {\text{s}}$ , the speed of particle will be(Given : $\pi=3.14$ ) [27-Jun-2022-Shift-2]

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📖 Explanation

$x= sin \; (\pi t+\;\frac{\pi}{3}) m$ $\Rightarrow \;\frac{d x}{d t}=\pi \cos (\pi t+\;\frac{\pi}{3})$ $=\pi \cos (\pi+\;\frac{\pi}{3}) \;\text{ at }\; t=1 s$ $=-\;\frac{\pi}{2} {\text{m}} / {\ s}$ or $|\;\frac{d x}{d t}{{}/{}|} =157 {\text{cm}} / {\ s}$

Q11JEE Main 2022MCQ4MElectrostatics

If a charge $q$ is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be : [27-Jun-2022-Shift-2]

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📖 Explanation

Flux passing through flat surface = Flux passing through curved surface. $\;\text{ So, }\; \varphi =\;\frac{q}{2 \epsilon_0}$ Remark : Electric flux through flat surface is zerobut no option is given, option is available forelectric flux passing through curved surface

Q12JEE Main 2022MCQ4MElectrostatics

Three identical charged balls each of charge $2 {\text{C}}$ are suspended from a common point $P$ by silk threads of $2 {\text{m}}$ each (as shown in figure). They form an equilateral triangle of side $1 {\text{m}}$ .The ratio of net force on a charged ball to the force between any two charged balls will be: [27-Jun-2022-Shift-2]

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📖 Explanation

$F_{\;\text{net }\;}$ on charge $3, F_1=\;{{\sqrt{3}} k q^2}/{1^2}$ Force between any 2 charges $F_2=\;\frac{k q^2}{1^2}$ So, $\;\frac{F_1}{F_2}={\sqrt{3}}$

Q13JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

Two long parallel conductors ${\text{S}}_1$ and ${\text{S}}_2$ are separated by a distance $10 {\text{cm}}$ and carrying currents of $4 {\text{A}}$ and $2 {\text{A}}$ respectively. The conductors are placed along $X$ -axis in $X-Y$ plane. There is a point $P$ located between the conductors (as shown in figure).A charge particle of $3 \pi$ coulomb is passing through the point ${\text{P}}$ with velocity $\vec{v}=(2 \hat{i}+3 \hat{j}) {\text{m}} / {\text{s}}$ ; where $\hat{i}$ and $\hat{j}$ represents unit vector along $x \& y$ axis respectively.The force acting on the charge particle is $4 \pi \times 10^{-5}(-x \hat{i}+2 \hat{j}) {\text{N}}$ . The value of ${\text{x}}$ is : [27-Jun-2022-Shift-2]

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📖 Explanation

$\;\text{ Field at }\; {\text{P}} \;\text{ is }\;= (\;\frac{\mu _0 \times i_1}{2 \pi r_1}-\;\frac{\mu _0 i_2}{2 \pi r_2}) (-\hat{k})$ $=- (\;\frac{\mu _0 4}{2 \pi \times 0.04}-\;\frac{\mu _0 \times 2}{2 \pi \times 0.06}) \hat{k}=-\;\frac{\mu _0 \times 200}{6 \pi} \hat{k}$ So, force $\vec{F}=\vec{q v} \times \vec{B}$ $=3 \pi(2 \hat{i}+3 \hat{j}) \times (- (\;\frac{\mu _0 \times 200}{6 \pi} \hat{k}) )$ $=3 \pi (\;\frac{200 \mu _0}{3 \pi} \hat{j}-\;\frac{100 \mu _0}{\pi} \hat{i})$ $=200 \mu _0 \hat{j}-300 \mu _0 \hat{i}$ $=4 \pi \times 10^{-5}(2 \hat{j}-3 \hat{i})$ So, $x=3$

Q14JEE Main 2022MCQ4MAlternating Current

If $L, C$ and $R$ are the self inductance, capacitance and resistance respectively, which of the following does not have the dimension of time? [27-Jun-2022-Shift-2]

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📖 Explanation

$(\;\frac{L}{C})$ does not have dimension of time. ${\text{RC}}, \;{ {\text{L}}}/{ {\text{R}}}$ are time constant while $\sqrt{ {\text{LC}}}$ is reciprocal of angular frequency or having dimension of time. $U=\;\frac{1}{2} L i^2=\;\frac{1}{2} C V^2$ So, $[\;\frac{L}{C}] =\;\frac{V^2}{i^2}=R^2$ is not the dimension of time.

Q15JEE Main 2022MCQ4MElectromagnetic Waves

Given below are two statements:Statement I : A time varying electric field is a source of changing magnetic field and vice-versa. Thus a disturbance in electric or magnetic field creates EM waves.Statement II : In a material medium, the EM wave travels with speed $v=\;\frac{1}{\sqrt{\mu _0 E_0}}$ . In the light of the above statements, choose the correct answer from the options given below. [27-Jun-2022-Shift-2]

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📖 Explanation

In a material medium speed of light is given by $v=\;{1}/{{\sqrt{\epsilon_0 \epsilon_r \mu _0 \mu _r}}}$ . So statement 2 is false.

Q16JEE Main 2022MCQ4MRay Optics and Optical Instruments

A convex lens has power P. It is cut into two halves along its principal axis. Further one piece (out of the two halves) is cut into two halves perpendicular to the principal axis (as shown in figures). Choose the incorrect option for the reported pieces. [27-Jun-2022-Shift-2]

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📖 Explanation

We know $P=\;\frac{1}{f}=(\mu -1) (\;\frac{1}{R_1}-\;\frac{1}{R_2})$ $L_1: \;\frac{1}{f_1}=(\mu -1) (\;\frac{1}{R_1}-\;\frac{1}{R_2}) =P_1=(\mu -1) (\;\frac{2}{R}) =P$ $L_2: \;\frac{1}{f_2}=(\mu -1) (\;\frac{1}{R_1}) =P_2=\;\frac{\mu -1}{R}$ $L_3: \;\frac{1}{f_3}=(\mu -1) (-\;\frac{1}{R_2}) =P_3=\;\frac{\mu -1}{R}$

Q17JEE Main 2022MCQ4MWaves

If a wave gets refracted into a denser medium, then which of the following is true? [27-Jun-2022-Shift-2]

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Q18JEE Main 2022MCQ4MAtoms and Nuclei

Given below are two statements :Statement I : In hydrogen atom, the frequency of radiation emitted when an electron jumps from lower energy orbit (E $. {\text{E}}_1)$ to higher energy orbit $( {\text{E}}_2)$ , is given as $h f= {\text{E}}_1- {\text{E}}_2$ Statement II : The jumping of electron from higher energy orbit $( {\text{E}}_2)$ to lower energy orbit $( {\text{E}}_1)$ is associated with frequency of radiation given as $f=(E_2-E_1) / h$ This condition is Bohr's frequency condition.In the light of the above statements, choose the correct answer from the options given below : [27-Jun-2022-Shift-2]

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📖 Explanation

Radiation is not emitted but absorbed when an electron jumps from low energy to high energy.Also, $E_2-E_1$ is the energy of photon $\Rightarrow {\text{E}}_2- {\text{E}}_1= {\text{hf}}$ $\Rightarrow f=\;\frac{E_2-E_1}{h}$

Q19JEE Main 2022MCQ4MSemiconductor Electronics

For a transistor to act as a switch, it must be operated in [27-Jun-2022-Shift-2]

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Q20JEE Main 2022MCQ4MElectromagnetic Waves

We do not transmit low frequency signal to long distances because-(a) The size of the antenna should be comparable to signal wavelength which is unreal solution for a signal of longer wavelength.(b) Effective power radiated by a long wavelength baseband signal would be high.(c) We want to avoid mixing up signals transmitted by different transmitter simultaneously.(d) Low frequency signal can be sent to long distances by superimposing with a high frequency wave as well.Therefore, the most suitable option will be : [27-Jun-2022-Shift-2]

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Q21JEE Main 2022NAT4MLaws of Motion

A mass of $10 {\text{kg}}$ is suspended vertically by a rope of length $5 {\text{m}}$ from the roof. A force of $30 {\text{N}}$ is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $\theta=\tan ^{-1}(x \times 10^{-1})$ . The value of $x$ is____(Given, $g=10 {\text{m}} / {\text{s}}^2$ ) [27-Jun-2022-Shift-2]

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📖 Explanation

$T \cos \theta=m g$ $T \cos \theta=100 N$ $T sin \;\theta=30$ $\Rightarrow \;\frac{T sin \;\theta}{T \cos \theta}=\;\frac{30}{100}$ $\Rightarrow \tan \theta=\;\frac{3}{10}$ $\therefore x=3$

Q22JEE Main 2022NAT4MRotational Motion

A rolling wheel of $12 {\text{kg}}$ is on an inclined plane at position ${\text{P}}$ and connected to a mass of $3 {\text{kg}}$ through a string of fixed length and pulley as shown in figure. Consider PR as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom ${\text{Q}}$ of the inclined plane ${\text{PQ}}$ will be $\;\frac{1}{2} \sqrt{x g h} {\text{m}} / {\text{s}}$ . The value of ${\text{x}}$ is_____ [27-Jun-2022-Shift-2]

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📖 Explanation

For rolling wheel ${[12 g sin \;\alpha-3 g sin \;\alpha] \times R= (2 \times 12 R^{2+}3 R^2) \times \;\frac{a}{R}}$ $\Rightarrow \;\frac{9 g sin \;\alpha}{27}=a$ $\Rightarrow a=\;\frac{g sin \;\alpha}{3}$ $\therefore v=\sqrt{2 \times \;\frac{g sin \;\alpha}{3} \times \;\frac{h}{ sin \;\alpha}}=\sqrt{\;\frac{2}{3} g h}$ $=\;\frac{1}{2} \times \sqrt{\;\frac{8}{3} g h}$ $\therefore x=\;\frac{8}{3}=2.67$

Q23JEE Main 2022NAT4MThermodynamics

A diatomic gas $(\gamma =1.4)$ does $400 {\text{J}}$ of work when it is expanded isobarically. The heat given to the gas in the process is_____ J. [27-Jun-2022-Shift-2]

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📖 Explanation

${\text{W}}= {\text{nR}} \Delta {\text{T}}=400 {\text{J}}$ $\therefore \Delta {\text{Q}}= {\text{nC}}_{ {\text{P}}} \Delta {\text{T}}$ $=n \times \;\frac{7}{2} R \times \Delta T=\;\frac{7}{2} \times (400)=1400$

Q24JEE Main 2022NAT4MOscillations

A particle executes simple harmonic motion. Its amplitude is $8 {\text{cm}}$ and time period is $6 {\text{s}}$ . The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is s. [27-Jun-2022-Shift-2]

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📖 Explanation

${\text{A}}=8 {\text{cm}}$ ${\text{T}}=6 {\ s}$ $A \cos (\;\frac{2 \pi t}{T}) =\;\frac{A}{2}$ $\Rightarrow \;\frac{2 \pi t}{T}=\;\frac{\pi}{3}$ or $t=\;\frac{T}{6}=1 s$

Q25JEE Main 2022NAT4MElectrostatic Potential and Capacitance

A parallel plate capacitor is made up of stair like structure with a plate area ${\text{A}}$ of each stair and that is connected with a wire of length ${\text{b}}$ , as shown in the figure. The capacitance of the arrangement is $\;\frac{x}{15} \;\frac{E_0 A}{b}$ . The value of ${\text{x}}$ is_____ [27-Jun-2022-Shift-2]

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📖 Explanation

The circuit is equivalent to 3 capacitors in parallel as shown $C_{e q}=\;\frac{\epsilon_0 A}{b} (1+\;\frac{1}{3}+\;\frac{1}{5}) =\;\frac{23}{15} \;\frac{\epsilon_0 A}{b}$ $\Rightarrow x=23$

Q26JEE Main 2022NAT4MCurrent Electricity

The current density in a cylindrical wire of radius $r=4.0 {\text{mm}}$ is $1.0 \times 10^6 {\text{A}} / {\text{m}}^2$ . The current through the outer portion of the wire between radial distances $\;\frac{r}{2}$ and ${\text{r}}$ is ${\text{x}} \pi {\text{A}}$ ; where ${\text{x}}$ is____ [27-Jun-2022-Shift-2]

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📖 Explanation

$i=A \times j$ $=\pi (R^{2-}\;\frac{R^2}{4}) j$ $=\;\frac{3 \pi R^2}{4} \times j$ $=\;\frac{3 \pi \times (4 \times 10^{-3}) ^2}{4} \times 1.0 \times 10^6$ $=12 \pi$

Q27JEE Main 2022NAT4MCurrent Electricity

In the given circuit 'a' is an arbitrary constant. The value of $m$ for which the equivalent circuit resistance is minimum, will be $\sqrt{\;\frac{x}{2}}$ . The value of $x$ is _____ [27-Jun-2022-Shift-2]

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📖 Explanation

$R_{n e t}=\;\frac{m a}{3}+\;\frac{a}{2 m}$ $=a [\;\frac{m}{3}+\;\frac{1}{2 m}-\;{2}/{{\sqrt{6}}}+\;{2}/{{\sqrt{6}}}]$ $=a [ (\sqrt{\;\frac{m}{3}}-\;{1}/{{\sqrt{2 m}}}) ^{2+}\sqrt{\;\frac{2}{3}}]$ This will be minimum when $\sqrt{\;\frac{m}{3}}=\;{1}/{{\sqrt{2 m}}}$ $\;\text{ or }\; m=\sqrt{\;\frac{3}{2}}$ $\;\text{ so }\; x=3$

Q28JEE Main 2022NAT4MMagnetic Effects of Current and Magnetism

A deuteron and a proton moving with equal kinetic energy enter into a uniform magnetic field at right angle to the field. If $r_d$ and ${\text{r}}_p$ are the radii of their circular paths respectively, then the ratio $\;\frac{r_d}{r_p}$ will be $\sqrt{x}: 1$ where ${\text{x}}$ is_____ [27-Jun-2022-Shift-2]

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📖 Explanation

$R=\;{{\sqrt{2 m K}}}/{q B}$ So, $\;\frac{r_d}{r_p}=\;{{\sqrt{m_d}} / q_d}/{{\sqrt{m_p}} / q_p}$ $={\sqrt{2}}$ So $x=2$

Q29JEE Main 2022NAT4MElectromagnetic Induction

A metallic rod of length $20 {\text{cm}}$ is placed in North-South direction and is moved at a constant speed of $20 {\text{m}} / {\text{s}}$ towards East. The horizontal component of the Earth's magnetic field at that place is $4 \times 10^{-3} {\text{T}}$ and the angle of dip is $45^{\circ}$ . The emf induced in the rod is ______ $m V$ [27-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$E=B l v$ $=4 \times 10^{-3} \times \;\frac{20}{100} \times 20$ Volts $=16 {\text{mV}}$

Q30JEE Main 2022NAT4MSemiconductor Electronics

The cut-off voltage of the diodes (shown in figure) in forward bias is $0.6 {\text{V}}$ . The current through the resister of $40 Ω$ is_____ ${\text{mA}}$ . [27-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{D}}_1$ : conducting $D_2$ : open circuit $i=\;\frac{1-0.6}{60+40} A$ $=\;\frac{0.4}{100} A$ $\Rightarrow i=4 {\text{mA}}$

Chemistry30 questions

Q31JEE Main 2022MCQ4MStates of Matter

Which amongst the given plots is the correct plot for pressure (p) vs density (d) for an ideal gas? [27-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

From ideal gas equation we know, ${\text{PV}}= {\text{nRT}}$ $\Rightarrow P V=\;\frac{W}{M} R T$ $\Rightarrow P=\;\frac{W}{V} \cdot \;\frac{R T}{M}$ $\Rightarrow P=d \cdot \;\frac{R T}{M} [ \because d=\;\frac{W}{V}]$ For a fixed amount of gas at a fixed temperature, ${\text{M}}$ and ${\text{T}}$ is constant. $\therefore P \propto d$ So graph between pressure $(P)$ and density $(d)$ is a straight line.Also, $\;\frac{P}{d}=\;\frac{R T}{M}$ $\Rightarrow \;\frac{P}{d} \propto T \;\text{ [as }\; {\text{R}}, {\text{M}}=\;\text{ constant }\;]$ $\therefore$ When temperature ${\text{T}}$ increases then slope of graph between ${\text{P}}$ and ${\text{d}}$ increases.As $T_3>T_2>T_1$ then slope of $T_3$ will be highest then $T_2$ and then $T_1$ .So, graph (B) is the right graph.

Q32JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

Identify the incorrect statement for ${ PCl}_5$ from the following. [27-Jun-2022-Shift-2]

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📖 Explanation

${\text{PCl}}_5$ All three equatorial bonds in a plane ${\ sp}^3 {\text{d}}$ hybridizationTrigonal bipyramidalAxial bonds are weaker than equatorial bonds.

Q33JEE Main 2022MCQ4Md and f Block Elements

Statement I : Leaching of gold with cyanide ion in absence of air / ${\text{O}}_2$ leads to cyano complex of ${\text{Au}}( {\text{III}})$ .Statement II : Zinc is oxidized during the displacement reaction carried out for gold extraction.In the light of the above statements, choose the correct answer from the options given below. [27-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Leaching of gold with cyanide ion is done in presence of air/ ${\text{O}}_2$ leading to cyano complex $[ {\text{Au}}( {\text{CN}})_2] ^{-}$ where ${\text{Au}}$ is in $+1$ oxidation state. $4 {\text{Au}}( {\ s})+8 {\text{CN}}^{-}( {\text{aq}})\;+2 {\text{H}}_2 {\text{O}}( {\text{aq}})+ {\text{O}}_2( {\text{g}}) \longrightarrow 4 [ {\text{Au}}( {\text{CN}})_2] _{ {\text{aq}}}^{-}+4 {\text{OH}}^{-}$ $2 [ {\text{Au}}( {\text{CN}})_2] _{( {\text{aq}})}^{-}\;+ {\text{Zn}}( {\ s})$ $\;\longrightarrow [ {\text{Zn}}( {\text{CN}})_4] _{( {\text{aq}})}^{2-}+2 {\text{Au}}( {\ s})$ Zinc is oxidised from (0) to $+2$ oxidation state during displacement reaction carried out for gold extraction.

Q34JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

The correct order of increasing intermolecular hydrogen bond strength is [27-Jun-2022-Shift-2]

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📖 Explanation

Due to the high difference in electronegativity of ${\text{H}}$ and ${\text{N}}$ the ${\text{H}}$ -bond strength of ${\text{NH}}_3$ is highest. There is no ${\text{H}}$ bond in ${\text{CH}}_4$ . ${\text{CH}}_4< {\text{HCN}}< {\text{NH}}_3$

Q35JEE Main 2022MCQ4MClassification of Elements

The correct order of increasing ionic radii is [27-Jun-2022-Shift-2]

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📖 Explanation

For isoelectronic speciesIonic radii $\propto \;\frac{(-) \;\text{ ve charge }\;}{(+) \;\text{ ve charge }\;}$ Hence, correct order of ionic radii is ${\text{Mg}}^{2+}< {\text{Na}}^{+}< {\text{F}}^{-}< {\text{O}}^{2-}< {\text{N}}^{3-}$

Q36JEE Main 2022MCQ4Mp Block Elements

The gas produced by treating an aqueous solution of ammonium chloride with sodium nitrite is [27-Jun-2022-Shift-2]

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📖 Explanation

${\text{N}}_2$ gas is produced by treating an aqueous solution of ammonium chloride with sodium nitrite. ${\text{NH}}_4 {\text{Cl}}+ {\text{NaNO}}_2 \to {\text{NaCl}}+ {\text{N}}_2( {\text{g}}) ↑+2 {\text{H}}_2 {\text{O}}$

Q37JEE Main 2022MCQ4Mp Block Elements

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.Assertion A : Fluorine forms one oxoacid.Reason R : Fluorine has smallest size amongst all halogens and is highly electronegative.In the light of the above statements, choose the most appropriate answer from the options given below : [27-Jun-2022-Shift-2]

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Q38JEE Main 2022MCQ4Md and f Block Elements

In $3 {\text{d}}$ series, the metal having the highest ${\text{M}}^{2+} / {\text{M}}$ standard electrode potential is [27-Jun-2022-Shift-2]

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📖 Explanation

${\text{Cr}}^{+2} / {\text{Cr}} \to -0.90 {\text{V}}$ ${\text{Fe}}^{+2} / {\text{Fe}} \to -0.44 {\text{V}}$ ${\text{Cu}}^{+2} / {\text{Cu}} \to +0.34 {\text{V}}$ ${\text{Zn}}^{+2} / {\text{Zn}} \to -0.76 {\text{V}}$ So Ans. is ${\text{Cu}}^{+2} / {\text{Cu}}$

Q39JEE Main 2022MCQ4Md and f Block Elements

The ' $f$ ' orbitals are half and completely filled, respectively in lanthanide ions[Given : Atomic no. Eu, 63; ${\text{Sm}}, 62 ; {\text{Tm}}, 69 ; {\text{Tb}}, 65 ; {\text{Yb}}, 70 ; {\text{Dy}}, 66]$ [27-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$T b \to 4 f^9 6 s^2$ ${\text{Tb}}^{+4} \to 4 {\text{f}}^7$ ${\text{Yb}} \to 4 f^{14} 6 {\ s}^2$ ${\text{Yb}}^{+2} \to 4 {\text{f}}^{14}$ Hence, the pair ${\text{Tb}}^{+4}$ and ${\text{Yb}}^{+2}$ have half filled and completely filled f subshells respectively.

Q40JEE Main 2022MCQ4MCoordination Compounds

Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers : ${\text{Mn}}=25 ; {\text{Fe}}=26$ )A. $[ {\text{FeF}}_6]^{3-}$ B. $[ {\text{Fe}}( {\text{CN}})_6]^{3-}$ C. $[ {\text{MnCl}}_6]^{3-}$ (high spin)D. $[ {\text{Mn}}( {\text{CN}})_6]^{3-}$ Choose the correct answer from the options given below : [27-Jun-2022-Shift-2]

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📖 Explanation

Coordination Compound Number of unpaired ${\text{e}}^{-}( {\text{n}})$ Magnetic moment $(\mu )( {\text{B}} . {\text{M}})$ A. $[ {\text{FeF}}_6] ^{3-}- {\text{d}}^5$ 5 $5.91$ B. $[ {\text{Fe}}( {\text{CN}})_6] ^{3-}- {\text{d}}^5$ 1 $1.73$ C. $[ {\text{MnCl}}]_6] ^{3-}- {\text{d}}^4$ 4 $4.89$ D. $[ {\text{Mn}}( {\text{CN}})_6] ^{3-}- {\text{d}}^4$ 2 $2.82$ Hence, correct order of magnetic moment is $2< 4< 3< 1$

Q41JEE Main 2022MCQ4Mp Block Elements

On the surface of polar stratospheric clouds, hydrolysis of chlorine nitrite gives A and B while its reaction with its reaction with ${\text{HCl}}$ produces ${\text{B}}$ and ${\text{C}}$ . A, ${\text{B}}$ and ${\text{C}}$ are, respectively [27-Jun-2022-Shift-2]

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📖 Explanation

On the surface of polar stratospheric clouds, hydrolysis of chlorine nitrate as ${\text{ClONO}}_2+ {\text{H}}_2 {\text{O}} \longrightarrow \;{{ {\text{HOCl}}}}_{{ {\text{A}}}}+\;{{ {\text{HNO}}_3}}_{{ {\text{B}}}}$ ${\text{ClONO}}_2+ {\text{HCl}} \longrightarrow \;{{ {\text{Cl}}_2}}_{{ {\text{C}}}}+ {\text{HNO}}_3$ Hence A, B and ${\text{C}}$ are ${\text{HOCl}}, {\text{HNO}}_3$ and ${\text{Cl}}_2$ respectively.

Q42JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Which of the following is most stable? [27-Jun-2022-Shift-2]

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Q43JEE Main 2022MCQ4MHydrocarbons

What will be the major product of following sequence of reactions? $\text{n}-\text{Bu}-\equiv {\to }^{ (\text{i}) \; \text{n}-\text{BuLi}, \text{n}-C_5 \text{H}_11 \text{Cl}}_{(\text{ii})\; \text{Lindlar cat}, \text{H}_2}$ [27-Jun-2022-Shift-2]

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Q44JEE Main 2022MCQ4MHaloalkanes and Haloarenes

Product 'A' of following sequence of reactions is $\text{ Ethylbenzene }\; {\to }_{(b) \text{Cl}_2, \Delta }_{ (c) \;\; \text{alc}. {\text{KOH}} } ^{(a)\;\; \text{Br}_2, \text{Fe}} {\'\text{A}\'} _{\text{Major Product}}$ [27-Jun-2022-Shift-2]

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Q45JEE Main 2022MCQ4MSalt Analysis

Choose the correct answer from the options given below: List-1 List-II (A) I. $B r_2$ in $C S_2$ (B) II. ${\text{Na}}_2 {\text{Cr}}_2 {\text{O}}_7 / {\text{H}}_2 {\text{SO}}_4$ (C) III. $\text{Zn}$ (D) IV. ${\text{CHCl}}_3 / {\text{NaOH}}$ [27-Jun-2022-Shift-2]

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Q46JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Decarboxylation of all six possible forms of diaminobenzoic acids ${\text{C}}_6 {\text{H}}_3( {\text{NH}}_2)_2 {\text{COOH}}$ yields three products ${\text{A}}, {\text{B}}$ and ${\text{C}}$ . Three acids give a product ' $A$ ', two acids gives a product ' $B$ ' and one acid give a product ' $C$ '. The melting point of product ' $C$ ' is [27-Jun-2022-Shift-2]

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Q47JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Which is true about Buna-N? [27-Jun-2022-Shift-2]

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Q48JEE Main 2022MCQ4MBiomolecules

Given below are two statements.Statement I : Maltose has two $\alpha$ -D-glucose units linked at ${\text{C}}_1$ and ${\text{C}}_4$ and is a reducing sugar.Statement II : Maltose has two monosaccharides : $\alpha$ -D-glucose and $\beta$ -D-glucose linked at ${\text{C}}_1$ and ${\text{C}}_6$ and it is a non-reducing sugar.In the light of the above statements, choose the correct answer from the options given below : [27-Jun-2022-Shift-2]

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📖 Explanation

Maltose is composed of two $\alpha$ -D-glucose units in which $C_1$ of one glucose unit and $C_4$ of second glucose unit are linked.

Q49JEE Main 2022MCQ4MSalt Analysis

Choose the correct answer from the options given below : List-I List-II (A) Antipyretic I. Reduces pain (B) Analgesic II. Reduces stress (C) Tranquilizer III. Reduces fever (D) Antacid IV. Reduces acidity (stomach) [27-Jun-2022-Shift-2]

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Q50JEE Main 2022MCQ4MSalt Analysis

Choose the correct answer from the options given below : List-I (Anion) List-II (A) ${\text{CO}}_3^{2-}$ I. Colourless gas which turns lead acetate paper black. (B) $\text{S}^{2-}$ II. Colourless gas which turns acidified potassium dichromate solution green. (C) ${\text{SO}}_3^{2-}$ III. Brown fumes which turns acidified KI solution containing starch blue. (D) ${\text{NO}}_2^{-}$ IV. Colourless gas evolved with brisk effervescence, which turns lime water milky. [27-Jun-2022-Shift-2]

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📖 Explanation

${\text{CO}}_3^{2-}$ : On action of dil sulphuric acid, ${\text{CO}}_2$ gas is released which turns lime water milky. ${\text{S}}^{2-}$ : On action of dil sulphuric acid, ${\text{H}}_2 {\text{S}}$ gas is released which turns lead acetate paper black. ${\text{SO}}_3^{2-}$ : On action of dil ${\text{H}}_2 {\text{SO}}_4, {\text{SO}}_2$ gas is evolved which turns acidified potassium dichromate solution green. ${\text{NO}}_2^{-}$ : On action of dil ${\text{H}}_2 {\text{SO}}_4, {\text{NO}}_2$ gas is evolved which turns ${\text{KI}}$ solution containg starch blue.

Q51JEE Main 2022NAT4MOrganic Chemistry - Some Basic Principles

$116 {\text{g}}$ of a substance upon dissociation reaction, yields $7.5 {\text{g}}$ of hydrogen, $60 {\text{g}}$ of oxygen and $48.5 {\text{g}}$ of carbon. Given that the atomic masses of ${\text{H}}, {\text{O}}$ and ${\text{C}}$ are 1, 16 and 12, respectively. The data agrees with how many formulae of the following?A. ${\text{CH}}_3 {\text{COOH}}$ , B. ${\text{HCHO}},$ $\text{C} . {\text{CH}}_3 {\text{OOCH}}_3$ , D. ${\text{CH}}_3 {\text{CHO}}$ [27-Jun-2022-Shift-2]

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📖 Explanation

$\% {\text{H}}=\;\frac{7.5}{116} \times 100=6.5$ $\% {\text{O}}=\;\frac{60}{116} \times 100=51.7$ $\% {\text{C}}=\;\frac{48.5}{116} \times 100=41.8$ Relative atomicities $= {\text{H}} \Rightarrow 6.5$ ${\text{O}} \Rightarrow \;\frac{51.7}{16}=3.25$ ${\text{C}} \Rightarrow \;\frac{41.8}{12}=3.5$ Emperically formula is approx.. ${\text{CH}}_2 {\text{O}}$ (A) ${\text{C}}_2 {\text{H}}_4 {\text{O}}_2$ (B) ${\text{CH}}_2 {\text{O}}$ relate to this formula.

Q52JEE Main 2022NAT4MStructure of Atom

Consider the following set of quantum numbers. n 1 $\text{m}_1$ A. 3 3 3 B. 3 2 2 C. 2 1 +1 D. 2 2 +2 The number of correct sets of quantum numbers is __________. [27-Jun-2022-Shift-2]

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📖 Explanation

For $A$ ,Given ${\text{n}}=3$ and ${\text{l}}=3$ but we know maximum value of ${\text{I}}= {\text{n}}-1$ . $\therefore$ I can't be equal to $n$ .So, Set A of quantum numbers is not possible.For B,Given $n=3, l=2, m=-2$ Here, $l=2$ which follow the rule $l=n-1$ .And we know possible value of $m$ is $-I$ to $+I$ .here possible value of $m=-2$ to $+2$ $\therefore$ This Set B is valid set of quantum numbers. For C,Given $n=2, I=1, m=+1$ Here $I=1$ which follows the rule $I=n-1$ .For $I=1$ possible value of $m=-1$ to $+1$ Here $m=+1$ . So value of $m$ is valid. $\therefore$ Set ${\text{C}}$ is valid set of quantum numbers.For D,Given $n=2, I=2, m=+2$ ${\text{I}}=2$ does not follow the rule ${\text{I}}= {\text{n}}-1$ rule. $\therefore$ Set $D$ is not valid set of quantum numbers.

Q53JEE Main 2022NAT4MCoordination Compounds

BeO reacts with HF in presence of ammonia to give $[A]$ which on thermal decomposition produces [B] and ammonium fluoride. Oxidation state of Be in $[ {\text{A}}]$ is___ [27-Jun-2022-Shift-2]

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Q54JEE Main 2022NAT4MChemical Thermodynamics

When 5 moles of He gas expand isothermally and reversibly at $300 {\text{K}}$ from 10 litre to 20 litre, the magnitude of the maximum work obtained is____ ${\text{J}} .$ [nearest integer] (Given : ${\text{R}}=8.3 {\text{J}} {\text{K}}^{-1} {\text{mol}}^{-1}$ and $\log 2=0.3010$ ) [27-Jun-2022-Shift-2]

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📖 Explanation

${\text{W}}_{\;\text{rev }\;}\;=-2.303 {\text{nRT}} \log _{10} (\;{ {\text{V}}_2}/{ {\text{V}}_1})$ $\;=-2.303 \times 5 \times 8.3 \times 300 \times \log _{10} (\;\frac{20}{10})$ $\;≃-8630 {\text{J}}$

Q55JEE Main 2022NAT4MSolutions

A solution containing $2.5 \times 10^{-3} {\text{kg}}$ of a solute dissolved in $75 \times 10^{-3} {\text{kg}}$ of water boils at $373.535 {\text{K}}$ . The molar mass of the solute is _____ ${\text{g}} {\text{mol}}{ }^{-1}$ . [nearest integer] (Given : ${\text{K}}_{ {\text{b}}}( {\text{H}}_2 {\text{O}})=0.52 {\text{K}} {\text{kg}} {\text{mol}}^{-1}$ and boiling point of water $=373.15 {\text{K}}$ ) [27-Jun-2022-Shift-2]

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📖 Explanation

$W_{\;\text{solute }\;}=2.5 \times 10^{-3} {\text{kg}}$ ${\text{W}}_{\;\text{solvent }\;}=75 \times 10^{-3} {\text{kg}}$ $\Delta {\text{T}}_{ {\text{b}}}=373.535-373.15$ $=0.385 {\text{K}}$ ${\text{K}}_{ {\text{b}}} ( {\text{H}}_2 {\text{O}}) =0.52 {\text{K}} {\text{kg}} {\text{mol}}^{-1}$ $\Delta {\text{T}}_{ {\text{b}}}=\;{ {\text{K}}_{ {\text{b}}} \times 10^3 \times {\text{W}}_{\;\text{solute }\;}}/{ {\text{M}}_{\;\text{solute }\;} \times {\text{W}}_{\;\text{solvent }\;}}$ ${\text{M}}_{\;\text{solute }\;}=\;\frac{0.52 \times 10^3 \times 2.5 \times 10^{-3}}{75 \times 10^{-3} \times 0.385}$ $=45.02$ $\approx 45$

Q56JEE Main 2022NAT4MIonic Equilibrium

${\text{pH}}$ value of $0.001 {\text{M}} {\text{NaOH}}$ solution is___ [27-Jun-2022-Shift-2]

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📖 Explanation

$[ {\text{OH}}^{-}] =0.001=10^{-3} {\text{M}}$ $[ {\text{H}}^{+}] \$ [ {\text{OH}}^{-}]$ =10^{-14}[ {\text{H}}^{+}] =10^{-11}{\text{pH}}=-\log [ {\text{H}}^{+}]=-\log (10^{-11}){\text{pH}}=11$

Q57JEE Main 2022NAT4MElectrochemistry

For the reaction taking place in the cell : ${\text{Pt}}( {\text{s}})| {\text{H}}_2( {\text{g}})| {\text{H}}^{+}( {\text{aq}}) \| {\text{Ag}}^{+}( {\text{aq}}) \mid {\text{Ag}}( {\ s})$ ${\text{E}}_{\;\text{cell }\;}^o=+0.5332 {\text{V}} .$ The value of $\Delta _f G^{\circ}$ is ____ ${\text{kJ}} {\text{mol}}^{-1}$ . (in nearestinteger) [27-Jun-2022-Shift-2]

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📖 Explanation

${\text{H}}_2 | {\text{H}}^{+} {{}/{}\| \|} {\text{Ag}}^{+}{{}/{}|} {\text{Ag}}$ Anode Cathode # At anode, oxidation occur ${\text{H}}_2 \to 2 {\text{H}}^{+}+2 {\text{e}}^{-}$ # At cathode, reduction occur $2 {\text{Ag}}^{+}+2 {\text{e}}^{-} \to 2 {\text{Ag}}$ Adding equation (1) and (2), we get ${\text{n}}=2$ , where ${\text{n}}=$ cancelled out electronNow, $\Delta G^{\circ}=-n F E_{c e l l}^o$ $=-2 \times 96500 \times 0.5332$ $=-102907.6$ $=-102.9 {\text{kJ}} / {\text{mol}}$ $=-103 {\text{kJ}} / {\text{mol}}$

Q58JEE Main 2022NAT4MChemical Kinetics

It has been found that for a chemical reaction with rise in temperature by $9 {\text{K}}$ the rate constant gets doubled. Assuming a reaction to be occurring at $300 {\text{K}}$ , the value of activation energy is found to be _____ ${\text{kJ}} {\text{mol}}^{-1}$ . [nearest integer](Given $\ln 10=2.3, {\text{R}}=8.3 {\text{JK}}^{-1} {\text{mol}}^{-1}, \log 2=0.30$ ) [27-Jun-2022-Shift-2]

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📖 Explanation

$T_1=300 {\text{K}}$ (Rate constant) ${\text{K}}_2=2 {\text{K}}_1$ , on increase temperature by $9 {\text{K}}$ ${\text{T}}_2=309 {\text{K}}$ ${\text{Ea}}=?$ $\log \;{ {\text{K}}_2}/{ {\text{K}}_1}=\;{ {\text{Ea}}}/{2.3 {\text{R}}} [\;{ {\text{T}}_2- {\text{T}}_1}/{ {\text{T}}_2 \cdot {\text{T}}_1}]$ $\log 2=\;{ {\text{Ea}}}/{2.3 \times 8.3} [\;\frac{9}{309 \times 300}]$ ${\text{Ea}}=\;\frac{0.3 \times 309 \times 300 \times 2.3 \times 8.3}{9}$ $=58988.1 {\text{J}} / {\text{mole}}$ $≃ 59 {\text{kJ}} / {\text{mole}}$

Q59JEE Main 2022NAT4MChemical Equilibrium

If the initial pressure of a gas is $0.03$ atm, the mass of the gas adsorbed per gram of the adsorbent is ____ $\times 10^{-2} {\text{g}}$ [27-Jun-2022-Shift-2]

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📖 Explanation

Given that $\log {\text{K}}=$ intercept $=0.602=\log 4$ $\therefore {\text{K}}=4$ $\;\text{ Slope }\;=\;\frac{1}{n}=1$ and initial pressure $=0.03 {\text{atm}}$ $\;\frac{x}{m}=K(p)^{1 / n}=4 \times 0.03=0.12=12 \times 10^{-2}$ mass of gas absorbed per gm of adsorbent $=12 \times 10^{-2} {\text{g}}$

Q60JEE Main 2022NAT4MPractical Organic Chemistry

$0.25 {\text{g}}$ of an organic compound containing chlorine gave $0.40 {\text{g}}$ of silver chloride in Carius estimation. The percentage of chlorine present in the compound is_____ [in nearest integer](Given : Molar mass of ${\text{Ag}}$ is $108 {\text{g}} {\text{mol}}^{-1}$ and that of ${\text{Cl}}$ is $35.5 {\text{g}} {\text{mol}}^{-1}$ ) [27-Jun-2022-Shift-2]

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📖 Explanation

Given, weight of organic compound $=0.25 {\text{g}}$ Moles of ${\text{AgCl}}=\;\frac{0.4}{M}$ Molecular mass of ${\text{AgCl}}(M)=143.5 {\text{gm}}$ $\therefore \;\text{ Moles of }\; {\text{AgCl}}=\;\frac{0.4}{143.5}$ $\therefore \;\text{ Mass of }\; {\text{Cl}}=\;\frac{0.4}{143.5} \times 35.5$ Mass $\%$ of ${\text{Cl}}$ in the organic compound $=\;\frac{\;\frac{35.5 \times 0.4}{143.5}}{0.25} \times 35.5$ $=39.58$ $≃ 40$

Mathematics30 questions

Q61JEE Main 2022MCQ4MComplex Numbers

The number of points of intersection of $|z-(4+3 i)|=2$ and $|z|+|z-4|=6, {\text{z}} \in {\text{C}}$ , is [27-Jun-2022-Shift-2]

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📖 Explanation

$C_1:|z-(4+3 i)|=2 \;\text{ and }\; C_2:|z|+|z-4|=6, z \in C$ ${\text{C}}_1$ represents a circle with centre $(4,3)$ and radius 2 and ${\text{C}}_2$ represents a ellipse with focii at $(0,0)$ and $(4,0)$ and length of major axis $=6$ , and length of semi-major axis $2 {\sqrt{5}}$ and $(4,2)$ lies inside the both $C_1$ and $C_2$ and $(4,3)$ lies outside the $C_2$ $\therefore$ number of intersection points $=2$

Q62JEE Main 2022MCQ4MDifferentiation

Let $f(x)=\begin{vmatrix}a & -1 & 0 \\ a x & a & -1 \\ a x^2 & a x & a\end{vmatrix}, a \in R$ . Then the sum of the squares of all the values of $a$ , for which $2 f^{'}(10)-f^{'}(5)+100=0$ , is [27-Jun-2022-Shift-2]

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📖 Explanation

$f(x)={\begin{vmatrix}a & -1 & 0 \\ a x & a & -1 \\ a x^2 & a x & a\end{vmatrix}}, a \in R$ $f(x)=a (a^{2+}a x) +1 (a^2 x+a x^2)$ $=a(x+a)^2$ $f^{'}(x)=2 a(x+a)$ Now, $2 f^{'}(10)-f^{'}(5)+100=0$ $\Rightarrow 2 \cdot 2 a(10+a)-2 a(5+a)+100=0$ $\Rightarrow 2 a(a+15)+100=0$ $\Rightarrow a^{2+}15 a+50=0$ $\Rightarrow a=-10,-5$ $\therefore$ Sum of squares of values of $a=125$

Q63JEE Main 2022MCQ4MComplex Numbers

Let for some real numbers $\alpha$ and $\beta, a=\alpha-i \beta$ . If the system of equations $4 i x+(1+i) y=0$ and $8(\cos \;\frac{2 \pi}{3}+i \sin \;\frac{2 \pi}{3}) x+\overline{a} y=0$ has more than one solution, then $\;\frac{\alpha}{\beta}$ is equal to [27-Jun-2022-Shift-2]

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📖 Explanation

Given $a=\alpha-i \beta$ and $4 i x+(1+i) y=0 . . . . . .$ . (i) $8 (\cos \;\frac{2 \pi}{3}+i sin \;\;\frac{2 \pi}{3}) x+\overline{a} y=0$ By (i) $\;\frac{x}{y}=\;\frac{-(1+i)}{4 i} . . . . . .$ (iii)By (ii) $\;\frac{x}{y}=\;{-\overline{a}}/{8 (\;\frac{-1}{2}+\;{{\sqrt{3 i}}}/{2}) } . . . .$ Now by (iii) and (iv) $\;\frac{1+i}{4 i}=\;{\overline{a}}/{4(-1+{\sqrt{3}} i)}$ $\Rightarrow \overline{a}=({\sqrt{3}}-1)+({\sqrt{3}}+1) i$ $\Rightarrow \alpha+i \beta=({\sqrt{3}}-1)+({\sqrt{3}}+1) i$ $\therefore \;\frac{\alpha}{\beta}=\;{{\sqrt{3}}-1}/{{\sqrt{3}}+1}=2-{\sqrt{3}}$

Q64JEE Main 2022MCQ4MMatrices and Determinants

Let ${\text{A}}$ and ${\text{B}}$ be two $3 \times 3$ matrices such that $\text{AB}=I$ and $|A|=\;\frac{1}{8}$ . Then $|\text{adj}(\text{B} \; \text{adj}(2\text{A}))|$ is equal to [27-Jun-2022-Shift-2]

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📖 Explanation

$A$ and $B$ are two matrices of order $3 \times 3$ .and $A B=I$ , $|A|=\;\frac{1}{8}$ Now, $|A||B|=1$ $|B|=8$ $\therefore \mid {\text{adj}} (B({\text{adj}}(2 A))|=| B(a d j(2 A)){{}/{}|} ^2.$ $=|B|^2|{\text{adj}}(2 A)|^2$ $=2^6|2 A|^{2 \times 2}$ $=2^6 \cdot 2^{12} \cdot \;\frac{1}{2^{12}}=64$

Q65JEE Main 2022MCQ4MSequences and Series

Let $S=2+\;\frac{6}{7}+\;\frac{12}{7^2}+\;\frac{20}{7^3}+\;\frac{30}{7^4}+. . . . .$ Then $4 S$ is equal to [27-Jun-2022-Shift-2]

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📖 Explanation

$S=2+\;\frac{6}{7}+\;\frac{12}{7^2}+\;\frac{20}{7^3}+\;\frac{30}{7^4}+. . . . . . . . .$ (i) $\;\frac{1}{7} S=\;\frac{2}{7}+\;\frac{6}{7^2}+\;\frac{12}{7^3}+\;\frac{20}{7^4}+. . . . . . . . .$ (ii)(i) - (ii) $\;\frac{6}{7} S=2+\;\frac{4}{7}+\;\frac{6}{7^2}+\;\frac{8}{7^3}+. . . . . . . . . . . .$ (iii) $\;\frac{6}{7^2} S=\;\frac{2}{7}+\;\frac{4}{7^2}+\;\frac{6}{7^3}+. . . . . . . . . . . .$ (iv)(iii) - (iv) $(\;\frac{6}{7}) ^2 S=2+\;\frac{2}{7}+\;\frac{2}{7^2}+\;\frac{2}{7^3}+. . . . . .$ . $=2 [\;\frac{1}{1-\;\frac{1}{7}}] =2 (\;\frac{7}{6})$ $\therefore 4 S=8 (\;\frac{7}{6}) ^3= (\;\frac{7}{3}) ^3$

Q66JEE Main 2022MCQ4MSequences and Series

If $a_1, a_2, a_3 . . . . . .$ and $b_1, b_2, b_3 . . . . . .$ are A.P., and $a_1=2, a_{10}=3, a_1 b_1=1=a_{10} b_{10}$ , then $a_4 b_4$ is equal to - [27-Jun-2022-Shift-2]

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📖 Explanation

$a_1, a_2, a_3 . . .$ are in A.P. (Let common difference is $d_1$ ) $b_1, b_2, b_3 . . .$ are in A.P. (Let common difference is $d_2$ )and $a_1=2, a_{10}=3, a_1 b_1=1=a_{10} b_{10}$ $\because a_1 b_1=1$ $\therefore b_1=\;\frac{1}{2}$ $a_{10} b_{10}=1$ $\therefore b_{10}=\;\frac{1}{3}$ Now, $a_{10}=a_1+9 d_1 \Rightarrow d_1=\;\frac{1}{9}$ $b_{10}=b_1+9 d_2 \Rightarrow d_2=\;\frac{1}{9} [\;\frac{1}{3}-\;\frac{1}{2}] =-\;\frac{1}{54}$ Now, $a_4=2+\;\frac{3}{9}=\;\frac{7}{3}$ ${\text{b}}_4=\;\frac{1}{2}-\;\frac{3}{54}=\;\frac{4}{9}$ $\therefore a_4 b_4=\;\frac{28}{27}$

Q67JEE Main 2022MCQ4MApplication of Derivatives

If ${\text{m}}$ and ${\text{n}}$ respectively are the number of local maximum and local minimum points of the function $f(x)=int_0^{x^2} \;\frac{t^{2-}5 t+4}{2+e^t} d t$ , then the ordered pair $( {\text{m}}, {\text{n}})$ is equal to [27-Jun-2022-Shift-2]

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📖 Explanation

$f(x)=int_{0}^{x^2} \;\frac{t^{2-}5 t+4}{2+e^t} \text{dt}$ $f^{'}(x)=2 x (\;\frac{x^{4-}5 x^{2+}4}{2+e^{x^2}}) =0$ $x=0$ , or $(x^{2-}4) (x^{2-}1) =0$ $x=0, x=\pm 2, \pm 1$ Now, $f^{'}(x)=\;\frac{2 x(x+1)(x-1)(x+2)(x-2)}{ (e^{x^2}+2) }$ ${\text{f}}^{'}( {\text{x}})$ changes sign from positive to negative at ${\text{x}}=-1,1$ So, number of local maximumpoints = 2 ${\text{f}}^{'}( {\text{x}})$ changes sign from negative to positive at ${\text{x}}=-2,0,2$ So, number of local minimumpoints = 3 $\therefore {\text{m}}=2, {\text{n}}=3$

Q68JEE Main 2022MCQ4MDefinite Integrals

Let ${\text{f}}$ be a differentiable function in $(0, \;\frac{\pi}{2})$ . If $int_{\cos x}^{1} t^2 f(t) d t=\sin ^3 x+\cos x$ , then $\;\frac{1}{\sqrt{3}} f^{'}(\;\frac{1}{\sqrt{3}})$ is equal to [27-Jun-2022-Shift-2]

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📖 Explanation

$int_{\cos x}^{1} t^2 f(t) \text{dt}= sin \;^3 x+\cos x$ $\Rightarrow sin \;x \cos ^2 x f(\cos x)=3 sin \;^2 x \cos x- sin \;x$ $\Rightarrow f(\cos x)=3 \tan x-\ sec ^2 x$ $\Rightarrow f^{'}(\cos x) \cdot (- sin \;x)=3 \ sec ^2 x-2 \ sec ^2 x \tan x$ Put $\cos x=\;{1}/{{\sqrt{3}}}$ , $\therefore f^{'} (\;{1}/{{\sqrt{3}}}) (-\;{{\sqrt{2}}}/{{\sqrt{3}}}) =9-6 {\sqrt{2}}$ $\;{1}/{{\sqrt{3}}} f^{'} (\;{1}/{{\sqrt{3}}}) =6-\;{9}/{{\sqrt{2}}}$

Q69JEE Main 2022MCQ4MDefinite Integrals

The integral $int_0^1 \;{1}/{7^{[\;\frac{1}{x}]}} d x$ , where [. ] denotes the greatest integer function, is equal to [27-Jun-2022-Shift-2]

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📖 Explanation

$int_0^1 \;{1}/ {{7} ^{ [\frac{1}{x}]}} \text{dx}, \;\text{ let }\; \;\frac{1}{x}=t$ $\;\frac{-1}{x^2} \text{dx}=d t$ $=int_{\infty }^{1} \;\frac{1}{-t^2 7^{[t]}} \text{dt}=int_{1}^{\infty } \;\frac{1}{t^2 7^{[t]}} \text{dt}$ $=int_{1}^{2} \;\frac{1}{7 t^2} \text{dt}+int_{2}^{3} \;\frac{1}{7^2 t^2} \text{dt}+. . .$ $=\;\frac{1}{7} [-\;\frac{1}{t}] _1^{2+}\;\frac{1}{7^2} [\;\frac{-1}{t}] _2^{3+}\;\frac{1}{7^3} [-\;\frac{1}{t}] _2^{3+}. . .$ $=sum_{n=1}^{\infty } \;\frac{1}{7^n} (\;\frac{1}{n}-\;\frac{1}{n+1})$ $=sum_{n=1}^{\infty } \;\frac{ (\frac{1}{7}) ^n}{n}-7 sum_{n=1}^{\infty } \;\frac{ (\frac{1}{7}) ^{n+1}}{n+1}$ $=-\log (1-\;\frac{1}{7}) +7 \log (1-\;\frac{1}{7}) +1$ $=1+6 \log \;\frac{6}{7}$

Q70JEE Main 2022MCQ4MDifferential Equations

If the solution curve of the differential equation $((\tan ^{-1} y)-x) d y=(1+y^2) d x$ passes through the point $(1,0)$ , then the abscissa of the point on the curve whose ordinate is $\tan (1)$ , is [27-Jun-2022-Shift-2]

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📖 Explanation

$( (\tan ^{-1} y) -x) d y= (1+y^2) \text{dx}$ $\;\frac{d x}{d y}+\;\frac{x}{1+y^2}=\;\frac{\tan ^{-1} y}{1+y^2}$ $\;\text{ I. F. }\;=e^{\int \;\frac{1}{1+y^2} d y}=e^{\tan ^{-1} y}$ $\therefore \;\text{ Solution }\;$ $\;\text{ x. }\; e^{\tan ^{-1} y}=\int \;{e^{\tan ^{-1} y} \tan ^{-1} y}/{1+y^2} d y$ $\;\text{ Let }\; e^{\tan ^{-1} y}=t$ $\;\frac{e^{\tan -1} y}{1+y^2}=d t$ $=x e^{\tan ^{-1} y}=\int \ln t \text{dt}=t \ln t-t+c$ $\therefore =x e^{\tan ^{-1} y}=e^{\tan ^{-1} y} \tan ^{-1} y-e^{\tan ^{-1} y}+c . . . . . \;\text{ (i) }\;$ $\because \;\text{ It passes through }\;(1,0) \Rightarrow c=2$ $\;\text{ Now put }\; y=\tan 1 \;\text{, then }\;$ $e x=e-e+2$ $$ $\Rightarrow x=\;\frac{2}{e}$

Q71JEE Main 2022MCQ4MParabola

If the equation of the parabola, whose vertex is at $(5,4)$ and the directrix is $3 x+y-29=0$ , is $x^{2+}a y^{2+}b x y+c x+d y+k=0$ , then $a+b+c+d+k$ is equal to [27-Jun-2022-Shift-2]

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📖 Explanation

Given vertex is $(5,4)$ and directrix $3 x+y-29=0$ Let foot of perpendicular of $(5,4)$ on directrix is $( {\text{x}}_1, {\text{y}}_1)$ $\;\frac{x_1-5}{3}=\;\frac{y_1-4}{1}=\;\frac{-(-10)}{10}$ $\therefore (x_1, y_1) \equiv (8,5)$ So, focus of parabola will be $S=(2,3)$ Let $P(x, y)$ be any point on parabola, then $(x-2)^{2+}(y-3)^2=\;\frac{(3 x+y-29)^2}{10}$ $\Rightarrow 10 (x^{2+}y^{2-}4 x-6 y+13) =9 x^{2+}y^{2+}841+6 x y-58 y-174 x$ $\Rightarrow x^{2+}9 y^{2-}6 x y+134 x-2 y-711=0$ and given parabola $x^{2+}a y^{2+}b x y+c x+d y+k=0$ $\therefore a=9, b=-6, c=134, d=-2, {\text{k}}=-711$ $\therefore a+b+c+d+k=-576$

Q72JEE Main 2022MCQ4MCircles

The set of values of ${\text{k}}$ , for which the circle $C: 4 x^{2+}4 y^{2-}12 x+8 y+k=0$ lies inside the fourth quadrant and the point $(1,-\;\frac{1}{3})$ lies on or inside the circle $C$ , is [27-Jun-2022-Shift-2]

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📖 Explanation

$C: 4 x^{2+}4 y^{2-}12 x+8 y+k=0$ $\because (1,-\;\frac{1}{3})$ lies on or inside the ${\text{C}}$ then $4+\;\frac{4}{9}-12-\;\frac{8}{3}+k \le 0$ $\Rightarrow k \le \;\frac{92}{9}$ Now, circle lies in $4^{\;\text{th }\;}$ quadrant centre $\equiv (\;\frac{3}{2},-1)$ $\therefore r< 1 \Rightarrow \sqrt{\;\frac{9}{4}+1-\;\frac{k}{4}}< 1$ $\Rightarrow \;\frac{13}{4}-\;\frac{k}{4}< 1$ $\Rightarrow \;\frac{k}{4}>\;\frac{9}{4}$ $\Rightarrow k>9$ $\therefore k \in (9, \;\frac{92}{9})$

Q73JEE Main 2022MCQ4MThree Dimensional Geometry

Let the foot of the perpendicular from the point $(1,2,4)$ on the line $\;\frac{x+2}{4}=\;\frac{y-1}{2}=\;\frac{z+1}{3}$ be $P$ . Then the distance of $P$ from the plane $3 x+4 y+12 z+23=0$ is [27-Jun-2022-Shift-2]

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📖 Explanation

$L: \;\frac{x+2}{4}=\;\frac{y-1}{2}=\;\frac{z+1}{3}=t$ Let $P=(4 t-2,2 t+1,3 t-1)$ $\because P$ is the foot of perpendicular of $(1,2,4)$ $\therefore 4(4 t-3)+2(2 t-1)+3(3 t-5)=0$ $\Rightarrow 29 t=29 \Rightarrow t=1$ $\therefore P=(2,3,2)$ Now, distance of $P$ from the plane $3 x+4 y+12 z+23=0$ , is $|\;{6+12+24+23}/{{\sqrt{9+16+144}}}{{}/{}|} =\;\frac{65}{13}=5$

Q74JEE Main 2022MCQ4MThree Dimensional Geometry

The shortest distance between the lines $\;\frac{x-3}{2}=\;\frac{y-2}{3}=\;\frac{z-1}{-1}$ and $\;\frac{x+3}{2}=\;\frac{y-6}{1}=\;\frac{z-5}{3}$ , is [27-Jun-2022-Shift-2]

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📖 Explanation

$L_1: \;\frac{x-3}{2}=\;\frac{y-2}{3}=\;\frac{z-1}{-1}$ $L_2: \;\frac{x+3}{2}=\;\frac{y-6}{1}=\;\frac{z-5}{3}$

\;\text{ Now, }\; \vec{p} \times \vec{q}= |\begin{array}{ccc}\hat{i}\;\hat{j}\;\hat{k}$ $2\;3\;-1$ $2\;1\;3\end{array}{{}/{}|} =10 \hat{i}-8 \hat{j}-4 \hat{k}

$\;\text{ and }\; \vec{a_2}-\vec{a}_1=6 \hat{i}-4 \hat{j}-4 \hat{k}$ $\therefore S . D .= |\;{60+32+16}/{{\sqrt{100+64+16}}}{{}/{}|} =\;{108}/{{\sqrt{180}}}=\;{18}/{{\sqrt{5}}}$

Q75JEE Main 2022MCQ4MVector Algebra

Let $\vec{a}$ and $\vec{b}$ be the vectors along the diagonals of a parallelogram having area $2 \sqrt{2}$ . Let the angle between $\vec{a}$ and $\vec{b}$ be acute, $|\vec{a}|=1$ , and $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$ . If $\vec{c}=2 \sqrt{2}(\vec{a} \times \vec{b})-2 \vec{b}$ , then an angle between $\vec{b}$ and $\vec{c}$ is [27-Jun-2022-Shift-2]

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📖 Explanation

$\because \vec{a}$ and $\vec{b}$ be the vectors along the diagonals of a parallelogram having area $2 {\sqrt{2}}$ . $\therefore \;\frac{1}{2} \mid \vec{a} \times \vec{b \mid }=2 {\sqrt{2}}$ $|\vec{a}||\vec{b}| sin \;\theta=4 {\sqrt{2}}$ $\Rightarrow \vec{\mid b} \mid sin \;\theta=4 {\sqrt{2}} . . . . \;\text{ (i) }\;$ $\;\text{ and }\; \vec{a} \cdot \vec{b}|=| \vec{a} \times \vec{b} \mid$ $|\vec{a}||\vec{b}| \cos \theta=|\vec{a}||\vec{b}| sin \;\theta$ $\Rightarrow \tan \theta=1$ $\therefore \theta=\;\frac{\pi}{4}$ By (i) $\vec{|b|}=8$ Now $\vec{c}=2 {\sqrt{2}} \vec{(a} \times \vec{b})-2 \vec{b}$ $\Rightarrow \vec{c} \cdot \vec{b}=-2|\vec{b}|^2=-128$ and $\vec{c} \cdot \vec{c}=8|\vec{a} \times \vec{b}|^{2+}4|\vec{b}|^2$ $\Rightarrow |c|^2=8.32+4.64$ $\Rightarrow |\vec{c}|=16 {\sqrt{2}}$ From (ii) and (iii) $|\vec{|c|}| \vec{b \mid } \cos \alpha=-128$ $\Rightarrow \cos \alpha=\;{-1}/{{\sqrt{2}}}$ $\alpha=\;\frac{3 \pi}{4}$

Q76JEE Main 2022MCQ4MStatistics

The mean and variance of the data $4,5,6,6,7,8, x, y$ , where $x< y$ , are 6 and $\;\frac{9}{4}$ respectively. Then $x^{4+}y^2$ is equal to [27-Jun-2022-Shift-2]

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📖 Explanation

$\;\text{ Mean }\;=\;\frac{4+5+6+6+7+8+x+y}{8}=6$ $\therefore x+y=12 . . . . \;\text{ (i) }\;$ And variance $=\;\frac{2^{2+}1^{2+}0^{2+}0^{2+}1^{2+}2^{2+}(x-6)^{2+}(y-6)^2}{8}$ $=\;\frac{9}{4}$ $\therefore (x-6)^{2+}(y-6)^2=8 . . . . \;\text{ (ii) }\;$ From (i) and (ii) $x=4$ and $y=8$ $\therefore x^{4+}y^2=320$

Q77JEE Main 2022MCQ4MProbability

If a point ${\text{A}}( {\text{x}}, {\text{y}})$ lies in the region bounded by the $y$ -axis, straight lines $2 y+x=6$ and $5 x-6 y=30$ , then the probability that $y< 1$ is [27-Jun-2022-Shift-2]

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📖 Explanation

The required probability $=\;\frac{\;\text{ Area of Region PQCAP }\;}{\;\text{ Area of Region ABCA }\;}$ $=\;\frac{\;\frac{1}{2} \times 8 \times 6-\frac{1}{2} \times 2 \times 4}{\frac{1}{2} \times 8 \times 6}$ $=\;\frac{5}{6}$

Q78JEE Main 2022MCQ4MInverse Trigonometric Functions

The value of $\cot (sum_{n=1}^{50} \tan ^{-1}(\;\frac{1}{1+n+n^2}))$ is [27-Jun-2022-Shift-2]

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📖 Explanation

$\cot (sum_{n=1}^{50} \tan ^{-1} (\;\frac{1}{1+n+n^2}) )$ $=\cot (sum_{n=1}^{50} \tan ^{-1} (\;\frac{(n+1)-n}{1+(n+1) n}) )$ $=\cot (sum_{n=1}^{50} (\tan ^{-1}(n+1)-\tan ^{-1} n) .$ $=\cot (\tan ^{-1} 51-\tan ^{-1} 1)$ $=\cot (\tan ^{-1} (\;\frac{51-1}{1+51}) )$ $=\cot (\cot ^{-1} (\;\frac{52}{50}) )$ $=\;\frac{26}{25}$

Q79JEE Main 2022MCQ4MTrigonometric Ratios and Identities

$\alpha=\sin 36^{\circ}$ is a root of which of the following equation? [27-Jun-2022-Shift-2]

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📖 Explanation

$\alpha= sin \;36^{\circ}=x$ (say) $\therefore x=\;{\sqrt{10-2 {\sqrt{5}}}}/{4}$ $\Rightarrow 16 x^2=10-2 {\sqrt{5}}$ $\Rightarrow (8 x^{2-}5) ^2=5$ $\Rightarrow 16 x^{4-}80 x^{2+}20=0$ $\therefore 4 x^{4-}20 x^{2+}5=0$

Q80JEE Main 2022MCQ4MSets and Relations

Which of the following statement is a tautology? [27-Jun-2022-Shift-2]

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📖 Explanation

$\because ((∼ q) ∧ p) ∨(p ∨(∼ p))$ $=(∼ q ∧ p) ∨ t(t \;\text{ is tautology }\;)$ $\equiv t$ $\therefore$ option (C) is correct.

Q81JEE Main 2022NAT4MFunctions

Let $S=\{1,2,3,4,5,6,7,8,9,10\}$ . Define $f: S \to S$ as $f(n)=\{\text{table} 2 n , \text{, if }\; n=\text{1,},\text{2,},\text{3,},\text{4,},\text{5} ; 2 n-11 , \text{, if }\; n=\text{6,} ,\text{7,},\text{8,},\text{9,},\text{10,}$ Let ${\text{g}}: {\text{S}} \to {\text{S}}$ be a function such that ${\text{fog}}(n)=\{\text{table} n+1 , \;\text{ , if n is odd }\; ; n-1 , \text{, if n is even }$ Then $g(10) g(1)+g(2)+g(3)+g(4)+g(5))$ is equal to [27-Jun-2022-Shift-2]

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📖 Explanation

$\because f(n)={\begin{cases}2 n & n=1 2 3 4 5 \\ 2 n-11 & n=6 7 8 9 10}\end{cases}$ $\therefore f(1)=2, f(2)=4, . . . . . ., f(5)=10$ $\;\text{ and }\; f(6)=1, f(7)=3, f(8)=5, . . . . . . ., f(10)=9$ Now, $f(g(n))={\begin{cases}n+1 & \\ \\ \text{ if }\\ \\ n \\ \\ \text{ is odd }\\ \\ \\ n-1 & \\ \\ \text{ if }\\ \\ n \\ \\ \text{ is even }\\ \\ }\end{cases}$ $f(g(10))=9 \;\; \Rightarrow g(10)=10$ $f(g(1))=2 \;\; \Rightarrow g(1)=1$ $\;\text{. }\; f(g(2))=1 \;\; \Rightarrow g(2)=6$ $\therefore f(g(3))=4 \;\; \Rightarrow g(3)=2$ $f(g(4))=3 \;\; \Rightarrow g(4)=7$ $f(g(5))=6 \;\; \Rightarrow g(5)=3$ $\therefore g(10) g(1)+g(2)+g(3)+g(4)+g(5))=190$ $$

Q82JEE Main 2022NAT4MQuadratic Equation and Inequalities

Let $\alpha, \beta$ be the roots of the equation $x^{2-}4 \lambda x+5=0$ and $\alpha, \gamma$ be the roots of the equation $x^{2-}(3 \sqrt{2}+2 \sqrt{3}) x+7+3 \lambda \sqrt{3}=0, \lambda >0$ . If $\beta+\gamma =3 \sqrt{2}$ , then $(\alpha+2 \beta+\gamma )^2$ is equal to___ [27-Jun-2022-Shift-2]

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📖 Explanation

$\because \alpha, \beta$ are roots of $x^{2-}4 \lambda x+5=0$ $\therefore \alpha+\beta=4 \lambda$ and $\alpha \beta=5$ Also, $\alpha, \gamma$ are roots of $x^{2-}(3 {\sqrt{2}}+2 {\sqrt{3}}) x+7+3 {\sqrt{3}} \lambda =0, \lambda >0$ $\therefore \alpha+\gamma =3 {\sqrt{2}}+2 {\sqrt{3}}, \alpha \gamma =7+3 {\sqrt{3}} \lambda$ $\because \alpha$ is common root $\therefore \alpha^{2-}4 \lambda \alpha+5=0$ and $\alpha^{2-}(3 {\sqrt{2}}+2 {\sqrt{3}}) \alpha+7+3 {\sqrt{3}} \lambda =0$ From (i) - (ii) : we get $\alpha=\;{2+3 {\sqrt{3}} \lambda }/{3 {\sqrt{2}}+2 {\sqrt{3}}-4 \lambda }$ $\because \beta+\gamma =3 {\sqrt{2}}$ $\therefore 4 \lambda +3 {\sqrt{2}}+2 {\sqrt{3}}-2 \alpha=3 {\sqrt{2}}$ $\Rightarrow 3 {\sqrt{2}}=4 \lambda +3 {\sqrt{2}}+2 {\sqrt{3}}-\;{4+6 {\sqrt{3}} \lambda }/{3 {\sqrt{2}}+2 {\sqrt{3}}-4 \lambda }$ $\Rightarrow 8 \lambda ^{2+}3({\sqrt{3}}+2 {\sqrt{2}}) \lambda -4-3 {\sqrt{6}}=0$ $\therefore \lambda =\;{6 {\sqrt{2}}-3 {\sqrt{2}} \pm \sqrt{9(11-4 {\sqrt{6}})+32(4+3 {\sqrt{6}})}}/{16}$ $\therefore \lambda ={\sqrt{2}}$ $\therefore (\alpha+2 \beta+\gamma )^2=(\alpha+\beta+\beta+\gamma )^2$ $=(4 {\sqrt{2}}+3 {\sqrt{2}})^2$ $=(7 {\sqrt{2}})^2=98$

Q83JEE Main 2022NAT4MPermutations and Combinations

Let $A$ be a matrix of order $2 \times 2$ , whose entries are from the set $\{0,1,2,3,4,5\}$ . If the sum of all the entries of $A$ is a prime number $p, 2< p< 8$ , then the number of such matrices $A$ is____ [27-Jun-2022-Shift-2]

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📖 Explanation

$\because$ Sum of all entries of matrix A must be prime $p$ such that $2< p< 8$ then sum of entries may be 3,5 or 7 .If sum is 3 then possible entries are $(0,0,0,3),(0,0,1,2)$ or $(0,1,1,1)$ . $\therefore$ Total number of matrices $=4+4+12=20$ If sum of 5 then possible entries are $(0,0,0,5),(0,0,1,4),(0,0,2,3),(0,1,1,3),(0,1,2,2)$ and $(1,1,1,2)$ . $\therefore$ Total number of matrices $=4+12+12+12+12+4=56$ If sum is 7 then possible entries are $(0,0,2,5),(0,0,3,4),(0,1,1,5),(0,3,3,1),(0,2,2,3),(1,1,1,4),(1,2,2,2),(1,1,2,3)$ and $(0,1,2,4)$ .Total number of matrices with sum $7=104$ $\therefore$ Total number of required matrices $=20+56+104$ = 180

Q84JEE Main 2022NAT4MBinomial Theorem

If the sum of the coefficients of all the positive powers of $x$ , in the Binomial expansion of $(x^n+\;\frac{2}{x^5})^7$ is 939 , then the sum of all the possible integral values of $n$ is____ [27-Jun-2022-Shift-2]

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📖 Explanation

Given, Binomial expression is $= (x^n+\;\frac{2}{x^5}) ^7$ $\therefore$ General term $T_{r+1}={ }^7 C_r \cdot (x^n) ^{7-r} \cdot (\;\frac{2}{x^5}) ^r$ $={ }^7 C_r \cdot x^{7 n-n r-5 r} \cdot 2^r$ For positive power of ${\text{x}}$ , $7 n-n r-5 r>0$ $\Rightarrow 7 n>r(n+5)$ $\Rightarrow r< \;\frac{7 n}{n+5}$ As $r$ represent term of binomial expression so $r$ is always integer.Given that sum of coefficient is 939. When $r=0$ ,sum of coefficient $={ }^7 C_0 \cdot 2^0=1$ when $r=1$ ,sum of coefficient $={ }^7 C_0 \cdot 2^{0+}{ }^7 C_1 \cdot 2^1=1+14=15$ when $r=2$ ,sum of coefficient $={ }^7 C_0 \cdot 2^{0+}{ }^7 C_1 \cdot 2^{1+}{ }^7 C_2 \cdot 2^2$ $=1+14+84$ $=99$ when $r=3$ ,sum of coefficient $={ }^7 C_0 \cdot 2^{0+}{ }^7 C_1 \cdot 2^{1+}{ }^7 C_2 \cdot 2^{2+}{ }^7 C_3 \cdot 2^3$ $=1+14+84+280$ $=379$ when $r=4$ ,sum of coefficient $={ }^7 C_0 \cdot 2^{0+}{ }^7 C_1 \cdot 2^{1+}{ }^7 C_2 \cdot 2^{2+}{ }^7 C_3 \cdot 2^{3+}{ }^7 C_4 \cdot 2^4$ $=1+14+84+280+560$ $=939$ To get value of $r=4$ , value of $\;\frac{7 n}{n+5}$ should be between 4 and 5 . $\therefore 4< \;\frac{7 n}{n+5}< 5$ $\Rightarrow 4 n+20< 7 n< 5 n+25$ $\therefore 4 n+20< 7 n$ $\Rightarrow 3 n>20$ $\Rightarrow n>\;\frac{20}{3}$ $\Rightarrow n>6.66$ and $7 n< 5 n+25$ $\Rightarrow 2 n< 25$ $\Rightarrow n< 12.5$ $\therefore 6.66< n< 12.5$ $\therefore$ Possible integer values of $n=7,8,9,10,11,12$ $\therefore$ Sum of values of $n=7+8+9+10+11+12$ $=57$

Q85JEE Main 2022NAT4MLimits, Continuity and Differentiability

Let $[ {\text{t}}]$ denote the greatest integer $\le {\text{t}}$ and $\{ {\text{t}}\}$ denote the fractional part of ${\text{t}}$ . The integral value of $\alpha$ for which the left hand limit of the function $f(x)=[1+x]+\;{\alpha^{2[x]+\{x\}}+[x]-1}/{2[x]+\{x\}}$ at ${\text{x}}=0$ is equal to $\alpha-\;\frac{4}{3}$ , is____ [27-Jun-2022-Shift-2]

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📖 Explanation

$f(x)=[1+x]+\;{a^{2[]]+\{x\}}+[x]-1}/{2[x]+\{x\}}$ $\lim _{x \to 0^{-}} f(x)=\alpha-\;\frac{4}{3}$ $\Rightarrow \lim _{x \to 0^{-}} 1+[x]+\;\frac{\alpha^{x+[x]}+[x]-1}{x+[x]}=\alpha-\;\frac{4}{3}$ $\Rightarrow \lim _{h \to 0^{-}} 1-1+\;\frac{\alpha^{-h-1}-1-1}{-h-1}=\alpha-\;\frac{4}{3}$ $\therefore \;\frac{\alpha^{-1}-2}{-1}=\alpha-\;\frac{4}{3}$ $\Rightarrow 3 \alpha^{2-}10 \alpha+3=0$ $\therefore \alpha=3$ or $\;\frac{1}{3}$ $\because \alpha$ in integer, hence $\alpha=3$

Q86JEE Main 2022NAT4MDifferentiation

If $y(x)=(x^x)^x, x>0$ , then $\;\frac{d^2 x}{d y^2}+20$ at $x=1$ is equal to____ [27-Jun-2022-Shift-2]

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📖 Explanation

$\because y(x)= (x^x) ^x$ $\therefore y=x^{x^2}$ $\therefore \;\frac{d y}{d x}=x^2 \cdot x^{x^{2-}1}+x^{x^2} \ln x \cdot 2 x$ $\therefore \;\frac{d x}{d y}=\;\frac{1}{x^{x^{2+}1}(1+2 \ln x)} ⋯ \cdot ( {\text{i}})$ Now, $\;\frac{d^2 x}{d x}=\;\frac{d}{d x} ( (x^{x^{2+}1}(1+2 \ln x)) ^{-1}) \cdot \;\frac{d x}{d y}$ $=\;\frac{-x (x^{x^{2+}1}(1+2 \ln x)) ^{-2} \cdot x^{x^2}(1+2 \ln x) (x^{2+}2 x^2 \ln x+3) }{x^{x^2}(1+2 \ln x)}$ $=\;\frac{-x^{x^2}(1+2 \ln x) (x^{3+}3+2 x^2 \ln x) }{ (x^{x^2}(1+2 \ln x)) ^3}$ $\;\frac{d^2 x}{d y^2(a t x=1)}=-4$ $\therefore \;\frac{d^2 x}{d y^2(a t x=1)}+20=16$

Q87JEE Main 2022NAT4MArea Under The Curves

If the area of the region $\{(x, y): x^{\;\frac{2}{3}}+y^{\;\frac{2}{3}} \le 1, x+y \ge 0, y \ge 0\}$ is ${\text{A}}$ , then $\;\frac{256 A}{\pi}$ is equal to_____ [27-Jun-2022-Shift-2]

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📖 Explanation

$\therefore$ Area of shaded region $=int_{{- (\;\frac{1}{2}) ^{\;\frac{3}{2}}}^0 ( (1-x^{\;\frac{2}{3}}) }^{\;\frac{3}{2}}+x) \text{dx}+int_{0^1 (1-x^{\;\frac{2}{3}}) }^{\;\frac{3}{2}} \text{dx}$ $=int_{{- (\;\frac{1}{2}) ^{\;\frac{3}{2}}}^0 (1-x^{\;\frac{2}{3}}) }^{\;\frac{3}{2}} \text{dx}+int_{{- (\;\frac{1}{2}) }^{\;\frac{3}{2}}}^0 x \text{dx}$ Let $x= sin \;^3 \theta$ $\therefore \text{dx}=3 sin \;^2 \theta \cos \theta d \theta$ $=int_{{-\;\frac{\pi}{4}}}^{\;\frac{\pi}{2}} 3 sin \;^2 \theta \cos ^4 \theta d \theta+ (0+\;\frac{1}{16})$ $=\;\frac{9 \pi}{64}+\;\frac{1}{16}-\;\frac{1}{16}=\;\frac{36 \pi}{256}=A$ $\therefore \;\frac{256 A}{\pi}=36$

Q88JEE Main 2022NAT4MDifferential Equations

Let $y=y(x)$ be the solution of the differential equation $(1-x^2) d y=(x y+(x^{3+}2) \sqrt{1-x^2}) d x,-1< x< 1$ , and $y(0)=0 .$ If $int_{\;\frac{-1}{2}}^{\;\frac{1}{2}} \sqrt{1-x^2} y(x) d x=k$ , then ${\text{k}}^{-1}$ is equal to____ [27-Jun-2022-Shift-2]

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📖 Explanation

$(1-x^2) d y= (x y+ (x^{3+}2) {\sqrt{1-x^2}}) \text{dx}$ $\therefore \;\frac{d y}{d x}-\;\frac{x}{1-x^2} y=\;{x^{3+}3}/{{\sqrt{1-x^2}}}$ $\therefore \;\text{ I. F. }\;=e^{\int-\;\frac{x}{1-x^2} \text{dx}}={\sqrt{1-x^2}}$ Solution isy. ${\sqrt{1-x^2}}=\int (x^{3+}3) \text{dx}$ $y \cdot {\sqrt{1-x^2}}=\;\frac{x^4}{4}+3 x+c$ $\because y(0)=0 \Rightarrow c=0$ $\therefore y(x)=\;{x^{4+}12 x}/{4 {\sqrt{1-x^2}}}$ $\therefore int_{{\;\frac{-1}{2}}}^{\;\frac{1}{2}} {\sqrt{1-x^2}} y(x) \text{dx}=int_{{\;\frac{-1}{2}}}^{\;\frac{1}{2}} (\;\frac{x^{4+}12 x}{4}) \text{dx}=int_{0}^{\;\frac{1}{2}} \;\frac{x^4}{2} \text{dx}$ $\therefore k=\;\frac{1}{320}$ $\therefore =k^{-1}=320$

Q89JEE Main 2022NAT4MCircles

Let a circle $C$ of radius 5 lie below the $x$ -axis. The line $L_1: 4 x+3 y+2=0$ passes through the centre $P$ of the circle $C$ and intersects the line $L_2=3 x-4 y-11=0$ at $Q$ . The line $L_2$ touches $C$ at the point $Q$ . Then the distance of $P$ from the line $5 x-$ $12 y+51=0$ is____ [27-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$L_1: 4 x+3 y+2=0$ $L_2: 3 x-4 y-11=0$ Since circle $C$ touches the line $L_2$ at $Q$ intersection point $Q$ of $L_1$ and $L_2$ , is $(1,-2)$ $\because P$ lies of $L_1$ $\therefore P (x,-\;\frac{1}{3}(2+4 x))$ Now, $P Q=5 \Rightarrow (x-1)^{2+} (\;\frac{4 x+2}{3}-2) ^2=25$ $\Rightarrow (x-1)^2 [1+\;\frac{16}{9}] =25$ $\Rightarrow (x-1)^2=9$ $\Rightarrow x=4,-2$ $\because$ Circle lies below the ${\text{x}}$ -axis $\therefore y=-6$ $P(4,-6)$ Now distance of $P$ from $5 x-12 y+51=0$ $= |\;\frac{20+72+51}{13}{{}/{}|} =\;\frac{143}{13}=11$

Q90JEE Main 2022NAT4MProbability

Let $S=\{E_1, E_2, . . . . . . . . ., E_8\}$ be a sample space of a random experiment such that $P(E_n)=\;\frac{n}{36}$ for every $n=1,2, . . . 8$ . Then the number of elements in the set $\{A ⊆ S: P(A) \ge \;\frac{4}{5}\}$ is___ [27-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Here $P (E_n) =\;\frac{n}{36}$ for ${\text{n}}=1,2,3, . . . . . ., 8$ Here $P(A)=\;\frac{\;\text{ Any possible \sum of }\;(1,2,3, . . ., 8)(=a \;\text{ say }\;)}{36}$ $\because \;\frac{a}{36} \ge \;\frac{4}{5}$ $\therefore a \ge 29$ If one of the number from $\{1,2, . . . . . ., 8\}$ is left then total $a \ge 29$ by 3 ways.Similarly by leaving terms more 2 or 3 we get 16 more combinations. $\therefore$ Total number of different set A possible is $16+3=19$

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