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26 Jun 2022 Shift 1

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Physics30 questions

Q1JEE Main 2022MCQ4MUnits and Measurements

An expression for a dimensionless quantity ${\text{P}}$ is given by $P=\;\frac{\alpha}{\beta} \log _e(\;\frac{k t}{\beta x})$ ; where $\alpha$ and $\beta$ are constants, ${\text{x}}$ is distance; ${\text{k}}$ is Boltzmann constant and ${\text{t}}$ is the temperature. Then the dimensions of $\alpha$ will be : [26-Jun-2022-Shift-1]

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📖 Explanation

${[P]=\;\frac{[\alpha]}{[\beta]} \Rightarrow 1 \;\frac{[\alpha]}{[\beta]} }$ $\Rightarrow [\alpha]=[\beta] . . . . . .(1)$ Now, $[\;\frac{k t}{\beta t}]=1$ We know, E $=\;\frac{3}{2} k T(K T G)$ $\therefore [$ Energy] $=[ {\text{KT}}]$ $\therefore [\beta x]=[k T]$ ${[\beta]=\;\frac{M L^2 T-2}{[x]}=\;\frac{M L^2 T^{-2}}{L} }$ $\therefore [\alpha]=[\beta]=M L T{ }^{-2}$

Q2JEE Main 2022MCQ4MLaws of Motion

A person is standing in an elevator. In which situation, he experiences weight loss? [26-Jun-2022-Shift-1]

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📖 Explanation

${\text{N}}_1= {\text{mg}}$ ${\text{N}}_2= {\text{mg}}+ {\text{ma}}$ ${\text{N}}_3= {\text{mg}}- {\text{ma}}$ ${\text{N}}_4= {\text{mg}}$ ${\text{N}}_5= {\text{mg}}$

Q3JEE Main 2022MCQ4MMotion in a Straight Line

An object is thrown vertically upwards. At its maximum height, which of the following quantity becomes zero? [26-Jun-2022-Shift-1]

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Q4JEE Main 2022MCQ4MCircular Motion

A ball is released from rest from point ${\text{P}}$ of a smooth semi-spherical vessel as shown in figure. The ratio of the centripetal force and normal reaction on the ball at point ${\text{Q}}$ is ${\text{A}}$ while angular position of point ${\text{Q}}$ is $\alpha$ with respect to point ${\text{P}}$ . Which of the following graphs represent the correct relation between ${\text{A}}$ and $\alpha$ when ball goes from ${\text{Q}}$ to ${\text{R}}$ ? [26-Jun-2022-Shift-1]

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📖 Explanation

$h=R \sin \alpha$ $v=\sqrt{2 g h}=\sqrt{2 g R \sin \alpha}$ $a_c=\;\frac{v^2}{R}=\;\frac{2 g R \sin \alpha}{R}=2 g \sin \alpha$ Now, at Q; $N-m g \sin \alpha=m a_c$ $N=m \times 2 g \sin \alpha+m g \sin \alpha$ $=3 m g \sin \alpha$ $\;\frac{M a_c}{N}=\;\frac{2 m g \sin \alpha}{3 m g \sin \alpha}=\;\frac{2}{3}= {\text{constant}}$

Q5JEE Main 2022MCQ4MRotational Motion

A thin circular ring of mass $M$ and radius ${\text{R}}$ is rotating with a constant angular velocity 2 rads ${ }^{-1}$ in a horizontal plane about an axis vertical to its plane and passing through the center of the ring. If two objects each of mass $m$ be attached gently to the opposite ends of a diameter of ring, the ring will then rotate with an angular velocity (in rads ${ }^{-1}$ ). [26-Jun-2022-Shift-1]

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📖 Explanation

$I_1 \omega _1=I_2 \omega _2$ $M R^2 \omega _1=(M R^{2+}2 m R^2) \omega _2$ $\omega _2=(\;\frac{M}{M+2 m}) \omega _1$ $\omega _2=2(\;\frac{M}{M+2 m})$

Q6JEE Main 2022MCQ4MGravitation

The variation of acceleration due to gravity $( {\text{g}})$ with distance ({\text{r}}) $from the center of the earth is correctly represented by :(Given$ {\text{R}}=$ radius of earth) [26-Jun-2022-Shift-1]

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📖 Explanation

For $r< R g=\;\frac{G m r}{R^3}={\text{Cr}}(C=$ Constant $)$ For $r>R g=\;\frac{G m}{r^2}=\;\frac{C^{'}}{r^2}(C^{'}=.$ Constant $)$ For the above equations the best suited graph is as given in option $(A)$

Q7JEE Main 2022MCQ4MThermodynamics

The efficiency of a Carnot's engine, working between steam point and ice point, will be : [26-Jun-2022-Shift-1]

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📖 Explanation

$η=1-\;\frac{T_2}{T_1}=1-\;\frac{273}{373}$ $=0.26809$ $\approx 26.81 \%$

Q8JEE Main 2022MCQ4MOscillations

Time period of a simple pendulum in a stationary lift is ' $T$ '. If the lift accelerates with $\;\frac{g}{6}$ vertically upwards then the time period will be :(Where $g=$ acceleration due to gravity) [26-Jun-2022-Shift-1]

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📖 Explanation

$T^{'}=2 \pi \sqrt{\;{I}/{g_{\;\text{eff }\;}}}$ $T^{'}=2 \pi \sqrt{\;\frac{I}{g+\;\frac{g}{6}}}=2 \pi \sqrt{\;\frac{6 l}{7 g}}$ $\Rightarrow T^{'}=\sqrt{\;\frac{6}{7}} T$

Q9JEE Main 2022MCQ4MThermodynamics

A thermally insulated vessel contains an ideal gas of molecular mass ${\text{M}}$ and ratio of specific heats $1.4$ . Vessel is moving with speed $v$ and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by : $( {\text{R}}=$ universal gas constant) [26-Jun-2022-Shift-1]

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📖 Explanation

Let there be ${\text{n}}$ moles of gas ${\text{E}}_{\;\text{loss }\;}= {\text{E}}_{ {\text{gain}}}$ $\;\frac{1}{2}(n M) v^2=n C_v \Delta T$ $\;\frac{1}{2} M v^2=C_v \Delta T$ here, $\gamma =1.4=\;\frac{7}{5}$ i.e. diatomic gas $\therefore C_v=\;\frac{5 R}{2}$ Now, $\;\frac{1}{2} M v^2=\;\frac{5 R}{2} \Delta T$ $\Delta T=\;\frac{M v^2}{5 R}$

Q10JEE Main 2022MCQ4MElectrostatic Potential and Capacitance

Two capacitors having capacitance ${\text{C}}_1$ and ${\text{C}}_2$ respectively are connected as shown in figure. Initially, capacitor ${\text{C}}_1$ is charged to a potential difference ${\text{V}}$ volt by a battery. The battery is then removed and the charged capacitor ${\text{C}}_1$ is now connected to uncharged capacitor ${\text{C}}_2$ by closing the switch ${\text{S}}$ . The amount of charge on the capacitor ${\text{C}}_2$ , after equilibrium, is : [26-Jun-2022-Shift-1]

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📖 Explanation

$V_{\;\text{common }\;}=\;\frac{C_1 V}{C_1+C_2}$ $\Rightarrow$ Charge on capacitor $C_2$

Q11JEE Main 2022MCQ4MElectrostatics

Given below two statements : One is labelled as Assertion (A) and other is labelled as Reason (R).Assertion (A) : Non-polar materials do not have any permanent dipole moment.Reason (R): When a non-polar material is placed in an electric field, the centre of the positive charge distribution of it's individual atom or molecule coincides with the centre of the negative charge distribution.In the light of above statements, choose the most appropriate answer from the options given below. [26-Jun-2022-Shift-1]

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Q12JEE Main 2022MCQ4MElectromagnetic Induction

The magnetic flux through a coil perpendicular to its plane is varying according to the relation $\varphi =(5 t^{3+}4 t^{2+}2 t-5)$ Weber. If the resistance of the coil is 5 ohm, then the induced current through the coil at $t=2 {\text{s}}$ will be, [26-Jun-2022-Shift-1]

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📖 Explanation

$\varphi =5 {\text{t}}^{3+}4 {\text{t}}^{2+}2 {\text{t}}-5$ $| {\text{e}}|=\;{ {\text{d}} \varphi }/{ {\text{dt}}}=15 {\text{t}}^{2+}8 {\text{t}}+2$ ${\text{At}} {\text{t}}=2,| {\text{e}}|=15 \times 2^{2+}8 \times 2+2$ $\Rightarrow {\text{e}}=78 {\text{V}}$ $\Rightarrow {\text{I}}=\;{ {\text{e}}}/{ {\text{R}}}=\;\frac{78}{5}=15.60$

Q13JEE Main 2022MCQ4MCurrent Electricity

An aluminium wire is stretched to make its length, $0.4 \%$ larger. The percentage change in resistance is : [26-Jun-2022-Shift-1]

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📖 Explanation

$R=\;\frac{\rho l}{A}$ Also volume will remain constanti.e., ${\text{Al}}=$ constant $\Rightarrow A \propto \;\frac{1}{l}$ $\therefore R \propto l^2$ $\;\frac{\Delta R}{R}=2 \;\frac{\Delta l}{l}=0.8$

Q14JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

A proton and an alpha particle of the same velocity enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and proton is : [26-Jun-2022-Shift-1]

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📖 Explanation

$R=\;\frac{m v}{q B}$ $\;{ {\text{R}}_alpha}/{ {\text{R}}_{ {\text{P}}}}=\;{ {\text{M}}_alpha}/{ {\text{M}}_{ {\text{P}}}} \times \;{ {\text{q}}_{ {\text{P}}}}/{ {\text{q}}_alpha}$ $\;{ {\text{R}}_alpha}/{ {\text{R}}_{ {\text{P}}}}=\;\frac{4}{1} \times \;\frac{1}{2}=2$

Q15JEE Main 2022MCQ4MElectromagnetic Waves

If Electric field intensity of a uniform plane electromagnetic wave is given as $E=-301.6 \sin (k z-\omega t) \hat{a}_x+452.4 \sin (k z-\omega t) \hat{a}_y \;\frac{V}{m}$ . Then magnetic intensity ' ${\text{H}}^{'}$ of this wave in ${\text{Am}}^{-1}$ will be:[Given : Speed of light in vacuum $c=3 \times 10^8 {\text{ms}}^{-1}$ , Permeability of vacuum $.\mu _0=4 \pi \times 10^{-7} {\text{NA}}^{-2}]$ [26-Jun-2022-Shift-1]

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📖 Explanation

$\vec{E}=-301.6 \sin (k z-\omega t) \hat{a}_x+452.4 \sin (k z-\omega t) \hat{a}_y$ $E_{0 x}=301.6$ $E_{0 y}=+452.4$ $E_0=\sqrt{E_{0 x}^{2+}E_{0 y}^2}$ Now, $\;\frac{E_0}{B_0}=C \Rightarrow B_0=\;\frac{E_0}{C}=\;{\sqrt{E_{0 x}^{2+}E_{0 y}^2}}/{C}$ Also, $\hat{B}=\hat{C} \times \hat{E} \Rightarrow \hat{k} \times \;{(E_{0 x} \hat{i}+E_{0 y} \hat{j})}/{\sqrt{E_{0 x}^{2+}E_{0 y}^2}}$ $\hat{B}=\;{-E_{0 x} \hat{j}-E_{0 y} \hat{i}}/{\sqrt{E_{0 x}^{2+}E_{0 y}^2}}$ $\vec{B}=-\;\frac{E_{0 x}}{C} \sin (k z-\omega t) \hat{a}_y-\;\frac{E_{0 y}}{C} \sin (k z-\omega t) \hat{a}_x$ $\vec{H}=\;\frac{\vec{B}}{\mu _0}$ $\vec{H}=-\;\frac{E_{0 x}}{\mu _0 C} \sin (k z-\omega t) \hat{a}_y-\;\frac{E_{0 y}}{\mu _0 C} \sin (k z-\omega t) \hat{a}_x$

Q16JEE Main 2022MCQ4MWave Optics

In free space, an electromagnetic wave of $3 {\text{GHz}}$ frequency strikes over the edge of an object of size $\;\frac{\lambda }{100}$ , where $\lambda$ is the wavelength of the wave in free space. The phenomenon, which happens there will be : [26-Jun-2022-Shift-1]

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📖 Explanation

$\;{ {\text{a}}}/{\lambda }=\;\frac{1}{100}$ For reflection size of obstacle must be much larger than wavelength, for diffraction size should be order of wavelength.Since the object is of size $\;\frac{\lambda }{100}$ , much smaller than wavelength, so scattering will occur.

Q17JEE Main 2022MCQ4MDual Nature of Matter and Radiation

An electron with speed $v$ and a photon with speed $c$ have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are $E_e$ and $p_e$ and that of photon are $E_{p h}$ and $p_{p h}$ respectively. Which of the following is correct? [26-Jun-2022-Shift-1]

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📖 Explanation

$\lambda =\;\frac{h}{p} \Rightarrow p=\;\frac{h}{\lambda }$ Now, A/Q, $\;\frac{h}{P_e}=\;{h}/{P_{\;\text{photon }\;}}$ $\Rightarrow P_e=P_{\;\text{photon }\;} . . . . . . .$ . (i)Now, $K_e=\;\frac{1}{2} M v^2=\;\frac{P v}{2}$ $K_{p h}=m c^2=P c . . .$ . (ii) $\;\frac{K_e}{K_{e q}}=\;\frac{v}{2 c}$

Q18JEE Main 2022MCQ4MAtoms and Nuclei

How many alpha and beta particles are emitted when Uranium ${ }_{92} {\text{U}}^{238}$ decays to lead ${ }_{82} {\text{Pb}}^{206}$ ? [26-Jun-2022-Shift-1]

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📖 Explanation

${ }_{92} U^{238} \longrightarrow { }_{82} {\text{Pb}}^{206}+x({ }_2 H e^4)+{ }_y(-1 \beta^0)$ $238=206+4 x+0$ $\Rightarrow 4 x=32 \Rightarrow x=8$ also, $92=82+2 x-y$ $y=82+16-92=6$

Q19JEE Main 2022MCQ4MSemiconductor Electronics

The I-V characteristics of a p-n junction diode in forward bias is shown in the figure. The ratio of dynamic resistance, corresponding to forward bias voltage of $2 {\text{V}}$ and $4 {\text{V}}$ respectively, is : [26-Jun-2022-Shift-1]

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📖 Explanation

$\therefore R=\;\frac{\Delta v}{\Delta i}$ Now, dynamic resistance at $V=2 V$ is $R_2=\;\frac{0.1}{5 \times 10^{-3}} Ω$ $=20 Ω$ Similarly, $R_4=\;\frac{0.2}{50 \times 10^{-3}}=4 Ω$ $\;\frac{R_2}{R_4}=\;\frac{5}{1}$

Q20JEE Main 2022MCQ4MElectromagnetic Waves

Choose the correct statement for amplitude modulation : [26-Jun-2022-Shift-1]

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Q21JEE Main 2022NAT4MMotion in a Plane

A fighter jet is flying horizontally at a certain altitude with a speed of $200 {\text{ms}}^{-1}$ . When it passes directly overhead an anti-aircraft gun, a bullet is fired from the gun, at an angle $\theta$ with the horizontal, to hit the jet. If the bullet speed is $400 {\text{m}} / {\text{s}}$ , the value of $\theta$ will be____________° [26-Jun-2022-Shift-1]

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📖 Explanation

To hit the jet both should have same horizontal component of velocity.To hit the jet $400 \cos \theta=200$ $\Rightarrow \cos \theta=\;\frac{1}{2}$ $\Rightarrow \theta=60^{\circ}$

Q22JEE Main 2022NAT4MMotion in a Straight Line

A ball of mass $0.5 {\text{kg}}$ is dropped from the height of $10 {\text{m}}$ . The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ____m. $[$ Use $g=10$ $. {\text{m}} / {\text{s}}^2]$ [26-Jun-2022-Shift-1]

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📖 Explanation

Let at ${\text{h}}$ height its velocity becomes $10 {\text{m}} / {\text{s}}$ (as given in question) $\Rightarrow v^{2-}u^2=2(-10) \times (-h)$ $\Rightarrow v^2=20 h$ $\Rightarrow 100=20 h$ $\Rightarrow h=5 m$ $h^{'}=10-5=5 m$

Q23JEE Main 2022NAT4MProperties of Matter

The elastic behaviour of material for linear stress and linear strain, is shown in the figure. The energy density for a linear strain of $5 \times 10^{-4}$ is_____ ${\text{kJ}} / {\text{m}}^3$ . Assume that material is elastic upto the linear strain of $5 \times 10^{-4}$ . [26-Jun-2022-Shift-1]

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📖 Explanation

slope of strain - stress curve given by $=\;\frac{10^{-10}}{20}$ for strain of $5 \times 10^{-4}$ stress is given by $5 \times 10^{-4}=\;\frac{10^{-10}}{20} \times$ stressstress $=10^8 {\text{N}} / {\text{m}}^2$ Energy density $=\;\frac{1}{2} \times$ strain $\times$ stress $=\;\frac{1}{2} \times 5 \times 10^{-4} \times 10^8$ $=25000 {\text{J}} / {\text{m}}^3$ $=25 {\text{kJ}} / {\text{m}}^3$

Q24JEE Main 2022NAT4MProperties of Matter

The elongation of a wire on the surface of the earth is $10^{-4} {\text{m}}$ . The same wire of same dimensions is elongated by $6 \times 10^{-5} {\text{m}}$ on another planet. The acceleration due to gravity on the planet will be ___ ${\text{ms}}^{-2}$ . (Take acceleration due to gravity on the surface of earth $=10 {\text{ms}}^{-2}$ ) [26-Jun-2022-Shift-1]

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📖 Explanation

on earth, $\Delta l=10^{-4} {\text{m}}$ on other planet $\Delta l^{'}=6 \times 10^{-5} {\text{m}}$ $\Delta l=\;\frac{F l}{A y} \Rightarrow \;\frac{\Delta l^{'}}{\Delta l}=\;{\;\frac{F^{'} l}{A y}}/{\frac{F l}{A y}}=\;\frac{m g^{'}}{m g}$ $\Rightarrow g^{'}=\;\frac{\Delta l^{'}}{\Delta l} \times g$ $=\;\frac{6 \times 10^{-5}}{10^{-4}} \times 10$ $\Rightarrow g^{'}=6 {\text{m}} / {\text{s}}^2$

Q25JEE Main 2022NAT4MElectromagnetic Induction

A $10 Ω, 20 {\text{mH}}$ coil carrying constant current is connected to a battery of $20 {\text{V}}$ through a switch. Now after switch is opened current becomes zero in $100 \mu$ s. The average e.m.f. induced in the coil is ______V. [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Initially current, $I_0=\;\frac{20}{10}=2 A$ (when initially switch closed) average emf induced in coil $=\;\frac{L d i}{d t}$ $\Rightarrow e_{a v g}=\;\frac{20 \times 10^{-3} \times (2-0)}{100 \times 10^{-6}}$ $e_{a v g}=400 {\text{V}}$

Q26JEE Main 2022NAT4MRay Optics and Optical Instruments

A light ray is incident, at an incident angle $\theta_1$ , on the system of tow plane mirrors $M_1$ and $M_2$ having an inclination angle $75^{\circ}$ between them (as shown in figure). After reflecting from mirror $M_1$ it gets reflected back by the mirror $M_2$ with an angle of reflection $30^{\circ}$ . The total deviation of the ray will be_____ degree. [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

On first reflection angel of deviation is $90^{\circ}$ and on second reflection angle of deviation is $120^{\circ}$ So total deviation is $\delta =90^{\circ}+120^{\circ}=210^{\circ}$

Q27JEE Main 2022NAT4MUnits and Measurements

In a vernier callipers, each ${\text{cm}}$ on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be _____ $\times 10^{-2} {\text{mm}}$ [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$1 {\text{MSD}}=\;\frac{1}{20} {\text{cm}}$ $\because 10 {\text{VSD}}=9 {\text{MSD}}$ $1 {\text{VSD}}=\;\frac{9}{10} \times \;\frac{1}{20} {\text{cm}}=\;\frac{9}{200} \times 10 {\text{mm}}=0.45 {\text{mm}}$ Now, $1 {\text{MSD}}=\;\frac{1}{20} \times 10 {\text{mm}}=0.50 {\text{mm}}$ ${\text{LC}}=(0.50-0.45) {\text{mm}}=0.05 {\text{mm}}$ $=5 \times 10^{-2} {\text{mm}}$

Q28JEE Main 2022NAT4MSemiconductor Electronics

As per the given circuit, the value of current through the battery will be____ A. [26-Jun-2022-Shift-1]

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📖 Explanation

${\text{R}}_{ {\text{eq}}}=16 \| 16+2=(8+2) Ω$ $=10 Ω$ ${\text{I}}_0=10 / 10$ $=1 {\text{A}}$

Q29JEE Main 2022NAT4MAlternating Current

A $110 {\text{V}}, 50 {\text{Hz}}, {\text{AC}}$ source is connected in the circuit (as shown in figure). The current through the resistance $55 Ω$ , at resonance in the circuit, will be _____A. [26-Jun-2022-Shift-1]

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📖 Explanation

$\;\frac{1}{Z}=\sqrt{(\;\frac{1}{X_L}-\;\frac{1}{X_C})^2}$ At resonance, ${\text{X}}_{ {\text{L}}}= {\text{X}}_{ {\text{C}}} \& {\text{Z}} \to \infty$ $\therefore {\text{Z}}_{\;\text{total circuit }\;} \to \infty$ i.e, ${\text{I}}=0$

Q30JEE Main 2022NAT4MProperties of Matter

An ideal fluid of density $800 {\text{kgm}}^{-3}$ , flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to $\;\frac{a}{2}$ . The pressure difference between the wide and narrow sections of pipe is $4100 {\text{Pa}}$ . At wider section, the velocity of fluid is $\;\frac{\sqrt{x}}{6} {\text{ms}}^{-1}$ for ${\text{x}}=$ (Given ${\text{g}}=10 {\text{ms}}^{-2}$ ) [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

From Bernoulli's equation $P_1+\;\frac{1}{2} \rho v_1^{2+}\rho g h_1=P_2+\;\frac{1}{2} \rho v_2^{2+}\rho g h_2$ $P_1-P_2+\rho g(h_1-h_2)=\;\frac{1}{2} \rho (v_2^{2-}v_1^2) . . . . . . \;\text{ (i) }\;$ Also, from equation of continuity $A_1 v_1=A_2 v_2$ $A v_1=\;\frac{A}{2} v_2$ $v_2=2 v_1 . . . . . . \;\text{ (ii) }\;$ $\;\text{ put equation (ii) \in (i), }\;$ $4100 \times 800 \times 10 \times 1=\;\frac{1}{2} \times 800 \times (4 v_1^{2-}v_1^2)$ $4100+8000=400 \times 3 v_1^2$ $v_1^2=\;\frac{12100}{3 \times 400}=\;\frac{121}{12}$ $v_1=\sqrt{\;\frac{121}{12}}$ Now, $\;\frac{\sqrt{x}}{6}=\sqrt{\;\frac{121}{12}}$ $\;\frac{x}{36}=\;\frac{121}{12}$ $x=121 \times 3=363$ $\therefore x=363$

Chemistry30 questions

Q31JEE Main 2022MCQ4MSolutions

A commercially sold conc. ${\text{HCl}}$ is $35 \% {\text{HCl}}$ by mass. If the density of this commercial acid is $1.46 {\text{g}} / {\text{mL}}$ , the molarity of this solution is:(Atomic mass : ${\text{Cl}}=35.5 {\text{amu}}, {\text{H}}=1 {\text{amu}}$ ) [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

35% ${\text{HCl}}$ by mass means in $100 {\text{gm}} {\text{HCl}}$ solution $35 {\text{gm}} {\text{HCl}}$ present.Now, volume of $100 {\text{gm}} {\text{HCl}}$ solution $=\;\frac{100}{1.46} {\text{ml}}$ $=\;\frac{\;\frac{100}{1.46}}{1000} {\text{I}}$ Moles of ${\text{HCl}}=\;\frac{35}{36.5}$ Now, molarity $=\;\frac{\;\text{ moles of solute }\;}{\;\text{ volume of solution }\;(\;\text{ \in }\; L)}$ $=\;\frac{\;\frac{35}{36.5}}{\frac{100}{1.46}}=14$

Q32JEE Main 2022MCQ4MStates of Matter

An evacuated glass vessel weighs $40.0 {\text{g}}$ when empty, $135.0 {\text{g}}$ when filled with a liquid of density $0.95 {\text{g}} {\text{mL}}-1$ and $40.5 {\text{g}}$ when filled with an ideal gas at $0.82$ atm at $250 {\text{K}}$ . The molar mass of the gas in ${\text{g}} {\text{mol}}^{-1}$ is :(Given : ${\text{R}}=0.082 {\text{L}} {\text{atm}} {\text{K}}^{-1} {\text{mol}}^{-1}$ ) [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Weight of empty glass vessel $=40 {\text{gm}}$ Weight of glass vessel filled with liquid $=135 {\text{gm}}$ $\therefore$ Weight of liquid $=135-40=95 {\text{gm}}$ Given density of liquid $=0.95 {\text{gm}} {\text{ml}}^{-1}$ $\therefore$ Volume of liquid $=\;\frac{95}{0.95}=100 {\text{ml}}$ Weight of glass filled with ideal gas $=40.5 {\text{gm}}$ $\therefore$ Weight of gas $=40.5-40=0.5 {\text{g}}$ Let the Molar mass $= {\text{M}}$ $\therefore$ Moles of gas $=\;\frac{0.5}{M}$ $\therefore$ Now applying ideal gas equation, ${\text{pV}}= {\text{nRT}}$ $\Rightarrow 0.82 \times \;\frac{100}{1000}=\;\frac{0.5}{M} \times 0.082 \times 250$ $\Rightarrow M=0.5 \times 250=125 {\text{g}} / {\text{mol}}$

Q33JEE Main 2022MCQ4MStructure of Atom

If the radius of the $3^{\;\text{rd }\;}$ Bohr's orbit of hydrogen atom is $r_3$ and the radius of $4^{\;\text{th }\;}$ Bohr's orbit is $r_4$ . Then : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

We know, $r=r_0 \times \;\frac{n^2}{z}$ For hydrogen atom, $\therefore r_3=r_0 \times \;\frac{3^2}{1}$ $\Rightarrow r_0 \times \;\frac{r_3}{9}$ $\;\text{ and }\; r_4=r_0 \times \;\frac{4^2}{1}$ $=\;\frac{r_3}{9} \times 16$ $=\;\frac{16}{9} r_3$

Q34JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

Consider the ions/molecule ${\text{O}}_2^{+}, {\text{O}}_2, {\text{O}}_2^{-}, {\text{O}}_2^{2-}$ For increasing bond order the correct option is : [26-Jun-2022-Shift-1]

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📖 Explanation

(1) Bond strength $\propto$ Bond order(2) Bond length $\propto \;\frac{1}{\;\text{ Bond order }\;}$ (3) Bond order $=\;\frac{1}{2}[ {\text{N}}_{ {\text{b}}}- {\text{N}}_{ {\text{a}}}]$ ${\text{N}}_{ {\text{b}}}=$ No of electrons in bonding molecular orbital $N_a=$ No of electrons in anti bonding molecular orbital(4) upto 14 electrons, molecular orbital configuration isHere ${\text{N}}_{ {\text{a}}}=$ Anti bonding electron $=4$ and ${\text{N}}_{ {\text{b}}}=10$ (5) After 14 electrons to 20 electrons molecular orbital configuration is - - -Here $N_a=10$ and ${\text{N}}_b=10$ In ${\text{O}}$ atom 8 electrons present, so in ${\text{O}}_2, 8 \times 2=16$ electrons present.Then in $O_2^{+}$ no of electrons $=15$ in $O_2^{-}$ no of electrons $=17$ in $O_2^{2-}$ no of electrons $=18$ $\therefore$ Molecular orbital configuration of ${\text{O}}_2$ (16 electrons) is $\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \sigma_{2 p_z^2} \pi_{2 p_x^2}=\pi_{2 p_y^2} \pi_{2 p_x^1}^*=\pi_{2 p_y^1}^*$ $\therefore {\text{N}}_{ {\text{a}}}=6$ $N_b=10$ $\therefore \;\; B O=\;\frac{1}{2}[10-6]=2$ Molecular orbital configuration of ${\text{O}}_2^{+}$ ( 15 electrons) is $\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \sigma_{2 p_z^2} \pi_{2 p_x^2}=\pi_{2 p_y^2} \pi_{2 p_x^1}^*=\pi_{2 p_y^o}^*$ $\therefore \;\; {\text{N}}_{ {\text{b}}}=10$ ${\text{N}}_{ {\text{a}}}=5$ $\therefore \;\; {\text{BO}}=\;\frac{1}{2}[10-5]=2.5$ Molecular orbital configuration of $O_2^{-}$ (17 electrons) is $\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \sigma_{2 p_z^2} \pi_{2 p_x^2}=\pi_{2 p_y^2} \pi_{2 p_x^2}^*=\pi_{2 p_y^1}^*$ $\therefore \;\; {\text{N}}_{ {\text{b}}}=10$ $N_a=7$ $\therefore \;\; B O=\;\frac{1}{2}[10-7]=1.5$ Molecular orbital configuration of ${\text{O}}_2^{2-}$ (18 electrons) is $\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \sigma_{2 p_z^2} \pi_{2 p_x^2}=\pi_{2 p_y^2} \pi_{2 p_x^2}^*=\pi_{2 p_y^2}^*$ $\therefore \;\; {\text{N}}_{ {\text{b}}}=10$ ${\text{N}}_{ {\text{a}}}=8$ $\therefore \;\; {\text{BO}}=\;\frac{1}{2}[10-8]=1$ So, correct order of Bond order is $O_2^{2-}< O_2^{-}< O_2< O_2^{+}$

Q35JEE Main 2022MCQ4MElectrochemistry

The $(\;\frac{∂ E}{∂ T})_P$ of different types of half cells are as follows: A B C D $1\times 10^{-4}$ $2\times 10^{-4}$ $0.1 \times 10^{-4}$ $0.2 \times 10^{-4}$ (Where $E$ is the electromotive force)Which of the above half cells would be preferred to be used as reference electrode? [26-Jun-2022-Shift-1]

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📖 Explanation

A cell with less variation in EMF with temperature is preferred as a reference electrode because it can be used for a wider range of temperatures without much derivation from standard value so a cell with less $(\;\frac{∂ E}{∂ T})_P$ is preferred.

Q36JEE Main 2022MCQ4Mp Block Elements

Choose the correct stability order of group 13 elements in their $+1$ oxidation state. [26-Jun-2022-Shift-1]

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📖 Explanation

Due to inert pair effect, stability of $+3$ oxidation state decreases and that of $+1$ oxidation state increases for (down the group) group 13 elements.So, the correct order of stability of group 13 elements in their $+1$ oxidation state is ${\text{Al}}< {\text{Ga}}< {\text{In}}< {\text{TI}}$ .

Q37JEE Main 2022MCQ4MRedox Reactions

Given below are two statements :Statement I : According to Ellingham diagram, any metal oxide with higher $\Delta {\text{G}}^{\circ}$ is more stable than the one with lower $\Delta {\text{G}}^{\circ}$ .Statement II : The metal involved in the formation of oxide placed lower in the Ellingham diagram can reduce the oxide of a metal placed higher in the diagram.In the light of the above statements, choose the most appropriate answer from the options given below : [26-Jun-2022-Shift-1]

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📖 Explanation

Ellingham diagram is plot of $\Delta {\text{G}} v {\text{v}} T$ . The criterion for the feasibility of a thermal reduction is that at a given temperature Gibbs energy change of a reaction must be negative. The change in Gibbs energy, $\Delta {\text{G}}$ for any process at any specified temperature is given by the equation $\Delta {\text{G}}=\Delta {\text{H}}- {\text{T}} \Delta {\text{S}}$ where $\Delta {\text{H}}=$ enthalpy change and $\Delta {\text{S}}=$ entropy changeAccording to the ellingham diagram, any metal oxide with higher $\Delta G^{\circ}$ has a tendency of getting reduced by the metal whose metal oxide has lower value of $\Delta {\text{G}}^{\circ}$ .

Q38JEE Main 2022MCQ4MElectrochemistry

Consider the following reaction: $2 {\text{HSO}}_4^{-}(a q) { \to } ^{(1) \text{Electrolysis}}_{(2) \;\text{ Hydrolysis} } 2 {\text{HSO}}_4^{-}+2 {\text{H}}^{+}+ {\text{A}}$ The dihedral angle in product ${\text{A}}$ in its solid phase at $110 {\text{K}}$ is : [26-Jun-2022-Shift-1]

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📖 Explanation

$2 {\text{HSO}}_4^{-}(a q) { \to } ^{(1) \text{Electrolysis}}_{(2) \;\text{ Hydrolysis} } 2 {\text{HSO}}_4^{-}+2 {\text{H}}^{+}+ \text{H}_2\text{O}_2_{(\text{A})}$ A should be ${\text{H}}_2 {\text{O}}_2$ Structure of ${\text{H}}_2 {\text{O}}_2$ is solid phase

Q39JEE Main 2022MCQ4MClassification of Elements

The correct order of melting point is : [26-Jun-2022-Shift-1]

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📖 Explanation

MP details ${\text{Be}}\;1560 {\text{K}}$ ${\text{Mg}}\;924 {\text{K}}$ ${\text{Ca}}\;1124 {\text{K}}$ ${\text{Sr}}\;1062 {\text{K}}$

Q40JEE Main 2022MCQ4Mp Block Elements

The correct order of melting points of hydrides of group 16 elements is : [26-Jun-2022-Shift-1]

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📖 Explanation

Hydride M.P. $H_2O$ 273K $H_2 S$ 188K $H_2Se$ 208K $H_2Te$ 222K

Q41JEE Main 2022MCQ4Mp Block Elements

Consider the following reaction: ${\text{A}}+$ alkali $\to {\text{B}}$ (Major Product)If $B$ is an oxoacid of phosphorus with no ${\text{P}}- {\text{H}}$ bond, then $A$ is : [26-Jun-2022-Shift-1]

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📖 Explanation

Red ${\text{P}}_4+$ Alkali $\to {\text{H}}_4 {\text{P}}_2 {\text{O}}_6$ (No P-H bond) Structure of ${\text{H}}_4 {\text{P}}_2 {\text{O}}_6=$ $HO-P^{||}^{O}_{|}_{\;\;\;OH}-P^{||}^{O}_{|}_{\;\;\;OH}-OH$ No P-H bond

Q42JEE Main 2022MCQ4Mp Block Elements

Polar stratospheric clouds facilitate the formation of : [26-Jun-2022-Shift-1]

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📖 Explanation

Polar stratospheric clouds provide surface on which hydrolysis of ${\text{ClONO}}_2$ takes place to form ${\text{HOCl}}$ (Hypochlorous acid) ${\text{ClONO}}_2( {\text{g}})+ {\text{H}}_2 {\text{O}}( {\text{g}}) \to {\text{HOCl}}( {\text{g}})+ {\text{HNO}}_3( {\text{g}})$

Q43JEE Main 2022MCQ4MPractical Organic Chemistry

Given below are two statements :Statement I : In 'Lassaigne's Test', when both nitrogen and sulphur are present in an organic compound, sodium thiocyanate is formed.Statement II : If both nitrogen and sulphur are present in an organic compound, then the excess of sodium used in sodium fusion will decompose the sodium thiocyanate formed to give ${\text{NaCN}} {\text{and}} {\text{Na}}_2 {\text{S}}$ .In the light of the above statements, choose the most appropriate answer from the options given below : [26-Jun-2022-Shift-1]

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📖 Explanation

Both statement I & statement II are correct ${\text{NaSCN}}+2 {\text{Na}} \to {\text{NaCN}}+ {\text{Na}}_2 {\text{S}}$

Q44JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

$( {\text{C}}_7 {\text{H}}_5 {\text{O}}_2)_2 {\longrightarrow }^{h v}[ {\text{X}}] \to 2 {\text{C}}_6 {\text{H}}_5+2 {\text{CO}}_2$ Consider the above reaction and identify the intermediate ' $X$ ' [26-Jun-2022-Shift-1]

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Q45JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

Consider the above reaction sequence and identify the product B. [26-Jun-2022-Shift-1]

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Q46JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Which will have the highest enol content? [26-Jun-2022-Shift-1]

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Q47JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Among the following structures, which will show the most stable enamine formation? $( {\text{Where}} {\text{Me}} { {\text{is}}}- {\text{CH}}_3)$ [26-Jun-2022-Shift-1]

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Q48JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Which of the following sets are correct regarding polymer?(A) Copolymer : Buna-S(B) Condensation polymer : Nylon-6,6(C) Fibres : Nylon-6,6(D) Thermosetting polymer : Terylene(E) Homopolymer : Buna-NChoose the correct answer from given options below : [26-Jun-2022-Shift-1]

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📖 Explanation

(A) Buna-S $- \;\;$ Copolymer(B) Nylon-6,6 $\;\; - \;\;$ Condensation polymer(C) Nylon-6,6 - Fibre(D) Terylene $\;\; - \;\;$ Thermoplastic(E) Buna-N - Copolymer $\therefore A, B$ and $C$ are correct.

Q49JEE Main 2022MCQ4MBiomolecules

A chemical which stimulates the secretion of pepsin is: [26-Jun-2022-Shift-1]

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Q50JEE Main 2022MCQ4MSalt Analysis

Which statement is not true with respect to nitrate ion test? [26-Jun-2022-Shift-1]

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📖 Explanation

Brown ring test ${\text{NO}}_3^{-}+3 {\text{Fe}}^{+2}+4 {\text{H}}^{+} \to {\text{NO}}+3 {\text{Fe}}^{+3}+2 {\text{H}}_2 {\text{O}}$ ${[ {\text{Fe}}( {\text{H}}_2 {\text{O}})_6]^{2+} {\text{NO}} \to [ {\text{Fe}}( {\text{H}}_2 {\text{O}})_5 {\text{NO}}]^{2+}_{\text{Brown} \text{ring}} {\text{H}}_2 {\text{O}}}$

Q51JEE Main 2022NAT4MChemical Thermodynamics

For complete combustion of methanol ${\text{CH}}_3 {\text{OH}}( {\text{I}})+\;\frac{3}{2} {\text{O}}_2( {\text{g}}) \to {\text{CO}}_2( {\text{g}})+2 {\text{H}}_2 {\text{O}}( {\text{I}})$ the amount of heat produced as measured by bomb calorimeter is $726 {\text{kJ}} {\text{mol}}^{-1}$ at $27^{\circ} {\text{C}}$ . The enthalpy of combustion for the reaction is $- {\text{x}} {\text{kJ}} {\text{mol}}^{-1}$ , where $x$ is____ (Nearest integer)(Given : ${\text{R}}=8.3 {\text{JK}}^{-1} {\text{mol}}^{-1}$ ) [26-Jun-2022-Shift-1]

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📖 Explanation

${\text{CH}}_3 {\text{OH}}( {\text{l}})+\;\frac{3}{2} {\text{O}}_2( {\text{g}}) \to {\text{CO}}_2( {\text{g}})+2 {\text{H}}_2 {\text{O}}( {\text{l}})$ $\Delta {\text{H}}=\Delta {\text{U}}+\Delta {\text{n}}_{ {\text{g}}} {\text{RT}}$ $=-726 {\text{kJ}}+(\;\frac{-1}{2}) \times 8.3 \times 300$ $≃-727 {\text{kJ}} {\text{mol}}^{-1}$

Q52JEE Main 2022NAT4MSolutions

A $0.5$ percent solution of potassium chloride was found to freeze at $-0.24^{\circ} {\text{C}}$ . The percentage dissociation of potassium chloride is____ (Nearest integer)(Molal depression constant for water is $1.80 {\text{K}} {\text{kg}} {\text{mol}}^{-1}$ and molar mass of ${\text{KCl}}^{\;\text{is }\;} 74.6 {\text{g}} {\text{mol}}^{-1}$ ) [26-Jun-2022-Shift-1]

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📖 Explanation

$\Delta {\text{T}}_{ {\text{f}}}= {\text{iK}}_{ {\text{b}}} {\text{m}}$ ${\text{i}}=\;\frac{0.24 \times 99.5 \times 74.6}{1.80 \times 0.5 \times 1000}$ $=1.98$ $\alpha=\;{ {\text{i}}-1}/{ {\text{n}}-1}=\;\frac{0.98}{1}=0.98=98 \%$

Q53JEE Main 2022NAT4MIonic Equilibrium

$50 {\text{mL}}$ of $0.1 {\text{M}} {\text{CH}}_3 {\text{COOH}}$ is being titrated against $0.1 {\text{M}} \; {\text{NaOH}}$ . When $25 {\text{mL}}$ of ${\text{NaOH}}$ has been added, the ${\text{pH}}$ of the solution will be ____ $\times 10^{-2}$ . (Nearest integer)(Given : $. {\text{pK}}_{ {\text{a}}}( {\text{CH}}_3 {\text{COOH}})=4.76)$ $\log 2=0.30$ $\log 3=0.48$ $\log 5=0.69$ $\log 7=0.84$ $\log 11=1.04$ [26-Jun-2022-Shift-1]

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📖 Explanation

${\text{CH}}_3 {\text{COOH}}+ {\text{NaOH}} \to {\text{CH}}_3 {\text{COONa}}+ {\text{H}}_2 {\text{O}}$ After adding $25 {\text{ml}}$ of ${\text{NaOH}}$ volume of mixture $=50+25=75 {\text{ml}}$ Initially,Number of millimole of ${\text{NaOH}}=25 \times 0.1=2.5 {\text{mm}}$ Number of millimole of ${\text{CH}}_3 {\text{COOH}}=50 \times 0.1=5 {\text{mm}}$ After nutrilisation,Millimole of ${\text{NaOH}}=0$ Millimole of ${\text{CH}}_3 {\text{COOH}}=5-2.5=2.5 {\text{mm}}$ Millimole of ${\text{CH}}_3 {\text{COONa}}=2.5$ After nutrilisation,Concentration of ${\text{CH}}_3 {\text{COOH}}=[ {\text{CH}}_3 {\text{COOH}}]=\;\frac{5-2.5}{75}=\;\frac{1}{30}$ Concentration of ${\text{CH}}_3 {\text{COONa}}=[ {\text{CH}}_3 {\text{COONa}}]=\;\frac{2.5}{75}=\;\frac{1}{30}$ $P^H=P^{ {\text{Ka}}}+\log \;{[ {\text{CH}}_3 {\text{COONa}}]}/{[ {\text{CH}}_3 {\text{COOH}}]}$ $=4.76+\log \;\frac{\;\frac{1}{30}}{\frac{1}{30}}$ $=4.76+\log (1)$ $=4.76+0$ $=4.76$ $=4.76 \times 10^{-2}$

Q54JEE Main 2022NAT4MChemical Kinetics

A flask is filled with equal moles of $A$ and $B$ . The half lives of $A$ and $B$ are 100 s and 50 s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of $B$ is____ s.(Given : $\ln 2=0.693$ ) [26-Jun-2022-Shift-1]

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📖 Explanation

${\text{k}}_{ {\text{A}}}=\;\frac{\ln 2}{100} ; {\text{k}}_{ {\text{B}}}=\;\frac{\ln 2}{50}$ ${\text{A}}_{ {\text{t}}}= {\text{A}}_0 \times {\text{e}}^{- {\text{k}}_{ {\text{A}}} {\text{t}}}$ ${\text{A}}_{ {\text{t}}}= {\text{A}}_0 \times {\text{e}}^{(\;\frac{-\ln 2}{100} \times {\text{t}})}$ ${\text{B}}_{ {\text{t}}}= {\text{B}}_0 \times {\text{e}}^{(\;\frac{-\ln 2}{50} \times {\text{t}})}$ ${\text{A}}_0= {\text{B}}_0$ $\& {\text{A}}_{ {\text{t}}}=4 {\text{B}}_{ {\text{t}}}$ ${\text{e}}^{-\;\frac{\ln 2}{100} \times {\text{t}}}=4 \times {\text{e}}^{-\;\frac{\ln 2}{50} \times {\text{t}}}$ ${\text{e}}^{\;\frac{\ln 2}{100} \times {\text{t}}}=4$ ${\text{e}}^{\;\frac{\ln 2}{100} \times {\text{t}}}=4$ $\;\frac{\ln 2}{100} \times {\text{t}}=\ln 4=2 \ln 2$ ${\text{t}}=200 {\sec}$

Q55JEE Main 2022NAT4MStates of Matter

$2.0 {\text{g}}$ of ${\text{H}}_2$ gas is adsorbed on $2.5 {\text{g}}$ of platinum powder at $300 {\text{K}}$ and $1 {\bar}$ pressure. The volume of the gas adsorbed per gram of the adsorbent is ___ ${\text{mL}} .$ [26-Jun-2022-Shift-1]

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📖 Explanation

${\text{PV}}= {\text{nRT}}$ ${\text{V}}=\;\frac{2 \times 0.083 \times 300}{2 \times 1}=24.9$ litre $\therefore$ Volume of the gas adsorbed per gram of the adsorbent $=\;\frac{24.9}{2.5}=9.96 {\text{L}}$ $=9960 {\text{ml}}$

Q56JEE Main 2022NAT4Md and f Block Elements

The spin-only magnetic moment value of the most basic oxide of vanadium among ${\text{V}}_2 {\text{O}}_3, {\text{V}}_2 {\text{O}}_4$ and ${\text{V}}_2 {\text{O}}_5$ is_____ B.M. (Nearest integer) [26-Jun-2022-Shift-1]

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📖 Explanation

The most basic oxide among ${\text{V}}_2 {\text{O}}_3, {\text{V}}_2 {\text{O}}_4$ and ${\text{V}}_2 {\text{O}}_5$ is ${\text{V}}_2 {\text{O}}_3$ ${\text{V}}_2 {\text{O}}_3= {\text{V}}^{+3}( {\text{d}}^2)$ Magnetic moment $=\sqrt{2(2+2)}=\sqrt{8}$ $=2.83 \approx 3$

Q57JEE Main 2022NAT4MCoordination Compounds

The spin-only magnetic moment value of an octahedral complex among ${\text{CoCl}}_3 \cdot 4 {\text{NH}}_3, {\text{NiCl}}_2 \cdot 6 {\text{H}}_2 {\text{O}}$ and ${\text{PtCl}}_4 \cdot 2 {\text{HCl}}$ , which upon reaction with excess of ${\text{AgNO}}_3$ gives 2 moles of ${\text{AgCl}}$ is ____B.M. (Nearest integer) [26-Jun-2022-Shift-1]

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📖 Explanation

${\text{CoCl}}_3 \cdot 4 {\text{NH}}_3 \;{{\longrightarrow }^{ {\text{AgNO}}}_{\;\text{ excess }\;_3}}[ {\text{Co}}( {\text{NH}}_3)_4 \cdot {\text{Cl}}_2]+ {\text{AgCl}}$ ${\text{NiCl}}_2 \cdot 6 {\text{H}}_2 {\text{O}} \;{{\longrightarrow }^{ {\text{AgNO}}}_{\;\text{ excess }\;_3}}[ {\text{Ni}}( {\text{H}}_2 {\text{O}})_6]^{2+}+2 {\text{AgCl}}$ ${\text{PtCl}}_4 \cdot 2 {\text{HCl}} \longrightarrow [ {\text{PtCl}}_6]^{4-}+ {\text{No}} {\text{AgCl}} {\text{ppt}}$ $[ {\text{Ni}}^{+2}( {\text{H}}_2 {\text{O}})_6]= {\text{d}}^8= {\text{t}} 2_{ {\text{g}}} {\text{e}}_{ {\text{g}}}^2=2$ unpaired electrons Magnetic moment $=\sqrt{2(2+2)}$ $2 \sqrt{8} \approx 3$

Q58JEE Main 2022NAT4MSome Basic Concepts of Chemistry

On complete combustion $0.30 {\text{g}}$ of an organic compound gave $0.20 {\text{g}}$ of carbon dioxide and $0.10 {\text{g}}$ of water. The percentage of carbon in the given organic compound is ____(Nearest integer) [26-Jun-2022-Shift-1]

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📖 Explanation

$C_x H_y+(x+\;\frac{y}{4}) O_2 \to x C O_2+\;\frac{y}{2} {\text{H}}_2 {\text{O}}$ Given organic compound ${\text{C}}_{ {\text{x}}} {\text{H}}_{ {\text{y}}}=0.3 {\text{gm}}$ Produced carbon dioxide $( {\text{CO}}_2)=0.2 {\text{gm}}$ Produced water $( {\text{H}}_2 {\text{O}})=0.1 {\text{gm}}$ Moles of ${\text{CO}}_2=\;\frac{0.2}{44}$ $\therefore$ Moles of $C$ atom $=\;\frac{0.2}{44}$ $\therefore$ Mass of $C$ atom $=\;\frac{0.2}{44} \times 12=0.0545$ Moles of ${\text{H}}_2 {\text{O}}=\;\frac{0.1}{18}$ $\therefore$ Moles of ${\text{H}}$ atoms $=\;\frac{0.1}{18} \times 2$ $\therefore$ Mass of ${\text{H}}$ atoms $=\;\frac{0.1 \times 2}{18} \times 1=0.0111$ $\therefore \%$ of $C$ atom $=\;\frac{0.0545}{0.3} \times 100=18 \%$

Q59JEE Main 2022NAT4MAlcohols, Phenols and Ethers

Compound 'P' on nitration with dil. ${\text{HNO}}_3$ yields two isomers $( {\text{A}})$ and $( {\text{B}})$ . These isomers can be separated by steam distillation. Isomers (A) and (B) show the intramolecular and intermolecular hydrogen bonding respectively. Compound (P) on reaction with conc. ${\text{HNO}}_3$ yields a yellow compound ' ${\text{C}}$ ', a strong acid. The number of oxygen atoms is present in compound ' ${\text{C}}$ '_____ [26-Jun-2022-Shift-1]

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Q60JEE Main 2022NAT4MBiomolecules

The number of oxygens present in a nucleotide formed from a base, that is present only in RNA is___ [26-Jun-2022-Shift-1]

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Mathematics30 questions

Q61JEE Main 2022MCQ4MFunctions

Let $f(x)=\;\frac{x-1}{x+1}, x \in R-\{0,-1,1\}$ . If $f^{n+1}(x)=f(f^n(x))$ for all ${\text{n}} \in {\text{N}}$ , then $f^6(6)+f^7(7)$ is equal to : [26-Jun-2022-Shift-1]

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📖 Explanation

Given, $f(x)=\;\frac{x-1}{x+1}$ Also given, $f^{n+1}(x)=f(f^n(x)) . . . .(1)$ $\therefore \;\text{ For }\; n=1$ $f^{1+1}(x)=f(f^1(x))$ $\Rightarrow f^2(x)=f(f(x))$ $=f(\;\frac{x-1}{x+1})$ $=\;\frac{\;\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}$ $=\;\frac{\;\frac{x-1-x-1}{x+1}}{\frac{x-1+x+1}{x+1}}$ $=\;\frac{-2}{2 x}$ $=-\;\frac{1}{x}$ From equation (1), when $n=2$ $f^{2+1}(x)=f(f^2(x))$ $\Rightarrow f^3(x)=f(f^2(x))$ $=f(-\;\frac{1}{x})$ $=\;\frac{-\frac{1}{x}-1}{-\;\frac{1}{x}+1}$ $=\;\frac{\;\frac{-1-x}{-1}}{\frac{-1+x}{x}}$ $=\;\frac{-1-x}{-1+x}=\;\frac{-(x+1)}{x-1}$ Similarly, $f^4(x)=f(f^3(x))$ $=f(\;\frac{-x+1}{x-1})$ $=\;\frac{\;\frac{-(x+1)}{x-1}-1}{\frac{-(x+1)}{x-1}+1}$ $=\;\frac{\;\frac{x-1-x+1}{x-1}}{\frac{-x-1+x-1}{x-1}}$ $=\;\frac{-2 x}{-2}=x$ $\therefore f^5(x)=f(f^4(x))$ $=f(x)$ $=\;\frac{x-1}{x+1}$ $f^6(x)=f(f^5(x))$ $=f(\;\frac{x-1}{x+1})$ $=-\;\frac{1}{x}$ (Already calculated earlier) $f^7(x)=f(f^6(x))$ $=f(-\;\frac{1}{x})$ $=\;\frac{-\frac{1}{x}-1}{-\;\frac{1}{x}+1}$ $=\;\frac{-(x+1)}{x-1}$ $\therefore f^6(6)=-\;\frac{1}{6}$ and $f^7(7)=\;\frac{-(7+1)}{7-1}=-\;\frac{8}{6}$ So, $f^6(6)+f^7(7)$ $=-\;\frac{1}{6}-\;\frac{8}{6}$ $=-\;\frac{3}{2}$

Q62JEE Main 2022MCQ4MComplex Numbers

Let $A=\{z \in C:|\;\frac{z+1}{z-1}|< 1\}$ and $B=\{z \in C: \arg (\;\frac{z-1}{z+1})=\;\frac{2 \pi}{3}\}$ . Then ${\text{A}} ∩ {\text{B}}$ is : [26-Jun-2022-Shift-1]

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📖 Explanation

Set A $\Rightarrow |\;{ {\text{z}}+1}/{ {\text{z}}-1}|< 1$ $\Rightarrow | {\text{z}}+1|< | {\text{z}}-1|$ $\Rightarrow ( {\text{x}}+1)^{2+} {\text{y}}^2< ( {\text{x}}-1)^{2+} {\text{y}}^2$ $\Rightarrow {\text{x}}< 0$ $\Rightarrow \arg (\;{ {\text{z}}-1}/{ {\text{z}}+1})=\;\frac{2 \pi}{3}$ $\Rightarrow \tan ^{-1}(\;{ {\text{y}}}/{ {\text{x}}-1})-\tan ^{-1}(\;{ {\text{y}}}/{ {\text{x}}+1})=\;\frac{2 \pi}{3}$ $\Rightarrow {\text{x}}^{2+} {\text{y}}^{2+}\;{2 {\text{y}}}/{\sqrt{3}}-1=0$ ${\text{A}} ∩ {\text{B}}$ $\Rightarrow$ Centre $(0,-\;\frac{1}{\sqrt{3}})$

Q63JEE Main 2022MCQ4MMatrices and Determinants

Let $A$ be a $3 \times 3$ invertible matrix. If $|{\text{adj}}(24 A)|=|{\text{adj}}(3 {\text{adj}}(2 A))|$ , then $|A|^2$ is equal to : [26-Jun-2022-Shift-1]

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📖 Explanation

We know, $|{\text{adj}} A|=|A|^{n-1}$ Now, $|a d j 24 A|=|a d j 3(a d j 2 A)|$ $\Rightarrow |24 A|^{3-1}=|3 a d j 2 A|^{3-1}$ $\Rightarrow |24 A|^2=|3 a d j 2 A|^2$ Also, we know, $|K A|=K^n|A|$ $\Rightarrow ((24)^2)^2|A|^2=((3)^3)^2|a d j 2 A|^2$ $\Rightarrow (24)^6|A|^2=3^6 \cdot (|2 A|^{3-1})^2$ $\Rightarrow (24)^6|A|^2=3^6 \cdot |2 A|^4$ $\Rightarrow (24)^6|A|^2=3^6 \cdot (2^3)^4 \cdot |A|^4$ $\Rightarrow 3^6 \cdot 8^6 \cdot |A|^2=3^6 \cdot 8^4 \cdot |A|^4$ $\Rightarrow 8^2=|A|^2$ $\Rightarrow |A|^2=64=2^6$

Q64JEE Main 2022MCQ4MMatrices and Determinants

The ordered pair $(a, b)$ , for which the system of linear equations $3 x-2 y+z=b$ $5 x-8 y+9 z=3$ $2 x+y+a z=-1$ has no solution, is : [26-Jun-2022-Shift-1]

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📖 Explanation

$\begin{vmatrix}3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 & a\end{vmatrix}=0 \Rightarrow -14 a-42=0 \Rightarrow a=-3$ Now 3 (equation (1)) - (equation (2)) $-2$ (equation (3)) is $3(3 x-2 y+z-b)-(5 x-8 y+9 z-3)-2(2 x+y+a z+1)=0$ $\Rightarrow -3 b+3-2=0 \Rightarrow b=\;\frac{1}{3}$ So for no solution $a=-3$ and $b \ne \;\frac{1}{3}$

Q65JEE Main 2021MCQ4MBinomial Theorem

The remainder when $(2021)^{2023}$ is divided by 7 is : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$(2021)^{2023}$ $=(2016+5)^{2023}[$ here 2016 is divisible by 7 $]$ $={ }^{2023} C_0(2016)^{2023}+. . . . . . . . .+{ }^{2023} C_{2022}(2016)(5)^{2022}+{ }^{2023} C_{2023}(5)^{2023}$ $=2016\$ [^{2023} C_0 ⋅(2016)^{2022}+. . . . . . .+{ }^{2023} C_{2022} ⋅(5)^{2022}]$+(5)^{2023}=2016 \lambda +(5)^{2023}=7 \times 288 \lambda +(5)^{2023}=7 K+(5)^{2023} . . . . . .(1) $Now,$ (5)^{2023}=(5)^{2022} \cdot 5=(5^3)^{674} \cdot 5=(125)^{674} \cdot 5=(126-1)^{674} \cdot 5=(126-1)^{674} \cdot 5=5[{ }^{674} {\text{C}}0(126)^{674}+. . . . . . . . . .-{ }^{674} {\text{C}}{673}(126)+{ }^{674} {\text{C}}_{674}]=5 \times 126$[{ }^{674} {\text{C}}0(126)^{673}+. . . . . . .-{ }^{674} {\text{C}}{673}]$+5=5.7 .18$[{ }^{674} {\text{C}}0(126)^{673}+. . . . . . .-{ }^{674} {\text{C}}{673}]$+5=7 \lambda +5 $Replacing$ (5)^{2023} $in equation (1) with$ 7 \lambda +5 $, we get,$ (2021)^{2023}=7 {\text{K}}+7 \lambda +5=7( {\text{K}}+\lambda )+5\therefore $Remainer$ =5$

Q66JEE Main 2022MCQ4MLimits, Continuity and Differentiability

$\lim _{x \to \;\frac{1}{\sqrt{2}}} \;\frac{\sin (\cos ^{-1} x)-x}{1-\tan (\cos ^{-1} x)}$ is equal to : [26-Jun-2022-Shift-1]

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📖 Explanation

$\lim _{x \to \;\frac{1}{\sqrt{2}}} \;\frac{\sin (\cos ^{-1} x)-x}{1-\tan (\cos ^{-1} x)}$ Let $\cos ^{-1} x=t$ $\Rightarrow x=\cos t$ When $x \to \;\frac{1}{\sqrt{2}}$ , then $t \to \cos ^{-1}(\;\frac{1}{\sqrt{2}}) \to \;\frac{\pi}{4}$ $\therefore \lim _{t \to \;\frac{\pi}{4}} \;\frac{\sin t-\cos t}{1-\tan (t)}$ $=\lim _{t \to \;\frac{\pi}{4}} \;\frac{\sin t-\cos t}{1-\;\frac{\sin t}{\cos t}}$ $=\lim _{t \to \;\frac{\pi}{4}} \;\frac{(\sin t-\cos t)(\cos t)}{(\cos t-\sin t)}$ $=\lim _{t \to \;\frac{\pi}{4}}-\cos t$ $=-\lim _{t \to \;\frac{\pi}{4}} \cos t$ $=-\;\frac{1}{\sqrt{2}}$

Q67JEE Main 2022MCQ4MLimits, Continuity and Differentiability

Let ${\text{f}}, {\text{g}}: {\text{R}} \to {\text{R}}$ be two real valued functions defined as $f(x)=\begin{cases}-\text{|}x+3\text{|} & x< 0 \\ e^x & \ \\ \ \\ x \ge 0$ and $g(x)=\{\text{table} x^{2+}k_1 x , , \ \\ \ \\ x < 0 \\ 4 x+k_2 & \ \\ \ \\ x \ge 0$ where $k_1$ and $k_2$ are real constants\end{cases} If (gof) is differentiable at $x=0$ , then (gof) $(-4)+(g o f)(4)$ is equal to : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\because$ gof is differentiable at $x=0$ So R.H.D = L.H.D $\;\frac{d}{d x}(4 e^x+k_2)=\;\frac{d}{d x}((-|x+3|)^{2-}k_1|x+3|)$ $\Rightarrow 4=6-k_1 \Rightarrow k_1=2$ Also $f(f(0^{+}))=g(f(0^{-}))$ $\Rightarrow 4+k_2=9-3 k_1 \Rightarrow k_2=-1$ Now $g(f(-4))+g(f(4))$ $=g(-1)+g(e^4)=(1-k_1)+(4 e^{4+}k_2)$ $=4 e^{4-}2$ $=2(2 e^{4-}1)$

Q68JEE Main 2022MCQ4MApplication of Derivatives

The sum of the absolute minimum and the absolute maximum values of thefunction $f(x)= 3 x-x^{2+}2|-x$ in the interval $[-1,2]$ is : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$f(x)=\begin{cases}x^{2-}4 x-2 & ∀ x \in (-1 \\ \\ \frac{3-\sqrt{17}}{2}) \\ -x^{2+}2 x+2 & ∀ x \in (\\ \\ \frac{3-\sqrt{17}}{2} & 2)\end{cases}$ $f^{'}(x)$ when $x \in (-1, \;\frac{3-\sqrt{17}}{2})$ $f^{'}(x)=2 x-4=0 \Rightarrow x=2$ $f^{'}( {\text{x}})=2( {\text{x}}-2) \;\; \Rightarrow f^{'}( {\text{x}})$ is always $↓$ $f(2)=2$ $f(-1)=3$ $f(\;\frac{3-\sqrt{17}}{2})=\;\frac{\sqrt{17}-3}{2}$ $f^{'}( {\text{x}})$ when ${\text{x}} \in (\;\frac{3-\sqrt{17}}{2}, 2)$ $f^{'}( {\text{x}})=-2 {\text{x}}+2$ $f^{'}(x)=-2(x-1)$ $f^{'}(x)=0$ when $x=1$ $f(1)=3$ absolute minimum value $=\;\frac{\sqrt{17}-3}{2}$ absolute maximum value $=3$ $\;\text{ Sum }\;=\;\frac{\sqrt{17}-3}{2}+3=\;\frac{\sqrt{17}+3}{2}$

Q69JEE Main 2022MCQ4MApplication of Derivatives

Let $S$ be the set of all the natural numbers, for which the line $\;\frac{x}{a}+\;\frac{y}{b}=2$ is a tangent to the curve $(\;\frac{x}{a})^n+(\;\frac{y}{b})^n=2$ at the point $(a, b), a b \ne 0$ . Then : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$(\;\frac{x}{a})^n+(\;\frac{y}{b})^n=2$ Differentiating both sides with respect to $x$ , we get $\Rightarrow n \cdot (\;\frac{x}{a})^{n-a} \cdot \;\frac{1}{a}+n \cdot (\;\frac{y}{b})^{n-1} \cdot \;\frac{1}{b} \cdot \;\frac{d y}{d x}=0$ $\Rightarrow \;\frac{d y}{d x}=-\;\frac{\;\frac{1}{a} \cdot (\frac{x}{a})^{n-1}}{\frac{1}{b}(\;\frac{y}{b})^{n-1}}$ $=-\;\frac{b}{a}(\;\frac{x b}{y a})^{n-1}$ Now, $\;\frac{d y}{d x}$ at $( {\text{a}}, {\text{b}})=-\;\frac{b}{a}[\;\frac{a b}{b a}]^{n-1}=-\;\frac{b}{a}$ Equation of tangent at $(a, b)$ is, $(y-b)=-\;\frac{b}{a}(x-a)$ $\Rightarrow \;\frac{y-b}{b}=-\;\frac{x-a}{a}$ $\Rightarrow \;\frac{y}{b}-1=-\;\frac{x}{a}+1$ $\Rightarrow \;\frac{x}{a}+\;\frac{y}{b}=2$ $\therefore$ It is tangent for all value of $n$ .

Q70JEE Main 2022MCQ4MArea Under The Curves

The area bounded by the curve $y=| x^{2-}9|$ and the line $y=3$ is : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$y=3$ and $y=|x^{2-}9|$ Intersect in first quadrant at $x=\sqrt{6}$ and $x=\sqrt{12}$ Required area $=2[\;\frac{2}{3}(6 \times \sqrt{6})+int_{\sqrt{6}}^3(3-(9-x^2)) d x+int_3^{\sqrt{12}}(3-(x^{2-}9)) d x]$ $=2[4 \sqrt{6}+.(\;\frac{x^3}{3}-6 x)|_{\sqrt{6}} ^{3+}.(12 x-\;\frac{x^3}{3})|_3 ^{\sqrt{12}}]$ $=2[4 \sqrt{6}+(4 \sqrt{6}-9)+(8 \sqrt{12}-27)]$ $=2[8 \sqrt{6}+16 \sqrt{3}-36]=8[2 \sqrt{6}+4 \sqrt{3}-9]$

Q71JEE Main 2022MCQ4MStraight Lines and Pair of Straight Lines

Let $R$ be the point $(3,7)$ and let $P$ and $Q$ be two points on the line $x+y=5$ such that $P Q R$ is an equilateral triangle. Then the area of $\triangle {\text{PQR}}$ is : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Let, side of triangle $=a$ . $h=\;\frac{|3+7-5|}{\sqrt{1^{2+}1^2}}$ $=\;\frac{5}{\sqrt{2}}$ From figure, $h=a \sin 60^{\circ}$ $\Rightarrow a=\;\frac{2 h}{\sqrt{3}}$ $=\;\frac{2}{\sqrt{3}} \times \;\frac{5}{\sqrt{2}}$ $=\;\frac{10}{\sqrt{6}}$ $\therefore \;\text{ Area }\;=\;\frac{3}{4}(\;\frac{10}{\sqrt{6}})^2$ $=\;\frac{25}{2 \sqrt{3}}$

Q72JEE Main 2022MCQ4MCircles

Let $C$ be a circle passing through the points $A(2,-1)$ and $B(3,4)$ . The line segment $A B$ s not a diameter of $C$ . If $r$ is the radius of $C$ and its centre lies on the circle $(x-5)^{2+}(y-1)^2=\;\frac{13}{2}$ , then $r^2$ is equal to: [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Equation of perpendicular bisector of $A B$ is $y-\;\frac{3}{2}=-\;\frac{1}{5}(x-\;\frac{5}{2}) \Rightarrow x+5 y=10$ Solving it with equation of given circle, $(x-5)^2(\;\frac{10-x}{5}-1)^2=\;\frac{13}{2}$ $\Rightarrow (x-5)^2(1+\;\frac{1}{25})=\;\frac{13}{2}$ $\Rightarrow x-5=\pm \;\frac{5}{2} \Rightarrow x=\;\frac{5}{2} \;\text{ or }\; \;\frac{15}{2}$ But $x \ne \;\frac{5}{2}$ because ${\text{AB}}$ is not the diameter.So, centre will be $(\;\frac{15}{2}, \;\frac{1}{2})$ Now, $r^2=(\;\frac{15}{2}-2)^{2+}(\;\frac{1}{2}+1)^2$ $=\;\frac{65}{2}$

Q73JEE Main 2022MCQ4MParabola

Let the normal at the point on the parabola ${\text{y}}^2=6 {\text{x}}$ pass through the point $(5,-8)$ . If the tangent at ${\text{P}}$ to the parabola intersects its directrix at the point $Q$ , then the ordinate of the point $Q$ is : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Let $P(a t^2, 2 a t)$ where $a=\;\frac{3}{2}$ ${\text{T}}: {\text{yt}}= {\text{x}}+ {\text{at}}^2$ So point ${\text{Q}}$ is $(-a, a t-\;\frac{a}{t})$ $N: y=-t x+2 a t+a t^3$ passes through $(5,-8)$ $-8=-5 t+3 t+\;\frac{3}{2} t^3$ $\Rightarrow 3 t^{3-}4 t+16=0$ $\Rightarrow (t+2)(3 t^{2-}6 t+8)=0$ $\Rightarrow {\text{t}}=2$ So ordinate of point $Q$ is $-\;\frac{9}{4}$ .

Q74JEE Main 2022MCQ4MThree Dimensional Geometry

If the two lines $l_1: \;\frac{x-2}{3}=\;\frac{y+1}{-2}, z=2$ and $l_2: \;\frac{x-1}{1}=\;\frac{2 y+3}{\alpha}=\;\frac{z+5}{2}$ are perpendicular, then an angle between the lines ${\text{I}}_2$ and $l_3: \;\frac{1-x}{3}=\;\frac{2 y-1}{-4}=\;\frac{z}{4}$ is : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\because L_1$ and $L_2$ are perpendicular, so $3 \times 1+(-2)(\;\frac{\alpha}{2})+0 \times 2=0$ $\Rightarrow \alpha=3$ Now angle between $I_2$ and $I_3$ , $\cos \theta=\;{1(-3)+\frac{\alpha}{2}(-2)+2(4)}/{\sqrt{1+\;\frac{\alpha^2}{4}+4 \sqrt{9+4+16}}}$ $\Rightarrow \cos \theta=\;\frac{2}{\frac{29}{2}} \Rightarrow \theta=\cos ^{-1}(\;\frac{4}{29})=\sec ^{-1}(\;\frac{29}{4})$

Q75JEE Main 2022MCQ4MThree Dimensional Geometry

Let the plane $2 x+3 y+z+20=0$ be rotated through a right angle about its line of intersection with the plane $x-3 y+5 z=8$ . If the mirror image of the point $(2,-\;\frac{1}{2}, 2)$ in the rotated plane is $B(a, b, c)$ , then : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Consider the equation of plane, $P:(2 x+3 y+z+20)+\lambda (x-3 y+5 z-8)=0$ $P:(2+\lambda ) x+3(3-3 \lambda ) y+1(1+5 \lambda ) z+(20-8 \lambda )=0$ $\because$ Plane ${\text{P}}$ is perpendicular to $2 x+3 y+z+20=0$ So, $4+2 \lambda +9-9 \lambda +1+5 \lambda =0$ $\Rightarrow \lambda =7$ $P: 9 x-18 y+36 z-36=0$ or $P: x-2 y+4 z=4$ If image of $(2,-\;\frac{1}{2}, 2)$ in plane $P$ is $(a, b, c)$ then $\;\frac{a-2}{1}=\;\frac{b+\frac{1}{2}}{-2}=\;\frac{c-2}{4}$ and $(\;\frac{a+2}{2})-2(\;\frac{b-\frac{1}{2}}{2})+4(\;\frac{c+2}{2})=4$ Clearly $a=\;\frac{4}{3}, b=\;\frac{5}{6}$ and $c=-\;\frac{2}{3}$ So, $a: b: c=8: 5:-4$

Q76JEE Main 2022MCQ4MVector Algebra

If $\vec{a} \cdot \vec{b}=1, \vec{b} \cdot \vec{c}=2$ and $\vec{c} \cdot \vec{a}=3$ , then the value of $[\vec{a} \times (\vec{b} \times \vec{c}), \vec{b} \times (\vec{c} \times \vec{a}), \vec{c} \times (\vec{b} \times \vec{a})]$ is : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\because \vec{a} \times (\vec{b} \times \vec{c})=3 \vec{b}-\vec{c}=\vec{u}$ $\vec{b} \times (\vec{c} \times \vec{a})=\vec{c}-2 \vec{a}=\vec{v}$ $\vec{c} \times (\vec{b} \times \vec{a})=3 \vec{b}-2 \vec{a}=\vec{w}$ $\therefore \vec{u}+\vec{v}=\vec{w}$ So, vectors $\vec{u, v}$ and $\vec{w}$ are coplanar, hence their Scalar triple product will be zero.

Q77JEE Main 2022MCQ4MProbability

Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Coin is tossed 5 times, so $n=5$ Let, $p=$ probability of getting heads $q=$ probability of getting tails. $\therefore p+q=1 . . . . . .(1)$ $\therefore$ Probability of getting 4 heads $={ }^5 C_4 \cdot p^4 \cdot q$ And probability of getting 5 heads $={ }^5 C_5 \cdot p^5$ Given, ${ }^5 C_4 \cdot p^4 \cdot q={ }^5 C_5 \cdot p^5$ $\Rightarrow 5 q=p . . . . . . \;\text{ (2) }\;$ From equation (1) and (2), we get, $5 q+q=1$ $\Rightarrow 6 q=1$ $\Rightarrow q=\;\frac{1}{6}$ $\therefore p=1-\;\frac{1}{6}=\;\frac{5}{6}$ Now, probability of getting atmost two heads $=p(x=0)+p(x=1)+p(x=2)$ $p(x=0)=$ Getting zero head in 5 trials $={ }^5 C_0 \cdot p^0 \cdot q^5$ $p(x=1)=$ Getting one head in 5 trials $={ }^5 C_1 \cdot p^1 \cdot q^4$ $p(x=2)=$ Getting two heads in 5 trials $={ }^5 C_2 \cdot p^2 \cdot q^3$ $={ }^5 {\text{C}}_0 \cdot q^{5+}{ }^5 {\text{C}}_1 \cdot p q^{4+}{ }^5 {\text{C}}_2 \cdot {\text{p}}^2 q^3$ $=(\;\frac{1}{6})^{5+}5 \cdot \;\frac{5}{6} \cdot (\;\frac{1}{6})^{4+}10 \cdot (\;\frac{5}{6})^2 \cdot (\;\frac{1}{6})^3$ $=\;\frac{1+25+250}{6^5}=\;\frac{276}{6^5}=\;\frac{46}{6^4}$

Q78JEE Main 2022MCQ4MStatistics

The mean of the numbers $a, b, 8,5,10$ is 6 and their variance is $6.8$ . If ${\text{M}}$ is the mean deviation of the numbers about the mean, then $25 {\text{M}}$ is equal to : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\sigma^2=\;\frac{Σ_{i=1}^{5} (x_i-\overline{x}) ^2}{n}$ $\;\text{ Mean }\;=6$ $\;{ {\text{a}}+ {\text{b}}+8+5+10}/{5}=6 \Rightarrow a+b=7$ $a+b=7$ ${\text{b}}=7- {\text{a}}$ $6.8=\;\frac{(a-6)^{2+}(b-6)^{2+}(8-6)^{2+}(5-6)^{2+}(10-6)^2}{5}$ $34=(a-6)^{2+}(7-a-6)^{2+}4+1+18$ $a^{2-}7 a+12=0 \Rightarrow a=4 \;\text{ or }\; a=3$ $a=4 \;\; a=3$ ${\text{b}}=3 \;\; {\text{b}}=4$ $M=\;\frac{Σ_{i=1}^{5}{| x_i-x |}}{n}$ $M=\;\frac{|a-6|+|b-6|+|8-6|+|5-6|+|10-6|}{5}$ $\;\text{ when }\; {\text{a}}=3, {\text{b}}=4 \;\; \;\text{ when }\; {\text{a}}=4, {\text{b}}=3$ ${\text{M}}=\;\frac{3+2+2+1+4}{5} \;\; {\text{M}}=\;\frac{2+3+2+1+7}{5}$ $M=\;\frac{12}{5} \;\; M=\;\frac{12}{5} \Rightarrow 25 {\text{M}}=25 \times \;\frac{12}{5}=60$

Q79JEE Main 2022MCQ4MApplication of Derivatives

Let $f(x)=2 \cos ^{-1} x+4 \cot ^{-1} x-3 x^{2-}2 x+10, x \in [-1,1]$ . If $[ {\text{a}}, {\text{b}}]$ is the range of the function ${\text{f}}$ , then $4 {\text{a}}- {\text{b}}$ is equal to : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$f(x)=2 \cos ^{-1} x+4 \cot ^{-1} x-3 x^{2-}2 x+10 ∀ x \in [-1,1]$ $\Rightarrow f^{'}(x)=-\;\frac{2}{\sqrt{1-x^2}}-\;\frac{4}{1+x^2}-6 x-2< 0 ∀ x \in [-1,1]$ So $f(x)$ is decreasing function and range of $f(x)$ is $[f(1), f(-1)]$ , which is $[\pi+5,5 \pi+9$ ]Now $4 a-b=4(\pi+5)-(5 \pi+9)$ $=11-\pi$

Q80JEE Main 2022MCQ4MSets and Relations

Let $\Delta , ∇ \in \{∧, V\}$ be such that ${\text{p}} ∇ {\text{q}} \Rightarrow (( {\text{p}} \Delta {\text{q}}) ∇ {\text{r}})$ is a tautology. Then (p $∇ {\text{q}}) \Delta {\text{r}}$ is logically equivalent to : [26-Jun-2022-Shift-1]

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📖 Explanation

Case-I If $\Delta \equiv ∇ \equiv ∧$ $(p ∧ q) \to ((p ∧ q) ∧ r)$ it can be false if $r$ is false,so not a tautologyCase-II If $\Delta \equiv ∇ \equiv V$ $( {\text{p}} ∨ {\text{q}}) \to (( {\text{p}} ∨ {\text{q}}) ∨ {\text{r}}) \equiv \;\text{ tautology }\;$ then $(p ∨ q) ∨ r \equiv (p \Delta r) ∨ q$ Case-III if $\Delta =∨, ∇=∧$ then $( {\text{p}} ∧ {\text{q}}) \to \{( {\text{p}} ∨ {\text{q}}) ∧ {\text{r}}\}$ Not a tautology $($ Check ${\text{p}} \to {\text{T}}, {\text{q}} \to {\text{T}}, {\text{r}} \to {\text{F}})$ Case-IV if $\Delta =∧, ∇=∨$ $( {\text{p}} ∧ {\text{q}}) \to \{( {\text{p}} ∧ {\text{q}}) ∨ {\text{r}}\}$ Not a tautology

Q81JEE Main 2022NAT4MQuadratic Equation and Inequalities

The sum of the cubes of all the roots of the equation $x^{4-}3 x^{3-}2 x^{2+}3 x+1=0 \;\text{ is }\;$ ____ [26-Jun-2022-Shift-1]

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📖 Explanation

$x^{4-}3 x^{3-}x^{2-}x^{2+}3 x+1=0$ $(x^{2-}1)(x^{2-}3 x-1)=0$ Let the root of $x^{2-}3 x-1=0$ be $\alpha$ and $\beta$ and other two roots of given equation are 1 and $-1$ So sum of cubes of roots $=1^{3+}(-1)^{3+}\alpha^{3+}\beta^3$ $=(\alpha+\beta)^{3-}3 \alpha \beta(\alpha+\beta)$ $=(3)^{3-}3(-1)(3)$ $=36$

Q82JEE Main 2022NAT4MPermutations and Combinations

There are ten boys $B_1, B_2, . . . . . ., B_{10}$ and five girls $G_1, G_2, . . . . . . ., G_5$ in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both $B_1$ and $B_2$ together should not be the members of a group, is___ [26-Jun-2022-Shift-1]

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📖 Explanation

Number of ways when $B_1$ and $B_2$ are not together $=$ Total number of ways of selecting 3 boys $- {\text{B}}_1$ and ${\text{B}}_2$ are together $={ }^{10} {\text{C}}_3-{ }^8 {\text{C}}_1$ $=\;\frac{10.9 .8}{1.2 .3}-8$ $=112$ Number of ways to select 3 girls $={ }^5 C_3=10$ $\therefore$ Total number of ways $=112 \times 10=1120$

Q83JEE Main 2022NAT4MEllipse

Let the common tangents to the curves $4(x^{2+}y^2)=9$ and $y^2=4 x$ intersect at the point Q. Let an ellipse, centered at the origin 0 , has lengths of semi-minor and semi-major axes equal to $O Q$ and 6, respectively. If e and I respectively denote the eccentricity and the length of the latus rectum of this ellipse, then $\;\frac{l}{e^2}$ is equal to____ [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Let $y=m x+c$ is the common tangentSo $c=\;\frac{1}{m}=\pm \;\frac{3}{2} \sqrt{1+m^2} \Rightarrow m^2=\;\frac{1}{3}$ So equation of common tangents will be $y=\pm \;\frac{1}{\sqrt{3}} x \pm \sqrt{3}$ , which intersects at $Q(-3,0)$ Major axis and minor axis of ellipse are 12 and 6.So eccentricity $e^2=1-\;\frac{1}{4}=\;\frac{3}{4}$ and length of latus rectum $=\;\frac{2 b^2}{a}=3$ Hence, $\;\frac{l}{e^2}=\;\frac{3}{3 / 4}=4$

Q84JEE Main 2022NAT4MDefinite Integrals

Let $f(x)=\max \{|x+1|,|x+2|, . . . .,|x+5|\}$ . Then $int_{-6}^{0} f(x) d x$ is equal to___ [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

For $|x+1|$ critical point, ${\text{x}}+1=0 \Rightarrow {\text{x}}=-1$ For $|x+2|$ critical point, ${\text{x}}+2=0 \Rightarrow {\text{x}}=-2$ For $|x+3|$ critical point, ${\text{x}}+3=0 \Rightarrow {\text{x}}=-3$ For $|x+4|$ critical point, ${\text{x}}+4=0 \Rightarrow {\text{x}}=-4$ For $|x+5|$ critical point, ${\text{x}}+5=0 \Rightarrow {\text{x}}=-5$ Here maximum function is represent by the dotted line. $\therefore$ Point of intersection $A$ of line $y=-x-1$ and $y=x+5$ : $-x-1=x+5$ $\Rightarrow 2 x=-6$ $\Rightarrow x=-3$ $\therefore y=-3+5=2$ $\therefore \;\text{ Point }\; A=(-3,2)$ $\therefore int_{-6}^0 f(x) d x$ $=int_{-6}^{-3}(-x-1) d x+int_{-3}^0(x+5) d x$ $=(-\;\frac{x^2}{2}-x)_{-6}^{-3}+\$ [;\frac{x^2}{2}+5 x]_{-3}^0=[(-;\frac{9}{2}+3)-(-;\frac{36}{2}+6)]+[0-(;\frac{9}{2}-15)]=(-;\frac{3}{2}+12)+;\frac{21}{2}=;\frac{21}{2} =21$

Q85JEE Main 2022NAT4MDifferential Equations

Let the solution curve $y=y(x)$ of the differential equation $(4+x^2) d y-2 x(x^{2+}3 y+4) d x=0$ pass through the origin. Then ${\text{y}}(2)$ is equal to____ [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$(4+x^2) d y-2 x(x^{2+}3 y+4) d x=0$ $\Rightarrow \;\frac{d y}{d x}=(\;\frac{6 x}{x^{2+}4}) y+2 x$ $\Rightarrow \;\frac{d y}{d x}-(\;\frac{6 x}{x^{2+}4}) y=2 x$ $\;\text{ I. F. }\;=e^{-3 \ln (x^{2+}4)}=\;\frac{1}{(x^{2+}4)^3}$ $\;\text{ So }\; \;\frac{y}{(x^{2+}4)^3}=\int \;\frac{2 x}{(x^{2+}4)^3} d x+c$ $\Rightarrow y=-\;\frac{1}{2}(x^{2+}4)+c(x^{2+}4)^3$ When ${\text{x}}=0, {\text{y}}=0$ gives $c=\;\frac{1}{32}$ ,So, for $x=2, y=12$

Q86JEE Main 2022NAT4MTrigonometric Ratios and Identities

If $\sin ^2(10^{\circ}) \sin (20^{\circ}) \sin (40^{\circ}) \sin (50^{\circ}) \sin (70^{\circ})=\alpha-\;\frac{1}{16} \sin (10^{\circ})$ , then $16+\alpha^{-1}$ is equal to____ [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$(\sin 10^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ}) \cdot (\sin 10^{\circ} \cdot \sin 20^{\circ} \cdot \sin 40^{\circ})$ $=(\;\frac{1}{4} \sin 30^{\circ}) \cdot [\;\frac{1}{2} \sin 10^{\circ}(\cos 20^{\circ}-\cos 60^{\circ})]$ $=\;\frac{1}{16}[\sin 10^{\circ}(\cos 20^{\circ}-\;\frac{1}{2})]$ $=\;\frac{1}{32}[2 \sin 10^{\circ} \cdot \cos 20^{\circ}-\sin 10^{\circ}]$ $=\;\frac{1}{32}[\sin 30^{\circ}-\sin 10^{\circ}-\sin 10^{\circ}]$ $=\;\frac{1}{64}-\;\frac{1}{64} \sin 10^{\circ}$ Clearly, $\alpha=\;\frac{1}{64}$ Hence $16+\alpha^{-1}=80$

Q87JEE Main 2022NAT4MSets and Relations

Let $A=\{n \in N$ : H.C.F. $(n, 45)=1\}$ andLet $B=\{2 k: k \in \{1,2, . . . . . ., 100\}\}$ . Then the sum of all the elements of $A ∩ B$ is______ [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Sum of all elements of $A ∩ B=2$ [Sum of natural numbers upto 100 which are neither divisible by 3 nor by 5 ] $=2[\;\frac{100 \times 101}{2}-3(\;\frac{33 \times 34}{2})-5(\;\frac{20 \times 21}{2})+15(\;\frac{6 \times 7}{2})]$ $=10100-3366-2100+630$ $=5264$

Q88JEE Main 2022NAT4MDefinite Integrals

The value of the integral $\;\frac{48}{\pi^4} int_0^\pi(\;\frac{3 \pi x^2}{2}-x^3) \;\frac{\sin x}{1+\cos ^2 x} d x \;\text{ is equal to }\;$ ____ [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$I=\;\frac{48}{\pi^4} int_0^\pi\$ [(;\frac{pi}{2}-x)^{3-};\frac{3 pi^2}{4}(;\frac{pi}{2}-x)+;\frac{pi^3}{4}]$ ;\frac{\sin x d x}{1+\cos ^2 x} $Using$ int_a^b f(x) d x=int_a^b f(a+b-x) d xI=;\frac{48}{\pi^4} int_0^\pi$[-(;\frac{pi}{2}-x)^{3+};\frac{3 pi^4}{4}(;\frac{pi}{2}-x)+;\frac{pi^3}{4}]$ ;\frac{\sin x d x}{1+\cos ^2 x} $Adding these two equations, we get$ 2 I=;\frac{48}{\pi^4} int_0^\pi ;\frac{\pi^3}{2} \cdot ;\frac{\sin x d x}{1+\cos ^2 x}\Rightarrow I=;\frac{12}{\pi}$[-\tan ^{-1}(\cos x)]_0^pi=;\frac{12}{pi} ⋅ ;\frac{pi}{2}=6$

Q89JEE Main 2022NAT4MSequences and Series

Let $A=sum_{i=1}^{10} sum_{j=1}^{10} \min \{i, j\}$ and $B=sum_{i=1}^{10} sum_{j=1}^{10} \max \{i, j\}$ . Then ${\text{A}}+ {\text{B}}$ is equal to [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{A}}=sum_{ {\text{i}}=1}^{10} sum_{ {\text{j}}=1}^{10} \min \{ {\text{i}}, {\text{j}}\}$ ${\text{B}}=sum_{ {\text{i}}=1}^{10} sum_{ {\text{j}}=1}^{10} \max \{ {\text{i}}, {\text{j}}\}$ ${\text{A}}=sum_{ {\text{j}}=1}^{10} \min ( {\text{i}}, 1)+\min ( {\text{j}}, 2)+. . . \min ( {\text{i}}, 10)$ $= {(1+1+1+. . .+1)}_{19 \;\text{ \times }\;}+ {(2+2+2 . . .+2)}_{17 \;\text{ \times }\;}+ {(3+3+3 . . .+3)}_{15 \;\text{ \times }\;}$ $+. . .(1) 1$ times ${\text{B}}=sum_{ {\text{j}}=1}^{10} \max ( {\text{i}}, 1)+\max ( {\text{j}}, 2)+. . . \max ( {\text{i}}, 10)$ $= {(10+10+. . .+10)}_{19 \;\text{ \times }\;}+{(9+9+. . .+9)}_{17 \;\text{ \times }\;}+. . .+11$ times ${\text{A}}+ {\text{B}}=20(1+2+3+. . .+10)$ $=20 \times \;\frac{10 \times 11}{2}=10 \times 110=1100$

Q90JEE Main 2022NAT4MDifferential Equations

Let $S=(0,2 \pi)-\{\;\frac{\pi}{2}, \;\frac{3 \pi}{4}, \;\frac{3 \pi}{2}, \;\frac{7 \pi}{4}\}$ . Let $y=y(x), {\text{x}} \in {\text{S}}$ , be the solution curve of the differential equation $\;\frac{d y}{d x}=\;\frac{1}{1+\sin 2 x}, y(\;\frac{\pi}{4})=\;\frac{1}{2}$ . If the sum of abscissas of all the points of intersection of the curve ${\text{y}}= {\text{y}}( {\text{x}})$ with the curve $y=\sqrt{2} \sin x$ is $\;\frac{k \pi}{12}$ , then ${\text{k}}$ is equal to____ [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{d y}{d x}=\;\frac{1}{1+\sin 2 x}$ $\Rightarrow d y=\;\frac{\sec ^2 x d x}{(1+\tan x)^2}$ $\Rightarrow y=-\;\frac{1}{1+\tan x}+c$ When $x=\;\frac{\pi}{4}, y=\;\frac{1}{2}$ gives ${\text{c}}=1$ So $y=\;\frac{\tan x}{1+\tan x} \Rightarrow y=\;\frac{\sin x}{\sin x+\cos x}$ Now, $y=\sqrt{2} \sin x \Rightarrow \sin x=0$ or $\sin x+\cos x=\;\frac{1}{\sqrt{2}}$ $\sin x=0$ gives $x=\pi$ only.and $\sin x+\cos x=\;\frac{1}{\sqrt{2}} \Rightarrow \sin (x+\;\frac{\pi}{4})=\;\frac{1}{2}$ So $x+\;\frac{\pi}{4}=\;\frac{5 \pi}{6}$ or $\;\frac{13 \pi}{6} \Rightarrow x=\;\frac{7 \pi}{12}$ or $\;\frac{23 \pi}{12}$ Sum of all solutions $=\pi+\;\frac{7 \pi}{12}+\;\frac{23 \pi}{12}=\;\frac{42 \pi}{12}$ Hence, $k=42$ .

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