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25 Jun 2022 Shift 2

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Physics30 questions

Q1JEE Main 2022MCQ4MMotion in a Plane

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.Assertion A: Two identical balls A and B thrown with same velocity 'u' at two different angles with horizontal attained the same range ${\text{R}}$ . IF $A$ and $B$ reached the maximum height $h_1$ and ${\text{h}}_2$ respectively, then $R=4 \sqrt{h_1 h_2}$ Reason R : Product of said heights. $h_1 h_2=(\;\frac{u^2 \sin ^2 \theta}{2 g}) \cdot (\;\frac{u^2 \cos ^2 \theta}{2 g})$ Choose the correct answer : [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$h_1=\;\frac{u^2 \sin ^2 \theta}{2 g}$ $h_2=\;\frac{u^2 \cos ^2 \theta}{2 g}$ $\therefore \sqrt{h_1 h_2}=\;\frac{u^2 \sin \theta \cos \theta}{2 g}$ $=\;\frac{R}{4}$ $\Rightarrow R=4 \sqrt{h_1 h_2}$

Q2JEE Main 2022MCQ4MMotion in a Straight Line

Two buses $P$ and $Q$ start from a point at the same time and move in a straight line and their positions are represented by $X_P(t)=\alpha t+\beta t^2$ and $X_Q(t)=f t-t^2$ At what time, both the buses have same velocity? [25-Jun-2022-Shift-2]

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📖 Explanation

$X_P=\alpha t+\beta t^2$ $X_Q=f t-t^2$ $\therefore V_P=\alpha+2 \beta t$ $V_Q=f-2 t$ $\because V_P=V_Q$ $\Rightarrow \alpha+2 \beta t=f-2 t$ $\Rightarrow t=\;\frac{f-\alpha}{2(1+\beta)}$

Q3JEE Main 2022MCQ4MCircular Motion

A disc with a flat small bottom beaker placed on it at a distance ${\text{R}}$ from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity $\omega$ . The coefficient of static friction between the bottom of the beaker and the surface of the disc is $\mu$ . The beaker will revolve with the disc if : [25-Jun-2022-Shift-2]

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📖 Explanation

To move together $\omega ^2 R \le \mu g$ $\Rightarrow R \le \;\frac{\mu g}{\omega ^2}$

Q4JEE Main 2022MCQ4MProperties of Matter

A solid metallic cube having total surface area $24 {\text{m}}^2$ is uniformly heated. If its temperature is increased by $10^{\circ} {\text{C}}$ , calculate the increase in volume of the cube. $(.$ Given $.\alpha=5.0 \times 10^{-4}{ }^{\circ} {\text{C}}^{-1})$ [25-Jun-2022-Shift-2]

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📖 Explanation

$6 \times l^2=24$ $\Rightarrow l=2 {\text{m}}$ $\therefore \;\frac{\Delta V}{V}=3 \times \;\frac{\Delta l}{l}$ $\Rightarrow \Delta V=3 \times (\alpha \Delta T) \times V$ $=3 \times 5 \times 10^{-4} \times 10 \times (8)$ $=120 \times 10^{-3} {\text{m}}^3$ $=120 \times 10^{-3} \times 10^6 {\text{cm}}^3$ $=1.2 \times 10^5 {\text{cm}}^3$

Q5JEE Main 2022MCQ4MThermodynamics

A copper block of mass $5.0 {\text{kg}}$ is heated to a temperature of $500^{\circ} {\text{C}}$ and is placed on a large ice block. What is the maximum amount of ice that can melt? $[Specific heat of copper : $0.39 {\text{J}} {\text{g}}^{-1}{ }^{\circ} {\text{C}}^{-1}$ and latent heat of fusion of water : $335 {\text{J}} {\text{g}}^{-1}$ ]$ [25-Jun-2022-Shift-2]

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📖 Explanation

$m L=\Delta Q=m s \Delta T$ $\Rightarrow m=\;\frac{5 \times 0.39 \times 10^3 \times 500}{335}$ $=2.9 {\text{kg}}$

Q6JEE Main 2022MCQ4MThermodynamics

The ratio of specific heats $(\;\frac{C_P}{C_V})$ in terms of degree of freedom (f) is given by : [25-Jun-2022-Shift-2]

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📖 Explanation

$\;\frac{C_P}{C_V}=\gamma$ $C_V=(\;\frac{f}{2}) R$ and $C_P-C_V=R$ $\Rightarrow \;\frac{C_P}{C}=\;\frac{1+f / 2}{f / 2}=1+\;\frac{2}{f}$

Q7JEE Main 2022MCQ4MCircular Motion

For a particle in uniform circular motion, the acceleration $\vec{a}$ at any point $P(R, \theta)$ on the circular path of radius ${\text{R}}$ is (when $\theta$ is measured from the positive $x$ -axis and $v$ is uniform speed) : [25-Jun-2022-Shift-2]

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📖 Explanation

As the particle in uniform circular motion experiences only centripetal acceleration of magnitude $\omega ^2 {\text{R}}$ or $\;\frac{v^2}{R}$ directed towards centre so from diagram, $\vec{a}=\;\frac{v^2}{R} \cos \theta(-\hat{i})+\;\frac{v^2}{R} \sin (-\hat{j})$

Q8JEE Main 2022MCQ4MElectrostatic Potential and Capacitance

Two metallic plates form a parallel plate capacitor. The distance between the plates is ' $d$ '. A metal sheet of thickness $\;\frac{d}{2}$ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor? [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$C_{e q}=\;\frac{\epsilon_0 A}{d-\;\frac{d}{2}+\;\frac{d}{2 k}}=\;\frac{\epsilon_0 A}{\frac{d}{2}}=\;\frac{2 \epsilon_0 A}{d}$ If $C=\;\frac{\epsilon_0 A}{d}$ $\Rightarrow C_{e q}=2 C \;\text{ or }\; \;{C_{\;\text{new }\;}}/{C_{\;\text{old }\;}}=\;\frac{2}{1}$

Q9JEE Main 2022MCQ4MCurrent Electricity

Two cells of same emf but different internal resistances $r_1$ and $r_2$ are connected in series with a resistance $R$ . The value of resistance ${\text{R}}$ , for which the potential difference across second cell is zero, is : [25-Jun-2022-Shift-2]

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📖 Explanation

$I=\;\frac{2 \epsilon}{R+r_1+r_2}$ As per the question, $\;\frac{2 \epsilon}{R+r_1+r_2} \times r_2-\epsilon=0$ $\Rightarrow R=r_2-r_1$

Q10JEE Main 2022MCQ4MMagnetism and Matter

Given below are two statements :Statement I : Susceptibilities of paramagnetic and ferromagnetic substances increase with decrease in temperature.Statement II : Diamagnetism is a result of orbital motions of electrons developing magnetic moments opposite to the applied magnetic field.Choose the correct answer from the options given below : [25-Jun-2022-Shift-2]

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📖 Explanation

Statement $I$ is true as susceptibility of ferromagnetic and paramagnetic materials is inversely related to temperature.Statement II is true as because of orbital motion of electrons the diamagnetic material is able to oppose external magnetic field.

Q11JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

A long solenoid carrying a current produces a magnetic field ${\text{B}}$ along its axis. If the current is doubled and the number of turns per ${\text{cm}}$ is halved, the new value of magnetic field will be equal to [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$B=\mu _0 n i$ Now $i \to 2 i$ And $n \to \;\frac{n}{2}$ $B^{'}=\mu _0 \;\frac{n}{2} \times 2 i=\mu _0 n i=B$

Q12JEE Main 2022MCQ4MAlternating Current

A sinusoidal voltage $V(t)=210 \sin 3000 {\text{t}}$ volt is applied to a series $L C R$ circuit in which $L=10 {\text{mH}}, {\text{C}}=25 \mu {\text{F}}$ and ${\text{R}}$ $=100 Ω$ . The phase difference $(Φ)$ between the applied voltage and resultant current will be : [25-Jun-2022-Shift-2]

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📖 Explanation

$X_L=3000 \times 10 \times 10^{-3}=30 Ω$ $X_C=\;\frac{1}{3000 \times 25} \times 10^6=\;\frac{40}{3} Ω$ So $X_L-X_C=30-\;\frac{40}{3}=\;\frac{50}{3} Ω$ $\tan \theta=\;\frac{X_L-X_C}{R}=\;\frac{50 / 3}{100}=\;\frac{1}{6}$ So $\theta=\tan ^{-1}(0.17)$

Q13JEE Main 2022MCQ4MElectromagnetic Waves

The electromagnetic waves travel in a medium at a speed of $2.0 \times 10^8 {\text{m}} / {\text{s}}$ . The relative permeability of the medium is $1.0$ . The relative permittivity of the medium will be : [25-Jun-2022-Shift-2]

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📖 Explanation

$n=\;\frac{c}{v}=\;\frac{3}{2}$ $\sqrt{\in \mu }=n$ So $\in =\;\frac{9}{4}=2.25$

Q14JEE Main 2022MCQ4MWave Optics

The interference pattern is obtained with two coherent light sources of intensity ratio $4: 1$ . And the ratio $\;\frac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }}$ is $\;\frac{5}{x}$ . Then, the value of $x$ will be equal to : [25-Jun-2022-Shift-2]

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📖 Explanation

$\;\frac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }}=\;\frac{I_1+I_2+2 \sqrt{I_1 I_2}+I_1+I_2-2 \sqrt{I_1 I_2}}{I_1+I_2+2 \sqrt{I_1 I_2}-I_1-I_2+2 \sqrt{I_1 I_2}}$ $=\;\frac{2(I_1+I_2)}{4 \sqrt{I_1 I_2}}$ $=\;{(\frac{I_1}{I_2}+1)}/{2 \sqrt{\;\frac{I_1}{I_2}}}=\;\frac{4+1}{2 \times 2}=\;\frac{5}{4}$ So $x=4$

Q15JEE Main 2022MCQ4MWave Optics

A light whose electric field vectors are completely removed by using a good polaroid, allowed to incident on the surface of the prism at Brewster's angle. Choose the most suitable option for the phenomenon related to the prism. [25-Jun-2022-Shift-2]

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Q16JEE Main 2022MCQ4MDual Nature of Matter and Radiation

A proton, a neutron, an electron and an $\alpha$ particle have same energy. If $\lambda _{ {\text{p}}}, \lambda _{ {\text{n}}}, \lambda _{ {\text{e}}}$ and $\lambda _{ {\text{a}}}$ are the de Broglie's wavelengths of proton, neutron, electron and $\alpha$ particle respectively, then choose the correct relation from the following: [25-Jun-2022-Shift-2]

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📖 Explanation

de Broglie wavelength $\lambda =\;\frac{h}{p}$ $\Rightarrow \lambda =\;\frac{h}{\sqrt{2 m K}}$ Where ${\text{K}}$ : kinetic energy $\Rightarrow$ For same ${\text{K}}, \lambda \propto \;\frac{1}{\sqrt{m}}$ Since $m_alpha>m_n>m_p>m_e$ $\Rightarrow \lambda _alpha< \lambda _n< \lambda _p< \lambda _e$

Q17JEE Main 2022MCQ4MAtoms and Nuclei

Which of the following figure represents the variation of $l_n(\;\frac{R}{R_0})$ with $l_n A$ (if ${\text{R}}=$ radius of a nucleus and ${\text{A}}=$ its mass number) [25-Jun-2022-Shift-2]

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📖 Explanation

We know that $R=R_0 A^{1 / 3}$ $\Rightarrow {\ln (\;\frac{R}{R_0})}_{╰──▾──╯}_{y}={1/3}_{⏟}_{m} {\;\;\ln (A)}_{⏟}_{x}$ $\Rightarrow$ Straight line

Q18JEE Main 2022MCQ4MSemiconductor Electronics

Identify the logic operation performed by the given circuit: [25-Jun-2022-Shift-2]

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📖 Explanation

According to the circuit, $Y=(A^{'}+B^{'})^{'}$ $\Rightarrow Y=A B$ $\Rightarrow \;\text{ AND gate }\;$

Q19JEE Main 2022MCQ4MElectromagnetic Waves

List - I List -II A. Facsimile I. Static Document Image B. Guided media Channel II. Local Broadcast Radio C. Frequency Modulation III. Rectangular wave D. Digital Signal IV. Optical Fiber Choose the correct answer from the following options: [25-Jun-2022-Shift-2]

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Q20JEE Main 2022MCQ4MCurrent Electricity

If ${\text{n}}$ represents the actual number of deflections in a converted galvanometer of resistance ${\text{G}}$ and shunt resistance ${\text{S}}$ . Then the total current I when its figure of merit is $K$ will be: [25-Jun-2022-Shift-2]

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📖 Explanation

$\Rightarrow \;\frac{S}{S+G} i=n K$ $\Rightarrow i=\;\frac{n K(S+G)}{S}$

Q21JEE Main 2022NAT4MUnits and Measurements

For $z=a^2 x^3 y^{\;\frac{1}{2}}$ , where 'a' is a constant. If percentage error in measurement of ' $x$ ' and ' $y$ ' are $4 \%$ and $12 \%$ respectively, then the percentage error for 'z' will be ______ $\%$ [25-Jun-2022-Shift-2]

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📖 Explanation

$\%$ error in $z=3 \times 4+\;\frac{1}{2} \times 12$ $=12+6=18 \%$

Q22JEE Main 2022NAT4MCircular Motion

A curved in a level road has a radius $75 {\text{m}}$ . The maximum speed of a car turning this curved road can be $30 {\text{m}} / {\text{s}}$ without skidding. If radius of curved road is changed to $48 {\text{m}}$ and the coefficient of friction between the tyres and the road remains same, then maximum allowed speed would be ____ ${\text{m}} / {\text{s}}$ [25-Jun-2022-Shift-2]

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📖 Explanation

$\because v=\sqrt{\mu g r}$ $\Rightarrow \;\frac{v_1}{v_2}=\sqrt{\;\frac{r_1}{r_2}}$ $\Rightarrow \;\frac{30}{v_2}=\sqrt{\;\frac{75}{48}}=\sqrt{\;\frac{25}{16}}=\;\frac{5}{4}$ $\Rightarrow V_2=24 {\text{m}} / {\text{s}}$

Q23JEE Main 2022NAT4MLaws of Motion

A block of mass $200 {\text{g}}$ is kept stationary on a smooth inclined plane by applying a minimum horizontal force ${\text{F}}=$ $\sqrt{x} {\text{N}}$ as shown in figure.The value of $x=____________$ [25-Jun-2022-Shift-2]

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📖 Explanation

$F \times \;\frac{1}{2}=0.2 \times 10 \times \;\frac{\sqrt{3}}{2}$ $\Rightarrow F=2 \sqrt{3}$ $\Rightarrow F=\sqrt{12} {\text{N}}$ $\therefore x=12$

Q24JEE Main 2022NAT4MRotational Motion

Moment of Inertia (M.I.) of four bodies having same mass 'M' and radius ' $2 R^{'}$ are as follows: $I_1=$ M.I. of solid sphere about its diameter ${\text{I}}_2=$ M.I. of solid cylinder about its axis $I_3=$ M.I. of solid circular disc about its diameter ${\text{I}}_4=$ M.I. of thin circular ring about its diameterIf $2(I_2+I_3)+I_4=x . I_1$ , then the value of $x$ will be [25-Jun-2022-Shift-2]

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📖 Explanation

$2(\;\frac{1}{2}+\;\frac{1}{4}) \times M(2 R)^{2+}\;\frac{1}{2} M(2 R)^2=x \;\frac{2}{5} M(2 R)^2$ $\Rightarrow 1+\;\frac{1}{2}+\;\frac{1}{2}=x \times \;\frac{2}{5}$ $\Rightarrow x=5$

Q25JEE Main 2022NAT4MGravitation

Two satellites $S_1$ and $S_2$ are revolving in circular orbits around a planet with radius $R_1=3200 {\text{km}}$ and $R_2=800 {\text{km}}$ respectively. The ratio of speed of satellite $S_1$ to be speed of satellite $S_2$ in their respective orbits would be $\;\frac{1}{x}$ where $x=$ [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$v=\sqrt{\;\frac{G M}{R}}$ $\Rightarrow \;\frac{v_1}{v_2}=\sqrt{\;\frac{R_2}{R_1}}$ $\;\frac{v_2}{v_1}=\sqrt{\;\frac{3200}{800}}=2$ $\Rightarrow \;\frac{v_1}{v_2}=\;\frac{1}{2}$ $x=2$

Q26JEE Main 2022NAT4MThermodynamics

When a gas filled in a closed vessel is heated by raising the temperature by $1^{\circ} {\text{C}}$ , its pressure increases by $0.4 \%$ . The initial temperature of the gas is ____ ${\text{K}} .$ [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$P V=n R T$ So $\;\frac{d P}{P} \times 100=\;\frac{d T}{T} \times 100$ $0.4=\;\frac{1}{T} \times 100$ $\Rightarrow T=250 {\text{K}}$

Q27JEE Main 2022NAT4MElectrostatic Potential and Capacitance

27 identical drops are charged at $22 {\text{V}}$ each. They combine to form a bigger drop. The potential of the bigger drop will be_____ V. [25-Jun-2022-Shift-2]

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📖 Explanation

Let the charge on one drop is $q$ and its radius is $r$ .So for one drop $V=\;\frac{k q}{r}$ For 27 drops merged new charge will be $Q=27 q$ and new radius is $R=3 r$ So new potential is $V^{'}=\;\frac{k Q}{R}=9 \;\frac{k q}{r}=9 \times 22 {\text{V}}$ $=198 {\text{V}}$

Q28JEE Main 2022NAT4MCurrent Electricity

The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be _____ $\%$ . [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Volume is constant so on length doubledArea is halved so $R=\rho \;\frac{l}{A}$ and $R^{'}=\rho \;\frac{2 l}{\frac{A}{2}}=4 \rho \;\frac{l}{A}=4 R$ So percentage increase will be $R \%=\;\frac{4 R-R}{R} \times 100=300 \%$

Q29JEE Main 2022NAT4MAlternating Current

In a series ${\text{LCR}}$ circuit, the inductance, capacitance and resistance are ${\text{L}}=100 {\text{mH}}, {\text{C}}=100 \mu {\text{F}}$ and ${\text{R}}=10 Ω$ respectively. They are connected to an ${\text{AC}}$ source of voltage $200 {\text{V}}$ and frequency of $50 {\text{Hz}}$ . The approximate value of current in the circuit will be A. [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$Z=\sqrt{R^{2+}(x_L+x_C)^2}$ $=\sqrt{10^{2+}[10 \pi-\;\frac{100}{\pi}]^2} Ω$ $≃ 10 Ω$ $\Rightarrow$ Current $=\;\frac{220}{10} {\text{A}}=22 {\text{A}}$

Q30JEE Main 2022NAT4MSemiconductor Electronics

In an experiment of CE configuration of $n-p-n$ transistor, the transfer characteristics are observed as given in figure.If the input resistance is $200 Ω$ and output resistance is $60 Ω$ , the voltage gain in this experiment will be_____ [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Voltage gain $=\;\frac{I_C R_0}{I_B R_i}$ $=\;\frac{(10 m A)(60 Ω)}{(200 \mu A)(200 Ω)}$ $\Rightarrow$ Voltage gain $=15$

Chemistry30 questions

Q31JEE Main 2022MCQ4MStructure of Atom

The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is :$[Given: The threshold frequency of platinum is $1.3 × 10^{15} {\text{s}}^{-1}$ and ${\text{h}}=6.6 × 10^{-34} {\text{J}} {\text{s}}$ .]$ [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

The minimum energy possessed by photons will be equal to the work function of the metal. $\therefore {\text{E}}_{\min }= {\text{h}} ν_0 {\text{J}}$ $=6.6 \times 10^{-34} \times 1.3 \times 10^{15}$ $=8.58 \times 10^{-19} {\text{J}}$

Q32JEE Main 2022MCQ4MChemical Thermodynamics

At $25^{\circ} {\text{C}}$ and $1 {\text{atm}}$ pressure, the enthalpy of combustion of benzene (I) and acetylene ( $. {\text{g}})$ are $-3268 {\text{kJ}} {\text{mol}}^{-1}$ and $-1300 {\text{kJ}}$ $\text{mol}^{-1}$ , respectively. The change in enthalpy for the reaction $3 {\text{C}}_2 {\text{H}}_2( {\text{g}}) \to {\text{C}}_6 {\text{H}}_6( {\text{I}})$ , is [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\Delta {\text{H}}=Σ \Delta {\text{H}}_{\;\text{Combustion }\;} \;\;$ (Reactant) $-Σ \Delta {\text{H}}_{\;\text{Combustion }\;}$ (Product) $=3 \times (-1300)-[-3268]$ $=-632 {\text{kJ}} {\text{mol}}^{-1}$

Q33JEE Main 2022MCQ4MSolutions

Solute A associates in water. When $0.7 {\text{g}}$ of solute ${\text{A}}$ is dissolved in $42.0 {\text{g}}$ of water, it depresses the freezing point by $0.2^{\circ} {\text{C}}$ . The percentage association of solute A in water, is :$[Given : Molar mass of $A=93 {\text{g}} {\text{mol}}^{-1}$ . Molal depression constant of water is $1.86 {\text{K}} {\text{kg}} {\text{mol}}^{-1}$ .]$ [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\Delta {\text{T}}= {\text{i}} {\text{k}}_{ {\text{f}}} \times {\text{m}}$ $0.2= {\text{i}} \times 1.86 \times \;\frac{0.7}{93} \times \;\frac{1000}{42}$ $i=\;\frac{0.2 \times 93 \times 6}{1.86 \times 100}$ ${\text{i}}=0.60$ $2 {\text{A}} ⇌ {\text{A}}_2$ $1-\alpha \;\; \;\frac{\alpha}{2}$ ${\text{i}}=1-\alpha+\;\frac{\alpha}{2}$ ${\text{i}}=1-\;\frac{\alpha}{2}$ $1-\;\frac{\alpha}{2}=0.60$ $1-0.60=\;\frac{\alpha}{2}$ $\alpha=0.80$

Q34JEE Main 2022MCQ4MIonic Equilibrium

The ${\text{K}}_{\text{sp}}$ for bismuth sulphide $(\text{Bi}_2 {\text{S}}_3)$ is $1.08 \times 10^{-73}$ . The solubility of ${\text{Bi}}_2 {\text{S}}_3$ in ${\text{mol}} {\text{L}}^{-1}$ at $298 {\text{K}}$ is [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{Bi}}_2 {\text{S}}_3 ⇌ \;{2 {\text{Bi}}^{3+}}_{2 {\text{s}}}+\;{3 {\text{S}}^{2-}}_{3 {\text{s}}}$ ${\text{K}}_{ { sp}}=(2 { s})^2(3 { s})^3=108 { s}^5$ $108 { s}^5=108 \times 10^{-75}$ ${ s}=1.0 \times 10^{-15} {\text{mol}} / {\text{L}}$

Q35JEE Main 2022MCQ4MBiomolecules

Match List I with List II List - I List -II A. Zymase I. Stomach B. Diastase II. Yeast C. Urease III. Malt D. Pepsin IV. Soyabean Choose the correct answer from the options given below: [25-Jun-2022-Shift-2]

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Q36JEE Main 2022MCQ4MClassification of Elements

The correct order of electron gain enthalpies of ${\text{Cl}}, {\text{F}}$ , Te and ${\text{Po}}$ is [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{Te}} \to -190 {\text{kJ}} {\text{mol}}^{-1}$ ${\text{Po}} \to -174 {\text{kJ}} {\text{mol}}^{-1}$ ${\text{F}} \to -333 {\text{kJ}} {\text{mol}}^{-1}$ ${\text{Cl}} \to -349 {\text{kJ}} {\text{mol}}^{-1}$ Hence, correct order is ${\text{Cl}}> {\text{F}}> {\text{Te}}> {\text{Po}}$

Q37JEE Main 2022MCQ4MElectrochemistry

Given below are two statements.Statement I : During electrolytic refining, blister copper deposits precious metals.Statement II : In the process of obtaining pure copper by electrolysis method, copper blister is used to make the anode.In the light of the above statements, choose the correct from the options given below. [25-Jun-2022-Shift-2]

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Q38JEE Main 2022MCQ4MIonic Equilibrium

Given below are two statements one is labelled as Assertion A and the other is labelled as Reason R :Assertion A : The amphoteric nature of water is explained by using Lewis acid/base concept.Reason R: Water acts as an acid with ${\text{NH}}_3$ and as a base with ${\text{H}}_2 {\text{S}}$ .In the light of the above statements choose the correct answer from the options given below : [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

The amphoteric nature of water is explained by using Bronsted-Lowry acid base concept $\;{ {\text{H}}_2 {\text{O}}}_{\;\text{ (acid) }\;}+ {\text{NH}}_3- {\text{OH}}^{-}+ {\text{NH}}_4^{+}$ $\;{ {\text{H}}_2 {\text{O}}}_{\;\text{ (base) }\;}+ {\text{H}}_2 {\text{S}} \;\; {\text{H}}_3 {\text{O}}^{+}+ {\text{HS}}^{-}$ Hence, ${\text{A}}$ is false but ${\text{R}}$ is true

Q39JEE Main 2022MCQ4MElectrochemistry

The correct order of reduction potentials of the following pairs isA. ${\text{Cl}}_2 / {\text{Cl}}^{-}$ B. $1^2 / 1^{-}$ C. ${\text{Ag}}^{+} / {\text{Ag}}$ D. ${\text{Na}}^{+} / {\text{Na}}$ E. ${\text{Li}}^{+} / {\text{Li}}$ Choose the correct answer from the options given below. [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{E}}_{ {\text{C}}_2 / {\text{CI}}}^{\circ}=+1.36 {\text{V}}$ ${\text{E}}_{ {\text{I}}_2 / {\text{I}}^{-}}^0=+0.54 {\text{V}}$ ${\text{E}}_{ {\text{Ag}}^{+} / {\text{Ag}}}^{\circ}=+0.80 {\text{V}}$ ${\text{E}}_{ {\text{Na}}^{+} / {\text{Na}}}^{\circ}=-2.71 {\text{V}}$ ${\text{E}}_{ {\text{L}}^{+} / {\text{Li}}}=-3.05 {\text{V}}$

Q40JEE Main 2022MCQ4Mp Block Elements

The number of bridged oxygen atoms present in compound B formed from the following reactions is ${\text{Pb}}( {\text{NO}}_3)_2 {\to }^{673 {\text{K}}} {\text{A}}+ {\text{PbO}}+ {\text{O}}_2$ ${\text{A}} {\to }^{\;\text{ Dimerise }\;} {\text{B}}$ [25-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$2 {\text{Pb}}( {\text{NO}}_3)_2 {\longrightarrow }^{673 {\text{K}}} 4 {\text{AO}}_2+2 {\text{PbO}}+ {\text{O}}_2$ Hence no bridged oxygen atom is present in ${\text{N}}_2 {\text{O}}_4$ Hence no bridged oxygen atom is present in ${\text{N}}_2 {\text{O}}_4$

Q41JEE Main 2022MCQ4Md and f Block Elements

The metal ion (in gaseous state) with lowest spin-only magnetic moment value is [25-Jun-2022-Shift-2]

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Q42JEE Main 2022MCQ4MBiomolecules

Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R.Assertion A: Polluted water may have a value of $B O D$ of the order of $17 {\text{ppm}}$ .Reason ${\text{R}}$ : BOD is a measure of oxygen required to oxidise both the bio-degradable an nonbiodegradable organic material in water.In the light of the above statements, choose the most appropriate answer from the options given below. [25-Jun-2022-Shift-2]

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Q43JEE Main 2022MCQ4MPractical Organic Chemistry

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.Assertion A : A mixture contains benzoic acid and napthalene. The pure benzoic acid can be separated out by the use of benzene.Reason ${\text{R}}$ : Benzoic acid is soluble in hot water.In the light of the above statements, choose the most appropriate answer from the options given below. [25-Jun-2022-Shift-2]

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Q44JEE Main 2022MCQ4MPractical Organic Chemistry

During halogen test, sodium fusion extract is boiled with concentrated ${\text{HNO}}_3$ to [25-Jun-2022-Shift-2]

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Q45JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

Amongst the following, the major product of the given chemical reaction is [25-Jun-2022-Shift-2]

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Q46JEE Main 2022MCQ4MHaloalkanes and Haloarenes

In the given reaction 'A' can be [25-Jun-2022-Shift-2]

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Q47JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Which of the following conditions or reaction sequence will NOT give acetophenone as the major product? [25-Jun-2022-Shift-2]

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Q48JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

The major product formed in the following reaction, is [25-Jun-2022-Shift-2]

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Q49JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Which of the following ketone will NOT give enamine on treatment with secondary amines? [where t-Bu is $. {\text{C}}( {\text{CH}}_3)_3]$ [25-Jun-2022-Shift-2]

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Q50JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

An antiseptic dettol is a mixture of two compounds ' $A$ ' and ' $B$ ' where A has $6 \pi$ electrons and $B$ has $2 \pi$ electrons. What is 'B'? [25-Jun-2022-Shift-2]

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📖 Explanation

Dettol is mixture of Chloroxylenol (Compound A) It has $6 \pi {\text{e}}^{-}$ and Terpineol (Compound B) It has $2 \pi {\text{e}}^{-}$

Q51JEE Main 2022NAT4MBiomolecules

A protein ' $A$ ' contains $0.30 \%$ of glycine (molecular weight 75 ). The minimum molar mass of the protein ' $A$ ' is ____ $\times 10^3 {\text{g}} {\text{mol}}^{-1}$ [nearest integer] [25-Jun-2022-Shift-2]

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📖 Explanation

Let, molar mass of protein ${\text{A}}= {\text{x}}$ Protein A contains $0.30 \%$ glycine $\therefore \;\frac{x \times 0.3}{100}=75$ $\Rightarrow {\text{x}}=25000=25 \times 10^3$

Q52JEE Main 2022NAT4MStates of Matter

A rigid nitrogen tank stored inside a laboratory has a pressure of $30 {\text{atm}}$ at $06: 00 {\text{am}}$ when the temperature is $27^{\circ} {\text{C}}$ . At 03:00 pm, when the temperature is $45^{\circ}$ , the pressure in the tank will be ____atm. [nearest integer] [25-Jun-2022-Shift-2]

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📖 Explanation

A nitrogen tank of fixed volume used where number of moles of nitrogen is fixed. $\therefore {\text{V}}=$ constant $n=$ constant $R=$ constantFrom ideal gas equation, ${\text{PV}}= {\text{nRT}}$ $\Rightarrow {\text{P}} \propto {\text{T}}[ {\text{As}} {\text{V}}, {\text{n}}, {\text{R}}=$ constant $]$ Here, initially $P_1=30$ atm, $T_1=300 {\text{K}}$ Finally, $P_2=?, T_2=318 {\text{K}}$ $\therefore \;\frac{P_1}{P_2}=\;\frac{T_1}{T_2}$ $\Rightarrow \;\frac{30}{P_2}=\;\frac{300}{318}$ $\Rightarrow P_2=31.8 ≃ 32$

Q53JEE Main 2022NAT4MChemical Bonding and Molecular Structure

Amongst ${\text{BeF}}_2, {\text{BF}}_3, {\text{H}}_2 {\text{O}}, {\text{NH}}_3, {\text{CCl}}_4$ and ${\text{HCl}}$ , the number of molecules with non-zero net dipole moment is____ [25-Jun-2022-Shift-2]

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📖 Explanation

${\text{BeF}}_2, {\text{BF}}_3$ and ${\text{CCl}}_4 \Rightarrow \mu _{\;\text{net }\;}=0$ ${\text{H}}_2 {\text{O}}, {\text{NH}}_3$ and ${\text{HCl}} \Rightarrow \mu _{\;\text{net }\;} \ne 0$

Q54JEE Main 2022NAT4MChemical Kinetics

At $345 {\text{K}}$ , the half life for the decomposition of a sample of a gaseous compound initially at $55.5 {\text{kPa}}$ was 340 s. When the pressure was $27.8 {\text{kPa}}$ , the half life was found to be $170 {\text{s}}$ . The order of the reaction is _____. [integer answer] [25-Jun-2022-Shift-2]

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📖 Explanation

${\text{t}}_{1 / 2} \times \;{1}/{[ {\text{P}}_0]^{ {\text{n}}-1}}$ $\;{ {\text{t}}_1}/{ {\text{t}}_2}=\;{( {\text{P}}_2)^{ {\text{n}}-1}}/{( {\text{P}}_1)^{ {\text{n}}-1}}$ $\;\frac{340}{170}=(\;\frac{27.8}{55.5})^{ {\text{n}}-1}$ $\Rightarrow 2=\;{1}/{(2)^{ {\text{n}}-1}}$ ${\text{n}}=0$

Q55JEE Main 2022NAT4MElectrochemistry

A solution of ${\text{Fe}}_2( {\text{SO}}_4)_3$ is electrolyzed for ' $x$ ' min with a current of $1.5 {\text{A}}$ to deposit $0.3482 {\text{g}}$ of Fe. The value of $x$ is [nearest integer]Given : $1 {\text{F}}=96500 {\text{C}} {\text{mol}}^{-1}$ Atomic mass of ${\text{Fe}}=56 {\text{g}} {\text{mol}}^{-1}$ [25-Jun-2022-Shift-2]

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📖 Explanation

${\text{Fe}}^{3+}+3 {\text{e}}^{-} \longrightarrow {\text{Fe}}$ $3 {\text{F}} \longrightarrow 1$ mole Fe is deposited For $56 {\text{g}} \longrightarrow 3 \times 96500$ (required charge) For $\text{lg} \longrightarrow \;\frac{3 \times 96500}{56}$ (required charge) For $0.3482 {\text{g}} \longrightarrow \;\frac{3 \times 96500}{56} \times 0.3482$ $=1800.06$ ${\text{Q}}= {\ it}$ $1800.06=1.5 {\text{t}}$ ${\text{t}}=20 {\min}$

Q56JEE Main 2022NAT4Mp Block Elements

Consider the following reactions: ${\text{PCl}}_3+ {\text{H}}_2 {\text{O}} \to {\text{A}}+ {\text{HCl}}$ ${\text{A}}+ {\text{H}}_2 {\text{O}} \to {\text{B}}+ {\text{HCl}}$ The number of ionisable protons present in the product $B$ is [25-Jun-2022-Shift-2]

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📖 Explanation

${\text{PCl}}_3+ {\text{H}}_2 {\text{O}} \;{\to }^{\text{Pattial}}_{\text{hydrolysis}} {\text{PCl}}_2( {\text{OH}}) \;\text{ (or) }\; {\text{PCl}}( {\text{OH}})_2+$ ${\text{HCl}}$ no. of ionisable protons in ${\text{B}}=2$

Q57JEE Main 2022NAT4MCoordination Compounds

Amongst ${\text{FeCl}}_3 \cdot 3 {\text{H}}_2 {\text{O}}, {\text{K}}_3[ {\text{Fe}}( {\text{CN}})_6]$ and $[ {\text{Co}}( {\text{NH}}_3)_6] {\text{Cl}}_3$ , the spin-only magnetic moment value of the innerorbital complex that absorbs light at shortest wavelength is_____ B.M. [nearest integer] [25-Jun-2022-Shift-2]

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📖 Explanation

$[ {\text{Fe}}( {\text{H}}_2 {\text{O}})_3 {\text{Cl}}_3], \;\; { {\text{K}}_3[ {\text{Fe}}( {\text{CN}})_6],[ {\text{Co}}( {\text{NH}}_3)_6] {\text{Cl}}_3}_{{\;\text{inner orbital complexes }\;}$ ${\text{K}}_3[ {\text{Fe}}( {\text{CN}})_6]$ has more value of $\Delta _0$ than that of $[ {\text{Co}}( {\text{NH}}_3)_6] {\text{Cl}}_3 ;$ as $\overline{ {\text{C}}} {\text{N}}$ is stronger ligand.More $\Delta _0 \Rightarrow$ smaller value of absorbed $\lambda$ Spin only magnetic moment $(\mu )=\sqrt{3} {\text{BM}}$ $=1.732 {\text{BM}}$ Rounding off $\Rightarrow 2$

Q58JEE Main 2022NAT4MOrganic Chemistry - Some Basic Principles

The Novolac polymer has mass of $963 {\text{g}}$ . The number of monomer units present in it are [25-Jun-2022-Shift-2]

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📖 Explanation

Monomer unit of Novolac is its molecular mass is 124 amu.Upon considering molecular weight of polymer as 963 amu (In question its given as 963 gram) Now if during formation of Novolac, $( {\text{n}}-1)$ unit of water are removed then ${\text{n}} \times 124=963+[18 \times ( {\text{n}}-1)]$ ${\text{n}}=9$

Q59JEE Main 2022NAT4MBiomolecules

How many of the given compounds will give a positive Biuret test ?Glycine, Glycylalanine, Tripeptide, Biuret [25-Jun-2022-Shift-2]

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Q60JEE Main 2022NAT4MIonic Equilibrium

The neutralization occurs when $10 {\text{mL}}$ of $0.1 {\text{M}}$ acid 'A' is allowed to react with $30 {\text{mL}}$ of $0.05 {\text{M}}$ base ${\text{M}}( {\text{OH}})_2$ . The basicity of the acid ' $A$ ' is[{\text{M}}$ is a metal] [25-Jun-2022-Shift-2]

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📖 Explanation

${\text{Acid}}_{0.1 \text{M}}_{10ml} + {\text{Base}}_{\text{M}(\text{OH})_2}_{0.05 \text{M}}_{30 \text{ml}} \longrightarrow \text{Salt} + {\text{H}}_2 {\text{O}}$ at equivalence point equivalent of acid $=$ equivalent of base $0.1 \times 10 \times {\text{n}}=30 \times 0.05 \times 2$ ${\text{n}}=3$

Mathematics30 questions

Q61JEE Main 2022MCQ4MSets and Relations

Let $A=\{x \in R:| x+1|< 2\}$ and $B=\{x \in R:| x-1| \ge 2\}$ . Then which one of the following statements is NOT true? [25-Jun-2022-Shift-2]

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📖 Explanation

$A=(-3,1)$ and $B=(-\infty ,-1] ∪[3, \infty )$ So, $A-B=(-1,1)$ $B-A=(-\infty ,-3] ∪[3, \infty )=R-(-3,3)$ $A ∩ B=(-3,-1]$ and $A ∪ B=(-\infty , 1) ∪[3, \infty )=R-[1,3)$ So Option B is not True

Q62JEE Main 2022MCQ4MQuadratic Equation and Inequalities

Let ${\text{a}}, {\text{b}} \in {\text{R}}$ be such that the equation $a x^{2-}2 b x+15=0$ has a repeated root $\alpha$ . If $\alpha$ and $\beta$ are the roots of the equation $x^{2-}2 b x+21=0$ , then $\alpha^{2+}\beta^2$ is equal to : [25-Jun-2022-Shift-2]

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📖 Explanation

$a x^{2-}2 b x+15=0$ has repeated root so $b^2=15 a$ and $\alpha=\;\frac{15}{b}$ $\because \alpha$ is a root of $x^{2-}2 b x+21=0$ So $\;\frac{225}{b^2}=9 \Rightarrow b^2=25$ Now $\alpha^{2+}\beta^2=(\alpha+\beta)^{2-}2 \alpha \beta=4 b^{2-}42=100-42=58$

Q63JEE Main 2022MCQ4MComplex Numbers

Let ${\text{z}}_1$ and ${\text{z}}_2$ be two complex numbers such that $\overline{z}_1=i \overline{z}_2$ and $\arg (\;\frac{z_1}{\overline{z}_2})=\pi$ . Then [25-Jun-2022-Shift-2]

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📖 Explanation

$\because \;\frac{z_1}{z_2}=-i \Rightarrow z_1=-i z_2$ $\Rightarrow \arg (z_1)=-\;\frac{\pi}{2}+\arg (z_2) . . . . . . \;\text{ (i) }\;$ Also $\arg (z_1)-\arg (\overline{z}_2)=\pi$ $\Rightarrow \arg (z_1)+\arg (z_2)=\pi . . . . \;\text{ (ii) }\;$ From (i) and (ii), we get $\arg (z_1)=\;\frac{\pi}{4}$ and $\arg (z_2)=\;\frac{3 \pi}{4}$

Q64JEE Main 2022MCQ4MMatrices and Determinants

The system of equations $-k x+3 y-14 z=25$ $-15 x+4 y-k z=3$ $-4 x+y+3 z=4$ is consistent for all ${\text{k}}$ in the set [25-Jun-2022-Shift-2]

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📖 Explanation

The system may be inconsistent if $\begin{vmatrix}-k & 3 & -14 \\ -15 & 4 & -k \\ -4 & 1 & 3\end{vmatrix}=0 \Rightarrow k=\pm 11$ Hence if system is consistent then $k \in R-\{11,-11\}$ .

Q65JEE Main 2022MCQ4MLimits, Continuity and Differentiability

$\lim _{x \to \;\frac{\pi}{2}}(\tan ^2 x((2 \sin ^2 x+3 \sin x+4)^{\;\frac{1}{2}}-(\sin ^2 x+6 \sin x+2)^{\;\frac{1}{2}}))$ is equal to [25-Jun-2022-Shift-2]

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📖 Explanation

$\lim _{x \to \;\frac{\pi}{2}} \tan ^2 x\{\sqrt{2 \sin ^2 x+3 \sin x+4}-\sqrt{\sin ^2 x+6 \sin x+2}\}$ $=\lim _{x \to \;\frac{\pi}{2}} \;\frac{\tan ^2 x(\sin ^2 x-3 \sin x+2)}{\sqrt{2 \sin ^2 x+3 \sin x+4}+\sqrt{\sin ^2 x+6 \sin x+2}}$ $=\;\frac{1}{6} \lim _{x \to \;\frac{\pi}{2}} \;\frac{(1-\sin x)(2-\sin x)}{\cos ^2 x} \cdot \sin ^2 x$ $=\;\frac{1}{6} \lim _{x \to \;\frac{\pi}{2}} \;\frac{(2-\sin x) \sin ^2 x}{1+\sin x}$ $=\;\frac{1}{12}$

Q66JEE Main 2022MCQ4MArea Under The Curves

The area of the region enclosed between the parabolas $y^2=2 x-1$ and $y^2=4 x-$ 3 is [25-Jun-2022-Shift-2]

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📖 Explanation

Area of the shaded region $=2 int_0^1(\;\frac{y^{2+}3}{4}-\;\frac{y^{2+}1}{2}) d y$ $=2 int_0^1(\;\frac{1}{4}-\;\frac{y^2}{4}) d y$ $=2[\;\frac{1}{4}-\;\frac{1}{12}]=\;\frac{1}{3}$

Q67JEE Main 2022MCQ4MBinomial Theorem

The coefficient of ${\text{x}}^{101}$ in the expression $(5+x)^{500}+x(5+x)^{499}+x^2(5+x)^{498}+. . .+x^{500}, {\text{x}}>0$ , is [25-Jun-2022-Shift-2]

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📖 Explanation

Given, $(5+x)^{500}+x(5+x)^{499}+x^2(5+x)^{498}+. . . . . . x^{500}$ This is a G.P. with first term $(5+x)^{500}$ Common ratio $=\;\frac{x(5+x)^{499}}{(5+x)^{500}}=\;\frac{x}{5+x}$ and 501 terms present. $\therefore \;\text{ Sum }\;=\;\frac{(5+x)^{500}((\frac{x}{5+x})^{501}-1)}{\frac{x}{5+x}-1}$ $=\;{(5+x)^{500}(\frac{x^{501}-(5+x)^{501}}{(5+x)^{501}})}/{\frac{x-5-x}{5+x}}$ $=\;{\;\frac{x^{501}-(5+x)^{501}}{5+x}}/{\frac{-5}{5+x}}$ $=\;\frac{1}{5}((5+x)^{501}-x^{501})$ Coefficient of ${\text{x}}^{101}$ in $(5+x)^{501}$ is $={ }^{501} C_{101} \cdot 5^{400}$ $\therefore \ln \;\frac{1}{5}((5+x)^{500}-x^{501}) \;\text{ coefficient of }\; x^{101} \;\text{ is }\;=\;\frac{1}{5} \cdot { }^{501} C_{101} \cdot 5^{400}$ $={ }^{501} C_{101} \cdot 5^{399}$

Q68JEE Main 2022MCQ4MSequences and Series

The sum $1+2 \cdot 3+3 \cdot 3^{2+}. . . . . . . . .+10.3^9$ is equal to [25-Jun-2022-Shift-2]

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📖 Explanation

Let $S=1.3^{0+}2.3^{1+}3.3^{2+}. . . . . .+10.3^9$ $3 S=1.3^{1+}2.3^{2+}. . . . . . . . .+10.3^{10}$ $-2 S=(1.3^{0+}1.3^{1+}1.3^{2+}. . . . . .+1.3^9)-10.3^{10}$ $\Rightarrow S=\;\frac{1}{2}[10.3^{10}-\;\frac{3^{10}-1}{-3-1}]$ $\Rightarrow S=\;\frac{19.3^{10}+1}{4}$

Q69JEE Main 2022MCQ4MThree Dimensional Geometry

Let $p$ be the plane passing through the intersection of the planes $\vec{r} \cdot (\hat{i}+3 \hat{j}-\hat{k})=5$ and $\vec{r} \cdot (2 \hat{i}-\hat{j}+\hat{k})=3$ , and the point $(2,1,-2)$ . Let the position vectors of the points $X$ and $Y$ be $\hat{i}-2 \hat{j}+4 \hat{k}$ and $5 \hat{i}-\hat{j}+2 \hat{k}$ respectively. Then the points [25-Jun-2022-Shift-2]

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📖 Explanation

Let the equation of required plane $\pi:(x+3 y-z-5)+\lambda (2 x-y+z-3)=0$ $\because (2,1,-2)$ lies on it so, $2+\lambda (-2)=0$ $\Rightarrow \lambda =1$ Hence, $\pi: 3 x+2 y-8=0$ $\because \pi_x=-9, \pi_y=5, \pi_{x+y}=4$ $\pi_{x-y}=-22$ and $\pi_{y-x}=6$ Clearly $X$ and $Y$ are on opposite sides of plane $\pi$

Q70JEE Main 2022MCQ4MCircles

A circle touches both the $y$ -axis and the line $x+y=0$ . Then the locus of its center is : [25-Jun-2022-Shift-2]

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📖 Explanation

Let the centre be $( {\text{h}}, {\text{k}})$ $\;\text{ So, }\;|h|=|\;\frac{h+k}{\sqrt{2}}|$ $\Rightarrow 2 h^2=h^{2+}k^{2+}2 h k$ Locus will be $x^{2-}y^2=2 x y$

Q71JEE Main 2022MCQ4MApplication of Derivatives

Water is being filled at the rate of $1 {\text{cm}}^3 / {\sec}$ in a right circular conical vessel (vertex downwards) of height $35 {\text{cm}}$ and diameter $14 {\text{cm}}$ . When the height of the water level is $10 {\text{cm}}$ , the rate (in ${\text{cm}}^2 / {\sec}$ ) at which the wet conical surface area of the vessel increases is [25-Jun-2022-Shift-2]

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📖 Explanation

$\because V=\;\frac{1}{3} \pi r^2 h$ and $\;\frac{r}{h}=\;\frac{7}{35}=\;\frac{1}{5}$ $\Rightarrow V=\;\frac{1}{75} \pi h^3$ $\;\frac{d V}{d t}=\;\frac{1}{25} \pi h^2 \;\frac{d h}{d t}=1$ $\Rightarrow \;\frac{d h}{d t}=\;\frac{25}{\pi h^2}$ Now, $S=\pi r l=\pi(\;\frac{h}{5}) \sqrt{h^{2+}\;\frac{h^2}{25}}=\;\frac{\pi}{25} \sqrt{26} h^2$ $\Rightarrow \;\frac{d S}{d t}=\;\frac{2 \sqrt{26} \pi h}{25} \cdot \;\frac{d h}{d t}=\;\frac{2 \sqrt{26}}{h}$ $\Rightarrow \;\frac{d S}{d t_{(h=10)}}=\;\frac{\sqrt{26}}{5}$

Q72JEE Main 2022MCQ4MDefinite Integrals

If $b_n=int_0^{\;\frac{\pi}{2}} \;\frac{\cos ^2 n x}{\sin x} d x, n \in N$ , then [25-Jun-2022-Shift-2]

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📖 Explanation

$b_n-b_{n-1}=int_0^{\;\frac{\pi}{2}} \;\frac{\cos ^2 n x-\cos ^2(n-1) x}{\sin x} d x$ $=int_0^{\;\frac{\pi}{2}} \;\frac{-\sin (2 n-1) x \cdot \sin x}{\sin x} d x$ $=.\;\frac{\cos (2 n-1) x}{2 n-1}|_0 ^{\pi / 2}=-\;\frac{1}{2 n-1}$ So, $b_3-b_2, b_4-b_3, b_5-b_4$ are in H.P. $\Rightarrow \;\frac{1}{b_3-b_2}, \;\frac{1}{b_4-b_3}, \;\frac{1}{b_5-b_4}$ are in A.P. with common difference $-2$ .

Q73JEE Main 2022MCQ4MDifferential Equations

If $y=y(x)$ is the solution of the differential equation $2 x^2 \;\frac{d y}{d x}-2 x y+3 y^2=0$ such that $y (e)=\;\frac{e}{3}$ , then ${\text{y}}(1)$ is equal to [25-Jun-2022-Shift-2]

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📖 Explanation

$2 x^2 \;\frac{d y}{d x}-2 x y+3 y^2=0$ $\Rightarrow 2 x(x d y-y d x)+3 y^3 d x=0$ $\Rightarrow 2(\;\frac{x d y-y d x}{y^2})+3 \;\frac{d x}{x}=0$ $\Rightarrow -\;\frac{2 x}{y}+3 \ln x=C$ $\because y(e)=\;\frac{e}{3} \Rightarrow -6+3=C \Rightarrow C=-3$ Now, at $x=1,-\;\frac{2}{y}+0=-3$ $y=\;\frac{2}{3}$

Q74JEE Main 2022MCQ4MApplication of Derivatives

If the angle made by the tangent at the point $( {\text{x}}_0, {\text{y}}_0)$ on the curve $x=12(t+\sin t \cos t), y=12(1+\sin t)^2, 0< t< \;\frac{\pi}{2}$ , with the positive $x$ -axis is $\;\frac{\pi}{3}$ , then ${\text{y}}_0$ is equal to: [25-Jun-2022-Shift-2]

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📖 Explanation

$\because \;\frac{d y}{d x}=\;\frac{24(1+\sin t) \cos t}{12(1+\cos 2 t)}=\;\frac{1+\sin t}{\cos t}=\tan (\;\frac{\pi}{4}+\;\frac{t}{2})$ $\because \;\frac{d y}{d x_{(x_0, y_0)}}=\sqrt{3}=\tan (\;\frac{\pi}{4}+\;\frac{t}{2})$ $\Rightarrow t=\;\frac{\pi}{6}$ So, $y_{0(\;\text{ at }\; t=\;\frac{\pi}{6})}=12(1+\sin \;\frac{\pi}{6})^2=27$

Q75JEE Main 2022MCQ4MTrigonometric Ratios and Identities

The value of $2 \sin (12^{\circ})-\sin (72^{\circ})$ is : [25-Jun-2022-Shift-2]

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📖 Explanation

$2 \sin 12^{\circ}-\sin 72^{\circ}$ $=\sin 12^{\circ}+(-2 \cos 42^{\circ} \cdot \sin 30^{\circ})$ $=\sin 12^{\circ}-\cos 42^{\circ}$ $=\sin 12^{\circ}-\sin 48^{\circ}$ $=2 \sin 18^{\circ} \cdot \cos 30^{\circ}$ $=-2(\;\frac{\sqrt{5}-1}{4}) \cdot \;\frac{\sqrt{3}}{2}$ $=\;\frac{\sqrt{3}(1-\sqrt{5})}{4}$

Q76JEE Main 2022MCQ4MProbability

A biased die is marked with numbers $2,4,8,16,32,32$ on its faces and the probability of getting a face with mark $n$ is $\;\frac{1}{n}$ . If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48 , is : [25-Jun-2022-Shift-2]

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📖 Explanation

There are only two ways to get sum 48 , which are $(32,8,8)$ and $(16,16,16)$ So, required probability $=3(\;\frac{2}{32} \cdot \;\frac{1}{8} \cdot \;\frac{1}{8})+(\;\frac{1}{16} \cdot \;\frac{1}{16} \cdot \;\frac{1}{16})$ $=\;\frac{3}{2^{10}}+\;\frac{1}{2^{12}}$ $=\;\frac{13}{2^{12}}$

Q77JEE Main 2022MCQ4MSets and Relations

The negation of the Boolean expression $((∼ q) ∧ p) \Rightarrow ((∼ p) ∨ q)$ is logically equivalent to : [25-Jun-2022-Shift-2]

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📖 Explanation

$\;\text{ Let }\; S:((∼ q) ∧ p) \Rightarrow ((∼ p) ∨ q)$ $\Rightarrow S: ∼((∼ q) ∧ p) ∨((∼ p) ∨ q)$ $\Rightarrow S:(q ∨(∼ p)) ∨((∼ p) ∨ q)$ $\Rightarrow S:(∼ p) ∨ q$ $\Rightarrow S: p \Rightarrow q$ So, negation of $S$ will be $∼(p \Rightarrow q)$

Q78JEE Main 2022MCQ4MApplication of Derivatives

If the line $y=4+k x, k>0$ , is the tangent to the parabola $y=x-x^2$ at the point $P$ and $V$ is the vertex of the parabola, then the slope of the line through $P$ and $V$ is : [25-Jun-2022-Shift-2]

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📖 Explanation

$\because$ Line $y=k x+4$ touches the parabola $y=x-x^2$ So, $k x+4=x-x^2 \Rightarrow x^{2+}(k-1) x+4=0$ has only one root $(k-1)^2=16 \Rightarrow k=5$ or $-3$ but $k>0$ So, $k=5$ .And hence $x^{2+}4 x+4=0 \Rightarrow x=-2$ So, ${\text{P}}(-2,-6)$ and ${\text{V}}$ is $(\;\frac{1}{2}, \;\frac{2}{4})$ Slope of $P V=\;\frac{\;\frac{1}{4}+6}{\frac{1}{2}+2}=\;\frac{5}{2}$

Q79JEE Main 2022MCQ4MInverse Trigonometric Functions

The value of $\tan ^{-1}(\;\frac{\cos (\frac{15 \pi}{4})-1}{\sin (\;\frac{\pi}{4})})$ is equal to : [25-Jun-2022-Shift-2]

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📖 Explanation

$\tan ^{-1}(\;\frac{\cos (\frac{15 \pi}{4})-1}{\sin \;\frac{\pi}{4}})$ $=\tan ^{-1}(\;{\;\frac{1}{\sqrt{2}}-1}/{\frac{1}{\sqrt{2}}})$ $=\tan ^{-1}(1-\sqrt{2})=-\tan ^{-1}(\sqrt{2}-1)$ $=-\;\frac{\pi}{8}$

Q80JEE Main 2022MCQ4MEllipse

The line $y=x+1$ meets the ellipse $\;\frac{x^2}{4}+\;\frac{y^2}{2}=1$ at two points $P$ and $Q$ . If $r$ is the radius of the circle with $P Q$ as diameter then $(3 r)^2$ is equal to: [25-Jun-2022-Shift-2]

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📖 Explanation

Let point $(a, a+1)$ as the point of intersection of line and ellipse.So, $\;\frac{a^2}{4}+\;\frac{(a+1)^2}{2}=1 \Rightarrow a^{2+}2(a^{2+}2 a+1)=4$ $\Rightarrow 3 a^{2+}4 a-2=0$ If roots of this equation are $\alpha$ and $\beta$ .So, $P(\alpha, \alpha+1)$ and $Q(\beta, \beta+1)$ $P Q=4 r^2=(\alpha-\beta)^{2+}(\alpha-\beta)^2$ $\Rightarrow 9 r^2=\;\frac{9}{4}(2(\alpha-\beta)^2)$ $=\;\frac{9}{2}[(\alpha+\beta)^{2-}4 \alpha \beta]$ $=\;\frac{9}{2}[(-\;\frac{4}{3})^{2+}\;\frac{8}{3}]$ $=\;\frac{1}{2}[16+24]=20$

Q81JEE Main 2022NAT4MMatrices and Determinants

Let $A=(\text{table} 2 , -2 ; 1 , -1)$ and $B=(\text{table} -1 , 2 ; -1 , 2)$ . Then the number of elements in the set $\{(n, m): n, m \in \{1,2, . . . . . . . . ., 10\}.$ and $.nA^n+m B^m=1\}$ is___ [25-Jun-2022-Shift-2]

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📖 Explanation

$A^2=\begin{bmatrix}2 & -2 \\ 1 & -1\end{bmatrix}\begin{bmatrix}2 & -2 \\ 1 & -1\end{bmatrix}=\begin{bmatrix}2 & -2 \\ 1 & -1\end{bmatrix}=A$ $\Rightarrow A^K=A, K \in I$ $B^2=\begin{bmatrix}-1 & 2 \\ -1 & 2\end{bmatrix}\begin{bmatrix}-1 & 2 \\ -1 & 2\end{bmatrix}=\begin{bmatrix}-1 & 2 \\ -1 & 2\end{bmatrix}=B$ $\;\text{ So, }\; B^K=B, K \in I$ $n A^n+m B^m=n A+m B$ $=\begin{bmatrix}2 n-2 n \\ n-n\end{bmatrix}+\begin{bmatrix}-m & 2 m \\ -m & 2 m\end{bmatrix}$ $=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ So, $2 n-m=1,-n+m=0,2 m-n=1$ So, $(m, n)=(1,1)$

Q82JEE Main 2022NAT4MLimits, Continuity and Differentiability

Let $f(x)=[2 x^{2+}1]$ and $g(x)=\begin{cases}2 x-3 & x< 0 \\ 2 x+3 & x \ge 0\end{cases}$ , where [t] is the greatest integer $\le {\text{t}}$ . Then, in the open interval $(-1,1)$ , the number of points where fog is discontinuous is equal to____ [25-Jun-2022-Shift-2]

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📖 Explanation

${\text{f}}( {\text{g}}( {\text{x}}))=\$ [2 {\text{g}}^2( {\text{x}})]$+1=\begin{cases}{[2(2 x-3)^2]$+1 \ x< 0} \ {[2(2 x+3)^2]$+1 \ x \ge 0}\end{cases}\therefore $fog is discontinuous whenever$ 2(2 x-3)^2 $or$ 2(2 x+3)^2 $belongs to integer except$ x=0 $.$ \therefore 62$ points of discontinuity.

Q83JEE Main 2022NAT4MDefinite Integrals

The value of $b>3$ for which $12 int_3^b \;\frac{1}{(x^{2-}1)(x^{2-}4)} d x=\log _e(\;\frac{49}{40})$ , is equal to [25-Jun-2022-Shift-2]

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📖 Explanation

$I=\int \;\frac{1}{(x^{2-}1)(x^{2-}4)} d x$ $=\;\frac{1}{3} \int(\;\frac{1}{x^{2-}4}-\;\frac{1}{x^{2-}1}) d x$ $=\;\frac{1}{3}(\;\frac{1}{4} \ln |\;\frac{x-2}{x+2}|-\;\frac{1}{2} \ln |\;\frac{x-1}{x+1}|)+C$ $12 I=\ln |\;\frac{x-2}{x+2}|+2 \ln |\;\frac{x-1}{x+1}|+C$ $12 int_3^b \;\frac{d x}{(x^{2-}4)(x^{2-}1)}$ $=\ln (\;\frac{b-2}{b+2})-2 \ln (\;\frac{b-1}{b+1})-(\ln (\;\frac{1}{5})-2 \ln (\;\frac{1}{2}))$ $=\ln ((\;\frac{b-2}{b+2}) \cdot \;\frac{(b+1)^2}{(b-1)^2})-(\ln \;\frac{4}{5})$ $\;\text{ So, }\; \;\frac{49}{40}=\;\frac{(b-2)}{(b+2)} \;\frac{(b+1)^2}{(b-1)^2} \cdot \;\frac{5}{4}$ $\Rightarrow b=6$

Q84JEE Main 2022NAT4MBinomial Theorem

If the sum of the co-efficient of all the positive even powers of $x$ in the binomial expansion of $(2 x^{3+}\;\frac{3}{x})^{10}$ is $5^{10}-\beta .3^9$ , then $\beta$ is equal to____ [25-Jun-2022-Shift-2]

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📖 Explanation

Given, Binomial Expansion $(2 x^{3+}\;\frac{3}{x})^{10}$ General term $T_{r+1}={ }^{10} C_r \cdot (2 x^3)^{10-r} \cdot (\;\frac{3}{x})^r$ $={ }^{10} C_r \cdot 2^{10-r} \cdot 3^r \cdot x^{30-3 r} \cdot x^{-r}$ $={ }^{10} C_r \cdot 2^{10-r} \cdot 3^r \cdot x^{30-4 r}$ For positive even power of $x, 30-4 r$ should be even and positive.For $r=0,30-4 \times 0=30$ (even and positive)For $r=1,30-4 \times 1=26$ (even and positive)For $r=2,30-4 \times 2=22$ (even and positive)For $r=3,30-4 \times 3=18$ (even and positive)For $r=4,30-4 \times 4=14$ (even and positive)For $r=5,30-4 \times 5=10$ (even and positive)For $r=6,30-4 \times 6=6$ (even and positive) For $r=7,30-4 \times 7=2$ (even and positive)For $r=8,30-4 \times 8=-2$ (even but not positive)So, for $r=1,2,3,4,5,6$ and 7 we can get positive even power of $x$ . $\therefore$ Sum of coefficient for positive even power of $x$ $={ }^{10} C_0 \cdot 2^{10} \cdot 3^{0+}{ }^{10} C_1.2^9 \cdot 3^{1+}{ }^{10} C_2 \cdot 2^8 \cdot 3^{2+}{ }^{10} C_3 \cdot 2^7 \cdot 3^{3+}{ }^{10} C_4 \cdot 2^6 \cdot 3^{4+}{ }^{10} C_5 \cdot 2^5 \cdot 3^{5+}{ }^{10} C_6 \cdot 2^4 \cdot 3^{6+}{ }^{10} C_7 \cdot 2^3 \cdot 3^7$ $={ }^{10} C_{10} \cdot 2^{10} \cdot 3^{0+}{ }^{10} C_1.2^9 \cdot 3^{1+}. . . . . .+{ }^{10} C_{10} \cdot 2^0 \cdot 3^{10}-[{ }^{10} C_8 \cdot 2^2 \cdot 3^{8+}{ }^{10} C_9 \cdot 2 \cdot 3^{9+}{ }^{10} C_{10} \cdot 2^0 \cdot 3^{10}]$ $=(2+3)^{10}-[45.4 .3^{8+}10.2 .3^{9+}1.1 .3^{10}]$ $=5^{10}-[60 \times 3^{9+}20 \cdot 3^{9+}3 \cdot 3^9]$ $=5^{10}-(60+20+3) 3^9$ $=5^{10}-83.3^9$ $\therefore \beta=83$

Q85JEE Main 2022NAT4MStatistics

If the mean deviation about the mean of the numbers $1,2,3$ , ${\text{n}}$ , where ${\text{n}}$ is odd, is $\;\frac{5(n+1)}{n}$ , then ${\text{n}}$ is equal to___ [25-Jun-2022-Shift-2]

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📖 Explanation

Mean $=\;\frac{n \frac{(n+1)}{2}}{n}=\;\frac{n+1}{2}$ M.D. $=\;\frac{2(\frac{n-1}{2}+\frac{n-3}{2}+\frac{n-5}{2}+. . . 0)}{n}=\;\frac{5(n+1)}{n}$ $\Rightarrow ((n-1)+(n-3)+(n-5)+. . . 0)=5(n+1)$ $\Rightarrow (\;\frac{n+1}{4}) \cdot (n-1)=5(n+1)$ So, $n=21$

Q86JEE Main 2022NAT4MVector Algebra

Let $\vec{b}=\hat{i}+\hat{j}+\lambda \hat{k}, \lambda \in {\text{R}}$ . If $\vec{a}$ is a vector such that $\vec{a} \times \vec{b}=13 \hat{i}-\hat{j}-4 \hat{k}$ and $\vec{a} \cdot \vec{b}+21=0$ , then $(\vec{b}-\vec{a}) \cdot (\hat{k}-\hat{j})+(\vec{b}+\vec{a}) \cdot (\hat{i}-\hat{k})$ is equal to____ [25-Jun-2022-Shift-2]

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📖 Explanation

Let $\vec{a}=x \hat{i}=y \hat{j}+z \hat{k}$ So, $\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 1 & 1 & \lambda \end{vmatrix}=\hat{i}(\lambda y-z)+\hat{j}(z-\lambda x)+\hat{k}(x-y)$ $\Rightarrow \lambda y-z=13, z-\lambda x=-1, x-y=-4$ and $x+y+\lambda z=-21$ $\Rightarrow$ Clearly, $\lambda =3, x=-2, y=2$ and $z=-7$ So, $\vec{b}-\vec{a}=3 \hat{i}-\hat{j}+10 \hat{k}$ and $\vec{b}+\vec{a}=-\hat{i}+3 \hat{j}-4 \hat{k}$ $\Rightarrow (\vec{b}-\vec{a}) \cdot (\hat{k}-\hat{j})+(\vec{b}+\vec{a}) \cdot (\hat{i}-\hat{k})=11+3=14$

Q87JEE Main 2022NAT4MPermutations and Combinations

The total number of three-digit numbers, with one digit repeated exactly two times, is___ [25-Jun-2022-Shift-2]

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📖 Explanation

$C-1$ : All digits are non-zero ${ }^9 C_2 \cdot 2 \cdot \;\frac{3 !}{2}=216$ $C-2:$ One digit is 0 $0,0, x \Rightarrow { }^9 C_1 .1=9$ $0, x, x \Rightarrow { }^9 C_1 .2=18$ Total $=216+27=243$

Q88JEE Main 2022NAT4MApplication of Derivatives

Let $f(x)=|(x-1)(x^{2-}2 x-3)|+x-3, x \in R$ . If ${\text{m}}$ and ${\text{M}}$ are respectively the number of points of local minimum and local maximum of $f$ in the interval $(0,4)$ , then $m+M$ is equal to____ [25-Jun-2022-Shift-2]

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📖 Explanation

$f(x)=|(x-1)(x+1)(x-3)|+(x-3)$ $f(x)=\begin{cases}(x-3)(x^2) & 3 \le x \le 4 \\ (x-3)(2-x^2) & 1 \le x< 3 \\ (x-3)(x^2) & 0< x< 1\end{cases}$ $f^{'}(x)=\begin{cases}3 x^{2-}6 x & 3< x< 4 \\ -3 x^{2+}6 x+2 & 1< x< 3 \\ 3 x^{2-}6 x & 0< x< 1\end{cases}$ $f^{'}(3^{+})>0 \;\; f^{'}(3^{-})< 0 \to \;\text{ Minimum }\;$ $f^{'}(1^{+})>0 \;\; f^{'}(1^{-})< 0 \to \;\text{ Minimum }\;$ $x \in (1,3) f^{'}(x)=0 \;\text{ at one point }\; \to \;\text{ Maximum }\;$ $x \in (3,4) f^{'}(x) \ne 0$ $x \in (0,1) f^{'}(x) \ne 0$ So, 3 points.

Q89JEE Main 2022NAT4MHyperbola

Let the eccentricity of the hyperbola $\;\frac{x^2}{a^2}-\;\frac{y^2}{b^2}=1$ be $\;\frac{5}{4}$ . If the equation of the normal at the point $(\;\frac{8}{\sqrt{5}}, \;\frac{12}{5})$ on the hyperbola is $8 \sqrt{5} x+\beta y=\lambda$ , then $\lambda -\beta$ is equal to____ [25-Jun-2022-Shift-2]

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📖 Explanation

$\;\frac{x^2}{a^2}-\;\frac{y^2}{b^2}=1(e=\;\frac{5}{4})$ So, $b^2=a^2(\;\frac{25}{16}-1) \Rightarrow b=\;\frac{3}{4} a$ Also $(\;\frac{8}{\sqrt{5}}, \;\frac{12}{5})$ lies on the given hyperbolaSo, $\;\frac{64}{5 a^2}-\;\frac{144}{25(\;\frac{9 a^2}{16})}=1 \Rightarrow a=\;\frac{8}{5}$ and $b=\;\frac{6}{5}$ Equation of normal $\;\frac{64}{25}(\;{x}/{\frac{8}{\sqrt{5}}})+\;\frac{36}{25}(\;\frac{y}{\frac{12}{5}})=4$ $\Rightarrow \;\frac{8}{5 \sqrt{5}} x+\;\frac{3}{5} y=4$ $\Rightarrow 8 \sqrt{5} x+15 y=100$ $\;\text{ So, }\; \beta=15 \;\text{ and }\; \lambda =100$ Gives $\lambda -\beta=85$

Q90JEE Main 2022NAT4MThree Dimensional Geometry

Let ${\text{I}}_1$ be the line in ${\text{xy}}$ -plane with ${\text{x}}$ and ${\text{y}}$ intercepts $\;\frac{1}{8}$ and $\;\frac{1}{4 \sqrt{2}}$ respectively, and ${\text{I}}_2$ be the line in $z x$ -plane with $x$ and $z$ intercepts $-\;\frac{1}{8}$ and $-\;\frac{1}{6 \sqrt{3}}$ respectively. If $d$ is the shortest distance between the line ${\text{I}}_1$ and ${\text{I}}_2$ , then ${\text{d}}^{-2}$ is equal to____ [25-Jun-2022-Shift-2]

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📖 Explanation

$\;\frac{x-\frac{1}{8}}{\frac{1}{8}}=\;{y}/{-\;\frac{1}{4 \sqrt{2}}}=\;\frac{z}{0} \;\; \;\; {\text{L}}_1$ or $\;\frac{x-\frac{1}{8}}{1}=\;\frac{y}{-\sqrt{2}}=\;\frac{z}{0} . . . . . .$ (i)Equation of ${\text{L}}_2$ $\;\frac{x+\frac{1}{8}}{-6 \sqrt{3}}=\;\frac{y}{0}=\;\frac{z}{8} . . . . . .$ (ii) $d=|\;{{(c-\vec{a}) \cdot \vec{(b} \times \vec{d})}^\frac{\to }{|\vec{b} \times \vec{d}|}|$ $=\;{(\frac{1}{4} \hat{i}) \cdot (4 \sqrt{2} \hat{i}+4 \hat{j}+3 \sqrt{6 \hat{k}})}/{\sqrt{(4 \sqrt{2})^{2+}4^{2+}(3 \sqrt{6})^2}}$ $=\;\frac{\sqrt{2}}{\sqrt{32+16+54}}=\;\frac{1}{\sqrt{51}}$ $d^{-2}=51$

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