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25 Jun 2022 Shift 1

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Physics30 questions

Q1JEE Main 2022MCQ4MUnits and Measurements

If $Z=\;\frac{A^2 B^3}{C^4}$ , then the relative error in $Z$ will be : [25-Jun-2022-Shift-1]

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📖 Explanation

$Z=\;\frac{A^2 B^3}{C^4}$ $\therefore \;\frac{\Delta Z}{Z}=\;\frac{2 \Delta A}{A}+3 \times \;\frac{\Delta B}{B}+\;\frac{4 \Delta C}{C}$

Q2JEE Main 2022MCQ4MUnits and Measurements

$\vec{A}$ is a vector quantity such that $|\vec{A}|=$ non-zero constant. Which of the following expression is true for $\vec{A}$ ? [25-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\vec{A} \times \vec{A}=A \times A \times \sin 0^{\circ}$ $=0$

Q3JEE Main 2022MCQ4MMotion in a Plane

Which of the following relations is true for two unit vector $\hat{A}$ and $\hat{B}$ making an angle $\theta$ to each other? [25-Jun-2022-Shift-1]

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📖 Explanation

$\because |\hat{A}-\hat{B}|=2 \sin (\;\frac{\theta}{2})$ and, $|\hat{A}+\hat{B}|=2 \cos (\;\frac{\theta}{2})$ $\Rightarrow \;\frac{|\hat{A}-\hat{B}|}{|\hat{A}+\hat{B}|}=\tan (\;\frac{\theta}{2})$

Q4JEE Main 2022MCQ4MRotational Motion

If force $\vec{F}=3 \hat{i}+4 \hat{j}-2 \hat{k}$ acts on a particle position vector $2 \hat{i}+\hat{j}+2 \hat{k}$ then, the torque about the origin will be : [25-Jun-2022-Shift-1]

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📖 Explanation

$\vec{\tau }=\vec{r} \times \vec{F}$ $=(2 \hat{i}+\hat{j}+2 \hat{k}) \times (3 \hat{i}+4 \hat{j}-2 \hat{k})$ $=-10 \hat{i}+10 \hat{j}+5 \hat{k}$

Q5JEE Main 2022MCQ4MGravitation

The height of any point $P$ above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point $P$ will be: (Given $g=$ acceleration due to gravity at the surface of earth). [25-Jun-2022-Shift-1]

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📖 Explanation

$h=2 R$ $\therefore g^{'}=\;\frac{G M}{(R+h)^2}$ $=\;\frac{G M}{9 R^2}$ $=\;\frac{g}{9}$

Q6JEE Main 2022MCQ4MProperties of Matter

The terminal velocity $(v_t)$ of the spherical rain drop depends on the radius $(r)$ of the spherical rain drop as : [25-Jun-2022-Shift-1]

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📖 Explanation

$6 \pi η v_t r=\;\frac{4}{3} \pi r^3(\rho -\sigma) g$ $\Rightarrow v_t=C r^2$ where ${\text{C}}$ is a constant or $v_t \propto r^2$

Q7JEE Main 2022MCQ4MThermodynamics

The relation between root mean square speed $(v_{r m s})$ and most probable sped $(v_p)$ for the molar mass $M$ of oxygen gas molecule at the temperature of $300 {\text{K}}$ will be: [25-Jun-2022-Shift-1]

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📖 Explanation

$v_{r m s}=\sqrt{\;\frac{3 R T}{M}}$ $v_p=\sqrt{\;\frac{2 R T}{M}}$ $\Rightarrow v_{r m s}=\sqrt{\;\frac{3}{2}} v_p$

Q8JEE Main 2022MCQ4MElectrostatics

In the figure, a very large plane sheet of positive charge is shown. $P_1$ and $P_2$ are two points at distance I and $2 \mid$ from the charge distribution. If $\sigma$ is the surface charge density, then the magnitude of electric fields ${\text{E}}_1$ and ${\text{E}}_2$ at $P_1$ and $P_2$ respectively are : [25-Jun-2022-Shift-1]

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📖 Explanation

For an infinite charged plane $E=\;\frac{\sigma}{2 \epsilon_0}$ for any value of ${\text{I}}$ $\Rightarrow E_1=E_2=\;\frac{\sigma}{2 \epsilon_0}$

Q9JEE Main 2022MCQ4MAlternating Current

Choose the correct answer from the options given below : List - I List -II (A) AC generator (I) Detects the presence of current in the circuit (B) Galvanometer (II) Converts mechanical energy into electrical energy (C) Transformer (III) Works on the principle of resonance in AC circuit (D) Metal detector (IV) Changes an alternating voltage for smaller or greater value [25-Jun-2022-Shift-1]

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Q10JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

A long straight wire with a circular cross-section having radius ${\text{R}}$ , is carrying a steady current ${\text{I}}$ . The current ${\text{I}}$ is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance $r(r< R)$ from its centre will be : [25-Jun-2022-Shift-1]

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📖 Explanation

$\int \overline{B} \cdot \overline{d l}=\mu _0 I_{i n}$ $\Rightarrow B \times 2 \pi r=\;\frac{\mu _0 I}{\pi R^2} \times \pi r^2$ $\Rightarrow B \propto r$

Q11JEE Main 2022MCQ4MAlternating Current

If wattless current flows in the AC circuit, then the circuit is : [25-Jun-2022-Shift-1]

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Q12JEE Main 2022MCQ4MElectromagnetic Waves

The electric field in an electromagnetic wave is given by ${\text{E}}=56.5 \sin \omega ( {\text{t}}- {\text{x}} / {\text{c}}) {\text{NC}}^{-1}$ . Find the intensity of the wave if it is propagating along $x$ -axis in the free space.(Given : $\epsilon_0=8.85 \times 10^{-12} {\text{C}}^2 {\text{N}}^{-1} {\text{m}}^{-2}$ ) [25-Jun-2022-Shift-1]

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📖 Explanation

$I=\;\frac{1}{2} \epsilon_0 E_0^2 c$ $=\;\frac{1}{2} 8.5 \times 10^{-12} \times (56.5)^2 \times 3 \times 10^8$ $=4.24 {\text{W}} / {\text{m}}^2$

Q13JEE Main 2022MCQ4MWave Optics

The two light beams having intensities I and 91 interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\pi / 2$ at point ${\text{P}}$ and $\pi$ at point ${\text{Q}}$ . Then the difference between the resultant intensities at ${\text{P}}$ and ${\text{Q}}$ will be : [25-Jun-2022-Shift-1]

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📖 Explanation

$I_P=I+9 I+2 \sqrt{I \times 9 I} \cos \;\frac{\pi}{2}=10 I$ $I_Q=I+9 I+2 \sqrt{I \times 9 I} \cos \pi=4 I$ So, $I_P-I_Q=6 I$

Q14JEE Main 2022MCQ4MRay Optics and Optical Instruments

A light wave travelling linearly in a medium of dielectric constant 4, incidents on the horizontal interface separating medium with air. The angle of incidence for which the total intensity of incident wave will be reflected back into the same medium will be :(Given : relative permeability of medium $\mu _r=1$ ) [25-Jun-2022-Shift-1]

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📖 Explanation

$n=\sqrt{K \mu }=2(n \Rightarrow$ refractive index $)$ So for TIR $\theta>\sin ^{-1}(\;\frac{1}{n})$ $\theta>30$ Only option is $60^{\circ}$

Q15JEE Main 2022MCQ4MDual Nature of Matter and Radiation

Given below are two statements :Statement I : Davisson-Germer experiment establishes the wave nature of electrons.Statement II : If electrons have wave nature, they can interfere and show diffraction.In the light of the above statements choose the correct answer from the option given below : [25-Jun-2022-Shift-1]

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Q16JEE Main 2022MCQ4MAtoms and Nuclei

The ratio for the speed of the electron in the $3^{\;\text{rd }\;}$ orbit of ${\text{He}}^{+}$ to the speed of the electron in the $3^{\;\text{rd }\;}$ orbit of hydrogen atom will be : [25-Jun-2022-Shift-1]

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📖 Explanation

We know that $v \propto \;\frac{Z}{n}$ $\Rightarrow$ Required ratio $=\;\frac{\;\frac{2}{3}}{\frac{1}{3}}$ $=2: 1$

Q17JEE Main 2022MCQ4MSemiconductor Electronics

The photodiode is used to detect the optical signals. These diodes are preferably operated in reverse biased mode because : [25-Jun-2022-Shift-1]

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Q18JEE Main 2022MCQ4MElectromagnetic Waves

A signal of $100 {\text{THz}}$ frequency can be transmitted with maximum efficiency by : [25-Jun-2022-Shift-1]

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📖 Explanation

Optical fibres supports frequency of electromagnetic waves in the range $10^{14} {\text{Hz}}$ to $10^{15} {\text{Hz}}$ .

Q19JEE Main 2022MCQ4MRay Optics and Optical Instruments

The difference of speed of light in the two media $A$ and $B(v_A-v_B)$ is $2.6 \times 10^7 {\text{m}} / {\text{s}}$ . If the refractive index of medium $B$ is $1.47$ , then the ratio of refractive index of medium $B$ to medium $A$ is : (Given : speed of light in vacuum $. {\text{c}}=3 \times 10^8 {\text{ms}}^{-1})$ [25-Jun-2022-Shift-1]

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📖 Explanation

Speed of light in a medium $=\;\frac{c}{n}$ $\Rightarrow$ According to given information, $\;\frac{c}{n_A}-\;\frac{c}{n_B}=2.6 \times 10^7$ $\Rightarrow \;\frac{n_B}{n_A}-1=\;\frac{2.6 \times 10^7}{3 \times 10^8} \times n_B$ $\Rightarrow \;\frac{n_B}{n_A} ≃ 1.13$

Q20JEE Main 2022MCQ4MCurrent Electricity

A teacher in his physics laboratory allotted an experiment to determine the resistance (G) of a galvanometer. Students took the observations for $\;\frac{1}{3}$ deflection in the galvanometer. Which of the below is true for measuring value of ${\text{G}}$ ? [25-Jun-2022-Shift-1]

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📖 Explanation

The circuit for the given situation is:Since $G$ and $S$ are in parallel, $\Rightarrow \;\frac{i}{3} \times {\text{G}}=\;\frac{2 i}{3} \times {\text{S}}$ $\Rightarrow {\text{G}}=2 {\text{S}}$ $\Rightarrow G$ equals twice the value of shunt resistance.

Q21JEE Main 2022NAT4MLaws of Motion

A uniform chain of $6 {\text{m}}$ length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is $0.5$ , the maximum length of the chain hanging from the table is_____ m. [25-Jun-2022-Shift-1]

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📖 Explanation

$(x) g \lambda =\mu (6-x) g \lambda$ where ${\text{x}}$ is length of hanging part $\Rightarrow x=3-0.5 x$ $\Rightarrow x=2 {\text{m}}$

Q22JEE Main 2022NAT4MWork, Power and Energy

A $0.5 {\text{kg}}$ block moving at a speed of $12 {\text{ms}}^{-1}$ compresses a spring through a distance $30 {\text{cm}}$ when its speed is halved. The spring constant of the spring will be ____ ${\text{Nm}}^{-1}$ [25-Jun-2022-Shift-1]

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📖 Explanation

$\;\frac{1}{2} m V^2=\;\frac{1}{2} k x^{2+}\;\frac{1}{2} m(\;\frac{v}{2})^2$ $\Rightarrow \;\frac{3}{8} m v^2=\;\frac{1}{2} k x^2$ $\Rightarrow k=\;\frac{3}{4} \times \;\frac{1}{2} \times \;\frac{144}{9} \times 100$ $=600$ $\Rightarrow 600$

Q23JEE Main 2022NAT4MProperties of Matter

The velocity of upper layer of water in a river is $36 {\text{kmh}}^{-1}$ . Shearing stress between horizontal layers of water is $10^{-3} {\text{Nm}}^{-2}$ . Depth of the river is m. (Co-efficient of viscosity of water is _____ $10^{-2} {\text{Pa}} . {\text{s}}$ ) [25-Jun-2022-Shift-1]

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📖 Explanation

$F=-η A \;\frac{d u}{d x}$ $\Rightarrow 10^{-3}=10^{-2} \times \;\frac{10}{h}$ $\Rightarrow h=\;\frac{10^{-1}}{10^{-3}} {\text{m}}=100 {\text{m}}$ $\Rightarrow (100)$

Q24JEE Main 2022NAT4MThermodynamics

A steam engine intakes $50 {\text{g}}$ of steam at $100^{\circ} {\text{C}}$ per minute and cools it down to $20^{\circ} {\text{C}}$ . If latent heat of vaporization of steam is $540 {\text{cal}} {\text{g}}^{-1}$ , then the heat rejected by the steam engine per minute is _____ $\times$ $10^3 {\text{cal}}$ (Given : specific heat capacity of water : $1 {\text{cal}} {\text{g}}^{-1}{ }^{\circ} {\text{C}}^{-1}$ ) [25-Jun-2022-Shift-1]

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📖 Explanation

$\Delta {\text{Q}}_{ {\text{rej}}}=50 \times 540+50 \times 1 \times (100-20)$ $=50 \times [540+80]$ $=50 \times 620$ $=31000 {\text{cal}}$ $=31 \times 10^3 {\text{cal}}$

Q25JEE Main 2022NAT4MWaves

The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is $20 {\text{cm}}$ . The length of the open organ pipe is ______cm. [25-Jun-2022-Shift-1]

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📖 Explanation

$2 \times (\;\frac{V}{2 L_0})=(\;\frac{V}{4 L_c})$ $\Rightarrow L_0=4 L_c$ $=4 \times 20$ $=80 {\text{cm}}$

Q26JEE Main 2022NAT4MElectrostatic Potential and Capacitance

The equivalent capacitance between points $A$ and $B$ in below shown figure will be _______ $\mu {\text{F}}$ [25-Jun-2022-Shift-1]

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📖 Explanation

$C_{e q}=\;\frac{(3 \times 8) \times 8}{(3 \times 8)+8}$ $=\;\frac{24 \times 8}{32}$ $=6 \mu {\text{F}}$

Q27JEE Main 2022NAT4MCurrent Electricity

A resistor develops $300 {\text{J}}$ of thermal energy in $15 {\text{s}}$ , when a current of $2 {\text{A}}$ is passed through it. If the current increases to $3 {\text{A}}$ , the energy developed in $10 {\text{s}}$ is_____J. [25-Jun-2022-Shift-1]

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📖 Explanation

$300=I^2 R \times 15$ $\Rightarrow R=5 Ω$ Now $I_2^2 R t_2$ $=9 \times 5 \times 10$ $=450 {\text{J}}$

Q28JEE Main 2022NAT4MCurrent Electricity

The total current supplied to the circuit as shown in figure by the $5 {\text{V}}$ battery is _______A. [25-Jun-2022-Shift-1]

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📖 Explanation

$\Rightarrow I=\;\frac{5}{2.5}=2 A$

Q29JEE Main 2022NAT4MElectromagnetic Induction

The current in a coil of self inductance $2.0 {\text{H}}$ is increasing according to ${\text{I}}=2 \sin ( {\text{t}}^2) {\text{A}}$ . The amount of energy spent during the period when current changes from 0 to ${\text{A}}$ is ______J. [25-Jun-2022-Shift-1]

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📖 Explanation

$U=\;\frac{1}{2} L I^2$ $=\;\frac{1}{2} 2 \times 2^2=4 {\text{J}}$

Q30JEE Main 2022NAT4MMotion in a Plane

A force on an object of mass $100 {\text{g}}$ is $(10 \hat{i}+5 \hat{j}) {\text{N}}$ . The position of that object at ${\text{t}}=2 {\text{s}}$ is $(a \hat{i}+b \hat{j}) {\text{m}}$ after starting from rest. The value of $\;\frac{a}{b}$ will be_____ [25-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\vec{F}=m \vec{a}$ $\Rightarrow \vec{a}=100 \hat{i}+50 \hat{j}$ So, $\vec{S}=\;\frac{1}{2} \vec{a} t^2$ $\;\frac{1}{2}(100 \hat{i}+50 \hat{j}) 2^2$ $=200 \hat{i}+100 \hat{j} {\text{m}}$ so a $=200 {\text{m}}$ and ${\text{b}}=100 {\text{m}}$ so $\;\frac{a}{b}=2$

Chemistry30 questions

Q31JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

Bonding in which of the following diatomic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron?(A) NO(B) ${\text{N}}_2$ (C) ${\text{O}}_2$ (D) ${\text{C}}_2$ (E) $B_2$ Choose the most appropriate answer from the options given below: [25-Jun-2022-Shift-1]

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📖 Explanation

If an electron is removed from the anti-bonding orbital, then it will tend to increase the bond order. The HOMO in ${\text{NO}}$ and ${\text{O}}_2$ is antibonding molecular orbital.Hence, in ${\text{NO}}$ and ${\text{O}}_2$ bond order will increase on loss of electron.

Q32JEE Main 2022MCQ4MSolutions

Incorrect statement for Tyndall effect is : [25-Jun-2022-Shift-1]

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Q33JEE Main 2022MCQ4MStructure of Atom

The pair, in which ions are isoelectronic with ${\text{Al}}^{3+}$ is : [25-Jun-2022-Shift-1]

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📖 Explanation

${\text{O}}^{2-}, {\text{Mg}}^{2+}$ and ${\text{Al}}^{3+}$ are isoelectronic. All have 10 electrons.

Q34JEE Main 2022MCQ4MCoordination Compounds

Leaching of gold with dilute aqueous solution of ${\text{NaCN}}$ in presence of oxygen gives complex [A], which on reaction with zinc forms the elemental gold and another complex [B]. [A] and [B], respectively are : [25-Jun-2022-Shift-1]

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📖 Explanation

In the metallurgy of gold $4 {\text{Au}}+8 {\text{CN}}^{-}+2 {\text{H}}_2 {\text{O}}+ {\text{O}}_2 \to 4[ {\text{Au}}( {\text{CN}})_2]^{-}+4 {\text{OH}}^{-}$ $2[ {\text{Au}}( {\text{CN}})_2]^{-}+ {\text{Zn}} \to [ {\text{Zn}}( {\text{CN}})_4]^{2-}+2 {\text{Au}}$

Q35JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

Number of electron deficient molecules among the following ${\text{PH}}_3, {\text{B}}_2 {\text{H}}_6, {\text{CCl}}_4, {\text{NH}}_3, {\text{LiH}}$ and ${\text{BCl}}_3$ is [25-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Electron deficient species have less than 8 electrons (or two electrons for ${\text{H}}$ ) in their valence (incomplete octet). ${\text{B}}_2 {\text{H}}_6, {\text{BCl}}_3$ have incomplete octet.

Q36JEE Main 2022MCQ4MElectrochemistry

Which one of the following alkaline earth metal ions has the highest ionic mobility in its aqueous solution? [25-Jun-2022-Shift-1]

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📖 Explanation

In aqueous solution, the ionic mobility is inversely proportional to the charge density on the ion. Hence ${\text{Sr}}^{2+}$ has highest ionic mobility in water.

Q37JEE Main 2022MCQ4MCoordination Compounds

White precipitate of ${\text{AgCl}}$ dissolves in aqueous ammonia solution due to formation of : [25-Jun-2022-Shift-1]

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Q38JEE Main 2022MCQ4Md and f Block Elements

Cerium (IV) has a noble gas configuration. Which of the following is correct statement about it? [25-Jun-2022-Shift-1]

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📖 Explanation

Cerium exists in two different oxidation state $+3,+4$ ${\text{Ce}}^{+4}+ {\text{e}}^{-} \to {\text{Ce}}^{3+}\; {\text{E}}^0=+1.61 {\text{V}}$ ${\text{Ce}}^{+3}+3 {\text{e}}^{-} \to {\text{Ce}}\; {\text{E}}^0=-2.336 {\text{V}}$ It shows ${\text{Ce}}^{+4}$ acts as a strong oxidising agent & accepts electron.

Q39JEE Main 2022MCQ4Md and f Block Elements

Among the following, which is the strongest oxidizing agent? [25-Jun-2022-Shift-1]

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📖 Explanation

Strongest oxidising agent have highest reduction potential value ${\text{E}}^{\circ}{}_{ {\text{Fe}}^{3+} \mid {\text{Fe}}^{2+}}=+0.77 {\text{V}}\;$ ${\text{E}}^{\circ} {}_{ {\text{Ti}}^{3+} \mid {\text{Ti}}^{2+}}=-0.37 {\text{V}}$ ${\text{E}}^{\circ} {}_{ {\text{Mn}}^{+3} \mid {\text{Mn}}^{+2}}=+1.57 {\text{V}}\;$ ${\text{E}}^{\circ} {}_{ {\text{Cr}}^{3+} \mid {\text{Cr}}^{2+}}=-0.41 {\text{V}}$ ${\text{Mn}}^{+3}$ is the best oxidising agent among the given series.

Q40JEE Main 2022MCQ4MBiomolecules

The eutrophication of water body results in: [25-Jun-2022-Shift-1]

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Q41JEE Main 2022MCQ4MPractical Organic Chemistry

Phenol on reaction with dilute nitric acid, gives two products. Which method will be most efficient for large scale separation? [25-Jun-2022-Shift-1]

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Q42JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

In the following structures, which on is having staggered conformation with maximum dihedral angle? [25-Jun-2022-Shift-1]

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Q43JEE Main 2022MCQ4MHydrocarbons

[25-Jun-2022-Shift-1]

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Q44JEE Main 2022MCQ4MHaloalkanes and Haloarenes

The IUPAC name of ethylidene chloride is : [25-Jun-2022-Shift-1]

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📖 Explanation

Ethylidene chloride is ${\text{CH}}_3- {\text{CHCl}}_2$ , its IUPAC name is 1,1-Dichloromethane.

Q45JEE Main 2022MCQ4MHaloalkanes and Haloarenes

The major product in the reaction [25-Jun-2022-Shift-1]

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Q46JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

The intermediate X, in the reaction : [25-Jun-2022-Shift-1]

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Q47JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

In the following reaction : The compounds A and B respectively are : [25-Jun-2022-Shift-1]

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Q48JEE Main 2022MCQ4MAmines

The reaction of $\text{R}-\text{C}_{||}_{\text{O}}-\text{NH}_2$ with bromine and $\text{KOH}$ gives $\text{RNH}_2$ as the end product. Which one of the following is the intermediate product formed in this reation? [25-Jun-2022-Shift-1]

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Q49JEE Main 2022MCQ4MSolutions

Using very little soap while washing clothes, does not serve the purpose of cleaning of clothes, because: [25-Jun-2022-Shift-1]

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Q50JEE Main 2022MCQ4MBiomolecules

Which one of the following is an example of artificial sweetner? [25-Jun-2022-Shift-1]

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Q51JEE Main 2022NAT4MSome Basic Concepts of Chemistry

The number of ${\text{N}}$ atoms in $681 {\text{g}}$ of ${\text{C}}_7 {\text{H}}_5 {\text{N}}_3 {\text{O}}_6$ is ${\text{x}} \times 10^{21}$ . The value of ${\text{x}}$ is $( {\text{N}}_{ {\text{A}}}=6.02 \times 10^{23} {\text{mol}}^{-1})$ (Nearest Integer) [25-Jun-2022-Shift-1]

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📖 Explanation

Molar mass of ${\text{C}}_7 {\text{H}}_5 {\text{N}}_3 {\text{O}}_6$ $=12 \times 7+5+14 \times 3+16 \times 6$ $=227 {\text{gm}}$ Given mass of ${\text{C}}_7 {\text{H}}_5 {\text{N}}_3 {\text{O}}_6=681 {\text{gm}}$ $\therefore$ Moles of $C_7 H_5 N_3 O_6=\;\frac{681}{227}=3$ In one molecule of $C_7 H_5 N_3 O_6, 3 {\text{N}}$ atoms present. $\therefore$ In 1 mole $C_7 H_5 N_3 O_6, 3$ moles of ${\text{N}}$ atoms presents. $\therefore$ In 3 moles of $C_7 H_5 N_3 O_6, 3 \times 3=9$ moles of ${\text{N}}$ atoms presents.We know, 1 mole of ${\text{N}}$ atoms $=6.02 \times 10^{23} {\text{N}}$ atoms. $\therefore 9$ moles of ${\text{N}}$ atoms $=9 \times 6.02 \times 10^{23}=54.18 \times 10^{23}=5418 \times 10^{21}$

Q52JEE Main 2022NAT4MSolid State

The distance between ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions in solid ${\text{NaCl}}$ of density $43.1 {\text{g}} {\text{cm}}^{-3}$ is _____ $\times 10^{-10} {\text{m}}$ . (Nearest Integer)(Given : ${\text{N}}_{ {\text{A}}}=6.02 \times 10^{23} {\text{mol}}^{-1}$ ) [25-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\rho =\;\frac{Z \times M}{a^3 \times N_A}$ $43.1=\;\frac{4 \times 58.5}{a^3 \times 6.02 \times 10^{23}}$ $a^3=0.9 \times 10^{-23}$ $=9 \times 10^{-24}$ $a=2.08 \times 10^{-8} {\text{cm}}$ $=2.08 \times 10^{-10} {\text{m}}$ for ${\text{NaCl}}$ , distance between ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}=\;\frac{a}{2}$ $=1.04 \times 10^{-10} {\text{m}}$

Q53JEE Main 2022NAT4MStructure of Atom

The longest wavelength of light that can be used for the ionisation of lithium atom (Li) in its ground state is $x \times 10^{-8} {\text{m}}$ . The value of $x$ is___ (Nearest Integer).(Given : Energy of the electron in the first shell of the hydrogen atom is $-2.2$ $\times 10^{-18} {\text{J}} ; {\text{h}}=6.63 \times 10^{-34} {\text{Js}}$ and ${\text{c}}=3 \times 10^8 {\text{ms}}^{-1}$ ) [25-Jun-2022-Shift-1]

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📖 Explanation

Bohr model is not valid for lithium atom (Li) as Bohr model is valid for only single electronic species, so it would be valid for ${\text{Li}}^{+2}$ but not ${\text{Li}}$ atom.

Q54JEE Main 2022NAT4MChemical Thermodynamics

The standard entropy change for the reaction $4 {\text{Fe}}( {\text{s}})+3 {\text{O}}_2( {\text{g}}) \to 2 {\text{Fe}}_2 {\text{O}}_3( {\text{s}})$ is $-550 {\text{J}} {\text{K}}^{-1}$ at $298 {\text{K}}$ .$[Given : The standard enthalpy change for the reaction is $-165 {\text{kJ}} {\text{mol}}^{-1}$ ]$.The temperature in ${\text{K}}$ at which the reaction attains equilibrium is____(Nearest Integer) [25-Jun-2022-Shift-1]

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📖 Explanation

$\Delta {\text{G}}=\Delta {\text{H}}- {\text{T}} \Delta {\text{S}}=0$ at equilibrium $\Rightarrow -165 \times 10^{3-} {\text{T}} \times (-505)=0$ $\Rightarrow {\text{T}}=300 {\text{K}}$

Q55JEE Main 2022NAT4MSolutions

$1 {\text{L}}$ aqueous solution of ${\text{H}}_2 {\text{SO}}_4$ contains $0.02 {\text{m}} {\text{mol}} {\text{H}}_2 {\text{SO}}_4 .50 \%$ of this solution is diluted with deionized water to give $1 {\text{L}}$ solution (A). In solution (A), $0.01 {\text{m}} {\text{mol}}$ of ${\text{H}}_2 {\text{SO}}_4$ are added. Total ${\text{m}}$ mols of ${\text{H}}_2 {\text{SO}}_4$ in the final solution is ____ $\times 10^3 {\text{m}} {\text{mols}}$ [25-Jun-2022-Shift-1]

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📖 Explanation

${\text{n}}_{ {\text{H}}_2 {\text{SO}}_4}$ in ${\text{Sol}}^{ {\text{n}}} {\text{A}}=50 \%$ of original solution $=0.01 {\text{m}} {\text{mol}}$ ${\text{n}}_{ {\text{H}}_2 {\text{SO}}_4}$ in Final solution $=0.01+0.01$ $=0.02 {\text{m}} {\text{mol}}$ $=0.00002 \times 10^3 {\text{m}} {\text{mol}}$ The answer 0

Q56JEE Main 2022NAT4MChemical Thermodynamics

The standard free energy change $(\Delta {\text{G}}^{\circ})$ for $50 \%$ dissociation of ${\text{N}}_2 {\text{O}}_4$ into ${\text{NO}}_2$ at $27^{\circ} {\text{C}}$ and 1 atm pressure is $-x {\text{J}} {\text{mol}}^{-1}$ . The value of $x$ is_______ (Nearest Integer)[Given : $. {\text{R}}=8.31 {\text{JK}}^{-1} {\text{mol}}^{-1}, \log 1.33=0.1239 \ln 10=2.3]$ [25-Jun-2022-Shift-1]

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📖 Explanation

${\text{k}}_{ {\text{P}}}=\;\frac{(\frac{1}{1.5} \times 1)^2}{(\;\frac{0.5}{1.5} \times 1)}=\;\frac{1}{0.75}=\;\frac{100}{75}$ $=1.33$ $\Delta {\text{G}}^0=- {\text{RT}} ℓ {\text{nk}}_{ {\text{P}}}$ $=-8.31 \times 300 \times \ln (1.33)=-710.45 {\text{J}} / {\text{mol}}$ $=-710 {\text{J}} / {\text{mol}}$

Q57JEE Main 2022NAT4MElectrochemistry

In a cell, the following reactions take place $\text{table} {\text{Fe}}^{2+} \to {\text{Fe}}^{3+}+e^{-} , E^o{}_{F e^{3+} / {\text{Fe}}^{2+}}=0.77 {\text{V}} ; 2 {\text{I}}^{-} \to {\text{I}}_2+2 e^{-} , E^o{}_{I_2 / I^{-}}=0.54 {\text{V}}$ The standard electrode potential for the spontaneous reaction in the cell is $x \times 10^{-2} {\text{V}} 298 {\text{K}}$ . The value of ${\text{x}}$ is ___(Nearest Integer) [25-Jun-2022-Shift-1]

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📖 Explanation

$\;{ {\text{Fe}}}_{\;\text{ Cathode }\;^{+3}}+\;{ {\text{I}}}_{\;\text{ anode }\;^{-}} \longrightarrow {\text{I}}_2+ {\text{Fe}}^{+2}$ $E_{\;\text{cell }\;}^{\circ}=E_{\;\text{cathode }\;}^{\circ}-E_{\;\text{anode }\;}^{\circ}$ $=0.77-0.54$ $=0.23 {\text{V}}$ $=23 \times 10^{-2} {\text{V}}$

Q58JEE Main 2022NAT4MChemical Kinetics

For a given chemical reaction $\gamma _1 {\text{A}}+\gamma _2 {\text{B}} \to \gamma _3 {\text{C}}+\gamma _4 {\text{D}}$ Concentration of $C$ changes from $10 {\text{mmol}} {\text{dm}}^{-3}$ to $20 {\text{mmol}} {\text{dm}}^{-3}$ in 10 seconds. Rate of appearance of $D$ is $1.5$ times the rate of disappearance of $B$ which is twice the rate of disappearance $A$ . The rate of appearance of $D$ has been experimentally determined to be $9 {\text{mmol}} {\text{dm}}^{-3} {\text{s}}^{-1}$ . Therefore, the rate of reaction is ___ ${\text{mmol}} {\text{dm}}^{-3} {\text{s}}^{-1} \cdot ($ Nearest Integer) [25-Jun-2022-Shift-1]

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📖 Explanation

Rate $=\;\frac{1}{r_1}(\;\frac{-d[A]}{d t})=\;\frac{1}{r_2}(\;\frac{-d[B]}{d t})=\;\frac{1}{r_3}(\;\frac{-d[C]}{d t})$ $=\;\frac{1}{r_4}(\;\frac{d[D]}{d t})$ $\;{ {\text{d}}[ {\text{D}}]}/{ {\text{dt}}}=\;{ {\text{r}}_4}/{ {\text{r}}_2}(\;{- {\text{d}}[ {\text{B}}]}/{ {\text{dt}}})$ $\;\frac{r_4}{r_2}=\;\frac{3}{2}$ $\;{- {\text{d}}[ {\text{B}}]}/{ {\text{dt}}}=\;{ {\text{r}}_2}/{ {\text{r}}_1}(\;{- {\text{d}}[ {\text{A}}]}/{ {\text{dt}}}) \Rightarrow \;{ {\text{r}}_2}/{ {\text{r}}_1}=2$ $r_2=2 r_1$ ${\text{r}}_4=1.5 {\text{r}}_2=3 {\text{r}}_1$ $\;{ {\text{d}}[ {\text{C}}]}/{ {\text{dt}}}=1 {\text{m}} \cdot {\text{mol}} {\text{dm}}^{-3} {\sec}^{-1}$ $\;{ {\text{d}}[ {\text{D}}]}/{ {\text{dt}}}=\;{ {\text{r}}_4}/{r_3} \cdot \;{ {\text{d}}[ {\text{C}}]}/{ {\text{dt}}} \Rightarrow \;{ {\text{r}}_4}/{ {\text{r}}_3}=9$ ${\text{r}}_4=9 {\text{r}}_3=3 {\text{r}}_1$ $\Rightarrow {\text{r}}_1=3 {\text{r}}_3$ $3 {\text{r}}_3 {\text{A}}+6 {\text{r}}_3 {\text{B}} \to {\text{r}}_3 {\text{C}}+9 {\text{r}}_3 {\text{D}}$ $\therefore$ rate of reaction $=\;\frac{1}{9} \times 9 {\text{m}} \cdot {\text{mol}} {\text{dm}}^{-3} {\sec}^{-1}$ $=1 {\text{m}} \cdot {\text{mol}} {\text{dm}}{ }^{-3} \sec ^{-1}$

Q59JEE Main 2022NAT4MCoordination Compounds

If $[ {\text{Cu}}( {\text{H}}_2 {\text{O}})_4]^{2+}$ absorbs a light of wavelength $600 {\text{nm}}$ for ${\text{d}}- {\text{d}}$ transition, then the value of octahedral crystal field splitting energy for $[ {\text{Cu}}( {\text{H}}_2 {\text{O}})_6]^{2+}$ will be ____ $\times 10^{-21} {\text{J}} .$ [Nearest Integer](Given : $h=6.63 \times 10^{-34} {\text{Js}}$ and ${\text{c}}=3.08 \times 10^8 {\text{ms}}^{-1}$ ) [25-Jun-2022-Shift-1]

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📖 Explanation

$[ {\text{Cu}}( {\text{H}}_2 {\text{O}})_4]^{2+}$ is tetrahedral $[ {\text{Cu}}( {\text{H}}_2 {\text{O}})_6]^{2+}$ is octahedral $\because \Delta _{ {\text{t}}}=\;\frac{4}{9} \times \Delta _0$ $\Delta _{ {\text{t}}}=\;\frac{6.63 \times 10^{-34} \times 3.08 \times 10^8}{600 \times 10^{-9}}$ $\Delta _0=\;\frac{9}{4} \times \;\frac{6.63 \times 10^{-34} \times 3.08 \times 10^8}{600 \times 10^{-9}} \approx 765 \times 10^{-21} {\text{J}}$

Q60JEE Main 2022NAT4MHydrocarbons

Number of grams of bromine that will completely react with $5.0 {\text{g}}$ of pent-1-ene is $\times 10^{-2} {\text{g}} .$ (Atomic mass of ${\text{Br}}=80 {\text{g}} / {\text{mol}}$ ) [Nearest Integer] [25-Jun-2022-Shift-1]

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📖 Explanation

Molar mass of ${\text{C}}_5 {\text{H}}_{10}=12 \times 5+10=70 {\text{gm}}$ Given mass of ${\text{C}}_5 {\text{H}}_{10}=5 {\text{gm}}$ $\therefore$ Moles of ${\text{C}}_5 {\text{H}}_{10}=\;\frac{5}{70}$ From reaction,1 mole of ${\text{C}}_5 {\text{H}}_{10}$ reacts with 1 mole of ${\text{Br}}_2$ $\therefore \;\frac{5}{70}$ moles of ${\text{C}}_5 {\text{H}}_{10}$ reacts with $\;\frac{5}{70}$ moles of ${\text{Br}}_2$ $\therefore$ Reacted ${\text{Br}}_2=\;\frac{5}{70} \times 160 {\text{gm}}$ $=11.428 {\text{gm}}$ $=1142.8 \times 10^{-2} {\text{gm}}$ $≃ 1143 \times 10^{-2} {\text{gm}}$

Mathematics30 questions

Q61JEE Main 2022MCQ4MCircles

Let a circle C touch the lines $L_1: 4 x-3 y+K_1=0$ and $L_2=4 x-3 y+K_2=0, K_1, K_2 \in R$ . If a line passing through the centre of the circle ${\text{C}}$ intersects ${\text{L}}_1$ at $(-1,2)$ and ${\text{L}}_2$ at $(3,-6)$ , then the equation of the circle ${\text{C}}$ is : [25-Jun-2022-Shift-1]

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📖 Explanation

Co-ordinate of centre $C \equiv (\;\frac{3+(-1)}{2}, \;\frac{-6-2}{2}) \equiv (1,-2)$ $L_1$ is passing through $A$ $\Rightarrow -4-6+K_1=0$ $\Rightarrow K_1=10$ $L_2$ is passing through $B$ $\Rightarrow 12+18+K_2=0$ $\Rightarrow K_2=-30$ Equation of $L_1: 4 x-3 y+10=0$ Equation of $L_2=4 x-3 y-30=0$ Diameter of circle $=|\;\frac{10+30}{\sqrt{4^{2+}(-3)^2}}|=8$ $\Rightarrow$ Radius $=4$ Equation of circle $(x-1)^{2+}(y+2)^2=16$

Q62JEE Main 2022MCQ4MDefinite Integrals

The value of $int_0^\pi \;\frac{e^{\cos x} \sin x}{(1+\cos ^2 x)(e^{\cos x}+e^{-\cos x})} \text{dx}$ is equal to: [25-Jun-2022-Shift-1]

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📖 Explanation

$int_0^\pi \;\frac{e^{\cos x} \sin x}{(1+\cos ^2 x)(e^{\cos x}+e^{-\cos x})} d x$ $\;\text{ Let }\; \cos x=t$ $\sin x d x=d t$ $=int_1^{-1} \;\frac{-e^t d t}{(1+t^2)(e^t+e^{-t})}$ $I=int_{-1}^1 \;\frac{e^t}{(1+t^2)(e^t+e^{-t})} d t . . . . . . . \;\text{ (i) }\;$ $I=int_{-1}^1 \;\frac{e^{-t}}{(1+t^2)(e^{-t}+e^t)} d t . . . . \;\text{ (ii) }\;$ Adding (i) and (ii) Adding (i) and (ii) $2 I=int_{-1}^1 \;\frac{d t}{1+t^2}$ $2 I=.\tan ^{-t}|_{-1} ^1$ $2 I=\;\frac{\pi}{4}-(-\;\frac{\pi}{4})$ $2 I=\;\frac{\pi}{2}$ $I=\;\frac{\pi}{4}$

Q63JEE Main 2022MCQ4MTrigonometric Ratios and Identities

Let $a, b$ and $c$ be the length of sides of a triangle ${\text{ABC}}$ such that $\;\frac{a+b}{7}=\;\frac{b+c}{8}=\;\frac{c+a}{9}$ . If ${\text{r}}$ and ${\text{R}}$ are the radius of incircle and radius of circumcircle of the triangle $A B C$ , respectively, then the value of $\;\frac{R}{r}$ is equal to: [25-Jun-2022-Shift-1]

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📖 Explanation

$\;\frac{a+b}{7}=\;\frac{b+c}{8}=\;\frac{c+a}{9}=\lambda$ $a+b=7 \lambda$ $b+c=8 \lambda$ $c+a=9 \lambda$ $a+b+c=12 \lambda$ $\therefore a=4 \lambda , b=3 \lambda , c=5 \lambda$ $S=\;\frac{4 \lambda +3 \lambda +5 \lambda }{2}=6 \lambda$ $\Delta =\sqrt{S(s-a)(s-b)(s-c)}=\sqrt{(6 \lambda )(2 \lambda )(3 \lambda )(\lambda )}=6 \lambda ^2$ $R=\;\frac{a b c}{4 \Delta }=\;\frac{(4 \lambda )(3 \lambda )(5 \lambda )}{4(6 \lambda ^2)}=\;\frac{5}{2} \lambda$ $r=\;\frac{\Delta }{s}=\;\frac{6 \lambda ^2}{6 \lambda }=\lambda$ $\;\frac{R}{r}=\;\frac{\;\frac{5}{2} \lambda }{\lambda }=\;\frac{5}{2}$

Q64JEE Main 2022MCQ4MFunctions

Let ${\text{f}}: {\text{N}} \to {\text{R}}$ be a function such that $f(x+y)=2 f(x) f(y)$ for natural numbers ${\text{x}}$ and ${\text{y}}$ . If ${\text{f}}( 1)$ $=2$ , then the value of $\alpha$ for which $sum_{k=1}^{10} f(\alpha+k)=\;\frac{512}{3}(2^{20}-1)$ holds, is : [25-Jun-2022-Shift-1]

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📖 Explanation

Given, $f(x+y)=2 f(x) f(y)$ and $f(1)=2$ For $x=1$ and $y=1$ , $f(1+1)=2 f(1) f(1)$ $\Rightarrow f(2)=2(f(1))^2=2(2)^2=2^3$ For $x=1, y=2$ $f(1+2)=2 f(1) y(2)$ $\Rightarrow f(3)=2 \cdot 2 \cdot 2^3=2^5$ For $x=1, y=3$ $f(1+3)=2 f(1) f(3)$ $\Rightarrow f(4)=2 \cdot 2 \cdot 2^5=2^7$ For $x=1, y=4$ $f(1+4)=2 f(1) f(4)$ $\Rightarrow f(5)=2.2 \cdot 2^7=2^9 . . . .$ Also given $sum_{k=1}^{10} f(\alpha+k)=\;\frac{512}{3}(2^{20}-1)$ $\Rightarrow f(\alpha+1)+f(\alpha+2)+f(\alpha+3)+. . .+f(\alpha+10)=\;\frac{512}{3}(2^{20}-1)$ $\Rightarrow f(\alpha+1)+f(\alpha+2)+f(\alpha+3)+. . .+f(\alpha+10)=\;\frac{2^9((2^2)^{10}-1)}{2^{2-}1}$ This represent a G.P with first term $=2^9$ and common ratio $=2^2$ $\therefore$ First term $=f(\alpha+1)=2^9 . . . .(2)$ From equation $(1), f(5)=2^9$ $\therefore$ From (1) and (2), we get $f(\alpha+1)=2^9=f(5)$ $\Rightarrow f(\alpha+1)=f(5)$ $\Rightarrow f(\alpha+1)=f(4+1)$ Comparing both sides we get, $\alpha=4$

Q65JEE Main 2022MCQ4MMatrices and Determinants

Let $A$ be a $3 \times 3$ real matrix such that $A(\text{table} 1 ; 1 ; 0)=(\text{table} 1 ; 1 ; 1 ) ;$ $A(\text{table} 1 ; 0 ; 1)=(\text{table} -1 ; 0 ; 1)$ and $A(\text{table} 0 ; 0 ; 1)=(\text{table} 1 ; 1 ; 2)$ If $X=(x_1, x_2, x_3)^T$ and I is an identity matrix of order 3 , then the system $(\text{A}-\2\text{I}) \text{X}=(\text{table} 4 ; 1 ; 1)$ has : [25-Jun-2022-Shift-1]

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📖 Explanation

Let $A=\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$ $A=\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}=\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$ $\Rightarrow \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$ $=\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$ \Rightarrow $\text{table} a+b=1 ; d+e=1;g+h=0$ $A=\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}=\begin{bmatrix}-1 \\ 0 \\ 1\end{bmatrix}$ $\Rightarrow \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}$ $=\begin{bmatrix}-1 \\ 0 \\ 1\end{bmatrix}$ \Rightarrow $\text{table} a+c=-1; d+f=1;g+i=0$ $A=\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}=\begin{bmatrix}1 \\ 1 \\ 2\end{bmatrix}$ $\Rightarrow \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$ $=\begin{bmatrix}1 \\ 1 \\ 2\end{bmatrix}$ \Rightarrow ${ \text{table} rc=1; f=1; i=2;$ Solving will get $a=-2, b=3, c=1, d=-1, e=2, f=1, g=-1, h=1, i=2$ $A=\begin{bmatrix}-2 & 3 & 1 \\ -1 & 2 & 1 \\ -1 & 1 & 2\end{bmatrix} \Rightarrow A=2 I$ $=\begin{bmatrix}-4 & 3 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 0\end{bmatrix}$ $(A-2 I) x=\begin{bmatrix}4 \\ 1 \\ 1\end{bmatrix}$ $\Rightarrow -4 x_1+3 x_2+x_3=4 . . . . \;\text{ (i) }\;$ $-x_1+x_3=1$ $-x_1+x_2=1 . . . \;\text{ (ii) }\;$ $\;\text{ So }\; 3( {\text{iii}})+( {\text{ii}})=( {\text{i}})$ $\therefore$ Infinite solution

Q66JEE Main 2022MCQ4MDifferentiation

Let ${\text{f}}: {\text{R}} \to {\text{R}}$ be defined as $f(x)=x^{3+}x-5$ . If ${\text{g}}( {\text{x}})$ is a function such that $f(g(x))=x, ∀^{'} x^{'} \in R$ , then $g^{'}(63)$ is equal to [25-Jun-2022-Shift-1]

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📖 Explanation

$f(x)=3 x^{2+}1$ ${\text{f}}^{'}( {\text{x}})$ is bijective functionand $f(g(x))=x \Rightarrow g(x)$ is inverse of ${\text{f}}( {\text{x}})$ $g(f(x))=x$ $g^{'}(f(x)) \cdot f^{'}(x)=1$ $g^{'}(f(x))=\;\frac{1}{3 x^{2+}1}$ Put $x=4$ we get $g^{'}(63)=\;\frac{1}{49}$

Q67JEE Main 2022MCQ4MSets and Relations

Consider the following two propositions: $P 1: ∼(p \to ∼ q)$ $P 2:(p ∧ ∼ q) ∧((∼ p) ∨ q)$ If the proposition $p \to ((∼ p) ∨ q)$ is evaluated as FALSE, then : [25-Jun-2022-Shift-1]

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📖 Explanation

p q ∼p ∼q ∼pvq p→ (∼pvq) p→ ∼q ∼(p→ ∼q) p ∧ ∼q $p_2$ T T F F T T F T F F T F F T F F T F T F F T T F T T T F F F F F T T T T T F F F ${\text{p}} \to (∼ {\text{p}} ∨ {\text{q}})$ is ${\text{F}}$ when ${\text{p}}$ is true ${\text{q}}$ is falseFrom table ${\text{P}} 1$ & ${\text{P}} 2$ both are false

Q68JEE Main 2022MCQ4MSequences and Series

If $\;\frac{1}{2 \cdot 3^{10}}+\;\frac{1}{2^2 \cdot 3^9}+. . .+\;\frac{1}{2^{10} \cdot 3}=\;\frac{K}{2^{10} \cdot 3^{10}}$ , then the remainder when ${\text{K}}$ is divided by 6 is : [25-Jun-2022-Shift-1]

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📖 Explanation

$\;\frac{1}{2 \cdot 3^{10}}+\;\frac{1}{2^2 \cdot 3^9}+. . .+\;\frac{1}{2^{10} \cdot 3}=\;\frac{K}{2^{10} \cdot 3^{10}}$ $\Rightarrow \;\frac{1}{2 \cdot 3^{10}}\$ [;\frac{(\frac{3}{2})^{10}-1}{\frac{3}{2}-1}]$=;\frac{K}{2^{10} \cdot 3^{10}}=;\frac{3^{10}-2^{10}}{2^{10} \cdot 3^{10}}=;\frac{K}{2^{10} \cdot 3^{10}} \Rightarrow K=3^{10}-2^{10} $Now$ K=(1+2)^{10}-2^{10}={ }^{10} C_0+{ }^{10} C_1 2+{ }^{10} C_2 2^{3+}. . .+{ }^{10} C_{10} 2^{10}-2^{10}={ }^{10} C_0+{ }^{10} C_1 2+6 \lambda +{ }^{10} C_9 \cdot 2^9=1+20+5120+6 \lambda =5136+6 \lambda +5=6 \mu +5\lambda , \mu \in N\therefore $remainder$ =5$

Q69JEE Main 2022MCQ4MLimits, Continuity and Differentiability

Let ${\text{f}}( {\text{x}})$ be a polynomial function such that $f(x)+f^{'}(x)+f^{' '}(x)=x^{5+}64$ . Then, the value of $\lim _{x \to 1} \;\frac{f(x)}{x-1}$ is equal to: [25-Jun-2022-Shift-1]

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📖 Explanation

$\lim _{x \to 1} \;\frac{f(x)}{x-1}$ $f(x)+f^{'}(x)+f^{' '}(x)=x^{5+}64$ Let $f(x)=x^{5+}a x^{4+}b x^{3+}c x^{2+}d x+e$ $f^{'}(x)=5 x^{4+}4 a x^{3+}3 b x^{2+}2 c x+d$ $f^{' '}(x)=20 x^{3+}12 a x^{2+}6 b x+2 c$ $x^5(a+5) x^{4+}(b+4 a+20) x^{3+}(c+3 b+12 a) x^{2+}(d+2 c+6 b) x+e+d+2 c=x^{5+}64$ $\Rightarrow a+5=0$ $b+4 a+20=0$ $c+3 b+12 a=0$ $d+2 c+6 b=0$ $e+d+2 c=64$ $\therefore a=-5, b=0, c=60, d=-120, e=64$ $\therefore f(x)=x^{5-}5 x^{4+}60 x^{2-}120 x+64$ Now, $\lim _{x \to 1} \;\frac{x^{5-}5 x^{4+}60 x^{2-}120 x+64}{x-1}$ is ( $\;\frac{0}{0}$ form)By L' Hospital rule $\lim _{x \to 1} \;\frac{5 x^{4-}20 x^{3+}120 x-120}{1}$ $=-15$

Q70JEE Main 2022MCQ4MProbability

Let ${\text{E}}_1$ and ${\text{E}}_2$ be two events such that the conditional probabilities $P(E_1 \mid E_2)=\;\frac{1}{2}, P(E_2 \mid E_1)=\;\frac{3}{4}$ and $P(E_1 ∩ E_2)=\;\frac{1}{8}$ . Then : [25-Jun-2022-Shift-1]

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📖 Explanation

$P(\;\frac{E_1}{E_2})=\;\frac{1}{2} \Rightarrow \;\frac{P(E_1 ∩ E_2)}{P(E_2)}=\;\frac{1}{2}$ $P(\;\frac{E_2}{E_1})=\;\frac{3}{4} \Rightarrow \;\frac{P(E_2 ∩ E_1)}{P(E_1)}=\;\frac{3}{4}$ $P(E_1 ∩ E_2)=\;\frac{1}{8}$ $P(E_2)=\;\frac{1}{4}, P(E_1)=\;\frac{1}{6}$ (A) $P(E_1 ∩ E_2)=\;\frac{1}{8}$ and $P(E_1) \cdot P(E_2)=\;\frac{1}{24}$ $\Rightarrow P(E_1 ∩ E_2) \ne P(E_1) \cdot P(E_2)$ (B) $P(E_1^{'} ∩ E_2^{'})=1-P(E_1 ∪ E_2)$ $=1-[\;\frac{1}{4}+\;\frac{1}{6}-\;\frac{1}{8}]=\;\frac{17}{24}$ $P(E_1^{'})=\;\frac{3}{4} \Rightarrow P(E_1^{'}) P(E_2)=\;\frac{3}{24}$ $\Rightarrow P(E_1^{'} ∩ E_2^{'}) \ne P(E_1^{'}) . P(E_2)$ (C) $P(E_1 ∩ E_2^{'})=P(E_1)-P(E_1 ∩ E_2)$ $=\;\frac{1}{6}-\;\frac{1}{8}=\;\frac{1}{24}$ $P(E_1) \cdot P(E_2)=\;\frac{1}{24}$ $\Rightarrow P(E_1 ∩ E_2^{'})=P(E_1) \cdot P(E_2)$ (D) $P(E_1^{'} ∩ E_2)=P(E_2)-P(E_1 ∩ E_2)$ $=\;\frac{1}{4}-\;\frac{1}{8}=\;\frac{1}{8}$ $P(E_1) P(E_2)=\;\frac{1}{24}$ $\Rightarrow P(E_1^{'} ∩ E_2) \ne P(E_1) \cdot P(E_2)$

Q71JEE Main 2022MCQ4MMatrices and Determinants

Let $A=\begin{bmatrix}0 & -2 \\ 2 & 0\end{bmatrix}$ . If ${\text{M}}$ and ${\text{N}}$ are two matrices given by $M=sum_{k=1}^{10} A^{2 k}$ and $N=sum_{k=1}^{10} A^{2 k-1}$ then ${\text{MN}}^2$ is : [25-Jun-2022-Shift-1]

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📖 Explanation

$A=\begin{bmatrix}0 & -2 \\ 2 & 0\end{bmatrix}$ $A^2=\begin{bmatrix}0 & -2 \\ 2 & 0\end{bmatrix}\begin{bmatrix}0 & -2 \\ 2 & 0\end{bmatrix}=\begin{bmatrix}-4 & 0 \\ 0 & -4\end{bmatrix}=-4 I$ $M=A^{2+}A^{4+}A^{6+}. . . . . .+A^{20}$ $=-4 I+16 I-64 I+. . . . .$ upto 10 terms $=-I[4-16+64 . . . .+$ upto 10 terms $]$ $=-I .4\$ [;\frac{(-4)^{10}-1}{-4-1}]$=;\frac{4}{5}(2^{20}-1) IN=A^{1+}A^{3+}A^{5+}. . . . .+A^{19}=A-4 A+16 A+. . . . . $upto 10 terms$ =A(;\frac{(-4)^{10}-1}{-4-1})=-(;\frac{2^{20}-1}{5}) AN^2=;\frac{(2^{20}-1)^2}{2^5} A^2=;\frac{-4}{24}(2^{20}-1)^2 IM N^2=;\frac{-16}{125}(2^{20}-1)^3 I=K I ;; (K \ne \pm 1)(M N^2)^T=(K I)^T=K I\therefore {\text{A}}$ is correct

Q72JEE Main 2022MCQ4MApplication of Derivatives

Let $g:(0, \infty ) \to R$ be a differentiable function such that $\int(\;\frac{x(\cos x-\sin x)}{e^x+1}+\;\frac{g(x)(e^x+1-x e^x)}{(e^x+1)^2}) d x=\;\frac{x g(x)}{e^x+1}+c$ , for all ${\text{x}}>0$ , where ${\text{c}}$ is an arbitrary constant. Then : [25-Jun-2022-Shift-1]

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📖 Explanation

$\int(\;\frac{x(\cos x-\sin x)}{e^x+1}+\;\frac{g(x)(e^x+1-x e^x)}{(e^x+1)^2}) d x=\;\frac{x g(x)}{e^x+1}+c$ On differentiating both sides w.r.t. $x$ , we get $(\;\frac{x(\cos x-\sin x)}{e^x+1}+\;\frac{g(x)(e^x+1-x e^x.}{(e^x+1)^2})$ $=\;\frac{(e^x+1)(g(x)+x g^{'}(x))-e^x \cdot x \cdot g(x)}{(e^x+1)^2}$ $(e^x+1) x(\cos x-\sin x)+g(x)(e^x+1-x e^x)$ $=(e^x+1)(g(x)+x g^{'}(x))-e^x \cdot x \cdot g(x)$ $\Rightarrow {\text{g}}^{'}( {\text{x}})=\cos {\text{x}}-\sin {\text{x}}$ $\Rightarrow {\text{g}}( {\text{x}})=\sin {\text{x}}+\cos {\text{x}}+ {\text{C}}$ ${\text{g}}( {\text{x}})$ is increasing in $(0, \pi / 4)$ $g^{' '}(x)=-\sin x-\cos x< 0$ $\Rightarrow {\text{g}}^{'}( {\text{x}})$ is decreasing functionlet $h(x)=g(x)+g^{'}(x)=2 \cos x+C$ $\Rightarrow h^{'}(x)=g^{'}(x)+g^{' '}(x)=-2 \sin x< 0$ $\Rightarrow {\text{h}}$ is decreasing let $\varphi (x)=g(x)-g^{'}(x)=2 \sin x+C$ $\Rightarrow \varphi ^{'}( {\text{x}})= {\text{g}}^{'}( {\text{x}})- {\text{g}}^{' '}( {\text{x}})=2 \cos {\text{x}}>0$ $\Rightarrow \varphi$ is increasing

Q73JEE Main 2022MCQ4MApplication of Derivatives

Let $f: R \to R$ and $g: R \to R$ be two functions defined by $f(x)=\log _e(x^{2+}1)-e^{-x}+1$ and $g(x)=\;\frac{1-2 e^{2 x}}{e^x}$ . Then, for which of the following range of $\alpha$ , the inequality $f(g(\;\frac{(\alpha-1)^2}{3}))>f(g(\alpha-\;\frac{5}{3}))$ holds ? [25-Jun-2022-Shift-1]

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📖 Explanation

$f(x)=\log _e(x^{2+}1)-e^{-x}+1$ $f^{'}(x)=\;\frac{2 x}{x^{2+}1}+e^{-x}$ $=\;\frac{2}{x+\;\frac{1}{x}}+e^{-x}>0 ∀ x \in R$ $g(x)=e^{-x}-2 e^x$ $g^{'}(x)--e^{-x}-2 e^x< 0 \;\; ∀ x \in R$ $\Rightarrow f( {\text{x}})$ is increasing and g $( {\text{x}})$ is decreasing function. $f(g(\;\frac{(\alpha-1)^2}{3}))>f(g(\alpha-\;\frac{5}{3}))$ $\Rightarrow \;\frac{(\alpha-1)^2}{3}< \alpha-\;\frac{5}{3}$ $=\alpha^{2-}5 \alpha+6< 0$ $=(\alpha-2)(\alpha-3)< 0$ $=\alpha \in (2,3)$

Q74JEE Main 2022MCQ4MVector Algebra

Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \widehat{k} a_i>0, i=1,2,3$ be a vector which makes equal angles with the coordinate axes OX, OY and OZ. Also, let the projection of $\vec{a}$ on the vector $3 \hat{i}+4 \hat{j}$ be 7. Let $\vec{b}$ be a vector obtained by rotating $\vec{a}$ with $90^{\circ}$ . If $\vec{a}, \vec{b}$ and $x$ -axis are coplanar, then projection of a vector $\vec{b}$ on $3 \hat{i}+4 \hat{j}$ is equal to: [25-Jun-2022-Shift-1]

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📖 Explanation

$\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma =1 \Rightarrow \cos ^2 \alpha=\;\frac{1}{3} \Rightarrow \cos \alpha=\;\frac{1}{\sqrt{3}}$ $\vec{a}=\;\frac{\lambda }{3}(\hat{i}+\hat{j}+\widehat{k}), \lambda >0$ $\;\frac{\lambda }{\sqrt{3}} \;\frac{(\hat{i}+\hat{j}+\widehat{k}) \cdot (3 \hat{i}+4 \hat{j})}{\sqrt{3^{2+}4^2}}=7$ $\Rightarrow \;\frac{\lambda }{\sqrt{3}}(3+4)=7 \times 5$ $\therefore \lambda =5 \sqrt{3}$ $\vec{a}=5(\hat{i}+\hat{j}+\widehat{k})$ Let $\vec{b}=p \hat{i}+q \hat{j}+r \widehat{k}$ $\vec{a} \cdot \vec{b}=0$ and $[\vec{a} b \hat{i}]=0$ $\Rightarrow p+q+r=0$ $\&\begin{vmatrix}p & q & r \\ 1 & 1 & 1 \\ 1 & 0 & 0\end{vmatrix}=0 \Rightarrow q=r$ $\vec{b}=-2 r \hat{i}+r \hat{j}+r \widehat{k}$ $\vec{b}=r(-2 \hat{i}+\hat{j}+\widehat{k})$ $\;\text{ Now }\;|\vec{a}|=|\vec{b}|$ $5 \sqrt{3}=|r| \sqrt{b} \Rightarrow |r|=\;\frac{5}{\sqrt{2}}$ $\Rightarrow \;\text{ Projection of }\; \vec{b} \;\text{ on }\; 3 \hat{i}+4 \hat{j}=|\;\frac{\vec{b} \cdot (3 \hat{i}+4 \hat{j})}{\sqrt{3^{2+}4^2}}|$ $=|r| \;\frac{(-6+4)}{5}=|\;\frac{-2 r}{5}|$ $\;\text{ Projection }\;=\;\frac{2}{5} \times \;\frac{5}{\sqrt{2}}=\sqrt{2}$ $\therefore B$ is correct.

Q75JEE Main 2022MCQ4MDifferential Equations

Let $y=y(x)$ be the solution of the differential equation $(x+1) y^{'}-y=e^{3 x}(x+1)^2$ , with $y(0)=\;\frac{1}{3}$ . Then, the point $x=-\;\frac{4}{3}$ for the curve $y=y(x)$ is : [25-Jun-2022-Shift-1]

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📖 Explanation

$(x+1) \;\frac{d y}{d x}-y=e^{3 x}(x+1)^2$ $\;\frac{d y}{d x}-\;\frac{y}{x+1}=e^{3 x}(x+1)$ If $e^{-\int \;\frac{1}{x+1} x}=e^{-\log (x+1)}=\;\frac{1}{x+1}$ $\therefore y(\;\frac{1}{x+1})=\int \;\frac{e^{3 x}(x+1)}{x+1} d x$ $\;\frac{y}{x+1}=\int e^{3 x} d x$ $\;\frac{y}{x+1}=\;\frac{e^{3 x}}{3}+c$ $\because y(0)=\;\frac{1}{3}$ $\;\frac{1}{3}=\;\frac{1}{3}+c$ $\therefore c=0$ So : $y=\;\frac{e^{3 x}}{3}(x+1)$ $y^{'}=e^{3 x}(x+1)+\;\frac{e^{3 x}}{3}=e^{3 x}(x+\;\frac{4}{3})$ $y^{' '}=3 e^{3 x}(x+\;\frac{4}{3})+e^{3 x}=e^{3 x}(3 x+5)$ $y^{'}=0$ at $x=\; {\frac{-4}{3}} \& y''=e^{-4}(1)>0$ at $x=\;\frac{-4}{3}$ $\Rightarrow x=\;\frac{-4}{3}$ is point of local minima

Q76JEE Main 2022MCQ4MStraight Lines and Pair of Straight Lines

If $y=m_1 x+c_1$ and $y=m_2 x+c_2, m_1 \ne m_2$ are two common tangents of circle $x^{2+}y^2=2$ and parabola $y^2= {\text{x}}$ , then the value of $8 |m_1 m_2|$ is equal to : [25-Jun-2022-Shift-1]

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📖 Explanation

Let tangent to $y^2=x$ be $y=m x+\;\frac{1}{4 m}$ For it being tangent to circle. $|\;\frac{\;\frac{1}{4} m}{\sqrt{1+m^2}}|=\sqrt{2}$ $\Rightarrow 32 m^{4+}32 m^{2-}1=0$ $\Rightarrow m^2=\;\frac{-32 \pm \sqrt{(32)^{2+}4(32)}}{64}$ $\Rightarrow 8 m_1 m_2=-4+3 \sqrt{2}$

Q77JEE Main 2022MCQ4MThree Dimensional Geometry

Let $Q$ be the mirror image of the point $P(1,0,1)$ with respect to the plane $S: x+y+z=5$ . If a line $L$ passing through $(1,-1,-1)$ , parallel to the line $P Q$ meets the plane $S$ at $R$ , then $Q R^2$ is equal to : [25-Jun-2022-Shift-1]

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📖 Explanation

As $L$ is parallel to $P Q$ d.r.s of $S$ is $⟨ 1,1,1⟩$ $\therefore L \equiv \;\frac{x-1}{1}=\;\frac{y+1}{1}=\;\frac{z+1}{1}$ Point of intersection of $L$ and $S$ be $\lambda$ $\Rightarrow (\lambda +1)+(\lambda -1)+(\lambda -1)=S$ $\Rightarrow \lambda =2$ $\therefore R \equiv (3,1,1)$ $\;\text{ Let }\; Q(\alpha, \beta, \gamma )$ $\Rightarrow \;\frac{\alpha-1}{1}=\;\frac{\beta}{1}=\;\frac{\gamma -1}{1}=\;\frac{-2(-3)}{3}$ $\Rightarrow \alpha=3, \beta=2, \gamma =3$ $\Rightarrow Q \equiv (3,2,3)$ $(Q R)^2=0^{2+}(1)^{2+}(2)^2=5$

Q78JEE Main 2022MCQ4MDifferential Equations

If the solution curve $y=y(x)$ of the differential equation $y^2 d x+(x^{2-}x y+y^2) d y=0$ , which passes through the point (1, 1) and intersects the line $y=\sqrt{3} x$ at the point $(\alpha, \sqrt{3} \alpha)$ , then value of $\log _e(\sqrt{3} \alpha)$ is equal to : [25-Jun-2022-Shift-1]

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📖 Explanation

$\;\frac{d y}{d x}=\;\frac{y^2}{x y-x^{2-}y^2}$ Put $y=v x$ we get $v+x \;\frac{d v}{d x}=\;\frac{v^2}{v-1-v^2}$ $\Rightarrow x \;\frac{d v}{d x}=\;\frac{v^{2-}v^{2+}v+v^3}{v-1-v^2}$ $\Rightarrow \int \;\frac{v-1-v^2}{v(1+v^2)} d v=\int \;\frac{d x}{x}$ $\tan ^{-1}(\;\frac{y}{x})-\ln (\;\frac{y}{x})=\ln x+c$ As it passes through $(1,1)$ $c=\;\frac{\pi}{4}$ $\Rightarrow \tan ^{-1}(\;\frac{y}{x}) \ln (\;\frac{y}{x})=\ln x+\;\frac{\pi}{4}$ Put $y=\sqrt{3} x$ we get $\Rightarrow \;\frac{\pi}{3}-\ln \sqrt{3}=\ln x+\;\frac{\pi}{4}$ $\Rightarrow \ln x=\;\frac{\pi}{12}-\ln \sqrt{3}=\ln \alpha$ $\therefore \ln (\sqrt{3} \alpha)=\ln \sqrt{3}+\ln \alpha$ $=\ln \sqrt{3}+\;\frac{\pi}{12}-\ln \sqrt{3}=\;\frac{\pi}{12}$

Q79JEE Main 2022MCQ4MParabola

Let $x=2 t, y=\;\frac{t^2}{3}$ be a conic. Let ${\text{S}}$ be the focus and ${\text{B}}$ be the point on the axis of the conic such that $S A ⟂ B A$ , where A is any point on the conic. If ${\text{k}}$ is the ordinate of the centroid of the $\Delta S A B$ , then $\lim _{t \to 1} k$ is equal to: [25-Jun-2022-Shift-1]

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📖 Explanation

$x=2 t, y=\;\frac{2}{3}$ $t \to 1 A \equiv (2, \;\frac{1}{3})$ Given conic is $x^2=12 y \Rightarrow S \equiv (0,3)$ Let $B \equiv (0, \beta)$ Given $S A ⟂ B A$ $(\;\frac{\;\frac{1}{3}}{2-3})(\;\frac{\beta-\frac{1}{3}}{-2})=-1$ $\Rightarrow (\beta-\;\frac{1}{3}) \;\frac{1}{3}=-2$ $\Rightarrow \beta=\;\frac{1}{3}(\;\frac{-17}{3})$ Ordinate of centroid $=k=\;\frac{\beta+\frac{1}{3}+3}{3}$ $=\;\frac{\;\frac{-17}{9}+\frac{10}{3}}{3}=\;\frac{13}{18}$

Q80JEE Main 2022MCQ4MComplex Numbers

Let a circle ${\text{C}}$ in complex plane pass through the points $z_1=3+4 i, z_2=4+3 i$ and $z_3=5 i$ . If $z(\ne z_1)$ is a point on ${\text{C}}$ such that the line through $z$ and $z_1$ is perpendicular to the line through $z_2$ and $z_3$ , then $\arg (z)$ is equal to: [25-Jun-2022-Shift-1]

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📖 Explanation

$z_1=3+4 i, z_2=4+3 i$ and $z_3=5 i$ Clearly, $C \equiv x^{2+}y^2=25$ Let $z(x, y)$ $\Rightarrow (\;\frac{y-4}{x-3})(\;\frac{2}{-4})=-1$ $\Rightarrow y=2 x-2 \equiv L$ $\therefore {\text{z}}$ is intersection of ${\text{C}} \& {\text{L}}$ $\Rightarrow z \equiv (\;\frac{-7}{5}, \;\frac{-24}{5})$ $\therefore {\text{Arg}}(z)=-\pi+\tan ^{-1}(\;\frac{24}{7})$

Q81JEE Main 2022NAT4MBinomial Theorem

Let ${\text{C}}_{ {\text{r}}}$ denote the binomial coefficient of ${\text{x}}^{ {\text{r}}}$ in the expansion of $(1+x)^{10}$ . If for $\alpha, \beta \in {\text{R}}, C_1+3.2 C_2+5.3 C_3+. . . . . .$ upto 10 terms $=\;\frac{\alpha \times 2^{11}}{2^\beta-1}(C_0+\;\frac{C_1}{2}+\;\frac{C_2}{3}+. . ..$ upto 10 terms $)$ then the value of $\alpha+\beta$ is equal to [25-Jun-2022-Shift-1]

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📖 Explanation

$(1+ {\text{x}})^{10}= {\text{C}}_0+ {\text{C}}_1 {\text{x}}+ {\text{C}}_2 {\text{x}}^{2+}. . . . . .+ {\text{C}}_{10} {\text{x}}^{10}$ Differentiating $10(1+ {\text{x}})^9= {\text{C}}_1+2 {\text{C}}_2 {\text{x}}+3 {\text{C}}_3 {\text{x}}^{2+}. . . .+10 {\text{C}}_{10} {\text{x}}^9$ replace $x \to x^2$ $10(1+x^2)^9=C_1+2 C_2 x^{2+}3 C_3 x^{4+}. . . .+10 C_{10} x^{18}$ $10 \cdot x(1+x^2)^9=C_1 x+2 C_2 x^{3+}3 C_3 x^{5+}. . . .+10 C_{10} x^{19}$ Differentiating $10((1+x^2)^9 \cdot 1+x \cdot 9(1+x^2)^8 2 x)$ $= {\text{C}}_1 {\text{x}}+2 {\text{C}}_2 \cdot 3 {\text{x}}^{3+}3 \cdot 5 \cdot {\text{C}}_3 {\text{x}}^{4+}. . . .+10 \cdot 19 {\text{C}}_{10} {\text{x}}^{18}$ putting ${\text{x}}=1$ $10(2^{9+}18 \cdot 2^8)$ $= {\text{C}}_1+3 \cdot 2 \cdot {\text{C}}_2+5 \cdot 3 \cdot {\text{C}}_3+. . . .+19 \cdot 10 \cdot {\text{C}}_{10}$ ${\text{C}}_1+3 \cdot 2 \cdot {\text{C}}_2+. . . . . .+19 \cdot 10 \cdot {\text{C}}_{10}$ $=10 \cdot 2^9 \cdot 10=100 \cdot 2^9$ ${\text{C}}_0+\;{ {\text{C}}_1}/{2}+\;{ {\text{C}}_2}/{3}+. . . .+\;{ {\text{C}}_9}/{11}_{10^{\;\text{th }\;} \text{term}}+\;{ {\text{C}}_{10}}/{11}_{11^{\;\text{th }\;} \text{term}}=\;\frac{2^{11}-1}{11}$ ${\text{C}}_0+\;{ {\text{C}}_1}/{2}+\;{ {\text{C}}_2}/{3}+. . . . .+\;{ {\text{C}}_9}/{11}=\;\frac{2^{11}-2}{11}$ Now, $100 \cdot 2^9=\;\frac{\alpha \cdot 2^{11}}{2^\beta-1}(\;\frac{2^{11}-2}{11})$ Eqn. of form ${\text{y}}= {\text{k}}(2^{ {\text{x}}}-1)$ It has infinite solutions even if we take $x, y \in N$ .

Q82JEE Main 2022NAT4MPermutations and Combinations

The number of 3-digit odd numbers, whose sum of digits is a multiple of 7 , is____ [25-Jun-2022-Shift-1]

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📖 Explanation

For odd numbers, unit place shall be $1,3,5,7$ or 9 . $x{x} y x{y}, x{x} y x{y}, x{x} y x{y}, x{x} y x{z}, x{x} y x{9}$ are the type of numbers.If $x y 1$ then $x+y=6,13,20 ⋯$ Total number $=6+6+0+⋯=12$ If $x$ y 3 then $x+y=4,11,18, ⋯$ Total number $=4+8+1+0 ⋯=13$ Similarly for $x$ y 5 , we have $x+y=2,9,16, ⋯$ Total number $=2+9+3=14$ for $x$ y 7 we have $x+y=0,7,14, ⋯$ Total number $=0+7+5=12$ waysAnd for $x$ y 9 we have $x+y=5,12,19 . . .$ Total number $=5+7+0 ⋯=12$ ways $\therefore$ Total odd numbers whose sum of digits is a multiple of 7 is 63 .

Q83JEE Main 2022NAT4MVector Algebra

Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$ , where $\vec{a \mid }|=4,| \vec{b \mid }=$ ,3 and $\theta \in (\;\frac{\pi}{4}, \;\frac{\pi}{3})$ . Then $|(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})|^{2+}4(\vec{a} \cdot \vec{b})^2$ is equal to__ [25-Jun-2022-Shift-1]

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📖 Explanation

$|(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})|^{2+}4(\vec{a} \cdot \vec{b})^2$ $\Rightarrow |\vec{a} \times \vec{a}+\vec{a} \times \vec{b}-\vec{b} \times \vec{a}-\vec{b} \times \vec{b}|^{2+}4(\vec{a} \cdot \vec{b})^2$ $\Rightarrow |2(\vec{a} \times \vec{b})|^{2+}4(\vec{a} \cdot \vec{b})^2$ $\Rightarrow 4(\vec{a} \times \vec{b})^{2+}(\vec{a} \cdot \vec{b})^2$ $.\Rightarrow 4{|\vec{a}|}^2 \;{|\vec{b}|}^2=4.16 .9=576$

Q84JEE Main 2022NAT4MCircles

Let the abscissae of the two points ${\text{P}}$ and ${\text{Q}}$ be the roots of $2 x^{2-}r x+p=0$ and the ordinates of ${\text{P}}$ and ${\text{Q}}$ be the roots of $x^{2-}s x-q=0$ . If the equation of the circle described on ${\text{PQ}}$ as diameter is $2(x^{2+}y^2)-11 x-14 y-22=0$ , then $2 r+s-2 q+p$ is equal to [25-Jun-2022-Shift-1]

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📖 Explanation

Let $P(x_1, y_1) \& Q(x_2, y_2)$ $\therefore$ Roots of $2 x^{2-}r x+p=0$ are $x_1, x_2$ and roots of $x^{2-}s x-q=0$ are $y_1, y_2$ $\therefore$ Equation of circle $\equiv (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$ $\Rightarrow x^{2-}(x_1+x_2) x+x_1 x_2+y^{2-}(y_1+y_2) y+y_1 y_2=0$ $\Rightarrow x^{2-}\;\frac{r}{2} x+\;\frac{p}{2}+y^{2+}s y-q=0$ $\Rightarrow 2 x^{2+}2 y^{2-}r x+2 s y+p-2 q=0$ Compare with $2 x^{2+}2 y^{2-}11 x-14 y-22=0$ We get $r=11, s=7, p-2 q=-22$ $\Rightarrow 2 r+s+p-2 q=22+7-22=7$

Q85JEE Main 2022NAT4MTrigonometric Ratios and Identities

The number of values of $x$ in the interval $(\;\frac{\pi}{4}, \;\frac{7 \pi}{4})$ for which $14 \text{cosec}^2 x-2 \sin ^2 x=21-4 \cos ^2 x$ holds, is____ [25-Jun-2022-Shift-1]

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📖 Explanation

$\;\frac{14}{\sin ^2 x}-2 \sin ^2 x=21-4(1-\sin ^2 x)$ Let $\sin ^2 x=t$ $\Rightarrow 14-2 t^2=21 t-4 t+4 t^2$ $\Rightarrow 6 t^{2+}17 t-14=0$ $\Rightarrow 6 t^{2+}21 t-4 t-14=0$ $\Rightarrow 3 t(2 t+7)-2(2 t+7)=0$ $\Rightarrow \sin 2 x=\;\frac{2}{3}$ or $-\;\frac{7}{3}($ rejected $)$ $\Rightarrow \sin x=\pm \sqrt{\;\frac{2}{3}}$

Q86JEE Main 2022NAT4MSequences and Series

For a natural number n, let $\alpha_n=19^n-12^n$ . Then, the value of $\;\frac{31 \alpha_9-\alpha_{10}}{57 \alpha_8}$ is___ [25-Jun-2022-Shift-1]

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📖 Explanation

$\alpha_n=19^n-12^n$ Let equation of roots $12 \& 19$ i.e. $x^{2-}31 x+228=0$ $\Rightarrow (31-x)=\;\frac{228}{x}$ (where ${\text{x}}$ can be 19 or 12 $)$ $\therefore \;\frac{31 \alpha_9-\alpha_{10}}{57 \alpha_8}=\;\frac{31(19^{9-}12^9)-(19^{10}-12^{10})}{57(19^{8-}12^8)}$ $=\;\frac{19^9(31-19)-12^9(31-12)}{57(19^{8-}12^8)}$ $=\;\frac{228(19^{8-}12^8)}{57(19^{8-}12^8)}=4$

Q87JEE Main 2022NAT4MFunctions

Let $f: R \to R$ be a function defined by $f(x)=(2(1-\;\frac{x^{25}}{2})(2+x^{25}))^{\;\frac{1}{50}}$ . If the function $g(x)=f(f(f(x))+f(f(x))$ , then the greatest integer less than or equal to $g(1)$ is___ [25-Jun-2022-Shift-1]

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📖 Explanation

Given, $f(x)=(2(1-\;\frac{x^{25}}{2})(2+x^{25}))^{\;\frac{1}{50}}$ and $g(x)=f(f(f(x)))+f(f(x))$ $\therefore g(1)=f(f(f(1)))+f(f(1))$ Now, $f(1)=(2(1-\;\frac{1^{25}}{2})(2+1^{25}))^{\;\frac{1}{50}}$ $=(2(1-\;\frac{1}{2})(2+1))^{\;\frac{1}{50}}$ $=(3)^{\;\frac{1}{50}}$ $\therefore f(f(1))=f(3^{\;\frac{1}{50}})$ $=(2(1-\;{(3^{\frac{1}{50}})^{25}}/{2})(2+(3^{\;\frac{1}{50}})^{25}))^{\;\frac{1}{50}}$ $=(2(1-\;{3^{\frac{1}{2}}}/{2})(2+3^{\;\frac{1}{2}}))^{\;\frac{1}{50}}$ $=(2 \times (\;\frac{2-\sqrt{3}}{2})(2+\sqrt{3}))^{\;\frac{1}{50}}$ $=[(2-\sqrt{3})(2+\sqrt{3})]^{\;\frac{1}{50}}$ $=(4-3)^{\;\frac{1}{50}}$ $=1^{\;\frac{1}{50}}=1$ Now, $f(f(f(1)))=f(1)=3^{\;\frac{1}{50}}$ $\therefore g(1)=f(f(f(1)))+f(f(1))$ $=3^{\;\frac{1}{50}}+1$ Now, greatest integer less than or equal to $g(1)$ $=[g(1)]$ $=[3^{\;\frac{1}{50}}+1]$ $=\$ [3^{;\frac{1}{50}}]$+[1]=[1.02]+1=1+1=2$

Q88JEE Main 2022NAT4MThree Dimensional Geometry

Let the lines $L_1: \vec{r}=\lambda (\hat{i}+2 \hat{j}+3 \widehat{k}), \lambda \in R$ $L_2: \vec{r}=(\hat{i}+3 \hat{j}+\widehat{k})+\mu (\hat{i}+\hat{j}+5 \widehat{k}) ; \mu \in R$ intersect at the point $S$ . If a plane $a x+b y-z+d=0$ passes through $S$ and is parallel to both the lines $L_1$ and $L_2$ , then the value of $a+b+d$ is equal to [25-Jun-2022-Shift-1]

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📖 Explanation

As plane is parallel to both the lines we have d.r's of normal to the plane as $< 7,-2,-1>$ $({\text{from}}\begin{vmatrix}\hat{i} & \hat{j} & \widehat{k} \\ 1 & 2 & 3 \\ 1 & 1 & 5\end{vmatrix}=7 \hat{i}-\hat{j}(2)+\widehat{k}(-1))$ Also point of intersection of lines is $2 \hat{i}+4 \hat{j}+6 \hat{k}$ $\therefore$ Equation of plane is $7(x-2)-2(y-4)-1(z-6)=0$ $\Rightarrow 7 x-2 y-z=0$ $a+b+d=7-2+0=5$

Q89JEE Main 2022NAT4MPermutations and Combinations

Let $A$ be a $3 \times 3$ matrix having entries from the set $\{-1,0,1\}$ . The number of all such matrices $A$ having sum of all the entries equal to 5 , is [25-Jun-2022-Shift-1]

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📖 Explanation

Case-I: $1 \to 7$ timesand $-1 \to 2$ timesnumber of possible marrix $=\;\frac{9 !}{7 ! 2 !}=36$ Case-II: $1 \to 6$ times, $-1 \to 1$ timesand $0 \to 2$ timesnumber of possible marrix $=\;\frac{9 !}{6 ! 2 !}=252$ Case-III: $1 \to 5$ timesand $0 \to 4$ timesnumber of possible marrix $=\;\frac{9 !}{5 ! 4 !}=126$ Hence total number of all such matrix $A=414$

Q90JEE Main 2022NAT4MSequences and Series

The greatest integer less than or equal to the sum of first 100 terms of the sequence $\;\frac{1}{3}, \;\frac{5}{9}, \;\frac{19}{27}, \;\frac{65}{81}$ , . is equal to [25-Jun-2022-Shift-1]

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📖 Explanation

$S=\;\frac{1}{3}+\;\frac{5}{9}+\;\frac{19}{27}+\;\frac{65}{81}+. . .$ $=sum_{r=1}^{100}(\;\frac{3^r-2^r}{3^r})$ $=100-\;\frac{2}{3} \;\frac{(1-(\frac{2}{3})^{100})}{1 / 3}$ $=98+2(\;\frac{2}{3})^{100}$ $\therefore [S]=98$

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