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27 Jun 2022 Shift 1

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Physics30 questions

Q1JEE Main 2022MCQ4MMotion in a Plane

A projectile is launched at an angle ' $\alpha$ ' with the horizontal with a velocity $20 {\text{ms}}^{-1}$ . After $10 {\text{s}}$ , its inclination with horizontal is ' $\beta$ '. The value of $\tan \beta$ will be : $( {\text{g}}=10 {\text{ms}}^{-2})$ . [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$v_y=20 \times \sin \alpha-10 \times 10$ $v_x=20 \cos \alpha$ $\therefore \tan \beta=\;\frac{v_y}{v_x}=\;\frac{20 \sin \alpha-100}{20 \cos \alpha}$ $=\tan \alpha-5 \sec \alpha$

Q2JEE Main 2022MCQ4MMotion in a Plane

A girl standing on road holds her umbrella at $45^{\circ}$ with the vertical to keep the rain away. If she starts running without umbrella with a speed of $15 \sqrt{2} {\text{kmh}}^{-1}$ , the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is : [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

From graph, $v_{R G}=15 \sqrt{2} \tan 45^{\circ}$ $=15 \sqrt{2}$ $=\;\frac{30}{\sqrt{2}}$

Q3JEE Main 2022MCQ4MUnits and Measurements

A silver wire has a mass $(0.6 \pm 0.006) {\text{g}}$ , radius $(0.5 \pm 0.005) {\text{mm}}$ and length $(4 \pm 0.04) {\text{cm}}$ . The maximum percentage error in the measurement of its density will be : [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\rho =\;\frac{m}{V}=\;\frac{m}{\pi r^2 \times l}$ $\therefore \%$ error in $\rho = (\;\frac{0.006}{0.6}+2 \times \;\frac{0.005}{0.5}+\;\frac{0.04}{4}) \times 100$ $=4 \%$

Q4JEE Main 2022MCQ4MLaws of Motion

A system of two blocks of masses $m=2 {\text{kg}}$ and $M=8 {\text{kg}}$ is placed on a smooth table as shown in figure. The coefficient of static friction between two blocks is $0.5$ . The maximum horizontal force ${\text{F}}$ that can be applied to the block of mass ${\text{M}}$ so that the blocks move together will be : [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\therefore a_{\max }=\mu {\text{g}}$ $=0.5 \times 9.8=4.9 {\text{m}} / {\ s}^2$ $\therefore {\text{F}}_{\max }=(8+2) \times 4.9=49 {\text{N}}$

Q5JEE Main 2022MCQ4MCentre of Mass and Collision

Two blocks of masses $10 {\text{kg}}$ and $30 {\text{kg}}$ are placed on the same straight line with coordinates $(0,0) {\text{cm}}$ and $(x, 0) {\text{cm}}$ respectively. The block of $10 {\text{kg}}$ is moved on the same line through a distance of $6 {\text{cm}}$ towards the other block. The distance through which the block of $30 {\text{kg}}$ must be moved to keep the position of centre of mass of the system unchanged is : [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

For COM to remain unchanged, $m_1 x_1=m_2 x_2$ $\Rightarrow 10 \times 6=30 \times x_2$ $\Rightarrow x_2=2 {\text{cm}}$ towards $10 {\text{kg}}$ block.

Q6JEE Main 2022MCQ4MCurrent Electricity

A $72 Ω$ galvanometer is shunted by a resistance of $8 Ω$ . The percentage of the total current which passes through the galvanometer is : [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

From the given setup $y \times R_G=(x-y) (R_S)$ $\Rightarrow y \times 72=(x-y) \times 8$ $\Rightarrow 9 y=x-y$ $\Rightarrow y=\;\frac{x}{10} \;\text{ or }\; 10 \% \;\text{ of }\; {\text{x}}$

Q7JEE Main 2022MCQ4MGravitation

Given below are two statements :Statement I: The law of gravitation holds good for any pair of bodies in the universe.Statement II : The weight of any person becomes zero when the person is at the centre of the earth.In the light of the above statements, choose the correct answer from the options given below. [27-Jun-2022-Shift-1]

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Q8JEE Main 2022MCQ4MCentre of Mass and Collision

What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass?(Assume the collision to be head-on elastic collision) [27-Jun-2022-Shift-1]

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📖 Explanation

For a head on elastic collision $v_2=\;\frac{m u_1}{m+5 m}+\;\frac{m u_1}{m+5 m}$ $=\;\frac{2 u_1}{6} \;\text{ or }\; \;\frac{u_1}{3}$ Initial kinetic energy of first mass $=\;\frac{1}{2} m u_1^2$ Final kinetic energy of second mass $=\;\frac{1}{2} \times 5 m (\;\frac{u_1}{3}) ^2$ $=\;\frac{5}{9} (\;\frac{1}{2} m u_1^2)$ $\Rightarrow$ kinetic energy transferred $=55 \%$ of initial kinetic energy of first colliding mass

Q9JEE Main 2022MCQ4MProperties of Matter

The velocity of a small ball of mass ' $m$ ' and density $d_1$ , when dropped in a container filled with glycerin, becomes constant after some time. If the density of glycerin is ${\text{d}}_2$ , then the viscous force acting on the ball, will be : [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Viscous force acting on the ball will be equal and opposite to net of weight and buoyant force $\Rightarrow F_0=\;\frac{4}{3} \pi r^3 d_1 g-\;\frac{4}{3} \pi r^3 d_2 g$ $=\;\frac{4}{3} \pi r^3 d_1 g (1-\;\frac{d_2}{d_1})$ $=m g (1-\;\frac{d_2}{d_1})$

Q10JEE Main 2022MCQ4MMagnetism and Matter

The susceptibility of a paramagnetic material is 99 . The permeability of the material in Wb/A-m, is :$[Permeability of free space $µ_0=4 pi × 10^{-7} {\text{Wb}} / {\text{A}}- {\text{m}}$ ]$ [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\mu _{ {\text{r}}}= {\text{x}}+1$ $=99+1=100$ $\Rightarrow \mu =\mu _{ {\text{r}}} \mu _0=100 \times 4 \mu \times 10^{-7} {\text{Wb}} / {\text{Am}}$ $=4 \mu \times 10^{-5} {\text{Wb}} / {\text{Am}}$

Q11JEE Main 2022MCQ4MAlternating Current

The current flowing through an ac circuit is given by $I=5 \sin (120 \pi t) A$ How long will the current take to reach the peak value starting from zero? [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\omega =120 \pi$ $\Rightarrow T=\;\frac{1}{60} {\ sec}$ The current will take its peak value in $\;\frac{T}{4}$ timeSo $t=\;\frac{T}{4}$ $=\;\frac{1}{240} s$

Q12JEE Main 2022MCQ4MElectromagnetic Waves

List - I List - II (a) Ultraviolet rays (i) Study crystal structure (b) Microwaves (ii) Greenhouse effect (c) Infrared rays (iii) Sterilizing surgical instrument (d) X-rays (iv) Radar system Choose the correct answer from the options given below: [27-Jun-2022-Shift-1]

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Q13JEE Main 2022MCQ4MDual Nature of Matter and Radiation

An $\alpha$ particle and a carbon 12 atom has same kinetic energy ${\text{K}}$ . The ratio of their de-Broglie wavelengths $(\lambda _alpha: \lambda _{C 12})$ is : [27-Jun-2022-Shift-1]

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📖 Explanation

$K_alpha=K_C$ $\;\frac{p_alpha^2}{2 m_alpha}=\;\frac{p_C^2}{2 m_C}$ $\;\frac{p_alpha}{p_C}=\sqrt{\;\frac{m_alpha}{m_C}}$ So $\;\frac{\lambda _alpha}{\lambda _C}=\;\frac{h / p_alpha}{h / p_C}=\sqrt{\;\frac{m_C}{m_alpha}}$ So $\;\frac{\lambda _alpha}{\lambda _C}={\sqrt{3}}$

Q14JEE Main 2022MCQ4MElectrostatics

A force of $10 {\text{N}}$ acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be. [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

E between two plates is $\;\frac{\sigma}{\epsilon_0}$ and due to one plate is $\;\frac{\sigma}{2 \epsilon_0}$ so the force will be halvedSo new force ${\text{F}}=5 {\text{N}}$

Q15JEE Main 2022MCQ4MOscillations

The displacement of simple harmonic oscillator after 3 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is : [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Time taken by the harmonic oscillator to move from mean position to half of amplitude is $\;\frac{T}{12}$ So, $\;\frac{T}{12}=3$ ${\text{T}}=36 {\ sec} .$

Q16JEE Main 2022MCQ4MWaves

An observer moves towards a stationary source of sound with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be: [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$f^{'}=f_0 [\;\frac{v-v_0}{v-v_s}]$ $\Rightarrow f^{'}=f_0 [\;\frac{v+\frac{v}{5}}{v}]$ $\Rightarrow f^{'}=\;\frac{6 f_0}{5}$ $\Rightarrow \% \;\text{ change }\;=20$

Q17JEE Main 2022MCQ4MRay Optics and Optical Instruments

Consider a light ray travelling in air is incident into a medium of refractive index $\sqrt{2 n}$ . The incident angle is twice that of refracting angle. Then, the angle of incidence will be : [27-Jun-2022-Shift-1]

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📖 Explanation

According to the law, $1 \times sin \;\theta={\sqrt{2 n}} \times sin \; (\;\frac{\theta}{2})$ $\Rightarrow \cos \;\frac{\theta}{2}=\sqrt{\;\frac{n}{2}}$ $\Rightarrow \theta=2 \cos ^{-1} (\sqrt{\;\frac{n}{2}})$

Q18JEE Main 2022MCQ4MAtoms and Nuclei

A hydrogen atom in its ground state absorbs $10.2 {\text{eV}}$ of energy. The angular momentum of electron of the hydrogen atom will increase by the value of :(Given, Planck's constant $=6.6 \times 10^{-34} {\text{Js}}$ ). [27-Jun-2022-Shift-1]

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📖 Explanation

$-13.6+10.2=\;\frac{-13.6}{n^2}$ $\Rightarrow \;\frac{13.6}{n^2}=3.4$ $\Rightarrow n=2$ $\Rightarrow \Delta L=2 \times \;\frac{h}{2 \lambda }-1 \times \;\frac{h}{2 \lambda }$ $=\;\frac{h}{2 \lambda }$ $\Rightarrow \Delta L ≃ 1.05 \times 10^{-34} {\text{Js}}$

Q19JEE Main 2022MCQ4MSemiconductor Electronics

Identify the correct Logic Gate for the following output (Y) of two inputs A and B. [27-Jun-2022-Shift-1]

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📖 Explanation

A B Y 001011101110 $\Rightarrow Y=(AB)'$

Q20JEE Main 2000MCQ4MThermodynamics

A mixture of hydrogen and oxygen has volume $2000 {\text{cm}}^3$ , temperature $300 {\text{K}}$ , pressure $100 {\text{kPa}}$ and mass $0.76 {\text{g}}$ . The ratio of number of moles of hydrogen to number of moles of oxy in the mixture will be:[Take gas constant $. {\text{R}}=8.3 {\text{JK}}^{-1} {\text{mol}}^{-1}]$ [27-Jun-2022-Shift-1]

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📖 Explanation

$P_1 V=n_1 R T$ $P_2 V=n_2 R T$ $\Rightarrow (100 {\text{kPa}}) {\text{V}}= ( {\text{n}}_1+ {\text{n}}_2) {\text{RT}}$ $\Rightarrow n_1+n_2=\;{(100 {\text{kPa}}) (2000 {\text{cm}}^3) }/{8.3 \times 300}$ Also, $n_1 \times 2+n_2 \times 32=0.76$ Solving (1) and (2), ${\text{n}}_1=0.06$ ${\text{n}}_2=0.02$ $\Rightarrow \;\frac{n_1}{n_2}=3$

Q21JEE Main 2022NAT4MThermodynamics

In a carnot engine, the temperature of reservoir is $527^{\circ} {\text{C}}$ and that of sink is $200 {\text{K}}$ . If the work done by the engine when it transfers heat from reservoir to sink is $12000 {\text{kJ}}$ , the quantity of heat absorbed by the engine from reservoir is ____ $\times$ $10^6 {\text{J}} .$ [27-Jun-2022-Shift-1]

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📖 Explanation

$η=1-\;\frac{T_2}{T_1}$ $=1-\;\frac{200}{800}=\;\frac{3}{4}$ $\therefore η=\;\frac{W}{Q_1}$ $\Rightarrow \;\frac{3}{4}=\;\frac{12000 \times 10^3}{Q_1}$ $\Rightarrow Q_1=16 \times 10^6 {\text{J}}$

Q22JEE Main 2022NAT4MAlternating Current

A $220 {\text{V}}, 50 {\text{Hz}}$ AC source is connected to a $25 {\text{V}}, 5 {\text{W}}$ lamp and an additional resistance R in series (as shown in figure) to run the lamp at its peak brightness, then the value of ${\text{R}}$ (in ohm) will be__ [27-Jun-2022-Shift-1]

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📖 Explanation

$R_b=\;\frac{(25)^2}{5}=125 Ω$ $I_{r m s}=\sqrt{\;\frac{5}{125}}=\;\frac{1}{5} {\text{A}}$ $\Rightarrow \;\frac{220}{R+125}=\;\frac{1}{5}$ $\Rightarrow R=1100-125$ $=975 Ω$

Q23JEE Main 2022NAT4MWave Optics

In Young's double slit experiment the two slits are $0.6 {\text{mm}}$ distance apart. Interference pattern is observed on a screen at a distance $80 {\text{cm}}$ from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be____ ${\text{nm}}$ . [27-Jun-2022-Shift-1]

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📖 Explanation

$y=\;\frac{d}{2}$ $\therefore \Delta x=y \;\frac{d}{D}$ $\Rightarrow \;\frac{d^2}{2 D}=\;\frac{\lambda }{2}$ $\Rightarrow \lambda =\;\frac{ (0.6 \times 10^{-3}) ^2}{0.8}$ $=450 {\text{nm}}$

Q24JEE Main 2022NAT4MAtoms and Nuclei

A beam of monochromatic light is used to excite the electron in ${\text{Li}}^{++}$ from the first orbit to the third orbit. The wavelength of monochromatic light is found to be $x \times 10^{-10} {\text{m}}$ . The value of $x$ is____[Given hc $=1242 {\text{eV}} {\text{nm}}]$ [27-Jun-2022-Shift-1]

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📖 Explanation

${\text{E}}( {\in } {\text{eV}})=13.6 \times 9 (1-\;\frac{1}{9})$ $=13.6 \times 8 {\text{eV}}$ $\Rightarrow \lambda =\;\frac{12420}{13.6 \times 8} \;{A}^{o}$ $=114.15 \;{A}^{o}$

Q25JEE Main 2022NAT4MCurrent Electricity

A cell, shunted by a $8 Ω$ resistance, is balanced across a potentiometer wire of length $3 {\text{m}}$ . The balancing length is $2 {\text{m}}$ when the cell is shunted by $4 Ω$ resistance. The value of internal resistance of the cell will be____ $Ω$ . [27-Jun-2022-Shift-1]

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📖 Explanation

$\;\frac{\epsilon_1 8}{r_1+8}=3 c$ $\;\frac{\epsilon_1 4}{r_1+4}=2 c$ $\Rightarrow \;\frac{2 (r_1+4) }{r_1+8}=\;\frac{3}{2}$ $\Rightarrow r_1=8 Ω$

Q26JEE Main 2022NAT4MCurrent Electricity

The current density in a cylindrical wire of radius $4 {\text{mm}}$ is $4 \times 10^6 {\text{Am}}^{-2}$ . The current through the outer portion of the wire between radial distances $\;\frac{R}{2}$ and ${\text{R}}$ is____ $\pi {\text{A}}$ [27-Jun-2022-Shift-1]

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📖 Explanation

$i=A \times j$ $=\pi (R^{2-}\;\frac{R^2}{4}) j$ $=\;\frac{3 \pi R^2}{4} \times j$ $=\;\frac{3 \pi \times (4 \times 10^{-3}) ^2}{4} \times 4 \times 10^6$ $=48 \pi$

Q27JEE Main 2022NAT4MElectrostatic Potential and Capacitance

A capacitor of capacitance $50 {\text{pF}}$ is charged by $100 {\text{V}}$ source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is____ nJ. [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Electrical energy lost $=\;\frac{1}{2} (\;\frac{1}{2} C V^2)$ $=\;\frac{1}{2} \times \;\frac{1}{2} \times 50 \times 10^{-12} \times (100)^2$ $=\;\frac{500}{4} {\text{nJ}}$ $=125 {\text{nJ}}$

Q28JEE Main 2022NAT4MElectromagnetic Waves

The height of a transmitting antenna at the top of a tower is $25 {\text{m}}$ and that of receiving antenna is, $49 {\text{m}}$ . The maximum distance between them, for satisfactory communication in LOS (Line-Of-Sight) is ${\text{K}} \sqrt{5} \times 10^2 {\text{m}}$ . The value of K is____(Assume radius of Earth is $64 \times 10^{+5} {\text{m}}$ ) [Calculate upto nearest integer value] [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$d={\sqrt{2 h_t R_e}}+{\sqrt{2 \times h_R R_e}}$ $={\sqrt{2 \times 25 \times 64 \times 10^5}}+{\sqrt{2 \times 49 \times 64 \times 10^5}}$ $=8000 {\sqrt{5}}+11200 {\sqrt{5}} {\text{m}}$ $=19200 {\sqrt{5}} {\text{m}}$ $=192 {\sqrt{5}}+10^2 {\text{m}}$

Q29JEE Main 2022NAT4MProperties of Matter

The area of cross-section of a large tank is $0.5 {\text{m}}^2$ . It has a narrow opening near the bottom having area of cross-section $1 {\text{cm}}$ . A load of $25 {\text{kg}}$ is applied on the water at the top in the tank. Neglecting the speed of water in the tank, the velocity of the water, coming out of the opening at the time when the height of water level in the tank is $40 {\text{cm}}$ above the bottom, will be ____ ${\text{cms}}^{-1} \cdot \$ [. $Take$ {\text{g}}=10 {\text{ms}}^{-2} $]\$ [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

By Bernoulli's theorem: $\;\frac{250}{0.5}+\rho g h=\;\frac{1}{2} \rho v^2$ $\Rightarrow {\text{v}}=3 {\text{m}} / {\ s}$ $\Rightarrow {\text{v}}=300 {\text{cm}} / {\ s}$

Q30JEE Main 2022NAT4MCentre of Mass and Collision

A pendulum of length $2 {\text{m}}$ consists of a wooden bob of mass $50 {\text{g}}$ . A bullet of mass $75 {\text{g}}$ is fired towards the stationary bob with a speed $v$ . The bullet emerges out of the bob with a speed $\;\frac{v}{3}$ and the bob just completes the vertical circle. The value of $v$ is _____ ${\text{ms}}^{-1}$ . (if $g=10 {\text{m}} / {\text{s}}^2$ ). [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$v_{b o b}={\sqrt{5 g l}}={\sqrt{5 \times 10 \times 2}}=10 {\text{m}} / {\ s}$ Conserving momentum : $75 \times v=75 \times \;\frac{v}{3}+50 \times 10$ $\Rightarrow 50 v=50 \times 10$ $\Rightarrow v=10 {\text{m}} / {\ s}$

Chemistry30 questions

Q31JEE Main 2022MCQ4MSolutions

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).Assertion (A) : At $10^{\circ} {\text{C}}$ , the density of a $5 {\text{M}}$ solution of ${\text{KCl}}$ $[atomic masses of ${\text{K}} \& {\text{Cl}}$ are $39 \& 35.5 {\text{g}} {\text{mol}}^{-1}$ respectively]$, is ' $x$ ' $g {\text{ml}}^{-1}$ . The solution is cooled to $-21^{\circ} {\text{C}}$ . The molality of the solution will remain unchanged.Reason (R): The molality of a solution does not change with temperature as mass remains unaffected with temperature.In the light of the above statements, choose the correct answer from the options given below : [27-Jun-2022-Shift-1]

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Q32JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

Based upon VSEPR theory, match the shape (geometry) of the molecules in List-I with the molecules in List-II and select the most appropriate option. List - I(Shape) List - II(Molecules) (A) T-shaped (I) XeF (B) Trigonal planar (II) SF (C) Square planar (III) ClF (D) See-saw (IV) BF [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

To determine the molecular shape using VSEPR theory, we count the total number of electron pairs (bond pairs + lone pairs) around the central atom.

(III) $ClF_3$ : The central atom Cl has 7 valence electrons, 3 bond pairs with F atoms, and 2 lone pairs $(7 - 3)/2 = 2$ . With 5 total electron pairs ( $AX_3E_2$ ), the geometry is T-shaped. So, (A) matches (III).
(IV) $BF_3$ : The central atom B has 3 valence electrons, 3 bond pairs with F, and 0 lone pairs. With 3 total electron pairs ( $AX_3$ ), the geometry is Trigonal planar. So, (B) matches (IV).
(I) $XeF_4$ : The central atom Xe has 8 valence electrons, 4 bond pairs with F, and 2 lone pairs $(8 - 4)/2 = 2$ . With 6 total electron pairs ( $AX_4E_2$ ), the geometry is Square planar. So, (C) matches (I).
(II) $SF_4$ : The central atom S has 6 valence electrons, 4 bond pairs with F, and 1 lone pair $(6 - 4)/2 = 1$ . With 5 total electron pairs ( $AX_4E_1$ ), the geometry is See-saw. So, (D) matches (II).

Therefore, the correct matching is (A)-(III), (B)-(IV), (C)-(I), and (D)-(II).

Q33JEE Main 2022MCQ4MChemical Thermodynamics

Choose the correct answer from the options given below: List - I List - II (A) Spontaneous process (I) $\Delta H< 0$ (B) Process with $\Delta P=0, \Delta T=0$ (II) $\Delta G_{T, P}< 0$ (C) $\Delta H_{\;\text{reaction }\;}$ (III) Isothermal and isobaric process (D) Exothermic Process (IV) [Bond energies of molecules in reactants]- [Bond energies of product molecules [27-Jun-2022-Shift-1]

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📖 Explanation

(A) For a spontaneous process $\Delta {\text{G}}_{ {\text{T}}, {\text{P}}}< 0$ (B) $\Delta {\text{P}}=0 \to$ Isobaric process, $\Delta {\text{T}}=0 \to$ Isothermal process(C) $\Delta {\text{H}}_{\;\text{reaction }\;}=(Σ$ Bond energies of reactants $)-(Σ$ bond energies of products $)$ (D) $\Delta {\text{H}}< 0$ is for exothermic reaction

Q34JEE Main 2022MCQ4MChemical Thermodynamics

Choose the correct answer from the options given below: List - I List - II (A) Lyophilic colloid (I) Liquid-liquid colloid (B) Emulsion (II) Protective colloid (C) Positively charged colloid (III) $\text{FeCl}_3 + \text{NaOH}$ (D) Negatively charged colloid (IV) $\text{FeCl}_3$ + hot water [27-Jun-2022-Shift-1]

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📖 Explanation

(A) Protective colloids are lyophilic colloids(B) Emulsions are liquid in liquid colloidal solutions(C) ${\text{FeCl}}_3$ + hot water forms positively charged colloidal solution of hydrated ferric oxide.(D) ${\text{FeCl}}_3+ {\text{NaOH}}$ forms negatively charged colloidal solution due to preferential adsorption of ${\text{OH}}^{-}$ ions

Q35JEE Main 2022MCQ4MClassification of Elements

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).Assertion (A) : The ionic radii of ${\text{O}}^{2-}$ and ${\text{Mg}}^{2+}$ are same.Reason (R): Both ${\text{O}}^{2-}$ and ${\text{Mg}}^{2+}$ are isoelectronic species.In the light of the above statements, choose the correct answer from the options given below. [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Correct order of ionic radii: ${\text{O}}^{-2}> {\text{Mg}}^{+2}$ This is because among isoelectronic species, the size of anions are greater than the size of cations. Statement (II) is correct as both ${\text{O}}^{-2}$ and ${\text{Mg}}^{+2}$ are isoelectronic.

Q36JEE Main 2022MCQ4Md and f Block Elements

Choose the correct answer from the options given below : List - I List - II (A) Concentration of gold ore (I) Aniline (B) Leaching of alumina (II) NaOH (C) Froth stabiliser (III) $\text{SO}_2$ (D) Blister copper (IV) NaCN [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Gold is concentrated by cyanidationLeaching of alumina is done by ${\text{NaOH}}$ Froth stabiliser is anilineBlister copper has condensed ${\text{SO}}_2$ on the surface

Q37JEE Main 2022MCQ4Mp Block Elements

Addition of ${\text{H}}_2 {\text{SO}}_4$ to ${\text{BaO}}_2$ produces: [27-Jun-2022-Shift-1]

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📖 Explanation

${\text{BaO}}_2 \cdot 8 {\text{H}}_2 {\text{O}}+ {\text{H}}_2 {\text{SO}}_4 \to {\text{BaSO}}_4+ {\text{H}}_2 {\text{O}}_2+ {\text{H}}_2 {\text{O}}$ This is a common method to prepare hydrogenperoxide

Q38JEE Main 2022MCQ4Mp Block Elements

${\text{BeCl}}_2$ reacts with ${\text{LiAlH}}_4$ to give: [27-Jun-2022-Shift-1]

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📖 Explanation

$2 {\text{BeCl}}_2+ {\text{LiAlH}}_4 \to 2 {\text{BeH}}_2+ {\text{LiCl}}+ {\text{AlCl}}_3$ The above reaction using ${\text{LiAlH}}_4$ is an important preparation method for production of hydrides.

Q39JEE Main 2022MCQ4Mp Block Elements

List - I (Si-Compounds) List - II (Si-Polymeric/Other Products (A) $(\text{CH}_3)_4\text{Si}$ (I) Chain Silicone (B) $(\text{CH}_3)\text{Si}(\text{OH})_3$ (II) Dimeric Silicone (C) $(\text{CH}_3)_2\text{Si}(\text{OH})_2$ (III) Silane (D) $(\text{CH}_3)_3\text{Si}(\text{OH})$ (IV) 2D - Silicone Choose the correct answer from the options given below : [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$( {\text{CH}}_3) _4 {\text{Si}}$ is a silane. $( {\text{CH}}_3) {\text{Si}}( {\text{OH}})_3$ polymerise to form 2D silicone. $( {\text{CH}}_3) _2 {\text{Si}}( {\text{OH}})_2$ polymerise to form chain silicone. $( {\text{CH}}_3) _3 {\text{Si}}( {\text{OH}})$ form dimer $( {\text{CH}}_3) _3 {\text{Si}}- {\text{O}}- {\text{Si}} ( {\text{CH}}_3) _3$

Q40JEE Main 2022MCQ4Mp Block Elements

Heating white phosphorus with conc. ${\text{NaOH}}$ solution gives mainly: [27-Jun-2022-Shift-1]

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📖 Explanation

${\text{P}}_4+3 {\text{NaOH}}+3 {\text{H}}_2 {\text{O}} \to 3 {\text{NaH}}_2 {\text{PO}}_2+ {\text{PH}}_3$

Q41JEE Main 2022MCQ4MCoordination Compounds

Which of the following will have maximum stabilization due to crystal field? [27-Jun-2022-Shift-1]

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📖 Explanation

The given complexes are: $[ {\text{Ti}} ( {\text{H}}_2 {\text{O}}) _6] ^{3+}, [ {\text{Co}} ( {\text{H}}_2 {\text{O}}) _6] ^{2+}, [ {\text{Co}}( {\text{CN}})_6] ^{-3}, [ {\text{Cu}} ( {\text{NH}}_3) _4] ^{2+}$ ${\text{CN}}^{-}$ is the strongest ligand among the given complexes CFSE value for the $[ {\text{Co}}( {\text{CN}})_6] ^{-3}$ complex will be highest as it has $d^6$ configuration with a CFSE value of $-2.40 \Delta _0+2 {\text{P}}$ , where $P$ represents pairing energy and $\Delta _0$ represents splitting energy in octahedral field.The value of $\Delta _0$ is high for cyanide complexes.

Q42JEE Main 2022MCQ4MRedox Reactions

Given below are two Statements:Statement I: Classical smog occurs in cool humid climate. It is a reducing mixture of smoke, fog and sulphur dioxide.Statement II : Photochemical smog has components, ozone, nitric oxide, acrolein, formaldehyde, PAN etc.In the light of the above statements, choose the most appropriate answer from the options given below. [27-Jun-2022-Shift-1]

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Q43JEE Main 2022MCQ4MPractical Organic Chemistry

Which of the following is structure of a separating funnel? [27-Jun-2022-Shift-1]

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Q44JEE Main 2022MCQ4MHydrocarbons

'A' and 'B' respectively are: $\text{A} {\to }^{(1) \;\text{O}_3}_{(2)\text{Zn}-\text{H}_2O} \text{Ethane-1,2-dicarbaldehyde + Glyoxal/Oxaldehyde}$ $\text{B} {\to }^{(1)\;\text{O}_3}_{(2)\text{Zn}-\text{H}_2O} \text{5-oxohexanal}$ [27-Jun-2022-Shift-1]

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Q45JEE Main 2022MCQ4MHaloalkanes and Haloarenes

The major product of the following reaction is : [27-Jun-2022-Shift-1]

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📖 Explanation

It is bimolecular nucleophilic substitution $( {\text{SN}}^2)$ which occur at benzylic carbon by inversion in contiguration. This reaction cannot undergo substitution at benzene ring.

Q46JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Which of the following reactions will yield benzaldehyde as a product? (A)(B)(C)(D) [27-Jun-2022-Shift-1]

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Q47JEE Main 2022MCQ4MAmines

Given below are two statements :Statement I : In Hofmann degradation reaction, the migration of only an alkyl group takes place from carbonyl carbon of the amide to the nitrogen atom.Statement II : The group is migrated in Hofmann degradation reaction to electron deficient atom.In the light of the above statements, choose the most appropriate answer from the options given below: [27-Jun-2022-Shift-1]

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Q48JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Choose the correct answer from the options given below: List - I (Polymer) List - II(Used in) (A) Bakelite (I) Radio and television cabinets (B) Glyptal (II) Electrical switches (C) PVC (III) Paints and Lacquers (D) Polystyrene (IV) Water pipes [27-Jun-2022-Shift-1]

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Q49JEE Main 2022MCQ4MBiomolecules

L-isomer of a compound 'A' $( {\text{C}}_4 {\text{H}}_8 {\text{O}}_4)$ gives a positive test with $[ {\text{Ag}}( {\text{NH}}_3)_2]^{+}$ . Treatment of 'A' with acetic anhydride yields triacetate derivative. Compound ' $A$ ' produces an optically active compound $(B)$ and an optically inactive compound (C) on treatment with bromine water and ${\text{HNO}}_3$ respectively. Compound $( {\text{A}})$ is : [27-Jun-2022-Shift-1]

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Q50JEE Main 2022MCQ4MPractical Organic Chemistry

List - I List - II (A) (I) Dishwashing powder (B) (II) Toothpaste (C) ${\text{C}}_{17} {\text{H}}_{35} {\text{COO}}^{-} {\text{Na}}^{+}+ {\text{Na}}_2 {\text{CO}}_3+$ Rosinate (III) Laundry soap (D) ${\text{CH}}_3( {\text{CH}}_2)_{16} {\text{COO}}( {\text{CH}}_2 {\text{CH}}_2 {\text{O}})_n {\text{CH}}_2 {\text{CH}}_2 {\text{OH}}$ (IV) Hair conditioner Choose the correct answer from the options given below : [27-Jun-2022-Shift-1]

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📖 Explanation

(A) $[ {\text{CH}}_3 ( {\text{CH}}_2) _{15}- {\text{N}} ( {\text{CH}}_3) _3] ^{+} {\text{Br}}^{-}$ is cationic detergents used in hair conditioner (B) is anionic detergent used in tooth pastes.(C) ${\text{C}}_{17} {\text{H}}_{35} {\text{COO}}^{-} {\text{Na}}^{+}+ {\text{Na}}_2 {\text{CO}}_3+$ Rosin ate is used as laundary soap(D) ${\text{CH}}_3 ( {\text{CH}}_2) _{16} {\text{COO}} ( {\text{CH}}_2 {\text{CH}}_2 {\text{O}}) _{ {\text{N}}} {\text{CH}}_2 {\text{CH}}_2 {\text{OH}}$ is nonionic detergents formed from stearic acid and poly ethylene glycol used as liquid dishwashing detergents

Q51JEE Main 2022NAT4MSolid State

Metal deficiency defect is shown by ${\text{Fe}}_{0.93} {\text{O}}$ . In the crystal, some ${\text{Fe}}^{2+}$ cations are missing and loss of positive charge is compensated by the presence of ${\text{Fe}}^{3+}$ ions. The percentage of ${\text{Fe}}^{2+}$ ions in the ${\text{Fe}}_{0.93} {\text{O}}$ crystals is ___(Nearest integer) [27-Jun-2022-Shift-1]

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📖 Explanation

${\text{Fe}}_{0.93} {\text{O}}$ Let the number of ${\text{O}}^{-2}$ ions be 100 and the number of ${\text{Fe}}^{+2}$ ions be ${\text{X}}$ The number of ${\text{Fe}}^{+3}$ ions be $(93- {\text{X}})$ $\therefore X(2)+(93-X) 3=200$ $279-X=200$ $X=79$ $\therefore \;\; \% \;\text{ of }\; F^{+2} \;\text{ ions }\;=\;\frac{79}{93} \times 100$ $≃ 85 \%$

Q52JEE Main 2022NAT4MStructure of Atom

If the uncertainty in velocity and position of a minute particle in space are, $2.4 \times 10^{-26}( {\text{m}} {\text{s}}^{-1})$ and $10^{-7}( {\text{m}})$ respectively. The mass of the particle in ${\text{g}}$ is____ (Nearest integer)(Given : $h=6.626 \times 10^{-34} {\text{Js}}$ ) [27-Jun-2022-Shift-1]

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📖 Explanation

We know from hisenberg uncertainty principle $\Delta x \cdot \Delta p=\;\frac{h}{4 \pi}$ $\Rightarrow \Delta x \cdot m \Delta v=\;\frac{h}{4 \pi}$ Given, $\Delta {\text{x}}=10^{-7} {\text{m}}$ $\Delta {\text{x}}=2.4 \times 10^{-26} {\text{m}} / {\ s}$ ${\text{h}}=6.626 \times 10^{-34} {\text{Js}}$ $\therefore 10^{-7} \times m \times 2.4 \times 10^{-26}=\;\frac{6.626 \times 10^{-34}}{4 \pi}$ $\Rightarrow 2.4 m=\;\frac{6.626}{4 \pi \times 10}$ $\Rightarrow {\text{m}}=0.022 {\text{kg}}$ $\Rightarrow {\text{m}}=22 {\text{gm}}$

Q53JEE Main 2022NAT4MSolutions

$2 {\text{g}}$ of a non-volatile non-electrolyte solute is dissolved in $200 {\text{g}}$ of two different solvents ${\text{A}}$ and ${\text{B}}$ whose ebullioscopic constants are in the ratio of $1: 8$ . The elevation in boiling points of $A$ and $B$ are in the ratio $\;\frac{x}{y}(x: y)$ . The value of $y$ is_____ (Nearest integer) [27-Jun-2022-Shift-1]

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📖 Explanation

$\Delta T b=k b m$ $\;\frac{ (\Delta T_b) _A}{ (\Delta T_b) _B}=\;\frac{ (k_b) _A}{ (k_b) _B}$ $=\;\frac{1}{8}=\;\frac{x}{y}$ $\therefore y=8$

Q54JEE Main 2022NAT4MChemical Equilibrium

$2 {\text{NOCl}}( {\text{g}}) ⇌ 2 {\text{NO}}( {\text{g}})+ {\text{Cl}}_2( {\text{g}})$ In an experiment, $2.0$ moles of ${\text{NOCl}}$ was placed in a one-litre flask and the concentration of ${\text{NO}}$ after equilibrium established, was found to be $0.4 {\text{mol}} / {\text{L}}$ . The equilibrium constant at $30^{\circ} {\text{C}}$ is______ $\times 10^{-4}$ [27-Jun-2022-Shift-1]

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📖 Explanation

$\;2 {\text{NOCl}}( {\text{g}}) ⇌\;2 {\text{NO}}( {\text{g}})+ {\text{Cl}}_2( {\text{g}})$ At $\; t=0:$ $2$ $0$ $0$ At $\; t=t_{\;\text{eq }\;}:$ $2-2x$ $2x$ $0$ Given that at equilibrium, concentration of $N O=0.4 {\text{mol}} / {\text{L}}$ $\therefore 2 x=0.4$ $\Rightarrow x=0.2$ $\therefore$ Concentration of ${\text{NOCl}}$ at equilibrium, $[ {\text{NOCl}}]_{ {\text{eq}}}=2-2 \times 0.2=1.6$ and $[ {\text{NO}}]_{ {\text{eq}}}=0.4$ and $[ {\text{Cl}}_2] _{ {\text{eq}}}=0.2$ We know, $K_C=\;{[N O]^2 [ {\text{Cl}}_2] }/{[ {\text{NOCl}}]^2}$ $=\;\frac{[0.4]^2[0.2]}{[1.6]^2}$ $\Rightarrow K_C=12.5 \times 10^{-3}$ $\Rightarrow K_C=125 \times 10^{-4}$

Q55JEE Main 2022NAT4MElectrochemistry

The limiting molar conductivities of ${\text{Nal}}, {\text{NaNO}}_3$ and ${\text{AgNO}}_3$ are $12.7,12.0$ and $13.3 {\text{mS}} {\text{m}}^2 {\text{mol}}^{-1}$ , respectively (all at $25^{\circ} {\text{C}}$ ). The limiting molar conductivity of ${\text{Agl}}$ at this temperature is ____ $m S\;\; m^2 m o l^{-1}$ [27-Jun-2022-Shift-1]

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📖 Explanation

Given(1) $\lambda _m^{\infty }( {\text{NaI}})=12.7 {\text{mSm}}^2 {\text{mol}}^{-1}$ (2) $\lambda _{ {\text{m}}}^{\infty } ( {\text{NaNO}}_3) =12.0 {\text{mSm}}^2 {\text{mol}}^{-1}$ (3) $\lambda _{ {\text{m}}}^{\infty } ( {\text{AgNO}}_3) =13.3 {\text{mSm}}^2 {\text{mol}}^{-1}$ $\lambda _{ {\text{m}}}^{\infty }( {\text{AgI}})=(1)+(3)-(2)$ $=12.7+13.3-12.0$ $=26.0-12.0$ $\lambda _{ {\text{m}}}^{\infty }( {\text{AgI}})=14.0$

Q56JEE Main 2022NAT4MChemical Kinetics

The rate constant for a first order reaction is given by the following equation: $\ln k=33.24-\;\frac{2.0 \times 10^4 K}{T}$ The activation energy for the reaction is given by ____ ${\text{kJ}} {\text{mol}}^{-1}$ . (In nearest integer) (Given : ${\text{R}}=8.3 {\text{JK}}^{-1} {\text{mol}}^{-1}$ ) [27-Jun-2022-Shift-1]

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📖 Explanation

$\ln {\text{k}}=\ln {\text{A}}-\;{ {\text{E}}_{ {\text{A}}}}/{ {\text{RT}}}$ Given: $\ln k=33.24-\;{2.0 \times 10^4}/{ {\text{T}}}$ $\therefore \;\text{ on comparing }\; \;{ {\text{E}}_{ {\text{A}}}}/{ {\text{R}}}=2.0 \times 10^4$ $\therefore {\text{E}}_{ {\text{A}}}=2.0 \times 10^4 \times {\text{R}}$ $\Rightarrow {\text{E}}_{ {\text{A}}}=2.0 \times 10^4 \times 8.3 {\text{J}}$ $\Rightarrow {\text{E}}_{ {\text{A}}}=16.6 \times 10^4 {\text{J}}=166 {\text{kJ}}$

Q57JEE Main 2022NAT4Md and f Block Elements

The number of statements correct from the following for Copper (at. no. 29) is/are____(A) ${\text{Cu}}( {\text{II}})$ complexes are always paramagnetic.(B) ${\text{Cu}}( {\text{I}})$ complexes are generally colourless(C) ${\text{Cu}}( {\text{I}})$ is easily oxidized(D) In Fehling solution, the active reagent has ${\text{Cu}}( {\text{I}})$ [27-Jun-2022-Shift-1]

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📖 Explanation

(A) ${\text{Cu}}$ (II) complexes are always paramagnetic as they have one unpaired electron due to $d^9$ configuration of ${\text{Cu}}( {\text{II}})$ (B) ${\text{Cu}}( {\text{I}})$ complexes are generally colourless due to $d^{10}$ configuration.(C) ${\text{Cu}}( {\text{I}})$ is easily oxidised to ${\text{Cu}}^{+2}$ in aqueous solution $2 {\text{Cu}}^{+} \to {\text{Cu}}^{+2}+ {\text{Cu}}$ ${\text{Cu}}^{+1}$ disproportionates to ${\text{Cu}}^{+2}$ and ${\text{Cu}}$ ( $E_{\;\text{cell }\;}^{\circ}>0$ for this cell reaction in aqueous solution)In Fehling's solution, active reagent has ${\text{Cu}}( {\text{II}})$ which is reduced to ${\text{Cu}}( {\text{I}})$ on reaction with aldehydes.

Q58JEE Main 2022NAT4Md and f Block Elements

Acidified potassium permanganate solution oxidises oxalic acid. The spin-only magnetic moment of the manganese product formed from the above reaction is ______B.M. (Nearest integer) [27-Jun-2022-Shift-1]

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📖 Explanation

${\text{KMnO}}_4 \;\text{ (acidic medium) }\;+ {\text{H}}_2 {\text{C}}_2 {\text{O}}_4 \to {\text{CO}}_2+ {\text{Mn}}^{+2}$ ${\text{Mn}}^{+2}$ has 5 unpaired electrons $\therefore \;\text{ Spin only magnetic moment }\;={\sqrt{5(5+2)}}$ $={\sqrt{5 \times 7}}$ $={\sqrt{35}}$ $≃ 5.92 \;\text{ B.M. }\;$ $≃ 6 \;\text{ B.M. }\;$

Q59JEE Main 2022NAT4MSome Basic Concepts of Chemistry

Two elements $A$ and $B$ which form $0.15$ moles of $A_2 B$ and $A B_3$ type compounds. If both $A_2 B$ and $A B_3$ weigh equally, then the atomic weight of $A$ is_____ times of atomic weight of $B$ . [27-Jun-2022-Shift-1]

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📖 Explanation

Let atomic weight of $A=x$ and atomic weight of $B=y$ $\therefore$ Molar mass of $A_2 B=2 x+y$ and molar mass of ${\text{AB}}_3= {\text{x}}+3 {\text{y}}$ Now, weight of $0.15$ moles of $A_2 B=(2 x+y) 0.15$ and weight of $0.15$ moles of $A B_3=(x+3 y) 0.15$ Given, weight of $0.15$ moles of $A_2 B$ and $A B_3$ are equal. $\therefore (2 x+y) 0.15=(x+3 y) 0.15$ $\Rightarrow x=2 y$ $\Rightarrow \;\frac{x}{y}=2$ $\therefore$ Atomic weight of $A$ is 2 times of atomic weight of $B$ .

Q60JEE Main 2022NAT4MHydrocarbons

Total number of possible stereoisomers of dimethyl cyclopentane is____ [27-Jun-2022-Shift-1]

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Mathematics30 questions

Q61JEE Main 2022MCQ4MComplex Numbers

The area of the polygon, whose vertices are the non-real roots of the equation $\overline{z}=i z^2$ is : [27-Jun-2022-Shift-1]

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📖 Explanation

$\overline{z}=i z^2$ Let $z=x+i y$ $x-i y=i(x^{2-}y^{2+}2 x i y)$ $x-i y=i(x^{2-}y^2)-2 x y$ $\therefore x=-2 y x \;\text{ or }\; x^{2-}y^2=-y$ $x=0 \;\text{ or }\; y=-\;\frac{1}{2}$ Case - I $x=0$ $-y^2=-y$ $y=0,1$ Case - II $y=-\;\frac{1}{2}$ $\Rightarrow x^{2-}\;\frac{1}{4}=\;\frac{1}{2} \Rightarrow x=\pm \;\frac{\sqrt{3}}{2}$ $x=\{0, i, \;\frac{\sqrt{3}}{2}-\;\frac{i}{2}, \;\frac{-\sqrt{3}}{2}-\;\frac{i}{2}\}$ $\;\text{ Area of polygon }\;=\;\frac{1}{2}\begin{vmatrix}0 & 1 & 1 \\ \\ \\ \frac{\sqrt{3}}{2} & \\ \\ \frac{-1}{2} & 1 \\ \\ \\ \frac{-\sqrt{3}}{2} & \\ \\ \frac{-1}{2} & 1\end{vmatrix}$ $=\;\frac{1}{2}|-\sqrt{3}-\;\frac{\sqrt{3}}{2}|=\;\frac{3 \sqrt{3}}{4}$

Q62JEE Main 2022MCQ4MMatrices and Determinants

Let the system of linear equations $x+2 y+z=2$ $\alpha x+3 y-z=\alpha$ $-\alpha x+y+2 z=-\alpha$ be inconsistent. Then $\alpha$ is equal to: [27-Jun-2022-Shift-1]

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📖 Explanation

$x+2 y+z=2$ $\alpha x+3 y-z=\alpha$ $-\alpha x+y+2 z=-\alpha$ $\Delta ={\begin{vmatrix}1 & 2 & 1 \\ \alpha & 3 & -1 \\ -\alpha & 1 & 2\end{vmatrix}}=1(6+1)-2(2 \alpha-\alpha)+1(\alpha+3 \alpha)$ $=7+2 \alpha$ $\Delta =0 \Rightarrow \alpha=-\;\frac{7}{2}$ $\Delta _1={\begin{vmatrix}2 & 2 & 1 \\ \alpha & 3 & -1 \\ -\alpha & 1 & 2\end{vmatrix}}=14+2 \alpha \ne 0 \;\text{ for }\; \alpha=-\;\frac{7}{2}$ $\therefore$ For no solution $\alpha=-\;\frac{7}{2}$

Q63JEE Main 2022MCQ4MSequences and Series

$x=sum_{n=0}^{\infty } a^n, y=sum_{n=0}^{\infty } b^n, z=sum_{n=0}^{\infty } c^n$ , where a, b, c are in A.P. and $| {\text{a}}|< 1,| {\text{b}}|< 1,| {\text{c}}|< 1$ , abc $\ne 0$ , then : [27-Jun-2022-Shift-1]

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📖 Explanation

$x=sum_{n=0}^{\infty } a^n=\;\frac{1}{1-a} ; y=sum_{n=0}^{\infty } b^n=\;\frac{1}{1-b} ; z=sum_{n=0}^{\infty } c^n=\;\frac{1}{1-c}$ Now, $a, b, c \to A P$ $1-a, 1-b, 1-c \to A P$ $\;\frac{1}{1-a}, \;\frac{1}{1-b}, \;\frac{1}{1-c} \to H P$ ${\text{x}}, {\text{y}}, {\text{z}} \to {\text{HP}}$ $\therefore \;\frac{1}{x}, \;\frac{1}{y}, \;\frac{1}{z} \to A P$

Q64JEE Main 2022MCQ4MDifferential Equations

Let $\;\frac{d y}{d x}=\;\frac{a x-b y+a}{b x+c y+a}$ , where $a, b, c$ are constants, represent a circle passing through the point $(2,5)$ . Then the shortest distance of the point $(11,6)$ from this circle is : [27-Jun-2022-Shift-1]

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📖 Explanation

$\;\frac{d y}{d x}=\;\frac{a x-b y+a}{b x+c y+a}$ $=b x d y+c y d y+a d y=a x d x-b y d x+a d x$ $=c y d y+a d y-a x d x-a d x+b(x d y+y d x)=0$ $=c \int y d y+a \int x d x-a \int d x+b \int d(x y)=0$ $=\;\frac{c y^2}{2}+a y-\;\frac{a x^2}{2}-a x+b x y=k$ $=a x^{2-}c y^{2+}2 a x-2 a y-2 b x y=k$ Above equation is circle $\Rightarrow {\text{a}}=- {\text{c}}$ and ${\text{b}}=0$ $a x^{2+}a y^{2+}2 a x-2 a y=k$ $\Rightarrow x^{2+}y^{2+}2 x-2 y=\lambda \;\; [\lambda =\;\frac{k}{a}]$ Passes through $(2,5)$ $4+25+4-10=\lambda \Rightarrow \lambda =23$ Circle $\equiv x^{2+}y^{2+}2 x-2 y-23=0$ Centre $(-1,1) r=\sqrt{(-1)^{2+}1^{2+}23}=5$ Shortest distance of $(11,6)=\sqrt{12^{2+}5^2}-5$ $=13-5$ $=8$

Q65JEE Main 2022MCQ4MLimits, Continuity and Differentiability

Let a be an integer such that $\lim _{x \to 7} \;\frac{18-[1-x]}{[x-3 a]}$ exists, where [t] is greatest integer $\le {\text{t}}$ . Then a is equal to : [27-Jun-2022-Shift-1]

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📖 Explanation

$\lim _{x \to 7} \;\frac{18-[1-x]}{[x-3 a]}$ exist $\& a \in I$ . $=\lim _{x \to 7} \;\frac{17-[-x]}{[x]-3 a}$ exist $R H L=\lim _{x \to 7^{+}} \;\frac{17-[-x]}{[x]-3 a}=\;\frac{25}{7-3 a}[a \ne \;\frac{7}{3}]$ $L H L=\lim _{x \to 7^{-}} \;\frac{17-[-x]}{[x]-3 a}=\;\frac{24}{6-3 a}[a \ne 2]$ For limit to exist $L H L=R H L$ $\;\frac{25}{7-3 a}=\;\frac{24}{6-3 a}$ $\Rightarrow \;\frac{25}{7-3 a}=\;\frac{8}{2-a}$ $\therefore a=-6$

Q66JEE Main 2022MCQ4MApplication of Derivatives

The number of distinct real roots of $x^{4-}4 x+1=0$ is: [27-Jun-2022-Shift-1]

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📖 Explanation

$f(x)=x^{4-}4 x+1=0$ $f^{'}(x)=4 x^{3-}4$ $=4(x-1)(x^{2+}1+x)$

Q67JEE Main 2022MCQ4MApplication of Derivatives

The lengths of the sides of a triangle are $10+x^2, 10+x^2$ and $20-2 x^2$ . If for $x=k$ , the area of the triangle is maximum, then $3 k^2$ is equal to: [27-Jun-2022-Shift-1]

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📖 Explanation

$C D=\sqrt{(10+x^2)^{2-}(10-x^2)^2}=2 \sqrt{10}|x|$ Area $=\;\frac{1}{2} \times C D \times A B=\;\frac{1}{2} \times 2 \sqrt{10}|x|(20-2 x^2)$ $A=\sqrt{10}|x|(10-x^2)$ $\;\frac{d A}{d x}=\sqrt{10} \;\frac{|x|}{x}(10-x^2)+\sqrt{10}|x|(-2 x)=0$ $\Rightarrow 10-x^2=2 x^2$ $3 x^2=10$ $x=k$ $3 k^2=10$

Q68JEE Main 2022MCQ4MDifferential Equations

If $\cos ^{-1}(\;\frac{y}{2})=\log _e(\;\frac{x}{5})^5,|y|< 2$ , then : [27-Jun-2022-Shift-1]

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📖 Explanation

$\cos ^{-1}(\;\frac{y}{2})=\log _e(\;\frac{x}{5})^5 \;\; |y|< 2$ Differentiating on both side $-\;{1}/{\sqrt{1-(\;\frac{y}{2})^2}} \times \;\frac{y^{'}}{2}=\;\frac{5}{\frac{x}{5}} \times \;\frac{1}{5}$ $\;\frac{-x y^{'}}{2}=5 \sqrt{1-(\;\frac{y}{2})^2}$ Square on both side $\;\frac{x^2 y^{' 2}}{4}=25(\;\frac{4-y^2}{4})$ Diff on both side $2 x y^{' 2}+2 y^{'} y^{' '} x^2=-25 \times 2 y y^{'}$ $x y^{'}+y^{' '} x^{2+}25 y=0$

Q69JEE Main 2022MCQ4MIndefinite Integrals

If $\int \;\frac{(x^{2+}1) e^x}{(x+1)^2} d x=f(x) e^x+C$ , where ${\text{C}}$ is a constant, then $\;\frac{d^3 f}{d x^3}$ at $x=1$ is equal to : [27-Jun-2022-Shift-1]

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📖 Explanation

$I=\int \;\frac{e^x(x^{2+}1)}{(x+1)^2} d x=f(x) e^x+c$ $=\int \;\frac{e^x(x^{2-}1+1+1)}{(x+1)^2} d x$ $=\int e^x\$ [;\frac{x-1}{x+1}+;\frac{2}{(x+1)^2}]$ d x=e^x(;\frac{x-1}{x+1})+c\therefore f(x)=;\frac{x-1}{x+1}f(x)=1-;\frac{2}{x+1}f^{'}(x)=2(;\frac{1}{x+1})^2f^{' '}(x)=-4(;\frac{1}{x+1})^3f^{' ' '}(x)=;\frac{12}{(x+1)^4} $for$ x=1f^{' ' '}(1)=;\frac{12}{2^4}=;\frac{12}{16}=;\frac{3}{4}$

Q70JEE Main 2022MCQ4MDefinite Integrals

The value of the integral $int_{-2}^2 \;\frac{ x^{3+}x|}{(e^{x|x|}+1)} d x$ is equal to: [27-Jun-2022-Shift-1]

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📖 Explanation

$I=int_{-2}^2 \;\frac{|x^{3+}x|}{e^{x|x|}+1} d x . . . .( {\text{i}})$ $I=int_{-2}^2 \;\frac{|x^{3+}x|}{e^{-x|x|}+1} d x . . . .( {\text{ii}})$ $2 I=int_{-2}^2|x^{3+}x| d x$ $2 I=2 int_0^2(x^{3+}x) d x$ $I=int_0^2(x^{3+}x) d x$ $.=\;\frac{x^4}{4}+\;\frac{x^2}{2}]_0^2$ $=(\;\frac{16}{4}+\;\frac{4}{2})-0$ $=4+2=6$

Q71JEE Main 2022MCQ4MDifferential Equations

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

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📖 Explanation

$\;\frac{d y}{d x}+\;\frac{2^{x-y}(2^y-1)}{2^x-1}=0, {\text{x}}, {\text{y}}>0, {\text{y}}(1)=1$ $\;\frac{d y}{d x}=-\;\frac{2^x(2^y-1)}{2^y(2^z-1)}$ $\int \;\frac{2^y}{2^y-1} d y=-\int \;\frac{2^x}{2^x-1} d x$ $=\;\frac{\log _e(2^y-1)}{\log _{e^2}}=-\;\frac{\log _e(2^x-1)}{\log _e 2}+\;\frac{\log _e c}{\log _e 2}$ $=|(2^y-1)(2^x-1)|=c$ $\because y(1)=1$ $\therefore c=1$ $=|(2^y-1)(2^x-1)|=1$ $\;\text{ For }\; x=2$ $|(2^y-1) 3|=1$ $2^y-1=\;\frac{1}{3} \Rightarrow 2 y=\;\frac{4}{3}$ Taking log to base 2. $\therefore y=2-\log _2 3$

Q72JEE Main 2022MCQ4MStraight Lines and Pair of Straight Lines

In an isosceles triangle $A B C$ , the vertex $A$ is $(6,1)$ and the equation of the base $B C$ is $2 x+y=4$ . Let the point $B$ lie on the line $x+$ $3 {\text{y}}=7$ . If $(\alpha, \beta)$ is the centroid of $\triangle {\text{ABC}}$ , then $15(\alpha+\beta)$ is equal to: [27-Jun-2022-Shift-1]

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📖 Explanation

$\text{table} 2 x+y=4 ; 2 x+6 y=14\} y=2, x=3$ ${\text{B}}(1,2)$ Let ${\text{C}}( {\text{k}}, 4-2 {\text{k}})$ Now $A B^2=A C^2$ $5^{2+}(-1)^2=(6-k)^{2+}(-3+2 k)^2$ $\Rightarrow 5 k^{2-}24 k+19=0$ $(5 k-19)(k-1)=0 \Rightarrow k=\;\frac{19}{5}$ $C(\;\frac{19}{5},-\;\frac{18}{5})$ Centroid $(\alpha, \beta)$ $\alpha=\;\frac{6+1+\frac{19}{5}}{3}=\;\frac{18}{5}$ $\beta=\;\frac{1+2-\frac{18}{5}}{3}=-\;\frac{1}{5}$ Now $15(\alpha+\beta)$ $15(\;\frac{17}{5})=51$

Q73JEE Main 2022MCQ4MEllipse

Let the eccentricity of an ellipse $\;\frac{x^2}{a^2}+\;\frac{y^2}{b^2}=1, a>b$ , be $\;\frac{1}{4}$ . If this ellipse passes through the point $(-4 \sqrt{\;\frac{2}{5}}, 3)$ , then $a^{2+}b^2$ is equal to : [27-Jun-2022-Shift-1]

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📖 Explanation

$\;\frac{x^2}{a^2}+\;\frac{y^2}{b^2}=1$ $\Rightarrow \;{(-4 \sqrt{\frac{2}{5}})^2}/{a^2}+\;\frac{32}{b^2}=1$ $\Rightarrow \;\frac{32}{5 a^2}+\;\frac{9}{b^2}=1 . . . . \;\text{ (i) }\;$ $a^2(1-e^2)=b^2$ $a^2(1-\;\frac{1}{16})=b^2$ $15 a^2=16 b^2 \Rightarrow a^2=\;\frac{16 b^2}{15}$ From (i) $\;\frac{6}{b^2}+\;\frac{9}{b^2}=1 \Rightarrow b^2=15 \& a^2=16$ $a^{2+}b^2=15+16=31$

Q74JEE Main 2022MCQ4MThree Dimensional Geometry

If two straight lines whose direction cosines are given by the relations $l+m-n=0,3 l^{2+}m^{2+}c n l=0$ are parallel, then the positive value of ${\text{c}}$ is : [27-Jun-2022-Shift-1]

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📖 Explanation

$l+m-n=0 \Rightarrow n=l+m$ $3 l^{2+}m^{2+}c n l=0$ $3 l^{2+}m^{2+}c l(l+m)=0$ $=(3+c) l^{2+}c l m+m^2=0$ $=(3+c)(\;\frac{l}{m})^{2+}c(\;\frac{l}{m})+1=0$ $\therefore$ Lines are parallel $D=0$ $c^{2-}4(3+c)=0$ $c^{2-}4 c-12=0$ $(c-4)(c+3)=0$ $c=4($ as $c>0)$

Q75JEE Main 2022MCQ4MVector Algebra

Let $\vec{a}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+2 \hat{k}$ . Then the number of vectors $\vec{b}$ such that $\vec{b} \times \vec{c}=\vec{a}$ and $|\vec{b}| \in \{1,2, . . . . . ., 10\}$ is : [27-Jun-2022-Shift-1]

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📖 Explanation

$\vec{a}=\hat{i}+\hat{j}-\hat{k}$ $\vec{c}=2 \hat{i}-3 \hat{j}+2 \hat{k}$ Now, $\vec{b} \times \vec{c}=\vec{a}$ $\vec{c} \cdot \vec{(b} \times \vec{c)}=\vec{c} \cdot \vec{a}$ $\vec{c} \cdot \vec{a}=0$ $\Rightarrow (\hat{i}+\hat{j}-\hat{k})(2 \hat{i}-3 \hat{j}+2 \hat{k})=0$ $=2-3-2=0$ $\Rightarrow -3=0$ (Not possible) $\Rightarrow$ No possible value of $\vec{b}$ is possible.

Q76JEE Main 2022MCQ4MProbability

Five numbers $x_1, x_2, x_3, x_4, x_5$ are randomly selected from the numbers $1,2,3, . . . . . . ., 18$ and are arranged in the increasing order $(x_1< x_2< x_3< x_4< x_5)$ . The probability that $x_2=7$ and $x_4=11$ is : [27-Jun-2022-Shift-1]

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📖 Explanation

No. of ways to select and arrange ${\text{x}}_1, {\text{x}}_2, {\text{x}}_3, {\text{x}}_4, {\text{x}}_5$ from $1,2,3 . . . . . . . . . . . . . . . . . . .$ ${\text{n}}( {\text{s}})={ }^{18} {\text{C}}_5$ ${\text{n}}( {\text{E}})={ }^6 {\text{C}}_1 \times { }^3 {\text{C}}_1 \times { }^7 {\text{C}}_1$ $P(E)=\;\frac{6 \times 3 \times 7}{{ }^{18} C_5}$ $\;\frac{1}{17 \times 4}=\;\frac{1}{68}$

Q77JEE Main 2022MCQ4MProbability

Let ${\text{X}}$ be a random variable having binomial distribution ${\text{B}}(7, {\text{p}})$ . If ${\text{P}}( {\text{X}}=3)=5 {\text{P}}( {\text{x}}=4)$ , then the sum of the mean and the variance of $X$ is : [27-Jun-2022-Shift-1]

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📖 Explanation

${\text{B}}(7, {\text{p}})$ ${\text{n}}=7 \;\; {\text{p}}= {\text{p}}$ given $P(x=3)=5 P(x=4)$ ${ }^7 C_3 \times p^3(1-p)^4=5 .{ }^7 C_4 p^4(1-p)^3$ $\;\frac{{ }^7 C_3}{5 \times { }^7 C_4}=\;\frac{p}{1-p}$ $1- {\text{p}}=5 {\text{p}}$ $6 p=1$ $p=\;\frac{1}{6} \Rightarrow q=\;\frac{5}{6}$ $n=7$ Mean $=n p=7 \times \;\frac{1}{6}=\;\frac{7}{6}$ ${\text{Var}}= {\text{npq}}=7 \times \;\frac{1}{6} \times \;\frac{5}{6}=\;\frac{35}{36}$ Sum $=\;\frac{7}{6}+\;\frac{35}{36}$ $=\;\frac{42+35}{36}$ $=\;\frac{77}{36}$

Q78JEE Main 2022MCQ4MTrigonometric Ratios and Identities

The value of $\cos (\;\frac{2 \pi}{7})+\cos (\;\frac{4 \pi}{7})+\cos (\;\frac{6 \pi}{7})$ is equal to: [27-Jun-2022-Shift-1]

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📖 Explanation

$\cos \;\frac{2 \pi}{7}+\cos \;\frac{4 \pi}{7}+\cos \;\frac{6 \pi}{7}$ $=\;\frac{\sin 3(\frac{\pi}{7})}{\sin \;\frac{\pi}{7}} \cos \;\frac{(\frac{2 \pi}{7}+\frac{6 \pi}{7})}{2}$ $=\;\frac{\sin (\frac{3 \pi}{7}) \cdot \cos (\frac{4 \pi}{7})}{\sin (\;\frac{\pi}{7})}$ $=\;\frac{2 \sin \frac{4 \pi}{7} \cos \frac{4 \pi}{7}}{2 \sin \;\frac{\pi}{7}}$ $=\;\frac{\sin (\frac{3 \pi}{7})}{2 \sin \;\frac{\pi}{7}}=\;\frac{-\sin \frac{\pi}{7}}{2 \sin \;\frac{\pi}{7}}=\;\frac{-1}{2}$

Q79JEE Main 2022MCQ4MInverse Trigonometric Functions

$\sin ^{-1}(\sin \;\frac{2 \pi}{3})+\cos ^{-1}(\cos \;\frac{7 \pi}{6})+\tan ^{-1}(\tan \;\frac{3 \pi}{4})$ is equal to: [27-Jun-2022-Shift-1]

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📖 Explanation

$\sin ^{-1}(\sin \;\frac{2 \pi}{3})+\cos ^{-1}(\cos \;\frac{7 \pi}{6})+\tan ^{-1} \tan (\;\frac{3 \pi}{4})$ $\sin ^{-1} \sin (\;\frac{2 \pi}{3})=\pi-\;\frac{2 \pi}{3}=\;\frac{\pi}{3}$ $\cos ^{-1}(\cos \;\frac{2 \pi}{6})=2 \pi-\;\frac{7 \pi}{6}=\;\frac{5 \pi}{6}$ $\tan ^{-1} \tan (\;\frac{3 \pi}{4})=\;\frac{3 \pi}{4}-\pi=\;\frac{-\pi}{4}$ $\sin ^{-1}(\sin \;\frac{2 \pi}{3})+\cos ^{-1} \cos \;\frac{7 \pi}{6}+\tan ^{-1} \tan \;\frac{3 \pi}{4}$ $=\;\frac{11 \pi}{12}$

Q80JEE Main 2022MCQ4MSets and Relations

The boolean expression $(∼(p ∧ q)) ∨ q$ is equivalent to : [27-Jun-2022-Shift-1]

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📖 Explanation

p q $p ∧ q$ $~p ∧ q$ $(∼(p ∧ q)) ∨ q$ $p ∨ q$ $p \to q$ $p \to (p ∨ q)$ T T T F T T T T T F F T T T F T F F F T T T T T F F F T T F T T Tautology Tautology

Q81JEE Main 2022NAT4MSequences and Series

Let ${\text{f}}: {\text{R}} \to {\text{R}}$ be a function defined by $f(x)=\;\frac{2 e^{2 x}}{e^{2 x}+e}$ . Then $f(\;\frac{1}{100})+f(\;\frac{2}{100})+f(\;\frac{3}{100})+. . .+f(\;\frac{99}{100})$ is equal to____ [27-Jun-2022-Shift-1]

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📖 Explanation

Given, $f(x)=\;\frac{2 e^{2 x}}{e^{2 x}+e}$ $\therefore f(1-x)=\;\frac{2 e^{2(1-x)}}{e^{2(1-s)}+e}$ $=\;{2 \cdot \frac{e^2}{e^{2 x}}}/{\frac{e^2}{e^{2 \pi}}+e}$ $=\;\frac{2 e^2}{e^{2+}e^{2 x} \cdot e}$ $=\;\frac{2 e^2}{e(e+e^{2 x})}$ $=\;\frac{2 e}{e+e^{2 \Delta }}$ $\therefore f(x)+f(1-x)=\;\frac{2 e^{2 x}}{e^{2 x}+e}+\;\frac{2 e}{e^{2 x}+e}$ $=\;\frac{2(e^{2 x}+e)}{e^{2 x}+e}$ $=2 . . . . . .$ (1)Now, $f(\;\frac{1}{100})+f(\;\frac{99}{100})$ $=f(\;\frac{1}{100})+f(1-\;\frac{1}{100})$ $=2$ [as $f(x)+f(1-x)=2$ ] $f(\;\frac{2}{100})+f(1-\;\frac{2}{100})=2$ ⁞ $f(\;\frac{49}{100})+f(1-\;\frac{49}{100})=2$ $\therefore$ Total sum $=49 \times 2$ Remaining term $=f(\;\frac{50}{100})=f(\;\frac{1}{2})$ Put $x=\;\frac{1}{2}$ in equation (1), we get $f(\;\frac{1}{2})+f(1-\;\frac{1}{2})=2$ $\Rightarrow 2 f(\;\frac{1}{2})=2$ $\Rightarrow f(\;\frac{1}{2})=1$ $\therefore \;\text{ Sum }\;=49 \times 2+1=99$

Q82JEE Main 2022NAT4MQuadratic Equation and Inequalities

If the sum of all the roots of the equation $e^{2 x}-11 e^x-45 e^{-x}+\;\frac{81}{2}=0$ is $\log _e p$ , then $p$ is equal to____ [27-Jun-2022-Shift-1]

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📖 Explanation

Let $e^x=t$ then equation reduces to $t^{2-}11 t-\;\frac{45}{t}+\;\frac{81}{2}=0$ $\Rightarrow 2 t^{3-}22 t^{2+}81 t-45=0 . . . . . \;\text{ (i) }\;$ if roots of $e^{2 x t}-11 e^x-45 e^{-x}+\;\frac{81}{2}=0$ are $\alpha, \beta, \gamma$ then roots of (i) will be $e^{\alpha_1} e^{\alpha_2} e^{\alpha 3}$ using product of roots $e^{\alpha_1+\alpha_2+\alpha_3}=45$ $\Rightarrow \alpha_1+\alpha_2+\alpha_3=\ln 45 \Rightarrow p=45$

Q83JEE Main 2022NAT4MMatrices and Determinants

The positive value of the determinant of the matrix A, whose ${\text{Adj}}({\text{Adj}}(A))=(\text{table} 14 , 28 , -14 ; -14 , 14 , 28 ; 28 , -14 , 14)$ , is____ [27-Jun-2022-Shift-1]

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📖 Explanation

$|{\text{adj}}({\text{adj}}(A))|=|A|^{2^2}=|A|^4$ $\therefore |A|^4=\begin{vmatrix}14 & 28 & -14 \\ -14 & 14 & 28 \\ 28 & -14 & 14\end{vmatrix}$ $=(14)^3\begin{vmatrix}1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1\end{vmatrix}$ $=(14)^3(3-2(-5)-1(-1))$ $|A|^4=(14)^4 \Rightarrow |A|=14$

Q84JEE Main 2022NAT4MPermutations and Combinations

The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is [27-Jun-2022-Shift-1]

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📖 Explanation

$x_1+x_2+x_3+x_4+x_5+x_6=11$ $x_1, x_6 \ge 0, \;\; x_2, x_3, x_4, x_5 \ge 2$ $x_2=t_1+2$ $x_3=t_3+2$ $x_4=t_4+2$ $x_5=t_5+2$ $x_1, t_2, t_3, t_4, t_5, x_6 \ge 0$ No. of solutions $={ }^{6+3-1} C_3={ }^8 C_3=56$

Q85JEE Main 2022NAT4MBinomial Theorem

If the coefficient of ${\text{x}}^{10}$ in the binomial expansion of $(\;{\sqrt{x}}/{5^{\;\frac{1}{4}}}+\;{\sqrt{5}}/{x^{\;\frac{1}{3}}})^{60}$ is $5^k . l$ , where I, ${\text{k}} \in {\text{N}}$ and ${\text{I}}$ is co-prime to 5 , then ${\text{k}}$ is equal to [27-Jun-2022-Shift-1]

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📖 Explanation

Given Binomial Expansion $=(\;{\sqrt{x}}/{5^{\;\frac{1}{4}}}+\;{\sqrt{x}}/{5^{\;\frac{1}{3}}})^{60}$ $\therefore$ General term $T_{r+1}={ }^{60} C_r \cdot (\;\frac{x^{1 / 2}}{5^{1 / 4}})^{60-r} \cdot (\;\frac{5^{1 / 2}}{x^{1 / 3}})^r$ $={ }^{60} C_r \cdot 5^{(\;\frac{r}{4}-15+\;\frac{r}{2})} \cdot x^{(30-\;\frac{r}{2}-\;\frac{r}{3})}$ $={ }^{60} C_r \cdot 5^{(\;\frac{3 r-60}{4})} \cdot x^{(\;\frac{180-5 r}{6})}$ For $x^{10}$ term, $\;\frac{180-5 r}{6}=10$ $\Rightarrow 5 r=120$ $\Rightarrow r=24$ $\therefore \;\text{ Coefficient of }\; x^{10}={ }^{60} C_{24} \cdot 5(\;\frac{3 \times 24-60}{4})$ $={ }^{60} C_{24} \cdot 5^3$ $=\;\frac{60 !}{24 ! 36 !} \cdot 5^3$ It is given that, $\;\frac{60 !}{24 ! 36 !} \cdot 5^3=5^k \cdot l . . . . . . \;\text{ (1) }\;$ Also given that, I is coprime to 5 means I can't be multiple of 5 . So we have to find all the factors of 5 in $60 !, 24$ ! and 36 ![Note : Formula for exponent or degree of prime number in n!.Exponent of $p$ in $n !=⌈\;\frac{n}{p}⌋+⌈\;\frac{n}{p^2}⌋+\$ [;\frac{n}{p^3}]$+. . . . $. until 0 comeshere$ p $is a prime number. ]$ \therefore $Exponent of 5 in 60 !$ =⌈;\frac{60}{5}⌋+⌈;\frac{60}{5^2}⌋+⌈;\frac{60}{5^3}⌋+. . . . . .=12+2+0+. . . .=14 $Exponent of 5 in 24 !$ =⌈;\frac{24}{5}⌋+⌈;\frac{24}{5^2}⌋+⌈;\frac{24}{5^3}⌋+. . . . . .=4+0+0 . . . . . .=4 $Exponent of 5 in 36 !$ =⌈;\frac{36}{5}⌋+⌈;\frac{36}{5^2}⌋+⌈;\frac{36}{5^3}⌋+. . . . . .=7+1+0 . . . . . .=8\therefore $From equation (1), exponent of 5 overall$ ;\frac{5^{14}}{5^4 \cdot 5^8} \cdot 5^3=5^k\Rightarrow 5^5=5^k\Rightarrow k=5$

Q86JEE Main 2022NAT4MArea Under The Curves

Let $A_1=\{(x, y):| x| \le y^2,|x|+2 y \le 8\}$ and $A_2=\{(x, y):| x|+|y| \le k\}$ . If $27(.$ Area $. {\text{A}}_1)=5$ (Area ${\text{A}}_2$ ), then ${\text{k}}$ is equal to : [27-Jun-2022-Shift-1]

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📖 Explanation

Required area (above $x$ -axis) $A_1=2 int_0^4(\;\frac{8-x}{2}-\sqrt{x}) d x$ $=2(16-\;\frac{16}{4}-\;\frac{8}{3 / 2})=\;\frac{40}{3}$ and $A_2=4(\;\frac{1}{2} \cdot k^2)=2 k^2$ $\therefore 27 \cdot \;\frac{40}{3}=5 \cdot (2 k^2)$ $\Rightarrow {\text{k}}=6$

Q87JEE Main 2022NAT4MSequences and Series

If the sum of the first ten terms of the series $\;\frac{1}{5}+\;\frac{2}{65}+\;\frac{3}{325}+\;\frac{4}{1025}+\;\frac{5}{2501}+. . .$ is $\;\frac{m}{n}$ , where $m$ and $n$ are co-prime numbers, then $m+n$ is equal to [27-Jun-2022-Shift-1]

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📖 Explanation

$T_r=\;\frac{r}{(2 r^2)^{2+}1}$ $=\;\frac{r}{(2 r^{2+}1)^{2-}(2 r)^2}$ $=\;\frac{1}{4} \;\frac{4 r}{(2 r^{2+}2 r+1)(2 r^{2-}2 r+1)}$ $S_{10}=\;\frac{1}{4} sum_{r=1}^{10}(\;\frac{1}{(2 r^{2-}2 r+1)}-\;\frac{1}{(2 r^{2+}2 r+1)})$ $=\;\frac{1}{4}[1-\;\frac{1}{5}+\;\frac{1}{5}-\;\frac{1}{13}+. . .+\;\frac{1}{181}-\;\frac{1}{221}]$ $\Rightarrow S_{10}=\;\frac{1}{4} \cdot \;\frac{220}{221}=\;\frac{55}{221}=\;\frac{m}{n}$ $\therefore m+n=276$

Q88JEE Main 2022NAT4MCircles

A rectangle R with end points of one of its sides as $(1,2)$ and $(3,6)$ is inscribed in a circle. If the equation of a diameter of the circle is $2 x-y+4=0$ , then the area of $R$ is [27-Jun-2022-Shift-1]

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📖 Explanation

As slope of line joining $(1,2)$ and $(3,6)$ is 2 given diameter is parallel to side $\therefore a=\sqrt{(3-1)^{2+}(6-2)^2}=\sqrt{20}$ and $b / 2=\;\frac{4}{\sqrt{5}} \Rightarrow b=\;\frac{8}{\sqrt{5}}$ $\;\text{ Area }\;=a b=2 \sqrt{5} \cdot \;\frac{8}{\sqrt{5}}=16$

Q89JEE Main 2022NAT4MCircles

A circle of radius 2 unit passes through the vertex and the focus of the parabola $y^2=2 x$ and touches the parabola $y=(x-\;\frac{1}{4})^{2+}\alpha$ , where $\alpha>0 .$ Then $(4 \alpha-8)^2$ is equal to___ [27-Jun-2022-Shift-1]

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📖 Explanation

Let the equation of circle be $x(x-\;\frac{1}{2})+y^{2+}\lambda y=0$ $\Rightarrow x^{2+}y^{2-}\;\frac{1}{2} x+\lambda y=0$ $\;\text{ Radius }\;=\sqrt{\;\frac{1}{16}+\;\frac{\lambda ^2}{4}}=2$ $\Rightarrow \lambda ^2=\;\frac{63}{4}$ $\Rightarrow (x-\;\frac{1}{4})^{2+}(y+\;\frac{\lambda }{2})^2=4$ $\because$ This circle and parabola $y-\alpha=(x-\;\frac{1}{4})^2$ touch each other, so $\alpha=-\;\frac{\lambda }{2}+2$ $\Rightarrow \alpha-2=-\;\frac{\lambda }{2}$ $\Rightarrow (\alpha-2)^2=\;\frac{\lambda ^2}{4}=\;\frac{63}{16}$ $\Rightarrow (4 \alpha-8)^2=63$

Q90JEE Main 2022NAT4MThree Dimensional Geometry

Let the mirror image of the point $(a, b, c)$ with respect to the plane $3 x-4 y+12 z+19=0$ be $(a-6, \beta, \gamma )$ . If $a+b+c=5$ , then $7 \beta$ $-9 \gamma$ is equal to [27-Jun-2022-Shift-1]

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📖 Explanation

$\;\frac{x-a}{3}=\;\frac{y-b}{-4}=\;\frac{z-c}{12}=\;\frac{-2(3 a-4 b+12 c+19)}{3^{2+}(-4)^{2+}12^2}$ $\;\frac{x-a}{3}=\;\frac{y-b}{-4}=\;\frac{z-c}{12}=\;\frac{-6 a+8 b-24 c-38}{169}$ $(x, y, z) \equiv (a-6, \beta, \gamma )$ $\;\frac{(a-6)-a}{3}=\;\frac{\beta-b}{-4}=\;\frac{\gamma -c}{12}=\;\frac{-6 a+8 b-24 c-38}{169}$ $\;\frac{\beta-b}{-4}=-2 \Rightarrow \beta=8+b$ $\;\frac{\gamma -c}{12}=-2 \Rightarrow \gamma =-24+c$ $\;\frac{-6 a+8 b-24 c-38}{169}=-2$ $\Rightarrow 3 a-4 b+12 c=150$ $a+b+c=5$ $3 a+3 b+3 c=15$ Applying (1) - (2) $-7 b+9 c=135$ $7 b-9 c=-135$ $7 \beta-9 \gamma =7(8+b)-9(-24+c)$ $=56+216+7 b-9 c$ $=56+216-135=137$

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