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24 Jun 2022 Shift 1

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Physics30 questions

Q1JEE Main 2022MCQ4MProperties of Matter

The bulk modulus of a liquid is $3 \times 10^{10} {\text{Nm}}^{-2}$ . The pressure required to reduce the volume of liquid by $2 \%$ is : [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\because B=\;\frac{\Delta P}{(-\;\frac{\Delta V}{V})}$ $\Rightarrow \Delta P=3 \times 10^{10} \times (0.02)$ $=6 \times 10^8 {\text{N}} / {\text{m}}^2$

Q2JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R).Assertion (A): In an uniform magnetic field, speed and energy remains the same for a moving charged particle.Reason (R): Moving charged particle experiences magnetic force perpendicular to its direction of motion. [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Magnetic force $\vec{F} ⟂ \vec{v}$ $\Rightarrow W_b=0$ $\Rightarrow \Delta K E=0$ and speed remains constant.

Q3JEE Main 2022MCQ4MCurrent Electricity

Two identical cells each of emf $1.5 {\text{V}}$ are connected in parallel across a parallel combination of two resistors each of resistance $20 Ω$ . A voltmeter connected in the circuit measures $1.2 {\text{V}}$ . The internal resistance of each cell is : [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{1.5 \times 10}{10+\;\frac{r}{2}}=1.2$ $\Rightarrow r=5 Ω$

Q4JEE Main 2022MCQ4MUnits and Measurements

Identify the pair of physical quantities which have different dimensions: [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$[S]=\;\frac{[C]}{[m] \times [\Delta T]}$ and, $[L]=\;\frac{[Q]}{[m]}$ $\Rightarrow$ They have different dimensions.

Q5JEE Main 2022MCQ4MMotion in a Plane

A projectile is projected with velocity of $25 {\text{m}} / {\text{s}}$ at an angle $\theta$ with the horizontal. After $t$ seconds its inclination with horizontal becomes zero. If ${\text{R}}$ represents horizontal range of the projectile, the value of $\theta$ will be :$[use ${\text{g}}=10 {\text{m}} / {\text{s}}^2$ ]$ [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$t=\;\frac{25 \sin \theta}{g}$ and, $R=\;\frac{(25)^2(2 \sin \theta \cos \theta)}{g}$ $\Rightarrow R=\;\frac{25 \times 25 \times 2}{g} \times \;\frac{g t}{25} \times \cos \theta$ $\Rightarrow R=50 t \cos \theta$ $\therefore \tan \theta=\;\frac{g t}{25} \times \;\frac{50 t}{R}$ $=\;\frac{20 t^2}{R}$ $\Rightarrow \theta=\cot ^{-1}(\;\frac{R}{20 t^2})$

Q6JEE Main 2022MCQ4MWork, Power and Energy

A block of mass $10 {\text{kg}}$ starts sliding on a surface with an initial velocity of $9.8 {\text{ms}}^{-1}$ . The coefficient of friction between the surface and block is $0.5$ . The distance covered by the block before coming to rest is:$[use ${\text{g}}=9.8 {\text{ms}}^{-2}$ ]$ [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$S=\;\frac{u^2}{2 a}=\;\frac{u^2}{2(\mu g)}$ $=\;\frac{(9.8)^2}{2 \times 0.5 \times (9.8)}$ $=\;\frac{9.8}{1}$ $=9.8 {\text{m}}$

Q7JEE Main 2022MCQ4MCircular Motion

A boy ties a stone of mass $100 {\text{g}}$ to the end of a $2 {\text{m}}$ long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of $80 {\text{N}}$ . If the maximum speed with which the stone can revolve is $\;\frac{K}{\pi} {\text{rev}} . / {\min}$ . The value of ${\text{K}}$ is :(Assume the string is massless and unstretchable) [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$T=m \omega ^2 r$ $\Rightarrow 80=0.1 \times (2 \pi \times \;\frac{K}{\pi} \times \;\frac{1}{60})^2 \times 2$ $\Rightarrow \;\frac{800}{2}=\;\frac{K^2}{900}$ $\Rightarrow K=30 \times 20=600$

Q8JEE Main 2022MCQ4MElectrostatics

A vertical electric field of magnitude $4.9 \times 10^5 {\text{N}} / {\text{C}}$ just prevents a water droplet of a mass $0.1 {\text{g}}$ from falling. The value of charge on the droplet will be :(Given : $g=9.8 {\text{m}} / {\text{s}}^2$ ) [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Since the droplet is at rest $\Rightarrow \;\text{ Net force }\;=0$ $\Rightarrow {\text{mg}}= {\text{qE}}$ $\Rightarrow q=\;\frac{m g}{E}=2 \times 10^{-9} {\text{C}}$

Q9JEE Main 2022MCQ4MWork, Power and Energy

A particle experiences a variable force $\vec{F}=(4 x \hat{i}+3 y^2 \hat{j})$ in a horizontal $x-y$ plane. Assume distance in meters and force is newton. If the particle moves from point $(1,2)$ to point $(2,3)$ in the $x-y$ plane, then Kinetic Energy changes by : [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$W=\int \vec{F} \cdot \vec{d r}$ $=int_1^2 4 x d x+int_2^3 3 y^2 d y$ $=\$ [2 x^2]_1^{2+}$[y^3]_2^3=2 \times 3+(27-8)=25 {\text{J}}$

Q10JEE Main 2022MCQ4MGravitation

The approximate height from the surface of earth at which the weight of the body becomes $\;\frac{1}{3}$ of its weight on the surface of earth is :[Radius of earth ${\text{R}}=6400 {\text{km}}$ and $\sqrt{3}=1.732$ ] [24-Jun-2022-Shift-1]

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📖 Explanation

$M^{'}=\;\frac{M}{3} g$ $g^{'}=\;\frac{g}{3}$ $g^{'}=g (\;{ {\text{R}}}/{ {\text{R}}+ {\text{h}}}) ^2=\;\frac{g}{3}$ $\;{ {\text{R}}}/{ {\text{R}}+ {\text{h}}}=\;{1}/{{\sqrt{3}}}$ ${\text{h}}=({\sqrt{3}}-1) {\text{R}}$ $=(1.732-1) 6400$ ${\text{h}}=4685 {\text{km}}$

Q11JEE Main 2022MCQ4MAlternating Current

A resistance of $40 Ω$ is connected to a source of alternating current rated $220 {\text{V}}, 50 {\text{Hz}}$ . Find the time taken by the current to change from its maximum value to the rms value : [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$I=I_0 \cos (\omega t)$ say $\Rightarrow$ At maximum $\omega t_1=0$ or $t_1=0$ Then at rms value $I=I_0 / \sqrt{2}$ $\Rightarrow \omega t_2=\pi / 4$ $\Rightarrow \omega (t_2-t_1)=\pi / 4$ $\Delta t=\;\frac{\pi}{4 \omega }=\;\frac{\pi T}{4 \times 2 \pi}$ $=\;\frac{1}{400} s \;\text{ or }\; 2.5 {\text{ms}}$

Q12JEE Main 2022MCQ4MWaves

The equations of two waves are given by: $y_1=5 \sin 2 \pi(x-v t) {\text{cm}}$ $y_2=3 \sin 2 \pi(x-v t+1.5) {\text{cm}}$ These waves are simultaneously passing through a string. The amplitude of the resulting wave is: [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$y_1=5 \sin (2 \pi x-2 \pi v t)$ $y_2=3 \sin (2 \pi x-2 \pi v t+3 \pi)$ $\Rightarrow$ Phase difference $=3 \pi$ $\Rightarrow A_{n e t}=\sqrt{A_1^{2+}A_2^{2+}2 A_1 A_2 \cos (3 \pi)}$ $\Rightarrow A_{n e t}=2 {\text{cm}}$

Q13JEE Main 2022MCQ4MElectromagnetic Waves

A plane electromagnetic wave travels in a medium of relative permeability $1.61$ and relative permittivity 6.44. If magnitude of magnetic intensity is $4.5 \times 10^{-2} {\text{Am}}^{-1}$ at a point, what will be the approximate magnitude of electric field intensity at that point?(Given : Permeability of free space $\mu _0=4 \pi \times 10^{-7} {\text{NA}}^{-2}$ , speed of light in vacuum ${\text{c}}=3 \times 10^8 {\text{ms}}^{-} 1)$ [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$H=4.5 \times 10^{-2}$ So $B=\mu _0 \mu {\text{H}}$ Thus $E=\;\frac{c}{n} B$ (where ${\text{n}} \Rightarrow$ refractive index)So $E=\;\frac{3 \times 10^8 \times 4 \pi \times 10^{-7} \times 1.61 \times 4.5 \times 10^{-2}}{\sqrt{1.61 \times 6.44}}$ $E=8.48$

Q14JEE Main 2022MCQ4MAtoms and Nuclei

Choose the correct option from the following options given below :In the ground state of Rutherford's model electrons are in stable equilibrium. While in Thomson's [24-Jun-2022-Shift-1]

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Q15JEE Main 2022MCQ4MAtoms and Nuclei

Nucleus A is having mass number 220 and its binding energy per nucleon is $5.6 {\text{MeV}}$ . It splits in two fragments ' $B$ ' and ' $C$ ' of mass numbers 105 and 115 . The binding energy of nucleons in ' $B$ ' and ' $C$ ' is $6.4 {\text{MeV}}$ per nucleon. The energy Q released per fission will be : [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${ }^{220} {\text{A}} \to { }^{105} {\text{B}}+{ }^{115} {\text{C}}$ $\Rightarrow {\text{Q}}=[105 \times 6.4+115 \times 6.4]-[220 \times 5.6] {\text{MeV}}$ $\Rightarrow {\text{Q}}=176 {\text{MeV}}$

Q16JEE Main 2022MCQ4MElectromagnetic Waves

A baseband signal of $3.5 {\text{MHz}}$ frequency is modulated with a carrier signal of $3.5 {\text{GHz}}$ frequency using amplitude modulation method. What should be the minimum size of antenna required to transmit the modulated signal? [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$v_c=3.5 \times 10^9 {\text{Hz}}$ $\therefore \lambda =\;\frac{c}{v_c}=\;\frac{3 \times 10^8}{3.5 \times 10^9}$ $\therefore$ Size of antenna $=\;\frac{\lambda }{4}$ $=\;\frac{8.57 \times 10^{-2}}{4}$ $=21.4 {\text{mm}}$

Q17JEE Main 2022MCQ4MThermodynamics

A Carnot engine whose heat sinks at $27^{\circ} {\text{C}}$ , has an efficiency of $25 \%$ . By how many degrees should the temperature of the source be changed to increase the efficiency by $100 \%$ of the original efficiency? [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Initially : $\;\frac{1}{4}=1-\;\frac{300}{T_H}$ $\Rightarrow T_H=400 K$ Finally : Efficiency becomes $\;\frac{1}{2}$ $\Rightarrow \;\frac{1}{2}=1-\;\frac{300}{T_H^{'}}$ $\Rightarrow T_H^{'}=600 K$ $\Rightarrow$ Temperature of the source increases by $200^{\circ} {\text{C}}$ .

Q18JEE Main 2022MCQ4MElectrostatic Potential and Capacitance

A parallel plate capacitor is formed by two plates each of area $30 \pi {\text{cm}}^2$ separated by $1 {\text{mm}}$ . A material of dielectric strength $3.6 \times 10^7 {\text{Vm}}^{-1}$ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is $7 \times 10^{-6} {\text{C}}$ , the value of dielectric constant of the material is : $[$ Use $.\;\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 {\text{Nm}}^2 {\text{C}}^{-2}]$ [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Field inside the dielectric $=\;\frac{\sigma}{k \epsilon_0}$ According to the given information, $\;\frac{\sigma}{k \epsilon_0}=3.6 \times 10^7$ $\Rightarrow \;\frac{\;\frac{Q}{A}}{k \epsilon_0}=3.6 \times 10^7$ $\Rightarrow k=2.33$

Q19JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

The magnetic field at the centre of a circular coil of radius $r$ , due to current I flowing through it, is $B$ . The magnetic field at a point along the axis at a distance $\;\frac{r}{2}$ from the centre is : [24-Jun-2022-Shift-1]

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📖 Explanation

$B=\;\frac{\mu _0 I}{2 r}$ $B_a=\;\frac{\mu _0 I r^2}{2(r^{2+}\;\frac{r^2}{4})}$ $\Rightarrow \;\frac{B_a}{B}=(\;\frac{2}{\sqrt{5}})^3$ $\Rightarrow B_a=(\;\frac{2}{\sqrt{5}})^3 B$

Q20JEE Main 2022MCQ4MProperties of Matter

Two metallic blocks $M_1$ and $M_2$ of same area of cross-section are connected to each other (as shown in figure). If the thermal conductivity of ${\text{M}}_2$ is ${\text{K}}$ then the thermal conductivity of ${\text{M}}_1$ will be :[Assume steady state heat conduction] [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Thermal current is same so $\;\frac{d Q}{d t}=\;\frac{\Delta T_1}{\frac{I_1}{K_1 A}}=\;\frac{\Delta T_2}{\frac{I_2}{K_2 A}}$ $\;\text{ or }\; \;\frac{20}{16} \times K^{'}=\;\frac{80}{8} \times K$ $\Rightarrow K^{'}=8 K$

Q21JEE Main 2022NAT4MThermodynamics

$0.056 {\text{kg}}$ of Nitrogen is enclosed in a vessel at a temperature of $127^{\circ} {\text{C}}$ . Th amount of heat required to double the speed of its molecules is _____ ${\text{k}}$ cal.Take ${\text{R}}=2 {\text{cal}} {\text{mole}}^{-1} {\text{K}}^{-1}$ ) [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Because the vessel is closed, it will be an isochoric process.To double the speed, temperature must be 4 times $(∨ \alpha \sqrt{T})$ So, $T_f=1600 {\text{K}}, T_i=400 {\text{K}}$ number of moles are $\;\frac{56}{28}=2$ $\;\text{ so }\; Q=n C v \Delta T=2 \times \;\frac{5}{2} \times 2 \times 1200$ $=12000 {\text{cal}}=12 {\text{K}} {\text{cal}}$

Q22JEE Main 2022NAT4MRay Optics and Optical Instruments

Two identical thin biconvex lens of focal length $15 {\text{cm}}$ and refractive index $1.5$ are in contact with each other. The space between the lenses is filled with a liquid of refractive index $1.25$ . The focal length of the combination is _____cm. [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{1}{f_l}=(\;\frac{\mu _e}{\mu _m}-1)(\;\frac{1}{R_1}-\;\frac{1}{R_2})$ $\;\text{ here }\;|R_1|=|R_2|=R$ $\Rightarrow \;\frac{1}{f_{l_1}}=(1.5-1)(\;\frac{2}{R})=\;\frac{1}{15}$ $\Rightarrow \;\frac{1}{R}=\;\frac{1}{15} \;\text{ or }\; R=15 {\text{cm}}$ for the concave lens made up of liquid $\;\frac{1}{f_{l_2}}=(1.25-1)(-\;\frac{2}{R})=-\;\frac{1}{30} {\text{cm}}$ now for equivalent lens $\;\frac{1}{f_e}=\;\frac{2}{f_{l_1}}+\;\frac{1}{f_{l_2}}$ $=\;\frac{2}{15}-\;\frac{1}{30}=\;\frac{3}{30}=\;\frac{1}{10}$ or $f_e=10 {\text{cm}}$

Q23JEE Main 2022NAT4MSemiconductor Electronics

A transistor is used in common-emitter mode in an amplifier circuit. When a signal of $10 {\text{mV}}$ is added to the base-emitter voltage, the base current changes by $10 \mu {\text{A}}$ and the collector current changes by $1.5 {\text{mA}}$ . The load resistance is $5 {\text{k}} Ω$ . The voltage gain of the transistor will be____ [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$R_B=\;\frac{10 \times 10^{-3}}{10 \times 10^{-6}}$ $=10^3 Ω$ $\therefore A_v=(\;\frac{\Delta I_C}{\Delta I_B}) \times (\;\frac{R_C}{R_B})$ $=\;\frac{1.5 \times 10^{-3}}{10 \times 10^{-6}} \times \;\frac{5 \times 10^3}{1 \times 10^3}$ $=\;\frac{1.5 \times 5}{10} \times (1000)$ $=750$

Q24JEE Main 2022NAT4MAlternating Current

As shown in the figure an inductor of inductance $200 {\text{mH}}$ is connected to an AC source of emf $220 {\text{V}}$ and frequency $50 {\text{Hz}}$ . The instantaneous voltage of the source is $0 {\text{V}}$ when the peak value of current is $\;\frac{\sqrt{a}}{\pi} {\text{A}}$ . The value of $a$ is _____ [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$I_{r m s}=\;\frac{V_{r m s}}{z}$ $z=X_2=\omega _2$ $=2 \pi \times 50 \times \;\frac{200}{1000}$ $=20 \pi$ $\therefore I_{r m s}=\;\frac{220}{20 \pi}=\;\frac{11}{\pi}$ $\therefore I_{p e a k}=\sqrt{2} \times \;\frac{11}{\pi}$ $=\;\frac{\sqrt{2 \times 121}}{\pi}$ $=\;\frac{\sqrt{242}}{\pi}$

Q25JEE Main 2022NAT4MWave Optics

Sodium light of wavelengths $650 {\text{nm}}$ and $655 {\text{nm}}$ is used to study diffraction at a single slit of aperture $0.5 {\text{mm}}$ . The distance between the slit and the screen is $2.0 {\text{m}}$ . The separation between the positions of the first maxima of diffraction pattern obtained in the two cases is ______ $\times 10^{-5}$ ${\text{m}} .$ [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Position of $1^{\;\text{st }\;}$ maxima is $\;\frac{3}{2} \;\frac{\lambda D}{a}$ $\Rightarrow$ According to given values, required separation $=\;\frac{3}{2} \times (655 {\text{nm}}-650 {\text{nm}}) \times \;{2 m}/{0.5 {\text{mm}}}$ $\Rightarrow$ Required separation $=3 \times 10^{-5} {\text{m}}$ .

Q26JEE Main 2022NAT4MDual Nature of Matter and Radiation

When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is $v_{1}$ . When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes $v_2$ . If $v_2=x v_1$ , the value of $x$ will be [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Let us say that work function is $\varphi$ $\Rightarrow 2 \varphi =\varphi +\;\frac{1}{2} m v_1^2 . . . . . . \;\text{ (1 }\;$ $\;\text{ and }\; 5 \varphi =\varphi +\;\frac{1}{2} m v_2^2 . . . .(2$ From (1) and (2) $\;\frac{v_2^2}{v_1^2}=\;\frac{4}{1} \;\text{ or }\; \;\frac{v_2}{v_1}=2$

Q27JEE Main 2022NAT4MMotion in a Straight Line

From the top of a tower, a ball is thrown vertically upward which reaches the ground in $6 {\text{s}}$ . A second ball thrown vertically downward from the same position with the same speed reaches the ground in $1.5 {\text{s}}$ . A third ball released, from the rest from the same location, will reach the ground in s. [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Based on the situation $h=-u t_1+\;\frac{1}{2} g t_1^2 \to \;\text{ throwing up ...... (i) }\;$ $h=u t_2+\;\frac{1}{2} g t_2^2 \to \;\text{ throwing up ...... (ii) }\;$ $h=\;\frac{1}{2} g t^2 \to \;\text{ dropping }\; . . . . . . . . . \;\text{ (iii) }\;$ $\;\text{ and }\; 0=u(t_1-t_2)-\;\frac{1}{2} g(t_1-t_2)^2 . . . . . . \;\text{ ( iv)}$ solving above equations $t=\sqrt{t_1 t_2}$ $\Rightarrow t=\sqrt{6 \times 1.5}=3 {\text{s}}$

Q28JEE Main 2022NAT4MWork, Power and Energy

A ball of mass $100 {\text{g}}$ is dropped from a height $h=10 {\text{cm}}$ on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance $\;\frac{h}{2}$ . The spring constant is ${\text{Nm}}^{-1}$ . (Use ${\text{g}}=10 {\text{ms}}^{-2}$ ) [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$m g(h+\;\frac{h}{2})=\;\frac{1}{2} k(\;\frac{h}{2})^2$ $\Rightarrow 0.1 \times 10 \times (0.15)=\;\frac{1}{2} k(0.05)^2$ $\Rightarrow k=120 {\text{N}} / {\text{m}}$

Q29JEE Main 2022NAT4MCurrent Electricity

In a potentiometer arrangement, a cell gives a balancing point at $75 {\text{cm}}$ length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is $3: 2$ , the difference in the balancing length of the potentiometer wire in above two cases will be ___ ${\text{cm}} .$ [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

At balancing point, we know that emf is proportional to the balancing length. i.e.,emf $\propto$ balancing lengthNow, let the emf's be $3 \epsilon$ and $2 \epsilon$ . $\Rightarrow 3 \epsilon= {\text{k}}(75) . . . . .(1)$ and $2 \epsilon= {\text{k}}( {\text{I}}) . . . . . .(2)$ $\Rightarrow {\text{I}}=50 {\text{cm}}$ $\Rightarrow$ Difference is $(75-50) {\text{cm}}=25 {\text{cm}}$ .

Q30JEE Main 2022NAT4MRotational Motion

A metre scale is balanced on a knife edge at its centre. When two coins, each of mass $10 {\text{g}}$ are put one on the top of the other at the $10.0 {\text{cm}}$ mark the scale is found to be balanced at $40.0 {\text{cm}}$ mark. The mass of the metre scale is found to be $x \times 10^{-2} {\text{kg}}$ . The value of $x$ is__ [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

If $\lambda$ is the mass per unit length of the scale then $0.02 \times (30) \times 10+\lambda 40 \times 20 \times 10=\lambda 60 \times 30 \times 10$ $0.006=\lambda 10$ $\;\text{ or }\; 100 \lambda =0.06 {\text{kg}}$ $=6 \times 10^{-2} {\text{kg}}$ $\Rightarrow x=6$

Chemistry30 questions

Q31JEE Main 2022MCQ4MSome Basic Concepts of Chemistry

If a rocket runs on a fuel $( {C}_{15} {H}_{30})$ and liquid oxygen, the weight of oxygen required and ${CO}_{2}$ released for every litre of fuel respectively are :(Given : density of the fuel is $0.756 {g} / {mL}$ ) [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${C}_{15} {H}_{30}+\;\frac{45}{2} {O}_{2} \to 15 {CO}_{2}+15 {H}_{2} {O}$ Given, volume of fuel $=1 {L}=1000 {ml}$ And density of fuel $=0.756 {g} / {ml}$ We know, $d=\;\frac{w}{v}$ $\Rightarrow 0.756=\;\frac{w}{1000}$ $\Rightarrow {W}=756 {gm}$ $\therefore$ weight of fuel $=756 {gm}$ Molar mass of ${C}_{15} {H}_{30}=15 \times 12+30=210$ $\therefore$ Moles of ${C}_{15} {H}_{30}=\;\frac{756}{210}$ From equation you can see, 1 mole of ${C}_{15} {H}_{30}$ react with $\;\frac{45}{2}$ mole of ${O}_{2}$ $\therefore \;\frac{756}{210}$ moles of ${C}_{15} {H}_{30}$ react with $\;\frac{45}{2} \times \;\frac{756}{210}$ moles of ${O}_{2}$ $\therefore$ Moles of ${O}_{2}$ required $=\;\frac{45}{2} \times \;\frac{756}{210}$ $\therefore$ Mass of ${O}_{2}$ required $=\;\frac{45}{2} \times \;\frac{756}{210} \times 32=$ $2592 g$ Also,From 1 mole of ${C}_{15} {H}_{30} 15$ moles of ${CO}_{2}$ formed $\therefore$ From $\;\frac{756}{210}$ moles of ${C}_{15} {H}_{30} 15 \times \;\frac{756}{210}$ moles of ${CO}_{2}$ formed $\therefore$ Moles of ${CO}_{2}$ formed $=15 \times \;\frac{756}{210}$ $\therefore$ Mass of ${CO}_{2}$ formed $=15 \times \;\frac{756}{210} \times 44=$ $2376 {g}$

Q32JEE Main 2022MCQ4MStructure of Atom

Consider the following pairs of electrons(A) $( {a}) {n}=3, l=1, {m}_{1}=1, {m}_{ {s}}=+\;\frac{1}{2}$ (b) $n=3, l=2, m_{1}=1, m_{s}=+\;\frac{1}{2}$ (B) (a) ${n}=3, l=2, {m}_{1}=-2, {m}_{ {s}}=-\;\frac{1}{2}$ (b) ${n}=3, l=2, {m}_{1}=-1, {m}_{ {s}}=-\;\frac{1}{2}$ (C) $( {a}) {n}=4, l=2, {m}_{1}=2, {m}_{ {s}}=+\;\frac{1}{2}$ (b) ${n}=3, l=2, {m}_{1}=2, {m}_{ {s}}=+\;\frac{1}{2}$ The pairs of electrons present in degenerate orbitals is/are : [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

For degenerate orbitals, only the value of ${\text{m}}$ must be different. The value of $(n+1)$ must be the same.Hence, the pair of electrons with quantum numbers given in (B) are degenerate.

Q33JEE Main 2022MCQ4MCoordination Compounds

List-1 List -II (A) $[PtCl_4]^{2-}$ I $sp^3d$ (B) $BrF_5$ II $d^2sp^3$ (C) $PCl_5$ III $dsp^2$ (D) $[CoNH_{3}{}_{6}]^{3+}$ IV $sp^3d^2$ Choose the most appropriate answer from the options given below : [24-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

To determine the hybridization, we analyze the central atom of each species.

A) $[PtCl_4]^{2-}$ : Platinum is $Pt^{2+}$ ( $5d^8$ configuration). Since ligands are considered strong-field for 4d and 5d metals, the electrons pair up, leaving one d-orbital empty. The hybridization is $dsp^2$ (square planar).
B) $BrF_5$ : Bromine has 7 valence electrons. It forms 5 sigma bonds with fluorine and has 1 lone pair. The steric number is $5+1=6$ , leading to $sp^3d^2$ hybridization (square pyramidal).
C) $PCl_5$ : Phosphorus has 5 valence electrons, forming 5 sigma bonds with chlorine and has no lone pairs. The steric number is $5+0=5$ , resulting in $sp^3d$ hybridization (trigonal bipyramidal).
D) $[Co(NH_3)_6]^{3+}$ : Cobalt is $Co^{3+}$ ( $3d^6$ configuration). $NH_3$ is a strong-field ligand, causing electron pairing. This vacates two inner d-orbitals, leading to $d^2sp^3$ hybridization (octahedral).

Matching the lists: (A) matches with (III), (B) with (IV), (C) with (I), and (D) with (II).

Q34JEE Main 2022MCQ4MChemical Equilibrium

For a reaction at equilibrium ${A}( {g}) ⇌ {B}( {g})+\;\frac{1}{2} {C}( {g})$ the relation between dissociation constant (K), degree of dissociation $(\alpha)$ and equilibrium pressure $( {p})$ is given by: [24-Jun-2022-Shift-1]

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📖 Explanation

$\text{table} ,{A}( {g}), ⇌, {B}( {g}),+,\;\frac{1}{2} {C}( {g}), \text{Total} \text{Moles} ; At t=0:, 1,, 0,,0,1;At t=t_\text{req}:, 1-\alpha,, \alpha,,\alpha/2,1;\text{Mole} \text{Fraction}: , \;\frac{\alpha}{1+\;\frac{\alpha}{2}} ,, (\;\frac{1-a}{1+\;\frac{a}{2}}),,P(\;\frac{2}{2}) P$ Now, $K_{P} \;\text{ or }\; K=\;{P_{B} \times (P_{C})^{\frac{1}{2}}}/{P_{A}}$ $=\;{(\frac{\alpha}{1+\frac{\alpha}{2}}) P \times [(\frac{\frac{\alpha}{2}}{1+\frac{\alpha}{2}}) P]^{\frac{1}{2}}}/{(\;\frac{1-\alpha}{1+\;\frac{\alpha}{2}}) P}$ $=\;{(\frac{2 \alpha}{2+\alpha}) P \times [(\frac{\alpha}{2+\alpha}) P]^{\frac{1}{2}}}/{(\;\frac{2(1-\alpha)}{2+\alpha}) P}$ $=\;\frac{\alpha}{1-\alpha} \times (\;\frac{\alpha P}{2+\alpha})^{\;\frac{1}{2}}$ $=\;{\alpha^{\frac{3}{2}} \cdot P^{\frac{1}{2}}}/{(1-\alpha)(2+\alpha)^{\;\frac{1}{2}}}$

Q35JEE Main 2022MCQ4MStates of Matter

Given below are two statements :Statement I : Emulsions of oil in water are unstable and sometimes they separate into two layers on standing.Statement II : For stabilisation of an emulsion, excess of electrolyte is added.In the light of the above statements, choose the most appropriate answer from the options given below : [24-Jun-2022-Shift-1]

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Q36JEE Main 2022MCQ4Mp Block Elements

Given below are the oxides: ${Na}_{2} {O}, {As}_{2} {O}_{3}, {N}_{2} {O}, {NO}$ and ${Cl}_{2} {O}_{7}$ Number of amphoteric oxides is : [24-Jun-2022-Shift-1]

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📖 Explanation

Oxides ${\text{Na}}_2 {\text{O}} \longrightarrow$ Basic ${\text{As}}_2 {\text{O}}_3 \longrightarrow$ Amphoteric ${\text{N}}_2 {\text{O}} \longrightarrow$ Neutral ${\text{NO}} \longrightarrow$ Neutral ${\text{Cl}}_2 {\text{O}}_7 \longrightarrow$ AcidicHence, only one amphoteric oxide is present.

Q37JEE Main 2022MCQ4Md and f Block Elements

List-I List -II (A) Sphalerite (I) $\text{FeCO}_3$ (B) Calamine (II) $\text{PbS}$ (C) Galena (III) $\text{ZnCO}_3$ (D) Siderite (IV) $\text{ZnS}$ Choose the most appropriate answer from the options given below : [24-Jun-2022-Shift-1]

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📖 Explanation

This question requires matching common mineral ores with their chemical formulas. Let's identify the chemical formula for each ore listed.

Sphalerite is the primary ore of zinc, with the chemical formula $\text{ZnS}$ . Thus, (A) matches (IV).
Calamine is an ore of zinc composed of zinc carbonate, $\text{ZnCO}_3$ . Thus, (B) matches (III).
Galena is the main ore of lead, chemically known as lead(II) sulfide, $\text{PbS}$ . Thus, (C) matches (II).
Siderite is an ore of iron, which is iron(II) carbonate, $\text{FeCO}_3$ . Thus, (D) matches (I).
The correct set of matches is (A)-(IV), (B)-(III), (C)-(II), and (D)-(I).

Q38JEE Main 2022MCQ4Mp Block Elements

The highest industrial consumption of molecular hydrogen is to produce compounds of element : [24-Jun-2022-Shift-1]

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📖 Explanation

Hydrogen combines with nitrogen to produce Ammonia in Haber's process. ${\text{N}}_2( {\text{g}})+3 {\text{H}}_2( {\text{g}}) ⇌ 2 {\text{NH}}_3( {\text{g}})$ In this process, iron oxide is used with small amounts of ${\text{K}}_2 {\text{O}}$ and ${\text{Al}}_2 {\text{O}}_3$ to increase the rate of attainment of equilibrium.Optimum conditions for the production of ammonia are a pressure of 200 atm and a temperature of $700 {\text{K}}$ .Earlier, iron was used as a catalyst with molybdenum as a promoter in this reaction.

Q39JEE Main 2022MCQ4MClassification of Elements

Which of the following statements are correct?(A) Both ${LiCl}$ and ${MgCl}_{2}$ are soluble in ethanol.(B) The oxides ${Li}_{2} {O}$ and ${MgO}$ combine with excess of oxygen to give superoxide.(C) LiF is less soluble in water than other alkali metal fluorides.(D) ${Li}_{2} {O}$ is more soluble in water than other alkali metal oxides.Choose the most appropriate answer from the options given below : [24-Jun-2022-Shift-1]

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📖 Explanation

(A) Both ${\text{LiCl}}$ and ${\text{MgCl}}_2$ are soluble in ethanol(B) Li and Mg do not form superoxide(C) LiF has high lattice energy(D) ${\text{Li}}_2 {\text{O}}$ is least soluble in water than another alkali metal oxides

Q40JEE Main 2022MCQ4Mp Block Elements

Identify the correct statement for ${B}_{2} {H}_{6}$ from those given below :(A) In ${B}_{2} {H}_{6}$ , all B-H bonds are equivalent.(B) In ${B}_{2} {H}_{6}$ , there are four 3-centre-2-electron bonds.(C) ${B}_{2} {H}_{6}$ is a Lewis acid.(D) ${B}_{2} {H}_{6}$ can be synthesized from both ${BF}_{3}$ and ${NaBH}_{4}$ .(E) ${B}_{2} {H}_{6}$ is a planar molecule.Choose the most appropriate answer from the options given below : [24-Jun-2022-Shift-1]

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📖 Explanation

It has two 3-centre-2-electron bonds and four 2-centre-2-electron bonds.Hence, all ${\text{B}}- {\text{H}}$ bonds are not equivalent.It is an electron deficient compound as the octet of boron is incomplete.Hence, it can behave as a Lewis acid.It can be synthesized from both ${\text{BF}}_3$ and ${\text{NaBH}}_4$ $2 {\text{BF}}_3+6 {\text{NaH}} {\longrightarrow }^{450 K} {\text{B}}_2 {\text{H}}_6+6 {\text{NaF}}$ $2 {\text{NaBH}}_4+ {\text{I}}_2 \to {\text{B}}_2 {\text{H}}_6+2 {\text{NaI}}+ {\text{H}}_2$ It is a non-planar molecule.Hence, only Statements (C) and (D) are correct.

Q41JEE Main 2022MCQ4Mp Block Elements

The most stable trihalide of nitrogen is: [24-Jun-2022-Shift-1]

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📖 Explanation

The stability of trihalides decreases down the group due to weakening of ${\text{N}}- {\text{X}}$ bond and inability of ${\text{N}}$ to accommodate large sized halogen atoms $( {\text{Cl}}, {\text{Br}}, {\text{l}})$ around it.

Q42JEE Main 2022MCQ4MBiomolecules

Which one of the following elemental forms is not present in the enamel of the teeth? [24-Jun-2022-Shift-1]

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📖 Explanation

${\text{P}}^{+3}$ is not present is enamel of teeth.Th compound present is $[3 {\text{Ca}}_3( {\text{PO}}_4)_2 \cdot {\text{CaF}}_2]$ Which contains ${\text{Ca}}^{+2}, {\text{P}}^{+5} \& {\text{F}}^{-}$

Q43JEE Main 2022MCQ4MHaloalkanes and Haloarenes

In the given reaction sequence, the major product $'C'$ is : $\text{C}_8\text{H}_10 {\to }^{ \text{HNO}_3} _{\text{H}_2\text{SO}_4} {A} {\to }^{Br_2}_{\Delta } {B} {\to }^{\text{alcoholic}} _{\text{KOH}}C$ [24-Jun-2022-Shift-1]

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📖 Explanation

${\text{C}}_8 {\text{H}}_{10} {\text{DU}}=9-5=4$

Q44JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Two statements are given below:Statement I : The melting point of monocarboxylic acid with even number of carbon atoms is higher than that of with odd number of carbon atoms acid immediately below and above it in the series.Statement II : The solubility of monocarboxylic acids in water decreases with increase in molar mass.Choose the most appropriate option : [24-Jun-2022-Shift-1]

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Q45JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Which of the following is an example of conjugated diketone? [24-Jun-2022-Shift-1]

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Q46JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

[24-Jun-2022-Shift-1]

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Q47JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Which of the following is an example of polyester? [24-Jun-2022-Shift-1]

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📖 Explanation

Polyesters are formed by condensation reaction between alcohols and carboxylic acid.Poly- $\beta$ -hydroxybutyrate-co- $\beta$ -hydroxy valerate (PHBV) is a polymer obtained by condensation reaction of 3 hydroxybutanoic acid with 3-hydroxypentanoic acid.

Q48JEE Main 2022MCQ4MBiomolecules

A polysaccharide ' ${\text{X}}$ ' on boiling with dil ${\text{H}}_2 {\text{SO}}_4$ at $393 {\text{K}}$ under 2-3 atm pressure yields ' ${\text{Y}}$ '. ' ${\text{Y}}$ ' on treatment with bromine water gives gluconic acid. 'X' contains $\beta$ -glycosidic linkages only. Compound ' $X$ ' is : [24-Jun-2022-Shift-1]

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📖 Explanation

Cellulose contains $\beta$ -glycosidic linkages only. Structure of celluloseOn boiling with dil. ${\text{H}}_2 {\text{SO}}_4$ at $393 {\text{K}}$ under 2-3 atm, ' ${\text{X}}$ ' forms glucose, which given gluconic acid on treatment with bromine water.

Q49JEE Main 2022MCQ4MBiomolecules

Which of the following is not a broad spectrum antibiotic? [24-Jun-2022-Shift-1]

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Q50JEE Main 2022MCQ4MSalt Analysis

During the qualitative analysis of salt with cation ${\text{y}}^{2+}$ , addition of a reagent $( {\text{X}})$ to alkaline solution of the salt gives a bright red precipitate. The reagent $( {\text{X}})$ and the cation $( {\text{y}}^{2+})$ present respectively are : [24-Jun-2022-Shift-1]

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📖 Explanation

On addition of dimethylglyoxime to alkaline solution of ${\text{Ni}}^{+2}$ , a bright red ppt. is obtained. ${\text{Ni}}^{+2}+2 {\text{dmg}} \to [ {\text{Ni}}( {\text{dmg}})_2]^{+2}($ Bright red ppt)

Q51JEE Main 2022NAT4MSolid State

Atoms of element $X$ form hcp lattice and those of element $Y$ occupy $\;\frac{2}{3}$ of its tetrahedral voids. The percentage of element $X$ in the lattice is____ (Nearest integer) [24-Jun-2022-Shift-1]

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📖 Explanation

Since $X$ occupies hop lattice, Number of particles of type $X$ in a unit cell $=6$ Number of particles of type $Y=\;\frac{2}{3} \times 12=8$ $\therefore$ Percentage of element $X=\;\frac{6}{14} \times 100$ $=\;\frac{300}{7}$ $=42.85$ $≃ 43 \%$

Q52JEE Main 2022NAT4MChemical Thermodynamics

$2 {\text{O}}_3( {\text{g}}) ⇌ 3 {\text{O}}_2( {\text{g}})$ At $300 {\text{K}}$ , ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is $(-) \underline{\hspace{3em}} {\text{J}} {\text{mol}}^{-1}$ . (Nearest integer)[Given : In $1.35=0.3$ and $. {\text{R}}=8.3 {\text{JK}}^{-1} {\text{mol}}^{-1}]$ [24-Jun-2022-Shift-1]

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📖 Explanation

${2O_3(g)}_{1-x} ⇌ {3O_2(g)}_{{3x}/2}$ Given, $x=0.5$ $\therefore {\text{k}}_{ {\text{p}}}=\;\frac{[3(0.5)]^3 \times 1}{[2]^3 \times (0.5)^2 \times 1.25}$ $\therefore {\text{k}}_{ {\text{p}}}=\;\frac{27}{8} \times \;\frac{0.5}{1.25}=1.35$ $\Delta {\text{G}}^{\circ}=-2.303 {\text{RT}} \log {\text{k}}_{ {\text{p}}}$ $=-2.303 \times 8.3 \times 300 \log 1.35$ $=-8.3 \times 300 \ln (1.35)$ $=-747 {\text{J}} {\text{mol}}^{-1}$

Q53JEE Main 2022NAT4MSolutions

The osmotic pressure of blood is $7.47$ bar at $300 {\text{K}}$ . To inject glucose to a patient intravenously, it has to be isotonic with blood. The concentration of glucose solution in ${\text{gL}}^{-1}$ is(Molar mass of glucose $=180 {\text{g}} {\text{mol}}^{-1}, {\text{R}}=0.083 {\text{L}} {\bar}^{-1} {\text{mol}}^{-1}$ ) _____(Nearest integer) [24-Jun-2022-Shift-1]

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📖 Explanation

$7.47= {\text{C}} \times 0.083 \times 300$ $(\pi= {\text{CRT}})$ (Where ${\text{C}}$ represents the concentration of glucose solution and $\pi$ represents osmotic pressure) ${\text{C}}=\;\frac{7.47}{0.083 \times 300}( {\text{mol}} {\text{L}}^{-1})$ which in ${\text{gm}} / {\text{L}}=\;\frac{7.47}{0.083 \times 300} \times 180$ $=54 {\text{gm}} / 1$

Q54JEE Main 2022NAT4MElectrochemistry

The cell potential for the following cell ${\text{Pt}}| {\text{H}}_2( {\text{g}})| {\text{H}}^{+}( {\text{aq}}) \| {\text{Cu}}^{2+}(0.01 {\text{M}}) \mid {\text{Cu}}( {\text{s}})$ is $0.576 {\text{V}}$ at $298 {\text{K}}$ . The ${\text{pH}}$ of the solution is _____(Nearest integer)(Given : ${E^o}_{C u^{2+} / C u}=0.34 {\text{V}}$ and $\;\frac{2.303 R T}{F}=0.06 {\text{V}}$ ) [24-Jun-2022-Shift-1]

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📖 Explanation

$E_{\;\text{cell }\;}=E_{\;\text{cell }\;}^{0-}\;\frac{0.06}{2} \log \;{[ {\text{H}}^{⊕}]^2}/{[ {\text{Cu}}^{+2}]\$ }0.576=0.34-0.03 \log ;{[ {\text{H}}^{⊕}]^2}/{[0.01]}0.576-0.34=-0.03 \log $[ {\text{H}}^{⊕}]^{2+}0.03 \log (0.01)=0.06 {\text{pH}}-0.06{\text{pH}} ≃ 4.93 ≃ 5$

Q55JEE Main 2022NAT4MChemical Kinetics

The rate constants for decomposition of acetaldehyde have been measured over the temperature range $700-1000 {\text{K}}$ . The data has been analysed by plotting $\ln {\text{k}} v {\text{s}} \;\frac{10^3}{T}$ graph. The value of activation energy for the reaction is ${\text{kJ}} {\text{mol}}^{-1}$ . _________(Nearest integer)(Given : ${\text{R}}=8.31 {\text{JK}}^{-1} {\text{mol}}^{-1}$ ) [24-Jun-2022-Shift-1]

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📖 Explanation

$\ln {\text{k}}=\ln {\text{A}}-\;{ {\text{E}}_{ {\text{a}}}}/{ {\text{RT}}}$ $\therefore$ Slope of the graph $=-\;\frac{E_a}{R \times 10^3}=-18.5$ $\therefore E_a=18.5 \times 8.31 \times 1000 ≃ 154 {\text{kJ}} {\text{mol}}^{-1}$

Q56JEE Main 2022NAT4Md and f Block Elements

The difference in oxidation state of chromium in chromate and dichromate salts is ____ [24-Jun-2022-Shift-1]

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📖 Explanation

Chromate ion $\to {\text{CrO}}_4^{2-}$ , oxidation state of ${\text{Cr}}=+6$ Dichromate ion $\to {\text{CrO}}_7^{2-}$ , oxidation state of ${\text{Cr}}=+6$ $\therefore$ Difference in oxidation state $=$ zero

Q57JEE Main 2022NAT4MCoordination Compounds

In the cobalt-carbonyl complex : $[ {\text{CO}}_2( {\text{CO}})_8]$ , number of ${\text{Co}}- {\text{Co}}$ bonds is "X" and terminal ${\text{CO}}$ ligands is "Y". $X+Y=$ [24-Jun-2022-Shift-1]

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📖 Explanation

$x=1$ $y=6$ $\therefore x+y=7$

Q58JEE Main 2022NAT4MPractical Organic Chemistry

A $0.166 {\text{g}}$ sample of an organic compound was digested with conc. ${\text{H}}_2 {\text{SO}}_4$ and then distilled with ${\text{NaOH}}$ . The ammonia gas evolved was passed through $50.0 {\text{mL}}$ of $0.5 {\text{N}} {\text{H}}_2 {\text{SO}}_4$ . The used acid required $30.0 {\text{mL}}$ of $0.25 {\text{N}} {\text{NaOH}}$ for complete neutralization. The mass percentage of nitrogen in the organic compound is [24-Jun-2022-Shift-1]

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📖 Explanation

Millimoles of used acid $=\;\frac{30 \times 0.25}{2}$ Millimoles of ${\text{NH}}_3=30 \times 0.25=7.5$ Mass $\%$ of nitrogen $=\;\frac{7.5}{0.166} \times 10^{-3} \times 14 \times 100 ≃ 63 \%$

Q59JEE Main 2022NAT4MOrganic Chemistry - Some Basic Principles

Number of electrophilic centres in the given compound is [24-Jun-2022-Shift-1]

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📖 Explanation

Given compounds : Number of electrophilic centres $=3$

Q60JEE Main 2022NAT4MHydrocarbons

The major product $'A'$ of the following given reaction has_____ ${sp}^2$ hybridized carbon atoms. $\text{2,7-Dimethyl-2, 6-octadiene} {\longrightarrow }^{\text{H}}^{+} A _{\text{Major} \text{product}}$ [24-Jun-2022-Shift-1]

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📖 Explanation

Number of ${ sp}^2$ hybridised carbon atoms $=2$

Mathematics30 questions

Q61JEE Main 2022MCQ4MComplex Numbers

Let ${A}=\{z \in C: 1 \le |z-(1+i)| \le 2\}$ and ${B}=\{z \in {A}:|z-(1-i)|=1\}$ . Then, ${B}$ : [24-Jun-2022-Shift-1]

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📖 Explanation

Let, $z=x+i y$ Given, $1 \le |z-(1+i)| \le 2$ $\Rightarrow 1 \le |x+i y-1-i| \le 2$ $\Rightarrow 1 \le |(x-1)+i(y-1)| \le 2$ $\Rightarrow 1 \le \sqrt{(x-1)^{2+}(y-1)^2} \le 2$ It represent two concentric circle both have center at $(1,1)$ and radius 1 and 2 . Also given, $|z-(1-i)|=1$ $\Rightarrow |x+i y-1+i|=1$ $\Rightarrow |(x-1)+i(y+1)|=1$ $\Rightarrow \sqrt{(x-1)^{2+}(y+1)^2}=1$ This represent a circle with center at $(1,-1)$ and radius $=1$ .

Q62JEE Main 2022MCQ4MSequences and Series

The remainder when $3^{2022}$ is divided by 5 is : [24-Jun-2022-Shift-1]

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📖 Explanation

$3^{2022}$ $=(3^2)^{1011}$ $=(9)^{1011}$ $=(10-1)^{1011}$ $={ }^{1011} C_0(10)^{1011}+. . . . . .+{ }^{1011} C_{1010} \cdot (10)^{1-}{ }^{1011} C_{1011}$ $=10\$ [{ }^{1011} C_0(10)^{1010}+. . . . . .+{ }^{1011} C_{1010}]$-1=10 K-1 $[As$ 10[{ }^{1011} C_0 \cdot (10)^{1010}+. . . . . .+{ }^{1011} C_{1010}] $is multiple of 10]$ =10 K+5-5-1=10 K-5+5-1=5(2 K-1)+4\therefore $Unit digit$ =4$ when divided by 5.

Q63JEE Main 2022MCQ4MApplication of Derivatives

The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is: [24-Jun-2022-Shift-1]

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📖 Explanation

We know,Surface area of balloon $(s)=4 \pi r^2$ $\therefore \;\frac{d s}{d t}=\;\frac{d}{d t}(4 \pi r^2)$ $\Rightarrow \;\frac{d s}{d t}=4 \pi(2 r) \times \;\frac{d r}{d t}$ $\Rightarrow \;\frac{d s}{d t}=8 \pi r \times \;\frac{d r}{d t}$ Given that, surface area of balloon is increasing in constant rate. $\therefore \;\frac{d s}{d t}=$ constant $= {\text{k}}$ (Assume) $\Rightarrow k=8 \pi r \cdot \;\frac{d r}{d t}$ $\Rightarrow \int k d t=\int 8 \pi r d r$ $\Rightarrow k t=8 \pi \times \;\frac{r^2}{2}+c$ $\Rightarrow k t=4 \pi r^{2+}c . . . . .(1)$ Given at $t=0$ , radius $r=3$ So, $0=4 \pi(3^2)+c$ $\Rightarrow c=-36 \pi$ $\therefore$ Equation (1) becomes $k t=4 \pi r^{2-}36 \pi$ Also given, at $t=5$ , radius $r=7$ $\therefore k(5)=4 \pi(7)^{2-}36$ $\Rightarrow k=32 \pi$ $\therefore$ Equation (1) is $32 \pi t=4 \pi r^{2-}36 \pi$ Now at $t=9$ $\Rightarrow 32 \pi(9)=4 \pi r^{2-}36 \pi$ $\Rightarrow 8 \times 9=r^{2-}9$ $\Rightarrow r^2=81 \Rightarrow r=9$

Q64JEE Main 2022MCQ4MProbability

Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag $A$ is $\;\frac{6}{11}$ , then $n$ is equal to [24-Jun-2022-Shift-1]

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📖 Explanation

$P(1 R$ and $1 B)=P(A) \cdot P(\;\frac{1 R 1 B}{A})+P(B) \cdot P(\;\frac{1 R 1 B}{B})$ $=\;\frac{1}{2} \cdot \;\frac{{ }^3 C_1 \cdot { }^1 C_1}{{ }^6 C_2}+\;\frac{1}{2} \cdot \;\frac{{ }^2 C_1 \cdot { }^3 C_1}{{ }^{n+5} C_2}$ $P(\;\frac{1 R 1 B}{A})=\;\frac{\;\frac{1}{2} \cdot \frac{3}{15}}{\frac{1}{2} \cdot \;\frac{3}{15}+\;\frac{1}{2} \cdot \;\frac{6 \cdot 2}{(n+5)(n+4)}}=\;\frac{6}{11}$ $\Rightarrow \;\frac{\;\frac{1}{10}}{\frac{1}{10}+\;\frac{6}{(n+5)(n+4)}}=\;\frac{6}{11}$ $\Rightarrow \;\frac{11}{10}=\;\frac{6}{10}+\;\frac{36}{(n+5)(n+4)}$ $\Rightarrow \;\frac{5}{10 \times 36}=\;\frac{1}{(n+5)(n+4)}$ $\Rightarrow n^{2+}9 n-52=0$ $\Rightarrow n=4$ is only possible value

Q65JEE Main 2022MCQ4MCircles

Let $x^{2+}y^{2+}A x+B y+C=0$ be a circle passing through $(0,6)$ and touching the parabola $y=x^2$ at $(2 , 4)$ . Then $A+C$ is equal to [24-Jun-2022-Shift-1]

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📖 Explanation

For tangent to parabola $y=x^2$ at $(2,4)$ $.\;\frac{d y}{d x}|_{(2,4)}=4$ Equation of tangent is $y-4=4(x-2)$ $\Rightarrow 4 x-y-4=0$ Family of circle can be given by $(x-2)^{2+}(y-4)^{2+}\lambda (4 x-y-4)=0$ As it passes through $(0,6)$ $2^{2+}2^{2+}\lambda (-10)=0$ $\Rightarrow \lambda =\;\frac{4}{5}$ Equation of circle is $(x-2)^{2+}(y-4)^{2+}\;\frac{4}{5}(4 x-y-4)=0$ $\Rightarrow (x^{2+}y^{2-}4 x-8 y+20)+(\;\frac{16}{5} x-\;\frac{4}{5} y-\;\frac{16}{5})=0$ $A=-4+\;\frac{16}{5}, C=20-\;\frac{16}{5}$ $\;\text{ So, }\; A+C=16$

Q66JEE Main 2022MCQ4MMatrices and Determinants

The number of values of $\alpha$ for which the system of equations: $x+y+z=\alpha$ $\alpha x+2 \alpha y+3 z=-1$ $x+3 \alpha y+5 z=4$ is inconsistent, is [24-Jun-2022-Shift-1]

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📖 Explanation

$\Delta ={\begin{vmatrix}1 & 1 & 1 \\ \alpha & 2 \alpha & 3 \\ 1 & 3 \alpha & 5\end{vmatrix}}$ $=1(10 \alpha-9 \alpha)-1(5 \alpha-3)+1(3 \alpha^{2-}2 \alpha)$ $=\alpha-5 \alpha+3+3 \alpha^{2-}2 \alpha$ $=3 \alpha^{2-}6 \alpha+3$ For inconsistency $\Delta =0$ i.e. $\alpha=1$ Now check for $\alpha=1$ $x+y+z=1 \;\; . . .(i)$ $x+2 y+3 z=-1 \;\; . . .(i i)$ $\;\text{ By (ii) }\; \times 2-\;\text{ (i) }\; \times 1$ $x+3 y+5 z=-3$ so equations are inconsistent for $\alpha=1$

Q67JEE Main 2022MCQ4MQuadratic Equation and Inequalities

If the sum of the squares of the reciprocals of the roots $\alpha$ and $\beta$ of the equation $3 x^{2+}\lambda x-1=0$ is 15 , then $6(\alpha^{3+}\beta^3)^2$ is equal to: [24-Jun-2022-Shift-1]

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📖 Explanation

$3 x^{2+}\lambda x-1=0$ Given, two roots are $\alpha$ and $\beta$ . $\therefore$ Sum of roots $=\alpha+\beta=\;\frac{-\lambda }{3}$ And product of roots $=\alpha \beta=\;\frac{-1}{3}$ Given that,Sum of square of reciprocal of roots $\alpha$ and $\beta$ is 15 . $\therefore \;\frac{1}{\alpha^2}+\;\frac{1}{\beta^2}=15$ $\Rightarrow \;\frac{\alpha^{2+}\beta^2}{\alpha^2 \beta^2}=15$ $\Rightarrow \;\frac{(\alpha+\beta)^{2-}2 \alpha \beta}{(\alpha \beta)^2}=15$ $\Rightarrow \;\frac{\;\frac{\lambda ^2}{9}+2 \times \frac{1}{3}}{\frac{1}{9}}=15$ $\Rightarrow \;\frac{\;\frac{\lambda ^{2+}6}{9}}{\frac{1}{9}}=15$ $\Rightarrow \lambda ^{2+}6=15$ $\Rightarrow \lambda ^2=9$ Now, $6(\alpha^{3+}\beta^3)^2$ $=6\{(\alpha+\beta)(\alpha^{2+}\beta^{2-}\alpha \beta)\}^2$ $=6(\alpha+\beta)^2\$ [(alpha+beta)^{2-}2 alpha beta-alpha beta]^2=6(;\frac{-λ}{3})^2$[(;\frac{-\lambda }{3})^{2-}3 \cdot ;\frac{-1}{3}]^2=6 \times ;\frac{\lambda ^2}{9} \times [;\frac{\lambda ^2}{9}+1]=6 \times ;\frac{9}{9} \times [;\frac{9}{9}+1]^2=6 \times (2)^2=6 \times 4=24$

Q68JEE Main 2022MCQ4MInverse Trigonometric Functions

The set of all values of $k$ for which $(\tan ^{-1} x)^{3+}(\cot ^{-1} x)^{3-}k \pi^3, x \in R_u$ is the interval : [24-Jun-2022-Shift-1]

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📖 Explanation

$(\tan ^{-1} x)^{3+}(\cot ^{-1} x)^3=k \pi^3$ Let $f(t)=t^{3+}(\;\frac{\pi}{2}-t)^3$ Where $t=\tan ^{-1} x ; x \in (-\;\frac{\pi}{2}, \;\frac{\pi}{2})$ $=t^{3+}(\;\frac{\pi}{2})^{3-}\;\frac{3 \pi^2 t}{4}+\;\frac{3 \pi}{2} t^{2-}t^3$ $f(t)=\;\frac{3 \pi}{2} t^{2-}\;\frac{3 \pi^2}{4} \cdot t+\;\frac{\pi^3}{8}$ This is a quadratic equation of ${\text{t}}$ .Here, coefficient of $t^2$ term is $\;\frac{3 \pi}{2}$ which is $>0$ . $\therefore$ It is a upward parabola.Now, $f^{'}(t)=3 \pi t-\;\frac{3 \pi^2}{4}$ $f^{' '}(t)=3 \pi>0$ $\therefore 3 \pi t-\;\frac{3 \pi^2}{4}=0$ $\Rightarrow t=\;\frac{\pi}{4} \;\text{ (minima) }\;$ $\therefore$ vertex of graph at $\;\frac{\pi}{4}$ $\therefore$ Minimum value at $\;\frac{\pi}{4}$ and maximum value at $-\;\frac{\pi}{2}$ . $\therefore f(\;\frac{\pi}{4})=\;\frac{\pi^3}{64}+(\;\frac{\pi}{2}-\;\frac{\pi}{4})^3=\;\frac{\pi^3}{32}$ $f(-\;\frac{\pi}{2})=-\;\frac{\pi^3}{8}+\pi^3$ $=\;\frac{7 \pi^3}{8}$ $\therefore k \pi^3 \in [\;\frac{\pi^3}{32}, \;\frac{7 \pi^3}{8})$ $\Rightarrow k \in [\;\frac{1}{32}, \;\frac{7}{8})$

Q69JEE Main 2022MCQ4MMatrices and Determinants

Let $S=\{\sqrt{n}: 1 \le n \le 50$ and $n$ is odd $\}$ .Let ${\text{a}} \in S$ and $A=\begin{bmatrix}1 & 0 & a \\ -1 & 1 & 0 \\ -a & 0 & 1\end{bmatrix}$ .If $sum_{ a \in S} {\det}({\text{adj}} A)=100 \lambda$ , then $\lambda$ is equal to : ion: [24-Jun-2022-Shift-1]

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📖 Explanation

Given, $A=\begin{bmatrix}1 & 0 & a \\ -1 & 1 & 0 \\ -a & 0 & 1\end{bmatrix}_{3 \times 3}$ ${\text{S}}=\{\sqrt{n}: 1 \le {\text{n}} \le 50$ and ${\text{n}}$ is odd $\}$ $\therefore S=\{1, \sqrt{3}, \sqrt{5}, \sqrt{7}, . . ., \sqrt{49}\}$ We know, $|{\text{adj}} A|=|A|^{n-1}$ Here, ${\text{n}}=$ order of matrix.Here, $n=3$ $\therefore |a d j A|=|A|^{3-1}=|A|^2$ Now, $|A|=\begin{vmatrix}1 & 0 & a \\ -1 & 1 & 0 \\ -a & 0 & 1\end{vmatrix}$ $=1(1-0)-0+a(0-(-a))$ $=a^{2+}1$ $\therefore |a d j A|=|A|^2=(a^{2+}1)^2$ Now, $sum_{a \in S} {\det}(a d j A)$ $=sum_{a \in S}(a^{2+}1)^2$ $=(1^{2+}1)^{2+}((\sqrt{3})^{2+}1)^{2+}((\sqrt{5})^{2+}1)^{2+}. . .+((\sqrt{49})^{2+}1)^2$ $=(1^{2+}1)^{2+}(3+1)^{2+}(5+1)^{2+}. . .+(49+1)^2$ $=2^{2+}4^{2+}6^{2+}. . .+50^2$ $=2^2(1^{2+}2^{2+}3^{2+}. . . .+25^2)$ $=4 . \;\frac{25.26 .51}{6}=100.221$ $\therefore 100 K=100.221$ $\Rightarrow K=221$

Q70JEE Main 2022MCQ4MApplication of Derivatives

For the function $f(x)=4 \log _e(x-1)-2 x^{2+}4 x+5, x>1$ , which one of the following is NOT correct? [24-Jun-2022-Shift-1]

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📖 Explanation

$f(x)=4 \log _e(x-1)-2 x^{2+}4 x+5, x>1$ $f^{'}(x)=\;\frac{4}{x-1}-4(x-1)$ For $1< x< 2 \Rightarrow f^{'}(x)>0$ For $x>2 \Rightarrow f^{'}(x)< 0$ (option $A$ is correct) $f(x)=-1$ has two solution (option $B$ is correct) $f(e)>0$ $f(e+1)< 0$ $f(e) \cdot f(e+1)< 0($ option $D$ is correct) $f^{'}(e)-f^{' '}(2)=\;\frac{4}{e-1}-4(e-1)+8>0$

Q71JEE Main 2022MCQ4MApplication of Derivatives

If the tangent at the point $( {x}_1, {y}_1)$ on the curve $y-x^{3+}3 x^{2+}5$ passes through the origin, then $( {x}_1.$ , $y_1$ ) does NOT lie on the curve: [24-Jun-2022-Shift-1]

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📖 Explanation

Given curve, $y=x^{3+}3 x^{2+}5$ Slope of tangent, $\;\frac{d y}{d x}=3 x^{2+}6 x$ Slope of line joined by $(x_1, y_1)$ and $(0,0)$ is $=\;\frac{y_0-0}{x_1-0}$ $\therefore \;\frac{y_1}{x_1}=3 x_1^{2+}6 x_1$ $\Rightarrow y_1=3 x_1^{3+}6 x_1^2 . . . . . . \;\text{ (1) }\;$ Point $(x_1, y_1)$ is on the line, $\therefore y_1=x_1^{3+}3 x_1^{2+}5 . . . .(2)$ $\therefore 3 x_1^{3+}6 x_1^2=x_1^{3+}3 x_1^{2+}6$ $\Rightarrow 2 x_1^{3+}3 x_1^{2-}5=0$ $x=1$ satisfy the equation $\;\text{ Now, }\; y_1=x_1^{3+}3 x_1^{2+}5$ $=1+3+5$ $=9$ $\therefore ( {\text{x}}_1, {\text{y}}_1)=(1,9)$ Option D does not satisfy point $(1,9)$ .

Q72JEE Main 2022MCQ4MApplication of Derivatives

The sum of absolute maximum and absolute minimum values of the function $f(x)- 2 x^{2+}3 x-2|+\sin x \cos x$ in the interval $[0,1]$ is : [24-Jun-2022-Shift-1]

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📖 Explanation

$f(x)=|(2 x-1)(x+2)|+\;\frac{\sin 2 x}{2}$ $0 \le x< \;\frac{1}{2} \;\; f(x)=(1-2 x)(x+2)+\;\frac{\sin 2 x}{2}$ $f^{'}(x)=-4 x-3+\cos 2 x < 0$ For $x \ge \;\frac{1}{2}: \;\; f^{'}(x)=4 x+3+\cos 2 x>0$ So, minima occurs at $x=\;\frac{1}{2}$ $.f(x)|_{\min } \;\; =|2(\;\frac{1}{2})^{2+}\;\frac{3}{2}-2|+\sin (\;\frac{1}{2}) \cdot \cos (\;\frac{1}{2})$ $\;\; =\;\frac{1}{2} \sin 1$ So, maxima is possible at $x=0$ or $x=1$ Now checking for $x=0$ and $x=1$ , we can see it attains its maximum value at $x=1$ ${c} .f(x)|_{\max }=|2+3-2|+\;\frac{\sin 2}{2}$ $=3+\;\frac{1}{2} \sin 2$ Sum of absolute maximum and minimum value $=3+\;\frac{1}{2}(\sin 1+\sin 2)$

Q73JEE Main 2022MCQ4MSequences and Series

If $\{{\text{a}}_i\}{}_{i=1}^{\text{n}}$ , where ${\text{n}}$ is an even integer, is an arithmetic progression with common difference1 , and $sum_{i=1}^{ {\text{n}}} {\text{a}}_i=192, sum_{i=1}^{ {\text{n}} / 2} {\text{a}}_{2 i}=120$ , then ${\text{n}}$ is equal to : [24-Jun-2022-Shift-1]

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📖 Explanation

$sum_{i=1}^n a_i=192$ $\Rightarrow a_1+a_2+a_3+. . . . . .+a_n=192$ $\Rightarrow \;\frac{n}{2}[a_1+a_n]=192$ $\Rightarrow a_1+a_n=\;\frac{384}{n} . . . . . . \;\text{ (1) }\;$ $\;\text{ Now, }\; sum_{i=1}^{\;\frac{n}{2}} a_{2 i}=120$ $\Rightarrow a_2+a_4+a_6+. . . . .+a_n=120$ Here total $\;\frac{n}{2}$ terms present. $\therefore \;\frac{\;\frac{n}{2}}{2}[a_2+a_n]=120$ $\Rightarrow \;\frac{n}{4}[a_1+1+a_n]=120$ $\Rightarrow a_1+a_n+1=\;\frac{480}{n} . . . . .$ Subtracting (1) from (2), we get $1=\;\frac{480}{n}-\;\frac{384}{n}$ $\Rightarrow 1=\;\frac{96}{n}$ $\Rightarrow {\text{n}}=96$

Q74JEE Main 2022MCQ4MDifferential Equations

If $x=x(y)$ is the solution of the differential equation $y \;\frac{d x}{d y}=2 x+y^3(y+1) e^y, x(1 )=0 ;$ then ${x}( {e})$ is equal to: [24-Jun-2022-Shift-1]

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📖 Explanation

$\;\frac{d x}{d y}-\;\frac{2 x}{y}=y^2(y+1) e^y$ $\;\text{ If }\;=e^{\int-\;\frac{2}{y} d y}=e^{-2 \ln y}=\;\frac{1}{y^2}$ Solution is given by $\;x \cdot \;\frac{1}{y^2}=\int y^2(y+1) e^y \cdot \;\frac{1}{y^2} d y$ $\Rightarrow \;\frac{x}{y^2}=\int(y+1) e^y d y$ $\Rightarrow \;\frac{x}{y^2}=y e^y+c$ $\Rightarrow x=y^2(y e^y+c) \;\text{ at, }\; y=1, x=0$ $\Rightarrow 0=1(1 \cdot e^{1+}c) \Rightarrow c=-e \;\text{ at }\; y=e$ $x \;\; =e^2(e \cdot e^e-e)$

Q75JEE Main 2022MCQ4MHyperbola

Let $\lambda x-2 y=\mu$ be a tangent to the hyperbola $a^2 x^{2-}y^2=b^2$ . Then $(\;\frac{\lambda }{a})^{2-}(\;\frac{\mu }{b})^2$ is equal to: [24-Jun-2022-Shift-1]

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📖 Explanation

$\;\frac{x^2}{(\;\frac{b^2}{a^2})}-\;\frac{y^2}{b^2}=1$ Tangent in slope form $\Rightarrow y=m x \pm \sqrt{\;\frac{b^2}{a^2} m^{2-}b^2}$ i.e., same as $y=\;\frac{\lambda x}{2}-\;\frac{\mu }{2}$ Comparing coefficients, $m=\;\frac{\lambda }{2}, \;\frac{b^2}{a^2} m^{2-}b^2=\;\frac{\mu ^2}{4}$ Eliminating $m, \;\frac{b^2}{a^2} \cdot \;\frac{\lambda ^2}{4}-b^2=\;\frac{\mu ^2}{4}$ $\Rightarrow \;\frac{\lambda ^2}{a^2}-\;\frac{\mu ^2}{b^2}=4$

Q76JEE Main 2022MCQ4MVector Algebra

Let ${ {\text{a}}^{\^}, { {\text{b}}^{\^}$ be unit vectors. If ${ \vec{\text{c}}$ be a vector such that the angle between ${ {\text{a}}^{\^}$ and ${ \vec{\text{c}}$ is $\;\frac{\pi}{12}$ , and ${ {\text{b}}^{\^}={ \vec{\text{c}}+2({ \vec{\text{c}} \times { {\text{a}}^{\^})$ , then $|6 { \vec{\text{c}}|^2$ is equal to : [24-Jun-2022-Shift-1]

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📖 Explanation

$\because \;\; \hat{b}=\vec{c}+2(\vec{c} \times \hat{a})$ $\Rightarrow \;\; \hat{b} \cdot \vec{c}=|\vec{c}|^2$ $\therefore \;\; \hat{b}-\vec{c}=2(\vec{c} \times \vec{a})$ $\Rightarrow \;\; |\hat{b}|^{2+}|\vec{c}|^{2-}2 \hat{b} \cdot \vec{c}=4|\vec{c}|^2|\vec{a}|^2 \sin ^2 \;\frac{\pi}{12}$ $\Rightarrow \;\; 1+|\vec{c}|^{2-}2|c|^2=4|\vec{c}|^2(\;\frac{\sqrt{3}-1}{2 \sqrt{2}})^2$ $\Rightarrow \;\; 1=|\vec{c}|^2(3-\sqrt{3})$ $\Rightarrow \;\; 36|\vec{c}|^2=\;\frac{36}{3-\sqrt{3}}=6(3+\sqrt{3})$

Q77JEE Main 2022MCQ4MProbability

If a random variable $X$ follows the Binomial distribution $B(33, p)$ such that $3 P(X=0)=P(X=1)$ , then the value of $\;\frac{P(X=15)}{P(X=18)}-\;\frac{P(X=16)}{P(X=17)}$ is equal to: [24-Jun-2022-Shift-1]

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📖 Explanation

$3 P(X=0)=P(X=1)$ $3 \cdot { }^n C_0 P^0(1-P)^n={ }^n C_1 P^1(1-P)^{n-1}$ $\;\frac{3}{n}=\;\frac{P}{1-P} \Rightarrow \;\frac{1}{11}=\;\frac{P}{1-P}$ $\Rightarrow 1-P=11 P$ $\Rightarrow P=\;\frac{1}{12}$ $\;\frac{P(X=15)}{P(X=18)}-\;\frac{P(X=16)}{P(X=17)}$ $\Rightarrow \;\frac{{ }^{33} C_{15} P^{15}(1-P)^{18}}{{ }^{33} C_{18} P^{18}(1-P)^{15}}-\;\frac{{ }^{33} C_{16} P^{16}(1-P)^{17}}{{ }^{33} C_{17} P^{17}(1-P)^{16}}$ $\Rightarrow (\;\frac{1-P}{P})^{3-}(\;\frac{1-P}{P})$ $\Rightarrow \;\; 11^{3-}11=1320$

Q78JEE Main 2022MCQ4MFunctions

The domain of the function $f(x)=\;\frac{\cos ^{-1}(\frac{x^{2-}5 x+6}{x^{2-}9})}{\log _e(x^{2-}3 x+2)} \;\text{ is : }\;$ [24-Jun-2022-Shift-1]

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📖 Explanation

$-1 \le \;\frac{x^{2-}5 x+6}{x^{2-}9} \le 1$ and $x^{2-}3 x+2>0, \ne 1$ $\;\frac{(x-3)(2 x+1)}{x^{2-}9} \ge 0 \mid \;\frac{5(x-3)}{x^{2-}9} \ge 0$ The solution to this inequality is $x \in [\;\frac{-1}{2}, \infty )-\{3\}$ for $x^{2-}3 x+2>0$ and $\ne 1$ $x \in (-\infty , 1) ∪(2, \infty )-\{\;\frac{3-\sqrt{5}}{2}, \;\frac{3+\sqrt{5}}{2}\}$ Combining the two solution sets (taking intersection) $x \in [-\;\frac{1}{2}, 1) ∪(2, \infty )-\{\;\frac{3-\sqrt{5}}{2}, \;\frac{3+\sqrt{5}}{2}\}$

Q79JEE Main 2022MCQ4MTrigonometric Ratios and Identities

Let $S=\{\theta \in [-\pi, \pi]-\{\pm \;\frac{\pi}{2}\}: \sin \theta \tan \theta+\tan \theta=\sin 2 \theta\}$ .If $T=sum_{\theta \in S} \cos 2 \theta$ , then ${T}+ {n}( {S})$ is equal to: [24-Jun-2022-Shift-1]

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📖 Explanation

$\tan \theta(\sin \theta+1)-\sin 2 \theta=0$ $\tan \theta(\sin \theta+1-2 \cos ^2 \theta)=0$ $\Rightarrow \tan \theta=0 \;\text{ or }\; 2 \sin ^2 \theta+\sin \theta-1=0$ $\Rightarrow (2 \sin \theta+1)(\sin \theta-1)=0$ $\Rightarrow \sin \theta=\;\frac{-1}{2} \;\text{ or }\; 1$ But, $\sin \theta=1$ not possible $\theta=0, \pi,-\pi,-\;\frac{\pi}{6}, \;\frac{-5 \pi}{6}$ ${\text{n}}( {\text{S}})=5$ $T=\sum \cos 2 \theta=\cos 0^{\circ}+\cos 2 \pi+\cos (-2 \pi)+\cos (-\;\frac{5 \pi}{3})+\cos (-\;\frac{\pi}{3})$ $=4$

Q80JEE Main 2022MCQ4MSets and Relations

The number of choices for $\Delta \in \{∧, ∨, \Rightarrow , ⇔\}$ , such that $(p \Delta q) \Rightarrow ((p \Delta ∼ q) ∨((∼ p) \Delta q))$ is a tautology, is : [24-Jun-2022-Shift-1]

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📖 Explanation

Then $(p \Delta ∼ q) ∨(∼ p \Delta q)$ becomes $(p ∨ ∼ q) ∨(∼ p ∨ q)$ which is always true, so $x$ becomes a tautology.Case-IIWhen $\Delta$ is same as $∧$ Then $(p ∧ q) \Rightarrow (p ∧ ∼ q) ∨(∼ p ∧ q)$ If $p ∧ q$ is $T$ , then $(p ∧ ∼ q) ∨(∼ p ∧ q)$ is $F$ so $x$ cannot be a tautology.Case-IIIWhen $\Delta$ is same as $\Rightarrow$ Then $(p \Rightarrow ∼ q) ∨(∼ p \Rightarrow q)$ is same at $(∼ p ∨ ∼ q) ∨(p ∨ q)$ , which is always true, so $x$ becomes a tautology.Case-IVWhen $\Delta$ is same as $⇔$ Then $(p ⇔ q) \Rightarrow (p ⇔ ∼ q) ∨(∼ p ⇔ q)$ $p ⇔ q$ is true when $p$ and $q$ have same truth values, then $p ⇔ ∼ q$ and $∼ p ⇔ q$ both are false. Hence $x$ cannot be a tautology.

Q81JEE Main 2022NAT4MPermutations and Combinations

The number of one-one functions $f:\{a, b, c, d\} \to \{0,1,2, . . . . . ., 10\}$ such that $2 f(a)-f(b)+3 f(c)+f(d)=0$ is [24-Jun-2022-Shift-1]

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📖 Explanation

Given one-one function $f:\{a, b, c, d\} \to \{0,1,2, . . . 10\}$ $\;\text{ and }\; 2 f(a)-f(b)+3 f(c)+f(d)=0$ $\Rightarrow 3 f(c)+2 f(a)+f(d)=f(b)$ Case I:(1) Now let $f(c)=0$ and $f(a)=1$ then $3 \times 0+2 \times 1+f(d)=f(b)$ $\Rightarrow 2+f(d)=f(b)$ Now possible value of $f(d)=2,3,4,5,6,7$ , and 8 . ${\text{f}}( {\text{d}})$ can't be 9 and 10 as if $f(d)=9$ or 10 then $f(b)=2+9=11$ or $f(b)=2+10=12$ , which is not possible as here any function's maximum value can be 10 . $\therefore$ Total possible functions when $f(c)=0$ and $f(a)=1$ are $=7$ (2) When $f(c)=0$ and $f(a)=2$ then $3 \times 0+2 \times 2+f(d)=f(b)$ $\Rightarrow 4+f(d)=f(b)$ $\therefore$ possible value of $f(d)=1,3,4,5,6$ $\therefore$ Total possible functions in this case $=5$ (3) When $f(c)=0$ and $f(a)=3$ then $3 \times 0+2 \times 3+f(d)=f(b)$ $\Rightarrow 6+f(d)=f(b)$ $\therefore$ Possible value of $f(d)=1,2,4$ $\therefore$ Total possible functions in this case $=3$ (4) When $f(c)=0$ and $f(a)=4$ then $3 \times 0+2 \times 4+f(d)=f(b)$ $\Rightarrow 8+f(d)=f(b)$ $\therefore$ Possible value of $f(d)=1,2$ $\therefore$ Total possible functions in this case $=2$ (5) When $f(c)=0$ and $f(a)=5$ then $3 \times 0+2 \times 5+f(d)=f(b)$ $\Rightarrow 10+f(d)=f(b)$ Possible value of $f(d)$ can be 0 but $f(c)$ is already zero. So, no value to $f(d)$ can satisfy. $\therefore$ No function is possible in this case. $\therefore$ Total possible functions when $f(c)=0$ and $f(a)=1,2,3$ and 4 are $=7+5+3+2=17$ Case II:(1) When $f(c)=1$ and $f(a)=0$ then $3 \times 1+2 \times 0+f(d)=f(b)$ $\Rightarrow 3+f(d)=f(b)$ $\therefore$ Possible value of $f(d)=2,3,4,5,6,7$ $\therefore$ Total possible functions in this case $=6$ (2) When $f(c)=1$ and $f(a)=2$ then $3 \times 1+2 \times 2+f(d)=f(b)$ $\Rightarrow 7+f(d)=f(b)$ $\therefore$ Possible value of $f(d)=0,3$ $\therefore$ Total possible functions in this case $=2$ (3) When $f(c)=1$ and $f(a)=3$ then $3 \times 1+2 \times 3+f(d)=f(b)$ $\Rightarrow 9+f(d)=f(b)$ $\therefore$ Possible value of $f(d)=0$ $\therefore$ Total possible functions in this case $=1$ $\therefore$ Total possible functions when $f(c)=1$ and $f(a)=0,2$ and 3 are $=6+2+1=9$ Case III:(1) When $f(c)=2$ and $f(a)=0$ then $3 \times 2+2 \times 0+f(d)=f(b)$ $\Rightarrow 6+f(d)=f(b)$ $\therefore$ Possible values of $f(d)=1,3,4$ $\therefore$ Total possible functions in this case $=3$ (2) When $f(c)=2$ and $f(a)=1$ then, $3 \times 2+2 \times 1+f(d)=f(b)$ $\Rightarrow 8+f(d)=f(b)$ $\therefore$ Possible values of $f(d)=0$ $\therefore$ Total possible function in this case $=1$ $\therefore$ Total possible functions when $f(c)=2$ and $f(a)=0,1$ are $=3+1=4$ Case IV:(1) When $f(c)=3$ and $f(a)=0$ then $3 \times 3+2 \times 0+f(d)=f(b)$ $\Rightarrow 9+f(d)=f(b)$ $\therefore$ Possible values of $f(d)=1$ $\therefore$ Total one-one functions from four cases $=17+9+4+1=31$

Q82JEE Main 2022NAT4MPermutations and Combinations

In an examination, there are 5 multiple choice questions with 3 choices, out of which exactly one is correct. There are 3 marks for each correct answer, $-2$ marks for each wrong answer and 0 mark if the question is not attempted. Then, the number of ways a student appearing in the examination gets 5 marks is___ [24-Jun-2022-Shift-1]

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📖 Explanation

Let student marks $x$ correct answers and $y$ incorrect. So $3 x-2 y=5$ and $x+y \le 5$ where $x, y \in {\text{W}}$ Only possible solution is $(x, y)=(3,2)$ Students can mark correct answers by only one choice but for an incorrect answer, there are two choices. So total number of ways of scoring 5 marks $={ }^5 C_3(1)^3 \cdot (2)^2=40$

Q83JEE Main 2022NAT4MApplication of Derivatives

Let $A(\;\frac{3}{\sqrt{a}}, \sqrt{a}), a>0$ , be a fixed point in the xy-plane. The image of ${A}$ in ${y}$ -axis be ${B}$ and the image of $B$ in $x$ -axis be $C$ . If $D(3 \cos \theta, a \sin \theta)$ is a point in the fourth quadrant such that the maximum area of $\triangle A C D$ is 12 square units, then a is equal to____ [24-Jun-2022-Shift-1]

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📖 Explanation

Clearly $B$ is $(-\;\frac{3}{\sqrt{a}},+\sqrt{a})$ and $C$ is $(-\;\frac{3}{\sqrt{a}},-\sqrt{a})$ Area of $\triangle A C D=\;\frac{1}{2}\begin{vmatrix}\\ \\ \frac{3}{\sqrt{a}} & \sqrt{a} & 1 \\ -\\ \\ \frac{3}{\sqrt{a}} & -\sqrt{a} & 1 \\ 3 \cos \theta & a \sin \theta & 1\end{vmatrix}$ $\Rightarrow \;\; \Delta =\begin{vmatrix}0 & 0 & 1 \\ -\\ \\ \frac{3}{\sqrt{a}} & -\sqrt{a} & 1 \\ 3 \cos \theta & a \sin \theta & 1\end{vmatrix}$ $\Rightarrow \;\; \Delta =|3 \sqrt{a} \sin \theta+3 \sqrt{a} \cos \theta|=3 \sqrt{a}|\sin \theta+\cos \theta|$ $\Rightarrow \;\; \Delta _{\max }=3 \sqrt{a} \cdot \sqrt{2}=12 \Rightarrow a=(2 \sqrt{2})^2=8$

Q84JEE Main 2022NAT4MThree Dimensional Geometry

Let a line having direction ratios $1,-4,2$ intersect the lines $\;\frac{x-7}{3}=\;\frac{y-1}{-1}=\;\frac{z+2}{1}$ and $\;\frac{x}{2}=\;\frac{y-7}{3}=\;\frac{z}{1}$ at the points $A$ and $B$ . Then $(A B)^2$ is equal to__ [24-Jun-2022-Shift-1]

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📖 Explanation

Let $A(3 \lambda +7,-\lambda +1, \lambda -2)$ and $B(2 \mu , 3 \mu +7, \mu )$ So, DR's of $A B \propto 3 \lambda -2 \mu +7,-(\lambda +3 \mu +6), \lambda -\mu -2$ Clearly $\;\frac{3 \lambda -2 \mu +7}{1}=\;\frac{\lambda +3 \mu +6}{4}=\;\frac{\lambda -\mu -2}{2}$ $\Rightarrow 5 \lambda -3 \mu =-16$ And $\lambda -5 \mu =10$ From (i) and (ii) we get $\lambda =-5, \mu =-3$ So, $A$ is $(-8,6,-7)$ and $B$ is $(-6,-2,-3)$ $A B=\sqrt{4+64+16} \Rightarrow (A B)^2=84$

Q85JEE Main 2022NAT4MLimits, Continuity and Differentiability

The number of points where the function $f(x)=\begin{cases}\text{|} 2 x^{2-}3 x-7\text{|} & \ \\ \text{ if }\ \\ x \le -1 \\ {[4 x^{2-}1]\$ } & \ \ \text{ if }\ \ -1< x< 1 \ \text{|}x+1\text{|}+\text{|}x-2\text{|} & \ \ \text{ if }\ \ x ≥ 1\end{cases} $[t] denotes the greatest integer$ ≤ {t}$ , is discontinuous is [24-Jun-2022-Shift-1]

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📖 Explanation

$\because f(-1)=2$ and $f(1)=3$ For $x \in (-1,1),(4 x^{2-}1) \in [-1,3)$ hence $f(x)$ will be discontinuous at $x=1$ and alsowhenever $4 x^{2-}1=0,1$ or 2 $\Rightarrow x=\pm \;\frac{1}{2}, \pm \;\frac{1}{\sqrt{2}}$ and $\pm \;\frac{\sqrt{3}}{2}$ So there are total 7 points of discontinuity.

Q86JEE Main 2022NAT4MDefinite Integrals

Let $f(\theta)=\sin \theta+int_{-\pi / 2}^{\pi / 2}(\sin \theta+ {\text{t}} \cos \theta) f( {\text{t}}) {\text{dt}}$ . Then the value of $|int_{0}^{\pi / 2} f(\theta) {\text{d}} \theta|$ is [24-Jun-2022-Shift-1]

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📖 Explanation

$f(\theta)=\sin \theta(1+int_{-\pi / 2}^{\pi / 2} f(t) d t)+\cos \theta(int_{-\pi / 2}^{\pi / 2} t f(t) d t)$ Clearly $f(\theta)=a \sin \theta+b \cos \theta$ Where $a=1+int_{-\pi / 2}^{\pi / 2}(a \sin t+b \cos t) d t \Rightarrow a=1+2 b \;\; . . .(i)$ and $b=int_{-\pi / 2}^{\pi / 2}(a t \sin t+b t \cos t) d t \Rightarrow b=2 a \;\; . . .(i i)$ from (i) and (ii) we get $a=-\;\frac{1}{3}$ and $b=-\;\frac{2}{3}$ So $f(\theta)=-\;\frac{1}{3}(\sin \theta+2 \cos \theta)$ $\Rightarrow |int_0^{\pi / 2} f(\theta) d \theta|=\;\frac{1}{3}(1+2 \times 1)=1$

Q87JEE Main 2022NAT4MDefinite Integrals

Let ${\text{Max}}_{0 \le x \le 2}\{\;\frac{9-x^2}{5-x}\}=\alpha$ and ${\text{Min}}_{0 \le x \le 2}\{\;\frac{9-x^2}{5-x}\}=\beta$ .If $int_{\beta-\;\frac{8}{3}}^{2 \alpha-1} {\text{Max}}\{\;\frac{9-x^2}{5-x}, x\} {\text{d}} x=\alpha_1+\alpha_2 \log _{ {\text{e}}}(\;\frac{8}{15})$ then $\alpha_1+\alpha_2$ is equal to [24-Jun-2022-Shift-1]

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📖 Explanation

Let $f(x)=\;\frac{x^{2-}9}{x-5} \Rightarrow f^{'}(x)=\;\frac{(x-1)(x-9)}{(x-5)^2}$ So, $\alpha=f(1)=2$ and $\beta=\min (f(0), f(2))=\;\frac{5}{3}$ Now, $int_{-1}^3 \max \{\;\frac{x^{2-}9}{x-5}, x\} d x=int_{-1}^{9 / 5} \;\frac{x^{2-}9}{x-5} d x+int_{9 / 5}^3 x d x$ $=int_{-1}^{9 / 5}(x+5+\;\frac{16}{x-5}) d x+.\;\frac{x^2}{2}|_{9 / 5} ^3$ $=\;\frac{28}{25}+14+16 \ln (\;\frac{8}{15})+\;\frac{72}{25}=18+16 \ln (\;\frac{8}{15})$ Clearly $\alpha_1=18$ and $\alpha_2=16$ , so $\alpha_1+\alpha_2=34$ .

Q88JEE Main 2022NAT4MParabola

If two tangents drawn from a point $(\alpha, \beta)$ lying on the ellipse $25 x^{2+}4 y^2=1$ to the parabola $y^2=4 x$ are such that the slope of one tangent is four times the other, then the value of $(10 \alpha+5)^{2+}(16 \beta^{2+}50)^2$ equals [24-Jun-2022-Shift-1]

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📖 Explanation

$\because (\alpha, \beta)$ lies on the given ellipse, $25 \alpha^{2+}4 \beta^2=1 \;\; . . .(i)$ Tangent to the parabola, $y=m x+\;\frac{1}{m}$ passes through $(\alpha, \beta)$ . So, $\alpha m^{2-}\beta m+1=0$ has roots $m_1$ and $4 m_1$ $m_1+4 m_1=\;\frac{\beta}{\alpha}$ and $m_1 \cdot 4 m_1=\;\frac{1}{\alpha}$ Gives that $4 \beta^2=25 \alpha \;\; . . .(i i)$ from (i) and (ii) $25(\alpha^{2+}\alpha)=1 \;\; . . .(i i i)$ Now, $(10 \alpha+5)^{2+}(16 \beta^{2+}50)^2$ $=25(2 \alpha+1)^{2+}2500(2 \alpha+1)^2$ $=2525(4 \alpha^{2+}4 \alpha+1)$ from equation (iii) $=2525(\;\frac{4}{25}+1)$ $=2929$

Q89JEE Main 2022NAT4MArea Under The Curves

Let $S$ be the region bounded by the curves $y=x^3$ and $y^2=x$ . The curve $y=2|x|$ divides $S$ into two regions of areas ${\text{R}}_1, {\text{R}}_2$ . If $\max \{ {\text{R}}_1, {\text{R}}_2\}= {\text{R}}_2$ , then $\;\frac{R_2}{R_1}$ is equal to [24-Jun-2022-Shift-1]

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📖 Explanation

$C_1: y=x^3$ $C_2: y^2=x$ and $C_3=y=2|x|$ $C_1$ and $C_2$ intersect at $(1,1)$ $C_2$ and $C_3$ intersect at $(\;\frac{1}{4}, \;\frac{1}{2})$ Clearly $R_1=int_0^{1 / 4}(\sqrt{x}-2 x) d x=\;\frac{2}{3}(\;\frac{1}{8})-\;\frac{1}{16}=\;\frac{1}{48}$ and $R_1+R_2=int_0^1(\sqrt{x}-x^3) d x=\;\frac{2}{3}-\;\frac{1}{4}=\;\frac{5}{12}$ So, $\;\frac{R_1+R_2}{R_1}=\;\frac{5 / 12}{1 / 48} \Rightarrow 1+\;\frac{R_2}{R_1}=20$ $\Rightarrow \;\frac{R_2}{R_1}=19$

Q90JEE Main 2022NAT4MVector Algebra

If the shortest distance between the lines $\vec{r}=(-\hat{i}+3 \hat{k})+\lambda (\hat{i}-a \hat{j})$ and $\vec{r}=(-\hat{j}+2 \hat{k})+\mu (\hat{i}-\hat{j}+\hat{k})$ is $\sqrt{\;\frac{2}{3}}$ , then the integral value of a is equal to [24-Jun-2022-Shift-1]

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📖 Explanation

$\vec{b}_1 \times \vec{b}_2=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & -a & 0 \\ 1 & -1 & 1\end{vmatrix}=-a \hat{i}-\hat{j}+(a-1) \hat{k}$ $\vec{a}_1-\vec{a}_2=-\hat{i}+\hat{j}+\hat{k}$ Shortest distance $=|\;\frac{(\vec{a}_1-\vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|}|$ $\Rightarrow \;\; \sqrt{\;\frac{2}{3}}=\;\frac{2(a-1)}{\sqrt{a^{2+}1+(a-1)^2}}$ $\Rightarrow 6(a^{2-}2 a+1)=2 a^{2-}2 a+2$ $\Rightarrow \;\; (a-2)(2 a-1)=0 \Rightarrow a=2$ because $a \in z$ .

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