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26 Jul 2022 Shift 1

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Physics30 questions

Q1JEE Main 2022MCQ4MLaws of Motion

Three masses $M=100 {\text{kg}}, {\text{m}}_1=10 {\text{kg}}$ and ${\text{m}}_2=20 {\text{kg}}$ are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force ${\text{F}}$ is applied on the system so that the mass ${\text{m}}_2$ moves upward with an acceleration of $2 {\text{ms}}^{-2}$ . The value of ${\text{F}}$ is : $(.$ Take $. {\text{g}}=10 {\text{ms}}^{-2})$ [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

In frame of block of mass M moving with acceleration a $m_1 a-T=2 m_1 \Rightarrow 10 a-T=20$ $T-m_2 g=m_2 2 \Rightarrow T-200=40 \Rightarrow T=240 . . . . . .$ (ii) $\Rightarrow$ From equation 1 and $210 {\text{a}}=260$ or ${\text{a}}=26 {\text{m}} / {\text{s}}^2$ for block $F(M+m_2) a=120 \times 26$ $=3120 N$

Explanation Figure 2

Explanation Figure 3

Q2JEE Main 2022MCQ4MElectromagnetic Waves

A radio can tune to any station in $6 {\text{MHz}}$ to $10 {\text{MHz}}$ band. The value of corresponding wavelength bandwidth will be: [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$v_1=6 \times 10^6 {\text{Hz}}$ $\Rightarrow \lambda _1=\;\frac{3 \times 10^8}{6 \times 10^6}=50 {\text{m}}$ $v_2=10 \times 10^6 {\text{Hz}}$ $\Rightarrow \lambda _2=\;\frac{3 \times 10^8}{10 \times 10^6}=30 {\text{m}}$ $\Rightarrow$ Wavelength band with $=|\lambda _1-\lambda _2|=20 {\text{m}}$

Q3JEE Main 2022MCQ4MAtoms and Nuclei

The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minute. 10 minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is : (Take $\log _{10} 1.88=0.274)$ [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$A_0=4250$ $A=2250=A_0 e^{-\lambda t}$ $\Rightarrow \;\frac{2250}{4250}=e^{-\lambda t}$ $\Rightarrow \lambda (10)=\ln (\;\frac{4250}{2250})$ $\lambda (10)=0.636$ $\lambda =0.063$

Q4JEE Main 2022MCQ4MDual Nature of Matter and Radiation

A parallel beam of light of wavelength $900 {\text{nm}}$ and intensity $100 {\text{Wm}}^{-2}$ is incident on a surface perpendicular to the beam. The number of photons crossing $1 {\text{cm}}^2$ area perpendicular to the beam in one second is : [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\lambda =900 {\text{nm}}$ ${\text{I}}=100 {\text{W}} / {\text{m}}^2$ $A=10^{-4}$ $\Rightarrow {\text{P}}=10^{-2} {\text{W}}$ $\Rightarrow$ Number of photons incident per second $=\;\frac{10^{-2} \lambda }{h c}$ $=\;\frac{9 \times 10^{-11} \times 10^2}{6.63 \times 10^{-34} \times 3 \times 10^8} ≃ 4.5 \times 10^{16}$

Q5JEE Main 2022MCQ4MWave Optics

In Young's double slit experiment, the fringe width is $12 {\text{mm}}$ . If the entire arrangement is placed in water of refractive index $\;\frac{4}{3}$ , then the fringe width becomes (in mm): [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$B=12 \times 10^{-3}$ $\beta^{'}=\;\frac{\beta}{\mu }=\;\frac{12 \times 10^{-3}}{\frac{4}{3}}$ $=9 \times 10^{-3} {\text{m}}=9 {\text{mm}}$

Q6JEE Main 2022MCQ4MElectromagnetic Waves

The magnetic field of a plane electromagnetic wave is given by : ${ \vec{\text{B}}=2 \times 10^{-8} \sin (0.5 \times 10^3 x+1.5 \times 10^{11} {\text{t}}) \hat{j} {\text{T}}$ The amplitude of the electric field would be : [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Speed of light $c=\;\frac{\omega }{k}=\;\frac{1.5 \times 10^{11}}{0.5 \times 10^3}=3 \times 10^8 {\text{m}} / {\sec}$ So, $E_0=B_0 c$ $=2 \times 10^{-8} \times 3 \times 10^8$ $=6 {\text{V}} / {\text{m}}$ Direction will be along z-axis.

Q7JEE Main 2022MCQ4MAlternating Current

In a series $L R$ circuit $X_L=R$ and power factor of the circuit is $P_1$ . When capacitor with capacitance $C$ such that $X_L=X_C$ is put in series, the power factor becomes $P_2$ . The ratio $\;\frac{P_1}{P_2}$ is: [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$P_1=\cos \varphi =\;\frac{1}{\sqrt{2}}(X_L=R)$ $P_2=\cos \varphi ^{'}=1$ (will become resonance circuit)So, $\;\frac{P_1}{P_2}=\;\frac{1}{\sqrt{2}}$

Q8JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

A charge particle is moving in a uniform magnetic field $(2 \hat{i}+3 \hat{j}) {\text{T}}$ . If it has an acceleration of $(\alpha \hat{i}-4 \hat{j}) {\text{m}} / {\text{s}}^2$ , then the value of $\alpha$ will be : [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

As magnetic force is perpendicular to magnetic fieldSo, $\vec{F} \cdot \vec{B}$ must be 0So, $2 \alpha-12=0$ $\alpha=6$

Q9JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

${\text{B}}_X$ and ${\text{B}}_{ {\text{Y}}}$ are the magnetic fields at the centre of two coils ${\text{X}}$ and ${\text{Y}}$ respectively each carrying equal current. If coil $X$ has 200 turns and $20 {\text{cm}}$ radius and coil $Y$ has 400 turns and $20 {\text{cm}}$ radius, the ratio of $B_X$ and $B_Y$ is : [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$B=\;\frac{\mu _0 N I}{2 R}$ $\;\frac{B_X}{B_Y}=\;\frac{N_x R_y}{N_y R_x}$ $=\;\frac{200 \times 20}{400 \times 20}=\;\frac{1}{2}$

Q10JEE Main 2022MCQ4MCurrent Electricity

The current I in the given circuit will be : [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

The grouping of resistance is a wheatstone bridgeSo, $R_{\;\text{net }\;}=4 Ω$ So, $i=\;{V}/{R_{\;\text{net }\;}}=10 {\text{A}}$

Explanation Figure 2

Q11JEE Main 2022MCQ4MElectrostatic Potential and Capacitance

The total charge on the system of capacitors $C_1=1 \mu {\text{F}}, C_2=2 \mu {\text{F}}, {\text{C}}_3=4 \mu {\text{F}}$ and ${\text{C}}_4=3 \mu {\text{F}}$ connected in parallel is : (Assume a battery of $20 {\text{V}}$ is connected to the combination) [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Equivalent $C=\sum C_i$ $=10 \mu {\text{F}}$ $\Rightarrow \;\text{ Charge }\; Q=C V$ $=200 \mu C$

Q12JEE Main 2022MCQ4MOscillations

When a particle executes Simple Hormonic Motion, the nature of graph of velocity as a function of displacement will be : [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Let $x=A \sin \omega t$ $\Rightarrow v=A \omega \cos \omega t$ $\Rightarrow v=\pm \omega \sqrt{A^{2-}x^2}$ $\Rightarrow \;\frac{v^2}{\omega ^2}+x^2=A^2$ $\Rightarrow \;\text{ Ellipse }\;$

Q13JEE Main 2022MCQ4MThermodynamics

$7 {\text{mol}}$ of a certain monoatomic ideal gas undergoes a temperature increase of $40 {\text{K}}$ at constant pressure. The increase in the internal energy of the gas in this process is : (Given ${\text{R}}=8.3 {\text{JK}}^{-1} {\text{mol}}^{-1}$ ) [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\Delta U=n C_v \Delta T$ $=7 \times \;\frac{3 R}{2} \times 40$ $=3486 {\text{J}}$

Q14JEE Main 2022MCQ4MThermodynamics

A monoatomic gas at pressure ${\text{P}}$ and volume ${\text{V}}$ is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be : [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$P V^\gamma =$ constant $\Rightarrow P V^\gamma =(P^{'})(\;\frac{v}{8})^\gamma \;\text{ where }\; \gamma =5 / 3$ $\Rightarrow P^{'}=32 P$

Q15JEE Main 2022MCQ4MProperties of Matter

A water drop of radius $1 {\text{cm}}$ is broken into 729 equal droplets. If surface tension of water is 75 dyne/ ${\text{cm}}$ , then the gain in surface energy upto first decimal place will be : (Given $\pi=3.14$ ) [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$729 \times \;\frac{4}{3} \pi r^3=\;\frac{4}{3} \pi R^3$ $\Rightarrow R=9 r . . . . . .(1)$ $\Delta U=S \times \Delta A . . . . . .(2)$ $\Rightarrow \Delta U=S \times \{-4 \pi R^{2+}729 \times 4 \pi r^2\}$ $=S \times 4 \pi\{729 r^{2-}81 r^2\}$ $=7.5 \times 10^{-4} J$

Q16JEE Main 2022MCQ4MGravitation

The percentage decrease in the weight of a rocket, when taken to a height of $32 {\text{km}}$ above the surface of earth will, be : ( Radius of earth $=6400 {\text{km}}$ ) [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\because g=\;\frac{G M}{r^2}$ $\Rightarrow \;\frac{\Delta g}{g}=2 \;\frac{\Delta r}{r}$ $\Rightarrow \;\frac{\Delta g}{g} \times 100=2 \times \;\frac{32}{6400} \times 100 \%=1 \%$ $\Rightarrow \%$ decrease in weight $=1 \%$

Q17JEE Main 2022MCQ4MOscillations

As per the given figure, two blocks each of mass $250 {\text{g}}$ are connected to a spring of spring constant $2 {\text{Nm}}^{-1}$ . If both are given velocity $v$ in opposite directions, then maximum elongation of the spring is: [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\because$ Loss in $K E=$ Gain in spring energy $\Rightarrow \;\frac{1}{2} m v^2 \times 2=\;\frac{1}{2} k x_m^2$ $\Rightarrow 2 \times \;\frac{1}{4} \times v^2=2 \times x_m^2$ $\Rightarrow x_m=\sqrt{\;\frac{v^2}{4}}=\;\frac{v}{2}$

Q18JEE Main 2022MCQ4MLaws of Motion

A monkey of mass $50 {\text{kg}}$ climbs on a rope which can withstand the tension (T) of $350 {\text{N}}$ . If monkey initially climbs down with an acceleration of $4 {\text{m}} / {\text{s}}^2$ and then climbs up with an acceleration of $5 {\text{m}} / {\text{s}}^2$ . Choose the correct option $(g=10 {\text{m}} / {\text{s}}^2)$ . [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$T_{\;\text{down }\;}=50 \times (10-4)$ $=50 \times 6$ $=300 {\text{N}}$ ${\text{T}}_{\;\text{up }\;}=50 \times (10+5)$ $=50 \times 15$ $=750 {\text{N}}$ $\Rightarrow$ Rope will break while climbing up.

Q19JEE Main 2022MCQ4MMotion in a Plane

Two projectiles thrown at $30^{\circ}$ and $45^{\circ}$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is : [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\because t_a=\;\frac{u \sin \theta}{g}$ $\Rightarrow \;\frac{u_1 \sin (30^{\circ})}{g}=\;\frac{u_1 \sin (45^{\circ})}{g}$ $\Rightarrow \;\frac{u_1}{u_2}=\;{\;\frac{1}{\sqrt{2}}}/{\frac{1}{2}}=\;\frac{\sqrt{2}}{1}$

Q20JEE Main 2022MCQ4MUnits and Measurements

A screw gauge of pitch $0.5 {\text{mm}}$ is used to measure the diameter of uniform wire of length $6.8 {\text{cm}}$ , the main scale reading is $1.5 {\text{mm}}$ and circular scale reading is 7 . The calculated curved surface area of wire to appropriate significant figures is : [Screw gauge has 50 divisions on its circular scale] [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Least count $=\;\frac{0.5}{50} {\text{mm}}=0.01 {\text{mm}}$ $\therefore$ Diameter, ${\text{d}}=1.5 {\text{mm}}+7 \times 0.01$ $=1.57 {\text{mm}}$ $\therefore$ Surface area $=(2 \pi r) \times l$ $=\pi d l$ $=3.142 \times \;\frac{1.57}{10} \times 6.8 {\text{cm}}^2$ $=3.354 {\text{cm}}^2=3.4 {\text{cm}}^2$

Q21JEE Main 2022NAT4MMotion in a Plane

If the initial velocity in horizontal direction of a projectile is unit vector $\hat{i}$ and the equation of trajectory is $y=5 x(1-x)$ . The $y$ component vector of the initial velocity is ________ $\hat{j}$ . Take $.g=10 {\text{m}} / {\text{s}}^2)$ [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$y=5 x-5 x^2$ $y=x \tan \theta-\;\frac{1}{2} \;\frac{g x^2}{v^2}$ $\tan \theta=5=\;\frac{u_y}{u_x}$ $\Rightarrow u_y=5$

Q22JEE Main 2022NAT4MRotational Motion

A disc of mass $1 {\text{kg}}$ and radius $R$ is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be $4 \sqrt{\;\frac{x}{3 R}} {\text{rad}} s^{-1}$ where $x=$ ________. $(g=10 {\text{ms}}^{-2})$ [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Loss in P.E. = Gain in K.E. $2 m g R=\;\frac{1}{2}\$ [;\frac{1}{2} m R^{2+}m R^2]$ w^22 m g R=;\frac{1}{2} \times ;\frac{3}{2} m R^2 w^2w^2=;\frac{8 g}{3 R}w=\sqrt{;\frac{8 g}{3 R}}=4 \sqrt{;\frac{g}{2 \times 3 R}}\Rightarrow x=;\frac{g}{2}=5$

Explanation Figure 1

Q23JEE Main 2022NAT4MProperties of Matter

In an experiment to determine the Young's modulus of wire of a length exactly $1 {\text{m}}$ , the extension in the length of the wire is measured as $0.4 {\text{mm}}$ with an uncertainty of $\pm 0.02 {\text{mm}}$ when a load of $1 {\text{kg}}$ is applied. The diameter of the wire is measured as $0.4 {\text{mm}}$ with an uncertainty of $\pm 0.01 {\text{mm}}$ . The error in the measurement of Young's modulus $(\Delta Y)$ is found to be $x \times 10^{10} {\text{Nm}}^{-2}$ . The value of $x$ is ________ $(.$ take $.g=10 {\text{ms}}^{-2})$ [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{F / A}{l / L}=Y, A=\pi D^2$ $\;\frac{\Delta Y}{Y}=\;\frac{\Delta F}{F}+\;\frac{2 \Delta D}{D}+\;\frac{\Delta l}{e}+\;\frac{\Delta L}{L}$ $=2 \times \;\frac{0.01}{0.4}+\;\frac{0.02}{0.4}$ $=\;\frac{0.04}{0.4}=\;\frac{1}{10}$ $Y=\;\frac{F l}{A \Delta l}$ $=\;\frac{10 \times 1}{\pi(0.1 m m)^2 \times 0.4 m m}$ $=1.988 \times 10^{11}$ $\approx 2 \times 10^{11}$ $\;\frac{\Delta y}{y}=\;\frac{1}{10}$ $\Delta y=\;\frac{y}{10}=2 \times 10^{10}$

Q24JEE Main 2022NAT4MWaves

When a car is approaching the observer, the frequency of horn is $100 {\text{Hz}}$ . After passing the observer, it is $50 {\text{Hz}}$ . If the observer moves with the car, the frequency will be $\;\frac{x}{3} {\text{Hz}}$ where $x=$ __________. [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$100=v_0 \;\frac{v}{v-v_c}$ $50=v_0 \;\frac{v}{v+v_c}$ $2=\;\frac{v+v_c}{v-v_c}$ $2 v-2 v_c=v+v_c$ $v_c=\;\frac{v}{3}$ $100=v_0 \;\frac{v \times 3}{2 v} \Rightarrow v_0=\;\frac{200}{3}=\;\frac{x}{3}$ $\Rightarrow x=200$

Q25JEE Main 2022NAT4MElectrostatic Potential and Capacitance

A composite parallel plate capacitor is made up of two different dielectric materials with different thickness $(t_1.$ and $.t_2)$ as shown in figure. The two different dielectric materials are separated by a conducting foil ${\text{F}}$ . The voltage of the conducting foil is V. [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{C_1}{C_2}=\;\frac{3 \times t_2}{t_1 \times 4}=\;\frac{3}{2}$ $\;\frac{q}{C_1}=v_1, \;\frac{q}{C_2}=v_2$ $\;\frac{v_1}{v_2}=\;\frac{C_2}{C_1}=\;\frac{2}{3}$

Explanation Figure 2

Q26JEE Main 2022NAT4MCurrent Electricity

Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length $l_1$ is $40 {\text{cm}}$ . Now an unknown resistance $x$ is connected in series with ${\text{P}}$ and new balancing length is found to be $80 {\text{cm}}$ measured from the same end. Then the value of $x$ will be _________ $Ω$ . [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{P}{40}=\;\frac{Q}{60} . . . . . .(1)$ $\;\frac{P+x}{80}=\;\frac{Q}{20} . . . .(2)$ $\;\frac{P}{P+x} \times \;\frac{80}{40}=\;\frac{20}{60}$ $\;\frac{4}{4+x} \times 2=\;\frac{1}{3}$ $24=4+x$ $x=20$

Q27JEE Main 2022NAT4MAlternating Current

The effective current I in the given circuit at very high frequencies will be A. [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Equivalent circuit will be $I=\;\frac{220}{5}=44 {\text{A}}$

Explanation Figure 2

Q28JEE Main 2022NAT4MRay Optics and Optical Instruments

The graph between $\;\frac{1}{u}$ and $\;\frac{1}{v}$ for a thin convex lens in order to determine its focal length is plotted as shown in the figure. The refractive index of lens is $1.5$ and its both the surfaces have same radius of curvature $R$ . The value of $R$ will be _________ cm. (where $u=$ object distance, $v=$ image distance) [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$f=10 {\text{cm}}$ $\;\frac{1}{f}=(\mu -1)(\;\frac{1}{R}-\;\frac{1}{-R})$ $\;\frac{1}{10}=\;\frac{1.5-1}{1} \times \;\frac{2}{R}$ $\;\frac{1}{10}=\;\frac{1}{R}$ $R=10 {\text{cm}}$

Q29JEE Main 2022NAT4MAtoms and Nuclei

In the hydrogen spectrum, $\lambda$ be the wavelength of first transition line of Lyman series. The wavelength difference will be "a $\lambda$ ' between the wavelength of $3^{ {\text{rd}}}$ transition line of Paschen series and that of $2^{\;\text{nd }\;}$ transition line of Balmer series where $a=$ _________. [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{1}{\lambda }=R_H(\;\frac{1}{1^2}-\;\frac{1}{2^2})$ $\;\frac{1}{\lambda _3}=R_H(\;\frac{1}{3^2}-\;\frac{1}{6^2})$ $\;\frac{1}{\lambda _2}=R_H(\;\frac{1}{2^2}-\;\frac{1}{4^2})$ $\therefore \lambda _3-\lambda _2=a \lambda$ $a=5$

Q30JEE Main 2022NAT4MSemiconductor Electronics

In the circuit shown below, maximum zener diode current will be___ ${\text{mA}}$ . [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$i_s=\;\frac{60}{4 \times 10^3}=15 \times 10^{-3}=15 {\text{mA}}$ $i_L=\;\frac{60}{10 \times 10^3}=6 {\text{mA}}$ $I_z=i_s-i_L=9 {\text{mA}}$

Explanation Figure 2

Chemistry30 questions

Q31JEE Main 2022MCQ4Md and f Block Elements

Match List-I with List-II : List I (Compound) List II (Shape) (A) ${\text{BrF}}_5$ (I)bent (B) $[ {\text{CrF}}_6]^{3-}$ (II) square pyramidal (C) ${\text{O}}_3$ (III) trigonal bipyramidal (D) ${\text{PCl}}_5$ (IV) octahedral Choose the correct answer from the options given below : [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

(A) ${\text{BrF}}_5$ - square pyramidal(B) $[ {\text{CrF}}_6]^{3-}$ - octahedral(C) ${\text{O}}_3-$ bent(D) ${\text{PCl}}_5$ - trigonal bipyramidal

Q32JEE Main 2022MCQ4Md and f Block Elements

Match List - I with List - II. List I (Processes / Reactions) List II (Catalyst) (A) $2 {\text{SO}}_2( {\text{g}})+ {\text{O}}_2( {\text{g}}) \to 2 {\text{SO}}_3( {\text{g}})$ (I) ${\text{Fe}}( {\text{s}})$ (B) $4 {\text{NH}}_3( {\text{g}})+5 {\text{O}}_2( {\text{g}}) \to 4 {\text{NO}}( {\text{g}})+6 {\text{H}}_2 {\text{O}}( {\text{g}})$ (II) ${\text{Pt}}( {\text{s}})- {\text{Rh}}( {\text{s}})$ (C) ${\text{N}}_2( {\text{g}})+3 {\text{H}}_2( {\text{g}}) \to 2 {\text{NH}}_3( {\text{g}})$ (III) ${\text{V}}_2 {\text{O}}_5$ (D) Vegetable oil(I) $+ {\text{H}}_2 \to$ Vegetable ghee(s) (IV) ${\text{Ni}}( {\text{s}})$ Choose the correct answer from the options given below: [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

(A) $2 {\text{SO}}_2( {\text{g}})+ {\text{O}}_2( {\text{g}}) {\longrightarrow }^{ {\text{V}}_2 {\text{O}}_5} 2 {\text{SO}}_3$ (B) $4 {\text{NH}}_3( {\text{g}})+5 {\text{O}}_2( {\text{g}}) {\longrightarrow }^{ {\text{Pt}}( {\text{s}})- {\text{Rh}}( {\text{s}})} 4 {\text{NO}}( {\text{g}})+6 {\text{H}}_2 {\text{O}}( {\text{g}})$ (C) ${\text{N}}_2( {\text{g}})+3 {\text{H}}_2( {\text{g}}) {\longrightarrow }^{ {\text{Fe}}( {\text{s}})} 2 {\text{NH}}_3( {\text{g}})$ (D) Vegetable oil(I) $+ {\text{H}}_2 {\longrightarrow }^{ {\text{Ni}}( {\text{s}})}$ Vegetable ghee(s)

Q33JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

Given two statements below : Statement I : In ${\text{Cl}}_2$ molecule the covalent radius is double of the atomic radius of chlorine. Statement II : Radius of anionic species is always greater than their parent atomic radius. Choose the most appropriate answer from options given below : [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

Q34JEE Main 2022MCQ4Mp Block Elements

Refining using liquation method is the most suitable for metals with: [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

Q35JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Which of the following can be used to prevent the decomposition of ${\text{H}}_2 {\text{O}}_2$ ? [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Urea is used as a stabilizer for the storage of ${\text{H}}_2 {\text{O}}_2$ .

Q36JEE Main 2022MCQ4Mp Block Elements

Reaction of ${\text{BeCl}}_2$ with ${\text{LiAlH}}_4$ gives : (A) ${\text{AlCl}}_3$ (B) ${\text{BeH}}_2$ (C) ${\text{LiH}}$ (D) ${\text{LiCl}}$ (E) ${\text{BeAlH}}_4$ Choose the correct answer from options given below: [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$2 {\text{BeCl}}_2+ {\text{LiAlH}}_4 \to 2 {\text{BeH}}_2+ {\text{LiCl}}+ {\text{AlCl}}_3$

Q37JEE Main 2022MCQ4Mp Block Elements

Borazine, also known as inorganic benzene, can be prepared by the reaction of 3-equivalents of "X" with 6-equivalents of "Y". "X" and "Y", respectively are : [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$3 {\text{B}}_2 {\text{H}}_6+6 {\text{NH}}_3 \to \;{2 {\text{B}}}_{\;\text{ (Borazine) }\;_3 {\text{N}}_3 {\text{H}}_6}$

Q38JEE Main 2022MCQ4MRedox Reactions

Which of the given reactions is not an example of disproportionation reaction? [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$2 {\text{NO}}_2+ {\text{H}}_2 {\text{O}} \to {\text{HNO}}_3^{+5}+ {\text{HNO}}_2^{+3}$ : Disproportionation ${\text{MnO}}_4^{-}+4 {\text{H}}^{+}+3 {\text{e}}^{-} \to {\text{MnO}}_2+2 {\text{H}}_2 {\text{O}}$ : reduction $3 {\text{MnO}}_4^{2-}+4 {\text{H}}^{+} \to 2 {\text{MnO}}_4^{-}+ {\text{MnO}}_2^{+4}+2 {\text{H}}_2 {\text{O}}$ : Disproportionation

Q39JEE Main 2022MCQ4MRedox Reactions

The dark purple colour of ${\text{KMnO}}_4$ disappears in the titration with oxalic acid in acidic medium. The overall change in the oxidation number of manganese in the reaction is : [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

In acidic medium, ${\text{MnO}}_4^{-} \to {\text{Mn}}^{+2}$ change in ox. no. $=5$

Q40JEE Main 2022MCQ4MHydrocarbons

${ {\text{C}}^{∙}}+ {\text{CH}}_4 \to {\text{A}}+ {\text{B}}$ $A$ and $B$ in the above atmospheric reaction step are : [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{Cl}}^{∙}+ {\text{CH}}_4 \longrightarrow ˙{C^{∙} H}_3+ {\text{HCl}}$

Q41JEE Main 2022MCQ4MPractical Organic Chemistry

Which technique among the following, is most appropriate in separation of a mixture of $100 {\text{mg}}$ of $p$ nitrophenol and picric acid? [26-Jul-2022-Shift-1]

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Q42JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

The difference in the reaction of phenol with bromine in chloroform and bromine in water medium is due to : [26-Jul-2022-Shift-1]

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Q43JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Which of the following compounds is not aromatic? [26-Jul-2022-Shift-1]

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📖 Explanation

[10] Annulene, although follow $(4 n+2) \pi$ electron rule, but it is non-aromatic due to its non planar nature. It is nonplanar due to repulsion of ${\text{C}}- {\text{H}}$ bonds present inside the ring.

Q44JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

The products formed in the following reaction, ${\text{A}}$ and ${\text{B}}$ are [26-Jul-2022-Shift-1]

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📖 Explanation

${\text{NaBH}}_4$ does not reduce carboxylic acid.

Explanation Figure 6

Q45JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Which reactant will give the following alcohol on reaction with one mole of phenyl magnesium bromide $( {\text{PhMgBr}})$ followed by acidic hydrolysis? ${\text{Ph}}- {\text{C}}^{|}^{\text{Ph}}_{|}_{\text{CH}_3}- {\text{OH}}$ [26-Jul-2022-Shift-1]

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Q46JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

The major product of the following reaction is [26-Jul-2022-Shift-1]

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Q47JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

The correct stability order of the following diazonium salt is [26-Jul-2022-Shift-1]

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📖 Explanation

Since diazonium ion is a cation hence it is stabilized by electron donating groups and destabilized by electron withdrawing group.Hence Stability order should be ${\text{A}}> {\text{C}}> {\text{D}}> {\text{B}}$ .

Explanation Figure 2

Q48JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Stearic acid and polyethylene glycol react to form which one of the following soap/s detergents ? [26-Jul-2022-Shift-1]

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Q49JEE Main 2022MCQ4MBiomolecules

Which one of the following is a reducing sugar? [26-Jul-2022-Shift-1]

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📖 Explanation

If any sugar is having free $- {\text{OH}}$ group at anomeric carbon then it will be a reducing sugar

Explanation Figure 5

Q50JEE Main 2022MCQ4MAmines

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Experimental reaction of ${\text{CH}}_3 {\text{Cl}}$ with aniline and anhydrous ${\text{AlCl}}_3$ does not give $o$ and $p$ methylaniline. Reason (R): The $- {\text{NH}}_2$ group of aniline becomes deactivating because of salt formation with anhydrous ${\text{AlCl}}_3$ and hence yields $m$ -methyl aniline as the product. In the light of the above statements, choose the most appropriate answer from the options given below : [26-Jul-2022-Shift-1]

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📖 Explanation

Aniline does not undergo Friedel Craft reaction because the reagent $AlCl_3$ being electron deficient acts as a Lewis acid.

Explanation Figure 1

Q51JEE Main 2022NAT4MSome Basic Concepts of Chemistry

Chlorophyll extracted from the crushed green leaves was dissolved in water to make $2 {\text{L}}$ solution of ${\text{Mg}}$ of concentration $48 {\text{ppm}}$ . The number of atoms of ${\text{Mg}}$ in this solution is $x \times 10^{20}$ atoms. The value of $x$ is _________. (Nearest Integer) (Given : Atomic mass of ${\text{Mg}}$ is $24 {\text{g}} {\text{mol}}^{-1} ; {\text{N}}_{ {\text{A}}}=6.02 \times 10^{23} {\text{mol}}^{-1}$ ) [26-Jul-2022-Shift-1]

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📖 Explanation

In $2 {\text{L}} \to 96 {\text{mg}}$ of ${\text{Mg}}$ Number of atoms of ${\text{Mg}}=\;\frac{96 \times 10^{-3}}{24} \times {\text{N}}_{ {\text{A}}}$ $\;=4 \times 10^{-3} \times 6 \times 10^{23}$ $\;=24 \times 10^{20}$ $\;x=24$

Q52JEE Main 2022NAT4MStates of Matter

A mixture of hydrogen and oxygen contains $40 \%$ hydrogen by mass when the pressure is $2.2$ bar. The partial pressure of hydrogen is bar. (Nearest Integer) [26-Jul-2022-Shift-1]

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📖 Explanation

$40 \% {\text{w}} / {\text{w}}$ hydrogen gas is given in mixture of ${\text{H}}_2$ and oxygen.Wt. of ${\text{H}}_2=40 {\text{g}}$ Wt. of ${\text{O}}_2=60 {\text{g}}$ $χ_{ {\text{H}}_2}=\;{ {\text{n}}_{ {\text{H}}_2}}/{ {\text{n}}_{ {\text{H}}_2}+ {\text{n}}_{ {\text{O}}_2}}$ $=\;\frac{\;\frac{40}{2}}{\frac{40}{2}+\;\frac{60}{32}}$ $=\;\frac{20}{20+1.875}$ $=\;\frac{20}{21.875}=0.914$ ${\text{P}}_{ {\text{H}}_2}=χ_{ {\text{H}}_2} \times {\text{P}}_{ {\text{T}}}$ $=0.914 \times 2.2$ $=2.01 ≃ 2 {\bar}$

Q53JEE Main 2022NAT4MStructure of Atom

The wavelength of an electron and a neutron will become equal when the velocity of the electron is $x$ times the velocity of neutron. The value of $x$ is ________.(Nearest Integer) (Mass of electron is $9.1 \times 10^{-31} {\text{kg}}$ and mass of neutron is $1.6 \times 10^{-27} {\text{kg}}$ ) [26-Jul-2022-Shift-1]

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📖 Explanation

$\lambda _e=\;\frac{h}{m_e \times V_e}, \;\; \lambda _N=\;\frac{h}{m_N \times V_N}$ $\lambda _e=\lambda _N$ When $V_e=x V_N$ $\;\frac{1}{m_e V_e}=\;\frac{1}{m_N \times V_N}$ $\;\frac{m_N}{m_e}=\;\frac{V_e}{V_N}=x$ $x=\;\frac{1.6 \times 10^{-27}}{9.1 \times 10^{-31}}$ $=0.17582 \times 10^4$ $≃ 1758$

Q54JEE Main 2022NAT4MChemical Thermodynamics

$2.4 {\text{g}}$ coal is burnt in a bomb calorimeter in excess of oxygen at $298 {\text{K}}$ and $1 {\text{atm}}$ pressure. The temperature of the calorimeter rises from $298 {\text{K}}$ to $300 {\text{K}}$ . The enthalpy change during the combustion of coal is $-x {\text{kJ}} {\text{mol}}^{-1}$ . The value of $x$ is ________. (Nearest Integer) (Given : Heat capacity of bomb calorimeter $20.0 {\text{kJ}} {\text{K}}^{-1}$ . Assume coal to be pure carbon) [26-Jul-2022-Shift-1]

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📖 Explanation

${\text{Q}}($ Heat evolved $)=-\;{ {\text{C}}_{ {\text{system}}} \Delta {\text{T}}}/{ {\text{n}}}$ ${\text{n}}_{\;\text{coal }\;}=\;\frac{2.4}{12}$ ${\text{Q}}=\;\frac{-20(300-298)}{0.2}$ $Q=-200 {\text{kJ}} / {\text{mol}}$ $x=200$

Q55JEE Main 2022NAT4MSolutions

When $800 {\text{mL}}$ of $0.5 {\text{M}}$ nitric acid is heated in a beaker, its volume is reduced to half and $11.5 {\text{g}}$ of nitric acid is evaporated. The molarity of the remaining nitric acid solution is $x \times 10^{-2} {\text{M}}$ . (Nearest integer) (Molar mass of nitric acid is $63 {\text{g}} {\text{mol}}^{-1}$ ) [26-Jul-2022-Shift-1]

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📖 Explanation

${\text{m}}$ moles of ${\text{HNO}}_3=800 \times 0.5$ Moles of ${\text{HNO}}_3=400 \times 10^{-3}=0.4$ molesWeight of ${\text{HNO}}_3=0.4 \times 63 {\text{g}}=25.2 {\text{g}}$ Remaining acid $=25.2-11.5=13.7 {\text{g}}$ $M \;\; =\;\frac{13.7 \times 1000}{400 \times 63}$ $\;\; =\;\frac{137}{252}=0.54$ $\;\; =54 \times 10^{-2}$

Q56JEE Main 2022NAT4MChemical Equilibrium

At $298 {\text{K}}$ , the equilibrium constant is $2 \times 10^{15}$ for the reaction : ${\text{Cu}}( {\text{s}})+2 {\text{Ag}}^{+}( {\text{aq}}) ⇌ {\text{Cu}}^{2+}( {\text{aq}})+2 {\text{Ag}}( {\text{s}})$ The equilibrium constant for the reaction $\;\frac{1}{2} {\text{Cu}}^{2+}( {\text{aq}})+ {\text{Ag}}( {\text{s}}) ⇌ \;\frac{1}{2} {\text{Cu}}( {\text{s}})+ {\text{Ag}}^{+}( {\text{aq}})$ is $x \times 10^{-8}$ . The value of $x$ is ________. (Nearest Integer) [26-Jul-2022-Shift-1]

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📖 Explanation

${\text{K}}_{ {\text{eq}}}^{'}=\;{1}/{\sqrt{ {\text{K}}_{ {\text{eq}}}}}=\;{1}/{\sqrt{2 \times 10^{15}}}= {\text{X}} \times 10^{-8}$ $\Rightarrow \;\frac{1}{\sqrt{20}} \times \;\frac{1}{10^7}= {\text{x}} \times 10^{-8}$ $\Rightarrow \;\frac{1}{\sqrt{20}} \times 10^{-7}= {\text{x}} \times 10^{-8}$ $\Rightarrow {\text{x}}=\;\frac{\sqrt{10}}{\sqrt{2}}=\sqrt{5}=2.236$ $≃ 2.24$

Q57JEE Main 2022NAT4MElectrochemistry

The amount of charge in ${\text{F}}$ (Faraday) required to obtain one mole of iron from ${\text{Fe}}_3 {\text{O}}_4$ is ______. (Nearest Integer) [26-Jul-2022-Shift-1]

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📖 Explanation

For ${\text{Fe}}_3 {\text{O}}_4$ , $x=\;\frac{+8}{3}$ where $x$ is oxidation state of Fe. ${\text{Fe}}_3 {\text{O}}_4+8 {\text{H}}^{+}+8 {\text{e}}^{-} \longrightarrow 3 {\text{Fe}}+4 {\text{H}}_2 {\text{O}}$ Charge required $=\;\frac{8}{3} \times {\text{F}}=\;{8 {\text{F}}}/{3} ≃ 3 {\text{F}}$

Q58JEE Main 2022NAT4MChemical Kinetics

For a reaction ${\text{A}} \to 2 {\text{B}}+ {\text{C}}$ the half lives are $100 {\text{s}}$ and $50 {\text{s}}$ when the concentration of reactant ${\text{A}}$ is $0.5$ and $1.0 {\text{mol}} {\text{L}}^{-1}$ respectively. The order of the reaction is _______. (Nearest Integer) [26-Jul-2022-Shift-1]

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📖 Explanation

$t_{1 / 2} \propto \;\frac{1}{(a_0)^{n-1}}$ ${\text{t}}_{1 / 2}=100 {\sec} \;\; {\text{a}}_0=0.5$ ${\text{t}}_{1 / 2}=50 {\sec} \;\; {\text{a}}_0=1$ $\;\frac{100}{50}=(\;\frac{1}{0 \cdot 5})^{n-1}$ $(2)=(2)^{n-1}$ ${\text{n}}-1=1$ $n=2$

Q59JEE Main 2022NAT4MCoordination Compounds

The difference between spin only magnetic moment values of $[ {\text{Co}}( {\text{H}}_2 {\text{O}})_6] {\text{Cl}}_2$ and $[ {\text{Cr}}( {\text{H}}_2 {\text{O}})_6] {\text{Cl}}_3$ is ________. [26-Jul-2022-Shift-1]

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📖 Explanation

${\text{Co}} \to 4 s^2 3 d^7$ ${\text{H}}_2 {\text{O}}$ is weak field ligand. ${\text{Co}}^{+2} \to 3 d^7$ ${\text{n}}=3 \;\;$ µ_1 ;; =√{n( {\n}+2)} ;; =\sqrt{15} ;\text{ B.M. }; ${\text{Cr}} \to 4 s^1 3 d^5$ ${\text{Co}}^{+3} \to 3 d^3$ ${\text{n}}=3 \;\; \mu _2=\sqrt{15} \;\text{ B.M. }\;$ $\mu _1-\mu _2=0$

Q60JEE Main 2022NAT4MHydrocarbons

In the presence of sunlight, benzene reacts with ${\text{Cl}}_2$ to give product, ${\text{X}}$ . The number of hydrogens in ${\text{X}}$ is _________. [26-Jul-2022-Shift-1]

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Mathematics30 questions

Q61JEE Main 2022MCQ4MFunctions

Let ${\text{f}}: {\text{R}} \to {\text{R}}$ be a continuous function such that $f(3 x)-f(x)=x$ . If $f(8)=7$ , then $f(14)$ is equal to: [26-Jul-2022-Shift-1]

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📖 Explanation

$f(3 x)-f(x)=x . . . . . . \;\text{ (1) }\;$ $x \to \;\frac{x}{3}$ $f(x)-f(\;\frac{x}{3})=\;\frac{x}{3} . . . . . .(2)$ Again $x \to \;\frac{x}{3}$ $f(\;\frac{x}{3})-f(\;\frac{x}{9})=\;\frac{x}{3^2} . . . . . .$ Similarly $f(\;\frac{x}{3^{n-2}})-f(\;\frac{x}{3^{n-1}})=\;\frac{x}{3^{n-1}} . . . .(n)$ Adding all these and applying $n \to \infty$ $\lim _{n \to \infty }(f(3 x)-f(\;\frac{x}{3^{n-1}}))=x(1+\;\frac{1}{3}+\;\frac{1}{3^2}+. . .)$ $f(3 x)-f(0)=\;\frac{3 x}{2}$ Putting $x=\;\frac{8}{3}$ $f(8)-f(0)=4$ $\Rightarrow f(0)=3$ Putting $x=\;\frac{14}{3}$ $f(14)-3=7 \Rightarrow f(14)=0$

Q62JEE Main 2022MCQ4MComplex Numbers

Let 0 be the origin and ${\text{A}}$ be the point $z_1=1+2 i$ . If ${\text{B}}$ is the point $z_2, {\text{Re}}(z_2) < 0$ , such that ${\text{OAB}}$ is a right angled isosceles triangle with ${\text{OB}}$ as hypotenuse, then which of the following is NOT true? [26-Jul-2022-Shift-1]

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📖 Explanation

$\;\frac{z_2-0}{(1+2 i)-0}=\;\frac{|O B|}{|O A|} e^{\;\frac{i \pi}{4}}$ $\Rightarrow \;\frac{z_2}{1+2 i}=\sqrt{2} e^{\;\frac{i \pi}{4}}$ OR $z_2=(1+2 i)(1+i)$ $=-1+3 i$ $\arg z_2=\pi-\tan ^{-1} 3$ $|z_2|=\sqrt{10}$ $z_1-2 z_2=(1+2 i)+2-6 i=3-4 i$ $\arg (z_1-2 z_2)=-\tan ^{-1} \;\frac{4}{3}$ $|2 z_1-z_2|=|2+4 i+1-3 i|=|3+i|=\sqrt{10}$

Explanation Figure 1

Q63JEE Main 2022MCQ4MMatrices and Determinants

If the system of linear equations. $8 x+y+4 z=-2$ $x+y+z=0$ $\lambda x-3 y=\mu$ has infinitely many solutions, then the distance of the point $(\lambda , \mu ,-\;\frac{1}{2})$ from the plane $8 x+y+4 z+2=0$ is : [26-Jul-2022-Shift-1]

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📖 Explanation

$\Delta ={\begin{vmatrix}8 & 1 & 4 \\ 1 & 1 & 1 \\ \lambda & -3 & 0\end{vmatrix}}$ $=8(3)-1(-\lambda )+4(-3-\lambda )$ $=24+\lambda -12-4 \lambda$ $=12-3 \lambda$ So for $\lambda =4$ , it is having infinitely many solutions. $\Delta _x={\begin{vmatrix}-2 & 1 & 4 \\ 0 & 1 & 1 \\ \mu & -3 & 0\end{vmatrix}}$ $=-2(3)-1(-\mu )+4(-\mu )$ $=-6-3 \mu =0$ For $\mu =-2$ Distance of $(4,-2, \;\frac{-1}{2})$ from $8 x+y+4 z+2=0$ $=\;\frac{32-2-2+2}{\sqrt{64+1+16}}=\;\frac{10}{3}$ units

Q64JEE Main 2022MCQ4MMatrices and Determinants

Let $A$ be a $2 \times 2$ matrix with ${\det}(A)=-1$ and ${\det}((A+I)({\text{Adj}}(A)+I))=4$ . Then the sum of the diagonal elements of $A$ can be : [26-Jul-2022-Shift-1]

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📖 Explanation

$|(A+I)(a d j A+I)|=4$ $\Rightarrow |A \text{adj} A+A+ \text{adj} A+I|=4$ $\Rightarrow |(A) I+A+\text{adj} A+I|=4$ $|A|=-1 \Rightarrow |A+\text{adj} A|=4$ $A=\begin{bmatrix}a & b \\ c & d\end{bmatrix} a d j A=\begin{bmatrix}a & -b \\ -c & d\end{bmatrix}$ $\Rightarrow {\begin{vmatrix}(a+d) & 0 \\ 0 & (a+d)\end{vmatrix}}=4$ $\Rightarrow a+d=\pm 2$

Q65JEE Main 2022MCQ4MArea Under The Curves

The odd natural number $a$ , such that the area of the region bounded by $y=1, y=3, x=0, x=y^a$ is $\;\frac{364}{3}$ , is equal to: [26-Jul-2022-Shift-1]

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📖 Explanation

$a$ is a odd natural number and $|int_1^3 y^a d y|=\;\frac{364}{3}$ $\Rightarrow |\;\frac{1}{a+1}(y^{a+1})_1^3|=\;\frac{364}{3}$ $\Rightarrow \;\frac{3^{a+1}-1}{a+1}=\pm \;\frac{364}{3}$ Solving with (-) sign, $\;\frac{3^{a+1}-1}{a+1}=\;\frac{364}{3} \Rightarrow (a=5)$ Solving with $(+)$ sign, $\;\frac{3^{a+1}-1}{a+1}=\;\frac{-364}{3}$ , No a exist $\therefore (a=5)$

Q66JEE Main 2022MCQ4MSequences and Series

Consider two G.Ps. $2,2^2, 2^3, . . . .$ . and $4,4^2, 4^3, . . .$ . of 60 and $n$ terms respectively. If the geometric mean of all the $60+n$ terms is $(2)^{\;\frac{225}{8}}$ , then $sum_{k=1}^n k(n-k)$ is equal to : [26-Jul-2022-Shift-1]

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📖 Explanation

Given G.P's $2,2^2, 2^3, . . . .60$ terms $4,4^2, . . . . n$ termsNow, G.M $=2^{\;\frac{225}{8}}$ $(2.2^2 . . . 4.4^2 . . .)^{\;\frac{1}{60+n}}=2^{\;\frac{225}{8}}$ $(2^{\;\frac{n^{2+}n+1830}{6+n}})=2^{\;\frac{225}{8}}$ $\Rightarrow \;\frac{n^{2+}n+1830}{60+n}=\;\frac{225}{8}$ $\Rightarrow 8 n^{2-}217 n+1140=0$ $n=\;\frac{57}{8}, 20, \;\text{ so }\; n=20$ $\therefore sum_{k=1}^{20} k(20-k)=20 \times \;\frac{20 \times 21}{2}-\;\frac{20 \times 21 \times 41}{6}$ $=\;\frac{20 \times 21}{2}[20-\;\frac{41}{3}]=1330$

Q67JEE Main 2022MCQ4MLimits, Continuity and Differentiability

If the function $f(x)=\begin{cases}\\ \\ \frac{\log _e(1-x+x^2)+\log _e(1+x+x^2)}{\sec x-\cos x} & { ,} \\ \\ \\ \\ x \in (\\ \\ \frac{-\pi}{2} & \\ \\ \frac{\pi}{2})-\{0\} \\ \\ k & {,} \\ \\ \\ \\ x=0\end{cases}$ is continuous at ${\text{x}}=0$ , then ${\text{k}}$ is equal to: [26-Jul-2022-Shift-1]

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📖 Explanation

$f(x)=\begin{cases}\\ \\ \frac{\log _e(1-x+x^2)+\log _e(1+x+x^2)}{\sec x-\cos x} & x \in (\\ \\ \frac{-\pi}{2} & \\ \\ \frac{\pi}{2})-\{0\} \\ k & x=0\end{cases}$ for continuity at $x=0$ $\lim _{x \to 0} f(x)=k$ $\therefore k=\lim _{x \to 0} \;\frac{\log _e(x^{4+}x^{2+}1)}{\sec x-\cos x}(\;\frac{0}{0} \;\text{ form }\;)$ $=\lim _{x \to 0} \;\frac{\cos x \log _e(x^{4+}x^{2+}1)}{\sin ^2 x}$ $=\lim _{x \to 0} \;\frac{\log _e(x^{4+}x^{2+}1)}{x^2}$ $=\lim _{x \to 0} \;\frac{\ln (1+x^{2+}x^4)}{x^{2+}x^4} \cdot \;\frac{x^{2+}x^4}{x^2}$ $=1$

Q68JEE Main 2022MCQ4MLimits, Continuity and Differentiability

If $f(x)=\begin{cases}x+a {,} & \\ \\ \\ \\ x \le 0 \\ {|x-4|} {,} & \\ \\ \\ \\ x > 0\end{cases}$ and $g(x)=\begin{cases}x+1 {,} & x < 0 \\ (x-4)^{2+}b {,} & \\ \\ \\ \\ x \ge 0\end{cases}$ are continuous on ${\text{R}}$ , then $(g of)(2)+(f o g)(-2)$ is equal to : [26-Jul-2022-Shift-1]

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📖 Explanation

$\because f(x)$ and $g(x)$ are continuous on ${\text{R}}$ $\therefore a=4$ and $b=1-16=-15$ then $(g \circ f)(2)+(f o g)(-2)$ $=g(2)+f(-1)$ $=-11+3=-8$

Q69JEE Main 2022MCQ4MLimits, Continuity and Differentiability

Let $f(x)=\begin{cases}x^{3-}x^{2+}10 x-7 & x \le 1 \\ -2 x+\log _2(b^{2-}4) & x > 1\end{cases}$ . Then the set of all values of $b$ , for which $f(x)$ has maximum value at $x=1$ , is : [26-Jul-2022-Shift-1]

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📖 Explanation

$f(x)=\begin{cases}x^{3-}x^{2+}10 x-7 & x \le 1 \\ -2 x+\log _2(b^{2-}4) & x>1\end{cases}$ If $f(x)$ has maximum value at $x=1$ then $f(1+) \le f(1)$ $-2+\log _2(b^{2-}4) \le 1-1+10-7$ $\log _2(b^{2-}4) \le 5$ $0< b^{2-}4 \le 32$ (i) $b^{2-}4>0 \Rightarrow b \in (-\infty ,-2) ∪(2, \infty )$ (ii) $b^{2-}36 \le 0 \Rightarrow b \in [-6,6]$ Intersection of above two sets $b \in [-6,-2) ∪(2,6]$

Q70JEE Main 2022MCQ4MApplication of Derivatives

If $a=\lim _{n \to \infty } sum_{k=1}^{n} \;\frac{2 n}{n^{2+}k^2}$ and $f(x)=\sqrt{\;\frac{1-\cos x}{1+\cos x}}, x \in (0,1)$ , then : [26-Jul-2022-Shift-1]

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📖 Explanation

$a=\lim _{n \to \infty } sum_{k=1}^n \;\frac{2 n}{n^{2+}k^2}$ $=\lim _{n \to \infty } \;\frac{1}{n} sum_{k=1}^n \;\frac{2}{1+(\;\frac{k}{n})^2}$ $a=int_0^1 \;\frac{2}{1+x^2} d x=2 \tan ^{-1} x int_0^1=\;\frac{\pi}{2}$ $f(x)=\sqrt{\;\frac{1-\cos x}{1+\cos x}}, x \in (0,1)$ $f(x)=\;\frac{1-\cos x}{\sin x}={\text{cosec}} x-\cot x$ $f^{'}(x)={\text{cosec}}{ }^2 x-{\text{cosec}} x \cot x$ $\text{table} f(\;\frac{a}{2})\;=f(\;\frac{\pi}{4})=\sqrt{2}-1 ;f^{'}(\;\frac{a}{2})=f^{'}(\;\frac{\pi}{4})=2-\sqrt{2} \} f^{'}(\;\frac{a}{2})=\sqrt{2} \cdot f(\;\frac{a}{2})$

Q71JEE Main 2022MCQ4MDifferential Equations

If $\;\frac{d y}{d x}+2 y \tan x=\sin x, 0 < x < \;\frac{\pi}{2}$ and $y(\;\frac{\pi}{3})=0$ , then the maximum value of $y(x)$ is : [26-Jul-2022-Shift-1]

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📖 Explanation

$\;\frac{d y}{d x}+2 y \tan x=\sin x$ which is a first order linear differential equation.Integrating factor (I. F.) $=e^{\int 2 \tan x d x}$ $=e^{2 \ln |\sec x|}=\sec ^2 x$ Solution of differential equation can be written as $y \cdot \sec ^2 x=\int \sin x \cdot \sec ^2 x d x=\int \sec x \cdot \tan x d x$ $y \sec ^2 x=\sec x+C$ $y(\;\frac{\pi}{3})=0,0=\sec \;\frac{\pi}{3}+C \Rightarrow C=-2$ $y=\;\frac{\sec x-2}{\sec ^2 x}=\cos x-2 \cos ^2 x$ $=\;\frac{1}{8}-2(\cos x-\;\frac{1}{4})^2$ $y_{\max }=\;\frac{1}{8}$

Q72JEE Main 2022MCQ4MCircles

A point $P$ moves so that the sum of squares of its distances from the points $(1,2)$ and $(-2,1)$ is 14 . Let $f(x, y)=0$ be the locus of ${\text{P}}$ , which intersects the $x$ -axis at the points ${\text{A}}, {\text{B}}$ and the $y$ -axis at the points ${\text{C}}, {\text{D}}$ . Then the area of the quadrilateral ${\text{ACBD}}$ is equal to : [26-Jul-2022-Shift-1]

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📖 Explanation

Let point $P:(h, k)$ $(h-1)^{2+}(k-2)^{2+}(h+2)^{2+}(k-1)^2=14$ $2 h^{2+}2 k^{2+}2 h-6 k-4=0$ Locus of $P: x^{2+}y^{2+}x-3 y-2=0$ Intersection with ${\text{x}}$ -axis, $x^{2+}x-2=0$ $\Rightarrow x=-2,1$ Intersection with y-axis, $y^{2-}3 y-2=0$ $\Rightarrow y=\;\frac{3 \pm \sqrt{17}}{2}$ Area of the quadrilateral ${\text{ACBD}}$ is $=\;\frac{1}{2}(|x_1|+|x_2|)(|y_1|+|y_2|)$ $=\;\frac{1}{2} \times 3 \times \sqrt{17}=\;\frac{3 \sqrt{17}}{2}$

Q73JEE Main 2022MCQ4MHyperbola

Let the tangent drawn to the parabola $y^2=24 x$ at the point $(\alpha, \beta)$ is perpendicular to the line $2 x+2 y=5$ . Then the normal to the hyperbola $\;\frac{x^2}{\alpha^2}-\;\frac{y^2}{\beta^2}=1$ at the point $(\alpha+4, \beta+4)$ does NOT pass through the point: [26-Jul-2022-Shift-1]

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📖 Explanation

Any tangent to $y^2=24 x$ at $(\alpha, \beta)$ $\beta y=12(x+\alpha)$ Slope $=\;\frac{12}{\beta}$ and perpendicular to $2 x+2 y=5$ $\Rightarrow \;\frac{12}{\beta}=1 \Rightarrow \beta=12, \alpha=6$ Hence hyperbola is $\;\frac{x^2}{36}-\;\frac{y^2}{144}=1$ and normal is drawn at $(10,16)$ Equation of normal $\;\frac{36 . x}{10}+\;\frac{144 \cdot y}{16}=36+144$ $\Rightarrow \;\frac{x}{50}+\;\frac{y}{20}=1$ This does not pass though $(15,13)$ out of given option.

Q74JEE Main 2022MCQ4MThree Dimensional Geometry

The length of the perpendicular from the point $(1,-2,5)$ on the line passing through $(1,2,4)$ and parallel to the line $x+y-z=0=x-2 y+3 z-5$ is : [26-Jul-2022-Shift-1]

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📖 Explanation

The line $x+y-z=0=x-2 y+3 z-5$ is parallel to the vector $\vec{b}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 1 & -2 & 3\end{vmatrix}=(1,4,-3)$ Equation of the line through $P(1,2,4)$ and parallel to $\vec{b}$ $\;\frac{x-1}{1}=\;\frac{y-2}{-4}=\;\frac{z-4}{-3}$ Let $N \equiv (\lambda +1,-4 \lambda +2,-3 \lambda +4)$ $\vec{Q N}=(\lambda ,-4 \lambda +4,-3 \lambda -1)$ $\vec{Q N}$ is perpendicular to $\vec{b}$ $\Rightarrow (\lambda ,-4 \lambda +4,-3 \lambda -1) \cdot (1,4,-3)=0$ $\Rightarrow \lambda =\;\frac{1}{2}$ Hence $\vec{Q N}=(\;\frac{1}{2}, 2, \;\frac{-5}{2})$ and $|\vec{Q N}|=\sqrt{\;\frac{21}{2}}$

Explanation Figure 1

Q75JEE Main 2022MCQ4MVector Algebra

Let ${ \vec{\text{a}}=\alpha \hat{i}+\hat{j}-\hat{k}$ and ${ \vec{\text{b}}=2 \hat{i}+\hat{j}-\alpha \hat{k}, \alpha > 0$ . If the projection of ${ \vec{\text{a}} \times { \vec{\text{b}}$ on the vector $-\hat{i}+2 \hat{j}-2 \hat{k}$ is 30 , then $\alpha$ is equal to : [26-Jul-2022-Shift-1]

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📖 Explanation

Given : $\vec{a}=(\alpha, 1,-1)$ and $\vec{b}=(2,1,-\alpha)$ $\vec{c}=\vec{a} \times \vec{b}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ \alpha & 1 & -1 \\ 2 & 1 & -\alpha\end{vmatrix}$ $=(-\alpha+1) \hat{i}+(\alpha^{2-}2) \hat{j}+(\alpha-2) \hat{k}$ Projection of $\vec{c}$ on $\vec{d}=-\hat{i}+2 \hat{j}-2 \hat{k}$ $.=|\vec{c} \cdot \;\frac{\vec{d}}{|d|}|=30 \;\text{ Given }\;\}$ $\Rightarrow =|\;\frac{\alpha-1-4+2 \alpha^{2-}2 \alpha+4}{\sqrt{1+4+4}}|=30$ On solving $\alpha=\;\frac{-13}{2}$ (Rejected as $.\alpha>0)$ and $\alpha=7$

Q76JEE Main 2022MCQ4MProbability

The mean and variance of a binomial distribution are $\alpha$ and $\;\frac{\alpha}{3}$ respectively. If ${\text{P}}(X=1)=\;\frac{4}{243}$ , then ${\text{P}}(X=4$ or 5 $)$ is equal to : [26-Jul-2022-Shift-1]

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📖 Explanation

Given, mean $=n p=\alpha$ .and variance $=n p q=\;\frac{\alpha}{3}$ $\Rightarrow q=\;\frac{1}{3}$ and $p=\;\frac{2}{3}$ $P(X=1)=n \cdot p^1 \cdot q^{n-1}=\;\frac{4}{243}$ $\Rightarrow n \cdot \;\frac{2}{3} \cdot (\;\frac{1}{3})^{n-1}=\;\frac{4}{243}$ $\Rightarrow n=6$ $P(X=4$ or 5 $)={ }^6 C_4 \cdot (\;\frac{2}{3})^4 \cdot (\;\frac{1}{3})^{2+}{ }^6 C_5 \cdot (\;\frac{2}{5})^5 \cdot \;\frac{1}{3}$ $=\;\frac{16}{27}$

Q77JEE Main 2022MCQ4MProbability

Let ${\text{E}}_1, {\text{E}}_2, {\text{E}}_3$ be three mutually exclusive events such that ${\text{P}}( {\text{E}}_1)=\;{2+3 {\text{p}}}/{6}, {\text{P}}( {\text{E}}_2)=\;{2- {\text{p}}}/{8}$ and ${\text{P}}( {\text{E}}_3)=\;{1- {\text{p}}}/{2}$ . If the maximum and minimum values of ${\text{p}}$ are ${\text{p}}_1$ and ${\text{p}}_2$ , then $( {\text{p}}_1+ {\text{p}}_2)$ is equal to [26-Jul-2022-Shift-1]

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📖 Explanation

$0 \le \;\frac{2+3 P}{6} \le 1 \Rightarrow P \in [-\;\frac{2}{3}, \;\frac{4}{3}]$ $0 \le \;\frac{2-P}{8} \le 1 \Rightarrow P \in [-6,2]$ $0 \le \;\frac{1-P}{2} \le 1 \Rightarrow P \in [-1,1]$ $0< P(E_1)+P(E_2)+P(E_3) \le 1$ $0< \;\frac{13}{12}-\;\frac{P}{8} \le 1$ $P \in [\;\frac{2}{3}, \;\frac{26}{3}]$ Taking intersection of all $P \in [\;\frac{2}{3}, 1)$ $P_1+P_2=\;\frac{5}{3}$

Q78JEE Main 2022MCQ4MTrigonometric Ratios and Identities

Let $S=\{\theta \in [0,2 \pi]: 8^{2 \sin ^2 \theta}+8^{2 \cos ^2 \theta}=16\}$ . Then $n(s)+sum_{\theta \in S}(\sec (\;\frac{\pi}{4}+2 \theta) {\text{cosec}}(\;\frac{\pi}{4}+2 \theta))$ is equal to: [26-Jul-2022-Shift-1]

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📖 Explanation

$S=\{\theta \in [0,2 \pi]: 8^{2 \sin ^2 \theta}+8^{2 \cos ^2 \theta}=16\}$ Now apply AM $\ge$ GM for $8^{2 \sin ^2 \theta}, 8^{2 \cos ^2 \theta}$ $\;\frac{8^{2 \sin ^2 \theta}+8^{2 \cos ^2 \theta}}{2} \ge (8^{2 \sin ^2 \theta+2 \cos ^2 \theta})^{\;\frac{1}{2}}$ $8 \ge 8$ $\Rightarrow 8^{2 \sin ^2 \theta}=8^{2 \cos ^2 \theta}$ $\;\text{ or }\; \sin ^2 \theta=\cos ^2 \theta$ $\therefore \theta=\;\frac{\pi}{4}, \;\frac{3 \pi}{4}, \;\frac{5 \pi}{4}, \;\frac{7 \pi}{4}$ $n(S)+sum_{\theta \in S} \sec (\;\frac{\pi}{4}+2 \theta) \cos e c(\;\frac{\pi}{4}+2 \theta)$ $4+sum_{\theta \in S} \;\frac{2}{2 \sin (\;\frac{\pi}{4}+2 \theta) \cos (\;\frac{\pi}{4}+2 \theta)}$ $=4+sum_{\theta \in S} \;\frac{2}{\sin (\;\frac{\pi}{2}+4 \theta)}=4+2 sum_{\theta \in S} {\text{cosec}}(\;\frac{\pi}{2}+4 \theta)$ $=4+2[{\text{cosec}}(\;\frac{\pi}{2}+\pi) {\text{cosec}}(\;\frac{\pi}{2}+3 \pi)+{\text{cosec}}(\;\frac{\pi}{2}+5 \pi)+{\text{cosec}}(\;\frac{\pi}{2}+7 \pi)]$ $=4+2[-{\text{cosec}} \;\frac{\pi}{2}-\cos e c \;\frac{\pi}{2}-\cos e c \;\frac{\pi}{2}-{\text{cosec}} \;\frac{\pi}{2}]$ $=4-2(4)$ $=4-8$ $=-4$

Q79JEE Main 2022MCQ4MInverse Trigonometric Functions

$\tan (2 \tan ^{-1} \;\frac{1}{5}+\sec ^{-1} \;\frac{\sqrt{5}}{2}+2 \tan ^{-1} \;\frac{1}{8})$ is equal to : [26-Jul-2022-Shift-1]

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📖 Explanation

$\tan (2 \tan ^{-1} \;\frac{1}{5}+\sec ^{-1} \;\frac{\sqrt{5}}{2}+2 \tan ^{-1} \;\frac{1}{8})$ $=\tan (2 \tan ^{-1}(\;\frac{\;\frac{1}{5}+\frac{1}{8}}{1-\;\frac{1}{5} \cdot \;\frac{1}{8}})+\sec ^{-1} \;\frac{\sqrt{5}}{2})$ $=\tan [2 \tan ^{-1} \;\frac{1}{3}+\tan ^{-1} \;\frac{1}{2}]$ $=\tan [\tan ^{-1} \;\frac{\;\frac{2}{3}}{1-\;\frac{1}{9}}+\tan ^{-1} \;\frac{1}{2}]$ $=\tan [\tan ^{-1} \;\frac{3}{4}+\tan ^{-1} \;\frac{1}{2}]$ $=\tan \$ [\tan ^{-1} ;\frac{;\frac{3}{4}+\frac{1}{2}}{1-;\frac{3}{8}}]$=\tan [\tan ^{-1} ;\frac{;\frac{5}{4}}{\frac{5}{8}}]=\tan $[\tan ^{-1} 2]$=2$

Q80JEE Main 2022MCQ4MSets and Relations

The statement $(∼( {\text{p}} ⇔ ∼ {\text{q}})) ∧ {\text{q}}$ is : [26-Jul-2022-Shift-1]

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📖 Explanation

$∼(p ⇔ ∼ q) ∧ q$ $=(p ⇔ q) ∧ q$ p q p ⟷ q (p → q)∧ q (p → q) (p → q) ∧q (p → q) ∧ p T T T T T T T T F F F F F F F T F F T T F F F T F T F F $\therefore (∼(p ⇔ ∼ q)) ∧ q$ is equivalent to $(p \Rightarrow q) ∧ p$ .

Q81JEE Main 2022NAT4MQuadratic Equation and Inequalities

If for some ${\text{p}}, {\text{q}}, {\text{r}} \in R$ , not all have same sign, one of the roots of the equation $( {\text{p}}^{2+} {\text{q}}^2) x^{2-}2 {\text{q}}( {\text{p}}+ {\text{r}}) x+ {\text{q}}^{2+} {\text{r}}^2=0$ is also a root of the equation $x^{2+}2 x-8=0$ , then $\;{ {\text{q}}^{2+} {\text{r}}^2}/{ {\text{p}}^2}$ is equal to ________. [26-Jul-2022-Shift-1]

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📖 Explanation

$\therefore \alpha+\beta>0$ and $\alpha \beta>0$ Also, it has a common root with $x^{2+}2 x-8=0$ $\therefore$ The common root between above two equations is 4 . $\Rightarrow 16(p^{2+}q^2)-8 q(p+r)+q^{2+}r^2=0$ $\Rightarrow (16 p^{2-}8 p q+q^2)+(16 q^{2-}8 q r+r^2)=0$ $\Rightarrow (4 p-q)^{2+}(4 q-r)^2=0$ $\Rightarrow q=4 p \;\text{ and }\; r=16 p$ $\therefore \;\frac{q^{2+}r^2}{p^2}=\;\frac{16 p^{2+}256 p^2}{p^2}=272$

Explanation Figure 1

Q82JEE Main 2022NAT4MPermutations and Combinations

The number of 5-digit natural numbers, such that the product of their digits is 36 , is _______. [26-Jul-2022-Shift-1]

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📖 Explanation

Factors of $36=2^2 \cdot 3^2 \cdot 1$ Five-digit combinations can be $(1,2,2,3,3)(1,4,3,3,1),(1,9,2,2,1)$ $(1,4,9,11)(1,2,3,6,1)(1,6,6,1,1)$ i.e., total numbers $\;\frac{5 !}{2 ! 2 !}+\;\frac{5 !}{2 ! 2 !}+\;\frac{5 !}{2 ! 2 !}+\;\frac{5 !}{3 !}+\;\frac{5 !}{2 !}+\;\frac{5 !}{3 ! 2 !}$ $=(30 \times 3)+20+60+10=180$

Q83JEE Main 2022NAT4MSequences and Series

The series of positive multiples of 3 is divided into sets : $\{3\},\{6,9,12\},\{15,18,21,24,27\}, . . .$ Then the sum of the elements in the $11 ^{\;\text{th }\;}$ set is equal to __________. [26-Jul-2022-Shift-1]

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📖 Explanation

$\{3\times 1\}_{1-\text{term}},\{3\times 2,3\times 3,3\times 4\}_{3 \text{terms}}, {\{3\times 5,3\times 6,3\times 7,3\times 8 ,3\times 9\}}_{5-\text{terms}}, . . .$ $\therefore 11^{\;\text{th }\;}$ set will have $1+(10) 2=21$ termAlso upto $10^{\;\text{th }\;}$ set total $3 \times k$ type terms will be $1+3+5+. . . . . .+19=100-$ term $\therefore$ Set $11=\{3 \times 101,3 \times 102, . . . . . . 3 \times 121\}$ $\therefore$ Sum of elements $=3 \times (101+102+. . .+121)$ $=\;\frac{3 \times 222 \times 21}{2}=6993$

Q84JEE Main 2022NAT4MApplication of Derivatives

The number of distinct real roots of the equation $x^5(x^{3-}x^{2-}x+1)+x(3 x^{3-}4 x^{2-}2 x+4)-1=0 \;\text{ is }\;$ ____________. [26-Jul-2022-Shift-1]

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📖 Explanation

$x^{8-}x^{7-}x^{6+}x^{5+}3 x^{4-}4 x^{3-}2 x^{2+}4 x-1=0$ $\Rightarrow x^7(x-1)-x^5(x-1)+3 x^3(x-1)-x(x^{2-}1)+2 x(1-x)+(x-1)=0$ $\Rightarrow (x-1)(x^{7-}x^{5+}3 x^{3-}x(x+1)-2 x+1)=0$ $\Rightarrow (x-1)(x^{7-}x^{5+}3 x^{3-}x^{2-}3 x+1)=0$ $\Rightarrow (x-1)(x^5(x^{2-}1)+3 x(x^{2-}1)-1(x^{2-}1))=0$ $\Rightarrow (x-1)(x^{2-}1)(x^{5+}3 x-1)=0$ $\therefore x=\pm 1$ are roots of above equation and $x^{5+}3 x-1$ is a monotonic term hence vanishs at exactly one value of $x$ other than 1 or $-1$ . $\therefore 3$ real roots.

Q85JEE Main 2022NAT4MBinomial Theorem

If the coefficients of $x$ and $x^2$ in the expansion of $(1+x)^{ {\text{p}}}(1-x)^{ {\text{q}}}, {\text{p}}, {\text{q}} \le 15$ , are $-3$ and $-5$ respectively, then the coefficient of $x^3$ is equal to _________. [26-Jul-2022-Shift-1]

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📖 Explanation

Coefficient of ${\text{x}}$ in $(1+x)^p(1-x)^q$ ${ }^p C_0{ }^q C_1+{ }^p C_1{ }^q C_0=-3 \Rightarrow p-q=-3$ Coefficient of ${\text{x}}^2$ in $(1+x)^p(1-x)^q$ ${ }^p C_0{ }^q C_2-{ }^p C_1{ }^q C_1+{ }^p C_2{ }^q C_0=-5$ $\;\frac{q(q-1)}{2}-p q+\;\frac{p(q-1)}{2}=-5$ $\;\frac{q^{2-}q}{2}-(q-3) q+\;\frac{(q-3)(q-4)}{2}=-5$ $\Rightarrow q=11, p=8$ Coefficient of ${\text{x}}^3$ in $(1+x)^8(1-x)^{11}$ $=-{ }^{11} C_3+{ }^8 C_1{ }^{11} C_2-{ }^8 C_2{ }^{11} C_1+{ }^8 C_3=23$

Q86JEE Main 2022NAT4MDefinite Integrals

If ${\text{n}}(2 {\text{n}}+1) int_0^1(1-x^{ {\text{n}}})^{2 {\text{n}}} {\text{d}} x=1177 int_0^1(1-x^{ {\text{n}}})^{2 {\text{n}}+1} {\text{d}} x$ , then ${\text{n}} \in N$ is equal to ________. [26-Jul-2022-Shift-1]

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📖 Explanation

$int_0^1(1-x^n)^{2 n+1} d x=int_0^1 1 \cdot (1-x^n)^{2 n+1} d x$ $=\$ [(1-x^n)^{2 n+1} ⋅ x]_0^{1-}int_0^1 x ⋅(2 n+1)(1-x^n)^{2 n} ⋅-n x^{n-1} d x=n(2 n+1) int_0^1(1-(1-x^n))(1-x^n)^{2 n} d x=n(2 n+1) int_0^1(1-x^n)^{2 n} d x-n(2 n+1) int_0^1(1-x^n)^{2 n+1} d x(1+n(2 n+1)) int_0^1(1-x^n)^{2 n+1} d x=n(2 n+1) int_0^1(1-x^n)^{2 n} d x(2 n^{2+}n+1) int_0^1(1-x^n)^{2 n+1} d x=1177 int_0^1(1-x^n)^{2 n+1} d x∴ 2 n^{2+}n+1=11772 n^{2+}n-1176=0∴ n=24 $or$ -;\frac{49}{2}∴ n=24$

Q87JEE Main 2022NAT4MDifferential Equations

Let a curve $y=y(x)$ pass through the point $(3,3)$ and the area of the region under this curve, above the $x$ -axis and between the abscissae 3 and $x( > 3)$ be $(\;\frac{y}{x})^3$ . If this curve also passes through the point $(\alpha, 6 \sqrt{10})$ in the first quadrant, then $\alpha$ is equal to __________. [26-Jul-2022-Shift-1]

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📖 Explanation

$int_3^x f(x) d x=(\;\frac{f(x)}{x})^3$ $x^3 \cdot int_3^x f(x) d x=f^3(x)$ Differentiate w.r.t. $x$ $x^3 f(x)+3 x^2 \cdot \;\frac{f^3(x)}{x^3}=3 f^2(x) f^{'}(x)$ $\Rightarrow 3 y^2 \;\frac{d y}{d x}=x^3 y+\;\frac{3 y^3}{x}$ $3 x y \;\frac{d y}{d x}=x^{4+}3 y^2$ $\;\text{ Let }\; y^2=t$ $\;\frac{3}{2} \;\frac{d t}{d x}=x^{3+}\;\frac{3 t}{x}$ $\;\frac{d t}{d x}-\;\frac{2 t}{x}=\;\frac{2 x^3}{3}$ $\;\text{ I. }\; F .=. e^{\int-\;\frac{2}{x} d x}=\;\frac{1}{x^2}$ Solution of differential equationt. $\;\frac{1}{x^2}=\int \;\frac{2}{3} x d x$ $\;\frac{y^2}{x^2}=\;\frac{x^2}{3}+C$ $y^2=\;\frac{x^4}{3}+C x^2$ Curve passes through $(3,3) \Rightarrow C=-2$ $y^2=\;\frac{x^4}{3}-2 x^2$ Which passes through $(\alpha, 6 \sqrt{10})$ $\;\frac{\alpha^{4-}6 \alpha^2}{3}=360$ $\alpha^{4-}6 \alpha^{2-}1080=0$ $\alpha=6$

Q88JEE Main 2022NAT4MStraight Lines and Pair of Straight Lines

The equations of the sides ${\text{AB}}, {\text{BC}}$ and ${\text{CA}}$ of a triangle ${\text{ABC}}$ are $2 x+y=0, x+ {\text{p}} y=15 {\text{a}}$ and $x-y=3$ respectively. If its orthocentre is $(2, a),-\;\frac{1}{2} < {\text{a}} < 2$ , then ${\text{p}}$ is equal to _________. [26-Jul-2022-Shift-1]

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📖 Explanation

Slope of $A H=\;\frac{a+2}{1}$ Slope of $B C=-\;\frac{1}{p}$ $\therefore p=a+2 . . . . . . \;\text{ (i) }\;$ Coordinate of $C=(\;\frac{18 p-30}{p+1}, \;\frac{15 p-33}{p+1})$ Slope of $H C=\;\frac{\;\frac{15 p-33}{p+3}-a}{\frac{18 p-33}{p+1}-2}=\;\frac{15 p-33-(p-2)(p+1)}{18 p-30-2 p-2}$ $=\;\frac{16 p-p^{2-}31}{16 p-32}$ $\because \;\frac{16 p-p^{2-}31}{16 p-32} \times -2=-1$ $\therefore p^{2-}8 p+15=0$ $\therefore p=3 \;\text{ or }\; 5$ But if $p=5$ then $a=3$ not acceptable $\therefore p=3$

Explanation Figure 1

Q89JEE Main 2022NAT4MParabola

Let the function $f(x)=2 x^{2-}\log _{ {\text{e}}} x, x > 0$ , be decreasing in $(0, {\text{a}})$ and increasing in $( {\text{a}}, 4)$ . A tangent to the parabola $y^2=4 a x$ at a point ${\text{P}}$ on it passes through the point $(8 {\text{a}}, 8 {\text{a}}-1)$ but does not pass through the point $(-\;\frac{1}{a}, 0)$ . If the equation of the normal at $P$ is : $\;\frac{x}{\alpha}+\;\frac{y}{\beta}=1$ , then $\alpha+\beta$ is equal to _________. [26-Jul-2022-Shift-1]

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📖 Explanation

$\delta ^{'}(x)=\;\frac{4 x^{2-}1}{x}$ so $f(x)$ is decreasing in $(0, \;\frac{1}{2})$ and increasing in $(\;\frac{1}{2}, \infty ) \Rightarrow a=\;\frac{1}{2}$ Tangent at $y^2=2 x \Rightarrow y=, x+\;\frac{1}{2 m}$ It is passing through $(4,3)$ $3=4 m+\;\frac{1}{2 m} \Rightarrow m=\;\frac{1}{2}$ or $\;\frac{1}{4}$ So tangent may be $y=\;\frac{1}{2} x+1$ or $y=\;\frac{1}{4} x+2$ But $y=\;\frac{1}{2} x+1$ passes through $(-2,0)$ so rejected.Equation of normal $y=-4 x-2(\;\frac{1}{2})(-4)-\;\frac{1}{2}(-4)^3$ or $y=-4 x+4+32$ or $\;\frac{x}{9}+\;\frac{y}{36}=1$

Q90JEE Main 2022NAT4MThree Dimensional Geometry

Let $Q$ and $R$ be two points on the line $\;\frac{x+1}{2}=\;\frac{y+2}{3}=\;\frac{z-1}{2}$ at a distance $\sqrt{26}$ from the point $P(4,2,7)$ . Then the square of the area of the triangle $P Q R$ is _________. [26-Jul-2022-Shift-1]

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📖 Explanation

$L: \;\frac{x+1}{2}=\;\frac{y+2}{3}=\;\frac{2-1}{2}$ Let $T(2 t-1,3 t-2,2 t+1)$ $\because P T ⟂^r Q R$ $\therefore 2(2 t-5)+3(3 t-4)+2(2 t-6)=0$ $17 t=34$ $\therefore t=2$ So $T(3,4,5)$ $\therefore P T=\sqrt{1+4+4}=3$ $\therefore Q T=\sqrt{26-9}=\sqrt{17}$ $\therefore$ Area of $\triangle P Q R=\;\frac{1}{2} \times 2 \sqrt{17} \times 3=3 \sqrt{17}$ $\therefore$ Square of $a r(\triangle P Q R)=153$

Explanation Figure 1

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