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28 Jul 2022 Shift 2

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Physics30 questions

Q1JEE Main 2022MCQ4MUnits and Measurements

Consider the efficiency of carnot's engine is given by $η=\;\frac{\alpha \beta}{\ sin \theta} \log e \;\frac{\beta x}{k T}$ , where $\alpha$ and $\beta$ are constants. If ${\text{T}}$ is temperature, ${\text{k}}$ is Boltzmann constant, $\theta$ is angular displacement and ${\text{x}}$ has the dimensions of length. Then, choose the incorrect option: [28-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

(A) $[\beta]=[\;\frac{k T}{x}]=[\;\frac{E}{x}]=\$ [M L T^{-2}]$=[F] $(B)$ [\alpha \beta]=[M^0 L^0 T^0][\alpha]^{-1}=[\beta]=[;\frac{k T}{x}] $So$ [\alpha]^{-1}[x]=[k T]=[M L^2 T^{-2}] $(C)$ η \ sin \theta=\alpha \beta $So$ [η \ sin \theta]=[\alpha \beta][η]=[M^0 L^0 T^0] $it is dimensionless quantity(D)$ [\alpha] \ne [\beta]$

Q2JEE Main 2022MCQ4MMotion in a Plane

At time $t=0$ a particle starts travelling from a height $7^{} \hat{z} {\text{cm}}$ in a plane keeping $z$ coordinate constant. At any instant of time it's position along the $\hat{x}$ and $\hat{y}$ directions are defined as $3 {\text{t}}$ and $5 {\text{t}}^3$ respectively. At ${\text{t}}=1 {\ s}$ acceleration of the particle will be [28-Jul-2022-Shift-2]

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📖 Explanation

$x=3 t \Rightarrow a_x=0$ $y=5 t^3 \Rightarrow a_y=30 t$ $\vec{a}(t=1)=30 \hat{y}$

Q3JEE Main 2022MCQ4MProperties of Matter

A pressure-pump has a horizontal tube of cross sectional area $10 {\text{cm}}^2$ for the outflow of water at a speed of $20 {\text{m}} / {\ s}$ . The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is :$[given: density of water $=1000 {\text{kg}} / {\text{m}}^3$ ]$ [28-Jul-2022-Shift-2]

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📖 Explanation

$F_w=\rho A v^2$ $=10^3 \times 10 \times 10^{-4} \times 20 \times 20$ $=400 N$

Q4JEE Main 2022MCQ4MLaws of Motion

A uniform metal chain of mass $m$ and length 'L' passes over a massless and frictionless pulley. It is released from rest with a part of its length 'l' is hanging on one side and rest of its length ' ${\text{L}}-l$ ' is hanging on the other side of the pully. At a certain point of time, when $l=\;\frac{L}{x}$ , the acceleration of the chain is $\;\frac{g}{2}$ . The value of ${\text{x}}$ is ________. [28-Jul-2022-Shift-2]

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📖 Explanation

$a=\;\frac{m_2-m_1}{m_1+m} g=\;\frac{g}{2}$ $\Rightarrow m_2=3 m_1$ $(L-l)=3 l$ $L=4 l$ $l=\;\frac{L}{4}$

Q5JEE Main 2022MCQ4MWork, Power and Energy

A bullet of mass $200 {\text{g}}$ having initial kinetic energy $90 {\text{J}}$ is shot inside a long swimming pool as shown in the figure. If it's kinetic energy reduces to $40 {\text{J}}$ within $1 {\ s}$ , the minimum length of the pool, the bullet has to travel so that it completely comes to rest is [28-Jul-2022-Shift-2]

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📖 Explanation

$\;\frac{1}{2} m x^2=90$ $\Rightarrow \;\frac{1}{2} \times 0.2 \times x^2=90$ $x^2=900$ $x=30 {\text{m}} / {\ s}$ $\;\frac{1}{2} m v^2=40 \Rightarrow v=\;\frac{2}{3} \times 30=20 {\text{m}} / {\ s}$ $20=30-a \times 1 \Rightarrow a=-10 {\text{m}} / {\ s}^2$ $0-x^2=2 a s$ $s=\;\frac{x^2}{-2 a}=\;\frac{30 \times 30}{2 \times 10}$ $=45 {\text{m}}$

Q6JEE Main 2022MCQ4MGravitation

Assume there are two identical simple pendulum clocks. Clock $-1$ is placed on the earth and Clock - 2 is placed on a space station located at a height ${\text{h}}$ above the earth surface. Clock - 1 and Clock - 2 operate at time periods $4 {\ s}$ and $6 s$ respectively. Then the value of $h$ is___(consider radius of earth $R_E=6400 {\text{km}}$ and ${\text{g}}$ on earth $10 {\text{m}} / {\ s}^2$ ) [28-Jul-2022-Shift-2]

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📖 Explanation

$T \propto \sqrt{1 / g}$ $\Rightarrow \;\frac{T_1}{T_2}=\sqrt{\;\frac{g_2}{g_1}}=\;\frac{R}{R+h}$ $\;\frac{4}{6}=\;\frac{R}{R+h}$ $\Rightarrow h=R / 2$ $=3200 {\text{km}}$

Q7JEE Main 2022MCQ4MProperties of Matter

Consider a cylindrical tank of radius $1 {\text{m}}$ is filled with water. The top surface of water is at $15 {\text{m}}$ from the bottom of the cylinder. There is a hole on the wall of cylinder at a height of $5 {\text{m}}$ from the bottom. A force of $5 \times 10^5 {\text{N}}$ is applied an the top surface of water using a piston. The speed of ifflux from the hole will be : (given atmospheric pressure ${\text{P}}_{ {\text{A}}}=1.01 \times 10^5 {\text{Pa}}$ , density of water $\rho _{ {\text{W}}}=1000 {\text{kg}} / {\text{m}}^3$ and gravitational acceleration ${\text{g}}=10 {\text{m}} / {\ s}^2$ ) [28-Jul-2022-Shift-2]

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📖 Explanation

By Bernoulli's theorem, $\;\frac{5 \times 10^5}{\pi(1)^2}+\rho g(10)=1.01 \times 10^{5+}\;\frac{1}{2} \rho (v)^2$ $\Rightarrow v^2=200+\;\frac{10^6}{1000 \pi}-202$ $\Rightarrow v ≃ 17.8 {\text{m}} / {\ s}$

Q8JEE Main 2022MCQ4MThermodynamics

A vessel contains $14 {\text{g}}$ of nitrogen gas at a temperature of $27^{\circ} {\text{C}}$ . The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be :Take ${\text{R}}=8.32 {\text{J}} {\text{mol}}^{-1} {\text{k}}^{-1}$ [28-Jul-2022-Shift-2]

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📖 Explanation

$n=0.5$ $T=300$ For ${\text{v}}_{ {\text{rms}}}$ to be doubled ${\text{T}}^{'}=4 \times 300=1200$ $\Rightarrow$ Heat transferred $=(0.5)(\;\frac{5}{2})(8.32)(900)$ $=9360 {\text{J}}$

Q9JEE Main 2022MCQ4MElectrostatic Potential and Capacitance

A slab of dielectric constant ${\text{K}}$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\;\frac{3}{4} {\text{d}}$ , where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :(Given ${\text{C}}_0=$ capacitance of capacitor with air as medium between plates.) [28-Jul-2022-Shift-2]

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📖 Explanation

$C_0=\;\frac{\epsilon_0 A}{d}$ $C=\;\frac{\epsilon_0 A}{d-\;\frac{3 d}{4}+\;\frac{3 d}{4 K}}=\;\frac{4 \epsilon_0 A K}{3 d+K d}$ $=\;\frac{4 K C_0}{3+K}$

Q10JEE Main 2022MCQ4MElectrostatics

A uniform electric field ${\text{E}}=(8 {\text{m}} / {\text{e}}) {\text{V}} / {\text{m}}$ is created between two parallel plates of length $1 {\text{m}}$ as shown in figure, (where ${\text{m}}=$ mass of electron and ${\text{e}}=$ charge of electron). An electron enters the field symmetrically between the plates with a speed of $2 {\text{m}} / {\ s}$ . The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be________. [28-Jul-2022-Shift-2]

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📖 Explanation

$E=\;\frac{8 m}{e} {\text{V}} / {\text{m}}$ $l=1 {\text{m}}$ $v_x=2 {\text{m}} / {\ s}$ $a_y=-8 {\text{m}} / {\ s}^2$ $t=\;\frac{l}{v_x}=\;\frac{1}{2} s$ $\Rightarrow |v_y|=4 {\text{m}} / {\ s}$ $\Rightarrow$ angle of deviation $=\theta$ $\tan \theta=\;\frac{v_y}{v_x}$ $\theta=\tan ^{-1}(\;\frac{4}{2})=\tan ^{-1}(2)$

Q11JEE Main 2022MCQ4MCurrent Electricity

Given below are two statements :Statement I : A uniform wire of resistance $80 Ω$ is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be $5 Ω$ .Statement II: Two resistances $2 {\text{R}}$ and $3 {\text{R}}$ are connected in parallel in a electric circuit. The value of thermal energy developed in $3 R$ and $2 R$ will be in the ratio $3: 2$ .In the light of the above statements, choose the most appropriate answer from the option given below [28-Jul-2022-Shift-2]

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📖 Explanation

Statement I: $R_{1 \;\text{ part }\;}=\;\frac{80}{4}=20 Ω$ $\Rightarrow R_{e f f}=\;\frac{20}{4}=5 Ω$ Statement II : Ratio $=\;\frac{\;\frac{(\Delta V)^2}{3 R}}{\frac{\Delta \Delta V)^2}{2 R}}$ $=\;\frac{2}{3}$

Q12JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

A triangular shaped wire carrying $10 {\text{A}}$ current is placed in a uniform magnetic field of $0.5 {\text{T}}$ , as shown in figure. The magnetic force on segment ${\text{CD}}$ is $($ Given ${\text{BC}}= {\text{CD}}= {\text{BD}}=5 {\text{cm}}$ . [28-Jul-2022-Shift-2]

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📖 Explanation

$\vec{F}=\vec{i l} \times \vec{B}$ $=i l B \ sin 60^{\circ}$ $=10 \times \;\frac{5}{100} \times 0.5 \times \;\frac{\sqrt{3}}{2}$ $=0.2165 {\text{N}}$

Q13JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

The magnetic field at the center of current carrying circular loop is $B_1$ . The magnetic field at a distance of $\sqrt{3}$ times radius of the given circular loop from the center on its axis is $B_2$ . The value of $B_1 / B_2$ will be [28-Jul-2022-Shift-2]

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📖 Explanation

$B_1=\;\frac{\mu _0 i}{2 R}$ $B_2=\;{\mu _0 i R^2}/{2(R^{2+}x^2)^{\;\frac{3}{2}}}$ $\Rightarrow \;\frac{B_1}{B_2}=\;\frac{1}{R^3}(R^{2+}x^2)^{\;\frac{3}{2}}$ $=\;\frac{1}{R^3}(8 R^3)$ $=8$

Q14JEE Main 2022MCQ4MAlternating Current

A transformer operating at primary voltage $8 {\text{kV}}$ and secondary voltage $160 {\text{V}}$ serves a load of $80 {\text{kW}}$ . Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be [28-Jul-2022-Shift-2]

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📖 Explanation

$V_1 i_1=V_2 i_2=80 {\text{kW}}$ $\Rightarrow i_1=10 A$ and $i_2=\;\frac{80 \times 1000}{160}=500 {\text{A}}$ $\Rightarrow R_1=\;\frac{V_1}{i_1}=800 Ω$ and $R_2=\;\frac{160}{500}=0.32 Ω$

Q15JEE Main 2022MCQ4MElectromagnetic Waves

Sun light falls normally on a surface of area $36 {\text{cm}}^2$ and exerts an average force of $7.2 \times 10^{-9} {\text{N}}$ within a time period of 20 minutes. Considering a case of complete absorption, the energy flux of incident light is [28-Jul-2022-Shift-2]

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📖 Explanation

$\;{ {\text{I}}}/{ {\text{C}}} \times$ area $=$ force $\;{ {\text{I}}}/{ {\text{C}}} \times 36 \times 10^{-4}=7.2 \times 10^{-9}$ ${\text{I}}=\;\frac{7.2 \times 10^{-9} \times 3 \times 10^8}{36 \times 10^{-9} \times 10}$ $=\;\frac{6 \times 10^{-1}}{10^{-3}}$ ${\text{I}}=6 \times 10^2 \;{ {\text{w}}}/{ {\text{m}}^2}$ $=0.06 \;{ {\text{w}}}/{ {\text{cm}}^2}$

Q16JEE Main 2022MCQ4MRay Optics and Optical Instruments

The power of a lens (biconvex) is $1.25 {\text{m}}^{-1}$ in particular medium. Refractive index of the lens is $1.5$ and radii of curvature are $20 {\text{cm}}$ and $40 {\text{cm}}$ respectively. The refractive index of surrounding medium: [28-Jul-2022-Shift-2]

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📖 Explanation

$P=\;\frac{\mu _2}{f}=(\mu _1-\mu _2)(\;\frac{1}{R_1}-\;\frac{1}{R_2}) \;\;$ (For this formula refer to NCERT Part-2, Chapter-9, solved example 8 $)$ $(\mu _1.$ is refractive index of lens and $\mu _2$ is of surrounding medium) $1.25=(1.5-\mu _2)(\;\frac{1}{0.2}+\;\frac{1}{0.4})$ $\;\frac{1.25 \times 0.08}{0.6}=(1.5-\mu _2)$ $\Rightarrow \mu _2=\;\frac{4}{3}$

Q17JEE Main 2022MCQ4MDual Nature of Matter and Radiation

Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted, in the two cases respectively, will be [28-Jul-2022-Shift-2]

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📖 Explanation

$\;\frac{1}{2} m v_1^2=5 \varphi -\varphi$ And, $\;\frac{1}{2} m v_2^2=10 \varphi -\varphi$ $\Rightarrow (\;\frac{v_1}{v_2})^2=\;\frac{4}{9}$ $\Rightarrow \;\frac{v_1}{v_2}=\;\frac{2}{3}$

Q18JEE Main 2022MCQ4MAtoms and Nuclei

A radioactive sample decays $\;\frac{7}{8}$ times its original quantity in 15 minutes. The half-life of the sample is [28-Jul-2022-Shift-2]

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📖 Explanation

$N=\;{N_0}/{2^{\;\frac{t}{T_{1 / 2}}}}$ $\Rightarrow 2^{\;\frac{t}{T_{1 / 2}}}=\;\frac{N_0}{N}=\;\frac{N_0}{(\;\frac{N_0}{8})}=8$ $\Rightarrow \;\frac{t}{T_{1 / 2}}=3$ $\Rightarrow T_{1 / 2}=\;\frac{15}{3}=5 {\min}$

Q19JEE Main 2022MCQ4MSemiconductor Electronics

An n.p.n transistor with current gain $\beta=100$ in common emitter configuration is shown in figure. The output voltage of the amplifier will be [28-Jul-2022-Shift-2]

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📖 Explanation

$\;{ {\text{V}}_{\;\text{out }\;}}/{ {\text{V}}_{\;\text{\in }\;}}=\beta \;{ {\text{R}}_{\;\text{out }\;}}/{ {\text{R}}_{\;\text{\in }\;}}$ ${\text{v}}_{\;\text{out }\;}=\;\frac{100 \times 10 \times 10^3}{10^3} \times 10^{-3}$ $=1 {\text{V}}$

Q20JEE Main 2022MCQ4MElectromagnetic Waves

A FM Broadcast transmitter, using modulating signal of frequency $20 {\text{kHz}}$ has a deviation ratio of 10 . The Bandwidth required for transmission is : [28-Jul-2022-Shift-2]

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📖 Explanation

GivenFM broadcastModulating frequency $=20 {\text{kHz}}= {\text{f}}$ Deviation ratio $=\;\frac{\;\text{ frequency deviation }\;}{\;\text{ modulating frequency }\;}=\;{\Delta {\text{f}}}/{ {\text{f}}}$ $\Rightarrow$ frequency deviation $-\Delta {\text{f}}= {\text{f}} \times 10$ $=20 {\text{kHz}} \times 10=200 {\text{kHz}}$ $\Rightarrow$ Bandwidth $=2( {\text{f}}+\Delta {\text{f}})$ $=2(20+200) {\text{kHz}}$ $=440 {\text{kHz}}$

Q21JEE Main 2022NAT4MMotion in a Straight Line

A ball is thrown vertically upwards with a velocity of $19.6 {\text{ms}}^{-1}$ from the top of a tower. The ball strikes the ground after $6 {\ s}$ . The height from the ground up to which the ball can rise will be $(\;\frac{k}{5}) {\text{m}}$ . The value of ${\text{k}}$ is (use $. {\text{g}}=9.8 {\text{m}} / {\ s}^2)$ [28-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{t}}_{ {\text{a}}}=\;{ {\text{u}}}/{ {\text{g}}}=\;\frac{19.6}{9.8}=2 {\ s}$ ${\text{t}}_{ {\text{d}}}=6-2 {\ s}=\sqrt{\;{2 {\text{h}}_{\max }}/{ {\text{g}}}}$ $\Rightarrow {\text{h}}_{\max }=\;\frac{16 \times 9.8}{2}=\;\frac{392}{5}$

Q22JEE Main 2022NAT4MCentre of Mass and Collision

The distance of centre of mass from end $A$ of a one dimensional rod $(A B)$ having mass density $\rho =\rho _0(1-\;\frac{x^2}{L^2}) {\text{kg}} / {\text{m}}$ and length $L$ (in meter) is $\;\frac{3 L}{\alpha} {\text{m}}$ . The value of $\alpha$ is_____ (where ${\text{x}}$ is the distance from end ${\text{A}}$ ) [28-Jul-2022-Shift-2]

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📖 Explanation

${\text{dm}}=\lambda \cdot {\text{dx}}=\lambda _0(1-\;{ {\text{x}}^2}/{ℓ^2})$ $X_{ {\text{cm}}}=\;{\int {\text{xdm}}}/{\int {\text{dm}}_{ℓ}}$ $=\;{\lambda _0 int_{0}^{ℓ}(x-\frac{x^3}{ℓ^2}) {\text{dx}}}/{int_{0}^{ℓ} \lambda _0(1-\;\frac{x^2}{ℓ^2}) {\text{dx}}}=\;\frac{\;\frac{ℓ^2}{2}-\frac{ℓ^4}{4 ℓ^2}}{ℓ-\;\frac{ℓ^3}{3 ℓ^2}}=\;\frac{3 ℓ}{8}$

Q23JEE Main 2022NAT4MProperties of Matter

A string of area of cross-section $4 {\text{mm}}^2$ and length $0.5 {\text{m}}$ is connected with a rigid body of mass $2 {\text{kg}}$ . The body is rotated in a vertical circular path of radius $0.5 {\text{m}}$ . The body acquires a speed of $5 {\text{m}} / {\ s}$ at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is ____ $\times 10^{-5}$ (use Young's modulus $10^{11} {\text{N}} / {\text{m}}^2$ and ${\text{g}}=10 {\text{m}} / {\ s}^2$ ) [28-Jul-2022-Shift-2]

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📖 Explanation

Strain $=$ F/AY $=\;\frac{m g+\frac{m v^2}{R}}{A Y}$ $=\;\frac{20+\frac{2(5)^2}{0.5}}{3 \times 10^{-6} \times 10^{11}}=30 \times 10^{-5}$

Q24JEE Main 2022NAT4MThermodynamics

At a certain temperature, the degrees of freedom per molecule for gas is 8 . The gas performs $150 {\text{J}}$ of work when it expands under constant pressure. The amount of heat absorbed by the gas will be ________ J. [28-Jul-2022-Shift-2]

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📖 Explanation

${\text{W}}= {\text{nR}} \Delta {\text{T}}=150 {\text{J}}$ $Q=(\;{ {\text{f}}}/{2}+1) {\text{nR}} \Delta {\text{T}}=(\;\frac{8}{2}+1) 150=750 {\text{J}}$

Q25JEE Main 2022NAT4MOscillations

The potential energy of a particle of mass $4 {\text{kg}}$ in motion along the $x$ -axis is given by ${\text{U}}=4(1-\cos 4 x) {\text{J}}$ . The time period of the particle for small oscillation $(\ sin \theta ≃ \theta)$ is $(\;\frac{\pi}{K}) s$ . The value of ${\text{K}}$ is ________. [28-Jul-2022-Shift-2]

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📖 Explanation

$U=4(1-\cos 4 x)$ $F=-\;\frac{d U}{d x}=-4(+\ sin 4 x) 4=-16 \ sin (4 x)$ For small $\theta$ $\ sin \theta \approx \theta$ ${\text{F}}=-64 {\text{x}}$ $a=-64 x / m=-16 x$ $\omega ^2=16$ ${\text{T}}=\;\frac{2 \pi}{\omega }=\;\frac{\pi}{2}$

Q26JEE Main 2022NAT4MCurrent Electricity

An electrical bulb rated $220 {\text{V}}, 100 {\text{W}}$ , is connected in series with another bulb rated $220 {\text{V}}, 60 {\text{W}}$ . If the voltage across combination is $220 {\text{V}}$ , the power consumed by the $100 {\text{W}}$ bulb will be about _______ W. [28-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{R}}_1=\;{ {\text{V}}^2}/{ {\text{P}}}=\;\frac{220^2}{100}=484$ ${\text{R}}_2=\;{ {\text{V}}^2}/{ {\text{P}}}=\;\frac{220^2}{60}=484(\;\frac{10}{6})$ ${\text{I}}=\;\frac{220}{484+484 \times \;\frac{10}{6}}$ ${\text{P}}_1= {\text{I}}^2 {\text{R}}_1=14.06 {\text{W}}$

Q27JEE Main 2022NAT4MElectromagnetic Induction

For the given circuit the current through battery of $6 {\text{V}}$ just after closing the switch 'S' will be ________ A. [28-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Just after closing the switch ${\text{S}}$ , inductor behaves like an open circuit. $I=\;\frac{6}{2+4}=1 {\text{A}}$

Q28JEE Main 2022NAT4MRay Optics and Optical Instruments

An object ' $O$ ' is placed at a distance of $100 {\text{cm}}$ in front of a concave mirror of radius of curvature $200 {\text{cm}}$ as shown in the figure. The object starts moving towards the mirror at a speed $2 {\text{cm}} / {\ s}$ . The position of the image from the mirror after $10 {\ s}$ will be at ____ ${\text{cm}}$ . [28-Jul-2022-Shift-2]

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📖 Explanation

After $10 {\ sec}$ . $u=-80 {\text{cm}}$ $f=-100 {\text{cm}}$ $\;{1}/{ {\text{v}}}+\;{1}/{ {\text{u}}}=\;{1}/{ {\text{f}}}$ ${\text{v}}=400 {\text{cm}}$

Q29JEE Main 2022NAT4MRay Optics and Optical Instruments

In an experiment with a convex lens, The plot of the image distance $(v^{'})$ against the object distance $(\mu ^{'})$ measured from the focus gives a curve $v^{'} \mu ^{'}=225$ . If all the distances are measured in ${\text{cm}}$ . The magnitude of the focal length of the lens is ____ ${\text{cm}}$ [28-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{ru}}= {\text{f}}^2$ (by Newton's formula) ${\text{f}}^2=225$ ${\text{f}}=15 {\text{cm}}$

Q30JEE Main 2022NAT4MUnits and Measurements

In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of $0.5 {\ s}$ is measured from time of 100 oscillation with a watch of $1 {\ s}$ resolution. If measured value of length is $10 {\text{cm}}$ known to $1 {\text{mm}}$ accuracy, The accuracy in the determination of ${\text{g}}$ is found to be $x \%$ . The value of $x$ is________. [28-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$T=2 \pi \sqrt{\;\frac{l}{g}}$ $\;\frac{d g}{g} \times 100=\;\frac{2 d T}{T} \times 100+\;\frac{d l}{l} \times 100$ $=2 \times \;\frac{1}{50} \times 100+\;\frac{1}{100} \times 100=5 \%$

Chemistry30 questions

Q31JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

Given below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason $R$ Assertion A : Zero orbital overlap is an out of phase overlap. Reason R: It results due to different orientation / direction of approach of orbitals. In the light of the above statements, choose the correct answer from the options given below [28-Jul-2022-Shift-2]

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Q32JEE Main 2022MCQ4MClassification of Elements

The correct decreasing order for metallic character is [28-Jul-2022-Shift-2]

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📖 Explanation

Metallic character increases top to bottom in group and decreases left to right in a period. ${\text{Mg}}$ is from second group it will be less metallic than ${\text{Na}}$ . Be comes above ${\text{Mg}}$ hence less metallic than ${\text{Mg}}$ . Si is more metallic than phosphorous.

Q33JEE Main 2022MCQ4MChemical Thermodynamics

Given below are two statements: One is labelled as Assertion AA and the other is labelled as Reason ${\text{RR}}$ Assertion A: The reduction of a metal oxide is easier if the metal formed is in liquid state than solid state. Reason RR : The value of $\Delta G^{⊖} \Delta {\text{G}}$ becomes more on negative side as entropy is higher in liquid state than solid state. In the light of the above statements, choose the most appropriate answer from the options given below [28-Jul-2022-Shift-2]

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Q34JEE Main 2022MCQ4MIonic Equilibrium

The products obtained during treatment of hard water using Clark's method are: [28-Jul-2022-Shift-2]

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📖 Explanation

In Clark's method lime water is used ${\text{Ca}}( {\text{HCO}}_3)_2+2 {\text{Ca}}( {\text{OH}})_2 \to 2 {\text{CaCO}}_3+2 {\text{H}}_2 {\text{O}}$ ${\text{Mg}}( {\text{HCO}}_3)_2+2 {\text{Ca}}( {\text{OH}})_2 \to 2 {\text{CaCO}}_3+ {\text{Mg}}( {\text{OH}})_2+2 {\text{H}}_2 {\text{O}}$

Q35JEE Main 2022MCQ4MClassification of Elements

Statement I: An alloy of lithium and magnesium is used to make aircraft plates. Statement II : The magnesium ions are important for cell-membrane integrity. In the light the above statements, choose the correct answer from the options given below [28-Jul-2022-Shift-2]

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Q36JEE Main 2022MCQ4Mp Block Elements

White phosphorus reacts with thionyl chloride to give [28-Jul-2022-Shift-2]

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📖 Explanation

$\;{{ {\text{P}}_4}}_{\;\text{ phosphorous }\;}+\;{{8 {\text{SOCl}}_2}}_{\;\text{ Thionyl chloride }\;} \longrightarrow 4 {\text{PCl}}_3+4 {\text{SO}}_2+2 {\text{S}}_2 {\text{Cl}}_2$

Q37JEE Main 2022MCQ4Mp Block Elements

Concentrated ${\text{HNO}}_3 {\text{HNO}} 3$ reacts with lodine to give [28-Jul-2022-Shift-2]

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📖 Explanation

${\text{I}}_2+10 {\text{HNO}}_{3(\;\text{ conc }\;)} \Rightarrow 2 {\text{HIO}}_3+10 {\text{NO}}_2+4 {\text{H}}_2 {\text{O}}$

Q38JEE Main 2022MCQ4MStructure of Atom

Which of the following pair is not isoelectronic species?(At. no. Sm, 62; Er, 68; Yb, 70; Lu, 71; Eu, 63; Tb, 65; Tm, 69) [28-Jul-2022-Shift-2]

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📖 Explanation

Species having same number of electrons are isoelectronic $\text{table} {\text{Tb}} \longrightarrow 65 {\text{Tb}}^{+2} \longrightarrow 63 \;\text{ electrons }\; ; {\text{Tm}} \longrightarrow 69 {\text{Tm}}^{+4} \longrightarrow 65 \;\text{ electrons }\;\} \text{not isoelectronic}$ $\text{table} {\text{Sm}}^{+2} \longrightarrow 60 \;\text{ electrons }\; ; {\text{Er}}^{+3} \longrightarrow 65 \;\text{ electrons }\;\} \text{not isoelectronic }$

Q39JEE Main 2022MCQ4MRedox Reactions

Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason ${\text{R}}$ Assertion A : Permanganate titrations are not performed in presence of hydrochloric acid.Reason R : Chlorine is formed as a consequence of oxidation of hydrochloric acid.In the light of the above statements, choose the correct answer from the options given below [28-Jul-2022-Shift-2]

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📖 Explanation

$2 {\text{KMnO}}_4+16 {\text{HCl}} \to 2 {\text{MnCl}}_2+2 {\text{KCl}}+8 {\text{H}}_2 {\text{O}}+ {\text{Cl}}_2$ ${\text{HCl}}$ is not used in the process of titration because it reacts with the $( {\text{KMnO}}_4)$ that is used in the process and gets oxidized.

Q40JEE Main 2022MCQ4MCoordination Compounds

Match List I with List II List - I (Complex) List - II (Hybridization) (A) ${\text{Ni}}( {\text{CO}})_4$ (I) $sp^3$ (B) $[ {\text{Ni}}( {\text{CN}})_4]^{2-}$ (II) ${sp^3} {d^2}$ (C) ${[ {\text{Co}}(C N)_6]^{3-}}$ (III) $\text{d}^2\ sp^3$ (D) $[ {\text{CoF}}_6]^{3-}$ (IV) $\text{dsp}^2$ Choose the correct answer from the options given below: [28-Jul-2022-Shift-2]

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📖 Explanation

${\text{Ni}}( {\text{CO}})_4$ Hybridisation ${\ sp}^3$ $[ {\text{Ni}}( {\text{CN}})_4]^{2-}$ Hybridisation dspe2 $[ {\text{Co}}( {\text{CN}})_6]^{3-}$ Hybridisation ${\text{d}}^2 {\ sp}^3$ $[ {\text{Co}}( {\text{F}})_6]^{3-}$ Hybridisation ${\ sp}^3 {\text{d}}^2$

Q41JEE Main 2022MCQ4MChemical Thermodynamics

Dinitrogen and dioxygen, the main constituents of air do not react with each other in atmosphere to form oxides of nitrogen because [28-Jul-2022-Shift-2]

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📖 Explanation

${\text{N}}_2$ is unreactive, its reaction with oxides is endothermic and require very high temperature. ${\text{N}}_2+ {\text{O}}_2 {⇋}^{(1483-2000 {\text{K}})} 2 {\text{NO}}$ (Endothermic and feasible at high temperature)

Q42JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

The major product in the given reaction is [28-Jul-2022-Shift-2]

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Q43JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Arrange the following in increasing order of reactivity towards nitration A. p-xylene B. bromobenzene C. mesitylene D. nitrobenzene E. benzene Choose the correct answer from the options given below [28-Jul-2022-Shift-2]

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Q44JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

Compound I is heated with Conc. HI to give a hydroxy compound A which is further heated with Zn dust to give compound B. Identify A and B. [28-Jul-2022-Shift-2]

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Q45JEE Main 2022MCQ4MAmines

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$ Assertion A : Aniline on nitration yields ortho, meta & para nitro derivatives of aniline.Reason R : Nitrating mixture is a strong acidic mixture.In the light of the above statements, choose the correct answer from the options given below [28-Jul-2022-Shift-2]

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Q46JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Match List I with List II List - I (Polymer) List - II (Nature) (I)Thermosetting polymer (II)Fibers (III)Elastomer (IV)Thermoplastic polymerChoose the correct answer from the options given below : [28-Jul-2022-Shift-2]

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Q47JEE Main 2022MCQ4MBiomolecules

Two statements in respect of drug-enzyme interaction are given belowStatement I : Action of an enzyme can be blocked only when an inhibitor blocks the active site of the enzyme.Statement II : An inhibitor can form a strong covalent bond with the enzyme.In the light of the above statements, choose the correct answer from the options given below [28-Jul-2022-Shift-2]

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📖 Explanation

Action of an enzyme can be altered by a number of factors like temperature, ${\text{pH}}$ , presence of activators and coenzymes and presence of inhibitors and poisons.Inhibitors or poisons interact with the active functional groups on the enzyme surface and often reduce or completely destroy the catalytic activity of the enzymes.

Q48JEE Main 2022MCQ4MPractical Organic Chemistry

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason ${\text{R}}$ Assertion A : Thin layer chromatography is an adsorption chromatography.Reason R: A thin layer of silica gel is spread over a glass plate of suitable size in thin layer chromatography which acts as an adsorbent.In the light of the above statements, choose the correct answer from the options given below [28-Jul-2022-Shift-2]

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📖 Explanation

Thin layer chromoatography (TLC) is another type of adsorption chromatography, which involve sepration of substance of a mixture ovel a thin layer of an adsorbent coated on glass plate.A thin layer (about $0.2 {\text{mm}}$ thick) of an adsorbent (silica gel) or (Alumina) in spread overa glass plate of suitable size. Hence Assertion (A) is correct and Reason (R) is correct explanation of $(A)$

Q49JEE Main 2022MCQ4MBiomolecules

The formulas of $A$ and $B$ for the following reaction sequence are [28-Jul-2022-Shift-2]

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Q50JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Find out the major product for the above reaction. [28-Jul-2022-Shift-2]

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Q51JEE Main 2022NAT4MSome Basic Concepts of Chemistry

$2 {\text{L}}$ of $0.2 {\text{M}} {\text{H}}_2 {\text{SO}}_4$ is reacted with $2 {\text{L}}$ of $0.1 {\text{M}} {\text{NaOH}}$ solution, the molarity of the resulting product ${\text{Na}}_2 {\text{SO}}_4$ in the solution is _______ millimolar. (Nearest integer) [28-Jul-2022-Shift-2]

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📖 Explanation

${\text{H}}_2 {\text{SO}}_4+2 {\text{NaOH}} \to {\text{Na}}_2 {\text{SO}}_4+2 {\text{H}}_2 {\text{O}}$ $0.4 {\text{mol}} \;\; \;\; 0.2 {\text{mol}} \;\;\;\; -$ $0.3 {\text{mol}} \;\;\;\;\;\; -\;\;\;\;\;\; \;\; 0.1 {\text{mol}}$ Molarity of ${\text{Na}}_2 {\text{SO}}_4$ is $\;\frac{0.1}{4}=0.025 {\text{M}}$ $=25 {\text{mM}}$

Q52JEE Main 2022NAT4MSolid State

Metal ${\text{M}}$ crystallizes into a fcc lattice with the edge length of $4.0 \times 10^{-8} {\text{cm}}$ . The atomic mass of the metal is ${\text{g}} /$ mol. (Nearest integer)( Use : ${\text{N}}_{ {\text{A}}}=6.02 \times 10^{23} {\text{mol}}^{-1}$ , density of metal, ${\text{M}}=9.03 {\text{g}} {\text{cm}}^{-3}$ ) [28-Jul-2022-Shift-2]

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📖 Explanation

$\rho \;\; =\;\frac{Z M}{N_A a^3} \Rightarrow M=\;\frac{9.03 \times 6.02 \times 10^{23} \times (4 \times 10^{-8})^3}{4}$ $\;\; =\;\frac{9.03 \times 6.02 \times 64 \times 10^{-1}}{4}$ $\;\; =86.9 {\text{g}} {\text{mol}}^{-1}$ $\;\; \approx 87 {\text{g}} {\text{mol}}^{-1}$

Q53JEE Main 2022NAT4MStructure of Atom

If the wavelength for an electron emitted from ${\text{H}}$ -atom is $3.3 \times 10^{-10} {\text{m}}$ , then energy absorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is times. (Nearest integer)$[Given : ${\text{h}}=6.626 × 10^{-34} {\text{J}} {\ s}$ ]$Mass of electron $=9.1 \times 10^{-31} {\text{kg}}$ [28-Jul-2022-Shift-2]

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📖 Explanation

$\lambda =\;{ {\text{h}}}/{ {\ mv}}$ $\Rightarrow {\ mv}=\;{ {\text{h}}}/{\lambda }=\;{6.626 \times 10^{-34} {\text{kg}} { {\text{m}}^2}/{ {\ sec}^2} \times {\ sec}}/{3.3 \times 10^{-10} {\text{m}}}$ ${\ mv}=\;\frac{6.626 \times 10^{-24}}{3.3}=2 \times 10^{-24} {\text{kg}} {\text{m}} {\ sec}^{-1}$ Kinetic energy $=\;\frac{1}{2} m v^2$ $=\;{( {\ mv})^2}/{2 {\text{m}}}$ $=\;{(2 \times 10^{-24})^2}/{2 \times 9.1 \times 10^{-31} {\text{kg}}}$ $\text{table} =2.18 \times 10^{-18} {\text{J}} ; =21.8 \times 10^{-19} {\text{J}}$ Total energy absorbed $=$ lonization energy $+$ Kinetic energy $=(21.76+21.8) \times 10^{-19}$ $=43.56 \times 10^{-19} {\text{J}}$ $\approx 2$ times of $21.76 \times 10^{-19} {\text{J}}$

Q54JEE Main 2022NAT4MSolutions

A gaseous mixture of two substances $A$ and $B$ , under a total pressure of $0.8 {\text{atm}}$ is in equilibrium with an ideal liquid solution. The mole fraction of substance $A$ is $0.5$ in the vapour phase and $0.2$ in the liquid phase. The vapour pressure of pure liquid ${\text{A}}$ is atm. (Nearest integer) [28-Jul-2022-Shift-2]

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📖 Explanation

Given that $X_A=0.2, Y_A=0.5, P_T=0.8 {\text{atm}}$ We know that $P_A=Y_A \times P_T$ $P_A=0.5 \times 0.8=0.4$ Now $P_A=X_A \times P_A^{\circ} \Rightarrow P_A^{\circ}=\;\frac{0.4}{0.2}=2 {\text{atm}}$

Q55JEE Main 2022NAT4MChemical Equilibrium

At $600 {\text{K}}, 2 {\text{mol}}$ of ${\text{NO}}$ are mixed with $1 {\text{mol}}$ of ${\text{O}}_2$ . $2 {\text{NO}}_{( {\text{g}})}+ {\text{O}}_2( {\text{g}}) ⇄ 2 {\text{NO}}_2( {\text{g}})$ The reaction occurring as above comes to equilibrium under a total pressure of $1 {\text{atm}}$ . Analysis of the system shows that $0.6 {\text{mol}}$ of oxygen are present at equilibrium. The equilibrium constant for the reaction is ____(Nearest integer) [28-Jul-2022-Shift-2]

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📖 Explanation

$2 {\text{NO}}( {\text{g}})+ {\text{O}}_2( {\text{g}}) ⇌ 2 {\text{NO}}_2( {\text{g}})$ at intial $2$ $1$ $0$ at equilibrium $2-0.8$ $0.6$ $0.8$ Partial pressure of ${\text{NO}}( {\text{g}})=\;\frac{1.2}{2.6} \times 1$ Partial pressure of ${\text{O}}_2( {\text{g}})=\;\frac{0.6}{2.6}$ Partial pressure of ${\text{NO}}_2( {\text{g}})=\;\frac{0.8}{2.6}$ ${\text{K}}_{ {\text{p}}}=\;{( {\text{P}}_{ {\text{NO}}_2})^2}/{( {\text{P}}_{ {\text{NO}}})^2( {\text{P}}_{ {\text{O}}_2})} \;\; =\;\frac{0.8 \times 0.8 \times 2.6}{1.2 \times 1.2 \times 0.6}$ $\;\; =1.925$ $\;\; \approx 2$

Q56JEE Main 2022NAT4MOrganic Chemistry - Some Basic Principles

A sample of $0.125 {\text{g}}$ of an organic compound when analyzed by Duma's method yields $22.78 {\text{mL}}$ of nitrogen gas collected over ${\text{KOH}}$ solution at $280 {\text{K}}$ and $759 {\text{mm}} {\text{Hg}}$ . The percentage of nitrogen in the given organic compound is ____(Nearest integer)Given :(a) The vapour pressure of water of $280 {\text{K}}$ is $14.2 {\text{mm}} {\text{Hg}}$ .(b) ${\text{R}}=0.082 {\text{L}}$ atm ${\text{K}}^{-1} {\text{mol}}^{-1}$ [28-Jul-2022-Shift-2]

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📖 Explanation

$\text{table} P_{\;\text{actual }\;}=759-14.2=744.8 {\text{mmHg}} ; {\text{n}}_{ {\text{N}}_2}=\;\frac{744.8 \times 22.78}{760 \times 0.0821 \times 280 \times 1000} ; =0.000971 {\text{mol}}$ Mass of ${\text{N}}_2=0.02719 {\text{gm}}$ Percentage of nitrogen $=\;\frac{0.0271}{0.125} \times 100=21.75 ≃ 22$

Q57JEE Main 2022NAT4MRedox Reactions

On reaction with stronger oxidizing agent like ${\text{KIO}}_4$ , hydrogen peroxide oxidizes with the evolution of ${\text{O}}_2$ . The oxidation number of ${\text{I}}$ in ${\text{KIO}}_4$ changes to____ [28-Jul-2022-Shift-2]

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📖 Explanation

${\text{KIO}}_4+ {\text{H}}_2 {\text{O}}_2 \longrightarrow {\text{KIO}}_3+ {\text{H}}_2 {\text{O}}+ {\text{O}}_2$

Q58JEE Main 2022NAT4MChemical Kinetics

For a reaction, given below is the graph of $\ln k$ vs $\;\frac{1}{T}$ . The activation energy for the reaction is equal to____ (nearest integer)(Given : ${\text{R}}=2 {\text{cal}} {\text{K}}^{-1} {\text{mol}}^{-1}$ ) [28-Jul-2022-Shift-2]

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📖 Explanation

The relationship between the rate constant ( $k$ ), activation energy ( $E_a$ ), and temperature ( $T$ ) is given by the Arrhenius equation, $\ln k = \ln A - \frac{E_a}{RT}$ . A plot of $\ln k$ versus $\frac{1}{T}$ yields a straight line with a slope of $m = -\frac{E_a}{R}$ .

From the given graph, we can calculate the slope using the y-intercept $(0, 20)$ and the x-intercept $(5, 0)$ .
Slope ( $m$ ) = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 20}{5 - 0} = -4$ .

Often in such plots, the x-axis is scaled by 1000 for convenience (i.e., the x-axis represents $\frac{1000}{T}$ ). In this case, the slope of the line is $m = -\frac{E_a}{1000R}$ .
Equating the calculated slope:
$-4 = -\frac{E_a}{1000R}$
$E_a = 4 \times 1000 \times R$

Substituting the given value of $R = 2 \text{ cal K}^{-1} \text{mol}^{-1}$ :
$E_a = 4000 \times 2 = 8000 \text{ cal/mol}$
$E_a = 8 \text{ kcal/mol}$ . The activation energy is 8.

Q59JEE Main 2022NAT4MSolid State

Among the following the number of curves not in accordance with Freundlich adsorption isotherm is ________. [28-Jul-2022-Shift-2]

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📖 Explanation

$\;{ {\text{X}}}/{ {\text{m}}}= {\text{KP}}^{\;\frac{1}{n}}$ $\log \;{ {\text{x}}}/{ {\text{m}}}=\;{1}/{ {\text{n}}} \log {\text{p}}+\log {\text{k}}$

Q60JEE Main 2022NAT4MChemical Thermodynamics

Among the following the number of state variables is _______. Internal energy (U) Volume (V) Heat (q) Enthalpy $( {\text{H}})$ [28-Jul-2022-Shift-2]

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📖 Explanation

State variables are internal energy $(U)$ , Volume $( {\text{V}})$ and Enthalpy $( {\text{H}})$ .

Mathematics30 questions

Q61JEE Main 2022MCQ4MQuadratic Equation and Inequalities

Let $S=\{x \in [-6,3]-\{-2,2\}: \;\frac{|x+3|-1}{|x|-2} \ge 0\}$ and $T=\{x \in ℤ: x^{2-}7|x|+9 \le 0\}$ Then the number of elements in ${\text{S}} ∩ {\text{T}}$ is : [28-Jul-2022-Shift-2]

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📖 Explanation

$|x^2|-7|x|+9 \le 0$ $\Rightarrow |x| \in [\;\frac{7-\sqrt{13}}{2}, \;\frac{7+\sqrt{13}}{2}]$ As $x \in Z$ So, $x$ can be $\pm 2, \pm 3, \pm 4, \pm 5$ Out of these values of $x$ , $x=3,-4,-5$ satisfy ${\text{S}}$ as well $n(S ∩ T)=3$

Q62JEE Main 2022MCQ4MQuadratic Equation and Inequalities

Let $\alpha, \beta$ be the roots of the equation $x^{2-}\sqrt{2} x+\sqrt{6}=0$ and $\;\frac{1}{\alpha^2}+1, \;\frac{1}{\beta^2}+1$ be the roots of the equation $x^{2+}a x+b=0$ . Then the roots of the equation $x^{2-}(a+b-2) x+(a+b+2)=0$ are : [28-Jul-2022-Shift-2]

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📖 Explanation

$\alpha+\beta=\sqrt{2}, \alpha \beta=\sqrt{6}$ $\;\frac{1}{\alpha^2}+1+\;\frac{1}{\beta^2}+1=2+\;\frac{\alpha^{2+}\beta^2}{6}$ $=2+\;\frac{2-2 \sqrt{6}}{6}=-a$ $(\;\frac{1}{\alpha^2}+1)(\;\frac{1}{\beta^2}+1)=1+\;\frac{1}{\alpha^2}+\;\frac{1}{\beta^2}+\;\frac{1}{\alpha^2 \beta^2}$ $=\;\frac{7}{6}+\;\frac{2-2 \sqrt{6}}{6}=b$ $\Rightarrow a+b=\;\frac{-5}{6}$ So, equation is $x^{2+}\;\frac{17 x}{6}+\;\frac{7}{6}=0$ OR $6 x^{2+}17 x+7=0$ Both roots of equation are - ve and distinct

Q63JEE Main 2022MCQ4MMatrices and Determinants

Let ${\text{A}}$ and ${\text{B}}$ be any two $3 \times 3$ symmetric and skew symmetric matrices respectively. Then which of the following is NOT true? [28-Jul-2022-Shift-2]

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📖 Explanation

(A) $M=A^{4-}B^4$ $M^T=(A^{4-}B^4)^T=(A^T)^{4-}(B^T)^4$ $=A^{4-}(-B)^4=A^{4-}B^4=M$ (B) $M=A B-B A$ $M^T=(A B-B A)^T=(A B)^T-(B A)^T$ $=B^T A^T-A^T B^T$ $=-B A-A(-B)$ $=A B-B A=M$ (C) $M=B^{5-}A^5$ $M^T=(B^T)^{5-}(A^T)^5=-(B^{5+}A^5) \ne -M$ (D) $M=A B+B A$ $M^T=(A B)^T+(B A)^T$ $=B^T A^T+A^T B^T=-B A-A B=-M$

Q64JEE Main 2022MCQ4MFunctions

Let $f(x)=a x^{2+}b x+c$ be such that $f(1)=3, f(-2)=\lambda$ and $f(3)=4$ . If $f(0)+f(1)+f(-2)+f(3)=14$ , then $\lambda$ is equal to : [28-Jul-2022-Shift-2]

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📖 Explanation

$f(1)=a+b+c=3 . . . .$ . (i) $f(3)=9 a+3 b+c=4 . . .$ . (ii) $f(0)+f(1)+f(-2)+f(3)=14$ ${\text{OR}} c+3+(4 a-2 b+c)+4=14$ OR $4 a-2 b+2 c=7 . . .$ . (iii)From (i) and (ii) $8 a+2 b=1 . . .$ . (iv)From (iii) $-(2) \times$ (i) $\Rightarrow 2 a-4 b=1 . . . . .( {\text{v}})$ From (iv) and (v) $a=\;\frac{1}{6}, b=\;\frac{-1}{6}$ and $c=3$ $f(-2)=4 a-2 b+c$ $=\;\frac{4}{6}+\;\frac{2}{6}+3=4$

Q65JEE Main 2022MCQ4MLimits, Continuity and Differentiability

The function $f: ℝ \to ℝ$ defined by $f(x)=\lim _{n \to \infty } \;\frac{\cos (2 \pi x)-x^{2 n} \ sin (x-1)}{1+x^{2 n+1}-x^{2 n}}$ is continuous for all ${\text{x}}$ in : [28-Jul-2022-Shift-2]

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📖 Explanation

$f(x)=\lim _{n \to \infty } \;\frac{\cos (2 \pi x)-x^{2 n} \ sin (x-1)}{1+x^{2 n+1}-x^{2 n}}$ For $|x|< 1, f(x)=\cos 2 \pi x$ , continuous function $|x|>1, f(x)=\lim _{n \to \infty } \;{\;\frac{1}{x^{2 n}} \cos 2 \pi x-\ sin (x-1)}/{\frac{1}{x^{2 n}}+x-1}$ $=\;\frac{-\ sin (x-1)}{x-1}$ , continuousFor $|x|=1, f(x)=\begin{cases}1 & \\ \\ \text{ if }\\ \\ x=1 \\ -(1+\\ sin 2) & \\ \\ \text{ if }\\ \\ x=-1\end{cases}$ Now, $\lim _{x \to 1^{+}} f(x)=-1, \lim _{x \to 1^{-}} f(x)=1$ , so discontinuous at $x=1$ $\lim _{x \to 1^{+}} f(x)=1, \lim _{x \to -1^{-}} f(x)=-\;\frac{\ sin 2}{2}$ , so discontinuous at $x=-1$ $\therefore f(x)$ is continuous for all $x \in R-\{-1,1\}$

Q66JEE Main 2022MCQ4MApplication of Derivatives

The function $f(x)=x {\text{e}}^{x(1-x)}, x \in ℝ$ , is : [28-Jul-2022-Shift-2]

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📖 Explanation

$f(x)=x e^{x(1-x)}, x \in R$ $f^{'}(x)=x e^{x(1-x)} \cdot (1-2 x)+e^{x(1-x)}$ $=e^{x(1-x)}[x-2 x^{2+}1]$ $=-e^{x(1-x)}[2 x^{2-}x-1]$ $=-e^{x(1-x)}(2 x+1)(x-1)$ $\therefore f(x)$ is increasing in $(-\;\frac{1}{2}, 1)$ and decreasing in $(-\infty ,-\;\frac{1}{2}) ∪(1, \infty )$

Q67JEE Main 2022MCQ4MApplication of Derivatives

The sum of the absolute maximum and absolute minimum values of the function $f(x)=\tan ^{-1}(\ sin x-\cos x)$ in the interval $[0, \pi]$ is : [28-Jul-2022-Shift-2]

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📖 Explanation

$f(x)=\tan ^{-1}(\ sin x-\cos x), \;\; [0, \pi]$ Let $g(x)=\ sin x-\cos x$ $=\sqrt{2} \ sin (x-\;\frac{\pi}{4})$ and $x-\;\frac{\pi}{4} \in [\;\frac{-\pi}{4}, \;\frac{3 \pi}{4}]$ $\therefore g(x) \in [-1, \sqrt{2}]$ and $\tan ^{-1} x$ is an increasing function $\therefore f(x) \in [\tan ^{-1}(-1), \tan ^{-1} \sqrt{2}]$ $\in [-\;\frac{\pi}{4}, \tan ^{-1} \sqrt{2}]$ $\therefore$ Sum of $f_{\max }$ and $f_{\min }=\tan ^{-1} \sqrt{2}-\;\frac{\pi}{4}$ $=\cos ^{-1}(\;\frac{1}{\sqrt{3}})-\;\frac{\pi}{4}$

Q68JEE Main 2022MCQ4MDifferentiation

Let $x(t)=2 \sqrt{2} \cos t \sqrt{\ sin 2 t}$ and $y(t)=2 \sqrt{2} \ sin t \sqrt{\ sin 2 t}, t \in (0, \;\frac{\pi}{2})$ Then $\;\frac{1+(\frac{d y}{d x})^2}{\frac{d^2 y}{d x^2}}$ at $t=\;\frac{\pi}{4}$ is equal to: [28-Jul-2022-Shift-2]

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📖 Explanation

$x=2 \sqrt{2} \cos t \sqrt{\ sin 2 t}, y=2 \sqrt{2} \ sin t \sqrt{\ sin 2 t}$ $\therefore \;\frac{d x}{d t}=\;\frac{2 \sqrt{2} \cos 3 t}{\sqrt{\ sin 2 t}}, \;\frac{d y}{d t}=\;\frac{2 \sqrt{2} \ sin 3 t}{\sqrt{\ sin 2 t}}$ $\therefore \;\frac{d y}{d x}=\tan 3 t,({\text{at}} t=\;\frac{\pi}{4}, \;\frac{d y}{d x}=-1)$ and $\;\frac{d^2 y}{d x^2}=3 \ sec ^2 3 t \cdot \;\frac{d t}{d x}=\;\frac{3 \ sec ^2 3 t \cdot \sqrt{\ sin 2 t}}{2 \sqrt{2} \cos 3 t}$ $(.$ At $.t=\;\frac{\pi}{4}, \;\frac{d^2 y}{d x^2}=-3)$ $\therefore \;\frac{1+(\frac{d y}{d x})^2}{\frac{d^2 y}{d x^2}}=\;\frac{2}{-3}=\;\frac{-2}{3}$

Q69JEE Main 2022MCQ4MDefinite Integrals

Let $I_n(x)=int_{0}^{x }\;\frac{1}{(t^{2+}5)^n} d t, n=1,2,3, . . .$ Then : [28-Jul-2022-Shift-2]

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📖 Explanation

$I_n(x)=int_{0}^{x }\;\frac{1}{(t^{2+}5)^n} d t$ $=int_{0}^{x }\;\frac{1}{⏟{2}_I+5)^n} \times int_{I I}^{I} d t$ $=.\;\frac{t}{(t^{2+}5)^n}|_0 ^x-int_{0}^{x }\;\frac{-2 n t}{(t^{2+}5)^{n+1}} \times t d t$ $=\;\frac{x}{(x^{2+}5)^n}+int_{0}^{x }2 n(\;\frac{t^{2+}5-5}{(t^{2+}5)^{n+1}}) d t$ $I_n(x)=\;\frac{x}{(x^{2+}5)^n}+2 n I_n(x)-10 n I_{n+1}(x)$ $10 n I_{n+1}(x)-(2 n-1) I_n(x)=x I_n^{'}(x)$ For $n=5$ $50 I_6(x)-9 I_5(x)=x I_5^{'}(x)$

Q70JEE Main 2022MCQ4MArea Under The Curves

The area enclosed by the curves $y=\log _e(x+ {\text{e}}^2), x=\log _e(\;\frac{2}{y})$ and $x=\log _{ {\text{e}}} 2$ , above the line $y=1$ is: [28-Jul-2022-Shift-2]

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📖 Explanation

According to NTA, the required region $A_2$ which is shaded in crossed lines and comes out to be $A_2=int_{1}^{2}(\ln \;\frac{2}{y}-e^y+e^2) d y=1+e-\ln 2$ But according to us the required region $A_1$ comes out to be shaded in parallel lines, which can be obtained as $A_1=int_{0}^{\ln 2}(\ln (x+e^2)-2 e^{-x}) d x$ $=.\{(x+e^2) \ln (x+e^2)-x+2 e^{-x}\}|_0 ^{\ln 2}$ $=(\ln 2+e^2) \ln (\ln 2+e^2)-\ln 2+1$ $-2 e^{2-}2$ $=(\ln 2+e^2) \ln (\ln 2+e^2)-\ln 2-2 e^{2-}1$ Not given in any option.The region asked in the question is bounded by three curves $y=\ln (x+e^2)$ $x=\ln (\;\frac{2}{y})$ $x=\ln 2$ There is only one region which satisfies above requirement and which also lies above line $y=1$ Line $y=1$ may or may not be the boundary of the region.

Q71JEE Main 2022MCQ4MDifferential Equations

Let $y=y(x)$ be the solution curve of the differential equation $\;\frac{d y}{d x}+\;\frac{1}{x^{2-}1} y=(\;\frac{x-1}{x+1})^{1 / 2}, x>1$ passing through the point $(2, \sqrt{\;\frac{1}{3}})$ . Then $\sqrt{7} y(8)$ is equal to: [28-Jul-2022-Shift-2]

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📖 Explanation

$\;\frac{d y}{d x}+\;\frac{1}{x^{2-}1} y=\sqrt{\;\frac{x-1}{x+1}}, x>1$ Integrating factor I.F. $=e^{\int \;\frac{1}{x^{2-}1} d x}=e^{\;\frac{1}{2} \ln |\;\frac{x-1}{x+1}|}$ $=\sqrt{\;\frac{x-1}{x+1}}$ Solution of differential equation $y \sqrt{\;\frac{x-1}{x+1}}=\int \;\frac{x-1}{x+1} d x=\int(1-\;\frac{2}{x+1}) d x$ $y \sqrt{\;\frac{x-1}{x+1}}=x-2 \ln |x+1|+C$ Curve passes through $(2, \sqrt{\;\frac{1}{3}})$ $\;\frac{1}{\sqrt{3}} \times \;\frac{1}{\sqrt{3}}=2-2 \ln 3+C$ $C=2 \ln 3-\;\frac{5}{3}$ $y(8) \times \;\frac{\sqrt{7}}{3}=8-2 \ln 9+2 \ln 3-\;\frac{5}{3}$ $\sqrt{7} \cdot y(8)=19-6 \ln 3$

Q72JEE Main 2022MCQ4MDifferential Equations

The differential equation of the family of circles passing through the points $(0,2)$ and $(0,-2)$ is : [28-Jul-2022-Shift-2]

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📖 Explanation

Family of circles passing through the points $(0,2)$ and $(0,-2)$ $x^{2+}(y-2)(y+2)+\lambda x=0, \lambda \in R$ $x^{2+}y^{2+}\lambda x-4=0 . . . . . . \;\text{ (1) }\;$ Differentiate w.r.t x $2 x+2 y \;\frac{d y}{d x}+\lambda =0 . . . . . . \;\text{ (2) }\;$ Using (1) and (2), eliminate $\lambda$ $x^{2+}y^{2-}(2 x+2 y \;\frac{d y}{d x}) x-4=0$ $2 x y \;\frac{d y}{d x}+x^{2-}y^{2+}4=0$

Q73JEE Main 2022MCQ4MCircles

Let the tangents at two points ${\text{A}}$ and ${\text{B}}$ on the circle $x^{2+} {\text{y}}^{2-}4 x+3=0$ meet at origin ${\text{O}}(0,0)$ . Then the area of the triangle ${\text{OAB}}$ is : [28-Jul-2022-Shift-2]

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📖 Explanation

$x^{2+}y^{2-}4 x+3=0$ $\Rightarrow (x-2)^{2+}y^2=1$ $A O=\sqrt{(O C)^{2-}(A C)^2}$ $=\sqrt{4-1}=\sqrt{3}$ $\ sin \theta=\;\frac{1}{2} \Rightarrow \theta=\;\frac{\pi}{6}$ Also, $A O=B O$ Area of $\Delta O A B=\;\frac{1}{2} \cdot O A \cdot O B \ sin 60^{\circ}$ $=\;\frac{1}{2} \times \sqrt{3} \cdot \sqrt{3} \cdot \;\frac{\sqrt{3}}{2}=\;\frac{3 \sqrt{3}}{4}$

Q74JEE Main 2022MCQ4MHyperbola

Let the hyperbola $H: \;\frac{x^2}{a^2}-\;\frac{y^2}{b^2}=1$ pass through the point $(2 \sqrt{2},-2 \sqrt{2})$ . A parabola is drawn whose focus is same as the focus of ${\text{H}}$ with positive abscissa and the directrix of the parabola passes through the other focus of ${\text{H}}$ . If the length of the latus rectum of the parabola is e times the length of the latus rectum of ${\text{H}}$ , where e is the eccentricity of ${\text{H}}$ , then which of the following points lies on the parabola? [28-Jul-2022-Shift-2]

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📖 Explanation

$H: \;\frac{x^2}{a^2}-\;\frac{y^2}{b^2}=1$ Focus of parabola : $(a e, 0)$ Directrix : $x=-a e$ .Equation of parabola $\equiv y^2=4 a e x$ Length of latus rectum of parabola $=4 a e$ Length of latus rectum of hyperbola $=\;\frac{2 . b^2}{a}$ as given, $4 a e=\;\frac{2 b^2}{a} \cdot e$ $2=\;\frac{b^2}{a^2} . . . . . .$ (i) $\because$ H passes through $(2 \sqrt{2},-2 \sqrt{2}) \Rightarrow \;\frac{8}{a^2}-\;\frac{8}{b^2}=1 . . . . . .$ . (ii)From (i) and (ii) $a^2=4$ and $b^2=8 \Rightarrow e=\sqrt{3}$ $\Rightarrow$ Equation of parabola is $y^2=8 \sqrt{3} x$ .

Q75JEE Main 2022MCQ4MThree Dimensional Geometry

Let the lines $\;\frac{x-1}{\lambda }=\;\frac{y-2}{1}=\;\frac{z-3}{2}$ and $\;\frac{x+26}{-2}=\;\frac{y+18}{3}=\;\frac{z+28}{\lambda }$ be coplanar and ${\text{P}}$ be the plane containing these two lines. Then which of the following points does NOT lie on P? [28-Jul-2022-Shift-2]

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📖 Explanation

$L_1: \;\frac{x-1}{\lambda }=\;\frac{y-2}{1}=\;\frac{z-3}{2} \;\text{, }\;$ through a point $\vec{a}_1 \equiv (1,2,3)$ parallel to $\vec{b_1} \equiv (\lambda , 1,2)$ $L_2: \;\frac{x+26}{-2}=\;\frac{y+18}{3}=\;\frac{z+28}{\lambda }$ through a point $\vec{a}_2=(-26,-18,-28)$ parallel to $\vec{b_2}=(-2,3,1)$ If lines are coplanar then, $(\vec{a}_2-\vec{a}_1) \cdot \vec{b_1} \times \vec{b_2}=0$ If lines are coplanar then, $(\vec{a_2}-\vec{a_1}) \cdot \vec{b_1} \times \vec{b_2}=0$ $\Rightarrow \begin{vmatrix}27 & 20 & 31 \\ \lambda & 1 & 2 \\ -2 & 3 & \lambda \end{vmatrix}=0 \Rightarrow \lambda =3$ Vector normal to the required plane $\vec{n}=\vec{b_1} \times \vec{b_2}$ $\Rightarrow \vec{n}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ -2 & 3 & 3\end{vmatrix}=-3 \hat{i}-13 \hat{j}+11 \hat{k}$ Equation of plane $\equiv ((x-1),(y-2),(z-3)) \cdot (-3,-13,11)=0$ $\Rightarrow 3 x+13 y-11 z+4=0$ Checking the option gives $(0,4,5)$ does not lie on the plane.

Q76JEE Main 2022MCQ4MThree Dimensional Geometry

A plane $P$ is parallel to two lines whose direction ratios are $-2,1,-3$ and $-1,2,-2$ and it contains the point $(2,2,-2)$ . Let ${\text{P}}$ intersect the co-ordinate axe at the points ${\text{A}}, {\text{B}}, {\text{C}}$ making the intercepts $\alpha, \beta, \gamma$ . If ${\text{V}}$ is the volume of the tetrahedron ${\text{OABC}}$ , where ${\text{O}}$ is the origin, and ${\text{p}}=\alpha+\beta+\gamma$ , then the ordered pair $( {\text{V}}, {\text{p}})$ is equal to: [28-Jul-2022-Shift-2]

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📖 Explanation

Let $\vec{a}_1=(-2,1,-3)$ and $\vec{a}_2=(-1,2,-2)$ Vector normal to plane $\overline{n}=\vec{a}_1 \times \vec{a}_2$ $\overline{n}=(4,-1,-3)$ Plane through $(2,2,-2)$ and normal to $\overline{n}$ $(x-2, y-2, z+2) \cdot (4,-1,-3)=0$ $\Rightarrow 4 x-y-3 z=12$ $\Rightarrow \;\frac{x}{3}+\;\frac{y}{-12}+\;\frac{z}{-4}=1$ Intercepts $\alpha, \beta, \gamma$ are $3,-12,-4$ $P=\alpha+\beta+\gamma =-13$ $V=\;\frac{1}{6} \times 3 \times 12 \times 4=24$

Q77JEE Main 2022MCQ4MVector Algebra

Let $S$ be the set of all ${\text{a}} \in R$ for which the angle between the vectors $\vec{u}=a(\log _e b) \hat{i}-6 \hat{j}+3 \hat{k}$ and $\vec{v}=(\log _e b) \hat{i}+2 \hat{j}+2 a(\log _e b) \hat{k},(b>1)$ is acute. Then ${\text{S}}$ is equal to : [28-Jul-2022-Shift-2]

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📖 Explanation

$\vec{u}=a(\log _e b) \hat{i}-6 \hat{j}+3 \hat{k}$ $\vec{v}=(\log _e b) \hat{i}+2 \hat{j}+2 a(\log _e b) \hat{k}$ For acute angle $\vec{u} \cdot \vec{v}>0$ $\Rightarrow a(\log _e b)^{2-}12+6 a(\log _e b)>0$ $\because b>1$ Let $\log _e b=t \Rightarrow t>0$ as $b>1$ $a t^{2+}6 a t-12>0 \;\; ∀ t>0$ $\Rightarrow a \in \varphi$

Q78JEE Main 2022MCQ4MTrigonometric Ratios and Identities

A horizontal park is in the shape of a triangle ${\text{OAB}}$ with ${\text{AB}}=16$ . A vertical lamp post ${\text{OP}}$ is erected at the point ${\text{O}}$ such that $∠ {\text{PAO}}=∠ {\text{PBO}}=15^{\circ}$ and $∠ {\text{PCO}}=45^{\circ}$ , where ${\text{C}}$ is the midpoint of ${\text{AB}}$ . Then $( {\text{OP}})^2$ is equal to [28-Jul-2022-Shift-2]

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📖 Explanation

$O P=O A \tan 15=O B \tan 15 . . . . . .$ . (i) $O P=O C \tan 45 \Rightarrow O P=O C . . . . . .$ . (ii) $O A=O B . . . . . .$ (iii) $O C^{2+}8^2=O A^2$ $O P^{2+}64=O P^2(\;\frac{\sqrt{3}+1}{\sqrt{3}-1})^2$ $64=O P^2[\;\frac{(\sqrt{3}+1)^{2-}(\sqrt{3-1})^2}{(\sqrt{3}-1)^2}]$ $=O P^2(\;\frac{4 \sqrt{3}}{(\sqrt{3}-1)^2})$ $O P^2=\;\frac{64(\sqrt{3}-1)^2}{4 \sqrt{3}}=\;\frac{32}{\sqrt{3}}(2-\sqrt{3})$

Q79JEE Main 2022MCQ4MProbability

Let ${\text{A}}$ and ${\text{B}}$ be two events such that $P(B \mid A)=\;\frac{2}{5}, P(A \mid B)=\;\frac{1}{7}$ and $P(A ∩ B)=\;\frac{1}{9} \cdot$ Consider(S1) $P(A^{'} ∪ B)=\;\frac{5}{6}$ ,(S2) $P(A^{'} ∩ B^{'})=\;\frac{1}{18}$ Then [28-Jul-2022-Shift-2]

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📖 Explanation

$P(A / B)=\;\frac{1}{7} \Rightarrow \;\frac{P(A ∩ B)}{P(B)}=\;\frac{1}{7}$ $\Rightarrow P(B)=\;\frac{7}{9}$ $P(B / A)=\;\frac{2}{5} \Rightarrow \;\frac{P(A ∩ B)}{P(A)}=\;\frac{2}{5}$ $P(A)=\;\frac{5}{2} \cdot \;\frac{1}{9}=\;\frac{5}{18}$ $S 2: P(A^{'} ∩ B^{'})=\;\frac{1}{18}$ $S 1:$ and $P(A^{'} ∪ B)=\;\frac{1}{9}+\;\frac{6}{9}+\;\frac{1}{18}=\;\frac{5}{6}$ .

Q80JEE Main 2022MCQ4MSets and Relations

Let ${\text{p}}$ : Ramesh listens to music. $q$ : Ramesh is out of his village. ${\text{r}}$ : It is Sunday. ${\ s}$ : It is Saturday.Then the statement "Ramesh listens to music only if he is in his village and it is Sunday or Saturday" can be expressed as [28-Jul-2022-Shift-2]

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📖 Explanation

${\text{p}}$ : Ramesh listens to music $q$ : Ramesh is out of his village $r$ : It is Sunday ${\ s}$ : It is Saturday $p \to q$ conveys the same $p$ only if $q$ Statement "Ramesh listens to music only if he is in his village and it is Sunday or Saturday" $p \Rightarrow ((∼ q) ∧(r ∨ s))$

Q81JEE Main 2022NAT4MBinomial Theorem

Let the coefficients of the middle terms in the expansion of $(\;\frac{1}{\sqrt{6}}+\beta x)^4,(1-3 \beta x)^2$ and $(1-\;\frac{\beta}{2} x)^6, \beta>0$ , respectively form the first three terms of an A.P. If ${\text{d}}$ is the common difference of this A.P. , then $50-\;\frac{2 d}{\beta^2}$ is equal to _________. [28-Jul-2022-Shift-2]

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📖 Explanation

Coefficients of middle terms of given expansions are ${ }^4 C_2 \;\frac{1}{6} \beta^2,{ }^2 C_1(-3 \beta),{ }^6 C_3(\;\frac{-\beta}{2})^3$ form an A.P. $\therefore 2.2(-3 \beta)=\beta^{2-}\;\frac{5 \beta^3}{2}$ $\Rightarrow -24=2 \beta-5 \beta^2$ $\Rightarrow 5 \beta^{2-}2 \beta-24=0$ $\Rightarrow 5 \beta^{2-}12 \beta+10 \beta-24=0$ $\Rightarrow \beta(5 \beta-12)+2(5 \beta-12)=0$ $\beta=\;\frac{12}{5}$ $d=-6 \beta-\beta^2$ $\therefore 50-\;\frac{2 d}{\beta^2}=50-2 \;\frac{(-6 \beta-\beta^2)}{\beta^2}=50+\;\frac{12}{\beta}+2=57$

Q82JEE Main 2022NAT4MPermutations and Combinations

A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168 , then ${\text{b}}+3 {\text{g}}$ is equal to ________. [28-Jul-2022-Shift-2]

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📖 Explanation

${ }^b C_3 \cdot { }^g C_2=168$ $\Rightarrow \;\frac{b(b-1)(b-2)}{6} \cdot \;\frac{g(g-1)}{2}=168$ $\Rightarrow b(b-1)(b-2) \;\; g(g-1)=2^5 \cdot 3^2 \cdot 7$ $\Rightarrow b(b-1)(b-2) \;\; g(g-1)=6.7 .8 .3 .2$ $\therefore b=8$ and $g=3$ $\therefore b+3 g=17$

Q83JEE Main 2022NAT4MEllipse

Let the tangents at the points ${\text{P}}$ and ${\text{Q}}$ on the ellipse $\;\frac{x^2}{2}+\;\frac{y^2}{4}=1$ meet at the point $R(\sqrt{2}, 2 \sqrt{2}-2)$ . If ${\text{S}}$ is the focus of the ellipse on its negative major axis, then ${\text{SP}}^{2+} {\text{SQ}}^2$ is equal to ________. [28-Jul-2022-Shift-2]

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📖 Explanation

$E \equiv \;\frac{x^2}{2}+\;\frac{y^2}{4}=1$ $T \equiv y=m x \pm \sqrt{2 m^{2+}4}$ $\;\;\;\; \;\; ↓(\sqrt{2}, 2 \sqrt{2}-2)$ $\Rightarrow (2 \sqrt{2}-2-m \sqrt{2})=\pm \sqrt{2 m^{2+}4}$ $\Rightarrow 2 m^{2-}2 m \sqrt{2}(2 \sqrt{2-2})+4(3-2 \sqrt{2})=2 m^{2+}4$ $\Rightarrow -2 \sqrt{2} m(2 \sqrt{2}-2)=4-12+8 \sqrt{2}$ $\Rightarrow -4 \sqrt{2} m(\sqrt{2}-1)=8(\sqrt{2}-1)$ $\Rightarrow m=-\sqrt{2} \;\text{ and }\; m \to \infty$ $\therefore$ Tangents are $x=\sqrt{2}$ and $y=-\sqrt{2} x+\sqrt{8}$ $\therefore P(\sqrt{2}, 0)$ and $Q(1, \sqrt{2})$ and $S=(0,-\sqrt{2})$ $\therefore (P S)^{2+}(Q S)^2=4+9=13$

Q84JEE Main 2022NAT4MBinomial Theorem

If $1+(2+{ }^{49} C_1+{ }^{49} C_2+. . .+{ }^{49} C_{49})({ }^{50} C_2+{ }^{50} C_4+. . .+{ }^{50} C_{50})$ is equal to $2^{ {\text{n}}} \cdot {\text{m}}$ , where ${\text{m}}$ is odd, then ${\text{n}}+ {\text{m}}$ is equal to _______. [28-Jul-2022-Shift-2]

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📖 Explanation

$l=1+(1+{ }^{49} C_0+{ }^{49} C_1+. . .+{ }^{49} C_{49})({ }^{50} C_2+{ }^{50} C_4+. . .+{ }^{50} C_{50})$ As ${ }^{49} C_0+{ }^{49} C_1+. . .+{ }^{49} C_{49}=2^{49}$ and ${ }^{50} C_0+{ }^{50} C_2+. . .+{ }^{50} C_{50}=2^{49}$ $\Rightarrow { }^{50} C_2+{ }^{50} C_4+. . .+{ }^{50} C_{50}=2^{49}-1$ $\therefore l=1+(2^{49}+1)(2^{49}-1)$ $=2^{98}$ $\therefore m=1$ and $n=98$ $m+n=99$

Q85JEE Main 2022NAT4MParabola

Two tangent lines $l_1$ and $l_2$ are drawn from the point $(2,0)$ to the parabola $2 {\text{y}}^2=-x$ . If the lines $l_1$ and $l_2$ are also tangent to the circle $(x-5)^{2+}y^2=r$ , then $17 r$ is equal to________. [28-Jul-2022-Shift-2]

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📖 Explanation

Given : $y^2=\;\frac{-x}{2}$ ${c} T \equiv y=m x-\;\frac{1}{8 m}$ $↓(2,0)$ $\Rightarrow m^2=\;\frac{1}{16} \Rightarrow m=\pm \;\frac{1}{4}$ Tangents are $y=\;\frac{1}{4} x-\;\frac{1}{2}, y=\;\frac{-x}{4}+\;\frac{1}{2}$ $4 y=x-2$ and $4 y+x=2$ If these are also tangent to circle then $d_c=r$ $\Rightarrow |\;\frac{5-2}{\sqrt{17}}|=\sqrt{r} \Rightarrow r=(\;\frac{3}{\sqrt{17}})^2$ $\Rightarrow 17 r=17 \cdot \;\frac{9}{17}=9$

Q86JEE Main 2022NAT4MSequences and Series

$\;\frac{6}{3^{12}}+\;\frac{10}{3^{11}}+\;\frac{20}{3^{10}}+\;\frac{40}{3^9}+. . .+\;\frac{10240}{3}=2^n \cdot m$ , where ${\text{m}}$ is odd, then ${\text{m}} . {\text{n}}$ is equal to_______. [28-Jul-2022-Shift-2]

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📖 Explanation

$\;\frac{1}{3^{12}}+5(\;\frac{2^0}{3^{12}}+\;\frac{2^1}{3^{11}}+\;\frac{2^2}{3^{10}}+. . . . .+\;\frac{2^{11}}{3})=2^n \cdot m$ $\Rightarrow \;\frac{1}{3^{12}}+5(\;\frac{1}{3^{12}} \;\frac{((6)^{2-}1)}{(6-1)})=2^n \cdot m$ $\Rightarrow \;\frac{1}{3^{12}}+\;\frac{5}{5}(\;\frac{1}{3^{12}} \cdot 2^{12} \cdot 3^{12}-\;\frac{1}{3^{12}})=2^n \cdot m$ $\Rightarrow \;\frac{1}{3^{12}}+2^{12}-\;\frac{1}{3^{12}}=2^n \cdot m$ $\Rightarrow 2^n \cdot m=2^{12}$ $\Rightarrow m=1$ and $n=12$ $m \cdot n=12$

Q87JEE Main 2022NAT4MTrigonometric Ratios and Identities

Let $S=[-\pi, \;\frac{\pi}{2})-\{-\;\frac{\pi}{2},-\;\frac{\pi}{4},-\;\frac{3 \pi}{4}, \;\frac{\pi}{4}\}$ . Then the number of elements in the set $\mid A=\{\theta \in S: \tan \theta(1+\sqrt{5} \tan (2 \theta))=\sqrt{5}-\tan (2 \theta)\}$ is _______. [28-Jul-2022-Shift-2]

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📖 Explanation

Let $\tan \alpha=\sqrt{5}$ $\therefore \tan \theta=\;\frac{\tan \alpha-\tan 2 \theta}{1+\tan \alpha \tan 2 \theta}$ $\therefore \tan \theta=\tan (\alpha-2 \theta)$ $\alpha-2 \theta=n \pi+\theta$ $\Rightarrow 3 \theta=\alpha-n \pi$ $\Rightarrow \theta=\;\frac{\alpha}{3}-\;\frac{n \pi}{3} \;\; ; n \in Z$ If $\theta \in [-\pi, \pi / 2]$ then $n=0,1,2,3,4$ are acceptable $\therefore 5$ solutions.

Q88JEE Main 2022NAT4MComplex Numbers

Let ${\text{z}}=a+i b, b \ne 0$ be complex numbers satisfying $z^2=\overline{z} \cdot 2^{1-|z|}$ . Then the least value of $n \in N$ , such that $z^n=(z+1)^n$ , is equal to ________. [28-Jul-2022-Shift-2]

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📖 Explanation

$\because z^2=\overline{z} \cdot 2^{1-|z|} . . . . . . \;\text{ (1) }\;$ $\Rightarrow |z|^2=|\overline{z}| \cdot 2^{1-|z|}$ $\Rightarrow |z|=2^{1-|z|}$ $\because b \ne 0 \Rightarrow |z| \ne 0$ $\therefore |z|=1 . . . . . . \;\text{ (2) }\;$ $\because z=a+i b \;\text{ then }\; \sqrt{a^{2+}b^2}=1 . . . . . . \;\text{ (3) }\;$ Now again from equation (1), equation (2), equation (3) we get : $a^{2-}b^{2+}i 2 a b=(a-i b) 2^0$ $\therefore a^{2-}b^2=a \;\text{ and }\; 2 a b=-b$ $\therefore a=-\;\frac{1}{2}$ and $b=\pm \;\frac{\sqrt{3}}{2}$ $z=-\;\frac{1}{2}+\;\frac{\sqrt{3}}{2} i$ or $z=-\;\frac{1}{2}-\;\frac{\sqrt{3}}{2} i$ $z^n=(z+1)^n \Rightarrow (\;\frac{z+1}{z})^n=1$ $(1+\;\frac{1}{z})^n=1$ $(\;\frac{1+\sqrt{3} i}{2})=1$ , then minimum value of $n$ is 6 .

Q89JEE Main 2022NAT4MProbability

A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let ${\text{X}}$ be the number of white balls, among the drawn balls. If $\sigma^2$ is the variance of ${\text{X}}$ , then $100 \sigma^2$ is equal to_______. [28-Jul-2022-Shift-2]

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📖 Explanation

$X=$ Number of white ball drawn $P(X=0)=\;\frac{{ }^6 C_3}{{ }^{10} C_3}=\;\frac{1}{6}$ $P(X=1)=\;\frac{{ }^6 C_2 \times { }^4 C_1}{{ }^{10} C_3}=\;\frac{1}{2}$ $P(X=2)=\;\frac{{ }^6 C_1 \times { }^4 C_2}{{ }^{10} C_3}=\;\frac{3}{10}$ $\;\text{ and }\; P(X=3)=\;\frac{{ }^6 C_0 \times { }^4 C_3}{{ }^{10} C_3}=\;\frac{1}{30}$ $\;\text{ Variance }\;=\sigma^2=\sum P_i X_i^{2-}(\sum P_i X_i)^2$ $\sigma^2=\;\frac{1}{2}+\;\frac{12}{10}+\;\frac{3}{10}-(\;\frac{1}{2}+\;\frac{6}{10}+\;\frac{1}{10})^2$ $=\;\frac{56}{100}$ $100 \sigma^2=56$

Q90JEE Main 2022NAT4MDefinite Integrals

The value of the integral $int_{0}^{\;\frac{\pi}{2}} 60 \;\frac{\ sin (6 x)}{\ sin x} d x$ is equal to________. [28-Jul-2022-Shift-2]

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📖 Explanation

$I=int_{0}^{\;\frac{\pi}{2}} 60 \cdot \;\frac{\ sin 6 x}{\ sin x} d x$ $=60 \cdot 2 int_{0}^{\;\frac{\pi}{2}}(3-4 \ sin ^2 x)(4 \cos ^2 x-3) \cos x d x$ $=120 int_{0}^{\;\frac{\pi}{2}}(3-4 \ sin ^2 x)(1-4 \ sin ^2 x) \cos x d x$ Let $\ sin x=t \Rightarrow \cos x d x=d t$ $=120 int_{0}^{1}(3-4 t^2)(1-4 t^2) d t$ $=120 int_{0}^{1}(3-16 t^{2+}16 t^4) d t$ $=120\$ [3 t-;\frac{16 t^3}{3}+;\frac{16 t^5}{5}]_0^1=104$

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