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25 Jul 2021 Shift 2

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Physics30 questions

Q1JEE Main 2021MCQ4MMotion in a Straight Line

The relation between time $t$ and distance $x$ for a moving body is given as ${t}={mx}^{2}+{nx}$ , where ${m}$ and ${n}$ are constants. The retardation of the motion is: (When v stands for velocity)

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📖 Explanation

${t}={mx}^{2}+{n} {x}$ $\frac{1}{{v}}=\frac{{dt}}{{dx}}=2 {m} {x}+{n}$ ${v}=\frac{1}{2 {mx}+{n}}$ $\frac{{d} v}{{dt}}=-\frac{2 {m}}{(2{m} {x}+{n})^{2}}(\frac{{d} {x}}{{dt}})$ ${a}=-(2 {m}) {v}^{3}$

Q2JEE Main 2021MCQ4MOscillations

In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.

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📖 Explanation

$K=\frac{1}{2} m \omega^{2}(A^{2}-x^{2})$ $=\frac{1}{2} m \omega^{2}(A^{2}-\frac{A^{2}}{4})$ $=\frac{1}{2} m \omega^{2}(\frac{3 A^{2}}{4})$ $K=\frac{3}{4}(\frac{1}{2} m \omega^{2} A^{2})$

Q3JEE Main 2021MCQ4MLaws of Motion

A force $\vec{F}=(40 {\hat{i}}+10 {\hat{j}}) {N}$ acts on a body of mass $5 {kg}$ . If the body starts from rest, its position vector ${\vec{r}}$ at time ${t}=10 {s}$ , will be :

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📖 Explanation

${{d} {\vec{v}}}/{{dt}}={\vec{a}}={{\vec{F}}}/{{m}}=(8 \hat{i}+2 {\hat{j}}) {m} / {s}^{2}$ ${{d} {\vec{r}}}/{{dt}}={\vec{v}}=(8 {t} {\hat{i}}+2 {t} {\hat{j}}) {m} / {s}$ ${\vec{r}}=(8 {\hat{i}}+2 {\hat{j}}) \frac{{t}^{2}}{2} {m}$ At ${t}=10 {sec}$ ${\vec{r}}=[(8 \hat{i}+2 {\hat{j}}) 50] {m}$ $\Rightarrow\vec{r}=(400 {\hat{i}}+100 \hat{j}) {m}$

Q4JEE Main 2021MCQ4MRay Optics and Optical Instruments

A prism of refractive index $\mu$ and angle of prism ${A}$ is placed in the position of minimum angle of deviation. If minimum angle of deviation is also ${A}$ , then in terms of refractive index

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📖 Explanation

$\mu={\sin (\frac{{A}+\delta_{\min }}{2})}/{\sin (\frac{{A}}{2})}$ $\mu={\sin (\frac{{A}+{A}}{2})}/{\sin (\frac{{A}}{2})}$ $\mu={\sin {A}}/{\sin \frac{{A}}{2}}=2 \cos \frac{{A}}{2}$ ${A}=2 \cos ^{-1}(\frac{\mu}{2})$

Q5JEE Main 2021MCQ4MThermodynamics

A heat engine has an efficiency of $\frac{1}{6}$ . When the temperature of sink is reduced by $62°{C}$ , its efficiency get doubled. The temperature of the source is :

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📖 Explanation

$\eta =1-{{T}_{{L}}}/{{T}_{{H}}} \........$ (i) $2 \eta =1-{({T}_{{L}}-62)}/{{T}_{{H}}}=1-{{T}_{{L}}}/{{T}_{{H}}}+{\frac{62}{T}_{{H}}}$ $\Rightarrow\eta ={62}/{{T}_{{H}}} \Rightarrow \frac{1}{6}={62}/{{T}_{{H}}} \$ $\Rightarrow {T}_{{H}}=6 \\times 62=372 {K}$ ${\text{In}}°{C} \Rightarrow372-273=99°{C}$

Q6JEE Main 2021MCQ4MCurrent Electricity

In the given potentiometer circuit arrangement, the balancing length ${AC}$ is measured to be $250 {cm}$ . When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes $400 {cm}$ . The ratio of the emf of two cells, $\frac{\epsilon_{1}}{\epsilon_{2}}$ is :

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📖 Explanation

${E}_{1}={k} l_{1}$ .........(i) ${E}_{1}+{E}_{2}={k} l_{2}$ ........(ii) $\frac{{E}_{1}}{{E}_{1}+{E}_{2}}=\frac{l_{1}}{l_{2}}=\frac{250}{400}=\frac{5}{8}$ $8 {E}_{1}=5 {E}_{1}+5 {E}_{2}$ $3 {E}_{1}=5 {E}_{2}$ $\frac{E_{1}}{E_{2}}=\frac{5}{3}$

Q7JEE Main 2021MCQ4MMagnetic Effects of Current and Magnetism

Two ions having same mass have charges in the ratio 1 : 2. They are projected normally in a uniform magnetic field with their speeds in the ratio 2 : 3. The ratio of the radii of their circular trajectories is :

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📖 Explanation

${R}=\frac{{mv}}{{qB}} \Rightarrow\frac{{R}_{1}}{{R}_{2}}={\frac{{mv}_{1}}{{q}_{1} {B}}}/{\frac{{mv}_{2}}{{q}_{2} {B}}}$ $=\frac{{v}_{1}}{{q}_{1}} \\times \frac{{q}_{2}}{{v}_{2}}=\frac{{q}_{2}}{{q}_{1}} \\times \frac{{v}_{1}}{{v}_{2}}$ $=\frac{2}{1} \\times (\frac{2}{3})=\frac{4}{3}$

Q8JEE Main 2021MCQ4MAlternating Current

A $10 \Omega$ resistance is connected across $220 {V}-50 {Hz}$ AC supply. The time taken by the current to change from its maximum value to the rms value is:

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📖 Explanation

$\Rightarrow{i}={i}_{0} \sin \omega{t}$ When ${i}={i}_{0}$ ${i}_{0}={i}_{0} \sin \omega{t}_{1} \Rightarrow\omega{t}_{1}=\frac{\pi}{2} \......$ (i) When ${i}=\frac{{i}_{0}}{\sqrt{2}}$ $\frac{{i}_{0}}{\sqrt{2}}={i}_{0} \sin \omega{t}_{2} \Rightarrow\omega{t}_{2}=\frac{\pi}{4} \.......$ (ii) Time taken by current from maximum value to ${rm} {s}$ value $\Rightarrow({t}_{1}-{t}_{2})=\frac{\pi}{2 \omega}-\frac{\pi}{4 \omega}$ $=\frac{\pi}{4 \omega}=\frac{\pi}{4 \\times 2 \pi{f}}$ $=\frac{1}{8 \\times 50}$ $=\frac{1}{400} {sec}$ $=2.5 {ms}$

Q9JEE Main 2021MCQ4MMotion in a Straight Line

A balloon was moving upwards with a uniform velocity of 10 m/s. An object of finite mass is dropped from the balloon when it was at a height of 75 m from the ground level. The height of the balloon from the ground when object strikes the ground was around : (takes the value of g as 10 m/s $^2$ )

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📖 Explanation

Object is projected as shown so as per motion under gravity ${S}={ut}+\frac{1}{2} {at}^{2}$ $-75=+10 t+\frac{1}{2}(-10) t^{2} \$ $\Rightarrow t=5 {sec}$ Object takes ${t}=5 {s}$ to fall on ground Height of balloon from ground ${H}=75+{ut}$ $=75+10 \\times 5=125 {m}$

Q10JEE Main 2021MCQ4MElectrostatic Potential and Capacitance

If ${q}_{{f}}$ is the free charge on the capacitor plates and ${q}_{{b}}$ is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge ${q}_{{b}}$ can be expressed as :

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📖 Explanation

When a dielectric is inserted in a capacitor Due to free charge $\vec{E}={\vec{E}}_{0}$ only After dielectric ${E}'=\frac{{E}_{0}}{{k}}$ ${q}_{{B}}={q}_{{f}}(1-\frac{1}{{k}})$

Q11JEE Main 2021MCQ4MGravitation

Consider a planet in some solar system which has a mass double the mass of earth and density equal to the average density of earth. If the weight of an object on earth is $W$ , the weight of the same object on that planet will be :

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📖 Explanation

Let the mass of planet be $m_{p}$ and that of earth be $m$ .Given, $m_{p}=2 m$ Density of planet $(\rho _{p})=$ Density ofearth $(\rho )$ As we know that,Weight of object on earth is W $w=m g \;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$ where, $g$ is acceleration due to gravity.and $\;\; g=\;\frac{G m}{R^{2}}$ where, $\;\; G=$ universal gravitationalconstant, $R \;\; =\;\text{ radius of earth }\;$ $\;\text{ and }\; \;\; m_{e} \;\; =\;\text{ mass of earth. }\;$ $\Rightarrow g \;\; =\;\frac{G \rho \frac{4}{3} \pi R^{3}}{R^{2}}$ $\Rightarrow g \;\; =G \rho \;\frac{4}{3} \pi R$ Putting this value in Eq. (i), we get $w=m G \rho \;\frac{4}{3} \pi R \;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$ As densities of planets earth is same.i.e., $\;\; \rho _{p}=\rho$ $\Rightarrow \;\frac{m_{p}}{\frac{4}{3} \pi R_{p}^{3}} \;\; =\;\frac{m}{\frac{4}{3} \pi R^{3}}$ $\Rightarrow \;\; R_{p} \;\; =(\;\frac{m_{p}}{m} R^{3})^{1 / 3}$ $\;\; =(\;\frac{2 m}{m} R^{3})^{1 / 3}$ $\Rightarrow \;\; R_{p} \;\; =(2)^{1 / 3} R$ Weight on planet, $w_{p}=\;\frac{4}{3}(2)^{1 / 3} \pi m G \rho R \;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (iii)$ Dividing Eq. (iii) by Eq. (ii), we get $\;\frac{w_{p}}{w} \;\; =\;\frac{\;\frac{4}{3}(2)^{1 / 3} \pi m G \rho R}{m \rho G \;\frac{4}{3} \pi R}$ $\Rightarrow \;\; w_{p} \;\; =2^{1 / 3} w$

Q12JEE Main 2021MCQ4MElectrostatics

Two ideal electric dipoles ${A}$ and ${B}$ , having their dipole moment ${p}_{1}$ and ${p}_{2}$ respectively are placed on a plane with their centres at ${O}$ as shown in the figure. At point ${C}$ on the axis of dipole ${A}$ , the resultant electric field is making an angle of $37°$ with the axis. The ratio of the dipole moment of ${A}$ and ${B}, \frac{{p}_{1}}{{p}_{2}}$ is : $($ take $\sin 37°=\frac{3}{5}$ )

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📖 Explanation

$\tan 37°=\frac{3}{4}={\frac{{kP}_{2}}{{r}^{3}}}/{\frac{2 {kP}_{1}}{{r}^{3}}}=\frac{{P}_{2}}{2 {P}_{1}}=\frac{3}{4}$ $\frac{{P}_{2}}{{P}_{1}}=\frac{3}{2}$ $\frac{{P}_{1}}{{P}_{2}}=\frac{2}{3}$

Q13JEE Main 2021MCQ4MProperties of Matter

Two spherical soap bubbles of radii $r_{1}$ and $r_{2}$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to :

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📖 Explanation

no. of moles is conserved ${n}_{1}+{n}_{2}={n}_{3}$ ${P}_{1} {V}_{1}+{P}_{2} {V}_{2}={P}_{3} {V}$ $\frac{4 {S}}{{r}_{1}}(\frac{4}{3} \pi{r}_{1}^{3})+\frac{4 {S}}{{r}_{2}}(\frac{4}{3} \pi{r}_{2}^{3})$ $=\frac{4 {S}}{{r}_{3}}(\frac{4}{3} \pi{r}_{3}^{3})$ ${r}_{1}^{2}+{r}_{2}^{2}={r}_{3}^{2}$ ${r}_{3}=\sqrt{{r}_{1}^{2}+{r}_{2}^{2}}$

Q14JEE Main 2021MCQ4MUnits and Measurements

The force is given in terms of time ${t}$ and displacement $x$ by the equation ${F}={A} \cos {Bx}+{C} \sin {Dt}$ The dimensional formula of $\frac{{AD}}{{B}}$ is :

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📖 Explanation

$[{A}]=[{ML} {T}^{-2}]$ $[{B}]=[{L}^{-1}]$ $[{D}]=[{T}^{-1}]$ $[\frac{{AD}}{{B}}]=\frac{[{MLT}^{-2}]\$ $[{T}^{-1}]$}{[{L}^{-1}]$}[\frac{{D}}{{B}}]=[{ML}^{2} {T}^{-3}]$

Q15JEE Main 2021MCQ4MCurrent Electricity

The given potentiometer has its wire of resistance 10Ω. When the sliding contact is in the middle of the potentiometer wire, the potential drop across 2Ω resistor is :

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📖 Explanation

$\frac{20-{V}_{0}}{5}+\frac{0-{V}_{0}}{5}+\frac{20-{V}_{0}}{2}=0$ $4+10=\frac{2 {V}_{0}}{5}+\frac{{V}_{0}}{2}$ $14=\frac{4 {V}_{0}+5 {V}_{0}}{10}$ ${V}_{0}=\frac{140}{9} {Volt}$ Potential difference across $2 \Omega$ resistor is $20-{V}_{0}$ That is $(20-\frac{140}{9})$ Volt Hence answer is $(\frac{40}{9})$ Volt

Q16JEE Main 2021MCQ4MDual Nature of Matter and Radiation

An electron moving with speed $v$ and a photon moving with speed c, have same D-Broglie wavelength. The ratio of kinetic energy of electron to that of photon is :

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📖 Explanation

$\lambda_{{e}}=\lambda_{{Ph}}$ ${{h}}/{{p}_{{e}}}={{h}}/{{p}_{{ph}}}$ $\sqrt{2 {mk}_{{e}}}={{E}_{{ph}}}/{{c}}$ $2 {mk}_{{e}}={({E}_{{ph}})^{2}}/{{c}^{2}}$ ${{k}_{{e}}}/{{E}_{{ph}}}={{E}_{{ph}}}/{{c}^{2}}(\frac{1}{2 {m}})$ $={{p}_{{ph}}}/{{c}}(\frac{1}{2 {m}})$ $={{p}_{{e}}}/{{c}}(\frac{1}{2 {m}})$ $=\frac{{mv}}{{c}} \frac{1}{2 {m}}$ $=\frac{{v}}{2 {c}}$

Q17JEE Main 2021MCQ4MMotion in a Straight Line

The instantaneous velocity of a particle moving in a straight line is given as $v=\alphat+\betat^{2}$ , where $\alpha$ and $\beta$ are constants. The distance travelled by the particle between $1 {s}$ and $2 {s}$ is :

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📖 Explanation

${V}=\alpha{t}+\beta{t}^{2}$ $\frac{{ds}}{{dt}}=\alpha{t}+\beta{t}^{2}$ $\int_{{S}_{1}} ^{{s}_{2}} {ds}=\int_{1} ^{2}(\alpha{t}+\beta{t}^{2}) {dt}$ ${S}_{2}-{S}_{1}=\$ [\frac{\alpha {t}^{2}}{2}+\frac{\beta{t}^{3}}{3}]_{1}^{2} $As particle is not changing direction So distance$ = $displacement. Distance$ =[\frac{\alpha[4-1]}{2}+\frac{\beta[8-1]}{3}]=\frac{3}{2} \alpha+\frac{7}{2} \beta$

Q18JEE Main 2021MCQ4MRay Optics and Optical Instruments

A ray of light entering from air into a denser medium of refractive index $\frac{4}{3}$ , as shown in figure.The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle $\theta$ should be equal to :

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📖 Explanation

At maximum angle $\theta$ ray at point B goes in gazing emergence, at all less values of $\theta$ , TIR occurs. At point ${B}$ $\frac{4}{3} \\times \sin \theta"=1 \\times \sin 90°$ $\theta"=\sin ^{-1}(\frac{3}{4})$ $\theta'=(\frac{\pi}{2}-\theta")$ At point ${A}$ $1 \\times \sin \theta=\frac{4}{3} \\times \sin \theta'$ $\sin \theta=\frac{4}{3} \\times \sin (\frac{\pi}{2}-\theta")$ $\sin \theta=\frac{4}{3} \cos [\cos ^{-1} \frac{\sqrt{7}}{4}]$ $\sin \theta=\frac{4}{3} \\times \frac{\sqrt{7}}{4}$ $\theta=\sin ^{-1}(\frac{\sqrt{7}}{3})$

Q19JEE Main 2021MCQ4MDual Nature of Matter and Radiation

When radiation of wavelength $\lambda$ is incident on a metallic surface, the stopping potential of ejected photoelectrons is $4.8 {V}$ . If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes $1.6 {V}$ . The threshold wavelength of the metal is :

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📖 Explanation

${V}_{{s}}={h} v-\phi$ $4.8=\frac{{hc}}{\lambda}-\phi$ .........(i) $1.6=\frac{{hc}}{2 \lambda}-\phi$ .......(ii) Using above equation (i) - (ii) $3.2=\frac{{hc}}{\lambda}-\frac{{hc}}{2 \lambda}$ $3.2=\frac{{hc}}{2 \lambda}$ .......(iii) $[\lambda=\frac{{hc}}{6.4}]$ Put in equation (ii) $\phi=1.6$ ${{hc}}/{\lambda_{{th}}}=1.6$ $\lambda_{{th}}=\frac{{hc}}{1.6}$ $=(\frac{{hc}}{6.4}) \\times 4=4 \lambda$

Q20JEE Main 2021MCQ4MUnits and Measurements

Two vectors $\vec{X}$ and $\vec{Y}$ have equal magnitude. The magnitude of $({\vec{X}}-\vec{Y})$ is ${n}$ times the magnitude of $({\vec{X}}+{\vec{Y}})$ . The angle between $\vec{X}$ and $\vec{Y}$ is :

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📖 Explanation

Given ${X}={Y}$ $\sqrt{X^{2}+Y^{2}-2 \\times Y \cos \theta}$ $=n \sqrt{X^{2}+Y^{2}+2 \\times Y \cos \theta}$ Square both sides $2 X^{2}(1-\cos \theta)=n^{2} \.2 X^{2}(1+\cos \theta)$ $1-\cos \theta=n^{2}+n^{2} \cos \theta$ $\cos \theta=\frac{1-{n}^{2}}{1+{n}^{2}}$ $\theta=\cos ^{-1}[\frac{n^{2}-1}{-n^{2}-1}]$

Q21JEE Main 2021NAT4MThermodynamics

A system consists of two types of gas molecules ${A}$ and ${B}$ having same number density $2 \\times 10^{25} / {m}^{3}$ . The diameter of ${A}$ and ${B}$ are $10\Å$ and $5 \Å$ respectively. They suffer collision at room temperature. The ratio of average distance covered by the molecule ${A}$ to that of ${B}$ between two successive collision is ______ $\\times 10^{-2}$

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📖 Explanation

$\because$ mean free path $\lambda=\frac{1}{\sqrt{2} \pi{d}^{2} {n}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{{d}_{2}^{2} {n}_{2}}{{d}_{1}^{2} {n}_{1}}$ $=(\frac{5}{10})^{2}=0.25=25 \\times 10^{-2}$

Q22JEE Main 2021NAT4MDual Nature of Matter and Radiation

A light beam of wavelength $500 {nm}$ is incident on a metal having work function of $1.25 {eV}$ , placed in a magnetic field of intensity ${B}$ . The electrons emitted perpendicular to the magnetic field ${B}$ , with maximum kinetic energy are bent into circular arc of radius $30 {cm}$ . The value of ${B}$ is _______ $\\times 10^{-7} {T}$ Given ${hc}=20 \\times 10^{-26} {J}-{m}$ , mass of electron $=9 \\times 10^{-31} {kg}$

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📖 Explanation

By photoelectric equation $\frac{{hc}}{\lambda}-\phi={k}_{\max }$ ${k}_{\max }=\frac{1240}{500}-1.25 \approx 1.25$ ${r}={\sqrt{2 {mk}}}/{{eB}}$ ${B}={\sqrt{2 {mk}}}/{{er}}$ $=125 \\times 10^{-7} {T}$

Q23JEE Main 2021NAT4MElectromagnetic Waves

A message signal of frequency 20 kHz and peak voltage of 20 volt is used to modulate a carrier wave of frequency 1 MHz and peak voltage of 20 volt. The modulation index will be :

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📖 Explanation

Modulation index $\mu=\frac{A_{m}}{A_{c}}=\frac{20}{20}=1$

Q24JEE Main 2021NAT4MCurrent Electricity

A $16 \Omega$ wire is bend to form a square loop. A $9 {V}$ supply having internal resistance of $1 \Omega$ is connected across one of its sides. The potential drop across the diagonals of the square loop is _________ $\\times 10^{-1} {V}$

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📖 Explanation

here assume current as By KVL in outer loop $9-12 {i}-4 {i}=0$ $16 {i}=9$ $8 {i}=\frac{9}{2}=4.5$ $=45 \\times 10^{-1}$

Q25JEE Main 2021NAT4MAlternating Current

Two circuits are shown in the figure (a) & (b). At a frequency of _______ rad/s the average power dissipated in one cycle will be same in both the circuits.

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📖 Explanation

For figure (a) ${P}_{{\text{avg}}}={{v}_{{\text{rms}}}^{2}}{{R}}$ ${{v}_{{\text{rms}}}^{2}}/{{Z}^{2}} \\times {R}={{v}_{{\text{rms}}}^{2}}{{R}} \\times 1$ ${R}^{2}={Z}^{2}$ $25=(\sqrt{({x}_{{C}}-{x}_{{L}})^{2}+5^{2}})^{2}$ $25=({x}_{{c}}-{x}_{{L}})^{2}+25$ ${x}_{{c}}={x}_{{L}} \Rightarrow\frac{1}{\omega{C}}=\omega{L}$ $\omega^{2}=\frac{1}{{LC}}=\frac{10^{6}}{0.1 \\times 40}$ $\omega=500$

Q26JEE Main 2021NAT4MAtoms and Nuclei

From the given data, the amount of energy required to break the nucleus of aluminium ${ }_{13}^{27} {Al}$ is _____ ${x} \\times 10^{-3} {J}$ Mass of neutron $=1.00866 {u}$ Mass of proton $=1.00726 {u}$ Mass of Aluminium nucleus $=27.18846 {u}$ (Assume 1 u corresponds to $x$ J of energy) (Round off to the nearest integer)

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📖 Explanation

$\Deltam=({Zm}_{{P}}+({A}-{Z}) {m}_{{n}})-{M}_{{A} l}$ $=(13 \\times 1.00726+14 \\times 1.00866)$ $-27.18846$ $=27.21562-27.18846$ $=0.02716 {u}$ ${E}=27.16 {x} \\times 10^{-3} {J}$

Q27JEE Main 2021NAT4MWork, Power and Energy

A force of ${F}=(5 {y}+20) {\hat{j}} {N}$ acts on a particle. The workdone by this force when the particle is moved from ${y}=0 {m}$ to ${y}=10 {m}$ is ______ J.

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📖 Explanation

${F}=(5 {y}+20) {\hat{j}}$ $\omega=\intF d y=\int_{0} ^{10}(5 y+20) d y$ $=(\frac{5 y^{2}}{2}+20 y)_{0}^{10}$ $=\frac{5}{2} \\times 100+20 \\times 10$ $=250+200=450 {J}$

Q28JEE Main 2021NAT4MRotational Motion

A solid disc of radius $20 {cm}$ and mass $10 {kg}$ is rotating with an angular velocity of 600 rpm, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in $10 { s}$ is _____ $\pi\\times 10^{-1} {Nm}$ .

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📖 Explanation

$\tau =\frac{\Delta{L}}{\Delta{t}}={{I}(\omega_{{f}}-\omega_{{i}})}/{\Delta{t}}$ $\tau ={\frac{{mR}^{2}}{2} \\times [0-\omega]}/{\Delta{t}}$ $=\frac{10 \\times (20 \\times 10^{-2})^{2}}{2} \\times \frac{600 \\times \pi}{30 \\times 10}$ $=0.4 \pi=4 \pi\\times 10^{-2}$

Q29JEE Main 2021NAT4MSemiconductor Electronics

In a semiconductor, the number density of intrinsic charge carriers at $27°{C}$ is $1.5 \\times 10^{16} / {m}^{3}$ . If the semiconductor is doped with impurity atom, the hole density increases to $4.5 \\times 10^{22} / {m}^{3}$ . The electron density in the doped semiconductor is_______ $\\times 10^{9} / {m}^{3}$

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📖 Explanation

$n_{e} n_{h}=n_{i}^{2}$ $n_{e}=\frac{n_{i}^{2}}{n_{h}}=\frac{(1.5 \\times 10^{16})^{2}}{4.5 \\times 10^{22}}$ $=\frac{1.5 \\times 1.5 \\times 10^{32}}{4.5 \\times 10^{22}}$ $5 \\times 10^{9} / {m}^{3}$

Q30JEE Main 2021NAT4MAtoms and Nuclei

The nuclear activity of a radioactive element becomes $(\frac{1}{8})^{\text{th }}$ of its initial value in 30 years. The half-life of radioactive element is ________ years.

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📖 Explanation

${A}={A}_{0} {e}^{-\lambdat}$ $\frac{{A}_{0}}{8}={A}_{0} {e}^{-\lambda{t}} \Rightarrow\lambda{t}=\ln 8$ $\lambda{t}=3 \ln 2$ $\frac{\ln 2}{\lambda}=\frac{{t}}{3}=\frac{30}{3}=10$ years

Chemistry30 questions

Q31JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

In the following the correct bond order sequence is:

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📖 Explanation

${O}_{2}$ (16 electrons) $\sigma_{1 {s}}^{2}, \sigma_{1 {s}}^{* 2}, \sigma_{2 {s}}^{2}, \sigma_{2 {s}}^{* 2}, \sigma_{2 {p_z}}^{2}$ $\pi_{2 p_{{x}}}^{2}=\pi_{2 {py}}^{2}, \pi_{2 {p}_{{x}}}^{* 1}=\pi_{2 {py}}^{* 1}, \sigma_{2 {p}_{{z}}}^{*}$ Bond order of ${O}_{2} \Rightarrow2$ Bond order of ${O}_{2}^{-} \Rightarrow1.5$ Bond order of ${O}_{2}^{2-} \Rightarrow1$ Bond order of ${O}_{2}^{+} \Rightarrow 2.5$

Q32JEE Main 2021MCQ4MBiomolecules

A biodegradable polyamide can be made from:

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Q33JEE Main 2021MCQ4Mp Block Elements

Match List I with List II : List-I Elements List-IIProperties (a) Li (i) Poor water solubility of $I^{-}$ salt (b) Na (ii) Most abundant element in cell fluid (c) K (iii) Bicarbonate salt used in fire extinguisher (d) Cs (iv) Carbonate salt decomposeseasily on heatingChoose the correct answer from the options given below :

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📖 Explanation

(a) ${C}_{{S}} {I}$ salt is poor water soluble due to it's low hydration energy (b) ${NaHCO}_{3}$ is used in fire extinguisher (c) ${K}$ is most abundant element in cell fluid (d) ${Li}_{2} {CO}_{3}$ decomposes easily due to high covalent character caused by small size ${Li}^{+}$ cation.

Q34JEE Main 2021MCQ4MCoordination Compounds

Which one of the following metal complexes is most stable?

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📖 Explanation

Complex $[{Co}({en})_{3}] {Cl}_{2}$ is most stable complex among the given complex compounds because more number of chelate rings are present in this complex as compare to others. (1) $[{Co}({en})({NH}_{3})_{4}] {Cl}_{2}$ 1 chelate ring (2) $[{Co}({en})_{3}] {Cl}_{2}$ 3 chelate ring (3) $[{Co}({en})_{2}({NH}_{3})_{2}] {Cl}_{2}$ 2 chelate ring (4) $[{Co}({NH}_{3})_{6}] {Cl}_{2}$ 0 chelate ring

Q35JEE Main 2021MCQ4Md and f Block Elements

Match List I with List II : (Both having metallurgical terms) List-I List-II (a) Concentration of Ag ore (i) Reverberatory furnace (b) Blast furnace (ii) Pig iron (c) Blister copper (iii) Leaching with dilute NaCN solution (d) Froth floatation method (iv) Sulfide ores Choose the correct answer from the options givenbelow :

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Q36JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

The ionic radii of ${F}^{-}$ and ${O}^{2-}$ respectively are $1.33$ $\Å$ and $1.4 \Å$ , while the covalent radius of ${N}$ is $0.74 \Å$ The correct statement for the ionic radius of ${N}^{3-}$ from the following is :

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📖 Explanation

${F}^{-}, {O}^{2-}$ and ${N}^{3-}$ all are isoelectronic species in which ${N}^{3-}$ have least number of protons due to which it's size increases as least nuclear attraction is experienced by the outer shell electrons. Size order ${N}^{3-}>{O}^{2-}>{F}^{-}$

Q37JEE Main 2021MCQ4MHaloalkanes and Haloarenes

The correct decreasing order of densities of the following compounds is :

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Q38JEE Main 2021MCQ4MAmines

Consider the above reaction, the Product "P" is :

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Q39JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

A reaction of benzonitrile with one equivalent $CH_3MgBr$ followed by hydrolysis produces a yellow liquid "P". The compound "P" will give positive_____.

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Q40JEE Main 2021MCQ4Md and f Block Elements

The spin only magnetic moments (in BM) for free ${Ti}^{3+}, {V}^{2+}$ and ${Sc}^{3+}$ ions respectively are (At.No. Sc : 21, Ti : 22, V: 23)

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📖 Explanation

$\text{table} ,, \mu=\sqrt{{n}({n}+2)} {BM}; {T} 1^{+3}=[{Ar}] 3 {d}^{1}, {n}=1 , \mu=1.73 {BM}; {V}^{+2}=[{Ar}] 3 {d}^{3} , {n}=3 , \mu=3.87 {BM} ;{Sc}^{+3}=[{Ar}] 3 {d}^{0} 4 {s}^{0} , {n}=0 , \mu=0$

Q41JEE Main 2021MCQ4MBiomolecules

Which one of the following is correct structure for cytosine ?

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Q42JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

Identify the species having one $\pi$ -bond and maximum number of canonical forms from the following:

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📖 Explanation

Among ${SO}_{3}, {O}_{2}, {SO}_{2}$ and ${CO}_{3}^{2-}$ , only ${O}_{2}$ and ${CO}_{3}^{2-}$ has only one $\pi$ -bond

Q43JEE Main 2021MCQ4Md and f Block Elements

Which one of the following metals forms interstitial hydride easily ?

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Q44JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Maleic anhydride Maleic anhydride can be prepared by :

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Q45JEE Main 2021MCQ4Mp Block Elements

Given below are two statements : Statement I : Chlorofluoro carbons breakdown by radiation in the visible energy region and release chlorine gas in the atmosphere which then reacts with stratospheric ozone. Statement II : Atmospheric ozone reacts with nitric oxide to give nitrogen and oxygen gases, which add to the atmosphere. For the above statements choose the correct answer from the options given below :

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📖 Explanation

Statement (1) CFCs are broken down by powerful UV radiation and releases chlorine free radical which reacts with ozone and start chain reaction. ${CF}_{2} {Cl}_{2({g})} {\to }^{{UV}} {{C}^{\bullet }} {l}_{({g})}+{{C}^{\bullet }} {F}_{2} {Cl}_{({g})}$ ${{C}}^{\bullet } {l}_{({g})}+{O}_{3({g})} \\to {{C}^{\bullet }} {lO}_{({g})}+{O}_{2({g})}$ ${{C}^{\bullet }} {O}_{({g})}+{O}_{({g})} \\to {{C}^{\bullet }} {l}_{({g})}+{O}_{2({g})}$ Statement (2) Atmosphere ozone reacts with nitric oxide to produce nitrogen dioxide and oxygen. ${NO}_{({g})}+{O}_{3({g})} \\to {NO}_{2({g})}+{O}_{2({g})}$

Q46JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Consider the above reaction sequence, Product "A" and Product "B" formed respectively are :

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Q47JEE Main 2021MCQ4MSolutions

Match List I with List II : List-I Example of colloids List-II Classification (a) Cheese (i) dispersion of liquid in liquid (b) Pumice stone (ii) dispersion of liquid in gas (c) Hair cream (iii) dispersion of gas in solid (d) Cloud (iv) dispersion of liquid in solid Choose the most appropriate answer from the options given below

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Q48JEE Main 2021MCQ4MAmines

What is the major product "P" of the following reaction ?

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Q49JEE Main 2021MCQ4MRedox Reactions

Identify the process in which change in the oxidation state is five :

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📖 Explanation

${MnO}_{4}^{-}+5e \\to {Mn}^{2+}$

Q50JEE Main 2021MCQ4MOrganic Chemistry - Some Basic Principles

Which among the following is the strongest acid ?

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Q51JEE Main 2021NAT4MChemical Thermodynamics

A system does 200 J of work and at the same time absorbs 150 J of heat. The magnitude of the change in internal energy is __________ J. (Nearest integer)

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📖 Explanation

${w}=-200 {J}, {q}=+150:$ $\Delta{U}={q}+{w}$ $\Delta{U}=150-200=-50 {J}:$ magnitude $=50 {J}=|\Delta{U}|$

Q52JEE Main 2021NAT4MStructure of Atom

An accelerated electron has a speed of $5 \\times 10^{6} {ms}^{-1}$ with an uncertainty of $0.02 \%$ . The uncertainty in finding its location while in motion is $x \\times 10^{-9} {m}$ . The value of $x$ is __________ $.$ (Nearest integer) [Use mass of electron $=9.1 \\times 10^{-31} {kg}$ , ${h}=6.63 \\times 10^{-34} {Js}, \pi=3.14]$

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📖 Explanation

$\Deltav=\frac{0.02}{100} \\times 5 \\times 10^{6}=10^{3} {m} / {s}$ $\Delta{x} \.\Delta{v}=\frac{{h}}{4 \pi{m}}$ ${x} \\times 10^{-9} \\times 10^{3}$ $=\frac{6.63 \\times 10^{-34}}{4 \\times 3.14 \\times 9.1 \\times 10^{-31}}$ ${x} \\times 10^{-9} \\times 10^{3}=0.058 \\times 10^{-3}$ ${x}=\frac{0.058 \\times 10^{-6}}{10^{-9}}=58$

Q53JEE Main 2021NAT4Md and f Block Elements

Number of electrons present in 4f orbital of $Ho^{3+}$ ion is __________ . (Given Atomic No. of Ho = 67)

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📖 Explanation

${Ho}=[{Xe}] 4 {f}^{11} 6 {s}^{2}$ ${Ho}^{3+}=[{Xe}] 4 {f}^{10}$ so number of ${e}^{-}$ present in $4 {f}$ is 10 .

Q54JEE Main 2021NAT4MHaloalkanes and Haloarenes

Consider the above chemical reaction. The total number of stereoisomers possible for Product 'P' is __________ .

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Q55JEE Main 2021NAT4MChemical Kinetics

For a chemical reaction ${A} \\to {B}$ , it was found that concentration of ${B}$ is increased by $0.2 {mol} {L}^{-1}$ in 30 min. The average rate of the reaction is _________ $\\times 10^{-1} {mol} {L}^{-1} {h}^{-1}$ . (in nearest integer)

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📖 Explanation

$\text{table} A ,\to ,B;t =0,, 0;t =30\min,, 0.2\text{M}$ Av. rate of reaction $=-\frac{\Delta[{A}]}{\Delta{t}}=\frac{\Delta[{B}]}{\Delta{t}}=\frac{(0.2-0)}{\frac{1}{2}}$ $=0.4=4 \\times 10^{-1} {mol} / {L} \\times {hr}$

Q56JEE Main 2021NAT4MSome Basic Concepts of Chemistry

The number of significant figures in 0.00340 is __________ .

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Q57JEE Main 2021NAT4MIonic Equilibrium

Assuming that ${Ba}({OH})_{2}$ is completely ionised in aqueous solution under the given conditions the concentration of ${H}_{3} {O}^{+}$ ions in $0.005 {M}$ aqueous solution of ${Ba}({OH})_{2}$ at $298 {K}$ is __________ $\\times 10^{-12} {mol} {L}^{-1}$ . (Nearest integer)

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📖 Explanation

${Ba}({OH})_{2} \\to {Ba}^{+2}+2 {OH}^{-}$ $↓$ $2 \\times 0.005=0.01=10^{-2}$ At $298 {K}:$ in aq. solution $[{H}_{3} {O}^{+}]\$ [{OH}^{-}]$=10^{-14}[{H}_{3} {O}^{+}]=\frac{10^{-14}}{10^{-2}}=10^{-12}$

Q58JEE Main 2021NAT4MPractical Organic Chemistry

0.8 g of an organic compound was analysed by Kjeldahl's method for the estimation of nitrogen. If the percentage of nitrogen in the compound was found to be 42%, then __________ mL of 1 M $H_2SO_4$ would have been neutralized by the ammonia evolved during the analysis.

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📖 Explanation

Organic compound : $0.8 {gm}$ wt. of ${N}=(\frac{42}{100} \\times 0.8) {gm}$ mole of ${N}=\frac{42 \\times 0.8}{100 \\times 14}=\frac{2.4}{100} {mol}$ moles of ${NH}_{3}=\frac{2.4}{100}$ $\text{table} 2 {NH}_{3},+,{H}_{2} {SO}_{4} ,\\to ,({NH}_{4})_{2} {SO}_{4} ;\frac{2.4}{100} \text{mole} ,,\frac{1.2}{100} \text{mole}$ $\frac{1.2}{100}=1 \\times {V}(l)$ $\Rightarrow{V}_{{H}_{2} {SO}_{4}}=\frac{1.2}{100} l=12 {m} l$

Q59JEE Main 2021NAT4MSolutions

When $3.00 {g}$ of a substance ' ${X}'$ is dissolved in $100 {g}$ of ${CCl}_{4}$ , it raises the boiling point by $0.60 {K}$ . The molar mass of the substance ' $X$ ' is ________ g $m o l^{-1}$ . (Nearest integer). $[$ Given ${K}_{{b}}$ for ${CCl}_{4}$ is $5.0 {K} {kg} {mol}^{-1}]$

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📖 Explanation

$\Delta{T}_{{b}} ={K}_{{b}} \\times \text{molality }$ $0.60 =5 \\times (\frac{3 / {M}}{100 / 100})$ ${M} =250$

Q60JEE Main 2021NAT4MStates of Matter

An LPG cylinder contains gas at a pressure of 300 ${kPa}$ at $27°{C}$ . The cylinder can withstand the pressure of $1.2 \\times 10^{6} {Pa}$ . The room in which the cylinder is kept catches fire. The minimum temperature at which the bursting of cylinder will take place is __________ ${ }°{C}$ . (Nearest integer)

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📖 Explanation

$\frac{{P}_{1}}{{T}_{1}}=\frac{{P}_{2}}{{T}_{2}} \Rightarrow\frac{300 \\times 10^{3}}{300}$ $=\frac{1.2 \\times 10^{6}}{{T}_{2}}$ $\Rightarrow{T}_{2}=1200 {K}$ $T_2=927°C$

Mathematics30 questions

Q61JEE Main 2021MCQ4MBinomial Theorem

The sum of all those terms which are rational numbers in the expansion of $(2^{1 / 3}+3^{1 / 4})^{12}$ is:

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📖 Explanation

${T}_{{r}+1}={ }^{12} {C}_{{r}}(2^{1 / 3})^{{r}} \.(3^{1 / 4})^{12-{r}}$ ${T}_{{r}+1}$ will be rational number when ${r}=0,3,6,9,12$ $\& \ {r}=0,4,8,12$ $\Rightarrow{r}=0,12$ ${T}_{1}+{T}_{13}=1 \\times 3^{3}+1 \\times 2^{4} \\times 1$ $=24+16=43$

Q62JEE Main 2021MCQ4MStatistics

The first of the two samples in a group has 100 items with mean 15 and standard deviation $3 .$ If the whole group has 250 items with mean $15.6$ and standard deviation $\sqrt{13.44}$ , then the standard deviation of the second sample is:

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📖 Explanation

$\text{table} {n}_{1}=100,{m}=250;\overline{{X}}_{1}=15,\overline{{X}}=15.6;{V}_{1}({x})=9 , {\text{Var}}({x})=13.44$ $\sigma^{2}=\frac{{n}_{1} \sigma_{1}^{2}+{n}_{2} \sigma_{2}^{2}}{{n}_{1}+{n}_{2}}+\frac{{n}_{1} {n}_{2}}{({n}_{1}+{n}_{2})^{2}}$ $(\ov {{x}}_{1}-\ov {{x}}_{2})^{2}$ ${n}_{2}=150, \overline{{x}}_{2}=16, {V}_{2}({x})=\sigma_{2}$ $13.44=\frac{100 \\times 9+150 \\times \sigma_{2}^{2}}{250}$ $+\frac{100 \\times 150}{(250)^{2}} \\times 1$ $\Rightarrow\sigma_{2}^{2}=16 \Rightarrow\sigma_{2}=4$

Q63JEE Main 2021MCQ4MLimits, Continuity and Differentiability

If $f(x)=\{\text{table} \int_{0} ^{x}(5+|1-t|) d t{,}, x>2 ;5 x+1{, }, x \\le 2\$ , then

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📖 Explanation

$f({x})=\int_{0} ^{1}(5+(1-{t})) {dt}+\int_{0} ^{x}(5+({t}-1)) {dt}$ $=6-\frac{1}{2}+(4 t+\frac{t^{2}}{2})_{1} ^{x}$ $=\frac{11}{2}+4 x+\frac{x^{2}}{2}-4-\frac{1}{2}$ $=\frac{x^{2}}{2}+4 x+1$ $f(2^{+})=2+8+1=11$ $f(2)=f(2)=5 \\times 2+1=11$ $\Rightarrow$ continuous at $x=2$ Clearly differentiable at $x=1$ ${Lf}'(2)=5$ ${Rf}'(2)=6$ $\Rightarrow$ not differentiable at $x=2$

Q64JEE Main 2021MCQ4MBinomial Theorem

If the greatest value of the term independent of ' ${x}$ ' in the expansion of $(x \sin \alpha+a \frac{\cos \alpha}{x})^{10}$ is $\frac{10 !}{(5 !)^{2}}$ ,then the value of 'a' is equal to:

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📖 Explanation

${T}_{{r}+1}={ }^{10} {C}_{{r}}({x} \sin \alpha)^{10-{r}}(\frac{{a} \cos \alpha}{{x}})^{{r}}$ ${r}=0,1,2, \....., 10$ ${T}_{r+1}$ will be independent of ${x}$ when $10-2 {r}=0 \Rightarrow{r}=5$ ${T}_{6}={ }^{10} {C}_{5}({x} \sin \alpha)^{5} \\times (\frac{{a} \cos \alpha}{{x}})^{5}$ $={ }^{10} {C}_{5} \\times {a}^{5} \\times \frac{1}{2^{5}}(\sin 2 \alpha)^{5}$ will be greatest when $\sin 2 \alpha=1$ $\Rightarrow{ }^{10} {C}_{5}\frac{{a}^{5}}{2^{5}}={ }^{10} {C}_{5} \$ $\Rightarrow {a}=2$

Q65JEE Main 2021MCQ4MSets and Relations

Consider the statement "The match will be played only if the weather is good and ground is not wet". Select the correct negation from the following:

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📖 Explanation

${p}$ : weather is food ${q}$ : ground is not wet $\sim ({p} \wedge {q}) \equiv \sim {p} \vee \sim {q}$ $\equiv$ weather is not good or ground is wet

Q66JEE Main 2021MCQ4MTrigonometric Ratios and Identities

The value of $\cot \frac{\pi}{24}$ is:

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📖 Explanation

$\cot \theta=\frac{1+\cos 2 \theta}{\sin 2 \theta}$ $={1+(\frac{\sqrt{3}+1}{2 \sqrt{2}})}/{(\frac{\sqrt{3}-1}{2 \sqrt{2}})}$ $\theta=\frac{\pi}{24}$ $\Rightarrow\cot (\frac{\pi}{24})={1+(\frac{\sqrt{3}+1}{2 \sqrt{2}})}/{(\frac{\sqrt{3}-1}{2 \sqrt{2}})}$ $=\frac{(2 \sqrt{2}+\sqrt{3}+1)}{(\sqrt{3}-1)} \\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$ $=\frac{2 \sqrt{6}+2 \sqrt{2}+3+\sqrt{3}+\sqrt{3}+1}{2}$ $=\sqrt{6}+\sqrt{2}+\sqrt{3}+2$

Q67JEE Main 2021MCQ4MLimits, Continuity and Differentiability

The lowest integer which is greater than $(1+\frac{1}{10^{100}})^{10^{100}}$ is ________.

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📖 Explanation

Let ${P}=(1+\frac{1}{10^{100}})^{10^{100}}$ Let ${x}=10^{100}$ $\Rightarrow{P}=(1+\frac{1}{{x}})^{{x}}$ $\Rightarrow{P}=1+({x})(\frac{1}{{x}})+\frac{({x})({x}-1)}{2!} \. \frac{1}{{x}^{2}}$ $+\frac{(x)(x-1)(x-2)}{3!} \.\frac{1}{x^{3}}+\.........$ (upto $10^{100}+1$ terms $)$ $\Rightarrow{P}=1+1+(\frac{1}{2!}-\frac{1}{2 !{x}^{2}})$ $+(\frac{1}{3!}-\......)+{so}$ on $\Rightarrow {P}=2+($ Positive value less then $\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\.....)$ Also e $=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+......$ $\Rightarrow\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+....={e}-2$ $\Rightarrow{P}=2+($ positive value less then ${e}-2)$ $\Rightarrow{P} \in(2,3)$ $\Rightarrow$ least integer value of ${P}$ is 3

Q68JEE Main 2021MCQ4MDefinite Integrals

The value of the integral $\int_{-1} ^{1} \log (x+\sqrt{x^{2}+1}) d x$ is:

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📖 Explanation

Let $I=\int_{-1} ^{1} \log (x+\sqrt{x^{2}+1}) d x$ $\because\log (x+\sqrt{x^{2}+1})$ is an odd function $\therefore {I}=0$

Q69JEE Main 2021MCQ4MVector Algebra

Let ${a}, {b}$ and ${c}$ be distinct positive numbers. If the vectors $a \hat{i}+{{a}}\hat{j}+{c} {\hat{k}}, {\hat{i}}+\hat{k}$ and ${c} {\hat{i}}+{{c}\hat{j}}+{b} {\hat{k}}$ are co-planar, then ${c}$ is equal to:

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📖 Explanation

Because vectors are coplanar Hence $\begin{vmatrix}a & a & c \\ 1 & 0 & 1 \\ c & c & b\end{vmatrix}=0$ $\Rightarrow{c}^{2}={ab} \Rightarrow{c}=\sqrt{{ab}}$

Q70JEE Main 2021MCQ4MSequences and Series

If $[{x}]$ be the greatest integer less than or equal to ${x}$ , then $\sum_{{n}=8} ^{100}[{(-1)^{{n}} {n}}/{2}]$ is equal to:

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📖 Explanation

$\sum_{{n}=8} ^{100}[{(-1)^{{n}} .{n}}/{2}]$ $=4-5+5-6+6+\....-50$ $+50=4$

Q71JEE Main 2021MCQ4MMatrices and Determinants

The number of distinct real roots of $\begin{vmatrix}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{vmatrix}=0$ in the interval $-\frac{\pi}{4} \\le {x} \\le \frac{\pi}{4}$ is:

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📖 Explanation

$\begin{vmatrix}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{vmatrix}=0,$ $-\frac{\pi}{4} \\le {x} \\le \frac{\pi}{4}$ Apply $: {R}_{1} \\to {R}_{1}-{R}_{2}\ \& \ {R}_{2} \\to {R}_{2}-{R}_{3}$ $\begin{vmatrix}\sin x-\cos {x} & \cos x-\sin {x} & 0 \\ 0 & \sin {x}-\cos {x} & \cos x-\sin x \\ \cos x & \cos x & \sin x\\end{vmatrix}=0$ $(\sin x-\cos x)^{2}$ $\begin{vmatrix}1 & -1 & 0 \\ 0 & 1 & -1 \\ \cos x & \cos x & \sin x\end{vmatrix}=0$ $(\sin x-\cos x)^{2}(\sin x+2 \cos x)=0$ $\therefore {x}=\frac{\pi}{4}$

Q72JEE Main 2021MCQ4MVector Algebra

If $|{\vec{a}}|=2,|{{\vec{b}}}|=5$ and $|{{\vec{a}}} \\times {\vec{b}}|=8$ , then $|{\vec{a}} \.{\vec{b}}|$ is equal to :

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📖 Explanation

$|{\vec{a}}|=2,|{{\vec{b}}}|=5$ $|{{\vec{a}}} \\times {\vec{b}}|=|{\vec{a}}| |{{\vec{b}}}| \sin \theta=\\pm 8$ $\sin \theta=\\pm \frac{4}{5}$ $\therefore \vec{a} \.{b}v=|\vec{a}||\vec{b}| \cos \theta$ $=10 \.(\\pm \frac{3}{5})=\\pm 6$ $|{a} ^{\to }\.\vec{b}|=6$

Q73JEE Main 2021MCQ4MQuadratic Equation and Inequalities

The number of real solutions of the equation, $x^{2}-|x|-12=0$ is:

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📖 Explanation

$|x|^{2}-|x|-12=0$ $(|x|+3)(|x|-4)=0$ $|x|=4 \Rightarrowx=\\pm 2$

Q74JEE Main 2021MCQ4MFunctions

Consider function $f: {A} \\to {B}$ and $g: {B} \\to {C}({A}, {B}, {C} \subseteq{R})$ such that (gof) $^{-1}$ exists, then:

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📖 Explanation

$\therefore$ (gof) $^{-1}$ exist $\Rightarrow$ gof is bijective $\Rightarrow ' f'$ must be one-one and 'g' must be ONTO

Q75JEE Main 2021MCQ4MMatrices and Determinants

If ${P}=\begin{bmatrix}1 & 0 \\ 1 / 2 & 1\end{bmatrix}$ , then ${P}^{50}$ is:

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📖 Explanation

$P=\begin{bmatrix}1 & 0 \\ 1 / 2 & 1\end{bmatrix}$ $P^2=\begin{bmatrix}1 & 0 \\ 1 / 2 & 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ 1 / 2 & 1\end{bmatrix}$ $=\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$ $P^3=\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ 1 / 2 & 1\end{bmatrix}$ $=\begin{bmatrix}1 & 0 \\ 3 / 2 & 1\end{bmatrix}$ $P^4=\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$ $=\begin{bmatrix}1 & 0 \\ 2 & 1\end{bmatrix}$ ∴ $P^50=\begin{bmatrix}1 & 0 \\ 25 & 1\end{bmatrix}$

Q76JEE Main 2021MCQ4MProbability

Let $x$ be a random variable such that the probability function of a distribution is given by ${P}({X}=0)=\frac{1}{2}, {P}({X}= {j})={1}/{3^{{j}}}$ $({j}=1,2,3, \......, \\infty )$ , Then the mean of the distribution and ${P}({X}$ is positive and even) respectively are:

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📖 Explanation

mean $=\sum{x}_{1} {p}_{{i}}=\sum_{{r}=0} ^{\\infty } {r} \.{1}/{3^{{r}}}=\frac{3}{4}$ ${p}({x}$ is even ) $=\frac{1}{3^{2}}+\frac{1}{3^{4}}+\....\\infty$ $=\frac{\frac{1}{9}}{1-\frac{1}{9}}=\frac{1 / 9}{8 / 9}=\frac{1}{8}$

Q77JEE Main 2021MCQ4MEllipse

If a tangent to the ellipse $x^{2}+4 y^{2}=4$ meets the tangents at the extremities of its major axis at ${B}$ and ${C}$ , then the circle with ${BC}$ as diameter passes through the point :

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📖 Explanation

$\frac{x^{2}}{4}+\frac{y^{2}}{1}=1$ Equation of tangent is $(\cos \theta) x+2 \sin \thetay=2$ ${B}(-2, \frac{1+\cos \theta}{\sin \theta}), \;\;\;{C}(2, \frac{1-\cos \theta}{\sin \theta})$ ${B}(-2, \cot \frac{\theta}{2}) \;\;\;{C}(2, \tan \frac{\theta}{2})$ Equation of circle is $(x+2)(x-2)+(y-\cot \frac{\theta}{2})$ $(y-\tan \frac{\theta}{2})=0$ $x^{2}-4+y^{2}-(\tan \frac{\theta}{2}+\cot \frac{\theta}{2})$ $y+1=0$ so, $(\sqrt{3}, 0)$ satisfying option (1)

Q78JEE Main 2021MCQ4MStraight Lines and Pair of Straight Lines

Let the equation of the pair of lines, ${y}={px}$ and $y=q x$ , can be written as $(y-p x)(y-q x)=0$ Then the equation of the pair of the angle bisectors of the lines $x^{2}-4 x y-5 y^{2}=0$ is:

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📖 Explanation

$\frac{x^{2}-y^{2}}{1-(-5)}=\frac{x y}{-2}$ $\frac{x^{2}-y^{2}}{6}=\frac{x y}{-2}$ $\Rightarrowx^{2}-y^{2}=-3 x y$ $\Rightarrowx^{2}+3 x y-y^{2}=0$

Q79JEE Main 2021MCQ4MPermutations and Combinations

If ${ }^{{n}} {P}_{{r}}={ }^{{n}} {P}_{{r}+1}$ and ${ }^{{n}} {C}_{{r}}={ }^{{n}} {C}_{{r}-1}$ , then the value of ${r}$ is equal to:

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📖 Explanation

${ }^{{n}} {P}_{{r}}={ }^{{n}} {P}_{{rtl}} \Rightarrow\frac{{n} !}{({n}-{r}) !}=\frac{{n} !}{({n}-{r}-1) !}$ $\Rightarrow({n}-{r})=1$ ........(1) ${ }^{{n}} {C}_{{r}}={ }^{{n}} {C}_{{r}-1}$ $\Rightarrow\frac{{n} !}{{r} !({n}-{r}) !}=\frac{{n} !}{({r}-1) !({n}-{r}+1) !}$ $\text{r}\frac{1}{r(n-r) !}=\frac{1}{(n-r+1)(n-r) !}$ $\Rightarrow{n}-{r}+1={r}$ $\Rightarrow{n}+1=2 {r}$ .......(2) (1) $\Rightarrow2 {r}-1-{r}=1 \Rightarrow{r}=2$

Q80JEE Main 2021MCQ4MDifferential Equations

Let $y=y(x)$ be the solution of the differential equation $x d y=(y+x^{3} \cos x) d x$ with $y(\pi)=0$ , then ${y}(\frac{\pi}{2})$ is equal to:

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📖 Explanation

$x d y=(y+x^{3} \cos x) d x$ $x d y=y d x+x^{3} \cos x d x$ $\frac{x d y-y d x}{x^{2}}=\frac{x^{3} \cos x d x}{x^{2}}$ $\frac{d}{d x}(\frac{y}{x})=\intx \cos x d x$ $\Rightarrow \frac{y}{x}=x \sin x-\int1 \.\sin x d x$ $\frac{y}{x}=x \sin x+\cos x+C$ $\Rightarrow0=-1+C \RightarrowC=1, x=\pi, y=0$ so $\frac{y}{x}=x \sin x+\cos x+1$ $y=x^{2} \sin x+x \cos x+x$ $\;\; x=\frac{\pi}{2}$ $y(\frac{\pi}{2})=\frac{\pi^{2}}{4}+\frac{\pi}{2}$

Q81JEE Main 2021NAT4MBinomial Theorem

Let ${n} \in {N}$ and $[{x}]$ denote the greatest integer less than or equal to $x$ . If the sum of $(n+1)$ terms ${ }^{{n}} {C}_{0}, 3 \.{ }^{{n}}{C}_{1}, 5 \.{ }^{{n}} {C}_{2}, 7 \.{ }^{{n}} {C}_{3}, \..... .$ is equal to $2^{100} \.101$ ,then $2[\frac{{n}-1}{2}]$ is equal to _________.

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📖 Explanation

$1 { }^{{n}} {C}_{0}+3 \.{ }^{{n}}{C}_{1}+ 5 \.{ }^{{n}} {C}_{2} +\.....$ $+(2 {n}+1) \.{ }^{{n}} {C}_{{n}}$ ${T}_{{r}}=(2 {r}+1)^{{n}} {C}_{{r}}$ ${S}=\sum{T}_{{r}}$ ${S}=\sum(2 {r}+1)^{{n}} {C}_{{r}}=\sum2 {r}^{{n}} {C}_{{r}}$ $+\sum^{{n}} {C}_{{r}}$ ${S}=2({n} \. 2^{{n}-1})+2^{{n}}$ $=2^{{n}}({n}+1)$ $2^{{n}}({n}+1)=2^{100} \.101 \$ $\Rightarrow {n}=100$ $2[\frac{{n}-1}{2}]=2[\frac{99}{2}]=98$

Q82JEE Main 2021NAT4MLimits, Continuity and Differentiability

Consider the function $f(x)= \begin{cases}\frac{P(x)}{\sin (x-2)}{,} & x \\ne 2 \\ 7 { , } & x=2$ Where P(x) is a polynomial such that P"(x) is always a constant and P(3) = 9\end{cases} If f(x) is continuous at x = 2, then P(5) is equal to ______.

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📖 Explanation

$f(x)= \begin{cases}\frac{P(x)}{\sin (x-2)}{,} & x \\ne 2 \\ 7 { , } & x=2$ ${P}"({x})=$ const\end{cases} $\Rightarrow{P}({x})$ is a 2 degree polynomial $f({x})$ is cont. at ${x}=2$ $f(2^{+})=f(2^{-})$ $\lim _{x \\to 2^{+}} \frac{P(x)}{\sin (x-2)}=7$ $\lim _{x \\to 2^{+}} \frac{(x-2)(a x+b)}{\sin (x-2)}=7$ $\Rightarrow2 a+b=7$ $P(x)=(x-2)(a x+b)$ $P(3)=(3-2)(3 a+b)=9$ $\Rightarrow 3 a+b=9$ ${a}=2, {b}=3$ ${P}(5)=(5-2)(2.5+3)$ $=3.13=39$

Q83JEE Main 2021NAT4MCircles

The equation of a circle is ${\text{Re}}(z^{2})+2({Im}(z))^{2}+2 {Re}(z)=0$ , where $z=x+iy.$ A line which passes through the center of the given circle and the vertex of the parabola, ${x}^{2}-6 {x}-{y}+13=0$ , has y-intercept equal to ____.

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📖 Explanation

Equation of circle is $(x^{2}-y^{2})+2 y^{2}+2 x=0$ $x^{2}+y^{2}+2 x=0$ Centre : $(-1,0)$ Parabola $: x^{2}-6 x-y+13=0$ $(x-3)^{2}=y-4$ Vertex : $(3,4)$ Equation of line $\equiv y-0=\frac{4-0}{3+1}(x+1)$ ${y}={x}+1$ $y$ -intercept $=1$

Q84JEE Main 2021NAT4MApplication of Derivatives

If a rectangle is inscribed in an equilateral triangle of side length $2 \sqrt{2}$ as shown in the figure, then the square of the largest area of such a rectangle is _______.

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📖 Explanation

In $\Delta{DBF}$ $\tan 60°=\frac{2 {b}}{2 \sqrt{2}-l} \$ $\Rightarrow {b}=\frac{\sqrt{3}(2 \sqrt{2}-l)}{2}$ ${A}=$ Area of rectangle $=l\\times {b}$ ${A}=l\\times \frac{\sqrt{3}}{2}(2 \sqrt{2}-l)$ $\frac{{d} {A}}{{d} l}=\frac{\sqrt{3}}{2}(2 \sqrt{2}-l)$ $-\frac{l\.\sqrt{3}}{2}=0$ $l=\sqrt{2}$ ${A}=l\\times {b}=\sqrt{2} \\times \frac{\sqrt{3}}{2}(\sqrt{2})$ $=\sqrt{3}$ $\Rightarrow{A}^{2}=3$

Q85JEE Main 2021NAT4MVector Algebra

If $(\vec{a}+3 \vec{b})$ is perpendicular to $(7 \vec{a}-5 \vec{b})$ and $({\vec{a}}-4 \vec{b})$ is perpendicular to $(7 {\vec{a}}-2 {{b} }^{\to })$ , then the angle between a and $\vec{b}$ (in degrees) is ________.

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📖 Explanation

$(\vec{a}+3 \vec{b}) ⊥ (7 \vec{a}-5 \vec{b})$ $(\vec{a}+3 \vec{b}) . (7 \vec{a}-5 \vec{b})=0$ $7{|\vec{a}|}^{2-}15{| \vec{b}|}^2 +16 \vec{a}. \vec{b}=0$ ......(1) $({\vec{a}}-4 \vec{b}) .(7 {\vec{a}}-2 {{b} }^{\to })=0$ $7{|\vec{a}|}^{2+}8{| \vec{b}|}^2 -30 \vec{a}. \vec{b}=0$ ......(2) from (1) & (2) ${|\vec{a}|}={| \vec{b}|}$ $\cos \theta={| \vec{b}|}/{2{| \vec{a}|}}\;\;\therefore \theta=60°$

Q86JEE Main 2021NAT4MDifferential Equations

Let a curve $y=f(x)$ pass through the point $(2,(\log _{{e}} 2)^{2})$ and have slope ${2 {y}}/{{x} \log _{{c}} {x}}$ for all positivereal value of $x$ . Then the value of $f(e)$ is equalto _______.

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📖 Explanation

$y'=\frac{2 y}{x ln x}$ $\Rightarrow \frac{d y}{y}=\frac{2 d x}{x ln x}$ $\Rightarrowl$ nlyl $=2 ln|ln x|+C$ put $x=2, y=(ln 2)^{2}$ $\Rightarrowc=0$ $\Rightarrowy=(ln x)^{2}$ $\Rightarrowf(e)=1$

Q87JEE Main 2021NAT4MQuadratic Equation and Inequalities

If ${a}+{b}+{c}=1, {ab}+{bc}+{ca}=2$ and ${abc}=3$ , then the value of $a^{4}+b^{4}+c^{4}$ is equal to _________.

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📖 Explanation

$a^{2}+b^{2}+c^{2}=(a+b+c)^{2}$ $-2 \suma b=-3$ $(a b+b c+c a)^{2}=\sum(a b)^{2}$ $+2 a b c \suma$ $\Rightarrow\sum(a b)^{2}=-2$ $a^{4}+b^{4}+c^{4}=(a^{2}+b^{2} +c^{2})^{2}$ $-2 \sum(a b)^{2}$ $=9-2(-2)=13$

Q88JEE Main 2021NAT4MProbability

A fair coin is tossed n-times such that the probability of getting at least one head is at least 0.9. Then the minimum value of n is _______.

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📖 Explanation

${P}($ Head $)=\frac{1}{2}$ $1-{P}($ All tail $) \\ge 0.9$ $1-(\frac{1}{2})^{{n}} \\ge 0.9$ $\Rightarrow(\frac{1}{2})^{{n}} \\le \frac{1}{10}$ $\Rightarrow{n}_{\min }=4$

Q89JEE Main 2021NAT4MBinomial Theorem

If the co-efficient of $x^{7}$ and $x^{8}$ in the expansion of $(2+\frac{x}{3})^{n}$ are equal, then the value of $n$ is equal to _______.

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📖 Explanation

${ }^{{n}} {C}_{7} 2^{{n}-7} \frac{1}{3^{7}}={ }^{{n}} {C}_{8} 2^{{n}-8} \frac{1}{3^{8}}$ $\Rightarrow{n}-7=48 \Rightarrow {n}=55$

Q90JEE Main 2021NAT4MThree Dimensional Geometry

If the lines $\frac{x-k}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+3}{1}$ are co-planar, then the value of ${k}$ is _______.

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📖 Explanation

$\begin{vmatrix}{k}+1 & 4 & 6 \\ 1 & 2 & 3 \\ 3 & 2 & 1\end{vmatrix}=0$ $({k}+1)[2-6]-4[1-9]+6[2-6]=0$ ${k}=1$

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