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20 Jul 2021 Shift 2

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Physics30 questions

Q1JEE Main 2021MCQ4MWork, Power and Energy

If the Kinetic energy of a moving body becomes four times its initial Kinetic energy, then the percentage change in its momentum will be :

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📖 Explanation

${K}_{2}=4 {K}_{1}$ $\frac{1}{2} {mv}_{2}^{2}=4 \frac{1}{2} {mv}_{1}^{2}$ ${v}_{2}=2 {v}_{1}$ ${P}={mv}$ ${P}_{2}={mv}_{2}=2 {mv}_{1}$ ${P}_{1}={mv}_{1}$ $\%$ change $=\frac{\Delta{P}}{{P}_{1}} \\times 100$ $=\frac{2 {mv}_{1}-{mv}_{1}}{{mv}_{1}} \\times 100=100 \%$

Q2JEE Main 2021MCQ4MMotion in a Straight Line

A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time $t_{1}$ . If he remains stationary on a moving escalator then the escalator takes him up in time $t_{2}$ . The time taken by him to walk up on the moving escalator will be:

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📖 Explanation

${L}=$ Length of escalator ${V}_{{b} / {esc}}=\frac{{L}}{{t}_{1}}$ When only escalator is moving. ${V}_{{esc}}=\frac{{L}}{{t}_{2}}$ when both are moving ${V}_{{b} / {g}}={V}_{{b} / {csc}}+{V}_{{csc}}$ ${V}_{{b} / {g}}=\frac{{L}}{{t}_{1}}+\frac{{L}}{{t}_{2}}$ $\Rightarrow[{t}={{L}}/{{V}_{{b} / {g}}}=\frac{{t}_{1} {t}_{2}}{{t}_{1}+{t}_{2}}]$

Q3JEE Main 2021MCQ4MGravitation

A satellite is launched into a circular orbit of radius R around earth, while a second satellite is launched into a circular orbit of radius 1.02 R. The percentage difference in the time periods of the two satellites is :

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📖 Explanation

${T}^{2} \propto{R}^{3}$ ${T}={kR}^{3 / 2}$ $\frac{{dT}}{{T}}=\frac{3}{2} \frac{{dR}}{{R}}$ $=\frac{3}{2} \\times 0.02=0.03$ $\%$ Change $=3 \%$

Q4JEE Main 2021MCQ4MElectromagnetic Waves

With what speed should a galaxy move outward with respect to earth so that the sodium-D line at wavelength 5890 Å is observed at 5896 Å ?

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📖 Explanation

${f}={f}_{0} \sqrt{\frac{1+\beta}{1-\beta}} \;\;\; \beta=\frac{{v}}{{c}}$ $\frac{{f}}{{f}_{0}} \sqrt{\frac{1+\beta}{1-\beta}}$ $(1+\frac{\Deltaf}{f_{0}})^{2}=(1+\beta)(1-\beta)^{-1}$ $\beta$ is small compared to 1 $(1+\frac{2 \Deltaf}{f_{0}})=(1+2 \beta)$ $\beta=\frac{\Delta{f}}{{f}_{0}}=\frac{{v}}{{c}}$ ${v}=6 \\times \frac{{c}}{5890}=305.6 {km} / {s}$

Q5JEE Main 2021MCQ4MProperties of Matter

The length of a metal wire is $l_{1}$ , when the tension in it is ${T}_{1}$ and is $l_{2}$ when the tension is ${T}_{2}$ . The natural length of the wire is :

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📖 Explanation

${T}_{1}={k}(l_{1}-l_{0})$ ${T}_{2}={k}(l_{2}-l_{0})$ $\frac{{T}_{1}}{{T}_{2}}=\frac{l_{1}-l_{0}}{l_{2}-l_{0}}$ $\frac{{T}_{1} l_{2}-{T}_{2} l_{1}}{{T}_{1}-{T}_{2}}=l_{0}$

Q6JEE Main 2021MCQ4MElectromagnetic Waves

In an electromagnetic wave the electric field vector and magnetic field vector are given as ${\vec{E}}={E}_{0} {\hat{i}}$ and ${\vec{B}}={B}_{0} {\hat{k}}$ respectively. The direction of propagation of electromagnetic wave is along:

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📖 Explanation

Direction of propagation $={\vec{E}} \\times {{\vec{B}}}={\hat{i}} \\times {\hat{k}}=-{\hat{j}}$

Q7JEE Main 2021MCQ4MAlternating Current

For a series LCR circuit with ${R}=100 \Omega$ , ${L}=0.5 {mH}$ and ${C}=0.1 {pF}$ connected across $220 {V}-50 {Hz}$ AC supply, the phase angle between current and supplied voltage and the nature of the circuit is :

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📖 Explanation

${R}=100 \Omega$ ${X}_{{L}}=\omega{L}=50 \pi\\times 10^{-3}$ $X_{C}=\frac{1}{\omegaC}=\frac{10^{11}}{100 \pi}$ ${X}_{{C}} \> >{X}_{{L}}$ $\&\| X_{C}-X_{L}|>>R$

Q8JEE Main 2021MCQ4MThermodynamics

Which of the following graphs represent the behavior of an ideal gas ? Symbols have their usual meaning.

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📖 Explanation

${PV}={nRT}$ ${PV} \propto{T}$ Straight line with positive slope (nR)

Q9JEE Main 2021MCQ4MOscillations

A particle is making simple harmonic motion along the X-axis. If at a distances ${x}_{1}$ and ${x}_{2}$ from the mean position the velocities of the particle are $v_{1}$ and $v_{2}$ respectively. The time period of its oscillation is given as:

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📖 Explanation

${v}^{2}=\omega^{2}({A}^{2}-{x}^{2})$ ${A}^{2}={x}_{1}^{2}+\frac{{v}_{1}^{2}}{\omega^{2}}={x}_{2}^{2}+\frac{{v}_{2}^{2}}{\omega^{2}}$ $\omega^{2}=\frac{{v}_{2}^{2}-{v}_{1}^{2}}{{x}_{1}^{2}-{x}_{2}^{2}}$ ${T}=2 \pi\sqrt{\frac{{x}_{2}^{2}-{x}_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}}$

Q10JEE Main 2021MCQ4MDual Nature of Matter and Radiation

An electron having de-Broglie wavelength $\lambda$ is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is :

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📖 Explanation

$\lambda=\frac{{h}}{{mv}}$ kinetic energy, $\frac{{P}^{2}}{2 {m}}=\frac{{h}^{2}}{2 {m} \lambda^{2}}={{hc}}/{\lambda_{{c}}}$ $\lambda_{{c}}=\frac{2 {m} \lambda^{2} {c}}{{h}}$

Q11JEE Main 2021MCQ4MRotational Motion

A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is :

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📖 Explanation

$\frac{1}{2} {I} \omega^{2}=\frac{1}{2} \\times \frac{1}{2} {mv}^{2}$ ${I}=\frac{1}{2} {m} {R}^{2}$ Body is solid cylinder

Q12JEE Main 2021MCQ4MUnits and Measurements

If time $(t)$ , velocity $(v)$ , and angular momentum $(l)$ are taken as the fundamental units. Then thedimension of mass $({m})$ in terms of $t, v$ and $l$ is :

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📖 Explanation

${m} \propto{t}^{{a}} {v}^{{b}} l^{{c}}$ ${m} \propto[{T}]^{{a}}\$ [{LT}^{-1}]^{{b}}$[{ML}^{2} {T}^{-1}]^{{c}}{M}^{1} {L}^{0} {T}^{0}={M}^{{c}} {L}^{{b}+2 {c}} {T}^{{a}-{b}-{c}} $comparing powers$ {c}=1, {b}=-2, {a}=-1{m} \propto{t}^{-1} {v}^{-2} l^{1}$

Q13JEE Main 2021MCQ4MThermodynamics

The correct relation between the degrees offreedom $f$ and the ratio of specific heat $\gamma$ is :

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📖 Explanation

$\gamma=1+\frac{2}{{f}}$ ${f}=\frac{2}{\gamma-1}$

Q14JEE Main 2021MCQ4MAtoms and Nuclei

For a certain radioactive process the graph between In R and t(sec) is obtained as shown in the figure. Then the value of half life for the unknown radioactive material is approximately :

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📖 Explanation

${R}={R}_{0} {e}^{-\lambda t}$ $l{n} {R}=l {n} {R}_{0}-\lambda {t}$ $-\lambda$ is slope of straight line $\lambda=\frac{3}{20}$ ${t}_{1 / 2}=\frac{l{n} 2}{\lambda}=4.62$

Q15JEE Main 2021MCQ4MGravitation

Consider a binary star system of star A and star B with masses ${m}_{{A}}$ and ${m}_{{B}}$ revolving in a circular orbit of radii $r_{A}$ and $r_{B}$ , respectively. If $T_{A}$ and $T_{B}$ are the time period of star ${A}$ and star ${B}$ , respectively, then:

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📖 Explanation

${T}_{{A}}={T}_{{B}}($ since $\omega_{{A}}=\omega_{{B}})$

Q16JEE Main 2021MCQ4MMagnetism and Matter

At an angle of $30°$ to the magnetic meridian, the apparent dip is $45°$ . Find the true dip:

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📖 Explanation

A $\tan \delta=\tan \delta' \cos \theta$ $=\tan 45°\cos 30°$ $\tan \delta=1 \\times \frac{\sqrt{3}}{2}$ $\delta=\tan ^{-1}(\frac{\sqrt{3}}{2})$

Q17JEE Main 2021MCQ4MWork, Power and Energy

A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time ' $t$ ' is proportional to:

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📖 Explanation

${P}=$ constant $\frac{1}{2} {mv}^{2}={Pt}$ $\Rightarrow v \propto\sqrt{t}$ $\frac{{dx}}{{dt}}={C} \sqrt{{t}}$ C = constant by integration. $x=C {t^{\frac{1}{2}+1}}/{\frac{1}{2}+1}$ $x \proptot^{3 / 2}$

Q18JEE Main 2021MCQ4MMotion in a Plane

Two vectors ${\vec{P}}$ and ${\vec{Q}}$ have equal magnitudes. If the magnitude of ${\vec{P}}+{\vec{Q}}$ is $n$ times the magnitude of $\vec{P}-{\vec{Q}}$ , then angle between $\vec{P}$ and ${\vec{Q}}$ is :

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📖 Explanation

$|{\vec{P}}|=|{\vec{Q}}|={x}$ .......(i) $|{\vec{P}}+{\vec{Q}}|={n}|{\vec{P}}-\vec{Q}|$ ${P}^{2}+{Q}^{2}+2 {PQ} \cos \theta$ $={n}^{2}({P}^{2}+{Q}^{2}-2 {P} {Q} \cos \theta)$ Using (i) in above equation $\cos \theta=\frac{{n}^{2}-1}{1+{n}^{2}}$ $\theta=\cos ^{-1}(\frac{{n}^{2}-1}{{n}^{2}+1})$

Q19JEE Main 2021MCQ4MProperties of Matter

Two small drops of mercury each of radius ${R}$ coalesce to form a single large drop. The ratio of total surface energy before and after the change is :

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📖 Explanation

$\frac{4}{3} \piR^{3}+\frac{4}{3} \piR^{3}=\frac{4}{3} \piR^{3}$ $R'=2^{\frac{1}{3}} R$ .......(i) $A_{i}=2[4 \piR^{2}]$ $A_{f}=4 \piR'^{2}$ $\frac{U_{i}}{U_{f}}=\frac{A_{i}}{A_{f}}=\frac{2 R^{2}}{2^{2 / 3} R^{2}}=2^{1 / 3}$

Q20JEE Main 2021MCQ4MMagnetism and Matter

The magnetic susceptibility of a material of a rod is 499. Permeability in vacuum is $4\pi\\times 10^{-7} {H} / {m}$ . Absolute permeability of the material of the rod is :

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📖 Explanation

$\mu=\mu_{0}(1+{x}_{{m}})$ $=4 \pi\\times 10^{-7} \\times 500$ $=2 \pi\\times 10^{-4} {H} / {m}$

Q21JEE Main 2021NAT4MSemiconductor Electronics

A zener diode having zener voltage $8 {V}$ and power dissipation rating of $0.5 {W}$ is connected across a potential divider arranged with maximum potential drop across zener diode is as shown in the diagram. The value of protective resistance $R_{p}$ is $........ \Omega$ .

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📖 Explanation

${P}={Vi}$ $0.5=8 {i}$ ${i}=\frac{1}{16} {A}$ ${E}=20=8+{i} {R}_{{P}}$ ${R}_{{P}}=12 \\times 16=192 \Omega$

Q22JEE Main 2021NAT4MLaws of Motion

A body of mass ' $m$ ' is launched up on a rough inclined plane making an angle of $30°$ with the horizontal. The coefficient of friction between the body and plane is $\frac{\sqrt{x}}{5}$ if the time of ascent is half of the time of descent. The value of $x$ is _______.

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📖 Explanation

${t}_{{a}}=\frac{1}{2} {t}_{{d}}$ $\sqrt{{2 {s}}/{{a}_{{a}}}}=\frac{1}{2} \sqrt{{2 {s}}/{{a}_{{d}}}}$ ......(i) ${a}_{{a}}={g} \sin \theta+\mu{g} \cos \theta$ $=\frac{{g}}{2}+\frac{\sqrt{3}}{2} \mu{g}$ $a_{{d}}=g \sin \theta-\mug \cos \theta$ $=\frac{{g}}{2}-\frac{\sqrt{3}}{2} \mu{g}$ using the above values of $a_{a}$ and $a_{d}$ and putting ineqution (i) we will gate $\mu=\frac{\sqrt{3}}{5}$

Q23JEE Main 2021NAT4MCurrent Electricity

In the given figure switches $S_{1}$ and $S_{2}$ are in open condition. The resistance across ab when the switches ${S}_{1}$ and ${S}_{2}$ are closed is_______ $\Omega$ .

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📖 Explanation

when switch ${S}_{1}$ and ${S}_{2}$ are closed $\frac{12 \\times 6}{12+6}+2+\frac{6 \\times 12}{6+12}$ $\frac{72}{18}+2+\frac{72}{18}=4+2+4=10 \Omega$

Q24JEE Main 2021NAT4MRotational Motion

Two bodies, a ring and a solid cylinder of same material are rolling down without slipping an inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of mass at the bottom of the inclined plane of the ring to that of the cylinder is ${\sqrt{{x}}}/{2}$ . Then, the value of ${x}$ is _____.

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📖 Explanation

I in both cases is about point of contact Ring ${mgh}=\frac{1}{2} {I} \omega^{2}$ ${mgh}=\frac{1}{2}(2 {m} {R}^{2}) {{v}_{{R}}^{2}}/{{R}^{2}}$ ${v}_{{R}}=\sqrt{{gh}}$ Solid cylinder ${mgh}=\frac{1}{2} {I} \omega^{2}$ ${mgh}=\frac{1}{2}(\frac{3}{2} {mR}^{2}) {{v}_{{C}}^{2}}/{{R}^{2}}$ ${v}_{{C}}=\sqrt{\frac{4 {gh}}{3}}$ ${{v}_{{R}}}/{{v}_{{C}}}=\frac{\sqrt{3}}{2}$

Q25JEE Main 2021NAT4MSemiconductor Electronics

For the forward biased diode characteristics shown in the figure, the dynamic resistance at ${I}_{{D}}=3 {mA}$ will be _______ $\Omega$ .

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📖 Explanation

${R}_{{d}}=\frac{{dV}}{{di}}={1}/{\frac{{di}}{{dv}}}$ $={1}/{\frac{5-1 \\times 10^{-3}}{0.75-0.65}}$ $\frac{100}{4}=25 \Omega$

Q26JEE Main 2021NAT4MAlternating Current

A series LCR circuit of ${R}=5 \Omega, {L}=20 {mH}$ and ${C}=0.5 \mu{F}$ is connected across an AC supply of $250 {V}$ , having variable frequency. The power dissipated at resonance condition is _______ $\\times 10^{2} { W}$ .

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📖 Explanation

$X_{L}=X_{C}$ (due to resonance) $Z=R$ so $i_{\text{rms }}={V}{Z}=\frac{V}{R}$ $\frac{V^{2}}{R}=\frac{250 \\times 250}{5}=125 \\times 10^{2} {W}$

Q27JEE Main 2021NAT4MThermodynamics

One mole of an ideal gas at $27°{C}$ is taken from ${A}$ to ${B}$ as shown in the given ${PV}$ indicator diagram. The work done by the system will be ______ $\\times 10^{-1} {J}$ . [Given: ${R}=8.3 {J} / {mole} {K}, \ln 2=0.6931]$ (Round off to the nearest integer)

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📖 Explanation

Process of isothermal ${W}={nRT} l{n}(\frac{{V}_{2}}{{V}_{1}})$ $=1 \\times 8.3 \\times 300 \\times \ln 2$ $=17258 \\times 10^{-1} {J}$

Q28JEE Main 2021NAT4MDual Nature of Matter and Radiation

A certain metallic surface is illuminated by monochromatic radiation of wavelength $\lambda$ . The stopping potential for photoelectric current for this radiation is $3 {V}_{0}$ . If the same surface is illuminated with a radiation of wavelength $2 \lambda$ , the stopping potential is $V_{0}$ . The threshold wavelength of this surface for photoelectric effect is________ $\lambda$ .

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📖 Explanation

${KE}=\frac{{hc}}{\lambda}-\phi{hc}$ ${e}(3{V}_{0})=\frac{{hc}}{\lambda_{0}}-\phi$ ......(i) ${eV}_{0}=\frac{{hc}}{2 \lambda_{0}}-\phi$ .........(ii) Using (i) & (ii) $\phi=\frac{{hc}}{4 \lambda_{0}}={{hc}}/{\lambda_{{t}}}$ $\lambda_{{t}}=4 \lambda_{0}$

Q29JEE Main 2021NAT4MRotational Motion

A body rotating with an angular speed of 600 rpm is uniformly accelerated to 1800 rpm in 10 sec. The number of rotations made in the process is___.

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📖 Explanation

Given, initial angular speed, $\omega_{0}=600 {rpm}=10 {rps}$ Final angular speed, $\omega_{ {F}}=1800 {rpm}=30 {rps}$ Time, $t=10 {s}$ $\because$ We know that $\omega_{F}=\omega_{0}+\alpha t$ $\Rightarrow \;\; 30=10+\alpha \times 10$ $\Rightarrow \;\; \;\frac{30-10}{10}=\alpha$ $\Rightarrow \;\; \alpha=\;\frac{20}{10}=2 {rps}^{2}$ Angular displacement is calculated by the rotational equation of motion as $\theta=\omega_{0} t+\;\frac{1}{2} \alpha t^{2}$ where, $\theta$ is the angular displacement. $\Rightarrow \theta=10 \times 10+\;\frac{1}{2} \times 2 \times (10)^{2}$ $\Rightarrow \theta=100+100$ $\Rightarrow \theta=200 {rad}$ Number of rotations $=\;\frac{\theta}{2 \pi}=\;\frac{200}{2 \pi}=31.8 \approx 32$

Q30JEE Main 2021NAT4MAtoms and Nuclei

A radioactive substance decays to $(\frac{1}{16})^{\text{th }}$ of its initial activity in 80 days. The half life of the radioactive substance expressed in days is ______.

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📖 Explanation

${N}_{0}{ \to }^{t^{1/2}}\frac{{N}_{0}}{2} { \to }^{t^{1/2}}\frac{{N}_{0}}{4}{ \to }^{t^{1/2}}\frac{{N}_{0}}{8} { \to }^{t^{1/2}}\frac{{N}_{0}}{16}$ $4 \\times {t}_{1 / 2}=80$ ${t}_{1/ 2}=20$ days

Chemistry30 questions

Q31JEE Main 2021MCQ4MOrganic Chemistry - Some Basic Principles

CHEMISTRY SECTION-A Which one of the following pairs of isomers is an example of metamerism ?

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Q32JEE Main 2021MCQ4MAmines

In the above reactions, product ${A}$ and product ${B}$ respectively are:

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Q33JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

The major product (P) in the following reaction is :

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Q34JEE Main 2021MCQ4Mp Block Elements

The single largest industrial application of dihydrogen is :

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📖 Explanation

Informative, according to ncert uses of di hydrogen. In fact ${NH}_{3}$ largest production in used to manufacture nitrogenous fertilisers.

Q35JEE Main 2021MCQ4Md and f Block Elements

Consider two chemical reactions (A) and (B) that take place during metallurgical process : (A) ${ZnCO}_{3({s})} \ { \to }^ {\Delta} {ZnO}_{({s})}+{CO}_{2({g})}$ (B) $2 {ZnS}_{(s)}+3 {O}_{2({g})} \ { \to }^ {\Delta} 2 {ZnO}_{({s})}+2 {SO}_{2({g})}$ The correct option of names given to them respectively is :

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📖 Explanation

(A) ${ZnCO}_{3}({S}) \ { \to }^ {\Delta} {ZnO}({s})+{CO}_{2}({g})$ Heating in absence of oxygen in calcination. (B) $2 {Zns}({s})+3 {O}_{2}({g}) \\to 2 {ZnO}({g})+2 {SO}_{2}({g})$ heating in presence of oxygen in roasting Hence (A) is calcination while (B) in roasting.

Q36JEE Main 2021MCQ4MIonic Equilibrium

A solution is $0.1 {M}$ in ${Cl}^{-}$ and $0.001 {M}$ in ${CrO}_{4}^{2-}$ . Solid ${AgNO}_{3}$ is gradually added to it Assuming that the addition does not change in volume and ${K}_{\text{sp }}({AgCl})=1.7 \\times 10^{-10} {M}^{2}$ and ${K}_{{sp}}({Ag}_{2} {CrO}_{4})=1.9 \\times 10^{-12} {M}^{3}$ Select correct statement from the following:

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📖 Explanation

(i) $[{Ag}^{+}]$ required to ppt ${AgCl}({s})$ ${Ksp}={IP}=\$ [{Ag}^{+}]$$[{Cl}^{-}]$=1.7 \times 10^{-10}[{Ag}^{+}]=1.7 \times 10^{-9} $(ii)$ [{Ag}^{+}] $required to ppt$ {Ag}{2} {CrO}{4}({s}){Ksp}={IP}=[{Ag}+]^{2}$[{CrO}_{4}^{-2}]$=1.9 \times 10^{-12}[{Ag}^{+}]=4.3 \times 10^{-5}[{Ag}^{+}] $required to ppt$ {AgCl} $is low so$ {AgCl} $will$ {ppt}1^{{st}}$

Q37JEE Main 2021MCQ4MClassification of Elements

Outermost electronic configuration of a group 13 element, ${E}$ , is $4 {s}^{2}, 4 {p}^{1}$ . The electronic configuration of an element of p-block period-five placed diagonally to element, E is :

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📖 Explanation

The element ${E}$ is Ga and the diagonal element of $5^{\text{th }}$ period is ${ }_{50} {Sn}$ having outer electronic configuration will be $[{Kr}] 5 {s}^{2} 4 {d}^{10} 5 {p}^{2}$ .

Q38JEE Main 2021MCQ4MHydrocarbons

Metallic sodium does not react normally with :

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Q39JEE Main 2021MCQ4MCoordination Compounds

Spin only magnetic moment of an octahedral complex of ${Fe}^{2+}$ in the presence of a strong field ligand in BM is :

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📖 Explanation

In presence of SFL $\Delta_{0}>{P}$ means pairing occurs therefore For ${Fe}^{+2} \\to 3 {d}^{6}$ $\therefore$ No of unpaired ${e}^{-}({s})=0$ $\therefore \mu=\sqrt{n(n+2)} B M=0$ $[{n}=$ No of unpaired ${e}^{-}({s})]$ In ${NiCl}_{2} {Ni}^{+2}$ is having configuration $3 {d}^{8}$ $\therefore$ Number of unpaired electron $=2$ After formation of oxidised product $[{Ni}({CN})_{6}]^{-2} {Ni}^{+4}$ is obtained ${Ni}^{+4} \Rightarrow3 {d}^{6}$ and ${CN}^{-}$ is strong field ligand $\therefore$ number of unpaired electrons $=0$ $\therefore$ The charge is $2-0=2$

Q40JEE Main 2021MCQ4MCoordination Compounds

Which one of the following species doesn't have a magnetic moment of $1.73 {BM}$ , (spin only value)?

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📖 Explanation

Species must not contain single unpaired (1) ${O}_{2}^{+} \\to \sigma_{1 {s}}^{2} < \sigma_{1 {s}}^{* 2} < \sigma_{2 {s}}^{2} < \sigma_{2 {s}}^{* 2} < \sigma_ {2 {p}}^{2}$ $< \pi_{2 {px}}^{2} =\pi_{2 {py}}^{2} < \pi_{2 {px}}^{*_{1}}=\pi_{2 {py}}^{*}$ unpaired ${e}^{-}=1 \therefore \mu=1.73 {BM}$ (1) ${Cu}^{+} {I}^{-} {Cu}^{+} \\to [{Ar}] 3 {d}^{10} \therefore$ unpaired ${e}^{-}=0$ ${I}^{-} \\to [{Xe}] \therefore$ unpaired ${e}^{-}=0$ therefore $\mu=0$ 3. $[{Cu}({NH}_{3})_{4}] {Cl}_{2}$ ${Cu} \\to [{A}] 3 {d}^{3} \therefore$ unpaired $=1 \therefore \mu=1.73 {BM}$ $4 . {O}_{2}^{-} \\to {d}$ $\sigma_{1 {s}}^{2} < \sigma_{1 {s}}^{* 2} < \sigma_{2 {s}}^{2} < \sigma_{2 {s}}^{* 2} < \sigma_{2 {px}}^{2} < \pi_{2 {px}}^{2}$ $=\pi_{2 {py}}^{2} < \pi_{2 {px}}^{*_{1}}=\pi_{2 {py}}^{*}$ $(11 e^{-})$ $\therefore$ unpaired $\therefore \mu=1.73 {BM}$

Q41JEE Main 2021MCQ4MBiomolecules

Which one of the following statements is not true about enzymes ?

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Q42JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

The hybridisations of the atomic orbitals of nitrogen in ${NO}_{2}^{-}, {NO}_{2}^{+}$ and ${NH}_{4}^{+}$ respectively are.

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Q43JEE Main 2021MCQ4MAlcohols, Phenols and Ethers

Bakelite is a cross-linked polymer of formaldehyde and :

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Q44JEE Main 2021MCQ4MHydrocarbons

Benzene on nitration gives nitrobenzene in presence of ${HNO}_{3}$ and ${H}_{2} {SO}_{4}$ mixture, where :

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Q45JEE Main 2021MCQ4MAmines

Consider the above reaction, compound B is :

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Q46JEE Main 2021MCQ4MHaloalkanes and Haloarenes

Major product P of above reaction, is :

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Q47JEE Main 2021MCQ4Md and f Block Elements

${Cu}^{2+}$ salt reacts with potassium iodide to give

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📖 Explanation

$2 {Cu}^{+2}+4 {I}^{-} \\to {Cu}_{2} {I}_{2}({s})+{I}_{2}$ $2 {Cu}^{+2}+3 {I}^{-} \\to 2 {CuI}+{I}_{2}$

Q48JEE Main 2021MCQ4MPractical Organic Chemistry

In Carius method, halogen containing organic compound is heated with fuming nitric acid in the presence of :

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📖 Explanation

Organic compound is heated with fuming nitric acid in the presence of silver nitrate in carius method. Lunar caustic $({AgNO}_{3})$ is used as reagent hare to distinguish ${Cl}^{-}, {Br}$ and ${I}^{-}$ respectively as follows. ${Cl}^{-}({aq}) \ { \to }^ {{AgNO}_{3}} {AgCl} \downarrow _{{ppt}}$ white ${Br}^{-}({aq}) \ { \to }^ {{AgNO}_{3}} {AgBr} \downarrow {{ppt}}$ pale yellow ${I}^{-}({aq}) \ { \to }^ {{AgNO}_{3}} {AgI} \downarrow {{ppt}}$ Dark yellow

Q49JEE Main 2021MCQ4MBiomolecules

Which one of the following gases is reported to retard photosynthesis?

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📖 Explanation

According to NCERT only ${NO}_{2}$ from the given options can retard the photosynthesis process in plants.

Q50JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

The correct order of their reactivity towards hydrolysis at room temperature is :

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Q51JEE Main 2021NAT4MChemical Thermodynamics

For a given chemical reaction ${A} \\to {B}$ at $300 {K}$ the free energy change is $-49.4 {kJ} {mol}^{-1}$ and the enthalpy of reaction is $51.4 {kJ} {mol}^{-1}$ . The entropy change of the reaction is ${J} {K}^{-1} {mol}^{-1}$ .

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📖 Explanation

Given chemical reaction: ${A} \ { \to }_{{T} 300 {K}} {B}[\Delta{G}]_{{P}, {T}}=-49.4$ kJ/mol $\Delta{H}_{{rxn}}=51.4$ kJ/mol $\Delta{S}_{{rxn}}=?$ $\Rightarrow$ From the relation $[\Delta{G}]_{{P}, {T}}=\Delta{H}-{T} \Delta{S}$ $\Rightarrow\Delta{S}_{{rxn}}={\Delta{H}_{{rn}}-[\Delta{G}]_{{P}, {T}}}/{{T}}$ $=\frac{[51.4-(-49.4)] \\times 1000}{300} \frac{{J}}{{mol} {K}}$ $\Rightarrow\Delta{S}_{{rxn}}=336 \frac{{J}}{{mol} {K}}$

Q52JEE Main 2021NAT4MStructure of Atom

The wavelength of electrons accelerated from rest through a potential difference of $40 {kV}$ is ${x} \\times 10^{-12} {m} .$ The value of ${x}$ is ________ $.$ (Nearest integer) Given : Mass of electron $=9.1 \\times 10^{-31} {kg}$ Charge on an electron $=1.6 \\times 10^{-19} {C}$ Planck's constant $=6.63 \\times 10^{-34} {J} {S}$

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📖 Explanation

De-broglie-wave length of electron: $\lambda_{{e}}={{h}}/{\sqrt{2 {m}({KE})}}$ $\{\because{e}^{-}$ is accelerated from rest $\Rightarrow{KE}={q} \\times {V}$ $\lambda={{h}}/{\sqrt{2 {mqv}}}$ $={6.63 \\times 10^{-34}}/{\sqrt{2 \\times 1.6 \\times 10^{-19} \\times 9.1 \\times 10^{-31} \\times 40 \\times 10^{3}}}$ $=0.614 \\times 10^{-11} {m}$ $=6.14 \\times 10^{-12} {m}$ Nearest integer $=6$

Q53JEE Main 2021NAT4MSolutions

The vapour pressures of ${A}$ and ${B}$ at $25°{C}$ are $90 {mm} {Hg}$ and $15 {mm} {Hg}$ respectively. If ${A}$ and ${B}$ are mixed such that the mole fraction of ${A}$ in themixture is $0.6$ , then the mole fraction of ${B}$ in the vapour phase is ${x} \\times 10^{-1}$ . The value of ${x}$ is ________.(Nearest integer)

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📖 Explanation

Given ${P}°_{{A}} =90 {mm} {Hg}$ , at $25°{C}$ ${P}°_{{B}} =15 {mm} \ {Hg}$ and $\text{table} {X}_{{A}}=0.6 ;{X}_{{B}}=0.4 \} {P}_{{T}}$ $={X}_{{A}} {P}°_{{A}} +{X}_{{B}} {P}_{{B}}°$ $=(0.6 \\times 90)+(0.4 \\times 15)$ $=54+6=60 {mm}$ Now mol fraction of B in the vapour phase i.e. $Y_{B}=\frac{P_{B}}{P_{T}}=\frac{X_{B} P°_{B} }{60}$ $=0.1=1 \\times 10^{-1}$ therefore: ${x}=1$

Q54JEE Main 2021NAT4MSome Basic Concepts of Chemistry

$4 {g}$ equimolar mixture of ${NaOH}$ and ${Na}_{2} {CO}_{3}$ contains ${x} {g}$ of ${NaOH}$ and y ${g}$ of ${Na}_{2} {CO}_{3}$ . The value of $x$ is_______ g. (Nearest integer)

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📖 Explanation

Total mass $=4 {g}$ Now ${NaOH}: {a} \ {mol} \;\;\;\;\; {W}_{{NaOH}}+{W}_{{Na}_{2}CO_3}=4$ ${Na}_{2} {CO}_{3}: '{a}'{mol}$ $\;\;\;\Rightarrow40 {a}+106 {a}=4$ $\Rightarrow {a}=\frac{4}{146} {mol}$ $\Rightarrow$ therefore mass of ${NaOH}$ is $: \frac{4}{146} \\times 40 {g}$ $=1.095 \ \approx 1$

Q55JEE Main 2021NAT4MPractical Organic Chemistry

When $0.15 {g}$ of an organic compound was analyzed using Carius method for estimation of bromine, $0.2397 {g}$ of ${AgBr}$ was obtained. The percentage of bromine in the organic compound is ________.(Nearest integer) [Atomic mass : Silver $=108$ , Bromine $=80$ ]

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📖 Explanation

Moles of Br = Moles of AgBr obtained $\Rightarrow$ Mass of ${Br}=\frac{0.2397}{188} \\times 80 {g}$ therefore % Br in the organic compound $={W_{{Br}}}/{{W}_{{T}}} \\times 100$ $=\frac{0.2397 \\times 80}{188 \\times 0.15} \\times 100=0.85 \\times 80$ $= 68$ ⇒ Nearest integer is '68'

Q56JEE Main 2021NAT4MSolutions

$100 {ml}$ of $0.0018 \%({w} / {v})$ solution of ${Cl}^{-}$ ion was the minimum concentration of ${Cl}^{-}$ required to precipitate a negative sol in one ${h}$ . The coagulating value of ${Cl}^{-}$ ion is _______. (Nearest integer)

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📖 Explanation

The coagulating value is the minimum concentration of an electrolyte in millimoles per liter (mmol/L) required to cause coagulation.
Given concentration is $0.0018 \% (\text{w/v})$ for $\text{Cl}^{-}$ ions. This means $0.0018 \text{ g}$ of $\text{Cl}^{-}$ is present in $100 \text{ ml}$ of solution.

First, convert the concentration to g/L:
$C_{\text{g/L}} = \frac{0.0018 \text{ g}}{100 \text{ mL}} \times \frac{1000 \text{ mL}}{1 \text{ L}} = 0.018 \text{ g/L}$
Next, convert g/L to mmol/L. The molar mass of $\text{Cl}^{-}$ is approximately $35.5 \text{ g/mol}$ .
$C_{\text{mol/L}} = \frac{0.018 \text{ g/L}}{35.5 \text{ g/mol}} \approx 0.000507 \text{ mol/L}$
Finally, convert mol/L to mmol/L:
$C_{\text{mmol/L}} = 0.000507 \text{ mol/L} \times 1000 \text{ mmol/mol} \approx 0.507 \text{ mmol/L}$
Rounding to the nearest integer, the coagulating value is $1$ .

The final answer is $\boxed{1}$ .

Q57JEE Main 2021NAT4MChemical Kinetics

${PCl}_{5}({g}) \\to {PCl}_{3}({g})+{Cl}_{2}({g})$ In the above first order reaction the concentration of ${PCl}_{5}$ reduces from initial concentration 50 mol ${L}^{-1}$ to 10 mol ${L}^{-1}$ in 120 minutes at $300 {~K}$ . The rate constant for the reaction at $300 {K}$ is ${x} \\times 10^{-2} {min}^{-1}$ . The value of ${x}$ is _______. $[$ Given $\log 5=0.6989]$

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📖 Explanation

${PCl}_{5}({g}) { \to }^{I \text{order}} _{300K}{PCl}_{3}({g})+{Cl}_{2}({g})$ $\text{table} {t}=0 , 50 {M} ;{t}=120 {min} , 10 {M}$ $\Rightarrow{K}=\frac{2.303}{{t}} \log {[{A}_{0}]}/{[{A}_{{t}}]}$ $\Rightarrow{K}=\frac{2.303}{120} \log \frac{50}{10}$ $\Rightarrow{K}=\frac{2.303}{120} \\times 0.6989$ $=0.013413 {min}^{-1}$ $=1.3413 \\times 10^{-2} {min}^{-1}$ $1.34 \Rightarrow$ Nearest integer $=1$

Q58JEE Main 2021NAT4MSolid State

Diamond has a three dimensional structure of C atoms formed by covalent bonds. The structure of diamond has face centred cubic lattice where 50% of the tetrahedral voids are also occupied by carbon atoms. The number of carbon atoms present per unit cell of diamond is _______.

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Q59JEE Main 2021NAT4MCoordination Compounds

An aqueous solution of ${NiCl}_{2}$ was heated with excess sodium cyanide in presence of strong oxidizing agent to form $[{Ni}({CN})_{6}]^{2-}$ . The total change in number of unpaired electrons on metal centre is _______.

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📖 Explanation

$[{Ni}({CN})_{6}]^{2-}$ ${Ni}^{+4} \\to {d}^{6}$ strong field ligand ↑↓ ↑↓ ↑↓ Pairing will be there zero unpaired electron ${NiCl}_{2} \\to {Ni}^{2+} \\to {d}^{8}$ ↑↓ ↑↓ ↑↓ ↑ ↑ → two unpaired ${e}^{-}$ Change = 2

Q60JEE Main 2021NAT4MElectrochemistry

Potassium chlorate is prepared by electrolysis of ${KCl}$ in basic solution as shown by following equation. $6 {OH}^{-}+{Cl}^{-} \\to {ClO}_{3}^{-}+3 {H}_{2} {O}+6 {e}^{-}$ A current of $x$ A has to be passed for $10 {h}$ to produce $10.0 {g}$ of potassium chlorate. the value of ${x}$ is ______ $.$ (Nearest integer) (Molar mass of ${KClO}_{3}=122.6 {g} \ { mol}^{-1}$ , ${F}=96500 {C})$

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📖 Explanation

${W}=\frac{{E}}{{F}} \\times {I} \\times {t}$ $10=\frac{122.6}{96500 \\times 6} \\times {x} \\times 10 \\times 3600$ ${X}=1.311$

Mathematics30 questions

Q61JEE Main 2021MCQ4MBinomial Theorem

For the natural numbers ${m}, {n}$ , if $(1-y)^{m}(1+y)^{n}=1+a_{1} y+a_{2} y^{2}+\......$ $+a_{m+n} y^{m+n}$ and $a_{1}=a_{2}=10$ , then the value of $(m+n)$ is equal to:

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📖 Explanation

$(1-y)^{{m}}(1+{y})^{{n}}$ Coefficient of $y(a_{1})=1 .{ }^{n} C_{1}+{ }^{m} C_{1}(-1)$ $={n}-{m}=10$ .......(1) Coefficient of ${y}^{2}({a}_{2})$ $=1 .{ }^{{n}} {C}_{2}-{ }^{{m}} {C}_{1} \.{ }^{{n}} {C}_{1} .+1 .{ }^{{m}} {C}_{2}=10$ $=\frac{{n}({n}-1)}{2}-{m} \.{n}+\frac{{m}({m}-1)}{2}=10$ ${m}^{2}+{n}^{2}-2 {mn}-({n}+{m})=20$ $({n}-{m})^{2}-({n}+{m})=20$ ${n}+{m}=80$ .......(2) By equation (1) & (2) ${m}=35, {n}=45$

Q62JEE Main 2021MCQ4MInverse Trigonometric Functions

The value of $\tan (2 \tan ^{-1}(\frac{3}{5})+\sin ^{-1}(\frac{5}{13}))$ is equalto :

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📖 Explanation

${\tan ^{-1} \frac{3}{5}+\tan ^{-1} \frac{3}{5}} +\tan ^{-1} \frac{5}{12}$ $\tan ^{-1} \frac{\frac{6}{5}}{1-\frac{9}{25}}={\tan ^{-1} \frac{15}{8}+\tan ^{-1} \frac{5}{12}}$ $\tan ^{-1} \frac{\frac{15}{8}+\frac{5}{12}}{1-\frac{15}{8} \. \frac{5}{12}}=\tan ^{-1} \frac{220}{21}$ $\tan (\tan ^{-1} \frac{220}{21})=\frac{220}{21}$

Q63JEE Main 2021MCQ4MCircles

Let ${r}_{1}$ and ${r}_{2}$ be the radii of the largest and smallest circles, respectively, which pass through the point $(-4,1)$ and having their centres on the circumference of the circle $x^{2}+y^{2}+2 x+4 y-4=0$ . If $\frac{{r}_{1}}{{r}_{2}}={a}+{b} \sqrt{2}$ , then ${a}+{b}$ is equal to :

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📖 Explanation

Centre of smallest circle is A Centre of largest circle is ${B}$ $r_{2}=|C P-C A|=3 \sqrt{2}-3$ $r_{1}=C P+C B=3 \sqrt{2}+3$ $\frac{r_{1}}{r_{2}}=\frac{3 \sqrt{2}+3}{3 \sqrt{2}-3}=\frac{(3 \sqrt{2}+3)^{2}}{9}$ $=(\sqrt{2}+1)^{2}=3+2 \sqrt{2}$ $a=3, b=2$

Q64JEE Main 2021MCQ4MSets and Relations

Consider the following three statements : (A) If 3 + 3 = 7 then 4 + 3 = 8. (B) If 5 + 3 = 8 then earth is flat. (C) If both (A) and (B) are true then 5 + 6 = 17. Then, which of the following statements is correct ?

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Q65JEE Main 2021MCQ4MThree Dimensional Geometry

The lines $x=a y-1=z-2$ and $x=3 y-2=b z-2,(a b \\ne 0)$ are coplanar, if :

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📖 Explanation

$\frac{x+1}{a}=y=\frac{z-1}{a}$ $\frac{x+2}{3}=y=\frac{z}{3 / b}$ lines are Co-planar $\begin{vmatrix}{a} & 1 & {a} \\ 3 & 1 & \frac{3}{{b}} \\ -1 & 0 & -1\end{vmatrix}=0 \$ $\Rightarrow -(\frac{3}{{b}}-{a})-1({a}-3)=0$ ${a}-\frac{3}{{b}}-{a}+3=0$ ${~b}=1, {a} \in{R}-\{0\}$

Q66JEE Main 2021MCQ4MDefinite Integrals

If $[{x}]$ denotes the greatest integer less than or equal to ${x}$ , then the value of the integral $\int_{-\pi/ 2} ^{\pi/ 2}[[x]-\sin x] {d} x$ is equal to :

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📖 Explanation

$I=\int{\frac{-\pi}{2}} ^{\frac{\pi}{2}}([x]+[-\sin x]) d x$ .......(1) $I=\int{\frac{-\pi}{2}} ^{\frac{\pi}{2}}([-x]+[\sin x]) d x$ ......(2) (King property) $2 {I}=\int{\frac{-\pi}{2}} ^{\frac{\pi}{2}}({[{x}]+[-{x}]}_{-1})$ $+({[\sin {x}]+[-\sin {x}]} _{-1}) {dx}$ $2 {I}=\int{\frac{-\pi}{2}} ^{\frac{\pi}{2}}(-2) {dx}=-2(\pi)$ ${I}=-\pi$

Q67JEE Main 2021MCQ4MComplex Numbers

If the real part of the complex number $(1-\cos \theta+2 i \sin \theta)^{-1}$ is $\frac{1}{5}$ for $\theta\in(0, \pi)$ , then the value of the integral $\int_{0} ^{\theta} \sin x d x$ is equal to:

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📖 Explanation

${z}=\frac{1}{1-\cos \theta+2 {i} \sin \theta}$ $=\frac{2 \sin ^{2} \frac{\theta}{2}-2 {i} \sin \theta}{(1-\cos \theta)^{2}+4 \sin ^{2} \theta}$ $=\frac{\sin \frac{\theta}{2}-2 {i} \cos \frac{\theta}{2}}{4 \sin \frac{\theta}{2}(\sin ^{2} \frac{\theta}{2}+4 \cos ^{2} \frac{\theta}{2})}$ ${Re}(z)=\frac{1}{2(\sin ^{2} \frac{\theta}{2}+4 \cos ^{2} \frac{\theta}{2})}=\frac{1}{5}$ $\sin \frac{2 \theta}{2}+4 \cos ^{2} \frac{\theta}{2}=\frac{5}{2}$ $1-\cos ^{2} \frac{\theta}{2}+4 \cos \frac{\theta}{2}=\frac{5}{2}$ $3 \cos ^{2} \frac{\theta}{2}=\frac{3}{2}$ $\cos ^{2} \frac{\theta}{2}=\frac{1}{2}$ $\frac{\theta}{2}={n} \pi\\pm \frac{\pi}{4}$ $\theta=2 {n} \pi\\pm \frac{\pi}{2}$ $\theta=2 {n} \pi\\pm \frac{\pi}{2}$ $\theta\in(0, \pi)$ $\theta=\frac{\pi}{2}$ $\int^{\frac{\pi}{2}}_{0} \sin \theta{d} \theta-[-\cos \theta]_{0}^{\frac{\pi}{2}}$ $=-(0-1)$ $=1$

Q68JEE Main 2021MCQ4MFunctions

Let ${f}: {R}-\{\frac{\alpha}{6}\} \\to {R}$ be defined by ${f}({x})=\frac{5 {x}+3}{6 {x}-\alpha}$ . Then the value of $\alpha$ for which $(f o f)(x)=x$ , for all ${x} \in{R}-\{\frac{\alpha}{6}\}$ , is

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📖 Explanation

$f(x)=\frac{5 x+3}{6 x-\alpha}=y$ $5 x+3=6 x y-\alphay$ $x(6 y-5)=\alphay+3$ $x=\frac{\alphay+3}{6 y-5}$ $f^{-1}(x)=\frac{\alpha x+3}{6 x-5}$ fo ${f}({x})={x}$ ${f}({x})={f}^{-1}({x})$ From eq $^{{n}}$ (i) & (ii) Clearly $(\alpha=5)$

Q69JEE Main 2021MCQ4MDefinite Integrals

If $f: {R} \\to {R}$ is given by $f({x})={x}+1$ , then the value of $\lim _{n \\to \\infty } \frac{1}{n}[f(0)+f(\frac{5}{n})+f(\frac{10}{n})+\....... .$ $+f(\frac{5(n-1)}{n})]$ is:

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📖 Explanation

$I=\sum_{r=0} ^{n-1} f(\frac{5 r}{n}) \frac{1}{n}$ $I=\int_{0} ^{1} f(5 x) d x$ $I=\int_{0} ^{1} (5 x+1) d x$ $I=\$ [\frac{5 x^{2}}{2}+x]_{0}^{1}I=\frac{5}{2}+1=\frac{7}{2}$

Q70JEE Main 2021MCQ4MProbability

Let ${A}, {B}$ and ${C}$ be three events such that the probability that exactly one of ${A}$ and ${B}$ occurs is $(1-{k})$ , the probability that exactly one of ${B}$ and ${C}$ occurs is $(1-2 {k})$ , the probability that exactly one of ${C}$ and ${A}$ occurs is $(1-{k})$ and the probability of all A, B and C occur simultaneously is ${k}^{2}$ , where $0 < {k} < 1$ . Then the probability that at least one of A, B and C occur is :

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📖 Explanation

${P}(\overline{{A}} \cap{B})+{P}({A} \cap \overline{{B}})=1-{k}$ ${P}(\overline{{A}} \cap{C})+{P}({A} \cap\overline{{C}})=1-2 {k}$ ${P}(\overline{{B}} \cap{C})+{P}({B} \cap\overline{{C}})=1-{k}$ ${P}({A} \cap {B} \cap{C})={k}^{2}$ ${P}({A})+{P}({B})-2 {P}({A} \cap{B})=1-{k}$ .......(1) ${P}({B})+{P}({C})-2 {P}({B} \cap{C})=1-{k}$ ........(2) ${P}({C})+{P}({A})-2 {P}({A} \cap{C})=1-2 {k}$ ........(3) (1) $+(2)+(3)$ ${P}({A})+{P}({B})+{P}({C})-{P}({A} \cap{B})$ $-{P}({B} \cap{C})-{P}({C} \cap{A})=\frac{-4 {k}+3}{2}$ So ${P}({A} \cup{B} \cup{C})=\frac{-4 {k}+3}{2}+{k}^{2}$ ${P}({A} \cup {B} \cup{C})=\frac{2 {k}^{2}-4 {k}+3}{2}$ $=\frac{2({k}-1)^{2}+1}{2}$ ${P}({A} \cup{B} \cup{C})>\frac{1}{2}$

Q71JEE Main 2021MCQ4MApplication of Derivatives

The sum of all the local minimum values of the twice differentiable function $f: {R} \\to {R}$ defined by $f({x})={x}^{3}-3 {x}^{2}-\frac{3 f"(2)}{2} {x}+f"({1})$ is :

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📖 Explanation

$f(x)=x^{3}-3 x^{2}-\frac{3}{2} f"(2) x+f"(1)$ .......(i) ${f}'({x})=3 {x}^{2}-6 {x}-\frac{3}{2} {f}"(2)$ ......(ii) ${f}"({x})=6 {x}-6 ........ .$ (iii) Now is $3^{\text{rd }}$ equation $f"(2)=12-6=6$ $f"(11=0)$ Use (ii) $f'(x)=3 x^{2}-6 x-\frac{3}{2} f"(2)$ $f'(x)=3 x^{2}-6 x-\frac{3}{2} \\times 6$ $f'(x)=3 x^{2}-6 x-9$ $f'(x)=0$ $3 x^{2}-6 x-9=0$ $\Rightarrow{x}=-1 & 3$ Use (iii) $f"(x)=6 x-6$ ${f}"(-1)=-12 < 0$ maxima $f"(3)=12>0$ minima. Use (i) $f(x)=x^{3}-3 x^{2}-\frac{3}{2} f"(2) x+f"(1)$ $f(x)=x^{3}-3 x^{2}-\frac{3}{2} \\times 6 \8x+0$ $f(x)=x^{3}-3 x^{2}-9 x$ $f(3)=27-27-9 \\times 3=-27$

Q72JEE Main 2021MCQ4MTrigonometric Ratios and Identities

Let in a right angled triangle, the smallest angle be $\theta$ . If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then $\sin \theta$ is equal to:

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📖 Explanation

$∠{A}=\theta$ $∠{B}=90-\theta$ ${a}=$ smallest side ${c}^{2}={a}^{2}+{b}^{2}$ $\frac{1}{{a}^{2}}=\frac{1}{{b}^{2}}+\frac{1}{{c}^{2}}$ $\frac{{b}^{2} {c}^{2}}{{a}^{2}}={b}^{2}+{c}^{2}$ Use ${a}=2 {R} \sin {A}=2 {R} \sin \theta$ ${b}=2 {R} \sin {B}=2 {R} \sin (90-\theta)$ $=2 {R} \cos \theta$ ${c}=2 {R} \sin {C}=2 \sin 90°=2 {R}$ $\frac{4 {R}^{2} \cos ^{2} \theta}{4 {R}^{2} \sin ^{2} \theta}=4 {R}^{2} \cos ^{2} \theta+4 {R}^{2}$ $\cos ^{2} \theta=\sin ^{2} \theta\cos ^{2} \theta+\sin ^{2} \theta$ $1-\sin ^{2} \theta=\sin ^{2} \theta(1-\sin ^{2} \theta)+\sin ^{2} \theta$ $\sin ^{2} \theta=\frac{3-\sqrt{5}}{2}$ $\Rightarrow \sin \theta=\frac{\sqrt{5}-1}{2}$

Q73JEE Main 2021MCQ4MDifferential Equations

Let $y=y(x)$ satisfies the equation $\frac{d y}{d x}-{|A|}=0$ ,for all $x>0$ , where $A=\begin{bmatrix}y & \sin x & 1 \\ 0 & -1 & 1 \\ 2 & 0 & \frac{1}{x}\end{bmatrix} .$ If ${y}(\pi)=\pi+2$ , then the value of ${y}(\frac{\pi}{2})$ is :

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📖 Explanation

$|{A}|=-\frac{{y}}{{x}}+2 \sin {x}+2$ $\frac{{dy}}{{dx}}={|{A}|}$ $\frac{d y}{d x}=-\frac{y}{x}+2 \sin x+2$ $\frac{d y}{d x}+\frac{y}{x}=2 \sin x+2$ I.F. $=e^{\int\frac{1}{x} d x}=x$ $\Rightarrowy x=\intx(2 \sin x+2) d x$ $x y=x^{2}-2 x \cos x+2 \sin x+c$ .......(i) Now $x=\pi, y=\pi+2$ Use in (i) ${c}=0$ Now (i) be comes $x y=x^{2}-2 x \cos x+2 \sin x$ put ${x}=\pi/ 2$ $\frac{\pi}{2} {y}=(\frac{\pi}{2})^{2}-2 \.\frac{\pi}{2} \cos \frac{\pi}{2}+2 \sin \frac{\pi}{2}$ $\frac{\pi}{2} {y}=\frac{\pi^{2}}{4}+2$

Q74JEE Main 2021MCQ4MThree Dimensional Geometry

Consider the line $L$ given by the equation $\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1} .$ Let $Q$ be the mirror image of the point $(2,3,-1)$ with respect to $L$ . Let a plane $P$ be such that it passes through ${Q}$ , and the line ${L}$ is perpendicular to P. Then which of the following points is on the plane P ?

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📖 Explanation

Plane $\mathrm{p}$ is $\perp ^{{r}}$ to line $\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1}$ & passes through pt. $(2,3)$ equation of plane ${p}$ $2(x-2)+1(y-3)+1(z+1)=0$ $2 x+y+z-6=0$ pt $(1,2,2)$ satisfies above equation

Q75JEE Main 2021MCQ4MStatistics

If the mean and variance of six observations 7, $10,11,15, {a}, {b}$ are 10 and $\frac{20}{3}$ , respectively,then the value of $|{a}-{b}|$ is equal to :

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📖 Explanation

$10=\frac{7+10+11+15+a+b}{6}$ $\Rightarrow{a}+{b}=17$ .......(i) $\frac{20}{3}=\frac{7^{2}+10^{2}+11^{2}+15^{2}+{a}^{2}+{b}^{2}}{6}$ $-10^{2}$ ${a}^{2}+{b}^{2}=145$ ......(ii) Solve (i) and (ii) ${a}=9, { b}=8$ or ${a}=8, {b}=9$ $|a-b|=1$

Q76JEE Main 2021MCQ4MDefinite Integrals

Let $g(t)=\int_{-\pi/ 2} ^{\pi/ 2} \cos (\frac{\pi}{4} t+f(x)) d x$ , where $f({x})=\log _{{e}}({x}+\sqrt{{x}^{2}+1}), {x} \in{R}$ . Then which one of the following is correct?

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📖 Explanation

$g(t)=int_{-\pi/ 2} ^{\pi/ 2}(\cos \frac{\pi}{4} t+f(x)) d x$ $g(t)=\pi\cos \frac{\pi}{4} t+int_{-\pi/ 2} ^{\pi/ 2}f(x) d x$ $g(t)=\pi\cos \frac{\pi}{4} t$ $g(1)=\frac{\pi}{\sqrt{2}}, g(0)=\pi$

Q77JEE Main 2021MCQ4MParabola

Let ${P}$ be a variable point on the parabola ${y}=4 {x}^{2}+1$ . Then, the locus of the mid-point of the point ${P}$ and the foot of the perpendicular drawn from the point ${P}$ to the line ${y}={x}$ is :

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📖 Explanation

$\frac{{K}-{C}}{{h}-{C}}=-1$ ${C}=\frac{{h}+{K}}{2}$ $R=(\frac{x+C}{2}, \frac{y+C}{2})$ ${R}=(\frac{{x}}{2}+\frac{{h}}{4}+\frac{{K}}{4}, \frac{{y}}{2}+\frac{{h}}{4}+\frac{{k}}{4})$ ${h}=\frac{{x}}{2}+\frac{{h}}{4}+\frac{{K}}{4}$ ${K}=\frac{{y}}{2}+\frac{{h}}{4}+\frac{{K}}{4}$ $\Rightarrow{x}=\frac{3 {h}}{2}-\frac{{K}}{2}, {y}=\frac{3 {K}}{2}-\frac{{h}}{2}$ $Y=4 x^{2}+1$ $(\frac{3 {k}-{h}}{2})=4(\frac{3 {h}-{k}}{2})^{2}+1$

Q78JEE Main 2021MCQ4MMatrices and Determinants

The value of ${k} \in{R}$ , for which the following system of linear equations $3 x-y+4 z=3$ $x+2 y-3 z=-2$ $6 x+5 y+k z=-3$ has infinitely many solutions, is :

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📖 Explanation

$\begin{vmatrix}3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & {K}\end{vmatrix}=0$ $\Rightarrow3(2 {K}+15)+{K}+18-28=0$ $\Rightarrow 7 {K}+35=0 \Rightarrow{K}=-5$

Q79JEE Main 2021MCQ4MSequences and Series

If sum of the first 21 terms of the series $\log _{9 / 2} x+\log _{9^{1 / 3}} x+\log _{9 / 4} x+\..... .$ where $x>0$ is 504 , then ${x}$ is equal to

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📖 Explanation

${s}=2 \log _{9} {x}+3 \log _{9} {x}^{+} \..... .+22 \log _{9} {x}$ ${s}=\log _{9} \\times (2+3+\...... .+22)$ ${s}=\log _{9} {x}\{\frac{21}{2}(2+22)\}$ Given $252 \log _{9} {x}=504$ $\Rightarrow\log _{9} {x}=2 \Rightarrow{x}=81$

Q80JEE Main 2021MCQ4MVector Algebra

In a triangle ${ABC}$ , if $|{{BC}}^{\to }|=3,|{\vec{CA}}|=5$ and $|{\vec{BA}}|=7$ , then the projection of the vector ${{BA}}^{\to }$ on ${{BC}}^{\to }$ is equal to

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📖 Explanation

Projection of ${\vec{BA}}$ on $\vec{BC}$ is equal to $=|{B\vec{A}}| \cos \angle {ABC}$ $=7|\frac{7^{2}+3^{2}-5^{2}}{2 \\times 7 \\times 3}|=\frac{11}{2}$

Q81JEE Main 2021NAT4MMatrices and Determinants

Let ${A}=\{{a}_{{ij}}\}$ be a $3 \\times 3$ matrix, where $a_{i j}=\begin{cases}(-1)^{j-1} & \text{if } i < j{,} \\ 2 & \text{if }i=j {,} \\ (-1)^{i+j} & \text{if }i>j{ ,}$ then ${det}(3 {Adj}(2 {A}^{-1}))$ is equal to ______\end{cases}

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📖 Explanation

${A}=\begin{bmatrix}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{bmatrix}$ $|{A}|=4$ $|3 {adj}(2 {A}^{-1})|=|3.2^{2} {adj}({A}^{-1})|$ $=12^{3}|{adj}({A}^{-1})|=12^{3}|{A}^{-1}|^{2}$ $=\frac{12^{3}}{|{A}|^{2}}=\frac{12^{3}}{16}=108$

Q82JEE Main 2021NAT4MLogarithms

The number of solutions of the equation $\log _{(x+1)}(2 x^{2}+7 x+5)+\log _{(2 x+5)}(x+1)^{2}$ $-4=0$ $x>0$ , is

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📖 Explanation

$\log _{(x+1)}(2 x^{2}+7 x+5)+\log _{(2 x+5)}(x+1)^{2}$ $-4=0$ $\log _{(x+1)}(2 x+5)(x+1)+2 \log _{(2 x+5)}(x+1)$ $=4$ $\log _{(x+1)}(2 x+5)+1+2 \log _{(2 x+5)}(x+1)$ $=4$ Put $\log _{(x+1)}(2 x+5)=t$ $t+\frac{2}{t}=3 \Rightarrowt^{2}-3 t+2=0$ $t=1,2$ $\log _{(x+1)}(2 x+5)=1 \&$ $\;\;\log _{(x+1)}(2 x+5)=2$ $x+1=2 x+3$ $\&\;\;\;\;\;\;\; 2 x+5=(x+1)^{2}$ ${x}=-4$ (rejected) $\;\;\;\;\;{x}^{2}=4 \Rightarrow{x}=2,-2$ (rejected) So, $x=2$ No. of solution $=1$

Q83JEE Main 2021NAT4MDifferential Equations

Let a curve $y=y(x)$ be given by the solution of the differential equation $\cos (\frac{1}{2} \cos ^{-1}({e}^{-{x}})) {dx}$ $=\sqrt{{e}^{2 {x}}-1} {dy}$ If it intersects $y$ -axis at ${y}=-1$ , and the intersection point of the curve with $x$ -axis is $(\alpha, 0)$ , then ${e}^{\alpha}$ is equal to ________.

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📖 Explanation

$\cos (\frac{1}{2} \cos ^{-1}({e}^{-{x}})) {dx}$ $=\sqrt{{e}^{2 {x}}-1} {dy}$ Put $\cos ^{-1}({e}^{-{x}}) \theta, \theta\in[0, \pi]$ $\cos \theta={e}^{-{x}} \Rightarrow2 \cos ^{2} {\theta}{2}-1={e}^{-{x}}$ $\cos \frac{\theta}{2}=\sqrt{{{e}^{-{x}}+1}/{2}}=\sqrt{{{e}^{{x}}+1}/{2 {c}^{{x}}}}$ $\sqrt{{{e}^{{x}}+1}/{2 {c}^{{x}}}} {dx}=\sqrt{{e}^{2 {x}}-1} {dy}$ $\frac{1}{\sqrt{2}} \int{{dx}}/{\sqrt{{e}^{{x}}} \sqrt{{e}^{{x}}-1}}=\int{dy}$ Put ${e}^{{x}}={t}, \frac{{dt}}{{dx}}={e}^{{x}}$ $\frac{1}{\sqrt{2}} \int{{dt}}/{{e}^{{x}} \sqrt{{e}^{{x}}}\sqrt{{e}^{{x}}-1}}=\int{dy}$ $\int{{dt}}/{{t} \sqrt{{t}^{2}-{t}}}=\sqrt{2} {y}$ Put ${t}=\frac{1}{{z}}, \frac{{dt}}{{dz}}=-\frac{1}{{z}^{2}}$ $\int{-\frac{{dz}}{{z}^{2}}}/{\frac{1}{{z}} \sqrt{\frac{1}{{z}^{2}}-\frac{1}{{z}}}}=\sqrt{2} /{y}$ $-\int{{dz}}/{\sqrt{1-{z}}}=\sqrt{2} {y}$ $\frac{-2(1-{z})^{1 / 2}}{-1}=\sqrt{2} {y}+{c}$ $2(1-\frac{1}{{t}})^{1 / 2}=\sqrt{2} {y}+{c}$ $2(1-{e}^{-{x}})^{1 / 2}=\sqrt{2 {y}}+{c} {\ \to } ^{ {(0,-1)}}$ $\Rightarrow {c}=\sqrt{2}$ $2(1-{e}^{-{x}})^{1 / 2}=\sqrt{2}({y}+1)$ , passes through $(\alpha, 0)$ $2(1-{e}^{-\alpha})^{1 / 2}=\sqrt{2}$ $\sqrt{1-{e}^{-\alpha}}=\frac{1}{\sqrt{2}} \Rightarrow1-{e}^{-\alpha}=\frac{1}{2}$ ${e}^{-\alpha}=\frac{1}{2} \Rightarrow{e}^{\alpha}=2$

Q84JEE Main 2021NAT4MVector Algebra

For ${p}>0$ , a vector ${{\vec{v}}}_{2}=2 {\hat{i}}+({p}+1) {\hat{j}}$ is obtained by rotating the vector ${\vec{v}}_{1}=\sqrt{3} {p} {\hat{i}}+\hat{j}$ by an angle $\theta$ about origin in counter clockwise direction. If $\tan \theta=\frac{(\alpha\sqrt{3}-2)}{(4 \sqrt{3}+3)}$ , then the value of $\alpha$ is equal to _________.

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📖 Explanation

$|{\vec{V}}_{1}|=|{{\vec{V}}_{2}}|$ $3 {P}^{2}+1=4+({P}+1)^{2}$ $2 {P}^{2}-2 {P}-4=0 \Rightarrow{P}^{2}-{P}-2=0$ ${P}=2,-1$ (rejected) $\cos \theta={{{\vec{V}}_{1}} \.{\vec{V}_{2}}}/{|{\vec{V}_{1}}||{\vec{V}_{2}}|}$ $={2 \sqrt{3} {P}+({P}+1)}/{\sqrt{({P}+1)^{2}+4}\sqrt{3 {P}^{2}+1}}$ $\cos \theta=\frac{4 \sqrt{3}+3}{\sqrt{13} \sqrt{13}}=\frac{4 \sqrt{3}+3}{13}$ $\tan \theta={\sqrt{112-24 \sqrt{3}}}/{4 \sqrt{3}+3}=\frac{6 \sqrt{3}-2}{4 \sqrt{3}+3}$ $=\frac{\alpha\sqrt{3}-2}{4 \sqrt{3}+3}$ $\Rightarrow\alpha=6$

Q85JEE Main 2021NAT4MStraight Lines and Pair of Straight Lines

Consider a triangle having vertices ${A}(-2,3), {B}(1,9)$ and ${C}(3,8)$ . If a line ${L}$ passing through the circum-centre of triangle ${ABC}$ , bisects line ${BC}$ ,and intersects $y$ -axis at point $(0, \frac{\alpha}{2})$ , then the value of real number alpha is ________.

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📖 Explanation

$(\sqrt{50})^{2}=(\sqrt{45})^{2}+(\sqrt{5})^{2}$ $\angle {B}=90°$ Circum-center $=(\frac{1}{2}, \frac{11}{2})$ Mid point of ${BC}=(2, \frac{17}{2})$ Line $:(y-\frac{11}{2})=2(x-\frac{1}{2}) \$ $\Rightarrow y=2 x+\frac{9}{2}$ Passing though $(0, \frac{\alpha}{2})$ $\frac{\alpha}{2}=\frac{9}{2} \Rightarrow\alpha=9$

Q86JEE Main 2021NAT4MApplication of Derivatives

If the point on the curve $y^{2}=6 x$ , nearest to the point $(3, \frac{3}{2})$ is $(\alpha, \beta)$ , then $2(\alpha+\beta)$ is equal to _______.

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📖 Explanation

$P \ \equiv (\frac{3}{2} t^{2}, 3 t)$ Normal at point P $t x+y=3 t+\frac{3}{2} t^{3}$ Passes through $(3, \frac{3}{2})$ $\Rightarrow3 t+\frac{3}{2}=3 t+\frac{3}{2} t^{3}$ ${P} \ \equiv (\frac{3}{2}, 3)=(\alpha, \beta)$ $\Rightarrow{t}^{3}=1 \Rightarrow{t}=1$ $2(\alpha+\beta)=2(\frac{3}{2}+3)=9$

Q87JEE Main 2021NAT4MLimits, Continuity and Differentiability

Let a function ${g}:[0,4] \\to {R}$ be defined as $g(x)= \begin{cases}\max _{0 \\le t \\le x}\{t^{3}-6 t^{2}+9 t-3\}{,} & 0 \\le x \\le 3 \\ 4-x {,} & 3 < x \\le 4$ then the number of points in the interval $(0 4)$ where ${g}({x})$ is NOT differentiable is _______\end{cases}

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📖 Explanation

$f(x)=x^{3}-6 x^{2}+9 x-3$ $f'(x)=3 x^{2}-12 x+9$ $=3(x-1)(x-3)$ $f(1)=1 f(3)=-3$ $g(x)=[\text{table} f(x) , 0 \\le x \\le 1 ; 0 , 1 \\le x \\le 3;-1 , 3 < x \\le 4$ ${g}({x})$ is continuous $g'(x)=[\text{table} 3(x-1)(x-3) ,0 \\le x \\le 1 ; 0 , 1 \\le x \\le 3; -1 , 3 < x \\le 4$ $g(x)$ is non-differentiable at $x=3$

Q88JEE Main 2021NAT4MSequences and Series

For ${k} \in{N}$ , let $\frac{1}{\alpha(\alpha+1)(\alpha+2) ...... .(\alpha+20)}$ $=\sum_{{K}=0} ^{20} {{A}_{{k}}}/{\alpha+{k}}$ ,where $\alpha > 0$ . Then the value of $100(\frac{{A}_{14}+{A}_{15}}{{A}_{13}})^{2}$ is equal to ________.

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📖 Explanation

$\frac{1}{\alpha(\alpha+1)(\alpha+2) ...... .(\alpha+20)}$ $=\sum_{{K}=0} ^{20} {{A}_{{k}}}/{\alpha+{k}}$ ${A}_{14}=\frac{1}{(-14)(-13)_{\..... .}(-1)(1) \.... .(6)}$ $=\frac{1}{14 ! \.6 !}$ ${A}_{15}=\frac{1}{(-15)(-14) \...... .(-1)(1) \..... .(5)}$ $=\frac{1}{15 ! \.5 !}$ ${A}_{13}=\frac{1}{(-13) \...... .(-1)(1) \..... .(7)}$ $=\frac{-1}{13 ! \. 7 !}$ $\frac{{A}_{14}}{{A}_{13}}=\frac{1}{14 ! \.6 !} \\times -13 ! \\times 7 !$ $=\frac{-7}{14}=-\frac{1}{2}$ $\frac{{A}_{15}}{{A}_{13}}=-\frac{1}{15 ! \\times 5 !} \\times -13 ! \\times 7 !$ $=\frac{42}{15 \\times 14}=\frac{1}{5}$ $100(\frac{{A}_{14}}{{A}_{13}}+\frac{{A}_{15}}{{A}_{13}})^{2}$ $=100(-\frac{1}{2}+\frac{1}{5})^{2}=9$

Q89JEE Main 2021NAT4MSequences and Series

Let $\{a_{n}\}_{n=1}^{\\infty }$ be a sequence such that $a_{1}=1, a_{2}=1$ and ${a}_{{n}+2}=2 {a}_{{n}+1}+{a}_{{n}}$ for all ${n} \\ge 1$ . Then the value of $47 \sum_{{n}=1} ^{\\infty } {{a}_{{n}}}/{2^{3 {n}}}$ is equal to ________.

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📖 Explanation

$a_{n+2}=2 a_{n+1}+a_{n}$ , let $sum_{{n}=1} ^{\\infty } \frac{a_{n}}{8^{n}}=P$ Divide by $8^{\text{n }}$ we get $\frac{a_{n+2}}{8^{n}}=\frac{2 a_{n+1}}{8^{n}}+\frac{a_{n}}{8^{n}}$ $\Rightarrow64 {{a}_{{n}+2}}/{8^{{n}+2}}={16 {a}_{{n}+1}}/{8^{{n}+1}}+{{a}_{{n}}}/{8^{{n}}}$ $64sum_{{n}=1} ^{\\infty } {{a}_{{n}+2}}/{8^{{n}+2}}=16 sum_{{n}=1} ^{\\infty } {{a}_{{n}+1}}/{8^{{n}+1}}$ $+sum_{{n}=1} ^{\\infty } {{a}_{{n}}}/{8^{{n}}}$ $64({P}-\frac{{a}_{1}}{8}-\frac{{a}_{2}}{8^{2}})=16({P}-\frac{{a}_{1}}{8})+{P}$ $\Rightarrow64({P}-\frac{1}{8}-\frac{1}{64})=16({P}-\frac{1}{8})+{P}$ $64 P-8-1=16 {P}-2+{P}$ $47 {P}=7$

Q90JEE Main 2021NAT4MLimits, Continuity and Differentiability

If $\lim _{x \\to 0} \frac{\alpha x e^{x}-\beta\log _{e}(1+x)+\gammax^{2} e^{-x}}{x \sin ^{2} x}$ $=10, \alpha, \beta, \gamma\in {R}$ then the value of $\alpha+\beta+\gamma$ is _______.

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📖 Explanation

$\lim _{x \\to 0} {\alpha x(1+x+\frac{x^{2}}{2})-\beta(x-\frac{x^{2}}{2}+\frac{x^{3}}{3})+\gammax^{2}(1-x)}/{x^{3}}$ $\lim _{x \\to 0} \frac{x(\alpha-\beta)+x^{2}(\alpha+\frac{\beta}{2}+\gamma)+x^{3}(\frac{\alpha}{2}-\frac{\beta}{3}-\gamma)}{x^{3}}=10$ For limit to exist $\alpha-\beta=0, \alpha+{\beta}{2}+\gamma=0$ $\frac{\alpha}{2}-\frac{\beta}{3}-\gamma=10 \.......$ (i) $\beta=\alpha, \gamma=-3 \frac{\alpha}{2}$ Put in (i) $\frac{\alpha}{2}-\frac{\alpha}{3}+\frac{3 \alpha}{2}=10$ $\frac{\alpha}{6}+\frac{3 \alpha}{2}=10$ $\Rightarrow\frac{\alpha+9 \alpha}{6}=10$ $\Rightarrow\alpha=6$ $\alpha=6, \beta=6, \gamma=-9$ $\alpha+\beta+\gamma=3$

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