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Physics30 questions

Q1JEE Main 2021MCQ4MAlternating Current

In a circuit consisting of a capacitance and a generator with alternating emf ${E}_{{g}}={E}_{{g}_{0}} \sin \omega{t}, {V}_{{C}}$ and ${I}_{{C}}$ are the voltage and current. Correct phasor diagram for such circuit is :

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📖 Explanation

In capacitor, current lead voltage by $\frac{\pi}{2}$

Q2JEE Main 2021MCQ4MCurrent Electricity

A Copper (Cu) rod of length $25 {cm}$ and crosssectional area $3 {mm}^{2}$ is joined with a similar Aluminium (Al) rod as shown in figure. Find the resistance of the combination between the ends A and B. (Take Resistivity of Copper $=1.7 \\times 10^{-8} \Omega{m}$ Resistivity of Aluminium $=2.6 \\times 10^{-8} \Omega{m}$ )

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📖 Explanation

${R}=\frac{{R}_{1} {R}_{2}}{{R}_{1}+{R}_{2}}=\frac{l}{{A}} \.\frac{\rho _{1} \rho _{2}}{\rho _{1}+\rho _{2}}$ ${R}=\frac{25 \\times 10^{-2}}{3 \\times 10^{-6}} \\times \frac{1.7 \\times 2.6 \\times 10^{-16}}{4.3 \\times 10^{-8}}$ ${R}=0.858 {m} \Omega$

Q3JEE Main 2021MCQ4MUnits and Measurements

What will be the projection of vector ${\vec{A}}=\hat{i}+{\hat{j}}+{\hat{k}}$ on vector ${{\vec{B}}}=\hat{i}+\hat{j}$ ?

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📖 Explanation

$(A \cos \theta) \hat{B}=A(\frac{\vec{A} \.\vec{B}}{A B}) \hat{B}=\frac{{A} ^{\to }\.{B} ^{\to }}{B} \hat{B}$ $=\frac{2}{\sqrt{2}}(\frac{\hat{i}+\hat{j}}{\sqrt{2}})=\hat{i}+\hat{j}$

Q4JEE Main 2021MCQ4MWork, Power and Energy

A porter lifts a heavy suitcase of mass $80 {kg}$ and at the destination lowers it down by a distance of $80 {cm}$ with a constant velocity. Calculate the workdone by the porter in lowering the suitcase. $($ take $g=9.8 {ms}^{-2})$

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📖 Explanation

${W}_{\text{Porter }}+{W}_{{mg}}=\Delta{K} . {E} .=0$ ${W}_{\text{Porter }}=-{W}_{{mg}}=-{mgh}$ $=-80 \\times 9.8 \\times .8=-627.2 { J}$

Q5JEE Main 2021MCQ4MOscillations

${T}_{0}$ is the time period of a simple pendulum at aplace. If the length of the pendulum is reduced to $\;\frac{1}{16}$ times of its initial value, the modified timeperiod is :

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📖 Explanation

${T}_{0}=2 \pi\sqrt{\frac{l}{{g}}}$ New time period ${T}=2 \pi\sqrt{\frac{l/ 16}{{ g}}}=\frac{2 \pi}{4} \sqrt{\frac{l}{{g}}}$ ${T}=\frac{{T}_{0}}{4}$

Q6JEE Main 2021MCQ4MRay Optics and Optical Instruments

A ray of light passes from a denser medium to a rarer medium at an angle of incidence i. The reflected and refracted rays make an angle of 90° with each other. The angle of reflection and refraction are respectively r and r'. The critical angle is given by :

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📖 Explanation

$r+r'+90°=180°\Rightarrowr'=90-r=90-i$ $n_{1} \sin i=n_{2} \sin r'=n_{2} \sin (90-i)$ $n_{1} \sin i=n_{2} \cos i \Rightarrow\tan i=\frac{n_{2}}{n_{1}}$ Now $\sin C=\frac{n_{2}}{n_{1}}=\tan i$ $\RightarrowC=\sin ^{-1}(\tan i)=\sin ^{-1}(\tan r)$

Q7JEE Main 2021MCQ4MMagnetism and Matter

Statement I : The ferromagnetic property depends on temperature. At high temperature, ferromagnet becomes paramagnet. Statement II : At high temperature, the domain wall area of a ferromagnetic substance increases. In the light of the above statements, choose the most appropriate answer from the options given below :

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Q8JEE Main 2021MCQ4MCentre of Mass and Collision

A bullet of ' $4 {g}$ ' mass is fired from a gun of mass $4 {kg}$ . If the bullet moves with the muzzle speed of $50 {ms}^{-1}$ , the impulse imparted to the gun and velocity of recoil of gun are:

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📖 Explanation

By momentum conservation $4 \\times 10^{-3}(50-{v})-4 {v}=0$ ${v}=\frac{4 \\times 10^{-3} \\times 50}{4+4 \\times 10^{-3}} \ \approx 0.05 {ms}^{-1}$ Impulse ${J}={mv}=4 \\times .05=0.2 {kgms}^{-1}$

Q9JEE Main 2021MCQ4MMagnetism and Matter

Choose the correct option :

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📖 Explanation

If apparent dip circle is at an angle alpha with true dip circle then As $\cos \alpha < 1$ Hence true dip $(\phi)$ is less than apparent ${\text{dip}}(\phi')$

Q10JEE Main 2021MCQ4MRotational Motion

Consider a situation in which a ring, a solid cylinder and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and having identical diameter. The correct statement for this situation is:-

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📖 Explanation

$a={g \sin \theta}{1+\frac{I}{m R^{2}}}$ $I_{\text{ring }}>I_{\text{solid cylinder }}>I_{\text{solid sphere }}$ $\Rightarrowa_{\text{ring }} < a_{\text{solid cylinder }} < a_{\text{solid sphere }}$ $\Rightarrowv_{\text{ring }} < v_{\text{solid cylinder }} < {v}_{\text{solid sphere }}$

Q11JEE Main 2021MCQ4MSemiconductor Electronics

Consider a situation in which reverse biased current of a particular P-N junction increases when it is exposed to a light of wavelength ≤ 621 nm. During this process, enhancement in carrier concentration takes place due to generation of hole-electron pairs. The value of band gap is nearly.

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📖 Explanation

Band gap $=\frac{h c}{\lambda_{0}}$ $\lambda_{0}$ ; threshold wavelength Band gap $=\frac{1242 e v-n m}{621 n m}=2 {eV}$

Q12JEE Main 2021MCQ4MAtoms and Nuclei

A nucleus with mass number 184 initially at rest emits an $\alpha$ -particle. If the $Q$ value of the reaction is $5.5 {MeV}$ , calculate the kinetic energy of the $\alpha-$ particle.

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📖 Explanation

$\frac{1}{2}(4 m) v^{2}+\frac{1}{2}(180 m)(\frac{4 v}{180})^{2}=5.5 {MeV}$ $\Rightarrow \frac{1}{2} 4 m v^{2}\$ [1+45(\frac{4}{180})^{2}]$=5.5 {MeV}\Rightarrow {K} . {E}{\alpha}=\frac{5.5}{1+45 .(\frac{4}{180})^{2}} {MeV}{K} . {E}{\alpha}=5.38 {MeV}$

Q13JEE Main 2021MCQ4MDual Nature of Matter and Radiation

An electron of mass ${m}_{{e}}$ and a proton of mass ${m}_{{P}}$ are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is :-

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📖 Explanation

${KE}={e} \Delta{V}$ $\lambda_{e}={{h}}/{\sqrt{2 {m}_{e}(e \Delta{V})}}$ $\lambda_{{P}}={{h}}/{\sqrt{2 {m}_{{p}}(e \Delta{V})}}$ \Rightarrow ${\lambda_{e}}/{\lambda_{{P}}}=\sqrt{{{m}_{{p}}}/{{m}_{e}}}$

Q14JEE Main 2021MCQ4MAlternating Current

Match List-I with List-II : List-I List-II (a) $\omega{L}>\frac{1}{\omega{C}}$ (i) Current is in phase with emf (b) $\omega{L}=\frac{1}{\omega{C}}$ (ii) Current lags behind the applied emf (c) $\omega{L}<\frac{1}{\omega{C}}$ (iii) Maximum current occurs (d) Resonant frequency (iv) Current leads the emfChoose the correct answer from the options given below :

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📖 Explanation

(a) For ${x}_{{L}}>{x}_{{C}}$ , voltage leads the current (ii) (b) For $x_{{L}}=x_{C}$ , voltage $\&$ current are in same phase (i)(c) For ${x}_{{L}} < {x}_{{C}}$ , current leads the voltage (iv) (d) For resonant frequency ${x}_{{L}}={x}_{{C}}$ , current is maximum (iii)

Q15JEE Main 2021MCQ4MElectromagnetic Waves

What should be the height of transmitting antenna and the population covered if the television telecast is to cover a radius of $150 {km}$ ? The average population density around the tower is $2000 / {km}^{2}$ and the value of ${R}_{{e}}=6.5 \\times 10^{6} {m}$ .

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📖 Explanation

Radius covered $r=\sqrt{2 {RH}_{{T}}}$ $150 {km}=\sqrt{2 \\times (6.5 \\times 10^{6} {m}) {H}_{{T}}}$ $(150 {km} \\times 10^{3})^{2}=2 6.5 \\times 10^{6} {H}_{{T}}$ ${H}_{{T}}=1731 {m}$ Population covered $=(\pi{r}^{2})(2000 / {km}^{2})$ $=3.14 \\times (150)^{2} \\times 2000=1413 \\times 10^{5}$

Q16JEE Main 2021MCQ4MThermodynamics

What will be the average value of energy for a monoatomic gas in thermal equilibrium at temperature ${T}$ ?

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📖 Explanation

As per Equi-partition law : Each degree of freedom contributes $\frac{1}{2} {k}_{{B}} {T}$ Average Energy In monoatomic gas D.O.F. $=3$ $\Rightarrow$ Average energy $=3 \\times \frac{1}{2} {k}_{{B}} {T}=\frac{3}{2} {k}_{{B}} {T}$

Q17JEE Main 2021MCQ4MElectromagnetic Waves

Intensity of sunlight is observed as $0.092 {Wm}^{-2}$ at a point in free space. What will be the peak value of magnetic field at that point? $(\epsilon_{0}=8.85 \\times 10^{-12} {C}^{2} {N}^{-1} {m}^{-2})$

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📖 Explanation

$I_{\text{avg }}=\frac{B_{0}^{2} C}{2 \mu_{0}}\ \&\ \frac{1}{\mu_{0}}=\in_{0} C^{2}$ $I=\frac{B_{0}^{2}}{2} \in_{0} C^{3}$ $B_{0}=\sqrt{\frac{2 I}{\in_{0} C^{3}}}$ $B_{0}=2.77 \\times 10^{-8} {T}$

Q18JEE Main 2021MCQ4MOscillations

The motion of a mass on a spring, with spring constant K is as shown in figure. The equation of motion is given by ${x}({t})=$ Asin\omega t+ Bcos\omegatwith $\omega=\sqrt{\frac{K}{m}}$ Suppose that at time ${t}=0$ , the position of mass is ${x}(0)$ and velocity ${v}(0)$ , then its displacement can also be represented as $x(t)=C \cos (\omegat-\phi)$ , where $C$ and $\phi$ are :

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📖 Explanation

${x}={A} \sin \omega{t}+{B} \cos \omega{t}$ ${v}=\frac{{dx}}{{dt}}={A} \omega\cos \omega{t}-{B} \omega\sin \omega{t}$ At ${t}=0, {x}(0)={B}$ ${v}(0)={A} \omega$ ${x}={A} \sin \omega{t}+{B} \sin (\omega{t}+90°)$ $A_{\text{net }}=\sqrt{A^{2}+B^{2}}$ $\tan \alpha=\frac{B}{A} \Rightarrow\cot \alpha=\frac{A}{B}$ $\Rightarrowx=\sqrt{A^{2}+B^{2}} \sin (\omegat+\alpha)$ $\Rightarrowx=\sqrt{A^{2}+B^{2}} \cos (\omegat-(90-\alpha))$ $x=C \cos (\omegat-\phi)$ $\RightarrowC=\sqrt{A^{2}+B^{2}}$ $C=\sqrt{\frac{[v(0)]^{2}}{\omega^{2}}+[x(0)]^{2}}$ $\phi=90-\alpha$ $\tan \alpha=\cos \alpha=\frac{A}{B}$ $\Rightarrow \tan \phi=\frac{v(0)}{x(0) \.\omega}$ $\phi=\tan ^{-1}(\frac{{v}(0)}{{x}(0) \omega})$

Q19JEE Main 2021MCQ4MElectrostatics

An electric dipole is placed on $x$ -axis in proximity to a line charge of linear charge density $3.0 \\times 10^{-6} {C} / {m}$ . Line charge is placed on z-axis and positive and negative charge of dipole is at a distance of $10 {mm}$ and $12 {mm}$ from the origin respectively. If total force of $4 {N}$ is exerted on the dipole, find out the amount of positive or negative charge of the dipole.

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📖 Explanation

$r=10 m m, x=2$ , $|\vec{F}_{q}|=\frac{2 k \lambda}{r} \.q$ $|\vec{F}_{-q}|=\frac{2 k \lambda}{r+x} \.q$ $\Rightarrow |\vec{F}_{n e t}|=\frac{2 k \lambda q}{r}-\frac{2 k \lambda q}{r+x}$ $|\vec{F}_{n e t}|=\frac{2 k \lambdaq \.x}{r(r+x)}$ $4=\frac{2 \\times 9 \\times 10^{9} \\times 3 \\times 10^{-6} \\times q \\times 2 m m}{10 {mm} .12 {mm}}$ $\Rightarrow q=4.44 \muC$

Q20JEE Main 2021MCQ4MGravitation

A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height h is ______ S.

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📖 Explanation

Applying energy conservation from (1) to (2) $\frac{1}{2} m \.(\frac{2 G M}{R_{e}})-\frac{G M m}{R_{e}}$ $=\frac{1}{2} m v^{2}-\frac{G M m}{R+r}$ $\Rightarrow \frac{1}{2} m v^{2}=\frac{G M m}{R+r}$ $v=\sqrt{\frac{2 G M}{R+r}}=\frac{d r}{d t}$ $\sqrt{2 G M} \int_{0} ^{t} d t=\int_{R_{e}} ^{R_{e}+h}(\sqrt{R+r}) d r$ $\sqrt{2 G M} \.t=\frac{2}{3}\$ [(R+r)^{3 / 2}]{R{e}}^{R_{e}+h}t=\frac{2}{3} \sqrt{\frac{R_{e}^{3}}{2 G M}}[(1+\frac{h}{R_{e}})^{3 / 2}-1]\frac{G M}{R_{e}^{2}}=gt=1/3\sqrt{{{2R}{{e}}}/{ {g}}}[(1+{{h}}/{{R}{{e}}})^{3 / 2}-1]$

Q21JEE Main 2021NAT4MSemiconductor Electronics

In a given circuit diagram, a $5 {V}$ zener diode along with a series resistance is connected across a $50 { V}$ power supply. The minimum value of the resistance required, if the maximum zener current is $90 {mA}$ will be _____ $\Omega$ .

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📖 Explanation

Voltage across ${R}_{{L}}=5 {V}$ $\Rightarrow{i}_{2}={5}/{{R}_{{L}}}$ Also voltage across ${R}=50-5=45$ volt ${By} {v}={i} {R} \Rightarrow{R}=\frac{{v}}{{i}}={45}/{{i}_{{i}}+{i}_{2}}$ $R={45}/{90 m A+\frac{5}{R_{L}}}$ Current in zener diode is maximum when ${R}_{{L}} \\to \\infty$ $({i}_{2} \\to 0$ and ${i}_{{i}}={i})$ So ${R}=\frac{45}{90 {mA}}=500 \Omega$

Q22JEE Main 2021NAT4MCentre of Mass and Collision

The position of the centre of mass of a uniform semi-circular wire of radius 'R' placed in ${x}-{y}$ plane with its centre at the origin and the line joining its ends as ${x}$ -axis is given by $(0, \frac{{xR}}{\pi})$ . Then, the value of $|{x}|$ is __________.

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📖 Explanation

COM of semi-circular ring is at $\frac{2 {R}}{\pi}$ Distance from centre ⇒ x = 2

Q23JEE Main 2021NAT4MCurrent Electricity

In an electric circuit, a call of certain emf provides a potential difference of $1.25 {V}$ across a load resistance of $5 \Omega$ . However, it provides a potential difference of $1 {V}$ across a load resistance of $2 \Omega$ .The emf of the cell is given by $\frac{x}{10} V$ . Then the value of x is ________.

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📖 Explanation

Terminal voltage ${v}={i} {R}=\frac{{ER}}{{R}+{r}}$ $1^{{st}} \\to 1.25=\frac{{E}(5)}{5+{r}}$ ........(i) $2^{\text{nd }} \\to 1=\frac{E(2)}{2+r} \......$ (ii) By (i) and (ii) ${r}=1 \Omega, {E}=\frac{3}{2} {V}=\frac{15}{10} {\text{volt}}$ $\Rightarrow{x}=15$

Q24JEE Main 2021NAT4MElectrostatics

The total charge enclosed in an incremental volume of $2 \\times 10^{-9} {m}^{3}$ located at the origin is ${nC}$ , if electric flux density of its field is found as ${D}={e}^{-{x}} \sin y \hat{i}-{e}^{-{x}} \cos y \hat{j}+2 {z} \hat{k} {C} / {m}^{2}$ .

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📖 Explanation

Electric flux density $({{D}})=\frac{\text{charge }}{\text{Area }} \\times {\hat{r}}=\frac{{Q}}{4 \pi{r}^{2}} {\hat{r}}=\in_{0}(\frac{{Q}}{4 \pi\in_{0} {r}^{2}} {\hat{r}})$ $\Rightarrow{\vec{E}}={{\vec{D}}}/{\in_{0}}$ $={e^{-{x}} \sin y {\hat{i}}-e^{-{x}} \cos y {\hat{j}}+2 z \hat{k}}/{\in_{0}}$ Also by Gauss's law $\frac{\rho }{\in_{0}}=(\frac{\ 𝜕}{\ 𝜕{x}} \hat{i}+\frac{\ 𝜕}{\ 𝜕{y}}{\hat{j}}+\frac{\ 𝜕}{\ 𝜕z} {\hat{k}}) \.{\vec{E}}=(\frac{\ 𝜕}{\ 𝜕{x}} \hat{i}+\frac{\ 𝜕}{\ 𝜕{y}} \hat{j}+\frac{\ 𝜕}{\ 𝜕{z}} {\hat{k}}) \.{{\vec{D}}}/{\in_{0}}$ $\Rightarrow\rho =\frac{\ 𝜕}{\ 𝜕x}(e^{-x} \sin y)+\frac{\ 𝜕}{\ 𝜕y}(-e^{-x} \cos y)$ $+\frac{\ 𝜕}{\ 𝜕 z}(2 z)$ $\rho =-e^{-x} \sin y+e^{-x} \sin y+2$ At origin $\rho =-e^{-0} \sin 0+e^{-0} \sin 0+2$ $\rho =2 {C} / {m}^{3}$ Charge $=\rho \\times$ volume $=2 \\times 2 \\times 10^{-9}=4 \\times 10^{-9}=$ $4 {nC}$

Q25JEE Main 2021NAT4MMotion in a Plane

Three particles ${P}, {Q}$ and ${R}$ are moving along thevectors ${\vec{A}}={\hat{i}}+\hat{j}, {\vec{B}}=\hat{j}+{{k}} ^{\^}$ and ${\vec{C}}=-\hat{i}+\hat{j}$ respectively. They strike on a point and start tomove in different directions. Now particle ${P}$ ismoving normal to the plane which contains vector $\vec{A}$ and $\vec{B}$ . Similarly particle ${Q}$ is moving normalto the plane which contains vector ${\vec{A}}$ and ${\vec{C}}$ . Theangle between the direction of motion of ${P}$ and ${Q}$ is $\cos ^{-1}(\frac{1}{\sqrt{x}}) .$ Then the value of $x$ is ______.

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📖 Explanation

Direction of $P\;\;\; \hat{v}_{1}=\\pm \frac{\vec{A} \\times \vec{B}}{|{A} ^{\to }\\times \vec{B}|}=\\pm \frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}$ Direction of ${Q}\;\;\; {\hat{v}}_{2}=\\pm {{{\vec{A}}} \\times {\vec{C}}}/{|{\vec{A}} \\times {\vec{C}}|}=\\pm {2 {\hat{k}}}/{2}=\\pm {\hat{k}}$ Angle between $\hat{v}_{1}$ and $\hat{v}_{2}$ ${{\hat{v}}_{1} \.\hat{v}_{2}}/{|\hat{v}_{1}||{\hat{v}}_{2}|}=\frac{\\pm 1 / \sqrt{3}}{(1)(1)}=\\pm \frac{1}{\sqrt{3}}$ $\Rightarrow{x}=3$

Q26JEE Main 2021NAT4MRotational Motion

The centre of a wheel rolling on a plane surface moves with a speed $v_{0}$ . A particle on the rim of the wheel at the same level as the centre will be moving at a speed $\sqrt{x} v_{0}$ . Then the value of $x$ is _______.

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📖 Explanation

For no slipping ${v}_{0}=\omega{R}$ Now $v_{{A}}=v_{B}=\sqrt{v_{0}^{2}+(\omegaR)^{2}}$ $=\sqrt{2} {v}_{0}$ \Rightarrow ${x}=2$

Q27JEE Main 2021NAT4MRay Optics and Optical Instruments

A ray of light passing through a prism $(\mu=\sqrt{3})$ suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then, the angle of prism is ________ (in degrees)

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📖 Explanation

At minimum deviation ${r}_{1}={r}_{2}=\frac{{A}}{2}$ Also given ${i}=2 {r}_{1}={A}$ Now $1 \.\sin {i}=\sqrt{3} \sin {r}_{1}$ $1 \sin A=\sqrt{3} \sin \frac{A}{2}$ $\Rightarrow 2 \sin \frac{{A}}{2} \cos \frac{{A}}{2}=\sqrt{3} \sin \frac{{A}}{2}$ $\Rightarrow \cos \frac{A}{2}=\frac{\sqrt{3}}{2} \Rightarrow\frac{A}{2}=30°$ $\Rightarrow {A}=60°$

Q28JEE Main 2021NAT4MProperties of Matter

The area of cross-section of a railway track is $0.01 {m}^{2}$ . The temperature variation is $10°{C}$ . Coefficient of linear expansion of material of track is $10^{-5} /{ }°{C}$ . The energy stored per meter in the track is _______ ${J} / {m}$ . (Young's modulus of material of track is $10^{11} {Nm}^{-2}$ )

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📖 Explanation

Elastic energy $=\frac{Y}{2}(\text{ strain })^{2} \\times$ Area $\\times$ length \Rightarrow Elastic energy per unit length $=\frac{Y}{2}(\text{ strain})^{2} \\times$ Area $($ strain $=\frac{\Deltal}{l}=\alpha\Delta{T}=10^{-5} \\times 10=10^{-4})$ $=\frac{10^{11}}{2} \\times (10^{-4})^{2} \\times 10^{-2}=5 {J} / {m}$

Q29JEE Main 2021NAT4MUnits and Measurements

Three students ${S}_{1}, {S}_{2}$ and ${S}_{3}$ perform an experiment for determining the acceleration due to gravity (g) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observationsare as shown in the table. Student No. Length of pendulum (cm No. of oscillations (n) Total time for n oscillations Time period (s) 1. 64.0 8 128.0 16.0 2. 64.0 4 64.0 16.0 3. 20.0 4 36.0 9.0 (Least count of length = 0.1 m least count for time = 0.1 s) If $E_1, E_2$ and $E_3$ are the percentage errors in 'g' for students 1, 2 and 3 respectively, then the minimum percentage error is obtained by student no._____.

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📖 Explanation

${T}=2 \pi\sqrt{\frac{l}{{g}}} \Rightarrow{g}=\frac{4 \pi^{2} l}{{T}^{2}}$ $\frac{\Deltag}{g}=\frac{\Deltal}{l}+\frac{2 \Delta{T}}{{T}}$ $\Delta{T}=\frac{\text{least count of time }(\Delta{T}_{0})}{\text{number of oscillations(n) }}$ $\frac{\Deltag}{g}=\frac{\Deltal}{l}+\frac{2 \Delta{T}_{0}}{{n} {T}}$ As $\Deltal$ and $\Delta{T}_{0}$ are same for all observations so ${\Delta{g}}/g$ is minimum for highest value of $l, {n}$ and ${T}$ \Rightarrow Minimum percentage error in ${g}$ is for student number-1

Q30JEE Main 2021NAT4MThermodynamics

In 5 minutes, a body cools from 75°C to 65°C at room temperature of 25°C. The temperature of body at the end of next 5 minutes is _______°C

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📖 Explanation

By newton's law of cooling (with approximation) $\frac{\Delta{T}}{\Deltat}=-{C}({T}_{\text{arg }}-{T}_{{s}})$ $1^{\text{st }} \frac{-10°{C}}{5 {min}}=-{C}(70°{C}-25°{C})$ $\Rightarrow {C}=\frac{2}{45} {min}^{-1}$ $2^{\text{nd }} \frac{{T}-65}{5 {min}}=-{C}(\frac{{T}+65}{2}-25)$ $=-(\frac{2}{45})(\frac{{T}+15}{2})$ $\Rightarrow 9({T}-65)=-({T}+15)$ ⇒ $10 {T}=570$ \Rightarrow ${T}=57°{C}$

Chemistry30 questions

Q31JEE Main 2021MCQ4MSolutions

The water having more dissolved $O_2$ is :

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📖 Explanation

On heating concentration of $O_2$ in water decreases. So boiling water and water at 80°C having less $O_2$ concentration. Polluted water also having less $O_2$ concentration. So water at 4°C having maximum $O_2$ concentration.

Q32JEE Main 2021MCQ4MClassification of Elements

Which one of the following statements for D.I. Mendeleeff, is incorrect?

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Q33JEE Main 2021MCQ4MPractical Organic Chemistry

Which purification technique is used for high boiling organic liquid compound (decomposes near its boiling point)?

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Q34JEE Main 2021MCQ4MAlcohols, Phenols and Ethers

Which of the following compounds will provide a tertiary alcohol on reaction with excess of ${CH}_{3} {MgBr}$ followed by hydrolysis?

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Q35JEE Main 2021MCQ4MOrganic Chemistry - Some Basic Principles

Which of the following compounds does not exhibit resonance?

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📖 Explanation

${CH}_{3}- {CH}_{2}-CH=CH- {CH}_{2} { -NH}_{2}$ No conjugation thus resonance is not possible.

Q36JEE Main 2021MCQ4MClassification of Elements

Match List-I with List-II List-I (Elements) List-II (Properties) (a) Ba (i) Organic solvent soluble compounds (b) Ca (ii) Outer electronic configuration $6s^2$ (c) Li (iii) Oxalate insoluble in water (d) Na (iv) Formation of very strong monoacidic base Choose the correct answer from the options given below :

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📖 Explanation

(a) 'Ba' having outer electronic configuration $6s^2$ - . (b) $CaC_2O_4$ is water insoluble (c) 'Li' is soluble in organic solvents (d) NaOH is strong Monoacidic base among given.

Q37JEE Main 2021MCQ4MAmines

In the chemical reactions given above $A$ and $B$ respectively are:

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Q38JEE Main 2021MCQ4MStructure of Atom

Isotope(s) of hydrogen which emits low energy $\beta^{-}$ particles with $t_½$ value > 12 years is/are

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📖 Explanation

${ }_{1}^{1} {H}$ and ${ }_{1}^{2} {H}$ are stable while ${ }_{1}^{3} {H}$ is radioactive.

Q39JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

Match List-I with List-II : List-I (Species) List-II (Hybrid Orbitals) (a) $SF_4$ (i) $sp^3d^2$ (b) $IF_5$ (ii) $d^2sp^3$ (c) $NO_2^{+}$ (iii) $sp^3d$ (d) $NH_4^{+}$ (iv) $sp^3$ (v) sp Choose the correct answer from the options given below :

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📖 Explanation

(a) ${SF}_{4}-{sp}^{3} {d}$ hybridisation (b) ${IF}_{5}-{sp}^{3} {d}^{2}$ hybridisation (c) ${NO}_{2}^{+}-$ sp hybridisation (d) ${NH}_{4}^{+}-{sp}^{3}$ hybridisation

Q40JEE Main 2021MCQ4MSolutions

When silver nitrate solution is added to potassium iodide solution then the sol produced is :

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📖 Explanation

${{AgNO}_{3}({aq} .)}_{(\text{drop} \text{by} \text{drop})}+{{KI}({aq} .) }_{\text{excess}}\\to { {AgI} / {I}^{-}}_{\text{Sol}}$

Q41JEE Main 2021MCQ4MOrganic Chemistry - Some Basic Principles

Which of the following molecules does not show stereo isomerism ?

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Q42JEE Main 2021MCQ4Mp Block Elements

Given below are the statements about diborane (a) Diborane is prepared by the oxidation of ${NaBH}_{4}$ with ${I}_{2}$ (b) Each boron atom is in sp $^{2}$ hybridized state (c) Diborane has one bridged 3 centre-2-electron bond (d) Diborane is a planar molecule The option with correct statement(s) is -

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📖 Explanation

Diborane is prepared by the reaction of ${NaBH}_{4}$ with ${I}_{2}$ . $2 {NaBH}_{4}+{I}_{2} \\to {B}_{2} {H}_{6}+2 {NaI}+{H}_{2}$ In diborane, 'B' is sp $^{3}$ hybrid, it is Non-planar and two $3 {c}-2 {e}^{-}$ bonds are present.

Q43JEE Main 2021MCQ4Mp Block Elements

Which one of the following group-15 hydride is the strongest reducing agent?

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📖 Explanation

Among $15^{\text{th }}$ group hydrides, ${BiH}_{3}$ is strongest reducing agent.

Q44JEE Main 2021MCQ4MOrganic Chemistry - Some Basic Principles

Match List-I with List-II : List-I List-II (a) Chloroprene (b) Neoprene (c) Acrylonitrile (d) Isoprene (iv) $CH_2=CH-CN$ Choose the correct answer from the options given below :

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Q45JEE Main 2021MCQ4Md and f Block Elements

The set having ions which are coloured and paramagnetic both is -

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📖 Explanation

$\text{table} {Cu}^{2+}:[{Ar}] 3 {d}^{9} 4 {s}^{0} ; {Cr}^{3+}:[{Ar}] 3 {d}^{3} 4 {s}^{0} ; {Sc}^{+}:[{Ar}] 3 {d}^{1} 4 {s}^{1} ]$ All are coloured and paramagnetic due to presence of unpaired electrons

Q46JEE Main 2021MCQ4MBiomolecules

Thiamine and pyridoxine are also known respectively as :

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📖 Explanation

Vitamine- ${B}_{1}$ is also known as Thiamine while vitamin B-6 is known as Pyridoxine

Q47JEE Main 2021MCQ4MClassification of Elements

Sulphide ion is soft base and its ores are common for metals. (a) Pb (b) Al (c) Ag (d) Mg Choose the correct answer from the options given below :

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📖 Explanation

Pb and Ag commonly exist in the form of sulphideore like PbS (galena) and $Ag_2S$ (Argentite) 'Al' is mainly found in the form of oxide orewhereas 'Mg' is found in the form of halide ore.

Q48JEE Main 2021MCQ4MAlcohols, Phenols and Ethers

An organic compound $A (C_6H_6O)$ gives dark green colouration with ferric chloride. On treatment with $CHCl_3$ and KOH, followed by acidification gives compound B. Compound B can also be obtained from compound C on reaction with pyridinium chlorochromate (PCC). Identify A, B and C .

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📖 Explanation

Compound A with the formula $C_6H_6O$ gives a dark green colour with $FeCl_3$ , which is a characteristic test for phenols. Thus, A is phenol. The reaction of phenol (A) with $CHCl_3$ and KOH is the Reimer-Tiemann reaction, which results in the ortho-formylation of phenol to produce 2-hydroxybenzaldehyde, also known as salicylaldehyde (B). Compound B can also be formed from C by reaction with pyridinium chlorochromate (PCC). PCC is a mild oxidizing agent that oxidizes primary alcohols to aldehydes. Since B is an aldehyde (salicylaldehyde), C must be the corresponding primary alcohol, which is 2-hydroxybenzyl alcohol.

Q49JEE Main 2021MCQ4MAmines

Which one of the following reactions does not occur ?

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📖 Explanation

(1) Aniline is lewis base give acid base reaction with ${AlCl}_{3}$ and form Anilinium ion (2) Anilinium ion has strongest deactivated ring so further Friedel craft Alkylation not occurs.

Q50JEE Main 2021MCQ4MSolutions

Which one of the following $0.06 {M}$ aqueous solutions has lowest freezing point?

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📖 Explanation

${T}_{{f}}-{T}_{{f}}'={i} {K}_{{f}} {m}$ For minimum ${T}_{{f}}'$ 'i' should be maximum. ${Al}_{2}({SO}_{4})_{3} \;\;\;{i}=5$ ${C}_{6} {H}_{12} {O}_{6} \;\;\;{i}=1$ ${KI} \;\;\;\;\;\;\;\;\;\;\;\;{i}=2$ ${K}_{2} {SO}_{4} \;\;\;\;\;\;\;{i}=3$

Q51JEE Main 2021NAT4MCoordination Compounds

The total number of unpaired electrons present in $[{Co}({NH}_{3})_{6}] {Cl}_{2}$ and $[{Co}({NH}_{3})_{6}] {Cl}_{3}$ is

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📖 Explanation

$[{Co}({NH}_{3})_{6}] {Cl}_{2}$ ${Co}^{2+}:[{Ar}] 3 {d}^{7} 4 {s}^{0} 4 {p}^{0}$ For this complex $\Delta_{0} <$ P.E., so pairing of electron does not take place. ${sp}^{3} {d}^{2}$ hybridisation Total 3 unpaired electrons are present. $[{Co}({NH}_{3})_{6}] {Cl}_{3}$ ${Co}^{3+}:[{Ar}] 3 {d}^{6} 4 {s}^{0} 4 {p}^{0}$ ${d}^{2} {sp}^{3}$ hybridisation ${NH}_{3}$ acts as SFL because $\Delta_{0}>{P} . {E}$ So here all electrons becomes paired.

Q52JEE Main 2021NAT4MOrganic Chemistry - Some Basic Principles

Methylation of 10 g of benzene gave 9.2 g of toluene. Calculate the percentage yield of toluene ______. (Nearest integer)

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📖 Explanation

${C}_{6} {H}_{6}+{CH}_{3} {Cl} \\to {C}_{6} {H}_{5} {CH}_{3}+{HCl}$ $10/78\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(10/78\times 92)gm\Rightarrow$ ${{A}_{{y}}}/{{T}_{{y}}}=\%$ yield $=\frac{9.2}{920} \\times 78 \\times 100 \Rightarrow 78 \%$

Q53JEE Main 2021NAT4MHydrocarbons

The number of acyclic structural isomers (including geometrical isomers) for pentene are ____

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Q54JEE Main 2021NAT4MElectrochemistry

Assume a cell with the following reaction ${Cu}_{(s)}+2 {Ag}^{+}(1 \\times 10^{-3} {M}) \\to {Cu}^{2+}(0.250 {M})$ $+2 {Ag}_{({s})}$ ${E}^{⊝}_{\text{cell }} =2.97 {V}$ ${E}_{\text{cell }}$ for the above reaction is __________ V. (Nearest integer) $[$ Given : $\log 2.5=0.3979, {T}=298 {K}]$

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📖 Explanation

${E}={E}°-\frac{0.059}{2} \log \frac{[{Cu}^{+2}]\$ }{[{Ag}^{+}]^{2}}=2.97-\frac{0.059}{2} \log \frac{0.25}{(10^{-3})^{2}}=2.81 {V}$

Q55JEE Main 2021NAT4MChemical Equilibrium

Value of $K_{p}$ for the equilibrium reaction ${N}_{2} {O}_{4} \⇌2 {NO}_{2({g})}$ at $288 {K}$ is $47.9 .$ The ${K}_{{C}}$ for this reaction at same temperature is _______. (Nearest integer) $({R}=0.083 {L}.$ bar ${K}^{-1} {mol}^{-1})$

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📖 Explanation

${K}_{{C}}={{K}_{{P}}}/{{RT}}=\frac{47.9}{0.083 \\times 288}=2$

Q56JEE Main 2021NAT4MChemical Thermodynamics

If the standard molar enthalpy change for combustion of graphite powder is $-2.48 \\times 10^{2} {kJ} {mol}^{-1}$ ,the amount of heat generated on combustion of $1 {g}$ . of graphite powder is _______ ${kJ}$ . (Nearest integer)

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📖 Explanation

$1 {mol}$ graphite $=12 {gm} {C}$ Ans. $=\frac{248}{12}=20.67 {kJ} / {gm}$ heat evolved

Q57JEE Main 2021NAT4MSolid State

A copper complex crystallising in a CCP lattice with a cell edge of $0.4518 {nm}$ has been revealed by employing X-ray diffraction studies. The density of a copper complex is found to be $7.62 {g} {cm}^{-3}$ . The molar mass of copper complex is ______ ${g} {mol}^{-1}$ (Nearest integer) $[$ Given $: {N}_{{A}}=6.022 \\times 10^{23} {mol}^{-1}]$

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📖 Explanation

${d}(\frac{{gm}}{{cc}})={4 \\times {{M}}/{{N}_{{A}}}}/{({a} {cm})^{3}}$ $7.62=\frac{4 \\times {M} / 6.022 \\times 10^{23}}{(0.4518 \\times 10^{-7} {cm})^{3}} \$ $\Rightarrow {M}=105.8 {g} / {mol}$

Q58JEE Main 2021NAT4MStructure of Atom

Number of electrons that Vanadium (Z = 23) has in p-orbitals is equal to ______

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📖 Explanation

${ }_{23} {V}: 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{3} 4 {s}^{2}$ Number of electrons in p-orbitals is equal to $12.00$

Q59JEE Main 2021NAT4MChemical Kinetics

${N}_{2} {O}_{5({g})} \\to 2 {NO}_{2({g})}+\frac{1}{2} {O}_{2({g})}$ In the above first order reaction the initial concentration of ${N}_{2} {O}_{5}$ is $2.40 \\times 10^{-2} {mol} {L}^{-1}$ at 318 ${K}$ . The concentration of ${N}_{2} {O}_{5}$ after 1 hour was $1.60$ $\\times 10^{-2} {mol} {L}^{-1}$ . The rate constant of the reaction at $318 {K}$ is s _______ $\\times 10^{-3} {min}^{-1} .$ (Nearest integer) $[$ Given : $\log 3=0.477, \log 5=0.699]$

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📖 Explanation

${K}=\frac{2.303}{{t}} \log {[{N}_{2} {O}_{5}]_{0}}/{[{N}_{2} {O}_{5}]_{{t}}}$ $=\frac{2.303}{60} \log \frac{2.4}{1.6}=6.76 \\times 10^{-3} {min}^{-1} \$ $\approx 7 \\times 10^{-3} {min}^{-1}$

Q60JEE Main 2021NAT4MSolutions

If the concentration of glucose $({C}_{6} {H}_{12} {O}_{6})$ in blood is $0.72 {g} {L}^{-1}$ , the molarity of glucose in blood is ______ $\\times 10^{-3} {M} .$ (Nearest integer) [Given: Atomic mass of ${C}=12, {H}=1, {O}=16 {u}]$

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📖 Explanation

$[$ Glucose ] $=\frac{{C}({gm} / l)}{{M}({gm} / {mol})}=\frac{0.72}{180}$ $=4 \\times 10^{-3} {M}$

Mathematics30 questions

Q61JEE Main 2021MCQ4MThree Dimensional Geometry

Let $L$ be the line of intersection of planes ${\vec{r}} \.({\hat{i}}-{\hat{j}}+2 {\hat{k}})=2$ and $\vec{r} \.(2 \hat{i}+\hat{j}-\hat{k})=2$ . If ${P}(\alpha, \beta, \gamma)$ is the foot of perpendicular on ${L}$ from the point $(1,2,0)$ , then thevalue of $35(\alpha+\beta+\gamma)$ is equal to :

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📖 Explanation

$P_{1}: x-y+2 z=2$ $P_{2}=2 x+y-3=2$ Let line of Intersection of planes ${P}_{1}$ and ${P}_{2}$ cuts ${xy}$ plane in point $Q$ . $\Rightarrow$ z-coordinate of point $Q$ is zero $\Rightarrow \text{table} x-y=2; \text{and }2 x+y=2 \} \$ $\Rightarrow x=\frac{4}{3}, y=\frac{-2}{3}$ $\Rightarrow {Q}(\frac{4}{3},\frac{-2}{3}, 0)$ Vector parallel to the line of intersection ${\vec{a}}=\begin{vmatrix}\hat{i} & {\hat{j}} & {\hat{k}} \\ 1 & -1 & 2 \\ 2 & 1 & -1\end{vmatrix}$ $=-{\hat{i}}+5 \hat{j}+3\hat{k}$ Equation of Line of intersection $\frac{x-\frac{4}{3}}{-1}=\frac{y+\frac{2}{3}}{5}=\frac{z-0}{3}=\lambda($ say $)$ Let coordinates of foot of perpendicular be ${F}(-\lambda+\frac{4}{3}, 5 \lambda-\frac{2}{3}, 3 \lambda)$ ${{PF}}^{\to }=(-\lambda+\frac{1}{3}) \hat{i}+(5 \lambda-\frac{8}{3}) \hat{j}+(3 \lambda) \hat{k}$ ${\vec{PF}} \.{\vec{a}}=0$ $\Rightarrow\lambda-\frac{1}{3}+25 \lambda\frac{-40}{3}+9 \lambda=0$ $\Rightarrow35 \lambda=\frac{41}{3} \Rightarrow\lambda=\frac{41}{105}$ Now, $\alpha=-\lambda+\frac{4}{3}, \beta=5 \lambda-\frac{2}{3}, \gamma=3 \lambda$ $\Rightarrow\alpha+\beta+\gamma=7 \lambda+\frac{2}{3}$ $=7(\frac{41}{105})+\frac{2}{3}$ $=\frac{51}{15}$ $\Rightarrow 35(\alpha+\beta+\gamma)=\frac{51}{15} \\times 35=119$

Q62JEE Main 2021MCQ4MSequences and Series

Let ${S}_{{n}}$ denote the sum of first n-terms of an arithmetic progression. If ${S}_{10}=530, {S}_{5}=140$ , then ${S}_{20}-{S}_{6}$ is equal to :

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📖 Explanation

${S}_{10}=530 \Rightarrow\frac{10}{2}\{2 {a}+9 {d}\}=530$ $\Rightarrow2 {a}+9 {d}=106 \........ .(1)$ and ${S}_{5}=140 \Rightarrow \frac{5}{2}\{2 {a}+4 {d}\}=140$ $\Rightarrow2 {a}+4 {d}=56 \...... .(2)$ $\Rightarrow5 {d}=50 \Rightarrow{d}=10 \Rightarrow{a=8}$ Now, ${S}_{20}-{S}_{6}=\frac{20}{2}\{2 {a}+19 {d}\}-\frac{6}{2}\{2 {a}+5 {d}\}$ $=14 {a}+175 {d}$ $=(14 \\times 8)+(175 \\times 10)$ $=1862$

Q63JEE Main 2021MCQ4MApplication of Derivatives

Let ${f}: {R} \\to {R}$ be defined as ${f}({x})=\begin{cases}-\frac{4}{3} {x}^{3}+2 {x}^{2}+3 {x} & {x}>0 \\ 3 {xe}^{{x}} & { ,} {x} \\le 0$ \end{cases} Then $f$ is increasing function in the interval

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📖 Explanation

${f'}({x}) \{\text{table} -{4} {x}^{3}+4 {x} +3 , {x}>0 ; 3 { e}^{{x}}(1+x) , {x} \\le 0$ For $x>0, f'(x)=-4 x^{2}+4 x+3$ ${f}({x})$ is increasing in $(-\frac{1}{2}, \frac{3}{2})$ For $x \\le 0, f'(x)=3 e^{x}(1+x)$ ${f}'({x})>0 \forall {x} \in (-1,0)$ $\Rightarrow{f}({x})$ is increasing in $(-1,0)$ So, in complete domain, ${f}({x})$ is increasing in $(-1, \frac{3}{2})$

Q64JEE Main 2021MCQ4MDifferential Equations

Let ${y}={y}({x})$ be the solution of the differential equation ${\text{cosec}}^{2} x d y+2 d x=(1+y \cos 2 x)$ ${\text{cosec}}^{2} x d x$ , with $y(\frac{\pi}{4})=0 .$ Then, the value of $(y(0)+1)^{2}$ is equal to:

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📖 Explanation

$\frac{d y}{d x}+2 \sin ^{2} x=1+y \cos 2 x$ $\Rightarrow\frac{d y}{d x}+(-\cos 2 x) y=\cos 2 x$ I.F. $=e^{\int-\cos 2 x d x}=e^{-\frac{\sin 2 x}{2}}$ Solution of D.E. $y(e^{-\frac{\sin 2 x}{2}})=\int(\cos 2 x)(e^{-\frac{\sin 2 x}{2}})$ $d x+c$ $\Rightarrowy(e^{-\frac{\sin 2 x}{2}})=-e^{-\frac{\sin 2 x}{2}}+c$ Given $y(\frac{\pi}{4})=0$ $\Rightarrow0=-{e}^{-1 / 2}+{c} \Rightarrow{c}={e}^{-1 / 2}$ $\Rightarrow{y}({e}^{-\frac{\sin 2 x}{2}})=-{e}^{-\frac{\sin 2 x}{2}}+{e}^{-1 / 2}$ at ${x}=0$ ${y}=-1+{e}^{-1 / 2}$ $\Rightarrow{y}(0)=-1+{e}^{-1 / 2} \$ $\Rightarrow ({y}(0)+1)^{2}={e}^{-1}$

Q65JEE Main 2021MCQ4MProbability

Four dice are thrown simultaneously and the numbers shown on these dice are recorded in $2 \\times 2$ matrices. The probability that such formed matrices have all different entries and are non-singular, is:

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📖 Explanation

$A=\begin{vmatrix}a & b \\ c & d\end{vmatrix} \;\;\;\;|A|=a d-b c$ Total case $=6^{4}$ For non-singular matrix $|{A}| \\ne 0 \Rightarrow{ad}-{bc} \\ne 0$ $\Rightarrow{ad} \\ne {bc}$ And $a, b, c, d$ are all different numbers in the set $\{1,2,3,4,5,6\}$ Now for ${ad}={bc}$ (i) $6 \\times 1=2 \\times 3$ $\Rightarrow \text{table} {a}=6{,} {b}=2{,} {c}=3{, }{d}=1 ; \text{or } {a}=1{, }{b}=2{,} {c}=3{, }{d}=6 ;:;:;: \}$ 8 such cases (ii) $6 \\times 2=3 \\times 4$ $\Rightarrow \text{table} {a}=6{,} {b}=3{,} {c}=4{,} { d}=2 ;\text{or }{a}=2{, }{b}=3{, }{c}=4{,} {d}=6 ;:;:;: \}$ 8 such cases favourable cases $={ }^{6} {C}_{4!}-16$ required probability $=\frac{{ }^{6} {C}_{4}4!-16}{6^{4}}=\frac{43}{162}$

Q66JEE Main 2021MCQ4MVector Algebra

Let a vector $\vec{a}$ be coplanar with vectors ${\vec{b}}=2 \hat{i}+\hat{j}+{\hat{k}}$ and $\vec{c}=\hat{i}-\hat{j}+\hat{k}$ . If $\vec{a}$ is perpendicular to ${\vec{d}}=3 \hat{i}+2 {\hat{j}}+6 {\hat{k}}$ , and $|\vec{a}|=\sqrt{10}$ . Then a possible value of $[\vec{a}\vec{b}\vec{c}]+[\vec{a}\vec{b}\vec{d}]+[\vec{a}\vec{c}\vec{d}]$ equal to:

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📖 Explanation

${\vec{a}}=\lambda\vec{b}+\mu{\vec{c}}={\hat{i}}(2 \lambda+\mu)+\hat{j}(\lambda-\mu)+{\hat{k}}(\lambda+\mu)$ $\vec{a}.\vec{d}=0=3(2 \lambda+\mu)+2(\lambda-\mu)+6(\lambda+\mu)$ $\Rightarrow14 \lambda+7 \mu=0 \Rightarrow\mu=-2 \lambda$ $\Rightarrow{\vec{a}}=(0) {\hat{i}}-3 \lambda{\hat{j}}+(-\lambda) {\hat{k}}$ $\Rightarrow|{\vec{a}}|=\sqrt{10}|\lambda|=\sqrt{10} \Rightarrow|\lambda|=1$ $\Rightarrow\lambda=1$ or $-1$ $[{\vec{a}} {\vec{b}} {\vec{c}}]=0$ $[{\vec{a}}{{\vec{b}}}{{\vec{c}}}]+[{{\vec{a}}} {{\vec{b}}}{{\vec{d}}}]+[{{\vec{a}}} {\vec{c}} {{\vec{d}}}]$ $=[{{\vec{a}}} {\vec{b}}+{\vec{c}} {{\vec{d}}}]$ $=\begin{vmatrix}0 & -3 \lambda & \lambda \\ 3 & 0 & 2 \\ 3 & 2 & 6\end{vmatrix}$ $=3 \lambda(12)+\lambda(6)=42 \lambda=-42$

Q67JEE Main 2021MCQ4MDefinite Integrals

If $\int_ {0} ^{100 \pi} {\sin ^{2} x}/{{e}^{(\frac{{x}}{\pi}[\frac{x}{\pi}])}} {dx}=\frac{\alpha\pi^{3}}{1+4 \pi^{2}}, \alpha\in {R}$ where $[{x}]$ is the greatest integer less than or equal to $x$ , then the value of $\alpha$ is :

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📖 Explanation

$I=int_ {0} ^{100 \pi} \frac{\sin ^{2} x}{ e^{(x / \pi}\}} d x=100 int_ {0} ^{ \pi} \frac{\sin ^{2} x}{{e}^{x / \pi}} d x$ $100 int_ {0} ^{ \pi} e^{-x / \pi} \frac{(1-\cos 2 x)}{2} d x$ $=50\{ int_ {0} ^{ \pi}{e}^{-x / \pi} d x- int_ {0} ^{ \pi} {e}^{-x / \pi} \cos 2 x d x\}$ $I_{1}= int_ {0} ^{ \pi} e^{-x / \pi} d x=\$ [-\pie^{-x / \pi}]{0}^{\pi}=\pi(1-e^{-1})I{2}= int_ {0} ^{ \pi} e^{-x / \pi} \cos 2 x d x=-\pi{e}^{-x / \pi} \cos 2 {x}]{0}^{\pi}-\int-\pi{e}^{-x / \pi}(-2 \sin 2 {x}) {dx}=\pi(1-{e}^{-1})-2 \piint {0} ^{ \pi} {e}^{-x / \pi} \sin 2 {x} {dx}=\pi(1-{e}^{-1})-2 \pi{-\pi{e}^{-x / \pi} \sin 2 {x}]{0}^{\pi}-int {0} ^{ \pi} -\pi{e}^{-x / \pi} 2 \cos 2 {xdx}}=\pi(1-{e}^{-1})-4 \pi^{2} {I}{2}\Rightarrow{I}{2}=\frac{\pi(1-{e}^{-1})}{1+4 \pi^{2}}\therefore {I}=50{\pi(1-{e}^{-1})-\frac{\pi(1-{e}^{-1})}{1+4 \pi^{2}}}=\frac{200(1-{e}^{-1}) \pi^{3}}{1+4 \pi^{2}}$

Q68JEE Main 2021MCQ4MVector Algebra

Let three vectors ${{\vec{a}}},{{\vec{b}}}$ and ${\vec{c}}$ be such that ${\vec{a}} \\times {\vec{b}}={\vec{c}},{\vec{b}} \\times {\vec{c}}={\vec{a}}$ and $|{\vec{a}}|=2 .$ Then which one of the following is not true?

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📖 Explanation

(1) $\vec{a} \\times (({\vec{b}}+\vec{c}) \\times ({\vec{b}}-{\vec{c}}))=\vec{0}$ $=\vec{a}(-\vec{b} \\times \vec{c}+\vec{c} \\times \vec{b})$ $=-2(\vec{a} \\times (b ^{\to }\\times \vec{c}))$ $=-2(\vec{a}\times \vec{a})\vec{0}$ (2) Projection of $\vec{a}$ on $({b} ^{\to }\\times \vec{c})$ $={a ^{\to }\.(\vec{b} \\times \vec{c})}/{|{\vec{b}} \\times \vec{c}|}=\frac{a ^{\to }\.\vec{a}}{|\vec{a}|}={|\vec{a}|}=2$ (3) $$$\text{table} \vec{a} ,\vec{b},{c} ]+\begin{bmatrix}{c} ^{\to } & {a} ^{\to } & \vec{b}\end{bmatrix}=2$$\text{table} \vec{a} ,\vec{b},{c} ]=2\vec{a}.(\vec{b}\times \vec{c})=2\vec{a} .{\vec{a}}=2|\vec{a}|^{2}=8 $(4)$ \vec{a} \times \vec{b}=\vec{c} $and$ \vec{b} \times \vec{c}=\vec{a}\Rightarrow\vec{a}, \vec{b}, \vec{c} $are mutually$ \perp $vectors.$ \therefore |{a} ^{\to }\times \vec{b}|=|\vec{c}| \Rightarrow|\vec{a}|| \vec{b}|=|\vec{c}|\Rightarrow|\vec{b}|={|\vec{c}|} / 2 $Also,$ |{b} ^{\to }\times \vec{c}|=|{a}| \Rightarrow|\vec{b} | |\vec{c}|=2 \Rightarrow|\vec{c}|=2\ &\ |\vec{b}|=1|\vec{a}+\vec{b}-2 \vec{c}|^{2}=(3 \vec{a}+{\vec{b}}-2 \vec{c}).(3 \vec{a}+\vec{b}-2 \vec{c})=9|{\vec{a}}|^{2}+|\vec{b}|^{2}+4|\vec{c}|^{2}=(9 \times 4)+1+(4 \times 4)=36+1+16=53$

Q69JEE Main 2021MCQ4MMatrices and Determinants

The values of $\lambda$ and $\mu$ such that the system of equations $x+y+z=6,3 x+5 y+5 z=26,$ ${x}+2 {y}+\lambda{z}=\mu$ has no solution, are :

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📖 Explanation

$x+y+z=6$ .....(i) $3 x+5 y+5 z=26$ .......(ii) $x+2 y+\lambdaz=\mu$ ........(iii) $5 \\times ({i})-({ii}) \Rightarrow2 {x}=4 \$ $\Rightarrow {x}=2$ ∴ from (i) and (iii) $y+z=4$ .......(iv) $2 {y}+\lambda{z}=\mu-2$ $({v})-2 \\times ({iv})$ $\Rightarrow(\lambda-2) {z}=\mu-10$ $\Rightarrow{z}=\frac{\mu-10}{\lambda-2} \ \& \ {y}=4-\frac{\mu-10}{\lambda-2}$ $\therefore$ For no solution $\lambda=2$ and $\mu\\ne 10$ .

Q70JEE Main 2021MCQ4MThree Dimensional Geometry

If the shortest distance between the straight lines $3({x}-1)=6({y}-2)=2({z}-1)$ and $4({x}-2)=2({y}-\lambda)=({z}-3), \lambda\in {R}$ is $\frac{1}{\sqrt{38}}$ , then the integral value of $\lambda$ is equal to :

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📖 Explanation

${L}_{1}: \frac{({x}-1)}{2}=\frac{({y}-2)}{1}=\frac{({z}-1)}{3}$ ${\vec{r}_{1}}=2 {\hat{i}}+{\hat{j}}+3 \hat{k}$ ${L}_{2}: \frac{({x}-2)}{1}=\frac{{y}-\lambda}{2}=\frac{{z}-3}{4}$ ${\vec{r}_{2}}={\hat{i}}+2 {\hat{j}}+4 {\hat{k}}$ Shortest distance $=$ Projection of ${\vec{a}}$ on ${\vec{r} }_{1} \times {\vec{r}}_{2}$ $={|{\vec{a}} \.({\vec{r}}_{1} \\times {\vec{r}}_{2})|}/{|{\vec{r}}_{1} \\times {\vec{r}}_{2}|}$ ${|{\vec{a}} \.({\vec{r}}_{1} \\times {\vec{r}}_{2})|}=\begin{vmatrix}1 & \lambda-2 & 2 \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{vmatrix}$ $={|14-5 \lambda|}$ $|{\vec{r}}_{1} \\times {\vec{r}}_{2}|=\sqrt{38}$ $\therefore \frac{1}{\sqrt{38}}=\frac{|14-5 \lambda|}{\sqrt{38}}$ $\Rightarrow|14-5 \lambda|=1$ $\Rightarrow14-5 \lambda=1$ or $14-5 \lambda=-1$ $\Rightarrow\lambda=\frac{13}{5}$ or 3 $\therefore$ Integral value of $\lambda=3$

Q71JEE Main 2021MCQ4MSets and Relations

Which of the following Boolean expressions is not a tautology?

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📖 Explanation

(1) $({p} \\to {q}) \vee (\sim {q} \\to {p})$ $=(\sim p \vee q) \vee (q \vee p)$ $=(\sim {p} \vee {p}) \vee {q}$ $=t \vee q=t$ (2) $({q} \\to {p}) \vee (\sim {q} \\to {p})$ $=(\sim q \vee p) \vee (q \vee p)$ $=(\sim q \vee q) \vee p$ $=t \vee p=t$ (3) $({p} \\to \sim {q}) \vee (\sim {q} \\to {p})$ $=(\sim p \vee \sim q) \vee (q \vee p)$ $=(\sim {p} \vee {p}) \vee (\sim {q} \vee {q})$ $=t \vee t=t$ (4) $(\sim {q} \\to {q}) \vee (\sim {q} \\to {p})$ $=(p \vee q) \vee (q \vee p)$ $=(p \vee p) \vee (q \vee p)$ $={p} \vee {q}$ Which is not a tautology.

Q72JEE Main 2021MCQ4MMatrices and Determinants

Let ${A}=[{a}_{{i} j}.$ ] be a real matrix of order $3 \\times 3$ , such that ${a}_{{i} 1}+{a}_{{i} 2}+{a}_{{i} 3}=1$ , for ${i}=1,2,3 .$ Then, the sum of all the entries of the matrix ${A}^{3}$ is equal to :

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📖 Explanation

$A=\begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\\end{bmatrix}$ Let $x=\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$ $A X =\begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\\end{bmatrix}$ $=\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$ $\Rightarrow{AX}={X}$ Replace $X$ by AX ${A}^{2} {X}={AX}={X}$ Replace $X$ by AX $A^{3} X=A X=X$ Let $A^{3}=\begin{bmatrix}x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ z_{1} & z_{2} & z_{3}\end{bmatrix}$ $A^{3}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}=\begin{bmatrix}x_{1}+x_{2}+x_{3} \\ y_{1}+y_{2}+y_{3} \\ z_{1}+z_{2}+z_{3}\end{bmatrix}$ $=\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$ Sum of all the element $=3$

Q73JEE Main 2021MCQ4MFunctions

Let $[{x}]$ denote the greatest integer less than or equal to $x$ . Then, the values of $x \inR$ satisfying the equation $[{e}^{{x}}]^{2}+\$ [{e}^{{x}}+1]$-3=0$ lie in the interval :

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📖 Explanation

$[{e}^{{x}}]^{2}+\$ [{e}^{{x}}+1]$-3=0\Rightarrow$[{e}^{{x}}]^{2}+$[{e}^{{x}}]$+1-3=0 $Let$ [{e}^{{x}}]={t}\Rightarrow{t}^{2}+{t}-2=0\Rightarrow{t}=-2,1[{e}^{{x}}]=-2 $(Not possible) or$ [{e}^{{x}}]=1 ;;;;\therefore 1 \≤ {e}^{{x}} < 2 $\Rightarrow$ \ln (1) \≤ x < \ln (2) $\Rightarrow$ 0 \≤ x < \ln (2) $\Rightarrow$ x \in[0, \ln 2)$

Q74JEE Main 2021MCQ4MCircles

Let the circle ${S}: 36 {x}^{2}+36 {y}^{2}-108 {x}+120 {y}+{C}=0$ be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, $x-2 y=4$ and $2 x-y=5$ lies inside the circle ${S}$ , then :

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📖 Explanation

${S}: 36 {x}^{2}+36 {y}^{2}-108 {x}+120 {y}$ $+{C}=0$ $\Rightarrowx^{2}+y^{2}-3 x+\frac{10}{3} y+\frac{C}{36}=0$ Centre $\equiv (-g,-f) \equiv (\frac{3}{2}, \frac{-10}{6})$ radius $=r=\sqrt{\frac{9}{4}+\frac{100}{36}-\frac{C}{36}}$ Now, $\Rightarrow {r} < \frac{3}{2}$ $\Rightarrow\frac{9}{4}+\frac{100}{36}-\frac{{C}}{36} < \frac{9}{4}$ $\Rightarrow {C}>100$ ........(1) Now point of intersection of $x-2 y=4$ and $2 {x}-{y}=5$ is $(2,-1)$ , which lies inside the circle ${S}$ . $\therefore {S}(2,-1) < 0$ $\Rightarrow(2)^{2}+(-1)^{2}-3(2)+\frac{10}{3}(-1)$ $+\frac{{C}}{36} < 0$ $\Rightarrow4+1-6-\frac{10}{3}+\frac{{C}}{36} < 0$ ${C} < 156$ ..........(2) From $(1) &(2)$ $100 < {C} < 156$ Ans.

Q75JEE Main 2021MCQ4MComplex Numbers

Let ${n}$ denote the number of solutions of the equation ${z}^{2}+3 \ov {{z}}=0$ , where ${z}$ is a complex number. Then the value of $\sum_{{k}=0} ^{\\infty } {1}/{{n}^{{k}}}$ is equal to

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📖 Explanation

${z}^{2}+3 \overline{{z}}=0$ Put $z=x+i y$ $\Rightarrowx^{2}-y^{2}+2 i x y+3(x-i y)=0$ $\Rightarrow(x^{2}-y^{2}+3 x)+i(2 x y-3 y)$ $=0+i 0$ $\therefore {x}^{2}-{y}^{2}+3 {x}=0$ ........(1) $2 x y-3 y=0$ ........(2) $x=\frac{3}{2}, y=0$ Put $x=\frac{3}{2}$ in equation (1) $\frac{9}{4}-y^{2}+\frac{9}{2}=0$ $y^{2}=\frac{27}{4} \Rightarrowy=\\pm \frac{3 \sqrt{3}}{2}$ $\therefore (x, y)=(\frac{3}{2}, \frac{3 \sqrt{3}}{2}),(\frac{3}{2}, \frac{-3 \sqrt{3}}{2})$ Put $y=0 \Rightarrowx^{2}-0+3 x=0$ $x=0,-3$ $\therefore (x, y)=(0,0),(-3,0)$ $\therefore$ No of solutions $=n=4$ $\sum_{K=0} ^{\\infty }(\frac{1}{n^{k}})=\sum_{K=0} ^{\\infty }(\frac{1}{4^{k}})$ $=\frac{1}{1}+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\........ .$ $=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$

Q76JEE Main 2021MCQ4MTrigonometric Ratios and Identities

The number of solutions of $\sin ^{7} x+\cos ^{7} x=1$ , $x \ \in[0,4 \pi]$ is equal to

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📖 Explanation

$\sin ^{7} x \\le \sin ^{2} x \\le \lambda 1$ .......(1) and $\cos ^{7} x \\le \cos ^{2} x \\le 1$ ........(2) also $\sin ^{2} x+\cos ^{2} x=1$ $\Rightarrow$ equality must hold for (1) & (2) $\Rightarrow\sin ^{7} x=\sin ^{2} x \ \&\ \cos ^{7}=\cos ^{2} x$ $\Rightarrow\sin x=0 \& \cos x=1$ or $\cos x=0 \ \&\ \sin x=1$ $\Rightarrow{x}=0,2 \pi, 4 \pi, \frac{\pi}{2}, \frac{5 \pi}{2}$ $\Rightarrow5$ solutions

Q77JEE Main 2021MCQ4MFunctions

If the domain of the function $f(x)={\cos ^{-1} \sqrt{x^{2}-x+1}}/{\sqrt{\sin ^{-1}(\frac{2 x-1}{2})}}$ is the interval $(\alpha, \beta]$ , then $\alpha+\beta$ is equal to $:$

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📖 Explanation

${O} \\le {x}^{2}-{x}+1 \\le 1$ $\Rightarrow{x}^{2}-{x} \\le 0$ $\Rightarrow{x} \in[0,1]$ Also, $0 < \sin ^{-1}(\frac{2 x-1}{2}) \\le \frac{\pi}{2}$ $\Rightarrow0 < \frac{2 x-1}{2} \\le 1$ $\Rightarrow 0 < 2 {x}-1 \\le 2$ $1 < 2 {x} \\le 3$ $\frac{1}{2} < {x} \\le \frac{3}{2}$ Taking intersection $x \in(\frac{1}{2}, 1]$ $\Rightarrow\alpha=\frac{1}{2}, \beta=1$ $\Rightarrow\alpha+\beta=\frac{3}{2}$

Q78JEE Main 2021MCQ4MLimits, Continuity and Differentiability

Let $f: {R} \\to {R}$ be defined as $f(x)=\{\text{table} \frac{x^{3}}{(1-\cos 2 x)^{2}} \log _{c}(\frac{1+2 x e^{-2 x}}{(1-x e^{-x})^{2}}) {,}, x \\ne 0; \alpha {, },x=0$ If $f$ is continuous at $x=0$ , then $\alpha$ is equal to :

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📖 Explanation

For continuity $\lim _{x \\to 0} \frac{x^{3}}{4 \sin ^{4} x}(ln(1+2 x e^{-2 x})-2 ln(1-x e^{-x}))$ $=\alpha$ $\lim _{x \\to 0} \frac{1}{4 x}\$ [2 x e^{-2 x}+2 x e^{-x}]$=\alpha=\frac{1}{4}(4)=\alpha=1$

Q79JEE Main 2021MCQ4MHyperbola

Let a line ${L}: 2 {x}+{y}={k}, {k}>0$ be a tangent to the hyperbola ${x}^{2}-{y}^{2}=3 .$ If ${L}$ is also a tangent to the parabola ${y}^{2}=\alpha{x}$ , then $\alpha$ is equal to $:$

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📖 Explanation

Tangent to hyperbola of Slope ${m}=-2$ (given) $y=-2 x \\pm \sqrt{3(3)}$ $(y=m x \\pm \sqrt{a^{2} m^{2}-b^{2}})$ $\Rightarrow {y}+2 {x}=\\pm 3 \Rightarrow2 {x}+{y}=3 ({k}>0)$ For parabola ${y}^{2}=\alpha{x}$ ${y}={m} {x}+\frac{\alpha}{4 {m}}$ $\Rightarrow{y}=-2 {x}+\frac{\alpha}{-8}$ $\Rightarrow\frac{\alpha}{-8}=3$ $\Rightarrow\alpha=-24 .$

Q80JEE Main 2021MCQ4MEllipse

Let ${E}_{1}: \frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1, {a}>{b} .$ Let ${E}_{2}$ be another ellipse such that it touches the end points of major axis of ${E}_{1}$ and the foci of ${E}_{2}$ are the end points of minor axis of ${E}_{1}$ . If ${E}_{1}$ and ${E}_{2}$ have same eccentricities, then its value is:

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📖 Explanation

${e}^{2}=1-\frac{{b}^{2}}{{a}^{2}}$ ${e}^{2}=1-\frac{{a}^{2}}{{c}^{2}}$ $\Rightarrow\frac{{b}^{2}}{{a}^{2}}=\frac{{a}^{2}}{{c}^{2}}$ $\Rightarrow{c}^{2}=\frac{{a}^{4}}{{b}^{2}} \Rightarrow{c}=\frac{{a}^{2}}{{b}}$ Also ${b}={ce}$ $\Rightarrow{c}=\frac{{b}}{{e}}$ $\frac{{b}}{{e}}=\frac{{a}^{2}}{{b}}$ $\Rightarrow{e}=\frac{{b}^{2}}{{a}^{2}}=1-{e}^{2}$ $\Rightarrow{e}^{2}+{e}-1=0$ $\Rightarrow {e}=\frac{-1+\sqrt{5}}{2}$

Q81JEE Main 2021NAT4MPermutations and Combinations

Let ${A}=\{0,1,2,3,4,5,6,7\} .$ Then the number of bijective functions $f: {A} \\to {A}$ such that $f(1)+f(2)=3-f(3)$ is equal to

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📖 Explanation

$f(1)+f(2)=3-f(3)$ $\Rightarrowf(1)+f(2)=3+f(3)=3$ The only possibility is : $0+1+2=3$ $\Rightarrow$ Elements $1,2,3$ in the domain can be mapped with $0,1,2$ only. So number of bijective functions. $={3}! \\times 5!=720$

Q82JEE Main 2021NAT4MPermutations and Combinations

If the digits are not allowed to repeat in any number formed by using the digits 0, 2, 4, 6, 8, then the number of all numbers greater than 10,000 is equal to _______.

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📖 Explanation

$=4 \\times 4 \\times 3 \\times 2=96$

Q83JEE Main 2021NAT4MMatrices and Determinants

Let ${A}=\begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}$ . Then the number of $3 \\times 3$ matrices ${B}$ with entries from the set $\{1,2,3,4,5\}$ and satisfying ${AB}={BA}$ is ______.

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📖 Explanation

Let matrix ${B}=\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$ \because ${AB}={BA}$ $\begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$ $=\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix} \begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}$ $\begin{bmatrix}d & e & f \\ a & b & c \\ g & h & i\end{bmatrix}=\begin{bmatrix}b & a & c \\ e & d & f \\ h & g & i\end{bmatrix}$ $\Rightarrowd=b, e=a, f=c, g=h$ $\therefore$ Matrix ${B}=\begin{bmatrix}{a} & {b} & {c} \\ b & a & c \\ g & g & i\end{bmatrix}$ No. of ways of selecting ${a}, {b}, {c}, {g}, {i}$ $=5 \\times 5 \\times 5 \\times 5 \\times 5$ $=5^{5}=3125$ $\therefore$ No. of Matrices ${B}=3125$

Q84JEE Main 2021NAT4MStatistics

Consider the following frequency distribution : Class : 0-6 6-12 12-18 18-24 24-30 Frequency : a b 12 9 5 If mean $=\frac{309}{22}$ and median $=14$ , then the value $(a-b)^{2}$ is equal to _______.

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📖 Explanation

Class Frequency $x_i$ $f_ix_i$ 0-6 a 3 3a 6-12 b 9 9b 12-18 12 15 180 18-24 9 21 189 24-30 5 27 135 N=(26+a+b) (504+3a+9b)Mean $=\frac{3 a+9 b+180+189+135}{a+b+26}$ $=\frac{309}{22}$ $\Rightarrow66 a+198 b+11088=309 a+309 b$ $+8034$ $\Rightarrow243 {a}+111 {b}=3054$ $\Rightarrow81 {a}+37 {b}=1018 \\to (1)$ Now, Median $=12+{\frac{{a}+{b}+26}{2}-({a}+{b})}/{12} \\times 6=14$ $\Rightarrow\frac{13}{2}-(\frac{{a}+{b}}{4})=2$ $\Rightarrow\frac{{a}+{b}}{4}=\frac{9}{2}$ $\Rightarrow{a}+{b}=18 \\to (2)$ From equation $(1) &(2)$ ${a}=8, {b}=10$ $\therefore ({a}-{b})^{2}=(8-10)^{2}$

Q85JEE Main 2021NAT4MSets and Relations

The sum of all the elements in the set $\{{n} \in\{1,2, \......., 100\} \|$ H.C.F. of ${n}$ and 2040 is 1 $\}$ is equal to _______.

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📖 Explanation

$2040=2^{3} \\times 3 \\times 5 \\times 17$ ${n}$ should not be multiple of $2,3,5$ and 17 . Sum of all ${n}=(1+3+5 \.......+99)-(3+9+15+$ $21+\........ .+99)-(5+25+35+55$ $+65+85+95)$ $-(17)$ $=2500-\frac{17}{2}(3+99)-365-17$ $=2500-867-365-17$ $=1251$

Q86JEE Main 2021NAT4MArea Under The Curves

The area (in sq. units) of the region bounded by the curves $x^{2}+2 y-1=0, y^{2}+4 x-4=0$ and $y^{2}-4 x-4=0$ in the upper half plane is _______.

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📖 Explanation

Required Area (shaded) $=2[\int_{0} ^{2}(\frac{4-y^{2}}{4}) d y-\int_{0} ^{1}(\frac{1-x^{2}}{2}) d x]$ $=2[\frac{4}{3}-\frac{1}{3}]=(2)$

Q87JEE Main 2021NAT4MLimits, Continuity and Differentiability

Let $f: {R} \\to {R}$ be a function defined as $f({x})=\begin{cases}3(1-\frac{|{x}|}{2}) & \text{ if } {|{x}|} \\le 2 \\ 0 & \text{if } {|{x}|}>2$ Let $g: {R} \\to {R}$ be given by $g(x)=f(x+2)-f(x-2)$ \end{cases} If ${n}$ and ${m}$ denote the number of points in ${R}$ where ${g}$ is not continuous and not differentiable,respectively, then ${n}+{m}$ is equal to _______.

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📖 Explanation

$f(x-2) \{\text{table} \frac{3 x}{2} , -4 \\le x \\le -2;-\frac{3 x}{2}, -2 < x \\le 0; 0 , x \in(-\\infty ,-4) \cup(0,+\\infty )\$ $f(x-2) \{\text{table} \frac{3 x}{2} , 0 \\le x \\le 2;-\frac{3 x}{2}+6, 2 \le x \\le 4; 0 , x \in(-\\infty ,0) \cup(4,+\\infty )\$ $g(x)=f(x+2)-f(x-2) \{\text{table} {3 x}{2}+6 , -4 \\le x \\le -2 ; -\frac{3 x}{2} , -2 < x < 2 ; \frac{3 x}{2}-6 , 2 \\le x \\le 4; 0 , x \in(-\\infty ,-4) \cup(4, \ \infty )$ ${n}=0$ ${m}=4 \Rightarrow({n}+{m}=4)$

Q88JEE Main 2021NAT4MBinomial Theorem

If the constant term, in binomial expansion of $(2 {x}^{{r}}+\frac{1}{{x}^{2}})^{10}$ is 180 , then ${r}$ is equal to _______.

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📖 Explanation

$(2 {x}^{{r}}+\frac{1}{{x}^{2}})^{10}$ General term $={ }^{10} {C}_{{R}}(2 {x}^{2})^{10-{R}} {x}^{-2 {R}}$ $\Rightarrow2^{10-{R}{ }^{10}} {C}_{{R}}=180 ...... . .(1)$ $\&\ (10-{R}) {r}-2 {R}=0$ ${r}=\frac{2 {R}}{10-{R}}$ ${r}=\frac{2({R}-10)}{10-{R}}+\frac{20}{10-{R}}$ $\Rightarrow{r}=-2+\frac{20}{10-{R}} \........ .$ (2) ${R}=8$ or 5 reject equation (1) not satisfied At ${R}=8$ $2^{10-{R}} ^{ 10}{C}_{{R}}=180 \Rightarrow{r}=8$

Q89JEE Main 2021NAT4MDifferential Equations

Let ${y}={y}({x})$ be the solution of the differential equation $((x+2) e^{(\frac{y+1}{x+2})}+(y+1)) d x$ $=(x+2) d y,$ $y(1)=1 .$ If the domain of $y=y(x)$ is an open interval $(\alpha, \beta)$ , then $|\alpha+\beta|$ is equal to _______.

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📖 Explanation

${y}+1={Y} \Rightarrow{dy}={d} {Y}$ ${x}+2={X} \Rightarrow{d} {x}={d} {X}$ $\Rightarrow({Xe}^{\frac{{Y}}{{X}}}+{Y}) {dX}={XdY}$ $\Rightarrow{XdY}-{YdX}={Xe}^{{Y} / {x}} {dX}$ $\Rightarrow{d}(\frac{{Y}}{{X}}) {e}^{-\frac{{Y}}{{x}}}=\frac{{dX}}{{X}}$ $-{e}^{-{Y} / {x}}=l|{X}|+{c}$ $(3,2) \\to -{e}^{-2 / 3}=l|3|+{c}$ $-{e}^{-\frac{{Y}}{{x}}}=l{n}|{X}|-{e}^{-\frac{2}{3}}-l{n} 3$ $e^{-\frac{Y}{x}}=e^{2 / 3}+ln 3-l n|X|>0$ $\ln |{X}| < ({e}^{2 / 3}+l{n} 3)$ Let $\lambda=({e}^{2 / 3}+l{n} 3)$ $|x+2| < {e}^{\lambda}$ $-{e}^{\lambda} < {x}+2 < {e}^{\lambda}$ $-{e}^{\lambda}-2 < {x} < {e}^{\lambda}-2$ $\alpha+\beta=-4 \Rightarrow|\alpha+\beta|=4$ Although ${x}=-2$ should be excluded from domain but according to the given problem it will the most appropriate solution.

Q90JEE Main 2021NAT4MFunctions

The number of elements in the set $\{{n} \in\{1,2,3, \......, 100\}$ $(11)^{{n}}>(10)^{{n}}+(9)^{{n}}\}$ is _______.

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📖 Explanation

$11^{{n}}>10^{{n}}+9^{{n}}$ $\Rightarrow11^{{n}}-9^{{n}}>10^{{n}}$ $\Rightarrow(10+1)^{{n}}-(10-1)^{{n}}>10^{{n}}$ $\Rightarrow\{{ }^{{n}} {C}_{1} \.10^{{n}-1}+{ }^{{n}} {C}_{3} 10^{{n}-0}+{ }^{{n}} {C}_{5} 10^{{n}-5}$ $+\...... \}>10^{{n}}$ $\Rightarrow2 {n} \.10^{{n}-1}+2\{{ }^{{n}} {C}_{3} 10^{{n}-3}+{ }^{{n}} {C}_{5} 10^{{n}-5}$ $+.......\}>10^{{n}}$ ........(1) For ${n}=5$ $10^{5}+2\{{ }^{5} {C}_{3} 10^{2}+{ }^{5} {C}_{5}\}>10^{5}$ (True) For ${n}=6,7,8, \.......100$ $2 {n} 10^{{n}-1}>10^{{n}}$ $\Rightarrow2 {n} 10^{n-1}+2\{{ }^{{n}} {C}_{3} 10^{{n}-3}+{ }^{{n}} {C}_{5} 10^{{n}-5}$ $+\........\}>10°$ $\Rightarrow11^{{n}}-9^{{n}}>10^{{n}}$ For ${n}=5,6,7, \.......100$ For ${n}=4$ , Inequality (1) is not satisfied $\Rightarrow$ Inequality does not hold good for ${N}=1,2,3,4$ So, required number of elements $=96$

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