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Physics30 questions

Q1JEE Main 2021MCQ4MUnits and Measurements

If $e$ is the electronic charge, $c$ is the speed of light in free space and $h$ is Planck's constant, the quantity $\;\frac{1}{4 \pi \epsilon_{0}} \;\frac{|e|^{2}}{h c}$ has dimensions of

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📖 Explanation

Dimensional formula of $[e]=[I T]$ $[h] \;\; =[M^{1} L^{2} T^{-1}]$ $[c] \;\; =[M^{0} L^{1} T^{-1}]$ $[\;\frac{1}{4 \pi \epsilon_{0}}] \;\; =[M^{1} L^{3} T^{-4} I^{-2}]$ where, $\;\frac{1}{4 \pi \epsilon_{0}}$ is Coulomb's constant. Therefore, dimensional formula of $\;\frac{1}{4 \pi \epsilon_{0}} \;\frac{| {e}|^{2}}{h c} \;\; =\$ [ {M}^{1} {L}^{3} {T}^{-4} {I}^{-2}]$ \cdot ;\frac{[ {IT}]^{2}}{[ {M}^{1} {L}^{2} {T}^{-1}]$$[ {M}^{0} {LT}^{-1}]$};; =$[ {M}^{0} {L}^{0} {T}^{0} {I}^{0}]$ ;\text{ or };[ {M}^{0} {L}^{0} {T}^{0}]$

Q2JEE Main 2021MCQ4MMotion in a Straight Line

A stone is dropped from the top of a building. When it crosses a point $5 {m}$ below the top, another stone starts to fall from a point $25 {m}$ below the top. Both stones reach the bottom of building simultaneously. The height of the building is

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📖 Explanation

Let the total height of building be $x$ . $T A \;\; =5 m$ $T B \;\; =25 m$ $\therefore \;\; A G=x-5$ and $B C=x-25$ For initial conditions, from second equation of motion under gravity, $s=u t+1 / 2 g t^{2}$ where, $g=10 {ms}^{-2}$ $\therefore \;\; 5=0+1 / 2 \times 10 t^{2} \Rightarrow t=1 {s}$ Now, by first equation of motion under gravity, $v_{A} \;\; =u+g t$ $\;\; =0+10=10 {ms}^{-1}$ From second equation of motion, $x-5=v_{A} t+1 / 2 g t^{2}$ . . . (i) Similarly, $x-25=1 / 2 {gt}^{2}$ Put the above value in Eq. (i), we get $x-5 \;\; =10 t+x-25$ $20 \;\; =10 t \Rightarrow t=2 {s}$ Put the value of $t$ in Eq. (i), we get $x-5 \;\; =10 \times 2+1 / 2 \times 10 \times 4$ $x-5 \;\; =20+20$ $x \;\; =45 {m}$

Q3JEE Main 2021MCQ4MRotational Motion

A sphere of radius $a$ and mass $m$ rolls along a horizontal plane with constant speed $v_{0}$ . It encounters an inclined plane at angle $\theta$ and climbs upwards. Assuming that it rolls without slipping, how far up the sphere will travel?

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📖 Explanation

Given, radius of sphere is a, mass of sphere is $m$ , horizontal speed of sphere is $v_{0}$ . Moment of inertia of solid sphere, $I=2 / 5 {ma}^{2}$ Let, $h$ be the height of inclined plane and $/$ be the length of inclined plane.By using law of conservation of energy, Total energy at $1=$ Total energy at 2 $\therefore \;\; {PE}_{1}+ {KE}_{1}= {PE}_{2}+ {KE}_{2}$ $\Rightarrow \;\; 0+1 / 2 m v_{0}^{2}+1 / 2 / \omega^{2}=m g h$ . . . (i) As, $\;\; v_{0}=a \omega \Rightarrow \omega=\;\frac{v_{0}}{a}$ $\therefore \;\; \;\frac{1}{2} m v_{0}^{2}+\;\frac{1}{2}(\;\frac{2}{5} m a^{2}) \;\frac{v_{0}^{2}}{a^{2}}=m g h$ $\Rightarrow \;\; \;\frac{1}{2} m v_{0}^{2}(1+2 / 5)=m g h$ $\Rightarrow \;\; \;\frac{7 v_{0}^{2}}{10}=g h$ . . . (ii) As, $\;\; \sin \theta=h / l$ $\Rightarrow \;\; h=I \sin \theta$ Put this value in Eq. (ii), we get $\Rightarrow$ $\;\frac{7}{10} v_{0}^{2}=g l \sin \theta \Rightarrow l=\;\frac{7 v_{0}^{2}}{10 g \sin \theta}$

Q4JEE Main 2021MCQ4MOscillations

The point $A$ moves with a uniform speed along the circumference of a circle of radius $0.36 {m}$ and covers $30^{\circ}$ in $0.1 {s}$ . The perpendicular projection $P$ from $A$ on the diameter $M N$ represents the simple harmonic motion of $P$ . The restoration force per unit mass when $P$ touches $M$ will be

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📖 Explanation

Given, radius of circle, $R=0.36 {m}$ Angular distance, $\theta=30^{\circ}=\pi / 6 {rad}$ Let $/$ be the arc length. $\because \;\; I=R \theta$ $\Rightarrow \;\; I=\;\frac{36}{100} \times \;\frac{\pi}{6}=\;\frac{6 \pi}{100} {m}$ A s, speed on circular track $(v)=Arclength(l) / Time(t)$ $\Rightarrow \;\; v=\;\frac{6 \pi}{100 \times 0.1}=\;\frac{6 \pi}{10} {ms}^{-1}$ If $F$ be the restoration force and $a_{t}$ be the radial acceleration $(=v^{2} / R)$ , then $F \;\; =m a_{r}$ $a_{r} \;\; =\;\frac{{F}}{m}=\;\frac{v^{2}}{R}=(\;\frac{6 \pi}{10})^{2} \times \;\frac{100}{36}$ $\;\; =\;\frac{36 \times 9.87}{100} \times \;\frac{100}{36}=9.87 {N}$

Q5JEE Main 2021MCQ4MThermodynamics

Thermodynamic process is shown below on a $p$ -V diagram for one mole of an ideal gas. If $V_{2}=2 V_{1}$ , then the ratio of temperature $\;\frac{T_{2}}{T_{1}}$ is

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📖 Explanation

Here, $p_{1}$ and $p_{2}, T_{1}$ and $T_{2}, V_{1}$ and $V_{2}$ are initial and final pressures, temperatures and volumes, respectively. Given, $\;\; V_{2}=2 V_{1}$ $p V^{1 / 2}=\;\text{ constant }\;$ From graph, $\gamma=$ adiabatic constant $=1 / 2$ $\Rightarrow \;\; p_{1} V_{1}^{1 / 2}=p_{2} V_{2}^{1 / 2} \Rightarrow \;\frac{p_{1}}{p_{2}}=(\;\frac{2 V_{1}}{V_{1}})^{1 / 2}=2^{1 / 2}$ Also, $\;\; p^{1-\gamma} T^{\gamma}=$ constant (for adiabatic process) $\Rightarrow \;\; p_{1}^{1-\gamma} T_{1}^{\gamma}=p_{2}^{1-\gamma} T_{2}^{\gamma}$ $\Rightarrow \;\; (\;\frac{p_{1}}{p_{2}})^{1-\gamma}=(\;\frac{T_{2}}{T_{1}})^{\gamma} \Rightarrow (\;\frac{p_{1}}{p_{2}})^{\;\frac{1-\gamma}{\gamma}}=\;\frac{T_{2}}{T_{1}}$ $\therefore \;\; T_{2} / T_{1}=(2^{1 / 2})^{\;\frac{1-0.5}{0.5}}=\sqrt{2}$

Q6JEE Main 2021MCQ4MThermodynamics

Given below are two statements: Statement I In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution. Statement II In a diatomic molecule, the rotational energy at a given temperature equals the translational kinetic energy for each molecule. In the light of the above statements, choose the correct answer from the options given below.

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📖 Explanation

According to Statement $1 :$ In diatomic molecule the rotational energy at a given temperature obeys Maxwell's distribution is correct. But, according to Statement $2:$ In diatomic molecule, the rotational energy at a given temperature equals translational energy for each molecule is false. Because kinetic energy $( {KE})$ of gaseous molecule is $\;\frac{f}{2} K_{B} T$ . where, $f$ is the degree of freedom, and $f($ diatomic atom $)=5=[3$ (translational) and 2 (rotational)] Therefore, translational ${KE}$ of gas $=\;\frac{3}{2} K_{B} T$ . . . (i) and rotational ${KE}$ of gas $=2 / 2 K_{B} T=K_{B} T$ . . . (ii) $\because$ Eq. (i) is not equal to Eq. (ii) Hence, option (c) is the correct.

Q7JEE Main 2021MCQ4MOscillations

Two identical springs of spring constant $2 k$ are attached to a block of mass $m$ and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is

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📖 Explanation

Let spring constants of two springs be $k_{1}$ and $k_{2} .$ Since, two springs are connected in parallel connection and parallel equivalent spring constant, $k_{\;\text{eq }\;}=k_{1}+k_{2}$ $\Rightarrow \;\; k_{ {eq}}=2 k+2 k=4 k$ As, time period, $T=2 \pi \sqrt{\;{m}/{k_{\;\text{eq }\;}}}$ $\Rightarrow \;\; T=2 \pi \sqrt{\;\frac{m}{4 k}}=\;\frac{2 \pi}{2} \sqrt{\;\frac{m}{k}}=\pi \sqrt{\;\frac{m}{k}}$

Q8JEE Main 2021MCQ4MOscillations

$Y=A \sin (\omega t+\varphi _{0})$ is the time-displacement equation of SHM. At $t=0$ , the displacement of the particle is $Y=\;\frac{A}{2}$ and it is moving along negative $x$ -direction. Then, the initial phase angle $\varphi _{0}$ will be

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📖 Explanation

Given, displacement-time equation, $Y=A \sin (\omega t+\varphi _{0})$ Here, $A$ is amplitude, $\omega$ is angular frequency, $t$ is time taken and $\varphi _{0}$ is the phase constant. At $t=0, Y=A / 2$ $\therefore \;Y=A / 2=A \sin (0+\varphi _{0})=A \sin \varphi _{0}$ $\Rightarrow \;\sin \varphi _{0}=1 / 2$ $\Rightarrow \;\varphi _{0}=\pi / 6$

Q9JEE Main 2021MCQ4MElectrostatics

A charge $q$ is placed at one corner of a cube as shown in figure. The flux of electrostatic field $E$ through the shaded area is

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📖 Explanation

Given, charge $q$ is at one of the corner of the cube. $\therefore$ Contribution of $q$ in cube will be $q_{\;\text{enclosed }\;}=q / 8$ As, only 3 faces of cube is allowing the flux lines to pass through it. $\therefore$ Flux $(\varphi )=\;{q_{\;\text{enclosed }\;}}/{\epsilon_{0}}=\;\frac{1}{3} \;\frac{q / 8}{\epsilon_{0}}=\;\frac{q}{24 \epsilon_{0}}$

Q10JEE Main 2021MCQ4MElectrostatic Potential and Capacitance

An electron with kinetic energy $K_{1}$ enters between parallel plates of a capacitor at an angle $\alpha$ with the plates. It leaves the plates at angle $\beta$ with kinetic energy $K_{2}$ . Then, the ratio of kinetic energies $K_{1}: K_{2}$ will be

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📖 Explanation

The given situation can be shown as belowLet, $v_{1}$ and $v_{2}$ be the incoming and outgoing velocities of electron into the capacitor and out of the capacitor, respectively. Since, electric field is along $X$ -axis, hence electric force on electron along $Y$ -axis, $(F_{y})=0$ $\therefore$ Change in momentum along $Y$ -axis, $\Delta p_{y}=0$ i.e. $\;\; p_{1}=p_{2}$ (along $Y$ -axis) $\Rightarrow \;\; m_{1} v_{1} \cos \alpha=m_{2} v_{2} \cos \beta$ $\Rightarrow \;\; v_{1} / v_{2}=\cos \beta / \cos \alpha$ $\because$ Kinetic energy $(K)=1 / 2 {mv}^{2}$ If mass is same, $K \propto v^{2}$ $\therefore \;\; \;\frac{K_{1}}{K_{2}}=(\;\frac{v_{1}}{v_{2}})^{2}=(\;\frac{\cos \beta}{\cos \alpha})^{2}=\;\frac{\cos ^{2} \beta}{\cos ^{2} \alpha}$

Q11JEE Main 2021MCQ4MMagnetism and Matter

In a ferromagnetic material, below the Curie temperature, a domain is defined as

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📖 Explanation

Since, in ferromagnetic material, with increase in temperature susceptibility decreases, $\therefore$ Ferromagnetic material below Curie temperature will show saturation magnetisation. Hence, option (b) is the correct i.e. domain is defined as a macroscopic region with saturation magnetisation.

Q12JEE Main 2021MCQ4MAlternating Current

An L-C-R circuit contains resistance of $110 Ω$ and a supply of $220 {V}$ at $300 {rad} / {s}$ angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by $45^{\circ}$ . If on the other hand, only inductor is removed the current leads by $45^{\circ}$ with the applied voltage. The rms current flowing in the circuit will be

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📖 Explanation

Given, resistance, $R=110 Ω$ Supply voltage $( {V})=220 {V}$ and angular frequency $(\omega)=300 {rads}^{-1}$ $\because$ Current lag and lead by same angle. $\therefore$ Circuit is in resonance i.e. $X_{L}=X_{C}$ As, $\;\; Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}$ $=\sqrt{(110)^{2}}+(X_{L}-X_{L})^{2}=110$ $\;\;[ \because X_{L}=X_{C}]$ and $I_{ {ms}}=\;{V_{ {ms}}}/{Z}=\;\frac{220}{110}=2 {A}$

Q13JEE Main 2021MCQ4MDual Nature of Matter and Radiation

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength $491 {nm}$ is $0.710 {V}$ . When the incident wavelength is changed to a new value, the stopping potential is $1.43 {V}$ . The new wavelength is

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📖 Explanation

Given, stopping potential $(V_{1})=0.710 {V}$ Incident wavelength of electrons $(\lambda_{1})=491 {nm}$ $=491 \times 10^{-9} {m}$ Let $\lambda_{2}$ will be the new incident wavelength of electrons. Stopping potential $(V_{2})=1.43 {V}$ As, energy $(E)=\;\frac{1240}{\lambda_{1}}=\varphi _{0}+ {eV}$ where, $\varphi _{0}$ is work-function and $V$ is applied potential $\therefore \;\; E_{1}=\;\frac{1240}{491}=\varphi _{0}+0.71$ - $1.43$ . . . (i) and $E_{2}=\;\frac{1240}{\lambda_{2}}=\varphi _{0}+1.43$ . . . (ii) Now, subtracting Eqs. (i) from (ii), we get $\;E_{2}-E_{1} \;\; =1240(\;\frac{1}{\lambda_{2}}-\;\frac{1}{491})=0.72$ $\Rightarrow \;\;\frac{1}{\lambda_{2}} \;\; =0.00058+0.00204=0.00262$ $\therefore \;\lambda_{2} \;\; =381.7 {nm}$ $\;\lambda_{2} \;\; =382 {nm}$

Q14JEE Main 2021MCQ4MWave Optics

Consider the diffraction pattern obtained from the sunlight incident on a pinhole of diameter $0.1 \mu {m}$ . If the diameter of the pinhole is slightly increased, it will affect the diffraction pattern such that

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📖 Explanation

Given, diameter of pinhole, $a=0.1 \mu {m}=0.1 \times 10^{-6} {m}$ $\because$ Path difference $(\Delta x)=a \sin \varphi =n \lambda$ . . . (i) where, $\varphi$ is the phase difference and $\lambda$ be the wavelength. As, $\;\; I=4 I_{0} \cos ^{2} \varphi$ and $\;\; \sin \varphi =\;\frac{n \lambda}{a}$ [from Eq. (i)] If $a$ increases $↔ \sin \varphi$ or $\varphi$ decreases As $\varphi$ decreases $↔ \cos \varphi$ increases $\therefore$ Intensity increases. Hence, on decreasing diameter of pinhloe, the size of diffraction pattern decreases and intensity increases.

Q15JEE Main 2021MCQ4MDual Nature of Matter and Radiation

An electron of mass $m_{e}$ and a proton of mass $m_{p}=1836 m_{e}$ are moving with the same speed. The ratio of their de-Broglie wavelength $\;{\lambda_{\;\text{electron }\;}}/{\lambda_{\;\text{proton }\;}}$ will be

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📖 Explanation

Given, mass of proton $(m_{p})$ is 1836 times the mass of electron $(m_{e})$ and velocity of proton $(v_{p})$ is equal to velocity of electron $(v_{e})$ . As, wavelength $(\lambda)=\;\frac{h}{p}=\;\frac{h}{m v}$ where, $h$ is Planck's constant and $p$ is momentum. $\therefore \;\; \lambda \propto \;\frac{1}{m}$ $\Rightarrow \;\; \;\frac{\lambda_{e}}{\lambda_{\rho }}=\;\frac{m_{p}}{m_{e}}=\;\frac{1836 m_{e}}{m_{e}}=1836$

Q16JEE Main 2021MCQ4MAtoms and Nuclei

The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from $n=2$ to $n=1$ state is

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📖 Explanation

Given, electron is moving from $n=2$ to $n=1$ . From Bohr's hydrogen spectrum (Rydberg formula) $\;\frac{1}{\lambda}=R[\;\frac{1}{n_{f}^{2}}-\;\frac{1}{n_{i}^{2}}]$ where, $\lambda=$ wavelength, $R=1.097 \times 10^{7} {m}^{-1}\;\;\text{ (Rydberg's constant) }\;$ $\Rightarrow \;\; \;\frac{1}{\lambda} \;\; =1.097 \times 10^{7}[\;\frac{1}{1^{2}}-\;\frac{1}{2^{2}}]$ $\;\; =1.097 \times 10^{7}[\;\frac{1}{1}-\;\frac{1}{4}]=\;\frac{3}{4} \times 1.097 \times 10^{7}$ $\lambda \;\; =\;\frac{4}{3 \times 1.097 \times 10^{7}}=1.215 \times 10^{-7}$ $\;\; =121.5 \times 10^{-9} ∼eq 121.8 {nm}$

Q17JEE Main 2021MCQ4MElectromagnetic Waves

If a message signal of frequency $f_{m}$ is amplitude modulated with a carrier signal of frequency $f_{c}$ and radiated through an antenna, the wavelength of the corresponding signal in air is

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📖 Explanation

Let $\lambda$ be the wavelength of carrier signal. Since, frequency $(f_{c})=\;\frac{\;\text{ speed }\;(c)}{\;\text{ wavelength }\;(\lambda)}$ $\therefore \;\; \lambda=\;\frac{c}{f_{c}}$

Q18JEE Main 2021MCQ4MSemiconductor Electronics

For extrinsic semiconductors when doping level is increased,

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📖 Explanation

In case of $n$ -type semiconductor, the energy level diagram will beIn case of $n$ -type semiconductor $n > p$ , so the fermi level will go upward. Similarly, in case of $p$ -type semiconductor $p > n$ , so the fermi level will go downward.

Q19JEE Main 2021MCQ4MSemiconductor Electronics

List-I List-II A. Rectifier 1. Used either forstepping up or stepping down the AC voltage B. Stabiliser 2. Used to convert ACvoltage into $D C$ voltage c. Transformer 3. Used to remove anyripple in the rectifiedoutput voltage D. Filter 4. Used for constantoutput voltage evenwhen the input voltageor load current changeChoose the correct answer from the options given below.

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📖 Explanation

Rectifier devices are used to convert $A C$ to $D C$ . Stabiliser is an electronic device which gives constant output even if input changes. Transformer is used to step up or down the AC voltage. Filter device which removes ripple in rectified output. So, the correct match is ${A} \to 2, {B} \to 4, {C} \to 1$ and ${D} \to 3$

Q20JEE Main 2021MCQ4MSemiconductor Electronics

The truth table for the followng logic circuit is

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📖 Explanation

Here $A$ and $B$ be the input and $Y$ be the output. $\because Y \;\; =\overline{A \overline{B}}+\overline{A B}=\overline{A \overline{B}} \cdot \overline{\overline{A} B}=(\overline{A}+\overline{\overline{B}}) \cdot (\overline{\overline{A}}+\overline{B})$ $\;\; =(\overline{A}+B)(A+\overline{B})=\overline{A} A+\overline{A B}+A B+B \overline{B}$ $\;\; =0+\overline{A B}+A B+0=A B+\overline{A B}$ According to the truth table $A$ $B$ $\overline{A}$ $\overline{B}$ $AB$ $\overline{AB}$ $Y=A B+\overline{A} \overline{B}$ 0 0 1 1 0 1 1 0 1 1 0 0 0 0 1 0 0 1 0 0 0 1 1 0 0 1 0 1

Q21JEE Main 2021NAT4MLaws of Motion

Two particles having masses $4 {g}$ and $16 {g}$ respectively are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is $n: 2$ . The value of $n$ will be ............ .

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📖 Explanation

Given, mass of particle $A, m_{A}=4 {g}$ Mass of particle $B, m_{B}=16 {g}$ Kinetic energy of $A$ and $B$ is same i.e. $\;\; {KE}_{A}= {KE}_{B}$ As, kinetic energy $( {KE})=p^{2} / 2 m$ where, $p$ is momentum and $m$ is mass. $\therefore \;\; \;\frac{(p_{A})^{2}}{m_{A}}=\;\frac{(p_{B}^{2})}{m_{B}} \Rightarrow \;\frac{p_{A}^{2}}{4}=\;\frac{p_{B}^{2}}{16} \Rightarrow \;\frac{p_{A}}{p_{B}}=\;\frac{1}{2}$ $\because$ linear momentum is $n: 2$ . $\therefore \;\; n=1$

Q22JEE Main 2021NAT4MGravitation

The initial velocity $v_{i}$ required to project a body vertically upward from the surface of the Earth to reach a height of $10 R$ , where $R$ is the radius of the Earth, may be described in terms of escape velocity $v_{e}$ such that $v_{i}=\sqrt{\;\frac{x}{y}} \times v_{e} .$ The value of $x$ will be ........... .

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📖 Explanation

By using law of conservation of energy, Energy on the surface of earth $(E_{\;\text{surface }\;})=$ Energy at height $(h=10 R)$ $\Rightarrow \;\frac{-G M m}{R}+\;\frac{1}{2} m v_{i}^{2}=\;\frac{-G M m}{R+10 R}+0=\;\frac{-G M m}{11 R}$ $\Rightarrow \;\; 1 / 2 m v_{i}^{2}=\;\frac{11}{11} \;\frac{G M m}{R}-\;\frac{G M m}{11 R}$ $\Rightarrow \;\; \;\frac{1}{2} m v_{i}^{2}=\;\frac{10 G M m}{11 R} \Rightarrow v_{i}^{2}=\;\frac{20 G M}{11 R}$ $\;\;\; v_{i}=\sqrt{\;\frac{10}{11}} v_{e} \;\; ( \because v_{e}=\sqrt{\;\frac{2 G M}{R}}=\;\text{ escape velocity) }\;$ $\Rightarrow \;\; x=10$

Q23JEE Main 2021NAT4MWaves

The percentage increase in the speed of transverse waves produced in a stretched string, if the tension is increased by $4 \%$ , will be ............ %.

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📖 Explanation

Error in tension is $4 \%$ . Speed of transverse wave, $v=\sqrt{\;\frac{\;\text{ Tension }\;(T)}{\;\text{ Mass per unit length }\;(\mu )}}$ On squaring both side $\Rightarrow \;\; v^{2}=\;\frac{T}{\mu }$ By using the concept of % error calculation $\Rightarrow \;\; \;\frac{2 \Delta v}{v}=\;\frac{\Delta T}{T} \Rightarrow \;\frac{\Delta v}{v}=\;\frac{1}{2} \;\frac{\Delta T}{T}$ $\therefore \;\; \;\frac{\Delta v}{v} \times 100=\;\frac{1}{2} \;\frac{\Delta T}{T} \times 100=\;\frac{4}{2}=2 \%$

Q24JEE Main 2021NAT4MLaws of Motion

If $P \times Q=Q \times P$ , the angle between $P$ and $Q$ is $\theta(0^{\circ} < \theta < 360^{\circ})$ . The value of $\theta$ will be ........... $^{\circ}$ .

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📖 Explanation

Given, $P \times Q=Q \times P$ $\Rightarrow \;\; P \times Q=-P \times Q \Rightarrow 2(P \times Q)=0$ or $P \times Q=0$ It is possible only, if $P=0$ or $Q=0$ , angle between them is $180^{\circ}$ . $\therefore \;\; \theta=180^{\circ}$

Q25JEE Main 2021NAT4MThermodynamics

A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by $52 {K}$ , its efficiency is doubled. The temperature in kelvin of the source will be ............ .

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📖 Explanation

Given, initial efficiency $(η_{h})=1 / 4$ $\Rightarrow \;\; η_{ {h}}=1-\;\frac{T_{2}}{T_{1}}=\;\frac{1}{4} \Rightarrow \;\frac{T_{2}}{T_{1}}=1-\;\frac{1}{4}=\;\frac{3}{4}$ When temperature of sink is reduced by $50 {K}$ , $\;η_{2} \;\; =2 η_{1}=1-\;\frac{T_{2}-52}{T_{1}}$ $\Rightarrow \;2 \times \;\frac{1}{4} \;\; =1-\;\frac{T_{2}}{T_{1}}+\;\frac{52}{T_{1}}$ $\Rightarrow \;\;\frac{1}{2} \;\; =1-\;\frac{3}{4}+\;\frac{52}{T_{1}} \Rightarrow \;\frac{1}{2}-\;\frac{1}{4}=\;\frac{52}{T_{1}}$ $\Rightarrow \;\;\frac{2}{4} \;\; =\;\frac{52}{T_{1}}$ $\Rightarrow \;T_{1} \;\; =104 {K}$

Q26JEE Main 2021NAT4MElectrostatics

Two small spheres each of mass $10 {mg}$ are suspended from a point by threads $0.5 {m}$ long. They are equally charged and repel each other to a distance of $0.20 {m}$ . The charge on each of the sphere is $\;\frac{a}{21} \times 10^{-8} {C}$ . The value of $a$ will be ............. . $[Given, $g=10 {ms}^{-2}$ ]$

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📖 Explanation

Given, mass of each spheres, $m=10 {mg}=10 \times 10^{-3} {g}$ Length of thread $(l)=0.5 {m}$ Separation between charges, $d=0.2 {m}$ Charge of each sphere, $q=\;\frac{a}{21} \times 10^{-8} C$ Acceleration due to gravity $(g)=10 {ms}^{-2}$ The situation can be shown as below,Taking component of tension (T) $T \cos \theta \;\; =m g$ . . . (i) $T \sin \theta \;\; =q E=\;\frac{k \mid q^{2}}{d^{2}}$ . . . (ii) $\sin \theta=\;\frac{d / 2}{l}=\;\frac{0.1}{0.5}=\;\frac{1}{5}$ $\cos \theta=\sqrt{1-\sin ^{2} \theta}=\sqrt{1-1 / 25}=\;\frac{\sqrt{24}}{5}$ $\therefore \;\; \tan \theta=\;\frac{\sin \theta}{\cos \theta}=\;\frac{k q^{2}}{d^{2} m g} \;\;$ [using Eqs. (i) and (ii)] $\Rightarrow \;\; \;\frac{1 / 5}{\sqrt{24} / 5}=\;\frac{k q^{2}}{d^{2} m g}$ $\Rightarrow \;\; \;\frac{1}{\sqrt{24}}=\;\frac{9 \times 10^{9} \times q^{2}}{(0.2)^{2} \times 10 \times 10^{-3} \times 10}$ $\Rightarrow \;\; q=\sqrt{\;\frac{(0.2)^{2} \times 10^{-1}}{\sqrt{24} \times 9 \times 10^{9}}} \Rightarrow q=3 \times 10^{-7} {C}$ $\Rightarrow \;\; \;\frac{a}{21} \times 10^{-8}=30 \times 10^{-8}$ $a=630$

Q27JEE Main 2021NAT4MElectrostatics

Two identical conducting spheres with negligible volume have $2.1 {nC}$ and $-0.1 {nC}$ charges, respectively. They are brought into contact and then separated by a distance of $0.5 {m}$ . The electrostatic force acting between the spheres is ____ $\times 10^{-9} {N}$ . $[Given, $4 pi epsilon_{0}=\;\frac{1}{9 × 10^{9}}$ SI unit]$

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📖 Explanation

According to Coulomb's law, the force of attraction or repulsion is directly proportional to the product of distance between them and inversely proportional to the square of the distance between them. ${\text{F}}= {\text{K}} \;\frac{q_1 q_2}{r^2}$ Here we have, $\text{bo} {q}_1$ is the charge of the first mass, $\text{bo} {q}_2$ is the charge of the second mass, $\text{bo} {r}$ is the distance and ${\text{K}}$ is the proportionally constant. Given: Charge of first sphere ${\text{q}}_1=2.1 {\text{nc}}$ and Charge of second sphere $q_2=-0.1 {\text{NC}}$ when they connect with each other the charge becomes, $q=\;\frac{q_2-q_1}{2}$ $\Rightarrow {\text{q}}=\;\frac{2.1-0.1}{2}$ $\Rightarrow {\text{q}}=1 {\text{nC}}$ sigma According to Coulomb's law, we have; ${\text{F}}=\;\frac{1}{4 \pi E_o} \;\frac{q_1 q_2}{r^2}$ Now, on putting the given values we have; $F=9 \times 10^9 \times \;\frac{1 \times 10^{-9} \times 1 \times 10^{-9}}{0.5^2}$ $F=9 \times 10^9 \times \;\frac{10^{-18}}{0.25}$ $\Rightarrow F=36 \times 10^{-9} {\text{N}}$ $\;\text{ Hence, }\; F=36 \times 10^{-9} {\text{N}}$

Q28JEE Main 2021NAT4MElectromagnetic Waves

The peak electric field produced by the radiation coming from the $8 {W}$ bulb at a distance of $10 {m}$ is $\;\frac{x}{10} \sqrt{\;\frac{\mu _{0} c}{\pi}} \;\frac{ {V}}{ {m}}$ . The efficiency of the bulb is $10 \%$ and it is a point source. The value of $x$ is ............ .

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📖 Explanation

Given, radiation power $(P)=8 {W}$ Distance $(d)=10 {m}$ $\because$ Intensity $(I)=\;\frac{1}{2} subset \epsilon_{0} E^{2}=\;\frac{\;\text{ Power }\;(P)}{\;\text{ Area }\;(A)}$ . . . (i) where, $c=\;{1}/{\sqrt{\mu _{0} \epsilon_{0}}}=$ speed of light in vacuum ... (ii) and $E=$ electric field From Eq. (ii), we get $\epsilon_{0}=\;\frac{1}{\mu _{0} c^{2}}$ Put this value of $\epsilon_{0}$ in Eq. (i), we get $I \;\; =\;\frac{1}{2} C \;\frac{1}{\mu _{0} C^{2}} E^{2}=\;\frac{P}{A} \Rightarrow \;\frac{1}{2} \;\frac{1}{\mu _{0} C} E^{2}=\;\frac{P}{A}$ $\Rightarrow \;\; E \;\; =\sqrt{\;\frac{2 P \mu _{0} C}{A}}=\sqrt{\;\frac{2 P \mu _{0} C}{4 \pi d^{2}}}=\sqrt{\;\frac{2 P}{4 d^{2}}} \sqrt{\;\frac{\mu _{0} C}{\pi}}$ $\;\; =\sqrt{\;\frac{2 \times 8}{4 \times 100}}=\sqrt{\;\frac{\mu _{0} C}{\pi}}=\;\frac{2}{10} \sqrt{\;\frac{\mu _{0} C}{\pi}} {V} / {m}$ $\therefore \;\; x \;\; =2$

Q29JEE Main 2021NAT4MCurrent Electricity

A current of $6 {A}$ enters one corner $P$ of an equilateral triangle $P Q R$ having three wires of resistance $2 Ω$ each and leaves by the corner $R$ . The currents $i$ , in ampere is ............ .

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📖 Explanation

Let resistances be $R_{1}, R_{2}, R_{3}$ and $R_{4}$ and $I_{1}$ current is passing through $R_{4}$ as shown in figure $\therefore I_{2}=(6-I_{1})$ is passing through $R_{2}$ As, same current is flowing through $R_{4}$ and $R_{3}$ . $\therefore R_{4}$ and $R_{3}$ are in series.and series equivalent resistance, $R_{\;\text{eq }\;}=R_{4}+R_{3}$ $\therefore \;\; R_{\;\text{eq }\;}=2+2=4 Ω$ Voltage through $R_{ {eq}}$ and $R_{2}$ will be same. $\Rightarrow \;\; I_{1} R_{ {eq}}=I_{2} R_{2} \Rightarrow I_{1} 4=(6-I_{1}) 2$ $\Rightarrow \;\; 2 I_{1}=6-I_{1} \Rightarrow I_{1}=2 {A}$

Q30JEE Main 2021NAT4MDual Nature of Matter and Radiation

The wavelength of an X-ray beam is $10 Å$ . The mass of a fictitious particle having the same energy as that of the X-ray photons is $\;\frac{x}{3} h {kg}$ . The value of $x$ is ............ ( $h$ = Planck's constant)

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📖 Explanation

Given, wavelength of X-rays, $\lambda=10 Å=10 \times 10^{-10} {m}$ Speed of light in free space, $c=3 \times 10^{8} {m} / {s}$ Since, energy $(E)=\;\frac{h c}{\lambda}=m c^{2}$ . . . (i) $\Rightarrow \;\; m=\;\frac{h}{d \lambda}=\;\frac{h}{3 \times 10^{8} \times 10 \times 10^{-10}}=\;\frac{h}{3 \times 10^{-1}}=\;\frac{10 h}{3} {kg}$ $\therefore \;\; x=10$

Chemistry30 questions

Q31JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

Which among the following species has unequal bond lengths ?

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📖 Explanation

Xe $(s p^{3} d^{2}$ -hybridised) Geometry : Octahedral Shape : Square planar It has four equivalent ${Xe}- {F}$ bonds in the same plane.Si ( $s p^{3}$ -hybridised) Geometry and shape : Tetrahedral It has four equivalent ${Si}- {F}$ bonds. ${S}( {sp}^{3} d$ -hybridised $)$ Geometry: Trigonal bipyramidal Shape : see-saw Axial $S-F$ bonds are longer than equatorial bonds.B (sp $^{3}$ -hybridised) Geometry and shape : Tetrahedral It has four equivalent ${B}- {F}$ bonds. Hence, among given species only ${SF}_{4}$ has unequal bond lengt

Q32JEE Main 2021MCQ4MIonic Equilibrium

The solubility of ${Ca}( {OH})_{2}$ in water is $[Given: The solubility product of ${Ca}( {OH})_{2}$ \in water $=5.5 × 10^{-6}$ ]$

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📖 Explanation

Let, solubility of ${Ca}( {OH})_{2}$ in pure water $=S {mol} / {L}$ $\; {Ca}( {OH})_{2} ⇌ {Ca}^{2+}+2 {OH}^{-}$ $\;S {mol} / {L} \;\; 2 \times S( {mol} / {L})$ $= \;\; {[ {Ca}^{2+}]\$ $[ {OH}^{-}]^{2}=S \times (2 S)^{2}=4 S^{3}( {mol} / {L}) } $The expression of$ K_{s p} $can also be written as,$ K_{s p} ;; =x^{x} \cdot y^{y} \cdot S^{x+y};; =1^{1} \cdot 2^{2} \cdot S^{1+2};; =4 S^{3}K_{s p} ;; =x^{x} \cdot y^{y} \cdot S^{x+y};; =1^{1} \cdot 2^{2} \cdot S^{1+2};; =4 S^{3} ;; [ \because ;\text{ For }; {Ca}( {OH}){2}: x=1, y=2]x $and$ y $are the coefficients of cations and anions respectively$ S ;; =(;\frac{K{s p}}{4})^{1 / 3}=(;\frac{5.5 \times 10^{-6}}{4})^{1 / 3};; =1.11 \times 10^{-2} {mol} / {L}$

Q33JEE Main 2021MCQ4MSolutions

Which one of the following statements is false for hydrophilic sols ?

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📖 Explanation

Statement (c) is false whereas other statements are true. Corrected statement is as follows: Viscosity of hydrophilic sols is higher than that of the dispersion medium, i.e. ${H}_{2} {O}$ , because there is a high concentration of dispersed phase in water.

Q34JEE Main 2021MCQ4Mp Block Elements

The correct order of bond dissociation enthalpy of halogens is

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📖 Explanation

Among halogens $( {F}_{2}, {Cl}_{2}, {Br}_{2}$ and ${I}_{2})$ , bond dissociation enthalpy ( $\Delta _{\;\text{diss }\;} H^{\circ}$ ) of ${I}_{2}$ , is minimum because of larger size of l-atom there is a steric repulsion between bonded l-atoms, which makes I- I bond weakest. Whereas, smaller size and highest electronegativity of F-atom cause highest electron density on ${F}$ -atom of ${F}_{2}$ molecule. As a result, ${F}- {F}$ bond becomes weaker due to electrostatic repulsion between bonded F-atoms. Thus, the order of $\Delta _{\;\text{diss }\;} H^{\circ}$ (in ${kJ} {mol}^{-1}$ ) is ${{Cl}- {Cl}}_{242.6} > {{Br}- {Br}}_{192.3} > {{F}- {F} }_{(158.8)\text{Electrostatic}\text{repulsion}} > { {I}- {I}}_{151.1 (\text{Steric}\text{repulsion})}$

Q35JEE Main 2021MCQ4Mp Block Elements

The method used for the purification of indium is

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Q36JEE Main 2021MCQ4MHydrocarbons

Water does not produce ${CO}$ on reacting with

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📖 Explanation

Water (steam) can produce ${CO}$ on reacting with ${CH}_{4}, {C}$ and ${C}_{3} {H}_{8}$ as, ${CH}_{4}+ {H}_{2} {O} {\to }_{\text{Ni}}^{1270 {K}} {CO}+3 {H}_{2}$ ${C}+ {H}_{2} {O} {\to }^{1270 {K}} {CO}+ {H}_{2}$ ${C}_{3} {H}_{8}+3 {H}_{2} {O} {\to }_{\text{Ni}}^{1270 {K}} 3 {CO}+7 {H}_{2}$ But, water on reaction with ${CO}_{2}$ produces carbonic acid $( {H}_{2} {CO}_{3})$ , not CO. ${CO}_{2}+ {H}_{2} {O} \longrightarrow {H}_{2} {CO}_{3}$

Q37JEE Main 2021MCQ4Mp Block Elements

Given below are two statements. Statement I $\alpha$ and $\beta$ -forms of sulphur can change reversibly between themselves with slow heating or slow cooling. Statement II At room temperature, the stable crystalline form of sulphur is monoclinic sulphur. In the light of the above statements, choose the correct answer from the options given below.

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📖 Explanation

Two crystalline allotropic forms of sulphur $( {S}_{\alpha}$ and ${S}_{\beta})$ can change reversibly between themselves with slow heating (above 369 K) or cooling. ${{S}_{\alpha}}_{(\text{Rhombic})} {⇌}_{\text{Cooling}}^{ > 369 {K}}{ {S}_{\beta}}_{(\text{Monoclinic})}$ So, statement 1 is true. At room temperature or at standard conditions of pressure and temperature, rhombic sulphur is the thermodynamically most stable crystalline allotrope of sulphur $(\Delta _{f} H^{\circ}=0)$ . So, statement II is false.

Q38JEE Main 2021MCQ4Md and f Block Elements

The major components of German silver are

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📖 Explanation

The major components of German silver are ${Cu}: 25-30 \%, {Zn}: 25-30 \%$ and ${Ni}: 40-30 \%$ .

Q39JEE Main 2021MCQ4MCoordination Compounds

In which of the following order the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment? (i) $[ {FeF}_{6}]^{3-} \;\;$ (ii) $[ {Co}( {NH}_{3})_{6}]^{3+}$ (iii) $[ {NiCl}_{4}]^{2-} \;\;$ (iv) $[ {Cu}( {NH}_{3})_{4}]^{2+}$

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📖 Explanation

Spin only magnetic moment, $\mu =\sqrt{n(n+2)} {BM}$ where, $n=$ number of unpaired electrons and $\mu \propto n$ . (i) $[ {FeF}_{6}]^{3-} \Rightarrow {Fe}^{3+}=(3 d^{5})+6 \times {F}^{-}$ (weak field ligand). Thus, pairing of electron does not take place.(ii) $[ {Co}( {NH}_{3})_{6}]^{3+} \Rightarrow {Co}^{3+}=(3 d^{6})+6 \times {NH}_{3}$ (strong field ligand). Thus, pairing of electron takes place.(iii) $[ {NiCl}_{4}]^{2-} \Rightarrow {Ni}^{2+}=(3 d^{8})+4 \times {Cl}^{-}$ (weak field ligand). Thus, pairing of electron takes place.(iv) $[ {Cu}( {NH}_{3})_{4}]^{2+} \Rightarrow {Cu}^{2+}(3 d^{9})+4 \times {NH}_{3}$ (strong field ligand). Thus, pairing of electron takes place.So, the decreasing order of $\mu$ is $\;{\mu _{ {i}}_{(n=5)}} > \;{\mu _{ {iii}}_{(n=2)}} > \;{\mu _{ {iv}}_{(n=1)} > } \;{\mu _{ {ii}}_{(n=0)}}$

Q40JEE Main 2021MCQ4MIonic Equilibrium

Given below are two statements. Statement I The ${pH}$ of rain water is normally $∼ 5.6$ . Statement II If the ${pH}$ of rain water drops below $5.6$ , it is called acid rain. In the light of the above statements, choose the correct answer from the options given below.

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📖 Explanation

The ${pH}$ of rain water is normally $5.6$ . If it ${pH}$ drop below $5.6$ , it is called acid rain. $\therefore$ Both statements are correct.

Q41JEE Main 2021MCQ4MPractical Organic Chemistry

Which of the following compound is added to the sodium extract before addition of silver nitrate for testing of halogens?

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📖 Explanation

Nitric acid is added to sodium extract before addition of silver nitrate for testing halogens. Because it decomposes ${NaCN}$ and ${Na}_{2} {S}$ or else they interfere in the test. The reaction are as follows : ${NaCN}+ {HNO}_{3}\;\longrightarrow {NaNO}_{3}+ {HCN} ↑$ ${ {Na}_{2} {S}}_{\text{Sodium} \text{extract}}+ {{HNO}_{3}_{\text{Nitric} \text{acid}}}\;\longrightarrow {NaNO}_{3}+ {H}_{2} {S} ↑$

Q42JEE Main 2021MCQ4MHydrocarbons

The major product of the following reaction is

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Q43JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

The major product of the following reaction is ${CH}_{3} {CH}_{2} {CH}= {CH}_{2} {\to }_{\text{Rh} \text{catalyst}}^{ {H}_{2} / {CO}}$

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📖 Explanation

The major product of the reaction is ${CH}_{3} {CH}_{2} {CH}_{2} {CH}_{2} {CHO}$ . Here, electrophilic addition of ${H}-{ {C}^{+}}= {O}$ (formylation) take place to the alkene through Markownikoff addition.

Q44JEE Main 2021MCQ4MOrganic Chemistry - Some Basic Principles

The correct sequence of reagents used in the preparation of 4-bromo-2-nitroethyl benzene from benzene is

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📖 Explanation

Benzene on reaction with ${CH}_{3} {COCl} / {AlCl}_{3}$ gives acetophenone which on reduction with ${Zn}- {Hg} / {HCl}$ gives ethyl benzene. Bromination of ethyl benzene give 4-bromo-ethylbenzene which upon nitration gives 4-bromo-2-nitroethyl benzene.$[-l-effect of Br directs ${NO}_{2}^{-}$ at its meta-position]$ So, option (b) is the correct sequence of reagents used.

Q45JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

The correct order of acid character of the following compounds is Options

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📖 Explanation

Acidity of phenol (I) is weaker than any carboxylic acid. Electron withdrawing nature $(-R,-I)$ , of $- {NO}_{2}$ group at para position increases acidic strength of (II), whereas $+$ l effect of $- {CH}_{3}$ group at para position decreases acidic strength of (IV). So, the order of acid character is II > III > IV > I.

Q46JEE Main 2021MCQ4MAlcohols, Phenols and Ethers

What is ' $X$ ' in the given reaction?

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📖 Explanation

When ethylene glycol is heated with oxalic acid at $210^{\circ} {C}$ , first we get an unstable cyclic-diester (ethylene oxalate) which readily decarboxylate to give ethylene as the major product.

Q47JEE Main 2021MCQ4MAmines

Correct statement about the given chemical reaction is

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📖 Explanation

During nitration, in strongly acidic medium, aniline gets protonated to form the anilinium ion i.e, $( {C}_{6} {H}_{5}- {N} {H}_{3})$ which is meta-directing as $- {N} {H}_{3}$ is an electron withdrawing ( $-l$ ) group. As a result, we get meta-nitro aniline $(B)$ as the major product.

Q48JEE Main 2021MCQ4MAmines

Carbylamine test is used to detect the presence of primary amino group in an organic compounD. Which of the following compound is formed when this test is performed with aniline?

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📖 Explanation

Aniline on carbylamine reaction produces a foul smelling gas, phenyl isocyanide $( {C}_{6} {H}_{5} {NC})$ .Carbylamine test is used to detect aliphatic and aromatic primary amines.

Q49JEE Main 2021MCQ4MBiomolecules

Which of the following is correct structure of $\alpha$ -anomer of maltose?

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📖 Explanation

Maltose is a disaccharide which is made of two $\alpha$ -D-glucose units in which ${C}_{1}$ (anomeric carbon) of one glucose (I) is linked to ${C}_{4}$ of another glucose unit (II).

Q50JEE Main 2021MCQ4MCoordination Compounds

Given below are two statements. Statement I The identification of ${Ni}^{2+}$ is carried out by dimethyl glyoxime in the presence of ${NH}_{4} {OH}$ . Statement II The dimethyl glyoxime is a bidentate neutral ligand. In the light of the above statements, choose the correct answer from the options given below.

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📖 Explanation

Both statements are true. Dimethyl glyoxime (DMG) is a neutral bidentate ligand $( {DMG}^{-})$ . ${Ni}^{2+}$ ion is identified with DMG in presence of ${NH}_{4} {OH}$ to give a red ppt. Its reaction is as follows.

Q51JEE Main 2021NAT4MSolutions

Consider titration of ${NaOH}$ solution versus $1.25 {M}$ oxalic acid solution. At the end point following burette readings were obtained. (i) $4.5 {mL}$ (ii) $4.5 {mL}$ (iii) $4.4 {mL}$ (iv) $4.4 {mL}$ (v) $4.4 {mL}$ If the volume of oxalic acid taken was $10.0 {mL}$ , then the molarity of the ${NaOH}$ solution is .......... M. (Rounded off to the nearest integer)

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📖 Explanation

Average burette reading $=$ Volume of ${NaOH}$ solution $(V_{1})$ $\;\; =\;\frac{4.5+4.5+4.4+4.4+4.4}{5}$ $\;\; =4.44 {mL}$ Strength of ${NaOH}$ solution $=S_{1}( {M})($ say $)=S_{1}( {N})$ Volume of oxalic acid solution $(V_{2})=10 {mL}$ Strength of oxalic acid solution $(S_{2})=1.25 {M}$ $=1.25 \times 2 {N}$ So, $V_{1} S_{1}=V_{2} S_{2}$ ( $\because$ Law of equivalence) $\Rightarrow \;\; S_{1} \;\; =\;\frac{V_{2} S_{2}}{V_{1}}=\;\frac{10 \times (1.25 \times 2)}{4.44}=5.63 {N}$ $\;∼eq 6 {M}=6 {M}$ Note $n$ -factor of ${NaOH}=1$ $n$ -factor of ${H}_{2} {C}_{2} {O}_{4}=2$ (oxalic acid) $N=M \times n$

Q52JEE Main 2021NAT4MSolid State

The unit cell of copper corresponds to a face centered cube of edge length $3.596 Å$ with one copper atom at each lattice point. The calculated density of copper in ${kg} / {m}^{3}$ is ......... [Molar mass of ${Cu}=63.54 {g}$ ;

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📖 Explanation

Density of copper, $d=\;\frac{Z \times M}{a^{3} \times N_{A}}$ Given, $Z \;\; =4, \;\text{ for fcc lattice, }\;$ $M \;\; =63.54 {g} {mol}^{-1}$ $\;\; =63.54 \times 10^{-3} {kg} {mol}^{-1}$ $a \;\; =3.596 Å=3.596 \times 10^{-10} {m},$ $N_{A} \;\; =6.022 \times 10^{23} {mol}^{-1}$ $M \;\; =63.54 {g} {mol}^{-1}$ $\;\; =63.54 \times 10^{-3} {kg} {mol}^{-1},$ $a \;\; =3.596 Å=3.596 \times 10^{-10} {m},$ $N_{A} \;\; =6.022 \times 10^{23} {mol}^{-1}$ On putting given values, we get $\Rightarrow \;\; d \;\; =\;\frac{4 \times (63.54 \times 10^{-3})}{(3.596 \times 10^{-10})^{3} \times (6.022 \times 10^{23})} {kg} / {m}^{3}$ $\;\; =9076.26 ∼eq 9077 {kg} / {m}^{3}$

Q53JEE Main 2021NAT4MStructure of Atom

Electromagnetic radiation of wavelength $663 {nm}$ is just sufficient to ionise the atom of metal $A$ . The ionisation energy of metal $A$ in ${kJ} {mol}^{-1}$ is ......... .(Rounded off to the nearest integer) $\;{[h=6.63 \times 10^{-34} {J}- {s}, {c}=3.00 \times 10^{8} {ms}^{-1},}$ $\;N_{ {A}}=6.02 \times 10^{23} {mol}^{-1}]$

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📖 Explanation

Energy of EMR = IE of the metal (A) $\;\; =h {v}=\;\frac{h c}{\lambda} {atom}^{-1}=-\;\frac{h c}{\lambda} \times {N}_{A} {mol}^{-1}$ $\;\; =\;\frac{(6.63 \times 10^{-34}) \times (3 \times 10^{8}) \times (6.02 \times 10^{23})}{(663 \times 10^{-9})} {J} {mol}^{-1}$ $\;{[ \because \lambda=663 {nm}=663 \times 10^{-9} {m}]\$ };; =180600 {J} {mol}^{-1}=180.6 {kJ} {mol}^{-1} ∼eq 180 {kJ} {mol}^{-1}$

Q54JEE Main 2021NAT4MChemical Thermodynamics

Five moles of an ideal gas at $293 {K}$ is expanded isothermally from an initial pressure of $2.1 {MPa}$ to $1.3 {MPa}$ against at constant external pressure $4.3 {MPa}$ . The heat transferred in this process is ......... kJ ${mol}^{-1}$ . (Rounded off to the nearest integer) $[R=8.314 {J} {mol}^{-1} {K}^{-1}]$

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📖 Explanation

The gas performs isothermal irreversible work ( $W$ ). where, $\Delta U=0$ (change in internal energy) From, 1st law of thermodynamics, $\Rightarrow \;\; \Delta U\;=\Delta Q+W$ $\Rightarrow \;\; 0\;=\Delta Q+W$ $\Rightarrow \;\; \Delta Q\;=-W$ $\;\text{ Now, }\; \;\; W\;=-p_{\;\text{ext }\;}(V_{2}-V_{1})$ $\;=-p_{\;\text{ext }\;}(\;\frac{n R T}{p_{2}}-\;\frac{n R T}{p_{1}})=-p_{\;\text{ext }\;} \times n R T(\;\frac{1}{p_{2}}-\;\frac{1}{p_{1}})$ Given, $p_{\;\text{ext }\;}=4.3 {MPa}, p_{1}=2.1 {MPa}, p_{2}=1.3 {MPa}$ , $n=5 {mol}, T \;\; =293 {K} \;\text{ and }\; R=8.314 {J} {mol}^{-1} {K}^{-1}$ $\;\; =-4.3 \times 5 \times 8.314 \times 293(\;\frac{1}{1.3}-\;\frac{1}{2.1})$ $\;\; =-15347.70 {J} {mol}^{-1}$ $\;\; =-15.347 {kJ} {mol}^{-1} ∼eq-15 {kJ} {mol}^{-1}$ $\Rightarrow \;\; \Delta Q \;\; =15 {kJ} {mol}^{-1}$

Q55JEE Main 2021NAT4MSolutions

If a compound $A B$ dissociates to the extent of $75 \%$ in an aqueous solution, the molality of the solution which shows a $2.5 {K}$ rise in the boiling point of the solution is .......... molal. (Rounded off to the nearest integer) $[K_{b}=0.52 {K} {kg} {mol}^{-1}]$

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📖 Explanation

As $A B$ is a binary electrolyte, $\therefore \;\; A B ⇌ A^{+}+B^{-}, n=2$ $\;\; i=1+\alpha(n-1)=1+\;\frac{75}{100}(2-1)=1.75$ $\;\text{ Given, }\; \;\;$ $\Delta T_{b} \;\; =2.5 {K}$ $K_{b} \;\; =0.52 {Kkg} {mol}^{-1}$ $\Delta T_{b} \;\; =K_{b} \times m \times i$ $m \;\; =\;\frac{\Delta T_{b}}{K_{b} \times i}=\;\frac{2.5}{0.52 \times 1.75}$ $\therefore \;\; \;\; =2.74 ∼ 3 {mol} / {kg}$

Q56JEE Main 2021NAT4MElectrochemistry

Copper reduces ${NO}_{3}^{-}$ into ${NO}$ and ${NO}_{2}$ depending upon the concentration of ${HNO}_{3}$ in solution. (Assuming fixed $[ {Cu}^{2+}]$ and $p_{ {NO}}=p_{ {NO}_{2}}$ ), the ${HNO}_{3}$ concentration at which the thermodynamic tendency for reduction of ${NO}_{3}^{-}$ into ${NO}$ and ${NO}_{2}$ by copper is same is $10^{x} {M}$ . The value of $2 x$ is ............ .(Rounded off to the nearest integer) $[$ Given, $E_{ {Cu}^{2+} / {Cu}}^{\circ}=0.34 {V}, E_{ {NO}_{3}^{-} / {NO}}^{\circ}=0.96 {V}$ $E_{ {NO}_{3}^{\circ} / {NO}_{2}}^{\circ} \;\; =0.79 {V} \;\text{ and at }\; 298 {K}, \;\frac{R T}{F}$ $(2.303) \;\; =0.059]$

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📖 Explanation

Cell-l $( {HNO}_{3} \longrightarrow {NO})$ $3 {Cu}+2 {NO}_{3}^{-}+8 {H}^{+} \longrightarrow 3 {Cu}^{2+}+2 {NO}+4 {H}_{2} {O}$ $Q_{1}=\;{[ {Cu}^{2+}]^{3} \times (p_{ {NO}})^{2}}/{[ {NO}_{3}^{-}]^{2} \times \$ [ {H}^{+}]^{8}}\because ;; E_{1}^{\circ}=0.96-(-0.34)=1.3 {V}E_{1}=1.3-;\frac{0.059}{6} \log Q_{1} $Cell-ll$ ( {HNO}{3} → {NO}{2}){Cu}+2 {NO}{3}^{-}+4 {H}^{+} ⟶ {Cu}^{2+}+2 {NO}{2}+2 {H}{2} {O}Q{2}=;{[ {Cu}^{2+}]$ \times (p_{ {NO}{2}})^{2}}/{[ {NO}{3}^{-}]^{2} \times $[ {H}^{+}]^{4}}E_{2}^{\circ}=0.79-(-0.34) {V}=1.13 {V}E_{2}=1.13-;\frac{0.059}{2} \log Q_{2}\because ;; E_{2} $Now,$ E_{1}=E_{2}1.3-;\frac{0.059}{6} \log Q_{1}=1.13-;\frac{0.059}{2} \log Q_{2}0.17=;\frac{0.059}{6}[\log Q_{1}-3 \log Q_{2}]=;\frac{0.059}{6} \log ;\frac{Q_{1}}{Q_{2}}=;\frac{0.059}{6} \log ;{[ {Cu}^{2+}]^{3} × (p_{ {NO}})^{2}}/{[ {NO}{3}^{-}]^{2} × $[ {H}^{+}]^{8}} \times ;{[ {NO}{3}^{-}]^{6} \times $[ {H}^{+}]^{12}}/{[ {Cu}^{2+}]^{3} × (p_{ {NO}{2}})^{6}}=;\frac{0.059}{6} \log ;{[ {H}^{+}]^{4} × $[ {NO}{3}^{-}]^{4}}/{(p_{ {NO}{2}})^{4}} ;; [ \because p{ {NO}}=p_{ {NO}{2}}]=;\frac{0.059}{6} \log ;{[ {HNO}{3}]^{4}}/{(p_{ {NO}{2}})^{4}} $Now,$ ;; p{ {NO}{2}} \equiv [ {HNO}{3}] $So,$ ;; 0.17=;\frac{0.059}{6} \log [ {HNO}{3}]^{8}=;\frac{0.059}{6} \times 8 \log [ {HNO}{3}]\log [ {HNO}{3}] ;; =2.16{[ {HNO}{3}] } ;; =10^{2.16} {M}=10^{x} {M}\therefore ;; x ;; =2.16\Rightarrow ;; 2 x ;; =2 \times 2.16=4.32 ∼eq 4$

Q57JEE Main 2021NAT4MChemical Kinetics

The rate constant of a reaction increases by five times on increase in temperature from $27^{\circ} {C}$ to $52^{\circ} {C}$ . The value of activation energy in ${kJ} {mol}^{-1}$ is ........ (Rounded off to the nearest integer) $[R=8.314 {J} {K}^{-1} {mol}^{-1}]$

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📖 Explanation

$T_{1}=(273+27)=300 {K}, T_{2}=(273+52)=325 {K}$ Given, temperature coefficient of the reaction, $\alpha_{T} \;\; =\;\frac{K_{325}}{K_{300}}=5$ $\log \;{K_{T_{2}}}/{K_{T_{1}}} \;\; =\;\frac{E_{a}}{2.303 R} \times (\;\frac{T_{2}-T_{1}}{T_{1} T_{2}})$ $\log \;\frac{K_{325}}{K_{300}} \;\; =\;\frac{E_{a}}{2.303 \times 8.314}(\;\frac{325-300}{300 \times 325})$ $\log 5 \;\; =\;\frac{E_{a}}{2.303 \times 8319} \times \;\frac{25}{300 \times 325}$ $E_{a} \;\; =52194.78 {Jmol}^{-}$ $\;\; =52.194 {kJ} {mol}^{-1}$ $\;\; =52 {kJ} {mol}^{-1}$

Q58JEE Main 2021NAT4MClassification of Elements

Among the following, number of metal(s) which can be used as electrodes in the photoelectric cell is (i) ${Li}$ (ii) ${Na}$ (ii) ${Rb}$ (iv) Cs

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📖 Explanation

Given metals $( {Li}, {Na}, {Rb}, {Cs})$ are alkali metals (Group 1). Their ionisation energy (IE) decreases down the group. Here, Cs has lowest IE, i.e. it has lowest value of threshold energy $(E_{0})$ or work function $(W)$ . Hence, it can be used as electrode in photoelectric cell.

Q59JEE Main 2021NAT4Md and f Block Elements

The spin only magnetic moment of a divaler ion in aqueous solution (atomic number $=29$ ) is ......... BM.

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📖 Explanation

$Z=29[ {Cu}] {\longrightarrow }^{-2 e^{-}} {Cu}^{2+}=[ {Ar}] 3 d^{9}$ Number of unpaired electron, $n=1$ $\therefore$ Spin only magnetic moment, $\mu \;\; =\sqrt{n(n+2)} {BM}=\sqrt{1(1+2)} {BM}=\sqrt{3} {BM}$ $\;\; =1.73 {BM}=2 {BM}$

Q60JEE Main 2021NAT4MAldehydes, Ketones and Carboxylic Acids

The number of compound(s) given below which contain(s) - ${COOH}$ group is .......... (i) Sulphanilic acid (ii) Picric acid (iii) Aspirin (iv) Ascorbic acid

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📖 Explanation

Structure of given compounds are as followsOnly one compound (iii) contains - ${COOH}$ group.

Mathematics30 questions

Q61JEE Main 2021MCQ4MMatrices and Determinants

If for the matrix, $A=\begin{bmatrix}1 & -\alpha \\ \alpha & \beta\end{bmatrix}, A A^{T}=I_{2}$ , then the value of $\alpha^{4}+\beta^{4}$ is

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📖 Explanation

Given, $A=\begin{bmatrix}1 & -\alpha \\ \alpha & \beta\end{bmatrix}$ $\Rightarrow \;\; A^{T}=\begin{bmatrix}1 & \alpha \\ -\alpha & \beta\end{bmatrix}$ Given, $A A^{T}=I_{2}$ i.e. $\;{\begin{bmatrix}1 & -\alpha \\ \alpha & \beta\end{bmatrix}\begin{bmatrix}1 & \alpha \\ -\alpha & \beta\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}}$ $\;{\begin{bmatrix}1+\alpha^{2} & \alpha-\alpha \beta \\ \alpha-\alpha \beta & \alpha^{2}+\beta^{2}\end{bmatrix}\$ =\begin{bmatrix}1 & 0 \ 0 & 1\end{bmatrix}} $Equating these matrices,$ 1=alpha^{2}+1 ;\text{ gives, }; alpha=0alpha(1-beta)=0alpha^{2}+beta^{2}=1 $Put$ alpha=0 $\in$ alpha^{2}+beta^{2}=1 $, we get$ 0+beta^{2}=1 $, gives$ beta=\pm 1 $where we take$ beta^{4}=1∴ ;; alpha^{4}+beta^{4}=0^{4}+1=1$

Q62JEE Main 2021MCQ4MMatrices and Determinants

Let $A$ be a $3 \times 3$ matrix with $det(A)=4$ . Let $R_{i}$ denote the ith row of $A$ . If a matrix $B$ is obtained by performing the operation $R_{2} \to 2 R_{2}+5 R_{3}$ on $2 A$ , then det (B) is equal to

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📖 Explanation

Let $A=\begin{bmatrix}a & b & c \\ d & \\ \\ \text{ e }\\ \\ f \\ \\ \\ \text{ g }\\ \\ h & i\end{bmatrix}$ Then, $2 A=\begin{bmatrix}2 a & 2 b & 2 c \\ 2 d & 2 {e} & 2 f \\ 2 g & 2 h & 2 i\end{bmatrix}$ Now, perform the operation $R_{2} \to 2 R_{2}+5 R_{3}$ on $2 A$ , we get $B=\begin{bmatrix}2 a & 2 b & 2 c \\ 4 d+10 g & 4 e+10 h & 4 f+10 i \\ 2 g & 2 h & 2 i\end{bmatrix}$ Using property of invariance to calculate $|B|$ , apply $R_{2} \to R_{2}-5 R_{3}$ $|B| \;\; =\begin{bmatrix}2 a & 2 b & 2 c \\ 4 d & 4 {e} & 4 f \\ 2 g & 2 h & 2 i\end{bmatrix}=2 \times 4 \times 2\begin{bmatrix}a & b & c \\ d & {e} & f \\ g & h & i\end{bmatrix} \;\; [ \because det(A)=4]$ $\;\; =16 \times det(A)$ $\;\; =16 \times 4=64$

Q63JEE Main 2021MCQ4MMatrices and Determinants

The following system of linear equations $\;2 x+3 y+2 z=9$ $\;3 x+2 y+2 z=9$ $x-y+4 z=8$

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📖 Explanation

The given system of equations is non-homogeneous and it can be written as, $\;{\begin{bmatrix}2 & 3 & 2 \\ 3 & 2 & 2 \\ 1 & -1 & 4\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}9 \\ 9 \\ 8\end{bmatrix}}$ $\;\text{ i.e,. }\; A X=B$ $\;\text{ Now, }\;|A|=2(8+2)-3(12-2)+2(-3-2)$ $\;\; =20-30-10=-20 \ne 0$ $\because |A| \ne 0$ , then this system have unique solution.

Q64JEE Main 2021MCQ4MDefinite Integrals

If $I_{n}=int_{\pi / 4}^{\pi / 2} \cot ^{n} x d x$ , then

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📖 Explanation

$I_{n}=int_{\;\frac{\pi}{4}}^{\;\frac{\pi}{2}} \cot ^{n} x d x=int_{\;\frac{\pi}{4}}^{\;\frac{\pi}{2}} \cot ^{n-2} x(\cot ^{2} x) d x$ $I_{n}=int_{\;\frac{\pi}{4}}^{\;\frac{\pi}{2}} \cot ^{n-2} x cosec^{2} x d x-int_{\;\frac{\pi}{4}}^{\;\frac{\pi}{2}} \cot ^{n-2} x d x$ $I_{n}+I_{n-2}=int_{\;\frac{\pi}{4}}^{\;\frac{\pi}{2}} \cot ^{n-2} x \cdot cosec^{2} x d x$ Now, let $\cot x=t$ , then $cosec^{2} x d x=-d t$ , limit will be $I_{n}+I_{n-2} \;\; =int_{1}^{0}-t^{n-2} d t$ $\;=\;\frac{-(t)^{n-1}}{n-1}]_{1}^{0}=-\{\;\frac{0}{n-1}-\;\frac{(1)^{n-1}}{n-1}\}$ $I_{n}+I_{n-2}=\;\frac{1}{n-1}$ Now, put $n=4$ $\Rightarrow \;\; I_{2}+I_{4}=\;\frac{1}{3}$ , then $\;\frac{1}{I_{2}+I_{4}}=3$ . . . (i) Put $n=5$ $\Rightarrow \;\; I_{5}+I_{3}=\;\frac{1}{4}$ , then $\;\frac{1}{I_{3}+I_{5}}=4$ . . . (ii) Put $n=6$ $\Rightarrow \;\; I_{6}+I_{4}=\;\frac{1}{5}$ , then $\;\frac{1}{I_{4}+I_{6}}=5$ . . . (iii) Here, from Eqs. (i), (ii) and (iii), we conclude $\;\frac{1}{I_{2}+I_{4}}, \;\frac{1}{I_{3}+I_{5}}$ and $\;\frac{1}{I_{4}+I_{6}}$ are in AP with common difference $1 .$

Q65JEE Main 2021MCQ4MSequences and Series

A function $f(x)$ is given by $f(x)=\;\frac{5^{x}}{5^{x}+5}$ , then the sum of the series $f(\;\frac{1}{20})+f(\;\frac{2}{20})+f(\;\frac{3}{20})+. . .+f(\;\frac{39}{20})$ is equal to

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📖 Explanation

Given, $f(x)=\;\frac{5^{x}}{5^{x}+5}$ , then, $f(2-x)=\;\frac{5^{2-x}}{5^{2-x}+5}$ $\Rightarrow \;\; f(2-x)=\;\frac{5}{5^{x}+5}$ This gives, $f(x)+f(2-x)=\;\frac{5^{x}+5}{5^{x}+5}=1$ $\Rightarrow f(\;\frac{1}{20})+f(2-\;\frac{1}{20})=f(\;\frac{1}{20})+f(\;\frac{39}{20})=1$ Similarly, $f(\;\frac{2}{20})+f(\;\frac{38}{20})=1 \;\text{ and so on, }\;$ $\therefore \;\; f(\;\frac{1}{20})+f(\;\frac{2}{20})+. . .+f(\;\frac{38}{20})+f(\;\frac{39}{20})$ $=1+1+. . .+1+f(\;\frac{20}{20})=19+f(1)=19+\;\frac{1}{2}=\;\frac{39}{2}$

Q66JEE Main 2021MCQ4MQuadratic Equation and Inequalities

Let $\alpha$ and $\beta$ be the roots of $x^{2}-6 x-2=0$ . If $a_{n}=\alpha^{n}-\beta^{n}$ for $n \ge 1$ , then the value of $\;\frac{a_{10}-2 a_{8}}{3 a_{9}}$ is

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📖 Explanation

We have, $x^{2}-6 x-2=0$ Given, $\alpha$ and $\beta$ are roots of above quadratic equation, then $\;\alpha^{2}-6 \alpha-2=0$ $\;\beta^{2}-6 \beta-2=0$ Also, given $a_{n}=\alpha^{n}-\beta^{n}$ , then $\;\frac{a_{10}-2 a_{8}}{3 a_{9}}$ $\;\; =\;\frac{(\alpha^{10}-\beta^{10}-2(\alpha^{8}-\beta^{8})}{3(\alpha^{9}-\beta^{9})}=\;\frac{\alpha^{10}-2 \alpha^{8}-\beta^{10}+2 \beta^{8}}{3(\alpha^{9}-\beta^{9})}$ $\;\; =\;\frac{\alpha^{8}(\alpha^{2}-2)-\beta^{8}(\beta^{2}-2)}{3(\alpha^{9}-\beta^{9})}$ $\;\; =\;\frac{\alpha^{8}(6 \alpha)-\beta^{8}(6 \beta)}{3(\alpha^{9}-\beta^{9})}=\;\frac{6 \alpha^{9}-6 \beta^{9}}{3(\alpha^{9}-\beta^{9})}=\;\frac{6(\alpha^{9}-\beta^{9})}{3(\alpha^{9}-\beta^{9})}$ $\;\; =2$

Q67JEE Main 2021MCQ4MApplication of Derivatives

The minimum value of $f(x)=a^{a^{x}}+a^{1-a^{x}}$ , where $a, x \in R$ and $a > 0$ , is equal to

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📖 Explanation

We already know, Arithmetic mean $\ge$ Geometric mean, Let us take AM and GM of two terms $a^{a^{x}}$ and $a^{1-a^{x}}$ , $\Rightarrow A M=\;{a^{a^{x}}+a^{1-a^{x}}}/{2}$ and $G M=\sqrt{a^{a^{x}} \cdot a^{1-a^{x}}}$ $\because A M \ge G M \Rightarrow \;{a^{a^{x}}+a^{1-a^{x}}}/{2} \ge \sqrt{a^{a^{x}} \cdot a^{1-a^{x}}} \Rightarrow a^{a^{x}}+a^{1-a^{x}} \ge 2 \sqrt{a^{1}}$ $\therefore$ Minimum value of $f(x)=a^{a^{x}}+a^{1-a^{x}}$ is $2 \sqrt{a}$ .

Q68JEE Main 2021MCQ4MIndefinite Integrals

The integral $\int \;{e^{3 \log _{e} 2 x}+5 e^{2 \log _{e} 2 x}}/{e^{4 \log _{e} x}+5 e^{3 \log _{e} x}-7 e^{2 \log _{e} x}} \text{dx}$ , $x > 0$ , is equal to (where, $c$ is a constant of integration)

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📖 Explanation

$I=\int \;{e^{3 \log _{e}(2 x)}+5 e^{2 \log _{e}(2 x)}}/{e^{4 \log _{e}(x)}+5 e^{3 \log _{e}(x)}-7 e^{2 \log _{e}(x)}} d x$ $=\int \;{e^{\log _{e}(2 x)^{3}}+5 e^{\log _{e}(2 x)^{2}}}/{e^{\log _{e} x^{4}}+5 e^{\log _{e}(x)^{3}}-7 e^{\log _{e}(x)^{2}}} d x$ $[using property $a \log x=\log x^{a}$ ]$ $=\int \;\frac{8 x^{3}+5(2 x)^{2}}{x^{4}+5(x)^{3}-7 x^{2}} d x \;\;$ [using property $a^{\log _{a} x}=x]$ $\;\; =\int \;\frac{8 x^{3}+20 x^{2}}{x^{4}+5 x^{3}-7 x^{2}} d x=\int \;\frac{4 x^{2}(2 x+5)}{x^{2}(x^{2}+5 x-7)} d x$ $\;\; =\int \;\frac{4(2 x+5)}{x^{2}+5 x+7} d x$ Let $x^{2}+5 x-7=t$ , then $(2 x+5) d x=d t$ $I=\int \;\frac{4 d t}{t}=4 \log _{e} t+c$ Put $t=x^{2}+5 x-7$ $I=4 \log _{ {e}}|x^{2}+5 x-7|+C$

Q69JEE Main 2021MCQ4MComplex Numbers

If $\alpha, \beta \in R$ are such that $1-2 i($ here $i^{2}=-1)$ is a root of $z^{2}+\alpha z+\beta=0$ , then $(\alpha-\beta)$ is equal to

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📖 Explanation

Given, root of $z^{2}+\alpha z+\beta=0$ is $1-2 i$ . Since, it is quadratic equation and one root is complex in nature, its another root is complex conjugate. $\therefore$ Two roots are $1-2 i$ and $1+2 i$ . Now, sum of roots $=-\;\frac{\alpha}{1}=-\alpha$ $=(1-2 i)+(1+2 i)=2$ Gives, $\alpha=-2$ Product of roots $=\;\frac{\beta}{1}=\beta$ Gives, $\beta=5=(1-2 i)(1+2 i)=1+4=5$ Gives, $\beta=5$ $\therefore \;\; \alpha-\beta=-2-5=-7$

Q70JEE Main 2021MCQ4MEllipse

If the curve $x^{2}+2 y^{2}=2$ intersects the line $x+y=1$ at two points $P$ and $Q$ , then the angle subtended by the line segment $P Q$ at the origin is

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📖 Explanation

Curve $x^{2}+2 y^{2}=2$ intersect the line $x+y=1$ at points $P$ and $Q$ . First we have to find any common relation between these two curves. Use substitution for the same as follows, $x^{2}+2 y^{2} \;\; =2$ . . . (i) $x+y \;\; =1, \;\text{ then }\;(x+y)^{2}=1^{2}$ $\Rightarrow \;\; x^{2}+y^{2}+2 x y=1$ . . . (ii) We can write Eq. (i) as, $x^{2}+2 y^{2}-2(1)^{2}=0$ $\Rightarrow \;\; x^{2}+2 y^{2}-2(x+y)^{2}=0$ $\Rightarrow x^{2}+2 y^{2}-2 x^{2}-2 y^{2}-4 x y=0$ $\Rightarrow \;\; -x^{2}-4 x y=0 \Rightarrow -x(x+4 y)=0$ Gives, $x=0$ and $x+4 y=0$ or $y=\;\frac{-1}{4} x$ Draw the line $y=\;\frac{-1}{4} x$ on graph and take arbitrary point (any one) as follows, From given graph, $\tan \theta=\;\frac{1}{4} \Rightarrow \theta=\tan ^{-1}(\;\frac{1}{4})$ We have two lines, $y=-\;\frac{1}{4} x$ and $x=0$ (i.e. $Y$ - axis). Thus, any line joining these two curves makes an angle $\;\frac{\pi}{2}+\theta$ at origin. $\therefore$ Answer is $\;\frac{\pi}{2}+\tan ^{-1}(\;\frac{1}{4})$ .

Q71JEE Main 2021MCQ4MApplication of Derivatives

The shortest distance between the line $x-y=1$ and the curve $x^{2}=2 y$ is

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📖 Explanation

Let $(x, y)$ be any arbitrary point on curve $x^{2}=2 y$ and find the tangent line equation at this point, such that tangent line at $(x, y)$ is parallel to line $x-y=1$ . To find tangent equation, differentiate the following equation so that we can find slope, $x^{2}-2 y \;\; =0$ $2 x-2 \;\frac{d y}{d x} \;\; =0 \;\text{ gives }\; \;\frac{d y}{d x}=x$ Slope $($ say $m_{1})=x$ Also, slope of line $x-y=1$ or $y=x-1$ is 1 (say $m_{2}$ ). Since, $x-y=1$ and tangent line is parallel, therefore, their slope be equal. Hence, $m_{1}=m_{2}$ gives, $x=1$ Put $x=1$ in Eq. (i), we get $y=1 / 2$ Thus, $(x, y)=(1, \;\frac{1}{2})$ Perpendicular distance between line $x-y=1$ and point $(1, \;\frac{1}{2})$ is given as, $P \;\; =|\;{(1)(1)+(\frac{1}{2})(-1)-1}/{\sqrt{(1)^{2}+(-1)^{2}}}|$ $\;\; =|\;\frac{-1}{2 \sqrt{2}}|$ $\;\; =\;\frac{1}{2 \sqrt{2}} \;\; \;\; \;\; \therefore \;\text{ using }\;$ ( $\therefore$ using perpendicular distance formula)

Q72JEE Main 2021MCQ4MHyperbola

A hyperbola passes through the foci of the ellipse $\;\frac{x^{2}}{25}+\;\frac{y^{2}}{16}=1$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is

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📖 Explanation

Given, ellipse is $\;\frac{x^{2}}{25}+\;\frac{y^{2}}{16}=1$ or $\;\; \;\frac{x^{2}}{(5)^{2}}+\;\frac{y^{2}}{(4)^{2}}=1$ Compare it with standard equation $\;\frac{x^{2}}{a^{2}}+\;\frac{y^{2}}{b^{2}}=1$ , we get $a=5, b=4$ Now, focus of ellipse $=(\pm C, 0)$ where $C=\sqrt{a^{2}-b^{2}}$ Put the values of $a$ and $b$ , we get $C=\sqrt{5^{2}-4^{2}}=\sqrt{25-16}=\sqrt{9}$ $\therefore$ Focus $=(\pm 3,0)$ According to question, hyperbola passes through the focus of ellipse. Let equation of hyperbola be $\;\frac{x^{2}}{a^{2}}-\;\frac{y^{2}}{b^{2}}=1$ Since, it passes through $(\pm 3,0)$ , we get $\;\frac{(\pm 3)^{2}}{a^{2}}-\;\frac{0}{b^{2}}=1$ , gives $a=\pm 3$ or $a^{2}=9$ Also, given that product of eccentricities is $1 .$ Now, (Eccentricity of ellipse ) (Eccentricity of hyperbola) $=1$ $\Rightarrow \;\; (\sqrt{1-\;\frac{16}{25}})(\sqrt{1+\;\frac{b^{2}}{9}})=1$ (using formula of eccentricity of ellipse and hyperbola) $\Rightarrow \;\; (\sqrt{\;\frac{9}{25}})(\sqrt{1+\;\frac{b^{2}}{9}})=1$ Squaring on both sides, $1+\;\frac{b^{2}}{9}=\;\frac{25}{9}$ $b^{2}=16$ Thus, equation of hyperbola is $\;\frac{x^{2}}{9}-\;\frac{y^{2}}{16}=1$ .

Q73JEE Main 2021MCQ4MThree Dimensional Geometry

A plane passes through the points $A(1,2,3), B(2,3,1)$ and $C(2,4,2)$ . If 0 is the origin and $P$ is $(2,-1,1)$ , then the projection of OP on this plane is of length[2021 25 Feb Shift 2]

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📖 Explanation

Refer diagram, the normal vector be $n$ and it is perpendicular to both $A B$ and $A C$ ${AB} \times {AC}= {n}$ Now, $A(1,2,3), B(2,3,1)$ and $C(2,4,2)$ Then, $A B=(2-1) \hat{i}+(3-2) \hat{j}+(1-3) \hat{k}$ $\;\; =\hat{i}+\hat{j}-2 \hat{k}$ $\;\; -1) \hat{i}+(4-2) \hat{j}+(2-3) \hat{k}$ ${AC} \;\; =(2-1) { {i}^{∧}+(4$ $\;\; ={ {i}^{∧}+2 { {j}^{∧}-{ {k}^{∧}$ Now, $A B \times A C=\begin{vmatrix}{i}^{∧} & {j}^{∧} & {k}^{∧} \\ 1 & 1 & -2 \\ 1 & 2 & -1\end{vmatrix}$ $\;\; =\hat{i}(-1+4)-\hat{j}(-1+2)+\hat{k}(2-1)$ $\;\; =3 \hat{i}-\hat{j}+\hat{k}$ $n \;\; =3 \hat{i}-\hat{j}+\hat{k}$ Let $P$ be any point on normal vector and ${O}$ be origin. Then refer the diagram, projection of $O P$ on plane have length $O M$ . $\;\text{ OP }\; \;\; =2 \hat{i}-\hat{j}+\hat{k} \;\text{ and }\; n=3 \hat{i}-\hat{j}+\hat{k}$ $\;\text{ OP. }\; n \;\; =|O P||n| \cos \theta$ $(6+1+1) \;\; =\sqrt{4+1+1}(\sqrt{9+1+1}) \cos \theta$ $8 \;\; =\sqrt{6} \sqrt{11} \cos \theta \Rightarrow \cos \theta=\;\frac{8}{\sqrt{66}}$ $\;\text{ Again, }\; \sin \theta \;\; =\;\frac{|O M|}{|O P|}, \;\text{ gives }\;|O M|=\sin \theta \cdot |O P|$ $\Rightarrow \;\; |O M| \;\; =\sqrt{1-\cos ^{2} \theta|O P|}$ $\;=\sqrt{1-\;\frac{64}{66}} \sqrt{4+1+1} \;\; \;\text{ (use }\; \cos \theta=8 \sqrt{66})$ $\;\; =\sqrt{\;\frac{2}{66}} \cdot \sqrt{6}$ $\therefore \;\; |O M|=\sqrt{\;\frac{2}{11}}$

Q74JEE Main 2021MCQ4MDefinite Integrals

$\lim _{n \to \infty }\$ [;\frac{1}{n}+;\frac{n}{(n+1)^{2}}+;\frac{n}{(n+2)^{2}}+. . .+;\frac{n}{(2 n-1)^{2}}]$ ;\text{ is };$ equal to

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📖 Explanation

(b) $\;\text{ Let }\; L \;\; =\lim _{n \to \infty }[\;\frac{1}{n}+\;\frac{n}{(n+1)^{2}}+\;\frac{n}{(n+2)^{2}}+. . .]$ $\;\text{ or }\; L \;\; =\lim _{n \to \infty }\$ [;\frac{n}{(2 n-1)^{2}}]$ ;\frac{n}{(n+0)^{2}}+;\frac{n}{(n+1)^{2}}+. . .+;; =\lim _{n \to \infty }[;\frac{n}{(n+n-1)^{2}}+;\frac{n}{(n+n)^{2}}-;\frac{n}{(n+n)^{2}}];; -\lim {n \to \infty }[;\frac{n}{(n+n)^{2}}];=\lim {n \to \infty } sum{r=0}^{n} ;\frac{n}{(n+1)^{2}}+. . .+;\frac{n}{(n+n)^{2}}];; =\lim {n \to \infty } sum{r=0}^{n} ;\frac{n}{(n+r)^{2}}-0(since, \lim {n \to \infty } ;\frac{1}{n}=0) $Now, for solving limit summation, we integrate it using some replacement.$ L=\lim {n \to \infty } sum{r=0}^{n} ;\frac{1}{r(1+r / n)} $Take$ ;\frac{r}{n} $as$ x $and$ ;\frac{1}{n} $as$ d x $. Lower limit is obtained by putting$ r=0 $in$ ;\frac{r}{n} $, we get Lower limit$ =0 $Upper limit is obtained by putting$ r=n $in$ ;\frac{r}{n} $, we get Upper limit = 1$ \therefore ;; L=int{0}^{1} ;\frac{1}{(1+x)^{2}} d x=;\frac{-1}{(1+x)}]{0}^{1}=-(;\frac{1}{2}-1)=;\frac{1}{2}\therefore ;; L=;\frac{1}{2}$

Q75JEE Main 2021MCQ4MProbability

In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarian and the remaining 140 are non-smokers and vegetarian. Their chances of getting a particular chest disorder are $35 \%, 20 \%$ and $10 \%$ , respectively. A person is chosen from the group at random and is found to be suffering from the chest disorder. The probability that the selected person is a smoker and non-vegetarian is

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📖 Explanation

Let ' $A$ ' be the event of smokers and non-vegetarian. Let ' $B$ ' be the event of smokers and vegetarian. Let ' $C$ ' be the event of non-smokers and vegetarian. Let ' $E$ ' be the event of chest disorders. According to question, $P(A)=\;\frac{160}{400} \;\; =\;\frac{2}{5}, P(B)=\;\frac{100}{400}=\;\frac{1}{4}, P(C)=\;\frac{140}{400}=\;\frac{7}{20}$ $P(\;\frac{E}{A})=35 \% \;\; =\;\frac{35}{100}, P(\;\frac{E}{B})=\;\frac{20}{100}, P(\;\frac{E}{C})=\;\frac{10}{100}$ $\;\text{ Thus, }\; P(\;\frac{A}{E}) \;\; =\;\frac{P(A) \cdot P(\frac{E}{A})}{P(A) \cdot P(\;\frac{E}{A})+P(B) P(\;\frac{E}{B})+P(C) P(\;\frac{E}{C})}$ $\;\; =\;\frac{\;\frac{2}{5} \cdot \frac{35}{100}}{\frac{2}{5} \cdot \;\frac{35}{100}+\;\frac{1}{4} \cdot \;\frac{20}{100}+\;\frac{7}{20} \cdot \;\frac{10}{100}}$ $\;\; =\;\frac{14}{\frac{14}{100}+\;\frac{5}{100}+\;\frac{2}{100}}$

Q76JEE Main 2021MCQ4MProbability

Let $A$ be a set of all 4-digit natural numbers whose exactly one digit is 7 . Then, the probability that a randomly chosen element of $A$ leaves remainder 2 when divided by 5 is

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📖 Explanation

Here, we have four digit natural numbers, then total cases will exclude those number which contain zero at thousands place. Hence, total cases will be $=({ }^{4} C_{1} \times 9 \times 9 \times 9)-({ }^{3} C_{1} \times 9 \times 9)=2673$ Again, only those numbers will have remainder 2 when divided by 5 either they have 2 at its unit place or 7 at its unit place. When unit digit is 2 , then total number of four digit numbers will be $=({ }^{3} C_{1} \times 9 \times 9)-({ }^{2} C_{1} \times 9)=225$ When unit digit is 7 , then total number of four digit numbers will be $=8 \times 9 \times 9=648$ Now, total favourable cases $=225+648=873$ Required probability $=\;\frac{\;\text{ Total favourable cases }\;}{\;\text{ Total number of cases }\;}$ $=\;\frac{873}{2673}=\;\frac{97}{297}$

Q77JEE Main 2021MCQ4MTrigonometric Ratios and Identities

If $0 < x, y < \pi$ and $\cos x+\cos y-\cos (x+y)=\;\frac{3}{2}$ , then $\sin x+\cos y$ is equal to

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📖 Explanation

Given, $\cos x+\cos y-\cos (x+y)=\;\frac{3}{2}$ $\Rightarrow 2 \cos (\;\frac{x+y}{2}) \cos (\;\frac{x-y}{2})-\$ [2 \cos ^{2}(;\frac{x+y}{2})-1]$=;\frac{3}{2} $[Use formula,$ ;\cos a+\cos b=2 \cos (;\frac{a+b}{2}) \cos (;\frac{a-b}{2});\cos 2 x=2 \cos ^{2} x-1]\Rightarrow 2 \cos (;\frac{x+y}{2}) \cos (;\frac{x-y}{2})-2 \cos ^{2}(;\frac{x+y}{2});; =;\frac{3}{2}-1=;\frac{1}{2}\Rightarrow 4 \cos (;\frac{x+y}{2}) \cos (;\frac{x-y}{2})-4 \cos ^{2}(;\frac{x+y}{2})=;\frac{1}{2} \times 2=1=\cos ^{2} ;\frac{(x-y)}{2}+\sin ^{2} ;\frac{(x-y)}{2}\Rightarrow [\cos (;\frac{x-y}{2})-2 \cos (;\frac{x+y}{2})]^{2}+\sin ^{2}(;\frac{x-y}{2})=0\Rightarrow \sin (;\frac{x-y}{2})=0 $and$ \cos (;\frac{x-y}{2})-2 \cos (;\frac{x+y}{2})=0\Rightarrow x=y $and$ \cos 0-2 \cos x=0 $Gives,$ \cos x=;\frac{1}{2}=\cos y\therefore \sin x=\sqrt{1-\cos ^{2} x}=\sqrt{1-;\frac{1}{4}}=;\frac{\sqrt{3}}{2}\therefore \sin x+\cos y=;\frac{\sqrt{3}}{2}+;\frac{1}{2}=;\frac{1+\sqrt{3}}{2}$

Q78JEE Main 2021MCQ4MFunctions

Let $x$ denote the total number of one-one functions from a set $A$ with 3 elements to a set $B$ with 5 elements and $y$ denote the total number of one-one functions from the set $A$ to the set $A \times B$ . Then,

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📖 Explanation

$x=\{f: A \to B, f$ is one - one $\}$ $y=\{g: A \to A \times B, g$ is one one $\}$ Number of elements in $A=3$ i.e. $|A|=3$ Similarly, $|B|=5$ Then, $|A \times B|=|A| \times |B|=3 \times 5=15$ Now, number of one-one function from $A$ to $B$ will be ${ }^{5} P_{3}=\;\frac{5 !}{(5-3) !}=\;\frac{5 !}{2 !}=5 \times 4 \times 3=60$ $\therefore \;\; x=60$ Now, number of one-one function from $A$ to $A \times B$ will be $\;\; ={ }^{15} P_{3}=\;\frac{15 !}{(15-3) !}=\;\frac{15 !}{12 !}$ $\;\; =15 \times 14 \times 13=2730$ $y=2730$ Thus, $2 \times (2730)=91 \times (60)$ $2 y=91 x$

Q79JEE Main 2021MCQ4MInverse Trigonometric Functions

$cosec[2 \cot ^{-1}(5)+\cos ^{-1}(\;\frac{4}{5})]$ is equal to

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📖 Explanation

$cosec[2 \cot ^{-1}(5)+\cos ^{-1}(4 / 5)]$ $=cosec\$ [2 \tan ^{-1}(;\frac{1}{5})+\cos ^{-1}(;\frac{4}{5})]$ ;; [ \because \tan ^{-1} x=\cot ^{-1}(;\frac{1}{x})]=cosec[\tan ^{-1}(;\frac{2(\frac{1}{5})}{1-(;\frac{1}{5})^{2}})+\cos ^{-1}(;\frac{4}{5})[[ \therefore 2 \tan ^{-1} \theta=\tan ^{-1}(;\frac{2 \theta}{1-\theta^{2}})]=cosec(\tan ^{-1} ;\frac{5}{12}+\cos ^{-1} ;\frac{4}{5}) $Let$ \tan ^{-1}(;\frac{5}{12})=x $, then$ \tan x=;\frac{5}{12} $gives$ \sin x=;\frac{5}{13}, \cos x=;\frac{12}{13} $Let$ \cos ^{-1}(;\frac{4}{5})=y $, then$ \cos y=;\frac{4}{5} $gives,$ \sin y=;\frac{3}{5} $Now,$ cosec(x+y)=;\frac{1}{\sin (x+y)}=;\frac{1}{\sin x \cos y+\cos x \sin y}=;\frac{1}{(;\frac{5}{13})(;\frac{4}{5})+(;\frac{12}{13})(;\frac{3}{5})}=;\frac{65}{56}$

Q80JEE Main 2021MCQ4MSets and Relations

The contrapositive of the statement; "If you will work, you will earn money" is

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📖 Explanation

Contrapositive of $p \to q$ is $∼ q \to ∼ p$ $p:$ you will work $\Rightarrow ∼ p:$ you will not work $q$ : you will earn money $\Rightarrow ∼ q$ : you will not earn money Then, $∼ q \to ∼ p:$ if you will not earn money, you will not work

Q81JEE Main 2021NAT4MLimits, Continuity and Differentiability

A function $f$ is defined on $[-3,3]$ as

f(x)= \begin{cases}\min \{|x|, 2-x^{2}\},\;-2 \le x \le 2 \\ {[|x|]}\;, 2 < |x| \le 3\end{cases}

where, $[x]$ denotes the greatest integer $\le x$ . The number of points, where $f$ is not differentiable in $(-3,3)$ is .......... .

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📖 Explanation

For this particular problem, try to draw graph in the region $(-3,3)$ , it will be as follows, Thus, points of discontinuity will be at $-2,2$ because the curve breaks at these points and at $-1,0,1$ because curve has sharp points. $\therefore$ Point of discontinuity are $-2,-1,0,1,2$ i.e. 5 points.

Q82JEE Main 2021NAT4MDifferential Equations

If the curve $y=y(x)$ represented by the solution of the differential equation $(2 x y^{2}-y) d x+x d y=0$ , passes through the intersection of the lines $2 x-3 y=1$ and $3 x+2 y=8$ , then $|y(1)|$ is equal to ........... .

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📖 Explanation

Given, $(2 x y^{2}-y) d x+x d y=0$ $\Rightarrow \;\; \;\frac{d y}{d x}-\;\frac{y}{x}=-2 y^{2}$ $\Rightarrow \;\; \;\frac{-1}{y^{2}} \;\frac{d y}{d x}+\;\frac{1}{x y}=2$ ...(i) $[divide by $y^{2}$ ]$ Let $\;\frac{1}{y}=v$ , then $-\;\frac{1}{y^{2}} \cdot \;\frac{d y}{d x}=\;\frac{d v}{d x}$ , putting in Eq. (i) $\;\frac{d v}{d x}+v(\;\frac{1}{x})=2$ (this is a linear form) Now, integrating factor (IF) $=e^{\int \;\frac{1}{x} d x}=e^{\log x}=x$ $\therefore \;\; (I F) v=\int 2 .(I F) d x=\int 2 x d x=2 \;\frac{x^{2}}{2}+C$ $\therefore \;\; (I A) v=x^{2}+C$ Put $v=\;\frac{1}{y}$ , this gives $x^{2}+c=\;\frac{x}{y}$ Now, first find point of intersection of lines $2 x-3 y=1$ and $3 x=-2 y+8$ by elimination method, we get $x=2$ , $y=1$ $\therefore$ The curve $x^{2}+c=\;\frac{1}{y}$ passes through $(2,1)$ . Put $x=2, y=1$ , we get $c=-2$ $\;\frac{x}{y}=x^{2}-2$ or $y=\;\frac{x}{x^{2}-2}$ Put $x=1$ , we get $y(1)=\;\frac{1}{1-2}=-1$ $\therefore$ $|y(1)|=1$

Q83JEE Main 2021NAT4MSequences and Series

The total number of two digit numbers ' $n$ ', such that $3^{n}+7^{n}$ is a multiple of 10 , is .......... .

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📖 Explanation

We may write, $7^{n}=(10-3)^{n}$ or $7^{n}=10 k+(-3)^{n}$ (using expansion) Let $n=$ even $=2 t, t \in N$ Then, $3^{n}=3^{2 t}=9^{t}=(10-1)^{t}$ $=10 p+(-1)^{t}=10 p \pm 1$ If $n=$ even, then $7^{n}+3^{n}$ will never be multiple of 10 . This implies $n=$ odd $n \;\; =11,13,15, . . . 99 \;\; \;\text{ (since, }\; n \;\text{ is two digit) }\;$ $10 \; < n < 100$ Total possible ' $n$ ' are 45 .

Q84JEE Main 2021NAT4MLimits, Continuity and Differentiability

If $\lim _{x \to 0} \;\frac{a x-(e^{4 x}-1)}{a x(e^{4 x}-1)}$ exists and is equal to $b$ , then the value of $a-2 b$ is ......... .

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📖 Explanation

$\lim _{x \to 0} \;\frac{a x-(e^{4 x}-1)}{a x(e^{4 x}-1)}=L$ (say) $[\;\frac{0}{0}$ form $]$ Apply L - Hospital rule, $L=\lim _{x \to 0} \;\frac{a-4 e^{4 x}}{a(e^{4 x}-1)+a x(4 e^{4 x})}$ [Limit exist everywhere except $a=4]$ Again, apply L-Hospital rule, $L \;\; =\lim _{x \to 0} \;\frac{-16 e^{4 x}}{a(4 e^{4 x})+a(4 e^{4 x})+a x(16 e^{4 x})}$ $\;\; =\;\frac{-16}{4 a+4 a}=\;\frac{-2}{a}$ $\;\; =\;\frac{-2}{4}=\;\frac{-1}{2}$ $L \;\; =\lim _{x \to 0} \;\frac{-16 e^{4 x}}{a(4 e^{4 x})+a(4 e^{4 x})+a x(16 e^{4 x})}$ $\;\; =\;\frac{-16}{4 a+4 a}=\;\frac{-2}{a}$ $\;\; =\;\frac{-2}{4}=\;\frac{-1}{2}$ Given, $\;\; L=b \Rightarrow \;\frac{-2}{a}=\;\frac{-1}{2}=b$ Then, $a-2 b=4-2(\;\frac{-1}{2})=4+1=5$

Q85JEE Main 2021NAT4MApplication of Derivatives

If the curves $x=y^{4}$ and $x y=k$ cut at right angles, then $(4 k)^{6}$ is equal to ............ .

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📖 Explanation

If the curves cut at right angle, then product of slopes will be $-1$ . First curve $x=y^{4}$ Differentiate it, we get $1 \;\; =4 y^{3} \;\frac{d y}{d x}$ $\;\frac{d y}{d x} \;\; =\;\frac{1}{4 y^{3}}$ $\Rightarrow \;\; \;\text{ Slope of first curve }\;(m_{1}) \;\; =\;\frac{1}{4 y_{1}^{3}}$ [at point $(x_{1}, y_{1})$ ] Second curve $x y=k$ Differentiate it, $0=x \;\frac{d y}{d x}+y$ $\Rightarrow \;\; \;\frac{d y}{d x}=\;\frac{-y}{x}$ Slope of second curve $(m_{2})=\;\frac{-y_{1}}{x_{1}}$ [at $(x_{1}, y_{1})$ ] $\Rightarrow \;\; m_{1} \cdot m_{2}=-1$ $\Rightarrow \;\; \;\frac{1}{4 y_{1}^{3}}(\;\frac{-y_{1}}{x_{1}})=-1 \Rightarrow \;\frac{-1}{4 y_{1}^{2} x_{1}}=-1$ $\Rightarrow \;\; \;\frac{-1}{4(y_{1})^{6}}=-1$ $\Rightarrow \;\; y_{1}^{6}=\;\frac{1}{4}$ Also, $x_{1} y_{1}=k$ , using $x_{1}=y_{1}^{4}$ , we get $k=y_{1}^{5}$ or $k^{6}=(y_{1})^{30}$ $\because \;\; y_{1}^{6}=\;\frac{1}{4}$ , then $y_{1}^{30}=(\;\frac{1}{4})^{5}$ $\therefore (4 k)^{6}=4^{6} \cdot k^{6}=4^{6}(y_{1})^{30}=4^{6}(\;\frac{1}{4})^{5}=4$ $\therefore (4 k)^{6}=4$

Q86JEE Main 2021NAT4MDefinite Integrals

The value of $int_{-2}^{2}|3 x^{2}-3 x-6| d x$ is .......... .

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📖 Explanation

$int_{-2}^{2}|3 x^{2}-3 x-6| d x=I$ (say) $I=3 int_{-2}^{2}|x^{2}-x-2| d x$ $\;\; =3[int_{-2}^{-1}(x^{2}-x-2) d x+int_{-1}^{2}(-x^{2}+x+2) d x]$ $\;\; =3[(\;\frac{x^{3}}{3}-\;\frac{x^{2}}{2}-2 x)^{-1}-(\;\frac{x^{3}}{3}-\;\frac{x^{2}}{2}-2 x)^{2}]$ $\;\; =19$

Q87JEE Main 2021NAT4MBinomial Theorem

If the remainder when $x$ is divided by 4 is 3 , then the remainder when $(2020+x)^{2022}$ is divided by 8 is ........ .

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📖 Explanation

Given, when $x$ is divided by 4 , the remainder is 3 . Let $x=4 p+3$ , then $(2020+x)^{2022}= \;\; (2020+4 p+3)^{2022}$ $= \;\; (2024+4 p-1)^{2022}$ $= \;\; (4 k-1)^{2022}( \because 2024 \;\text{ is divisible by 4) }\;$ $= \;\; { }^{2022} C_{0}(4 K)^{2022}(-1)^{0}+{ }^{2022} C_{1}(4 K)^{2021}$ $\;(-1)^{1}+. . . .+{ }^{2022} C_{2022}(4 A)^{0}(-1)^{2022}$ On expansion $(2020+x)^{2022}$ , we get the form of $8 \lambda+1 .$ Since, each terms have 2022 and $4 k$ , so if we take 2 common from 2022 we get 8. Thus, each term have 8 in common. Hence, remainder is $1 .$

Q88JEE Main 2021NAT4MThree Dimensional Geometry

A line ' $l$ ' passing through origin is perpendicular to the lines $\;l_{1}: r=(3+t) \hat{i}+(-1+2 t) \hat{j}+(4+2 t) \hat{k}$ $\;l_{2}: r(3+2 s) \hat{i}+(3+2 s) \hat{j}+(2+s) \hat{k}$ If the coordinates of the point in the first octant on ' $I_{2}$ ' at a distance of $\sqrt{17}$ from the point of intersection of ' $l$ ' and ' $l_{1}$ ' are $(a, b, c)$ , then $18(a+b+c)$ is equal to ........... .

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📖 Explanation

Let $L_{1} \Rightarrow \;\frac{x-3}{1}=\;\frac{y-3}{2}=\;\frac{z-4}{2}=u$ (say) $\Rightarrow$ Direction ratios of $\;\; L_{1}=1,2,2$ $L_{2} \Rightarrow \;\frac{x-3}{2}=\;\frac{y-3}{2}=\;\frac{z-2}{1}=v \;\text{ (say) }\;$ Direction ratios of $L_{2}=2,2,1$ Line $L$ passing through origin is perpendicular to $L_{1}$ and $L_{2}$ . Hence, direction ratios of $L$ is parallel to $(L_{1} \times L_{2})$ . $\Rightarrow \;\; (-2,3,-2)$ Equation of $L \Rightarrow \;\frac{x}{2}=\;\frac{y}{-3}=\;\frac{z}{2}=\lambda$ (say) Solve $L$ and $L_{1}$ , we get $(2 \lambda,-3 \lambda, 2 \lambda)=(\mu +3,2 \mu -1,2 \mu +4)$ Gives, $\lambda=1, \mu =-1$ So, intersection point $P(2,-3,2)$ . Let $Q(2 v+3,2 v+3, v+2)$ be required point on $L_{2}$ . Now, $P Q=\sqrt{17}$ (given) $\;\text{ Now, }\; P Q=\sqrt{17} \;\text{ (given) }\;$ $P Q \;\; =\sqrt{(2 v+1)^{2}+(2 v+6)^{2}+(v)^{2}}$ $\;\; =\sqrt{17}$ $\Rightarrow (2 v+1)^{2} \;+(2 v+6)^{2}+v^{2}=17 \;\; \;\text{ (squaring on both sides) }\;$ $\Rightarrow \;\; 9 v^{2}+28 v+20=0$ On solving, we get $v=-2$ (rejected), $\;\frac{-10}{9}$ (accepted) $\therefore Q$ is $(\;\frac{7}{9}, \;\frac{7}{9}, \;\frac{8}{9})$ $\therefore 18(a+b+c)=18(\;\frac{7}{9}+\;\frac{7}{9}+\;\frac{8}{9})=44$

Q89JEE Main 2021NAT4MCircles

A line is a common tangent to the circle $(x-3)^{2}+y^{2}=9$ and the parabola $y^{2}=4 x$ . If the two points of contact $(a, b)$ and $(c, d)$ are distinct and lie in the first quadrant, then $2(a+c)$ is equal to . . . . . . . .

🚀 Solve in Practice Mode

📖 Explanation

Given, circle $\Rightarrow (x-3)^{2}+y^{2}=9$ Parabola $\Rightarrow y^{2}=4 x$ Let equation of common tangent be $\;y=m x+\;\frac{a}{m}$ $\;y=m x+\;\frac{1}{m}$ $\Rightarrow m^{2} x-m y+1=0$ $\therefore$ Perpendicular from $(3,0)$ to line $=3$ $\Rightarrow \;\; |\;{(3 m^{2}-0+1)}/{\sqrt{m^{2}+m^{4}}}|=3$ $\Rightarrow \;\; \;\; (3 m^{2}+1)^{2}=9(m^{2}+m^{4}) \Rightarrow m=\pm \;\frac{1}{\sqrt{3}}$ Tangent is $y=\;\frac{1}{\sqrt{3}} x+\sqrt{3}$ $\Rightarrow \;\; m=\;\frac{1}{\sqrt{3}}$ or $\;\; y=\;\frac{(-1)}{\sqrt{3}} \times (-\sqrt{3})\;\;$ (rejected) For parabola, point of contact is $(c, d)=(\;\frac{a}{m^{2}}, \;\frac{2 a}{n})$ $\therefore \;\; (c, d)=(3,2 \sqrt{3})$ Again, solve circle and line equation, we get $(x-3)^{2}+[(\;\frac{1}{\sqrt{3}} x+\sqrt{3})]^{2} \;\; =9$ $\Rightarrow \;\; x^{2}+9-6 x+\;\frac{1}{3} x^{2}+3+2 x \;\; =9$ $\;\frac{4}{3} x^{2}-4 x+3 \;\; =0$ $\Rightarrow \;\; x \;\; =\;\frac{3}{2}=a$ $\therefore \;\; 2(a+c) \;\; =2(\;\frac{3}{2}+3)=9$

Q90JEE Main 2021NAT4MVector Algebra

Let $a=\hat{i}+\alpha \hat{j}+3 \hat{k}$ and $b=3 \hat{i}-\alpha \hat{j}+\hat{k}$ . If the area of the parallelogram whose adjacent sides are represented by the vectors $a$ and $b$ is $8 \sqrt{3}$ square units, then $a \cdot b$ is equal to ............ .

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📖 Explanation

Area of parallelogram $=|a \times b|$ $\;\; =|(\hat{i}+\alpha \hat{j}+3 \hat{k}) \times (3 \hat{i}-\alpha \hat{j}+\hat{k})|$ $(64)(3) \;=16 \alpha^{2}+64+16 \alpha^{2} \;\text{ (given, area }\;=8 \sqrt{3})$ $\;\; =|(\hat{i}+\alpha \hat{j}+3 \hat{k}) \times (3 \hat{i}-\alpha \hat{j}+\hat{k})|$ $(64)(3) \;\; =16 \alpha^{2}+64+16 \alpha^{2} \;\text{ (given, area }\;=8 \sqrt{3} \;\text{ ) }\;$ $\;\;\text{ (squaring on both sides) }\;$ $\Rightarrow \;\; \alpha^{2}=4$ Now, a. b $=3-\alpha^{2}+3$ $=6-\alpha^{2}=6-4=2$

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