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27 Jul 2021 Shift 1

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Physics30 questions

Q1JEE Main 2021MCQ4MCurrent Electricity

In the given figure, a battery of emf $E$ is connected across a conductor ${PQ}$ of length $' l'$ and different area of cross-sections having radii ${r}_{1}$ and ${r}_{2}$ {r}{2} < {r}{1})$.Choose the correct option as one moves from P to Q :

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📖 Explanation

Current is constant in conductor ${i}=$ constant Resistance of element ${dR}=\frac{\rho {dx}}{\pi{r}^{2}}$ ${dV}={idR}=\frac{{i} \rho {d} {x}}{\pi{r}^{2}}$ ${E}=\frac{{d} {V}}{{d} {x}}=\frac{{i} \rho }{\pi{r}^{2}}$ $\& \ V_{{d}}=\frac{{eE} \tau }{{m}}$ $\therefore {V}_{{d} \ \propto } {E}$ → ${E} \ \propto \frac{1}{{r}^{2}}$ if ${r}$ decreases, ${E}$ will increase $\therefore {V}_{{d}}$ will increase

Q2JEE Main 2021MCQ4MAtoms and Nuclei

The number of molecules in one litre of an ideal gas at $300 {K}$ and 2 atmospheric pressure with mean kinetic energy $2 \\times 10^{-9} {J}$ per molecules is :

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📖 Explanation

${KE}=\frac{3}{2} {k} {T}$ ${PV}={{N}}/{{N}_{{A}}}{RT}$ ${N}=\frac{{PV}}{{kT}}$ $={N}=1.5 \\times 10^{11}$

Q3JEE Main 2021MCQ4MElectrostatic Potential and Capacitance

The relative permittivity of distilled water is 81 . The velocity of light in it will be : $($ Given $\mu_{{r}}=1)$

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📖 Explanation

${V}={{c}}/{\sqrt{\mu_{{r}} \epsilon_{{r}}}}$ $=3.33 \times 10^{7} {m} /{s}$

Q4JEE Main 2021MCQ4MRotational Motion

List-I List-II (a) Moment of Inertia(MI) of the rod (length L, Mass M, about an axis ⊥ to the rod passing through the midpoint) (i) $8 ML^2/3$ (b) Moment of Inertia(MI) of the rod (length L, Mass 2M, about an axis ⊥ to the rod passing through one of its end) (ii) $ML^2/3$ (c) Moment of Inertia(MI) of the rod (length 2L, Mass M, about an axis ⊥ to the rod passing through its midpoint) (iii) $ML^2/12$ (d) Moment of Inertia(MI) of the rod (Length 2L, Mass 2M, about an axis ⊥ to the rod passing through one of its end) (iv) $2 ML^2/3$ Choose the correct answer from the options given below :

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Q5JEE Main 2021MCQ4MCentre of Mass and Collision

Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. The masses of A, B and C are m, 2 m and 2 m respectively. A moves towards B with a speed of 9 m/s and makes an elastic collision with it. Thereafter B makes a completely inelastic collision with C. All motions occur along same straight line. The final speed of C is :

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📖 Explanation

Collision between A and B ${m} \\times 9={mv}_{1}+2 {m} {v}_{2}$ (from momentum conservation) ${e}=1=\frac{{v}_{2}-{v}_{1}}{9}$ $\Rightarrow{v}_{2}=6 {m} / {sec}, {v}_{1}=-3 {m} / {sec}$ collision between B and $C$ $2m \times 6 = 4mv$ (from momentum conservation) $v = 3$ m/s

Q6JEE Main 2021MCQ4MElectrostatic Potential and Capacitance

A capacitor of capacitance ${C}=1 \mu{F}$ is suddenly connected to a battery of 100 volt through a resistance ${R}=100 \Omega.$ The time taken for the capacitor to be charged to get $50 {V}$ is : $[$ Take $\ln 2=0.69]$

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📖 Explanation

${V}={V}_{0}(1-{e}^{-\frac{{t}}{{RC}}})$ $50=100(1-{e}^{-\frac{{t}}{{RC}}})$ ${t}=0.69 \\times 10^{-4} {sec}$

Q7JEE Main 2021MCQ4MElectrostatic Potential and Capacitance

In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be : (Given area of plate = A)

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📖 Explanation

$\frac{1}{C_{e f f}}=\frac{d}{K \epsilon_{0} A}+\frac{2 d}{3 K \epsilon_{0} A}+\frac{3 d}{5 K \epsilon_{0} A}$ $C_{\text{eff }}=\frac{15 K \epsilon_{0} A}{34 d}$

Q8JEE Main 2021MCQ4MRotational Motion

The figure shows two solid discs with radius R and r respectively. If mass per unit area is same for both, what is the ratio of MI of bigger disc around axis AB (Which is ⊥ to the plane of the disc and passing through its centre) of MI of smaller disc around one of its diameters lying on its plane? Given 'M' is the mass of the larger disc. (MI stands for moment of inertia)

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📖 Explanation

Ratio of moment of inertia $=\frac{\frac{1}{2} {MR}^{2}}{\frac{1}{4} {mr}^{2}}$ $=\frac{2 \sigma\piR^{2} R^{2}}{\sigma\pir^{2} r^{2}}=\frac{2 R^{4}}{r^{4}}$

Q9JEE Main 2021MCQ4MWave Optics

In Young's double slit experiment, if the source of light changes from orange to blue then :

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Q10JEE Main 2021MCQ4MThermodynamics

In the reported figure, there is a cyclic process ABCDA on a sample of 1 mol of a diatomic gas. The temperature of the gas during the process ${A} \\to {B}$ and ${C} \\to {D}$ are ${T}_{1}$ and ${T}_{2}({T}_{1}>{T}_{2})$ respectively. Choose the correct option out of the following for work done if processes ${BC}$ and ${DA}$ are adiabatic.

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📖 Explanation

Work done in adiabatic process $=\frac{-{nR}}{\gamma-1}({T}_{{f}}-{T}_{{i}})$ $\therefore {W}_{{AD}}=\frac{-{nR}}{\gamma-1}({T}_{2}-{T}_{1})$ and ${W}_{{BC}}=\frac{-{nR}}{\gamma-1}$ {T}{2}-{T}{1})\therefore {W}{{AD}}={W}{{BC}}$

Q11JEE Main 2021MCQ4MCentre of Mass and Collision

Assertion ${A}:$ If ${A}, {B}, {C}, {D}$ are four points on a semi-circular arc with centre at 'O' such that $|{\vec{AB}}|=|{\vec{BC}}|=|{\vec{CD}}|$ , then ${{AB}}^{\to }+{\vec{AC}}+{\vec{AD}}=4 {\vec{AO}}+{\vec{OB}}+{{OC}}^{\to }$ Reason R : Polygon law of vector addition yields ${{AB}}^{\to }+{{BC}}^{\to }+{\vec{CD}}+{\vec{AD}}=2{{AO}}^{\to }$ In the light of the above statements, choose the most appropriate answer from the options given below :

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Q12JEE Main 2021MCQ4MCentre of Mass and Collision

A light cylindrical vessel is kept on a horizontal surface. Area of base is A. A hole of crosssectional area 'a' is made just at its bottom side. The minimum coefficient of friction necessary to prevent sliding the vessel due to the impact force of the emerging liquid is (a < < A) :

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📖 Explanation

For no sliding ${f} \\ge {\rho av}^{2}$ $\mu{mg} \\ge \rho {av}^{2}$ $\mu\rho {Ahg} \\ge \rho {a} 2 {gh}$ $\mu\\ge \frac{2 {a}}{{A}}$ Option (3)

Q13JEE Main 2021MCQ4MOscillations

A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy ${E}$ . At any instant, its kinetic energy is $\frac{3 {E}}{4}$ then its displacement 'y' is given by:

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📖 Explanation

${E}=\frac{1}{2} {Ka}^{2}$ $\frac{3 {E}}{4}=\frac{1}{2} {K}({a}^{2}-{y}^{2})$ $\frac{3}{4} \\times \frac{1}{2} {Ka}^{2}=\frac{1}{2} {K}({a}^{2}-{y}^{2})$ ${y}^{2}={a}^{2}-\frac{3 {a}^{2}}{4}$ ${y}=\frac{{a}}{2}$

Q14JEE Main 2021MCQ4MAtoms and Nuclei

If ' $f$ denotes the ratio of the number of nuclei decayed $({N}_{{d}})$ to the number of nuclei at ${t}=0({N}_{0})$ then for a collection of radioactive nuclei, the rate of change of $f'$ with respect to time is given as : $[\lambda$ is the radioactive decay constant]

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📖 Explanation

${N}={N}_{0} {e}^{-\lambdat}$ ${N}_{{d}}={N}_{0}-{N}$ ${N}_{{d}}={N}_{0}(1-{e}^{-\lambdat})$ ${{N}_{{d}}}/{{N}_{0}}={f}=1-{e}^{-\lambda{t}}$ $\frac{{df}}{{dt}}=\lambda{e}^{-\lambda{t}}$

Q15JEE Main 2021MCQ4MElectrostatic Potential and Capacitance

Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be :

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📖 Explanation

${V}_{{C}}=\frac{2 {CV}+{CV}}{{KC}+2 {C}}$ $=\frac{3 {V}}{{K}+2}$

Q16JEE Main 2021MCQ4MMotion in a Straight Line

A ball is thrown up with a certain velocity so that it reaches a height ' $h$ '. Find the ratio of the two different times of the ball reaching $\frac{{h}}{3}$ in both the directions.

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📖 Explanation

${u}=\sqrt{2 {gh}}$ Now, ${S}=\frac{{h}}{3} \;\;\;\;\;{a}=-g$ ${S}={ut}+\frac{1}{2} {at}^{2}$ $\frac{{h}}{3}=\sqrt{2 {gh}} {t}+\frac{1}{2}(-{g}) {t}^{2}$ $t^{2}(\frac{g}{2})-\sqrt{2 g h} t+\frac{h}{3}=0$ From quadratic equation $t_{1}, t_{2}={\sqrt{2 g h} \\pm \sqrt{2 g h-\frac{4 g}{2} \frac{h}{3}}}/{g}$ $\frac{{t}_{1}}{{t}_{2}}={\sqrt{2 {gh}}-\sqrt{\frac{4 {gh}}{3}}}/{\sqrt{2 {gh}}+\sqrt{\frac{4 {gh}}{3}}}$ $=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

Q17JEE Main 2021MCQ4MAlternating Current

A $0.07 {H}$ inductor and a $12 \Omega$ resistor are connected in series to a $220 {V}, 50 {Hz}$ ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take $\pi$ as $\frac{22}{7}$ $]$

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📖 Explanation

$\phi=\tan ^{-1}({X_{{L}}}/{{R}})$ ${X}_{{L}}=\omega {L}$ ${X}_{{L}}=2 \\times \frac{22}{7} \\times 50 \\times 0.07=22 \Omega$ $\phi=\tan ^{-1}(\frac{22}{12}) \;\;\;{R}=12 \Omega$ $\phi=\tan ^{-1}(\frac{11}{6})$ ${Z}=\sqrt{{X}_{{L}}^{2}+{R}^{2}}=25.059$ ${I}=\frac{{V}}{{Z}}=\frac{220}{25.059}=8.77 {A}$

Q18JEE Main 2021MCQ4MElectrostatics

Two identical tennis balls each having mass 'm' and charge ' $q$ ' are suspended from a fixed point by threads of length $'l '$ . What is the equilibrium separation when each thread makes a small angle $' \theta'$ with the vertical ?

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📖 Explanation

${T\cos} \theta=m g$ ${T\sin} \theta=\frac{{kq}^{2}}{{x}^{2}}$ $\tan \theta=\frac{{kq}^{2}}{{x}^{2} {mg}}$ as $\tan \theta\approx \sin \theta\approx \frac{{x}}{2 {L}}$ $\frac{{x}}{2 { L}}=\frac{{Kq}^{2}}{{x}^{2} {mg}}$ ${x}=(\frac{{q}^{2} {L}}{2 \pi\epsilon_{0} {mg}})^{1 / 3}$

Q19JEE Main 2021MCQ4MUnits and Measurements

Assertion A : If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is 5 mm and there are 50 total divisions on circular scale, then least count is 0.001 cm. Reason R : Least Count $=\frac{\text{Pitch }}{\text{Total divisionson circular scale }}$ In the light of the above statements, choose the most appropriate answer from the options given below :

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📖 Explanation

Least Count $=\frac{\text{Pitch }}{\text{Total divisionson circular scale }}$ In 5 revolution, distance travel, 5 mm In 1 revolution, it will travel 1 mm. So least count $=\frac{1}{50}=0.02$

Q20JEE Main 2021MCQ4MThermodynamics

A body takes 4 min. to cool from 61° C to 59°C. If the temperature of the surroundings is 30°C, the time taken by the body to cool from 51°C to 49° C is :

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📖 Explanation

$\frac{\Delta{T}}{\Delta{t}}={K}({T}_{{t}}-{T}_{{s}})$ ${T}_{{t}}=$ average temp ${T}_{{S}}=$ surrounding temp. $\frac{61-59}{4}={K}(\frac{61+59}{2}-30)$ .........(1) $\frac{51-49}{{t}}={K}(\frac{51+49}{2}-30)$ .........(2) Divide (1) & (2) $\frac{{t}}{4}=\frac{60-30}{50-30}=\frac{30}{20}$ so, ${t}=6$ minutes

Q21JEE Main 2021NAT4MCurrent Electricity

Consider an electrical circuit containing a two way switch ' ${S}'$ . Initially ${S}$ is open and then ${T}_{1}$ isconnected to ${T}_{2}$ . As the current in ${R}=6 \Omega$ attains amaximum value of steady state level, ${T}_{1}$ is disconnected from ${T}_{2}$ and immediately connectedto ${T}_{3}$ . Potential drop across ${r}=3 \Omega$ resistorimmediately after ${T}_{1}$ is connected to ${T}_{3}$ is _____V. (Round off to the Nearest Integer)

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📖 Explanation

When ${T}_{1}$ and ${T}_{2}$ are connected, then the steady state current in the inductor ${I}=\frac{6}{6}=1 {A}$ When ${T}_{1}$ and ${T}_{3}$ are connected then current through inductor remains same. So potential difference across $3 \Omega$ ${V}={Ir}=1 \\times 3=3 \text{volt }$

Q22JEE Main 2021NAT4MGravitation

Suppose two planets (spherical in shape) of radii R and 2R, but mass M and 9 M respectively have a centre to centre separation 8 R as shown in the figure. A satellite of mass 'm' is projected from the surface of the planet of mass 'M' directly towards the centre of the second planet. The minimum speed 'v' required for the satellite to reach the surface of the second planet is $\sqrt{a/7 \text{GM}/ R$ then the value of 'a' is ________. [Given : The two planets are fixed in their position]

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📖 Explanation

Assume that at a distance $x$ from the planet of mass $M$ , the net gravitational field becomes zero. $\therefore \;\;\frac{G M}{x^{2}}=\;\frac{G \times 9 M}{(8 R-x)^{2}}$ $\Rightarrow \;\frac{1}{x^{2}}=\;\frac{9}{(8 R-x)^{2}}$ $\Rightarrow (\;\frac{1}{x})=(\;\frac{3}{8 R-x})^{2} \Rightarrow \;\frac{1}{x}=\;\frac{3}{8 R-x}$ $\Rightarrow 3 x=8 R-x$ $\Rightarrow 4 x=8 R$ $\Rightarrow x=2 R \;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$ Now, a satellite should be projected in such a way that its covers a minimum distance of $2 R$ . $\therefore \;\frac{1}{2} m v^{2} \;\; - \;\frac{G M m}{R}-\;\frac{G(9 M) m}{7 R}$ $\;\; =\;\frac{-G M m}{2 R}-\;\frac{G(9 M) m}{6 R}$ where, $m$ is the mass of satellite. $\Rightarrow \;\; \;\frac{1}{2} v^{2}=\;\frac{2 G M}{7 R} \Rightarrow v=\sqrt{\;\frac{4}{7} \;\frac{G M}{R}} \;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$ According to question, the minimum speed $v$ required for the satellite to reach the surface of the second planet is $\sqrt{\;\frac{a G M}{7 R}}$ . So, on comparing it with Eq. (ii), we can write $a=4$

Q23JEE Main 2021NAT4MAtoms and Nuclei

In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius $0.5 \Å$ . If the speed of electron is $2.2 \\times 16^{6}{m} / {s}$ , then the current associated with the electron will be _______ $\\times 10^{-2} {mA} .[$ Take $\pi$ as $\frac{22}{7}$ ]

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📖 Explanation

${I}=\frac{{e}}{{T}}=\frac{{e} \omega}{2 \pi}=\frac{{eV}}{2 \pi{r}}$ ${I}=\frac{1.6 \\times 10^{-19} \\times 2.2 \\times 10^{6} \\times 7}{2 \\times 22 \\times 0.5 \\times 10^{-10}}$ $=1.12 {mA}$ $112 \\times 10^{-2} {mA}$

Q24JEE Main 2021NAT4MAtoms and Nuclei

A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance 200 µF is first charged and later connected with resistor 'R'. If the ratio of charge on capacitor to the activity of radioactive sample is fixed with respect to time then the value of 'R' should be ______Ω.

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📖 Explanation

${T}_{{m}}=30 {ms}$ ${C}=200 \mu{F}$ $\frac{{q}}{{N}}={{Q}_{0} {e}^{-t / {RC}}}/{{N}_{0} {e}^{-\lambda{t}}}=\frac{{Q}_{0}}{{N}_{0}} {e}^{t(\lambda-\frac{1}{{RC}})}$ Since ${q} / {N}$ is constant hence $\lambda=\frac{1}{{RC}}$ ${R}=\frac{1}{\lambda{C}}={{T}_{{m}}}/{{C}}=\frac{30 \\times 10^{-3}}{200 \\times 10^{-6}}$ $=150 \Omega$

Q25JEE Main 2021NAT4MDual Nature of Matter and Radiation

A particle of mass $9.1 \\times 10^{-31} {kg}$ travels in a medium with a speed of $10^{6} {m} / {s}$ and a photon of aradiation of linear momentum $10^{-27} {kg} {m} / {s}$ travelsin vacuum. The wavelength of photon is _______ times the wavelength of the particle.

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📖 Explanation

For photon $\lambda_{1}=\frac{{h}}{{P}}=\frac{6.6 \\times 10^{-34}}{10^{-27}}$ For particle $\lambda_{2}=\frac{{h}}{{mv}}=\frac{6.6 \times 10^{-34}}{9.1 \\times 10^{-31} \\times 10^{6}}$ $\therefore \frac{\lambda_{1}}{\lambda_{2}}=910$

Q26JEE Main 2021NAT4MRay Optics and Optical Instruments

A prism of refractive index $n_{1}$ and another prism of refractive index ${n}_{2}$ are stuck together (as shown in the figure). ${n}_{1}$ and ${n}_{2}$ depend on $\lambda$ , the wavelength of light, according to the relation ${n}_{1}=1.2+\frac{10.8 \\times 10^{-14}}{\lambda^{2}}$ and ${n}_{2}=1.45+\frac{1.8 \\times 10^{-14}}{\lambda^{2}}$ The wavelength for which rays incident at any angle on the interface ${BC}$ pass through without bending at that interface will be ________ ${nm}$ .

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📖 Explanation

For no bending, ${n}_{1}={n}_{2}$ $1.2+\frac{10.8 \\times 10^{-14}}{\lambda^{2}}=1.45+\frac{1.8 \\times 10^{-4}}{\lambda^{2}}$ On solving, $9 \\times 10^{-14}=25 \lambda^{2}$ $\lambda=6 \\times 10^{-7}$ $\lambda=600 {nm}$

Q27JEE Main 2021NAT4MProperties of Matter

A stone of mass $20 {g}$ is projected from a rubber catapult of length $0.1 {m}$ and area of cross section $10^{-6} {m}^{2}$ stretched by an amount $0.04 {m}$ . The velocity of the projected stone is __________ ${m} / {s}$ (Young's modulus of rubber $=0.5 \\times 10^{9} {N} / {m}^{2}$ )

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📖 Explanation

By energy conservation $\frac{1}{2} \.\frac{{YA}}{{L}} \.{x}^{2}=\frac{1}{2} {mv}^{2}$ $\frac{0.5 \\times 10^{9} \\times 10^{-6} \\times (0.04)^{2}}{0.1}$ $=\frac{20}{1000} {v}^{2}$ $\therefore {v}^{2}=400$ ${v}=20 {m} / {s}$

Q28JEE Main 2021NAT4MSemiconductor Electronics

A transistor is connected in common emitter circuit configuration, the collector supply voltage is 10 V and the voltage drop across a resistor of 1000 Ω in the collector circuit is 0.6 V. If the current gain factor (beta) is 24, then the base current is _______ µA. (Round off to the Nearest Integer)

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📖 Explanation

$\beta={{I}_{{C}}}/{{I}_{{B}}}=24 ;$ ${R}_{{C}}=1000$ $\Delta{V}=0.6$ ${I}_{{C}}=\frac{0.6}{1000}$ ${I}_{{C}}=6 \\times 10^{-4}$ ${I}_{{B}}={{I}_{{C}}}/{\beta}=\frac{6 \\times 10^{-4}}{24}$ $=25 \mu{A}$

Q29JEE Main 2021NAT4MWave Optics

The amplitude of upper and lower side bands of A.M. wave where a carrier signal with frequency 11.21 MHz, peak voltage 15 V is amplitude modulated by a 7.7 kHz sine wave of 5V amplitude are $a/10V$ and $b/10V$ respectively. Then the value of $a/ b$ is ___________

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📖 Explanation

$\frac{{a}}{10}=\frac{{b}}{10}={\mu{A}_{{C}}}/{2}$ $\Rightarrow\frac{{a}}{{b}}=1$

Q30JEE Main 2021NAT4MMagnetic Effects of Current and Magnetism

In a uniform magnetic field, the magnetic needle has a magnetic moment $9.85 \\times 10^{-2} {A} / {m}^{2}$ and moment of inertia $5 \\times 10^{-6} {kg} {m}^{2}$ . If it performs 10 complete oscillations in 5 seconds then themagnitude of the magnetic field is _______ ${m} {T}$ [Take $\pi^{2}$ as $9.85]$

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📖 Explanation

${T}=2 \pi\sqrt{\frac{{I}}{{MB}}}$ ${B}=80 \\times 10^{-4}=8 {mT}$

Chemistry30 questions

Q31JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Which one of the following compounds will give orange precipitate when treated with 2,4-dinitrophenyl hydrazine ?

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Q32JEE Main 2021MCQ4MElectrochemistry

The product obtained from the electrolytic oxidation of acidified sulphate solutions, is :

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📖 Explanation

Electrolysis of concentrated solution of acidified sulphate solution yields ${H}_{2} {S}_{2} {O}_{8}$

Q33JEE Main 2021MCQ4MSolid State

The parameters of the unit cell of a substance are ${a}=2.5, {b}=3.0, {c}=4.0, \alpha=90°,$ $\beta=120° \gamma=90°.$ The crystal system of the substance is :

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📖 Explanation

${a} \\ne {b} \\ne {c}$ and $\alpha=\gamma=90°\ne \beta$ are parameters of monoclinic unit cell.

Q34JEE Main 2021MCQ4Mp Block Elements

The oxidation states of ' ${P}'$ in ${H}_{4} {P}_{2} {O}_{7}, {H}_{4} {P}_{2} {O}_{5}$ and ${H}_{4} {P}_{2} {O}_{6}$ , respectively, are:

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📖 Explanation

Oxidation state of ${P}$ in ${H}_{4} {P}_{2} {O}_{7}, {H}_{4} {P}_{2} {O}_{5}$ and ${H}_{4} {P}_{2} {O}_{6}$ is $5,3\ \&\ 4$ respectively ${H}_{4} {P}_{2} {O}_{7}$ $2 x+4(+1)+7(-2)=0$ ${x}=+5$ ${H}_{4} {P}_{2} {O}_{5}$ $2 x+4(+1)+5(-2)=0$ ${x}=+3$ ${H}_{4} {P}_{2} {O}_{6}$ $2 x+4(+1)+6(-2)=0$ ${x}=+4$

Q35JEE Main 2021MCQ4MChemical Kinetics

For a reaction of order ${n}$ , the unit of the rate constant is :

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📖 Explanation

Rate $={k}[{A}]^{{n}}$ comparing units $\frac{(m o l / l)}{\sec }={k}(\frac{{mol}}{l})^{{n}}$ $\Rightarrow{k}={mol}^{(1-{n})} l^{({n}-1)} {s}^{-1}$

Q36JEE Main 2021MCQ4MAmines

Given below are two statements : Statement I : Aniline is less basic than acetamide. Statement II : In aniline, the lone pair of electrons on nitrogen atom is delocalised over benzene ring due to resonance and hence less available to a proton. Choose the most appropriate option ;

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Q37JEE Main 2021MCQ4MCoordination Compounds

The type of hybridisation and magnetic property of the complex $[{MnCl}_{6}]^{3-}$ , respectively, are:

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📖 Explanation

$[{MnCl}_{6}]^{3-}$ Paramagnetic and having 4 unpaired electrons.

Q38JEE Main 2021MCQ4MCoordination Compounds

The number of geometrical isomers found in the metal complexes $[{PtCl}_{2}({NH}_{3})_{2}]$ , $[{Ni}({CO})_{4}], \;\;\;[{Ru}({H}_{2} {O})_{3} {Cl}_{3}]$ and $[{CoCl}_{2}({NH}_{3})_{4}]^{+}$ respectively, are :

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Q39JEE Main 2021MCQ4MSome Basic Concepts of Chemistry

Which one of the following statements is NOT correct ?

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📖 Explanation

Eutrophication leads to decrease in oxygen level of water. $3^{\text{rd}}$ statement is incorrect

Q40JEE Main 2021MCQ4MStructure of Atom

Given below are two statements : Statement I : Rutherford's gold foil experiment cannot explain the line spectrum of hydrogen atom. Statement II : Bohr's model of hydrogen atom contradicts Heisenberg's uncertainty principle. In the light of the above statements, choose the most appropriate answer from the options given below :

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Q41JEE Main 2021MCQ4MChemical Equilibrium

Presence of which reagent will affect the reversibility of the following reaction, and change it to a irreversible reaction: ${CH}_{4}+{I}_{2} {⇌}_{\text{ Reversible }} ^{{hv}} {CH}_{3}-{I}+{HI}$

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📖 Explanation

lodination of alkane is reversible reaction. It can be irreversible in the presence of strong oxidising agent like conc. $HNO_3$ or conc. $HIO_3$

Q42JEE Main 2021MCQ4MBiomolecules

Which one among the following chemical tests is used to distinguish monosaccharide from disaccharide ?

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Q43JEE Main 2021MCQ4MBiomolecules

Match List-I with List-II : List-I (Drug) List-II (Class of Drug) (a) Furacin (i) Antibiotic (b) Arsphenamine (ii) Tranquilizers (c) Dimetone (iii) Antiseptic (d) Valium (iv) Synthetic antihistaminesChoose the most appropriate match :

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Q44JEE Main 2021MCQ4MChemical Thermodynamics

The statement that is INCORRECT about Ellingham diagram is

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Q45JEE Main 2021MCQ4MAlcohols, Phenols and Ethers

Consider the above reaction and identify the Product P :

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Q46JEE Main 2021MCQ4MBiomolecules

The compound 'A' is a complementary base of __________ in DNA stands.

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Q47JEE Main 2021MCQ4MOrganic Chemistry - Some Basic Principles

Staggered and eclipsed conformers of ethane are :

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Q48JEE Main 2021MCQ4MIonic Equilibrium

Match List - I with List - II : List-I (Drug) List-II (Class of Drug) (a) NaOH (i) Acidic (b) $Be(OH)_3$ (ii) Basic (c) $Ca(OH)_3$ (iii) Amphoteric (d) $B(OH)_3$ (e) $Al(OH)_3$ Choose the most appropriate answer from the options given below

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📖 Explanation

${NaOH} \\to$ Basic ${Be}({OH})_{2} \\to$ Amphoteric ${Ca}({OH})_{2} \\to$ Basic ${B}({OH})_{3} \\to$ Acidic ${Al}({OH})_{3} \\to$ Amphoteric

Q49JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

The correct order of stability of given carbocation is :

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Q50JEE Main 2021MCQ4Mp Block Elements

Given below are two statements : One is labelled as Assertion A and the other labelled as Reason R. Assertion A : Lithium halides are some what covalent in nature. Reason R : Lithium possess high polarisation capability. In the light of the above statements, choose the most appropriate answer from the options given below:

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Q51JEE Main 2021NAT4MIonic Equilibrium

The density of ${\text{NaOH}}$ solution is $1.2 {g} {cm}^{-3}$ . The molality of this solution is ________ ${m}$ . (Round off to the Nearest Integer) [Use : Atomic masses : ${Na}: 23.0 {u} \;\;{O}: 16.0 {u}$ ${H}: 1.0 {u}$ Density of ${H}_{2} {O}: 1.0 {g} {cm}^{-3}]$

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📖 Explanation

Consider $1 l$ solution mass of solution $=(1.2 \\times 1000) {g}$ $=1200 {gm}$ Neglecting volume of ${NaOH}$ Mass of water $=1000 {gm}$ $\Rightarrow$ Mass of ${NaOH}=(1200-1000) {gm}$ $=200 {gm}$ $\Rightarrow$ Moles of ${NaOH}=\frac{200 {g}}{50 {g} / {mol}}=5 {mol}$ $\Rightarrow{\text{molality}}=\frac{5 {mol}}{1 {kg}}=5 {m}$

Q52JEE Main 2021NAT4MChemical Equilibrium

${CO}_{2}$ gas adsorbs on charcoal following Freundlich adsorption isotherm. For a given amount of charcoal, the mass of ${CO}_{2}$ adsorbed becomes 64 times when the pressure of ${CO}_{2}$ is doubled. The value of ${n}$ in the Freundlich isotherm equation is _________ $\\times 10^{-2}$ . (Round off to the Nearest Integer)

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📖 Explanation

Freundlich isotherm. ; $\frac{{x}}{{m}}={k} \.{p}^{\frac{1}{{n}}}$ Substituting values ; $(\frac{64}{1})=(2)^{\frac{1}{n}} \$ $\Rightarrow {n}=\frac{1}{6}=0.166$ $\cong 17 \\times 10^{-2}$

Q53JEE Main 2021NAT4MIonic Equilibrium

The conductivity of a weak acid HA of concentration $0.001 {mol} {L}^{-1}$ is $2.0 \\times 10^{-5} {S} {cm}^{-1}$ . If $\Lambda_{{m}}^{{o}}({HA})=190 {S} {cm}^{2} {mol}^{-1},$ the ionization constant $({K}_{{a}})$ of ${HA}$ is equal to ________ $\\times 10^{-6}$ . (Round off to the Nearest Integer)

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📖 Explanation

$\Lambda_{{m}}=1000 \\times \frac{{K}}{{M}}$ $=1000 \\times \frac{2 \\times 10^{-5}}{0.001}=20 {S} {cm}^{2} {mol}^{-1}$ $\Rightarrow\alpha={\Lambda_{{m}}}/{\Lambda_{{m}}^{\\infty }}=\frac{20}{190}=(\frac{2}{19})$ ${HA} \⇌ {H}^{+}+{A}^{-}$ $0.001(1-\alpha) 0.001 \alpha0.001 \alpha$ $\Rightarrow {k}_{{a}}=0.001(\frac{\alpha^{2}}{1-\alpha})$ $=\frac{0.001 \\times (\frac{2}{19})^{2}}{1-(\frac{2}{19})}$ $=12.3 \\times 10^{-6}$

Q54JEE Main 2021NAT4MSolutions

$1.46 {g}$ of a biopolymer dissolved in a $100 {mL}$ water at $300 {K}$ exerted an osmotic pressure of $2.42 \\times 10^{-3}$ bar The molar mass of the biopolymer is _____ $-10^{4} {g}$ ${mol}^{-1}$ . (Round off to the Nearest Integer) [Use : ${R}=0.083 {L}$ bar ${mol}^{-1} {K}^{-1}]$

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📖 Explanation

$\pi= CRT$ $\pi=$ osmotic pressure ${C}=$ molarity ${T}=$ Temperature of solution let the molar mass be M gm / mol $2.42 \\times 10^{-3}$ bar $=$ ${(\frac{1.46 {g}}{{Mgm} / {mol}})}/{0.1 l} \\times (\frac{0.083 l-{bar}}{{mol}-{K}})$ $\\times (300 {K})$ $\Rightarrow{M}=15.02 \\times 10^{4} {g} / {mol}$

Q55JEE Main 2021NAT4MHaloalkanes and Haloarenes

An organic compound is subjected to chlorination to get compound A using 5.0 g of chlorine. When 0.5 g of compound A is reacted with $AgNO_3$ [Carius Method], the percentage of chlorine in compound A is _______ when it forms 0.3849 g of AgCl. (Round off to the Nearest Integer) (Atomic masses of Ag and Cl are 107.87 and 35.5 respectively)

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📖 Explanation

${n}_{{c} l}$ in compound $={n}_{{AgCl}}=\frac{0.3849 {g}}{(107.87+35.5)} {g} / {mol}$ \Rightarrow mass of chlorine $={n}_{{Cl}} \\times 35.5=0.0953 {gm}$ \Rightarrow $\%$ wt of chlorine $=\frac{0.0953}{0.5} \\times 100$ $= 19.06%$

Q56JEE Main 2021NAT4MCoordination Compounds

The number of geometrical isomers possible in triamminetrinitrocobalt (III) is X and in trioxalatochromate (III) is Y. Then the value of X + Y is _______.

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📖 Explanation

Triamminetrinitrocobalt(III) $\\to [{Co}({NO}_{2})_{3}({NH}_{3})_{3}]$ trioxalatochromate(III) ion $\\to [{Cr}({C}_{2} {O}_{4})_{3}]^{3-}$ $[{Co}({NO}_{2})_{3}({NH}_{3})_{3}]$ $X + Y = 2 + 0 = 2.0$

Q57JEE Main 2021NAT4MAmines

In gaseous triethyl amine the "-C-N-C-" bond angle is ________ degree.

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Q58JEE Main 2021NAT4MChemical Thermodynamics

For water at $100°{C}$ and 1 bar, $\Delta_{\text{vap }} H-\Delta_{\text{vap }} U=\ ─── \\times 10^{2} { J} \ { mol}^{-1} .$ (Round off to the Nearest Integer) [Use : ${R}=8.31 {J} {mol}^{-1} {K}^{-1}]$ [Assume volume of ${H}_{2} {O}({l})$ is much smaller than volume of ${H}_{2} {O}({g}) .$ Assume ${H}_{2} {O}({g})$ treated as an ideal gas]

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📖 Explanation

${H}_{2} {O}_{(l)} \⇌{H}_{2} {O}_{({v})}$ $\Delta{H}=\Delta{U}+\Delta{n}_{{g}} {RT}$ for 1 mole waters ; $\Delta{n}_{{g}}=1$ $\therefore \Delta{n}_{{g}} {RT}=1 {mol} \\times 8.31 {J} / {mol}$ $-{k} \\times 373 {K}$ $=3099.63 {J} \cong 31 \\times 10^{2} {J}$

Q59JEE Main 2021NAT4MChemical Equilibrium

${PCl}_{5} \⇌{PCl}_{3}+{Cl}_{3} \;\;\;{K}_{{c}}=1.844$ $3.0$ moles of ${PCl}_{5}$ is introduced in a $1 {L}$ closed reaction vessel at $380 {K}$ . The number of moles of ${PCl}_{5}$ at equilibrium is _______ $\\times 10^{-3}$ (Round off to the Nearest Integer)

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📖 Explanation

${PCl}_{5(g)} \⇌{PCl}_{3(g)}+{Cl}_{2(g)} \;\;\;{K}_{{c}}=1.844$ $\Rightarrow\frac{[{PCl}_{3}][{Cl}_{2}$$}{[{PCl}_{5}]}=\frac{{x}^{2}}{3-{x}}=1.844$ $\Rightarrow{x}^{2}+1.844-5.532=0$ $\Rightarrow {x}={-1.844+\sqrt{(1.844)^{2}+4 \\times 5.532}}/{2}$ $\cong 1.604$ $\Rightarrow$ Moles of ${PCl}_{5}=3-1.604 \cong 1.396$

Q60JEE Main 2021NAT4MChemical Bonding and Molecular Structure

The difference between bond orders of ${CO}$ and ${NO}^{\oplus }$ is $\frac{{x}}{2}$ where ${x}=$ _______. (Round off to the Nearest Integer)

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📖 Explanation

Bond order of ${\text{CO}}=3$ $\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2$ B. O. $=\;\frac{10-4}{2}=3$ ${\text{NO}}^{+}:$ B. O. $=\;\frac{8-2}{2}=3$ Difference $=3-3=0$

Mathematics30 questions

Q61JEE Main 2021MCQ4MStatistics

If the mean and variance of the following data: $6,10,7,13, {a}, 12, {b}, 12$ are 9 and $\frac{37}{4}$ respectively, then $(a-b)^{2}$ is equal to:

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📖 Explanation

Mean $=\frac{6+10+7+13+{a}+12+{b}+12}{8}=9$ $60+{a}+{b}=72$ ${a}+{b}=12$ .......(1) variance $={\sum{x}_{{i}}^{2}}/{{n}}-({\sum{x}_{{i}}}/{{n}})^{2}=\frac{37}{4}$ $\sum{x}_{{i}}^{2}=6^{2}+10^{2}+7^{2}$ $+13^{2}+{a}^{2}+{b}^{2}+12^{2}+12^{2}$ $={a}^{2}+{b}^{2}+642$ $\frac{{a}^{2}+{b}^{2}+642}{8}-(9)^{2}=\frac{37}{4}$ $\frac{{a}^{2}+{b}^{2}}{8}+\frac{321}{4}-81=\frac{37}{4}$ $\frac{{a}^{2}+{b}^{2}}{8}=81+\frac{37}{4}-\frac{321}{4}$ $\frac{{a}^{2}+{b}^{2}}{8}=81-71$ $\therefore {a}^{2}+{b}^{2}=80$ .......(2) From $(1) {a}^{2}+{b}^{2}+2 {ab}=144$ $80+2 {ab}=144 \;\;\therefore 2 {ab}=64$ $({a}-{b})^{2}={a}^{2}+{b}^{2}-2 {ab}$ $=80-64=16$

Q62JEE Main 2021MCQ4MLimits, Continuity and Differentiability

The value of $\lim _{n \\to \\infty } \frac{1}{n} \sum_{j=1} ^{n} \frac{(2 j-1)+8 n}{(2 j-1)+4 n}$ is equal to :

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📖 Explanation

$\lim _{n \\to \\infty } \frac{1}{n} \sum_{j=1} ^{n} \frac{(\frac{2 j}{n}-\frac{1}{n}+8)}{(\frac{2 j}{n}-\frac{1}{n}+4)}$ $\int_{0} ^{1} \frac{2 x+8}{2 x+4} d x=int_{0} ^{1} \ d x+int_{0} ^{1} \frac{4}{2 x+4} d x$ $=1+4{ \frac{1}{2}(\ln |2 x+4|)|}_{0} ^{1}$ $=1+2 \ln(\frac{3}{2})$

Q63JEE Main 2021MCQ4MVector Algebra

Let $\vec{a}={\hat{i}}+\hat{j}+2 {\hat{k}}$ and ${{\vec{b}}}=-\hat{i}+2 \hat{j}+3 \hat{k}$ . Then the vector product $({\vec{a}}+{\vec{b}}) \\times (({\vec{a}} \\times (({\vec{a}}-{{\vec{b}}}) \\times {\vec{b}}))$ $\\times {\vec{b}})$ is equal to:

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📖 Explanation

$\vec{a}={\hat{i}}+\hat{j}+2 {\hat{k}}$ ${{\vec{b}}}=-\hat{i}+2 \hat{j}+3 \hat{k}$ . $\vec{a}+\vec{b}=3 \hat{j}+5 \hat{k} ;$ ${a} ^{\to }\.\vec{b}=-1+2+6=7$ $((\vec{a} \\times ((\vec{a}-\vec{b}) \\times {\vec{b}})) \\times \vec{b})$ $(({a} ^{\to }\\times ({\vec{a}} \\times \vec{b}-{\vec{b}} \\times \vec{b})) \\times \vec{b})$ $(\vec{a} \\times ({\vec{a}} \\times {\vec{b}}-0)) \\times \vec{b}$ $(\vec{a} \\times ({a} ^{\to }\\times \vec{b})) \\times \vec{b}$ $(({a} ^{\to }\.{\vec{b}}) {\vec{a}}-({\vec{a}} \.{\vec{a}}) {\vec{b}}) \\times {\vec{b}}$ $(\vec{a} \.\vec{b}) \vec{a} \\times \vec{b}-({a} ^{\to }\.\vec{a})(\vec{b} \\times \vec{b})$ $({a} ^{\to }\.{\vec{b}})(\vec{a} \\times {\vec{b}})$ $\vec{a} \\times \vec{b}=\begin{vmatrix}i ^{\^} & j ^{\^} & k ^{\^} \\ 1 & 1 & 2 \\ -1 & 2 & 3\end{vmatrix}$ $=-\hat{i}-5 \hat{j}+3 \hat{k}$ $\therefore 7(-\hat{i}-5 \hat{j}+3 \hat{k})$ $(\vec{a}+\vec{b}) \\times (7(-\hat{i}-5 \hat{j}+3 \hat{k}))$ $7(0 \hat{i}+3 \hat{j}+5 \hat{k}) \\times (-\hat{i}-5 \hat{j}+3 \hat{k})$ $\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & 5 \\ -1 & -5 & 3\end{vmatrix}$ $\Rightarrow34 \hat{i}-(5) \hat{j}+(3 \hat{k})$ $\Rightarrow34 \hat{i}-5 \hat{j}+3 \hat{k}$ $\therefore 7(34 \hat{i}-5 \hat{j}+3 \hat{k})$

Q64JEE Main 2021MCQ4MDefinite Integrals

The value of the definite integral $\int_{-\frac{\pi}{4}} ^{\frac{\pi}{4}} \frac{d x}{(1+e^{x \cos x})(\sin ^{4} x+\cos ^{4} x)}$ is equal to:

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📖 Explanation

$\int_{-\frac{\pi}{4}} ^{\frac{\pi}{4}} \frac{d x}{(1+e^{x \cos x})(\sin ^{4} x+\cos ^{4} x)}$ .........(1) Using $\int_{a} ^{b} f(x) d x=\int_{a} ^{b} f(a+b-x) d x$ $I=int_{-\frac{\pi}{4}} ^{\frac{\pi}{4}}\frac{d x}{(1+e^{-x \cos x})(\sin ^{4} x+\cos ^{4} x)}$ ........(2) Add (1) and (2) 2I $=int_{-\frac{\pi}{4}} ^{\frac{\pi}{4}} \frac{d x}{\sin ^{4} x+\cos ^{4} x}$ $2 I=2 int_{0} ^{\frac{\pi}{4}}\frac{d x}{\sin ^{4} x+\cos ^{4} x}$ $I= int_{0} ^{\frac{\pi}{4}} \frac{(1+\tan ^{2} x) \sec ^{2} x}{\tan ^{4} x+1} d x$ $I= int_{0} ^{\frac{\pi}{4}}(1+\frac{1}{\tan ^{2} x}) \sec ^{2} \frac{x}{(\tan x-\frac{1}{\tan x})^{2}+2} d x$ $\tan x-\frac{1}{\tan x}=t$ $(1+\frac{1}{\tan ^{2} x}) \sec ^{2} x d x=d t$ $I=\int_{-\\infty } ^{0} \frac{d t}{t^{2}+2}=\$ [\frac{1}{\sqrt{2}} \tan ^{-1}(\frac{t}{\sqrt{2}})]_{-\∞ }^{0}{I}=0-\frac{1}{\sqrt{2}}(-\frac{\pi}{2})=\frac{\pi}{2 \sqrt{2}}$

Q65JEE Main 2021MCQ4MComplex Numbers

Let ${C}$ be the set of all complex numbers. Let ${S}_{1}=\{{z} \in{C}|| {z}-3-2 {i}|^{2}=8\}$ ${S}_{2}=\{{z} \in{C} \|{Re}({z}) \\ge 5\}$ and ${S}_{3}=\{{z} \in{C}|| {z}-\ov {{z}} \\ge 8\}$ Then the number of elements in ${S}_{1} \cap{S}_{2} \cap{S}_{3}$ is equal to

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📖 Explanation

${S}_{1}:| {z}-3-2 {i}|^{2}=8$ $|{z}-3-2 {i}|=2 \sqrt{2}$ $({x}-3)^{2}+({y}-2)^{2}=(2 \sqrt{2})^{2}$ ${S}_{2}: {x} \\ge 5$ ${S}_{3}:| {z}-\overline{{z}}| \\ge 8$ $|2 {iy}| \\ge 8$ $2|{y}| \\ge 8 \therefore {y} \\ge 4, {y} \\le -4$ $n{S}_{1} \cap{S}_{2} \cap{S}_{3}=1$

Q66JEE Main 2021MCQ4MArea Under The Curves

If the area of the bounded region ${R}=\{({x}, {y}): \max \{0, \log _{{e}} {x}\} \\le {y} \\le 2^{{x}},$ $\frac{1}{2} \\le {x} \\le 2\}$ is, $\alpha(\log _{{e}} 2)^{-1}+\beta(\log _{{e}} 2)+\gamma$ , then the value of $(\alpha+\beta-2 \gamma)^{2}$ is equal to :

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📖 Explanation

${R}=\{({x}, {y}): \max \{0, \log _{{e}} {x}\} \\le {y} \\le 2^{{x}},$ $\frac{1}{2} \\le {x} \\le 2\}$ $\int_{\frac{1}{2}} ^{2} 2^{x} d x-\int_{1} ^{2} ln x d x$ $\Rightarrow\$ [\frac{2^{x}}{\ln 2}]{1 / 2}^{2}-[x \ln x-x]{1}^{2}\Rightarrow{(2^{2})-2^{1 / 2}}/{\log _{{e}} 2}-(2 \ln 2-1)\Rightarrow{(2^{2}-\sqrt{2})}/{\log _{{e}} 2}-2 \ln 2+1\therefore \alpha=2^{2}-\sqrt{2}, \beta=-2, \gamma=1\Rightarrow(\alpha+\beta+2 \gamma)^{2}\Rightarrow(2^{2}-\sqrt{2}-2-2)^{2}\Rightarrow(\sqrt{2})^{2}=2$

Q67JEE Main 2021MCQ4MEllipse

A ray of light through $(2,1)$ is reflected at a point ${P}$ on the ${y}$ -axis and then passes through the point $(5,3)$ . If this reflected ray is the directrix of an ellipse with eccentricity $\frac{1}{3}$ and the distance of the nearer focus from this directrix is $\frac{8}{\sqrt{53}}$ , then the equation of the other directrix can be:

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📖 Explanation

Equation of reflected Ray $y-1=\frac{2}{7}(x+2)$ $7 y-7=2 x+4$ $2 x-7 y+11=0$ Let the equation of other directrix is $2 x-7 y+\lambda$ Distance of directrix from Focub $\frac{{a}}{{e}}-{ae}=\frac{8}{\sqrt{53}}$ $3 a-\frac{a}{3}=\frac{8}{\sqrt{53}}$ or $a=\frac{3}{\sqrt{53}}$ Distance from other focus $\frac{{a}}{{e}}+{ae}$ $3 {a}+\frac{{a}}{3}=\frac{10 {a}}{3}=\frac{10}{3} \\times \frac{3}{\sqrt{53}}$ $=\frac{10}{\sqrt{53}}$ Distance between two directrix $=\frac{2 {a}}{{e}}$ $=2 \\times 3 \\times \frac{3}{\sqrt{53}}=\frac{18}{\sqrt{53}}$ $|\frac{\lambda-11}{\sqrt{53}}|=\frac{18}{\sqrt{53}}$ $\lambda-11=18$ or $-18$ $\lambda=29$ or $-7$ $2 x-7 y-7=0$ or $2 x-7 y+29=0$

Q68JEE Main 2021MCQ4MBinomial Theorem

If the coefficients of $x^{7}$ in $(x^{2}+\frac{1}{b x})^{11}$ and $x^{-7}$ in $(x-\frac{1}{b x^{2}})^{11}, b \\ne 0$ , are equal, then the value of $b$ is equal to:

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📖 Explanation

Coefficient of ${x}^{7}$ in $({x}^{2}+\frac{1}{{bx}})^{11}$ ${ }^{11} C_{r}(x^{2})^{11-r} \.(\frac{1}{b x})^{r}$ ${ }^{11} {C}_{{r}} {x}^{22-3 {r}} \.{1}/{{b}^{{r}}}$ $22-3 r=7$ ${r}=5$ $\therefore { }^{11} {C}_{5} \.\frac{1}{{b}^{5}} \.{x}^{7}$ Coefficient of $x^{-7}$ in $(x-\frac{b}{b x^{2}})^{11}$ ${ }^{11} {C}_{{r}}({x})^{11-{r}} \.(-\frac{1}{{bx}^{2}})^{{r}}$ ${ }^{11} {C}_{{r}} {x}^{11-3 {r}} \.{(-1)^{{r}}}/{{b}^{{r}}}$ $11-3 {r}=-7 \therefore {r}=6$ ${ }^{11} {C}_{6} \.\frac{1}{{b}^{6}} {x}^{-7}$ ${ }^{11} {C}_{5} \.\frac{1}{{b}^{5}}={ }^{11} {C}_{6} \. \frac{1}{{b}^{6}}$ Since ${b} \\ne 0 \;\;\therefore {b}=1$

Q69JEE Main 2021MCQ4MSets and Relations

The compound statement $(P \vee Q) \wedge (\sim P) \RightarrowQ$ is equivalent to:

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📖 Explanation

Using Truth Table $P$ $Q$ $P∨Q$ $P$ $(P∨Q)∧P$ $(P∨Q)∧∼P\to Q$ T T T F F T T F T F F T F T T T T T F F F T F T $P$ $Q$ $∼Q$ $P∧ ∼Q$ $P\to Q$ $∼(P\to Q)$ T T F F T F T F T T F T F F T F T F $\sim (P \\to Q)$ ${P} \wedge \sim {Q}$ $\sim ({P} \\to {Q}) \⇔ {P} \wedge \sim {Q}$ F F T T T T F F T F F T

Q70JEE Main 2021MCQ4MTrigonometric Ratios and Identities

If $\sin \theta+\cos \theta=\frac{1}{2}$ , then16 $(\sin (2 \theta)+\cos (4 \theta)+\sin (6 \theta))$ is equal to:

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📖 Explanation

$\sin \theta+\cos \theta=\frac{1}{2}$ $\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta=\frac{1}{4}$ $\sin 2 \theta=-\frac{3}{4}$ Now : $\cos 4 \theta=1-2 \sin ^{2} 2 \theta$ $=1-2(-\frac{3}{4})^{2}$ $=1-2 \\times \frac{9}{16}=-\frac{1}{8}$ $\sin 6 \theta=3 \sin 2 \theta-4 \sin ^{3} 2 \theta$ $=(3-4 \sin ^{2} 2 \theta) \. \sin 2 \theta$ $=[3-4(\frac{9}{16})] \.(-\frac{3}{4})$ $\Rightarrow[\frac{3}{4}] \\times (-\frac{3}{4})$ $=-\frac{9}{16}$ $16[\sin 2 \theta+\cos 4 \theta+\sin 6 \theta]$ $16(-\frac{3}{4}-\frac{1}{8}-\frac{9}{16})=-23$

Q71JEE Main 2021MCQ4MMatrices and Determinants

Let ${A}=\begin{bmatrix}1 & 2 \\ -1 & 4\end{bmatrix} .$ If ${A}^{-1}=\alpha{I}+\beta {A}, \alpha, \beta \ \in {R}, {I}$ is a $2 \\times 2$ identity matrix, then $4(\alpha-\beta)$ is equal to :

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📖 Explanation

${A}=\begin{bmatrix}1 & 2 \\ -1 & 4\\end{bmatrix},{|{A}|}=6$ ${A}^{-1}=\frac{{adj} {A}}{|{A}|}=\frac{1}{6}\begin{bmatrix}4 & -2 \\ 1 & 1\\end{bmatrix}$ $=\begin{bmatrix}\frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6}\end{bmatrix}$ $\begin{bmatrix}\frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6}\end{bmatrix}=\begin{bmatrix}\alpha & 0 \\ 0 & \alpha\end{bmatrix}$ $+\begin{bmatrix}\beta & 2 \beta \\ -\beta & 4 \beta\end{bmatrix}$ $\text{table} \alpha+\beta=\frac{2}{3} ; \beta=-\frac{1}{6} \} \$ $\Rightarrow \alpha=\frac{2}{3}+\frac{1}{6}=\frac{5}{6}$ $4(\alpha-\beta)=4(1)=4$

Q72JEE Main 2021MCQ4MLimits, Continuity and Differentiability

Let ${f}:(-\frac{\pi}{4}, \frac{\pi}{4}) \\to {R}$ be defined as $f(x)=\{\text{table} (1+|\sin x|)^{\frac{3 a}{|\sin x|}} ,{ , },-\frac{\pi}{4} < x < 0 ; b , {,}, x=0;e^{\cot 4 x / \cot 2 x} ,{,} , 0 < x < \frac{\pi}{4}$ If ${f}$ is continuous at ${x}=0$ , then the value of $6 {a}+{b}^{2}$ is equal to:

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📖 Explanation

$\lim _{x \\to 0} f(x)=b$ $\lim _{x \\to 0^{+}} x e^{\frac{\cot 4 x}{\cot 2 x}}=e^{\frac{1}{2}}=b$ $\lim _{x \\to 0^{-}}(1+|\sin x|)^{\frac{3 a}{{|\sin} x \|}}=e^{3 a}=e^{\frac{1}{2}}$ $\lim _{x \\to 0^{-}}(1+|\sin x|)^{\frac{3 a}{|\sin x \|}}=e^{3 a}=e^{\frac{1}{2}}$ $a=\frac{1}{6} \Rightarrow6 a=1$ $(6 a+b^{2})=(1+e)$

Q73JEE Main 2021MCQ4MDifferential Equations

Let ${y}={y}({x})$ be solution of the differential equation $\log _{e}(\frac{d y}{d x})=3 x+4 y$ , with $y(0)=0$ If ${y}(-\frac{2}{3} \log _{{e}} 2)=\alpha\log _{{e}} 2$ , then the value of $\alpha$ is equal to:

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📖 Explanation

$\frac{d y}{d x}=e^{3 x} \.e^{4 y} \Rightarrow\inte^{-4 y} d y$ $=\int e^{3 x} d x$ $\frac{e^{-4 y}}{-4}={e^{3 x}}{3}+C \Rightarrow-\frac{1}{4}-\frac{1}{3}=C \$ $\Rightarrow C=-\frac{7}{12}$ $\frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}-\frac{7}{12} \$ $\Rightarrow e^{-4 y}=\frac{4 e^{3 x}-7}{-3}$ $e^{4 y}=\frac{3}{7-4 e^{3 x}} \Rightarrow4 y=\ln(\frac{3}{7-4 e^{3 x}})$ $4 y=\ln(\frac{3}{6})$ when $x=-\frac{2}{3} \ln 2$ $y=\frac{1}{4} \ln(\frac{1}{2})=-\frac{1}{4} \text{l}{n} 2$

Q74JEE Main 2021MCQ4MThree Dimensional Geometry

Let the plane passing through the point $(-1,0,-2)$ and perpendicular to each of the planes $2 {x}+{y}-{z}=2$ and $x-y-z=3$ be $a x+b y+c z+8=0$ . Then the value of ${a}+{b}+{c}$ is equal to:

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📖 Explanation

Normal of req. plane $(2 \hat{i}+{\hat{j}}-{\hat{k}}) \\times ({\hat{i}}-{\hat{j}}-{\hat{k}})$ $=-2 \hat{i}+{\hat{j}}-3 {\hat{k}}$ Equation of plane $-2(x+1)+1(y-0)-3(z+2)=0$ $-2 x+y-3 z-8=0$ $2 x-y+3 z+8=0$ ${a}+{b}+{c}=4$

Q75JEE Main 2021MCQ4MCircles

Two tangents are drawn from the point ${P}(-1,1)$ to the circle $x^{2}+y^{2}-2 x-6 y+6=0$ . If these tangents touch the circle at points ${A}$ and ${B}$ , and if ${D}$ is a point on the circle such that length of the segments ${AB}$ and ${AD}$ are equal, then the area of the triangle ABD is equal to:

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📖 Explanation

$\Delta{ABD}=\frac{1}{2} \\times 2 \\times 4$ $=4$

Q76JEE Main 2021MCQ4MLimits, Continuity and Differentiability

Let ${f}: {R} \\to {R}$ be a function such that ${f}(2)=4$ and ${f}'(2)=1 .$ Then, the value of $\lim _{x \\to 2} \frac{x^{2} f(2)-4 f(x)}{x-2}$ is equal to :

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📖 Explanation

Apply L'Hopital Rule $\lim _{x \\to 2}(\frac{2 x f(2)-4 f'(x)}{1})$ $=\frac{4(4)-4}{1}=12$

Q77JEE Main 2021MCQ4MCircles

Let ${P}$ and ${Q}$ be two distinct points on a circle which has center at ${C}(2,3)$ and which passes through origin ${O}$ . If ${OC}$ is perpendicular to both the line segments ${CP}$ and ${CQ}$ , then the set $\{{P}, {Q}\}$ is equal to

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📖 Explanation

$\tan \theta=-\frac{2}{3}$ Using symmetric from of line ${P}, {Q}:(2 \\pm \sqrt{13} \cos \theta, 3 \\pm \sqrt{13} \sin \theta)$ $(2 \\pm \sqrt{13} \.(-\frac{3}{\sqrt{13}}),$ $3 \\pm \sqrt{13}(\frac{2}{\sqrt{13}}))$ $(-1,5)\ \& \ (5,1)$

Q78JEE Main 2021MCQ4MQuadratic Equation and Inequalities

Let $\alpha, \beta$ be two roots of the equation ${x}^{2}+(20)^{1 / 4} {x}+(5)^{1 / 2}=0$ . Then $\alpha^{8}+\beta^{8}$ is equal to

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📖 Explanation

$\;\; x^{2+}(20)^{\;\frac{1}{4}} \cdot x+(5)^{\;\frac{1}{2}}=0$ . $\text{roots} \alpha \; \& \;\beta .$ $\alpha+\beta=-(20)^{\;\frac{1}{4}}$ $\alpha \beta=(5)^{\;\frac{1}{2}} .$ $\alpha^{8+}\beta^8= (\alpha^4) ^{2+} (\beta^4) ^2$ $= (\alpha^{4-}\beta^4) ^{2+}2(\alpha \beta)^4 . \;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$ $\Rightarrow \;\; (\alpha+\beta)^2= (\alpha^{2+}\beta^2) +2 \alpha \beta .$ $\Rightarrow \;\; (20)^{\;\frac{1}{2}}= (\alpha^{2+}\beta^2) +2 \cdot 5^{\;\frac{1}{2}}$ $\Rightarrow 2 \cdot (5)^{\;\frac{1}{2}}= (\alpha^{2+}\beta^2) +2 \cdot 5^{\;\frac{1}{2}}$ $\Rightarrow \;\; 0 = (\alpha^{2+}\beta^2)$ From eqn (1) $\alpha^{8+}p^8\;= ( (\alpha^{2+}p^2) \cdot (\alpha^{2-}\beta^2) ) ^{2+}2 \cdot (5)^{1/2}$ $\;=0+2 \times 5^2$ $\;=2 \times 25$ $\;=50 \;\text{ (Ans) }\;$

Q79JEE Main 2021MCQ4MProbability

The probability that a randomly selected 2 -digit number belongs to the set $\{{n} \in{N}:(2^{{n}}-2).$ is a multiple of 3 $\}$ is equal to

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📖 Explanation

Total number of cases $={ }^{90} {C}_{1}=90$ Now, $2^{{n}}-2=(3-1)^{{n}}-2$ ${ }^{n} C_{0} 3^{n}-{ }^{n} C_{1} \. 3^{n-1}+\.....$ $+(-1)^{n-1} \.{ }^{n} C_{n-1} 3+(-1)^{n} \.{ }^{n} C_{n}-2$ $3(3^{{n}-1}-{n} 3^{{n}-2}+\......+(-1)^{{n}-1} \. {n})+(-1)^{{n}}-2$ $(2^{{n}}-2)$ is multiply of 3 only when ${n}$ is odd Req. Probability $=\frac{45}{90}=\frac{1}{2}$

Q80JEE Main 2021MCQ4MCircles

Let ${A}=\{({x}, {y}) \in{R} \\times {R} \|2 {x}^{2} +2 {y}^{2}-2 {x}$ $-2 {y}=1\}$ ${B}=\{({x}, {y}) \in{R} \\times {R} \|4 {x}^{2}+4 {y}^{2}-16 {y}$ $+7=0\}$ and ${C}=\{({x}, {y}) \in{R} \\times {R} \|{x}^{2}+{y}^{2}-4 {x}-2 {y}$ $+5 \\le {r}^{2}\}$ Then the minimum value of $|{r}|$ such that ${A} \cup{B} \subseteq{C}$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

${S}_{1}: {x}^{2}+{y}^{2}-{x}-{y}-\frac{1}{2}=0 ;$ ${C}_{1}(\frac{1}{2}, \frac{1}{2})$ ${r}_{1}=\sqrt{\frac{1}{4}+\frac{1}{4}+\frac{1}{2}}=1$ ${S}_{2}: {x}^{2}+{y}^{2}-4 {y}+\frac{7}{4}=0 ;$ ${C}_{2}:(0,2)$ ${r}_{2}=\sqrt{4-\frac{7}{4}}=\frac{3}{2}$ ${ S}_{3}: {x}^{2}+{y}^{2}-4 {x}-2 {y}+5-{r}^{2}=0$ ${C}_{3}:(2,1)$ ${r}_{3}=\sqrt{4+1-5+{r}^{2}}=|{r}|$ ${C}_{1} {C}_{3}=\sqrt{\frac{5}{2}}$ $\sqrt{\frac{5}{2}} \\le { |r-1|} \$ $\Rightarrow \text{table} r \\le 1+\sqrt{\frac{5}{2}}; r \\ge \frac{3}{2}+\sqrt{5} \}$ ${C}_{2} {C}_{3}=\sqrt{5} \\le |{r}-\frac{3}{2}|$ $\text{table} {r}-\frac{3}{2} \\ge \sqrt{5} ; {r}-\frac{3}{2} \\le -\sqrt{5}\ \}$

Q81JEE Main 2021NAT4MMatrices and Determinants

For real numbers $\alpha$ and $\beta$ , consider the following system of linear equations : $x+y-z=2, x+2 y+\alphaz=1,$ $2 x-y+z=\beta$ If the system has infinite solutions, then $\alpha+\beta$ is equal to ________ .

🚀 Solve in Practice Mode

📖 Explanation

For infinite solutions $\Delta=\Delta_{1}=\Delta_{2}=\Delta_{3}=0$ $\Delta=\begin{vmatrix}1 & 1 & -1 \\ 1 & 2 & \alpha \\ 2 & -1 & 1\end{vmatrix}=0$ $\Delta=\begin{vmatrix}3 & 0 & 0 \\ 1 & 2 & \alpha \\ 2 & -1 & 1\end{vmatrix}=0$ $\Delta=3(2+\alpha)=0$ $\Rightarrow\alpha=-2$ $\Delta_{2}=\begin{vmatrix}1 & 2 & -1 \\ 1 & 1 & -2 \\ 2 & \beta & 1\end{vmatrix}=0$ $1(1+2 \beta)-2(1+4)-(\beta-2)=0$ $\beta-7=0$ $\beta=7$ $\therefore \alpha+\beta=5$ Ans.

Q82JEE Main 2021NAT4MVector Algebra

Let ${\vec{a}}={\hat{i}}+{\hat{j}}+{\hat{k}},\vec{b}$ and $\vec{c}=\hat{j}-\hat{k}$ be three vectors such that $\vec{a} \\times \vec{b}=\vec{c}$ and $\vec{a} \. \vec{b}=1$ . If the length of projection vector of the vector $\vec{b}$ on the vector $\vec{a} \\times {\vec{c}}$ is $l$ , then the value of $3 l^{2}$ is equal to _______.

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📖 Explanation

${a} ^{\to }\\times \vec{b}=\vec{c}$ Take Dot with $\vec{c}$ $({{a}} ^{\to }\\times {\vec{b}}) {\vec{c}}=|\vec{c}|^{2}=2$ Projection of ${\vec{b}}$ or ${{\vec{a}}} \\times {{\vec{c}}}=\text{l}$ ${|{\vec{b}} \.({{\vec{a}}} \times {\vec{c}})|}/{|{\vec{a}} \\times {\vec{c}}|}=\text{l}$ $\therefore \text{l}=\frac{2}{\sqrt{6}} \Rightarrow\text{l}^{2}=\frac{4}{6}$ $3 \text{l}^{2}=2$

Q83JEE Main 2021NAT4MLogarithms

If $\log _{3} 2, \log _{3}(2^{x}-5), \log _{3}(2^{x}-\frac{7}{2})$ are in an arithmetic progression, then the value of ${x}$ is equal to _________.

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📖 Explanation

$2 \log _{3}(2^{x}-5)=\log _{3} 2+\log _{3}(2^{x}-\frac{7}{2})$ Let $2^{{x}}={t}$ $\log _{3}(t-5)^{2}=\log _{3} 2(t-\frac{7}{2})$ $(t-5)^{2}=2 t-7$ ${t}^{2}-12 {t}+32=0$ $(t-4)(t-8)=0$ $\Rightarrow2^{{x}}=4$ or $2^{{x}}=8$ ${X}=2$ (Rejected) Or ${x}=3$

Q84JEE Main 2021NAT4MLogarithms

Let the domain of the function $f(x)=\log _{4}(\log _{5}(\log _{3}(18 x-x^{2}-77)))$ be $(a, b)$ Then the value of the integral $\int_{a} ^{b} \frac{\sin ^{3} x}{(\sin ^{3} x+\sin ^{3}(a+b-x))} d x$ is equal to ________.

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📖 Explanation

For domain $\log _{5}(\log _{3}(18 x-x^{2}-77))>0$ $\log _{3}(18 x-x^{2}-77)>1$ $18 x-x^{2}-77>3$ $x^{2}-18 x+80 < 0$ $x \ \in(8,10)$ $\Rightarrowa=8$ and $b=10$ $I=\int_{a} ^{b} \frac{\sin ^{3} x}{\sin ^{3} x+\sin ^{3}(a+b-x)} d x$ $I=int_{a} ^{b} \frac{\sin ^{3}(a+b-x)}{\sin ^{3} x+\sin ^{3}(a+b-x)}$ $2 I=(b-a) \RightarrowI=\frac{b-a}{2}$ $(\becausea=8$ and $b=10)$ $I=\frac{10-8}{2}=1$

Q85JEE Main 2021NAT4MTrigonometric Ratios and Identities

Let ${f}({x})=\begin{vmatrix}\sin ^{2} {x} & -2+\cos ^{2} {x} & \cos 2 {x} \\ 2+\sin ^{2} {x} & \cos ^{2} {x} & \cos 2 {x} \\ \sin ^{2} {x} & \cos ^{2} {x} & 1+\cos 2 {x}\\end{vmatrix},$ ${x} \in[0, \pi]$ Then the maximum value of ${f}({x})$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

$\begin{vmatrix}-2 & -2 & 0 \\ 2 & 0 & -1 \\ \sin ^{2} x & \cos ^{2} x & 1+\cos 2 x\end{vmatrix}$ $( {R}_{1} \\to {R}_{1}-{R}_{2} \ & \ {R}_{2} \\to {R}_{2}-{R}_{3} )$ $-2(\cos ^{2} x)+2(2+2 \cos 2 x+\sin ^{2} x)$ $4+4 \cos 2 x-2(\cos ^{2} x-\sin ^{2} x)$ $f(x)=4+{2 \cos 2 x} _{\max =1}$ $f(x)_{\max }=4+2=6$

Q86JEE Main 2021NAT4MDifferential Equations

Let ${F}:[3,5] \\to {R}$ be a twice differentiable function on $(3,5)$ such that ${F}({x})={e}^{-{x}} \int_{3} ^{{x}}(3 {t}^{2}+2 {t}+4 {F}'({t})) {dt}$ If $F'(4)=\frac{\alpha {e}^{\beta}-224}{({e}^{\beta}-4)^{2}}$ , then $\alpha+\beta$ is equal to _________.

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📖 Explanation

$F(3)=0$ ${e}^{{x}} {F}({x})=\int_{3} ^{{x}}(3 {t}^{2}+2 {t}+4 {F}'({t})) {dt}$ ${e}^{{x}} {F}({x})+{e}^{{x}} {F}'({x})=3 {x}^{2}+2 {x}+4 {F}'({x})$ $(e^{x}-4) \frac{d y}{d x}+e^{x} y=(3 x^{2}+2 x)$ $\frac{d y}{d x}+\frac{e^{x}}{(e^{x}-4)} y=\frac{(3 x^{2}+2 x)}{(e^{x}-4)}$ $y e^{\int \frac{e^{x}}{(e^{x}-4)} d x}=\int\frac{(3 x^{2}+2 x)}{(e^{x}-4)} e^{\int\frac{e^{x}}{e^{x}-4} d x} d x$ $y \.(e^{x}-4)=\int(3 x^{2}+2 x) d x+c$ $y(e^{x}-4)=x^{3}+x^{2}+c$ Put $x=3 \Rightarrow c=-36$ $F(x)=\frac{(x^{3}+x^{2}-36)}{(e^{x}-4)}$ $F'(x)=\frac{(3 x^{2}+2 x)(e^{x}-4)-(x^{3}+x^{2}-36) e^{x}}{(e^{x}-4)^{2}}$ Now put value of ${x}=4$ we will get $\alpha=12 \ \&\ \beta=4$

Q87JEE Main 2021NAT4MThree Dimensional Geometry

Let a plane P pass through the point $(3,7,-7)$ and contain the line, $\frac{x-2}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ . If distance of the plane ${P}$ from the origin is ${d}$ , then ${d}^{2}$ is equal to _________.

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📖 Explanation

${{BA}}^{\to }=(\hat{i}+4 {\hat{j}}-5 \hat{k})$ ${\vec{BA}}=(\hat{i}+4 \hat{j}-5 \hat{k})$ ${\vec{BA}} \\times \vec{l}={\vec{n}}=$ $\begin{vmatrix}{\hat{i}} & \hat{j} & \hat{k} \\ -3 & 2 & 1 \\ 1 & 4 & -5\end{vmatrix}$ ${a\hat{i}}+{{b}\hat{j}}+{c} {\hat{k}}=-14 \hat{i}-{\hat{j}}(14)+\hat{k}(-14)$ ${a}=1, {b}=1, {c}=1$ Plane is $({x}-2)+({y}-3)+({z}+2)=0$ ${x}+{y}+{z}-3=0$ ${d}=\sqrt{3} \Rightarrow {d}^{2}=3$

Q88JEE Main 2021NAT4MFunctions

Let ${S}=\{1,2,3,4,5,6,7\} .$ Then the number of possible functions ${f}: {S} \\to {S}$ such that ${f}({m} \.{n})={f}({m}) \.{f}({n})$ for every ${m}, {n} \in{S}$ and ${m} \.{n} \in{S}$ is equal to ______.

🚀 Solve in Practice Mode

📖 Explanation

$F(mn) = f(m). f(n)$ Put $m = 1 f(n) = f(1). f(n) \Rightarrow f(1) = 1$ Put $m = n = 2$ $f(4)=f(2) . f(2)\begin{cases}f(2)=1 \Rightarrowf(4)=1 \\ \text{or } \\ f(2)=2 \Rightarrowf(4)=4$ Put m = 2 & n = 3 $f(6)=f(2)\end{cases} f(3)\{\text{table} \text{when }f(2)=1 ; f(3)=1 \text{to }7;,; f(2)=2 f(3)=1 \text{or } 2 \text{or } 3$ ${f}(5), {f}(7)$ can take any value Total $=(1 \\times 1 \\times 7 \\times 1 \\times 7 \\times 1 \\times 7)$ $+(1 \\times 1 \\times 3 \\times 1 \\times 7 \\times 1 \\times 7)$ $= 490$

Q89JEE Main 2021NAT4MDifferential Equations

If ${y}={y}({x}), {y} \in[0, \frac{\pi}{2})$ is the solution of the differential equation $\sec y \frac{d y}{d x}-\sin (x+y)-\sin (x-y)=0$ , with $y(0)=0$ then $5 y'(\frac{\pi}{2})$ is equal to _________.

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📖 Explanation

$\sec y \frac{d y}{d x}=2 \sin x \cos y$ $\sec ^{2} y d y=2 \sin x d x$ $\tan y=-2 \cos x+c$ $c=2$ $\tan y=-2 \cos x+2 \Rightarrow$ at $x=\frac{\pi}{2}$ $\tan y=2$ $\sec ^{2} y \frac{d y}{d x}=2 \sin x$ $5 \frac{d y}{d x}=2$

Q90JEE Main 2021NAT4MLimits, Continuity and Differentiability

Let ${f}:[0,3] \\to {R}$ be defined by $f(x)=\min \{x-[x], 1+[x]-x\}$ where $[{x}]$ is the greatest integer less than or equal to $x$ . Let $P$ denote the set containing all $x \in[0,3]$ where f is discontinuous, and $Q$ denote the set containing all ${x} \in(0,3)$ where ${f}$ is not differentiable. Then the sum of number of elements in ${P}$ and ${Q}$ is equal to _________.

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📖 Explanation

Non differentiable at ${x}=\frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}$

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