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25 Jul 2021 Shift 1

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Physics30 questions

Q1JEE Main 2021MCQ4MThermodynamics

For a gas ${C}_{{P}}-{C}_{{V}}={R}$ in a state ${P}$ and ${C}_{{P}}-{C}_{{V}}=1.10 {R}$ in a state ${Q}, {T}_{{P}}$ and ${T}_{{Q}}$ are the temperatures in two different states ${P}$ and ${Q}$ respectively. Then

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📖 Explanation

${C}_{{P}}-{C}_{{V}}={R}$ for ideal gas and gas behaves as ideal gas at high temperature so ${T}_{{P}}>{T}_{{Q}}$

Q2JEE Main 2021MCQ4MRotational Motion

Given below are two statements: one is labelled as Assertion ${A}$ and the other is labelled as Reason ${R}$ . Assertion A: Moment of inertia of a circular disc of mass ' ${M}$ ' and radius 'R' about ${X}, {Y}$ axes (passing through its plane) and Z-axis which is perpendicular to its plane were found to be $I_{x}, I_{y}$ and ${I}_{{z}}$ respectively. The respective radii of gyration about all the three axes will be the same. Reason R : A rigid body making rotational motion has fixed mass and shape. In the light of the above statements, choose the most appropriate answer from the options given below:

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📖 Explanation

${I}_{{z}}={I}_{{x}}+{I}_{{y}}$ (using perpendicular axis theorem) $\& \ {I}={mk}^{2}({K}:$ radius of gyration $)$ so ${mK}_{{z}}^{2}={m} {K}_{{x}}^{2}+{mK}_{{y}}^{2}$ ${K}_{{z}}^{2}={K}_{{x}}^{2}+{K}_{{y}}^{2}$ so radius of gyration about axes $x, y\ \&\ z$ won't be same hense asseration ${A}$ is not correct reason ${R}$ is correct statement (property of a rigid body)

Q3JEE Main 2021MCQ4MDual Nature of Matter and Radiation

What should be the order of arrangement of de-Broglie wavelength of electron $(\lambda_{{e}})$ , an $\alpha$ -particle $(\lambda_{\alpha})$ and proton $(\lambda_{{p}})$ given that all have the same kinetic energy ?

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📖 Explanation

$\lambda=\frac{{h}}{{p}}={{h}}/{\sqrt{2 {mE}}} \propto{1}/{\sqrt{{m}}}$ ${m}_{\alpha}>{m}_{{p}}>{m}_{{e}}$ so $\lambda_{{e}}>\lambda_{{p}}>\lambda_{\alpha}$

Q4JEE Main 2021MCQ4MSemiconductor Electronics

Identify the logic operation carried out.

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Q5JEE Main 2021MCQ4MDual Nature of Matter and Radiation

A particle of mass $4 {M}$ at rest disintegrates into two particles of mass ${M}$ and $3 {M}$ respectively having non zero velocities. The ratio of de-Broglie wavelength of particle of mass ${M}$ to that of mass $3 {M}$ will be :

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📖 Explanation

$\lambda=\frac{{h}}{{p}}$ both the particles will move with momentum same in magnitude & opposite in direction. So De-Broglie wavelength of both will be same i.e. ratio 1 : 1

Q6JEE Main 2021MCQ4MAtoms and Nuclei

Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as : (where $\lambda$ is the decay constant)

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📖 Explanation

$\frac{3 {N}_{0}}{4}={N}_{0} {e}^{-\lambdat_{1}}$ $\frac{{N}_{0}}{2}={N}_{0} {e}^{-\lambdat_{2}}$ $\ln (3 / 4)=-\lambda{t}_{1} ..........$ (i) $\ln (1 / 2)=-\lambda{t}_{2}$ ..........(ii) $\ln (3 / 4)-\ln (1 / 2)=\lambda({t}_{2}-{t}_{1})$ ........(iii) $\Delta{t}=\frac{\ln (3 / 2)}{\lambda}$

Q7JEE Main 2021MCQ4MMotion in a Plane

Match List I with List II. Choose the correct answer from the options given below :

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📖 Explanation

(a) ${\vec{C}}={\vec{A}}+{\vec{B}}$ Option (iv) (b) ${\vec{A}}={\vec{B}}+{\vec{C}}={\vec{C}}+{\vec{B}}$ Option (iii) (c) ${\vec{B}}=\vec{A}+{\vec{C}}$ Option (i) (d) ${\vec{A}}+{\vec{B}}+\vec{C}=0$ Option (ii)

Q8JEE Main 2021MCQ4MElectrostatic Potential and Capacitance

A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as : $\epsilon({x})=\epsilon_{0}+{kx}$ , for $(0 < {x} \\le \frac{{d}}{2})$ $\epsilon({x})=\epsilon_{0}+{k}({d}-{x})$ , for $(\frac{{d}}{2} \\le {x} \\le {d})$

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📖 Explanation

Taking an element of width ${dx}$ at a distance ${x}({x} < {d} / 2)$ from left plate ${dc}=\frac{(\epsilon_{0}+{kx}) {A}}{{dx}}$ Capacitance of half of the capacitor $\frac{1}{{C}}=\int_{0} ^{{d} / 2} \frac{1}{{dc}}=\frac{1}{{A}} int_{0} ^{{d} / 2} \frac{{dx}}{\epsilon_{0}+{kx}}$ $\frac{1}{{C}}=\frac{1}{{kA}} \ln (\frac{\epsilon_{0}+{kd} / 2}{\epsilon_{0}})$ Capacitance of second half will be same $C_{e q}=\frac{C}{2}={k A}/{2 \ln (\frac{2 \epsilon_{0}+k d}{2 \epsilon_{0}})}$

Q9JEE Main 2021MCQ4MThermodynamics

A monoatomic ideal gas, initially at temperature ${T}_{1}$ is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature ${T}_{2}$ by releasing the piston suddenly. If $l_{1}$ and $l_{2}$ are the lengths of the gas column, before and after the expansion respectively, then the value of $\frac{{T}_{1}}{{T}_{2}}$ will be :

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📖 Explanation

${PV}^{{r}}={\text{const}}$ ${TV}^{{r}-1}={\text{const}} .$ ${T}(l)^{\frac{5}{3}-1}={\text{const}}$ $\frac{{T}_{1}}{{ T}_{2}}=(\frac{l_{2}}{l_{1}})^{2 / 3}$

Q10JEE Main 2021MCQ4MRay Optics and Optical Instruments

A ray of laser of a wavelength 630 nm is incident at an angle of 30° at the diamond-air interface. It is going from diamond to air. The refractive index of diamond is 2.42 and that of air is 1. Choose the correct option.

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📖 Explanation

$\sin \theta_{{C}}=\frac{1}{\mu}=\frac{1}{2 \mu_{2}} < \sin \theta_{{C}}$ $\sin \theta>\sin \theta_{{C}}$ $\theta>\theta_{C}$ Total internal reflection will happen

Q11JEE Main 2021MCQ4MProperties of Matter

Two wires of same length and radius are joined end to end and loaded. The Young's modulii of the materials of the two wires are $Y_{1}$ and $Y_{2}$ . The combination behaves as a single wire then its Young's modulus is :

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📖 Explanation

In series combination $\Delta{l}=l_{1}+l_{2}$ ${Y}=\frac{{F} / {A}}{\Deltal/ l} \Rightarrow\Deltal=\frac{{F} l}{{AY}}$ $\Rightarrow\Deltal\propto\frac{l}{{Y}}$ Equivalent length of rod after joing is $=2 l$ As, lengths are same and force is also same in series $\Deltal=\Deltal_{1}+\Deltal_{2}$ ${l_{{cq}}}/{{Y}_{{cq}}}=\frac{l}{{Y}_{1}}+\frac{l}{{Y}_{2}} \$ $\Rightarrow \frac{2 l}{{Y}}=\frac{l}{{Y}_{1}}+\frac{l}{{Y}_{2}}$ $\therefore {Y}=\frac{2 {Y}_{1} {Y}_{2}}{{Y}_{1}+{Y}_{2}}$

Q12JEE Main 2021MCQ4MAtoms and Nuclei

The half-life of ${ }^{198} {Au}$ is 3 days. If atomic weight of ${ }^{198} {Au}$ is $198 {g} /$ mol then the activity of $2 {mg}$ of ${ }^{198} {Au}$ is [in disintegration/second] :

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📖 Explanation

${A}=\lambda{N}$ $\lambda=\frac{\ln 2}{{t}_{1 / 2}}=\frac{\ln 2}{3 \\times 24 \\times 60 \\times 60} {sec}^{-1}$ $=2.67 \\times 10^{-6} {sec}^{-1}$ ${N}=$ Number of atoms in $2 {mg} \ {Au}$ $=\frac{2 \\times 10^{-3}}{198} \\times 6 \\times 10^{23}$ $=6.06 \\times 10^{15}$ ${A}=\lambda{N}=1.618 \\times 10^{13}$ $=16.18 \\times 10^{12} {dps}$

Q13JEE Main 2021MCQ4MCentre of Mass and Collision

Two billiard balls of equal mass 30 g strike a rigid wall with same speed of 108 kmph (as shown) but at different angles. If the balls get reflected with the same speed then the ratio of the magnitude of impulses imparted to ball 'a' and ball 'b' by the wall along 'X' direction is :

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📖 Explanation

Impulse = change in momentum Ball (a) $|{\Delta{p}}^{\to }|=2 {\mu}={J}_{1}$ Ball (b) $|{{\Deltap}}^{\to }|=2 {\mu} \cos 45°={J}_{2}$ $\frac{{J}_{1}}{{J}_{2}}=\frac{1}{\cos 45°}=\sqrt{2}$

Q14JEE Main 2021MCQ4MWave Optics

In the Young's double slit experiment, the distance between the slits varies in time as ${d}({t})={d}_{0}+{a}_{0} \sin\omegat;$ where ${d}_{0}, \omega$ and ${a}_{0}$ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

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📖 Explanation

Fringe Width, $\beta=\frac{\lambda{D}}{{d}}$ $\beta_{\max } \Rightarrow{d}_{\min }$ and $\beta_{\min } \Rightarrow{d}_{\max }$ ${d}={d}_{0}+{a}_{0} \sin \omega {t}$ ${d}_{\max }={d}_{0}+{a}_{0}$ and ${d}_{\min }={d}_{0}-{a}_{0}$ $\therefore \beta_{\min }=\frac{\lambda{D}}{{d}_{0}+{a}_{0}}$ and $\therefore \beta_{\max }=\frac{\lambda{D}}{{d}_{0}-{a}_{0}}$ $\beta_{\max }-\beta_{\min }=\frac{\lambda{D}}{{d}_{0}-{a}_{0}}-\frac{\lambda{D}}{{d}_{0}+{a}_{0}}$ $=\frac{2 \lambda{Da}_{0}}{{d}_{0}^{2}-{a}_{0}^{2}}$

Q15JEE Main 2021MCQ4MThermodynamics

Two different metal bodies A and B of equal mass are heated at a uniform rate under similar conditions. The variation of temperature of the bodies is graphically represented as shown in the figure. The ratio of specific heat capacities is :

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📖 Explanation

$(\frac{\DeltaQ}{\Deltat})_{A}=(\frac{\DeltaQ}{\Deltat})_{B}$ $m S_{A}(\frac{\DeltaT}{\Deltat})_{{A}}=m S_{B}(\frac{\DeltaT}{\Deltat})_{{B}}$ $\frac{S_{A}}{S_{B}}=\frac{(\frac{\DeltaT}{\Deltat})_{A}}{(\frac{\DeltaT}{\Deltat})_{B}}$ $=\frac{90 / 6}{120 / 3}=\frac{15}{40}=\frac{3}{8}$

Q16JEE Main 2021MCQ4MElectromagnetic Waves

A linearly polarized electromagnetic wave in vacuum is $\mathrm{E}=3.1 \cos \left[(1.8) \mathrm{z}-\left(5.4 \times 10^{6}\right) \mathrm{t}\right]\$ \hat{\mathrm{i}} \mathrm{N} / \mathrm{C} $is incident normally on a perfectly reflecting wall at$ \mathrm{z}=\mathrm{a}$. Choose the correct option

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Q17JEE Main 2021MCQ4MCurrent Electricity

In the given figure, there is a circuit of potentiometer of length ${AB}=10 {m} .$ The resistance per unit length is $0.1 \Omega$ per ${cm}$ . Across ${AB}$ , a battery of emf ${E}$ and internal resistance ' ${r}'$ is connected. The maximum value of emf measured by this potentiometer is:

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📖 Explanation

Max. voltage that can be measured by this potentiometer will be equal to potential drop across ${AB}$ ${R}_{{AB}}=10 \\times 0.1 \\times 100=100 {ohm} .$ ∴ ${V}_{{AB}}=\frac{6}{20+100} \\times 100$ $=6 \\times \frac{100}{120}=5 {V}$

Q18JEE Main 2021MCQ4MElectromagnetic Waves

In amplitude modulation, the message signal ${V}_{{m}}({t})=10 \sin (2 \pi\\times 10^{5} {t})$ volts and Carrier signal ${V}_{{C}}({t})=20 \sin (2 \pi\\times 10^{7} {t})$ volts The modulated signal now contains the message signal with lower side band and upper side band frequency, therefore the bandwidth of modulated signal is $\alpha\ {kHz}$ . The value of $\alpha$ is :

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📖 Explanation

Bandwidth $=2 \\times {f}_{{m}}$ $=2 \\times 10^{5} {HZ}=200 {KHZ}$

Q19JEE Main 2021MCQ4MMotion in a Straight Line

Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at $4^{\text{th }}$ second after its fall to the next droplet is $34.3 {m}$ . At what rate the droplets are coming from the tap ? (Take ${g}=9.8 {m} / {s}^{2}$ )

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📖 Explanation

In 4 sec. $1^{\text{st }}$ drop will travel $\Rightarrow \frac{1}{2} \\times (9.8) \\times (4)^{2}=78.4 {m}$ $\therefore 2^{\text{nd }}$ drop would have travelled $\Rightarrow78.4-34.3=44.1 {m}$ . Time for $2^{\text{nd }}$ drop $\frac{1}{2}(9.8) {t}^{2}=44.1$ $t=3 {sec}$ $\therefore$ each drop have time gap of $1 {sec}$ $\therefore 1$ drop per sec

Q20JEE Main 2021MCQ4MGravitation

The minimum and maximum distances of a planet revolving around the Sun are $x_{1}$ and $x_{2}$ . If the minimum speed of the planet on its trajectory is $\mathrm{v}_{0}$ then its maximum speed will be :

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📖 Explanation

Angular momentum conservation equation ${v}_{0} {x}_{2}={v}_{1} {x}_{1}$ ${v}_{1}=\frac{{v}_{0} {x}_{2}}{{x}_{1}}$

Q21JEE Main 2021NAT4MCentre of Mass and Collision

A body of mass 2 kg moving with a speed of 4 m/s. makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its initial speed. The speed of the two body centre of mass is $x/10$ m / s . Then the value of x is_______.

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📖 Explanation

${p}_{{i}}={p}_{{f}}$ $2 \\times 4=2 \\times 1+{m}_{2} \\times {v}_{2}$ ${m}_{2} {v}_{2}=6$ .......(i) by coefficient of restitution $1=\frac{{v}_{2}-1}{4} \Rightarrow{v}_{2}=5 {m} / {s}$ by (i) ${m}_{2} \\times 5=6$ ${m}_{2}=1.2 {kg}$ ${v}_{{cm}}=\frac{{m}_{1} {v}_{1}+{m}_{2} {v}_{2}}{{m}_{1}+{m}_{2}}$ ${v}_{{cm}}=\frac{2 \\times 1+1.2 \\times 5}{2+1.2}$ $=\frac{8}{3.2}=\frac{25}{10}$ ${x}=25$

Q22JEE Main 2021NAT4MUnits and Measurements

Student A and Student B used two screw gauges of equal pitch and 100 equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is 0.322 cm. The absolute value of the difference between the final circular scale readings observed by the students A and B is ________. [Figure shows position of reference 'O' when jaws of screw gauge are closed] Given pitch = 0.1 cm.

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📖 Explanation

For (A) Reading = MSR + CSR + Error 0.322 = 0.300 + CSR + 5 × LC 0.322 = 0.300 + CSR + 0.005 CSR = 0.017 For B Reading = MSR + CSR + Error 0.322 = 0.200 + CSR + 0.092 CSR = 0.030 Difference = 0.030 - 0.017 = 0.013 cm Division on circular scale $=\frac{0.013}{0.001}=13$

Q23JEE Main 2021NAT4MElectromagnetic Induction

An inductor of $10 {mH}$ is connected to a $20 {V}$ battery through a resistor of $10 {k} \Omega$ and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after $1 \mu{s}$ is $\frac{{x}}{100} {mA} .$ Then ${x}$ is equal to ______. (Take ${e}^{-1}=0.37$ )

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📖 Explanation

${I}_{\max }=\frac{{V}}{{R}}=\frac{20 {V}}{10 {K} \Omega}=2 {mA}$ For LR $-$ decay circuit ${I}={I}_{\max } {e}^{-{R} / {L}}$ ${I}=2 {mAe}^{\frac{-10 \\times 10^{3} \\times 1 \\times 10^{-6}}{10 \\times 10^{-3}}}$ ${I}=2 {mA}\ {e}^{-1}$ ${I}=2 \\times 0.37 {mA}$ ${I}=\frac{74}{100} {mA}$ ${x}=74$

Q24JEE Main 2021NAT4MElectromagnetic Induction

A circular conducting coil of radius $1 {m}$ is being heated by the change of magnetic field ${\vec{B}}$ passing perpendicular to the plane in which the coil is laid. The resistance of the coil is $2 \mu\Omega$ . The magnetic field is slowly switched off such that its magnitude changes in time as ${B}=\frac{4}{\pi} \\times 10^{-3} {T}(1-\frac{{t}}{100})$ The energy dissipated by the coil before the magnetic field is switched off completely is ${E}=────\ {mJ}$

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📖 Explanation

$\phi={\vec{B}} \.{\vec{S}}$ $\phi=\frac{4}{\pi} \\times 10^{-3}(1-\frac{t}{100}) \.\piR^{2}$ $\phi=4 \\times 10^{-3} \\times (1)^{2}(1-\frac{{t}}{100})$ $\epsilon=\frac{-{d} \phi}{{dt}}$ $\epsilon=\frac{-{d}}{{dt}}(4 \\times 10^{-3}(1-\frac{{t}}{100}))$ $\epsilon=4 \\times 10^{-3}(\frac{1}{100})=4 \\times 10^{-5} {V}$ When ${B}=0$ $1-\frac{t}{100}=0$ ${t}=100 {sec}$ Heat $=\frac{\epsilon^{2}}{{R}} {t}$ Heat $=\frac{(4 \\times 10^{-5})^{2}}{2 \\times 10^{-6}} \\times 100 {J}$ Heat $=\frac{16 \\times 10^{-10} \\times 100}{2 \\times 10^{-6}} {J}$ Heat $=0.08 {J}$ Heat $=80 {mJ}$

Q25JEE Main 2021NAT4MOscillations

In the reported figure, two bodies A and B of masses 200 g and 800 g are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be _______ rad/s when k = 20 N/m.

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📖 Explanation

$\omega=\sqrt{{{k}_{{eq}}}/{\mu}}$ $\mu=$ reduced mass springs are in series connection ${k}_{{eq}}=\frac{{k}_{1} {k}_{2}}{{k}_{1}+{k}_{2}}$ ${k}_{{eq}}=\frac{{k} \\times 4 {k}}{5 {k}}=\frac{4 {k}}{5}$ ${k}_{{eq}}=\frac{4 \\times 20}{5}{N} / {m}=16 {N} / {m}$ $\mu=\frac{{m}_{1} {m}_{2}}{{m}_{1}+{m}_{2}}=\frac{0.2 \\times 0.8}{0.2+0.8}$ $=0.16 {kg}$ $\omega=\sqrt{\frac{16}{0.16}}=\sqrt{100}=10$

Q26JEE Main 2021NAT4MMagnetism and Matter

The value of aluminium susceptibility is $2.2 \\times 10^{-5}$ . The percentage increase in the magnetic field if space within a current carrying toroid is filled with aluminium is $\frac{x}{10^{4}}$ . Then the value of $x$ is ______.

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📖 Explanation

${B}=\mu\.({H}+{I})$ ${B}=\mu\.{H}(1+\frac{{I}}{{H}})$ ${B}={B}_{0}(1+{x})$ ${B}-{B}_{0}={B}_{0} {x}$ $\frac{{B}-{B}_{0}}{{B}_{0}}={x}$ $\frac{{B}-{B}_{0}}{{B}_{0}} \\times 100=100 {x}$ $=2.2 \\times 10^{-3}=\frac{22}{10^{4}}$

Q27JEE Main 2021NAT4MOscillations

A particle of mass $1 {mg}$ and charge ${q}$ is lying at the mid-point of two stationary particles kept at a distance '2 ${m}$ ' when each is carrying same charge 'q'. If the free charged particle is displaced from its equilibrium position through distance ' ${x}'$ $({x} < < 1 {m})$ . The particle executes SHM. Its angular frequency of oscillation will be _____ $\\times 10^{5} {\text{rad}} / {s}$ if $q^{2}=10 C^{2}$

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📖 Explanation

Net force on free charged particle $F=\frac{k q^{2}}{(d+x)^{2}}-\frac{k q^{2}}{(d-x)^{2}}$ $F=-k q^{2}[\frac{4 d x}{(d^{2}-x^{2})^{2}}]$ ${a}=-\frac{4 {kq}^{2} {d}}{{m}}(\frac{{x}}{{d}^{4}})$ ${a}=-(\frac{4 {kq}^{2}}{{md}^{3}}) {x}$ So, angular frequency $\omega=\sqrt{\frac{4 {kq}^{2}}{{md}^{3}}}$ $\omega=\sqrt{\frac{4 \\times 9 \\times 10^{9} \\times 10}{1 \\times 10^{-6} \\times 1^{3}}}$ $\omega=6 \\times 10^{8} {rad} / {sec}$

Q28JEE Main 2021NAT4MCurrent Electricity

An electric bulb rated as 200 W at 100 V is used in a circuit having 200 V supply. The resistance 'R' that must be put in series with the bulb so that the bulb delivers the same power is _____ Ω

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📖 Explanation

Power, ${P}={{V}^{2}}/{{R}_{{B}}}$ ${R}_{{B}}=\frac{{V}^{2}}{{P}}=\frac{100 \\times 100}{200}$ ${R}_{{B}}=50 \Omega$ To produce same power, same voltage (i.e. $100 {V}$ ) should be across the bulb Hence, ${R}={R}_{{B}}$ ${R}=50 \Omega$

Q29JEE Main 2021NAT4MWork, Power and Energy

A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 50 cm long. The speed of bob, when the length makes an angle of 60° to the vertical will be (g = 10 m/s $^{2}$ ) ______ m/s.

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📖 Explanation

Applying work energy theorem: ${W}_{{g}}+{w}_{{T}}=\Delta{K}$ $-{mgl}(1-\cos 60°)=\frac{1}{2} {mv}^{2}-\frac{1}{2} {\mu}^{2}$ $v^{2}=u^{2}-2 g l(1-\cos 60°)$ ${v}^{2}=9-2 \\times 10 \\times 0.5(\frac{1}{2})$ ${v}^{2}=4$ ${v}=2 {m} / {s}$

Q30JEE Main 2021NAT4MRotational Motion

A particle of mass 'm' is moving in time 't' on a trajectory given by ${\vec{r}}=10 \alpha{t}^{2} {\hat{i}}+5 \beta({t}-5) \hat{j}$ Where alpha and beta are dimensional constants. The angular momentum of the particle becomes the same as it was for t = 0 at time t = ______seconds.

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📖 Explanation

${\vec{r}}=10 \alpha{t}^{2} {\hat{i}}+5 \beta({t}-5) \hat{j}$ ${\vec{v}}=20 \alpha\hat{i}+5 \beta\hat{j}$ ${\vec{L}}={m}({\vec{r}} \\times {\vec{v}})$ $={m}\$ [10 \alpha{t}^{2} \hat{i}+5 \beta({t}-5) \hat{j}]$ $ $\times [20 \alphat \hat{i}+5 \beta{\hat{j}}]$ ${\vec{L}}={m}[50 \alpha\beta{t}^{2} \hat{k}-100 \alpha\beta({t}^{2}-5 {t}) \hat{k}]$ At ${t}=0, {\vec{L}}=\vec{0}$ $50 \alpha\beta{t}^{2}-100 \alpha\beta({t}^{2}-5 {t})=0$ ${t}-2({t}-5)=0$ ${t}=10 {sec}$

Chemistry30 questions

Q31JEE Main 2021MCQ4MBiomolecules

is a repeating unit for :

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📖 Explanation

Phenol $+H-C^{∥}^{O}-H{\to }^{H^{+}} { \text{Novalac} }_{(\text{Given}\text{stucture})\text{Linear} \text{polymer}}$ ${\to }^{\text{Polymerisation}} \text{Bakelite}_{(3\text{D}\text{structures})}$ Novolac is low molecular weight polymer derived from phenol and formaldehyde. They are related to bakelite, which is more highly cross-linked.

Q32JEE Main 2021MCQ4MCoordination Compounds

Which one of the following species responds to an external magnetic field?

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📖 Explanation

$[{Fe}({H}_{2} {O})_{6}]^{3+}$ ${Fe}^{3+}:[{Ar}] 3 {d}^{5}$ Hybridisation : ${sp}^{3} {d}^{2}$ Magnetic nature : Paramagnetic (so this complex response to external magnetic field) 2. $[{Ni}({CN})_{4}]^{2-}$ ${Ni}^{2+}:[{Ar}] 3 {d}^{8}$ Hybridisation: dsp $^{2}$ Magnetic nature : diamagnetic 3. $[{Co}({CN})_{6}]^{3-}$ ${Co}^{3+}:[{Ar}] 3 {d}^{6}$ Hybridisation: ${d}^{2} {sp}^{3}$ Magnetic nature : diamagnetic 4. $[{Ni}({CO})_{4}]$ ${Ni}:[{Ar}] 3 {d}^{8} 4{s}^{2}$ Hybridisation : sp $^{3}$ Magnetic nature : diamagnetic

Q33JEE Main 2021MCQ4MAlcohols, Phenols and Ethers

Consider the above reaction, the major product 'P' is:

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Q34JEE Main 2021MCQ4MSolutions

Sodium stearate ${CH}_{3}({CH}_{2})_{16} {COO}^{-} {Na}^{+}$ is ananionic surfactant which forms micelles in oil. Choose the correct statement for it from thefollowing:

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📖 Explanation

Forms spherical micelles with $CH_3(CH_2)_16$ group pointing towards the centre of sphere

Q35JEE Main 2021MCQ4MBiomolecules

The water soluble protein is :

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Q36JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

At $298.2 {K}$ the relationship between enthalpy of bond dissociation (in ${kJ} {mol}^{-1}$ ) for hydrogen $({E}_{{H}})$ and its isotope, deuterium $({E}_{{D}})$ , is best described by:

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📖 Explanation

Enthalpy of bond dissociation (kJ/mole) at 298.2K For , hydrogen = 435.88 For , Deuterium = 443.35 $\therefore {E}_{{H}} \simeq {E}_{{D}}-7.5$

Q37JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Consider the given reaction, the product 'X' is:

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Q38JEE Main 2021MCQ4MHaloalkanes and Haloarenes

The given reaction can occur in the presence of : (a) Bromine water (b) ${Br}_{2}$ in ${CS}_{2}, 273 {K}$ (c) ${Br}_{2} / {FeBr}_{3}$ (d) ${Br}_{2}$ in ${CHCl}_{3}, 273 {K}$ Choose the correct answer from the options givenbelow :

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Q39JEE Main 2021MCQ4MAmines

Given below are two statements, one is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (R) : Gabriel phthalimide synthesis cannot be used to prepare aromatic primary amines. Reason : Aryl halides do not undergo nucleophilic substitution reaction. In the light of the above statements, choose the correct answer from the options given below :

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Q40JEE Main 2021MCQ4MChemical Kinetics

For the following graphs, Choose from the options given below, the correct one regarding order of reaction is :

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📖 Explanation

For zero order reaction's rate $={K}[\text{Reactant }]^{0}$ $\Rightarrow{r}={k}$ Reactant concentration after time ${t} \\to$ ${C}_{{t}}={C}_{0} {e}^{-{kt}}$

Q41JEE Main 2021MCQ4MAmines

Which one of the products of the following reactions does not react with Hinsberg reagent to form sulphonamide?

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Q42JEE Main 2021MCQ4MClassification of Elements

The ionic radii of ${K}^{+}, {Na}^{+}, {Al}^{3+}$ and ${Mg}^{2+}$ are in the order :

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📖 Explanation

${Al}^{3+}, {Mg}^{2+}$ and ${Na}^{+}$ are isoelectronic ionic species. For monoatomic ionic isoelectronic species as positive charge increases ionic size decreases. The order of size of ${Na}^{+} \ \& { K}^{+}$ is ${Na}^{+} < {K}^{+}$ , $\therefore$ order of ionic radii is : ${Al}^{3+} < {Mg}^{2+} < {Na}^{+} < {K}^{+}$

Q43JEE Main 2021MCQ4Mp Block Elements

Which one of the following compounds of Group-14 elements is not known?

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📖 Explanation

$[{SiCl}_{6}]^{2-}$ does not exist due to steric crowding of surrounding atoms.

Q44JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

Which one among the following resonating structures is not correct?

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Q45JEE Main 2021MCQ4MIonic Equilibrium

Given below are two statements : Statement I : None of the alkaline earth metal hydroxides dissolve in alkali. Statement II : Solubility of alkaline earth metal hydroxides in water increases down the group. In the light of the above statements, choose the most appropriate answer from the options given below :

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📖 Explanation

Statement-I is incorrect ${Be}({OH})_{2}$ dissolve in alkali due to it's amphoteric nature. Statement-II is correct Solubility of alkaline earth metal hydroxide in water increases down the group due to rapid decreases in lattice energy as compared to hydration energy.

Q46JEE Main 2021MCQ4Md and f Block Elements

The correct order of following $3 {d}$ metal oxides, according to their oxidation numbers is : (a) ${CrO}_{3}$ (b) ${Fe}_{2} {O}_{3}$ (c) ${MnO}_{2}$ (d) ${V}_{2} {O}_{5}$ (e) ${Cu}_{2} {O}$

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📖 Explanation

(a) ${{Cr}^{+6}O}_{3}$ (b) ${Fe}_{2}^{+3} {O}_{3}$ (c) ${{Mn}^{+4}O}_{2}$ (d) ${V}_{2}^{+5} {O}_{5}$ (e) ${Cu}_{2}^{+1} {O}$ So order of oxidation state $a > d > c > b > e$

Q47JEE Main 2021MCQ4MHaloalkanes and Haloarenes

Which one of the following chemical agent is not being used for dry-cleaning of clothes?

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📖 Explanation

${CO}_{2}, {CCl}_{4}$ and ${Cl}_{2} {C}={CCl}_{2}$ are used as dry cleaning agents for clothes. ${H}_{2} {O}_{2}$ is used as bleaching agent in laundry. Carbon tetrachloride was commonly blended with other solvents for use as a dry cleaning solvent. Now, because of its high toxicity and tendency to corrode equipment, carbon tetrachloride is no longer used as a drycleaning solvent.

Q48JEE Main 2021MCQ4MAmines

Which one of the following compounds will liberate ${\text{CO}}_{2}$ , when treated with ${\text{NaHCO}}_{3} ?$

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Q49JEE Main 2021MCQ4MCoordination Compounds

In the leaching of alumina from bauxite, the ore expected to leach out in the process by reacting with ${NaOH}$ is :

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📖 Explanation

In bauxite impurities of ${Fe}_{2} {O}_{3}, {TiO}_{2}$ and ${SiO}_{2}$ are present, ${Fe}_{2} {O}_{3}$ and ${TiO}_{2}$ are basic oxides therefore does not reacts with or dissolve in ${NaOH}$ whereas ${SiO}_{2}$ is acidic oxide it gets dissolve in ${NaOH}$ , hence leach out ${SiO}_{2}+2 {NaOH} \\to {Na}_{2} {SiO}_{3}({aq} .)$ $+{H}_{2} {O}$

Q50JEE Main 2021MCQ4MHydrocarbons

An organic compound 'A' ${C}_{4} {H}_{8}$ on treatment with ${KMnO}_{4} / {H}^{+}$ yields compound ' ${B}' {C}_{3} {H}_{6} {O}$ . Compound 'A' also yields compound 'B' an ozonolysis. Compound 'A' is :

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Q51JEE Main 2021NAT4MHydrocarbons

SECTION-BThe number of sigma bonds in ${H}_{3} {C}-{C}_{|}_{H}={CH}-{C} \equiv {C}-{H}$ is _________ .

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Q52JEE Main 2021NAT4MCoordination Compounds

Three moles of ${AgCl}$ get precipitated when one mole of an octahedral co-ordination compound with empirical formula ${CrCl}_{3} .3 {NH}_{3} \.3 {H}_{2} {O}$ reacts with excess of silver nitrate. The number of chloride ions satisfying the secondary valency of the metal ion is _____.

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📖 Explanation

Mole of AgCl precipitated is equal the mole of $Cl^{- }$ present in ionization sphere. Since none of $Cl^{-}$ is present in the co-ordination sphere. Therefore answer is zero.

Q53JEE Main 2021NAT4MPhotoelectric Effect

A source of monochromatic radiation of wavelength $400 {nm}$ provides $1000 {J}$ of energy in 10 seconds. When this radiation falls on the surface of sodium, $x \\times 10^{20}$ electrons are ejected per second. Assume that wavelength $400 {nm}$ is sufficient for ejection of electron from the surface of sodium metal. The value of ${x}$ is _______. (Nearest integer) $({h}=6.626 \\times 10^{-34} {Js})$

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📖 Explanation

Total energy provided by Source per second $=\frac{1000}{10}=100 {J}$ Energy required to eject electron $=\frac{{hc}}{\lambda}$ $=\frac{6.626 \\times 10^{-34}}{400 \\times 10^{-9}} \\times 3 \\times 10^{8}$ Number of electrons ejected $={100}/{\frac{6.626 \\times 10^{-34} \\times 3 \\times 10^{8}}{400 \\times 10^{-9}}}$ $=\frac{400 \\times 10^{-7} \\times 10^{26}}{6.626 \\times 3}$ $=\frac{40 \\times 10^{-20}}{6.626 \\times 3}$ $=2.01 \\times 10^{20}$

Q54JEE Main 2021NAT4MSolutions

${CO}_{2}$ gas is bubbled through water during a soft drink manufacturing process at $298 {K}$ . If ${CO}_{2}$ exerts a partial pressure of $0.835$ bar then ${x} {m} {\text{mol}}$ of ${CO}_{2}$ would dissolve in $0.9 {L}$ of water. The value of $x$ is ________ $.$ (Nearest integer) (Henry's law constant for ${CO}_{2}$ at $298 {K}$ is $1.67 \\times 10^{3}$ bar $)$

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📖 Explanation

From Henry's law ${P}_{{\text{gas}}}={K}_{{H}} \.{X}_{{\text{gas}}}$ $0.835=1.67 \\times 10^{3} \\times \frac{{n}({CO}_{2})}{\frac{0.9 \\times 1000}{18}}$ ${n}({CO}_{2})=0.025$ Millimoles of ${CO}_{2}=0.025 \\times 1000=25$

Q55JEE Main 2021NAT4MChemical Equilibrium

For the reaction ${A}+{B} \⇌2 {C}$ the value of equilibrium constant is 100 at $298 {K}$ . If the initial concentration of all the three species is $1 {M}$ each, then the equilibrium concentration of ${C}$ is ${x} \\times 10^{-1} {M}$ . The value of ${x}$ is ________ . (Nearest integer)

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📖 Explanation

$\text{table} {A},+,{B}, \⇌,2 {C};1,,1,,1;-x,,-x,,2x$ $───────────────$ $\text{table} 1-x,,1-x,,1+2x$ ${K}={[{C}]_{{eq}}^{2}}/{[{A}]_{{eq}}[{B}]_{{cq}}}=\frac{(1+2 {x})^{2}}{(1-{x})(1-{x})}$ $100=(\frac{1+2 {x}}{1-{x}})^{2}$ $(\frac{1+2 {x}}{1-{x}})=10$ $x=\frac{3}{4}$ $[$ C]e $_{q .}=1+2 x$ $=1+2(\frac{3}{4})$ $=2.5 {M}$ $25 \\times 10^{-1} {M}$

Q56JEE Main 2021NAT4MElectrochemistry

Consider the cell at $25°{C}$ ${Zn} \|{Zn}^{2+}({aq}),(1 {M}) ∥{Fe}^{3+}({aq}), {Fe}^{2+}({aq})) {Pt}({s})$ The fraction of total iron present as ${Fe}^{3+}$ ion at the cell potential of $1.500 {V}$ is ${x} \\times 10^{-2}$ . The value of ${x}$ is _______ $\.$ (Nearest integer) (Given : $E°_{{Fe}^{3+/ {Fe}^{2+}} }=0.77 {V}, {E}°_{{Zn}^{2+} / {Zn}} =-0.76 {V})$

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📖 Explanation

${Zn} \\to {Zn}^{2+}+2 {e}^{-}$ $2 {Fe}^{3+} \\to 2 {e}^{-}+2 {e}^{2+}$ $─────────────────$ ${Zn}+2 {Fe}^{3+} \\to {Zn}^{2+}+2 {Fe}^{2+}$ $─────────────────$ ${E}°_{\text{cell }} =0.77-(0.76)$ $= 1.53 V$ $1.50=1.53-\frac{0.06}{2} \log (\frac{{Fe}^{2+}}{{Fe}^{3+}})^{2}$ $\log (\frac{{Fe}^{2+}}{{Fe}^{3+}})=\frac{0.03}{0.06}=\frac{1}{2}$ $\frac{[{Fe}^{2+}]\$ }{[{Fe}^{3+}]$}=10^{1 / 2}=\sqrt{10}\frac{[{Fe}^{3+}]$}{[{Fe}^{2+}]$}=\frac{1}{\sqrt{10}}\frac{[{Fe}^{3+}]$}{[{Fe}^{2+}]$+$[{Fe}^{3+}]$}=\frac{1}{1+\sqrt{10}}=\frac{1}{4.16}=0.2402=24 \times 10^{-2}$

Q57JEE Main 2021NAT4MChemical Thermodynamics

At $298 {K}$ , the enthalpy of fusion of a solid $({X})$ is $2.8 {~kJ} {mol}^{-1}$ and the enthalpy of vaporisation of the liquid $(X)$ is $98.2 {kJ} {mol}^{-1}$ . The enthalpy of sublimation of the substance $({X})$ in ${kJ} {mol}^{-1}$ is ________ . (in nearest integer)

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📖 Explanation

$\Delta{H}_{{\text{sub}}}=\Delta{H}_{\text{fivs. }}+\Delta{H}_{\text{vap. }}$ $=2.8+98.2$ $=101 {kJ} / {mol}$

Q58JEE Main 2021NAT4MSome Basic Concepts of Chemistry

A home owner uses $4.00 \\times 10^{3} {m}^{3}$ of methane $({CH}_{4})$ gas, $($ assume ${CH}_{4}$ is an ideal gas) in a year to heat his home. Under the pressure of $1.0$ atm and $300 {K}$ , mass of gas used is ${x} \\times 10^{5} {g}$ . The value of $x$ is _______ $.$ (Nearest integer) $($ Given ${R}=0.083 {L} {atm} {K}^{-1} {mol}^{-1})$

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📖 Explanation

${n}({CH}_{4})=\frac{{PV}}{{RT}}$ $=\frac{1 \\times 4 \\times 10^{3} \\times 1000}{0.083 \\times 300}$ Weight of ${CH}_{4}$ $=\frac{40 \\times 16 \\times 10^{5}}{0.083 \\times 300} {gm}$ $=25.7 \\times 10^{5} {gm}$

Q59JEE Main 2021NAT4MRedox Reactions

When $10 {mL}$ of an aqueous solution of ${Fe}^{2+}$ ions was titrated in the presence of dil ${H}_{2} {SO}_{4}$ using diphenylamine indicator, $15 {mL}$ of $0.02 {M}$ solution of ${K}_{2} {Cr}_{2} {O}_{7}$ was required to get the end point. The molarity of the solution containing ${Fe}^{2+}$ ions is ${x} \\times 10^{-2} {M}$ . The value of ${x}$ is _______.(Nearest integer)

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📖 Explanation

milli-equivalents of ${Fe}^{2+}=$ milli-equivalents of ${K}_{2} {Cr}_{2} {O}_{7}$ ${M} \\times 10 \\times 1=0.02 \\times 15 \\times 6$ ${M}=0.18=18 \\times 10^{-2} {M}$

Q60JEE Main 2021NAT4MHydrocarbons

Consider the complete combustion of butane, the amount of butane utilized to produce $72.0 {g}$ of water is _______ $\\times 10^{-1} {g}$ . (in nearest integer)

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📖 Explanation

${C}_{4} {H}_{10}+\frac{13}{2} {O}_{2} \\to 4 {CO}_{2}+5 {H}_{2} {O}$ Moles of ${H}_{2} {O}=\frac{72}{18}=4$ Moles of ${C}_{4} {H}_{10}$ used $=\frac{1}{5} \\times 4$ Weight of ${C}_{4} {H}_{10}$ used $=\frac{4}{5} \\times 58$ $=46.4 {gm}$

Mathematics30 questions

Q61JEE Main 2021MCQ4MThree Dimensional Geometry

A spherical gas balloon of radius 16 meter subtends an angle 60° at the eye of the observer A while the angle of elevation of its center from the eye of A is 75°. Then the height (in meter) of the top most point of the balloon from the level of the observer's eye is :

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📖 Explanation

O → centre of sphere P,Q → point of contact of tangents from A Let T be top most point of balloon & R be foot of perpendicular from O to ground. From triangle ${OAP}, {OA}=16 {\text{cosec}} 30°=32$ From triangle ${ABO}, {OR}={OA} \sin 75°=32 \frac{(\sqrt{3}+1)}{2 \sqrt{2}}$ So level of top most point $={OR}+{OT}$ $=8(\sqrt{6}+\sqrt{2}+2)$

Q62JEE Main 2021MCQ4MDifferentiation

Let ${f}({x})=3 \sin ^{4} {x}+10 \sin ^{3} {x}+6 \sin ^{2} {x}-3$ , ${x} \in[-\frac{\pi}{6}, \frac{\pi}{2}] .$ Then, ${f}$ is :

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📖 Explanation

${f}({x})=3 \sin ^{4} {x}+10 \sin ^{3} {x}+6 \sin ^{2} {x}-3$ , ${x} \in[-\frac{\pi}{6}, \frac{\pi}{2}] .$ $f'(x)=12 \sin ^{3 } x \cos x+30 \sin ^{2} x \cos x$ $+12 \sin x \cos x$ $=6 \sin x \cos x(2 \sin ^{2} x+5 \sin x+2)$ $=6 \sin x \cos x(2 \sin x+1)(\sin +2)$ decreasing in $(-\frac{\pi}{6}, 0)$

Q63JEE Main 2021MCQ4MSequences and Series

Let ${S}_{{n}}$ be the sum of the first ${n}$ terms of an arithmetic progression. If ${S}_{3 {n}}=3 {S}_{2 {n}}$ , then the value of ${{S}_{4 {n}}}/{{S}_{2 {n}}}$ is :

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📖 Explanation

Let a be first term and d be common diff. of this A.P. Given $S_{3 n}=3 S_{2 n}$ $\Rightarrow\frac{3 n}{2}[2 a+(3 n-1) d]$ $=3 \frac{2 n}{2}[2 a+(2 n-1) d]$ $\Rightarrow2 a+(3 n-1) d=4 a+(4 n-2) d$ $\Rightarrow2 a+(n-1) d=0$ Now $\frac{S_{4 n}}{S_{2 n}}=\frac{\frac{4 n}{2}[2 a+(4 n-1) d]}{\frac{2 n}{2}[2 a+(2 n-1) d]}$ $=\frac{2[{2 a+(n-1) d} +3 n d]}{[{2 a+(n-1) d} +n d]}$ $=\frac{6 n d}{\text{nd }}=6$

Q64JEE Main 2021MCQ4MHyperbola

The locus of the centroid of the triangle formed by any point ${P}$ on the hyperbola $16 {x}^{2}-9 {y}^{2}+32 {x}+36 {y}-164=0$ , and its foci is :

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📖 Explanation

Given hyperbola is $16(x+1)^{2}-9(y-2)^{2}$ $=164+16-36=144$ $\Rightarrow\frac{(x+1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1$ Eccentricity, ${e}=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$ $\Rightarrow$ foci are $(4,2)$ and $(-6,2)$ Let the centroid be $({h}, {k})$ & ${A}(\alpha, \beta)$ be point on hyperbola So ${h}=\frac{\alpha-6+4}{3}, {k}=\frac{\beta+2+2}{3}$ $\Rightarrow\alpha=3 {h}+2, \beta=3 {k}-4$ $(\alpha, \beta)$ lies on hyperbola so $16(3 h+2+1)^{2}-9(3 k-4-2)^{2}$ $=144$ $\Rightarrow144({h}+1)^{2}-81({k}-2)^{2}=144$ $\Rightarrow16(h^{2}+2 h+1)-9(k^{2}-4 k$ $+4)=16$ $\Rightarrow16 x^{2}-9 y^{2}+32 x$ $+36 y-36=0$

Q65JEE Main 2021MCQ4MVector Algebra

Let the vectors $(2+{a}+{b}) {\hat{i}}+({a}+2 {b}+{c}) {\hat{j}}-({b}+{c}) {\hat{k}},$ $(1+{b}) {\hat{i}}+2 {{b}}-{b} {\hat{k}}$ and $(2+b) \hat{i}+2 b \hat{j}+(1-b) {k} ^{\^}$ $\;\;a, b, c, \in{R}$ be co-planar. Then which of the following is true?

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📖 Explanation

If the vectors are co-planar, $\begin{vmatrix}a +b +2 & a +2b+ c & -b- c \\ b +1 & 2b & - b \\ b+ 2 & 2b & 1- b\end{vmatrix}=0$ Now ${R}_{3} \\to {R}_{3}-{R}_{2}, {R}_{1} \\to {R}_{1}-{R}_{2}$ So $\begin{vmatrix}{a}+1 & {a}+{c} & -{c} \\ {b}+1 & 2 {b} & -{b} \\ 1 & 0 & 1\end{vmatrix}=0$ $=({a}+1) 2 {b}-({a}+{c})(2 {b}+1)-{c}(-2 {b})$ $=2 {ab}+2 {b}-2 {ab}-{a}-2 {bc}-{c}+2 {bc}$ $=2 {b}-{a}-{c}=0$

Q66JEE Main 2021MCQ4MLimits, Continuity and Differentiability

Let ${f}: {R} \\to {R}$ be defined as $f(x)= \begin{cases}\frac{\lambda|x^{2}-5 x+6|}{\mu(5 x-x^{2}-6)}{,} & x < 2 \\ e^{\frac{\tan (x-2)}{x-[x]}} {,} & x>2 \\ \mu {, } & x=2$ where $[{x}]$ is the greatest integer less than or equal to $x$ \end{cases} If $f$ is continuous at $x=2$ , then $\lambda+\mu$ is equal to:

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📖 Explanation

$\lim _{x \\to 2^{+}} f(x)=\lim _{x \\to 2^{+}} e^{\frac{\tan (x-2)}{x-2}}=e^{1}$ $\lim _{x \\to 2^{-}} f(x)=\lim _{x \\to 2^{-}} \frac{-\lambda(x-2)(x-3)}{\mu(x-2)(x-3)}$ $=-\frac{\lambda}{\mu}$ For continuity $\mu=e=-\frac{\lambda}{\mu} \Rightarrow\mu=e, \lambda=-e^{2}$ $\lambda+\mu=e(-e+1)$

Q67JEE Main 2021MCQ4MDefinite Integrals

The value of the definite integral $\int_{\pi/ 24}^ {5 \pi/ 24} \frac{d x}{1+\sqrt^{3}{\tan 2 x}}$ is

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📖 Explanation

Let $I=\int_{\pi/ 24}^ {5 \pi/ 24} \frac{(\cos 2 x)^{1 / 3}}{(\cos 2 x)^{1 / 3}+(\sin 2 x)^{1 / 3}} d x$ .......(i) $\RightarrowI=\int_{\pi/ 24}^ {5 \pi/ 24} {(\cos \{2(\frac{\pi}{4}-x)\})^{\frac{1}{3}}}/{(\cos \{2(\frac{\pi}{4}-x)\})^{\frac{1}{3}}+(\sin \{2(\frac{\pi}{4}-x)\})^{\frac{1}{3}}} d x$ $\{\int_{a} ^{b} f(x) d x=\int_{a} ^{b} f(a+b-x) d x\}$ So $I=\int_{\pi/ 24}^ {5 \pi/ 24} \frac{(\sin 2 x)^{1 / 3}}{(\sin 2 x)^{1 / 3}+(\cos 2 x)^{1 / 3}} d x .$ .......(ii) Hence $2 {I}=int_{\pi/ 24}^ {5 \pi/ 24} {dx}$ $[({i})+({ii})]$ $\Rightarrow2 {I}=\frac{4 \pi}{24} \Rightarrow{I}=\frac{\pi}{12}$

Q68JEE Main 2021MCQ4MDifferential Equations

If ${b}$ is very small as compared to the value of ${a}$ , so that the cube and other higher powers of $\frac{{b}}{{a}}$ can be neglected in the identity $\frac{1}{a-b}+\frac{1}{a-2 b}+\frac{1}{a-3 b}+\.......$ $+\frac{1}{a-n b}=\alphan+\betan^{2}+\gamman^{3}$ then the value of $\gamma$ is :

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📖 Explanation

$(a-b)^{-1}+(a-2 b)^{-1}+\......$ $+(a-n b)^{-1}$ $=\frac{1}{a} \sum_{r=1} ^{n}(1-\frac{r b}{a})^{-1}$ $=\frac{1}{a} \sum_{r=1} ^{n}\{(1+\frac{r b}{a}+\frac{r^{2} b^{2}}{a^{2}})+($ terms to be neglected $)\}$ $=\frac{1}{a}[n+\frac{n(n+1)}{2} \.\frac{b}{a}+\frac{n(n+1)(2 n+1)}{6} \.\frac{b^{2}}{a^{2}}]$ $=\frac{1}{a}[n^{3}(\frac{b^{2}}{3 a^{2}})+\.......]$ So $\gamma=\frac{b^{2}}{3 a^{3}}$

Q69JEE Main 2021MCQ4MDifferential Equations

Let $y=y(x)$ be the solution of the differentialequation $\frac{{dy}}{{dx}}=1+{x} {e}^{{y}-{x}},-\sqrt{2} < {x} < \sqrt{2}, {y}(0)=0$ then, the minimum value of $y(x), x \in(-\sqrt{2}, \sqrt{2})$ is equal to:

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📖 Explanation

$\frac{d y-d x}{e^{y-x}}=x d x$ $\Rightarrow \frac{d y-d x}{e^{y-x}}=x d x$ $\Rightarrow-e^{x-y}=\frac{x^{2}}{2}+c$ At $x=0, y=0 \Rightarrowc=-1$ $\Rightarrowe^{x-y}=\frac{2-x^{2}}{2}$ $\Rightarrowy=x-ln(\frac{2-x^{2}}{2})$ $\Rightarrow \frac{d y}{d x}=1+\frac{2 x}{2-x^{2}}=\frac{2+2 x-x^{2}}{2-x^{2}}$ So minimum value occurs at ${x}=1-\sqrt{3}$ $y(1-\sqrt{3})=(1-\sqrt{3})-\ln(\frac{2-(4-2 \sqrt{3})}{2})$ $=(1-\sqrt{3})-\ln (\sqrt{3}-1)$

Q70JEE Main 2021MCQ4MSets and Relations

The Boolean expression $({p} \Rightarrow{q}) \wedge ({q} \Rightarrow\sim {p})$ is equivalent to

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📖 Explanation

$({p} \\to {q}) \wedge ({q} \\to \sim {p})$ $\equiv (\sim p \vee q) \wedge (\sim q \vee \sim p) \{p \\to q \equiv \sim p \vee q\}$ $\equiv (\sim {p} \vee {q}) \wedge (\sim {p} \vee \sim {q}) \{$ commutative property $\}$ $\equiv ∼{p} \vee ({q} \wedge \sim {q}) \{$ distributive property $\}$ $\equiv \sim {p}$

Q71JEE Main 2021MCQ4MArea Under The Curves

The area (in sq. units) of the region, given by the set $\{({x}, {y}) \in{R} \\times {R} \|{x} \\ge 0,2 {x}^{2} \\le {y} \\le 4-2 {x}\}$ is :

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📖 Explanation

Required area $=\int_{0}^{1}(4-2 x-2 x^{2}) d x$ $={4 x-x^{2}-\frac{2 x^{3}}{3}|}_{0} ^{1}$ $=4-1-\frac{2}{3}=\frac{7}{3}$

Q72JEE Main 2021MCQ4MTrigonometric Ratios and Identities

The sum of all values of $x$ in $[0,2 \pi]$ , for which $\sin x+\sin 2 x+\sin 3 x+\sin 4 x=0$ , is equal to

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📖 Explanation

$(\sin x+\sin 4 x)+(\sin 2 x+\sin 3 x)=0$ $\Rightarrow2 \sin \frac{5 x}{2}\{\cos \frac{3 x}{2}+\cos \frac{x}{2}\}=0$ $\Rightarrow2 \sin \frac{5 x}{2}\{2 \cos x \cos \frac{x}{2}\}=0$ $2 \sin \frac{5 x}{2}=0 \Rightarrow\frac{5 X}{2}$ $=0, \pi, 2 \pi, 3 \pi, 4 \pi, 5 \pi$ $\Rightarrowx=0, \frac{2 \pi}{5}, \frac{4 \pi}{5}, \frac{6 \pi}{5}, \frac{8 \pi}{5}, 2 \pi$ $\cos \frac{x}{2}=0 \Rightarrow\frac{x}{2}=\frac{\pi}{2} \Rightarrowx=\pi$ $\cos x=0 \Rightarrowx=\frac{\pi}{2}, \frac{3 \pi}{2}$ So sum $=6 \pi+\pi+2 \pi=9 \pi$

Q73JEE Main 2021MCQ4MSets and Relations

Let ${g}: {N} \\to {N}$ be defined as $g(3 n+1)=3 n+2$ $g(3 n+2)=3 n+3$ $g(3 n+3)=3 n+1$ , for all $n \\ge 0$ Then which of the following statements is true?

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📖 Explanation

${g}: {N} \\to {N}$ $g(3 n+1)=3 n+2$ $g(3 n+2)=3 n+3$ $g(3 n+3)=3 n+1$ $g(x)=[\text{table} x+1 ,x =3 k+1 ; x+1 , x=3 k+2 ;x-2 , x=3 k+3$ $g(g(x))=[\text{table} x+2 , x=3 k+1;x-1 , x=3 k+2; x-1 , x=3 k+3$ $g(g(g(x)))=[\text{table} x , x=3 k+1; x , x=3 k+2 ; x , x=3 k+3$ If $f: {N} \\to {N}, f$ is a one-one function such that $f({g}({x}))=f({x}) \text{r} {g}({x})={x}$ , which is not the caseIf ${f} f: {N} \\to {N} f$ is an onto function such that $f(g(x))=f(x)$ , one possibility is $f(x)=[\text{table} {n} , {x}=3 {n}+1;{n} , {x}=3 {n}+2; {n} , {x}=3 {n}+3$ ${n} \in{N}_{0}.$ Here $f({x})$ is onto, also $f({g}({x}))=f({x}) \forall {x} \in{N}$

Q74JEE Main 2021MCQ4MLimits, Continuity and Differentiability

Let ${f}:[0, \\infty ) \\to [0, \\infty )$ be defined as ${f}({x})=\int_{0} ^{{x}}[{y}] {dy}$ where $[{x}]$ is the greatest integer less than or equal to ${x}$ . Which of the following is true?

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📖 Explanation

$f:[0, \\infty ) \\to [0, \\infty ), f({x})=\int_{0} ^{{x}}[{y}] {dy}$ Let ${x}={n}+f, f \in (0,1)$ So $f({x})=0+1+2+\.......+({n}-1)$ $+\int_{{n}} ^{{n}+f} {n} {dy}$ $f({x})=\frac{{n}({n}-1)}{2}+{n} f$ $=\frac{[{x}]([{x}]-1)}{2}+[{x}]\{{x}\}$ Note $\lim _{x \\to {n}^{+}} f({x})=\frac{{n}({n}-1)}{2}, \lim _{x \\to {n}^{-}} f({x})$ $=\frac{({n}-1)({n}-2)}{2}+({n}-1)$ $=\frac{{n}({n}-1)}{2}$ $f({x})=\frac{{n}({n}-1)}{2}({n} \in {N}_{0})$ so $f(x)$ is cont. $\forall x \\ge 0$ and diff. except at integer points

Q75JEE Main 2021MCQ4MMatrices and Determinants

The values of a and ${b}$ , for which the system of equations $2 x+3 y+6 z=8$ $x+2 y+a z=5$ $3 x+5 y+9 z=b$ has no solution, are :

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📖 Explanation

$D=\begin{vmatrix}2 & 3 & 6 \\ 1 & 2 & {a} \\ 3 & 5 & 9\end{vmatrix}=3-{a}$ ${D}=\begin{vmatrix}2 & 3 & 8 \\ 1 & 2 & 5 \\ 3 & 5 & {b}\end{vmatrix}={b}-13$ If ${a}=3, {b} \\ne 13$ , no solution.

Q76JEE Main 2021MCQ4MProbability

Let 9 distinct balls be distributed among 4 boxes, ${B}_{1}, {B}_{2}, {B}_{3}$ and ${B}_{4}$ . If the probability than ${B}_{3}$ contains exactly 3 balls is ${k}(\frac{3}{4})^{9}$ then ${k}$ lies in the set :

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📖 Explanation

required probability $=\frac{{ }^{9} {C}_{3} \.3^{6}}{4^{9}}$ $=\frac{{ }^{9} {C}_{3}}{27} \.(\frac{3}{4})^{9}$ $=\frac{28}{9} \.(\frac{3}{4})^{9} \Rightarrow{k}=\frac{28}{9}$ Which satisfies $|{x}-3| < 1$

Q77JEE Main 2021MCQ4MParabola

Let a parabola ${P}$ be such that its vertex and focus lie on the positive $x$ -axis at a distance 2 and 4 units from the origin, respectively. If tangents are drawn from ${O}(0,0)$ to the parabola ${P}$ which meet ${P}$ at ${S}$ and ${R}$ , then the area (in sq. units) of $\Delta{SOR}$ is equal to:

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📖 Explanation

Clearly RS is latus-rectum $\ \because {VF}=2={a}$ $\therefore {RS}=4 {a}=8$ Now ${OF}=2 {a}=4$ $\Rightarrow$ Area of triangle ${ORS}=16$

Q78JEE Main 2021MCQ4MFunctions

The number of real roots of the equation ${e}^{6 {x}}-{e}^{4 {x}}-2 {e}^{3 {x}}-12 {e}^{2 {x}}+{e}^{{x}}+1=0$ is:

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📖 Explanation

${e}^{6 {x}}-{e}^{4 {x}}-2 {e}^{3 {x}}-12 {e}^{2 {x}}+{e}^{{x}}+1=0$ $\Rightarrow({e}^{3 {x}}-1)^{2}-{e}^{{x}}({e}^{3 {x}}-1)=12 {e}^{2 {x}}$ $({e}^{3 {x}}-1)^{2}({e}^{{x}}-{e}^{-{x}}-{e}^{-2 {x}})=12$ $\Rightarrow{{e}^{{x}}-{e}^{-{x}}-{e}^{-2 {x}}}={12}/{{e}^{3 {x}}-1}$ ⇒ No. of real roots = 2

Q79JEE Main 2021MCQ4MEllipse

Let an ellipse ${E}: \frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1, {a}^{2}>{b}^{2}$ , passes through $(\sqrt{\frac{3}{2}}, 1)$ and has eccentricity $\frac{1}{\sqrt{3}}$ . If a circle, centered at focus ${F}(\alpha, 0), \alpha>0$ , of ${E}$ and radius $\frac{2}{\sqrt{3}}$ , intersects ${E}$ at two points ${P}$ and ${Q}$ , then ${PQ}^{2}$ is equal to :

🚀 Solve in Practice Mode

📖 Explanation

$\frac{3}{2 {a}^{2}}+\frac{1}{{b}^{2}}=1$ and $1-\frac{{b}^{2}}{{a}^{2}}=\frac{1}{3}$ $\Rightarrowa^{2}=3 b^{2}=3$ $\Rightarrow\frac{x^{2}}{3}+\frac{y^{2}}{2}=1$ ........(i) Its focus is $(1,0)$ Now, eqn of circle is $(x-1)^{2}+y^{2}=\frac{4}{3}$ .......(ii) Solving (i) and (ii) we get $y=\\pm \frac{2}{\sqrt{3}}, x=1$ $\Rightarrow{PQ}^{2}=(\frac{4}{\sqrt{3}})^{2}=\frac{16}{3}$

Q80JEE Main 2021MCQ4MThree Dimensional Geometry

Let the foot of perpendicular from a point ${P}(1,2,-1)$ to the straight line ${L}: \frac{{x}}{1}=\frac{{y}}{0}=\frac{{z}}{-1}$ be ${N}$ . Let a line be drawn from P parallel to the plane $x+y+2 z=0$ which meets $L$ at point $Q$ . If $\alpha$ is the acute angle between the lines ${PN}$ and ${PQ}$ , then $\cos \alpha$ is equal to _________.

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📖 Explanation

${\vec{PN}} \.({\hat{i}}-{\hat{k}})=0$ $\Rightarrow{N}(1,0,-1)$ Now, ${{PQ}}^{\to } \.(\hat{i}+\hat{j}+2 {\hat{k}})=0$ $\Rightarrow\mu=-1$ $\Rightarrow{Q}(-1,0,1)$ ${\vec{PN}}=2 {\hat{j}}$ and ${\vec{PQ}}=2 {\hat{i}}+2 {\hat{j}}-2 {\hat{k}}$ $\Rightarrow\cos \alpha=\frac{1}{\sqrt{3}}$

Q81JEE Main 2021NAT4MDifferential Equations

Let ${y}={y}({x})$ be solution of the following differential equation ${e}^{y} \frac{{dy}}{{dx}}-2 {e}^{y} \sin {x}+\sin {x} \cos ^{2} {x}=0, {y}(\frac{\pi}{2})=0$ If ${y}(0)=\log _{e}(\alpha+\beta{e}^{-2})$ , then $4(\alpha+\beta)$ is equal to ________.

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📖 Explanation

Let ${e}^{{y}}={t}$ $\Rightarrow\frac{{dt}}{{dx}}-(2 \sin {x}) {t}=-\sin {x} \cos ^{2} {x}$ I.F. $={e}^{2 {\text{cosx}}}$ $\Rightarrow{t} \.{e}^{2 \cos {x}}=\int{e}^{2 \cos x} \.(-\sin x \cos ^{2} {x}) {dx}$ $\Rightarrow{e}^{{y}} \.{e}^{2 \cos {x}}=\int{e}^{2 {z}} \.{z}^{2} {d} z, {z}={e}^{2 \cos {x}}$ at $x=\frac{\pi}{2}, y=0 \RightarrowC=\frac{3}{4}$ $\Rightarrow {e}^{y}=\frac{1}{2} \cos ^{2} x-\frac{1}{2} \cos x+\frac{1}{4}+\frac{3}{4} \.{e}^{-2 \cos x}$ $\Rightarrowy=\log [\frac{\cos ^{2} x}{2}-\frac{\cos x}{2}+\frac{1}{4}+\frac{3}{4} {e}^{-2 \cos x}]$ Put $x=0$ $\Rightarrowy=\log \$ [\frac{1}{4}+\frac{3}{4} {e}^{-2}]$ $ $\Rightarrow \alpha=\frac{1}{4}, \beta=\frac{3}{4}$

Q82JEE Main 2021NAT4MSequences and Series

If the value of $(1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\....... . \text{upto }\\infty )^{\log _{(0.25)}(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\...... \text{upto }\\infty )}$ is $l$ , then $l^{2}$ is equal to ________.

🚀 Solve in Practice Mode

📖 Explanation

$l=(1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\......)^{\log _{025}(\frac{1}{3}+\frac{1}{3^{2}}+\......)}$ ${S}=1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\....$ $\frac{{S}}{3}=\frac{1}{3}+\frac{2}{3^{2}}+\frac{6}{3^{3}}+\...... .$ $-\;\;\;-\;\;\;\;\;\;\;\;-\;\;\;\; -$ $────────────────$ $\frac{2 {S}}{3}=1+\frac{1}{3}+\frac{4}{3^{2}}+\frac{4}{3^{3}}+\.....$ $\frac{2 {S}}{3}=\frac{4}{3}+\frac{4}{3^{2}}+\frac{4}{3^{3}}+\......$ ${S}=\frac{3}{2}(\frac{4 / 3}{1-1 / 3})=3$ Now $l=(3)^{\log _{0.2}(\frac{1 / 3}{1-1 / 3})}$ $l=3^{\log _{(144)(\frac{1}{2})}}=3^{1 / 2}=\sqrt{3}$ $\Rightarrowl^2=2$

Q83JEE Main 2021NAT4MStatistics

Consider the following frequency distribution : class : 10-20 20-30 30-40 40-50 50-60 Frequency : alpha 110 54 30 betaIf the sum of all frequencies is 584 and median is 45 , then $|\alpha-\beta|$ is equal to _________.

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📖 Explanation

$\because$ Sum of frequencies $=584$ $\Rightarrow\alpha+\beta=390$ Now, Median is at $\frac{584}{2}=292^{\text{th }}$ $\because$ Median $=45$ (lies in class $40-50)$ $\Rightarrow\alpha+110+54+15=292$ $\Rightarrow\alpha=113, \beta=277$ $\Rightarrow|\alpha-\beta|=164$

Q84JEE Main 2021NAT4MVector Algebra

Let ${\vec{p}}=2 {\hat{i}}+3 \hat{j}+\hat{k}$ and ${{\vec{q}}}={\hat{i}}+2 \hat{j}+{\hat{k}}$ be two vectors. If a vector ${\vec{r}}=(\alpha\hat{i}+\beta{\hat{j}}+\gamma{\hat{k}})$ is perpendicular to each of the vectors $({\vec{p}}+{\vec{q}})$ and $({\vec{p}}-{\vec{q}})$ , and $|{\vec{r}}|=\sqrt{3}$ , then $|\alpha|+|\beta|+|\gamma|$ is equal to _______.

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📖 Explanation

${\vec{p}}=2 {\hat{i}}+3 \hat{j}+\hat{k}$ (Given) ${{\vec{q}}}={\hat{i}}+2 \hat{j}+{\hat{k}}$ Now $({\vec{p}}+{\vec{q}})\times$ $({\vec{p}}-{\vec{q}})=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 1 \\ 1 & 1 & 0\end{vmatrix}$ $=-2\hat{i}-2\hat{j}-2\hat{k}$ $\vec{r}=\pm \sqrt3(({\vec{p}}+{\vec{q}})\times ({\vec{p}}-{\vec{q}}))/{|({\vec{p}}+{\vec{q}})\times ({\vec{p}}-{\vec{q}})|}$ $\pm {\sqrt3(-2\hat{i}-2\hat{j}-2\hat{k})}/{\sqrt{2^{2+}2^{2+}2^2}$ $\vec{r}=\pm (-\hat{i}-\hat{j}-\hat{k})$ According to question $\vec{r}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$ So $|\alpha|=1,|\beta|=1,|\gamma|=1$ $\Rightarrow|\alpha|+|\beta|+|\gamma|=3$

Q85JEE Main 2021NAT4MBinomial Theorem

The ratio of the coefficient of the middle term in the expansion of $(1+x)^{20}$ and the sum of the coefficients of two middle terms in expansion of $(1+x)^{19}$ is ________.

🚀 Solve in Practice Mode

📖 Explanation

Coeff. of middle term in $(1+x)^{20}={ }^{20} C_{10}$ & Sum of Coeff. of two middle terms in $(1+x)^{19}={ }^{19} C_{9}+{ }^{19} C_{10}$ So required ratio $=\frac{{ }^{20} {C}_{10}}{{ }^{19} {C}_{9}+{ }^{19} {C}_{10}}=\frac{{ }^{20} {C}_{10}}{{ }^{20} {C}_{10}}=1$

Q86JEE Main 2021NAT4MMatrices and Determinants

Let ${M}=\{{A}=(\text{table} {a} ,{b} ;{c} , {d} ): {a}, {b}, {c}, {d} \in\{\\pm 3, \\pm 2, \\pm 1,0\}\}$ . Define ${f}: {M} \\to {Z}$ , as ${f}({A})={\det}({A})$ , for all ${A} \in{M}$ where ${Z}$ is set of all integers. Then the number of $A \inM$ such that $f(A)=15$ is equal to ________.

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📖 Explanation

$|{A}|={ad}-{bc}=15$ where $a, b, c, d \in\{\\pm 3, \\pm 2, \\pm 1,0\}$ Case I ad $=9\ \&\ {bc}=-6$ For ad possible pairs are $(3,3),(-3,-3)$ For bc possible pairs are $(3,-2),(-3,2),(-2,3),(2,-3)$ So total matrix $=2 \\times 4=8$ Case II $a d=6\ \&\ b c=-9$ Similarly total matrix $=2 \\times 4=8$ $\Rightarrow$ Total such matrices are $=16$

Q87JEE Main 2021NAT4MPermutations and Combinations

There are 5 students in class 10, 6 students in class 11 and 8 students in class 12. If the number of ways, in which 10 students can be selected from them so as to include at least 2 students from each class and at most 5 students from the total 11 students of class 10 and 11 is 100 k, then k is equal to _______.

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📖 Explanation

$\text{table} \text{Class}, 10^{th}, 11^{th },12^{th} ;\text{Total} \text{student} ,5, 6 ,8;,2 ,3 ,5,\Rightarrow{ }^{5} {C}_{2} \\times { }^{6} {C}_{3} \\times { }^{8} {C}_{5};\text{Number} \text{of} \text{selection}, 2, 2, 6,\Rightarrow { }^{5} {C}_{2} \\times { }^{6} {C}_{2} \\times { }^{8} {C}_{6};,3,2,5,\Rightarrow { }^{5} {C}_{3} \\times { }^{6} {C}_{2} \\times { }^{8} {C}_{5}$ $\Rightarrow$ Total number of ways $=23800$ According to question $100 {K}=23800$ $\Rightarrow{K}=238$

Q88JEE Main 2021NAT4MQuadratic Equation and Inequalities

If $\alpha, \beta$ are roots of the equation ${x}^{2}+5(\sqrt{2}) {x}+10=0, \alpha>\beta$ and ${P}_{{n}}=\alpha^{{n}}-\beta^{{n}}$ for each positive integer ${n}$ , then the value of $(\frac{{P}_{17} {P}_{20}+5 \sqrt{2} {P}_{17} {P}_{19}}{{P}_{18} {P}_{19}+5 \sqrt{2} {P}_{18}^{2}})$ is equal to ________.

🚀 Solve in Practice Mode

📖 Explanation

$x^{2}+5 \sqrt{2} x+10=0$ $\ \ & {p}_{{n}}=\alpha^{{n}}-\beta^{{n}}$ (Given) Now $\frac{{P}_{17} {P}_{20}+5 \sqrt{2} {P}_{17} {P}_{19}}{{P}_{18} {P}_{19}+5 \sqrt{2} {P}_{18}^{2}}$ $=\frac{{P}_{17}({P}_{20}+5 \sqrt{2} {P}_{19})}{{P}_{18}({P}_{19}+5 \sqrt{2} {P}_{18})}$ $\frac{{P}_{17}(\alpha^{20}-\beta^{20}+5 \sqrt{2}(\alpha^{19}-\beta^{19}))}{{P}_{18}(\alpha^{19}-\beta^{19}+5 \sqrt{2}(\alpha^{18}-\beta^{18}))}$ $\frac{{P}_{17}(\alpha^{19}(\alpha+5 \sqrt{2})-\beta^{19}(\beta+5 \sqrt{2}))}{{P}_{18}(\alpha^{18}(\alpha+5 \sqrt{2})-\beta^{18}(\beta+5 \sqrt{2}))}$ Since $\alpha+5 \sqrt{2}=-10 / \alpha......$ (1) $\&\ \beta+5 \sqrt{2}=-10 / \beta........$ (2) Now put there values in above expression $=-\frac{10 P_{17} P_{18}}{-10 P_{18} P_{17}}=1$

Q89JEE Main 2021NAT4MBinomial Theorem

The term independent of ' ${x}'$ in the expansion of $(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}})^{10}$ , where $x \\ne 0,1$ is equal to ________.

🚀 Solve in Practice Mode

📖 Explanation

$((x^{1 / 3}+1)-(\frac{x^{1 / 2}+1}{x^{1 / 2}}))^{10}$ $=(x^{1 / 3}-\frac{1}{x^{1 / 2}})^{10}$ Now General Term ${T}_{{r}+1}={ }^{10} {C}_{{r}}({x}^{1 / 3})^{10-{r}} \.(-\frac{1}{{x}^{1 / 2}})^{{r}}$ For independent term ${10-{r}}{3}-\frac{{r}}{2}=0 \Rightarrow {r}=4$ $\Rightarrow{T}_{5}={ }^{10} {C}_{4}=210$

Q90JEE Main 2021NAT4MMatrices and Determinants

Let ${S}=\{{n} \in{N} \|(\text{table} 0 , {i} ;1, 0 )^{{n}}(\text{table} {a} ,{b} ; {c} , {d} )=(\text{table} {a}, {b} ; {c} ,{d}) \forall {a}, {b}, {c}, {d} \in{R}\}$ where ${i}=\sqrt{-1}$ . Then the number of 2 -digit numbers in the set ${S}$ is

🚀 Solve in Practice Mode

📖 Explanation

Let $X=(\text{table} {a} ,{b} ; {c} , {d} )\ &\ A=(\text{table} 0 , {i} ;1, 0 )^{{n}}$ $\Rightarrow{AX}={IX}$ $\Rightarrow{A}={I}$ $\Rightarrow(\text{table} 0 , {i} ;1, 0 )^{{n}}={I}$ $\RightarrowA^{8}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ \Rightarrow n is multiple of 8 So number of 2 digit numbers in the set ${S}=11(16,24,32, ......., 96)$

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