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20 Jul 2021 Shift 1

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Physics30 questions

Q1JEE Main 2021MCQ4MCurrent Electricity

The value of current in the 6Ω resistance is :

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📖 Explanation

Applying KCL at point P, $\frac{V-0}{6}+\frac{V-90}{5}+\frac{V-140}{20}=0$ $\Rightarrow10 {V}+12 {V}-1080+3 {V}-420=0$ $\Rightarrow{V}=60$ $\therefore$ current in $6 \Omega={{V}-0}{6}=10 {A}$ Hence option 3.

Q2JEE Main 2021MCQ4MCircular Motion

The normal reaction 'N' for a vehicle of $800 {kg}$ mass, negotiating a turn on a $30°$ banked road at maximum possible speed without skidding is _______ $\\times 10^{3} {kg} {m} / {s}^{2}$

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📖 Explanation

At ${v}_{\max }, {f}$ will be limiting in nature. $\therefore$ Balancing force in vertical direction, $N \cos 30°-m g-\muN \cos 60°=0$ $\RightarrowN[\cos 30°-\mu\cos 60°]={mg}$ $\therefore {N}=\frac{800 \\times 10}{(0.87-0.1)} \approx 10.2 \\times 10^{3} {kg} {m} / {s}^{2}$ Hence option 1 .

Q3JEE Main 2021MCQ4MAtoms and Nuclei

A radioactive material decays by simultaneous emissions of two particles with half lives of 1400 years and 700 years respectively. What will be the time after the which one third of the material remains ? (Take ln 3 = 1.1)

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📖 Explanation

Given $\lambda_{1}=\frac{l{n} 2}{700} /$ year, $\lambda_{2}=\frac{l{n} 2}{1400} /$ year $\therefore \lambda_{{net}}=\lambda_{1}+\lambda_{2}=\ln 2[\frac{1}{700}+\frac{1}{1400}]$ $=\frac{3 l{n} 2}{1400} / {\text{year}}$ Now, Let initial no. of radioactive nuclei be No. $\therefore \frac{{N}_{0}}{3}={N}_{0} {e}^{-\lambda_{{net}}}$ $\Rightarrowl{n} \frac{1}{3}=-\lambda_{{net}} {t}$ $\Rightarrow1.1=\frac{3 \\times 0.693}{1400} {t} \Rightarrow{t} \approx 740$ years Hence option 4

Q4JEE Main 2021MCQ4MLaws of Motion

A steel block of $10 {kg}$ rests on a horizontal floor as shown. When three iron cylinders are placed on it as shown, the block and cylinders go down with an acceleration $0.2 {m} / {s}^{2}$ . The normal reaction ${R}'$ by the floor if mass of the iron cylinders are equal and of $20 {kg}$ each, is ______ N. [Take ${g}=10 {m} / {s}^{2}$ and $\mu_{s}=0.2]$

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📖 Explanation

Writing force equation in vertical direction ${Mg}-{N}={Ma}$ $\Rightarrow70 {g}-{N}=70 \\times 0.2$ $\Rightarrow{N}=70[{g}-0.2]=70 \\times 9.8$ $\therefore {N}=686$ Newton Note : Since there is no compressive normal from the sides, hence friction will not act. Hence option 2.

Q5JEE Main 2021MCQ4MAlternating Current

AC voltage $V(t)=20$ sinomegat of frequency $50 {Hz}$ is applied to a parallel plate capacitor. The separation between the plates is $2 {mm}$ and the area is $1 {m}^{2}$ . The amplitude of the oscillating displacement current for the applied AC voltage is _______. $[$ Take $\epsilon_{0}=8.85 \\times 10^{-12} {F} / {m}]$

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📖 Explanation

From the given information, ${C}=\frac{\epsilon_{0} {A}}{{d}}=\frac{\in_{0} \\times 1}{2 \\times 10^{-3}} {F}$ $\therefore X_{{C}}=\frac{1}{\omega{C}} =\frac{2 \\times 10^{-3}}{2 \\times 50 \pi\\times \in_{0}}$ $=\frac{2 \\times 10^{-3}}{25 \\times 4 \pi \in_{0}} \Omega$ $\therefore {X}_{{C}}=\frac{2 \\times 10^{-3}}{25} \\times 9 \\times 10^{9}$ $=\frac{18}{25} \\times 10^{6} \Omega$ $\therefore {i}_{0}={{V}_{0}}/{{X}_{{C}}}=\frac{20 \\times 25}{18} \\times 10^{-6} {A}$ $=27.47 \mu{A} .$ The value of amplitude of displacement current will be same as value of amplitude of conventional current. Hence option 3.

Q6JEE Main 2021MCQ4MRay Optics and Optical Instruments

Region I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of the image from the surface is :

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📖 Explanation

$\frac{\mu_{2}}{{v}}-\frac{\mu_{1}}{{u}}=\frac{\mu_{2}-\mu_{1}}{{R}}$ $\frac{1.4}{{v}}-\frac{1.25}{-40}=\frac{1.4-1.25}{-25}$ $\frac{1.4}{{v}}=-\frac{0.15}{25}-\frac{1.25}{40}$ ${v}=-37.58 {cm}$ Hence option (4)

Q7JEE Main 2021MCQ4MGravitation

A person whose mass is 100 kg travels from Earth to Mars in a spaceship. Neglect all other objects in sky and take acceleration due to gravity on the surface of the Earth and Mars as 10 m/s $^2$ and 4 m/s $^2$ respectively. Identify from the below figures, the curve that fits best for the weight of the passenger as a function of time.

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📖 Explanation

At neutral point g = 0 so graph (C) is correct Given,Acceleration due to gravity on Earth, $g_{E}=10 {m} / {s}^{2}$ and acceleration due to gravity on Mars, $g_{M}=4 {m} / {s}^{2}$ We know that, $g_{E}$ (at height $h)=g_{\;\text{Earth }\;}(1-\;\frac{2 h}{R})$ $\therefore$ Weight at Earth, $m g_{E}=100 \times 10=1000 {N}$ As the spaceship moves far away from Earth, the value of $g_{E}$ decreases to zero at a point where $g_{E}+g_{M}=0$ and hence weight will also be zero.This point is called neutral point and is shown by graph (c) in the given figure.Then, $g_{E}$ increases $4 {ms}^{-2}$ at Mars surface and weight becomes $400 {N}$ which is also exhibited by graph (c).

Q8JEE Main 2021MCQ4MThermodynamics

The amount of heat needed to raise the temperature of 4 moles of a rigid diatomic gas from 0°C to 50°C when no work is done is _______. (R is the universal gas constant)

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📖 Explanation

$\Delta{Q}=\Delta{U}+\Delta{W}$ Here $\Delta{W}=0$ $\DeltaQ=\DeltaU=n C_{V} \DeltaT$ $\Delta{Q}=4 \\times \frac{5 {R}}{2}(50)=500 {R}$ Hence option (4).

Q9JEE Main 2021MCQ4MMotion in a Plane

If $\vec{A}$ and $\vec{B}$ are two vectors satisfying the relation ${\vec{A}} \.{\vec{B}}=|{\vec{A}} \\times \vec{B}| .$ Then the value of $|{\vec{A}}-{\vec{B}}|$ will be

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📖 Explanation

${\vec{A}} \.\vec{B}=|\vec{A} \\times {\vec{B}}|$ ${AB} \cos \theta={AB} \sin \theta \Rightarrow \theta=45°$ $|\vec{A}-{\vec{B}}|=\sqrt{{A}^{2}+{B}^{2}-2 {AB} \cos 45°}$ $=\sqrt{{A}^{2}+{B}^{2}-\sqrt{2} {AB}}$ Hence option (4).

Q10JEE Main 2021MCQ4MMagnetic Effects of Current and Magnetism

A deuteron and an alpha particle having equal kinetic energy enter perpendicular into a magnetic field. Let ${r}_{{d}}$ and ${r}_{\alpha}$ be their respective radii of circular path. The value of $\frac{r_{d}}{r_{\alpha}}$ is equal to :

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📖 Explanation

${r}=\frac{{mv}}{{qB}}={\sqrt{2 {mk}}}/{{qB}}$ ${{r}_{{d}}}/{{r}_{\alpha}}=\sqrt{{{m}_{{d}}}/{{m}_{\alpha}}} {{q}_{\alpha}}/{{q}_{{d}}}$ $=\sqrt{\frac{2}{4}}(\frac{2}{1})=\sqrt{2}$ Hence option (2).

Q11JEE Main 2021MCQ4MAtoms and Nuclei

A nucleus of mass ${M}$ emits $\gamma$ -ray photon of frequency ' $v' .$ The loss of internal energy by the nucleus is : [Take 'c' as the speed of electromagnetic wave]

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📖 Explanation

Energy of $\gamma$ ray $[{E}_{\gamma}]={hv}$ Momentum of $\gamma$ ray $[{P}_{\gamma}]=\frac{{h}}{\lambda}=\frac{{h} v}{{C}}$ Total momentum is conserved. ${\vec{P}}_{\gamma}+\vec{P}_{{Nu}}=0$ Where ${\vec{P}}_{{Nu}}=$ Momentum of decayed nuclei $\Rightarrow{P}_{\gamma}={P}_{{Nu}}$ $\Rightarrow\frac{{hv}}{{C}}={P}_{{Nu}}$ $\Rightarrow$ K.E. of nuclei $=\frac{1}{2} {Mv}^{2}={({P}_{{Nu}})^{2}}/{2 {M}}=\frac{1}{2 {M}}[\frac{{h} v}{{C}}]^{2}$ Loss in internal energy $={E}_{\gamma}+{K} \.{E}_{{Nu}}$ $=h v+\frac{1}{2 M}[\frac{h v}{C}]^{2}$ $={hv}[1+\frac{{h} v}{2 {MC}^{2}}]$

Q12JEE Main 2021MCQ4MElectrostatics

A certain charge $Q$ is divided into two parts $q$ and $({Q}-{q}) .$ How should the charges ${Q}$ and ${q}$ be divided so that ${q}$ and $({Q}-{q})$ placed at a certain distance apart experience maximum electrostatic repulsion?

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📖 Explanation

${F}_{{q}}=\frac{{kq}({Q}-{q})}{{L}^{2}}=\frac{{k}}{{L}^{2}}({q} {Q}-{q}^{2})$ $\frac{{dF}}{{dq}}=0$ when force is maximum $\frac{{dF}}{{dq}}=\frac{{k}}{{L}^{2}}[{Q}-2 {q}]=0$ $\Rightarrow{Q}-2 {q}=0 \Rightarrow{Q}=2 {q}$

Q13JEE Main 2021MCQ4MCurrent Electricity

A current of $5 {A}$ is passing through a non-linear magnesium wire of cross-section $0.04 {m}^{2}$ . At every point the direction of current density is at an angle of $60°$ with the unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is : (Resistivity of magnesium $\rho =44 \\times 10^{-8} \Omega{m}$ )

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📖 Explanation

${I}={\vec{J}} \. {\vec{A}}={J} {A} \cos (\theta)$ $5={J}(\frac{4}{100}) \\times \cos (60)$ ${J}=5 \\times 50=250 {A} / {m}^{2}$ Now, ${\vec{E}}=\rho \.{\vec{J}}$ $=44 \\times 10^{-8} \\times 250=11 \\times 10^{-5} {V} / {m}$

Q14JEE Main 2021MCQ4MThermodynamics

Consider a mixture of gas molecule of types A, B and ${C}$ having masses ${m}_{{A}} < {m}_{{B}} < {m}_{{C}}$ . The ratio of their root mean square speeds at normal temperature and pressure is :

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📖 Explanation

$V_{{RMS}}=\sqrt{\frac{3 {RT}}{{M}}}$ ${m}_{{A}} < {m}_{{B}} < {m}_{{C}}$ $\Rightarrow {v}_{{A}}>{v}_{{B}}>{v}_{{C}}$ $\Rightarrow {1}/{{v}_{{A}}} < {1}/{{v}_{{B}}} < {1}/{{v}_{{C}}}$

Q15JEE Main 2021MCQ4MMotion in a Plane

A butterfly is flying with a velocity $4 \sqrt{2} {m} / {s}$ in North-East direction. Wind is slowly blowing at $1 {m} / {s}$ from North to South. The resultant displacement of the butterfly in 3 seconds is :

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📖 Explanation

$\vec{V}_{{BW}}=4 \sqrt{2} \cos 45 \hat{i}+4 \sqrt{2} \sin 45 \hat{j}$ $=4 {\hat{i}}+4 \hat{j}$ ${{V}}^{ \to }_{{w}}=-\hat{j}$ ${\vec{V}}_{{B}}=\vec{V}_{{BW}}+{\vec{V}}_{{w}}=4 {\hat{i}}+3 {\hat{j}}$ ${\vec{S}}_{{B}}={\vec{V}}_{{B}} \\times {t}=(4 \hat{i}+3 {\hat{j}}) \\times 3$ $=12 {\hat{i}}+9 \hat{j}$ $|{\vec{S}}_{{B}}|=\sqrt{(12)^{2}+(9)^{2}}=15 {m}$

Q16JEE Main 2021MCQ4MProperties of Matter

The value of tension in a long thin metal wire has been changed from ${T}_{1}$ to ${T}_{2}$ . The lengths of the metal wire at two different values of tension ${T}_{1}$ and ${T}_{2}$ are $l_{1}$ and $l_{2}$ respectively. The actual length of the metal wire is :

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📖 Explanation

$Y=\frac{F L}{A \DeltaL}$ $\RightarrowY=\frac{T_{1} l_{0}}{A(l_{1}-l_{0})}=\frac{T_{2} l_{0}}{{A}(l_2-l_0)}$ $1=\frac{T_{1}(l_{2}-l_{0})}{T_{2}(l_{1}-l_{0})}$ $T_{2} l_{1}-T_{2} l_{0}=T_{1} l_{2}-T_{1} l_{0}$ $(T_{1}-T_{2}) l_{0}=T_{1} l_{2}-T_{2} l_{1}$ $l_0=(\frac{{T}_{1} l_{2}-{T}_{2} l_{1}}{{T}_{1}-{T}_{2}})$

Q17JEE Main 2021MCQ4MSemiconductor Electronics

For the circuit shown below, calculate the value of $I_z:$

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📖 Explanation

${I}=\frac{50}{1000}=50 {mA}$ ${I}=\frac{50}{2000}=25 {mA}$ ${I}_{{Z}}={I}_{1000}-{I}_{2000}$ $=50-25=25 {mA}$

Q18JEE Main 2021MCQ4MElectromagnetic Induction

The arm PQ of a rectangular conductor is moving from x = 0 to x = 2b outwards and then inwards from x = 2b to x = 0 as shown in the figure. A uniform magnetic field perpendicular to the plane is acting from x = 0 to x = b. Identify the graph showing the variation of different quantities with distance :

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📖 Explanation

As rod moves in field area increases upto x = b then field is absent and again flux is generated on return journey from x = b to x = 0. Thus plot A for flux. $\Rightarrow{e}=-\frac{{d} \phi}{{dt}} \Rightarrow$ curve ${B}$ for ${emf}$ $\Rightarrow$ Power dissipated $=$ vi $\Rightarrow$ curve $C$ for power dissipated

Q19JEE Main 2021MCQ4MThermodynamics

The entropy of any system is given by ${S}=\alpha^{2} \beta \ln [\frac{\mu{k} {R}}{{J} \beta^{2}}+3]$ where $\alpha$ and $\beta$ are the constants. $\mu, {J}, {k}$ and ${R}$ are no. of moles, mechanical equivalent of heat, Boltzmann constant and gas constant respectively. $[\text{Take }{S}=\frac{{dQ}}{{T}}]$ Choose the incorrect option from the following :

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📖 Explanation

${S}=\alpha^{2} \beta \ln (\frac{\mu{KR}}{{J} \beta^{2}}+3)$ ${S}=\frac{{Q}}{{T}}=$ joulek $/ {k}$ $[\alpha^{2} \beta]=$ Joule $/ {k}$ ${PV}={nRT} \;\;\;\$ [\frac{\mu{KR}}{{J} \beta^{2}}]$=1{R}=\frac{\text{Joule }}{{K}}\Rightarrow{R}=\frac{\text{Joule }}{{K}}, {K}=\frac{\text{Joule }}{{R}}\Rightarrow\beta=(\frac{{Joule}}{{K}})\alpha^{2} \beta=(\frac{{J}_{0} {c} l {e}}{{K}})\Rightarrow\alpha=$ dimensionless

Q20JEE Main 2021MCQ4MDual Nature of Matter and Radiation

The radiation corresponding to $3 \\to 2$ transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of $5 \\times 10^{-4} {T}$ . Assume that the radius of the largest circular path followed by these electrons is $7 {mm}$ , the work function of the metal is: (Mass of electron $=9.1 \\times 10^{-31} {kg}$ )

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📖 Explanation

$1.51 ───── n = 3$ $3.4─────n = 3$ $13.6─────n = 1$ $3 \\to 2 \Rightarrow 1.89 {eV}$ $5 \\times 10^{-4} {T} \;\;\;\;\;{r}=7 {mm}$ ${r}=\frac{{mv}}{{qB}} \Rightarrow{mv}={qrB}$ $\Rightarrow{E}=\frac{{P}^{2}}{2 {m}}=\frac{({qRB})^{2}}{2 {m}}$ $={(1.6 \\times 10^{-19} \\times 7 \\times 10^{-3} \\times 5 \\times 10^{-4})^{2}}/{2 \\times 9.1 \\times 10^{-31} {\text{Joule}}}$ $=\frac{3136 \\times 10^{-52}}{18.2 \\times 10^{-31} \\times 1.6 \\times 10^{-19}} \\times {eV}$ $=1.077 {eV}$ We know work function $=$ energy incident $({KE})_{\text{electron }}$ $\phi=1.89-1.077=0.813 {eV}$

Q21JEE Main 2021NAT4MMotion in a Plane

In a spring gun having spring constant $100 {N} / {m}$ a small ball 'B' of mass $100 {g}$ is put in its barrel (as shown in figure) by compressing the spring through $0.05 {m}$ . There should be a box placed at a distance 'd' on the ground so that the ball falls in it. If the ball leaves the gun horizontally at a height of $2 {m}$ above the ground. The value of ${d}$ is ${m}$ . $({g}=10 {m} / {s}^{2})$

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📖 Explanation

$\frac{1}{2} {kx}^{2}=\frac{1}{2} {mv}^{2}$ ${Kx}^{2}={mv}^{2}$ ${v}={x} \sqrt{\frac{{k}}{{m}}}=0.05 \sqrt{\frac{100}{0.1}}$ $=0.05 \\times 10 \sqrt{10}$ ${v}=0.5 \sqrt{10}$ From ${h}=\frac{1}{2} {gt}^{2}$ ${t}=\sqrt{\frac{2 {h}}{{g}}}=\sqrt{\frac{2 \\times 2}{10}}=\frac{2}{\sqrt{10}}$ $\therefore {d}={vt}=0.5 \sqrt{10} \\times \frac{2}{\sqrt{10}}=1 {m}$

Q22JEE Main 2021NAT4MAlternating Current

In an LCR series circuit, an inductor $30 {mH}$ and a resistor $1 \Omega$ are connected to an AC source of angular frequency $300 {rad} / {s}$ . The value of capacitance for which, the current leads the voltage by $45°$ is $\frac{1}{x} \\times 10^{-3} {F}$ . Then the value of $x$ is ______.

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📖 Explanation

$\tan \phi=\frac{x_{C}-x_{L}}{R}$ $\tan 45=\frac{x_{C}-x_{L}}{R}$ $x_{C}-x_{L}=R$ $\frac{1}{\omegaC}-\omegaL=R$ $\frac{1}{\omegaC}-300 \\times 0.03=1$ $\frac{1}{\omegaC}=10$ $C=\frac{1}{10 \omega}=\frac{1}{10 \\times 300}$ $C=\frac{1}{3} \\times 10^{-3}$ $X=3$

Q23JEE Main 2021NAT4MWaves

The amplitude of wave disturbance propagating in the positive $x$ -direction is given by $y=\frac{1}{(1+x)^{2}}$ at time ${t}=0$ and ${y}=\frac{1}{1+({x}-2)^{2}}$ at ${t}=1 {s}$ , where ${x}$ and $y$ are in meres. The shape of wave does not change during the propagation. The velocity of the wave will be ____ ${m} / {s}$ .

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📖 Explanation

At ${t}=0, {y}=\frac{1}{1+{x}^{2}}$ At time ${t}={t}, {y}=\frac{1}{1+({x}-{vt})^{2}}$ At $t=1, y=\frac{1}{1+( x-v)^{2}}$ .......(i) At ${t}=1, {y}=\frac{1}{1+({x}-2)^{2}}$ ........(ii) Comparing (i) & (ii) ${v}=2 {m} / {s}$

Q24JEE Main 2021NAT4MElectrostatics

A body having specific charge $8 \mu{C} / {g}$ is resting on a frictionless plane at a distance $10 {cm}$ from the wall (as shown in the figure). It starts moving towards the wall when a uniform electric field of $100 {V} / {m}$ is applied horizontally towards the wall. If the collision of the body with the wall is perfectly elastic, then the time period of the motion will be _____ ${s}$ .

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📖 Explanation

$F = ma$ $qE = ma$ $a=\frac{q E}{m}$ Now $d=\frac{1}{2} a t^{2}$ $t=\sqrt{\frac{2 d}{a}}$ $t=\sqrt{{2 {d}}/{(\frac{{qE}}{{m}})}}$ ${t}=\sqrt{{2 \\times 0.1}/{(\frac{8 \\times 10^{-6}}{10^{-3}}) \\times 100}}$ $=\frac{1}{2}$ $\therefore$ Time period $=2 {t}=1 {sec}$

Q25JEE Main 2021NAT4MThermodynamics

In the reported figure, heat energy absorbed by a system in going through a cyclic process is ______ piJ.

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📖 Explanation

For complete cyclic process $\Delta{U}=0$ $\therefore$ from $\Delta{Q}=\Delta {U}+{W}$ $=0+{W}$ $\Delta{Q}={W}$ $=$ Area $=\pi{r}_{1} \.{r}_{2}$ $=\pi\\times (10 \\times 10^{3}) \\times (10 \\times 10^{-3})$ $\Delta{Q}=100 \pi$

Q26JEE Main 2021NAT4MRotational Motion

A circular disc reaches from top to bottom of an inclined plane of length ' $L' .$ When it slips down the plane, it takes time ' ${t}_{1}'$ . When it rolls down the plane, it takes time $t_{2}$ . The value of $\frac{t_{2}}{t_{1}}$ is $\sqrt{\frac{3}{x}} .$ The value of x will be ______.

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📖 Explanation

If disk slips on inclined plane, then it's acceleration $a_{1}={g\sin} \theta$ $L=\frac{1}{2} a_{1} t_{1}^{2}$ $\Rightarrowt_{1}=\sqrt{\frac{2 L}{a_{1}}}$ .......(i) If disk rolls on inclined plane, its acceleration, ${a}_{2}={{g} \sin \theta}/{1+\frac{{I}}{{m} {R}^{2}}}$ ${a}_{2}={{g} \sin \theta}/{1+\frac{{m} {R}^{2}}{2 {m} {R}^{2}}}$ ${a}_{2}=\frac{2}{3} {g} \sin \theta$ Now ${L}=\frac{1}{2} {a}_{2} \.{t}_{2}^{2}$ $\Rightarrow{t}_{2}=\sqrt{\frac{2 {L}}{{a}_{2}}}$ ......(ii) Now $\frac{{t}_{2}}{{t}_{1}}=\sqrt{\frac{{a}_{1}}{{a}_{2}}}=\sqrt{\frac{3}{2}}$ $\Rightarrow{x}=2$

Q27JEE Main 2021NAT4MCentre of Mass and Collision

A rod of mass ${M}$ and length ${L}$ is lying on a horizontal frictionless surface. A particle of mass 'm' travelling along the surface hits at one end of the rod with a velocity 'u' in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses $(\frac{{m}}{{M}})$ is $\frac{1}{{x}} .$ The value of ' $x$ ' will be ______.

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📖 Explanation

Just after collision From momentum conservation, ${P}_{{i}}^{0}={P}_{{f}}$ ${\mu}={Mv} \.......$ (i) From angular momentum conservation about ${O}$ , $m u \.\frac{L}{2}=\frac{M L^{2}}{12} \omega$ $\Rightarrow\omega=\frac{6 {\mu}}{{ML}}$ ........(ii) From e $=\frac{\text{R.V.S }}{\text{R.V.A }}$ $1={{V}+\frac{\omega{L}}{2}}/{{u}}$ ${v}+\frac{\omega {L}}{2}={u}$ ${v}+\frac{3 {\mu}}{{M}}={u}$ $\frac{{\mu}}{{M}}+\frac{3 {\mu}}{{M}}={u}$ $\frac{4 {\mu}}{{M}}={u}$ $\frac{{m}}{{M}}=\frac{1}{4}$ ${X}=4$

Q28JEE Main 2021NAT4MRay Optics and Optical Instruments

An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification '6', gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to ______ cm.

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📖 Explanation

For simple microscope, ${m}=1+\frac{{D}}{{f}_{0}}$ $6=1+\frac{{D}}{{f}_{0}}$ $5=\frac{25}{{f}_{0}}$ ${f}_{0}=5 {cm}$ For compound microscope, ${m}={l\.{D}}/{{f}_{0} \.{f}_{{e}}}$ $12={60 \\times 25}/{5 \.{f}_{{e}}}$ ${f}_{{e}}=25 {cm}$

Q29JEE Main 2021NAT4MWaves

The frequency of a car horn encountered a change from 400 Hz to 500 Hz. When the car approaches a vertical wall. If the speed of sound is 330 m/s. Then the speed of car is ______ km/h.

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📖 Explanation

Wall as an observer Frequency received by wall ${f}_{1}=f_{0}(\frac{{C}}{{C}-{V}})$ Again wall as a source Frequency received by observer on car ${f}_{2}={f}_{1}(\frac{{C}+{V}}{{C}})$ ${f}_{2}={f}_{0}(\frac{{C}+{V}}{{C}-{V}})$ $500=400(\frac{{C}+{V}}{{C}-{V}})$ $\frac{5}{4}=\frac{{C}+{V}}{{C}-{V}}$ ${C}=9 {V}$ ${V}=\frac{{C}}{9}=\frac{330}{9} {m} / {s}$ ${V}=\frac{330}{9} \\times \frac{18}{5}=132$ km / hr

Q30JEE Main 2021NAT4MElectromagnetic Waves

A carrier wave ${V}_{{C}}({t})=160 \sin (2 \pi\\times 10^{6} {t})$ volts is made to vary between $V_{\max }=200 {V}$ and ${V}_{\min }=120 {V}$ by a message signal ${V}_{{m}}({t})={A}_{{m}} \sin (2 \pi\\times 10^{3} {t})$ volts. The peak voltage ${A}_{{m}}$ of the modulating signal is _______.

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📖 Explanation

Maximum amplitude ${A}_{\max }={A}_{{m}}+{A}_{{C}}$ $\Rightarrow{V}_{\max }={V}_{{m}}+{V}_{{C}}$ $200={V}_{{m}}+160$ ${V}_{{m}}=40$ $\therefore$ Peak voltage ${A}_{{m}}=40$

Chemistry30 questions

Q31JEE Main 2021MCQ4MCoordination Compounds

According to the valence bond theory the hybridization of central metal atom is ${dsp}^{2}$ for which one of the following compounds?

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Q32JEE Main 2021MCQ4MBiomolecules

The correct structure of Rhumann's Purple, the compound formed in the reaction of ninhydrin with proteins is :

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Q33JEE Main 2021MCQ4MPractical Organic Chemistry

Green chemistry in day-to-day life is in the use of:

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📖 Explanation

Chlorine gas was used earlier for bleaching paper. These days, hydrogen peroxide $({H}_{2} {O}_{2})$ with suitable catalyst. Tetra chlroroethene $({Cl}_{2} {C}={CCl}_{2})$ was earlier used as solvent for dry cleaning. The compound contaminates the ground water and is also a suspected carcinogen. Replacement of halogenated solvent by liquid ${CO}_{2}$ will result in less harm to groundwater. Hence given statement (4) is correct.

Q34JEE Main 2021MCQ4MCoordination Compounds

The correct order of intensity of colors of the compounds is:

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📖 Explanation

$\text{table} ,[{NiCl}_{4}]^{2-},>,[{Ni}({H}_{2} {O})_{6}]^{2+},>,[{Ni}({CN})_{4}]^{2-};\text{Splitting}\text{energy} \text{order} ,\Delta_{{t}},<,\Delta_{0},<,\Delta_{{sq}} ;\text{absorbed}\text{energy} \text{order},[{NiCl}_{4}]^{2-},<,[{Ni}({H}_{2} {O})_{6}]^{2+},<,[{Ni}({CN})_{4}]^{2-};\text{intensity} \text{of}\text{colour} \text{of}\ compound,[{NiCl}_{4}]^{2-} ,>,[{Ni}({H}_{2} {O})_{6}]^{2+},>,[{Ni}({CN})_{4}]^{2-}$

Q35JEE Main 2021MCQ4MClassification of Elements

The set in which compounds have different natureis :

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📖 Explanation

${B}({OH})_{3}$ acidic and ${H}_{3} {PO}_{3}$ acidic 2) ${B}({OH})_{3}$ acidic and ${Al}({OH})_{3}$ amphoteric 3) ${NaOH}$ basic and ${Ca}({OH})_{2}$ basic 4) ${Be}({OH})_{2}$ amphoteric and ${Al}({OH})_{3}$ amphoteric

Q36JEE Main 2021MCQ4MRedox Reactions

The species given below that does NOT show disproportionation reaction is :

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📖 Explanation

In ${BrO}_{4}^{\⊝}, {Br}$ is in highest oxidation state $(+7)$ , So it cannot oxidise further hence it cannot show disproportionation reaction.

Q37JEE Main 2021MCQ4MSolid State

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Sharp glass edge becomes smooth on heating it upto its melting point. Reason R : The viscosity of glass decreases on melting. Choose the most appropriate answer from the options given below.

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Q38JEE Main 2021MCQ4MBiomolecules

Orlon fibres are made up of :

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Q39JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

Given below are two statements : One is labelled as Assertion A and other is labelled as Reason R. Assertion A : The dihedral angles in $H_2O_2$ in gaseous phase is 90.2° and in solid phase is 111.5°. Reason R : The change in dihedral angle in solid and gaseous phase is due to the difference in the intermolecular forces. Choose the most appropriate answer from the options given below for A and R.

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📖 Explanation

(a) ${H}_{2} {O}_{2}$ structure in gas phase, dihedral angle is $111.5°$ . (b) ${H}_{2} {O}_{2}$ structure in solid phase at $110 {K}$ , dihedral angle is $90.2°$ . Hence given statement (A) is not correct But statement (B) is correct.

Q40JEE Main 2021MCQ4Mp Block Elements

Chemical nature of the nitrogen oxide compound obtained from a reaction of concentrated nitric acid and $P_4O_10$ (in 4 : 1 ratio) is :

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📖 Explanation

$4 {HNO}_{3}+{P}_{4} {O}_{10}$ $↓$ $2 { N}_{2} {O}_{5}+({HPO}_{3})_{4}$ Ans. ${N}_{2} {O}_{5}$ is acidic in nature.

Q41JEE Main 2021MCQ4MSalt Analysis

An inorganic Compound 'X' on treatment with concentrated ${H}_{2} {SO}_{4}$ produces brown fumes and gives dark brown ring with ${FeSO}_{4}$ in presence of concentrated ${H}_{2} {SO}_{4}$ . Also Compound ' ${X}'$ gives precipitate 'Y', when its solution in dilute ${HCl}$ is treated with ${H}_{2} {S}$ gas. The precipitate ' ${Y}'$ on treatment with concentrated ${HNO}_{3}$ followed by excess of ${NH}_{4} {OH}$ further gives deep blue coloured solution, Compound ' ${X}'$ is:

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📖 Explanation

${{NO}_{3}^{-}}_{X(\text{Anion})}+{{H}_{2} {SO}_{4}}_{(\text{Conc}.)} \\to {{NO}_{2} \uparrow }_{\text{Brown} \text{fumes}}+{H}_{2} {O}$ $\text{table} {FeSO}_{4},+,{H}_{2} {SO}_{4},+,{NO}_{3}^{-} ;{\text{Sol}}^{{n}} ,, \text{conc. }, ,{X}$

Q42JEE Main 2021MCQ4MOrganic Chemistry - Some Basic Principles

Among the given species the Resonance stabilised carbocations are:

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Q43JEE Main 2021MCQ4MClassification of Elements

A s-block element (M) reacts with oxygen to form an oxide of the formula $MO_2$ . The oxide is pale yellow in colour and paramagnetic. The element (M) is:

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📖 Explanation

(A) $2 {Mg}+{O}_{2} \\to 2 {MgO}$ (Diamagnetic) (B) $2 {Na}+{O}_{2} \\to {Na}_{2} {O}$ (Diamagnetic) $2 {Na}+{O}_{2} _{(\text{excess})}\\to {Na}_{2} {O}_{2}$ (Diamagnetic) (C) $2 {Ca}+{O}_{2} \\to 2 {CaO}$ (Diamagnetic) ${Ca}+{O}_{2} \\to {CaO}_{2}$ (Diamagnetic) (D) ${K}+{O}_{2} _{(\text{excess})}\\to {KO}_{2}$ (Paramagnetic)

Q44JEE Main 2021MCQ4MAlcohols, Phenols and Ethers

In the given reaction 3-Bromo-2, 2 -dimethyl butane ${\to }^{{C}_{2} {H}_{3} {OH}} {\text{'A' }}_{\text{(Major Product) }}$ Product A is:

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Q45JEE Main 2021MCQ4Md and f Block Elements

The metal that can be purified economically by fractional distillation method is:

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Q46JEE Main 2021MCQ4MAmines

Compound A is converted to B on reaction with $CHCl_3$ and KOH. The compound B is toxic and can be decomposed by C. A, B and C respectively are :

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Q47JEE Main 2021MCQ4MSolutions

The conditions given below are in the context of observing Tyndall effect in colloidal solutions: (A) The diameter of the colloidal particles is comparable to the wavelength of light used. (B) The diameter of the colloidal particles is much smaller than the wavelength of light used. (C) The diameter of the colloidal particles is much larger than the wavelength of light used. (D) The refractive indices of the dispersed phase and the dispersion medium are comparable. (E) The dispersed phase has a very different refractive index from the dispersion medium. Choose the most appropriate conditions from the options given below:

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Q48JEE Main 2021MCQ4MBiomolecules

Identify the incorrect statement from the following

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📖 Explanation

Amylose is a linear chain polymer of $\alpha$ -D-glucose while amylopectine is branched chain polymer of $\alpha-D$ glucose

Q49JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Which among the above compound/s does/do not form Silver mirror when treated with Tollen's reagent?

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Q50JEE Main 2021MCQ4MHydrocarbons

For above chemical reactions, identify the correct statement from the following:

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Q51JEE Main 2021NAT4MChemical Bonding and Molecular Structure

The number of lone pairs of electrons on the central I atom in ${I}_{3}^{-}$ is _________.

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📖 Explanation

${I}_{3}^{-}:$ The number of lone pairs of electron on the central atom is 3.

Q52JEE Main 2021NAT4MIonic Equilibrium

$250 {mL}$ of $0.5 {M} {NaOH}$ was added to $500 {mL}$ of $1 {M}$ HCl. The number of unreacted HCl molecules in the solution after complete reaction is _______ $\\times 10^{21} .$ (Nearest integer) $({N}_{{A}}=6.022 \\times 10^{23})$

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📖 Explanation

We known that no. of moles $={V}_{\text{litre }} \\times$ Molarity & No. of millimoles $={V}_{{ml}} \\times$ Molarity so millimoles of ${NaOH}=250 \\times 0.5=125$ Millimoles of ${HCl}=500 \\times 1=500$ Now reaction is $\text{table} ,{NaOH},+,{HCl}, \\to ,{NaCl},+,{H}_{2} {O};t = 0, 125,, 500,, 0,, 0; t = t, 0,, 375,, 125 ,,125$ so millimoles of ${HCl}$ left $=375$ Moles of ${HCl}=375 \\times 10^{-3}$ No. of HCl molecules $=6.022 \\times 10^{23} \\times 375 \\times 10^{-3}$ $=225.8 \\times 10^{21}$ $\simeq 226 \\times 10^{21}=226$

Q53JEE Main 2021NAT4MStructure of Atom

The Azimuthal quantum number for the valence electrons of $Ga^{+}$ ion is _______. (Atomic number of Ga = 31)

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📖 Explanation

${Ga}^{+}: {Is}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{10} 4 {s}^{2}$ The azimuthal quantum number for the valenceelectrons (4s-subshell) of ${Ga}^{+}$ ion is zero(0).

Q54JEE Main 2021NAT4MCoordination Compounds

The spin-only magnetic moment value for thecomplex $[{Co}({CN})_{6}]^{4-}$ is _______ ${BM}$ . [At. no. of ${Co}=27]$

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📖 Explanation

$[{Co}({CN})_{6}]^{4-}$ $x+6 \\times (-1)=-4$ $x=+2$ ${Co}^{2+}:[{Ar}] 3 {d}^{7}$ and ${CN}^{-}$ is a strong field ligand which can pair electron of central atom. It has one unpaired electron (n) in 4 d-subshell. So spin only magnetic moment $(\mu)=$ $\sqrt{n(n+2)}$ B.M where ${n}=$ number of unpaired electrons. $\mu=\sqrt{3}$ B.M $\mu=1.73 {BM}$

Q55JEE Main 2021NAT4MChemical Equilibrium

$2 {SO}_{2}({g})+{O}_{2}({g}) \⇌2 {SO}_{3}({g})$ In an equilibrium mixture, the partial pressures are ${P}_{{SO}_{3}}=43 {kPa} ; {P}_{{O}_{2}}=530 {Pa}$ and ${P}_{{SO}_{2}}=45 {kPa}$ . The equilibrium constant ${K}_{{P}}=$ ______ $\\times 10^{-2} .$ (Nearest integer)

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📖 Explanation

$2 {SO}_{2}({g})+{O}_{2}({g}) =2 {SO}_{3}({g})$ ${K}_{{P}}={({pSO}_{3({g})})^{2}}/{{pSO} 2({g})} \\times {pO}_{2({g})}$ $=\frac{43 \\times 43}{45 \\times 45} \\times 530 {Pa}^{-1}$ $=172.28 \\times 10^{-5} {Pa}^{-1}$ $=172.28 {atm}$ $=17228 \\times 10^{-2} {atm}$

Q56JEE Main 2021NAT4MAldehydes, Ketones and Carboxylic Acids

The number of nitrogen atoms in a semicarbazone molecule of acetone is _________.

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Q57JEE Main 2021NAT4MAlcohols, Phenols and Ethers

To synthesise 1.0 mole of 2-methylpropan-2-ol from Ethylethanoate ________ equivalents of $CH_3MgBr$ reagent will be required. (Integer value)

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Q58JEE Main 2021NAT4MChemical Kinetics

The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute after preparation, $10 \%$ of the virus is inactivated. The rate constant for viral inactivation is __________ $\\times 10^{-3} {min}^{-1}$ . (Nearest integer) [Use : $\ln 10=2.303 ; \log _{10} 3=0.477$ ;property of logarithm $: \log x^{y}=y \log x]$

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📖 Explanation

As the unit of rate constant is $\min ^{-1}$ so it must be a first order reaction ${K} \\times {t}=2.303 \log {A}_{0} / {A}_{{t}}$ in 1 min $10 \%$ is in activated so tabing ${A}_{0}=100 \;\;\; {A}_{{t}}=90$ in $1 {min}$ So ${K} \\times 1=2.303 \\times \log \frac{100}{90}$ $=2.303 \\times (\log 10-2 \log 3)$ $=2.303 \\times (1-2 \\times 0.477)$ $=0.10593$ $=105.93 \\times 10^{-3}$ $\approx 106$

Q59JEE Main 2021NAT4MBiomolecules

An average person needs about $10000 {kJ}$ energy per day. The amount of glucose (molar mass $=180.0 {~g} {mol}^{-1}$ ) needed to meet this energy requirement is ________ ${g}$ . (Use: $\Delta_{{C}} {H}($ glucose $)=-2700 {kJ} {mol}^{-1}$ )

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📖 Explanation

1 mole glucose give $2700 {kJ}$ energy so mole of glucose needed for $10^{5} {kJ}$ energy $=\frac{10000}{2700}=370$ moles wt. of glucose $=3.10 \\times 180$ $= 666.666$ $\approx 667$ gm ${Y_{\text{Benzene }}}/{Y_{\text{M.B }}}=\frac{P°_{{B}} X_{{B}}}/{P°_{{MB}} X_{{MB}}}$ $=\frac{70 \\times 1}{20 \\times 1}=\frac{7}{2}$ ${Y}_{\text{Benzene}}=\frac{7}{9}=77.77 \\times 10^{-2}$ $=78 \\times 10^{-12}$

Q60JEE Main 2021NAT4MSolutions

At 20°C, the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at 20°C above an equimolar mixture of benzene and methyl benzene is ________ $\\times 10^{-2}$ . (Nearest integer)

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📖 Explanation

${P}°_{{B}} =40$ ${P}°_{{T}} =20$ ${K}_{{B}}=0.5={K}_{{T}}$ Now ${y}_{{B}}={{K}_{{B}} {P}°_{{B}} }/{{K}_{{B}} {P}°_{{B}} +{K}_{{T}} {P}°_{{T}} }$ $=\frac{70 \\times 0.5}{70 \\times 0.5+20 \\times 0.5}$

Mathematics30 questions

Q61JEE Main 2021MCQ4MSets and Relations

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

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📖 Explanation

$p$ $q$ $∼p$ $∼q$ $p∧∼q$ $q∨∼p$ $({p} \wedge \sim q) \Rightarrow(q \vee \sim p)$ $p\Rightarrow q$ T F F T T F F F F T T F F T T T T T F F F T T T F F T T F T T T $\therefore ({p} \wedge \sim q) \Rightarrow(q \vee \sim p)$ $\equiv p\Rightarrow q$ So, option (2) is correct.

Q62JEE Main 2021MCQ4MDefinite Integrals

Let a be a positive real number such that $\int_{0} ^{a} {e}^{{x}-[{x}]} {dx}=10 {e}-9$ where $[{x}]$ is the greatest integer less than or equal to $x$ . Then a is equal to :

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📖 Explanation

$a > 0$ Let ${n} \\le {a} < {n}+1, {n} \ \in{W}$ $\text{table} \therefore {a}=,[{a}],+\{{a}\};,⇓,⇓ ;, \text{G}.\text{I}.\text{F} , \text{Fractional} \text{part}$ Here $[{a}]={n}$ Now, $\int_{0} ^{a} {e}^{x-[x]} d x=10 e-9$ $\Rightarrow \int_{0} ^{{n}} {e}^{\{{x}\}} {dx}+\int_{{n}} ^{{a}} {e}^{{x}-[{x}]} {d} {x}=10 {e}-9$ $\therefore n \int_{0} ^{1} e^{x} d x+\int_{n} ^{a} e^{x-n} d x=10 e-9$ $\Rightarrow{n}({e}-1)+({e}^{{a}-{n}}-1)=10 {e}-9$ $\therefore {n}=0$ and $\{{a}\}=\log _{{e}} 2$ So, $a=[a]+\{a\}=(10+\log _{e} 2)$ $\Rightarrow$ Option (2) is correct.

Q63JEE Main 2021MCQ4MStatistics

The mean of 6 distinct observations is $6.5$ and their variance is $10.25$ . If 4 out of 6 observations are 2 ,4,5 and 7, then the remaining two observationsare:

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📖 Explanation

Let other two numbers be a, (21-a) Now, $10.25$ $=\frac{(4+16+25+49+{a}^{2}+(21-{a})^{2})}{6}$ $-(6.5)^{2}$ (Using formula for variance) $\Rightarrow6(10.25)+6(6.5)^{2}=94+{a}^{2}+(21-{a})^{2}$ $\Rightarrowa^{2}+(21-a)^{2}=221$ $\therefore {a}=10$ and $(21-{a})=21-10=11$ So, remaining two observations are 10,11 . $\Rightarrow$ Option (1) is correct.

Q64JEE Main 2021MCQ4MDefinite Integrals

The value of the integral $\int_{-1} ^{1} \log _{e}(\sqrt{1-x}+\sqrt{1+x}) d x$ is equal to :

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📖 Explanation

Let $I=2 \int_{0} ^{1} {\ln (\sqrt{1-x}+\sqrt{1+x})} 1 d x$ $\therefore I=2[(x \.\ln (\sqrt{1-x}+\sqrt{1-x}))_{0}^{1}$ $- \int_{0} ^{1} x \.(\frac{1}{\sqrt{1-x}+\sqrt{1+x}}) \.(\frac{1}{2 \sqrt{1+x}}-\frac{1}{2 \sqrt{1-x}}) d x$ $=2(\ln \sqrt{2}-0)-\frac{2}{2} \ \int_{0} ^{1} {x \sqrt{1-x}-\sqrt{1+x} d x}/{(\sqrt{1-x}+\sqrt{1+x}) \sqrt{1-x^{2}}}$ $=(\log _{{e}} 2)-\int_{0} ^{1} {{x} \.(2-2 \sqrt{1-{x}^{2}})}/{-2 {x} \sqrt{1-{x}^{2}}} {dx}$ (After rationalisation) $=(\log _{{e}} 2)+int_{0} ^{1}({1-\sqrt{1-{x}^{2}}}/{\sqrt{1-{x}^{2}}}) {d} {x}$ $=(\log _{{e}} 2)+(\sin ^{-1} {x})_{0}^{1}-1$ $=\log _{{e}} 2+(\frac{\pi}{2}-0)-1$ $\therefore {I}=(\log _{{e}} 2)+\frac{\pi}{2}-1$ $\Rightarrow$ Option (3) is correct.

Q65JEE Main 2021MCQ4MQuadratic Equation and Inequalities

If $\alpha$ and $\beta$ are the distinct roots of the equation $x^{2}+(3)^{1 / 4} x+3^{1 / 2}=0$ , then the value of $\alpha^{96}(\alpha^{12}-1)+\beta^{96}(\beta^{12}-1)$ is equal to :

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📖 Explanation

As, $(\alpha^{2}+\sqrt{3})=-(3)^{1 / 4} \.\alpha$ $\Rightarrow(\alpha^{4}+2 \sqrt{3} \alpha^{2}+3)=\sqrt{3} \alpha^{2}$ (On squaring) $\therefore (\alpha^{4}+3)=(-) \sqrt{3} \alpha^{2}$ $\Rightarrow\alpha^{8}+6 \alpha^{4}+9=3 \alpha^{4}$ (Again squaring) $\therefore \alpha^{8}+3 \alpha^{4}+9=0$ $\Rightarrow\alpha^{8}=-9-3 \alpha^{4}$ (Multiply by $\alpha^{4}$ ) So, $\alpha^{12}=-9 \alpha^{4}-3 \alpha^{8}$ $\therefore \alpha^{12}=-9 \alpha^{4}-3(-9-3 \alpha^{4})$ $\Rightarrow\alpha^{12}=-9\alpha^{4}+27+9 \alpha^{4}$ Hence, $\alpha^{12}=(27)^{2}$ $\Rightarrow (\alpha^{12})^{8}=(27)^{8}$ $\Rightarrow\alpha^{96}=(3)^{24}$ Similarly $\beta^{96}=(3)^{24}$ $\therefore \alpha^{96}(\alpha^{12}-1)+\beta^{96}(\beta^{12}-1)=(3)^{24} \\times 52$ $\Rightarrow$ Option (3) is correct.

Q66JEE Main 2021MCQ4MMatrices and Determinants

Let ${A}=\begin{bmatrix}2 & 3 \\ {a} & 0\end{bmatrix}, {a} \in{R}$ be written as ${P}+{Q}$ where ${P}$ is a symmetric matrix and ${Q}$ is skew symmetric matrix. If ${\det}(Q)=9$ , then the modulus of the sum of all possible values of determinant of ${P}$ is equal to :

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📖 Explanation

${A}=\begin{bmatrix}2 & 3 \\ {a} & 0\end{bmatrix}, {a} \in{R}$ and $P=\frac{A+A^{T}}{2}=\begin{bmatrix}2 & \frac{3+a}{2} \\ \frac{a+3}{2} & 0\end{bmatrix}$ and $P=\frac{A-A^{T}}{2}=\begin{bmatrix}0 & \frac{3-a}{2} \\ \frac{a-3}{2} & 0\end{bmatrix}$ As, ${\det}(Q)=9$ $\Rightarrow(a-3)^{2}=36$ $\Rightarrowa=3 \\pm 6$ $\therefore a=9,-3$ $=0-\frac{(a-3)^{2}}{4}=0$ , for $a=-3$ $=0-\frac{({a}-3)^{2}}{4}=-\frac{1}{4}(12)(12)$ , for ${a}=9$ $\therefore$ Modulus of the sum of all possible values ofdet. $({P})=|-36|+|0|=36$ Ans. $\Rightarrow$ Option (1) is correct

Q67JEE Main 2021MCQ4MComplex Numbers

If $z$ and $\omega$ are two complex numbers such that $|z \omega|=1$ and $\arg (z)-\arg (\omega)=\frac{3 \pi}{2}$ , then $\arg (\frac{1-2 \overline{z} \omega}{1+3 \overline{z} \omega})$ is : (Here $\arg ({z})$ denotes the principal argument of complex number z)

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📖 Explanation

As $|z \omega|=1$ $\Rightarrow$ If $|z|=r$ , then $|\omega|=\frac{1}{r}$ Let $\arg (z)=\theta$ $\therefore \arg (\omega)=(\theta-\frac{3 \pi}{2})$ So, $z=r e^{i \theta}$ $\Rightarrow\overline{{Z}}={re}^{{i}(-\theta)}$ $\omega=\frac{1}{{r}} {e}^{{i}(\theta-\frac{3 \pi}{2})}$ Now, consider $\frac{1-2 \overline{z} \omega}{1+3 \overline{z} \omega}={1-2 {e}^{{i}(-\frac{3 \pi}{2})}}/{1+3 {e}^{(-\frac{3 \pi}{2})}}=(\frac{1-2 {i}}{1+3 {i}})$ $=\frac{(1-2 {i})(1-3 {i})}{(1+3 {i})(1-3 {i})}=-\frac{1}{2}(1+{i})$ $\therefore$ prin $\arg ({1-2 \overline{{z}} \omega}/{1+3 \overline{{z}} \omega})$ $=$ prin $\arg ({1-2 \overline{{z}} \omega}/{1+3 \overline{{z}} \omega})$ $=(-\frac{1}{2}(1+{i}))$ $=-(\pi-\frac{\pi}{4})=\frac{-3 \pi}{4}$ So, option (2) is correct.

Q68JEE Main 2021MCQ4MTrigonometric Ratios and Identities

If in a triangle ${ABC}, {AB}=5$ units, $\angle {B}=\cos ^{-1}(\frac{3}{5})$ and radius of circumcircle of $\Delta {ABC}$ is 5 units,then the area (in sq. units) of $\Delta{ABC}$ is :

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📖 Explanation

As, $\cos B=\frac{3}{5} \RightarrowB=53°$ ${\text{As}}, {R}=5 \Rightarrow\frac{{c}}{\sin {c}}=2 {R}$ $\Rightarrow\frac{5}{10}=\sin c \RightarrowC=30°$ Now, $\frac{{b}}{\sin {B}}=2 {R} \Rightarrow{b}=2(5)(\frac{4}{5})=8$ Now, by cosine formula $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}$ $\Rightarrow\frac{3}{5}=\frac{{a}^{2}+25-64}{2(5) {a}}$ $\Rightarrow{a}^{2}-6 {a}-3 {g}=0$ $\therefore {a}=\frac{6 \\pm \sqrt{192}}{2}=\frac{6 \\pm 8 \sqrt{3}}{2}$ $\Rightarrow3+4 \sqrt{3} ($ Reject ${a}=3-4 \sqrt{3})$ Now, $\Delta=\frac{{abc}}{4 {R}}=\frac{(3+4 \sqrt{3})(8)(5)}{4(5)}$ $=2(3+4 \sqrt{3})$ $\Rightarrow\Delta=(6+8 \sqrt{3})$ $\Rightarrow$ Option (3) is correct.

Q69JEE Main 2021MCQ4MFunctions

Let $[{x}]$ denote the greatest integer $\\le {x}$ , where ${x} \in{R}$ . If the domain of the real valued function ${f}({x})=\sqrt{\frac{|[{x}]|-2}{|[{x}]|-3}}$ is $(-\\infty , a) \cup[{b}, {c}) \cup[4, \\infty ), {a }< {b} < {c}$ , then the value of ${a}+{b}+{c}$ is:

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📖 Explanation

For domain, $\frac{|[x]|-2}{|[x]|-3} \\ge 0$ Case I : When $|[x]|-2 \\ge 0$ and $|[x]|-3>0$ $\therefore {x} \in(-\\infty ,-3) \cup[4, \\infty )$ ........(1) Case II : When $|[x]|-2 \\ge 0$ and $|[x]|-3 < 0$ $\therefore {x} \in[-2,3)$ .......(2) So, from (1) and (2) we get Domain of function $=(-\\infty ,-3) \cup[-2,3) \cup[4, \\infty )$ $\therefore (a+b+c)=-3+(-2)+3$ $=-2(a < b < c)$ $\Rightarrow$ Option (3) is correct.

Q70JEE Main 2021MCQ4MDifferential Equations

Let $y=y(x)$ be the solution of the differential equation $x \tan (\frac{y}{x}) d y=(y \tan (\frac{y}{x})-x) d x$ $-1 \\le {x} \\le 1, {y}(\frac{1}{2})=\frac{\pi}{6} .$ Then the area of the region bounded by the curves $x=0, x=\frac{1}{\sqrt{2}}$ and $y=y(x)$ in the upper half plane is:

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📖 Explanation

We have $\frac{d y}{d x}=\frac{x(\frac{y}{x} \.\tan \frac{y}{x}-1)}{x \tan \frac{y}{x}}$ $\therefore \frac{{dy}}{{dx}}=\frac{{y}}{{x}}-\cot (\frac{{y}}{{x}})$ Put $\frac{{y}}{{x}}={v}$ $\Rightarrow{y}={vn}$ $\therefore \frac{{dy}}{{dx}}={v}+\frac{{ndv}}{{dx}}$ Now, we get $v+n \frac{d v}{d x}=v-\cot (v)$ $\Rightarrow\int(\tan ) {dv}=-\int\frac{{dx}}{{x}}$ $\therefore \ln |\sec (\frac{y}{x})|=-ln |x|+c$ ${\text{As}}(\frac{1}{2})=(\frac{{y}}{{x}}) \Rightarrow{C}=0$ $\therefore \sec (\frac{{y}}{{x}})=\frac{1}{{x}}$ $\Rightarrow\cos (\frac{{y}}{{x}})={x}$ $\therefore y=x \cos ^{-1}(x)$ So, required bounded area $=\int_{0} ^{1 / \sqrt{2}} {{x}}(\cos ^{-1} {x}) {d} {x}=(\frac{\pi-1}{8})$ $\therefore$ option (1) is correct.

Q71JEE Main 2021MCQ4MBinomial Theorem

The coefficient of ${x}^{256}$ in the expansion of $(1-x)^{101}(x^{2}+x+1)^{100}$ is:

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📖 Explanation

$(1-x)^{100} \.(x^{2}+x+1)^{100} \.(1-x)$ $=((1-x)(x^{2}+x+1))^{100}(1-x)$ $=(1^{3}-x^{3})^{100}(1-x)$ $=(1-x^{3})^{100}(1-x)$ $={(1-x^{3})^{100}} -{x(1-x^{3})^{100}}$ Required coefficient $(-1) \\times (-{ }^{100} {C}_{85})$ $={ }^{100} {C}_{85}={ }^{100} {C}_{15}$

Q72JEE Main 2021MCQ4MMatrices and Determinants

Let ${A}=[{a}_{{i} j}]$ be a $3 \\times 3$ matrix, where $a_{i j}=\{ \text{table} 1 , {, }, \text{if }{ i=j } ;-x , {, }, \text{if }{|i-j|}=1 ;2 x+1 , {, }, \text{otherwise }$ Let a function ${f}: {R} \\to {R}$ be defined as ${f}({x})={\det}({A})$ Then the sum of maximum and minimum values of f on $\mathbf{R}$ is equal to:

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📖 Explanation

$A=\begin{bmatrix}1 & -x & 2 x+1 \\ -x & 1 & -x \\ 2 x+1 & -x & 1\end{bmatrix}$ $|A|=4 x^{3}-4 x^{2}-4 x=f(x)$ $f'(x)=4(3 x^{2}-2 x-1)=0$ $\Rightarrowx=1 ; x=\frac{-1}{3}$ $\therefore {f}(1)=-4 ; f(-\frac{1}{3})=\frac{20}{27}$ Sum $=-4+\frac{20}{27}=-\frac{88}{27}$

Q73JEE Main 2021MCQ4MVector Algebra

Let ${\vec{a}}=2 \hat{i}+\hat{j}-2 \hat{k}$ and ${{\vec{b}}}=\hat{i}+\hat{j}$ . If ${\vec{c}}$ is a vector such that $\vec{a} \.\vec{c}=|\vec{c}|,|\vec{c}-\vec{a}|=2 \sqrt{2}$ and the angle between $(\vec{a} \\times {\vec{b}})$ and $\vec{c}$ is $\frac{\pi}{6}$ , then the value of $|({{\vec{a}}} \\times \vec{b}) \\times \vec{c}|$ is

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📖 Explanation

$|{\vec{a}}|=3={a} ; {\vec{a}} \.{\vec{c}}={c}$ Now $|{{\vec{c}}}-{\vec{a}}|=2 \sqrt{2}$ $\Rightarrow{c}^{2}+{a}^{2}-2 {\vec{c}} \.{\vec{a}}=8$ $\Rightarrow {c}^{2}+9-2({c})=8$ $\Rightarrow{c}^{2}-2 {c}+1=0 \Rightarrow{c}=1=|{\vec{c}}|$ Also, ${\vec{a}} \\times {{\vec{b}}}=2 {\hat{i}}-2 {\hat{j}}+{\hat{k}}$ Given $({\vec{a}} \\times {\vec{b}})={|{\vec{a}} \\times {\vec{b}} \|} {|{{\vec{c}}}|} \sin \frac{\pi}{6}$ $=(3)(1)(1 / 2)$ $=3 / 2$

Q74JEE Main 2021MCQ4MInverse Trigonometric Functions

The number of real roots of the equation $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{4}$ is :

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📖 Explanation

$\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{4}$ For equation to be defined, $x^{2}+x \\ge 0$ $\Rightarrowx^{2}+x+1 \\ge 1$ $\therefore$ only possibility that the equation is defined $x^{2}+x=0 \;\;\;\Rightarrowx=0 ; x=--1$ None of these values satisfy $\therefore$ No of roots $=0$

Q75JEE Main 2021MCQ4MDifferential Equations

Let ${y}={y}({x})$ be the solution of the differential equation ${e}^{{x}} \sqrt{1-{y}^{2}} {dx}+(\frac{{y}}{{x}}) {dy}=0,$ ${y}(1)=-1$ .Then the value of $({y}(3))^{2}$ is equal to:

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📖 Explanation

${e}^{{x}} \sqrt{1-{y}^{2}} {d} {x}+\frac{{y}}{{x}} {dy}=0$ $\Rightarrow{e}^{{x}} \sqrt{1-{y}^{2}} {d} {x}+\frac{-{y}}{{x}} {dy}$ $\Rightarrow\int {-{y}}/{\sqrt{1-{y}^{2}}} {dy}=\int {{e}^{{x}}} {{x}} {dx}$ $\Rightarrow\sqrt{1-{y}^{2}}={e}^{{x}}({x}-1)+{c}$ Given $:$ At ${x}=1, {y}=-1$ $\Rightarrow0=0+{c} \Rightarrow{c}=0$ $\therefore \sqrt{1-{y}^{2}}={e}^{{x}}({x}-1)$ At ${x}=3$ $1-{y}^{2}=({e}^{3} 2)^{2} \Rightarrow{y}^{2}=1-4 {e}^{6}$

Q76JEE Main 2021MCQ4MApplication of Derivatives

Let 'a' be a real number such that the function ${f}({x})={ax}^{2}+6 {x}-15, {x} \in{R}$ is increasing in $(-\\infty , \frac{3}{4})$ and decreasing in $(\frac{3}{4}, \\infty )$ . Then the function $g(x)=a x^{2}-6 x+15, x \in{R}$ has a:

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📖 Explanation

$\frac{-B}{2 A}=\frac{3}{4}$ $\Rightarrow\frac{-(6)}{2 a}=\frac{3}{4}$ $\Rightarrowa=\frac{-6 \\times 4}{6} \Rightarrowa=-4$ $\therefore g(x)=4 x^{2}-6 x+15$ Local max. at $x=\frac{-B}{2 A}=-\frac{(-6)}{2(-4)}$ $={-3}/4$

Q77JEE Main 2021MCQ4MLimits, Continuity and Differentiability

Let a function $f: {R} \\to {R}$ be defined as $f(x)= \begin{cases}\sin x-e^{x} & \text{if }x \\le 0 \\ a+[-x] & \text{if } 0 < x < 1 \\ 2 x-b & \text{ if }x \\ge 1$ Where $[{x}]$ is the greatest integer less than or equal to ${x}$ \end{cases} If $f$ is continuous on ${R}$ , then $({a}+{b})$ is equal to:

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📖 Explanation

Continuous at ${x}=0$ ${f}(0^{+})={f}(0^{-}) \Rightarrow{a}-1=0-{e}^{0}$ $\Rightarrow a=0$ Continuous at ${x}=1$ ${f}(1^{+})={f}(1^{-})$ $\Rightarrow2(1)-{b}={a}+(-1)$ $\Rightarrow{b}=2-{a}+1 \Rightarrow{b}=3$ $\therefore {a}+{b}=3$

Q78JEE Main 2021MCQ4MProbability

Words with or without meaning are to be formed using all the letters of the word EXAMINATION. The probability that the letter ${M}$ appears at the fourth position in any such word is:

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📖 Explanation

AAEIIMNNOTX --------M---------- Total words with ${M}$ at fourth Place $=\frac{10 !}{2 ! 2 ! 2 !}$ Total words $=\frac{11 !}{2 ! 2 ! 2 !}$ Required probability $=\frac{10 !}{11 !}=\frac{1}{11}$

Q79JEE Main 2021MCQ4MProbability

The probability of selecting integers ${a} \in[-5,30]$ such that $x^{2}+2(a+4) x-5 a+64>0$ , for all ${x} \in{R}$ , is:

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📖 Explanation

${D} < 0$ $\Rightarrow 4(a+4)^{2}-4(-5 a+64) < 0$ $\Rightarrowa^{2}+16+8 a+5 a-64 < 0$ $\Rightarrow{a}^{2}+13 {a}-48 < 0$ $\Rightarrow (a+16)(a-3) < 0$ $\Rightarrow{a} \in(-16,3)$ $\therefore$ Possible a : $\{-5,-4, ......, 3\}$ $\therefore$ Required probability $=\frac{8}{36}$ $=2/9$

Q80JEE Main 2021MCQ4MParabola

Let the tangent to the parabola ${S}: {y}^{2}=2 {x}$ at the point ${P}(2,2)$ meet the ${x}$ -axis at ${Q}$ and normal at it meet the parabola $S$ at the point $R$ . Then the area (in sq. units) of the triangle PQR is equal to:

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📖 Explanation

Tangent at $P: y(2)=2(1 / 2)(x+2)$ $\Rightarrow2 y=x+2$ $\therefore Q=(-2,0)$ Normal at $P: y-2=-\frac{(2)}{2 \.1 / 2}(x-2)$ $\Rightarrowy-2=-2(x-2)$ $\Rightarrowy=6-2 x$ $\therefore$ Solving with $y^{2}=2 x \RightarrowR(\frac{9}{2}-3)$ $\therefore {Ar}(\Delta{PQR})=\frac{1}{2}\begin{vmatrix}2 & 2 & 1 \\ -2 & 1 & 1 \\ \frac{9}{2} & 3- & 1\end{vmatrix}$ $\therefore \frac{25}{2}$ sq.units

Q81JEE Main 2021NAT4MVector Algebra

Let ${\vec{a}}, \vec{b}, {\vec{c}}$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $\theta$ , with the vector ${\vec{a}}+\vec{b}+\vec{c}$ . Then $36 \cos ^{2} 2 \theta$ is equal to _______.

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📖 Explanation

$|\vec{a}+\vec{b}+\vec{c}|^2={|\vec{a}|}^{2+}{|\vec{b}|}^{2+}{|\vec{c}|}^2$ $+2(\vec{a}.\vec{b}+\vec{a}.\vec{c}+\vec{b}.\vec{c})=3$ $\Rightarrow |\vec{a}+\vec{b}+\vec{c}|=\sqrt3$ $\vec{a}(\vec{a}+\vec{b}+\vec{c})=|\vec{a}+|\vec{a}+\vec{b}+\vec{c}|$ $\cos\theta$ $\Rightarrow1=\sqrt{3} \cos \theta$ $\Rightarrow\cos 2 \theta=-\frac{1}{3}$ $\Rightarrow36 \cos ^{2} 2 \theta=4$

Q82JEE Main 2021NAT4MMatrices and Determinants

Let ${A}=(\text{table} 1 , -1 ,0 ; 0 , 1 , -1 ; 0 , 0 ,1 )$ and ${B}=7 {A}^{20}-20 {A}^{7}+2 {I}$ ,where ${I}$ is an identity matrix of order $3 \\times 3$ . If ${B}=[{b}_{{ij}}]$ , then ${b}_{13}$ is equal to _________.

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📖 Explanation

Let $A=(\text{table} 1 , -1 ,0 ; 0 , 1 , -1 ; 0 , 0 ,1 )=I+C$ where ${I}=(\text{table} 1 , 0 , 0;0 ,1 , 0;0 , 0, 1 ),$ ${C}=(\text{table} 0 , -1 , 0 ;0 , 0 , -1 ; 0, 0 , 0 )$ ${C}^{2}=(\text{table} 0 , 0 , 1;0 ,0 , 0;0 , 0, 0)$ , $C^{3}=(\text{table} 0 , 0 , 0;0 ,0 , 0;0 , 0, 0)$ $={C}^{4}={C}^{5}=......$ ${B}=7 {A}^{20}-20 {A}^{7}+2 {I}$ $=7({I}+{C})^{20}-20({I}+{C})^{7}+2 {I}$ $=7({I}+20 {C}+{ }^{20} {C}_{2} {C}^{2})-20({I}+7 {C}$ $+{ }^{7} {C}_{2} {C}^{2})+2 {I}$ ${\text{So}}$ ${b}_{13}=7 \\times { }^{20} {C}_{2}-20 \\times { }^{7} {C}_{2}=910$

Q83JEE Main 2021NAT4MThree Dimensional Geometry

Let ${P}$ be a plane passing through the points $(1,0,1),(1,-2,1)$ and $(0,1,-2)$ . Let a vector $\vec{a}=\alpha{\hat{i}}+\beta{\hat{j}}+\gamma{\hat{k}}$ be such that $\vec{a}$ is parallel to theplane ${P}$ , perpendicular to $(\hat{i}+2 {\hat{j}}+3 {\hat{k}})$ and ${{a}} ^{\to }\.(\hat{i}+\hat{j}+2 \hat{k})=2$ , then $(\alpha-\beta+\gamma)^{2}$ equals _______.

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📖 Explanation

Equation of plane : $\begin{vmatrix}x-1 & y-0 & z-1 \\ 1-1 & 2 & 1-1 \\ 1-0 & 0-1 & 1+2\end{vmatrix}=0$ $\Rightarrow3 x-z-2=0$ ${\vec{a}}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k} ∥$ to $3 x-z-2=0$ $\Rightarrow3 \alpha-8=0$ .......(1) $\vec{a} \perp \hat{i}+2 \hat{j}+3 \hat{k}$ $\Rightarrow\alpha+2 \beta+38=0$ .......(2) $\vec{a} \.(\hat{i}+\hat{j}+2 \hat{k})=0$ $\Rightarrow\alpha+\beta+28=2$ .......(3) on solving $1,2 \ \&\ 3$ $\alpha=1, \beta=-5,8=3$ So $(\alpha-\beta+8)=81$

Q84JEE Main 2021NAT4MBinomial Theorem

The number of rational terms in the binomial expansion of $(4^{\frac{1}{4}}+5^{\frac{1}{6}})^{120}$ is ________.

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📖 Explanation

$(4^{\frac{1}{4}}+5^{\frac{1}{6}})^{120}$ ${T}_{{r}+1}={ }^{120} {C}_{{r}}(2^{1 / 2})^{120-{r}}(5)^{{r} / 6}$ for rational terms ${r}=6 \lambda\;\;\;0 \\le {r} \\le 120$ so total no of forms are 21 .

Q85JEE Main 2021NAT4MThree Dimensional Geometry

If the shortest distance between the lines ${\vec{r}}_{1}=\alpha{\hat{i}}+2 \hat{j}+2 {\hat{k}}+\lambda(\hat{i}-2 {\hat{j}}+2 {\hat{k}}),$ $\lambda\in{R}, \alpha>0$ and $\overrightarrow{\mathrm{r}_{2}}=-4 {\hat{i}}-{\hat{k}}+\mu(3 {\hat{i}}-2 \hat{j}-2 {\hat{k}}), \mu\in{R}$ is 9, then $\alpha$ is equal to ________.

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📖 Explanation

If ${\vec{r}}={{\vec{a}}}+\lambda{\vec{b}}$ and ${{\vec{r}}}=\vec{c}+\lambda{\vec{d}}$ then shortest distance between two lines is $L=\frac{(\vec{a}-\vec{c}) \.(\vec{b} \\times \vec{d})}{|b \\times d|}$ $\therefore \vec{a}-\vec{c}=((\alpha+4) \hat{i}+2 \hat{j}+3 {\hat{k}})$ ${\vec{b} \\times {\vec{d}}}/{|{b} \\times {d}|}={(2 \hat{i}+2 {\hat{j}}+\hat{k})}/{3}$ $\therefore ((\alpha+4) \hat{i}+2 \hat{j}+3 \hat{k}) \.\frac{(2 \hat{i}+2 \hat{j}+\hat{k})}{3}$ $=9$ or $\alpha=6$

Q86JEE Main 2021NAT4MArea Under The Curves

Let ${T}$ be the tangent to the ellipse ${E}: {x}^{2}+4 {y}^{2}=5$ at the point ${P}(1,1) .$ If the area of the region bounded by the tangent $T$ , ellipse $E$ , lines $x=1$ and ${x}=\sqrt{5}$ is $\alpha\sqrt{5}+\beta+\gamma\cos ^{-1}(\frac{1}{\sqrt{5}})$ , then $|\alpha+\beta+\gamma|$ is equal to __________.

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📖 Explanation

Tangent at ${P}: {x}+4 {y}=5$ Required Area $=\int_{1} ^{\sqrt{5}}(\frac{5-x}{4}-{\sqrt{5-x^{2}}}/{2}) d x$ $=\$ [\frac{5 x}{4}-\frac{x^{2}}{8}-\frac{x}{4} \sqrt{5-x^{2}}-\frac{5}{2} \sin ^{-1} \frac{x}{\sqrt{5}}]_{1}^{\sqrt{5}}=\frac{5}{4} \sqrt{5}-\frac{5}{4}-\frac{5}{4} \cos ^{-1}(\frac{1}{\sqrt{5}}) $It we assume$ \alpha, \beta, \gamma, \in{Q} $(Not given \in question) then$ \alpha=\frac{5}{4}, \beta=-\frac{5}{4}\ &\ \gamma=-\frac{5}{4}|\alpha+\beta+\gamma|=1.25$

Q87JEE Main 2021NAT4MMatrices and Determinants

Let ${a}, {b}, {c}, {d}$ be in arithmetic progression with common difference $\lambda$ . If $\begin{vmatrix}x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c\end{vmatrix}=2$ then value of $\lambda^{2}$ is equal to ___________.

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📖 Explanation

$\begin{vmatrix}x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c\end{vmatrix}=2$ ${C}_{2} \\to {C}_{2}-{C}_{3}$ $\Rightarrow\begin{vmatrix}{x}-2 \lambda & \lambda & {x}+{a} \\ {x}-1 & \lambda & {x}+{b} \\ {x}+2 \lambda & \lambda & {x}+{c}\\end{vmatrix}=2$ ${R}_{2} \\to {R}_{2}-{R}_{1}, \;\;\;\;{R}_{3} \\to {R}_{3}-{R}_{1}$ $\Rightarrow \lambda\begin{vmatrix}{x}-2 \lambda & 1 & {x}+{a} \\ 2 \lambda-1 & 0 & \lambda \\ 4 \lambda & 0 & 2 \lambda\end{vmatrix}=2$ $\Rightarrow1(4 \lambda^{2}-4 \lambda^{2}+2 \lambda)=2$ $\Rightarrow \lambda^2=1$

Q88JEE Main 2021NAT4MPermutations and Combinations

There are 15 players in a cricket team, out ofwhich 6 are bowlers, 7 are batsmen and 2 arewicketkeepers. The number of ways, a team of 11players be selected from them so as to include atleast 4 bowlers, 5 batsmen and 1 wicketkeeper,is ________.

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📖 Explanation

15: Players 6: Bowlers 7: Batsman 2: Wicket keepers Total number of ways for: at least 4 bowlers, 5 batsman & 1 wicket keeper $={ }^{6} {C}_{4}({ }^{7} {C}_{6} \\times { }^{2} {C}_{1}+{ }^{7} {C}_{5} \\times { }^{2} {C}_{2})+{ }^{6} {C}_{5} \$ $\times { }^{7} {C}_{5} \\times { }^{2} {C}_{1}$ $=777$

Q89JEE Main 2021NAT4MParabola

Let $y=m x+c, m>0$ be the focal chord of $y^{2}=-64 x$ , which is tangent to $(x+10)^{2}+y^{2}=4$ Then, the value of $4 \sqrt{2}({m}+{c})$ is equal to ______.

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📖 Explanation

$y^{2}=-64 x$ focus : $(-16,0)$ ${y}={mx}+{c}$ is focal chord $\Rightarrow{c}=16 {m}.........$ (1) $y=m x+c$ is tangent to $(x+10)^{2}+y^{2}=4$ $\Rightarrow{y}={m}({x}+10) \\pm 2 \sqrt{1+{m}^{2}}$ $\Rightarrow{c}=10 {m} \\pm 2 \sqrt{1+{m}^{2}}$ $\Rightarrow16 {m}=10 {m} \\pm 2 \sqrt{1+{m}^{2}}$ $\Rightarrow6 {m}=2 \sqrt{1+{m}^{2}} \;\;\;\;({m}>0)$ $\Rightarrow9 {m}^{2}=1+{m}^{2}$ $\Rightarrow{m}=\frac{1}{2 \sqrt{2}} \ & {c}=\frac{8}{\sqrt{2}}$ $4 \sqrt{2}({m}+{c})=4 \sqrt{2}(\frac{17}{2 \sqrt{2}})=34$

Q90JEE Main 2021NAT4MLimits, Continuity and Differentiability

If the value of $\lim _{x \\to 0}(2-\cos x \sqrt{\cos 2 x})^{(\frac{x+2}{x^{2}})}$ is equal to ${e}^{{a}}$ , then a is equal to _______.

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📖 Explanation

$\lim _{x \\to 0}(2-\cos x \sqrt{\cos 2 x})^{(\frac{x+2}{x^{2}})}$ form: $1^{\\infty }$ $={e}^{\lim _{x \\to 0}$ {1-\cos x \sqrt{\cos 2 {x}}}/{x^{2}}) \× (x+2)} $Now$ \lim {x \→ 0} \frac{1-\cos x \sqrt{\cos 2 x}}{x^{2}}=\lim{x \→ 0} {\sin x \sqrt{\cos 2 x}-\cos x \× \frac{1}{2 \sqrt{\cos 2 x}} \× (-2 \sin 2 x)}/{2 x} $(by$ {L}' $Hospital Rule)$ {\lim t}{x \→ 0} \frac{\sin x \cos 2 x+\sin 2 x .\cos x}{2 x}=\frac{1}{2}+1=\frac{3}{2} $So,$ e^{{\lim t}{x \→ 0}(\frac{1-\cos x \sqrt{\cos 2 x}}{x^{2}})(x+2)}=e^{\frac{3}{2} \× 2}=e^{3}\Rightarrowa=3$

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