Previous Year Paper

24 Feb 2021 Shift 1

90 Questions360 Marks180 MinTake as Timed Mock →

🎯 Practice smarter, not harder

Practice 24 Feb 2021 Shift 1 questions with full analytics — free

✓ Attempt tracking & analytics
✓ Timer & exam-mode simulation
✓ Adaptive difficulty (Easy → Hard)
✓ Bookmarks, notes & mistake review

Just sign in to unlock everything. Free for all students.

🚀 Start Practice Mode Sign in free →

Physics30 questions

Q1JEE Main 2021MCQ4MUnits and Measurements

The work done by a gas molecule in an isolated system is given by, $W=\alpha \beta^{2} e^{-\;\frac{x^{2}}{\alpha k T}}$ , where $x$ is the displacement, $k$ is the Boltzmann constant and $T$ is the temperature, $\alpha$ and $\beta$ are constants. Then, the dimensions of $\beta$ will be 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

Given, work done, $W=\alpha \cdot \beta^{2} e^{-\;\frac{x^{2}}{a k T}}$ where, $k$ is Boltzmann constant, $T$ is temperature and $x$ is displacement. We know that, $\;\frac{x^{2}}{\alpha k T}$ is a dimensionless quantity. $\therefore \;\; \$ [;\frac{x^{2}}{alpha k T}]$=$[ {M}^{0} {L}^{0} {T}^{0}]$ \Rightarrow [\alpha]=;\frac{[x^{2}]$}{[ {k}][T]}\Rightarrow ;; [alpha]=;\frac{[ {L}^{2}]$}{[ {k}][T]} $Since, dimensions of$ k $are$ [k]=[M^{1} L^{2} T^{-2} K^{-1}] $... (i) Dimensions of temperature are$ [T]=[K] $Substituting Eqs. (ii) and (iii) in Eq. (i), we get$ [\alpha] ;; =;\frac{[L^{2}]$}{[M^{1} L^{2} T^{-2} K^{-1}]$[K]}[\alpha] ;; =[M^{-1} T^{2}] $According to dimensional analysis,$ {[W]};=[\alpha \beta^{2}]\Rightarrow ;; $[beta^{2}]$;=;\frac{[W]}{[\alpha]}\Rightarrow ;; $[beta^{2}]$;=;\frac{[M^{1} L^{2} T^{-2}]$}{[M^{-1} T^{2}]$}=[M^{2} L^{2} T^{-4}]\Rightarrow ;; [\beta];=[M L T^{-2}]$

Q2JEE Main 2021MCQ4MGravitation

Two stars of masses $m$ and $2 m$ at a distance $d$ rotate about their common centre of mass in free space. The period of revolution is 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

The given situation is shown below The gravitational force between these two stars provide the required centripetal force for rotation in a circle about their common centre. Assuming 2 mat origin, the centre of mass of the system lies at $x=\;\frac{2 m \times 0+m \times d}{2 m+m}=\;\frac{d}{3}$ Hence, $F_{G}=F_{C}$ where, $F_{C}$ is gravitational force between them and $F_{C}$ is centripetal force. $\Rightarrow \;\; \;\frac{G m_{1} m_{2}}{r^{2}} \;\; =2 m \omega^{2} x$ $\Rightarrow \;\; \;\frac{G(2 m)(m)}{d^{2}} \;\; =2 m \omega^{2} \times \;\frac{d}{3} \Rightarrow \omega^{2}=\;\frac{3 G m}{d^{3}}$ $\omega \;\; \omega \;\; =\sqrt{\;\frac{3 G m}{d^{3}}}$ We know that, $\;\omega=\;\frac{2 \pi}{T}$ $\;T=\;\frac{2 \pi}{\omega}$ $\Rightarrow$ $T \;\; =\;{2 \pi}/{\sqrt{\;\frac{3 G m}{d^{3}}}}$ $\;\; =2 \pi \sqrt{\;\frac{d^{3}}{3 G m}}$ [using Eq. (i)]

Q3JEE Main 2021MCQ4MGravitation

Four identical particles of equal masses $1 {kg}$ made to move along the circumference of a circle of radius $1 {m}$ under the action of their own mutual gravitational attraction. The speed of each particle will be 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

Given, $m=1 {kg}, R=1 {m}$ We know that, $F \;\; =\;\frac{G m_{1} m_{2}}{r^{2}}$ $\because \;\; F_{1} \;\; =\;\frac{G m m}{(2 R)^{2}}=\;\frac{G m^{2}}{4 R^{2}}$ $\;\text{ and }\; \;\; F_{2} \;\; =\;\frac{G m m}{(\sqrt{2} R)^{2}}=\;\frac{G m^{2}}{2 R^{2}}$ Net force on one particle, $F_{\;\text{net }\;} \;\; =F_{1}+F_{2} \cos 45^{\circ}+F_{2} \cos 45^{\circ}$ $\;\; =F_{1}+2 F_{2} \cos 45^{\circ}$ $\;\; =\;\frac{G m^{2}}{4 R^{2}}+2(\;\frac{G m^{2}}{2 R^{2}}) \cdot \;\frac{1}{\sqrt{2}}$ $\;\; =\;\frac{G m^{2}}{4 R^{2}}+\;\frac{G m^{2}}{\sqrt{2} R^{2}}$ $\;\; =\;\frac{G m^{2}}{R^{2}}[\;\frac{1}{4}+\;\frac{1}{\sqrt{2}}]$ As the gravitational force provides the necessary centripetal force, so $F_{\;\text{net }\;}=F_{C}=\;\frac{m v^{2}}{R}$ Here, $F_{C}=$ centripetal force. $\Rightarrow \;\frac{G m^{2}}{R^{2}}[\;\frac{1}{4}+\;\frac{1}{\sqrt{2}}] \;\; =\;\frac{m v^{2}}{R}$ $\Rightarrow \;\; v \;\; =\;\frac{1}{2} \sqrt{\;\frac{G m}{R}(1+2 \sqrt{2})}$ $\Rightarrow \;\; v \;\; =\;\frac{1}{2} \sqrt{G(1+2 \sqrt{2})}$

Q4JEE Main 2021MCQ4MRotational Motion

Moment of inertia (MI) of four bodies, having same mass and radius, are reported as $I_{1}=$ Ml of thin circular ring about its diameter, $I_{2}=$ MI of circular disk about an axis perpendicular to the disk and going through the centre, $I_{3}= {Ml}$ of solid cylinder about its axis and $I_{4}=$ Ml of solid sphere about its diameter. Then,24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

Let $M$ and $R$ be the mass and radius of four bodies. Then, as per question, their moment of inertia are $I_{1}=\;\frac{M R^{2}}{2}, I_{2}=\;\frac{M R^{2}}{2}, I_{3}=\;\frac{M R^{2}}{2}$ $I_{4}=\;\frac{2}{5} M R^{2}$ Clearly, $I_{1}=I_{2}=I_{3} > I_{4}$

Q5JEE Main 2021MCQ4MCircular Motion

Consider two satellites $S_{1}$ and $S_{2}$ with periods of revolution $1 {h}$ and $8 {h}$ respectively, revolving around a planet in circular orbits. The ratio of angular velocity of satellite $S_{1}$ to the angular velocity of satellite $S_{2}$ is 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

Given, period of revolution of first satellite, $T_{1}=1 {h}$ Period of revolution of second satellite, $T_{2} \;\; =8 {h}$ $\;\frac{T_{1}}{T_{2}} \;\; =\;\frac{1}{8}$ We know that, $\omega=\;\frac{2 \pi}{T}$ $\Rightarrow$ $\omega \propto \;\frac{1}{T}$ $\because \;\; \;\frac{\omega_{1}}{\omega_{2}}=\;\frac{T_{2}}{T_{1}} \Rightarrow \;\frac{\omega_{1}}{\omega_{2}}=\;\frac{8}{1}$ or $\;\; \omega_{1}: \omega_{2}=8: 1$

Q6JEE Main 2021MCQ4MProperties of Matter

Each side of a box made of metal sheet in cubic shape is $a$ at room temperature $T$ , the coefficient of linear expansion of the metal sheet is $\alpha$ . The metal sheet is heated uniformly, by a small temperature $\Delta T$ , so that its new temperature is $T+\Delta T$ . Calculate the increase in the volume of the metal box. 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

We know that, $\gamma=3 \alpha$ . ... (i) where, $\alpha$ is the coefficient of linear expansion and $\gamma$ is the coefficient of volume expansion. We know that, $\;\frac{\Delta V}{V}\;=\gamma \Delta T\;$ $\;\frac{\Delta V}{V}\;=3 \alpha \Delta T\; \;\; [\;\text{ from Eq. (i)] }\;$ $\Rightarrow \;\; \Delta V\;=3 a^{3} \alpha \Delta T\;{[ \because \;\text{ volume of cube }\;=a^{3}]\$ }$

Q7JEE Main 2021MCQ4MProperties of Matter

If $Y, K$ and $η$ are the values of Young's modulus, bulk modulus and modulus of rigidity of any material, respectively. Choose the correct relation for these parameters. 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

We know that, $Y=3 K(1-2 \sigma)$ $\Rightarrow \;\; \sigma=\;\frac{1}{2}(1-\;\frac{Y}{3 K})$ Also, $\;\; Y=2 η(1+\sigma)$ $\Rightarrow \;\; \sigma=\;\frac{Y}{2 η}-1$ ... (ii) On comparing Eqs. (i) and (ii), we get $(1-\;\frac{Y}{3 K}) \;\frac{1}{2}=\;\frac{Y}{2 η}-1$ On solving, we get $K=\;\frac{η Y}{9 η-3 Y} {N} / {m}^{2}$

Q8JEE Main 2021MCQ4MMotion in a Straight Line

If the velocity-time graph has the shape $A M B$ , what would be the shape of the corresponding acceleration-time graph ? 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

From the graph for first line, the slope is negative and intercept is positive. So, equation of line is $v \;\; =-m t+c$ $a_{1} \;\; =\;\frac{d v}{d t}=-m$ Similarly, for second line, the slope is positive and intercept is negative, so equation of line is $v \;\; =m t-c$ $a_{2} \;\; =\;\frac{d v}{d t}=m$ $\therefore$ The corresponding acceleration-time graph as shown below Hence, option (a) is correct.

Q9JEE Main 2021MCQ4MThermodynamics

$n$ mole of a perfect gas undergoes a cyclic process $A B C A$ (see figure) consisting of the following processes. $A \to B$ : Isothermal expansion at temperature $T$ , so that the volume is doubled from $V_{1}$ to $V_{2}=2 V_{1}$ and pressure changes from $p_{1}$ to $p_{2}$ . $B \to C:$ Isobaric compression at pressure $p_{2}$ to initial volume $V_{1}$ . $C \to A$ : Isochoric change leading to change of pressure from $p_{2}$ to $p_{1}$ . Total work done in the complete cycle $A B C A$ is 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

As, $A B$ is isothermal process, so work done by isothermal process is given by $\;W_{A B}=n R T \ln (\;\frac{V_{2}}{V_{1}})$ $\;W_{A B}=n R T \ln (\;\frac{2 V_{1}}{V_{1}})=n R T \ln 2$ As, $B C$ is isobaric process, so work done by isobaric process is given by $W_{B C} \;\; =p \Delta V=p_{2}(V_{1}-V_{2})$ $\;\; =p_{2}(\;\frac{V_{2}}{2}-V_{2})=-\;\frac{p_{2} V_{2}}{2}$ $\;\; =\;\frac{-n R T}{2}$ As, $C A$ is isochoric process, so work done by isochoric process is given by, $W_{C A}=0 \;\; [ \because \Delta V=0]$ Total work done in the cycle $A B C A$ , $W_{A B C A} \;\; =W_{A B}+W_{B C}+W_{C A}$ $\;\; =n R T \ln 2+(\;\frac{-n R T}{2})+0$ $\;\; =n R T(\ln 2-\;\frac{1}{2})$

Q10JEE Main 2021MCQ4MThermodynamics

Match List-I with List-II. List-I List-II A. Isothermal 1. Pressure constant B. Isochoric 2. Temperature constant C. Adiabatic C. Adiabatic D. Isobaric 4. Heat content is constantChoose the correct answer from the options given below.24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

We know that, in isothermal process, $\Delta T=0$ In isochoric process, $\Delta V=0$ In adiabatic process, $\Delta Q=0$ In isobaric process, $\Delta p=0$ So, the correct match is, ${A} \to 2, {B} \to 3, {C} \to 4, {D} \to 1$ .

Q11JEE Main 2021MCQ4MOscillations

In the given figure, a mass $M$ is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is $k$ . The mass oscillates on a frictionless surface with time period $T$ and amplitude $A$ . When the mass is in equilibrium position as shown in the figure, another mass $m$ is gently fixed upon it. The new amplitude of oscillation will be 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

Given, initial amplitude $=A$ Velocity at mean position, $v=A \omega$ Applying conservation of momentum at mean position, we get $M_{1} v_{1} \;\; =M_{2} v_{2}$ $M A \omega \;\; =(M+m) v^{'}$ $\Rightarrow \;\; v^{'} \;\; =\;\frac{M A \omega}{M+m}=\;{M A \sqrt{\frac{k}{M}}}/{M+m}$ $\;\; v^{'} \;\; =A^{'} \omega^{'}=A^{'} \sqrt{\;\frac{k}{M+m}}$ $\therefore \;\; A^{'} \;\; =\;{M A \sqrt{\frac{k}{M}}}/{M+m} \times \sqrt{\;\frac{M+m}{k}}$ $A^{'} \;\; =\sqrt{\;\frac{M}{M+m}} A$

Q12JEE Main 2021MCQ4MCurrent Electricity

A cell $E_{1}$ of emf $6 {V}$ and internal resistance $2 Ω$ is connected with another cell $E_{2}$ of emf $4 {V}$ and internal resistance $8 Ω$ (as shown in the figure). The potential difference across points $X$ and $Y$ is $?$ 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

The circuit can be shown as below The current through the circuit, $I=\;\frac{\epsilon_{1}-\epsilon_{2}}{r_{1}+r_{2}}=\;\frac{6-4}{10}=\;\frac{1}{5} {A}$ $\therefore$ Potential difference across points $X$ and $Y$ is $V_{X X}=E_{2}+I r_{2}=4+\;\frac{1}{5} \times 8=5.6 {V}$

Q13JEE Main 2021MCQ4MCurrent Electricity

A current through a wire depends on time as $i=\alpha_{0} t+\beta t^{2}$ , where $\alpha_{0}=20 {A} / {s}$ and $\beta=8 {As}^{-2}$ . Find the charge crossed through a section of the wire in $15 {s}$ . 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

Given, $\;\; i=\alpha_{0} t+\beta t^{2}$ where, $\alpha_{0}=20 {A} / {s}, \beta=8 {As}^{2}$ We know that, $\;\; i=\;\frac{d q}{d t}$ $\Rightarrow \;\;\frac{d q}{d t}=i=\alpha_{0} t+\beta t^{2}=20 t+8 t^{2}$ $\Rightarrow \;d q=(20 t+8 t^{2}) d t$ On integrating both sides, we get $int_{0}^{q} d q=int_{0}^{15}(20 t+8 t^{2}) d t$ $q=\$ [;\frac{20 t^{2}}{2}+;\frac{8 t^{3}}{3}]_{0}^{15}=10 × (15)^{2}+;\frac{8}{3} × (15)^{3}q=11250 {C}$

Q14JEE Main 2021MCQ4MElectrostatic Potential and Capacitance

Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

Given, $C_{1}=C_{2}=C$ When both capacitors are connected in series, their equivalent capacitance will be $\;\frac{1}{C_{s}}=\;\frac{1}{C}+\;\frac{1}{C}=\;\frac{2}{C}$ $\;C_{s}=\;\frac{C}{2}$ $\Rightarrow \;\; C_{s}=\;\frac{C}{2}$ When both capacitors are connected in parallel, their equivalent capacitance will be $C_{p}=C+C=2 C$ $\therefore$ The ratio of equivalent capacitance in series and parallel combination is $\;\frac{C_{s}}{C_{p}} \;\; =\;\frac{C / 2}{2 C}=\;\frac{1}{4}$ $\Rightarrow \;\; C_{s}: C_{p} \;\; =1: 4$

Q15JEE Main 2021MCQ4MElectrostatics

A cube of side $a$ has point charges $+Q$ located at each of its vertices except at the origin, where the charge is $-Q$ . The electric field at the centre of cube is 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

We can replace $-Q$ charge at origin by $+Q$ and $-2 Q .$ Now, due to $+Q$ charge at every corner of cube, electric field at centre of cube is zero. So, net electric field at centre is only due to $-2 Q$ charge at origin. Vector form of electric field strength, $E=\;\frac{K q r}{r^{3}}$ Here, position vector, $r=\;\frac{a}{2}(\hat{x}+\hat{y}+\hat{z})$ $\Rightarrow \;\; |r|=\sqrt{(\;\frac{a}{2})^{2}+(\;\frac{a}{2})^{2}+(\;\frac{a}{2})^{2}}=\;\frac{\sqrt{3} a}{2}$ $\Rightarrow \;\; E=\;\frac{1}{4 \pi \epsilon_{0}} \times \;\frac{(-2 Q) a / 2}{(\;\frac{a}{2} \sqrt{3})^{3}} \cdot (\hat{x}+\hat{y}+\hat{z})$ $E=\;{-2 Q(\hat{x}+\hat{y}+\frac\hat{z})}{3 \sqrt{3} \pi a^{2} \epsilon_{0}}$

Q16JEE Main 2021MCQ4MSemiconductor Electronics

If an emitter current is changed by $4 {mA}$ , the collector current changes by $3.5 {mA}$ . The value of $\beta$ will be 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

Given, emitter current, $I_{E}=4 {mA}$ Collector current, $I_{C}=3.5 {mA}$ Current gain in common base amplifier, $\;\alpha=\;\frac{I_{C}}{I_{E}}$ $\;\alpha=\;\frac{3.5}{4}=\;\frac{7}{8}$ Also, current gain in common emitter amplifier, $\;\beta=\;\frac{\alpha}{1-\alpha}$ $\;\beta=\;\frac{7 / 8}{1-7 / 8}$ $\;\beta=7$

Q17JEE Main 2021MCQ4MAtoms and Nuclei

In the given figure, the energy levels of hydrogen atom have been shown alongwith some transitions marked $A, B, C, D$ and $E$ . The transitions $A, B$ and $C$ respectively represent 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

In transition $A$ , hydrogen atom comes from higher energy state $n=\infty$ to lower energy state $n=1$ . Hence, transition $A$ represents series limit of Lyman series. In transition $B$ , hydrogen atom comes from higher energy state $n=5$ to lower energy state $n=2$ . Hence, transition $B$ represents 3 rd line of Balmer series. In transition $C$ , hydrogen atom comes from higher energy state $n=5$ to lower energy state $n=3$ . Hence, transition C represents 2 nd line of Paschen series. Hence, option (d) is correct.

Q18JEE Main 2021MCQ4MDual Nature of Matter and Radiation

Given below are two statements : Statement I Two photons having equal linear momenta have equal wavelengths. Statement II If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease. In the light of the above statements, choose the correct answer from the options given below.24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

As we know, $\lambda=\;\frac{h}{p}=\;\frac{h}{\sqrt{2 m K}}$ If linear momenta of two photons are equal, then their wavelengths is also equal. Also, if the wavelength is decreased, then the momentum |and energy of photon will increase. Hence, option(c) is correct.

Q19JEE Main 2021MCQ4MRay Optics and Optical Instruments

The focal length $f$ is related to the radius of curvature $r$ of the spherical convex mirror by 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

For convex mirror, the focal length $(f)$ and radius of curvature $(r)$ are related as $f=+\;\frac{r}{2}$ .

Q20JEE Main 2021MCQ4MWave Optics

In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern. 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

Given, amplitude $\propto$ width of slit $\Rightarrow \;\; A_{2}=3 A_{1}$ We know that, $\;\frac{I_{\max }}{I_{\min }}=\;{(\sqrt{I_{1}}+\sqrt{I_{2}})^{2}}/{(\sqrt{I_{1}}-\sqrt{I_{2}})^{2}}$ $\because \;\;$ Intensity, $l \propto A^{2}$ $\therefore \;\; \;\frac{I_{\max }}{I_{\min }}=\;\frac{(A_{1}+A_{2})^{2}}{(|A_{1}-A_{2}|)^{2}}$ $=(\;\frac{A_{1}+3 A_{1}}{|A_{1}-3 A_{1}|)})^{2}=(\;\frac{4 A_{1}}{2 A_{1}})^{2}=\;\frac{4}{1}$ $\therefore \;\; I_{\;\text{max }\;}: I_{\;\text{min }\;}=4: 1$

Q21JEE Main 2021NAT4MLaws of Motion

The coefficient of static friction between a wooden block of mass $0.5 {kg}$ and a vertical rough wall is $0.2$ . The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall will be ....... N. $[Take, $g=10 {ms}^{-2}$ ]$24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

Given, coefficient of static friction, $\mu _{s}=0.2$ Various forces acting on block are shown belowFrictional force $\le m g$ $\Rightarrow \;\; N \times 0.2 \le 5$ $\Rightarrow \;\; N \le 25$ $\therefore \;\;$ Magnitude of horizontal force, $F=N=25 {N}$

Q22JEE Main 2021NAT4MWave Optics

An unpolarised light beam is incident on the polariser of a polarisation experiment and the intensity of light beam emerging from the analyser is measured as 100 lumens. Now, if the analyser is rotated around the horizontal axis (direction of light) by $30^{\circ}$ in clockwise direction, the intensity of emerging light will be .............. lumens.24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

Given, $I_{0}=100$ lumens When analyser is rotated through an angle $\theta$ , the intensity of light will becomes $I \;\; =I_{0} \cos ^{2} \theta=100 \times \cos ^{2} 30^{\circ}$ $\;\; =100 \times (\;\frac{\sqrt{3}}{2})^{2}=75 \;\text{ lumens }\;$

Q23JEE Main 2021NAT4MCentre of Mass and Collision

A ball with a speed of $9 {m} / {s}$ collides with another identical ball at rest. After the collision, the direction of each ball makes an angle of $30^{\circ}$ with the original direction. The ratio of velocities of the balls after collision is $x: y$ , where $x$ is ................ .24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

The situation is shown below Using conservation of linear momentum in $y$ -direction, $\;\text{ As, }\;\;p_{i}\;=p_{f}$ $\;\text{ and }\;\;p_{i}\;=0$ $\Rightarrow \;p_{f}\;=m v_{1} \sin 30^{\circ}-m v_{2} \sin 30^{\circ}$ $\;0\;=m \times \;\frac{1}{2} v_{1}-m \times \;\frac{1}{2} v_{2}$ $\Rightarrow \;v_{1}\;=v_{2} \;\text{ or }\; v_{1}: v_{2}=1: 1$ $\;\text{ Since, }\; v_{1}: v_{2}=\;x: y\;\;\text{ (gi ven) }\;$ $\therefore \;x=1$

Q24JEE Main 2021NAT4MProperties of Matter

A hydraulic press can lift $100 {kg}$ when a mass $m$ is placed on the smaller piston. It can lift ............ ${kg}$ when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times keeping the same mass $m$ on the smaller piston.24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

According to Pascal's law, $\;\frac{F_{1}}{A_{1}}=\;\frac{F_{2}}{A_{2}}$ Initially, $\;\; \;\frac{100 {g}}{A_{1}}=\;\frac{m g}{A_{2}}$ . . . (i) Finally, $\;\; \;\frac{M g}{16 A_{1}}=\;{m g}/{(\;\frac{A_{2}}{16})}$ . . . (ii) On dividing Eqs. (i) by (ii), we get $\;\frac{100 \times 16}{M}=\;\frac{1}{16}$ $M=25600 {kg}$

Q25JEE Main 2021NAT4MLaws of Motion

An inclined plane is bent in such a way that the vertical cross-section is given by $y=\;\frac{x^{2}}{4}$ where, $y$ is in vertical and $x$ in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction $\mu =0.5$ , the maximum height in ${cm}$ at which a stationary block will not slip downward is .............. ${cm}$ .24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

The graph for given equation is shown belowAt maximum height, the slope of tangent drawn, $\tan \theta\;=\;\frac{d y}{d x}=\;\frac{2 x}{4}=\;\frac{x}{2}$ $\Rightarrow \;\; 0.5\;=\;\frac{x}{2} \;\; [ \therefore y=\;\frac{x^{2}}{4}]$ $\Rightarrow \;\; x\;=1 {m}$ $\therefore \;\; y\;=\;\frac{x^{2}}{4}=\;\frac{1}{4}=0.25 {m}=25 {cm}$

Q26JEE Main 2021NAT4MAlternating Current

A resonance circuit having inductance and resistance $2 \times 10^{-4} {H}$ and $6.28 Ω$ respectively oscillates at $10 {MHz}$ frequency. The value of quality factor of this resonator is $[$ Take, $\pi=3.14]$ 24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

Given, $L=2 \times 10^{-4} {H}, R=6.28 Ω$ , $f_{0}=10 {MHz}=10 \times 10^{6} {Hz}$ $\therefore$ Quality factor $=\omega_{0} \;\frac{L}{R}=2 \pi f_{0} \;\frac{L}{R}$ $=2 \pi \times 10 \times 10^{6} \times \;\frac{2 \times 10^{-4}}{6.28}$ $=2 \times 10^{3}=2000$

Q27JEE Main 2021NAT4MElectromagnetic Waves

An audio signal $v_{m}=20 \sin 2 \pi(1500 t)$ amplitude modulates a carrier $v_{c}=80 \sin 2 \pi(100000 t)$ The value of per cent modulation is ........... .24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

Given, audio signal, $V_{m}=20 \sin 2 \pi(1500 t)$ . . . (i) Carrier signal, $V_{c}=80 \sin 2 \pi(100000 t)$ . . . (ii) We know that, modulation index, $m_{f}=\;\frac{A_{m}}{A_{c}}$ From Eqs. (i) and (ii), we get $A_{m}=20, A_{c}=80$ Percentage of modulation index, $m_{f}=\;\frac{A_{m}}{A_{c}} \times 100=\;\frac{20}{80} \times 100=25 \%$

Q28JEE Main 2021NAT4MSemiconductor Electronics

In connection with the circuit drawn below, the value of current flowing through $2 {k} Ω$ resistor is ........ $\times 10^{-4} {A}$ .24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

Given, resistance, $R=2 {k} Ω=2 \times 10^{3} Ω$ In Zener breakdown, $i \;\; =\;\frac{V}{R}=\;\frac{5}{2 \times 10^{3}}=2.5 \times 10^{-3}$ $\because \;\; x \times 10^{-4}= \;\; 25 \times 10^{-4}$ $\therefore \;\; x \;\; =25$

Q29JEE Main 2021NAT4MElectromagnetic Waves

An electromagnetic wave of frequency $5 {GHz}$ , is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are 2 . Its velocity in this medium is $. . . . . . \times 10^{7} {m} / {s}$ .24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

Given, $\mu _{r}=\epsilon_{r}=2$ where, $\mu _{r}$ is relative permeability, $\epsilon_{r}$ is relative permittivity. Speed of electromagnetic wave $v$ is given by $v=\;\frac{C}{n}$ where, $n=$ refractive index $=\sqrt{\mu _{r} \epsilon_{r}}=\sqrt{4}=2$ $\Rightarrow \;\; v=\;\frac{3 \times 10^{8}}{2}=15 \times 10^{7} {m} / {s}$ $\because \;\; x \times 10^{7}=15 \times 10^{7}$ $\Rightarrow \;\; x=15$

Q30JEE Main 2021NAT4MSemiconductor Electronics

A common transistor radio set requires $12 {V}( {DC})$ for its operation. The DC source is constructed by using a transformer and a rectifier circuit, which are operated at $220 {V}$ (AC) on standard domestic AC supply. The number of turns of secondary coil are 24 , then the number of turns of primary are ............ .24 Feb 2021 Shift1

🚀 Solve in Practice Mode

📖 Explanation

In a transformer, $\;\frac{N_{p}}{N_{s}}=\;\frac{V_{p}}{V_{s}}$ where, $N_{p}=$ number of turns in primary circuit, $N_{s}=$ number of turns in secondary circuit $=24$ , $V_{p}=$ potential of primary circuit $=220 {V}$ and $V_{s}=$ potential of secondary circuit $=12 {V}$ $\Rightarrow \;\;\frac{N_{p}}{24} \;\; =\;\frac{220}{12}$ $\Rightarrow \;N_{p} \;\; =440$

Chemistry30 questions

Q31JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

Which of the following are isostructural pairs? A. ${SO}_{4}^{2-}$ and ${CrO}_{4}^{2-}$ B. ${SiCl}_{4}$ and ${TiCl}_{4}$ C. ${NH}_{3}$ and ${NO}_{3}^{-}$ D. ${BCl}_{3}$ and ${BrCl}_{3}$

🚀 Solve in Practice Mode

📖 Explanation

Isostructural compounds are those compounds which have same structure as well as same hybridisation. Formula for find hybridisation : $H=(l p+\sigma+$ coordinate bond $)$ ${Ip} \;\; =\;\text{ Lone pair }\;$ ${bp} \;\; =\;\text{ Bond pair }\;$ $\sigma \;\; =\;\text{ Sigma bond }\;$ S. No. Molecule H (Steric number) Hybridisation B.P. L.P. Shape (A) ${SO}_{4}^{2-}$ 4 $s p^{3}$ 4 0 ${CrO}_{4}^{2-}$ 4 $s p^{3}$ 4 0 (B) ${SiCl}_{4}$ 4 $s p^{3}$ 4 0 ${TiCl}_{4}$ 4 $s p^{3}$ 4 0 (C) ${NH}_{3}$ 4 $s p^{3}$ 3 1 ${NO}_{3}^{-}$ 4 $s p^{2}$ 3 0 (D) ${BCl}_{3}$ 3 $s p^{2}$ 3 0 ${BrCl}_{3}$ 5 $s p^{3} d$ 3 2 If number of sigma bond $(\sigma)$ , co-ordinate bond and lone pair are same for given pairs, they are isostructural. Hence, ${SO}_{4}^{2-}, {CrO}_{4}^{2-}, {SiCl}_{4}$ and ${TiCl}_{3}$ have tetrahedral shape and are isostructural.

Q32JEE Main 2021MCQ4MSolutions

In Freundlich adsorption isotherm, slope of $A B$ line is

🚀 Solve in Practice Mode

📖 Explanation

Given graph is Freundlich adsorption isotherm. According to equation, $\;\frac{x}{m}=k p^{\;\frac{1}{n}}$ $. . .$ (i) where, $\;\frac{x}{m}=$ amount of gas $k$ and $n=$ constants depending upon nature of adsorbate and absorbent. Now, taking log both side in Eq. (i), we get $\log \;\frac{x}{m}=\log k+\;\frac{1}{n} \log p$ For straight line, comparing above equation to $y=m x+c$ ${SO}$ , $m=$ slope $=\;\frac{1}{n}$ hence, $0 \le \;\frac{1}{n} \le 1$ It value lies in between 0 to 1 . Hence, $\;\frac{1}{n}$ with $(\;\frac{1}{n}=0$ to 1 $)$ .

Q33JEE Main 2021MCQ4MClassification of Elements

Consider the elements ${Mg}, {Al}, {S}, {P}$ and ${Si}$ , the correct increasing order of their first ionisation enthalpy is

🚀 Solve in Practice Mode

📖 Explanation

On moving left to right in a period of the periodic table, lonisation energy (I.E.) increases due to increase in effective nuclear charge i.e. $Z_{\;\text{eff }\;}$ . But due to extra stability of fully filled and half-filled electronic configuration of ${Mg}$ and ${P}$ required ionisation enthalpy is more from neighbouring elements. i.e. First ionisation enthalpy order is ${Al} < {Mg} < {Si} < {S} < {P}$ .

Q34JEE Main 2021MCQ4Md and f Block Elements

Which of the following ore is concentrated using group 1 cyanide salt?

🚀 Solve in Practice Mode

📖 Explanation

The chemical formulas of given ore in options are as follows Sphalerite ore: ZnS Calamine ore : ${ZnCO}_{3}$ Siderite ore : ${FeCO}_{3}$ Malachite ore : ${Cu}( {OH})_{2} \cdot {CuCO}_{3}$ Sphalerite ore containing ${ZnS}$ is concentrated using group is cyanide salt ({NaCN}$ ) as a depressant by froth flotation method.

Q35JEE Main 2021MCQ4MRedox Reactions

(A) ${HOCl}+ {H}_{2} {O}_{2} \longrightarrow {H}_{3} {O}^{+}+ {Cl}^{-}+ {O}_{2}$ (B) ${I}_{2}+ {H}_{2} {O}_{2}+2 {OH}^{-} \longrightarrow 2 {I}^{-}+2 {H}_{2} {O}+ {O}_{2}$ Choose the correct option.

🚀 Solve in Practice Mode

📖 Explanation

In equation (A), ${HOCl}$ undergoes reduction in presence of ${H}_{2} {O}_{2}$ . Here, oxidation state of ${Cl}$ changes from $+1$ to $-1$ (i.e. reduces) $\therefore {H}_{2} {O}_{2}$ act as reducing agent ${I}_{2}$ reduces to ${T}$ in presence of ${H}_{2} {O}_{2}$ . In equation (B), Here oxidation state of iodine decreases (from 0 to $-1$ ) $\therefore {H}_{2} {O}_{2}$ act as reducing agent in both the equations.

Q36JEE Main 2021MCQ4Mp Block Elements

${Al}_{2} {O}_{3}$ was leached with alkali to get $X$ . The solution of $X$ on passing of gas $Y$ , forms $Z$ . $X$ , $Y$ and $Z$ respectively are

🚀 Solve in Practice Mode

📖 Explanation

${Al}_{2} {O}_{3}$ (aluminium oxide) was leached with alkali to get ${Na}[ {Al}( {OH})_{4}](X)$ (sodium aluminate). The solution of sodium aluminate when passed through gas $Y$ , i.e. carbon dioxide $( {CO}_{2})$ forms $( {Al}_{2} {O}_{3} \cdot {xH}_{2} {O})( {Z})$ . Complete reactions are as follows ${{Al}_{2} {O}_{3}}_{\text{Aluminium} \text{oxide}}+{2 {NaOH}( {aq})}_{\text{Sodium} \text{hydroxide}}+3 {H}_{2} {O}(l) \longrightarrow$ $2 {Na}[ {Al}_{(\text{X})}( {OH})_{4}]( {aq}) {\longrightarrow }^{+ {CO}_{2}( {Y})} { {Al}_{2} {O}_{3} \cdot {xH}_{2} {O}}_{(\text{Z})}+{2 {NaHCO}_{3}}_{\text{Sodium} \text{bicarbonate}}$ Hence, $X \;\; = {Na}[ {Al}( {OH})_{4}]$ $Y \;\; = {CO}_{2}$ $Z \;\; = {Al}_{2} {O}_{3} \cdot x {H}_{2} {O}$

Q37JEE Main 2021MCQ4MElectrochemistry

The electrode potential of $M^{2+} / M$ of $3 d$ -series elements shows positive value for

🚀 Solve in Practice Mode

📖 Explanation

In the electrode potential series, only copper have positive value for electrode potential because copper has lower tendency than hydrogen to form ions. So, if standard hydrogen electrode $(E_{\;\text{Cell }\;}=0)$ is connected to copper half-cell, the copper with be relatively less negative or less number of electrons. $\;E_{( {Cu}^{2+} / {Cu})}^{\circ}=+0.34 {V} ; \;\; E_{( {Fe}^{2+} / {Fe})}^{\circ}=-0.41 {V}$ $\;E_{ {Co}^{\circ} / 2 / {Co}}^{\circ}=-0.28 {V} ; \;\; E_{( {Zn}^{2+} / {Zn})}^{\circ}=-0.76 {V}$ $\therefore$ Electrode potential of ${Cu} E_{ {Cu}^{+2} / {Cu}}^{\circ}$ show positive value.

Q38JEE Main 2021MCQ4Md and f Block Elements

The major components in Gun metal are

🚀 Solve in Practice Mode

📖 Explanation

'Gun metal' also called as red brass (in USA). It is greyish in colour and originally used for making guns. The major components of 'Gun metal' are ${Cu}=87 \%$ ${Zn}=3 \%$ ${Sn}=10 \%$ It is also used in manufacturing of gears and bearing that are to be subjected to heavy loads and low speeds.

Q39JEE Main 2021MCQ4MHydrocarbons

The gas released during anaerobic degradation of vegetation may lead to

🚀 Solve in Practice Mode

📖 Explanation

Methane $( {CH}_{4})$ gas is evolved due to anaerobic degradation of vegetation which causes global warming and cancer. It is a heat trapping gas that forces the planet to warm drastically and quickly.

Q40JEE Main 2021MCQ4MAlcohols, Phenols and Ethers

Which of the following compound gives pink colour on reaction with phthalic anhydride in conc. ${H}_{2} {SO}_{4}$ followed by treatment with ${NaOH}$ ?.

🚀 Solve in Practice Mode

📖 Explanation

Correct answer isFirstly phthalic anhydride in presence of conc. ${H}_{2} {SO}_{4}$ undergoes protonation to give an intermediate carbocation. This carbocation reacts with 2-propylphenol in presence of ${NaOH}$ to give pink colour compound. Mechanism2-propylphenol gives pink colour on reaction with phthalic anhydride in conc. ${H}_{2} {SO}_{4}$ followed by treatment with ${NaOH}$ .

Q41JEE Main 2021MCQ4MHydrocarbons

What is the major product formed by HI or reaction with

🚀 Solve in Practice Mode

📖 Explanation

The major product formed is 2 -iodo- 2,3 -dimethylbutane. Steps involved in the reaction are as follows: Step 1 it is electrophilic addition reaction, $\pi$ -bond of alkene attack ${H}^{+}$ ion of ${Hl}$ and form more stable carbocation. Step 2 Formation of more stable $3^{\circ}$ carbocation take place by 1,2 shifting of Me group. Step 3 Direct addition of $\text{I}^{⊖}$ ion.

Q42JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Which of the following reagent is used for the following reaction ? ${CH}_{3} {CH}_{2} {CH}_{3} {\longrightarrow }^{?} {CH}_{3} {CH}_{2} {CHO}$

🚀 Solve in Practice Mode

📖 Explanation

Molybdenum oxide $( {Mo}_{2} {O}_{3})$ is used for oxidising alkanes to aldehyde. It used to manufacture molybdenum metal, which serves as an additive to steel and corrosive resistant alloys. ${CH}_{3}- {{CH}_{2}}_{\text{Propane}}- {CH}_{3} {\to }^{ {Mo}_{2} {O}_{3}}_{\text{Oxidising} \text{agent}} {CH}_{3}- {{CH}_{2}}_{\text{Propanal}}- {CHO}$

Q43JEE Main 2021MCQ4MHaloalkanes and Haloarenes

The product formed in the first step of the reaction of ${CH}_{3}- {CH}_{2}- {CH}^{|}^{\text{Br}}- {CH}_{2}- {CH}_{|}_{\text{Br}}- {CH}_{3}$ with excess ${Mg} / {Et}_{2} {O}( {Et}= {C}_{2} {H}_{5})$ is

🚀 Solve in Practice Mode

📖 Explanation

Here, in first step only one mole of ${Mg} / {Et}_{2} {O}$ attacks on bromine and form two $2 {MgBr}$ in the first step.On further moving in the reaction, two ${MgBr}$ are eliminated to form alkene in respective positions.

Q44JEE Main 2021MCQ4MHaloalkanes and Haloarenes

What is the final product (major) ' $A$ ' in the given reaction ?

🚀 Solve in Practice Mode

📖 Explanation

Steps involved in this reaction are as follows Step ${AHCl} \longrightarrow {H}^{+}+ {Cl}^{-}$ ${H}^{+}$ attacks on $- {OH}$ (lone pair) and formed ${OH}_{2}^{⊕}$ ion. Here, ${OH}_{2}^{⊕}$ is the good leaving group.Step 2 Elimination reaction always give minor product than substitution reaction.

Q45JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Identify products $A$ and $B$ .

🚀 Solve in Practice Mode

📖 Explanation

In first step, ${KMnO}_{4}$ (dil.) help to break alkene in ketone and aldehyde and in second step, ${CrO}_{3}$ is used for selective over oxidation of aldehyde only. ${CrO}_{3}$ will form aldehyde to acid in product $(B)$ .

Q46JEE Main 2021MCQ4MAmines

In the following reaction the reason why meta-nitro product also formed is

🚀 Solve in Practice Mode

📖 Explanation

Aniline on protonation forms anilinium ion, which is meta-directing. So, a considerable amount of meta product is formed alongwith o-nitroaniline and $p$ -nitroaniline. Nitrating mixture is mixture of conc. ${HNO}_{3}$ and a conc. ${H}_{2} {SO}_{4}$ . When aniline is reacted with nitrating mixture ortho, meta and para nitroanilines are obtained. Aniline being basic in nature, reacts with acids to form anilinium ion which is meta directing.

Q47JEE Main 2021MCQ4MAmines

$' A'$ and ' $B$ ' in the following reaction are

🚀 Solve in Practice Mode

📖 Explanation

Step 1 In step $1, {NaNO}_{2}+ {HCl}, 0-5^{\circ} {C}$ used for diazotisation reaction. It will form diazonium salt. Further, it will react with KCN to form cyanobenzene. Step 2 In step $2, {SnCl}_{2}$ and ${HCl}$ is a Stephen's reduction reagent. Cyanobenzene reduced to benzaldehyde by ${SnCl}_{2} / {HCl}$ .

Q48JEE Main 2021MCQ4MBiomolecules

Match List-I with List-II. List-I(Monomer Unit) List-II(Polymer) A. Caprolactum 1. Natural rubber B. 2-chloro buta-1,3-diene 2. Buna- ${N}$ C. Isoprene 3. Nylon-6 D. Acrylonitrile 4. NeopreneChoose the correct answer from the options given below

🚀 Solve in Practice Mode

Q49JEE Main 2021MCQ4MBiomolecules

Out of the following, which type of interaction is responsible for the stabilisation of $\alpha$ -helix structure of proteins?

🚀 Solve in Practice Mode

📖 Explanation

$\alpha$ -helix structure of proteins is stabilised by H-bonds between hydrogen of $- {NH}$ and oxygen of $- {C}= {O}$ group is amino acids.

Q50JEE Main 2021MCQ4MSalt Analysis

Given below are two statements. Statement I Colourless cupric metaborate is reduced to cuprous metaborate in a luminous flame. Statement II Cuprous metaborate is obtained by heating boric anhydride and copper sulphate in a non-luminous flame. In the light of the above statements, choose the most appropriate answer from the options given below.

🚀 Solve in Practice Mode

📖 Explanation

(i) In presence of luminous flame, blue cupric metaborate is reduced to colourless cuprous metaborate. $2 {Cu}( {BO}_{2})_{2}+2 {NaBO}_{2}+ {C} {\to }^{\;\text{ Luminous }\;}_{\text{flame}} 2 {CuBO}_{2}+ {Na}_{2} {B}_{4} {O}_{7}+ {CO}$ If non-luminous flame is present in reaction, cupric metaborate is obtained by heating boric anhydride with copper sulphate. ${CuSO}_{4}+ {B}_{2} {O}_{3} {\to }^{\;\text{ Non-luminous }\;} _{\text{flame}} {{Cu}( {BO})_{2}}_{\text{Cupric} \text{metaborate}}+ {SO}_{3}$ So, both statements are false.

Q51JEE Main 2021NAT4MSolutions

$4.5 {g}$ of compound $A(M W=90)$ was used to make $250 {mL}$ of its aqueous solution. The molarity of the solution in $M$ is $x \times 10^{-1}$ . The value of $x$ is ......... (Rounded off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

Given, weight of compound $A=4.5 {g}$ Molecular weight of compound $A=90 {g} / {mol}$ Volume of solution (in ${mL}$ ) $=250 {mL}$ Now, molarity is defined as number of moles of solute or compound A divided by volume of solution (in L). $M \;\; =\;\frac{\;\text{ Number of moles of solute }\;(n)}{\;\text{ Volume of solution }\;}$ $\;\; =\;{\;\frac{4.5}{\frac{90}{250}}}/{1000}=0.2 \;\text{ or }\;=2 \times 10^{-1} {M}$ $\therefore \;\; n \;\; =\;\frac{\;\text{ Weight of solute (compound }\; A \;\text{ ) }\;}{\;\text{ Molecular weight of solute (compound }\; A \;\text{ ) }\;}$ $\;\text{ Hence, }\; x \times 10^{-1} \mu$ $x=2$

Q52JEE Main 2021NAT4MSolid State

The coordination number of an atom in a body centered cubic structure is ............ .[Assume that the lattice is made up of atoms.]

🚀 Solve in Practice Mode

Q53JEE Main 2021NAT4MStructure of Atom

A proton and ${a} {Li}^{3+}$ nucleus are accelerated by the same potential. If $\lambda_{ {Li}}$ and $\lambda_{ {P}}$ denote the de-Broglie wavelengths of ${Li}^{3+}$ and proton respectively, then the value of $\;\frac{\lambda L i}{\lambda_{P}}$ is $x \times 10^{-1}$ . The value of $x$ is ........... . (Rounded off to the nearest integer) (Mass of ${Li}^{3+}=8.3$ mass of proton)

🚀 Solve in Practice Mode

📖 Explanation

Given, mass of ${Li}^{3+}=8.3$ times of mass of proton formula, De-Broglie wavelength, $\lambda=\;\frac{h}{\sqrt{2 m q V}}$ Here, $h=$ Planck's constant $=6.624 \times 10^{-34} {J}- {s}$ $m=$ Mass of atom $q=$ Charge (or number of electrons) $\;\lambda_{ {Li}}=\;{h}/{\sqrt{2 m_{ {Li}}(3 {e}) v}}$ . . . (i) $\;\lambda_{ {P}}=\;{h}/{\sqrt{2 m_{ {P}}( {e}) {v}}}$ . . . (ii) Now, Eq. (i) divided by Eq. (ii), we get, $\;{\lambda_{ {Li}}}/{\lambda_{ {P}}}=\sqrt{\;{m_{ {P}}( {e}) {v}}/{m_{ {Li}}(3 {e}) {v}}}$ We know that $m_{ {Li}}=8.3 m_{p}$ $\;{\lambda_{ {Li}}}/{\lambda_{ {P}}}= \;\; \sqrt{\;{m_{ {P}} \times {e} \times V}/{8.3 m_{ {P}} \times 3 {e} \times V}}=\sqrt{\;\frac{1}{8.3 \times 3}}=\;\frac{1}{5}=0.2$ $\;2 \times 10^{-1}$ Hence, $x \times 10^{-1}, x=2$

Q54JEE Main 2021NAT4MChemical Equilibrium

For the reaction, $A(g) \longrightarrow B(g)$ , the value of the equilibrium constant at $300 {K}$ and 1 atm is equal to $100.0$ . The value of $\Delta , G$ for the reaction at $300 {K}$ and $1 {atm}$ in ${J} {mol}^{-1}$ is $-x R_{i}$ where $x$ is ......... (Rounded off to the nearest integer) $(R=8.31 {J} {mol}^{-1} {K}^{-1}$ and $\ln 10=2.3)$

🚀 Solve in Practice Mode

📖 Explanation

For a reaction, $A( {g}) \longrightarrow B( {g})$ Given, $K_{p}$ (equilibrium constant) $=100$ $\;\text{ Temperature }\;=300 {K}$ $\;\text{ Pressure }\;=1 {atm}$ Formula used, $\Delta G^{\circ}=-R T \ln K_{p}$ . . . (i) Here, $\Delta G^{\circ}=$ standard Gibb's free energy $R=$ gas constant $=8.31 {J} {mol}^{-1} {K}^{-1}$ Put value in Eq (i), we get $\Delta G^{\circ}=-R(300) \ln 100$ $\Delta G^{\circ}=-R(300)(2) \ln (10)$ $\because \ln (10)=2.3$ $\Delta G^{\circ}=-R(300)(2)(2.3)$ $\Delta G^{\circ}=-1380 R$ $\;\text{ Hence, }\; \Delta G^{\circ}=-x R$ $x=1380$

Q55JEE Main 2021NAT4MIonic Equilibrium

When $9.45 {g}$ of ${Cl} {CH}_{2} {COOH}$ is added to $500 {mL}$ of water, its freezing point drops by $0.5^{\circ} {C}$ . The dissociation constant of ${ClCH}_{2} {COOH}$ is $x \times 10^{-3}$ . The value of $x$ is ............. . (Rounded off to the nearest integer) $[K_{f({H}_{2} {O})}=1.86 {K} {kg} {mol}^{-1}]$

🚀 Solve in Practice Mode

📖 Explanation

${ClCH}_{2} {COOH} ⇌\; {ClCH}_{2} {COO}^{-}+ {H}^{+}$ Here, $i \;\; =\;\frac{\;\text{ Final moles }\;}{\;\text{ Initial moles }\;}$ $i \;\; =1-\alpha+\alpha+\alpha \Rightarrow i=1+\alpha$ Formula used for freezing point; $\Delta T_{f}=i K_{f} m$ Here, $K_{f}$ = freezing constant of ${H}_{2} {O}$ ${K}_{f}( {H}_{2} {O})=1.86 {Kkg} {mol}^{-1}$ $m=$ molarity $\Delta {T}_{f}=i K_{f} m$ $0.5=(1+\alpha)(1.86) \;\frac{(9.45 / 94.5)}{(500 / 1000)}$ $\;\frac{5}{3.72}=1+\alpha \Rightarrow \;\frac{5}{3.72}-1=\alpha$ $\Rightarrow \;\; \alpha=\;\frac{1.28}{3.72}=\;\frac{32}{93}=0.344$ Percentage of dissociation $=34.4 \%$

Q56JEE Main 2021NAT4MChemical Equilibrium

At $1990 {K}$ and $1 {atm}$ pressure, there are equal number of ${Cl}_{2}$ molecules and ${Cl}$ atoms in the reaction mixture. The value of $k_{p}$ for the reaction ${Cl}_{2}(g) ⇌ 2 {Cl}(g)$ under the above conditions is $x \times 10^{-1}$ . The value of $x$ is ........... . (Rounded off to the nearest integer)

🚀 Solve in Practice Mode

📖 Explanation

${Cl}_{2}( {g}) ⇌ 2 {Cl}( {g})$ Let, moles of both of ${Cl}_{2}$ and ${Cl}$ molecule be $x$ . Partial pressure of ${Cl}$ is, $p_{ {Cl}}=\;\frac{x}{2 x} \times 1=\;\frac{1}{2}$ Partial pressure of ${Cl}_{2}$ is $p_{ {Cl}_{2}}=\;\frac{x}{2 x} \times 1=\;\frac{1}{2}$ Now, $\;\; K_{p}=\;{(p_{ {C}})^{2}}/{p_{ {Cl}_{2}}} \Rightarrow K_{p}=\;\frac{(1 / 2)^{2}}{1 / 2}=\;\frac{1}{2}=0.5$ or $5 \times 10^{-1}$ Hence, $x \times 10^{-1}$

Q57JEE Main 2021NAT4MRedox Reactions

The reaction of sulphur in alkaline medium is given below ${S}_{8}( {s})+a {OH}^{-}(a q) \longrightarrow b {S}^{2-}(a q)+c {S}_{2} {O}_{3}^{2-}(a q)$ $+d {H}_{2} {O}(')$ The value of ' $a$ ' is ........... (Integer answer)

🚀 Solve in Practice Mode

📖 Explanation

The two half reaction, one separately are as follows ${S}_{8}+16 {e}^{-} \longrightarrow 8 {S}^{2-} \;\;$ (Reduction) $\;\frac{ {S}_{8}+12 {H}_{2} {O} \longrightarrow 4 {S}_{2} {O}_{3}^{2-}+24 {H}^{+}+16 {e}^{-}}{2 {S}_{8}+12 {H}_{2} {O} \longrightarrow 8 {S}^{2-}+4 {S}_{2} {O}_{3}^{2-}+24 {H}^{+}}$ For balancing in basic medium, Add an equal number of ${OH}^{-}$ that of ${H}^{+}$ , we get $2 {S}_{8}+12 {H}_{2} {O}+24 {OH}^{-} \longrightarrow 8 {S}^{2-}+4 {S}_{2} {O}_{3}^{2-}+24 {H}_{2} {O}$ $2 {S}_{8}+24 {OH}^{-} \longrightarrow 8 {S}^{2-}+4 {S}_{2} {O}_{3}^{2-}+12 {H}_{2} {O}$ or ${S}_{8}+12 {OH}^{-} \longrightarrow 4 {S}^{2-}+2 {S}_{2} {O}_{3}^{2-}+6 {H}_{2} {O}$ . . . (i) On comparing (i) with ${S}_{8}+ {aOH}^{-}( {aq}) \longrightarrow {bS}^{2-}( {aq})+ {c} {S}_{2} {O}_{3}^{2-}+ {d} {H}_{2} {O}$ We get, $a=12$ $b=4 ; \;\; c=2 ; \;\; d=6$

Q58JEE Main 2021NAT4MChemical Kinetics

Gaseous cyclobutene isomerises to butadiene in a first order process which has a ' $k$ ' value of $3.3 \times 10^{-4} {s}^{-1}$ at $153^{\circ} {C}$ . The time in minutes it takes for the isomerisation to proceed $40 \%$ to completion at this temperature is ........ (Rounded off to the nearest integer)

🚀 Solve in Practice Mode

📖 Explanation

It is a first order isomerisation reaction. Integrated rate law for 1 st order reaction is $k t=\ln \;\frac{[A]_{0}}{[A]_{t}}$ . . . (i) Here, $k=$ rate constant $[A]_{0}=$ initial concentration $[A]_{t}=$ concentration at time $' t '$ Given, $k=3.3 \times 10^{-4} {s}^{-1}$ $[A]_{0}=100 \Rightarrow [A]_{t}=100-40=60$ Put values in Eq. (i), we get $3.3 \times 10^{-4} {s}^{-1} \times t=\ln \;\frac{100}{60}$ $t=1547.95 {s}=25.79 {min} \;\; (1 {min}=60 {s}$ or $1 {s}=\;\frac{1}{60} {min})$ $=26$ minutes

Q59JEE Main 2021MCQ4MClassification of Elements

Number of amphoteric compounds among the following is

🚀 Solve in Practice Mode

📖 Explanation

An amphoteric compound is a molecule or ion that can react with both as an acid or as a base. ${BeO}=\;\text{ Amphoteric }\;$ ${BaO}=\;\text{ Basic }\;$ ${Be}( {OH})_{2}=\;\text{ Amphoteric }\;$ ${Sr}( {OH})_{2}=\;\text{ Basic }\;$ Both beryllium compound ${BeO}$ and ${Be}( {OH})_{2}$ are amphoteric in nature while compound ${BaO}$ and ${Sr}( {OH})_{2}$ are basic in nature, they form alkaline solution in ${H}_{2} {O}$ .

Q60JEE Main 2021NAT4MCoordination Compounds

The stepwise formation of $[ {Cu}( {NH}_{3})_{4}]^{2+}$ is given below ${Cu}^{2+}+ {NH}_{3} ⇌[ {Cu}( {NH}_{3})]^{2+}$ $\;{[ {Cu}( {NH}_{3})]^{2+}+ {NH}_{3} {⇌}^{ {K}_{2}}[ {Cu}( {NH}_{3})_{2}]^{2+}}$ $\;{[ {Cu}( {NH}_{3})_{2}]^{2+}+ {NH}_{3} {⇌}^{ {K}_{3}}[ {Cu}( {NH}_{3})]^{2+}}$ $[ {Cu}( {NH}_{3})_{3}]^{2+}+ {NH}_{3} {⇌}^{K_{4}}[ {Cu}( {NH}_{3})_{4}]^{2+}$ The value of stability constants $K_{1}, K_{2}, K_{3}$ and $K_{4}$ are $10^{4}, 1.58 \times 10^{3}, 5 \times 10^{2}$ and $10^{2}$ respectively. The overall equilibrium constants for dissociation of $[ {Cu}( {NH}_{3})]^{2+}]$ is $x \times 10^{-12}$ . The value of $x$ is ............. . (Rounded off to the nearest integer)

🚀 Solve in Practice Mode

📖 Explanation

Given, stability constant value, $k_{1}=10^{4}$ $k_{2}=1.58 \times 10^{3}$ $k_{3}=5 \times 10^{2}$ $k_{4}=10^{2}$ ${Cu}^{2+}+ {NH}_{3} {⇌}^{k_{1}}[ {Cu}( {NH}_{3})]^{2+}$ . . . (i) $[ {Cu}( {NH}_{3})]^{2+}+ {NH}_{3} {⇌}^{k_{2}}[ {Cu}( {NH}_{3})_{2}]^{2+}$ . . . (ii) $[ {Cu}( {NH}_{3})_{2}]^{2+}+ {NH}_{3} {⇌}^{k_{3}}[ {Cu}( {NH}_{3})_{3}]^{2+}$ . . . (iii) $[ {Cu}( {NH}_{3})_{3}]^{2+}+ {NH}_{3} {⇌}^{k_{4}}[ {Cu}( {NH}_{3})_{4}]^{2+}$ . . . (iv) On adding Eqs. (i), (ii), (iii) and (iv), we get ${Cu}^{2+}+4 {NH}_{3} \;{\text{stackrel}{k}}_{⇌{⇌}}[ {Cu}( {NH}_{3})_{4}]^{2+}$ $\therefore$ The overall reaction constant $(k)$ or equilibrium constant for formation of $[ {Cu}( {NH}_{3})_{4}]^{2+}$ is $\;k=k_{1} \times k_{2} \times k_{3} \times k_{4}$ $\;k=10^{4} \times 1.58 \times 10^{3} \times 5 \times 10^{2} \times 10^{2}$ $\;k=7.9 \times 10^{11}$ where, $k=$ equilibrium constant for formation of $[ {Cu}( {NH}_{3})_{4}]^{2+}$ So, equilibrium constant $k^{'}$ for dissociation of $[ {Cu}( {NH}_{3})_{4}]^{2+}$ is $\;\frac{1}{k}$ . $k^{'}=\;\frac{1}{k}=\;\frac{1}{7.9 \times 10^{11}}=1.26 \times 10^{-12}$ Hence, $k^{'} \;\; =x \times 10^{-12}$ $x \;\; =1.26$

Mathematics30 questions

Q61JEE Main 2021MCQ4MFunctions

Let $f: R \to R$ be defined as $f(x)=2 x-1$ and $g: R-\{1\} \to R$ be defined as $g(x)=\;\frac{x-\frac{1}{2}}{x-1}$ . Then, the composition function $f(g(x))$ is

🚀 Solve in Practice Mode

📖 Explanation

Given, $f(x)=2 x-1 ; f: R \to R$ $g(x) \;\; =\;\frac{x-1 / 2}{x-1} ; g: R-\{1\} \to R$ $f[g(x)] \;\; =2 g(x)-1$ $\;\; =2 \times (\;\frac{x-\frac{1}{2}}{x-1})-1=2 \times (\;\frac{2 x-1}{2(x-1)})-1$ $\;\; =\;\frac{2 x-1}{x-1}-1=\;\frac{2 x-1-x+1}{x-1}=\;\frac{x}{x-1}$ $\therefore \;\; f[g(x)] \;\; =1+\;\frac{1}{x-1}$ Now, draw the graph of $1+\;\frac{1}{x-1}$ , $\because$ Any horizontal line does not cut the graph at more than one points, so it is one-one and here, co-domain and range are not equal, so it is into.Hence, the required function is one-one into.

Q62JEE Main 2021MCQ4MQuadratic Equation and Inequalities

Let $p$ and $q$ be two positive numbers, such that $p+q=2$ and $p^{4}+q^{4}=272$ . Then, $p$ and $q$ are roots of the equation

🚀 Solve in Practice Mode

📖 Explanation

Given, $p$ and $q$ are positive numbers. $p+q=2$ . . . (i) $p^{4}+q^{4}=272$ $\Rightarrow \;\; (p^{2}+q^{2})^{2}-2 p^{2} q^{2}=272$ $\Rightarrow \;\; \$ [(p+q)^{2}-2 p q]^{2}-2 p^{2} q^{2}=272\Rightarrow ;; $[(2)^{2}-2 p q]^{2}-2 p^{2} q^{2}=272 $[from Eq. (i)]$ \Rightarrow ;; (4-2 p q)^{2}-2 p^{2} q^{2}=272\Rightarrow 16+4 p^{2} q^{2}-16 p q-2 p^{2} q^{2}=272\Rightarrow ;; 2 p^{2} q^{2}-16 p q-256=0\Rightarrow ;; p^{2} q^{2}-8 p q-128=0p q=;\frac{8 \pm \sqrt{64+4 \times 128}}{2 \times 1}=;\frac{8 \pm \sqrt{576}}{2}=;\frac{8 \pm 24}{2}\therefore ;; p q=16,-8 $Here,$ p q=-8 $is not possible as$ p $and$ q $are positive.$ \therefore ;; p q=16 $Now, the equation whose roots are$ p $and$ q $is$ x^{2}-(p+q) x+p q=0x^{2}-2 x+16=0$

Q63JEE Main 2021MCQ4MMatrices and Determinants

The system of linear equations $\;3 x-2 y-k z=10$ $\;2 x-4 y-2 z=6$ $\;x+2 y-z=5 m$ is inconsistent, if

🚀 Solve in Practice Mode

📖 Explanation

Given, $3 x-2 y-k z=10$ $2 x-4 y-2 z \;\; =6$ $x+2 y-z \;\; =5 m$ Condition for inconsistency $\Rightarrow \Delta =0$ and atleast one of the $\Delta _{x^{'}} \Delta _{y}, \Delta _{z}$ is non-zero. $\;\text{ Now, }\; \Delta \;\; =\begin{vmatrix}3 & -2 & -k \\ 2 & -4 & -2 \\ 1 & 2 & -1\end{vmatrix}$ $\;\; =3(4+4)+2(-2+2)-k(4+4)$ $\;\; =24-8 k$ Now, $\Delta =0$ $\therefore \;\; 24-8 k=0$ $\Rightarrow \;\; 8 k=24 \Rightarrow k=\;\frac{24}{8}=3$ $\Delta _{x}=\begin{vmatrix}10 & -2 & -k \\ 6 & -4 & -2 \\ 5 m & 2 & -1\end{vmatrix}$ $=10(4+4)+2(-6+10 m)-3(12+20 m)$ $=80-12+20 m-36-60 m$ $=32-40 m=8(4-5 m)$ $\Delta _{y}=\begin{vmatrix}3 & 10 & -k \\ 2 & 6 & -2 \\ 1 & 5 m & -1\end{vmatrix}$ $=3(-6+10 m)-10(-2+2)-3(10 m-6)$ $=-18+30 m-30 m+18$ $=0$ $\Delta _{z}=\begin{vmatrix}3 & -2 & 10 \\ 2 & -4 & 6 \\ 1 & 2 & 5 m\end{vmatrix}$ $\;\; =3(-20 m-12)+2(10 m-6)+10(4+4)$ $\;\; =-60 m-36+20 m-12+80$ $\;\; =-40 m+32=8(4-5 m)$ Here, either $\Delta _{x}$ or $\Delta _{z} \ne 0$ $\;\Rightarrow \;8(4-5 m)\;\ne 0$ $\Rightarrow \;m \ne 4 / 5$ Hence, the required values are $k=3 ; m \ne \;\frac{4}{5}$ .

Q64JEE Main 2021MCQ4MBinomial Theorem

The value of $-{ }^{15} C_{1}+2 \cdot { }^{15} C_{2}-3 \cdot { }^{15} C_{3}+. . .$ $-15 \cdot { }^{15} C_{15}+{ }^{14} C_{1}+{ }^{14} C_{3}+{ }^{14} C_{5}+. . .+{ }^{14} C_{11}$ is

🚀 Solve in Practice Mode

📖 Explanation

Given, $(-{ }^{15} C_{1}+2 \cdot { }^{15} C_{2}-3 \cdot { }^{15} C_{3}+. . .-15 \cdot { }^{15} C_{15})$ $\;\; +({ }^{14} C_{1}+{ }^{14} C_{3}+{ }^{14} C_{5}+. . .+{ }^{14} C_{11})$ Let $S_{1}=-{ }^{15} C_{1}+2 \cdot { }^{15} C_{2}-3 \cdot { }^{15} C_{3}+. . .-15 \cdot { }^{15} C_{15}$ $\;\; =sum_{r=1}^{15}(-1)^{r} \cdot r \cdot { }^{15} C_{r}=15 sum_{r=1}^{15}(-1)^{r}{ }^{14} C_{r-1}$ $\;\; =15(-{ }^{14} C_{0}+{ }^{14} C_{1}-{ }^{14} C_{2}+. . .-{ }^{14} C_{14})$ $\;\; =15(0)=0$ $S_{2}={ }^{14} C_{1}+{ }^{14} C_{3}+{ }^{14} C_{5}+. . .+{ }^{14} C_{11}$ $\;\; =({ }^{14} C_{1}+{ }^{14} C_{3}+. . .+{ }^{14} C_{13})-{ }^{14} C_{13}$ $\;\; =2^{13}-14$ Now, the required value is $(-{ }^{15} C_{1}\;+2 \cdot { }^{15} C_{2}-3 \cdot { }^{15} C_{3}+. . .-15 \cdot { }^{15} C_{15})$ $\;+({ }^{14} C_{1}+{ }^{14} C_{3}+. . .+{ }^{14} C_{11})$ $\;\; =S_{1}+S_{2}=0+2{ }^{13}-14=2^{13}-14$

Q65JEE Main 2021MCQ4MSequences and Series

If $e^{(\cos ^{2} x+\cos ^{4} x+\cos ^{6} x+. . . \infty ) \log _{e} 2}$ satisfies the equation $t^{2}-9 t+8=0$ , then the value of $\;\frac{2 \sin x}{\sin x+\sqrt{3} \cos x},(0 < x < \;\frac{\pi}{2})$ is

🚀 Solve in Practice Mode

📖 Explanation

Given, $"\; t^{2}-9 t+8=0, \;\text{ is satisfied by }\;$ $\;\text{ Now, }\; a^{locos^{2} x+\cos ^{4} x+\cos ^{6} x+. . . \infty ) \log _{c} 2}=b^{\log _{c} a}$ $\therefore \;\; e^{(\cos ^{2} x+\cos ^{4} x+. . . \infty ) \log 2}$ $=2^{(\cos ^{2} x+\cos ^{4} x+\infty ) \log _{e} e}=2^{\cos ^{2} x+\cos ^{4} x+. . . . . . \infty }$ $\;\text{ Here, }\; \cos ^{2} x+\cos ^{4} x+. . . \infty \;\text{ are \in GP, }\;$ $\;\text{ where }\; a=\cos ^{2} x$ $\therefore \;\; r=\cos ^{2} x< 1$ $\therefore \;\; S_{\infty }=\;\frac{a}{1-r}$ $\therefore \;\; S_{\infty }=\;\frac{\cos ^{2} x}{1-\cos ^{2} x}=\;\frac{\cos ^{2} x}{\sin ^{2} x}=\cot ^{2} x$ $\therefore \;\; \cos ^{2} x+\cos ^{4} x+. . . \infty =\cot ^{2} x$ $\;\text{ Now, }\; 2^{\cos ^{2} x+\cos ^{4} x+. . . \infty }=2^{\cot ^{2} x}$ $\;\text{ Now, roots of equation }\; t^{2}-9 t+8=0, \;\text{ are }\;$ $\;\;\; (t-1)(t-8)=0$ $\;\;\; \;\; t=1,8$ $\Rightarrow \;\; 2^{\cot ^{2} x}=1 \;\text{ or }\; 8$ $\Rightarrow \;\; \cot ^{2} x=0 \;\text{ or }\; \cot ^{2} x=3 \Rightarrow \cot x=0 \;\text{ or }\; \cot x=\sqrt{3}$ \But here, $0 < x < \;\frac{\pi}{2}$ . $\therefore \cot x=0$ not possible. Hence, $\cot x=\sqrt{3}$ is only possible value. Now, $\;\frac{2 \sin x}{\sin x+\sqrt{3} \cos x}$ Dividing numerator and denominator by $\sin x$ , we get $\;\;\frac{2}{1+\sqrt{3} \cot x}$ $= \;\; \;\frac{2}{1+\sqrt{3} \times \sqrt{3}}$ $= \;\; \;\frac{2}{1+3}=\;\frac{2}{4}=\;\frac{1}{2}$ $[ \because \cot x=\sqrt{3}]$

Q66JEE Main 2021MCQ4MLimits, Continuity and Differentiability

$\lim _{x \to 0} \;{int_{0}^{x^{2}}(\sin \sqrt{t}) d t}/{x^{3}}$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Given, $\lim _{x \to 0} \;{int_{0}^{x^{2}}(\sin \sqrt{t}) d t}/{x^{3}}$ $\because$ It is of the form $\;\frac{0}{0}$ . By differentiating numerator and denominator, $\lim _{x \to 0} \;{\sin \sqrt{x^{2}} \cdot 2 x}/{3 x^{2}} \;\; =\lim _{x \to 0} \;\frac{\sin x \cdot 2 x}{3 x^{2}}$ $\;\; =\;\frac{2}{3} \lim _{x \to 0} \;\frac{\sin x}{x}=\;\frac{2}{3}(1)=\;\frac{2}{3}$

Q67JEE Main 2021MCQ4MApplication of Derivatives

The function $f(x)=\;\frac{4 x^{3}-3 x^{2}}{6}-2 \sin x+(2 x-1) \cos x$

🚀 Solve in Practice Mode

📖 Explanation

Given, $f(x)=\;\frac{4 x^{3}-3 x^{2}}{6}-2 \sin x+(2 x-1) \cos x$ $f^{'}(x)=\;\frac{12 x^{2}-6 x}{6}-2 \cos x+(2 x-1)(-\sin x)+\cos x(2)$ $=(2 x^{2}-x)-2 \cos x-2 x \sin x+\sin x+2 \cos x$ $=2 x^{2}-x-2 x \sin x+\sin x$ $=2 x(x-\sin x)-1(x-\sin x)$ $f^{'}(x)=(2 x-1)(x-\sin x)$ for $x > 0, x-\sin x > 0$ $x < 0, x-\sin x < 0$ for $x \in(-\infty , 0] \cup[\;\frac{1}{2}, \infty ), f^{'}(x) \ge 0$ for $x \in[0, \;\frac{1}{2}], f^{'}(x) \le 0$ Hence, $f(x)$ increases in $[\;\frac{1}{2}, \infty )$ .

Q68JEE Main 2021MCQ4MPermutations and Combinations

A scientific committee is to be formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and double the number of foreigners as Indians. Then, the number of ways, the committee can be formed, is

🚀 Solve in Practice Mode

📖 Explanation

Given, Number of Indians $=6$ Number of foreigners $=8$ Committee of atleast 2 Indians and double number of foreigners is to be formed. Hence, the required cases are $\;(2 I, 4 F)+(3 I, 6 F)+(4 I, 8 F)$ $= \;\; { }^{6} C_{2} \times { }^{8} C_{4}+{}^{6} C_{3} \times { }^{8} C_{6}+{ }^{6} C_{4} \times { }^{8} C_{8}$ $= \;\; (15 \times 70)+(20 \times 28)+(15 \times 1)$ $= \;\; 1050+560+15=1625$

Q69JEE Main 2021MCQ4MLimits, Continuity and Differentiability

If $f: R \to R$ is a function defined by $f(x)=[x-1] \cos (\;\frac{2 x-1}{2}) \pi$ , where [.] denotes the greatest integer function, then $f$ is

🚀 Solve in Practice Mode

📖 Explanation

Given, $f(x)=[x-1] \cos (\;\frac{2 x-1}{2}) \pi$ where $[\cdot ]$ is greatest integer function and $f: R \to R$ $\because$ It is a greatest integer function then|we need to check its continuity at $x \in I$ except these it is continuous. Let, $x=n$ where $n \in I$ $\;\text{ Then }\; {LHL} \;\; =\lim _{x \to n^{-}}[x-1] \cos (\;\frac{2 x-1}{2}) \pi$ $\;\; =(n-2) \cos (\;\frac{2 n-1}{2}) \pi=0$ ${RHL} \;\; =\lim _{x \to n^{+}}[x-1] \cos (\;\frac{2 x-1}{2}) \pi$ $\;\; =(n-1) \cos (\;\frac{2 n-1}{2}) \pi=0$ and $f(n)=0$ . Here, $\lim _{x \to n^{-}} f(x)=\lim _{x \to n^{+}} f(x)=f(n)$ $\therefore$ It is continuous at every integers. Therefore, the given function is continuous for all real $x$ .

Q70JEE Main 2021MCQ4MIndefinite Integrals

If $\int \;\frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x=a \sin ^{-1}(\;\frac{\sin x+\cos x}{b})+c$ , where $c$ is a constant of integration, then the ordered pair $(a, b)$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Given, $\int \;\frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x$ $=a \sin ^{-1}(\;\frac{\sin x+\cos x}{b})+c$ Put, $\sin x+\cos x=t$ Also, $\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=t^{2}$ $\Rightarrow$ $2 \sin x \cos x=t^{2}-1$ $\Rightarrow$ $\sin 2 x=(t^{2}-1)$ and $(\cos x-\sin x) d x=d t$ Now, $\int \;\frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x=\int \;{d x}/{\sqrt{8-(t^{2}-1)}}$ $=\int \;{d t}/{\sqrt{9-t^{2}}}=\sin ^{-1}(\;\frac{t}{3})+c$ Now, according to question, $\int \;\frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x=a \sin ^{-1}(\;\frac{\sin x+\cos x}{b})+c$ $\Rightarrow \sin ^{-1}(\;\frac{t}{3})+c=a \sin ^{-1} \;\frac{(\sin x+\cos x)}{b}+c$ $\Rightarrow 1 \sin ^{-1}(\;\frac{\sin x+\cos x}{3})+c=a \sin ^{-1}(\;\frac{\sin x+\cos x}{b})+c$ $( \because t=\sin x+\cos x)$ $\therefore \;\; a=1$ $\;b=3$ Hence, $(a, b)=(1,3)$

Q71JEE Main 2021MCQ4MArea Under The Curves

The area (in sq. units) of the part of the circle $x^{2}+y^{2}=36$ , which is outside the parabola $y^{2}=9 x$ , is

🚀 Solve in Practice Mode

📖 Explanation

Given, equation of circle $\Rightarrow x^{2}+y^{2}=36$ Equation of parabola $\Rightarrow y^{2}=9 x$ We have to find area of shaded region.Required area $\;\; =\pi r^{2}-2[int_{0}^{3} \sqrt{9 x} d x+int_{3}^{6} \sqrt{36-x^{2}} d x]$ $\;\; =36 \pi-12 \sqrt{3}-2|\;\frac{x}{2} \sqrt{36-x^{2}}+18 \sin ^{-1}(\;\frac{x}{6})|_{3}^{6}$ $\;\; =36 \pi-12 \sqrt{3}-2[9 \pi-(\;\frac{9 \sqrt{3}}{2}+3 \pi)]$ $\;\; =(24 \pi-3 \sqrt{3})$

Q72JEE Main 2021MCQ4MDifferential Equations

The population $P=P(t)$ at time $t$ of a certain species follows the differential equation $\;\frac{d P}{d t}=0.5 P-450$ . If $P(0)=850$ , then the time at which population becomes zero is

🚀 Solve in Practice Mode

📖 Explanation

Given, $\;\frac{d P}{d t}=0.5 P-450$ and $\;\; P(0)=850$ $\because \;\; \;\frac{d P}{d t}=\;\frac{1}{2} P-450=\;\frac{P-900}{2}$ $\Rightarrow \;\; int_{0}^{1} \;\frac{d P}{P-900}=int_{0}^{1} \;\frac{d t}{2}$ $\Rightarrow \;\; [\log _{e}|P-900|]_{0}^{t}=\{\;\frac{t}{2}\}_{0}^{t}$ $\Rightarrow \log _{e}|P(t)-900|-\log _{e}|P(0)-900|=\;\frac{t}{2}$ $\Rightarrow \;\; \log _{e}|P(t)-900|-\log _{e} 50=\;\frac{t}{2}$ Let at $t=t_{1}, P(t)=0$ Hence, $\log _{e}|P(t)-900|-\log _{e} 50=t_{1} / 2$ $\therefore \;\; t_{1}=2 \log _{e} 18$

Q73JEE Main 2021MCQ4MStraight Lines and Pair of Straight Lines

A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this line on the coordinate axes is $\;\frac{1}{4}$ . Three stones $A, B$ and $C$ are placed at the points $(1,1),(2,2)$ and $(4,4)$ , respectively. Then, which of these stones is / are on the path of the man?

🚀 Solve in Practice Mode

📖 Explanation

Given, position of $A=(1,1)$ Position of $B=(2,2)$ Position of $C=(4,4)$ Let $x$ -intercept be $a$ and $y$ -intercept be $b$ . Equation of line traced is $\;\frac{x}{a}+\;\frac{y}{b}=1$ This is the equation of path, let a point $(h, k)$ lie on this path. Then $\;\frac{h}{a}+\;\frac{k}{b}=1$ Also, AM of reciprocal of $a$ and $b=\;\frac{1}{4}$ $\therefore$ $\;\frac{\;\frac{1}{a}+\frac{1}{b}}{2}=\;\frac{1}{4}$ $\;\frac{1}{a}+\;\frac{1}{b}=\;\frac{1}{2}$ On comparing Eqs. (i) and (ii), we get $(h, k)=(2,2)$ Hence, the required stone is $B(2,2)$ .

Q74JEE Main 2021MCQ4MParabola

The locus of the mid-point of the line segment joining the focus of the parabola $y^{2}=4 a x$ to a moving point of the parabola, is another parabola whose directrix is

🚀 Solve in Practice Mode

📖 Explanation

Given, equation of parabola $\Rightarrow y^{2}=4 a x$ Focus $=S(a, 0)$ Let any point on the parabola be $P( {at}^{2}, 2 {at})$ . and let the mid-point of $P S$ be $M(h, k)$ . $\; \therefore \;h=\;\frac{a t^{2}+a}{2} ;\;k=\;\frac{2 a t+0}{2}$ $\Rightarrow \;t^{2}=\;\frac{2 h-a}{a} ; t \;\; =\;\frac{k}{a}$ $\;\Rightarrow \;\; t^{2} \;\; =\;\frac{k^{2}}{a^{2}}$ $\;\text{ Now, }\; \;\frac{2 h-a}{a}=\;\frac{k^{2}}{a^{2}}$ $\Rightarrow \;\; 2 h-a=\;\frac{k^{2}}{a} \Rightarrow k^{2}=a(2 h-a)$ $\therefore \;\text{ Locus of }\;(h, k) \;\text{ is }\; y^{2}=a(2 x-a)$ $\;y^{2}=2 a(x-\;\frac{a}{2})$ $\therefore$ The directrix of this parabola is $x-\;\frac{a}{2}=-\;\frac{a}{2} \Rightarrow x=0$

Q75JEE Main 2021MCQ4MApplication of Derivatives

If the tangent to the curve $y=x^{3}$ at the point $P(t, t^{3})$ meets the curve again at $Q$ , then the ordinate of the point which divides $P Q$ internally in the ratio $1: 2$ is

🚀 Solve in Practice Mode

📖 Explanation

Given, curve $\Rightarrow y=x^{3}$ $P(t, t^{3})$ Equation of tangent at $P(t, t^{3})$ $(y-t^{3})=3 t^{2}(x-t)$ From Eqs. (i) and (ii), $x^{3}-t^{3} \;\; =3 t^{2}(x-t)$ $\Rightarrow \;(x-t)(x^{2}+t^{2}+x t)=3 t^{2}(x-t)$ $\Rightarrow \;x^{2}+x t-2 t^{2}=0$ $\Rightarrow \;(x-t)(x+2 t)=0$ $\Rightarrow \;x=t \;\text{ or }\; x=-2 t$ This is not possible. Now, the coordinate of $Q=(x, y)=(-2 t,(-2 t)^{3})$ $\therefore \;\; Q=(-2 t,-8 t^{3})$ $\therefore$ Ordinate of the point dividing $P Q$ in the ratio $1: 2$ is $\;\frac{2 t^{3}+(-8 t^{3})}{1+2}=-2 t^{3}$

Q76JEE Main 2021MCQ4MThree Dimensional Geometry

The equation of the plane passing through the point $(1,2,-3)$ and perpendicular to the planes $3 x+y-2 z=5$ and $2 x-5 y-z=7$ , is

🚀 Solve in Practice Mode

📖 Explanation

Given, equation of planes are $\;3 x+y-2 z=5$ $\;2 x-5 y-z=7$ and point $(1,2,-3)$ . Normal vector of required plane $=\hat{i}(-1-10)-\hat{j}(-3+4)+\hat{k}(-15-2)$ $=-11 \hat{i}-\hat{j}-17 \hat{k}$ Now, the equation of plane passing through $(1,2,-3)$ having normal vector $(-11 \hat{i}-\hat{j}-17 \hat{k})$ is $-[11(x-1)+(y-2)+17(z+3)] \;\; =0$ $11 x+y+17 z+38 \;\; =0$

Q77JEE Main 2021MCQ4MThree Dimensional Geometry

The distance of the point $(1,1,9)$ from the point of intersection of the line $\;\frac{x-3}{1}=\;\frac{y-4}{2}$ $=\;\frac{z-5}{2}$ and the plane $x+y+z=17$ is

🚀 Solve in Practice Mode

📖 Explanation

Given, $P(1,1,9)$ . Equation of plane $x+y+z \;\; =17$ $\;\text{ Equation of line }\; \Rightarrow \;\frac{x-3}{1} \;\; =\;\frac{y-4}{2}=\;\frac{z-5}{2}$ $\Rightarrow \;\; \;\frac{x-3}{1}=\;\frac{y-4}{2}=\;\frac{z-5}{2}=\lambda$ (let) $\Rightarrow \;\; x=\lambda+3 ; y=2 \lambda+4 ; z=2 \lambda+5$ $\therefore$ The point we have is $(\lambda+3,2 \lambda+4,2 \lambda+5) .$ $\because$ This point lies on the plane $x+y+z=17 .$ $\therefore \;\; \lambda+3+2 \lambda+4+2 \lambda+5=17$ $\Rightarrow \;\; \;\; \lambda=1$ $\therefore$ The coordinate of point is $(4,6,7)$ $\therefore$ Required distance between $(1,1,9)$ and $(4,6,7)$ is $\;\; =\sqrt{(4-1)^{2}+(6-1)^{2}+(7-9)^{2}}$ $\;\; =\sqrt{9+25+4}=\sqrt{38}$

Q78JEE Main 2021MCQ4MProbability

An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is

🚀 Solve in Practice Mode

📖 Explanation

Given, $P$ (odd number 2 times) $=P($ even number 3 times) $\Rightarrow \;\; { }^{n} C_{2}(\;\frac{1}{2})^{n}={ }^{n} C_{3}(\;\frac{1}{2})^{n}$ where, $n=$ Number of times the die is thrown. $\Rightarrow \;\; \;\frac{n !}{2 !(n-2) !} \times (\;\frac{1}{2})^{n}=\;\frac{n !}{3 !(n-3) !} \times (\;\frac{1}{2})^{n}$ $\Rightarrow \;\; \;\frac{3 ! \times (n-3) !}{2 ! \times (n-2) !}=1$ $\Rightarrow \;\; \;\frac{3 \times 2 ! \times (n-3) !}{2 ! \times (n-2)(n-3) !}=1 \Rightarrow 3=(n-2)$ $\Rightarrow \;\; n=5$ $\therefore$ Probability of getting an odd number, odd number of times $\;\; =P(1)+P(3)+P(5)$ $\;\; ={ }^{5} C_{1}(\;\frac{1}{2})^{5}+{ }^{5} C_{3}(\;\frac{1}{2})^{5}+{ }^{5} C_{5}(\;\frac{1}{2})^{5}$ $\;\; =\;\frac{16}{2^{5}}=\;\frac{16}{32}=\;\frac{1}{2}$

Q79JEE Main 2021MCQ4MTrigonometric Ratios and Identities

Two vertical poles are $150 {m}$ apart and the height of one is three times that of the other If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in metres) is

🚀 Solve in Practice Mode

📖 Explanation

Given, distance between both poles $=150 {m}$ Let $A B=h$ (smaller pole) and $D E=3 h$ (larger pole) $∠ A C B=\theta$ Then, $∠ E C D=90^{\circ}-\theta$ $[ \because$ angles are complementary] In $\;\; \Delta A C B, \tan \theta=\;\frac{h}{75}$ Also, in $\Delta E C D, \tan \theta=\;\frac{75}{3 h}$ $\Rightarrow \;\; \;\frac{h}{75}=\;\frac{75}{3 h}$ $\Rightarrow \;\; h^{2}=\;\frac{(75)^{2}}{3}$ $\therefore \;\; h=\sqrt{\;\frac{75 \times 75}{3}}=25 \sqrt{3} {m}$ $\therefore$ Required height $=25 \sqrt{3} {m}$

Q80JEE Main 2021MCQ4MSets and Relations

The statement among the following that is a tautology is

🚀 Solve in Practice Mode

📖 Explanation

Given, $[A ∧(A \to B)] \to B$ $\;\; =A ∧(∼ A ∨ B) \to B$ $\;\; =[(A ∧ ∼ A) ∨(A ∧ B)] \to B$ $\;\; =(A ∧ B) \to B$ $\;\; =∼ A ∨ ∼ B ∨ B$ $\;\; =t$ Hence, $[A ∧(A \to B)] \to B$ is a tautology.

Q81JEE Main 2021NAT4MComplex Numbers

If the least and the largest real values of $\alpha$ , for which the equation $z+\alpha z-1|+2 i=0$ $(z \in C$ and $i=\sqrt{-1})$ has a solution, are $p$ and $q$ respectively, then $4(p^{2}+q^{2})$ is equal to .................... .

🚀 Solve in Practice Mode

📖 Explanation

Given, $\alpha_{\;\text{least }\;}=p$ $\alpha_{\max }=q$ Equation given is $z+\alpha|z-1|+2 i=0 ; z \in C$ and $i=\sqrt{-1}$ Let $z=x+i y$ Then, $z+\alpha|z-1|+2 i=0$ $\Rightarrow x+i y+\alpha \sqrt{(x-1)^{2}+y^{2}}+2 i=0$ $\Rightarrow (x+\alpha \sqrt{(x-1)^{2}+y^{2}})+i(y+2)=0$ $\therefore \;\; y+2=0$ and $x+\alpha \sqrt{(x-1)^{2}+y^{2}}=0$ $\;\; y=-2$ and $x^{2}=\alpha^{2}(x^{2}+1-2 x+y^{2})$ $\;\; x^{2}=\alpha^{2}(x^{2}-2 x+5)$ $\;\; \alpha^{2}=\;\frac{x^{2}}{x^{2}-2 x+5}$ $\therefore \;\; \alpha^{2} \in[0, \;\frac{5}{4}]$ $\therefore \;\; \alpha \in[-\;\frac{\sqrt{5}}{2}, \;\frac{\sqrt{5}}{2}]$ Now, $4(p^{2}+q^{2})=4[(\alpha_{\;\text{least }\;})^{2}+(\alpha_{\max })^{2}]$ $\;\; =4[(-\;\frac{\sqrt{5}}{2})^{2}+(\;\frac{\sqrt{5}}{2})^{2}]$ $\Rightarrow \;\; (\;\frac{5}{4}+\;\frac{5}{4}]=10$

Q82JEE Main 2021NAT4MProbability

Let $B_{i}(i=1,2,3)$ be three independent events in a sample space. The probability that only $B_{1}$ occur is $\alpha$ , only $B_{2}$ occurs is $\beta$ and only $B_{3}$ occurs is $\gamma$ . Let $P$ be the probability that none of the events $B_{i}$ occurs and these 4 probabilities satisfy the equations $(\alpha-2 \beta) P=\alpha \beta$ and $(\beta-3 \gamma) P=2 \beta \gamma($ All the probabilities are assumed to lie in the interval $(0,1)$ ). Then, $\;\frac{P(B_{1})}{P(B_{3})}$ is equal to ......... .

🚀 Solve in Practice Mode

📖 Explanation

Given, $B_{1}, B_{2}$ and $B_{3}$ are three independent events. Let $x, y, z$ be the probability of $B_{1}, B_{2}, B_{3}$ , respectively. $P(\;\text{ only }\; B_{1} \;\text{ occur }\;)=\alpha$ $P(B_{1}) \cdot P(\overline{B}_{2}) \cdot P(\overline{B}_{3})=\alpha$ $\Rightarrow \;\; x \cdot (1-y) \cdot (1-z)=\alpha$ $P(\;\text{ only }\; B_{2} \;\text{ occur }\;)=\beta\;$ $\;P(\overline{B}_{1}) \cdot P(B_{2}) \cdot P(\overline{B}_{3}) \;\; =\beta$ $\Rightarrow \;(1-x) \cdot y \cdot (1-z) \;\; =\beta$ $\;P(\;\text{ only }\; B_{3} \;\text{ occur }\;)=\gamma$ $\Rightarrow \;P(\overline{B}_{1}) \cdot P(\overline{B}_{2}) \cdot P(B_{3})=\gamma$ $\Rightarrow \;\; (1-x) \cdot (1-y) \cdot z=\gamma$ $\;P(\;\text{ none occur }\;)=P$ $\;P(\overline{B}_{1}) \cdot P(\overline{B}_{2}) \cdot P(\overline{B}_{3})=P$ $\Rightarrow \;\; (1-x) \cdot (1-y) \cdot (1-z)=P$ Now, we have given relations $(\alpha-2 \beta) P=\alpha \beta$ $\Rightarrow [x(1-y)(1-z)-2 y(1-x)(1-z)](1-x)(1-y)(1-z)$ $\;\; =x \cdot (1-y)(1-z) \cdot y(1-x)(1-z)$ $\Rightarrow \;\; (1-z)[x(1-y)-2 y(1-x)]=x \cdot y \cdot (1-z)$ $\Rightarrow \;\; \;\; x-x y-2 y+2 x y=x y$ $\Rightarrow \;\; \;\; x=2 y$ Similarly, on solving the second relation, $(\beta-3 \gamma) P=2 \beta \gamma$ by putting $\beta, \gamma$ and $P$ , we get $y=3 z \;\; \;\; x=2 \times 3 z$ From Eqs. (i) and (ii), we get $\Rightarrow \;\; \;\; x=6 z \Rightarrow \;\frac{x}{z}=6$ Now, $\;\frac{P(B_{1})}{P(B_{3})}=\;\frac{x}{z}=6$ $\therefore$ Required answer is $6 .$

Q83JEE Main 2021NAT4MMatrices and Determinants

Let $P=\begin{bmatrix}3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0\end{bmatrix}$ , where $\alpha \in R$ . Suppose $Q=[q_{i j}]$ is a matrix satisfying $P Q=k l_{3}$ for some non-zero $k \in R$ . If $q_{23}=-k / 8$ and $|Q|=\;\frac{k^{2}}{2}$ , then $\alpha^{2}+k^{2}$ is equal to ............. .

🚀 Solve in Practice Mode

📖 Explanation

Given, $P=\begin{bmatrix}3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0\end{bmatrix}$ $\Rightarrow \;\; |P|=\begin{vmatrix}3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0\end{vmatrix}=(20+12 \alpha)$ According to the question, $P Q=k I_{3} \Rightarrow Q=k P^{-1} I_{3}$ Now, $Q=\;\frac{k}{|P|}(adj P) I_{3}$ $=\;\frac{k}{(20+12 \alpha)}\begin{bmatrix}5 \alpha & -10 & -\alpha \\ 3 \alpha & 6 & -3 \alpha-4 \\ -10 & 12 & -2\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$ $\because \;\; q_{23}=-\;\frac{k}{8}$ $\therefore \;\; \;\frac{k}{(20+12 \alpha)}(-3 \alpha-4)=-\;\frac{k}{8}$ $\Rightarrow \;\; 2(3 \alpha+4)=5+3 \alpha \Rightarrow 3 \alpha=-3$ $\Rightarrow$ $\alpha=-1$ Also, $|Q|=\;\frac{k^{3}|I|}{|P|}$ $\Rightarrow \;\; \;\frac{k^{2}}{2}=\;\frac{k^{3}}{20+12 \alpha} \Rightarrow 20+12 \alpha=2 k$ $\Rightarrow \;\; 2 k=20-12$ $\Rightarrow \;\; 2 k=8$ $\;\; k=4$ $\therefore$ Required value of $k^{2}+\alpha^{2}=4^{2}+(-1)^{2}=17$

Q84JEE Main 2021NAT4MMatrices and Determinants

Let $M$ be any $3 \times 3$ matrix with entries from the set $\{0,1,2\}$ . The maximum number of such matrices, for which the sum of diagonal elements of $M^{T} M$ is seven, is ........... .

🚀 Solve in Practice Mode

📖 Explanation

Given, $M$ is a $3 \times 3$ matrix. $\;\text{ Let }\; M=\begin{bmatrix}a & b & c \\ d & {e} & f \\ g & h & i\end{bmatrix}, \;\text{ then }\; M^{T}=\begin{bmatrix}{a} & d & g \\ b & {e} & h \\ c & f & i\end{bmatrix}$ Now, $M^{T} M=\begin{bmatrix}a & d & g \\ b & e & h \\ c & f & i\end{bmatrix} \times \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$ Sum of diagonal elements $=a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}+g^{2}+h^{2}+i^{2}=7$ According to the question, the entries are $\{0,1,2\} . \;\; [ \because M^{⊤} M=7]$ i.e. $\{a, b, c, . . ., h, i\}=\{0,1,2\}$ So, for Eq. (i) to be true, there are two cases. Case I When $7-1^{'}$ s are there and $2-0$ 's are there. $\Rightarrow \;\; { }^{9} C_{7} \times { }^{2} C_{2}=36$ ways of arrangements. Case II When $1-2$ is there, $3-1^{'}$ s and $5-0^{'}$ s are there. ${ }^{9} C_{1} \times { }^{8} C_{3} \times { }^{5} C_{5}=9 \times \;\frac{8 !}{3 ! 5 !} \times 1$ $=504$ ways of arrangements. $\therefore$ Total possible arrangements $=36+504=540$

Q85JEE Main 2021NAT4MSets and Relations

Let $A=\{n \in N: n$ is a 3 -digit number $\}$ $B=\{9 k+2: k \in N\}$ and $C=\{9 k+l: k \in N\}$ for some $/(0< l< 9)$ If the sum of all the elements of the set $A \cap(B \cup C)$ is $274 \times 400$ , then $/$ is equal to $. . . . . . . . . . . .$

🚀 Solve in Practice Mode

📖 Explanation

Given, $A=\{n \in N: n$ is a 3 -digit number $\}$ $B=\{9 k+2: k \in N\}$ $C=\{9 k+l: k \in N\}$ $\{101,109, . . . 992\}$ $\Rightarrow \;\; \;\text{ Sum }\;=\;\frac{100}{2}[101+992]=\;\frac{100 \times 1093}{2}$ $\;\text{ Similarly, }\; 3 \;\text{-digit number of the form }\; 9 k+5 \;\text{ is }\;$ $\;\;\;\; \;\frac{100}{2}[104+995]=\;\frac{100 \times 1099}{2}$ $\;\text{ Their \sum }\;=\;\frac{100 \times 1093}{2}+\;\frac{100 \times 1099}{2}$ $=100 \times 1096=400 \times 274$ Hence, we can say the value of $I=5$ as the second series of numbers obtained by set $C$ is of the form $9 k+5$ . $\therefore$ Required value of $I=5$

Q86JEE Main 2021NAT4MApplication of Derivatives

The minimum value of $\alpha$ for which the equation $\;\frac{4}{\sin x}+\;\frac{1}{1-\sin x}=\alpha$ has at least one solution in $(0, \;\frac{\pi}{2})$ is

🚀 Solve in Practice Mode

📖 Explanation

Given, equation $\;\frac{4}{\sin x}+\;\frac{1}{1-\sin x}=\alpha$ Let $f(x)=\;\frac{4}{\sin x}+\;\frac{1}{1-\sin x}$ $\Rightarrow \;\; y=\;\frac{4-3 \sin x}{\sin x(1-\sin x)}$ Let $\sin x=t$ when $t \in(0,1)$ as $x \in(0, \;\frac{\pi}{2})$ . $\begin{bmatrix}\because t_{1}=\sin 0=0 \\ t_{2}=\sin \\ \\ \frac{\pi}{2}=1\end{bmatrix}$ Now, $y=\;\frac{4-3 t}{t-t^{2}}$ $\;\frac{d y}{d t}=\;\frac{-3(t-t^{2})-(1-2 t)(4-3 t)}{(t-t^{2})^{2}}=0$ $\Rightarrow 3 t^{2}-3 t-(4-11 t+6 t^{2})=0$ $\Rightarrow \;\; 3 t^{2}-8 t+4=0$ $\Rightarrow \;\; 3 t^{2}-6 t-2 t+4=0$ $\Rightarrow \;\; 3 t(t-2)-2(t-2)=0$ $\Rightarrow \;\; (t-2)(3 t-2)=0$ $\therefore \;\; t-2=0$ or $3 t-2=0$ $t=2$ or $t=2 / 3$ $\sin x=2$ or $\sin x=\;\frac{2}{3}$ $x=\sin ^{-1}(\;\frac{2}{3})$ Also, $y=\;\frac{4-3 t}{t-t^{2}}=\;\frac{4-3 \times 2 / 3}{\frac{2}{3}-(\;\frac{2}{3})^{2}}$ $y=9$ $[ \because \sin \theta \le 1]$ $[ \because t=2 / 3]$ Now, required graph isFrom the graph, minimum value of $f(x) \ge 9$ $\alpha \ge 9$ $\therefore \alpha$ has minimum value 9 .

Q87JEE Main 2021NAT4MDefinite Integrals

If $int_{-a}^{a}(|x|+|x-2|) d x=22,(a>2)$ and $[x]$ denotes the greatest integer $\le x$ , then $int_{a}(x+[x]) d x$ is equal to ........ .

🚀 Solve in Practice Mode

📖 Explanation

Given, $int_{-a}^{a}(|x|+|x-2|) d x=22$ $\Rightarrow int_{-a}^{0}(-2 x+2) d x+int_{0}^{2}(x+2-x) d x+int_{2}^{a}(2 x-2) d x=22$ $\Rightarrow (x^{2}-2 x)|_{0} ^{-a}+(2 x)|_{0} ^{2}+(x^{2}-2 x)|_{2} ^{a}=22$ $\Rightarrow \;\; a^{2}+2 a+4+a^{2}-2 a-(4-4)=22$ $\Rightarrow \;\; 2 a^{2}=18$ $\Rightarrow \;\; a^{2}=9$ $\Rightarrow \;\; a=3$ $\therefore \;\; int_{a}^{-a} x+[x] d x=int_{3}^{-3}(x+[x]) d x$ $=-int_{-3}^{3}(x+[x]) d x$ $=-[-3-2-1+1+2]$ $=-[-3]=3$ $\therefore$ Required value is 3 .

Q88JEE Main 2021NAT4MCircles

If one of the diameters of the circle $x^{2}+y^{2}-2 x-6 y+6=0$ is a chord of another circle ' $C$ , whose centre is at $(2,1)$ , then its radius is

🚀 Solve in Practice Mode

📖 Explanation

Given, circle $\Rightarrow x^{2}+y^{2}-2 x-6 y+6=0$ Coordinate of $D=(1,3)$ $\;\text{ Radius }\; \;\; =r=\sqrt{1^{2}+3^{2}-6}$ $\;\; =\sqrt{4}=2 \;\text{ units }\;$ $C D \;\; =\sqrt{(2-1)^{2}+(1-3)^{2}}$ $\;\; =\sqrt{1+4}=\sqrt{5}$ Now, by using Pythagoras theorem in $\triangle A D C$ , $A C^{2}=A D^{2} \;+C D^{2}$ $\;\; =(2)^{2}+(\sqrt{5})^{2}$ $\;\; =4+5$ $A C^{2} \;\; =9$ $A C=\sqrt{9} \;\; =3$ $\therefore \;\text{ Required radius }\;=3$

Q89JEE Main 2021NAT4MVector Algebra

Let three vectors $a, b$ and $c$ be such that $c$ is coplanar with $a$ and $b$ , $a \cdot c=7$ and $b$ is perpendicular to $c$ , where $a=-\hat{i}+\hat{j}+\hat{k}$ and $b=2 \hat{i}+\hat{k}$ , then the value of $2|a+b+c|^{2}$ is .............. .

🚀 Solve in Practice Mode

📖 Explanation

Given, $c$ is co-planar with $a$ and $b$ $a \cdot c=7$ $b ⟂ c \Rightarrow b \cdot c=0$ $a \;\; =-\hat{i}+\hat{j}+\hat{k}$ $b \;\; =2 \hat{i}+\hat{k}$ $\;\text{ Now, }\; c \;\; =\lambda[b \times (a \times b)] \;\; [ \because c \;\text{ is coplanar with }\; a \;\text{ and }\; b]$ $\;\; =\lambda[(b \cdot b) a-(b \cdot a) b]$ $\;\; =\lambda[(\sqrt{5})^{2}(-\hat{i}+\hat{j}+\hat{k})-(-2+1)(2 \hat{i}+\hat{k})]$ $\;\; =\lambda[5(-\hat{i}+\hat{j}+\hat{k})+2 \hat{i}+\hat{k}]$ $\;\; =\lambda(-3 \hat{i}+5 \hat{j}+6 \hat{k})$ Now, $c \cdot a=7$ $\Rightarrow \;\; \lambda(-3 \hat{i}+5 \hat{j}+6 \hat{k}) \cdot (-\hat{i}+\hat{j}+\hat{k})=7$ $\Rightarrow \;\; 3 \lambda+5 \lambda+6 \lambda=7$ $\Rightarrow \;\; \lambda=\;\frac{7}{14}=\;\frac{1}{2}$ $\therefore 2|a+b+c|^{2}$ $\;\;\; =2|(\;\frac{-3}{2}-1+2) \hat{i}+(\;\frac{5}{2}+1) \hat{j}+(3+1+1) \hat{k}|^{2}$ [by putting $a, b, c$ ] $=2(\;\frac{1}{4}+\;\frac{49}{4}+25)=25+50=75$ $\therefore$ Required value is 75 .

Q90JEE Main 2021NAT4MLimits, Continuity and Differentiability

$\lim _{n \to \infty } \tan \{sum_{r=1}^{n} \tan ^{-1}(\;\frac{1}{1+r+r^{2}})\}$ is equal to ........... .

🚀 Solve in Practice Mode

📖 Explanation

Given, $\;\lim _{n \to \infty } \tan \{sum_{r=1}^{n} \tan ^{-1}(\;\frac{1}{1+r+r^{2}})\}$ $= \;\; \tan (\lim _{n \to \infty } sum_{r=1}^{n}\$ [\tan ^{-1}(r+1)-\tan ^{-1} r]$)= ;; \tan (\lim _{n \to \infty }(\tan ^{-1}(n+1)-;\frac{\pi}{4}))= ;; \tan (;\frac{\pi}{2}-;\frac{\pi}{4})= ;; \tan ;\frac{\pi}{4}=1 $Hence, the required value is$ 1 .$

← All JEE Main 2021 papers Take as Timed Mock →