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Physics30 questions

Q1JEE Main 2021MCQ4MRotational Motion

A triangular plate is shown below. A force $F=4 \hat{i}-3 \hat{j}$ is applied at point $P$ . The torque at point $P$ with respect to point $O$ and $Q$ are

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📖 Explanation

Given, $F=4 \hat{i}-3 \hat{j}$ Resolving the components of position vector in horizontal and vertical direction. $\therefore$ Horizontal component, $r_{x}=r \cos \theta$ $=10 \cos 60^{\circ}$ $[ \because \cos 60^{\circ}=\;\frac{1}{2}]$ $=10 \times \;\frac{1}{2}=5$ units Vertical component, $r_{y}=r \sin \theta$ $=10 \sin 60^{\circ}$ $[ \because \sin 60^{\circ}=\;\frac{\sqrt{3}}{2}]$ $=10 \times \;\frac{\sqrt{3}}{2}=5 \sqrt{3}$ units $\therefore$ Position vector w.r.t. $O, r_{O}=5 \hat{i}+5 \sqrt{3} \hat{j}$ and position vector w.r.t. Q, $r_{Q}=-5 \hat{i}+5 \sqrt{3} j$ Torque at point $P$ w.r.t. $O$ , $\tau _{O}=r_{O} \times F=(5 \hat{i}+5 \sqrt{3} \hat{j}) \times (4 \hat{i}-3 \hat{j})$ $=(-15-20 \sqrt{3}) {k}^{∧}=(15+20 \sqrt{3})(-{k}^{∧})$ Torque at point $P$ w.r.t. $Q$ , $\tau _{Q}=r_{Q} \times F=(4 \hat{i}-3 {j}^{∧}) \times (-5 {i}^{∧}+5 \sqrt{3} {j}^{∧})$ $=(-15+20 \sqrt{3}) \hat{k}=(15-20 \sqrt{3})(-\hat{k})$

Q2JEE Main 2021MCQ4MProperties of Matter

When two soap bubbles of radii $a$ and $b(b>a)$ coalesce, the radius of curvature of common surface is

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📖 Explanation

When two soap bubbles of radii a and $b(b>a)$ coalesce, it can be shown as followsPressure inside the bubble $1, p_{1}=p_{0}+\;\frac{4 T}{a}$ where, $p_{0}$ is the initial pressure, $T$ is the surface tension and $a$ is the radius of the bubble $1 .$ Similarly, pressure inside the bubble $2, p_{2}=p_{0}+\;\frac{4 T}{b}$ where, $b$ is the radius of the bubble 2 . Excess pressure at common surface is given by $p_{\;\text{ex }\;}=p_{1}-p_{2}$ Let, $r$ be the radius of common surface, then $\Rightarrow \;\; \;\frac{4 T}{r}=(p_{0}+\;\frac{4 T}{a})-(p_{0}+\;\frac{4 T}{b})$ $\Rightarrow \;\; \;\frac{4 T}{r}=p_{0}+\;\frac{4 T}{a}-p_{0}-\;\frac{4 T}{b}$ $\Rightarrow \;\; \;\frac{4 T}{r}=\;\frac{4 T}{a}-\;\frac{4 T}{b} \Rightarrow \;\frac{1}{r}=\;\frac{1}{a}-\;\frac{1}{b} \Rightarrow \;\frac{1}{r}=\;\frac{b-a}{a b}$ $\Rightarrow r=\;\frac{a b}{b-a}$ , which is the required expression for radius of curvature at common surface.

Q3JEE Main 2021MCQ4MThermodynamics

A polyatomic ideal gas has 24 vibrational modes. What is the value of $\gamma$ ?

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📖 Explanation

Since we know that, the heat capacity ratio $\gamma$ for an ideal gas can be related to the degrees of freedom of gas molecules $(f)$ by formula $\gamma=1+\;\frac{2}{f}$ .....(i) As each vibrational mode has 2 degrees of freedom, hence total vibrational degrees of freedom $=2 \times 24=48$ $\Rightarrow f=3$ (rotational) $+3$ (translational) $+48$ (vibrational) $\Rightarrow \;\; f=3+3+48 \Rightarrow f=54$ Now, put the value of $f$ in Eq. (i), we get $\gamma=1+\;\frac{2}{54} \Rightarrow \gamma=1+\;\frac{1}{27}$ $\Rightarrow \;\; \gamma=\;\frac{28}{27}$ or $\gamma=1.03$

Q4JEE Main 2021MCQ4MAtoms and Nuclei

If an electron is moving in the $n$ th orbit of the hydrogen atom, then its velocity $v_{n}$ for the $n$ th orbit is given as

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📖 Explanation

As we know that, the velocity of electron in the nth shell of hydrogen atom is given by $v_{n}=\;\frac{Z e^{2}}{2 \epsilon_{0} n h}$ where, $Z=$ atomic number, $n=$ order of orbit, $e=$ charge on an electron, $h=$ Planck's constant and $\epsilon_{0}=$ absolute permittivity of free space $\Rightarrow \;\; v_{n} \propto \;\frac{1}{n}$

Q5JEE Main 2021MCQ4MDual Nature of Matter and Radiation

An electron of mass $m$ and a photon have same energy $E$ . The ratio of wavelength of electron to that of photon is (c being the velocity of light)

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📖 Explanation

We know $\lambda=\;\frac{h}{p}$ ...(i) where, $\lambda=$ wavelength, $h=$ Planck's constant and $\;\; p=$ linear momentum ${AlsO}$ , kinetic energy $( {KE})=\;\frac{1}{2} m v^{2}=\;\frac{p^{2}}{2 m}$ $\Rightarrow \;\; p=\sqrt{2 m K E}$ ...(ii) From Eqs. (i) and (ii), we get $\lambda=\;\frac{h}{\sqrt{2 m K E}}$ or $\lambda=\;\frac{h}{\sqrt{2 m E}} \;\; [ \because {KE}=E]$ Wavelength of electron, $\lambda_{e}=\;\frac{h}{\sqrt{2 m E}}$ and wavelength of photon, $\lambda_{p}=\;\frac{h c}{E}$ $[ \because$ Both electron and photon have same energy. $]$ $\Rightarrow \;\; \;\frac{\lambda_{e}}{\lambda_{p}}=\;{\;\frac{h}{\sqrt{2 m E}}}/{\frac{h c}{E}} \Rightarrow \;\frac{\lambda_{e}}{\lambda_{p}}=\;\frac{1}{c}(\;\frac{E}{2 m})^{1 {/} 2}$

Q6JEE Main 2021MCQ4MProperties of Matter

Two identical metal wires of thermal conductivities $K_{1}$ and $K_{2}$ respectively are connected in series. The effective thermal conductivity of the combination is

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📖 Explanation

Two identical metal wires of thermal conductivities $K_{1}$ and $K_{2}$ respectively connected in series are represented as followsThe above figure can also be represented as Therefore, the effective thermal conductivity of the combination will be given by $R_{\;\text{eff }\;}=\;\frac{I}{K_{1} A}+\;\frac{I}{K_{2} A}=\;\frac{2 l}{K_{e q} A}$ $\Rightarrow \;\; \;{2 I}/{K_{ {eq}} A}=\;\frac{1}{A}(\;\frac{1}{K_{1}}+\;\frac{1}{K_{2}}) \Rightarrow \;{2 I}/{K_{ {eq}} A}=\;\frac{I}{A}(\;\frac{K_{1}+K_{2}}{K_{1} K_{2}})$ $\Rightarrow \;\; \;{2}/{K_{ {eq}}}=\;{K_{1}+K_{2}}/{K_{ {l}} K_{2}} \Rightarrow K_{ {eq}}=\;\frac{2 K_{1} K_{2}}{K_{1}+K_{2}}$

Q7JEE Main 2021MCQ4MUnits and Measurements

The vernier scale used for measurement has a positive zero error of $0.2 {mm}$ . If while taking a measurement, it was noted that ' 0 ' on the vernier scale lies between $8.5 {cm}$ and $8.6 {cm}$ , vernier coincidence is 6 , then the correct value of measurement is ......... ${cm}$ .

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📖 Explanation

Given, positive zero error $=0.2 {mm}=0.02 {cm}$ $[ \therefore$ Least count $L C=0.01 {cm}]$ Main scale reading $=8.5 {cm}$ Vernier scale reading $=$ Vernier scale coincidence $\times$ Least count $=6 \times 0.01=0.06 {cm}$ Final reading $=$ Main scale reading $+$ Vernier scale reading - Zero error $=8.5+0.06-0.02$ $=8.54 {cm}$

Q8JEE Main 2021MCQ4MAlternating Current

An AC current is given by $I=I_{1} \sin \omega t+I_{2} \cos \omega t$ . A hot wire ammeter will give a reading

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📖 Explanation

Given, $I=I_{1} \sin \omega t+I_{2} \cos \omega t$ ..(i) We know that rms value of current is given by $I_{ {rms}}=\sqrt{\;\frac{\int I^{2} {d} t}{T}}$ $\Rightarrow I_{ {rms}}=\sqrt{\;\frac{int_{0}^{T}(I_{1} \sin \omega t+I_{2} \cos \omega t)^{2} d t}{T}}$ using Eq. (i)] Squaring on both sides of the above equation, we get $\Rightarrow \;\; (I_{ {rms}})^{2}=int_{0}^{T} \;\frac{(I_{1} \sin \omega t+I_{2} \cos \omega t)^{2}}{T} d t$ $\Rightarrow \;\; (I_{ {rms}})^{2}=int_{0}^{T} \;\frac{I_{1}^{2}(\sin ^{2} \omega t+I_{2} \cos ^{2} \omega t+2 I_{1} I_{2} \sin \omega t \cos \omega t) d t}{T}$ $\Rightarrow \;\; I_{ {rms}}=\sqrt{\;\frac{l_{1}^{2}}{2}+\;\frac{I_{2}^{2}}{2}+0}$ $\Rightarrow I_{ {rms}}=\sqrt{\;\frac{l_{1}^{2}+I_{2}^{2}}{2}}$

Q9JEE Main 2021MCQ4MCircular Motion

A modern grand-prix racing car of mass $m$ is travelling on a flat track in a circular arc of radius $R$ with a speed $v$ . If the coefficient of static friction between the tyres and the track is $\mu _{s}$ , then the magnitude of negative lift $f_{L}$ acting downwards on the car is (Assume forces on the four tyres are identical and $g=$ acceleration due to gravity)

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📖 Explanation

We know that, static friction force, $f_{s}=\mu _{s} N$ where, $\mu _{s}$ is the coefficient of static friction and $N$ is the normal force acting on the body. As the car is travelling on a circular track, so centripetal force is also acting on it. $\Rightarrow \;\; f_{C}=\;\frac{m v^{2}}{R}$ In limiting condition, $f_{s}=f_{C} \Rightarrow \mu _{s} N=\;\frac{m v^{2}}{R}$ $\Rightarrow \;\; N=\;\frac{m v^{2}}{\mu _{s} R}$ ....(i) The magnitude of negative lift $f_{L}$ acting downwards on the car is given by $f_{L}=m g-N$ $=m g-\;\frac{m v^{2}}{\mu _{s} R}$ $\Rightarrow \;\; f_{L}=m(g-\;\frac{v^{2}}{\mu _{s} R}) \Rightarrow f_{L}=-m(\;\frac{v^{2}}{\mu _{s} R}-g)$ $\Rightarrow \;\; {|f_{L}|}=m(\;\frac{v^{2}}{\mu _{s} R}-g)$

Q10JEE Main 2021MCQ4MMotion in a Straight Line

A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If the total time elapsed is $t$ seconds, the total distance travelled is

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📖 Explanation

Consider 1st case, A car starts from rest at a constant acceleration $\alpha$ . It means Acceleration, $a_{1}=\alpha$ initial velocity, $u=0$ $v=u+a_{1} t \Rightarrow v_{0}=0+\alpha t_{1}$ $\Rightarrow \;\; v_{0}=\alpha t_{1}$ ...(i) Consider 2nd case, The same car decelerates after some time at a constant acceleration $\beta$ and comes to rest. It means Acceleration, $a_{2}=\beta$ Final velocity $=0$ $\Rightarrow \;\; v_{0}=\beta t_{2}$ ...(ii) According to question, $t_{1}+t_{2}=t$ $\Rightarrow \;\; v_{0}(\;\frac{1}{\alpha}+\;\frac{1}{\beta})=t$ [from Eqs. (i) & (ii)] $\Rightarrow \;\; v_{0}=\;\frac{\alpha \beta t}{\alpha+\beta}$ ...(iii) If we draw the velocity-time graph for the given situation, it will be as follows $\therefore$ Distance travelled $=$ Area of $v-t$ graph $=\;\frac{1}{2} \times t \times v_{0}=\;\frac{1}{2} \times t \times \;\frac{\alpha \beta t}{(\alpha+\beta)}$ [from Eq. (iii)] $=\;\frac{1}{2} \;\frac{\alpha \beta t^{2}}{(\alpha+\beta)}=\;\frac{\alpha \beta t^{2}}{2(\alpha+\beta)}$

Q11JEE Main 2021MCQ4MMagnetic Effects of Current and Magnetism

A solenoid of 1000 turns per metre has a core with relative permeability 500 . Insulated windings of the solenoid carry an electric current of $5 {A}$ . The magnetic flux density produced by the solenoid is (Permeability of free space $=4 \pi \times 10^{-7} \frac{H}{m}$ )

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📖 Explanation

Given, number of turns per unit length, $n=1000$ Relative magnetic permeability, $\mu _{t}=500$ Current flowing through solenoid, $I=5 {A}$ Magnetic permeability of free space, $\mu _{0}=4 \pi \times 10^{-7} \frac{H}{m}$ We know that, magnetic field inside the solenoid can be given as $B=\mu n I$ ....(i) where, $\mu =$ permeability of medium. As $\;\; \mu =\mu _{r} \mu _{0}$ ...(ii) From Eqs. (i) and (ii), we get $\therefore \;\; B=\mu _{r} \mu _{0} n l$ $=500 \times 4 \pi \times 10^{-7} \times 1000 \times 5$ $=100 \pi \times 10^{-7} \times 10^{5}$ $=\pi T$

Q12JEE Main 2021MCQ4MRotational Motion

A mass $M$ hangs on a massless rod of length $/$ which rotates at a constant angular frequency. The mass $M$ moves with steady speed in a circular path of constant radius. Assume that the system is in steady circular motion with constant angular velocity $\omega$ . The angular momentum of $M$ about point $A$ is $L_{A}$ which lies in the positive $z$ -direction and the angular momentum of $M$ about $B$ is $L_{B}$ . The correct statement for this system is

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📖 Explanation

It is given that the system is assumed to be in steady circular motion with constant angular velocity $\omega$ . We know that $L=m(r \times v)$ where, $L=$ angular momentum, $v=$ velocity of the particle, $m=$ mass of the particle and $r=$ radius of the circular path traced by the particle. Since, $L_{A}$ is the angular momentum of $M$ about point $A$ which lies in positive $z$ -direction. Therefore, with respect to point $A$ , we will get the direction of $L$ along positive $Z$ -axis and of constant magnitude of mvr. Since, $L_{B}$ is the angular momentum of $M$ about point $B$ , so with respect to point $B$ , we will get the constant magnitude of $L$ but its direction will be continuously changing. Hence, option (d) is correct.

Q13JEE Main 2021MCQ4MOscillations

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

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📖 Explanation

Potential energy of ${SHM}=\;\frac{1}{2} k x^{2}=\;\frac{1}{2} m \omega^{2} x^{2}$ where $m=$ mass of particle, $\omega=$ angular velocity and $x=$ displacement. Kinetic energy of ${SHM}=\;\frac{1}{2} m(\omega \sqrt{A^{2}-x^{2}})^{2}$ Here, $A$ is the amplitude of SHM. According to question, Potential energy of ${SHM}=$ Kinetic energy of ${SHM}$ $\;\frac{1}{2} m \omega^{2} x^{2}=\;\frac{1}{2} m(\omega \sqrt{A^{2}-x^{2}})^{2}$ $\Rightarrow \;\; \omega^{2} x^{2}=\omega^{2}(\sqrt{A^{2}-x^{2}})^{2} \Rightarrow \omega^{2} x^{2}=\omega^{2}(A^{2}-x^{2})$ $\Rightarrow \;\; \omega^{2} x^{2}=\omega^{2} A^{2}-\omega^{2} x^{2} \Rightarrow 2 \omega^{2} x^{2}=\omega^{2} A^{2}$ $\Rightarrow \;\; 2 x^{2}=A^{2} \Rightarrow x^{2}=\;\frac{A^{2}}{2}$ $\Rightarrow \;\; x=\pm \sqrt{\;\frac{A^{2}}{2}} \Rightarrow x=\pm \;\frac{A}{\sqrt{2}}$

Q14JEE Main 2021MCQ4MThermodynamics

A Carnot's engine working between $400 {K}$ and $800 {K}$ has a work output of $1200 {J}$ per cycle. The amount of heat energy supplied to the engine from the source in each cycle is

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📖 Explanation

Efficiency of Carnot heat engine is given by $η=1-\;\frac{T_{2}}{T_{1}}$ $=1-\;\frac{Q_{2}}{Q_{1}}=\;\frac{W}{Q_{1}}$ .....(i) where. $W=$ net work done by the gas $Q_{1}=$ heat absorbed by the gas, $Q_{2}=$ heat released by the gas, $T_{1}=$ temperature of hot reservoir, and $T_{2}=$ temperature of cold reservoir. Using Eq. (i), we can write $1-\;\frac{T_{2}}{T_{1}}=\;\frac{W}{Q_{1}} \Rightarrow \;\frac{T_{2}}{T_{1}}=1-\;\frac{W}{Q_{1}}$ $\Rightarrow \;\; \;\frac{400}{800}=1-\;\frac{W}{Q_{1}} \Rightarrow \;\frac{W}{Q_{1}}=1-\;\frac{400}{800}$ $\Rightarrow \;\; \;\frac{W}{Q_{1}}=1-\;\frac{1}{2}=\;\frac{1}{2} \Rightarrow \;\frac{W}{Q_{1}}=\;\frac{1}{2}$ $\Rightarrow \;\; Q_{1}=2 \times W=2 \times 1200[ \because W_{1}=1200 J]$ $=2400 {J}$

Q15JEE Main 2021MCQ4MRay Optics and Optical Instruments

The thickness at the centre of a plano convex lens is $3 {mm}$ and the diameter is $6 {cm}$ . If the speed of light in the material of the lens is $2 \times 10^{8} {ms}^{-1}$ , then the focal length of the lens is

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📖 Explanation

Given, thickness at the centre of plano-convex lens, $t=3 {mm}$ $=3 \times 10^{-3} {m}$ Diameter of plano-convex lens, $d=6 {cm}$ $\therefore$ Radius of plano-convex lens, $r=3 {cm}=3 \times 10^{-2} {m}$ Speed of light in lens material, $v=2 \times 10^{8} {ms}^{-1}$ Refractive index $=\;\frac{\;\text{ Speed of light \in air }\;}{\;\text{ Speed of light \in medium }\;}$ $\mu =\;\frac{3 \times 10^{8}}{2 \times 10^{8}}=\;\frac{3}{2}=1.5$ $\Rightarrow \;\; \mu =1.5$ We know that, $R^{2}=(R-t)^{2}+r^{2}$ where, $R$ is the radius of curvature of plano convex lens. $\Rightarrow \;\; R^{2}=R^{2}+t^{2}-2 R t+r^{2} \Rightarrow 2 R t=r^{2}+t^{2}$ As, $t$ is small, then $t^{2}$ will be very very small, so it can be neglected. $\Rightarrow \;\; 2 R t=r^{2}$ $\Rightarrow \;\; R=r^{2} {/} 2 t$ ....(i) From lens Maker's formula, we have $\;\frac{1}{f}=(\mu -1)(\;\frac{1}{R_{1}}-\;\frac{1}{R_{2}})$ Here $R_{1}=R \& R_{2}=\infty$ $\Rightarrow \;\; \;\frac{1}{f}=(\mu -1)(\;\frac{1}{R}-\;\frac{1}{\infty })$ $\Rightarrow \;\; \;\frac{1}{f}=\;\frac{\mu -1}{R}$ ...(ii) From Eqs. (i) and (ii), we get $\;\frac{1}{f}=\;\frac{\mu -1(2 t)}{r^{2}} \Rightarrow f=\;\frac{r^{2}}{(\mu -1) 2 t}$ Putting the given values in above equation, we get $f=\;\frac{(3 \times 10^{-2})^{2}}{(1.5-1) 2 \times 3 \times 10^{-3}} \Rightarrow f=0.3 {m}$ $f=30 {cm}$

Q16JEE Main 2021MCQ4MSemiconductor Electronics

The output of the given combination gates represents

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📖 Explanation

We can find the output of given combination of gates using Boolean algebra. Consider the given figureThe output $Y=\overline{A \cdot B}$ which is the expression for NAND gate. Alternate method We can find the output of given combination of gates using truth table. Consider the given figureAB $Y_{1}=A \cdot B$ $Y=\overline{A \cdot B}$ 0001010110011110The above truth table shows the output of NAND gate.

Q17JEE Main 2021MCQ4MWork, Power and Energy

A boy is rolling a $0.5 {kg}$ ball on the frictionless floor with the speed of $20 {ms}^{-1}$ . The ball gets deflected by an obstacle on the way. After deflection it moves with $5 \%$ of its initial kinetic energy. What is the speed of the ball now ?

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📖 Explanation

Given, mass of rolling ball, $m=0.5 {kg}$ Speed of ball, $u=20 {ms}^{-1}$ Before deflection, Initial kinetic energy, ${KE}_{1}=\;\frac{1}{2} m u^{2}$ $=\;\frac{1}{2} \times 0.5 \times (20)^{2}$ $=\;\frac{1}{2} \times 0.5 \times 400=\;\frac{1}{2} \times \;\frac{1}{2} \times 400=100 {J}$ It is given in the question that after deflection the ball moves with $5 \%$ of its initial kinetic energy ${KE}_{f}=5 \%$ of ${KE}_{ {i}} \Rightarrow {KE}_{f}=\;\frac{5}{100} \times 100=5 {J}$ If the final speed of the ball is $v {ms}^{-1}$ , then ${KE}_{f}=\;\frac{1}{2} m v^{2}$ $\Rightarrow \;\; 5=\;\frac{1}{2} \times 0.5 \times v^{2} \Rightarrow 10=0.5 \times v^{2}$ $\Rightarrow \;\; v^{2}=\;\frac{10}{0.5}=\;\frac{100}{5} \Rightarrow v^{2}=20$ $\Rightarrow \;\; v=\sqrt{20}=4.47 {ms}^{-1}$ $\Rightarrow \;\; v=4.47 {ms}^{-1}$

Q18JEE Main 2021MCQ4MAtoms and Nuclei

Which level of the single ionized carbon has the same energy as the ground state energy of hydrogen atom?

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📖 Explanation

Since we know that, energy of hydrogen atom is $E=-13.6 \;\frac{Z^{2}}{n^{2}}$ where, $Z$ is the atomic number and $n$ is the energy state. $\because$ For hydrogen atom, $Z=1$ and $n=1$ $\Rightarrow \;\; E=\;\frac{-13.6(1)^{2}}{1^{2}}$ $\Rightarrow \;\; E=-13.6 {eV}$ ...(i) For carbon atom, $Z=6, n=$ ? $\Rightarrow \;\; E=\;\frac{-13.6(6)^{2}}{n^{2}}$ ....(ii) According to question, carbon has same energy as the ground stat energy of hydrogen atom. $\therefore$ From Eqs. (i) and (ii), we get $-13.6=\;\frac{-13.6(6)^{2}}{n^{2}} \Rightarrow n^{2}=6^{2}$ $\Rightarrow \;\; n=6$

Q19JEE Main 2021MCQ4MThermodynamics

Two ideal polyatomic gases at temperatures $T_{1}$ and $T_{2}$ are mixed so that there is no loss of energy. If $f_{1}$ and $f_{2}, m_{1}$ and $m_{2}, n_{1}$ and $n_{2}$ be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is

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📖 Explanation

According to question, two ideal atomic gases at temperatures $T_{1}$ and $T_{2}$ are mixed. Let the final temperature of this mixture be $T$ . As per question there is no loss of energy, it means $\Delta U=0$ ...(i) As, we know, $\Delta U=\;\frac{f_{1} n_{1} R \Delta T}{2}+\;\frac{f_{2} n_{2} R \Delta T}{2}$ ....(ii) From Eqs. (i) and (ii), we get $\;\frac{f_{1} n_{1} R \Delta T}{2}+\;\frac{f_{2} n_{2} R \Delta T}{2}=0$ $\Rightarrow \;\; f_{1} n_{1} R \Delta T+f_{2} n_{2} R \Delta T=0$ $\Rightarrow \;\; f_{1} n_{1} R(T_{1}-T)+f_{2} n_{2} R(T_{2}-T)=0$ $\Rightarrow \;\; f_{1} n_{1}(T_{1}-T)+f_{2} n_{2}(T_{2}-T)=0$ $\Rightarrow \;\; f_{1} n_{1} T_{1}-f_{1} n_{1} T+f_{2} n_{2} T_{2}-f_{2} n_{2} T=0$ $\Rightarrow \;\; f_{1} n_{1} T+f_{2} n_{2} T=f_{1} n_{1} T_{1}+f_{2} n_{2} T_{2}$ $\Rightarrow T(f_{1} n_{1}+f_{2} n_{2})=f_{1} n_{1} T_{1}+f_{2} n_{2} T_{2}$ $T=\;\frac{f_{1} n_{1} T_{1}+f_{2} n_{2} T_{2}}{f_{1} n_{1}+f_{2} n_{2}}$

Q20JEE Main 2021MCQ4MCurrent Electricity

A current of $10 {A}$ exists in a wire of cross sectional area of $5 {mm}^{2}$ with a drift velocity of $2 \times 10^{-3} {ms}^{-1}$ . The number of free electrons in each cubic metre of the wire is

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📖 Explanation

Given, current, $I=10 {A}$ Cross-sectional area, $A=5 {mm}^{2}=5 \times 10^{-6} {m}^{2}$ Drift velocity, $v_{d}=2 \times 10^{-3} {ms}^{-1}$ The value of current flowing through a conductor can be given by $I=n e A v_{d}$ ....(i) where, $n=$ number of free electrons and ${e}=$ charge on an electron Putting all the given values in Eq. (i) we get $10=n \times 1.6 \times 10^{-19} \times 5 \times 10^{-6} \times 2 \times 10^{-3}$ $\Rightarrow \;\; n=\;\frac{10}{1.6 \times 10^{-19} \times 5 \times 10^{-6} \times 2 \times 10^{-3}}$ $=0.625 \times 10^{28}=625 \times 10^{25}$

Q21JEE Main 2021NAT4MElectromagnetic Waves

For VHF signal broadcasting, ........ ${km}^{2}$ of maximum service area will be covered by an antenna tower of height $30 {m}$ , if the receiving antenna is placed at ground. Let radius of the Earth be $6400 {km}$ . (Round off to the nearest integer). (Take $\pi$ as $3.14$ )

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📖 Explanation

Given, height of antenna tower, $h=30 {m}$ $=30 \times 10^{-3} {km}$ Radius of earth, $R=6400 {km}$ We know that, $d=\sqrt{2 R h}$ ....(i) where, $d=$ distance up to which signals can be transmitted Area, $A=\pi d^{2}$ ...(ii) From Eqs. (i) and (ii), we get $A=\pi(\sqrt{2 R h})^{2}=2 \pi R h$ $=2 \times 3.14 \times 6400 \times 30 \times 10^{-3}=1205.76 {km}^{2}$ $=1206 {km}^{2}$

Q22JEE Main 2021NAT4MRotational Motion

The angular speed of truck wheel is increased from 900 rpm to $2460 {rpm}$ in $26 {s}$ . The number of revolutions by the truck engine during this time is ......... (Assuming the acceleration to be uniform).

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📖 Explanation

Initial angular speed, $\omega_{0}=900 {rpm}=900 \times \;\frac{2 \pi}{60}=30 \pi \frac{rad}{s}$ Final angular speed, $\omega=2460 {rpm}$ $=2460 \times \;\frac{2 \pi}{60}=41 \times 2 \pi=82 \pi \frac{rad}{s}$ We know that, $\omega=\omega_{0}+\alpha t \Rightarrow \omega-\omega_{0}=\alpha t$ $\Rightarrow \;\; 82 \pi-30 \pi=\alpha \times 26 \Rightarrow 52 \pi=26 \alpha$ $\Rightarrow \;\; \alpha=2 \pi \frac{rad}{s}^{2}$ Also we know that, $\theta=\omega_{0} t+\;\frac{1}{2} \alpha t^{2}$ $\Rightarrow \;\; 2 \pi n=30 \pi \times 26+\;\frac{1}{2} \times 2 \pi \times (26)^{2}$ $\Rightarrow 2 n=30 \times 26+(26)^{2} \Rightarrow n=\;\frac{30 \times 26+(26)^{2}}{2}$ $=\;\frac{26(30+26)}{2}=13 \times 56=728$

Q23JEE Main 2021NAT4MCurrent Electricity

The equivalent resistance of series combination of two resistors is $s$ . When they are connected in parallel, the equivalent resistance is $p$ . If $s=n p$ , then the minimum value for $n$ is .......... (Round off to the nearest integer)

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📖 Explanation

Let two resistors have resistances $R_{1}$ and $R_{2}$ , respectively. As per question, equivalent resistance of series combination is $s$ $\Rightarrow$ $s=R_{1}+R_{2}$ .....(i) and equivalent resistance of parallel combination is $p$ $\Rightarrow \;\; p=\;\frac{R_{1} R_{2}}{R_{1}+R_{2}}$ ....(ii) According to the question, $s=n p$ ....(iii) From Eqs. (i),(ii) and (iii), we get $R_{1}+R_{2}=n \;\frac{R_{1} R_{2}}{(R_{1}+R_{2})} \Rightarrow n(R_{1} R_{2})=(R_{1}+R_{2})^{2}$ $\Rightarrow n=\;\frac{(R_{1}+R_{2})^{2}}{R_{1} R_{2}}$ For $n$ to be minimum, $R_{1}=R_{2}=R$ $\Rightarrow \;\; n=\;\frac{(R+R)^{2}}{R \cdot R}=\;\frac{(2 R)^{2}}{R^{2}}=\;\frac{4 R^{2}}{R^{2}} \Rightarrow n=4$

Q24JEE Main 2021NAT4MElectrostatic Potential and Capacitance

Four identical rectangular plates with length, $I=2 {cm}$ and breadth, $b=\;\frac{3}{2} {cm}$ are arranged as shown in figure. The equivalent capacitance between $A$ and $C$ is $\;\frac{x \epsilon_{0}}{d}$ . The value of $x$ is ......... (Round off to the nearest integer)

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📖 Explanation

The given figure can be shown with capacitors as Let $C_{0}$ be the capacitance of each capacitor. In the above figure capacitance $C_{1}$ is in series combination with equivalent of parallel combination of capacitance $C_{2}$ and $C_{3}$ . $C_{e q}^{'}=C_{2}+C_{3} \Rightarrow C_{e q}^{'}=C_{0}+C_{0}=2 C_{0}$ $\Rightarrow \;\; \;{1}/{C_{ {eq}}}=\;\frac{1}{C_{1}}+\;{1}/{C_{ {eq}}^{'}}=\;\frac{1}{C_{0}}+\;\frac{1}{2 C_{0}}=\;\frac{3}{2 C_{0}}$ $\Rightarrow \;\; \;\; C_{\;\text{eq }\;}=\;\frac{2 C_{0}}{3}$ ....(i) As, $C_{0}=\;\frac{\epsilon_{0} A}{d}$ ...(ii) From Eqs. (i) and (ii), we get $C_{\;\text{eq }\;}=\;\frac{2}{3}(\;\frac{\epsilon_{0} A}{d}) \Rightarrow C_{\;\text{eq }\;}=\;\frac{2 \epsilon_{0}}{3 d}(I \times b)=\;\frac{2 \epsilon_{0}}{3 d}(2 \times \;\frac{3}{2})$ $\Rightarrow \;\; C_{\;\text{eq }\;}=\;\frac{2 \epsilon_{0}}{d}$ ...(iii) According to question, the equivalent capacitance between $A$ and $C$ is $\;\frac{x \epsilon_{0}}{d}$ . So, comparing it with Eq. (iii), we get $x=2$

Q25JEE Main 2021NAT4MGravitation

The radius in kilometre to which the present radius of Earth $(R=6400 {km})$ to be compressed so that the escape velocity is increased 10 times is

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📖 Explanation

We know that, $v_{e}=\sqrt{\;\frac{2 {Gm}}{R_{1}}}$ ...(i) where, $v_{e}=$ escape velocity, $G=$ gravitational constant, $R=$ radius of earth and $m=$ mass of the body. Now, the escape velocity is increased to 10 times, $\Rightarrow \;\; 10 v_{e}=\sqrt{\;\frac{2 G m}{R_{2}}}$ ...(ii) Dividing Eq. (ii) by Eq. (i), we get $\Rightarrow$ $10=\sqrt{\;\frac{R_{1}}{R_{2}}} \Rightarrow 100=\;\frac{R_{1}}{R_{2}}$ $\Rightarrow \;\; R_{2}=\;\frac{R_{1}}{100}=\;\frac{6400}{100} \Rightarrow R_{2}=64 {km}$

Q26JEE Main 2021NAT4MOscillations

Consider two identical springs each of spring constant $k$ and negligible mass compared to the mass $M$ as shown. Fig.1 shows one of them and Fig.2 shows their series combination. The ratios of time period of oscillation of the two SHM is $\;\frac{T_{b}}{T_{a}}=\sqrt{x}$ , where value of $x$ is ......... (Round off to the nearest integer)

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📖 Explanation

Time period of Fig. 1 can be given as $T_{a}=2 \pi \sqrt{\;\frac{M}{k}}$ where, $M$ is mass of the suspended object and $k$ is the force constant. In Fig. 2, both the springs are in series combination, therefore its time period can be given as $T_{b}=2 \pi \sqrt{\;{M}/{k_{ {eq}}}}=2 \pi \sqrt{\;\frac{M}{k {/} 2}} \;\; ( \because k_{ {eq}}=\;\frac{k \times k}{k+k})$ Now, $\;\frac{T_{b}}{T_{a}}=\;{2 \pi \sqrt{\frac{M}{k {/} 2}}}/{2 \pi \sqrt{\;\frac{M}{k}}}$ $\Rightarrow \;\; \;\frac{T_{b}}{T_{a}}=\sqrt{2}$ ...(i) According to question, the ratio of time period of oscillation of two ${SHM}$ is $T_{b} {/} T_{a}=\sqrt{x}$ , so on comparing it with Eq. (i) we can say, $x=2$

Q27JEE Main 2021NAT4MRotational Motion

The following bodies, 1. a ring 2. a disc 3. a solid cylinder 4. a solid sphere of same mass $m$ and radius $R$ are allowed t roll down without slipping simultaneously from the top of the inclined plane. The boo which will reach first at the bottom of the inclined plane is .......... . (Mark the body as per their respective numbering given in the question)

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📖 Explanation

For rolling without slipping on an inclined plane, we can write $R m g \sin \theta=(m K^{2}+m R^{2}) \alpha$ $\Rightarrow \;\; R m g \sin \theta=m(K^{2}+R^{2}) \alpha$ $\Rightarrow \;\; \alpha=\;\frac{R g \sin \theta}{K^{2}+R^{2}} \Rightarrow \;\frac{\alpha}{R}=\;{g \sin \theta}/{1+\;\frac{K^{2}}{R^{2}}}$ $\Rightarrow \;\; a=\;{g \sin \theta}/{1+\;\frac{K^{2}}{R^{2}}}$ $[ \because a=\;\frac{\alpha}{R}] . . .$ (i) Time period, $t=\sqrt{\;\frac{2 s}{a}}$ ...(ii) From Eqs. (i) and (ii), we get $t=\sqrt{\;\frac{2 s}{g \sin \theta}(1+\;\frac{k^{2}}{R^{2}})}$ For least time acceleration a should be maximum and $K$ should be minimum and we know that $K$ is least for solid sphere. So, time will be least for sphere. It means the body which will reach first at the bottom of the inclined plane is 4 , i.e. solid sphere.

Q28JEE Main 2021NAT4MElectrostatic Potential and Capacitance

A parallel plate capacitor whose capacitance $C$ is $14 {pF}$ is charged by a battery to a potential difference $V=12 {V}$ between its plates. The charging battery is now disconnected and a porcelain plate with $K=7$ is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of ......... pJ. (Assume no friction)

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📖 Explanation

Given, capacitance of capacitor, $C=14 {pF}$ Potential difference, $V=12 {V}$ Energy of capacitor, $E_{i}=\;\frac{1}{2} C V^{2}$ $=\;\frac{1}{2} \times 14 \times 12 \times 12=1008 {pJ}$ Now, the charging battery is disconnected and a porcelain plate with $K=7$ is inserted between the plates $\therefore \;\; E_{f}=\;\frac{E_{i}}{7} \Rightarrow f=\;\frac{1008}{7} {pj} \Rightarrow f=144 {pj}$ Mechanical energy with which the plate would oscillate back and forth between the plates will be $=(1008-144) {pJ}=864 {pJ}$

Q29JEE Main 2021NAT4MLaws of Motion

Two blocks $(m=0.5 {kg}$ and $M=4.5 {kg})$ are arranged on a horizontal frictionless table as shown in figure. The coefficient of static friction between the two blocks is $\;\frac{3}{7}$ . Then, the maximum horizontal force that can be applied on the larger block so that the blocks move together is ...... ${N}$ . (Round off to the nearest integer. Take, $g=9.8 {ms}^{-2}$ )

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📖 Explanation

When both the blocks move together as a system, then acceleration of this system will be given as $a=\;\frac{F}{m+M}$ ...(i) Frictional force on mass, $f=m a$ ...(ii) From Eqs. (i) and (ii), we get $f=m(\;\frac{F}{m+M})$ For no slipping, $f \le \mu m g \;\; [ \because \mu$ being the coefficient of static friction $\Rightarrow \;\; m(\;\frac{F}{m+M}) \le \mu m g \Rightarrow F \le \mu (m+M) g$ $\therefore \;\; F_{\max }=\;\frac{3}{7}(0.5+4.5) \times 9.8 \Rightarrow F_{\max }=21 {N}$

Q30JEE Main 2021NAT4MElectromagnetic Waves

If $2.5 \times 10^{-6} {N}$ average force is exerted by a light wave on a non-reflecting surface of $30 {cm}^{2}$ area during 40 min of time span, the energy flux of light just before it falls on the surface is ......W / ${cm}^{2}$ . (Round off to the nearest integer. Assume complete absorption and normal incidence conditions are there.)

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📖 Explanation

Given, average force, $F=2.5 \times 10^{-6} {N}$ Area, $A=30 {cm}^{2}$ Time, $t=40 {min}$ We know that, $F=\;\frac{I A}{c}$ where, $I=$ energy flux of light and $c=$ speed of light in air. $\Rightarrow \;\; I=\;\frac{F C}{A}=\;\frac{2.5 \times 10^{-6} \times 3 \times 10^{8}}{30}=25 \frac{W}{cm}^{2}$

Chemistry30 questions

Q31JEE Main 2021MCQ4MBiomolecules

With respect to drug-enzyme interaction, identify the wrong statement

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Q32JEE Main 2021MCQ4MOrganic Chemistry - Some Basic Principles

Which of the following is an aromatic compound?

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📖 Explanation

is an aromatic compound $(6 \pi e^{-})$ . Aromatic compound possess following characteristics: (i) Planarity (ii) Complete delocalisation of the $\pi$ -electrons in the ring. (iii) Presence of $(4 n+2) \pi$ -electrons in the ring, where $n$ is an integer $(n=0,1,2 . . .)$ . That compound follow all aromatic characteristics.

Q33JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

The product $A$ in the above reaction is

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Q34JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

A central atom in a molecule has two lone pairs of electrons and forms three single bonds. The shape of this molecule is

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📖 Explanation

2 lone pair $+3$ bond pair Stearic number $=5$ So, general formula is $A B_{3} L_{2}$ . (Here, $L$ = number of lone pair.) Hybridisation $=s p^{3} d .$ So, the shape is T- shapeNote Geometry of the compound having $s p^{3} d$ hybridisation should to tetragonal bipyramidal (TBP) according to VSEPR theory. But due to two lone pair, geometry is distorted. In TBP lone pair occupy equatorial position.

Q35JEE Main 2021MCQ4Md and f Block Elements

Given below are two statements. Statement I Potassium permanganate on heating at $573 {K}$ forms potassium manganate. Statement II Both potassium permanganate and potassium manganate are tetrahedral and paramagnetic in nature. In the light of the above statements, choose the most appropriate answer from the options given below

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📖 Explanation

Potassium permanganate on heating at $573 {K}$ forms potassium manganate. ${KMnO}^{+7}_{}_{4}_{\text{permanganate(Diamagnetic)}} {\to }^{573 {K}} {K}_{2} {MnO}_{4}^{+6}_{\text{Potassium manganate Paramagnetic)}}+ {MnO}_{2}+ {O}_{2}$ Both potassium permanganate and potassium manganate are tetrahedral. Manganese has four sigma bonds forming tetrahedral structure. ${KMnO}_{4} ;+1+x-8=0$ $x=+7$ ${K}_{2} {MnO}_{4} ;+2+x-8=0$ $x=+6$ So, ${Mn}$ is present in $+7$ oxidation state in ${KMnO}_{4}$ and $+6$ oxidation state in ${K}_{2} {MnO}_{4}$ . ${ }_{25} {Mn}=[ {Ar}] 4 s^{2} 3 d^{5}$ ${Mn}^{+7}=[ {Ar}] 4 s^{0} 3 d^{0}$ (Diamagnetic) ${Mn}^{+6}=[ {Ar}] 4 s^{1}$ (Paramagnetic) So, potassium permanganate is diamagnetic and potassium manganate is paramagnetic. Statement I is true but statement II is false.

Q36JEE Main 2021MCQ4MBiomolecules

Which of the following is correct structure of tyrosine?

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Q37JEE Main 2021MCQ4MHaloalkanes and Haloarenes

The above reaction requires which of the following reaction conditions?

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📖 Explanation

When chlorobenzene is treated with ${NaOH}$ at $623 {K}$ and 300 atm sodium phenoxide is formed by nucleophilic substitution.Nucleophilic substitution is not preferred in aromatic compounds. So, to do such reactions high temperature and pressure are required.

Q38JEE Main 2021MCQ4Mp Block Elements

The absolute value of the electron gain enthalpy of halogens satisfies

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📖 Explanation

(c) Order of electron gain enthalpy (absolute value) ${Cl}> {F}> {Br}>1$ ElementFClBrI $\Delta H_{e}(k J {/}$ Mole $)$ $-328$ $- 349$ $- 325$ $-295$ Absolute value of electron gain enthalpy decreases down the group in a periodic table due to increase in size. But magnitude of electron gain enthalpy of second period element ${F}$ Due to small size of ' $F$ ' there is high electronic repulsion between the existing and approaching electron. Hence, magnitude of electron gain enthalpy decreases.

Q39JEE Main 2021MCQ4MChemical Bonding and Molecular Structure

Which of the following compound cannot act as a Lewis base?

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📖 Explanation

${PCl}_{5}$ has no lone-pair on the central atom but has empty $d$ -orbital in valence shell, so that it can accept a pair of electrons (from a Lewis acid Species which accepts a pair of electron in their vacant orbital e.g. ${BF}_{3}, {AlCl}_{3}, {PCl}_{5}$ . Lewis base Species which can donate a pair of electron, e.g. ${ {P^{..}Cl}}_{3}, { {N^{..}H}_{3}}, { {N^{..}F}_{3}}, { :{{Cl}^{..}F}_{3}}, { {S^{..}F}_{4}} .$

Q40JEE Main 2021MCQ4MSome Basic Concepts of Chemistry

Reducing smog is a mixture of

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📖 Explanation

Classical smog (reducing smog) occurs in cool humid climate, it is a mixture of smoke, fog and ${SO}_{2}$ . It is a reducing mixture/smog. Reducing smog $=$ smoke $+$ fog $+ {SO}_{2}$ Note Photochemical smog occurs in warm, dry and sunny climate. The main components of the photochemical smog result from the action of sunlight on unsaturated hydrocarbons and nitrogen oxides produced by automobiles and factories. Photochemical smog has high concentration of oxidising agents and is therefore, called as oxidising smog.

Q41JEE Main 2021MCQ4MAmines

Hofmann bromamide degradation of benzamide gives product $A$ , which upon heating with ${CHCl}_{3}$ and ${NaOH}$ gives product $B$ . The structures of $A$ and $B$ are

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📖 Explanation

Hofmann bromamide degradation of benzamide gives product aniline $(A)$ , which upon heating with ${CHCl}_{3}$ and ${NaOH}$ gives product benzene isocyanide $(B)$ , by carbylamine reaction. Hofmann bromamide degradation method is for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. The amine formed contains one carbon less than that present in the amide. Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances.Note Secondary and tertiary amines do not show this reaction. This reaction is known as carbylamine reaction or isocyanide test and is used as a test for primary amines.

Q42JEE Main 2021MCQ4MAldehydes, Ketones and Carboxylic Acids

Mesityl oxide is a common name of

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Q43JEE Main 2021MCQ4MAmines

Which of the following reaction is an example of ammonolysis ?

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📖 Explanation

Reaction of alkyl halide with ammonia is called ammonolysis reaction. ${C}_{6} {H}_{5} {CH}_{2} {Cl} {\longrightarrow }^{ {NH}_{3}} {C}_{6} {H}_{5} {CH}_{2} {NH}_{2}$ Carbon-halogen bond in alkyl or benzyl halides can be easily cleaved by a nucleophile. Hence, an alkyl or benzyl halide on reaction with an ethanolic solution of ammonia undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino $(- {NH}_{2})$ group. This process of cleavage of the $C-X$ bond by ammonia molecule is known as ammonolysis.

Q44JEE Main 2021MCQ4MHydrocarbons

Product $A$ is

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📖 Explanation

Alkene undergoes electrophilic addition with HBr to give a major product by Markownikoff rule by ${S}_{N} 1$ mechanism. Step 1. Addition of electrophile to form carbocation. $2^{\circ}$ carbocation undergoes methyl shift to form more stable $3^{\circ}$ carbocation. Step 2. Attack of nucleophile on carbocation.

Q45JEE Main 2021MCQ4MSolutions

A colloidal system consisting of a gas dispersed in a solid is called a/an

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Q46JEE Main 2021MCQ4MClassification of Elements

The incorrect statement(s) about heavy water is (are) (A) used as a moderator in nuclear reactor. (B) obtained as a by-product in fertilizer industry. (C) used for the study of reaction mechanism. (D) has a higher dielectric constant than water. Choose the correct answer from the options given below:

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Dielectric constant of water is greater than heavy water $( {D}_{2} {O})$ . A. It is widely used as moderator in nuclear reactors. B. It can be prepared as a by-product in some fertiliser industries. C. It is used in exchange reactions for the study of the reaction mechanism. D. SpeciesDielectric constant $(C^2{/}NM^2)$ $D_2O$ $78.06$ $H_2O$ $78.39$

Q47JEE Main 2021MCQ4MElectrochemistry

The correct order of conductivity of ions in water is

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📖 Explanation

Correct order of conductivity of ions in water is ${Cs}^{+}> {Rb}^{+}> {K}^{+}> {Na}^{+}$ ${Cs}^{+}( {aq})$ has lower hydrated radius so its electrical conductivity is higher. Extent of hydration depends on charge density on the ion. As the size of gaseous ion decreases, it get more hydrated in water and hence, the size of aqueous ion increases. When this bulky ion move in solution, it experience greater resistance and hence lower conductivity. Size of gaseous ion: ${Cs}^{+}> {Rb}^{+}> {K}^{+}> {Na}^{+}$ Size of aqueous ion: ${Cs}^{+}< {Rb}^{+}< {K}^{+}< {Na}^{+}$ Conductivity: ${Cs}^{+}> {Rb}^{+}> {K}^{+}> {Na}^{+}$

Q48JEE Main 2021MCQ4Md and f Block Elements

What is the spin-only magnetic moment value (BM) of a divalent metal ion with atomic number 25 , in it's aqueous solution?

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📖 Explanation

Atomic number 25 is for manganese $(M n)$ . Configuration of ${Mn}=[ {Ar}] 4 s^{2} 3 d^{5}$ In aqueous medium, ${Mn}^{2+}=[ {Ar}] 3 d^{5}$ Total number of unpaired electrons $=5$ $\mu$ (magnetic moment $)=\sqrt{n}(n+2)$ BM where, $n=$ number of unpaired electron $\therefore \mu =\sqrt{5(5+2)}=\sqrt{35} {BM}=5.92 {BM}$

Q49JEE Main 2021MCQ4MPractical Organic Chemistry

Given below are two statements. Statement I Retardation factor $(R_{f})$ can be measured in metre/centimetre. Statement II $R_{f}$ value of a compound remains constant in all solvents. Choose the most appropriate answer from the options given below

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📖 Explanation

$R_{f}=\;\frac{\;\text{ Distance moved by the substance from base line }\;(x)}{\;\text{ Distance moved by the solvent from base line }\;(y)}$ $R_{f}$ (Retardation factor is dimensionless) Different compounds are differently adsorbed in various. So, $R_{f}$ value of a compound varies with solvent and it is not constant. Both statements I and II are false.

Q50JEE Main 2021MCQ4MChemical Thermodynamics

The point of intersection and sudden increase in the slope, in the diagram given below, respectively, indicates.

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📖 Explanation

At intersection point $\Delta G=0$ , which represents the equilibrium point and sudden increase in slope is due to melting or boiling point of the metal oxide. According to Gibb's free rule, $\Delta G=\Delta H-T \Delta S$ where, $\Delta H=$ change in enthalpy $\Delta G \;\; =\;\text{ change \in Gibb's free energy }\;$ $\Delta S \;\; =\;\text{ change \in entropy }\;$ We can see that $\Delta G$ is dependent on $\Delta S$ . So, when the phase changes $(s \to I$ or $I \to$ g) the entropy increases and hence, slope increases.

Q51JEE Main 2021NAT4Mp Block Elements

The reaction of white phosphorus on boiling with alkali in inert atmosphere resulted in the formation of product $A$ . The reaction of $1 {mol}$ of ${A}$ with excess of ${AgNO}_{3}$ in aqueous medium gives mol(s) of Ag (Round off to the nearest integer).

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📖 Explanation

${P}_{4}+3 {NaOH}+3 {H}_{2} {O} \longrightarrow {PH}_{3}+{3 {NaH}_{2} {PO}_{2} }_{( A)_{\text{Sodiumhypophosphite}} }$ 1 mole of ${NaH}_{2} {PO}_{2}$ reacts with excess of ${AgNO}_{3}$ in aqueous medium to give 4 moles of ${Ag}$ . ${NaH}_{2} {PO}_{2}+4 {AgNO}_{3}+2 {H}_{2} {O} \longrightarrow 4 {Ag}+ {H}_{3} {PO}_{4}+ {NaNO}_{3}+3 {HNO}_{3}$ Hence, 1 mole produce 4 mole of ${Ag}$ .

Q52JEE Main 2021NAT4MIonic Equilibrium

$0.01$ moles of a weak acid ${HA}(K_{a}=2.0 \times 10^{-6})$ is dissolved in $1.0 {L}$ of $0.1 {M} {HCl}$ solution. The degree of dissociation of ${HA}$ is ...... $\times 10^{-5}$ (Round off to the nearest integer). [Neglect volume change on adding HA. Assume degree of dissociation $≪< 1$ ]

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📖 Explanation

Given, $[ {HA}]=0.01$ $[ {HCl}]=0.1 {M}$ When strong acid ({HCl}) $is completely dissociated,$ {HCl}{0.1} ⟶ {{H}^{+}( {aq})}{0.1} +{ {Cl}^{-}( {aq})}{0.1} $For weak acid, dissociation is very less,$ \text{table} ,{H} A, ⇌ , {H}^{+}( {aq}),+, {A}^{-}( {aq});t=0,0.01,,0.1,,0;t=t{;\text{eq };},0.01 -0.01 - alpha,,0.1+0.01 alpha,,0.01 alphaK_{ {a}}=;\frac{[ {H}^{+}]$$[A^{-}]$}{[ {H} A]}=2 \times 10^{-6} \Rightarrow ;\frac{(0.1+0.01 \alpha)(0.01 \alpha)}{(0.01-0.01 \alpha)}=2 \times 10^{-6} $As$ 0.01 \alpha< < 0.1[ {H}^{+}]=0.1 $and$ 0.01 \alpha< < 0.01[ {HA}]=0.01\therefore 2 \times 10^{-6}=;\frac{(0.1)(0.01 \alpha)}{0.01} \Rightarrow \alpha=2 \times 10^{-5}\Rightarrow ;; x=2$

Q53JEE Main 2021NAT4MStructure of Atom

A certain orbital has $n=4$ and $m_{l}=-3$ . The number of radial nodes in this orbital is (Round off to the nearest integer).

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📖 Explanation

Given, $n=4, m_{l}=-3$ , so $l=3$ Possible value of $I=0,1,2,3$ As $m=-3$ is possible only for $I=3$ For $l=3, m_{l}=-3,-2,-1,0,1,2,3$ So, this is $4 f$ -orbital. Number of radial nodes $=n-l-1$ $=4-3-1$ $=0$ So, there is no radial node in $4 f$ orbital.

Q54JEE Main 2021NAT4MOrganic Chemistry - Some Basic Principles

In the above reaction $3.9 {g}$ of benzene on nitration gives $4.92 {g}$ of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is ......... % (Round off to the nearest integer). (Given, atomic mass ${C}: 12.0 {u}, {H}: 1.0 {u}$ , ${O}: 16.0 {u}, {N}: 14.0 {u}$ )

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📖 Explanation

${C}_{6} {H}_{6} {\to }^{ {HNO}_{3}}_{ {H}_{2} {SO}_{4}} {C}_{6} {H}_{5} {NO}_{2}$ ( Molar mass $=123)$ Moles of ${C}_{6} {H}_{6}=(\;\frac{3.9}{78})=0.05$ So, moles of ${C}_{6} {H}_{5} {NO}_{2}$ formed should be $\;\frac{3.9}{78}$ . Mass of ${C}_{6} {H}_{5} {NO}_{2}=$ Moles $\times$ Molar mass $=\;\frac{3.9}{78} \times 123=6.15 {g}$ By conserving moles of carbon, mole or ${C}_{6} {H}_{5} {NO}_{2}$ . $\%$ yield $=\;\frac{\;\text{ Weight actually formed }\;}{\;\text{ Weight theoretical }\;} \times 100$ $=\;\frac{4.92}{6.15} \times 100=80 \%$

Q55JEE Main 2021NAT4MSolutions

The mole fraction of a solute in a 100 molal aqueous solution ......... $\times 10^{-2}$ (Round off to the nearest integer). [Given, atomic masses ${H}: 1.0 {u}, {O}: 16.0 {u}$ ]

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📖 Explanation

Given, molality $=100$ $\Rightarrow 100$ moles of solute in $1 {kg}$ of solvent Mole fraction of solute $(χ_{\;\text{solute }\;})=\;\frac{\;\text{ Moles of solute }\;}{\;\text{ Total moles }\;}$ $=\;{n_{\;\text{Solute }\;}}/{n_{\;\text{Solute }\;}+n_{\;\text{Solvent }\;}}$ $n_{\;\text{solvent }\;}=\;\frac{1000}{18}=55.5$ $χ_{\;\text{solute }\;}=\;\frac{100}{100+55.5}=\;\frac{100}{155.5}=0.643$ $χ_{\;\text{solute }\;}=64.3 \times 10^{-2} \Rightarrow 64 \times 10^{-2}$

Q56JEE Main 2021NAT4MChemical Kinetics

For a certain first order reaction $32 \%$ of the reactant is left after $570 {s}$ . The rate constant of this reaction is ...... $\times 10^{-3} {s}^{-1}$ . (Round off to the nearest integer). [Given, $\log _{10} 2=0.301$ , In $10=2.303$ ]

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📖 Explanation

For first order reaction, $k \times t=\ln \;\frac{A_{0}}{A_{t}}$ ....(i) where, rate $=$ rate constant $t=\;\text{ time }\;$ Initial conc. of reactant, $A_{0}=100$ Conc. of reactant after ' $t$ ' time, $A_{t}=32$ Put all values in equation (i) $t=$ time $A_{t}=32$ $k \times 570=\ln \;\frac{100}{32}$ $k \times 570=2.303(\log 10^{2}-\log 2^{5})$ $\Rightarrow \;\; k=\;\frac{2.303}{570}(2-5 \log 2)$ $=\;\frac{2.303}{570}[2-(5 \times 0.3)]$ $=\;\frac{2.303}{570} \times 0.5$ $k=2 \times 10^{-3} {s}^{-1}$

Q57JEE Main 2021NAT4MChemical Thermodynamics

The standard enthalpies of formation of ${Al}_{2} {O}_{3}$ and ${CaO}$ are $-1675 {kJ} {mol}{ }^{-1}$ and $-625 {kJ} {mol}^{-1}$ respectively. For the reaction, $3 {CaO}+2 {Al} \longrightarrow 3 {Ca}+ {Al}_{2} {O}_{3}$ the standard reaction enthalpy $\Delta _{t} {H}^{\circ} . . . . . . . . {kJ}$ . (Round off to the nearest integer).

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📖 Explanation

Given, $\Delta _{f} H_{ {Al}_{2} {O}_{3}}^{ {o}}=-1675 \frac{kJ}{mol}$ $\Delta _{f} H_{ {CaO}}^{\circ}=-625 \frac{kJ}{mol}$ To find $\Delta _{r} H^{\circ}$ for the reaction $3 {CaO}+2 {Al} \longrightarrow 3 {Ca}+ {Al}_{2} {O}_{3} ; \Delta _{r} {H}^{\circ}=?$ $\Delta _{r} H^{\circ}=sum_{\Delta _{f}} H_{\;\text{product }\;}^{\circ}-sum_{\Delta _{f}} H_{\;\text{Reactant }\;}^{\circ}$ $=\{\Delta _{f} H_{ {Al}_{2} {O}_{3}}^{\circ}+3 \Delta _{f} H_{ {Ca}}^{\circ}\}-\{3 \Delta _{f} H_{ {CaO}}^{\circ}+2 \Delta _{f} H_{ {Al}}^{\circ}\}$ $=\Delta H_{f}^{\circ}( {Al}_{2} {O}_{3})-3 \times \Delta H_{f}^{a}( {CaO})$ We know, $\Delta _{f} H^{\circ}$ for elemental state $=0$ $\Rightarrow \Delta _{f} H_{ {Ca}}^{ {o}}=0, \Delta _{f} H_{ {Al}}^{ {o}}=0$ Putting the value, $\Delta _{t} H^{\circ}=\{-1675+0\}-\{3(-625)+0\}$ $=-1675+1875$ $=200 {kJ}$

Q58JEE Main 2021NAT4MRedox Reactions

$15 {mL}$ of aqueous solution of ${Fe}^{2+}$ in acidic medium completely reacted with $20 {mL}$ of $0.03 {M}$ aqueous ${Cr}_{2} {O}_{7}^{2-}$ . The molarity of the ${Fe}^{2+}$ solution is ...... $\times 10^{-2} {M}$ (Round off to the nearest integer).

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📖 Explanation

${ {Cr}_{2} {O}_{7}^{2-}}_{+6}+ {Fe}^{2+}_{+2} \longrightarrow {Cr}^{3+}_{+3}+ {Fe}^{3+}_{+3}$ Dichromate $( {Cr}_{2} {O}_{7}^{2-})$ oxidised ${Fe}^{2+}$ to ${Fe}^{3+}$ and itself get reduced to ${Cr}^{3+}$ . Valency factor of ${Cr}_{2} {O}_{7}^{2-}=6$ as ${Cr}$ gets reduced from $+6$ to $+3$ From law of equivalence, Milliequivalent of ${Cr}_{2} {O}_{7}^{2-}=$ Milliequivalent of ${Fe}^{2+}$ $M_{1} V_{1} n_{1}=M_{2} V_{2} n_{2}$ $(M_{1} \times V_{1}) \times 6=(M_{2} \times V_{2}) \times 1$ $(0.03 \times 20) \times 6=(M_{2} \times 15) 1$ $\Rightarrow \;\; \;\; M_{2}=0.24$ molar $=24 \times 10^{-2}$ molar

Q59JEE Main 2021NAT4MSolutions

The oxygen dissolved in water exerts a partial pressure of $20 {kPa}$ in the vapour above water. The molar solubility of oxygen in water is ..... $\times 10^{-5} {mol} {} {dm}^{-3}$ . (Round off to the nearest integer). $[Given, Henry's law constant $(K_{H})$ $=8.0 × 10^{4} {kPa}$ for ${O}_{2}$ , density of water with dissolved oxygen $=1.0 {kg} {dm}^{-3}$ ]$.

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📖 Explanation

Given, partial pressure of ${O}_{2}=20 {kPa}$ $K_{ {H}}($ Henry's constant $)=8 \times 10^{4} {kPa}$ From Henry's law, $p(g)=[K_{H}] χ_{O_{2}}$ where, $χ_{ {O}_{2}}=$ solubility of oxygen $20 \times 10^{3}=(8 \times 10^{4} \times 10^{3}) χ_{o_{2}}$ $\Rightarrow \;\; χ_{ {O}_{2}}=\;\frac{20}{8 \times 10^{4}}$ Solubility $=2.5 \times 10^{-4}=25 \times 10^{-5}$

Q60JEE Main 2021NAT4MStates of Matter

The pressure exerted by a non-reactive gaseous mixture of $6.4 {g}$ of methane and $8.8 {g}$ of carbon dioxide in a $10 {L}$ vessel at $27^{\circ} {C}$ is ...... kPa (Round off to the nearest integer) (Assume gases are ideal, $R=8.314 {J} {} {mol}^{-1} {K}^{-1}$ Atomic mass, C: $12.0 {u}, {H}: 1.0 {u}, {O}: 16.0 {u}$ )

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📖 Explanation

Given, mass of methane $=6.4 {g}$ mass of ${CO}_{2}=8.8 {g}$ volume of container $=10 {L}$ temperature $=27^{\circ} {C}$ To find, $p_{\;\text{total }\;}=$ Total pressure Moles of ${CH}_{4}=\;\frac{\;\text{ Given mass }\;}{\;\text{ Molar mass }\;}$ $=\;\frac{6.4}{16}=0.4 {mol}$ Moles of ${CO}_{2}=\;\frac{8.8}{44}=0.2 {mol}$ $n_{\;\text{total }\;}=0.4+0.2=0.6 {mol}$ $p_{\;\text{total }\;}=\;{n_{\;\text{total }\;} R T}/{V}=\;\frac{0.6 \times 8.314 \times 300}{10}$ $=149.652$ $=150 {kPa}$

Mathematics30 questions

Q61JEE Main 2021MCQ4MFunctions

The inverse of $y=5^{\log x}$ is

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📖 Explanation

$y=5^{\log x}$ Taking log on both sides, $\log y=\log x \cdot \log 5$ $\Rightarrow \;\; \;\frac{1}{\log 5}=\;\frac{\log x}{\log y} \Rightarrow \;\frac{1}{\log 5}=\log _{y} x$ $x=y^{\;\frac{1}{\log 5}}$

Q62JEE Main 2021MCQ4MVector Algebra

Let $a=2 \hat{i}-3 \hat{j}+4 \hat{k}$ and $b=7 \hat{i}+\hat{j}-6 \hat{k}$ . If $r \times a=r \times b, r \cdot (\hat{i}+2 \hat{j}+\hat{k})=-3$ , then. $r \cdot (2 \hat{i}-3 \hat{j}+\hat{k})$ is equal to

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📖 Explanation

$a=2 \hat{i}-3 \hat{j}+4 \hat{k}$ $b=7 \hat{i}+\hat{j}-6 \hat{k}$ If $r \times a=r \times b$ $\Rightarrow \;\; r=\lambda(a-b)=\lambda(5 \hat{i}+4 \hat{j}-10 k)$ Now, $r \cdot (\hat{i}+2 \hat{j}+\hat{k})=-3$ $\Rightarrow \;\; \lambda(5 \hat{i}+4 \hat{j}+10 \hat{k}) \cdot (\hat{i}+2 \hat{j}+\hat{k})=-3$ $\Rightarrow \;\; \lambda(5+8+10)=-3$ $\Rightarrow \;\; \lambda=+1$ $\therefore r=-5 \hat{i}-4 \hat{j}+10 \hat{k}$ ${So}, r \cdot (2 \hat{i}-3 \hat{j}+\hat{k})$ $=(-5 \hat{i}-4 \hat{j}+10 \hat{k})(2 \hat{i}-3 \hat{j}+\hat{k})$ $=-10+12+10$ $=12$

Q63JEE Main 2021MCQ4MStraight Lines and Pair of Straight Lines

In a $\triangle P Q R$ , the coordinates of the points $P$ and $Q$ are $(-2,4)$ and $(4,-2)$ , respectively. If the equation of the perpendicular bisector of $P R$ is $2 x-y+2=0$ , then the centre of the circumcircle of the $\triangle P Q R$ is

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📖 Explanation

Let $O$ be the centre of the circumcircle.And $T$ be the mid-point of $P R$ . Since $O T$ is perpendicular bisector of $P R$ . So, equation of $O T$ is given as $2 x-y+2=0$ Let $S$ be the mid-point of $P Q$ . Now, $S$ will be $(\;\frac{-2+4}{2}, \;\frac{4-2}{2})=(1,1)$ and $O S$ will be perpendicular bisector of $P Q$ . Equation of $O S \Rightarrow \;\frac{y-1}{x-1}=\;\frac{-1}{m_{P Q}}$ $\because m_{P Q}=\;\frac{-2-4}{4+2}=-1$ $\therefore$ Equation of $O S=y-1=1(x-1)$ $\Rightarrow \;\; y=x$ Now, coordinates of $O$ will be the intersection of lines $O S$ and $O T$ . $\begin{cases}y=x \\ 2 x-y+2=0\end{cases}$ $\Rightarrow \;\; 2 x-x+2=0 \Rightarrow x=-2$ $\therefore \;\; y=-2$ $O=(-2,-2)$

Q64JEE Main 2021MCQ4MMatrices and Determinants

The system of equations $k x+y+z=1$ , $x+k y+z=k$ and $x+y+z k=k^{2}$ has no solution, if $k$ is equal to

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📖 Explanation

$k x+y+z=1$ $x+k y+z=k$ $x+y+k z=k^{2}$ For this set of equations to have no solution, $\Delta =0$ $\begin{vmatrix}k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k\end{vmatrix}=k(k^{2}-1)-1(k-1)+(1-k)$ $=k^{3}-k-k+1+1-k=k^{3}-3 k+2$ Now, $\Delta =0$ $\Rightarrow \;\; k^{3}-3 k+2=0 \Rightarrow (k-1)(k^{2}+k-2)=0$ $\Rightarrow \;\; (k-1)(k-1)(k+2)=0$ $\therefore k=1,-2$ $\text{table}, x+y+z=1;\;\text{ If }\; k=1, x+y+z=1; ,x+y+z=1\}$ There are same equations and they will have infinite solutions. So, $K=-2$

Q65JEE Main 2021MCQ4MInverse Trigonometric Functions

If $\cot ^{-1}(\alpha)=\cot ^{-1} 2+\cot ^{-1} 8+\cot ^{-1} 18+\cot ^{-1} 32+. . . . .$ upto 100 terms, then $\alpha$ is

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📖 Explanation

$\cot ^{-1} \alpha=\cot ^{-1} 2+\cot ^{-1} 8+\cot ^{-1} 18+\cot ^{-1} 32 . . .$ upto 100 terms Let $T_{n}$ be the $n$ th term of $\cot ^{-1} \alpha$ $\Rightarrow \;\; T_{n}=\cot ^{-1}(2 n^{2})=\tan ^{-1}(\;\frac{1}{2 n^{2}})=\tan ^{-1}[\;\frac{(2 n+1)-(2 n-1)}{1+(2 n+1)(2 n-1)}]$ $\because \;\; \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}(\;\frac{x-y}{1+x y})$ $\therefore \tan ^{-1}\{\;\frac{(2 n+1)-(2 n-1)}{1+(2 n+1)(2 n-1)}\}=\tan ^{-1}(2 n+1)-\tan ^{-1}(2 n-1)$ $T_{1}=\tan ^{-1} 3-\tan ^{-1} 1$ $T_{2}=\tan ^{-1} 5-\tan ^{-1} 3$ $T_{3}=\tan ^{-1} 7-\tan ^{-1} 5$ : $T_{99}=\tan ^{-1} 199-\tan ^{-1} 197$ $T_{100}=\tan ^{-1} 201-\tan ^{-1} 199$ $sum_{n=1}^{100} T_{r}=\tan ^{-1} 201-\tan ^{-1} 1$ $=\tan ^{-1}(\;\frac{201-1}{1+201.1})=\tan ^{-1}(\;\frac{200}{202})$ $\Rightarrow \;\cot ^{-1}(\alpha)=\tan ^{-1}(\;\frac{200}{202})$ $\Rightarrow \;\cot ^{-1}(\alpha)=\cot ^{-1}(\;\frac{202}{200})$ $\therefore \;\alpha=\;\frac{202}{200}=\;\frac{101}{100}=1.01$

Q66JEE Main 2021MCQ4MThree Dimensional Geometry

The equation of the plane which contains the $Y$ -axis and passes through the point $(1,2,3)$ is

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📖 Explanation

Equation of plane passing through a point $(x_{1}, y_{1}, z_{1})$ is $a(x-x_{1})+b(y-y_{1})+c(z-z_{1})=0$ Here, $(x_{1}, y_{1}, z_{1})=(1,2,3)$ ${So}, a(x-1)+b(y-2)+c(z-3)=0$ Now, $Y$ -axis lies on it. Direction ratio of $Y$ -axis is $(0,1,0)$ . Normal vector to the plane $=a \hat{i}+b \hat{j}+c \hat{k}$ So, the normal vector of the plane will be perpendicular to direction ratio of $Y$ -axis. $a \cdot 0+b \cdot 1+c \cdot 0=0 \Rightarrow b=0$ Equation of plane becomes $a(x-1)+c(z-3)=0$ Now, $x=0, z=0$ also satisfies the equation. $a(0-1)+c(0-3)=0$ $\Rightarrow \;\; {t} \;\; - {a}-3 c=0 \Rightarrow {a}=-3 c$ So, $-3 c(x-1)+c(z-3)=0$ $-3 x+3+z-3=0$ $[ {as}, {C} \ne 0]$ $\Rightarrow \;\; 3 x-z=0$

Q67JEE Main 2021MCQ4MMatrices and Determinants

If $A=(\text{table} 0 , \sin \alpha ; \sin \alpha , 0)$ and $det(A^{2}-\;\frac{1}{2} f)=0$ , then a possible value of $\alpha$ is

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📖 Explanation

$A=(\text{table} 0 , \sin \alpha ; \sin \alpha , 0)$ and $\det(A^{2}-\;\frac{1}{2} I)=0$ $\therefore \;\; A^{2}=(\text{table} 0 , \sin \alpha ; \sin \alpha , 0)(\text{table} 0 , \sin \alpha ; \sin \alpha , 0)$ $A^{2}=(\text{table} \sin ^{2} \alpha , 0 ; 0 , \sin ^{2} \alpha)$ $\Rightarrow \;\; \;\frac{I}{2}=(\text{table} 1{/}2 , 0 ; 0 , 1 {/} 2)$ $(A^{2}-\;\frac{1}{2})=(\text{table} \sin ^{2} \alpha-1 {/}2 , 0 ; 0 , \sin ^{2} \alpha-1 {/}2)$ $\therefore \;\; \det(A^{2}-\;\frac{1}{2} I)={\begin{vmatrix}\sin ^{2} \alpha-1{/} 2 & 0 \\ 0 & \sin ^{2} \alpha-1 {/} 2\end{vmatrix}}$ $(\sin ^{2} \alpha-\;\frac{1}{2})^{2}=0$ $\sin \alpha=\pm \;\frac{1}{\sqrt{2}} \Rightarrow \alpha=\;\frac{\pi}{4}$ is one possibility.

Q68JEE Main 2021MCQ4MSets and Relations

If the Boolean expression $(p \Rightarrow q) ⇔(q^{*}(∼ p))$ is a tautology, then the Boolean expression $(p^{*}(∼ q))$ is equivalent to

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📖 Explanation

The Boolean expression $(p \Rightarrow q) ⇔(q^{*}(∼ p))$ is a tautology. Making the truth table for this pqp→ q $q*~p$ $~q$ $-q ∧ p$ $∼(∼ q ∧ p)$ TTTTFFTTFFFTTFFTTTFFTFFTTTFT $\therefore \;\; ∼(∼ q ∧ p)=q ∨ ∼ p=∼ p ∨ q$ $\therefore *$ is equivalent to \vee . So, $\;\; p^{*} ∼ q=p ∨ ∼ q=∼ q p$ $=q \Rightarrow p$

Q69JEE Main 2021MCQ4MProbability

Two dices are rolled. If both dices have six faces numbered 1, 2, 3, 5, 7 and 11 , then the probability that the sum of the numbers on the top faces is less than or equal to 8 is

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📖 Explanation

Six faces have numbers $\{1,2,3,5,7,11\}$ . Sum of the numbers on top faces of both the sides is less than or equal to 8 Then possible outcomes $(1,1),(1,2),(1,3),(1,5),(1,7)\;\to 5$ $(2,1),(2,2),(2,3),(2,5)\;\to 4$ $(3,1),(3,2),(3,3),(3,5)\;\to 4$ $(5,1),(5,2),(5,3)\;\to 3$ $(7,1)\;\to 1$ $n(S)=5+4+4+3+1=17$ Total number of outcomes $n(T)=6 \times 6=36$ $\therefore \;\;$ Probability $=\;\frac{17}{36}$

Q70JEE Main 2021MCQ4MBinomial Theorem

If the fourth term in the expansion of $(x+x^{\log _{2} x})^{7}$ is 4480 , then the value of $x$ , where $x \in N$ is equal to

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📖 Explanation

$(x+x^{\log 2 x})^{7}$ $T_{4}=4480$ $T_{r}={ }^{n} C_{r}(x^{\log _{2} x})^{r} x^{(n-r)}$ $T_{4}={ }^{7} C_{3}(x^{\log _{2} x})^{3} x^{4}=35 x^{3 \log _{2} x} x^{4}$ $T_{4}=4480$ $35 x^{4+3 \log _{2} x}=4480$ $x^{4+3 \log 2 x}=128=2^{7}$ Taking log on both sides, $\log _{2} x^{4 \pm 3 \log _{2} x}=\log _{2} 2^{7}$ $\Rightarrow (4+3 \log _{2} x) \log _{2} x=7$ $4 \log _{2} x+3(\log _{2} x)^{2}=7$ Let $\log _{2} x=y$ $\Rightarrow \;\; 3 y^{2}+4 y-7=0 \Rightarrow (y-1)(3 y+7)=0$ $\Rightarrow \;\; y=1$ or $y=\;\frac{-7}{3} \Rightarrow \log _{2} x=1$ $\Rightarrow$ $x=2$

Q71JEE Main 2021MCQ4MSets and Relations

In a school, there are three types of games to be played. Some of the students play two types of games, but none play all the three games. Which Venn diagram can justify the above statement?

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Q72JEE Main 2021MCQ4MInverse Trigonometric Functions

The sum of possible values of $x$ for $\tan ^{-1}(x+1)+\cot ^{-1}(\;\frac{1}{x-1})=\tan ^{-1}(\;\frac{8}{31})$ is

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📖 Explanation

$\tan ^{-1}(x+1)+\cot ^{-1}(\;\frac{1}{x-1})=\tan ^{-1}(\;\frac{8}{31})$ $\Rightarrow \tan ^{-1}(x+1)+\tan ^{-1}(x-1)=\tan ^{-1}(8 / 31)$ $\Rightarrow \;\; \tan ^{-1}[\;\frac{(x+1)+(x-1)}{1-(x+1)(x-1)}]=\tan ^{-1}(\;\frac{8}{31})$ $\Rightarrow \;\; \tan ^{-1}\$ [;\frac{2 x}{1-(x^{2}-1)}]$=\tan ^{-1}(;\frac{8}{31})\Rightarrow ;\frac{2 x}{2-x^{2}}=;\frac{8}{31}\Rightarrow 62 x=16-8 x^{2}\Rightarrow 8 x^{2}+62 x-16=0\Rightarrow 2(x+8)(4 x-1)=0\Rightarrow x=-8,1 / 4 $But at$ x=;\frac{1}{4}{LHS} \Rightarrow \tan ^{-1}(1+;\frac{1}{4})+\cot ^{-1}(;\frac{1}{\frac{1}{4}-1})=\tan ^{-1} ;\frac{5}{4}+\cot ^{-1}(;\frac{-4}{3})={{\tan^{-1}5/4}{{╰──▾──╯}{<; \pi/2}}+{\cot^{-1}({-4}/3)}{{╰──▾──╯}{>;\pi/2}}}{{╰─────────▾────────╯}{>;\pi/2}}\therefore {LHS}>\pi {/} 2 $RHS$ ={\tan^{-1}(8/31)}{╰────▾────╯}{< ;\pi/2 } $As, LHS$ >\pi {/} 2 $and RHS$ < \pi {/} 2 $So,$ x=-8 $is the only solution.$ x=;\frac{-32}{4}$ is correct option.

Q73JEE Main 2021MCQ4MComplex Numbers

The area of the triangle with vertices $A(z), B(i z)$ and $C(z+i z)$ is

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📖 Explanation

Area of triangle whose vertices are $A(z), B(i z), C(z+i z)$ Area of the triangle $=\;\frac{1}{2}{|z||i z|}=\;\frac{1}{2}{|z|}^{2}$

Q74JEE Main 2021MCQ4MCircles

The line $2 x-y+1=0$ is a tangent to the circle at the point $(2,5)$ and the centre of the circle lies on $x-2 y=4$ . Then, the radius of the circle is

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📖 Explanation

Given, $2 x-y+1=0$ is a tangent to the circle at $(2,5)$ So, normal at $(2,5)$ will be $\;\frac{(y-5)}{(x-2)}=\;\frac{-1}{2}$ \Rightarrow $2 y-10 \;\; =-x+2$ $x+2 y \;\; =12$ Now, it is also given that centre lies on $x-2 y=4$ . So, coordinates of centre will be the solution of $\{\text{table} x+2 y=12; x-2 y=4$ $\Rightarrow x=8, y=2$ Radius will be the distance between $(8,2)$ and $(2,5)$ $r^{2}=(8-2)^{2}+(2-5)^{2}$ $\Rightarrow \;r^{2}\;=36+9$ $\Rightarrow \;r\;=\sqrt{45}=3 \sqrt{5}$

Q75JEE Main 2021MCQ4MPermutations and Combinations

Team $A$ consists of 7 boys and $n$ girls and Team $B$ has 4 boys and 6 girls. If a total of 52 single matches can be arranged between these two teams, when a boy plays against a boy and a girl plays against a girl, then $n$ is equal to

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📖 Explanation

BoysGirlsTeam A7nTeam B46Number of matches between Team $A$ and Team $B$ when a boy play against a boy $({ }^{7} C_{1} \times { }^{4} C_{1})=28$ Similarly, number of matches between Team $A$ and Team $B$ when a girl play against a girl $({ }^{n} C_{1} \times { }^{6} C_{1})=6 n$ According to question, $28+6 n=52$ $6 n=24$ $n=4$

Q76JEE Main 2021MCQ4MQuadratic Equation and Inequalities

The value of $4+\;{1}/{5+\;{1}/{4+\;\frac{1}{5+\;\frac{1}{4+. . . \infty }}}}$ is

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📖 Explanation

Let $x=4+\;{1}/{5+\;{1}/{4+\;\frac{1}{5+\;\frac{1}{4+. . . \infty }}}}$ $\Rightarrow \;\; x=4+\;\frac{1}{5+\;\frac{1}{x}}$ $\Rightarrow \;\; x=4+\;\frac{x}{5 x+1}$ $\Rightarrow \;\; (x-4)(5 x+1)=x \Rightarrow 5 x^{2}-19 x-4=x$ $\Rightarrow \;\; 5 x^{2}-20 x-4=0$ $\Rightarrow \;\; x=\;\frac{20 \pm \sqrt{400+80}}{10}$ $\Rightarrow \;\; x=\;\frac{20 \pm \sqrt{480}}{10}$ $\Rightarrow \;\; x=2 \pm \sqrt{\;\frac{480}{100}}=2 \pm \;\frac{2}{5} \sqrt{30}$ $\because \;\; x ≮ 0$ So, $x=2+\;\frac{2}{5} \sqrt{30}$

Q77JEE Main 2021MCQ4MCircles

Choose the incorrect statement about the wo circles whose equations are given below $x^{2}+y^{2}-10 x-10 y+41=0$ and $x^{2}+y^{2}-16 x-10 y+80=0$

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📖 Explanation

$C_{1} \Rightarrow x^{2}+y^{2}-10 x-10 y+41=0$ $(x-5)^{2}+(y-5)^{2}+41=25+25$ $\Rightarrow$ Centre $=(5,5)$ and radius $=3$ $\Rightarrow \;\; (x-5)^{2}+(y-5)^{2}=3^{2}$ $C_{2} \Rightarrow x^{2}+y^{2}-16 x-10 y+80=0$ $\Rightarrow \;\; (x-8)^{2}+(y-5)^{2}+80=64+25$ $\Rightarrow \;\; (x-8)^{2}+(y-5)^{2}=3^{2}$ $\Rightarrow$ Centre $=(8,5)$ and radius $=3$ Now, distance between centres $=\sqrt{(8-5)^{2}+(5-5)^{2}}=3$ Average radii $=\;\frac{3+3}{2}=3$ $\therefore$ Option (a) is correct. $C_{1}(8,5)=(8-5)^{2}+(5-5)^{2}-9=0$ $C_{2}(5,5)=(5-8)^{2}+(5-5)^{2}-9=0$ Centers of each other lies on circumference of each other. Hence, (b) is incorrect statement.

Q78JEE Main 2021MCQ4MDefinite Integrals

Which of the following statements is correct for the function $g(\alpha)$ for $\alpha \in R$ , such that $g(\alpha)=int_{\pi / 6}^{\pi / 3} \;\frac{\sin ^{\alpha} x}{\cos ^{\alpha} x+\sin ^{\alpha} x} d x$

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📖 Explanation

$g(\alpha)=int_{\pi {/} 6}^{\pi {/}3} \;\frac{\sin ^{\alpha} x}{\cos ^{\alpha} x+\sin ^{\alpha} x} d x$ ...(i) Applying $int_{a}^{b} f(x) d x=int_{a}^{b} f(a+b-x) d x$ $g(\alpha)=int_{\pi {/} 6}^{\pi {/}3} \;\frac{\sin ^{\alpha}(\pi{/} 2-x)}{\cos ^{\alpha}(\;\frac{\pi}{2}-x)+\sin ^{\alpha}(\;\frac{\pi}{2}-x)} d x$ $g(\alpha)=int_{\pi{/} 6}^{\pi / 3} \;\frac{\cos ^{\alpha} x}{\cos ^{\alpha} x+\sin ^{\alpha} x} d x$ .....(ii) Adding Eqs. (i) and (ii), $2 g(\alpha)=int_{\pi {/}6}^{\pi{/} 3} \;\frac{\sin ^{\alpha} x+\cos ^{\alpha} x}{\sin ^{\alpha} x+\cos ^{\alpha} x} d x$ $2 g(\alpha)=int_{\pi{/} 6}^{\pi {/}3} 1 \cdot d x=[x]_{\pi{/}6}^{\pi{/}3}=\;\frac{\pi}{3}-\;\frac{\pi}{6}=\;\frac{\pi}{6}$ $\therefore \;\; g(\alpha)=\;\frac{\pi}{12}$ $g(\alpha)$ is constant function. $\therefore$ It is even function.

Q79JEE Main 2021MCQ4MDifferential Equations

Which of the following is true for $y(x)$ that satisfies the differential equation $\;\frac{d y}{d x}=x y-1+x-y ; y(0)=0$

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📖 Explanation

$\;\frac{d y}{d x}=x y-1+x-y$ $\;\frac{d y}{d x}=x(y+1)-1(y+1) \Rightarrow \;\frac{d y}{d x}=(x-1)(y+1)$ $\int \;\frac{d y}{y+1}=\int(x-1) d x$ $\log _{e}(y+1)=\;\frac{x^{2}}{2}-x+c$ $[ \because \log _{e} 1=0]$ $\log _{e} 1=0+C \Rightarrow C=0$ $y(1) \Rightarrow \log _{e}(y+1)=\;\frac{1}{2}-1$ $y+1=e^{-1{/}2}$ $y(1)=-1+e^{-1 {/} 2}$

Q80JEE Main 2021MCQ4MLimits, Continuity and Differentiability

The value of $\lim _{x \to 0^{+}} \;\frac{\cos ^{-1}(x-[x]^{2}) \cdot \sin ^{-1}(x-[x]^{2})}{x-x^{3}}$ , where $[x]$ denotes the greatest integer $\le x$ is

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📖 Explanation

$\lim _{x \to 0^{+}} \;\frac{\cos ^{-1}(x-[x]^{2}) \cdot \sin ^{-1}(x-[x]^{2})}{x-x^{3}}$ $x \to 0+h$ $=\lim _{h \to 0} \;\frac{\cos ^{-1}(h-0) \cdot \sin ^{-1}(h-0)}{h-h^{3}}$ $=\lim _{h \to 0} \;\frac{\cos ^{-1} h \cdot \sin ^{-1} h}{h(1-h)(1+h)}$ $=\lim _{h \to 0}(\;\frac{\sin ^{-1} h}{h})\$ [;\frac{\cos ^{-1} h}{(1-h)(1+h)}]$=1 \cdot ;\frac{\pi}{2}{RHL}=;\frac{\pi}{2}$

Q81JEE Main 2021NAT4MApplication of Derivatives

The maximum value of $z$ in the following equation $z=6 x y+y^{2}$ , where $3 x+4 y \le 100$ and $4 x+3 y \le 75$ for $x \ge 0$ and $y \ge 0$ is

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📖 Explanation

$z=6 x y+y^{2}$ $3 x+4 y \le 100$ $4 x+3 y \le 75$ $x, y \ge 0$ $z=y(6 x+y)$ z will be maximum at the corner points. $x \le \;\frac{75-3 y}{4}$ $z=y(6 x+y)$ $z \le y[6(\;\frac{75-3 y}{4})+y]$ $z \le \;\frac{1}{2}(225-7 y^{2})$ $(225 y-7 y^{2})$ is a quadratic in $y$ whose maximum value is $\;\frac{-D}{4 a}$ . Here, $D=\;\frac{(225)^{2}-4 \cdot 0(-7)}{4(-7)}$ $\therefore \;\; z \le \;\frac{(225)^{2}}{2 \cdot 4 \cdot 7}=\;\frac{50625}{56} ≃ 904$

Q82JEE Main 2021NAT4MLimits, Continuity and Differentiability

If the function $f(x)=\;\frac{\cos (\sin x)-\cos x}{x^{4}}$ is continuous at each point in its domain and $f(0)=\;\frac{1}{k}$ , then $k$ is ..........

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📖 Explanation

$f(x)=\;\frac{\cos (\sin x)-\cos x}{x^{4}}$ As, $f(x)$ is continuous everywhere, so $f(0)=\lim _{x \to 0} f(x)$ $=\lim _{x \to 0} \;\frac{\cos (\sin x)-\cos x}{x^{4}}$ On expanding the numerator and only identifying the coefficient of $x^{4}$ will give us the required limit. $\cos (\sin x)=(1-\;\frac{\sin ^{2} x}{2}+\;\frac{\sin ^{4} x}{24})$ $=1-\;\frac{1}{2}(x-\;\frac{x^{3}}{6})^{2}+\;\frac{1}{24}(x)$ $=1-\;\frac{1}{2}(x^{2}-\;\frac{x^{4}}{3})+\;\frac{x^{4}}{24}$ $=1-\;\frac{x^{2}}{2}+\;\frac{x^{4}}{6}+\;\frac{x^{4}}{24}$ and $\cos x=1-\;\frac{x^{2}}{2}+\;\frac{x^{4}}{24}$ $\therefore \lim _{x \to 0} \;\frac{\cos (\sin x)-\cos x}{x^{4}}$ $=\lim _{x \to 0} \;{(1-\frac{x^{2}}{2}+\frac{x^{4}}{6}+\frac{x^{4}}{24})-(1-\frac{x^{2}}{2}+\frac{x^{4}}{24})}/{x^{4}}$ $=1 {/} 6$ $\therefore \;\; f(0)=1{/} 6=1 {/} k$ Hence, $k=6$ .

Q83JEE Main 2021NAT4MInverse Trigonometric Functions

If $f(x)=\sin (\cos ^{-1}(\;\frac{1-2^{2 x}}{1+2^{2 x}}))$ and its first derivative with respect to $x$ is $-\;\frac{b}{a} \log _{e} 2$ when $x=1$ , where $a$ and $b$ are integers, then the minimum value of $|a^{2}-b^{2}|$ is ......... .

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📖 Explanation

$f(x)=\sin (\cos ^{-1}(\;\frac{1-2^{2 x}}{1+2^{2 x}}))$ Let $2^{2 x}$ be $\tan ^{2} \theta$ $\therefore \;\; f(x)=\sin [\cos ^{-1}(\;\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta})]$ $=\sin \$ [\cos ^{-1}(\cos 2 theta)]$=\sin 2 \theta=;\frac{2 \tan \theta}{1+\tan ^{2} \theta}=;\frac{2 \cdot 2^{x}}{1+2^{2 x}}f(x)=2 \cdot (;\frac{2^{x}}{1+2^{2 x}})f^{'}(x)=2 \cdot [;\frac{(1+2^{2 x})(2^{x} \log 2)-2^{x}(2^{2 x} \log 2 \cdot 2)}{(1+2^{2 x})^{2}}]f^{'}(1)=2(;\frac{5 \cdot 2 \log 2-2 \cdot 8 \log 2}{5^{2}})=(-;\frac{12}{25}) \log 2=;\frac{-b}{a} \log {e} 2\Rightarrow ;; a=25 $and$ b=12\therefore |a^{2}-b^{2}|{\min }=|25^{2}-12^{2}|=481$

Q84JEE Main 2021NAT4MProbability

Let there be three independent events $E_{1}, E_{2}$ and $E_{3}$ . The probability that only $E_{1}$ occurs is $\alpha$ , only $E_{2}$ occurs is $\beta$ and only $E_{3}$ occurs is $\gamma$ . Let $p$ denote the probability of none of events occur that satisfies the equations $(\alpha-2 \beta) p=\alpha \beta$ and $(\beta-3 \gamma) p=2 \beta \gamma$ . All the given probabilities are assumed to lie in the interval $(0,1)$ . Then, $\;\frac{\;\text{ probability of occurrence of }\; E_{1}}{\;\text{ probability of occurrence of }\; E_{3}}$ is equal to ..........

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📖 Explanation

Let $P(E_{1})=x, P(E_{2})=y, P(E_{3})=z$ Then, $(1-x)(1-y)(1-z)=P$ ......(A) Probability of only $E_{1}=P(E_{1} \cap \overline{E}_{2} \cap \overline{E}_{3})$ $\alpha=x(1-y)(1-z)$ .....(B) Similarly, $\beta=(1-x) y(1-z)$ .....(C) $\gamma=(1-x)(1-y) z$ ......(D) On dividing Eq. (A) by Eq. (B), $\;\frac{P}{\alpha}=\;\frac{1-x}{x} \Rightarrow x P=\alpha-\alpha x$ $\Rightarrow \;\; x(P+\alpha)=\alpha \Rightarrow x=\;\frac{\alpha}{\alpha+P}$ Similarly, $y=\;\frac{\beta}{\beta+P}$ $z=\;\frac{\gamma}{\gamma+P}$ $\;\frac{P(E_{1})}{P(E_{3})}=\;\frac{x}{Z}=(\;\frac{\alpha}{\alpha+P}) \times (\;\frac{\gamma+P}{\gamma})$ $\;\frac{x}{z}=(\;\frac{\alpha}{\gamma})(\;\frac{\gamma+P}{\alpha+P})=(\;\frac{1+P / \gamma}{1+P / \alpha})$ Also given, $P=\;\frac{\alpha \beta}{\alpha-2 \beta}=\;\frac{2 \beta \gamma}{\beta-3 \gamma}$ $\alpha P-2 \beta P=\alpha \beta$ $\Rightarrow \;\; \;\frac{\alpha P}{\alpha+2 P}=\beta$ ....(i) $\beta P-3 P \gamma=2 \beta \gamma$ $\beta(P-2 Y)=3 P Y$ $\Rightarrow \;\; \beta=\;\frac{3 P \gamma}{P-2 \gamma}$ ....(ii) From Eqs. (i) and (ii), $\;\frac{\alpha P}{\alpha+2 P}=\;\frac{3 P \gamma}{P-2 \gamma}$ $\Rightarrow \;\; \;\frac{1}{1+2(\;\frac{P}{\alpha})}=\;\frac{3}{(\;\frac{P}{\gamma})-2}$ $\Rightarrow \;\; (P / \gamma)-2=3+6(P / \alpha)$ $\Rightarrow \;\; \;\frac{P}{\gamma}=5+6(\;\frac{P}{\alpha})$ $\;\frac{x}{z}=\;\frac{1+P / \gamma}{1+P / \alpha}=\;\frac{1+5+6(P / \alpha)}{1+P / \alpha}$ $=6(\;\frac{1+P / \alpha}{1+P / \alpha})=6$

Q85JEE Main 2021NAT4MVector Algebra

If $a=\alpha \hat{j}+\beta \hat{j}+3 \hat{k}, b=-\beta \hat{i}-\alpha \hat{j}-\hat{k}$ and $c=\hat{i}-2 \hat{j}-\hat{k}$ such that $a \cdot b=1$ and $b . c=-3$ , then $\;\frac{1}{3}[(a \times b) \cdot c]$ is equal to

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📖 Explanation

$a=⟨\alpha, \beta, 3⟩$ $b=⟨-\beta,-\alpha,-1⟩$ $c=⟨ 1,-2,-1⟩$ $a \cdot b=1$ $-\alpha \beta-\alpha \beta-3=1$ $\alpha \beta=-2$ $b \cdot c=-3$ $-\beta+2 \alpha+1=-3$ $\beta-2 \alpha=4$ $\Rightarrow \;\; \beta-2(\;\frac{-2}{\beta})=4$ $\Rightarrow \;\; \beta^{2}+4=4 \beta \Rightarrow \beta^{2}-4 \beta+4=0$ $\Rightarrow \;\; (\beta-2)^{2}=0 \Rightarrow \beta=2$ $\alpha \beta=-2 \Rightarrow \alpha \cdot 2=-2$ $\Rightarrow \;\; \alpha=-1$ Hence, $\;\; \;\frac{1}{3}[(a \times b) \cdot c]$ $a=⟨-1,2,3⟩$ $b=⟨-2,1,-1⟩$ $c=⟨ 1,-2,-1⟩$ $(a \times b)=\begin{vmatrix}i^{∧} & j^{∧} & k^{∧} \\ -1 & 2 & 3 \\ -2 & 1 & -1\end{vmatrix}=-5 \hat{i}-7 \hat{j}+3 \hat{k}$ $(a \times b) \cdot c=(-5 \hat{i}-7 \hat{j}+3 \hat{k})(\hat{i}-2 \hat{j}-\hat{k})$ $=-5+14-3=6$ $\therefore \;\; \;\frac{1}{3}[(a \times b) \cdot c]=\;\frac{1}{3} \times 6=2$

Q86JEE Main 2021NAT4MMatrices and Determinants

If $A=\begin{bmatrix}2 & 3 \\ 0 & -1\end{bmatrix}$ , then the value of $\det(A^{4})+\det[A^{10}-\text{Adj}(2 A)^{10}]$ is equal to $. . . . .$

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📖 Explanation

If $A=\begin{bmatrix}2 & 3 \\ 0 & -1\end{bmatrix}$ $det(A^{4})+det[A^{10}-[Adj(2 A)]^{10}]$ $A \cdot A \;\; =\begin{bmatrix}2 & 3 \\ 0 & -1\end{bmatrix}\begin{bmatrix}2 & 3 \\ 0 & -1\end{bmatrix}=\begin{bmatrix}4 & 3 \\ 0 & 1\end{bmatrix}$ $A \;\; =\begin{bmatrix}2 & 3 \\ 0 & -1\end{bmatrix}, A^{2}=\begin{bmatrix}4 & 3 \\ 0 & 1\end{bmatrix}$ $A^{3} \;\; =A^{2} \cdot A=\begin{bmatrix}4 & 3 \\ 0 & 1\end{bmatrix}\begin{bmatrix}2 & 3 \\ 0 & -1\end{bmatrix}=\begin{bmatrix}8 & 9 \\ 0 & -1\end{bmatrix}$ $A^{4} \;\; =A^{3} \cdot A$ $=\begin{bmatrix}8 & 9 \\ 0 & -1\end{bmatrix}\begin{bmatrix}2 & 3 \\ 0 & -1\end{bmatrix}=\begin{bmatrix}16 & 15 \\ 0 & 1\end{bmatrix}$ $A^{n}=\begin{bmatrix}2^{n} & 2^{n}-(-1)^{n} \\ 0 & (-1)^{n}\end{bmatrix}$ $A^{10}=\$ \begin{bmatrix}2^{10} & 2^{10}-1 \ 0 & 1\end{bmatrix}$ \Rightarrow 2 A=\begin{bmatrix}4 & 6 \ 0 & -2\end{bmatrix}adj(2 A)=\begin{bmatrix}-2 & -6 \ 0 & 4\end{bmatrix}adj(2 A)=-2(\text{table} 1 , 3 ; 0 , -2)\begin{bmatrix}\because x=(\text{table} a & b \ c & d) \ adj(x)=(\text{table} d & -b \ -c & a)\end{bmatrix}[adj(2 A)]^{2}=4(\text{table} 1 , 3 ; 0 , -2)(\text{table} 1 , 3 ; 0 , -2)=4(\text{table} 1 , -3 ; 0 , 4)[adj(2 A)]^{3}=4(\text{table} 1 , -3 ; 0 , 4) \times (-2)(\text{table} 1 , 3 ; 0 , -2)=-8(\text{table} 1 , 9 ; 0 , -8)[adj(2 A)]^{3}=4(\text{table} 1 , -3 ; 0 , 4) \times (-2)(\text{table} 1 , 3 ; 0 , -2)=-8(\text{table} 1 , 9 ; 0 , -8)[adj(2 A)]^{n}=(-2)^{n}\begin{bmatrix}1 & (-1)^{n}$[2^{n}-(-1)^{n}\end{bmatrix}$ ; 0 , (-1)^{n} 2^{n}][adj(2 A)]^{10}=2^{10}(\text{table} 1 , -(2^{10}-1) ; 0 , 2^{10}) $Now,$ A^{10}-[adj(2 A)]^{10}=$\begin{bmatrix}2^{10} & 2^{10}-1 \ 0 & 1\end{bmatrix}$-\begin{bmatrix}2^{10} & -(2^{20}-2^{10}) \ 0 & 2^{20}\end{bmatrix}=\begin{bmatrix}0 & 2 \cdot 2^{10}-2^{20}-1 \ 0 & 1-2^{20}\end{bmatrix}det|A^{10}-adj(2 A)^{10}|=0\therefore ;; det(A^{4})+det$[A^{10}-adj(2 A)^{10}]$=(16)^{4}+0=16$

Q87JEE Main 2021NAT4MDefinite Integrals

If [⋅] represents the greatest integer function, then the value of $|int_{0}^{\sqrt{\pi {/} 2}}\$ [x^{2}-\cos x]$ d x|$ is .........

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📖 Explanation

Let $I=int_{0}^{\sqrt{\pi {/} 2}}[[x^{2}]\$ -\cos x] d xI=int_{0}^{1} 0 d x+int_{1}^{\sqrt{pi {/} 2}} d x-int_{0^{\sqrt{pi {/} 2}} d x\Rightarrow ;; I=0+\sqrt{;\frac{pi}{2}}-1-;\frac{pi}{2}=-1|I|=|-1|^{2}=1$

Q88JEE Main 2021NAT4MCircles

The minimum distance between any two points $P_{1}$ and $P_{2}$ while considering point $P_{1}$ on one circle and point $P_{2}$ on the other circle for the given circles equations $x^{2}+y^{2}-10 x-10 y+41=0$ $x^{2}+y^{2}-24 x-10 y+160=0$ is ......... .

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📖 Explanation

$C_{1} \Rightarrow x^{2}+y^{2}-10 x-10 y+41=0$ $\Rightarrow \;\; (x-5)^{2}+(y-5)^{2}=3^{2}$ $C_{2} \Rightarrow x^{2}+y^{2}-24 x-10 y+160=0$ $\Rightarrow \;\; (x-12)^{2}+(y-5)^{2}=3^{2}$ $\therefore$ Minimum distance $=1$ unit.

Q89JEE Main 2021NAT4MThree Dimensional Geometry

If the equation of the plane passing through the line of intersection of the planes $2 x-7 y+4 z-3=0,3 x-5 y+4 z+11=0$ and the point $(-2,1,3)$ is $a x+b y+c z-7=0$ , then the value of $2 a+b+c-7$ is ......... .

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📖 Explanation

Equation of the plane passing through the line of intersections of planes $2 x-7 y+4 z-3=0$ and $3 x-5 y+4 z+11=0$ is $(2 x-7 y+4 z-3)+\lambda(3 x-5 y+4 z+11)=0$ Since this plane passes thought the point $(-2,1,3)$ $\therefore (-4-7+12-3)+\lambda(-6-5+12+11)=0$ $-2+12 \lambda=0 \Rightarrow \lambda=1 {/} 6$ $\therefore$ Equation of plane is $(2 x-7 y+4 z-3)+\;\frac{1}{6}(3 x-5 y+4 z+11)=0$ $15 x-47 y+28 z-7=0$ $\therefore a=15, b=-47, c=28$ $2 a+b+c-7=30-47+28-7=4$

Q90JEE Main 2021NAT4MBinomial Theorem

If $(2021)^{3762}$ is divided by 17 , then the remainder is

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📖 Explanation

$(2021)^{3762}$ $2021=(17 \times 119-2) \Rightarrow (17 \lambda-2)$ $(2021)^{3762}=(17 \lambda-2)^{3762}=C_{0}(17 \lambda)^{3762}-C_{1}(17 \lambda)^{3761}$ $2^{1}+. . . C_{n} 2^{3762}$ Now, $(2021)^{3762}$ will be divisible by 17 all the terms except the last one for last one. $\therefore \;\; (2021)^{3762}=17 \mu -2^{3762}$ $=17 \mu -2^{2}(2^{3760})=17 \mu -4(16)^{235}$ $=17 \mu -4 \cdot (17-1)^{235}$ $(17-1)^{235}=(-1)(1-17)^{235}$ $=-(C_{0}-C_{1} 17+C_{2} 17^{2}-. . .)$ $=-C_{0}+17 \gamma=-1+17 \gamma$ $17 \mu -4(17-1)^{235}=17 \mu -4[-1+17 \gamma]$ $=17(\mu -4 \gamma)+4$ $\therefore \;\; (2021)^{3762}=17 k+4$ Hence, 4 is the remainder.

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