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29 Jul 2022 Shift 2

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Physics30 questions

Q1JEE Main 2022MCQ4MElectrostatics

Two identical metallic spheres ${\text{A}}$ and ${\text{B}}$ when placed at certain distance in air repel each other with a force of ${\text{F}}$ . Another identical uncharged sphere ${\text{C}}$ is first placed in contact with ${\text{A}}$ and then in contact with ${\text{B}}$ and finally placed at midpoint between spheres ${\text{A}}$ and ${\text{B}}$ . The force experienced by sphere ${\text{C}}$ will be: [29-Jul-2022-Shift-2]

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📖 Explanation

Let two identical spheres have charge q. And distance between them $=r$ $\therefore$ Force between the spheres $(F)=\;\frac{k q^2}{r^2}$ Now when an identical uncharged sphere $C$ comes in contact with $A$ , charge $q$ on sphere A get's divided equally to both sphere.So, both sphere ${\text{A}}$ and ${\text{C}}$ have charge $=\;\frac{q}{2}$ Now, C get's in contact with B. So their total charge $(q+\;\frac{q}{2})$ gets divided equally.So, charge on both $B$ and $C$ is $=\;\frac{q+\frac{q}{2}}{2}=\;\frac{3 q}{4}$ Now, ${\text{C}}$ is place midpoint between ${\text{A}}$ and ${\text{B}}$ .Repulsion force between A and C, $F_{A C}=\;\frac{k (\frac{q}{2}) (\frac{3 q}{4}) }{ (\;\frac{r}{2}) ^2}=\;\frac{4 k \times 3 q^2}{8 r^2}$ Repulsion force between B and C, $F_{B C}=\;\frac{k (\frac{3 q}{4}) (\frac{3 q}{4}) }{ (\;\frac{r}{2}) ^2}=\;\frac{4 k \times 9 q^2}{16 r^2}$ $\therefore$ Net force on C, $F_{n e t}=F_{B C}-F_{A C}$ $=\;\frac{4 k \times 9 q^2}{16 r^2}-\;\frac{4 k \times 3 q^2}{8 r^2}$ $=\;\frac{k q^2}{r^2} (\;\frac{9}{4}-\;\frac{3}{2})$ $=\;\frac{k q^2}{r^2} \times \;\frac{3}{4}$ $=\;\frac{3}{4} F [.$ as $F=\;\frac{k q^2}{r^2}]$

Q2JEE Main 2022MCQ4MUnits and Measurements

Match List I with List II. List I List II A.Torque l. ${\text{Nms}}^{-1}$ B.Stress II. $\;\; {\text{J}} {\text{kg}}^{-1}$ C.Latent Heat III. ${\text{Nm}}$ D.Power IV. ${\text{Nm}}^{-2}$ Choose the correct answer from the options given below : [29-Jul-2022-Shift-2]

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📖 Explanation

Torque $\to {\text{Nm}}$ Stress $\to N / {\text{m}}^2$ Latent heat $\to {\text{J}} / {\text{kg}}$ Power $\to {\text{Nm}} / {\ s}$ A-III, B-IV, C-II, D-I

Q3JEE Main 2022MCQ4MElectrostatic Potential and Capacitance

Two identical thin metal plates has charge $q_1$ and $q_2$ respectively such that $q_1 > q_2$ . The plates were brought close to each other to form a parallel plate capacitor of capacitance ${\text{C}}$ . The potential difference between them is: [29-Jul-2022-Shift-2]

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📖 Explanation

Charge on the left surface of plate ${\text{A}}=\;\frac{\;\text{ Total charge }\;}{2}$ $=\;\frac{q_1+q_2}{2}$ Let right surface of plate ${\text{A}}$ has charge $=x$ And total charge on plate $A=q_1$ $\therefore q_1=$ Charge on left surface of plate ${\text{A}}+$ Charge on right surface of plate ${\text{A}}$ $=\;\frac{q_1+q_2}{2}+x$ $\Rightarrow x=q_1-\;\frac{q_1+q_2}{2}$ $=\;\frac{2 q_1-q_1-q_2}{2}$ $=\;\frac{q_1-q_2}{2}$ Let potential difference between two plates $=v$ For capacitor we know, $q=C V$ $\therefore \;\frac{q_1-q_2}{2}=C V$ $\Rightarrow V=\;\frac{q_1-q_2}{2 C}$

Q4JEE Main 2022MCQ4MCurrent Electricity

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Alloys such as constantan and manganin are used in making standard resistance coils. Reason R: Constantan and manganin have very small value of temperature coefficient of resistance. In the light of the above statements, choose the correct answer from the options given below. [29-Jul-2022-Shift-2]

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Q5JEE Main 2022MCQ4MCurrent Electricity

${\text{A}} 1 {\text{m}}$ long wire is broken into two unequal parts ${\text{X}}$ and ${\text{Y}}$ . The ${\text{X}}$ part of the wire is streched into another wire ${\text{W}}$ . Length of $W$ is twice the length of $X$ and the resistance of $W$ is twice that of $Y$ . Find the ratio of length of ${\text{X}}$ and ${\text{Y}}$ . [29-Jul-2022-Shift-2]

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📖 Explanation

$\;{ {\text{R}}_{ {\text{X}}}}/{ {\text{R}}_{ {\text{Y}}}}=\;{ℓ_{ {\text{X}}}}/{ℓ_{ {\text{Y}}}}$ When wire is stretched to double of its length, then resistance becomes 4 times ${\text{R}}_{ {\text{W}}}=4 {\text{R}}_{ {\text{X}}}=2 {\text{R}}_{ {\text{Y}}}$ $\;{ {\text{R}}_{ {\text{X}}}}/{ {\text{R}}_{ {\text{Y}}}}=\;\frac{1}{2}$ So. $\;\frac{ℓ_x}{ℓ_y}=\;\frac{1}{2}$

Q6JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

A wire $X$ of length $50 {\text{cm}}$ carrying a current of $2 {\text{A}}$ is placed parallel to a long wire $Y$ of length $5 {\text{m}}$ . The wire ${\text{Y}}$ carries a current of $3 {\text{A}}$ . The distance between two wires is $5 {\text{cm}}$ and currents flow in the same direction. The force acting on the wire $Y$ is [29-Jul-2022-Shift-2]

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📖 Explanation

Length of wire $Y$ is very high compared to length of wire $X$ . So we can assume length of wire $Y$ is infinite compare to wire $X$ .Magnetic field due to wire $y$ from $5 {\text{cm}}$ from wire towards wire $X$ is, $B=\;\frac{\mu _0 I_1}{2 \pi d}=\;\frac{\mu _0 \times 3}{2 \pi \times 0.05}$ And direction of $B$ is outward.Now force on wire $X$ due to wire $Y$ , $F_{Y X}=I_2 B l$ $=2 \times \;\frac{\mu _0 \times 3}{2 \pi \times 0.05} \times 0.5$ $=1.2 \times 10^{-5} {\text{N}}$ Current flowing in both the wire in same direction so both wire will attract each other with equal force. $\therefore F_{Y X}=F_{X Y}$ $\therefore F_{X Y}=1.2 \times 10^{-5} N$ , attractive force towards wire $X$ .

Q7JEE Main 2022MCQ4MMotion in a Straight Line

A juggler throws balls vertically upwards with same initial velocity in air. When the first ball reaches its highest position, he throws the next ball. Assuming the juggler throws ${\text{n}}$ balls per second, the maximum height the balls can reach is [29-Jul-2022-Shift-2]

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📖 Explanation

The juggler throws ${\text{n}}$ balls per second. $\therefore$ Interval between balls $=\;\frac{1}{n}$ seconds At maximum height velocity of first ball $v=0$ Juggler throws all balls with same initial velocity $=u$ For 1 st ball, $u_1=u, v_1=0, a=-g, S=H_{\max }$ Using formula, $v^2=u^{2+}2 a s$ $0=u^{2-}2 g H_{\max }$ $\Rightarrow H_{\max }=\;\frac{u^2}{2 g} . . . . . . \;\text{ (1) }\;$ And using formula, $v=u+a t$ $\Rightarrow 0=u-g t$ $\Rightarrow t=\;\frac{u}{g}$ $\therefore$ Time taken by ball 1 to reach maximum height $(H_{\max }) =\;\frac{u}{g}$ According to the question, $t=\;\frac{1}{n}$ $\Rightarrow \;\frac{u}{g}=\;\frac{1}{n}$ $\Rightarrow u=\;\frac{g}{n} . . . . . .(2)$ Putting value of $u$ in equation (1), we get $H_{\max }=\;\frac{ (\frac{g}{n}) ^2}{2 g}$ $=\;\frac{g^2}{2 n^2 g}$ $=\;\frac{g}{2 n^2}$

Q8JEE Main 2022MCQ4MAlternating Current

A circuit element ${\text{X}}$ when connected to an a.c. supply of peak voltage $100 {\text{V}}$ gives a peak current of $5 {\text{A}}$ which is in phase with the voltage. A second element ${\text{Y}}$ when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by $\;\frac{\pi}{2}$ . If ${\text{X}}$ and ${\text{Y}}$ are connected in series to the same supply, what will be the rms value of the current in ampere? [29-Jul-2022-Shift-2]

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📖 Explanation

Element ${\text{X}}$ should be resistive with, $R=\;\frac{100}{5}=20 Ω$ Element $Y$ should be inductive with, $X_L=\;\frac{100}{5}=20 Ω$ When ${\text{X}}$ and ${\text{Y}}$ are connector in series, $Z={\sqrt{20^{2+}20^2}}=20 {\sqrt{2}} Ω$ $I=\;\frac{100}{Z}=\;{100}/{20 {\sqrt{2}}}=\;{5}/{{\sqrt{2}}}$ $i_{ {\text{rms}}}=\;{1}/{{\sqrt{2}}} I$ $=\;\frac{5}{2}$

Q9JEE Main 2022MCQ4MWave Optics

An unpolarised light beam of intensity $2 I_0$ is passed through a polaroid ${\text{P}}$ and then through another polaroid $Q$ which is oriented in such a way that its passing axis makes an angle of $30^{\circ}$ relative to that of $P$ . The intensity of the emergent light is [29-Jul-2022-Shift-2]

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📖 Explanation

${\text{I}}_1=\;\frac{1}{2} (2 {\text{I}}_0) = {\text{I}}_0$ ${\text{I}}_2= {\text{I}}_1 \cos ^2 30^{\circ}$ $= {\text{I}}_0 \cdot \;\frac{3}{4}=\;{3 {\text{I}}_0}/{4}$

Q10JEE Main 2022MCQ4MElectrostatic Potential and Capacitance

An $\alpha$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be: [29-Jul-2022-Shift-2]

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📖 Explanation

We know, ${\text{Momentum}}(p)={\sqrt{2 m E_k}}$ and $E_k=q V_{a c c}$ $\therefore p=\sqrt{2 m q V_{a c c}}$ Both $\alpha$ particle and proton are passed through same potential difference. $\therefore (V_{a c c}) _alpha= (V_{a c c}) _p=v$ $\therefore p_alpha={\sqrt{2 m_alpha q_alpha v}}$ $p_p={\sqrt{2 m_p q_p v}}$ $\therefore \;\frac{p_alpha}{p_p}=\sqrt{\;\frac{m_alpha q_alpha}{m_p q_p}}$ $=\sqrt{\;\frac{4 m_p \times 2 e}{m_p \times e}}$ $=\sqrt{\;\frac{8}{1}}$ $=\;{2 {\sqrt{2}}}/{1}$

Q11JEE Main 2022MCQ4MAtoms and Nuclei

Read the following statements: (A) Volume of the nucleus is directly proportional to the mass number. (B) Volume of the nucleus is independent of mass number. (C) Density of the nucleus is directly proportional to the mass number. (D) Density of the nucleus is directly proportional to the cube root of the mass number. (E) Density of the nucleus is independent of the mass number. Choose the correct option from the following options. [29-Jul-2022-Shift-2]

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📖 Explanation

We know,Radius of nucleus, $r=r_0 A^{\;\frac{1}{3}}$ where, ${\text{A}}=$ mass number $\therefore \;\text{ Volume }\;(v)=\;\frac{4}{3} \pi r^3$ $=\;\frac{4}{3} \pi r_0^3 \cdot A$ $\therefore v \propto A$ $\therefore$ Volume is directly proportional to mass number.We know, ${\text{density}}(d)=\;\frac{m}{v}$ $=\;\frac{z m_p+(A-z) m_N}{\frac{4}{3} \pi r_0^3 (A^{1 / 3}) ^3}$ $=\;\frac{z m_p+A m_p-z m_p}{\frac{4}{3} \pi r_0^3 A} [ \because m_p ≃ m_N]$ $=\;\frac{A m_p}{\frac{4}{3} \pi r_0^3 A}$ $=\;\frac{m_p}{\frac{4}{3} \pi r_0^3}$ $\therefore$ Density of nucleus is independent of mass number.

Q12JEE Main 2022MCQ4MGravitation

An object of mass $1 {\text{kg}}$ is taken to a height from the surface of earth which is equal to three times the radius of earth. The gain in potential energy of the object will be [If, ${\text{g}}=10 {\text{ms}}^{-2}$ and radius of earth $=6400 {\text{km}}]$ [29-Jul-2022-Shift-2]

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📖 Explanation

$\Delta U=U_f-U_i$ $=-\;\frac{G M m}{4 R}+\;\frac{G M m}{R}$ $=\;\frac{3 G M m}{4 R}=\;\frac{3}{4} m g R$ $=48 {\text{MJ}}$

Q13JEE Main 2022MCQ4MMotion in a Straight Line

A ball is released from a height ${\text{h}}$ . If $t_1$ and $t_2$ be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between $t_1$ and $t_2$ . [29-Jul-2022-Shift-2]

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📖 Explanation

For first $\;\frac{h}{2}$ distance, $\;\frac{h}{2}=\;\frac{1}{2} g t_1^2$ $\Rightarrow h=g t_1^2$ For total distance h, $h=\;\frac{1}{2} g (t_1+t_2) ^2$ $\Rightarrow g t_1^2=\;\frac{1}{2} g (t_1+t_2) ^2$ $\Rightarrow 2 t_1^2= (t_1+t_2) ^2$ $\Rightarrow {\sqrt{2}} t_1=t_1+t_2$ $\Rightarrow ({\sqrt{2}}-1) t_1=t_2$

Q14JEE Main 2022MCQ4MLaws of Motion

Two bodies of masses $m_1=5 {\text{kg}}$ and $m_2=3 {\text{kg}}$ are connected by a light string going over a smooth light pulley on a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass ${\text{m}}_1$ will be : $[.$ Take $. {\text{g}}=10 {\text{ms}}^{-2}]$ [29-Jul-2022-Shift-2]

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📖 Explanation

Force on equilibrium, $m_2 g=m_1 g sin \;\theta$ Normal force on ${\text{m}}_1, N=m_1 g \cos \theta$ $\Rightarrow \;\frac{N}{m_2 g}=\cot \theta$ $\Rightarrow N=3 \times 10 \times \cot \theta=3 \times 10 \times \;\frac{4}{3} \;\; ( \because sin \;\theta=\;\frac{3}{5})$ $\Rightarrow N=40 {\text{N}}$

Q15JEE Main 2022MCQ4MWork, Power and Energy

If momentum of a body is increased by $20 \%$ , then its kinetic energy increases by [29-Jul-2022-Shift-2]

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📖 Explanation

Let, initial momentum of body $(p_i) =p$ $\therefore$ Final momentum $(p_f) =p_i+20 \%$ of $p_i$ $=p+0.2 p$ $=1.2 p$ We know,Kinetic energy $(E)=\;\frac{p^2}{2 m}$ $\therefore E_i=\;\frac{p^2}{2 m}$ and $E_f=\;\frac{(1.2 p)^2}{2 m}=\;\frac{1.44 p^2}{2 m}$ $\therefore \%$ Change in kinetic energy $=\;\frac{E_f-E_i}{E_i} \times 100$ $=\;\frac{\;\frac{1.44 p^2}{2 m}-\frac{p^2}{2 m}}{\frac{p^2}{2 m}} \times 100$ $=\;\frac{\;\frac{p^2}{2 m}(1.44-1)}{\frac{p^2}{2 m}} \times 100=0.44 \times 100=44$

Q16JEE Main 2022MCQ4MRotational Motion

The torque of a force $5 \hat{i}+3 \hat{j}-7 \hat{k}$ about the origin is $\tau$ . If the force acts on a particle whose position vector is $2 i+2 j+k$ , then the value of $\tau$ will be [29-Jul-2022-Shift-2]

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📖 Explanation

$\vec{\tau } \;\; ={\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 5 & 3 & -7\end{vmatrix}}$ $\;\; =\hat{i}(-14-3)+\hat{j}(5+14)+\hat{k}(6-10)$ $\;\; =-17 \hat{i}+19 \hat{j}-4 \hat{k}$

Q17JEE Main 2022MCQ4MThermodynamics

A thermodynamic system is taken from an original state ${\text{D}}$ to an intermediate state ${\text{E}}$ by the linear process shown in the figure. Its volume is then reduced to the original volume from ${\text{E}}$ to ${\text{F}}$ by an isobaric process. The total work done by the gas from $D$ to $E$ to $F$ will be [29-Jul-2022-Shift-2]

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📖 Explanation

${\text{W}}_{ {\text{DE}}}=\;\frac{1}{2}(600+300) 3 {\text{J}}$ $=1350 {\text{J}}$ ${\text{W}}_{ {\text{EF}}}=-300 \times 3=-900 {\text{J}}$ ${\text{W}}_{ {\text{DEF}}}=450 {\text{J}}$

Q18JEE Main 2022MCQ4MMagnetism and Matter

The vertical component of the earth's magnetic field is $6 \times 10^{-5} {\text{T}}$ at any place where the angle of dip is $37^{\circ}$ . The earth's resultant magnetic field at that place will be (Given $\tan 37^{\circ}=\;\frac{3}{4}$ ) [29-Jul-2022-Shift-2]

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📖 Explanation

$\therefore sin \;37^{\circ}=\;\frac{B_v}{B_{n e t}}$ $\Rightarrow B_v=B_{n e t} sin \;37^{\circ}$ $\Rightarrow B_{n e t}=\;\frac{B_v}{ sin \;37^{\circ}}$ $=\;\frac{6 \times 10^{-5}}{\frac{3}{5}}$ $=10 \times 10^{-5} {\text{T}}$ $=1 \times 10^{-4} {\text{T}}$

Q19JEE Main 2022MCQ4MThermodynamics

The root mean square speed of smoke particles of mass $5 \times 10^{-17} {\text{kg}}$ in their Brownian motion in air at NTP is approximately. $[Given ${\text{k}}=1.38 × 10^{-23} {\text{JK}}^{-1}$ ]$ [29-Jul-2022-Shift-2]

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📖 Explanation

At NTP, $T=298 {\text{K}}$ $v_{ {\text{rms}}} \;\; =\sqrt{\;\frac{3 R T}{M}}$ $\;\; =\sqrt{\;\frac{3 k N_A \times 298}{5 \times 10^{-17} \times N_A}}$ $≃ 15 {\text{mm}} / {\ s}$

Q20JEE Main 2022MCQ4MRay Optics and Optical Instruments

Light enters from air into a given medium at an angle of $45^{\circ}$ with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of $15^{\circ}$ from its original direction. The refractive index of the medium is: [29-Jul-2022-Shift-2]

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📖 Explanation

Let, refractive index of medium $=\mu$ $\therefore r+15^{\circ}=45^{\circ}$ $\Rightarrow r=30^{\circ}$ Using Snell's law, $\;\text{ 1. }\; sin \;45^{\circ}= sin \;30^{\circ} \times \mu$ $\Rightarrow \;{1}/{{\sqrt{2}}}=\;\frac{1}{2} \times \mu$ $\Rightarrow \mu =\;{2}/{{\sqrt{2}}}={\sqrt{2}}=1.141$

Q21JEE Main 2022NAT4MCircular Motion

A tube of length $50 {\text{cm}}$ is filled completely with an incompressible liquid of mass $250 {\text{g}}$ and closed at both ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity $x \sqrt{F} {\text{rads}}^{-1}$ . If ${\text{F}}$ be the force exerted by the liquid at the other end then the value of $x$ will be ________. [29-Jul-2022-Shift-2]

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📖 Explanation

${\text{F}}=\int( {\text{dm}}) \omega ^2 {\text{x}}$ $=int_{0}^{ {\text{L}}} (\;{ {\text{m}}}/{ {\text{L}}} {\text{dx}}) \omega ^2 {\text{x}}$ $=\;{ {\text{m}}}/{ {\text{L}}} \omega ^2 \;{ {\text{L}}^2}/{2}$ $=\;{ {\text{m}} \omega ^2 {\text{L}}}/{2}$ $\omega =\sqrt{\;{2}/{ {\text{mL}}}} \sqrt{ {\text{F}}}$ $=\sqrt{\;\frac{2}{0.25 \times 0.5}} \sqrt{ {\text{F}}}$ $={\sqrt{16}} \sqrt{ {\text{F}}}$ $=4 \sqrt{ {\text{F}}}$

Q22JEE Main 2022NAT4MElectromagnetic Waves

Nearly $10 \%$ of the power of a $110 {\text{W}}$ light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of $1 {\text{m}}$ from the bulb to a distance of $5 {\text{m}}$ is $a \times 10^{-2} {\text{W}} / {\text{m}}^2$ . The value of 'a' will be ________. [29-Jul-2022-Shift-2]

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📖 Explanation

$P^{'}=10 \%$ of $110 W$ $=\;\frac{10}{100} \times 110 {\text{W}}$ $=11 {\text{W}}$ ${\text{I}}_1- {\text{I}}_2=\;{ {\text{P}}^{'}}/{4 \pi \pi_1^2}-\;{ {\text{P}}^{'}}/{4 \pi \pi_2^2}$ $=\;\frac{11}{4 \pi} [\;\frac{1}{1}-\;\frac{1}{25}]$ $=\;\frac{11}{4 \pi} \times \;\frac{24}{25}$ $=\;\frac{264}{\pi} \times 10^{-2}=84 \times 10^{-2} {\text{W}} / {\text{m}}^2$

Q23JEE Main 2022NAT4MCircular Motion

A metal wire of length $0.5 {\text{m}}$ and cross-sectional area $10^{-4} {\text{m}}^2$ has breaking stress $5 \times 10^8 {\text{Nm}}^{-2}$ . A block of $10 {\text{kg}}$ is attached at one end of the string and is rotating in a horizontal circle. The maximum linear velocity of block will be _______ ${\text{ms}}^{-1}$ [29-Jul-2022-Shift-2]

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📖 Explanation

${\text{T}}=\;\frac{ {\ mv}^2}{ℓ}=\;{10 \times {\text{v}}^2}/{0.5}=20 {\text{v}}^2$ ${\text{T}}_{\max }=$ Breaking stress $\times$ Area $=5 \times 10^8 \times 10^{-4}=5 \times 10^4$ $20 {\text{V}}^2=5 \times 10^4$ ${\text{V}}=\sqrt{\;\frac{1}{4} \times 10^4}=50 {\text{m}} / {\ s}$

Q24JEE Main 2022NAT4MProperties of Matter

The velocity of a small ball of mass $0.3 {\text{g}}$ and density $8 {\text{g}} / {\text{cc}}$ when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is $1.3 {\text{g}} / {\text{cc}}$ , then the value of viscous force acting on the ball will be $x \times 10^{-4} {\text{N}}$ , The value of $x$ is _______. $[use ${\text{g}}=10 {\text{m}} / {\text{s}}^2$ ]$ [29-Jul-2022-Shift-2]

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📖 Explanation

$F_{ {\text{V}}}+F_B=m g(v= {\text{constant}})$ $F_V=m g-F_B$ $=\rho _{ {\text{B}}} {\text{Vg}}-\rho _{ {\text{L}}} {\text{Vg}}$ $= (\rho _{ {\text{B}}}-\rho _{ {\text{L}}}) {\text{Vg}}$ $=(8-1.3) \times 10^{+3} \times \;\frac{0.3 \times 10^{-3}}{8 \times 10^3} \times 10$ $=\;\frac{6.7 \times 0.3}{8} \times 10^{-2} \;\; ( {\text{g}}=10)$ $=\;\frac{67 \times 3}{8} \times 10^{-4}=25.125 \times 10^{-4}$

Q25JEE Main 2022NAT4MSemiconductor Electronics

A modulating signal $2 \sin (6.28 \times 10^6) t$ is added to the carrier signal $4 \sin (12.56 \times 10^9) t$ for amplitude modulation. The combined signal is passed through a non-linear square law device. The output is then passed through a band pass filter. The bandwidth of the output signal of band pass filter will be _______ MHz. [29-Jul-2022-Shift-2]

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📖 Explanation

${\text{W}}_{ {\text{C}}}=12.56 \times 10^9$ $W_m=6.25 \times 10^6$ After amplitude modulationBandwidth frequency $=\;\frac{2 W_m}{2 \pi}=\;\frac{2 \times 6.28}{2 \pi} \times 10^6=2 {\text{MHz}}$

Q26JEE Main 2022NAT4MProperties of Matter

The speed of a transverse wave passing through a string of length $50 {\text{cm}}$ and mass $10 {\text{g}}$ is $60 {\text{ms}}^{-1}$ . The area of cross-section of the wire is $2.0 {\text{mm}}^2$ and its Young's modulus is $1.2 \times 10^{11} {\text{Nm}}^{-2}$ . The extension of the wire over its natural length due to its tension will be $x \times 10^{-5} {\text{m}}$ . The value of $x$ is _______. [29-Jul-2022-Shift-2]

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📖 Explanation

${\text{V}}_{ {\text{w}}}=\sqrt{\;{ {\text{T}}}/{\mu }}$ $60=\sqrt{\;{ {\text{T}}}/{10 \times 10^{-3}} \times 0.5}$ ${\text{T}}=\;\frac{(60)^2 \times 10^{-2}}{0.5}=72 {\text{N}}$ $\Delta ℓ=\;{ {\text{F}} ℓ}/{ {\text{AY}}}=\;\frac{72 \times 0.5}{2 \times 10^{-6} \times 1.2 \times 10^{11}}$ $=\;\frac{72 \times 5}{24} \times 10^{-5}=15 \times 10^{-5}$

Q27JEE Main 2022NAT4MOscillations

The metallic bob of simple pendulum has the relative density 5 . The time period of this pendulum is $10 {\text{s}}$ . If the metallic bob is immersed in water, then the new time period becomes $5 \sqrt{x} {\text{s}}$ . The value of $x$ will be ______. [29-Jul-2022-Shift-2]

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📖 Explanation

${\text{mg}}^{'}= {\text{mg}}- {\text{F}}_{ {\text{B}}}$ ${\text{g}}^{'}=\;{ {\text{mg}}- {\text{F}}_{ {\text{B}}}}/{ {\text{F}}_{ {\text{B}}}}$ $=\;{\rho _{ {\text{B}}} {\text{Vg}}-\rho _{ {\text{w}}} {\text{Vg}}}/{\rho _{ {\text{B}}} {\text{V}}}$ $= (\;{\rho _{ {\text{B}}}-\rho _{ {\text{w}}}}/{\rho _{ {\text{B}}}}) {\text{g}}$ $=\;\frac{5-1}{5} \times {\text{g}}$ $=\;\frac{4}{5} {\text{g}}$ We know, $T=2 \pi \sqrt{\;{ℓ}/{ {\text{g}}}}$ $\;{ {\text{T}}^{'}}/{ {\text{T}}}=\sqrt{\;{ {\text{g}}}/{ {\text{g}}^{'}}}=\sqrt{\;{ {\text{g}}}/{5} {\text{g}}}=\sqrt{\;\frac{5}{4}}$ ${\text{T}}^{'}= {\text{T}} \sqrt{\;\frac{5}{4}}=\;\frac{10}{2} {\sqrt{5}}$ ${\text{T}}^{'}=5 {\sqrt{5}}$

Q28JEE Main 2022NAT4MSemiconductor Electronics

A $8 {\text{V}}$ Zener diode along with a series resistance ${\text{R}}$ is connected across a $20 {\text{V}}$ supply (as shown in the figure). If the maximum Zener current is $25 {\text{mA}}$ , then the minimum value of ${\text{R}}$ will be _______ $Ω$ [29-Jul-2022-Shift-2]

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📖 Explanation

$R$ will be minimum when $R_L$ is infinitely large, so $R_{\;\text{Zener }\;}=\;\frac{8}{25 \times 10^{-3}}=320 Ω$ So $\;{R}/{R_{\;\text{Zener }\;}}=\;\frac{12}{8}$ $R=\;\frac{12}{8} \times 320=480 Ω$

Q29JEE Main 2022NAT4MAtoms and Nuclei

Two radioactive materials ${\text{A}}$ and ${\text{B}}$ have decay constants $25 \lambda$ and $16 \lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $B$ to that of $A$ will be "e" after a time $\;\frac{1}{a \lambda }$ . The value of $a$ is _______. [29-Jul-2022-Shift-2]

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📖 Explanation

$N_A=N_0 e^{-25 \lambda t}$ $N_B=N_0 e^{-16 \lambda t}$ $\;\frac{N_B}{N_A}=e=e^{9 \lambda t}$ $t=\;\frac{1}{9 \lambda }$

Q30JEE Main 2022NAT4MOscillations

A capacitor of capacitance $500 \mu {\text{F}}$ is charged completely using a dc supply of $100 {\text{V}}$ . It is now connected to an inductor of inductance $50 {\text{mH}}$ to form an LC circuit. The maximum current in LC circuit will be ________ A. [29-Jul-2022-Shift-2]

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📖 Explanation

At steady state charge stored on the capacitor, $q_{\max }=C V$ $=500 \times 10^{-6} \times 100$ $=5 \times 10^{-2} C$ Energy stored in the capacitor, $U_{\max }=\;\frac{q_{\max }^2}{2 C}$ Now, when electrostatic energy of capacitor converted to magnetic field energy then all energy of capacitor is transferrd to the inductor. $\therefore$ Maximum energy stored in the inductor $U_{L \max }=\;\frac{1}{2} L I_{\max }^2$ $\therefore \;\frac{1}{2} L I_{\max }^2=\;\frac{q_{\max }^2}{2 C}$ $\Rightarrow I_{\max }=\;{q_{\max }}/{{\sqrt{L C}}}$ $=\;{5 \times 10^{-2}}/{\sqrt{50 \times 10^{-3} \times 500 \times 10^{-6}}}$ $=\;\frac{5 \times 10^{-2}}{5 \times 10^{-3}}$ $=10 {\text{A}}$

Chemistry30 questions

Q31JEE Main 2022MCQ4MSome Basic Concepts of Chemistry

Consider the reaction $4 {\text{HNO}}_3(1)+3 {\text{KCl}}( {\text{s}}) \to {\text{Cl}}_2( {\text{g}})+ {\text{NOCl}}( {\text{g}})+2 {\text{H}}_2 {\text{O}}( {\text{g}})+3 {\text{KNO}}_3( {\text{s}})$ The amount of ${\text{HNO}}_3$ required to produce $110.0 {\text{g}}$ of ${\text{KNO}}_3$ is (Given: Atomic masses of ${\text{H}}, {\text{O}}, {\text{N}}$ and ${\text{K}}$ are $1,16,14$ and 39 , respectively.) [29-Jul-2022-Shift-2]

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📖 Explanation

$4 {\text{HNO}}_3( {\text{l}})+3 {\text{KCl}}( {\ s}) \longrightarrow {\text{Cl}}_2( {\text{g}})+ {\text{NOCl}}( {\text{g}})+$ $2 {\text{H}}_2 {\text{O}}( {\text{g}})+3 {\text{KNO}}_3( {\ s})$ $\because 110 {\text{g}}$ of ${\text{KNO}}_3 \Rightarrow$ moles of ${\text{KNO}}_3=\;\frac{110}{101}$ $=1.089 {\text{mol}}$ As, 4 mole of ${\text{HNO}}_3$ produces $3 {\text{mol}}$ of ${\text{KNO}}_3$ .Hence, the moles of ${\text{HNO}}_3$ required to produce $1.089$ moles of ${\text{KNO}}_3=\;\frac{4}{3} \times 1.089=1.452 {\text{mol}}$ Hence, mass of ${\text{HNO}}_3$ required is $1.452 \times 63$ $=91.5 {\text{g}}$

Q32JEE Main 2022MCQ4MStructure of Atom

Given below are the quantum numbers for 4 electrons. A. ${\text{n}}=3,1=2, {\text{m}}_1=1, {\text{m}}_{ {\text{s}}}=+1 / 2$ B. ${\text{n}}=4,1=1, {\text{m}}_1=0, {\text{m}}_{ {\text{s}}}=+1 / 2$ C. ${\text{n}}=4,1=2, {\text{m}}_1=-2, {\text{m}}_{ {\text{s}}}=-1 / 2$ D. ${\text{n}}=3,1=1, {\text{m}}_1=-1, {\text{m}}_{ {\text{s}}}=+1 / 2$ The correct order of increasing energy is [29-Jul-2022-Shift-2]

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📖 Explanation

Energy of the sub-shell is given by, $(n+l)$ rule. ${\text{A}} \Rightarrow 3 {\text{d}} \Rightarrow {\text{n}}+1=5$ $B \Rightarrow 4 {\text{p}} \Rightarrow {\text{n}}+\lambda =5$ ${\text{C}} \Rightarrow 4 {\text{d}} \Rightarrow {\text{n}}+ℓ \Rightarrow 6$ ${\text{D}} \Rightarrow 3 {\ s} \Rightarrow ( {\text{n}}+ℓ)=4$ ${\text{D}}< {\text{A}}< {\text{B}}< {\text{C}}$

Q33JEE Main 2022MCQ4MChemical Thermodynamics

${C}( {\text{s}})+ {\text{O}}_2( {\text{g}}) \to {\text{CO}}_2( {\text{g}})+400 {\text{k}} . {\text{J}}$ ${C}( {\text{s}})+\;\frac{1}{2} {\text{O}}_2( {\text{g}}) \to {\text{CO}}( {\text{g}})+100 {\text{kJ}}$ When coal of purity $60 \%$ is allowed to burn in presence of insufficient oxygen, $60 \%$ of carbon is converted into ' ${\text{CO}}$ ' and the remaining is converted into ' ${\text{CO}}_2$ '. The heat generated when $0.6 {\text{kg}}$ of coal is burnt is _______. [29-Jul-2022-Shift-2]

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📖 Explanation

Weight of coal $=0.6 {\text{kg}}=600 {\text{gm}}$ $\therefore 60 \%$ of it is carbonSo weight of carbon $=600 \times \;\frac{60}{100}=360 {\text{g}}$ $\therefore$ moles of carbon $=\;\frac{360}{12}=30$ moles $\therefore$ Heat generated $=12 \times 400+18 \times 100=6600 {\text{kJ}}$

Q34JEE Main 2022MCQ4MIonic Equilibrium

$200 {\text{mL}}$ of $0.01 {\text{MHCl}}$ is mixed with $400 {\text{mL}}$ of $0.01 {\text{MH}}_2 {\text{SO}}_4$ . The ${\text{pH}}$ of the mixture is _______. Given: $\log 2=0.30, \log 3=0.48, \log 5=0.70, \log 7=0.84, \log 11=1.04$ [29-Jul-2022-Shift-2]

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📖 Explanation

${[ {\text{H}}^{+}]\$ } ;; =;\frac{0.01 × 200+2 × 0.01 × 400}{600};; =;\frac{0.01+2 × 0.01 × 2}{3};; =;\frac{0.01+0.04}{3};; =;\frac{5}{3} × 10^{-2}{\text{pH}} ;; =-\log [ {\text{H}}^{+}];; =-\log (;\frac{5}{3} × 10^{-2});; =- [\log ;\frac{5}{3}+\log 10^{-2}];; =-[\log 5-\log 3-2];; =-0.7+0.48+2;; =2.48-0.7;; =1.78$

Q35JEE Main 2022MCQ4MStates of Matter

Given below are the critical temperatures of some of the gases: Gas Critical temperature (K) ${\text{He}}$ 5.2 ${\text{CH}}_4$ 190.0 ${\text{CO}}_2$ 304.2 ${\text{NH}}_3$ 405.5The gas showing least adsorption on a definite amount of charcoal is [29-Jul-2022-Shift-2]

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Q36JEE Main 2022MCQ4Mp Block Elements

In liquation process used for tin $( { Sn})$ , the metal [29-Jul-2022-Shift-2]

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Q37JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

Given below are two statements. Statement I: Stannane is an example of a molecular hydride. Statement II : Stannane is a planar molecule. In the light of the above statement, choose the most appropriate answer from the options given below. [29-Jul-2022-Shift-2]

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📖 Explanation

Stannane or tin hydride is an inorganic compound with formula ${\text{SnH}}_4$ $\therefore$ It is a non-planar molecule.

Q38JEE Main 2022MCQ4Mp Block Elements

Portland cement contains ' $X$ ' to enhance the setting time. What is ' $X$ '? [29-Jul-2022-Shift-2]

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📖 Explanation

Gypsum $( {\text{CaSO}}_4 \cdot 2 {\text{H}}_2 {\text{O}})$ is used to enhance setting time in portland cement.

Q39JEE Main 2022MCQ4MSalt Analysis

When borax is heated with ${\text{CoO}}$ on a platinum loop, blue coloured bead formed is largely due to [29-Jul-2022-Shift-2]

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📖 Explanation

${\text{Na}}_2 {\text{B}}_4 {\text{O}}_7 \;{\longrightarrow }^{\Delta } 2 {\text{NaBO}}_2+ {\text{B}}_2 {\text{O}}_3$ ${\text{B}}_2 {\text{O}}_3+ {\text{CoO}} \to {\text{Co}} ( {\text{BO}}_2) _2$

Q40JEE Main 2022MCQ4Md and f Block Elements

Which of the following $3 {\text{d}}$ -metal ion will give the lowest enthalpy of hydration $(\Delta _{\;\text{hyd }\;} {\text{H}})$ when dissolved in water? [29-Jul-2022-Shift-2]

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📖 Explanation

Ion $\Delta {\text{H}}_{ {\text{Hyd}}}^{ {\text{O}}}( {\text{kJ}} / {\text{mole}})$ ${\text{Cr}}^{2+}$ $-1925$ ${\text{Mn}}^{2+}$ $-1862$ ${\text{Fe}}^{2+}$ $-1998$ ${\text{Co}}^{2+}$ $-2079$

Q41JEE Main 2022MCQ4MCoordination Compounds

Octahedral complexes of copper(II) undergo structural distortion (Jahn-Teller). Which one of the given copper (II) complexes will show the maximum structural distortion? (en - ethylenediamine; $. {\text{H}}_2 {\text{N}}_{-} {\text{CH}}_2- {\text{CH}}_2- {\text{NH}}_2)$ [29-Jul-2022-Shift-2]

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📖 Explanation

According to John teller any nonlinear molecular system in a degenerate electronic state will be unstable and will undergo some kind of distortion which will lower its symmetry and energy and split the degenerate state.In case of octahedral $d^9$ configuration, the last electron may occupy either $d^2$ or $d_{x^2 y^2}$ orbitals of $e_g$ set.If it occupies ${\text{dz}}^2$ orbital most of the electron density will be concentrated between the metal and the two ligands on the ${\text{z}}$ axis. Thus there will be greater electrostatic repulsion associated with these ligands than with the other four on xy plane.The Jahn Teller effect is mostly observed in octahedral environments. The considerable distortions are usually observed in high spin ${\text{d}}^4$ , low spin ${\text{d}}^7$ and ${\text{d}}^9$ configuration.

Q42JEE Main 2022MCQ4Mp Block Elements

Dinitrogen is a robust compound, but reacts at high altitudes to form oxides. The oxide of nitrogen that can damage plant leaves and retard photosynthesis is [29-Jul-2022-Shift-2]

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📖 Explanation

High concentration of ${\text{NO}}_2$ can damage plant leaves and retard photosynthesis.

Q43JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Correct structure of $\gamma$ -methylcyclohexane carbaldehyde is [29-Jul-2022-Shift-2]

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📖 Explanation

$\gamma$ -Methyl cyclohexane carbaldehyde

Q44JEE Main 2022MCQ4MHaloalkanes and Haloarenes

Compound 'A' undergoes following sequence of reactions to give compound ' $B$ '. The correct structure and chirality of compound ' $B$ ' is [where Et is $- {\text{C}}_2 {\text{H}}_5$ ] [29-Jul-2022-Shift-2]

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Q45JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Given below are two statements. Statement 1: The compound is optically active. Statement II : is mirror image of above compound ${\text{A}}$ . In the light of the above statement, choose the most appropriate answer from the options given below. [29-Jul-2022-Shift-2]

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Q46JEE Main 2022MCQ4MHydrocarbons

When enthanol is heated with conc. ${\text{H}}_2 {\text{SO}}_4$ , a gas is produced. The compound formed, when this gas is treated with cold dilute aqueous solution of Baeyer's reagent, is [29-Jul-2022-Shift-2]

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📖 Explanation

${\text{CH}}_3- {\text{CH}}_2- {\text{OH}} \;{\;\text{ Conc. }\; {\text{H}}_2 {\text{SO}}_4}/{\Delta }▶ {\text{CH}}_2= {\text{CH}}_2 \;\frac{\;\text{ Cold dil solution of }\;}{\;\text{ Bayers reagent }\;} ▶ {\text{OH}}- {\text{CH}}_2- {\text{CH}}_2- {\text{OH}}($ Glycol $)$

Q47JEE Main 2022MCQ4MAmines

The Hinsberg reagent is [29-Jul-2022-Shift-2]

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Q48JEE Main 2022MCQ4MBiomolecules

Which of the following is NOT a natural polymer? [29-Jul-2022-Shift-2]

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Q49JEE Main 2022MCQ4MBiomolecules

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Amylose is insoluble in water. Reason R: Amylose is a long linear molecule with more than 200 glucose units. In the light of the above statements, choose the correct answer from the options given below. [29-Jul-2022-Shift-2]

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📖 Explanation

Amylose is a linear polymer formed by combination of $\alpha-D$ glucose through 1 , 4- glycosidic linkage.It is water soluble.So, assertion is incorrect.

Q50JEE Main 2022MCQ4MIonic Equilibrium

A compound ' $X$ ' is a weak acid and it exhibits colour change at ${\text{pH}}$ close to the equivalence point during neutralization of ${\text{NaOH}}$ with ${\text{CH}}_3 {\text{COOH}}$ . Compound ' $X$ ' exists in ionized form in basic medium. The compound ' $X$ ' is [29-Jul-2022-Shift-2]

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Q51JEE Main 2022NAT4MStates of Matter

'x' ${\text{g}}$ of molecular oxygen $( {\text{O}}_2)$ is mixed with $200 {\text{g}}$ of neon $( {\text{Ne}})$ . The total pressure of the nonreactive mixture of ${\text{O}}_2$ and ${\text{Ne}}$ in the cylinder is 25 bar. The partial pressure of ${\text{Ne}}$ is 20 bar at the same temperature and volume. The value of ' $x$ ' is. _______. [Given: Molar mass of ${\text{O}}_2-32 {\text{g}}^{ {\text{mol}}} {\text{mol}}^{-1}$ . Molar mass of $. {\text{Ne}}=20 {\text{g}} {\text{mol}}^{-1}]$ [29-Jul-2022-Shift-2]

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📖 Explanation

$O_2+N e$ $\;\text{ Xgm 200gm }\;$ $P_{\;\text{total }\;}=25 \;\text{ bar }\; ; P_{ {\text{Ne}}}=2 0$ ${\text{P}}_{ {\text{O}}_2}+ {\text{P}}_{ {\text{Ne}}}=25$ ${\text{P}}_{ {\text{O}}_2}=25-20=5 \;\text{ bar }\;$ $5=\;{\;{ {\text{x}}}/{32}}/{{ {\text{x}}}/{32}+\;\frac{200}{20}} \times 25$ $\;\frac{1}{5}=\;\frac{\;\frac{x}{32}}{\frac{x}{32}+10}$ $\;\frac{1}{5}=\;\frac{x \times 32}{32(x+320)}$ $5 x=x+320$ $4 x=320$ $x=\;\frac{320}{4}=80 {\text{gm}}$

Q52JEE Main 2022NAT4MChemical Bonding and Molecular Structure

Consider, ${\text{PF}}_5, {\text{BrF}}_5, {\text{PCl}}_3, {\text{SF}}_6,[ {\text{ICl}}_4]^{-}, {\text{ClF}}_3$ and ${\text{IF}}_5$ . Amongst the above molecule(s)/ion(s), the number of molecule(s)/ion(s) having $sp^3d^2$ hybridisation is ________. [29-Jul-2022-Shift-2]

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Q53JEE Main 2022NAT4MSolutions

$1.80 {\text{g}}$ of solute A was dissolved in $62.5 {\text{cm}}^3$ of ethanol and freezing point of the solution was found to be $155.1 {\text{K}}$ . The molar mass of solute ${\text{A}}$ is ${\text{g}} {\text{mol}}-1$ ______. $[Given : Freezing point of ethanol is $156.0 {\text{K}}$ . Density of ethanol is $0.80 {\text{g}} {\text{cm}}^{-3}$ . Freezing point depression constant of ethanol is $2.00 {\text{K}} {\text{kg}} {\text{mol}}^{-1}$ ]$ [29-Jul-2022-Shift-2]

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📖 Explanation

Mass of solvent $=d x v=0.8 \times 62.5=50 {\text{gram}}$ $\Delta T_f=k_f \times m$ $0.9=2 [\;{1.8 \times 1000}/{M_{\;\text{Solute }\;} \times 50}]$ $M_{\;\text{Solute }\;}= (\;\frac{2 \times 1.8 \times 1000}{0.9 \times 50}) =80$

Q54JEE Main 2022NAT4MElectrochemistry

For a cell, ${\text{Cu}}( {\text{s}})| {\text{Cu}}^{2+}(0.001 {\text{M}}) \| {\text{Ag}}^{+}(0.01 {\text{M}})| {\text{Ag}}( {\text{s}})$ the cell potential is found to be $0.43 {\text{V}}$ at $298 {\text{K}}$ . The magnitude of standard electrode potential for ${\text{Cu}}^{2+} / {\text{Cu}}$ is _______ $\times 10^{-2} {\text{V}}$ [Given : $E_{A g^{+} / A g}^{Θ}=0.80 {\text{V}}$ and $.\;\frac{2.303 R R T}{F}=0.06 {\text{V}}]$ [29-Jul-2022-Shift-2]

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📖 Explanation

Anode $: {\text{Cu}}( {\ s}) \longrightarrow {\text{Cu}}^{2+}( {\text{aq}})+2 {\text{e}}^{-}$ Cathode: $[ {\text{Ag}}^{+}+ {\text{e}}^{-} \longrightarrow {\text{Ag}}( {\ s})] 2$ Cus(s) $+2 {\text{Ag}}^{+}( {\text{aq}}) \longrightarrow {\text{Cu}}^{2+}( {\text{aq}})+2 {\text{Ag}}( {\ s})$ $E_{\;\text{cell }\;}=E_{\;\text{cell }\;}^{0-}\;\frac{0.06}{2} \log \;{[ {\text{Cu}}^{2+}]\$ }/{[ {\text{Ag}}^{+}]$ ^2}0.43= {\text{E}}{;\text{cell };}^{0-};\frac{0.06}{2} \log (;\frac{10^{-3}}{ (10^{-2}) ^2})0.43=E{;\text{cell };}^{0-}0.03 \log 10{\text{E}}{;\text{cell };}^0=0.46 {\text{V}}{\text{E}}{;\text{cell };}^0= {\text{E}}{ {\text{Ag}}^{+} / {\text{Ag}}}^{0-} {\text{E}}{ {\text{Cu}}^{2+} / {\text{Cu}}}^0{\text{E}}_{ {\text{Cu}}^{2+} / {\text{Cu}}^0}=(0.80-0.46)=0.34 {\text{V}}=34 \times 10^{-2}$

Q55JEE Main 2022NAT4MChemical Kinetics

Assuming $1 \mu {\text{g}}$ of trace radioactive element ${\text{X}}$ with a half life of 30 years is absorbed by a growing tree. The amount of $X$ remaining in the tree after 100 years is _______ $\times 10^{-1} \mu {\text{g}}$ . [Given : $\ln 10=2.303 ; \log 2=0.30]$ [29-Jul-2022-Shift-2]

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📖 Explanation

$t=\;\frac{1}{\lambda } \ln (\;\frac{a}{a-x})$ $\Rightarrow 100= (\;\frac{30}{\ln 2}) [\ln (\;\frac{1}{w}) ]$ $\Rightarrow [\;\frac{100 \times \log 2}{30}] =\log (\;\frac{1}{w})$ $\Rightarrow 1=\log (\;\frac{1}{w})$ $\Rightarrow \;\frac{1}{w}=10$ So $w=0.1 \mu g$

Q56JEE Main 2022NAT4MCoordination Compounds

Sum of oxidation state (magnitude) and coordination number of cobalt in ${\text{Na}} \cdot [ {\text{Co}}( {\text{bpy}}) {\text{Cl}}_4]$ is ________. [29-Jul-2022-Shift-2]

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📖 Explanation

${\text{Na}} [ {\text{Co}}( {\text{bpy}}) {\text{Cl}}_4]$ Oxidation state of cobalt $=+3$ Coordination number of cobalt $=6$ [As bpy is bidentate]So, sum $=9$

Q57JEE Main 2022NAT4MChemical Bonding and Molecular Structure

Consider the following sulphur based oxoacids. ${\text{H}}_2 {\text{SO}}_3, {\text{H}}_2 {\text{SO}}_4, {\text{H}}_2 {\text{S}}_2 {\text{O}}_8$ and ${\text{H}}_2 {\text{S}}_2 {\text{O}}_7$ . Amongst these oxoacids, the number of those with peroxo $( {\text{O}}- {\text{O}})$ bond is _______. [29-Jul-2022-Shift-2]

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Q58JEE Main 2022NAT4MAlcohols, Phenols and Ethers

A $1.84 {\text{mg}}$ sample of polyhydric alcoholic compound 'X' of molar mass $92.0 {\text{g}} / {\text{mol}}$ gave $1.344 {\text{mL}}$ of ${\text{H}}_2$ gas at STP. The number of alcoholic hydrogens present in compound ' ${\text{X}}$ ' is ________. [29-Jul-2022-Shift-2]

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📖 Explanation

Volume of ${\text{H}}_2$ gas $=1.344 {\text{mL}}$ Mole of ${\text{H}}_2$ gas $=\;\frac{1.344}{22400}=6 \times 10^{-5}$ No of ${\text{H}}$ atoms per molecule of ${\text{H}}_2=2$ .No. of moles of organic compound $=\;\frac{1.84 \times 10^{-3}}{92}=2 \times 10^{-5}$ No. of $- {\text{OH}}$ (hydroxyl group in one molecule) $=\;\frac{6 \times 10^{-5}}{2 \times 10^{-5}}=3$

Q59JEE Main 2022NAT4MOrganic Chemistry - Some Basic Principles

The number of stereoisomers formed in a reaction of $(\pm ) {\text{Ph}}( {\text{C}}= {\text{O}}) {\text{C}}( {\text{OH}})( {\text{CN}}) {\text{Ph}}$ with ${\text{HCN}}$ is ________. [where ${\text{Ph}}$ is $- {\text{C}}_6 {\text{H}}_5$ ] [29-Jul-2022-Shift-2]

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Q60JEE Main 2022NAT4MHaloalkanes and Haloarenes

The number of chlorine atoms in bithionol is ________. [29-Jul-2022-Shift-2]

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Mathematics30 questions

Q61JEE Main 2022MCQ4MComplex Numbers

If $z \ne 0$ be a complex number such that $|z-\;\frac{1}{z}|=2$ , then the maximum value of $|z|$ is: [29-Jul-2022-Shift-2]

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📖 Explanation

We know, $|| z_1|-| z_2|| \le |z_1+z_2{{}/{}|} \le |z_1{{}/{}|} + |z_2{{}/{}|}$ $\therefore || z |-\;\frac{1}{|z|}{{}/{}|} \le |z-\;\frac{1}{z}{{}/{}|}$ $\Rightarrow || z |-\;\frac{1}{|z|}{{}/{}|} \le 2 [\;\text{ Given }\; |z-\;\frac{1}{z}{{}/{}|} =2]$ $\Rightarrow |\;\frac{|z|^{2-}1}{|z|}{{}/{}|}\le 2$ $\Rightarrow -2 \le \;\frac{|z|^{2-}1}{|z|} \le 2$ $\therefore \;\frac{|z|^{2-}1}{|z|} \le 2$ $\Rightarrow |z|^{2-}1 \le 2|z|$ $\Rightarrow |z|^{2-}2|z|-1 \le 0$ $\Rightarrow |z|^{2-}2|z|+1-2 \le 0$ $\Rightarrow (|z|-1)^{2-}2 \le 0$ $\Rightarrow -{\sqrt{2}} \le |z|-1 \le {\sqrt{2}}$ $\Rightarrow 1-{\sqrt{2}} \le |z| \le 1+{\sqrt{2}} . . . . .(1)$ or $-2 \le \;\frac{|z|^{2-}1}{|z|}$ $\Rightarrow |z|^{2-}1 \le -2|z|$ $\Rightarrow |z|^{2+}2|z|-1 \le 0$ $\Rightarrow |z|^{2+}2|z|+1-2 \le 0$ $\Rightarrow (|z|+1)^{2-}2 \le 0$ $\Rightarrow -{\sqrt{2}} \le |z|+1 \le +{\sqrt{2}}$ $\Rightarrow -{\sqrt{2}}-1 \le |z| \le {\sqrt{2}}-1 . . . . . . \;\text{ (2) }\;$ From (1) and (2) we get,Maximum value of $|z|={\sqrt{2}}+1$ and minimum value of $|z|=-{\sqrt{2}}-1$

Q62JEE Main 2022MCQ4MMatrices and Determinants

Which of the following matrices can NOT be obtained from the matrix $\begin{bmatrix}-1 & 2 \\ 1 & -1\end{bmatrix}$ by a single elementary row operation? [29-Jul-2022-Shift-2]

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📖 Explanation

$A={\begin{bmatrix}-1 & 2 \\ 1 & -1\end{bmatrix}}$ (1) ${\text{R}}_1 \to {\text{R}}_1+ {\text{R}}_2 ;{\begin{bmatrix}0 & 1 \\ 1 & -1\end{bmatrix}}$ possible(2) ${\text{R}}_1 ↔ {\text{R}}_2 ;{\begin{bmatrix}1 & -1 \\ -1 & 2\end{bmatrix}}$ possible(3) Option is not possible(4) $R_2 \to R_2+2 R_1 ;{\begin{bmatrix}-1 & 2 \\ -1 & 3\end{bmatrix}}$ possible

Q63JEE Main 2022MCQ4MMatrices and Determinants

If the system of equations $x+y+z=6$ $2 x+5 y+\alpha z=\beta$ $x+2 y+3 z-14$ has infinitely many solutions, then $\alpha+\beta$ is equal to [29-Jul-2022-Shift-2]

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📖 Explanation

Given, $x+y+z=6 . . . . .(1)$ $2 x+5 y+\alpha z=\beta . . . . .(2)$ $x+2 y+3 z=14 . . . . . . \;\text{ (3) }\;$ System of equation have infinite many solutions. $\therefore \Delta _x=\Delta _y=\Delta _z=0 \;\text{ and }\; \Delta =0$ $\;\text{ Now, }\; \Delta ={\begin{vmatrix}1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3\end{vmatrix}}=0$ $C_1 \to C_1-C_3$ $C_2 \to C_2-C_3$ $\Rightarrow {\begin{vmatrix}0 & 0 & 1 \\ 2-\alpha & 5-\alpha & \alpha \\ -2 & -1 & 3\end{vmatrix}}=0$ $\Rightarrow -2+\alpha+10-2 \alpha=0$ $\Rightarrow 8-\alpha=0$ $\Rightarrow \alpha=8$ Now, $x+y+z=6$ $2 x+5 y+8 z=\beta$ $x+2 y+3 z=14$ $\therefore \Delta _x={\begin{vmatrix}6 & 1 & 1 \\ \beta & 5 & 8 \\ 14 & 2 & 3\end{vmatrix}}=0$ $C_1 \to C_1-6 C_3$ $C_2 \to C_2-C_3$ $\Rightarrow {\begin{vmatrix}0 & 0 & 1 \\ \beta-48 & -3 & 8 \\ -4 & -1 & 3\end{vmatrix}}=0$ $\Rightarrow -\beta+48-12=0$ $\Rightarrow \beta=36$ $\therefore \alpha+\beta=8+36=44$

Q64JEE Main 2022MCQ4MLimits, Continuity and Differentiability

Let the function $f(x)=\begin{cases}\\ \\ \frac{\log _e(1+5 x)-\log _e(1+a x x)}{x} & { \\ } \\ \\ \text{ if }\\ \\ x \ne 0 \\ 10 & { \\ } \\ \\ \text{ if }\\ \\ x=0\end{cases}$ be continuous at $x=0$ . Then $\alpha$ is equal to [29-Jul-2022-Shift-2]

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📖 Explanation

$f(x)$ is continuous at $x=0$ $\therefore f(0)=\lim _{x \to 0} f(x)$ $\Rightarrow 10=\lim _{x \to 0} \;\frac{\log _e(1+5 x)-\log _e(1+\alpha x)}{x}$ $=\lim _{x \to 0} \;\frac{\log (1+5 x)}{5 x} \times 5-\;\frac{\log _e(1+\alpha x)}{\alpha x} \times \alpha$ $=1 \times 5-\alpha$ $\Rightarrow \alpha=5-10=-5$

Q65JEE Main 2022MCQ4MDefinite Integrals

If $[t]$ denotes the greatest integer $\le t$ , then the value of $int_0^1\$ [2 x- |3 x^{2-}5 x+2|+1]$ {\text{d}} x$ is: [29-Jul-2022-Shift-2]

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📖 Explanation

$I=int_{0}^{1} \$ [2 x-13 x^{2-}3 x-2 x+2 ∣+1]$ \text{dt}I=int_{0}^{1}[2 x-|(3 x-2)(x-1)|] \text{dt}+int_{0}^{1} 1 \text{dt}I=int_{0}^{2 / 3} $[ (2 x- (3 x^{2-}5 x+2) ) ]$ \text{dt}+int_{2 / 3}^{1} (2 x+ (3 x^{2-}5 x+2) ) \text{dt}+1I=int_{0}^{2 / 3} $[-3 x^{2+}7 x-2]$ \text{dt}+int_{2 / 3}^{1} (3 x^{2-}3 x+2) \text{dt}+1int_{0}^{\alpha}(-2) \text{dt}+int_{\alpha}^{1 / 3}(-1) \text{dt}+int_{1 / 3}^{\beta} 0 \text{dt}+int_{\beta}^{2 / 3} 1 . \text{dt}=-2 \alpha- (;\frac{1}{3}-\alpha) +;\frac{2}{3}-\beta=-\alpha-\beta+;\frac{1}{3}y=3 x^{2-}3 x+2 $When$ {\text{x}} \in (;\frac{2}{3}, 1)3 x^{2-}3 x+2 \in (;\frac{4}{3}, 2)[3 x^{2-}3 x+2] =1\therefore int_{2 / 3}^{1} $[3 x^{2-}3 x+2]$ \text{dt}=1 (1-;\frac{2}{3}) =;\frac{1}{3} $Hence$ {\text{I}}= (;\frac{1}{3}-(\alpha+\beta)) + (;\frac{1}{3}) +1=;\frac{5}{3}- (;{7-{\sqrt{37}}}/{6}+;{7-{\sqrt{13}}}/{6})=;\frac{-2}{3}+;{{\sqrt{37}}+{\sqrt{13}}}/{6}=;{{\sqrt{37}}+{\sqrt{13}}-4}/{6}$

Q66JEE Main 2022MCQ4MSequences and Series

Let $\{a_n\}_{n=0}^{\infty }$ be a sequence such that $a_0-a_1-0$ and $a_{n+2}=3 a_{n+1}-2 a_n+1, ∀ n \ge 0$ . Then $a_{25} a_{23}-2 a_{25} a_{22}-2 a_{23} a_{24}+4 a_{22} a_{24}$ is equal to [29-Jul-2022-Shift-2]

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📖 Explanation

$a_0=0, a_1=0$ $a_{ {\text{n}}+2}=3 a_{ {\text{n}}+1}-2 a_{ {\text{n}}+1}: {\text{n}} \ge 0$ $a_{ {\text{n}}+2}-a_{ {\text{n}}+1}=2 (a_{ {\text{n}}+1}- {\text{a}}_{ {\text{n}}}) +1$ ${\text{n}}=0 \;\; a_2-a_1=2 (a_1-a_0) +1$ ${\text{n}}=1 \;\; a_3-a_2=2 (a_2-a_1) +1$ ${\text{n}}=2 \;\; a_4-a_3=2 (a_3-a_2) +1$ ${\text{n}}= {\text{n}} \;\; a_{ {\text{n}}+2}-a_{ {\text{n}}+1}=2 (a_{ {\text{n}}+1}-a_{ {\text{n}}}) +1$ $(a_{ {\text{n}}+2}-a_1) -2 (a_{ {\text{n}}+1}-a_0) -( {\text{n}}+1)=0$ $a_{ {\text{n}}+2}=2 a_{ {\text{n}}+1}+( {\text{n}}+1)$ ${\text{n}} \to {\text{n}}-2$ $a_{ {\text{n}}}-2 a_{ {\text{n}}-1}= {\text{n}}-1$ $\;\text{ Now }\; {\text{a}}_{25} {\text{a}}_{23}-2 {\text{a}}_{25} {\text{a}}_{22}-2 {\text{a}}_{23} {\text{a}}_{24}+4 {\text{a}}_{22} {\text{a}}_{24}$ $= (a_{25}-2 a_{24}) (a_{23}-2 a_{22}) =(24)(22)=528$

Q67JEE Main 2022MCQ4MSequences and Series

$sum_{r=1}^{20}(r^{2+}1)(r !)$ is equal to [29-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Given, $sum_{r=1}^{20} (r^{2+}1) (r !)$ Let, $f(r)= (r^{2+}1) (r !)$ $= (r^2) (r !)+r !$ $=r(r r !)+r !$ $=r[(r+1-1) r !]+r !$ $=r[(r+1) r !-r !]+r !$ $=r[(r+1) !-(r !)]+r !$ $=r(r+1) !-r(r !)+r !$ $=(r+2-2)(r+1) !-r(r !)+r !$ $=(r+2)(r+1) !-2(r+1) !-[(r+1-1)(r !)]+r !$ $=(r+2) !-2(r+1) !-(r+1) !+r !+r !$ $=(r+2) !-3(r+1) !+2 r !$ $=[(r+2) !-(r+1) !]-2[(r+1) !-r !]$ $\therefore sum_{r=1}^{20} f(r)$ $=sum_{r=1}^{20}[(r+2) !-(r+1) !]-2 sum_{r=1}^{20}[(r+1) !-r !]$ $=[(22 !+21 !+20 !+. . . .+4 !+3 !)-(21 !+20 !+19 !+. . . .+3 !+2 !]-2[(21 !+20 !+. . .+3 !+2 !)-(20 !+19 !+. . . . . . 1 !)]$ $=[(22 !)-(2 !)]-2[(21) !-(1 !)]$ $=22 !-2 !-2 \cdot (21) !+2.1 !$ $=22 !-2 \cdot (21) !$

Q68JEE Main 2022MCQ4MIndefinite Integrals

For $I(x)-\int \;\frac{\sec ^2 x-2022}{\sin ^{2022} x} d x$ , if $I(\;\frac{\pi}{4})-2^{1011}$ , then [29-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Given, $I(x)=\int \;\frac{\ sec ^2 x-2022}{ sin \;^{2022} x} \text{dt}$ $=\int \;\frac{\ sec ^2 x}{ sin \;^{2022} x} \text{dt}-\int \;\frac{2022}{ sin \;^{2022} x} \text{dt}$ $=\int \;\frac{1}{ sin \;^{2022} x} \cdot \ sec ^2 x \text{dt}-\int \;\frac{2022}{ sin \;^{202} x} \text{dt}$ $=\;\frac{1}{ sin \;^{2022} x} \cdot \tan x-\int (\;\frac{-2022}{ sin \;^{2023} x} \cdot \cos x \cdot \tan x) \text{dt}-\int \;\frac{2022}{ sin \;^{2022} x} \text{dt}+C$ $=\;\frac{\tan x}{ sin \;^{2022} x}+\int (\;\frac{2022}{ sin \;^{2023} x} \cdot \cos x \cdot \;\frac{ sin \;x}{\cos x}) \text{dt}-\int \;\frac{2022}{ sin \;^{2022} x} \text{dt}+C$ $=\;\frac{\tan x}{ sin \;^{2022} x}+\int \;\frac{2022}{ sin \;^{2022} x} \text{dt}-\int \;\frac{2022}{ sin \;^{2022} x} \text{dt}$ $=\;\frac{\tan x}{ sin \;^{2022} x}+C$ Given, $I (\;\frac{\pi}{4}) =2^{1011}$ $\therefore I (\;\frac{\pi}{4}) =\;\frac{\tan (\frac{\pi}{4}) }{ ( sin \;\;\frac{\pi}{4}) ^{2022}}+C$ $\Rightarrow 2^{1011}=\;{1}/{ (\;{1}/{{\sqrt{2}}}) ^{2022}}+C$ $\Rightarrow C=2^{1011}-2^{1011}=0$ $\therefore I(x)=\;\frac{\tan x}{ sin \;^{2022} x}$ $\therefore I (\;\frac{\pi}{3}) =\;\frac{\tan \frac{\pi}{3}}{ ( sin \;\;\frac{\pi}{3}) ^{2022}}=\;{{\sqrt{3}}}/{ (\;{{\sqrt{3}}}/{2}) ^{2022}}$ $I (\;\frac{\pi}{6}) =\;{\;{1}/{{\sqrt{3}}}}/{ (\;\frac{1}{2}) ^{2022}}=\;{1}/{{\sqrt{3}}} \times (2)^{2022}$ From option $(A)$ , $3^{1010} \cdot I (\;\frac{\pi}{3}) -I (\;\frac{\pi}{6})$ $=3^{1010} \cdot {\sqrt{3}} \cdot (\;{2}/{{\sqrt{3}}}) ^{2022}-\;{(2)^{2022}}/{{\sqrt{3}}}$ $=3^{1010} \cdot {\sqrt{3}} \times \;\frac{2^{2022}}{3^{1011}}-\;{2^{2022}}/{{\sqrt{3}}}$ $=\;{2^{2022}}/{{\sqrt{3}}}-\;{2^{2022}}/{{\sqrt{3}}}=0$

Q69JEE Main 2022MCQ4MDifferential Equations

If the solution curve of the differential equation $\;\frac{d y}{d x}=\;\frac{x+y-2}{x-y}$ passes through the points $(2,1)$ and $( {\text{k}}+1,2), {\text{k}} > 0$ , then [29-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{d y}{d x}=\;\frac{x+y-2}{x-y}=\;\frac{(x-1)+(y-1)}{(x-1)-(y-1)}$ Let $x-1=X, y-1=Y$ $\;\frac{d Y}{d X}=\;\frac{X+Y}{X-Y}$ Let $Y=t X \Rightarrow \;\frac{d Y}{d X}=t+X \;\frac{d t}{d X}$ $t+X \;\frac{d t}{d X}=\;\frac{1+t}{1-t}$ $X \;\frac{d t}{d X}=\;\frac{1+t}{1-t}-t=\;\frac{1+t^2}{1-t}$ $\int \;\frac{1-t}{1+t^2} \text{dt}=\int \;\frac{d X}{X}$ $\tan ^{-1} t-\;\frac{1}{2} \ln (1+t^2) =\ln |X|+c$ $\tan ^{-1} (\;\frac{y-1}{x-1}) -\;\frac{1}{2} \ln (1+ (\;\frac{y-1}{x-1}) ^2) =\ln |x-1|+c$ Curve passes through $(2,1)$ $0-0=0+c \Rightarrow c=0$ If $(k+1,2)$ also satisfies the curve $\tan ^{-1} (\;\frac{1}{k}) -\;\frac{1}{2} \ln (\;\frac{1+k^2}{k^2}) =\ln k$ $2 \tan ^{-1} (\;\frac{1}{k}) =\ln (1+k^2)$

Q70JEE Main 2022MCQ4MDifferential Equations

Let $y=y(x)$ be the solution curve of the differential equation $\;\frac{d y}{d x}+(\;\frac{2 x^{2+}11 x+13}{x^{3+}6 x^{2+}11 x+6}) y=\;\frac{(x+3)}{x+1}, x > -1$ , which passes through the point $(0,1)$ . Then $y(1)$ is equal to: [29-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{d y}{d x}+ (\;\frac{2 x^{2+}11 x+13}{x^{3+}6 x^{2+}11 x+6}) y=\;\frac{(x+3)}{x+1}, x>-1$ Integrating factor I.F. $=e^{\int \;\frac{2 x^{2+}11 x+13}{x^{3+}6 x^{2+}11 x+6} \text{dt}}$ Let $\;\frac{2 x^{2+}11 x+13}{(x+1)(x+2)(x+3)}=\;\frac{A}{x+1}+\;\frac{B}{x+2}+\;\frac{C}{x+3}$ $A=2, B=1, C=-1$ I.F. $=e^{(2 \ln |x+1|+\ln |x+2|-\ln |x+3|)}$ $=\;\frac{(x+1)^2(x+2)}{x+3}$ Solution of differential equation $y \cdot \;\frac{(x+1)^2(x+2)}{x+3}=\int(x+1)(x+2) \text{dt}$ $y \;\frac{(x+1)^2(x+2)}{x+3}=\;\frac{x^3}{3}+\;\frac{3 x^2}{2}+2 x+c$ Curve passes through $(0,1)$ $1 \times \;\frac{1 \times 2}{3}=0+c \Rightarrow c=\;\frac{2}{3}$ So, $y(1)=\;\frac{\;\frac{1}{3}+\frac{3}{2}+2+\frac{2}{3}}{\frac{ (2^2 \times 3) }{4}}=\;\frac{3}{2}$

Q71JEE Main 2022MCQ4MStraight Lines and Pair of Straight Lines

Let $m_1, m_2$ be the slopes of two adjacent sides of a square of side a such that $a^{2+}11 a+3(m_1^{2+}m_2^2)=220$ . If one vertex of the square is $(10(\cos \alpha-\sin \alpha), 10(\sin \alpha+\cos \alpha))$ , where $\alpha \in (0, \;\frac{\pi}{2})$ and the equation of one diagonal is $(\cos \alpha-\sin \alpha) x+(\sin \alpha+\cos \alpha) y-10$ , then $72(\sin ^4 \alpha+\cos ^4 \alpha)+a^{2-}3 a+13$ is equal to: [29-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$m_1 m_2=-1$ $a^{2+}11 a+3 (m_1^{2+}\;\frac{1}{m_1^2}) =220$ Eq. of AC ${\text{AC}}=(\cos \alpha- sin \;\alpha)+( sin \;\alpha+\cos \alpha) {\text{y}}=10$ ${\text{BD}}=( sin \;\alpha-\cos \alpha) {\text{x}}+( sin \;\alpha-\cos \alpha) {\text{y}}=0$ $(10(\cos \alpha- sin \;\alpha), 10( sin \;\alpha-\cos \alpha))$ Slope of ${\text{AC}}= (\;\frac{ sin \;\alpha-\cos \alpha}{ sin \;\alpha+\cos \alpha}) =\tan \theta= {\text{M}}$ Eq. of line making an angle $\pi_4$ with AC ${\text{m}}_1, {\text{m}}_2=\;{ {\text{m}} \pm \tan \frac{\pi}{4}}/{1 \pm {\text{m}} \tan \;\frac{\pi}{4}}$ $=\;\frac{m+1}{1-m}$ or $\;\frac{m-1}{1+m}$ $\;\frac{\;\frac{ sin \;\alpha-\cos \alpha}{ sin \;\alpha+\cos \alpha}+1}{1- (\;\frac{ sin \;\alpha-\cos \alpha}{ sin \;\alpha+\cos \alpha}) }, \;\frac{\;\frac{ sin \;\alpha-\cos \alpha}{ sin \;\alpha+\cos \alpha}-1}{1+\;\frac{ sin \;\alpha-\cos \alpha}{ sin \;\alpha+\cos \alpha}}$ $m_1, m_2=\tan \alpha, \cot \alpha$ mid point of ${\text{AC}} \& {\text{BD}}$ $= {\text{M}}(5(\cos \alpha- sin \;\alpha), 5(\cos \alpha+ sin \;\alpha))$ B $(10(\cos \alpha- sin \;\alpha), 10(\cos \alpha+ sin \;\alpha))$ ${\text{a}}= {\text{AB}}={\sqrt{2}} {\text{BM}}={\sqrt{2}}(5 {\sqrt{2}})=10$ $a=10$ $\because a^{2+}11 a+3 (m_1^{2+}\;\frac{1}{m_1 2}) =220$ $100+110+3 (\tan ^2 \alpha+\cot ^2 \alpha) =220$ Hence $\tan ^2 \alpha=3, \tan ^2 \alpha=\;\frac{1}{3} \Rightarrow \alpha=\;\frac{\pi}{3}$ or $\;\frac{\pi}{6}$ Now $72 ( sin \;^4 \alpha+\cos ^4 \alpha) +a^{2-}3 a+13$ $=72 (\;\frac{9}{16}+\;\frac{1}{16}) +100-30+13$ $=72 (\;\frac{5}{8}) +83=45+83=128$

Q72JEE Main 2022MCQ4MFunctions

The number of elements in the set $S=\{x \in ℝ: 2 \cos (\;\frac{x^{2+}x}{6})=4^x+4^{-x}\}$ is: [29-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$2 \cos (\;\frac{x^{2+}x}{6}) =4^x+4^{-x}$ L.H.S $\le 2$ . & R.H.S. $\ge 2$ Hence L.H.S $=2 &$ R.H.S $=2$ $2 \cos (\;\frac{x^{2+}x}{6}) =2 \;\; 4^x+4^{-x}=2$ Check $x=0$ Possible hence only one solution.

Q73JEE Main 2022MCQ4MCircles

Let ${\text{A}}(\alpha,-2), {\text{B}}(\alpha, 6)$ and ${\text{C}}(\;\frac{\alpha}{4},-2)$ be vertices of a $\triangle {\text{ABC}}$ . If $(5, \;\frac{\alpha}{4})$ is the circumcentre of $\triangle {\text{ABC}}$ , then which of the following is NOT correct about $\triangle {\text{ABC}}$ ? [29-Jul-2022-Shift-2]

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📖 Explanation

Circumcentre of $\triangle A B C$ $= (\;\frac{\alpha+\frac{\alpha}{4}}{2}, \;\frac{6-2}{2})$ $= (\;\frac{5 \alpha}{8}, 2)$ $= (5, \;\frac{\alpha}{4})$ $\Rightarrow \alpha=8$ ${\text{area}}(\triangle A B C)=\;\frac{1}{2} \cdot \;\frac{3 \alpha}{4} \times 8=24$ sq. units Perimeter $=8+\;\frac{3 \alpha}{4}+\sqrt{8^{2+} (\;\frac{3 \alpha}{4}) ^2}$ $=8+6+10=24$ Circumradius $=\;\frac{10}{2}=5$ $r=\;\frac{\Delta }{s}=\;\frac{24}{12}=2$

Q74JEE Main 2022MCQ4MThree Dimensional Geometry

Let $Q$ be the foot of perpendicular drawn from the point $P(1,2,3)$ to the plane $x+2 y+z=14$ . If $R$ is a point on the plane such that $∠ P R Q=60^{\circ}$ , then the area of $\triangle P Q R$ is equal to: [29-Jul-2022-Shift-2]

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📖 Explanation

Length of perpendicular $P Q= |\;{1+4+3-14}/{{\sqrt{6}}}{{}/{}|} ={\sqrt{6}}$ ${\text{QR}}=( {\text{PQ}}) \cot 60^{\circ}={\sqrt{2}}$ $\therefore$ Area of $\triangle {\text{PQR}}=\;\frac{1}{2}( {\text{PQ}})( {\text{QR}})={\sqrt{3}}$

Q75JEE Main 2022MCQ4MThree Dimensional Geometry

If $(2,3,9),(5,2,1),(1, \lambda , 8)$ and $(\lambda , 2,3)$ are coplanar, then the product of all possible values of $\lambda$ is: [29-Jul-2022-Shift-2]

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📖 Explanation

$\because A(2,3,9), B(5,2,1), C(1, \lambda , 8)$ and ${\text{D}}(\lambda , 2,3)$ are coplanar. $\therefore \$ [ {\text{ABACAD}}^{ → }]$=0{\begin{vmatrix}3 & -1 & -8 \ -1 & \lambda -3 & -1 \ \lambda -2 & -1 & -6\end{vmatrix}}=0\Rightarrow [-6(\lambda -3)-1]-8(1-(\lambda -3)(\lambda -2))+(6+(\lambda -2)=0\Rightarrow 3(-6 \lambda +17)-8 (-\lambda ^{2+}5 \lambda -5) +(\lambda +4)=8\Rightarrow 8 \lambda ^{2-}57 \lambda +95=0\therefore \lambda _1 \lambda _2=;\frac{95}{8}$

Q76JEE Main 2022MCQ4MProbability

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is: [29-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

A : Drown ball from boy II is blackB : Red ball transferred $P (\;\frac{B}{A}) =\;\frac{P(A ∩ B)}{P(A)}$ $=\;\frac{\;\frac{3}{9} \times \frac{5}{10}}{\frac{3}{9} \times \;\frac{5}{10}+\;\frac{4}{9} \times \;\frac{6}{10}+\;\frac{3}{9} \times \;\frac{5}{10}}$ $=\;\frac{15}{15+24+15}=\;\frac{15}{54}=\;\frac{5}{18}$

Q77JEE Main 2022MCQ4MComplex Numbers

Let ${\text{S}}=\{z=x+i y:| z-1+i| \ge z|,|z| < 2,|z+i|=|z-1|\}$ . Then the set of all values of $x$ , for which $u=2 x+i y \in {\text{S}}$ for some $y \in ℝ$ , is [29-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

We are given that $u=2x+iy$ is an element of the set $S$ . Therefore, $u$ must satisfy the three conditions defining $S$ . The third condition, $|u+i|=|u-1|$ , geometrically represents the perpendicular bisector of the segment connecting $(1,0)$ and $(0,-1)$ . Substituting $u=2x+iy$ , we get $|2x+i(y+1)|=|(2x-1)+iy|$ . Squaring and simplifying yields the linear relationship $y=-2x$ .

Now, we substitute $y=-2x$ into the other two conditions. The first condition, $|u-1+i| \ge |u|$ , simplifies to $y \ge 2x-1$ . With $y=-2x$ , this becomes $-2x \ge 2x-1$ , which leads to $4x \le 1$ or $x \le \frac{1}{4}$ .

The second condition, $|u|<2$ , becomes $4x^{2+}y^2<4$ . Substituting $y=-2x$ gives $4x^{2+}(-2x)^2<4$ , which simplifies to $8x^2<4$ or $x^2<\frac{1}{2}$ . This implies $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$ .

To find the set of all possible values for $x$ , we must find the intersection of the two derived conditions: $x \le \frac{1}{4}$ and $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$ . Combining these results gives the interval $-\frac{1}{\sqrt{2}} < x \le \frac{1}{4}$ .

Q78JEE Main 2022MCQ4MVector Algebra

Let $\vec{a}, \vec{b}, \vec{c}$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and $(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c})+(\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a})+(\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b})=168$ , then $|\vec{a}|+|\vec{b}|+|\vec{c}|$ is equal to: [29-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$|\vec{a}||\vec{b}||\vec{c}|=14$ $\vec{a} ∧ \vec{b}=\vec{b} ∧ \vec{c}=\vec{c} ∧ \vec{a}=\theta=\;\frac{2 \pi}{3}$ So, $\vec{a} \cdot \vec{b}=-\;\frac{1}{2} a b, \vec{b} \cdot \vec{c}=-\;\frac{1}{2} b c, \vec{a} \cdot \vec{c} .=-\;\frac{1}{2} {\text{ac}}$ (let) $(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c})-(\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{b})$ $=\;\frac{1}{4} a b^2 c+\;\frac{1}{2} a b^2 c=\;\frac{3}{4}{a b^2} c$ Similarly $(\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a})$ $=\;\frac{3}{4} a b c^2$ $(\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b})$ $=\;\frac{3}{4} a^2 b c$ $168$ $=\;\frac{3}{4} a b c(a+b+c)$ So, $(a+b+c)=16$

Q79JEE Main 2022MCQ4MFunctions

The domain of the function $f(x)=\sin ^{-1}(\;\frac{x^{2-}3 x+2}{x^{2+}2 x+7})$ is: [29-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$f(x)= sin \;^{-1} (\;\frac{x^{2-}3 x+2}{x^{2+}2 x+7})$ $-1 \le \;\frac{x^{2-}3 x+2}{x^{2+}2 x+7} \le 1$ $\;\frac{x^{2-}3 x+2}{x^{2+}2 x+7} \le 1$ $x^{2-}3 x+2 \le x^{2+}2 x+7$ $5 x \ge -5$ $x \ge -1$ And $\;\frac{x^{2-}3 x+2}{x^{2+}2 x+7} \ge -1$ $x^{2-}3 x+2 \ge -x^{2-}2 x-7$ $2 x^{2-}x+9 \ge 0$ $x \in R$ (i) $∩$ (ii)Domain $\in [-1, \infty )$

Q80JEE Main 2022MCQ4MSets and Relations

The statement $(p \Rightarrow q) ∨(p \Rightarrow r)$ is NOT equivalent to [29-Jul-2022-Shift-2]

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📖 Explanation

(A) $(p ∧(∼ r)) \Rightarrow q$ $∼(p ∧ ∼ r) ∨ q$ $\equiv (∼ p ∨ r) ∨ q$ $\equiv ∼ p ∨(r ∨ q)$ $\equiv p \to (q ∨ r)$ $\equiv (p \Rightarrow q) ∨(p \Rightarrow r)$ (C) $p \Rightarrow (q ∨ r)$ $\equiv ∼ p ∨(q ∨ r)$ $\equiv (∼ p ∨ q) ∨(∼ p ∨ r)$ $\equiv (p \to q) ∨(p \to r)$ (D) $(p ∧ ∼ q) \Rightarrow r$ $\equiv p \Rightarrow (q ∨ r)$ $\equiv (p \Rightarrow q) ∨(p \Rightarrow r)$

Q81JEE Main 2022NAT4MProbability

The sum and product of the mean and variance of a binomial distribution are $82.5$ and 1350 respectively. Then the number of trials in the binomial distribution is _______. [29-Jul-2022-Shift-2]

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📖 Explanation

Let two roots of a quadratic equation are mean $=n p$ and variance $=n p q$ .Given $n p+n p q=82.5$ and $n p(n p q)=1350$ $\therefore$ Quadratic equation is $x^{2-}82.5 x+1350=0$ $\Rightarrow x^{2-}22.5 x-60 x+1350=0$ $\Rightarrow x-(x-22.5)-60(x-22.5)=0$ Mean $=60$ and Variance $=22.5$ $n p=60, n p q=22.5$ $\Rightarrow q=\;\frac{9}{24}=\;\frac{3}{8}, p=\;\frac{5}{8}$ $\therefore \;\; n \;\frac{5}{8}=60 \;\; \Rightarrow n=96$

Q82JEE Main 2022NAT4MQuadratic Equation and Inequalities

Let $\alpha, \beta(\alpha > \beta)$ be the roots of the quadratic equation $x^{2-}x-4=0$ . If $P_n=\alpha^n-\beta^n, n \in {\text{N}}$ , then $\;\frac{P_{15} P_{16}-P_{14} P_{16}-P_{15}^{2+}P_{14} P_{15}}{P_{13} P_{14}}$ is equal to [29-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\alpha$ and $\beta$ are the roots of the quadratic equation $x^{2-}x-4=0$ . $\therefore \alpha$ and $\beta$ are satisfy the given equation. $\alpha^{2-}\alpha-4=0$ $\Rightarrow \alpha^{n+1}-\alpha^n-4 \alpha^{n-1}=0 . . . . . . . (i)$ and $\beta^{2-}\beta-4=0$ $\Rightarrow \beta^{n+1}-\beta^n-4 \beta^{n-1}=0 \;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)$ Substituting (2) from (1), we get, $(\alpha^{n+1}-\beta^{n+1}) - (\alpha^n-\beta^n) -4 (\alpha^{n-1}-\beta^{n-1}) =0$ $\Rightarrow P_{n+1}-P_n-4 P_{n-1}=0$ $\Rightarrow P_{n+1}=P_n+4 P_{n-1}$ $\Rightarrow P_{n+1}-P_n=4 P_{n-1}$ $\;\text{ For }\; n=14, P_{15}-P_{14}=4 P_{13}$ $\;\text{ For }\; n=15, P_{16}-P_{15}=4 P_{14}$ $\;\text{ Now, }\; \;\frac{P_{15} P_{16}-P_{14} P_{16}-P_{15}^{2+}P_{14} P_{15}}{P_{13} P_{14}}$ $=\;\frac{P_{16} (P_{15}-P_{14}) -P_{15} (P_{15}-P_{14}) }{P_{13} P_{14}}$ $=\;\frac{ (P_{15}-P_{14}) (P_{16}-P_{15}) }{P_{13} P_{14}}$ $=\;\frac{ (4 P_{13}) (4 P_{14}) }{P_{13} P_{14}}$ $=16$

Q83JEE Main 2022NAT4MMatrices and Determinants

Let $X=\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$ and $A-\begin{bmatrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{bmatrix}$ . For ${\text{k}} \in N$ , if $X^{'} A^k X-33$ , then ${\text{k}}$ is equal to ________. [29-Jul-2022-Shift-2]

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📖 Explanation

$X={\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}} ; A={\begin{bmatrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{bmatrix}}$ $X^{ {\text{T}}} A^{ {\text{K}}} X=33$ ${{\begin{bmatrix}1 & 1 & 1\end{bmatrix}}{\begin{bmatrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{bmatrix}}^{ {\text{k}}}{\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}}=33 }$ ${{\begin{bmatrix}1 & 1 & 1\end{bmatrix}}{\begin{bmatrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{bmatrix}}{\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}}=33 }$ As $A^2={\begin{bmatrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{bmatrix}}{\begin{bmatrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{bmatrix}}={\begin{bmatrix}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}}$ $A^4={\begin{bmatrix}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}}{\begin{bmatrix}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}}={\begin{bmatrix}1 & 0 & 12 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}}$ ${\text{A}}^8={\begin{bmatrix}1 & 0 & 24 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}}$ $A^{10}={\begin{bmatrix}1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}}{\begin{bmatrix}1 & 0 & 24 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}}={\begin{bmatrix}1 & 0 & 30 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}}$ for ${\text{K}} \to {\text{Even}} {\text{A}}^{ {\text{K}}}={\begin{bmatrix}1 & 0 & 3 {\text{K}} \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}}$ ${\text{X}}^{ {\text{T}}} {\text{A}}^{ {\text{K}}} {\text{X}}=33$ (This is not correct) ${{\begin{bmatrix}1 & 1 & 1\end{bmatrix}}{\begin{bmatrix}1 & 0 & 3 {\text{K}} \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}}{\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}}}$ $={\begin{bmatrix}1 & 1 & 3 {\text{K}}+1\end{bmatrix}}{\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}}=[3 {\text{K}}+3]$ $\therefore 3 {\text{K}}+3=33 \therefore {\text{K}}=10$ But it should be dropped as 33 is not matrixIf $K$ is odd ${\text{X}}^{ {\text{T}}} {\text{A}}^{ {\text{K}}} {\text{X}}=33$ ${\text{X}}^{ {\text{T}}} {\text{AA}}^{ {\text{K}}-1} {\text{X}}=33$ ${{\begin{bmatrix}1 & 1 & 1\end{bmatrix}}{\begin{bmatrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{bmatrix}}{\begin{bmatrix}1 & 0 & 3 {\text{k}}-3 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}}{\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}}=33}$ ${{\begin{bmatrix}-1 & 3 & 8\end{bmatrix}}{\begin{bmatrix}3 {\text{k}}-2 \\ 1 \\ 1\end{bmatrix}}=[33]}$ ${[-3 {\text{k}}+13]=[33]}$ ${\text{k}}=20 / 3 \;\text{ (not possible) }\;$

Q84JEE Main 2022NAT4MPermutations and Combinations

The number of natural numbers lying between 1012 and 23421 that can be formed using the digits $2,3,4,5,6$ (repetition of digits is not allowed) and divisible by 55 is _______. [29-Jul-2022-Shift-2]

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📖 Explanation

4 digit numbersFor divisibility by 55 , no. should be div. by 5 and 11 bothAlso, for divisibility by 11 $a+c=b+5$ for $b=1 \;\; a=2, \;\; c=4$ ${\text{a}}=4, \;\; {\text{c}}=2$ for $b=2 \;\; a=3, \;\; c=4$ ${\text{a}}=4, \;\; {\text{c}}=3$ for ${\text{b}}=3 \;\; {\text{a}}=6, \;\; {\text{c}}=2$ $a=2, \;\; c=6$ $\therefore 6$ possible four digit no.s are div. by 55(II) 5 digit number is not possible

Q85JEE Main 2022NAT4MBinomial Theorem

If $sum_{k=1}^{10} K^2(10_{C_K})^2=22000 L$ , then $L$ is equal to _______. [29-Jul-2022-Shift-2]

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📖 Explanation

Given, $sum_{k=1}^{10} k^2 ({ }^{10} C_k) ^2=2200 L$ $\Rightarrow sum_{k=1}^{10} (k \cdot { }^{10} C_k) ^2=22000 L$ $\Rightarrow sum_{k=1}^{10} (k \cdot \;\frac{10}{k} \cdot { }^9 C_{k-1}) ^2=22000 L$ $\Rightarrow sum_{k=1}^{10} (10 \cdot { }^9 C_{k-1}) ^2=22000 L$ $\Rightarrow 100 \cdot sum_{k=1}^{10} ({ }^9 C_{k-1}) ^2=22000 L$ $\Rightarrow 100 ( ({ }^9 C_0) ^{2+} ({ }^9 C_1) ^{2+}. . .+ ({ }^9 C_9) ^2) =22000 L$ $\Rightarrow 100 ({ }^{18} C_9) =22000 L$ [Note : $({ }^n C_1) ^{2+} ({ }^n C_2) ^{2+}. . .+ ({ }^n C_n) ^2={ }^{2 n} C_n]$ $\Rightarrow 100 \times \;\frac{18 !}{9 ! 9 !}=22000 L$ $\Rightarrow L=221$

Q86JEE Main 2022NAT4MLimits, Continuity and Differentiability

If $[t]$ denotes the greatest integer $\le t$ , then the number of points, at which the function $f(x)=4 |{}/{}2 x+3|+9[x+\;\frac{1}{2}]-12[x+20]$ is not differentiable in the open interval $(-20,20)$ , is ________. [29-Jul-2022-Shift-2]

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📖 Explanation

$f(x)=4|2 x+3|+9 [x+\;\frac{1}{2}] -12[x+20]$ $=4|2 x+3|+9 [x+\;\frac{1}{2}] -12[x]-240$ $f(x)$ is non differentiable at $x=-\;\frac{3}{2}$ and $f(x)$ is discontinuous at $\{-19,-18, . . ., 18,19\}$ as well as $\{-\;\frac{39}{2},-\;\frac{37}{2}, . . .,-\;\frac{3}{2},-\;\frac{1}{2}, \;\frac{1}{2}, . . ., \;\frac{39}{2}\}$ ,at same point they are also non differentiable $\therefore$ Total number of points of non differentiability $=39+40$ $=79$

Q87JEE Main 2022NAT4MApplication of Derivatives

If the tangent to the curve $y=x^{3-}x^{2+}x$ at the point $(a, b)$ is also tangent to the curve $y=5 x^{2+}2 x-25$ at the point $(2,-1)$ , then $|2 a+9 b|$ is equal to ________. [29-Jul-2022-Shift-2]

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📖 Explanation

Slope of tangent to curve $y=5 x^{2+}2 x-25$ $=m= (\;\frac{d y}{d x}) _{{\text{at}}(2,-1)}=22$ $\therefore \;\;$ Equation of tangent $: y+1=22(x-2)$ $\therefore \;\; y=22 x-45 \;\text{. }\;$ Slope of tangent to $y=x^{3-}x^{2+}x$ at point $(a, b)$ $=3 a^{2-}2 a+1$ $3 a^{2-}2 a+1=22$ $3 a^{2-}2 a-21=0$ $\therefore \;\; a=3 \;\text{ or }\;-\;\frac{7}{3}$ Also $b=a^{3-}a^{2+}a$ Then $(a, b)=(3,21)$ or $(-\;\frac{7}{3},-\;\frac{151}{9})$ . $(-\;\frac{7}{3},-\;\frac{151}{9})$ does not satisfy the equation of tangent $\therefore \;\; a=3, b=21$ $\therefore |2 a+9 b|=195$

Q88JEE Main 2022NAT4MCircles

Let $A B$ be a chord of length 12 of the circle $(x-2)^{2+}(y+1)^2=\;\frac{169}{4}$ . If tangents drawn to the circle at points $A$ and $B$ intersect at the point $P$ , then five times the distance of point $P$ from chord $A B$ is equal to ________. [29-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Here $A M=B M=6$ $O M=\sqrt{ (\;\frac{13}{2}) ^{2-}6^2}=\;\frac{5}{2}$ $sin \;\theta=\;\frac{12}{13}$ In $\triangle P A O$ : $\;\frac{P O}{O A}=\ sec \theta$ $P O=\;\frac{13}{2} \cdot \;\frac{13}{5}=\;\frac{169}{10}$ $\therefore P M=\;\frac{169}{10}-\;\frac{5}{2}=\;\frac{144}{10}=\;\frac{72}{5}$ $\therefore 5 P M=72$

Q89JEE Main 2022NAT4MVector Algebra

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}+\vec{b}|^2=|\vec{a}|^{2+}2|\vec{b}|^2, \vec{a} \cdot \vec{b}=3$ and $|\vec{a} \times \vec{b}|^2=75$ . Then $|\vec{a}|^2$ is equal to _______. [29-Jul-2022-Shift-2]

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📖 Explanation

$\because |\vec{a}+˙{b}|^2=|\vec{a}|^{2+}2|b|^2$ or $|\vec{a}|^{2+}|\vec{b}|^{2+}2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2+}2|\vec{b}|^2$ $\therefore |\vec{b}|^2=6$ Now $|\vec{a} \times \vec{b}|^2=|\vec{a}|^2|\vec{b}|^{2-}(\vec{a} \cdot \vec{b})^2$ $75=|\vec{a}|^2 \cdot 6-9$ $\therefore |\vec{a}|^2=14$

Q90JEE Main 2022NAT4MEllipse

Let $S=\{(x, y) \in ℕ \times ℕ: 9(x-3)^{2+}16(y-4)^2 \le 144\}$ and $T=\{(x, y) \in ℝ \times ℝ:(x-7)^{2+}(y-4)^2 \le 36\}$ . Then $n(S ∩ T)$ is equal to _________. [29-Jul-2022-Shift-2]

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📖 Explanation

$S: \;\frac{(x-3)^2}{16}+\;\frac{(y-4)^2}{9} \le 1 ; x, y \in \{1,2,3, . . . . . .\}$ $T:(x-7)^{2+}(y-4)^2 \le 36 x, y \in R$ Let $x-3=x: y-4=y$ ${\text{S}}: \;{ {\text{x}}^2}/{16}+\;{ {\text{y}}^2}/{9} \le 1 ; {\text{x}} \in \{-2,-1,0,1, . . . . . .\}$ ${\text{T}}:( {\text{x}}-4)^{2+} {\text{y}}^2 \le 36 ; {\text{y}} \in \{-3,-2,-1,0, . . . . . .$ . ${\text{S}} ∩ {\text{T}}=(-2,0),(-1,0), . . . . .(4,0) \to (7)$ $(-1,1),(0,1), . . . . . . .(3,1) \to (5)$ $(-1,-1),(0,-1), . . . . . . .(3,-1) \to (5)$ $(-1,2),(0,2),(1,2),(2,2) \to (4)$ $(-1,-2),(0,-2),(1,-2),(2,-2) \to (4)$ $(0,3)(0,-3) \to (2)$

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