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27 Jul 2022 Shift 2

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Physics30 questions

Q1JEE Main 2022MCQ4MUnits and Measurements

An expression of energy density is given by $u=\;\frac{\alpha}{\beta} \sin (\;\frac{\alpha x}{k t})$ , where $\alpha, \beta$ are constants, $x$ is displacement, $k$ is Boltzmann constant and ${\text{t}}$ is the temperature. The dimensions of $\beta$ will be : [27-Jul-2022-Shift-2]

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📖 Explanation

$u=\;\frac{\alpha}{\beta} \ sin (\;\frac{\alpha x}{k t})$ ${[\alpha]=[\;\frac{k t}{x}]=\;\frac{[\;\text{ Energy }\;]}{[\;\text{ Dis }\; \tan c e]} }$ ${[\beta]=\;\frac{[\alpha]}{[u]} }$ $=\;\frac{[\;\text{ Energy }\;] /[\;\text{ Distan }\; c e]}{[\;\text{ Energy }\;] / \;\text{ Volume }\;]}$ $=[L^2]$

Q2JEE Main 2022MCQ4MMotion in a Plane

A body of mass $10 {\text{kg}}$ is projected at an angle of $45^{\circ}$ with the horizontal. The trajectory of the body is observed to pass through a point $(20,10)$ . If ${\text{T}}$ is the time of flight, then its momentum vector, at time ${\text{t}}=\;{ {\text{T}}}/{\sqrt{2}}$ , is [Take $. {\text{g}}=10 {\text{m}} / {\text{s}}^2]$ __________. [27-Jul-2022-Shift-2]

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📖 Explanation

$m=10 {\text{kg}}$ $\theta=45^{\circ}$ $y=x \tan \theta(1-\;\frac{x}{R})$ $\Rightarrow 10=20(1-\;\frac{20}{R})$ $\Rightarrow R=40$ $40=\;\frac{u^2}{10} \Rightarrow u=20$ $\Rightarrow T=\;{20 \times 20 \times \frac{1}{\sqrt{2}}}/{10}=\;\frac{4}{\sqrt{2}} s \Rightarrow t=2 s$ at $t=2, \vec{v}=(10 \sqrt{2} \hat{i})+(10 \sqrt{2}-2 \times 10)$ $\Rightarrow \vec{p}=10[10 \sqrt{2} \hat{i}+(10 \sqrt{2}-20) \hat{j}]$ $=100 \sqrt{2} \hat{i}+(100 \sqrt{2}-200) \hat{j}$

Q3JEE Main 2022MCQ4MLaws of Motion

A block of mass $M$ slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is $\theta$ . The magnitude of the contact force will be : [27-Jul-2022-Shift-2]

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📖 Explanation

${\text{N}}= {\text{Mg}} \cos \theta$ ${\text{f}}= {\text{Mg}} \ sin \theta$ ${\text{R}}=\sqrt{ {\text{N}}^{2+} {\text{f}}^2}$ ${\text{R}}= {\text{Mg}}$

Q4JEE Main 2022MCQ4MLaws of Motion

A block 'A' takes $2 {\text{s}}$ to slide down a frictionless incline of $30^{\circ}$ and length 'I', kept inside a lift going up with uniform velocity 'v'. If the incline is changed to $45^{\circ}$ , the time taken by the block, to slide down the incline, will be approximately : [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{a}}= {\text{g}} \ sin \theta$ $ℓ=\;\frac{1}{2} {\text{g}} \ sin 30^{\circ}(2)^2$ $ℓ=\;\frac{1}{2} {\text{g}} \ sin 45^{\circ} {\text{t}}^2$ $(\;\frac{1}{2})(4)=\;\frac{1}{\sqrt{2}} {\text{t}}^2 \Rightarrow {\text{t}}=\sqrt{2 \sqrt{2}} ≃ 1.68$

Q5JEE Main 2022MCQ4MMotion in a Straight Line

The velocity of the bullet becomes one third after it penetrates $4 {\text{cm}}$ in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at $(4+x) {\text{cm}}$ inside the block. The value of $x$ is : [27-Jul-2022-Shift-2]

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📖 Explanation

$(\;{ {\text{V}}}/{3})^2= {\text{V}}^{2-}2 {\text{a}}(4) \Rightarrow {\text{a}}=\;{8 {\text{V}}^2}/{9(8)}=\;{ {\text{V}}^2}/{9}$ $0= {\text{V}}^{2-}2 {\text{a}}(4+ {\text{x}})$ $\Rightarrow \;\; {\text{V}}^2=2(\;{ {\text{V}}^2}/{9})(4+ {\text{x}})$ $4.5=4+ {\text{x}}$ ${\text{x}}=0.5$

Q6JEE Main 2022MCQ4MGravitation

A body of mass ${\text{m}}$ is projected with velocity $\lambda v_{ {\text{e}}}$ in vertically upward direction from the surface of the earth into space. It is given that $v_{ {\text{e}}}$ is escape velocity and $\lambda < 1$ . If air resistance is considered to be negligible, then the maximum height from the centre of earth, to which the body can go, will be : ({\text{R}}$ : radius of earth) [27-Jul-2022-Shift-2]

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📖 Explanation

$-\;{ {\text{GMm}}}/{ {\text{R}}}+\;\frac{1}{2} {\text{m}} \lambda ^2 {\text{V}}_{ {\text{e}}}^2=-\;{ {\text{GMm}}}/{ {\text{h}}}$ $-\;{ {\text{GMm}}}/{ {\text{R}}}+\;\frac{1}{2} \lambda ^2 \;{2 {\text{GMm}}}/{ {\text{R}}}=-\;{ {\text{GMm}}}/{ {\text{h}}}$ $\;{\lambda ^2}/{ {\text{R}}}-\;{1}/{ {\text{R}}}=\;{-1}/{ {\text{h}}}$ $\;{1}/{ {\text{h}}}=\;{1-\lambda ^2}/{ {\text{R}}}$ ${\text{h}}=\;{ {\text{R}}}/{1-\lambda ^2}$

Q7JEE Main 2022MCQ4MProperties of Matter

A steel wire of length $3.2 {\text{m}}( {\text{Y}}_{ {\text{s}}}=2.0 \times 10^{11} {\text{Nm}}^{-2})$ and a copper wire of length $4.4 {\text{m}}( {\text{Y}}_{ {\text{c}}}=1.1 \times 10^{11} {\text{Nm}}^{-2})$ , both of radius $1.4 {\text{mm}}$ are connected end to end. When stretched by a load, the net elongation is found to be $1.4 {\text{mm}}$ . The load applied, in Newton, will be: ( Given $.\pi=\;\frac{22}{7})$ [27-Jul-2022-Shift-2]

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📖 Explanation

$\Delta ℓ_1+\Delta ℓ_2=\Delta ℓ$ $\;{ {\text{F}} ℓ_1}/{ {\text{A}}_1 {\text{y}}_1}+\;{ {\text{F}} ℓ_2}/{ {\text{A}}_2 {\text{y}}_2}=\Delta ℓ$ ${\text{F}}=\;{\Delta ℓ}/{{ℓ_1}/{ {\text{A}}_1 {\text{y}}_1}+\;{ℓ_2}/{ {\text{A}}_2 {\text{y}}_2}}=1.54 \times 10^2=154$

Q8JEE Main 2022MCQ4MThermodynamics

In $1^{\;\text{st }\;}$ case, Carnot engine operates between temperatures $300 {\text{K}}$ and $100 {\text{K}} . \ln 2^{\;\text{nd }\;}$ case, as shown in the figure, a combination of two engines is used. The efficiency of this combination (in $2^{\;\text{nd }\;}$ case) will be : [27-Jul-2022-Shift-2]

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📖 Explanation

First case $: η=1-\;\frac{100}{300}=\;\frac{2}{3}$ Second case $: η_{\;\text{net }\;}=η_1+η_2-η_1 η_2$ $η_1=1-\;\frac{200}{300}=\;\frac{1}{3}$ $η_2=1-\;\frac{100}{200}=\;\frac{1}{2}$ $η_{\;\text{net }\;}=\;\frac{1}{3}+\;\frac{1}{2}-\;\frac{1}{6}=\;\frac{2}{3}$ $η($ first case $)=η($ second case $)$

Q9JEE Main 2022MCQ4MThermodynamics

Which statements are correct about degrees of freedom ? (A) A molecule with $n$ degrees of freedom has $n^2$ different ways of storing energy. (B) Each degree of freedom is associated with $\;\frac{1}{2}$ RT average energy per mole. (C) A monatomic gas molecule has 1 rotational degree of freedom where as diatomic molecule has 2 rotational degrees of freedom. (D) ${\text{CH}}_4$ has a total of 6 degrees of freedom. Choose the correct answer from the options given below : [27-Jul-2022-Shift-2]

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Q10JEE Main 2022MCQ4MElectrostatics

A charge of $4 \mu {\text{C}}$ is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be : [27-Jul-2022-Shift-2]

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📖 Explanation

${\text{F}}=\;{ {\text{Kq}}(4- {\text{q}})}/{ {\text{d}}^2}$ $\;{ {\text{dF}}}/{ {\text{dq}}}=\;{ {\text{K}}}/{ {\text{d}}^2}[4-2 {\text{q}}]=0$ ${\text{q}}=2$

Q11JEE Main 2022MCQ4MCurrent Electricity

(A) The drift velocity of electrons decreases with the increase in the temperature of conductor. (B) The drift velocity is inversely proportional to the area of cross-section of given conductor. (C) The drift velocity does not depend on the applied potential difference to the conductor. (D) The drift velocity of electron is inversely proportional to the length of the conductor. (E) The drift velocity increases with the increase in the temperature of conductor. Choose the correct answer from the options given below : [27-Jul-2022-Shift-2]

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📖 Explanation

Drift velocity $=(\;\frac{e \tau }{m}) E$ ${\text{v}}_{ {\text{d}}}=(\;{ {\text{e}} \tau }/{ {\text{m}}})(\;{\Delta {\text{V}}}/{ℓ})$ $\Delta {\text{V}}=$ Potential difference applied across the wireAs temperature increases, relaxation time decreases, hence ${\text{V}}_{ {\text{d}}}$ decreases.As per formula, ${\text{V}}_{ {\text{d}}} \propto \;\frac{1}{ℓ}$ ${\text{v}}_{ {\text{d}}}=\;{ {\text{I}}}/{\;\text{ neA }\;}$ , as it is not mentioned that current is at steady state neither it is mentioned that ${\text{n}}$ is constant for given conductor. So it can't be said that ${\text{v}}_{ {\text{d}}}$ is inversely proportional to ${\text{A}}$ . ${\text{I}}=n e A v_d=\;\frac{V}{R}=\;{ {\text{V}}}/{\rho ℓ} {\text{A}}$ ${\text{v}}_{ {\text{d}}}=\;{ {\text{V}}}/{\rho ℓ \;\text{ ne }\;} \;\; ( {\text{E}}=\;{ {\text{V}}}/{ℓ})$ ${\text{v}}_{ {\text{d}}}=\;{ {\text{eE}} \tau }/{ {\text{m}}}$ $\tau$ decrease with temperature increase.First and fourth statements are correct.

Q12JEE Main 2022MCQ4MMagnetism and Matter

A compass needle of oscillation magnetometer oscillates 20 times per minute at a place ${\text{P}}$ of ${\text{dip}} 30^{\circ}$ . The number of oscillations per minute become 10 at another place ${\text{Q}}$ of $60^{\circ}$ dip. The ratio of the total magnetic field at the two places $(B_Q: B_P)$ is : [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$T=2 \pi \sqrt{\;{ {\text{I}}}/{ {\text{B}}_{ {\text{H}}} {\text{M}}}}$ $T_1=3 \ sec =2 \pi \sqrt{\;\frac{I}{(B_P \cos 30^{\circ}) M}}$ ${\text{T}}_2=6 {\ sec}=2 \pi \sqrt{\;{ {\text{I}}}/{( {\text{B}}_{ {\text{Q}}} \cos 60^{\circ}) {\text{M}}}}$ $\;\frac{3}{6}=\sqrt{\;{1}/{( {\text{B}}_{ {\text{P}}} \;\frac{\sqrt{3}}{2})} \times ( {\text{B}}_{ {\text{Q}}} / 2)}$ $\;\frac{3}{6}=\sqrt{(\;{B_{ {\text{Q}}}}/{\sqrt{3} B_P})}$ $\;\frac{\sqrt{3}}{4}=\;{B_{ {\text{Q}}}}/{ {\text{B}}_{ {\text{P}}}}$ $B_Q: B_P=\sqrt{3}: 4$

Q13JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

A cyclotron is used to accelerate protons. If the operating magnetic field is $1.0 {\text{T}}$ and the radius of the cyclotron 'dees' is $60 {\text{cm}}$ , the kinetic energy of the accelerated protons in ${\text{MeV}}$ will be : $[use ${\text{m}}_{ {\text{p}}}=1.6 × 10^{-27} {\text{kg}}, {\text{e}}=1.6 × 10^{-19} {\text{C}}$ ]$ [27-Jul-2022-Shift-2]

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📖 Explanation

Kinetic energy of electron in cyclotron $=[\;{ {\text{q}}^2 {\text{B}}^2 {\text{r}}_0^2}/{2 {\text{m}}}]$ $=18 {\text{MeV}}$

Q14JEE Main 2022MCQ4MAlternating Current

A series ${\text{LCR}}$ circuit has ${\text{L}}=0.01 {\text{H}}, {\text{R}}=10 Ω$ and ${\text{C}}=1 \mu {\text{F}}$ and it is connected to ac voltage of amplitude $( {\text{V}}_{ {\text{m}}}) 50 {\text{V}}$ . At frequency $60 \%$ lower than resonant frequency, the amplitude of current will be approximately : [27-Jul-2022-Shift-2]

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📖 Explanation

Resonant frequency, $\omega _0=\;{1}/{\sqrt{ {\text{LC}}}}=10^4 {\text{rad}} / {\ sec}$ $\omega ^{'}=.4 \times 10^4=4000 {\text{rad}} / {\ sec}$ ${\text{i}}_0=\;{ {\text{V}}_0}/{\sqrt{ {\text{R}}^{2+}( {\text{X}}_{ {\text{C}}}^{'}- {\text{X}}_{ {\text{L}}}^{'})^2}}=238 {\text{mA}}$

Q15JEE Main 2022MCQ4MElectromagnetic Waves

Identify the correct statements from the following descriptions of various properties of electromagnetic waves. (A) In a plane electromagnetic wave electric field and magnetic field must be perpendicular to each other and direction of propagation of wave should be along electric field or magnetic field. (B) The energy in electromagnetic wave is divided equally between electric and magnetic fields. (C) Both electric field and magnetic field are parallel to each other and perpendicular to the direction of propagation of wave. (D) The electric field, magnetic field and direction of propagation of wave must be perpendicular to each other. (E) The ratio of amplitude of magnetic field to the amplitude of electric field is equal to speed of light. Choose the most appropriate answer from the options given below : [27-Jul-2022-Shift-2]

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📖 Explanation

In an EM wave:1. $\vec{E} ⟂ \vec{B}$ 2. $\vec{V} \equiv \vec{E} \times \vec{B}$ 3. Energy is equally divided4. $|\vec{V}|=|\vec{E}| /|\vec{B}|$

Q16JEE Main 2022MCQ4MWave Optics

Two coherent sources of light interfere. The intensity ratio of two sources is $1: 4$ . For this interference pattern if the value of $\;\frac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }}$ is equal to $\;\frac{2 \alpha+1}{\beta+3}$ , then $\;\frac{\alpha}{\beta}$ will be : [27-Jul-2022-Shift-2]

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📖 Explanation

$\;{ {\text{I}}_1}/{ {\text{I}}_2}=\;\frac{1}{4}$ ${\text{I}}_2=4 {\text{I}}_1$ ${\text{I}}_{\max }= {\text{I}}_1+4 {\text{I}}_1+2 \sqrt{ {\text{I}}_1 4 {\text{I}}_1}=9 {\text{I}}_1$ ${\text{I}}_{\min }= {\text{I}}_1+4 {\text{I}}_1-2 \sqrt{ {\text{I}}_1 4 {\text{I}}_1}= {\text{I}}_1$ $\therefore \;{9 {\text{I}}_1+ {\text{I}}_1}/{9 {\text{I}}_1- {\text{I}}_1}=\;\frac{10}{8}=\;\frac{5}{4}=\;\frac{2 \alpha+1}{\beta+1}$ $\alpha=2 \;\; \beta=1$ $\therefore \;\frac{\alpha}{\beta}=\;\frac{2}{1}=2$

Q17JEE Main 2022MCQ4MDual Nature of Matter and Radiation

With reference to the observations in photo-electric effect, identify the correct statements from below (A) The square of maximum velocity of photoelectrons varies linearly with frequency of incident light. (B) The value of saturation current increases on moving the source of light away from the metal surface. (C) The maximum kinetic energy of photo-electrons decreases on decreasing the power of LED (light emitting diode) source of light. (D) The immediate emission of photo-electrons out of metal surface can not be explained by particle nature of light/electromagnetic waves. (E) Existence of threshold wavelength can not be explained by wave nature of light / electromagnetic waves. Choose the correct answer from the options given below : [27-Jul-2022-Shift-2]

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📖 Explanation

$\because \;\frac{1}{2} m v_m^2=h ν-\varphi$ $\Rightarrow v_m^2$ varies linearly with frequency.And, threshold wavelength can be explained by particle nature of light.

Q18JEE Main 2022MCQ4MAtoms and Nuclei

The activity of a radioactive material is $6.4 \times 10^{-4}$ curie. Its half life is 5 days. The activity will become $5 \times 10^{-6}$ curie after : [27-Jul-2022-Shift-2]

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📖 Explanation

${\text{A}}_0=6.4 \times 10^{-4}$ Curie ${\text{T}}_{1 / 2}=5 \;\text{ days }\;=\;\frac{\ln 2}{\lambda }$ ${\text{A}}= {\text{A}}_0 {\text{e}}^{-\lambda {\text{t}}}$ $5 \times 10^{-6}=6.4 \times 10^{-4} {\text{e}}^{-\lambda {\text{t}}}$ $\;\frac{5}{6.4} \times 10^{-2}= {\text{e}}^{-\lambda {\text{t}}}$ $7.8 \times 10^{-3}= {\text{e}}^{-\lambda t}$ $\log (7.8 \times 10^{-3})=-\lambda {\text{t}} \ln {\text{e}}$ $\ln (7.8 \times 10^{-3})=-\;{\lambda {\text{n}} 2}/{5} \cdot {\text{t}}$ $\therefore \;\frac{5 \times 4.853}{0.693}= {\text{t}}=35 \;\text{ days }\;$

Q19JEE Main 2022MCQ4MSemiconductor Electronics

For a constant collector-emitter voltage of $8 {\text{V}}$ , the collector current of a transistor reached to the value of $6 {\text{mA}}$ from $4 {\text{mA}}$ , whereas base current changed from $20 \mu {\text{A}}$ to $25 \mu {\text{A}}$ value. If transistor is in active state, small signal current gain (current amplification factor) will be : [27-Jul-2022-Shift-2]

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📖 Explanation

$\beta=\;\frac{\Delta I_C}{\Delta I_B}$ $=\;\frac{(6-4) \times 10^{-3}}{(25-20) \times 10^{-6}}$ $=\;\frac{2}{5} \times 10^3$ $=400$

Q20JEE Main 2022MCQ4MElectromagnetic Waves

A square wave of the modulating signal is shown in the figure. The carrier wave is given by $C(t)=5 \sin (8 \pi t)$ Volt. The modulation index is : [27-Jul-2022-Shift-2]

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📖 Explanation

Modulation Index $\mu =\;\frac{A_m}{A_C}=\;\frac{1}{5}=0.2$ $A_m=$ amp. of modulating signal $A_C=$ amp. of carrier wave

Q21JEE Main 2022NAT4MProperties of Matter

In an experiment to determine the Young's modulus, steel wires of five different lengths $(1,2,3,4$ , and $5 {\text{m}})$ but of same cross section $(2 {\text{mm}}^2)$ were taken and curves between extension and load were obtained. The slope (extension/load) of the curves were plotted with the wire length and the following graph is obtained. If the Young's modulus of given steel wires is $x \times 10^{11} {\text{Nm}}^{-2}$ , then the value of $x$ is _________. [27-Jul-2022-Shift-2]

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📖 Explanation

Slope $=\;{\Delta {\text{l}} / {\text{w}}}/{ {\text{L}}}=\;{\Delta {\text{l}} / {\text{L}}}/{ {\text{W}}}=\;{1}/{ {\text{YA}}}$ $\Rightarrow {\text{Y}}=\;{1}/{(\;\text{ slope }\;) {\text{A}}}$ ${\text{Y}}=\;\frac{1}{2 \times 10^{-6}(0.25 \times 10^{-5})}$ ${\text{Y}}=2 \times 10^{11} {\text{N}} / {\text{m}}^2$

Q22JEE Main 2022NAT4MCurrent Electricity

In the given figure of meter bridge experiment, the balancing length $A C$ corresponding to null deflection of the galvanometer is $40 {\text{cm}}$ . The balancing length, if the radius of the wire ${\text{AB}}$ is doubled, will be _________ ${\text{cm}}$ . [27-Jul-2022-Shift-2]

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📖 Explanation

Even if the radius of wire is doubled, the balancing point would not change as $\;\frac{x}{l-x}=\;\frac{R_1}{R_2}$ , which is not including a term of area.

Q23JEE Main 2022NAT4MRay Optics and Optical Instruments

A thin prism of angle $6^{\circ}$ and refractive index for yellow light $( {\text{n}}_{ {\text{Y}}}) 1.5$ is combined with another prism of angle $5^{\circ}$ and ${\text{n}}_Y=1.55$ . The combination produces no dispersion. The net average deviation $(\delta )$ produced by the combination is $(\;\frac{1}{x})^0$ . The value of $x$ is ________. [27-Jul-2022-Shift-2]

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📖 Explanation

$\delta _{\;\text{net }\;}=\delta _1+\delta _2$ $=|(\mu _1-1) A_1-(\mu _2-1) A_2|$ $=|3^{\circ}-2.75^{\circ}|$ $\delta _{n e t}=\;\frac{1^{\circ}}{4}$ $\Rightarrow x=4$

Q24JEE Main 2022NAT4MElectromagnetic Induction

A conducting circular loop is placed in $X-Y$ plane in presence of magnetic field ${ \vec{\text{B}}=(3 {\text{t}}^3 \hat{j}+3 {\text{t}}^2 \hat{k})$ in SI unit. If the radius of the loop is $1 {\text{m}}$ , the induced emf in the loop, at time, ${\text{t}}=2 {\text{s}}$ is ${\text{n}} \pi {\text{V}}$ . The value of ${\text{n}}$ is ________. [27-Jul-2022-Shift-2]

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📖 Explanation

$\varphi ={ \vec{\text{B}} \cdot { \vec{\text{A}}$ $=(3 {\text{t}}^3 { {\text{j}}^{\^}+3 {\text{t}}^2 { {\text{k}}^{\^}) \cdot (\pi(1)^2 { {\text{k}}^{\^})$ $\varphi =3 {\text{t}}^2 \pi$ $\epsilon_{ {\text{IND}}}=|\;{ {\text{d}} \varphi }/{ {\text{dt}}}|=6 {\text{t}} \pi$ at ${\text{t}}=2, \epsilon_{ {\text{IND}}}=12$

Q25JEE Main 2022NAT4MElectrostatic Potential and Capacitance

As show in the figure, in steady state, the charge stored in the capacitor is _______ $\times 10^{-6} {\text{C}}$ . [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{q}}= {\text{C}} {\text{V}}_{100 Ω}$ $=(1.1 \times 10^{-6})(\;{10}/{ {\text{R}}+ {\text{r}}} {\text{R}})$ $=1.1 \times 10^{-6}(\;\frac{10}{110} \times 100)$ $=10 \mu {\text{C}}$

Q26JEE Main 2022NAT4MElectrostatic Potential and Capacitance

A parallel plate capacitor with width $4 {\text{cm}}$ , length $8 {\text{cm}}$ and separation between the plates of $4 {\text{mm}}$ is connected to a battery of $20 {\text{V}}$ . A dielectric slab of dielectric constant 5 having length $1 {\text{cm}}$ , width $4 {\text{cm}}$ and thickness $4 {\text{mm}}$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be _______ $E_0 {\text{J}}$ . (Where $E_0$ is the permittivity of free space) [27-Jul-2022-Shift-2]

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📖 Explanation

${\text{C}}_{ {\text{eff}}}=[\;\frac{\epsilon_0(7 \times 4)}{4 / 10}+\;\frac{5 \epsilon_0(1 \times 4)}{4 / 10}] \times 10^{-2}$ ${\text{C}}_{ {\text{eff}}}=1.2 \epsilon_0$ Energy $=\;\frac{1}{2} {\text{C}}_{ {\text{eff}}} {\text{V}}^2$ $=\;\frac{1}{2}(1.2) \epsilon_0(20)(20)=240 \epsilon_0$

Q27JEE Main 2022NAT4MWaves

A wire of length $30 {\text{cm}}$ , stretched between rigid supports, has it's $n^{\;\text{th }\;}$ and $(n+1)^{\;\text{th }\;}$ harmonics at 400 ${\text{Hz}}$ and $450 {\text{Hz}}$ , respectively. If tension in the string is $2700 {\text{N}}$ , it's linear mass density is _________ ${\text{kg}} / {\text{m}}$ . [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;{ {\text{nv}}}/{0.6}=400 \& \;{( {\text{n}}+1) {\text{v}}}/{0.6}=450$ $\Rightarrow [\;{0.6 \times 400}/{ {\text{v}}}+1] \;{ {\text{v}}}/{0.6}=450$ $\Rightarrow = {\text{v}}=30$ $\Rightarrow \sqrt{\;{ {\text{T}}}/{\mu }}=30$ $\Rightarrow \;\frac{2700}{\mu }=900=\mu =3$

Q28JEE Main 2022NAT4MProperties of Matter

A spherical soap bubble of radius $3 {\text{cm}}$ is formed inside another spherical soap bubble of radius 6 ${\text{cm}}$ . If the internal pressure of the smaller bubble of radius $3 {\text{cm}}$ in the above system is equal to the internal pressure of the another single soap bubble of radius $r {\text{cm}}$ . The value of $r$ is _________. [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{P}}_2- {\text{P}}_0=\;{4 {\text{T}}}/{6} \; \& {\text{P}}_1- {\text{P}}_2=\;{4 {\text{T}}}/{3}$ $\Rightarrow {\text{P}}_1- {\text{P}}_0=\;{4 {\text{T}}}/{2}=2$

Q29JEE Main 2022NAT4MRotational Motion

A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should be unbinding the strings to achieve a speed of $4 {\text{ms}}^{-1}$ , is ________ cm. $(.$ take $g=10 {\text{ms}}^{-2}$ ) [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

From energy conservation ${\text{mgh}}=\;\frac{1}{2} {\ mv}^{2+}\;\frac{1}{2} {\text{I}} \omega ^2$ ${\text{mgh}}=\;\frac{1}{2} {\ mv}^{2+}\;\frac{1}{2} \;{ {\text{mR}}^2}/{2} \omega ^2$ $10 {\text{h}}=\;\frac{16}{2}+\;\frac{16}{4} \Rightarrow {\text{h}}=1.2 {\text{m}}=120 {\text{cm}}$

Q30JEE Main 2022NAT4MLaws of Motion

Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane $A B$ along its surface with a velocity just sufficient to carry it to the top Point $B$ at a height $10 {\text{m}}$ . After reaching the Point ${\text{B}}$ the block slides down on inclined plane ${\text{BC}}$ . Time it takes to reach to the point $C$ from point $A$ is $t(\sqrt{2}+1)$ s. The value of $t$ is _______. (use ${\text{g}}=10 {\text{m}} / {\text{s}}^2$ ) [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

From E.C. $=\;\frac{1}{2} {\ mv}_0^2= {\text{mgh}}$ ${\text{v}}_0=10 \sqrt{2}$ For ${\text{A}} \to {\text{B}}$ at ${\text{B}}, \;\; {\text{v}}=0$ ${\text{a}}=- {\text{g}} \ sin 45^{\circ}=\;\frac{-10}{\sqrt{2}}$ ${\text{v}}= {\text{u}}+ {\text{at}}_1 \Rightarrow 0=10 \sqrt{2}-\;\frac{10}{\sqrt{2}} {\text{t}}_1 \Rightarrow {\text{t}}_1=2 {\ sec}$ For $B \to C$ ${\ s}=u {\text{t}}_2+\;\frac{1}{2} {\text{at}}_2^2$ $\;\frac{10}{\ sin 30^{\circ}}=\;\frac{1}{2}(10 \ sin 30^{\circ}) {\text{t}}_2^2$ $t_2=2 \sqrt{2}$ So total time ${\text{T}}= {\text{t}}_1+ {\text{t}}_2$ ${\text{T}}= {\text{t}}_1+ {\text{t}}_2$ $=2 \sqrt{2}+2$ $=2(\sqrt{2}+1) {\ sec}$

Chemistry30 questions

Q31JEE Main 2022MCQ4MStructure of Atom

The correct decreasing order of energy for the orbitals having, following set of quantum numbers : (A) $n=3, I=0, m=0$ (B) $n=4, I=0, m=0$ (C) $n=3, I=1, m=0$ (D) $n=3, l=2, m=1$ is : [27-Jul-2022-Shift-2]

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📖 Explanation

(A) ${\text{n}}+ℓ=3+0=3$ (B) ${\text{n}}+ℓ=4+0=4$ (C) ${\text{n}}+ℓ=3+1=4$ (D) ${\text{n}}+ℓ=3+2=5$ Higher ${\text{n}}+ℓ$ value, higher the energy & if same ${\text{n}}+ℓ$ value, then higher ${\text{n}}$ value, higher the energy. Thus : ${\text{D}}> {\text{B}}> {\text{C}}> {\text{A}}$ .

Q32JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

Match List - I with List - II. List I List II (A) $ψ_{ {\text{MO}}}=ψ_{ {\text{A}}}-ψ_{ {\text{B}}}$ (I)Dipole moment (B) $\;\; \mu =Q \times r$ (II)Bonding molecular orbital (C) $\;{ {\text{N}}_{ {\text{b}}}- {\text{N}}_{ {\text{a}}}}/{2}$ (III)Anti-bonding molecular orbital (D) $ψ_{ {\text{MO}}}=ψ_{ {\text{A}}}+ψ_{ {\text{B}}}$ (IV)Bond orderChoose the correct answer from the options given below : [27-Jul-2022-Shift-2]

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📖 Explanation

$Ψ_A-Ψ_B=Ψ_{M O}$ is anti-boding molecular orbital $\mu =Q \times r$ -is dipole moment $\;\frac{N_b-N_a}{2}=$ bond order $Ψ_A+Ψ_B=Ψ_{M O}$ is bonding molecular orbital.

Q33JEE Main 2022MCQ4MIonic Equilibrium

The plot of ${\text{pH}}$ -metric titration of weak base ${\text{NH}}_4 {\text{OH}}$ vs strong acid ${\text{HCl}}$ looks like : [27-Jul-2022-Shift-2]

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📖 Explanation

${\text{NH}}_4 {\text{OH}}$ is a weak base and ${\text{HCl}}$ is a strong acid.With the addition of ${\text{HCl}}$ to ${\text{NH}}_4 {\text{OH}}, {\text{pH}}$ of solution will decrease gradually.So, the correct graph should be

Q34JEE Main 2022MCQ4MElectrochemistry

Given below are two statements : Statement I : For KI, molar conductivity increases steeply with dilution Statement II : For carbonic acid, molar conductivity increases slowly with dilution In the light of the above statements, choose the correct answer from the options given below: [27-Jul-2022-Shift-2]

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Q35JEE Main 2022MCQ4MSolutions

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A) : Dissolved substances can be removed from a colloidal solution by diffusion through a parchment paper. Reason $( {\text{R}})$ : Particles in a true solution cannot pass through parchment paper but the colloidal particles can pass through the parchment paper. In the light of the above statements, choose the correct answer from the options given below : [27-Jul-2022-Shift-2]

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Q36JEE Main 2022MCQ4MClassification of Elements

Outermost electronic configurations of four elements ${\text{A}}, {\text{B}}, {\text{C}}, {\text{D}}$ are given below : (A) $3 s^2$ (B) $3 s^2 3 p^1$ (C) $3 s^2 3 p^3$ (D) $3 s^2 3 p^4$ The correct order of first ionization enthalpy for them is : [27-Jul-2022-Shift-2]

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📖 Explanation

Orbitals with fully filled and half-filled electronic configuration are stable, and require more energy for ionizationElements with greater electronegativity require more energy for ionisationHence the correct order is $C>D>A>B$

Q37JEE Main 2022MCQ4MClassification of Elements

An element $A$ of group 1 shows similarity to an element $B$ belonging to group 2 . If $A$ has maximum hydration enthalpy in group 1 then $B$ is : [27-Jul-2022-Shift-2]

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📖 Explanation

${\text{Li}}^{+} \to$ Maximum hydration enthalpy in group 1 due to small size.So ' $B$ ' is ${\text{Mg}}$ .

Q38JEE Main 2022MCQ4Mp Block Elements

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A) : Boron is unable to form ${\text{BF}}_6^{3-}$ . Reason $( {\text{R}})$ : Size of ${\text{B}}$ is very small. In the light of the above statements, choose the correct answer from the options given below : [27-Jul-2022-Shift-2]

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📖 Explanation

The outer most shell of Boron is 2 and its maximum covalency is 4 .Therefore, boron cannot form ${\text{BF}}_6^{3-}$ .Hence Assertion is correctBoron is the first element of group-13 of modern periodic table. It is very small in size.But it does not provide correct explanation of Assertion

Q39JEE Main 2022MCQ4MRedox Reactions

In neutral or alkaline solution, ${\text{MnO}}_4^{-}$ oxidises thiosulphate to : [27-Jul-2022-Shift-2]

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📖 Explanation

$8 {\text{MnO}}_4^{-}+3 {\text{S}}_2 {\text{O}}_3^{2-}+ {\text{H}}_2 {\text{O}} \;\frac{\;\text{ neutral or }\;}{\;\text{ alk. solution }\;} 8 {\text{MnO}}_2+6 {\text{SO}}_4^{2-}+2 {\text{OH}}^{-}$

Q40JEE Main 2022MCQ4MCoordination Compounds

Low oxidation state of metals in their complexes are common when ligands: [27-Jul-2022-Shift-2]

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📖 Explanation

Ligands like :CO, are sigma donor and $\pi$ -acceptor and they make stronger bond with lower oxidation state metal ion, in this case back bonding is more effective

Q41JEE Main 2022MCQ4MSome Basic Concepts of Chemistry

Given below are two statements: Statement I: The non bio-degradable fly ash and slag from steel industry can be used by cement industry. Statement II : The fuel obtained from plastic waste is lead free. In the light of the above statements, choose the most appropriate answer from the options given below: [27-Jul-2022-Shift-2]

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Q42JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

The structure of ${\text{A}}$ in the given reaction is : [27-Jul-2022-Shift-2]

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Q43JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

Major product B' of the following reaction sequence is : [27-Jul-2022-Shift-2]

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Q44JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Match List - I with List - II. List I List II (A) (I)Gatterman Koch reaction (B) (II)Etard reaction (C) (III)Stephen reaction (D) (IV)Rosenmund reactionChoose the correct answer from the options given below : [27-Jul-2022-Shift-2]

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Q45JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Match List - I with List - II. List I (Polymer) List II (Monomer) (A)Neoprene (I)Acrylonitrile (B)Teflon (II)Chloroprene (C)Acrilan (III)Tetrafluoroethene (D)Natural rubber (IV)IsopreneChoose the correct answer from the options given below : [27-Jul-2022-Shift-2]

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Q46JEE Main 2022MCQ4MAmines

An organic compound ' A ' contains nitrogen and chlorine. It dissolves readily in water to give a solution that turns litmus red. Titration of compound ${ }^{'} {\text{A}}^{'}$ with standard base indicates that the molecular weight of ${ }^{'} {\text{A}}^{'}$ is $131 \pm 2$ . When a sample of ${ }^{'} {\text{A}}^{'}$ is treated with aq. ${\text{NaOH}}$ , a liquid separates which contains ${\text{N}}$ but not ${\text{Cl}}$ . Treatment of the obtained liquid with nitrous acid followed by phenol gives orange precipitate. The compound ' ${\text{A}}^{'}$ is : [27-Jul-2022-Shift-2]

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Q47JEE Main 2022MCQ4MBiomolecules

Match List - I with Match List - II. List I List II (A)Glucose + HI (I)Gluconic acid (B)Glucose + ${\text{Br}}_2$ water (II)Glucose pentacetate (C)Glucose + acetic anhydride (III)Saccharic acid (D)Glucose $+ {\text{HNO}}_3$ (IV)HexaneChoose the correct answer from the options given below: [27-Jul-2022-Shift-2]

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Q48JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Which of the following enhances the lathering property of soap? [27-Jul-2022-Shift-2]

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Q49JEE Main 2022MCQ4MPractical Organic Chemistry

Match List - I with List - II. List I (Mixture) List II (Purification Process) (A)Chloroform & Aniline (I)Steam distillation (B)Benzoic acid & Napthalene (II)Sublimation (C)Water & Aniline (III)Distillation (D)Napthalene & Sodium chloride (IV)CrystallisationChoose the correct answer from the options given below : [27-Jul-2022-Shift-2]

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📖 Explanation

(A) Chloroform + Aniline $\to$ (III) Distillation(B) Benzoic acid $+$ Napthalene $\to$ (IV) Crystallisation(C) Water + Aniline $\to$ (I) Steam distillation(D) Napthalene + Sodium chloride $\to$ (II) Sublimation

Q50JEE Main 2022MCQ4MCoordination Compounds

${\text{Fe}}^{3+}$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of : [27-Jul-2022-Shift-2]

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📖 Explanation

$4 {\text{Fe}}^{3+}+3[ {\text{Fe}}( {\text{CN}})_6]^{-4} \longrightarrow {\text{Fe}}_4[ {\text{Fe}}( {\text{CN}})_6]_3$

Q51JEE Main 2022NAT4MSolutions

The normality of ${\text{H}}_2 {\text{SO}}_4$ in the solution obtained on mixing $100 {\text{mL}}$ of $0.1 {\text{M}} {\text{H}}_2 {\text{SO}}_4$ with $50 {\text{mL}}$ of $0.1 {\text{M}} {\text{NaOH}}$ ________ is $\times 10^{-1} {\text{N}}$ . (Nearest Integer) [27-Jul-2022-Shift-2]

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📖 Explanation

No. of equivalents of ${\text{H}}_2 {\text{SO}}_4=100 \times 0.1 \times 2=20$ No. of equivalents of ${\text{NaOH}}=50 \times 0.1=5$ No. of equivalents of ${\text{H}}_2 {\text{SO}}_4$ left $=20-5=15$ $\Rightarrow 150 \times x=15$ $x=\;\frac{1}{10}=0.1 {\text{N}}=1 \times 10^{-1} {\text{N}}$

Q52JEE Main 2022NAT4MStates of Matter

For a real gas at $25^{\circ} {\text{C}}$ temperature and high pressure ( $99 {\bar}$ ) the value of compressibility factor is 2 , so the value of Vander Waal's constant 'b' should be ________ $\times 10^{-2} {\text{L}} {\text{mol}}^{-1}$ (Nearest integer) (Given ${\text{R}}=0.083 {\text{L}}$ bar ${\text{K}}^{-1} {\text{mol}}^{-1}$ ) [27-Jul-2022-Shift-2]

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📖 Explanation

For real gas under high pressure ${\text{Z}}=1+\;{ {\text{Pb}}}/{ {\text{RT}}}\;\Rightarrow {\text{b}}=\;{ {\text{RT}}}/{ {\text{P}}}$ $\;\; =\;\frac{0.083 \times 298}{99}$ $\;\; =0.25 \times 10^{-2} {\text{L}} {\text{mol}}^{-1}$

Q53JEE Main 2022NAT4MChemical Thermodynamics

A gas (Molar mass $=280 {\text{g}} {\text{mol}}^{-1}$ ) was burnt in excess ${\text{O}}_2$ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from $298.0 {\text{K}}$ to $298.45 {\text{K}}$ . If the heat capacity of calorimeter is $2.5 {\text{kJ}} {\text{K}}^{-1}$ and enthalpy of combustion of gas is $9 {\text{kJ}} {\text{mol}}^{-1}$ then amount of gas burnt is ________ g. (Nearest Integer) [27-Jul-2022-Shift-2]

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📖 Explanation

Let $x g$ is burnt $\;\text{ moles }\;=\;\frac{x}{280}$ heat released by $\;\frac{x}{280}$ mole $=2.5 \times 0.45 {\text{kJ}}$ heat released by 1 mole $=\;\frac{2.5 \times 0.45 \times 280}{x} {\text{kJ}}$ $\Delta {\text{H}}=\Delta {\text{U}}+\Delta {\text{ngRT}}$ $\Delta {\text{H}} ≃ \Delta {\text{U}}$ $9=\;\frac{2.5 \times 280 \times 0.45}{x}$ $x=35 g$

Q54JEE Main 2022NAT4MSolutions

When a certain amount of solid ${\text{A}}$ is dissolved in $100 {\text{g}}$ of water at $25^{\circ} {\text{C}}$ to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is $23.76 {\text{mmHg}}$ . The number of moles of solute ${\text{A}}$ added is _______ . (Nearest Integer) [27-Jul-2022-Shift-2]

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📖 Explanation

$\because$ Diliute solution given: $\;{ {\text{P}}^{0-} {\text{P}}_{ {\text{S}}}}/{ {\text{P}}^0} ∼ \;{{ }^{ {\text{n}}} \;\text{ solute }\;}/{{ }^{ {\text{n}}} \;\text{ solvent }\;}$ $\;{ {\text{P}}^{0-} {\text{P}}^0 / 2}/{ {\text{P}}^0}=\;{{ }^{ {\text{n}}} \;\text{ solute }\;}/{{ }^{ {\text{n}}} \;\text{ solvent }\;}$ ${ }^n$ Solute $∼ \;\frac{{ }^n \;\text{ solvent }\;}{2}=\;\frac{100}{18 \times 2}=2.78 {\text{mol}}$ More accurate approach: $\;{ {\text{P}}^{0-} {\text{P}}_{ {\text{S}}}}/{ {\text{P}}_{ {\text{S}}}}=\;{{ }^{ {\text{n}}} \;\text{ Solute }\;}/{{ }^n \;\text{ solvent }\;}$ $\;{ {\text{P}}^{0-} {\text{P}}^0 / 2}/{ {\text{P}}^0 / 2}=\;{{ }^{ {\text{n}}} \;\text{ solute }\;}/{{ }^n \;\text{ solvent }\;}$ ${ }^{ {\text{n}}}$ Solute $={ }^{ {\text{n}}}$ solvent $=\;\frac{100}{18}=5.55 {\text{mol}}$

Q55JEE Main 2022NAT4MChemical Kinetics

$[A] _{\text{Reactant}} \;\; \to \;\; [B]_{\text{Product}}$ If formation of compound $[ {\text{B}}]$ follows the first order of kinetics and after 70 minutes the concentration of $[ {\text{A}}]$ was found to be half of its initial concentration. Then the rate constant of the reaction is $x \times 10^{-6} {\text{s}}^{-1}$ . The value of $x$ is ________.(Nearest Integer) [27-Jul-2022-Shift-2]

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📖 Explanation

${\text{K}}=\;{0.693}/{ {\text{t}}_{1 / 2}}=\;\frac{0.693}{70 \times 60}$ $=\;\frac{6930}{7 \times 6} \times 10^{-6}$ $=165 \times 10^{-6} {\ s}^{-1}$

Q56JEE Main 2022NAT4Md and f Block Elements

Among the following ores Bauxite, Siderite, Cuprite, Calamine, Haematite, Kaolinite, Malachite, Magnetite, Sphalerite, Limonite, Cryolite, the number of principal ores if iron is _________. [27-Jul-2022-Shift-2]

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📖 Explanation

Bauxite $- {\text{AlO}}_{ {\text{x}}}( {\text{OH}})_{3-2 {\text{x}}}($ where $0< {\text{x}}< 1)$ $✔$ Siderite $- {\text{FeCO}}_3$ Cuprite $- {\text{Cu}}_2 {\text{O}}$ Calamine - ${\text{ZnCO}}_3$ $✔ {\text{Fe}}_2 {\text{O}}_3$ Haematite- $- {\text{Al}}_2( {\text{OH}}_4)_4 {\text{Si}}_2 {\text{O}}_5$ Kaolinite $- {\text{CuCO}}_3 \cdot {\text{Cu}}_2( {\text{OH}})_2$ Malachite $- {\text{Fe}}_3 {\text{O}}_4$ $✔$ Magnetite $- {\text{ZnS}}_2$ Sphalerite $- {\text{Fe}}_2 {\text{O}}_3 \cdot 3 {\text{H}}_2 {\text{O}}$ $✔$ Limonite $- {\text{Na}}_3 {\text{AlF}}_6$

Q57JEE Main 2022NAT4MRedox Reactions

The oxidation state of manganese in the product obtained in a reaction of potassium permanganate and hydrogen peroxide in basic medium is ________. [27-Jul-2022-Shift-2]

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📖 Explanation

$2 {\text{KMnO}}_4+3 {\text{H}}_2 {\text{O}}_2 {\longrightarrow }^{\;\text{ basic medium }\;} 2 {\text{MnO}}_2+3 {\text{O}}_2+2 {\text{H}}_2 {\text{O}}+2 {\text{KOH}}$

Q58JEE Main 2022NAT4MChemical Bonding and Molecular Structure

The number of molecule(s) or ion(s) from the following having non-planar structure is _________. ${\text{NO}}_3^{-}, {\text{H}}_2 {\text{O}}_2, {\text{BF}}_3, {\text{PCl}}_3, {\text{XeF}}_4, {\text{SF}}_4, {\text{XeO}}_3, {\text{PH}}_4^{+}, {\text{SO}}_3,[ {\text{Al}}( {\text{OH}})_4]^{-}$ [27-Jul-2022-Shift-2]

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📖 Explanation

${\text{NO}}_3^{⊖} \to$ Trigonal planar (Planar) ${\text{H}}_2 {\text{O}}_2 \to$ Open book (Non-planar) ${\text{BF}}_3 \to$ Trigonal planar (Planar) ${\text{PCl}}_3 \to$ Pyramidal (Non-planar) ${\text{XeF}}_4 \to$ Square planar (Planar) ${\text{SF}}_4 \to$ See-Saw (Non-planar) ${\text{XeO}}_3 \to$ Pyramidal (Non-planar) ${\text{PH}}_4^{⊕} \to$ Tetrahedral (Non-planar) ${\text{SO}}_3 \to$ Trigonal planar (Planar) $[ {\text{Al}}( {\text{OH}})_4]^{-} \to$ Tetrahedral (Non-planar)

Q59JEE Main 2022NAT4MCoordination Compounds

The spin only magnetic moment of the complex present in Fehling's reagent is _________ B.M. (Nearest integer). [27-Jul-2022-Shift-2]

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📖 Explanation

Fehling solution is a complex of ${\text{Cu}}^{++}$ ${\text{Cu}}^{++}=3 {\text{d}}^9$ No. of unpaired ${\text{e}}^{-}=1$ $\;\text{ M.M }\;=\sqrt{1(1+2)}=\sqrt{3}=1.73 {\text{BM}}$

Q60JEE Main 2022NAT4MAldehydes, Ketones and Carboxylic Acids

In the above reaction, $5 {\text{g}}$ of toluene is converted into benzaldehyde with $92 \%$ yield. The amount of benzaldehyde produced is ________ $\times 10^{-2} {\text{g}}$ . (Nearest integer) [27-Jul-2022-Shift-2]

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📖 Explanation

Moles $=\;\frac{5}{92}$ Moles of benzaldehyde produced $=\;\frac{5}{92} \times 0.92=0.05$ $\therefore$ Mass of benzaldehyde formed $=0.05 \times 106$ $=5.3 {\text{g}}$ $=530 \times 10^{-2}$

Mathematics30 questions

Q61JEE Main 2022MCQ4MFunctions

The domain of the function $f(x)=\sin ^{-1}\$ [2 x^{2-}3]$+\log _2(\log _{;\frac{1}{2}}(x^{2-}5 x+5)) $, where$ [ {\text{t}}]$ is the greatest integer function, is : [27-Jul-2022-Shift-2]

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📖 Explanation

$-1 \le 2 x^{2-}3< 2$ or $2 \le 2 x^2< 5$ or $1 \le x^2< \;\frac{5}{2}$ $x \in (-\sqrt{\;\frac{5}{2}},-1] ∪[1, \sqrt{\;\frac{5}{2}})$ $\log _{\;\frac{1}{2}}(x^{2-}5 x+5)>0$ $0< x^{2-}5 x+5< 1$ $x^{2-}5 x+5>0 \& x^{2-}5 x+4< 0$ $x \in (-\infty , \;\frac{5-\sqrt{5}}{2}) ∪(\;\frac{5+\sqrt{5}}{2}, \infty )$ $\& x \in (-\infty , 1) ∪(4, \infty )$ Taking intersection $x \in (1, \;\frac{5-\sqrt{5}}{2})$

Q62JEE Main 2022MCQ4MComplex Numbers

Let $S$ be the set of all $(\alpha, \beta), \pi < \alpha, \beta < 2 \pi$ , for which the complex number $\;\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely imaginary and $\;\frac{1+i \cos \beta}{1-2 i \cos \beta}$ is purely real. Let $Z_{\alpha \beta}=\sin 2 \alpha+i \cos 2 \beta,(\alpha, \beta) \in S$ . Then $sum_{(\alpha, \beta) \in S}(i Z_{\alpha \beta}+\;\frac{1}{i Z_{\alpha \beta}})$ is equal to : [27-Jul-2022-Shift-2]

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📖 Explanation

$\because \;\frac{1-i \ sin \alpha}{1+2 i \ sin \alpha}$ is purely imaginary $\therefore \;\frac{1-i \ sin \alpha}{1+2 i \ sin \alpha}+\;\frac{1+i \ sin \alpha}{1-2 i \ sin \alpha}=0$ $\Rightarrow 1-2 \ sin ^2 \alpha=0$ $\therefore \alpha=\;\frac{5 \pi}{4}, \;\frac{7 \pi}{4}$ and $\;\frac{1+i \cos \beta}{1-2 i \cos \beta}$ is purely real $\;\frac{1+i \cos \beta}{1-2 i \cos \beta}-\;\frac{1-i \cos \beta}{1+2 i \cos \beta}=0$ $\Rightarrow \cos \beta=0$ $\therefore \beta=\;\frac{3 \pi}{2}$ $\therefore S=\{(\;\frac{5 \pi}{2}, \;\frac{3 \pi}{2}),(\;\frac{7 \pi}{4}, \;\frac{3 \pi}{2})\}$ $Z_{\alpha \beta}=1-i$ and $Z_{\alpha \beta}=-1-i$ $\therefore sum_{(\alpha, \beta) \in S}(i Z_{\alpha \beta}+\;\frac{1}{i \overline{Z}_{\alpha \beta}})=i(-2 i)+\;\frac{1}{i}[\;\frac{1}{1+i}+\;\frac{1}{-1+i}]$ $=2+\;\frac{1}{i} \;\frac{2 i}{-2}=1$

Q63JEE Main 2022MCQ4MQuadratic Equation and Inequalities

If $\alpha, \beta$ are the roots of the equation $x^{2-}(5+3^{\sqrt{\log _3 5}}-5^{\sqrt{\log _5 3}})+3(3^{(\log _3 5)^{\;\frac{1}{3}}}-5^{(\log _5 3)^{\;\frac{2}{3}}}-1)=0$ then the equation, whose roots are $\alpha+\;\frac{1}{\beta}$ and $\beta+\;\frac{1}{\alpha}$ , is : [27-Jul-2022-Shift-2]

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📖 Explanation

$3^{\sqrt{\log _3 5}}-5^{\sqrt{\log _5 3}}=3^{\sqrt{\log _3 5}}-(3^{\log _3 5})^{\sqrt{\log _5 3}}$ $3^{(\log _3 5)^{\;\frac{1}{3}}}-5^{(\log _5 3)^{\;\frac{2}{3}}}=5^{(\log _5 3)^{\;\frac{2}{3}}}-5^{(\log _5 3)^{\;\frac{2}{3}}}=0$ Note: In the given equation ' $x$ ' is missing.So $x^{2-}5 x+3(-1)=0{}^{╱}^{\;\;\;\;\;\;\alpha}_{╲}_{\;\;\;\;\;\;\beta}$ $\alpha+\beta+\;\frac{1}{\alpha}+\;\frac{1}{\beta}=(\alpha+\beta)+\;\frac{\alpha+\beta}{\alpha \beta}$ $=5-\;\frac{5}{3}=\;\frac{10}{3}$ $(\alpha+\;\frac{1}{\beta})(\beta+\;\frac{1}{\alpha})=2+\alpha \beta+\;\frac{1}{\alpha \beta}=2-3-\;\frac{1}{3}=\;\frac{-4}{3}$ So Equation must be option (B).

Q64JEE Main 2022MCQ4MMatrices and Determinants

Let $A=(\text{table} 4 , -2 ; \alpha , \beta)$ . If ${\text{A}}^{2+}\gamma {\text{A}}+18 {\text{I}}= {\text{O}}$ , then ${\det}( {\text{A}})$ is equal to ________. [27-Jul-2022-Shift-2]

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📖 Explanation

Characteristic equation of $A$ is given by $|A-\lambda I|=0$ $\begin{vmatrix}4-\lambda & -2 \\ \alpha & \beta-\lambda \end{vmatrix}=0$ $\Rightarrow \lambda ^{2-}(4+\beta) \lambda +(4 \beta+2 \alpha)=0$ $\;\text{ So, }\; A^{2-}(4+\beta) A+(4 \beta+2 \alpha) I=0$ $|A|=4 \beta+2 \alpha=18$

Q65JEE Main 2022MCQ4MLimits, Continuity and Differentiability

If for ${\text{p}} \ne {\text{q}} \ne 0$ , the function $f(x)=\;{\;^{7}\sqrt{ {\text{p}}(729+x)}-3}/{\;^{3}\sqrt{729+ {\text{q}} x}-9}$ is continuous at $x=0$ , then : [27-Jul-2022-Shift-2]

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📖 Explanation

$f(x)=\;\frac{\;^{7}\sqrt{p(729+x)}-3}{\;^{3}\sqrt{729+q x}-9}$ for continuity at $x=0, \lim _{x \to 0} f(x)=f(0)$ Now, $\therefore \lim _{x \to 0} f(x)=\lim _{x \to 0} \;\frac{\;^{7}\sqrt{p(729+x)}-3}{\;^{3}\sqrt{729+q x}-9}$ $\Rightarrow p=3$ (To make indeterminant form)So, $\lim _{x \to 0} f(x)=\lim _{x \to 0} \;{(3^{7+}3 x)^{\frac{1}{7}}-3}/{(729+q x)^{\;\frac{1}{3}}-9}$ $=\lim _{x \to 0} \;{3\$ [(1+\frac{x}{3^6})^{\frac{1}{7}}-1]$}/{9$[(1+;\frac{q}{729} x)^{;\frac{1}{3}}-1]$}=;\frac{1}{3} \cdot ;\frac{;\frac{1}{7} \cdot \frac{1}{3^6}}{\frac{1}{3} \cdot ;\frac{q}{729}}\therefore f(0)=;\frac{1}{7 q}\therefore $Option$ (B)$ is correct.

Q66JEE Main 2022MCQ4MDefinite Integrals

Let $f(x)=2+|x|-|x-1|+|x+1|, x \in R$ . Consider (S1): $f^{'}(-\;\frac{3}{2})+f^{'}(-\;\frac{1}{2})+f^{'}(\;\frac{1}{2})+f^{'}(\;\frac{3}{2})=2$ (S2) : $int_{-2}^2 f(x) {\text{d}} x=12$ Then, [27-Jul-2022-Shift-2]

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📖 Explanation

$f(x)=2+|x|-|x-1|+|x+1|, x \in R$ $\therefore f(x)=\begin{cases}-x & {, \\ \\ } , x< -1 \\ x+2 & {, \\ \\ } , -1 \le x< 0 \\ 3 x+2 & {, \\ \\ } , 0 \le x< 1 \\ x+4 & {, \\ \\ } , x \ge 1\end{cases}$ $\therefore f^{'}(-\;\frac{3}{2})+f^{'}(-\;\frac{1}{2})+f^{'}(\;\frac{1}{2})+f^{'}(\;\frac{3}{2})=-1+1+3+1=4$ $\;\text{ and }\; int_{-2}^{2} f(x) d x=int_{-2}^{-1} f(x) d x+int_{-1}^{0} f(x) d x+int_{0^1 f(x) d x+int_1}^{2 }f(x) d x$ $=\$ [-;\frac{x^2}{2}]2^{-1}+$[;\frac{(x+2)^2}{2}]{-1}^{0+}$[;\frac{(3 x+2)^2}{6}]_0^{1+}$[;\frac{(x+4)^2}{2}]_1^2=;\frac{3}{2}+;\frac{3}{2}+;\frac{7}{2}+;\frac{11}{2}=;\frac{24}{2}=12\therefore $ Only (S2) is correct

Q67JEE Main 2022MCQ4MSequences and Series

Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5 . Let the sum of its first five terms be $\;\frac{98}{25}$ . Then the sum of the first 21 terms of an AP, whose first term is $10 a r, n^{\;\text{th }\;}$ term is ${\text{a}}_{ {\text{n}}}$ and the common difference is $10 {\text{ar}}^2$ , is equal to : [27-Jul-2022-Shift-2]

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📖 Explanation

Let first term of G.P. be a and common ratio is $r$ Then, $\;\frac{a}{1-r}=5 . . . . . .$ (i) $a \;\frac{(r^{5-}1)}{(r-1)}=\;\frac{98}{25} \Rightarrow 1-r^5=\;\frac{98}{125}$ $\therefore r^5=\;\frac{27}{125}, r=(\;\frac{3}{5})^{\;\frac{3}{5}}$ $\therefore \;\text{ Then, }\; S_{21}=\;\frac{21}{2}[2 \times 10 a r+20 \times 10 a r^2]$ $=21[10 a r+10.10 a r^2]$ $=21 a_{11}$

Q68JEE Main 2022MCQ4MArea Under The Curves

The area of the region enclosed by $y \le 4 x^2, x^2 \le 9 y$ and $y \le 4$ , is equal to : [27-Jul-2022-Shift-2]

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📖 Explanation

$y \le 4 x^2, x^2 \le 9 y, y \le 4$ So, required area $A=2 int_{0}^{4}(3 \sqrt{y}-\;\frac{1}{2} \sqrt{y}) d y$ $=2 \cdot \;\frac{5}{2} \;\; \$ [;\frac{2}{3} y^{;\frac{3}{2}}]_0^4=;\frac{10}{3} ;; $[4^{;\frac{3}{2}}-0]$=;\frac{80}{3}$

Q69JEE Main 2022MCQ4MDefinite Integrals

$int_0^2( |2 x^{2-}3 x|+[x-\;\frac{1}{2}]) {\text{d}} x$ , where $[ {\text{t}}]$ is the greatest integer function, is equal to: [27-Jul-2022-Shift-2]

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📖 Explanation

$int_{0}^{2}|2 x^{2-}3 x| d x+int_{0}^{2}[x-\;\frac{1}{2}] d x$ $=int_{0}^{3 / 2}(3 x-2 x^2) d x+int_{3 / 2}^{2}(2 x^{2-}3 x) d x+int_{0}^{1 / 2}-1 d x+int_{1 / 2}^{3 / 2} 0 d x+int_{3 / 2}^{2} 1 d x$ $=(\;\frac{3 x^2}{2}-\;\frac{2 x^3}{3})|_0 ^{3 / 2}+.(\;\frac{2 x^3}{3}-\;\frac{3 x^2}{2})|_{3 / 2} ^{2-}\;\frac{1}{2}+\;\frac{1}{2}$ $=(\;\frac{27}{8}-\;\frac{27}{12})+(\;\frac{16}{3}-6-\;\frac{27}{12}+\;\frac{27}{8})$ $=\;\frac{19}{12}$

Q70JEE Main 2022MCQ4MDifferential Equations

Consider a curve $y=y(x)$ in the first quadrant as shown in the figure. Let the area ${\text{A}}_1$ is twice the area ${\text{A}}_2$ . Then the normal to the curve perpendicular to the line $2 x-12 y=15$ does NOT pass through the point. [27-Jul-2022-Shift-2]

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📖 Explanation

$A_1+A_2=x y-8 \& A_1=2 A_2$ $A_1+\;\frac{A_1}{2}=x y-8$ $A_1=\;\frac{2}{3}(x y-8)$ $int_{4}^{x }f(x) d x=\;\frac{2}{3}(x f(x)-8)$ Differentiate w.r.t. x $f(x)=\;\frac{2}{3}\{x f^{'}(x)+f(x)\}$ $\;\frac{2}{3} x f^{'}(x)=\;\frac{1}{3} f(x)$ $2 \int \;\frac{f^{'}(x)}{f(x)} d x=\int \;\frac{d x}{x}$ $2 \ln f(x)=\ln x+\ln c$ $f^2(x)=c x$ Which passes through $(4,2)$ $4=c \times 4 \Rightarrow c=1$ Equation of required curve $y^2=x$ Equation of normal having slope $(-6)$ is $y=-6 x-2(\;\frac{1}{4})(-6)-\;\frac{1}{4}(-6)^3$ $y=-6 x+57$ Which does not pass through $(10,-4)$

Q71JEE Main 2022MCQ4MStraight Lines and Pair of Straight Lines

The equations of the sides ${\text{AB}}, {\text{BC}}$ and ${\text{CA}}$ of a triangle ${\text{ABC}}$ are $2 x+y=0, x+ {\text{p}} y=39$ and $x-y=3$ respectively and ${\text{P}}(2,3)$ is its circumcentre. Then which of the following is NOT true? [27-Jul-2022-Shift-2]

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📖 Explanation

Intersection of $2 x+y=0$ and $x-y=3: A(1,-2)$ Equation of perpendicular bisector of $A B$ is $x-2 y=-4$ Equation of perpendicular bisector of AC is $x+y=5$ Point ${\text{B}}$ is the image of ${\text{A}}$ in line $x-2 y+4=0$ which is obtained as $B(\;\frac{-13}{5}, \;\frac{26}{5})$ Similarly vertex $C:(7,4)$ Equation of line $B C: x+8 y=39$ So, $p=8$ $A C=\sqrt{(7-1)^{2+}(4+2)^2}=6 \sqrt{2}$ Area of triangle $A B C=32.4$

Q72JEE Main 2022MCQ4MCircles

A circle $C_1$ passes through the origin ${\text{O}}$ and has diameter 4 on the positive $x$ -axis. The line $y=2 x$ gives a chord ${\text{OA}}$ of circle ${\text{C}}_1$ . Let ${\text{C}}_2$ be the circle with ${\text{OA}}$ as a diameter. If the tangent to ${\text{C}}_2$ at the point ${\text{A}}$ meets the $x$ -axis at ${\text{P}}$ and $y$ -axis at ${\text{Q}}$ , then ${\text{QA}}: {\text{AP}}$ is equal to: [27-Jul-2022-Shift-2]

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📖 Explanation

Equation of $C_1$ $x^{2+}y^{2-}4 x=0$ Intersection with $y=2 x$ $x^{2+}4 x^{2-}4 x=0$ $5 x^{2-}4 x=0$ $\Rightarrow x=0, \;\frac{4}{5}$ $y=0, \;\frac{8}{5}$ $A:(\;\frac{4}{5}, \;\frac{8}{5})$ Tangent of ${\text{C}}_2$ at $A(\;\frac{4}{5}, \;\frac{8}{5})$ $x+2 y=4 \Rightarrow P:(4,0), Q:(0,2)$ $Q A: A P=1: 4$

Q73JEE Main 2022MCQ4MParabola

If the length of the latus rectum of a parabola, whose focus is $(a, a)$ and the tangent at its vertex is $x+y=a$ , is 16 , then $|a|$ is equal to : [27-Jul-2022-Shift-2]

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📖 Explanation

Equation of tangent at vertex: $L \equiv x+y-a=0$ Focus : $F \equiv (a, a)$ Perpendicular distance of $L$ from $F$ $=|\;\frac{a+a-a}{\sqrt{2}}|=|\;\frac{a}{\sqrt{2}}|$ Length of latus rectum $=4|\;\frac{a}{\sqrt{2}}|$ Given 4. $|\;\frac{a}{\sqrt{2}}|=16$ $\Rightarrow |a|=4 \sqrt{2}$

Q74JEE Main 2022MCQ4MThree Dimensional Geometry

If the length of the perpendicular drawn from the point $P(a, 4,2), {\text{a}} > 0$ on the line $\;\frac{x+1}{2}=\;\frac{y-3}{3}=\;\frac{z-1}{-1}$ is $2 \sqrt{6}$ units and $Q(\alpha_1, \alpha_2, \alpha_3)$ is the image of the point ${\text{P}}$ in this line, then ${\text{a}}+sum_{i=1}^3 \alpha_i$ is equal to : [27-Jul-2022-Shift-2]

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📖 Explanation

$\because {\text{PR}}$ is perpendicular to given line, so $2(2 \lambda -1-a)+3(3 \lambda -1)-1(-\lambda -1)=0$ $\Rightarrow a=7 \lambda -2$ Now, $\because P R=2 \sqrt{6}$ $\Rightarrow (-5 \lambda +1)^{2+}(3 \lambda -1)^{2+}(\lambda +1)^2=24$ $\Rightarrow 5 \lambda ^{2-}2 \lambda -3=0 \Rightarrow \lambda =1 \;\text{ or }\;-\;\frac{3}{5}$ $\because a>0$ so $\lambda =1$ and $a=5$ Now $sum_{i=1}^{3} \alpha_i=2$ (Sum of co-ordinate of $. {\text{R}})-($ Sum of coordinates of ${\text{P}})$ $=2(7)-11=3$ $a+sum_{i=1}^{3} \alpha_i=5+3=8$

Q75JEE Main 2022MCQ4MThree Dimensional Geometry

If the line of intersection of the planes $a x+b y=3$ and $a x+b y+c z=0, {\text{a}} > 0$ makes an angle $30^{\circ}$ with the plane $y-z+2=0$ , then the direction cosines of the line are : [27-Jul-2022-Shift-2]

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📖 Explanation

$P_1: a x+b y+0 z=3$ , normal vector $: \vec{n}_1=(a, b, 0)$ $P_2: a x+b y+c z=0$ , normal vector $: \vec{n}_2=(a, b, c)$ Vector parallel to the line of intersection $=\vec{n}_1 \times \vec{n}_2$ $\vec{n}_1 \times \vec{n}_2=(b c,-a c, 0)$ Vector normal to $0 . x+y-z+2=0$ is $\vec{n}_3=(0,1,-1)$ Angle between line and plane is $30^{\circ}$ $\Rightarrow |\;\frac{0-a c+0}{\sqrt{b^2 c^{2+}c^2 a^2} \sqrt{2}}|=\;\frac{1}{2}$ $\Rightarrow a^2=b^2$ Hence, $\vec{n}_1 \times \vec{n}_2=(a c,-a c, 0)$ Direction ratios $=(\;\frac{1}{\sqrt{2}},-\;\frac{1}{\sqrt{2}}, 0)$

Q76JEE Main 2022MCQ4MProbability

Let ${\text{X}}$ have a binomial distribution ${\text{B}}( {\text{n}}, {\text{p}})$ such that the sum and the product of the mean and variance of ${\text{X}}$ are 24 and 128 respectively. If $P(X > n-3)=\;\frac{k}{2^n}$ , then ${\text{k}}$ is equal to : [27-Jul-2022-Shift-2]

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📖 Explanation

Mean $=n p=16$ Variance $=n p q=8$ $\Rightarrow q=p=\;\frac{1}{2}$ and $n=32$ $P(x>n-3)=p(x=n-2)+p(x=n-1)+p(x=n)$ $=({ }^{32} C_2+{ }^{32} C_1+{ }^{32} C_0) \cdot \;\frac{1}{2^n}$ $=\;\frac{529}{2^n}$

Q77JEE Main 2022MCQ4MProbability

A six faced die is biased such that $3 \times {\text{P}}($ a prime number $)=6 \times {\text{P}}($ a composite number $)=2 \times {\text{P}}$ . Let $X$ be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of ${\text{X}}$ is : [27-Jul-2022-Shift-2]

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📖 Explanation

Let ${\text{P}}($ a prime number $)=\alpha$ $P($ a composite number $)=\beta$ and ${\text{P}}(1)=\gamma$ $\because 3 \alpha=6 \beta=2 \gamma =k$ (say)and $3 \alpha+2 \beta+\gamma =1$ $\Rightarrow k+\;\frac{k}{3}+\;\frac{k}{2}=1 \Rightarrow k=\;\frac{6}{11}$ Mean $=n p$ where $n=2$ and $p=$ probability of getting perfect square $=P(1)+P(4)=\;\frac{k}{2}+\;\frac{k}{6}=\;\frac{4}{11}$ So, mean $=2 \cdot (\;\frac{4}{11})=\;\frac{8}{11}$

Q78JEE Main 2022MCQ4MTrigonometric Ratios and Identities

The angle of elevation of the top $P$ of a vertical tower $P Q$ of height 10 from a point $A$ on the horizontal ground is $45^{\circ}$ . Let $R$ be a point on $A Q$ and from a point $B$ , vertically above $R$ , the angle of elevation of ${\text{P}}$ is $60^{\circ}$ . If $∠ {\text{BAQ}}=30^{\circ}, {\text{AB}}= {\text{d}}$ and the area of the trapezium ${\text{PQRB}}$ is $\alpha$ , then the ordered pair $( {\text{d}}, \alpha)$ is : [27-Jul-2022-Shift-2]

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📖 Explanation

Let $B R=x$ $\;\frac{x}{d}=\;\frac{1}{2} \Rightarrow x=\;\frac{d}{2}$ $\;\frac{10-x}{10-x \sqrt{3}}=\sqrt{3} \Rightarrow 10-x=10 \sqrt{3}-3 x$ $2 x=10(\sqrt{3}-1)$ $x=5(\sqrt{3}-1)$ $d=2 x=10(\sqrt{3}-1)$ $\alpha=\;\frac{1}{2}(x+10)(10-x \sqrt{3})={\text{Area}}( {\text{PQRB}})$ $=\;\frac{1}{2}(5 \sqrt{3}-5+10)(10-5 \sqrt{3}(\sqrt{3}-1))$ $=\;\frac{1}{2}(5 \sqrt{3}+5)(10-15+5 \sqrt{3})-\;\frac{1}{2}(75-25)=25$

Q79JEE Main 2022MCQ4MSequences and Series

Let $S=\{\theta \in (0, \;\frac{\pi}{2}): sum_{m=1}^{9} \sec (\theta+(m-1) \;\frac{\pi}{6}) \sec (\theta+\;\frac{m \pi}{6})=-\;\frac{8}{\sqrt{3}}\}$ . Then [27-Jul-2022-Shift-2]

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📖 Explanation

$s=\{0 \in (0, \;\frac{\pi}{2}): sum_{m=1}^{9} \ sec (\theta+(m-1) \;\frac{\pi}{6}) \ sec (\theta+\;\frac{m \pi}{6})=-\;\frac{8}{\sqrt{3}}\} .$ $sum_{m=1}^{9} \;\frac{1}{\cos (\theta+(m-1) \;\frac{\pi}{6})} \cos (\theta+m \;\frac{\pi}{6})$ $\;\frac{1}{\ sin (\;\frac{\pi}{6})} sum_{m=1}^{9} \;\frac{\ sin [(\theta+\frac{m \pi}{6})-(\theta+(m-1) \frac{\pi}{6})]}{\cos (\theta+(m-1) \;\frac{\pi}{6}) \cos (\theta+\;\frac{\pi}{6})}$ $=2 sum_{m=1}^{9}[\tan (\theta+\;\frac{m \pi}{6})-\tan (\theta+(m-1) \;\frac{\pi}{6})]$ Now, $\text{table} m=1 , 2[\tan (\theta+\;\frac{\pi}{6})-\tan (\theta)] ; m=2 , 2[\tan (\theta+\;\frac{2 \pi}{6})-\tan (\theta+\;\frac{\pi}{6})]$ . . . $m=9 \;\; 2[\tan (\theta+\;\frac{9 \pi}{6})-\tan (\theta+8 \;\frac{\pi}{6})]$ $\therefore =2[\tan (\theta+\;\frac{3 \pi}{2})-\tan \theta]=\;\frac{-8}{\sqrt{3}}$ $=-2[\cot \theta+\tan \theta]=\;\frac{-8}{\sqrt{3}}$ $=-\;\frac{2 \times 2}{2 \ sin \theta \cos \theta}=\;\frac{-8}{\sqrt{3}}$ $=\;\frac{1}{\ sin 2 \theta}=\;\frac{2}{\sqrt{3}}$ $\Rightarrow \ sin 2 \theta=\;\frac{\sqrt{3}}{2}$ $2 \theta=\;\frac{\pi}{3}$ $2 \theta=\;\frac{2 \pi}{3}$ $\theta=\;\frac{\pi}{6}$ $\theta=\;\frac{\pi}{3}$ $\sum \theta_i=\;\frac{\pi}{6}+\;\frac{\pi}{3}=\;\frac{\pi}{2}$

Q80JEE Main 2022MCQ4MSets and Relations

If the truth value of the statement $(P ∧(∼ R)) \to ((∼ R) ∧ Q)$ is ${\text{F}}$ , then the truth value of which of the following is ${\text{F}}$ ? [27-Jul-2022-Shift-2]

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📖 Explanation

${\text{X}} \Rightarrow {\text{Y}}$ is a falsewhen $X$ is true and $Y$ is falseSo, ${\text{P}} \to {\text{T}}, {\text{Q}} \to {\text{F}}, {\text{R}} \to {\text{F}}$ (A) ${\text{P}} ∨ {\text{Q}} \to ∼ {\text{R}}$ is ${\text{T}}$ (B) ${\text{R}} ∨ {\text{Q}} \to ∼ {\text{P}}$ is ${\text{T}}$ (C) $∼( {\text{P}} ∨ {\text{Q}}) \to ∼ {\text{R}}$ is ${\text{T}}$ $( {\text{D}}) ∼( {\text{R}} ∨ {\text{Q}}) \to ∼ {\text{P}}$ is ${\text{F}}$

Q81JEE Main 2022NAT4MMatrices and Determinants

Consider a matrix $A=\begin{bmatrix}\alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma ^2 \\ \beta+\gamma & \gamma +\alpha & \alpha+\beta\end{bmatrix}$ , where $\alpha, \beta, \gamma$ are three distinct natural numbers. If $\;{{\det}({\text{adj}}({\text{adj}}({\text{adj}}({\text{adj}} A))))}/{(\alpha-\beta)^{16}(\beta-\gamma )^{16}(\gamma -\alpha)^{16}}=2^{32} \times 3^{16}$ , then the number of such 3 - tuples $(\alpha, \beta, \gamma )$ is _________. [27-Jul-2022-Shift-2]

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📖 Explanation

${\det}(A)=\begin{vmatrix}\alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma ^2 \\ \beta+\gamma & \gamma +\alpha & \alpha+\beta\end{vmatrix}$ $R_3 \to R_3+R_1$ $\Rightarrow (\alpha+\beta+\gamma )\begin{vmatrix}\alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma ^2 \\ 1 & 1 & 1\end{vmatrix}$ $\therefore {\det}(A)=(\alpha+\beta+\gamma )(\alpha-\beta)(\beta-\gamma )(\gamma -\alpha)$ Also, ${\det}(a d j(a d j(a d j(a d j(A)))))$ $=({\det}(A))^{2^4}=({\det}(A)^{16}.$ $\therefore \;\frac{(\alpha+\beta+\gamma )^{16}(\alpha-\beta)^{16}(\beta-\gamma )^{16}(\gamma -\alpha)^{16}}{(\alpha-\beta)^{16}(\beta-\gamma )^{16}(\gamma -\alpha)^{16}}=(4.13)^{16}$ $\Rightarrow \alpha+\beta+\gamma =12$ $\Rightarrow (\alpha, \beta, \gamma ) {\text{distinct}}$ natural triplets $={ }^{11} C_2-1-{ }^3 C_2(4)=55-1-12$ =42

Q82JEE Main 2022NAT4MPermutations and Combinations

The number of functions $f$ , from the set ${\text{A}}=\{x \in N: x^{2-}10 x+9 \le 0\}$ to the set ${\text{B}}=\{ {\text{n}}^2: {\text{n}} \in N\}$ such that $f(x) \le (x-3)^{2+}1$ , for every $x \in {\text{A}}$ , is _________. [27-Jul-2022-Shift-2]

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📖 Explanation

$A=\{x \in N, x^{2-}10 x+9 \le 0\}$ $=\{1,2,3, . . ., 9\}$ $B=\{1,4,9,16, . . . . . .\}$ $f(x) \le (x-3)^{2+}1$ $f(1) \le 5, f(2) \le 2, . . . . . . . . . . f(9) \le 37$ $x=1$ has 2 choices $x=2$ has 1 choice $x=3$ has 1 choice $x=4$ has 1 choice $x=5$ has 2 choices $x=6$ has 3 choices $x=7$ has 4 choices $x=8$ has 5 choices $x=9$ has 6 choices $\therefore$ Total functions $=2 \times 1 \times 1 \times 1 \times 2 \times 3 \times 4 \times 5 \times 6=1440$

Q83JEE Main 2022NAT4MBinomial Theorem

Let for the $9^{\;\text{th }\;}$ term in the binomial expansion of $(3+6 x)^{ {\text{n}}}$ , in the increasing powers of $6 x$ , to be the greatest for $x=\;\frac{3}{2}$ , the least value of ${\text{n}}$ is ${\text{n}}_0$ . If ${\text{k}}$ is the ratio of the coefficient of $x^6$ to the coefficient of $x^3$ , then ${\text{k}}+ {\text{n}}_0$ is equal to: [27-Jul-2022-Shift-2]

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📖 Explanation

$(3+6 x)^n=3^n(1+2 x)^n$ If $T_9$ is numerically greatest term $\therefore T_8 \le T_9 \le T_{10}$ ${ }^n C_7 3^{n-7}(6 x)^7 \le { }^n C_8 3^{n-8}(6 x)^8 \ge { }^n C_9 3^{n-9}(6 x)^9$ $\Rightarrow \;\frac{n !}{(n-7) ! 7 !} 9 \le \;\frac{n !}{(n-8) ! 8 !} 3 \cdot (6 x) \ge \;\frac{n !}{(n-9) ! 9 !}(6 x)^2$ $\Rightarrow {\;\frac{9}{(n-7)(n-8)}} _{⏟}\le {\;\frac{18(\frac{3}{2})}{(n-8) 8} \ge \;\frac{36}{9.8} \;\frac{9}{4}} _{⏟}$ $72 \le 27(n-7) \;\text{ and }\; 27 \ge 9(n-8)$ $\;\frac{29}{3} \le n \;\text{ and }\; n \le 11$ $\therefore n_0=10$ $\;\text{ For }\;(3+6 x)^{10}$ $T_{r+1}={ }^{10} C_r$ $3^{10-r}(6 x)^r$ For coeff. of $x^6$ $r=6 \Rightarrow { }^{10} C_6 3^4 \cdot 6^6$ For coeff. of $x^3$ $r=3 \Rightarrow { }^{10} C_3 3^7 \cdot 6^3$ $\therefore k=\;\frac{{ }^{10} C_6}{{ }^{10} C_3} \cdot \;\frac{3^4 \cdot 6^6}{3^7 \cdot 6^3}=\;\frac{10 ! 7 ! 3 !}{6 ! 4 ! 10 !} \cdot 8$ $\Rightarrow k=14$ $\therefore k+n_0=24$

Q84JEE Main 2022NAT4MSequences and Series

$\;\frac{2^{3-}1^3}{1 \times 7}+\;\frac{4^{3-}3^{3+}2^{3-}1^3}{2 \times 11}+\;\frac{6^{3-}5^{3+}4^{3-}3^{3+}2^{3-}1^3}{3 \times 15}+⋯+\;\frac{30^{3-}29^{3+}28^{3-}27^{3+}. . .+2^{3-}1^3}{15 \times 63}$ is equal to ________. [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$T_n=\;\frac{sum_{k=1}^{n}\$ [(2 k)^{3-}(2 k-1)^3]$}{n(4 n+3)}=;\frac{sum_{k=1}^{n} 4 k^{2+}(2 k-1)^{2+}2 k(2 k-1)}{n(4 n+3)}=;\frac{sum_{k=1}^{n}(12 k^{2-}6 k+1)}{n(4 n+3)}=;\frac{2 n(2 n^{2+}3 n+1)-3 n^{2-}3 n+n}{n(4 n+3)}=;\frac{n^2(4 n+3)}{n(4 n+3)}=n\therefore T_n=nS_n=sum_{n=1}^{15} T_n=;\frac{15 \times 16}{2}=120$

Q85JEE Main 2022NAT4MApplication of Derivatives

A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semivertical angle is $\tan ^{-1} \;\frac{3}{4}$ . Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is _________. [27-Jul-2022-Shift-2]

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📖 Explanation

$v=\;\frac{1}{3} \pi r^2 h . . . . . \;\text{ (i) }\;$ And $\tan \theta=\;\frac{3}{4}=\;\frac{r}{h} . . . . . .$ . (ii)i.e. if $h=4, r=3$ $v=\;\frac{1}{3} \pi r^2(\;\frac{4 r}{3})$ $\;\frac{d v}{d t}=\;\frac{4 \pi}{9} 3 r^2 \;\frac{d r}{d t} \Rightarrow 6=\;\frac{4 \pi}{3}(9) \;\frac{d r}{d t}$ $\Rightarrow \;\frac{d r}{d t}=\;\frac{1}{2 \pi}$ Curved area $=\pi r \sqrt{r^{2+}h^2}$ $=\pi r \sqrt{r^{2+}\;\frac{16 r^2}{9}}$ $=\;\frac{5}{3} \pi r^2$ $\;\frac{d A}{d t}=\;\frac{10}{3} \pi r \;\frac{d r}{d t}$ $=\;\frac{10}{3} \pi \cdot 3 \cdot \;\frac{1}{2 \pi}$ $=5$

Q86JEE Main 2022NAT4MDifferentiation

For the curve $C:(x^{2+}y^{2-}3)+(x^{2-}y^{2-}1)^5=0$ , the value of $3 y^{'}-y^3 y^{' '}$ , at the point $(\alpha, \alpha), \alpha > 0$ , on ${\text{C}}$ , is equal to ________. [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\because C:(x^{2+}y^{2-}3)+(x^{2-}y^{2-}1)^5=0$ for point $(\alpha, \alpha)$ $\alpha^{2+}\alpha^{2-}3+(\alpha^{2-}\alpha^{2-}1)^5=0$ $\therefore \alpha=\sqrt{2}$ On differentiating $(x^{2+}y^{2-}3)+(x^{2-}y^{2-}1)^5=0$ we get $x+y y^{'}+5(x^{2-}y^{2-}1)^4(x-y y^{'})=0$ When $x=y=\sqrt{2}$ then $y^{'}=\;\frac{3}{2}$ Again on differentiating eq. (i) we get : $1+(y^{'})^{2+}y y^{' '}+20(x^{2-}y^{2-}1)(2 x-2 y y^{'})(x-y^{'} y)+5(x^{2-}y^{2-}1)^4(1-y^{' 2}-y y^{' '})=0$ For $x=y=\sqrt{2}$ and $y^{'}=\;\frac{3}{2}$ we get $y^{' '}=-\;\frac{23}{4 \sqrt{2}}$ $\therefore 3 y^{'}-y^3 y^{' '}=3 . \;\frac{3}{2}-(\sqrt{2})^3 \cdot (-\;\frac{23}{4 \sqrt{2}})=16$

Q87JEE Main 2022NAT4MDefinite Integrals

Let $f(x)=\min \{[x-1],[x-2], . . .,[x-10]\}$ where $[ {\text{t}}]$ denotes the greatest integer $\le {\text{t}}$ . Then $int_0^{10} f(x) {\text{d}} x+int_0^{10}(f(x))^2 {\text{d}} x+int_0^{10}|f(x)| {\text{d}} x$ is equal to ________. [27-Jul-2022-Shift-2]

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📖 Explanation

$\because f(x)=\min \{[x-1],[x-2], . . . . . .,[x-10]\}=[x-10]$ Also $|f(x)|=\begin{cases}-f(x){, \\ \\ } , \\ \\ \text{ if }\\ \\ x \le 10 \\ f(x){, \\ \\ } , \\ \\ \text{ if }\\ \\ x \ge 10\end{cases}$ $\therefore int_{0}^{10} f(x) d x+int_{0}^{10}(f(x))^2 d x+int_{0}^{10}(-f(x)) d x$ $=int_{0}^{10}(f(x))^2 d x$ $=10^{2+}9^{2+}8^{2+}. . . . . .+1^2$ $=\;\frac{10 \times 11 \times 21}{6}=385$

Q88JEE Main 2022NAT4MDifferential Equations

Let ${\text{f}}$ be a differentiable function satisfying $f(x)=\;\frac{2}{\sqrt{3}} int_0^{\sqrt{3}} f(\;\frac{\lambda ^2 x}{3}) {\text{d}} \lambda , x > 0$ and $f(1)=\sqrt{3}$ . If $y=f(x)$ passes through the point $(\alpha, 6)$ , then $\alpha$ is equal to __________. [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\because f(x)=\;\frac{2}{\sqrt{3}} int_{0}^{\sqrt{3}} f(\;\frac{\lambda ^2 x}{3}) d \lambda , x>0$ On differentiating both sides w.r.t., ${\text{x}}$ , we get $f^{'}(x)=\;\frac{2}{\sqrt{3}} int_{0}^{\sqrt{3}} \;\frac{\lambda ^2}{3} f^{'}(\;\frac{\lambda ^2 x}{3}) d \lambda$ $f^{'}(x)=\;\frac{1}{\sqrt{3}} int_{0}^{\sqrt{3}} \lambda \cdot \;\frac{2 \lambda }{3} f^{'}(\;\frac{\lambda ^2 x}{3}) d \lambda$ $\therefore \sqrt{3} f^{'}(x)=\$ [;\frac{λ}{x} ⋅ f(;\frac{λ^2 x}{3})]0^{\sqrt{3}}-int{0}^{\sqrt{3}} ;\frac{1}{x} f(;\frac{λ^2 x}{3}) d x\sqrt{3} x f^{'}(x)=\sqrt{3} f(x)-;\frac{\sqrt{3}}{2} f(x)x f^{'}(x)=;\frac{f(x)}{2} $On integrating we get$ : \ln y=;\frac{1}{2} \ln x+\ln c\because f(1)=\sqrt{3} $then$ c=\sqrt{3}∴ (alpha, 6) $lies on$ ∴ y=\sqrt{3 x}∴ 6=\sqrt{3 alpha} \Rightarrow alpha=12$

Q89JEE Main 2022NAT4MApplication of Derivatives

A common tangent ${\text{T}}$ to the curves ${\text{C}}_1: \;\frac{x^2}{4}+\;\frac{y^2}{9}=1$ and $C_2: \;\frac{x^2}{42}-\;\frac{y^2}{143}=1$ does not pass through the fourth quadrant. If ${\text{T}}$ touches ${\text{C}}_1$ at $(x_1, y_1)$ and ${\text{C}}_2$ at $(x_2, y_2)$ , then $2 x_1+x_2|$ is equal to ________. [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Equation of tangent to ellipse $\;\frac{x^2}{4}+\;\frac{y^2}{9}=1$ and given slope ${\text{m}}$ is : $y=m x+\sqrt{4 m^{2+}9} . . .$ . (i)For slope m equation of tangent to hyperbola is : $y=m x+\sqrt{42 m^{2-}143}$ Tangents from (i) and (ii) are identical then $4 m^{2+}9=42 m^{2-}143$ $\therefore m=\pm 2(+2 \;\text{ is not acceptable }\;)$ $\therefore m=-2 .$ Hence, $x_1=\;\frac{8}{5}$ and $x_2=\;\frac{84}{5}$ $\therefore |2 x_1+x_2|=|\;\frac{16}{5}+\;\frac{84}{5}|=20$

Q90JEE Main 2022NAT4MVector Algebra

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors such that ${ \vec{\text{a}} \times { \vec{\text{b}}=4 ^{}{ \vec{\text{c}}, { \vec{\text{b}} \times { \vec{\text{c}}=9^{} { \vec{\text{a}}$ and ${ \vec{\text{c}} \times { \vec{\text{a}}=\alpha { \vec{\text{b}}, \alpha>0$ If $|\vec{a}|+|\vec{b}|+|\vec{c}|=\;\frac{1}{36}$ , then $\alpha$ is equal to________. [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Given, $\vec{a} \times \vec{b}=4 \cdot \vec{c} . . . . . \;\text{ (i) }\;$ $\vec{b} \times \vec{c}=9 \cdot \vec{a} . . . . . \;\text{ (ii) }\;$ $\vec{c} \times \vec{a}=\alpha \cdot \vec{b} . . . \;\text{. (iii) }\;$ Taking dot products with ${c, \vec{a}, \vec\vec{b}$ we get $\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0$ Hence, $\;\text{ (i) }\; \Rightarrow |\vec{a}| \cdot |\vec{b}|=4 . \vec{c \mid } . . . . . \;\text{ (iv) }\;$ (ii) $\Rightarrow \vec{|b|} \cdot |\vec{c}|=9 \cdot |\vec{a}| . . .$ . (v)(iii) $\Rightarrow \vec{c \mid } \cdot |\vec{a}|=\alpha \cdot \vec{|b|} . . .$ (vi) Multiplying (iv), (v) and (vi) $\Rightarrow |\vec{a}| \cdot \vec{\mid b}|\cdot | \vec{c} \mid =36 \alpha . . . . . \;\text{ (vii) }\;$ Dividing (vii) by (iv) $\Rightarrow |\vec{c}|^2=9 \alpha \Rightarrow \mid \vec{c}=3 \sqrt{\alpha} . . .$ . (viii)Dividing (vii) by (v) $\Rightarrow |\vec{a}|^2=4 \alpha \Rightarrow |\vec{a}|=2 \sqrt{\alpha}$ Dividing (viii) by (vi) $\Rightarrow \overrightarrow{|\vec{b}|^2}=36 \Rightarrow \vec{b} \mid =6$ Now, as given, $3 \sqrt{\alpha}+2 \sqrt{\alpha}+6=\;\frac{1}{36} \Rightarrow \sqrt{\alpha}=\;\frac{-43}{36}$

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