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11 Apr 2023 Shift 1

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Physics30 questions

Q1JEE Main 2023MCQ4MGravitation

The radii of two planets 'A' and 'B' are 'R' and '4R' and their densities are $\rho$ and $\rho / 3$ respectively. The ratio of acceleration due to gravity at their surfaces $( {\text{g}}_A: {\text{g}}_B)$ will be : [11-Apr-2023 shift 1]

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📖 Explanation

$\;g=\;\frac{4 \pi}{3} {\text{GR}} \delta$ $\;g \propto \delta R$ $\;\;\frac{g_A}{g_B}=\;\frac{\delta _A R_A}{\delta _B \cdot R_B}=\;\frac{\delta \cdot R}{\frac{\delta }{3} \cdot 4 R}=\;\frac{3}{4}$

Q2JEE Main 2023MCQ4MCircular Motion

A coin placed on a rotating table just slips when it is placed at a distance of $1 {\text{cm}}$ from the center. If the angular velocity of the table in halved, it will just slip when placed at a distance of _______ from the centre : [11-Apr-2023 shift 1]

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📖 Explanation

${fr}= {\text{m}\omega ^2r}$ $\;\mu {\text{mg}}= {\text{m}} \omega ^2 {\text{r}}= {\text{const}}$ $\;\omega ^2 {\text{r}}= {\text{const}}$ $\;\omega _1^2 {\text{r}}_1=\omega _2^2 {\text{r}}_2$ $\;\omega ^2(1)= (\;\frac{\omega }{2}) ^2 r_2$ $\;r_2=4 {\text{cm}}$

Q3JEE Main 2023MCQ4MThermodynamics

Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and third contains uranium hexafluoride (polyatomic). Arrange these on the basis of their root mean square speed $( {\text{v}}_{ {\text{rms}}})$ and choose the correct answer from the options given below : [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{VRMS}}=\sqrt{\;{\gamma {\text{RT}}}/{ {\text{m}}}}$ $\;\gamma =1+\;\frac{2}{f} \;\; \;\text{ so }\; \text{bo} {r}_{\;\text{monochromic }\;} > \text{bo} {r}_{\;\text{diatomic }\;} > \text{bo} {r}_{\;\text{poly }\;} \;\text{. }\;$ $\;V_{\;\text{mono }\;} > V_{\;\text{diatomic }\;} > V_{\;\text{poly. }\;}$ $\;\;\text{ Ans.(1) }\;$ $\;$

Q4JEE Main 2023MCQ4MAtoms and Nuclei

Two radioactive elements $A$ and $B$ initially have same number of atoms. The half life of $A$ is same as the average life of B. If $\lambda _{ {\text{A}}}$ and $\lambda _{ {\text{B}}}$ are decay constants of ${\text{A}}$ and ${\text{B}}$ respectively, then choose the correct relation from the given options. [11-Apr-2023 shift 1]

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📖 Explanation

$\;\text{ Now }\;\;\;\frac{\ln (2)}{\lambda _A}=\;\frac{1}{\lambda _B}\;\Rightarrow \lambda _{ {\text{A}}}=\lambda _{ {\text{B}}} \cdot \ln (2)$

Q5JEE Main 2023MCQ4MAlternating Current

As per the given graph, choose the correct representation for curve $A$ and curve $B$ . $\{.$ Where ${\text{X}}_{ {\text{C}}}=$ reactance of pure capacitive circuit connected with ${\text{A}}$ .C. source ${\text{X}}_{ {\text{L}}}=$ reactance of pure inductive circuit connected with A.C. source ${\text{R}}=$ impedance of pure resistive circuit connected with A.C. source. ${\text{Z}}=$ impedance of the LCR series circuit [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{X}}_{ {\text{L}}}= {\text{W}}_{ {\text{L}}}=2 \pi {\text{fL}}$ $\; {\text{X}}_{ {\text{C}}}=\;{1}/{ {\text{Wc}}}=\;{1}/{2 \pi {\text{fc}}}$ $\; {\text{R}}=\;\text{ const. }\;$ $\; {\text{A}} \to {\text{X}}_{ {\text{C}}}$ $\; {\text{B}} \to {\text{X}}_{ {\text{L}}}$

Q6JEE Main 2023MCQ4MElectromagnetic Waves

A transmitting antenna is kept on the surface of the earth. The minimum height of receiving antenna required to receive the signal in line of sight at $4 {\text{km}}$ distance from it is $x \times 10^{-2} {\text{m}}$ . The value of ${\text{x}}$ is (Let, radius of earth $R=6400 {\text{km}}$ ) [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{d}}=\sqrt{2 {\text{R}} \cdot {\text{h}}}$ $\; {\text{d}}^2=2 {\text{Rh}}$ $\;(4)^2=2 \times 6400 \times {\text{h}}$ $\;\;\frac{16}{2 \times 6400}= {\text{h}}=\;\frac{1}{800} {\text{km}}$ $\; {\text{h}}=\;\frac{1000}{800}=\;\frac{5}{4} {\text{m}}$ $\; {\text{x}} \times 10^{-2}=\;\frac{5}{4}$ $\; {\text{x}}=\;\frac{500}{4}=125 \;\; \;\text{ Ans. Option }\; \to (1)$

Q7JEE Main 2023MCQ4MSemiconductor Electronics

The logic performed by the circuit shown in figure is equivalent to : [11-Apr-2023 shift 1]

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📖 Explanation

$\;Y=Y=\overline{(\overline{a}+\overline{b})}$ $\;Y=Y=\overline{\overline{(a \cdot b})}$ $\;Y=a \cdot b$ Ans. $\to$ AND gate Option $\to 3$

Q8JEE Main 2023MCQ4MElectromagnetic Waves

The electric field in an electromagnetic wave is given as ${ {\text{E}}}^{\to }=20 sin \;\omega ( {\text{t}}-\;{ {\text{x}}}/{ {\text{c}}}) { {\text{j}}}^{\to } {\text{NC}}^{-1}$ where $\omega$ and ${\text{c}}$ are angular frequency and velocity of electromagnetic wave respectively. the energy contained in a volume of $5 \times 10^{-4} {\text{m}}^3$ will be (Given $\epsilon_0=8.85 \times 10^{-12} {\text{c}}^2 / {\text{Nm}}^2$ ) [11-Apr-2023 shift 1]

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📖 Explanation

${ {\text{E}}}^{\to }=20 sin \; {\text{w}} [ {\text{t}}-\;{ {\text{x}}}/{ {\text{c}}}]$ $E_o 2_o$ $\;\;\text{ Energy density }\;=\;\frac{1}{2} \epsilon_{ {\text{o}}} E_{ {\text{o}}}^2$ $\;=\;\frac{1}{2} \times 8.85 \times 10^{-12} \times 400$ $\;=200 \times 8.85 \times 10^{-12} \times 5 \times 10^{-4}$ $\;=8.85 \times 10^{-12} \times 10^{-4} \times 1000$ $\;\;\text{ Energy }\;=8.85 \times 10^{-13} {\text{J}}$

Q9JEE Main 2023MCQ4MCurrent Electricity

Two identical heater filaments are connected first in parallel and then in series. At the same applied voltage, the ratio of heat produced in same time for parallel to series will be : [11-Apr-2023 shift 1]

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📖 Explanation

${\text{H}}_1=\;{ {\text{V}}^2}/{( {\text{R}} / 2)} {\text{t}}=\;{2 {\text{V}}^2}/{ {\text{R}}}$ . . . (1) ${\text{H}}_2=\;{ {\text{V}}^2}/{2 {\text{R}}} {\text{t}}$ $\;{ {\text{H}}_1}/{ {\text{H}}_2}= (\;{2 {\text{V}}^2 {\text{t}}}/{ {\text{R}}}) \times \;{2 {\text{R}}}/{ {\text{V}}^2 {\text{t}}}=\;\frac{4}{1}$

Q10JEE Main 2023MCQ4MElectrostatic Potential and Capacitance

A parallel plate capacitor of capacitance $2 {\text{F}}$ is charged to a potential ${\text{V}}$ , The energy stored in the capacitor is ${\text{E}}_1$ . The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is ${\text{E}}_2$ . The ratio ${\text{E}}_2 / {\text{E}}_1$ is : [11-Apr-2023 shift 1]

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📖 Explanation

${\text{C}}=2 {\text{F}}$ ${\text{E}} 1=\;\frac{1}{2} {\text{CV}}^2$ . . . (1)Now $\; {\text{V}}_{ {\text{C}}}=\;{ {\text{C}}_1 {\text{V}}_1+ {\text{C}}_2 {\text{V}}_2}/{ {\text{C}}_1+ {\text{C}}_2}$ $\; {\text{V}}_{ {\text{C}}}=\;{ {\text{CV}}+ {\text{O}}}/{2 {\text{C}}}=\;{ {\text{V}}}/{2}$ $\; {\text{E}}_2= {\text{CV}}_{ {\text{C}}}{ }^2= {\text{C}} \cdot \;{ {\text{V}}^2}/{4}$ . . . (2) $\;\;{ {\text{E}}_2}/{ {\text{E}}_1}=\;{ ({ {\text{CV}}^2}/{4}) }/{ (\;{ {\text{CV}}^2}/{2}) }=\;\frac{2}{1} \;\; \;\text{ option }\; \to (4)$

Q11JEE Main 2023MCQ4MLaws of Motion

An average force of $125 {\text{N}}$ is applied on a machine gun firing bullets each of mass $10 {\text{g}}$ at the speed of $250 {\text{m}} / {\ s}$ to keep it in position. The number of bullets fired per second by the machine gun is : [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{F}}=125 {\text{N}}$ $\; {\text{F}}=\;{ {\text{dp}}}/{ {\text{dt}}} \;\; {\text{n}} \to \;\text{ No. of bullets }\;$ $\; {\text{F}}=\;{ {\text{d}}( {\text{nmv}})}/{ {\text{dt}}}= {\ mv} \;{ {\text{dn}}}/{ {\text{dt}}}$ $\;125=\;{10 {\text{n}}}/{1000} \times 250 \times \;{ {\text{dn}}}/{ {\text{dt}}}$ $\;\;\frac{125 \times 1000}{2500}=\;{ {\text{dn}}}/{ {\text{dt}}}$ $\;\;{ {\text{dn}}}/{ {\text{dt}}}=50$ option $\to$ (4)

Q12JEE Main 2023MCQ4MOscillations

The variation of kinetic energy $( {\text{KE}})$ of a particle executing simple harmonic motion with the displacement $( {\text{x}})$ starting from mean position to extreme position (A) is given by [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{K}} \cdot {\text{E}}= {\text{T}} \cdot {\text{E}}- {\text{P}} \cdot {\text{E}}$ $\; {\text{K}} \cdot {\text{E}}=\;\frac{1}{2} {\text{KA}} 2-\;\frac{1}{2} {\text{Kx}}^2$ Graph ${\text{b}} / {\text{w}} {\text{K}} \cdot {\text{E}}$ and ${\text{x}}$ will be parabola Option $\to$ (1)

Q13JEE Main 2023MCQ4MMotion in a Straight Line

From the ${\text{v}}- {\text{t}}$ graph shown, the ratio of distance to displacement in $25 {\ s}$ of motion is: [11-Apr-2023 shift 1]

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📖 Explanation

Displacement $=$ Area of graph with sign $\;\;\text{ Displacement }\;= (\;\frac{1}{2} \times 10 \times 5) +(10 \times 5)+ (\;\frac{1}{2} \times 5 \times 30) + (\;\frac{1}{2} \times 5 \times 20) -\;\frac{1}{2}(5)(20)$ $\;=25+50+75+50-50$ $\;=150 {\text{m}}$ Distance $\to$ Area of graph with positive valueDistance $=25+50+75+50=250$ $\;\frac{\;\text{ Distance }\;}{\;\text{ Displacement }\;}=\;\frac{250}{150}=\;\frac{5}{3} \;\; \;\text{ option }\; \to (3)$

Q14JEE Main 2023MCQ4MThermodynamics

On a temperature scale ' ${\text{X}}$ ', the boiling point of water is $65^{\circ} {\text{X}}$ and the freezing point is $-15^{\circ} {\text{X}}$ . Assume that the ${\text{X}}$ scale is linear. The equivalent temperature corresponding to $-95^{\circ} {\text{X}}$ on the Farenheit scale would be : [11-Apr-2023 shift 1]

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📖 Explanation

$\;\;\frac{X_T-X_L}{X_H-X_L}=\;\frac{T_F-32}{212-32}$ $\;\;\frac{-95^{\circ}- (-15^{\circ}) }{65^{\circ}- (-15^{\circ}) }=\;\frac{T_F-32}{180}$ $\;\;\frac{-80^{\circ}}{80^{\circ}}=\;\frac{T_F-32}{180^{\circ}}$ $\;-180= {\text{T}}_{ {\text{F}}}-32$ $\; {\text{T}}_{ {\text{F}}}=-180+32=-148^{\circ} {\text{F}}$ Ans. option $\to$ (2)

Q15JEE Main 2023MCQ4MMagnetism and Matter

The free space inside a current carrying toroid is filled with a material of susceptibility $2 \times 10^{-2}$ . The percentage increase in the value of magnetic field inside the toroid will be [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{X}}=2 \times 10^{-2}$ $\;\mu {\text{r}}=1+ {\text{x}}=1+0.02=1.02$ Bo $\to$ magnetic field due to magnetic material ${\text{B}}_{ {\text{m}}} \to$ magnetic field due to magnetic material ${\text{B}}_{ {\text{m}}}=\mu _{ {\text{r}}} {\text{B}}_0$ $\Delta {\text{B}}=\;{ {\text{B}}_{ {\text{m}}}- {\text{B}}_0}/{ {\text{B}}_0} \times 100=\;{\mu _{ {\text{r}}} {\text{B}}_0- {\text{B}}_0}/{ {\text{B}}_0} \times 100$ $\Delta {\text{B}} \%=\;{( {\text{X}}+1)-1}/{1} \times 100= {\text{X}} \times 100$ $\Delta {\text{B}} \%=2 \times 10^{-2} \times 100=2 \%$ Ans. Option (3)

Q16JEE Main 2023MCQ4MRay Optics and Optical Instruments

The critical angle for a denser-rarer interface is $45^{\circ}$ . The speed of light in rarer medium is $3 \times 10^8 {\text{m}} / {\ s}$ . The speed of light in the denser medium is : [11-Apr-2023 shift 1]

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📖 Explanation

$\;{\text{Sin}} {\text{ic}}=\;{\mu _{ {\text{r}}}}/{\mu _{ {\text{d}}}} \;\; \Rightarrow sin \;45^{\circ}=\;{\mu _{ {\text{r}}}}/{\mu _{ {\text{d}}}}$ $\;\Rightarrow \;{\mu _{ {\text{r}}}}/{\mu _{ {\text{d}}}}={\sqrt{2}}$ . . . (1)We know $\; {\text{V}} \propto \;\frac{1}{\mu } \Rightarrow \;{ {\text{V}}_{ {\text{d}}}}/{ {\text{V}}_{ {\text{r}}}}=\;{\mu _{ {\text{r}}}}/{\mu _{ {\text{d}}}}$ $\;=\;{ {\text{V}}_{ {\text{d}}}}/{3 \times 10^8}=\;{1}/{{\sqrt{2}}}$ $\; {\text{V}}_{ {\text{d}}}=\;{3}/{{\sqrt{2}}} \times 10^8=3 \times 0.7 \times 10^8$ $\; {\text{V}}_{ {\text{d}}}=2.12 \times 10^8 {\text{m}} / {\ sec}$ Ans. Option (1)

Q17JEE Main 2023MCQ4MUnits and Measurements

Given below are two statements :Statements I : Astronomical unit $( {\text{Au}})$ , Parsec $( {\text{Pc}})$ and Light year (ly) are units for measuring astronomical distances.Statements II : ${\text{Au}}<$ Parsec $( {\text{Pc}})<$ lyIn the light of the above statements, choose the most appropriate answer from the options given below : [11-Apr-2023 shift 1]

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📖 Explanation

A.V., Par sec and light year are the unit of distance Light year $\to$ distance travelled by light in one year $1 {\text{ly}}=9.5 \times 10^{15} {\text{m}}$ parcec $=3.262$ light year A.V. $=1.58 \times 10^{-5}$ light year A. V. $< 1 {\text{y}} <$ Parsec. Statement I correct and statement II incorrect.

Q18JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

The current sensitivity of moving coil galvanometer is increased by $25 \%$ .This increase is achieved only by changing in the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be : [11-Apr-2023 shift 1]

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📖 Explanation

$\tau = {\text{mB}}\; {\text{A}}=\;\text{ area of coil }\;$ ${\text{K}} \theta= {\text{IANB}}\; {\text{B}}=\;\text{ magnetic field }\;$ $\;{\theta}/{ {\text{I}}}=\;{ {\text{ANB}}}/{ {\text{K}}}\;\;\text{ Currect senstivity }\;$ $1.25 (\;{\theta}/{ {\text{I}}}) _2= (\;{\theta}/{ {\text{I}}}) _1\;. . . . . . . . . .(1)$ $1.25 [\;{ {\text{AN}}_2 {\text{B}}}/{ {\text{K}}}] = [\;{ {\text{AN}}_1 {\text{B}}}/{ {\text{K}}}]$ $1.25=\;{ {\text{N}}_1}/{ {\text{N}}_2}=\;\frac{5}{4}\;. . . . . .(2)$ $\Rightarrow {\text{R}}=\;{\delta ℓ}/{ {\text{a}}}=\;\text{ const. }\;\;$ $\Rightarrow ℓ= {\text{a}}$ Voltage sensitivity $=\;\frac{\theta}{V}=\;\frac{\theta}{ℝ}=\;{\;\text{ Current sensitivity }\;}/{ {\text{R}}}$ ${\text{R}}=$ constantVoltage sensitivity $\propto$ current sensitivity

Q19JEE Main 2023MCQ4MDual Nature of Matter and Radiation

A metallic surface is illuminated with radiation of wavelength $\lambda$ , the stopping potential is $V_0$ . If the same surface is illuminated with radiation of wavelength $2 \lambda$ , the stopping potential becomes $\;\frac{V_0}{4}$ . The threshold wavelength for this metallic surface will be [11-Apr-2023 shift 1]

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📖 Explanation

${\text{E}}= {\text{K}} \cdot {\text{E}}+\varphi _0$ Now $\;{ {\text{hc}}}/{\lambda }= {\text{ev}}_0+\varphi _0$ . . . (1) And $\;{ {\text{hc}}}/{2 \lambda }=\;{ {\text{eV}}_0}/{4}+\varphi _0$ . . . (2) $\;(2) \times 4$ . . . (1) $\;\;{2 {\text{hc}}}/{\lambda }-\;{ {\text{hc}}}/{\lambda }=0+ (4 \varphi _0-\varphi _0)$ $\;\;{ {\text{hc}}}/{\lambda }=3 \varphi _0$ $\;\;{ {\text{hc}}}/{\lambda }=3 \;{ {\text{hc}}}/{\lambda _0}$ $\;\lambda _0=3 \lambda$

Q20JEE Main 2023MCQ4MThermodynamics

$1 {\text{kg}}$ of water at $100^{\circ} {\text{C}}$ is converted into steam at $100^{\circ} {\text{C}}$ by boiling at atmospheric pressure. The volume of water changes from $1.00 \times 10^{-3} {\text{m}}^3$ as a liquid to $1.671 {\text{m}}^3$ as steam. The change in internal energy of the system during the process will be(Given latent heat of vaporisation $=2257 {\text{kJ}} / {\text{kg}}$ , Atmospheric pressure $=1 \times 10^5 {\text{Pa}}$ ) [11-Apr-2023 shift 1]

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📖 Explanation

Change in volume at constant pressure and temp $\to$ $\Delta {\text{V}}=\; {\text{V}}_2- {\text{V}}_1=1.671-0.001$ $\Delta {\text{V}}=\;1.67 {\text{m}}^3 \;\; . . . . . .(1)$ $\Delta {\text{Q}}=\;\Delta {\text{U}}+ {\text{w}}$ ${\text{mL}}_{ {\text{v}}}=\;\Delta {\text{U}}+ (1.013 \times 10^5) (1.67)$ $\;\Delta {\text{U}}=(2257-170) 10^3$ $\;\Delta {\text{U}}=2090 {\text{kJ}} \;\text{ (approx.) }\; \;\; \;\text{ Ans. Option }\; \to 4$

Q21JEE Main 2023NAT4MRay Optics and Optical Instruments

The radius of curvature of each surface of a convex lens having refractive index 1.8 is $20 {\text{cm}}$ . The lens is now immersed in a liquid of refractive index 1.5. The ratio of power of lens in air to its power in the liquid will be $x: 1$ . The value of $x$ is ________. [11-Apr-2023 shift 1]

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📖 Explanation

$\;\;{1}/{ {\text{f}}}= (\;{\mu _{ℓ}}/{\mu _{ {\text{m}}}}-1) (\;{1}/{ {\text{R}}}-\;{1}/{- {\text{R}}})$ $\; {\text{P}}_1=\;{2}/{ {\text{R}}} (\;\frac{1.8}{1}-1]$ $\; {\text{P}}_1=\;{2}/{ {\text{R}}}(0.8)=\;{1.6}/{ {\text{R}}}$ . . . (1)Now, $\; {\text{P}}_2=\;{2}/{ {\text{R}}} [\;\frac{1.8}{1.5}-1]$ $\; {\text{P}}_2=\;{2}/{ {\text{R}}} [\;\frac{0.3}{1.5}] =\;{2}/{ {\text{R}}} \times \;\frac{1}{5}=\;{2}/{5 {\text{R}}}$ $\;{ {\text{P}}_{\;\text{air }\;}}/{ {\text{P}}_{\;\text{liquid }\;}}=\;{ {\text{P}}_1}/{ {\text{P}}_2}=\;{ ({1.6}/{ {\text{R}}}) }/{ (\;{0.4}/{ {\text{R}}}) }=\;\frac{4}{1}$ Ans. $\to 4$

Q22JEE Main 2023NAT4MElectromagnetic Induction

The magnetic field B crossing normally a square metallic plate of area $4 {\text{m}}^2$ is changing with time as shown in figure. The magnitude of induced emf in the plate during $t=2 s$ to $t=4 s$ , is _______ ${\text{mV}}$ . [11-Apr-2023 shift 1]

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📖 Explanation

${\text{emf}}\;=\;{ {\text{d}} \varphi }/{ {\text{dt}}}$ ${\text{Emf}}\;=\;{ {\text{dBA}}}/{ {\text{dt}}}=\;{ {\text{AdB}}}/{ {\text{dt}}}$ ${\text{Emf}}\;=4 \cdot \;\text{ Slope of B.t curve }\;$ $\;=4 \cdot [\;\frac{8-4}{4-2}] =4 \times 2$ $\;\text{ Emf }\;\;=8 {\text{Volt}}$

Q23JEE Main 2023NAT4MProperties of Matter

The length of a wire becomes $l_1$ and $l_2$ when $100 {\text{N}}$ and $120 {\text{N}}$ tensions are applied respectively. If $10 l_2=11 l_l$ , the natural length of wire will be $\;\frac{1}{x} l_1$ . Here the value of $x$ is ________. [11-Apr-2023 shift 1]

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📖 Explanation

${\text{F}}= {\text{kx}}$ ${\text{F}}=\;{ {\text{yA}}}/{ℓ_0} \cdot {\text{x}} \;\text{. }\;$ Sol when ${\text{F}}=100 {\text{N}}$ $100= {\text{k}} (ℓ_1-ℓ_0)$ . . . (1)When ${\text{F}}=120 {\text{N}}$ $120= {\text{K}} ( (ℓ_1-ℓ_0) .$ Given that $10 ℓ_2\;=11 ℓ_1$ $ℓ_2\;=1.1 ℓ_1$ So $\;\; 120= {\text{K}} (1.1 ℓ_1-ℓ_0)$ . . . (2) Now (2) $\\ (1)$ $\;\frac{120}{100}=\;{ {\text{K}} (1.1 ℓ_1-ℓ_0) }/{ {\text{K}} (ℓ_1-ℓ_0) }$ $1.2=\;\frac{1.1 ℓ_1-ℓ_0}{ℓ_1-ℓ_0}$ $1.2 ℓ_1-1.2 ℓ_0=1.1 ℓ_1-ℓ_0$ $0.1 ℓ_1=0.2 ℓ_0$ $ℓ_0=\;\frac{ℓ_1}{2}$ So ${\text{x}}=2$ Ans.

Q24JEE Main 2023NAT4MAtoms and Nuclei

A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is ${\text{x}} \times 10^{15}$ ${\text{Hz}}$ . The value of ${\text{x}}$ is _________.(Given ${\text{h}}=4.25 \times 10^{-15} {\text{eVs}}$ ) [11-Apr-2023 shift 1]

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📖 Explanation

Total emission lines $=6$ (given $)$ So electron absorbed energy and jump from $n=1$ to ${\text{n}}=4$ $\Delta {\text{E}} 13.6 \$ [;\frac{ℓ}{ℓ^2}-;\frac{1}{4^2}]$ {\text{ev}};=13.6 [1-;\frac{1}{16}] {\text{ev}};\Delta {\text{E}}= {\text{hf}};12.75=4.5 \times 10^{-15},; {\text{f}}=;\frac{12.75}{4.25} \times 10^{15}=3 \times 10^{15} {\text{Hz}}; {\text{x}}=3 ;\text{ Ans. };$

Q25JEE Main 2023NAT4MWork, Power and Energy

A force ${ {\text{F}}}^{\to }=(2+3 {\text{x}}) { {\text{i}}^{\^}}$ acts on a particle in the ${\text{x}}$ direction where ${\text{F}}$ is in newton and ${\text{x}}$ is in meter. The work done by this force during a displacement from $x=0$ to $x=4 {\text{m}}$ , is ________ mJ. [11-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;{ {\text{F}}}^{\to }=(2+3 {\text{x}}) {\text{i}}$ $\; {\text{w}}=int_{0}^{4} {\text{F}} \cdot {\text{dx}}=int_{0}^{4}(2+3 {\text{x}}) \cdot {\text{dx}}$ $\; {\text{w}}= (2 {\text{x}}+\;{3 {\text{x}}^2}/{2}) ^4=(8+24)$ $\; {\text{w}}=32 {\text{J}}$

Q26JEE Main 2023NAT4MElectrostatics

As shown in the figure, a configuration of two equal point charges $( {\text{q}}_0=+2 \mu {\text{C}})$ is placed on an inclined plane. Mass of each point charge is $20 {\text{g}}$ . Assume that there is no friction between charge and plane. For the system of two point charges to be in equilibrium (at rest) the height ${\text{h}}= {\text{x}} \times 10^{-3} {\text{m}}$ . The value of ${\text{x}}$ is ________. (Take $\;\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 {\text{Nm}}^2 {\text{C}}^{-2}, {\text{g}}=10 {\text{ms}}^{-2}$ ) [11-Apr-2023 shift 1]

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📖 Explanation

Point charge on equilibrium is at rest. $\;\;\text{ So }\; \;\; {\text{F}}_{ {\text{e}}}= {\text{mg}} sin \;\theta$ $\;\;{ {\text{kq}}_0 \cdot {\text{q}}_0}/{ {\text{r}}^2}= {\text{mg}} sin \;30^{\circ}$ $\;\;{ {\text{kq}}_0^2}/{ (\;{ {\text{h}}}/{ sin \;30^{\circ}}) ^2}=\;{ {\text{mg}}}/{2}$ $\;\;{9 \times 10^9 \times (2 \times 10^{-6}) ^2}/{4 {\text{h}}^2}=\;\frac{20 \times 10^{-3} \times 10}{2}$ $\;\;{9 \times 4 \times 10^9 \times 10^{-12}}/{4 {\text{h}}^2}=10^{-1}$ $\; {\text{h}}^2=9 \times 10^{-2} \;\; \;\text{ Ans. }\;$ $\; {\text{h}}=0.3 {\text{m}}=300 \times 10^{-3} {\text{m}}$ $\; {\text{x}}=300 \;\; \;\text{ Ans. }\;$

Q27JEE Main 2023NAT4MRotational Motion

A solid sphere of mass $500 {\text{g}}$ and radius $5 {\text{cm}}$ is rotated about one of its diameter with angular speed of $10 {\text{rad}}$ ${\ s}^{-1}$ . If the moment of inertia of the sphere about its tangent is ${\text{x}} \times 10^{-2}$ times its angular momentum about the diameter. Then the value of ${\text{x}}$ will be ________. [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{I}}_1=\;\frac{2}{5} {\text{mR}}^2$ $\; {\text{I}}_2=\;\frac{2}{5} {\text{mR}}^{2+} {\text{mR}}^2=\;\frac{7}{5} {\text{mR}}^2$ Angular moment about diameter is ${\text{L}}_{ {\text{com}}}= {\text{I}}_1 {\text{w}}=\;\frac{2}{5} {\text{mR}}^2 {\text{w}}$ Now, $\;\;{ {\text{I}}_2}/{ {\text{L}}_{\;\text{com }\;}}=\;{\;\frac{7}{5} {\text{mR}}^2}/{\frac{2}{5} {\text{mR}}^2 {\text{w}}}=\;\frac{7}{2} {\text{w}}$ $\;\;{ {\text{I}}_2}/{ {\text{L}}_{\;\text{com }\;}}=\;\frac{7}{2 \times 10}=\;\frac{7}{20}$ Now $\;\frac{7}{20}= {\text{x}} \times 10^{-2}$ $\;x=\;\frac{7}{20} \times 100$ $\;x=35$ Ans.

Q28JEE Main 2023NAT4MWaves

The equation of wave is given by $Y=10^{-2} sin \;2 \pi(160 t-0.5 x+\pi / 4)$ where ${\text{x}}$ and ${\text{Y}}$ are in ${\text{m}}$ and ${\text{t}}$ in ${\ s}$ . The speed of the wave is ________ ${\text{km}} {\text{h}}^{-1}$ . [11-Apr-2023 shift 1]

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📖 Explanation

$Y=10^{-2} sin \;2 \pi(160 t-0.5 x+\pi / 4)$ Speed of wave $=\;{ {\text{w}}}/{ {\text{k}}}=\;\frac{160}{0.5}=320 {\text{m}} / {\ sec}=320 \times \;\frac{18}{5}=1152 {\text{km}} / {\text{h}} \;\;$ Ans.

Q29JEE Main 2023NAT4MCurrent Electricity

In the circuit diagram shown in figure given below, the current flowing through resistance $3 Ω$ is $\;\frac{x}{3} A$ . The value of $x$ is ________ [11-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Req. }\;=0.5+1+4.5+ (\;\frac{3.6}{9})$ $\;\;\text{ Req. }\;=6+2=8 Ω$ $\; {\text{I}}=\;\frac{8-4}{8}=\;\frac{1}{2} {\text{A}}=0.5 {\text{A}}$ $\; {\text{I}}_1: {\text{I}}_2=\;\frac{1}{3}: \;\frac{1}{6}$ $\; {\text{I}}_1: {\text{I}}_2=2: 1$ $\;\;\text{ and }\; {\text{I}}_1+ {\text{I}}_2=0.5 {\text{A}}$ $\; {\text{I}}_1=\;\frac{2}{3} \times 0.5=\;\frac{1}{3} {\text{A}}$ $\;\;\text{ So }\; \;\frac{1}{3}=\;{ {\text{x}}}/{3} \Rightarrow {\text{x}}=1$

Q30JEE Main 2023NAT4MMotion in a Plane

A projectile fired at $30^{\circ}$ to the ground is observed to be at same height at time $3 {\ s}$ and $5 {\ s}$ after projection, during its flight. The speed of projection of the projectile is _______ ${\text{ms}}^{-1}$ . $($ Given ${\text{g}}=10 {\text{ms}}^{-2}$ ) [11-Apr-2023 shift 1]

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📖 Explanation

Time of flight $=5+3=8 {\ sec}$ .Now, $\;\; T=\;\frac{2 u sin \;\theta}{g}$ $\;8=\;{2 {\text{u}} \cdot sin \;30^{\circ}}/{10}$ $\;\Rightarrow {\text{u}}=80 {\text{m}} / {\ sec}$ Ans.

Chemistry30 questions

Q31JEE Main 2023MCQ4MCoordination Compounds

Which of the following complex has a possibility to exist as meridional isomer? [11-Apr-2023 shift 1]

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📖 Explanation

$[ {\text{MA}}_3 {\text{B}}_3]$ type of compound exists as facial and meridional isomer.

Q32JEE Main 2023MCQ4MBiomolecules

L-isomer of tetrose ${\text{X}} ( {\text{C}}_4 {\text{H}}_8 {\text{O}}_4)$ gives positive schiff's test and has two chiral carbons. On acetylation. ' ${\text{X}}$ ' yields triacetate. ' ${\text{X}}$ ' undergoes following reactions $'A' {◀─────}^{ {\text{HNO}}_3} {\text{X}}^{'} {\to }^{ {\text{NaBH}}_4} { '{\text{B}}'}_{\text{Chiral} \text{compound}}$ ' ${\text{X}}$ ' is [11-Apr-2023 shift 1]

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📖 Explanation

(x) gives positive schiff's test due $- {\text{CHO}}$ group (x) is L-tetrose.

Q33JEE Main 2023MCQ4MClassification of Elements

Match list I with list II: List I List II A. ${\text{K}}$ I. Thermonuclear ractions B. ${\text{KCl}}$ II. Fertilizer C. ${\text{KOH}}$ III. Sodium potassium pump D. Li IV. Absorbent of ${\text{CO}}_2$ Choose the correct answer from the options given below: [11-Apr-2023 shift 1]

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📖 Explanation

${\text{K}}^{+}$ -Sodium-Potassium Pump ${\text{KCl}}$ -Fertiliser ${\text{KOH}}-$ absorber of ${\text{CO}}_2$ ${\text{Li}}$ - used in thermonuclear reactions

Q34JEE Main 2023MCQ4Mp Block Elements

For compound having the formula ${\text{GaAlCl}}_4$ , the correct option form the following is [11-Apr-2023 shift 1]

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📖 Explanation

Gallous tetrachloro aluminate ${\text{Ga}}^{+} {\text{AlCl}}_4^{-}$ $2 {\text{Ga}}+ {\text{Ga}}^{+} {\text{Cl}}_4^{-}+2 {\text{Al}}_2 {\text{Cl}}_6 {\to }^{190^{\circ}} 4 {\text{Ga}}^{+} {\text{AlCl}}_4^{-}$ Structure of ${\text{Ga}}^{+} {\text{AlCl}}_4^{-}$ ${\text{Ga}}$ is cationic part of salt ${\text{GaAlCl}}_4$ .

Q35JEE Main 2023MCQ4MPractical Organic Chemistry

Thin layer chromatography of a mixture shows the following observation :The correct order of elution in the silica gel column chromatography is [11-Apr-2023 shift 1]

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📖 Explanation

According to the observation, ${\text{A}}$ is more mobile and interacts with the mobile phase more than ${\text{C}}$ , and ${\text{C}}$ is more drawn to the mobile phase than $B$ .Hence, the correct order of elution in the silica gel column chromatography is - ${\text{B}} < {\text{C}} < {\text{A}}$

Q36JEE Main 2023MCQ4MSalt Analysis

When a solution of mixture having two inorganic salts was treated with freshly prepared ferrous sulphate in acidic medium, a dark brown ring was formed whereas on treatment with neutral ${\text{FeCl}}_3$ . it gave deep red colour which disappeared on boiling and a brown red ppt was formed. The mixture contains [11-Apr-2023 shift 1]

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Q37JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

The polymer X-consists of linear molecules and is closely packed. It prepared in the presence of triethylaluminium and titranium tetrachloride under low pressure. The polymer ${\text{X}}$ is- [11-Apr-2023 shift 1]

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📖 Explanation

Ethene undergoes addition polymerisation to high density polythene in the presence of catalyst such as AlEt 3 and ${\text{TiCl}}_4$ (Ziegler - Natta catalyst) at a temperature of $333 {\text{K}}$ to $343 {\text{K}}$ and under a pressure of $6-7$ atmosphere.

Q38JEE Main 2023MCQ4MChemical Bonding and Molecular Structure

Match list I with list II List I Species List II Geometry/ Shape A. ${\text{H}}_3 {\text{O}}^{+}$ I. Tetrahedral B. Acetylide anion II. Linera C. ${\text{NH}}_4^{+}$ III. Pyramidal D. ${\text{ClO}}_2^{-}$ IV. BentChoose correct answer from the options given below: [11-Apr-2023 shift 1]

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Q39JEE Main 2023MCQ4Mp Block Elements

Given below are two statement :Statement I : Methane and steam passed over a heated Ni catalyst produces hydrogen gasStatement II : Sodium nitrite reacts with ${\text{NH}}_4 {\text{Cl}}$ to give ${\text{H}}_2 {\text{O}}, {\text{N}}_2$ and ${\text{NaCl}}$ In the light of the above statements, choose the most appropriate answer from the options given below: [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{CH}}_4( {\text{g}})+ {{\text{H}}_2 {\text{O}}( {\text{g}}) }_{\text{Steam}}\;{ {\text{Ni}}}/{1270 {\text{K}}} \to {\text{CO}}( {\text{g}})+3 {\text{H}}_2( {\text{g}})$ $\; {\text{NaNO}}_2 \;\text{ (aq) }\;+ {\text{NH}}_4 {\text{Cl}}(\;\text{ aq }\;) \to {\text{N}}_2( {\text{g}})+ {\text{NaCl}}( {\text{aq}})+2 {\text{H}}_2 {\text{O}}(ℓ)$

Q40JEE Main 2023MCQ4MCoordination Compounds

The set which does not have ambidentate ligand (s) is [11-Apr-2023 shift 1]

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📖 Explanation

${\text{NO}}_2^{-} {\text{NCS}}^{-}$ are ambidentate ligandEDTA Ethylene diamine tetra acetate

Q41JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

Arrange the following compounds in increasing order of rate of aromatic electrophilic substitution reaction [11-Apr-2023 shift 1]

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📖 Explanation

Benzene becomes more reactive towards EAS when any substituent raises the electron density.Correct order ${\text{c}} < {\text{a}} < {\text{b}} < {\text{d}}$

Q42JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

Where ${\text{Nu}}=$ NucleophileFind out the correct statement from the options given below for the above 2 reactions. [11-Apr-2023 shift 1]

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📖 Explanation

Electron Donating group ${\text{S}}_{ {\text{N}}}{ }^1 {\text{Mech}} .: {\text{I}}^{\;\text{st }\;}$ orderElectron withdrawing group ${\text{S}}_{ {\text{N}}^2}$ Mech: $2^{\;\text{nd }\;}$ order

Q43JEE Main 2023MCQ4MAmines

o-Phenylenediamine ${\to }^{ {\text{HNO}}_2}$ ' ${\text{X}}$ ' Major Product ' ${\text{X}}$ ' is [11-Apr-2023 shift 1]

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Q44JEE Main 2023MCQ4MClassification of Elements

For elements B, C, N, Li, Be, O and F, the correct order of first ionization enthalpy is [11-Apr-2023 shift 1]

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📖 Explanation

First I.E. $\; {\text{F}} > {\text{N}} > {\text{O}} > {\text{C}} > {\text{Be}} > {\text{B}}> {\text{Li}}$ $\; {\text{Li}}-520 {\text{kJ}} / {\text{mol}}$ $\; {\text{Be}}-899 {\text{kJ}} / {\text{mol}}$ $\; {\text{B}}-801 {\text{kJ}} / {\text{mol}}$ $\; {\text{C}}-1086 {\text{kJ}} / {\text{mol}}$ $\; {\text{N}}-1402 {\text{kJ}} / {\text{mol}}$ $\; {\text{O}}-1314 {\text{kJ}} / {\text{mol}}$ $\; {\text{F}}-1681 {\text{kJ}} / {\text{mol}}$

Q45JEE Main 2023MCQ4Md and f Block Elements

In the extraction process of copper, the product obtained after carrying out the reactions(i) $2 {\text{Cu}}_2 {\text{S}}+3 {\text{O}}_2 \to 2 {\text{Cu}}_2 {\text{O}}+2 {\text{SO}}_2$ (ii) $2 {\text{Cu}}_2 {\text{O}}+ {\text{Cu}}_2 {\text{S}} \to 6 {\text{Cu}}+ {\text{SO}}_2$ is called [11-Apr-2023 shift 1]

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📖 Explanation

$\;2 {\text{Cu}}_2 {\text{S}}+3 {\text{O}}_2 \to 2 {\text{Cu}}_2 {\text{O}}+3 {\text{SO}}_2$ $\;2 {\text{Cu}}_2 {\text{O}}+ {\text{Cu}}_2 {\text{S}} \to { 6 {\text{Cu}}+ {\text{SO}}_2}_{\text{Blister} \text{copper}}$ Due to evolution of ${\text{SO}}_2$ , the solidified copper formed has a blistered look and is referred to as blister copper.

Q46JEE Main 2023MCQ4MIonic Equilibrium

$25 {\text{mL}}$ of silver nitrate solution (1M) is added dropwise to $25 {\text{mL}}$ of potassium iodide $(1.05 {\text{M}})$ solution. The ion(s) present in very small quantity in the solution is/are [11-Apr-2023 shift 1]

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📖 Explanation

On adding ${\text{AgNO}}_3$ into ${\text{KI}}, {\text{AgI}}$ will form and solubility of ${\text{AgI}}$ is very low. So, $[ {\text{Ag}}^{+}]$ and $[Γ]$ will be present in very small quantity.

Q47JEE Main 2023MCQ4MSome Basic Concepts of Chemistry

Given below are two statements:Statement I : If BOD is $4 {\text{ppm}}$ and dissolved oxygen is $8 {\text{ppm}}$ , it is a good quality water.Statement II : If the concentration of zinc and nitrate salts are $5 {\text{ppm}}$ each, than it can be good quality water.In the light of the above statements choose the most appropriate answer from the options given below: [11-Apr-2023 shift 1]

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📖 Explanation

Clean water would have BOD value of less than $5 {\text{ppm}}$ .Maximum limit of ${\text{Zn}}$ in clean water $=5.0 {\text{ppm}}$ or ${\text{mg}} {\text{dm}}^{-3}$ Maximum limit of ${\text{NO}}_3{ }^{-}$ in clean water $=50 {\text{ppm}}$ or ${\text{mg}} {\text{dm}}^{-3}$

Q48JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

' ${\text{A}}$ ' and ' ${\text{B}}$ ' in the above reactions are : [11-Apr-2023 shift 1]

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Q49JEE Main 2023MCQ4MStructure of Atom

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R: Assertion A : In the photoelectric effect electrons are ejected from the metal surface as soon as the beam of light of frequency greater than threshold frequency strikes the surface. Reason R: When the photon of any energy strikes an electron in the atom transfer of energy from the photon to the electron takes place. In the light of the above statements, choose the most appropriate answer from the options given below: [11-Apr-2023 shift 1]

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Q50JEE Main 2023MCQ4MCoordination Compounds

The complex that dissolves in water is [11-Apr-2023 shift 1]

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📖 Explanation

${\text{Fe}}_4 [ {\text{Fe}}( {\text{CN}})_6] _3$ Prussian Blue-water insoluble ${\text{K}}_3 [ {\text{Co}} ( {\text{NO}}_2) _6]$ very poorly water soluble $( {\text{NH}}_4) _3 [ {\text{As}} ( {\text{MO}}_3 {\text{O}}_{10}) _4]$ water insoluble ammonium arseno molybdate $[ {\text{Fe}}_3( {\text{OH}})_2( {\text{OAc}})_6] {\text{Cl}}$ is water soluble.

Q51JEE Main 2023NAT4MChemical Thermodynamics

Solid fuel used in rocket is a mixture of ${\text{Fe}}_2 {\text{O}}_3$ and ${\text{Al}}$ (in ratio $1: 2$ ) the heat evolved $( {\text{KJ}})$ per gram of the mixture is ________ (Nearest integer)Givne $\Delta {\text{H}}_{ {\text{f}}}^{⊖} ( {\text{Al}}_2 {\text{O}}_3) =-1700 {\text{KJ}} {\text{mol}}^{-1}$ $\Delta {\text{H}}_{ {\text{f}}}^{⊖} ( {\text{Fe}}_2 {\text{O}}_3) =-840 {\text{KJ}} {\text{mol}}^{-1}$ [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{Fe}}_2 {\text{O}}_3+2 {\text{Al}} \to {\text{Al}}_2 {\text{O}}_3+2 {\text{Fe}}$ $\Delta {\text{H}}_{ {\text{r}}}\;= (\Delta {\text{H}}_{ {\text{f}}}) {\text{Al}}_2 {\text{O}}_3-\Delta {\text{H}}_{ {\text{f}}}^{\circ} ( {\text{Fe}}_2 {\text{O}}_3)$ $\;=-1700-(-840)$ $\;=-860 {\text{kJ}}$ $\; {\text{Fe}}_2 {\text{O}}_3 & {\text{Al}} \to 1: 2$ ${\text{Fe}}_2 {\text{O}}_3=1 {\text{mole}}\;=(2 \times 25+48)$ $\;=112+48=160 {\text{gm}}$ ${\text{Al}}=2 {\text{mole}}=2 \times 27=54 {\text{gm}}$ Total mass $=160+54=214 {\text{gm}}$ Heat evolved per gm $=\;\frac{-860}{214} {\text{kJ}}=-4.01 \approx 4 {\text{kJ}}$

Q52JEE Main 2023NAT4MChemical Kinetics

${\text{KClO}}_3+6 {\text{FeSO}}_4+3 {\text{H}}_2 {\text{SO}}_4 \to {\text{KCl}}+3 {\text{Fe}}_2 ( {\text{SO}}_4) _3+3 {\text{H}}_2 {\text{O}}$ The above reaction was studied at $300 {\text{K}}$ by monitoring the concentration of ${\text{FeSO}}_4$ in which initial concentration was $10 {\text{M}}$ and after half an hour became $8.8 {\text{M}}$ . The rate of production of ${\text{Fe}}_2 ( {\text{SO}}_4) _3$ is ________ $\times 10^{-6} {\text{mol}} {\text{L}}^{-1} {\ s}^{-1}$ [11-Apr-2023 shift 1]

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📖 Explanation

$\;{-\Delta {\text{FeSO}}_4}/{\Delta {\text{t}}}=\;\frac{10-8.8}{30 \times 60}=\;\frac{1.2}{1800}$ From given equation : $-\;\frac{1}{6} \;{\Delta {\text{FeSO}}_4}/{\Delta {\text{t}}}=\;\frac{1}{3} \times (\;\text{ Rate of production of }\; {\text{Fe}}_2 ( {\text{SO}}_4) _3)$ Rate of production of ${\text{Fe}}_2 ( {\text{SO}}_4) _3=\;\frac{3}{6} \times \;\frac{1.2}{1800}$ $\;=\;\frac{1}{3} \times 10^{-3}$ $\;=\;\frac{1000}{3} \times 10^{-6}$ $\;=333.33 \times 10^{-6}$

Q53JEE Main 2023NAT4MSolutions

$0.004 \; {\text{M}} {\text{K}}_2 {\text{SO}}_4$ solution is isotonic with $0.01 {\text{M}}$ glucose solution. Percentage dissociation of ${\text{K}}_2 {\text{SO}}_4$ is (Nearest integer) [11-Apr-2023 shift 1]

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📖 Explanation

For isotonic solution $\;( {\text{ic}})_{\;\text{glucose }\;}=( {\text{ic}})_{ {\text{K}}_2 {\text{SO}}_4}$ $\;0.01= {\text{i}}(0.004)$ $\; {\text{i}}=\;\frac{0.01}{0.004}=\;\frac{10}{4}=\;\frac{5}{2}$ $\;1+( {\text{n}}-1) \alpha=\;\frac{5}{2}$ $\;1+(3-1) \alpha=\;\frac{5}{2} \;\; ( \because {\text{n}}=3 \;\text{ for }\; {\text{K}}_2 {\text{SO}}_4)$ $\;2 \alpha=\;\frac{3}{2}$ $\;\alpha=\;\frac{3}{4} \to 75 \%$

Q54JEE Main 2023NAT4MOrganic Chemistry - Some Basic Principles

The number of hyperconjugation structures involved to stabilize carbocation formed in the above reaction is ________ [11-Apr-2023 shift 1]

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Q55JEE Main 2023NAT4MChemical Equilibrium

A mixture of 1 mole of ${\text{H}}_2 {\text{O}}$ and 1 mole of ${\text{CO}}$ is taken in a 10 litre container and heated to $725 {\text{K}}$ . At equilibrium $40 \%$ of water by mass reacts with carbon monoxide according to the equation : ${\text{CO}}( {\text{g}})+ {\text{H}}_2 {\text{O}}( {\text{g}}) ⇌$ ${\text{CO}}_2( {\text{g}})+ {\text{H}}_2( {\text{g}})$ . The equilibrium constant ${\text{K}}_{ {\text{e}}} \times 10^2$ for the reaction is ________ (Nearest integer) [11-Apr-2023 shift 1]

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📖 Explanation

${ {\text{CO}}( {\text{g}})}_{1 \text{mole}}+{ {\text{H}}_2 {\text{O}}( {\text{g}}) }_{1 \text{mole}} ⇌ {\text{CO}}_2( {\text{g}})+ {\text{H}}_2( {\text{g}})$ $\; {\text{K}}_{ {\text{c}}}=\;\frac{0.4 \times 0.4}{0.6 \times 0.6}=\;\frac{4}{9}$ $\; {\text{K}}_{ {\text{c}}} \times 10^2=\;\frac{4}{9} \times 100=\;\frac{400}{9}=44.44 \approx 44$

Q56JEE Main 2023NAT4MSolid State

An atomic substance $A$ of molar mass $12 {\text{g}} {\text{mol}}^{-1}$ has a cubic crystal structure with edge length of $300 {\pm}$ . The no. of atoms present in one unit cell of $A$ is ______ (Nearest integer)Given the density of $A$ is $3.0 {\text{g}} {\text{mL}}^{-1}$ and ${\text{N}}_{ {\text{A}}}=6.02 \times 10^{23} {\text{mol}}^{-1}$ [11-Apr-2023 shift 1]

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📖 Explanation

${\text{d}}\;=\;{\;{ {\text{Z}}}/{ {\text{N}}_{ {\text{A}}}} \times {\text{M}}}/{ {\text{a}}^3}$ $3\;=\;{ {\text{Z}}}/{6.02 \times 10^{23}} \times \;\frac{12}{ (300 \times 10^{-10}) ^3}$ ${\text{Z}}\;=\;\frac{3 \times 6.02 \times 27 \times 10^6 \times 10^{-30} \times 10^{23}}{12}$ $\;=40.635 \times 10^{-1}=4.0635 \approx 4$

Q57JEE Main 2023NAT4MAldehydes, Ketones and Carboxylic Acids

The ratio ${\text{x}} / {\text{y}}$ on completion of the above reaction is [11-Apr-2023 shift 1]

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📖 Explanation

$\therefore x=2$ mole $\;\frac{x}{y}=\;\frac{2}{1}=2$

Q58JEE Main 2023NAT4MCoordination Compounds

The ratio of spin-only magnetic moment values $\mu _{\;\text{eff }\;} [ {\text{Cr}}( {\text{CN}})_6] ^{3-} / \mu _{\;\text{eff }\;} [ {\text{Cr}} ( {\text{H}}_2 {\text{O}}) _6] ^{3+}$ is ________. [11-Apr-2023 shift 1]

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📖 Explanation

Spin magnetic moment of $[ {\text{Cr}}( {\text{CN}})_6] ^{3-} ( {\text{t}}_{2 {\text{g}}}^3 {\text{eg}}_{ {\text{g}}}^0)$ $\mu _1={\sqrt{3(3+2)}}={\sqrt{15}} {\text{BM}}$ Spin magnetic moment of $[ {\text{Cr}} ( {\text{H}}_2 {\text{O}}) _6] ^{3+} ( {\text{t}}_{2 {\text{g}}}^3 {\text{eg}}_{ {\text{g}}}^0)$ $\;\mu _2={\sqrt{3(3+2)}}={\sqrt{15}} {\text{BM}}$ $\;\;\frac{\mu _1}{\mu _2}=\;{{\sqrt{51}}}/{{\sqrt{51}}}=1$

Q59JEE Main 2023NAT4MElectrochemistry

In an electrochemical reaction of lead, at standard temperature, if ${\text{E}}_{ ( {\text{pb}}^{2+} / {\text{pb}}) }^{\circ}= {\text{m}}$ volt and ${\text{E}}_{ ( {\text{pb}}^{+} / {\text{Pb}}_{ {\text{b}}}) }^{\circ}= {\text{n}}$ volt, then the value of ${\text{E}}_{ ( {\text{Pb}}^{2+} / {\text{Pb}}^{+}) }^{\circ}$ is given by ${\text{m}}- {\text{xn}}$ . The value of ${\text{x}}$ is ________ (Nearest integer) [11-Apr-2023 shift 1]

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Q60JEE Main 2023NAT4MSolutions

A solution of sugar is obtained by mixing $200 {\text{g}}$ of its $25 \%$ solution and $500 {\text{g}}$ of its $40 \%$ solution (both by mass). The mass percentage of the resulting sugar solution is _______ (Nearest integer) [11-Apr-2023 shift 1]

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📖 Explanation

$( {\text{I}}) \to$ Mass of sugar $=200 \times \;\frac{25}{100}=50 {\text{gm}}$ Mass of solution $=200 {\text{gm}}$ Solution (II) $\to$ Mass of solution $=500 {\text{gm}}$ Mass of sugar $=\;\frac{40}{100} \times 500=200 {\text{gm}}$ $\;\text{ Final }\; \% {\text{w}} / {\text{w}}\;=\;\frac{\;\text{ Total mass of sugar }\;}{\;\text{ Total mass of solution }\;} \times 100$ $\;=\;\frac{50+200}{200+500} \times 100=\;\frac{250}{7}$ $\;=35.71 \% \approx 36$

Mathematics30 questions

Q61JEE Main 2023MCQ4MSequences and Series

Let ${\text{x}}_1, {\text{x}}_2, . . . . ., {\text{x}}_{100}$ be in an arithmetic progression, with ${\text{x}}_1=2$ and their mean equal to 200 . If ${\text{y}}_{ {\text{i}}}= {\text{i}} ( {\text{x}}_{ {\text{i}}}- {\text{i}})$ , $1 \le i \le 100$ , then the mean of $y_1, y_2, . . . . . y_{100}$ is : [11-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Mean }\;=200$ $\;\Rightarrow \;{\;\frac{100}{2}(2 \times 2+99 {\text{d}})}/{100}=200$ $\;\Rightarrow 4+99 {\text{d}}=400$ $\;\Rightarrow {\text{d}}=4$ $\;y_i=i(x i-1)$ $\;=i(2+(i-1) 4-i)=3 i^{2-}2 i$ $\;\;\text{ Mean }\;=\;\frac{Σ y_i}{100}$ $\;=\;\frac{1}{100} sum_{{i=1}}^{{100}} 3 i^{2-}2 i$ $\;=\;\frac{1}{100} \{\;\frac{3 \times 100 \times 101 \times 201}{6}-\;\frac{2 \times 100 \times 101}{2}\}$ $\;=101 \{\;\frac{201}{2}-1\} =101 \times 99.5$ $\;=10049.50$

Q62JEE Main 2023MCQ4MTrigonometric Ratios and Identities

The number of elements in the set $S= \{\theta \in [0,2 \pi]: 3 \cos ^4 \theta-5 \cos ^2 \theta-2 sin \;^6 \theta+2=0\}$ is : [11-Apr-2023 shift 1]

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📖 Explanation

$\;3 \cos ^4 \theta-5 \cos ^2 \theta-2 sin \;^6 \theta+2=0$ $\;\Rightarrow 3 \cos ^4 \theta-3 \cos ^2 \theta-2 \cos ^2 \theta-2 sin \;^6 \theta+2=0$ $\;\Rightarrow 3 \cos ^4 \theta-3 \cos ^2 \theta+2 sin \;^2 \theta-2 sin \;^6 \theta=0$ $\;\Rightarrow 3 \cos ^2 \theta (\cos ^2 \theta-1) +2 sin \;^2 \theta ( sin \;^4 \theta-1) =0$ $\;\Rightarrow -3 \cos ^2 \theta sin \;^2 \theta+2 sin \;^2 \theta (1+ sin \;^2 \theta) \cos ^2 \theta-1$ $\;\Rightarrow sin \;^2 \theta \cos ^2 \theta (2+2 sin \;^2 \theta-3) =0$ $\;\Rightarrow sin \;^2 \theta \cos ^2 \theta (2 sin \;^2 \theta-1) =0$ $\;( {\text{C}} 1) sin \;^2 \theta=0 \to 3 \;\text{ solution; }\; \theta=\{0, \pi, 2 \pi\}$ $\;\;\text{ (C2) }\; \cos ^2 \theta=0 \to 2 \;\text{ solution; }\; \theta= \{\;\frac{\pi}{2}, \;\frac{3 \pi}{2}\}$ (C3) $sin \;^2 \theta=\;\frac{1}{2} \to 4$ solution; $\theta= \{\;\frac{\pi}{4}, \;\frac{3 \pi}{4}, \;\frac{5 \pi}{4}, \;\frac{7 \pi}{4}\}$ No. of solution $=9$

Q63JEE Main 2023MCQ4MDefinite Integrals

The value of the integral ${\int}_{-\log _e 2}^{\log _e 2} e^x (\log _e (e^x+\sqrt{1+e^{2 x}}) )$ is equal to : [11-Apr-2023 shift 1]

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📖 Explanation

$I={\int}_{-\ln 2}^{\ln 2} e^x (\ln (e^x+\sqrt{1+e^{2 x}}) ) \text{dx}$ Put ${\text{e}}^{ {\text{x}}}= {\text{t}} \Rightarrow {\text{e}}^{ {\text{x}}} {\text{dx}}= {\text{dt}}$ $I=int_{1 / 2}^{2} \ln (t+{\sqrt{1+t^2}}) \text{dt}$ Applying integration by parts. $\;= \$ [t \ln ( {\text{t}}+\sqrt{1+ {\text{t}}^2}) ]$ {;\frac{1}{2}}^{2-}int{1 / 2}^{2} ;{ {\text{t}}}/{ {\text{t}}+\sqrt{1+ {\text{t}}^2}} (1+;{2 {\text{t}}}/{2 \sqrt{1+ {\text{t}}^2}}) \text{dt};=2 \ln (2+{\sqrt{5}})-;\frac{1}{2} \ln (;{1+{\sqrt{5}}}/{2}) -int_{1 / 2}^{2} ;{ {\text{t}}}/{\sqrt{1+ {\text{t}}^2}} d {\text{t}};=2 \ln (2+{\sqrt{5}})-;\frac{1}{2} \ln (;{1+{\sqrt{5}}}/{2}) -;{{\sqrt{5}}}/{2};=\ln (;{(2+{\sqrt{5}})^2}/{ ( (;{{\sqrt{5+1}}}/{2}) ^{;\frac{1}{2}}.}) -;{{\sqrt{5}}}/{2}$

Q64JEE Main 2023MCQ4MProbability

Let $S= \{M= [a_{i j}] , a_{i j} \in \{0,1,2\}, 1 \le i, j \le 2\}$ be a sample space and $A=\{M \in S: M$ is invertible $\}$ be an event. Then ${\text{P}}( {\text{A}})$ is equal to : [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{M}}{\begin{bmatrix}{\text{a}} & {\text{b}} \\ {\text{c}} & {\text{d}}\end{bmatrix}}, \;\text{ where }\; {\text{a}}, {\text{b}}, {\text{c}}, {\text{d}}, \in \{0,1,2\}$ $\; {\text{n}}( {\ s})=3^4=81$ $\;\;\text{ we first bound p }\;(\overline{ {\text{A}}})$ $\;| {\text{m}}|=0 \Rightarrow {\text{ad}}= {\text{bc}}$ $\; {\text{ad}}= {\text{bc}}=0 \Rightarrow \;\text{ no. of }\;( {\text{a}}, {\text{b}}, {\text{c}}, {\text{d}})= (3^{2-}2^2) ^2=25$ $\; {\text{ad}}= {\text{bc}}=1 \Rightarrow \;\text{ no. of }\;( {\text{a}}, {\text{b}}, {\text{c}}, {\text{d}})=1^2=1$ $\; {\text{ad}}= {\text{bc}}=2 \Rightarrow \;\text{ no. of }\;( {\text{a}}, {\text{b}}, {\text{c}}, {\text{d}})=2^2=4$ $\; {\text{ad}}= {\text{bc}}=4 \Rightarrow \;\text{ no. of }\;( {\text{a}}, {\text{b}}, {\text{c}}, {\text{d}})=1^2=1$ $\;: {\text{P}}(\overline{ {\text{A}}})=\;\frac{31}{81} \Rightarrow {\text{p}}( {\text{A}})=\;\frac{50}{81}$

Q65JEE Main 2023MCQ4MApplication of Derivatives

Let $f:[2,4] \to ℝ$ be a differentiable function such that $(x \log _e x) f^{'}(x)+ (\log _e x) f(x)+f(x) \ge 1, x \in [2,4]$ with $f(2)=\;\frac{1}{2}$ and $f(4)=\;\frac{1}{4}$ . Consider the following two statements : (A) : $f(x) \le 1$ , for all $x \in [2,4]$ (B) : $f(x) \ge \;\frac{1}{8}$ , for all $x \in [2,4]$ Then, [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{x}} ℓ {\text{nxf}} {\text{f}}^{'}( {\text{x}})+\ln {\text{xf}}( {\text{x}})+ {\text{f}}( {\text{x}}) \ge 1, {\text{x}} \in [2,4]$ $\;\;\text{ And }\; f(2)=\;\frac{1}{2}, f(4)=\;\frac{1}{4}$ $\;\;\text{ Now }\; x \ln x, \;\frac{d y}{d x}+(\ln +1) y \ge 1$ $\;\;\frac{d}{d x}(y \cdot x \ln x) \ge 1$ $\;\;\frac{d}{d x}(f(x) \cdot x \ln x) \ge 1$ $\;\Rightarrow \;\frac{d}{d x}(x \ln x f(x)-x) \ge 0, x \in [2,4]$ $\;\Rightarrow \;\text{ The function }\; g(x)=x \ln x f(x)-x \;\text{ is increasing \in }\;$ $\;{[2,4]}$ $\;\;\text{ And }\; g(2)=2 \ln 2 f(2)-2=\ln 2-2$ $\; {\text{g}}(2)=4 \ln 4 f(4)-4=\ln 4-4$ $\;=2(\ln 2-2)$ $\;\;\text{ Now }\; g(2) \le g(x) \le g(4)$ $\;{\text{Ln}} 2-2 \le x \ln x f(x)-x \le 2(\ln 2-2)$ $\;\;\frac{\ln 2-2}{x \ln x}+\;\frac{1}{\ln x} \le f(x) \le \;\frac{2(\ln 2-2)}{x \ln x}+\;\frac{1}{\ln x}$ $\;\;\text{ Now for }\; {\text{x}} \in [2,4]$ $\;\;\frac{2(\ln 2-2)}{x \ln x}+\;\frac{1}{\ln x} < \;\frac{2(\ln 2-2)}{2 \ln 2}+\;\frac{1}{\ln 2}=1-\;\frac{1}{\ln 2} < 1$ $\;\Rightarrow {\text{f}}( {\text{x}}) \le 1 \;\text{ for }\; {\text{x}} \in [2,4]$ $\;\;\frac{\ln 2-2}{x \ln x}+\;\frac{1}{\ln x} \ge \;\frac{\ln 2-2}{4 \ln 4}+\;\frac{1}{\ln 4}=\;\frac{1}{8}+\;\frac{1}{2 \ln 2} > \;\frac{1}{8}$ $⇔ {\text{f}}( {\text{x}}) \ge \;\frac{1}{8}$ for ${\text{x}} \in [2,4]$ Hence both ${\text{A}}$ and ${\text{B}}$ are true.

Q66JEE Main 2023MCQ4MMatrices and Determinants

Let $A$ be a $2 \times 2$ matrix with real entries such that $A^{'}=\alpha A+I$ , where $a \in ℝ-\{-1,1\}$ . If ${\det} (A^{2-}A) =4$ , then the sum of all possible values of $\alpha$ is equal to : [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{A}}^{ {\text{T}}}=\alpha {\text{A}}+ {\text{I}}$ $\; {\text{A}}=\alpha {\text{A}}^{ {\text{T}}}+ {\text{I}}$ $\; {\text{A}}=\alpha(\alpha {\text{A}}+ {\text{I}})+ {\text{I}}$ $\; {\text{A}}=\alpha^2 {\text{A}}+(\alpha+1) {\text{I}}$ $\; {\text{A}} (1-\alpha^2) =(\alpha+1) {\text{I}}$ $\; {\text{A}}=\;{ {\text{I}}}/{1-\alpha} . . .(1)$ $\;| {\text{A}}|=\;\frac{1}{(1-\alpha)^2} . . .(2)$ $\;{| {\text{A}}^{2-} {\text{A}} |}=| {\text{A}}|| {\text{A}}-1| . . .(3)$ $\; {\text{A}}- {\text{I}}=\;{ {\text{I}}}/{ {\text{I}}-\alpha}- {\text{I}}=\;\frac{\alpha}{1-\alpha} {\text{I}}$ $\;| {\text{A}}- {\text{I}}|= (\;\frac{\alpha}{1-\alpha}) ^2 . . .(4)$ $\;\;\text{ Now }\;{| {\text{A}}^{2-} {\text{A}} |}=4$ $\;| {\text{A}} \| {\text{A}}- {\text{I}}|=4$ $\;\Rightarrow \;\frac{1}{(1-\alpha)^2} \;\frac{\alpha^2}{ (1-\alpha^2) }=4$ $\;\Rightarrow \;\frac{\alpha}{(1-\alpha)^2}= \pm 2$ $\;\Rightarrow 2(1-\alpha)^2= \pm \alpha$ $\; ( {\text{C}}_1) 2(1-\alpha)^2=\alpha\;\;\;$ $\;\;\;(C_2) 2(1-\alpha)^3=-\alpha$ $\alpha_1+\alpha_2=\;\frac{5}{2}$

Q67JEE Main 2023MCQ4MLogarithms

The number of integral solutions $x$ of $\log _{ (x+\;\frac{7}{2}) } (\;\frac{x-7}{2 x-3}) ^2 \ge 0$ is : [11-Apr-2023 shift 1]

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📖 Explanation

$\log _{x+\;\frac{7}{2}} (\;\frac{x-7}{2 x-3}) ^2 \ge 0$ Feasible region: $x+\;\frac{7}{2} > 0 \Rightarrow x > -\;\frac{7}{2}$ And $x+\;\frac{7}{2} \ne 1 \Rightarrow x \ne \;\frac{-5}{2}$ Taking intersection: ${\text{x}} \in (\;\frac{-7}{2}, \infty ) - \{-\;\frac{5}{2}, \;\frac{3}{2}, 7\}$ Now $\log _a b \ge 0$ if $a>1$ and $b \ge 1$ ${\text{a}} \in (0,1)$ and ${\text{b}} \in (0,1)$ ${\text{C}}- {\text{I}} ; {\text{x}}+\;\frac{7}{2} > 1$ and $(\;{ {\text{x}}-7}/{2 {\text{x}}-3}) ^2 \ge 1$ $\;x > -\;\frac{5}{2} ;(2 x-3)^{2-}(x-7)^2 \le 0$ $\;(2 x-3+x-7)(2 x-3-x+7) \le 0$ $\;(3 x-10)(x+4) \le 0$ $\;x \in [-4, \;\frac{10}{3}]$ Intersection: ${\text{x}} \in (\;\frac{-5}{2}, \;\frac{10}{3}]$ $\;C-Π: x+\;\frac{7}{2} \in (0,1) \;\text{ and }\; (\;\frac{x-7}{2 x-3}) ^2 \in (0,1)$ $\;0 < x+\;\frac{7}{2} < 1 ; (\;\frac{x-7}{2 x-3}) ^2 < 1$ $\;-\;\frac{7}{2} < x < \;\frac{-5}{2} ;(x-7)^2 < (2 x-3)^2$ $\;x \in (-\infty ,-4) ∪ (\;\frac{10}{3}, \infty )$ No common values of ${\text{x}}$ .Hence intersection with feasible regionWe get ${\text{x}} \in (\;\frac{-5}{2}, \;\frac{10}{3}] - \{\;\frac{3}{2}\}$ Integral value of ${\text{x}}$ are $\{-2,-1,0,1,2,3\}$ No. of integral values $=6$

Q68JEE Main 2023MCQ4MVector Algebra

For any vector $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$ , with $10{| a_{ {\text{i}}} |} < 1, i=1,2,3$ , consider the following statements :(A) : $\max \{{| {\text{a}}_1 |},{| {\text{a}}_2 |},{| {\text{a}}_3 |}\} \le |{ {\text{a}}}^{\to }|$ (B) : $|\vec{a}| \le 3 \max \{{| {\text{a}}_1 |},{| {\text{a}}_2 |},{| {\text{a}}_3 |}\}$ [11-Apr-2023 shift 1]

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📖 Explanation

Without loss of generalityLet ${| a_1 |} \le {| a_2 |} \le {| a_3 |}$ $\;|{ {\text{a}}}^{\to }|^2={| { {\text{a}}}^{\to }_1 |}^{2+}{| { {\text{a}}}^{\to }_2 |}^{2+}{| { {\text{a}}}^{\to }_3 |}^2 \ge ( {\text{a}}_3) ^2$ $\;\Rightarrow |{ {\text{a}}}^{\to }| \ge {| {\text{a}}_3 |}=\max \{{| {\text{a}}_1 |},{| {\text{a}}_2 |},{| {\text{a}}_3 |}\}$ $A$ is true $\;|\vec{a}|^2={| a_1 |}^{2+}{| a_2 |}^{2+}{| a_3 |}^2 \le {| a_3 |}^{2+}{| a_3 |}^{2+}{| a_3 |}^2$ $\;\Rightarrow |\vec{a}|^2 \le 3{| a_3 |}^2$ $\;\Rightarrow |\vec{a}| \le {\sqrt{3}}{| \vec{a}_3 |}={\sqrt{3}} \max \{{| a_1 |},{| a_2 |},{| a_3 |}\}$ $\le 3 \max \{{| a_1 |},{| a_2 |},{| a_3 |}\}$ (2) is true

Q69JEE Main 2023MCQ4MPermutations and Combinations

The number of triplets $(x, y, z)$ , where $x, y, z$ are distinct non negative integers satisfying $x+y+z=15$ , is : [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{x}}+ {\text{y}}+ {\text{z}}=15$ $\;\;\text{ Total no. solution }\;={ }^{15+3-1} {\text{C}}_3=136 . . . \;\text{ (1) }\;$ $\;\;\text{ Let }\; {\text{x}}= {\text{y}} \ne {\text{z}}$ $\;2 {\text{x}}+ {\text{z}}=15 \Rightarrow {\text{z}}=15-2 {\text{t}}$ $\;\Rightarrow {\text{r}} \in \{0,1,2, . . . 7\}-\{5\}$ $\; \therefore 7 \;\text{ solutions }\;$ $\; \therefore \;\text{ there are }\; 21 \;\text{ solutions \in which exactly }\;$ $\;\;\text{ Two of }\; {\text{x}}, {\text{y}}, {\text{z}} \;\text{ are equal ... (2) }\;$ $\;\;\text{ There is one solution \in which }\; {\text{x}}= {\text{y}}= {\text{z}} . . . \;\text{ (3) }\;$ $\;\;\text{ Required answer }\;=136-21-1=144$

Q70JEE Main 2023MCQ4MStatistics

Let sets $A$ and $B$ have 5 elements each. Let mean of the elements in sets $A$ and $B$ be 5 and 8 respectively and the variance of the elements in sets $A$ and $B$ be 12 and 20 respectively. A new set $C$ of 10 elements is formed by subtracting 3 from each element of A and adding 2 to each element of B. Then the sum of the mean and variance of the elements of ${\text{C}}$ is _______. [11-Apr-2023 shift 1]

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📖 Explanation

$\;\omega A= \{a_1, a_2, a_3, a_4, a_5\}$ $\;B= \{b_1, b_2, b_3, b_4, b_5\}$ $\;\;\frac{Σ_{i=1}^{5} a_i^2}{5}- (\;\frac{Σ_{i=1}^{5} a_i}{5}) ^2=12, \;\frac{Σ_{i=1}^{5} b_i^2}{5}- (\;\frac{Σ_{i=1}^{5} b_i}{5}) ^2=20$ $\;\Rightarrow sum_{i=1}^{5} a_i^2=185, \;\; sum_{i=1}^{5} b_i^2=420$ Now, ${\text{C}}= \{ {\text{C}}_1, {\text{C}}_2, . . . {\text{C}}_{10}\}$ $\; \therefore \;\text{ Mean of }\; {\text{C}}, \overline{ {\text{C}}}=\;{ (Σ {\text{a}}_{ {\text{i}}}-15) + (Σ {\text{b}}_{ {\text{i}}}-10) }/{10}$ $\;\overline{ {\text{C}}}=\;\frac{10+50}{10}=6$ $\; \therefore \sigma^2=\;{Σ_{{ {\text{i}}=1}}^{{10}} {\text{C}}_{ {\text{i}}}^2}/{10}=(\overline{ {\text{C}}})^2$ $\;=\;{Σ ( {\text{a}}_{ {\text{i}}}-3) ^{2+}Σ ( {\text{b}}_{ {\text{i}}}-2) ^{2+}}/{10}-(6)^2$ $\;=\;{Σ {\text{a}}_{ {\text{i}}}^{2+}Σ {\text{b}}_{ {\text{i}}}^{2-}6 Σ {\text{a}}_{ {\text{i}}}+4 Σ {\text{b}}_{ {\text{i}}}+65}/{10}-36$ $\;=\;\frac{185+420-150+160+65}{10}-36$ $\;=32$ $\; \therefore \;\text{ Mean }\;+ \;\text{ Variance }\;=\overline{ {\text{C}}}+\sigma^2=6+32=38$

Q71JEE Main 2023MCQ4MArea Under The Curves

Area of the region $\{(x, y): x^{2+}(y-2)^2 \le 4, x^2 \ge 2 y\}$ is : [11-Apr-2023 shift 1]

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📖 Explanation

$x^{2+}(y-2)^2 \le 2^2 \;\text{ and }\; x^2 \ge 2 y$ Solving circle and parabola simultaneously: $\;2 y+y^{2-}4 y+4=4$ $\;y^{2-}2 y=0$ $\;y=0,2$ Put $y=2$ in $x^2=2 y \to x= \pm 2$ $\Rightarrow (2,2) \;\text{ and }\;(-2,2)$ $=2 \times 2-\;\frac{1}{4} \cdot \pi \cdot 2^2=4-\pi$ $\;\;\text{ Required area }\;=2 [int_{0}^{2} \;\frac{x^2}{2} \text{dx}-(4-\pi)]$ $\;=2 [\;\frac{4}{3}+\pi-4]$ $\;=2 [\pi-\;\frac{8}{3}]$ $\;=2 \pi-\;\frac{16}{6}$

Q72JEE Main 2023MCQ4MStraight Lines and Pair of Straight Lines

Let $R$ be a rectangle given by the line $x=0, x=2, y=0$ and $y=5$ . Let $A(\alpha, 0)$ and $B(0, \beta), \alpha \in [0,2]$ and $\beta \in [0,5]$ , be such that the line segment $A B$ divides the area of the rectangle $R$ in the ratio $4: 1$ . Then, the midpoint of $A B$ lies on $a$ : [11-Apr-2023 shift 1]

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📖 Explanation

$\;{{\text{ar}}( {\text{OPQR}})}/{{\text{or}}( {\text{OAB}})}=\;\frac{4}{1}$ Let ${\text{M}}$ be the mid-point of ${\text{AB}}$ . $\;M(h, k) \equiv (\;\frac{\alpha}{2}, \;\frac{\beta}{2})$ $\;\Rightarrow \;\frac{10-\frac{1}{2} \alpha \beta}{\frac{1}{2} \alpha \beta}=4$ $\;\Rightarrow \;\frac{5}{2} \alpha \beta=10 \Rightarrow \alpha \beta=4$ $\;\Rightarrow (2 h)(2 {\text{K}})=4$ $\therefore$ Locus of ${\text{M}}$ is xy $=1$ Which is a hyperbola.

Q73JEE Main 2023MCQ4MVector Algebra

Let $\vec{a}$ be a non-zero vector parallel to the line of intersection of the two places described by ${ {\text{i}}^{\^}}+{ {\text{j}}^{\^}}, { {\text{i}}^{\^}}+{ {\text{k}}^{\^}}$ and $\hat{i}-\hat{j}, \hat{j}-\hat{k}$ . If $\theta$ is the angle between the vector $\vec{a}$ and the vector $\vec{b}=2 \hat{i}-2 \hat{j}+\hat{k}$ and $\vec{a} \cdot \vec{b}=6$ , then ordered pair $(\theta,|\vec{a} \times \vec{b}|)$ is equal to : [11-Apr-2023 shift 1]

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📖 Explanation

${ {\text{n}}}^{\to }_1$ and ${ {\text{n}}}^{\to }_2$ are normal vector to the plane ${ {\text{i}}^{\^}}+{ {\text{j}}^{\^}}, { {\text{i}}^{\^}}+{ {\text{k}}^{\^}}$ and ${ {\text{i}}^{\^}}-{ {\text{j}}^{\^}}, { {\text{i}}^{\^}}-{ {\text{k}}^{\^}}$ respectively ${ {\text{n}}}^{\to }_1={\begin{vmatrix}{ {\text{i}}^{\^}} & { {\text{j}}^{\^}} & { {\text{j}}^{\^}} \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{vmatrix}}={ {\text{i}}^{\^}}-{ {\text{j}}^{\^}}-{ {\text{k}}^{\^}}$ ${ {\text{n}}}^{\to }_2={\begin{vmatrix}{ {\text{i}}^{\^}} & { {\text{j}}^{\^}} & { {\text{j}}^{\^}} \\ 1 & -1 & 0 \\ 1 & 0 & -1\end{vmatrix}}={ {\text{i}}^{\^}}+{ {\text{j}}^{\^}}+{ {\text{k}}^{\^}}$ ${ {\text{a}}}^{\to }=\lambda {| { {\text{n}}}^{\to }_2 \times { {\text{n}}}^{\to }_2 |}$ $=\lambda {\begin{vmatrix}\hat{i} & \hat{j} & \hat{j} \\ 1 & -1 & -1 \\ 1 & 1 & -1\end{vmatrix}}=\lambda (-2 \hat{j}+2 \hat{k})$ $\;{ {\text{a}}}^{\to } \cdot { {\text{b}}}^{\to }=\lambda |0+4+2|=6$ $\;\Rightarrow \lambda =1$ $\;\vec{\alpha}=-2 { {\text{j}}^{\^}}+2 { {\text{k}}^{\^}}$ $\;\cos \theta=\;{{ {\text{a}}}^{\to } \cdot { {\text{b}}}^{\to }}/{| {\text{a}}|| {\text{b}}|}$ $\;\cos \theta=\;{6}/{2 {\sqrt{2}} \times 3}=\;{1}/{{\sqrt{2}}}$ $\;\theta=\;\frac{\pi}{4}$ $\;\;\text{ Now }\;|{ {\text{a}}}^{\to } \cdot { {\text{b}}}^{\to }|^{2+}|{ {\text{a}}}^{\to } \times { {\text{b}}}^{\to }|^2=| {\text{a}}|^2| {\text{b}}|^2$ $\;36+{| { {\text{a}}}^{\to } \times {\text{b}}^2 |}=8 \times 9=72$ $\;|{ {\text{a}}}^{\to } \times {\text{b}}|^2=36$ $\;|{ {\text{a}}}^{\to } \times { {\text{b}}}^{\to }|=6$

Q74JEE Main 2023MCQ4MComplex Numbers

Let ${\text{w}}_1$ be the point obtained by the rotation of ${\text{z}}_1=5+4 {\text{i}}$ about the origin through a right angle in the anticlockwise direction, and ${\text{w}}_2$ be the point obtained by the rotation of ${\text{z}}_2=3+5 {\text{i}}$ about the origin through a right angle in the clockwise direction. Then the principal argument of ${\text{w}}_1- {\text{w}}_2$ is equal to : [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{W}}_1=z_{ {\text{i}}} {\text{i}}=(5+4 {\text{i}}) {\text{i}}=-4+5 {\text{i}} . . . \;\text{ (i) }\;$ $\; {\text{W}}_1=z_2(- {\text{i}})=(3+5 {\text{i}})(- {\text{i}})=5-3 {\text{i}} . . .$ (2) $\; {\text{W}}_1- {\text{W}}_2=-9+8 {\text{i}}$ $\;\;\text{ Principal argument }\;=\pi-\tan ^{-1} (\;\frac{8}{9})$

Q75JEE Main 2023MCQ4MEllipse

Consider ellipse $E_k: k x^{2+}k^2 y^2=1, k=1,2, . . ., 20$ . Let $C_k$ be the circle which touches the four chords joining the end points (one on minor axis and another on major axis) of the ellipse ${\text{E}}_{ {\text{k}}}$ . If ${\text{r}}_{ {\text{k}}}$ is the radius of the circle ${\text{C}}_{ {\text{k}}}$ then the value of $sum_{{ {\text{k}}=1}}^{{20}} \;{1}/{ {\text{r}}_{ {\text{k}}}^2}$ is [11-Apr-2023 shift 1]

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📖 Explanation

$K {\text{Kx}}^{2+} {\text{K}}^2 {\text{y}}^2=1$ $\;\frac{x^2}{1 / K}+\;{y^2}/{1 / {\text{K}}^2}=1$ NowEquation of $\; {\text{A}}_1 {\text{B}}_2 ; \;{ {\text{x}}}/{1 / \sqrt{ {\text{K}}}}+\;{ {\text{y}}}/{1 / {\text{K}}}=1 \Rightarrow \sqrt{ {\text{K}}} {\text{x}}+ {\text{Ky}}=1$ $\; {\text{r}}_{ {\text{K}}}=⟂ {\text{r}} \;\text{ distance of }\;(0,0) \;\text{ from line }\; {\text{A}}_1 {\text{B}}_1$ $\; {\text{r}}_{ {\text{K}}}={| \;{(0+0-1)}/{\sqrt{ {\text{K}}+ {\text{K}}^2}} |}=\;{1}/{\sqrt{ {\text{K}}+ {\text{K}}^2}}$ $\;\;{1}/{ {\text{r}}_{ {\text{K}}}^2}= {\text{K}}+ {\text{K}}^2 \Rightarrow sum_{{ {\text{k}}=1}}^{{20}} \;{1}/{ {\text{r}}_{ {\text{K}}}^2}=sum_{{ {\text{K}}=1}}^{{20}} ( {\text{K}}+ {\text{K}}^2)$ $\;=sum_{{ {\text{K}}=1}}^{{20}} {\text{K}}+sum_{{ {\text{K}}=1}}^{{20}} {\text{K}}^2$ $\;=\;\frac{20 \times 21}{2}+\;\frac{20.21 .41}{6}$ $\;=210+10 \times 7 \times 41$ $\;=210+2870$ $\;=3080$

Q76JEE Main 2023MCQ4MThree Dimensional Geometry

If the equation of the plane that contains the point $(-2,3,5)$ and is perpendicular to each of the planes $2 x+4 y+5 z$ $=8$ and $3 x-2 y+3 z=5$ is $\alpha x+\beta y+\gamma z+97=0$ then $\alpha+\beta+\gamma =$ : [11-Apr-2023 shift 1]

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📖 Explanation

The equation of plane through $(-2,3,5)$ is $a(x+2)+b(y-3)+c(z-5)=0$ it is perpendicular to $2 x+4 y+5 z=8 & 3 x-2 y+3 z=5$ $\; \therefore 2 {\text{a}}+4 {\text{b}}+5 {\text{c}}=0$ $\;3 {\text{a}}-2 {\text{b}}+3 {\text{c}}=0$ $\; \therefore \;{ {\text{a}}}/{{\begin{vmatrix}4 & 5 \\ -2 & 3\end{vmatrix}}}=\;{-b}/{{\begin{vmatrix}2 & 5 \\ 3 & 3\end{vmatrix}}}=\;{ {\text{c}}}/{{\begin{vmatrix}2 & 4 \\ 3 & -2\end{vmatrix}}}$ $\;\Rightarrow \;{ {\text{a}}}/{22}=\;{ {\text{b}}}/{9}=\;{ {\text{c}}}/{-16}$ $\therefore$ Equation of plane is $\;22(x+2)+9(y-3)-16(z-5)=0$ $\;\Rightarrow 22 x+9 y-16 z+97=0$ $\;\;\text{ Comparing with }\; \alpha x+\beta y+\gamma x+97=0$ $\;\;\text{ We get }\; \alpha+\beta+\gamma =22+9-16=15$

Q77JEE Main 2023MCQ4MSets and Relations

An organization awarded 48 medals in event ' ${\text{A}}$ ', 25 in event ' ${\text{B}}$ ' and 18 in event ' ${\text{C}}$ '. If these medals went to total 60 men and only five men got medals in all the three events, then, how many received medals in exactly two of three events ? [11-Apr-2023 shift 1]

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📖 Explanation

$\;|A|=48$ $\;|B|=25$ $\;|C|=18$ $\;|A ∪ B ∪ C|=60 \;\text{ [Total] }\;$ $\;|A ∩ B ∩ C|=5$ $\;| {\text{A}} ∩ {\text{B}} ∩ {\text{C}}|=\sum| {\text{A}}|-\sum| {\text{A}} ∩ {\text{B}}|+| {\text{A}} ∩ {\text{B}} ∩ {\text{C}}|$ $\;\Rightarrow \sum| {\text{A}} ∩ {\text{B}}|=48+25+18+5-60$ $\;=36$ No. of men who received exactly 2 medals $\;\Rightarrow \sum| {\text{A}} ∩ {\text{B}}|-3| {\text{A}} ∩ {\text{B}} ∩ {\text{C}}|$ $\;=36-15$ $\;=21$

Q78JEE Main 2023MCQ4MDifferential Equations

Let $y=y(x)$ be a solution curve of the differential equation. $(1-x^2 y^2) \text{dx}=y \text{dx}+x d y$ . If the line $x=1$ intersects the curve $y=y(x)$ at $y=2$ and the line $x=2$ intersects the curve $y=y(x)$ at $y=\alpha$ , then a value of $\alpha$ is : [11-Apr-2023 shift 1]

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📖 Explanation

$\; (1-x^2 y^2) \text{dx}=y \text{dx}+x d y, y(1)=2$ $\;y(2)=\infty$ $\;d x=\;\frac{d(x y)}{1-(x y)^2}$ $\;\int \text{dx}=\int \;\frac{d(x y)}{1-(x y)^2}$ $\;x=\;\frac{1}{2} \ln {| \;\frac{1+x y}{1-x y} |}+C$ $\;\;\text{ Put }\; x=1 \;\text{ and }\; y=2:$ $\;1=\;\frac{1}{2} \ln {| \;\frac{1+2}{1-2} |}+C$ $\;C=1-\;\frac{1}{2} \ln 3$ Now put ${\text{x}}=2$ : $\;2=\;\frac{1}{2} \ln {| \;\frac{1+2 \alpha}{1-2 \alpha} |}+1-\;\frac{1}{2} \ln 3$ $\;1+\;\frac{1}{2} \ln 3=\;\frac{1}{2}{| \;\frac{1+2 \alpha}{1-2 \alpha} |}$ $\;2+\ln 3={| \;\frac{1+2 \alpha}{1-2 \alpha} |}$ $\;{| \;\frac{1+2 \alpha}{1-2 \alpha} |}=3 {\text{e}}^2$ $\;\;\frac{1+2 \alpha}{1-2 \alpha}=3 {\text{e}}^2,-3 {\text{e}}^2$ $\;\;\frac{1+2 \alpha}{1-2 \alpha}=3 {\text{e}}^2 \Rightarrow \alpha=\;{3 {\text{e}}^{2-}1}/{2 (3 {\text{e}}^{2+}1) }$ And $\;\frac{1+2 \alpha}{1-2 \alpha}=-3 e^2 \Rightarrow \alpha=\;\frac{3 e^{2+}1}{2 (3 e^{2-}1) }$

Q79JEE Main 2023MCQ4MThree Dimensional Geometry

Let $(\alpha, \beta, \gamma )$ be the image of the point $P(2,3,5)$ in the plane $2 x+y-3 z=6$ . Then $\alpha+\beta+\gamma$ is equal to : [11-Apr-2023 shift 1]

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📖 Explanation

$\;\frac{\alpha-2}{2}=\;\frac{\beta-3}{1}=\;\frac{\gamma -5}{-3}=-2 (\;\frac{2 \times 2+3-3 \times 5-6}{2^{2+}1^{2+}1-3^2}) =2$ $\;\frac{\alpha-2}{2}=2\;\beta-3=2\;\gamma -5=-6$ $\alpha=6\;\beta=5\;\gamma =-1$ $\alpha+\beta+\gamma =10$

Q80JEE Main 2023MCQ4MLimits, Continuity and Differentiability

Let $f(x)= \$ [x^{2-}x]$ +|-x+[x]| $, where$ x \in ℝ $and$ [t] $denotes the greatest integer less than or equal to$ t $. Then,$ {\text{f}}$ is : [11-Apr-2023 shift 1]

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📖 Explanation

Here $f(x)=[x(x-1)]+\{x\}$ ${\text{f}} (0^{+}) =-1+0=-1\; {\text{f}} (1^{+}) =0+0=0$ ${\text{f}}(0)=0\; {\text{f}}(1)=0$ $\; {\text{f}} (1^{-}) =-1+1=0$ $\therefore {\text{f}}( {\text{x}})$ is continuous at ${\text{x}}=1$ , discontinuous at ${\text{x}}=0$

Q81JEE Main 2023NAT4MBinomial Theorem

The number of integral terms in the expansion of $(3^{\;\frac{1}{2}}+5^{\;\frac{1}{4}}) ^{680}$ is equal to : [11-Apr-2023 shift 1]

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📖 Explanation

The number of integral term in the expression of $(3^{\;\frac{1}{2}}+5^{\;\frac{1}{4}}) ^{680}$ is equal to $\;\;\text{ General term }\;={ }^{680} {\text{C}}_{ {\text{r}}} (3^{\;\frac{1}{2}}) ^{680- {\text{r}}} (5^{\;\frac{1}{4}}) ^{ {\text{r}}}$ $\;={ }^{680} {\text{C}}_{ {\text{r}}} 3^{\;{680- {\text{r}}}/{2}} 5^{\;{ {\text{r}}}/{4}}$ Values' s of $r$ , where $\;\frac{r}{4}$ goes to integer $r=0,4,8,12, . . . . . . 680$ All value of $r$ are accepted for $\;\frac{680-r}{2}$ as well so No of integral terms $=171$ .

Q82JEE Main 2023NAT4MSets and Relations

The number of ordered triplets of the truth values of $p, q$ and $r$ such that the truth value of the statement $(p$ $∨ q) ∧(p ∨ r) \Rightarrow (q ∨ r)$ is True, is equal to ________. [11-Apr-2023 shift 1]

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Q83JEE Main 2023NAT4MMatrices and Determinants

Let $A={\begin{bmatrix}0 & 1 & 2 \\ {\text{a}} & 0 & 3 \\ 1 & {\text{c}} & 0\end{bmatrix}}$ , where ${\text{a}}, {\text{c}} \in ℝ$ . If ${\text{A}}^3= {\text{A}}$ and the positive value of a belongs to the interval $( {\text{n}}-1, {\text{n}}]$ , where ${\text{n}} \in ℕ$ , then ${\text{n}}$ is equal to ______. [11-Apr-2023 shift 1]

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📖 Explanation

$A\;={\begin{bmatrix}0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0\end{bmatrix}}$ $A^3\;=A$ $A^2\;={\begin{bmatrix}0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0\end{bmatrix}}{\begin{bmatrix}0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0\end{bmatrix}}$ $A^2\;={\begin{bmatrix}a+2 & 2 c & 3 \\ 3 & a+3 c & 2 a \\ a c & 1 & 2+3 c\end{bmatrix}}$ $A^3\;={\begin{bmatrix}a+2 & 2 c & 3 \\ 3 & a+3 c & 2 a \\ a c & a & 2+3 c\end{bmatrix}}{\begin{bmatrix}0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0\end{bmatrix}}$ $A^3\;={\begin{bmatrix}2 a c+3 & a+2+3 c & 2 a+4+6 c \\ a(a+3 c)+2 a & 3+2 a c & 6+3 a+9 c \\ a+2+3 c & a c+c(2+3 c) & 2 a c+3\end{bmatrix}}$ $\;\;\text{ Given }\; A^3=A$ $\;2 a c+3=0 . . .(1) \;\text{ and }\; a+2+3 c=1$ $\;a+1+3 c=0$ $\;a+1-\;\frac{9}{2 a}=0$ $\;2 a^{2+}2 a-9=0$ $\;f(1) < 0, f(2) > 0$ $\;a \in (1,2]$ $\;n=2$

Q84JEE Main 2023NAT4MDefinite Integrals

For ${\text{m}}, {\text{n}} > 0$ , let $\alpha( {\text{m}}, {\text{n}})=int_{0}^{2} {\text{t}}^{ {\text{m}}}(1+3 {\text{t}})^{ {\text{n}}} {\text{dt}}$ . If $11 \alpha(10,6)+18 \alpha(11,5)= {\text{p}}(14)^6$ , then ${\text{p}}$ is equal to _______. [11-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\alpha( {\text{m}}, {\text{n}})=int_{0}^{2} {\text{t}}^{ {\text{m}}}(1+3 {\text{t}})^{ {\text{n}}} {\text{dt}}$ $\;\;\text{ If }\; 11 \alpha(10,6)+18 \alpha(11,5)= {\text{p}}(14)^6 \;\text{ then }\; {\text{P}}$ $\;=11 int_{0}^{2} \;{t^{10}}/{ {\text{II}}} \;{(1+3 {\text{t}})^6}/{ {\text{I}}}+10 \int^2 {\text{t}}^{11}(1+3 {\text{t}})^5 {\text{dt}}$ $\;=11 \$ [(1+3 {\text{t}})^6 ⋅ ;\frac{t^{11}}{11}-int 6(1+3 {\text{t}})^5 ⋅ 3 ;\frac{t^{11}}{11}]$ 0^{2+}18 int{0}^{2} t^{11}(1+3 t)^5 \text{dt};= (t^{11}(1+3 t)^6) _0^2;=2^{11}(7)^6;=2^5(14)^6;=32(14)^6$

Q85JEE Main 2023NAT4MSequences and Series

Let ${\text{S}}= {\text{S}}=109+\;\frac{108}{5}+\;\frac{107}{5^2}+. . . .+\;\frac{2}{5^{107}}+\;\frac{1}{5^{108}}$ . Then the value of $(16 {\text{S}}-(25)^{-54})$ is equal to _______. [11-Apr-2023 shift 1]

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📖 Explanation

$\;S=109+\;\frac{108}{5}+\;\frac{107}{5^2} . . . . . .+\;\frac{1}{5^{108}}$ $\;\;{\;\frac{S}{5}=\frac{109}{5}+\frac{108}{5^2} . . . . . . \cdot \frac{2}{5^{108}}+\frac{1}{5^{109}}}/{{4 {\text{S}}}/{5}=109-\;\frac{1}{5}-\;\frac{1}{5^2} . . . . . .-\;\frac{1}{5^{108}}-\;\frac{1}{5^{109}}}$ $\;=109- (\;\frac{1}{5} \;{ (1-\frac{1}{5^{109}}) ) }/{ (1-\;\frac{1}{5}) })$ $\;=109-\;\frac{1}{4} (1-\;\frac{1}{5^{109}})$ $\;=109-\;\frac{1}{4}+\;\frac{1}{4} \times \;\frac{1}{5^{109}}$ $\; {\text{S}}=\;\frac{5}{4} (109-\;\frac{1}{4}+\;\frac{1}{4.5^{109}})$ $\;16 {\text{S}}=20 \times 109-5+\;\frac{1}{5^{108}}$ $\;16 {\text{S}}-(25)^{-54}=2180-5=2175$

Q86JEE Main 2023NAT4MHyperbola

Let ${\text{H}}_{ {\text{n}}}: \;{ {\text{x}}^2}/{1+ {\text{n}}}-\;{ {\text{y}}^2}/{3+ {\text{n}}}=1, {\text{n}} \in {\text{N}}$ . Let ${\text{k}}$ be the smallest even value of ${\text{n}}$ such that the eccentricity of ${\text{H}}_{ {\text{k}}}$ is a rational number. If $l$ is the length of the latus rectum of ${\text{H}}_{ {\text{k}}}$ , then $21 l$ is equal to _______. [11-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{Hn}} \Rightarrow \;{ {\text{x}}^2}/{1+ {\text{n}}}-\;{ {\text{y}}^2}/{3+ {\text{n}}}=1$ $\;e=\sqrt{1+\;\frac{b^2}{a^2}}=\sqrt{1+\;\frac{3+n}{1+n}}=\sqrt{\;\frac{2 n+4}{n+1}}$ $\;e=\sqrt{\;\frac{2 n+4}{n+1}}$ $\; {\text{n}}=48 \;\text{ (smallest even value for which }\; {\text{e}} \in {\text{Q}} \;\text{ ) }\;$ $\; {\text{e}}=\;\frac{10}{7}$ $\; {\text{a}}^2= {\text{n}}+1 \;\; {\text{b}}^2= {\text{n}}+3$ $\;=49 \;\; =51$ $\;1=\;\text{ length of }\; L R=\;\frac{2 b^2}{a}$ $\; {\text{L}}=2 \cdot \;\frac{51}{7}$ $\;1=\;\frac{102}{7}$ $\;21 ℓ=306$ $\;$

Q87JEE Main 2023NAT4MBinomial Theorem

The mean of the coefficients of $x, x^2, . . . x^7$ in the binomial expansion of $(2+x)^9$ is ________. [11-Apr-2023 shift 1]

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📖 Explanation

Coefficient of ${\text{x}}={ }^9 {\text{C}}_1 2^8$ Coef. ${\text{x}}^2={ }^9 {\text{C}}_2 2^7$ Coef. ${\text{x}}^7={ }^9 {\text{C}}_7 \cdot 2^2$ Mean $=\;{{ }^9 {\text{C}}_1 \cdot 2^{8+}{ }^9 {\text{C}}_2 \cdot 2^7 . . .+{ }^9 {\text{C}}_7 \cdot 2^2}/{7}$ $=\;{(1+2)^{9-}{ }^9 {\text{C}}_0 \cdot 2^{9-}{ }^9 {\text{C}}_8 \cdot 2^{1-}{ }^9 {\text{C}}_9}/{7}$ $=\;\frac{3^{9-}2^{9-}18-1}{7}$ $=\;\frac{19152}{7}=2736$

Q88JEE Main 2023NAT4MQuadratic Equation and Inequalities

If $a$ and $b$ are the roots of the equation $x^{2-}7 x-1=0$ , then the value of $\;\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}$ is equal to ________. [11-Apr-2023 shift 1]

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📖 Explanation

By newton's theorem $\; {\text{S}}_{ {\text{n}}+2}-7 {\text{S}}_{ {\text{n}}+1}- {\text{S}}_{ {\text{n}}}=0$ $\; {\text{S}}_{21}-7 {\text{S}}_{20}- {\text{S}}_{19}=0$ $\; {\text{S}}_{20}-7 {\text{S}}_{19}- {\text{S}}_{18}=0$ $\; {\text{S}}_{19}-7 {\text{S}}_{18}- {\text{S}}_{17}=0$ $\;\;{ {\text{S}}_{21}+ {\text{S}}_{17}}/{ {\text{S}}_{19}}=\;{ {\text{S}}_{21}+ ( {\text{S}}_{19}-7 {\text{S}}_{18}) }/{ {\text{S}}_{19}}$ $\;=\;{50 {\text{S}}_{19}+ ( {\text{S}}_{21}-7 {\text{S}}_{20}) }/{ {\text{S}}_{19}}$ $\;=51 \cdot \;{ {\text{S}}_{19}}/{ {\text{S}}_{19}}=51$

Q89JEE Main 2023NAT4MPermutations and Combinations

In an examination, 5 students have been allotted their seats as per their roll numbers. The number of ways, in which none of the students sits on the allotted seat, is _______. [11-Apr-2023 shift 1]

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📖 Explanation

Derangement of 5 students $\; {\text{D}}_5=5 ! (1-\;\frac{1}{1 !}+\;\frac{1}{2 !}-\;\frac{1}{3 !}+\;\frac{1}{4 !}-\;\frac{1}{5 !})$ $\;=120 (\;\frac{1}{2}-\;\frac{1}{6}+\;\frac{1}{24}-\;\frac{1}{120})$ $\;=60-20+5-1$ $\;=40+4$ $\;=44$

Q90JEE Main 2023NAT4MThree Dimensional Geometry

Let a line $l$ pass through the origin and be perpendicular to the lines $1_1: { {\text{r}}}^{\to }={ {\text{i}}^{\^}}-11 { {\text{j}}^{\^}}-7 { {\text{k}}^{\^}}+\lambda { {\text{i}}^{\^}}+2 { {\text{j}}^{\^}}+3 { {\text{k}}^{\^}}, \lambda \in ℝ$ and $1_2: { {\text{r}}}^{\to }=-{ {\text{i}}^{\^}}+{ {\text{k}}^{\^}}+\mu 2 { {\text{i}}^{\^}}+2 { {\text{j}}^{\^}}+{ {\text{k}}^{\^}}, \mu \in ℝ$ .If ${\text{P}}$ is the point of intersection of $l$ and $l_1$ , and ${\text{Q}}(\alpha, \beta, \gamma )$ is the foot of perpendicular from ${\text{P}}$ on $l_2$ , then $9(\alpha+\beta+\gamma )$ is equal to ______. [11-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Let }\; ℓ=(0 { {\text{i}}^{\^}}+0 { {\text{j}}^{\^}}+0 { {\text{k}}^{\^}})+\gamma (a \hat{i}+b { {\text{j}}^{\^}}+c \hat{k})$ $\;=\gamma (a \hat{i}+b { {\text{j}}^{\^}}+c { {\text{k}}^{\^}})$ $\;a \hat{i}+b { {\text{j}}^{\^}}+c { {\text{k}}^{\^}}={\begin{vmatrix}{ {\text{i}}^{\^}} & { {\text{j}}^{\^}} & { {\text{k}}^{\^}} \\ 1 & 2 & 3 \\ 2 & 2 & 1\end{vmatrix}}$ $\;={ {\text{i}}^{\^}}(2-6)-{ {\text{j}}^{\^}}(1-6)+{ {\text{k}}^{\^}}(2-4)$ $\;=-4 { {\text{i}}^{\^}}-5 { {\text{j}}^{\^}}-2 { {\text{k}}^{\^}}$ $\;ℓ=\gamma (-4 { {\text{i}}^{\^}}+5 { {\text{j}}^{\^}}-2 { {\text{k}}^{\^}})$ ${\text{P}}$ is intersection of $ℓ$ and $ℓ_1$ $-4 \gamma =1+\lambda , 5 \gamma =-11+2 \lambda ,-2 \gamma =-7+3 \lambda$ By solving these equation $\gamma =-1, {\text{P}}(4,-5,2)$ Let ${\text{Q}}(-1+2 \mu , 2 \mu , 1+\mu )$ $\;{ {\text{PQ}}}^{\to } \cdot (2 { {\text{i}}^{\^}}+2 { {\text{j}}^{\^}}+{ {\text{k}}^{\^}})=0$ $\;-2+4 \mu +4 \mu +1+\mu =0$ $\;9 \mu =1$ $\;\mu =\;\frac{1}{9}$ $\; {\text{Q}} (\;\frac{-7}{9}, \;\frac{2}{9}, \;\frac{10}{9})$ $\;9(\alpha+\beta+\gamma )=9 (\;\frac{-7}{9}+\;\frac{2}{9}+\;\frac{10}{9})$ $\;=5$

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