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29 Jan 2023 Shift 1

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Physics30 questions

Q1JEE Main 2023MCQ4MUnits and Measurements

Match List I with List II : List-I (PhysicalQuantity) List-II (Dimensional Formula) A Pressure gradient I $[ {\text{M}}^0 {\text{L}}^2 {\text{T}}^{-2}]$ B Energy density II $[ {\text{M}}^1 {\text{L}}^{-1} {\text{T}}^{-2}]$ C Electric Field III $[ {\text{M}}^1 {\text{L}}^{-2} {\text{T}}^{-2}]$ D Latent heat IV $[ {\text{M}}^1 {\text{L}}^1 {\text{T}}^{-3} {\text{A}}^{-1}]$ Choose the correct answer from the options given below: [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\;\text{ Pressure gradient }\;=\;{ {\text{dp}}}/{ {\text{dx}}}=\;{[ {\text{ML}}^{-1} {\text{T}}^{-2}]\$ }/{[ {\text{L}}]};= [M^1 {\text{L}}^{-2} {\text{T}}^{-2}];;\text{ Energy density };=;\frac{;\text{ energy };}{;\text{ volume };}=;{[ {\text{ML}}^2 {\text{T}}^{-2}]$ }/{[ {\text{L}}^3]$ };= [ {\text{M}}^1 {\text{L}}^{-1} {\text{T}}^{-2}];;\text{ Electric field };=;\frac{;\text{ Force };}{;\text{ charg e };}=;{[ {\text{MLT}}^{-2}]$ }/{[ {\text{A}} \cdot {\text{T}}]};= [ {\text{M}}^1 {\text{L}}^1 {\text{T}}^{-3} {\text{A}}^{-1}];;\text{ Latent heat };=;\frac{;\text{ heat };}{;\text{ mass };}=;{[ {\text{ML}}^2 {\text{T}}^{-2}]$ }/{[ {\text{M}}]};= [ {\text{M}}^0 {\text{L}}^2 {\text{T}}^{-2}];$

Q2JEE Main 2023MCQ4MElectrostatics

In a cuboid of dimension $2 {\text{L}} \times 2 {\text{L}} \times {\text{L}}$ , a charge ${\text{q}}$ is placed at the centre of the surface ' $S$ ' having area of $4 {\text{L}}^2$ . The flux through the opposite surface to ' $S$ ' is given by [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\varphi =\;\frac{Q / \epsilon_0}{6}$ Flux passing through shaded face $=\;{ {\text{q}}}/{6 \epsilon_0}$

Q3JEE Main 2023MCQ4MCurrent Electricity

Ratio of thermal energy released in two resistor $R$ and 3R connected in parallel in an electric circuit is : [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{H}}=\;{ {\text{V}}^2}/{ {\text{R}}} \times {\text{t}}$ $\;\;{ {\text{H}}_1}/{ {\text{H}}_2}=\;{\;{ {\text{V}}^2 {\text{t}}}/{ {\text{R}}}}/{{ {\text{V}}^2 {\text{t}}}/{3 {\text{R}}}}=3: 1$

Q4JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

A single current carrying loop of wire carrying current I flowing in anticlockwise direction seen from +ve ${\text{z}}$ direction and lying in xy plane in shown in figure. The plot of $\hat{j}$ component of magnetic field (By) at a distance ' $a$ ' (less than radius of the coil) and on yz plane vs ${\text{z}}$ coordinate look like [29-Jan-2023 Shift 1]

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📖 Explanation

$B_y=0$ in plane of coil ${\text{B}}_{ {\text{y}}}$ is opposite of each other in $- {\text{z}}$ and $+ {\text{z}}$ positions.

Q5JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

The magnitude of magnetic induction at mid-point ${\text{O}}$ due to current arrangement as shown in Fig will be : [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Magnetic field due to current in BC and ET are outward at point ' ${\text{O}}$ ' ${\text{B}}_0=\;{\mu _0 {\text{i}}}/{4 \pi {\text{r}}}+\;{\mu _0 {\text{i}}}/{4 \pi {\text{r}}}=\;{\mu _0 {\text{i}}}/{2 \pi {\text{r}}}=\;{\mu _0 {\text{i}}}/{\pi {\text{a}}}$

Q6JEE Main 2023MCQ4MElectromagnetic Induction

Find the mutual inductance in the arrangement, when a small circular loop of wire of radius ' $R$ ' is placed inside a large square loop of wire of side ${\text{L}}$ $( {\text{L}} ≫ {\text{R}})$ . The loops are coplanar and their centres coincide : [29-Jan-2023 Shift 1]

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📖 Explanation

$\;\varphi = {\text{Mi}}$ $\;\varphi =( {\text{BA}})$ $\;\varphi =\pi {\text{R}}^2 (4 \;\frac{\mu _0}{4 \pi} \;{ {\text{i}}}/{ (\;{ {\text{L}}}/{2}) } {\sqrt{2}})$ $\;\Rightarrow {\text{M}}=\;{2 {\sqrt{2}} \mu _0 {\text{R}}^2}/{ {\text{L}}}$

Q7JEE Main 2023MCQ4MElectromagnetic Waves

Which of the following are true?A. Speed of light in vacuum is dependent on the direction of propagation.B. Speed of light in a medium in independent of the wavelength of light.C. The speed of light is independent of the motion of the source.D. The speed of light in a medium is independent of intensity.Choose the correct answer from the option given below : [29-Jan-2023 Shift 1]

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Q8JEE Main 2023MCQ4MWave Optics

In a Young's double slit experiment, two slits are illuminated with a light of wavelength $800 {\text{nm}}$ . The line joining ${\text{A}}_1 {\text{P}}$ is perpendicular to ${\text{A}}_1 {\text{A}}_2$ as shown in the figure. If the first minimum is detected at $P$ , the value of slits separation ' $a$ ' will be :The distance of screen from slits $D=5 {\text{cm}}$ [29-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{A}}_2 {\text{P}}- {\text{A}}_1 {\text{P}}=\;\frac{\lambda }{2} \;\text{ (Condition of minima) }\;$ $\;\sqrt{ {\text{D}}^{2+} {\text{a}}^2}- {\text{D}}=\;\frac{\lambda }{2}$ $\; {\text{D}} (1+\;{ {\text{a}}^2}/{ {\text{D}}^2}) ^{1 / 2}- {\text{D}}=\;\frac{\lambda }{2}$ $\; {\text{D}} (1+\;\frac{1}{2} \times \;{ {\text{a}}^2}/{ {\text{D}}^2}) - {\text{D}}=\;\frac{\lambda }{2}$ $\;\;{ {\text{a}}^2}/{2 {\text{D}}}=\;\frac{\lambda }{2} \Rightarrow {\text{a}}=\sqrt{\lambda . {\text{D}}}$ $\;=\sqrt{800 \times 10^{-6} \times 50}$ $\; {\text{a}}=0.2 {\text{mm}}$

Q9JEE Main 2023MCQ4MMotion in a Plane

A stone is projected at angle $30^{\circ}$ to the horizontal. The ratio of kinetic energy of the stone at point of projection to its kinetic energy at the highest point of flight will be : [29-Jan-2023 Shift 1]

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📖 Explanation

$\;{K E_{\;\text{POP }\;}}/{K E_{\;\text{top }\;}}=\;\frac{\;\frac{1}{2} M(u)^2}{\frac{1}{2} M (u \cos 30^{\circ}) ^2}=\;\frac{4}{3}$

Q10JEE Main 2023MCQ4MLaws of Motion

A block of mass $m$ slides down the plane inclined at angle $30^{\circ}$ with an acceleration $\;{ {\text{g}}}/{4}$ . The value of coefficient of kinetic friction will be : [29-Jan-2023 Shift 1]

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📖 Explanation

${\text{Mg}} sin \;30^{\circ}-\mu {\text{mg}} \cos 30^{\circ}= {\text{ma}}$ $\;\frac{g}{2}-\;{{\sqrt{3}}}/{2} \cdot \mu g=\;\frac{g}{4}$ $\;\;{{\sqrt{3}}}/{2} \mu =\;\frac{1}{4}$ $\;\mu =\;{1}/{2 {\sqrt{3}}}$

Q11JEE Main 2023MCQ4MCircular Motion

A car is moving on a horizontal curved road with radius $50 {\text{m}}$ . The approximate maximum speed of car will be, if friction between tyres and road is 0.34. $[.$ Take $g=10 {\text{ms}}^{-2}]$ [29-Jan-2023 Shift 1]

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📖 Explanation

$f_s=\;{m v^2}/{ {\text{r}}}$ For maximum speed in safe turning, $\; {\text{f}}_5= {\text{f}}_5 \max =\mu {\text{mg}}$ $\; {\text{v}}_{\max } \;\text{ (for safe turning }\;=\sqrt{\mu {\mu g}}$ $\;={\sqrt{0.34 \times 50 \times 10}} \approx 13 {\text{m}} / {\ s}$

Q12JEE Main 2023MCQ4MGravitation

Two particles of equal mass ' $m$ ' move in a circle of radius ' $r$ ' under the action of their mutual gravitational attraction. The speed of each particle will be : [29-Jan-2023 Shift 1]

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📖 Explanation

$\;\;{G {\text{m}}^2}/{4 {\text{r}}^2}=\;{ {\ mv}^2}/{ {\text{r}}}$ $\;v=\sqrt{\;{ {\text{Cm}}}/{4 {\text{r}}}}$

Q13JEE Main 2023MCQ4MProperties of Matter

Surface tension of a soap bubble is $2.0 \times 10^{-2} {\text{Nm}}^{-1}$ .Work done to increase the radius of soap bubblefrom $3.5 {\text{cm}}$ to $7 {\text{cm}}$ will be : [Take $\pi=\;\frac{22}{7}$ ] [29-Jan-2023 Shift 1]

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📖 Explanation

Surface area of soap bubble $=2 \times 4 \pi R^2$ Work done $=$ change in surface energy $\times {\text{T}}_{ {\text{S}}}$ $= {\text{T}}_{ {\text{S}}} \times 8 \pi \times ( {\text{R}}_2^{2-} {\text{R}}_1^2)$ $=2 \times 10^{-2} \times 8 \times \;\frac{22}{7} \times 49 \times \;\frac{3}{4} \times 10^{-4}$ $=18.48 \times 10^{-4} {\text{J}}$

Q14JEE Main 2023MCQ4MThermodynamics

Given below are two statements. One is labelled as Assertion $A$ and the other is labelled as Reason R. Assertion ${\text{A}}$ : If ${\text{dQ}}$ and ${\text{dW}}$ represent the heat supplied to the system and the work done on the system respectively. Then according to the first law of thermodynamics $d Q=d U-d W$ .Reason R : First law of thermodynamics is based on law of conservation of energy.In the light of the above statements, choose the correct answer from the option given below : [29-Jan-2023 Shift 1]

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📖 Explanation

First law of thermodynamics is based on law of conservation of energy and it can be written as ${\text{dQ}}= {\text{dU}}- {\text{dW}}$ .where ${\text{dW}}$ is work done on the system

Q15JEE Main 2023MCQ4MThermodynamics

A bicycle tyre is filled with air having pressure of $270 {\text{kPa}}$ at $27^{\circ} {\text{C}}$ . The approximate pressure of the air in the tyre when the temperature increases to $36^{\circ} {\text{C}}$ is [29-Jan-2023 Shift 1]

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📖 Explanation

Taking volume constant : $\;\frac{P_1}{T_1}=\;\frac{P_2}{T_2}$ $\Rightarrow P_2=\;{P_1}/{ {\text{T}}_1} \times {\text{T}}_2=\;\frac{270 \times (309)}{300}$ $=278 {\text{kPa}}$

Q16JEE Main 2023MCQ4MWaves

A person observes two moving trains, 'A' reaching the station and ' ${\text{B}}$ ' leaving the station with equal speed of $30 {\text{m}} / {\ s}$ . If both trains emit sounds with frequency $300 {\text{Hz}}$ , (Speed of sound : $330 {\text{m}} / {\ s}$ ) approximate difference of frequencies heard by the person will be : [29-Jan-2023 Shift 1]

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📖 Explanation

$\;f_1=300 (\;\frac{330-0}{330-(-30)}) =275$ $\;f_2=300 (\;\frac{330-0}{330-(30)}) =330$ $\;\Delta {\text{f}}=330-275=55 {\text{Hz}} .$

Q17JEE Main 2023MCQ4MElectromagnetic Waves

If the height of transmitting and receiving antennas are $80 {\text{m}}$ each, the maximum line of sight distance will be :Given : Earth's radius $=6.4 \times 10^6 {\text{m}}$ . [29-Jan-2023 Shift 1]

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📖 Explanation

Maximum line of sight distance between two antennas, ${\text{d}}_{ {\text{M}}}=\sqrt{2 {\text{Rh}}_{ {\text{T}}}}+\sqrt{2 {\text{R}} \cdot {\text{h}}_{ {\text{R}}}}$ ${\text{d}}_{ {\text{M}}}=2 \times {\sqrt{2 \times 6.4 \times 10^6 \times 80}}=64 {\text{km}}$

Q18JEE Main 2023MCQ4MDual Nature of Matter and Radiation

The threshold wavelength for photoelectric emission from a material is $5500 Å$ . Photoelectrons will be emitted, when this material is illuminated with monochromatic radiation from aA. $75 {\text{W}}$ infra-red lampB. $10 {\text{W}}$ infra-red lampC. $75 {\text{W}}$ ultra-violet lampD. $10 {\text{W}}$ ultra-violet lampChoose the correct answer from the options given below : [29-Jan-2023 Shift 1]

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📖 Explanation

$\lambda < 5500 Å$ for photoelectric emission $\lambda _{ {\text{uv}}}< 5500 Å$

Q19JEE Main 2023MCQ4MAtoms and Nuclei

If a radioactive element having half-life of $30 {\min}$ is undergoing beta decay, the fraction of radioactive element remains undecayed after $90 {\min}$ . will be : [29-Jan-2023 Shift 1]

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📖 Explanation

$\;\;{ {\text{N}}}/{ {\text{N}}_0}= (\;\frac{1}{2}) ^{ {\text{t}} / t \;\frac{1}{2}}= (\;\frac{1}{2}) ^{\;\frac{90}{30}}$ $\;\;{ {\text{N}}}/{ {\text{N}}_0}= (\;\frac{1}{2}) ^3=\;\frac{1}{8}$

Q20JEE Main 2023MCQ4MSemiconductor Electronics

Which of the following statement is not correct in the case of light emitting diodes?A. It is a heavily doped ${\text{p}}- {\text{n}}$ junction.B. It emits light only when it is forward biased.C. It emits light only when it is reverse biased.D. The energy of the light emitted is equal to or slightly less than the energy gap of the semiconductor used.Choose the correct answer from the options given below : [29-Jan-2023 Shift 1]

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Q21JEE Main 2023NAT4MAtoms and Nuclei

A radioactive element ${ }_{92}^{242} {\text{X}}$ emits two $\alpha$ -particles, one electron and two positrons. The product nucleus is represented by ${ }_{p}^{234} {\text{Y}}$ . The value of ${\text{P}}$ is _______ [29-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{P}}=92-2-2+1-1-1$ $\; {\text{P}}=92-5$ $\; {\text{P}}=8 7$

Q22JEE Main 2023NAT4MWaves

Two simple harmonic waves having equal amplitudes of $8 {\text{cm}}$ and equal frequency of $10 {\text{Hz}}$ are moving along the same direction. The resultant amplitude is also $8 {\text{cm}}$ . The phase difference between the individual waves is degree ________. [29-Jan-2023 Shift 1]

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📖 Explanation

$\;2 {\text{A}} \cos (\;\frac{\Delta \varphi }{2}) = {\text{A}}$ $\;\cos (\;\frac{\Delta \varphi }{2}) =\;\frac{1}{2}$ $\;\;\frac{\Delta \varphi }{2}=60^{\circ}$

Q23JEE Main 2023NAT4MThermodynamics

A body cools from $60^{\circ} {\text{C}}$ to $40^{\circ} {\text{C}}$ in 6 minutes. If, temperature of surroundings is $10^{\circ} {\text{C}}$ . Then, after the next 6 minutes, its temperature will be ${ }^{\circ} {\text{C}}$ . [29-Jan-2023 Shift 1]

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📖 Explanation

By average form of Newton's law of cooling $\;\frac{20}{6}= {\text{k}}(50-10)$ . . . (i) $\;\frac{40-T}{6}=K (\;\frac{40+T}{2}-10)$ . . . (ii)From equation (i) and (ii) $\;\;{20}/{40- {\text{T}}}=\;{40}/{10+ {\text{T}} / 2}$ $\;10+\;{ {\text{T}}}/{2}=80-2 {\text{T}}$ $\;\;{5 {\text{T}}}/{2}=70 \Rightarrow {\text{T}}=28^{\circ} {\text{C}}$

Q24JEE Main 2023NAT4MRotational Motion

A solid sphere of mass $2 {\text{kg}}$ is making pure rolling on a horizontal surface with kinetic energy $2240 {\text{J}}$ . The velocity of centre of mass of the sphere will be _______ ${\text{ms}}^{-1}$ . [29-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{KE}}=\;\frac{1}{2} {\ mv}^{2+}\;\frac{1}{2} {\text{I}} \omega ^2$ $\;2240=\;\frac{1}{2} 2( {\text{v}})^{2+}\;\frac{1}{2} \;\frac{2}{5}(2) {\text{R}}^2 \cdot (\;{ {\text{v}}}/{ {\text{R}}}) ^2$ $\;2240= {\text{v}}^{2+}\;\frac{2}{5} {\text{v}}^2$ $\;\Rightarrow {\text{v}}=40 {\text{m}} / {\ s}$

Q25JEE Main 2023NAT4MWork, Power and Energy

A $0.4 {\text{kg}}$ mass takes $8 {\ s}$ to reach ground when dropped from a certain height ' $P$ ' above surface of earth. The loss of potential energy in the last second of fall is J. $[Take ${\text{g}}=10 {\text{m}} / {\ s}^2$ ]$ [29-Jan-2023 Shift 1]

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📖 Explanation

Displacement is $8^{\;\text{th }\;} {\ sec}$ . $\; {\text{S}}_8=0+\;\frac{1}{2} \times 10 \times (2 \times 8-1)$ $\; {\text{S}}_8=5 \times 15$ $\;\Delta {\text{U}}=0.4 \times 10 \times 5 \times 15$ $\;\Delta {\text{U}}=20 \times 15=300$

Q26JEE Main 2023NAT4MLaws of Motion

A tennis ball is dropped on to the floor from a height of $9.8 {\text{m}}$ . It rebounds to a height $5.0 {\text{m}}$ . Ball comes in contact with the floor for $0.2 {\ s}$ . The average acceleration during contact is _______ ${\text{ms}}^{-2}$ .$[Given ${\text{g}}=10 {\text{ms}}^{-2}$ ]$ [29-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{v}}_{ {\text{i}}}=\sqrt{2 {\text{gh}}_{ {\text{i}}}}$ $\;={\sqrt{2 \times 10 \times 9.8}} ↓$ $\;=14 {\text{m}} / {\ s} ↓$ $\; {\text{v}}_{ {\text{f}}}=\sqrt{2 {\text{gh}}_{ {\text{f}}}}$ $\;={\sqrt{2 \times 10 \times 5}} ↑$ $\;=10 {\text{m}} / {\ s} ↑$ $\; |{ \vec{\text{a}}_{\;\text{avg }\;}|= |\;{\Delta { \vec{\text{v}}}/{\Delta {\text{t}}}|=\;\frac{24}{0.2}=120 {\text{m}} / {\ s}^2$

Q27JEE Main 2023NAT4MElectrostatics

A point charge $q_1=4 q_0$ is placed at origin. Another point charge ${\text{q}}_2=- {\text{q}}_0$ is placed at ${\text{x}}=12 {\text{cm}}$ . Charge of proton is ${\text{q}}_0$ . The proton is placed on ${\text{x}}$ -axis so that the electrostatic force on the proton in zero. In this situation,the position of the proton from the origin is _______ ${\text{cm}}$ . [29-Jan-2023 Shift 1]

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📖 Explanation

$\;\;{ {\text{q}}_0}/{ {\text{x}}^2}=\;{4 {\text{q}}_0}/{( {\text{x}}+12)^2}$ $\; {\text{x}}+12=2 {\text{x}}$ $\; {\text{x}}=12$ Distance from origin $=x+12=24 {\text{cm}}$ .

Q28JEE Main 2023NAT4MCurrent Electricity

In a metre bridge experiment the balance point in obtained if the gaps are closed by $2 Ω$ and $3 Ω$ . A shunt of $X Ω$ is added to $3 Ω$ resistor to shift the balancing point by $22.5 {\text{cm}}$ . The value of $X$ is _______ [29-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\frac{2}{ (\;\frac{3 x}{3+x}) }=\;\frac{40+22.5}{60-22.5}=\;\frac{62.5}{37.5}=\;\frac{5}{3}$ $\;\;\frac{6}{5}=\;\frac{3 x}{3+x}$ $\;6+2 x=5 x \Rightarrow x=2$

Q29JEE Main 2023NAT4MElectromagnetic Induction

A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field $B=0.8 {\text{T}}$ . When released the radius of the loop starts shrinking at a constant rate of $2 {\text{cm}}^{-1}$ . The induced emf in the loop at an instant when the radius of the loop is $10 {\text{cm}}$ will _______be ${\text{mV}}$ . [29-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\text{ EMF }\;=\;{ {\text{d}}}/{ {\text{dt}}} (B \pi r^2)$ $\;=2 {\text{B}} \pi {\text{r}} \;{ {\text{dr}}}/{ {\text{dt}}}=2 \times \pi \times 0.1 \times 0.8 \times 2 \times 10^{-2}$ $\;=2 \pi \times 1.6=10.06 \;\text{ [round off }\; 10.06=10]$

Q30JEE Main 2023NAT4MWave Optics

As shown in figures, three identical polaroids $P_1$ , $P_2$ and $P_3$ are placed one after another. The pass axis of $P_2$ and $P_3$ are inclined at angle of $60^{\circ}$ and $90^{\circ}$ with respect to axis of $P_1$ . The source $S$ has an intensity of $256 \;{ {\text{W}}}/{ {\text{m}}^2}$ . The intensity of light at point $O$ is $-----\;{W}/{ {\text{m}}^2}$ . [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

By first polaroid P1 intensity will be halved then P2 and P3 will make intensity $\cos ^2 (60^{\circ})$ and $\cos ^2 (30^{\circ})$ times respectively. $\;\text{ Intensity out }\;=\;\frac{256}{2} \times \;\frac{1}{4} \times (\;{{\sqrt{3}}}/{2}) ^2=\;\frac{256 \times 3}{2 \times 4 \times 4}=24$

Chemistry30 questions

Q31JEE Main 2023MCQ4MStates of Matter

For $1 {\text{mol}}$ of gas, the plot of ${\text{pV}}$ vs ${\text{p}}$ is shown below. ${\text{p}}$ is the pressure and ${\text{V}}$ is the volume of the gas.What is the value of compressibility factor at point ${\text{A}}$ ? [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

For 1 mole of real gas ${\text{PV}}= {\text{ZRT}}$ from graph PV for real gas is less than PV for ideal gas at point ${\text{A}}$ ${\text{Z}}< 1$ ${\text{Z}}=1-\;{ {\text{a}}}/{ {\text{V}}_{ {\text{m}}} {\text{RT}}}$

Q32JEE Main 2023MCQ4MStructure of Atom

The shortest wavelength of hydrogen atom in Lyman series is $\lambda$ . The longest wavelength in Balmer series of ${\text{He}}^{+}$ is [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

For ${\text{H}}: \;\frac{1}{\lambda }= {\text{R}}_{ {\text{H}}} \times 1^2 (\;\frac{1}{1^2}-\;\frac{1}{\infty ^2})$ . . . (1) $\;{1}/{\lambda _{ {\text{He}}^{+}}}= {\text{R}}_{ {\text{H}}} \times 2^2 \times (\;\frac{1}{4}-\;\frac{1}{9})$ . . . (2)From (1) & (2) $\;{\lambda _{ {\text{He}}^{+}}}/{\lambda }=\;\frac{9}{5}$ $\;\lambda _{ {\text{He}}^{+}}=\lambda \times \;\frac{9}{5}$ $\;\lambda _{ {\text{He}}^{+}}=\;\frac{9 \lambda }{5}$

Q33JEE Main 2023MCQ4MSolutions

Which of the following salt solutions would coagulate the colloid soloution formed when ${\text{FeCl}}_3$ is added to ${\text{NaOH}}$ solution, at the fastest rate? [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Sol. Formed is negatively charged solution, therefore ${\text{Al}}^{3+}$ has highest coagulating power

Q34JEE Main 2023MCQ4MChemical Bonding and Molecular Structure

The bond dissociation energy is highest for [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Bond energy of ${\text{F}}_2$ less than ${\text{Cl}}_2$ due to lone pair lone pair repulsions.Bond energy order ${\text{Cl}}_2 > {\text{Br}}_2 > {\text{F}}_2 > {\text{I}}_2$

Q35JEE Main 2023MCQ4MCoordination Compounds

The reaction representing the Mond process for metal refining is [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Mond's process uses: ${\text{Ni}}+4 {\text{CO}} \to [ {\text{Ni}}( {\text{CO}})_4]$

Q36JEE Main 2023MCQ4MSolid State

Which of the given compounds can enhance the efficiency of hydrogen storage tank? [29-Jan-2023 Shift 1]

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Q37JEE Main 2023MCQ4MChemical Thermodynamics

The correct order of hydration enthalpies is(A) ${\text{K}}^{+}$ (B) ${\text{Rb}}^{+}$ (C) ${\text{Mg}}^{2+}$ (D) ${\text{Cs}}^{+}$ (E) ${\text{Ca}}^{2+}$ Choose the correct answer from the options given below: [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Hydration enthalpies:(i) ${\text{K}}^{+} > {\text{Rb}}^{+} > {\text{Cs}}^{+}:( {\text{A}}) >$ (B) $>$ (D)(ii) ${\text{Mg}}^{+2} > {\text{Ca}}^{+2}:( {\text{C}}) > ( {\text{E}})$ Option (D) $( {\text{C}}) > ( {\text{E}}) > ( {\text{A}}) > ( {\text{B}})>( {\text{D}})$

Q38JEE Main 2023MCQ4MChemical Bonding and Molecular Structure

The magnetic behaviour of ${\text{Li}}_2 {\text{O}}, {\text{Na}}_2 {\text{O}}_2$ and ${\text{KO}}_2$ , respectively, are [29-Jan-2023 Shift 1]

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📖 Explanation

${\text{Li}}_2 {\text{O}} \to {\text{O}}^{2-} \to$ diamagnetic ${\text{Na}}_2 {\text{O}}_2 \to {\text{O}}_2{ }^{2-} \to$ diamagnetic ${\text{KO}}_2 \to {\text{O}}_2^{-} \to$ paramagnetic

Q39JEE Main 2023MCQ4Mp Block Elements

"A" obtained by Ostwald's method involving air oxidation of ${\text{NH}}_3$ , upon further air oxidation produces "B". "B" on hydration forms an oxoacid of Nitrogen along with evolution of "A". The oxoacid also produces "A" and gives positive brown ring test [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$4 {\text{NH}}_3+5 {\text{O}}_2 \;{\longrightarrow }^{\Delta } {4 {\text{NO}}}_{(\text{A})}+6 {\text{H}}_2 {\text{O}}$ $2 {\text{NO}}+ {\text{O}}_2 \longrightarrow {2 {\text{NO}}_2 }_{(\text{B})}$

Q40JEE Main 2023MCQ4MElectrochemistry

The standard electrode potential $( {\text{M}}^{3+} / {\text{M}}^{2+})$ for ${\text{V}}$ , ${\text{Cr}}, {\text{Mn}} & {\text{Co}}$ are $-0.26 {\text{V}},-0.41 {\text{V}},+1.57 {\text{V}}$ and $+1.97 {\text{V}}$ , respectively. The metal ions which can liberate ${\text{H}}_2$ from a dilute acid are [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Metal cation with $(-)$ value of reduction potential $( {\text{M}}^{+3} / {\text{M}}^{+2})$ or with $(+)$ value of oxidation potential $( {\text{M}}^{+2} / {\text{M}}^{+3})$ will liberate ${\text{H}}_2$ Therefore they will reduce ${\text{H}}^{+}$ i.eV ${\text{eV}}^{+2}$ and ${\text{Cr}}^{+2}$

Q41JEE Main 2023MCQ4MRedox Reactions

Correct statement about smog is [29-Jan-2023 Shift 1]

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📖 Explanation

Photochemical smog has high concentration of oxidising agents ${\text{NO}}_2$ is produced from ${\text{NO}}$ and ${\text{O}}_3$ in the presence of sunlightClassical smog contain smoke, fog and ${\text{SO}}_2$ and it is known as reducing smog, as chemically it is reducing mixture

Q42JEE Main 2023MCQ4MCoordination Compounds

Chiral complex from the following is:Here en $=$ ethylene diamine [29-Jan-2023 Shift 1]

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Q43JEE Main 2023MCQ4MHaloalkanes and Haloarenes

Identify the correct order for the given property for following compoundsChoose the correct answer from the option given below :- [29-Jan-2023 Shift 1]

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Q44JEE Main 2023MCQ4MAlcohols, Phenols and Ethers

The increasing order of ${\text{pK}}_{ {\text{a}}}$ for the following phenols is [29-Jan-2023 Shift 1]

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📖 Explanation

Order of acidity for following phenol is- $M$ and $- {\text{I}}$ increases acidity $+ {\text{M}}$ and $+ {\text{I}}$ decreases acidity

Q45JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

Match List I with List II. List-I Reaction List-II Reagents A Hoffmann Degradation I Conc. ${\text{KOH}}, \Delta$ B Clemenson reduction II ${\text{CHCl}}_3, {\text{NaOH}} / {\text{H}}_3 {\text{O}}^{+}$ C Cannizaro reaction III ${\text{Br}}_2, {\text{NaOH}}$ D Reimer-Tiemann reaction IV ${\text{Zn}}- {\text{Hg}} / {\text{HCl}}$ [29-Jan-2023 Shift 1]

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📖 Explanation

Reaction Reagents used A HoffmannDegradation ${\text{Br}}_2, {\text{NaOH}}$ B Clemensonreduction ${\text{Zn}}- {\text{Hg}} / {\text{HCl}}$ C Cannizaro reaction Conc. ${\text{KOH}}, \Delta$ D Reimer-Tiemannreaction ${\text{CHCl}}_3, {\text{NaOH}} / {\text{H}}_3 {\text{O}}^{+}$

Q46JEE Main 2023MCQ4MAmines

The major product ' ${\text{P}}$ ' for the following sequence| of reactions is: [29-Jan-2023 Shift 1]

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Q47JEE Main 2023MCQ4MSalt Analysis

During the borax bead test with ${\text{CuSO}}_4$ , a blue green colour of the bead was observed in oxidising flame due to the formation of [29-Jan-2023 Shift 1]

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📖 Explanation

Blue green colour is due to formation of ${\text{Cu}} ( {\text{BO}}_2) _2$ ${\text{CuSO}}_4 \;{\longrightarrow }^{\Delta } {\text{CuO}}+ {\text{SO}}_3$ ${\text{CuO}}+ {\text{B}}_2 {\text{O}}_3 \to {\text{Cu}} ( {\text{BO}}_2) _2$

Q48JEE Main 2023MCQ4MBiomolecules

Match List I with List II List-I Antimicrobials List-II Names A Narrow SpectrumAntibiotic I Furacin B Antiseptic II Sulphur dioxide C Disinfectants III Penicillin-G D Broad spectrum antibiotic IV Chloramphenicol [29-Jan-2023 Shift 1]

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Q49JEE Main 2023MCQ4MBiomolecules

Number of cyclic tripeptides formed with 2 amino acids $A$ and $B$ is: [29-Jan-2023 Shift 1]

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Q50JEE Main 2023MCQ4MPractical Organic Chemistry

Compound that will give positive Lassaigne's test for both nitrogen and halogen is [29-Jan-2023 Shift 1]

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📖 Explanation

${\text{CH}}_3 {\text{NH}}_2 \cdot {\text{HCl}}{\to } ^{{\text{Na}}}_{\;\text{ fusion }\;} {\text{NaCN}}$ and ${\text{NaCl}}$ ${\text{NaCN}}$ gives +ve test for nitrogen and ${\text{NaCl}}$ gives +ve test for halogen

Q51JEE Main 2023NAT4MIonic Equilibrium

Millimoles of calcium hydroxyide required to produce $100 {\text{mL}}$ of the aqueous solution of ${\text{pH}} 12$ is $x \times 10^{-1}$ . The value of $x$ is _______(Nearest integer).Assume complete dissociation. [29-Jan-2023 Shift 1]

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$\; \because {\text{pH}}=12$ $\; \therefore \$ [ {\text{H}}^{+}]$ =10^{-12} {\text{M}}; \therefore $[ {\text{OH}}^{-}]$ =10^{-2} {\text{M}}; \therefore [ {\text{Ca}}( {\text{OH}})_2] =5 \times 10^{-3} {\text{M}};5 \times 10^{-3}=;{;\text{ milli moles of Ca};( {\text{OH}})_2}/{100 {\text{mL}}};;\text{ milli moles of }; {\text{Ca}}( {\text{OH}})_2=5 \times 10^{-1}$

Q52JEE Main 2023NAT4MChemical Bonding and Molecular Structure

The number of molecules or ions from the following, which do not have odd number of electrons are ________.(A) ${\text{NO}}_2$ (B) ${\text{ICl}}_4^{-}$ (C) ${\text{BrF}}_3$ (D) ${\text{ClO}}_2$ (E) ${\text{NO}}_2^{+}$ (F) ${\text{NO}}$ [29-Jan-2023 Shift 1]

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📖 Explanation

${\text{ICl}}_4^{-}, {\text{BrF}}_3$ and ${\text{NO}}_2^{+}$ do not have odd number of ${\text{e}}^{-}$

Q53JEE Main 2023NAT4MChemical Equilibrium

Consider the following reaction approaching equilibrium at $27^{\circ} {\text{C}}$ and $1 {\text{atm}}$ pressure ${\text{A}}+ {\text{B}} {⇌}^{ {\text{K}}_{ {\text{f}}}=10^3} _{ {\text{K}}_{ {\text{r}}}=10^2} {\text{C}}+ {\text{D}}$ The standard Gibb's energy change $(\Delta _{ {\text{r}}} G^{\circ})$ at $27^{\circ} {\text{C}}$ is (-) _______ ${\text{kJ}} {\text{mol}}^{-1}$ (Nearest integer).(Given : ${\text{R}}=8.3 {\text{J}} {\text{K}}^{-1} {\text{mol}}^{-1}$ and $\ln 10=2.3$ ) [29-Jan-2023 Shift 1]

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📖 Explanation

$\; \because \Delta {\text{G}}^{\circ}=- {\text{RT}} \ln {\text{K}}_{ {\text{eq}}}$ $\;\;\text{ and }\; {\text{K}}_{ {\text{eq}}}=\;{ {\text{K}}_{ {\text{f}}}}/{ {\text{K}}_{ {\text{b}}}}$ $\; \therefore {\text{K}}_{ {\text{eq}}}=\;\frac{10^3}{10^2}=10$ $\; \therefore \Delta {\text{G}}=- {\text{RT}} \ln 10$ $\;\Rightarrow -(8.3 \times 300 \times 2.3)=-5.7 {\text{kJ}} {\text{mole}}=6 {\text{kJ}}$ $\; {\text{mole}}^{-1}(\;\text{ nearest integer) }\;$ $\;\;\text{ Ans }\;=6$

Q54JEE Main 2023NAT4MIonic Equilibrium

Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at $100.15^{\circ} {\text{C}}$ . When $0.2 {\text{mol}}$ of ${\text{NaCl}}$ is added to the resulting solution, it was observed that the solution froze at $-0.8^{\circ} {\text{C}}$ . The solutbility product of ${\text{PbCl}}_2$ formed is ________ $\times 10^{-6}$ at $298 {\text{K}}$ . (Nearest integer)Given : ${\text{K}}_{ {\text{b}}}=0.5 {\text{K}} {\text{kg}} {\text{mol}}^{-1}$ and ${\text{K}}_{ {\text{f}}}=1.8 {\text{kg}} {\text{mol}}^{-1}$ .Assume molality to be equal to molarity in all cases. [29-Jan-2023 Shift 1]

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Let a mole ${\text{Pb}} ( {\text{NO}}_3) _2$ be added ${{\text{Pb}} ( {\text{NO}}_3) _2}_{\text{a}} \to { {\text{Pb}}^{2+}}_{\text{a}}+{2 {\text{NO}}_3^{-}}_{2a}$ $\Delta {\text{T}}_{ {\text{b}}}=0.15=0.5[3 {\text{a}}] \Rightarrow {\text{a}}=0.1$ $\; {\text{Pb}}_{( {\text{aq}})}^{2+}+\;2 {\text{Cl}}_{( {\text{aq}})}^{-} \to {\text{PbCl}}_2( {\ s})$ ${\text{t}}=0\;0.1\;0.2$ ${\text{t}}=\infty \;(0.1- {\text{x}})\;(0.2-2 {\text{x}})$ In final solution $\;\Delta {\text{T}}_{ {\text{f}}}=0.8=1.8 [\;{0.3-3 {\text{x}}+0.2+0.2}/{1}]$ $\;\Rightarrow {\text{x}}=\;\frac{2.3}{27}$ $\;\Rightarrow {\text{K}}_{\;\text{sp }\;}= (0.1-\;\frac{2.3}{27}) (0.2-\;\frac{4.6}{27}) ^2=13 \times 10^{-6}$

Q55JEE Main 2023NAT4MChemical Equilibrium

Water decomposes at $2300 {\text{K}}$ ${\text{H}}_2 {\text{O}}( {\text{g}}) \to {\text{H}}_2( {\text{g}})+\;\frac{1}{2} {\text{O}}_2( {\text{g}})$ The percent of water decomposing at $2300 {\text{K}}$ and 1 bar is ________ (Nearest integer).Equilibrium constant for the reaction is $2 \times 10^{-3}$ at $2300 {\text{K}}$ [29-Jan-2023 Shift 1]

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$\; {\text{H}}_2 {\text{O}}( {\text{g}}) ⇌ {\text{H}}_2( {\text{g}})+\;\frac{1}{2} {\text{O}}_2( {\text{g}})$ $\; {\text{P}}_0[1-\alpha] \;\; {\text{P}}_0 \alpha \;\; \;{ {\text{P}}_0 \alpha}/{2} \;\; \;\text{ partial pr. at eq. }\;$ $\; {\text{P}}_0 [1+\;\frac{\alpha}{2}] =1$ . . . (i) $\; {\text{K}}_{ {\text{p}}}=\;{ ( {\text{P}}_{ {\text{H}}_2}) ( {\text{P}}_{ {\text{O}}_2}) ^{1 / 2}}/{ {\text{P}}_{ {\text{H}}_2 {\text{O}}}}$ $\;\;{ ( {\text{P}}_0 \alpha) ({ {\text{P}}_0 \alpha}/{2}) ^{1 / 2}}/{ {\text{P}}_0[1-\alpha]}=2 \times 10^{-3}$ since $\alpha$ is negligible w.r.t 1 so $P_0=1$ and $1-\alpha \approx 1$ $\;\;{\alpha {\sqrt{\alpha}}}/{{\sqrt{2}}}=2 \times 10^{-3}$ $\;\alpha^{3 / 2}=2^{3 / 2} \times 10^{-3}$ $\;\alpha=2^{3 / 2 \times 2 / 3} \times 10^{-3 \times 2 / 3}$ $\;\alpha=2 \times 10^{-2} \;\; \% \alpha=2 \%$

Q56JEE Main 2023NAT4MElectrochemistry

Following figure shows dependence of molar conductance of two electrolytes on concentration. \wedge ${\text{m}}^{\0}$ is the limiting molar conductivityThe number of Incorrect statement(s) from the following is ________ (A) $Lambda {\text{m}}$ for electrolyte $A$ is obtained by extrapolation(B) For electrolyte B, vx $Lambda {\text{m}}$ vs $\sqrt{ {\text{c}}}$ graph is a straight line with intercept equal to $Lambda \;{ {\text{m}}}^{0}$ (C) At infinite dilution, the value of degree of dissociation approach zero for electrolyte B.(D) $Lambda {\text{m}}$ for any electrolyte ${\text{A}}$ or ${\text{B}}$ can be calculated using $\lambda ^0$ for individual ions. [29-Jan-2023 Shift 1]

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Q57JEE Main 2023NAT4MChemical Kinetics

For certain chemical reaction $X \to Y$ , the rate of formation of product is plotted against the time as shown in the figure. The number of Correct statement/s from the following is ________(A) Over all order of this reaction is one(B) Order of this reaction can't be determined(C) In region-I and III, the reaction is of first and zero order respectively(D) In region-II, the reaction is of first order(E) In region-II, the order of reaction is in the range of $0.1$ to $0.9$ . [29-Jan-2023 Shift 1]

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Q58JEE Main 2023NAT4MCoordination Compounds

The sum of bridging carbonyls in ${\text{W}}( {\text{CO}})_6$ and ${\text{Mn}}_2$ $( {\text{CO}})_{10}$ is _______. [29-Jan-2023 Shift 1]

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📖 Explanation

$[( {\text{CO}})_5 {\text{Mn}}- {\text{Mn}}( {\text{CO}})_5]$

Q59JEE Main 2023NAT4MPractical Organic Chemistry

Following chromatogram was developed by adsorption of compound ' ${\text{A}}$ ' on a $6 {\text{cm}}$ TLC glass plate. Retardation factor of the compound ' ${\text{A}}$ ' is $\times 10^{-1}$ . [29-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{R}}_{ {\text{f}}}=\;\frac{\;\text{ Distance moved by the substance frombase line }\;}{\;\text{ Distance move dby the solvent frombase line }\;}$ $\;=\;{3.0 {\text{cm}}}/{5.0 {\text{cm}}}=0.6 \;\text{ or }\; 6 \times 10^{-1}$

Q60JEE Main 2023NAT4MHydrocarbons

$17 {\text{mg}}$ of a hydrocarbon (M.F. ${\text{C}}_{10} {\text{H}}_{16}$ ) takes up $8.40 {\text{mL}}$ of the ${\text{H}}_2$ gas measured at $0^{\circ} {\text{C}}$ and 760 ${\text{mm}}$ of ${\text{Hg}}$ . Ozonolysis of the same hydrocarbon yields ${\text{CH}}_3- {\text{C}}_{||}_{\text{O}}- {\text{CH}}_3-, {\text{H}}- {\text{C}}_{||}_{\text{O}}- {\text{H}}, {\text{H}}- {\text{C}}_{||}_{\text{O}}- {\text{CH}}_2- {\text{CH}}_2- {\text{C}}_{||}_{\text{O}}- {\text{C}}_{||}_{\text{O}}- {\text{H}}$ The number of double bond/s present in the hydrocarbon is ________. [29-Jan-2023 Shift 1]

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📖 Explanation

Moles of hydrocarbon $=\;\frac{17 \times 10^{-3}}{136}=1.25 \times 10^{-4}$ Mole of ${\text{H}}_2$ gas $\;\Rightarrow 1 \times \;\frac{8.40}{1000}= {\text{n}} \times 0.0821 \times 273$ $\;\Rightarrow {\text{n}}=3.75 \times 10^{-4}$ Hydrogen molecule used for 1 molecule of hydrocarbon is 3 $=\;\frac{3.75 \times 10^{-4}}{1.25 \times 10^{-4}}=3$

Mathematics30 questions

Q61JEE Main 2023MCQ4MFunctions

The domain of $f(x)=\;\frac{\log _{(x+1)}(x-2)}{e^{2 \log _e x}-(2 x+3)}, x \in R$ is [29-Jan-2023 Shift 1]

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📖 Explanation

$\;x-2 > 0 \Rightarrow x >2$ $\;x+1 > 0 \Rightarrow x > -1$ $\;x+1 \ne 1 \Rightarrow x \ne 0 \;\text{ and }\; x > 0$ Denominator $\;x^{2-}2 x-3 \ne 0$ $\;(x-3)(x+1) \ne 0$ $\;x \ne -1,3$ So Ans $(2, \infty )-\{3\}$

Q62JEE Main 2023MCQ4MFunctions

Let $f: R \to R$ be a function such that $f(x)=\;\frac{x^{2+}2 x+1}{x^{2+}1}$ . Then [29-Jan-2023 Shift 1]

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📖 Explanation

$\;f(x)=\;\frac{(x+1)^2}{x^{2+}1}=1+\;\frac{2 x}{x^{2+}1}$ $\;f(x)=1+\;\frac{2}{x+\;\frac{1}{x}}$

Q63JEE Main 2023MCQ4MComplex Numbers

For two non-zero complex number $z_1$ and $z_2$ , if ${\text{Re}} (z_1 z_2) =0$ and ${\text{Re}} (z_1+z_2) =0$ , then which of the following are possible ?(A) ${\text{Im}} ( {\text{z}}_1) > 0$ and ${\text{Im}} ( {\text{z}}_2) > 0$ (B) ${\text{Im}} ( {\text{z}}_1) < 0$ and ${\text{Im}} ( {\text{z}}_2) > 0$ (C) ${\text{Im}} ( {\text{z}}_1) > 0$ and ${\text{Im}} ( {\text{z}}_2) < 0$ (D) ${\text{Im}} ( {\text{z}}_1) < 0$ and ${\text{Im}} ( {\text{z}}_2) < 0$ Choose the correct answer from the options given below : [29-Jan-2023 Shift 1]

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📖 Explanation

${\text{z}}_1= {\text{x}}_1+ {\text{iy}}_1$ $z_2=x_2+y_2$ ${\text{Re}} (z_1 z_2) =x_1 x_2-y_1 y_2=0$ ${\text{Re}} (z_1+z_2) =x_1+x_2=0$ ${\text{x}}_1 & {\text{x}}_2$ are of opposite sign $y_1 & y_2$ are of opposite sign

Q64JEE Main 2023MCQ4MQuadratic Equation and Inequalities

Let $\lambda \ne 0$ be a real number. Let $\alpha, \beta$ be the roots of the equation $14 x^{2-}31 x+3 \lambda =0$ and $\alpha, \gamma$ be the roots of the equation $35 x^{2-}53 x+4 \lambda =0$ . Then $\;\frac{3 \alpha}{\beta}$ and $\;\frac{4 \alpha}{\gamma }$ are the roots of the equation : [29-Jan-2023 Shift 1]

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📖 Explanation

$14 x^{2-}31 x+3 \lambda =0$ $\alpha+\beta=\;\frac{31}{14} . . . .(1)$ and $\alpha \beta=\;\frac{3 \lambda }{14}$ $\;35 x^{2-}53 x+4 \lambda =0$ $\;\alpha+\gamma =\;\frac{53}{35} . . .(3) \;\text{ and }\; \alpha \gamma =\;\frac{4 \lambda }{35} . . .$ $\;\;\frac{(2)}{(4)} \Rightarrow \;\frac{\beta}{\gamma }=\;\frac{3 \times 35}{4 \times 14}=\;\frac{15}{8} \Rightarrow \beta=\;\frac{15}{8} \gamma$ $\;\;\text{ (1) }\;-(3) \Rightarrow \beta-\gamma =\;\frac{31}{14}-\;\frac{53}{35}=\;\frac{155-106}{70}=\;\frac{7}{10}$ $\;\;\frac{15}{8} \gamma -\gamma =\;\frac{7}{10} \Rightarrow \gamma =\;\frac{4}{5}$ $\;\Rightarrow \beta=\;\frac{15}{8} \times \;\frac{4}{5}=\;\frac{3}{2}$ $\;\Rightarrow \alpha=\;\frac{31}{14}-\beta=\;\frac{31}{14}-\;\frac{3}{2}=\;\frac{5}{7}$ $\;\Rightarrow \lambda =\;\frac{14}{3} \alpha \beta=\;\frac{14}{3} \times \;\frac{5}{7} \times \;\frac{3}{2}=5$ so, sum of roots $\;\frac{3 \alpha}{\beta}+\;\frac{4 \alpha}{\gamma }= (\;\frac{3 \alpha \gamma +4 \alpha \beta}{\beta \gamma })$ $\;=\;\frac{ (3 \times \frac{4 \lambda }{35}+4 \times \frac{3 \lambda }{14}) }{\beta \gamma }=\;\frac{12 \lambda (14+35)}{14 \times 35 \beta \gamma }$ $\;=\;\frac{49 \times 12 \times 5}{490 \times \;\frac{3}{2} \times \;\frac{4}{5}}=5$ Product of roots $=\;\frac{3 \alpha}{\beta} \times \;\frac{4 \alpha}{\gamma }=\;\frac{12 \alpha^2}{\beta \gamma }=\;\frac{12 \times \frac{25}{49}}{\frac{3}{2} \times \;\frac{4}{5}}=\;\frac{250}{49}$ So, required equation is $x^{2-}5 x+\;\frac{250}{49}=0$ $\Rightarrow 49 x^{2-}245 x+250=0$

Q65JEE Main 2023MCQ4MMatrices and Determinants

Consider the following system of questions $\;\alpha x+2 y+z=1$ $\;2 \alpha x+3 y+z=1$ $\;3 x+\alpha y+2 z=\beta$ For some $\alpha, \beta \in ℝ$ . Then which of the following is NOT correct. [29-Jan-2023 Shift 1]

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📖 Explanation

$\;D={\begin{vmatrix}\alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & 2\end{vmatrix}}=0 \Rightarrow \alpha=-1,3$ $\;D_x={\begin{vmatrix}2 & 1 & 1 \\ 3 & 1 & 1 \\ \alpha & 2 & \beta\end{vmatrix}}=0 \Rightarrow \beta=2$ $\;D_y={\begin{vmatrix}\alpha & 1 & 1 \\ 2 \alpha & 1 & 1 \\ 3 & 2 & \beta\end{vmatrix}}=0$ $\;D_z={\begin{vmatrix}\alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & \beta\end{vmatrix}}=0$ $\;\beta=2, \alpha=-1$ $\;\alpha=-1, \beta=2 \;\text{ Infinite solution }\;$

Q66JEE Main 2023MCQ4MMatrices and Determinants

Let $\alpha$ and $\beta$ be real numbers. Consider a $3 \times 3$ matrix $A$ such that $A^2=3 A+\alpha I$ . If $A^4=21 A+\beta I$ , then [29-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{A}}^2=3 {\text{A}}+\alpha {\text{I}}$ $\; {\text{A}}^3=3 {\text{A}}^{2+}\alpha {\text{A}}$ $\; {\text{A}}^3=3(3 {\text{A}}+\alpha {\text{I}})+\alpha {\text{A}}$ $\; {\text{A}}^3=9 {\text{A}}+\alpha {\text{A}}+3 \alpha {\text{I}}$ $\; {\text{A}}^4=(9+\alpha) {\text{A}}^{2+}3 \alpha {\text{A}}$ $\;=(9+\alpha)(3 {\text{A}}+\alpha {\text{I}})+3 \alpha {\text{A}}$ $\;= {\text{A}}(27+6 \alpha)+\alpha(9+\alpha)$ $\;\Rightarrow 27+6 \alpha=21 \Rightarrow \alpha=-1$ $\;\Rightarrow \beta=\alpha(9+\alpha)=-8$

Q67JEE Main 2023MCQ4MLimits, Continuity and Differentiability

Let $x=2$ be a root of the equation $x^{2+}p x+q=0$ and ${\text{f}}( {\text{x}})={\begin{cases}\\ \\ {1-\cos ( {\text{x}}^{2-}4 p {\text{p}}+ {\text{q}}^{2+}8 {\text{q}}+16) }/{( {\text{x}}-2 {\text{p}})^4} & {\text{x}} \ne 2 {\text{p}} \\ 0 & {\text{x}}=2 {\text{p}}}\end{cases}$ Then $\lim _{x \to 22^{+}}[f(x)]$ where [. ] denotes greatest integer function, is [29-Jan-2023 Shift 1]

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📖 Explanation

$\;\lim _{x \to 2 p^{+}} (\;\frac{1-\cos (x^{2-}4 p x+q^{2+}8 q+16) }{ (x^{2-}4 p x+q^{2+}8 q+16) ^2}) (\;\frac{ (x^{2-}4 p x+q^{2+}8 q+16) ^2}{(x-2 p)^2})$ $\;\lim _{h \to 0} \;\frac{1}{2} (\;\frac{(2 p+h)^{2-}4 p(2 p+h)+q^{2+}82+16}{h^2}) ^2=\;\frac{1}{2}$ $\;\;\text{ Using L'Hospital's }\;$ $\;\lim _{x \to 2 {\text{p}}^{+}}[f(x)]=0$

Q68JEE Main 2023MCQ4MDefinite Integrals

Let $\;\; f(x)=x+\;\frac{a}{\pi^{2-}4} sin \;x+\;\frac{b}{\pi^{2-}4} \cos x$ , $x \in ℝ$ be a function which satisfies $f(x)=x+int_{0}^{\pi / 2} sin \;(x+y) f(y) d y$ . Then $( {\text{a}}+ {\text{b}})$ is equal to [29-Jan-2023 Shift 1]

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📖 Explanation

$f(x)=x+int_{0}^{\pi / 2}( sin \;x \cos y+\cos x sin \;y) f(y) d y$ $f(x)=x+int_{0}^{\pi / 2}((\cos y f(y) d y) sin \;x+( sin \;y f(y) d y) \cos x)$ On comparing with $f(x)=x+\;\frac{a}{\pi^{2-}4} sin \;x+\;\frac{b}{\pi^{2-}4} \cos x, x \in ℝ$ then $\;\Rightarrow \;\frac{a}{\pi^{2-}4}=int_{0}^{\pi / 2} \cos y f(y) d y$ . . . (2) $\;\Rightarrow \;\frac{b}{\pi^{2-}4}=int_{0}^{\pi / 2} sin \;y f(y) d y$ . . . (3)Add (2) and (3) $\;\;\frac{a+b}{\pi^{2-}4}=int_{0}^{\pi / 2}( sin \;y+\cos y) f(y) d y$ . . . (4) $\;\;\frac{a+b}{\pi^{2-}4}=int_{0}^{\pi / 2}( sin \;y+\cos y) f (\;\frac{\pi}{2}-y) d y$ . . . (5) $\;\;\text{ Add (4) and (5) }\;$ $\;\;\frac{2(a+b)}{\pi^{2-}4}=int_{0}^{x / 2}( sin \;y+\cos y) (\;\frac{\pi}{2}+\;\frac{(a+b)}{\pi^{2-}4}( sin \;y+\cos y)) d y$ $\;=\pi+\;\frac{a+b}{\pi^{2-}4} (\;\frac{\pi}{2}+1)$ $\;(a+b)=-2 \pi(\pi+2)$ $\;$

Q69JEE Main 2023MCQ4MArea Under The Curves

Let $A= \{(x, y) \in ℝ^2: y \ge 0,2 x \le y \le {\sqrt{4-(x-1)^2}}\}$ and ${\text{B}}= \{( {\text{x}}, {\text{y}}) \in ℝ \times ℝ: 0 \le {\text{y}} \le \min \{2 {\text{x}}, \sqrt{4-( {\text{x}}-1)^2}\} \}$ Then the ratio of the area of ${\text{A}}$ to the area of ${\text{B}}$ is [29-Jan-2023 Shift 1]

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📖 Explanation

$y^{2+}(x-1)^2=4$ shaded portion $=$ circular $( {\text{OABC}})$ $\;-{\text{Ar}}(\Delta {\text{OAB}})$ $\;=\;\frac{\pi(4)}{4}-\;\frac{1}{2}(2)(1)$ $A=(\pi-1)$ Area ${\text{B}}={\text{Ar}}(\triangle {\text{AOB}})+$ Area of arc of circle $( {\text{ABC}})$ $=\;\frac{1}{2}(1)(2)+\;\frac{\pi(2)^2}{4}=\pi+1$ $\;{ {\text{A}}}/{ {\text{B}}}=\;\frac{\pi-1}{\pi+1}$

Q70JEE Main 2023MCQ4MArea Under The Curves

Let $\Delta$ be the area of the region $\{(x, y) \in ℝ^2: x^{2+}y^2 \le 21, y^2 \le 4 x, x \ge 1\}$ . Then $\;\frac{1}{2} (\Delta -21 sin \;^{-1} \;{2}/{{\sqrt{7}}})$ is equal to [29-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\text{ Area }\; 2 int_{1}^{3} 2 {\sqrt{x}} \text{dx}+2 int_{3}^{{\sqrt{21}}} {\sqrt{21-x^2 \text{dx}}}$ $\;\Delta =\;\frac{8}{3}(3 {\sqrt{3}}-1)+21 sin \;^{-1} (\;{2}/{{\sqrt{7}}}) -6 {\sqrt{3}}$ $\;\;\frac{1}{2} (\Delta -21 sin \;^{-1} (\;{2}/{{\sqrt{7}}}) ) =\;{2 {\sqrt{3}}-\frac{8}{3}}/{2}$ $\;={\sqrt{3}}-\;\frac{4}{3}$

Q71JEE Main 2023MCQ4MStraight Lines and Pair of Straight Lines

A light ray emits from the origin making an angle $30^{\circ}$ with the positive ${\text{x}}$ -axis. After getting reflected by the line $x+y=1$ , if this ray intersects $x$ -axis at $Q$ , then the abscissa of $Q$ is [29-Jan-2023 Shift 1]

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📖 Explanation

Slope of reflected ray $=\tan 60^{\circ}={\sqrt{3}}$ Line ${\text{y}}=\;{ {\text{x}}}/{{\sqrt{3}}}$ intersect ${\text{y}}+ {\text{x}}=1$ at $(\;{{\sqrt{3}}}/{{\sqrt{3}}+1}, \;{1}/{{\sqrt{3}}+1})$ Equation of reflected ray is $y-\;{1}/{{\sqrt{3}}+1}={\sqrt{3}} (x-\;{{\sqrt{3}}}/{{\sqrt{3}}+1})$ Put ${\text{y}}=0 \Rightarrow {\text{x}}=\;{2}/{3+{\sqrt{3}}}$

Q72JEE Main 2023MCQ4MStraight Lines and Pair of Straight Lines

Let ${\text{B}}$ and ${\text{C}}$ be the two points on the line ${\text{y}}+ {\text{x}}=0$ such that ${\text{B}}$ and ${\text{C}}$ are symmetric with respect to the origin. Suppose ${\text{A}}$ is a point on ${\text{y}}-2 {\text{x}}=2$ such that $\triangle {\text{ABC}}$ is an equilateral triangle. Then, the area of the $\triangle {\text{ABC}}$ is [29-Jan-2023 Shift 1]

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📖 Explanation

At $A \;\; x=y$ $Y-2 x=2$ $(-2,-2)$ Height from line $x+y=0$ ${\text{h}}=\;{4}/{{\sqrt{2}}}$ Area of $\Delta =\;{{\sqrt{3}}}/{4} \;{ {\text{h}}^2}/{ sin \;^2 60}=\;{8}/{{\sqrt{3}}}$

Q73JEE Main 2023MCQ4MCircles

Let the tangents at the points ${\text{A}}(4,-11)$ and ${\text{B}}(8,-5)$ on the circle ${\text{x}}^{2+} {\text{y}}^{2-}3 {\text{x}}+10 {\text{y}}-15=0$ , intersect at the point ${\text{C}}$ . Then the radius of the circle, whose centre is ${\text{C}}$ and the line joining ${\text{A}}$ and ${\text{B}}$ is its tangent, is equal to [29-Jan-2023 Shift 1]

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📖 Explanation

Equation of tangent at ${\text{A}}(4,-11)$ on circle is $\;\Rightarrow 4 x-11 y-3 (\;\frac{x+4}{2}) +10 (\;\frac{y-11}{2}) -15=0$ $\;\Rightarrow 5 x-12 y-152=0 . . . . . . \;\text{ (1) }\;$ Equation of tangent at B $(8,-5)$ on circle is $\;\Rightarrow 8 x-5 y-3 (\;\frac{x+8}{2}) +10 (\;\frac{y-5}{2}) -15=0$ $\;\Rightarrow 13 x-104=0 \Rightarrow x=8$ put in (1) $\Rightarrow {\text{y}}=\;\frac{28}{3}$ ${\text{r}}= |\;{3.8+\frac{2.28}{3}-34}/{{\sqrt{13}}}|=\;{2 {\sqrt{13}}}/{3}$

Q74JEE Main 2023MCQ4MDefinite Integrals

Let $[ {\text{x}}]$ denote the greatest integer $\le {\text{x}}$ . Consider the function ${\text{f}}( {\text{x}})=\max \{ {\text{x}}^2, 1+[ {\text{x}}]\}$ . Then the value of the integral $int_{0}^{2} f(x) \text{dx}$ is: [29-Jan-2023 Shift 1]

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📖 Explanation

$\;A=int_{0}^{1} 1 \cdot \text{dx}+int_{1}^{{\sqrt{2}}} 2 \text{dx}+int_{{{\sqrt{2}}}}^{2} x^2 \text{dx}$ $\;=1+2 {\sqrt{2}}-2+\;\frac{8}{3}-\;{2 {\sqrt{2}}}/{3}$ $\;=\;\frac{5}{3}+\;{4 {\sqrt{2}}}/{3}$

Q75JEE Main 2023MCQ4MVector Algebra

If the vectors $\vec{a}=\lambda \hat{i}+\mu \hat{j}+4 \hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}-2 \hat{k}$ and $\vec{c}=2 \hat{i}+3 \hat{j}+\hat{k}$ are coplanar and the projection of $\vec{a}$ on the vector $\vec{b}$ is ${\sqrt{54}}$ units, then the sum of all possible values of $\lambda +\mu$ is equal to [29-Jan-2023 Shift 1]

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📖 Explanation

${\begin{vmatrix}\lambda & \mu & 4 \\ -2 & 4 & -2 \\ 2 & 3 & 1\end{vmatrix}}=0$ $\lambda (10)-\mu (2)+4(-14)=0$ $10 \lambda -2 \mu =56$ $5 \lambda -\mu =28$ . . . (1) ${\vec{\text{a}} \cdot \vec{\text{b}}}/{|\vec{\text{b}}|}={\sqrt{54}}$ $\;\;{-2 \lambda +4 \mu -8}/{{\sqrt{24}}}={\sqrt{54}}$ $\;-2 \lambda +4 \mu -8={\sqrt{54 \times 24}}$ . . . (2)By solving equation (1) & (2) $\Rightarrow \lambda +\mu =24$

Q76JEE Main 2023MCQ4MProbability

Fifteen football players of a club-team are given 15 T-shirts with their names written on the backside. If the players pick up the ${\text{T}}$ -shirts randomly, then the probability that at least 3 players pick the correct ${\text{T}}$ -shirt is [29-Jan-2023 Shift 1]

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📖 Explanation

Required probability $=1-\;\frac{D_{(15)}+{ }^{15} C_1 \cdot D_{(14)}+{ }^{15} C_2 D_{(13)}}{15 !}$ Taking ${\text{D}}_{(15)}$ as $\;\frac{15 \;\text{ ! }\;}{e}$ ${\text{D}}_{(14)}$ as $\;\frac{14 !}{e}$ ${\text{D}}_{(13)}$ as $\;\frac{13 !}{e}$ We get, $1- (\;\frac{\;\frac{15 !}{e}+15 \cdot \frac{14 !}{e}+\frac{15 \times 14}{2} \times \frac{13 !}{e}}{15 !})$ $=1- (\;\frac{1}{e}+\;\frac{1}{e}+\;\frac{1}{2 e}) =1-\;\frac{5}{2 e} \approx .08$

Q77JEE Main 2023MCQ4MDifferentiation

Let $f(\theta)=3 ( sin \;^4 (\;\frac{3 \pi}{2}-\theta) + sin \;^4(3 \pi+\theta)) -2 (1- sin \;^2 2 \theta)$ and $S= \{\theta \in [0, \pi]: f^{'}(\theta)=-\;{{\sqrt{3}}}/{2}\}$ . If $4 \beta=sum_{\theta \in S} \theta$ , then $f(\beta)$ is equal to [29-Jan-2023 Shift 1]

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📖 Explanation

$\;f(\theta)=3 ( sin \;^4 (\;\frac{3 \pi}{2}-\theta) + sin \;^4(3 x+\theta)) -2 (1- sin \;^2 2 \theta)$ $\;S= \{\theta \in [0, \pi]: f^{'}(\theta)=-\;{{\sqrt{3}}}/{2}\}$ $\;\Rightarrow {\text{f}}(\theta)=3 (\cos ^4 \theta+ sin \;^4 \theta) -2 \cos ^2 2 \theta$ $\;\Rightarrow {\text{f}}(\theta)=3 (1-\;\frac{1}{2} sin \;^2 2 \theta) -2 \cos ^2 2 \theta$ $\;\Rightarrow {\text{f}}(\theta)=3-\;\frac{3}{2} sin \;^2 2 \theta-2 \cos ^2 \theta$ $\;=\;\frac{3}{2}-\;\frac{1}{2} \cos ^2 2 \theta=\;\frac{3}{2}-\;\frac{1}{2} (\;\frac{1+\cos 4 \theta}{2})$ $\;f(\theta)=\;\frac{5}{4}-\;\frac{\cos 4 \theta}{4}$ $\;f^{'}(\theta)= sin \;4 \theta$ $\;\Rightarrow {\text{f}}^{'}(\theta)= sin \;4 \theta=-\;{{\sqrt{3}}}/{2}$ $\;\Rightarrow 4 \theta= {\text{n}} \pi+(-1)^{ {\text{n}}} \;\frac{\pi}{3}$ $\;\Rightarrow \theta=\;{ {\text{n}} \pi}/{4}+(-1)^{ {\text{n}}} \;\frac{\pi}{12}$ $\;\Rightarrow \theta=\;\frac{\pi}{12}, (\;\frac{\pi}{4}-\;\frac{\pi}{12}) , (\;\frac{\pi}{2}+\;\frac{\pi}{12}) , (\;\frac{3 \pi}{4}-\;\frac{\pi}{12})$ $\;\Rightarrow 4 \beta=\;\frac{\pi}{4}+\;\frac{\pi}{2}+\;\frac{3 \pi}{4}=\;\frac{3 \pi}{2}$ $\;\Rightarrow \beta=\;\frac{3 \pi}{8} \Rightarrow {\text{f}}(\beta)=\;\frac{5}{4}-\;\frac{\cos \frac{3 \pi}{2}}{4}=\;\frac{5}{4}$

Q78JEE Main 2023MCQ4MSets and Relations

If ${\text{p}}, {\text{q}}$ and ${\text{r}}$ are three propositions, then which of the following combination of truth values of $p, q$ and $r$ makes the logical expression $\{(p ∨ q) ∧((∼ p) ∨ r)\} \to ((∼ q) ∨ r)$ false ? [29-Jan-2023 Shift 1]

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📖 Explanation

Option (3) $( {\text{p}} ∨ {\text{q}}) ∧(∼ {\text{q}} ∨ {\text{r}}) \to (∼ {\text{p}} ∨ {\text{r}})$ will be False.

Q79JEE Main 2023MCQ4MProbability

There rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable ${\text{X}}$ denote the number of rotten apples. If $\mu$ and $\sigma^2$ represent mean and variance of $X$ , respectively, then $10 (\mu ^{2+}\sigma^2)$ is equal to [29-Jan-2023 Shift 1]

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📖 Explanation

$\;\sum x P(x)=\;\frac{6}{2}=\mu$ $\;\sigma^2=\sum x^2 P(x)-\mu ^2$ $\;\sigma^{2+}\mu ^2=0+\;\frac{1}{2}+\;\frac{12}{10}+\;\frac{9}{30}=2$ $\;10 (\sigma^{2+}\mu ^2) =20 \;\text{ Ans. }\;$

Q80JEE Main 2023MCQ4MDifferential Equations

Let $y=f(x)$ be the solution of the differential equation $y(x+1) \text{dx}-x^2 d y=0, y(1)=e$ . Then $\lim _{x \to 0^{+}} f(x)$ is equal to [29-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\frac{x+1}{x^2} \text{dx}=\;\frac{d y}{y}$ $\;\ln x-\;\frac{1}{x}=\ln y+c$ $\;(1, {\text{e}})$ $\; {\text{c}}=-2$ $\;\ln x-\;\frac{1}{x}=\ln y-2$ $\;y=e^{\ln x}-\;\frac{1}{x}+2$ $\;\lim _{x \to 0^{+}} e^{\ln x-1}-\;\frac{1}{x}+2$ $\;=e^{-\infty }$ $\;=0$

Q81JEE Main 2023NAT4MThree Dimensional Geometry

Let the co-ordinates of one vertex of $\triangle {\text{ABC}}$ be ${\text{A}}(0,2, \alpha)$ and the other two vertices lie on the line $\;{ {\text{x}}+\alpha}/{5}=\;{ {\text{y}}-1}/{2}=\;\frac{z+4}{3}$ . For $\alpha \in ℤ$ , if the area of $\triangle {\text{ABC}}$ is 21 sq. units and the line segment ${\text{BC}}$ has length $2 {\sqrt{21}}$ units, then $\alpha^2$ is equal to _______. [29-Jan-2023 Shift 1]

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📖 Explanation

A. $( {\text{O}}_1 2, \alpha)$ $\;{\sqrt{(2 \alpha+5)^{2+}(2 \alpha+20)^{2+}(2 \alpha-5)^2}}={\sqrt{21}} {\sqrt{38}}$ $\;\Rightarrow 12 \alpha^{2+}80 \alpha+450=798$ $\;\Rightarrow 12 \alpha^{2+}80 \alpha-348=0$ $\;\Rightarrow \alpha=3 \Rightarrow \alpha^2=9$

Q82JEE Main 2023NAT4MThree Dimensional Geometry

Let the equation of the plane P containing the line $x+10=\;\frac{8-y}{2}=z$ be $a x+b y+3 z=2(a+b)$ and the distance of the plane ${\text{P}}$ from the point $(1,27,7)$ be c. Then ${\text{a}}^{2+} {\text{b}}^{2+} {\text{c}}^2$ is equal to ________. [29-Jan-2023 Shift 1]

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📖 Explanation

The line $\;\frac{x+10}{1}=\;\frac{y-8}{-2}=\;\frac{z}{1}$ have a point $(-10,8,0)$ with d. r. $(1,-2,1)$ $\because$ the plane $a x+b y+3 z=2(a+b)$ $\Rightarrow b=2 a$ $\&$ dot product of d.r.'s is zero $\therefore a-2 b+3=0$ $\therefore {\text{a}}=1 & {\text{b}}=2$ Distance from $(1,27,7)$ is $\;c=\;{1+54+21-6}/{{\sqrt{14}}}=\;{70}/{{\sqrt{14}}}=5 {\sqrt{14}}$ $\; \therefore {\text{a}}^{2+} {\text{b}}^{2+} {\text{c}}^2=1+4+350$ $\;=355$

Q83JEE Main 2023NAT4MSequences and Series

Suppose ${\text{f}}$ is a function satisfying $f(x+y)=f(x)+f(y)$ for all $x, y \in ℕ$ and $f(1)=\;\frac{1}{5}$ . If $sum_{n=1}^{m} \;\frac{f(n)}{n(n+1)(n+2)}=\;\frac{1}{12}$ , then $m$ is equal to ________ [29-Jan-2023 Shift 1]

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📖 Explanation

$\; \because f(1)=\;\frac{1}{5} \therefore f(2)=f(1)+f(1)=\;\frac{2}{5}$ $\;f(2)=\;\frac{2}{5} \;\; f(3)=f(2)+f(1)=\;\frac{3}{5}$ $\;f(3)=\;\frac{3}{5}$ $\; \therefore sum_{n=1}^{m} \;\frac{f(n)}{n(n+1)(n+2)}$ $\;=\;\frac{1}{5} sum_{n=1}^{m} (\;\frac{1}{n+1}-\;\frac{1}{n+2})$ $\;=\;\frac{1}{5} (\;\frac{1}{2}-\;\frac{1}{3}+\;\frac{1}{3}-\;\frac{1}{4}+. . . .+\;\frac{1}{m+1}-\;\frac{1}{m+2})$ $\;=\;\frac{1}{5} (\;\frac{1}{2}-\;\frac{1}{m+2}) =\;\frac{m}{10(m+2)}=\;\frac{1}{12}$ $\; \therefore m=10$

Q84JEE Main 2023NAT4MSequences and Series

Let $a_1, a_2, a_3, . . .$ . be a GP of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24 , then $a_1 a_9+a_2 a_4 a_9+a_5+a_7$ is equal to ______. [29-Jan-2023 Shift 1]

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📖 Explanation

$\;a_4 \cdot a_6=9 \Rightarrow (a_5) ^2=9 \Rightarrow a_5=3$ $\;\& a_5+a_7=24 \Rightarrow a_5+a_5 r^2=24 \Rightarrow (1+r^2) =8 \Rightarrow r={\sqrt{7}}$ $\;\Rightarrow a=\;\frac{3}{49}$ $\;\Rightarrow a_1 a_9+a_2 a_4 a_9+a_5+a_7=9+27+3+21=60$

Q85JEE Main 2023NAT4MVector Algebra

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero non-coplanar vectors. Let the position vectors of four points $A$ , $B, \;\; C$ and $D$ be $\vec{a}-\vec{b}+\vec{c}, \;\; \lambda \vec{a}-3 \vec{b}+4 \vec{c}$ , $-\vec{a}+2 \vec{b}-3 \vec{c}$ and $2 \vec{a}-4 \vec{b}+6 \vec{c}$ respectively. If $\vec{A B}$ , ${ \vec{\text{AC}}$ and ${ \vec{\text{AD}}$ are coplanar, then $\lambda$ is :Official Ans. by NTA [29-Jan-2023 Shift 1]

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📖 Explanation

$\;\overline{A B}=(\lambda -1) \overline{a}-2 \overline{b}+3 \overline{c}$ $\;\overline{A C}=2 \overline{a}+3 \overline{b}-4 \overline{c}$ $\;\overline{A D}=\overline{a}-3 \overline{b}+5 \overline{c}$ $\;{\begin{vmatrix}\lambda -1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5\end{vmatrix}}=0$ $\;\Rightarrow (\lambda -1)(15-12)+2(-10+4)+3(6-3)=0$ $\;\Rightarrow (\lambda -1)=1 \Rightarrow \lambda =2$

Q86JEE Main 2023NAT4MPermutations and Combinations

If all the six digit numbers ${\text{x}}_1 {\text{x}}_2 {\text{x}}_3 {\text{x}}_4 {\text{x}}_5 {\text{x}}_6$ with $0 < {\text{x}}_1 < {\text{x}}_2 < {\text{x}}_3 < {\text{x}}_4 < {\text{x}}_5 < {\text{x}}_6$ are arranged in the increasing order, then the sum of the digits in the $72^{\;\text{th }\;}$ number is _______. [29-Jan-2023 Shift 1]

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📖 Explanation

1 2 $={ }^7 {\text{C}}_4=35$ 1 3 $={ }^6 {\text{C}}_4=15$ 1 4 $={ }^5 {\text{C}}_4=5$ 1 5 $={ }^4 {\text{C}}_4=1$ 2 3 $={ }^6 {\text{C}}_4=15$ 71 words $\;245678 \to 72^{\;\text{th }\;} \;\text{ word }\;$ $\;2+4+5+6+7+8=32$

Q87JEE Main 2023NAT4MLimits, Continuity and Differentiability

Let $f: ℝ \to ℝ$ be a differentiable function that satisfies the relation ${\text{f}}( {\text{x}}+ {\text{y}})= {\text{f}}( {\text{x}})+ {\text{f}}( {\text{y}})-1, ∀ {\text{x}}$ , $y \in ℝ$ . If $f^{'}(0)=2$ , then $|f(-2)|$ is equal to ______ [29-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{f}}( {\text{x}}+ {\text{y}})= {\text{f}}( {\text{x}})+ {\text{f}}( {\text{y}})-1$ $\;f^{'}(x)=\lim _{h \to 0} \;\frac{f(x+h)-f(x)}{h}$ $\;f^{'}(x)=\lim _{h \to 0} \;\frac{f(h)-f(0)}{h}=f^{'}(0)=2$ $\;f^{'}(x)=2 \Rightarrow d y=2 \text{dx}$ $\; {\text{y}}=2 {\text{x}}+ {\text{C}}$ $\; {\text{x}}=0, {\text{y}}=1, {\text{c}}=1$ $\; {\text{y}}=2 {\text{x}}+1$ $\;|f(-2)|=|-4+1|=|-3|=3$

Q88JEE Main 2023NAT4MBinomial Theorem

If the co-efficient of $x^9$ in $(\alpha x^{3+}\;\frac{1}{\beta x}) ^{11}$ and the co-efficient of $x^{-9}$ in $(\alpha x-\;\frac{1}{\beta x^3}) ^{11}$ are equal, then $(\alpha \beta)^2$ is equal to _______. [29-Jan-2023 Shift 1]

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📖 Explanation

Coefficient of ${\text{x}}^9$ in $(\alpha x^{3+}\;\frac{1}{\beta x}) ={ }^{11} C_6 \cdot \;\frac{\alpha^5}{\beta^6}$ $\because$ Both are equal $\; \therefore \;\frac{11}{C_6} \cdot \;\frac{\alpha^5}{\beta^6}=-\;\frac{11}{C_5} \cdot \;\frac{\alpha^6}{\beta^5}$ $\;\Rightarrow \;\frac{1}{\beta}=-\alpha$ $\;\Rightarrow \alpha \beta=-1$ $\;\Rightarrow (\alpha \beta)^2=1$

Q89JEE Main 2023NAT4MBinomial Theorem

Let the coefficients of three consecutive terms in the binomial expansion of $(1+2 x)^{ {\text{n}}}$ be in the ratio $2: 5: 8$ . Then the coefficient of the term, which is in the middle of these three terms, is _______. [29-Jan-2023 Shift 1]

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📖 Explanation

$\;t_{r+1}={ }^n C_r(2 x)^r$ $\;\Rightarrow \;{{ }^n C_{r-1}(2)^{ {\text{r}}-1}}/{{ }^n C_r(2)^{ {\text{r}}}}=\;\frac{2}{5}$ $\Rightarrow \;\frac{\;\frac{n !}{(r-1) !(n-r+1) !}}{\frac{n !(2)}{r !(n-r) !}}=\;\frac{2}{5}$ $\;\Rightarrow \;\frac{r}{n-r+1}=\;\frac{4}{5} \Rightarrow 5 r=4 n-4 r+4$ $\;\Rightarrow 9 r=4(n+1)$ . . . (1) $\;\Rightarrow \;{{ }^n C_r(2)^{ {\text{r}}}}/{{ }^{ {\text{n}}} C_{r+1}(2)^{ {\text{r}}+1}}=\;\frac{5}{8}$ $\;\Rightarrow \;{\;{ {\text{n}} !}/{ {\text{r}} !( {\text{n}}- {\text{r}}) !}}/{\frac{n !}{(r+1) !(n-r-1) !}}=\;\frac{5}{4} \Rightarrow \;\frac{r+1}{n-r}=\;\frac{5}{4}$ $\;\Rightarrow 4 r+4=5 n-5 r \Rightarrow 5 n-4=9 r . . .(2)$ From (1) and (2) $\Rightarrow 4 n+4=5 n-4 \Rightarrow n=8$ (1) $\Rightarrow r=4$ so, coefficient of middle term is ${ }^8 {\text{C}}_4 2^4=16 \times \;\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=16 \times 70=1120$

Q90JEE Main 2023NAT4MPermutations and Combinations

Five digit numbers are formed using the digits 1,2 , $3,5,7$ with repetitions and are written in descending order with serial numbers. For example, the number 77777 has serial number 1. Then the serial number of 35337 is _______. [29-Jan-2023 Shift 1]

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📖 Explanation

No of 5 digit numbers starting with digit 1 $=5 \times 5 \times 5 \times 5=625$ No of 5 digit numbers starting with digit 2 $=5 \times 5 \times 5 \times 5=625$ No of 5 digit numbers starting with 31 $=5 \times 5 \times 5=125$ No of 5 digit numbers starting with 32 $=5 \times 5 \times 5=125$ No of 5 digit numbers starting with 33 $=5 \times 5 \times 5=125$ No of 5 digit numbers starting with 351 $=5 \times 5=25$ No of 5 digit numbers starting with 352 $=5 \times 5=25$ No of 5 digit numbers starting with 3531 $=5$ No of 5 digit numbers starting with 3532 $=5$ Before 35337 will be 4 numbers,So rank of 35337 will be 1690 So, in descending order serial number will be $3125-1690+1=1436$

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