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25 Jan 2023 Shift 1

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Physics30 questions

Q1JEE Main 2023MCQ4MDual Nature of Matter and Radiation

Electron beam used in an electron microscope, when accelerated by a voltage of $20 {\text{kV}}$ . has a de-Broglie wavelength of $\lambda _o$ . If the voltage is increased to $40 {\text{kV}}$ . then the de-Broglie wavelength associated with the electron beam would be: [25-Jan-2023 Shift 1]

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📖 Explanation

When electron is accelerated through potential difference $V$ , then $\;\;\text{ K.E. }\;= {\text{eV}}$ $\;\Rightarrow \lambda =\;{ {\text{h}}}/{\sqrt{2 {\text{m}}( {\text{KE}})}}=\;{ {\text{h}}}/{\sqrt{2 {\text{meV}}}}$ $\; \therefore \;\; \lambda \alpha \;{1}/{\sqrt{ {\text{V}}}}$ $\; \therefore \;\; \;\frac{\lambda }{\lambda _0}=\sqrt{\;\frac{20}{40}}$ $\; \therefore \;\; \lambda =\;{\lambda _0}/{{\sqrt{2}}}$

Q2JEE Main 2023MCQ4MRotational Motion

An object of mass $8 {\text{kg}}$ is hanging from one end of a uniform rod ${\text{CD}}$ of mass $2 {\text{kg}}$ and length $1 {\text{m}}$ pivoted at its end ${\text{C}}$ on a vertical wall as shown in figure. It is supported by a cable ${\text{AB}}$ such that the system is in equilibrium. The tension in the cable is :(Take ${\text{g}}=10 {\text{m}} / {\ s}^2$ ) [25-Jan-2023 Shift 1]

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📖 Explanation

Taking torque about point ${\text{C}}$ $\;\;{ {\text{T}}}/{2} \times 60=20 \times 50+80 \times 100$ $\;\Rightarrow 3 {\text{T}}=100+800$ $\;\Rightarrow =300 {\text{N}}$

Q3JEE Main 2023MCQ4MThermodynamics

A Carnot engine with efficiency $50 \%$ takes heat from a source at $600 {\text{K}}$ . In order to increase the efficiency to $70 \%$ , keeping the temperature of sink same, the new temperature of the source will be : [25-Jan-2023 Shift 1]

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📖 Explanation

Initially $η=\;\frac{1}{2}$ But $\;η=1-\;{ {\text{T}}_2}/{ {\text{T}}_1}$ $\; \therefore \;\frac{1}{2}=1-\;{ {\text{T}}_2}/{600}$ $\;\Rightarrow \;{ {\text{T}}_2}/{600}=\;\frac{1}{2}$ $\;\;\text{ Now efficiency is increased to }\; 70 \% \;\text{ and }\; {\text{T}}_2=300$ $\; {\text{K}}, \;\text{ Let temp of source }\; {\text{T}}_1= {\text{T}}$ $\;\Rightarrow \;\frac{7}{10}=1-\;{300}/{ {\text{T}}}$ $\;\Rightarrow \;{300}/{ {\text{T}}}=1-\;\frac{7}{10}$ $\;\Rightarrow \;{300}/{ {\text{T}}}=\;\frac{3}{10} \;\; \therefore {\text{T}}_2=300 {\text{K}}$

Q4JEE Main 2023MCQ4MGravitation

${\text{T}}$ is the time period of simple pendulum on the earth's surface. Its time period becomes x ${\text{T}}$ when taken to a height ${\text{R}}$ (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be: [25-Jan-2023 Shift 1]

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📖 Explanation

At surface of earth time period ${\text{T}}=2 \pi \sqrt{\;{ℓ}/{ {\text{g}}}}$ At height ${\text{h}}= {\text{R}}$ $\; {\text{g}}^{'}=\;{ {\text{g}}}/{ (1+\;{ {\text{h}}}/{ {\text{R}}}) ^2}=\;{ {\text{g}}}/{4}$ $\; {\text{xT}}=2 \pi \sqrt{\;{ℓ}/{( {\text{g}} / 4)}}$ $\;\Rightarrow {\text{xT}}=2 \times 2 \pi \sqrt{\;{ℓ}/{ {\text{g}}}}$ $\;\Rightarrow {\text{xT}}=2 {\text{T}} \Rightarrow {\text{x}}=2$

Q5JEE Main 2023MCQ4MOscillations

Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is $100 {\text{g}}$ . The time period of the motion of the particle will be (approximately) (take $g=10 {\text{ms}}^{-2}$ ,radius of earth $=6400 {\text{km}}$ ) [25-Jan-2023 Shift 1]

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📖 Explanation

Let at some time particle is at a distance ${\text{x}}$ from centre of Earth, then at that position field ${\text{E}}=\;{ {\text{GM}}}/{ {\text{R}}^3} {\text{x}}$ $\therefore$ Acceleration of particle ${ \vec{\text{a}}=-\;{ {\text{GM}}}/{ {\text{R}}^3} { \vec{\text{x}}$ $\;\Rightarrow \omega =\sqrt{\;{ {\text{GM}}}/{ {\text{R}}^3}}=\sqrt{\;{ {\text{g}}}/{ {\text{R}}}}$ $\;\;\text{ Now }\; {\text{T}}=\;\frac{2 \pi}{\omega }=2 \pi \sqrt{\;{ {\text{R}}}/{ {\text{g}}}}$ $\;\Rightarrow {\text{T}}=2 \times 3.14 \times \sqrt{\;\frac{6400 \times 10^3}{10}}$ $\;=2 \times 3.14 \times 800 {\ sec} \approx 1 \;\text{ hour }\; 24 \;\text{ minutes }\;$

Q6JEE Main 2023MCQ4MMotion in a Straight Line

A car travels a distance of ' $x$ ' with speed $V_1$ and then same distance ' $x$ ' with speed $V_2$ in the same direction. The average speed of the car is: [25-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\text{ Average velocity }\;=\;\frac{\;\text{ Total displacement }\;}{\;\text{ Total time }\;}$ $\;=\;{ {\text{x}}+ {\text{x}}}/{{ {\text{x}}}/{ {\text{v}}_1}+\;{ {\text{x}}}/{ {\text{v}}_2}}=\;{2 {\text{v}}_1 {\text{v}}_2}/{ {\text{v}}_1+ {\text{v}}_2}$

Q7JEE Main 2023MCQ4MElectrostatic Potential and Capacitance

A parallel plate capacitor has plate area $40 {\text{cm}}^2$ and plates separation $2 {\text{mm}}$ . The space between the plates is filled with a dielectric medium of a thickness $1 {\text{mm}}$ and dielectric constant 5 . The capacitance of the system is : [25-Jan-2023 Shift 1]

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📖 Explanation

This can be seen as two capacitors in series combination so $\;\;{1}/{ {\text{C}}_{ {\text{eq}}}}=\;{1}/{ {\text{C}}_1}+\;{1}/{ {\text{C}}_2}$ $\;=\;{1}/{{ {\text{K}} \in _0 {\text{A}}}/{ {\text{t}}}}+\;{1}/{{E_0 {\text{A}}}/{ {\text{d}}- {\text{t}}}}$ $\;=\;{ {\text{t}}}/{ {\text{K}} \in _0 {\text{A}}}+\;{ {\text{d}}- {\text{t}}}/{E_0 {\text{A}}}$ $\;=\;\frac{1 \times 10^{-3}}{5 E_0 \times 40 \times 10^{-4}}+\;\frac{1 \times 10^{-3}}{E_0 40 \times 10^{-4}}$ $\;\;{1}/{ {\text{C}}_{ {\text{eq}}}}=\;\frac{1}{20 E_0}+\;\frac{1}{4 E_0}$ $\; {\text{C}}_{ {\text{eq}}}=\;\frac{20 \times 4 E_0}{24}=\;\frac{10 E_0}{3} {\text{F}}$

Q8JEE Main 2023MCQ4MThermodynamics

The root mean square velocity of molecules of gas is [25-Jan-2023 Shift 1]

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📖 Explanation

The rms speed of a gas molecule is $\; {\text{V}}_{ {\text{RMS}}}=\sqrt{\;{3 {\text{RT}}}/{ {\text{M}}}}$ $\; {\text{V}}_{ {\text{RMS}}} \propto \sqrt{ {\text{T}}}$

Q9JEE Main 2023MCQ4MUnits and Measurements

Match List I with List II List - I List - II A. Surface tension I. $\;\; {\text{Kg}} {\text{m}}^{-1} {\ s}^{-1}$ B. Pressure II. ${\text{Kg}} {\text{ms}}^{-1}$ C. Viscosity III. ${\text{Kg}} {\text{m}}^{-1} {\ s}^{-2}$ D. Impulse IV. ${\text{Kg}} {\ s}^{-2}$ Choose the correct answer from the options given below : [25-Jan-2023 Shift 1]

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📖 Explanation

$\;\text{ (A) Surface Tension }\; =\;\frac{F}{ℓ}\;=\;{ {\text{MLT}}^{-2}}/{ {\text{L}}}= {\text{ML}}^{-1} {\text{T}}^{-2}$ $\;= {\text{Kgs}}^{-2}( {\text{IV}})$ $\;\;\text{ (B) Pressure }\;=\;\frac{F}{A}=\;{ {\text{MLT}}^{-2}}/{ {\text{L}}^2}$ $\;= {\text{kg}} {\text{m}}^{-1} {\ s}^{-2} \;\text{ (III) }\;$ (C) $\;\text{ Viscosity }\;\;==\;{ {\text{F}}}/{ {\text{A}} (\;{ {\text{dV}}}/{ {\text{dz}}}) }=\;{ {\text{MLT}}^{-2}}/{ {\text{L}}^2 (\;{ {\text{LT}}^{-1}}/{ {\text{L}}}) }$ $\;= {\text{ML}}^{-1} {\text{T}}^{-1}= {\text{kg}} {\text{m}}^{-1} {\ s}^{-1}( {\text{I}})$ $\;\text{ (D) }\; \;\text{ Impulse }\;\;=\int {\text{Fdt}}= {\text{MLT}}^{-2} \times {\text{T}}$ $\;= {\text{MLT}}^{-1}= {\text{Kgms}}^{-1} \;\text{ (II) }\;$ So A-(IV), B-(III), C-(I), D- (II)

Q10JEE Main 2023MCQ4MAlternating Current

In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes ${\text{x}}$ times its initial resonant frequency $\omega _0$ . The value of ${\text{x}}$ is: [25-Jan-2023 Shift 1]

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📖 Explanation

The resonance frequency of LC oscillations circuit is $\;\omega _0=\;{1}/{\sqrt{ {\text{LC}}}}$ $\; {\text{L}} \to 2 {\text{L}}$ $\; {\text{C}} \to 8 {\text{C}}$ $\;\omega =\;{1}/{\sqrt{2 {\text{L}} \times 8 {\text{C}}}}=\;{1}/{4 \sqrt{ {\text{LC}}}}$ $\;\omega =\;\frac{\omega _0}{4}$ So $x=\;\frac{1}{4}$

Q11JEE Main 2023MCQ4MAtoms and Nuclei

The ratio of the density of oxygen nucleus $({ }_8^{16} {\text{O}})$ and helium nucleus $({ }_2^4 {\text{He}})$ is [25-Jan-2023 Shift 1]

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📖 Explanation

Nuclear density is independent of mass numberAs nuclear density $=\;\frac{A u}{\frac{4}{3} \pi R^3}$ Also, ${\text{R}}= {\text{R}}_0 {\text{A}}^{\;\frac{1}{3}}$ And ${\text{R}}^3= {\text{R}}_0^3 {\text{A}}$ $\Rightarrow$ Nuclear density $=\;{ {\text{Au}}}/{\frac{4}{3} \pi {\text{R}}_0^3 {\text{A}}}$ Nuclear density $=\;{3 {\text{u}}}/{4 \pi {\text{R}}_0^3}$ $\Rightarrow$ Nuclear density is independent of A

Q12JEE Main 2023MCQ4MSemiconductor Electronics

A message signal of frequency $5 {\text{kHz}}$ is used to modulate a carrier signal of frequency 2 ${\text{MHz}}$ . The bandwidth for amplitude modulation is: [25-Jan-2023 Shift 1]

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📖 Explanation

GivenSignal frequency $f_m=5 {\text{kHz}}$ Carrier wave frequency $f_c=2 {\text{MHz}}$ ${\text{f}}_{ {\text{c}}}=2000 {\text{KHz}}$ The resultant signal will have band width of frequency given by $\;{ [ ( {\text{f}}_{ {\text{c}}}+ {\text{f}}_{ {\text{m}}}) - ( {\text{f}}_{ {\text{c}}}- {\text{f}}_{ {\text{m}}}) ] }$ $\;\Rightarrow [(2000+5)-(2000-5)] {\text{kHz}}$ $\;\Rightarrow 10 {\text{kHz}}$

Q13JEE Main 2023MCQ4MElectromagnetic Waves

All electromagnetic wave is transporting energy in the negative ${\text{z}}$ direction. At a certain point and certain time the direction of electric field of the wave is along positive ${\text{y}}$ direction. What will be the direction of the magnetic field of the wave at that point and instant? [25-Jan-2023 Shift 1]

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📖 Explanation

As , poynting vector ${ \vec{\text{S}}={ \vec{\text{E}} \times { \vec{\text{H}}$ Given energy transport $=$ negative ${\text{z}}$ directionElectric field = positive $y$ direction $(-{ {\text{k}}^{\^})=(+{ {\text{j}}^{\^}) \times [{ {\text{i}}^{\^}]$ Hence according to vector cross product magnetic field should be positive ${\text{x}}$ direction.

Q14JEE Main 2023MCQ4MWave Optics

In Young's double slits experiment, the position of $5^{\;\text{th }\;}$ bright fringe from the central maximum is $5 {\text{cm}}$ . The distance between slits and screen is $1 {\text{m}}$ and wavelength of used monochromatic light is $600 {\text{nm}}$ . The separation between the slits is: [25-Jan-2023 Shift 1]

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📖 Explanation

Given ${\text{D}}\;=1 {\text{m}}$ $\lambda \;=600 \times 10^{-9} {\text{m}}$ ${\text{n}}\;=5$ $\;\;\text{ As }\; {\text{y}}_{ {\text{nth}}}=\;{ {\text{n}} \lambda {\text{D}}}/{ {\text{d}}}$ $\;\Rightarrow \;{5 \times 600 \times 10^{-9} \times 1}/{ {\text{d}}}=5 \times 10^{-2}$ $\;\Rightarrow {\text{d}}=\;\frac{5 \times 600 \times 10^{-9} \times 1}{5 \times 10^{-2}}=60 \times 10^{-6} {\text{m}}$ $\;\Rightarrow {\text{d}}=60 \mu {\text{m}}$

Q15JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

Math List I with List IIChoose the correct answer from the option given below: [25-Jan-2023 Shift 1]

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📖 Explanation

(A) $\; {\text{B}}_{ {\text{ab}}}=\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}} \;\text{ (out of the plane) }\;$ $\; {\text{B}}_{ {\text{bcd}}}=\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}}(2 \pi) \;\text{ (\in the plane) }\;$ $\; {\text{B}}_{ {\text{de}}}=\;\frac{\mu _0}{4pi} \;{ {\text{I}}}/{ {\text{r}}} \;\text{ (out of the plane) }\;$ Hence magnetic field at ${\text{O}}$ is $\; {\text{B}}_0=-\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}}+\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}}(2 \pi)-\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}}$ $\; {\text{B}}_0=\;\frac{\mu _0}{2 \pi} \;{ {\text{I}}}/{{\text{r}}}(\pi-1) . . . . . . . . \;\text{ (III) }\;$ (B) $\; {\text{B}}_{ {\text{ab}}}=\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}} \;\text{ (out of the plane) }\;$ $\; {\text{B}}_{ {\text{bcd}}}=\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}}(\pi) \;\text{ (out of the plane) }\;$ $\; {\text{B}}_{ {\text{de}}}=\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}} \;\text{ (out of theplane) }\;$ Hence magnetic field at ${\text{O}}$ is $\; {\text{B}}_0=\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}}+\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}}(\pi)+\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}}$ $\; {\text{B}}_0=\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}}(\pi+2) . . . \;\text{. (I) }\;$ (C) ${\text{B}}_{ {\text{ab}}}=\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}}$ (in the plane) ${\text{B}}_{ {\text{bcd}}}=\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}}(\pi)$ (in the plane) ${\text{B}}_{ {\text{de}}}=0$ (at the axis)Hence magnetic field at ${\text{O}}$ is ${\text{B}}_0=\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}}(1+\pi) . . .$ (D) ${\text{B}}_{ {\text{ab}}}=0$ (at the axis) ${\text{B}}_{ {\text{bcd}}}=\;\frac{\mu _0}{4 \pi} \;{ {\text{I}}}/{ {\text{r}}}(\pi)$ (out of the plane) ${\text{B}}_{ {\text{de}}}=0$ (at the axis)Hence magnetic field at ${\text{O}}$ is ${\text{B}}_0=\;{\mu _0 {\text{I}}}/{4 {\text{r}}} . . .$

Q16JEE Main 2023MCQ4MSemiconductor Electronics

Given below are two statements : one is labeled as Assertion A and the other is labeled as Reason ${\text{R}}$ Assertion A: Photodiodes are used in forward bias usually for measuring the light intensity.Reason R: For a p-n junction diode, at applied voltage ${\text{V}}$ the current in the forward bias is more than the current in the reverse bias for $| {\text{V}}_{ {\text{z}}}{{}/{}|} > \pm {\text{V}} \ge | {\text{V}}_0{{}/{}|}$ where ${\text{V}}_{ {\text{o}}}$ is the threshold voltage and ${\text{V}}_{ {\text{z}}}$ is the breakdown voltage.In the light of the above statements, choose the correct answer from the options given below [25-Jan-2023 Shift 1]

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📖 Explanation

Theory basedPhotodiodes are operated in reverse bias condition.For P-N junction current in forward bias (for ${\text{V}} \ge {\text{V}}_0$ ) is always greater than current in reverse bias $(.$ for ${\text{V}} \le {\text{V}}_{ {\text{Z}}}$ ) .Hence Assertion if false but Reason is true

Q17JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube $2 {\text{m}}$ long and $0.2 {\text{m}}$ in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is: [25-Jan-2023 Shift 1]

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📖 Explanation

Magnetic field at centre inside the solenoid is given by ${\text{B}}=\mu _0 {\text{nI}}$ So magnetic intensity at centre $\; {\text{H}}=\;{ {\text{B}}}/{\mu _0}= {\text{nI}}= (\;\frac{1200}{2}) (2)$ $\; {\text{H}}=1.2 \times 10^3 {\text{Am}}^{-1}$

Q18JEE Main 2023MCQ4MCurrent Electricity

A uniform metallic wire carries a current $2 {\text{A}}$ . when $3.4 {\text{V}}$ battery is connected across it. The mass of uniform metallic wire is $8.92 \times 10^{-3} {\text{kg}}$ . density is $8.92 \times 10^3 {\text{kg}} / {\text{m}}^3$ and resistivity is $1.7 \times 10^{-8} Ω- {\text{m}}$ . The length of wire is : [25-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{I}}=2 {\text{A}}$ $\;\Delta {\text{V}}=3.4 {\text{V}}$ Using Ohm's Law $\; {\text{R}}=\;\frac{3.4}{2}=1.7 Ω$ $\;1.7=\;{\rho {\text{L}}}/{ {\text{A}}}$ $\; {\text{L}}=\;{1.7( {\text{A}})}/{\rho }$ $\; {\text{M}}=(\;\text{ density volume }\;)$ $\;\;\text{ Volume }\;=\;\frac{8.92 \times 10^{-3}}{8.92 \times 10^3}=10^{-6}$ $\; {\text{L}}^2=\;\frac{1.7}{\rho } (10^{-6}) =\;\frac{1.7}{1.7} \times 10^2$ $\; {\text{L}}=10 {\text{m}}$

Q19JEE Main 2023MCQ4MThermodynamics

A bowl filled with very hot soup cools from $98^{\circ} {\text{C}}$ to $86^{\circ} {\text{C}}$ in 2 minutes when the room temperature is $22^{\circ} {\text{C}}$ . How long it will take to cool from $75^{\circ} {\text{C}}$ to $69^{\circ} {\text{C}}$ ? [25-Jan-2023 Shift 1]

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📖 Explanation

$\;\;{\Delta {\text{Q}}}/{\Delta {\text{t}}}=- {\text{K}} ( {\text{T}}- {\text{T}}_0)$ $\;\;{\Delta {\text{Q}}}/{\Delta {\text{t}}}=- {\text{K}} ( {\text{T}}_{ {\text{avg}}}- {\text{T}}_0)$ (i) $\;{ {\text{ms}} \times 12}/{2}=- {\text{K}} (\;\frac{98+86}{2}-22)$ $\;6=-\;{ {\text{K}}}/{ {\text{ms}}} [\;\frac{98+86}{2}-22]$ $\;6=-\;{ {\text{K}}}/{ {\text{ms}}}[70] . . . . . . . . . . \;\text{ (i) }\;$ (ii) $\;{ {\text{ms}} \times 6}/{\Delta {\text{t}}}=- {\text{K}} (\;\frac{75+69}{2}-22)$ $\;{6}/{\Delta {\text{t}}}=-\;{ {\text{K}}}/{ {\text{ms}}}(50)$ ............. (II)(ii) $\div$ (i) $\;\;{6}/{\Delta {\text{t}}(6)}=\;\frac{50}{70}$ $\;\Delta {\text{t}}=\;\frac{7}{5}=1.4 {\min}$

Q20JEE Main 2023MCQ4MCircular Motion

A car is moving with a constant speed of 20 ${\text{m}} / {\ s}$ in a circular horizontal track of radius 40 ${\text{m}}$ . A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be : (Take $g=10$ ${\text{m}} / {\ s}^2)$ [25-Jan-2023 Shift 1]

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📖 Explanation

${\text{T}} \cos \theta= {\text{mg}}$ ${\text{T}} sin \;\theta=\;{ {\ mv}^2}/{ {\text{R}}}$ $\tan \theta=\;{ {\text{v}}^2}/{ {\text{Rg}}}$ $\;\tan \theta=\;\frac{20^2}{40 \times 10}$ $\;\tan \theta=1$ $\;\Rightarrow \theta=\;\frac{\pi}{4}$

Q21JEE Main 2023NAT4MRay Optics and Optical Instruments

A ray of light is incident from air on a glass plate having thickness ${\sqrt{3}} {\text{cm}}$ and refractive index ${\sqrt{2}}$ . The angle of incidence of a ray is equal to the critical angle for glass-air interface. The lateral displacement of the ray when it passes through the plate is $\times 10^{-}$ ${ }^2 {\text{cm}}$ . (given $sin \;15^{\circ}=0.26$ ) [25-Jan-2023 Shift 1]

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📖 Explanation

$\; sin \; {\text{c}}=\;{1}/{{\sqrt{2}}}$ $\; {\text{c}}=45^{\circ}$ $\; sin \; {\text{c}}=\mu sin \;\theta$ $\;\;{1}/{{\sqrt{2}}}={\sqrt{2}} sin \;\theta$ $\;\theta=30^{\circ}$ Lateral displacement: $\;x=t sin \;(i-r) \ sec r$ $\;x={\sqrt{3}} sin \; (45^{\circ}-30^{\circ}) \ sec 30^{\circ}$ $\;x={\sqrt{3}}(0.26) (\;{2}/{{\sqrt{3}}})$ $\;X=0.52 {\text{cm}}$ $\;x=52 \times 10^{-2} {\text{cm}}$

Q22JEE Main 2023NAT4MCurrent Electricity

In the given circuit, the equivalent resistance between the terminal ${\text{A}}$ and ${\text{B}}$ is_______ $Ω$ . [25-Jan-2023 Shift 1]

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📖 Explanation

Both $4 Ω$ resistance gets short.Remove the resistors that have no current. $\; {\text{R}}_{ {\text{eq}}}=3+(2 \| 2)+6$ $\; {\text{R}}_{ {\text{eq}}}=3+1+6$ $\; {\text{R}}_{ {\text{eq}}}=10 Ω$

Q23JEE Main 2023NAT4MProperties of Matter

As shown in the figure, in an experiment to determine Young's modulus of a wire, the extension-load curve is plotted. The curve is a straight line passing through the origin and makes an angle of $45^{\circ}$ with the load axis. The length of wire is $62.8 {\text{cm}}$ and its diameter is 4 ${\text{mm}}$ . The Young's modulus is found to be ${\text{x}} \times$ $10^4 {\text{Nm}}^{-2}$ .The value of $x$ is _______. [25-Jan-2023 Shift 1]

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📖 Explanation

From graph: $\; {\text{F}}=\Delta {\text{L}}$ $\; {\text{Y}}=\;{ {\text{FL}}}/{ {\text{A}} \Delta {\text{L}}}$ $\; {\text{Y}}=\;{ {\text{L}}}/{ {\text{A}}}$ $\; {\text{Y}}=\;\frac{62.8 \times 10^{-2}}{\pi (2 \times 10^{-3}) ^2}$ $\;Y=5 \times 10^4 {\text{N}} / {\text{m}}^2$

Q24JEE Main 2023NAT4MWork, Power and Energy

An object of mass ' $m$ ' initially at rest on a smooth horizontal plane starts moving under the action of force ${\text{F}}=2 {\text{N}}$ . In the process of its linear motion, the angle $\theta$ (as shown in figure) between the direction of force and horizontal varies as $\theta= {\text{kx}}$ , where ${\text{k}}$ is a constant and ${\text{x}}$ is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be ${\text{E}}=\;{ {\text{n}}}/{ {\text{k}}} sin \;\theta$ . The value of ${\text{n}}$ is _______. [25-Jan-2023 Shift 1]

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📖 Explanation

${\text{F}} \cos \theta= {\text{ma}}$ $\;2 \cos ( {\text{kx}})=\;{ {\ mvdv}}/{ {\text{dx}}}$ $\;int_{0}^{ {\text{v}}} {\text{vdv}}=2 int_{0}^{ {\text{x}}} \cos ( {\text{kx}}) {\text{dx}}$ $\;\;\frac{ {\ mv}^2}{2}=\;{2}/{ {\text{k}}} sin \; {\text{kx}}$ $\;\;\text{ K.E. }\;=\;{2}/{ {\text{k}}} sin \;\theta$ $\;n=2$

Q25JEE Main 2023NAT4MAtoms and Nuclei

The wavelength of the radiation emitted is $\lambda _0$ when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be $\;\frac{20}{x} \lambda _0$ . The value of $x$ is _______. [25-Jan-2023 Shift 1]

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📖 Explanation

___________n=4 ___________n=3 ___________ n=2 ___________n=1 Second excited state $\to$ first excited state $\; {\text{n}}=3 \to {\text{n}}=2$ $\;\;{ {\text{hc}}}/{\lambda _0}=13.6 (\;\frac{1}{2^2}-\;\frac{1}{3^2}) . . . (\text{i})$ Third excited state $\to$ second orbit ${\text{n}}=4 \to {\text{n}}=2$ $\;\frac{h c}{ (20 \lambda _0 / x) }=13.6 (\;\frac{1}{2^2}-\;\frac{1}{4^2}) \;\text{. }\;$ . . . (ii)(ii) $\div$ (i) $\;\;\frac{x}{20}=\;\frac{\;\frac{1}{2^2}-\frac{1}{4^2}}{\frac{1}{2^2}-\;\frac{1}{3^2}}$ $\;x=27$

Q26JEE Main 2023NAT4MRotational Motion

${\text{I}}_{ {\text{CM}}}$ is moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. ${\text{I}}_{ {\text{AB}}}$ is it's moment of inertia about an axis ${\text{AB}}$ perpendicular to plane and parallel to axis ${\text{CM}}$ at a distance $\;\frac{2}{3} {\text{R}}$ from center. Where ${\text{R}}$ is the radius of the disc. The ratio of ${\text{I}}_{ {\text{AB}}}$ and ${\text{I}}_{ {\text{CM}}}$ is $x: 9$ . The value of $x$ is ________. [25-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{I}}_{ {\text{cm}}}=\;{ {\text{mR}}^2}/{2}$ $\; {\text{I}}_{ {\text{AB}}}=\;{ {\text{mR}}^2}/{2}+ {\text{m}} (\;{2 {\text{R}}}/{3}) ^2=\;\frac{17}{18} {\text{mR}}^2$ $\;\;{ {\text{I}}_{ {\text{AB}}}}/{ {\text{I}}_{ {\text{cm}}}}=\;\frac{17}{9} \Rightarrow {\text{x}}=17$

Q27JEE Main 2023NAT4MWaves

The distance between two consecutive points with phase difference of $60^{\circ}$ in a wave of frequency $500 {\text{Hz}}$ is $6.0 {\text{m}}$ . The velocity with which wave is traveling is ________ ${\text{km}} / {\ s}$ [25-Jan-2023 Shift 1]

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📖 Explanation

$\;\Delta \varphi =\;\frac{2 \pi}{\lambda } \Delta {\text{x}}$ $\;\;\frac{\pi}{3}=\;\frac{2 \pi}{\lambda }(6 {\text{m}})$ $\Rightarrow \;\; \lambda =36 {\text{m}}$ $\; {\text{V}}= {\text{f}} \lambda =(500 {\text{Hz}})(36 {\text{m}})$ $=\;18000 {\text{m}} / {\ s}=18 {\text{km}} / {\ s}$

Q28JEE Main 2023NAT4MElectrostatics

A uniform electric field of $10 {\text{N}} / {\text{C}}$ is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy $0.5 {\text{eV}}$ . The length of each plate is $10 {\text{cm}}$ . The angle $(\theta)$ of deviation of the path of electron as it comes out of the field is ______ (in degree). [25-Jan-2023 Shift 1]

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📖 Explanation

$\;0.5 {\text{e}}=\;\frac{1}{2} {\ mv}_{ {\text{x}}}^2 \Rightarrow {\text{v}}_{ {\text{x}}}=\sqrt{\;{ {\text{e}}}/{ {\text{m}}}}$ $\;\;\text{ Along }\; x {\text{L}}= {\text{v}}_{ {\text{x}}} {\text{t}}=\sqrt{\;{ {\text{e}}}/{ {\text{m}}}} {\text{t}}$ $\;\;\text{ Along y }\; {\text{v}}_{ {\text{y}}}=\;{ {\text{eE}}}/{ {\text{m}}} {\text{t}}$ $\;{\text{dividing}} \;{ {\text{v}}_{ {\text{y}}}}/{ {\text{L}}}= {\text{E}} \sqrt{\;{ {\text{e}}}/{ {\text{m}}}}= {\text{Ev}}_{ {\text{x}}}$ $\;\Rightarrow {\text{Tan}} \theta=\;{ {\text{v}}_{ {\text{y}}}}/{ {\text{v}}_{ {\text{x}}}}= {\text{E}} \times {\text{L}}=10 \times 0.1=1$ $\;\theta=45^{\circ}$ $\;$

Q29JEE Main 2023NAT4MAlternating Current

An LCR series circuit of capacitance $62.5 {\text{nF}}$ and resistance of $50 Ω$ . is connected to an A.C. source of frequency $2.0 {\text{kHz}}$ . For maximum value of amplitude of current in circuit, the value of inductance is ________ ${\text{mH}}$ .(take $\pi^2=10$ ) [25-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{f}}=\;{1}/{2 \pi \sqrt{ {\text{LC}}}}$ $\;2000 {\text{Hz}}=\;{1}/{2 \pi \sqrt{ {\text{L}} \times 62.5 \times 10^{-9}}}$ $\; {\text{L}}=\;\frac{1}{4 \pi^2 \times 2000^2 \times 62.5 \times 10^{-9}}=0.1 {\text{H}}=100 {\text{mH}}$

Q30JEE Main 2023NAT4MRotational Motion

If ${ \vec{\text{P}}=3 { {\text{i}}^{\^}+{\sqrt{3}} { {\text{j}}^{\^}+2 { {\text{k}}^{\^}$ and ${ \vec{\text{Q}}=4 { {\text{i}}^{\^}+{\sqrt{3}} { {\text{j}}^{\^}+2.5 { {\text{k}}^{\^}$ then, The unit vector in the direction of ${ \vec{\text{P}} \times { \vec{\text{Q}}$ is $\;\frac{1}{x}({\sqrt{3}} \hat{i}+\hat{j}-2 {\sqrt{3}} \hat{k})$ . The value of $x$ is [25-Jan-2023 Shift 1]

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Chemistry30 questions

Q31JEE Main 2023MCQ4MHaloalkanes and Haloarenes

The compound which will have the lowest rate towards nucleophilic aromatic substitution on treatment with ${\text{OH}}^{-}$ is [25-Jan-2023 Shift 1]

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Q32JEE Main 2023MCQ4MBiomolecules

The variation of the rate of an enzyme catalyzed reaction with substrate concentration is correctly represented by graph (A)(B)(C)(D) [25-Jan-2023 Shift 1]

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Q33JEE Main 2023MCQ4MAmines

Identify the product formed (A and E) [25-Jan-2023 Shift 1]

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Q34JEE Main 2023MCQ4MSalt Analysis

Choose the correct answer from the options given below: List I List II Elements Colour imparted to the flame A. K I. Brick Red B. Ca II. Violet C. Sr III. Apple Green D. Ba IV. Crimson Red Choose the correct answer from the options given below: [25-Jan-2023 Shift 1]

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Q35JEE Main 2023MCQ4Mp Block Elements

Reaction of thionyl chloride with white phosphorus forms a compound $[ {\text{A}}]$ , which on hydrolysis gives [B], a dibasic acid. [A] and [B] are respectively [25-Jan-2023 Shift 1]

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📖 Explanation

${\text{P}}_4+8 {\text{SOCl}}_2 \to \;{{4 {\text{PCl}}_3}}_{{[ {\text{A}}]}}+4 {\text{SO}}_2+2 {\text{S}}_2 {\text{Cl}}_2$ ${\text{PCl}}_3+3 {\text{H}}_2 {\text{O}} \to { {\text{H}}_3 {\text{PO}}_3}_{[\text{B}]}+3 {\text{HCl}}$

Q36JEE Main 2023MCQ4MSolid State

A cubic solid is made up of two elements ${\text{X}}$ and ${\text{Y}}$ . Atoms of ${\text{X}}$ are present on every alternate corner and one at the center of cube. $Y$ is at $\;\frac{1}{3}^{\;\text{rd }\;}$ of the total faces. The empirical formula of the compound is [25-Jan-2023 Shift 1]

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📖 Explanation

$\;X_{4 \times \;\frac{1}{8}+1 \times 1} Y_{6 \times \;\frac{1}{3} \times \;\frac{1}{2}}$ $\;\Rightarrow \;\; {\text{X}}_{\;\frac{1}{2}+1} \;\; {\text{Y}}_1$ $\;\Rightarrow \;\; {\text{X}}_{\;\frac{2}{3}} \;\; {\text{Y}}_1$ $\;\Rightarrow \;\; {\text{X}}_{1.5} \;\; {\text{Y}}_1$ $\;\Rightarrow {\text{X}}_3 \;\; {\text{Y}}_2$ $\;$

Q37JEE Main 2023MCQ4MStructure of Atom

The radius of the $2^{\;\text{nd }\;}$ orbit of ${\text{Li}}^{2+}$ is ${\text{x}}$ . The expected radius of the $3^{\;\text{rd }\;}$ orbit of ${\text{Be}}^{3+}$ is [25-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{Li}}^{2+}$ $\; {\text{r}}_2= {\text{x}}= {\text{k}} \times \;\frac{2^2}{3}=\;{4 {\text{k}}}/{2}$ $\;\;{ {\text{y}}}/{ {\text{x}}}=\;\frac{9}{4} \times \;\frac{3}{4}=\;\frac{27}{16}$ $\; {\text{y}}=\;\frac{27}{16} {\text{x}}$ $\; {\text{Be}}^{3+}$ $\; {\text{r}}_3= {\text{y}}= {\text{k}} \times \;\frac{3^2}{4}$

Q38JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

Which of the following conformations will be the most stable? [25-Jan-2023 Shift 1]

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Q39JEE Main 2023MCQ4MBiomolecules

Match items of Row I with those of Row II.Row I :Row II : (i) $\alpha$ -D-(-) Fructofuranose. (ii) $\beta$ -D-(-) Fructofuranose (iii) $\alpha$ -D-(-) Glucopyranose. (iv) $\beta$ -D-(-) Glucopyranose Correct match is [25-Jan-2023 Shift 1]

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📖 Explanation

Represents $\alpha$ -D-(+) GlucopyranoseRepresents $\beta$ -D- $(+)$ GlucopyranoseRepresents $\beta$ -D-(-) FructofuranoseRepresents $\beta$ -D- $(-)$ Fructofuranose (from the given options best answer is D)

Q40JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason $R$ :Assertion A : Acetal/Ketal is stable in basic medium.Reason $R$ : The high leaving tendency of alkoxide ion gives the stability to acetal/ketal in basic medium.In the light of the above statements, choose the correct answer from the options given below: [25-Jan-2023 Shift 1]

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📖 Explanation

For Assertion :Acetal and ketals are basically ethers hence they must be stable in basic medium but should break down in acidic medium.Hence assertion is correct.For reason: Alkoxide ion $( {\text{RO}}^{-})$ is not considered a good leaving group hence reason must be false.

Q41JEE Main 2023MCQ4Mp Block Elements

Inert gases have positive electron gain enthalpy. Its correct order is [25-Jan-2023 Shift 1]

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📖 Explanation

Element $\Delta {\text{egH}}[ {\text{KJ}} / {\text{mol}}]$ He +48 Ne +116 Kr +96 Xe +77From NCERTSo, order is ${\text{Ne}} > {\text{Kr}} > {\text{Xe}} > {\text{He}}$

Q42JEE Main 2023MCQ4Md and f Block Elements

Which one of the following reactions does not occur during extraction of copper ? [25-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{CuFeS}}_2+ {\text{O}}_2 \;{\to }^{\;\text{ Partial roasting }\;}$ $\; {\text{Cu}}_2 {\text{S}}+ {\text{FeO}}+ {\text{SO}}_2+\;{ {\text{FeS}}}_{\;\text{ very small }\;}+\;{ {\text{Cu}}_2 {\text{O}}_{\;\text{ very small }\;}}$ $\; {\text{Cu}}_2 {\text{S}}+ {\text{O}}_2 \to {\text{Cu}}_2 {\text{O}}+ {\text{SO}}_2$ $\; {\text{FeS}}+ {\text{O}}_2 \to {\text{FeO}}+ {\text{SO}}_2$ $\; {\text{FeO}}+ {\text{SiO}}_2 \to {\text{FeSiO}}_3$ No formation of calcium silicate $( {\text{CaSiO}}_3)$ in extraction of ${\text{Cu}}$ .

Q43JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

The correct sequence of reagents for the preparation of ${\text{Q}}$ and ${\text{R}}$ is : [25-Jan-2023 Shift 1]

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Q44JEE Main 2023MCQ4MAmines

The correct order in aqueous medium of basic strength in case of methyl substituted amines is : [25-Jan-2023 Shift 1]

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📖 Explanation

In aqueous medium basic strength is dependent on electron density on nitrogen as well as solvation of cation formed after accepting ${\text{H}}^{+}$ . After considering all these factors overall basic strength order is ${\text{Me}}_2 {\text{NH}} > {\text{MeNH}}_2 > {\text{Me}}_3 {\text{N}} > {\text{NH}}_3$

Q45JEE Main 2023MCQ4MSolutions

' 25 volume' hydrogen peroxide means [25-Jan-2023 Shift 1]

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📖 Explanation

Volume $=11.35 \times {\text{M}}$ Strength $\; {\text{M}}=\;\frac{25}{11.35} {\text{M}}$ $\; {\text{g}} / {\text{L}}=25 \times 34 / 11.35$ $\;=74.889$

Q46JEE Main 2023MCQ4MBiomolecules

Which of the following statements is incorrect for antibiotics? [25-Jan-2023 Shift 1]

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Q47JEE Main 2023MCQ4Mp Block Elements

Compound ${\text{A}}$ reacts with ${\text{NH}}_4 {\text{Cl}}$ and forms a compound B. Compound B reacts with ${\text{H}}_2 {\text{O}}$ and excess of ${\text{CO}}_2$ to form compound ${\text{C}}$ which on passing through or reaction with saturated ${\text{NaCl}}$ solution forms sodium hydrogen carbonate. Compound A. B and C, are respectively. [25-Jan-2023 Shift 1]

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📖 Explanation

${{\text{Ca}}( {\text{OH}})_2}_{(\text{A})} +2 {\text{NH}}_4 {\text{Cl}}\;{\longrightarrow }{2 {\text{NH}}_3}_{(\text{B})}+ {\text{CaCl}}_2+2 {\text{H}}_2 {\text{O}}$ $\;{ {\text{NH}}_3}_{\;\text{ (B) }\;}+ {\text{H}}_2 {\text{O}}+\;{ {\text{CO}}_2}_{\;\text{ (exc) }\;} \longrightarrow \;{ {\text{NH}}_4}_{\;\text{ (C) }\;} {\text{HCO}}_3$ ${ {\text{NH}}_4 {\text{HCO}}_3}_{(\text{C})}+ {\text{NaCl}} \longrightarrow {\text{NaHCO}}_3 ↓+ {\text{NH}}_4 {\text{Cl}}$

Q48JEE Main 2023MCQ4Mp Block Elements

Some reactions of ${\text{NO}}_2$ relevant to photochemical smog formation areIdentify ${\text{A}}, {\text{B}}, {\text{X}}$ and ${\text{Y}}$ [25-Jan-2023 Shift 1]

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Q49JEE Main 2023MCQ4MSalt Analysis

Match the List-I with List-II : Cations Group reaction ${\text{P}} \to {\text{Pb}}^{2+}, {\text{Cu}}^{2+}$ (i) ${\text{H}}_2 {\text{S}}$ gas in presence of dilute ${\text{HCl}}$ ${\text{Q}} \to {\text{Al}}^{3+}, {\text{Fe}}^{3+}$ (ii) $( {\text{NH}}_4) _2 {\text{CO}}_3$ in presence of ${\text{NH}}_4 {\text{OH}}$ ${\text{R}} \to {\text{Co}}^{2+}, {\text{Ni}}^{2+}$ (iii) ${\text{NH}}_4 {\text{OH}}$ in presence of ${\text{NH}}_4 {\text{CI}}$ ${\text{S}} \to {\text{Ba}}^{2+}, {\text{Ca}}^{2+}$ (iv) ${\text{H}}_2 {\text{S}}$ in presence of ${\text{NH}}_4 {\text{OH}}$ [25-Jan-2023 Shift 1]

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📖 Explanation

Cations Group No. Group reagent ${\text{Pb}}^{+2}, {\text{Cu}}^{+2}$ II ${\text{H}}_2 {\text{S}}( {\text{g}})$ in presence of dilHCl ${\text{Al}}^{+3}, {\text{Fe}}^{+3}$ III ${\text{NH}}_4 {\text{OH}}$ in presence of ${\text{NH}}_4 {\text{Cl}}$ ${\text{CO}}^{+2}, {\text{Ni}}^{+2}$ IV ${\text{H}}_2 {\text{S}}$ in presence of ${\text{NH}}_4 {\text{OH}}$ ${\text{Ba}}^{+2}, {\text{Ca}}^{+2}$ V $( {\text{NH}}_4) _2 {\text{CO}}_3$ in presence of ${\text{NH}}_4 {\text{OH}}$

Q50JEE Main 2023MCQ4MAlcohols, Phenols and Ethers

In the cumene to phenol preparation in presence of air, the intermediate is [25-Jan-2023 Shift 1]

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Q51JEE Main 2023NAT4MChemical Thermodynamics

An athlete is given $100 {\text{g}}$ of glucose $( {\text{C}}_6 {\text{H}}_{12} {\text{O}}_6)$ for energy. This is equivalent to $1800 {\text{kJ}}$ of energy. The $50 \%$ of this energy gained is utilized by the athlete for sports activities at the event. In order to avoid storage of energy, the weight of extra water he would need to perspire is _______ ${\text{g}}$ (Nearest integer)Assume that there is no other way of consuming stored energy.Given : The enthalpy of evaporation of water is 45 ${\text{kJ}} {\text{mol}}^{-1}$ Molar mass of ${\text{C}}, {\text{H}} & {\text{O}}$ are $12.1$ and $16 {\text{g}} {\text{mol}}^{-1}$ . [25-Jan-2023 Shift 1]

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📖 Explanation

${\text{C}}_6 {\text{H}}_{12} {\text{O}}_6( {\ s})+6 {\text{O}}_2 \to 6 {\text{CO}}_2( {\text{g}})+6 {\text{H}}_2 {\text{O}}( {\text{l}})$ Extra energy used to convert ${\text{H}}_2 {\text{O}}( {\text{l}})$ into ${\text{H}}_2 {\text{O}}( {\text{l}})$ into ${\text{H}}_2 {\text{O}}( {\text{g}})$ $=\;\frac{1800}{2}=900 {\text{kJ}}$ $\;\Rightarrow \;\; 900= {\text{n}}_{ {\text{H}}_2 {\text{O}}} \times 45$ $\; {\text{n}}_{ {\text{H}}_2 {\text{O}}}=\;\frac{900}{45}=20 {\text{mole}}$ $\; {\text{W}}_{ {\text{H}}_2 {\text{O}}}=20 \times 18=360 {\text{g}}$

Q52JEE Main 2023NAT4MChemical Equilibrium

A litre of buffer solution contains $0.1$ mole of each of ${\text{NH}}_3$ and ${\text{NH}}_4 {\text{Cl}}$ . On the addition of $0.02$ mole of ${\text{HCl}}$ by dissolving gaseous ${\text{HCl}}$ , the ${\text{pH}}$ of the solution is found to be ________ $\times 10^{-3}($ Nearest integer) $[.$ Given : ${\text{pK}}_{ {\text{b}}} ( {\text{NH}}_3) =4.745$ $\log 2=0.301$ $\log 3=0.477$ ${\text{T}}=298 {\text{K}}]$ [25-Jan-2023 Shift 1]

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📖 Explanation

In resultant solution ${\text{n}}_{ {\text{NH}}_3}=\;0.1-0.02=0.08$ ${\text{n}}_{ {\text{NH}}_4 {\text{Cl}}}\;= {\text{n}}_{ {\text{NH}}_4^{+}}=0.1+0.02=0.12$ ${\text{pOH}}=\; {\text{pK}} {\text{b}}_{ {\text{b}}}+\log \;{[ {\text{NH}}_4^{+}]\$ }/{ [ {\text{NH}}_3] };=4.745+\log ;\frac{0.12}{0.08};=4.745+\log ;\frac{3}{2};=4.745+0.477-0.301; {\text{pOH}}=4.921; {\text{pH}}=14- {\text{pH}};=9.079$

Q53JEE Main 2023NAT4MSolutions

The osmotic pressure of solutions of PVC in cyclohexanone at $300 {\text{K}}$ are plotted on the graph.The molar mass of PVC is _________ ${\text{g}}\; {\text{mol}}^{-1}$ (Nearest integer)(Given : ${\text{R}}=0.083 {\text{L}} {\text{atm}} {\text{K}}^{-1} {\text{mol}}^{-1}$ ) [25-Jan-2023 Shift 1]

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📖 Explanation

$\;\pi= {\text{M}}^{'} {\text{RT}}= (\;{ {\text{W}} / {\text{M}}}/{ {\text{V}}}) {\text{RT}}$ $\;\Rightarrow \;\; \pi= (\;{ {\text{W}}}/{ {\text{V}}}) (\;{1}/{ {\text{M}}}) {\text{RT}}= {\text{C}} (\;{ {\text{RT}}}/{ {\text{M}}})$ $\;\Rightarrow \;\; \;{\pi}/{ {\text{C}}}=\;{ {\text{RT}}}/{ {\text{M}}} \ne {\text{f}}( {\text{c}})$ If we assume graph between $\;{\pi}/{ {\text{C}}}$ and ${\text{C}}$ Assuming $\pi$ vs $C$ graph $\; \;\; \;\text{ Slope }\;=\;{ {\text{RT}}}/{ {\text{M}}}=\;{0.083 \times 300}/{ {\text{M}}}=6 \times 10^{-4}$ $\; \therefore \;\; {\text{M}}=\;\frac{0.083 \times 300}{6 \times 10^{-4}}=\;\frac{830 \times 300}{6}$ $\;=41,500 {\text{gm}} / {\text{mole}}$

Q54JEE Main 2023NAT4Md and f Block Elements

How many of the following metal ions have similar value of spin only magnetic moment in gaseous state?________ (Given: Atomic number: V, $23 ; {\text{Cr}}, 24 ; {\text{Fe}}, 26$ ; ${\text{Ni}}, 28)$ ${\text{V}}^{3+} \cdot {\text{Cr}}^{3+}, {\text{Fe}}^{2+}, {\text{Ni}}^{3+}$ [25-Jan-2023 Shift 1]

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📖 Explanation

$\mu _s=\sqrt{ {\text{n}}( {\text{n}}+2)} B M \;\; ( {\text{n}}=$ no. of unpaired electrons $)$ ${\text{V}}^{3+}:[ {\text{Ar}}] 3 {\text{d}}^2 4 {\ s}^0\;2$ ${\text{Cr}}^{3+}:[ {\text{Ar}}] 3 {\text{d}}^3 4 {\ s}^0\;3$ ${\text{Fe}}^{2+}:[ {\text{Ar}}] 3 {\text{d}}^6 4 {\ s}^0\;4$ ${\text{Ni}}^{3+}:[ {\text{Ar}}] 3 {\text{d}}^7 4 {\ s}^0\;3$ ${\text{Cr}}^{3+} &\; {\text{Ni}}^{3+}$ have same value of $\mu _{ {\ s}}$

Q55JEE Main 2023NAT4MSome Basic Concepts of Chemistry

The density of a monobasic strong acid (Molar mass $24.2 {\text{g}} {\text{mol}}$ ) is $1.21 {\text{kg}} {\text{L}}$ . The volume of its solution required for the complete neutralization of $25 {\text{mL}}$ of $0.24 {\text{M}} {\text{NaOH}}$ is ________ x $10^{-2} {\text{mL}}$ (Nearest integer) [25-Jan-2023 Shift 1]

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📖 Explanation

$\;\text{ millimole of }\; {\text{NaOH}}=0.24 \times 25$ $\; \therefore \;\; \;\text{ millimole of acid }\;=0.24 \times 25$ $\;\Rightarrow \;\; \;\text{ mass of acid }\;=0.24 \times 25 \times 24.2 {\text{mg}}$ for pure acid, ${\text{V}}=\;{ {\text{w}}}/{ {\text{d}}} ;( {\text{d}}=1.21 {\text{kg}} / {\text{L}}=1.21 {\text{g}} / {\text{ml}})$ $\therefore {\text{V}}=\;\;\frac{0.24 \times 25 \times 24.2}{1.12} \times 10^{-3}$ $\;=120 \times 10^{-3} {\text{ml}}$ $\;=12 \times 10^{-2} {\text{ml}}$

Q56JEE Main 2023NAT4MChemical Kinetics

For the first order reaction ${\text{A}} \to {\text{B}} {\text{Br}}$ the half life is $30 {\text{mm}}$ . The time taken for $75 \%$ completion of the reaction is _______ mm. (Nearest mteger)Given : $\log 2=0.3010$ $\log 3=0.4771$ $\log 5=0.6989$ [25-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{t}}_{1 / 2}= {\text{T}}_{50}=30 {\min}$ $\; {\text{T}}_{75}=2 {\text{t}}_{1 / 2}=30 \times 2=60 {\min}$

Q57JEE Main 2023NAT4MChemical Bonding and Molecular Structure

The total number of lone pairs of electrons on oxygen atoms of ozone is _______ [25-Jan-2023 Shift 1]

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📖 Explanation

(Total no, of lone pairs on oxygen atoms $=6$

Q58JEE Main 2023NAT4MOrganic Chemistry - Some Basic Principles

In sulphur estimation. $0.471 {\text{g}}$ of an organic compound gave $1.4439 {\text{g}}$ of barium sulphate.The percentage of sulphur in the compound is ________ (Nearest Integer)(Given: Atomic mass Ba: $137 {\text{u}}: {\text{S}}: 32 {\text{u}}, {\text{O}}: 16 {\text{u}}$ ) [25-Jan-2023 Shift 1]

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📖 Explanation

$\;\text{ \%sulphur }\;\;=\;\frac{32}{233} \times \;{\;\text{ weight of }\; {\text{BaSO}}_4 \;\text{ formed }\;}/{\;\text{ weight of organic compound }\;} \times 100$ $\;=\;\frac{32}{233} \times \;\frac{1.4439}{0.471} \times 100$ $\;=42.10$ Nearest integer 42

Q59JEE Main 2023NAT4MCoordination Compounds

The number of paramagnetic species from the following is ______. $\;{ [ {\text{Ni}}( {\text{CN}})_4] ^{2-}, [ {\text{Ni}}( {\text{CO}})_4] , [ {\text{NiCl}}_4] ^{2-}}$ $\;{ [ {\text{Fe}}( {\text{CN}})_6] ^{4-}, [ {\text{Cu}} ( {\text{NH}}_3) _4] ^{2+}}$ $\;{ [ {\text{Fe}}( {\text{CN}})_6] ^{3-} \;\text{ and }\; [ {\text{Fe}} ( {\text{H}}_2 {\text{O}}) _6] ^{2+}}$ [25-Jan-2023 Shift 1]

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Q60JEE Main 2023NAT4MElectrochemistry

Consider the cell ${\text{Pt}}( {\ s}) | {\text{H}}_2( {\ s})( {\text{latm}}) {\text{H}}^{+} ( {\text{aq}}, \$ [ {\text{H}}^{+}]$ =1){\text{Fe}}^{3+}( {\text{aq}}), {\text{Fe}}^{2+}( {\text{aq}}) \mid {\text{Pt}}( {\ s}) $Given:$ {\text{E}}{ {\text{Fe}}^{3+/ / {\text{Fe}}^{2+}}}^{\circ}=0.771 {\text{V}} $and$ {\text{E}}{ {\text{H}}^{+} / ;\frac{1}{2} {\text{H}}_2}^{\circ}=0 {\text{V}}, {\text{T}}=298 {\text{K}} $If the potential of the cell is$ 0.712 {\text{V}} $the ratio of concentration of$ {\text{Fe}}^{2+} $to$ {\text{Fe}}^{2+}$ is ________ (Nearest integer) [25-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\frac{1}{2} {\text{H}}_2( {\text{g}})+ {\text{Fe}}^{3+}( {\text{aq}} .) \longrightarrow {\text{H}}^{+}( {\text{aq}})+ {\text{Fe}}^{2+}( {\text{aq}} .)$ $\; {\text{E}}= {\text{E}}^{ {\text{o}}}-\;\frac{0.059}{1} \log \;{[ {\text{Fe}}^{2+}]\$ }/{[ {\text{Fe}}^{3+}]$ };\Rightarrow ;; 0.712=(0.771-0)-;\frac{0.059}{1} \log ;{[ {\text{Fe}}^{2+}]$ }/{[ {\text{Fe}}^{3+}]$ };\Rightarrow ;; \log ;{[ {\text{Fe}}^{2+}]$ }/{[ {\text{Fe}}^{3+}]$ }=;\frac{(0.771-0712)}{0.059}=1;\Rightarrow ;; ;{[ {\text{Fe}}^{2+}]$ }/{[ {\text{Fe}}^{3+}]$ }=10$

Mathematics30 questions

Q61JEE Main 2023MCQ4MProbability

Let $M$ be the maximum value of the product of two positive integers when their sum is 66 . Let the sample space $S= \{x \in Z: x(66-x) \ge \;\frac{5}{9} M\}$ and the event ${\text{A}}=\{ {\text{x}} \in {\text{S}}: {\text{x}}$ is a multiple of 3 $\}$ . Then ${\text{P}}( {\text{A}})$ is equal to [25-Jan-2023 Shift 1]

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📖 Explanation

$\;M=33 \times 33$ $\;x(66-x) \ge \;\frac{5}{9} \times 33 \times 33$ $\;11 \le x \le 55$ $\; {\text{A}}:\{12,15,18, . . . .54\}$ $\; {\text{P}}( {\text{A}})=\;\frac{15}{45}=\;\frac{1}{3}$

Q62JEE Main 2023MCQ4MVector Algebra

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non zero vectors such that ${ \vec{\text{b}} \cdot { \vec{\text{c}}=0$ and ${ \vec{\text{a}} \times ({ \vec{\text{b}} \times { \vec{\text{c}})=\;{{ \vec{\text{b}}-{ \vec{\text{c}}}/{2}$ . If ${ \vec{\text{d}}$ be a vector such that $\vec{b} \cdot \vec{d}=\vec{a} \cdot \vec{b}$ , then $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d})$ is equal to [25-Jan-2023 Shift 1]

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Q63JEE Main 2023MCQ4MDifferential Equations

Let $y=y(x)$ be the solution curve of the differential equation $\;\frac{d y}{d x}=\;\frac{y}{x} (1+x y^2 (1+\log _e x) )$ , $x > 0, y(1) =3$ . Then $\;\frac{y^2(x)}{9}$ is equal to : [25-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\frac{d y}{d x}-\;\frac{y}{x}=y^3 (1+\log _e x)$ $\;\;\frac{1}{y^3} \;\frac{d y}{d x}-\;\frac{1}{x y y^2}=1+\log _e x$ $\;\;\text{ Let }\;-\;{1}/{ {\text{y}}^2}= {\text{t}} \Rightarrow \;{2}/{ {\text{y}}^3} \;{ {\text{dy}}}/{ {\text{dx}}}=\;{ {\text{dt}}}/{ {\text{dx}}}$ $\; \therefore \;{ {\text{dt}}}/{ {\text{dx}}}+\;{2 {\text{t}}}/{ {\text{x}}}=2 (1+\log _{ {\text{e}}} {\text{x}})$ $\;\;\text{ I.F. }\;=e^{\int \;\frac{2}{x} \text{dx}}=x^2$ $\;\;\frac{-x^2}{y^2}=\;\frac{2}{3} ( (1+\log _e x) x^{3-}\;\frac{x^3}{3}) +C$ $\;y(1)=3$ $\;\;\frac{y^2}{9}=\;\frac{x^2}{5-2 x^3 (2+\log _e x^3) }$ $\;x d y=y \text{dx}+x y^3 (1+\log _e x) \text{dx}$ $\;\;\frac{x d y-y \text{dx}}{y^3}=x (1+\log _e x) \text{dx}$ $\;-\;\frac{x}{y} d (\;\frac{x}{y}) =x^2 (1+\log _e x) \text{dx}$ $\;- (\;\frac{x}{y}) ^2=2 \int x^2 (1+\log _e x) \text{dx}$

Q64JEE Main 2023MCQ4MLimits, Continuity and Differentiability

The value of ${\text{Lim}}_{n \to \infty } \;{1+2-3+4+5-6+. . .+(3 n-2)+(3 n-1)-3 n}/{\sqrt{2 n^{4+}4 n+3-{\sqrt{n^{4+}5 n+4}}}}$ is : [25-Jan-2023 Shift 1]

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📖 Explanation

$\;{\text{Lim}}_{n \to \infty } \;{0+3+6+9+. . . . n \;\text{ terms }\;}/{{\sqrt{2 n^{4+}4 n+3}}-{\sqrt{n^{4+}5 n+4}}}$ $\;{\text{Lim}}_{n \to \infty } \;{3 n(n-1)}/{2 ({\sqrt{2 n^{4+}4 n+3}}-{\sqrt{n^{4+}5 n+4}}) }$ $\;=\;{3}/{2({\sqrt{2}}-1)}=\;\frac{3}{2}({\sqrt{2}}+1)$

Q65JEE Main 2023MCQ4MCircles

The points of intersection of the line $a x+b y=0$ , $(a \ne b)$ and the circle $x^{2+}y^{2-}2 x=0$ are ${\text{A}}(\alpha, 0)$ and ${\text{B}}(1, \beta)$ . The image of the circle with ${\text{AB}}$ as a diameter in the line ${\text{x}}+ {\text{y}}+2=0$ is : [25-Jan-2023 Shift 1]

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📖 Explanation

Only possibility $\alpha=0, \beta=1$ $\therefore$ equation of circle ${\text{x}}^{2+} {\text{y}}^{2-} {\text{x}}- {\text{y}}=0$ Image of circle in $x+y+2=0$ is $x^{2+}y^{2+}5 x+5 y+12=0$

Q66JEE Main 2023MCQ4MStatistics

The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12 . If the new mean of the marks is $10.2$ . then their new variance is equal to: [25-Jan-2023 Shift 1]

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📖 Explanation

$sum_{{ {\text{i}}=1}}^{{ {\text{n}}}} {\text{x}}_{ {\text{i}}}=10 {\text{n}}$ $sum_{{ {\text{i}}=1}}^{{ {\text{n}}}} {\text{x}}_{ {\text{i}}}-8+12=(10.2) {\text{n}} \;\; \therefore {\text{n}}=20$ $\;\text{ Now }\; \;{sum_{{ {\text{i}}=1}}^{{20}} {\text{x}}_{ {\text{i}}}^2}/{20}-(10)^2=4 \Rightarrow sum_{{ {\text{i}}=1}}^{{20}} {\text{x}}_{ {\text{i}}}^2=2080$ $\;{sum_{{ {\text{i}}=1}}^{{20}} {\text{x}}_{ {\text{i}}}^{2-}8^{2+}12^2}/{20}-(10.2)^2$ $=108-104.04=3.96$

Q67JEE Main 2023MCQ4MDifferentiation

Let $y(x)=(1+x) (1+x^2) (1+x^4) (1+x^8) (1+x^{16}) .$ Then $y^{'}-y^{' '}$ at $x=-1$ is equal to [25-Jan-2023 Shift 1]

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📖 Explanation

$\;y=\;\frac{1-x^{32}}{1-x} \Rightarrow y-x y=1-x^{32}$ $\;y^{'}-x y^{'}-y=-32 x^{31}$ $\;y^{' '}-x y^{' '}-y^{'}-y^{'}=-(32)(31) x^{30}$ $\;\;\text{ at }\; x=-1 \Rightarrow y^{'}-y^{' '}=496$

Q68JEE Main 2023MCQ4MVector Algebra

The vector $\vec{a}=-\hat{i}+2 \hat{j}+\hat{k}$ is rotated through a right angle, passing through the $y$ -axis in its way and the resulting vector is $\vec{b}$ . Then the projection of $3 \vec{a}+{\sqrt{2}} \vec{b}$ on $\vec{c}=5 \hat{i}+4 \hat{j}+3 \hat{k}$ is [25-Jan-2023 Shift 1]

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Q69JEE Main 2023MCQ4MApplication of Derivatives

The minimum value of the function $f(x)=int_{0}^{2} e {}^{}^{|x-t|} \text{dt}$ is is [25-Jan-2023 Shift 1]

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📖 Explanation

For ${\text{x}} \le 0$ $f(x)=int_{0}^{2} e^{t-x} \text{dt}=e^{-x} (e^{2-}1)$ For $0< x< 2$ $f(x)=int_{0}^{x} e^{x-t} \text{dt}+int_{x}^{2} e^{t-x} \text{dt}=e^x+e^{2-x}-2$ For ${\text{x}} \ge 2$ $f(x)=int_{0}^{2} e^{x-t} \text{dt}=e^{x-2} (e^{2-}1)$ For ${\text{x}} \le 0, {\text{f}}( {\text{x}})$ is $↓$ and ${\text{x}} \ge 2, {\text{f}}( {\text{x}})$ is $↑$ $\therefore$ Minimum value of ${\text{f}}( {\text{x}})$ lies in ${\text{x}} \in (0,2)$ Applying A.M $\ge$ G.Mminimum value of $f(x)$ is $2(e-1)$

Q70JEE Main 2023MCQ4MThree Dimensional Geometry

Consider the lines ${\text{L}}_1$ and ${\text{L}}_2$ given by $\;L_1: \;\frac{x-1}{2}=\;\frac{y-3}{1}=\;\frac{z-2}{2}$ $\;L_2: \;\frac{x-2}{1}=\;\frac{y-2}{2}=\;\frac{z-3}{3}$ A line ${\text{L}}_3$ having direction ratios $1,-1,-2$ , intersects ${\text{L}}_1$ and ${\text{L}}_2$ at the points ${\text{P}}$ and ${\text{Q}}$ respectively. Then the length of line segment PQ is [25-Jan-2023 Shift 1]

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📖 Explanation

Let $P=(2 \lambda +1, \lambda +3,2 \lambda +2)$ Let $Q=(\mu +2,2 \mu +2,3 \mu +3)$ $\;\Rightarrow \;\frac{2 \lambda -\mu -1}{1}=\;\frac{\lambda -2 \mu +1}{-1}=\;\frac{2 \lambda -3 \mu -1}{-2}$ $\;\Rightarrow \lambda =\mu =3 \Rightarrow {\text{P}}(7,6,8) \;\text{ and }\; {\text{Q}}(5,8,12)$ $\; {\text{PQ}}=2 {\sqrt{6}}$

Q71JEE Main 2023MCQ4MApplication of Derivatives

Let ${\text{x}}=2$ be a local minima of the function $f(x)=2 x^{4-}18 x^{2+}8 x+12, x \in (-4,4)$ . If $M$ is local maximum value of the function ${\text{f}}$ in $(-4,4)$ , then $M=$ [25-Jan-2023 Shift 1]

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📖 Explanation

$\;f^{'}(x)=8 x^{3-}36 x+8=4 (2 x^{3-}9 x+2)$ $\;f^{'}(x)=0$ $\; \therefore x=\;{{\sqrt{6}}-2}/{2}$ $\;\;\text{ Now }\;$ $\;f(x)= (x^{2-}2 x-\;\frac{9}{2}) (2 x^{2+}4 x-1) +24 x+7.5$ $\; \therefore f (\;{{\sqrt{6}}-2}/{2}) =M=12 {\sqrt{6}}-\;\frac{33}{2}$

Q72JEE Main 2023MCQ4MComplex Numbers

Let $z_1=2+3 i$ and $z_2=3+4 i$ . The set ${\text{S}}= \{ {\text{z}} \in {\text{C}}: | {\text{z}}- {\text{z}}_1{{}/{}|} ^{2-}z-z_2{{}/{}|} ^2=z_1- {\text{z}}_2{{}/{}|} ^2\}$ represents a [25-Jan-2023 Shift 1]

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📖 Explanation

$\; (( {\text{x}}-2)^{2+}( {\text{y}}-3)^2) - (( {\text{x}}-3)^{2-}( {\text{y}}-4)^2) =1+1$ $\;\Rightarrow {\text{x}}+ {\text{y}}=7$

Q73JEE Main 2023MCQ4MParabola

The distance of the point $(6,-2 {\sqrt{2}})$ from the common tangent ${\text{y}}= {\text{mx}}+ {\text{c}}, {\text{m}} > 0$ , of the curves ${\text{x}}=2 {\text{y}}^2$ and ${\text{x}}=1+ {\text{y}}^2$ is [25-Jan-2023 Shift 1]

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📖 Explanation

For ${\text{y}}^2=\;{ {\text{x}}}/{2}, {\text{T}}: {\text{y}}= {\text{mx}}+\;{1}/{8 {\text{m}}}$ For tangent to $y^{2+}1=x$ $\;\Rightarrow ( {\text{mx}}+\;{1}/{8 {\text{m}}}) ^{2+}1= {\text{x}}$ $\; {\text{D}}=0 \Rightarrow {\text{m}}=\;{1}/{2 {\sqrt{2}}}$ $\; \therefore {\text{T}}: {\text{x}}-2 {\sqrt{2}} {\text{y}}+1=0$ $\; {\text{d}}= |\;{6+8+1}/{{\sqrt{9}}}{{}/{}|} =5$

Q74JEE Main 2023MCQ4MMatrices and Determinants

Let $S_1$ and $S_2$ be respectively the sets of all $a \in R-\{0\}$ for which the system of linear equations $\;a x+2 a y-3 a z=1$ $\;(2 a+1) x+(2 a+3) y+(a+1) z=2$ $\;(3 a+5) x+(a+5) y+(a+2) z=3$ has unique solution and infinitely many solutions. Then [25-Jan-2023 Shift 1]

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📖 Explanation

$\;\Delta ={\begin{vmatrix}a & 2 a & -3 a \\ 2 a+1 & 2 a+3 & a+1 \\ 3 a+5 & a+5 & a+2\end{vmatrix}}$ $\;=a (15 a^{2+}31 a+36) =0 \Rightarrow a=0$ $\;\Delta \ne 0 \;\text{ for all }\; a \in R-\{0\}$ $\;\;\text{ Hence }\; S_1=R-\{0\} \;\; S_2=Φ$

Q75JEE Main 2023MCQ4MIndefinite Integrals

Let $f(x)=\int \;\frac{2 x}{ (x^{2+}1) (x^{2+}3) } \text{dx}$ .If ${\text{f}}(3)=\;\frac{1}{2} (\log _e 5-\log _e 6)$ , then ${\text{f}}(4)$ is equal to [25-Jan-2023 Shift 1]

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📖 Explanation

Put $x^2=t$ $\;\int \;\frac{d t}{(t+1)(t+3)}=\;\frac{1}{2} \int (\;\frac{1}{t+1}-\;\frac{1}{t+3}) \text{dt}$ $\;f(x)=\;\frac{1}{2} \ln (\;\frac{x^{2+}1}{x^{2+}3}) +C$ $\;f(3)=\;\frac{1}{2}(\ln 10-\ln 12)+C$ $\;\Rightarrow C=0$ $\;f(4)=\;\frac{1}{2} \ln (\;\frac{17}{19})$

Q76JEE Main 2023MCQ4MSets and Relations

The statement $( {\text{p}} ∧(∼ {\text{q}})) \Rightarrow ( {\text{p}} \Rightarrow (∼ {\text{q}}))$ is [25-Jan-2023 Shift 1]

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📖 Explanation

$\;( {\text{p}} ∧ ∼ {\text{q}}) \to ( {\text{p}} \to ∼ {\text{q}})$ $\;\equiv (∼( {\text{p}} ∧ ∼ {\text{q}})) ∨(∼ {\text{p}} ∨ ∼ {\text{q}})$ $\;\equiv (∼ {\text{p}} ∨ {\text{q}}) ∨(∼ {\text{p}} ∨ ∼ {\text{q}})$ $\;\equiv ∼ {\text{p}} ∨ {\text{t}} \equiv {\text{t}}$

Q77JEE Main 2023MCQ4MApplication of Derivatives

Let ${\text{f}}:(0,1) \to ℝ$ be a function defined by $f(x)=\;\frac{1}{1-e^{-x}}$ , and $g(x)=(f(-x)-f(x))$ . Consider two statements(I) ${\text{g}}$ is an increasing function in $(0,1)$ (II) $g$ is one-one in $(0,1)$ Then, [25-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{g}}( {\text{x}})= {\text{f}}(- {\text{x}})- {\text{f}}( {\text{x}})=\;{1+ {\text{e}}^{ {\text{x}}}}/{1- {\text{e}}^{ {\text{x}}}}$ $\;\Rightarrow {\text{g}}^{'}( {\text{x}})=\;{2 {\text{e}}^{ {\text{x}}}}/{ (1- {\text{e}}^{ {\text{x}}}) ^2} > 0$ $\;\Rightarrow {\text{g}} \;\text{ is increasing \in }\;(0,1)$ $\;\Rightarrow {\text{g}} \;\text{ is one-one \in }\;(0,1)$

Q78JEE Main 2023MCQ4MThree Dimensional Geometry

The distance of the point ${\text{P}}(4,6,-2)$ from the line passing through the point $(-3,2,3)$ and parallel to a line with direction ratios $3,3,-1$ is equal to : [25-Jan-2023 Shift 1]

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📖 Explanation

Equation of line is $\;{ {\text{x}}+3}/{3}=\;{ {\text{y}}-2}/{3}=\;{ {\text{z}}-3}/{-1}=\lambda$ ${\text{M}}(3 \lambda -3,3 \lambda +2,3-\lambda )$ D.R of ${\text{PM}}(3 \lambda -7,3 \lambda -4,5-\lambda )$ Since PM is perpendicular to line $\;\Rightarrow 3(3 \lambda -7)+3(3 \lambda -4)-1(5-\lambda )=0$ $\;\Rightarrow \lambda =2$ $\;\Rightarrow {\text{M}}(3,8,1) \Rightarrow {\text{PM}}={\sqrt{14}}$

Q79JEE Main 2023MCQ4MMatrices and Determinants

Let $x, y, z >1$ and $A={\begin{bmatrix}1 & \log _x y & \log _x z \\ \log _y x & 2 & \log _y z \\ \log _z x & \log _z y & 3\end{bmatrix}} .$ Then $|{\text{adj}} ({\text{adj}} A^2) {{}/{}|}$ is equal to [25-Jan-2023 Shift 1]

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📖 Explanation

$\;|A|=\;\frac{1}{\log x \cdot \log y \cdot \log z}{\begin{vmatrix}\log x & \log y & \log z \\ \log x & 2 \log y & \log z \\ \log x & \log y & 3 \log z\end{vmatrix}}=2$ $\;\Rightarrow |{\text{adj}} ({\text{adj}} A^2) {{}/{}|} = |A^2{{}/{}|} ^4=2^8$

Q80JEE Main 2023MCQ4MBinomial Theorem

If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1+x)^{10}$ , then $sum_{r=1}^{10} r^3 (\;\frac{a_r}{a_{r-1}}) ^2$ is equal to [25-Jan-2023 Shift 1]

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📖 Explanation

${\text{a}}_{ {\text{r}}}={ }^{10} {\text{C}}_{10- {\text{r}}}={ }^{10} {\text{C}}_{ {\text{r}}}$ $\Rightarrow sum_{{\text{r}}=1}^{10} r^3 (\;\frac{{ }^{10} C_r}{{ }^{10} C_{r-1}}) ^2$ $=sum_{{ {\text{r}}=1}}^{10} {\text{r}}^3 (\;{11- {\text{r}}}/{ {\text{r}}}) ^2$ $=sum_{{ {\text{r}}=1}}^{10} {\text{r}}(11- {\text{r}})^2$ $=sum_{{ {\text{r}}=1}}^{10} (121 {\text{r}}+ {\text{r}}^{3-}22 {\text{r}}^2) =1210$

Q81JEE Main 2023NAT4MPermutations and Combinations

Let ${\text{S}}=\{1,2,3,5,7,10,11\}$ . The number of nonempty subsets of ${\text{S}}$ that have the sum of all elements a multiple of 3 , is _______. [25-Jan-2023 Shift 1]

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📖 Explanation

Elements of the type $3 {\text{k}}=3$ Elements of the type $3 {\text{k}}+1=1,7,9$ Elements of the type $3 {\text{k}}+2=2,5,11$ Subsets containing one element $S_1=1$ Subsets containing two elements ${\text{S}}_2={ }^3 {\text{C}}_1 \times { }^3 {\text{C}}_1=9$ Subsets containing three elements $S_3={ }^3 C_1 \times { }^3 C_1+1+1=11$ Subsets containing four elements ${\text{S}}_4={ }^3 {\text{C}}_3+{ }^3 {\text{C}}_3+{ }^3 {\text{C}}_2 \times { }^3 {\text{C}}_2=11$ Subsets containing five elements ${\text{S}}_5={ }^3 {\text{C}}_2 \times { }^3 {\text{C}}_2 \times 1=9$ Subsets containing six elements ${\text{S}}_6=1$ Subsets containing seven elements ${\text{S}}_7=1$ $\Rightarrow \;\text{ \sum }\;=43$

Q82JEE Main 2023NAT4MFunctions

For some $a, b, c \in ℕ$ , let $f(x)=a x-3$ and $g(x)=x^b+c, x \in ℝ$ . If $(\;\text{ fog }\;)^{-1}(x)= (\;\frac{x-7}{2}) ^{1 / 3}$ then (fog) $(a c)+(g o f)$ (b) is equal to ________. [25-Jan-2023 Shift 1]

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📖 Explanation

Let fog $(x)=h(x)$ $\;\Rightarrow {\text{h}}^{-1}( {\text{x}})= (\;{ {\text{x}}-7}/{2}) ^{\;\frac{1}{3}}$ $\;\Rightarrow {\text{h}}( {\text{x}})= {\text{fog}}( {\text{x}})=2 {\text{x}}^{3+}7$ $\;\;\text{ fog }\;(x)=a (x^b+c) -3$ $\;\Rightarrow {\text{a}}=2, {\text{b}}=3, {\text{c}}=5$ $\;\Rightarrow \;\text{ fog }\;( {\text{ac}})={\text{fog}}(10)=2007$ $\; {\text{g}} ( {\text{f}}( {\text{x}})=(2 {\text{x}}-3)^{3+}5.$ $\;\Rightarrow {\text{gof}}( {\text{b}})={\text{gof}}(3)=32$ $\;\Rightarrow {\ \sum}=2039$

Q83JEE Main 2023NAT4MHyperbola

The vertices of a hyperbola ${\text{H}}$ are $(\pm 6,0)$ and its eccentricity is $\;{{\sqrt{5}}}/{2}$ . Let ${\text{N}}$ be the normal to ${\text{H}}$ at a point in the first quadrant and parallel to the line ${\sqrt{2}} x+y=2 {\sqrt{2}}$ . If ${\text{d}}$ is the length of the line segment of ${\text{N}}$ between ${\text{H}}$ and the ${\text{y}}$ -axis then ${\text{d}}^2$ is equal to _______. [25-Jan-2023 Shift 1]

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📖 Explanation

${\text{H}}: \;{ {\text{x}}^2}/{36}-\;\frac{y^2}{9}=1$ equation of normal is $6 x \cos \theta+3 y \cot \theta=45$ $\;\;\text{ slope }\;=-2 sin \;\theta=-{\sqrt{2}}$ $\;\Rightarrow \theta=\;\frac{\pi}{4}$ Equation of normal is ${\sqrt{2}} {\text{x}}+ {\text{y}}=15$ ${\text{P}}:( {\text{a}} \ sec \theta, b \tan \theta)$ $\Rightarrow {\text{P}}(6 {\sqrt{2}}, 3)$ and ${\text{K}}(0,15)$ $d^2=216$

Q84JEE Main 2023NAT4MQuadratic Equation and Inequalities

Let $S= \{\alpha: \log _2 (9^{2 \alpha-4}+13) -\log _2 (\;\frac{5}{2} \cdot 3^{2 \alpha-4}+1) =2\} \;\text{. }\;$ Then the maximum value of $\beta$ for which the equation $x^{2-}2 (sum_{\alpha \in s} \alpha) ^2 x+sum_{\alpha \in s}(\alpha+1)^2 \beta=0$ has real roots, is ______. [25-Jan-2023 Shift 1]

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📖 Explanation

$\;\log _2 (9^{2 \alpha-4}+13) -\log _2 (\;\frac{5}{2} \cdot 3^{2 \alpha-4}+1) =2$ $\;\Rightarrow \;\frac{9^{2 \alpha-4}+13}{\frac{5}{2} 3^{2 \alpha-4}+1}=4$ $sum_{\alpha \in S} \alpha=5$ and $sum_{{\alpha \in S}}(\alpha+1)^2{=25}$ $\;\Rightarrow x^2{-50} {\text{x}}+25 \beta=0 \;\text{ has real roots }\;$ $\;\Rightarrow \beta \le 25$ $\;\Rightarrow \beta_{\max }=25$

Q85JEE Main 2023NAT4MBinomial Theorem

The constant term in the expansion of $(2 x+\;\frac{1}{x^7}+3 x^2) ^5 \;\text{ is }\;$ ______. [25-Jan-2023 Shift 1]

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📖 Explanation

General term is $\sum \;{5 !(2 x)^{ {\text{n}}_1} ( {\text{x}}^{-7}) ^{ {\text{n}}_2} (3 {\text{x}}^2) ^{ {\text{n}}_3}}/{ {\text{n}}_{1} ! {\text{n}}_{2} ! {\text{n}}_{3} !}$ For constant term, ${\text{n}}_1+2 {\text{n}}_3=7 {\text{n}}_2$ $\& {\text{n}}_1+ {\text{n}}_2+ {\text{n}}_3=5$ Only possibility ${\text{n}}_1=1, {\text{n}}_2=1, {\text{n}}_3=3$ $\Rightarrow$ constant term $=1080$

Q86JEE Main 2023NAT4MSequences and Series

Let $A_1, A_2, A_3$ be the three A.P. with the same common difference ${\text{d}}$ and having their first terms as ${\text{A}}, {\text{A}}+1, {\text{A}}+2$ , respectively. Let a, b, c be the $7^{\;\text{th }\;}, 9^{\;\text{th }\;}, 17^{\;\text{th }\;}$ terms of $A_1, A_2, A_3$ , respectively such that ${\begin{vmatrix}{\text{a}} & 7 & 1 \\ 2 {\text{b}} & 17 & 1 \\ {\text{c}} & 17 & 1\end{vmatrix}}+70=0$ If $a=29$ , then the sum of first 20 terms of an AP whose first term is ${\text{c}}- {\text{a}}- {\text{b}}$ and common difference is $\;{ {\text{d}}}/{12}$ , is equal to ________. [25-Jan-2023 Shift 1]

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📖 Explanation

$\;{\begin{vmatrix}{\text{A}}+6 {\text{d}} & 7 & 1 \\ 2( {\text{A}}+1+8 {\text{d}}) & 17 & 1 \\ {\text{A}}+2+16 {\text{d}} & 17 & 1\end{vmatrix}}+70=0$ $\;\Rightarrow {\text{A}}=-7 \;\text{ and }\; {\text{d}}=6$ $\; \therefore {\text{c}}- {\text{a}}- {\text{b}}=20$ $\; {\text{S}}_{20}=495$

Q87JEE Main 2023NAT4MInverse Trigonometric Functions

If the sum of all the solutions of $\tan ^{-1} (\;\frac{2 x}{1-x^2}) +\cot ^{-1} (\;\frac{1-x^2}{2 x}) =\;\frac{\pi}{3}$ $-1 < x < 1, x \ne 0$ , is $\alpha-\;{4}/{{\sqrt{3}}}$ , then $\alpha$ is equal to _______. [25-Jan-2023 Shift 1]

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📖 Explanation

Case I : $x > 0$ $\;\tan ^{-1} \;\frac{2 x}{1-x^2}+\tan ^{-1} \;\frac{2 x}{1-x^2}=\;\frac{\pi}{3}$ $\;x=2-{\sqrt{3}}$ Case II : $x < 0$ $\;\tan ^{-1} \;\frac{2 x}{1-x^2}+\tan ^{-1} \;\frac{2 x}{1-x^2}+\pi=\;\frac{\pi}{3}$ $\;x=\;{-1}/{{\sqrt{3}}} \Rightarrow \alpha=2$

Q88JEE Main 2023NAT4MThree Dimensional Geometry

Let the equation of the plane passing through the line $x-2 y-z-5=0=x+y+3 z-5$ and parallel to the line ${\text{x}}+ {\text{y}}+2 {\text{z}}-7=0=2 {\text{x}}+3 {\text{y}}+ {\text{z}}-2$ be $a x+b y+c z=65$ . Then the distance of the point (a, b, c) from the plane $2 x+2 y-z+16=0$ is _______. [25-Jan-2023 Shift 1]

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📖 Explanation

Equation of plane is $\;(x-2 y-z-5)+b(x+y+3 z-5)=0$ $\;{\begin{vmatrix}1+b & -2+b & -1+3 b \\ 1 & 1 & 2 \\ 2 & 3 & 1\end{vmatrix}}=0$ $\;\Rightarrow b=12$ $\; \therefore \;\text{ plane is }\; 13 x+10 y+35 z=65$ Distance from given point to plane $=9$

Q89JEE Main 2023NAT4MPermutations and Combinations

Let ${\text{x}}$ and ${\text{y}}$ be distinct integers where $1 \le {\text{x}} \le 25$ and $1 \le y \le 25$ . Then, the number of ways of choosing ${\text{x}}$ and ${\text{y}}$ , such that ${\text{x}}+ {\text{y}}$ is divisible by 5 , is _______. [25-Jan-2023 Shift 1]

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📖 Explanation

$x+y=5 \lambda$ Cases: Total $=120$

Q90JEE Main 2023NAT4MArea Under The Curves

It the area enclosed by the parabolas $P_1: 2 y=5 x^2$ and $P_2: x^{2-}y+6=0$ is equal to the area enclosed by $P_1$ and $y=\alpha x, \alpha>0$ , then $\alpha^3$ is equal to _______. [25-Jan-2023 Shift 1]

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📖 Explanation

Abscissa of point of intersection of $2 {\text{y}}=5 {\text{x}}^2$ and $y=x^{2+}6$ is $\pm 2$ $\;\;\text{ Area }\;=2 int_{0}^{2} (x^{2+}6-\;\frac{5 x^2}{2}) \text{dx}=int_{0}^{\;\frac{2 \alpha}{5}} (\alpha x-\;\frac{5 x^2}{2}) \text{dx}$ $\;\Rightarrow int_{0}^{\;\frac{2 \alpha}{5}} (\alpha x-\;\frac{5 x^2}{2}) \text{dx}=16$ $\;\Rightarrow \alpha^3=600$

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