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24 Jan 2023 Shift 1

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Physics30 questions

Q1JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

Two long straight wires ${\text{P}}$ and ${\text{Q}}$ carrying equal current $10 {\text{A}}$ each were kept parallel to each other at $5 {\text{cm}}$ distance. Magnitude of magnetic force experienced by $10 {\text{cm}}$ length of wire $P$ is $F_{1}$ . If distance between wires is halved and currents on them are doubled, force $F_{2}$ on $10 {\text{cm}}$ length of wire ${\text{P}}$ will be : [24-Jan-2023 Shift 1]

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📖 Explanation

Force per unit length between two parallel straight $\;\;\text{ wires }\;=\;{\mu _0 {\text{i}}_1 {\text{i}}_2}/{2 \pi {\text{d}}}$ $\;\;\frac{F_1}{F_2}=\;{\;{\mu _0(10)^2}/{2 \pi(5 {\text{cm}})}}/{{\mu _0(20)^2}/{2 \pi (\;{5 {\text{cm}}}/{2}) }}=\;\frac{1}{8}$ $\;\Rightarrow {\text{F}}_2=8 {\text{F}}_1$ $\;$

Q2JEE Main 2023MCQ4MLaws of Motion

Given below are two statements :Statement-I : An elevator can go up or down with uniform speed when its weight is balanced with the tension of its cable.Statement-II : Force exerted by the floor of an elevator on the foot of a person standing on it is more than his/her weight when the elevator goes down with increasing speed.In the light of the above statements, choose the correct answer from the options given below : [24-Jan-2023 Shift 1]

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📖 Explanation

Statement-1When elevator is moving with uniform speed ${\text{T}}= {\text{F}}_{ {\text{g}}}$ Statement-2When elevator is going down with increasing speed, its acceleration is downward.Hence ${\text{W}}- {\text{N}}=\;{ {\text{W}}}/{ {\text{g}}} \times {\text{a}}$ ${\text{N}}= {\text{W}} (1-\;{ {\text{a}}}/{ {\text{g}}})$ i.e. less than weight.

Q3JEE Main 2023MCQ4MDual Nature of Matter and Radiation

From the photoelectric effect experiment, following observations are made. Identify which of these are correctA. The stopping potential depends only on the work function of the metal.B. The saturation current increases as the intensity of incident light increases.C. The maximum kinetic energy of a photo electron depends on the intensity of the incident light.D. Photoelectric effect can be explained using wave theory of light.Choose the correct answer from the options given below: [24-Jan-2023 Shift 1]

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📖 Explanation

(A) Stopping potential depends on both frequency of light and work function.(B) Saturation current $\propto$ intensity of light(C) Maximum KE depends on frequency(D) Photoelectric effect is explained using particle theory

Q4JEE Main 2023MCQ4MGravitation

The weight of a body at the surface of earth is 18 ${\text{N}}$ . The weight of the body at an altitude of 3200 ${\text{km}}$ above the earth's surface is (given, radius of earth ${\text{R}}_{ {\text{e}}}=6400 {\text{km}}$ ) [24-Jan-2023 Shift 1]

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📖 Explanation

Acceleration due to gravity at height ${\text{h}}$ ${\text{g}}^{'}=\;{ {\text{g}}}/{ [1+\;{ {\text{h}}}/{ {\text{R}}}] ^2}$ So weight at given height ${\text{mg}}^{'}=\;{ {\text{mg}}}/{ [1+\;{ {\text{h}}}/{ {\text{R}}}] ^2}=\;\frac{18}{ [1+\;\frac{1}{2}] ^2}=8 {\text{N}}$

Q5JEE Main 2023MCQ4MProperties of Matter

A $100 {\text{m}}$ long wire having cross-sectional area $6.25$ $\times 10^{-4} {\text{m}}^{2}$ and Young's modulus is $10^{10} {\text{Nm}}^{-2}$ is subjected to a load of $250 {\text{N}}$ , then the elongation in the wire will be : [24-Jan-2023 Shift 1]

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📖 Explanation

Elongation in wire $\delta =\;{ {\text{F}} ℓ}/{ {\text{AY}}}$ $\;\delta =\;\frac{250 \times 100}{6.25 \times 10^{-4} \times 10^{10}}$ $\;\delta =4 \times 10^{-3} {\text{m}}$

Q6JEE Main 2023MCQ4MThermodynamics

$1 {\text{g}}$ of a liquid is converted to vapour at $3 \times 10^{5} {\text{Pa}}$ pressure. If $10 \%$ of the heat supplied is used for increasing the volume by $1600 {\text{cm}}^{3}$ during this phase change, then the increase in internal energy in the process will be : [24-Jan-2023 Shift 1]

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📖 Explanation

Work done $= {\text{P}} \Delta {\text{V}}$ $=3 \times 10^5 \times 1600 \times 10^{-6}$ $=480 {\text{J}}$ Only $10 \%$ of heat is used in work done.Hence $\Delta {\text{Q}}=4800 {\text{J}}$ The rest goes in internal energy, which is $90 \%$ of heat.Change in internal energy $=0.9 \times 4800=4320 {\text{J}}$

Q7JEE Main 2023MCQ4MElectromagnetic Waves

A modulating signal is a square wave, as shown in the figure. If the carrier wave is given as ${\text{c}}( {\text{t}})=2 sin \;(8 \pi {\text{t}})$ volts, the modulation index is : [24-Jan-2023 Shift 1]

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📖 Explanation

Modulation index $\;=\;\frac{\;\text{ Amplitude of mod ulating signal }\;}{\;\text{ Amplitude of carrier wave }\;}$ $\;\mu =\;\frac{1}{2}$

Q8JEE Main 2023MCQ4MLaws of Motion

As per given figure, a weightless pulley $P$ is attached on a double inclined frictionless surface. The tension in the string (massless) will be (if ${\text{g}}=$ $10 {\text{m}} / {\ s}^{2}$ ) [24-Jan-2023 Shift 1]

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📖 Explanation

$4 g sin \;60^{\circ}-T=4 a$ . . . (1) $4 {\text{g}} sin \;60^{\circ}$ ${\text{T}}- {\text{g}} sin \;30^{\circ}= {\text{a}}$ . . . (2)Solving (1) and (2) we get. $\;20 {\sqrt{3}}-T=4 T-20$ $\;T=4({\sqrt{3}}+1) N$

Q9JEE Main 2023MCQ4MSemiconductor Electronics

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R Assertion A : Photodiodes are preferably operated in reverse bias condition for light intensity measurement.Reason R : The current in the forward bias is more than the current in the reverse bias for a ${\text{p}}- {\text{n}}$ junction diode.In the light of the above statement, choose the correct answer from the options given below : [24-Jan-2023 Shift 1]

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Q10JEE Main 2023MCQ4MElectromagnetic Waves

If $\vec{E}$ and ${ \vec{\text{K}}$ represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by : $(\omega -$ angular frequency) : [24-Jan-2023 Shift 1]

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📖 Explanation

Magnetic field vector will be in the direction of $\;{ {\text{K}}}^{∧} \times \;{ {\text{E}}}^{∧}$ magnitude of $B=\;\frac{E}{C}=\;\frac{K}{\omega } E$ Or $\;\; { \vec{\text{B}}=\;\frac{1}{\omega }({ \vec{\text{K}} \times { \vec{\text{E}})$

Q11JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

A circular loop of radius $r$ is carrying current I A. The ratio of magnetic field at the centre of circular loop and at a distance $r$ from the center of the loop on its axis is : [24-Jan-2023 Shift 1]

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📖 Explanation

Magnetic field due to current carrying circular loop on its axis is given as $\;{\mu _0 {\text{ir}}^2}/{2 ( {\text{r}}^{2+} {\text{x}}^2) ^{3 / 2}}$ At centre, $x=0, B_1=\;{\mu _0 {\text{i}}}/{2 {\text{r}}}$ $\;\;\text{ At }\; x=r, B_2=\;{\mu _0 {\text{i}}}/{2 \times 2 {\sqrt{2}} r}$ $\;\;\frac{B_1}{B_2}=2 {\sqrt{2}}$

Q12JEE Main 2023MCQ4MWaves

A travelling wave is described by the equation ${\text{y}}( {\text{x}}, {\text{t}})=[0.05 sin \;(8 {\text{x}}-4 {\text{t}}] {\text{m}}$ The velocity of the wave is : [all the quantities are in SI unit] [24-Jan-2023 Shift 1]

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📖 Explanation

From the given equation ${\text{k}}=8 {\text{m}}^{-1}$ and $\omega =4 {\text{rad}} / {\ s}$ Velocity of wave $=\;{\omega }/{ {\text{k}}}$ ${\text{v}}=\;\frac{4}{8}=0.5 {\text{m}} / {\ s}$

Q13JEE Main 2023MCQ4MCurrent Electricity

As shown in the figure, a network of resistors is connected to a battery of $24 {\text{V}}$ with an internal resistance of $3 Ω$ . The currents through the resistors ${\text{R}}_{4}$ and ${\text{R}}_{5}$ are ${\text{I}}_{4}$ and ${\text{I}}_{5}$ respectively. The values of ${\text{I}}_{4}$ and ${\text{I}}_{5}$ are : $(24 {\text{V}}, 3 Ω)$ [24-Jan-2023 Shift 1]

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📖 Explanation

Equivalent resistance of circuit ${\text{R}}_{ {\text{eq}}}\;=3+1+2+4+2$ $\;=12 Ω$ Current through battery ${\text{i}}=\;\frac{24}{12}=2 {\text{A}}$ $\; {\text{I}}_4=\;{ {\text{R}}_5}/{ {\text{R}}_4+ {\text{R}}_5} \times 2=\;\frac{5}{20+5} \times 2=\;\frac{2}{5} {\text{A}}$ $\; {\text{I}}_5=2-\;\frac{2}{5}=\;\frac{8}{5} {\text{A}}$

Q14JEE Main 2023MCQ4MWave Optics

Given below are two statements :Statement I : If the Brewster's angle for the light propagating from air to glass is $\theta_{ {\text{B}}}$ , then Brewster's angle for the light propagating from glass to air is $\;\frac{\pi}{2}-\theta_{ {\text{B}}}$ .Statement II : The Brewster's angle for the light propagating from glass to air is $\tan ^{-1} (\mu _{ {\text{g}}})$ where $\mu _{ {\text{g}}}$ is the refractive index of glass.In the light of the above statements, choose the correct answer from the options given below : [24-Jan-2023 Shift 1]

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📖 Explanation

$\;\mu _{ {\text{a}}} sin \; {\text{i}}_1=\mu _{ {\text{g}}} sin \; (90- {\text{i}}_1)$ $\;\tan {\text{i}}_1=\;{\mu _{ {\text{g}}}}/{\mu _{ {\text{a}}}}$ When going from glass to air $\tan i_2=\;{\mu _{ {\text{a}}}}/{\mu _{ {\text{g}}}}=\cot i_1$ Hence ${\text{i}}_2=\;\frac{\pi}{2}- {\text{i}}_1$

Q15JEE Main 2023MCQ4MElectrostatics

If two charges ${\text{q}}_{1}$ and ${\text{q}}_{2}$ are separated with distance ' $d$ ' and placed in a medium of dielectric constant ${\text{K}}$ . What will be the equivalent distance between charges in air for the same electrostatic force ? [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{F}}=\;\frac{1}{ (4 \pi \epsilon_0) } \;{ {\text{q}}_1 {\text{q}}_2}/{ {\text{kd}}^2} \;\text{ (\in medium) }\;$ $\; {\text{F}}_{\;\text{Air }\;}=\;\frac{1}{4 \pi \epsilon_0} \;{ {\text{q}}_1 {\text{q}}_2}/{ {\text{d}}^{' 2}}$ $\; {\text{F}}= {\text{F}}_{ {\text{Air}}}$ $\;\;{ {\text{q}}_1 {\text{q}}_2}/{4 \pi \epsilon_0 {\text{kd}}^2}=\;{ {\text{q}}_1 {\text{q}}_2}/{4 \pi \epsilon_0 {\text{d}}^{' 2}}$ $\; {\text{d}}^{'}= {\text{d}} \sqrt{ {\text{k}}}$

Q16JEE Main 2023MCQ4MAtoms and Nuclei

Consider the following radioactive decay process ${ {}_{84}^{218} A\;{\longrightarrow }^{\alpha} A_{1} {\longrightarrow }^{\beta^{-}}A_{2} \;{\longrightarrow }^{\gamma } A_{3} \;{\longrightarrow }^{\alpha} A_{4}{\longrightarrow } ^{ \beta^{+}}A_{5} \;{\longrightarrow }^{\gamma } A_{6}$ The mass number and the atomic number ${\text{A}}_{6}$ are given by : [24-Jan-2023 Shift 1]

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📖 Explanation

$\;{ }_{84}^{218} {\text{A}} {\to }^{\alpha}{ }_{82}^{214} {\text{A}}_1 {\to }^{\beta^{-}}{ }_{83}^{214} {\text{A}}_2 {\to }^{\gamma }{ }_{83}^{214} {\text{A}}_3$ $\;{ }_{83}^{214} {\text{A}}_3 {\to }^{\alpha}{ }_{81}^{210} {\text{A}}_4 {\to }^{\beta^{+}}{ }_{80}^{210} {\text{A}}_5 {\to }^{\gamma }{ }_{80}^{210} {\text{A}}_6$

Q17JEE Main 2023MCQ4MThermodynamics

Given below are two statements :Statements I : The temperature of a gas is $-73^{\circ} {\text{C}}$ . When the gas is heated to $527^{\circ} {\text{C}}$ , the root mean square speed of the molecules is doubled.Statement II : The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules.In the light of the above statements, choose the correct answer from the options given below : [24-Jan-2023 Shift 1]

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📖 Explanation

Statement-I $\; {\text{T}}_1=-73^{\circ} {\text{C}}=200 {\text{K}}$ $\; {\text{T}}_2=527^{\circ} {\text{C}}=800 {\text{K}}$ $\;\;{ {\text{V}}_1}/{ {\text{V}}_2}=\;{\sqrt{{3 {\text{RT}}_1}/{ {\text{M}}}}}/{\sqrt{\;{3 {\text{RT}}_2}/{ {\text{M}}}}}=\sqrt{\;{ {\text{T}}_1}/{ {\text{T}}_2}}$ $\;=\sqrt{\;\frac{200}{800}}=\;\frac{1}{2}$ $\; {\text{V}}_2=2 {\text{V}}_1 \;\text{ (True) }\;$ Statement-II ${\text{PV}}= {\text{nRT}}$ Translational ${\text{KE}}=\;\frac{3}{2} {\text{nRT}}$ (False)

Q18JEE Main 2023MCQ4MMotion in a Plane

The maximum vertical height to which a man can throw a ball is $136 {\text{m}}$ . The maximum horizontal distance upto which he can throw the same ball is [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{H}}_{\max }=\;{ {\text{v}}^2}/{2 {\text{g}}}=136 {\text{m}}$ $\; {\text{R}}_{\max }=\;{ {\text{v}}^2}/{ {\text{g}}}=2 {\text{H}}_{\;\text{max }\;}$ $\;=2(136)$ $\;=272 {\text{m}}$

Q19JEE Main 2023MCQ4MElectromagnetic Induction

A conducting loop of radius $\;{10}/{{\sqrt{\pi}}} {\text{cm}}$ is placed perpendicular to a uniform magnetic field of $0.5 {\text{T}}$ . The magnetic field is decreased to zero in $0.5 {\ s}$ at a steady rate. The induced emf in the circular loop at $0.25 {\ s}$ is: [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{EMF}}=\;{ {\text{d}} \varphi }/{ {\text{dt}}}=\;{ {\text{BA}}-0}/{ {\text{t}}}$ $\; {\text{A}}=\pi {\text{r}}^2=\pi (\;\frac{0.1^2}{\pi}) =0.01$ $\; {\text{B}}=0.5$ $\; {\text{EMF}}=\;\frac{(0.5)(0.01)}{0.5}=0.01 {\text{V}}=10 {\text{mV}}$

Q20JEE Main 2023MCQ4MUnits and Measurements

Match List I with List II LIST I LIST II A. Planck's constant (h) I. $[ {\text{M}}^{1} {\text{L}}^{2} {\text{T}}^{-2}]$ B. Stopping potential (Vs) II. $[ {\text{M}}^{1} {\text{L}}^{1} {\text{T}}^{-1}]$ C. Work function (Ø) III. $[ {\text{M}}^{1} {\text{L}}^{2} {\text{T}}^{-1}]$ D. Momentum (p) IV. $[ {\text{M}}^{1} {\text{L}}^{2} {\text{T}}^{-3} {\text{A}}^{-1}]$ [24-Jan-2023 Shift 1]

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📖 Explanation

(A) Planck's constant ${\text{h}} v\;= {\text{E}}$ ${\text{h}}\;=\;{ {\text{E}}}/{v}=\;{ {\text{M}}^1 {\text{L}}^2 {\text{T}}^{-2}}/{ {\text{T}}^{-1}}= {\text{M}}^1 {\text{L}}^2 {\text{T}}^{-1}$ (B) ${\text{E}}= {\text{qV}}$ $V=\;\frac{E}{q}=\;\frac{M^1 L^2 T^{-2}}{A^1 T^1}=M^1 L^2 T^{-3} A^{-1}(I V)$ (C) $\varphi ($ work function $)=$ energy $= {\text{M}}^1 {\text{L}}^2 {\text{T}}^{-2}$ (D) Momentum (p) = F.t $\;= {\text{M}}^1 {\text{L}}^1 {\text{T}}^{-2} {\text{T}}^1$ $\;= {\text{M}}^1 {\text{L}}^1 {\text{T}}^{-1}$

Q21JEE Main 2023NAT4MWork, Power and Energy

A spherical body of mass $2 {\text{kg}}$ starting from rest acquires a kinetic energy of $10000 {\text{J}}$ at the end of $5^{\;\text{th }\;}$ second. The force acted on the body is ___ ${\text{N}}$ . [24-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\frac{1}{2} \times 2 \times v^2=10000$ $\;\Rightarrow {\text{v}}^2=10000$ $\;\Rightarrow {\text{v}}=100 {\text{m}} / {\ s}$ $\;\Rightarrow {\text{v}}= {\text{at}}= {\text{a}} \times 5=100$ $\;\Rightarrow {\text{a}}=20 {\text{m}} / {\ s}^2$ $\; {\text{F}}= {\text{ma}}=2 \times 20=40 {\text{N}}$ $\;$

Q22JEE Main 2023NAT4MOscillations

A block of mass $2 {\text{kg}}$ is attached with two identical springs of spring constant $20 {\text{N}} / {\text{m}}$ each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is $\;{\pi}/{\sqrt{ {\text{x}}}}$ in SI unit. The value of ${\text{x}}$ is_____ [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{F}}=-2 {\text{kx}}, {\text{a}}=-\;{2 {\text{kx}}}/{ {\text{m}}}, \omega =\sqrt{\;{2 {\text{k}}}/{ {\text{m}}}}=\sqrt{\;\frac{2 \times 20}{2}}$ $\;={\sqrt{20}} {\text{rad}} / {\ s}$ $\; {\text{T}}=\;\frac{2 \pi}{\omega }=\;{2 \pi}/{{\sqrt{20}}}=\;{\pi}/{{\sqrt{5}}}$ $\; {\text{x}}=5$

Q23JEE Main 2023NAT4MProperties of Matter

A hole is drilled in a metal sheet. At $27^{\circ} {\text{C}}$ , the diameter of hole is $5 {\text{cm}}$ . When the sheet is heated to $177^{\circ} {\text{C}}$ , the change in the diameter of hole is ${\text{d}} \times$ $10^{-3} {\text{cm}}$ . The value of ${\text{d}}$ will be __________if coefficient of linear expansion of the metal is $1.6 \times$ $10^{-5} /{ }^{\circ} {\text{C}}$ . [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{d}}_0 \;\text{ at }\; 27^{\circ} {\text{C}} & {\text{d}}_1 \;\text{ at }\; 177^{\circ} {\text{C}}$ $\; {\text{d}}_1= {\text{d}}_0(1+\alpha \Delta {\text{T}})$ $\; {\text{d}}_1- {\text{d}}_0=5 \times 1.6 \times 10^{-5} \times 150 {\text{cm}}$ $\;=12 \times 10^{-3} {\text{cm}}$

Q24JEE Main 2023NAT4MAlternating Current

In the circuit shown in the figure, the ratio of the quality factor and the band width is____s. [24-Jan-2023 Shift 1]

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📖 Explanation

$\;\Delta \omega =\;{ {\text{R}}}/{ {\text{L}}}$ $\; {\text{Q}}=\;\frac{\omega _0}{\Delta \omega }=\omega _0 \;{ {\text{L}}}/{ {\text{R}}}$ $\;\omega _0=\;{1}/{\sqrt{3 \times 27 \times 10^{-6}}}=\;\frac{1}{9 \times 10^{-3}}$ $\;\;{ {\text{Q}}}/{\Delta \omega }=\;{\omega _0 { {\text{L}}}/{ {\text{R}}}}/{{ {\text{R}}}/{ {\text{L}}}}=\omega _0 \;{ {\text{L}}^2}/{ {\text{R}}^2}=\sqrt{\;{1}/{ {\text{LC}}} \;{ {\text{L}}^2}/{ {\text{R}}^2}}$ $\;=\;\frac{1}{9 \times 10^{-3}} \times \;\frac{9}{100}=10 {\ s}$

Q25JEE Main 2023NAT4MCurrent Electricity

A hollow cylindrical conductor has length of $3.14$ ${\text{m}}$ , while its inner and outer diameters are $4 {\text{mm}}$ and $8 {\text{mm}}$ respectively. The resistance of the conductor is ${\text{n}} \times 10^{-3} Ω$ .If the resistivity of the material is $2.4 \times 10^{-8} Ω {\text{m}}$ . The value of ${\text{n}}$ is___ [24-Jan-2023 Shift 1]

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📖 Explanation

${\text{R}}=\rho \;{ℓ}/{ {\text{A}}}$ , the cross-sectional area is $\pi ( {\text{b}}^{2-} {\text{a}}^2)$ $\; {\text{R}}=\rho \;{ℓ}/{\pi ( {\text{b}}^{2-} {\text{a}}^2) }=\;\frac{2.4 \times 10^{-8} \times 3.14}{3.14 \times (4^{2-}2^2) \times 10^{-6}}$ $\;=2 \times 10^{-3} Ω$ $\;\to {\text{n}}=2$

Q26JEE Main 2023NAT4MRay Optics and Optical Instruments

As shown in the figure, a combination of a thin plano concave lens and a thin plano convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is $30 {\text{cm}}$ and refraction index of the material for both the lenses is $1.75$ . Both the lenses are placed at distance of 40 ${\text{cm}}$ from each other. Due to the combination, the image of the object is formed at distance ${\text{x}}=$ ___ ${\text{cm}}$ , from concave lens. [24-Jan-2023 Shift 1]

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📖 Explanation

$\;\;{1}/{ {\text{f}}_1}=(1.75-1) (-\;\frac{1}{30})$ $\;\Rightarrow {\text{f}}_1=-40 {\text{cm}}$ $\;\;{1}/{ {\text{f}}_2}=(1.75-1) (\;\frac{1}{30}) \Rightarrow {\text{f}}_2=40 {\text{cm}}$ Image from ${\text{L}}_1$ will be virtual and on the left of ${\text{L}}_1$ at focal length $40 {\text{cm}}$ . So the object for $L_2$ will be $80 {\text{cm}}$ from ${\text{L}}_2$ which is $2 {\text{f}}$ . Final image is formed at $80 {\text{cm}}$ from ${\text{L}}_2$ on the right.So ${\text{x}}=120$

Q27JEE Main 2023NAT4MRotational Motion

Solid sphere $A$ is rotating about an axis PQ. If the radius of the sphere is $5 {\text{cm}}$ then its radius of gyration about PQ will be ${\sqrt{x}} {\text{cm}}$ . The value of $x$ is___ [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{I}}_{ {\text{cm}}}=\;\frac{2}{5} {\text{MR}}^2$ $\; {\text{I}}_{ {\text{PQ}}}= {\text{I}}_{ {\text{cm}}}+ {\text{md}}^2$ $\; {\text{I}}_{ {\text{PQ}}}=\;\frac{2}{5} {\text{mR}}^{2+} {\text{m}}(10 {\text{cm}})^2$ For radius of gyration $\; {\text{I}}_{ {\text{PQ}}}= {\text{mk}}^2$ $\; {\text{k}}^2=\;\frac{2}{5} {\text{R}}^{2+}(10 {\text{cm}})^2$ $\;=\;\frac{2}{5}(5)^{2+}100$ $\;=10+100=110$ $\; {\text{k}}={\sqrt{110}} {\text{cm}}$ $\; {\text{x}}=110$

Q28JEE Main 2023NAT4MUnits and Measurements

Vectors $a {i}^{∧}+b {j}^{∧}+{k}^{∧}$ and $2 {i}^{∧}-3 {j}^{∧}+4 {k}^{∧}$ are perpendicular to each other when $3 a+2 b=7$ , the ratio of a to $b$ is $\;\frac{x}{2}$ . The value of $x$ is____ [24-Jan-2023 Shift 1]

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📖 Explanation

For two perpendicular vectors $\;(a {i}^{\wedge }+b {j}^{\wedge }+{k}^{\wedge }) \cdot (2 {i}^{\wedge }-3 {j}^{\wedge }+4 {k}^{\wedge })=0$ $\;2 a-3 b+4=0$ On solving, $2 a-3 b=-4$ Also given $3 a+2 b=7$ We get ${\text{a}}=1, {\text{b}}=2$ $\;\;{ {\text{a}}}/{ {\text{b}}}=\;{ {\text{x}}}/{2} \Rightarrow {\text{x}}=\;{2 {\text{a}}}/{ {\text{b}}}=\;\frac{2 \times 1}{2}$ $\Rightarrow {\text{x}}=1$

Q29JEE Main 2023NAT4MAtoms and Nuclei

Assume that protons and neutrons have equal masses. Mass of a nucleon is $1.6 \times 10^{-27} {\text{kg}}$ and radius of nucleus is $1.5 \times 10^{-15} {\text{A}}^{1 / 3} {\text{m}}$ . The approximate ratio of the nuclear density and water density is ${\text{n}} \times 10^{13}$ . The value of ${\text{n}}$ is____ [24-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\text{ density of nuclei }\;=\;\frac{\;\text{ mass of nuclei }\;}{\;\text{ volume of nuclei }\;}$ $\;\rho =\;{1.6 \times 10^{-27} {\text{A}}}/{\frac{4}{3} \pi (1.5 \times 10^{-15}) ^3 {\text{A}}}$ $\;=\;\frac{1.6 \times 10^{-27}}{14.14 \times 10^{-45}}=0.113 \times 10^{18}$ $\;\rho _{ {\text{w}}}=10^3$ $\;\;\frac{\rho }{\rho _w}=11.31 \times 10^{13}$ $\;$

Q30JEE Main 2023NAT4MElectrostatics

A stream of a positively charged particles having $\;{ {\text{q}}}/{ {\text{m}}}=2 \times 10^{11} \;{ {\text{C}}}/{ {\text{kg}}}$ and velocity ${ \vec{\text{v}}_{0}=3 \times 10^{7} { {\text{i}}^{∧} {\text{m}} / {\ s}$ is deflected by an electric field $1.8 { {\text{j}}^{∧} {\text{kV}} / {\text{m}}$ . The electric field exists in a region of $10 {\text{cm}}$ along ${\text{x}}$ direction. Due to the electric field, the deflection of the charge particles in the $y$ direction is ____ ${\text{mm}}$ . [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{a}}=\;{ {\text{F}}}/{ {\text{m}}}=\;{ {\text{qE}}}/{ {\text{m}}}= (2 \times 10^{11}) (1.8 \times 10^3)$ $\;=3.6 \times 10^{14} {\text{m}} / {\ s}^2$ $\;\;\text{ Time to cross plates }\;=\;{ {\text{d}}}/{ {\text{v}}}$ $\; {\text{t}}=\;\frac{0.10}{3 \times 10^7}$ $\; {\text{y}}=\;\frac{1}{2} {\text{at}}^2=\;\frac{1}{2} (3.6 \times 10^{14}) (\;\frac{0.01}{9 \times 10^{14}})$ $\;=0.2 \times 0.01$ $\;=0.002 {\text{m}}$ $\;=2 {\text{mm}}$

Chemistry30 questions

Q31JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

Compound $( {\text{X}})$ undergoes following sequence of reactions to give the Lactone $( {\text{Y}})$ . [24-Jan-2023 Shift 1]

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Q32JEE Main 2023MCQ4MHaloalkanes and Haloarenes

Assertion A : Hydrolysis of an alkyl chloride is a slow reaction but in the presence of NaI, the rate of the hydrolysis increases.Reason R: ${\text{I}}^{-}$ is a good nucleophile as well as a good leaving group.In the light of the above statements, choose the correct answer from the options given below. [24-Jan-2023 Shift 1]

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📖 Explanation

The rate of hydrolysis of alkyl chloride improves because of better Nucleophilicity of ${\text{I}}^{-}$ .

Q33JEE Main 2023MCQ4MChemical Bonding and Molecular Structure

Order of Covalent bond;A. ${\text{KF}}> {\text{KI}} ; {\text{LiF}}> {\text{KF}}$ B. ${\text{KF}}< {\text{KI}} ; {\text{LiF}}> {\text{KF}}$ C. ${\text{SnCl}}_{4}> {\text{SnCl}}_{2} ; {\text{CuCl}}> {\text{NaCl}}$ D. ${\text{LiF}}> {\text{KF}} ; {\text{CuCl}}< {\text{NaCl}}$ E. ${\text{KF}}< {\text{KI}} ; {\text{CuCl}}> {\text{NaCl}}$ [24-Jan-2023 Shift 1]

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📖 Explanation

According to Fajan's Rule,A. ${\text{KF}} > {\text{KI}}-$ False; ${\text{LiF}} > {\text{KF}}$ - TrueB. ${\text{KF}} < {\text{KI}}$ - True; ${\text{LiF}} > {\text{KF}}$ - TrueC. ${\text{SnCl}}_4 > {\text{SnCl}}_2-$ True; ${\text{CuCl}} > {\text{NaCl}}-$ TrueD. ${\text{LiF}}> {\text{KF}}$ - True; ${\text{CuCl}} < {\text{NaCl}}-$ FalseE. ${\text{KF}} < {\text{KI}}-$ True; ${\text{CuCl}} > {\text{NaCl}}-$ True

Q34JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

Increasing order of stability of the resonance structure is : A. B. C. D. [24-Jan-2023 Shift 1]

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📖 Explanation

No option is matching the correct answer.Order should be : ${\text{C}} < {\text{A}} < {\text{B}} < {\text{D}}$

Q35JEE Main 2023MCQ4Md and f Block Elements

The magnetic moment of a transition metal compound has been calculated to be $3.87$ B.M. The metal ion is [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{Cr}}^{+2}:[ {\text{Ar}}], 3 {\text{d}}^4, 4 {\ s}^0 {\text{n}}=4, \mu ={\sqrt{4(4+2)}}={\sqrt{24}}$ $\;=4.89 {\text{BM}}$ $\; {\text{Mn}}^{+2}:[ {\text{Ar}}], 3 {\text{d}}^5, 4 {\ s}^0 {\text{n}}=5, \mu ={\sqrt{5(5+2)}}={\sqrt{35}}$ $\;=5.91 {\text{BM}}$ $\; {\text{V}}^{+2}:[ {\text{Ar}}], 3 {\text{d}}^3, 4 {\ s}^0 {\text{n}}=3, \mu ={\sqrt{3(3+2)}}={\sqrt{15}}$ $\;=3.87 {\text{BM}}$ $\; {\text{Ti}}^{+2}:[ {\text{Ar}}], 3 {\text{d}}^2, 4 {\ s}^0 {\text{n}}=2, \mu ={\sqrt{2(2+2)}}={\sqrt{8}}$ $\;=2.82 {\text{BM}}$

Q36JEE Main 2023MCQ4Md and f Block Elements

Match List I with List II. LIST I LIST II A. Reverberatory furnace I. Pig Iron B. Electrolytic cell II. Aluminum C. Blast furnace III. Silicon D. Zone Refining furnace IV. Copper [24-Jan-2023 Shift 1]

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Q37JEE Main 2023MCQ4MStructure of Atom

It is observed that characteristic X-ray spectra of elements show regularity. When frequency to the power 'n' i.e. $v^{ {\text{n}}}$ of X-rays emitted is plotted against atomic number ' ${\text{Z}}$ ', following graph is obtained. Thevalue of ' $n$ ' is [24-Jan-2023 Shift 1]

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📖 Explanation

According to Henry Moseley ${\sqrt{v}} \alpha {\text{z}}- {\text{b}}$ So ${\text{n}}=\;\frac{1}{2}$

Q38JEE Main 2023MCQ4Mp Block Elements

Which of the Phosphorus oxoacid can create silver mirror from ${\text{AgNO}}_{3}$ solution ? [24-Jan-2023 Shift 1]

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📖 Explanation

Oxyacid having ${\text{P}}- {\text{H}}$ bond can reduce ${\text{AgNO}}_3$ toAg.

Q39JEE Main 2023MCQ4MCoordination Compounds

The primary and secondary valencies of cobalt respectively in $[ {\text{Co}} ( {\text{NH}}_{3}) _{5} {\text{Cl}}] {\text{Cl}}_{2}$ are : [24-Jan-2023 Shift 1]

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📖 Explanation

$[ {\text{Co}} ( {\text{NH}}_3) _5 {\text{Cl}}] {\text{Cl}}_2$ Oxidation number of ${\text{Co}}$ is $+3$ .So primary valency is 3 .It is an octahedral complex so secondary valency 6 or Co-ordination number 6 .

Q40JEE Main 2023MCQ4MSalt Analysis

An ammoniacal metal salt solution gives a brilliant red precipitate on addition of dimethylglyoxime. The metal ion is : [24-Jan-2023 Shift 1]

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📖 Explanation

${\text{Ni}}^{+2}+2 {\text{DMG}} {\to }^{ {\text{NH}}_3( {\text{aq}})} [ {\text{Ni}}( {\text{DMG}})_2]$

Q41JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

' ${\text{R}}$ ' formed in the following sequence of reaction is: [24-Jan-2023 Shift 1]

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Q42JEE Main 2023MCQ4MSome Basic Concepts of Chemistry

Choose the correct answer from the options given below : LIST I LIST II A. Chlorophyll I. ${\text{Na}}_2 {\text{CO}}_3$ B. Soda ash II. ${\text{CaSO}}_4$ C. Dentistry, Ornamental work III ${\text{Mg}}^{2+}$ D. Used in white washing IV. ${\text{Ca}}( {\text{OH}})_2$ [24-Jan-2023 Shift 1]

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📖 Explanation

Chlorophyll : ${\text{Mg}}^{+2}$ complexSoda ash : ${\text{Na}}_2 {\text{CO}}_3$ Dentistry, Ornamental work: ${\text{CaSO}}_4$ Used in white washing : ${\text{Ca}}( {\text{OH}})_2$

Q43JEE Main 2023MCQ4MSolutions

Statement I : For colloidal particles, the values of colligative properties are of small order as compared to values shown by true solutions at same concentration.Statement II : For colloidal particles, the potential difference between the fixed layer and the diffused layer of same charges is called the electrokinetic potential or zeta potential.In the light of the above statements, choose the correct answer from the options given below. [24-Jan-2023 Shift 1]

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Q44JEE Main 2023MCQ4Mp Block Elements

Reaction of ${\text{BeO}}$ with ammonia and hydrogen fluoride gives 'A' which on thermal decomposition gives ${\text{BeF}}_{2}$ and ${\text{NH}}_{4} {\text{F}}$ . What is 'A' ? [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{BeO}}+2 {\text{NH}}_3+4 {\text{HF}} \to ( {\text{NH}}_4) _2 {\text{BeF}}_4+ {\text{H}}_2 {\text{O}}$ $\; ( {\text{NH}}_4) _2 {\text{BeF}}_4 {\to }^{\Delta } {\text{BeF}}_2+ {\text{NH}}_4 {\text{F}}$

Q45JEE Main 2023MCQ4MAlcohols, Phenols and Ethers

'A' and 'B' formed in the following set of reactions are : [24-Jan-2023 Shift 1]

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Q46JEE Main 2023MCQ4MHydrocarbons

In the following given reaction ' ${\text{A}}$ ' is [24-Jan-2023 Shift 1]

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Q47JEE Main 2023MCQ4MStates of Matter

Decreasing order of the hydrogen bonding in following forms of water is correctly represented byA. Liquid waterB. IceC. Impure water [24-Jan-2023 Shift 1]

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📖 Explanation

Ice $>$ Liquid water $>$ Impure waterDue to impurity extent of ${\text{H}}$ -Bonding decreases.

Q48JEE Main 2023MCQ4MBiomolecules

Given below are two statements:Statement I : Noradrenaline is a neurotransmitter.Statement II : Low level of noradrenaline is not the cause of depression in human.In the light of the above statements, choose the correct answer from the options given below [24-Jan-2023 Shift 1]

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Q49JEE Main 2023MCQ4MSolutions

In the depression of freezing point experimentA. Vapour pressure of the solution is less than that of pure solventB. Vapour pressure of the solution is more than that of pure solventC. Only solute molecules solidify at the freezing pointD. Only solvent molecules solidify at the freezing point [24-Jan-2023 Shift 1]

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Q50JEE Main 2023MCQ4MHaloalkanes and Haloarenes

Which of the following is true about freons? [24-Jan-2023 Shift 1]

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Q51JEE Main 2023NAT4MIonic Equilibrium

The dissociation constant of acetic is ${\text{x}} \times 10^{-5}$ . When $25 {\text{mL}}$ of $0.2 {\text{M}} \; {\text{CH}}_{3} {\text{COONa}}$ solution is mixed with $25 {\text{mL}}$ of $0.02 \; {\text{M}} \; {\text{CH}}_{3} {\text{COOH}}$ solution, the ${\text{pH}}$ of the resultant solution is found to be equal to 5 . The value of $x$ is___ [24-Jan-2023 Shift 1]

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📖 Explanation

Buffer of ${\text{HOAc}}$ and ${\text{NaOAc}}$ $\; {\text{pH}}= {\text{pKa}}+\log \;\frac{0.1}{0.01}$ $\;5= {\text{pKa}}+1$ $\; {\text{pKa}}=4$ $\; {\text{Ka}}=10^{-4}$ $\; {\text{x}}=10$

Q52JEE Main 2023NAT4MSome Basic Concepts of Chemistry

$5 {\text{g}}$ of ${\text{NaOH}}$ was dissolved in deionized water to prepare a $450 {\text{mL}}$ stock solution. What volume (in ${\text{mL}}$ ) of this solution would be required to prepare $500 {\text{mL}}$ of $0.1 {\text{M}}$ solution?Given : Molar Mass of ${\text{Na}}, {\text{O}}$ and ${\text{H}}$ is 23, 16 and 1 ${\text{g}} {\text{mol}}^{-1}$ respectively [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{M}}=\;\frac{5}{40} \times \;\frac{1000}{450}$ $\; {\text{M}}_1 {\text{V}}_1= {\text{M}}_2 {\text{V}}_2$ $\; (\;\frac{5}{40} \times \;\frac{1000}{450}) \times {\text{V}}_1=0.1 \times 500$ $\; {\text{V}}_1=180$

Q53JEE Main 2023NAT4MStructure of Atom

If wavelength of the first line of the Paschen series of hydrogen atom is $720 {\text{nm}}$ , then the wavelength of the second line of this series is _____ ${\text{nm}}$ (Nearest integer) [24-Jan-2023 Shift 1]

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📖 Explanation

$\;\;{1}/{ (\lambda _1) _{ {\text{P}}}}= {\text{R}}_{ {\text{H}}} {\text{Z}}^2 (\;\frac{1}{9}-\;\frac{1}{16})$ $\;\;{1}/{ (\lambda _2) _{ {\text{P}}}}= {\text{R}}_{ {\text{H}}} {\text{Z}}^2 (\;\frac{1}{9}-\;\frac{1}{25})$ $\;\;{ (\lambda _2) _{ {\text{P}}}}/{ (\lambda _1) _{ {\text{P}}}}=\;\frac{\;\frac{7}{16 \times 9}}{\frac{16}{25 \times 9}}=\;\frac{25 \times 7}{16 \times 16}$ $\; (\lambda _2) _{ {\text{P}}}=\;\frac{25 \times 7}{16 \times 16} \times 720$ $\; (\lambda _2) _{ {\text{P}}}=492 {\text{nm}}$

Q54JEE Main 2023NAT4MChemical Kinetics

The number of correct statement/s from the following is___A. Larger the activation energy, smaller is the value of the rate constant.B. The higher is the activation energy, higher is the value of the temperature coefficient.C. At lower temperatures, increase in temperature causes more change in the value of ${\text{k}}$ than at higher temperature.D. A plot of $\ln {\text{k}}$ vs $\;{1}/{ {\text{T}}}$ is a straight line with slope equal to $-\;\frac{E a}{R}$ [24-Jan-2023 Shift 1]

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📖 Explanation

$\text{bo} {A}: {\text{k}}=A e^{-\;{ {\text{Ea}}}/{R T}}$ As Ea increases ${\text{k}}$ decreasesB : Temperature coefficient $=\;{ {\text{k}}_{ {\text{T}}+10}}/{ {\text{k}}_{ {\text{T}}}}$ Option (C) is wrong. $\Delta {\text{k}}$ may be greater or lesser depending on temperature. ${\text{D}}: \ln {\text{k}}=\ln {\text{A}}-\;{ {\text{Ea}}}/{ {\text{RT}}}$

Q55JEE Main 2023NAT4MElectrochemistry

At $298 {\text{K}}$ , a 1 litre solution containing $10 {\text{mmol}}$ of ${\text{Cr}}_{2} {\text{O}}_{7}{ }^{2-}$ and $100 {\text{mmol}}$ of ${\text{Cr}}^{3+}$ shows a pH of $3.0$ .Given : ${\text{Cr}}_{2} {\text{O}}_{7}^{2-} \to {\text{Cr}}^{3+} ; {\text{E}}^{0}=1.330 {\text{V}}$ and $\;{2.303 {\text{RT}}}/{ {\text{F}}}=0.059 {\text{V}}$ The potential for the half cell reaction is ${\text{x}} \times 10^{-3}$ ${\text{V}}$ . The value of $x$ is____ [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{Cr}}_2 {\text{O}}_7^{2-}+14 {\text{H}}^{+}+6 {\text{e}}^{-} \to 2 {\text{Cr}}^{3+}+7 {\text{H}}_2 {\text{O}}$ $\; {\text{E}}=1.33-\;\frac{0.059}{6} \log \;\frac{(0.1)^2}{ (10^{-2}) (10^{-3}) ^{14}}$ $\; {\text{E}}=1.33-\;\frac{0.059}{6} \times 42=0.917$ $\; {\text{E}}=917 \times 10^{-3}$ $\; {\text{x}}=917$

Q56JEE Main 2023NAT4MSolid State

When ${\text{Fe}}_{0.93} {\text{O}}$ is heated in presence of oxygen, it converts to ${\text{Fe}}_{2} {\text{O}}_{3}$ . The number of correct statement/s from the following is___A. The equivalent weight of ${\text{Fe}}_{0.93} {\text{O}}$ is $\text{Molecular weight}/0.79$ . B. The number of moles of ${\text{Fe}}^{2+}$ and ${\text{Fe}}^{3+}$ in 1 mole of ${\text{Fe}}_{0.93} {\text{O}}$ is $0.79$ and $0.14$ respectively.C. ${\text{Fe}}_{0.93} {\text{O}}$ is metal deficient with lattice comprising of cubic closed packed arrangement of ${\text{O}}^{2-}$ ions.D. The % composition of ${\text{Fe}}^{2+}$ and ${\text{Fe}}^{3+}$ in ${\text{Fe}}_{0.93} {\text{O}}$ is $85 \%$ and $15 \%$ respectively. [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{A}}: {\text{Fe}}_{0.93} {\text{O}} \to {\text{Fe}}_2 {\text{O}}_3$ $\; {\text{nf}}= (3-\;\frac{200}{93}) \times 0.93$ $\; {\text{nf}}=0.79$ $\; {\text{B}}: 2 {\text{x}}+(0.93- {\text{x}}) \times 3=2$ $\; {\text{x}}=0.79$ $\; {\text{Fe}}^{2+}=0.79, {\text{Fe}}^{3+}=0.21$ $\; {\text{C}}: {\text{Fact}}^{ {\text{D}}}: \% \% {\text{Fe}}^{2+}=\;\frac{0.79}{0.93} \times 100=85 \% ; {\text{Fe}}^{3+}=15 \%$

Q57JEE Main 2023NAT4MCoordination Compounds

The d-electronic configuration of $[ {\text{CoCl}}_4] ^{2-}$ in tetrahedral crystal field is ${\text{e}}^{ {\text{m}}} {\text{t}}_2{ }^{ {\text{n}}}$ . Sum of ' ${\text{m}}$ ' and 'number of unpaired electrons is_____ [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{Co}}^{2+}: 3 {\text{d}}^7 4 {\ s}^0, {\text{Cl}}^{-}:$ Configuration $e^4 {\text{t}}_2{ }^3: {\text{m}}=4$ Number of unpaired electrons $=3$ So, answer $=7$

Q58JEE Main 2023NAT4MChemical Thermodynamics

For independent process at 300 K. Process $\Delta {\text{H}} / {\text{kJ}} {\text{mol}}^{-1}$ $\Delta {\text{S}} / {\text{J}} {\text{K}}^{-1}$ ${\text{A}}$ $-25$ $-80$ B $-22$ 40 C 25 $-50$ D 22 20 The number of non-spontaneous process from the following is____ [24-Jan-2023 Shift 1]

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📖 Explanation

$\;\Delta {\text{G}}=\Delta {\text{H}}- {\text{T}} \Delta {\text{S}}$ $\; {\text{A}}: \Delta {\text{G}} ( {\text{J}} {\text{mol}}^{-1}) =-25 \times 10^{3+}80 \times 300:- {\text{ve}}$ $\; {\text{B}}: \Delta {\text{G}} ( {\text{J}} {\text{mol}}^{-1}) =-22 \times 10^{3-}40 \times 300:- {\text{ve}}$ $\; {\text{C}}: \Delta {\text{G}} ( {\text{J}} {\text{mol}}^{-1}) =25 \times 10^{3+}300 \times 50:+ {\text{ve}}$ $\; {\text{D}}: \Delta {\text{G}} ( {\text{J}} {\text{mol}}^{-1}) =22 \times 10^{3-}20 \times 300:+ {\text{ve}}$ Processes ${\text{C}}$ and ${\text{D}}$ are non-spontaneous.

Q59JEE Main 2023NAT4MBiomolecules

Uracil is base present in RNA with the following structure. $\%$ of ${\text{N}}$ in uracil is___Given :Molar mass ${\text{N}}=14 {\text{g}} {\text{mol}}^{-1} ; {\text{O}}=16 {\text{g}} {\text{mol}}^{-1} ; {\text{C}}=$ $12 {\text{g}} {\text{mol}}^{-1} ; {\text{H}}=1 {\text{g}} {\text{mol}}^{-1}$ ; [24-Jan-2023 Shift 1]

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📖 Explanation

Mol. Wt of ${\text{C}}_4 {\text{N}}_2 {\text{H}}_4 {\text{O}}_2=112$ $\% {\text{N}}=\;\frac{28}{112} \times 100=25 \%$

Q60JEE Main 2023NAT4MHaloalkanes and Haloarenes

Number of moles of ${\text{AgCl}}$ formed in the following reaction is [24-Jan-2023 Shift 1]

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Mathematics30 questions

Q61JEE Main 2023MCQ4MThree Dimensional Geometry

The distance of the point $(7,-3,-4)$ from the plane passing through the points $(2,-3,1),(-1,1,-2)$ and $(3,-4,2)$ is : [24-Jan-2023 Shift 1]

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📖 Explanation

Equation of Plane is $={\begin{vmatrix}x-2 & y+3 & z-1 \\ -3 & 4 & -3 \\ 4 & -5 & 4\end{vmatrix}}=0$ ${\text{x}}- {\text{z}}-1=0$ Distance of $P(7,-3,-4)$ from Plane is ${\text{d}}= |\;{7+4-1}/{{\sqrt{2}}}|=5 {\sqrt{2}}$

Q62JEE Main 2023MCQ4MLimits, Continuity and Differentiability

$\lim _{t \to 0} (1^{\;\frac{1}{ sin \;^2 t}}+2^{\;\frac{1}{ sin \;^2 t}}+. . . .+n^{\;\frac{1}{ sin \;^2 t}}) ^{ sin \;^2 t}$ is equal to [24-Jan-2023 Shift 1]

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📖 Explanation

$\;\lim _{t \to 0} (1^{{\text{cosec}}^2 t}+2^{{\text{cosec}}^2 t}+. . . . . . . .+n^{{\text{cosec}}^2 t}) ^{ sin \;^2 t}$ $\;=\lim _{t \to 0} n ( (\;\frac{1}{n}) ^{{\text{cosec}}^2 t}+ (\;\frac{2}{n}) ^{{\text{cosec}}^2 t}+. . . . . . . .+1) ^{ sin \;^2 t}$ $\;=n$

Q63JEE Main 2023MCQ4MVector Algebra

Let $\;\; { \vec{\text{u}}={ {\text{i}}^{∧}-{ {\text{j}}^{∧}-2 {\text{k}}, { \vec{\text{v}}=2 { {\text{i}}^{∧}+{ {\text{j}}^{∧}- {\text{k}}, { \vec{\text{v}} \cdot { \vec{\text{w}}=2 \;\;$ and ${ \vec{\text{v}} \times { \vec{\text{w}}={ \vec{\text{u}}+\lambda { \vec{\text{v}}$ . Then ${ \vec{\text{u}} \cdot { \vec{\text{w}}$ is equal to [24-Jan-2023 Shift 1]

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📖 Explanation

$\;{ \vec{\text{u}}=(1,-1,-2), { \vec{\text{v}}=(2,1,-1), { \vec{\text{v}} \cdot { \vec{\text{w}}=2$ $\;{ \vec{\text{v}} \times { \vec{\text{w}}={ \vec{\text{u}}+\lambda { \vec{\text{v}} . . . . . . . . . . . . . . . \cdot (1)$ $\;\;\text{ Taking dot with }\; { \vec{\text{w}} \;\text{ \in }\;(1)$ $\;{ \vec{\text{W}} \cdot ({ \vec{\text{v}} \times { \vec{\text{w}})={ \vec{\text{u}} \cdot { \vec{\text{w}}+\lambda { \vec{\text{v}} \cdot { \vec{\text{w}}$ $\;\Rightarrow 0={ \vec{\text{u}} \cdot { \vec{\text{w}}+2 \lambda$ $\;\;\text{ Taking dot with }\; { \vec{\text{v}} \;\text{ \in }\;(1)$ $\;{ \vec{\text{v}} \cdot ({ \vec{\text{v}} \times { \vec{\text{w}})={ \vec{\text{u}} \cdot { \vec{\text{v}}+\lambda { \vec{\text{v}} \cdot { \vec{\text{v}}$ $\;\Rightarrow 0=(2-1+2)+\lambda (6)$ $\;\lambda =-\;\frac{1}{2}$ $\;\Rightarrow { \vec{\text{u}} \cdot { \vec{\text{w}}=-2 \lambda =1$

Q64JEE Main 2023MCQ4MBinomial Theorem

The value $\sum _{ \ r =0}^{22} { }^{22}{\text{C}}_{\text{r}} { }^{23}{\text{C}}_{\text{r}}$ is [24-Jan-2023 Shift 1]

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📖 Explanation

$sum_{{ {\text{r}}=0}}^{22}{ }^{22} {\text{C}}_{ {\text{r}}} \cdot { }^{23} {\text{C}}_{ {\text{r}}}=sum_{r=0}^{22}{ }^{22} {\text{C}}_{ {\text{r}}} \cdot { }^{23} {\text{C}}_{23- {\text{r}}}$ $={ }^{45} {\text{C}}_{23}$

Q65JEE Main 2023MCQ4MParabola

Let a tangent to the curve $y^2=24 x$ meet the curve $x y=2$ at the points $A$ and $B$ . Then the mid points of such line segments $A B$ lie on a parabola with the [24-Jan-2023 Shift 1]

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📖 Explanation

$\;y^2=24 x$ $\;a=6$ $\;x y=2$ $\;A B \equiv t y=x+6 t^2$ . . . (1) $\; {\text{AB}} \equiv T=S_1$ $\;k x+h y=2 h k . . . (2)$ From (1) and (2) $\;---$ $\;\Rightarrow \;\text{ then locus is }\; y^2=-3 x$ Therefore directrix is $4 x=3$

Q66JEE Main 2023MCQ4MMatrices and Determinants

Let ${\text{N}}$ denote the number that turns up when a fair die is rolled. If the probability that the system of equations $x+y+z=1$ $2 x+N y+2 z=2$ $3 x+3 y+N z=3$ has unique solution is $-$ , then the sum of value of ${\text{k}}$ and all possible values of ${\text{N}}$ is [24-Jan-2023 Shift 1]

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📖 Explanation

$\;x+y+z=1$ $\;2 x+N y+2 z=2$ $\;3 x+3 y+N z=3$ $\;\Delta ={\begin{vmatrix}1 & 1 & 1 \\ 2 & N & 2 \\ 3 & 3 & N\end{vmatrix}}$ $\;=(N-2)(N-3)$ For unique solution $\Delta \ne 0$ So ${\text{N}} \ne 2,3$ $P($ system has unique solution $)=-$ So ${\text{k}}=4$ Therefore sum $=4+1+4+5+6=20$

Q67JEE Main 2023MCQ4MInverse Trigonometric Functions

$\tan ^{-1} (\;{1+{\sqrt{3}}}/{3+{\sqrt{3}}}) +\ sec ^{-1} (\sqrt{\;{8+4 {\sqrt{3}}}/{6+3 {\sqrt{3}}}})$ is equal to [24-Jan-2023 Shift 1]

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📖 Explanation

$\;\tan ^{-1} (\;{1+{\sqrt{3}}}/{3+{\sqrt{3}}}) +\ sec ^{-1} (\sqrt{\;{8+4 {\sqrt{3}}}/{6+3 {\sqrt{3}}}})$ $\;=\tan ^{-1} (\;{1}/{{\sqrt{3}}}) +\ sec ^{-1} (\;{2}/{{\sqrt{3}}}) =\;\frac{\pi}{3}$

Q68JEE Main 2023MCQ4MStraight Lines and Pair of Straight Lines

Let PQR be a triangle. The points ${\text{A}}, {\text{B}}$ and ${\text{C}}$ are on the sides ${\text{QR}}, {\text{RP}}$ and PQ respectively such that $\;{ {\text{QA}}}/{ {\text{AR}}}=\;{ {\text{RB}}}/{ {\text{BP}}}=\;{ {\text{PC}}}/{ {\text{CQ}}}=\;\frac{1}{2}$ . Then $\;{\;\text{ Area }\;(\triangle {\text{PQR}})}/{\;\text{ Area }\;(\triangle {\text{ABC}})}$ is equal to [24-Jan-2023 Shift 1]

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📖 Explanation

Let $P$ is $\vec{0}$ , Q is $\vec{q}$ and $R$ is $\vec{r}$ $A$ is $\;\frac{2 \vec{q}+\vec{r}}{3}, B$ is $\;\frac{2 \vec{r}}{3}$ and $C$ is $\;\frac{\vec{q}}{3}$ Area of $\triangle {\text{PQR}}$ is $=\;\frac{1}{2}|{ \vec{\text{q}} \times { \vec{\text{r}}|$ Area of $\triangle {\text{ABC}}$ is $\;\frac{1}{2}|{ \vec{\text{AB}} \times { \vec{\text{AC}}|$ ${ \vec{\text{AB}}=\;{{ \vec{\text{r}}-2 { \vec{\text{q}}}/{3}, { \vec{\text{AC}}=\;{-{ \vec{\text{r}}-{ \vec{\text{q}}}/{3}$ Area of $\triangle {\text{ABC}}=\;\frac{1}{6}|{ \vec{\text{q}} \times { \vec{\text{r}}|$ $\;{\;\text{ Area }\;(\triangle {\text{PQR}})}/{{\text{Area}}(\triangle {\text{ABC}})}=3$

Q69JEE Main 2023MCQ4MMatrices and Determinants

If ${\text{A}}$ and ${\text{B}}$ are two non-zero ${\text{n}} \times {\text{n}}$ matrics such that $A^{2+}B=A^2 B$ , then [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{A}}^{2+} {\text{B}}= {\text{A}}^2 {\text{B}}$ $\; ( {\text{A}}^{2-} {\text{I}}) ( {\text{B}}- {\text{I}})= {\text{I}}$ $\; {\text{A}}^{2+} {\text{B}}= {\text{A}}^2 {\text{B}}$ $\; {\text{A}}^2( {\text{B}}- {\text{I}})= {\text{B}}$ $\; {\text{A}}^2= {\text{B}}( {\text{B}}- {\text{I}})^{-1}$ $\; {\text{A}}^2= {\text{B}} ( {\text{A}}^{2-} {\text{I}})$ $\; {\text{A}}^2=B A^{2-} {\text{B}}$ $\; {\text{A}}^{2+}B=B A^2$ $\; {\text{A}}^2 {\text{B}}= {\text{BA}}^2$

Q70JEE Main 2023MCQ4MDifferential Equations

Let $y=y(x)$ be the solution of the differential equation ${\text{x}}^3 {\text{dy}}+( {\text{xy}}-1) {\text{dx}}=0, {\text{x}}>0$ , $y (\;\frac{1}{2}) =3-e \cdot$ Then $y(1)$ is equal to [24-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\frac{d y}{d x}=\;\frac{1-x y}{x^3}=\;\frac{1}{x^3}-\;\frac{y}{x^2}$ $\;\;\frac{d y}{d x}+\;\frac{y}{x^2}=\;\frac{1}{x^3}$ $\;\;\text{ If }\;=e^{\int \;\frac{1}{x^2} \text{dx}}=e^{-\;\frac{1}{x}}$ $\;y \cdot e^{-\;\frac{1}{x}}=\int e^{-\;\frac{1}{x}} \cdot \;\frac{1}{x^3} \text{dx} (\;\text{ put }\;-\;\frac{1}{x}=t)$ $\;y \cdot e^{-\;\frac{1}{x}}=-\int e^t \cdot t \text{dt}$ $\;y=\;\frac{1}{x}+1+C e^{\;\frac{1}{x}}$ Where ${\text{C}}$ is constantPut $x=\;\frac{1}{2}$ $\;3- {\text{e}}=2+1+ {\text{Ce}}^2$ $\; {\text{C}}=-\;{1}/{ {\text{e}}}$ $\;y(1)=1$

Q71JEE Main 2023MCQ4MArea Under The Curves

The area enclosed by the curves $y^{2+}4 x=4$ and ${\text{y}}-2 {\text{x}}=2$ is : [24-Jan-2023 Shift 1]

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📖 Explanation

$\;y^{2+}4 x=4$ $\;y^2=-4(x-1)$ $\;A=int_{-4}^{2} (\;\frac{4-y^2}{4}-\;\frac{y-2}{2}) d y=9$

Q72JEE Main 2023MCQ4MMatrices and Determinants

Let $\alpha$ be a root of the equation $(a-c) x^{2+}(b-a) x+(c-b)=0$ where $a, b, c$ are distinct real numbers such that the matrix ${\begin{bmatrix}\alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c\end{bmatrix}\$ } $is singular. Then the value of$ ;\frac{(a-c)^2}{(b-a)(c-b)}+;\frac{(b-a)^2}{(a-c)(c-b)}+;\frac{(c-b)^2}{(a-c)(b-a)}$ is [24-Jan-2023 Shift 1]

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📖 Explanation

$\;\Delta =0={\begin{vmatrix}\alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ {\text{a}} & {\text{b}} & {\text{c}}\end{vmatrix}}$ $\;\Rightarrow \alpha^2( {\text{c}}- {\text{b}})-\alpha( {\text{c}}- {\text{a}})+( {\text{b}}- {\text{a}})=0$ It is singular when $\alpha=1$ $\;\;\frac{(a-c)^2}{(b-a)(c-b)}+\;\frac{(b-a)^2}{(a-c)(c-b)}+\;\frac{(c-b)^2}{(a-c)(b-a)}$ $\;\;\frac{(a-b)^{3+}(b-c)^{3+}(c-a)^3}{(a-b)(b-c)(c-a)}$ $\;=3 \;\frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=3$

Q73JEE Main 2023MCQ4MThree Dimensional Geometry

The distance of the point $(-1,9,-16)$ from the plane $2 x+3 y-z=5$ measured parallel to the line $\;{ {\text{x}}+4}/{3}=\;\frac{2-y}{4}=\;\frac{z-3}{12}$ is [24-Jan-2023 Shift 1]

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📖 Explanation

Equation of line $\;{ {\text{x}}+1}/{3}=\;{ {\text{y}}-9}/{-4}=\;{ {\text{z}}+16}/{12}$ G.P on line $(3 \lambda -1,-4 \lambda +9,12 \lambda -16)$ point of intersection of line $\&$ plane $6 \lambda -2-12 \lambda +27-12 \lambda +16=5$ $\lambda =2$ Point $(5,1,8)$ Distance $={\sqrt{36+64+576}}=26$

Q74JEE Main 2023MCQ4MLogarithms

For three positive integers ${\text{p}}, {\text{q}}, {\text{r}}, {\text{x}}^{ {\text{pq}} {\text{q}}^2}= {\text{y}}^{ {\text{qr}}}= {\text{z}}^{ {\text{p}}^2 {\text{r}}}$ and $r=p q+1$ such that $3,3 \log _y x, 3 \log _z y, 7 \log _x z$ are in A.P. with common difference $\;\frac{1}{2}$ . Then ${\text{r}}- {\text{p}}- {\text{q}}$ is equal to [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{pq}}^2=\log _{ {\text{x}}} \lambda$ $\; {\text{qr}}=\log _{ {\text{y}}} \lambda$ $\; {\text{p}}^2 {\text{r}}=\log _{ {\text{z}}} \lambda$ $\;\log _{ {\text{y}}} {\text{x}}=\;{ {\text{qr}}}/{ {\text{pq}}^2}=\;{ {\text{r}}}/{ {\text{pq}}} . . . . (1)$ $\;\log _{ {\text{x}}} {\text{z}}=\;{ {\text{pq}}^2}/{ {\text{p}}^2 {\text{r}}}=\;{ {\text{q}}^2}/{ {\text{pr}}} . . . .$ . . . . . . (2) $\;\log _{ {\text{z}}} {\text{y}}=\;{ {\text{p}}^2 {\text{r}}}/{ {\text{qr}}}=\;{ {\text{p}}^2}/{ {\text{q}}} . . . . . . (3)$ $\;3, \;{3 {\text{r}}}/{ {\text{pq}}}, \;{3 {\text{p}}^2}/{ {\text{q}}}, \;{7 {\text{q}}^2}/{ {\text{pr}}} \;\text{ \in A.P }\;$ $\;\;{3 {\text{r}}}/{ {\text{pq}}}-3=\;\frac{1}{2}$ $\; {\text{r}}=\;\frac{7}{6} {\text{pq}}$ . . . . . . (4) $\; {\text{r}}= {\text{pq}}+1$ $\; {\text{pq}}=6$ . . . . . . (5) $\; {\text{r}}=7 . . . . . . (6)$ $\;\;{3 {\text{p}}^2}/{ {\text{q}}}=4$ After solving ${\text{p}}=2$ and ${\text{q}}=3$

Q75JEE Main 2023MCQ4MComplex Numbers

Let ${\text{p}}, {\text{q}} \in ℝ$ and $(1-{\sqrt{3}} {\text{i}})^{200}=2^{199}( {\text{p}}+ {\text{iq}})$ , ${\text{i}}={\sqrt{-1}}$ Then ${\text{p}}+ {\text{q}}+ {\text{q}}^2$ and ${\text{p}}- {\text{q}}+ {\text{q}}^2$ are roots of the equation. [24-Jan-2023 Shift 1]

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📖 Explanation

$\;(1-{\sqrt{3}} {\text{i}})^{200}=2^{199}( {\text{p}}+ {\text{iq}})$ $\;2^{200} (\cos \;\frac{\pi}{3}- {\text{i}} sin \;\;\frac{\pi}{3}) ^{200}=2^{199}( {\text{p}}+ {\text{iq}})$ $2 (-\;\frac{1}{2}- {\text{i}} \;{{\sqrt{3}}}/{2}) = {\text{p}}+ {\text{iq}}$ $\; {\text{p}}=-1, {\text{q}}=-{\sqrt{3}}$ $\;\alpha= {\text{p}}+ {\text{q}}+ {\text{q}}^2=2-{\sqrt{3}}$ $\;\beta=p-q+q^2=2+{\sqrt{3}}$ $\;\; =4$ $\;\; =1$ equation $x^{2-}4 x+1=0$

Q76JEE Main 2023MCQ4MSets and Relations

The relation ${\text{R}}=\{( {\text{a}}, {\text{b}}): {\gcd}( {\text{a}}, {\text{b}})=1,2 {\text{a}} \ne {\text{b}}, {\text{a}}, {\text{b}} \in ℤ\}$ is: [24-Jan-2023 Shift 1]

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📖 Explanation

Reflexive : $(a, a) \;\; {\gcd}$ of $(a, a)=1$ Which is not true for every a $E {\text{Z}}$ .Symmetric:Take $a=2, b=1 \;\; {\gcd}(2,1)=1$ Also $2 a=4 \ne b$ Now when ${\text{a}}=1, {\text{b}}=2 \;\; {\gcd}(1,2)=1$ Also now $2 {\text{a}}=2= {\text{b}}$ Hence $a=2 b$ $\Rightarrow {\text{R}}$ is not SymmetricTransitive:Let ${\text{a}}=14, {\text{b}}=19, {\text{c}}=21$ ${\gcd}( {\text{a}}, {\text{b}})=1$ ${\gcd}(b, c)=1$ ${\gcd}( {\text{a}}, {\text{c}})=7$ Hence not transitive ${\text{R}}$ is neither symmetric nor transitive.

Q77JEE Main 2023MCQ4MSets and Relations

The compound statement $(∼( {\text{P}} ∧ {\text{Q}})) ∨((∼ {\text{P}}) ∧ {\text{Q}}) \Rightarrow ((∼ {\text{P}}) ∧(∼ {\text{Q}}))$ is equivalent to [24-Jan-2023 Shift 1]

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📖 Explanation

Let $\;\; \text{bo} {r}=(∼( {\text{P}} ∧ {\text{Q}})) ∨((∼ {\text{P}}) ∧ {\text{Q}}) ; \;\; {\ s}=$ $((∼ {\text{P}}) ∧(∼ {\text{Q}}))$ Option (A) : $((∼ {\text{P}}) ∨ {\text{Q}}) ∧((∼ {\text{Q}}) ∨ {\text{P}})$ is equivalent to (not of only $P) ∧($ not of only ${\text{Q}})$ $=($ Both P, Q) and (neither P nor Q)

Q78JEE Main 2023MCQ4MLimits, Continuity and Differentiability

Let $f(x)={\begin{cases}x^2 sin \\ \\ (\\ \\ \frac{1}{x}) & {, \\ \\ } x \ne 0 \\ 0 & {, \\ \\ } x=0}\end{cases}$ ; Then at $x=0$ [24-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\text{ Continuity of }\; f(x): f (0^{+}) =h^2 \cdot sin \;\;\frac{1}{h}=0$ $\;f (0^{-}) =(-h)^2 \cdot sin \; (\;\frac{-1}{h}) =0$ $\;f(0)=0$ $f(x)$ is continuous $\;f^{'} (0^{+}) =\lim _{h \to 0} \;\frac{f(0+h)-f(0)}{h}=\;\frac{h^2 \cdot sin \; (\frac{1}{h}) -0}{h}=0$ $\;f^{'} (0^{-}) =\lim _{h \to 0} \;\frac{f(0-h)-f(0)}{-h}=\;\frac{h^2 \cdot sin \; (\frac{1}{-h}) -0}{-h}=0$ $f(x)$ is differentiable. $f^{'}(x)=2 x \cdot sin \; (\;\frac{1}{x}) +x^2 \cdot \cos (\;\frac{1}{x}) \cdot \;\frac{-1}{x^2}$ $f^{'}(x)={\begin{cases}2 x \cdot sin \\ \\ (\\ \\ \frac{1}{x}) -\cos (\\ \\ \frac{1}{x}) & x \ne 0 \\ 0 & x=0}\end{cases}$ (x) is not continuous (as $\cos (\;\frac{1}{x})$ is highly oscillating at ${\text{x}}=0$ )

Q79JEE Main 2023MCQ4MFunctions

The equation ${\text{x}}^{2-}4 {\text{x}}+[ {\text{x}}]+3= {\text{x}}[ {\text{x}}]$ , where $[ {\text{x}}]$ denotes the greatest integer function, has: [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{x}}^{2-}4 {\text{x}}+[ {\text{x}}]+3= {\text{x}}[ {\text{x}}]$ $\;\Rightarrow {\text{x}}^{2-}4 {\text{x}}+3= {\text{x}}[ {\text{x}}]-[ {\text{x}}]$ $\; \;\; ( {\text{x}}-1)( {\text{x}}-3)=[ {\text{x}}] .( {\text{x}}-1)$ $\;\Rightarrow {\text{x}}=1 \;\text{ or }\; {\text{x}}-3=[ {\text{x}}]$ $\;\Rightarrow {\text{x}}-[ {\text{x}}]=3$ $\; \;\; \{ {\text{x}}\}=3 \;\text{ (Not Possible) }\;$ Only one solution $x=1$ in $(-\infty , \infty )$

Q80JEE Main 2023MCQ4MProbability

Let $Ω$ be the sample space and $A ⊆ Ω$ be an event. Given below are two statements :(S1) : If ${\text{P}}( {\text{A}})=0$ , then ${\text{A}}=\varphi$ $( {\text{S}} 2)$ : If ${\text{P}}( {\text{A}})=1$ , then ${\text{A}}=Ω$ Then [24-Jan-2023 Shift 1]

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📖 Explanation

$Ω=$ sample space ${\text{A}}=$ be an event $A= \{\;\frac{1}{2}\} , Ω=[0,1]$ If $P(A)=0 \;\; {\text{A}} \ne \varphi$ If ${\text{P}}(\overline{ {\text{A}}})=1 \Rightarrow \overline{ {\text{A}}} \ne Ω$ Then both statement are false

Q81JEE Main 2023NAT4MEllipse

Let ${\text{C}}$ be the largest circle centred at $(2,0)$ and inscribed in the ellipse $=\;\frac{x^2}{36}+\;\frac{y^2}{16}=1$ .If $(1, \alpha)$ lies on ${\text{C}}$ , then $10 \alpha^2$ is equal to____ [24-Jan-2023 Shift 1]

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📖 Explanation

Equation of normal of ellipse $\;\frac{x^2}{36}+\;\frac{y^2}{16}=1$ at any point ${\text{P}}(6 \cos \theta, 4 sin \;\theta)$ is $3 \ sec \theta x-2 {\text{cosec}} \theta y=10$ this normal is also the normal of the circle passing through the point $(2,0)$ So, $6 \ sec \theta=10$ or $sin \;\theta=0$ (Not possible) $\cos \theta=\;\frac{3}{5}$ and $sin \;\theta=\;\frac{4}{5}$ so point ${\text{P}}= (\;\frac{18}{5}, \;\frac{16}{5})$ So the largest radius of circle ${\sqrt{320}}$ So the equation of circle $(x-2)^{2+}y^2=\;\frac{64}{5}$ Passing it through $(1, \alpha)$ Then $\alpha^2=\;\frac{59}{5}$ $10 \alpha^2=118$

Q82JEE Main 2023NAT4MBinomial Theorem

Suppose $sum_{ {\text{r}}=0}^{2023} {{\text{r}}^2 \;\;^{2023} {\text{C}}_{ {\text{r}}}=2023 \times \alpha \times 2} ^{2022}$ . Then the value of $\alpha$ is [24-Jan-2023 Shift 1]

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📖 Explanation

using result $sum_{{ {\text{r}}=0}}^{ {\text{n}}}$ ${\text{r}}^2{ }^{ {\text{n}}} {\text{C}}_{ {\text{r}}}= {\text{n}}( {\text{n}}+1) \cdot 2^{ {\text{n}}-2}$ Then $sum_{r=0}^{2023}$ ${\text{r}}^2\;\;{ }^{2023} {\text{C}}_{ {\text{r}}}=2023 \times 2024 \times 2^{2021}$ $=2023 \times \alpha \times 2^{2022} \;\text{ So, }\;$ $\alpha=1012$

Q83JEE Main 2023NAT4MDefinite Integrals

The value of $12 int_{0}^{3}{| x^{2-}3 x+2 |} \text{dx}$ is [24-Jan-2023 Shift 1]

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📖 Explanation

$\;12 int_{0}^{3} |x^{2-}3 x+2| \text{dx}$ $\;=12 int_{0}^{3} | (x-\;\frac{3}{2}) ^{2-}\;\frac{1}{4}| \text{dx}$ If ${\text{x}}-\;\frac{3}{2}= {\text{t}}$ ${\text{dx}}= {\text{dt}}$ $\;=24 int_{0}^{3 / 2} | {\text{t}}^{2-}\;\frac{1}{4}| {\text{dt}}$ $\;=24 \$ [-int_{0}^{1 / 2} ( {\text{t}}^{2-};\frac{1}{4}) {\text{dt}}+int_{1 / 2}^{3 / 2} ( {\text{t}}^{2-};\frac{1}{4}) {\text{dt}}]$ =22$

Q84JEE Main 2023NAT4MPermutations and Combinations

The number of 9 digit numbers, that can be formed using all the digits of the number 123412341 so that the even digits occupy only even places, is__ [24-Jan-2023 Shift 1]

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📖 Explanation

Even digits occupy at even places $\;\frac{4 !}{2 ! 2 !} \times \;\frac{5 !}{2 ! 3 !}=\;\frac{24 \times 120}{4 \times 12}=60$

Q85JEE Main 2023NAT4MQuadratic Equation and Inequalities

Let $\lambda \in ℝ$ and let the equation $E$ be $| {\text{x}}|^{2-}2| {\text{x}}|+|\lambda -3|=0$ . Then the largest element in the set ${\text{S}}=$ $\{ {\text{x}}+\lambda : {\text{x}}$ is an integer solution of ${\text{E}}\}$ is [24-Jan-2023 Shift 1]

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📖 Explanation

$\;| {\text{x}}|^{2-}2| {\text{x}}|+|\lambda -3|=0$ $\;| {\text{x}}|^{2-}2| {\text{x}}|+|\lambda -3|-1=0$ $\;(| {\text{x}}|-1)^{2+}|\lambda -3|=1$ At $\lambda =3, {\text{x}}=0$ and 2 ,at $\lambda =4$ or 2 , then ${\text{x}}=1$ or $-1$ So maximum value of $x+\lambda =5$

Q86JEE Main 2023NAT4MPermutations and Combinations

A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is [24-Jan-2023 Shift 1]

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📖 Explanation

For at most two language courses $={ }^5 {\text{C}}_2 \times { }^7 {\text{C}}_3+{ }^5 {\text{C}}_1 \times { }^7 {\text{C}}_4+{ }^7 {\text{C}}_5=546$

Q87JEE Main 2023NAT4MEllipse

Let a tangent to the Curve $9 x^{2+}16 y^2=144$ intersect the coordinate axes at the points ${\text{A}}$ and ${\text{B}}$ . Then, the minimum length of the line segment $A B$ is [24-Jan-2023 Shift 1]

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📖 Explanation

Equation of tangent at point ${\text{P}}(4 \cos \theta, 3 sin \;\theta)$ is $\;{ {\text{x}} \cos \theta}/{4}+\;{ {\text{y}} sin \;\theta}/{3}=1$ So $A$ is $(4 \ sec \theta, 0)$ and point ${\text{B}}$ is $(0,3 {\text{cosec}} \theta)$ Length ${A B}=\sqrt{16 sec ^2 \theta+9 {\text{cosec}}^2theta}$ $={\sqrt{25+16 \tan ^2 \theta+9 \cot ^2 \theta}} \ge 7$

Q88JEE Main 2023NAT4MDefinite Integrals

The value of $\;\frac{8}{\pi} int_{0}^{\;\frac{\pi}{2}} \frac{(\cos x)^{2023}}{( sin \;x)^{2023}+(\cos x)^{2023}} \text{dx}$ is____ [24-Jan-2023 Shift 1]

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📖 Explanation

$I=\;\frac{8}{\pi} int_{0}^{\;\frac{\pi}{2}} \frac{(\cos x)^{2023}}{( sin \;x)^{2023}+(\cos x)^{2023}} \text{dx}$ . . . . . (1) $\;\;\text{ Using }\; int_{0}^{a} f(x) \text{dx}=int_{0}^{a} f(a-x) \text{dx}$ $\;I=\;\frac{8}{\pi} int_{0}^{\;\frac{\pi}{2}} \;\frac{( sin \;x)^{2023}}{( sin \;x)^{2023}+(\cos x)^{2023}}\text{dx}$ . . . . . (2) $\;\;\text{ Adding }\;(1) &(2)$ $\;I I=\;\frac{8}{\pi} int_{0}^{\;\frac{\pi}{2}} 1 \text{dx}$

Q89JEE Main 2023NAT4MThree Dimensional Geometry

The shortest distance between the lines $\;\frac{x-2}{3}=\;\frac{y+1}{2}=\;\frac{z-6}{2}$ and $\;\frac{x-6}{3}=\;\frac{1-y}{2}=\;\frac{z+8}{0}$ is equal to [24-Jan-2023 Shift 1]

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📖 Explanation

$=\;{\begin{vmatrix}4 & 2 & -14 \\ 3 & 2 & 2 \\ 3 & -2 & 0\end{vmatrix}}/ {\begin{vmatrix}\\ \\ {i} ^{\^} & \\ \\ \hat{j} & {\text{k}} \\ 3 & 2 & 2 \\ 3 & -2 & 0 \\ \end{vmatrix}}$ $=\;\frac{16+12+168}{|-4 \hat{i}+6 \hat{j}-12 k|}=\;\frac{196}{14}=14$

Q90JEE Main 2023NAT4MSequences and Series

The $4^{\;\text{th }\;}$ term of GP is 500 and its common ratio is $\;{1}/{ {\text{m}}}, {\text{m}} \in {\text{N}}$ . Let ${\text{S}}_{ {\text{n}}}$ denote the sum of the first ${\text{n}}$ terms of this GP. If ${\text{S}}_6> {\text{S}}_5+1$ and ${\text{S}}_7< {\text{S}}_6+\;\frac{1}{2}$ , then the number of possible values of ${\text{m}}$ is [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{T}}_4=500 \;\text{ where }\; {\text{a}}=\;\text{ first term, }\;$ $\; {\text{r}}=\;\text{ common ratio }\;=\;{1}/{ {\text{m}}}, {\text{m}} \in {\text{N}}$ $\;a r^3=500$ $\;\;{ {\text{a}}}/{ {\text{m}}^3}=500$ $\;S_n-S_{n-1}=a r^{n-1}$ $\; {\text{S}}_6 > {\text{S}}_5+1 \;\; \;\text{ and }\; {\text{S}}_7- {\text{S}}_6 < \;\frac{1}{2}$ $\; {\text{S}}_6- {\text{S}}_5 > 1 \;\; \;{ {\text{a}}}/{ {\text{m}}^6} < \;\frac{1}{2}$ $\;a r^5 > 1 \;\; {\text{m}}^3 > 10^3$ $\;\;{500}/{ {\text{m}}^2}>1 \;\; {\text{m}}>10$ $\; {\text{m}}^2 < 500 \;\text{......(1) }\;$ $\;\;\text{ From (1) and (2) }\;$ $\; {\text{m}}=11,12,13 . . . . . . . . . . . . ., 22$ $\;$ So number of possible values of ${\text{m}}$ is 12

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