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30 Jan 2023 Shift 1

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Physics30 questions

Q1JEE Main 2023MCQ4MCurrent Electricity

The charge flowing in a conductor changes with time as ${\text{Q}}( {\text{t}})=\alpha {\text{t}}-\beta {\text{t}}^{2+}\gamma {\text{t}}^3$ . Where $\alpha, \beta$ and $\gamma$ are constants. Minimum value of current is : [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;Q= (\alpha t-\beta t^{2+}\gamma t^3)$ $\;i=\;\frac{d Q}{d t}= (\alpha-2 \beta t+3 \gamma t^2)$ $\;\;\frac{d i}{d t}=(3 \gamma t-2 \beta)=0$ $\;\Rightarrow t=\;\frac{\beta}{3 \gamma }$ $\;i= (\alpha-2 \beta t+3 \gamma t^2) = (\alpha-\;\frac{\beta^2}{3 \gamma })$

Q2JEE Main 2023MCQ4MThermodynamics

The pressure $( {\text{P}})$ and temperature $( {\text{T}})$ relationship of an ideal gas obeys the equation ${\text{PT}}^2=$ constant. The volume expansion coefficient of the gas will be: [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{PT}}^2=$ constant, ${\text{Using}} {\text{PV}}= {\text{nRT}}$ ${\text{P}}=\;{ {\text{nRT}}}/{ {\text{V}}}$ ${\text{PT}}^2=\;{ {\text{nRT}}}/{ {\text{V}}} \times {\text{T}}^2=$ constant $\Rightarrow {\text{T}}^3= {\text{KV}}$ So, $\;{ {\text{d}}}/{ {\text{dT}}}( {\text{KV}})=3 {\text{T}}^2$ $\Rightarrow \;{ {\text{KdV}}}/{ {\text{dT}}}=3 {\text{T}}^2$ $\Rightarrow {\text{dV}}=\;{3 {\text{T}}^2}/{ {\text{K}}} {\text{dT}}$ ${\text{dV}}= {\text{V}} \gamma {\text{dT}}$ $\Rightarrow \gamma {\text{V}}=\;{3 {\text{T}}^2}/{ {\text{K}}} \;\; \Rightarrow \gamma =\;{3 {\text{T}}^2}/{ {\text{KV}}}=\;{3 {\text{T}}^2}/{ {\text{T}}^3}=\;{3}/{ {\text{T}}}$

Q3JEE Main 2023MCQ4MRay Optics and Optical Instruments

A person has been using spectacles of power-1.0 diopter for distant vision and a separate reading glass of power $2.0$ diopters. What is the least distance of distinct vision for this person: [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\;{1}/{ {\text{v}}}-\;{1}/{ {\text{u}}}=\;{1}/{ {\text{f}}}$ $\; {\text{P}}=2 {\text{D}}=2 {\text{m}}^{-1}$ $\;\Rightarrow \;{1}/{ {\text{f}}}=\;\frac{2}{100} {\text{cm}}^{-1}$ $\;\;{1}/{ {\text{V}}}- (-\;\frac{1}{25}) =\;\frac{2}{100}$ $\;\Rightarrow \;{1}/{ {\text{V}}}=\;\frac{1}{50}-\;\frac{1}{25}$ $\;\Rightarrow V=-50 {\text{cm}}$

Q4JEE Main 2023MCQ4MCentre of Mass and Collision

As per the given figure, a small ball P slides down the quadrant of a circle and hits the other ball ${\text{Q}}$ of equal mass which is initially at rest. Neglecting the effect of friction and assume the collision to be elastic, the velocity of ball ${\text{Q}}$ after collision will be $: ( {\text{g}}=10 {\text{m}} / {\ s}^2)$ [30-Jan-2023 Shift 1]

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📖 Explanation

The velocities will be interchanged after collision.Speed of $P$ just before collision $=\sqrt{2 {\text{gh}}}$ $={\sqrt{2 \times 10 \times 0.2}}=2 {\text{m}} / {\ s}$

Q5JEE Main 2023MCQ4MProperties of Matter

Choose the correct relationship between Poisson ratio $(\sigma)$ . bulk modulus $( {\text{K}})$ and modulus of rigidity $(η)$ of a given solid object: [30-Jan-2023 Shift 1]

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📖 Explanation

$\;Y=3 η(1+\sigma)$ $\;Y=3 K(1-\sigma)$ $\;\Rightarrow 2 η(1+\sigma)=3 K(1-2 \sigma)$ $\;\Rightarrow \sigma= (\;\frac{3 K-2 η}{6 K+2 η})$

Q6JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

The magnetic moments associated with two closely wound circular coils ${\text{A}}$ and ${\text{B}}$ of radius ${\text{r}}_{ {\text{A}}}=10 {\text{cm}}$ and $r_B=20 {\text{cm}}$ respectively are equal if: (Where ${\text{N}}_A, {\text{I}}_{ {\text{A}}}$ and ${\text{N}}_{ {\text{B}}}, {\text{I}}_{ {\text{B}}}$ are number of turn and current of $A$ and $B$ respectively) [30-Jan-2023 Shift 1]

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📖 Explanation

${\text{M}}=$ NIA $\; {\text{M}}_{ {\text{A}}}= {\text{M}}_{ {\text{B}}}$ $\therefore \; {\text{N}}_{ {\text{A}}} {\text{I}}_{ {\text{A}}} {\text{A}}_{ {\text{A}}}= {\text{N}}_{ {\text{B}}} {\text{I}}_{ {\text{B}}} {\text{A}}_{ {\text{B}}}$ $\therefore \; {\text{N}}_{ {\text{A}}} {\text{I}}_{ {\text{A}}} \pi(0.1)^2= {\text{N}}_{ {\text{B}}} {\text{I}}_{ {\text{B}}} \pi(0.2)^2$ $\therefore \; {\text{N}}_{ {\text{A}}} {\text{I}}_{ {\text{A}}}=4 {\text{N}}_{ {\text{B}}} {\text{I}}_{ {\text{B}}}$

Q7JEE Main 2023MCQ4MDual Nature of Matter and Radiation

A small object at rest, absorbs a light pulse of power $20 {\text{mW}}$ and duration $300 {\text{ns}}$ . Assuming speed of light as $3 \times 10^8 {\text{m}} / {\ s}$ . the momentum of the object becomes equal to : [30-Jan-2023 Shift 1]

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📖 Explanation

Momentum $=\;{\;\text{ Energy }\;}/{ {\text{C}}}$ $=\;{\;\text{ Power }\; \times \;\text{ time }\;}/{ {\text{C}}}$ $=\;{ (20 \times 10^{-3} {\text{w}}) (300 \times 10^{-9} {\ s}) }/{3 \times 10^8 {\text{m}} / {\ s}}$ $=2 \times 10^{-17} {\text{kg}}- {\text{m}} / {\ s}$

Q8JEE Main 2023MCQ4MAtoms and Nuclei

Speed of an electron in Bohr's $7^{\;\text{th }\;}$ orbit for Hydrogen atom is $3.6 \times 10^6 {\text{m}} / {\ s}$ . The corresponding speed of the electron in $3^{\;\text{rd }\;}$ orbit, in ${\text{m}} / {\ s}$ is : [30-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{V}}_{ {\text{n}}} \propto \;{ {\text{Z}}}/{ {\text{n}}}$ $\; {\text{Z}}=1, \therefore {\text{V}}_{ {\text{n}}} \propto \;{1}/{ {\text{n}}}$ $\therefore \;{ {\text{V}}_3}/{ {\text{V}}_7}=\;\frac{7}{3}$ $\; \therefore {\text{V}}_3=\;\frac{7}{3} {\text{V}}_7$ $\;\; =\;\frac{7}{3} \times 3.6 \times 10^6 {\text{m}} / {\ s}$ $\;\; =8.4 \times 10^6 {\text{m}} / {\ s}$

Q9JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

A massless square loop, of wire of resistance $10 Ω$ . supporting a mass of I g. hangs vertically with one of its sides in a uniform magnetic field of $10^3 {\text{G}}$ , directed outwards in the shaded region. A dc voltage ${\text{V}}$ is applied to the loop. For what value of V. the magnetic force will exactly balance the weight of the supporting mass of $1 {\text{g}}$ ?(If sides of the loop $=10 {\text{cm}}, {\text{g}}=10 {\text{ms}}^{-2}$ ) [30-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{F}}_{ {\text{m}}}= {\text{mg}}$ $\; \therefore \;\; \;\text{ ILB }\;= {\text{mg}}$ $\; \therefore \;\; (\;{ {\text{V}}}/{ {\text{R}}}) {\text{LB}}= {\text{mg}}$ $\; \therefore \;\; {\text{V}}=\;{ {\text{mgR}}}/{ {\text{LB}}}$ $\;=\;{ (1 \times 10^{-3} {\text{kg}}) (10 {\text{m}} / {\ s}^2) (10 Ω)}/{(0.1 {\text{m}}) (10^3 \times 10^{-4} {\text{T}}) }=10 {\text{V}}$

Q10JEE Main 2023MCQ4MElectrostatics

Two isolated metallic solid spheres of radii ${\text{R}}$ and $2 {\text{R}}$ are charged such that both have same charge density $\sigma$ . The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $\sigma^{'}$ . The ratio $\;\frac{\sigma^{'}}{\sigma}$ is: [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\frac{Q_1^{'}}{4 \pi \epsilon_0 R}=\;\frac{Q_2^{'}}{4 \pi \epsilon_0(2 R)}$ $\therefore \;\; Q_2^{'}=2 Q_1^{'}$ $\;Q_1^{'}+Q_2^{'}=Q_1+Q_2$ $\therefore \;\;\frac{Q_2^{'}}{2}+Q_2^{'}=20 \pi R^2 \sigma$ $\;\;\frac{3}{2} Q_2^{'}=20 \pi R^2 \sigma$ $\therefore \;\;\frac{Q_2^{'}}{4 \pi(2 R)^2}=\;\frac{2}{3} \cdot \;\frac{20 \pi R^2 \sigma}{16 \pi R^2}$ $\therefore \;\;\frac{\sigma^{'}}{\sigma}=\;\frac{5}{6}$

Q11JEE Main 2023MCQ4MThermodynamics

Heat is given to an ideal gas in an isothermal process.A. Internal energy of the gas will decrease.B. Internal energy of the gas will increase.C. Internal energy of the gas will not change.D. The gas will do positive work.E. The gas will do negative work.Choose the correct answer from the options given below: [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\text{ Sol. }\;\; {\text{dQ}}= {\text{dU}}+ {\text{dW}} \Rightarrow {\text{dU}}= {\text{nC}}_{ {\text{V}}} {\text{dT}}$ $\; {\text{dU}}=0 \;\; \;\text{ (for isothermal) }\;$ $\therefore \;\; {\text{U}}=\;\text{ constant }\;$ $\;\;\text{ Also }\; {\text{dQ}} > 0 \;\; \;\text{ (supplied) }\;$ $\;\;\text{ Hence }\; {\text{dW}} > 0$

Q12JEE Main 2023MCQ4MElectrostatics

Electric field in a certain region is given by $\vec{E}= (\;\frac{A}{x^2} {i}^{\wedge }+\;\frac{B}{y^3} {j}^{\wedge })$ . The SI unit of $A$ and $B$ are : [30-Jan-2023 Shift 1]

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📖 Explanation

$\;{ \vec{\text{E}}=\;{ {\text{A}}}/{ {\text{x}}^2} \;{ {\text{i}}}^{\^}+\;{ {\text{B}}}/{ {\text{y}}^3} \;{ {\text{j}}}^{\^}$ $\;{[\;{ {\text{A}}}/{ {\text{x}}^2}]\$ = {\text{NC}}^{-1} \Rightarrow [ {\text{A}}]= {\text{Nm}}^2 {\text{C}}^{-1}};{[;{ {\text{B}}}/{ {\text{y}}^3}]$ = {\text{NC}}^{-1} \Rightarrow [ {\text{B}}]= {\text{Nm}}^3 {\text{C}}^{-1}}$

Q13JEE Main 2023MCQ4MSemiconductor Electronics

The output waveform of the given logical circuit for the following inputs $A$ and $B$ as shown below, is: [30-Jan-2023 Shift 1]

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📖 Explanation

$\;(\overline{ {\text{A}} \cdot {\text{A}}})=\overline{ {\text{A}}}$ $\;\overline{ {\text{B}} \cdot {\text{B}}}=\overline{ {\text{B}}}$ $\;(\overline{\overline{ {\text{A}}} \cdot \overline{ {\text{B}}}})= {\text{A}}+ {\text{B}}$ OR Gate

Q14JEE Main 2023MCQ4MProperties of Matter

The height of liquid column raised in a capillary tube of certain radius when dipped in liquid A vertically is, $5 {\text{cm}}$ . If the tube is dipped in a similar manner in another liquid $B$ of surface tension and density double the values of liquid $A$ , the height of liquid column raised in liquid $B$ would be _______ ${\text{m}}$ . [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\; {\text{h}}=\;{2 {\text{S}} \cos \theta}/{ {\text{r} \rho g}}$ $\therefore \;\; \;{ {\text{h}}_1}/{ {\text{h}}_2}\;=\;{ {\text{S}}_1}/{ {\text{S}}_2} \;\frac{\rho _2}{\rho _1}$ $\;{5}/{ {\text{h}}_2}\;= [\;\frac{1}{2}] [\;\frac{2}{1}] \Rightarrow {\text{h}}_2=5 {\text{cm}}=0.05 {\text{m}}$ {Info about angle of contact not there so most appropriate is 3 $\}$

Q15JEE Main 2023MCQ4MSemiconductor Electronics

A sinusoidal carrier voltage is amplitude modulated. The resultant amplitude modulated wave has maximum and minimum amplitude of $120 {\text{V}}$ and $80 {\text{V}}$ respectively. The amplitude of each sideband is : [30-Jan-2023 Shift 1]

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📖 Explanation

${\text{A}}_{ {\text{c}}}+ {\text{A}}_{ {\text{m}}}=120$ ${\text{A}}_{ {\text{c}}}- {\text{A}}_{ {\text{m}}}=80$ $\therefore \overline{ {\text{A}}_c=100}$ ${\text{A}}_{ {\text{m}}}=20$ Modulation index $=\;\frac{20}{100}=\;\frac{1}{5}$ Amplitude of each sideband $= {\text{A}}_{ {\text{c}}} \;\frac{\;\text{ (mod ulation index) }\;}{2}$ $=100 \times \;\frac{1}{10}=10$ volt

Q16JEE Main 2023MCQ4MAlternating Current

In a series $L R$ circuit with ${\text{X}}_{ {\text{L}}}= {\text{R}}$ . power factor is $P_1$ . If a capacitor of capacitance $C$ with $X_C=X_L$ is added to the circuit the power factor becomes $P_2$ . The ratio of $P_1$ to $P_2$ will be : [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{P}}=\;{ {\text{R}}}/{ {\text{Z}}} \Rightarrow {\text{P}}_1=\;{ {\text{R}}}/{\sqrt{ {\text{R}}^{2+} {\text{X}}_{ {\text{L}}}^2}}=\;{ {\text{R}}}/{ {\text{R}} {\sqrt{2}}} (\;\text{ as }\; {\text{X}}_{ {\text{L}}}= {\text{R}})$ $\; {\text{P}}_1=\;{1}/{{\sqrt{2}}}$ $\; {\text{P}}_2=\;{ {\text{R}}}/{\sqrt{ {\text{R}}^{2+} ( {\text{X}}_{ {\text{L}}}- {\text{X}}_{ {\text{L}}}) ^2}}= {\text{P}}_2=1$ $\;\;{ {\text{P}}_1}/{ {\text{P}}_2}=\;{1}/{{\sqrt{2}}}$

Q17JEE Main 2023MCQ4MGravitation

If the gravitational field in the space is given as $(-\;{ {\text{K}}}/{ {\text{r}}^2})$ . Taking the reference point to be at ${\text{r}}=2 {\text{cm}}$ with gravitational potential ${\text{V}}=10 {\text{J}} / {\text{kg}}$ . Find the gravitational potential at ${\text{r}}=3 {\text{cm}}$ in ${\text{SI}}$ unit (Given, that ${\text{K}}=6 {\text{J}} {\text{cm}} / {\text{kg}}$ ) [30-Jan-2023 Shift 1]

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📖 Explanation

$\;-\;{ {\text{dV}}}/{ {\text{dr}}}=-\;{ {\text{k}}}/{ {\text{r}}^2} \Rightarrow int_{10}^{ {\text{V}}} {\text{dV}}=int_{2}^{3} \;{ {\text{k}}}/{ {\text{r}}^2} {\text{dr}}$ $\; {\text{V}}-10= {\text{k}} [\;\frac{1}{2}-\;\frac{1}{3}]$ $\; {\text{V}}-10=\;{ {\text{k}}}/{6} \Rightarrow {\text{V}}=11 \;\text{ volts }\;$

Q18JEE Main 2023MCQ4MCentre of Mass and Collision

A ball of mass $200 {\text{g}}$ rests on a vertical post of height $20 {\text{m}}$ . A bullet of mass $10 {\text{g}}$ , travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance $30 {\text{m}}$ and the bullet at a distance of $120 {\text{m}}$ from the foot of the post. The value of initial velocity of the bullet will be (if ${\text{g}}=10 {\text{m}} / {\ s}^2$ ) : [30-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{v}}_1=\;{30}/{\sqrt{\;{2 {\text{h}}}/{ {\text{g}}}}}, \;\; {\text{v}}_2=\;{120}/{\sqrt{\;{2 {\text{h}}}/{ {\text{g}}}}}$ $\;(0.01) {\text{u}}=(0.2) \;{30 \sqrt{ {\text{g}}}}/{\sqrt{2 {\text{h}}}}+(0.01) \;{120 \sqrt{ {\text{g}}}}/{\sqrt{2 {\text{h}}}}$ $\; {\text{u}}=300+60=360 {\text{ms}}^{-1}$

Q19JEE Main 2023MCQ4MProperties of Matter

Choose the correct answer from the options given below: [30-Jan-2023 Shift 1]

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📖 Explanation

$\;{ {\text{dx}}}/{ {\text{dt}}}=$ slope $\ge 0$ always increasing(A - II) $\;{ {\text{dx}}}/{ {\text{dt}}} < 0$ ; and at ${\text{t}} \to \infty \;{ {\text{dx}}}/{ {\text{dt}}} \to 0$ (B - IV) $\;{ {\text{dx}}}/{ {\text{dt}}} > 0$ for first half $\;{ {\text{dx}}}/{ {\text{dt}}} < 0$ for second half.(C - III) $\;{ {\text{dx}}}/{ {\text{dt}}}=$ constant(D - I)

Q20JEE Main 2023MCQ4MLaws of Motion

The figure represents the momentum time $(p-t)$ curve for a particle moving along an axis under the influence of the force. Identify the regions on the graph where the magnitude of the force is maximum and minimum respectively ?If $(t_3-t_2) < t_1$ . [30-Jan-2023 Shift 1]

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📖 Explanation

$|\;{ {\text{d}} { \vec{\text{p}}}/{ {\text{dt}}}|=|{ \vec{\text{F}}| \Rightarrow \;{ {\text{d}} { \vec{\text{p}}}/{ {\text{dt}}}=$ Slope of curvemax slope (c) $\min$ slope (b)

Q21JEE Main 2023NAT4MOscillations

The general displacement of a simple harmonic oscillator is ${\text{x}}= {\text{A}} sin \;\omega {\text{t}}$ . Let ${\text{T}}$ be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when $t=\;\frac{T}{\beta}$ . The value of $\beta$ is _______. [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$x=A sin \;(\omega t)$ ${\text{U}}_{( {\text{x}})}=\;\frac{1}{2} {\text{kx}}^2$ $\;{ {\text{dU}}}/{ {\text{dt}}}=\;\frac{1}{2} {\text{k}} 2 {\text{x}} \;{ {\text{dx}}}/{ {\text{dt}}}$ $= {\text{kA}}^2 \omega sin \;\omega {\text{t}} \cos \omega {\text{t}} \times \;\frac{2}{2}$ $\; (\;{ {\text{dU}}}/{ {\text{dt}}}) _{\max }=\;{ {\text{kA}}^2 \omega }/{2}( sin \;2 \omega {\text{t}})_{\max }$ $\;2 \omega {\text{t}}=\;\frac{\pi}{2} \Rightarrow {\text{t}}=\;\frac{\pi}{4} \omega =\;{ {\text{T}}}/{8} \Rightarrow \beta=8$

Q22JEE Main 2023NAT4MElectrostatic Potential and Capacitance

A capacitor of capacitance $900 \mu {\text{F}}$ is charged by a $100 {\text{V}}$ battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as $x \times 10^{-2} {\text{J}}$ . The value of $x$ is ______. [30-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{C}}=900 \mu {\text{F}}$ $\; {\text{Q}}= {\text{CV}}=900 \times 10^{-6} \times 100=9 \times 10^{-2}=90 {\text{MC}}$ NowCommon potential will be developed across both capacitors by ${\text{kVL}}$ Total charge on left plates of capacitors should be conserved. $\therefore \;\; 90 {\text{mc}}+0=2 {\text{cv}}_0$ $\; {\text{cv}}_0=45 {\text{mc}}$ Heat dissipated $= {\text{U}}_{ {\text{i}}}- {\text{U}}_{ {\text{f}}}$ [Change in energy stored in the capacitors] $\;=\;\frac{1}{2} \;{(90 {\text{mc}})^2}/{900 \mu {\text{F}}}-2 \times \;\frac{1}{2} \;{(45 {\text{mc}})^2}/{900 \mu {\text{F}}} [ {\text{U}}=\;{ {\text{Q}}^2}/{2 {\text{c}}}]$ $\;=\;\frac{1}{2 \times 900 \times 10^{-6}}(8100-4050) \times 10^{-6}$ $\;=2.25 {\text{Joule}}$ OR $\;\;\text{ Heat }\;=\;\frac{1}{2} \;{ {\text{C}}_1 {\text{C}}_2}/{ {\text{C}}_1+ {\text{C}}_2} ( {\text{V}}_1- {\text{V}}_2) ^2$ $\;=\;\frac{1}{2} \;{ {\text{C}}^2}/{2 {\text{C}}}(100-0)^2$ $\;=\;\frac{1}{2} \;\frac{900 \times 10^{-6}}{2} \times 10^4=\;\frac{9}{4} \;\text{ Joule }\;=2.25 \;\text{ Joule }\;$

Q23JEE Main 2023NAT4MWave Optics

In Young's double slit experiment, two slits $S_1$ and $S_2$ are ' $d$ ' distance apart and the separation from slits to screen is D (as shown in figure). Now if two transparent slabs of equal thickness $0.1 {\text{mm}}$ but refractive index $1.51$ and $1.55$ are introduced in the path of beam $(\lambda =4000 Å)$ from $S_1$ and $S_2$ respectively. The central bright fringe spot will shift by _______ number of fringes. [30-Jan-2023 Shift 1]

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📖 Explanation

Path difference at ${\text{P}}$ be $\Delta {\text{x}}$ $\Delta {\text{x}}\;= (\mu _2-\mu _1) {\text{t}}$ $\;=(1.55-1.51) 0.1 {\text{mm}}$ $\;=0.04 \times 10^{-4}$ $\Delta {\text{x}}\;=4 \times 10^{-6}=4 \mu {\text{m}}$ ${\text{y}}\;=\;{\Delta {\text{xD}}}/{ {\text{d}}}=4 \times 10^{-6} \;{ {\text{D}}}/{ {\text{d}}}$ y is the distance of central maxima from geometric center $\;\text{ fringe width }\;$ (beta) $=\;\frac{\lambda D}{d}=4 \times 10^{-6} {\text{m}} \;{ {\text{D}}}/{ {\text{d}}}=4 \mu {\text{m}} \;{ {\text{D}}}/{ {\text{d}}}$ Central bright fringe spot will shift by ' $x$ ' $\;\;\text{ Number of shift }\;=\;{ {\text{y}}}/{\beta}$ $\;=\;{4 \times 10^{-6} {\text{D}} / {\text{d}}}/{4 \times 10^{-7} {\text{D}} / {\text{d}}}=10 \;\text{ Ans }\;$

Q24JEE Main 2023NAT4MCurrent Electricity

In the following circuit, the magnitude of current ${\text{I}}_1$ , is ______ A. [30-Jan-2023 Shift 1]

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📖 Explanation

Junction law at A, $\;\frac{x-(y+5)}{1}+\;\frac{x-2}{2}+\;\frac{x-0}{2}=0$ . . . (1)Junction law at B, $\;\frac{y+5-x}{1}+\;\frac{y-0}{1}+\;\frac{y-2}{1}=0$ . . . (2)On solving equation (1) and Equation (2) ${\text{x}}=3$ & ${\text{y}}=0$ At ${\text{D}}$ junction $I_1\;=i_1+i_2$ $I_1\;=\;\frac{y-0}{1}+\;\frac{x-0}{2}$ $\;=\;\frac{0-0}{1}+\;\frac{3-0}{2}$ $I_1\;=1.5 {\text{A}}$

Q25JEE Main 2023NAT4MMotion in a Straight Line

A horse rider covers half the distance with $5 {\text{m}} / {\ s}$ speed. The remaining part of the distance was travelled with speed $10 {\text{m}} / {\ s}$ for half the time and with speed $15 {\text{m}} / {\ s}$ for other half of the time. The mean speed of the rider averaged over the whole time of motion is $x / 7 {\text{m}} / {\ s}$ . The value of $x$ is [30-Jan-2023 Shift 1]

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📖 Explanation

${\text{t}}_{ {\text{AB}}}=\;{ {\text{x}}}/{5 {\text{m}} / {\ s}}$ In motion ${\text{BC}}$ ${\text{x}}= {\text{d}}_1+ {\text{d}}_2$ where ${\text{d}}_1 & {\text{d}}_2$ we the distance travelled with 10 ${\text{m}} / {\ s}$ and $15 {\text{m}} / {\ s}$ respectively in equal time intervals $\;\frac{\;\text{ 't' }\;}{2}$ each $\; {\text{d}}_1=\;{10 {\text{t}}}/{2}, {\text{d}}_2=\;{15 {\text{t}}}/{2}$ $\; {\text{d}}_1+ {\text{d}}_2= {\text{x}}=\;{ {\text{t}}}/{2}(10+15)=\;{25 {\text{t}}}/{2}$ $\;< {\text{v}}>=\;{2 {\text{x}}}/{{ {\text{x}}}/{5}+\;{2 {\text{x}}}/{25}}=\;\frac{2 \times 25}{5+2}=\;\frac{50}{7} {\text{m}} / {\ s}$ Ans. : 50

Q26JEE Main 2023NAT4MDual Nature of Matter and Radiation

A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of $24 {\text{W}}$ The radius of curvature of hemisphere is $10 {\text{cm}}$ and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is ________ $\times 10^{-8} {\text{N}}$ [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\text{ Force }\;=\int {\text{PdA}} \cos \theta$ $\;=\;{2 {\text{I}}}/{ {\text{C}}} \int {\text{dA}} \cos \theta=\;{2 {\text{I}}}/{ {\text{C}}} \pi {\text{R}}^2=2 \;{ {\text{p}}_0}/{4 \pi {\text{R}}^2} \cdot \;{\pi {\text{R}}^2}/{ {\text{C}}}$ $\;=\;{ {\text{p}}_0}/{2 {\text{C}}}=\;\frac{24}{2 \times 3 \times 10^8}=4 \times 10^{-8} {\text{N}} \;\text{ (Ans) }\;$

Q27JEE Main 2023NAT4MElectromagnetic Induction

As per the given figure, if $\;{ {\text{dI}}}/{ {\text{dt}}}=-1 {\text{A}} / {\ s}$ then the value of ${\text{V}}_{ {\text{AB}}}$ at this instant will be _______ V. [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\;{ {\text{dI}}}/{ {\text{dt}}}=-1 \;{ {\text{A}}}/{ {\ sec}}$ $\;V_A-I R-L \;\frac{d I}{d t}-12=V_B$ $\;V_A-2 \times 12-6(-1)-12=V_B$ $\;V_A-V_B=36-6=30 \;\text{ volt }\;$

Q28JEE Main 2023NAT4MUnits and Measurements

In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by $0.5 {\text{mm}}$ on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible while $46^{\;\text{th }\;}$ division the circular scale coincide with the reference line. The diameter of the wire is _______ $\times 10^{-2} {\text{mm}}$ . [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\text{ Least count }\;=\;\frac{\;\text{ Pitch }\;}{\;\text{ No. of circular divisions }\;}$ $\;=\;{0.5 {\text{mm}}}/{100}$ $\;\;\text{ Least count }\;=5 \times 10^{-3} {\text{mm}}$ $\;\;\text{ Positive Error }\;= {\text{MSR}}+ {\text{CSR}}( {\text{LC}})$ $\;=0 {\text{mm}}+6 (5 \times 10^{-3} {\text{mm}})$ Reading of Diameter $=$ MSR $+$ CSR $( {\text{LC}})-$ Positive zero error $\;=4 \times 0.5 {\text{mm}}+ (46 (5 \times 10^{-3}) ) -6 (5 \times 10^{-3}) {\text{mm}}$ $\;=2 {\text{mm}}+40 \times 5 \times 10^{-3} {\text{mm}}=2.2 {\text{mm}} \;\text{ (Ans.) }\;$

Q29JEE Main 2023NAT4MRay Optics and Optical Instruments

In an experiment for estimating the value of focal length of converging mirror, image of an object placed at $40 {\text{cm}}$ from the pole of the mirror is formed at distance $120 {\text{cm}}$ from the pole of the mirror. These distances are measured with a modified scale in which there are 20 small divisions in $1 {\text{cm}}$ . The value of error in measurement of focal length of the mirror is $1 / {\text{K}}$ ${\text{cm}}$ . The value of ${\text{K}}$ is ________. [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\;{1}/{ {\text{v}}}+\;{1}/{ {\text{u}}}=\;{1}/{ {\text{f}}}$ $\;\;\frac{-1}{120}-\;\frac{1}{40}=\;{1}/{ {\text{f}}}, \;\; {\text{f}}=-30 {\text{cm}}$ Now, $\;\frac{-1}{v^2} d v-\;\frac{1}{u^2} d u=-\;\frac{1}{f^2} d f$ Also ${\text{dv}}= {\text{du}}=\;\frac{1}{20} {\text{cm}}$ $\therefore \;\; \;\frac{\;\frac{1}{20}}{(120)^2}+\;\frac{\;\frac{1}{20}}{(40)^2}=\;{ {\text{df}}}/{(30)^2}$ On solving $\; {\text{df}}=\;\frac{1}{32} {\text{cm}}$ $\therefore \;\; {\text{k}}\;=32$

Q30JEE Main 2023NAT4MRotational Motion

A thin uniform rod of length $2 {\text{m}}$ . cross sectional area ' ${\text{A}}$ ' and density ' ${\text{d}}$ ' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity $\omega$ . If value of $\omega$ in terms of its rotational kinetic energy $E$ is $\sqrt{\;\frac{\alpha E}{A d}}$ then the value of $\alpha$ is ________. [30-Jan-2023 Shift 1]

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📖 Explanation

$\;( {\text{KE}})_{\;\text{Rotational }\;}=\;\frac{1}{2} {\text{I}} \omega ^2= {\text{E}}$ $\; {\text{E}}=\;\frac{1}{2} \;{ {\text{m}} ℓ^2}/{12} \omega ^2$ $\; {\text{E}}=\;\frac{1}{2} \;{ {\text{dA}} ℓ^3}/{12} \omega ^2$ $\; {\text{E}}=\;{ {\text{dA}}(2)^3}/{24} \omega ^2$ $\;\sqrt{\;{3 {\text{E}}}/{ {\text{dA}}}}=\omega$ $\;\alpha=3 \;\text{ Ans. }\;$

Chemistry30 questions

Q31JEE Main 2023MCQ4MPractical Organic Chemistry

Which of the following compounds would give the following set of qualitative analysis ?(i) Fehling's Test : Positive(ii) ${\text{Na}}$ fusion extract upon treatment with sodium nitroprusside gives a blood red colour but not [30-Jan-2023 Shift 1]

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Q32JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

What is the correct order of acidity of the protons marked A-D in the given compounds ? [30-Jan-2023 Shift 1]

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Q33JEE Main 2023MCQ4MBiomolecules

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).Assertion (A) : Ketoses give Seliwanoff's test faster than Aldoses.Reason (R) : Ketoses undergo $\beta$ -elimination followed by formation of furfural.In the light of the above statements, choose the correct answer from the options given below : [30-Jan-2023 Shift 1]

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Q34JEE Main 2023MCQ4Md and f Block Elements

In the extraction of copper, its sulphide ore is heated in a reverberatory furnace after mixing with silica to : [30-Jan-2023 Shift 1]

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📖 Explanation

The copper ore contains iron, it is mixed with silica before heating in reverberatory furnace. ${\text{FeO}}$ slags off as ${\text{FeSiO}}_3$ . ${\text{FeO}}+ {\text{SiO}}_2 \to {\text{FeSiO}}_3$

Q35JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

Amongst the following compounds, which one is an antacid ? [30-Jan-2023 Shift 1]

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Q36JEE Main 2023MCQ4MHydrocarbons

The major products ' $A^{\;\text{' }\;}$ and ' $B$ ', respectively, are [30-Jan-2023 Shift 1]

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Q37JEE Main 2023MCQ4MAmines

Benzyl isocyanide can be obtained by :Choose the correct answer from the options given below : [30-Jan-2023 Shift 1]

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📖 Explanation

Benzyl isocyanide can be synthesized via two main pathways, which were likely presented as options A and B in the original question.

Carbylamine Reaction: This method involves the reaction of a primary amine, benzylamine ( $C_6H_5CH_2NH_2$ ), with chloroform ( $CHCl_3$ ) and a strong base like alcoholic KOH. This reaction is specific for the synthesis of isocyanides from primary amines.
$C_6H_5CH_2NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5CH_2NC + 3KCl + 3H_2O$
Reaction with Silver Cyanide: Treating benzyl chloride ( $C_6H_5CH_2Cl$ ) with silver cyanide ( $AgCN$ ) yields benzyl isocyanide. The $AgCN$ compound has a more covalent character, so the nitrogen atom's lone pair attacks the electrophilic carbon of the benzyl group, forming the isocyanide.
$C_6H_5CH_2Cl + AgCN \rightarrow C_6H_5CH_2NC + AgCl$

Since both of these are valid and standard methods for preparing benzyl isocyanide, the correct choice combines these two reactions.

Q38JEE Main 2023MCQ4MStates of Matter

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).Assertion (A) : In expensive scientific instruments, silica gel is kept in watch-glasses or in semipermeable membrane bags.Reason (R): Silica gel adsorbs moisture from air via adsorption, thus protects the instrument from water corrosion (rusting) and / or prevents malfunctioning. In the light of the above statements, choose the correct answer from the options given below : [30-Jan-2023 Shift 1]

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Q39JEE Main 2023MCQ4MHaloalkanes and Haloarenes

Match List I with List II [30-Jan-2023 Shift 1]

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Q40JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

Caprolactam when heated at high temperature in presence of water, gives [30-Jan-2023 Shift 1]

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Q41JEE Main 2023MCQ4MClassification of Elements

The alkaline earth metal sulphate(s) which are readily soluble in water is/are:(A) ${\text{BeSO}}_4$ (B) ${\text{MgSO}}_4$ (C) ${\text{CaSO}}_4$ (D) ${\text{SrSO}}_4$ (E) ${\text{BaSO}}_4$ Choose the correct answer from the options given below: [30-Jan-2023 Shift 1]

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📖 Explanation

Due to high hydration energy ${\text{Be}}^{2+}$ and ${\text{Mg}}^{2+}$ , ${\text{BeSO}}_4$ and ${\text{MgSO}}_4$ are readily soluble in water.

Q42JEE Main 2023MCQ4MCoordination Compounds

Which of the following is correct order of ligand field strength? [30-Jan-2023 Shift 1]

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📖 Explanation

The increasing order of field strength of ligands (according to spectrochemical series) ${\text{S}}^{2-} < {\text{C}}_2 {\text{O}}_4^{2-} < {\text{NH}}_3 < {\text{en}} < {\text{CO}}$

Q43JEE Main 2023MCQ4Mp Block Elements

Formation of photochemical smog involves the following reaction in which ${\text{A}}, {\text{B}}$ and ${\text{C}}$ are respectively.(i) ${\text{NO}}_2 {\to }^{\;\text{ hu }\;} {\text{A}}+ {\text{B}}$ (ii) ${\text{B}}+ {\text{O}}_2 \to {\text{C}}$ (iii) ${\text{A}}+ {\text{C}} \to {\text{NO}}_2+ {\text{O}}_2$ Choose the correct answer from the options given below: [30-Jan-2023 Shift 1]

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📖 Explanation

${\text{NO}}_{2 {\text{g}}} {\to }^{ {\text{hv}}} {{\text{NO}}_{ {\text{g}}}}_{(\text{A})}+ {{\text{O}}_{ {\text{g}}}}_{(\text{B})}$ $\;{{ {\text{O}}_{ {\text{g}}}}}_{{( {\text{g}})}}+ {\text{O}}_{2 {\text{g}}} ⇌ \; {\text{O}}_{3 {\text{g}}} _{( {\text{C}})}$ $\;{{ {\text{NO}}_{ {\text{g}}}}}_{{( {\text{A}})}}+\;{ {\text{O}}_{3 {\text{g}}_{\;\text{ (C) }\;}}} \to {\text{NO}}_{2 {\text{g}}}+ {\text{O}}_{2 {\text{g}}}$

Q44JEE Main 2023MCQ4MSalt Analysis

During the qualitative analysis of ${\text{SO}}_3^{2-}$ using dilute ${\text{H}}_2 {\text{SO}}_4, {\text{SO}}_2$ gas is evolved which turns ${\text{K}}_2 {\text{Cr}}_2 {\text{O}}_7$ solution (acidified with dilute ${\text{H}}_2 {\text{SO}}_4$ ): [30-Jan-2023 Shift 1]

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📖 Explanation

${\text{Cr}}_2 {\text{O}}_7^{2-}+ {\text{SO}}_3^{2-} {\to }^{ {\text{H}}^{+}} \;{{ {\text{Cr}}^{3+}}}_{{ {\text{Green}}}}+ {\text{SO}}_4^{2-}$

Q45JEE Main 2023MCQ4MCoordination Compounds

To inhibit the growth of tumours, identify the compounds used from the following:(A) EDTA(B) Coordination Compounds of Pt(C) D-Penicillamine(D) Cis - PlatinChoose the correct answer from the option given below: [30-Jan-2023 Shift 1]

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📖 Explanation

Cis - Platin is used in chemotherapy to inhibits the growth of tumors. (cis $[ {\text{Pt}} ( {\text{NH}}_3) _2 {\text{Cl}}_2] )$

Q46JEE Main 2023MCQ4MSalt Analysis

In the wet tests for identification of various cations by precipitation, which transition element cation doesn't belong to group IV in qualitative inorganic analysis? [30-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{Zn}}^{2+}, {\text{Co}}^{2+}, {\text{Ni}}^{2+}= {\text{IV}}^{\;\text{th }\;} \;\text{ Group }\;$ $\; {\text{Fe}}^{3+}= {\text{III}}^{\;\text{rd }\;} \;\text{ Group }\;$

Q47JEE Main 2023MCQ4MClassification of Elements

Match List I with List II LIST-I (molecules/ions) LIST-II (No. of lone pairs of ${\text{e}}^{-}$ on central atom) (A) $\;\; {\text{IF}}_7$ I. Three (B) ${\text{ICl}}_4^{-}$ II. One (C) $\;\; {\text{XeF}}_6$ III. Two (D) ${\text{XeF}}_2$ IV. ZeroChoose the correct answer from the options given below: [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\; {\text{IF}}_7$ zero lone pair ${\text{ICl}}_4^{-}$ two lone pair $\;\; {\text{XeF}}_6$ one lone pair ${\text{XeF}}_2$ three lone pair

Q48JEE Main 2023MCQ4MChemical Bonding and Molecular Structure

For ${\text{OF}}_2$ molecule consider the following:(A) Number of lone pairs on oxygen is 2 .(B) FOF angle is less than $104.5^{\circ}$ .(C) Oxidation state of ${\text{O}}$ is $-2$ .(D) Molecule is bent ' ${\text{V}}$ ' shaped.(E) Molecular geometry is linear.Correct options are: [30-Jan-2023 Shift 1]

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📖 Explanation

Two lone pair one oxygen- Molecule is ' $v$ ' shaped- Bond angle is less than $104.5^{\circ} (102^{\circ})$ - ${\text{O}} \cdot {\text{S}} \cdot$ of ' ${\text{O}}$ ' is $+2$

Q49JEE Main 2023MCQ4Mp Block Elements

Lithium aluminium hydride can be prepared from the reaction of [30-Jan-2023 Shift 1]

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📖 Explanation

$8 {\text{LiH}}+ {\text{Al}}_2 {\text{Cl}}_6 \to 2 {\text{LiAlH}}_4+6 {\text{LiCl}}$

Q50JEE Main 2023MCQ4MClassification of Elements

Match List - I with List - II LIST-I (Atomic number) LIST-II (Block of periodictable) (A) 37 I. p-block (B) 78 II. d-block (C) 52 III. f-block (D) 65 IV. s-blockChoose the correct answer from the options given below: [30-Jan-2023 Shift 1]

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Q51JEE Main 2023NAT4MElectrochemistry

Consider the cell ${\text{Pt}}_{( {\ s})} | {\text{H}}_2( {\text{g}}, 1 {\text{atm}})| {\text{H}}^{+}( {\text{aq}}, 1 {\text{M}})$ ${\text{Fe}}^{3+}( {\text{aq}}), {\text{Fe}}^{2+}( {\text{aq}}) \mid {\text{Pt}}( {\ s})$ When the potential of the cell is $0.712 {\text{V}}$ at $298 {\text{K}}$ , the ratio $[ {\text{Fe}}^{2+}] / [ {\text{Fe}}^{3+}]$ is _______.(Nearest integer)Given: ${\text{Fe}}^{3+}+ {\text{e}}^{-}= {\text{Fe}}^{2+}, {\text{E}}^{\circ} {\text{Fe}}^{3+}, {\text{Fe}}^{2+} \mid {\text{Pt}}=0.771$ $\;{2.303 {\text{RT}}}/{ {\text{F}}}=0.06 {\text{V}}$ [30-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{Pt}}_{( {\ s})} | {\text{H}}_2( {\text{g}}, 1 {\text{latm}})| {\text{H}}^{+}( {\text{aq}}, 1 {\text{M}}) \| {\text{Fe}}^{3+}( {\text{aq}}), {\text{Fe}}^{2+}( {\text{aq}}) \mid {\text{Pt}}( {\ s})$ $\;\;\text{ at anode }\; {\text{H}}_2 \to 2 {\text{H}}^{+}+2 {\text{e}}^{-}$ $\;\;\text{At cathode }\; {\text{Fe}}_{ {\text{aq}}}^{3+}+ {\text{e}}^{-} \to {\text{Fe}}_{ {\text{aq}}}^{2+}$ $\; {\text{E}}^{\circ}= {\text{E}}_{ {\text{H}}_2 \mid {\text{H}}^{+}}^{\circ}+ {\text{E}}_{ {\text{Fe}}^{\circ +} \mid {\text{Fe}}^{2+}}=0 \cdot 771 {\text{V}}$ $\; {\text{E}}= {\text{E}}^{\circ}-\;\frac{0 \cdot 06}{1} \log \;{ {\text{Fe}}^{2+}}/{ {\text{Fe}}^{3+}}$ $\;0 \cdot 712=(0+0 \cdot 771)-\;\frac{0 \cdot 06}{1} \log \;{ {\text{Fe}}^{2+}}/{ {\text{Fe}}^{3+}}$ $\;\log \;{ {\text{Fe}}^{2+}}/{ {\text{Fe}}^{3+}}=\;\frac{0 \cdot 059}{0 \cdot 06} \approx 1$ $\;\;{ {\text{Fe}}^{2+}}/{ {\text{Fe}}^{3+}}=10$

Q52JEE Main 2023NAT4MSolutions

A $300 {\text{mL}}$ bottle of soft drink has $0.2 {\text{M}} {\text{CO}}_2$ dissolved in it. Assuming ${\text{CO}}_2$ behaves as an ideal gas, the volume of the dissolved ${\text{CO}}_2$ at STP is _______ ${\text{mL}}$ . (Nearest integer)Given: At STP, molar volume of an ideal gas is $22.7 {\text{L}} {\text{mol}}^{-1}$ [30-Jan-2023 Shift 1]

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📖 Explanation

Mole of ${\text{CO}}_2=0.2 {\text{M}} \times (300 \times 10^{-3}) {\text{L}}$ $=0.06$ MoleVolume of $0.06$ mole ${\text{CO}}_2$ at S.T.P $\;=0.06 \times 22.7$ $\;=1.362 {\text{L}}$

Q53JEE Main 2023NAT4MSolutions

A solution containing $2 {\text{g}}$ of a non-volatile solute in $20 {\text{g}}$ of water boils at $373.52 {\text{K}}$ . The molecular mass of the solute is ________ ${\text{gmol}}^{-1}$ . (Nearest integer)Given, water boils at $373 {\text{K}}, {\text{K}}_{ {\text{b}}}$ for water $=0.52 {\text{K}} {\text{kg}} {\text{mol}}^{-1}$ [30-Jan-2023 Shift 1]

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📖 Explanation

$\Delta {\text{T}}_{ {\text{b}}}=373.52-373$ $=0.52$ $\Delta {\text{T}}_{ {\text{b}}}= {\text{Kb}} \cdot {\text{m}}$ $0.52=0.52 \times \;\frac{2}{\;\text{ Molar Mass }\;} \times \;\frac{1}{20 \times 10^{-3}}$ Molar Mass $=100 {\text{g}} / {\text{mol}}$

Q54JEE Main 2023NAT4MChemical Kinetics

If compound A reacts with ${\text{B}}$ following first order kinetics with rate constant $2.011 \times 10^{-3} {\ s}^{-1}$ . The time taken by A (in seconds) to reduce from $7 {\text{g}}$ to $2 {\text{g}}$ will be ________. (Nearest Integer) $[\log 5=0.698, \log 7=0.845, \log 2=0.301]$ [30-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{A}}+ {\text{B}} \to {\text{P}}$ $\; {\text{t}}= {\text{t}}= {\text{t}} \;\; 2 {\text{g}}$ $\;\;\text{ at constant volume }\;$ $\; \;\;\;\; {\text{t}}\;=\;{2.303}/{ {\text{K}}} \log \;{[ {\text{A}}]_0}/{[ {\text{A}}]_{ {\text{t}}}}$ $\;=\;\frac{2 \cdot 303}{2 \cdot 011 \times 10^{-3}} \log \;\frac{7}{2}$ $\;=\;\frac{2 \cdot 303 \times 0 \cdot 544}{2 \cdot 011 \times 10^{-3}}$ $\;=622.989$ $\;\; \approx 623$

Q55JEE Main 2023NAT4MStructure of Atom

The energy of one mole of photons of radiation of frequency $2 \times 10^{12} {\text{Hz}}$ in ${\text{Jmol}}^{-1}$ is _______.(Nearest integer)(Given: ${\text{h}}=6.626 \times 10^{-34} {\text{Js}}$ ${\text{N}}_{ {\text{A}}}=6.022 \times 10^{23} {\text{mol}}^{-1})$ [30-Jan-2023 Shift 1]

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📖 Explanation

For one photon ${\text{E}}= {\text{hv}}$ For one mole photon, ${\text{E}}\;=6.023 \times 10^{23} \times 6.626 \times 10^{-34} \times 2 \times 10^{12}$ $\;=798.16 {\text{J}}$ $\;\; \approx 798 {\text{J}}$

Q56JEE Main 2023NAT4MRedox Reactions

The number of electrons involved in the reduction of permanganate to manganese dioxide in acidic medium is ________. [30-Jan-2023 Shift 1]

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📖 Explanation

${\text{MnO}}_4^{-}+4 {\text{H}}^{+}+3 {\text{e}}^{-} \to {\text{MnO}}_2+2 {\text{H}}_2 {\text{O}}$

Q57JEE Main 2023NAT4MChemical Thermodynamics

When 2 litre of ideal gas expands isothermally into vacuum to a total volume of 6 litre, the change in internal energy is ______ J. (Nearest integer) [30-Jan-2023 Shift 1]

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📖 Explanation

For ideal gas $U=f(T)$ and for isothermal process, $\Delta {\text{U}}=0$

Q58JEE Main 2023NAT4MIonic Equilibrium

$600 {\text{mL}}$ of $0.01 {\text{M}} {\text{HCl}}$ is mixed with $400 {\text{mL}}$ of $0.01 {\text{M}} {\text{H}}_2 {\text{SO}}_4$ . The ${\text{pH}}$ of the mixture is _______ $\times 10^{-2}$ . (Nearest integer)[Given $\log 2=0.30, \;\; \log 3=0.48$ $\log 5=0.69 \;\; \log 7=0.84$ $\log 11=1.04]$ [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\text{ Total milimoles of }\; {\text{H}}^{+}\;=(600 \times 0.01)+(400 \times 0.01 \times 2)$ $\;=14$ ${[ {\text{H}}^{+}]\$ };=;\frac{14}{1000}=14 × 10^{-3}{\text{pH}};=3-\log 14;=1.86;=186 × 10^{-2}$

Q59JEE Main 2023NAT4MAlcohols, Phenols and Ethers

A trisubstituted compound ' ${\text{A}}$ ', ${\text{C}}_{10} {\text{H}}_{12} {\text{O}}_2$ gives neutral ${\text{FeCl}}_3$ test positive. Treatment of compound ' ${\text{A}}$ ' with ${\text{NaOH}}$ and ${\text{CH}}_3 {\text{Br}}$ gives ${\text{C}}_{11} {\text{H}}_{14} {\text{O}}_2$ , with hydroiodic acid gives methyl iodide and with hot conc. ${\text{NaOH}}$ gives a compound $B, {\text{C}}_{10} {\text{H}}_{12} {\text{O}}_2$ . Compound 'A' also decolorises alkaline ${\text{KMnO}}_4$ . The number of $\pi$ bond/s present in the compound ' ${\text{A}}$ ' is _______. [30-Jan-2023 Shift 1]

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Q60JEE Main 2023NAT4MSolutions

Some amount of dichloromethane $( {\text{CH}}_2 {\text{Cl}}_2)$ is added to $671.141 {\text{mL}}$ of chloroform $( {\text{CHCl}}_3)$ to prepare $2.6 \times 10^{-3} {\text{M}}$ solution of ${\text{CH}}_2 {\text{Cl}}_2( {\text{DCM}})$ . The concentration of ${\text{DCM}}$ is ppm _______ (by mass).Given: Atomic mass: ${\text{C}}=12 ; {\text{H}}: 1 ; {\text{Cl}}=35.5$ density of ${\text{CHCl}}_3=1.49 {\text{g}} {\text{cm}}^{-3}$ [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\text{ Molarity }\;=\;\frac{\;\text{ mole }\;}{\;\text{ volume }\;}$ $\;2.6 \times 10^{-3}=\;\frac{x / 85}{0.67141}$ $\; {\text{x}}=0.148 {\text{g}}$ $\;\;\text{ conc. Fo DCM \in ppm }\;=\;\frac{0.148}{1.49 \times 671.141} \times 10^6$ $\;=148 {\text{ppm}}$ $\;$

Mathematics30 questions

Q61JEE Main 2023MCQ4MMatrices and Determinants

Let ${\text{A}}={(\text{table} {\text{m}} , {\text{n}} ; {\text{p}} , {\text{q}})}, {\text{d}}=| {\text{A}}| \ne 0| {\text{A}}- {\text{d}}({\text{Adj}} {\text{A}})|=0$ Then [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\text{ Sol. }\; \;\; {\text{A}}={\begin{bmatrix}{\text{m}} & {\text{n}} \\ {\text{p}} & {\text{q}}\end{bmatrix}}, \;\; | {\text{A}}- {\text{d}}({\text{adj}} {\text{A}})|=0$ $\;\Rightarrow |A-d({\text{adj}} A)|= |{\begin{bmatrix}m & n \\ p & q\end{bmatrix}}-d{\begin{bmatrix}q & -n \\ -p & m\end{bmatrix}}|$ $\;={\begin{vmatrix}{\text{m}}- {\text{qd}} & {\text{n}}(1+ {\text{d}}) \\ {\text{p}}(1+ {\text{d}}) & {\text{q}}- {\text{md}}\end{vmatrix}}=0$ $\;\Rightarrow \;\; ( {\text{m}}- {\text{qd}})( {\text{q}}- {\text{md}})- {\text{np}}(1+ {\text{d}})^2=0$ $\;\Rightarrow \;\; m q-m^2 d-q^2 d+m q d^{2-}n p(1+d)^2=0$ $\;\Rightarrow \;\; (m q-n p)+d^2(m q-n p)-d (m^{2+}q^{2+}2 n p) =0$ $\;\Rightarrow \;\; {\text{d}}+ {\text{d}}^{3-} {\text{d}} (( {\text{m}}+ {\text{q}})^{2-}2 {\text{d}}) =0$ $\;\Rightarrow \;\; 1+d^2=( {\text{m}}+ {\text{q}})^{2-}2 d$ $\;\Rightarrow \;\; (1+ {\text{d}})^2=( {\text{m}}+ {\text{q}})^2$ $\;$

Q62JEE Main 2023MCQ4MThree Dimensional Geometry

The line $l_1$ passes through the point $(2,6,2)$ and is perpendicular to the plane $2 x+y-2 z=10$ . Then the shortest distance between the line $l_1$ and the line $\;{ {\text{x}}+1}/{2}=\;{ {\text{y}}+4}/{-3}=\;{ {\text{z}}}/{2}$ is : [30-Jan-2023 Shift 1]

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📖 Explanation

Line $ℓ$ , is given by $L_1: \;\frac{x-2}{2}=\;\frac{y-6}{1}=\;\frac{z-2}{-2}$ Given, ${\text{L}}_2: \;{ {\text{x}}+1}/{2}=\;{ {\text{y}}+4}/{-3}=\;{ {\text{z}}}/{2}$ $\;\;\text{ Shortest distance }\;= |\;{{ \vec{\text{AB}} \cdot { \vec{\text{MN}}}/{ {\text{MN}}}|$ $\;\overline{ {\text{AB}}}=3 \;{ {\text{i}}}^{\^}+10 \;{ {\text{j}}}^{\^}+2 \;{ {\text{k}}}^{\^}$ $\;\overline{ {\text{MN}}}={\begin{vmatrix}\\ \\ { {\text{i}}}^{\^} & \\ \\ { {\text{j}}}^{\^} & \\ \\ { {\text{k}}}^{\^} \\ 2 & 1 & -2 \\ 2 & -3 & 2\end{vmatrix}}=-4 \;{ {\text{i}}}^{\^}-8 \;{ {\text{j}}}^{\^}-8 \;{ {\text{k}}}^{\^}$ $\; {\text{MN}}={\sqrt{16+64+64}}=12$ $\; \therefore \;\; \;\text{ Shortest distance }\;= |\;\frac{-12-80-16}{12}|=9$ $\therefore \;\;$ Option (4) is correct.

Q63JEE Main 2023MCQ4MProbability

If an unbiased die, marked with $-2,-1,0,1,2,3$ on its faces, is through five times, then the probability that the product of the outcomes is positive, is : [30-Jan-2023 Shift 1]

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📖 Explanation

Either all outcomes are positive or any two are negative. $\;\;\text{ Now, }\; {\text{p}}= {\text{P}}(\;\text{ positive }\;)=\;\frac{3}{6}=\;\frac{1}{2}$ $\; {\text{q}}= {\text{p}}(\;\text{ negative }\;)=\;\frac{2}{6}=\;\frac{1}{3}$ Required probability $\;={ }^5 {\text{C}}_5 (\;\frac{1}{2}) ^{5+}{ }^5 {\text{C}}_2 (\;\frac{1}{3}) ^2 (\;\frac{1}{2}) ^{3+}{ }^5 {\text{C}}_4 (\;\frac{1}{3}) ^4 (\;\frac{1}{2}) ^1$ $\;=\;\frac{521}{2592}$ $\therefore$ Option (2) is correct.

Q64JEE Main 2023MCQ4MMatrices and Determinants

Let the system of linear equations $\;x+y+k z=2$ $\;2 x+3 y-z=1$ $\;3 x+4 y+2 z=k$ have infinitely many solutions. Then the system $\;( {\text{k}}+1) {\text{x}}+(2 {\text{k}}-1) {\text{y}}=7$ $\;(2 {\text{k}}+1) {\text{x}}+( {\text{k}}+5) {\text{y}}=10 \;\text{ has: }\;$ [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\; {\begin{vmatrix}1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2\end{vmatrix}}=0$ $\Rightarrow \;\; 1(10)-1(7)+k(-1)-0$ $\Rightarrow \;\; k=3$ For ${\text{k}}=3,2^{\;\text{nd }\;}$ system is $4 x+5 y=7$ . . . (1)and $7 x+8 y=10$ . . . (2)Clearly, they have a unique solution(2) $-(1) \Rightarrow 3 x+3 y=3$ $\Rightarrow x+y=1$

Q65JEE Main 2023MCQ4MTrigonometric Ratios and Identities

If $\tan 15^{\circ}+\;\frac{1}{\tan 75^{\circ}}+\;\frac{1}{\tan 105^{\circ}}+\tan 195^{\circ}=2 {\text{a}}$ then the value of $(a+\;\frac{1}{a})$ is : [30-Jan-2023 Shift 1]

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📖 Explanation

Option (1) $\;\tan 15^{\circ}=2-{\sqrt{3}}$ $\;\;\frac{1}{\tan 75^{\circ}}=\cot 75^{\circ}=2-{\sqrt{3}}$ $\;\;\frac{1}{\tan 105^{\circ}}=\cot (105^{\circ}) =-\cot 75^{\circ}={\sqrt{3}}-2$ $\;\tan 195^{\circ}=\tan 15^{\circ}=2-{\sqrt{3}}$ $\therefore \;\; 2(2-{\sqrt{3}})=2 {\text{a}} \Rightarrow {\text{a}}=2-{\sqrt{3}}$ $\Rightarrow \;\; {\text{a}}+\;{1}/{ {\text{a}}}=4$

Q66JEE Main 2023MCQ4MFunctions

Suppose ${\text{f}}: {\text{R}} \to (0, \infty )$ be a differentiable function such that $5 {\text{f}}( {\text{x}}+ {\text{y}})= {\text{f}}( {\text{x}}) \cdot {\text{f}}( {\text{y}}), ∀ {\text{x}}, {\text{y}} \in {\text{R}}$ . If $f(3)=320$ , then $sum_{n=0}^{5} f(n)$ is equal to: [30-Jan-2023 Shift 1]

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📖 Explanation

Option (3) $\;5 {\text{f}}( {\text{x}}+ {\text{y}})= {\text{f}}( {\text{x}}) \cdot {\text{f}}( {\text{y}})$ $\;5 {\text{f}}(0)= {\text{f}}(0)^2 \Rightarrow {\text{f}}(0)=5$ $\;5 {\text{f}}( {\text{x}}+1)= {\text{f}}( {\text{x}}) \cdot {\text{f}}(1)$ $\Rightarrow \;\; \;{ {\text{f}}( {\text{x}}+1)}/{ {\text{f}}( {\text{x}})}=\;{ {\text{f}}(1)}/{5}$ $\Rightarrow \;\; \;{ {\text{f}}(1)}/{ {\text{f}}(0)} \cdot \;{ {\text{f}}(2)}/{ {\text{f}}(1)} \cdot \;{ {\text{f}}(3)}/{ {\text{f}}(2)}= (\;{ {\text{f}}(1)}/{5}) ^3$ $\Rightarrow \;\; \;\frac{320}{5}=\;{( {\text{f}}(1))^3}/{5^3} \Rightarrow {\text{f}}(1)=20$ $\therefore \;\; 5 {\text{f}}( {\text{x}}+1)=20 \cdot {\text{f}}( {\text{x}}) \Rightarrow {\text{f}}( {\text{x}}+1)=4 {\text{f}}( {\text{x}})$ $\; \;\; sum_{{ {\text{n}}=0}}^{5} {\text{f}}( {\text{n}})=5+5.4+5.4^{2+}5.4^{3+}5.4^{4+}5.4^5$ $\;=\;\frac{5 \$ [4^{6-}1]$ }{3}=6825$

Q67JEE Main 2023MCQ4MSequences and Series

If $a_n=\;\frac{-2}{4 n^{2-}16 n+15}$ , then $a_1+a_2+. . . . . .+a_{25}$ is equal to : [30-Jan-2023 Shift 1]

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📖 Explanation

Option (3)If ${\text{a}}_{ {\text{n}}}=\;{-2}/{4 {\text{n}}^{2-}16 {\text{n}}+15}$ then ${\text{a}}_1+ {\text{a}}_2+. . . . . . . . . {\text{a}}_{25}$ $\;\Rightarrow \;\; sum_{n=1}^{25} a_n=\sum \;\frac{-2}{4 n^{2-}16 n+15}$ $\;\; =\sum \;\frac{-2}{4 n^{2-}6 n-10 n+15}$ $\;\; =\sum \;\frac{-2}{2 n(2 n-3)-5(2 n-3)}$ $\;\; =\sum \;\frac{-2}{(2 n-3)(2 n-5)}$ $\;\; =\sum \;\frac{1}{2 n-3}-\;\frac{1}{2 n-5}$ $\;\; =\;\frac{1}{47}-\;\frac{1}{(-3)}$ $\;\; =\;\frac{50}{141}$

Q68JEE Main 2023MCQ4MBinomial Theorem

If the coefficient of $x^{15}$ in the expansion of $(a x^{3+}\;{1}/{b x^{\;\frac{1}{3}}}) ^{15}$ is equal to the coefficient of $x^{-15}$ in the expansion of $(a x^{\;\frac{1}{3}}-\;\frac{1}{b x^3}) ^{15}$ , where $a$ and $b$ are positive real numbers, then for each such ordered pair (a, b) : [30-Jan-2023 Shift 1]

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📖 Explanation

Option (2) $\;\;\text{ Coefficient Of }\; x^{15} \;\text{ \in }\; (a x^{3+}\;\frac{1}{b x^{1 / 3}}) ^{15}$ $\;T_{r+1}={ }^{15} C_r (a x^3) ^{15-r} (\;\frac{1}{b x^{1 / 3}}) ^r$ $\;45-3 r-\;\frac{r}{3}=15$ $\;30=\;\frac{10 r}{3}$ $\;r=9$ Coefficient of $x^{15}={ }^{15} C_9 a^6 b^{-9}$ Coefficient of $x^{-15}$ in $(a x^{1 / 3}-\;\frac{1}{b x^3}) ^{15}$ $T_{r+1}={ }^{15} C_r (a x^{1 / 3}) ^{15-r} (-\;\frac{1}{b x^3}) ^r$ $\;5-\;\frac{r}{3}-3 r=-15$ $\;\;\frac{10 r}{3}=20$ $\;r=6$ $\;\;\text{ Coefficient }\;={ }^{15} {\text{C}}_6 {\text{a}}^9 \times {\text{b}}^{-6}$ $\Rightarrow \;\; \;{ {\text{a}}^9}/{ {\text{b}}^6}=\;{ {\text{a}}^6}/{ {\text{b}}^9}$ $\Rightarrow \;\; {\text{a}}^3 {\text{b}}^3=1 \Rightarrow {\text{ab}}=1$

Q69JEE Main 2023MCQ4MVector Algebra

If ${ \vec{\text{a}}, { \vec{\text{b}}, { \vec{\text{c}}$ are three non-zero vectors and $\;{ {\text{n}}}^{\^}$ is a unit vector perpendicular to ${ \vec{\text{c}}$ such that ${ \vec{\text{a}}=\alpha { \vec{\text{b}}-\;{ {\text{n}}}^{\^},(\alpha \ne 0)$ and ${ \vec{\text{b}} \cdot { \vec{\text{c}}=12$ , then $|{ \vec{\text{c}} \times ({ \vec{\text{a}} \times { \vec{\text{b}})|$ is equal to : [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\;{ {\text{n}}}^{\^} ⟂ { \vec{\text{c}} \;\; { \vec{\text{a}}=\alpha { \vec{\text{b}}-{ \vec{\text{n}}$ $\;{ \vec{\text{b}} \cdot { \vec{\text{c}}=12$ $\;\vec{a} \cdot { \vec{\text{c}}=\alpha(\vec{b} \cdot { \vec{\text{c}})-{ \vec{\text{n}} \cdot { \vec{\text{c}}$ $\;\vec{a} \cdot { \vec{\text{c}}=\alpha({ \vec{\text{b}} \cdot \overline{ {\text{c}}})$ $\;|{ \vec{\text{c}} \times ({ \vec{\text{a}} \times { \vec{\text{b}})|=|({ \vec{\text{c}} \cdot { \vec{\text{b}}) { \vec{\text{a}}-({ \vec{\text{c}} \cdot { \vec{\text{a}}) { \vec{\text{b}}|$ $\;=|({ \vec{\text{c}} \cdot { \vec{\text{b}}) { \vec{\text{a}}-\alpha({ \vec{\text{b}} \cdot { \vec{\text{c}}) { \vec{\text{b}}|$ $\;=|({ \vec{\text{c}} \cdot { \vec{\text{b}})||{ \vec{\text{a}}-\alpha { \vec{\text{b}}|$ $\;=12 \times (|\vec{n}|)$ $\;=12 \times 1$ $\;=12$ $\;$

Q70JEE Main 2023MCQ4MApplication of Derivatives

The number of points on the curve $y=54 x^{5-}$ $135 x^{4-}70 x^{3+}180 x^{2+}210 x$ at which the normal lines are parallel to ${\text{x}}+90 {\text{y}}+2=0$ is : [30-Jan-2023 Shift 1]

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📖 Explanation

Normal of line is parallel to line $x+90 y+2=0$ $\; {\text{m}}_{ {\text{N}}}=-\;\frac{1}{90}$ $\;- (\;{ {\text{dx}}}/{ {\text{dy}}}) _{ (x_1 y_1) }=-\;\frac{1}{90} \Rightarrow (\;{ {\text{dy}}}/{ {\text{dx}}}) _{ (x_1 y_1) }=90$ Now, $\;\;\frac{d y}{d x}=270 x^{4-}540 x^{3-}210 x^{2+}360 x+210=90$ $\;\Rightarrow x=1,2, \;\frac{-2}{3}, \;\frac{-1}{3}$ (4) normals

Q71JEE Main 2023MCQ4MCircles

Let $y=x+2,4 y=3 x+6$ and $3 y=4 x+1$ be three tangent lines to the circle $( {\text{x}}- {\text{h}})^{2+}( {\text{y}}- {\text{k}})^2= {\text{r}}^2$ . Then ${\text{h}}+ {\text{k}}$ is equal to : [30-Jan-2023 Shift 1]

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📖 Explanation

${\text{L}}_1: {\text{y}}= {\text{x}}+2, {\text{L}}_2: 4 {\text{y}}=3 {\text{x}}+6, {\text{L}}_3: 3 {\text{y}}=4 {\text{x}}+1$ Bisector of lines ${\text{L}}_2 & {\text{L}}_3$ $\;\;\frac{4 x-3 y+1}{5}=\pm (\;\frac{3 x-4 y+6}{5})$ $\;\;\text{ (+) }\; 4 x-3 y+1=3 x-4 y+6$ $\;x+y=5$ Centre lies on Bisector of $4 x-3 y+1=0 &$ $\;\text{ (0) }\; 3 x-4 y+6=0$ $\Rightarrow {\text{h}}+ {\text{k}}=5$

Q72JEE Main 2023MCQ4MDifferential Equations

Let the solution curve ${\text{y}}= {\text{y}}( {\text{x}})$ of the differential equation $\;\; \;\frac{d y}{d x}-\;{3 x^5 \tan ^{-1} (x^3) }/{ (1+x^6) ^{\;\frac{3}{2}}} y=2 x$ $\exp \;{x^{3-}\tan ^{-1} x^3}/{{\sqrt{(1+x)^6}}}$ pass through the origin. Then ${\text{y}} (1)$ is equal to : [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\frac{d y}{d x}+ (\;\frac{-3 x^5 \tan ^{-1} x^3}{ (1+x^6) ^{3 / 2}}) y=2 e^{ \{\;{x-\tan x}/{{\sqrt{1+x^6}}}\} }$ $\;\text{ I.F. }\;\;=e^{\int \;\frac{-3 x^5 \tan ^{-1} x^3}{ (1+x^6) ^{3 / 2}} \text{dx}}$ $\;=e^{\;{\tan ^{-1} x^{3-}x^3}/{{\sqrt{1+x^6}}}}$ Solution of differential equation $\;y \cdot e^{\;{\tan ^{-1} x^{3-}x^3}/{{\sqrt{1+x^6}}}}=\int 2 x e^{ (\;{x^{3-}\tan ^{-1} x^3}/{{\sqrt{1+x^6}}}) } \cdot e^{ (\;{\tan ^{-1} (x^3) -x^3}/{{\sqrt{1+x^6}}}) } \text{dx}$ $\; \;\; =\int 2 x \text{dx}+c$ $\;y \cdot e^{\;{\tan ^{-1} x^{3-}x^3}/{{\sqrt{1+x^6}}}}=x^{2+}c$ Also it passes through origin ${\text{c}}=0$ $\;y(1) \cdot e^{\;{\tan ^{-1}(1)-1}/{{\sqrt{2}}}}=1$ $\;y(1) \cdot e^{\;{\;\frac{\pi}{4}-1}/{{\sqrt{2}}}}=1$ $\;y(1) \cdot e^{\;{\pi-4}/{4 {\sqrt{2}}}}=1$ $\;y(1)=\;{1}/{e^{\;{\pi-4}/{4 {\sqrt{2}}}}}=e^{\;{4-\pi}/{4 {\sqrt{2}}}}$

Q73JEE Main 2023MCQ4MVector Algebra

Let a unit vector ${ {\text{OP}}^{◠}}$ make angle $\alpha, \beta, \gamma$ with the positive directions of the co-ordinate axes ${\text{OX}}, {\text{OY}}$ , ${\text{OZ}}$ respectively, where $\beta \in (0, \;\frac{\pi}{2}) { {\text{OP}}^{◠}}$ is perpendicular to the plane through points (1, 2, 3), $(2,3,4)$ and $(1,5,7)$ , then which one of the following is true ? [30-Jan-2023 Shift 1]

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📖 Explanation

Equation of plane :- $\;{\begin{vmatrix}{\text{x}}-1 & {\text{y}}-2 & {\text{z}}-3 \\ 1 & 1 & 1 \\ 0 & 3 & 4\end{vmatrix}}=0$ $\;\Rightarrow [ {\text{x}}-1]-4[ {\text{y}}-2]+3[ {\text{z}}-3]=0$ $\;\Rightarrow {\text{x}}-4 {\text{y}}+3 {\text{z}}=2$ D.R's of normal of plane $< 1,-4,3>$ D.C's of $\;\; ⟨\pm \;{1}/{{\sqrt{26}}}, ∓ \;{4}/{{\sqrt{26}}}, \pm \;{3}/{{\sqrt{26}}}⟩$ $\;\cos \beta=\;{4}/{{\sqrt{26}}}$ $\;\cos \alpha=\;{-1}/{{\sqrt{26}}} \;\; \;\;\;\;\frac{\pi}{2} < \alpha < \pi$ $\;\cos \gamma =\;{-3}/{{\sqrt{26}}} \;\; \;\;\;\;\frac{\pi}{2} < \gamma < \pi$

Q74JEE Main 2023MCQ4MDefinite Integrals

If [t denotes the greatest integer $\le 1$ , then the value of of $\;\frac{3(e-1)^2}{e} int_{1}^{2} x^2 e^{[x]+ \$ [x^3]$ } \text{dx}$ is : [30-Jan-2023 Shift 1]

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📖 Explanation

$int_{1}^{2} {\text{x}}^2 {\text{e}}^{[ {\text{x}}^3]\$ +1} \text{dx}{\text{x}}^3= {\text{t}}3 {\text{x}}^2 {\text{dx}}= {\text{dt}}=;{ {\text{e}}}/{3} int_{1}^{8} {\text{e}}^{[ {\text{t}}]} {\text{dt}}=;{ {\text{e}}}/{3} {int_{1}^{2} {\text{e}} {\text{dt}}+int_{2}^{3} {\text{e}}^2 {\text{dt}}+. . . . . . . . .+int_{7}^{8} {\text{e}}^7 {\text{dt}}}=;{ {\text{e}}}/{3} ( {\text{e}}+ {\text{e}}^{2+}. . . . . . . . .+ {\text{e}}^7)=;{ {\text{e}}^2}/{3} (1+ {\text{e}}+. . . . . . . . .+ {\text{e}}^6) =;{ {\text{e}}^2}/{3} ;{ ( {\text{e}}^{7-}1) }/{( {\text{e}}-1)};{3( {\text{e}}-1)}/{ {\text{e}}} int_{1}^{2} {\text{x}}^2 × {\text{e}}^{[ {\text{x}}]+ $[ {\text{x}}^3]$ } {\text{dx}}=;{3}/{ {\text{e}}}( {\text{e}}-1) × ;{ {\text{e}}^2}/{3} ;{ ( {\text{e}}^{7-}1) }/{( {\text{e}}-1)}= {\text{e}} ( {\text{e}}^{7-}1)= {\text{e}}^{8-} {\text{e}}$

Q75JEE Main 2023MCQ4MApplication of Derivatives

If ${\text{P}}( {\text{h}}, {\text{k}})$ be point on the parabola ${\text{x}}=4 {\text{y}}^2$ , which is nearest to the point $Q(0,33)$ , then the distance of $P$ from the directrix of the parabola ${\text{y}}^2=4( {\text{x}}+ {\text{y}})$ is equal to : [30-Jan-2023 Shift 1]

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📖 Explanation

Equation of normal $\;y=-t x+2 a t+a t^3$ $\;y=-t x+\;\frac{2}{16} t+\;\frac{1}{16} t^3$ It passes through $(0,33)$ $\;33=\;\frac{t}{8}+\;\frac{t^3}{16}$ $\;t^{3+}2 t-528=0$ $\;(t-8) (t^{2+}8 t+66) =0$ $\;t=8$ $\;P (a t^2, 2 a t) = (\;\frac{1}{16} \times 64,2 \times \;\frac{1}{16} \times 8) =(4,1)$ Parabola : $\;y^2=4(x+y)$ $\;\Rightarrow y^{2-}4 y=4 x$ $\;\Rightarrow (y-2)^2=4(x+1)$ Equation of directix :- $\;x+1=-1$ $\;x=-2$ Distance of point $=6$ Ans. : (4)

Q76JEE Main 2023MCQ4MStraight Lines and Pair of Straight Lines

A straight line cuts off the intercepts ${\text{OA}}= {\text{a}}$ and ${\text{OB}}= {\text{b}}$ on the positive directions of ${\text{x}}$ -axis and ${\text{y}}-$ axis respectively. If the perpendicular from origin ${\text{O}}$ to this line makes an angle of $\;\frac{\pi}{6}$ with positive direction of $y$ -axis and the area of $\triangle O A B$ is $\;\frac{98}{3} {\sqrt{3}}$ , then ${\text{a}}^{2-} {\text{b}}^2$ is equal to: [30-Jan-2023 Shift 1]

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📖 Explanation

Equation of straight line : $\;{ {\text{x}}}/{ {\text{a}}}+\;{ {\text{y}}}/{ {\text{b}}}=1$ Or $x \cos \;\frac{\pi}{3}+y sin \;\;\frac{\pi}{3}=p$ $\;\frac{x}{2}+\;{y {\sqrt{3}}}/{2}=p$ $\;\frac{x}{3 p}+\;\frac{y}{2 p}=1$ Comparing both : $a=2 p, b=\;{2 p}/{{\sqrt{3}}}$ Now area of $\triangle {\text{OAB}}=\;\frac{1}{2} \cdot {\text{ab}}=\;\frac{98}{3} \cdot {\sqrt{3}}$ $\;\frac{1}{2} \cdot 2 {\text{p}} \cdot \;{2 {\text{p}}}/{{\sqrt{3}}}=\;\frac{98}{3} \cdot {\sqrt{3}}$ ${\text{p}}^2=49$ ${\text{a}}^{2-} {\text{b}}^2=4 {\text{p}}^{2-}\;{4 {\text{p}}^2}/{3}=\;\frac{2}{3} 4 {\text{p}}^2$ $=\;\frac{8}{3} \cdot 49=\;\frac{392}{3}$

Q77JEE Main 2023MCQ4MBinomial Theorem

The coefficient of ${\text{x}}^{301}$ in $(1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+. . . . .+x^{500}$ is: [30-Jan-2023 Shift 1]

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📖 Explanation

$\;(1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+. . . .+x^{500}$ $\;=(1+x)^{500} \cdot \{\;\frac{1- (\frac{x}{1+x}) ^{501}}{1-\;\frac{x}{1+x}}\}$ $\;=(1+x)^{500} \;\frac{ ((1+x)^{501}-x^{501}) }{(1+x)^{501}} \cdot (1+x)$ $\;=(1+x)^{501}-x^{501}$ Coefficient of $x^{301}$ in $(1+x)^{501}-x^{501}$ is given by ${ }^{501} {\text{C}}_{301}={ }^{501} {\text{C}}_{200}$

Q78JEE Main 2023MCQ4MSets and Relations

Among the statements:(S1) $\;\; (( {\text{p}} ∨ {\text{q}}) \Rightarrow {\text{r}}) ⇔( {\text{p}} \Rightarrow {\text{r}})$ (S2) $\;\; (( {\text{p}} ∨ {\text{q}}) \Rightarrow {\text{r}}) ⇔(( {\text{p}} \Rightarrow {\text{r}}) ∨( {\text{q}} \Rightarrow {\text{r}}))$ [30-Jan-2023 Shift 1]

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📖 Explanation

${\text{S}}_1 \equiv (( {\text{p}} ∨ {\text{q}}) \Rightarrow {\text{r}}) ⇔( {\text{p}} \Rightarrow {\text{r}})$ $S_2 \equiv (p ∨ q) \Rightarrow r ⇔((p \Rightarrow r) ∨(q \Rightarrow r))$ S2 $\to$ not a tautology

Q79JEE Main 2023MCQ4MSets and Relations

The minimum number of elements that must be added to the relation ${\text{R}}=\{( {\text{a}}, {\text{b}}),( {\text{b}}, {\text{c}})\}$ on the set $\{ {\text{a}}, {\text{b}}, {\text{c}}\}$ so that it becomes symmetric and transitive is: [30-Jan-2023 Shift 1]

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📖 Explanation

For Symmetric $(a, b),(b, c) \in R$ $\Rightarrow (b, a),(c, b) \in R$ For Transitive $(a, b),(b, c) \in R$ $\Rightarrow (a, c) \in R$ Now1. Symmetric $\therefore (a, c) \in R \Rightarrow (c, a) \in R$ 2. Transitive $\; \therefore (a, b),(b, a) \in R$ $\;\Rightarrow (a, a) \in R &(b, c),(c, b) \in R$ $\;\Rightarrow (b, b) &(c, c) \in R$ $\therefore$ Elements to be added ${\{\text{table} (b, a),(c, b),(a, c),(c, a) ;{, \;}(a, a),(b, b),(c, c) \}}$ Number of elements to be added $=7$

Q80JEE Main 2023MCQ4MLogarithms

If the solution of the equation $\log _{\cos x} \cot x+4 \log _{ sin \;x} \tan x=1, x \in (0, \;\frac{\pi}{2}) , \;\;$ is $sin \;^{-1} (\;{\alpha+{\sqrt{\beta}}}/{2})$ , where $\alpha, \beta$ are integers, then $\alpha+\beta$ is equal to: [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\log _{\cos x} \cot x+4 \log _{ sin \;x} \tan x=1$ $\;\Rightarrow \;\frac{\ln \cos x-\ln sin \;x}{\ln \cos x}+4 \;\frac{\ln sin \;x-\ln \cos x}{\ln sin \;x}=1$ $\;\Rightarrow (\ln sin \;x)^{2-}4(\ln sin \;x)(\ln \cos x)+4(\ln \cos x)^2=1$ $\;\Rightarrow \ln sin \;x=2 \ln \cos x$ $\;\Rightarrow sin \;^2 x+ sin \;x-1=0 \Rightarrow sin \;x=\;{-1+{\sqrt{5}}}/{2}$ $\; \therefore \alpha+\beta=4$ Correct option (4)

Q81JEE Main 2023NAT4MPermutations and Combinations

Let $S=\{1,2,3,4,5,6\}$ . Then the number of oneone functions $f: S \to P(S)$ , where $P(S)$ denote the power set of S, such that $f(n) ⊂ f(m)$ where ${\text{n}} < {\text{m}}$ is ________. [30-Jan-2023 Shift 1]

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📖 Explanation

Let $S=\{1,2,3,4,5,6\}$ , then the number of one-one functions, ${\text{f}}: {\text{S}} \cdot {\text{P}}( {\text{S}})$ , where ${\text{P}}( {\text{S}})$ denotes the power set of ${\text{S}}$ , such that ${\text{f}}( {\text{n}}) < {\text{f}}( {\text{m}})$ where ${\text{n}} < {\text{m}}$ is $\;n(S)=6$ $\;P(S)={\begin{cases}\varphi & \{1\}\end{cases} . .\{6\},\{1,2\}, . . ., ; \{5,6\}, . . .,\{1,2,3,4,5,6\} \}}$ $\; \;\; -64 \;\text{ elements }\;$ case $-1$ $f(6)=S$ i.e. 1 option, $f(5)=$ any 5 element subset A of S i.e. 6 options, $f(4)=$ any 4 element subset B of $A$ i.e. 5 options, ${\text{f}}(3)=$ any 3 element subset ${\text{C}}$ of ${\text{B}}$ i.e. 4 options, ${\text{f}}(2)=$ any 2 element subset D of ${\text{C}}$ i.e. 3 options, $f(1)=$ any 1 element subset $E$ of $D$ or empty subset i.e. 3options,Total functions $=1080$ Case - 2 $f(6)=$ any 5 element subset $A$ of $S$ i.e. 6 options, $f(5)=$ any 4 element subset B of A i.e. 5 options,f' (4) = any 3 element subset $C$ of B i.e. 4 options, $f(3)=$ any 2 element subset $D$ of $C$ i.e. 3 options,f' $(2)=$ any 1 element subset $E$ of D i.e. 2 options, $f(1)=$ empty subset i.e. 1 optionTotal functions $=720$ Case $-3$ ${\text{f}}(6)= {\text{S}}$ $f(5)=$ any 4 element subset A of' S i.e. 15 options, $f(4)=$ any 3 element subset B of A i.e. 4 options, $f(3)=$ any 2 element subset $C$ of B i.e. 3 options, $f(2)=$ any 1 element subset D of C i.e. 2 options, $f(1)=$ empty subset i.e. 1 option Total functions $=360$ Case $-4$ ${\text{f}}(6)= {\text{S}}$ $f(5)=$ any 5 element subset $A$ of $S$ i.e. 6 options, $f(4)=$ any 3 element subset B of A i.e. 10 options, $f(3)=$ any 2 element subset $C$ of B i.e. 3 options, $f(2)=$ any 1 element subset $D$ of $C$ i.e. 2 options, $f(1)=$ empty subset i.e. 1 optionTotal functions $=360$ Case $-5$ ${\text{f}}(6)= {\text{S}}$ $f(5)=$ any 5 element subset A of S i.e. 6 options, $f(4)=$ any 4 element subset B of A i.e. 5 options, $f(3)=$ any 2 element subset $C$ of B i.e. 6 options, $f(2)=$ any 1 element subset $D$ of $C$ i.e. 2 options, ${\text{f}}(1)=$ empty subset i.e. 1 optionTotal functions $=360$ Case - 6 ${\text{f}}(6)= {\text{S}}$ $f(5)=$ any 5 element subset $A$ of $S$ i.e. 6 options, $f(4)=$ any 4 element subset B of A i.e. 5 options, $f(3)=$ any 3 element subset $C$ of B i.e. 4 options, $f(2)=$ any 1 element subset $D$ of $C$ i.e. 3 options, $f(1)=$ empty subset i.e. 1 optionTotal functions $=360$ $\therefore$ Number of such functions $=3240$

Q82JEE Main 2023NAT4MArea Under The Curves

Let $\alpha$ be the area of the larger region bounded by the curve ${\text{y}}^2=8 {\text{x}}$ and the lines ${\text{y}}= {\text{x}}$ and ${\text{x}}=2$ , which lies in the first quadrant. Then the value of $3 \alpha$ is equal to _______. [30-Jan-2023 Shift 1]

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📖 Explanation

$\;y=x$ $\;\ y^2=8 x$ Solving it $\;x^2=8 x$ $\therefore \;\; x=0,8$ $\therefore \;\; y=0,8$ ${\text{x}}=2$ will intersect occur at $y^2=16 \Rightarrow \;\; y=\pm 4$ $\therefore \;\;$ Area of shaded $\;=int_{2}^{8}({\sqrt{8 x}}-x) \text{dx}=\int{2^8(2 {\sqrt{2}} {\sqrt{x}}-x) \text{dx}$ $= \$ [2 {\sqrt{2}} ⋅ ;\frac{x^{3 / 2}}{3 / 2}-;\frac{x^2}{2}]$ _0^8;= (;{4 {\sqrt{2}}}/{3} \cdot 2^{9 / 2}-32) - (;{4 {\sqrt{2}}}/{3} \cdot 2^{9 / 2}-2);=;\frac{128}{3}-32-;\frac{16}{3}+2=;\frac{112-90}{3}=;\frac{22}{3}= {\text{A}}; \therefore 3 {\text{A}}=22$

Q83JEE Main 2023NAT4MThree Dimensional Geometry

If $\lambda _1 < \lambda _2$ are two values of $\lambda$ such that the angle between the planes $P_1: \overline{ {\text{r}}}(3 \;{ {\text{i}}}^{\^}-5 \;{ {\text{j}}}^{\^}+\;{ {\text{k}}}^{\^})=7$ and $P_2: \vec{r} \cdot (\lambda \hat{i}+\hat{j}-3 \hat{k})=9$ is $sin \;^{-1} (\;{2 {\sqrt{6}}}/{5})$ , then the square of the length of perpendicular from the point $(38 \lambda _1, 10 \lambda _2, 2)$ to the plane $P_1$ is _______. [30-Jan-2023 Shift 1]

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📖 Explanation

$P_1={ \vec{\text{r}} \cdot (3 \;{ {\text{i}}}^{\^}-5 \;{ {\text{j}}}^{\^}+\;{ {\text{k}}}^{\^})=7$ ${\text{P}}_2={ \vec{\text{r}} \cdot (\lambda \;{ {\text{i}}}^{\^}+\;{ {\text{j}}}^{\^}-3 \;{ {\text{k}}}^{\^})=9$ $\;\theta= sin \;^{-1} (\;{2 {\sqrt{6}}}/{5})$ $\;\Rightarrow sin \;\theta=\;{2 {\sqrt{6}}}/{5}$ $\therefore \cos \theta=\;\frac{1}{5}$ $\cos \theta=\;{{ \vec{\text{r}} \cdot { \vec{\text{r}}}/{ |\overline{ {\text{r}}}_1| |{ \vec{\text{r}}_2|}$ $=\;{(3 i-5 j+K)(\lambda i+j-3 K)}/{{\sqrt{35}} \cdot {\sqrt{\lambda ^{2+}10}}}$ $\;\frac{1}{5}= |\;{3 \lambda -8}/{{\sqrt{35}} \cdot {\sqrt{\lambda ^{2+}10}}}|$ Square $\Rightarrow \;\frac{1}{25}=\;\frac{9 \lambda ^{2+}64-48 \lambda }{35 (\lambda ^{2+}10) }$ $\Rightarrow 19 \lambda ^{2-}120 \lambda +125=0$ $\;\Rightarrow 19 \lambda ^{2-}95 \lambda -25 \lambda +125=0$ $\;\Rightarrow x=5, \;\frac{25}{19}$ Perpendicular distance of point $(38 \lambda _1, 10 \lambda _2, 2) \equiv (50,50,2)$ from plane $P_1$ $=\;{|3 \times 50-5 \times 50+2-7|}/{{\sqrt{35}}}=\;{105}/{{\sqrt{35}}}$ $\;\text{ Square }\;=\;\frac{105 \times 105}{35}=315$

Q84JEE Main 2023NAT4MComplex Numbers

Let ${\text{z}}=1+ {\text{i}}$ and ${\text{z}}_1=\;{1+ {\text{i}} \overline{ {\text{z}}}}/{\overline{ {\text{z}}}(1- {\text{z}})+\;{1}/{ {\text{z}}}}$ . Then $\;\frac{12}{\pi}$ $\arg ( {\text{z}}_1)$ is equal to _______. [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;z_1=\;\frac{1+i}{\overline{z}(1-z)+\;\frac{1}{z}}$ $\;z_1=\;\frac{1+i(1-i)}{(1-i)(1-1-i)+\;\frac{1}{1+i}}$ $\;=\;\frac{1+i-i^2}{(1-i)(-i)+\;\frac{1-i}{2}}$ $\;=\;\frac{2+i}{-3 i-1}=\;\frac{4+2 i}{-3 i-1}$ $\;=\;\frac{-(4+2 i)(3 i-1)}{(3 i)^{2-}(1)^2}$ $\;{\text{Arg}} (z_1) =\;\frac{3 \pi}{4}$ $\; \therefore \;\frac{12}{\pi} \arg (z_1) =\;\frac{12}{\pi} \times \;\frac{3 \pi}{4}=9$

Q85JEE Main 2023NAT4MLimits, Continuity and Differentiability

$\lim _{x \to 0} \;\frac{48}{x^4} int_{0}^{x} \;\frac{t^3}{t^{6+}1} \text{dt}$ is equal to _______. [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$48 \lim _{x \to 0} \;\frac{int_{0}^{x} \frac{t^3}{t^{6+}1} \text{dt}}{x^4} (\;\frac{0}{0})$ Applying L'Hospitals Rule $\;48 \lim _{x \to 0} \;\frac{x^3}{x^{6+}1} \times \;\frac{1}{4 x^3}$ $=\;12$

Q86JEE Main 2023NAT4MStatistics

The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted a and $b$ are respectively mean and variance of remaining 6 observation, then $a+3 b-5$ is equal to _______. [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\;\frac{x_1+x_2+. . . . .+x_7}{7}=8$ $\;\;\frac{x_1+x_2+x_3 . . .+x_6+14}{7}=8$ $\;\Rightarrow {\text{x}}_1+ {\text{x}}_2+. . . .+ {\text{x}}_6=42$ $\; \therefore \;{ {\text{x}}_1+ {\text{x}}_2 . . . .+ {\text{x}}_6}/{6}=\;\frac{42}{6}=7= {\text{a}}$ $\;\;{\sum {\text{x}}_{ {\text{i}}}^2}/{7}-8^2=16$ $\;Σ {\xi}^2=560$ $\;\Rightarrow x_1^{2+}x_2^{2+}. . .+x_6^2=364$ $\; {\text{b}}=\;{ {\text{x}}_1^{2+} {\text{x}}_2^{2+}. . .+ {\text{x}}_6^2}/{6}-7^2$ $\;=\;\frac{364}{6}-49$ $\; {\text{b}}=\;\frac{70}{6}$ $\;a+3 b-5=7+3 \times \;\frac{70}{6}-5$ $\;=37$ $\;$

Q87JEE Main 2023NAT4MThree Dimensional Geometry

If the equation of the plane passing through the point $(1,1,2)$ and perpendicular to the line ${\text{x}}-3 {\text{y}}+2 {\text{z}}-1=0 \;\; 4 {\text{x}}- {\text{y}}+ {\text{z}}$ is ${\text{Ax}}+ {\text{By}}+ {\text{Cz}}=1$ , then $140( {\text{C}}- {\text{B}}+ {\text{A}})$ is equal to _______. [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;x-3 y+2 z-1=0$ $\;4 x-y+z=0$ $\;\overline{ {\text{n}}}_1 \times \overline{ {\text{n}}}_2={\begin{vmatrix}\\ \\ { {\text{i}}}^{\^} & \\ \\ { {\text{j}}}^{\^} & \\ \\ { {\text{k}}}^{\^} \\ 1 & -3 & 2 \\ 4 & -1 & 1\end{vmatrix}}$ $\;=-\;{ {\text{i}}}^{\^}+7 \;{ {\text{j}}}^{\^}+11 \;{ {\text{k}}}^{\^}$ Dre of normal to the plane is $-1,7,11$ Equation of plane : $\;-1(x-1)+7(y-1)+11(z-2)=0$ $\;-x+7 y+11 z=28$ $\;\;\frac{-1}{28} x+\;\frac{7 y}{28}+\;\frac{11 z}{28}=1$ $\;A x+B y+C z=1$ $140(C-B+A)=140 (\;\frac{11}{28}-\;\frac{7}{28}-\;\frac{1}{28})$ $=140 \times \;\frac{3}{28}=15$

Q88JEE Main 2023NAT4MSequences and Series

Let $\;\; sum_{n=0}^{\infty } \;\frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}=a e+\;\frac{b}{e}+c$ , where $a, b, c \in ℤ$ and $e=sum_{n=0}^{\infty } \;\frac{1}{n}$ Then $a^{2-}b+c$ is equal to _______. [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;sum_{n=0}^{\infty } \;\frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}$ $=\;sum_{n=0}^{\infty } \;\frac{1}{(n-3) !}+sum_{n=0}^{\infty } \;\frac{3}{(n-2) !}$ $\;+sum_{n=0}^{\infty } \;\frac{1}{(n-1) !}+sum_{n=0}^{\infty } \;\frac{1}{(2 n-1) !}-sum_{n=0}^{\infty } \;\frac{1}{(2 n) !}$ $\;=e+3 e+e+\;\frac{1}{2} (e-\;\frac{1}{e}) -\;\frac{1}{2} (e+\;\frac{1}{e})$ $\;=5 e-\;\frac{1}{e}$ $\therefore \;a^{2-}b+c=26$

Q89JEE Main 2023NAT4MPermutations and Combinations

Number of 4-digit numbers (the repetition of digits is allowed) which are made using the digits 1, 2, 3 and 5 , and are divisible by 15 , is equal to _______ [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

For number to be divisible by 15 , last digit should be 5 and sum of digits must be divisible by 3 . Possible combinations are 1 2 1 5 Numbers $=3$ 2 2 3 5 Numbers $=3$ 3 3 1 5 Numbers $=3$ 1 1 5 5 Numbers $=3$ 2 3 5 5 Numbers $=3$ 3 3 5 5 Total Numbers $=21$

Q90JEE Main 2023NAT4MFunctions

Let $f^1(x)=\;\frac{3 x+2}{2 x+3}, x \in R- \{\;\frac{-3}{2}\}$ For ${\text{n}} \ge 2$ , define ${\text{f}}^{ {\text{n}}}( {\text{x}})= {\text{f}}^1 0 {\text{f}}^{ {\text{n}}-1}( {\text{x}})$ .If ${\text{f}}^5( {\text{x}})=\;{ {\text{ax}}+ {\text{b}}}/{ {\text{bx}}+ {\text{a}}}, {\gcd}( {\text{a}}, {\text{b}})=1$ , then ${\text{a}}+ {\text{b}}$ is equal to ________. [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;f^1(x)=\;\frac{3 x+2}{2 x+3}$ $\;\Rightarrow f^2(x)=\;\frac{13 x+12}{12 x+13}$ $\;\Rightarrow f^3(x)=\;\frac{63 x+62}{62 x+63}$ $\; \therefore f^5(x)=\;\frac{1563 x+1562}{1562 x+1563}$ $\;a+b=3125$

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