Previous Year Paper

1 Feb 2023 Shift 1

90 Questions360 Marks180 MinTake as Timed Mock →

🎯 Practice smarter, not harder

Practice 1 Feb 2023 Shift 1 questions with full analytics — free

✓ Attempt tracking & analytics
✓ Timer & exam-mode simulation
✓ Adaptive difficulty (Easy → Hard)
✓ Bookmarks, notes & mistake review

Just sign in to unlock everything. Free for all students.

🚀 Start Practice Mode Sign in free →

Physics30 questions

Q1JEE Main 2023MCQ4MSemiconductor Electronics

Match the List I with List II List I List II A. IntrinsicSemiconductor I. Fermi-level near conduction band B. n-typesemiconductor II. Fermi-level at middle C. p-typesemiconductor III. Fermi-level near valence band D. Metals IV. Fermi-level inside conduction bandChoose the correct answer from the options given below: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

Q2JEE Main 2023MCQ4MMotion in a Straight Line

An object moves with speed $v_{1}, v_{2}$ , and $v_{3}$ along a line segment ${\text{AB}}, {\text{BC}}$ and ${\text{CD}}$ respectively as shown in figure. Where ${\text{AB}}= {\text{BC}}$ and ${\text{AD}}=3 {\text{AB}}$ , then average speed of the object will be : [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{AB}}=x$ ${\text{BC}}=x$ $2 x+ {\text{CD}}=3 x$ $\; {\text{CD}}=x$ $\;⟨ {\text{v}}⟩=\;{3 x}/{\frac{x}{v_{1}}+\;\frac{x}{v_{2}}+\;\frac{x}{v_{3}}}=\;\frac{3 v_{1} v_{2} v_{3}}{v_{2} v_{3}+v_{1} v_{3}+v_{1} v_{2}}$

Q3JEE Main 2023MCQ4MGravitation

Given below are two statements :Statement-I: Acceleration due to gravity is different at different places on the surface of earth.Statement-II: Acceleration due to gravity increases as we go down below the earth's surface. In the light of the above statements, choose the correct answer from the options given below [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$g_{e f f}=g-\omega ^{2} R_{e} sin \;^{2} \theta, \theta \to$ co-latitude angle $g_{\;\text{eff }\;}=g (1-\;\frac{d}{R_{e}}) , d$ here depth

Q4JEE Main 2023MCQ4MAlternating Current

Match the List I with List II List I List II A. ${\text{AC}}$ generator I. Presence of both ${\text{L}}$ and ${\text{C}}$ B. Transformer II. ElectromagneticInduction C. Resonancephenomenonto occur III. Quality factor D. Sharpness of resonance IV. Mutual Inductance Choose the correct answer from the options given below: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

Q5JEE Main 2023MCQ4MElectromagnetic Waves

Match the List-I with List-II: List I List II A. Microwaves I. Radio active decay of the nucleus B. Gamma rays II. Rapid acceleration and deceleration of electron in aerials C. Radio waves III. Inner shell electrons D. X-rays IV. Klystron valveChoose the correct answer from the options given below: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

Q6JEE Main 2023MCQ4MGravitation

If earth has a mass nine times and radius twice to the of a planet ${\text{P}}$ . Then $\;\frac{v_e}{3} {\sqrt{x}} {\text{ms}}^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of ${\text{P}}$ , where $v_e$ is escape velocity on earth. The value of $x$ is [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;v_{\;\text{(escape) plant }\;}=\sqrt{\;\frac{2 G M_P}{R_P}}$ $\;=\sqrt{\;\frac{2 G (\frac{M_e}{9}) }{ (\;\frac{R_e}{2}) }}=\;{v_e {\sqrt{2}}}/{3} \therefore x=2$

Q7JEE Main 2023MCQ4MWave Optics

' $n$ ' polarizing sheets are arranged such that each makes an angle $45^{\circ}$ with the proceeding sheet. An unpolarized light of intensity $I$ is incident into this arrangement. The output intensity is found to be $\;\frac{I}{64}$ . The value of $n$ will be: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

After passing through first sheet $I_1=\;\frac{I}{2}$ After passing through second sheet $I_2=I_1 \cos ^2 (45^{\circ}) =\;\frac{I}{4}$ After passing through $n^{\;\text{th }\;}$ sheet $\;I_{ {\text{n}}}=\;{I}/{2^{ {\text{n}}}}=\;\frac{I}{64}$ $\;n=6$

Q8JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

Find the magnetic field at the point ${\text{P}}$ in figure. The curved portion is a semicircle connected to two long straight wires. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$B_{ {\text{P}}}= (\;\frac{\mu _0 i}{4 r}+\;\frac{\mu _0 i}{4 \pi r}) =\;\frac{\mu _0 i}{2 r} (\;\frac{1}{2}+\;\frac{1}{2 \pi})$

Q9JEE Main 2023MCQ4MElectromagnetic Waves

Which of the following frequencies does not belong to FM broadcast. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

FM broadcast range is $88 {\text{MHz}}$ to $108 {\text{MHz}}$

Q10JEE Main 2023MCQ4MWaves

A steel wire with mass per unit length $7.0 \times 10^{-3} {\text{kg}}$ ${\text{m}}^{-1}$ is under tension of $70 {\text{N}}$ . The speed of transverse waves in the wire will be: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$v=\sqrt{\;\frac{T}{\mu }}=\sqrt{\;\frac{70}{70 \times 10^{-3}}}=100 {\text{m}} / {\ s}$

Q11JEE Main 2023MCQ4MMotion in a Plane

A child stands on the edge of the cliff $10 {\text{m}}$ above the ground and throws a stone horizontally with an initial speed of $5 {\text{ms}}^{-1}$ . Neglecting the air resistance, the speed with which the stone hits the ground will be ______ ${\text{ms}}^{-1}$ (given, ${\text{g}}=10 {\text{ms}}^{-2}$ ). [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{v}}_{ {\text{y}}}={\sqrt{2 g h}}={\sqrt{200}}$ $\;v_{\;\text{net }\;}={\sqrt{25+200}}=15 {\text{m}} / {\ s}$

Q12JEE Main 2023MCQ4MDual Nature of Matter and Radiation

A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of $\lambda$ . An alpha particle having certain kinetic energy has the same de-Brogle wavelength $\lambda$ . The ratio of kinetic energy of proton and that of alpha particle is: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;K E=\;\frac{p^2}{2 m}=\;\frac{h^2}{2 m \lambda ^2}$ $\;\;\frac{K E_p}{K E_alpha}=\;\frac{m_alpha}{m_p}=4: 1$

Q13JEE Main 2023MCQ4MThermodynamics

A sample of gas at temperature $T$ is adiabatically expanded to double its volume. The work done by the gas in the process is $(.$ given, $\gamma =\;\frac{3}{2})$ : [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;T_1 V_1^{\gamma -1}=T_2 V_2^{\gamma -1}$ $\;T V^{1 / 2}=T_2(2 V)^{1 / 2}$ $\;T_2=\;{T}/{{\sqrt{2}}}$ $\;W=\;\frac{R (T_1-T_2) }{\gamma -1}=\;{R (T-{T}/{{\sqrt{2}}}) }/{\frac{1}{2}}=R T(2-{\sqrt{2}})$

Q14JEE Main 2023MCQ4MCurrent Electricity

The equivalent resistance between $A$ and $B$ of the network shown in figure: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Wheat stone bridge is in balanced condition. $\;\;\frac{1}{R_{e q}}=\;\frac{1}{4 R}+\;\frac{1}{8 R}$ $\;R_{e q}=\;\frac{8 R}{3}$

Q15JEE Main 2023MCQ4MElectrostatics

Let $\sigma$ be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields in three different region $E_{ {\text{I}}}, E_{ {\text{II}}}$ and $E_{ {\text{III}}}$ are: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Assuming RHS to be $\hat{n}$ $\;\vec{E}_{ {\text{I}}}=\;\frac{\sigma}{2 E_0}(-\hat{n})+\;\frac{\sigma}{2 E_0}(-\hat{n})=-\;\frac{\sigma}{E_0}\hat{n}$ $\;\vec{E}_{I I}=0$ $\;\vec{E}_{I I I}=\;\frac{\sigma}{2 E_0}(\hat{n})+\;\frac{\sigma}{2 E_0}(\hat{n})=\;\frac{\sigma}{E_0}(\hat{n})$

Q16JEE Main 2023MCQ4MProperties of Matter

A mercury drop of radius $10^{-3} {\text{m}}$ is broken into 125 equal size droplets. Surface tension of mercury is $0.45 {\text{Nm}}^{-1}$ . The gain in surface energy is: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Sol. Initial surface energy $=0.45 \times 4 \pi (10^{-3}) ^2$ $\;\;\frac{4}{3} \pi (10^{-3}) ^3=125 \times \;\frac{4 \pi}{3} R_{\;\text{new }\;}^3$ $\therefore \;10^{-3}=5 R_{\;\text{new }\;}$ $\therefore \;R_{\;\text{new }\;}=\;\frac{10^{-3}}{5} {\text{m}}$ So, final surface energy $=0.45 \times 125 \times 4 \pi (\;\frac{10^{-3}}{5}) ^2$ Increase in energy $=0.45 \times 4 \pi \times (10^{-3}) ^2 [\;\frac{125}{25}-1]$ $=4 \times 0.45 \times 4 \pi \times 10^{-6}$ $=2.26 \times 10^{-5} {\text{J}}$

Q17JEE Main 2023MCQ4MAtoms and Nuclei

The mass of proton, neutron and helium nucleus are respectively $1.0073 {\text{u}}, 1.0087 {\text{u}}$ and $4.0015 {\text{u}}$ . The binding energy of helium nucleus is: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\;\text{ B.E of Helium }\;= (2 m_P+2 m_N-m_{H e}) c^2$ $\;=28.4 {\text{MeV}}$

Q18JEE Main 2023MCQ4MUnits and Measurements

$(P+\;\frac{a}{V^2}) (V-b)=R T$ represents the equation of state of some gases. Where $P$ is the pressure, $V$ is the volume, $T$ is the temperature and $a, b, R$ are the constants. The physical quantity, which has dimensional formula as that of $\;\frac{b^2}{a}$ , will be : [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;{[b]=[V]}$ $\; \;\;\;\; \$ [;\frac{a}{b^2}]$ =[P] ;; \therefore ;; $[;\frac{b^2}{a}]$ =;\frac{1}{[P]}=;\frac{1}{[B]}=[K]$

Q19JEE Main 2023MCQ4MThermodynamics

The average kinetic energy of a molecule of the gas is [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Basic theoryTranslational $K . E$ on average of a molecule is $\;\frac{3}{2}$ KT which is independent of nature, pressure and volume.

Q20JEE Main 2023MCQ4MLaws of Motion

A block of mass $5 {\text{kg}}$ is placed at rest on a table of rough surface. Now, if a force of $30 {\text{N}}$ is applied in the direction parallel to surface of the table, the block slides through a distance of $50 {\text{m}}$ in an interval of time $10 {\ s}$ . Coefficient of kinetic friction is (given, ${\text{g}}=10 {\text{ms}}^{-2}$ ): [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;S=u t+\;\frac{1}{2} a t^2$ $\;50=0+\;\frac{1}{2} \times a \times 100$ $\;a=1 {\text{m}} / {\ s}^2$ $\;F-\mu m g=m a$ $\;30-\mu \times 50=5 \times 1$ $\;50 \mu =25$ $\;\mu =\;\frac{1}{2}$

Q21JEE Main 2023NAT4MMagnetic Effects of Current and Magnetism

A charge particle of $2 \mu {\text{C}}$ accelerated by a potential difference of $100 {\text{V}}$ enters a region of uniform magnetic field of magnitude $4 {\text{mT}}$ at right angle to the direction of field. The charge particle completes semicircle of radius $3 {\text{cm}}$ inside magnetic field. The mass of the charge particle is ______ $\;\; \times 10^{-18} {\text{kg}}$ . [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$r=\;\frac{m v}{q B}=\;{\sqrt{2 {\text{km}}}}/{q B}, m=\;\frac{r^2 q^2 B^2}{2 k}$ ${\text{m}}=\;\frac{\;\frac{1}{100} \times \frac{3}{100} \times 2 \times 2 \times 4 \times 10^{-3} \times 4 \times 10^{-3} \times 10^{-12}}{2 \times (100) \times 10^{-6}}$ $\;=144 \times 10^{-18} {\text{kg}}$

Q22JEE Main 2023NAT4MCurrent Electricity

In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf $1.5 {\text{V}}$ is found to be $60 {\text{cm}}$ . If this cell is replaced by another cell of emf $E$ . the length-of null point increases by $40 {\text{cm}}$ . The value of $E$ is $\;\frac{x}{10} {\text{V}}$ . The value of $x$ is ________. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\;\frac{E_1}{E_2}=\;\frac{l_1}{l_2}$ $\;\;\frac{1.5}{E_2}=\;\frac{60}{60+40}=\;\frac{6}{10}=\;\frac{3}{5}$ $\;E_2=\;\frac{5}{2}=\;\frac{x}{10}$ $\;x=25$

Q23JEE Main 2023NAT4MWork, Power and Energy

A small particle moves to position $5{i}^{\wedge }-2 {j}^{\wedge }+{k}^{\wedge }$ from its initial position $2 {i}^{\wedge }+3 {j}^{\wedge }-4{k}^{\wedge }$ under the action of force $5 {i}^{\wedge }+2 {j}^{\wedge }+7 {k} ^{\wedge } {\text{N}}$ . The value of work done will be ______ J. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;W=\vec{F} \cdot (\vec{r}_f-\vec{r}_1)$ $\;=(5 {i}^{\wedge }+2 {j}^{\wedge }+7 {k}^{\wedge }) \cdot ((5 {i}^{\wedge }-2 {j}^{\wedge }+{k}^{\wedge })-(2 {i}^{\wedge }+3 {j}^{\wedge }-4{k}^{\wedge }))$ $\;W=40 {\text{J}}$

Q24JEE Main 2023NAT4MAtoms and Nuclei

A light of energy $12.75 {\text{eV}}$ is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is $\;\frac{x}{\pi} \times 10^{-17} {\text{eVs}}$ . The value of $x$ is _______ (use $h=4.14 \times 10^{-15} {\text{eVs}}, c=3 \times 10^8$ ${\text{ms}}^{-1})$ [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

In the ground state energy $=-13.6 {\text{eV}}$ So energy $\; \;\; \;{-13.6 {\text{eV}}}/{n^2}=-13.6+12.75$ $\;\;{-13.6 {\text{eV}}}/{n^2}=-0.85$ $\;n={\sqrt{16}}$ $\; \;\;\;\; n=4$ $\;\;\text{ Angular momentum }\;=\;\frac{n h}{2 \pi}=\;\frac{4 h}{2 \pi}=\;\frac{2 h}{\pi}$ $\;\;\text{ Angular momentum }\;=\;\frac{2}{\pi} \times 4.14 \times 10^{-15}$ $\; \;\;\;\; =\;\frac{828 \times 10^{-17}}{\pi} {\text{eVs}}$

Q25JEE Main 2023NAT4MProperties of Matter

A certain pressure ${ }^{'} P^{'}$ is applied to 1 litre of water and 2 litre of a liquid separately. Water gets compressed to $0.01 \%$ whereas the liquid gets compressed to $0.03 \%$ . The ratio of Bulk modulus of water to that of the liquid is $\;\frac{3}{x}$ . The value of $x$ is _______. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;B_{\;\text{water }\;}=\;\frac{-\Delta P}{ (\;\frac{\Delta V}{V}) }=\;\frac{-\Delta P}{\frac{0.01}{100}}$ $\;B_{\;\text{liquid }\;}=\;\frac{-\Delta P}{\frac{0.03}{100}}$ $\;\;{B_{\;\text{water }\;}}/{B_{\;\text{liquid }\;}}=3$ $\;x=1$

Q26JEE Main 2023NAT4MElectrostatics

Two equal positive point charges are separated by a distance $2 a$ . The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge $q_0$ becomes maximum is $\;{a}/{{\sqrt{x}}}$ . The value of $x$ is _______. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$F=\;\frac{2 K q q_0 x}{ (x^{2+}a^2) ^{3 / 2}}$ For $F$ to be maximum $\;{ {\text{d}} F}/{ {\text{d}} x}=0$ $x=\;{a}/{{\sqrt{2}}}$

Q27JEE Main 2023NAT4MRay Optics and Optical Instruments

A thin cylindrical rod of length $10 {\text{cm}}$ is placed horizontally on the principle axis of a concave mirror of focal length $20 {\text{cm}}$ . The rod is placed in a such a way that mid point of the rod is at $40 {\text{cm}}$ from the pole of mirror. The length of the image formed by the mirror will be $\;\frac{x}{3} {\text{cm}}$ . The value of $x$ is _______. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;U_A=-45 {\text{cm}}, {\text{f}}=-20 {\text{cm}}$ $\;V_A=\;\frac{-45 \times (-20)}{-45-(-20)}=\;\frac{-900}{25}=-36 {\text{cm}}$ $\;\;\text{ And }\; U_B=-35 {\text{cm}}$ $\; \therefore V_B=\;\frac{-35 \times (-20)}{-35-(-20)}=\;\frac{700}{-15}$ $\; \therefore V_A-V_B=\;\text{ length of image }\;$ $\;= (-36+\;\frac{140}{3}) {\text{cm}}$ $\;=\;\frac{-108+140}{3} {\text{cm}}$ $\;=\;\frac{32}{3} {\text{cm}}$ $\; \therefore x=32$

Q28JEE Main 2023NAT4MRotational Motion

A solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60 {\text{cm}}$ . If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is ________ ${\text{ms}}^{-1}$ . (Given ${\text{g}}=10 {\text{ms}}^{-2}$ ) [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$v=\sqrt{\;\frac{2 g h}{1+\;\frac{k^2}{R^2}}}$ Where $h=60 sin \;30^{\circ}=30 {\text{cm}}$ $k^2=\;\frac{R^2}{2}$ $v=2 {\text{ms}}^{-1}$

Q29JEE Main 2023NAT4MOscillations

The amplitude of a particle executing SHM is 3 ${\text{cm}}$ . The displacement at which its kinetic energy will be $25 \%$ more than the potential energy is:_______ cm. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;K E=P E+\;\frac{P E}{4}$ $\;K E=\;\frac{5}{4} P E$ $\;\;\frac{1}{2} m \omega ^2 (A^{2-}x^2) =\;\frac{5}{4} \times \;\frac{1}{2} m \omega ^2 x^2$ $\;{[ {\text{v}}=\omega \sqrt{A^{2-} {\text{x}}^2}]\$ };A^{2-}x^2=;\frac{5}{4} x^2;;\frac{9 x^2}{4}=A^2;x=;\frac{2}{3} A; ∴ x=;\frac{2}{3} × 3 {\text{cm}};x=2 {\text{cm}}$

Q30JEE Main 2023NAT4MAlternating Current

A series LCR circuit is connected to an ac source of $220 {\text{V}}, 50 {\text{Hz}}$ . The circuit contain a resistance $R=100 Ω$ and an inductor of inductive reactance $X_L=79.6 Ω$ . The capacitance of the capacitor needed to maximize the average rate at which energy is supplied will be _______ $\mu {\text{F}}$ . [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

To maximize the average rate at which energy supplied i.e. power will be maximum.So in LCR circuit power will be maximum at the condition of resonance and in resonance condition $X_L\;=X_C$ $79.6\;=\;\frac{1}{\omega C}$ $\therefore C\;=\;\frac{1}{2 \pi \times 50 \times 79.6}$ $\therefore C\;=40 \mu {\text{F}}$

Chemistry30 questions

Q31JEE Main 2023MCQ4MSolid State

Which of the following represents the lattice structure of ${\text{A}}_{0.95} {\text{O}}$ containing ${\text{A}}^{2+}, {\text{A}}^{3+}$ and ${\text{O}}^{2-}$ ions? $⊙ {\text{A}}^{2+} ⊙ {\text{A}}^{3+} ⊙ {\text{O}}^{2-}$ [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Applying electrical neutrality principle in metal defficiency defect. $3 {\text{A}}^{2+}$ are replaced by $2 {\text{A}}^{3+}$ , thus one vacant site per pair of ${\text{A}}^{3+}$ is created

Q32JEE Main 2023MCQ4MBiomolecules

The correct representation in six membered pyranose form for the following sugar $[ {\text{X}}]$ is [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

Q33JEE Main 2023MCQ4Md and f Block Elements

Highest oxidation state of ${\text{Mn}}$ is exhibited in ${\text{Mn}}_2 {\text{O}}_7$ . The correct statements about ${\text{Mn}}_2 {\text{O}}_7$ are(A) ${\text{Mn}}$ is tetrahedrally surrounded by oxygen atoms(B) ${\text{Mn}}$ is octahedrally surrounded by oxygen atoms(C) Contains Mn-O-Mn bridge(D) Contains Mn-Mn bond.Choose the correct answer from the options given below [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

Q34JEE Main 2023MCQ4MAlcohols, Phenols and Ethers

Decreasing order of dehydration of the following alcohols is [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

Q35JEE Main 2023MCQ4MStates of Matter

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason $\text{bo} {R}$ . Assertion A: Amongst ${\text{He}}, {\text{Ne}}, {\text{Ar}}$ and ${\text{Kr}}$ ; $1 {\text{g}}$ of activated charcoal adsorbs more of ${\text{Kr}}$ . Reason $\text{bo} {R}$ : The critical volume ${\text{V}}_{ {\text{c}}} ( {\text{cm}}^3 {\text{mol}}^{-1})$ and critical pressure $P_c( {\text{atm}})$ is highest for Krypton but the compressibility factor at critical point $Z_c$ is lowest for Krypton.In the light of the above statements, choose the correct answer from the options given below. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Adsorption $\propto$ vanderwaal attraction forces ${\text{Z}}_{ {\text{c}}}=\;\frac{3}{8}$ for all real gases

Q36JEE Main 2023MCQ4MAmines

In the following reaction, ' $A$ ' is'A' Major product. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Initially lone pair electron of $- {\text{NH}}_2$ attack on electrophilic carbon, after then lone pair electron of oxygen attacks leading to formation of cyclic compound.

Q37JEE Main 2023MCQ4MBiomolecules

Match List I with List II List I List II A. Tranquilizers I. Anti blood clotting B. Aspirin II. Salvarsan C. Antibiotic III. Antidepressant drugs D. Antiseptic IV. SoframicineChoose the correct answer from the options given below: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

Q38JEE Main 2023MCQ4Mp Block Elements

Given below are two statements:Statement I: Chlorine can easily combine with oxygen to from oxides: and the product has a tendency to explode.Statement II: Chemical reactivity of an element can be determined by its reaction with oxygen and halogens.In the light of the above statements, choose the correct answer from the options given below. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Chlorine oxides, ${\text{Cl}}_2 {\text{O}}, {\text{ClO}}_2, {\text{Cl}}_2 {\text{O}}_6$ and ${\text{Cl}}_2 {\text{O}}_7$ are highly reactive oxidising agents and tend to explode.

Q39JEE Main 2023MCQ4MChemical Bonding and Molecular Structure

Resonance in carbonate ion $( {\text{CO}}_3^{2-})$ isWhich of the following is true? [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

Q40JEE Main 2023MCQ4MHaloalkanes and Haloarenes

Identify the incorrect option from the following: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

In alcoholic ${\text{KOH}}$ , elimination reaction takes place.

Q41JEE Main 2023MCQ4MCoordination Compounds

A solution of ${\text{FeCl}}_3$ when treated with ${\text{K}}_4 [ {\text{Fe}}( {\text{CN}})_6]$ gives a prussiun blue precipitate due to the formation of [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

Q42JEE Main 2023MCQ4MCoordination Compounds

Which of the following are the example of double salt?(A) ${\text{FeSO}}_4 \cdot ( {\text{NH}}_4) _2 {\text{SO}}_4 \cdot 6 {\text{H}}_2 {\text{O}}$ (B) ${\text{CuSO}}_4 \cdot 4 {\text{NH}}_3 \cdot {\text{H}}_2 {\text{O}}$ (C) ${\text{K}}_2 {\text{SO}}_4 \cdot {\text{Al}}_2 ( {\text{SO}}_4) _3 \cdot 24 {\text{H}}_2 {\text{O}}$ (D) ${\text{Fe}}( {\text{CN}})_2 \cdot 4 {\text{KCN}}$ Choose the correct answer. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Double salt contain's two or more types of salts. ${\text{CuSO}}_4 .4 {\text{NH}}_3 \cdot {\text{H}}_2 {\text{O}}$ and ${\text{Fe}}( {\text{CN}})_2 .4 {\text{KCN}}$ are complex compounds.

Q43JEE Main 2023MCQ4MCoordination Compounds

Which of the following complex will show largest splitting of d-orbitals? [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\overline{ {\text{C}} N}$ is a strong field ligand so maximum splitting in d orbitals take place.

Q44JEE Main 2023MCQ4MSome Basic Concepts of Chemistry

How can photochemical smog be controlled? [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

Q45JEE Main 2023MCQ4Mp Block Elements

Match List I with List II List I List II A. Slaked lime I. ${\text{NaOH}}$ B. Dead burnt plaster II. ${\text{Na}}_2 {\text{CO}}_3 \cdot 10 {\text{H}}_2 {\text{O}}$ C. Caustic soda III. ${\text{Na}}_2 {\text{CO}}_3 \cdot 10 {\text{H}}_2 {\text{O}}$ D. Washing soda IV. ${\text{CaSO}}_4$ Choose the correct answer form the options given below: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

Q46JEE Main 2023MCQ4MClassification of Elements

Choose the correct statement(s):A. Beryllium oxide is purely acidic in nature.B. Beryllium carbonate is kept in the atmosphere of ${\text{CO}}_2$ .C. Beryllium sulphate is readily soluble in water.D. Beryllium shows anomalous behavior.Choose the correct answer from the options given below: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

A. Beryllium oxide is amphoteric in nature.B. Beryllium carbonate is kept in the atmosphere of ${\text{CO}}_2$ because it is thermally less stable.C. Beryllium sulphate is readily soluble in water due to high degree of hydration.D. Beryllium shows anomalous behaviour due to small size, high ionization energy and high value of $\varphi$ (polarising power).

Q47JEE Main 2023MCQ4MChemical Thermodynamics

Given below are two statements: one is labelled as Assertion $\text{bo} {A}$ and the other is labelled as Reason $\text{bo} {R}$ Assertion A: In an Ellingham diagram, the oxidation of carbon to carbon monoxide shows a negative slope with respect to temperature.Reason R: CO tends to get decomposed at higher temperature.In the light of the above statements, choose the correct answer from the options given below [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$2 {\text{C}}( {\ s})+ {\text{O}}_2( {\text{g}}) \to 2 {\text{CO}}( {\text{g}})$ $\Delta _{ {\text{r}}} {\text{S}}^0$ is $+ {\text{ve}}, \Delta _{ {\text{r}}} {\text{G}}^0=\Delta _{ {\text{r}}} {\text{H}}^{0-} {\text{T}} \Delta _{ {\text{r}}} {\text{S}}^0$ ; thus slope is negativeAs temperature increases $\Delta _{ {\text{r}}} {\text{G}}^0$ becomes more negative thus it has lower tendency to get decomposed.

Q48JEE Main 2023MCQ4MHydrocarbons

But-2-yne is reacted separately with one mole of Hydrogen as shown below: ${\text{B}} _{-}{←}_{\text{liq} {\text{NH}}_3}^{ {\text{Na}}} {\text{NH}}_3- {\text{C}} \equiv {\text{C}}- {\text{CH}}_3 {\to }_{ \Delta}^{ {\text{Pd}} / {\text{C}}} {{\text{A}}_{-}$ Identify the incorrect statements from the options given below:A. A is more soluble than B.B. The boiling point & melting point of ${\text{A}}$ are higher and lower than B respectively.C. A is more polar than ${\text{B}}$ because dipole moment of ${\text{A}}$ is zero.D. ${\text{Br}}_2$ adds easily to ${\text{B}}$ than ${\text{A}}$ . [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Incorrect statements are ${\text{C}}$ and ${\text{D}}$ only

Q49JEE Main 2023MCQ4MSome Basic Concepts of Chemistry

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason $\text{bo} {R}$ Assertion A: Hydrogen is an environment friendly fuel.Reason R: Atomic number of hydrogen is 1 and it is a very light element.In the light of the above statements, choose the correct answer from the options given below [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

Q50JEE Main 2023MCQ4MPractical Organic Chemistry

Match List I and List II List I Test List II Functional group /Class of Compound A. Molisch's Test I. Peptide B. Biuret Test II. Carbohydrate C. Carbylamine Test III. Primary amine D. Schiff s Test IV. AldehydeChoose the correct answer from the options given below: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

Q51JEE Main 2023NAT4MSolutions

The density of $3 {\text{M}}$ solution of ${\text{NaCl}}$ is $1.0 {\text{g}} {\text{mL}}^{-1}$ .Molality of the solution is ______ $\times 10^{-2} {\text{m}}$ (Nearest integer).Given: Molar mass of ${\text{Na}}$ and ${\text{Cl}}$ is 23 and $35.5 {\text{g}}$ ${\text{mol}}^{-1}$ respectively. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{m}}=\;{1000 \times {\text{M}}}/{1000 \times {\text{d}}- {\text{M}} \times {\text{M}} . {\text{W}} \;\text{ of solute }\;}$ $\;=\;\frac{1000 \times 3}{1000 \times 1-(3 \times 58.5)}=3.64$ $\;=364 \times 10^{-2}$

Q52JEE Main 2023NAT4MStructure of Atom

Electrons in a cathode ray tube have been emitted with a velocity of $1000 {\text{ms}}^{-1}$ . The number of following statements which is/are true about the emitted radiation is _______.Given : ${\text{h}}=6 \times 10^{-34} {\text{Js}}, {\text{m}}_{ {\text{e}}}=9 \times 10^{-31} {\text{kg}}$ .(A) The deBroglie wavelength of the electron emitted is $666.67 {\text{nm}}$ .(B) The characteristic of electrons emitted depend upon the material of the electrodes of the cathode ray tube.(C) The cathode rays start from cathode and move towards anode.(D) The nature of the emitted electrons depends on the nature of the gas present in cathode ray tube. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

(A) ${\text{V}}_{ {\text{e}}}=1000 {\text{m}} / {\ s} ; {\text{h}}=6 \times 10^{-34} {\text{Js}}$ ; $\; {\text{m}}_{ {\text{e}}}=9 \times 10^{-31} {\text{kg}}$ $\;\lambda =\;{ {\text{h}}}/{ {\ mv}}=\;\frac{6 \times 10^{-34}}{9 \times 10^{-31} \times 1000}=666.67 \times 10^{-9} {\text{m}}$ $\;=666.67 {\text{nm}}$ (B) The characteristic of electrons emitted is independent of the material of the electrodes of the cathode ray tube.(C) The cathode rays start from cathode and move towards anode.(D) The nature of the emitted electrons is independent on the nature of the gas present in cathode ray tube.

Q53JEE Main 2023NAT4MRedox Reactions

Sum of oxidation states of bromine in bromic acid and perbromic acid is _______. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{HBrO}}_3$ (Bromic acid)Ox. State of ${\text{Br}}=+5$ ${\text{HBrO}}_4$ (per bromic acid)OX. State of ${\text{Br}}=+7$ Sum of Ox. State $=12$

Q54JEE Main 2023NAT4MElectrochemistry

At what ${\text{pH}}$ , given half cell ${\text{MnO}}_4^{-}(0.1 {\text{M}}) \mid {\text{Mn}}^{2+}$ (0.001 M) will have electrode potential of $1.282$ ${\text{V}}$ ? _______ (Nearest Integer)Given $E_{ {\text{MnO}}_4^{-} / {\text{Mn}}^{2+}}^o=1.54 {\text{V}}, \;\frac{2.303 R T}{F}=0.059 {\text{V}}$ [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{MnO}}_4^{-}+8 {\text{H}}^{+}+5 {\text{e}}^{-} ⇌ {\text{Mn}}^{2+}+4 {\text{H}}_2 {\text{O}}$ $\; {\text{E}}= {\text{E}}^{\circ}-\;\frac{0.059}{5} \log \;{[ {\text{Mn}}^{2+}]\$ }/{[ {\text{MnO}}_4^{-}]$ $[ {\text{H}}^{+}]$ ^8};1.282=1.54-;\frac{0.059}{5} \log ;{10^{-3}}/{10^{-1} \times $[ {\text{H}}^{+}]$ ^8};;\frac{0.258 \times 5}{0.059}=\log ;{10^{-2}}/{[ {\text{H}}^{+}]$ ^8}\Rightarrow ;; 21.86=-2+8 {\text{pH}}∴ ;; {\text{pH}}=2.98;≃ 3$

Q55JEE Main 2023NAT4MOrganic Chemistry - Some Basic Principles

Number of isomeric compounds with molecular formula ${\text{C}}_9 {\text{H}}_{10} {\text{O}}$ which (i) do not dissolve in ${\text{NaOH}}$ (ii) do not dissolve in ${\text{HCl}}$ . (iii) do not give orange precipitate with 2, 4 - DNP (iv) on hydrogenation give identical compound with molecular formula ${\text{C}}_9 {\text{H}}_{12} {\text{O}}$ is ________. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

As per the language of given question, the best possible isomeric structure is ${\text{Ph}}- {\text{CH}}= {\text{CH}}- {\text{O}}- {\text{CH}}_3$ (cis and trans). So, the answer is 2.

Q56JEE Main 2023NAT4MChemical Equilibrium

(i) ${\text{X}}( {\text{g}}) ⇌ {\text{Y}}( {\text{g}})+ {\text{Z}}( {\text{g}}) {\text{K}}_{ {\text{p}} 1}=3$ (ii) ${\text{A}}( {\text{g}}) ⇌ 2 {\text{B}}( {\text{g}}) \;\; {\text{K}}_{ {\text{p}} 2}=1$ If the degree of dissociation and initial concentration of both the reactants $X( {\text{g}})$ and ${\text{A}}( {\text{g}})$ are equal, then the ratio of the total pressure at equilibrium $(\;\frac{p_1}{p_2})$ is equal to $x: 1$ . The value of $x$ is ________ (Nearest integer) [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\; {\text{x}}( {\text{g}}) ⇌ {\text{y}}( {\text{g}})+ {\text{z}}( {\text{g}}) \;\; {\text{k}}_{ {\text{p}}_1}=3$ Initial moles $\;\; {\text{n}} \;\; - \;\; -$ at equilibrium ${\text{n}}-\alpha {\text{n}} \;\; \alpha {\text{n}} \;\; \alpha {\text{n}}$ $\; {\text{k}}_{ {\text{p}}_1}=\;{ (\frac{\alpha}{1+\alpha} \times {\text{p}}_1) ^2}/{\frac{1-\alpha}{1+\alpha} {\text{p}}_1}$ $\;3=\;{\alpha^2 \times {\text{p}}_1}/{1-\alpha^2}$ $\; {\text{A}}( {\text{g}}) ⇌\;2 {\text{B}}( {\text{g}})\; {\text{k}}_{ {\text{p}}_2}=1$ $\;\text{ Initial mole }\;\; {\text{n}}\;-\;$ $\;\text{ at equilibrium }\;\; {\text{x}}-\alpha {\text{n}}\;2 \alpha {\text{n}}\; {\text{p}}_{\;\text{total }\;}= {\text{p}}_2$ $\; {\text{k}}_{ {\text{p}}_2}=\;{ (\frac{2 \alpha}{1+\alpha} \times {\text{p}}_2) ^2}/{\frac{1-\alpha}{1+\alpha} \times {\text{p}}_2}$ $\;1=\;{4 \alpha^2 \times {\text{p}}_2}/{1-\alpha^2}$ $\;\;{ {\text{k}}_{ {\text{p}}_1}}/{ {\text{k}}_{ {\text{p}}_2}}=\;{ {\text{p}}_1}/{4 {\text{p}}_2}$ $\;\;\frac{3}{1}=\;{ {\text{p}}_1}/{4 {\text{p}}_2} \;\; \therefore {\text{p}}_1: {\text{p}}_2=12: 1$ $\;x=12$ $\;$

Q57JEE Main 2023NAT4MOrganic Chemistry - Some Basic Principles

The total number of chiral compound/s from the following is ________. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

Q58JEE Main 2023NAT4MChemical Kinetics

${\text{A}}$ and ${\text{B}}$ are two substances undergoing radioactive decay in a container. The half life of ${\text{A}}$ is $15 {\min}$ and that of ${\text{B}}$ is $5 {\min}$ . If the initial concentration of B is 4 times that of $A$ and they both start decaying at the same time, how much time will it take for the concentration of both of them to be same? ________ $\min$ . [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$[ {\text{A}}]_{ {\text{t}}}=[ {\text{A}}]_0 {\text{e}}^{- {\text{kt}}}$ For $A: L e t[A]_t$ be $y$ and $[A]_0$ be $x ; k=\;\frac{\ln 2}{t_{1 / 2}}=$ $\;\frac{\ln 2}{15 \min }$ ${\text{y}}= {\text{xe}} {\text{e}}^{- {\text{kt}}}$ $=x e^{- (\;\frac{\ln 2}{15}) t}$ For $\text{bo} {B}:[ {\text{B}}]_{ {\text{t}}}=[ {\text{B}}]_0 {\text{e}}^{- {\text{kt}}}$ ${\text{Let}}[B]_{ {\text{t}}}= {\text{y}} ;[ {\text{B}}]_0=4 {\text{x}} ; {\text{k}}=\;{\ln 2}/{ {\text{t}}_{1 / 2}}=\;\frac{\ln 2}{5 \min }$ $y=4 x e^{- (\;\frac{\ln 2}{5}) t}$ $\;\Rightarrow \;\; x e^{- (\;\frac{\ln 2}{15}) {\text{t}}}=4 x e^{- (\;\frac{\ln 2}{5}) {\text{t}}}$ $\; {\text{e}}^{ {\text{t}} (\;\frac{\ln 2}{5}-\;\frac{\ln 2}{15}) }=4$ $\; {\text{t}} \times [\;\frac{\ln 2}{5}-\;\frac{\ln 2}{15}] =\ln 4$ $\; {\text{t}} \times \ln 2 [\;\frac{1}{5}-\;\frac{1}{15}] =2 \ln 2$ $\; {\text{t}}=15 {\min}$

Q59JEE Main 2023NAT4MChemical Thermodynamics

At $25^{\circ} {\text{C}}$ , the enthalpy of the following processes are given: $\; {\text{H}}_2( {\text{g}})+ {\text{O}}_2( {\text{g}}) \to 2 {\text{OH}}( {\text{g}}) \Delta {\text{H}}^0=78 {\text{kJ}} {\text{mol}}^{-1}$ $\; {\text{H}}_2( {\text{g}})+1 / 2 {\text{O}}_2( {\text{g}}) \to {\text{H}}_2 {\text{O}}( {\text{g}}) \Delta {\text{H}}^0=-242 {\text{kJ}} {\text{mol}}^{-1}$ $\; {\text{H}}_2( {\text{g}}) \to 2 {\text{H}}( {\text{g}}) \Delta {\text{H}}^0=436 {\text{kJ}} {\text{mol}}^{-1}$ $\;1 / 2 {\text{O}}_2( {\text{g}}) \to {\text{O}}( {\text{g}}) \Delta {\text{H}}^0=249 {\text{kJ}} {\text{mol}}^{-1}$ What would be the value of ${\text{X}}$ for the following reaction? _______ (Nearest integer) ${\text{H}}_2 {\text{O}}( {\text{g}}) \to {\text{H}}( {\text{g}})+ {\text{OH}}( {\text{g}}) \Delta {\text{H}}^{ {\text{o}}}= {\text{X}} {\text{kJ}} {\text{mol}}^{-1}$ [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$2 {\text{H}}_2 {\text{O}}( {\text{g}}) \to 2 {\text{H}}_2( {\text{g}})+ {\text{O}}_2( {\text{g}})\;+(242 \times 2) {\text{kJ}} {\text{mol}}^{-1}$ ${\text{H}}_2( {\text{g}})+ {\text{O}}_2( {\text{g}}) \to 2 {\text{OH}}\;+78 {\text{kJ}} {\text{mol}}^{-1}$ ${\text{H}}_2( {\text{g}}) \to 2 {\text{H}}\;+436 {\text{kJ}} {\text{mol}}^{-1}$ $2 {\text{H}}_2 {\text{O}} \to 2 {\text{H}}+2 {\text{OH}}\;+998 {\text{kJ}} {\text{mol}}^{-1}$ $\;$ ${\text{H}}_2 {\text{O}} \to {\text{H}}+ {\text{OH}} \;\; 998 \times \;\frac{1}{2}=+499 {\text{kJ}} {\text{mol}}^{-1}$

Q60JEE Main 2023NAT4MSolutions

$25 {\text{mL}}$ of an aqueous solution of ${\text{KCl}}$ was found to require $20 {\text{mL}}$ of $1 {\text{M}} \; {\text{AgNO}_3}$ solution when titrated using ${\text{K}}_2 {\text{CrO}}_4$ as an indicator. What is the depression in freezing point of ${\text{KCl}}$ solution of the given concentration? ________ (Nearest integer).(Given : ${\text{K}}_{ {\text{f}}}=2.0 {\text{K}} {\text{kg}} {\text{mol}}^{-1}$ )Assume1) $100 \%$ ionization and2) density of the aqueous solution as $1 {\text{g}} {\text{mL}}^{-1}$ [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

At equivalence point, mmole of ${\text{KCl}}=$ mmole of ${\text{AgNO}}_3$ $=20$ mmoleVolume of solution $=25 {\text{ml}}$ Mass of solution $=25 {\text{gm}}$ Mass of solvent $=25$ - mass of solute $=25- [20 \times 10^{-3} \times 74.5]$ $=23.51 {\text{gm}}$ Molality of ${\text{KCl}}=\;{\;\text{ mole of }\; {\text{KCl}}}/{\;\text{ mass of solvent \in }\; {\text{kg}}}$ $=\;\frac{20 \times 10^{-3}}{23.51 \times 10^{-3}}=0.85$ $\;\;\text{ i of }\; {\text{KCl}}=2(100 \% \;\text{ ionisation }\;)$ $\;\Delta {\text{T}}_{ {\text{f}}}= {\text{i}} \times {\text{K}}_{ {\text{f}}} \times {\text{m}}$ $\;=2 \times 2 \times 0.85$ $\;=3.4$ $\;≃ 3$

Mathematics30 questions

Q61JEE Main 2023MCQ4MLimits, Continuity and Differentiability

$\lim _{n \to \infty } (\;\frac{1}{1+n}+\;\frac{1}{2+n}+\;\frac{1}{3+n}+. . .+\;\frac{1}{2 n})$ is equal to :- [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\lim _{n \to \infty } (\;\frac{1}{1+n}+. . . .+\;\frac{1}{n+n}) =\lim _{n \to \infty } sum_{r=1}^{n} \;\frac{1}{n+r}$ $\;\; =\lim _{n \to \infty } sum_{r=1}^{n} \;\frac{1}{n} (\;\frac{1}{1+\;\frac{r}{n}})$ $\;\; =int_{0}^{1} \;\frac{1}{1+x} \text{dx}= [\ln (1+x]_0^1=\ln 2.$

Q62JEE Main 2023MCQ4MSets and Relations

The negation of the expression $q ∨((∼ q) ∧ p)$ is equivalent to [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;∼(q ∨((∼ q) ∧ p))$ $\;=∼ q ∧ ∼((∼ q) ∧ p)$ $\;=∼ q ∧(q ∨ ∼ p)$ $\;=(∼ q ∧ q) ∨(∼ q ∧ ∼ p)$ $\;=(∼ q ∧ ∼ p)$

Q63JEE Main 2023MCQ4MProbability

In a binomial distribution ${\text{B}}( {\text{n}}, {\text{p}})$ , the sum and product of the mean & variance are 5 and 6 respectively, then find $6(n+p-q)$ is equal to :- [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;n p+n p q=5, n p \cdot n p q=6$ $\;n p(1+q)=5, n^2 p^2 q=6$ $\;n^2 p^2(1+q)^2=25, n^2 p^2 q=6$ $\;\;\frac{6}{q}(1+q)^2=25$ $\;6 q^{2+}12 q+6=25 q$ $\;6 q^{2-}13 q+6=0$ $\;6 q^{2-}9 q-4 q+6=0$ $\;(3 q-2)(2 q-3)=0$ $\;q=\;\frac{2}{3}, \;\frac{3}{2}, q=\;\frac{2}{3} \;\text{ is accepted }\;$ $\;p=\;\frac{1}{3} \Rightarrow n \cdot \;\frac{1}{3}+n \cdot \;\frac{1}{3} \cdot \;\frac{2}{3}=5$ $\;\;\frac{3 n+2 n}{9}=5$ $\;n=9$ So $6(n+p-q)=6 (9+\;\frac{1}{3}-\;\frac{2}{3}) =52$

Q64JEE Main 2023MCQ4MSequences and Series

The sum to 10 terms of the series $\;\frac{1}{1+1^{2+}1^4}+\;\frac{2}{1+2^{2+}2^4}+\;\frac{3}{1+3^{2+}3^4}+. . .$ . is:- [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{T}}_{ {\text{r}}}=\;{ ( {\text{r}}^{2+} {\text{r}}+1) - ( {\text{r}}^{2-} {\text{r}}+1) }/{2 ( {\text{r}}^{4+} {\text{r}}^{2+}1) }$ $\Rightarrow {\text{T}}_{ {\text{r}}}=\;\frac{1}{2} [\;{1}/{ {\text{r}}^{2-} {\text{r}}+1}-\;{1}/{ {\text{r}}^{2+} {\text{r}}+1}]$ $T_1=\;\frac{1}{2} [\;\frac{1}{1}-\;\frac{1}{3}]$ $T_2=\;\frac{1}{2} [\;\frac{1}{3}-\;\frac{1}{7}]$ $T_3=\;\frac{1}{2} [\;\frac{1}{7}-\;\frac{1}{13}]$ ${\text{T}}_{10}=\;\frac{1}{2} [\;\frac{1}{91}-\;\frac{1}{111}]$ $\Rightarrow sum_{{\text{r}}=1}^{10} {\text{T}}_{\text{r}}=\;\frac{1}{2} [1-\;\frac{1}{111}] =\;\frac{55}{111}$

Q65JEE Main 2023MCQ4MBinomial Theorem

The value of $\;\frac{1}{1 ! 50 !}+\;\frac{1}{3 ! 48 !}+\;\frac{1}{5 ! 46 !}+. . . .+\;\frac{1}{49 ! 2 !}+\;\frac{1}{51 ! 1 !}$ is [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$sum_{{ {\text{r}}=1}}^{26} \;{1}/{(2 {\text{r}}-1) !(51-(2 {\text{r}}-1)) !}=sum_{{\text{r}}=1}^{26} { }^{51} {\text{C}}_{(2 {\text{r}}-1)} \;\frac{1}{51 !}$ $=\;\frac{1}{51 !} \{{ }^{51} C_1+{ }^{51} C_3+. . . .+{ }^{51} C_{51}\} =\;\frac{1}{51 !} (2^{50})$

Q66JEE Main 2023MCQ4MStraight Lines and Pair of Straight Lines

If the orthocentre of the triangle, whose vertices are $(1,2),(2,3)$ and $(3,1)$ is $(\alpha, \beta)$ , then the quadratic equation whose roots are $\alpha+4 \beta$ and $4 \alpha+\beta$ , is [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Here ${\text{mBH}} \times {\text{mAC}}=-1$ $\; (\;\frac{\beta-3}{\alpha-2}) (\;\frac{1}{-2}) =-1$ $\;\beta-3=2 \alpha-4$ $\;\beta=2 \alpha-1$ $\; {\text{m}}_{ {\text{AH}}} \times {\text{m}}_{ {\text{BC}}}=-1$ $\Rightarrow \;\; (\;\frac{\beta-2}{\alpha-1}) (-2)=-1$ $\Rightarrow \;\; 2 \beta-4=\alpha-1$ $\Rightarrow \;\; 2(2 \alpha-1)=\alpha+3$ $\Rightarrow \;\; 3 \alpha=5$ $\;\alpha=\;\frac{5}{3}, \beta=\;\frac{7}{3} \Rightarrow {\text{H}} (\;\frac{5}{3}, \;\frac{7}{3})$ $\;\alpha+4 \beta=\;\frac{5}{3}+\;\frac{28}{3}=\;\frac{33}{3}=11$ $\;\beta+4 \alpha=\;\frac{7}{3}+\;\frac{20}{3}=\;\frac{27}{3}=9$ $\; {\text{x}}^{2-}20 {\text{x}}+99=0$

Q67JEE Main 2023MCQ4MTrigonometric Ratios and Identities

For a triangle ${\text{ABC}}$ , the value of $\cos 2 {\text{A}}+\cos 2 {\text{B}}+\cos 2 {\text{C}}$ is least. If its inradius is 3 and incentre is $M$ , then which of the following is NOT correct? [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

If $\cos 2 {\text{A}}+\cos 2 {\text{B}}+\cos 2 {\text{C}}$ is minimum then ${\text{A}}=$ ${\text{B}}= {\text{C}}=60^{\circ}$ So $\triangle {\text{ABC}}$ is equilateralNow in-radias $r=3$ So in $\triangle {\text{MBD}}$ we have $\;{\text{Tan}} 30^{\circ}=\;\frac{M D}{B D}=\;\frac{r}{a / 2}=\;\frac{6}{a}$ $\;1 / {\sqrt{3}}=\;\frac{1}{a}=a=6 {\sqrt{3}}$ Perimeter of $\triangle {\text{ABC}}=18 {\sqrt{3}}$ Area of $\triangle {\text{ABC}}=\;{{\sqrt{3}}}/{4} a^2=27 {\sqrt{3}}$

Q68JEE Main 2023MCQ4MStraight Lines and Pair of Straight Lines

The combined equation of the two lines $a x+b y+c=0$ and $a^{'} x+b^{'} y+c^{'}=0$ can be written as $(a x+b y+c) (a^{'} x+b^{'} y+c^{'}) =0$ The equation of the angle bisectors of the lines represented by the equation $2 x^{2+}x y-3 y^2=0$ is [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Equation of the pair of angle bisector for the homogenous equation $a x^{2+}2 h x y+b^2=0$ is given as $\;\frac{x^{2-}y^2}{a-b}=\;\frac{x y}{h}$ Here ${\text{a}}=2, {\text{h}}=1 / 2 & {\text{b}}=-3$ Equation will become $\;\;\frac{x^{2-}y^2}{2-(-3)}=\;\frac{x y}{1 / 2}$ $\;x^{2-}y^2=10 x y$ $\;x^{2-}y^{2-}10 x y=0$

Q69JEE Main 2023MCQ4MThree Dimensional Geometry

The shortest distance between the lines $\;\frac{x-5}{1}=\;\frac{y-2}{2}=\;\frac{z-4}{-3}$ and $\;\frac{x+3}{1}=\;\frac{y+5}{4}=\;\frac{z-1}{-5}$ is [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Shortest distance between two lines $\;\frac{x-x_1}{a_1}=\;\frac{y-y_1}{a_2}=\;\frac{z-z_1}{a_3} &$ $\;\frac{x-x_2}{b_1}=\;\frac{y-y_2}{b_2}=\;\frac{z-z_2}{b_3}$ is given as $\;{{\begin{vmatrix}x_1-x_2 & y_1-y_2 & z_1-z_2 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3\end{vmatrix}}}/{{\sqrt{ (a_1 b_3-a_3 b_2) ^{2+} (a_1 b_3-a_3 b_1) ^{2+} (a_1 b_2-a_2 b_1) ^2}}}$ $\;{{\begin{vmatrix}5-(3) & 2-(-5) & 4-1 \\ 1 & 2 & -3 \\ 1 & 4 & -5\end{vmatrix}}}/{{\sqrt{(-10+12)^{2+}(-5+3)^{2+}(4-2)^2}}}$ $\;{{\begin{vmatrix}8 & 7 & 3 \\ 1 & 2 & -3 \\ 1 & 4 & -5\end{vmatrix}}}/{{\sqrt{(2)^{2+}(2)^{2+}(2)^2}}}$ $\;=\;{|8(-10+12)-7(-5+3)+3(4-2)|}/{{\sqrt{4+4+4}}}$ $\;=\;{16+14+6}/{{\sqrt{12}}}=\;{36}/{{\sqrt{12}}}=\;{36}/{2 {\sqrt{3}}}$ $\;=\;{18}/{{\sqrt{3}}}=6 {\sqrt{3}}$

Q70JEE Main 2023MCQ4MMatrices and Determinants

Let ${\text{S}}$ denote the set of all real values of $\lambda$ such that the system of equations $\;\lambda x+y+z=1$ $\;x+\lambda y+z=1$ $\;x+y+\lambda z=1$ is inconsistent, then $sum_{\lambda \in S}$ $(|\lambda |^2 + |\lambda |)$ is equal to [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;{\begin{vmatrix}\lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{vmatrix}}=0$ $\;(\lambda +2){\begin{vmatrix}1 & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{vmatrix}}=0$ $\;(\lambda +2) \$ [1 (λ^{2-}1) -1(λ-1)+(1-λ)]$ =0;(\lambda +2) [ (\lambda ^{2-}2 \lambda +1) =0.;(\lambda +2)(\lambda -1)^2=0 \Rightarrow \lambda =-2, \lambda =1 $at$ \lambda =1 $system has infinite solution, for inconsistent$ \lambda =-2 $so$ \sum (|-2|^{2+}|-2|) =6$

Q71JEE Main 2023MCQ4MQuadratic Equation and Inequalities

Let ${\text{S}}= \{ {\text{x}}: {\text{x}} \in ℝ.$ and $({\sqrt{3}}+{\sqrt{2}})^{ {\text{x}}^{2-}4}+({\sqrt{3}}-{\sqrt{2}})^{ {\text{x}}^{2-}4}=10\}$ .Then ${\text{n}}( {\text{S}})$ is equal to [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Sol. Let $({\sqrt{3}}+{\sqrt{2}})^{ {\text{x}}^{2-}4}= {\text{t}}$ $\; {\text{t}}+\;{1}/{ {\text{t}}}=10$ $\Rightarrow \;\; {\text{t}}=5+2 {\sqrt{6}}, 5-2 {\sqrt{6}}$ $\Rightarrow \;\; ({\sqrt{3}}+{\sqrt{2}})^{ {\text{x}}^{2-}4}=5+2 {\sqrt{6}}, 5-2 {\sqrt{6}}$ $\Rightarrow \;\; {\text{x}}^{2-}4=2,-2 \;\; \;\text{ or }\; {\text{x}}^2=6,2$ $\Rightarrow \;\; {\text{x}}=\pm {\sqrt{2}}, \pm {\sqrt{6}}$

Q72JEE Main 2023MCQ4MInverse Trigonometric Functions

Let ${\text{S}}$ be the set of all solutions of the equation $\cos ^{-1}(2 x)-2 \cos ^{-1} ({\sqrt{1-x^2}}) =\pi, \;\; x \in [-\;\frac{1}{2}, \;\frac{1}{2}]$ .Then $sum_{x \in {\text{S}}} 2 sin \;^{-1} ( {\text{x}}^{2-}1)$ is equal to [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\cos ^{-1}(2 x)=\pi+2 \cos ^{-1} ({\sqrt{1-x^2}})$ $\;\;\text{ LHS }\;=[0, \pi]$ For equation to be meaningful $\cos ^{-1} 2 x=\pi$ and $\cos ^{-1} ({\sqrt{1-x^2}}) =0$ $x=\;\frac{-1}{2}$ and $x=0$ which is not possible $\therefore {\text{x}} \in \varphi$ Now $Σ(x)=0$ $\therefore$ Sum over empty set is always 0

Q73JEE Main 2023MCQ4MComplex Numbers

If the center and radius of the circle $|\;\frac{z-2}{z-3}|=2$ are respectively $(\alpha, \beta)$ and $\gamma$ , then $3(\alpha+\beta+\gamma )$ is equal to [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;{\sqrt{(x-2)^{2+}y^2}}=2 {\sqrt{(x-3)^{2+}y^2}}$ $\;=x^{2+}y^{2-}4 x+4=4 x^{2+}4 y^{2-}24 x+36$ $\;=3 x^{2+}3 y^{2-}20 x+32=0$ $\;=x^{2+}y^{2-}\;\frac{20}{3} x+\;\frac{32}{3}=0$ $\;=(\alpha, \beta)= (\;\frac{10}{3}, 0)$ $\;\gamma =\sqrt{\;\frac{100}{9}-\;\frac{32}{3}}=\sqrt{\;\frac{4}{9}}=\;\frac{2}{3}$ $\;3(\alpha, \beta, \gamma )=3 (\;\frac{10}{3}+\;\frac{2}{3})$ $\;=12$

Q74JEE Main 2023MCQ4MDifferential Equations

If ${\text{y}}= {\text{y}}( {\text{x}})$ is the solution curve of the differential equation $\;{ {\text{dy}}}/{ {\text{dx}}}+ {\text{y}} \tan {\text{x}}= {\text{x}} \ sec {\text{x}}, \;\; 0 \le {\text{x}} \le \;\frac{\pi}{3}$ , $y(0)=1$ , then $y (\;\frac{\pi}{6})$ is equal to [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Here I.F. $=\ sec {\text{x}}$ Then solution of D.E : $\;y(\ sec x)=x \tan x-\ln (\ sec x)+c$ $\;\;\text{ Given }\; y(0)=1 \Rightarrow c=1$ $\; \therefore y(\ sec x)=x \tan x-\ln (\ sec x)+1$ $\;\;\text{ At }\; x=\;\frac{\pi}{6}, y=\;\frac{\pi}{12}+\;{{\sqrt{3}}}/{2} \ln \;{{\sqrt{3}}}/{2}+\;{{\sqrt{3}}}/{2}$

Q75JEE Main 2023MCQ4MSets and Relations

Let ${\text{R}}$ be a relation on $ℝ$ , given by ${\text{R}}=\{( {\text{a}}, {\text{b}}): 3 {\text{a}}-3 {\text{b}}+{\sqrt{7}}$ is an irrational number $\}$ . Then ${\text{R}}$ is [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Check for reflexivity:As 3(a-a) $+{\sqrt{7}}={\sqrt{7}}$ which belongs to relation so relation is reflexiveCheck for symmetric:Take ${\text{a}}=\;{{\sqrt{7}}}/{3}, b=0$ Now $(a, b) \in R$ but $(b, a) ∉ R$ As $3(b-a)+{\sqrt{7}}=0$ which is rational so relation is not symmetric.Check for Transitivity:Take $( {\text{a}}, {\text{b}})$ as $(\;{{\sqrt{7}}}/{3}, 1)$ $\&(b, c)$ as $(1, \;{2 {\sqrt{7}}}/{3})$ So now $( {\text{a}}, {\text{b}}) \in {\text{R}} &( {\text{b}}, {\text{c}}) \in {\text{R}}$ but $( {\text{a}}, {\text{c}}) ∉ {\text{R}}$ which means relation is not transitive

Q76JEE Main 2023MCQ4MThree Dimensional Geometry

Let the image of the point ${\text{P}}(2,-1,3)$ in the plane $x+2 y-z=0$ be $Q$ . Then the distance of the plane $3 x+2 y+z+29=0$ from the point $Q$ is [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

eq. of line PM $\;\frac{x-2}{1}=\;\frac{y+1}{2}=\;\frac{z-3}{-1}=\lambda$ any point on line $=(\lambda +2,2 \lambda -1,-\lambda +3)$ for point ' $m$ ' $(\lambda +2)+2(2 \lambda -1)-(3-\lambda )=0$ $\lambda =\;\frac{1}{2}$ Point ${\text{m}} (\;\frac{1}{2}+2,2 \times \;\frac{1}{2}-1, \;\frac{-1}{2}+3)$ $= (\;\frac{5}{2}, 0, \;\frac{5}{2})$ For Image ${\text{Q}}(\alpha, \beta, \gamma )$ $\;\;\frac{\alpha+2}{2}=\;\frac{5}{2}, \;\frac{\beta-1}{2}=0,$ $\;\;\frac{\gamma +3}{2}=\;\frac{5}{2}$ $\;Q:(3,1,2)$ $\;d= |\;{3(3)+2(1)+2+29}/{{\sqrt{3^{2+}2^{2+}1^2}}}|$ $\;d=\;{42}/{{\sqrt{14}}}=3 {\sqrt{14}}$

Q77JEE Main 2023MCQ4MApplication of Derivatives

Let $\;\; f(x)={\begin{vmatrix}1+ sin \\ \\ ^2 x & \cos ^2 x & sin \\ \\ 2 x \\ sin \\ \\ ^2 x & 1+\cos ^2 x & sin \\ \\ 2 x \\ sin \\ \\ ^2 x & \cos ^2 x & 1+ sin \\ \\ 2 x\end{vmatrix}}$ , ${\text{x}} \in [\;\frac{\pi}{6}, \;\frac{\pi}{3}]$ . If $\alpha$ a $\beta$ respectively are the maximum and the minimum values of $f$ , then [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{C}}_1 \to {\text{C}}_1+ {\text{C}}_2+ {\text{C}}_3$ ${\text{f}}( {\text{x}})={\begin{vmatrix}2+ sin \\ \\ 2 x & \cos ^2 x & sin \\ \\ 2 x \\ 2+ sin \\ \\ 2 x & 1+\cos ^2 x & sin \\ \\ 2 x \\ 2+ sin \\ \\ 2 x & \cos ^2 x & 1+ sin \\ \\ 2 x\end{vmatrix}}$ $f(x)=(2+ sin \;2 x){\begin{vmatrix}1 & \cos ^2 x & sin \\ \\ 2 x \\ 1 & 1+\cos ^2 x & sin \\ \\ 2 x \\ 1 & \cos ^2 x & 1+ sin \\ \\ 2 x\end{vmatrix}}$ $\;R_2 \to R_2-R_1$ $\;R_3 \to R_3-R_1$ $f(x)=2+ sin \;2 x){\begin{vmatrix}1 & \cos ^2 x & sin \\ \\ 2 x \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{vmatrix}}$ $\;=(2+ sin \;2 x)(1)=2+ sin \;2 x$ $\;= sin \;2 x \in [\;{{\sqrt{3}}}/{2}, 1]$ Hence $2+ sin \;2 x \in [2+\;{{\sqrt{3}}}/{2}, 3]$

Q78JEE Main 2023MCQ4MApplication of Derivatives

Let $f(x)=2 x+\tan ^{-1} x$ and $g(x)=\log _e ({\sqrt{1+x^2}}+x)$ , ${\text{x}} \in [0,3]$ . Then [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{f}}( {\text{x}})=2 {\text{x}}+\tan ^{-1} {\text{x}}$ and ${\text{g}}( {\text{x}})=\ln ({\sqrt{1+x^2}}+x)$ and ${\text{x}} \in [0,3]$ ${\text{g}}^{'}( {\text{x}})=\;{1}/{{\sqrt{1+x^2}}}$ Now, $0 \le x \le 3$ $0 \le x^2 \le 9$ $1 \le 1+x^2 \le 10$ So, $\;\; 2+\;\frac{1}{10} \le f^{'}(x) \le 3$ $\;\frac{21}{10} \le f^{'}(x) \le 3 \;\text{ and }\; \;{1}/{{\sqrt{10}}} \le g^{'}(x) \le 1$ option (4) is incorrectFrom above, ${\text{g}}^{'}( {\text{x}})< {\text{f}}^{'}( {\text{x}}) ∀ {\text{x}} \in [0,3]$ Option (1) is incorrect. ${\text{f}}^{'}( {\text{x}}) & {\text{g}}^{'}( {\text{x}})$ both positive so ${\text{f}}( {\text{x}}) & {\text{g}}( {\text{x}})$ both are increasingSo, $\;\; \max ( {\text{f}}( {\text{x}}).$ at ${\text{x}}=3$ is $6+\tan ^{-1} 3$ ${\text{Max}}(g(x)$ at $x=3$ is $\ln (3+{\sqrt{10}})$ And $6+\tan ^{-1} 3 > \ln (3+{\sqrt{10}})$ Option (2) is correct

Q79JEE Main 2023MCQ4MStatistics

The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are $1,3,5$ , then the sum of cubes of the remaining two observations is [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\;\frac{1+3+5+a+b}{5}=5$ $\;a+b=16 . . . . . .(1)$ $\;\sigma^2=\;\frac{\sum x_1^2}{5}- (\;\frac{\sum x}{5}) ^2$ $\;8=\;\frac{1^{2+}3^{2+}5^{2+}a^{2+}b^2}{5}-25$ $\;a^{2+}b^2=130$ . . . (2) $\;\;\text{ by }\;(1),(2)$ $\;a=7, b=9$ $\;\;\text{ or }\; a=9, b=7$

Q80JEE Main 2023MCQ4MArea Under The Curves

The area enclosed by the closed curve $C$ given by the differential equation $\;\frac{d y}{d x}+\;\frac{x+a}{y-2}=0, y(1)=0$ is $4 \pi$ .Let ${\text{P}}$ and ${\text{Q}}$ be the points of intersection of the curve ${\text{C}}$ and the ${\text{y}}$ -axis. If normals at ${\text{P}}$ and ${\text{Q}}$ on the curve ${\text{C}}$ intersect ${\text{x}}$ -axis at points ${\text{R}}$ and ${\text{S}}$ respectively, then the length of the line segment ${\text{RS}}$ is [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\;\frac{d y}{d x}+\;\frac{x+a}{y-2}=0$ $\;\;\frac{d y}{d x}=\;\frac{x+a}{2-y}$ $\;(2- {\text{y}}) {\text{dy}}=( {\text{x}}+ {\text{a}}) {\text{dx}}$ $\;2 y \;\frac{-y}{2}=\;\frac{x^2}{2}+ {\text{ax}}+ {\text{c}}$ $\; {\text{a}}+ {\text{c}}=-\;\frac{1}{2} \;\text{ as }\; {\text{y}}(1)=0$ $\; {\text{X}}^{2+} {\text{y}}^{2+}2 {\text{ax}}-4 {\text{y}}-1-2 {\text{a}}=0$ $\;\pi {\text{r}}^2=4 \pi$ $\; {\text{r}}^2=4$ $\;4={\sqrt{a^{2+}4+1+2 a}}$ $\;( {\text{a}}+1)^2=0$ $P, Q=(0,2 \pm {\sqrt{3}})$ Equation of normal at ${\text{P}}, {\text{Q}}$ are ${\text{y}}-2={\sqrt{3}}( {\text{x}}-1)$ $\; {\text{y}}-2=-{\sqrt{3}}(x-1)$ $\; {\text{R}}= (1-\;{2}/{{\sqrt{3}}}, 0)$ $\; {\text{S}}= (1+\;{2}/{{\sqrt{3}}}, 0)$ $\; {\text{RS}}=\;{4}/{{\sqrt{3}}}=4 \;{{\sqrt{3}}}/{3}$

Q81JEE Main 2023NAT4MSequences and Series

Let $a_1=8, a_2, a_3, . . . . a_n$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170 , then the product of its middle two terms is ______. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;a_1+a_2+a_3+a_4=50$ $\;\Rightarrow 32+6 d=50$ $\;\Rightarrow d=3$ $\;\;\text{ and, }\; a_{n-3}+a_{n-2}+a_{n-1}+a_n=170$ $\;\Rightarrow 32+(4 n-10) \cdot 3=170$ $\;\Rightarrow n=14$ $\;a_7=26, a_8=29$ $\;\Rightarrow a_7 \cdot a_8=754$

Q82JEE Main 2023NAT4MVector Algebra

${\text{A}}(2,6,2), {\text{B}}(-4,0, \lambda ), {\text{C}}(2,3,-1)$ and ${\text{D}}(4,5,0)$ , $|\lambda | \le 5$ are the vertices of a quadrilateral ${\text{ABCD}}$ . If its area is 18 square units, then $5-6 \lambda$ is equal to ______. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{A}}(2,6,2) \;\; {\text{B}}(-4,0, \lambda ), {\text{C}}(2,3,-1) {\text{D}}(4,5,0)$ $\;\;\text{ Area }\;=\;\frac{1}{2}|\vec{B D} \times \vec{A C}|=18$ $\;\vec{A C} \times \vec{B D}={\begin{bmatrix}{i}^{∧} & j & k \\ 0 & -3 & -3 \\ 8 & 5 & -\lambda \end{bmatrix}\$ };=(3 λ+15) ;\hat{i}-j(-24)+k(-24);\vec{A C} × \vec{B D}=(3 λ+15) ;\hat{i}+24 j-24 k;={\sqrt{(3 λ+15)^{2+}(24)^{2+}(24)^2}}=36;=λ^{2+}10 λ+9=0;=λ=-1,-9;|λ| ≤ 5 \Rightarrow λ=-1;5-6 λ=5-6(-1)=11$

Q83JEE Main 2023NAT4MPermutations and Combinations

The number of 3 -digit numbers, that are divisible by either 2 or 3 but not divisible by 7 is _______. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Divisible by $2 \to 450$ Divisible by $3 \to 300$ Divisible by $7 \to 128$ Divisible by $2 & 7 \to 64$ Divisible by $3 & 7 \to 43$ Divisible by $2 & 3 \to 150$ Divisible by $2,3 & 7 \to 21$ $\therefore$ Total numbers $=450+300-150-64-43+21=514$

Q84JEE Main 2023NAT4MBinomial Theorem

The remainder when $19^{200}+23^{200}$ is divided by 49 , is _______. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;(21+2)^{200}+(21-2)^{200}$ $\;\Rightarrow 2 [^{100} {\text{C}}_0 21^{200}+200 {\text{C}}_2 21^{198} \cdot 2^{2+}. . . . .+{ }^{200} {\text{C}}_{198}.$ $\;21^2 \cdot 2^{198}+2^{200}]$ $\;\Rightarrow 2 \$ [49 {\text{I}}_1+2^{200}]$ =49 {\text{I}}_1+2^{201};;\text{ Now };, 2^{201}=(8)^{67}=(1+7)^{67}=49 {\text{I}}_2+{ }^{67} {\text{C}}_0{ }^{67} {\text{C}}_1 \cdot 7=;49 {\text{I}}_2+470=49 {\text{I}}_2+49 \times 9+29$

Q85JEE Main 2023NAT4MDefinite Integrals

If $int_{0}^{1} (x^{21}+x^{14}+x^7) (2 x^{14}+3 x^{7+}6) ^{1 / 7} \text{dx}=\;\frac{1}{l}{(11)}^{m / n}$ where $1, {\text{m}}, {\text{n}} \in ℕ, {\text{m}}$ and ${\text{n}}$ are coprime then ${\text{l}}+ {\text{m}}+ {\text{n}}$ is equal to ________. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\int (x^{20}+x^{13}+x^6) (2 x^{21}+3 x^{14}+6 x^7) ^{1 / 7} \text{dx}$ $\;2 {\text{x}}^{21}+3 {\text{x}}^{14}+6 {\text{x}}^7= {\text{t}}$ $\;42 ( {\text{x}}^{20}+ {\text{x}}^{13}+ {\text{x}}^6) {\text{dx}}= {\text{dt}}$ $\;\;\frac{1}{42} int_{0}^{11} t^{\;\frac{1}{7}} \text{dt}= (\;{t^{\frac{8}{7}}}/{\frac{8}{7}} \times \;\frac{1}{42}) _0^{11}$ $\; \;\; =\;\frac{1}{48} (t^{\;\frac{8}{7}}) _0^{11}=\;\frac{1}{48}(11)^{8 / 7}$ $\;l=48, {\text{m}}=8, {\text{n}}=7$ $\;l+ {\text{m}}+ {\text{n}}=63$

Q86JEE Main 2023NAT4MDifferentiation

If $f(x)=x^{2+}g^{'}(1) x+g^{' '}(2)$ and $g(x)=f(1) x^{2+}x f^{'}(x)+f^{' '}(x),$ then the value of $f(4)-g(4)$ is equa to ________. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{f}}( {\text{x}})= {\text{x}}^{2+} {\text{g}}^{'}(1) {\text{x}}+ {\text{g}}^{' '}(2)$ $\; {\text{f}}^{'}( {\text{x}})=2 {\text{x}}+ {\text{g}}^{'}(1)$ $\; {\text{f}}^{' '}( {\text{x}})=2$ $\; {\text{g}}( {\text{x}})= {\text{f}}(1) {\text{x}}^{2+} {\text{x}} \$ [2 {\text{x}}+ {\text{g}}^{'}(1)]$ +2; {\text{g}}^{'}( {\text{x}})=2 {\text{f}}(1) {\text{x}}+4 {\text{x}}+ {\text{g}}^{'}(1); {\text{g}}^{' '}( {\text{x}})=2 {\text{f}}(1)+4; {\text{g}}^{' '}( {\text{x}})=0;2 {\text{f}}(1)+4=0; {\text{f}}(1)=-2-2=1+ {\text{g}}^{'}(1)= {\text{g}}^{'}(1)=-3;\text{ So, };; {\text{f}}^{'}( {\text{x}})=2 {\text{x}}-3; {\text{f}}( {\text{x}})= {\text{x}}^{2-}3 {\text{x}}+ {\text{c}}{\text{c}}=0{\text{f}}( {\text{x}})= {\text{x}}^{2-}3 {\text{x}}{\text{g}}( {\text{x}})=-3 {\text{x}}+2{\text{f}}(4)- {\text{g}}(4)=14$

Q87JEE Main 2023NAT4MVector Algebra

Let $\vec{v}=\alpha \hat{i}+2 j-3 k, \vec{w}=2 \alpha \hat{i}+j-k$ , and ${ \vec{\text{u}}$ be a vector such that $|\vec{u}|=\alpha>0$ . If the minimum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is $-\alpha {\sqrt{3401}}$ , and ${|\vec{\text{u}} . \;{ {\text{i}}}^{\^}|}^2=\;{{\text{m}}}/{ {\text{n}}}$ where ${\text{m}}$ and ${\text{n}}$ are coprime natural numbers, then ${\text{m}}+ {\text{n}}$ is equal to _______. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$[\vec{u} \vec{v} \vec{w}]=\vec{u} \cdot (\vec{v} \times \vec{w})$ $\;\min .(|u||\vec{v} \times \vec{w}| \cos \theta)=-\alpha {\sqrt{3401}}$ $\;\Rightarrow \;\; \cos \theta=-1$ $\;|u|=\alpha \;\text{ (Given) }\;$ $\;|\vec{v} \times \vec{w}|={\sqrt{3401}}$ $\;\vec{v} \times \vec{w}={\begin{vmatrix}\\ \\ {i} & j & k \\ \alpha & 2 & -3 \\ 2 \alpha & 1 & -1\end{vmatrix}}$ $\;\vec{v} \times \vec{w}=\;{i}-5 \alpha j-3 \alpha k$ $\;|\vec{v} \times \vec{w}|={\sqrt{1+25 \alpha^{2+}9 \alpha^2}}^{\^}^{\^}={\sqrt{3401}}$ $\;34 \alpha^2=3400$ $\;\alpha^2=100$ $\;\alpha=10$ $\;\;\text{ (as }\; \alpha > 0 \;\text{ ) }\;$ $\;$ So $\;\vec{u}=\lambda (\;{i}-5 \alpha j-3 \alpha k)$ $\;\vec{u}={\sqrt{\lambda ^{2+}25 \alpha^2 \lambda ^{2+}9 \alpha^2 \lambda }}^{\^}$ $\;\alpha^2=\lambda ^2 (1+25 \alpha^{2+}9 \alpha^2)$ $\;100=\lambda ^2(1+34 \times 100)$ $\;\lambda ^2=\;\frac{100}{3401}=\;\frac{m}{n}$

Q88JEE Main 2023NAT4MPermutations and Combinations

The number of words, with or without meaning, that can be formed using all the letters of the word ASSASSINATION so that the vowels occur together, is ________. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Vowels : A,A,A,I,I,OConsonants : S,S,S,S,N,N,T $\because$ Total number of ways in which vowels come together $=\;\frac{x{8}}{⌊ 4 \mid 2} \times \;\frac{x{6}}{⌊x{3} L 2}=50400$

Q89JEE Main 2023NAT4MArea Under The Curves

Let $A$ be the area bounded by the curve $y=x| x-3|$ , the $x$ -axis and the ordinates $x=-1$ and $x=2$ . Then $12 {\text{A}}$ is equal to _______. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$A=int_{-1}^{0} (x^{2-}3 x) \text{dx}+int_{0}^{2} (3 x-x^2) \text{dx}$ $\Rightarrow \;\; {\text{A}}=\;{ {\text{x}}^3}/{3}-\;{{3 {\text{x}}^2}/{2}}| ^0_{-1}+\;{3 {\text{x}}^2}/{2}-\;{ {\text{x}}^3}/{3}|_0 ^2$ $\;\Rightarrow \;\; {\text{A}}=\;\frac{11}{6}+\;\frac{10}{3}=\;\frac{31}{6}$ $\; \therefore \;\; 12 {\text{A}}=62$

Q90JEE Main 2023NAT4MDifferential Equations

Let $f: ℝ \to ℝ$ be a differentiable function such that $f^{'}(x)+f(x)=int_{0}^{2} f(t) \text{dt}$ . If $f(0)=e^{-2}$ , then $2 f(0)-f(2)$ is equal to _______. [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\;\frac{d y}{d x}+y=k$ $\;y \cdot e^x=k \cdot e^x+c$ $\;f(0)=e^{-2}$ $\;\Rightarrow \;\; {\text{c}}= {\text{e}}^{-2}- {\text{k}}$ $\; \therefore \;\; {\text{y}}= {\text{k}}+ ( {\text{e}}^{-2}- {\text{k}}) {\text{e}}^{- {\text{x}}}$ $\Rightarrow \;\; k=e^{-2}-1$ $\therefore \;\; {\text{y}}= ( {\text{e}}^{-2}-1) + {\text{e}}^{- {\text{x}}}$ $\; {\text{f}}(2)=2 {\text{e}}^{-2}-1, {\text{f}}(0)= {\text{e}}^{-2}$ $\;2 {\text{f}}(0)- {\text{f}}(2)=1$

← All JEE Main 2023 papers Take as Timed Mock →