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10 Apr 2023 Shift 1

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Physics30 questions

Q1JEE Main 2023MCQ4MElectrostatic Potential and Capacitance

The equivalent capacitance of the combination shown is [10-Apr-2023 shift 1]

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📖 Explanation

Capacitor (3) & (4) are short ckt $\therefore {\text{C}}_1 & {\text{C}}_2$ are in parallel ${\text{C}}_{ {\text{eq}}}= {\text{C}}+ {\text{C}}=2 {\text{C}}$

Q2JEE Main 2023MCQ4MThermodynamics

Match List I with List II : List I List II (A) 3 Translational degrees of freedom (I) Monoatomic gases (B) 3 Translational, 2 rotational degrees of freedoms (II) Polyatomic gases (C) 3 Translational, 2 rotational and 1 vibrational degrees of freedom (III) Rigid diatomic gases (D) 3 Translational, 3 rotational and more than one vibrational degrees of freedom (IV) Nonrigid diatomic gasesChoose the correct answer from the options given below : [10-Apr-2023 shift 1]

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📖 Explanation

Fact Based Type of gas No of degree of freedom 1 Monoatomic 3 (Translational) 2. Diatomic + rigid 3 (Translational +2 Rotational $=5)$ 3. Diatomic + non - rigid 3 (Trans) +2 (Rotational) +1 (vibrational) 4. Polyatomic 3 (Trans) +2 (Rotational) + more than 1 (vibrational)

Q3JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

Given below are two statements :Statements I : If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double.Statements II : Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratioIn the light of the above statements, choose the correct answer from the options given below : [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{I}}=\;{ {\text{k}} \theta}/{\;\text{ NBA }\;}$ $\; {\text{C}} \cdot {\text{S}}=\;{\theta}/{ {\text{I}}}=\;{ {\text{NBA}}}/{ {\text{K}}}$ $\; {\text{N}} \to 2 {\text{N}} \;\; {\text{C}} \cdot {\text{S}} \to 2 {\text{CS}}$ But V.S. $=\;{\theta}/{ {\text{V}}}=\;{ {\text{NBA}}}/{ {\text{KR}}}$ ${\text{N}} \to 2 {\text{NC}} \cdot {\text{S}} \to 2 {\text{CS}}$ But V.S. $=\;{\theta}/{ {\text{v}}}=\;{\theta}/{ {\text{IR}}}=\;{ {\text{NBA}}}/{ {\text{RK}}}$ As ${\text{N}} \to 2 {\text{N}}, {\text{R}} \to 2 {\text{R}}$ So V.S = constant

Q4JEE Main 2023MCQ4MAlternating Current

Given below are two statements:Statement I : Maximum power is dissipated in a circuit containing an inductor, a capacitor and a resistor connected in series with an ${\text{AC}}$ source, when resonance occursStatement II : Maximum power is dissipated in a circuit containing pure resistor due to zero phase difference between current and voltage.In the light of the above statements, choose the correct answer from the options given below : [10-Apr-2023 shift 1]

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📖 Explanation

Power is more when total impendence of ckt in minimum $\; {\text{Z}}=\sqrt{ {\text{R}}^{2+} ( {\text{X}}_{ {\text{L}}}- {\text{X}}_{ {\text{C}}}) ^2}$ $\; \because {\text{X}}_{ {\text{L}}}= {\text{X}}_{ {\text{C}}} \;\text{ (conductor of resonance) }\;$ $\; \therefore {\text{Z}}_{\min }= {\text{R}} \;\; \therefore {\text{V}} & {\text{I}} \;\text{ \in same phase }\;$

Q5JEE Main 2023MCQ4MMotion in a Plane

The range of the projectile projected at an angle of $15^{\circ}$ with horizontal is $50 {\text{m}}$ . If the projectile is projected with same velocity at an angle of $45^{\circ}$ with horizontal, then its range will be [10-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ So }\; {\text{R}}=\;{ {\text{u}}^2 sin \;(2 \times 15)}/{ {\text{g}}}=\;{ {\text{u}}^2}/{2 {\text{g}}}=\;\text{ So }\; \Rightarrow \;{ {\text{u}}^2}/{ {\text{g}}}=100$ $\; {\text{R}}^{'}=\;{ {\text{u}}^2 sin \;(2 \times 45)}/{ {\text{g}}}=\;{ {\text{u}}^2}/{ {\text{g}}}=100 {\text{m}}$

Q6JEE Main 2023MCQ4MCentre of Mass and Collision

A particle of mass ${\text{m}}$ moving with velocity $v$ collides with a stationary particle of mass $2 {\text{m}}$ . After collision, they stick together and continue to move together with velocity [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{p}} {\text{i}}= {\text{pf}} \Rightarrow {\ mv}+2 {\text{m}}(0)=3 {\text{m}} ( {\text{v}}^{'})$ $\; {\text{v}}^{'}=\;{ {\text{v}}}/{3}$

Q7JEE Main 2023MCQ4MGravitation

Two satellites of masses ${\text{m}}$ and $3 {\text{m}}$ revolve around the earth in circular orbits of radii ${\text{r}} & 3 {\text{r}}$ respectively. The ratio of orbital speeds of the satellites respectively is [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{v}}=\sqrt{\;{ {\text{GM}}}/{ {\text{r}}}} \Rightarrow {\text{v}} \times \;{1}/{\sqrt{ {\text{r}}}} ; {\text{M}}=\;\text{ mass of earth, }\; r=\;\text{ radius of earth }\;$ $\;\;{ {\text{v}}_1}/{ {\text{v}}_2}=\sqrt{\;{ {\text{r}}_2}/{ {\text{r}}_1}}=\sqrt{\;{3 {\text{r}}}/{ {\text{r}}}}={\sqrt{3}}$

Q8JEE Main 2023MCQ4MGravitation

Assuming the earth to be a sphere of uniform mass density, the weight of a body at a depth $d=\;\frac{R}{2}$ from the surface of earth, if its weight on the surface of earth is $200 {\text{N}}$ , will be : [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{mg}}=200 {\text{N}}$ $\; {\text{g}}^{'}= {\text{g}} (1-\;{ {\text{d}}}/{ {\text{R}}}) = {\text{g}} (1-\;{ {\text{R}}}/{2 \times {\text{R}}}) =\;{ {\text{g}}}/{2}$ $\;\;\text{ weight }\;= {\text{mg}}^{'}=\;{ {\text{mg}}}/{2}=\;\frac{200}{2}=100 {\text{N}}$

Q9JEE Main 2023MCQ4MDual Nature of Matter and Radiation

The de Broglie wavelength of a molecule in a gas at room temperature $(300 {\text{K}})$ is $\lambda _1$ . If the temperature of the gas is increased to $600 {\text{K}}$ , then the de Broglie wavelength of the same gas molecule becomes [10-Apr-2023 shift 1]

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📖 Explanation

$\;\lambda =\;{ {\text{h}}}/{\sqrt{3 {\text{mK}}( {\text{T}})}}$ $\;\;\frac{\lambda _1}{\lambda _2}=\sqrt{\;{ {\text{T}}_2}/{ {\text{T}}_1}}$ $\;\lambda _2=\lambda _1 \sqrt{\;{ {\text{T}}_1}/{ {\text{T}}_2}}$ $\;=\lambda _1 \sqrt{\;\frac{300}{600}}=\;{\lambda _2}/{{\sqrt{2}}}$

Q10JEE Main 2023MCQ4MUnits and Measurements

A physical quantity $P$ is given as $P=\;{a^2 b^3}/{c {\sqrt{d}}}$ The percentage error in the measurement of ${\text{a}}, {\text{b}}, {\text{c}}$ and ${\text{d}}$ are $1 \%, 2 \%, 3 \%$ and $4 \%$ respectively. The percentage error in the measurement of quantity ${\text{P}}$ will be [10-Apr-2023 shift 1]

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📖 Explanation

$\;\;{ {\text{dP}}}/{ {\text{P}}} \times 100= (2 \;{ {\text{da}}}/{ {\text{a}}}+3 \;{ {\text{db}}}/{ {\text{b}}}+\;{ {\text{dc}}}/{ {\text{c}}}+\;\frac{1}{2} \;{ {\text{d}}( {\text{d}})}/{ {\text{d}}}) \times 100$ $\;=2 \times 1+3 \times 2+3+\;\frac{1}{2} \times 4$ $\;=2+6+3+2$ $\;=13 \%$

Q11JEE Main 2023MCQ4MThermodynamics

Consider two containers A and B containing monoatomic gases at the same Pressure (P), Volume (V) and Temperature (T). The gas in A is compressed isothermally to $\;\frac{1}{8}$ of its original volume while the gas in B is compressed adiabatically to $\;\frac{1}{8}$ of its original volume. The ratio of final pressure of gas in B to that of gas in A is [10-Apr-2023 shift 1]

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📖 Explanation

By Isothermal Process for (A) $\; {\text{P}}_1 {\text{V}}_1= {\text{P}}_2 {\text{V}}_2$ $\; {\text{PV}}= {\text{P}}_2 \;{ {\text{V}}}/{8}$ $\; {\text{P}}_2=8 {\text{P}}$ For B adiabatically $\gamma _{\;\text{mono }\;}=\;\frac{5}{3}$ $\; {\text{P}}_1 {\text{V}}_1^\gamma = {\text{P}}_2 {\text{V}}_2^\gamma$ $\; {\text{PV}}^{5 / 3}= {\text{P}}_2 (\;{ {\text{V}}}/{8}) ^{5 / 3}$ $\; {\text{P}}_2=(8)^{5 / 3} {\text{P}}$ $\;\;{ {\text{P}}_2}/{ {\text{P}}_1}=\;{8^{5 / 3}}/{8 {\text{P}}}=(8)^{\;\frac{2}{3}}=4$

Q12JEE Main 2023MCQ4MProperties of Matter

Given below are two statements :Statements I : Pressure in a reservoir of water is same at all points at the same level of water.Statements II : The pressure applied to enclosed water is transmitted in all directions equally. In the light of the above statements, choose the correct answer from the options given below : [10-Apr-2023 shift 1]

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Q13JEE Main 2023MCQ4MMotion in a Straight Line

The positon-time graphs for two students ${\text{A}}$ and B returning from the school to their homes are shown in figure.(A) A lives closer to the school(B) B lives closer to the school(C) A takes lesser time to reach home(D) A travels faster than $B$ (E) ${\text{B}}$ travels faster than ${\text{A}}$ Choose the correct answer from the options given below : [10-Apr-2023 shift 1]

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📖 Explanation

(A) and (E) onlySlope of $A=V_A$ Slope of $B=V_B$ (slope $)_B > (\;\text{ slope }\;)_A$ ${\text{V}}_{ {\text{B}}} > {\text{V}}_{ {\text{A}}}$ $\therefore {\text{t}}_{ {\text{B}}} < {\text{t}}_{ {\text{A}}}$

Q14JEE Main 2023MCQ4MElectromagnetic Waves

The energy of an electromagnetic wave contained in a small volume oscillates with [10-Apr-2023 shift 1]

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📖 Explanation

double the frequency of the wave $\; {\text{E}}= {\text{E}}_0 sin \;( {\text{wt}}- {\text{kx}})$ $\;\;\text{ Energy density }\;=\;\frac{1}{2} \epsilon_0 {\text{E}}_{ {\text{net}}}$ $\;=\;\frac{1}{2} \epsilon_0 {\text{E}}_0^2 sin \;^2( {\text{wt}}- {\text{kx}})$ $\;=\;\frac{1}{4} \epsilon_0 {\text{E}}_0^2(1-\cos (2 {\text{wt}}-2 {\text{kx}}))$

Q15JEE Main 2023MCQ4MCurrent Electricity

The equivalent resistance of the circuit shown below between points a and ${\text{b}}$ is : [10-Apr-2023 shift 1]

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📖 Explanation

$\;\;\frac{1}{R_{a b}}=\;\frac{1}{16}+\;\frac{1}{8}+\;\frac{1}{8}$ $\;=\;\frac{1}{R_{a b}}=\;\frac{1+2+2}{16}=\;\frac{5}{16}$ $\;R_{a b}=\;\frac{16}{5}=3.2$

Q16JEE Main 2023MCQ4MElectromagnetic Waves

A carrier wave of amplitude $15 {\text{V}}$ modulated by a sinusoidal base band signal of amplitude $3 {\text{V}}$ . The ratio of maximum amplitude to minimum amplitude in an amplitude modulated wave is [10-Apr-2023 shift 1]

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📖 Explanation

$\;V_C=15$ $\;V_m=3$ $\;V_{\max }=15+3=18$ $\;V_{\min }=15-3=12$ $\;V_{\max }=\;\frac{18}{12}=\;\frac{3}{2}=3: 2$

Q17JEE Main 2023MCQ4MOscillations

A particle executes S.H.M. of amplitude A along $x$ -axis. At $t=0$ , the position of the particle is $x=\;\frac{A}{2}$ and it moves along positives $x$ -axis. The displacement of particle in time $t$ is ${\text{x}}= {\text{A}} sin \;(\omega t+\delta )$ , then the value $\delta$ will be [10-Apr-2023 shift 1]

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📖 Explanation

$\;{\text{Cos}} \theta=\;{ {\text{A}}}/{2 \times {\text{A}}}=\;\frac{1}{2}$ $\;\theta=\;\frac{\pi}{3}$ $\;\delta =\;\frac{\pi}{2}-\;\frac{\pi}{3}=\;\frac{\pi}{6}$

Q18JEE Main 2023MCQ4MAtoms and Nuclei

The angular momentum for the electron in Bohr's orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, them the change in angular momentum will be [10-Apr-2023 shift 1]

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📖 Explanation

Angular momentum $=\;{ {\text{nh}}}/{2 \pi}$ $\; {\text{n}}=1, {\text{L}}_1=\;{ {\text{h}}}/{2 \pi}= {\text{L}}$ $\; {\text{n}}=2, {\text{L}}_2=\;{2 {\text{h}}}/{2 \pi}=2 {\text{L}}$ $\;\Delta {\text{L}}=2 {\text{L}}- {\text{L}}= {\text{L}}$

Q19JEE Main 2023MCQ4MRay Optics and Optical Instruments

An object is placed at a distance of $12 {\text{cm}}$ in front of a plane mirror. The virtual and erect image is formed by the mirror. Now the mirror is moved by $4 {\text{cm}}$ towards the stationary object. The distance by which the position of image would be shifted, will be [10-Apr-2023 shift 1]

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📖 Explanation

$8 {\text{cm}}$ towards mirrorImage will be shifted $8 {\text{cm}}$ towards mirror.

Q20JEE Main 2023MCQ4MSemiconductor Electronics

A zener diode of power rating $1.6 {\text{W}}$ is be used as voltage regulator. If the zener diode has a breakdown of $8 {\text{V}}$ and it has to regulate voltage fluctuating between $3 {\text{V}}$ and $10 {\text{V}}$ . The value of resistance ${\text{R}}_8$ for safe operation of diode will be [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{I}}_{ {\text{t}}}=\;{ {\text{P}}_{ {\text{t}}}}/{ {\text{V}}_{ {\text{t}}}}=\;\frac{1.6}{8}=0.2 {\text{A}}$ $\; {\text{R}}_{ {\ s}}=\;{10-8}/{ {\text{I}}}$ $\; {\text{R}}_{ {\ s}}=\;\frac{2}{0.2}=10 Ω$

Q21JEE Main 2023NAT4MWave Optics

Unpolarised light of intensity $32 {\text{Wm}}^{-2}$ passes through the combination of three polaroids such that the pass axis of the last polaroid is perpendicular to that of the pass axis of first polaroid. If intensity of emerging light is 3 ${\text{Wm}}^{-2}$ , then the angle between pass axis of first two polaroids is ________. [10-Apr-2023 shift 1]

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📖 Explanation

${\text{I}}_{\;\text{net }\;}=3=\;\frac{32}{8}( sin \;2 \theta)^2\;=\;{ {\text{I}}_0}/{2} \cos ^2 \theta sin \;^2 \theta$ ${\text{Sin}}(2 \theta)=\;{{\sqrt{3}}}/{2} \Rightarrow 2 \theta=60 & 120^{\circ}\;=\;{ {\text{I}}_0}/{8}( sin \;2 \theta)^2$ $\theta=30^{\circ} & 60^{\circ}\;$

Q22JEE Main 2023NAT4MRotational Motion

If the earth suddenly shrinks to $\;\frac{1}{64}$ th of its original volume with its mass remaining the same, the period of rotation of earth becomes $\;\frac{24}{x} h$ . The value of ${\text{x}}$ is _______. [10-Apr-2023 shift 1]

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📖 Explanation

By AMC $\;\;\frac{2}{5} {\text{MR}}^2 \omega _1^2=\;\frac{2}{5} {\text{M}} (\;{ {\text{R}}}/{4}) ^2 \omega _2$ $\;\;\frac{\omega _1}{\omega _2}=\;\frac{1}{16}=\;\frac{T_2}{T_1}=\;\frac{T_2}{24}$ $\;T_2=\;\frac{24}{16} \;\; \therefore {\text{x}}=16 \;\text{ Ans. }\;$

Q23JEE Main 2023NAT4MElectrostatic Potential and Capacitance

Three concentric spherical metallic shells ${\text{X}}, {\text{Y}}$ and ${\text{Z}}$ of radius ${\text{a}}, {\text{b}}$ and ${\text{c}}$ respectively $[ {\text{a}} < {\text{b}} < {\text{c}}]$ have surface charge densities $\sigma,-\sigma$ and $\sigma$ , respectively. The shells ${\text{X}}$ and ${\text{Z}}$ are at same potential. If the radii of ${\text{X}}$ & ${\text{Y}}$ are $2 {\text{cm}}$ and $3 {\text{cm}}$ , respectively. The radius of shell ${\text{Z}}$ is ________ ${\text{cm}}$ . [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{q}}_{ {\text{x}}}=\sigma 4 \pi {\text{a}}^2$ $\; {\text{q}}_{ {\text{y}}}=-\sigma 4 \pi {\text{b}}^2$ $\; {\text{q}}_{ {\text{z}}}=\sigma 4 \pi {\text{c}}^2$ Potential of $y$ $\;\;{ {\text{q}}_{ {\text{x}}}}/{4 \pi \epsilon_0 {\text{a}}}+\;{ {\text{q}}_{ {\text{y}}}}/{4 \pi \epsilon_0 {\text{b}}}+\;{ {\text{q}}_{ {\text{z}}}}/{4 \pi \epsilon_0 {\text{c}}}=\;{ {\text{q}}_{ {\text{x}}}}/{4 \pi \epsilon_0 {\text{c}}}+\;{ {\text{q}}_{ {\text{y}}}}/{4 \pi \epsilon_0 {\text{c}}}+\;{ {\text{q}}_z}/{4 \pi \epsilon_0 {\text{c}}}$ $\;\;{\sigma 4 \pi {\text{a}}^2}/{ {\text{a}}}-\;{\sigma 4 \pi {\text{b}}^2}/{ {\text{b}}}+\;{\sigma 4 \pi {\text{c}}^2}/{ {\text{c}}}=4 \pi \sigma \;{ ( {\text{a}}^{2-} {\text{b}}^{2+} {\text{c}}^2) }/{ {\text{c}}}$ $\; {\text{c}}( {\text{a}}- {\text{b}}+ {\text{c}})= {\text{a}}^{2-} {\text{b}}^{2+} {\text{c}}^2$ $\; {\text{c}}( {\text{a}}- {\text{b}})+ {\text{c}}^2=( {\text{a}}+ {\text{b}})( {\text{a}}- {\text{b}})$ $\; {\text{c}}( {\text{a}}- {\text{b}})=( {\text{a}}+ {\text{b}})( {\text{a}}- {\text{b}})$ $\; {\text{c}}= {\text{a}}+ {\text{b}}=2+3$ $\; {\text{c}}=5 {\text{cm}} \;\text{ Ans. }\;$

Q24JEE Main 2023NAT4MWaves

A transverse harmonic wave on a string is given by $y(x, t)=5 sin \;(6 {\text{t}}+0.003 {\text{x}})$ where ${\text{x}}$ and ${\text{y}}$ are in ${\text{cm}}$ and ${\text{t}}$ in sec. The wave velocity is _______ ${\text{ms}}^{-1}$ . [10-Apr-2023 shift 1]

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📖 Explanation

$v=\;{ {\text{w}}}/{ {\text{k}}}=\;\frac{6}{.003 \times 10^2}=\;\frac{6}{.3}=\;\frac{60}{3}=20 {\text{m}} / {\ s}$

Q25JEE Main 2023NAT4MCurrent Electricity

10 resistors each of resistance $10 Ω$ can be connected in such as to get maximum and minimum equivalent resistance. The ratio of maximum and minimum equivalent resistance will be _______. [10-Apr-2023 shift 1]

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📖 Explanation

$\;R_{\max } \Rightarrow \;\text{ \in series }\; \Rightarrow 10 R=10 \times 10=100 Ω$ $\;R_{\max } \Rightarrow \;\text{ \in parallel }\;=\;\frac{R}{10}=\;\frac{10}{10}=1 Ω$ $\;\;\frac{R_{\max }}{R_{\min }}=\;\frac{100}{1}=100 \;\text{ Ans. }\;$ $\;R_{\min } \Rightarrow \;\frac{100}{1}=100 \;\text{ Ans. }\;$

Q26JEE Main 2023NAT4MAtoms and Nuclei

The decay constant for a radioactive nuclide is $1.5 \times 10^{-5} {\ s}^{-1}$ . Atomic weight of the substance is $60 {\text{g}} {\text{mole}}^{-1}$ , $( {\text{N}}_{ {\text{A}}}=6 \times 10^{23})$ . The activity of $1.0 \mu {\text{g}}$ of the substance is _______ $\times 10^{10} {\text{Bq}}$ . [10-Apr-2023 shift 1]

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📖 Explanation

No. of moles $=\;\frac{1 \times 10^{-6}}{60}=\;\frac{10^{-7}}{6}$ No. of atom $=n(N A)=\;\frac{10^{-7}}{6} \times 6 \times 10^{23}=10^{16}$ at $( {\text{t}}=0) {\text{A}}_0= {\text{No}} \lambda =10^{16} \times 1.5 \times 10^{-5}=15 \times 10^{10} {\text{Bq}}$

Q27JEE Main 2023NAT4MProperties of Matter

Two wires each of radius $0.2 {\text{cm}}$ and negligible mass, one made of steel and the other made of brass are loaded as shown in the figure. The elongation of the steel wire is ______ $\times 10^{-6} {\text{m}}$ . [Young's modulus for steel $=2 \times$ $10^{11} {\text{Nm}}^{-2}$ and ${\text{g}}=10 {\text{ms}}^{-2}$ [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{T}}_2= {\text{T}}_1+20=20+11.4$ $\; {\text{T}}_2=31.4$ $\therefore$ Elongation in steel wire $\;\Delta {\text{L}}=\;{ {\text{T}}_2 {\text{L}}}/{ {\text{AY}}}$ $\;=\;\frac{31.4 \times 1.6}{\pi (0.2 \times 10^{-2}) ^2 \times 2 \times 10^{11}}$ $\;=2 \times 10^{-5}$ $\;\Delta {\text{L}}=20 \times 10^{-6} {\text{m}}$

Q28JEE Main 2023NAT4MWork, Power and Energy

A closed circular tube of average radius $15 {\text{cm}}$ , whose inner walls are rough, is kept in vertical plane. A block of mass $1 {\text{kg}}$ just fit inside the tube. The speed of block is $22 {\text{m}} / {\ s}$ , when it is introduced at the top of tube. After completing five oscillations, the block stops at the bottom region of tube. The work done by the tube on the block is _______ J. (Given : ${\text{g}}=10 {\text{m}} / {\ s}^2$ ) [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{R}}_{\;\text{arg }\;}=15 {\text{cm}}=.15 {\text{m}}$ $\;\;\text{ By WET }\;$ $\; {\text{W}}_{ {\text{f}}}+ {\text{W}}_{\;\text{gravity }\;}=\Delta {\text{K}}= {\text{Kf}}- {\text{Ki}}$ $\; {\text{W}}_{ {\text{f}}}+10 \times .3=0-\;\frac{1}{2} \times 1 \times (22)^2$ $\; {\text{W}}_{ {\text{f}}}=-3-\;\frac{484}{2}=3-242=-245$ Work by friction $=-245$ By NTA $(+245)$

Q29JEE Main 2023NAT4MElectromagnetic Induction

A $1 {\text{m}}$ long metal rod ${\text{XY}}$ completes the circuit as shown in figure. The plane of the circuit is perpendicular to the magnetic field of flux density $0.15 {\text{T}}$ . If the resistance of the circuit is $5 Ω$ , the force needed to move the rod in direction, as indicated, with a constant speed of $4 {\text{m}} / {\ s}$ will be _______ $10^{-3} {\text{N}}$ . [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{F}}= {\text{I}} ℓ {\text{B}}= (\;{ {\text{e}}}/{ {\text{R}}}) ℓ {\text{B}}=\;{( {\text{Bv}} ℓ) {\text{B}} ℓ}/{ {\text{R}}}=\;{ {\text{B}}^2 ℓ^2 v}/{ {\text{R}}}$ $\;=\;\frac{(\cdot 15)^2 \times (1)^2 \times 4}{5}=180 \times 10^{-4}$ $\;=18 \times 10^{-3}=18 \;\text{ Ans. }\;$

Q30JEE Main 2023NAT4MMagnetic Effects of Current and Magnetism

The current required to be passed through a solenoid of $15 {\text{cm}}$ length and 60 turns in order to demagnetize a bar magnet of magnetic intensity $2.4 \times 10^3 {\text{Am}}^{-1}$ is _______ A. [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{H}}=2.4 \times 10^3 {\text{A}} / {\text{m}}$ $\; {\text{H}}= {\text{nI}}=\;{ {\text{N}}}/{ℓ} {\text{I}}$ $\; {\text{I}}=\;{ {\text{H}} ℓ}/{ {\text{N}}}=\;\frac{2.4 \times 10^3 \times 15 \times 10^{-2}}{60}$ $\; {\text{I}}=6 {\text{A}}$

Chemistry30 questions

Q31JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

The major product ' ${\text{P}}$ ' formed in the given reaction is [10-Apr-2023 shift 1]

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Q32JEE Main 2023MCQ4MSalt Analysis

Prolonged heating is avoided during the preparation of ferrous ammonium sulphate to [10-Apr-2023 shift 1]

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Q33JEE Main 2023MCQ4MHaloalkanes and Haloarenes

Choose the correct answer from the options given below: [10-Apr-2023 shift 1]

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📖 Explanation

(A) ${\text{S}}_{ {\text{N}}} 2 \to$ for ${\text{S}}_{ {\text{N}}} 2$ Reaction $1^{\circ} > 2^{\circ} > 3^{\circ}$ (B) ${\text{S}}_{ {\text{N}}} 1 \to$ reactivity $\times$ Stability of Carbocation formed(C) Electrophilic Substitution reaction ${\text{rate}} \times \;\frac{1}{E W G}$ (D) Nucleophilic substitution :- rate $\times$ no. of EWG atloched at benzons

Q34JEE Main 2023MCQ4MChemical Thermodynamics

Given (A) $2 {\text{CO}}( {\text{g}})+ {\text{O}}_2( {\text{g}}) \to 2 {\text{CO}}_2( {\text{g}}) \Delta {\text{H}}_1^{ {\text{o}}}=- {\text{x}} {\text{kJ}} {\text{mol}}^{-1}$ (B) ${\text{C}}$ (graphite) $+ {\text{O}}_2( {\text{g}}) \to {\text{CO}}_2( {\text{g}}) \;\; \Delta {\text{H}}_2^{\circ}=- {\text{y}} {\text{KJ}} {\text{mol}}{ }^{-1}$ The $\Delta {\text{H}}^{\circ}$ for the reaction ${\text{C}} (graphite )+\;\frac{1}{2} {\text{O}}_2( {\text{g}}) \to {\text{CO}}( {\text{g}})$ is [10-Apr-2023 shift 1]

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📖 Explanation

$\;2 {\text{CO}}( {\text{g}})+ {\text{O}}_2( {\text{g}}) \to 2 {\text{CO}}_2( {\text{g}}) \;\; \Delta {\text{H}}_1^{\circ}= {\text{y}} {\text{kJ}} / {\text{mol}}$ ...........(2) $\; {\text{C}}(\;\text{ graphite }\;)+ {\text{O}}_2( {\text{g}}) \to {\text{CO}}_2( {\text{g}}) \;\; \Delta {\text{H}}_2^{\circ}=- {\text{y}} \;\; {\text{kJ}} / {\text{mol}} . . . . . . . . . \;\text{ (2) }\;$ _________________________________________________ $\; {\text{C}}(\;\text{ graphite }\;)+{ }^{1 / 2} {\text{O}}_2( {\text{g}}) \to {\text{CO}}( {\text{g}}) \;\; \Delta {\text{H}}_3^{\circ}=?$ $\;\Delta {\text{H}}_3^{\circ}= {\text{H}}_2^{\circ}-\;{ {\text{H}}_1^{\circ}}/{2}=- {\text{y}}-\;{- {\text{x}}}/{2}$ $\;\Delta {\text{H}}_3^{\circ}=\;{ {\text{x}}}/{2}- {\text{y}}=\;{ {\text{x}}-2 {\text{y}}}/{2}$

Q35JEE Main 2023MCQ4MPractical Organic Chemistry

Using column chromatography mixture of two compounds ' ${\text{A}}$ ' and ' ${\text{B}}$ ' was separated. 'A' eluted first, this indicates ' ${\text{B}}$ ' has [10-Apr-2023 shift 1]

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📖 Explanation

More Polar the compound, the more it will adhere to the adsorbent and the smaller the distance it will travel from baseline, and Lower its $R_f$ value. $B$ has Low $R_f$ value and strong Adsoption ${\text{R}}_{ {\text{f}}}=\;\frac{\;\text{ distance covered by substance from base line }\;}{\;\text{ total distance covered by solvent form base line }\;}$

Q36JEE Main 2023MCQ4Mp Block Elements

Lime reacts exothermally with water to give ' ${\text{A}}$ ' which has low solubility in water. Aqueous solution of ' ${\text{A}}$ ' is often used for the test of ${\text{CO}}_2$ . a test in which insoluble ${\text{B}}$ is formed. If ${\text{B}}$ is further reacted with ${\text{CO}}_2$ then soluble compound is formed. ' ${\text{A}}$ ' is [10-Apr-2023 shift 1]

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📖 Explanation

${\text{CaO}}+ {\text{H}}_2 {\text{O}} \to {\text{Ca}}( {\text{OH}})_2$ ${\text{A}}(\;\text{ less soluble }\;)$ ${\text{Ca}}( {\text{OH}})_2+ {\text{CO}}_2 \to {\text{CaCO}}_3+ {\text{H}}_2 {\text{O}}$ ${\text{B}}(\;\text{ insoluble }\;)$ $( {\text{CaCO}}_3)_{\text{B}}+ {\text{H}}_2 {\text{O}}+ {\text{CO}}_2 \to ({\text{Ca}} ( {\text{HCO}}_3) _2)_{\text{Soluble}}$

Q37JEE Main 2023MCQ4MSome Basic Concepts of Chemistry

Match list I with list II List I Industry List II Waste Generated (A) Steel plants (I) Gypsum (B) Thermal power plants (II) Fly ash (C) Fertilizer industries (III) Slag (D) Paper mills (IV) Bio-degradable wastes Choose the correct answer from the options given below [10-Apr-2023 shift 1]

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Q38JEE Main 2023MCQ4MAlcohols, Phenols and Ethers

Suitable reaction condition for preparation of Methyl phenyl ether is [10-Apr-2023 shift 1]

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Q39JEE Main 2023MCQ4MBiomolecules

The one that does not stabilize $2^{\circ}$ and $3^{\circ}$ structures of proteins is [10-Apr-2023 shift 1]

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📖 Explanation

FactThe main forces which stabilize the secondary and tertiary structure of proteins are $\to$ Hydrogen bonds $\to {\text{S}}- {\text{S}}$ Linkages $\to$ vanderwaals force $\to$ electrostatic force of attraction

Q40JEE Main 2023MCQ4MSome Basic Concepts of Chemistry

The compound which does not exist is [10-Apr-2023 shift 1]

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Q41JEE Main 2023MCQ4MSome Basic Concepts of Chemistry

Given below are two reactions, involved in the commercial production of dihydrogen $( {\text{H}}_2)$ . The two reactions are carried out at temperature " ${\text{T}}_1$ " and " ${\text{T}}_2$ ", respectively $\; {\text{C}}( {\ s})+ {\text{H}}_2 {\text{O}}( {\text{g}}) {\to }^{ {\text{T}}_1} {\text{CO}}( {\text{g}})+ {\text{H}}_2( {\text{g}})$ $\; {\text{CO}}( {\text{g}})+ {\text{H}}_2 {\text{O}}( {\text{g}}) \;{\to }^{ {\text{T}}_2} _{\;\text{ catalyst }\;} {\text{CO}}_2( {\text{g}})+ {\text{H}}_2( {\text{g}})$ The temperatures $T_1$ and $T_2$ are correctly related as [10-Apr-2023 shift 1]

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📖 Explanation

${\text{T}}_1=1270 {\text{K}} {\text{T}}_2=673 {\text{K}}$ ${\text{T}}_1 > {\text{T}}_2$ on the basis of data

Q42JEE Main 2023MCQ4MSolutions

The enthalpy change for the adsorption process and micelle formation respectively are [10-Apr-2023 shift 1]

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📖 Explanation

Adsorption $\to$ Exothermic $(\Delta {\text{H}}_{ {\text{ads}}}=- {\text{ve}})$ Micelle formation $\to$ Endothermic $(\Delta {\text{H}}_{\;\text{mic }\;}=+ {\text{ve}})$ $\Delta {\text{H}}_{\;\text{ads }\;} < {\text{O}} \;\text{ and }\; \Delta {\text{H}}_{\;\text{mic }\;} > {\text{O}}$

Q43JEE Main 2023MCQ4MChemical Bonding and Molecular Structure

The pair from the following pairs having both compounds with net non-zero dipole moment is [10-Apr-2023 shift 1]

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Q44JEE Main 2023MCQ4Md and f Block Elements

Which of the following is used as a stabilizer during the concentration of sulphide ores? [10-Apr-2023 shift 1]

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Q45JEE Main 2023MCQ4Md and f Block Elements

Which of the following statements are correct?(A) The ${\text{M}}^{3+} / {\text{M}}^{2+}$ reduction potential for iron is greater than manganese(B) The higher oxidation states of first row d-block elements get stabilized by oxide ion.(C) Aqueous solution of ${\text{Cr}}^{2+}$ can liberate hydrogen from dilute acid.(D) Magnetic moment of ${\text{V}}^{2+}$ is observed between 4.4-5.2 BM.Choose the correct answer from the options given below: [10-Apr-2023 shift 1]

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📖 Explanation

(A) The ${\text{M}}^{3+} / {\text{M}}^{2+}$ reduction potential for manganese is greater than iron(B) $\; {\text{E}}_{ {\text{Fe}}^{+3} / {\text{Fe}}^{+2}}^0=+0.77$ $\; {\text{E}}_{ {\text{Mn}}^{+3} / {\text{Mn}}^{+2}}^0=+1.57$ (C) ${\text{E}}_{ {\text{Cr}}^{+3} / {\text{Cr}}^{+2}}^0=-0.26$ $\therefore \;\; {\text{Cr}}^{2 ⊕}+ {\text{H}}^{⊕} \to {\text{Cr}}^{3 ⊕}+\;\frac{1}{2} {\text{H}}_2$ (D) ${\text{V}}^{2 ⊕}=3$ unpaired electron Magnetic Moment $=3.87$ B.M

Q46JEE Main 2023MCQ4MRedox Reactions

Given below are two statements :Statement I : Aqueous solution of ${\text{K}}_2 {\text{Cr}}_2 {\text{O}}_7$ is preferred as a primary standard in volumetric analysis over ${\text{Na}}_2 {\text{Cr}}_2 {\text{O}}_7$ aqueous solution.Statement II : ${\text{K}}_2 {\text{Cr}}_2 {\text{O}}_7$ has a higher solubility in water than ${\text{Na}}_2 {\text{Cr}}_2 {\text{O}}_7$ In the light of the above statements, choose the correct answer from the options given below: [10-Apr-2023 shift 1]

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📖 Explanation

(1) ${\text{K}}_2 {\text{Cr}}_2 {\text{O}}_7$ is used as primary standard. The concentration ${\text{Na}}_2 {\text{Cr}}_2 {\text{O}}_7$ changes in aq. solution.(2) It is less soluble than ${\text{Na}}_2 {\text{Cr}}_2 {\text{O}}_7$

Q47JEE Main 2023MCQ4MCoordination Compounds

The octahedral diamagnetic low spin complex among the following is [10-Apr-2023 shift 1]

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📖 Explanation

(1) Paramagnetic, High Spin & Tetrahedral(2) Paramagnetic, High Spin & Octahedral(3) Paramagnetic, High Spin & Octahedral(4) Diamagnetic, Low Spin & Octahedral $\; [ {\text{Co}} ( {\text{NH}}_3) _6] ^{3+}, {\text{CN}}=6 {\text{CN}}=6 \;\text{ (Octahedral }\;)$ $\; {\text{NH}}_3= {\text{SFL}}$ $\; {\text{Co}}^{+3}=[ {\text{Ar}}] 3 {\text{d}}^6$ Diamagnetic & Low spin complex

Q48JEE Main 2023MCQ4MAmines

Isomeric amines with molecular formula ${\text{C}}_8 {\text{H}}_{11} {\text{N}}$ given the following tests Isomer $( {\text{P}}) \Rightarrow$ Can be prepared by Gabriel phthalimide synthesis Isomer $(Q) \Rightarrow$ Reacts with Hinsberg's reagent to give solid insoluble in ${\text{NaOH}}$ Isomer $( {\text{R}}) \Rightarrow$ Reacts with ${\text{HONO}}$ followed by $\beta$ -naphthol in ${\text{NaOH}}$ to given red dye. Isomer $( {\text{P}}),( {\text{Q}})$ and $( {\text{R}})$ respectively are [10-Apr-2023 shift 1]

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📖 Explanation

${\text{P}}=$ Can be prepased by Gabriel phthalimide synthesis it should be i-amine ${\text{Q}}=$ React with Hinsberg's reagent and insoluble in ${\text{NaOH}}$ it should be $2^{\circ}$ -amine ${\text{R}}$ = React with ${\text{HNO}}_2$ followed by B-Napthol in ${\text{NaOH}}$ it give red dye it must be Aromatic Amine

Q49JEE Main 2023MCQ4MSome Basic Concepts of Chemistry

The number of molecules and moles in 2.8375 litres of ${\text{O}}_2$ at STP are respectively [10-Apr-2023 shift 1]

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📖 Explanation

Moles of ${\text{O}}_2 ( {\text{n}}_{ {\text{O}}_2}) =\;{\;\text{ Volume of }\; {\text{O}}_2}/{22.7}=0.125$ molesMolecules of ${\text{O}}_2=$ moles $\times {\text{N}}_{ {\text{A}}}$ $=0.125 \times 6.022 \times 10^{23}$ $=7.527 \times 10^{22}$ moleculesAns (1) $7.527 \times 10^{22}$ and $0.125 {\text{mole}}$

Q50JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

Match list I with List II List I polymer List II Type/Class (A) Nylon-2-Nylon-6 (I) Thermosetting polymer (B) Buna-N (II) Biodegradable polymer (C) Urea-Formaldehyde resin (III) Synthetic rubber (D) Dacron (IV) PolyesterChoose the correct answer from the options given below: [10-Apr-2023 shift 1]

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📖 Explanation

Fact Base(A) Nylon-2-Nylon- $6 \to$ It is $\alpha$ Biodegradable polymer

Q51JEE Main 2023NAT4MSolutions

If the degree of dissociation of aqueous solution of weak monobasic acid is determined to be 0.3 , then the observed freezing point will be ________ $\%$ higher than the expected/theoretical freezing point. (Nearest integer) [10-Apr-2023 shift 1]

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📖 Explanation

For mono basic acid $\to {\text{n}}=2$ $\; {\text{i}}=1+( {\text{n}}-1) \alpha=1+(2-1) 0.3$ $\; {\text{i}}=1.3$ $\;\% \;\text{ increase }\;=\;{ (\Delta {\text{T}}_{ {\text{f}}}) _{ {\text{obs}}}- (\Delta {\text{T}}_{ {\text{f}}}) _{ {\text{cal}}}}/{ (\Delta {\text{T}}_{ {\text{f}}}) _{ {\text{cal}}}} \times 100$ $\;=\;{ {\text{K}}_{ {\text{f}}} \times {\text{i}} \times {\text{m}}- {\text{K}}_{ {\text{f}}} \times {\text{m}}}/{ {\text{K}}_{ {\text{f}}} \times {\text{m}}} \times 100$ $\;=\;{ {\text{i}}-1}/{1} \times 100=30 \%$

Q52JEE Main 2023NAT4Mp Block Elements

In the following reactions, the total number of oxygen atoms in ${\text{X}}$ and ${\text{Y}}$ is _______ $\; {\text{Na}}_2 {\text{O}}+ {\text{H}}_2 {\text{O}} \to 2 {\text{X}}$ $\; {\text{Cl}}_2 {\text{O}}_7+ {\text{H}}_2 {\text{O}} \to 2 {\text{Y}}$ [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{Na}}_2 {\text{O}}+ {\text{H}}_2 {\text{O}} \to 2 {\text{NaOH}}$ $\; {\text{Cl}}_2 {\text{O}}_7+ {\text{H}}_2 {\text{O}} \to 2 {\text{HClO}}_4$ $\;1+4=5$

Q53JEE Main 2023NAT4MChemical Bonding and Molecular Structure

The sum of lone pairs present on the central atom of the interhalogen ${\text{IF}}_5$ and ${\text{IF}}_7$ is _______ [10-Apr-2023 shift 1]

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📖 Explanation

${\text{IF}}_5=1$ lone pair ${\text{IF}}_7=0$ lone pair $1+0=1$

Q54JEE Main 2023NAT4MChemical Bonding and Molecular Structure

The number of bent-shaped molecule/s from the following is _________ ${\text{N}}_3^{-}, {\text{NO}}_2^{-}, {\text{I}}_3^{-}, {\text{O}}_3, {\text{SO}}_2$ [10-Apr-2023 shift 1]

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📖 Explanation

${\text{N}}_3^{-}$ linear ${\text{NO}}_2^{-}$ bent ${\text{I}}_3^{-}$ linear ${\text{O}}_3$ bent ${\text{SO}}_2$ bent

Q55JEE Main 2023MCQ4MChemical Equilibrium

The number of correct statement/s involving equilibria in physical from the following is ________ [10-Apr-2023 shift 1]

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📖 Explanation

(A) is correct(B) for equilibrium $r_{ {\text{f}}}= {\text{r}}_{ {\text{b}}}$ $\Rightarrow (B)$ is correct(C) at equilibrium the value of parameters become constant of a given temperature and not equal $\Rightarrow ( {\text{C}})$ is incorrect(D) for a given solid solute and a liquid solvent solubility depends upon temperature only $\Rightarrow ( {\text{D}})$ is correct

Q56JEE Main 2023NAT4MStates of Matter

At constant temperature, a gas is at pressure of $940.3 {\text{mm}} {\text{Hg}}$ . The pressure at which its volume decreases by $40 \%$ is ________ ${\text{mm}} {\text{Hg}}$ . (Nearest integer) [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{P}}_{\;\text{initial }\;}=940.3 {\text{mm}} {\text{Hg}} \;\; \;\text{ Vinitial }\;=100 \;\text{ (Assume) }\;$ $\; {\text{P}}_{\;\text{final }\;}=?$ $\; {\text{P}}_{ {\text{i}}} {\text{V}}_{ {\text{i}}}= {\text{P}}_{ {\text{f}}} {\text{V}}_{ {\text{f}}}$ $\;940.3 \times 100= {\text{P}}_{ {\text{f}}} \times 60$ $\; {\text{P}}_{ {\text{f}}}=1567.16 {\text{mm}} \;\text{ of }\; {\text{Hg}}$ $\; {\text{P}}_{ {\text{f}}}=1567$

Q57JEE Main 2023NAT4MElectrochemistry

${\text{FeO}}_4^{2-} {\to }^{+2.2 {\text{V}}} {\text{Fe}}^{3+} {\to }^{+0.70 {\text{V}}} {\text{Fe}}^{2+} {\to }^{-0.45 {\text{V}}} {\text{Fe}}^0$ ${\text{E}}_{ {\text{FeO}}_4^{2-} / {\text{Fe}}^{2+}}^{\circ}$ is ${x} \times 10^{-3} {\text{V}}$ . The value of ${x}$ is _______ [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{FeO}}_4^{2-}+3 {\text{e}}^{⊖} \to {\text{Fe}}^{+3} \Delta {\text{G}}_1$ $\;\;{ {\text{Fe}}^{+3}+ {\text{e}}^{⊖} \to {\text{Fe}}^{+2} \Delta {\text{G}}_2}/{ {\text{FeO}}_4^{-2}+4 {\text{e}}^{Θ} \to {\text{Fo}}^{+2} \Delta {\text{G}}_3}$ $\;\Delta {\text{G}}_3=\Delta {\text{G}}_1+\Delta {\text{G}}_2$ $\;(-) 4 {\text{E}}_3^{\circ} {\text{F}}=(-) 3 \times 2.2 \times {\text{F}}+(-) 1 \times 0.7 \times {\text{f}}$ $\;4 {\text{E}}_3^{\circ}=6.6+0.7=7.3$ $\; {\text{E}}_3^{\circ}=\;\frac{7.3}{4}=1.825=1825 \times 10^{-3}$

Q58JEE Main 2023NAT4MChemical Kinetics

A molecule undergoes two independent first order reactions whose respective half lives are $12 {\min}$ and $3 {\min}$ . If both the reactions are occurring then the time taken for the $50 \%$ consumption of the reactant is _______ min. (Nearest integer) [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{k}}_{\;\text{eff }\;}= {\text{k}}_1+ {\text{k}}_2$ $\;\;{ℓ {\text{n}}^2}/{ {\text{t}}_{\;\text{eff }\;}}=\;{ℓ {\text{n}}^2}/{ {\text{t}}_1}+\;{ℓ {\text{n}}^2}/{ {\text{t}}_2}$ $\;\;{1}/{ {\text{t}}_{\;\text{eff }\;}}=\;\frac{1}{12}+\;\frac{1}{3}=\;\frac{1+4}{12}=5 / 12$ $\; {\text{t}}_{\;\text{eff }\;}=\;\frac{12}{5}=2.4=2$

Q59JEE Main 2023NAT4MStructure of Atom

The number of incorrect statement/s about the black body from the following is ________(A) Emit or absorb energy in the form of electromagnetic radiation(B) Frequency distribution of the emitted radiation depends on temperature(C) At a given temperature, intensity vs frequency curve passes through a maximum value(D) The maximum of the intensity vs frequency curve is at a higher frequency at higher temperature compared to that at lower temperature [10-Apr-2023 shift 1]

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Q60JEE Main 2023NAT4MCoordination Compounds

In potassium ferrocyanide, there are _______ pairs of electrons in the $t_{2 g}$ set of orbitals. [10-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{Fe}}^{+2}=[ {\text{Ar}}] 3 {\text{d}}^6$ $\; {\text{CN}}^{-}= {\text{SFL}}$ $t_{2 g}$ contain 6 electron so it become 3 pairs

Mathematics30 questions

Q61JEE Main 2023MCQ4MVector Algebra

An arc $P Q$ of a circle subtends a right angle at its centre $O$ . The mid point of the arc $P Q$ is $R$ . If $\vec{O P}={ {\text{u}}}^{\to }, \vec{O R}={ {\text{v}}}^{\to }$ and $\vec{O Q}=\alpha { {\text{u}}}^{\to }+\beta { {\text{v}}}^{\to }$ , then $\alpha, \beta^2$ are the roots of the equation : [10-Apr-2023 shift 1]

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📖 Explanation

Let $\vec{O P}={ {\text{u}}}^{\to }={ {\text{i}}^{\^}}$ ${ {\text{OQ}}}^{\to }={ {\text{q}}}^{\to }={ {\text{j}}^{\^}}$ $\because {\text{R}}$ is the mid point of ${ {\text{PQ}}}^{\to }$ Then $\vec{O R}=\vec{v}=\;{1}/{{\sqrt{2}}} \hat{i}+\;{1}/{{\sqrt{2}}} \hat{j}$ Now $\;{ {\text{OQ}}}^{\to }=\alpha { {\text{u}}}^{\to }+\beta { {\text{v}}}^{\to }$ $\;{ {\text{j}}^{\^}}=\alpha { {\text{i}}^{\^}}+\beta (\;{1}/{{\sqrt{2}}} { {\text{i}}^{\^}}+\;{1}/{{\sqrt{2}}} { {\text{j}}^{\^}})$ $\;\beta={\sqrt{2}}, \alpha+\;{\beta}/{{\sqrt{2}}}=0 \Rightarrow \alpha=-1$ Now equation $\;x^{2-} (\alpha+\beta^2) x+\alpha \beta^2=0$ $\;x^{2-}(-1+2) x+(-1)(2)=0$ $\;x^{2-}x-2=0$

Q62JEE Main 2023MCQ4MApplication of Derivatives

A square piece of tin of side $30 {\text{cm}}$ is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in ${\text{cm}}^2)$ is equal to : [10-Apr-2023 shift 1]

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📖 Explanation

Let the side of the square to be cut off be ${\text{x}} {\text{cm}}$ .Then, the length and breadth of the box will be $(30-2 {\text{x}}) {\text{cm}}$ each and the height of the box is ${\text{x}} {\text{cm}}$ therefore,The volume $V(x)$ of the box is given by $\;V(x)=x(30-2 x)^2$ $\;\;\frac{d v}{d x}=(30-2 x)^{2+}2 x \times (30-2 x)(-2)$ $\;0=(30-2 x)^{2-}4 x(30-2 x)$ $\;0=(30-2 x)[(30-2 x)-4 x]$ $\;0=(30-2 x)(30-6 x)$ $\;x=15,5$ $\;x \ne 15 \;\text{ (Not possible) }\;$ $\;\{ \therefore V=0\} \;\;$ (Not possible)Surface area without top of the box $=ℓ {\text{b}}+2( {\text{bh}}+ {\text{h}} ℓ)$ $\;=(30-2 x)(30-2 x)+2[(30-2 x) x+(30-2 x) x]$ $\;= [(30-2 x)^{2+}4\{(30-2 x) x\}.$ $\;= [(30-10)^{2+}4(5)(30-10)]$ $\;=400+400$ $\;=800 {\text{cm}}^2$

Q63JEE Main 2023MCQ4MVector Algebra

Let $O$ be the origin and the position vector of the point $P$ be $-\hat{i}-2 \hat{j}+3 \hat{k}$ . If the position vectors of the $A, B$ and $C$ are $-2 \hat{i}+\hat{j}-3 \hat{k}, 2 \hat{i}+4 \hat{j}-2 \hat{k}$ and $-4 \hat{i}+2 \hat{j}-\hat{k}$ respectively, then the projection of the vector $\vec{O P}$ on a vector perpendicular to the vectors ${ {\text{AB}}}^{\to }$ and ${ {\text{AC}}}^{\to }$ is : [10-Apr-2023 shift 1]

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📖 Explanation

Position vector of the point $P(-1,-2,3), A(-2,1,-3) B(2,4,-2)$ , and $C(-4,2,-1)$ Then $\vec{O P} \cdot \;\frac{\vec{A B} \times \vec{A C}}{\mid \vec{A B} \times \vec{A C}) \mid }$ $\;\vec{A B} \times \vec{A C}={\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 1 \\ -2 & 1 & 2\end{vmatrix}}$ $\;=\hat{i}(5)-\hat{j}(8+2)+\hat{k}(4+6)$ $\;=5 \hat{i}-10 \hat{j}+10 \hat{k}$ Now $\;\vec{O P} \cdot \;\frac{\vec{A B} \times \vec{A C}}{\mid \vec{A B} \times \vec{A C}) \mid }=(-{ {\text{i}}^{\^}}-2 \hat{j}+3 \hat{k}) \cdot \;{({ {\text{i}}^{\^}}-10 \hat{j}+10 \hat{k})}/{{\sqrt{(5)^{2+}(-10)^{2+}(10)^2}}}$ $\;=\;{-5+20+30}/{{\sqrt{25+100+100}}}$ $\;=\;{45}/{{\sqrt{225}}}=\;\frac{45}{15}=3$

Q64JEE Main 2023MCQ4MMatrices and Determinants

If $A$ is a $3 \times 3$ matrix and $|A|=2$ , then ${| 3 {\text{adj}} (|3 {\text{A}}| {\text{A}}^2) |}$ is equal to : [10-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Given }\;| {\text{A}}|=2$ $\;\;\text{ Now, }\;{| 3 {\text{adj}} (|3 {\text{A}}| {\text{A}}^2) |}$ $\;|3 {\text{A}}|=3^3 \cdot | {\text{A}}|$ $\;=3^3 \;\text{. }\;(2)$ $\;\;\text{ Adj. }\; (|3 A| A^2) ={\text{adj}} \{ (3^3 \cdot 2) A^2\}$ $\;= (2.3^3) ^2({\text{adj}} A)^2$ $\;=2^2 \cdot 3^6 \cdot ({\text{adjA}})^2$ $\;{| 3 {\text{adj}} (|3 A| A^2) |}={| 2^2 \cdot 3.3^6({\text{adj}} A)^2 |}$ $\;= (2^2 .3^7) ^3|{\text{adjA}}|^2$ $\;=2^6 \cdot 3^{21} (| {\text{A}}|^2) ^2$ $\;=2^6 \cdot 3^{21} (2^2) ^2$ $\;=2^{10} \cdot 3^{21}$ $\;=2^{10} \cdot 3^{10} \cdot 3^{11}$ $\;{| 3 {\text{adj}} (|3 {\text{A}}| {\text{A}}^2) |}=6^{10} \cdot 3^{11}$ $\;$

Q65JEE Main 2023MCQ4MThree Dimensional Geometry

Let two vertices of a triangle ${\text{ABC}}$ be $(2,4,6)$ and $(0,-2,-5)$ , and its centroid be $(2,1,-1)$ . If the image of the third vertex in the plane $x+2 y+4 z=11$ is $(\alpha, \beta, \gamma )$ , then $\alpha \beta+\beta \gamma +\gamma \alpha$ is equal to : [10-Apr-2023 shift 1]

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📖 Explanation

Given Two vertices of Triangle $A(2,4,6)$ and $B(0,-2,-5)$ and if centroid ${\text{G}}(2,1,-1)$ Let Third vertices be $( {\text{x}}, {\text{y}}, {\text{z}})$ Now $\;\frac{2+0+x}{3}=2, \;\frac{4-2+y}{3}=1, \;\frac{6-5+z}{3}=-1$ ${\text{x}}=4, {\text{y}}=1, \;\; {\text{z}}=-1$ Third vertices $C(4,1,-4)$ Now, Image of vertices $C(4,1,-4)$ in the given plane is $D(\alpha, \beta, \gamma )$ Now $\;\;\frac{\alpha-4}{1}=\;\frac{\beta-1}{2}=\;\frac{\gamma +4}{4}=-2 \;\frac{(4+2-16-11)}{1+4+16}$ $\;\;\frac{\alpha-4}{1}=\;\frac{\beta-1}{2}=\;\frac{\gamma +4}{4}=\;\frac{42}{21} \Rightarrow 2$ $\;\alpha=6, \beta=5, \gamma =4$ $\;\;\text{ Then }\; \alpha \beta+\beta \gamma +\gamma \alpha$ $\;=(6 \times 5)+(5 \times 4)+(4 \times 6)$ $\;=30+20+24$ $\;=74$

Q66JEE Main 2023MCQ4MSets and Relations

The negation of the statement : $( {\text{p}} ∨ {\text{q}}) ∧( {\text{q}} ∨(∼ {\text{r}}))$ is [10-Apr-2023 shift 1]

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📖 Explanation

$\;( {\text{p}} ∨ {\text{q}}) ∧( {\text{q}} ∨(∼ {\text{r}}))$ $\;∼[( {\text{p}} ∨ {\text{q}}) ∧( {\text{q}} ∨(∼ {\text{r}}))]$ $\;=∼( {\text{p}} ∨ {\text{q}}) ∧(∼ {\text{q}} ∧ {\text{r}})$ $\;=(∼ {\text{p}} ∧ ∼ {\text{q}}) ∨(∼ {\text{q}} ∧ {\text{r}})$ $\;=(∼ {\text{p}} ∨ {\text{r}}) ∧(∼ {\text{q}})$

Q67JEE Main 2023MCQ4MThree Dimensional Geometry

The shortest distance between the lines $\;\frac{x+2}{1}=\;\frac{y}{-2}=\;\frac{z-5}{2}$ and $\;\frac{x-4}{1}=\;\frac{y-1}{2}=\;\frac{z+3}{0}$ is : [10-Apr-2023 shift 1]

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📖 Explanation

$\;\frac{x+2}{1}=\;\frac{y}{-2}=\;\frac{z-5}{2}$ and $\;\frac{x-4}{1}=\;\frac{y-1}{2}=\;\frac{z+3}{0}$ $\;=\;{|-54|}/{|-4 { {\text{i}}^{\^}}+2 { {\text{j}}^{\^}}+4 {\text{k}}|}$ $\;=\;{54}/\sqrt{16+4+16}$ $\;=\;\frac{54}{6}$ $\;=9$

Q68JEE Main 2023MCQ4MBinomial Theorem

If the coefficient of $x^7$ in $(a x-\;\frac{1}{b x^2}) ^{13}$ and the coefficient of $x^{-5}$ in $(a x+\;\frac{1}{b x^2}) ^{13}$ are equal, then $a^4 b^4$ is equal to : [10-Apr-2023 shift 1]

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📖 Explanation

$\; (a x-\;{1}/{ {\text{bx}}^2}) ^{13}$ $\;\;\text{ We have, }\;$ $\; {\text{T}}_{ {\text{r}}+1}={ }^{ {\text{n}}} {\text{C}}_{ {\text{r}}}( {\text{p}})^{ {\text{n}}- {\text{r}}}( {\text{q}})^{ {\text{r}}}$ $\; {\text{T}}_{ {\text{r}}+1}={ }^{13} {\text{C}}_{ {\text{r}}}( {\text{ax}})^{13- {\text{r}}} (-\;{1}/{ {\text{bx}}{ }^2}) ^{ {\text{r}}}$ $\;={ }^{13} {\text{C}}_{ {\text{r}}}( {\text{a}})^{13- {\text{r}}} (-\;{1}/{ {\text{b}}}) ^{ {\text{r}}}( {\text{x}})^{13- {\text{r}}} \cdot ( {\text{x}})^{-2 {\text{r}}}$ $\;={ }^{13} {\text{C}}_{ {\text{r}}}( {\text{a}})^{13- {\text{r}}} (-\;{1}/{ {\text{b}}}) ^{ {\text{r}}}( {\text{x}})^{13-3 {\text{r}}}$ . . . (1) $\;\;\text{ Coefficient of }\; x^7$ $\;\Rightarrow 13-3 r=7$ $\;r=2$ $\;r \;\text{ \in equation (1) }\;$ $\; {\text{T}}_3={ }^{13} {\text{C}}_2( {\text{a}})^{13-2} (-\;{1}/{ {\text{b}}}) ^2( {\text{x}})^{13-6}$ $\;={ }^{13} {\text{C}}_2( {\text{a}})^{11} (\;{1}/{ {\text{b}}}) ^2( {\text{x}})^7$ Coefficient of $x^7$ is ${ }^{13} {\text{C}}_2 \;\frac{(a)^{11}}{b^2}$ Now, $(a x+\;\frac{1}{b x^2}) ^{13}$ $\; {\text{T}}_{ {\text{r}}+1}={ }^{13} {\text{C}}_{ {\text{r}}}( {\text{ax}})^{13- {\text{r}}} (\;{1}/{ {\text{bx}}^2}) ^{ {\text{r}}}$ $\;={ }^{13} {\text{C}}_{ {\text{r}}}( {\text{a}})^{13-r} (\;{1}/{ {\text{b}}}) ^{ {\text{r}}}( {\text{x}})^{13- {\text{r}}}( {\text{x}})^{-2 {\text{r}}}$ $\;={ }^{13} {\text{C}}_{ {\text{r}}}( {\text{a}})^{13- {\text{r}}} (\;{1}/{ {\text{b}}}) ^{ {\text{r}}}( {\text{x}})^{13-3 r}$ . . . (2) Coefficient of ${\text{x}}^{-5}$ $\;\Rightarrow 13-3 r=-5$ $\; {\text{r}}=6$ $r$ in equation $\; {\text{T}}_7={ }^{13} {\text{C}}_6( {\text{a}})^{13-6} (\;{1}/{ {\text{b}}}) ^6( {\text{x}})^{13-18}$ $\; {\text{T}}_7={ }^{13} {\text{C}}_6( {\text{a}})^7 (\;{1}/{ {\text{b}}}) ^6( {\text{x}})^{-5}$ Coefficient of ${\text{x}}^{-5}$ is ${ }^{13} {\text{C}}_6( {\text{a}})^7 (\;{1}/{ {\text{b}}}) ^6$ ATQCoefficient of $x^7=$ coefficient of ${\text{x}}^{-5}$ ${\text{T}}_3= {\text{T}}_7$ $\;{ }^{13} {\text{C}}_2 (\;{ {\text{a}}^{11}}/{ {\text{b}}^2}) ={ }^{13} {\text{C}}_6( {\text{a}})^7 (\;{1}/{ {\text{b}}}) ^6$ $\; {\text{a}}^4 \cdot {\text{b}}^4=\;{{ }^{13} {\text{C}}_6}/{{ }^{13} {\text{C}}_2}$ $\;=\;\frac{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 1}{13 \times 12 \times 6 \times 5 \times 4 \times 3}=22$

Q69JEE Main 2023MCQ4MCircles

A line segment ${\text{AB}}$ of length $\lambda$ moves such that the points ${\text{A}}$ and ${\text{B}}$ remain on the periphery of a circle of radius $\lambda$ . Then the locus of the point, that divides the line segment $A B$ in the ratio $2: 3$ , is a circle of radius : [10-Apr-2023 shift 1]

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📖 Explanation

Since ${\text{OAB}}$ form equilateral $\Delta$ $\; \therefore ∠ {\text{OAP}}=60^{\circ}$ $\; {\text{AP}}=\;\frac{2 \lambda }{5}$ $\;\cos 60^{\circ}=\;{ {\text{OA}}^{2+} {\text{AP}}^{2-} {\text{OP}}^2}/{2 {\text{OA}} \cdot {\text{AP}}}$ $\;\Rightarrow \;\frac{1}{2}=\;{\lambda ^{2+}\frac{4 \lambda ^2}{25}- {\text{OP}}^2}/{2 \lambda (\;\frac{2 \lambda }{5}) }$ $\;\Rightarrow \;\frac{2 \lambda ^2}{5}=\lambda ^{2+}\;\frac{4 \lambda ^2}{25}- {\text{OP}}^2$ $\;\Rightarrow {\text{OP}} 2=\;\frac{19 \lambda ^2}{25}$ $\;\Rightarrow {\text{OP}}=\;{{\sqrt{19}}}/{5} \lambda$ Therefore, locus of point $P$ is $\;{{\sqrt{19}}}/{5} \lambda$

Q70JEE Main 2023MCQ4MMatrices and Determinants

For the system of linear equations $\;2 x-y+3 z=5$ $\;3 x+2 y-z=7$ $\;4 x+5 y+\alpha z=\beta$ which of the following is NOT correct ? [10-Apr-2023 shift 1]

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📖 Explanation

Given $\;2 x-y+3 z=5$ $\;3 x+2 y-z=7$ $\;4 x+5 y+\alpha z=\beta$ $\;\Delta ={\begin{vmatrix}2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \alpha\end{vmatrix}}=7 \alpha+35$ $\;\Delta =7(\alpha+5)$ For unique solution $\Delta \ne 0$ $\alpha \ne -5$ For inconsistent & Infinite solution $\;\Delta =0$ $\;\alpha+5=0 \Rightarrow \alpha=-5$ $\;\Delta _1={\begin{vmatrix}5 & -1 & 3 \\ 7 & 2 & -1 \\ \beta & 5 & -5\end{vmatrix}}=-5(\beta-9)$ $\;\Delta _2={\begin{vmatrix}2 & 5 & 3 \\ 3 & 7 & -1 \\ 4 & \beta & -5\end{vmatrix}}=11(\beta-9)$ $\;\Delta _3={\begin{vmatrix}2 & -1 & 5 \\ 3 & 2 & 7 \\ 4 & 5 & \beta\end{vmatrix}}$ $\;\Delta _3=7(\beta-9)$ For Inconsistent system : -At least one $\Delta _1, \Delta _2 & \Delta _3$ is not zero $\alpha=-5, \beta=8$ option (A) True Infinite solution: $\Delta _1=\Delta _2=\Delta _3=0$ From here $\beta-9=0 \Rightarrow \beta=9$ $\alpha=-5 &$ option (D) True $\beta=9$ Unique solution $\alpha \ne -5, \beta=8 \to \;\text{ option }\;(C) \;\text{ True }\;$ Option (B) FalseFor Infinitely many solution $\alpha$ must be -5 .

Q71JEE Main 2023MCQ4MSequences and Series

Let the first term $a$ and the common ratio ${\text{r}}$ of a geometric progression be positive integers. If the sum of squares of its first three is 33033 , then the sum of these terms is equal to : [10-Apr-2023 shift 1]

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📖 Explanation

Let a, $a r, a r^2$ be three terms of GPGiven : $a^{2+}(a r)^{2+} (a r^2) ^2=33033$ $\; {\text{a}}^2 (1+ {\text{r}}^{2+} {\text{r}}^4) =11^2 .3 .7 .13$ $\;\Rightarrow {\text{a}}=11 \;\text{ and }\; 1+ {\text{r}}^{2+} {\text{r}}^4=3.7 .13$ $\;\Rightarrow {\text{r}}^2 (1+ {\text{r}}^2) =273-1$ $\;\Rightarrow {\text{r}}^2 ( {\text{r}}^{2+}1) =272=16 \times 17$ $\;\Rightarrow {\text{r}}^2=16$ $\; \therefore {\text{r}}=4 \;\; [ \because {\text{r}} > 0]$ Sum of three terms $=a+a r+a r^2=a (1+r+r^2)$ $\;=11(1+4+16)$ $\;=11 \times 21=231$

Q72JEE Main 2023MCQ4MThree Dimensional Geometry

Let $P$ be the point of intersection of the line $\;\frac{x+3}{3}=\;\frac{y+2}{1}=\;\frac{1-z}{2}$ and the plane $x+y+z=2$ . If the distance of the point $P$ from the plane $3 x-4 y+12 z=32$ is $q$ , then $q$ and $2 q$ are the roots of the equation : [10-Apr-2023 shift 1]

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📖 Explanation

$\;\;\frac{x+3}{3}=\;\frac{y+2}{1}=\;\frac{1-z}{2}=\lambda$ $\;x=3 \lambda -3, y=\lambda -2, z=1-2 \lambda$ ${\text{P}}(3 \lambda -3, \lambda -2,1-2 \lambda )$ will satisfy the equation of plane $x+y+z=2$ . $3 \lambda -3+\lambda -2+1-2 \lambda =2$ $2 \lambda -4=2$ $\lambda =3$ ${\text{P}}(6,1,-5)$ Perpendicular distance of $P$ from plane $3 x-4 y+12 z-32=0$ is $\;q={| \;{3(6)-4(1)+12(-5)-32}/{{\sqrt{9+16+144}}} |}$ $\;q=6$ $\;2 q=12$ Sum of roots $=6+12=18$ Product of roots $=6(12)=72$ $\therefore$ Quadratic equation having ${\text{q}}$ and $2 {\text{q}}$ as roots is ${\text{x}}^{2-}18 {\text{x}}+72$ .

Q73JEE Main 2023MCQ4MDifferential Equations

Let ${\text{f}}$ be a differentiable function such that $x^2 f(x)-x=4 int_{0}^{x} t f(t) \text{dt}, f(1)=\;\frac{2}{3}$ .. Then $18 \text{f}(3)$ is equal to : [10-Apr-2023 shift 1]

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📖 Explanation

$x^2 f(x)-x=4 int_{0}^{x} t f(t) \text{dt}$ Differentiate w.r.t. ${\text{x}}$ $\;x^2 f^{'}(x)+2 x f(x)-1=4 x f(x)$ $\;\;\text{ Let }\; y=f(x)$ $\;\Rightarrow x^2 \;\frac{d y}{d x}-2 x y-1=0$ $\;\;\frac{d y}{d x}-\;\frac{2}{x} y=\;\frac{1}{x^2}$ $\;\;\text{ I.F. }\;=e^{\int \;\frac{-2}{x} \text{dx}}=\;\frac{1}{x^2}$ Its solution is $\;\;\frac{y}{x^2}=\int \;\frac{1}{x^4} \text{dx}+C$ $\;\;\frac{y}{x^2}=\;\frac{-1}{3 x^3}+C$ $\; \because f(1)=\;\frac{2}{3} \Rightarrow y(1)=\;\frac{2}{3}$ $\;\Rightarrow \;\frac{2}{3}=-\;\frac{1}{3}+C$ $\;\Rightarrow C=1$ $\; \because y=-\;\frac{1}{3 x}+x^2$ $\;f(x)=-\;\frac{1}{3 x}+x^2$ $\;f(3)=-\;\frac{1}{9}+9=\;\frac{80}{9} \Rightarrow 18 f(3)=160$

Q74JEE Main 2023MCQ4MProbability

Let ${\text{N}}$ denote the sum of the numbers obtained when two dice are rolled. If the probability that $2^{ {\text{N}}} < {\text{N}}$ ! is $\;{ {\text{m}}}/{ {\text{n}}}$ , where ${\text{m}}$ and ${\text{n}}$ are coprime, then $4 {\text{m}}-3 {\text{n}}$ equal to : [10-Apr-2023 shift 1]

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📖 Explanation

$2^{ {\text{N}}} < {\text{N}}$ ! is satisfied for ${\text{N}} \ge 4$ Required probability $P(N \ge 4)=1- {\text{P}}( {\text{N}} < 4)$ ${\text{N}}=1$ (Not possible) ${\text{N}}=2(1,1)$ $\Rightarrow {\text{P}}( {\text{N}}=2)=\;\frac{1}{36}$ ${\text{N}}=3(1,2),(2,1)$ $\Rightarrow {\text{P}}( {\text{N}}=3)=\;\frac{2}{36}$ ${\text{P}}( {\text{N}}< 4)=\;\frac{1}{36}+\;\frac{2}{36}=\;\frac{3}{36}$ $\therefore {\text{P}}( {\text{N}} \ge 4)=1-\;\frac{3}{36}=\;\frac{33}{36}=\;\frac{11}{12}=\;{ {\text{m}}}/{ {\text{n}}}$ $\Rightarrow {\text{m}}=11, {\text{n}}=12$ $\therefore 4 m-3 n=4(11)-3(12)=8$

Q75JEE Main 2023MCQ4MIndefinite Integrals

If $I(x)=\int e^{ sin \;^2 x}(\cos x sin \;2 x- sin \;x) \text{dx}$ and $I(0)=1$ , then $I (\;\frac{\pi}{3})$ is equal to : [10-Apr-2023 shift 1]

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📖 Explanation

$\;I=\int ⏟{e^{ sin \;^2 x}}_{\;\text{II }\;} sin \;2 x ⏟{\cos x \text{dx}}_I-\int e^{ sin \;^2 x} sin \;x \text{dx}$ $\;=\cos x \int e^{ sin \;^2 x} sin \;2 x \text{dx}-\int ((- sin \;x) \int e^{ sin \;^2 x} sin \;2 x \text{dx}) \text{dx}-\int e^{ sin \;^2 x} sin \;x \text{dx}$ $\; sin \;^2 x=t$ $\; sin \;2 x \text{dx}=d t$ $\;=\cos x \int e^t \text{dt}+\int ( sin \;x \int e^t \text{dt}) \text{dx}-\int e^{ sin \;^2 x} sin \;x \text{dx}$ $\;=e^{ sin \;^2 x} \cos x+\int e^{ sin \;^2 x} sin \;x \text{dx}-\int e^{ sin \;^2 x} sin \;x \text{dx}$ $\; {\text{I}}= {\text{e}}^{ sin \;^2 x} \cos x+C$ $\; {\text{I}}(0)=1$ $\;\Rightarrow 1=1+C$ $\;\Rightarrow C=0$ $\; \therefore {\text{I}}= {\text{e}}^{ sin \;^2 x} \cos x$ $\; {\text{I}} (\;\frac{\pi}{3}) = {\text{e}}^{ sin \;^2 \;\frac{\pi}{3}} \cos \;\frac{\pi}{3}$ $\;=\;{ {\text{e}}^{\frac{3}{4}}}/{2}$

Q76JEE Main 2023MCQ4MTrigonometric Ratios and Identities

$96 \cos \;\frac{\pi}{33} \cos \;\frac{2 \pi}{33} \cos \;\frac{4 \pi}{33} \cos \;\frac{8 \pi}{33} \cos \;\frac{16 \pi}{33}$ is equal to : [10-Apr-2023 shift 1]

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📖 Explanation

$\;96 \cos \;\frac{\pi}{33} \cos \;\frac{2 \pi}{33} \cos \;\frac{2^2 \pi}{33} \cos \;\frac{2^3 \pi}{33} \cos \;\frac{2^4 \pi}{33}$ $\; \because \cos A \cos 2 A \cos 2^2 A . . . \cos 2^{ {\text{n}}-1} A=\;{ sin \; (2^{ {\text{n}}} {\text{A}}) }/{2^{ {\text{n}}} sin \; {\text{A}}}$ Here ${\text{A}}=\;\frac{\pi}{33}, {\text{n}}=5$ $=\;\frac{96 sin \; (2^5 \frac{\pi}{33}) }{2^5 sin \; (\;\frac{\pi}{33}) }$ $=\;\frac{96 sin \; (\frac{32 \pi}{33}) }{32 sin \; (\;\frac{\pi}{33}) }$ $=\;\frac{3 sin \; (\pi-\frac{\pi}{33}) }{ sin \; (\;\frac{\pi}{33}) }=3$

Q77JEE Main 2023MCQ4MComplex Numbers

Let the complex number $z=x+$ iy be such that $\;\frac{2 z-3 i}{2 z+i}$ is purely imaginary. If $x+y^2=0$ , then $y^{4+}y^{2-}y$ is equal to : [10-Apr-2023 shift 1]

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📖 Explanation

$\;z=x+i y$ $\;\;\frac{(2 z-3 i)}{2 z+i}=\;\text{ purely imaginary }\;$ $\;\;\text{ Means }\; {\text{Re}} (\;\frac{2 z-3 i}{2 z+i}) =0$ $\;\Rightarrow \;\frac{(2 z-3 i)}{(2 z+i)}=\;\frac{2(x+i y)-3 i}{2(x+i y)+i}$ $\;\; = \;\frac{2 x+2 y i-3 i}{2 x+i 2 y+i}$ $\;\; = \;\frac{2 x+i(2 y-3)}{2 x+i(2 y+1)} \times \;\frac{2 x-i(2 y+1)}{2 x-i(2 y+1)}$ $\;{\text{Re}} [\;\frac{2 z-3 i}{2 z+i}] =\;\frac{4 x^{2+}(2 y-3)(2 y+1)}{4 x^{2+}(2 y+1)^2}=0$ $\;\Rightarrow 4 x^{2+}(2 y-3)(2 y+1)=0$ $\;\Rightarrow 4 x^{2+} \$ [4 y^{2+}2 y-6 y-3]$ =0; \because x+y^2=0 \Rightarrow x=-y^2;\Rightarrow 4 (-y^2) ^{2+}4 y^{2-}4 y-3=0;\Rightarrow 4 y^{4+}4 y^{2-}4 y=3;\Rightarrow y^{4+}y^{2-}y=;\frac{3}{4}$Therefore, correct answer is option (4).

Q78JEE Main 2023MCQ4MFunctions

If $f(x)=\;\frac{ (\tan 1^{\circ}) x+\log _e(123)}{x \log _e(1234)- (\tan 1^{\circ}) }, x > 0$ , then the least value of $f(f(x))+f (f (\;\frac{4}{x}) )$ is: [10-Apr-2023 shift 1]

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📖 Explanation

$f(x)=\;\frac{(\tan 1) x+\log 123}{x \log 1234-\tan 1}$ Let ${\text{A}}=\tan 1, {\text{B}}=\log 123, {\text{C}}=\log 1234$ $\;f(x)=\;\frac{A x+B}{x C-A}$ $\;f(f(x))=\;\frac{A (\frac{A x+B}{x C-A}) +B}{C (\;\frac{A x+B}{C X-A}) -A}$ $\;=\;\frac{A^2 x+A B+x B C-A B}{A C x+B C-A C x+A^2}$ $\; \;\; =\;\frac{x (A^{2+}B C) }{ (A^{2+}B C.}=x$ $\;f(f(x))=x$ $\;f (f (\;\frac{4}{x}) ) =\;\frac{4}{x}$ $\;f(f(x))+f (f (\;\frac{4}{x}) )$ $\;A M \ge G M$ $x+\;\frac{4}{x} \ge 4$

Q79JEE Main 2023MCQ4MDifferential Equations

The slope of tangent at any point $(x, y)$ on a curve $y=y(x)$ is $\;\frac{x^{2+}y^2}{2 x y}, x > 0$ . If $y (2) =0$ , then a value of $y(8)$ is: [10-Apr-2023 shift 1]

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📖 Explanation

$\;\;\frac{d y}{d x}=\;\frac{x^{2+}y^2}{2 x y}$ $\;y=v x$ $\;y(2)=0$ $\;y(8)=?$ $\;\;\frac{d v}{d x}=v+x \;\frac{d v}{d x}$ $\;v+\;\frac{x d v}{d x}=\;\frac{x^{2+}v^2 x^2}{2 v x^2}$ $\;x \cdot \;\frac{d v}{d x}= (\;\frac{v^{2+}1}{2 v}-v)$ $\;\;\frac{2 v d v}{ (1-v^2) }=\;\frac{d x}{x}$ $\;-\ln (1-v^2) =\ln x+C$ $\;\ln x+\ln (1-v^2) =C$ $\;\ln \$ [x (1-;\frac{y^2}{x^2}) ]$ =C;\ln $[ (;\frac{x^{2-}y^2}{x}) ]$ =C;x^{2-}y^2=c x;y(2)=0 ;\text{ at }; x=2, y=0;4=2 C \Rightarrow C=2;x^{2-}y^2=2 x;;\text{ Hence, at }; x=8;64-y^2=16;y={\sqrt{48}}=4 {\sqrt{3}};y(8)=4 {\sqrt{3}}$Option (1)

Q80JEE Main 2023MCQ4MEllipse

Let the ellipse $E: x^{2+}9 y^2=9$ intersect the positive $x$ -and $y$ -axes at the points $A$ and $B$ respectively. Let the major axis of ${\text{E}}$ be a diameter of the circle ${\text{C}}$ . Let the line passing through ${\text{A}}$ and ${\text{B}}$ meet the circle ${\text{C}}$ at the point $P$ . If the area of the triangle with vertices $A, P$ and the origin $O$ is $\;\frac{m}{n}$ , where $m$ and $n$ are coprime, then ${\text{m}}- {\text{n}}$ is equal to : [10-Apr-2023 shift 1]

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📖 Explanation

Equation of line ${\text{AB}}$ or ${\text{AP}}$ is $\;\;\frac{x}{3}+\;\frac{y}{1}=1$ $\;x+3 y=3$ $\;x=(3-3 y)$ Intersection point of line AP & circle is ${\text{P}} ( {\text{x}}_0, {\text{y}}_0)$ $\;x^{2+}y^2=9 \Rightarrow (3-3 y)^{2+}y^2=9$ $\;\Rightarrow 3^2 (1+y^{2-}2 y) +y^2=9$ $\;\Rightarrow 5 y^{2-}9 y=0 \Rightarrow y(5 y-9)=0$ $\;\Rightarrow y=9 / 5$ Hence $x=3(1-y)=3 (1-\;\frac{9}{5}) =3 (\;\frac{-4}{5})$ $\;x=\;\frac{-12}{5}$ $\;P (x_0, y_0) = (\;\frac{-12}{5}, \;\frac{9}{5})$ Area of $\triangle {\text{AOP}}$ is $=\;\frac{1}{2} \times {\text{OA}} \times$ height $\;\text{ Height }\;=9 / 5, \;\; {\text{OA}}=3$ $=\;\frac{1}{2} \times 3 \times \;\frac{9}{5}=\;\frac{27}{10}=\;{ {\text{m}}}/{ {\text{n}}}$ Compare both side ${\text{m}}=27, {\text{n}}=10 \Rightarrow {\text{m}}- {\text{n}}=17$ Therefore, correct answer is option-D

Q81JEE Main 2023NAT4MPermutations and Combinations

Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple in a match, is 840 , then the total numbers of persons, who participated in the tournament, is _______. [10-Apr-2023 shift 1]

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📖 Explanation

Let number of couples $= {\text{n}}$ $\; \therefore { }^{ {\text{n}}} {\text{C}}_2 \times { }^{ {\text{n}}-2} {\text{C}}_2 \times 2=840$ $\;\Rightarrow {\text{n}}( {\text{n}}-1)( {\text{n}}-2)( {\text{n}}-3)=840 \times 2$ $\;=21 \times 40 \times 2$ $\;=7 \times 3 \times 8 \times 5 \times 2$ $\; {\text{n}}( {\text{n}}-1)( {\text{n}}-2)( {\text{n}}-3)=8 \times 7 \times 6 \times 5$ $\; \therefore {\text{n}}=8$ Hence, number of persons $=16$ .

Q82JEE Main 2023NAT4MQuadratic Equation and Inequalities

The number of elements in the set $\{n \in Z:{| n^{2-}10 n+19 |} < 6\}$ is [10-Apr-2023 shift 1]

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📖 Explanation

$-6 < {\text{n}}^{2-}10 {\text{n}}+19 < 6\;$ $\Rightarrow {\text{n}}^{2-}10 {\text{n}}+25 > 0 \;\text{ and }\;\; {\text{n}}^{2-}10 {\text{n}}+13 < 0$ $( {\text{n}}-5)^2 > 0\;5-3 {\sqrt{2}} < {\text{n}} < 5+3 {\sqrt{2}}$ ${\text{N}} \in {\text{Z}}-\{5\}\; {\text{n}}=\{2,3,4,5,6,7,8\}$ $. . .( {\text{i}})\;. . . \;\text{ (ii) }\;$ $\;\text{ From (i) }\; ∩(i i)\;$ ${\text{N}}=\{2,3,4,5,6,8,\}\;$ $\;\text{ Number of values of }\; {\text{n}}=6\;$

Q83JEE Main 2023NAT4MPermutations and Combinations

The number of permutations of the digits $1,2,3, . . . ., 7$ without repetition, which neither contain the string 153 nor the string 2467 , is _______. [10-Apr-2023 shift 1]

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📖 Explanation

Numbers are $1,2,3,4,5,6,7$ Numbers having string $(154)=(154), 2,3,6,7=5$ !Numbers having string $(2467)=(2467), 1,3,5=4$ !Number having string (154) and (2467) $\;=(154),(2467)=2 !$ $\;\;\text{ Now }\; n(154 ∪ 2467)=5 !+4 !-2 !$ $\;=120+24-2=142$ Again total numbers $=7 !=5040$ Now required numbers $= {\text{n}}$ (neither 154 nor 2467) $\;=5040-142$ $\;=4898$

Q84JEE Main 2023NAT4MLimits, Continuity and Differentiability

Let $f:(-2,2) \to ℝ$ be defined by $f(x)={\begin{cases}x[x] & -2 < x < 0 \\ (x-1)[x] & 0 \le x < 2}\end{cases}$ where $[x]$ denotes the greatest integer function. If ${\text{m}}$ and ${\text{n}}$ respectively are the number of points in $(-2,2)$ at which $y=|f(x)|$ is not continuous and not differentiable, then ${\text{m}}+ {\text{n}}$ is equal to _______. [10-Apr-2023 shift 1]

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📖 Explanation

${\text{f}}( {\text{x}})={\begin{cases}-2 {\text{x}} & -2< {\text{x}} < -1 \\ - {\text{x}} & -1 \le {\text{x}} < 0 \\ 0 & 0 \le {\text{x}} < 1 \\ {\text{x}}-1 & 1 \le {\text{x}} < 2 }\end{cases}$ Clearly $f(x)$ is discontinuous at $x=-1$ also non differentiable. $\therefore {\text{m}}=1$ Now for differentiability $f^{'}(x)={\begin{cases}-2 & -2 < x < -1 \\ -1 & -1 < x < 0 \\ 0 & 0 < x < 1 \\ -1 & 1 < x < 2 }\end{cases}$ Clearly $f(x)$ is non-differentiable at $x=-1,0,1$ Also, $|f(x)|$ remains same. $\; \therefore n=3$ $\; \therefore m+n=4$

Q85JEE Main 2023NAT4MParabola

Let a common tangent to the curves $y^2=4 x$ and $(x-4)^{2+}y^2=16$ touch the curves at the points $P$ and $Q$ . Then $( {\text{PQ}})^2$ is equal to ______. [10-Apr-2023 shift 1]

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📖 Explanation

$\;y^2=4 x$ $\;(x-4)^{2+}y^2=16$ Let equation of tangent of parabola ${\text{y}}= {\text{mx}}+1 / {\text{m}}$ Now equation 1 also touches the circle $\; \therefore {| \;{4 {\text{m}}+1 / {\text{m}}}/{\sqrt{1+ {\text{m}}^2}} |}=4$ $\;(4 {\text{m}}+1 / {\text{m}})^2=16+16 {\text{m}}^2$ $\;16 {\text{m}}^{4+}8 {\text{m}}^{2+}1=16 {\text{m}}^{2+}16 {\text{m}}^4$ $\;8 {\text{m}}^2=1$ $\; {\text{m}}^2=1 / 8 \;\; \{ {\text{m}}^4=0\} ( {\text{m}} \to \infty )$ For distinet points consider only ${\text{m}}^2=1 / 8$ . Point of contact of parabola $\; {\text{P}}(8,4 {\sqrt{2}})$ $\; \therefore {\text{PQ}}=\sqrt{ {\text{S}}_1} \Rightarrow ( {\text{PQ}})^2= {\text{S}}_1$ $\;=16+32-16=32$

Q86JEE Main 2023NAT4MStatistics

If the mean of the frequency distribution Class: 0-10 10-20 20-30 30-40 40-50 Frequency: 2 3 x 5 4is 28 , then its variance is _______. [10-Apr-2023 shift 1]

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📖 Explanation

C.I. f x $f_{ {\text{i}}} {\text{x}}_{ {\text{i}}}$ ${\text{x}}_{ {\text{i}}}^2$ 0-10 2 5 10 25 10-20 3 15 45 225 20-30 x 25 25x 625 30-40 5 35 175 1225 40-50 4 45 180 2025 $\;\overline{ {\text{x}}}=\;{Σ {\text{f}}_{ {\text{i}}} {\text{x}}_{ {\text{i}}}}/{ {\text{N}}}$ $\;28=\;{10+45+25 {\text{x}}+175+130}/{14+ {\text{x}}}$ $\;28 \times 14+28 {\text{x}}=410+25 {\text{x}}$ $\;\Rightarrow 3 {\text{x}}=410-392$ $\;\Rightarrow {\text{x}}=\;\frac{18}{3}=6$ $\; \therefore \;\text{ Variance }\;=\;{1}/{ {\text{N}}} \sum {\text{f}}_{ {\text{i}}} {\text{x}}_{ {\text{i}}}{ }^{2-}(\overline{ {\text{x}}})^2$ $\;=\;\frac{1}{20} 18700-(28)^2$ $\;=935-784=151$

Q87JEE Main 2023NAT4MBinomial Theorem

The coefficient of $x^7$ in $(1-x+2 x^3) ^{10}$ is _______. [10-Apr-2023 shift 1]

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📖 Explanation

$(1-x+2 x^3) ^{10}$ $\;T_{ {\text{n}}}=\;\frac{10 !}{a ! b ! c !}(-2 x)^b (x^3) ^c$ $\;=\;\frac{10 !}{a ! b ! c !}(-2)^b x^{b+3 c}$ $\;\Rightarrow b+3 c=7, a+b+c=10$ $\; \therefore \;\text{ Coefficient of }\; x^7=\;\frac{10 !}{3 ! 7 ! 0 !}(-1)^{7+}\;\frac{10 !}{5 ! 4 ! 1 !}(-1)^4(2)$ $\;+\;\frac{10 !}{7 ! 1 ! 2 !}(-1)^1(2)^2$ $\;=-120+2520-1440=960$

Q88JEE Main 2023NAT4MArea Under The Curves

Let $y=p(x)$ be the parabola passing through the points $(-1,0),(0,1)$ and $(1,0)$ . If the area of the region $\{(x, y):(x+1)^{2+}(y-1)^2 \le 1, y \le p(x)\}$ is $A$ , then $12(\pi-4 {\text{A}})$ is equal to _______. [10-Apr-2023 shift 1]

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📖 Explanation

There can be infinitely many parabolas through given points. Let parabola $x^2=-4 a(y-1)$ Passes through $(1,0)$ $\; \therefore {\text{b}}=-4 {\text{a}}(-1) \Rightarrow {\text{a}}=\;\frac{1}{4}$ $\; \therefore {\text{x}}^2=-( {\text{y}}-1)$ Now area covered by parabola $=int_{-1}^{0} (1-x^2) \text{dx}$ $\;= (x-\;\frac{x^3}{3}) _1^b=(0-0)- \{-1+\;\frac{1}{3}\}$ $\;=\;\frac{2}{3}$ Required Area $=$ Area of sector $-\{$ Area of square - Area covered by Parabola $\}$ $\;=\;\frac{\pi}{4}- \{1-\;\frac{2}{3}\}$ $\;=\;\frac{\pi}{4}-\;\frac{1}{3}$ $\; \therefore 12(\pi-4 A)=12 [\pi-4 (\;\frac{\pi}{4}-\;\frac{1}{3}) ]$ $\;=12 [\pi-\pi+\;\frac{4}{3}]$ $\;=16$

Q89JEE Main 2023NAT4MLogarithms

Let $a, b, c$ be three distinct positive real numbers such that $(2 a)^{\log _e a}=(b c)^{\log _e b}$ and $b^{\log _e 2}=a^{\log _c c}$ . Then $6 a+$ $5 {\text{bc}}$ is equal to _______. [10-Apr-2023 shift 1]

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📖 Explanation

$\;(2 a)^{\ln a}=(b c)^{\ln b} \;\; 2 a > 0, b c > 0$ $\;\ln a(\ln 2+\ln a)=\ln b(\ln b+\ln c)$ $\;\ln 2 \cdot \ln b=\ln c \cdot \ln a$ $\;\ln 2=\alpha, \ln a=x, \ln b=y, \ln c=z$ $\;\alpha y=x z$ $\;x(\alpha+x)=y(y+z)$ $\;\alpha=\;\frac{x z}{y}$ $\;x (\;\frac{x z}{y}+x) =y(y+z)$ $\;x^2(z+y)=y^2(y+z)$ $\;y+z=0 \;\text{ or }\; x^2=y^2 \Rightarrow x=-y$ $\;b c=1 \;\text{ or ab }\;=1$ $\;b c=1 \;\text{ or }\; a b=1$ $( {\text{a}}, {\text{b}}, {\text{c}})= (\;\frac{1}{2}, \lambda , \;\frac{1}{\lambda }) , \lambda \ne 1,2, \;\frac{1}{2}$ then $\;6 a+5 b c=3+5=8$ $\;\;\text{ (II) }\;(a, b, c)= (\lambda , \;\frac{1}{\lambda }, \;\frac{1}{2}) , \lambda \ne 1,2, \;\frac{1}{2}$ In this situation infinite answer are possibleSo, Bonus.

Q90JEE Main 2023NAT4MSequences and Series

The sum of all those terms, of the arithmetic progression $3,8,13, . . . ., 373$ , which are not divisible by 3 , is equal to _______. [10-Apr-2023 shift 1]

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📖 Explanation

A.P: $3,8,13 . . . . .373$ $\;T_n=a+(n-1) d$ $\;373=3+(n-1) 5$ $\;\Rightarrow n=\;\frac{370}{5}$ $\;\Rightarrow n=75$ $\;\;\text{ Now Sum }\;=\;{ {\text{n}}}/{2}[a+1]$ $\;=\;\frac{75}{2}[3+373]=14100$ Now numbers divisible by 3 are, $\;3,18,33 . . . . . . .363$ $\;363=3+( {\text{k}}-1) 15$ $\;\Rightarrow {\text{k}}-1=\;\frac{360}{15}=24 \Rightarrow {\text{k}}=25$ $\;\;\text{ Now, \sum }\;=\;\frac{25}{2}(3+363)=4575 {\ s}$ $\; \therefore \;\text{ req. \sum }\;=14100-4575$ $\;=9525$

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