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8 Apr 2023 Shift 2

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Physics30 questions

Q1JEE Main 2023MCQ4MProperties of Matter

A hydraulic automobile lift is designed to lift vehicles of mass $5000 {\text{kg}}$ . The area of cross section of the cylinder carrying the load is $250 {\text{cm}}^2$ . The maximum pressure the smaller piston would have to bear is $[Assume ${\text{g}}=10 {\text{m}} / {\ s}^2$ ]$ : [8-Apr-2023 shift 2]

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📖 Explanation

From pascal law same $\Delta {\text{P}}$ transmitted through out liquid $\;\Delta {\text{P}}=\;{ {\text{F}}}/{ {\text{A}}}=\;\frac{5000 \times 10}{250 \times 10^{-4}}$ $\;=2 \times 10^6 {\text{Pa}}$

Q2JEE Main 2023MCQ4MGravitation

The orbital angular momentum of a satellite is ${\text{L}}$ , when it is revolving in a circular orbit at height ${\text{h}}$ from earth surface. If the distance of satellite from the earth center is increased by eight times to its initial value, then the new angular momentum will be- [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{L}}= {\ mv} {\text{v}}_0 {\text{r}} ( {\text{v}}_0=\sqrt{\;{ {\text{GM}}}/{ {\text{h}}}})$ $\; {\text{L}}= {\text{m}} \sqrt{ {\text{GMh}}}$ $\; {\text{h}}^{'} \to {\text{h}}+8 {\text{h}}=9 {\text{h}}$ . . . (1) $\; {\text{L}}^{'}= {\text{m}} \sqrt{ {\text{GM}9h}}$ . . . (2) $\;\;{ {\text{L}}^{'}}/{ {\text{L}}^{'}}=3$ $\; {\text{L}}^{'}=3 {\text{L}}$

Q3JEE Main 2023MCQ4MElectromagnetic Waves

The waves emitted when a metal target is bombarded withhigh energy eléctrons are [8-Apr-2023 shift 2]

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Q4JEE Main 2023MCQ4MUnits and Measurements

Match List I with List II LIST - I LIST - II A. Torque I. ${\text{ML}}^{-2} {\text{T}}^{-2}$ B. Stress II. $\;\; {\text{ML}}^2 {\text{T}}^{-2}$ C. Pressure gradient III. $\text{MLL}^{-1} {\text{t}}^{-1}$ D. Coefficient of viscosity IV. ${\text{ML}}^{-1} {\text{T}}^{-2}$ Choose the correct answer from the options given below: [8-Apr-2023 shift 2]

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📖 Explanation

$\;{[\;\text{ Torque }\;]=\;\text{ F.L }\;}$ $\; {\text{MLT}}^{-2} \cdot {\text{L}}= {\text{ML}}^2 {\text{T}}^{-2}$ $\;{[\;\text{ Stress }\;]=\;{ {\text{F}}}/{ {\text{A}}}}$ $\;\;{M L T^{-2}}/{ {\text{L}}^2}= {\text{ML}}^{-1} {\text{T}}^{-2}$ $\;{[\;\text{ Pressure gradient }\;]=\;{\Delta {\text{P}}}/{\Delta {\text{L}}}=\;{ {\text{F}}}/{ {\text{A}}^2 . {\text{L}}}}$ $\;=\;{ {\text{MLT}}^{-2}}/{ {\text{L}}^3}$ $\;= {\text{ML}}^{-2} {\text{T}}^{-2}$ $\; {\text{F}}= {\text{nA}} \;{ {\text{dv}}}/{ {\text{dy}}}$ $\;η=\text{ML}^{-1} {\text{T}}^{-1}$

Q5JEE Main 2023MCQ4MMotion in a Straight Line

Give below are two statementsStatement I : Area under velocity- time graph gives the distance travelled by the body in a given time.Statement II : Area under acceleration- time graph is equal to the change in velocity- in the given time.In the light of given statement, choose the correct answer from the options given below. [8-Apr-2023 shift 2]

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📖 Explanation

${ {\text{v}}}^{\to }=\;{ {\text{d}} { {\ s}}^{\to }}/{ {\text{dt}}} \Rightarrow \int {\text{d}} { {\ s}}^{\to }=\int { {\text{v}}}^{\to } {\text{dt}}$ Area of $\vec{v}$ vs time gives displacement $\vec{a}=\;{ {\text{d}} { {\text{v}}}^{\to }}/{ {\text{dt}}} \Rightarrow \int {\text{d}} { {\text{v}}}^{\to }=\int { {\text{a}}}^{\to } {\text{dt}}$ Area of $\vec{a}$ vs t graph gives change in velocity

Q6JEE Main 2023MCQ4MElectromagnetic Waves

The power radiated from a linear antenna of length $l$ is proportional to (Given, $\lambda =$ Wavelength of wave): [8-Apr-2023 shift 2]

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📖 Explanation

The power ( $P$ ) radiated by an antenna is directly proportional to its radiation resistance ( $R_{rad}$ ), as given by the formula $P = I^2 R_{rad}$ . For a short linear antenna of length $l$ , which is much smaller than the wavelength $\lambda$ (i.e., $l \ll \lambda$ ), the radiation resistance is given by the standard formula $R_{rad} = 80 \pi^2 (\frac{l}{\lambda })^2$ . From this relationship, we can see that the radiation resistance is proportional to the square of the ratio of the antenna's length to the wavelength. Therefore, since $P \propto R_{rad}$ , the radiated power is proportional to $(\frac{l}{\lambda })^2$ .

Q7JEE Main 2023MCQ4MElectrostatic Potential and Capacitance

Electric potential at a point ' ${\text{P}}$ ' due to a point charge of $5 \times 10^{-9} {\text{C}}$ is $50 {\text{V}}$ . The distance of ' ${\text{P}}$ ' from the point charge is:(Assume, $\;\frac{1}{4 \pi \epsilon_0}=9 \times 10^{+9} {\text{Nm}}^2 {\text{C}}^{-2}$ ) [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{V}}=\;\frac{1}{4 \pi \epsilon_0} \;{ {\text{q}}}/{ {\text{r}}}$ $\;\Rightarrow {\text{r}}=\;\frac{9 \times 10^9 \times 5 \times 10^{-9}}{50}$ $\;\Rightarrow {\text{r}}=\;\frac{9}{10} \times 100 {\text{cm}}$ $\; {\text{r}}=90 {\text{cm}}$

Q8JEE Main 2023MCQ4MGravitation

The acceleration due to gravity at height ${\text{h}}$ above the earth if ${\text{h}}< < {\text{R}}$ (Radius of earth) is given by [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{g}}^{'}=\;{ {\text{GM}}}/{( {\text{R}}+ {\text{h}})^2}$ $\; {\text{g}}^{'}=\;{ {\text{GM}}}/{ {\text{R}}^2 (1+\;{ {\text{h}}}/{ {\text{R}}}) ^2}$ using binomial expansion & neglect higher order term $\Rightarrow {\text{g}}= {\text{g}} (1-\;{2 {\text{h}}}/{ {\text{R}}})$

Q9JEE Main 2023MCQ4MElectromagnetic Induction

An emf of $0.08 {\text{V}}$ is induced in a metal rod of length $10 {\text{cm}}$ held normal to a uniform magnetic field of $0.4 {\text{T}}$ , when moves with a velocity of: [8-Apr-2023 shift 2]

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📖 Explanation

$\;\epsilon= {\text{Blv}}$ $\;\Rightarrow 0.08= {\text{v}} \times 0.4 \times \;\frac{10}{100}$ $\;\Rightarrow {\text{v}}=2 {\text{m}} / {\ s}$

Q10JEE Main 2023MCQ4MThermodynamics

Work done by a Carnot engine operating between temperatures $127^{\circ} {\text{C}}$ and $27^{\circ} {\text{C}}$ is $2 {\text{kJ}}$ . The amount of heat transferred to the engine by the reservoir is: [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{n}}=1-\;\frac{300}{400}=\;\frac{1}{4}$ $\; {\text{n}}=\;{ {\text{w}}}/{ {\text{Q}}_1}=\;\frac{1}{4} \Rightarrow {\text{Q}}_1=8 {\text{kJ}}$

Q11JEE Main 2023MCQ4MWave Optics

The width of fringe is $2 {\text{mm}}$ on the screen in a double slits experiment for the light of wavelength of $400 {\text{nm}}$ . The width of the fringe for the light of wavelength $600 {\text{nm}}$ will be: [8-Apr-2023 shift 2]

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📖 Explanation

$\;\beta=\;{ {\text{D}} \lambda }/{ {\text{d}}}$ $\;\Rightarrow \beta \propto \lambda$ $\;\;\frac{\beta_1}{\beta_2}=\;\frac{\lambda _1}{\lambda _2} \Rightarrow \;\frac{2}{\beta}=\;\frac{400}{600}$ $\;\beta=3 {\text{mm}}$

Q12JEE Main 2023MCQ4MThermodynamics

The temperature at which the kinetic energy of oxygen molecules becomes double than its value at $27^{\circ} {\text{C}}$ is [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{KE}} \;\text{ of }\; {\text{O}}_2 \;\text{ molecules }\;=5 \times (\;\frac{1}{2} {\text{KT}})$ $\;( {\text{KE}})_{27^{\circ} {\text{C}}}=5 \times \;\frac{1}{2} {\text{k}}(27+273)=\;\frac{5}{2} {\text{k}} \times 300$ $\;( {\text{KE}})_{ {\text{T}}}=2 (\;\frac{5}{2} {\text{k}}) \times 300=\;\frac{5}{2} {\text{k}}(600)$ $\;\;\text{ i.e. }\; {\text{T}}=600 {\text{K}}$ $\;=600-273$ $\; {\text{T}}=327^{\circ} {\text{C}}$

Q13JEE Main 2023MCQ4MMagnetic Effects of Current and Magnetism

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason $\text{bo} {R}$ Assertion A : Electromagnets are made of soft iron.Reason R : Soft iron has high permeability and low retentivity.In the light of above, statements, chose the most appropriate answer from the options given below. [8-Apr-2023 shift 2]

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📖 Explanation

An electromagnet is a temporary magnet that is created by passing an electric current through a coil of wire wrapped around a core material. The core material should become strongly magnetized when the current is on and lose its magnetism when the current is off.

Assertion (A) is correct. Soft iron is the preferred material for the core of electromagnets.

Reason (R) is also correct and explains Assertion (A). Soft iron has high magnetic permeability ( $\mu_r$ ), which allows it to be strongly magnetized by a relatively weak external field (from the current in the coil), creating a powerful electromagnet. It also has low retentivity, which means it loses most of its magnetism once the external field is removed (the current is switched off). These two properties make soft iron ideal for creating controllable magnets.

Therefore, both statements are correct, and R is the correct explanation for A.

Q14JEE Main 2023MCQ4MMotion in a Plane

The trajectory of projectile, projected from the ground is given by ${\text{y}}= {\text{x}}-\;{ {\text{x}}^2}/{20}$ . Where ${\text{x}}$ and ${\text{y}}$ are measured in meter. The maximum height attained by the projectile will be. [8-Apr-2023 shift 2]

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📖 Explanation

$\;y=x-\;\frac{x^2}{20}$ $\; (\;\frac{d y}{d x}) =1-\;\frac{x}{10} \;\text{ for }\; y_{\max } ; \;\frac{d y}{d x}=0$ $\;x=10$ $\;y_{\max }=10-\;\frac{100}{20}=5 m$

Q15JEE Main 2023MCQ4MCentre of Mass and Collision

A bullet of mass $0.1 {\text{kg}}$ moving horizontally with speed $400 {\text{ms}}^{-1}$ hits a wooden block of mass $3.9 {\text{kg}}$ kept on a horizontal rough surface. The bullet gets embedded into the block and moves $20 {\text{m}}$ before coming to rest. The coefficient of friction between the block and the surface is $(.$ Given $g=10 {\text{m}} / {\ s}^2)$ [8-Apr-2023 shift 2]

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📖 Explanation

Apply momentum conservation just before and just after the collision $\;0.1 \times 400=(3.9+.1) {\text{u}}$ $\;\Rightarrow {\text{u}}=10 {\text{m}} / {\ s}$ $\;\Delta {\text{KE}}= {\text{w}}_{\;\text{all FORCE }\;}$ $\; \because {\text{f}}=\mu {\text{mg}} \;\text{ (kinetic friction) }\;$ $\;\Rightarrow 0-\;\frac{1}{2}(4)(10)^2=-\mu (4) {\text{g}} \times 20$ $\;\Rightarrow \mu =0.25$

Q16JEE Main 2023MCQ4MSemiconductor Electronics

For a given transistor amplifier circuit in CE configuration ${\text{V}}_{ {\text{CC}}}=1 {\text{V}}, {\text{R}}_{ {\text{C}}}=1 {\text{k}} Ω, {\text{R}}_{ {\text{b}}}=100 {\text{k}} Ω$ and $\beta=100$ . Value of base current $I_b$ is [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{V}}_{ {\text{cc}}}=1 {\text{V}}$ $\; {\text{R}}_{ {\text{c}}} {\text{I}}_{ {\text{c}}}=1$ $\; {\text{I}}_{ {\text{c}}}=\;\frac{1}{10^3} {\text{A}}=1 {\text{mA}}$ $\;\beta=\;{ {\text{I}}_{ {\text{c}}}}/{ {\text{I}}_beta}$ $\; {\text{I}}_beta=\;{ {\text{I}}_{ {\text{C}}}}/{\beta}$ $\;\; =1 \times 10^{-5} {\text{A}}$ $\;\; =10 \mu {\text{A}}$

Q17JEE Main 2023MCQ4MCircular Motion

For particle ${\text{P}}$ revolving round the centre ${\text{O}}$ with radius of circular path ${\text{r}}$ and angular velocity $\omega$ , as shown in below figure, the projection of ${\text{OP}}$ on the ${\text{x}}$ -axis at time ${\text{t}}$ is [8-Apr-2023 shift 2]

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📖 Explanation

$\theta=\omega {\text{t}}$ Angle from ${\text{x}}$ axis $=\omega {\text{t}}+\;\frac{\pi}{6}$ Projection of OP on $x$ axis $=r \cos (\omega t+\;\frac{\pi}{6})$

Q18JEE Main 2023MCQ4MAtoms and Nuclei

A radio active material is reduced to $1 / 8$ of its original amount in 3 days. If $8 \times 10^{-3} {\text{kg}}$ of the material is left after 5 days the initial amount of the material is [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{m}}= {\text{m}}_0 {\text{e}}^{-\lambda {\text{t}}}$ $\;\;{ {\text{m}}_0}/{8}= {\text{m}}_0 {\text{e}}^{-\lambda {\text{t}}}$ $\;-\ln 8=-\lambda {\text{t}}$ $\;=\lambda =\;\frac{\ln 8}{3} \;\text{ per day }\;$ $\; {\text{m}}= {\text{m}}_0 {\text{e}}^{-\lambda {\text{t}}}$ $\;8= {\text{m}}_0 {\text{e}}^{-\;\frac{\ln 8}{3} \times 5}$ $\;\Rightarrow 8= {\text{m}}_0 {\text{e}}^{-\;\frac{3 \ln 2}{3} \times 5}$ $\;8= {\text{m}}_0 {\text{e}}^{\ln 2^{-5}}$ $\;=8 {\text{m}}_0 (\;\frac{1}{2^5})$ $\; {\text{m}}_0=8 \times 2^5$ $\;=8 \times 32$ $\; {\text{m}}_0=256 {\text{gm}}^2$

Q19JEE Main 2023MCQ4MCurrent Electricity

The equivalent resistance between ${\text{A}}$ and ${\text{B}}$ as shown in figure is: [8-Apr-2023 shift 2]

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📖 Explanation

Potential different across all resistor is same So they are in parallel $\;\;\frac{1}{R}=\;\frac{1}{20}++\;\frac{1}{20}+\;\frac{1}{10}$ $\;R_{ {\text{eq}}}=5 {\text{k}} Ω$

Q20JEE Main 2023MCQ4MDual Nature of Matter and Radiation

In photoelectric effectA. The photocurrent is proportional to the intensity of the incident radiation.B. Maximum Kinetic energy with which photoelectrons are emitted depends on the intensity of incident light.C. Max K.E with which photoelectrons are emitted depends on the frequency of incident light.D. The emission of photoelectrons require a minimum threshold intensity of incident radiation.E. Max. K.E of the photoelectrons is independent of the frequency of the incident light.Choose the correct answer from the options given below: [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{hv}}=\varphi +( {\text{KE}})_{\max }$ $\;( {\text{KE}})_{\max }= {\text{hv}}-\varphi$

Q21JEE Main 2023NAT4MElectrostatic Potential and Capacitance

A $600 {\text{pF}}$ capacitor is charged by $200 {\text{V}}$ supply. It is then disconnected from the supply and is connected to another uncharged $600 {\text{pF}}$ capacitor. Electrostatic energy lost in the process is _______ $\mu {\text{J}}$ [8-Apr-2023 shift 2]

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📖 Explanation

$\;\;\text{ loss of strength }\;=\;\frac{1}{2} \;{ {\text{c}} \times {\text{c}}}/{ {\text{c}}+ {\text{c}}} ( {\text{v}}_1- {\text{v}}_2) ^2$ $\;=\;\frac{1}{2} \times \$ [;\frac{600 × 10^{-12}}{2}]$ \times (200)^2;=600 \times 10^{-12} \times 10^4=6 \times 10^{-6}=6 \mu {\text{J}}$

Q22JEE Main 2023NAT4MAlternating Current

A series combination of resistor of resistance $100 Ω$ , inductor of inductance $1 {\text{H}}$ and capacitor of capacitance $6.25 \mu {\text{F}}$ is connected to an ac source. The quality factor of the circuit will be _______. [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{Q}}=\;{1}/{ {\text{R}}} \sqrt{\;{ {\text{L}}}/{ {\text{C}}}}$ $\;=\;\frac{1}{100} \sqrt{\;\frac{1}{6.25 \times 10^{-6}}}$ $\;=4$

Q23JEE Main 2023NAT4MCurrent Electricity

The number density of free electrons in copper is nearly $8 \times 10^{28} {\text{m}}^{-3}$ . A copper wire has its area of cross section $=2 \times 10^{-6} {\text{m}}^2$ and is carrying a current of $3.2 {\text{A}}$ . The drift speed of the electrons is ________ $\times 10^{-6} {\text{ms}}^{-1}$ [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{I}}=\;\text{ neAvd }\;$ $\;\Rightarrow 3.2=8 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6} ( {\text{v}}_{ {\text{d}}})$ $\;\Rightarrow {\text{v}}_{ {\text{d}}}=\;\frac{1}{8 \times 10^{-6} \times 10^9}$ $\;\Rightarrow {\text{v}}_{ {\text{d}}}=125 \times 10^{-6} {\text{m}} / {\ s}$

Q24JEE Main 2023NAT4MRotational Motion

A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity $3 {\text{m}} / {\ s}$ (as shown in figure). Maximum height with respect to the initial position covered by it will be _______ ${\text{cm}} (.$ take, $g=10 {\text{m}} / {\ s}^2)$ [8-Apr-2023 shift 2]

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📖 Explanation

For pure rolling $( {\text{M}} . {\text{E}})_{ {\text{A}}}=( {\text{M}} . {\text{E}})_{ {\text{B}}}$ $\;\Rightarrow \;\frac{1}{2} {\ mv}_0^{2+}\;\frac{1}{2} \times (\;\frac{2}{3} {\text{mR}}^2) (\;{ {\text{v}}_0}/{ {\text{R}}}) ^2= {\text{mgH}}_{\max }$ $\;\Rightarrow {\text{H}}_{\max }=\;\frac{5}{6} \;{ {\text{v}}_0^2}/{ {\text{g}}}=\;\frac{5}{6} \times \;\frac{3^2}{10}=0.75 {\text{m}}$ $\;\Rightarrow {\text{H}}_{\max }=75 {\text{cm}}$

Q25JEE Main 2023NAT4MProperties of Matter

A steel rod of length $1 {\text{m}}$ and cross sectional area $10^{-4} {\text{m}}^2$ is heated from $0^{\circ} {\text{C}}$ to $200^{\circ} {\text{C}}$ without being allowed to extend or bend. The compressive tension produced in the rod is ________ of steel $=2 \times 10^{11} {\text{Nm}}^{-2}$ , coefficient of linear expansion $=10^{-5} {\text{K}}^{-1}$ ) $\times 10^4 {\text{N}}$ . (Given Young's modulus [8-Apr-2023 shift 2]

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📖 Explanation

Thermal stress $= {\text{Y}} \alpha \Delta {\text{T}}$ $\; {\text{F}}= {\text{YA}} \alpha \Delta {\text{T}}$ $\;=2 \times 10^{11} \times 10^{-4} \times 10^{-5} \times 200$ $\;=4 \times 10^4$ $\; {\text{x}}=4$

Q26JEE Main 2023NAT4MMagnetic Effects of Current and Magnetism

The ratio of magnetic field at the centre of a current carrying coil of radius ${\text{r}}$ to the magnetic field at distance ${\text{r}}$ from the centre of coil on its axis is ${\sqrt{x}}: 1$ . The value of $x$ is _______. [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{B}}_{\;\text{axis }\;}=\;{\mu _0 {\text{i}} {\text{R}}^2}/{2 ( {\text{R}}^{2+} {\text{x}}^2) ^{3 / 2}}$ $\;\;{ ( {\text{B}}_{\;\text{axis }\;}) {\text{x}}= {\text{R}}}/{ ( {\text{B}}_{\;\text{axis }\;}) {\text{x}}=0}=\;{\;{\mu _0 {\text{iR}}^2}/{2 ( {\text{R}}^{2+} {\text{R}}^2) ^{3 / 2}}}/{{\mu _0 {\text{iR}}^2}/{2 (R^2) ^{3 / 2}}}=\;{R^3}/{2^{3 / 2} {\text{R}}^3}=\;{1}/{{\sqrt{8}}}$ $\;\;{(B)_{ {\text{At}} \;\text{ centrs }\;}}/{(B)_{A t} x=R}=\;{{\sqrt{8}}}/{1}$ $\;x=8$

Q27JEE Main 2023NAT4MAtoms and Nuclei

The ratio of wavelength of spectral lines ${\text{H}}_alpha$ and ${\text{H}}_beta$ in the Balmer series is $\;{ {\text{x}}}/{20}$ . The value of ${\text{x}}$ is [8-Apr-2023 shift 2]

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📖 Explanation

The wavelength of a spectral line is determined by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$ . For the Balmer series, the final energy level is $n_f = 2$ . The ${\text{H}}_\alpha$ line corresponds to the transition from $n_i = 3$ , so its inverse wavelength is $\frac{1}{\lambda_\alpha} = R \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = \frac{5R}{36}$ . The ${\text{H}}_\beta$ line corresponds to the transition from $n_i = 4$ , so its inverse wavelength is $\frac{1}{\lambda_\beta} = R \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = \frac{3R}{16}$ . The ratio of the wavelengths is therefore $\frac{\lambda_\alpha}{\lambda_\beta} = \frac{1/\lambda_\beta}{1/\lambda_\alpha} = \frac{3R/16}{5R/36}$ . Simplifying this expression gives $\frac{3}{16} \times \frac{36}{5} = \frac{27}{20}$ . Comparing this to the given ratio $\frac{x}{20}$ , we find that $x = 27$ .

Q28JEE Main 2023NAT4MRay Optics and Optical Instruments

Two transparent media having refractive indices 1.0 and 1.5 are separated by a spherical refracting surface of radius of curvature $30 {\text{cm}}$ . The centre of curvature of surface is towards denser medium and a point object is placed on the principle axis in rarer medium at a distance of $15 {\text{cm}}$ from the pole of the surface. The distance of image from the pole of the surface is ______ ${\text{cm}}$ . [8-Apr-2023 shift 2]

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📖 Explanation

$\;\;{1.5}/{ {\text{v}}}-\;\frac{1}{(-15)}=\;\frac{1.5-1}{30}$ $\;\Rightarrow \;{1.5}/{ {\text{v}}}-\;\frac{1}{60}-\;\frac{1}{15}=\;\frac{1-4}{60}$ $\; {\text{V}}=-30 {\text{cm}}$ $\;=30 {\text{cm}}$

Q29JEE Main 2023NAT4MWaves

A guitar string of length $90 {\text{cm}}$ vibrates with a fundamental frequency of $120 {\text{Hz}}$ . The length of the string producing a fundamental frequency of $180 {\text{Hz}}$ will be ________ ${\text{cm}}$ . [8-Apr-2023 shift 2]

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📖 Explanation

$\text{bo} {f}\;=\;{ {\text{V}}}/{2 {\text{L}}}$ (Fundamental Frequency) $120\;=\;{ {\text{V}}}/{2 {\text{L}}}$ . . . (1) $180\;=\;{ {\text{V}}}/{2 {\text{L}}^{'}}$ . . . (2) $\;{ {\text{L}}^{'}}/{ {\text{L}}^{'}}\;=\;\frac{120}{180}$ ${\text{L}}^{'}\;=\;\frac{2}{3} \times 90$ $L^{'}\;=60 {\text{cm}}$

Q30JEE Main 2023NAT4MWork, Power and Energy

A body of mass $5 {\text{kg}}$ is moving with a momentum of $10 {\text{kg}} {\text{ms}}^{-1}$ . Now a force of $2 {\text{N}}$ acts on the body in the direction of its motion for $5 {\ s}$ . The increase in the Kinetic energy of the body is _______ J. [8-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\;( {\text{KE}})=\;{ {\text{P}}^2}/{2 {\text{M}}}$ $\;\Rightarrow \;\frac{1}{2} {\mu}^2=\;\frac{(10)^2}{2 \times 5}$ $\;=\;\frac{1}{2} \times 5 \times {\text{u}}^2=\;\frac{100}{10}$ $\;\;\text{ Initial speed }\; {\text{u}}=2 {\text{m}} / {\ s}$ $\;\Delta {\text{KE}}= {\text{w}}_{\;\text{all forces }\;}$ $\;={ {\text{F}}}^{\to } \cdot { {\text{S}}}^{\to } \;\; (\theta=0^{\circ})$ $\;= {\text{F}} ( {\text{ut}}+\;\frac{1}{2} {\text{at}}^2)$ $\;=2 \cdot [2 \times 5+\;\frac{1}{2} \times \;\frac{2}{5} \times 5^2]$ $\;=30 {\text{J}}$

Chemistry30 questions

Q31JEE Main 2023MCQ4MBiomolecules

The statement/s which are true about antagonists from the following is/are :A. They bind to the receptor siteB. Get transferred inside the cell for their actionC. Inhibit the natural communication of the bodyD. Mimic the natural messenger.Choose the correct answer from the options given below: [8-Apr-2023 shift 2]

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Q32JEE Main 2023MCQ4MChemical Kinetics

The correct reaction profile diagram for a positive catalyst reaction. [8-Apr-2023 shift 2]

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Q33JEE Main 2023MCQ4MClassification of Elements

Given below are two statements : One is labelled as Assertion A and other is labelled as Reason R Assertion A : Sodium is about 30 times as abundant as potassium in the oceans.Reason $\text{bo} {R}$ : Potassium is bigger in size than sodium.In the light of the above statements, choose the correct answer from the options given below [8-Apr-2023 shift 2]

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Q34JEE Main 2023MCQ4MChemical Kinetics

Which of these reactions is not a part of breakdown of ozone in stratosphere? [8-Apr-2023 shift 2]

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📖 Explanation

Ozone destruction ${\text{CF}}_2 {\text{Cl}}_2 {\to }^{\;\text{ hv }\;} {\text{C}} ^{∙}{\text{l}}+{\text{C}}^{∙} {\text{F}}_2 {\text{Cl}}( {\text{g}})$ $\; {\text{C}^{∙}\text{l}}+ {\text{O}}_3 \to {\text{Cl}} {\text{O}}^{∙}+ {\text{O}}_2$ $\; { {\text{ClO}}^{∙}}+{\text{O}}^{∙} \to {\text{Cl}}^{∙}+ {\text{O}}_2$

Q35JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

The correct IUPAC nomenclature for the following compound is : [8-Apr-2023 shift 2]

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Q36JEE Main 2023MCQ4MStructure of Atom

Henry Moseley studied characteristic X-ray spectra of elements. The graph which represents his observation correctly isGiven $v=$ frequency of ${\text{X}}$ -ray emitted ${\text{Z}}=$ atomic number [8-Apr-2023 shift 2]

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📖 Explanation

${\sqrt{v}} \alpha Z$

Q37JEE Main 2023MCQ4MBiomolecules

Match list I with list II LIST - I Coordination complex LIST - II Number of unpaired electrons A. $[ {\text{Cr}}( {\text{CN}})_6] ^{3-}$ I. 0 B. $[ {\text{Fe}} ( {\text{H}}_2 {\text{O}}) _6] ^{2+}$ II. 3 C. $[ {\text{Co}} ( {\text{NH}}_3) _6] ^{3+}$ III. 2 D. $[ {\text{Ni}} ( {\text{NH}}_3) _6] ^{2+}$ IV. 4Choose the correct answer from the options given below: [8-Apr-2023 shift 2]

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📖 Explanation

For option (A) ${\text{Cr}}^{+3}: 3 {\text{d}}^3$ ${\text{CN}}^{-} \to {\text{SFL}}$ $\Rightarrow$ No. of unpaired electrons $=3$ For option (B) ${\text{Fe}}^{+2}: 3 {\text{d}}^6$ ${\text{H}}_2 {\text{O}}: {\text{WFL}}$ No. of unpaired electrons $=4$ For option $( {\text{C}})$ ${\text{Co}}^{+3}: 3 {\text{d}}^6$ ${\text{NH}}_3: {\text{SFL}}$ No. of unpaired electrons $=0$ For option (D) ${\text{Ni}}^{+2}: 3 {\text{d}}^8$ ${\text{NH}}_3: {\text{SFL}}$ No. of unpaired electrons $=2$

Q38JEE Main 2023MCQ4MAldehydes, Ketones and Carboxylic Acids

Major product ' ${\text{P}}$ ' formed in the following reaction is : [8-Apr-2023 shift 2]

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Q39JEE Main 2023MCQ4MSolid State

For a good quality cement, the ratio of lime to the total of the oxides of ${\text{Si}}, {\text{Al}}$ and ${\text{Fe}}$ should be as close as to [8-Apr-2023 shift 2]

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📖 Explanation

$\;{\% {\text{CaO}}}/{\% {\text{SiO}}_2+\% {\text{Al}}_2 {\text{O}}_3+\% {\text{Fe}}_2 {\text{O}}_3}=1.9-2.1$ Option (1) is correct.

Q40JEE Main 2023MCQ4MBiomolecules

Match list I with list II LIST - I Natural amino acid LIST - II One letter code A. Glutamic acid I. Q B. Glutamine II. W C. Tyrosine III. E D. Tryptophan IV. YChoose the correct answer from the options given below: [8-Apr-2023 shift 2]

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Q41JEE Main 2023MCQ4MSome Basic Concepts of Chemistry

Which of the following have same number of significant figures?A. 0.00253B. 1. 0003C. 15.0D. 163Choose the correct answer from the options given below [8-Apr-2023 shift 2]

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📖 Explanation

$0.00253,15.0,163$ All have three significant figures.

Q42JEE Main 2023MCQ4MIonic Equilibrium

Given below are two statements :Statement I : Methyl orange is a weak acid.Statement II : The benzenoid form of methyl orange is more intense/deeply coloured than the quinonoid form. In the light of the above statement, choose the most appropriate answer from the options given below: [8-Apr-2023 shift 2]

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Q43JEE Main 2023MCQ4MOrganic Chemistry - Some Basic Principles

The descending order of acidity for the following carboxylic acid is -A. ${\text{CH}}_3 {\text{COOH}}$ B. ${\text{F}}_3 {\text{C}}- {\text{COOH}}$ C. ${\text{ClCH}}_2- {\text{COOH}}$ D. ${\text{BrCH}}_2- {\text{COOH}}$ Choose the correct answer from the options given below: [8-Apr-2023 shift 2]

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📖 Explanation

Acidity $\alpha$ stability of conjugate baseStability order ${\text{F}}_3 {\text{C}}- {\text{COO}}^{-} > {\text{F}}- {\text{CH}}_2- {\text{COO}}^{-} > {\text{Cl}}- {\text{CH}}_2- {\text{COO}}^{-} > {\text{Br}}- {\text{CH}}_2- {\text{COO}}^{-} > {\text{CH}}_3 {\text{COO}}^{-}$

Q44JEE Main 2023MCQ4MElectrochemistry

In Hall-Heroult process, the following is used for reducing ${\text{Al}}_2 {\text{O}}_3$ :- [8-Apr-2023 shift 2]

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📖 Explanation

In case of Hall's process, reduction of ${\text{Al}}_2 {\text{O}}_3$ to ${\text{Al}}$ can be done using graphite.

Q45JEE Main 2023MCQ4MStates of Matter

Arrange the following gases in increasing order of van der waals constant ' $a$ 'A. ArB. ${\text{CH}}_4$ C. ${\text{H}}_2 {\text{O}}$ D. ${\text{C}}_6 {\text{H}}_6$ Choose the correct options from the following [8-Apr-2023 shift 2]

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📖 Explanation

A $\alpha$ force of attraction vanderwal force depends on molecular size and molecular mass and there is H-bonding in water, so correct option will be ${\text{A}} < {\text{B}} < {\text{C}} < {\text{D}}$ .

Q46JEE Main 2023MCQ4MElectrochemistry

Given below are two statement :Statement I : In redox titration, the indicators used are sensitive to change in ${\text{pH}}$ of the solution.Statement II : In acid-base titration, the indicators used are sensitive to change in oxidation potential.In the light of the above statement, choose the most appropriate answer from the options given below [8-Apr-2023 shift 2]

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Q47JEE Main 2023MCQ4MChemical Kinetics

Which of the following can reduce decomposition of ${\text{H}}_2 {\text{O}}_2$ on exposure to light [8-Apr-2023 shift 2]

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📖 Explanation

Urea acts as a stabilizer in the decomposition of ${\text{H}}_2 {\text{O}}_2$

Q48JEE Main 2023MCQ4MHaloalkanes and Haloarenes

The correct order of reactivity of following haloarenes towards nucleophilic substitution with aqueous ${\text{NaOH}}$ is Choose the correct answer from the options given below: [8-Apr-2023 shift 2]

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📖 Explanation

Rate $\alpha$ EWG $\alpha \;\frac{1}{\;\text{ EDG }\;}$ ${\text{NO}}_2 \to -$ Meffect ${\text{OMe}} \to + {\text{M}}$ effect

Q49JEE Main 2023MCQ4MAlcohols, Phenols and Ethers

A compound ' ${\text{X}}$ ' when treated with phthalic anhydride in presence of concentrated ${\text{H}}_2 {\text{SO}}_4$ yields ' ${\text{Y}}$ ' ' ${\text{Y}}$ ' is used as an acid/base indicator. ' ${\text{X}}$ ' and ' ${\text{Y}}$ ' are respectively: [8-Apr-2023 shift 2]

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Q50JEE Main 2023MCQ4MAmines

The product $( {\text{P}})$ formed from the following multistep reaction is : [8-Apr-2023 shift 2]

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Q51JEE Main 2023NAT4MCoordination Compounds

The observed magnetic moment of the complex $[ {\text{Mn}}(x{ {\text{NCS}}})_6] ^{ {\text{x}}-}$ is $6.06 {\text{BM}}$ . The numerical value of ${\text{x}}$ is _______. [8-Apr-2023 shift 2]

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📖 Explanation

$[ {\text{Mn}}( {\text{NCS}})_6] ^{ {\text{x}-}}$ Number of unpaired electron $=5$ So, Mn must be in +2 oxidation state $( {\text{Mn}}^{+2})$ $\Rightarrow 2+(-6)=-x \;\; \Rightarrow -4=-x \;\; \Rightarrow x=4$

Q52JEE Main 2023NAT4MChemical Thermodynamics

For complete combustion of ethane, ${\text{C}}_2 {\text{H}}_4( {\text{g}})+3 {\text{O}}_2( {\text{g}}) \to 2 {\text{CO}}_2( {\text{g}})+2 {\text{H}}_2 {\text{O}}( {\text{l}})$ The amount of heat produced as measured in bomb calorimeter is $1406 {\text{KJ}} {\text{mol}}^{-1}$ at $300 {\text{K}}$ . The minimum value of ${\text{T}} \Delta {\text{S}}$ needed to reach equilibrium is _______ $(-)$ ${\text{KJ}}$ (Nearest integer)Given : ${\text{R}}=8.3 {\text{JK}}^{-1} {\text{mol}}^{-1}$ [8-Apr-2023 shift 2]

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📖 Explanation

$\Delta {\text{G}}=\Delta {\text{H}}- {\text{T}} \Delta {\text{S}}$ at equilibrium:- $\;\Delta {\text{G}}=0$ $\; {\text{T}} \Delta {\text{S}}=\Delta {\text{H}}=\Delta {\text{U}}+\Delta {\text{ngRT}}=-1406+(-2) \times 8.3 \times 300 \times 10^{-3}=-1410.98 \approx 1411$

Q53JEE Main 2023NAT4MIonic Equilibrium

The solubility product of ${\text{BaSO}}_4$ is _______ $1 \times 10^{-10}$ at $298 {\text{K}}$ . The solubility of ${\text{BaSO}}_4$ in $0.1 {\text{M}} {\text{K}}_2 {\text{SO}}_4$ (aq) solut is $\times 10^{-9} {\text{g}} {\text{L}}^{-1}$ (Nearest integer)Given: Molar mass of ${\text{BaSO}}_4$ is $233 {\text{g}} {\text{mol}}^{-1}$ [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{K}}_{\;\text{sp }\;}= {\text{x}}( {\text{x}}+0.1)=10^{-10}$ $\;0.1 {\text{x}}=10^{-10}$ $\; {\text{x}}=10^{-9} {\text{M}}$ $\; {\text{x}}(\;\text{ \in }\; {\text{g}} / {\text{l}})=233 \times 10^{-9}$

Q54JEE Main 2023NAT4MStructure of Atom

The number of atomic orbitals from the following having 5 radial nodes is _______ $7 {\ s}, 7 {\text{p}}, 6 {\ s}, 8 {\text{p}}, 8 {\text{d}}$ [8-Apr-2023 shift 2]

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📖 Explanation

No. of radial node $\;= {\text{n}}-ℓ-1$ $\;\;\text{ For }\; 6 {\text{S}} \to 6-0-1=5$ $\;7 {\text{P}} \to 7-1-1=5$ $\;8 {\text{d}} \to 8-2-1=5$

Q55JEE Main 2023MCQ4MElectrochemistry

The number of incorrect statement from the following is _______ [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{dG}}= {\text{vdp}}- {\ sdT}$ $\; {\text{dG}}=- {\ sdT}$ $\;\;{ {\text{dG}}}/{ {\text{dT}}}=- {\text{S}} \Rightarrow \;{ {\text{d}} \Delta {\text{G}}}/{ {\text{dT}}}=-\Delta {\text{S}}$ $\;\;{ {\text{dE}}^{\circ}}/{ {\text{dT}}}=\;{-\Delta {\text{S}}}/{- {\text{nF}}}$

Q56JEE Main 2023NAT4MSolutions

Coagulating value of the electrolytes ${\text{AlCl}}_3$ and ${\text{NaCl}}$ for ${\text{As}}_2 {\text{S}}_3$ are 0.09 and 50.04 respectively. The coagulating power of ${\text{AlCl}}_3$ is ${\text{x}}$ times the coagulating power of ${\text{NaCl}}$ . The value of ${\text{x}}$ is _______. [8-Apr-2023 shift 2]

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📖 Explanation

Coagulating power $\propto \;\frac{1}{\;\text{ coagulation value }\;}$ $\;{( {\text{CP}})_{ {\text{AlCl}}_3}}/{( {\text{CP}})_{ {\text{NaCl}}}}=\;\frac{50.04}{0.09}=556$

Q57JEE Main 2023NAT4MSolutions

If the boiling points of two solvents ${\text{X}}$ and ${\text{Y}}$ (having same molecular weights) are in the ratio $2: 1$ and their enthalpy of vaporizations are in the ratio $1: 2$ , then the boiling point elevation constant of ${\text{X}}$ is ${\text{m}}$ times the boiling point elevation constant of $Y$ . The value of ${\text{m}}$ is _______ (nearest integer) [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{K}}_{ {\text{b}}}=\;{ {\text{RT}}_{ {\text{b}}}^2 {\text{m}}}/{1000 \Delta {\text{H}}_{ {\text{vap}}}}$ $\;\;{ ( {\text{K}}_{ {\text{b}}}) _{ {\text{x}}}}/{ ( {\text{K}}_{ {\text{b}}}) _{ {\text{y}}}}=\;{ ( {\text{T}}_{ {\text{b}}}^2 {\text{M}}) _{ {\text{x}}}}/{ ( {\text{T}}_{ {\text{b}}}^2 {\text{M}}) _{ {\text{y}}}} \times \;{(\Delta {\text{H}})_{ {\text{y}}}}/{(\Delta {\text{H}})_{ {\text{x}}}}= (\;\frac{2}{1}) ^2 \times (\;\frac{2}{1}) =\;\frac{8}{1}$

Q58JEE Main 2023NAT4MChemical Bonding and Molecular Structure

The number of species from the following carrying a single lone pair on central atom Xenon is ________ ${\text{XeF}}_5{ }^{+}, {\text{XeO}}_3, {\text{XeO}}_2 {\text{F}}_2, {\text{XeF}}_5^{-}, {\text{XeO}}_3 {\text{F}}_2, {\text{XeOF}}_4, {\text{XeF}}_4$ [8-Apr-2023 shift 2]

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Q59JEE Main 2023NAT4MChemical Bonding and Molecular Structure

The ratio of sigma and $\pi$ bonds present in pyrophosphoric acid is _______ [8-Apr-2023 shift 2]

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📖 Explanation

$\;\frac{\sigma}{\pi}=\;\frac{12}{2}=6$ So, Answer is 6

Q60JEE Main 2023NAT4Md and f Block Elements

The sum of oxidation state of the metals in ${\text{Fe}}( {\text{CO}})_5, {\text{VO}}^{2+}$ and ${\text{WO}}_3$ is ______ [8-Apr-2023 shift 2]

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📖 Explanation

${\text{Fe}}^{(0)}( {\text{CO}})_5 \;\; {\text{V}}^{(+4)} {\text{O}}^{2+} \;\; {\text{W}^{(+6)}O}_3$ So, Sum of oxidation state $=0+4+6=10$

Mathematics30 questions

Q61JEE Main 2023MCQ4MComplex Numbers

Let $A= \{\theta \in (0,2 \pi): \;\frac{1+2 i sin \;\theta}{1-i sin \;\theta}.$ is purely imaginary $\}$ . Then the sum of the elements in $A$ is. [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{z}}=\;{1+2 {\text{i}} sin \;\theta}/{1- {\text{i}} sin \;\theta} \times \;{1+ {\text{i}} sin \;\theta}/{1+ {\text{i}} sin \;\theta}$ $\; {\text{z}}=\;{1-2 sin \;^2 \theta+ {\text{i}}(3 sin \;\theta)}/{1+ sin \;^2 \theta}$ $\;{\text{Re}}( {\text{z}})=0$ $\;\;\frac{1-2 sin \;^2 \theta}{1+ sin \;^2 \theta}=0$ $\; sin \;\theta=\;{ \pm 1}/{{\sqrt{2}}}$ $\; {\text{A}}= \{\;\frac{\pi}{4}, \;\frac{3 \pi}{4}, \;\frac{5 \pi}{4}, \;\frac{7 \pi}{4}\}$ $\;\;\text{ \sum }\;=4 \pi(\;\text{ Option }\; 3)$

Q62JEE Main 2023MCQ4MThree Dimensional Geometry

Let $P$ be the plane passing through the line $\;\frac{x-1}{1}=\;\frac{y-2}{-3}=\;\frac{z+5}{7}$ and the point $(2,4,-3)$ . If the image of the point $(-1,3,4)$ in the plane $P$ is $(\alpha, \beta, \gamma )$ then $\alpha+\beta+\gamma$ is equal to [8-Apr-2023 shift 2]

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📖 Explanation

Equation of plane is given by $\;{\begin{vmatrix}x-1 & y-2 & z+5 \\ 1 & 2 & 2 \\ 1 & -3 & 7\end{vmatrix}}=0$ $\;4 x-y-z=7$ $\;\;\frac{\alpha+1}{4}=\;\frac{\beta-3}{-1}=\;\frac{\gamma -4}{-1}=\;\frac{-2(-4-3-4-7)}{16+1+1}=2$ $\;\alpha=7, \beta=1, \gamma =2$ $\;\alpha+\beta+\gamma =10(\;\text{ Option }\; 3)$

Q63JEE Main 2023MCQ4MMatrices and Determinants

If ${\text{A}}={\begin{bmatrix}1 & 5 \\ \lambda & 10\end{bmatrix}}, {\text{A}}^{-1}=\alpha A+\beta I$ and $\alpha+\beta=-2$ , then $4 \alpha^{2+}\beta^{2+}\lambda ^2$ is equal to : [8-Apr-2023 shift 2]

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📖 Explanation

$| {\text{A}}- {\text{xI}}|=0 \Rightarrow {\begin{vmatrix}1- {\text{x}} & 5 \\ \lambda & 10- {\text{x}}\end{vmatrix}}=0 \Rightarrow {\text{x}}^{2-}11 {\text{x}}+10-5 \lambda =0$ $\Rightarrow (10-5 \lambda ) {\text{A}}^{-1}=- {\text{A}}+11 {\text{I}}$ $\therefore \alpha=\;\frac{-1}{10-5 \lambda } \;\; \;\text{ and }\; \;\; \beta=\;\frac{+11}{10-5 \lambda }$ $\alpha+\beta=-2 \Rightarrow \;\frac{10}{10-5 \lambda }=-2 \Rightarrow 10-5 \lambda =-5 \Rightarrow \lambda =3$ $\therefore \alpha=\;\frac{1}{5} \;\; & \;\; \beta=\;\frac{-11}{5}$ $\therefore 4 {\text{a}}^{2+}\beta^{2+}\lambda ^2=\;\frac{4}{25}+\;\frac{121}{25}+3^2=14 \;\text{ Ans. }\;$

Q64JEE Main 2023MCQ4MThree Dimensional Geometry

The area of the quadrilateral ${\text{ABCD}}$ with vertices ${\text{A}}(2,1,1), {\text{B}}(1,2,5), {\text{C}}(-2,-3,5)$ and ${\text{D}}(1,-6,-7)$ is equal to [8-Apr-2023 shift 2]

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📖 Explanation

$\;\;\text{ Vector Area }\;={ {\text{v}}}^{\to }$ $\;=\;\frac{1}{2} { {\text{AB}}}^{\to } \times { {\text{AC}}}^{\to }+\;\frac{1}{2} { {\text{AC}}}^{\to } \times { {\text{AD}}}^{\to }$ $\;=\;\frac{1}{2}({ {\text{AB}}}^{\to }-{ {\text{AD}}}^{\to }) \times { {\text{AC}}}^{\to }$ ${(\text{table} { {\text{AB}}}^{\to }=-{ {\text{i}}^{\^}}+{ {\text{j}}^{\^}}+4 { {\text{k}}^{\^}} ; { {\text{AD}}}^{\to }=-{ {\text{i}}^{\^}}-7 { {\text{j}}^{\^}}-8 { {\text{k}}^{\^}} ; { {\text{AC}}}^{\to }=-4 { {\text{i}}^{\^}}-4 { {\text{j}}^{\^}}+4 { {\text{k}}^{\^}})}$ $\;=\;\frac{1}{2}(8 { {\text{j}}^{\^}}+12 { {\text{k}}^{\^}}) \times (-4)({ {\text{i}}^{\^}}+{ {\text{j}}^{\^}}-{ {\text{k}}^{\^}})$ $\;=\;\frac{1}{2}{\begin{vmatrix}\hat{i} & { {\text{j}}^{\^}} & { {\text{k}}^{\^}} \\ 0 & 8 & 12 \\ 1 & 1 & -1\end{vmatrix}}$ $\;=(-2)(-20 { {\text{i}}^{\^}}+12 { {\text{j}}^{\^}}-8 { {\text{k}}^{\^}})$ $\;=8(5 { {\text{i}}^{\^}}-3 { {\text{j}}^{\^}}+2 { {\text{k}}^{\^}})$ $\; \therefore \;\text{ Area }\;=|{ {\text{v}}}^{\to }|=8 {\sqrt{25+9+4}}=8 {\sqrt{38}} \;\text{ Ans. }\;$ $\;$

Q65JEE Main 2023MCQ4MBinomial Theorem

$25^{190}-19^{190}-8^{190}+2^{190}$ is divisible by [8-Apr-2023 shift 2]

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📖 Explanation

$25^{190}-8^{190}$ is divisible by $25-8=17$ $19^{190}-2^{190}$ is divisible by $19-2=17$ $25^{190}-19^{190}$ is divisible by $25-19=6$ $8^{190}-2^{190}$ is divisible by $8-2=6$ L.C.M. of $1746=34$ $\therefore$ divisible by 34 but not by 14

Q66JEE Main 2023MCQ4MCircles

Let ${\text{O}}$ be the origin and OP and ${\text{OQ}}$ be the tangents to the circle ${\text{x}}^{2+} {\text{y}}^{2-}6 x+4 y+8=0$ at the points ${\text{P}}$ and ${\text{Q}}$ on it. If the circumcircle of the triangle OPQ passes through the point $(\alpha, \;\frac{1}{2})$ , then a value of $\alpha$ is. [8-Apr-2023 shift 2]

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📖 Explanation

Circumcircle of $\triangle {\text{OPQ}}$ $\;(x-0)(x-3)+(y-0)(y+2)=0$ $\;x^{2+}y^{2-}3 x+2 y=0$ passes through $(\alpha, \;\frac{1}{2})$ $\; \therefore \alpha^{2+}\;\frac{1}{4}-3 \alpha+1=0$ $\;\Rightarrow \alpha^{2-}3 \alpha+\;\frac{5}{4}=0 \Rightarrow 4 \alpha^{2-}12 \alpha+5=0$ $\;\Rightarrow 4 \alpha^{2-}10 \alpha-2 \alpha+5=0$ $\;(2 \alpha-1)(2 \alpha-5)=0 \therefore \alpha=\;\frac{1}{2}, \;\frac{5}{2} \;\text{ Ans. }\;$

Q67JEE Main 2023MCQ4MSequences and Series

Let ${\text{a}}_{ {\text{n}}}$ be the ${\text{n}}^{\;\text{th }\;}$ term of the series $5+8+14+23+35+50+. . .$ . and ${\text{Sn}}=sum_{{ {\text{k}}=1}}^{{ {\text{n}}}} {\text{a}}_K$ . Then ${\text{S}}_{30}- {\text{a}}_{40}$ is equal to [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{S}}_{ {\text{n}}}=5+8+14+23+35+50+. . .+ {\text{a}}_{ {\text{n}}}$ $\; {\text{S}}_{ {\text{n}}}=5+8+14+23+35+. . .+ {\text{a}}_{ {\text{n}}}$ ──────────────────────────── $\; {\text{O}}=5+3+6+9+12+15+. . .- {\text{a}}_{ {\text{n}}}$ $\; {\text{a}}_{ {\text{n}}}=5+(3+6+9+. . .( {\text{n}}-1) \;\text{ terms }\;)$ $\; {\text{a}}_{ {\text{n}}}=\;{3 {\text{n}}^{2-}3 {\text{n}}+10}/{2}$ $\; {\text{a}}_{40}=\;\frac{3(40)^{2-}3(40)+10}{2}=2345$ $\; {\text{S}}_{30}=\;{3 Σ_{{ {\text{n}}=1}}^{{30}} {\text{n}}^{2-}3 Σ_{{ {\text{n}}=1}}^{{30}} {\text{n}}+10 Σ_{{ {\text{n}}=1}}^{{30}} 1}/{2}$ $=\;\frac{\;\frac{3 \times 30 \times 31 \times 61}{6}-\frac{3 \times 30 \times 31}{2}+10 \times 30}{2}$ $\; {\text{S}}_{30}=13635$ $\; {\text{S}}_{30}- {\text{a}}_{40}=13635-2345$ $\;=11290(\;\text{ Option }\;(3))$

Q68JEE Main 2023MCQ4MLimits, Continuity and Differentiability

If $\alpha > \beta > 0$ are the roots of the equation $a x^{2+}b x+1=0$ , and $\lim _{x \to \;\frac{1}{\alpha}} (\;\frac{1-\cos (x^{2+}b x+a) }{2(1-a x)^2}) ^{\;\frac{1}{2}}=\;\frac{1}{k} (\;\frac{1}{\beta}-\;\frac{1}{\alpha})$ , then $k$ is equal to [8-Apr-2023 shift 2]

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📖 Explanation

$\; \therefore a x^{2+}b x+1=a(x-\alpha)(x-\beta) \therefore \alpha \beta=\;\frac{1}{a}$ $\; \therefore x^{2+}b x+a=a(1-\alpha x)(1-\beta x)$ $\; \therefore \lim _{x \to \;\frac{1}{\alpha}} \{\;\frac{1-\cos (x^{2+}b x+a) }{2(1-\alpha x)^2}\} ^{\;\frac{1}{2}}= \lim _{x \to \;\frac{1}{2}} \{\;\frac{1-\cos a(1-\alpha x)(1-\beta x)}{2\{a(1-\alpha x)(1-\beta x)\}^2} \cdot a^2(1-\beta x)^2\} ^{\;\frac{1}{2}}$ $\;\; = \$ [;\frac{1}{2} ⋅ ;\frac{1}{2} a^2 (1-;\frac{beta}{alpha}) ^2]$ ^{;\frac{1}{2}};; =;\frac{1}{2} ;\frac{1}{\alpha \beta} (1-;\frac{\beta}{\alpha}) =;\frac{1}{2} (;\frac{1}{\alpha \beta}-;\frac{1}{\alpha^2});; =;\frac{1}{2 \alpha} (;\frac{1}{\beta}-;\frac{1}{\alpha}) =;\frac{1}{k} (;\frac{1}{\beta}-;\frac{1}{\alpha}); \therefore k=2 \alpha ;\text{ Ans. };$

Q69JEE Main 2023MCQ4MPermutations and Combinations

If the number of words, with or without meaning, which can be made using all the letters of the word MATHEMATICS in which ${\text{C}}$ and ${\text{S}}$ do not come together, is $(6 !) {\text{k}}$ , is equal to [8-Apr-2023 shift 2]

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📖 Explanation

${\text{M}}_2 {\text{A}}_2 {\text{T}}_2 {\text{HEICS}}$ $=$ total words - when $C & S$ are together $\;\;\frac{⌊ 11}{⌊ 2⌊ 2⌊ 2}-\;\frac{⌊ 10}{⌊ 2⌊ 2≪corner 2} \times ⌊ 2$ $\;\;\frac{⌊x{10}}{⌊ 2⌊ 2≪corner 2} \times 9$ $\;=\;\frac{9 \times 10 \times 9 \times 8 \times 7}{8}⌊ 6$ $\;=5670⌊ 6$ $\;k=5670(\;\text{ Option 4) }\;$ $\;$

Q70JEE Main 2023MCQ4MMatrices and Determinants

Let $S$ be the set of all values of $\theta \in [-\pi, \pi]$ for which the system of linear equations $\;x+y+{\sqrt{3}} z=0$ $\;-x+(\tan \theta) y+{\sqrt{7}} z=0$ $\;x+y+(\tan \theta) z=0$ has non-trivial solution. Then $\;\frac{120}{\pi} sum_{\theta \in S} \theta$ is equal to [8-Apr-2023 shift 2]

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📖 Explanation

For non trivial solutions $\; {\text{D}}=0$ $\;{\begin{vmatrix}1 & 1 & {\sqrt{3}} \\ -1 & \tan \theta & {\sqrt{7}} \\ 1 & 1 & \tan \theta\end{vmatrix}}=0$ $\;\tan ^2 \theta-({\sqrt{3}}-1)-{\sqrt{3}}=0$ $\;\tan \theta={\sqrt{3}},-1$ $\;\theta= \{\;\frac{\pi}{3}, \;\frac{-2 \pi}{3}, \;\frac{-\pi}{4}, \;\frac{3 \pi}{4}\}$ $\;\;\frac{120}{\pi}(Σ \theta)=\;\frac{120}{\pi} \times \;\frac{\pi}{6}=20 \;\text{ (Option 1) }\;$

Q71JEE Main 2023MCQ4MThree Dimensional Geometry

For ${\text{a}}, {\text{b}} \in {\text{Z}}$ and $| {\text{a}}- {\text{b}}| \le 10$ , let the angle between the plane ${\text{P}}: {\text{ax}}+ {\text{y}}- {\text{z}}= {\text{b}}$ and the line $1: {\text{x}}-1= {\text{a}}- {\text{y}}= {\text{z}}$ +1 be $\cos ^{-1} (\;\frac{1}{3})$ . If the distance of the point $(6,-6,4)$ from the plane $P$ is $3 {\sqrt{6}}$ , then ${\text{a}}^{4+} {\text{b}}^2$ is equal to [8-Apr-2023 shift 2]

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📖 Explanation

$\;\theta=\cos ^{-1} \;\frac{1}{3} \therefore sin \;\theta=\sqrt{1-\;\frac{1}{9}}=\;{2 {\sqrt{2}}}/{3}$ $\; sin \;\theta=\;{a \cdot 1+1(-1)+(-1) \cdot 1}/{{\sqrt{a^{2+}1+1}} \cdot {\sqrt{3}}}=\;{2 {\sqrt{2}}}/{3}$ $\;\Rightarrow \{3(a-2)\}^2=24 (a^{2+}2)$ $\;\Rightarrow 3 (a^{2-}4 a+4) =8 a^{2+}16$ $\;\Rightarrow 5 a^{2+}12 a+4=0$ $\;\Rightarrow 5 a^{2+}10 a+2 a+4=0$ $\; \therefore a=-2, \;\frac{-2}{5} \because a \in z$ $\; \therefore a=-2$ Distance of $(6,-6,4)$ from $-2 {\text{x}}+ {\text{y}}- {\text{z}}- {\text{b}}=0$ is $3 {\sqrt{6}}$ $\; \therefore {| \;{-12-6-4-b}/{{\sqrt{4+1+1}}} |}=3 {\sqrt{6}}$ $\;\Rightarrow |b+22|=18 \therefore {\text{b}}=-40,-4$ $\; \because | {\text{a}}- {\text{b}}| \le 10$ $\; \therefore {\text{b}}=-4$ $\; \therefore {\text{a}}^{4+} {\text{b}}^2$ $\;=32 \;\text{ Ans. }\;$ $\;$

Q72JEE Main 2023MCQ4MVector Algebra

Let the vectors ${ {\text{u}}}^{\to }_1=\hat{i}+\hat{j}+a \hat{k}, \vec{\mu }_2=\hat{i}+b \hat{j}+\hat{k}$ and ${ {\text{u}}}^{\to }_3=c \hat{i}+\hat{j}+\hat{k}$ be coplanar. If the vectors ${ {\text{v}}}^{\to }_1=( {\text{a}}+ {\text{b}}) { {\text{i}}^{\^}}+ {\text{cj}}+ {\text{ck}}, { {\text{v}}}^{\to }_2= {\text{a}} { {\text{i}}^{\^}}+( {\text{b}}+ {\text{c}}) { {\text{j}}^{\^}}+ {\text{ak}} { {\text{k}}^{\^}}$ and ${ {\text{v}}}^{\to }_3=b \hat{i}+b \hat{j}+( {\text{c}}+ {\text{a}}) { {\text{k}}^{\^}}$ are also coplanar, then $6( {\text{a}}+ {\text{b}}+ {\text{c}})$ is equal to [8-Apr-2023 shift 2]

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📖 Explanation

${\begin{bmatrix}{ {\text{u}}}^{\to }_1 & { {\text{u}}}^{\to }_2 & { {\text{u}}}^{\to }_3\end{bmatrix}\$ }=0 ;; ∴ {\begin{vmatrix}1 & 1 & {\text{a}} \ 1 & {\text{b}} & 1 \ {\text{c}} & 1 & 1\end{vmatrix}}=0;\Rightarrow {\text{b}}-1+ {\text{c}}-1+ {\text{a}}(1- {\text{bc}})=0; ∴ {\text{abc}}= {\text{a}}+ {\text{b}}+ {\text{c}}-2{\begin{bmatrix}{ {\text{v}}}^{→ }_1 & { {\text{v}}}^{→ }_2 & { {\text{v}}}^{→ }_3\end{bmatrix}$}=0 ;; \therefore {\begin{vmatrix}{\text{a}}+ {\text{b}} & {\text{c}} & {\text{c}} \ {\text{a}} & {\text{b}}+ {\text{c}} & {\text{a}} \ {\text{b}} & {\text{b}} & {\text{c}}+ {\text{a}}\end{vmatrix}}=0;\Rightarrow -2 {\text{a}} ( {\text{ac}}- {\text{bc}}- {\text{c}}^2) +2 {\text{c}} ( {\text{a}}^{2+} {\text{ab}}- {\text{ac}}) =0;\Rightarrow -2 {\text{a}}^2 {\text{c}}+2 {\text{abc}}+2 {\text{ac}}^{2+}2 {\text{a}}^2 {\text{c}}+2 {\text{abc}}-2 {\text{ac}}^2=0;\Rightarrow 4 a b c=0 ;; \therefore a b c=0; \therefore {\text{a}}+ {\text{b}}+ {\text{c}}=2 ;; \therefore 6( {\text{a}}+ {\text{b}}+ {\text{c}})=12 ;\text{ Ans. };;$

Q73JEE Main 2023MCQ4MBinomial Theorem

The absolute difference of the coefficients of $x^{10}$ and $x^7$ in the expansion of $(2 x^{2+}\;\frac{1}{2 x}) ^{11}$ is equal to [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{T}}_{ {\text{r}}+1}={ }^{11} {\text{C}}_{ {\text{r}}} (2 {\text{x}}^2) ^{11- {\text{r}}} (\;{1}/{2 {\text{x}}}) ^{ {\text{r}}}$ $\;={ }^{11} {\text{C}}_{ {\text{r}}} 2{ }^{11-2 {\text{r}}} {\text{x}}^{22-3 {\text{r}}}$ $\;22-3 {\text{r}}=10 \;\; \;\text{ and }\; \;\; 22-3 {\text{r}}=7$ $\; {\text{r}}=4 \;\; \;\text{ and }\; \;\; {\text{r}}=5$ $\;\;\text{ Coefficient of }\; {\text{x}}^{10}={ }^{11} {\text{C}}_4 \cdot 2^3$ $\;\;\text{ Coefficient of }\; {\text{x}}^7={ }^{11} {\text{C}}_5 \cdot 2^1$ $\;\;\text{ difference }\;={ }^{11} {\text{C}}_4 \cdot 2^{3-}{ }^{11} {\text{C}}_5 \cdot 2$ $\;=\;\frac{11 \times 10 \times 9 \times 8}{24} \times 8-\;\frac{11 \times 10 \times 9 \times 8 \times 7}{120} \times 2$ $\;=11 \times 10 \times 3 \times 8-11 \times 3 \times 4 \times 7$ $\;=11 \times 3 \times 4 \times (20-7)$ $\;=11 \times 12 \times 13$ $\;=12(12-1)(12+1)$ $\;=12 (12^{2-}1)$ $\;=12^{3-}12(\;\text{ Option }\; 3)$

Q74JEE Main 2023MCQ4MSets and Relations

Let ${\text{A}}=\{1,2,3,4,5,6,7\}$ . Then the relation ${\text{R}}=\{( {\text{x}}, {\text{y}}) \in {\text{A}} \times {\text{A}}: {\text{x}}+ {\text{y}}=7\}$ is [8-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$R=\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}$

Q75JEE Main 2023MCQ4MProbability

If the probability that the random variable $X$ takes values $x$ is given by $P(X=x)=k(x+1) 3^{-x}, x=0,1,2,3, . . .$ , where $k$ is a constant, then $P(X \ge 2)$ is equal to [8-Apr-2023 shift 2]

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📖 Explanation

$\;sum_{{ {\text{x}}=0}}^{{\infty }} {\text{P}}( {\text{X}}= {\text{x}})=1$ $\; {\text{k}} (1+2 \cdot 3^{-1}+3 \cdot 3^{-2}+4 \cdot 3^{-3}+. . . \infty ) =1$ Let $\;\; {\ s}=1+\;\frac{2}{3}+\;\frac{3}{3^2}+\;\frac{4}{3^3}+. . . \infty$ $\;\;\;\;\;\frac{s}{3}=\;\frac{1}{3}+\;\frac{2}{3^2}+\;\frac{3}{3^2}+. . . \infty$ ───────────────────── $\;\;\;\;\frac{2 s}{3}=1+\;\frac{1}{3}+\;\frac{1}{3^2}+\;\frac{1}{3^2}+. . . \infty$ $\;\;\frac{2 s}{3}-\;\frac{1}{1-\;\frac{1}{3}}=\;\frac{3}{2}$ $s=\;\frac{9}{4}$ so ${\text{k}}=\;\frac{4}{9}$ ${\text{P}}( {\text{X}} \ge 2)\;=1- {\text{P}}( {\text{x}}=0)- {\text{P}}( {\text{x}}=1)$ $\;=1-\;\frac{4}{9} (1+\;\frac{2}{3})$ $\;=\;\frac{7}{27}(\;\text{ Option }\; 1)$

Q76JEE Main 2023MCQ4MIndefinite Integrals

The integral $\int ( (\;\frac{x}{2}) ^x+ (\;\frac{2}{x}) ^x) \log _2 x \text{dx}$ is equal to [8-Apr-2023 shift 2]

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📖 Explanation

$\;\int (x^x 2^{-x}+2^x x^{-x}) \log _2^x \text{dx}$ $\;\int (e^{x \ln x} \cdot e^{-x \ln 2}+e^{x \ln 2} \cdot e^{-x \ln x}) \text{dx}$ $\int (e^{x \ln x-x \ln 2}+e^{x \ln 2-x \ln x}) \;\frac{\ln x}{\ln 2} \text{dx}$ $\;\text{ let }\; \;\; x \ln x-x \ln 2=t$ $(\ln x+1-\ln 2) \text{dx}=d t$

Q77JEE Main 2023MCQ4MTrigonometric Ratios and Identities

The value of $36 (4 \cos ^2 9^{\circ}-1) (4 \cos ^2 27^{\circ}-1) (4 \cos ^2 81^{\circ}-1) (4 \cos ^2 243^{\circ}-1)$ is [8-Apr-2023 shift 2]

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📖 Explanation

$4 \cos ^2 \theta-1=4 (1- sin \;^2 \theta) -1=3-4 sin \;^2 \theta=\;\frac{ sin \;3 \theta}{ sin \;\theta}$ so given expression can be written as $\;36 \times \;\frac{ sin \;27^{\circ}}{ sin \;9^{\circ}} \times \;\frac{ sin \;81^{\circ}}{ sin \;27^{\circ}} \times \;\frac{ sin \;243^{\circ}}{ sin \;81^{\circ}} \times \;\frac{ sin \;729^{\circ}}{ sin \;243^{\circ}}$ $\;36 \times \;\frac{ sin \;729^{\circ}}{ sin \;9^{\circ}}=36$

Q78JEE Main 2023MCQ4MParabola

Let ${\text{A}}(0,1), {\text{B}}(1,1)$ and ${\text{C}}(1,0)$ be the mid-points of the sides of a triangle with incentre at the point ${\text{D}}$ . If the focus of the parabola $y^2=4 {\text{ax}}$ passing through $D$ is $(\alpha+\beta {\sqrt{3}}, 0)$ , where $\alpha$ and $\beta$ are rational numbers, then $\;\frac{\alpha}{\beta^2}$ is equal to [8-Apr-2023 shift 2]

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📖 Explanation

${\text{a}}= {\text{OP}}=2 \;\; {\text{b}}= {\text{OQ}}=2 \;\; {\text{c}}= {\text{PQ}}=2 {\sqrt{2}}$ $\; {\text{D}} (\;{4}/{2+2+2 {\sqrt{2}}}, \;{4}/{2+2+2 {\sqrt{2}}}) \equiv {\text{D}} (\;{2}/{2+{\sqrt{2}}}, \;{2}/{2+{\sqrt{2}}})$ $\; {\text{y}}^2=4 {\text{ax}} \Rightarrow (\;{2}/{2+{\sqrt{2}}}) ^2=4 {\text{a}} \cdot (\;{2}/{2+{\sqrt{2}}})$ $\; \therefore 4 {\text{a}}=\;{2}/{2+{\sqrt{2}}} \therefore {\text{a}}=\;\frac{1}{2} \cdot \;{2-{\sqrt{2}}}/{4-2}=\;\frac{1}{4}(2-{\sqrt{2}})$ $\; \therefore \alpha=\;\frac{2}{4}=\;\frac{1}{2} \;\; \beta=\;\frac{-1}{4}$ $\; \therefore \;\frac{\alpha}{\beta^2}=8 \;\text{ Ans. }\;$

Q79JEE Main 2023MCQ4MSets and Relations

The negation of $( {\text{p}} ∧(∼ {\text{q}})) ∨(∼ {\text{p}})$ is equivalent to [8-Apr-2023 shift 2]

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Q80JEE Main 2023MCQ4MStatistics

Let the mean and variance of 12 observations be $\;\frac{9}{2}$ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is $\;{ {\text{m}}}/{ {\text{n}}}$ , where $m$ and $n$ are coprime, then $m+n$ is equal to [8-Apr-2023 shift 2]

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📖 Explanation

$\;\;{Σ {\text{x}}}/{12}=\;\frac{9}{2}$ $\;\sum {\text{x}}=54$ $\;\;{Σ {\text{x}}^2}/{12}- (\;\frac{9}{2}) ^2=4$ $\;\sum {\text{x}}^2=291$ $\;\sum {\text{x}}_{\;\text{new }\;}=54-(9+10)+7+14=56$ $\;\sum {\text{x}}_{\;\text{new }\;}^2=291-(81+100)+49+196=355$ $\;\sigma_{\;\text{new }\;}^2=\;\frac{355}{12}- (\;\frac{56}{12}) ^2$ $\;\sigma_{\;\text{new }\;}^2=\;\frac{281}{36}=\;{ {\text{m}}}/{ {\text{n}}}$ $\; {\text{m}}+ {\text{n}}=317 \;\text{ Option (2) }\;$

Q81JEE Main 2023NAT4MPermutations and Combinations

Let ${\text{R}}=\{ {\text{a}}, {\text{b}}, {\text{c}}, {\text{d}}, {\text{e}}\}$ and ${\text{S}}=\{1,2,3,4\}$ . Total number of onto functions ${\text{f}}: {\text{R}} \to {\text{S}}$ such that ${\text{f}}( {\text{a}}) \ne 1$ , is equal to _______. [8-Apr-2023 shift 2]

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📖 Explanation

Total onto function $\;\frac{⨆ 5}{⌊ 3⌊} \times ⌊ 4=240$ Now when $f(a)=1$ $⌊ 4+\;\frac{⌊ 4}{⌊ 2⌊ 2} \times ⌊x{3}=24+36=60.$ so required ${\text{f}}^{ {\text{n}}}=240-60=180$

Q82JEE Main 2023NAT4MQuadratic Equation and Inequalities

Let $m$ and $n$ be the numbers of real roots of the quadratic equations $x^{2-}12 x+[x]+31=0$ and $x^{2-}5|x+2|-4$ $=0$ respectively, where $[x]$ denotes the greatest integer $\leq x$ . Then $m^{2+}m n+n^2$ is equal to [8-Apr-2023 shift 2]

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📖 Explanation

$\;x^{2-}12 x+[x]+31=0$ $\;\{x\}=x^{2-}11 x+31$ $\;0 \le x^{2-}11 x+31< 1$ $\;x^{2-}11 x+30n< 0$ $\;x \in (5,6)$ $\;\;\text{ so }\; \;\; [x]=5$ $\;x^{2-}12 x+5+31=0$ $\;x^{2-}12 x+36=0$ ${\text{x}}=6\;\;\text{ but }\; {\text{x}} \in (5,6)$ $\;\text{ so }\;\; {\text{x}} \in \varphi$ $\; {\text{m}}=0$ $\; {\text{x}}=\{7,-2,-3\}$ ${\text{n}}=3$ ${\text{m}^2}+ {mn}+ {\text{n}^2}= {\text{n}^2}=9$

Q83JEE Main 2023NAT4MThree Dimensional Geometry

Let $P_1$ be the plane $3 x-y-7 z=11$ and $P_2$ be the plane passing through the points $(2,-1,0),(2,0,-1)$ , and $(5,1$ , 1). If the foot of the perpendicular drawn from the point $(7,4,-1)$ on the line of intersection of the planes $P_1$ and $P_2$ is $(\alpha, \beta, \gamma )$ , then $\alpha+\beta+\gamma$ is equal to _______. [8-Apr-2023 shift 2]

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📖 Explanation

${\text{P}}_2$ is given by $\;{\begin{vmatrix}x-5 & y-1 & z-1 \\ 3 & 2 & 1 \\ 3 & 1 & 2\end{vmatrix}}=0$ $\;x-y-z=3$ DR of line intersection of ${\text{P}}_1 & {\text{P}}_2$ $\;{\begin{vmatrix}i & j & k \\ 1 & -1 & 1 \\ 3 & -1 & -7\end{vmatrix}}$ $\;+6 \hat{i}+4 \hat{j}+2 \hat{k}$ Let $z=0, \;\; x-y=3$ $3 x-y=112 x=8$ $x=4$ $y=1$ So Line is $\; \;\;\;\; \;\frac{x-4}{6}=\;\frac{y-1}{4}=\;\frac{z-0}{2}=r$ $\;(\alpha, \beta, \gamma )=(6 r+4,4 {\text{r}}+1,2 {\text{r}})$ $\;6(\alpha-7)+4(\beta-4)+2(\gamma +1)=0$ $\;6 \alpha-42+4 \beta-16+2 \gamma +2=0$ $\;36 {\text{r}}+24+16 {\text{r}}+4+4 {\text{r}}-56=0$ $\;56 {\text{r}}=28$ $\; {\text{r}}=\;\frac{1}{2}$ $\alpha+\beta+\gamma =12 r+5$ $\;\; =6+5=11$

Q84JEE Main 2023NAT4MFunctions

If domain of the function $\log _e (\;\frac{6 x^{2+}5 x+1}{2 x-1}) +\cos ^{-1} (\;\frac{2 x^{2-}3 x+4}{3 x-5})$ is $(\alpha, \beta) ∪(\gamma , \delta ]$ , then, $18 (\alpha^{2+}\beta^{2+}\gamma ^{2+}\delta ^2)$ is equal to [8-Apr-2023 shift 2]

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📖 Explanation

$\;\;\frac{6 x^{2+}5 x+1}{2 x-1} > 0$ $\;\;{(3 {\text{x}}+1)(2 {\text{x}}+1)}/{2 {\text{x}}-1} > 0$ $\; {\text{x}} \in [\;{-1}/{{\sqrt{2}}}, \;{1}/{{\sqrt{2}}}] ∪ (\;\frac{5}{3}, \infty )$ . . . (B) $\;x< \;\frac{5}{3}$ . . . (C) $\; {\text{A}} ∩ {\text{B}} ∩ {\text{C}} \equiv (\;\frac{-1}{2}, \;\frac{-1}{3}) ∪ (\;\frac{1}{2}, \;{1}/{{\sqrt{2}}}]$ $\;\;\text{ So }\; 18 (\alpha^{2+}\beta^{2+}\gamma ^{2+}\delta ^2) =18 (\;\frac{1}{4}+\;\frac{1}{9}+\;\frac{1}{4}+\;\frac{1}{2})$ $\;=18+2=20$

Q85JEE Main 2023NAT4MArea Under The Curves

Let the area enclosed by the lines $x+y=2, y=0 x=0$ and the curve $f(x)=\min \{x^{2+}\;\frac{3}{4}, 1+[x]\}$ where $[x]$ denotes the greatest integer $\le {\text{x}}$ , be ${\text{A}}$ . Then the value of $12 {\text{A}}$ is _______. [8-Apr-2023 shift 2]

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📖 Explanation

$\;int_{0}^{\;\frac{1}{2}} ( {\text{x}}^{2+}\;\frac{3}{4}) {\text{dx}}+\;\frac{1}{2} \times (\;\frac{3}{2}+\;\frac{1}{2}) \times 1$ $\;= \$ [;{ {\text{x}}^3}/{3}+;{3 {\text{x}}}/{4}]$ _0^{;\frac{1}{2}}+1; {\text{A}}=;\frac{1}{24}+;\frac{3}{8}+1;12 {\text{A}}=;\frac{1}{2}+;\frac{36}{8}+12;=;\frac{1}{2}+;\frac{9}{2}+12;=5+12;=17$

Q86JEE Main 2023NAT4MSequences and Series

Let $0 < z < y < x$ be three real numbers such that $\;\frac{1}{x}, \;\frac{1}{y}, \;\frac{1}{z}$ are in an arithmetic progression and $x, {\sqrt{2}} y, z$ are in a geometric progression. If $x y+y z+z x=\;{3}/{{\sqrt{2}}} x y z$ , then $3(x+y+z)^2$ is equal to _______. [8-Apr-2023 shift 2]

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📖 Explanation

$\;\;\frac{2}{y}=\;\frac{1}{x}+\;\frac{1}{z}$ $\;2 y^2=x z$ $\;\;\frac{2}{y}=\;\frac{x+z}{x z}=\;\frac{x+z}{2 y^2}$ $\;x+z=4 y$ $\;x y+y z+z x=\;{3}/{{\sqrt{2}}} x y z$ $\;y(x+z)+z x=\;{3}/{{\sqrt{2}}} x z \cdot y$ $\;4 y^{2+}2 y^2=\;{3}/{{\sqrt{2}}} y \cdot 2 y^2$ $\;6 y^2=3 {\sqrt{2}} y^3$ $\;y={\sqrt{2}}$ $\;x+y+z=5 y=5 {\sqrt{2}}$ $\;3(x+y+z)^2=3 \times 50=150$

Q87JEE Main 2023NAT4MDifferential Equations

Let the solution curve $x=x(y), 0< y< \;\frac{\pi}{2}$ , of the differential equation $(\log _e(\cos y)) ^2 \cos y \text{dx}- (1+3 x \log _e.$ (cos $y)) sin \;y d y=0$ satisfy $x (\;\frac{\pi}{3}) =\;{1}/{2 \log _{ {\text{c}}} 2}$ . If $x (\;\frac{\pi}{6}) =\;{1}/{\log _{ {\text{e}}} m-\log _{ {\text{c}}} n}$ , where $m$ and $n$ are co-prime, then $m n$ is equal to [8-Apr-2023 shift 2]

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📖 Explanation

${\text{Cos}} y \ln ^2 \cos y \text{dx}=(1+3 x \ln \cos y) sin \;y d y$ $\;\;\frac{d x}{d y}=\tan y (\;\frac{3 x}{\ln \cos y}+\;\frac{1}{\ln ^2 \cos y})$ $\;\;\frac{d x}{d y}- (\;\frac{3 \tan y}{\ln \cos y}) x=\;\frac{\tan y}{\ln ^2 \cos y}$ $\;\;\text{ If }\;=e^{-\;\frac{3}{\ln \cos y \cos y} {\text{dy}}}$ $\;\;\text{ ln cosy }\;= {\text{t}}$ $\;\;\frac{1}{\cos y}- sin \;y d y=d t$ $\;\;\text{ If }\;= {\text{e}}^{\;\frac{3}{t} \text{dt}}=e^{3 \ln t}=t^3=\ln ^3 \cos y$ $\;\;\text{ solution is }\; x \cdot \ln ^3 \cos y= \{\;\frac{ sin \;y}{\cos y} \cdot \ln \cos y d y+C.$ $\; \;\;\;\; x \ln ^3 \cos y=\;\frac{-\ln ^2 \cos y}{2}+C$ $\;x (\;\frac{\pi}{3}) =\;\frac{1}{2 \ln 2} {\ so} \;\frac{1}{2 \ln ^2} \times \ln ^3 (\;\frac{1}{2}) =-\;\frac{\ln ^3 (\frac{1}{2}) }{2}+C$ $\;\;\text{ C }\;=0 \;\; x \;\; \ln ^3 \;{{\sqrt{3}}}/{2}=-\;\frac{1}{2} \ln ^2 \;{{\sqrt{3}}}/{2}+0$ $\;y=\;\frac{\pi}{6} \;\;$ $\;x=-\;{1}/{2 \ln (\;{{\sqrt{3}}}/{2}) }$ $\;x=\;\frac{1}{\ln \;\frac{4}{3}}=\;\frac{1}{\ln 4-\ln 3}$ $\;m n=12$

Q88JEE Main 2023NAT4MDefinite Integrals

Let $[t]$ denote the greatest integer function. If $int_{0}^{2.4} \$ [x^2]$ {\text{dx}}=\alpha+\beta {\sqrt{2}}+\gamma {\sqrt{3}}+\delta {\sqrt{5}} $, then$ \alpha+\beta+\gamma +\delta $ is equal to _______. [8-Apr-2023 shift 2]

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📖 Explanation

$\;int_{0}^{1} 0 {\text{dx}}+int_{1}^{{\sqrt{2}}} 1 {\text{dx}}+int_{{\sqrt{2}}}^{{\sqrt{3}}} 2 {\text{dx}}+int_{{{\sqrt{3}}}}^{2} 3 {\text{dx}}+int_{2}^{{\sqrt{5}}} 4 {\text{dx}}+int_{{\sqrt{5}}}^{2 \cdot 4} 5 {\text{dx}}$ $\;{\sqrt{2}}-1+2({\sqrt{3}}-{\sqrt{2}})+3(2-{\sqrt{3}})+4({\sqrt{5}}-2)+5((2 \cdot 4)-{\sqrt{5}})$ $\;=9-{\sqrt{2}}-{\sqrt{3}}-{\sqrt{5}}$ $\;\alpha+\beta+\gamma +\delta =9-1-1-1=6$

Q89JEE Main 2023NAT4MParabola

The ordinates of the points ${\text{P}}$ and ${\text{Q}}$ on the parabola with focus (3.0) and directrix ${\text{x}}=-3$ are in the ratio $3: 1:$ . If ${\text{R}}(\alpha, \beta)$ is the point of intersection of the tangents to the parabola at $P$ and ${\text{Q}}$ , then $\;\frac{\beta^2}{\alpha}$ is equal to ________. [8-Apr-2023 shift 2]

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📖 Explanation

Parabola is ${\text{y}}^2=12 {\text{x}}$ Let ${\text{Q}} (3 {\text{t}}^2, 6 {\text{t}})$ so ${\text{P}} (27 {\text{t}}^2, 18 {\text{t}})$ ${\text{R}}(\alpha, \beta)\;= ( {\text{at}}_1 {\text{t}}_2, {\text{a}} ( {\text{t}}_1+ {\text{t}}_2) )$ $\;=(3 {\text{t}} \cdot 3 {\text{t}}, 3( {\text{t}}+3 {\text{t}}))$ ${\text{R}}(\alpha, \beta)\;= (9 {\text{t}}^2, 12 {\text{t}})$ ${\text{R}}(\alpha, \beta)= (9 t^2, 12 t)$ $\;\frac{\beta^2}{\alpha}=\;{(12 {\text{t}})^2}/{9 {\text{t}}^2}=\;\frac{144}{9}=16$

Q90JEE Main 2023NAT4MLimits, Continuity and Differentiability

Let $k$ and $m$ be positive real numbers such that the function $f(x)={\begin{cases}3 x^{2+}k {\sqrt{x+1}} & 0 < x < 1 \\ m x^{2+}k^2 & \ \\ \ \\ x \ge 1\}}$ is differentiable for all $x>0$ \end{cases} Then $\;\frac{8 f^{'}(8)}{f^{'} (\;\frac{1}{8}) }$ is equal to ________. [8-Apr-2023 shift 2]

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📖 Explanation

function is differentiable $∀ x < 0$ so $\; {\text{f}} (1^{-}) = {\text{f}}(1)$ . . . (1) $\;3+{\sqrt{2}} {\text{k}}= {\text{m}}+ {\text{k}}^2$ and $f_{+}^1 (1^{-}) =f_{-}^1 (1^{+})$ $\;2 {\text{m}}=6+\;{ {\text{k}}}/{2 {\sqrt{2}}}$ $\; {\text{m}}=3+\;{ {\text{k}}}/{4 {\sqrt{2}}}$ . . . (2) ${\text{k}}^{2+}3+\;{ {\text{k}}}/{4 {\sqrt{2}}}=3+{\sqrt{2}} {\text{k}}$ $\; {\text{k}}=\;{7}/{4 {\sqrt{2}}}, 0$ $\; {\text{m}}=3+\;\frac{7}{32}$ ${\text{m}}=\;\frac{103}{32}$ $\;=\;\frac{8 \times 2 \times 8 \times \frac{103}{32}}{\frac{16}{12}}$ $\;=103 \times 3=309$

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