Previous Year Paper

26 Jul 2022 Shift 2

90 Questions360 Marks180 MinTake as Timed Mock →

🎯 Practice smarter, not harder

Practice 26 Jul 2022 Shift 2 questions with full analytics — free

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

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

🚀 Start Practice Mode Sign in free →

Physics30 questions

Q1JEE Main 2022MCQ4MMotion in a Plane

Two projectiles are thrown with same initial velocity making an angle of $45^{\circ}$ and $30^{\circ}$ with the horizontal respectively. The ratio of their respective ranges will be : [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Let projection speed is $u$ ${\text{R}}_1=\;{ {\text{u}}^2 {\text{Sin}}(90^{\circ})}/{g} ; {\text{R}}_2=\;{ {\text{u}}^2 \sin (60^{\circ})}/{g}$ $\;{ {\text{R}}_1}/{ {\text{R}}_2}=\;\frac{2}{\sqrt{3}}$

Q2JEE Main 2022MCQ4MUnits and Measurements

In a Vernier Calipers, 10 divisions of Vernier scale is equal to the 9 divisions of main scale. When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and $4^{\;\text{th }\;}$ Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to $1 {\text{mm}}$ . While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and $6^{\;\text{th }\;}$ Vernier scale division exactly coincides with the main scale reading. The diameter of the spherical body will be : [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\text{ 1 M.S.D }\;=1 {\text{mm}}$ $\;\text{ 9 M.S.D }\;=10 {\text{V}} . S . D$ $\;\text{ 1 V.S.D }\;=0.9 \;\text{ M.S.D }\;=0.9 {\text{mm}}$ $\;\text{ L.C of vernier caliper }\;=1-0.9=0.1 {\text{mm}}=0.01 {\text{cm}}$ $\;\text{ zero error }\;=(10-4) \times 0.1 {\text{mm}}=-0.6 {\text{mm}}$ $\;\text{ Reading }\;=\;\text{ M.S.R }\;+\;\text{ V.S.R }\;-\;\text{ Zero error }\;$ $=3 {\text{cm}}+6 \times 0.01-[-0.06]$ $=3+0.06+0.06$ $=3.12 {\text{cm}}$ Nearest given answer in the options is $3.10$

Q3JEE Main 2022MCQ4MCentre of Mass and Collision

A ball of mass $0.15 {\text{kg}}$ hits the wall with its initial speed of $12 {\text{ms}}^{-1}$ and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is $100 {\text{N}}$ , calculate the time duration of the contact of ball with the wall. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${ \vec{\text{P}}_{ {\text{i}}}=0.15 \times 12({ {\text{i}}^{\^})$ ${ \vec{\text{P}}_{ {\text{f}}}=0.15 \times 12(-{ {\text{i}}^{\^})$ $|{\Delta \vec{\text{P}}|=3.6 {\text{kg}}- {\text{m}} / {\text{s}}$ $3.6= {\text{F}} \Delta {\text{t}}$ $3.6=100 \Delta {\text{t}}$ $\Delta {\text{t}}=0.036 {\sec}$

Q4JEE Main 2022MCQ4MWork, Power and Energy

A body of mass $8 {\text{kg}}$ and another of mass $2 {\text{kg}}$ are moving with equal kinetic energy. The ratio of their respective momentum will be : [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{K}} . {\text{E}}=\;{ {\text{P}}^2}/{2 {\text{m}}}$ ${\text{K}}_1=\;{ {\text{P}}_1^2}/{2(8)} ; {\text{K}}_2=\;{ {\text{P}}_2^2}/{2(2)}$ ${\text{K}}_1= {\text{K}}_2$ ${\text{So}}_0$ $4 {\text{P}}_2^2= {\text{P}}_1^2$ $\;{ {\text{P}}_1}/{ {\text{P}}_2}=2$

Q5JEE Main 2022MCQ4MElectrostatic Potential and Capacitance

Two uniformly charged spherical conductors $A$ and $B$ of radii $5 {\text{mm}}$ and $10 {\text{mm}}$ are separated by a distance of $2 {\text{cm}}$ . If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitudes of the electric fields at the surface of the sphere $A$ and $B$ will be : [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{V}}_{ {\text{A}}}= {\text{V}}_{ {\text{B}}}$ $\;{ {\text{KQ}}_{ {\text{A}}}}/{ {\text{R}}_{ {\text{A}}}}=\;{ {\text{KQ}}_{ {\text{B}}}}/{ {\text{R}}_{ {\text{B}}}}$ $\;{ {\text{Q}}_{ {\text{A}}}}/{ {\text{Q}}_{ {\text{B}}}}=\;{ {\text{R}}_{ {\text{A}}}}/{ {\text{R}}_{ {\text{B}}}}=\;\frac{1}{2}$ ${\text{E}}_{ {\text{A}}}=\;{ {\text{KQ}}_{ {\text{A}}}}/{ {\text{R}}_{ {\text{A}}}^2} ; {\text{E}}_{ {\text{B}}}=\;{ {\text{KQ}}_{ {\text{B}}}}/{ {\text{R}}_{ {\text{B}}}{ }^2}$ $\;{ {\text{E}}_{ {\text{A}}}}/{ {\text{E}}_{ {\text{B}}}}=\;{ {\text{Q}}_{ {\text{A}}}}/{ {\text{Q}}_{ {\text{B}}}} \times \;{ {\text{R}}_{ {\text{B}}}/{ }^2}{ {\text{R}}_{ {\text{A}}}{ }^2}=\;{ {\text{R}}_{ {\text{B}}}}/{ {\text{R}}_{ {\text{A}}}}=\;\frac{2}{1}$

Q6JEE Main 2022MCQ4MElectromagnetic Waves

The oscillating magnetic field in a plane electromagnetic wave is given by $B_y=5 \times 10^{-6} \sin 1000 \pi(5 x-4 \times 10^8 t) T$ . The amplitude of electric field will be : [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{B}}_0=5 \times 10^{-6}$ ${\text{v}}=$ Speed of wave $=\;\frac{4 \times 10^8}{5}=8 \times 10^7[ \therefore {\text{v}}=\;{ {\text{W}}}/{ {\text{k}}}]$ ${\text{E}}_0= {\text{vBB}}_0=40 \times 10^1$ $=4 \times 10^2 {\text{V}} / {\text{m}}$

Q7JEE Main 2022MCQ4MRay Optics and Optical Instruments

Light travels in two media $M_1$ and $M_2$ with speeds $1.5 \times 10^8 {\text{ms}}^{-1}$ and $2.0 \times 10^8 {\text{ms}}^{-1}$ respectively. The critical angle between them is : [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{v}}=\;{ {\text{c}}}/{ {\text{n}}}$ ${\text{n}}_{ {\text{d}}} \sin {\text{i}}_{ {\text{c}}}= {\text{n}}_{ {\text{r}}} \sin 90^{\circ}$ $\sin {\text{i}}_{ {\text{c}}}=\;{ {\text{n}}_{ {\text{r}}}}/{ {\text{n}}_{ {\text{d}}}}=\;{ {\text{V}}_{ {\text{d}}}}/{ {\text{V}}_{ {\text{r}}}}$ $\sin i_c=\;\frac{1.5 \times 10^8}{2 \times 10^8}=\;\frac{1.5}{2}$ $\sin i_c=\;\frac{3}{4}$ $\tan i_c=\;\frac{3}{\sqrt{4^{2-}3^2}} \Rightarrow \;\frac{3}{\sqrt{7}}$ $i_c=\tan ^{-1}(\;\frac{3}{\sqrt{7}})$

Q8JEE Main 2022MCQ4MGravitation

A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be : (Take radius of earth $=6400 {\text{km}}$ and ${\text{g}}=10 {\text{ms}}^{-2}$ ) [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{v}}_{ {\text{e}}}=\sqrt{\;{2 {\text{Gm}}}/{ {\text{R}}}}$ $\;{- {\text{GMm}}}/{ {\text{R}}}+\;\frac{1}{2} {\text{m}} \;{ {\text{V}}_{ {\text{e}}}/{ }^2}{9}=-\;{ {\text{GMm}}}/{ {\text{R}}+ {\text{h}}}$ $\;{ {\text{GM}}}/{ {\text{R}}+ {\text{h}}}=\;{ {\text{GM}}}/{ {\text{R}}}-\;{ {\text{V}}_{ {\text{e}}}^2}/{18}$ $\;{ {\text{GM}}}/{ {\text{R}}+ {\text{h}}}=\;{ {\text{GM}}}/{ {\text{R}}}-\;{ {\text{GM}}}/{9 {\text{R}}}$ $\;{ {\text{GM}}}/{ {\text{R}}+ {\text{h}}}=\;{8 {\text{GM}}}/{9 {\text{R}}}$ $\;{1}/{ {\text{R}}+ {\text{h}}}=\;{8}/{9 {\text{R}}}$ $9 {\text{R}}=8 {\text{R}}+8 {\text{h}}$ ${\text{h}}=\;{ {\text{R}}}/{8} \Rightarrow \;\frac{6400}{8} \Rightarrow 800 {\text{km}}$

Q9JEE Main 2022MCQ4MElectromagnetic Waves

The maximum and minimum voltage of an amplitude modulated signal are $60 {\text{V}}$ and $20 {\text{V}}$ respectively. The percentage modulation index will be : [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{V}}_{\max }=60$ ${\text{V}}_{\min }=20$ $\%$ modulation $=$ $(\;{ {\text{V}}_{\max }- {\text{V}}_{\min }}/{ {\text{V}}_{\max }+ {\text{V}}_{\min }}) 100 \Rightarrow (\;\frac{60-20}{60+20}) 100 \Rightarrow (\;\frac{40}{80}) 100$ $\Rightarrow 50 \%$

Q10JEE Main 2022MCQ4MDual Nature of Matter and Radiation

A nucleus of mass $M$ at rest splits into two parts having masses $\;\frac{M^{'}}{3}$ and $\;\frac{2 M^{'}}{3}(M^{'} < M)$ . The ratio of de Broglie wavelength of two parts will be : [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$|{ \vec{\text{P}}_1|=|{ \vec{\text{P}}_2|$ Here ${ \vec{\text{P}}$ is momentumSo $\lambda =\;\frac{h}{P}$ Hence both will have same de broglie wavelength.

Q11JEE Main 2022MCQ4MThermodynamics

An ice cube of dimensions $60 {\text{cm}} \times 50 {\text{cm}} \times 20 {\text{cm}}$ is placed in an insulation box of wall thickness $1 {\text{cm}}$ . The box keeping the ice cube at $0^{\circ} {\text{C}}$ of temperature is brought to a room of temperature $40^{\circ} {\text{C}}$ The rate of melting of ice is approximately : (Latent heat of fusion of ice is $3.4 \times 10^5 {\text{J}} {\text{kg}}^{-1}$ and thermal conducting of insulation wall is $.0.05 {\text{Wm}}^{-1 \circ } {\text{C}}^{-1})$ [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;{ {\text{dQ}}}/{ {\text{dt}}}=\;{ {\text{KA}} \Delta {\text{T}}}/{ℓ}$ $A=2(0.6 \times 0.5+0.5 \times 0.2+0.2 \times 0.6)$ $=2(0.3+0.1+0.12)$ $=2(0.4+0.12)$ $=2(0.52)$ $=1.04 {\text{m}}^2$ ${\text{R}}_{ {\text{th}}}=\;{ℓ}/{ {\text{KA}}} \Rightarrow \;\frac{1 \times 10^{-2}}{0.05 \times 1.04} \Rightarrow \;\frac{10^{-2}}{0.052}$ $\;{ {\text{dQ}}}/{ {\text{dt}}}=\;{\Delta {\text{T}}}/{ {\text{R}}_{ {\text{th}}}} \Rightarrow \;\frac{40 \times 0.052}{10^{-2}} \Rightarrow 2.08 \times 10^2 {\text{J}} / {\text{s}}$ $2.08 \times 10^2= {\text{m}} \times 3.4 \times 10^5$ ${\text{m}}=\;\frac{2.08}{3.4 \times 10^3} \Rightarrow 0.61 \times 10^{-3} {\text{kg}} / {\text{s}}$ $=61 \times 10^{-5} {\text{Kg}} / {\text{s}}$

Q12JEE Main 2022MCQ4MThermodynamics

A gas has $n$ degrees of freedom. The ratio of specific heat of gas at constant volume to the specific heat of gas at constant pressure will be : [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{C}}_{ {\text{v}}}=\;{ {\text{nR}}}/{2} \;\; {\text{C}}_{ {\text{p}}}=\;{( {\text{n}}+2) {\text{R}}}/{2}$ $\;{ {\text{C}}_{ {\text{v}}}}/{ {\text{C}}_{ {\text{p}}}}=\;{ {\text{n}}}/{ {\text{n}}+2}$

Q13JEE Main 2022MCQ4MWaves

A transverse wave is represented by $y=2 \sin (\omega t-k x) {\text{cm}}$ . The value of wavelength (in ${\text{cm}}$ ) for which the wave velocity becomes equal to the maximum particle velocity, will be : [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$y=2 \sin (\omega t-k x)$ Maximum particle velocity $= {\text{A}} \omega$ Wave velocity $=\;{\omega }/{ {\text{k}}}$ $\;{\omega }/{ {\text{k}}}= {\text{A}} \omega$ ${\text{k}}=\;{1}/{ {\text{A}}}=\;\frac{2 \pi}{\lambda }$ $\lambda =2 \pi {\text{A}}$ $=4 \pi {\text{cm}}$

Q14JEE Main 2022MCQ4MCurrent Electricity

A battery of $6 {\text{V}}$ is connected to the circuit as shown below. The current I drawn from the battery is: [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Balanced wheat stone bridge in circuit so there is no current in $5 Ω$ resistor so it can be removed from the circuit. ${\text{R}}_{ {\text{eq}}}=\;\frac{6 \times 12}{6+12}+2$ $=\;\frac{6 \times 12}{18}+2$ ${\text{R}}_{ {\text{eq}}}=6 Ω$ ${\text{I}}=\;{ {\text{V}}}/{ {\text{R}}_{ {\text{eq}}}}=\;\frac{6}{6}=1 {\text{Amp}}$ .

Q15JEE Main 2022MCQ4MElectrostatic Potential and Capacitance

A source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key ' $K$ ' is closed, the total energy stored across the combination is $E_1$ . Now key ' $K$ ' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $E_2$ . The ratio $E_1 / E_2$ will be : [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

(1) Switch is closed ${\text{C}}_{ {\text{eq}}}-2 {\text{C}}$ Energy ${\text{E}}_1=\;\frac{1}{2} {\text{C}}_{ {\text{eq}}} {\text{V}}^2$ $=\;\frac{1}{2} 2 {\text{C}} \times {\text{V}}^2$ ${\text{E}}_1= {\text{CV}}^2$ (ii) When switch is opened charge on right capacitor remain CV while potential on left capacitor remain sameDielectric ${\text{K}}=5$ ${\text{C}}^{'}= {\text{KC}}$ $C^{'}=5 C$ ${\text{E}}_2=\;\frac{1}{2}(5 {\text{C}}) {\text{V}}^{2+}\;{( {\text{CV}})^2}/{2(5 {\text{C}})}$ ${\text{E}}_2=\;{5 {\text{CV}}^2}/{2}+\;{ {\text{CV}}^2}/{10}$ ${\text{E}}_2=\;{13 {\text{CV}}^2}/{5}$ $\;\frac{E_1}{E_2}=\;\frac{C V^2}{\frac{13 C V^2}{5}}=\;\frac{5}{13}$ $\;{ {\text{E}}_1}/{ {\text{E}}_2}=\;\frac{5}{13}$

Q16JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

Two concentric circular loops of radii $r_1=30 {\text{cm}}$ and $r_2=50 {\text{cm}}$ are placed in ${\text{X}}- {\text{Y}}$ plane as shown in the figure. A current $I=7 {\text{A}}$ is flowing through them in the direction as shown in figure. The net magnetic moment of this system of two circular loops is approximately : [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q17JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

A velocity selector consists of electric field $\vec{E}=E \hat{k}$ and magnetic field $\vec{B}=B \hat{j}$ with $B=12 m T$ . The value of $E$ required for an electron of energy $728 {\text{e}} V$ moving along the positive $x$ -axis to pass undeflected is : (Given, mass of electron $=9.1 \times 10^{-31} {\text{kg}}$ ) [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${ \vec{\text{E}}= {\text{E}} { {\text{k}}^{\^} \;\; {\text{B}}=12 {\ mT}$ ${ \vec{\text{B}}= {\text{B}} { {\text{j}}^{\^} \;\;$ Energy $=728 {\ eV}$ Energy $=\;\frac{1}{2} {\text{m} v}^2$ $728 {\text{eV}}=\;\frac{1}{2} \times 9.1 \times 10^{-31} \times {\ v}^2$ $728 \times 1.6 \times 10^{-19}=\;\frac{1}{2} \times 9.1 \times 10^{-31} \times {\text{v}}^2$ ${\text{v}}=16 \times 10^6 {\text{m}} / {\text{s}}$ ${\text{E}}= {\text{vB}}$ ${\text{E}}=16 \times 10^6 \times 12 \times 10^{-3}$ ${\text{E}}=192 \times 10^3 {\text{V}} / {\ m}$

Q18JEE Main 2022MCQ4MLaws of Motion

Two masses $M_1$ and $M_2$ are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass $M_2$ is twice that of $M_1$ , the acceleration of the system is $a_1$ When the mass $M_2$ is thrice that of $M_1$ , the acceleration of the system is $a_2$ . The ratio $\;\frac{a_1}{a_2}$ will be : [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Case 1 ${\text{a}}=\;{ {\text{m}}_2 {\text{g}}- {\text{m}}_1 {\text{g}}}/{ {\text{m}}_1+ {\text{m}}_2}$ ${\text{M}}_2=2 {\text{m}}_1$ $a_1=\;\frac{2 m_1 g-m_1 g}{3 m_1}$ ${\text{a}}_1= {\text{g}} / 3$ Case $-2$ ${\text{M}}_2=3 {\text{m}}_1$ ${\text{a}}_2=\;{3 {\text{m}}_1 {\text{g}}- {\text{m}}_1 {\text{g}}}/{4 {\text{m}}_1}$ ${\text{a}}_2=\;{ {\text{g}}}/{2}$ $\;{ {\text{a}}_1}/{ {\text{a}}_2}=\;{\;{ {\text{g}}}/{3}}/{{ {\text{g}}}/{2}}=\;\frac{2}{3}$

Q19JEE Main 2022MCQ4MAtoms and Nuclei

Mass numbers of two nuclei are in the ratio of $4: 3$ . Their nuclear densities will be in the ratio of [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Radius of nucleus ${\text{R}}= {\text{R}}_0 {\text{A}}^{\;\frac{1}{3}}$ Density of nucleus $=\;\frac{\;\text{ Mass of nucleus }\;}{\;\text{ volume of nucleus }\;}$ $\rho =\;{ {\text{m}} \times {\text{A}}}/{\frac{4}{3} \pi {\text{R}}^3}$ Where ${\text{m}}:$ mass of proton or neutron $\rho =\;{ {\text{m}} \times {\text{A}}}/{\frac{4}{3} \pi {\text{R}}_0{ }^3 {\text{A}}}$ $\rho \propto {\text{A}}^0$ Hence density of nucleus is independent of mass number

Q20JEE Main 2022MCQ4MProperties of Matter

The area of cross section of the rope used to lift a load by a crane is $2.5 \times 10^{-4} {\text{m}}^2$ . The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross section of the rope should be : $(.$ take $.g=10 m s^{-2})$ [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Since breaking stress (Maximum lifting capacity) is the property of material so it will remain same. $\;\text{ breaking stress }\;=\;\frac{\;\text{ Maximum lifting capacity }\;}{\;\text{ Area of cross section of rope }\;}$ $\;\frac{10}{2.5 \times 10^{-4}}=\;{25}/{ {\text{A}}}$ ${\text{A}}= \;\; =625 \times 10^{-6}$ $= \;\; 6.25 \times 10^{-4} {\text{m}}^2$

Q21JEE Main 2022NAT4MMotion in a Plane

If $\vec{A}=(2 \hat{i}+3 \hat{j}-\hat{k}) {\text{m}}$ and $\vec{B}=(\hat{i}+2 \hat{j}+2 \hat{k}) {\text{m}}$ . The magnitude of component of vector $\vec{A}$ along vector $\vec{B}$ will be ${\text{m}}$ . [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${ \vec{\text{A}}=(2 { {\text{i}}^{\^}+3 { {\text{j}}^{\^}-{ {\text{k}}^{\^}) {\text{m}}$ and ${ \vec{\text{B}}=({ {\text{i}}^{\^}+2 { {\text{j}}^{\^}+2 { {\text{k}}^{\^}) {\text{m}}$ Component of $\vec{A}$ along $\vec{B}=\vec{A} \cdot \hat{B}$ $=\;{{ \vec{\text{A}} \cdot { \vec{\text{B}}}/{|{ \vec{\text{B}}|}=\;\frac{2+6-2}{\sqrt{1^{2+}2^{2+}2^2}}$ $=\;\frac{6}{3}=2$

Q22JEE Main 2022NAT4MRotational Motion

The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through the center will be ______m. Given, the length of the rod is $10 \sqrt{3} {\text{m}}$ . [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

The moment of inertia ( $I$ ) of a cylindrical rod about an axis perpendicular to its length and passing through its center is $I = \frac{ML^2}{12}$ , where $M$ is the mass and $L$ is the length. The radius of gyration ( $K$ ) is related to the moment of inertia by the formula $I = MK^2$ .

By equating these two expressions, we get $MK^2 = \frac{ML^2}{12}$ .
This simplifies to an expression for the radius of gyration in terms of length: $K^2 = \frac{L^2}{12}$ , or $K = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}}$ .

Given the length of the rod is $L = 10\sqrt{3}$ m.
Substituting the value of $L$ , we find the radius of gyration:
$K = \frac{10\sqrt{3}}{2\sqrt{3}} = \frac{10}{2} = 5$ m.

Q23JEE Main 2022NAT4MRay Optics and Optical Instruments

In the given figure, the face $A C$ of the equilateral prism is immersed in a liquid of refractive index ' $n$ '. For incident angle $60^{\circ}$ at the side $A C$ , the refracteel light beam just grazes along face $A C$ . The refractive index of the liquid $n=\;\frac{\sqrt{x}}{4}$ . The value of $x$ is ________. (Given refractive index of glass $=1.5$ ) [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Using snell's law at face AC $1.5 \sin 60^{\circ}= {\text{n}} \times \sin 90^{\circ}$ $1.5 \times \;\frac{\sqrt{3}}{2}= {\text{n}}=\;{\sqrt{ {\text{x}}}}/{4}$ $3 \sqrt{3}=\sqrt{x}$ $x=27$

Q24JEE Main 2022NAT4MAtoms and Nuclei

Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below : ${ }_1^2 X+{ }_1^2 X={ }_2^4 Y$ The binding energies per nucleon for $\;\frac{2}{1} X$ and ${ }_2^4 Y$ are $1.1 {\text{MeV}}$ and $7.6 {\text{MeV}}$ respectively. The energy released in this process is _________ ${\text{MeV}}$ [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Energy released in the given process $=$ Binding energy of product - Binding energy of reactants $=7.6 \times 4-(1.1 \times 2) \times 2$ $=30.4-4.4$ $=26 {\text{MeV}}$

Q25JEE Main 2022NAT4MProperties of Matter

A uniform heavy rod of mass $20 {\text{kg}}$ , cross sectional area $0.4 {\text{m}}^2$ and length $20 {\text{m}}$ is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is $x \times 10^{-9} {\text{m}}$ . The value of $x$ is ________. (Given, young modulus $Y=2 \times 10^{11} {\text{Nm}}^{-2}$ and ${\text{g}}=10 {\text{ms}}^{-2}$ ) [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$Y=\;\frac{T}{A} \;\frac{d x}{d y}$ ${\text{m}}=20 {\text{kg}}$ $A=\mathbf{0 . 4 \mathrm { m } ^ { 2 }}$ $1=20 {\text{m}}$ ${\text{Y}}=\;\frac{\;\text{ stress }\;}{\;\text{ strain }\;}$ $Y=\;{\;{ {\text{T}}}/{ {\text{A}}}}/{{ {\text{dy}}}/{ {\text{dx}}}}=\;{ {\text{T}}}/{ {\text{A}}} \cdot \;{ {\text{dx}}}/{ {\text{dy}}}$ $d y=\;\frac{T d x}{A Y}$ Tension at a distance ${\text{x}}$ from lower end $=\;{ {\text{mg}}}/{ℓ} {\text{x}}$ So. $int_0^{\Delta l} d {\text{y}}=int_0^{ℓ} \;{ {\text{mg}}}/{ℓ} {\text{x}} \;{ {\text{dx}}}/{ {\text{AY}}}$ $\Delta ℓ=\;{ {\text{mg}}}/{ℓ {\text{AY}}}\$ [;{ {\text{x}}^2}/{2}]_0^{ℓ}\Delta ℓ=;{ {\text{mg}} ℓ}/{2 {\text{AY}}}\Delta ℓ=;\frac{20 × 10 × 20}{2 × 0.4 × 2 × 10^{11}}2500 × 10^{-11}\Delta ℓ=25 × 10^{-9}= {\text{x}} × 10^{-9}{\text{x}}=25$

Q26JEE Main 2022NAT4MSemiconductor Electronics

The typical transfer characteristics of a transistor in CE configuration is shown in figure. A load resistor of $2 k Ω$ is connected in the collector branch of the circuit used. The input resistance of the transistor is $0.50 {\text{k}} Ω$ . The voltage gain of the transistor is ________. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Current gain in ${\text{C}}- {\text{E}}$ configuration $\Rightarrow \beta=\;{\Delta {\text{I}}_{ {\text{C}}}}/{\Delta {\text{I}}_{ {\text{B}}}}$ ${\text{R}}_{ {\text{C}}}=2 {\text{k}} Ω, {\text{R}}_{ {\text{B}}}=0.50 {\text{k}} Ω$ $\;\text{ Voltage gain }\;=\;{\Delta {\text{I}}_{ {\text{C}}} {\text{R}}_{ {\text{C}}}}/{\Delta {\text{I}}_{ {\text{B}}} {\text{R}}_{ {\text{B}}}}=\;\frac{5 \times 10^{-3}}{100 \times 10^{-6}} \times \;\frac{2}{0.5}$ $=\;\frac{10^{-2}}{5 \times 10^{-5}}=\;\frac{1000}{5}=200$

Q27JEE Main 2022NAT4MElectrostatics

Three point charges of magnitude $5 \mu {\text{C}}, 0.16 \mu {\text{C}}$ and $0.3 \mu {\text{C}}$ are located at the vertices $A, B, C$ of a right angled triangle whose sides are $A B=3 {\text{cm}}, B C=3 \sqrt{2} {\text{cm}}$ and $C A=3 {\text{cm}}$ and point $A$ is the right angle corner. Charge at point ${\text{A}}$ experiences _________ ${\text{N}}$ of electrostatic force due to the other two charges. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{F}}_1=\;{ {\text{k}} \times 5 \times 0.3 \times 10^{-12}}/{9 \times 10^{-4}}$ $=\;\frac{9 \times 10^9 \times 5 \times 0.3 \times 10^{-12}}{9 \times 10^{-4}}$ $=1.5 \times 10=15 {\text{N}}$ ${\text{F}}_2=\;\frac{9 \times 10^9 \times 5 \times 0.16 \times 10^{-12}}{9 \times 10^{-4}}=8 {\text{N}}$ force experienced by charge at ${\text{A}}=\sqrt{ {\text{F}}_1{ }^{2+} {\text{F}}_2{ }^2}$ $=\sqrt{15^{2+}8^2}$ $=\sqrt{289}=17 {\text{N}}$

Q28JEE Main 2022NAT4MElectromagnetic Induction

In a coil of resistance $8 Ω$ , the magnetic flux due to an external magnetic field varies with time as $\varphi =\;\frac{2}{3}(9-t^2)$ . The value of total heat produced in the coil, till the flux becomes zero, will be _________ $J .$ [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\varphi =\;\frac{2}{3}(9-t^2)=0$ $t=3 {\sec}$ $e=\;\frac{-d \varphi }{d t}=-\;\frac{2}{3}(0-2 t)=\;\frac{4 t}{3}$ Heat produced in $3 {\sec}=\int \;{ {\text{e}}^2}/{ {\text{r}}} {\text{dt}}=int_0^3 \;{16 {\text{t}}^2}/{9 \times 8} {\text{dt}}=2 {\text{J}}$

Q29JEE Main 2022NAT4MCurrent Electricity

A potentiometer wire of length $300 {\text{cm}}$ is connected in series with a resistance $780 Ω$ and a standard cell of emf $4 {\text{V}}$ . A constant current flows through potentiometer wire. The length of the null point for cell of emf $20 {\text{mV}}$ is found to be $60 {\text{cm}}$ . The resistance of the potentiometer wire is ________ $Ω$ [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Let resistance of potentiometers wire is ${\text{R}}$ $i=\;\frac{4}{R+780}$ Potential difference across ${\text{AB}}$ $=\;\frac{4 R}{R+780}$ Potential difference across AC $=\;\frac{4 R \times 60}{(R+780) \times 300}=\;\frac{4 R}{5(R+780)}$ This should be equal to $20 {\text{mV}}$ $\;{4 {\text{R}}}/{5( {\text{R}}+780)}=20 \times 10^{-3}=2 \times 10^{-2}$ $4 {\text{R}}=10^{-1}( {\text{R}}+780)$ $4 {\text{R}}=\;{ {\text{R}}}/{10}+78$ $4 {\text{R}}-\;{ {\text{R}}}/{10}=78$ $\;{39 {\text{R}}}/{10}=78$ ${\text{R}}=20 Ω$

Q30JEE Main 2022NAT4MOscillations

As per given figures, two springs of spring constants $k$ and $2 k$ are connected to mass $m$ . If the period of oscillation in figure (a) is $3 {\text{s}}$ , then the period of oscillation in figure (b) will be $\sqrt{x} s$ . The value of $x$ is ________. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

For figure (a) : ${\text{K}}_{\;\text{eq }\;}=\;{ {\text{K}} \times 2 {\text{K}}}/{ {\text{K}}+2 {\text{K}}}=\;{2 {\text{K}}}/{3}$ ${\text{T}}=2 \pi \sqrt{\;{ {\text{m}}}/{ {\text{K}}_{ {\text{eq}}}}}=2 \pi \sqrt{\;{ {\text{m}}}/{2 {\text{K}} / 3}}=2 \pi \sqrt{\;{3 {\text{m}}}/{2 {\text{K}}}}$ For figure (b): ${\text{K}}_{ {\text{eq}}}=3 {\text{K}}, {\text{T}}^{'}=2 \pi \sqrt{\;{ {\text{m}}}/{3 {\text{K}}}}$ $\;{ {\text{T}}^{'}}/{ {\text{T}}}=\sqrt{\;{ {\text{m}} \times 2 {\text{K}}}/{3 {\text{K}} \times 3 {\text{m}}}}=\;\frac{\sqrt{2}}{3}$ ${\text{T}}^{'}=\sqrt{2}$ ${\text{x}}=2$

Chemistry30 questions

Q31JEE Main 2022MCQ4MSome Basic Concepts of Chemistry

Hemoglobin contains $0.34 \%$ of iron by mass. The number of Fe atoms in $3.3 {\text{g}}$ of hemoglobin is (Given: Atomic mass of Fe is $56 {\text{u}}, {\text{N}}_{ {\text{A}}}=6.022 \times 10^{23} {\text{mol}}^{-1}$ .) [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

According to the question, $100 {\text{g}}$ of hemoglobin contains $0.34 {\text{g}}$ of iron $3.3 {\text{g}}$ of hemoglobin contains $\;\frac{0.34}{100} \times 3.3 {\text{g}}$ of ironmoles of ${\text{Fe}}=\;\frac{0.34 \times 3.3}{100 \times 56}=\;{ {\text{N}}}/{ {\text{N}}_{ {\text{A}}}}$ $N=\;\frac{0.34 \times 3.3 \times 6.022 \times 10^{23}}{100 \times 56}$ $=1.21 \times 10^{20}$

Q32JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

Arrange the following in increasing order of their covalent character. A. ${\text{CaF}}_2$ B. ${\text{CaCl}}_2$ C. ${\text{CaBr}}_2$ D. ${\text{CaI}}_2$ Choose the correct answer from the options given below. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

From Fajan's rule, for a given metal ion, as the size of anion increases, polarizability of anion increases and hence covalent character of the given ionic compound increases.Hence, the increasing order of covalent character is ${\text{CaF}}_2< {\text{CaCl}}_2< {\text{CaBr}}_2< {\text{Cal}}_2$

Q33JEE Main 2022MCQ4MIonic Equilibrium

Class XII students were asked to prepare one litre of buffer solution of ${\text{pH}} 8.26$ by their Chemistry teacher: The amount of ammonium chloride to be dissolved by the student in $0.2 {\text{M}}$ ammonia solution to make one litre of the buffer is (Given: ${\text{pK}}_{ {\text{b}}}( {\text{NH}}_3)=4.74$ Molar mass of ${\text{NH}}_3=17 {\text{g}} {\text{mol}}^{-1}$ Molar mass of ${\text{NH}}_4 {\text{Cl}}=53.5 {\text{g}} {\text{mol}}^{-1}$ ) [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

For basic Buffer, ${\text{pOH}}= {\text{pK}}_{ {\text{b}}}+\log \;\frac{[\;\text{ salt }\;]}{[\;\text{ Base }\;]}$ ${\text{pOH}}=14-8.26=5.74$ $5.74=4.74+\log \;{[ {\text{NH}}_4 {\text{Cl}}]}/{0.2}$ $[ {\text{NH}}_4 {\text{Cl}}]=2 {\text{M}}$ Moles of ${\text{NH}}_4 {\text{Cl}}=2 \times 1=2$ molesWeight of ${\text{NH}}_4 {\text{Cl}}=2 \times 53.5=107 {\text{g}}$

Q34JEE Main 2022MCQ4MChemical Kinetics

At $30^{\circ} {\text{C}}$ , the half life for the decomposition of ${\text{AB}}_2$ is $200 {\text{s}}$ and is independent of the initial concentration of ${\text{AB}}_2$ . The time required for $80 \%$ of the ${\text{AB}}_2$ to decompose is Given: $\log 2=0.30 \;\; \log 3=0.48$ [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Since, half-life is independent of the initial concentration of $A B_2$ . Hence, the reaction is "First Order". $k=\;\frac{2.303 \log 2}{t_{1 / 2}}$ $\;\frac{2.303 \log 2}{t_{1 / 2}}=\;\frac{2.303}{t} \log \;\frac{100}{(100-80)}$ $\;\frac{2.303 \times 0.3}{200}=\;\frac{2.303}{t} \log 5$ $t=467 {\text{s}}$

Q35JEE Main 2022MCQ4MSolutions

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Finest gold is red in colour, as the size of the particles increases, it appears purple then blue and finally gold. Reason R: The colour of the colloidal solution depends on the wavelength of light scattered by the dispersed particles. In the light of the above statements, choose the most appropriate answer from the options given below. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Finest gold sol is red in colour; as the size of particles increases, it appears purple, then blue and finally golden.The colour of colloidal solution depends on the wavelength of light scattered by the dispersed particles. The wavelength of light further depends on size and nature of the particles.Hence, Both $A$ and $\$ R $are true and$ R $is the correct explanation of$ A$

Q36JEE Main 2022MCQ4MClassification of Elements

The metal that has very low melting point and its periodic position is closer to a metalloid is [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{Al}} \to \;933 {\text{K}}$ ${\text{Ga}} \to \;303 {\text{K}}$ ${\text{In}} \to \;430 {\text{K}}$ ${\text{Se}} \to \;490 {\text{K}}$ Among the given elements, Gallium has the lowest melting point, Gallium is also close to a metalloid

Q37JEE Main 2022MCQ4Md and f Block Elements

The metal that is not extracted from its sulfide ore is [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Aluminium is not extracted from sulphide ore. It is usually extracted from bauxite ore, leaching of bauxite ore is done followed by electrolytic reduction. ${\text{Al}}$ is extracted from ${\text{Al}}_2 {\text{O}}_3 \cdot 2 {\text{H}}_2 {\text{O}}$ i.e., Bauxite ore

Q38JEE Main 2022MCQ4MRedox Reactions

The products obtained from a reaction of hydrogen peroxide and acidified potassium permanganate are [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$2 {\text{MnO}}_4^{-}+6 {\text{H}}^{+}+5 {\text{H}}_2 {\text{O}}_2 \longrightarrow 2 {\text{Mn}}^{2+}+8 {\text{H}}_2 {\text{O}}+5 {\text{O}}_2$ This reaction shows reducing action of ${\text{H}}_2 {\text{O}}_2$ in acidic medium.The products formed are ${\text{Mn}}^{2+}, {\text{H}}_2 {\text{O}}$ and ${\text{O}}_2$

Q39JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion ${\text{A}}$ : ${\text{LiF}}$ is sparingly soluble in water. Reason R: The ionic radius of ${\text{Li}}^{+}$ ion is smallest among its group members, hence has least hydration enthalpy. In the light of the above statements, choose the most appropriate answer from the options given below. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

LiF is sparingly soluble in water.The low solubility of ${\text{LiF}}$ in water is due to its high lattice enthalpy (Since ${\text{Li}}^{+}$ and ${\text{F}}^{-}$ are small in size). Also, due to small size of ${\text{Li}}^{+}$ , its hydration enthalpy is high.Hence, Assertion is true but Reason is false

Q40JEE Main 2022MCQ4Mp Block Elements

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion : Boric acid is a weak acid. Reason ${\text{R}}$ : Boric acid is not able to release ${\text{H}}^{+}$ ion on its own. It receives ${\text{OH}}^{-}$ ion from water and releases ${\text{H}}^{+}$ ion. In the light of the above statements, choose the most appropriate answer from the options given below. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Boric acid is a weak acid ${\text{H}}_3 {\text{BO}}_3+ {\text{H}}_2 {\text{O}} ⇌[ {\text{B}}( {\text{OH}})_4]^{⊖}+ {\text{H}}^{⊕}$ Boric acid is not able to release ${\text{H}}^{+}$ ion on its own. It receives ${\text{OH}}^{-}$ ion from water and releases ${\text{H}}^{+}$ ion as shown in the above reaction.Hence, Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$ .

Q41JEE Main 2022MCQ4MCoordination Compounds

The metal complex that is diamagnetic is (Atomic number: ${\text{Fe}}, 26 ; {\text{Cu}}, 29$ ) [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\Rightarrow {\text{K}}_3[ {\text{Cu}}( {\text{CN}})_4]$ is diamagnetic ${\text{Cu}}( {\text{I}}) \Rightarrow {\text{d}}^{10}$ configuration $\Rightarrow$ No unpaired electrons. $\Rightarrow {\text{K}}_2[ {\text{Cu}}( {\text{CN}})_4], {\text{K}}_3[ {\text{Fe}}( {\text{CN}})_4]$ and ${\text{K}}_4[ {\text{FeCl}}_6]$ are paramagnetic in nature

Q42JEE Main 2022MCQ4Md and f Block Elements

Match List I with List II. List I Pollutant List II Source (A) Microorganisms (I)Strip mining (B) Plant nutrients (II)Domestic sewage (C) Toxic heavy metals (III) Chemical fertilizer (D) Sediment (IV) Chemical factoryChoose the correct answer from the options given below : [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q43JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

The correct decreasing order of priority of functional groups in naming an organic Question: compound as per IUFAC system of nomenclature is [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$- {\text{SO}}_3 {\text{H}}>- {\text{COCl}}>- {\text{CONH}}_2>- {\text{CN}}$

Q44JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Which of the following is not an example of benzenoid compound? [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Benzenoid compounds are aromatic hydrocarbons that contain one or more benzene rings. Let's examine the options. Option A (triphenylene), Option C (naphthalene), and Option D (aniline) all contain at least one benzene ring in their structure, making them benzenoid compounds. In contrast, Option B is cyclooctatetraene ( $C_8H_8$ ), an eight-membered ring that does not contain a benzene ring. It is a non-benzenoid compound; in fact, due to its non-planar, tub-like shape, it is non-aromatic. Therefore, cyclooctatetraene is not an example of a benzenoid compound.

Q45JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

Hydrolysis of which compound will give carbolic acid? [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q46JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Consider the above reaction and predict the major product. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q47JEE Main 2022MCQ4MAmines

The correct sequential order of the reagents for the given reaction is [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q48JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Vulcanization of rubber is carried out by heating a mixture of [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q49JEE Main 2022MCQ4MBiomolecules

Animal starch is the other name of [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q50JEE Main 2022MCQ4MIonic Equilibrium

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R .$ Assertion A: Phenolphthalein is a ${\text{pH}}$ dependent indicator, remains colourless in acidic solution and gives pink colour in basic medium. Reason R: Phenolphthalein is a weak acid. It doesn't dissociate in basic medium. In the light of the above statements, choose the most appropriate answer from the options given below. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Phenolphthalein is a pH dependent indicator. It is a weak acid which is colourless in the acidic solution but gives pink colour in basic medium. The pink colour is due to its conjugate form. Therefore, assertion (A) is true but Reason (R) is false. Phenolphthalein dissociate in basic medium ${\text{HPh}}( {\text{aq}}) ⇌ {\text{H}}^{+}+ {\text{Ph}}^{-}$ (colourless) (Pink)

Q51JEE Main 2022NAT4MStates of Matter

A $10 {\text{g}}$ mixture of hydrogen and helium is contained in a vessel of capacity $0.0125 {\text{m}}^3$ at 6 bar and $27^{\circ} {\text{C}}$ . The mass of helium in the mixture is _______ g. (nearest integer) Given: ${\text{R}}=8.3 {\text{J}} {\text{K}}^{-1} {\text{mol}}^{-1}$ (Atomic masses of ${\text{H}}$ and ${\text{He}}$ are $1 {\text{u}}$ and $4 {\text{u}}$ , respectively) [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Number of moles of mixture of ${\text{H}}_2$ and ${\text{He}}$ $=\;\frac{P V}{R T}$ $=\;\frac{6 \times 10^5 \times 0.0125}{8.3 \times 300}=3$ Let the mass of He in $10 {\text{g}}$ mixture be ${\text{xg}}$ $\therefore \;\frac{x}{4}+\;\frac{10-x}{2}=3$ On solving $x=8 {\text{g}}$ $\therefore$ Mass of ${\text{He}}$ in the mixture $=8 {\text{g}}$

Q52JEE Main 2022NAT4MStructure of Atom

Consider an imaginary ion ${ }_{22}^{48} {\text{X}}^{3-}$ . The nucleus contains ' $a$ ' $\%$ more neutrons than the number of electrons in the ion. The value of ' $a$ ' is ________ . [nearest integer] [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Number of electrons in ${ }_{22}^{48} X^{3-}$ is 25 .Number of neutrons $=48-22=26$ . $\%$ increase in the number of neutrons over electrons $=(\;\frac{26-25}{25}) 100=4 \%$ $\therefore a=4$

Q53JEE Main 2022NAT4MChemical Thermodynamics

For the reaction ${H}_2 {\text{F}}_2( {\text{g}}) \to {\text{H}}_2( {\text{g}})+ {\text{F}}_2( {\text{g}})$ $\Delta U=-59.6 {\text{kJ}} {\text{mol}}^{-1} \;\text{ at }\; 27^{\circ} {\text{C}}$ The enthalpy change for the above reaction is $(-)$ _______ ${\text{kJ}} {\text{mol}}^{-1}$ [nearest integer] Given: ${\text{R}}=8.314 {\text{J}} {\text{K}}^{-1} {\text{mol}}^{-1}$ [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{H}}_2 {\text{F}}_2( {\text{g}}) \longrightarrow {\text{H}}_2( {\text{g}})+ {\text{F}}_2( {\text{g}})$ $\Delta {\text{U}}=-59.6 {\text{kJ}} {\text{mol}}^{-1}$ at $27^{\circ} {\text{C}}$ $\Delta {\text{H}}=\Delta {\text{U}}+\Delta {\text{n}}_g {\text{RT}}$ $=-59.6+\;\frac{1 \times 8.314 \times 300}{1000}$ $=-57.10 {\text{kJ}} {\text{mol}}^{-1}$

Q54JEE Main 2022NAT4MSolutions

The elevation in boiling point for 1 molal solution of non-volatile solute ${\text{A}}$ is $3 {\text{K}}$ . The depression in freezing point for 2 molal solution of ${\text{A}}$ in the same solvent is $6 K$ . The ratio of $K_b$ and $K_f$ i.e., $K_b / K_f$ is $1: X$ . The value of $X$ is [nearest integer] [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Molality of a solution of non volatile solute $(A)=1$ Elevation in boiling point is given by $\Delta {\text{T}}_{ {\text{b}}}= {\text{K}}_{ {\text{b}}} {\text{m}}$ $3= {\text{K}}_{ {\text{b}}} \times 1$ Molality of $(A)$ in the same solvent $=2$ Depression in freezing point is given by $\Delta {\text{T}}_{ {\text{f}}}= {\text{K}}_{ {\text{f}}} {\text{m}}$ $6= {\text{K}}_{ {\text{f}}} \times 2 \;\; . . .(2)$ $\;\text{ Dividing }\;(1) \;\text{ by }\;(2)$ $\;{ {\text{K}}_{ {\text{b}}}}/{ {\text{K}}_{ {\text{f}}}}=\;{1}/{ {\text{X}}}=\;\frac{1}{1}$ $\therefore \;\; {\text{X}}=1$

Q55JEE Main 2022NAT4MRedox Reactions

$20 {\text{mL}}$ of $0.02 {\text{M}}$ hypo solution is used for the titration of $10 {\text{mL}}$ of copper sulphate solution, in the presence of excess of ${\text{KI}}$ using starch as an indicator. The molarity of ${\text{Cu}}^{2+}$ is found to be _______ $\times 10^{-2} {\text{M}}$ [nearest integer] Given $: 2 {\text{Cu}}^{2+}+4 {\text{I}}^{-} \to {\text{Cu}}_2 {\text{I}}_2+ {\text{I}}_2$ ${\text{I}}_2+2 {\text{S}}_2 {\text{O}}_3^{2-} \to 2 {\text{I}}^{-}+ {\text{S}}_4 {\text{O}}_6^{2-}$ [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$2 {\text{Cu}}^{2+}+4 {\text{I}}^{-} \to {\text{Cu}}_2 {\text{I}}_2+ {\text{I}}_2$ ${\text{I}}_2+ {\text{S}}_2 {\text{O}}_3^{2-} \to 2 {\text{I}}^{-}+ {\text{S}}_4 {\text{O}}_6^{2-}$ Milliequivalents of hypo solution $=0.02 \times 20=0.4$ Milliequivalents of ${\text{Cu}}^{2+}$ in $10 {\text{mL}}$ solution =Milliequivalents of ${\text{I}}_2=$ Milliequivalents of hypo $=0.4$ Millimoles of ${\text{Cu}}^{2+}$ ions in $10 {\text{mL}}=0.4$ Molarity of ${\text{Cu}}^{2+}$ ions $=\;\frac{0.4}{10}=0.04 {\text{M}}$ $=4 \times 10^{-2} {\text{M}}$

Q56JEE Main 2022NAT4Mp Block Elements

The number of non-ionisable protons present in the product B obtained from the following reactions is ________. ${\text{C}}_2 {\text{H}}_5 {\text{OH}}+ {\text{PCl}}_3 \to {\text{C}}_2 {\text{H}}_5 {\text{Cl}}+ {\text{A}}$ ${\text{A}}+ {\text{PCl}}_3 \to {\text{B}}$ [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{C}}_2 {\text{H}}_5 {\text{OH}}+ {\text{PCl}}_3 \longrightarrow {\text{C}}_2 {\text{H}}_5 {\text{Cl}}+ {\text{H}}_3 {\text{PO}}_3$ ${\text{H}}_3 {\text{PO}}_3+ {\text{PCl}}_3 \longrightarrow {\text{H}}_4 {\text{P}}_2 {\text{O}}_5+ {\text{HCl}}$

Q57JEE Main 2022NAT4Md and f Block Elements

The spin-only magnetic moment value of the compound with strongest oxidizing ability among ${\text{MnF}}_4, {\text{MnF}}_3$ and ${\text{MnF}}_2$ is _______ B.M. [nearest integer] [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Hence ${\text{MnF}}_3 \Rightarrow$ strongest O.A $\mu =\sqrt{4(4+2)}=\sqrt{24}=4.89=5$

Q58JEE Main 2022NAT4MOrganic Chemistry - Some Basic Principles

Total number of isomers (including stereoisomers) obtained on monochlorination of methylcyclohexane is _________. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

Q59JEE Main 2022NAT4MAlcohols, Phenols and Ethers

A $100 {\text{mL}}$ solution of ${\text{CH}}_3 {\text{CH}}_2 {\text{MgBr}}$ on treatment with methanol produces $2.24 {\text{mL}}$ of a gas at STP. The weight of gas produced is _______ mg. [nearest integer] [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{n}} \;\; =\;\frac{2.24 \times 10^{-3}}{22.4}=10^{-4}$ ${\text{W}} \;\; = {\text{n}} \times {\text{M}}$ $\;\; =10^{-4} \times 30=3 {\text{mg}}$

Q60JEE Main 2022NAT4MBiomolecules

How many of the following drugs is/are example(s) of broad spectrum antibiotics? Ofloxacin, Penicillin G, Terpineol, Salvarsan. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Ofloxacin is the only broad-spectrum antibiotic given in the questionPenicilin - $G$ is a narrow-spectrum antibiotic.Salvarsan is mainly active against spirochete, a bacteria that causes syphilisTerpineol is an antiseptic.

Mathematics30 questions

Q61JEE Main 2022MCQ4MQuadratic Equation and Inequalities

The minimum value of the sum of the squares of the roots of $x^{2+}(3-a) x+1=2 a$ is: [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\alpha+\beta=a-3, \alpha \beta=1-2 a$ $\Rightarrow \alpha^{2+}\beta^2=(a-3)^{2-}2(1-2 a)$ $=a^{2-}6 a+9-2+4 a$ $=a^{2-}2 a+7$ $=(a-1)^{2+}6$ So, $\alpha^{2+}\beta^2 \ge 6$

Q62JEE Main 2022MCQ4MComplex Numbers

If $z=x+i y$ satisfies $|z|-2=0$ and $|z-i|-|z+5 i|=0$ , then [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$|z-i|=|z+5 i|$ So, ${\text{z}}$ lies on $⟂^r$ bisector of $(0,1)$ and $(0,-5)$ i.e., line $y=-2$ as $|z|=2$ $\Rightarrow z=-2 i$ $x=0$ and $y=-2$ so, $x+2 y+4=0$

Q63JEE Main 2022MCQ4MMatrices and Determinants

Let $A=\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$ and $B=\begin{bmatrix}9^2 & -10^2 & 11^2 \\ 12^2 & 13^2 & -14^2 \\ -15^2 & 16^2 & 17^2\end{bmatrix}$ , then the value of $A^{'} B A$ is: [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$A^{'} B A=\begin{bmatrix}1 & 1 & 1\end{bmatrix}\$ \begin{bmatrix}9^2 & -10^2 & 11^2 \ 12^2 & 13^2 & -14^2 \ -15^2 & 16^2 & 17^2\end{bmatrix}$ A=$\begin{bmatrix}9^{2+}12^{2-}15^2 & -10^{2+}13^{2+}16^2 & 11^{2-}14^{2+}17^2\end{bmatrix}$\begin{bmatrix}1 \ 1 \ 1\end{bmatrix}=[9^{2+}12^{2-}15^{2-}10^{2+}13^{2+}16^{2+}11^{2-}14^{2+}17^2]=[(9^{2-}10^2)+(11^{2+}12^2)+(13^{2-}14^2)+(16^{2-}15^2)+17^2]=[-19+265+(-27)+31+289]=[585-46]=[539]$

Q64JEE Main 2022MCQ4MBinomial Theorem

$sum_{{{{i, j=0 }_{ i \ne j}}}}^{n}{ }^n C_i{ }^n C_j$ is equal to [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\sum^{n}_{i, j=0,i \ne j}\;{}^nC_i\;\;{ }^n C_j=\sum^{n}_{i, j=0}{ }^n C_i\;{ }^n C_j-sum_{i=j} ^{n}\;{ }^n C_i\;\;{ }^n C_j$ $=\sum^{n}_{j=0} { }^n C_i sum_{j=0} ^{n} { }^n C_j-sum_{i=0}^{n} { }^n C_i C_i$ $=2^n \cdot 2^n-{ }^{2 n} C_n$ $=2^{2 n}-{ }^{2 n} C_n$

Q65JEE Main 2022MCQ4MApplication of Derivatives

Let ${\text{P}}$ and ${\text{Q}}$ be any points on the curves $(x-1)^{2+}(y+1)^2=1$ and $y=x^2$ , respectively. The distance between $P$ and $Q$ is minimum for some value of the abscissa of $P$ in the interval [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$y=m x+2 a+\;\frac{1}{m^2}$ (Equation of normal to $x^2=4 a y$ in slope form) through $(1,-1)$ . $4 m^{3+}6 m^{2+}1=0$ $\Rightarrow m ≃-1.6$ Slope of normal $≃ \;\frac{-8}{5}=\tan \theta$ $\Rightarrow \cos \theta ≃ \;\frac{-5}{\sqrt{89}}, \sin \theta ≃ \;\frac{8}{\sqrt{89}}$ $x_p=1+\cos \theta ≃ 1-\;\frac{5}{\sqrt{89}} \in (\;\frac{1}{4}, \;\frac{1}{2})$

Q66JEE Main 2022MCQ4MApplication of Derivatives

If the maximum value of $a$ , for which the function $f_a(x)=\tan ^{-1} 2 x-3 a x+7$ is non-decreasing in $(-\;\frac{\pi}{6}, \;\frac{\pi}{6})$ , is $\overline{a}$ , then $f_{\overline{a}}(\;\frac{\pi}{8})$ is equal to [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$f_a(x)=\tan ^{-1} 2 x-3 a x+7$ $\because f_a(x)$ is non-decreasing in $(-\;\frac{\pi}{6}, \;\frac{\pi}{6})$ $\therefore f_a^{'}(x) \ge 0 \Rightarrow \;\frac{2}{1+4 x^2}-3 a \ge 0$ $\Rightarrow 3 a \le \;\frac{2}{1+4 x^2}$ So, $a_{\max }=\;\frac{2}{3}(\;\frac{1}{1+4 \times \;\frac{\pi^2}{36}})=\;\frac{6}{9+\pi^2}=\overline{a}$ $\therefore f_{\overline{a}}(\;\frac{\pi}{8})=1-3 \;\frac{\pi}{8} \cdot \;\frac{6}{9+\pi^2}+7$ $=8-\;\frac{9 \pi}{4(9+\pi^2)}$

Q67JEE Main 2022MCQ4MLimits, Continuity and Differentiability

Let $\beta=\lim _{x \to 0} \;\frac{\alpha x-(e^{3 x}-1)}{\alpha x(e^{3 x}-1)}$ for some $\alpha \in ℝ$ . Then the value of $\alpha+\beta$ is : [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\beta=\lim _{x \to 0} \;\frac{\alpha x-(e^{3 x}-1)}{\alpha x(e^{3 x}-1)}, \alpha \in R$ $=\lim _{x \to 0} \;{\;\frac{\alpha}{3}-(\frac{e^{3 x}-1}{3 x})}/{\alpha x(\;\frac{e^{3 x-1}}{3 x})}$ So, $\alpha=3$ (to make independent form) $\beta=\lim _{x \to 0} \;{1-(\frac{e^{3 x}-1}{3 x})}/{3 x}=\;{1-\frac{(3 x+\frac{9 x^2}{2}+. . . . .)}{3 x}}/{3 x}$ $=\;\frac{-(\frac{9}{2} x^{2+}\frac{(3 x)^3}{31}+. . . . .)}{9 x^2}=\;\frac{-1}{2}$ $\therefore \alpha+\beta=3-\;\frac{1}{2}=\;\frac{5}{2}$

Q68JEE Main 2022MCQ4MDifferentiation

The value of $\log _e 2 \;\frac{d}{d x}(\log _{\cos x} {\text{cosec}} x)$ at $x=\;\frac{\pi}{4}$ is [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Let $f(x)=\log _{\cos x} {\text{cosec}} x$ $=\;{\log {\text{cosec}} x}/{\log \cos x}$ $\Rightarrow f^{'}(x)=\;{\log \cos x \cdot \sin x \cdot (-{\text{cosec}} x \cot x-\log {\text{cosec}} x \cdot \frac{1}{\cos x} \cdot -\sin x)}/{(\log \cos x)^2}$ $\;\text{ at }\; x=\;\frac{\pi}{4}$ $f^{'}(\;\frac{\pi}{4})=\;{-\log (\frac{1}{\sqrt{2}})+\log \sqrt{2}}/{(\log \;\frac{1}{\sqrt{2}})^2}=\;\frac{2}{\log \sqrt{2}}$ $\therefore \log _e 2 f^{'}(x) \;\text{ at }\; x=\;\frac{\pi}{4}=4$

Q69JEE Main 2022MCQ4MDefinite Integrals

$int_0^{20 \pi}(|\sin x|+|\cos x|)^2 d x$ is equal to [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$I=int_0^{20 \pi}(|\sin x|+|\cos x|)^2 d x$ $=20 int_0^\pi(1+|\sin 2 x|) d x$ $=40 int_0^{\;\frac{\pi}{2}}(1+\sin 2 x) d x$ $=.40(x-\;\frac{\cos 2 x}{2})|_0 ^{\;\frac{\pi}{2}}$ $=40(\;\frac{\pi}{2}+\;\frac{1}{2}+\;\frac{1}{2})=20(\pi+2)$

Q70JEE Main 2022MCQ4MDifferential Equations

Let the solution curve $y=f(x)$ of the differential equation $\;\frac{d y}{d x}+\;\frac{x y}{x^{2-}1}=\;\frac{x^{4+}2 x}{\sqrt{1-x^2}}, x \in (-1,1)$ pass through the origin. Then $int_{-\;\frac{\sqrt{3}}{2}}^{\;\frac{\sqrt{3}}{2}} f(x) d x$ is equal to [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{d y}{d x}+\;\frac{x y}{x^{2-}1}=\;\frac{x^{4+}2 x}{\sqrt{1-x^2}}$ which is first order linear differential equation.Integrating factor $(I . F)=.e^{\int \;\frac{x}{x^{2-}1} d x}$ $=e^{\;\frac{1}{2} \ln |x^{2-}1|}=\sqrt{|x^{2-}1|}$ $=\sqrt{1-x^2}$ $\because x \in (-1,1)$ Solution of differential equation $y \sqrt{1-x^2}=\int(x^{4+}2 x) d x=\;\frac{x^5}{5}+x^{2+}c$ Curve is passing through origin, $c=0$ $y=\;\frac{x^{5+}5 x^2}{5 \sqrt{1-x^2}}$ $int_{\;\frac{-\sqrt{3}}{2}}^{\;\frac{\sqrt{3}}{2}} \;\frac{x^{5+}5 x^2}{5 \sqrt{1-x^2}} d x=0+2 int_0^{\;\frac{\sqrt{3}}{2}} \;\frac{x^2}{\sqrt{1-x^2}} d x$ put $x=\sin \theta$ $d x=\cos \theta d \theta$ $I=2 int_0^{\;\frac{\pi}{3}} \;\frac{\sin ^2 \theta \cdot \cos \theta d \theta}{\cos \theta}$ $=int_0^{\;\frac{\pi}{3}}(1-\cos 2 \theta) d \theta$ $= (\theta-\;\frac{\sin 2 \theta}{2}) |{}_{0}^{\frac{\pi}{3}}$

Q71JEE Main 2022MCQ4MEllipse

The acute angle between the pair of tangents drawn to the ellipse $2 x^{2+}3 y^2=5$ from the point $(1,3)$ is [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$2 x^{2+}3 y^2=5$ Equation of tangent having slope $m$ . $y=m x \pm \sqrt{\;\frac{5}{2} m^{2+}\;\frac{5}{3}}$ which passes through $(1,3)$ $3=m \pm \sqrt{\;\frac{5}{2} m^{2+}\;\frac{5}{3}}$ $\;\frac{5}{2} m^{2+}\;\frac{5}{3}=9+m^{2-}6 m$ $\;\frac{3}{2} m^{2+}6 m-\;\frac{22}{3}=0$ $9 m^{2+}36 m-44=0$ $m_1+m_2=-4, m_1 m_2=-\;\frac{44}{9}$ $(m_1-m_2)^2=16+4 \times \;\frac{44}{9}=\;\frac{320}{9}$ Acute angle between the tangents is given by $\alpha=\tan ^{-1}|\;\frac{m_1-m_2}{1+m_1 m_2}|$ $=\tan ^{-1}|\;{\;\frac{8 \sqrt{5}}{3}}/{1-\;\frac{44}{9}}|$ $=\tan ^{-1}(\;\frac{24 \sqrt{5}}{35})$ $\alpha=\tan ^{-1}(\;\frac{24}{7 \sqrt{5}})$

Q72JEE Main 2022MCQ4MApplication of Derivatives

The equation of a common tangent to the parabolas $y=x^2$ and $y=-(x-2)^2$ is [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Equation of tangent of slope $m$ to $y=x^2$ $y=m x-\;\frac{1}{4} m^2$ Equation of tangent of slope $m$ to $y=-(x-2)^2$ $y=m(x-2)+\;\frac{1}{4} m^2$ If both equation represent the same line $\;\frac{1}{4} m^{2-}2 m=-\;\frac{1}{4} m^2$ $m=0,4$ So, equation of tangent $y=4 x-4$

Q73JEE Main 2022MCQ4MCircles

Let the abscissae of the two points $P$ and $Q$ on a circle be the roots of $x^{2-}4 x-6=0$ and the ordinates of ${\text{P}}$ and ${\text{Q}}$ be the roots of $y^{2+}2 y-7=0$ . If ${\text{PQ}}$ is a diameter of the circle $x^{2+}y^{2+}2 a x+2 b y+c=0$ , then the value of $(a+b-c)$ is __________. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Abscissae of ${\text{PQ}}$ are roots of $x^{2-}4 x-6=0$ Ordinates of $P Q$ are roots of $y^{2+}2 y-7=0$ and $P Q$ is diameter $\Rightarrow$ Equation of circle is $x^{2+}y^{2-}4 x+2 y-13=0$ But, given $x^{2+}y^{2+}2 a x+2 b y+c=0$ By comparison $a=-2, b=1, c=-13$ $\Rightarrow a+b-c=-2+1+13=12$

Q74JEE Main 2022MCQ4MHyperbola

If the line $x-1=0$ is a directrix of the hyperbola $k x^{2-}y^2=6$ , then the hyperbola passes through the point [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Given hyperbola : $\;\frac{x^2}{6 / k}-\;\frac{y^2}{6}=1$ Eccentricity $=e=\sqrt{1+\;\frac{6}{6 / k}}=\sqrt{1+k}$ Directrices : $x=\pm \;\frac{a}{e} \Rightarrow x=\pm \;\frac{\sqrt{6}}{\sqrt{k} \sqrt{k+1}}$ As given : $\;\frac{\sqrt{6}}{\sqrt{k} \sqrt{k+1}}=1$ $\Rightarrow k=2$ Here hyperbola is $\;\frac{x^2}{3}-\;\frac{y^2}{6}=1$ Checking the option gives $(\sqrt{5},-2)$ satisfies it.

Q75JEE Main 2022MCQ4MVector Algebra

A vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}, \hat{i}+\hat{j}$ and the plane determined by the vectors $\hat{i}-\hat{j}, \hat{i}+\hat{k}$ . The obtuse angle between $\vec{a}$ and the vector $\vec{b}=\hat{i}-2 \hat{j}+2 \hat{k}$ is [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

If $\vec{n_1}$ is a vector normal to the plane determined by $\hat{i}$ and $\hat{i}+\hat{j}$ then $\vec{n}_1=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ 1 & 1 & 0\end{vmatrix}=\hat{k}$ If $\vec{n}_2$ is a vector normal to the plane determined by $\hat{i}-\hat{j}$ and $\hat{i}+\hat{k}$ then $\overline{n}_2=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 0 & 1\end{vmatrix}=-\hat{i}-\hat{j}+\hat{k}$ Vector $\vec{a}$ is parallel to $\vec{n}_1 \times \vec{n}_2$ i.e. $\vec{a}$ is parallel to $\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 1 \\ -1 & -1 & 1\end{vmatrix}=\hat{i}-\hat{j}$ Given $\vec{b}=\hat{i}-2 \hat{j}+2 \hat{k}$ Cosine of acute angle between $\vec{a}$ and $\vec{b}=|\;\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot |\vec{b}|}|=\;\frac{1}{\sqrt{2}}$ Obtuse angle between $\vec{a}$ and $\vec{b}=\;\frac{3 \pi}{4}$

Q76JEE Main 2022MCQ4MInverse Trigonometric Functions

If $0 < x < \;\frac{1}{\sqrt{2}}$ and $\;\frac{\sin ^{-1} x}{\alpha}=\;\frac{\cos ^{-1} x}{\beta}$ , then the value of $\sin (\;\frac{2 \pi \alpha}{\alpha+\beta})$ is [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Let $\;\frac{\sin ^{-1} x}{\alpha}=\;\frac{\cos ^{-1} x}{\beta}=k \Rightarrow \sin ^{-1} x+\cos ^{-1} x=k(\alpha+\beta)$ $\Rightarrow \alpha+\beta=\;\frac{\pi}{2 k}$ Now, $\;\frac{2 \pi \alpha}{\alpha+\beta}=\;\frac{2 \pi \alpha}{\frac{\pi}{2 k}}=4 k \alpha=4 \sin ^{-1} x$ Here $\sin (\;\frac{2 \pi \alpha}{\alpha+\beta})=\sin (4 \sin ^{-1} x)$ Let $\sin ^{-1} x=\theta$ $\because x \in (0, \;\frac{1}{\sqrt{2}}) \Rightarrow \theta \in (0, \;\frac{\pi}{4})$ $\Rightarrow x=\sin \theta$ $\Rightarrow \cos \theta=\sqrt{1-x^2}$ $\Rightarrow \sin 2 \theta=2 x \cdot \sqrt{1-x^2}$ $\Rightarrow \cos 2 \theta=\sqrt{1-4 x^2(1-x^2)}=\sqrt{(2 x^{2-}1)^2}=1-2 x^2$ $\because (\cos 2 \theta>0.$ as $.2 \theta \in (0, \;\frac{\pi}{2}))$ $\Rightarrow \sin 4 \theta=2 \cdot 2 x \sqrt{1-x^2}(1-2 x^2)$ $=4 x \sqrt{1-x^2}(1-2 x^2)$

Q77JEE Main 2022MCQ4MSets and Relations

Negation of the Boolean expression $p ⇔(q \Rightarrow p)$ is [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$p ⇔(q \Rightarrow p)$ $∼(p ⇔(q ⇔ p))$ $\equiv p ⇔ ∼(q \Rightarrow p)$ $\equiv p ⇔(q ∧ ∼ p)$ $\equiv (p \Rightarrow (q ∧ ∼ p)) ∧((q ∧ ∼ p) \Rightarrow p))$ $\equiv (∼ p ∨(q ∧ ∼ p)) ∧((∼ q ∨ p) ∨ p))$ $\equiv ((∼ p ∨ q) ∧ ∼ p) ∧(∼ q ∨ p)$ $\equiv ∼ p ∧(∼ q ∨ p)$ $\equiv (∼ p ∧ ∼ q) ∨(∼ p ∧ p)$ $\equiv (∼ p ∧ ∼ q) ∨ c$ $\equiv (∼ p ∧ ∼ q)$

Q78JEE Main 2022MCQ4MProbability

Let $X$ be a binomially distributed random variable with mean 4 and variance $\;\frac{4}{3}$ . Then, $54 P(X \le 2)$ is equal to [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Mean $=4=\mu =n p$ Variance $=\sigma^2=n p(1-P)=\;\frac{4}{3}$ $4(1-P)=\;\frac{4}{3}$ $P=\;\frac{2}{3}$ $n \times \;\frac{2}{3}=4$ $n=6$ $P(X=k)={ }^n C_k P^k(1-P)^{n-k}$ $P(X \le 2)=P(X=0)+P(X=1)+P(X=2)$ $={ }^6 C_0 P^0(1-P)^{6+}{ }^6 C_1 P^1(1-P)^{5+}{ }^6 C_2 P^2(1-P)^4$ $={ }^6 C_0(\;\frac{1}{3})^{6+}{ }^6 C_1(\;\frac{2}{3})(\;\frac{1}{3})^{5+}{ }^6 C_2(\;\frac{2}{3})^2(\;\frac{1}{3})^4$ $=(\;\frac{1}{3})^6[1+12+60]=\;\frac{73}{3^6}$ $54 P(X \le 2)=\;\frac{73}{3^6} \times 54=\;\frac{146}{27}$

Q79JEE Main 2022MCQ4MIndefinite Integrals

The integral $\int \;{(1-\frac{1}{\sqrt{3}})(\cos x-\sin x)}/{(1+\;\frac{2}{\sqrt{3}} \sin 2 x)} d x$ is equal to [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$=\int \;{(1-\frac{1}{\sqrt{3}})(\cos x-\sin x)}/{(1+\;\frac{2}{\sqrt{3}} \sin 2 x)} \text{dx}$ $=\int \;{(\frac{\sqrt{3}-1}{\sqrt{3}}) \sqrt{2} \sin (\frac{\pi}{4}-x)}/{(\;\frac{2}{\sqrt{3}})(\sin \;\frac{\pi}{3}+\sin 2 x)} \text{dx}$ $=\int \;{\;\frac{(\sqrt{3}-1)}{\sqrt{2}} \sin (\frac{\pi}{4}-x)}/{(\sin \;\frac{\pi}{3}+\sin 2 x)} \text{dx}$ $=\int \;{\;\frac{\sqrt{3}-1}{2 \sqrt{2}} \sin (\frac{\pi}{4}-x)}/{\sin (\;\frac{\pi}{6}+x) \cos (\;\frac{\pi}{6}-x)} \text{dx}$ $=\;\frac{1}{2} \int \;\frac{2 \sin \frac{\pi}{12} \sin (\frac{\pi}{4}-x)}{\sin (\;\frac{\pi}{6}+x) \cos (\;\frac{\pi}{6}-x)} \text{dx}$ $=\;\frac{1}{2} \int \;\frac{\cos (\frac{\pi}{6}-x)-\cos (\frac{\pi}{3}-x)}{\sin (\;\frac{\pi}{6}+x) \cos (\;\frac{\pi}{6}-x)} \text{dx}$ $=\;\frac{1}{2}[\int {\text{cosec}}(\;\frac{\pi}{6}+x) d x-\int \sec (\;\frac{\pi}{6}-x) d x]$ $=\;\frac{1}{2}[\ln |\tan (\;\frac{\pi}{12}+\;\frac{x}{2})|-\int {\text{cosec}}(\;\frac{\pi}{3}-x) d x]$ $=\;\frac{1}{2}[\ln |\tan (\;\frac{\pi}{12}+\;\frac{x}{2})|-\ln |\;\frac{\pi}{6}+\;\frac{x}{2}|]+C$ $=\;\frac{1}{2} \ln |\;\frac{\tan (\frac{\pi}{2}+\frac{x}{2})}{\tan (\;\frac{\pi}{6}+\;\frac{x}{2})}|+C$

Q80JEE Main 2022MCQ4MArea Under The Curves

The area bounded by the curves $y= |x^{2-}1|$ and $y=1$ is [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Area $=2 int_0^{\sqrt{2}}(1-|x^{2-}1|) d x$ $2[int_0^1(1-(1-x^2)) d x+int_1^{\sqrt{2}}(2-x^2) d x]$ $.=2[[\;\frac{x^3}{3}]_0^{1+}\$ [2 x-;\frac{x^3}{3}]_1^{\sqrt{2}}]]=2(;\frac{4 \sqrt{2}-4}{3})=;\frac{8}{3}(\sqrt{2}-1)$

Q81JEE Main 2022NAT4MSets and Relations

Let $A=\{1,2,3,4,5,6,7\}$ and $B=\{3,6,7,9\}$ . Then the number of elements in the set $\{C ⊆ A: C ∩ B \ne \varphi \}$ is _______. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$A=\{7,2,3,4,5,6,7\}$ and $B=\{3,6,7,9\}$ Total subset of $A=2^7=128$ ${\text{C}} ∩ {\text{B}}=\varphi$ when set ${\text{C}}$ contains the element $1,2,4,5$ $\therefore {\text{S}}=\{ {\text{C}} ⊆ {\text{A}} ; {\text{C}} ∩ {\text{B}} \ne \varphi \}$ $=\;\text{ Total }\;-( {\text{C}} ∩ {\text{B}}=\varphi )$ $=128-2^4=112$

Q82JEE Main 2022NAT4MThree Dimensional Geometry

The largest value of $a$ , for which the perpendicular distance of the plane containing the lines $\vec{r}=(\hat{i}+\hat{j})+\lambda (\hat{i}+a \hat{j}-\hat{k})$ and $\vec{r}=(\hat{i}+\hat{j})+\mu (-\hat{i}+\hat{j}-a \hat{k})$ from the point $(2,1,4)$ is $\sqrt{3}$ , is _________. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;\text{ Normal to plane }\;=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & a & -1 \\ -1 & 1 & -a\end{vmatrix}$ $=\hat{i}(1-a^2)-\hat{j}(-a-1)+\hat{k}(1+a)$ $=(1-a) \hat{i}+\hat{j}+\hat{k}$ Distance from $(2,1,4)$ is $\sqrt{3}$ i.e. $\Rightarrow |\;\frac{(1-a)+0+4}{\sqrt{(1-a)^{2+}1+1}}|=\sqrt{3}$ $\Rightarrow 25+a^{2-}10 a=3 a^{2-}6 a+9$ $\Rightarrow 2 a^{2+}4 a-16=0$ $\Rightarrow a^{2+}2 a-8=0$ $a=2 \;\text{ or }\;-4$ $\therefore a_{\max }=2$

Q83JEE Main 2022NAT4MPermutations and Combinations

Numbers are to be formed between 1000 and 3000 , which are divisible by 4 , using the digits $1,2,3,4,5$ and 6 without repetition of digits. Then the total number of such numbers is _______. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Here $1^{\;\text{st }\;}$ digit is 1 or 2 onlyCase-IIf first digit is 1Then last two digits can be $24,32,36,52,56,64$ Case - IIIf first digit is 2 then last two digit can be $16,36,56,64$ Total ways $=12+18=30$ ways

Q84JEE Main 2022NAT4MSequences and Series

If $sum_{k=1}^{10} \;\frac{k}{k^{4+}k^{2+}1}=\;\frac{m}{n}$ , where ${\text{m}}$ and ${\text{n}}$ are co-prime, then $m+n$ is equal to ________. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$sum_{k=1}^{10} \;\frac{k}{k^{4+}k^{2+}1}$ $=\;\frac{1}{2}[sum_{k=1}^{10}(\;\frac{1}{k^{2-}k+1}-\;\frac{1}{k^{2+}k+1}).$ $=\;\frac{1}{2}[1-\;\frac{1}{3}+\;\frac{1}{3}-\;\frac{1}{7}+\;\frac{1}{7}-\;\frac{1}{13}+. . .+\;\frac{1}{91}-\;\frac{1}{111}]$ $=\;\frac{1}{2}[1-\;\frac{1}{111}]=\;\frac{110}{2.111}=\;\frac{55}{111}=\;\frac{m}{n}$ $\therefore m+n=55+111=166$

Q85JEE Main 2022NAT4MTrigonometric Ratios and Identities

If the sum of solutions of the system of equations $2 \sin ^2 \theta-\cos 2 \theta=0$ and $2 \cos ^2 \theta+3 \sin \theta=0$ in the interval $[0,2 \pi]$ is $k \pi$ , then $k$ is equal to ________. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Equation (1) $2 \sin ^2 \theta=1-2 \sin ^2 \theta$ $\Rightarrow \sin ^2 \theta=\;\frac{1}{4}$ $\Rightarrow \sin \theta=\pm \;\frac{1}{2}$ $\Rightarrow \theta=\;\frac{\pi}{6}, \;\frac{5 \pi}{6}, \;\frac{7 \pi}{6}, \;\frac{11 \pi}{6}$ Equation (2) $2 \cos ^2 \theta+3 \sin \theta=0$ $\Rightarrow 2 \sin ^2 \theta-3 \sin \theta-2=0$ $\Rightarrow 2 \sin ^2 \theta-4 \sin \theta+\sin \theta-2=0$ $\Rightarrow (\sin \theta-2)(2 \sin \theta+1)=0$ $\Rightarrow \sin \theta=\;\frac{-1}{2}$ $\Rightarrow \theta=\;\frac{7 \pi}{6}, \;\frac{11 \pi}{6}$ $\therefore$ Common solutions $=\;\frac{7 \pi}{6} ; \;\frac{11 \pi}{6}$ Sum of solutions $=\;\frac{7 \pi+11 \pi}{6}=\;\frac{18 \pi}{6}=3 \pi$ $\therefore k=3$

Q86JEE Main 2022NAT4MStatistics

The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If $\sigma$ is the standard deviation of the data after omitting the two wrong observations from the data, then $38 \sigma^2$ is equal to _________. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\mu \;\; =\;\frac{\sum x_i}{40}=30 \Rightarrow \sum x_i=1200$ $\sigma^2 \;\; =\;\frac{\sum x_i^2}{40}-(30)^2=25 \Rightarrow \sum x_i^2=37000$ After omitting two wrong observations $\sum y_i=1200-12-10=1178$ $\sum y_i^2=37000-144-100=36756$ Now $\sigma^2=\;\frac{\sum y_i^2}{38}-(\;\frac{\sum y_i}{38})^2$ $=\;\frac{36756}{38}-(\;\frac{1178}{38})^2=-31^2$ $=38 \sigma^2=36756-36518=238$

Q87JEE Main 2022NAT4MThree Dimensional Geometry

The plane passing through the line $L: l x-y+3(1-l) z=1, x+2 y-z=2$ and perpendicular to the plane $3 x+2 y+z=6$ is $3 x-8 y+7 z=4$ . If $\theta$ is the acute angle between the line $L$ and the $y$ -axis, then $415 \cos ^2 \theta$ is equal to ________. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$L: l x-y+3(1-l) z=1, x+2 y-z=2$ and plane containing the line $p: 3 x-8 y+7 z=4$ Let $\vec{n}$ be the vector parallel to ${\text{L}}$ . $\;\text{ then }\; \vec{n}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ l & -1 & 3(1-l) \\ 1 & 2 & -1\end{vmatrix}$ $=(6 l-5) \hat{i}+(3-2 l) \hat{j}+(2 l+1) \hat{k}$ $\because {\text{R}}$ containing ${\text{L}}$ $3(6 l-5)-8(3-2 l)+7(2 l+1)=0$ $18 l+16 l+14 l-15-24+7=0$ $\therefore l=\;\frac{32}{48}=\;\frac{2}{3}$ Let $\theta$ be the acute angle between $L$ and $y$ -axis $\therefore \cos \theta=\;{\;\frac{5}{3}}/{\sqrt{1+\;\frac{25}{9}+\;\frac{49}{9}}}=\;\frac{5}{\sqrt{83}}$ $\therefore 415 \cos ^2 \theta=125$

Q88JEE Main 2022NAT4MDifferential Equations

Suppose $y=y(x)$ be the solution curve to the differential equation $\;\frac{d y}{d x}-y=2-e^{-x}$ such that $\lim _{x \to \infty } y(x)$ is finite. If $a$ and $b$ are respectively the $x$ - and $y$ -intercepts of the tangent to the curve at $x=0$ , then the value of $a-4 b$ is equal to _________. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

IF $=e^{-x}$ $y \cdot e^{-x}=-2 e^{-x}+\;\frac{e^{-2 x}}{2}+C$ $\Rightarrow y=-2+e^{-x}+C e^x$ $\lim _{x \to \infty } y(x)$ is finite so $C=0$ $y=-2+e^{-x}$ $\Rightarrow \;\frac{d y}{d x}=-.e^{-x} \Rightarrow \;\frac{d y}{d x}|_{x=0}=-1$ Equation of tangent $y+1=-1(x-0)$ or $y+x=-1$ So $a=-1, b=-1$ $\Rightarrow a-4 b=3$

Q89JEE Main 2022NAT4MSequences and Series

Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ________. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$d_1=\;\frac{199-100}{2} ∉ I$ $d_2=\;\frac{199-100}{3}=33$ $d_3=\;\frac{199-100}{4} ∉ I$ $d_n=\;\frac{199-100}{i+1} \in I$ $d_i=33+11,9$ Sum of CD's $=33+11+9$ $=53$

Q90JEE Main 2022NAT4MMatrices and Determinants

The number of matrices $A=(\text{table} a , b ; c , d)$ , where $a, b, c, d \in \{-1,0,1,2,3, . . ., ., 10\}$ , such that $A=A^{-1}$ , is _________. [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\because A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ then $A^2=\begin{bmatrix}a^{2+}b c & b(a+d) \\ c(a+d) & b c+d^2\end{bmatrix}$ For ${\text{A}}^{-1}$ must exist $a d-b c \ne 0 . . . . . .$ (i)and $A=A^{-1} \Rightarrow A^2=I$ $\therefore a^{2+}b c=d^{2+}b c=1$ and $b(a+d)=c(a+d)=0$ Case ${\text{I}}$ : When ${\text{a}}= {\text{d}}=0$ , then possible values of $( {\text{b}}, {\text{c}})$ are $(1,1),(-1,1)$ and $(1,-1)$ and $(-1,1)$ .Total four matrices are possible.Case II: When ${\text{a}}=- {\text{d}}$ then $( {\text{a}}, {\text{d}})$ be $(1,-1)$ or $(-1,1)$ .Then total possible values of $(b, c)$ are $(12+11) \times 2=46$ . $\therefore$ Total possible matrices $=46+4=50$ .

← All JEE Main 2022 papers Take as Timed Mock →