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28 Jun 2022 Shift 2

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Physics30 questions

Q1JEE Main 2022MCQ4MUnits and Measurements

Velocity (v) and acceleration (a) in two systems of units 1 and 2 are related as $v_2=\;\frac{n}{m^2} v_1$ and $a_2=\;\frac{a_1}{m n}$ respectively. Here $m$ and $n$ are constants. The relations for distance and time in two systems respectively are: [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${[L]=\;\frac{[v^2]\$ }{[a]}};\text{ so }; ;\frac{ [v_2] ^2}{ [a_2] }=;\frac{[\frac{n}{m^2} v_1]$ ^2}{ [;\frac{a_1}{m_n}] };\frac{ [v_2] ^2}{ [a_2] }=;\frac{n^3}{m^3} ;\frac{ [v_1] ^2}{ [a_1] };\text{ or }; [L_2] =;\frac{n^3}{m^3} [L_1] $Similarly,$ [T]=;\frac{[v]}{[a]} $So,$ [T_2] =;\frac{n^2}{m} [T_1]$

Q2JEE Main 2022MCQ4MRotational Motion

A ball is spun with angular acceleration $\alpha=6 t^{2-}2 t$ where $t$ is in second and $\alpha$ is in rads $-2$ . At $t=0$ , the ball has angular velocity of 10 rads $^{-1}$ and angular position of 4 rad. The most appropriate expression for the angular position of the ball is [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\alpha=\;\frac{d \omega }{d t}=6 t^{2-}2 t$ $int_{0}^{\omega } d \omega =int_{0}^{t} (6 t^{2-}2 t) \text{dt}$ so $\omega =2 t^{3-}t^{2+}10$ and $\;\frac{d \theta}{d t}=2 t^{3-}t^{2+}10$ so $int_{4}^{\theta} d \theta=int_{0}^{t} (2 t^{3-}t^{2+}10) \text{dt}$ $\theta=\;\frac{t^4}{2}-\;\frac{t^3}{3}+10 t+4$

Q3JEE Main 2022MCQ4MWork, Power and Energy

A block of mass $2 {\text{kg}}$ moving on a horizontal surface with speed of $4 {\text{ms}}^{-1}$ enters a rough surface ranging from $x=0.5 {\text{m}}$ to $x=1.5 {\text{m}}$ . The retarding force in this range of rough surface is related to distance by $F=-k x$ where $k=12 {\text{Nm}}$ . The speed of the block as it just crosses the rough surface will be : [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$F=-12 x$ $m v \;\frac{d v}{d x}=-12 x$ $int_{4}^{v} v d v=-6 int_{0.5}^{1.5} x \text{dx}(m=2 {\text{kg}})$ $\;\frac{v^{2-}16}{2}=-6 [\;\frac{1.5^{2-}0.5^2}{2}]$ $\;\frac{v^{2-}16}{2}=-6$ $v=2 {\text{m}} / {\ sec}$

Q4JEE Main 2022MCQ4MRotational Motion

A $\sqrt{34} {\text{m}}$ long ladder weighing $10 {\text{kg}}$ leans on a frictionless wall. Its feet rest on the floor $3 {\text{m}}$ away from the wall as shown in the figure. If ${\text{E}}_{ {\text{f}}}$ and ${\text{F}}_{ {\text{w}}}$ are the reaction forces of the floor and the wall, then ratio of $F_w / F_f$ will be :(Use ${\text{g}}=10 {\text{m}} / {\text{s}}^2$ .) [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Taking torque from B $F_w \times 5=\;\frac{3}{2} m g$ $\Rightarrow F_w=\;\frac{3}{10} \times 10 \times 10$ $=30 N$ $N=m g=100 N$ and $f_r=F_w=30 {\text{N}}$ so $F_f={\sqrt{N^{2+}f_r^2}}={\sqrt{10900}}=10 {\sqrt{109}} N$ so $\;\frac{F_w}{F_f}=\;{3}/{{\sqrt{109}}}$

Q5JEE Main 2022MCQ4MWork, Power and Energy

Water falls from a $40 {\text{m}}$ high dam at the rate of $9 \times 10^4 {\text{kg}}$ per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydroelectric energy number of $100 {\text{W}}$ lamps, that can be lit, is : $(.$ Take $. {\text{g}}=10 {\text{ms}}^{-2})$ [28-Jun-2022-Shift-2]

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📖 Explanation

Total gravitational PE of water per second $=\;\frac{m g h}{T}$ $=\;\frac{9 \times 10^4 \times 10 \times 40}{3600}=10^4 {\text{J}} / {\ sec}$ $50 \%$ of this energy can be converted into electrical energy so total electrical energy $=\;\frac{10^4}{2}=5000$ WSo total bulbs lit can be $=\;{5000 {\text{W}}}/{100 {\text{W}}}=50$ bulbs

Q6JEE Main 2022MCQ4MGravitation

Two objects of equal masses placed at certain distance from each other attracts each other with a force of $F$ . If one-third mass of one object is transferred to the other object, then the new force will be : [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Let the masses are ${\text{m}}$ and distance between them is ${\text{I}}$ , then $F=\;\frac{G m^2}{l^2}$ .When $1 / 3^{\;\text{rd }\;}$ mass is transferred to the other then masses will be $\;\frac{4 m}{3}$ and $\;\frac{2 m}{3}$ . so new force will be $F^{'}=\;\frac{G \frac{4 m}{3} \times \frac{2 m}{3}}{l^2}=\;\frac{8}{9} \;\frac{G m^2}{l^2}=\;\frac{8}{9} F$

Q7JEE Main 2022MCQ4MProperties of Matter

A water drop of radius $1 \mu {\text{m}}$ falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is $1.8 \times 10^{-5} {\text{Nsm}}^{-2}$ and its density is negligible as compared to that of water $10^6 {\text{gm}}^{-3}$ . Terminal velocity of the water drop is(Take acceleration due to gravity $=10 {\text{ms}}^{-2}$ ) [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$6 \pi η r v=m g$ $6 \pi η r v=\;\frac{4}{3} \pi r^3 \rho g$ or $v=\;\frac{2}{9} \;\frac{\rho r^2 g}{η}=\;\frac{2}{9} \times \;\frac{10^3 \times (10^{-6}) ^2 \times 10}{1.8 \times 10^{-5}}$ $=123.4 \times 10^{-6} {\text{m}} / {\ s}$

Q8JEE Main 2022MCQ4MThermodynamics

A sample of an ideal gas is taken through the cyclic process ${\text{ABCA}}$ as shown in figure. It absorbs, $40 {\text{J}}$ of heat during the part $A B$ , no heat during $B C$ and rejects $60 {\text{J}}$ of heat during $C A$ . A work of $50 {\text{J}}$ is done on the gas during the part $B C$ . The internal energy of the gas at $A$ is $1560 {\text{J}}$ . The workdone by the gas during the part ${\text{CA}}$ is : [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\Delta {\text{U}}_{ {\text{AB}}}=40 {\text{J}}$ as process is isochoric. $\Delta {\text{U}}_{ {\text{BC}}}=+50 ( {\text{W}}_{ {\text{BC}}}=-50 {\text{J}})$ ${\text{U}}_{ {\text{C}}}= {\text{U}}_{ {\text{A}}}+\Delta {\text{U}}_{ {\text{AB}}}+\Delta {\text{U}}_{ {\text{BC}}}=1650$ For CA process, $Q_{C A}=-60 {\text{J}}$ $\Delta U_{C A}+W_{C A}=-60$ $-90+W_{C A}=-60$ $\Rightarrow W_{C A}=+30 {\text{J}}$ The graph given is inconsistent with the statement ${\text{BC}}$ may be adiabatic and ${\text{CA}}$ cannot be like isobaric as shown, as increasing volume while rejecting heat at same time.

Q9JEE Main 2022MCQ4MThermodynamics

What will be the effect on the root mean square velocity of oxygen molecules if the temperature is doubled and oxygen molecule dissociates into atomic oxygen? [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

As $v_{r m s}=\sqrt{\;\frac{3 R T}{M_0}}$ ${\text{T}}$ is doubled and oxygen molecule is dissociated into atomic oxygen molar mass is halved.So, $v_{r m s}^{'}=\sqrt{\;\frac{3 R T \times 2 T_0}{M_0 / 2}}=2 v_{r m s}$ So velocity of atomic oxygen is doubled.

Q10JEE Main 2022MCQ4MElectrostatics

Two point charges $A$ and $B$ of magnitude $+8 \times 10^{-6} {\text{C}}$ and $-8 \times 10^{-6} {\text{C}}$ respectively are placed at a distance ${\text{d}}$ apart. The electric field at the middle point ${\text{O}}$ between the charges is $6.4 \times 10^4 {\text{NC}}^{-1}$ . The distance ' ${\text{d}}$ ' between the point charges ${\text{A}}$ and $B$ is : [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Electric field at ${\text{P}}$ will be $E=\;\frac{k q}{(d / 2)^2} \times 2=\;\frac{8 k q}{d^2}$ So, $\;\frac{8 \times 9 \times 10^9 \times 8 \times 10^{-6}}{d^2}=6.4 \times 10^4$ So, $d=3 {\text{m}}$

Q11JEE Main 2022MCQ4MCurrent Electricity

Resistance of the wire is measured as $2 Ω$ and $3 Ω$ at $10^{\circ} {\text{C}}$ and $30^{\circ} {\text{C}}$ respectively. Temperature co-efficient of resistance of the material of the wire is : [28-Jun-2022-Shift-2]

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📖 Explanation

${\text{R}}_{10}=2= {\text{R}}_0(1+\alpha \times 10)$ ${\text{R}}_{30}=3= {\text{R}}_0(1+\alpha \times 30)$ On solving $\alpha=0.033 /{ }^{\circ} {\text{C}}$

Q12JEE Main 2022MCQ4MMagnetism and Matter

The space inside a straight current carrying solenoid is filled with a magnetic material having magnetic susceptibility equal to $1.2 \times 10^{-5}$ . What is fractional increase in the magnetic field inside solenoid with respect to air as medium inside the solenoid? [28-Jun-2022-Shift-2]

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📖 Explanation

${B^\vec{'}=\mu _0(1+X) n i$ (in the material) ${B^\vec{'}=\mu _0 n i$ (without material)So fractional increase is $\;\frac{B^{'}-B}{B}=X=1.2 \times 10^{-5}$

Q13JEE Main 2022MCQ4MMagnetic Effects of Current and Magnetism

Two parallel, long wires are kept $0.20 {\text{m}}$ apart in vacuum, each carrying current of $x {\text{A}}$ in the same direction. If the force of attraction per meter of each wire is $2 \times 10^{-6} {\text{N}}$ , then the value of ${\text{x}}$ is approximately : [28-Jun-2022-Shift-2]

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📖 Explanation

$\;\frac{d F}{d l}=2 \times 10^{-6} {\text{N}} / {\text{m}}=\;\frac{\mu _0 i_1 i_2}{2 \pi d}$ $2 \times 10^{-6}=\;\frac{2 \times 10^{-7} \times x^2}{0.2}$ $x={\sqrt{2}} ≃ 1.4$

Q14JEE Main 2022MCQ4MElectromagnetic Induction

A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be :(Assume the coil to be short circuited.) [28-Jun-2022-Shift-2]

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📖 Explanation

As number of turns are halved so length of wire is halved, and radius is doubled, then area will be 4 times the previous one if previous resistance is $R$ then new resistance is $\;\frac{R}{8}$ and if previous emf is $E$ then new emf will be $\;\frac{E}{2}$ so $P_i=\;\frac{E^2}{R}$ $P_f=\;\frac{(E / 2)^2}{R / 8}=\;\frac{2 E^2}{R}=2 P_i$

Q15JEE Main 2022MCQ4MElectromagnetic Waves

An EM wave propagating in $x$ -direction has a wavelength of $8 {\text{mm}}$ . The electric field vibrating $y$ -direction has maximum magnitude of $60 {\text{Vm}}^{-1}$ . Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum : [28-Jun-2022-Shift-2]

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📖 Explanation

In first 3 options speed of light is $3 \times 10^8 {\text{m}} / {\ sec}$ and in the fourth option it is $4 \times 10^8 {\text{m}} / {\ sec}$ .Using $E=C B$ We can check the option is $B$ .

Q16JEE Main 2022MCQ4MWave Optics

In Young's double slit experiment performed using a monochromatic light of wavelength $\lambda$ , when a glass plate ( $\mu =1.5)$ of thickness $x \lambda$ is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of $x$ will be : [28-Jun-2022-Shift-2]

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📖 Explanation

For the intensity to remain same the position must be of a maxima so path difference must be ${\text{n}} \lambda$ so $(1.5-1) \times \lambda =n \lambda$ $x=2 n(n=0,1,2 . . .)$ So, value of $x$ will be $x=0,2,4,6 . . .$

Q17JEE Main 2022MCQ4MDual Nature of Matter and Radiation

Let ${\text{K}}_1$ and ${\text{K}}_2$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $\lambda$ 1 and $\lambda _2$ , respectively are incident on a metallic surface. If $\lambda _1=3 \lambda _2$ then : [28-Jun-2022-Shift-2]

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📖 Explanation

$K_1=\;\frac{h c}{\lambda _1}-\varphi =\;\frac{h c}{3 \lambda _2}-\varphi$ and $K_2=\;\frac{h c}{\lambda _2}-\varphi$ from (i) and (ii) we can say $3 K_1=K_2-2 \varphi$ $K_1< \;\frac{K_2}{3}$

Q18JEE Main 2022MCQ4MAtoms and Nuclei

Following statements related to radioactivity are given below :(A) Radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions.(B) The number of un-decayed nuclei in the radioactive sample decays exponentially with time.(C) Slope of the graph of $\log _e$ (no. of undecayed nuclei) Vs. time represents the reciprocal of mean life time $(\tau )$ .(D) Product of decay constant $(\lambda )$ and half-life time $(T_{1 / 2})$ is not constant.Choose the most appropriate answer from the options given below : [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Radioactive decay is a random and spontaneous process it depends on unbalancing of nucleus. $N=N_0 e^{-\lambda t} . . . . .( {\text{B}})$ $\ln N=-\lambda t+\ln N_0$ So, slope $=-\lambda . . . .$ . (C) $t_{1 / 2}=\;\frac{\ln 2}{\lambda }$ So $t_{1 / 2} \times \lambda =\ln 2=$ Constant

Q19JEE Main 2022MCQ4MSemiconductor Electronics

In the given circuit the input voltage $V_{i n}$ is shown in figure. The cut-in voltage of $p$ -n junction diode $(D_1.$ or $.D_2)$ is $0.6 V$ . Which of the following output voltage $( {\text{V}}_0)$ waveform across the diode is correct? [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{Till}}|V| \le 0.6 {\text{V}}$ $|V_0{{}/{}|} =|V|$ So correct graph will be D.

Q20JEE Main 2022MCQ4MElectromagnetic Waves

Amplitude modulated wave is represented by $V_{A M}=10[1+0.4 \cos (2 \pi \times 10^4 t) \cos (2 \pi \times 10^7 t).$ . The total bandwidth of the amplitude modulated wave is : [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Bandwidth $=2 \times {\text{f}}_{ {\text{m}}}$ $=2 \times 10^4 {\text{Hz}}=20 {\text{kHz}}$

Q21JEE Main 2022NAT4MUnits and Measurements

A student in the laboratory measures thickness of a wire using screw gauge. The readings are $1.22 {\text{mm}}, 1.23 {\text{mm}}, 1.19 {\text{mm}}$ and $1.20 {\text{mm}}$ . The percentage error is $\;\frac{x}{121} \%$ . The value of ${\text{x}}$ is [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$I_{\;\text{mean }\;}=\;\frac{1.22+1.23+1.19+1.20}{4}=1.21$ $\Delta I_{\;\text{mean }\;}=\;\frac{0.01+0.02+0.02+0.01}{4}=0.015$ So $\% I=\;{\Delta I_{\;\text{mean }\;}}/{I_{\;\text{mean }\;}} \times 100=\;\frac{0.015}{1.21} \times 100=\;\frac{150}{121} \%$ $x=150$

Q22JEE Main 2022NAT4MSemiconductor Electronics

A zener of breakdown voltage $V_Z=8 {\text{V}}$ and maximum zener current, $I_{Z M}=10 {\text{mA}}$ is subjectd to an input voltage $V_i=10 {\text{V}}$ with series resistance ${\text{R}}=100 Ω$ . In the given circuit ${\text{R}}_{ {\text{L}}}$ represents the variable load resistance. The ratio of maximum and minimum value of $R_L$ is_____ [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Minimum value of $R_L$ for which the diode is shorted is $\;\frac{R_L}{R_L+100} \times 10=8 \Rightarrow R_L=400 Ω$ For maximum value of $R_L$ , current through diode is $10 {\text{mA}}$ So $i_R=i_{R_L}+I_{Z M}$ $\;\frac{2}{100}=\;\frac{8}{R_L}+10 \times 10^{-3}$ $10 \times 10^{-3}=\;\frac{8}{R_L}$ $R_L=800 Ω$ So $\;\frac{R_{L \max }}{R_{L \min }}=2$

Q23JEE Main 2022NAT4MWave Optics

In a Young's double slit experiment, an angular width of the fringe is $0.35^{\circ}$ on a screen placed at $2 {\text{m}}$ away for particular wavelength of $450 {\text{nm}}$ . The angular width of the fringe, when whole system is immersed in a medium of refractive index $7 / 5$ , is $\;\frac{1}{\alpha}$ . The value of $\alpha$ is____ [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Angular fringe width $\theta=\;\frac{\lambda }{D}$ So $\;\frac{\theta_1}{\lambda _1}=\;\frac{\theta_2}{\lambda _2}$ $\theta_2=\;{0.35^{\circ}}/{450 {\text{nm}}} \times \;{450 {\text{nm}}}/{715}=0.25^{\circ}=\;\frac{1}{4}$ So $\alpha=4$

Q24JEE Main 2022NAT4MAlternating Current

In the given circuit, the magnitude of ${\text{V}}_{ {\text{L}}}$ and ${\text{V}}_{ {\text{C}}}$ are twice that of ${\text{V}}_{ {\text{R}}}$ . Given that $f=50 {\text{Hz}}$ , the inductance of the coil is $\;\frac{1}{K \pi} {\text{mH}}$ . The value of ${\text{K}}$ is____ [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$V_L=2 V_R$ So $\omega L i=2 R i$ $\Rightarrow L=\;\frac{2 R}{\omega }=\;\frac{2 \times 5}{2 \pi \times 50}=\;\frac{1}{10 \pi} H=\;\frac{100}{\pi} H$ So $k=\;\frac{1}{100} ≃ 0$

Q25JEE Main 2022NAT4MCurrent Electricity

All resistances in figure are $1 Ω$ each. The value of current 'I' is $\;\frac{a}{5} {\text{A}}$ . The value of a is______ [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Let the current is ${\text{i}}$ Using Kirchhoff's law $i R+\;\frac{i}{2} R+\;\frac{i}{4} R+\;\frac{i}{8} R=3$ $i=\;\frac{3 \times 8}{15}=\;\frac{8}{5} A$ So $a=8$

Q26JEE Main 2022NAT4MElectrostatic Potential and Capacitance

A capacitor ${\text{C}}_1$ of capacitance $5 \mu {\text{F}}$ is charged to a potential of $30 {\text{V}}$ using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor ${\text{C}}_2$ of capacitance $10 \mu {\text{F}}$ as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor ${\text{C}}_2$ is ____ $\mu {\text{C}}$ [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Let the charge $q$ is flown in the circuit.So using Kirchhoff's law, $\;\frac{q}{10}=\;\frac{150-q}{5}$ $q=100 \mu C$

Q27JEE Main 2022NAT4MWaves

A tunning fork of frequency $340 {\text{Hz}}$ resonates in the fundamental mode with an air column of length $125 {\text{cm}}$ in a cylindrical tube closed at one end. When water is slowly poured in it, the minimum height of water required for observing resonance once again is _____cm.(Velocity of sound in air is $340 {\text{ms}}^{-1}$ ) [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Given $340=\;\frac{n}{4 \times 125} v$ $\Rightarrow n=5$ So $\lambda =100 {\text{cm}}$ So minimum height is $\;\frac{\lambda }{2}=50 {\text{cm}}$

Q28JEE Main 2022NAT4MProperties of Matter

A liquid of density $750 {\text{kgm}}^{-3}$ flows smoothly through a horizontal pipe that tapers in cross-sectional area from ${\text{A}}_1=1.2 \times 10$ $-2 {\text{m}}^2$ to ${\text{A}}_2=\;\frac{A_1}{2}$ . The pressure difference between the wide and narrow sections of the pipe is $4500 {\text{Pa}}$ . The rate of flow of liquid is _____ $\times 10^{-3} {\text{m}}^3 {\text{s}}^{-1}$ [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Using Bernoulli's equation $P_1+\;\frac{1}{2} \rho v^2=P_2+\;\frac{1}{2} \rho 4 v^2$ $\;\frac{3}{2} \rho v^2=P_1-P_2$ $\Rightarrow v=\sqrt{\;\frac{2 (P_1-P_2) }{3 \rho }}$ $=\sqrt{\;\frac{2 \times 4500}{3 \times 750}}=2 {\text{m}} / {\ sec}$ So $Q=A_1 v=24 \times 10^{-3} {\text{m}}^3 / {\ sec}$

Q29JEE Main 2022NAT4MRotational Motion

A uniform disc with mass $M=4 {\text{kg}}$ and radius ${\text{R}}=10 {\text{cm}}$ is mounted on a fixed horizontal axle as shown in figure. A block with mass $m=2 {\text{kg}}$ hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is ____ ${\text{N}} .$ $(.$ Take $. {\text{g}}=10 {\text{ms}}^{-2})$ [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$20-T=2 a$ and $0.1 \times {\text{T}}=0.02 \alpha=\;\frac{0.02 a}{0.1}$ ${\text{T}}=2 {\text{a}}$ $\Rightarrow {\text{a}}=5 {\text{m}} / {\ sec}^2$ So ${\text{T}}=10 {\text{N}}$

Q30JEE Main 2022NAT4MMotion in a Straight Line

A car covers $A B$ distance with first one-third at velocity $v_1 {\text{ms}}^{-1}$ , second one-third at ${\text{v}}_2 {\text{ms}}^{-1}$ and last one-third at ${\text{v}}_3 {\text{ms}}{ }^{-1}$ . If $v_3=3 v_1, v_2=2 v_1$ and $v_1=11 {\text{ms}}^{-1}$ then the average velocity of the car is ______ ${\text{ms}}^{-1}$ [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$v_{\;\text{mean }\;}=\;\frac{3 v_1 v_2 v_3}{v_1 v_2+v_2 v_3+v_3 v_1}$ $=\;\frac{3 \times 11 \times 22 \times 33}{11 \times 22+22 \times 33+33 \times 11}$ $=18 {\text{m}} / {\ sec}$

Chemistry30 questions

Q31JEE Main 2022MCQ4MSome Basic Concepts of Chemistry

Compound A contains $8.7 \%$ Hydrogen, $74 \%$ Carbon and $17.3 \%$ Nitrogen. The molecular formula of the compound is,Given : Atomic masses of ${\text{C}}, {\text{H}}$ and ${\text{N}}$ are 12,1 and 14 amu respectively.The molar mass of the compound $A$ is $162 {\text{g}} {\text{mol}}^{-1}$ . [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Mole ratio of ${\text{H}}, {\text{C}}$ and ${\text{N}}$ $=\;\frac{8.7}{1}: \;\frac{74}{12}: \;\frac{17.3}{14}$ $=8.7: 6.167: 1.23$ $=\;\frac{8.7}{1.23}: \;\frac{6.167}{1.23}: \;\frac{1.23}{1.23}$ $=7: 5: 1$ $\therefore \;\text{ Emperical formula }\;=C_5 H_7 N$ $\therefore \;\text{ Molecular formula }\;= (C_5 H_7 N) _n$ $\;\text{ Given molecular mass }\;=162$ $\;\text{ Molecular mass of }\; (C_5 H_7 N) _n$ $=(5 \times 12+7 \times 1+14) \times n$ $=(81) \times n$ $\therefore 81 \times n=162$ $\Rightarrow n=2$ $\therefore$ Molecular formula $= {\text{C}}_{10} {\text{H}}_{14} {\text{N}}_2$

Q32JEE Main 2022MCQ4MStructure of Atom

Consider the following statements :(A) The principal quantum number ' $n$ ' is a positive integer with values of ' $n$ ' $=1,2,3, . . .$ (B) The azimuthal quantum number 'I' for a given 'n' (principal quantum number) can have values as 'I' = 0, 1, 2,..... $n$ (C) Magnetic orbital quantum number 'm' for a particular 'l' (azimuthal quantum number) has (2l + 1 ) values.(D) $\pm 1 / 2$ are the two possible orientations of electron spin.(E) For $I=5$ , there will be a total of 9 orbitalWhich of the above statements are correct? [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

(A) Principle quantum number ${\text{n}}$ is a positive integer and it's possible values are ${\text{n}}=1,2,3 . . . . . .$ . $\therefore {\text{A}}$ is correct.(B) Azimuthal quantum number 'I' for a given ' $n$ ' can have values as $I=0,1,2 . . . . . . .(n-1)$ $\therefore$ Statement $B$ is wrong.(C) Magnetic orbital quantum number $m_l$ for particular I has values from $-I$ to $+1$ including zero means $2 {\text{l}}+1$ values. $\therefore$ Statement ${\text{C}}$ is correct.(D) $\pm \;\frac{1}{2}$ are the two possible orientation of electron spin. $\therefore$ Statement $D$ is correct.(E) For $I=5$ , there will be a total of 11 orbitals. $l=0 \Rightarrow 5$ subshell $l=1 \Rightarrow {\text{p}}$ subshell $l=2 \Rightarrow {\text{d}}$ subshell $l=3 \Rightarrow {\text{f}}$ subshell $l=4 \Rightarrow {\text{g}}$ subshell $l=5 \Rightarrow {\text{h}}$ subshellWe know,Number of orbital in any subshell $=2 l+1$ . $\therefore$ For h subshell, number of orbitals $=2 \times 5+1=11$

Q33JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

In the structure of ${\text{SF}}_4$ , the lone pair of electrons on ${\text{S}}$ is in. [28-Jun-2022-Shift-2]

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📖 Explanation

${\text{SF}}_4 \to {\ sp}^3 {\text{d}}$ hybridisation.The lone pair of electrons on ${\text{S}}$ is in an equatorial position and there are two lone pair-bond pair repulsions at $90^{\circ}$ .

Q34JEE Main 2022MCQ4MIonic Equilibrium

A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of $\;{[ {\text{CH}}_3 {\text{CH}}_2 {\text{COO}}^{-}]\$ }/{[ {\text{CH}}_3 {\text{CH}}_2 {\text{COOH}} ].} $required to make buffer isGiven :$ K_a( {\text{CH}}_3 {\text{CH}}_2 {\text{COOH}})=1.3 × 10^{-5}$ [28-Jun-2022-Shift-2]

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📖 Explanation

${\text{CH}}_3 {\text{CH}}_2 {\text{COOH}} ⇌ {\text{CH}}_3 {\text{CH}}_2 {\text{COO}}^{-}+ {\text{H}}^{+}$ From Henderson equation ${\text{pH}}= {\text{pK}}_{ {\text{a}}}+\log \;{[ {\text{CH}}_3 {\text{CH}}_2 {\text{COO}}^{-}]\$ }/{ [ {\text{CH}}_3 {\text{CH}}_2 {\text{COOH}}] }4=-\log 1.3 × 10^{-5}+\log ;{[ {\text{CH}}_3 {\text{CH}}_2 {\text{COO}}^{-}]$ }/{[ {\text{CH}}_3 {\text{CH}}_2 {\text{COOH}}^2]$ }-\log 10^{-4}=-\log 1.3 × 10^{-5}+\log ;{[ {\text{CH}}_3 {\text{CH}}_2 {\text{COO}}^{-}]$ }/{[ {\text{CH}}_3 {\text{CH}}_2 {\text{COOH}}^{-}]$ }-\log 10^{-4}=-\log 1.3 × 10^{-5} ;{ [ {\text{CH}}_3 {\text{CH}}_2 {\text{COOH}}] }/{[ {\text{CH}}_3 {\text{CH}}_2 {\text{COO}}^{-}]$ }10^{-4}=1.3 \times 10^{-5} ;{[ {\text{CH}}_3 {\text{CH}}_2 {\text{COOH}}^3]$ }/{[ {\text{CH}}_3 {\text{CH}}_2 {\text{COO}}^{-}]$ };{[ {\text{CH}}_3 {\text{CH}}_2 {\text{COO}}^{-}]$ }/{ [ {\text{CH}}_3 {\text{CH}}_2 {\text{COOH}}] }=0.13$

Q35JEE Main 2022MCQ4MSolutions

Choose the correct answer from the options given below: List-I List-II (A) negatively charged sol (I) ${\text{Fe}}_2 {\text{O}}_3 \cdot x {\text{H}}_2 {\text{O}}$ (B) macromolecular colloid (II) CdS sol (C) positively charged sol (III) Starch (D) Cheese (IV) a gel [28-Jun-2022-Shift-2]

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📖 Explanation

Negative charged sol $= {\text{CdS}}$ sol (II)Macromolecular colloid = starch (III)Positively charged sol $= {\text{Fe}}_2 {\text{O}}_3 \cdot {\text{xH}}_2 {\text{O}}$ (I)Cheese $=$ gel (IV)

Q36JEE Main 2022MCQ4MSolutions

Choose the correct answer from the options given below: List-I (Oxide) List-II (Nature) (A) ${\text{Cl}}_2 {\text{O}}_7$ (I) Amphoteric (B) ${\text{Na}}_2 {\text{O}}$ (II) Basic (C) ${\text{Al}}_2 {\text{O}}_3$ (III) Neutral (D) ${\text{N}}_2 {\text{O}}$ (IV) Acidic [28-Jun-2022-Shift-2]

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📖 Explanation

(A) ${\text{Cl}}_2 {\text{O}}_7 \to$ Acidic(B) ${\text{Na}}_2 {\text{O}} \to$ Basic(C) ${\text{Al}}_2 {\text{O}}_3 \to$ Amphoteric(D) ${\text{N}}_2 {\text{O}} \to$ NeutralOxides of metals are basic in nature whereas oxides of non-metals are acidic in nature. ${\text{N}}_2 {\text{O}}$ is a neutral oxide.

Q37JEE Main 2022MCQ4Md and f Block Elements

In the metallurgical extraction of copper, following reaction is used: ${\text{FeO}}+ {\text{SiO}}_2 \to {\text{FeSiO}}_3$ ${\text{FeO}}$ and ${\text{FeSiO}}_3$ respectively are. [28-Jun-2022-Shift-2]

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📖 Explanation

${\text{FeO}}+ {\text{SiO}}_2 \to {\text{FeSiO}}_3$ ${\text{FeO}}=\;\text{ Gangue and }\; {\text{FeSiO}}_3=\;\text{ Slag }\;$

Q38JEE Main 2022MCQ4MStructure of Atom

Hydrogen has three isotopes : protium $({ }^1 {\text{H}})$ , deuterium $({ }^2 {\text{H}}.$ or $. {\text{D}})$ and tritium $({ }^3 {\text{H}}.$ or $. {\text{T}})$ . They have nearly same chemical properties but different physical properties. They differ in [28-Jun-2022-Shift-2]

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📖 Explanation

${ }^1 {\text{H}},{ }^2 {\text{H}}$ and ${ }^3 {\text{H}}$ have same atomic number but their atomic masses are different.Isotopes have same atomic number (i.e. same number of protons)

Q39JEE Main 2022MCQ4MClassification of Elements

Among the following, basic oxide is : [28-Jun-2022-Shift-2]

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📖 Explanation

Since, oxides of metals are basic in nature. Hence ${\text{CaO}}$ is a basic oxide. ${\text{SO}}_3$ and ${\text{SiO}}_2$ are acidic oxides and ${\text{Al}}_2 {\text{O}}_3$ is a amphoteric oxide.

Q40JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

Among the given oxides of nitrogen; ${\text{N}}_2 {\text{O}}, {\text{N}}_2 {\text{O}}_3, {\text{N}}_2 {\text{O}}_4$ and ${\text{N}}_2 {\text{O}}_5$ , the number of compound/(s) having ${\text{N}}- {\text{N}}$ bond is : [28-Jun-2022-Shift-2]

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📖 Explanation

${\text{N}}_2 {\text{O}}, {\text{N}}_2 {\text{O}}_3$ and ${\text{N}}_2 {\text{O}}_4$ contain ${\text{N}}- {\text{N}}$ bond

Q41JEE Main 2022MCQ4Mp Block Elements

Which of the following oxoacids of sulphur contains "S" in two different oxidation states? [28-Jun-2022-Shift-2]

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📖 Explanation

In ${\text{H}}_2 {\text{S}}_2 {\text{O}}_3$ , sulphur exhibits two different oxidation states $+6$ and $-2$ .

Q42JEE Main 2022MCQ4MHydrocarbons

Correct statement about photo-chemical smog is : [28-Jun-2022-Shift-2]

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Q43JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

The correct IUPAC name of the following compound is : [28-Jun-2022-Shift-2]

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Q44JEE Main 2022MCQ4MAlcohols, Phenols and Ethers

The major product $(P)$ of the given reaction is(where, Me is $- {\text{CH}}_3$ ) [28-Jun-2022-Shift-2]

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📖 Explanation

(1) In compound A, positive charge on ${\text{O}}$ atom is not stable that is why $\;{ {\text{O}}}^{⊕} {\text{H}}_2$ get's removed.(2) In Carbocation (B), two $\alpha- {\text{H}}$ present and after 1, 2-methyl shift number of $\alpha- {\text{H}}$ becomes six so it becomes more stable as far carbocation when number of $\alpha$ -H increases stability increases.(3) Among product D, E and F, D is major product as in case of alkene most stable, alkene are those which have more $\alpha- {\text{H}}$ . And every reactant produce that product which is more stable.

Q45JEE Main 2022MCQ4MHaloalkanes and Haloarenes

In the above reaction 'A' is [28-Jun-2022-Shift-2]

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Q46JEE Main 2022MCQ4MAldehydes, Ketones and Carboxylic Acids

Isobutyraldehyde on reaction with formaldehyde and ${\text{K}}_2 {\text{CO}}_3$ gives compound ' ${\text{A}}$ '. Compound ' ${\text{A}}$ ' reacts with ${\text{KCN}}$ and yields compound ' $B$ ', which on hydrolysis gives a stable compound ' $C$ '. The compound ' $C$ ' is [28-Jun-2022-Shift-2]

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Q47JEE Main 2022MCQ4MAmines

With respect to the following reaction, consider the given statements : (A) o-Nitroaniline and p-nitroaniline are the predominant products.(B) p-Nitroaniline and m-nitroaniline are the predominant products.(C) ${\text{HNO}}_3$ acts as an acid.(D) ${\text{H}}_2 {\text{SO}}_4$ acts as an acid.Choose the correct option. [28-Jun-2022-Shift-2]

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Q48JEE Main 2022MCQ4MBiomolecules

Given below are two statements, one is Assertion (A) and other is Reason (R).Assertion (A) : Natural rubber is a linear polymer of isoprene called sis-polyisoprene with elastic properties.Reason $(R)$ : The cis-polyisoprene molecules consist of various chains held together by strong polar interactions with coiled structure.In the light of the above statements, choose the correct one from the options given below : [28-Jun-2022-Shift-2]

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Q49JEE Main 2022MCQ4MBiomolecules

When sugar ' ${\text{X}}$ ' is boiled with dilute ${\text{H}}_2 {\text{SO}}_4$ in alcoholic solution, two isomers ' $A$ ' and ' $B$ ' are formed. ' $A$ ' on oxidation with ${\text{HNO}} 3$ yields saccharic acid where as ' ${\text{B}}$ ' is laevorotatory. The compound ' ${\text{X}}$ ' is : [28-Jun-2022-Shift-2]

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Q50JEE Main 2022MCQ4MBiomolecules

The drug tegamet is : [28-Jun-2022-Shift-2]

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Q51JEE Main 2022NAT4MStates of Matter

$100 {\text{g}}$ of an ideal gas is kept in a cylinder of $416 {\text{L}}$ volume at $27^{\circ} {\text{C}}$ under $1.5$ bar pressure. The molar mass of the gas is _____ ${\text{g}} {\text{mol}}^{-1}$ . (Nearest integer)(Given : ${\text{R}}=0.083 {\text{L}} {\bar} {\text{K}}^{-1} {\text{mol}}^{-1}$ ) [28-Jun-2022-Shift-2]

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📖 Explanation

Given, Mass of ideal gas $=100 {\text{gm}}$ Let the molar mass of ideal gas $=M$ $\therefore$ Number of moles of gas $(n)=\;\frac{100}{M}$ Volume of cylinder $(V)=416 {\text{L}}$ Temperature $(T)=(27+273) K=300 {\text{K}}$ Pressure $(P)=1.5$ bar $R=0.083 {\text{L}} {\bar} {\text{K}}^{-1} {\text{mol}}^{-1}$ Using ideal gas equation, ${\text{PV}}= {\text{nRT}}$ $\Rightarrow 1.5 \times 416=\;\frac{100}{M} \times 0.083 \times 300$ $\Rightarrow {\text{M}}=4$

Q52JEE Main 2022NAT4MChemical Thermodynamics

For combustion of one mole of magnesium in an open container at $300 {\text{K}}$ and 1 bar pressure, $\Delta _C {\text{H}}^{Θ}=-601.70 {\text{kJ}}$ mol ${ }^{-1}$ , the magnitude of change in internal energy for the reaction is _____ kJ. (Nearest integer)(Given : ${\text{R}}=8.3 {\text{JK}}^{-1} {\text{mol}}^{-1}$ ) [28-Jun-2022-Shift-2]

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📖 Explanation

${\text{Mg}}( {\ s})+\;\frac{1}{2} {\text{O}}_2( {\text{g}}) \longrightarrow {\text{MgO}}( {\ s})$ $\Delta {\text{H}}=\Delta {\text{U}}+\Delta {\text{ngRT}}$ $\Delta {\text{ng}}=-\;\frac{1}{2}$ $-601.70=\Delta {\text{U}}-\;\frac{1}{2}(8.3)(300) \times 10^{-3}$ $\Delta {\text{U}}=-601.70+1.245$ $\Delta {\text{U}} ≃-600 {\text{kJ}}$ The magnitude of change in internal energy is $600 {\text{kJ}}$ .

Q53JEE Main 2022NAT4MSolutions

$2.5 {\text{g}}$ of protein containing only glycine $( {\text{C}}_2 {\text{H}}_5 {\text{NO}}_2)$ is dissolved in water to make $500 {\text{mL}}$ of solution. The osmotic pressure of this solution at $300 {\text{K}}$ is found to be $5.03 \times 10^{-3}$ bar. The total number of glycine units present in the protein is_____(Given : ${\text{R}}=0.083 {\text{L}} {\bar} {\text{K}}^{-1} {\text{mol}}^{-1}$ ) [28-Jun-2022-Shift-2]

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📖 Explanation

Since, $\pi= {\text{icRT}}$ $5.03 \times 10^{-3}=\;\frac{2.5}{M} \times \;\frac{1000}{500} \times 0.083 \times 300$ Molar mass of protein $=24751.5 {\text{g}} / {\text{mol}}$ Number of glycine units in protein $=\;\frac{24751.5}{75}$ $=330$

Q54JEE Main 2022NAT4MElectrochemistry

For the given reactions ${\text{Sn}}^{2+}+2 {\text{e}}^{-} \to {\text{Sn}}$ ${\text{Sn}}^{4+}+4 {\text{e}}^{-} \to {\text{Sn}}$ the electrode potentials are $; E_{S n^{2+} / S n}^o=-0.140 {\text{V}}$ and $E_{S n^{4+} / S n}^o=-0.010 {\text{V}}$ . The magnitude of standard electrode potential for $S n^{4+} / S n^{2+}$ i.e. $E_{S n^{4+} / S n^{2+}}^o$ is _____ $\times 10^{-2} {\text{V}}$ . (Nearest integer) [28-Jun-2022-Shift-2]

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📖 Explanation

${\text{Sn}} \longrightarrow {\text{Sn}}^{2+}+2 {\text{e}}^{-} \;\; {\text{E}}_1^0=0.140 {\text{V}}$ ${\text{Sn}}^{4+}+4 {\text{e}}^{-} \longrightarrow {\text{Sn}} \;\; {\text{E}}_2^0=0.010 {\text{V}}$ ${\text{Sn}}^{4+}+2 {\text{e}}^{-} \longrightarrow {\text{Sn}}^{2+} \;\; {\text{E}}_{\;\text{cell }\;}^0$ ${\text{E}}_{\;\text{cell }\;}^{ {\text{O}}}=\;{ {\text{n}}_2 {\text{E}}_2^{ {\text{o}}}+ {\text{n}}_1 {\text{E}}_1^0}/{ {\text{n}}}=\;\frac{4(0.010)+2(0.140)}{2}$ ${\text{E}}_{\;\text{cell }\;}^0=0.16 {\text{V}}=16 \times 10^{-2} {\text{V}}$

Q55JEE Main 2022NAT4MChemical Kinetics

A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is_______ (Nearest integer)(Given : antilog $0.125=1.333$ , antilog $0.693=4.93$ ) [28-Jun-2022-Shift-2]

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📖 Explanation

$\lambda =\;\frac{2.303}{t} \log \;\frac{A_0}{A}$ $\;\frac{0.693}{200}=\;\frac{2.303}{83} \log \;\frac{A_0}{A}$ $\;\frac{A}{A_0}=0.75$ Hence, percentage of original activity remaining after 83 days is $75 \%$

Q56JEE Main 2022NAT4MCoordination Compounds

${[ {\text{Fe}}( {\text{CN}})_6]^{4-}}$ ${[ {\text{Fe}}( {\text{CN}})_6]^{3-}}$ ${[ {\text{Ti}}( {\text{CN}})_6]^{3-}}$ ${[ {\text{Ni}}( {\text{CN}})_4]^{2-}}$ ${[ {\text{Co}}( {\text{CN}})_6]^{3-}}$ Among the given complexes, number of paramagnetic complexes is____ [28-Jun-2022-Shift-2]

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📖 Explanation

${ [ {\text{Fe}}( {\text{CN}})_6] ^{4-}}\;\;\text{ Diamagnetic }\;$ ${ [ {\text{Fe}}( {\text{CN}})_6] ^{3-} \;\; \;\text{ Paramagnetic (1 unpaired electron) }\;}$ ${ [ {\text{Ti}}( {\text{CN}})_6] ^{3-} \;\; \;\text{ Paramagnetic (1 unpaired electron) }\;}$ ${ [ {\text{Ni}}( {\text{CN}})_4] ^{2-} \;\; \;\text{ Diamagnetic }\;}$ ${ [ {\text{Co}}( {\text{CN}})_6] ^{3-} \;\; \;\text{ Diamagnetic }\;}$

Q57JEE Main 2022NAT4MCoordination Compounds

(a) ${\text{CoCl}}_3 \cdot 4 {\text{NH}}_3$ ,(b) ${\text{CoCl}}_3 .5 {\text{NH}}_3$ ,(c) ${\text{CoCl}}_3 \cdot 6 {\text{NH}}_3$ and(d) ${\text{CoCl}}( {\text{NO}}_3)_2 \cdot 5 {\text{NH}}_3$ .Number of complex(es) which will exist in cis-trans form is/are_______ [28-Jun-2022-Shift-2]

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📖 Explanation

${\text{CoCl}}_3 \cdot 4 {\text{NH}}_3 \Rightarrow [ {\text{Co}} ( {\text{NH}}_3) _4 {\text{Cl}}_2] {\text{Cl}}$ ${\text{CoCl}}_3 \cdot 5 {\text{NH}}_3 \Rightarrow [ {\text{Co}} ( {\text{NH}}_3) _5 {\text{Cl}}] {\text{Cl}}_2$ ${\text{CoCl}}_3 \cdot 6 {\text{NH}}_3 \Rightarrow [ {\text{Co}} ( {\text{NH}}_3) _6] {\text{Cl}}_3$ Only $[ {\text{Co}} ( {\text{NH}}_3) _4 {\text{Cl}}_2]$ can show geometrical isomerism. Hence can exist in cis-trans form.

Q58JEE Main 2022NAT4MOrganic Chemistry - Some Basic Principles

The complete combustion of $0.492 {\text{g}}$ of an organic compound containing ' ${\text{C}}$ ', ' ${\text{H}}$ ' and ' ${\text{O}}$ ' gives $0.793 {\text{g}}$ of ${\text{CO}}_2$ and $0.442 {\text{g}}$ of ${\text{H}}_2 {\text{O}}$ . The percentage of oxygen composition in the organic compound is_______ (nearest integer) [28-Jun-2022-Shift-2]

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📖 Explanation

Total organic compound $=0.492 {\text{gm}}$ Produced ${\text{CO}}_2=0.793 {\text{gm}}$ $\therefore$ Moles of ${\text{CO}}_2=\;\frac{0.793}{44}$ $\therefore$ Moles of ${\text{C}}$ atoms $=\;\frac{0.793}{44}$ $\therefore$ Weight of ${\text{C}}$ atoms $=\;\frac{0.793}{44} \times 12=0.216 {\text{g}}$ Produced ${\text{H}}_2 {\text{O}}=0.442 {\text{gm}}$ $\therefore$ Moles of ${\text{H}}_2 {\text{O}}=\;\frac{0.442}{18}$ $\therefore$ Moles of ${\text{H}}$ atoms $=\;\frac{0.442}{18}$ $\therefore$ Weight of ${\text{H}}$ atoms $=\;\frac{0.442}{18} \times 2=0.05 {\text{g}}$ $\therefore$ Weight of $O$ atoms $=0.492-(0.216+0.05)$ $=0.226 {\text{gm}}$ $\%$ by mass of oxygen in compound $=\;\frac{0.226}{0.492} \times 100=46 \%$

Q59JEE Main 2022NAT4MHaloalkanes and Haloarenes

The major product of the following reaction contains _____bromine atom(s). [28-Jun-2022-Shift-2]

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📖 Explanation

Here only carbon 12th and 13th is not taking part in the resonance. All other carbons are taking part in resonance and lone pair of ${\text{O}}$ is also taking part in the resonance.As because of resonance stability of compound increases Br will not attack to any carbon which is taking part in the resonance and decrease the stability of compound.So, Br can only attack 12 th carbon or 13 th carbon atom but 12th carbon is a alpha carbon and it can easily donate ${\text{H}}^{+}$ ion and add ${\text{Br}}^{-}$ ion and increase stability.

Q60JEE Main 2022NAT4MRedox Reactions

0.01 ${\text{M}} {\text{KMnO}}_4$ solution was added to $20.0 {\text{mL}}$ of $0.05 {\text{M}} {\text{Mohr}}$ s salt solution through a burette. The initial reading of $50 {\text{mL}}$ burette is zero. The volume of ${\text{KMnO}}_4$ solution left in the burette after the end point is _____ ${\text{mL}}$ . (nearest integer) [28-Jun-2022-Shift-2]

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📖 Explanation

Meq of oxidizing agent $=$ Meq of reducing agent $( {\text{M}} \times {\text{V}} \times {\text{n}}_{ {\text{F}}}) _{ {\text{KMnO}}_4}= ( {\text{M}} \times {\text{V}} \times {\text{n}}_{ {\text{F}}}) _{ {\text{Mohr}}^{'} \;\text{ salt }\;}$ $0.01 \times 20 \times 5=0.05 \times V \times 1$ Volume required $=20 {\text{ml}}$ Since initial volume of ${\text{KMnO}}_4$ in burette is $50 {\text{ml}}$ .Hence volume of ${\text{KMnO}}_4$ left in the burette after end point is $30 {\text{ml}}$ .

Mathematics30 questions

Q61JEE Main 2022MCQ4MSets and Relations

Let $R_1=\{(a, b) \in N \times N:| a-b| \le 13\}$ and $R_2=\{(a, b) \in N \times N:| a-b| \ne 13\}$ . Then on $N:$ [28-Jun-2022-Shift-2]

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📖 Explanation

$R_1=\{(a, b) \in N \times N:|a-b| \le 13\}$ and $R_2=\{(a, b) \in N \times N:|a-b| \ne 13\}$ In $R_1: \because |2-11|=9 \le 13$ $\therefore \;\; (2,11) \in R_1$ and $(11,19) \in R_1$ but $(2,19) ∉ R_1$ $\therefore \;\; R_1$ is not transitiveHence $R_1$ is not equivalenceIn $R_2:(13,3) \in R_2$ and $(3,26) \in R_2$ but $(13,26) ∉ R_2( \because |13-26|=13)$ $\therefore R_2$ is not transitiveHence $R_2$ is not equivalence.

Q62JEE Main 2022MCQ4MQuadratic Equation and Inequalities

Let $f(x)$ be a quadratic polynomial such that $f(-2)+f(3)=0$ . If one of the roots of $f(x)=0$ is $-1$ , then the sum of the roots of $f(x)=0$ is equal to : [28-Jun-2022-Shift-2]

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📖 Explanation

$\because x=-1$ be the roots of $f(x)=0$ $\therefore$ Let $f(x)=A(x+1)(x-1) . . . . . .$ (i)Now, $f(-2)+f(3)=0$ $\Rightarrow A[-1(-2-b)+4(3-b)]=0$ $b=\;\frac{14}{3}$ $\therefore$ Second root of $f(x)=0$ will be $\;\frac{14}{3}$ $\therefore$ Sum of roots $=\;\frac{14}{3}-1=\;\frac{11}{3}$

Q63JEE Main 2022MCQ4MPermutations and Combinations

The number of ways to distribute 30 identical candies among four children $C_1, C_2, C_3$ and $C_4$ so that $C_2$ receives at least 4 and at most 7 candies, $C_3$ receives at least 2 and at most 6 candies, is equal to: [28-Jun-2022-Shift-2]

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📖 Explanation

By multinomial theorem, no. of ways to distribute 30 identical candies among four children $C_1, C_2$ and ${\text{C}}_3, {\text{C}}_4$ $=\;\text{ Coefficient of }\; x^{30} \;\text{ \in }\; (x^{4+}x^{5+}. . . .+x^7) (x^{2+}x^{3+}. . . .+x^6) (1+x+x^2 . . . . .) ^2$ $=\;\text{ Coefficient of }\; x^{24} \;\text{ \in }\; \;\frac{ (1-x^4) }{1-x} \;\frac{ (1-x^5) }{1-x} \;\frac{ (1-x^{31}) ^2}{(1-x)^2}$ $=\;\text{ Coefficient of }\; x^{24} \;\text{ \in }\; (1-x^{4-}x^{5+}x^9) (1-x)^{-4}$ $={ }^{27} C_{24}-{ }^{23} C_{20}-{ }^{22} C_{19}+{ }^{18} C_{15}=430$

Q64JEE Main 2022MCQ4MBinomial Theorem

The term independent of $x$ in the expansion of $(1-x^{2+}3 x^3)(\;\frac{5}{2} x^{3-}\;\frac{1}{5 x^2})^{11}, x \ne 0$ is : [28-Jun-2022-Shift-2]

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📖 Explanation

General term of Binomial expansion $(\;\frac{5}{2} x^{3-}\;\frac{1}{5 x^2}) ^{11}$ is $T_{r+1}={ }^{11} C_r \cdot (\;\frac{5}{2} x^3) ^{11-r} \cdot (-\;\frac{1}{5 x^2}) ^r$ $={ }^{11} C_r \cdot (\;\frac{5}{2}) ^{11-r} \cdot (-\;\frac{1}{5}) ^r \cdot x^{33-5 r}$ In the term, $(1-x^{2+}3 x^3) (\;\frac{5}{2} x^{3-}\;\frac{1}{5 x^2}) ^{11}$ Term independent of ${\text{x}}$ is when $\;\text{ (1) }\; 33-5 r=0$ $\Rightarrow r=\;\frac{33}{5} ∉ \;\text{ integer }\;$ $\;\text{ (2) }\; 33-5 r=-2$ $\Rightarrow 5 r=35$ $\Rightarrow r=7 \in \;\text{ integer }\;$ $\;\text{ (3) }\; 33-5 r=-3$ $\Rightarrow 5 r=36$ $\Rightarrow r=\;\frac{36}{5} ∉ \;\text{ integer }\;$ $\therefore$ Only for $r=7$ independent of $x$ term possible. $\therefore$ Independent of ${\text{x}}$ term $=- ({ }^{11} C_7 (\;\frac{5}{2}) ^4 \cdot (-\;\frac{1}{5}) ^7)$ $=- (\;\frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} \cdot \;\frac{5^4}{2^4} \cdot -\;\frac{1}{5^7})$ $=\;\frac{11 \cdot 10 \cdot 3}{2^4 \cdot 5^3}$ $=\;\frac{11 \cdot 3}{2^3 \cdot 5^2}$ $=\;\frac{33}{200}$

Q65JEE Main 2022MCQ4MSequences and Series

If ${\text{n}}$ arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is $1: 7$ and a $+$ $n=33$ , then the value of $n$ is : [28-Jun-2022-Shift-2]

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📖 Explanation

${\text{a}}, {\text{A}}_1, {\text{A}}_2 . . . . . . . . . . . {\text{A}}_n, 100$ Let ${\text{d}}$ be the common difference of above A.P. then $\;\frac{a+d}{100-d}=\;\frac{1}{7}$ $\Rightarrow 7 a+8 d=100 . . . . . . . \;\text{ (i) }\;$ and $a+n=33$ and $100=a+(n+1) d$ $\Rightarrow 100=a+(34-a) \;\frac{(100-7 a)}{8}$ $\Rightarrow 800=8 a+7 a^{2-}338 a+3400$ $\Rightarrow 7 a^{2-}330 a+2600=0$ $\Rightarrow a=10, \;\frac{260}{7}, \;\text{ but }\; a \ne \;\frac{260}{7}$ $\therefore n=23$

Q66JEE Main 2022MCQ4MLimits, Continuity and Differentiability

Let ${\text{f}}, {\text{g}}: {\text{R}} \to {\text{R}}$ be functions defined by $f(x)=\begin{cases}{[x],} & x< 0 \\ {|1-x|,} & \\ \\ \\ \\ x \ge 0\end{cases}$ and $g(x)=\begin{cases}{e^x-x ,} & x< 0 \\ {(x-1)^{2-}1 ,} & \\ \\ \\ \\ x \ge 0\end{cases}$ where $[ {\text{x}}]$ denote the greatest integer less than or equal to $x$ . Then, the function fog is discontinuous at exactly : [28-Jun-2022-Shift-2]

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📖 Explanation

$f(x)={\begin{cases}{[x],} & x< 0 \\ {|1-x|,} & x \ge 0}$ and $g(x)={\{\text{table} e^x-x{,} , x< 0 \\ (x-1)^{2-}1{,} & x \ge 0}$ $f \circ g(x)={\{\text{table} {[g(x)],} , g(x)< 0 \\ {|1-g(x)|,} & g(x) \ge 0}\end{cases}$ $=\begin{cases}{|1+x-e^x|,} & x< 0 \\ 1{,} & x=0 \\ \$ [(x-1)^{2-}1]$ {,} & 0< x< 2 \ {|2-(x-1)^2|,} & x \ge 2 $So$ x=0 2$ are the two points where fog is discontinuous\end{cases}

Q67JEE Main 2022MCQ4MLimits, Continuity and Differentiability

Let ${\text{f}}: {\text{R}} \to {\text{R}}$ be a differentiable function such that $f(\;\frac{\pi}{4})=\sqrt{2}, f(\;\frac{\pi}{2})=0$ and $f^{'}(\;\frac{\pi}{2})=1$ and let $g(x)=int_x^{\pi / 4}(f^{'}(t) \sec t+\tan t \sec t f(t)) d t$ for $x \in [\;\frac{\pi}{4}, \;\frac{\pi}{2})$ . Then $\lim _{x \to (\;\frac{\pi}{2})^{-}} g(x)$ is equal to : [28-Jun-2022-Shift-2]

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📖 Explanation

Given: $f (\;\frac{\pi}{4}) ={\sqrt{2}}, f (\;\frac{\pi}{2}) =0$ and $f^{'} (\;\frac{\pi}{2}) =1$ $g(x)=int_{x^{\;\frac{\pi}{4}} (f}^{'}(t) \ sec t+\tan t \ sec t f(t)) \text{dt}$ $=[\ sec t+f(t)]_x^{\;\frac{\pi}{4}}=2-\ sec x f(x)$ Now, $\lim _{x \to \;\frac{\pi^{-}}{2}} g(x)=\lim _{h \to 0} g (\;\frac{\pi}{2}-h)$ $=\lim _{h \to 0} 2-({\text{cosec}} h) f (\;\frac{\pi}{2}-h)$ $=\lim _{h \to 0} [2-\;\frac{f (\frac{\pi}{2}-h) }{ sin \;h}]$ $=\lim _{h \to 0} [2+\;\frac{f^{'} (\frac{\pi}{2}-h) }{\cos h}]$ $=3$

Q68JEE Main 2022MCQ4MDefinite Integrals

Let ${\text{f}}: {\text{R}} \to {\text{R}}$ be a continuous function satisfying $f(x)+f(x+k)=n$ , for all $x \in R$ where $k>0$ and $n$ is a positive integer. If $I_1=int_0^{4 n k} f(x) d x$ and $I_2=int_{-k}^{3 k} f(x) d x$ , then : [28-Jun-2022-Shift-2]

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📖 Explanation

$f: R \to R \;\text{ and }\; f(x)+f(x+k)=n \;\; ∀ x \in R$ $x \to x+k$ $f(x+k)+f(x+2 k)=n$ $\therefore \;\; f(x+2 k)=f(x)$ So, period of $f(x)$ is $2 k$ Now, $I_1=int_{0}^{4 n k} f(x) \text{dx}=2 n int_{0}^{2 k} f(x) \text{dx}$ $=2 n [int_{0}^{k} f(x) \text{dx}+int_{k}^{2 k} f(x) \text{dx}]$ $x=t+k \Rightarrow \text{dx}=d t$ (in second integral) $=2 n [int_{0}^{k} f(x) \text{dx}+int_{0}^{k} f(t+k) \text{dt}]$ $=2 n^2 k$ Now, $I_2=int_{-k}^{3 k} f(x) \text{dx}=2 int_{0}^{2 k} f(x) \text{dx}$ $I_2=2(n k)$ $\therefore \;\; l_1+n l_2=4 n^2 k$

Q69JEE Main 2022MCQ4MArea Under The Curves

The area of the bounded region enclosed by the curve $y=3- |x-\;\frac{1}{2}|-|x+1|$ and the $x$ -axis is : [28-Jun-2022-Shift-2]

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📖 Explanation

$y={\begin{cases}2 x-\\ \\ \frac{7}{2} & x< -1 \\ \\ \\ \frac{3}{2} & -1 \le x \le \\ \\ \frac{1}{2} \\ \\ \\ \frac{5}{2}-2 x & x>\\ \\ \frac{1}{2} }\end{cases}$ $y=3- |x-\;\frac{1}{2}{{}/{}|} -|x+1|$ Area of shaded region (required area) $=\;\frac{1}{2} (3+\;\frac{3}{2}) \cdot \;\frac{3}{2}=\;\frac{27}{8}$

Q70JEE Main 2022MCQ4MDifferential Equations

Let $x=x(y)$ be the solution of the differential equation $2 y e^{x / y^2} d x+(y^{2-}4 x e^{x / y^2}) d y=0$ such that ${\text{x}}(1)=0$ . Then, ${\text{x}}( {\text{e}})$ is equal to : [28-Jun-2022-Shift-2]

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📖 Explanation

Given differential equation $2 y e^{\;\frac{x}{y^2}} \text{dx}+ (y^{2-}4 x e^{\;\frac{x}{y^2}}) d y=0, x(1)=0$ $\Rightarrow e^{\;\frac{x}{y^2}}[2 y \text{dx}-4 x d y]=-y^2 d y$ $\Rightarrow e^{\;\frac{x}{y^2}} \$ [;\frac{2 y^2 \text{dx}-4 x y d y}{y^4}]$ =;\frac{-1}{y} d y\Rightarrow 2 e^{;\frac{x}{y^2}} d (;\frac{x}{y^2}) =-;\frac{1}{y} d y\Rightarrow 2 e^{;\frac{x}{y^2}}=-\ln y+c . . . . . ;\text{ (i) }; $Now, using$ x(1)=0, c=2 $So, for$ x(e) $, Put$ y=e $in (i)$ 2 e^{;\frac{x}{e^2}}=-1+2\Rightarrow ;{}/{e^2}=\ln (;\frac{1}{2}) \Rightarrow x(e)=-e^2 \ln 2$

Q71JEE Main 2022MCQ4MDifferential Equations

Let the slope of the tangent to a curve ${\text{y}}= {\text{f}}( {\text{x}})$ at $( {\text{x}}, {\text{y}})$ be given by $2 \tan x(\cos x-y)$ . If the curve passes through the point $(\;\frac{\pi}{4}, 0)$ , then the value of $int_0^{\pi / 2} y d x$ is equal to: [28-Jun-2022-Shift-2]

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📖 Explanation

$\;\frac{d y}{d x}=2 \tan x(\cos x-y)$ $\Rightarrow \;\frac{d y}{d x}+2 \tan x y=2 sin \;x$ $I. F .=e^{\int 2 \tan x \text{dx}}=\ sec ^2 x$ $\therefore$ Solution of D.E. will be $y(x) \ sec ^2 x=\int 2 sin \;x \ sec ^2 x \text{dx}$ $y \ sec ^2 x=2 \ sec x+c$ $\because$ Curve passes through $(\;\frac{\pi}{4}, 0)$ $\therefore c=-2 {\sqrt{2}}$ $\therefore y=2 \cos x-2 {\sqrt{2}} \cos ^2 x$ $\therefore int_{0}^{\pi / 2} y \text{dx}=int_{0}^{\pi / 2} (2 \cos x-2 {\sqrt{2}} \cos ^2 x) \text{dx}$ $=2-2 {\sqrt{2}} \cdot \;\frac{\pi}{4}=2-\;{\pi}/{{\sqrt{2}}}$

Q72JEE Main 2022MCQ4MStraight Lines and Pair of Straight Lines

Let a triangle be bounded by the lines $L_1: 2 x+5 y=10 ; L_2:-4 x+3 y=12$ and the line $L_3$ , which passes through the point $P(2,3)$ , intersects $L_2$ at $A$ and $L_1$ at $B$ . If the point $P$ divides the line-segment $A B$ , internally in the ratio $1: 3$ , then the area of the triangle is equal to : [28-Jun-2022-Shift-2]

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📖 Explanation

$L_1: 2 x+5 y=10$ $L_2:-4 x+3 y=12$ Solving $L_1$ and $L_2$ we get $C \equiv (\;\frac{-15}{13}, \;\frac{32}{13})$ Now, Let $A (x_1, \;\frac{1}{3} (12+4 x_1) )$ and $B (x_2, \;\frac{1}{5} (10-2 x_2) )$ $\therefore \;\; \;\frac{3 x_1+x_2}{4}=2$ and $\;\frac{ (12+4 x_1) +\frac{10-2 x_2}{5}}{4}=3$ So, $3 x_1+x_2=8$ and $10 x_1-x_2=-5$ So, $(x_1, x_2) = (\;\frac{3}{13}, \;\frac{95}{13})$ $A= (\;\frac{3}{13}, \;\frac{56}{13})$ and $B= (\;\frac{95}{13}, \;\frac{-12}{13})$ $= |\;\frac{1}{2} (\;\frac{3}{13} (\;\frac{-44}{13}) \;\frac{-56}{13} (\;\frac{110}{13}) +1 (\;\frac{2860}{169}) ) {{}/{}|}$ $=\;\frac{132}{13}$ sq. units

Q73JEE Main 2022MCQ4MHyperbola

Let $a>0, b>0$ . Let e and I respectively be the eccentricity and length of the latus rectum of the hyperbola $\;\frac{x^2}{a^2}-\;\frac{y^2}{b^2}=1$ . Let e' and I' respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If $e^2=\;\frac{11}{14} l$ and $(e^{'})^2=\;\frac{11}{8} l^{'}$ , then the value of $77 a+44 b$ is equal to : [28-Jun-2022-Shift-2]

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📖 Explanation

$H: \;\frac{x^2}{a^2}-\;\frac{y^2}{b^2}=1$ , then $e^2=\;\frac{11}{14} l($ I be the length of ${\text{LR}})$ $\Rightarrow a^{2+}b^2=\;\frac{11}{7} b^2 a . . . . . \;\text{ (i) }\;$ and $e^{' 2}=\;\frac{11}{8} l^{'}$ (I' be the length of LR of conjugate hyperbola) $\Rightarrow a^{2+}b^2=\;\frac{11}{4} a^2 b$ By (i) and (ii) $7 a=4 b$ then by (i) $\;\frac{16}{49} b^{2+}b^2=\;\frac{11}{7} b^2 \cdot \;\frac{4 b}{7}$ $\Rightarrow 44 b=65 \;\text{ and }\; 77 a=65$ $\therefore 77 a+44 b=130$

Q74JEE Main 2022MCQ4MVector Algebra

Let $\vec{a}=\alpha \hat{i}+2 \hat{j}-\hat{k}$ and $\vec{b}=-2 \hat{i}+\alpha \hat{j}+\hat{k}$ , where $\alpha \in R$ . If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is $\sqrt{15(\alpha^{2+}4)}$ , then the value of $2{|{\vec{a}}|}^{2+}(\vec{a} \cdot \vec{b}){|{\vec{b}}|}^2$ is equal to : [28-Jun-2022-Shift-2]

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📖 Explanation

$\vec{a}=\alpha \hat{i}+2 \hat{j}-\hat{k}$ and $\vec{b}=-2 \hat{i}+\alpha \hat{j}+\hat{k}$ $\therefore \vec{a} \times \vec{b}={\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -1 \\ -2 & \alpha & 1\end{vmatrix}}=(2+\alpha) \hat{i}-(\alpha-2) \hat{j}+ (\alpha^{2+}4) \hat{k}$ Now $|\vec{a} \times \vec{b}|={\sqrt{15 (\alpha^{2+}4) }}$ $\Rightarrow (2+\alpha)^{2+}(\alpha-2)^{2+} (\alpha^{2+}4) ^2=15 (\alpha^{2+}4)$ $\Rightarrow \alpha^{4-}5 \alpha^{2-}36=0$ $\therefore \alpha=\pm 3$ Now, $2|\vec{a}|^{2+}(\vec{a}-\vec{b})|\vec{b}|^{-2}=2.14-14=14$

Q75JEE Main 2022MCQ4MParabola

If vertex of a parabola is $(2,-1)$ and the equation of its directrix is $4 x-3 y=21$ , then the length of its latus rectum is : [28-Jun-2022-Shift-2]

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📖 Explanation

Vertex of Parabola : $(2,-1)$ and directrix : $4 x-3 y=21$ Distance of vertex from the directrix $a= |\;{8+3-21}/{{\sqrt{25}}}{{}/{}|} =2$ $\therefore$ length of latus rectum $=4 a=8$

Q76JEE Main 2022MCQ4MThree Dimensional Geometry

Let the plane $a x+b y+c z=d$ pass through $(2,3,-5)$ and is perpendicular to the planes $2 x+y-5 z=10$ and $3 x+5 y-7 z=12$ . If $a, b, c, d$ are integers $d>0$ and $g c d(|a|,|b|,|c|, d)=1$ , then the value of $a+7 b+c$ $+20 d$ is equal to : [28-Jun-2022-Shift-2]

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📖 Explanation

Equation of pane through point $(2,3,-5)$ and perpendicular to planes $2 x+y-5 z=10$ and $3 x+5 y$ $-7 z=12$ is ${\begin{vmatrix}x-2 & y-3 & z+5 \\ 2 & 1 & -5 \\ 3 & 5 & -7\end{vmatrix}}=0$ $\therefore$ Equation of plane is $(x-2)(-7+25)-(y-3)$ $(-14+15)+(z+5) \cdot 7=0$ $\therefore 18 x-y+7 z+2=0$ $\Rightarrow 18 x-y+7 z=-2$ $\therefore -18 x+y-7 z=2$ On comparing with $a x+b y+c z=d$ where ${\text{d}}>0$ is ${\text{a}}=-18, {\text{b}}=1, {\text{c}}=-7, {\text{d}}=2$ $\therefore a+7 b+c+20 d=22$

Q77JEE Main 2022MCQ4MProbability

The probability that a randomly chosen one-one function from the set $\{a, b, c, d\}$ to the set $\{1,2,3,4,5\}$ satisfies $f(a)+2 f(b)$ $-f(c)=f(d)$ is : [28-Jun-2022-Shift-2]

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📖 Explanation

Number of one-one function from $\{a, b, c, d\}$ to set $\{1,2,3,4,5\}$ is ${ }^5 P_4=120 n(s)$ .The required possible set of value (f(a), $f(b), f(c), f(d))$ such that $f(a)+2 f(b)-f(c)=f(d)$ are $(5,3,2,1),(5,1,2,3),(4,1,3,5),(3,1,4,5),(5,4,3,2)$ and $(3,4,5,2)$ $\therefore n(E)=6$ $\therefore$ Required probability $=\;\frac{n(E)}{n(S)}=\;\frac{6}{120}=\;\frac{1}{20}$

Q78JEE Main 2022MCQ4MLimits, Continuity and Differentiability

The value of $\lim _{n \to \infty } 6 \tan \{sum_{r=1}^{n} \tan ^{-1}(\;\frac{1}{r^{2+}3 r+3})\}$ is equal to : [28-Jun-2022-Shift-2]

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📖 Explanation

$\lim _{n \to \infty } 6 \tan \{sum_{{r=1}^n \tan }^{-1} (\;\frac{1}{r^{2+}3 r+3}) \}$ $=\lim _{n \to \infty } 6 \tan \{sum_{{r=1}^n \tan }^{-1} (\;\frac{(r+2)-(r+1)}{1+(r+2)(r+1)}) \}$ $=\lim _{n \to \infty } 6 \tan \{sum_{{r=1}^n (\tan ^{-1}(r+2)-\tan }^{-1}(r+1)) \}$ $=\lim _{n \to \infty } 6 \tan \{\tan ^{-1}(n+2)-\tan ^{-1} 2\}$ $=6 \tan \{\;\frac{\pi}{2}-\cot ^{-1} (\;\frac{1}{2}) \}$ $=6 \tan (\tan ^{-1} (\;\frac{1}{2}) )$ $=3$

Q79JEE Main 2022MCQ4MVector Algebra

Let $\vec{a}$ be a vector which is perpendicular to the vector $3 \hat{i}+\;\frac{1}{2} \hat{j}+2 \hat{k}$ . If $\vec{a} \times (2 \hat{i}+\hat{k})=2 \hat{i}-13 \hat{j}-4 \hat{k}$ , then the projection of the vector $\vec{a}$ on the vector $2 \hat{i}+2 \hat{j}+\hat{k}$ is : [28-Jun-2022-Shift-2]

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📖 Explanation

Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$ and $\vec{a} \cdot (3 \hat{i}-\;\frac{1}{2} \hat{j}+2 \hat{k}) =0 \Rightarrow 3 a_1+\;\frac{a_2}{2}+2 a_3=0 . . . .$ . (i)and $\vec{a} \times (2 \hat{i}+\hat{k})=2 \hat{i}-13 \hat{j}-4 \hat{k}$ $\Rightarrow a_2 \hat{i}+ (2 a_3-a_1) \hat{j}-2 a_2 \hat{k}=2 \hat{i}-13 \hat{j}-4 \hat{k}$ $\therefore a_2=2 . . . .$ . (ii)and $a_1-2 a_3=13 . . . .$ . (iii)From eq. (i) and (iii) : $a_1=3$ and $a_3=-5$ $\therefore \vec{a}=3 \hat{i}+2 \hat{j}-5 \hat{k}$ $\therefore$ projection of $\vec{a}$ on $2 \hat{i}+2 \hat{j}+\hat{k}=\;\frac{6+4-5}{3}=\;\frac{5}{3}$

Q80JEE Main 2022MCQ4MTrigonometric Ratios and Identities

If $\cot \alpha=1$ and $\sec \beta=-\;\frac{5}{3}$ , where $\pi< \alpha< \;\frac{3 \pi}{2}$ and $\;\frac{\pi}{2}< \beta< \pi$ , then the value of $\tan (\alpha+\beta)$ and the quadrant in which $\alpha+\beta$ lies, respectively are : [28-Jun-2022-Shift-2]

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📖 Explanation

$\because \cot \alpha=1, \;\; \alpha \in (\pi, \;\frac{3 \pi}{2})$ then $\tan \alpha=1$ and $\ sec \beta=-\;\frac{5}{3}, \;\; \beta \in (\;\frac{\pi}{2}, \pi)$ then $\tan \beta=-\;\frac{4}{3}$ $\therefore \tan (\alpha+\beta)=\;\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}$ $=\;\frac{1-\frac{4}{3}}{1+\;\frac{4}{3}}$ $=-\;\frac{1}{7}$ $\alpha+\beta \in (\;\frac{3 \pi}{2}, 2 \pi)$ i.e. fourth quadrant

Q81JEE Main 2022NAT4MThree Dimensional Geometry

Let the image of the point ${\text{P}}(1,2,3)$ in the line $L: \;\frac{x-6}{3}=\;\frac{y-1}{2}=\;\frac{z-2}{3}$ be Q. Let ${\text{R}}(\alpha, \beta, \gamma )$ be a point that divides internally the line segment $P Q$ in the ratio $1: 3$ . Then the value of $22(\alpha+\beta+\gamma )$ is equal to___ [28-Jun-2022-Shift-2]

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📖 Explanation

The point dividing $P Q$ in the ratio $1: 3$ will be mid-point of $P &$ foot of perpendicular from $P$ on the line. $\therefore$ Let a point on line be $\lambda$ $\Rightarrow \;\frac{x-6}{3}=\;\frac{y-1}{2}=\;\frac{z-2}{3}=\lambda$ $\Rightarrow P^{'}(3 \lambda +6,2 \lambda +1,3 \lambda +2)$ as ${\text{P}}^{'}$ is foot of perpendicular $(3 \lambda +5) 3+(2 \lambda -1) 2+(3 \lambda -1) 3=0$ $\Rightarrow 22 \lambda +15-2-3=0$ $\Rightarrow \lambda =\;\frac{-5}{11}$ $\therefore P^{'} (\;\frac{51}{11}, \;\frac{1}{11}, \;\frac{7}{11})$ $\;\text{ Mid-point of }\; P P^{'} \equiv (\;\frac{\;\frac{51}{11}+1}{2}, \;\frac{\;\frac{1}{11}+2}{2}, \;\frac{\;\frac{7}{11}+3}{2})$ $\equiv (\;\frac{62}{22}, \;\frac{23}{22}, \;\frac{40}{22}) \equiv (\alpha, \beta, \gamma )$ $\Rightarrow 22(\alpha, \beta, \gamma )=62+23+40=125$

Q82JEE Main 2022NAT4MStatistics

Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62 , and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is_____ [28-Jun-2022-Shift-2]

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📖 Explanation

According to given data $\;\frac{sum_{i=1}^{7} (x_i-62) ^2}{7}=20$ $\Rightarrow sum_{i=1}^{7} (x_i-62) ^2=140$ So for any $x_i, (x_i-62) ^2 \le 140$ $\Rightarrow x_i>50 ∀ i=1,2,3, . . . . . . 7$ So no student is going to score less than 50.

Q83JEE Main 2022NAT4MCircles

If one of the diameters of the circle $x^{2+}y^{2-}2 \sqrt{2} x-6 \sqrt{2} y+14=0$ is a chord of the circle $(x-2 \sqrt{2})^{2+}(y-2 \sqrt{2})^2=r^2$ , then the value of $r^2$ is equal to_____ [28-Jun-2022-Shift-2]

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📖 Explanation

For $x^{2+}y^{2-}2 {\sqrt{2}} x-6 {\sqrt{2}} y+14=0$ $\;\text{ Radius }\;=\sqrt{({\sqrt{2}})^{2+}(3 {\sqrt{2}})^{2-}14}={\sqrt{6}}$ $\Rightarrow \;\text{ Diameter }\;=2 {\sqrt{6}}$ If this diameter is chord to $(x-2 {\sqrt{2}})^{2+}(y-2 {\sqrt{2}})^2=r^2 \;\text{ then }\;$ $\Rightarrow r^2=6+ (\sqrt{({\sqrt{2}})^{2+}({\sqrt{2}})^2}) ^2$ $\Rightarrow r^2=6+4=10$ $\Rightarrow r^2=10$

Q84JEE Main 2022NAT4MLimits, Continuity and Differentiability

If $\lim _{x \to 1} \;\frac{\sin (3 x^{2-}4 x+1)-x^{2+}1}{2 x^{3-}7 x^{2+}a x+b}=-2$ , then the value of $( {\text{a}}- {\text{b}})$ is equal to____ [28-Jun-2022-Shift-2]

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📖 Explanation

$\;\lim _{x \to 1} \;\frac{ (\frac{ sin \; (3 x^{2-}4 x+1) }{3 x^{2-}4 x+1}) (3 x^{2-}4 x+1) -x^{2+}1}{2 x^{3-}7 x^{2+}a x+b}=-2$ $\Rightarrow \lim _{x \to 1} \;\frac{3 x^{2-}4 x+1-x^{2+}1}{2 x^{3-}7 x^{2+}a x+b}=-2$ $\Rightarrow \lim _{x \to 1} \;\frac{2(x-1)^2}{2 x^{3-}7 x^{2+}a x+b}=-2$ So $f(x)=2 x^{3-}7 x^{2+}a x+b=0$ has $x=1$ as repeated root, therefore $f(1)=0$ and $f^{'}(1)=0$ gives $a+b+5$ and $a=8$ So, $a-b=11$

Q85JEE Main 2022NAT4MSequences and Series

Let for $n=1,2, . . . . . ., 50, S_n$ be the sum of the infinite geometric progression whose first term is $n^2$ and whose common ratio is $\;\frac{1}{(n+1)^2}$ . Then the value of $\;\frac{1}{26}+sum_{n=1}^{50}(S_n+\;\frac{2}{n+1}-n-1)$ is equal to____ [28-Jun-2022-Shift-2]

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📖 Explanation

$S_n=\;\frac{n^2}{1-\;\frac{1}{(n+1)^2}}=\;\frac{n(n+1)^2}{n+2}= (n^{2+}1) -\;\frac{2}{n+2}$ Now $\;\frac{1}{26}+sum_{n=1}^{50} (S_n+\;\frac{2}{n+1}-n-1)$ $=\;\frac{1}{26}+sum_{n=1}^{50} \{ (n^{2-}n) +2 (\;\frac{1}{n+1}-\;\frac{1}{n+2}) \}$ $=\;\frac{1}{26}+\;\frac{50 \times 51 \times 101}{6}-\;\frac{50 \times 51}{2}+2 (\;\frac{1}{2}-\;\frac{1}{52})$ $=1+25 \times 17(101-3)$ $=41651$

Q86JEE Main 2022NAT4MMatrices and Determinants

If the system of linear equations $2 x-3 y=\gamma +5 \;\text{, }\;$ $\alpha x+5 y=\beta+1$ , where $\alpha, \beta, \gamma \in {\text{R}}$ has infinitely many solutions then the value of $|9 \alpha+3 \beta+5 \gamma |$ is equal to_____ [28-Jun-2022-Shift-2]

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📖 Explanation

If $2 x-3 y=\gamma +5$ and $\alpha x+5 y=\beta+1$ have infinitely many solutions then $\;\frac{2}{\alpha}=\;\frac{-3}{5}=\;\frac{\gamma +5}{\beta+1}$ $\Rightarrow \alpha=-\;\frac{10}{3}$ and $3 \beta+5 \gamma =-28$ So $|9 \alpha+3 \beta+5 \gamma |=|-30-28|=58$

Q87JEE Main 2022NAT4MMatrices and Determinants

Let $A=(\text{table} \1+\text{i} , \1 ; \-\text{i} , \0)$ where $i=\sqrt{-1}$ . Then, the number of elements in the set $\{ {\text{n}} \in \{1,2, . . . . . ., 100\}: {\text{A}}^{ {\text{n}}}= {\text{A}}\}$ is____ [28-Jun-2022-Shift-2]

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📖 Explanation

$\therefore A^2={\begin{bmatrix}1+i & 1 \\ -i & 0\end{bmatrix}$ ${\begin{bmatrix}1+i & 1 \\ -1 & 0\end{bmatrix}}$ $={\begin{bmatrix}i & 1+i \\ 1-i & -i\end{bmatrix}}$ $A^4= {\begin{bmatrix}i & 1+i \\ 1-i & -i\end{bmatrix}} {\begin{bmatrix}i & 1+i \\ 1-i & -i\end{bmatrix}}=I$ So $A^5=A, A^9=A$ and so on.Clearly ${\text{n}}=1,5,9, . . . . . ., 97$ Number of values of $n=25$

Q88JEE Main 2022NAT4MComplex Numbers

Sum of squares of modulus of all the complex numbers $z$ satisfying $\overline{z}=i z^{2+}z^{2-}z$ is equal to_____ [28-Jun-2022-Shift-2]

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📖 Explanation

Let $z=x+i y$ So $2 x=(1+i) (x^{2-}y^{2+}2 x y i)$ $\Rightarrow 2 x=x^{2-}y^{2-}2 x y$ (i) and $x^{2-}y^{2+}2 x y=0$ From (i) and (ii) we get $x=0 \;\text{ or }\; y=-\;\frac{1}{2}$ When $x=0$ we get $y=0$ When $y=-\;\frac{1}{2}$ we get $x^{2-}x-\;\frac{1}{4}=0$ $\Rightarrow \;\; x=\;{-1 \pm {\sqrt{2}}}/{2}$ So there will be total 3 possible values of $z$ , which are $0, (\;{-1+{\sqrt{2}}}/{2}) -\;\frac{1}{2} i$ and $(\;{-1-{\sqrt{2}}}/{2}) -\;\frac{1}{2} i$ Sum of squares of modulus $=0+ (\;{{\sqrt{2}}-1}/{2}) ^{2+}\;\frac{1}{4}+ (\;{{\sqrt{2}}+1}/{2}) ^2=+\;\frac{1}{4}$ $=2$

Q89JEE Main 2022NAT4MPermutations and Combinations

Let $S=\{1,2,3,4\}$ . Then the number of elements in the set $\{f: S \times S \to S: f$ is onto and $f(a, b)=f(b, a) \ge a ∀(a, b) \in S \times$ ${\text{S}}\}$ is____ [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

There are 16 ordered pairs in $S \times S$ . We write all these ordered pairs in 4 sets as follows. $A=\{(1,1)\}$ $B=\{(1,4),(2,4),(3,4)(4,4),(4,3),(4,2),(4,1)\}$ $C=\{(1,3),(2,3),(3,3),(3,2),(3,1)\}$ $D=\{(1,2),(2,2),(2,1)\}$ All elements of set $B$ have image 4 and only element of $A$ has image 1 .All elements of set $C$ have image 3 or 4 and all elements of set $D$ have image 2 or 3 or 4 .We will solve this question in two cases. Case I: When no element of set $C$ has image 3 .Number of onto functions $=2$ (when elements of set $D$ have images 2 or 3 )Case II: When atleast one element of set $C$ has image 3 .Number of onto functions $= (2^{3-}1) (1+2+2)$ $=35$ Total number of functions $=37$

Q90JEE Main 2022NAT4MSets and Relations

The maximum number of compound propositions, out of ${\text{p}} ∨ {\text{r}} ∨ {\text{s}}, {\text{p}} ∨ {\text{r}} \vee ∼ {\text{s}}, {\text{p}} ∨ ∼ {\text{q}} ∨ {\text{s}}, ∼ {\text{p}} ∨ ∼ {\text{r}} ∨ {\text{s}}, ∼ {\text{p}} ∨ ∼ {\text{r}} ∨ ∼ {\text{s}}, ∼ {\text{p}} ∨ {\text{q}} ∨ ∼$ ${\text{s}}, {\text{q}} ∨ {\text{r}} ∨ ∼ {\text{s}}, {\text{q}} ∨ ∼ {\text{r}} ∨ ∼ {\text{s}}, ∼ {\text{p}} ∨ ∼ {\text{q}} ∨ ∼ {\text{s}}$ that can be made simultaneously true by an assignment of the truth values to p, q, ${\text{r}}$ and $s$ , is equal to___ [28-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

There are total 9 compound propositions, out of which 6 contain $∼ s$ . So if we assign $s$ as false, these 6 propositions will be true.In remaining 3 compound propositions, two contain $p$ and the third contains $∼ r$ . So if we assign $p$ and $r$ as true and false respectively, these 3 propositions will also be true.Hence maximum number of propositions that can be true are 9.

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