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27 Jul 2022 Shift 1

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Physics30 questions

Q1JEE Main 2022MCQ4MUnits and Measurements

A torque meter is calibrated to reference standards of mass, length and time each with $5 \%$ accuracy. After calibration, the measured torque with this torque meter will have net accuracy of : [27-Jul-2022-Shift-1]

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📖 Explanation

Dimensional formula for Torque $[\tau ]=[ {\text{ML}}^2 {\text{T}}^{-2}]$ NowPercentage error in torque $=\% \tau =\% {\text{M}}+2 \% {\text{L}}$ $2 \% {\text{T}}$ $\% \tau =25 \%$

Q2JEE Main 2022MCQ4MMotion in a Straight Line

A bullet is shot vertically downwards with an initial velocity of $100 {\text{m}} / {\text{s}}$ from a certain height. Within $10 {\text{s}}$ , the bullet reaches the ground and instantaneously comes to rest due to the perfectly inelastic collision. The velocity-time curve for total time ${\text{t}}=20 {\text{s}}$ will be: (Take $. {\text{g}}=10 {\text{m}} / {\text{s}}^2)$ . [27-Jul-2022-Shift-1]

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📖 Explanation

$|v_{10}|=(100+10 \times 10) {\text{m}} / {\text{s}}$ $v_{10}=-200 {\text{m}} / {\text{s}}$ and $v_0=-100 {\text{m}} / {\text{s}}$ from 10 s to 20 s velocity remains zero $\Rightarrow$ from $t=0$ s to $10 {\text{s}}$ velocity increases in magnitude linearly. $\Rightarrow$ graph given in option A fits correctly.

Q3JEE Main 2022MCQ4MWork, Power and Energy

Sand is being dropped from a stationary dropper at a rate of $0.5 {\text{kgs}}^{-1}$ on a conveyor belt moving with a velocity of $5 {\text{ms}}^{-1}$ . The power needed to keep the belt moving with the same velocity will be : [27-Jul-2022-Shift-1]

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📖 Explanation

$\;\frac{d m}{d t}=0.5 {\text{kg}} / {\text{s}}$ $v=5 {\text{m}} / {\text{s}}$ $F=\;\frac{v d m}{d t}=2.5 {\text{kg}} {\text{m}} / {\text{s}}^2$ $P=\overline{F} \cdot \overline{v}=(2.5)(5) {\text{W}}$ $=12.5 {\text{W}}$

Q4JEE Main 2022MCQ4MLaws of Motion

A bag is gently dropped on a conveyor belt moving at a speed of $2 {\text{m}} / {\text{s}}$ . The coefficient of friction between the conveyor belt and bag is $0.4$ . Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion, is : $[Take $g=10 {\text{m}} / {\text{s}}^{-2}$ ]$ [27-Jul-2022-Shift-1]

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📖 Explanation

$v=2 {\text{m}} / {\text{s}}$ $\mu =0.4$ ${\text{a}}=+(0.4)( {\text{g}})$ $=+4 {\text{m}} / {\text{s}}^2$ ${\text{v}}^{2-} {\text{u}}^2=2$ as $\Rightarrow (4)=2 \times (4)( {\text{s}})$ $s=0.5 {\text{m}}$

Q5JEE Main 2022MCQ4MWork, Power and Energy

Two cylindrical vessels of equal cross-sectional area $16 {\text{cm}}^2$ contain water upto heights $100 {\text{cm}}$ and $150 {\text{cm}}$ respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, is [Take, density of water $=10^3 {\text{kg}} / {\text{m}}^3$ and $. {\text{g}}=10 {\text{ms}}^{-2}]$ : [27-Jul-2022-Shift-1]

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📖 Explanation

$A=16 \times 10^{-4} {\text{m}}^2$ $E_{\;\text{\in }\;}=m_1 g \;\frac{H_1}{2}+m_2 g \;\frac{H_2}{2}$ $=\rho g \;\frac{A}{2}(H_1^{2+}H_2^2)=\rho g \;\frac{A}{2}(1^{2+}1.5^2)$ $E_{f i n}=\rho g \;\frac{A}{2}(2 H^2)=\rho g \;\frac{A}{2}(2 \times 1.25^2)$ $W=\rho g \;\frac{A}{2}(3.25-3.125)$ $=1 {\text{J}}$

Q6JEE Main 2022MCQ4MGravitation

Two satellites ${\text{A}}$ and ${\text{B}}$ , having masses in the ratio $4: 3$ , are revolving in circular orbits of radii $3 {\text{r}}$ and $4 {\text{r}}$ respectively around the earth. The ratio of total mechanical energy of $A$ to $B$ is : [27-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Given that $\;{ {\text{m}}_1}/{ {\text{m}}_2}=\;\frac{4}{3}, \;{ {\text{r}}_1}/{ {\text{r}}_2}=\;\frac{3}{4}$ Now ${\text{TE}}=\;\frac{1}{2} {\ mv}^{2+}(\;{- {\text{GMm}}}/{ {\text{r}}})$ but $\;{ {\ mv}^2}/{ {\text{r}}}=\;{ {\text{GMm}}}/{ {\text{r}}^2} \Rightarrow {\ mv}^2=\;{ {\text{GMm}}}/{ {\text{r}}}$ $\Rightarrow {\text{TE}}=-\;{ {\text{GMm}}}/{2 {\text{r}}} \propto \;{ {\text{m}}}/{ {\text{r}}}$ $\;{ {\text{TE}}_1}/{ {\text{TE}}_2}=\;{ {\text{m}}_1}/{ {\text{m}}_2} \cdot \;{ {\text{r}}_2}/{ {\text{r}}_1}=\;\frac{4}{3} \times \;\frac{4}{3}=\;\frac{16}{9}$

Q7JEE Main 2022MCQ4MThermodynamics

If $K_1$ and $K_2$ are the thermal conductivities, $L_1$ and $L_2$ are the lengths and $A_1$ and $A_2$ are the cross sectional areas of steel and copper rods respectively such that $\;\frac{K_2}{K_1}=9, \;\frac{A_1}{A_2}=2, \;\frac{L_1}{L_2}=2$ . Then, for the arrangement as shown in the figure, the value of temperature ${\text{T}}$ of the steel - copper junction in the steady state will be: [27-Jul-2022-Shift-1]

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📖 Explanation

$\;{ {\text{d}} \theta}/{ {\text{dt}}}=\;{ {\text{K}}_1 {\text{A}}_1}/{ {\text{l}}_1}( {\text{T}}_1- {\text{T}})=\;{ {\text{K}}_2 {\text{A}}_2}/{ {\text{l}}_2}( {\text{T}}- {\text{T}}_2)$ $\Rightarrow \;{450- {\text{T}}}/{ {\text{T}}-0}=\;{ {\text{K}}_2 {\text{A}}_2 {\text{l}}_1}/{ {\text{K}}_1 {\text{A}}_1 {\text{l}}_2}=9 \times \;\frac{1}{2} \times 2$ $\Rightarrow 450- {\text{T}}=9 {\text{T}} \Rightarrow {\text{T}}=45^{\circ} {\text{C}}$

Q8JEE Main 2022MCQ4MThermodynamics

Read the following statements : A. When small temperature difference between a liquid and its surrounding is doubled, the rate of loss of heat of the liquid becomes twice. B. Two bodies $P$ and $Q$ having equal surface areas are maintained at temperature $10^{\circ} {\text{C}}$ and $20^{\circ} {\text{C}}$ . The thermal radiation emitted in a given time by ${\text{P}}$ and ${\text{Q}}$ are in the ratio $1: 1.15$ . C. A Carnot Engine working between $100 {\text{K}}$ and $400 {\text{K}}$ has an efficiency of $75 \%$ . D. When small temperature difference between a liquid and its surrounding is quadrupled, the rate of loss of heat of the liquid becomes twice. Choose the correct answer from the options given below : [27-Jul-2022-Shift-1]

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📖 Explanation

Heat TransferA. by Newton's low of colling $\;{ {\text{d}} \theta}/{ {\text{dt}}}=\propto \Delta {\text{T}}$ B. ${\text{H}}=\;{ {\text{d}} \theta}/{ {\text{dt}}}=\sigma {\text{eAT}} {\text{T}}^4 \Rightarrow \;{ {\text{H}}_{ {\text{P}}}}/{ {\text{H}}_{ {\text{Q}}}}=(\;{ {\text{T}}_{ {\text{P}}}}/{ {\text{T}}_{ {\text{Q}}}})^4=(\;\frac{283}{293})^4$ ${\text{H}}_{ {\text{P}}}: {\text{H}}_{ {\text{Q}}}=1(1.03)^4=1:(1.03)^4=1: 1.15$ $\Rightarrow {\text{B}}$ is correctC. $η=1-\;\frac{100}{400}=\;\frac{3}{4}=75 \%$ D. is wrong as $\;{ {\text{d}} \theta}/{ {\text{dt}}} \propto \Delta {\text{T}}$

Q9JEE Main 2022MCQ4MThermodynamics

Same gas is filled in two vessels of the same volume at the same temperature. If the ratio of the number of molecules is $1: 4$ , then A. The r.m.s. velocity of gas molecules in two vessels will be the same. B. The ratio of pressure in these vessels will be $1: 4$ . C. The ratio of pressure will be $1: 1$ . D. The r.m.s. velocity of gas molecules in two vessels will be in the ratio of $1: 4$ . Choose the correct answer from the options given below : [27-Jul-2022-Shift-1]

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📖 Explanation

A. ${\text{V}}_{ {\text{Rms}}}=\sqrt{\;{3 {\text{RT}}}/{ {\text{M}}_{ {\text{w}}}}} \Rightarrow {\text{V}}_{ {\text{Rms}}}$ is sameB. $\;{ {\text{P}}_1}/{ {\text{P}}_2}=\;{ {\text{N}}_1}/{ {\text{N}}_2} \Rightarrow {\text{B}}$ is correct

Q10JEE Main 2022MCQ4MOscillations

Two identical positive charges $Q$ each are fixed at a distance of '2a' apart from each other. Another point charge $q_0$ with mass ' $m$ ' is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge ${\text{q}}_0$ executes ${\text{SHM}}$ . The time period of oscillation of charge ${\text{q}}_0$ will be : [27-Jul-2022-Shift-1]

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📖 Explanation

$(x< < a)(\alpha$ is acceleration $)$ $F_{n e t}=-(\;\frac{k q_0 Q}{(a-x)^2}-\;\frac{k Q q_0}{(a+x)^2})$ $m \alpha=-\;\frac{k q_0 Q}{a^4} 4 a x$ $\Rightarrow \alpha=-\;\frac{4 k q_0 Q}{m a^3} x$ $\;\text{ So, }\; T=2 \pi \sqrt{\;\frac{4 \pi \epsilon_0 m a^3}{4 q_0 Q}}$ $\;\text{ or }\; T=\sqrt{\;\frac{4 \pi^3 \epsilon_0 m a^3}{q_0 Q}}$

Q11JEE Main 2022MCQ4MCurrent Electricity

Two sources of equal emfs are connected in series. This combination is connected to an external resistance ${\text{R}}$ . The internal resistances of the two sources are $r_1$ and $r_2(r_1 > r_2)$ . If the potential difference across the source of internal resistance $r_1$ is zero, then the value of ${\text{R}}$ will be : [27-Jul-2022-Shift-1]

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📖 Explanation

${\text{I}}=\;{2 {\text{E}}}/{ {\text{r}}_1+ {\text{r}}_2+ {\text{R}}}$ ${\text{IR}}= {\text{E}}- {\text{Ir}}_2$ ${\text{I}}( {\text{R}}+ {\text{r}}_2)= {\text{E}}$ ${\text{I}}=\;{ {\text{E}}}/{ {\text{R}}+ {\text{r}}_2}$ $\;{2 {\text{E}}}/{ {\text{r}}_1+ {\text{r}}_2+ {\text{R}}}=\;{ {\text{E}}}/{ {\text{R}}+ {\text{r}}_2}$ $2 {\text{R}}+2 {\text{r}}_2= {\text{r}}_1+ {\text{r}}_2+ {\text{R}}$ ${\text{R}}= {\text{r}}_1- {\text{r}}_2$

Q12JEE Main 2022MCQ4MMagnetism and Matter

Two bar magnets oscillate in a horizontal plane in earth's magnetic field with time periods of $3 {\text{s}}$ and $4 {\text{s}}$ respectively. If their moments of inertia are in the ratio of $3: 2$ , then the ratio of their magnetic moments will be: [27-Jul-2022-Shift-1]

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📖 Explanation

${\text{T}}=2 \pi \sqrt{\;{ {\text{I}}}/{ {\text{MB}}_{ {\text{H}}}}}$ $\;{ {\text{T}}_1}/{ {\text{T}}_2}=\;{2 \pi \sqrt{{ {\text{I}}_1}/{ {\text{M}}_1 {\text{B}}_{ {\text{H}}}}}}/{2 \pi \sqrt{\;{ {\text{I}}_2}/{ {\text{M}}_2 {\text{B}}_{ {\text{H}}}}}}=\;\frac{3}{4}$ $\sqrt{\;{ {\text{I}}_1}/{ {\text{I}}_2} \times \;{ {\text{M}}_2}/{ {\text{M}}_1}}=\;\frac{3}{4}$ $\sqrt{\;{ {\text{I}}_1}/{ {\text{I}}_2}} \times \sqrt{\;{ {\text{M}}_2}/{ {\text{M}}_1}}=\;\frac{3}{4}$ $\sqrt{\;\frac{3}{2}} \times \sqrt{\;{ {\text{M}}_2}/{ {\text{M}}_1}}=\;\frac{3}{4}$ $\;\frac{3}{2} \times \;{ {\text{M}}_2}/{ {\text{M}}_1}=\;\frac{9}{16}$ $\;{ {\text{M}}_1}/{ {\text{M}}_2}=\;\frac{8}{3}$

Q13JEE Main 2022MCQ4MMagnetism and Matter

A magnet hung at $45^{\circ}$ with magnetic meridian makes an angle of $60^{\circ}$ with the horizontal. The actual value of the angle of dip is - [27-Jul-2022-Shift-1]

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📖 Explanation

$\tan \theta^{'}=\;\frac{\tan \theta}{\cos \alpha}$ $\theta^{'}=60^{\circ}$ $\alpha=45^{\circ}$ $\sqrt{3}=\;{\tan \theta}/{\frac{1}{\sqrt{2}}}$ $\tan \theta=\sqrt{\;\frac{3}{2}}$ $\theta=\tan ^{-1} \sqrt{\;\frac{3}{2}}$

Q14JEE Main 2022MCQ4MAlternating Current

A direct current of $4 {\text{A}}$ and an alternating current of peak value $4 {\text{A}}$ flow through resistance of $3 Ω$ and $2 Ω$ respectively. The ratio of heat produced in the two resistances in same interval of time will be : [27-Jul-2022-Shift-1]

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📖 Explanation

${\text{H}}_1= {\text{i}}^2 {\text{R}}_1 {\text{t}}\; {\text{H}}_2= {\text{i}}_{\;\text{rms }\;}^2 {\text{R}}_2 {\text{t}}\{ {\text{i}}_{\;\text{rms }\;}=\;{ {\text{i}}_0}/{\sqrt{2}}\}$ ${\text{H}}_1=16(3) {\text{t}} \;\; {\text{H}}_2=\;{ {\text{i}}_0^2}/{2} {\text{R}}_2 {\text{t}}$ ${\text{H}}_2=16 {\text{t}}$ ${\text{H}}_1: {\text{H}}_2=3: 1$

Q15JEE Main 2022MCQ4MElectromagnetic Waves

A beam of light travelling along $X$ -axis is described by the electric field $E_y=900 \sin \omega ( {\text{t}}-x / c)$ . The ratio of electric force to magnetic force on a charge q moving along $Y$ -axis with a speed of $3 \times 10^7 {\text{ms}}^{-1}$ will be : (Given speed of light $=3 \times 10^8 {\text{ms}}^{-1}$ ) [27-Jul-2022-Shift-1]

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📖 Explanation

${\text{E}}_{ {\text{y}}}=900 \sin (\omega {\text{t}}-\;{\omega {\text{x}}}/{ {\text{c}}})$ ${\text{E}}_0=900$ ${\text{F}}_{ {\text{E}}}= {\text{qE}}_0$ ${\text{F}}_{ {\text{B}}}= {\text{qvB}}_0$ $\;{ {\text{F}}_{ {\text{E}}}}/{ {\text{F}}_{ {\text{B}}}}=\;{ {\text{E}}_0}/{ {\text{vB}}_0}=\;{ {\text{c}}}/{ {\text{v}}}=\;\frac{3 \times 10^8}{3 \times 10^7}=10: 1$

Q16JEE Main 2022MCQ4MWave Optics

A microscope was initially placed in air (refractive index 1). It is then immersed in oil (refractive index 2). For a light whose wavelength in air is $\lambda$ , calculate the change of microscope's resolving power due to oil and choose the correct option. [27-Jul-2022-Shift-1]

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📖 Explanation

$\because$ Resolving power $=\;\frac{2 \mu \sin \theta}{1.22 \lambda }$ $\;\frac{P_1}{P_2}=\;\frac{\mu _1}{\mu _2} \times \;\frac{\mu _1}{\mu _2}$ $=(\;\frac{\mu _1}{\mu _2})^2$ $\Rightarrow \;\frac{P_1}{P_2}=\;\frac{1}{4}$ $\Rightarrow P_2=4 P_1$

Q17JEE Main 2022MCQ4MDual Nature of Matter and Radiation

An electron (mass ${\text{m}}$ ) with an initial velocity $\vec{v}=v_0 i(v_0 > 0)$ is moving in an electric field $\vec{E}=-E_0 \hat{i}(E_0 > 0)$ where $E_0$ is constant. If at ${\text{t}}=0$ de Broglie wavelength is $\lambda _0=\;\frac{h}{m v_0}$ , then its de Broglie wavelength after time $t$ is given by [27-Jul-2022-Shift-1]

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📖 Explanation

${\text{E}}=- {\text{E}}_0 { {\text{i}}^{\^}$ $\lambda _0=\;{ {\text{h}}}/{ {\ mv}}$ $v=v_0+\;{ {\text{eE}}_0 {\text{t}}}/{ {\text{m}}}$ $\lambda =\;\frac{h}{m v}=\;\frac{h}{m(v_0+\;\frac{e E_0}{m} t)}$ $\lambda ^{'}=\;{h}/{{\ mv}_0(1+\;{ {\text{eE}}}/{ {\ mv}_0} {\text{t}})}$ $\lambda ^{'}=\;{\lambda _0}/{(1+\;{ {\text{eE}}_0}/{ {\ mv}_0} {\text{t}})}$

Q18JEE Main 2022MCQ4MAtoms and Nuclei

What is the half-life period of a radioactive material if its activity drops to $1 / 16^{\;\text{th }\;}$ of its initial value in 30 years? [27-Jul-2022-Shift-1]

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📖 Explanation

${\text{A}}= {\text{A}}_0 {\text{e}}^{-\lambda {\text{t}}}$ $\Rightarrow -\lambda {\text{t}}=\ln (\;{ {\text{A}}}/{ {\text{A}}_0})$ $\Rightarrow -\;{ℓ {\text{n}} 2}/{ {\text{t}}_{1 / 2}} \times 30=ℓ {\text{n}}(\;\frac{1}{16})$ $\Rightarrow -\;{ℓ {\text{n}} 2}/{ {\text{t}}_{1 / 2}} \times 30=-4 ℓ {\text{n}} 2$ $\Rightarrow {\text{t}}_{1 / 2}=\;\frac{30}{4}=7.5 {\text{yrs}}$

Q19JEE Main 2022MCQ4MSemiconductor Electronics

A logic gate circuit has two inputs $A$ and $B$ and output $Y$ . The voltage waveforms of $A, B$ and $Y$ are shown below.The logic gate circuit is : [27-Jul-2022-Shift-1]

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Q20JEE Main 2022MCQ4MElectromagnetic Waves

At a particular station, the TV transmission tower has a height of $100 {\text{m}}$ . To triple its coverage range, height of the tower should be increased to [27-Jul-2022-Shift-1]

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📖 Explanation

Let d be range $d^2=(h+R)^{2-}R^2$ $=h^{2+}R^{2+}2 R H-R^2$ $d^2=h^{2+}2 R h$ as ${\text{R}}>≫>>$ h then ${\text{d}} \approx \sqrt{2 {\text{Rh}}} . . . \;\text{... (1) }\;$ Now, if coverage is to be increased 3 times $3 {\text{d}}=\sqrt{2 {\text{Rh}}^{'}} . . .$ (2)Divide 2 and $1 \;\frac{3 d}{d}=\sqrt{\;\frac{2 R h^{'}}{2 R h}}$ $9=\;\frac{h^{'}}{h}$ $9 h=h^{'}$ If ${\text{h}}=100 {\text{m}}$ then tower of height $900 {\text{m}}$ is required

Q21JEE Main 2022NAT4MCurrent Electricity

In a meter bridge experiment, for measuring unknown resistance ' $S$ ', the null point is obtained at a distance $30 {\text{cm}}$ from the left side as shown at point ${\text{D}}$ . If ${\text{R}}$ is $5.6 {\text{k}} Ω$ , then the value of unknown resistance 'S' will be _______ $Ω$ . [27-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\;{ {\text{S}}}/{30}=\;\frac{5.6 \times 10^3}{70}$ ${\text{S}}=\;\frac{3}{7} \times 5.6 \times 10^3=2400$

Q22JEE Main 2022NAT4MUnits and Measurements

The one division of main scale of Vernier callipers reads $1 {\text{mm}}$ and 10 divisions of Vernier scale is equal to the 9 divisions on main scale. When the two jaws of the instrument touch each other, the zero of the Vernier lies to the right of zero of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws, the zero of the Vernier scale lies in between $4.1 {\text{cm}}$ and $4.2 {\text{cm}}$ and $6^{\;\text{th }\;}$ Vernier division coincides scale division. The diameter of the bob will be ________ $\times 10^{-2} {\text{cm}}$ . [27-Jul-2022-Shift-1]

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📖 Explanation

$10 {\text{VSD}}=9 {\text{MSD}}$ $1 {\text{VST}}=.9 {\text{MSD}}$ L.C. $=.1 {\text{mm}}=.01 {\text{cm}}$ $+$ ve zero error $=.4 {\text{mm}}$ $=0.04 {\text{cm}}$ Negative zero error $=4.1 {\text{cm}}+6 \times .01$ $=4.12 {\text{cm}}$ $=412 \times 10^{-2} {\text{cm}}$

Q23JEE Main 2022NAT4MWave Optics

Two beams of light having intensities $I$ and $4 {\text{I}}$ interfere to produce a fringe pattern on a screen. The phase difference between the two beams are $\pi / 2$ and $\pi / 3$ at points ${\text{A}}$ and ${\text{B}}$ respectively. The difference between the resultant intensities at the two points is $x I$ . The value of $x$ will be ________. [27-Jul-2022-Shift-1]

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📖 Explanation

$\varphi _{ {\text{A}}}=\;\frac{\pi}{2}$ $\varphi _{ {\text{B}}}=\;\frac{\pi}{3}$ ${\text{I}}_{ {\text{A}}}= {\text{I}}+4 {\text{I}}+2 \sqrt{ {\text{I}}} \sqrt{4 {\text{I}}} \cos (\;\frac{\pi}{2})$ $=5 {\text{I}}+4 {\text{I}}(0)=5 {\text{I}}$ ${\text{I}}_{ {\text{B}}}= {\text{I}}+4 {\text{I}}+2 \sqrt{ {\text{I}}} \sqrt{4 {\text{I}}} \cos (60^{\circ})$ $=5 {\text{I}}+4 {\text{I}} \times \;\frac{1}{2}=7 {\text{I}}$ ${\text{I}}_{ {\text{B}}}- {\text{I}}_{ {\text{A}}}=7 {\text{I}}-5 {\text{I}}=2 {\text{I}},( {\text{x}}=2)$

Q24JEE Main 2022NAT4MAlternating Current

To light, a $50 {\text{W}}, 100 {\text{V}}$ lamp is connected, in series with a capacitor of capacitance $\;\frac{50}{\pi \sqrt{x}} \mu F$ , with $200 {\text{V}}, 50 {\text{Hz}} {\text{AC}}$ source. The value of $x$ will be _________. [27-Jul-2022-Shift-1]

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📖 Explanation

${\text{P}}=\;{ {\text{V}}^2}/{ {\text{R}}} \Rightarrow {\text{R}}=\;{ {\text{V}}^2}/{ {\text{P}}}$ $( {\text{V}}_{ {\text{R}}}) \;\; ( {\text{V}}_{ {\text{C}}})$ ${\text{R}}=\;\frac{100 \times 10^2}{50}= {\text{R}}=200 Ω$ ${\text{V}}_{ {\text{R}}}^{2+} {\text{V}}_{ {\text{C}}}^2= {\text{V}}^2$ $(100)^{2+} {\text{V}}_{ {\text{C}}}^2=(200)^2$ ${\text{i}}=\;\frac{100}{200}=\;\frac{1}{2} ; \;\; {\text{V}}^2=40000$ ${\text{V}}= {\text{I}} \times {\text{X}}_{ {\text{C}}} ; \;\; {\text{V}}_{ {\text{C}}}^2=30000$ ${\text{V}}_{ {\text{C}}}=100 \sqrt{3}$ ${\text{X}}_{ {\text{C}}}=200 \sqrt{3}$ $200 \sqrt{3}=\;{1}/{\omega {\text{C}}}$ $C=\;\frac{1}{20 \times 50 \times 20 \sqrt{3}}=\;\frac{50 \times 10^{-6}}{\sqrt{x}}$ $\sqrt{x}=50 \times 10^{-6} \times 100 \times 200 \sqrt{3}$ $X=3$

Q25JEE Main 2022NAT4MCurrent Electricity

A $1 {\text{m}}$ long copper wire carries a current of $1 {\text{A}}$ . If the cross section of the wire is $2.0 {\text{mm}}^2$ and the resistivity of copper is $1.7 \times 10^{-8} Ω {\text{m}}$ , the force experienced by moving electron in the wire is _______ $\times 10^{-23} {\text{N}}$ . (charge on electron $=1.6 \times 10^{-19} {\text{C}}$ ) [27-Jul-2022-Shift-1]

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📖 Explanation

$l=1 {\text{m}}$ ${\text{i}}=1 {\text{A}}$ Area $=2 \times 10^{-6}$ $\rho =1.7 \times 10^{-8}$ ${\text{R}}=\;{\rho ℓ}/{ {\text{A}}}=\;\frac{1.7 \times 10^{-8} \times 1}{2 \times 10^{-5}}=\;\frac{1.7}{2} \times 10^{-2}$ ${\text{v}}=\;\frac{1.7}{2} \times 10^{-2}$ ${\text{F}}=1.6 \times 10^{-19} \times \;\frac{1.7}{2} \times 10^{-2}$ $=1.36 \times 10^{-21}$ $=136 \times 10^{-23}$

Q26JEE Main 2022NAT4MElectrostatics

A long cylindrical volume contains a uniformly distributed charge of density $\rho {\text{Cm}}^{-3}$ . The electric field inside the cylindrical volume at a distance $x=\;\frac{2 \epsilon_0}{\rho } {\text{m}}$ from its axis is _______ ${\text{Vm}}^{-1}$ [27-Jul-2022-Shift-1]

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📖 Explanation

$\int {\text{Eds}\;cos} 0=\;{ {\text{q}}}/{\epsilon_0}$ $\Rightarrow {\text{E}} \cdot 2 \pi {\text{xh}}=\;{\rho \times \pi {\text{x}}^2 {\text{h}}}/{\epsilon_0}$ $\Rightarrow {\text{E}}=\;{\rho {\text{x}}}/{2 \epsilon_0}$ $\Rightarrow {\text{E}}=\;\frac{\rho }{2 \epsilon_0} \times \;\frac{2 \epsilon_0}{\rho }=1$

Q27JEE Main 2022NAT4MOscillations

A mass $0.9 {\text{kg}}$ , attached to a horizontal spring, executes SHM with an amplitude ${\text{A}}_1$ . When this mass passes through its mean position, then a smaller mass of $124 {\text{g}}$ is placed over it and both masses move together with amplitude $A_2$ . If the ratio $\;\frac{A_1}{A_2}$ is $\;\frac{\alpha}{\alpha-1}$ , then the value of $\alpha$ will be ________. [27-Jul-2022-Shift-1]

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📖 Explanation

$\;\frac{1}{2} {\text{kA}}^2=\;{ {\text{p}}^2}/{2 {\text{m}}}$ $\Rightarrow (\;{ {\text{A}}_1}/{ {\text{A}}_2})^2=\;{ {\text{m}}_2}/{ {\text{m}}_1}=\;\frac{1024}{900}$ $\Rightarrow \;{ {\text{A}}_1}/{ {\text{A}}_2}=\;\frac{32}{30}=\;\frac{16}{15}=\;\frac{16}{16-1}$ $\therefore \alpha=16$

Q28JEE Main 2022NAT4MProperties of Matter

A square aluminum (shear modulus is $25 \times 10^9 {\text{Nm}}^{-2}$ ) slab of side $60 {\text{cm}}$ and thickness $15 {\text{cm}}$ is subjected to a shearing force (on its narrow face) of $18.0 \times 10^4 {\text{N}}$ . The lower edge is riveted to the floor. The displacement of the upper edge is _______ $\mu {\text{m}}$ . [27-Jul-2022-Shift-1]

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📖 Explanation

$\;{ {\text{F}}}/{ {\text{A}}}=η \;{ {\text{x}}}/{ℓ} \Rightarrow \;{ {\text{F}} ℓ}/{ {\text{A}} η}= {\text{x}}$ $\Rightarrow {\text{x}}=\;\frac{18 \times 10^4 \times 60 \times 10^{-2}}{60 \times 10^{-2} \times 15 \times 10^{-2} \times 25 \times 10^9}$ $=48 \times 10^{-6} {\text{m}}=48 \mu {\text{m}}$

Q29JEE Main 2022NAT4MRotational Motion

A pulley of radius $1.5 {\text{m}}$ is rotated about its axis by a force $F=(12 {\text{t}}-3 {\text{t}}^2) N$ applied tangentially (while $t$ is measured in seconds). If moment of inertia of the pulley about its axis of rotation is $4.5 {\text{kg}} {\text{m}}$ , the number of rotations made by the pulley before its direction of motion is reversed, will be $\;\frac{K}{\pi}$ . The value of ${\text{K}}$ is ________. [27-Jul-2022-Shift-1]

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📖 Explanation

$\tau = {\text{I}} \alpha \Rightarrow (12 {\text{t}}-3 {\text{t}}^2) 1.5=4.5 \alpha$ $\Rightarrow \alpha=4 {\text{t}}- {\text{t}}^2$ $\Rightarrow \;{ {\text{d}} \omega }/{ {\text{dt}}}=4 {\text{t}}- {\text{t}}^2 \Rightarrow \omega =int_{0}^{ {\text{t}}}(4 {\text{t}}- {\text{t}}^2) {\text{dt}}$ $\Rightarrow \omega =2 {\text{t}}^{2-} {\text{t}}^3 / 3$

Q30JEE Main 2022NAT4MMotion in a Plane

A ball of mass ${\text{m}}$ is thrown vertically upward. Another ball of mass $2 {\text{m}}$ is thrown at an angle $\theta$ with the vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is $\;\frac{1}{x}$ . The value of ${\text{x}}$ is ________. [27-Jul-2022-Shift-1]

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📖 Explanation

$\therefore u_1=u_2 sin \;\theta$ $\;\frac{H_1}{H_2}=\;\frac{\;\frac{u_1^2}{2 g}}{u_2^2 \;\frac{ sin \;^2 \theta}{2 g}}$ $= (\;\frac{u_1}{u_2 sin \;\theta}) ^2=1$

Chemistry30 questions

Q31JEE Main 2022MCQ4MSolutions

$250 {\text{g}}$ solution of ${\text{D}}$ -glucose in water contains $10.8 \%$ of carbon by weight. The molality of the solution is nearest to (Given: Atomic Weights are, ${\text{H}}, 1 {\text{u}} ; {\text{C}}, 12 {\text{u}} ; {\text{O}}, 16 {\text{u}}$ ) [27-Jul-2022-Shift-1]

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📖 Explanation

${\text{C}}_6 {\text{H}}_{12} {\text{O}}_6 \to$ GlucoseWe know: $\;{\;\text{ mass of }\; {\text{C}}}/{\;\text{ mass of glucose }\;}=\;\frac{72}{180}$ Given: $\% {\text{C}}=10.8=\;{\;\text{ mass of }\; {\text{C}}}/{\;\text{ mass of solution }\;} \times 100$ $\;\frac{10.8 \times 250}{100}=$ mass of ${\text{C}} \Rightarrow$ Mass of ${\text{C}}=27 {\text{gm}}$ $\therefore$ mass of glucose $=67.5$ gm $\therefore$ moles of glucose $=0.375 {\text{moles}}$ Mass of solvent $=250-67.5 {\text{gm}}=182.5 {\text{gm}}$ $\therefore$ Molality $=\;\frac{0.375}{0.1825}=2.055 \approx 2.06$

Q32JEE Main 2022MCQ4MSolid State

Given below are two statements. Statement ${\text{l}}_2 {\text{O}}_2, {\text{Cu}}^{2+}$ , and ${\text{Fe}}^{3+}$ are weakly attracted by magnetic field and are magnetized in the same direction as magnetic field. Statement II: ${\text{NaCl}}$ and ${\text{H}}_2 {\text{O}}$ are weakly magnetized in opposite direction to magnetic field. In the light of the above statements, choose the most appropriate answer from the options given below. [27-Jul-2022-Shift-1]

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📖 Explanation

${\text{O}}_2, {\text{Cu}}^{2+}$ and ${\text{Fe}}^{3+}$ have 2,1 and 5 unpaired electrons respectively, so these are the paramagnetic species. Hence, they are attracted by magnetic field. ${\text{NaCl}}$ and ${\text{H}}_2 {\text{O}}$ are the diamagnetic species so they are repelled by the magnetic field.

Q33JEE Main 2022MCQ4MStructure of Atom

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Energy of $2 s$ orbital of hydrogen atom is greater than that of $2 s$ orbital of lithium. Reason R: Energies of the orbitals in the same subshell decrease with increase in the atomic number. In the light of the above statements, choose the correct answer from the options given below. [27-Jul-2022-Shift-1]

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📖 Explanation

As the atomic number increases then the potential energy of electrons present in same shell becomes more and more negative. And therefore total energy also becomes more negative. ${\text{E}}_{ {\text{total}}}=-13.6 \;{ {\text{Z}}^2}/{ {\text{n}}^2} {\text{eV}}$ $\therefore$ Energies of the orbitals in the same subshell decreases with increase in atomic number.

Q34JEE Main 2022MCQ4MStates of Matter

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Activated charcoal adsorbs ${\text{SO}}_2$ more efficiently than ${\text{CH}}_4$ . Reason R: Gases with lower critical temperatures are readily adsorbed by activated charcoal. In the light of the above statements, choose the correct answer from the options given below. [27-Jul-2022-Shift-1]

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📖 Explanation

More polar gases easily adsorbs on activated charcoal.And more polar gases has more (higher) critical temperature as compared to non-polar or less polar gases. $\therefore$ Gases with higher critical temperature adsorbed more.

Q35JEE Main 2022MCQ4MSolutions

Boiling point of a $2 \%$ aqueous solution of a non-volatile solute ${\text{A}}$ is equal to the boiling point of $8 \%$ aqueous solution of a non-volatile solute $B$ . The relation between molecular weights of $A$ and $B$ is [27-Jul-2022-Shift-1]

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📖 Explanation

For $A: 100 {\text{gm}}$ solution $\to 2 {\text{gm}}$ solute $A$ $\therefore$ Molality $=\;{2 / {\text{M}}_{ {\text{A}}}}/{0.098}$ For $B: 100 {\text{gm}}$ solution $\to 8 {\text{gm}}$ solute $B$ $\therefore \;\text{ Molality }\;=\;{8 / {\text{M}}_{ {\text{B}}}}/{0.092}$ $\because (\Delta {\text{T}}_{ {\text{B}}})_{ {\text{A}}}=(\Delta {\text{T}}_{ {\text{B}}})_{ {\text{B}}}$ $\therefore$ Molality of $A=$ Molality of $B$ $\therefore \;{2}/{0.098 {\text{M}}_{ {\text{A}}}}=\;{8}/{0.092 {\text{M}}_{ {\text{B}}}}$ $\;\frac{2}{98} \times \;\frac{92}{8}=\;\frac{M_A}{M_B}$ $\;\frac{1}{4.261}=\;{ {\text{M}}_{ {\text{A}}}}/{ {\text{M}}_{ {\text{B}}}}$ $\therefore {\text{M}}_{ {\text{B}}}=4.261 \times {\text{M}}_{ {\text{A}}}$

Q36JEE Main 2022MCQ4MClassification of Elements

The incorrect statement is [27-Jul-2022-Shift-1]

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📖 Explanation

On moving down in a group ionisation energy decrease $\therefore 1^{\;\text{st }\;}$ ionisation enthalpy order is ${\text{Li}}> {\text{Na}}> {\text{K}}$ ${\text{Zn}}$ has more ionisation energy as compared to ${\text{Ga}}$ because of their pseudo inert gas configuration.

Q37JEE Main 2022MCQ4Md and f Block Elements

Which of the following methods are not used to refine any metal? A. Liquation B. Calcination C. Electrolysis D. Leaching E. Distillation Choose the correct answer from the options given below: [27-Jul-2022-Shift-1]

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Q38JEE Main 2022MCQ4MRedox Reactions

Given below are two statements. Statement I: Hydrogen peroxide can act as an oxidizing agent in both acidic and basic conditions. Statement II: Density of hydrogen peroxide at $298 {\text{K}}$ is lower than that of ${\text{D}}_2 {\text{O}}$ . In the light of the above statements, choose the correct answer from the options given below. [27-Jul-2022-Shift-1]

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📖 Explanation

Depending on the nature of reducing agent ${\text{H}}_2 {\text{O}}_2$ can act as an oxidising agent in both acidic as well as basic medium.Density of ${\text{D}}_2 {\text{O}}=1.1 {\text{g}} / {\text{cc}}$ Density of ${\text{H}}_2 {\text{O}}_2=1.45 {\text{g}} / {\text{cc}}$

Q39JEE Main 2022MCQ4Mp Block Elements

Given below are two statements. Statement I: The chlorides of ${\text{Be}}$ and ${\text{Al}}$ have ${\text{Cl}}$ -bridged structure. Both are soluble in organic solvents and act as Lewis bases. Statement II: Hydroxides of Be and ${\text{Al}}$ dissolve in excess alkali to give beryllate and aluminate ions. In the light of the above statements, choose the correct answer from the options given below. [27-Jul-2022-Shift-1]

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📖 Explanation

${\text{Be}}_2 {\text{Cl}}_4$ is lewis acid and ${\text{Al}}_2 {\text{Cl}}_6$ has complete octet. ${\text{Be}}$ and ${\text{Al}}$ are amphoteric metals therefore dissolve in acid as well as alkaline solution and form beryllate and aluminate ions in excess alkali.

Q40JEE Main 2022MCQ4Mp Block Elements

Which oxoacid of phosphorous has the highest number of oxygen atoms present in its chemical formula? [27-Jul-2022-Shift-1]

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📖 Explanation

Pyrophosphorous acid $\to {\text{H}}_4 {\text{P}}_2 {\text{O}}_5$ .Hypophosphoric acid $\to {\text{H}}_4 {\text{P}}_2 {\text{O}}_6$ .Phosphoric acid $\to {\text{H}}_3 {\text{PO}}_4$ .Pyrophosphoric acid $\to {\text{H}}_4 {\text{P}}_2 {\text{O}}_7$ .

Q41JEE Main 2022MCQ4Md and f Block Elements

Given below are two statements. Statement I: Iron (III) catalyst, acidified ${\text{K}}_2 {\text{Cr}}_2 {\text{O}}_7$ and neutral ${\text{KMnO}}_4$ have the ability to oxidise ${\text{I}}^{-}$ to ${\text{I}}_2$ independently. Statement II: Manganate ion is paramagnetic in nature and involves ${\text{p}} \pi- {\text{p}} \pi$ bonding. In the light of the above statements, choose the correct answer from the options given below. [27-Jul-2022-Shift-1]

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📖 Explanation

Manganate ion ${\text{MnO}}_4^{2-}$ has tetrahedral structurehas only ${\text{d}} \pi- {\text{p}} \pi \pi$ -bonds. ${\text{Fe}}^{3+}$ is not used as a catalyst in the conversion of I' to ${\text{I}}_2$ by ${\text{K}}_2 {\text{Cr}}_2 {\text{O}}_7 \cdot {\text{K}}_2 {\text{Cr}}_2 {\text{O}}_7$ oxidise ${\text{I}}^{-}$ in acidic medium easily

Q42JEE Main 2022MCQ4MChemical Bonding and Molecular Structure

The total number of ${\text{Mn}}= {\text{O}}$ bonds in ${\text{Mn}}_2 {\text{O}}_7$ is ________. [27-Jul-2022-Shift-1]

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📖 Explanation

Structure of $\mathrm{Mn}_2 \mathrm{O}_7$ is as : $\therefore$ There are total $6 {\text{M}}= {\text{O}}$ bonds are present in ${\text{Mn}}_2 {\text{O}}_7$ compound.

Q43JEE Main 2022MCQ4MAmines

Match List I with List II. List I Pollutant List II Disease/sickness (A) Sulphate (>500 ppm) (I)Methemoglobinemia (B) Nitrate (>50 ppm) (II) Brown mottling of teeth (C) Lead (>50 ppb) (III) Laxative effect (D) Fluoride (>2 ppm) (IV) Kidney damageChoose the correct answer from the options given below: [27-Jul-2022-Shift-1]

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📖 Explanation

A. Sulphate ( $>500 {\text{ppm}})$ - Causes Laxative effect that leads to dehydrationB. Nitrate (>50 ppm) - Causes Methemoglobinemia, skin appears blueC. Lead ( $>50 {\text{ppb}})$ - It damage kidney and RBCD. Fluoride ( $>2 {\text{ppm}})$ - It Causes Brown mottling of teeth

Q44JEE Main 2022MCQ4MOrganic Chemistry - Some Basic Principles

Given below are two statements: one is labelled as Assertion ${\text{A}}$ and, the other is labelled as Reason R. Assertion A: [6] Annulene, [8] Annulene and cis-[10] Annulene, are respectively aromatic, not-aromatic and aromatic. Reason R: Planarity is one of the requirements of aromatic systems. In the light of the above statements, choose the most appropriate answer from the options given below. [27-Jul-2022-Shift-1]

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📖 Explanation

[6] Annulene is aromatic because it is planar.[8] Annulene and [10] Annulene are both not aromatic because they are not planar. So, Assertion (A) is not correct.Reason $(R)$ is correct because planarity is one of the requirements of aromatic system.

Q45JEE Main 2022MCQ4MHaloalkanes and Haloarenes

In the above reaction product $B$ is: [27-Jul-2022-Shift-1]

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📖 Explanation

Product $B$ is 4-iodomethylphenol.

Q46JEE Main 2022MCQ4MAmines

Match List I with List II. List I Polymers List II Commercial names (A) Phenol-formaldehyde resin (I)Glyptal (B) Copolymer of 1,3-butadiene and sterene (II) Novolac (C) Polyester of glycol and phthalic acid (III) Buna-S (D) Polyester of glycol and terephthalic acid (IV) DacronChoose the correct answer from the options given below: [27-Jul-2022-Shift-1]

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Q47JEE Main 2022MCQ4MBiomolecules

A sugar ' $X$ ' dehydrates very slowly under acidic condition to give furfural which on further reaction with resorcinol gives the coloured product after sometime. Sugar ' $X$ ' is [27-Jul-2022-Shift-1]

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Q48JEE Main 2022MCQ4MAmines

Match List I with List II. List I List II (A) (I)Anti-depressant (B) (II)550 times sweeter than cane sugar. (C) (III)Narcotic analgesic (D) (IV)AntisepticChoose the correct answer from the options given below: [27-Jul-2022-Shift-1]

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Q49JEE Main 2022MCQ4MPractical Organic Chemistry

In Carius method of estimation of halogen, $0.45 {\text{g}}$ of an organic compound gave $0.36 {\text{g}}$ of ${\text{AgBr}}$ . Find out the percentage of bromine in the compound. (Molar masses : ${\text{AgBr}}=188 {\text{g}} {\text{mol}}^{-1} ; {\text{Br}}=80 {\text{g}} {\text{mol}}^{-1}$ ) [27-Jul-2022-Shift-1]

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📖 Explanation

Mass of organic compound $=0.45 {\text{gm}}$ Mass of ${\text{AgBr}}$ obtained $=0.36 {\text{gm}}$ $\therefore$ Moles of ${\text{AgBr}}=\;\frac{0.36}{188}$ $\therefore$ Mass of Bromine $=\;\frac{0.36}{188} \times 80=0.1532 {\text{gm}}$ $\therefore \% {\text{Br}}$ in compound $=\;\frac{0.1532}{0.45} \times 100=34.04 \%$

Q50JEE Main 2022MCQ4MAmines

Match List I with List II. List I List II (A) Benzenesulphonyl chloride (I)Test for primary amines (B) Hoffmann bromamide reaction (II) Anti Saytzeff (C) Carbylamine reaction (III) Hinsberg reagent (D) Hoffmann orientation (IV) Known reaction of Isocyanates.Choose the correct answer from the options given below: [27-Jul-2022-Shift-1]

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📖 Explanation

(A) Benzene sulphonyl chloride is also known as Hinsberg reagent.(B) Hoffmann bromamide reaction involves conversion of amide to amine having one ${\text{C}}$ atom less. This reaction involves isocyanate as intermediate.(C) Carbylamine reaction is a test given by all primary amines.(D) Hoffmann orientation refers to the addition of molecules to unsymmetrical alkenes according to anti Saytzeff's rule.Correct match isA-III, B-IV, C-I, D-II

Q51JEE Main 2022NAT4MRedox Reactions

$20 {\text{mL}}$ of $0.02 {\text{M}}_2 {\text{Cr}}_2 {\text{O}}_7$ solution is used for the titration of $10 {\text{mL}}$ of ${\text{Fe}}^{2+}$ solution in the acidic medium. The molarity of ${\text{Fe}}^{2+}$ solution is _______ $\times 10^{-2} {\text{M}}$ . (Nearest Integer) [27-Jul-2022-Shift-1]

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📖 Explanation

Eq. of ${\text{K}}_2 {\text{Cr}}_2 {\text{O}}_7=$ Eq. of ${\text{Fe}}^{2+}$ $\Rightarrow$ (Molarity $\times$ volume $\times$ n.f) of ${\text{K}}_2 {\text{Cr}}_2 {\text{O}}_7=$ (molarity $\times$ volume $\times$ n.f ) of ${\text{Fe}}^{2+}$ $\Rightarrow 0.02 \times 20 \times 6= {\text{M}} \times 10 \times 1$ $\Rightarrow {\text{M}}=0.24$ $\Rightarrow$ Molarity $=24 \times 10^{-2}$

Q52JEE Main 2022NAT4MChemical Kinetics

$2 {\text{NO}}+2 {\text{H}}_2 \to {\text{N}}_2+2 {\text{H}}_2 {\text{O}}$ The above reaction has been studied at $800^{\circ} {\text{C}}$ . The related data are given in the table below Reaction serial number Initial Pressure of ${\text{H}}_2 / k P a$ Initial Pressureof $N O / k P a$ Initial rate $(\;\frac{-d p}{d t}) /(k P a / s)$ 1 65.6 40.0 0.135 2 65.6 20.1 0.033 3 38.6 65.6 0.214 4 19.2 65.6 0.106The order of the reaction with respect to ${\text{NO}}$ is [27-Jul-2022-Shift-1]

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📖 Explanation

Let the rate of reaction $(r)$ is as ${\text{r}}= {\text{K}}[ {\text{NO}}]^n[ {\text{H}}_2]^{ {\text{m}}}$ From $1^{\;\text{st }\;}$ data $0.135= {\text{K}}[40]^{ {\text{n}}} \cdot (65.6)^{ {\text{m}}} \;\; . . .(1)$ From $2^{\;\text{nd }\;}$ data $0.033= {\text{K}}(20.1)^{ {\text{n}}} \cdot (65.6)^{ {\text{m}}} \;\; . . .(2)$ On dividing equation (1) by equation (2) $\;\frac{0.135}{0.033}=(\;\frac{40}{20.1})^n$ $4=(2)^n$ $\therefore n=2$ $\therefore$ Order of reaction w.r.t. ${\text{NO}}$ is 2 .

Q53JEE Main 2022NAT4Mp Block Elements

Amongst the following, the number of oxide(s) which are paramagnetic in nature is ${\text{Na}}_2 {\text{O}}, {\text{KO}}_2, {\text{NO}}_2, {\text{N}}_2 {\text{O}}, {\text{ClO}}_2, {\text{NO}}, {\text{SO}}_2, {\text{Cl}}_2 {\text{O}}$ [27-Jul-2022-Shift-1]

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📖 Explanation

Paramagnetic species: ${\text{KO}}_2, {\text{NO}}_2, {\text{ClO}}_2, {\text{NO}}$ Diamagnetic species are: ${\text{Na}}_2 {\text{O}}, {\text{N}}_2 {\text{O}}, {\text{SO}}_2, {\text{Cl}}_2 {\text{O}}$

Q54JEE Main 2022NAT4MChemical Thermodynamics

The molar heat capacity for an ideal gas at constant pressure is $20.785 {\text{J}} {\text{K}}^{-1} {\text{mol}}^{-1}$ . The change in internal energy is $5000 {\text{J}}$ upon heating it from $300 {\text{K}}$ to $500 {\text{K}}$ . The number of moles of the gas at constant volume is _______. [ [Nearest integer] (Given: ${\text{R}}=8.314 {\text{J}} {\text{K}}^{-1} {\text{mol}}^{-1}$ ) [27-Jul-2022-Shift-1]

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📖 Explanation

${\text{C}}_{ {\text{p}}}=20.785 {\text{J}} {\text{K}}^{-1} {\text{mol}}^{-1}$ and $\Delta {\text{U}}= {\text{nCv}} \Delta {\text{T}}$ $\therefore \;\; {\text{nC}}_{ {\text{v}}}=\;\frac{5000}{200}=25$ and we know that $C_p-C_v=R$ $20.785-\;\frac{25}{n}=8.314$ $n=\;\frac{25}{(20.785-8.314)}=2$

Q55JEE Main 2022NAT4MChemical Bonding and Molecular Structure

According to MO theory, number of species/ions from the following having identical bond order is ________. ${\text{CN}}^{-}, {\text{NO}}^{+}, {\text{O}}_2, {\text{O}}_2^{+}, {\text{O}}_2^{2+}$ [27-Jul-2022-Shift-1]

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📖 Explanation

${\text{CN}}^{-}, {\text{NO}}^{+}$ and ${\text{O}}_2^{2+}$ have bond order of 3 ${\text{O}}_2$ has bond order of 2 ${\text{O}}_2^{+}$ has bond order of $2.5$ $\therefore 3$ species have similar bond order.

Q56JEE Main 2022NAT4MIonic Equilibrium

At $310 {\text{K}}$ , the solubility of ${\text{CaF}}_2$ in water is $2.34 \times 10^{-3} {\text{g}} / 100 {\text{mL}}$ . The solubility product of ${\text{CaF}}_2$ is _______ $\times 10^{-8}( {\text{mol}} / {\text{L}})^3$ . (Give molar mass : ${\text{CaF}}_2=78 {\text{g}} {\text{mol}}^{-1}$ ) [27-Jul-2022-Shift-1]

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📖 Explanation

${\text{CaF}}_2 {⇌}^{ {\ s}} \;{{ {\text{Ca}}^{2+}}}_{ {\ s}}+2 \;{{ {\text{F}}^{-}}}_{2 {\ s}}$ ${\text{K}}_{ {\ sp}}= {\ s}(2 {\ s})^2$ $\;\; =4 {\ s}^3$ Solubility $( {\ s})=2.34 \times 10^{-3} {\text{g}} / 100 {\text{mL}}$ $=\;\frac{2 \cdot 34 \times 10^{-3} \times 10}{78}$ mole $/$ lit $=3 \times 10^{-4} {\text{mole}} /$ lit $\therefore {\text{K}}_{ {\ sp}}=4 \times (3 \times 10^{-4})^3$ $=108 \times 10^{-12}$ $=0.0108 \times 10^{-8}( {\text{mole}} / \;\text{ lit }\;)^3$ $\therefore x \approx 0$

Q57JEE Main 2022NAT4MCoordination Compounds

The conductivity of a solution of complex with formula ${\text{CoCl}}_3( {\text{NH}}_3)_4$ corresponds to $1: 1$ electrolyte, then the primary valency of central metal ion is _______. [27-Jul-2022-Shift-1]

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📖 Explanation

In 1: 1 type of electrolyte the ions have $+1$ and $-1$ charge on them $\therefore$ Possible compound is $\to [ {\text{Co}}( {\text{NH}}_3)_4 {\text{Cl}}_2]^{+} {\text{Cl}}^{-}$ Oxidation state of central atom represents the total number of primary valency $\therefore$ Primary valency will be 3 .

Q58JEE Main 2022NAT4MRedox Reactions

In the titration of ${\text{KMnO}}_4$ and oxalic acid in acidic medium, the change in oxidation number of carbon at the end point is ________. [27-Jul-2022-Shift-1]

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📖 Explanation

$16 {\text{H}}^{+}+2 {\text{MnO}}_4^{-}+5 {\text{C}}_2 {\text{O}}_4^{2-} \to 10 {\text{CO}}_2+2 {\text{Mn}}^{2+}+8 {\text{H}}_2 {\text{O}}$ During titration of oxalic acid by ${\text{KMnO}}_4$ , oxalic acid converts into ${\text{CO}}_2$ . $\therefore$ Change in oxidation state of carbon $=1$

Q59JEE Main 2022NAT4MOrganic Chemistry - Some Basic Principles

Optical activity of an enantiomeric mixture is $+12.6^{\circ}$ and the specific rotation of $(+)$ isomer is $+30^{\circ}$ The optical purity is ________ $\%$ [27-Jul-2022-Shift-1]

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📖 Explanation

Optical purity $=\;\frac{\;\text{ Total rotation }\;}{\;\text{ Specific rotation }\;} \times 10$ $=\;\frac{12 \cdot 6}{30} \times 100$ $=42 \%$

Q60JEE Main 2022NAT4MAlcohols, Phenols and Ethers

In the following reactionThe $\%$ yield for reaction I is $60 \%$ and that of reaction II is $50 \%$ . The overall yield of the complete reaction is _______%. [nearest integer] [27-Jul-2022-Shift-1]

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📖 Explanation

Let initial moles of reactant taken $= {\text{n}}$ Total moles obtained for benzene sulphonic acid (with $\%$ yield $=60 %)=0.6 {\text{n}}$ Moles of benzene sulphonic acid before reaction ${\text{II}}=0.6 {\text{n}}$ Moles obtained for phenol (with % yield $=50 \%$ ) $=0.6 \times 0.5 {\text{n}}=0.3 {\text{n}}$ So over all % yield of complete reaction $=\;{0.3 {\text{n}}}/{ {\text{n}}} \times 100=30$

Mathematics30 questions

Q61JEE Main 2022MCQ4MSets and Relations

Let $R_1$ and $R_2$ be two relations defined on $ℝ$ by $a R_1 b ⇔ a b \ge 0$ and $a R_2 b ⇔ a \ge b$ Then, [27-Jul-2022-Shift-1]

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📖 Explanation

${\text{R}}_1=\{ {\text{xy}} \ge 0, {\text{x}}, {\text{y}} \in {\text{R}}\}$ For reflexive ${\text{x}} \times {\text{x}} \ge 0$ which is true.For symmetricIf ${\text{xy}} \ge 0 \Rightarrow {\text{yx}} \ge 0$ If $x=2, y=0$ and $z=-2$ Then ${\text{x}} \cdot {\text{y}} \ge 0 & {\text{y}} \cdot {\text{z}} \ge 0$ but ${\text{x}} \cdot {\text{z}} \ge 0$ is not true $\Rightarrow$ not transitive relation. $\Rightarrow {\text{R}}_1$ is not equivalence ${\text{R}}_2$ if ${\text{a}} \ge {\text{b}}$ it does not implies ${\text{b}} \ge {\text{a}}$ $\Rightarrow {\text{R}}_2$ is not equivalence relation

Q62JEE Main 2022MCQ4MFunctions

Let $f, g: ℕ-\{1\} \to ℕ$ be functions defined by $f(a)=\alpha$ , where $\alpha$ is the maximum of the powers of those primes $p$ such that $p^\alpha$ divides $a$ , and $g(a)=a+1$ , for all $a \in ℕ-\{1\}$ . Then, the function $f+g$ is [27-Jul-2022-Shift-1]

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📖 Explanation

$f, g: N-\{1\} \to N$ defined as $f(a)=\alpha$ , where $\alpha$ is the maximum power of those primes ${\text{p}}$ such that ${\text{p}}^\alpha$ divides a. $g(a)=a+1$ Now, $f(2)=1, \;\; g(2)=3 \;\; \Rightarrow \;\; (f+g)(2)=4$ $f(3)=1, \;\; g(3)=4 \;\; \Rightarrow \;\; (f+g)(3)=5$ $f(4)=2, \;\; g(4)=5 \Rightarrow (f+g)(4)=7$ $f(5)=1, \;\; g(5)=6 \Rightarrow (f+g)(5)=7$ $\because (f+g)(5)=(f+g)(4)$ $\therefore f+g$ is not one-oneNow, $\because f_{\min }=1, g_{\min }=3$ So, there does not exist any $x \in N-\{1\}$ such that $(f+g)(x)=1,2,3$ $\therefore f+g$ is not onto

Q63JEE Main 2022MCQ4MApplication of Derivatives

Let the minimum value $v_0$ of $v=|z|^{2+}|z-3|^{2+}|z-6 i|^2, z \in C$ is attained at $z=z_0$ . Then $| 2 z_0^{2-}\overline{z}_0^{3+}3 {}/{}|^{2+}v_0^2$ is equal to [27-Jul-2022-Shift-1]

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📖 Explanation

Let $z=x+i y$ $v=x^{2+}y^{2+}(x-3)^{2+}y^{2+}x^{2+}(y-6)^2$ $=(3 x^{2-}6 x+9)+(3 y^{2-}12 y+36)$ $=3(x^{2+}y^{2-}2 x-4 y+15)$ $=3[(x-1)^{2+}(y-2)^{2+}10]$ $v_{\;\text{min }\;} \;\text{ at }\; z=1+2 i=z_0 \;\text{ and }\; v_0=30$ $\;\text{ so }\;|2(1+2 i)^{2-}(1-2 i)^{3+}3|^{2+}900$ $=\mid 2(-3+4 i)-(1-8 i^{3-}6 i(1-2 i)+.3|^{2+}900.$ $=|-6+8 i-(1+8 i-6 i-12)+3|^{2+}900$ $=|8+6 i|^{2+}900$ $=|8+6 i|^{2+}900$ $=1000$

Q64JEE Main 2022MCQ4MMatrices and Determinants

Let $A=(\text{table} 1 , 2 ; -2 , -5)$ . Let $\alpha, \beta \in ℝ$ be such that $\alpha A^{2+}\beta A=2 I$ . Then $\alpha+\beta$ is equal to [27-Jul-2022-Shift-1]

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📖 Explanation

$A^2=\begin{bmatrix}1 & 2 \\ -2 & -5\end{bmatrix}\begin{bmatrix}1 & 2 \\ -2 & -5\end{bmatrix}=\begin{bmatrix}-3 & -8 \\ 8 & 21\end{bmatrix}$ $\alpha A^{2+}\beta A=\begin{bmatrix}-3 \alpha & -8 \alpha \\ 8 \alpha & 21 \alpha\end{bmatrix}+\begin{bmatrix}\beta & 2 \beta \\ -2 \beta & -5 \beta\end{bmatrix}$ $=\begin{bmatrix}-3 \alpha+\beta & -8 \alpha+2 \beta \\ 8 \alpha-2 \beta & 21 \alpha-5 \beta\end{bmatrix}=\begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}$ On Comparing $8 \alpha=2 \beta,-3 \alpha+\beta=2,21 \alpha-5 \beta=2$ $\Rightarrow \alpha=2, \beta=8$ So, $\alpha+\beta=10$

Q65JEE Main 2021MCQ4MBinomial Theorem

The remainder when $(2021)^{2022}+(2022)^{2021}$ is divided by 7 is [27-Jul-2022-Shift-1]

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📖 Explanation

$(2021)^{2022}+(2022)^{2021}$ $=(7 k-2)^{2022}+(7 k_1-1)^{2021}$ $=\$ [(7 k-2)^3]^{674}+(7 k_1)^{2021}-2021(7 k_1)^{2020}+. . .-1=(7 k_2-1)^{674}+(7 m-1)=(7 n+1)+(7 m-1)=7(m+n) $(multiple of 7$ )∴ $Remainder$ =0$

Q66JEE Main 2022MCQ4MSequences and Series

Suppose $a_1, a_2, . . ., a_n, .$ . be an arithmetic progression of natural numbers. If the ratio of the sum of first five terms to the sum of first nine terms of the progression is $5: 17$ and , $110 < a_{15} < 120$ , then the sum of the first ten terms of the progression is equal to [27-Jul-2022-Shift-1]

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📖 Explanation

$\because a_1, a_2, . . . . a_n$ be an A.P of natural numbers and $\;\frac{S_5}{S_9}=\;\frac{5}{17} \Rightarrow \;\frac{\;\frac{5}{2}[2 a_1+4 d]}{\frac{9}{2}[2 a_1+8 d]}=\;\frac{5}{17}$ $\Rightarrow 34 a_1+68 d=18 a_1+72 d$ $\Rightarrow 16 a_1=4 d$ $\therefore d=4 a_1$ And $110< a_{15}< 120$ $\therefore 110< a_1+14 d< 120 \Rightarrow 110< 57 a_1< 120$ $\therefore a_1=2( \because a_i \in N)$ $d=8$ $\therefore S_{10}=5[4+9 \times 8]=380$

Q67JEE Main 2022MCQ4MDefinite Integrals

Let $f: ℝ \to ℝ$ be a function defined as $f(x)=a \sin (\;\frac{\pi[x]}{2})+[2-x], a \in ℝ$ where $[t]$ is the greatest integer less than or equal to $t$ . If $\lim _{x \to -1} f(x)$ exists, then the value of $int_0^4 f(x) d x$ is equal to [27-Jul-2022-Shift-1]

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📖 Explanation

$f(x)=a \sin (\;\frac{\pi[x]}{2})+[2-x] a \in R$ Now, $\because \lim _{x \to -1} f(x) \;\text{ exist }\;$ $\therefore \lim _{x \to -1^{-}} f(x)=\lim _{x \to -1^{+}} f(x)$ $\Rightarrow a \sin (\;\frac{-2 \pi}{2})+3=a \sin (\;\frac{-\pi}{2})+2$ $\Rightarrow -a=1 \Rightarrow a=-1$ $\;\text{ Now, }\; int_0^4 f(x) d x=int_0^4(-\sin (\;\frac{\pi[x]}{2})+[2-x]) d x$ $=int_0^1 1 d x+int_1^{2-}1 d x+int_2^{3-}1 d x+int_3^4(1-2) d x$ $=1-1-1-1=-2$

Q68JEE Main 2022MCQ4MDefinite Integrals

Let $I=int_{\pi / 4}^{\pi / 3}(\;\frac{8 \sin x-\sin 2 x}{x}) d x$ . Then [27-Jul-2022-Shift-1]

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📖 Explanation

I comes out around $1.536$ which is not satisfied by any given options. $int_{\pi / 4}^{\pi / 3} \;\frac{8 x-2 x}{x} d x>I>int_{\pi / 4}^{\pi / 3} \;\frac{8 \sin x-2 x}{x} d x$ $\;\frac{\pi}{2}>I>int_{\pi / 4}^{\pi / 3}(\;\frac{8 \sin x}{x}-2) d x$ $\;\frac{\sin x}{x}$ is decreasing in $(\;\frac{\pi}{4}, \;\frac{\pi}{3})$ so it attains maximum at $x=\;\frac{\pi}{4}$ $I>int_{\pi / 4}^{\pi / 3}(\;\frac{8 \sin x / 3}{x / 3}-2) d x$ $I>\sqrt{3}-\;\frac{\pi}{6}$

Q69JEE Main 2022MCQ4MArea Under The Curves

The area of the smaller region enclosed by the curves $y^2=8 x+4$ and $x^{2+}y^{2+}4 \sqrt{3} x-4=0$ is equal to [27-Jul-2022-Shift-1]

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📖 Explanation

$\cos \theta=\;\frac{2 \sqrt{3}}{4}=\;\frac{\sqrt{3}}{2} \Rightarrow \theta=30^{\circ}$ Area of the required region $=\;\frac{2}{3}(4 \times \;\frac{1}{2})+4^2 \times \;\frac{\pi}{6}-\;\frac{1}{2} \times 4 \times 2 \sqrt{3}$ $=\;\frac{4}{3}+\;\frac{8 \pi}{3}-4 \sqrt{3}=\;\frac{1}{3}\{4-12 \sqrt{3}+8 \pi\}$

Q70JEE Main 2022MCQ4MDifferential Equations

Let $y=y_1(x)$ and $y=y_2(x)$ be two distinct solutions of the differential equation $\;\frac{d y}{d x}=x+y$ , with $y_1(0)=0$ and $y_2(0)=1$ respectively. Then, the number of points of intersection of $y=y_1(x)$ and $y=y_2(x)$ is [27-Jul-2022-Shift-1]

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📖 Explanation

$\;\frac{d y}{d x}=x+y$ Let $x+y=t$ $1+\;\frac{d y}{d x}=\;\frac{d t}{d x}$ $\;\frac{d t}{d x}-1=t \Rightarrow \int \;\frac{d t}{t+1}=\int d x$ $\ln |t+1|=x+C^{'}$ $|t+1|=C e^x$ $|x+y+1|=C e^x$ For $y_1(x), y_1(0)=0 \Rightarrow C=1$ For $y_2(x), y_2(0)=1 \Rightarrow C=2$ $y_1(x)$ is given by $|x+y+1|=e^x$ $y_2(x)$ is given by $|x+y+1|=2 e^x$ At point of intersection $e^x=2 e^x$ No solutionSo, there is no point of intersection of $y_1(x)$ and $y_2(x)$ .

Q71JEE Main 2022MCQ4MParabola

Let $P(a, b)$ be a point on the parabola $y^2=8 x$ such that the tangent at $P$ passes through the centre of the circle $x^{2+}y^{2-}10 x-14 y+65=0$ . Let $A$ be the product of all possible values of $a$ and $B$ be the product of all possible values of $b$ . Then the value of $A+B$ is equal to [27-Jul-2022-Shift-1]

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📖 Explanation

Centre of circle $x^{2+}y^{2-}10 x-14 y+65=0$ is at $(5,7)$ Let the equation of tangent to $y^2=8 x$ is $y t=x+2 t^2$ which passes through $(5,7)$ $7 t=5+2 t^2$ $\Rightarrow 2 t^{2-}7 t+5=0$ $t=1, \;\frac{5}{2}$ $A=2 \times 1^2 \times 2 \times (\;\frac{5}{2})^2=25$ $B=2 \times 2 \times 1 \times 2 \times 2 \times \;\frac{5}{2}=40$ $A+B=65$

Q72JEE Main 2022MCQ4MVector Algebra

Let $\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$ and $\vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$ be two vectors, such that $\vec{a} \times \vec{b}=-\hat{i}+9 \hat{i}+12 \hat{k}$ . Then the projection of $\vec{b}-2 \vec{a}$ on $\vec{b}+\vec{a}$ is equal to [27-Jul-2022-Shift-1]

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📖 Explanation

$\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}, \vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$ $\vec{a} \times \vec{b}=-\hat{i}+9 \hat{j}+12 \hat{k}$ $\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ \alpha & 1 & \beta \\ 3 & -5 & 4\end{vmatrix}=-\hat{i}+9 \hat{j}+12 \hat{k}$ $4+5 \beta=-1 \Rightarrow \beta=-1$ $-5 \alpha-3=12 \Rightarrow \alpha=-3$ $\vec{b}-2 \vec{a}=3 \hat{i}-5 \hat{j}+4 \hat{k}-2(-3 \hat{i}+\hat{j}-\hat{k})$ $\vec{b}-2 \vec{a}=9 \hat{i}-7 \hat{j}+6 \hat{k}$ $\vec{b}+\vec{a}=(3 \hat{i}-5 \hat{j}+4 \hat{k})+(-3 \hat{i}+\hat{j}-\hat{k})$ $\vec{b}+\vec{a}=-4 \hat{j}+3 \hat{k}$ Projection of $\vec{b}-2 \vec{a}$ on $\vec{b}+\vec{a}$ is $=\;\frac{(\vec{b}-2 \vec{a}) \cdot (\vec{b}+\vec{a})}{|\vec{b}+\vec{a}|}$ $=\;\frac{28+18}{5}=\;\frac{46}{5}$

Q73JEE Main 2022MCQ4MVector Algebra

Let $\vec{a}=2 \hat{i}-\hat{j}+5 \hat{k}$ and $\vec{b}=\alpha \hat{i}+\beta \hat{j}+2 \hat{k}$ . If $((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k}=\;\frac{23}{2}$ , then $|\vec{b} \times 2 \hat{j}|$ is equal to [27-Jul-2022-Shift-1]

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📖 Explanation

Given, $\vec{a}=2 \hat{i}-\hat{j}+5 \hat{k}$ and $\vec{b}=\alpha \hat{i}+\beta \hat{j}+2 \hat{k}$ Also, $((\vec{a} \times \vec{b}) \times i) \cdot \hat{k}=\;\frac{23}{2}$ $\Rightarrow ((\vec{a} \cdot \hat{i}) \vec{b}-(\vec{b} \cdot \hat{i}) \cdot \overline{a}) \cdot \hat{k}=\;\frac{23}{2}$ $\Rightarrow (2 \cdot \vec{b}-\alpha \cdot \vec{a}) \cdot \hat{k}=\;\frac{23}{2}$ $\Rightarrow 2 \cdot 2-5 \alpha=\;\frac{23}{2} \Rightarrow \alpha=\;\frac{-3}{2}$ $\;\text{ Now, }\;|\vec{b} \times 2 j|=|(\alpha \hat{i}+\beta \hat{j}+2 \hat{k}) \times 2 \hat{j}|$ $=|2 \alpha \hat{k}+0-4 \hat{i}|$ $=\sqrt{4 \alpha^{2+}16}$ $=\sqrt{4(\;\frac{-3}{2})^{2+}16}=5$

Q74JEE Main 2022MCQ4MProbability

Let $S$ be the sample space of all five digit numbers. It $p$ is the probability that a randomly selected number from $S$ , is a multiple of 7 but not divisible by 5 , then $9 p$ is equal to [27-Jul-2022-Shift-1]

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📖 Explanation

Among the 5 digit numbers,First number divisible by 7 is 10003 and last is 99995 . $\Rightarrow$ Number of numbers divisible by 7 . $=\;\frac{99995-10003}{7}+1$ $=12857$ First number divisible by 35 is 10010 and last is 99995. $\Rightarrow$ Number of numbers divisible by 35 $=\;\frac{99995-10010}{35}+1$ $=2572$ Hence number of number divisible by 7 but not by 5 $=12857-2572$ $=10285$ $9 P .=\;\frac{10285}{90000} \times 9$ $=1.0285$

Q75JEE Main 2022MCQ4MTrigonometric Ratios and Identities

Let a vertical tower $A B$ of height $2 h$ stands on a horizontal ground. Let from a point $P$ on the ground a man can see upto height $h$ of the tower with an angle of elevation $2 \alpha$ . When from $P$ , he moves a distance $d$ in the direction of $\vec{A P}$ , he can see the top $B$ of the tower with an angle of elevation $\alpha$ . If $d=\sqrt{7} h$ , then $\tan \alpha$ is equal to [27-Jul-2022-Shift-1]

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📖 Explanation

$\triangle A P M$ gives $\tan 2 \alpha=\;\frac{h}{x} . . . .$ . (i) $\triangle A Q B$ gives $\tan 2 \alpha=\;\frac{2 h}{x+d}=\;\frac{2 h}{x+h \sqrt{7}} . . . . . .$ (ii)From (i) and (ii) $\tan 2 \alpha=\;\frac{2 \cdot \tan 2 \alpha}{1+\sqrt{7} \cdot \tan 2 \alpha}$ Let $t=\tan \alpha$ $\Rightarrow t=\;\frac{2 \frac{2 t}{1-t^2}}{1+\sqrt{7} \cdot \;\frac{2 t}{1-t^2}}$ $\Rightarrow t^{2-}2 \sqrt{7} t+3=0$ $t=\sqrt{7}-2$

Q76JEE Main 2022MCQ4MSets and Relations

$(p ∧ r) ⇔(p ∧(∼ q))$ is equivalent to $(∼ p)$ when $r$ is [27-Jul-2022-Shift-1]

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📖 Explanation

p q ∼p ∼q p\wedge q p\wedge ∼q p\wedge q ⇔ p\wedge ∼q T T F F T F F T F F T F T F F T T F F F T F F T T F F T Clearly ${\text{p}} ∧ {\text{q}} ⇔ {\text{p}} ∧ ∼ {\text{q}} \equiv ∼ {\text{p}}$ $\therefore r=q$

Q77JEE Main 2022MCQ4MThree Dimensional Geometry

If the plane $P$ passes through the intersection of two mutually perpendicular planes $2 x+k y-5 z=1$ and $3 k x-k y+z=5, k < 3$ and intercepts a unit length on positive $x$ -axis, then the intercept made by the plane $P$ on the $y$ -axis is [27-Jul-2022-Shift-1]

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📖 Explanation

$P_1: 2 x+k y-5 z=1$ $P_2: 3 k x-k y+z=5$ $\because P_1 ⟂ P_2 \Rightarrow 6 k-k^{2+}5=0$ $\Rightarrow k=1,5$ $\because k< 3$ $\therefore k=1$ $P_1: 2 x+y-5 z=1$ $P_2: 3 x-y+z=5$ $P:(2 x+y-5 z-1)+\lambda (3 x-y+z-5)=0$ Positive $x$ -axis intercept $=1$ $\Rightarrow \;\frac{1+5 \lambda }{2+3 \lambda }=1$ $\Rightarrow \lambda =\;\frac{1}{2}$ $\therefore P: 7 x+y-4 z=7$ ${\text{y}}$ intercept $=7$

Q78JEE Main 2022MCQ4MArea Under The Curves

Let $A(1,1), B(-4,3), C(-2,-5)$ be vertices of a triangle $A B C, P$ be a point on side $B C$ , and $\Delta _1$ and $\Delta _2$ be the areas of triangles $A P B$ and $A B C$ , respectively. If $\Delta _1: \Delta _2=4: 7$ , then the area enclosed by the lines $A P, A C$ and the $x$ -axis is [27-Jul-2022-Shift-1]

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📖 Explanation

$\;\frac{\Delta _1}{\Delta _2}=\;\frac{\;\frac{1}{2} \times B P \times A H}{\frac{1}{2} \times B C \times A H}=\;\frac{4}{7}$ $P(\;\frac{-20}{7}, \;\frac{-11}{7})$ Line $A C: y-1=2(x-1)$ Intersection with $x$ -axis $=(\;\frac{1}{2}, 0)$ Line $A P: y-1=\;\frac{2}{3}(x-1)$ Intersection with $x$ -axis $(\;\frac{-1}{2}, 0)$ Vertices are $(1,1),(\;\frac{1}{2}, 0)$ and $(\;\frac{-1}{2}, 0)$ Area $=\;\frac{1}{2}$ sq. unit

Q79JEE Main 2022MCQ4MCircles

If the circle $x^{2+}y^{2-}2 g x+6 y-19 c=0, g, c \in ℝ$ passes through the point $(6,1)$ and its centre lies on the line $x-2 c y=8$ , then the length of intercept made by the circle on $x$ -axis is [27-Jul-2022-Shift-1]

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📖 Explanation

Circle $: x^{2+}y^{2-}2 g x+6 y-19 c=0$ It passes through $h(6,1)$ $\Rightarrow 36+1-12 g+6-19 c=0$ $=12 g+19 c=43 . . . . .(1)$ Line $x-2 c y=8$ passes through centre $\Rightarrow g+6 c=8 . . . . . . \;\text{ (2) }\;$ From (1) & (2) $g=2, c=1$ $C: x^{2+}y^{2-}4 x+6 y-19=0$ xint $=2 \sqrt{g^{2-}C}$ $=2 \sqrt{4+19}$ $=2 \sqrt{23}$

Q80JEE Main 2022MCQ4MDefinite Integrals

Let a function $f: ℝ \to ℝ$ be defined as : $f(x)=\begin{cases}int_{0}^{x}(5-{|t-3|}) d t & {, \ \\ } x>4 \\ x^{2+}b x {, \ \\ } , x \le 4$ where ${\text{b}} \in ℝ$ \end{cases} If $f$ is continuous at $x=4$ , then which of the following statements is NOT true? [27-Jul-2022-Shift-1]

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📖 Explanation

$\because f(x)$ is continuous at $x=4$ $\Rightarrow f(4^{-})=f(4^{+})$ $\Rightarrow 16+4 b=int_0^4(5-|t-3|) d t$ $=int_0^3(2+t) d t+int_3^4(8-t) d t$ $..=2 t+\;\frac{t^2}{2})_0^{3+}8 t-\;\frac{t^2}{3}]_3^4$ $=6+\;\frac{9}{2}-0+(32-8)-(24-\;\frac{9}{2})$ $16+4 b=15$ $\Rightarrow b=\;\frac{-1}{4}$ $\Rightarrow f(x)=\begin{cases}int_0^x{ 5} -\text{|t-3|} d t & x>4 \\ x^{2-}\\ \\ \frac{x}{4} & x \le 4\end{cases}$ $\Rightarrow f^{'}(x)=\begin{cases}5-\text{|x-3|} & x>4 \\ 2 x-\\ \\ \frac{1}{4} & x \le 4\end{cases}$ $\Rightarrow f^{'}(x)=\begin{cases}8-x & x>4 \\ 2 x-\\ \\ \frac{1}{4} & x \le 4\end{cases}$ $f^{'}(x)< 0=x \in (-\infty , \;\frac{1}{8}) ∪(8, \infty )$ $f^{'}(3)+f^{'}(5)=6-\;\frac{1}{4}=\;\frac{35}{4}$ $f^{'}(x)=0 \Rightarrow x=\;\frac{1}{8}$ have local minima

Q81JEE Main 2022NAT4MInverse Trigonometric Functions

For $k \in ℝ$ , let the solutions of the equation $\cos (\sin ^{-1}(x \cot (\tan ^{-1}(\cos (\sin ^{-1} x)))))=k, 0 < |{}/{}x| < \;\frac{1}{\sqrt{2}}$ be $\alpha$ and $\beta$ , where the inverse trigonometric functions take only principal values. If the solutions of the equation $x^{2-}b x-5=0$ are $\;\frac{1}{\alpha^2}+\;\frac{1}{\beta^2}$ and $\;\frac{\alpha}{\beta}$ , then $\;\frac{b}{k^2}$ is equal to ________. [27-Jul-2022-Shift-1]

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📖 Explanation

$\cos (\sin ^{-1}(x \cot (\tan ^{-1}(\cos (\sin ^{-1}))))=k.$ $\Rightarrow \cos (\sin ^{-1}(x \cot (\tan ^{-1} \sqrt{1-x^2})))=k$ $\Rightarrow \cos (\sin ^{-1}(\;\frac{x}{\sqrt{1-x^2}}))=k$ $\Rightarrow \;\frac{\sqrt{1-2 x^2}}{\sqrt{1-x^2}}=k$ $\Rightarrow \;\frac{1-2 x^2}{1-x^2}=k^2$ $\Rightarrow 1-2 x^2=k^{2-}k^2 x^2$ $\therefore x^{2-}(\;\frac{k^{2-}1}{k^{2-}2})=0\;\;^{╱^{\;\;\;\alpha}}_{╲{\;\;_\beta}}$ $\;\frac{1}{\alpha^2}+\;\frac{1}{\beta^2}=2(\;\frac{k^{2-}2}{k^{2-}1}) . . . . . .(1)$ and $\;\frac{\alpha}{\beta}=-1 . . . . . .$ (2) $\therefore 2(\;\frac{k^{2-}2}{k^{2-}1})(-1)=-5$ $\Rightarrow k^2=\;\frac{1}{3}$ and $b=S \cdot R=2(\;\frac{k^{2-}2}{k^{2-}1})-1=4$ $\therefore \;\frac{b}{k^2}=\;\frac{4}{\frac{1}{3}}=12$

Q82JEE Main 2022NAT4MStatistics

The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is ________. [27-Jul-2022-Shift-1]

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📖 Explanation

$\;\text{ Given }\; \;\frac{sum_{i=1}^{10} x_i}{10}=15 . . . . .$ $\Rightarrow sum_{i=1}^{10} x_i=150$ $\;\text{ and }\; \;\frac{sum_{i=1}^{10} x_i^2}{10}-15^2=15$ $\Rightarrow sum_{i=1}^{10} x_i^2=2400$ Replacing 25 by 15 we get $sum_{i=1}^9 x_i+25=150$ $\Rightarrow sum_{i=1}^9 x_i=125$ $\therefore$ Correct mean $=\;\frac{sum_{i=1}^9 x_i+15}{10}=\;\frac{125+15}{10}=14$ Similarly, $sum_{i=1}^2 x_i^2=2400-25^2=1775$ $\therefore$ Correct variance $=\;\frac{sum_{i=1}^9 x_i^{2+}15^2}{10}-14^2$ $=\;\frac{1775+225}{10}-14^2=4$ $\therefore$ Correct $S . D=\sqrt{4}=2$ .

Q83JEE Main 2022NAT4MThree Dimensional Geometry

Let the line $\;\frac{x-3}{7}=\;\frac{y-2}{-1}=\;\frac{z-3}{-4}$ intersect the plane containing the lines $\;\frac{x-4}{1}=\;\frac{y+1}{-2}=\;\frac{z}{1}$ and $4 a x-y+5 z-7 a=0=2 x-5 y-z-3, a \in ℝ$ at the point $P(\alpha, \beta, \gamma )$ . Then the value of $\alpha+\beta+\gamma$ equals ________. [27-Jul-2022-Shift-1]

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📖 Explanation

Equation of plane containing the line $4 a x-y+5 z-7 a=0=2 x-5 y-z-3$ can be written as $4 a x-y+5 z-7 a+\lambda (2 x-5 y-z-3)=0$ $(4 a+2 \lambda ) x-(1+5 \lambda ) y+(5-\lambda ) z-(7 z+3 \lambda )=0$ Which is coplanar with the line $\;\frac{x-4}{1}=\;\frac{y+1}{-2}=\;\frac{z}{1}$ $4(4 a+2 \lambda )+(1+5 \lambda )-(7 a+3 \lambda )=0$ $9 a+10 \lambda +1=0 . . . . .(1)$ $(4 a+2 \lambda ) 1+(1+5 \lambda ) 2+5-\lambda =0$ $4 a+11 \lambda +7=0 . . . . .(2)$ $a=1, \lambda =-1$ Equation of plane is $x+2 y+3 z-2=0$ Intersection with the line $\;\frac{x-3}{7}=\;\frac{y-2}{-1}=\;\frac{z-3}{-4}$ $(7 t+3)+2(-t+2)+3(-4 t+3)-2=0$ $-7 t+14=0$ $t=2$ So, the required point is $(17,0,-5)$ $\alpha+\beta+\gamma =12$

Q84JEE Main 2022NAT4MEllipse

An ellipse $E: \;\frac{x^2}{a^2}+\;\frac{y^2}{b^2}=1$ passes through the vertices of the hyperbola $H: \;\frac{x^2}{49}-\;\frac{y^2}{64}=-1$ . Let the major and minor axes of the ellipse $E$ coincide with the transverse and conjugate axes of the hyperbola $H$ , respectively. Let the product of the eccentricities of $E$ and $H$ be $\;\frac{1}{2}$ . If $l$ is the length of the latus rectum of the ellipse $E$ , then the value of $113 l$ is equal to ________. [27-Jul-2022-Shift-1]

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📖 Explanation

Vertices of hyperbola $=(0, \pm 8)$ As ellipse pass through it i.e., $0+\;\frac{64}{b^2}=1 \Rightarrow b^2=64 . . . . . . \;\text{ (1) }\;$ As major axis of ellipse coincide with transverse axis of hyperbola we have $b>a$ i.e. $e_E=\sqrt{1-\;\frac{a^2}{64}}=\;\frac{\sqrt{64-a^2}}{8}$ and $e_H=\sqrt{1+\;\frac{49}{64}}=\;\frac{\sqrt{113}}{8}$ $\therefore e_E \cdot e_H=\;\frac{1}{2}=\;\frac{\sqrt{64-a^2} \sqrt{113}}{64}$ $\Rightarrow (64-a^2)(113)=32^2$ $\Rightarrow a^2=64-\;\frac{1024}{113}$ L.R of ellipse $=\;\frac{2 a^2}{b}=\;\frac{2}{8}(\;\frac{113 \times 64-1024}{113})$ $=l=\;\frac{1552}{113}$ $\therefore 113 l=1552$

Q85JEE Main 2022NAT4MDifferential Equations

Let $y=y(x)$ be the solution curve of the differential equation $\sin (2 x^2) \log _e(\tan x^2) d y+(4 x y-4 \sqrt{2} x \sin (x^{2-}\;\frac{\pi}{4})) d x=0,0 < x < \sqrt{\;\frac{\pi}{2}}$ , which passes through the point $(\sqrt{\;\frac{\pi}{6}}, 1)$ . Then $|y(\sqrt{\;\frac{\pi}{3}})|$ is equal to ________. [27-Jul-2022-Shift-1]

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📖 Explanation

$\;\frac{d y}{d x}+y(\;\frac{4 x}{\sin (2 x^2) \ln (\tan x^2)})=\;\frac{4 \sqrt{2} x \sin (x^{2-}\frac{\pi}{4})}{\sin (2 x^2) \ln (\tan x^2)}$ $I . F=e^{\int \;\frac{4 x}{\sin (2 x^2) \ln (\tan x^2)} d x}$ $=e^{\ln \mid \ln (\tan x^2)}=\ln (\tan x^2)$ $\therefore \int d(y \cdot \ln (\tan x^2))=\int \;\frac{4 \sqrt{2} x \sin (x^{2-}\frac{\pi}{4})}{\sin (2 x^2)} d x$ $\Rightarrow y \ln (\tan x^2)=\ln |\;{\sec x^{2+}\tan x^2}/{{\text{cosec}} x^{2-}\cot x^2}|+C$ $\ln (\;\frac{1}{\sqrt{3}})=\ln (\;{\;\frac{3}{\sqrt{3}}}/{2-\sqrt{3}})+C$ $e=\ln (\;\frac{1}{\sqrt{3}})-\ln (\;\frac{\sqrt{3}}{2-\sqrt{3}})$ For $y(\sqrt{\;\frac{\pi}{3}})$ $y \ln (\sqrt{3})=\ln |\;{2+\sqrt{3}}/{\frac{2}{\sqrt{3}}+\;\frac{1}{\sqrt{3}}}|+\ln (\;\frac{1}{\sqrt{3}})-\ln (\;\frac{\sqrt{3}}{2 \sqrt{3}})$ $=\ln (2+\sqrt{3})+\ln (\;\frac{1}{\sqrt{3}})+\ln (\;\frac{1}{\sqrt{3}})-\ln (\;\frac{\sqrt{3}}{2-\sqrt{3}})$ $\Rightarrow y \ln \sqrt{3}=\ln (\;\frac{1}{\sqrt{3}})$ $\Rightarrow \;\frac{y}{2} \ln 3=-\;\frac{1}{2} \ln 3$ $\Rightarrow y=1$ $\therefore |y(\sqrt{\;\frac{\pi}{3}})|=1$

Q86JEE Main 2022NAT4MApplication of Derivatives

Let $M$ and $N$ be the number of points on the curve $y^{5-}9 x y+2 x=0$ , where the tangents to the curve are parallel to $x$ -axis and $y$ -axis, respectively. Then the value of $M+N$ equals ________. [27-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Here equation of curve is $y^{5-}9 x y+2 x=0 \;\text{..... (i) }\;$ On differentiating: $5 y^4 \;\frac{d y}{d x}-9 y-9 x \;\frac{d y}{d x}+2=0$ $\therefore \;\frac{d y}{d x}=\;\frac{9 y-2}{5 y^{4-}9 x}$ When tangents are parallel to $x$ -axis then $9 y-2=0$ $\therefore M=1 \;\text{. }\;$ For tangent perpendicular to ${\text{x}}$ -axis $5 y^{4-}9 x=0 . . . . . . \;\text{ (ii) }\;$ From equation (i) and (ii) we get only one point. $\therefore N=1 \;\text{. }\;$ $\therefore M+N=2$

Q87JEE Main 2022NAT4MQuadratic Equation and Inequalities

Let $f(x)=2 x^{2-}x-1$ and $S=\{n \in ℤ:| f(n)| \le 800\}$ . Then, the value of $sum_{n \in S} f(n)$ is equal to _______. [27-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\because |f(n)| \le 800$ $\Rightarrow -800 \le 2 n^{2-}n-1 \le 800$ $\Rightarrow 2 n^{2-}n-801 \le 0$ $\therefore n \in [\;\frac{-\sqrt{6409+1}}{4}, \;\frac{\sqrt{6409+1}}{4}]$ and $n \in z$ $\therefore n=-19,-18,-17, . . . . . . . . ., 19,20$ . $\therefore \sum(2 x^{2-}x-1)=2 \sum x^{2-}\sum x-\sum 1$ . $=2.2 \cdot (1^{2+}2^{2+}. . .+19^2)+2.20^{2-}20-40$ $=10620$

Q88JEE Main 2022NAT4MMatrices and Determinants

Let $S$ be the set containing all $3 \times 3$ matrices with entries from $\{-1,0,1\}$ . The total number of matrices $A \in S$ such that the sum of all the diagonal elements of $A^{ {\text{T}}} A$ is 6 is ________. [27-Jul-2022-Shift-1]

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📖 Explanation

Sum of all diagonal elements is equal to sum of square of each element of the matrix.i.e., $A=\begin{bmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{bmatrix}$ then $t_r(A \cdot A^T)$ $=a_1^{2+}a_2^{2+}a_3^{2+}b_1^{2+}b_2^{2+}b_3^{2+}c_1^{2+}c_2^{2+}c_3^2$ $\because a_i, b_i, c_i \in \{-1,0,1\}$ for $i=1,2,3$ $\therefore$ Exactly three of them are zero and rest are 1 or $-1$ .Total number of possible matrices ${ }^9 C_3 \times 2^6$ $=\;\frac{9 \times 8 \times 7}{6} \times 64$ $=5376$

Q89JEE Main 2022NAT4MEllipse

If the length of the latus rectum of the ellipse $x^{2+}4 y^{2+}2 x+8 y-\lambda =0$ is 4 , and $l$ is the length of its major axis, then $\lambda +l$ is equal to ________. [27-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Equation of ellipse is : $x^{2+}4 y^{2+}2 x+8 y-\lambda =0$ $(x+1)^{2+}4(y+1)^2=\lambda +5$ $\;\frac{(x+1)^2}{\lambda +5}+\;\frac{(y+1)^2}{(\;\frac{\lambda +5}{4})}=1$ Length of latus rectum $=\;\frac{2 \cdot (\frac{\lambda +5}{4})}{\sqrt{\lambda +5}}=4$ . $\therefore \lambda =59$ Length of major axis $=2 \cdot \sqrt{\lambda +5}=16=l$ $\therefore \lambda +l=75 \;\text{. }\;$

Q90JEE Main 2022NAT4MComplex Numbers

Let $S=\{z \in C: z^{2+}\overline{z}=0\}$ . Then $sum_{z \in S}({\text{Re}}(z)+{\text{Im}}(z))$ is equal to ________. [27-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\because z^{2+}\overline{z}=0$ Let $z=x+i y$ $\therefore x^{2-}y^{2+}2 i x y+x-i y=0$ $(x^{2-}y^{2+}x)+i(2 x y-y)=0$ $\therefore x^{2+}y^2=0$ and $(2 x-1) y=0$ if $x=+\;\frac{1}{2}$ then $y=\pm \;\frac{\sqrt{3}}{2}$ And if $y=0$ then $x=0,-1$ $\therefore z=0+0 i,-1+0 i, \;\frac{1}{2}+\;\frac{\sqrt{3}}{2} i, \;\frac{1}{2}-\;\frac{\sqrt{3}}{2} i$ $\therefore \sum(R_e(z)+m(z))=0$

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