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JEE MAIN 2020 - 05 SEP S1 2020

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CHEMISTRY25 questions

Q1JEE Main 2020MCQ4MRedox Reactions

The equation that represents the water-gas shift reaction is:

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📖 Explanation

(1) Water gas shift reaction ${CO}( {g})+ {H}_{2} {O}( {g})\ {───── ▸ }^{673K}_{\text{Catalyst}}$ ${CO}_{2}( { g})+ {H}_{2}( { g})$ (2) Water gas is produced by this reaction. ${CH}_4( {g})+ {H}_{2} {O}( {g})\ {───── ▸ }^{1270K}_{ \text{Ni}}$ ${CO} ( { g})+3 {H}_{2}( { g})$ (3) Water gas is produced by this reaction ${C } ( {s})+ {H}_{2} {O}( {g})\ {───── ▸ }^{1270K}$ ${CO} ( { g})+ {H}_{2}( { g})$ (4) producer gas is produced by this reaction. $2 {C } ( {s})+ {O} _{2} ( {g})+4N_2(g)\ {───── ▸ }^{1273K}$ $2 {CO} ( { g})+ 4{N}_{2}( { g})$

Q2JEE Main 2020MCQ4MChemical Equilibrium

Consider the following reaction ${N}_{2} {O}_{4}( { g}) ⇌ 2NO_2(g);\Delta{H}^{0}=+58 {kJ}$ For each of the following cases $(a, b),$ the direction in which the equilibrium shifts is: (a) Temperature is decreased (b) Pressure is increased by adding $N_2$ at constant T

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📖 Explanation

$\Delta{H}°>0 \;\;{T} \downarrow$ equation shifts back ward. ${N}_{2}$ is treated as inert gas in this case hence no effect on equilibrium.

Q3JEE Main 2020MCQ4MCoordination Compounds

The values of the crystal field stabilization energies for a high spin $d ^{6}$ metal ion in octahedral and tetrahedral fields, respectively, are :

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📖 Explanation

For high spin octahedral field ${CFSE}=(4) (-0.4 \Delta_{0} )+2 (0.6 \Delta_{0} )$ $=-0.4 \Delta_{0}$ For high spin tetrahedral field CFSE $=3 (-0.6 \Delta_{t} )+3 (0.4 \Delta_{t} )$ $=-0.6 \Delta_{t}$

Q4JEE Main 2020MCQ4MBiomolecules

Which of the following is not an essential amino acid:

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Q5JEE Main 2020MCQ4MAldehydes, Ketones and Carboxylic Acids

In the following reaction sequence the major products A and B are :

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Q6JEE Main 2020MCQ4MAldehydes, Ketones and Carboxylic Acids

The increasing order of the acidity of the α-hydrogen of the following compounds is :

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Q7JEE Main 2020MCQ4MChemical Thermodynamics

An Ellingham diagram provides information about:

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Q8JEE Main 2020MCQ4MAldehydes, Ketones and Carboxylic Acids

Which of the following derivatives of alcohols is unstable in an aqueous base ?

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Q9JEE Main 2020MCQ4Mp Block Elements

The structure of ${PCl}_{5}$ in the solid state is

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📖 Explanation

${PCl}_{5({ s})}$ exist as $[{PCl}_{4} ]^{+}$ and $[{PCl}_{6} ]$

Q10JEE Main 2020MCQ4MAmines

The most appropriate reagent for conversion of ${C}_{2} {H}_{5} {CN}$ into ${CH}_{3} {CH}_{2} {CH}_{2} {NH}_{2}$ is

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📖 Explanation

${CH}_{3}- {CH}_{2}-{C}\equiv {N}{─── ▸}^{?}$ ${CH}_{3}- {CH}_{2}- {CH}_{2}- {NH}_{2}$ ${CH}_{3}- {CH}_{2}-{C}\equiv {N}{─── ▸}^{ \text{LiAlH}_4}$ ${CH}_{3}- {CH}_{2}- {CH}_{2}- {NH}_{2}$

Q11JEE Main 2020MCQ4MStructure of Atom

The difference between the radii of $3 ^{\text{rd}}$ and $4^{ {\text{th} }}$ orbits of ${Li}^{2+}$ is $\Delta{R}_{1}$ . The difference between theradii of $3^{\text{rd}}$ and $4^{ {\text{th} }}$ orbits of ${He}^{+}$ is $\Delta{R}_{2}$ . Ratio $\Delta{R}_{1}: \Delta{R}_{2}$ is :

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📖 Explanation

$\frac{\Delta{R}_{1}}{\Delta{R}_{2}}={({r}_{4}- {r}_{3} )_{4^{2+}}}/{({r}_{4}- {r}_{3} )_{ {He}^{+}}}$ $={\frac{4^{2}}{3}- \frac{3^{2}}{3}}/{ \frac{4^{2}}{2}- \frac{3}{2}^{2}}= \frac{7 / 3}{7 / 2}= \frac{2}{3}$

Q12JEE Main 2020MCQ4MChemical Kinetics

A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives for A and ${B}$ are $300 { s}$ and $180 { s}$ , respectively. If the concentrations of ${A}$ and ${B}$ are equal initially, the time required for the concentration of A to be four times that of ${B}($ in ${s}):($ Use $\ln 2=0.693)$

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📖 Explanation

$[ {A}]_{ {t}}=4[ {B}]_{ {t}}$ $[ {A}]_{0} {e}^{-({ln}^{2} / 300 )^{1}}$ $=4[ {B}]_{0} {e}^{(-\ln 2 / 180) {t}}$ ${e}^{(\frac{\ln ^{2}}{180} \frac{\ln ^{2}}{300} )}=4$ $(\frac{\ln ^{2}}{180}- \frac{\ln ^{2}}{300} ) {t}=\ln 4$ $(\frac{1}{180}- \frac{1}{300}){t}=2$ $\Rightarrow {t}= \frac{2 \times 180 \\times 300}{120}=900 {\sec}$

Q13JEE Main 2020MCQ4MAmines

The increasing order of basicity of the following compounds is

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Q14JEE Main 2020MCQ4MSolutions

Identify the correct molecular picture showing that happens at the critical micellar concentration (CMC) of an aqueous solution of a surfactant.

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Q15JEE Main 2020MCQ4MBiomolecules

If a person is suffering from the deficiency of nor- adrenaline, what kind of drug can be suggested ?

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📖 Explanation

Anti depressant $\to$ drug which enhance the mood. Non adrenaline is neurotransmitter and its level is low in body due to some reason then person suffers from depression and in that situation anti depressant drug is required.

Q16JEE Main 2020MCQ4Md and f Block Elements

The correct electronic configuration and spinonly magnetic moment (BM) of Gd $^{3+}$ (Z = 64), respectively, are

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📖 Explanation

Electronic configuration of Gd $^{3+}$ is ${}_64Gd^{3+} = [Xe]4f^7$ [Xe] ↿ ↿ ↿ ↿ ↿ ↿ ↿ ${Gd}^{3+}$ having 7 unpaired electrons. Magnetic moment $(\mu)=\sqrt{ {n}( {n}+2)} {B} .{M}$ $\mu=\\sqrt{7(7+2)}$ B.M. = 7.9 B.M. n⇒ Number of unpaired electrons.

Q17JEE Main 2020MCQ4MSolutions

The condition that indicates a polluted environment is

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Q18JEE Main 2020MCQ4MStructure of Atom

In the sixth period, the orbitals that are filled are

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📖 Explanation

As per $( {n}+l)$ rule in $6^{ {\text{th} }}$ period, order of orbitals filling is $6 { s}, 4 {f}, 5 {d}, 6 {p} .$

Q19JEE Main 2020MCQ4MChemical Bonding and Molecular Structure

The potential energy curve for the ${H}_{2}$ molecule as a function of internuclear distance is :

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📖 Explanation

Potential energy curve for $H_2$ molecule is.

Q20JEE Main 2020MCQ4MSolid State

A diatomic molecule ${X}_{2}$ has a body-centred cubic (bec) structure with a cell edge of $300 {pm}$ . The density of the molecule is $6.17 { g} { cm}^{-3}$ . The number of molecules present in $200 { g}$ of ${X}_{2}$ is (Avogadro constant $({N}_{ {A}} )=6 \\times 10^{23} { mol}^{-1} )$

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📖 Explanation

${p}= {2 \times {{M}}/{ {N}_{ {A}}}}/{ {a}^{3}}$ $\Rightarrow 6.17= {2 \times {{M}}/{ {N}_{ {A}}}}/{ (3 \\times 10^{-8}{cm} )^{3}}$ $\Rightarrow {M} \simeq 50 {gm} / {mol}$ ${No}=\frac{{W}}{ {M}} \times {N}_{ {A}}= \frac{200}{50}\times {N}_{ {A}}$ $=4 {N}_{ {A}}$

Q21JEE Main 2020NAT4MElectrochemistry

an oxidation-reduction reaction in which 3 electrons are transferred has a $\Delta{G} °$ of $17.37 \ {kJ}$ $\ {mol}^{-1}$ at $25°{C}$ . The value of ${E}_{\text{cell}}°($ in ${V})$ is ________. $\\times 10^{-2}$ $(1 {F}=96,500 {C}\ {mol}^{-1} )$

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📖 Explanation

$\Delta{G}°=- {AFE}°$ $=-3 \\times 96500 \\times {E}°$ $\Rightarrow {E}°=-6 \\times 10^{-2} \ {V}$

Q22JEE Main 2020NAT4MHydrocarbons

The minimum number of moles of ${O}_{2}$ required for complete combustion of 1 mole of propane and 2 moles of butane is ______.

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📖 Explanation

${{C}_{3} {H}_{8}}_{1 \text{mole}}+ {{SO}_{2}}_{5 \text{mole}} \to 3 {Co}_{2}$ $+4 {H}_{2} {O}$ For 1 mole propane combustion 5 mole $O_2$ required ${C}_{4} {H}_{10}+ \frac{13}{2} {O}_{2} \to 4 {CO}_{2}$ $+5 {H}_{2}{O}$ $\text{table} 1 \text{mole} ,6.5 \text{mole};2 \text{mole} ,13 \text{mole}$ For 2 moles of butane 13 mole of $O_2$ is required total moles = 13 + 5 = 18

Q23JEE Main 2020NAT4MCoordination Compounds

The total number of coordination sites in ethylenediaminetetraacetate (EDTA $^{4-}$ ) is _______.

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📖 Explanation

(EDTA $^{4-}$ ) is hexadentate ligand, so its donation sites are six.

Q24JEE Main 2020NAT4MBiomolecules

The number of chiral carbon(s) present in peptide, Ile-Arg-Pro, is ______.

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📖 Explanation

A chiral carbon is a carbon atom bonded to four different groups. To find the total number, we analyze each amino acid in the peptide Ile-Arg-Pro.

Isoleucine (Ile): It has two chiral carbons. One is the α-carbon, and the other is the β-carbon within its sec-butyl side chain (-CH(CH₃)(CH₂CH₃)).
Arginine (Arg): It has one chiral carbon, which is its α-carbon. Its side chain (-(CH₂)₃NHC(NH)NH₂) does not have any chiral centers.
Proline (Pro): It has one chiral carbon, its α-carbon, which is part of the cyclic structure.

Therefore, the total number of chiral carbons in the tripeptide is the sum from each residue: $2$ (from Ile) + $1$ (from Arg) + $1$ (from Pro) = $4$ .

Q25JEE Main 2020NAT4MIonic Equilibrium

A soft drink was bottled with a partial pressure of ${CO}_{2}$ of 3 bar over the liquid at room temperature. The partial pressure of ${CO}_{2}$ over the solution approaches a value of 30 bar when $44 { g}$ of ${CO}_{2}$ is dissolved in $1 \ { kg}$ of water at room temperature. The approximate ${pH}$ of the soft drink is ________ $\times 10^{-1}$ . (First dissociation constant of ${H}_{2} {CO}_{3}=4.0 \times 10^{-7}$ ; $\log 2=0.3 ;$ density of the soft drink $=1 { g} {mL}^{-1} )$

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📖 Explanation

${P}_{ {CO}_{2}}={K}_{ {H}}\times {CO}_{2}$ $\frac{3}{30}={{K}_{ {H}}.{n}_{ {CO}_{2}}}/{ { K}_{ {H}} 1}$ $\Rightarrow {n}_{ {CO}_{2}=0.1}{mol}$ ${pH}= \frac{1}{2}({pka}_{1}-\log {c} )$ $= \frac{1}{2}(6.4 \\times 1)=3.7$ $\ {pH}=37 \\times 10^{-1}$

MATHEMATICS25 questions

Q26JEE Main 2020MCQ4MSequences and Series

If $3^{{2} \sin 2 \alpha -1},14$ and $3^{{4-2} \sin 2 \alpha }$ are the first three terms of an A.P. for some $\alpha ,$ then the sixth termof this A.P. is :

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📖 Explanation

Given that $3^{4-\sin 2 \alpha }+3^{{2} \sin 2 \alpha -1}=28$ Let $3^{{2} \sin 2 \alpha }=t$ $\frac{81}{t}+\frac{t}{3}=28$ ${t}=81,3$ $3^{{2} \sin 2 \alpha }=3^{1}, 3^{4}$ $2 \sin 2 \alpha =1,4$ sin $2 \alpha = \frac{1}{2}, 2($ rejected $)$ First term $a=3^{{2} \sin 2 \alpha -1}$ $a = 1$ Second term = 14 $\therefore$ common difference ${d}=13$ ${T}_{6}= {a}+5 {d}$ ${T}_{6}=1+5 \\times 13$ ${T}_{6}=66$

Q27JEE Main 2020MCQ4MLimits, Continuity and Differentiability

If the function ${f}( {x})=\{ \text{table} k_1(x-\pi )^{2-}1{,}, {x} \le { \pi } ;k_2\text{cosx}{,},x >\pi$ is twice differentiable, then the ordered pair $({k}_{1},{ k}_{2} )$ is equal to :

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📖 Explanation

f(x) is continuous and differentiable ${f} (\pi ^{-} )= {f}(\pi )= {f} (\pi ^{+}\ )$ $-1=- {k}_{2}$ $[ {k}_{2}=1]$ ${f}( {x})=\{ \text{table} 2k_1(x-\pi )-1 ,{;} , {x} \le { \pi }; -k_2\text{cosx},{;},x >\pi$ $f'(\pi ^{-} )=f'(\pi ^{+} )$ $0=0$ so, differentiable at $x=0$ $f"(x)=\{\text{table} 2k_1,{;},x\le \pi ;-k_2\cos x,{;},x>\pi$ $f"(\pi ^{-})=f"(\pi ^{+})$ $2 {k}_{1}= {k}_{2}$ $[{k}_{1}=\ \frac{1}{2}]$ $({k}_{1}, {k}_{2} )=(\frac{1}{2}, 1 )$

Q28JEE Main 2020MCQ4MParabola

If the common tangent to the parabolas, ${y}^{2}=4 {x}$ and ${x}^{2}=4 {y}$ also touches the circle ${x}^{2}+ {y}^{2}= {c}^{2},$ then ${c}$ is equal to :

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📖 Explanation

${y}= {mx}+ \frac{1}{ { m}}$ (tangent at ${y}^{2}=4 {x} )$ ${y}= {mx}- {m}^{2}$ (tangent at ${x}^{2}=4 {y} )$ $\frac{1}{ { m}}=- { m}^{2}$ (for common tangent) ${m}^{3}=-1$ ${m}=-1$ $[m =-1]$ $y=-x-1$ $x+y+1=0$ This line touches circle $\therefore$ apply ${p}= {r}$ $c=|\frac{0+0+1}{\\sqrt{2}} |= \frac{1}{\\sqrt{2}}$

Q29JEE Main 2020MCQ4MSets and Relations

The negation of the Boolean expression ${x} ↔ ∼ {y}$ is equivalent to :

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📖 Explanation

$p \leftrightarrow q \ \equiv (p \to q) \ ∧(q \to p)$ $x \leftrightarrow \sim y \ \equiv (x \to \sim y) \ ∧(\sim y \to x)$ $\ \because (p \to q \ \equiv \sim p \vee q)$ $x \leftrightarrow \sim y \ \equiv (\sim x \vee \sim y) ∧(y \vee x)$ $\sim (x \leftrightarrow \sim y) \ \equiv (x \wedge y) \vee (\sim x \wedge \sim y)$

Q30JEE Main 2020MCQ4MVector Algebra

If the volume of a parallelopiped, whose coterminus edges are given by the vectors ${\vec{a}}=\hat{i}+\hat{j}+n \hat{k},$ ${\vec{b}}=2\hat{i}+4 \hat{j}-{n} \hat{k}$ and ${\vec{c}}=\hat{i}+ {n}\hat{j}+3\hat{k} ({n} \ge 0),$ is $158{cu} .$ units, then :

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📖 Explanation

$v=\begin{bmatrix}\vec{a} & \vec{b} & \vec{c}\end{bmatrix}$ $158 =\begin{vmatrix}1 & 1 & n \\ 2 & 4 & -n \\ 1 & n & 3\end{vmatrix},n\ge 0$ $158=1 (12+n^{2} )-(6+n)$ $+n(2 n-4)$ $158=n^{2}+12-6-n$ $+2 n^{2}-4 n$ $3 n^{2}-5 n-152=0$ ${n}=8,- \frac{38}{6}$ (rejected) ${\vec{a}} .{\vec{c}}=1+ {n}+3 {n}$ $=1+4 {n}=33$ ${\vec{b}} .{\vec{c}}=2+4 {n}-3 {n}$ $=2+ {n}=10$

Q31JEE Main 2020MCQ4MDifferential Equations

If ${y}= {y}( {x})$ is the solution of the differential equation ${5+ {e}^{ {x}}}/{2+ {y}} .\frac{ {dy}}{ {dx}}+ {e}^ { {x}}=0$ satisfying ${y}(0)=1,$ then a value of ${y} (\log _{ {e}} 13 )$ is :

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📖 Explanation

${5+ {e}^{ {x}}}/{2+ {y}} .\frac{ {dy}}{ {dx}}=- {e}^ { {x}}$ $∫\frac{d y}{2+y}=∫\frac{-e^{x}}{e^{x}+5} d x$ $\ln (y+2)=-\ln (e^{x}+5 )+k$ $(y+2) (e^{x}+5 )=C$ $\becausey(0)=1$ $\RightarrowC=18$ ${y}+2={18}/{ {e}^{ {x}}+5}$ at ${x}=\ln 13$ ${y}+2= \frac{18}{13+5}=1$ $[y=-1]$

Q32JEE Main 2020MCQ4MSets and Relations

A survey shows that 73% of the persons working in an office like coffee, whereas 65% like tea. If x denotes the percentage of them, who like both coffee and tea, then x cannot be:

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📖 Explanation

$\ {C} \to$ person like coffee $\ {T} \to$ person like Tea ${n}( {C})=73$ ${n}( {T})=65$ ${n}( {C} ∪{T}) \le 100$ ${n}( {C})+ {n}( {T})- {n}( {C} ∩{T}) \le 100$ $73+65-x \le 100$ $x \ge 38$ $73-x \ge 0 \Rightarrowx \le 73$ $65-x \ge 0 \Rightarrow x \le 65$ $[38 \le x \le 65]$

Q33JEE Main 2020MCQ4MQuadratic Equation and Inequalities

The product of the roots of the equation $9 {x}^{2}-18| {x}|+5=0,$ is

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📖 Explanation

$9 {x}^{2}-18| {x}|+5=0$ $9| {x}|^{2}-15| {x}|-3| {x}|+5=0$ $(\because {x}^{2}=| {x}|^{2}$ $3| {x}|(3| {x}|-5)$ $-(3| {x}|-5)=0$ $| {x}|= \frac{1}{3}, \frac{5}{3}$ ${x}=±\frac{1}{3},±\frac{5}{3}$ Product of roots = $25/81$

Q34JEE Main 2020MCQ4MIndefinite Integrals

If $∫({e}^{2{x}}+2 {e}^{ {x}}- {e}^{- {x}}-1 ) {e}^{({e}^{ {x}}+ {e}^{-x} )} {d} {x}$ $=g(x) e^{ (e^{x}+e^{-x} )}+c,$ where $c$ is a constant of integration, then ${g}(0)$ is equal to :

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📖 Explanation

${e}^{2{x}}+2 {e}^{ {x}}- {e}^{- {x}}-1$ $= e^x (e^x + 1) -e^{-x }(e^x + 1) + e^x$ $= [(e^x + 1) (e^x -e^{-x}) + e^x ]$ so $I=∫(e^{x}+1)(e^{x}-e^{-x} ) e^{e^{x}+e^{-x}}$ $+∫ e^{x} . e^{e^{x}+e^{-x}} d x$ $=({e}^{ {x}}+1 ) {e}^{ {e}^{ {x}}+ {e}^{- {x}}}- ∫{e}^{ {x}} . {e}^{ {e}^{ {x}}+ {e}^{- {x}}}dx$ $+∫{e}^{ {x}} .{e}^{ {e}^{ {x}}+ {e}^{- {x}}}dx$ $=({e}^{ {x}}+1 ) {e}^{ {e}^{ {x}}+ {e}^{- {x}}}+C$ ∴ $g(x)=e^{x}+1 \Rightarrowg(0)=2$

Q35JEE Main 2020MCQ4MApplication of Derivatives

If the minimum and the maximum values of thefunction ${f}:[\frac{\pi }{4}, \frac{\pi }{2} ] \to {R},$ defined by : $f(\theta )=\begin{vmatrix}-\sin ^{2} \theta & -1-\sin ^{2} \theta & 1 \\ -\cos ^{2} \theta & -1-\cos ^{2} \theta & 1 \\ 12 & 10 & -2\end{vmatrix}$ are ${m}$ and ${M}$ respectively, then the ordered pair $( {m}, {M})$ is equal to :

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📖 Explanation

${C}_{3} \to {C}_{3}-({C}_{1}- {C}_{2} )$ $f(\theta )=\begin{vmatrix}-\sin ^{2} \theta & -1-\sin ^{2} \theta & 0 \\ -\cos ^{2} \theta & -1-\cos ^{2} \theta & 0 \\ 12 & 10 & -4\end{vmatrix}$ $=-4[(1+\cos ^{2} \theta ) \sin ^{2} \theta -\cos ^{2} \theta$ $(1+\sin ^{2} \theta )]$ $=-4 [\sin ^{2} \theta +\sin ^{2} \theta \cos ^{2} \theta -\cos ^{2} \theta$ $-\cos ^{2} \theta \sin ^{2} \theta ]$ $f(\theta )=4 \cos 2 \theta$ $\theta \in [\ \frac{\pi }{4}, \frac{\pi }{2} ]$ $2 \theta \in [ \frac{\pi }{2}, \pi ]$ $f(\theta \in [-4,0]$ $( { m},{M})=(-4,0)$

Q36JEE Main 2020MCQ4MMatrices and Determinants

Let $\lambda \in {R}$ . The system of linear equations $2 {x}_{1}-4 {x}_{2}+\lambda{x}_{3}=1$ $x_{1}-6 x_{2}+x_{3}=2$ $\lambdax_{1}-10 x_{2}+4 x_{3}=3$ is inconsistent for :

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📖 Explanation

$D=\begin{vmatrix}2 & -4 & \lambda \\ 1 & -6 & 1 \\ \lambda & -10 & 4\end{vmatrix}$ $=2(3 \lambda+2)(\lambda-3)$ $D_{1}=-2(\lambda-3)$ $D_{2}=-2(\lambda+1)(\lambda-3)$ $D_{3}=-2(\lambda-3)$ When $[{\lambda=3}$ , then ${D}= {D}_{1}= {D}_{2}= {D}_{3}=0$ $\Rightarrow$ Infinite many solution when $[\lambda=- \frac{2}{3} ]$ then ${D}_{1}, {D}_{2}, {D}_{3}$ none of them is zero so equations are inconsistant $\therefore [ \lambda=- \frac{2}{3}]$

Q37JEE Main 2020MCQ4MSequences and Series

If ${S}$ is the sum of the first 10 terms of the series $\tan ^{-1}(\frac{1}{3} )+\tan ^{-1}(\frac{1}{7} )+\tan ^{-1}(\frac{1}{13} )$ $+\tan ^{-1}(\frac{1}{21} )+......$ then $\tan ( {S})$ is equal to :

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📖 Explanation

$S=\tan ^{-1}(\frac{1}{3} )+\tan ^{-1}(\frac{1}{7} )$ $+\tan ^{-1}(\frac{1}{13} ) + ......$ ${S}=\tan ^{-1}({2-1}{1+1.2} )$ $+\tan ^{-1}(\frac{3-2}{1+2 \\times 3} )+\tan ^{-1}$ $(\frac{4-3}{1+3 \\times 4} )+....+\tan ^{-1}(\frac{11-10}{1+10 \times 11} )$ ${S}=(\tan ^{-1} 2-\tan ^{-1} 1 )+ (\tan ^{-1} 3$ $-\tan ^{-1} 2 )+$ $(\tan ^{-1} 4-\tan ^{-1} 3 )+.....$ $+ (\tan ^{-1}(11)-\tan ^{-1}(10) )$ ${S}=\tan ^{-1} 11-\tan ^{-1} 1$ $=\tan ^{-1}(\frac{11-1}{1+11} )$ $\tan ( {S})= \frac{11-1}{1+11 \\times 1}$ $= \frac{10}{12}= \frac{5}{6}$

Q38JEE Main 2020MCQ4MComplex Numbers

If the four complex numbers $z, \overline{z}, \overline{z}-2 {\text{Re}}(\overline{z})$ and $z-2 {\text{Re}}(z)$ represent the vertices of a square of side 4 units in the Argand plane, then $|z|$ is equal to : :

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📖 Explanation

Let $z=x+i y$ Length of side $=4$ $m {AB}=4$ $|z-\overline{z}|=4$ $|2 y|=4 ;|y|=2$ ${BC}=4$ $\| \overline{{z}}-(\overline{{z}}-2 {\text{Re}}(\overline{{z}})|=4$ $|2 x|=4 ;|x|=2$ $|z|=\\sqrt{x^{2}+y^{2}}$ $=\\sqrt{4+4}=2 \\sqrt{2}$

Q39JEE Main 2020MCQ4MApplication of Derivatives

If the point ${P}$ on the curve, $4 {x}^{2}+5 {y}^{2}=20$ is farthest from the point $Q(0,-4),$ then $P Q^{2}$ is equal to :

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📖 Explanation

Given ellipse is $\frac{x^{2}}{5}+\frac{y^{2}}{4}=1$ Let point $P$ is $(\\sqrt{5} \cos \theta , 2 \sin \theta )$ $( {PQ})^{2}=5 \cos ^{2} \theta +4(\sin \theta +2)^{2}$ $( {PQ})^{2}=\cos ^{2} \theta +16 \sin \theta +20$ $( {PQ})^{2}=-\sin ^{2} \theta +16 \sin \theta +21$ $=85-(\sin \theta -8)^{2}$ will be maximum when $\sin \theta =1$ $\Rightarrow ({PQ})^{2}{ }_{\max }=85-49=36$

Q40JEE Main 2020MCQ4MStatistics

The mean and variance of 7 observations are 8 and $16,$ respectively. If five observations are $2,4,10,12,14,$ then the absolute difference of the remaining two observations is :

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📖 Explanation

$\overline{x}= \frac{2+4+10+12+14+x+y}{7}$ $=8$ $x+y=14$ ....(i) $(\sigma )^{2}= {∑({x}_{ {i}} )^{2}}/{ {n}}-({\sum {x}_{ {i}}}/{ {n}} )^{2}$ $16= \frac{4+16+100+144+196+x^{2}+y^{2}}{7}-8^{2}$ $16+64= \frac{460+x^{2}+y^{2}}{7}$ $560=460+x^{2}+y^{2}$ $x^{2}+y^{2}=100$ ....(ii) Clearly by (i) and (ii), $|x-y|=2$

Q41JEE Main 2020MCQ4MThree Dimensional Geometry

If $( {a}, {b}, {c})$ is the image of the point (1,2,-3)in the line, $\frac{x+1}{2}= \frac{y-3}{-2}= \frac{z}{-1},$ then ${a}+ {b}+ {c}$ is equal to

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📖 Explanation

Line is $\frac{x+1}{2}= \frac{y-3}{-2}= \frac{z}{-1}=\lambda:$ Let point $R$ is $(2 \lambda-1,-2 \lambda+3,-\lambda)$ Direction ratio of ${PQ}\equiv (2 \lambda-2,-2 \lambda+1,3-\lambda)$ $\ {PQ}$ is $⊥{r}$ to line \Rightarrow $2(2 \lambda-2)-2(-2 \lambda+1)$ $-1(3-\lambda)=0$ $4 \lambda-4+4 \lambda-2-3+\lambda=0$ $9 \lambda=9\Rightarrow \lambda=1$ ⇒ Point ${R}$ is (1,1,-1) $\frac{a+1}{2}=1$ $a=1$ $\frac{b+2}{2}=1$ $b=0$ $\frac{c-3}{2}=-1$ $c=1$ \Rightarrow $a+b+c=2$

Q42JEE Main 2020MCQ4MDefinite Integrals

The value of $\int _ {-\pi / 2}^{\pi / 2} {1}/{1+ {e}^{\sin {x}}} {d} {x}$ is

🚀 Solve in Practice Mode

📖 Explanation

$I= ∫ _ {-\pi / 2}^{\pi / 2} {1}/{1+ {e}^{\sin {x}}} {d} {x}$ ....(1) Apply King property ${I}= ∫ _ {-\pi / 2}^{\pi / 2}{1}/{1+ {e}^{-\sin {x}}} {d} {x}$ $= ∫ _ {-\pi / 2}^{\pi / 2} {{e}^{\sin {x}}}/{1+ {e}^{\sin {x}}} {d} {x}$ ...(2) Add (1) & (2) $2 {I}=∫ _ {-\pi / 2}^{\pi / 2}{d} {x}=\pi$ ${I}= \frac{\pi }{2}$

Q43JEE Main 2020MCQ4MSequences and Series

If $2^{10}+2^{9} .3^{1}+2^{8.3^{2}}+.....+2.3^{9}$ $+3^{10} = {S}-2^{11}$ then ${S}$ is equal to :

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📖 Explanation

${a}=2^{10} ; {r}= \frac{3}{2} ; {n}=11$ (G.P.) ${S}'= (2^{10} ) \frac{((\frac{3}{2} )^{11}-1 )}{ \frac{3}{2}-1}$ $=2^{11}$ \frac{3^{11}}{2^{11}}-1 ){S}'=3^{11}-2^{11}= {S}-2^{11}( $Given$ )\therefore \ {S}=3^{11}$

Q44JEE Main 2020MCQ4MEllipse

If the co-ordinates of two points ${A}$ and ${B}$ are $(\\sqrt{7}, 0)$ and $(-\\sqrt{7}, 0)$ respectively and ${P}$ is any point on the conic, $9 {x}^{2}+16 {y}^{2}=144,$ then ${PA}+ {PB}$ is equal to :

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📖 Explanation

$\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ $a=4 ; b=3 ;$ $e=\sqrt{\frac{16-9}{16}}=\frac{\sqrt {7}}{4}$ A and $B$ are foci $\Rightarrow{PA}+ {PB}$ $=2 {a}=2 \\times 4=8$

Q45JEE Main 2020MCQ4MLimits, Continuity and Differentiability

If $\alpha$ is the positive root of the equation, $p(x)=x^{2}-x-2=0,$ then $\lim _{x \to \alpha ^{+}} \frac{\sqrt{1-\cos (p(x))}}{x+\alpha -4}$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

$x^{2}-x-2=0$ roots are $2 \ & -1$ $\Rightarrow \lim _{x \to 2^{+}}{\sqrt{1-\cos (x^{2}-x-2 )}}/{(x-2)}$ $=\lim_{x \to 2^{+}}{\\sqrt{2 \sin ^{2} \frac{(x^{2}-x-2 )}{2}}}/{(x-2)}$ $=\lim _{x \to 2^{+}}\frac{\\sqrt{2} \sin(\frac{(x-2)(x+1)}{2} )}{(x-2)}$ $=\frac{3}{\\sqrt{2}}$

Q46JEE Main 2020NAT4MProbability

Four fair dice are thrown independently 27 times. Then the expected number of times, at least two dice show up a three or a five, is ______.

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📖 Explanation

4 dice are independently thrown. Each die has probability to show 3 or 5 is ${p}= \frac{2}{6}= \frac{1}{3}$ $\therefore \ {q}=1- \frac{1}{3}= \frac{2}{3}$ (not showing 3 or 5) Experiment is performed with 4 dices independently. ∴ Their binomial distribution is $({q}+ {p})^{4}=( {q})^{4}+{ }^{4} {C}_{1} {q}^{3} {p}+{ }^{4} {C}_{2} {q}^{2} {p}^{2}$ $+{ }^{4} {C}_{3} {qp}^{3}+{ }^{4} {C}_{4} {p}^{4}$ $\therefore$ In one throw of each dice probability of showing 3 or 5 at least twice is $= {p}^{4}+{ }^{4} {C}_{3} {qp}^{3}+{ }^{4} {C}_{2 {q}^{2} {p}^{2}$ $= \frac{33}{81}$ $\therefore$ Such experiment performed 27 times $\therefore$ so expected out comes $= {np}$ $= \frac{33}{81} \\times 27$ $=11$

Q47JEE Main 2020NAT4MStraight Lines and Pair of Straight Lines

If the line, $2 {x}-{y}+3=0$ is at a distance $\frac{1}{\\sqrt{5}}$ and $\frac{2}{\\sqrt{5}}$ from the lines $4 m{x}-2 {y}+\alpha =0$ and $6 x-3 y+\beta =0,$ respectively, then the sum of all possible values of $\alpha$ and $\beta$ is _____.

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📖 Explanation

Apply distance between parallel line formula $4 x-2 y+\alpha =0$ $4 x-2 y+6=0$ $|\frac{\alpha -6}{255} |= \frac{1}{55}$ $|\alpha -6|=2 \Rightarrow \alpha =8,4$ sum $=12$ again $6 x-3 y+\beta =0$ $6 x-3 y+9=0$ $|\frac{\beta -9}{3 \\sqrt{5}} |= \frac{2}{\\sqrt{5}}$ $|\beta -9|=6 \Rightarrow \beta =15,3$ sum $=18$ sum of all values of $\alpha$ and $\beta$ is $=30$

Q48JEE Main 2020NAT4MBinomial Theorem

The natural number ${m}$ , for which the coefficient of $x$ in the binomial expansion of $(x^{m}+ {1} /{x^{2}} )^{22}$ is $1540,$ is _______.

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📖 Explanation

${T}_{ {r}+1}= {}^22C_r(x^m)^{22-r}(1/x^2) ^{ {r}}$ $={}^22 {C}_{ {r}} {x}^{22 { m}- {mr}-2 {r}}$ $={ }^{22} {C}_{ {r}} \ {x}$ $\because{}^22 {C}_{3}=22 {C}_{19}=1540$ $\therefore {r}=3$ or 19 $22 { m}- {mr}-2 {r}=1$ ${ m}= \frac{2 {r}+1}{22-5}$ ${r}=3, \ { m}= \frac{7}{19} ∉{N}$ ${r}=19,\ { m}= \frac{38+1}{22-19}= \frac{39}{3}=13$ ${ m}=13$

Q49JEE Main 2020NAT4MPermutations and Combinations

The number of words, with or without meaning, that can be formed by taking 4 letters at a time from the letters of the word 'SYLLABUS' such that two letters are distinct and two letters are alike, is ______.

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📖 Explanation

${S}_{2} {YL}_{2} {ABU}$ ABCC type words ${{ }^{2} {C}_{1}}_{\text{selection} \text{of} \text{two} \text{alike}\text{letters}}\times {{ }^{5} {C}_{2}}_{\text{selection} \text{of} \text{of} \text{two} \text{distinct}\text{letters}}$ $\times {4!/2!}_{\text{arrangement} \text{of} \text{selected} \text{letters}}$ $=240$

Q50JEE Main 2020NAT4MLimits, Continuity and Differentiability

Let $f(x)=x.[ \frac{x}{2} ],$ for $-10 < x < 10,$ where $[t]$ denotes the greatest integer function. Then the number of points of discontinuity of ${f}$ is equal to _______.

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📖 Explanation

${x} \in (-10,10)$ $\frac{x}{2} \in (-5,5) \to 9$ integers check continuity at ${x}=0$ $\text{table} f(0)=0;f(0^{-})=0;f(0^{+})=0\}$ continuous at x = 0 function will be distcontinuous when $\frac{x}{2}=\±4,\±3,\±2,\±1$ 8 points of discontinuity

PHYSICS25 questions

Q51JEE Main 2020MCQ4MMotion in a Plane

A balloon is moving up in air vertically above a point A on the ground. When it is at a height $\ {h}_{1},$ a girl standing at a distance $\ {d}$ (point $\ {B}$ ) from A (see figure) sees it at an angle $45°$ with respect to the vertical. When the balloon climbs up a further height $\ {h}_{2}$ , it is seen at an angle $60°$ with respect to the vertical if the girl moves further by a distance 2.464 d (point $C$ ). Then the height $\ {h}_{2}$ is (given tan $30°=0.5774 ):$

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📖 Explanation

$\frac{{h}_{1}}{ {d}}=\tan 45°\Rightarrow {h}_{1}= {d}$ ......(1) $\frac{{h}_{1}+ {h}_{2}}{ {d}+2.464 {d}}=\tan 30°$ $\Rightarrow ({h}_{1}+ {h}_{2} ) \times \sqrt{3}=3.46 \ {d}$ $({h}_{1}+ {h}_{2} )= \frac{3.46 {d}}{\\sqrt{3}}$ $\Rightarrow {d}+ {h}_{2}= \frac{3.46 {d}}{\\sqrt{3}}$ ${h}_{2}= {d}$

Q52JEE Main 2020MCQ4MWaves

In a resonance tube experiment when the tube is filled with water up to height of $17.0 \ {cm}$ frombottom, it resonates with a given tuning fork. When the water level is raised the nextresonance with the same tuning fork occurs at a height of $24.5 \ {cm} .$ If the velocity of sound in air is $330 \ {m} / {s},$ the tuning fork frequency is:

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📖 Explanation

$\Rightarrow \lambda=2 (l_{2}-l_{1}\ ) \Rightarrow 2 \\times (24.5-17)$ $\Rightarrow2 \\times 7.5=15 \ {cm}$ $\& \ {v}=\ {f} \lambda\Rightarrow 330=\lambda\times 15 \\times 10^{-2}$ $\lambda= \frac{330}{15} \\times 100 \$ $\Rightarrow \frac{1100 \\times 100}{5}$ $\Rightarrow2200 \ {Hz}$

Q53JEE Main 2020MCQ4MMotion in a Straight Line

A helicopter reises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is a height $\ {h}$ . The time taken by the packet to reach the ground is close to $[\ {g}$ is the acceleration due to gravity]:

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📖 Explanation

$v^{2}=u^{2}+2 a s$ $v^{2}=2 g h+2 g h$ $v=\\sqrt{4 g h}$ $\Rightarrow \sqrt{4 {gh}}=\\sqrt{2 {gh}}+ {gt}$ $\Rightarrow {t}=\sqrt{\frac{4 {h}}{ {g}}}-\sqrt{\frac{2 {h}}{ {g}}}$ $\Rightarrow 3.4 \sqrt{\frac{{h}}{ {g}}}$

Q54JEE Main 2020MCQ4MAtoms and Nuclei

Activities of three radioactive substances ${A}, \ {B}$ and $\ {C}$ are represented by the curves $\ {A}, \ {B}$ and C, in the figure. Then their half-lives $\ {T}_{ \frac{1}{2}}( {A}): {T}_{ \frac{1}{2}}( {B}): {T}_{ \frac{1}{2}}({C})$ are in the ratio :

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📖 Explanation

${R}= {R}_{0} {e}^{-\lambda{t}}$ In ${R}=\ln {R}_{0}-\lambda{t}$ $\lambda_{ {A}}= \frac{6}{10}\Rightarrow {T}_{ {A}}=\frac{10}{6} \ln 2$ $\lambda_{ {B}}= \frac{6}{5} \Rightarrow {T}_{ {B}}= \frac{5 l {n} 2}{6}$ $\lambda_{ {C}}= \frac{2}{5} \Rightarrow {T}_{ {C}}= \frac{5 l {n} 2}{2}$ $\frac{10}{6}: \frac{5}{6}: \frac{15}{6}:: 2: 1: 3$

Q55JEE Main 2020MCQ4MProperties of Matter

A hollow spherical shell at outer radius $\ {R}$ floats just submerged under the water surface. The inner radius of the shell is r. If the specific gravity of the shell material is $\ \frac{27}{8}$ w.r.t. water, the value of $\ {r}$ is :

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📖 Explanation

$\frac{4}{3} \pi ({R}^{3}- {r}^{3} ) \rho _{ {m}} {g}= \frac{4}{3} \pi \ {R}^{3} \rho _{ {w}} {g}$ $1-(\frac{{r}}{ {R}} )^{3}= \frac{8}{27}$ $\Rightarrow \frac{{r}}{ {R}}=(\frac{19}{27} )^{1 / 3}= \frac{19^{1 / 3}}{3}$ $=0.88≃\frac{8}{9}$

Q56JEE Main 2020MCQ4MThermodynamics

Three different processes that can occur in an ideal monoatomic gas are shown in the $\ {P}$ vs $\ {V}$ diagram. The paths are labelled as $\ {A} \to {B}, {A}\to {C}$ and $\ {A}\to {D} .$ The change ininternal energies during these process are taken as $\ {E}_{ {AB}}, {E}_{ {AC}}$ and ${E}_{ {AD}}$ and the workdone as ${W}_{ {AB}}$ ${W}_{ {AC}}$ and ${W}_{ {AD}}$ . The correct relation between these parametersare :

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📖 Explanation

(b) Temperature change $\Delta T$ is same for all three processes $A \to B ; A \to C$ and $A \to D$ $\Delta U=n C_{v} \Delta T=$ same $E_{A B}=E_{A C}=E_{A D}$ Work done, $W=P \times \Delta V$ $A B \to$ volume is increasing $\Rightarrow W_{A B}>0$ $A D \to$ volume is decreasing $\Rightarrow W_{A D}<0$ $A C \to$ volume is constant $\Rightarrow W_{A C}=0$

Q57JEE Main 2020MCQ4MElectrostatic Potential and Capacitance

A solid sphere of radius $\ {R}$ carries a charge $( {Q}+ {q})$ distributed uniformly over its volume. A very small point like piece of it of mass m gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge q. If it acquires a speed $\ {v}$ when it has fallen through a vertical height y (see figure), then : (assume the remaining portion to be spherical).

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📖 Explanation

$\frac{{kQq}}{ {R}}+ {mgy}$ $=\frac{{kQq}}{ {R}+ {y}}+ \frac{1}{2} {mv}^{2}$ ${v}^{2}=2 {gy}+ \frac{2 {kQqy}}{ {mR}( {R}+ {y})}$

Q58JEE Main 2020MCQ4MSemiconductor Electronics

With increasing biasing voltage of a photodiode, the photocurrent magnitude :

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Q59JEE Main 2020MCQ4MRay Optics and Optical Instruments

For a concave lens of focal length f, the relation between object and image distance $\ {u}$ and $\ {v},$ respectively, from its pole can best be represented by $(u=v$ is the reference line $):$

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📖 Explanation

${v}=\frac{{uf}}{ {u}+ {f}}$ Case-I If $v=u$ $\Rightarrow {f}+ {u}= {f}$ $\Rightarrow {u}=0$ Case-II If ${u}=\infty$ then $v=f$ Only option (4) satisfies this condition.

Q60JEE Main 2020MCQ4MCurrent Electricity

An electrical power line, having a total resistance of $2 \Omega,$ delivers $1 \ {kW}$ at $220 \ {V}$ . The efficiency of the transmission line is approximately:

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📖 Explanation

$\ {vi}=10^{3}$ $\ {i}=\ \frac{1000}{220}$ loss $= {i}^{2} {R}=(\frac{50}{11} )^{2} \\times 2$ efficiency $= \frac{1000}{1000+ {i}^{2} {R}} \\times 100=96 \%$

Q61JEE Main 2020MCQ4MWaves

Assume that the displacement(s) of air is proportional to the pressure difference $(\Delta{p})$ created by a sound wave. Displacement(s) further depends on the speed of sound (v), density of air $(\rho )$ and the frequency (f). If $\Delta{p}∼ 10 \ {Pa}, \ {v} \sim 300 {m} / {s},$ ${p} \sim 1 {kg} / {m}^{3}$ and ${f} \sim 1000 {Hz},$ then $\ {s}$ will be the order of (take multiplicative constant to be 1 )

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📖 Explanation

$\Delta{p}=\ {BkS}_{0}$ $=\ \rho {v}^{2}\times \frac{\omega }{ {v}} \times {S}_{0}$ $\Rightarrow {S}_{0}= \frac{\Delta{p}}{\ \rho {v} \omega }$ $\approx \frac{10}{1 \\times 300 \\times 1000} \ {m}$ $= \frac{1}{30} \ {mm} \approx \frac{3}{100} \ {mm}$

Q62JEE Main 2020MCQ4MThermodynamics

A bullet of mass $5 \ { g}$ , travelling with a speed of 210 $\ {m} / {s},$ strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is $0.030 \ {cal} /({g}-{ }°{C} )$ (1 cal $=4.2 \\times 10^{7}$ ergs) close to :

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📖 Explanation

$\frac{1}{2} {mv}^{2}\times \frac{1}{2}= {ms} \Delta{T}$ $\Delta{T}=\frac{{v}^{2}}{4 \\times 5}$ $= \frac{210^{2}}{4 \\times 30 \\times 4.200}$ $=87.5°{C}$

Q63JEE Main 2020MCQ4MThermodynamics

Number of molecules in a volume of $4 \ { cm}^{3}$ of a perfect monoatomic gas at some temperature $\ {T}$ and at a pressure of $2 \ { cm}$ of mercury is close to ? (Given, mean kinetic energy of a molecule (at T) is $4 \\times 10^{-14}$ erg, $\ {g}=980 { cm} / { s}^{2},$ density of mercury $\left.=13.6 { g} / { cm}^{3} )$

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📖 Explanation

${n}=\frac{{PV}}{ {RT}}, \frac{3}{2} \ {kT}=4 \\times 10^{-14}$ ${N}=\frac{{PV}}{ {RT}} \times {Na}$ $= \frac{2 \\times 13.6 \\times 980 \\times 4}{ \frac{8}{3} \\times 10^{-14}}$ $=3.99 \\times 10^{18}$

Q64JEE Main 2020MCQ4MMagnetic Effects of Current and Magnetism

A square loop of side $2 \ {a},$ and carrying current I, is kept in XZ plane with its centre at origin. A long wire carrying the same current I is placed parallel to the z-axis and passing through the point $(0, {b}, 0),( {b}>> {a}) .$ The magnitude of the torque on the loop about z-axis is given by:

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📖 Explanation

$\vec{\tau }={\vec{M}} \times {\vec{B}}$ $=4 {a}^{2} {I} \times \frac{\mu_{0} {I}}{2 \pi {b}}$

Q65JEE Main 2020MCQ4MUnits and Measurements

A physical quantity z depends on fourobservables $\ {a}, \ {b}, \ {c}$ and $\ {d},$ as $\ {z}={{a}^{2} \ {b}^{\frac{2}{3}}}/{\\sqrt{ {c}} {d}^{3}} .$ The percentage of error in the measurement of a, b, c and d $2 \%, 1.5 \%, 4 \%$ and $2.5 \%$ respectively. The percentage of error in z is:

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📖 Explanation

$\frac{\Delta Z}{Z}= \frac{2 \Delta a}{a}+ \frac{2}{3} \frac{\Delta b}{b}+ \frac{1}{2} \frac{\Deltac}{c}$ $+ \frac{3 \Deltad}{d}=14.5 \%$

Q66JEE Main 2020MCQ4MCurrent Electricity

A galvanometer of resistance $\ {G}$ is converted into a voltmeter of range $0-1 \ {V}$ by connecting a resistance $\ {R}_{1}$ in series with it. The additional resistance that should be connected in series with $\ {R}_{1}$ to increase the range of the voltmeter to $0-2 \ {V}$ will be :

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📖 Explanation

$\Rightarrow 1=\ {i}_{ {g}}({G}+ {R}_{1})$ ....(1) $\Rightarrow2=\ {i}_{ {g}}({R}_{1}+ {R}_{2}+ {G} )$ ....(2) (1) % (2) $\Rightarrow \frac{1}{2}=\ \frac{G+R_{1}}{G+R_{1}+R_{2}}$ $G+R_{1}+R_{2}=2 G+2 h_{1}$ $\ (R_{2}=G+R_{1} )$

Q67JEE Main 2020MCQ4MRotational Motion

A wheel is rotaing freely with an angular speed $\omega$ on a shaft. The moment of inertia of the wheel is I and the moment of inertia of the shaft is negligible. Another wheel of momet of inertia 3I initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is :

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📖 Explanation

By anglar momentum conservation $\omega {I}+3 {I} \\times 0=4 {I} \omega '$ $\Rightarrow \omega '= \frac{\omega }{4}$ $({KE})_{ {i}}=\frac{1}{2} {I} \omega ^{2}$ $({KE})_{{f}}={1}{2} \\times (4 {I})\times (\frac{\omega }{4})^{2}$ $=\frac{{I} \omega ^{2}}{8}$ $\Delta{KE}=\frac{3}{8} {I} \omega ^{2}$ fractional loss $=\frac{\Delta{KE}}{{KE}_{1}}={\frac{3}{8} {I} \omega ^{2}}/{{1/2}{ {I} \omega ^{2}}}= \frac{3}{4}$

Q68JEE Main 2020MCQ4MGravitation

The value of the acceleration due to gravity is ${g}_{1}$ at a height ${h}=\frac{{R}}{2}( {R}=$ radius of the earth $)$ from the surface of the earth. It is again equal to ${g}_{1}$ at a depth ${d}$ below the surface of the earth. The ratio $(\frac{{d}}{ {R}} )$ equals

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📖 Explanation

${g}_{1}= {\text{ GM}}/{({R}+\frac{{R}}{2} )^{2}}$ .....(1) ${g}_{2}=\frac{{GM}( {R}- {d})}{ {R}^{3}}$ ......(2) ${ g}_{1}= { g}_{2}$ ${{GM}}/{(\frac{3 {R}}{2} )^{2}}=\frac{{GM}( {R}- {d})}{ {R}^{3}}$ $\Rightarrow \frac{4}{9}=\frac{({R}- {d})}{ {R}}$ $4 {R}=9 {R}-9 {d}$ $5 {R}=9 {d} \Rightarrow \frac{{d}}{{R}}=\frac{5}{9}$

Q69JEE Main 2020MCQ4MElectromagnetic Waves

An electron is constrained to move along the y-axis with a speed of 0.1 c (c is the speed of light) in the presence of electromagnetic wave, whose electric field is ${\vec{E}}=30{{j} ^{\^}} \sin (1.5\times 10^{7} {t}$ $-5 \times 10^{-2} {x} ) {V} / {m}$ The maximum magnetic force experienced by the electron will be : (given ${c}=3 \times 10^{8} {ms}^{-1}$ and electron charge $=1.6\times 10^{-19} {C} )$

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📖 Explanation

⇒ $E={\vec{E}}=30{{j} ^{\^}} \sin (1.5\times 10^{7} {t}$ $-5 \times 10^{-2} {x} ) {V} / {m}$ $\Rightarrow {B}\Rightarrow {E} / {V}$ $\Rightarrow \frac{30}{1.5\times 10^{7}}\times 5 \\times 10^{-2}$ $\Rightarrow10^{-7}$ Tesla $\Rightarrow {F}_{ {mag}}= {q}({\vec{V}} \times {\vec{B}})=| {qVB}|$ $=1.6 \times 10^{-19}\times 0.1 \times 3$ $\times 10^{8}\times 10^{-7}$ $=4.8 \\times 10^{-19} \ {N}$

Q70JEE Main 2020MCQ4MElectrostatic Potential and Capacitance

Two capacitors of capacitances ${C}$ and $2 {C}$ are charged to potential differences ${V}$ and $2 {V},$ respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is:

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📖 Explanation

$\Rightarrow$ By conservation of charge ${q}_{ {i}}= {q}_{ {f}}$ ${Q}_{1}+ {Q}_{2}={q}_{1}+ {q}_{2}$ $4 {CV}-{CV}=( {C}+2 {C}) {V}_{ {C}}$ ${V}_{ {C}}= \frac{3 {CV}}{3 {C}} \Rightarrow {V}$ $\Rightarrow \frac{1}{2} \times (3 {C}) \times {V}_{{C}}^{2}$ $= \frac{1}{2}\times 3 {C} \times {V}^{2}= \frac{3}{2} {CV}^{2}$

Q71JEE Main 2020NAT4MElectromagnetic Induction

Two concentric circular coils, ${C}_{1}$ and ${C}_{2},$ are placed in the XY plane. ${C}_{1}$ has 500 turns, and a radius of $1 {-cm} . {C}_{2}$ has 200 turns and radius of $20{cm} . {C}_{2}$ carries a time dependent current ${I}( {t})=(5 {t}^{2}-2 {t}+3 )$ A where ${t}$ is in ${s}$ . The emf induced in ${C}_{1}$ (in ${mV}$ ), at the instant ${t}=1 { s}$ is $\frac{4}{x} .$ The value of $x$ is ______.

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📖 Explanation

${B}= \frac{\mu_{0} {NI}}{2 {R}}$ $\phi = \frac{\mu_{0} {NN}'{I}}{2 {R}} \pi {r}^{2}$ $\epsilon =\frac{{d} \phi }{\ {dt}}$ $= \frac{2 \pi \times 10^{-7}\times 10^{5}\times \pi \\times 10^{-4}}{0.2}$ $=8 \\times 10^{-4}=0.8 \ {mV}$

Q72JEE Main 2020NAT4MRotational Motion

A force ${\vec{F}}=( \hat{i}+2 \hat{j}+3 \hat{k}) {N}$ acts at a point $(4 \hat{i}+3\hat{j}-\hat{k}) {m}.$ Then the magnitude of torque about the point $( \hat{i}+2 \hat{j}+ \hat{k}) { m}$ will $b e$ $\sqrt{{x}} {N}- {m} .$ The value of ${x}$ is _______.

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📖 Explanation

${\tau }^{\to }=(\vec{r}_{2}-\vec{r}_{1} ) \times \vec{F}$ $=[(4 \hat{i}+3\hat{j}-\hat{k})-( \hat{i}+2 \hat{j}+3 \hat{k}) ]$ $\times { {F}}^{\to }$ $=(3\hat{i}+\hat{j}-2\hat{k}) \times (\hat{i}+2\hat{j}+3 \hat{k})$ $\tau =\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & 2 & 3\end{vmatrix}$ $=7 \hat{i}-11 \hat{j}+5 \hat{k}$ $| \vec{\tau }|=\\sqrt{195}$

Q73JEE Main 2020NAT4MDual Nature of Matter and Radiation

A beam of electrons of energy E scatters from a target having atomic spacing of $1 \Å$ . The first maximum intensity occurs at $\theta =60° .$ Then $\ {E}$ (in eV) is _______. (Planck constant ${h}=6.64\times 10^{-34} {Js}$ , $1 \ {eV}=1.6 \times 10^{-19} { J},$ electron mass ${m}=9.1 \times 10^{-31} { kg} )$

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📖 Explanation

$2 {d} \sin \theta =\lambda={{h}}/{\\sqrt{2 {mE}}}$ $2 \\times 10^{-10} \times \frac{\\sqrt{3}}{2}= {6.6 \\times 10^{-34}}/{\\sqrt{2 {mE}}}$ ${E}= \frac{1}{2}\times \frac{6.64^{2} \\times 10^{-48}}{9.1 \\times 10^{-31} \\times 3 \\times 1.6 \\times 10^{-19}}$ $=50.47$

Q74JEE Main 2020NAT4MCentre of Mass and Collision

A particle of mass $200 {MeV} / {c}^{2}$ collides with a hydrogen atom at rest. Soon after the collision the particle comes to rest, and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in ${eV}$ ) is $\frac{{N}}{4}$ . The value of ${N}$ is : (Given the mass of the hydrogen atom to be $1 {GeV} / {c}^{2} )$ _______.

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📖 Explanation

${mV}_{0}= {MV}= {p}$ $10.2=\frac{{p}^{2}}{2 { m}}-\frac{{p}^{2}}{2 {M}}$ $=\frac{{p}^{2}}{2 { m}} (1-\frac{{m}}{ {M}} )$ $=\frac{{p}^{2}}{2 { m}}(1-0.2)$ $\Rightarrow \frac{{p}^{2}}{2 { m}}= { K}= \frac{10.2}{0.8}$

Q75JEE Main 2020NAT4MRay Optics and Optical Instruments

A compound microscope consists of an objective lens of focal length $1 { cm}$ and an eye piece of focal length $5 {cm}$ with a separation of $10 { cm}$ The distance between an object and the objective lens, at which the strain on the eye is minimum is $\frac{{n}}{40} {cm} .$ The value of ${n}$ is _____.

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📖 Explanation

Final image at $\infty$ $\Rightarrow$ obj. for eye piece at $5 \ { cm}$ $\Rightarrow$ image for objective at $5 {cm}$ $\frac{1}{v}- \frac{1}{u}= \frac{1}{f}$ $\frac{1}{5}+\frac{1}{x}=1$ $\frac{1}{x}=1- \frac{1}{5}= \frac{4}{5}$ $\Rightarrow x= \frac{5}{4}$

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