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JEE MAIN 2020 - 05 SEP S2 2020

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CHEMISTRY25 questions

Q1JEE Main 2020MCQ4MChemical Kinetics

The rate constant (k) of a reaction is measured at different temperatures (T), and the data are plotted in the given figure. The activation energy of the reaction in ${kJ} {mol}^{-1}$ is : (R is gas constant)

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📖 Explanation

Slope $=- \frac{E_{a}}{R}$ $- \frac{10}{5}=- \frac{E_{a}}{R}$ $E_{a}=2 R$

Q2JEE Main 2020MCQ4MAlcohols, Phenols and Ethers

The major product of the following reaction is:

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Q3JEE Main 2020MCQ4MBiomolecules

The following molecule acts as an :

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Q4JEE Main 2020MCQ4MSolid State

An element crystallises in a face-centred cubic (fcc) unit cell with cell edge a. The distance between the centres of two nearest octahedral voids in the crystal lattice is

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📖 Explanation

distance between nearest octahedral voids(O.V.) $=\sqrt{(\frac{{a}}{2} )^{2}+(\frac{{a}}{2} )^{2}} \;\;\$ $\Rightarrow =\frac{{a}}{\\sqrt{2}}$

Q5JEE Main 2020MCQ4MElectrochemistry

The variation of molar conductivity with concentration of an electrolyte (X) in aqueous solution is shown in the given figure. The electrolyte X is :

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📖 Explanation

Its a weak electrolyte hence : $CH_3COOH$

Q6JEE Main 2020MCQ4MSolutions

The one that is NOT suitable for the removal of permanent hardness of water is :

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Q7JEE Main 2020MCQ4MStructure of Atom

The correct statement about probability density (except at infinite distance from nucleus) is :

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Q8JEE Main 2020MCQ4MAlcohols, Phenols and Ethers

The increasing order of boiling points of the following compounds is :

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Q9JEE Main 2020MCQ4MChemical Bonding and Molecular Structure

The compound that has the largest H-M-H bond angle (M=N, O, S, C), is :

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Q10JEE Main 2020MCQ4MOrganic Chemistry - Some Basic Principles

Among the following compounds, geometrical isomerism is exhibited by :

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Q11JEE Main 2020MCQ4MOrganic Chemistry - Some Basic Principles

Which one of the following polymers is not obtained by condensation polymerisation?

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Q12JEE Main 2020MCQ4MAmines

The final major product of the following reaction is :

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Q13JEE Main 2020MCQ4MChemical Bonding and Molecular Structure

Hydrogen peroxide, in the pure state, is :

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Q14JEE Main 2020MCQ4Mp Block Elements

Boron and silicon of very high purity can be obtained through :

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Q15JEE Main 2020MCQ4MSolutions

Lattice enthalpy and enthalpy of solution of ${NaCl}$ are $788 \ {kJ} \ {mol}^{-1}$ and $4 \ {kJ} \ {mol}^{-1},$ respectively. The hydration enthalpy of $\ {NaCl}$ is :

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📖 Explanation

$4=788+\Delta{H}$ $\Delta{H}=-784 \ {kJ}$

Q16JEE Main 2020MCQ4Mp Block Elements

Reaction of ammonia with excess ${Cl}_{2}$ gives:

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📖 Explanation

${NH}_{3}+{3 {Cl}_{2}}_{\text{excess}}{ \to }{NCl}_{3}$ $+3 {HCl}$

Q17JEE Main 2020MCQ4MClassification of Elements

The correct order of the ionic radii of ${O}^{2-}, {N}^{3-}, {F}^{-}, {Mg}^{2+}, {Na}^{+}$ and ${Al}^{3+} {is}:$

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📖 Explanation

Correct order of size for isoelectronic species. ${Al}^{3+} < {Mg}^{2+} < {Na}^{+} < {F}^{-}$ $< {O}^{2-} < {N}^{3-}$

Q18JEE Main 2020MCQ4MCoordination Compounds

Consider the complex ions, trans - $[Co(en)_2Cl_2]^{+} (A)$ and cis $-[Co(en)_2Cl_2]^{+} (B)$ . The correct statement regarding them is :

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📖 Explanation

(A) trans - $[Co(en)_2Cl_2 ] ^{+}$ \Rightarrow (A) is trans form and shows plane of symmetry which is optically inactive (not optically active) (B) cis- $[Co(en)_2Cl_2 ] ^{+}$ ⇒ (B) is cis form and does not shows plane of symmetry, hence it is optically active

Q19JEE Main 2020MCQ4MStates of Matter

Adsorption of a gas follows Freundlich adsorption isotherm. If ${x}$ is the mass of the gas adsorbed on mass ${m}$ of the adsorbent, the correct plot of $\frac{x}{m}$ versus $p$ is :

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📖 Explanation

$\frac{{X}}{ {m}}= {K} .{P}^{1 / {n}}$ $({T}_{2}> {T}_{1} )$

Q20JEE Main 2020MCQ4MHydrocarbons

The major product formed in the following reaction is : $CH_3 CH= CHCH(CH_3 )_2{ \to } ^{HBr}$

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Q21JEE Main 2020NAT4MBiomolecules

The number of chiral carbons present in sucrose is_______ .

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Q22JEE Main 2020NAT4MChemical Thermodynamics

For a dimerization reaction, $2{A}( {g}) \to {A}_{2}({g})$ at $298 {K}, \Delta{U}^{\ominus },=-20 {kJ} \ {mol}^{-1},$ $\Delta{S}^{\ominus }=-30{J}$ ${K}^{-1} {mol}^{-1},$ then the $\Delta{G}^{\ominus }$ will be _______ J.

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📖 Explanation

$\Delta{G}°=\Delta{H}°- {T} \Delta{S}°$ $= (\Delta{U}°+\Delta{n}_{ {g}} {RT} )- {T} \Delta{S}°$ $= [\{-20+(-1)) \frac{8.314}{1000} \\times 298 \}$ $- \frac{298}{1000} \times (-30) ] {kJ}$ $=-13.537572 \ {kJ}$ $=-13537.57$ Joule

Q23JEE Main 2020NAT4MChemical Equilibrium

For a reaction ${X}+ {Y} ⇌2 {Z}, 1.0 {mol}$ of ${X}, 1.5$ mol of Y and $0.5{mol}$ of ${Z}$ were taken in a 1 L vessel and allowed to react. At equilibrium, the concentration of $Z$ was $1.0{mol} {L}^{-1}$ . The equilibrium constant of the reaction is ______ $x/15 .$ The value of x is _______.

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📖 Explanation

$\text{table} ,X,+ ,Y,=, 2Z;t =0, 1 ,,1.5,, 0.5;\text{At} \text{eq}., 0.75 ,,1.25 ,,1$ ; ${K}_{ {\text{eq}.}}= \frac{1^{2}}{ \frac{3}{4}\times \frac{5}{4}}= \frac{16}{15}$

Q24JEE Main 2020NAT4MRedox Reactions

The volume, in ${mL},$ of $0.02 {M} {K}_{2} {Cr}_{2} {O}_{7}$ solutionrequired to react with 0.288 g of ferrous oxalate in acidic medium is ______. (Molar mass of ${Fe}=56 {g}\ {mol}^{-1}$ )

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📖 Explanation

${ {K}_{2} {Cr}_{2} {O}_{7}}_{n=6}+ {{FeC}_{2} {O}_{4}}_{n=3} {\to }$ ${Cr}^{+3}+ {Fe}^{+3}+ {CO}_{2}$ $\frac{0.02 \\times 6\times {V}( {mL})}{1000}$ $= \frac{0.288}{144} \\times 3$ \Rightarrow $V=50\text{ML}$

Q25JEE Main 2020NAT4MCoordination Compounds

Considering that $\Delta_{0}> {P},$ the magnetic moment (in BM) of $[{Ru}({H}_{2} {O} )_{6} ]^{2+}$ would be ______.

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📖 Explanation

Magnetic moment (in B.M.) of $[{Ru}({H}_{2} {O})_{6} ]^{2+}$ would be; while considering that $\Delta_{0}> {P}$ , ${Ru}_{(44)} ;[ {Kr}] 4 {d}^{7} 5 {s}^{1}$ (in ground state) $\Rightarrow$ In $\ {Ru}^{2+} \Rightarrow 4 {d}^{6}$ $\Rightarrow ({t}_{2} {g} )^{6}( {eg})^{0}$ $\Rightarrow$ Here number of unpaired electrons in ${Ru}^{2+}=({t}_{2} {g} )^{6}( {eg})^{0}=0$ and Hence $\mu_{ {m}}=\\sqrt{ {n}( {n}+2)} {B}.{M}$ $=[0\ {B.M}]$

MATHEMATICS25 questions

Q26JEE Main 2020MCQ4MMatrices and Determinants

If the system of linear equations $x+y+3 z=0$ $x+3 y+k^{2} z=0$ $3 x+y+3 z=0$ has a non-zero solution $( {x}, {y}, {z})$ for some ${k} \ \in {R},$ then ${x}+(\frac{{y}}{ {z}} )$ is equal to :

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📖 Explanation

$x+y+3 z=0$ .....(i) $x+3 y+k^{2} z=0$ ....(ii) $3 x+y+3 z=0$ .....(iii) $\begin{vmatrix}1 & 1 & 3 \\ 1 & 3 & k^2 \\ 3 & 1 & 3\end{vmatrix}=0$ $\Rightarrow9+3+3 k^{2}-27-k^{2}$ $-3=0$ $\Rightarrow {k}^{2}=9$ (i) $-$ (iii) $\Rightarrow -2 {x}=0 \Rightarrow {x}=0$ Now from $( {i}) \Rightarrow {y}+3 {z}=0$ $\Rightarrow \frac{y}{z}=-3$ $x+ \frac{y}{z}=-3$

Q27JEE Main 2020MCQ4MQuadratic Equation and Inequalities

If $\alpha$ and $\beta$ are the roots of the equation, $7 {x}^{2}-3 {x}-2=0,$ then the value of $\frac{\alpha }{1-\alpha ^{2}}+ \frac{\beta }{1-\beta ^{2}}$ is equal to :

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📖 Explanation

$7 x^{2}-3 x-2=0$ $\alpha +\beta = \frac{3}{7}$ $\alpha \beta = \frac{-2}{7}$ $\frac{\alpha }{1-\alpha ^{2}}+ \frac{\beta }{1-\beta ^{2}}$ $= \frac{\alpha +\beta -\alpha \beta (\alpha +\beta )}{1-\alpha ^{2}-\beta ^{2}+\alpha ^{2} \beta ^{2}}$ $= \frac{\frac{3}{7}+ \frac{2}{7}(\frac{3}{7} )}{1-(\alpha +\beta )^{2}+2 \alpha \beta +\alpha ^{2} \beta ^{2}}$ $= \frac{27}{16}$

Q28JEE Main 2020MCQ4MSequences and Series

If the sum of the first 20 terms of the series $\log _{ (7^{1 / 2} )} {x}+\log _{ (7^{1 / 3} )} {x}+\log _{ (7^{1/4} )} {x}$ $+.....$ is $460,$ then x is equal to :

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📖 Explanation

$460=\log _{7} x .(2+3+4+.. . .$ $+20+21)$ $\Rightarrow 460=\log _{7} x \.( \frac{21 \\times 22}{2}-1 )$ $\Rightarrow460=230 \.\log _{7} x$ $\Rightarrow\log _{7} x=2 \Rightarrowx=49$

Q29JEE Main 2020MCQ4MLimits, Continuity and Differentiability

$\lim _{x \to 0} {x (e^{ (\\sqrt{1+x^{2}+x^{4}}-1 ) / x}-1 )}/{\\sqrt{1+x^{2}+x^{4}}-1}$

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📖 Explanation

$\lim _{x \to 0} {x (e^{ (\\sqrt{1+x^{2}+x^{4}}-1 ) / x}-1 )}/{\\sqrt{1+x^{2}+x^{4}}-1}$ $\because\lim _{x \to 0} {\\sqrt{1+x^{2}+x^{4}}-1}{x}(\frac{0}{0}$ from $)$ $\lim _{x \to 0}{ (1+x^{2}+x^{4} )-1}/{x (\\sqrt{1+x^{2}+x^{4}}+1 }$ $\lim _{x \to 0}{x(1+x^{2} )}/{ (\\sqrt{1+x^{2}+x^{4}}+1 )}=0$ So $\lim _{x \to 0} {x (e^{({\sqrt{1+x^{2}+x^{4}}-1}/{x} )}-1 )}/{\\sqrt{1+x^{2}+x^{4}}-1}$ $\;( \frac{0}{0}$ from $)$ $\lim _{x \to 0} {e^{{\sqrt{1+x^{2}+x^{4}}-1}{x}}-1}/{({\sqrt{1+x^{2}+x^{4}}-1}/{x} )}=1$

Q30JEE Main 2020MCQ4MSequences and Series

If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243, then the sum of the first 50 terms of this G.P. is :

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📖 Explanation

Let first term = a > 0 Common ratio = r > 0 $ar + ar^2 + ar^3 = 3$ ....(i) $ar ^5 + ar^6 + ar ^7 = 243$ ....(ii) $r^4(ar + ar^2 + ar^3) = 243$ $r^4(3) = 243\Rightarrow r = 3$ as $r > 0$ from (1) $3a + 9a + 27a = 3$ $a=1/13$ ${S}_{50}=\frac{{a}({r}^{50}-1 )}{( {r}-1)}$ $= \frac{1}{26} (3^{50}-1 )$

Q31JEE Main 2020MCQ4MComplex Numbers

The value of $(\frac{-1+i \\sqrt{3}}{1-i} )^{30}$ is :

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📖 Explanation

$(\frac{-1+i \\sqrt{3}}{1-i} )^{30}=$ \frac{2 \omega }{1-i} )^{30}= \frac{2^{30}. \omega ^{30}}{ ((1-i)^{2} )^{30}}= \frac{2^{30} . 1}{ (1+i^{2}-2 i )^{15}}= \frac{2^{30}}{-2^{15} .{i}^{15}}=-2^{15} \ {i}$

Q32JEE Main 2020MCQ4MDifferentiation

The derivative of $\tan ^{-1}({\sqrt{1+ {x}^{2}}-1}/{ {x}} )$ with respect to $\tan ^{-1}({2 x \\sqrt{1-x^{2}}}/{1-2 x^{2}} )$ at $x= \frac{1}{2}$ is :

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📖 Explanation

Let $f=\tan ^{-1}({\sqrt{1+x^{2}}-1}/{x} )$ Put $x=\tan \Rightarrow\theta =\tan ^{-1} x$ $f=\tan ^{-1}(\frac{\sec \theta -1}{\tan \theta } )$ $f=\tan ^{-1}$ \frac{1-\cos \theta }{\sin \theta } )= \frac{\theta }{2}{f}= \frac{\tan ^{-1} {x}}{2}\Rightarrow \frac{{df}}{ {dx}}= \frac{1}{2 (1+ {x}^{2} )} $....(i) Let$ g=\tan ^{-1}({2 x \sqrt{1-x^{2}}}/{1-2 x^{2}} ) $Put$ x=\sin \theta \Rightarrow\theta =\sin ^{-1} x{g}=\tan ^{-1}(\frac{2 \sin \theta \cos \theta }{1-2 \sin ^{2} \theta } ){g}=\tan ^{-1}(\tan 2 \theta )=2 \theta {g}=2 \sin ^{-1} {x}\frac{{dg}}{{dx}}={2}/{\sqrt{1- {x}^{2}}} $....(ii)$ \frac{{df}}{ {dg}}= \frac{1}{2 (1+ {x}^{2} )} {\sqrt{1-\ {x}^{2}}}/{2} $at$ x= \frac{1}{2}(\frac{d f}{d g} )_{x= \frac{1}{2}}= \frac{\sqrt{3}}{10}$

Q33JEE Main 2020MCQ4MArea Under The Curves

The area (in sq. units) of the region ${A}=\{( {x}, {y}):( {x}-1)[ {x}] \le {y} \le 2 \\sqrt{ {x}},$ $0 \le {x} \le 2\}$ where [t] denotes the greatest integer function, is :

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📖 Explanation

$( {x}-1)[ {x}] \le {y} \le 2 \\sqrt{ {x}},$ $[ 0 \le {x} \le 2 ]$ Draw $y=2 \\sqrt{x} \Rightarrowy^{2}=4 x \;\; [{x \ge 0}]$ $y=(x-1)[x]$ $=\begin{cases}0 & 0\le x<1 \\ x-1 & 1\le x<2 \\ 2 & x=2$ ${A}=\int _{0} ^ {2} 2 \\sqrt{ {x}} {dx}-\frac{1}{2} 1 \.1$ ${ A}=2\end{cases}\$ [{{x}^{3 / 2}}{(3 / 2)} ]_{0}^{2}- \frac{1}{2}= \frac{8 \sqrt{2}}{3}- \frac{1}{2}$

Q34JEE Main 2020MCQ4MCircles

If the length of the chord of the circle, ${x}^{2}+{y}^{2}= {r}^{2}( {r}>0)$ along the line, ${y}-2 {x}=3$ is r, then ${r}^{2}$ is equal to :

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📖 Explanation

Let chord ${AB}= {r}$ $\because$ ΔAOM is right angled triangle $\therefore {OM}=\frac{{r} \\sqrt{3}}{2}=$ perpendicular distance of line AB from (0,0) $\frac{{r} \\sqrt{3}}{2}=|\frac{3}{\\sqrt{5}} |$ $\ {r}^{2}= \frac{12}{5}$

Q35JEE Main 2020MCQ4MApplication of Derivatives

If $x=1$ is a critical point of the function $f(x)= (3 x^{2}+a x-2-a ) e^{x},$ then :

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📖 Explanation

$f(x)= (3 x^{2}+a x-2-a ) e^{x}$ $f'(x)= (3 x^{2}+a x-2-a ) e^{x}$ $+e^{x}(6 x+a)$ $= {e}^{x} (3 {x}^{2}+ {x}(6+ {a})-2 )$ $f'(x)=0$ at $x=1$ $\Rightarrow3+(6+a)-2=0$ $a=-7$ $f'(x)=e^{x}(3 x^{2}-x-2 )$ $=e^{x}(x-1)(3 x+2)$ x = 1 is point of local minima ${x}= \frac{-2}{3}$ is point of local maxima

Q36JEE Main 2020MCQ4MStatistics

If the mean and the standard deviation of the data $3,5,7, {a}, {b}$ are 5 and 2 respectively, then a and $b$ are the roots of the equation :

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📖 Explanation

Mean = 5 $\frac{3+5+7+a+b}{5}=5$ $a+b=10$ ....(i) S.d. $=2\Rightarrow \sqrt{ \frac{\sum _{i=1} ^{5} (x_{i}-\overline{x} )^{2}}{5}}=2$ $(3-5)^{2}+(5-5)^{2}+(7-5)^{2}$ $+(a-5)^{2}+(b-5)^{2}=20$ $\Rightarrow4+0+4+(a-5)^{2}$ $+(b-5)^{2}=20$ $a^{2}+b^{2}-10(a+b)$ $+50=12$ $(a+b)^{2}-2 a b-100$ $+50=12$ $ab = 19$ .....(ii) Equation is $x^{2}-10 x+19=0$

Q37JEE Main 2020MCQ4MMatrices and Determinants

If $a+x=b+y=c+z+1,$ where $a, b, c, x$ ${y}, {z}$ are non-zero distinct real numbers, then $\begin{vmatrix}x & a+y & x+a \\ y & b+y & y+b \\ {z} & {c}+ {y} & {z}+ {c}\end{vmatrix}$ is equal to :

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📖 Explanation

$a+x=b+y=c+z+1$ $\begin{vmatrix}x & a+y & x+a \\ y & b+y & y+b \\ {z} & {c}+ {y} & {z}+ {c}\end{vmatrix}$ ${C}_{3}\to {C}_{3}- {C}_{1}$ $\begin{vmatrix}x & a+y & a \\ y & b+y & b \\ {z} & {c}+ {y} & {c}\end{vmatrix}$ ${C}_{2} \to {C}_{2}- {C}_{3}$ $\begin{vmatrix}x & y & a \\ y & y & b \\ {z} & {y} & {c}\end{vmatrix}$ ${R}_{3}\to {R}_{3}- {R}_{1}, {R}_{2} \to {R}_{2}- {R}_{1}$ $\\begin{vmatrix}x & y & a \\ y-x & 0 & b-a \\ z-x & 0 & c-a\end{vmatrix}$ $=(-y)[(y-x)(c-a)-(b-a)$ $(z-x)]$ $=(-y)[(a-b)(c-a)+(a-b)$ $(a-c-1)]$ $=(-y)[(a-b)(c-a)+(a-b)$ $(a-c)+b-a)$ $=-y(b-a)=y(a-b)$

Q38JEE Main 2020MCQ4MIndefinite Integrals

If $∫\frac{\cos \theta }{5+7 \sin \theta -2 \cos ^{2} \theta } {d} \theta$ $= {A} \log _{ {e}}{| {B}(\theta )|}+ {C}$ , where $\ {C}$ is a constant of integration, then $\frac{{B}(\theta )}{ {A}}$ can be :

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📖 Explanation

$∫\frac{\cos \theta }{5+7 \sin \theta -2 \cos ^{2} \theta }$ $∫\frac{\cos \theta {d} \theta }{3+7 \sin \theta +2 \sin ^{2} \theta }$ $\sin \theta =\ {t}$ , $\cos \theta {d} \theta = {dt}$ $∫\frac{d t}{2 t^{2}+7 t+3}=∫\frac{d t}{(2 t+1)(t+3)}$ $= \frac{1}{5} ∫(\frac{2}{2 t+1}- \frac{1}{t+3}){dt}$ $= \frac{1}{5} \ln |\frac{2 t+1}{t+3} |+C$ $= \frac{1}{5} \ln |\frac{2 \sin \theta +1}{\sin \theta +3}|+C$ ${A}= \frac{1}{5}$ and ${B}(\theta )= \frac{2 \sin \theta +1}{\sin \theta +3}$

Q39JEE Main 2020MCQ4MStraight Lines and Pair of Straight Lines

If the line ${y}= {mx}+ {c}$ is a common tangent to the hyperbola $\frac{x^{2}}{100}- \frac{y^{2}}{64}=1$ and the circle ${x}^{2}+ {y}^{2}=36$ , then which one of the following is true?

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📖 Explanation

y = mx + c is tangent to $\frac{x^{2}}{100}- \frac{y^{2}}{64}=1$ and ${x}^{2}+ {y}^{2}=36$ ${c}^{2}=100{m}^{2}-64$ । ${c}^{2}$ $=36 (1+{m}^{2} )$ \Rightarrow $100{m}^{2}-64=36+36 {m}^{2}$ ${m}^{2}= \frac{100}{64} \Rightarrow {m}=\±\frac{10}{8}$ ${c}^{2}=36 (1+\frac{100}{64} )=\frac{36 \\times 164}{64}$ $4 {c}^{2}=369$

Q40JEE Main 2020MCQ4MPermutations and Combinations

There are 3 sections in a question paper and each section contains 5 questions. A candidate has to answer a total of 5 questions, choosing at least one question from each section. Then the number of ways, in which the candidate can choose the questions, is:

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📖 Explanation

A B B [5] [5] [5] 1 2 2 2 1 2 2 2 1 1 1 3 1 3 1 3 1 1Total number of selection $=({ }^{5} {C}_{1}{ }^{5} {C}_{2}{ }^{5} {C}_{2}) \.3+({ }^{5} {C}_{1}{ }^{5} {C}_{1}{ }^{5} {C}_{3}) \.3$ $=5 \.10 \.10 \.3+5 \.5 \.10 \.3$ $=2250$

Q41JEE Main 2020MCQ4MThree Dimensional Geometry

If for some $\alpha \in {R},$ the lines ${L}_{1}: \frac{{x}+1}{2}=\frac{{y}-2}{-1}=\frac{{z}-1}{1}$ and ${L}_{2}: \frac{{x}+2}{\alpha }=\frac{{y}+1}{5-\alpha }=\frac{{z}+1}{1}$ are coplanar, then the line ${L}_{2}$ passes through the point

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📖 Explanation

${L}_{1}\equiv \frac{{x}+1}{2}=\frac{{y}-2}{-1}=\frac{{z}-1}{1}$ ${L}_{2}\equiv \frac{{x}+2}{\alpha }=\frac{{y}+1}{5-\alpha }=\frac{{z}+1}{1}$ Point ${A}(-1,2,1) {B}(-2,-1,-1)$ $\ \because {L}_{1}$ and ${L}_{2}$ are coplanar \Rightarrow $\begin{vmatrix}2 & -1 & 1 \\ \alpha & 5-\alpha & 1 \\ 1 & 3 & 2\end{vmatrix}=0$ $\alpha =-4$ ${L}_{2} \ \equiv \frac{{x}+2}{-4}=\frac{{y}+1}{9}=\frac{{z}+1}{1}$ Check options (2,-10,-2) lies on ${L}_{2}$

Q42JEE Main 2020MCQ4MDifferential Equations

Let $y=y(x)$ be the solution of the differential equation $\cos x {d y} / {d x}+2 y \sin x=\sin 2 x$ , ${x} \in (0, \frac{\pi }{2} ) .$ If ${y}(\pi / 3)=0,$ then ${y}(\pi / 4)$ is equal to :

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📖 Explanation

$\cos x {d y} / {d x}+2 y \sin x=\sin 2 x$ $\frac{d y}{d x}+\frac{2 \sin x}{\cos x} y=2 \sin x$ I.F. $={e}^{\int 2 \frac{\sin x}{\cos x} {dx}}$ $={e}^{{2} \ln \sec x}=\sec ^{2} x$ $y \.\sec ^{2} x=\int 2 \sin x \.\sec ^{2} x d x$ $y\ \sec ^{2} x=2 \int \tan x \sec x d x$ $y \; \sec ^{2} x=2 \sec x+c$ At $x=\frac{\pi }{3}, y=0$ $\Rightarrow0=2 \sec \frac{\pi }{3}+C$ $\Rightarrow C=-4$ $y\ \sec ^{2} x=2 \sec x-4$ Put $x=\frac{\pi }{4}$ $y \.2=2 \\sqrt{2}-4$ $y=\\sqrt{2}-2$

Q43JEE Main 2020MCQ4MApplication of Derivatives

Which of the following points lies on thetangent to the curve $x^{4} e^{y}+2 \\sqrt{y+1}=3$ at thepoint (1,0) $?$

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📖 Explanation

$x^{4} e^{y}+2 \\sqrt{y+1}=3$ d.w.r. to x $x^{4} {e}^{y} y'+{e}^{y} 4 x^{3}+\frac{2 y'}{2 \\sqrt{y+1}}=0$ at P(1, 0) ${y}_{{P}}'+4+{y}_{{P}}'=0$ $\Rightarrow {y}_{{P}}'=-2$ Tangent at ${P}(1,0)$ is $y-0=-2(x-1)$ $2 x+y=2$ (-2,6) lies on it

Q44JEE Main 2020MCQ4MSets and Relations

The statement $({p} \to ({q}\to {p})) \to ({p} \to ({p} \vee {q}))$ is:

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Q45JEE Main 2020MCQ4MTrigonometric Ratios and Identities

If $L=\sin ^{2}(\frac{\pi }{16})-\sin ^{2}(\frac{\pi }{8})$ and $M=\cos ^{2}(\frac{\pi }{16})-\sin ^{2}(\frac{\pi }{8}),$ then :

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📖 Explanation

$L=\sin ^{2}(\frac{\pi }{16})-\sin ^{2}(\frac{\pi }{8})$ $(\because \sin ^{2} \theta =\frac{1-\cos 2 \theta }{2})$ $\Rightarrow{L}=(\frac{1-\cos (\pi / 8)}{2})$ $-(\frac{1-\cos (\pi / 4)}{2})$ ${L}=\frac{1}{2}[\cos (\frac{\pi }{4})-\cos (\frac{\pi }{8})]$ ${L}=\frac{1}{2 \\sqrt{2}}-\frac{1}{2} \cos (\frac{\pi }{8})$ ${M}=\cos ^{2}(\frac{\pi }{16})-\sin ^{2}(\frac{\pi }{8})$ ${M}=\frac{1+\cos (\pi / 8)}{2}$ $-\frac{1-\cos (\pi / 4)}{2}$ ${M}=\frac{1}{2} \cos (\frac{\pi }{8})+\frac{1}{2 \\sqrt{2}}$

Q46JEE Main 2020NAT4MProbability

In a bombing attack, there is 50% chance that a bomb will hit the target. At least two independent hits are required to destroy the target completely. Then the minimum number of bombs, that must be dropped to ensure that there is at least 99% chance of completely destroying the target, is _______.

🚀 Solve in Practice Mode

📖 Explanation

${P}({H})=\frac{1}{2}$ ${P}(\overline{{H}})=\frac{1}{2}$ Let total 'n' bomb are required to destroy the target $1-{ }^{n} C_{n}(\frac{1}{2})^{n}-{ }^{n} C_{1}(\frac{1}{2})^{n}$ $\ge \frac{99}{100}$ $1-\frac{1}{2^{n}}-\frac{n}{2^{n}}\ge \frac{99}{100}$ $\frac{1}{100}\ge {{n}+1}/{2^{{n}}}$ Now check for value of ${n}$ $[n=11]$

Q47JEE Main 2020NAT4MPermutations and Combinations

Let ${A}=\{{a}, {b}, {c}\}$ and ${B}=\{1,2,3,4\}$ Then the number of elements in the set ${C}=\{{f}: {A} \to {B} \|2 \in {f}({A})$ and ${f}$ is not one-one $\}$ is ________.

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📖 Explanation

${C}=\{{f}: {A} \to {B} \|2 \in {f}({A})$ and f is not one-one} Case-I : If $f(x) = 2 ∀ x\in A$ then number of function = 1 Case-II : If f(x) = 2 for exactly two elements then total number of many-one function $={ }^{3} {C}_{2}{ }^{3} {C}_{1}=9$ Case-III : If f(x) = 2 for exactly one element then total number of many-one functions $={ }^{3} {C}_{1}{ }^{3} {C}_{1}=9$ Total = 19

Q48JEE Main 2020NAT4MBinomial Theorem

The coefficient of $x^{4}$ in the expansion of $(1+x+x^{2}+x^{3})^{6}$ in powers of $x,$ is _____.

🚀 Solve in Practice Mode

📖 Explanation

$(1+x+x^{2}+x^{3})^{6}=((1+x)(1+x^{2}))^{6}$ $=(1+x)^{6}(1+x^{2})^{6}$ $=\sum _{r=0} ^{6}{ }^{6} C_{r} x^{r} \sum _{r=0} ^{6}{ }^{6} C_{t} x^{2 t}$ $=\sum _{r=0} ^{6}\ \sum _{t=0}^{6}{ } ^{6} C_{r}{ }^{6} C_{t} x^{r+2 t}$ For coefficient of $x^{4} \Rightarrowr+2 t=4$ r t 0 2 2 1 4 0Coefficient of $x^{4}$ $={ }^{6} {C}_{0}{ }^{6} {C}_{2}+{ }^{6} {C}_{2}{ }^{6} {C}_{1}+{ }^{6} {C}_{4}{ }^{6} {C}_{0}$ $=120$

Q49JEE Main 2020NAT4MApplication of Derivatives

If the lines $x+y=a$ and $x-y=b$ touch the curve ${y}={x}^{2}-3 {x}+2$ at the points where the curve intersects the $x$ -axis, then $\frac{a}{b}$ is equal to _____.

🚀 Solve in Practice Mode

📖 Explanation

$y=x^{2}-3 x+2$ At $x$ -axis $y=0=x^{2}-3 x+2$ $x=1,2$ $\frac{d y}{d x}=2 x-3$ $A(1,0) B(2,0)$ $(\frac{{dy}}{{dx}})_{{x}=1}=-1$ and $(\frac{{dy}}{{dx}})_{{x}=2}=1$ $\# {x}+{y}={a} \Rightarrow \frac{{dy}}{{dx}}=-1$ So ${A}(1,0)$ lies on it $\Rightarrow1+0={a} \Rightarrow [{{a}=1}]$ $\# {x}-{y}={b} \Rightarrow\frac{{dy}}{{dx}}=1$ So ${B}(2,0)$ lies on it $2-0={b} \Rightarrow [{b}=2]$ $\ \frac{a}{b}=0.50$

Q50JEE Main 2020NAT4MVector Algebra

Let the vectors ${\vec{a}}, {\vec{b}}, {\vec{c}}$ be such that $|{\vec{a}}|=2,| {\vec{b}}|=4$ and $|\vec{c}|=4 .$ If the projection of ${\vec{b}}$ on ${\vec{a}}$ is equal to the projection of ${{\vec{c}}}$ on ${\vec{a}}$ and ${\vec{b}}$ is perpendicular to ${\vec{c}},$ then the value of $|{\vec{a}}+{\vec{b}}-{\vec{c}}|$ is ______.

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📖 Explanation

Projection of $\vec{b}$ on $\vec{a}=$ projection of $\vec{c}$ on $\vec{a}$ \Rightarrow $\frac{b^{\to }.a^{\to }}{|a^{\to }|}=\frac{c^{\to }.a^{\to }}{|a^{\to }| }$ $\Rightarrow b^{\to }.a^{\to }=c^{\to }.a^{\to }$ $\because \vec{b}$ is perpendicular to $\vec{c}\Rightarrow \vec{b}.\vec{c}=0$ Let $|\vec{a}+\vec{b}-\vec{c}|=k$ Square both sides ${k}^{2}=\vec{a}^{2}+\vec{b}^{2}+\vec{c}^{2}+2 \vec{a} {.\vec{b}}$ $-2\vec{a} .\vec{c}-2 \vec{b} .\vec{c}$ ${k}^{2}={{a} ^{\to }}^{2}+{{ b}^{\to }}^{2}+\vec{c}^{2}=36$ ${k}=6=|\vec{a}+\vec{b}-\vec{c}|$

PHYSICS25 questions

Q51JEE Main 2020MCQ4MCurrent Electricity

PHYSICS In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point B with respect to the point $A$ is :

🚀 Solve in Practice Mode

📖 Explanation

Let us asssume the potential at ${A}={V}_{ {A}}=0$ Now at junction C, According to KCL ${i}_{1}+ {i}_{3}= {i}_{2}$ $1 { A}+ {i}_{3}=2 { A}$ ${i}_{3}=2 { A}$ Now Analyse potential along ACDB $\ {v}_{ {A}}+1+ {i}_{3}(2)-2= {v}_{ {B}}$ $0+1+2(1)-2= {v}_{ {B}}$ ${v}_{ {B}}=3-2$ ${v}_{ {B}}=1$ Amp

Q52JEE Main 2020MCQ4MElectrostatic Potential and Capacitance

A parallel plate capacitor has plate of length 'l', width 'w' and separation of plates is 'd'. It is connected to a battery of emf ${V}$ . A dielectric slab of the same thickness 'd' and of dielectric constant ${k}=4$ is being inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored?

🚀 Solve in Practice Mode

📖 Explanation

Before inserting slab ${C}_{ {i}}= \frac{\epsilon _{0} {A}}{ {d}}$ ${C}_{ {i}}= \frac{\epsilon _{0} l{w}}{ {d}}$ After inserting dielectric slab ${C}_{ {f}}= {C}_{1}+ {C}_{2}$ ${C}_{ {f}}=\frac{{K} \epsilon _{0} { A}_{1}}{ {d}}+ \frac{\epsilon _{0} { A}_{2}}{ {d}}$ ${C}_{ {f}}=\frac{{K} \epsilon _{0} { wx}}{ {d}}+ \frac{\epsilon _{0} {w}(l- {x})}{ {d}}$ ${C}_{ {f}}=2 {C}_{ {i}} \Rightarrow \frac{ {K} \epsilon _{0} {wx}}{{x}}+\ \frac{\epsilon _{0} {w}(l - {x})}{ {d}}$ $= \frac{2 \epsilon _{0}l{w}}{ {d}}$ $4 x+l-x=2l$ $[x= \frac{l}{3}]$

Q53JEE Main 2020MCQ4MElectrostatics

Ten charges are placed on the circumference of a circle of radius ${R}$ with constant angular separation between successive charges. Alternate charges 1,3,5,7,9 have charge $(+ {q})$ each, while 2,4,6,8,10 have charge $(-q)$ each. The potential ${V}$ and the electric field ${E}$ at the centre of the circle are respectively: (Take ${V}=0$ at infinity)

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📖 Explanation

Potential of centre $= {V}=$ $∑(\frac{{kq}}{ {R}} )$ ${V}_{ {C}}=\frac{{K}(\sum {q})}{\ {R}}$ ${V}_{{C}}=\frac{{K}(0)}{ {R}}=0$ Electric field at centre ${\vec{E}}_{ {B}}=\sum {\vec{E}}$ Let $\mathrm{E}$ be electric field produced by each charge at the centre, then resultant electric field will be ${E}_{ {C}}=0,$ since equal electric field vectors are acting at equal angle so their resultant is equal to zero.

Q54JEE Main 2020MCQ4MMagnetism and Matter

An iron rod of volume $10^{-3} { m}^{3}$ and relative permeability 1000 is placed as core in a solenoid with 10 turns/cm. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be :

🚀 Solve in Practice Mode

📖 Explanation

${M}=\mu_{ {r}} {NiA}$ Here $\mu_{ {r}}=$ Relative permeability ${N}= {\text{No}} .$ of turns ${i}= {\text{Current}}$ ${A}=$ Aea of cross section ${M}=\mu_{ {r}} {NiA}=\mu_{ {r}} {n} l{i} {A}$ ${M}=\mu_{ {r}} {niV}$ $=1000(1000) 0.5 (10^{-3} )$ $=500=5\times 10^{2} \ {Am}^{2}$

Q55JEE Main 2020MCQ4MElectromagnetic Induction

An infinitely long straight wire carrying current I, one side opened rectangular loop and a conductor ${C}$ with a sliding connector are located in the same plane, as shown in the figure. The connector has length $l$ and resistance ${R}$ . It slides to the right with a velocity v. The resistance of the conductor and the self inductance of the loop are negligible. The induced current in the loop, as a function of separation ${r},$ between the connector and the straight wire is :

🚀 Solve in Practice Mode

📖 Explanation

${B}= \frac{\mu_{0} {i}}{2 \pi {r}}$ $\phi = \frac{\mu_{0} {i}}{2 \pi {r}}l{dr}$ $\Rightarrow \frac{{d} \phi }{ {dt}}= \frac{\mu_{0} {i}l}{2 \pi {r}} .\frac{{dr}}{{dt}}$ $\Rightarrow {e}= \frac{\mu_{0}}{2 \pi } .\frac{{iv}l}{ {r}}$ ${i}=\frac{{e}}{ {R}}= \frac{\mu_{0}}{2 \pi }.\frac{{iv} l}{ {Rr}}$

Q56JEE Main 2020MCQ4MGravitation

The acceleration due to gravity on the earth's surface at the poles is $\ {g}$ and angular velocity of the earth about the axis passing through the pole is $\omega .$ An object is weighed at the equator and at a height h above the poles by using a spring balance. If the weights are found to be same, then ${h}$ is $:({h} < < {R},$ where ${R}$ is the radius of the earth)

🚀 Solve in Practice Mode

📖 Explanation

${g}_{ {e}}= {g}- {R} \omega ^{2}$ ${g}_{2}= { g} (1- \frac{2 {h}}{ {R}} )$ ${g}_{2} ={ g}- \frac{2 {gh}}{ {R}}$ Now ${R} \omega ^{2}= \frac{2 {gh}}{ {R}}$ $h=\frac{{R}^{2} \omega ^{2}}{2 { g}}$

Q57JEE Main 2020MCQ4MWaves

Two coherent sources of sound, ${S}_{1}$ and ${S}_{2}$ , produce sound waves of the same wavelength, $\lambda=1 { m},$ in phase. ${S}_{1}$ and ${S}_{2}$ are placed $1.5 { m}$ apart (see fig.) A listener, located at L, directly in front of ${S}_{2}$ finds that the intensity is at a minimum when he is $2 { m}$ away from ${S}_{2}$ . The listener moves away from ${S}_{1}$ , keeping his distance from ${S}_{2}$ fixed. The adjacent maximum of intensity is observed when the listener is at a distance ${d}$ from ${S}_{1} .$ Then, ${d}$ is :

🚀 Solve in Practice Mode

📖 Explanation

Initially ${S}_{2} { L}=2 {m}$ ${S}_{1} {L}=\\sqrt{2^{2}+(3 / 2)^{2}}$ ${S}_{1}{L}= \frac{5}{2}=2.5{m}$ $\Delta{x}= +{S}_{1} { L}- {S}_{2} { L}=0.5 {m}$ So since $\lambda=1{m}\;\;\therefore \Delta {x}= \frac{\lambda}{2}$ So while listener moves away from ${S}_{1}$ Then, $\Delta{x} (= {S}_{1} { L}- {S}_{2}{L} )$ increases and hence, at $\Delta{x}=\lambda$ first maxima will appear. $\Deltax=\lambda=S_{1} L-S_{2} L$ $1= {d}-2 \Rightarrow {d}=3{m}$

Q58JEE Main 2020MCQ4MWaves

A driver in a car, approaching a vertical wall notices that the frequency of his car horn, has changed from $440 {Hz}$ to $480 \ {Hz},$ when it gets reflected from the wall. If the speed of sound in air is $345 {m} /{s},$ then the speed of the car is

🚀 Solve in Practice Mode

📖 Explanation

${f}_{1}=$ frequency heard by wall $= {f}_{ {s}}=({{v}_{ {s}}}/{ {v}_{ {s}}- {v}_{ {c}}} )$ ${f}_{2}=$ frequency heard by driver after reflection from wall ${f}_{2}=({{v}_{ {s}}+ {v}_{ {c}}}/{ {v}_{ {s}}} ) {f}_{1}$ $=({{v}_{ {s}}+ {v}_{ {c}}}/{ {v}_{ {s}}- {v}_{ {c}}} ) {f}_{0}$ $\frac{{f}_{2}}{ {f}_{0}}={{v}_{ {s}}- {v}_{ {c}}}/{ {v}_{ {s}}+ {v}_{ {c}}}$ $\frac{48}{44}= {{v}_{ {s}}- {v}_{ {c}}}/{ {v}_{ {s}}+ {v}_{ {c}}}$ $12({v}_{ {s}}+ {v}_{{c}} )=11({v}_{ {s}}- {v}_{ {c}} )$ $23 {v}_{ {c}}= {v}_{ {s}}$ ${v}_{ {c}}={{v}_{ {s}}}/{23}= \frac{345}{23}=15 { m} / { s}$ $= \frac{15 \\times 18}{5}=54{km} / {hr}$

Q59JEE Main 2020MCQ4MThermodynamics

In an adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The final pressure of the gas is found to be ${n}$ times the initial pressure. The value of ${n}$ is:

🚀 Solve in Practice Mode

📖 Explanation

In adiabatic process ${PV}^{\gamma }=$ constant ${P} (\frac{{m}}{\rho } )^{\gamma }= {\text{constant}}$ as mass is constant ${P} \propto\rho ^ \gamma$ $\frac{P_{f}}{P_{i}}=(\frac{\rho _{f}}{\rho _{i}} )^{\gamma }=(32)^{7 / 5}$ $=2^{7}=128$

Q60JEE Main 2020MCQ4MAtoms and Nuclei

A radioactive nucleus decays by two different processes. The half life for the first process is $10 {s}$ and that for the second is $100 { s}$ . the effective half life of the nucleus is close to:

🚀 Solve in Practice Mode

📖 Explanation

${1}/{ {T}_{\text{eff }}}= \frac{1}{ {T}_{1}}+\frac{1}{ {T}_{2}}$ ${T}_{ {\text{eff}}}=\frac{{T}_{1} {T}_{2}}{ {T}_{1}+ {T}_{2}}$ $= \frac{1000}{110}= \frac{100}{11}=9.09$ ${T}_{ {\text{eff}}} \cong 9$

Q61JEE Main 2020MCQ4MOscillations

A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with a time period ${T}_{1}$ and, (ii) back and forth in a direction perpendicular to its plane, with a period ${T}_{2}$ . the ratio $\frac{ {T}_{1}}{ {T}_{2}}$ will be :

🚀 Solve in Practice Mode

📖 Explanation

Moment of inertia in case (i) is I $_{1}$ Moment of inertia in case (ii) is ${I}_{2}$ ${I}_{1}=2 {MR}^{2}$ ${I}_{2}= \frac{3}{2} {MR}^{2}$ ${T}_{1}=2 \pi \sqrt{\frac{{I}_{1}}{ {Mgd}}} ;$ ${T}_{2}=2 \pi \sqrt{\frac{{I}_{2}}{ {Mgd}}}$ ${{T}_{1}}{ {T}_{2}}=\sqrt{\frac{{I}_{1}}{ {I}_{2}}}$ $=\sqrt{\frac{2 {MR}^{2}}{ \frac{3}{2} {MR}^{2}}}= \frac{2}{\\sqrt{3}}$

Q62JEE Main 2020MCQ4MElectrostatic Potential and Capacitance

In the circuit shown, charge on the $5 \mu{F}$ capacitor is :

🚀 Solve in Practice Mode

📖 Explanation

Let potential of point ${O}\;\;\;\; {v}_{0}=0$ Now, using junction analysis We can say, $q_{1}+q_{2}+q_{3}=0$ $2(x-6)+4(x-6)+5(x)=0$ ${x}= \frac{36}{11} \;\;\;{q}_{3}= \frac{36(5)}{11}= \frac{180}{11}$ ${q}_{3}=16.36 \mu{C}$

Q63JEE Main 2020MCQ4MProperties of Matter

In an experiment to verify Stokes law, a small spherical ball of radius ${r}$ and density $\rho$ falls under gravity through a distance ${h}$ in air before entering a tank of water. If the terminal velocity of the ball inside water is same as its velocity just before entering the water surface, then the value of h is proportional to : (ignore viscosity of air)

🚀 Solve in Practice Mode

📖 Explanation

After falling through h, the velocity be equal to terminal velocity $\sqrt{2 {gh}}= \frac{2}{9} \frac{{r}^{2} { g}}{\eta } (\rho _{l}-\rho )$ $\Rightarrow {h}= \frac{2}{81} \frac{{r}^{4} { g} (\rho _{l}-\rho )^{2}}{\eta ^{2}}$ $\Rightarrow {h} \propto{r}^{4}$

Q64JEE Main 2020MCQ4MProperties of Matter

Two different wires having lengths $L_{1}$ and $L_{2}$ , and respective temperature coefficient of linear expansion $\alpha _{1}$ and $\alpha _{2},$ are joined end-to-end. Then the effective temperature coefficient of linear expansion is :

🚀 Solve in Practice Mode

📖 Explanation

At ${T}°{C} \;\;\;{L}= {L}_{1}+ { L}_{2}$ At ${T}+\Delta{T} \;\;\;{L}'_{ {\text{eq}}}= {L}'_{1}+ { L}'+_{2}$ where $L_{1}'=L_{1} (1+\alpha _{1} \DeltaT )$ ${L}_{2}'= {L}_{2} (1+\alpha _{2} {\DeltaT} )$ ${L}_{{\text{eq}}}'=({L}_{1}+{L}_{2} ) (1+\alpha _{ {\text{avg}}} \Delta{T} )$ $\Rightarrow ({L}_{1}+{L}_{2}$ (1+\alpha { {\text{avg}}} {ST} )= {L}{1}+{L}{2}+{L}{1} \alpha {1} \Delta{T}+ {L}{2} \alpha {2} \Delta{T}\Rightarrow ({L}{1}+{L}_{2} ) \alpha { {\text{avg}}}= {L}{1} \alpha {1}+{L}{2} \alpha {2}\Rightarrow \alpha { {\text{avg}}}=\frac{{L}{1} \alpha {1}+{L}{2} \alpha {2}}{{L}{1}+{L}{2}}$

Q65JEE Main 2020MCQ4MUnits and Measurements

The quantities $x= {1}/{\\sqrt{\mu_{0} \in _{0}}}, y= \frac{E}{B}$ and ${z}= \frac{l}{ {CR}}$ are defined where ${C}$ -capacitance ${R}$ -Resistance, $l$ -length, $E$ -Electric field, ${B}$ -magnetic field and $\in _{0}, \mu_{0},$ -free space permittivity and permeability respectively. Then :

🚀 Solve in Practice Mode

📖 Explanation

${x}= {1}/{\\sqrt{\mu_{0} \in _{0}}}=$ speed $\Rightarrow [{x}]=[{L}^{1} {T}^{-1} ]$ ${y}=\frac{{E}}{ {B}}=$ speed $\Rightarrow [{y}]=[{L}^{1} {T}^{-1} ]$ ${z}= \frac{l}{ {RC}}= \frac{l}{\tau }$ $\Rightarrow [{z}]=[{L}^{1} {T}^{-1} ]$ So, ${x}, {y}, {z}$ all have the same dimensions.

Q66JEE Main 2020MCQ4MCurrent Electricity

A galvanometer is used in laboratory for detecting the null point in electrical experiments. If, on passing a current of $6 {mA}$ it produces a deflection of $2°,$ its figure of merit is close to :

🚀 Solve in Practice Mode

📖 Explanation

Figure of Merit $= {C}=\frac{{i}}{\theta }$ $= {C}= \frac{6 \\times 10^{-3}}{2}$ $=3 \\times 10^{-3} {Am}^{2}$

Q67JEE Main 2020MCQ4MMotion in a Straight Line

The velocity (v) and time (t) graph of a body in a straight line motion is shown in the figure. The point $S$ is at 4.333 seconds. The total distance covered by the body in 6 s is :

🚀 Solve in Practice Mode

📖 Explanation

${OS}=4+ \frac{1}{3}= \frac{13}{3}$ ${SD}=2- \frac{1}{3}= \frac{5}{3}$ Area of OABS is A $_{1}$ Area of ${SCD}$ is ${A}_{2}$ Distance $=|{A}_{1} |+|{A}_{2} |$ ${A}_{1}= \frac{1}{2}|\frac{13}{3}+1 ] 4=\frac{32}{3}$ ${A}_{2}= \frac{1}{2} \times \frac{5}{3} \times 2= \frac{5}{3}$ Distance $=|{A}_{1} |+|{A}_{2} |$ $= \frac{32}{3}+ \frac{5}{3}$ $= \frac{37}{3}$

Q68JEE Main 2020MCQ4MLaws of Motion

A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate $\frac{{dM}( {t})}{ {dt}}= {bv}^{2}( {t}),$ where ${v}( {t})$ is its instantaneous velocity. The instantaneous acceleration of the satellite is:

🚀 Solve in Practice Mode

📖 Explanation

$\frac{{dm}( {t})}{ {dt}}= {bv}^{2}$ ${F}_{\text{thast }}= \frac{v}{{dm}}/{ {dt}}$ Force on statellile $=-{\vec{v} }\frac{{dm}( {t})}{ {dt}}$ ${M}( {t}) {a}=- {v}({bv}^{2} )$ $[ {a}= {a} \frac{{bv}^{3}}{ {M}( {t})}]$

Q69JEE Main 2020MCQ4MSemiconductor Electronics

Two Zener diodes (A and B) having breakdownvoltages of $6 {V}$ and $4 {V}$ respectively, are connected as shown in athe circuit below. Theoutput voltage ${V}_{0}$ variation with input voltage linearly increasing with time, is given by : $({V}_{\text{input }}=0 {V}$ at ${t}=0$ ) (figures are qualitative)

🚀 Solve in Practice Mode

📖 Explanation

Till input voltage Reaches $4 {V}$ No zener is in Breakdown Region So ${V}_{0}= {V}_{ {i}}$ Then Now when ${V}_{ {i}}$ changes between $4 {V}$ to $6 {V}$ One Zener with $4 {V}$ will Breakdown are P.D. across This zener will become constant and Remaining Potential will drop. acro Resistance in series with $4 {V}$ Zener. Now current in circuit increases Abruptly and source must have an internal resistance due to which. Some potential will get drop across the source also so correct graph between ${V}_{0}$ and t. will beWe have to Assume some resistance in series with source.

Q70JEE Main 2020MCQ4MElectromagnetic Waves

The correct match between the entries in column I and column II are : I II Radiation Wavelength (a) Microwave (i) $100{m}$ (b) Gamma rays (ii) $10^{-15} {m}$ (c) A.M. radio waves (iii) $10^{-10} {m}$ (d) X-rays (iv) $10^{-3} {m}$

🚀 Solve in Practice Mode

📖 Explanation

Energes of given Radiation can have The following relation ${E}_{\gamma -\text{ Rays }}> {E}_{ {X}-\text{ Rays }}> {E}_{\text{microwave }}$ $> {E}_{\text{AM Radiowaves}}$ ∴ ${\lambda}_{\gamma -\text{ Rays }}> {\lambda}_{ {X}-\text{ Rays }}> {\lambda}_{\text{microwave }}$ $> {\lambda}_{\text{AM Radiowaves}}$ According To tres. (a) Microwave $\to 10^{-3} {m}$ (iv) (b) Gamma Rays $\to 10^{-15} {m}$ (ii) (c) AM Radio wave $\to 100 {m}$ (i) (d) X-Rays $\to 10^{-10}{m}$ (iii)

Q71JEE Main 2020NAT4MDual Nature of Matter and Radiation

The surface of a metal is illuminated alternately with photons of energies ${E}_{1}=4 {eV}$ and ${E}_{2}=2.5 {eV}$ respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is $2 .$ The work function of the metal in $( {eV})$ is ______.

🚀 Solve in Practice Mode

📖 Explanation

${E}_{1}=\phi + {K}_{1}$ ....(1) ${E}_{2}=\phi + {K}_{2}$ ......(2) ${E}_{1}- {E}_{2}= { K}_{1}- { K}_{2}$ Now $\frac{{V}_{1}}{ {V}_{2}}=2$ $\frac{{K}_{1}}{{K}_{2}}=4$ ${K}_{1}=4 {K}_{2}$ Now from equation (2) $\Rightarrow 4-2.5=4 {K}_{2}-{K}_{2}$ $1.5=3{K}_{2}$ ${K}_{2}=0.5 {eV}$ Now putting This Value in equation (2) $2.5=\phi +0.5 {eV}$ $\phi =2 {ev}$

Q72JEE Main 2020NAT4MThermodynamics

Nitrogen gas is at $300°{C}$ temperature. The temperature (in ${K}$ ) at which the ${\text{rms}}$ speed of a ${H}_{2}$ molecule would be equal to the rms speed of a nitrogen molecule, is _____. (Molar mass of ${N}_{2}$ gas $28{g}$ )

🚀 Solve in Practice Mode

📖 Explanation

${V}_{ {\text{rms}}} =\\sqrt{ \frac{3 {RT}}{ {M}}}$ ${V}_{{N}_{2}} = {V}_{ {H}_{2}}$ $\sqrt{{3 {RT}_{ {N}_{2}}}/{ {M}_{ {N}_{2}}}}=\sqrt{{3 {RT}_{ {H}_{2}}}/{ {M}_{ {H}_{2}}}}$ $\frac{573}{28}={{T}_{ {H}_{2}}}/{2}$ $\Rightarrow {T}_{ {H}_{2}}=40.928$

Q73JEE Main 2020NAT4MRotational Motion

A thin rod of mass $0.9 {kg}$ and length $1{m}$ is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of move $0.1 {kg}$ moving in a straight line with velocity $80 {m} / {s}$ hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in rad/s) of the rod immediately after the collision will be ______.

🚀 Solve in Practice Mode

📖 Explanation

${\vec{L}}_{ {i}}={\vec{L}}_{ {f}}$ ${mvL}= {I} \omega$ ${mvL}=(\frac{{ML}^{2}}{3}+{mL}^{2}) \omega$ $0.1 \\times 80 \\times 1$ $=(\frac{0.9 \\times 1^{2}}{3}+0.1 \\times 1^{2} ) \omega$ $8=(\frac{3}{10}+ \frac{1}{10} ) \omega$ $8= \frac{4}{10} \omega$ $\omega =20 {\text{rad}} {{\text{rad}}}/{{\sec}}$

Q74JEE Main 2020NAT4MWork, Power and Energy

A body of mass $2{kg}$ is driven by an engine delivering a constant power $1{J} / {s}$ . The body starts from rest and moves in a straight line. After 9 seconds, the body has moved a distance (in m)______.

🚀 Solve in Practice Mode

📖 Explanation

${P}= {mav}$ ${m} \frac{{dv}}{ {dt}} {v}= {P}$ $\int _{0} ^{v} v d v= \frac{P}{m} \int _{0} ^{t} d t$ $\frac{{v}^{2}}{2}=\frac{{Pt}}{ {m}}\Rightarrow {v}=(\frac{2 {Pt}}{ {m}} )^{1 / 2}$ $\frac{d x}{d t}=\\sqrt{ \frac{2 P}{m}} t^{1 / 2}$ $∫_{0} ^{x} d x=\sqrt{ \frac{2 P}{m}} ∫_{0} ^{t} t^{1 / 2} d t$ ${x}=\sqrt{\frac{2 {P}}{{m}}}\frac{{t}^{3/ 2}}{3 / 2}$ $=\sqrt{\frac{2 {P}}{ {m}}}\times \frac{2}{3} {t}^{3 / 2}$ $=\sqrt{\frac{2 \\times 1}{2}}\times {2}{3} \\times 9^{3 / 2}$ $= \frac{2}{3} \\times 27=18$

Q75JEE Main 2020NAT4MRay Optics and Optical Instruments

A prism of angle $A=1°$ has a refractive index $\mu=1.5 .$ A good estimate for the minimum angle of deviation (in degrees) is close to ${N} /$ 10. Value of ${N}$ is _______.

🚀 Solve in Practice Mode

📖 Explanation

$\delta_{\min }=(\mu-1)A$ $=(1.5-1) 1$ $=0.5$ $\delta_{\min }= \frac{5}{10}$ ${N}=5$

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