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JEE MAIN 2020 - 03 SEP S1 2020

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CHEMISTRY25 questions

Q1JEE Main 2020MCQ4MCoordination Compounds

The complex that can show optical activity is:

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Q2JEE Main 2020MCQ4MAlcohols, Phenols and Ethers

An organic compound [A], molecular formula ${C}_{10} {H}_{20} {O}_{2}$ was hydrolyzed with dilute sulphuricacid to give a carboxylic acid [B] and alcohol[C]. Oxidation of [C] with ${CrO}_{3}-{H}_{2} {SO}_{4}$ produced [B]. Which of the following structuresare not possible for [A]?

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Q3JEE Main 2020MCQ4MChemical Bonding and Molecular Structure

If the boiling point of $H_2O$ is 373 K, the boiling point of $H_2S$ will be :

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📖 Explanation

Boiling point of ${H}_{2} {S} <$ Boiling point of ${H}_{2} {O}$ $(213 K)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(373 K)$

Q4JEE Main 2020MCQ4MChemical Bonding and Molecular Structure

In a molecule of pyrophosphoric acid, thenumber of ${P}- {OH}, {P}= {O}$ and ${P}- {O}- {P}$ bonds $/$ moiety(ies) respectivey are:

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Q5JEE Main 2020MCQ4MChemical Kinetics

It is true that:

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Q6JEE Main 2020MCQ4MOrganic Chemistry - Some Basic Principles

Which of the following compounds produces an optically inactive compound on hydrogenation ?

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Q7JEE Main 2020MCQ4MSolutions

Henry's constant (in kbar) for four gases $\alpha , \beta$ $\gamma$ and $\delta$ in water at $298 \ {K}$ is given below: α β γ δ $K_H$ 50 2 $2\times 10^{-5}$ 0.5(density of water $=10^{3} \ {kg} \ {m}^{-3}$ at $298 \ {K}$ ) This table implies that :

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📖 Explanation

(1) ${P}_{\gamma }= {K}_{ {H}} {X}_{ {Y}}$ ${P}_{\gamma }=2 \\times 10^{-15} \\times \frac{55.5}{55.5+ \frac{1000}{18}}$ $=2 \\times 10^{-5} \ {K} \ {bar}$ $=2 \\times 10^{-2}$ bar (2) ${P}_{\delta}={K}_{ {H}} {X}_{\delta}$ ${P}_{\delta}=0.5\times \frac{55.5}{55.5+ \frac{1000}{18}}$ $=.249 {K} \ {bar}=249 \ {bar}$ (3) On increasing temperature solubility of gases decreases (4) $\ {K}_{\ {H}} \downarrow$ solubility $\uparrow$ and lowest $\ {K}_{ {H}}$ is for $\gamma$

Q8JEE Main 2020MCQ4MSolutions

Tyndall effect of observed when :

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Q9JEE Main 2020MCQ4Mp Block Elements

Thermal power plants can lead to :

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Q10JEE Main 2020MCQ4MCoordination Compounds

The electronic spectrum of $[Ti(H_2O)_6]^{3+}$ shows a single broad peak with a maximum at 20,300 cm $^{-1}$ . The crystal field stabilization energy (CFSE) of the complex ion, in kJ mol $^{-1}$ , is :

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📖 Explanation

$\ {CFSE}=0.4 \Delta_{0}$ $=0.4 \times \frac{20300}{83.7}$ $=97 {kJ} / {mol}$

Q11JEE Main 2020MCQ4MRedox Reactions

Aqua regia is used for dissolving noble metals (Au, Pt, etc). The gas evolved in this process is :

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📖 Explanation

$Au + HNO_3 + 4HCl \to HAuCl _4$ $+ NO + 2H _2 O$

Q12JEE Main 2020MCQ4MOrganic Chemistry - Some Basic Principles

The Kjeldahl method of Nitrogen estimation fails for which of the following reaction products ?

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📖 Explanation

Kjeldahl method is used for $\ {N}$ estimation Butnot given by 'Diazo' compounds

Q13JEE Main 2020MCQ4MHaloalkanes and Haloarenes

The mechanism of ${S}_{ {N}}^{1}$ reaction is given as : ${R}- {X} \to {{R}^{\oplus } {X}^{\ominus }}_{\text{Ion}\text{pair}}\to {{R}^{\oplus }∥ {X}^{\ominus } }_{\text{Solvent}\text{separated} \text{ion}\text{pair}}\to ^{{Y}^{\ominus }} {R}- {Y}+ {X}^{ {\ominus }}$ A student writes general characteristics based on the given mechanism as : (a) The reaction is favoured by weak nucleophiles (b) ${R}^{\oplus }$ would be easily formed if the substituents are bulky (c) The reaction is accompained by recemization (d) The reaction is favoured by non-polar solvents. Which observations are correct?

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📖 Explanation

${S}_{{N}}^{1}$ favours (a) The reaction is favoured by weak nucleophiles (b) ${R} ^{⊕}$ would be easily formed if the substituents are bulky (c) The reaction is accompained by recemization

Q14JEE Main 2020MCQ4MOrganic Chemistry - Some Basic Principles

Which one of the following compounds possesses the most acidic hydrogen ?

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Q15JEE Main 2020MCQ4MPractical Organic Chemistry

Glycerol is separated in soap industries by :

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Q16JEE Main 2020MCQ4MChemical Bonding and Molecular Structure

Of the species, NO, NO $^{ +}$ , NO $^{ 2+}$ , NO $^{ -}$ , the one with minimum bond strength is :

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📖 Explanation

Bond order of $\ {NO}^{2+}=2.5$ Bond order of $\ {NO}^{+}=3$ Bond order of $\ {NO}=2.5$ Bond order of $\ {NO}^{-}=2$ Bond order $\alpha$ bond strength.

Q17JEE Main 2020MCQ4MClassification of Elements

The atomic number of the element unnilennium is :

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📖 Explanation

$\text{table} 1,0,9;\text{un},\text{nil},\text{enn}$ Hence correct name→ unnilennium

Q18JEE Main 2020MCQ4MIonic Equilibrium

An acidic buffer is obtained on mixing :

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📖 Explanation

$\text{table} HCl ,+,CH _3COONa,\to , CH _3COOH,+, NaCl;10 \text{mili} \text{mol},,20\text{mili} \text{mol},,-,,-;-,,10\text{mili} \text{mol},,10\text{mili} \text{mol},,10\text{mili} \text{mol}$ So finaly we get mixture of $\ {CH}_{3} {COOH}+ {CH}_{3} {COONa}$ that will work like acidic buffer solution.

Q19JEE Main 2020MCQ4MElectrochemistry

Let ${C}_{ {NaCl}}$ and ${C}_{ {BaSO}_{4}}$ be the conductances (in ${S}$ ) measured for saturated aqueous solutions of ${NaCl}$ and ${BaSO}_{4},$ respectively, at a temperatureT. Which of the following is false?

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📖 Explanation

Dissolution of $\ {BaSO}_{4}$ is an endothermic reaction 50 on increasing temperature number of ions of $\ {BaSO}_{4}$ decrease so it's conduction also decrease.

Q20JEE Main 2020MCQ4MPractical Organic Chemistry

The antifertility drug 'Novestrol" can react with :

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📖 Explanation

gives (1) $Br _2 + H _2O$ test (2) Lucas test with $ZnCl _2 + HCl$ (3) $FeCl_3$ test of phenolic group.

Q21JEE Main 2020NAT4MSolutions

The volume strength of 8.9 M $H_2O_2$ solution calculated at 273 K and 1 atm is _____. (R=0.0821 L atm K $^{-1}$ mol $^{-1}$ ) (rounded off to thenearest integer)

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📖 Explanation

Volume strength of $\ {H}_{2} {O}_{2}$ at 1 atm 273 kelvin $= {M} \\times 11.2$ $=8.9 \\times 11.2=99.68$ Ans: 100

Q22JEE Main 2020NAT4MSolutions

The mole fraction of glucose $({C}_{6} {H}_{12} {O}_{6} )$ in anaqueous binary solution is $0.1 .$ The masspercentage of water in it, to the nearest integer,is _______.

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📖 Explanation

${X}_{{C}_{6} {H}_{12} {O}_{6}}=0.1$ Let total mole is 1 mol then mole of glucose will be 0.1 and mole of water will be 0.9 so mass % of water $= \frac{0.9 \\times 18}{0.1 \\times 180+0.9 \\times 18} \\times 100$ $=47.36$ Ans: 47

Q23JEE Main 2020NAT4MElectrochemistry

The photoelectric current from Na (work function, $w_{0}=2.3 \ {eV}$ ) is stopped by the output voltage of the cell ${Pt}( {s}) | {H}_{2}( {g}, 1 {bar}) | {HCl}( {aq} .{pH}=1)$ $| {AgCl}( {s})| {Ag}( {s})$ The pH of aq. HCl required to stop thephotoelectric current from ${K}({w}_{0}=2.25 {eV} ),$ all other conditions remaining the same, is ______. $\\times 10^{-2}$ (to the nearest integer). Given, $2.303\frac{{RT}}{ {F}}=0.06 {V} ;$ ${E}°_{ {AgCl} |{Ag}| {Cl}^{-}} =0.22 {V}$

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📖 Explanation

$\frac{1}{2} {H}_{2} \to {H}^{+}+ {e}^{\Theta }$ ${e}^{\Theta }+ {AgCl}_{( {s})} \to {Ag}_{( {s})}+ {Cl}^{\Theta }$ $────────────────────────$ $\frac{1}{2} {H}_{2}+ {AgCl}_{( {s})} \to {H}_{( {aq})}^{+}$ $+ {Ag}_{( {s})}+ {Cl}_{( {aq})}^{\Theta }$ ${E}=\epsilon ^{0}- \frac{.06}{1} \log {[{H}^{+}]\$ $[ {Cl}^{\Theta } ]$}/{ {P}{ {H}{2}}^{ \frac{1}{2}}}{E}=0.22-.06 \log {(10^{-1} ) (10^{-1})}/{1^{ \frac{1}{2}}}{E}=0.22+.12=.34 {volt}\Rightarrow $total energy of photon will be (for Na)$ =2.3+0.34=2.64 \ {eV}\Rightarrow $stopping potential required for$ \ {K}=2.64-2.25=0.39 $volt$ \ {E}=\epsilon ^{0}- \frac{.06}{1} \log {[ {H}^{+}]$$[{Cl}^{-} ]$}/{ {P}{ {H}{2}}^{ \frac{1}{2}}} $as$ [{H}^{+} ]=$[{Cl}^{Θ} ]$ {\text{so}}0.39=0.22-.06 \log $[{H}^{+} ]^2/1^{1/2}0.17=+.12 {pH}\ {pH}=1.4166 \Rightarrow1.42$

Q24JEE Main 2020NAT4MSolid State

An element with molar mass $2.7 \\times 10^{-2} \ {kgmol}^{-1}$ forms a cubic unit cell with edge length $405 \ {pm}$ . If its density is $2.7 \\times 10^{3} \ {kgm}^{-3},$ the radius of theelement is approximately ______ $\\times 10^{-12} \ {m}$ (to thenearest integer).

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📖 Explanation

$\ {d}={{z}({{M}}/{ {N}_{ {A}}} )}/{ {a}^{3}}$ $2.7 \\times 10^{3}= {z} {(\frac{2.7 \\times 10^{-2}}{6 \\times 10^{23}} )}/{ (405 \\times 10^{-12}\ )^{3}}$ $2.7 \\times 10^{3}$ $=\ {z}\frac{(2.7 \\times 10^{-2} )}{6 \\times 10^{23} (4.05 \\times 10^{-10} )^{3}}$ $2.7 \\times 10^{3}$ $=\ {z} \frac{(2.7 \\times 10^{-2}\ )}{6 \\times 10^{23} \\times 66.43 \\times 10^{-30}}$ $3.98=z$ ${z} \ \approx 4$ structure is $\ {fcc}$ $\frac{a}{\\sqrt{2}}=2 r$ $r=\ \frac{a}{2 \\sqrt{2}}=\frac{\sqrt{2} a}{4}$ $= \frac{1.414 \\times 405 \\times 10^{-12}}{4}$ $r=143.16 \\times 10^{-12}$

Q25JEE Main 2020NAT4MOrganic Chemistry - Some Basic Principles

The total number of monohalogenated organic products in the following (including stereoisomers) reaction is _____. $A_{(\text{simplest} \text{optically} \text{active} \text{alkene})}\xrightarrow{(i)H_2/Ni/\Delta}_{(ii)X_2/\Delta}$

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MATHEMATICS25 questions

Q26JEE Main 2020MCQ4MProbability

A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of $4 .$ Then the conditional probability that the score 4 has appeared atleast once is :

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📖 Explanation

A : Sum obtained is a multiple of 4 . $\ {A}=\{(1,3),(2,2),(3,1),(2,6),$ $(3,5),(4,4)(5,3),(6,2),(6,6)\}$ B : Score of 4 has appeared at least once. ${B}=\{(1,4),(2,4),(3,4),$ $(4,4),(5,4),(6,4),$ $(4,1),(4,2),(4,3),(4,5),(4,6)\}$ Required probability $= {P}(\frac{{B}}{ {A} })=\frac{{P}( {B} ∩{A})}{ {P}( {A})}$ $=\frac{ 1 / 36}{9 / 36}= \frac{1}{9}$

Q27JEE Main 2020MCQ4MThree Dimensional Geometry

The lines ${\vec{r}}=(\hat{i}-\hat{j})+l(2 \hat{i}+\hat{k})$ and $\vec{r}=(2\hat{i}-\hat{j})+ {m}(\hat{i}+\hat{j}-\hat{k})$

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📖 Explanation

${\vec{r}}=\hat{i}(1+2 l)+\hat{j}(-1)$ $+\hat{k}(l)$ ${\vec{r}}=\hat{i}(2+ {m})+\hat{j}( {m}-1)$ $+\hat{k}(- {m})$ For intersection $1+2l=2+ {m}$ ...... (i) $-1=\ {m}-1$ ...... (ii) $l=-\ {m}$ ...... (iii) from (ii) m = 0 from (iii)l = 0 These values of ${m}$ and $l$ do not satisfyequation (1) Hence the two lines do not intersect for anyvalues of $l$ and ${m}$ .

Q28JEE Main 2020MCQ4MThree Dimensional Geometry

The foot of the perpendicular drawn from thepoint (4,2,3) to the line joining the points(1,-2,3) and (1,1,0) lies on the plane :

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📖 Explanation

Equation of ${AB}={{r}}=(\hat{i}+\hat{j})+\lambda(3 \hat{j}-3 \hat{k})$ Let coordinates of $\ {M}=(1,(1+3 \lambda),-3 \lambda)$ . ${\vec{PM}}=-3 \hat{i}+(3 \lambda-1) \hat{j}$ $-3(\lambda+1) \hat{k}$ ${\vec{AB}}=3\hat{j}-3\hat{k}$ $\because {\vec{PM}} \perp {\vec{AB}} \Rightarrow {\vec{PM}} .{\vec{AB}}=0$ $\Rightarrow3(3 \lambda-1)+9(\lambda+1)=0$ $\Rightarrow \lambda=- \frac{1}{3}$ $\therefore \ {M}=(1,0,1)$ Clearly $\ {M}$ lies on $2 {x}+ {y}- {z}=1$

Q29JEE Main 2020MCQ4MHyperbola

A hyperbola having the transverse axis oflength $\\sqrt{2}$ has the same foci as that of the ellipse $3 x^{2}+4 y^{2}=12,$ then this hyperbola does notpass through which of the following points?

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📖 Explanation

Ellipse $: \frac{x^{2}}{4}+ \frac{y^{2}}{3}=1$ eccentricity $=\sqrt{1- \frac{3}{4}}= \frac{1}{2}$ $\therefore$ foci $=(\±1,0)$ for hyperbola, given $2 \ {a}=\\sqrt{2} \Rightarrow {a}= \frac{1}{\\sqrt{2}}$ $\therefore$ hyperbola will be $\frac{x^{2}}{1 / 2}- \frac{y^{2}}{b^{2}}=1$ eccentricity $=\\sqrt{1+2 {b}^{2}}$ $\therefore$ foci $=(±\sqrt{ \frac{1+2 {b}^{2}}{2}}, 0 )$ Ellipse and hyperbola have same foci $\Rightarrow \ \sqrt{\frac{1+2 b^{2}}{2}}=1$ $\Rightarrow \ b^{2}= \frac{1}{2}$ $\therefore$ Equation of hyperbola $: \frac{{x}^{2}}{1 / 2}-\frac{{y}^{2}}{1 / 2}=1$ $\Rightarrow \ x^{2}-y^{2}= \frac{1}{2}$ Clearly $(\sqrt{\frac{3}{2}}, \frac{1}{\\sqrt{2}} )$ does not lie on it.

Q30JEE Main 2020MCQ4MArea Under The Curves

The area (in sq. units) of the region $\{( {x}, {y}): 0\le {y} \le {x}^{2}+1,$ $0 \le {y}\le {x}+1,$ $\frac{1}{2} \le x \le 2 \}$ is :

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📖 Explanation

$\{( {x}, {y}): 0\le {y} \le {x}^{2}+1,$ $0 \le {y}\le {x}+1,$ $\frac{1}{2} \le x \le 2 \}$ Required area $=∫_{1 / 2} ^{1}({x}^{2}+1 ) {d} {x}+ \frac{1}{2}(2+3) \\times 1$ $= \frac{19}{24}+ \frac{5}{2}= \frac{79}{24}$

Q31JEE Main 2020MCQ4MSequences and Series

If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is :

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📖 Explanation

Sum of 1st 25 terms = sum of its next 15 terms $\Rightarrow ({T}_{1}+.... .+ {T}_{25})$ $=({T}_{26}+.... . .+ {T}_{40} )$ $\Rightarrow ({T}_{1}+.... .+ {T}_{40} )$ $=2 ({T}_{1}+....+ {T}_{25} )$ $\Rightarrow \frac{40}{2}[2 \\times 3+(39 {d})]$ $=2\times \frac{25}{2}[2 \\times 2+24 {d}]$ $\Rightarrow {d}= \frac{1}{6}$

Q32JEE Main 2020MCQ4MParabola

Let $\ {P}$ be a point on the parabola, $\ {y}^{2}=12 \ {x}$ and $\ {N}$ be the foot of the perpendicular drawn from $\mathrm{P}$ on the axis of the parabola. A line is nowdrawn through the mid-point $\ {M}$ of $\ {PN},$ parallelto its axis which meets the parabola at $Q .$ If they-intercept of the line $\ {NQ}$ is $\ \frac{4}{3},$ then :

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📖 Explanation

Let $P= (3 t^{2}, 6 t) ; N= (3 t^{2}, 0 )$ ${M}= (3 {t}^{2}, 3 {t} )$ Equation of ${MQ}: {y}=3 {t}$ $\therefore \ Q=(\frac{3}{4} t^{2}, 3 t )$ Equation of NQ ${y}= {3 {t}}/ {(\frac{3}{4} {t}^{2}-3 {t}^{2} )}({x}-3 {t}^{2} )$ ${y}$ -intercept of ${NQ}=4 {t}= \frac{4}{3}\Rightarrow {t}= \frac{1}{3}$ $\therefore \ {MQ}={9}{4} {t}^{2}=\ \frac{1}{4}$ $\ {PN}=6 \ {t}=2$

Q33JEE Main 2020MCQ4MStatistics

For the frequency distribution: $\text{table} \text{Variate} ( {x}):, {x}_{1} , {x}_{2} ,{x}_{3}..... {x}_{15} ;\text{Frequency}({f}): ,{f}_{1} ,{f}_{2},{f}_{3}....{f}_{15}$ where $0 < \ {x}_{1} < \ {x}_{2} < {x}_{3} < ....$ $< {x}_{15}=10$ and $∑_{i=1}^{ {15}} f_{i}>0,$ the standard deviation cannot be :

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📖 Explanation

$\ \because \sigma ^{2}\le \frac{1}{4}( {M}- {m})^{2}$ Where ${M}$ and ${m}$ are upper and lower boundsof values of any random variable. $\therefore \ \sigma ^{2} < \frac{1}{4}(10-0)^{2}$ $\Rightarrow \ 0 < \sigma <5$ $\therefore \ \sigma \ne 6$

Q34JEE Main 2020MCQ4MDefinite Integrals

$∫_{{-\pi } }^{{\pi }}{|\pi -| {x}|| } {d} {x}$ is equal to :

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📖 Explanation

$∫_{{-\pi } }^{{\pi }}{|\pi -| {x}|| } {d} {x}$ $=∫_{{0} }^{{\pi }}{|\pi -x|}dx$ $=2 ∫_{{0} }^{{\pi }}(\pi - {x}) {d} {x}$ $=2 \$ [\pi {x}-\frac{{x}^{2}}{2} ]_{0}^{\pi }=\pi ^{2}$

Q35JEE Main 2020MCQ4MQuadratic Equation and Inequalities

Consider the two sets: ${A}=\{ {m} \in {R}:$ both the roots of ${x}^{2}-( {m}+1) {x}+ {m}+4=0$ are real $\}$ and ${B}=[-3,5)$ . Which of the following is not true?

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📖 Explanation

${A}: {D} \ge 0$ $\Rightarrow \ ( {m}+1)^{2}-4( {m}+4) \ge 0$ $\Rightarrow {m}^{2}+2 {m}+1-4 {m}-16 \ge 0$ $\Rightarrow {m}^{2}-2 {m}-15 \ge 0$ $\Rightarrow ({m}-5)( {m}+3) \ge 0$ $\Rightarrow {m} \in (-\infty ,-3] \cup [5, \infty )$ $\therefore \ {A}=(-\infty ,-3] \cup [5, \infty )$ ${B}=[-3,5)$ ${A}- {B}=(-\infty ,-3) \cup [5, \infty )$ $A \cap B=\{-3\}$ $B-A=(-3,5)$ $A \cap B=R$

Q36JEE Main 2020MCQ4MDifferentiation

If ${y}^{2}+\log _{ {e}} (\cos ^{2} {x} )={y},$ $\;{x}\in (- \frac{\pi }{2}, \frac{\pi }{2} ),$ then :

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📖 Explanation

$\text{table} y^{2}+\ln (\cos ^{2} x )=y ,x \in (- \frac{\pi }{2}, {\pi }{2} ) ;\text{for} x=0 , y=0 \text{or} 1$ Differentiating wrt $x$ $\Rightarrow \ 2 y y'-2 \tan x=y'$ At $(0,0) \ y'=0$ At $(0,1) \ y'=0$ Differentiating wrt $x$ $2yy" +2({y}')^{2}-2 \sec ^{2} x= {y}"$ At $(0,0) \ {y}"=-2$ At $(0,1) \ {y}"=2$ ∴ $|y" (0) \|=2$

Q37JEE Main 2020MCQ4MApplication of Derivatives

The function, ${f}( {x})=(3 {x}-7) {x}^{2 / 3}, {x}\in {R},$ isincreasing for all x lying in :

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📖 Explanation

$f(x)=(3 x-7) x^{2 / 3}$ $\Rightarrow\ f(x)=3 x^{5 / 3}-7 x^{2 / 3}$ $\Rightarrow f(x)=5 x^{2 / 3}- \frac{14}{3 x^{1 / 3}}$ $= \frac{15 x-14}{3 x^{1 / 3}}>0$ $\therefore \ {f}( {x})>0 \forall$ ${x} \in (-\infty , 0) ∪( \frac{14}{15}, \infty )$

Q38JEE Main 2020MCQ4MSequences and Series

The value of $(2.1 {P}_{0}-3.2 {P}_{1}+4 .^{3} {P}_{2}-...$ up to $51^{ {\text{th} }}$ term $)+ (1 !-2 !+3 !-.....$ up to $51^{ {\text{th} }}$ term $)$ is equal to :

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📖 Explanation

${S}= (2 .{ }^{1} {p}_{0}-3.2 {p}_{1}$ $+4 .{ }^{3} {p}_{2} ....$ upto 51 terms $)$ $+(1 !+2 !+3 !....$ upto 51 terms $)$ $[\because {n} p_{ {n}-1}=\ {n} ! ]$ $\ {S}=(2 \\times 1 !-3 \\times 2 !+4 \$ $\times 3 ! \....+52.51 !)$ $+(1 !-2 !+3 ! ....(51) !)$ $=(2 !-3 !+4 ! .....+52 !)$ $+(1 !-2 !+3 !-4 !+....+(51) !)$ $=1 !+52 !$

Q39JEE Main 2020MCQ4MMatrices and Determinants

If $\Delta$ $=\begin{vmatrix}x-2 & 2 x-3 & 3 x-4 \\ 2 x-3 & 3 x-4 & 4 x-5 \\ 3 x-5 & 5 x-8 & 10 x-17\end{vmatrix}=$ ${Ax}^{3}+ {B} {x}^{2}+ {Cx}+ {D},$ then ${B}+ {C}$ is equal to :

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📖 Explanation

$\Delta$ $=\begin{vmatrix}x-2 & 2 x-3 & 3 x-4 \\ 2 x-3 & 3 x-4 & 4 x-5 \\ 3 x-5 & 5 x-8 & 10 x-17\end{vmatrix}$ ${Ax}^{3}+ {B} {x}^{2}+ {Cx}+ {D} .$ ${R}_{2} \to {R}_{2}- {R}_{1} \;\;\;{R}_{3}\to {R}_{3}- {R}_{2}$ $\Delta$ $=\begin{vmatrix}x-2 & 2 x-3 & 3 x-4 \\ x-1 & x-1 & x-1 \\ x-2 & 2( x-2) & 6( x-2)\end{vmatrix}$ $=(x-1)(x-2)$ $\begin{vmatrix}x-2 & 2x-3 & 3x-4 \\ 1 & 1 & 1 \\ 1 & 2 & 6\end{vmatrix}$ $=-3(x-1)^{2}(x-2)$ $=-3 x^{3}+12 x^{2}-15 x+6$ $\therefore \ B+C=12-15=-3$

Q40JEE Main 2020MCQ4MDifferential Equations

The solution curve of the differential equation, $(1+ {e}^{- {x}} ) (1+ {y}^{2} ) \frac{{dy}}{ {dx}}= {y}^{2},$ which passesthrough the point $(0,1),$ is :

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📖 Explanation

$(1+ {e}^{-x})(1+ {y}^{2} ) \frac{{dy}}{ {dx}}= {y}^{2}$ $\Rightarrow (1+ {y}^{-2} ) {dy}=({{e}^{ {x}}}/{1+ {e}^{ {x}}} ) {d} {x}$ $\Rightarrow({y}- \frac{1}{ {y}} )=\ln (1+ {e}^{{x}} )+ {c}$ $\therefore$ It passes through $(0,1)\Rightarrow {c}=-l{n} 2$ $\Rightarrow \ {y}^{2}=1+ {y} l{n} ({1+ {e}^{ {x}}}/{2} )$

Q41JEE Main 2020MCQ4MBinomial Theorem

If the number of integral terms in the expansionof $(3^{1 / 2}+5^{1 / 8})^{n}$ is exactly $33,$ then the least valueof $\ {n}$ is :

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📖 Explanation

${T}_{ {r}+1}={ }^{ {n}} {C}_{ {r}}(3)^{{ {n}- { {r}}}/{2}}(5)^{\frac{{r}}{8}} \;\;\;( {n}\ge {r})$ Clearly $\ {r}$ should be a multiple of 8 . $\because$ there are exactly 33 integral terms Possible values of $\ {r}$ can be $0,8,16,......, 32 \\times 8$ $\therefore$ least value of $\ {n}=256 .$

Q42JEE Main 2020MCQ4MQuadratic Equation and Inequalities

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

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📖 Explanation

$\alpha , \beta$ are roots of ${x}^{2}+ {px}+2=0$ $\Rightarrow \alpha ^{2}+p \alpha +2=0$ $\ \&\ \beta ^{2}+p \beta +2=0$ $\Rightarrow \frac{1}{\alpha }, \frac{1}{\beta }$ are roots of $2 {x}^{2}+ {px}+1=0$ But $\frac{1}{\alpha }, \frac{1}{\beta }$ are roots of $2 {x}^{2}+2 {qx}+1=0$ $\Rightarrow \ p=2 q$ Also $\alpha +\beta =- {p} \ \;\;\; \alpha \beta =2$ $(\alpha - \frac{1}{\alpha } ) (\beta - \frac{1}{\beta })(\alpha + \frac{1}{\beta } )$ $(\beta + \frac{1}{\alpha } )$ $=(\frac{\alpha ^{2}-1}{\alpha } )(\frac{\beta ^{2}-1}{\beta })$ $(\frac{\alpha \beta +1}{\beta })(\frac{\alpha \beta +1}{\alpha } )$ $= \frac{(- {p} \alpha -3)(- {p} \beta -3)(\alpha \beta +1)^{2}}{(\alpha \beta )^{2}}$ $= \frac{9}{4} ( {p} \alpha \beta +3 {p}(\alpha +\beta )+9)$ $= \frac{9}{4} (9- {p}^{2} )= \frac{9}{4} (9-4 {q}^{2} )$

Q43JEE Main 2020MCQ4MLimits, Continuity and Differentiability

Let $[ {t}]$ denote the greatest integer $\le {t}$ . If for some $\lambda\in {R}-\{0,1\}, \lim _{x \to 0}|\frac{1- {x}+| {x}|}{\lambda- {x}+[ {x}]}|= {L},$ then $\ {L}$ isequal to :

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📖 Explanation

LHL $: \lim _{x \to 0^{-}} | \frac{1-x-x}{\lambda-x-1} |= | \frac{1}{\lambda -1} |$ RHL $: \lim _ {x \to 0^{+}}|\frac{1-x+x}{\lambda-x+1} |=|\frac{1}{\lambda} |$ For existence of limit LHL = RHL $\Rightarrow \ \frac{1}{|\lambda-1|}= \frac{1}{|\lambda|}\Rightarrow \lambda= \frac{1}{2}$ $\therefore \ L= \frac{1}{|\lambda|}=2$

Q44JEE Main 2020MCQ4MInverse Trigonometric Functions

$2 \pi - (\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{16}{65} )$ is equal to:

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📖 Explanation

$2 \pi - (\sin ^{-1} (\frac{4}{5})+\sin ^{-1} (\frac{5}{13})$ $+\sin ^{-1} (\frac{16}{65} ))$ $=2 \pi - (\tan ^{-1}(\frac{4}{3} )+\tan ^{-1}(\frac{5}{12} )$ $+\tan ^{-1}(\frac{16}{63}))$ $=2 \pi - (\tan ^{-1}(\frac{63}{16} )+\tan ^{-1}(\frac{16}{63}))$ $=2 \pi - \frac{\pi }{2}= \frac{3 \pi }{2}$

Q45JEE Main 2020MCQ4MSets and Relations

The proposition ${p} \to \sim ( {p} \ ∧∼ {q})$ is equivalentto:

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📖 Explanation

${p} \to \sim ( {p} \ ∧∼ {q})$ $=\sim {p} \vee ∼({p} \ ∧∼{q})$ $=∼{p} \vee ∼{p} ∨{q}$ $=∼({p} \ ∧ {q}) ∨{q}$ $=∼ {p} ∨{q}$

Q46JEE Main 2020NAT4MMatrices and Determinants

Let ${A}= \begin{bmatrix}{x} & 1 \\ 1 & 0\end{bmatrix}, {x} \in {R}$ and ${A}^{4}=[{a}_{ {ij}}] .$ If $a_{11}=109,$ then $a_{22}$ is equal to _____.

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📖 Explanation

${A}= \begin{bmatrix}{x} & 1 \\ 1 & 0\end{bmatrix}$ ${A}^2= \begin{bmatrix}{x} & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}{x} & 1 \\ 1 & 0\end{bmatrix}$ $=\begin{bmatrix}x^{2+}1 & x \\ x & 1\end{bmatrix}$ $A^4=\begin{bmatrix}x^{2+}1 & x \\ x & 1\end{bmatrix}$ $\begin{bmatrix}x^{2+}1 & x \\ x & 1\end{bmatrix}$ $=\begin{bmatrix}(x^{2}+1 )^{2}+x^{2} & x (x^{2}+1 )+x \\ x (x^{2}+1 )+x & x^{2}+1\end{bmatrix}$ $a_{11}= (x^{2}+1 )^{2}+x^{2}=109$ $\Rightarrowx=\±3$ $a_{22}=x^{2}+1=10$

Q47JEE Main 2020NAT4MLimits, Continuity and Differentiability

If $\lim _{x \to 0} \{ \frac{1}{x^{8}} (1-\cos \frac{x^{2}}{2}-\cos \frac{x^{2}}{4}$ $+\cos \frac{x^{2}}{2} \cos \frac{x^{2}}{4})\}=2^{-k},$ then the value of $\ {k}$ is ________.

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📖 Explanation

$\lim _{x \to 0} \{ \frac{1}{x^{8}} (1-\cos \frac{x^{2}}{2}-\cos \frac{x^{2}}{4}$ $+\cos \frac{x^{2}}{2} \cos \frac{x^{2}}{4})\}=2^{-k}$ $\Rightarrow \lim _{x \to 0} { (1-\cos \frac{x^{2}}{2} )}/{4(\frac{x^{2}}{2} )^{2}} {(1-\cos \frac{x^{2}}{4} )}/{16(\frac{x^{2}}{4} )^{2}}$ $= \frac{1}{8} \times \frac{1}{32}=2^{-k}$ $\Rightarrow2^{-8}=2^{-k} \Rightarrowk=8$ .

Q48JEE Main 2020NAT4MCircles

The diameter of the circle, whose centre lies onthe line $x+y=2$ in the first quadrant and whichtouches both the lines $x=3$ and $y=2,$ is ______.

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📖 Explanation

$\because$ center lies on $\ {x}+ {y}=2$ and in 1 st quadrant center $=(\alpha , 2-\alpha )$ where $\alpha >0$ and $2-\alpha >0 \Rightarrow0 < \alpha < 2$ $\because \$ circle touches $x=3$ and $y=2$ $\Rightarrow3-\alpha |=|2-(2-\alpha )|=$ radius $\Rightarrow 3-\alpha |=|\alpha | \Rightarrow \alpha = \frac{3}{2}$ $\therefore$ radius $=\alpha$ $\Rightarrow$ Diameter $=2 \alpha =3$

Q49JEE Main 2020NAT4MLogarithms

The value of $(0.16)^{\log _{2.5}(\frac{1}{3}+ \frac{1}{3^{2}}+ \frac{1}{3^{3}}+.....\to \infty )}$ is equal to ______.

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📖 Explanation

$(0.16)^{\log _{2.5}(\frac{1}{3}+ \frac{1}{3^{2}}+ \frac{1}{3^{3}}+.....\to \infty )}$ $=(\frac{4}{25} )^{\log _{(\frac{5}{2})}(\frac{1}{2} )}$ $=(\frac{1}{2} )^{\log (\frac{5}{2})(\frac{4}{25} )}$ $=(\frac{1}{2} )^{-2}=4$

Q50JEE Main 2020NAT4MComplex Numbers

If $(\frac{1+i}{1-i} )^{ \frac{m}{2}}= (\frac{1+i}{i-1} )^{ \frac{n}{3}}$ $=1,(m, n \in N)$ then thegreatest common divisor of the least values of $\ {m}$ and $\ {n}$ is ______.

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📖 Explanation

$(\frac{1+i}{1-i} )^{ \frac{m}{2}}= (\frac{1+i}{i-1} )^{ \frac{n}{3}}=1$ $\Rightarrow (\frac{(1+i)^{2}}{2} )^{ {m} / 2}=(\frac{(1+i)^{2}}{-2} )^{ {n} / 3}=1$ $\Rightarrow ({i})^{ {m} / 2}=(- {i})^{ {n} / 3}=1$ $\Rightarrow \ \frac{{m}}{2}=4 {k}_{1}$ and $\frac{{n}}{3}=4 {k}_{2}$ $\Rightarrow \ {m}=8 {k}_{1}$ and ${n}=12 {k}_{2}$ Least value of $\ {m}=8$ and $\ {n}=12$ . $\therefore \ {GCD}=4$

PHYSICS25 questions

Q51JEE Main 2020MCQ4MUnits and Measurements

Using screw gauge of pitch $0.1 \ {cm}$ and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as:

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📖 Explanation

${LC}={{\text{pitch}}}/{ {CSD}}= \frac{0.1 {cm}}{50}$ $=0.002 {cm}$ So any measurement will be integral Multiple of LC. So ans. will be $2.124 \ {cm}$

Q52JEE Main 2020MCQ4MCurrent Electricity

Model a torch battery of length $l$ to be made up of a thin cylindrical bar of radius 'a' and a concentric thin cylindrical shell of radius 'b' filled in between with an electrolyte of resistivity $\rho$ (see figure). If the battery is connected to a resistance of value $\ {R},$ the maximum Joule heating in $\ {R}$ will take place for:-

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📖 Explanation

Maximum power in external resistance is generated when it is equal to internal resistance of battery. $P_{R}=(\frac{\in }{r+R}\ )^{2} R$ $\ {P}_{ {R}}$ is max. when ${r}= {R}$ $\int {dr}=∫_{ {a}} ^{ {b}} \frac{\rho {dr}}{2 \pi {r} l}$ $\Rightarrow {r}= \frac{\rho }{2 \pi l} \ln \frac{{b}}{\ {a}}$

Q53JEE Main 2020MCQ4MDual Nature of Matter and Radiation

When the wavelength of radiation falling on a metal is changed from $500 \ {nm}$ to $200 \ {nm},$ the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to :

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📖 Explanation

$\frac{3}{1}={\frac{{hc}}{200 \ {nm}}-\phi }/{\frac{{h} {c}}{500 \ {nm}}-\phi },$ ${hc}=1240 \ {eV}- {nm}$ On solving $\phi =0.61 \ {eV}$

Q54JEE Main 2020MCQ4MRotational Motion

Moment of inertia of a cylinder of mass M, length $\ {L}$ and radius $\ {R}$ about an axis passing through its centre and perpendicular to the axis of the cylinder is $\ {I}=\ {M}(\frac{{R}^{2}}{4}+\frac{{L}^{2}}{12}\ ).$ If such a cylinder is to be made for a given mass of material, the ratio L/R for it to have minimum possible I is :-

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📖 Explanation

$\ {I}=\ {M}(\frac{{R}^{2}}{4}+\frac{{L}^{2}}{12}\ )$ ......(1) as mass is constant $\Rightarrow {m}=\rho {V}={\text{constant}}$ $\ {V}=$ constant $\pi ^{2} \ {R} l=$ constant $\Rightarrow {R}^{2} \ {L}=$ constant $2 {RL}+ {R}^{2} \frac{{dL}}{\ {dR}}=0$ .....(2) From equation (1) $\frac{{dI}}{ {dR}}= {M}(\frac{2 {R}}{4}+ \frac{2 {L}}{12} \times \frac{{dL}}{ {dr}} )=0$ $\frac{{R}}{2}+\frac{{L}}{6} \frac{{dL}}{ {dR}}=0$ Substituting value of $\frac{{d} {L}}{ {d} {R}}$ from eqution (2) $\frac{{R}}{2}+\frac{{L}}{6}(\frac{-2 {L}}{\ {R}} )=0$ $\frac{{R}}{2}=\frac{{L}^{2}}{3 {R}}$ $\Rightarrow \frac{{L}}{ {R}}=\\sqrt{ \frac{3}{2}}$

Q55JEE Main 2020MCQ4MElectromagnetic Waves

The magnetic field of a plane electromagnetic wave is ${\vec{B}}=3 \\times 10^{-8} \sin [200 \pi ( {y}+ {ct})] \hat{i} {T}$ Where $c=3 \\times 10^{8} \ {ms}^{-1}$ is the speed of light. The corresponding electric field is:

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📖 Explanation

${\vec{B}}=3 \\times 10^{-8} \sin [200 \pi ( {y}+ {ct})] \hat{i} {T}$ ${E}_{0}= {CB}_{0}\Rightarrow {E}_{0}$ $=3 \\times 10^{8} \\times 3 \\times 10^{-8}$ $=9 V/M$ and direction of wave propagation is given as $( {\vec{E}} \times {\vec{B}}) ∥ {\vec{C}}$ $\hat{B}=\hat{i}\ \& \ \hat{C}=-\hat{j}$ so $\hat{E}=-\hat{k}$ $\therefore {\vec{E}}=E_0\sin [200 \pi ( {y}+ {ct})]{{(-k)}}^{\^} \$ $V/M$

Q56JEE Main 2020MCQ4MMagnetic Effects of Current and Magnetism

A charged particle carrying charge $1 \mu{C}$ is moving with velocity $(2\hat{i}+3 \hat{j}+4 \hat{k}) \ {ms}^{-1}$ . If an external magnetic field of $(5 {\hat{i}}+3 {\hat{j}}-6 {\hat{k}}) \\times 10^{-3} \ {T}$ exists in the region where the particle is moving then the force on the particle is ${\vec{F}} \\times 10^{-9} \ {N}$ . The vector ${\vec{F}}$ is :

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📖 Explanation

$\vec{F}=9(\vec{V} \times \vec{B})$ (Force on charge particle moving in magnetic field) $\vec{V}\times \vec{B}=(2\hat{i}+3 \hat{j}+4 \hat{k})$ $\times (5 {\hat{i}}+3 {\hat{j}}-6 {\hat{k}}) \\times 10^{-3}$ $(\text{table} \hat{i},\hat{j},\hat{k};2,3,4;5,3,-6)\times 10^{-3}$ $=[\hat{i}[-18-12]- \hat{j}[-12-20]$ $+\hat{k}[6-15]] \\times 10^{-3}$ $=[\hat{i}[-30]+\hat{j}[32]+\hat{k}[-9]]$ $\times 10^{-3}$ Force $=10^{-6}[-30 \hat{i}+32 \hat{j}-9 \hat{k}] \\times 10^{-3}$ $=10^{-9}[-30 \hat{i}+32\hat{j}-9 \hat{k}]$

Q57JEE Main 2020MCQ4MAlternating Current

A $750 \ {Hz}, 20 \ {V}$ (rms) source is connected to a resistance of $100 \ {a},$ an inductance of $0.1803 \ {H}$ and a capacitance of 10 $_{\mu}$ F all in series. The time in which the resistance (heat capacity $2 \ {J} /{ }°{C}$ ) will get heated by $10°{C}$ . (assume no loss of heat to the surroundings) is close to :

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📖 Explanation

$\ {f}=750 \ {Hz}, \ {V}_{ {\text{rms}}}=20 \ {V}$ $\ {R}=100 \ {a}, \ {L}=0.1803 \ {H}$ $\ {C}=10 \ {\mu} \ {F}, \ {S}=2 \ {J} /{ }°{C}$ ${Z}=\sqrt{{R}^{2}+({X}_{ {L}}- {X}_{ {C}} )^{2}}$ $=\sqrt{{R}^{2}+(\omega \ {L}-1 / \omega {C})^{2}}$ $=\sqrt{{R}^{2}+ (2 \pi {fL}- \frac{1}{2 \pi {fC}} )^{2}}$ Putting values $|Z|=834 \Omega$ In ${AC}$ power ${P}= {V}_{ {\text{rms}}} {i}_{ {\text{rms}}} \cos\phi$ $\cos \phi = \frac{R}{|Z|} \;\;\ i_{ {\text{rms}}}= {V_{ {\text{rms}}}}/{| {Z}|}$ $={{V}_{ {\text{rms}}}^{2} {R}}{(| {Z}|)^{2}}$ $=(\frac{20}{834} )^{2} \\times 100=0.0575 \ {J} / {s}$ ${H}= {Pt}= {S} {\Delta\theta }$ ${t}= \frac{2(10)}{0.0575}=348 \ {sec}$

Q58JEE Main 2020MCQ4MAtoms and Nuclei

In a radioactive material, fraction of active material remaining after time $t$ is $9 / 16 .$ The fraction that was remaining after $\ {t} / 2$ is :

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📖 Explanation

First order decay ${N}( {t})={N}_{0} {e}^{-\lambda} {t}$ Given ${{N}({t})} / {N}_{0}$ $=9 / 16= {e}^{-\lambda t}$ ${\text{Now}}, \;\;{N}( {t} / 2)= {N}_{0} {e}^{- {\lambda} {t} / 2}$ $\frac{{N}( {t} / 2)}{ {N}_{0}}=\sqrt{{e}^{-\lambda t}}=\\sqrt{9 / 16}$ ${N}(\ {t} / 2)=3 / 4 {N}_{0}$

Q59JEE Main 2020MCQ4MThermodynamics

A balloon filled with helium $(32°{C}$ and1.7 atm.) bursts. Immediately afterwards the expansion of helium can be considered as:

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Q60JEE Main 2020MCQ4MProperties of Matter

Pressure inside two soap bubbles are 1.01 and 1.02 atmosphere, respectively. The ratio of their volumes is :

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📖 Explanation

$\DeltaP_{1}=0.01=4 {T} / {R}_{1}$ .....(1) $\Delta{P}_{2}=0.02=4 {T} / {R}_{2}$ ......(2) Equation $(1)÷(2)$ $\frac{1}{2}= \frac{R_{2}}{R_{1}}$ $R_{1}=2 R_{2}$ $\frac{V_{1}}{V_{2}}= \frac{R_{1}^{3}}{R_{2}^{3}}= \frac{8 R_{2}^{3}}{R_{2}^{3}}=8$

Q61JEE Main 2020MCQ4MGravitation

A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth's radius $\ {R}_{ {e}} .$ By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so thatis become $\sqrt{\frac{3}{2}}$ times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is $\ {R},$ value of $\ {R}$ is :

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📖 Explanation

${V}_{0}=\sqrt{{{GM}}/{ {R}_{ {e}}}}$ ${-{GMm}}/{ {R}_{ {e}}}+ \frac{1}{2} {mv}^{2}$ $= \frac{- {GMm}}{ {R}_{\max }}+ \frac{1}{2} {mv'}^{ 2}$ ......(i) ${ {VR}_ {e}}= {V}' {R}_{\max }$ .....(ii) Solving (i) & (ii) $[{R}_{\max }=3 {R}_{ {e}}]$

Q62JEE Main 2020MCQ4MElectrostatic Potential and Capacitance

In the circuit shown in the figure, the total charge in $750 \mu{C}$ and the voltage across capacitor $\ {C}_{2}$ is $20 \ {V}$ . Then the charge on capacitor $\ {C}_{2}$ is :

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📖 Explanation

${q}_{3}=20 \\times 8=160 \mu{C}$ $\therefore {q}_{2}=750-160$ $=590 \mu{C}$

Q63JEE Main 2020MCQ4MElectromagnetic Induction

An elliptical loop having resistance $\ {R},$ of semi major axis a, and semi minor axis b is placed in a magnetic field as shown in the figure. If the loop is rotated about the x-axis with angular frequency $\omega ,$ the average power loss in the loop due to Joule heating is :

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📖 Explanation

$\in = \ {NAB} \omega \cos \omega {t}$ $P_{ {\text{avg}}}=< \frac{\in ^{2}}{\ {R}}>=<\ \frac{(\ {AB} \omega \cos \omega {t})^{2}}{\ {R}}>$ $=\frac{{A}^{2} {B}^{2} \omega ^{2}}{\ {R}}\frac{1}{2}= \frac{\pi ^{2} {a}^{2} {b}^{2} {B}^{2} \omega ^{2}}{2 {R}}$

Q64JEE Main 2020MCQ4MSemiconductor Electronics

When a diode is forward biased, it has a voltage drop of $0.5 \ {V}$ . The safe limit of current through the diode is $10 \ {mA}$ . If a battery of emf $1.5 \ {V}$ is used in the circuit, the value of minimum resistance to be connected in series with thediode so that the current does not exceed thesafe limit is :

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📖 Explanation

$1.5-0.5- {R} \\times 10 \\times 10^{-3}=0$ $\therefore {R}=100 \Omega$

Q65JEE Main 2020MCQ4MWaves

A uniform thin rope of length $12 \ {m}$ and mass $6 \ {kg}$ hangs vertically from a rigid support and a block of mass $2 \ {kg}$ is attached to its free end. A transverse short wavetrain of wavelength $6 \ {cm}$ is produced at the lower end of the rope. What is the wavelength of the wavetrain (in $\ {cm}$ ) when it reaches the top of the rope?

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📖 Explanation

${V} \propto \lambda \;\;\;{T}_{2}=8 {g}$ ${T}_{1}=2 {g}$ $\frac{V_{1}}{V_{2}}= \frac{\lambda_{1}}{\lambda_{2}}$ $\lambda_{2}=\frac{{V}_{2}}{ {V}_{1}} \lambda_{1}$ $=\sqrt{\frac{{T}_{2}}{{T}_{1}}} \times \lambda_{ {I}}$ $=\sqrt{\frac{8 {g}}{2 {g}}} \lambda_{1}$ $=2 \\times 6=12 \ {cm}$

Q66JEE Main 2020MCQ4MRotational Motion

A block of mass $\ {m}=1 \ {kg}$ slides with velocity $v=6 \ {m} / {s}$ on a frictionless horizontal surfaceand collides with a uniform vertical rod andsticks to it as shown. The rod is pivoted about $\ {O}$ and swings as a result of the collision making angle $\theta$ before momentarily coming to rest. If the rod has mass ${M}=2 \ {kg},$ and length $l=1 \ {m},$ the value of $\theta$ is approximately :(Take ${g}=10 {m} / {s}^{2} )$

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📖 Explanation

Angular momentum conservation ${mv} l= \frac{ {M} l^{2}}{3} \omega + {m} l^{2} \omega$ $\Rightarrow \omega = \frac{1 \\times 6 \\times 1}{\ \frac{2}{3}+1}= \frac{18}{5}$ Now using energy consevation $\frac{1}{2}({M} \frac{l^{2}}{3} ) \omega ^{2}+ \frac{1}{2}({m} l^{2} ) \omega ^{2}$ $=( {m}+ {M}) {r}_{ {cm}}(1-\cos \theta )$ $= ({m}+ {M}) ({{m} l+\frac{{M} l}{2}}/{ {m}+ {M}} )$ ${g}(1-\cos \theta )$ $\frac{5}{6} \\times (\frac{18}{5} )^{2}=20(1-\cos \theta )$ $\Rightarrow 1-\cos \theta = \frac{18}{5}\times \frac{3}{20}$ $\cos \theta =1- \frac{27}{50}$ $\cos \theta = \frac{23}{50} \Rightarrow \theta \ ≃ 63°$

Q67JEE Main 2020MCQ4MWave Optics

In a Young's double slit experiment, light of $500 \ {nm}$ is used to produce an interference pattern. When the distance between the slits is $0.05 \ {mm}$ the angular width (in degree) of the fringes formed on the distance screen is close to :

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📖 Explanation

$\Delta \theta _{0} =(\frac{\lambda}{ {d}}\times \frac{180}{\pi } )^{0}$ $=0.57°$

Q68JEE Main 2020MCQ4MMagnetic Effects of Current and Magnetism

Magnitude of magnetic field (in SI units) at thecentre of a hexagonal shape coil of side $10 \ {cm}$ ,50 turns and carrying current I (Ampere) inunits of $\frac{\mu_{0} {I}}{\pi }$ is :

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📖 Explanation

$\ {B}= \frac{6 \mu_{0} {I}}{4 \pi {a} \cos 30°} \\times 2 \sin 30° \\times 50$ $= \frac{\mu_{0} {I}}{\pi } \frac{150}{\\sqrt{3} {a}}= \frac{50 \\sqrt{3}}{0.1} \frac{\mu_{0} {I}}{\pi }$ $=500 \\sqrt{3} \frac{\mu_{0} {I}}{\pi }$

Q69JEE Main 2020MCQ4MThermodynamics

Consider a gas of triatomic molecules. Themolecules are assumed to the triangular andmade of massless rigid rods whose vertices areoccupied by atoms. The internal energy of a mole of the gas at temperature $\ {T}$ is :

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📖 Explanation

$\ {DOF}=3+3=6$ $\ {U}=\frac{{f}}{2} {nRT}=3 \ {RT}$

Q70JEE Main 2020MCQ4MElectrostatic Potential and Capacitance

Two isolated conducting spheres $\ {S}_{1}$ and $\ {S}_{2}$ ofradius $\ \frac{2}{3} \ {R}$ and $\ \frac{1}{3} \ {R}$ have $12 \mu{C}$ and $-3 \mu{C}$ charges, respectively, and are at a large distancefrom each other. They are now connected bya conducting wire. A long time after this is donethe charges on $\ {S}_{1}$ and $\ {S}_{2}$ are respectively:

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📖 Explanation

Now ${Q}_{1}+ {Q}_{2}= {Q}_{1}'+ {Q}_{2}'$ $=12 \mu{C}-3 \mu{C}=9 \mu{C}$ $\& \ {V}_{1}= {V}_{2}$ $\Rightarrow {{KQ}_{1}'}/{ \frac{2 {R}}{3}}={{KQ}_{2}'}/{\frac{{R}}{3}}$ $Q_{1}'=2 Q_{2}'$ $\Rightarrow 2 Q_{2}'+Q_{2}'=9 \mu C$ $\Rightarrow Q_{2}'=3 \mu C$ $\& \ {Q}_{1}'=6 \mu{C}$

Q71JEE Main 2020NAT4MProperties of Matter

A bakelite beaker has volume capacity of $500 \ {cc}$ at $30°{C}$ . When it is partially filled with $\ {V}_{ {m}}$ volume(at $30°$ ) of mercury, it is found that the unfilledvolume of the beaker remains constant astemperature is varied. If $\gamma _{(\text{beaker })}=6 \\times 10^{-6}{ }°{C}^{-1}$ and $\gamma _{(\text{mercury })}=1.5 \\times 10^{-4}{ }°{C}^{-1},$ where $\gamma$ is thecoefficient of volume expansion, then $\ {V}_{ {m}}$ (in cc) is close to _____.

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📖 Explanation

$[\Delta{V}=({V}_{0}- {V}_{ {m}} )$ After increasing temperature $\Delta{V}'=({{V}_{0}'- {V}_{ {m}}'})$ $\Delta{V}'=\Delta{V}$ ${V}_{0}- {V}_{ {m}}= {V}_{0} (1+\gamma _{ {b}} \Delta{T} )$ $- {V}_{ {m}}(1+\gamma _{ {M}} \Delta{T} )$ $\ {V}_{0} \gamma _{ {b}}=\ {V}_{ {m}} \gamma _{ {m}}$ ${V}_{ {m}}={{V}_{0} \gamma _{ {b}}}/{\gamma _{ {m}}}$ $= \frac{(500) (6 \\times 10^{-6} )}{ (1.5 \\times 10^{-4} )}$ $=20 \ {CC}$

Q72JEE Main 2020NAT4MWork, Power and Energy

A cricket ball of mass $0.15 \ {kg}$ is thrownvertically up by a bowling machine so that itrises to a maximum height of $20 \ {m}$ after leaving the machine. If the part pushing the ball applies a constant force $\ {F}$ on the ball and moveshorizontally a distance of $0.2 \ {m}$ whilelaunching the ball, the value of $\ {F}($ in $\ {N})$ is $({g}=10 {ms}^{-2}\ )$ _______.

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📖 Explanation

${W}_{ {F}}= \frac{1}{2} {mv}^{2}= {mgh}$ ${F}( {S})= {mgh}$ ${F}(0.2)=(0.15)(10)(20)$ ${F}=150 \ {N}$

Q73JEE Main 2020NAT4MProperties of Matter

When a long glass capillary tube of radius $0.015 \ {cm}$ is dipped in a liquid, the liquid rises to a heightof $15 \ {cm}$ within it. If the contact angle betweenthe liquid and glass to close to $0°,$ the surface tension of the liquid, in milliNewton $\ {m}^{-1},$ is $\ [\rho _{( {\text{liquid}})}=900 \ {kgm}^{-3},$ ${g}=10 \ {ms}^{-2} ]$ (Give answerin closest integer) ______.

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📖 Explanation

Capillary rise ${h}= {2 {S} \cos \theta }{\rho {gr}}$ ${S}= {\rho {grh}}{2 \cos \theta }$ $=\ \frac{(900)(10) (15 \\times 10^{-5} ) (15 \\times 10^{-2} )}{2}$ $\ {S}=1012.5 \\times 10^{-4}$ $\ {S}=101.25 \\times 10^{-3}=101.25 \ {mN} / {m}$ In question closest integer is asked so closest integer $=101.00$

Q74JEE Main 2020NAT4MRotational Motion

A person of $80 \ {kg}$ mass is standing on the rim of a circular platform of mass $200 \ {kg}$ rotating about its axis as 5 revolutions per minute (rpm). The person now starts moving towards the centre of the platform. What will be the rotational speed (in rpm) of the platform whenthe person reaches its centre ________.

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📖 Explanation

${L}_{ {i}}= {L}_{ {f}}$ $\ (80 {R}^{2}+ \frac{200 {R}^{2}}{2} ) \omega$ $= (0+ \frac{200 {R}^{2}}{2} ) \omega _{1}$ $180 \omega _{0}=100 \omega _{1}$ $\omega _{1}=1.8 \omega _{0}=1.8 \\times 5$ $=9 \ {\text{rpm}}$

Q75JEE Main 2020NAT4MRay Optics and Optical Instruments

An observer can see through a small hole onthe side of a jar (radius $15 \ {cm}$ ) at a point atheight of $15 \ {cm}$ from the bottom (see figure). The hole is at a height of $45 \ {cm} .$ When the jaris filled with a liquid up to a height of $30 \ {cm}$ the same observer can see the edge at thebottom of the jar. If the refractive index of the liquid $\ {N} / 100,$ where $\ {N}$ is an integer, the value of $\ {N}$ is _______.

🚀 Solve in Practice Mode

📖 Explanation

$\tan {r}= \frac{15}{30}=\ \frac{1}{2}$ $\sin {r}= \frac{1}{\\sqrt{5}}$ $1 \sin 45°=\mu \sin \ {r}$ $\frac{1}{\\sqrt{2}}=\mu(\frac{1}{\\sqrt{5}} )$ $[\mu=\\sqrt{ \frac{5}{2}}=1.581]$ $\frac{{N}}{100}=\mu$ $\ {N}=100 \mu$ $\ {N}=158.11$ So integer value of $\ {N}=15800$

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