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JEE MAIN 2020 - 04 SEP S2 2020

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CHEMISTRY25 questions

Q1JEE Main 2020MCQ4MChemical Equilibrium

If the equilibrium constant for ${A} ⇌{B}+\ {C}$ is ${K}_{ {eq}}^{(1)}$ and that of ${B}+ {C} ⇌{P}$ is ${K}_{ {eq}}^{(2)},$ the equilibrium constant for ${A} ⇌{P}$ is :-

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📖 Explanation

${A} ⇌{B}+\ {C} \;\; {K}_{ {eq}}^{(1)}$ $={[B][C]}/[A]$ .....(1) ${B}+ {C} ⇌{P} \;\;\; {K}_{ {eq}}^{(2)}$ $=[P]/{[B][C]}$ .....(2) For ${A} ⇌{P} {K}_{ {eq}}= \frac{[ {P}]}{[ {A}]}$ Multiplying equation (1) & (2) ${K}_{ {eq}}^{(1)} \times {K}_{ {eq}}^{(2)}= \frac{[ {P}]}{[ {A}]}= {K}_{ {eq}}$

Q2JEE Main 2020MCQ4MChemical Thermodynamics

Five moles of an ideal gas at 1 bar and $298 {K}$ is expanded into vacuum to double the volume. The work done is :-

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📖 Explanation

As the expansion is done in vaccum that is in absence of $\ {p}_{\text{ext}}$ so $\ {W}=$ zero

Q3JEE Main 2020MCQ4MClassification of Elements

The process that is NOT endothermic in nature is :-

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📖 Explanation

${H}_{( {g})}+ {e}^{-}\to {H} ^{-}$ is exothermic rest of all endothermic process.

Q4JEE Main 2020MCQ4MCoordination Compounds

The crystal Field stabilization Energy (CFSE)of $[{CoF}_{3}({H}_{2} {O} )_{3}](\Delta_{0} < {P} )$ is :-

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📖 Explanation

$[{CoF}_{3}({H}_{2} {O} )_{3}]\;\;\;(\Delta_{0} < {P} )$ Means all ligands behaves as weak field ligands $=[-0.4 \\times 4+0.6 \\times 2] \Delta_{0}$ $=[-1.6+1.2] \Delta_{0}$ $=[-0.4 \Delta_{0}\ ]$

Q5JEE Main 2020MCQ4MBiomolecules

The mechanism of action of "Terfenadine" (Seldane) is : -

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Q6JEE Main 2020MCQ4MClassification of Elements

An alkaline earth metal 'M' readily forms water soluble sulphate and water insoluble hydroxide. Its oxide MO is very stable to heat and does not have rock-salt structure. M is :-

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📖 Explanation

[Be] $BeSO_4$ is water soluble $Be(OH)_2$ is water insoluble BeO is stable to heat

Q7JEE Main 2020MCQ4MChemical Bonding and Molecular Structure

The reaction in which the hybridisation of the underlined atom is affected is :-

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📖 Explanation

$\text{table} {XeF}_{4},+, {SbF}_{5} ,\to ,[{XeF}_{3} ]^{+},[{SbF}_{6} ]^{-}; {sp}^{3} {d}^{2},, {sp}^{3} {d} , ,{sp}^{3} {d} , {sp}^{3} {d}^{2}$

Q8JEE Main 2020MCQ4MCoordination Compounds

The one that can exhibit highest paramagnetic behaviour among the following is :- gly = glycinato; bpy = 2, 2'-bipyridine

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📖 Explanation

$[{Co}( {OX})_{2}( {OH})_{2} ]^{-} \;\;\Delta_{0}> {P} [{S.F.L}]$ It has highest number of unpaired $e^{-}s$ . so it is most paramagnetic.

Q9JEE Main 2020MCQ4MHaloalkanes and Haloarenes

In the following reaction sequence, [C] is :-

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Q10JEE Main 2020MCQ4MSolutions

A sample of red ink (a colloidal suspension) is prepared by mixing eosin dye, egg white, HCHO and water. The component which ensures stability of the ink sample is :-

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Q11JEE Main 2020MCQ4Mp Block Elements

The processes of calcination and roasting in metallurgical industries, respectively, can lead to :-

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📖 Explanation

Due to industrial process ${SO}_{2}$ gas is released which is responsible for acid rain & global warming.

Q12JEE Main 2020MCQ4MHaloalkanes and Haloarenes

Which of the following compounds will form the precipitate with aq. $AgNO_3$ solution most readily?

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📖 Explanation

${R}- {x}+ {aq.} {AgNO}_{3}{\to }^{{R.D.S}}$ ${R}^{\oplus }+ { {Agx}}_{{( {PPT})}}$ (1) So rate of P.P.T formation of Agx depend's on stability of carbocation (R $^{+}$ ) In given question formed carbocation will beMost stable carbocation is (b) so give fastest P.P.T of AgBr with aq $AgNO _3$

Q13JEE Main 2020MCQ4MCoordination Compounds

The molecule in which hybrid MOs involve only one d-orbital of the central atom is :-

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📖 Explanation

$[Ni(CN)_4 ]^{ 2-}$ $dsp^2$ hybridisation.

Q14JEE Main 2020MCQ4MHaloalkanes and Haloarenes

Among the following compounds, which one has the shortest ${C}-\ {Cl}$ bond $?$

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Q15JEE Main 2020MCQ4MAlcohols, Phenols and Ethers

The major product [R] in the following sequence of reactions is :-

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📖 Explanation

Now :- (i) $HgSO_4/\text{dil}.H _2SO _4$ (ii) $NaBH_4$ is convert triple bond into ketone and formed ketone is reduced by $NaBH_4$ and convert into Alcohol.

Q16JEE Main 2020MCQ4Md and f Block Elements

The incorrect statement(s) among (a) - (c) is (are) :- (a) W(VI) is more stable than Cr(VI). (b) in the presence of HCl, permanganate titrations provide satisfactory results. (c) some lanthanoid oxides can be used as phosphors.

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📖 Explanation

$KMnO_4$ will not give satisfactory result when it is titrated by HCl.

Q17JEE Main 2020MCQ4MElectrochemistry

250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M $AgNO_3$ and 0.1 M AuCl. The solution was electrolyzed at 2 V by passing a current of 1 A for 15 minutes. The metal/metals electrodeposited will be :- $({E}°_{{Ag}^{+} / {Ag}} =0.80 {V},$ ${E}°_{ {Au}^{+} / {Au}} =1.69 {V}\ )$

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📖 Explanation

As voltage is '2V' so both Ag $^{+} \&$ Au $^{+}$ will reduce and their equal gm equivalent will reduce so gmeq $\ {Ag}=$ gmeq of $\ {Au}$ ${{Wt}_{ {Ag}}}/{ {E}_{{ \text{qwt}} _{Ag}}}={{Wt}_{{Au}}}/{ {E}_{{ {\text{qwt}}} _{Au}}}$ So ${{Wt}_{ {Ag}}}/{{Wt}_{{Au}}}={ {E}_{{ \text{qwt}} _{Ag}}}/{ {E}_{{ {\text{qwt}}} _{Au}}}$ $={{At} {wt}_{ {Ag}}}/{ {Atwt}_{ {Au}}}$

Q18JEE Main 2020MCQ4MAlcohols, Phenols and Ethers

The major product [B] in the following reactions is :- $CH _3-CH_2 -{CH}^{|}^{CH_3}-CH_2 -OCH_2 -CH_3$ ${\to }^{\text{HI}}_{\text{Heat}}[A]\text{alcohol}{\to }^{\text{H} _2\text{SO}_4}_{\Delta}[B]$

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Q19JEE Main 2020MCQ4MHydrocarbons

The major product [C] of the following reaction sequence will be :-

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Q20JEE Main 2020MCQ4MStructure of Atom

The shortest wavelength of $\ {H}$ atom is the Lyman series is $\lambda_{1} .$ The longest wavelength in the Balmer series of $\ {He}^{+}$ is :-

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📖 Explanation

As we know $\Delta{E}=\frac{{hc}}{\lambda}$ So $\lambda=\frac{{hc}}{\Delta{E}}\;\;\;$ forλ minimum i.e. shortest;ΔE = maximum for Lyman series ${n}=1 \&$ for $\Delta{E}_{\max }$ Transition must be form ${n}=\infty$ to ${n}=1$ So $\ \frac{1}{\lambda}= {R}_{ {H}} {Z}^{2}(\frac{1}{{n}_{1}^{2}}- \frac{1}{ {n}_{2}^{2}} )$ $\frac{1}{\lambda}= {R}_{ {H}} {Z}^{2}(1-0)$ $\frac{1}{\lambda}= {R} \\times (1)^{2}\Rightarrow \lambda_{1}=\frac{1}{ {R}}$ For longest wavelength $\Delta{E}=$ minimum for Balmer series $\ {n}=3$ to $\ {n}=2$ will have $\Delta{E}$ minimum for ${He}^{+} {Z}=2$ So $\frac{1}{\lambda_{2}}= {R}_{ {H}}\times {Z}^{2}(\frac{1}{ {n}_{1}^{2}}- \frac{1}{ {n}_{2}^{2}} )$ $\frac{1}{\lambda_{2}}= {R}_{ {H}}\times 4$ \frac{1}{4}-\frac{1}{9} )\frac{1}{\lambda_{2}}= {R}{ {H}}×\frac{5}{9}\lambda{2}=\lambda_{1} ×\frac{9}{5}$

Q21JEE Main 2020NAT4MSome Basic Concepts of Chemistry

A $100 \ {mL}$ solution was made by adding $1.43 {g}$ of $\ {Na}_{2} {CO}_{3}.{xH}_{2} {O} .$ The normality of the solution is 0.1 N. The value of $x$ is_____. (The atomic mass of Na is $23 {g} / {mol}$ ) :-

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📖 Explanation

Molar mass of ${Na}_{2} {CO}_{3}.{xH}_{2} {O}$ $\Rightarrow 23 \\times 2+12+48+18 x$ $\Rightarrow 46+12+48+18 x$ $\Rightarrow(106+18 x)$ Eqwt $=\ \frac{M}{2}=(53+9 x)$ As ${n}_{\text{factor }}$ in dissolution will be determined from net cationic or anionic charge; which is 2 so Eqwt $= \frac{M}{2}=53+9 x$ ${Gmeq}=\frac{{wt}}{\text{ Eqwt }}= \frac{1.43}{53+9 {x}}$ Normality $= {\text{Gmeq }}/{ {V}_{\text{litre }}}$ Normality $=0.1= {1.43}/{ \frac{53+9 {x}}{0.1}}$ As volume $=100 \ {ml}$ $= 0.1$ Litre So $10^{-2}= \frac{1.43}{53+9 \ {x}}$ $53+9 \ {x}=143$ $9x = 90$ $x = 10.00$

Q22JEE Main 2020NAT4MSolutions

The osmotic pressure of a solution of ${NaCl}$ is 0.10 atm and that of a glucose solution is 0.20 atm. The osmotic pressure of a solution formed by mixing $1 \ {L}$ of the sodium chloride solution with $2 \ {L}$ of the glucose solutionis $x \\times 10^{-3}$ atm. $x$ is _____. (nearest integer) :-

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📖 Explanation

Osmotic pressure $=\pi = {i} \times {C}\times {RT}$ For ${NaCl} {i}=2$ so $\pi _{ {NaCl}} {i} \times {C}_{ {NaCl}} \times {RT} \;\;{C}_{ {NaCl}}=$ conc. of ${NaCl}$ $0.1=2\times {C}_{ {NaCl}}\times {RT}$ ${C}_{ {NaCl}}= \frac{0.05}{ {RT}} \;\;\;{C}_{\text{glucose }}=$ conc. of glucose For glucose $i=1$ so $\pi _{\text{Glucose }}= {i} \times {C}_{\text{glucose }} \times {RT}$ $0.2=1\times {C}_{ {\text{glucose}}}\times {RT}$ ${C}_{\text{Glucose }}= \frac{0.2}{ {RT}} \;\;\; \eta _{ {NaCl}}=$ No. of moles ${NaCl}$ $\eta _{ {NaCl}}$ in $1 {L}= {C}_{ {NaCl}}\times {V}_{ {\text{Litre}}}$ $= \frac{0.05}{\ {RT}}\;\;\; \eta _{\text{glucose}}=$ No. of moles glucose $\eta _{\text{glucose }}$ in $2 {L}= {C}_{\text{glucose }} \times {V}_{\text{Litre }}$ $= \frac{0.4}{\ {RT}}$ ${V}_{\text{Total }}=1+2=3 \ {L}$ so Final conc. ${NaCl}= \frac{0.05}{3 {RT}}$ Final conc. glucose $= \frac{0.4}{3 {RT}}$ $\pi _{\text{Total }}=\pi _{ {NaCl}}+\pi _{\text{glucose }}$ $=[{i} \times {C}_{ {NaCl}}+ {C}_{ {glucose}} ] \times {RT}$ $=(\frac{2 \\times 0.05}{3 {RT}}+ \frac{0.4}{3 {RT}} ) \times {RT}$ $= \frac{0.5}{3} \ {\text{atm}}$ $=0.1666 \ {\text{atm}}$ $=166.6 \\times 10^{-3} {\text{atm}}$ $\Rightarrow 167.00 \\times 10^{-3} \ {\text{atm}}$ so $x=167.00$

Q23JEE Main 2020NAT4MBiomolecules

The number of chiral centres present in threonine is _________.

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Q24JEE Main 2020NAT4MRedox Reactions

Consider the following equations: $2 {Fe}^{2+}+ {H}_{2} {O}_{2} \to {x} {A}+ {y} {B}$ (in basic medium) $2 \ {MnO}_{4}^{-}+6 \ {H}^{+}+5 \ {H}_{2} {O}_{2}$ $\to {x}"{C}+ {y}"{D}+ {z}"{E}$ (in acidic medium) The sum of the stoichiometric coefficients ${x}, \ {y}, \ { {x}'}, {y}'$ and ${z}'$ for products $\ {A}, \ {B}, \ {C}, \ {D}$ and E, respectively, is _______.

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📖 Explanation

$[ {Fe}^{2+} \to {Fe}^{3+}+ {e}^{-} ] \\times 2$ ${H}_{2} {O}_{2}+2 {e}^{-} \to 2 {HO}^{\ominus }$ $────────────── ───────$ $2 {Fe}^{2+}+ {H}_{2} {O}_{2} \to 2 {Fe}^{3+}$ $+2 {HO}_{( {q} \omega )}^{\ominus }$ $x=2\;\;\; y=2$ $[8 H^{+}+M n O_{4}^{-}+5 e^{-} \to M n^{2+}$ $+4 H_{2} O]\times 2$ $[{H}_{2} {O}_{2} \to {O}_{2( {g})}+2 {H}^{+}$ $+2 {e}^{-} ] \\times 5$ $\Rightarrow 16 {H}^{+}+2 {MnO}_{4}^{-}+5 {H}_{2} {O}_{2}$ $\to 2 {Mn}^{2+}+8 {H}_{2} {O}+5 {O}_{2( {g})}+10 {H}^{+}$ \Rightarrow $6 {H}^{+}+2 {MnO}_{4}^{-}+5 {H}_{2} {O}_{2}$ $\to 2 {Mn}^{2+}+8 {H}_{2} {O}+5 {O}_{2({g})}$ $\text{So}\ x'=2 \;\; y'=8 \;\; z'=5$ so $x+y+x'+y'+z'$ $\Rightarrow \;2+2+2+8+5$ $\Rightarrow 19$

Q25JEE Main 2020NAT4MChemical Kinetics

The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from $27°{C}$ to $42°{C} .$ Its energy of activation in $\ {J} / {mol}$ is __________. ( Take $\ln 5=1.6094;$ ${R}=8.314 \ {J} \ {mol}^{-1} \ {K}^{-1}\ )$

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📖 Explanation

$\ {T}_{1}=300 \ {K} \;\;\;{T}_{2}=315 \ {K}$ As per question ${K}_{{T}_{2}}=5 {K}_{ {T}_{ {1}}}$ as molecules activated are increased five times so k will increases 5 times Now $\ln({{K}_{ {T}_{2}}}/{ {K}_{ {T}_{1}}} )= \frac{{Ea}}{ {R}}(\frac{1}{ {T}_{1}}- \frac{1}{ {T}_{2}} )$ $\ln 5=\frac{{Ea}}{ {R}}$ \frac{15}{300 \× 315} ) $So$ ;; {Ea}= \frac{1.6094 \× 8.314 \× 300 \× 315}{15}\ {Ea}=84297.47$ Joules/mole

MATHEMATICS25 questions

Q26JEE Main 2020MCQ4MLimits, Continuity and Differentiability

The function $f(x) =\{\text{table} \frac{\pi }{4}+\tan ^{-1} x {,} ,{|x| }\le 1 ; {1} /{2}(|x|-1) {,} ,{|x|}>1$ is :

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📖 Explanation

$f(x)=\begin{cases}\frac{\pi }{4}+\tan ^{-1} \ {x} & {x} \in (-\infty -1] \cup [1 \infty ) \\ -{({x}+1)}{2} & {x} \in (-1 0] \\ \ \frac{x-1}{2} & { ,} x \in (0 1)$ for continuity at $x=-1$ L\end{cases}H.L. $= \frac{\pi }{4}- \frac{\pi }{4}=0$ ${R} . {H.L.}=0$ so, continuous at $x=-1$ for continuity at $x=1$ L.H.L. $=0$ ${R.H.L.}= \frac{\pi }{4}+ \frac{\pi }{4}= \frac{\pi }{2}$ so, not continuous at $x=1$ For differentiability at $\ {x}=-1$ L.H.D. $=\ \frac{1}{1+1}= \frac{1}{2}$ ${R}.{H.D.}=- \frac{1}{2}$ so, non differentiable at $x=-1$

Q27JEE Main 2020MCQ4MSets and Relations

Let $U_{i=1}^{50} X_{i}= U_{i=1}^{n}Y_{i}=T,$ where each $X_{i}$ contains 10elements and each $Y_{i}$ contains 5 elements. Ifeach element of the set $\ {T}$ is an element ofexactly 20 of sets $\ {X}_{ {i}}$ 's and exactly 6 of sets ${Y}_{ {i}}'{s}$ then $\ {n}$ is equal to :

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📖 Explanation

${n}({X}_{{i}} )=10 . U_{i=1}^{50} {X}_{ {i}}= {T},$ $\Rightarrow \ {n}( {T})=500$ each element of $\ {T}$ belongs to exactly 20elements of $\ {X}_{ {i}} \Rightarrow \frac{500}{20}=25$ distinct elements so $\frac{5 {n}}{6}=25\Rightarrow {n}=30$

Q28JEE Main 2020MCQ4MQuadratic Equation and Inequalities

Let $\lambda\ne 0$ be in $\ {R} .$ If $\alpha$ and $\beta$ are the roots of the equation, $\ {x}^{2}- {x}+2 \lambda=0$ and $\alpha$ and $\gamma$ arethe roots of the equation, $3 x^{2}-10 x+27 \lambda=0$ then $\frac{\beta \gamma }{\lambda}$ is equal to :

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📖 Explanation

$\alpha +\beta =1, \alpha \beta =2 \lambda$ $\alpha +\beta =\ \frac{10}{3}, \;\;\alpha \gamma = \frac{27 \lambda}{3}=9 \lambda$ $\gamma -\beta =\ \frac{7}{3}$ $\frac{\gamma }{\beta }= \frac{9}{2}\Rightarrow \gamma = \frac{9}{2} \beta = \frac{9}{2} \times \frac{2}{3}$ $\Rightarrow \gamma =3$ $\frac{9}{2} \beta -\beta = \frac{7}{3}$ $\frac{9}{2} \beta = \frac{7}{3} \Rightarrow \beta = \frac{2}{3}$ $\alpha =1- \frac{2}{3}= \frac{1}{3}$ $2 \lambda=\ \frac{2}{9} \Rightarrow \lambda=\ \frac{1}{9}$ $\frac{\beta \gamma }{\lambda}=\frac{\frac{2}{3} \\times 3}{\ \frac{1}{9}}=18$

Q29JEE Main 2020MCQ4MDifferential Equations

The solution of the differential equation $\frac{d y}{d x}- \frac{y+3 x}{\log _{e}(y+3 x)}+3=0$ is :- (where $\mathrm{C}$ is a constant of integration.)

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📖 Explanation

$\ln ( {y}+3 {x})= {z}$ (let) $\frac{1}{ {y}+3 {x}} .(\frac{{dy}}{ {dx}}+3 )=\frac{{dz}}{ {dx}}$ ....(1) $\ \frac{d y}{d x}+3=\ \frac{y+3 x}{\ln(y+3 x)}$ (given) $\ \frac{d z}{d x}=\ \frac{1}{z}$ $\Rightarrow {z} {dz}= {dx} \Rightarrow \frac{{z}^{2}}{2}= {x}+ {C}$ $\Rightarrow \frac{1}{2} \ln ^{2}( {y}+3 {x})= {x}+ {C}$ $\Rightarrow {x}- \frac{1}{2}(\ln ( {y}+3 {x}))^{2}= {C}$

Q30JEE Main 2020MCQ4MSequences and Series

Let $a_{1}, a_{2} ....., a_{n}$ be a given A.P. whose common difference is an integer and $S_{n}=a_{1}+a_{2}+\ .....+a_{n}$ If $a_{1}=1, a_{n}=300$ and $15 \ \le n \ \le 50,$ then the ordered pair $({S}_{ {n}-4} {a}_{ {n}-4} )$ is equal to :

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📖 Explanation

$a_{n}=a_{1}+(n-1) d$ $\Rightarrow 300=1+( {n}-1) {d}$ $\Rightarrow ({n}-1) {d}=299=13 \\times 23$ since, $\ {n} \in [15,50]$ $\therefore {n}=24$ and $\ {d}=13$ $\ {a}_{ {n}-4}= {a}_{20}=1+19 \\times 13=248$ $\Rightarrow {a}_{ {n}-4}=248$ ${S}_{ {n}-4}= \frac{20}{2}\{1+248\}=2490$

Q31JEE Main 2020MCQ4MThree Dimensional Geometry

The distance of the point (1,-2,3) from theplane $\ {x}- {y}+ {z}=5$ measured parallel to the line $\frac{x}{2}= \frac{y}{3}= \frac{z}{-6}$ is :

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📖 Explanation

equation of line parallel to $\frac{x}{2}= \frac{y}{3}= \frac{z}{-6}$ passes through (1,-2,3) is $\frac{x-1}{2}= \frac{y+2}{3}= \frac{z-3}{-6}=r$ $x=2 r+1$ $y=3 r-2$ $z=-6 r+3$ So $2 r+1-3 r+2$ $-6 r+3=5$ \Rightarrow $-7 r+1=0$ $\ {r}= \frac{1}{7}$ ${x}= \frac{9}{7}, {y}= \frac{-11}{7}, {z}= \frac{15}{7}$ Distance is $=\sqrt{(\frac{9}{7}-1 )^{2}+ (2- \frac{11}{7} )^{2}+ (3- \frac{15}{7} )^{2}}$ $=\\sqrt{(\frac{2}{7} )^{2}+(\frac{3}{7} )^{2}+(\frac{6}{7} )^{2}}$ $= \frac{1}{7} \\sqrt{4+9+36}$ $= \frac{1}{7} \\sqrt{49}=1$

Q32JEE Main 2020MCQ4MLimits, Continuity and Differentiability

Let ${f}:(0, \infty ) \to (0, \infty )$ be a differentiable function such that $\ {f}(1)= {e}$ and $\lim _{t \to x} \frac{t^{2} f^{2}(x)-x^{2} f^{2}(t)}{t-x}=0$ If $\ {f}( {x})=1,$ then $\ {x}$ is equal to :

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📖 Explanation

$L=\lim _{t \to x} \frac{t^{2} f^{2}(x)-x^{2} f^{2}(t)}{t-x}$ using L.H. rule ${L}=\lim _{t \to x}\frac{2 {tf}^{2}( {x})- {x}^{2} \. 2 {f}'({t}) .m{f}( {t})}{1}$ $\Rightarrow {L}=2x{f}( {x})\ ({f}({x})- {x} {f}'( {x}) )=0$ (given) $\Rightarrow {f}( {x})= {xf}'({x}) \Rightarrow ∫\frac{{f}'( {x}) {d} {x}}{ {f}({x})}$ $=∫\frac{{d} {x}}{ {x}}$ \Rightarrow $\ln | {f}(x)|=\ln |x|+C$ \because $f(1)=e, x>0, f(x)>0$ $\Rightarrow {f}( {x})= {ex}, \;\;$ if ${f}( {x})=1 \Rightarrow {x}= \frac{1}{ {e}}$

Q33JEE Main 2020MCQ4MMatrices and Determinants

If the system of equations $x+y+z=2$ $2 x+4 y-z=6$ $3 x+2 y+\lambda z=\mu$ has infinitely many solutions, then :

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📖 Explanation

For infinite solutions $\Delta=\Delta_{ {x}}=\Delta_{ {y}}=\Delta_{ {z}}=0$ Now $\Delta=0 \Rightarrow\begin{vmatrix}1 & 1 & 1 \\ 2 & 4 & -1 \\ 3 & 2 & \lambda\end{vmatrix}$ $=0$ $\Rightarrow \lambda=\ \frac{9}{2}$ $\Delta_{x=0} \Rightarrow\begin{vmatrix}2 & 1 & 1 \\ 6 & 4 & -1 \\ \mu & 2 & -9/2\end{vmatrix}$ $=0$ $\Rightarrow \mu=5$ For $\lambda= \frac{9}{2}\ \&\ \mu=5, \Delta_{y}=\Delta_{z}=0$ Now check option $2 \lambda+\mu=14$

Q34JEE Main 2020MCQ4MApplication of Derivatives

The minimum value of $2^{\sin x}+2^{\cos x}$ is :-

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📖 Explanation

Usnign AM ≥ GM $\Rightarrow \frac{2^{\sin x}+2^{\cos x}}{2} \ge \\sqrt{2^{\sin x} \.2^{\cos x}}$ $\Rightarrow 2^{\sin x}+2^{\cos x} \ge 2^{1+ ( \frac{\sin x+\cos x}{2} )}$ $\Rightarrow \min (2^{\sin x}+2^{\cos x} )=2^{1- \frac{1}{\\sqrt{2}}}$

Q35JEE Main 2020MCQ4MDefinite Integrals

$\int _{\pi / 6} ^{\pi / 3} \tan ^{3} x. \sin ^{2} 3 x (2 \sec ^{2} x .\sin ^{2} 3 x$ $+3 \tan x. \sin 6 x ) {d} x$ is equal to :

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📖 Explanation

$I=\int _{\pi / 6} ^{\pi / 3}((2 \tan ^{3} x. \sec ^{2} x . \sin ^{4} 3 x )$ $+ (3 \tan ^{4} x.\sin ^{3} 3 x . \cos 3 x)) \ {d} x$ $\Rightarrow {I}= \frac{1}{2} \int _{\pi / 6} ^{\pi / 3}\ {d} ((\sin 3 {x})^{4}(\tan {x})^{4} )$ $\Rightarrow {I}= ((\sin 3 {x})^{4}(\tan {x})^{4} )_{\pi / 6}^{\pi / 3}$ $\Rightarrow {I}=- \frac{1}{18}$

Q36JEE Main 2020MCQ4MCircles

The circle passing through the intersection of the circles, $x^{2}+y^{2}-6 x=0$ and $x^{2}+y^{2}-4 y=0$ having its centre on the line, $2 {x}-3 {y}+12=0$ also passes through the point:

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📖 Explanation

Let ${S}$ be the circle pasing through point of intersection of ${S}_{1}\ \& \ {S}_{2}$ $\therefore \ {S}=\ {S}_{1}+\lambda {S}_{2}=0$ $\Rightarrow {S}:({x}^{2}+ {y}^{2}-6 {x} )$ $+\lambda( {x}^{2}+ {y}^{2}-4 {y})=0$ $\Rightarrow {S}: {x}^{2}+ {y}^{2}-(\frac{6}{1+\lambda} ) {x}$ $-(\frac{4 \lambda}{1+\lambda} ) {y}=0$ ...(1) Centre $(\frac{3}{1+\lambda}, \frac{2 \lambda}{1+\lambda} )$ lies on $2 x-3 y+12=0 \Rightarrow \lambda=-3$ put in $(1) \Rightarrow{S}: {x}^{2}+ {y}^{2}+3 {x}-6 {y}=0$ Now check options point (-3,6) lies on S.

Q37JEE Main 2020MCQ4MTrigonometric Ratios and Identities

The angle of elevation of a cloud $\ {C}$ from a point P, 200 m above a still lake is $30°$ . If the angle of depression of the image of $\ {C}$ in the lake from the point $\ {P}$ is $60°$ , then $\ {PC}$ (in $\ {m}$ ) is equal to :

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📖 Explanation

Let ${PA}= {x}$ For $\Delta{APC}$ $\ {AC}=\frac{{PA}}{\\sqrt{3}}=\frac{{x}}{\\sqrt{3}}$ $\ {AC}^{1}=\ {AB}+ {BC}^{1}$ ${AC}^{1}= {AB}+ {BC}$ ${AC}^{1}=400+\frac{{x}}{\\sqrt{3}}$ From $\Delta{C}^{1} {PA}: {AC}^{1}=\\sqrt{3} \ {PA}$ $\Rightarrow (400+\frac{{x}}{\\sqrt{3}} )=\\sqrt{3} {x}$ $\Rightarrow {x}=(200)(\\sqrt{3})$ from $\Delta{APC}: {PC}=\frac{2 {x}}{\\sqrt{3}} \Rightarrow {PC}=400$

Q38JEE Main 2020MCQ4MComplex Numbers

If a and b are real numbers such that $(2+\alpha )^{4}= {a}+ {b} \alpha ,$ where $\alpha = \frac{-1+ {i} \\sqrt{3}}{2},$ then $\ {a}+\ {b}$ is equal to :

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📖 Explanation

$\alpha =\omega \;\;\;\;(\omega ^{3}=1 )$ $\Rightarrow \ (2+\omega )^{4}=a+b \omega$ $\Rightarrow \ 2^{4}+4 \. 2^{3} \omega +6 \. 2^{2} \omega ^{3}$ $+4.2 \. \omega ^{3}+\omega ^{4}=a+b \omega$ $\Rightarrow \ 16+32 \omega +24 \omega ^{2}+8+\omega$ $=a+b \omega$ $\Rightarrow\ 24+24 \omega ^{2}+33 \omega$ $=a+b \omega$ $\Rightarrow \ -24 \omega +33 \omega =a+b \omega$ $\Rightarrow \ a=0, b=9$

Q39JEE Main 2020MCQ4MProbability

In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws a total of 6 before $\ {B}$ throws a total of 7 and $\ {B}$ wins the game if he throws a total of 7 before A throws a total of six The game stops as soon as either of the players wins. The probability of A winning the game is :

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📖 Explanation

${P}(6)= \frac{1}{6}, {P}(7)= \frac{5}{36}$ ${P}( {A})= {W}+ {FFW}$ $+ {FFFFW}+.....$ $= \frac{1}{6}+ \frac{5}{6} \times \frac{31}{36} \times \frac{1}{6}$ $+(\frac{5}{6} )^{2}(\frac{31}{36} )^{2} \frac{1}{6}+\......$ $=\frac{\frac{1}{6}}{1- \frac{155}{216}}= \frac{36}{61}$

Q40JEE Main 2020MCQ4MEllipse

Let $x=4$ be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\frac{1}{2}$ . If $\ {P}(1, \beta ), \beta >0$ is a point on this ellipse, then the equation of the normal to it at $\ {P}$ is :-

🚀 Solve in Practice Mode

📖 Explanation

Ellipse $: \frac{x^{2}}{a^{2}}+ \frac{y^{2}}{b^{2}}=1$ directrix $: x= \frac{a}{e}=4\ \&\ e= \frac{1}{2}$ $\Rightarrow {a}=2\ \&\ {b}^{2}= {a}^{2} (1- {e}^{2} )=3$ $\Rightarrow \$ Ellipse is $\frac{x^{2}}{4}+ \frac{y^{2}}{3}=1$ $P$ is $(1, \frac{3}{2} )$ Normal is : $\frac{4 x}{1}- \frac{3 y}{3 / 2}=4-3$ $\Rightarrow 4 x-2 y=1$

Q41JEE Main 2020MCQ4MLimits, Continuity and Differentiability

Contrapositive of the statement: 'If a function f is differentiable at a, then it is also continuous at a', is :-

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📖 Explanation

p = function is differantiable at a q = function is continuous at a contrapositive of statement p → q is $∼q \to ∼p$

Q42JEE Main 2020MCQ4MApplication of Derivatives

The area (in sq. units) of the largest rectangle ABCD whose vertices A and B lie on the x-axis and vertices $\ {C}$ and $\ {D}$ lie on the parabola, $\ {y}= {x}^{2}-1$ below the $\ {x}$ -axis, is :

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📖 Explanation

Area $( {A})=2 {t}.(1- {t}^{2} )$ $(0 < \ {t} < 1)$ $A=2 t-2 t^{3}$ $\ \frac{d A}{d t}=2-6 t^{2}$ $t=\ \frac{1}{\\sqrt{3}}$ $\Rightarrow {A}_{\max }= \frac{2}{\\sqrt{3}} (1- \frac{1}{3} )=\frac{4}{3 \\sqrt{3}}$

Q43JEE Main 2020MCQ4MBinomial Theorem

If for some positive integer $\ {n},$ the coefficients of three consecutive terms in the binomial expansion of $(1+x)^{n+5}$ are in the ratio $5: 10: 14,$ then the largest coefficient in this expansion is :-

🚀 Solve in Practice Mode

📖 Explanation

Let $\ {n}+5= {N}$ ${}^NC_{r-1}: {}^NC_{r }: {}^NC_{r+1}$ $= 5 : 10 : 14$ $\Rightarrow \ \frac{{}^NC_{r }}{{}^NC_{r-1 }}=\frac{{N}+1- {r}}{ {r}}=2$ $\Rightarrow \ \frac{{}^NC_{r +1}}{{}^NC_{r }}=\frac{{N} - {r}}{ {r+1}}=7/5$ $\Rightarrow \ r=4, N=11$ $\Rightarrow(1+x)^{11}$ Largest coefficient $={ }^{11} {C}_{6}=462$

Q44JEE Main 2020MCQ4MStraight Lines and Pair of Straight Lines

If the perpendicular bisector of the line segment joining the points $\ {P}(1,4)$ and $\ {Q}(\ {k}, 3)$ has $\ {y}-$ intercept equal to $-4,$ then a value of $\ {k}$ is :-

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📖 Explanation

Slope $= {m}=\frac{1}{1- {k}}$ Equation of $\perp$ bisector is $y+4=(k-1)(x-0)$ $\Rightarrow y+4=x(k-1)$ $\Rightarrow {7}{2}+4=\frac{{k}+1}{2}( {k}-1)$ $\Rightarrow \frac{15}{2}={{k}^{2}-1}{2}$ $\Rightarrow {k}^{2}=16\Rightarrow {k}=4,-4$

Q45JEE Main 2020MCQ4MMatrices and Determinants

Suppose the vectors $\ {x}_{1}, {x}_{2}$ and ${x}_{3}$ are the solutions of the system of linear equations, ${Ax}= {b}$ when the vector $\ {b}$ on the right side is equal to $\ {b}_{1}, \ {b}_{2}$ and $\ {b}_{3}$ respectively. If $x_1=\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix},x_2=\begin{bmatrix}0 \\ 2 \\ 1\end{bmatrix},$ $x_3=\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix},b_1=\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$ $b_2=\begin{bmatrix}0 \\ 2 \\ 0\end{bmatrix}$ and $b_3=\begin{bmatrix}0 \\ 0 \\ 2\end{bmatrix},$ then the determinant of A is equal to :-

🚀 Solve in Practice Mode

📖 Explanation

$\ {Ax}_{1}= {b}_{1}$ $\ {Ax}_{2}= {b}_{2}$ $\ {Ax}_{3}= {b}_{3}$ \Rightarrow $|A|=\begin{vmatrix}1 & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & 1\end{vmatrix}$ $=\begin{vmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{vmatrix}$ $\Rightarrow {| {A}|}= \frac{4}{2}=2$

Q46JEE Main 2020NAT4MPermutations and Combinations

A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is _____.

🚀 Solve in Practice Mode

📖 Explanation

Ways $={ }^{6} {C}_{4} . 1^{4} .3^{2}$ $=15 \\times 9$ $=135$

Q47JEE Main 2020NAT4MCircles

Let $\ {PQ}$ be a diameter of the circle $\ {x}^{2}+ {y}^{2}=9 .$ If $\alpha$ and $\beta$ are the lengths of the perpendiculars from $\ {P}$ and $\ {Q}$ on the straight line, $\ {x}+ {y}=2$ respectively, then the maximum value of $\alpha \beta$ is _______.

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📖 Explanation

Let ${P}(3 \cos \theta , 3 \sin \theta )$ $\ {Q}(-3 \cos \theta ,-3 \sin \theta )$ $\Rightarrow \alpha \beta =\frac{|(3 \cos \theta +3 \sin \theta )^{2}-4 |}{2}$ $\Rightarrow \alpha \beta = \frac{5+9 \sin 2 \theta }{2} \le 7$

Q48JEE Main 2020NAT4MDefinite Integrals

Let $\{x\}$ and $[x]$ denote the fractional part of $x$ and the greatest integer $\le {x}$ respectively of a real number x. If $∫_{0} ^ {n }\{x\} {d} x, ∫_{0} ^ {n}[x] \ {d} x$ and $10({n}^{2}- {n} t), {n}\in {N}, {n}>1)$ are three consecutive terms of a G.P., then $\ {n}$ is equal to ______.

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📖 Explanation

$∫_{0} ^ {n }\{x\} {d} x=n ∫_{0} ^ {1 }\{x\} {d} x$ $=n ∫_{0} ^ {1 } x d x=\ \frac{n}{2}$ $∫_{0} ^ {n }[x] d x=∫_{0} ^ {n }(x-\{x\}) d x$ $=\ \frac{n^{2}}{2}- \frac{n}{2}$ $\Rightarrow (\frac{{n}^{2}-{n}}{2} )^{2}=\frac{{n}}{2} \. 10 \. {n}(\ {n}-1)($ where ${n}>1)$ $\Rightarrow \frac{{n}-1}{4}=5 \Rightarrow {n}=21$

Q49JEE Main 2020NAT4MVector Algebra

If ${\vec{a}}=2 \hat{i}+\hat{j}+2 \hat{k}, \;\;$ then the value of ${| \hat{i} \\times ( \vec{a}\times \hat{i})|}^{2}+{| {j} ^{\^}\times ({a} ^{\to }\times \hat{j})|}^{2}$ $+{| \hat{k} \times (\vec{a}\times \hat{k})|}^{2}$ is equal to _______.

🚀 Solve in Practice Mode

📖 Explanation

$\sum |{\vec{a}}-({\vec{a}} .{i}) {i}|^{2}$ $\Rightarrow \sum (|\ \vec{a}|^{2}+({\vec{a}}.{i})^{2}$ $-2({\vec{a}} .{i})^{2} )$ $\Rightarrow 3|{\vec{a}}|^{2}-\sum ({\vec{a}} .{i})^{2}$ $\Rightarrow \ 2|{\vec{a}}|^{2}$ $\Rightarrow 18$

Q50JEE Main 2020NAT4MStatistics

If the variance of the following frequency distribution : Class : 10-20 20-30 30-40 Frequency : 2 x 2is 50, then x is equal to ____.

🚀 Solve in Practice Mode

📖 Explanation

∵ Variance is independent of shifting of origin $\text{table} x_i, : ,15 ,25 ,35,\text{or} ,-10,0,10;f_i,:,2,x,2,,2,x,2$ $\Rightarrow \$ Variance $(\sigma ^{2} )= {\sum {x}_{ {i}}^{2} {f}_{ {i}}}/{\sum {f}_{ {i}}}-({\vec{x}})^{2}$ $\Rightarrow \ 50= \frac{200+0+200}{ {x}+4}-0$ $\;\;\;\{\overline{{x}}=0\}$ $\Rightarrow \ 200+50 {x}=200+200$ $\Rightarrow \ {x}=4$

PHYSICS25 questions

Q51JEE Main 2020MCQ4MSemiconductor Electronics

PHYSICSIdentify the operation performed by the circuit given below :

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Q52JEE Main 2020MCQ4MRotational Motion

Consider two uniform discs of the samethickness and different radii $\ {R}_{1}=\ {R}$ and $\ {R}_{2}=\alpha {R}$ made of the same material. If the ratio of their moments of inertia $\ {I}_{1}$ and $\ {I}_{2}$ , respectively, about their axes is $\ {I}_{1}: {I}_{2}=1: 16$ then the value of $\alpha$ is :

🚀 Solve in Practice Mode

📖 Explanation

${I}_{1}=\frac{{MR}^{2}}{2}= \frac{\ \rho (\pi {R}^{2} ) {t} .{R}^{2}}{2}$ ${I} \ \propto {R}^{4}$ $\frac{{I}_{1}}{ {I}_{2}}=\frac{{R}_{1}^{4}}{ {R}_{2}^{4}}= \frac{1}{16}$ $\therefore \frac{ {R}_{1}}{ {R}_{2}}= \frac{1}{2}$

Q53JEE Main 2020MCQ4MElectrostatic Potential and Capacitance

A capacitor $\ {C}$ is fully charged with voltage $\ {V}_{0}$ . After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $\frac{\ {C}}{2} .$ The energy loss in the process after the charge is distributed between the two capacitors is :

🚀 Solve in Practice Mode

📖 Explanation

$\frac{{CV}_{0}- {q}}{ {C}}=\frac{{q}}{ {C} / 2}= \frac{2 {q}}{ {C}}$ ${V}_{0}= \frac{3 {q}}{ {C}} \Rightarrow {q}=\frac{{CV}_{0}}{3}$ ${U}_{{i}}= \frac{1}{2} {CV}_{0}^{2}$ ${U}_{{f}}={(\frac{2 {CV}_{0}}{3} )^{2}}/{2 {C}}+{(\frac{{CV}_{0}}{3} )^{2}}/{2(\frac{{C}}{2} )}$ $= \frac{1}{2} {CV}_{0}^{2}[\frac{4}{9}+ \frac{2}{9} ]$ $= \frac{1}{2} {CV}_{0}^{2}(\frac{2}{3} )$ Heat loss $= \frac{1}{2} {CV}_{0}^{2}-(\frac{2}{3} )(\frac{1}{2} {CV}_{0}^{2}\ )$ $= \frac{1}{6} {CV}_{0}^{2}$

Q54JEE Main 2020MCQ4MWork, Power and Energy

A person pushes a box on a rough horizontal platform surface. He applies a force of $200 \ {N}$ over a distance of $15 \ {m} .$ Thereafter, he getsprogressively tired and his applied force reduces linearly with distance of $100 \ {N}$ . The total distance through which the box has been moved is $30 \ {m} .$ What is the work done by the person during the total movement of the box?

🚀 Solve in Practice Mode

📖 Explanation

${F}=200 {N} \;\;\;$ for $0 \le {x} \le 15$ $=200- \frac{100}{15}( {x}-15)\;\;\;$ for $15\le {x} < 30$ ${W}=∫{Fdx}$ $=∫_{0} ^{15} 200 {dx}+∫_{15} ^{30} (300-\frac{100}{15} {x} ) {dx}$ $=200 \\times 15+300 \\times 15-\ \frac{100}{15}$ $\times \frac{(30^{2}-15^{2} )}{2}$ $=3000+4500-2250$ $=5250 \ {J}$

Q55JEE Main 2020MCQ4MElectromagnetic Waves

The electric field of a plane electromagnetic wave is given by ${\vec{E}}= {E}_{0} (\hat{x}+\hat{y}) \sin ( {kz}-\omega {t})$ Its magnetic field will be given by :

🚀 Solve in Practice Mode

📖 Explanation

${\vec{E}}= {E}_{0} (\hat{x}+\hat{y}) \sin ( {kz}-\omega {t})$ direction of propagation $=+\hat{k}$ $\hat{E}=\frac{\hat{i}+\hat{j}}{\\sqrt{2}}$ $\hat{k}=\hat{E}\times \hat{B}$ $\hat{k}=(\frac{\hat{i}+\hat{j}}{\\sqrt{2}} ) \times \hat{B} \;\;$ $\Rightarrow \hat{B}=\frac{-\hat{i}+\hat{j}}{\\sqrt{2}}$ $\therefore \vec{B}=\frac{{E}_{0}}{ {c}}({{-x}^{\^}}+\hat{y}) \sin ( {kz}-\omega {t})$

Q56JEE Main 2020MCQ4MAtoms and Nuclei

Find the bindng energy per nucleon for ${ }_{50}^{120} {Sn}$ . Mass of proton ${m}_{ {P}}=1.00783 \ {U},$ mass of neutron ${m}_{ {n}}=1.00867 \ {U}$ and mass of tin nucleus ${m}_{ {Sn}}=119.902199$ U. (take $1 {U}=931 {MeV})$

🚀 Solve in Practice Mode

📖 Explanation

B.E. $=[\Delta{m}].{c}^{2}$ ${M}_{\text{expected}}= {ZM}_{ {p}}+( {A}-{Z}) {M}_{ {n}}$ $=50[1.00783]+70[1.00867]$ ${M}_{\text{actual }}=119.902199$ B.E. $=[ 50[1.00783]+70[1.00867]$ $-119.902199] \times 931$ $=1020.56$ $\frac{ {BE}}{\text{ nucleon }}= \frac{1020.56}{120}=8.5 \ {MeV}$

Q57JEE Main 2020MCQ4MMotion in a Straight Line

A small ball of mass is thrown upward with velocity u from the ground. The ball experiences a resistive force $\ {mkv}^{2}$ where $\ {v}$ is its speed. The maximum height attained by the ball is :

🚀 Solve in Practice Mode

📖 Explanation

${ \vec{F}}= {mkv}^{2}- {mg}$ ${\vec{a}}= {{\vec{F}}}/{ {m}}=-[{kv}^{2}+ {g}]$ $\Rightarrow$ v. $\frac{{d} v}{ {dh}}=-[{kv}^{2}+ {g} ]$ $\Rightarrow ∫_{ {u}} ^{0}\frac{{v}.{d} {v}}{ {kv}^{2}+ {g}}=-∫_ {0} ^{ {H}} {dh}$ $\frac{1}{2 {K}} \ln \$ [{kv}^{2}+ {g} ]_{ {u}}^{0}=- {H}\Rightarrow \frac{1}{2 {K}} \ln $[\frac{{ku}^{2}+ {g}}{ {g}} ]$= {H}$

Q58JEE Main 2020MCQ4MRotational Motion

For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through $\ {O}$ (the centre of mass) and O' (corner point) is :

🚀 Solve in Practice Mode

📖 Explanation

${I}_{ {O}}=\frac{{M}}{12}[{L}^{2}+ {B}^{2} ]$ $=\frac{{M}}{12} [80^{2}+60^{2} {J} ]$ ${I}_{ {O}'}= {I}_{0}+ {Md}^{2}\{$ parallel axis theorem $\}$ $=\frac{{M}}{12} \$ [80^{2}+60^{2} ]$+\ {M}[50]^{2}{{I}{ {O}}}/{ {I}{ {O}'}}={{M} / 12 $[80^{2}+60^{2} ]$}/{\frac{{M}}{12} $[80^{2}+60^{2} ]$+ {M}[50]^{2}}= \frac{1}{4}$

Q59JEE Main 2020MCQ4MThermodynamics

Match the thermodynamic processes taking place in a system with the correct conditions. In the table : $\DeltaQ$ is the heat supplied, $\DeltaW$ is the work done and $\Delta{U}$ is change in internal energy of the system : Process Condition (I) Adiabatic (A) $\Delta{W}=0$ (II) Isothermal (B) $\DeltaQ=0$ (III) Isochoric (C) $\Delta{U} \ne 0, \Delta{W} \ne 0$ , $\Delta Q \ne 0$ (IV) Isobaric (D) $\Delta{U}=0$

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📖 Explanation

(I) Adiabatic process $\Rightarrow\DeltaQ=0$ No exchange of heat takes place with surroundings (II) Isothermal proess $\Rightarrow$ Temperature remains constant $(\Delta{T}=0)$ $\Delta{u}=\frac{{F}}{2} {nR} \Delta{T} \Rightarrow \Delta{u}=0$ No change in internal energy $[\Delta{u}=0]$ (III) Isochoric process Volume remains constant $\Delta{V}=0$ ${W}=∫{P.dV}=0$ Hence work done is zero. (IV) Isobaric process $\Rightarrow$ Pressure remains constant ${W}= {P.} \Delta{V} \ne 0$ $\Delta{u}=\frac{{F}}{2} {nR} \Delta{T}$ $=\frac{{F}}{2}[ {P} \Delta{V}] \ne 0$ $\Delta{Q}= {nC}_{ {p}} \Delta{T} \ne 0$

Q60JEE Main 2020MCQ4MMagnetism and Matter

A paramagnetic sample shows a net magnetisation of $6 \ {A} / {m}$ when it is placed in an external magnetic field of $0.4 \ {T}$ at a temperature of $4 \ {K}$ . When the sample is placed in an external magnetic field of $0.3 \ {T}$ at a temperature of $24 \ {K},$ then the magnetisation will be :

🚀 Solve in Practice Mode

📖 Explanation

For paramagnetic material According to curies law $\chi \propto\frac{1}{\ {T}}$ $\chi \propto\frac{1}{ {T}}\Rightarrow \chi _{1} {T}_{1}=\chi _{2} {T}_{2}$ $\Rightarrow \frac{6}{0.4} \\times 4=\frac{{I}}{0.3} \\times 24$ $\ {I}= \frac{0.3}{0.4}=0.75 {A} / {m}$

Q61JEE Main 2020MCQ4MCurrent Electricity

A series L-R circuit is connected to a battery of ${\text{emf}} \ {V}$ . If the circuit is switched on at $\ {t}=0$ ,then the time at which the energy stored in the inductor reaches $(\frac{1}{ {n}} )$ times of its maximum value, is :

🚀 Solve in Practice Mode

📖 Explanation

${U}_{\max }= \frac{1}{2} {L} {I}_{\max }^{2}$ ${i}= {I}_{\max } (1-{e}^{- {Rt} / {L}} )$ For ${U}$ to be $\frac{{U}_{\max }}{ {n}} ; {i}$ has to be ${{I}_{\max }}/{\\sqrt{ {n}}}$ ${I_{\max }}/{\\sqrt{ {n}}}= {I}_{\max } (1- {e}^{- {Rt} / {L}} )$ ${e}^{- {Rt} / {L}}=1- {1}/{\\sqrt{ {n}}}={\sqrt{{n}}-1}/{\sqrt{ {n}}}$ $-\frac{{Rt}}{ {L}}=\ln({{{n}}-1}/{\\sqrt{ {n}}})$ $\ {t}=\frac{{L}}{ {R}} \ln({{{n}}}/{\sqrt{{n}}-1} )$

Q62JEE Main 2020MCQ4MWaves

The driver of a bus approaching a big wall notices that the frequency of his bus's horn changes from $420 \ {Hz}$ to $490 \ {Hz},$ when he hears it after it gets reflected from the wall. Find the speed of the bus if speed of the sound is $330 \ {ms}^{-1}$

🚀 Solve in Practice Mode

📖 Explanation

$f_{1}=(\frac{330}{330-v_{B}} ) 420$ $f_{2}=(\frac{330+v_{0}}{330})$ $(\frac{330}{330-v_{B}} ) 420$ $490=(\frac{330+v_{B}}{330-v_{B}} ) 420$ $\frac{7}{6}= \frac{330+v_{B}}{330-v_{B}}$ $v_{B}= \frac{330}{13} \ {m} / \ {s}$ $= \frac{330}{13} \times \frac{18}{5} \approx 91 \ {km} / {hr}$

Q63JEE Main 2020MCQ4MWork, Power and Energy

Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density d. The area of the base of both vessels is $\ {S}$ but the height of liquid in one vessel is $\ {x}_{1}$ and in the other, $\ {x}_{2} .$ When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is :

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📖 Explanation

${U}_{ {i}}= (\rho {Sx}_{1} ) {g}.\frac{{x}_{1}}{2}$ $+ (\rho {Sx}_{2} ) {g} .\frac{{x}_{2}}{2}$ ${U}_{ {f}}= (\rho {Sx}_{ {f}} ) {g}.{{x}_{ {f}}}/{2} \\times 2$ By volume conservation ${Sx}_{1}+ {Sx}_{2}= {S} (2 {x}_{ {f}} )$ ${x}_{ {f}}=\frac{{x}_{1}+ {x}_{2}}{2}$ $\Delta{U}=\rho {Sg}[(\frac{{x}_{1}^{2}}{2}+\frac{{x}_{2}^{2}}{2} ) - {x}_{ {f}}^{2} ]$ $=\rho {Sg}[\frac{x_{1}^{2}}{2}+ \frac{x_{2}^{2}}{2}-(\frac{x_{1}+x_{2}}{2} )^{2} ]$ $= \frac{\rho {Sg}}{2}[\frac{x_{1}^{2}}{2}+ \frac{x_{2}^{2}}{2}-x_{1} x_{2} ]$ $= \frac{\rho {Sg}}{4}\ (x_{1}-x_{2} )^{2}$

Q64JEE Main 2020MCQ4MMagnetic Effects of Current and Magnetism

A circular coil has moment of inertia $0.8 \ {kg} \ {m}^{2}$ around any diameter and is carrying current to produce a magnetic moment of $20 \ {Am}^{2} .$ The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of $4 \ {T}$ is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by $60°$ will be:

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📖 Explanation

${I}_{ {\text{dia}}}=0.8 {kg} / {m}^{2}$ $\ {M}=20 \ {Am}^{2}$ ${U}_{ {i}}+ {K}_{ {i}}= {U}_{ {f}}+ {K}_{ {f}}$ $0+0=- {MB} \cos 30°+ \frac{1}{2} {I} \omega ^{2}$ $20 \\times 4\times \frac{\\sqrt{3}}{2}= \frac{1}{2}(0.8) \omega ^{2}$ $\omega =\\sqrt{100 \\sqrt{3}}=10(3)^{1 / 4}$

Q65JEE Main 2020MCQ4MElectrostatic Potential and Capacitance

A particle of charge q and mass $\ {m}$ is subjected to an electric field ${E}= {E}_{0} (1- {ax}^{2} )$ in the $\ {x}$ -direction, where a and $\ {E}_{0}$ are constants. Initially the particle was at rest at $\ {x}=0 .$ Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is :

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📖 Explanation

${E}= {E}_{0} (1- {ax}^{2} )$ ${W}=∫{qE} {dx}$ $= {qE}_{0} ∫_ {0} ^{ {x}_{0}} (1- {ax}^{2} ) {dx}$ $= {qE}_{0}[{x}_{0}-\frac{{ax}_{0}^{3}}{3} ]$ For $\Delta{KE}=0, \;\;{W}=0$ Hence ${x}_{0}=\sqrt{\frac{3}{\ {a}}}$

Q66JEE Main 2020MCQ4MProperties of Matter

A cube of metal is subjected to a hydrostatic pressure of 4 GPa. The percentage change in the length of the side of the cube is close to : (Given bulk modulus of metal, ${B}=8 \\times 10^{10} \ {Pa}$ )

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📖 Explanation

${B}=- {\Delta{P}}/{ \frac{\Delta{V}}{ {V}}}$ $|\frac{\Delta{V}}{{V}} |= \frac{\Delta{P}}{ {B}}$ $=\frac{4 \\times 10^{9}}{8 \\times 10^{10}}=\ \frac{1}{20}$ $\frac{\Deltal}{l}= \frac{1}{3} \times \frac{{\DeltaV}}{ {V}}=\frac{1}{60}$ Percentage change $= \frac{\Deltal}{l}\times 100 \%$ $= \frac{100}{60} \%=1.67 \%$

Q67JEE Main 2020MCQ4MCurrent Electricity

The value of current $i_{1}$ flowing from $\ {A}$ to $\ {C}$ in the circuit diagram is :

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📖 Explanation

Voltage across ${AC}=8 {V}$ ${R}_{ {AC}}=4+4=8 \Omega$ ${i}_{1}={{V}}/{\ {R}_{ {AC}}}= \frac{8}{8}=1 \ {A}$

Q68JEE Main 2020MCQ4MGravitation

A body is moving in a low circular orbit about a planet of mass $\ {M}$ and radius $\ {R} .$ The radius of the orbit can be taken to be $\ {R}$ itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is :

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📖 Explanation

$\ {V}_{\text{orbit }}=\sqrt{\frac{{GM}}{ {R}}}$ ${V}_{\text{escape }}=\sqrt{\frac{2 {GM}}{\ {R}}}$ ${{V}_{\text{orbit}}}/{ {V}_{\text{escape }}}= \frac{1}{\\sqrt{2}}$

Q69JEE Main 2020MCQ4MUnits and Measurements

If $I$ is moment of inertia, $F$ is force, $v$ is velocity, $E$ is energy and $L$ is length then, dimension of $\;\frac{ IFv ^{2}}{ EL ^{4}}$ will be:

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📖 Explanation

$\;\frac{ IFv ^{2}}{ EL ^{4}}$ $=\;\frac{( M ^{'} L ^{2})( M ^{'} L ^{'} T ^{-2})( L T ^{-1})^{2}}{( M ^{'} L ^{2} T ^{-2})( L ^{4})}$ $=\;\frac{ M ^{1} L ^{-2} T ^{-2}}{ L ^{3}}$ $= M ^{1} L ^{-1} T ^{-2}$

Q70JEE Main 2020MCQ4MDual Nature of Matter and Radiation

In a photoelectric effect experiment, the graph of stopping potential $\ {V}$ versus reciprocal of wavelength obtained is shown in the figure. As the intensity of incident radiation is increased :

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📖 Explanation

${eV}=\frac{{hc}}{\lambda}-\phi$ ${V}=(\frac{{hc}}{ {e}})(\frac{1}{\lambda} )-\phi$ Slope of the line in above equation and all other terms are independent of intensity. The graph does not change.

Q71JEE Main 2020NAT4MWave Optics

Orange light of wavelength $6000 \\times 10^{-10} \ {m}$ in illuminates a single slit of width $0.6 \\times 10^{-4} \ {m}$ . The maximum possible number of diffraction minima produced on both sides of the central maximum is _____.

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📖 Explanation

Condition for minimum, $\ {d} \sin \theta = {n} \lambda$ $\therefore \sin \theta =\frac{{n} \lambda}{\ {d}}<1$ ${n}<\frac{{d}}{\lambda}= \frac{6 \\times 10^{-5}}{6 \\times 10^{-7}}=100$ $\therefore$ Total number of minima on one side $=$ 99 Total number of minima $=198$ Correct Answer is 198

Q72JEE Main 2020NAT4MMotion in a Straight Line

The speed verses time graph for a particle is shown in the figure. The distance travelled (in $\ {m}$ ) by the particle during the time interval $\ {t}=0$ to $\ {t}=5 \ {s}$ will be _________ :

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📖 Explanation

Distance $=∫{v} \ {dt}$ Area under graph $= \frac{1}{2} \\times 5 \\times 8=20$

Q73JEE Main 2020NAT4MCurrent Electricity

Four resistances $40 \Omega, 60 \Omega, 90 \Omega$ and $110 \Omega$ make the arms of a quadrilateral ABCD. Across AC is a battery of emf $40 \ {V}$ and internal resistance negligible. The potential difference across BD is $\ {V}$ is ________.

🚀 Solve in Practice Mode

📖 Explanation

${i}_{1}= \frac{40}{40+60}=0.4$ $\ {i}_{2}= \frac{40}{90+110}= \frac{1}{5}$ $\ {v}_{\ {B}}+ {i}_{1}(40)- {i}_{2}(90)= {v}_{ {D}}$ $\ {v}_{ {B}}- {v}_{ {D}}= \frac{1}{5}(90)- \frac{4}{10} \\times 40$ ${v}_{ {B}}- {v}_{ {D}}=18-16=2$

Q74JEE Main 2020NAT4MRay Optics and Optical Instruments

The distance between an object and a screen is $100 \ {cm} .$ A lens can produce real image of the object on the screen for two different positions between the screen and the object. The distance between these two positions is $40 \ {cm} .$ If the power of the lens is close to $(\frac{{N}}{100} ) {D}$ where $\ {N}$ is an integer, the value of $\ {N}$ is _______.

🚀 Solve in Practice Mode

📖 Explanation

Using displacement method ${f}=\frac{{D}^{2}- {d}^{2}}{4 {D}}$ Here, $\ {D}=100 {cm}$ $\ {d}=40 {cm}$ ${f}= \frac{100^{2}-40^{2}}{4(100)}=21 {cm}$ ${P}= \frac{1}{ {f}}= \frac{100}{21} {D}$ $\frac{{N}}{100}= \frac{100}{21} {N}=476$

Q75JEE Main 2020NAT4MThermodynamics

The change in the magnitude of the volume of an ideal gas when a small additional pressure $\Delta{P}$ is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity $\Delta{T}$ at constant pressure. The initial temperature and pressure of the gas were $300 \ {K}$ and 2 atm respectively. If $|\Delta{T}|= {C}|\Delta{P}|$ then value of $\ {C}$ in $(\ {K} / {atm})$ is ______:

🚀 Solve in Practice Mode

📖 Explanation

$P V=n R T$ $P \DeltaV+V \DeltaP=0$ (for constant temp.) $P \DeltaV=n R \DeltaT$ (for constant pressure) $\DeltaT= \frac{P \DeltaV}{n R}$ $\Delta{P}=-\frac{{P} \Delta{V}}{ {V}} (\Delta{V}$ is same in both cases $)$ $\frac{\Delta{T}}{\Delta{P}}=\frac{{P} \Delta{V}}{ {nR}} \frac{{V}}{- {P} \Delta{V}}$ $= \frac{- {V}}{ {nR}} =-\frac{{T}}{ {P}}$ $( {PV}= {nRT})$ $$\frac{{V}}{ {nR}}=\frac{{T}}{ {P}} )|\ \frac{\Delta{T}}{\Delta{P}} |=|\frac{-300}{2} |=150$

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