Previous Year Paper

JEE MAIN 2020 - 04 SEP S1 2020

75 Questions300 Marks180 MinTake as Timed Mock →

🎯 Practice smarter, not harder

Practice JEE MAIN 2020 - 04 SEP S1 2020 questions with full analytics — free

✓ Attempt tracking & analytics
✓ Timer & exam-mode simulation
✓ Adaptive difficulty (Easy → Hard)
✓ Bookmarks, notes & mistake review

Just sign in to unlock everything. Free for all students.

🚀 Start Practice Mode Sign in free →

CHEMISTRY25 questions

Q1JEE Main 2020MCQ4Mp Block Elements

On heating, lead(II) nitrate gives a brown gas (A). The gas (A) on cooling changes to a colourless solid/liquid (B). (B) on heating with NO changes to a blue solid (C). The oxidation number of nitrogen in solid (C) is:

🚀 Solve in Practice Mode

📖 Explanation

${Pb}({NO}_{3}) _{2} {──── ▸}^{\Delta} {PbO}+2 {NO}_{2}_{\text{Brown} \text{gas}(\text{A})}$ $+ {1}/2{O}_{2}( {g})$ ${NO}_{2}( {g}) {──── ▸}^ { { \text{Cooling} }} {{N}_{2} {O}_{4}}_{(B)}$ ${N}_{2} {O}_{4}+ {NO} {──── ▸}^{\Delta}{{N}_{2} {O}_{3}}_{\text{Blue} \text{Solid}(C)}$ O.S. of nitrogen in ${N}_{2} {O}_{3}$ is +3 $\ {N}_{2} {O}_{3}\ 2 {x}+3(-2)=0$ $x=+3$

Q2JEE Main 2020MCQ4MAmines

Which of the following will react with ${CHCl}_{3}+$ alc. ${KOH} ?$

🚀 Solve in Practice Mode

📖 Explanation

Adenine and lysine Both have primary amine react with ${CHCl}_{3}+$ alc. ${KOH}$

Q3JEE Main 2020MCQ4MAlcohols, Phenols and Ethers

When neopentyl alcohol is heated with an acid, it slowly converted into an 85: 15 mixture of alkenes A and B, respectively. What are these alkenes?

🚀 Solve in Practice Mode

Q4JEE Main 2020MCQ4Md and f Block Elements

Among statements (a) -(d), the correct ones are: (a) Lime stone is decomposed to CaO during the extraction of iron from its oxides. (b) In the extraction of silver, silver is extracted as an anionic complex. (c) Nickel is purified by Mond's process. (d) $\ {Zr}$ and $\ {Ti}$ are purified by Van Arkel method.

🚀 Solve in Practice Mode

📖 Explanation

(a) ${CaCO}_{3} \xrightarrow{\Delta} {CaO}+ {CO}_{2}$ , In Blast furnace $\}$ lime stone (b) Ag form cyanide complex [Ag(CN) $_{2}]^{-}$ during cyaride process ${Ag} / {Ag}_{2} {S}+ {CN}^{\ominus } \to [{Ag}( {CN})_{2} ]^{-}$ (c) Ni is purified by mond's process (d) $\ {Zr}$ and $\ {Ti}$ are purified by van arkel method All (a), (b), (c), (d) are correct statements Thus correct option is (4)

Q5JEE Main 2020MCQ4MElectrochemistry

${{E}^{0}}_{ {Cu}^{2+}|{Cu}} =+0.34 {V}$ ${{E}^0}_{ {Zn}^{2+}|{Zn} }=-0.76 {V}$ Identify the incorrect statement from the options below for the above cell:

🚀 Solve in Practice Mode

📖 Explanation

$\;{E^{0}}_{\text{cell}}={E^{0}}_{C u^{2+} \mid C u}-{E^{0}}_{Z n^{2+} \mid Z n}$ $\;\; =0.34-(-0.76)$ $\;\; =1.10 {V}$ If ${E}_{ {ext}}< 1.1 {V}$ then ${Zn}$ dissolves at anode and copper deposits at Cathode. If ${E}_{\;\text{ext }\;}>1.1 {V}$ then ${Zn}$ deposited at zinc electrodes and ${Cu}$ deposits at ${Cu}$ electrode.

Q6JEE Main 2020MCQ4MAldehydes, Ketones and Carboxylic Acids

The IUPAC name of the following compound is :

🚀 Solve in Practice Mode

Q7JEE Main 2020MCQ4MChemical Equilibrium

For the equilibrium $\ {A} ⇌{B},$ the variation of the rate of the forward (a) and reverse (b) reaction with time is given by

🚀 Solve in Practice Mode

📖 Explanation

at equilibrium ${r}_{ {a}}= {r}_{ {b}}$

Q8JEE Main 2020MCQ4MHaloalkanes and Haloarenes

The decreasing order of reactivity of the following organic molecules towards AgNO $_{3}$ solution is :

🚀 Solve in Practice Mode

Q9JEE Main 2020MCQ4MAlcohols, Phenols and Ethers

An organic compound (A) (molecular formula ${C}_{6} {H}_{12} {O}_{2}$ ) was hydrolysed with dil. ${H}_{2} {SO}_{4}$ togive a carboxylic acid (B) and an alcohol (C). 'C' give white turbidity immediately when treated with anhydrous ${ZnCl}_{2}$ and conc. HCl. The organic compound (A) is :

🚀 Solve in Practice Mode

Q10JEE Main 2020MCQ4MSolutions

Match the following : (i) Foam (a) smoke (ii) Gel (b) cell fluid (iii) Aerosol (c) jellies (iv) Emulsion (d) rubber (e) froth (f) milk

🚀 Solve in Practice Mode

Q11JEE Main 2020MCQ4Md and f Block Elements

The elements with atomic numbers 101 and 104 belong to, respectively:

🚀 Solve in Practice Mode

Q12JEE Main 2020MCQ4MClassification of Elements

On combustion Li, Na and K in excess of air, the major oxides formed, respectively, are :

🚀 Solve in Practice Mode

📖 Explanation

Action of oxygen on alkali metals - Nature of oxides formed depends upon the nature of alkali metal $2 N a+O_{2} \to N a_{2} O_{2}($ peroxide $)$ $M+O_{2} \to M O_{2}($ super oxide $)(M=K, R b, C s)$ $4 L i+O_{2} \to 2 L i_{2} O($ mono oxide $)$ Li form $L i_{2} O$ with air Na form $Na _{2} O _{2}$ with air $K$ form $K O_{2}$ with air

Q13JEE Main 2020MCQ4MAldehydes, Ketones and Carboxylic Acids

$[\ {P}]$ on treatment with $\ {Br}_{2} / {FeBr}_{3}$ in $\ {CCl}_{4}$ produced a single isomer $\ {C}_{8} {H}_{7} {O}_{2}$ . Br while heating [P] with sodalime gave toluene. The compound [P] is :

🚀 Solve in Practice Mode

Q14JEE Main 2020MCQ4MCoordination Compounds

The number of isomers possible for $[{Pt}( {en})({NO}_{2} )_{2} ]$ is :

🚀 Solve in Practice Mode

📖 Explanation

$[Pt (en) (NO_2 )_2 ]$ \Rightarrow does not show G.I. as well as optical isomerismThis complex will have three linkage isomers as follows :- ${Pt}(\ {en})({NO}_{2} ) 2 ] \ {I}$ $[{Pt}(\ {en})({NO}_{2}\ )( {ONO}) ] \ {II}$ $[{Pt}(\ {en})( {ONO})_{2}]\ {III}$

Q15JEE Main 2020MCQ4MClassification of Elements

The ionic radii of $O_2 ^{-}, F ^{-}, Na ^{+}$ and $Mg^{2+}$ are in the order :

🚀 Solve in Practice Mode

📖 Explanation

$\text{table} ,O_2 ^{-}, F ^{-}, Na ^{+} ,Mg^{2+};z ,8 ,9, 11, 12;e^{ -} ,10, 10 ,10 ,10;z/e,0.8 ,0.9, 1.1 ,1.2$ as $\frac{{z}}{ {e}}$ ratio increases size decreases. Thus correct ionic radii order is $O_2 ^{-}> F ^{-} > Na ^{+} > Mg^{2+}$ Therefore correct option is (4)

Q16JEE Main 2020MCQ4MStructure of Atom

The region in the electromagnetic spectrum where the Balmer series lines appear is

🚀 Solve in Practice Mode

Q17JEE Main 2020MCQ4MChemical Bonding and Molecular Structure

The intermolecular potential energy for the molecules A, B, C and D given below suggests that :

🚀 Solve in Practice Mode

Q18JEE Main 2020MCQ4MBiomolecules

What are the functional groups present in the structure of maltose ?

🚀 Solve in Practice Mode

Q19JEE Main 2020MCQ4MChemical Thermodynamics

For one mole of an ideal gas, which of these statements must be true ? (a) U and H each depends only on temperature (b) Compressibility factor z is not equal to 1 (c) $C_{P,m} -C_{V,m} = R$ (d) $dU = C_V dT$ for any process

🚀 Solve in Practice Mode

📖 Explanation

For ideal Gas $\# {U}= {f}( {T}), {H}= {f}( {T})$ $\# {Z}=1$ $\# {C}_{ {P}}- {C}_{{V}}= {R}$ $\# {dU}= {C}_{{V}} {dT}$

Q20JEE Main 2020MCQ4MCoordination Compounds

The pair in which both the species have the same magnetic moment (spin only) is :

🚀 Solve in Practice Mode

📖 Explanation

Thus complex $[Cr(H _2O) _6 ] ^{2+}$ and $[Fe(H _2O)_6 ] ^{2+}$ have same no. of unpaired $\ {e}^{-}$ and hence same magnetic moment (spin only).

Q21JEE Main 2020NAT4MSome Basic Concepts of Chemistry

The mass of ammonia in grams produced when $2.8 \ {kg}$ of dinitrogen quantitatively reacts with $1 \ {kg}$ of dihydrogen is ______.

🚀 Solve in Practice Mode

📖 Explanation

$\text{table} N _2 ,+ ,3H_2, \to ,2NH_3; \frac{2.8}{28} {K} \ {mol} ,,\frac{1}{2} {K} \ {mol};= 0.1 K mol ,,0.5 K mol,, -;0,, 0.2 K mol ,,0.2 K mol$ $\text{mass} (NH_3) = 0.2 \times 17 Kg$ $= 3.4 Kg$ $= 3400 gm$

Q22JEE Main 2020NAT4MOrganic Chemistry - Some Basic Principles

The number of chiral centres present in [B] is _______.

🚀 Solve in Practice Mode

Q23JEE Main 2020NAT4MRedox Reactions

A $20.0 {mL}$ solution containing $0.2 \ {g}$ impure $\ {H}_{2} {O}_{2}$ reacts completely with $0.316 \ {g}$ of ${KMnO}_{4}$ in acid solution. The purity of ${H}_{2} \ {O}_{2}$ (in $\%$ ) is ______ (mol. wt. of ${H}_{2} {O}_{2}=34 ;$ mol. wt. of ${KMnO}_{4}=158 )$

🚀 Solve in Practice Mode

📖 Explanation

Eq of ${H}_{2} {O}_{2}= {Eq}$ of ${KMnO}_{4}$ $x \\times 2=\ \frac{0.316}{158} \\times 5$ $x=5 \\times 10^{-3} \ {mol}$ $\ {m}_{ {H}_{2} {O}_{2}}=5 \\times 10^{-3} \\times 34$ $=0.17 \ {gm}$ $\% \ {H}_{2} {O}_{2}= \frac{0.17}{0.2}$ $\\times 100=85$

Q24JEE Main 2020NAT4MChemical Kinetics

If $75 \%$ of a first order reaction was completed in 90 minutes, $60 \%$ of the same reaction would be completed in approximately (in minutes) ______. (Take $: \log 2=0.30 ; \log 2.5=0.40)$

🚀 Solve in Practice Mode

📖 Explanation

${t}_{0.75}=2\times \frac{\ln 2}{\ {k}}=90$ $\ {k}= \frac{\ln 2}{45} \ {\min}^{-1}$ $\ {kt}=\ln \frac{1}{1-0.6}=\ln 2.5$ $\frac{\ln 2}{45}\times {t}=\ln 2.5$ ${t} =45 \times \frac{\log 2.5}{\log 2}$ $=45\times \frac{0.4}{0.3}=60 \ {\min}$

Q25JEE Main 2020NAT4MSolutions

At 300 K, the vapour pressure of a solution containing 1 mole of n-hexane and 3 moles of n-heptane is 550 mm of Hg. At the same temperature, if one more mole of n-heptane is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. What is the vapour pressure in mm Hg of n-heptane in its pure state _________ ?

🚀 Solve in Practice Mode

📖 Explanation

$550= {P}_{ {A}}^{o}\times \frac{1}{4}+ {P}_{ {B}}^{ {o}} \times \frac{3}{4}$ $2200= {P}_{ {A}}^{{o}}+3 {P}_{ {B}}^{ {o}}$ ......(i) $2800= {P}_{ {A}}^{ {o}}+4 {P}_{ {B}}^{ {o}}$ .......(ii) $560= {P}_{ {A}}^{ {o}}\times \frac{1}{5}+ {P}_{ {B}}^{ {o}} \times \frac{4}{5}$ ${P}_{ {B}}^{ {o}}=600, {P}_{ {A}}^{ {o}}=400$

MATHEMATICS25 questions

Q26JEE Main 2020MCQ4MMatrices and Determinants

If $A=\begin{bmatrix}\cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta\end{bmatrix},$ $(\theta =\pi /24)$ and $A^5=\begin{bmatrix}a & b \\ c & d\end{bmatrix},$ where $i =\sqrt{-1}$ , then which one of the following is not true?

🚀 Solve in Practice Mode

📖 Explanation

$A^2=\begin{bmatrix}\cos 2\theta & i\sin 2\theta \\ i\sin 2\theta & \cos 2\theta\end{bmatrix}$ Similarly, $A^5=\begin{bmatrix}\cos 5\theta & i\sin 5\theta \\ i\sin5 \theta & \cos 5\theta\end{bmatrix}$ $= (\text{table} a,b;c,d)$ (1) $a^{2}+b^{2}$ $=\cos ^{2} 5 \theta -\sin ^{2} 5 \theta$ $=\cos 10 \theta =\cos 75°$ (2) $a^{2}-d^{2}$ $=\cos ^{2} 5 \theta -\cos ^{2} 5 \theta =0$ (3) $a^{2}-b^{2}$ $=\cos ^{2} 5 \theta +\sin ^{2} 5 \theta =1$ (4) $a^{2}-c^{2}$ $=\cos ^{2} 5 \theta +\sin ^{2} 5 \theta =1$

Q27JEE Main 2020MCQ4MFunctions

Let [t] denote the greatest integer $\le {t}$ . Then the equation in ${x},[ {x}]^{2}+2[ {x}+2]-7=0$ has :

🚀 Solve in Practice Mode

📖 Explanation

$[x]^{2}+2[x+2]-7=0$ $\Rightarrow[x]^{2}+2[x]+4-7=0$ $\Rightarrow[x]=1,-3$ $\Rightarrowx \in [1,2) \cup [-3,-2)$

Q28JEE Main 2020MCQ4MQuadratic Equation and Inequalities

Let $\alpha$ and $\beta$ be the roots of $x^{2}-3 x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^{2}-6 x+q=0 .$ If $\alpha$ $\beta , \gamma , \delta$ form a geometric progression. Then ratio $(2 q+p):(2 q-p)$ is :

🚀 Solve in Practice Mode

📖 Explanation

$x^{2}-3 x+p=0\begin{cases}\alpha \\ \beta$ $\alpha & \beta \gamma \delta$ in G\end{cases}P. $\alpha +\alpha r=3$ ....(1) $x^{2}-6 x+q=0\begin{cases}\gamma \\ \delta$ $\alpha r^{2}+\alpha r^{3}=6$ \end{cases}...(2) $(2) \÷(1)$ $r^{2}=2$ So, $\frac{2 q+p}{2 q-p}= \frac{2 r^{5}+r}{2 r^{5}-r}$ $= \frac{2 r^{4}+1}{2 r^{4}-1}= \frac{9}{7}$

Q29JEE Main 2020MCQ4MEllipse

Let $\frac{x^{2}}{a^{2}}+ \frac{y^{2}}{b^{2}}=1(a>b)$ be a given ellipse length of whose latus rectum is $10 .$ If its eccentricity is the maximum value of the function, $\phi ( {t})= \frac{5}{12}+ {t}- {t}^{2},$ then ${a}^{2}+ {b}^ {2}$ is equal to:

🚀 Solve in Practice Mode

📖 Explanation

$\frac{x^{2}}{a^{2}}+ \frac{y^{2}}{b^{2}}=1(a>b) ;$ $\frac{2 b^{2}}{a}=10 \Rightarrowb^{2}=5 a$ ......(i) Now, $\phi (t)= \frac{5}{12}+t-t^{2}$ $= \frac{8}{12}- (t- \frac{1}{2} )^{2}$ $\phi ( {t})_{\max }= \frac{8}{12}= \frac{2}{3}= {e}$ $\Rightarrow {e}^{2}=1-\frac{{b}^{2}}{ {a}^{2}}=4/9$ ...... (ii) $\Rightarrow {a}^{2}=81$ (from(i) $\&$ (ii)) So, $a^{2}+b^{2}$ $=81+45=126$

Q30JEE Main 2020MCQ4MVector Algebra

A triangle $\ {ABC}$ lying in the first quadrant has two vertices as $\ {A}(1,2)$ and $\ {B}(3,1)$ $\angle {BAC}=90°,$ and ${\text{ar}}(\Delta{ABC})=5 \\sqrt{5}$ sq. units, then the abscissa of the vertex $C$ is:

🚀 Solve in Practice Mode

📖 Explanation

$(\frac{{K}-2}{ {h}-1})(\frac{1-2}{3-1} )$ $=-1 \Rightarrow {K}=2 {h}$ .....(1) $\\sqrt{5}| {h}-1|=10$ $\ \because [\Delta{ABC}]=5 \\sqrt{5}$ $\Rightarrow \frac{1}{2}(\\sqrt{5}) \\sqrt{( {h}-1)^{2}+(\ {K}-2)^{2}}$ $=5 \\sqrt{5}$ .....(2) $\Rightarrow {h}=2 \\sqrt{5}+1(\ {h}>0)$

Q31JEE Main 2020MCQ4MDefinite Integrals

Let $f(x)=|x-2|$ and $g(x)=f(f(x)), x \in [0,4]$ . Then $∫_{0 }^{3}( {g}( {x})- {f}( {x})) {d} {x}$ is equal to :

🚀 Solve in Practice Mode

📖 Explanation

$∫_{0 }^{3}{g}( {x})- {f}({x})=∫_{0 }^{3}{|| {x}-2|-2|} {d} {x}$ $-∫_{0 }^{3}{|\ {x}-2| }{d} {x}$ $=(\frac{1}{2} \\times 2 \\times 2+1+\ \frac{1}{2} \\times 1 \\times 1 )$ $-(\frac{1}{2} \\times 2 \\times 2+ \frac{1}{2} \\times 1 \\times 1 )$ $= (2+1+ \frac{1}{2} )$ $- (2+ \frac{1}{2} )=1$

Q32JEE Main 2020MCQ4MSets and Relations

Given the following two statements: $({S}_{1} ):( {q} ∨{p}) \to ( {p} \leftrightarrow ∼{q})$ is a tautology $({S}_{2} ): ∼{q} \wedge (∼{p} \leftrightarrow {q})$ is a fallacy. Then :

🚀 Solve in Practice Mode

📖 Explanation

Let TV(r) denotes truth value of a statement r. ${\text{Now}},$ if ${TV}( {p})= {TV}( {q})= {T}$ \Rightarrow ${TV}({S}_{1} )= {F}$ Also, if $T V(p)=T \& \ T V(q)=F$ $\Rightarrow {TV}({S}_{2} )= {T}$

Q33JEE Main 2020MCQ4MHyperbola

Let $\ {P}(3,3)$ be a point on the hyperbola, $\frac{x^{2}}{a^{2}}- \frac{y^{2}}{b^{2}}=1 .$ If the normal to it at $P$ intesects the x-axis at (9,0) and e is its eccentricity, then the ordered pair $({a}^{2}, {e}^{2} )$ is equal to :

🚀 Solve in Practice Mode

📖 Explanation

Since, (3,3) lies on $\frac{x^{2}}{a^{2}}- \frac{y^{2}}{b^{2}}=1$ $\frac{9}{ {a}^{2}}- \frac{9}{ {b}^{2}}=1$ ....(1) Now, normal at (3,3) is ${y}-3=-\frac{{a}^{2}}{ {b}^{2}}( {x}-3)$ , which passes through $(9,0) \Rightarrow {b}^{2}=2 {a}^{2}$ ....(2) So, ${e}^{2}=1+\frac{{b}^{2}}{ {a}^{2}}=3$ Also, $a^{2}= \frac{9}{2}$ (from (i) & (ii)) Thus, $(a^{2}, e^{2} )=(\frac{9}{2}, 3 )$

Q34JEE Main 2020MCQ4MDefinite Integrals

Let $f(x)=∫\frac{\sqrt{x}}{(1+x)^{2}} d x$ $(x \ge 0) .$ Then $f(3)-f(1)$ is equal to :

🚀 Solve in Practice Mode

📖 Explanation

$f(x)=∫_{1} ^{3} \frac{\sqrt{x} d x}{(1+x)^{2}}$ $=∫_{1} ^{\sqrt3} \frac{t .2 t d t}{ (1+t^{2} )^{2}} ($ put $\sqrt{x}=t)$ $= (-\frac{{t}}{1+ {t}^{2}} )_{1}^{\\sqrt{3}}+ (\tan ^{-1} {t} )_{1}^{\\sqrt{3}} \;\;\;$ [Appling by parts $]$ $=-(\frac{\sqrt{3}}{4}- \frac{1}{2} )+ \frac{\pi }{3}- \frac{\pi }{4}$ $=\frac{\pi }{12}+ \frac{1}{2}-\frac{\sqrt{3}}{4}$

Q35JEE Main 2020MCQ4MSets and Relations

A survey shows that 63% of the people in a city read newspaper A whereas 76% read newspaper B. If x% of the people read both the newspapers, then a possible value of x can be:

🚀 Solve in Practice Mode

📖 Explanation

${n}({B}) \le {n}( {A} ∪{B}) \le {n}( {U})$ $\Rightarrow76 \le 76+63- {x} \ \le 100$ $\Rightarrow-63 \le -\ {x} \le -39$ $\Rightarrow63 \ge {x} \ge 39$

Q36JEE Main 2020MCQ4MComplex Numbers

Let ${u}= \frac{2 {z}+ {i}}{ {z}- {ki}}, {z}= {x}+ {iy}$ and ${k}>0 .$ If the curve represented by ${\text{Re}}( {u})+ {\text{Im}}( {u})=1$ intersects the $\ {y}$ -axis at the points $\ {P}$ and $\ {Q}$ where $\ {PQ}=5,$ then the value of $\ {k}$ is :

🚀 Solve in Practice Mode

📖 Explanation

${u}= \frac{2 {z}+ {i}}{ {z}- {ki}}$ $= \frac{2 x^{2}+(2 y+1)(y-k)}{x^{2}+(y-k)^{2}}$ $+i \frac{(x(2 y+1)-2 x(y-k))}{x^{2}+(y-k)^{2}}$ Since ${\text{Re}}( {u})+ {\text{Im}}( {u})=1$ $\Rightarrow 2 {x}^{2}+(2 {y}+1)( {y}- {k})$ $+ {x}(2 {y}+1)-2 {x}( {y}- {k})$ $= {x}^{2}+( {y}- {k})^{2}$ $\text{table} P(0,y_1);Q(0,y_2)\}\Rightarrow y^2 + y -k -k^2 = 0\{\text{table} {y}_{1}+ {y}_{2}=-1; {y}_{1} {y}_{2}=- {k}- {k}^{2}$ $\because {PQ}=5$ $\Rightarrow|{y}_{1}- {y}_{2} {I}=5 \Rightarrow {k}^{2}+ {k}-6=0$ $\Rightarrow {k}=-3,2$ ${\text{So}}, {k}=2( {k}>0)$

Q37JEE Main 2020MCQ4MApplication of Derivatives

Let $x_{0}$ be the point of local maxima of ${f}(x)= \vec{a} .( \vec{b} \times \vec{c}),$ where ${\vec{a}}= {x} \hat{i}-2 \hat{j}+3 \hat{k}$ , ${\vec{b}}=-2\hat{i}+ {x} \hat{j}-\hat{k}$ and ${\vec{c}}=7\hat{i}-2 \hat{j}+ {x} \hat{k}.$ Then the value of ${\vec{a}}.{\vec{b}}+{\vec{b}} .{\vec{c}}+{\vec{c}}.{\vec{a}}$ at ${x}= {x}_{0}$ is :

🚀 Solve in Practice Mode

📖 Explanation

${f}(x)={\vec{a}} .({\vec{b}} \times {\vec{c}})$ $=\begin{vmatrix}x & -2 & 2 \\ -2 & x & -1 \\ 7 & -2 & x\end{vmatrix}$ $=x^{3-}27x +26$ $f'(x)=3 x^{2}-27=0 \Rightarrowx=\±3$ and $"(-3)<0$ $\Rightarrow$ local maxima at $x=x_{0}=-3$ Thus, $\vec{a}=-3 \hat{i}-2 \hat{j}+3 \hat{k}$ , $\vec{b}=-2 \hat{i}-3 \hat{j}- \hat{k}$ , and $\vec{c}=7 \hat{i}-2 \hat{j}-3 \hat{k}$ \Rightarrow ${\vec{a}}.{\vec{b}}+{\vec{b}} .{\vec{c}}+{\vec{c}}.{\vec{a}}$ $=9-5-26=-22$

Q38JEE Main 2020MCQ4MStraight Lines and Pair of Straight Lines

Two vertical poles $\ {AB}=15 \ {m}$ and $\ {CD}=10 \ {m}$ are standing apart on a horizontal ground with points $\ {A}$ and $\ {C}$ on the ground. If $\ {P}$ is the point of intersection of $\ {BC}$ and $\ {AD},$ then the height of $\ {P}$ (in $\ {m}$ ) above the line $\ {AC}$ is :

🚀 Solve in Practice Mode

📖 Explanation

$\tan \theta = \frac{10}{x}= \frac{h}{x_{2}} \Rightarrowx_{2}= \frac{h x}{10}$ $\tan \phi = \frac{15}{x}= \frac{h}{x_{1}} \Rightarrowx_{1}= \frac{h x}{15}$ Now, $x_{1}+x_{2}=x= \frac{h x}{15}+ \frac{h x}{10}$ $\Rightarrow1=\ \frac{h}{10}+\ \frac{h}{15} \Rightarrow h=6$

Q39JEE Main 2020MCQ4MStatistics

The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is :

🚀 Solve in Practice Mode

📖 Explanation

$\overline{x}=10$ $\Rightarrow {\ov x}= \frac{63+a+b}{8}=10$ $\Rightarrow a+b=17$ .....(1) since, variance is independent of origin. So, we subtract 10 from each observation. So, $\sigma ^{2}=13.5$ $= \frac{79+(a-10)^{2}+(b-10)^{2}}{8}$ $-(10-10)^{2}$ $\Rightarrow {a}^{2}+ {b}^{2}-20( {a}+ {b})=-171$ $\Rightarrow {a}^{2}+ {b}^{2}=169$ From(i) $\&$ (ii) ; ${a}=12\ \&\ {b}=5$

Q40JEE Main 2020MCQ4MIndefinite Integrals

The integral $∫(\frac{{x}}{ {x} \sin {x}+\cos {x}} )^{2} {d} {x}$ is equal to : (where $\ {C}$ is a constant of integration)

🚀 Solve in Practice Mode

📖 Explanation

$∫(\frac{{x}}{ {x} \sin {x}+\cos {x}} )^{2} {d} {x}$ $=∫(\frac{{x}}{\cos {x}} ) .\frac{ {x} \cos {x} {d} {x}}{( {x} \sin {x}+\cos {x})^{2}}$ $= \frac{x}{\cos x}(- \frac{1}{x \sin x+\cos x} )$ $+∫( \frac{\cos x+x \sin x}{\cos ^{2} x}$ (\frac{1}{x \sin x+\cos x} ) d x=- \frac{x \sec x}{x \sin x+\cos }+∫ \sec ^{2} x d x=- \frac{x \sec x}{x \sin x+\cos }+\tan x+C$

Q41JEE Main 2020MCQ4MSequences and Series

If $1+ (1-2^{2} . 1 )+ (1-4^{2} . 3 )$ $+ (1-6^{2} . 5 )+......+(1-20^{2} .19 )$ $=\alpha -220 \beta ,$ then an ordered pair $(\alpha , \beta )$ is equal to :

🚀 Solve in Practice Mode

📖 Explanation

$1+ (1-2^{2} . 1 )+ (1-4^{2} . 3 )$ $+ (1-6^{2} . 5 )+......+(1-20^{2} .19 )$ $=\alpha -220 \beta$ $=11-(2^{2} .1+4^{2} .3+....$ $+20^{2} .19 )$ $=11-2^{2} . \sum _{{r}=1} ^{10} {r}^{2}(2 {r}-1)$ $=11-4(\frac{110^{2}}{2}-35 \\times 11 )$ $=11-220(103)$ $\Rightarrow \alpha =11, \beta =103$

Q42JEE Main 2020MCQ4MDifferential Equations

Let $y=y(x)$ be the solution of the differential equation, $x y'-y=x^{2}(x \cos x+\sin x),$ $x>0$ If $\ {y}(\pi )=\pi ,$ then $\ {y}"( \frac{\pi }{2} )+ {y}(\frac{\pi }{2} )$ is equal to :

🚀 Solve in Practice Mode

📖 Explanation

$x \frac{dy}{dx}-y=x^{2}(x \cos x+\sin x),$ $x>0$ $\frac{{dy}}{ {dx}}-\frac{{y}}{ {x}}= {x}({x} \cos{x}+\sin {x})$ $\Rightarrow \frac{{dy}}{ {dx}}+ {Py}= {Q}$ so, I.F. $= {e}^{∫- \frac{1}{ {x}} {dx}}= \frac{1}{|\ {x}|}$ $= \frac{1}{ {x}}( {x}>0)$ Thus, $\frac{{y}}{ {x}}=∫\frac{1}{ {x}}( {x}( {x} \cos {x}+\sin {x})) {d} {x}$ $\Rightarrow \frac{{y}}{ {x}}= {x} \sin {x}+ {C}$ $\because {y}(\pi )=\pi \Rightarrow {C}=1$ so, ${y}= {x}^{2} \sin {x}+{x}$ $\Rightarrow ({y})_{\pi / 2}= \frac{\pi ^{2}}{4}+\ \frac{\pi }{2}$ Also, $\frac{{dy}}{ {dx}}= {x}^{2} \cos {x}$ $+2 {x} \sin {x}+1$ $\Rightarrow \frac{{d}^{2} {y}}{ {dx}^{2}}=- {x}^{2} \sin {x}$ $+4 {x} \cos {x}+2 \sin {x}$ $\Rightarrow {\frac{{d}^{2} {y}}{ {dx}^{2}} |}_{\frac{\pi }{2}}=- \frac{\pi ^{2}}{4}+2$ Thus, ${y}_{(\frac{\pi }{2} )}+\frac{{d}^{2} {y}}{ {dx}^{2}(\frac{\pi }{2} )}= \frac{\pi }{2}+2$

Q43JEE Main 2020MCQ4MBinomial Theorem

The value of $\sum _{ {r}=0}^{20}{ }_{0}^{50-\ {r}} {C}_{6}$ is equal to :

🚀 Solve in Practice Mode

📖 Explanation

$\sum _{ {r}=0}^{20}{ }^{50-r} C_{6}={ }^{50} C_{6}+{ }^{49} C_{6}$ $+{ }^{48} C_{6}+....+{ }^{30} C_{6}$ $={ }^{50} {C}_{6}+{ }^{49} {C}_{6}+...+{ }^{31} {C}_{6}$ $+ ({ }^{30} {C}_{6} +{ }^{30} {C}_{7} )-{ }^{30} {C}_{7}$ $={ }^{50} {C}_{6}+{ }^{49} {C}_{6}+....$ $+ ({ }^{31} {C}_{6}+{ }^{31} {C}_{7} )-{ }^{30} {C}_{7}$ $={ }^{50} {C}_{6}+{ }^{50} {C}_{7}-{ }^{30} {C}_{7}$ $={ }^{51} {C}_{7}-{ }^{30} {C}_{7}$ $[{ }^{ {n}} {C}_{ {r}}+{ }^{ {n}} {C}_{{r}-1}={ }^{ {n}+1} {C}_{ {r}}]$

Q44JEE Main 2020MCQ4MApplication of Derivatives

Let $\ {f}$ be a twice differentiable function on $(1,6) .$ If $\ {f}(2)=8, {f}'(2)=5, {f}'( {x}) \ge 1$ and ${f}"( {x}) \ge 4,$ for all $\ {x} \in (1,6),$ then :

🚀 Solve in Practice Mode

📖 Explanation

Given, $f^{'}(x) \ge 1$ $\therefore ∫_{2}^{5} f^{'}(x) d x \ge ∫_{2}^{5} d x$ $\Rightarrow f(5)-f(2) \ge 3$ $\Rightarrow f(5)-8 \ge 3$ $\Rightarrow f(5) \ge 11 . . .(1)$ $\;\text{ Also, }\; f^{' '}(x) \ge 4$ $\therefore ∫_{2}^{5} f^{' '}(x) d x \ge ∫_{2}^{5} 4 d x$ $\Rightarrow f^{'}(5)-f^{'}(2) \ge 4(3)$ $\Rightarrow f^{'}(5)-5 \ge 12$ $\Rightarrow f^{'}(5) \ge 17 . . .(2)$ From (1) and (2), $\;f^{'}(5)+f^{'}(5) \ge 11+17$ $\Rightarrow f^{'}(5)+f^{'}(5) \ge 28$

Q45JEE Main 2020MCQ4MDifferentiation

If $(a+\\sqrt{2} b \cos x)(a-\\sqrt{2} b \cos y)$ $=a^{2}-b^{2}$ where $a>b>0,$ then $\ \frac{d x}{d y}$ at $$\frac{\pi }{4}, \frac{\pi }{4} )$ is :

🚀 Solve in Practice Mode

📖 Explanation

$(a+\\sqrt{2} b \cos x)(a-\\sqrt{2} b \cos y)$ $=a^{2}-b^{2}$ $\Rightarrow {a}^{2}-\\sqrt{2} {ab\cos} {y}$ $+\\sqrt{2}{ab} \cos {x}$ $-2 {b}^{2} \cos {x} \cos {y}= {a}^{2}- {b}^{2}$ Differentiating both sides : $0-\\sqrt{2} {ab}(-\sin {y} \frac{{dy}}{ {dx}} )+\\sqrt{2} {ab}(-\sin {x})$ $-2 b^{2} [\cos x (-\sin y \frac{d y}{d x} )$ $+\cos y(-\sin x) ]=0$ At $(\frac{\pi }{4}, \frac{\pi }{4} ):$ ab $\frac{d y}{d x}-a b-2 b^{2} (- \frac{1}{2} \frac{d y}{d x}- \frac{1}{2} )=0$ $\Rightarrow \frac{d x}{d y}= \frac{a b+b^{2}}{a b-b^{2}}$ $= \frac{a+b}{a-b} ; \;\;a, b>0$

Q46JEE Main 2020NAT4MMatrices and Determinants

If the system of equations $x-2 y+3 z=9$ $2 x+y+z=b$ $x-7 y+a z=24$ has infinitely many solutions, then $\ {a}-\ {b}$ is equalto _______.

🚀 Solve in Practice Mode

📖 Explanation

$D=\begin{vmatrix}1 & -2 & 3 \\ 2 & 1 & 1 \\ 1 & -7 & a\end{vmatrix}$ $=0\Rightarrow a=8$ also, $D_1=\begin{vmatrix}9 & -2 & 3 \\ b & 1 & 1 \\ 24 & -7 & 8\end{vmatrix}$ $0=\Rightarrow b=3$ hence, $a - b = 8 - 3 = 5$

Q47JEE Main 2020NAT4MProbability

The probability of a man hitting a target is $\frac{1}{10}$ . The least number of shots required, so that the probability of his hitting the target at least once is greater than $\frac{1}{4},$ is ______.

🚀 Solve in Practice Mode

📖 Explanation

We have, $1-$ (probability of all shots result irfailure ) $>\ \frac{1}{4}$ $\Rightarrow1-(\frac{9}{10} )^{n}> \frac{1}{4}$ $\Rightarrow \frac{3}{4}>(\frac{9}{10} )^{n} \Rightarrow {n} \ge 3$

Q48JEE Main 2020NAT4MLimits, Continuity and Differentiability

Suppose a differentiable function $\ {f}( {x})$ satisfies the identity $f(x+y)=f(x)+f(y)+x y^{2}$ $+x^{2} y$ for all real $\ {x}$ and $\ {y} .$ If $\lim _{\ {x} \to 0} \frac{{f}({x})}{ {x}}=1,$ then ${f}'(3)$ is equal to _______.

🚀 Solve in Practice Mode

📖 Explanation

since, $\lim _{\ {x} \to 0}\frac{f(x)}{x}$ exist $\Rightarrow f(0)=0$ Now, $f'(x)=\lim _{\ {h} \to 0}\frac{f(x+h)-f(x)}{h}$ $=\lim _{\ {h} \to 0}\frac{f(h)+x h^{2}+x^{2} h}{h}($ take $y=h)$ $=\lim_{\ {h} \to 0}\frac{f(h)}{h}+\lim_{\ {h} \to 0}(x h)+x^{2}$ $\Rightarrow f'(x)=1+0+x^{2}$ $\Rightarrowf'(3)=10$

Q49JEE Main 2020NAT4MBinomial Theorem

Let $(2 x^{2}+3 x+4 )^{10}=\sum _{r=0}^{20} a_{r} x^{r} .$ Then $\frac{a_{7}}{a_{13}}$ is equal to _______.

🚀 Solve in Practice Mode

📖 Explanation

Given $(2 x^{2}+3 x+4 )^{10}=\sum _{r=0}^{20} a_{r}\ x^{r}$ .... (1) replace ${x}$ by $\frac{2}{ {x}}$ in above identity :- $\frac{2^{10} (2 x^{2}+3 x+4 )^{10}}{x^{20}}$ $=\sum _{r=0}^{20}\frac{a_{r}\ 2^{r}}{x^{r}}$ $\Rightarrow2^{10}\sum _{r=0}^{20} a_{r}\ x^{r}=\sum _{r=0}^{20}a_{r} \ 2^{r} \ x^{(20-r)}$ (from (i)) now, comparing coefficient of $x^{7}$ from both sides (take ${r}=7$ in L.H.S. & $\ {r}=13$ in ${R} . \ {H.S.})$ $2^{10} {a}_{7}= {a}_{13} \ 2^{13} \Rightarrow \frac{{a}_{7}}{ {a}_{13}}=2^{3}=8$

Q50JEE Main 2020NAT4MThree Dimensional Geometry

If the equation of a plane P, passing through the intesection of the planes, $x+4 y-z+7=0$ and $3 x+y+5 z=8$ is $a x+b y+6 z=15$ for some $\ {a}, \ {b} \in {R},$ then the distance of the point $(3,2,$ -1 ) from the plane $P$ is ______.

🚀 Solve in Practice Mode

📖 Explanation

$D_1=\begin{vmatrix}-7 & 4 & -1 \\ 8 & 1 & 5 \\ 15 & b & 6\end{vmatrix}$ $=0\Rightarrow b=-3$ $D=\begin{vmatrix}1 & 4 & -1 \\ 3 & 1 & 5 \\ a & b & 6\end{vmatrix} =0$ $\Rightarrow21 {a}-8 {b}-66=0$ .... (1) $P: 2 x-3 y+6 z=15$ so required distance $= \frac{21}{7}=3$

PHYSICS25 questions

Q51JEE Main 2020MCQ4MWave Optics

A beam of plane polarised light of large cross sectional area and uniform intensity of 3.3 $\ {Wm}^{-2}$ falls normally on a polariser (cross sectional area $3 \\times 10^{-4} \ {m}^{2}$ ) which rotates about its axis with an angular speed of $31.4 \ {rad} / {s}$ . The energy of light passing through the polariser per revolution, is close to :

🚀 Solve in Practice Mode

📖 Explanation

Intensity, $\ {I}=3.3 \ {Wm}^{-2}$ Area, $A=3 \\times 10^{-4 }\ {m}^{2}$ Angular speed, $\omega =31.4 \ {rad} / {s}$ $\because<\cos ^{2} \theta >=\ \frac{1}{2},$ in one time period $\therefore$ Average energy $=\ {I}_{0} \ {A} \times \frac{1}{2}$ $= \frac{(3.3) (3 \\times 10^{-4} )}{2}$ $\ ≃ 5 \\times 10^{-4} \ {J}$

Q52JEE Main 2020MCQ4MThermodynamics

Match the ${C}_{ {p}} / {C}_{ {V}}$ ratio for ideal gases with different type of molecules: Molecular type ${C}_{ {p}} / {C}_{ {V}}$ (A) Monoatomic (I) $7 / 5$ (B) Diatomic rigidmolecules (II) $9 / 7$ (C) Diatomic non-rigidmolecules (III) $4 / 3$ (D) Triatomic rigidmolecules (IV) $5 / 3$

🚀 Solve in Practice Mode

📖 Explanation

$\gamma = \frac{C_{p}}{C_{v}}=1+ \frac{2}{f}$ where 'f' is degree of freedom (A) Monoatomic $\ {f}=3, \gamma =1+ \frac{2}{3}= \frac{5}{3}$ (B) Diatomic rigid molecules, ${f}=5, \gamma =1+\ \frac{2}{3}= \frac{7}{5}$ (C) Diatomic non-rigid molecules ${f}=7, \gamma =1+ \frac{2}{7}= {9}/ {7}$ (D) Triatomic rigid molecules ${f}=6, \gamma =1+ \frac{2}{6}= \frac{4}{3}$

Q53JEE Main 2020MCQ4MElectromagnetic Waves

Choose the correct option relating wavelengthsof differnet parts of electromagnetic wavespectrum :

🚀 Solve in Practice Mode

📖 Explanation

Information based $\lambda _{\text{radio waves }} > \lambda _{\text{micro waves}} > \lambda _{\text{visible }}$ $>\lambda _{\text{x-rays }}$

Q54JEE Main 2020MCQ4MProperties of Matter

A air bubble of radius $1 \ {cm}$ in water has an upward acceleration $9.8 \ {cm} \ {s}^{-2}$ . The density of water is $1 \ {gm} \ {cm}^{-3}$ and water offers negligible drag force on the bubble. The mass of the bubble is $( {g}=980 \ {cm} / \ {s}^{2} )$

🚀 Solve in Practice Mode

📖 Explanation

Volume ${V}= \frac{4 \pi }{3} \ {r}^{3}$ $= \frac{4 \pi }{3} \\times (1)^{3}=4.19 \ {cm}^{3}$ $\ {a}=9.8 \ {cm} / {s}^{2}$ $\ {B}-\ {mg}=\ {ma}$ ${m}=\frac{{B}}{ {g}+ {a}}$ ${m}=\frac{({V} \rho _{\omega } {g} )}{ {g}+ {a}}={{V} \rho _{\omega }}/{1+ \frac{ {a}}{ {g}}}$ $=\ \frac{(4.19) \\times 1}{1+\ \frac{9.8}{980}}$ $=\ \frac{4.19}{1.01}=4.15 \ {gm}$

Q55JEE Main 2020MCQ4MThermodynamics

Dimensional formula for thermal conductivity is (here $\ {K}$ denotes the temperature)

🚀 Solve in Practice Mode

📖 Explanation

$\because \frac{{d} \theta }{ {dt}}= {kA} \frac{{dT}}{ {dx}}$ ${k}={(\frac{{d} \theta }{ {dt}} )}/{ {A}(\frac{{dT}}{\ {dx}} )}$ $[ {k}]= \frac{[{ML}^{2} {T}^{-3} ]\$ }{[{L}^{2} ]$$[{KL}^{-1} ]$}=[{MLT}^{-3} {K}^{-1} ]$

Q56JEE Main 2020MCQ4MGravitation

On the x-axis and a dsitance $x$ from the origin, the gravitational field due to a mass distribution is given by $\frac{{Ax}}{({x}^{2}+ {a}^{2} )^{3 / 2}}$ in the $\ {x}$ -direction. The magnitude of gravitational potential on the $\ {x}$ -axis at a distance $\ {x},$ taking its value to be zero at infinity, is :

🚀 Solve in Practice Mode

📖 Explanation

Given ${E}_{ {G}}=\frac{{Ax}}{({x}^{2}+ {a}^{2} )^{3 / 2}}, {V}_{\infty }=0$ $∫_{{\ {V}_{\infty }} } ^{V_x}{dV}=-∫_{\infty }^ {\ {x}} {\vec{E}}_{ {G}}.{\vec{d}}_{ {x}}$ ${V}_{ {x}}- {V}_{\infty }=-∫_{\infty }^ {\ {x}} {{Ax}}/{{{x}^{2}+ {a}^{2} )^{3 / 2}} {d} {x}$ put $x^{2}+a^{2}=z$ $2 x d x=d z$ ${V}_{ {x}}-0=-∫_{\infty }^ {\ {x}} \frac{{A} {d} {z}}{2( {z})^{3 / 2}}=\$ [\frac{{A}}{ {z}^{1 / 2}} ]{∞ }^{ {x}}=$[\frac{{A}}{({x}^{2}+ {a}^{2} )^{1 / 2}} ]{\infty }^{ {x}}{V}_{ {x}}=\frac{{A}}{({x}^{2}+ {a}^{2} )^{1 / 2}}-0=\frac{{A}}{({x}^{2}+ {a}^{2} )^{1 / 2}}$

Q57JEE Main 2020MCQ4MMotion in a Plane

Starting from the origin at time $\ {t}=0,$ with initialvelocity $5\hat{j} \ {ms}^{-1},$ a particle moves in the $\ {x}-\ {y}$ plane with a constant acceleration of $(10 \hat{i}+4 \hat{j}) \ {ms}^{-2} .$ At time $\ {t},$ its coordinates are $(20 \ {m}, \ {y}_{0} \ {m} ) .$ The values of $\ {t}$ and $\ {y}_{0},$ are respectively:

🚀 Solve in Practice Mode

📖 Explanation

Given ${\vec{u}}=5\hat{j} {m} / {s}, {\vec{a}}=10 { \hat{i}}+4\hat{j}$ final coordinate $(20, \ {y}_{0} )$ in time $\ {t}$ ${S}_{ {x}}=4_{ {x}} {t}+ \frac{1}{2} {a}_{ {x}} {t}^{2}$ $20-0=0+ \frac{1}{2} \\times 10 \times {t}^{2}$ $\ {t}=2 \ {sec}$ $\ {S}_{ {y}}= {u}_{ {y}} \times {t}+ \frac{1}{2} {a}_{ {y}} {t}^{2}$ ${y}_{0}=5 \\times 2+ \frac{1}{2} 4 \\times 2^{2}=18 \ {m}$ 2 sec and $18 \ {m}$

Q58JEE Main 2020MCQ4MDual Nature of Matter and Radiation

Given figure shows few data points in a photo electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surfface is : (Plancks constant ${h}=6.62 \\times 10^{-34} \ {J} . \ {s} )$

🚀 Solve in Practice Mode

📖 Explanation

Graph of $\ {V}_{ {s}}$ and $\ {f}$ given $( {B} 5.5,0)$ ${h} v=\phi + {eV}_{ {s}}$ at B ${V}_{ {s}}=0,\;\; v=5.5$ $\Rightarrow \ {h} \\times 5.5 \\times 10^{14}=\phi$ $\phi = \frac{6.62 \\times 10^{-34} \\times 5.5 \\times 10^{14}}{1.6 \\times 10^{-19}} \ {eV}$ $=2.27 \ {eV}$

Q59JEE Main 2020MCQ4MMagnetic Effects of Current and Magnetism

A wire $\ {A},$ bent in the shape of an arc of a circle, carrying a current of $2 \ {A}$ and having radius $2 \ {cm}$ and another wire $\ {B},$ also bent in the shape of arc of a circle, carrying a current of $3 \ {A}$ and having radius of $4 \ {cm},$ are placed as shown in the figure. The ratio of the magnetic fields due to the wires $\ {A}$ and $\ {B}$ at the common centre $\ {O}$ is :

🚀 Solve in Practice Mode

📖 Explanation

Given ${i}_{ {A}}=2, {r}_{ {A}}=2 {cm},$ $\theta _{ {A}}=2 \pi - \frac{\pi }{2}= \frac{3 \pi }{2}$ ${i}_{ {B}}=3, {r}_{ {B}}=4 {cm},$ $\theta _{ {B}}=2 \pi - \frac{\pi }{3}= \frac{5 \pi }{3}$ ${B}= \frac{\mu_{0} {I} \theta }{4 \pi {R}}$ ${{B}_{ {A}}}/{ {B}_{ {B}}}={{I}_{ {A}}}/{ {I}_{ {B}}} \times {\theta _{ {A}} {R}_{ {B}}}/{\theta _{ {B}} {R}_{ {A}}}= \frac{6}{5}$

Q60JEE Main 2020MCQ4MWaves

For a transverse wave travelling along a straight line, the distance between two peaks (crests) is $5 \ {m},$ while the distance between one crest and one trough is $1.5 \ {m}$ . The possible wavelengths (in $\ {m}$ ) of the waves are :

🚀 Solve in Practice Mode

📖 Explanation

Given $\ {T}$ to $\ {C}\ 1.5 \ {m}$ $\ {C}$ to $\ {C}\ 5 \ {m}$ $\ {T}$ to $\ {C}= (2 {n}_{1}+1 ) \frac{\lambda}{2}$ $\ {C}$ to $\ {C}=\ {n}_{2} \lambda$ $\ \frac{1.5}{5}=\frac{(2 n_{1}+1 )}{2 n_{2}} \$ $\Rightarrow 3 n_{2}=10 n_{1}+5$ $\ {n}_{1}=1, \ {n}_{2}=5 \to \lambda=1$ $\ {n}_{1}=4, \ {n}_{2}=15 \to \lambda=1 / 3$ $\ {n}_{1}=7, \ {n}_{2}=25 \to \lambda=1 / 5$

Q61JEE Main 2020MCQ4MMagnetic Effects of Current and Magnetism

A small bar magnet placed with its axis at $30°$ with an external field of 0.06 T experiences a torque of 0.018 Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is :

🚀 Solve in Practice Mode

📖 Explanation

Torque on a bar magnet $: {I}= {MB} \sin \theta$ Here, $\theta =3o°, {I}=0.018 {N}- {m}, {B}=0.06 {T}$ $\Rightarrow0.018= {M} \\times 0.06 \\times \sin 30°$ $\Rightarrow 0.018= {M} \\times 0.06 \\times \frac{1}{2}$ $\Rightarrow {M}=0.6 {A}- {m}^{2}$ Now ${v}=- {MB} \cos \theta$ Position of stable equilibrium $(\theta =0°):$ ${u}_{ {i}}=- {MB}$ Position of unstable equilibrium $(\theta =180°):$ ${u}_{ {f}}= {MB}$ $\Rightarrow$ work done $: \Delta{U}$ $\Rightarrow {W}=2 \ {MB}$ $\Rightarrow{W}=2 \\times 0.6 \\times 0.06$ $\Rightarrow {W}=7.2 \\times 10^{-2} \ {J}$ option (4) is correct

Q62JEE Main 2020MCQ4MCentre of Mass and Collision

Particle $\ {A}$ of mass $\ {m}_{ {A}}=\frac{{m}}{2}$ moving along the x-axis with velocity $v_{0}$ collides elastically with another particle $\ {B}$ at rest having mass $\ {m}_{ {B}}=\frac{{m}}{3}$ . If both particles move along the x-axis after the collision, the change $\Delta\lambda$ in de-Brogliewavelength of particle $\ {A},$ in terms of its de-Broglie wavelength $(\lambda_{0} )$ before collision is :

🚀 Solve in Practice Mode

📖 Explanation

Applying momentum conservation $\frac{{m}}{2}\times {V}_{0}+\frac{{m}}{3} \\times (0)$ $=\frac{{m}}{2} {V}_{ {A}}+\frac{{m}}{3} {V}_{ {B}}$ $=\frac{{V}_{0}}{2}={{V}_{ {A}}}/{2}+{{V}_{ {B}}}/{3}$ .......(1) since, collision is elastic $(\mathrm{e}=1)$ ${e}=1={{V}_{ {B}}- {V}_{ {A}}}/{ {V}_{0}}$ $\Rightarrow {V}_{0}= {V}_{ {B}}- {V}_{ {A}}$ ....(2) On solving(1) $\ \&\ (2): {V}_{ {A}}=\frac{{V}_{0}}{5}$ Now, De-Broglie wavelength of A before collision: $\lambda_{0}={{h}}/{ {m}_{ {A}} {V}_{0}}={{h}}/{(\frac{{m}}{2} ) {V}_{0}}$ $\Rightarrow \lambda_{0}= \frac{2 {h}}{ {mV}_{0}}$ Final De-Broglie wavelength : $\lambda_{ {f}}={{h}}/{\ {m}_{ {A}} {V}_{0}}={{h}}/{\frac{{m}}{2} \\times \frac{ {V}_{0}}{5}}$ $\Rightarrow \lambda_{ {f}}= \frac{10 {h}}{\ {mV}_{0}}$ Now $\Delta\lambda=\lambda_{ {f}}-\lambda_{0}$ $\Delta \lambda= \frac{10 {h}}{ {mV}_{0}}- \frac{2 {h}}{ {mV}_{0}}$ $\Rightarrow \Delta\lambda= \frac{8 {h}}{ {mv}_{0}} \$ $\Rightarrow \Delta \lambda=4 \times \frac{2 {h}}{ {mv}_{0}}$ $\Rightarrow \Delta\lambda=4 \lambda_{0}$ option (1) is correct.

Q63JEE Main 2020MCQ4MElectromagnetic Induction

A small bar magnet is moved through a coil at constant speed from one end to the other. Which of the following series of observations wil be seen on the galvanometer G attached across the coil? Three positions shown describe :(a) the magnet's entry (b) magnet is completely inside and (c) magnet's exit.

🚀 Solve in Practice Mode

Q64JEE Main 2020MCQ4MCurrent Electricity

A battery of $3.0 \ {V}$ is connected to a resistor dissipating $0.5 \ {W}$ of power. If the terminal voltage of the battery is $2.5 \ {V},$ the power dissipated within the internal resistance is :

🚀 Solve in Practice Mode

📖 Explanation

${P}_{ {R}}=0.5 {W}$ $\Rightarrow {i}^{2} {R}=0.5 {W}$ Also, ${V}= {E}- {ir}$ $2.5=3- {ir}$ $\Rightarrow$ ir $=0.5$ Power dissipated across 'r' : ${P}_{ {r}}= {i}^{2} {r}$ Now iR $=2.5$ $\ {ir}=0.5$ On dividing $: \frac{{R}}{ {r}}=5$ Now $\frac{P_{R}}{P_{r}}= \frac{i^{2} R}{i^{2} r}\Rightarrow \frac{P_{R}}{P_{r}}=\ \frac{R}{r}$ $\Rightarrow \frac{P_{R}}{P_{r}}=5$ $\Rightarrow P_{ {r}}={{P}_{ {R}}}/{5}$ $\Rightarrow {P}_{ {r}}= \frac{0.50}{5}$ $\Rightarrow {P}_{ {r}}=0.10 {W}$ option (4) is correct.

Q65JEE Main 2020MCQ4MCentre of Mass and Collision

Blocks of masses ${m}, 2 {m}, 4 {m}$ and $8 {m}$ are arranged in a line on a frictionless floor. Another block of mass $\ {m},$ moving with speed $\ {v}$ along the same line (see figure) collides with mass $\ {m}$ in perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block of mass $8 \ {m}$ starts moving the total energy loss is p% of the original energy. Value of 'p' is close to :

🚀 Solve in Practice Mode

📖 Explanation

All collisions are perfectly inelastic, so after the final collision, all blocks are moving together. So let the final velocity be ${v}',$ so on applying momentum conservation: ${mv}=16 {m} {v}'\Rightarrow {v}'= {v} / 16$ Now initial energy ${E}_{ {i}}= \frac{1}{2} \ {mv}^{2}$ Final energy $: {E}_{ {f}}= \frac{1}{2} \\times 16 \ {m} \times (\frac{{v}}{16} )^{2}$ $\Rightarrow {E}_{ {f}}= \frac{1}{2} {m} \frac{{v}^{2}}{16}$ Energy loss : ${E}_{ {i}}- {E}_{ {f}}$ $\Rightarrow \frac{1}{2} {mv}^{2}- \frac{1}{2} {m} \frac{{v}^{2}}{16}$ $\Rightarrow \frac{1}{2} {mv}^{2} [1- \frac{1}{16} ] \Rightarrow \frac{1}{2} {mv}^{2}[{15} /{16} ]$ $\% {p}= \frac{\text{Energy loss}}{\text{ Original energy }} \\times 100$ $=\frac{ \frac{1}{2} {mv}^{2}[\frac{15}{16} ]}{ \frac{1}{2} {mv}^{2}} \\times 100$ $=93.75 \%$ $\Rightarrow$ Value of $\ {P}$ is close to 94.

Q66JEE Main 2020MCQ4MThermodynamics

The specific heat of water $=4200 {J} {kg}^{-1} {K}^{-1}$ and the latent heat of ice $=3.4\times 10^{5} \ {J} \ {kg}^{-1}$ 100 grams of ice at $0° {C}$ is placed in $200 \ {g}$ of water at $25°{C}$ . The amount of ice that will melt as the temperature of water reaches $0°{C}$ is close to (in grams):

🚀 Solve in Practice Mode

📖 Explanation

Here the water will provide heat for ice to melt therefore ${m}_{ {w}} {s}_{ {w}} \Delta\theta = {m}_{ {\text{ice}}} {L}_{ {\text{ice}}}$ $\ {m}_{ {\text{ice}}}=\ \frac{0.2 \\times 4200 \\times 25}{3.4 \\times 10^{5}}$ $=0.0617 \ {kg}$ $=61.7 \ {gm}$ Remaining ice will remain un-melted so correct answer is 1

Q67JEE Main 2020MCQ4MMotion in a Straight Line

A Tennis ball is released from a height h and after freely falling on a wooden floor it rebounds and reaches height $\frac{{h}}{2} .$ The velocity versus height of the ball during its motion may be represented graphically by : (graph are drawn schematically and on not to scale $)$

🚀 Solve in Practice Mode

📖 Explanation

Velocity at ground (means zero height) is non- zero therefore one is incorrect and velocity versus height is non-linear therefore two is also incorrect. $\ {v}^{2}=2 \ {gh}$ ${v} \frac{{dv}}{ {dh}}=2 {g}= {\text{const}.}$ $\frac{{dv}}{ {dh}}= \frac{\text{ constant }}{\ {v}}$ Here we can see slope is very high when velocity is low therefore at Maximum height the slope should be very large which is in option 3 and as velocity increases slope must decrease there for option 3 is correct.

Q68JEE Main 2020MCQ4MElectrostatic Potential and Capacitance

A two point charges $4 \ {q}$ and $-\ {q}$ are fixed on the $\ {x}$ -axis at $\ {x}=-\frac{{d}}{2}$ and $\ {x}=\frac{{d}}{2},$ respectively. If a third point charge 'q' is taken from the origin to $\ {x}=\ {d}$ along the semicircle as shown in the figure, the energy of the charge will:

🚀 Solve in Practice Mode

📖 Explanation

Potential of $-\ {q}$ is same as initial and final point of the path therefore potential due to $4 \ {q}$ will only change and as potential is decreasing the energy will decrease Decrease in potential energy $=\ {q}({V}_{ {i}}- {V}_{ {f}} )$ Decrease in potential energy $= {q} [ \frac{ {k} 4 {q}}{ {d} / 2}-\frac{{k} 4 {q}}{3 {d} / 2} ]= \frac{4 {q}^{2}}{3 \pi \epsilon _{0} {d}}$ Therefore correct answer is 3 .

Q69JEE Main 2020MCQ4MElectrostatics

Two charged thin infinite plane sheets of uniform surface charge density $\sigma _{+}$ and $\sigma _{-}$ where $|\sigma _{+} |> |\sigma _{-} |$ intersect at right angle. Which of the following best represents the electric field lines for this system :

🚀 Solve in Practice Mode

📖 Explanation

Thin infinite uniformly charged planes produces uniform electric field therefore option 2 and option 3 are obviously wrong. And as positive charge density is bigger in magnitude so its field along Y direction will be bigger than field of negative charge in $\ {X}$ direction and this is evident in option 1 so it is correct.

Q70JEE Main 2020MCQ4MSemiconductor Electronics

Take the breakdown voltage of the zener diode used in the given circuit as $6 \ {V}$ . For the input voltage shown in figure below, the time variation of the output voltage is : (Graphs drawn are schematic and not to scale)

🚀 Solve in Practice Mode

📖 Explanation

As there are two zener diodes in reverse polarity so if one is in forward bias the other will be in reverse bias and above $6 \ {V}$ the reversebias will too be in conduction mode. Thereforewhen voltage is more than $6 \ {V}$ the output will be constant. And when it is less than $6 \ {V}$ it will follow the input voltage so correct answer istwo.

Q71JEE Main 2020NAT4MRay Optics and Optical Instruments

In a compound microscope, the magnified virtual image is formed at a distance of $25 \ {cm}$ from the eye-piece. The focal length of its objective lens is $1 \ {cm} .$ If the magnification is 100 and the tube length of the microscope is $20 \ {cm},$ then the focal length of the eye-piecelens (in $\ {cm}$ ) is _______.

🚀 Solve in Practice Mode

📖 Explanation

for first lens $= \frac{1}{ {v}_{1}}- \frac{1}{- {x}}= \frac{1}{1} \Rightarrow {v}_{1}=\frac{{x}}{ {x}-1}$ also magnification $|{m}_{1} |=|\frac{{v}_{1}}{ {u}_{1}} |= \frac{1}{ {x}-1}$ for $2^{ {\text{nd} }}$ lens this is acting as object so ${u}_{2}=- (20- {v}_{1} )$ $=- (20-\frac{{x}}{ {x}-1})$ and $\ {v}_{2}=-25 {cm}$ angular magnification $|{m}_{ {A}}|=|\frac{{D}}{ {u}_{2}} |=\ \frac{25}{|{u}_{2} |}$ Total magnification ${m}= {m}_{1} {m}_{ {A}}=100$ $(\frac{1}{ {x}-1})({25}/{20-\frac{{x}}{ {x}-1}} )=100$ $\frac{25}{20( {x}-1)-{x}}=100$ $\Rightarrow 1=80( {x}-1)-4 {x}$ $\Rightarrow76 \ {x}=81\Rightarrow {x}=\frac{81}{76}$ $\Rightarrow {u}_{2}=- (20- \frac{81 / 76}{81 / 76-1} )=\ \frac{-19}{5}$ now by lens formula $\frac{1}{-25}- \frac{1}{-19 / 5}= {1}/{ {f}_{ {e}}}$ $\Rightarrow {f}_{ {e}}= \frac{25 \\times 19}{106} \approx 4.48 {cm}$

Q72JEE Main 2020NAT4MRotational Motion

ABC is a plane lamina of the shape of an equilateral triagnle. D, E are mid points of AB, $\ {AC}$ and $\ {G}$ is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through $\ {G}$ and perpendicular to the plane $\ {ABC}$ is $\ {I}_{0} .$ If part $\ {ADE}$ is removed, the moment of inertia of the remaining part about the same axis is $\frac{{NI}_{0}}{16}$ where $\ {N}$ is an integer. Value of $\ {N}$ is _____.

🚀 Solve in Practice Mode

📖 Explanation

Let side of triangle is a and mass is $\ {m}$ MOI of plate ABC about centroid ${I}_{0}=\frac{{m}}{3} ( (\frac{{a}}{2 \\sqrt{3}} )^{2} \\times 3\ )=\frac{{ma}^{2}}{12}$ triangle ADE is also an equilateral triangle of side a/2. Let moment of inertia of triangular plate ADE about it's centroid (G') is I $_{1}$ and mass is ${m}_{1}$ ${m}_{1}={{m}}/{\frac{\sqrt{3} {a}^{2}}{4}} \\times \frac{\sqrt{3}}{4}(\frac{{a}}{2} )^{2}=\frac{{m}}{4}$ ${I}_{1}=\frac{{m}_{1}}{12}(\frac{{a}}{2} )^{2}$ $=\frac{{m}}{4 \\times 12}\frac{{a}^{2}}{4}=\frac{{ma}^{2}}{192}$ distance ${GG}'=\frac{{a}}{\\sqrt{3}}-\frac{{a}}{2 \sqrt{3}}=\frac{{a}}{2 \\sqrt{3}}$ so MOI of part ADE about centroid G is ${I}_{2}= {I}_{1}+ {m}_{1}(\frac{{a}}{2 \\sqrt{3}} )^{2}$ $=\frac{{ma}^{2}}{192}+\frac{{m}}{4} .\frac{{a}^{2}}{12}$ $= \frac{5 \ {ma}^{2}}{192}$ now MOI of remaining part $=\frac{{ma}^{2}}{12}- \frac{5 {ma}^{2}}{192}$ $= \frac{11 {ma}^{2}}{12 \\times 16}=\ \frac{11 {I}_{0}}{16}$ \Rightarrow $N=11$

Q73JEE Main 2020NAT4MRotational Motion

A circular disc of mass $\ {M}$ and radius $\ {R}$ isrotating about its axis with angular speed $\omega _{1}$ . If another stationary disc having radius $\frac{{R}}{2}$ and same mass $\ {M}$ is dropped co-axially on to the rotating disc. Gradually both discs attain constant angular speed $\omega _{2}$ . The energy lost in the process is $\ {p} \%$ of the initial energy. Value of p is ______.

🚀 Solve in Practice Mode

📖 Explanation

Let moment of inertia of bigger disc is $I =\frac{{MR}^{2}}{2}$ $\Rightarrow$ MOI of small disc $\ {I}_{2}={{M}(\frac{{R}}{2} )^{2}}/{2}= \frac{{I}}{4}$ by angular momentum conservation ${I} \omega _{1}+\frac{{I}}{4}(0)= {I} \omega _{2}+\frac{{I}}{4} \omega _{2}$ $\Rightarrow \omega _{2}= \frac{4 \omega _{1}}{5}$ initial kinetic energy ${K}_{1}= \frac{1}{2} {I} \omega _{1}^{2}$ final kinetic energy $\ {K}_{2}$ $= \frac{1}{2}({I}+\frac{{I}}{4})(\frac{4 \omega _{1}}{5} )^{2}$ $= \frac{1}{2} {I} \omega _{1}^{2}(\frac{4}{5} )$ ${P} \%=\frac{{K}_{1}- {K}_{2}}{ {K}_{1}} \\times 100 \%$ $= \frac{1-4 / 5}{1} \\times 100=20 \%$

Q74JEE Main 2020NAT4MThermodynamics

A closed vessel contains $0.1 \ {\text{mole}}$ of a monoatomic ideal gas at 200 K. If 0.05 mole of the same gas at $400 \ {K}$ is added to it, the final equilibrium temperature (in $\ {K}$ ) of the gas in the vessel will be closed to ______.

🚀 Solve in Practice Mode

📖 Explanation

As work done on gas and heat supplied to the gas are zero, total internal energy of gases remain same $\ {u}_{1}+\ {u}_{2}=\ {u}_{1}'+\ {u}_{2}'$ $(0.1) {C}_{ {v}}(200)+(0.05) {C}_{ {v}}(400)$ $=(0.15) {C}_{ {v}} {T}$ ${T}= \frac{800}{3} {k}=266.67 \ {k}$

Q75JEE Main 2020NAT4MAtoms and Nuclei

In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is $304 \Å$ . Thecorresponding difference for the Paschan series in $\Å$ is : _______.

🚀 Solve in Practice Mode

📖 Explanation

$\lambda= {c}/ {(\frac{1}{n_{1}^{2}}- \frac{1}{n_{2}^{2}} )}$ for lyman series $\lambda_{1}= {c}/{ \frac{1}{1^{2}}- \frac{1}{\infty ^{2}}}$ $=c( {n}=\infty$ to $\ {n}=1)$ $\lambda_{2}={{c}}/{ \frac{1}{1^{2}}-\frac{1}{2^{2}}}= \frac{4 \ {c}}{3}$ $(\ {n}=2$ to $\ {n}=1)$ $\Delta \lambda=\lambda_{2}-\lambda_{1}=\frac{{c}}{3}$ $=304 \Å \Rightarrow {c}=912 \Å$ for paschen series $\lambda_{1}={{c}}/{ \frac{1}{3^{2}}- \frac{1}{\infty ^{2}}}$ $=9 \ {c}( {n}=\infty$ to $\ {n}=3)$ $\lambda_{2}={{c}}/{ \frac{1}{3^{2}}- \frac{1}{4^{2}}}$ $= \frac{144 \ {c}}{7}( {n}=4$ to ${n}=3)$ $\Delta\lambda=\lambda_{2}-\lambda_{1}= \frac{144 \ {c}}{7}-9 {c}$ $= \frac{81 {c}}{7}= \frac{81 \\times 912}{7}$ $=10553.14 \Å$

← All JEE Main 2020 papers Take as Timed Mock →