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JEE MAIN 2020 - 06 SEP S2 2020

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CHEMISTRY25 questions

Q1JEE Main 2020MCQ4MChemical Equilibrium

The value of ${K}_{{C}}$ is 64 at $800 {K}$ for the reaction ${\text{N}}_{2}({g})+3 {\text{H}}_{2}({g}) 2 ⇌{\text{NH}}_{3}({g})$ The value of ${\text{K}}_{{\text{C}}}$ for the following reaction is : ${\text{NH}}_{3}({g}) ⇌\frac{1}{2} {\text{N}}_{2}({g})+\frac{3}{2} {\text{H}}_{2}({g})$

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📖 Explanation

${\text{N}}_{2} +3 {\text{H}}_{2} 2 ⇌{\text{NH}}_{3} \to \text{K}_\text{C}=64$ $2 {\text{NH}}_{3}⇌ {\text{N}}_{2} +3 {\text{H}}_{2} \to \text{K}_\text{C}=1/64$ $\text{NH}_3 ⇌ 1/2\text{N}_2+3/2\text{H}_2\to \text{K}_\text{C}$ $=(1/64)^{1/2}=1/8$

Q2JEE Main 2020MCQ4Md and f Block Elements

The element that can be refined by distillation is :

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Q3JEE Main 2020MCQ4MOrganic Chemistry - Some Basic Principles

The correct match between Item-I and Item-II : Item-I Item-II (a) Natural rubber (I) 1, 3-butadiene + styrene (b) Neoprene (II) 1, 3-butadiene + acrylonitrile (c) Buna-N (III) Chloroprene (d) Buna-S (IV) Isoprene

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Q4JEE Main 2020MCQ4Md and f Block Elements

Mischmetal is an alloy consisting mainly of:

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📖 Explanation

Alloys of lanthanides with Fe are called Mischmetal, which consists of a lanthanoid metal $(∼ 95 \%)$ and iron $(∼5 \%)$ and traces of ${\text{S}}, {\text{C}}, {\text{Ca}}$ and Al.

Q5JEE Main 2020MCQ4Mp Block Elements

Reaction of an inorganic sulphite $\text{X}$ with dilute ${\text{H}}_{2} {\text{SO}}_{4}$ generates compound Y. Reaction of ${\text{Y}}$ with ${\text{NaOH}}$ gives ${\text{X}}$ . Further, the reaction of ${\text{X}}$ with $\text{Y}$ and water affords compound $\text{Z}$ . Y and Z, respectively, are:

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📖 Explanation

${\text{Na}_2\text{SO}_3}_{(\text{X})} {──────🢖}^{\text{dil}. \text{H}_2\text{SO}_4} {\text{SO}_2}_{(\text{Y})}{─────🢖}^{\text{NaOH}} {\text{Na}_2\text{SO}_3}_{(\text{X})}{────────🢖}^{\text{SO}_2+\text{H}_2\text{O}}{\text{NaHSO}_3}_{(\text{Z})}$

Q6JEE Main 2020MCQ4MOrganic Chemistry - Some Basic Principles

The IUPAC name of the following compound is :

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Q7JEE Main 2020MCQ4MElectrochemistry

Dihydrogen of high purity $(>99.95 \%)$ is obtained through:

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📖 Explanation

High purity $(>99.95 \%)$ dihydrogen is obtained by electrolyzing warm aqueous barium hydroxide solution between nickel electrodes.

Q8JEE Main 2020MCQ4MPractical Organic Chemistry

Test/MethodReagent(i) Lucas Test (a) $\text{C} _6 \text{H} _5\text{SO}_2\text{Cl}/\text{aq}. \text{KOH}$ (ii) Dumas method(b) ${\text{HNO}}_{3} / {\text{AgNO}}_{3}$ (iii) Kjeldahl's method(c) ${\text{CuO}}/ {\text{CO}}_{2}$ (iv) Hinsberg Test(d) Conc. HCl and ${\text{ZnCl}}_{2}$ (e) ${\text{H}}_{2} {\text{SO}}_{4}$

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📖 Explanation

Here is a step-by-step breakdown of matching each test/method to its correct reagent:

Lucas Test (i): This test is used to differentiate between primary, secondary, and tertiary alcohols. The reagent used is Lucas reagent, a solution of anhydrous zinc chloride ({\text{ZnCl}}_{2}$) in concentrated hydrochloric acid (Conc. HCl). Therefore, (i) matches with (d).
Dumas method (ii): This is a quantitative method for estimating nitrogen in an organic compound. The compound is heated with copper(II) oxide ({\text{CuO}} $) \in an atmosphere of carbon dioxide ({\text{CO}}_{2}$ ), and the evolved nitrogen gas is collected. Thus, (ii) matches with (c).
Kjeldahl's method (iii): This method also estimates nitrogen. The organic compound is digested with concentrated sulfuric acid ({\text{H}}{2} {\text{SO}}{4}$), converting the nitrogen into ammonium sulfate. Hence, (iii) matches with (e).
Hinsberg Test (iv): This test is used to distinguish between primary, secondary, and tertiary amines using benzenesulfonyl chloride (\text{C} _6 \text{H} 5\text{SO} 2\text{Cl}$) in the presence of aqueous alkali (KOH). So, (iv) matches with (a).

Combining these matches gives the sequence: (i)-(d), (ii)-(c), (iii)-(e), (iv)-(a).

Q9JEE Main 2020MCQ4Mp Block Elements

The reaction of ${\text{NO}}$ with ${\text{N}}_{2} {\text{O}}_{4}$ at $250 {\text{K}}$ gives :

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📖 Explanation

$2\text{NO}+\text{N}_2\text{O}_4 \to ^{250\text{K}}{2\text{N}_2\text{O}_3}_{\text{blue} \text{solid}}$

Q10JEE Main 2020MCQ4MElectrochemistry

For the given cell ; ${\text{Cu}}({\text{s}})|{\text{Cu}}^{2+}({\text{C}}_{1} {\text{M}}) \| {\text{Cu}}^{2+}({\text{C}}_{2} {\text{M}})| {\text{Cu}}({\text{s}})$ change in Gibbs energy ( $\Delta{\text{G}}$ ) is negative, if :

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📖 Explanation

For a reaction to be spontaneous, the change in Gibbs free energy ( $\Delta G$ ) must be negative. The relationship between $\Delta G$ and cell potential ( $E_{cell}$ ) is $\Delta G = -nFE_{cell}$ . Therefore, a negative $\Delta G$ requires a positive $E_{cell}$ .

This is a concentration cell, where the anode is the half-cell with the lower concentration and the cathode has the higher concentration. The overall reaction is ${\text{Cu}}^{2+}({\text{C}}_{2}) \rightarrow {\text{Cu}}^{2+}({\text{C}}_{1})$ . The Nernst equation for this cell is $E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF}\ln Q$ .

For a concentration cell, $E^{\circ}_{cell} = 0$ , and the reaction quotient is $Q = \frac{[\text{Products}]}{[\text{Reactants}]} = \frac{C_{1}}{C_{2}}$ .
Thus, $E_{cell} = -\frac{RT}{nF}\ln\left(\frac{C_{1}}{C_{2}}\right) = \frac{RT}{nF}\ln\left(\frac{C_{2}}{C_{1}}\right)$ .
For $E_{cell}$ to be positive, $\ln\left(\frac{C_{2}}{C_{1}}\right)$ must be positive, which means $\frac{C_{2}}{C_{1}} > 1$ , or $C_{2} > C_{1}$ .
Examining the options, only $C_{2} = \sqrt{2}C_{1}$ satisfies the condition $C_{2} > C_{1}$ because $\sqrt{2} \approx 1.414 > 1$ .

Q11JEE Main 2020MCQ4MSolid State

A crystal is made up of metal ions $'\text{M} _1 '$ ana $'\text{M}_{2}'$ and oxide ions. Oxide ions form a ccp lattice structure. The cation $'\text{M} _{1}'$ occupies $50 \%$ of octahedral voids and the cation $'\text{M} _{2}'$ occupies $12.5 \%$ of tetrahedral voids of oxide lattice. The oxidation numbers of $'\text{M} _{1}'$ and $'\text{M} _{2}'$ are, respectively :

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📖 Explanation

${\text{O}}^{-2}$ ions form ccp. ${\text{O}}_{4}_{(-8\text{charges})}$ ${\text{M}}_{1}=50 \%$ of ${\text{O}} . {\text{V}} . \Rightarrow \frac{50}{100} \times 4=2:({\text{M}}_{1})_{2}$ ${\text{M}}_{2}=12.5 \%$ of ${\text{T}.\text{V}} . \Rightarrow \frac{12.5}{100} \times 8=1:({\text{M}}_{2})_{1}$ So formula is : $({\text{M}}_{1})_{2}({\text{M}}_{2})_{1} {\text{O}}_{4}$ This must be neutral. Both metals must have +8 charge in total. From given options: $\begin{cases}\text{O}\end{cases}\text{N}.\text{of}, \text{M}_1 =+2;,\text{M}_2=+4\}$

Q12JEE Main 2020MCQ4MCoordination Compounds

For a ${d}^{4}$ metal ion in an octahedral field, the correct electronic configuration is:

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📖 Explanation

back pairing is not possible because pairing energy $>\Delta_{o}.$

Q13JEE Main 2020MCQ4MAmines

Which of the following compounds can be prepared in good yield by Gabriel phthalimide synthesis?

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📖 Explanation

Gabriel phthalimide synthesis is used for preparation of $1°$ Aliphatic amine

Q14JEE Main 2020MCQ4MAldehydes, Ketones and Carboxylic Acids

The correct match between Item-I (starting material) and Item-II (reagent) for the preparation of benzaldehyde is : Item-IItem-II(I) Benzene(P) ${\text{HCl}}$ and ${\text{SnCl}}_{2}, {\text{H}}_{3} {\text{O}}^{+}$ (II) Benzonitrile (Q) ${\text{H}}_{2}, {\text{Pd}}-{\text{BaSO}}_{4}, {\text{S}}$ and quinoline (III) Benzoyl Chloride (R) $\text{CO}, \text{HCl}$ and $\text{AlCl} _{3}$

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Q15JEE Main 2020MCQ4MSome Basic Concepts of Chemistry

The average molar mass of chlorine is $35.5 {g}$ ${\text{mol}}^{-1}$ . The ratio of $^{35} {\text{Cl}}$ to $^{37} {\text{Cl}}$ in naturally occurring chlorine is close to :

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📖 Explanation

$\text{table} ,^35\text{Cl},,^37\text{Cl};\text{let} \text{mole}\text{ratio},x,:,1$ Average Molar mass = 35.5 Av. molar mass $={{n}_{1} {\text{M}}_{1}+{n}_{2} {\text{M}}_{2}}/{({n}_{1}+{n}_{2})}$ $35.5=\frac{x \times 35+1 \times 37}{x+1}$ $x=3$

Q16JEE Main 2020MCQ4MBiomolecules

Which one of the following statements not true?

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Q17JEE Main 2020MCQ4MSolutions

A set of solutions is prepared using $180 {g}$ of water as a solvent and $10 {g}$ of different non-volatile solutes ${\text{A}}, {\text{B}}$ and ${\text{C}}$ . The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of ${\text{A}}=100 {\text{g}} {\text{mol}}^{-1} ; {\text{B}}=200 {\text{g}} {\text{mol}}^{-1};$ ${\text{C}}=10,000 {\text{g}} {\text{mol}}^{-1}]$

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📖 Explanation

Relative lowering of ${\text{V}} . {\text{P}} .={\Delta{\text{P}}}/{{\text{P}}^{0}}={x}_{{\text{solute} }}$ $({\Delta{\text{P}}}/{{\text{P}}^{0}})_{{\text{A}}}=\frac{\frac{10}{100}}{\frac{10}{100}+\frac{180}{18}}:({\Delta{\text{P}}}/{{\text{P}}^{0}})_{{\text{B}}}$ $=\frac{\frac{10}{200}}{\frac{10}{200}+\frac{180}{18}}$ $({\Delta{\text{P}}}/{{\text{P}}^{0}})_{{\text{C}}}=\frac{\frac{10}{10,000}}{\frac{10}{10,000}+\frac{180}{18}}:({\Delta{\text{P}}}/{{\text{P}}^{0}})_{{\text{A}}}>({\Delta{\text{P}}}/{{\text{P}}^{0}})_{{\text{B}}}>({\Delta{\text{P}}}/{{\text{P}}^{0}})_{{\text{C}}}$

Q18JEE Main 2020MCQ4MChemical Thermodynamics

For a reaction, $4 {\text{M}}({\text{s}})+{\text{nO}}_{2}({\text{g}}) \to 2 {\text{M}}_{2} {\text{O}}_{{n}}({\text{s}})$ the free energy change is plotted as a function of temperature. The temperature below which the oxide is stable could be inferred from the plot as the point at which :

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📖 Explanation

The stability of a compound is determined by the spontaneity of its formation reaction. A reaction is spontaneous when the Gibbs free energy change ( $\Delta G$ ) is negative. For the given formation of the metal oxide, the oxide ${\text{M}}_{2} {\text{O}}_{{n}}$ is stable as long as $\Delta G < 0$ . The oxide becomes unstable and will decompose back to its elements when the formation reaction is no longer spontaneous, which is when $\Delta G > 0$ . The transition point between stability and instability occurs at the equilibrium temperature where $\Delta G = 0$ . Therefore, the temperature below which the oxide is stable is found by identifying the point on the plot where $\Delta G$ changes from a negative value to a positive value.

Q19JEE Main 2020MCQ4MClassification of Elements

Match the following compounds (Column-I) with their uses (Column-II) : S.No. Column I S.No. Column II (I) $\text{Ca}(\text{OH})_2$ (A) casts of statues (II) $\text{NaCl}$ (B) white wash (III) $\text{Ca}\text{SO}_4 . 1/2 \text{H}_2\text{O}$ (C) antacid (IV) $\text{CaCO}_3$ (D) washing soda preparation

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📖 Explanation

Here is a step-by-step explanation of the correct matches:

(I) $\text{Ca}(\text{OH})_2$ (Calcium hydroxide): Also known as slaked lime, its suspension in water is used for whitewashing walls, which corresponds to (B) white wash.
(II) $\text{NaCl}$ (Sodium chloride): Common salt is a primary raw material for the industrial production of washing soda (\text{Na}_2\text{CO}_3$) through the Solvay process, matching with (D) washing soda preparation.
(III) $\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}$ (Plaster of Paris): This compound sets into a hard solid upon mixing with water, making it ideal for creating molds and (A) casts of statues.
(IV) $\text{CaCO}_3$ (Calcium carbonate): As a mild base, it effectively neutralizes excess acid in the stomach, so it is used as an (C) antacid.

Therefore, the correct sequence of matches is (I)-(B), (II)-(D), (III)-(A), (IV)-(C).

Q20JEE Main 2020MCQ4MHaloalkanes and Haloarenes

The increasing order of the boiling points of the major products ${\text{A}}, {\text{B}}$ and ${\text{C}}$ of the following reactions will be :

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📖 Explanation

The boiling points of isomeric haloalkanes decrease with increase in branching So order of B.P. is $A > C > B$ .

Q21JEE Main 2020NAT4MChemical Equilibrium

For Freundlich adsorption isotherm, a plot of $\log ({x} / {m})({y}$ -axis $)$ and $\log {p}$ (x-axis) gives a straight line. The intercept and slope for the line is 0.4771 and 2 , respectively. The mass of gas, adsorbed per gram of adsorbent if the initial pressure is 0.04 atm, is ________ $\times 10^{-4 }g$ $(\log 3=0.4771)$

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📖 Explanation

$\frac{x}{{m}}={\text{KP}}^{1 / {a}}$ $\log (\frac{{x}}{{m}})=\frac{1}{{n}} \log {\text{P}}+\log {\text{K}}$ $(\log 3=0.4771)$ slope $=\frac{1}{{n}}=2$ intercept $=\log {\text{K}}=0.4771$ ${\text{K}}=3$ mass of gas adsorbed per ${\text{gm}}$ of adsorbent $=\frac{{x}}{{m}}$ $\frac{x}{m}=3 \times (0.04)^{2}$ $=48 \times 10^{-4}$

Q22JEE Main 2020NAT4MAlcohols, Phenols and Ethers

A solution of phenol in chloroform when treated with aqueous NaOH gives compound ${\text{P}}$ as a major product. The mass percentage of carbon in ${\text{P}}$ is __________. (to the nearest integer) (Atomic mass : ${\text{C}}=12 ; {\text{H}}=1 ; {\text{O}}=16)$

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📖 Explanation

Molecular weight of ${\text{C}}_{7} {\text{H}}_{6} {\text{O}}_{2}=122$ $\% {\text{C}}=\frac{12 \times 7 \times 100}{122}$ $=68.85 \approx 69$

Q23JEE Main 2020NAT4MIonic Equilibrium

If the solubility product of ${\text{AB}}_{2}$ is $3.20 \times 10^{-11} {\text{M}}^{3},$ then the solubility of ${\text{AB}}_{2}$ in pure water is ________ $\times 10^{-4} {\text{mol}} {\text{L}}^{-1} .$ [Assuming that neither kind of ion reacts with water]

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📖 Explanation

${\text{AB}}_{2}({\text{s}}) ⇌{{\text{A}}_{({\text{aq}} .)}^{+2}}_{\text{S}}+{2 {\text{B}}_{({\text{aq}.})}^{-} }_{2\text{S}}: {\text{K}}_{{\text{s}\text{p}}}$ ${\text{K}}_{{\text{SP}}}={\text{S}}^{1} \times (2 {\text{s}})^{2}=4 {\text{s}}^{3}$ $3.2 \times 10^{-11}=4\times {\text{S}}^{3}$ ${\text{S}}=2 \times 10^{-4} {\text{M}} / {\text{L}}$

Q24JEE Main 2020NAT4MChemical Kinetics

The rate of a reaction decreased by 3.555 times when the temperature was changed from $40° {C}$ to $30°{C}$ . The activation energy (in ${\text{kJ}} {\text{mol}}^{-1}$ ) of the reaction is ____________. Take; ${\text{R}}=8.314 {\text{J}} {\text{mol}}^{-1} {\text{K}}^{-1}$ , $\ln3.555=1.268$

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📖 Explanation

${T}_{1}=303 {K} ; {T}_{2}=313 {K}$ $\ln ({{K}_{{T}_{2}}}/{{K}_{{T}_{1}}})={{E}_{{a}}}/{{R}}[\frac{1}{{T}_{1}}-\frac{1}{{T}_{2}}]$ ${{K}_{{T}_{2}}}/{{K}_{{T}_{1}}}=3.555$ ${\ln}(3.555)={{E}_{{a}}}/{8.314}[\frac{1}{303}-\frac{1}{313}]$ ${E}_{{a}}=99980.715$ ${E}_{{a}}=99.98 {{\text{kJ}}}/{{\text{mole}}}$

Q25JEE Main 2020NAT4MClassification of Elements

The atomic number of Unnilunium is _______

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MATHEMATICS25 questions

Q26JEE Main 2020MCQ4MApplication of Derivatives

The set of all real values of $\lambda$ for which the function $f(x)=(1-\cos ^{2} x) .(\lambda+\sin x),$ ${x} \in (-\frac{\pi }{2}, \frac{\pi }{2}),$ has exactly one maxima and exactly one minima, is :

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📖 Explanation

$f(x)=(1-\cos ^{2} x) .(\lambda+\sin x),$ ${x} \in (-\frac{\pi }{2}, \frac{\pi }{2})$ $f(x)=\lambda\sin ^{2} x+\sin ^{3} x$ $f'(x)=2 \lambda\sin x \cos x+3 \sin ^{2} x \cos x$ $f'(x)=\sin x \cos x(2 \lambda+3 \sin x)$ $\sin {x}=0, \frac{-2 \lambda}{3},(\lambda \ne 0)$ for exactlyone maxima & minima $\frac{-2 \lambda}{3}\in (-1,1)$ $\Rightarrow \lambda\in (\frac{-3}{2}, \frac{3}{2})$ $\lambda\in (-\frac{3}{2}, \frac{3}{2})-\{0\}$

Q27JEE Main 2020MCQ4MApplication of Derivatives

For all twice differentiable functions $f: {R} \to {R},$ with $f(0)=f(1)=f'(0)=0$

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📖 Explanation

$f(0)=f(1)=f'(0)=0$ Apply Rolles theorem on ${y}=f({x})$ in ${x} \in [0,1]$ $f(0)=f(1)=0$ $\Rightarrow f'(\alpha )=0$ where $\alpha \in (0,1)$ Now apply Rolles theorem on ${y}=f'({x})$ in ${x} \in [0, \alpha ]$ $f'(0)=f'(\alpha )=0$ and $f'({x})$ is continuous and differentiable $f'(0)=f'(\alpha )=0$ and $f'({x})$ is continuous and differentiable $\Rightarrow f"(\beta )=0$ for some $, \beta \in (0, \alpha ) \in (0,1)$ $\Rightarrow f"(x)=0$ for some $x \in\in (0,1)$

Q28JEE Main 2020MCQ4MApplication of Derivatives

If the tangent to the curve, ${y}=f({x})={x\log}_{{e}} {x}$ $(x>0)$ at a point $(c, f(c))$ is parallel to the line - segment joining the points (1,0) and $({e}, {e})$ , then $c$ is equal to :

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📖 Explanation

$f(x)=x \log _{e} x$ $f'(x)|_{(c, f(c))}=\frac{e-0}{e-1}$ $f'(x)=1+\log _{e} x$ $f'(x)|_{(c, f(c))}=1+\log _{e} c=\frac{e}{e-1}$ $\log _{e} c=\frac{e-(e-1)}{e-1}$ $={1}{e-1}$ $\Rightarrow c=e^{\frac{1}{e-1}}$

Q29JEE Main 2020MCQ4MSets and Relations

Consider the statement : "For an integer ${n},$ if ${n}^{3}-1$ is even, then ${n}$ is odd." The contrapositive statement of this statement is:

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📖 Explanation

Contrapositive of $({p}\to {q})$ is $∼{q} \to ∼{p}$ For an integer ${n},$ if ${n}$ is even then $({n}^{3}-1)$ is odd

Q30JEE Main 2020MCQ4MEllipse

If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity e of the ellipse satisfies :

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📖 Explanation

$\frac{{a}^{2}{x}}{{x}_{1}}-\frac{{b}^{2} {y}}{{y}_{1}}={a}^{2} {e}^{2}$ $\frac{{a}^{2} {x}}{{ae}}-\frac{{b}^{2} {y}}{{b}^{2}} .{a}={a}^{2} {e}^{2}$ $\frac{a x}{e}-a y=a^{2} e^{2}$ $\Rightarrow \frac{x}{e}-y=a e^{2}$ passes through $(0, {b})$ $-b=a e^{2} \Rightarrow b^{2}=a^{2} e^{4}$ $a^{2}(1-e^{2})=a^{2} e^{4}$ $\Rightarrow e^{4}+e^{2}=1$

Q31JEE Main 2020MCQ4MThree Dimensional Geometry

A plane ${\text{P}}$ meets the coordinate axes at ${\text{A}}, {\text{B}}$ and C respectively. The centroid of $\Delta {\text{ABC}}$ is given to be $(1,1,2) .$ Then the equation of the line through this centroid and perpendicular to the plane ${\text{P}}$ is :

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📖 Explanation

$x/a+y/b+z/c=1$ ${\text{A}} \equiv ({a}, 0,0), {\text{B}} \equiv (0, {b}, 0)$ , ${\text{C}} \equiv (0,0, {c})$ Centroid $\equiv (\frac{{a}}{3}, \frac{{b}}{3}, \frac{{c}}{3})=(1,1,2)$ $a=3, b=3, c=6$ Plane $: \frac{x}{3}+\frac{y}{3}+\frac{z}{6}=1$ $2 x+2 y+z=6$ line $⊥$ to the plane $({\text{DR}}$ of line $=2 \hat{i}+2 \hat{j}+\hat{k})$ $\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{1}$

Q32JEE Main 2020MCQ4MQuadratic Equation and Inequalities

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

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📖 Explanation

$\alpha$ and $\beta$ are the roots of the equation $4 x^{2}+2 x-1=0$ $4 \alpha ^{2}+2 \alpha =1 \Rightarrow \frac{1}{2}=2 \alpha ^{2}+\alpha$ ....(1) $\beta =-1/2-\alpha$ using equation(1) $\beta =-(2 \alpha ^{2}+\alpha )-\alpha$ $\beta =-2 \alpha ^{2}-2 \alpha$ $\beta =-2 \alpha (\alpha +1)$

Q33JEE Main 2020MCQ4MComplex Numbers

Let $z=x+$ iy be a non-zero complex number such that ${z}^{2}={i}|{z}|^{2},$ where ${i}=\sqrt{-1},$ then ${z}$ lies on the :

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📖 Explanation

${z}={x}+{iy}$ ${z}^{2}={i}|{z}|^{2}$ $(x+i y)^{2}=i(x^{2}+y^{2})$ $(x^{2}-y^{2})-i(x^{2}+y^{2}-2 x y)=0$ $(x-y)(x+y)-i(x-y)^{2}=0$ $(x-y)((x+y)-i(x-y))=0$ $\Rightarrow x=y$ $z$ lies on $y=x$

Q34JEE Main 2020MCQ4MSequences and Series

The common difference of the A.P. ${b}_{1}, {b}_{2}, ...$ ${b}_{{m}}$ is 2 more than the common difference of A.P. $a_{1}, a_{2}, ..., a_{n}$ . If $a_{40}=-159, a_{100}=-399$ and ${b}_{100}={a}_{70},$ then ${b}_{1}$ is equal to :

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📖 Explanation

$a_{1}, a_{2}, ..., a_{n} \to (C D=d)$ $b_{1}, b_{2}, ...., b_{m} \to (C D=d+2)$ $a_{40}=a+39 d=-159$ ...(1) $a_{100}=a+99 d=-399$ ...(2) Subtract $: 60 {d}=-240 \Rightarrow {d}=-4$ using equation (1) $a+39(-4)=-159$ $a=156-159=-3$ $a_{70}=a+69 d$ $=-3+69(-4)=-279=b_{100}$ ${b}_{100}=-279$ ${b}_{1}+99({d}+2)=-279$ ${b}_{1}-198=-279\Rightarrow {b}_{1}=-81$

Q35JEE Main 2020MCQ4MTrigonometric Ratios and Identities

The angle of elevation of the summit of a mountain from a point on the ground is $45° .$ After climbing up one ${\text{km}}$ towards the summit at an inclination of $30°$ from the ground, the angle of elevation of the summit is found to be $60° .$ Then the height (in ${\text{km}}$ ) of the summit from the ground is :

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📖 Explanation

$\sin 30°=x\Rightarrow x=\frac{1}{2}$ $\cos 30°=z \Rightarrow z=\frac{\sqrt{3}}{2}$ $\tan 45°=\frac{{h}}{{y}+{z}}$ $\Rightarrow {h}={y}+{z}$ $\tan 60°=\frac{{h}-{x}}{{y}}$ $\Rightarrow \tan 60°=\frac{{h}-{x}}{{h}-{z}}$ $\sqrt{3}({h}-{z})={h}-{x}$ $(\\sqrt{3}-1) {h}=\sqrt{3} {z}-{x}$ $\Rightarrow (\sqrt{3}-1) {h}=\frac{3}{2}-\frac{1}{2}$ $\Rightarrow {h}=\frac{1}{\sqrt{3}-1}$

Q36JEE Main 2020MCQ4MMatrices and Determinants

Let $\theta =\pi /5$ and $\text{A}=\begin{bmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$ . If $\text{B}=\text{A}+{\text{A}}^{4},$ then ${\det}({\text{B}}):$

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📖 Explanation

$\text{A}=\begin{bmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$ $\text{A}^2=\begin{bmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$ $\text{A}^2=\begin{bmatrix}\cos2\theta & \sin2\theta \\ -\sin2\theta & \cos2\theta\end{bmatrix}$ $\text{B}=\text{A}+\text{A}^4$ $=\begin{bmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$ $+\begin{bmatrix}\cos4\theta & \sin4\theta \\ -\sin4\theta & \cos4\theta\end{bmatrix}$ $\text{B}=\begin{bmatrix}(\cos\theta +\cos 4\theta ) & (\sin\theta +\sin4\theta ) \\ -(\sin\theta +\sin4\theta ) & (\cos\theta +\cos 4\theta )\end{bmatrix}$ $|{\text{B}}|=(\cos \theta +\cos 4 \theta )^{2}$ $+(\sin \theta +\sin 4 \theta )^{2}$ $|{\text{B}}|=2+2 \cos 3 \theta$ , when $\theta =\pi /5$ $|{\text{B}}|=2+2 \cos \frac{3 \pi }{5}=2(1-\sin 18)$ $|{B}|=2(1-\frac{\sqrt{5}-1}{4})$ $=2(\frac{5-\sqrt{5}}{4})=\frac{5-\sqrt{5}}{2}$

Q37JEE Main 2020MCQ4MFunctions

For a suitably chosen real constant a, let a function, $f: {\text{R}}-\{-{a}\} \to {\text{R}}$ be defined by $f(x)=\frac{a-x}{a+x} .$ Further suppose that for any real number ${x} \ne -{a}$ and $f({x}) \ne -{a},(f {o} f)({x})={x} .$ Then $f(-\frac{1}{2})$ is equal to :

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📖 Explanation

$f({x})=\frac{{a}-{x}}{{a}+{x}}$ , ${x} \in {R}-\{-{a}\} \to {R}$ $f(f({x}))=\frac{{a}-f({x})}{{a}+f({x})}$ $={{a}-(\frac{{a}-{x}}{{a}+{x}})}/{{a}+(\frac{{a}-{x}}{{a}+{x}})}$ $f(f(x))=\frac{(a^{2}-a)+x(a+1)}{(a^{2}+a)+x(a-1)}=x$ $\Rightarrow (a^{2}-a)+x(a+1)$ $=(a^{2}+a) x+x^{2}(a-1)$ $\Rightarrow a(a-1)+x(1-a^{2})$ $-x^{2}(a-1)=0$ $\Rightarrow a=1$ $f(x)=\frac{1-x}{1+x}$ $f(\frac{-1}{2})=\frac{1+\frac{1}{2}}{1-\frac{1}{2}}=3$

Q38JEE Main 2020MCQ4MLimits, Continuity and Differentiability

Let $f: {\text{R}} \to {\text{R}}$ be a function defined by $f(x)=\max \{x, x^{2}\} .$ Let $S$ denote the set of all points in ${\text{R}},$ where $f$ is not differentiable. Then :

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📖 Explanation

$f(x)=\max (x, x^{2})$ Non-differentiable at $x=0,1$ ${S}=\{0,1\}$

Q39JEE Main 2020MCQ4MArea Under The Curves

The area (in sq. units) of the region enclosed by the curves $y=x^{2}-1$ and $y=1-x^{2}$ is equal to :

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📖 Explanation

${y}={x}^{2}-1$ and ${y}=1-{x}^{2}$ $A=∫^{1}_{-1}((1-x^{2})-(x^{2}-1)) d x$ $A=∫^{1}_{-1}(2-2 x^{2}) d x$ $=4 ∫^{1}_{0}(1-x^{2}) d x$ $A=4(x-\frac{x^{3}}{3})_{0}^{1}$ $=4(\frac{2}{3})=\frac{8}{3}$

Q40JEE Main 2020MCQ4MProbability

The probabilities of three events ${\text{A}}, {\text{B}}$ and ${\text{C}}$ are given by ${\text{P}}({\text{A}})=0.6, {\text{P}}({\text{B}})=0.4$ and ${\text{P}}({\text{C}})=0.5$ If ${\text{P}}({\text{A}}∪ {\text{B}})=0.8$ , ${\text{P}} ({\text{A}} ∩ {\text{C}})=0.3$ , ${\text{P}}({\text{A}} ∩ {\text{B}} ∩ {\text{C}})=0.2$ , ${\text{P}}({\text{B}} ∩ {\text{C}})=\beta$ and ${\text{P}}({\text{A}}∪ {\text{B}}∪ {\text{C}})=\alpha$ where $0.85 \le \alpha \le 0.95,$ then $\beta$ lies in the interval:

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📖 Explanation

$\text{P}(\text{A}∪ \text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A} ∩ \text{B})$ $0.8=0.6+0.4-\text{P}(\text{A} ∩ \text{B})$ $\text{P}(\text{A} ∩ \text{B})=0.2$ $\text{P}(\text{A}∪\text{B}∪\text{C})$ $= \sum \text{P}(\text{A})-\sum \text{P}(\text{A} ∩ \text{B})$ $+\text{P}(\text{A}∩ \text{B} ∩\text{C})$ $\alpha =1.5-(0.2+0.3+\beta )+0.2$ $\alpha =1.2-\beta \in [0.85,0.95]$ (where $\alpha \in [0.85,0.95])$ $\beta \in [0.25,0.35]$

Q41JEE Main 2020MCQ4MBinomial Theorem

if the constant term in the binomial expansion of $(\sqrt{x}-\frac{k}{x^{2}})^{10}$ is $405,$ then $|k|$ equals :

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📖 Explanation

$(\sqrt{x}-\frac{k}{x^{2}})^{10}$ $T_{r+1}={ }^{10} C_{r}(\sqrt{x})^{10-r}(\frac{-k}{x^{2}})^{r}$ $T_{r+1}={ }^{10} C_{r} . x^{\frac{10-r}{2}} .(-k)^{r} .x^{-2 r}$ $T_{r+1}={ }^{10} C_{r} x^{\frac{10-5 r}{2}}(-k)^{r}$ Constant term $: \frac{10-5 {r}}{2}=0 \Rightarrow {r}=2$ ${T}_{3}={ }^{10} {C}_{2} .(-{k})^{2}=405$ ${k}^{2}=\frac{405}{45}=9$ ${k}=± 3\Rightarrow |{k}|=3$

Q42JEE Main 2020MCQ4MDefinite Integrals

The integral $∫^{2}_{1}{e}^{{x}} . {x}^{{x}}(2+\log _{{e}} {x}) {d} {x}$ equal :

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📖 Explanation

$∫^{2}_{1}{e}^{{x}} . {x}^{{x}}(2+\log _{{e}} {x}) {d} {x}$ $=∫^{2}_{1} {e}^{{x}}(2 {x}^{{x}}+{x}^{{x}} \log _{{e}} {x}) {d} {x}$ $=∫^{2}_{1} e^x(x^x_{f(x)} +{x^x(1+\log_e x)}_{f'(x)})dx$ $({e}^{{x}} .{x}^{{x}})_{1}^{2}=4 {e}^{2}-{e}$

Q43JEE Main 2020MCQ4MStraight Lines and Pair of Straight Lines

Let $L$ denote the line in the $x y$ -plane with $x$ and y intercepts as 3 and 1 respectively. Then the image of the point (-1,-4) in this line is :

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📖 Explanation

${L}: \frac{{x}}{3}+\frac{{y}}{1}=1$ $\Rightarrow {x}+3 {y}-3=0$ Image of point (-1,-4) $\frac{x+1}{1}=\frac{y+4}{3}=-2(\frac{-1-12-3}{10}$$$ \frac{x+1}{1}=\frac{y+4}{3}=\frac{16}{5}({x}, {y}) \equiv (\frac{11}{5}, \frac{28}{5})$

Q44JEE Main 2020MCQ4MDifferential Equations

If ${y}=(\frac{2}{\pi } {x}-1) {\text{cosec}} {x}$ is the solution of the differential equation, $\frac{d y}{d x}+p(x) y=\frac{2}{\pi } {\text{cosec}} x$ , $0 < x < \frac{\pi }{2},$ then the function ${p}({x})$ is equal to

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📖 Explanation

${y}=(\frac{2 {x}}{\pi }-1) {\text{cosec}} {x}$ ...(1) $\frac{d y}{d x}=\frac{2}{\pi } {\text{cosec}} x$ $-(\frac{2 x}{\pi }-1) {\text{cosec}} x \cot x$ $\frac{d y}{d x}={2 {\text{cosec}} x}/{\pi }-y \cot x$ using equation (1) $\frac{d y}{d x}+y \cot x={2 {\text{cosec}} x}/{\pi }$ $\frac{{dy}}{{dx}}+{p}({x}).{y}={2 {\text{cosec}} {x}}/{\pi } ,$ ${x} \in (0, \frac{\pi }{2})$ Compare : p(x) = cot x

Q45JEE Main 2020MCQ4MApplication of Derivatives

The centre of the circle passing through the point (0,1) and touching the parabola $y=x^{2}$ at the point (2,4) is :

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📖 Explanation

${y}={x}^{2}$ ${\frac{{dy}}{{dx}}|}_{{\text{P}}}=4$ $({y}-4)=4({x}-2)$ $4 {x}-{y}-4=0$ Circle $:(x-2)^{2}+(y-4)^{2}+\lambda(4 x-y-4)=0$ passes through (0, 1) $4+9+\lambda(-5)=0$ $\Rightarrow \lambda =\frac{13}{5}$ Circle $: x^{2}+y^{2}+x(4 \lambda-4)$ $+y(-\lambda-8)+(20-4 \lambda)=0$ Centre $:(2-2 \lambda, \frac{\lambda+8}{2})$ $\equiv (\frac{-16}{5}, \frac{53}{10})$

Q46JEE Main 2020NAT4MMatrices and Determinants

The sum of distinct values of $\lambda$ for which the system of equations $(\lambda-1) x+(3 \lambda+1) y+2 \lambdaz=0$ $(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$ $2 x+(3 \lambda+1) y+3(\lambda-1) z=0$ has non-zero solutions, is ___________.

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📖 Explanation

$(\lambda-1) x+(3 \lambda+1) y+2 \lambdaz=0$ $(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$ $2 x+(3 \lambda+1) y+3(\lambda-1) z=0$ $\begin{vmatrix}\lambda-1 & 3\lambda+1 & 2\lambda \\ \lambda-1 & 4\lambda-2 & \lambda+3 \\ 2 & 3\lambda+1 & 3\lambda-3\end{vmatrix}=0$ ${R}_{1}\to {R}_{1}-{R}_{2}$ & ${R}_{2}\to {R}_{2}-{R}_{3}$ $=\begin{vmatrix}0 & 3-\lambda & \lambda-3 \\ \lambda-3 & \lambda-3 & -2(\lambda-3) \\ 2 & 3\lambda+1 & 3\lambda-3\end{vmatrix}=0$ $=(\lambda-3)^2\begin{vmatrix}0 & -1 & 1 \\ 1 & 1 & -2 \\ 2 & 3\lambda+1 & 3\lambda-3\end{vmatrix}=0$ $(\lambda-3)^{2}[(3 \lambda+1)+(3 \lambda-1)]=0$ $6 \lambda(\lambda-3)^{2}=0\Rightarrow \lambda=0,3$ Sum $=3$

Q47JEE Main 2020NAT4MSequences and Series

Suppose that a function $f: {R}\to {R}$ satisfies $f(x+y)=f(x) f(y)$ for all $x, y \in R$ and $f(1)=3 .$ If $\sum _{i=1} ^{n} f(i)=363,$ then $n$ is equal to

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📖 Explanation

$f(x+y)=f(x) f(y)$ put $x=y=1$ → $f(2)=(f(1))^{2}=3^{2}$ put $x=2, y=1$ → $f(3)=(f(1))^{3}=3^{3}$ Similarly $f(x)=3^{x}$ $∑_{i=0}^nf(i)=363$ $\Rightarrow ∑_{i=0}^n 3^{i}=363$ $(3+3^{2}+...+3^{n})=363$ $\frac{3(3^{n}-1)}{2}=363$ $3^{n}-1=242 \Rightarrow 3^{n}=243$ $\Rightarrow n=5$

Q48JEE Main 2020NAT4MPermutations and Combinations

The number of words (with or without meaning) that can be formed from all the letters of the word "LETTER" in which vowels never come together is __________.

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📖 Explanation

LETTER vowels = EE, consonant = LTTR _ L _ T _ T _ R _ $\frac{4 !}{2 !} \times { }^{5} {C}_{2} \times \frac{2 !}{2 !}$ $=12 \times 10=120$

Q49JEE Main 2020NAT4MBinomial Theorem

Consider the data on x taking the values 0,2 , $4,8, ..., 2^{{n}}$ with frequencies ${ }^{{n}} {C}_{0},{ }^{{n}} {C}_{1},{ }^{{n}} {C}_{2}, ...$ ${ }^{n} C_{n}$ respectively. If the mean of this data is $\frac{728}{2^{n}},$ then $n$ is equal to ___________ .

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📖 Explanation

x0248 $2^n$ f $^nC_0$ $^nC_1$ $^nC_2$ $^nC_3$ $^nC_n$ Mean $={∑{x}_{{i}}\; f_{{i}}}/{∑f_{{i}}}={\sum _{r=0}^n 2^{{r}}\;{ }^{{n}} {C}_{{r}}}/{\sum _{r=1}^n { }^{{n}}{C}_{{r}}}$ Mean $={(1+2)^n-^nC_0}/2^n=728/2^n$ $\Rightarrow {3^n-1}/2^n =728/2^n$ $\Rightarrow 3^n=729\Rightarrow n=6$

Q50JEE Main 2020NAT4MVector Algebra

If ${{x^{\to }}}$ and ${{y^{\to }}}$ be two non-zero vectors such that $|{{x^{\to }}}+{{y^{\to }}}|=|{{x^{\to }}|$ and $2 {{x^{\to }}}+\lambda{{y^{\to }}}$ is perpendicular to ${{y^{\to }}},$ then the value of $\lambda$ is __________.

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📖 Explanation

$|{{x^{\to }}}+{{y^{\to }}}|=|{{x^{\to }}|$ $\sqrt{ |{{x^{\to }}}|^{2+}|{{y^{\to }}}|^{2+}2\vec{x}.\vec{y}}=|{{x^{\to }}|$ $|{{y^{\to }}|^{2+}2\vec{x}.\vec{y}=0$ ....(1) Now $(2\vec{x}+\lambda\vec{y}).\vec{y}=0$ $2\vec{x}.\vec{y} +\lambda|\vec{y}|^2=0$ from (1) $-|\vec{y}|^{2+}\lambda|\vec{y}|^2=0$ $(\lambda-1)|\vec{y}|^2=0$ given $|\vec{y}|\ne 0\Rightarrow \lambda=1$

PHYSICS25 questions

Q51JEE Main 2020MCQ4MGravitation

Two planets have masses ${M}$ and $16 {M}$ and their radii are a and 2a, respectively. The separation between the centers of the planets is 10a. A body of mass ${m}$ is fired from the surface of the larger planet towards the smaller planet along the line joining their centers. For the body to be able to reach at the surface of smaller planet, the minimum firing speed needed is :

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📖 Explanation

$\frac{{GM}}{{x}^{2}}=\frac{{G}(16 {M})}{(10 {a}-{x})^{2}}$ $\frac{1}{x}=\frac{4}{(10 a-x)}$ $\Rightarrow 4x=10a-x$ $\Rightarrow x=2a$ ....(i) COME $-\frac{{GMm}}{8 {a}}-\frac{{G}(16 {M}) {m}}{2 {a}}+{KE}$ $=-\frac{{GMm}}{2 {a}}-\frac{{G}(16 {M}) {m}}{8 {a}}$ ${KE}={GMm}[\frac{1}{8 {a}}+\frac{16}{2 {a}}-\frac{1}{2 {a}}-\frac{16}{8 {a}}]$ ${KE}={GMm}[\frac{1+64-4-16}{8 {a}}]$ $\frac{1}{2} {mv}^{2}={GMm}[\frac{45}{8 {a}}]$ ${\text{V}}=\sqrt{\frac{90 {GM}}{8 {a}}}$ ${\text{V}}=\frac{3}{2} \sqrt{\frac{ {GM}}{{a}}}$

Q52JEE Main 2020MCQ4MThermodynamics

Three rods of identical cross-section and lengths are made of three different materials of thermal conductivity ${K}_{1}, {K}_{2},$ and ${K}_{3},$ respectively. They are joined together at their ends to make a long rod (see figure). One end of the long rod is maintained at $100°{C}$ and the other at $0°{C}$ (see figure). If the joints of the rod are at $70° {C}$ and $20°{C}$ in steady state and there is no loss of energy from the surface of the rod, the correct relationship between ${K}_{1}, {K}_{2}$ and ${K}_{3}$ is :

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📖 Explanation

Rods are identical have same length $(l)$ and area of cross-section (A) Combination are in series, so heat current is same for all Rods $(\frac{\DeltaQ}{\Deltat})_{{\text{AB}}}=(\frac{\DeltaQ}{\Deltat})_{\text{BC}}$ $=(\frac{\DeltaQ}{\Deltat})_{\text{CD}}=$ Heat current ${(100-70) {\text{K}}_{1} {\text{A}}}/{l}$ $={(70-20) {\text{K}}_{2} {\text{A}}}/{l}$ $={(20-0) {\text{K}}_{3} {\text{A}}}/{l}$ $30 {\text{K}}_{1}=50 {\text{K}}_{2}=20 {\text{K}}_{3}$ $3 {\text{K}}_{1}=2 {\text{K}}_{3}$ $\frac{K_{1}}{K_{3}}=\frac{2}{3}=2: 3$ $5 {\text{K}}_{2}=2 {\text{K}}_{3}$ ${{\text{K}}_{2}}{{\text{K}}_{3}}=\frac{2}{5}=2: 5$

Q53JEE Main 2020MCQ4MElectromagnetic Waves

For a plane electromagnetic wave, the magnetic field at a point $x$ and time $t$ is ${\vec{\text{B}}}({x}, {t})=\$ [1.2 ×10^{-7} \sin (0.5 ×10^{3} {x}+1.5 ×10^{11} {t}) \hat{k}]$ {T} $The instantaneous electric field$ {\vec{\text{E}}} $corresponding to$ {{\vec{B}}} $is : (speed of light$ {c}=3 \times 10^{8} {\text{ms}}^{-1}$ )

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📖 Explanation

${{\\vec{E}}}$ and ${\vec{\text{B}}}$ are perpendicular for EM wave ${\text{E}}_{0}={\text{CB}}_{0}$ $=3 \times 10^{8} \times 1.2 \times 10^{-7}$ $=36$ Having same phase Propagation is along $-{x}$ -axis, ${\vec{\text{B}}}$ is along z-axis hence ${{\vec{E}}}$ must be along ${y}$ -axis.

Q54JEE Main 2020MCQ4MRay Optics and Optical Instruments

A double convex lens has power ${\text{P}}$ and same radii of curvature ${\text{R}}$ of both the surfaces. The radius of curvature of a surface of a plano-convex lens made of the same material with power $1.5 {\text{P}}$ is:

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📖 Explanation

${R}_{1}={R}_{2}={R}$ Power (P) Refractive index is assume $(\mu_{l})$ ${P}=\frac{1}{{f}}=(\mu_{l}-1)(\frac{2}{{R}})$ ....(i) $P'={1}{f'}=(\mu_{l}-1)(\frac{1}{R'})$ ....(ii) $P'=\frac{3}{2} P$ $(\mu_{l}-1)(\frac{1}{{R'}})=\mu\frac{3}{2}(\mu_{l}-1)(\frac{2}{{R}})$ $\therefore R'=R/3$

Q55JEE Main 2020MCQ4MCurrent Electricity

A circuit to verify Ohm's law uses ammeter and voltmeter in series or parallel connected correctly to the resistor. In the circuit :

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Q56JEE Main 2020MCQ4MLaws of Motion

A particle moving in the xy plane experiences a velocity dependent force ${{\vec{F}}}={k}({v}_{{y}} \hat{i}+{v}_{{x}} \hat{j}),$ where ${v}_{{x}}$ and ${v}_{{y}}$ are the ${x}$ and ${y}$ components of its velocity ${{\vec{v}}}$ . If ${\vec{a}}$ Ls the acceleration of the particle, then which of the following statements is true for the particle ?

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📖 Explanation

${{d} v_{{x}}}/{{dt}}=\frac{{k}}{{m}} {v}_{{y}}$ ${{d} {v}_{{y}}}/{{dt}}=\frac{{k}}{{m}} {v}_{{x}}$ ${{d} {v}_{{y}}}/{{d} {v}_{{x}}}={{v}_{{x}}}/{{v}_{{y}}}$ $\Rightarrow ∫{v}_{{y}} {d} {v}_{{y}}=∫{v}_{{x}} {d} {v}_{{x}}$ ${v}_{{y}}^{2}={v}_{{x}}^{2}+{C}$ ${v}_{{y}}^{2}-{v}_{{x}}^{2}={\text{constant}}$

Q57JEE Main 2020MCQ4MElectrostatics

Consider the force ${F}$ on a charge 'q' due to a uniformly charged spherical shell of radius ${R}$ carrying charge Q distributed uniformly over it. Which one of the following statements is true for ${F},$ if 'q' is placed at distance ${r}$ from the centre of the shell ?

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📖 Explanation

Inside the shell, $E=0$ hence ${F}=0$ Outside the shell, ${\text{E}}=\frac{1}{4 \pi \epsilon _{0}} {{\text{Q}}}/{{r}^{2}}$ hence ${\text{F}}=\frac{1}{4 \pi \epsilon _{0}} {{\text{Qq}}}/{{r}^{2}}$ for ${r} > {\text{R}}$

Q58JEE Main 2020MCQ4MAtoms and Nuclei

Given the masses of various atomic particles ${m}_{{p}}=1.0072 {u}, {m}_{{n}}=1.0087 {u}$ ${m}_{{e}}=0.000548 {u}, {m}_{{{\ov v}}}=0, {m}_{{d}}=2.0141 {u}$ where ${p} \equiv$ proton, ${n} \equiv$ neutron, ${e} \equiv$ electron, $\overline{{v}} \equiv$ antineutrino and ${d} \equiv$ deuteron. Which of the following process is allowed by momentum and energy conservation?

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📖 Explanation

Only in case-I, ${\text{M}}_{{\text{LHS}}} > {\text{M}}_{{\text{RHS}}}$ i.e. total mass on reactant side is greater then that on the product side. Hence it will only be allowed.

Q59JEE Main 2020MCQ4MElectrostatic Potential and Capacitance

Two identical electric point dipoles have dipole moments ${{\\vec{p}}}_{1}={\text{p}} \hat{i}$ and ${\vec{\text{p}}}_{2}=-{\text{p}} \hat{i}$ and are held on the $x$ axis at distance 'a' from each other. When released, they move along the $x$ -axis with the direction of their dipole moments remaining unchanged. If the mass of each dipole is 'm', their speed when they arc infinitely far apart is:

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📖 Explanation

Using energy conservation: ${\text{KE}}_{{i}}+{\text{PE}}_{{i}}={\text{KE}}_{{f}}+{\text{PE}}_{{f}}$ ${O}+{2 {\text{KP}}}/{{a}^{3}} \times {P}=\frac{1}{2} {mv}^{2} \times 2+0$ ${V}=\sqrt{\frac{2 {P}^{2}}{4 \pi \epsilon _{0} {a}^{3} {m}}}=\frac{{P}}{{a}} \sqrt{\frac{1}{2 \pi \epsilon _{0} {am}}}$

Q60JEE Main 2020MCQ4MCurrent Electricity

In the figure shown, the current in the $10 {\text{V}}$ battery is close to :

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📖 Explanation

${E}_{{eq}}=\frac{20 \times 10}{17}=\frac{200}{17}$ and ${R}_{{eq}}=\frac{7 \times 10}{17}=\frac{70}{17}$ ${I}=\frac{\frac{20}{17}-10}{4+\frac{70}{17}}=0.21 {\text{A}}$ from +ve to -ve terminal

Q61JEE Main 2020MCQ4MRotational Motion

The linear mass density of a thin rod ${\text{AB}}$ of length ${\text{L}}$ varies from ${\text{A}}$ to ${\text{B}}$ as $\lambda({x})=\lambda_{0}(1+{{x}} /{{\text{L}}}),$ where ${x}$ is the distance from A. If ${\text{M}}$ is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is:

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📖 Explanation

${I}=∫{r}^{2} {dm}=∫{x}^{2} \lambda{d} {x}$ ${I}= ∫_{0}^{\text{L}}{x}^{2} \lambda_{0}(1+\frac{{x}}{{L}}) {d} {x}$ ${I}=\lambda_{0} ∫_{0}^{\text{L}}({x}^{2}+{{x}^{3}}/{{\text{L}}}) {d} {x}$ ${I}={7 {\text{L}}^{3} \lambda_{0}}/{12}$ ....(i) ${M}= ∫_{0}^{\text{L}} \lambda {d} {x}= ∫_{0}^{\text{L}} \lambda_{0}(1+\frac{{x}}{{L}}) {d} {x}$ ${M}=\lambda_{0}({\text{L}}+{{\text{L}}}/{2})=\lambda_{0} {3 {\text{L}}}/{2}$ $\frac{2}{3} {\text{M}}=(\lambda_{0} {\text{L}})$ ....(ii) From (i) & (ii) ${I}=\frac{7}{12}(\frac{2}{3} {M}) {L}^{2}={7 {\text{ML}}^{2}}/{18}$

Q62JEE Main 2020MCQ4MUnits and Measurements

A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings $5.50 {\text{mm}}, 5.55 {\text{mm}}, 5.45 {\text{mm}} ; 5.65 {\text{mm}} .$ The average of these four readings is $5.5375 {\text{mm}}$ and the standard deviation of the data is $0.07395 {\text{mm}} .$ The average diameter of the pencil should therefore be recorded as :

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Q63JEE Main 2020MCQ4MOscillations

When a particle of mass ${m}$ is attached to a vertical spring of spring constant ${k}$ and released, its motion is described by ${y}({t})={y}_{0} \sin ^{2} \omega {t},$ where 'y' is measured from the lower end of unstretched spring. Then $\omega$ is :

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📖 Explanation

${y}={y}_{0} \sin ^{2} \omega {t}$ ${y}=\frac{{y}_{0}}{2}(1-\cos 2 \omega {t})$ ${y}-\frac{{y}_{0}}{2}=-\frac{{y}_{0}}{2} \cos 2 \omega {t}$ Amplitude : $\frac{{y}_{0}}{2}$ $\frac{{y}_{0}}{2}=\frac{{mg}}{{K}}$ $2 \omega =\sqrt{\frac{{K}}{{m}}}=\sqrt{\frac{2 {g}}{{y}_{0}}}$ $\omega =\sqrt{\frac{{g}}{2 {y}_{0}}}$

Q64JEE Main 2020MCQ4MThermodynamics

In a dilute gas at pressure ${P}$ and temperature ${T}$ , the mean time between successive collisions of a molecule varies with T as:

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📖 Explanation

${v}_{{\text{avg}}} \propto \sqrt{{T}}$ ${t}_{0}:$ mean time $\lambda:$ mean free path ${t}_{0}={\lambda}/{{v}_{{\text{avg} }}} \propto {1}/{\sqrt{{T}}}$

Q65JEE Main 2020MCQ4MProperties of Matter

A fluid is flowing through a horizontal pipe of varying cross-section, with speed ${v} {ms}^{-1}$ at a point where the pressure is P Pascal. At another point where pressure is $\;\frac{ {P}}{2}$ Pascal its speed is ${V} {ms}^{-1} .$ If the density of the fluid is $\rho \; {kg} \; {m}^{-3}$ and the flow is streamline, then ${V}$ is equal to :

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📖 Explanation

Applying Bernoulli's Equation ${P}_{1}+\frac{1}{2} \rho {v}_{1}^{2}+\rho {g} {y}_{1}$ $={P}_{2}+\frac{1}{2} \rho {v}_{2}^{2}+\rho {g} {y}_{2}$ ${P}+\frac{1}{2} \rho {v}^{2}=\frac{{P}}{2}+\frac{1}{2} \rho {\text{V}}^{2}$ $\frac{2 {P}}{2 \rho }+\frac{1}{2} \frac{\rho {v}^{2}}{\rho } \times 2={\text{V}}^{2}$ $\sqrt{\frac{{P}}{{\rho }}+{v}^{2}}=\text{V}$

Q66JEE Main 2020MCQ4MDual Nature of Matter and Radiation

Assuming the nitrogen molecule is moving with r.m.s. velocity at $400 {\text{K}},$ the de-Broglie wavelength of nitrogen molecule is close to : (Given : nitrogen molecule weight $: 4.64 \times 10^{-26} {\text{kg}}$ , $\text{Boltzman} \text{constant} : 1.38 \times 10^{-23} {\text{J}} / {\text{K}}$ , $\text{Planck}\text{constant} : 6.63 \times 10^{-34} {\text{J}} . {\text{s}})$

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📖 Explanation

${v}_{{\text{rms}}}=\sqrt{{3 {\text{KT}}}/{{m}}}$ ${m}\to$ mass of one molecule $($ in ${\text{kg}})=$ ${\text{molar mass }}/{{ \text{NA} }}$ de-Broglie wavelenth, $\lambda=\frac{{h}}{{mv}}$ given, ${v}={v}_{{\text{rms}}}$ $\lambda={{h}}/{{m} \sqrt{{3 {\text{KT}}}/{{m}}}}$ $={{h}}/{\sqrt{3 {\text{KTm}}}}$ $={6.63 \times 10^{-34}}/{\sqrt{3 \times 1.38 \times 10^{-23} \times 400 \times (\frac{28 \times 10^{-3}}{6.023 \times 10^{-23}})}}$ $\lambda=\frac{6.63 \times 10^{-11}}{2.77}=2.39 \times 10^{-11} {m}$ $\lambda=0.24 \Å$

Q67JEE Main 2020MCQ4MCentre of Mass and Collision

Particle A of mass ${m}_{1}$ moving with velocity $(\sqrt{3} \hat{i}+\hat{j}) \text{ms}^{-1}$ collides with another particle $B$ of mass ${m}_{2}$ which is at rest initially. Let ${\vec{v}}_{1}$ and ${{\\vec{V}}}_{2}$ be the velocities of particles ${\text{A}}$ and ${\text{B}}$ after collision respectively. If ${m}_{1}=2 {m}_{2}$ and after collision ${{\text{V}}}_{1}=(\hat{i}+\sqrt{3} \hat{j}) {\text{ms}}^{-1},$ the angle between ${{\\vec{V}}}_{1}$ and ${{\\vec{V}}}_{2}$ is :

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📖 Explanation

${{\vec{v}}}_{01}=(\sqrt{3} \hat{i}+\hat{j}) {\text{m}} / {\text{s}}$ ${\vec{v}}_{02}=\vec{0}$ ${m}_{1}=2 {m}_{2}$ After collision, $v_\vec{1}=(\hat{i}+\sqrt3\hat{j})\text{m}/\text{s}$ $v_\vec{2}=\?$ Applying conservation of linear momentum, ${m}_{1} {\vec{v}}_{01}+{m}_{2} {\vec{v}}_{02}={m}_{1} {{\vec{v}}}_{1}+{m}_{2} {{\vec{v}}}_{2}$ $2 {m}_{2}(\sqrt{3} \hat{i}+\hat{j})+0=2 {m}_{2}(\hat{i}+\sqrt{3} \hat{j})+{m}_{2} {{\vec{v}}}_{2}$ ${{v}}_{2}=2(\sqrt{3} \hat{i}+\hat{j})-2(\hat{i}+\sqrt{3} \hat{j})$ $=2(\sqrt{3} {\hat{i}}-{\hat{j}})+2({\hat{i}}-\sqrt{3} {\hat{j}})$ ${{\vec{v}}}_{2}=2(\sqrt{3}-1)(\hat{i}-\hat{j})$ for angle between ${{\vec{v}}}_{1} \& {{\vec{v}}}_{2}$ , $\cos \theta ={{{\vec{v}}}_{1}.{{\vec{v}}}_{2}}/{{{\vec{v}}}_{1} {{\vec{v}}}_{2}}$ $=\frac{2(\sqrt{3}-1)(1-\sqrt{3})}{2 \times 2 \sqrt{2}(\sqrt{3}-1)}$ $\cos \theta =\frac{1-\sqrt{3}}{2 \sqrt{2}}\Rightarrow \theta =105°$ or

Q68JEE Main 2020MCQ4MMagnetic Effects of Current and Magnetism

A charged particle going around in a circle can be considered to be a current loop. A particle of mass m carrying charge q is moving in a plane with speed ${v}$ under the influence of magnetic field ${{\\vec{B}}}$ . The magnetic moment of this moving particle :

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📖 Explanation

Magnetic moment, ${\text{M}}={\text{iA}}$ $M=(\frac{q}{T})\times \pi r^{2}$ $=\frac{q \pi r^{2}}{(\frac{2 \pi r}{v})}=\frac{q v r}{2}$ $M=\frac{q v}{2} \times \frac{v m}{q B}=\frac{mv^2}{2\text{B}}$ As we can see from the figure, direction of magnetic moment (M) is opposite to magnetic field. ${\vec{M}}=-\frac{{mv}^{2}}{2 {B}} \hat{B}$ $=-{{mv}^{2} {{\\vec{B}}}}/{2{\text{B}}^{2}}$

Q69JEE Main 2020MCQ4MMagnetic Effects of Current and Magnetism

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

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📖 Explanation

${F}={BI} 2 {a}=\frac{\mu_{0} {I}}{2 \pi {r}} {I} \times 2 {a}$ ${F}={\mu_{0} {I}^{2} {a}}/{\pi \sqrt{{b}^{2}+{a}^{2}}}$ $\tau ={F} \cos \theta \times 2 {a}$ $={\mu_{0} {I}^{2} {a}}/{\pi \sqrt{{b}^{2}+{a}^{2}}} \times {{b}}/{\sqrt{{b}^{2}+{a}^{2}}} \times 2 {a}$ $\tau =\frac{2 \mu_{0} {I}^{2} {a}^{2} {b}}{\pi ({a}^{2}+{b}^{2})}$ If ${b}>> a$ then $\tau =\frac{2 \mu_{0} {I}^{2} {a}^{2}}{\pi {b}}$ But among the given options (1) is most appropriate.

Q70JEE Main 2020MCQ4MMotion in a Plane

When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed ${v}$ , he sees that rain drops are coming at an angle $60°$ from the horizontal. On further increasing the speed of the car to $(1+\beta ){v},$ this angle changes to $45° .$ The value of $\beta$ is close to:

🚀 Solve in Practice Mode

📖 Explanation

Rain is falling vertically downwards. ${{\vec{v}}}_{{r} / {m}}={{\vec{v}}}_{{r}}-{{\vec{v}}}_{{m}}$ $\tan 60°={{v}_{{r}}}/{{v}_{{m}}}=\sqrt{3}$ ${v}_{{r}}={v}_{{m}} \sqrt{3}={v} \sqrt{3}$ Now, ${v}_{{m}}=(1+{\text{B}}) {v}$ and $\theta =45°$ $\tan 45°={{v}_{{r}}}/{{v}_{{m}}}=1$ ${v}_{{r}}={v}_{{m}}$ ${v} \sqrt{3}=(1+\beta ) {v}$ $\sqrt{3}=1+\beta$ $\Rightarrow \beta =\sqrt{3}-1=0.73$

Q71JEE Main 2020NAT4MThermodynamics

An engine operates by taking a monatomic ideal gas through the cycle shown in the figure. The percentage efficiency of the engine is close to ________.

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📖 Explanation

${\text{W}}_{{\text{ABCDA}}}=2 {\text{P}}_{0} {\text{V}}_{0}$ ${\text{Q}}_{{\in}}={\text{Q}}_{{\text{AB}}}+{\text{Q}}_{{\text{BC}}}$ ${\text{Q}}_{{\text{AB}}}={\text{nC}}({\text{T}}_{{\text{B}}}-{\text{T}}_{{\text{A}}})$ $={{\text{n}} 3 {\text{R}}}/{2}({\text{T}}_{{\text{B}}}-{\text{T}}_{{\text{A}}})$ $=\frac{3}{2}({\text{P}}_{{\text{B}}} {\text{V}}_{{\text{B}}}-{\text{P}}_{{\text{A}}} {\text{V}}_{{\text{A}}})$ $=\frac{3}{2}(3 {\text{P}}_{{\text{B}}} {\text{V}}_{0}={\text{P}}_{0} {\text{V}}_{0})$ $=3 {\text{P}}_{0} {\text{V}}_{0}$ ${\text{Q}}_{{\text{BC}}}={\text{nC}}_{{\text{P}}}({\text{T}}_{{\text{C}}}-{\text{T}}_{{\text{B}}})$ $={{\text{n}} 5 {\text{R}}}/{2}({\text{T}}_{{\text{C}}}-{\text{T}}_{{\text{B}}})$ $=\frac{5}{2}({\text{P}}_{{\text{C}}} {\text{V}}_{{\text{C}}}-{\text{P}}_{{\text{B}}} {\text{V}}_{{\text{B}}})$ $=\frac{5}{2}(6 {{\text{P}}}_{0} {\text{V}}_{0}-{3} {{\text{P}}}_{0} {\text{V}}_{0})$ $=\frac{15}{2} {\text{P}}_{0} {\text{V}}_{0}$ $\eta ={{\text{W}}}/{{\text{Q}}_{{\in}}} \times 100$ $={2 {\text{P}}_{0} {\text{V}}_{0}}/{3 {\text{P}}_{0} {\text{V}}_{0}+{15}{2} {\text{P}}_{0} {\text{V}}_{0}} \times 100$ $\eta =\frac{400}{21}=19.04 \approx 19$ $\eta =19$

Q72JEE Main 2020NAT4MCentre of Mass and Collision

The centre of mass of a solid hemisphere of radius $8 {\text{cm}}$ is ${\text{X}} {\text{cm}}$ from the centre of the flat surface. Then value of $x$ is

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📖 Explanation

$x=\frac{3 R}{8}=3 {cm}$ $x=3$

Q73JEE Main 2020NAT4MSemiconductor Electronics

The output characteristics of a transistor is shown in the figure. When ${\text{V}}_{{\text{CE}}}$ is $10 {\text{V}}$ and $\text{I}_{{\text{C}}}=4.0 {\text{mA}},$ then value of $\beta _{{\text{ac}}}$ is __________ .

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📖 Explanation

$\Delta{\text{I}}_{{\text{B}}}=(30-20)=10 \mu{\text{A}}$ $\Delta{\text{I}}_{{\text{C}}}=(4.5-3) {\text{mA}}=1.5 {\text{mA}}$ $\beta _{{\text{ac}}}={\Delta{\text{I}}_{{\text{C}}}}/{\Delta{\text{I}}_{{\text{B}}}}={1.5 {\text{mA}}}/{10 \mu{\text{A}}}=150$ $\beta _{{\text{ac}}}=150$

Q74JEE Main 2020NAT4MAlternating Current

In a scries LR circuit, power of $400 {\text{W}}$ is dissipated from a source of $250 {\text{V}}, 50 {\text{Hz}}$ . The power factor of the circuit is $0.8 .$ In order to bring the power factor to unity, a capacitor of value ${\text{C}}$ is added in series to the ${\text{L}}$ and ${\text{R}}$ . Taking the value of ${\text{C}}$ as $(\frac{{n}}{3 \pi }) \mu {\text{F}},$ then value of ${n}$ is

🚀 Solve in Practice Mode

📖 Explanation

${\text{P}}={{\text{E}}_{{\text{rms}}}^{2}}/{{\text{Z}}} \cos \phi$ $400={(250)^{2} \times 0.8}/{{\text{Z}}}$ ${\text{Z}}=25 \times 5=125$ ${\text{X}}_{{\text{L}}}=125 \sin \phi =125 \times 0.6=75$

Q75JEE Main 2020NAT4MWave Optics

A Young's double-slit experiment is performed using monochromatic light of wavelength $\lambda$ . The intensity of light at a point on the screen, where the path difference is $\lambda,$ is ${\text{K}}$ units. The intensity of light at a point where the path difference is A $\frac{\lambda}{6}$ is given by ${{\text{nK}}}/{12},$ where ${n}$ is an integer. The value of ${\text{n}}$ is

🚀 Solve in Practice Mode

📖 Explanation

${I}_{\max }={k}$ ${I}_{1}={I}_{2}={K} / 4$ $\Delta{x}=\lambda/ 6 \Rightarrow \Delta\phi =\pi / 3$ ${I}={I}_{1}+{I}_{2}+2 \sqrt{{I}_{1} {I}_{2}} \cos \phi$ ${I}=\frac{{K}}{4}+\frac{{K}}{4}+2 \times \frac{{K}}{4} \frac{1}{2}$ $=\frac{{K}}{2}+\frac{{K}}{4}=\frac{3 {K}}{4}=\frac{9 {K}}{12}$ $n=9$

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