Previous Year Paper

JEE MAIN 2020 - 08 JAN S1 2020

75 Questions300 Marks180 MinTake as Timed Mock →

🎯 Practice smarter, not harder

Practice JEE MAIN 2020 - 08 JAN 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 2020MCQ4MPractical Organic Chemistry

CHEM ISTRYA flask contains a mixture of isohexane and 3- methylphentane. One of the liquids boils at 63°C while the other boils at 60°C. What is the best way to seprate the two liquids and which one will be distilled out first?

🚀 Solve in Practice Mode

Q2JEE Main 2020MCQ4MClassification of Elements

The first ionization energy (in kJ/mol) of Na, Mg, Al and Si respectively, are :

🚀 Solve in Practice Mode

📖 Explanation

Elecronic configuration of $Na = [Ne] 3s^1$ $Mg = [Ne] 3s^2$ $Al = [Ne] 3s^2 3p ^{1}$ $Si = [Ne] 3s^2 3p^2$ So order of first ionisation energy is $Na < Mg > Al < Si$ kj/molNa < A1 < Mg < Si ( $IE_1$ order)

Q3JEE Main 2020MCQ4MAldehydes, Ketones and Carboxylic Acids

The most suitable reagent for the given conversion is :

🚀 Solve in Practice Mode

📖 Explanation

Most suitable reagent for given conversion is $B_2H_6$ (electrophilic reducing agent)

Q4JEE Main 2020MCQ4Md and f Block Elements

The third ionization enthalpy is minimum for :

🚀 Solve in Practice Mode

Q5JEE Main 2020MCQ4MOrganic Chemistry - Some Basic Principles

The predominant intermolecular forces present in ethyl acetate, a liquid, are :

🚀 Solve in Practice Mode

Q6JEE Main 2020MCQ4MSolutions

The strength of an aqueous NaOH solution is most accurately determined by titrating : (Note : consider that an appropriate indicator is used)

🚀 Solve in Practice Mode

Q7JEE Main 2020MCQ4MCoordination Compounds

The com plex that can show fac-and mer-isomers is :

🚀 Solve in Practice Mode

📖 Explanation

$[Ma_3b_3]$ type complex shows fac and mer isomerism. $[Co(NH_3)_3(NO_2)_3]$

Q8JEE Main 2020MCQ4MIonic Equilibrium

The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively :

🚀 Solve in Practice Mode

📖 Explanation

From the graph & dimensions salt is : $XY_2$ $[X] = 1 \times 10^{-3}M$ $[Y] = 2\times 10^{-3}M$ $XY_2_(s) ⇌{X^{2+}}_({aq.})+2{Y^{-}}_({aq.})$ $ksp = \$ [X^{2+}]$ $[Y^{-}]^2= (10^{-3}) (2 × 10^{-3})^2= 4× 10^{-9} M^3$

Q9JEE Main 2020MCQ4MHaloalkanes and Haloarenes

The decreasing order of reactivity towards dehydrohalogenation $(E_1)$ reaction of the following compounds is :

🚀 Solve in Practice Mode

Q10JEE Main 2020MCQ4MChemical Bonding and Molecular Structure

The number of bonds between sulphur and oxygen atoms in $S_2O_8^{2-}$ and the number of bonds between sulphur and sulphur atoms in rhombic sulphur, respectively, are :

🚀 Solve in Practice Mode

📖 Explanation

$S_2O_8^{2-}$ 8 bonds are present between sulphur and oxygen. (It is best answer in given options)Rhombic sulphur : 8 bonds are present between sulphur and sulphur atoms

Q11JEE Main 2020MCQ4MChemical Kinetics

The rate of a certain biochemical reaction at physiological temperature (T) occurs $10^6$ times faster with enzyme than without. The change in the activation energy upon adding enzyme is :

🚀 Solve in Practice Mode

📖 Explanation

$K=Ae^{-E_a/{RT}}$ $K'=Ae^{-{E _a}'/{RT}}=10^6K$ $Ae^{-{E _a}'/{RT}}=10^6Ae^{-E_a/{RT}}$ on solving we get $E_a'-E_a=-RT\ln10^6$

Q12JEE Main 2020MCQ4MBiomolecules

Which of the following statement is not true for glucose?

🚀 Solve in Practice Mode

Q13JEE Main 2020MCQ4MStates of Matter

A graph of vapour pressure and temperature for three different liquids X, Y and Z is shown below :The following inferences are made :(A) X has higher intermolecular interactions compared to Y.(B) X has lower intermolecular interactions compared to Y.(C) Z has lower intermolecular interactions compared to Y.The correct inference(s) is/are :

🚀 Solve in Practice Mode

Q14JEE Main 2020MCQ4MChemical Bonding and Molecular Structure

Among the gases (a) - (e), the gases that cause greenhouse effect are :(a) $CO_2$ (b) $H_2O$ (c) CFCs (d) $O_2$ (e) $O_3$

🚀 Solve in Practice Mode

📖 Explanation

$C O_2, H_2O, CFCs$ and $O_3$ are green house gases.

Q15JEE Main 2020MCQ4MSolutions

As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order :

🚀 Solve in Practice Mode

📖 Explanation

Since, $Fe(OH)_3$ is positively charged sol, hence, anionic charge will flocculate As per Hardy Schulze rules coagulation power of anion follows the order : $Fe(CN)_6^{3-} > CrO_4^{2-}$ $> Cl^{-} = B r^{-} = NO_3 ^{-}$ Higher the coagulation power lower will be its flocculation valuetherefore order will be : $Fe(CN)_6^{3-} < CrO_4^{2- }<$ $Cl^{-} = B r ^{-}= NO_3^{-}$

Q16JEE Main 2020MCQ4MOrganic Chemistry - Some Basic Principles

The major products A and B in the following reactions are :

🚀 Solve in Practice Mode

Q17JEE Main 2020MCQ4MStructure of Atom

For the Balmer series in the spectrum of Hatom, $\ov v=R_H\{1/n_1^{2-}1/n_2^2\},$ the correct statements among (I) and (IV) are : (I) As wavelength decreases, the lines in the series converge (II) The integer n, is equal to 2 (III) The lines of longest wavelength corresponds to $n_2 = 3$ (IV) The ionization energy of hydrogen can be calculated from wave number of these lines

🚀 Solve in Practice Mode

📖 Explanation

For balmer : $n_1 = 2$ , $n_2 = 3, 4, 5, ... \infty$ $\ov v=1/\lambda =R_H[1/2^{2-}1/n_2^2]$ $1/\lambda _\text{longest}=1/R_H[1/2^{2-}1/3^2]$

Q18JEE Main 2020MCQ4MAlcohols, Phenols and Ethers

Arrange the following compounds in increasing order of C-OH bond length :methanol, phenol, p-ethoxyphenol

🚀 Solve in Practice Mode

Q19JEE Main 2020MCQ4MAlcohols, Phenols and Ethers

The major product of the following reaction is :

🚀 Solve in Practice Mode

Q20JEE Main 2020MCQ4Mp Block Elements

When gypsum is heated to 393 K, it forms :

🚀 Solve in Practice Mode

📖 Explanation

${CaSO_4 . 2H_2O}_{\text{Gypsum}}\to ^{393K}$ ${CaSO_4 . 1/2H_2O}_{\text{Plaster} \text{of} \text{paris}}+{3/2H_2O}$

Q21JEE Main 2020NAT4MBiomolecules

The number o f chiral centres in penicillin is

🚀 Solve in Practice Mode

Q22JEE Main 2020NAT4MChemical Thermodynamics

The magnitude of work done by a gas that undergoes a reversible expansion along the path ABC shown in the figure is ______ .

🚀 Solve in Practice Mode

Q23JEE Main 2020NAT4MCoordination Compounds

The volume (in mL) of 0.125 M $AgNO_3$ , required to quantitatively precipitate chloride ions in 0.3 g of $[Co(NH_3)_6]Cl_3$ is ________. $^M[Co(NH_3)_6]Cl_3$ = 267.46 g/mol $^MAgNO_3$ = 169.87 g/mol

🚀 Solve in Practice Mode

📖 Explanation

Number of moles of $Cl^{-}$ precipitated in $[Co(NH_3)_6]Cl_3$ is equal to number of moles of $AgNO_3$ used. $0.3/267.46\times 3={0.125\times V}/1000$ where V is volume of $AgNO_3$ (in mL)V = 26.92 mL

Q24JEE Main 2020NAT4MElectrochemistry

What would be the electrode potential for the given half cell reaction at pH = 5 ?_______ $2H_2O\to O_2+4H^{+}+4e^{-}$ ; $E°_\text{red} = 1.23 V$ ( $R = 8.314 J mol^{-1} K ^{-1 }$ ; Temp = 298 K; oxygen under std. atm. pressure of 1 bar)

🚀 Solve in Practice Mode

📖 Explanation

$2H_2O\to O_2+4H^{+}+4e^{-}$ ; $E°_\text{red} = 1.23 V$ From nernst equation $E_\text{cell}=E°_\text{cell}-\frac{RT}{nF}\text{lnQ}$ at 1 bar & 298 K ${2.303RT}/F=0.059$ $pH = 5 \Rightarrow \$ [H^{+}]$ = 10^{-5} ME°\text{oxidation}= - 1.23 VoltE\text{cell}=-1.23-0.059/4\log$[H^{+}]^4E_\text{cell}=-1.23-0.059/4\log$[10^{-5}]^4= -1.23 + 0.059 \times 5$ = - 0.935 V

Q25JEE Main 2020NAT4MSolutions

Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve 10 ppm of iron in 100 kg of wheat is _____ . A tomic weight : Fe = 55.85 ; S = 32.0 ; O = 16.00

🚀 Solve in Practice Mode

📖 Explanation

$FeSO_4 .7H_2O$ (M = 277.85) $\text{ppm}=\frac{\text{wt}. \text{of} \text{Fe}}{\text{wt}. \text{of} \text{wheat}}\times 10^6$ let the wt. of salt be = w gmmoles $=w/277.85$ wt. of Fe $=W/277.85\times 55.85 \text{gm}$ $10={W/277.85\times 55.85 }/10^5\times 10^6$ $W=4.97$

MATHEMATICS25 questions

Q26JEE Main 2020MCQ4MEllipse

Let the line $y = mx$ and the ellipse $2x^2 + y^2 = 1$ intersect at a ponit P in the first quadrant. If the normal to this ellipse at P meets the co-ordinate axes at $(-1/{3\sqrt2},0)$ and (0,β), then β is equal to

🚀 Solve in Practice Mode

📖 Explanation

Any normal to the ellipse is ${x\sec\theta }/\sqrt2-y\text{cosec}\theta =-1/2$ $\Rightarrow x/ ({-\cos\theta }/\sqrt2)+y/({\sin\theta }/2)=1$ $\Rightarrow {\cos\theta }/\sqrt2=1/{3\sqrt2}$ and ${\sin\theta }/2=\beta$ $\Rightarrow \beta =\sqrt2/3$

Q27JEE Main 2020MCQ4MApplication of Derivatives

Let f : R → R be such that for all $x \in R (2^{1+x }+ 2^{1-x}), f(x)$ and $(3^x + 3^{-x})$ are in A.P., then the minimum value of $f(x)$ is

🚀 Solve in Practice Mode

📖 Explanation

$f(x)={2(2^x+2^{-x})+(3^x+3^{-x})}/2 \ge 3$ (A.M ≥ G.M)

Q28JEE Main 2020MCQ4MVector Algebra

Let the volume of a parallelopiped whose coterminous edges are given by $\vec{u}=\hat{i}+\hat{j}+\lambda \hat{k}$ , $\vec{v}=\hat{i}+\hat{j}+3\hat{k}$ and $\vec{w} = 2\hat{i} + j ^{\^}+ \hat{k}$ be 1 cu. unit. If θ be the angle between the edges $\vec{u}$ and $\vec{w}$ , then cosθ can be

🚀 Solve in Practice Mode

📖 Explanation

$\begin{vmatrix}1 & 1 & \lambda \\ 1 & 1 & 3 \\ 2 & 1 & 1\end{vmatrix}=1\Rightarrow \lambda =2,4$ Now, $\cos\theta =\frac{\vec{u}.\vec{w}}{|\vec{u}||\vec{w}|}$ $=5/{\sqrt6\sqrt6}$ or $7/{\sqrt6\sqrt18}$ $=5/6\text{or} 7/{6\sqrt3}$

Q29JEE Main 2020MCQ4MBinomial Theorem

If a,b and c are the greatest value of $^19C_p,^20C_q$ and $^21C_r$ respectively, then

🚀 Solve in Practice Mode

📖 Explanation

$a = ^19C_10, b = ^20C_10$ and $c = ^21C _10$ $\Rightarrow a=^19C_9,b=2(^19C_9)$ and $c=21/11(^20C_10)$ $\Rightarrow b =2a \text{and} c=21/11b=42/11a$ $\Rightarrow a:b:c=a:2a:42/11a$ $= 11 : 22 : 42$

Q30JEE Main 2020MCQ4MInverse Trigonometric Functions

Let $f(x)(\sin(\tan^{-1}x))+\sin (\cot^{-1}x))^{2-}1$ $,|x| > 1.$ If $\frac{dy}{dx}=1/2d/{dx}(\sin^{-1}(f(x)))$ and $y(\sqrt3)=\pi /6,$ then $y(-\sqrt3)$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Let $\tan^{-1}x=\theta$ , $\theta \in (-\pi /2,\pi /4) ∪(\pi /4,\pi /2)$ $f(x)=(\sin\theta +\cos\theta )^{2-}1$ $=\sin2\theta =\frac{2x}{1+x^2}$ Now, $\frac{dy}{dx}=1/2d/{dx}\sin^{-1}(\frac{2x}{1+x^2})$ $=-1/{1+x^2},{|x|} >1$ Since, we can integrate only in the continuous interval. So we have to take integral in two cases separtely namely for x < -1 and for x > 1. $\Rightarrow y=\begin{cases}-\tan^{-1}x+c_1{,} & x >1 \\ -\tan^{-1}+c_2{,} & x < -1$ So $c_1=\pi /2$ as $y(\sqrt3)=\pi /6$ But we cannot find $c_2$ as we do not have any other additional information for x < -1\end{cases} So, all of the given options may be correct as $c_2$ is unknown so, it should be bonus.

Q31JEE Main 2020MCQ4MLimits, Continuity and Differentiability

$\lim_{x\to 0}(\frac{3x^{2+}2}{7x^{2+}2})^{1/x^2}$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Required limit $=e^{\lim_{x\to 0}(\frac{3x^{2+}2}{7x^{2+}2}-1) {1/x^2}}$ $=e^{\lim_{x\to 0}(\frac{-4}{7x^{2+}2})}=1/e^2$

Q32JEE Main 2020MCQ4MStraight Lines and Pair of Straight Lines

Let two points be A(1,-1) and B(0,2). If a point P(x',y') be such that the area of APAB = 5 sq. units and it lies on the line, 3x + y - 4λ = 0, then a value of λ is

🚀 Solve in Practice Mode

📖 Explanation

$\ov {AB}:3x+y-2=0$ Also, $1/2\times \sqrt10\times h=5$ $\Rightarrow h=\sqrt10$ $\Rightarrow {|4\lambda -2|}/\sqrt10\Rightarrow \lambda =3,-2$

Q33JEE Main 2020MCQ4MStatistics

The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 resepectively. Each of these 10 observations is multiplied by p and then reduced by q, where p ≠ 0 and q ≠ 0. If the new mean and new s.d. become half of their original values, then q is equal to

🚀 Solve in Practice Mode

📖 Explanation

$20p - q = 10$ ...(i)and $2|p|=1\Rightarrow p±1/2$ ...(ii)So, $p=-1/2$ and $q=-20$

Q34JEE Main 2020MCQ4MDifferential Equations

Let $y = y(x)$ be a solution of the differential equation, $\sqrt{1-x^2}\frac{dy}{dx}+\sqrt{1-y^2}=0,{|x| < 1}$ . If $y(1/2)=\sqrt3/2,$ then $y(-1/\sqrt2)$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

(BONUS) $\frac{dy}{dx}=-\sqrt{1-y^2}/\sqrt{1-x^2}$ So, ${dy}/\sqrt{1-y^2}+{dx}/\sqrt{1-x^2}=0$ Integrating, $\sin ^{-1}x + \sin ^{-1}y = c$ So, $\pi /6+\pi /3=c$ Hence, $\sin^{-1}x+\sin^{-1}y=\pi /2$ Put $x=-1/\sqrt2,\sin^{-1}y={3\pi }/4$ (Not possible)

Q35JEE Main 2020MCQ4MComplex Numbers

If the equation, $x^2 + bx + 45 = 0$ (b ∈ R) has conjugate complex roots and they satisfy $|z + 1| = 2 \sqrt10$ , then

🚀 Solve in Practice Mode

📖 Explanation

Assuming z is a root of the given equation, $z={-b±i\sqrt{180-b^2}}/2$ So, $(1-b/2)^{2+}{180+b^2}/4=40$ $\Rightarrow -4b + 184 = 160 \Rightarrow b = 6$

Q36JEE Main 2020MCQ4MArea Under The Curves

For a > 0, let the curves $C_1 : y^2 = ax$ and $C_2 : x^2 = ay$ intersect at origin O and a point P. Let the line x = b(0 < b < a) intersect the chord OP and the x-axis at points Q and R, respectively. If the line $x=b$ bisects the area bounded by the curves, $C_1$ and $C_2$ , and the area of $\DeltaOQR=1/2$ , then 'a' satisfies the equation

🚀 Solve in Practice Mode

📖 Explanation

$∫^{b}_{0}(\sqrt{ax}-x^2/a)dx$ $=1/2\times {16(a/4)(a/4)}/3$ $\Rightarrow \$ [{2\text{sqrta}}/3x^{3/2}-x^3/{3a}]^b_0=a^2/6\Rightarrow {2\text{sqrta}}/3b^{3/2}=b^3/{3a}=a^2/6 $...(i)Also,$ 1/2×b^2=1/2\Rightarrow b=1 $So,$ {2\text{sqrta}}/3-1/{3a}=a^2/6\Rightarrow a^{3-}4a^{3/2}+2=0\Rightarrow a^6 + 4a^3 + 4 = 16a^3\Rightarrow a^6 - 12a^3 + 4 = 0$

Q37JEE Main 2020MCQ4MSets and Relations

Which one of the following is a tautology ?

🚀 Solve in Practice Mode

📖 Explanation

(1) $P \wedge (P\vee Q) \equiv P$ (2) $P\vee (P\wedge Q) \equiv P$ (3) $Q \to (P\wedge (P \to Q))$ $\equiv Q \to (P \wedge (~ P \vee Q)) \equiv Q \to (P \wedge Q )$ $\equiv (~ Q)\vee (P \wedge Q) \equiv (P \vee (~ Q))$ (4) $(P \wedge (P \to Q))\to Q$ $\equiv (P \wedge (~ P \vee Q)) \to Q$ $\equiv (P \wedge Q) \to Q$ $\equiv ((~ P) \vee (~ Q)) \vee Q$ $\equiv (~P)\vee t = t$

Q38JEE Main 2020MCQ4MParabola

The locus of a point which divides the line segment joining the point (0,-1) and a point on the parabola, $x^2 = 4y$ , internally in the ratio 1 : 2, is-

🚀 Solve in Practice Mode

📖 Explanation

$\Rightarrow 3h = 2t$ and $3k = t^2 - 2$ $\Rightarrow 3y=({3x}/2)^{2-}2$ $\Rightarrow 12y=9x^{2-}8$

Q39JEE Main 2020MCQ4MApplication of Derivatives

If c is a point at which Rolle's theorem holds for the function, $f(x)=\log_e(\frac{x^{2+}\alpha }{7x})$ in the interval [3,4], where $\alpha \in R$ , then $f"(c)$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

${9+\alpha }/21={16+\alpha }/28\Rightarrow \alpha =12$ Also, $f'(x)=\frac{7x}{x^{2+}12}\times \frac{x^{2-}12}{7x^2}$ $=\frac{x^{2-}12}{x(x^{2+}12)}$ Hence, $c=2\sqrt3$ Now, $f"(c)=1/12$

Q40JEE Main 2020MCQ4MMatrices and Determinants

For which of the following ordered pairs $(\mu ,\delta )$ , the system of linear equations $x + 2y + 3z = 1$ $3x + 4y + 5z = \mu$ $4x + 4y + 4z =\delta$ is inconsistent ?

🚀 Solve in Practice Mode

📖 Explanation

2 × (ii) - 2 × (i) - (iii) $0 = 2\mu - 2 - \delta$ $\Rightarrow \delta = 2(\mu - 1)$

Q41JEE Main 2020MCQ4MProbability

Let A and B be two independent events such that $P(A) = 1/3$ and $P(B) = 1/6$ Then, which of the following is TRUE ?

🚀 Solve in Practice Mode

📖 Explanation

Given that events A and B are independent, a fundamental property is that the occurrence of one does not affect the probability of the other. This means $P(A/B) = P(A)$ . This property also holds for the complements of the events. Therefore, if A and B are independent, then A and B' (the complement of B) are also independent.

Let's evaluate the expression in option C, $P(A/B')$ .
Using the definition of conditional probability:
$P(A/B') = \frac{P(A \cap B')}{P(B')}$
Since A and B' are independent, $P(A \cap B') = P(A) \cdot P(B')$ .
Substituting this into the equation, we get:
$P(A/B') = \frac{P(A) \cdot P(B')}{P(B')} = P(A)$
As we are given $P(A) = 1/3$ , it follows that $P(A/B') = 1/3$ .

Q42JEE Main 2020MCQ4MFunctions

The inverse function of $f(x)=\frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}},x \in (-1,1),$ is

🚀 Solve in Practice Mode

📖 Explanation

$f(x)=y=\frac{8^{4x}-1}{8^{4x}+1}=1-2/{8^{4x}+1}$ So, $8^{4x}+1=2/{1-y}\Rightarrow 8^{4x}=\frac{1+y}{1-y}$ $\Rightarrow x=\ln (\frac{1+y}{1-y})\times 1/{4\ln8}=f^{-1}(y)$ Hence, $f^{-1}(x)=1/4(\log_8e)\log_e(\frac{1+x}{1-x})$

Q43JEE Main 2020MCQ4MIndefinite Integrals

If $∫\frac{\cos xdx}{\sin^3x(1+\sin^6x)^{2/3}}$ $=f(x)(1+\sin^6x)^{1/\lambda }+c$ where c is a constant o f integration, then $\lambda f(\pi /3)$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

$∫\frac{\cos xdx}{\sin^3x(1+\sin^6x)^{2/3}}$ $=\frac{-6}{-6}∫\frac{\cos xdx}{\sin^7x(1/{\sin^6x}+1)^{2/3}}$ $=-1/6\times 3(1/{\sin^6x}+1)^{1/3}+c$ $=-1/2 \frac{(1+\sin^6x)^{1/3}}{\sin^x}+c$ Hence, $\lambda = 3$ and $f(x)=-1/{2\sin^2x}$ So, $\lambda f(\pi /3)=-2$ REMARK : Technically, this question should be marked as bonus. Because f(x) and λ cannot be found uniquely.

Q44JEE Main 2020MCQ4MThree Dimensional Geometry

The shortest distance between the lines ${x-3}/3=\frac{y-8}{-1}={z-3}/1$ and $\frac{x+3}{-3}={y+7}/2={z-6}/4$ is

🚀 Solve in Practice Mode

📖 Explanation

Shortest distance $={\begin{vmatrix}6 & 15 & -3 \\ 3 & -1 & 1 \\ -3 & 2 & 4\end{vmatrix}}/\sqrt{11\times 29-49}$ $=270/\sqrt270=3\sqrt30$

Q45JEE Main 2020MCQ4MApplication of Derivatives

Let $f (x) = x\cos^{-1}(-\sin|x|), x \in [-\pi /2,\pi /2],$ which of the following is true ?

🚀 Solve in Practice Mode

📖 Explanation

f(x) is an odd function.Now, if x ≥ 0, then $f(x) = x\cos^{-1}(-\sin x)$ $=x(\pi /2-\sin^{-1}(-\sin x))=x(\pi /2+x)$ Hence, $f(x)=\{\text{table} x(\pi /2+x){,},x\in [0,\pi /2];x(\pi /2-x){,},x\in [-\pi /2,0)$ So, $f'(x)=\{\text{table} \pi /2+2x {,},x\in [0,\pi /2); \pi /2-2x {,},x\in (-\pi /2,0)$

Q46JEE Main 2020NAT4MMatrices and Determinants

The number of all 3 × 3 matrices A, with enteries from the set {-1,0,1} such that the sum of the diagonal elements of $\text{AA}^\text{T}$ is 3, is

🚀 Solve in Practice Mode

📖 Explanation

trace $(\text{AA}^\text{T})= ∑a_\text{ij}^2=3$ Hence, number of such matrices $= ^9C_3\times 2^3 = 672.00$

Q47JEE Main 2020NAT4MQuadratic Equation and Inequalities

The least positive value of 'a' for which the equation $2x^2 + (a - 10)x +33/2= 2a$ has real roots is

🚀 Solve in Practice Mode

📖 Explanation

$D \ge 0$ $\Rightarrow (a - 10)^2 - 4 \times 2\times (33/2-2a) \ge 0$ $\Rightarrow a^2 - 4a - 3^2 > 0$ $\Rightarrow a\in (-\infty , 4] ⋃ [8,\infty )$

Q48JEE Main 2020NAT4MApplication of Derivatives

Let the normal at a point P on the curve $y^2 - 3x^2 + y + 10 = 0$ intersect the y-axis at $(0,3/2)$ . If m is the slope of the tangent at P to the curve, then $|m|$ is equal to

🚀 Solve in Practice Mode

📖 Explanation

Let P(α,β)So, $\beta ^{2-}3\alpha ^{2+}\beta +10=0$ ...(i)Now, $2yy' - 6x + y' = 0$ $\Rightarrow m=\frac{6\alpha }{2\beta +1}$ ...(ii)Also, ${\beta -3/2}/\alpha =-1/m$ $\Rightarrow \frac{2\beta -3}{2\alpha }=-(2\beta +1)/{6\alpha }$ (from (ii)) $\Rightarrow \beta =1\Rightarrow \alpha ^2=4$ (from (1))Hence, $|m|=12/3=4.00$

Q49JEE Main 2020NAT4MSequences and Series

The sum $∑^{20}_{k=1}(1+2+3+...+k)$ is

🚀 Solve in Practice Mode

📖 Explanation

$∑^{20}_{k=1}{k(k+1)}/2$ $=1/2∑^{20}_{k=1}{k(k+1)(k+2)-(k-1)k(k+1)}/3$ $=1/6\times 20\times 21\times 22=1540.00$

Q50JEE Main 2020NAT4MPermutations and Combinations

An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at the most three of them are red is ....

🚀 Solve in Practice Mode

📖 Explanation

The question does not mention that whether same coloured marbles are distinct or identical. So, assuming they are distinct our required answer $= ^12C_4 - ^5C_4 = 490$ And, if same coloured marbles are identical then required answer = (2 + 3 + 4 + 4)= 13

PHYSICS25 questions

Q51JEE Main 2020MCQ4MProperties of Matter

Consider a solid sphere of radius R and mass density $\rho (r)=\rho _0(1-r^2/R^2),0 < r \le R$ The minimum density of a liquid in which it will float is :

🚀 Solve in Practice Mode

📖 Explanation

In case of minimum density of liqued, sphere will be floating while completely submerged So mg = B $m=∫^{R}_{0} \rho (4\pi r^2dr)=B$ $=\rho _0∫^{R}_{0}(1-r^2/R^2)4\pi r^2dr=4/3\pi R^2\rho _lg$ On Solving $\rho _l={2\rho _0}/5$

Q52JEE Main 2020MCQ4MDual Nature of Matter and Radiation

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy $T_A$ eV end de-Broglie wavelength $\lambda _A$ . The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is $T_B = (T_A - 1.5) eV$ . If the de-Broglie wavelength of these photoelectrons $\lambda _B=2\lambda _A$ , then the work function of metal B is :

🚀 Solve in Practice Mode

📖 Explanation

$\lambda _B=2\lambda _A$ $\Rightarrow h/\sqrt{2T_Bm}={2h}/\sqrt{2T_Am}$ $T_A=4T_B$ ...(i) and $T_B=(T_A-1.5)\text{eV}$ ....(ii)from (i) and (ii) $3T_B=1.5\text{eV}$ $\Rightarrow T_B=0.5\text{eV}$ $\Rightarrow 4.5\text{eV}=ϕ_B$ $ϕ_B=4\text{eV}$

Q53JEE Main 2020MCQ4MCurrent Electricity

The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA. For a cell of emf 5V and internal resistance of 20Ω, the null point on it is found to be a 1000cm. The resistance of whole wire is :

🚀 Solve in Practice Mode

📖 Explanation

$5 = \lambda l$ where λ is potential gradient & L is total length of wire $5={\DeltaV}/Ll$ $\Delta={5\times L}/l=5\times 12/10=6\text{V}$ = 60 mA × RR = 100 Ω

Q54JEE Main 2020MCQ4MMagnetic Effects of Current and Magnetism

Photon with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of $10^12$ $m/s^2$ by an applied magnetic field (west to east). The value of magnetic field : (Rest mass of proton is $1.6 \times 10^{-27}$ kg) :

🚀 Solve in Practice Mode

📖 Explanation

$a={qvB}/m$ $B=\frac{ma}{qv}={ma\text{sqrtm}}/\sqrt{2k}$ $=\frac{m^{3/2}a}{e\sqrt{2k}}$ $={(1.6\times 10^{-27})^{3/2}\times 10^12}/{1.6\times 10^{-19}\sqrt{2\times 1\times 10^6\times 1.6\times 10^{-19}}}$ = 0.71 mT

Q55JEE Main 2020MCQ4MThermodynamics

The plot that depicts the behavior of the mean free time t (time between two successive collisions) for the molecules of an ideal gas, as a function of temperature (T), qualitatively, is: (Graphs are schematic and not drawn to scale)

🚀 Solve in Practice Mode

📖 Explanation

Mean free time $=\text{ Mean free path}/\text{Average speed}$ $={1/{\sqrt2\pi D^2n}}/{\sqrt{\frac{8RT}{\pi M_w}}}$ $t\propto 1/\text{sqrtT}$

Q56JEE Main 2020MCQ4MRotational Motion

Consider a uniform rod o f mass M = 4m and length l pivoted about its centre. A mass moving with velocity v making angle $\theta =\pi /4$ to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is :

🚀 Solve in Practice Mode

📖 Explanation

Let angular velocity o f the system after collision be $\omega$ . By conservation of angular momentum about the hinge : $m(v/\sqrt2)(l/2)=\$ [{4ml^2}/12+{ml^2}/4]$\omega $On solving$ \omega ={3\sqrt2}/7v/l$

Q57JEE Main 2020MCQ4MDual Nature of Matter and Radiation

The dimension of stopping potential $V_0$ in photoelectric effect in units of Planck's constant 'h', speed of light 'c' and Gravitational constant 'G' and ampere A is :

🚀 Solve in Practice Mode

📖 Explanation

$v_0 = h^xc ^yG^z A^w$ $\frac{ML^2T^{-2}}{AT}=(ML^2T^{-1})^x (LT^{-1})^y(M^{-1}L^{3}T^{-2})^zA^w$ $x = 0$ $z = -1$ $2\times 0 + y + 3x-1=2$ $y = 5$ $\Rightarrow v_0=h^0c^5G^{-1}A^{-1}$ So Bonus

Q58JEE Main 2020MCQ4MGravitation

Consider two solid spheres of radii $R_1 = 1m$ , $R_2 = 2m$ and masses $M_1$ and $M_2$ , respectively. The gravitational field due to sphere (1) and (2) are shown. The value of $M_1/M_2$ is :

🚀 Solve in Practice Mode

📖 Explanation

Gravitational field on the surface of a solid sphere $I_g={GM}/R^2$ By the graph ${GM_1}/1^2=2$ and ${GM_2}/2^2=3$ On solving $M_1/M_2=1/6$

Q59JEE Main 2020MCQ4MElectrostatics

In finding the electric field using Gauss Law the formula $|\vec{E}|=q_\text\frac{ene}{\epsilon _0|A|}$ is applicable. In the formula $s_0$ is permittivity of free space, A is the area of Gaussian surface and $q_\text{enc}$ is charge enclosed by the Gaussian surface. The equation can be used in which of the following situation?

🚀 Solve in Practice Mode

📖 Explanation

$|\vec{E}|$ should be constant on the surface and the surface should be equipotential.

Q60JEE Main 2020MCQ4MThermodynamics

A thermodynamic cycle xyzx is shown on a V-T diagram.The P-V diagram that best describes this cycle is : (Diagrams are schematic and not to scale)

🚀 Solve in Practice Mode

Q61JEE Main 2020MCQ4MCentre of Mass and Collision

The coordinates of centre of mass of a uniform flag shaped lamina (thin flat plate) of mass 4kg. (The coordinates of the same are shown in figure) are :

🚀 Solve in Practice Mode

📖 Explanation

$m_1 = 3kg,m_2 = 1kg$ Mass of plate-1 is assumed to be concentrated at (0.5, 1.5)Mass of plate-2 is assumed to be concentrated at (1.5, 2.5). $x_\text{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$ $={3\times 0.5+1\times 1.5}/4=0.75$ $y_\text{cm}=\frac{m_1y_1+m_2y_2}{m_1+m_2}$ $={3\times 1.5+1\times 2.5}/4=1.75$

Q62JEE Main 2020MCQ4MRay Optics and Optical Instruments

The magnifying power of a telescope with tube 60 cm is 5. What is the focal length of its eye piece ?

🚀 Solve in Practice Mode

📖 Explanation

$L = f_0 + f_e = 60 cm$ $M=f_0/f_e=5$ $\Rightarrow f_0=5f_e$ $\therefore 6f_e=60cm$ $f_e=10cm$

Q63JEE Main 2020MCQ4MAtoms and Nuclei

The graph which depicts the results of Rutherform gold foil experiment with α-particales is :0 : Scattering angleY : Number of scattered α-particles detected (Plots are schematic and not to scale)

🚀 Solve in Practice Mode

📖 Explanation

$Y \propto 1/(\sin\theta /2)^4$

Q64JEE Main 2020MCQ4MCircular Motion

A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length L The other end is fixed. The system is given an angular speed $\omega$ about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is :

🚀 Solve in Practice Mode

📖 Explanation

$kx=ml\omega ^{2+}mx\omega ^2$ $x=\frac{ml\omega ^2}{k-m\omega ^2}$

Q65JEE Main 2020MCQ4MRay Optics and Optical Instruments

The critical angle of a medium for a specific wavelength , if the medium has relative permittivity 3 and relative permeability $4/3$ for this wavelength, will be :

🚀 Solve in Practice Mode

📖 Explanation

$\sin\theta _C=1/\mu =1/\sqrt{3\times 4/3}$ $\theta _C=30°$

Q66JEE Main 2020MCQ4MProperties of Matter

A leak proof cylinder of length 1 m , made of a metal which has very low coefficient of expansion is floating vertically in water at 0°C such that its height above the water surface is 20 cm. When the temperature of water is increased to 4°C, the height of the cylinder above the water surface becomes 21 cm. The density of water at T = 4°C, relative to the density at T = 0°C is close to :

🚀 Solve in Practice Mode

📖 Explanation

$m = \rho _0A (80)$ ...(ii) $m = \rho _0A (79)$ ....(ii)

Q67JEE Main 2020MCQ4MSemiconductor Electronics

Boolean relation at the output stage-Y for the following circuit is :

🚀 Solve in Practice Mode

📖 Explanation

$\text{table} A,B,Y;0,0,1;1,0,0;0,1,0;1,1,0$

Q68JEE Main 2020MCQ4MElectrostatics

Three charged particle A, B and C with charges -4q, 2q and -2q are present on the circumference of a circle of radius d. the charged particles A, C and centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along x-direction is :

🚀 Solve in Practice Mode

📖 Explanation

$E_x={K(4q)}/R^2\cos30°$ $+{K(2q)}/R^2\cos30°$ $+{K(2q)}/R^2\cos30°$

Q69JEE Main 2020MCQ4MElectromagnetic Induction

At time t = 0 magnetic field of 100 Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 Gauss, in the next 5s, then induced EMF in the loop is :

🚀 Solve in Practice Mode

📖 Explanation

$\frac{dB}{dt}=100$ $A=16\times 4-4\times 2=56\text{cm}^2$ $\epsilon =\frac{dB}{dt}A=100\times 10^{-4}\times 56\times 10^{-4}$

Q70JEE Main 2020MCQ4MElectrostatic Potential and Capacitance

Effective capacitance of parallel combination of two capacitors $C_1$ and $C_2$ is 10 µF. When these capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor $C_2$ is 4 times that of $C_1$ . If these capacitors are connected in series, their effective capacitance will be :

🚀 Solve in Practice Mode

📖 Explanation

$C_1 + C _2 = 10$ $1/2C_2V^2=4\times 1/2C_1V^2$ $\therefore C_2=4C_1$ $\therefore C_1=2$ & $C_2=8$ For series combination $E_\text{eq}=\frac{C_1C_2}{C_1+C_2}=1.6$

Q71JEE Main 2020NAT4MCurrent Electricity

Four resistances o f 15Ω, 12Ω, 4Ω and 10Ω respectively in cyclic order to form Wheatstone's network. The resistance that is to be connected in parallel with the resistance of 10Q to balance the network is ______ Ω.

🚀 Solve in Practice Mode

📖 Explanation

Let the resistance to be connected is R.For balanced Wheatstone bridge $15\times 4-12\times \frac{10R}{10+R}$ $R=10\Omega$

Q72JEE Main 2020NAT4MRay Optics and Optical Instruments

A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is 30 cm and the refractive index of the lens material is 1.5, then the focal length of the lens (in cm) is_______ .

🚀 Solve in Practice Mode

📖 Explanation

Using Lens-Maker's formula : $1/f=(\mu -1)(1/R_1-1/R_2)$ $\Rightarrow 1/f=(1.5-1)(1/30-0)$ $f=60\text{cm}$

Q73JEE Main 2020NAT4MCentre of Mass and Collision

A body A, of mass m = 0.1 kg has an initial velocity of $3\hat{i} m s^{-1}$ . It collides elastically with another body, B of the same mass which has an initial velocity of $5\hat{j} m s^{-1}$ . After collision, A moves with a velocity $\vec{v} = 4(\hat{i} + \hat{j})$ . The energy of B after collision is written as $x/10J$ . The value of x is

🚀 Solve in Practice Mode

📖 Explanation

By conservation of linear momentum : $(0.1)(3\hat{i})+(0.1)(5\hat{j})$ $=(0.1)(4)(\hat{i}+\hat{j})+(0.1)\vec{v}$ $\Rightarrow \vec{v}=-\hat{i}+\hat{j}$ ∴ Speed of B after collision $|\vec{v}|=\sqrt2$ Now, kinetic energy $=1/2mv^{2-}1/2(0.1)(2)=1/10$ $\therefore x=1$

Q74JEE Main 2020NAT4MMotion in a Plane

A particle is moving along the x-axis with its coordinate with the time 't' given be $x(t) = 10 + 8t - 3t^2$ . Another particle is moving the y-axis with its coordinate as a function of time given by $y(t) = 5 - 8t^3$ . At $t = 1s$ , the speed of the second particle as measured in the frame of the first particle is given as $\text{sqrtv}$ . Then v (in m/s) is ____________.

🚀 Solve in Practice Mode

📖 Explanation

$x = 10 + 8t - 3t^2$ $v_x = 8 - 6t$ $(v_x)_{t=i }= 2\hat{i}$ $y = 5 - 8t^3$ $v_y = - 24t^2$ $(v_y)_{t=1}=-24\hat{j}$ Now $\text{sqrtv}=\sqrt{24^{2+}2^2}=\sqrt580$ $\therefore v=580 m^2/s^2$

Q75JEE Main 2020NAT4MWaves

A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is _________Hz.

🚀 Solve in Practice Mode

📖 Explanation

$v_s=\sqrt{{\gamma P}/ \rho }$ $v_\text{gas}/v_\text{air}=\sqrt{\rho _\text{air}/\rho _\text{gas}}$ $\Rightarrow v_\text{gas}/300=1/\sqrt2$ $\therefore v_\text{gas}=150\sqrt2$ Now $n_2-n_1=v_\text\frac{gas}{2l}$ $=\frac{150\sqrt2}{2(1)}=75\sqrt2$ $\Deltan=106.06\text{Hz}$

← All JEE Main 2020 papers Take as Timed Mock →