# JEE Main 2020 Question Paper with Solution (Jan 7th first shift)

## 7th January, Morning Session ( 9:30 am to 12:30 pm)

JEE Main 2020 is conducting by National Test Agency. Through this exam, students will get admission into B.tech /B.E courses. JEE Main 2020 exam will be conducted in 2 shifts; Morning shift ( 9:30 am to 12:30 pm) and Afternoon shift (2:30 pm to 5:30 pm). In this article, our experts share the exam analysis as per the student's feedback of the 7th January Morning session paper and JEE main 2020 question paper with solutions. Aspirants who appeared for the JEE Mains exam can check the 7th January, Morning session question paper and solutions here. With the help of the solutions, students who appeared for the JEE Main 2020 - Jan 7th first shift exam will be able to check their performance in the exam. As we know the JEE main 2020 exam is divided into 3 sections physics, chemistry, and mathematics. Below you can check JEE main 2020 exam question and solutions - subject wise.

Exam date and Shift Article URL

JEE Main 2020 exam - Chemistry ( 7th January first shift)

Q1. The relative strength of interionic/intermolecular forces in decreasing order is:

(a) Ion-dipole > dipole-dipole > Ion-ion
(b) dipole-dipole > ion-dipole > ion-ion
(c) ion-ion > ion-dipole > dipole-dipole
(d) ion-dipole > ion-ion > dipole-dipole

Solution

The correct order of intermolecular forces is:

ion-ion > ion-dipole > dipole-dipole

Therefore, Option(c) is correct.

Q2. The oxidation number of potassium in K2O , K2O2 and KO2, respectively is :

Solution

For K2O: 2x - 2 = 0
x = +1

For K2O2: 2x - 2 = 0 (In peroxide, the oxidation state of oxygen is -1)
x = +1

For KO2x - 2(1/2) = 0 (In supeoxide, the oxidation state of oxygen is -1/2)
x = +1

Q3. At 35oC, the vapour pressure of CS2 is 512mm Hg and that of acetone is 344mm Hg. A solution of CS2 in acetone has a total vapour pressure of 600mm Hg. The false statement among the following is:
(a) CS2 and acetone are less attracted to each other than to themselves.
(b) Heat must be absorbed in order to produce the solution at 35oC.
(c) Raoult's law is not obeyed by this system.
(d) A mixture of 100ml CS2 and 100ml acetone has a volume of <200ml.

Solution

We have:

Pobserved = 600Hg

We know:

Pcalc = PoAxA + PoBxB
= 512x + 344 - 344x
= 168x + 344
= 512 ( if x =1 for miximum value)

Thus, Pcalc < Pobserved
Thus, the process is deviated from Raoult's law, showing positive deviation from Raoult's law.

$\mathrm{\Delta H\: >\: 0}$, i.e, heat absorbed

Therefore, Option(d) is correct.

Q4. The atomic radius of Ag is closest to:
(a) Ni
(b) Cu
(c) Au
(d) Hg

Solution

The atomic radius of Ag is closest to Au.

Therefore, Option(c) is correct.

Q5. The dipole moments of CCl4, CHCl3 and CH4 are in the order:

Solution

Q6. In comparison to the zeolite process for the removal of permanent hardness, the synthetic resins method is:

(a) Less efficient as it exchanges only anions
(b) More efficient as it can exchange only cations.
(c) Less efficient as the resins cannot be generated
(d) More efficient as it can exchange both cations and anions

Solution

The synthetic resins method is more efficient as it can exchange both cations and anions.

Therefore, Option(d) is correct.

Q7. Among the following statements that which was not proposed by Dalton was:

(a) Matter consists of invisible atoms
(b) When gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same T and P.
(c) Chemical reactions involve the reorganization of atoms.These are neither created nor destroyed in a chemical reaction

Solution

Option(b) is correct as it was proposed by Gay Lussac's law of gaseous volume.

Therefore, Option(b) is correct.

Q8. The increasing order of pKb for the following compounds will be:

(a) NH2-CH=NH
(b)
(c) CH3NHCH3

Solution

$\mathrm{pK_{b}\: \propto\: \frac{1}{basicity}\: \propto\: acidity}$

Thus, the increasing order of basicity is:
(b) > (a) > (c)

pKorder is given as:

(c) > (a) > (b)

The conjugate acid of guanidine is most resonance stabilized followed by imidine.

Q9.  What is the product of following reaction:

Solution

The reactions occur as follows:

Q10. The number of orbitals associated with quantum numbers n = 5, ms = +1/2 is:

Solution

We have:

n - 5, ms = +1/2

Thus, values of l are from 0 to (n-1)

l = 0 to 4

Thus, values of l are 5s, 5p, 5d, 5f and 5g

Now, the total number of orbitals = n2 = 52 = 25

Q11. The purest form of commercial iron is:

Solution

The purest form of iron is wrought iron.

Q12. The theory that can completely /properly explain the nature of bonding in [Ni(CO)4]
(a) Werner's theory
(b) Crystal field theory
(c) Molecular Orbital theory
(d) Valence bond theory

Solution

The metal-carbonyl $\pi$ is formed by the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding $\pi$* orbital of carbon monoxide.

Therefore, Option(c) is correct.

Q13. The IUPAC name of the complex [Pt(NH3)2Cl(NH2CH3)]Cl is:

Solution

The IUPAC name of the given compound is:
Diamminechloromethylammineplatinum(II)chloride.

Q14. 1-methyl ethylene oxide when treated with an excess of HBr produces:

Solution

Q15.  Consider the following reaction:

The product 'X' is used:

(a) in proteins estimation as an alternative to ninhydrin
(c) in laboratory test for phenols
(d) in acid-base titration as an indicator

Solution

Q16. Match the following:

(i) Riboflavin                          (a) Beriberi
(ii) Thiamine                           (b) Scurvy
(iii) Pyridoxine                        (c) Cheilosis
(iv) Ascorbic acid                    (d) Convulsions

Solution

(i) Riboflavin              -          (c) Cheilosis
(ii) Thiamine              -          (a) Beriberi
(iii) Pyridoxine           -          (d) Convulsions
(iv) Ascorbic acid       -          (b) Scurvy

Q17. Given that the standard potentials(Eo) of Cu2+|Cu and Cu+|Cu are 0.34V and 0.522V respectively, the Eo of Cu2+|Cu+ is:

Solution

We have:

ΔG = -nFEo

-2xFx0.34V = -1xFxEo -1xFx0.522V

-Eo = -2x0.34V + 0.522V

Thus, Eo = 0.158V

Q18.  A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO3 to give fraction A. The leftover orgainc phase was extracted with dilute NaOH solution to give fraction. The final organic layer was labelled as fraction C. Fractions A, B and C contain respectively:

Solution

The reactions occur as follows:

Q19.  The electron gain enthalpy(in kJ/mol) of fluorine, chlorine, bromine and iodine respectively are:

Solution

The electron gain enthalpy values are given below:

Fluorine = -333kJ/mol
Chlorine = -348kJ/mol
Bromine = -324kJ/mol
Iodine = -295kJ/mlol

Thus the correct order is:

Cl > F > Br > I

Q20.  Consider the following reactions:

(a) $\mathrm{(CH_{3})_{3}CCH(OH)CH_{3}\: \overset{conc.\: H_{2}SO_{4}}{\longrightarrow}}$

(b) $\mathrm{(CH_{3})_{2}CHCH(Br)CH_{3}\: \overset{alc.\: KOH}{\longrightarrow}}$

(c) $\mathrm{(CH_{3})_{2}CHCH(Br)CH_{3}\: \overset{(CH_{3})_{3}O^{-}K^{+}}{\longrightarrow}}$

(d) $\mathrm{(CH_{3})_{2}COHCH_{2}CHO\: \overset{\Delta }{\longrightarrow}}$

Which of these reaction(s) will not produce Saytzeff product?

Solution

The reactions occur as follows:

Therefore, Option(c) is correct.

Q21.  Two solutions A and B each of 100L was made by dissolving 4g of NaOH and 9.8g of H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40L of solution A and 10L of solution B is:

Solution

For given solutions, we have:

Moles of NaOH  4/40 = 0.1 moles

Moles of H2SO4 = 9.8/98 = 0.1 moles

Molarity of NaOH = 0.1/100L
And molarity of H2SO4 = 0.1/100

Now, 40L of NaOH solution and 10L of H2SO4 solution are added, thus we get:

Total volume = 50L

Milliequivalents of NaOH = 40x(0.1/100)x1 = 0.04

Milliequivalents of H2SO4 = 10x(0.2/100)x2 = 0.02

Thus, Meq of NaOH left = 0.04 - 0.02 = 0.02

[OH-] = 4x10-4

pOH = -log[4x10-4]

pOH = -log4 - log10-4

pOH = -0.60 + 4 = 3.4

Further, we know:

pH = 14 - 3.4

pH = 10.5

Q22. During the nuclear explosion, one of the products of 90Sr was absorbed in the bones of a newly born bany in place of Ca, how much time is years is required to reduce it by 90% if it is not lost metabolically.
(Half-life of 90Sr is 6.93years).

Solution

We know:

$\\\mathrm{t\: =\: \frac{2.303}{k}log\frac{[R_{o}]}{[R]}}\\\\\mathrm{t\: =\: \frac{2.303}{k}log\frac{[R_{o}]}{[0.1R_{o}]}}$

$\\\mathrm{t\: =\: \frac{2.303}{k}log10}$

K is given as:
$\\\mathrm{k\: =\: \frac{0.693}{t_{\frac{1}{2}}}}$

$\\\mathrm{k\: =\: \frac{0.693}{6.93}\: =\: 0.1}$

Thus, t is given as:

$\\\mathrm{t\: =\: \frac{2.303}{0.1}}$

Thus, t = 23.03 years

Q23.  Chlorine reacts with hot and concentrated NaOH and produces compounds (X) and (Y). Compound (X) gives white precipitate with silver nitrate solution. The average bond order between Cl and O atoms in (Y) is:

Solution

We have:

$\mathrm{Cl_{2}\: +\: 6NaOH(hot\: and\: conc.)\: \rightarrow \: NaClO_{3}\: +\: 5NaCl\: +\: 3H_{2}O}$
Y                    X

$\mathrm{NaCl\: +\: AgNO_{3}\: \rightarrow \: AgCl(s)\: +\: NaNO_{3}(aq)}$
white(ppt)

The structure of NaClO-3 is given below:

The average bond order is given as:

B.O = 5/3 = 1.66

Q24. The number of chiral carbons in chloramphenicol is:

Solution

The structure of chloramphenicol is given below:

Thus, the number of chiral carbons in chloramphenicol is 2.

Q25.  For the reaction:

$\mathrm{A(l)\: \rightarrow \: 2B(g)}$

ΔU = 2.1 Kcal, ΔS = 20calK-1 at 300K. Hence ΔG in Kcal is:

Solution

We know:

$\mathrm{\Delta G\: =\: \Delta H\: -\: T\Delta S}$

$\mathrm{\Delta H\: =\: \Delta U\: +\: 2RT}$

Thus, we have:

$\mathrm{\Delta G\: =\: \Delta U\: +\: 2RT\: -\: T\Delta S}$

On putting the given values we get:

$\mathrm{\Delta G\: =\: 2.1\: +\: 2\, x\, 2\, x\, 300\, x\, 10^{-3}\: -\: 300\, x\, 20\, x\, 10^{-3}}$

$\mathrm{\Delta G\: =\: 2.1\: +\:4\, x\, 300\, x\, 10^{-3}\: -\: 300\, x\, 20\, x\, 10^{-3}}$

$\mathrm{\Delta G\: =\: 1200\, x\, 10^{-3}\: -\: 6000\, x\, 10^{-3}}$

$\mathrm{\Delta G\: =\:2.1\: +\: 1.2\: -\: 6}$

$\mathrm{\Delta G\: =\:3.3\: -\: 6}$

$\mathrm{\Delta G\: =\:-2.7Kcal}$

JEE Main 2020 exam - Mathematics ( 7th January first shift)

Q1.  if f(a+b+1-x) = f(x) , for all $x$ , where a and b are fixed positive real numbers, then

$\frac{1}{a+b}\int_{a}^{b} x(f(x)+f(x+1))dx$ is equal to:

Solution

$f(a+b+1-x)=f(x)$

$\text{put x =1+x}$

$f(a+b-x)=f(x+1)$

$I=\frac{1}{a+b}\int_{a}^{b}\left [x(f(x)+f(x+1)) \right ]dx$...........1

$I=\frac{1}{a+b}\int_{a}^{b}\left [(a+b-x)(f(a+b-x)+f(a+b-x+1)) \right ]dx$

$I=\frac{1}{a+b}\int_{a}^{b}\left [(a+b-x)(f(x+1)+f(x)) \right ]dx$..........2

$I=\frac{1}{a+b}\int_{a}^{b}\left [(a+b)(f(x+1)+f(x)) \right ]dx$

$I=\int_{a}^{b}\left [(f(x+1)+f(x)) \right ]dx$

$\begin{array}{l}{2 \mathrm{I}=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}+1) \mathrm{d} \mathrm{x}+\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}} \\ 2I={\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{a}+\mathrm{b}+1-\mathrm{x}) \mathrm{d} \mathrm{x}+\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}} \\ {2 \mathrm{I}=2 \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}}\end{array}$

Q2.  Let $\alpha$ and beta be two real roots of the equation $(k+1)\tan^2(x)-\sqrt{2}\lambda \tan(x)=(1-K),$ where K $\neq$1 and $\lambda$ are real numbers. if $\tan^{-1}(\alpha +\beta )=50$. Then value of $\lambda$ is :

Solution

$\tan \alpha + \tan{\beta } = \frac{\sqrt{2} \lambda}{1+k}\\\\\tan \alpha \times \tan{\beta } = \frac{k-1}{1+k}$      ...... given

since $\tan \alpha \& \tan \beta$ are the roots

$\frac{\tan \alpha + \tan{\beta }}{1-\tan{\alpha }\tan{\beta }} =50$

$\Rightarrow \frac{\frac{\sqrt{2}\lambda }{1+k}}{1- \frac{k-1}{k+1}} =50$

$\lambda = 10$

Q3. A total number of 6 digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear is :

Solution

5X6! / 2!

Q4. if $y=mx+4$ is a tangent to both the parabola, $y^2=4x \ \text{and}\ x^2=2by$ , then b is equal to :

Solution

\begin{align*} y&= mx-am^2 \\ y&=\frac{x}{4}+4\\ y=4x , y& = mx+\frac{1}{m}\\ \Rightarrow m&=\frac{1}{4} \\ y &= \frac{x}{4}+4 \end{align*}

also ,

$x^2=2by \\ y = mx -\frac{bm^2}{2}\\ y= \frac{x}{4}+4\\ \implies \frac{bm^2}{2}=-4\\$

since , $m= \frac{1}{4}$

Therefore , $b = \textbf{-128}$

Q5. Let $\alpha$ be a root of the equation $x^2+x+1=0$ and the matrix $A =\frac{1}{\sqrt{3}}\begin{bmatrix} 1& 1 &1 \\ 1& \alpha & \alpha ^2\\ 1& \alpha ^2 & \alpha ^4 \end{bmatrix}$ , then the matrix $A^{31}$ is equal to:

Solution

Given

$\\x^2+x+1=0\\\Rightarrow \alpha\;is\;the\;root\;of\;the\;eq\\\alpha=\omega,\;\;\omega^2\\\text{A}=\frac{1}{\sqrt3}\begin{bmatrix} 1 &1 &1 \\ 1& \omega &\omega^2 \\ 1 &\omega^2 &\omega \end{bmatrix}\\\text{A}^2=\frac{1}{3}\begin{bmatrix} 3 &0 &0 \\ 0& 0 &3 \\ 0 &3 &0 \end{bmatrix}$

$\\\text{A}^4=\frac{1}{3}\begin{bmatrix} 3 &0 &0 \\ 0& 0 &3 \\ 0 &3 &0 \end{bmatrix}\times \frac{1}{3}\begin{bmatrix} 3 &0 &0 \\ 0& 0 &3 \\ 0 &3 &0 \end{bmatrix}\\\text{A}^4=I\\\therefore A^{31}=A^{28}\cdot A^3=A^3$

Q6. if $y=y(x)$ is the solution of the differential equation, $e^y\left( \frac{dy}{dx}-1 \right )-e^x$, such that $y(0)=0$ then $y(1)$ is equal to :

Solution

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{e^y}{e^x} \right )=\frac{e^y\cdot e^x\frac{\mathrm{d} y}{\mathrm{d} x}-e^x\cdot e^y}{(e^x)^2}$

$\int \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{e^y}{e^x} \right )=\int 1\cdot dx$

$\left ( \frac{e^y}{e^x} \right )=x$

$\left ( e^y \right )=x{e^x}+c$

at y(0)=0

c=1

y(1)=1+ln2

Q7. if $y(\alpha)= \sqrt{2\left(\frac{\tan \alpha + \cot \alpha }{1+\tan^2 \alpha } \right )+\frac{1}{\sin^2 \alpha}} , \alpha \in \left[\frac{3\pi}{4}, \pi \right ]$

then $\frac{dy}{dx}$ at $\alpha = \frac{5\pi}{4}$   is :

Solution

$\\y(x)=\sqrt{2\frac{\tan x+\cot x}{1+\tan^2 x}+\frac{1}{\sin^2 x}}\\\\=\sqrt{2\frac{\frac{1}{\sin x\cos x}}{1+\frac{\sin^2 x}{\cos^2 x}}+\frac{1}{\sin^2 x}}\\\\=\sqrt{2\frac{\cos x}{\sin x}+\frac{1}{\sin^2 x}}\\\\ =\sqrt{\frac{2\cos x\sin x+1}{\sin^2 x}}\\\\ =\sqrt{\frac{(\cos x+\sin x)^2}{\sin^2 x}}\\\\ =\frac{(\cos x+\sin x)}{\sin x}$

$|1+\cot x|= \frac{\mathrm{d} }{\mathrm{d} x}(-1-\cot x)=\csc^2x$

put $x=\frac{5\pi}{6}$

we get 4

Q8.  let the function $f: [ -7, \infty] \rightarrow R$ be continuous on [ -7, 0] and differentiable on (-7, 0). If f(-7)=-3 and $f'(x) \leq 2$  for all $x \in (-7,0 )$ then for all such functions f , f(-1)+ f(0) lies in the interval :

Solution

Directly use $\\\frac{f(-1)-f(-7)}{-1+7}=f'(c)\;\;\;\;\;\&\;\;\;\;\;\;\;\frac{f(0)-f(-7)}{-1+7}=f'(c)\\\\\frac{f(-1)-f(-7)}{-1+7}=f'(2)\;\;\;\;\;\&\;\;\;\;\;\;\;\frac{f(0)-f(-7)}{0+7}=f'(2)$

we get

$f(-1)\leq 9\;\;\;\;\;\;\&\;\;\;\;\;\;\;\;f(0)\leq 11$

$f(-1)+f(0)\leq 20$

Q9.  A vector $\vec{a}= \alpha \hat{i}+ 2 \hat{j}+ \beta \hat{k}, ( \alpha , \beta \in R)$ lies in the plane of the vectors $\vec{b}= \hat{i}+ \hat{j}$ and $\vec{c}= \hat{i}- \hat{j}+ 4 \hat{k}$. If $\vec{a}$ bisects the angle between $\vec{b}$ and $\vec{c}$. Then

Solution

$\text { Given: } \vec{a}=\alpha \hat{i}+2 \hat{j}+\beta \hat{k}, \vec{b}=\hat{i}+\hat{j} ,\vec{c}=\hat i-\hat{j}+4\hat{k} \text { are coplanar. }$

$\begin{array}{l}{\Rightarrow\left[\begin{array}{lll}{\vec{a}} & {\vec{b}} & {\vec{c}}\end{array}\right]=0} \\ {\Rightarrow\left|\begin{array}{lll}{\alpha} & {2} & {\beta} \\ {1} & {1} & {0} \\ {1} & {-1} & {4}\end{array}\right|=0}\end{array}$

$\\\Rightarrow 4\alpha-8-2\beta=0\\\text { Also it is given that } \vec{a} \text { bisects the angle between } \vec{b} \text { and } \vec{c}\\$

$\Rightarrow \frac{\alpha+2}{|\vec{a}||\vec{b}|}=\frac{\alpha-2+4\beta}{|\vec{a}||\vec{c}|}$

$(\alpha,\beta)=(4,4)$

Q10.  If distance between the foci of an ellipse is 6 and the distance between directrices is 12, then the length of its latus rectum is:

Solution

given

$2ae = 6 \\ \therefore ae = 3$.......(1)

Also,

$\frac{2a}{e}=126 \\ \therefore \frac{a}{e}=63$.......(2)

from (1) and (2)

$e=\frac{1}{\sqrt{2}}, a = 3\sqrt{2}$

since , $b^2= a^2(1-e^2) \\$

Substitute the values of 'e' and 'a' in above equation .

$\Rightarrow b^2= 9 \\ \therefore b= \pm 3$

length of latus rectum = $\frac{2b^2}{a} =\frac{ 2 \times 9}{3\sqrt{2}}=3\sqrt{2}$

Q11. The greatest positive integer k, for which $49^k + 1$ is factor of the sum $49^{125} + 49^{124} +......49^2+49+1$, is :

Solution

\begin{align*} &1+49+49^2+......49^{125}\\ &\Rightarrow \frac{a(r^{n+1}-1)}{r-1}\\ &\Rightarrow \frac{1(49^{126}-1)}{48}\\ &\Rightarrow \frac{(49^{63}-1)(49^{63}+1)}{48}\\ \end{align*}

Hence, K=63

Q12. Let P be the plane passing through the pts. (2,1,0), (4,1,1) and (5,0,1) and R be any point ( 2,1,6). Then the image of R in the plane P is:

Solution

Points of plane P ( 2,1,0), (4,1,1)  and ( 5,0,1)  and point R ( 2,1,6) .

Then the image of R in the plane P is:

$\begin{vmatrix} x-2 &y-1 &z \\ 2&0 &1 \\ 3& -1& 1 \end{vmatrix} = (x-2)[0+1]-(y-1)[2-3]+ z[-2] \\ \\ \Rightarrow x-2+y-1-2z=0\\ \text{plane} , x+y-2z=3 \\ \text{image} = ( 1,1,-2)$

Q13 .  If the system of linear equation

\begin{align*} 2x+2ay+az&=0\\ 2x+3by+bz&=0\\ 2x+4cy+cz&=0 \end{align*}

Where a, b , c $\in R$ are non-zero and distinct; has a non-zero solution then:

Solution

For non zero solutions D = 0

$\\D=\begin{vmatrix} 2 & 2a &a \\ 2& 3b &b \\ 2& 4c &c \end{vmatrix}=0\\C_2\rightarrow C_2-2C_3\\D=\begin{vmatrix} 1 &0 &a \\ 1& b &b \\ 1 & 2c & c \end{vmatrix}=0\\$

on solving

$\\-bc+2ac-ab=0\\\\\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$

Q14. The area of the region enclosed by the circle $x^2+y^2=2$ which is not common to the region bounded by the parabola $y^2=x$ and the straight line $y=x$, is :

Solution

$\\x^2+y^2=2\Rightarrow r=\sqrt 2\\y^2=x\\y=x\\$

Q15. The logical statement $(p\Rightarrow q) \wedge(q\Rightarrow \sim p )$ is equivalent to :

Solution

$\begin{array}{c|c|ccc}{\mathbf{p}} & {\mathbf{q}} & {((\mathbf{p} \rightarrow \mathbf{q})} & {\mathbf{\wedge}(\mathbf{q} \rightarrow \neg \mathbf{p}))} & {} \\ \hline F & {F} & {} & {\mathbf{T}} \\ \hline F & {T} & {} & {\mathbf{T}} \\ \hline T & {F} & {} & {F} \\ \hline T & {T} & {} & {F}\end{array}$

which is equal to $\sim p$

Q16.  An unbiased coin is tossed 5 times . Suppose that a variable X is assigned the value k when k consecutive heads are obtained for k = 3,4,5 .Otherwise X takes the value -1. Then the expected value of X is :

Solution

\begin{align*} E(X)&=3 \times \left( \frac{1}{2}\right )^5\times 3 + 2\times 4 \times \left( \frac{1}{2}\right )^5+ 5 \times \left( \frac{1}{2}\right )^5- 18\times \left( \frac{1}{2}\right )^5\\ &= \left( \frac{1}{2}\right )^5 [9+8+5]- \frac{18}{32}\\ &= |\frac{4}{32}|=\frac{1}{8} \end{align*}

Q17.  Let $x^k+y^k = a^k ,\ ( a, k > 0 ) \ \text{and } \ \frac{dy}{dx}+\left( \frac{y}{x}\right )^\frac{1}{3}=0$, then 'k' is :

Solution

$\\x^k+y^k=a^k\;\;\;\;\;\;\;(a,k>0)\\\frac{dy}{dx}+\left (\frac{y}{x} \right )^{1/3}=0\\\Rightarrow \;kx^{k-1}+ky^{k-1}\cdot\frac{dy}{dx}=0\\\frac{dy}{dx}+\left (\frac{x}{y} \right )^{k-1}=0\\k-1=-1/3\\k=2/3$

Q18. if $Re\left[\frac{z-1}{2z+i} \right ]=1 , \ \text{ where} \ z=x+iy,$  then the point (x,y) lies on a:

Solution

$\\Re\left ( \frac{z-1}{2z+i} \right )=1\\put\;z=x+iy\\Re\left ( \frac{z-1}{2z+i} \right )=\frac{(x-1)2x+y(2y+1)}{4x^2+4y^2+1+4y}=1\\x^2+y^2+(3/2)y+x+1/2=0\\g=-1/2,\;\;f=-3/4\\radius=\sqrt5/4$

Q19. If $g(x)=x^2+x-1\ \text{and } gof(x)=4x^2 -10x+5 , \text{then} \ f(\frac{5}{4})$ is equal to :

Solution

\begin{align*} g(x)&=x^2+x-1 \\ gof(x)&= 4x^2-10x+5\\ f(x)&= ax+b\\ gof(x)&= (ax+b)^2+ax+b-1\\ gof(x)&= a^2x^2+b^2+2axb+ax+b-1 = 4x^2-10x+5\\ \implies a^2x^2&+x(2ab+a)+b^2+b-1=4x^2-10x+5\\ \end{align*}

Therefore , $a^2=4, \ 2ab+a = -10, \ b^2+b-1=5$

Solving the above equations,

$a=2, \ b = -3,$

Therefore,

$f(x)=2x-3 \\ f\left(\frac{5}{4}\right)= 2 \times \frac{5}{4}-3$

$=\frac{-1}{2}$

Q20.  Five numbers are in A.P whose sum is 25 and product is 2520. If one of these five numbers is $\frac{-1}{2}$, then the greatest number amongst them is :

Solution

Five terms are in AP

let terms are a-2d, a-d,a,a+d,a+2d

Sum of the terms

5a=25

a=5

Product of the terms

( a-2d) (a-d)(a+d)(a+2d)a

d= 11/2 and -11/2 satisfies the condition

so a+ 2d = 16

Q21. Let A(1,0)  , B( 6,2)  and $C(\frac{3}{2}, 6)$ be the vertices of a triangle ABC. If P is a point inside the $\Delta ABC$ such that triangles $\Delta APC , \Delta APB$ and $\Delta BPC$ have equal area. Then the length of the line segment PQ, where Q is the point $( -\frac{7}{6}, -\frac{1}{3})$

Solution

A(1,0)  B(6,2)    C(3/2,6)

Point P is the centroid of triangle ABC

P(17/6,8/3)

Distance between PQ is 5

Q22. If the variance of the first 'n' natural numbers is 10 and the variance of the first 'm' even natural numbers is 16, then m+n is equal to :

Solution

$\sigma _n=\frac{n^2-1}{12}=10 \\ \\ \sigma _{m(even)} = \frac{m^2-1}{3}=16$

Therefore n = 11

m= 7

$n+m = 18$

Q23.  $\lim_{x\rightarrow 2} \frac{3^x+3^{3-x}-12}{3^{\frac{-x}{2}}-3^{1-x}}$ is equal to :

Solution

$\lim_{x\rightarrow 2} \frac{3^x+3^{3-x}-12}{3^{\frac{-x}{2}}-3^{1-x}}$

$\Rightarrow \lim_{x\rightarrow 2} \frac{3^x+\frac{3^3}{3^x}-12}{3^{\frac{1}{x/2}}-\frac{3}{3^x}}$

$\Rightarrow \lim_{x\rightarrow 2} \frac{3^{2x}+3^3-3^x\cdot 12}{3^{x/2}-3}$
$\Rightarrow \frac{2 \ln3 \cdot 9^x-4\ln(3)\cdot 3^{x+1}}{\frac{1}{2}\ln(3)e^{\frac{x}{2}\ln\left(\frac{3x}{2} \right )}}$
$\Rightarrow \frac{4\cdot 9^x-8\cdot 3^{x+1}}{3x}$

for, put x=2 in the above equation, to get

$\Rightarrow \frac{4\cdot 9^2-8\cdot 3^{2+1}}{3\times 2}= 18$

Q24. If the sum of the coefficients of all even powers of 'x' in the product $(1+x+x^2+.......x^2n)(1-x+x^2-x^3+.......x^2 n )$ is  is 61 then 'n' is equal to:

Solution

$\text { Let } \left[\left(1-x+x^{2} \ldots \ldots\right)\left(1+x+x^{2} \ldots \ldots .)\right]=a_{0}+a_{1} x+a_{2} x^{2}+\ldots \ldots\right.$$\\\text{put } x=1 \\1(2 n+1) &=a_{0}+a_{1}+a_{2}+\ldots \ldots a_{2 n} \\ \text { put } x &=-1 \\(2 n+1) \times 1=& a_{0}-a_{1}+a_{2}+\ldots \ldots . a_{2 n} \\ \text { Form }(i)+(i i) & \\ 4 n+2 &=2\left(a_{0}+a_{2}+\ldots .\right) \\ &=2 \times 61 \\ & \Rightarrow 2 n+1=61 \Rightarrow n=30$

Q25. Let S be the set of points where the function $f(x)= \left | 2-\left | x-3 \right | \right |. \ x \in R ,$ is not differentiable . Then $\sum_{}^{x \in S}$ $f(f(x))$  is equal to :

Solution

$\\f(x)=|2-|x-3||\\f(f(x))=|2-||2-|x-3||-3||\\$

$\\S=\left \{ (1,0),(3,2),(5,0) \right \}\\\sum_{x\in S}f(f(x))=\sum_{x\in S}|2-||2-|x-3||-3||\\=3$

JEE Main 2020 exam - Physics ( 7th January first shift)

Q1. Which of the following gives a reversible operation?

Solution

For reversible operation, NOT gate is used. If an input is$A$ then output=$\bar{A}$. The following circuit represents the NOT gate

Q2. A 60 HP electric motor lifts an elevator having a maximum total load capacity of 1000 kg. If the frictional force on the elevator is 4000N , then the speed of the elevator at full load is close to: (1 HP=746 W, g=10 ms-2).

Solution

The elevator moves upward hence frictional force (F) opposes their motion downward.

so, net force act on elevator =mg+F

Fnet=(1000 x 10+4000)=14000 N

using power =Fnet x speed

So speed=$v= \frac{60*746}{14000}$=3.19

Q3.  An LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having dampling constant 'b', then the correct equivalence would be:-

a) $L\leftrightarrow m, C\leftrightarrow \frac{1}{K}, R\leftrightarrow b$

b) $L\leftrightarrow \frac{1}{b}, C\leftrightarrow \frac{1}{m}, R\leftrightarrow \frac{1}{K}$

c) $L\leftrightarrow K, C\leftrightarrow b, R\leftrightarrow m$

d) $L\leftrightarrow m, C\leftrightarrow K, R\leftrightarrow b$

Solution

For the spring-mass damping system, the governing equation is given by

$m \frac{dx^2}{dt^2}+b\frac{dx}{dt}+kx=0...............(1)$

For an LCR damped oscillator, the equation is

$L \frac{dQ^2}{dt^2}+R\frac{dQ}{dt}+\frac{Q}{C}=0...............(2)$

Comparing 1 and 2

$\\L\rightarrow m\\C\rightarrow \frac{1}{k}\\R\rightarrow b$

So the answer is option 1

Q4. As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a flywheel (disc) of radius r and mass m. When released from rest, the bob starts falling vertically. When is has covered a distance h, the angular speed of the wheel will be:-

Solution

$\\mg-T=ma\\T\times R=I\alpha\\T=\frac{mR^2}{2}\times\frac{a}{R}\times\frac{1}{R}\\T=\frac{ma}{2}\\mg=\frac{3ma}{2}\\a=\frac{2g}{3}\\Also,\ v=\sqrt{2as}=\sqrt{\frac{4gh}{3}}\\also, v=\omega R\\ \omega=\frac{v}{R}$

$\Rightarrow \sqrt{\frac{4gh}{3}}\times\frac{1}{R}=\sqrt{\frac{4gh}{3R^2}}$

Q5. Three-point particles of masses 1.0 kg, 1.5 kg and 2.5 kg are placed at three corners of a right triangle of sides 4.0 cm, 3.0 cm, and 5.0 cm as shown. The centre of mass of the system is at a point:

Solution:-

let m1=1 kg, m2=1.5 kg and m3=2.5 kg

x1=0, x2=3, x3=0 and y1=0, y2=0, y3=4

$x_{com}=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}$ and $y_{com}=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}$

Let point A be origin and mass m1=1.0 kg be at origin.

So, $x_{com}=\frac{1\times0+1.5\times3+2.5\times0}{1+1.5+2.5}=\frac{4.5}{5}=0.9$

and $y_{com}=\frac{1\times0+1.5\times0+2.5\times4}{1+1.5+2.5}=\frac{10}{5}=2$

so centre of mass of the system is at (0.9,2).

Q6. A parallel plate capacitor has plates of area A separated by distance d with a dielectric which has a dielectric constant that varies as

$k(x)=k(1+\alpha x)$  where x is the distance measured from one of the plates.

If $\alpha d << 1$ then the total capacitance of the system is best given by expression by

A) $\frac{A\varepsilon _0K}{d}(1+(\frac{\alpha d}{2})^2)$

B) $\frac{A\varepsilon _0K}{d}(1+(\frac{\alpha ^2d^2}{2}))$

C) $\frac{A\varepsilon _0K}{d}(1+ \alpha d)$

D) $\frac{A\varepsilon _0K}{d}(1+(\frac{\alpha d}{2}))$

Solution

If K filled between the plates -

$\dpi{100} {C}'=K\frac{\epsilon _{0}A}{d}={C}'=Ck$

Given $K=K_{0}+K_0\alpha x$

$V= -\int_{0}^{d} Edr= V=\int_{0}^{d}\frac{\sigma }{K\epsilon _{0}} dx$

$V= \frac{\sigma}{\varepsilon _{0}} \int_{0}^{d}\frac{1}{K+K \alpha x} dx= \frac{\sigma }{K \alpha \varepsilon _{0} }[ln(K+K \alpha d)-lnK]$

$V= \frac{\sigma }{K \alpha \varepsilon _{0} }ln (1+ \alpha d)$

For

$ln (1+ \alpha d)=\alpha d-\frac{(\alpha d)^2}{2}+\frac{(\alpha d)^3}{3}.......$

$V= \frac{\sigma }{K \alpha \varepsilon _{0} }ln (1+ \alpha d)$

$C= \frac{Q}{V}=\frac{\sigma A}{V}= \frac{\sigma A}{\frac{\sigma }{K\varepsilon _{0} \alpha }ln (1+ \alpha d)}$

here, $C_0=\frac{\varepsilon _{0}A}{d}$

$C=\frac{Kd\alpha }{ln\left ( 1+ \alpha d \right )}C_{0}=\frac{KdC_0\alpha }{\alpha d-\frac{(\alpha d)^2}{2}+\frac{(\alpha d)^3}{3}.....}=\frac{KdC_0\alpha }{(\alpha d)*(1-\frac{\alpha d}{2}+\frac{(\alpha d)^2}{3}...)}$

since $\alpha d << 1$

So

$C=\frac{KdC_0\alpha }{(\alpha d)*(1-\frac{\alpha d}{2} )}= \frac{KdC_0\alpha }{(\alpha d)}*(1-\frac{\alpha d}{2} )^{-1}=\frac{KdC_0\alpha }{(\alpha d)}*(1+\frac{\alpha d}{2} )=\frac{K\varepsilon _0A}{d}*(1+\frac{\alpha d}{2} )$

Q7. The time period of revolution of electron in its ground state orbit in a hydrogen atom is $1.6\times10^{-16}s$. The frequency of the electron in its first excited state (in s-1) is:-

Solution

Velocity of the nth orbit

$V_n \ \alpha \ \frac{z}{n}$

$V_n \ \alpha \ \frac{z}{n}$$r_n \ \alpha \ \frac{n^2}{z}$

$T=\frac{2\pi r}{V}\\ \Rightarrow T \ \alpha \ \frac{n^2*n}{z*z}=\frac{n^3}{z^2}\\ \Rightarrow f \ \alpha \ \frac{z^2}{n^3}$

$f_1=\frac{1}{1.6*10^{-16}} \ s^{-1}$

and

$\frac{f_1}{f_2}=(\frac{n_2}{n_1})^{3}=8\\ \Rightarrow f_2=\frac{f_1}{8}=7.8 *10^{14} \ s^{-1}$

Q8. The current I1 (in A) flowing through $1\Omega$ resistor in the following circuit is:-

Solution

$R_{AB}=\frac{1}{2}+2=\frac{5}{2}$

$R_{CD}=2$

$R_{eq}=\frac{2.5*2}{2.5+2}=\frac{10}{9}$

$I=\frac{V}{R_{eq}}=\frac{10}{9}$

As

$\Delta V_{AB}= \Delta V_{CD}\\ \Rightarrow 2.5*I_2 =2*I_3...(1)$

and $I_2+I_3=I=\frac{10}{9 }...(2)$

From equation (1) and (2)

$I_3=\frac{50}{81} , I_2=\frac{40}{81} ,$

and $I_1= \frac{I_2}{2}=\frac{20}{81} ,$

Q9. A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches a height of R(R=radius of r=earth), it ejects a rocket of mass $\frac{m}{10}$ so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G=gravitational constant; M= mass of the earth):

a) $\frac{3m}{8}\left [ u+\sqrt\frac{5GM}{6R} \right ]$

b) $\frac{m}{20}\left [ u-\sqrt\frac{2GM}{3R} \right ]$

c) $5m\left [ u^2-\frac{119}{200}\frac{GM}{R} \right ]$

d) $\frac{m}{20}\left [ u^2+\frac{113}{200}\frac{GM}{R} \right ]$

Solution

Before the rocket rejection

Apply energy conservation

$\frac{1}{2}mu^2-\frac{GMm}{R}= \frac{1}{2}mV^2-\frac{GMm}{2R}\\\Rightarrow \frac{1}{2}mV^2=\frac{1}{2}mu^2--\frac{GMm}{2R}\\ \Rightarrow V=\sqrt{u^2-\frac{GM}{R}}$

After the rocket rejection

Apply momentum conservation

Along y-axis

$mV=\frac{m}{10}V_3\Rightarrow V_3=10V$

Along x-axis

$\frac{9m}{10}V_1=\frac{m}{10}V_2\Rightarrow V_2=9V_1$

And since $V_1=V_{orbital}=\sqrt{\frac{GM}{2R}}$

So Kinetic energy of the rocket

$K=\frac{1}{2}*\frac{m}{10}*(V_2^2+V_3^2)=\frac{1}{2}*\frac{m}{10}*(\frac{81GM}{2R}+100(u^2-\frac{GM}{R}))\\ K=5m(u^2-\frac{119}{200}*\frac{GM}{R})$

Q10. A long solenoid of radius R carries a time(t) dependent current $I(t)=I_o(1-t)$. A ring of radius 2R is placed coaxially near its middle. During the time interval $0\leq t\leq 1$, the induced current (IR) and the induced EMF (VR) in the ring changes as:-

Solution

$B=\mu _0nI_0(t)(1-t)$

using $B_0=\mu _0nI_0$

SO $B=B_0t(1-t)$

Now considering solenoid as ideal solenoid extended up to infinite and ring as its centre

$\phi =BA_s=B_0t(1-t) A_s$

$e=-\frac{d \phi}{dt} =-B_0A_s (1-2t)$

Since induced emf is changing so current will also be changing

because $i=\frac{e}{R}$

So, since direction of emf is changing so direction of current is also changing.

And it will be zero at t=0.5 sec

Q11. The radius of gyration of a uniform rod of length l, about an axis passing through a point $\frac{l}{4}$ away from the centre of the rod and perpendicular to it, is:-

Solution

Moment of inertia of rod about an axis perpendicular through COM

$I_{COM}=\frac{ML^2}{12}$

And $I_N=I_{COM}+M(\frac{L}{4})^2=\frac{ML^2}{12}+M(\frac{L^2}{16})=\frac{7ML^2}{48}$

$K=\sqrt{\frac{I_N}{M}}=\sqrt{\frac{7ML^2}{48M}}=\frac{L}{4}*\sqrt{\frac{7}{3}}$

Q12. Two moles of an ideal gas with $C_p=\frac{5}{3}$ are mixed 3 moles of another ideal gas with $\frac{C_p}{C_v}=\frac{4}{3}$. The value of $\frac{C_p}{C_v}$ for the mixture is:-

Solution

For ideal gas:- $C_p-C_v=R$

For first case:-

$\frac{C_{p1}}{C_{v1}}=\frac{5}{3} \ and \ C_{p1}-C_{v1}=R$

$C_{p1}=\frac{5}{3}{C_{v1}} \ and \ \frac{5}{3}{C_{v1}} -C_{v1}=R\Rightarrow \frac{2}{3}C_{v1}=R\Rightarrow C_{v1}=\frac{3}{2}R$

So, $C_{p1}=\frac{5}{2}R$

For second case:-

$\frac{C_{p2}}{C_{v2}}=\frac{4}{3} \ and \ C_{p2}-C_{v2}=R$

$C_{p2}=\frac{4}{3}{C_{v2}} \ and \ \frac{4}{3}{C_{v2}}-C_{v2}=R\Rightarrow C_{v2}=3R$ and $C_{p2}=4R$

Now, $Y_{mix}=\frac{n_1C_{p1}+n_2C_{p2}}{n_1C_{v1}+n_2C_{v2}}=\frac{2\times\frac{5}{2}R+3\times4R}{2\times\frac{3}{2}R+3\times3R}= 1.417$

Q13.  A liter of dry air at STP expands adiabatically to a volume of 3l. If $\gamma=1.40$, then work done by air is $(3^{1.4}=4.6555)$ [take air to be an ideal gas]:-

Solution

$W_{adiabatic} = \frac{P_1V_1 - P_2V_2}{\gamma -1}$

At STP, Pressure = P1 = 100000 Pascal

Volume = V1 = 1 liter (As given in question)

So,

$\\ P_1V^{\gamma}_1 = P_2V^{\gamma}_2 \\ \\ \Rightarrow \ 1 \times (1)^{\gamma} = P_2 \times (3)^{1.4} \\ \\ P_2 = \frac{1}{4.65}\ atm \\$

$W_{adiabatic} = (\frac{1\ \times {1} - \frac{1}{4.65} \times {3}}{0.4}) \times 101.33 \\ \\ = 90.5 \ Joule$

Q14. If the magnetic field in a plane electromagnetic wave is given by $\vec B=3\times10^{-8} sin (1.6\times10^3x + 48\times10^{10}t)T$; then what will be the expression for the electric field?

Solution

given, $\vec B=3\times10^{-8} sin (1.6\times10^3x + 48\times10^{10}t)T$

$\left | \vec E \right |=BC=3\times10^{-8} sin (1.6\times10^3x + 48\times10^{10}t)\hat{j}\times (3\times10^8)\\ =9 sin (1.6\times10^3x + 48\times10^{10}t)T$

wave is propagating in -x direction, i.e., in - i direction.

the direction of EMW wave is in the direction of $\vec E\times \vec B$.

Since B is in j direction and EMW is in -i direction. Therefore E is in (k) direction.

Q15. A polarizer-analyzer set is adjusted such that the intensity of light coming out of the analyzer is just 10% of the original intensity. Assuming that the polarizer-analyzer set does not absorb any light, the angle by which the analyzer needs to be rotated further to reduce the output intensity to be zero is:

a) 45o

b) 71.6o

c) 90o

d) 18.4o

Solution

output intensity is given by $I=I_0Cos^2(\theta )$

Initial output intensity=10% of I_0

I.e $\frac{10I_0}{100}=I_0Cos^2(\theta )\Rightarrow \theta =71.57$

Final output intensity=O

means new angle is $90^0$

the angle by which the analyser needs to be rotated further is $90^0-\theta =18.4^0$

Q16. Speed of transverse wave of a straight wire (mass=6.0 g, length=60 cm and area of cross-section=1.0 mm2) is 90 ms-1. If the young's modulus of wire is $16\times10^{11} Nm^{-2}$, the extension of wire over its natural length is:

Solution

The linear mass density

$\\\mu=\frac{m}{l}=\frac{6\times10^{-3}}{60\times10^{-2}}\\=10^{-2}Kgm^{-1}\\given\ v=90ms^{-1}\\v=\sqrt{\frac{T}{\mu}}\\T=\mu v^2=81N\\Youngs\ modulus\\Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}$

$\\\Delta l=\frac{FL}{AY}=\frac{81\times0.6}{10^{-6}\times16\times10^{11}}=0.3mm$

Q17. Two infinite planes each with uniform surface charge density +$\sigma$ are kept in each such a way that the angle between them is 300. The electric field in the region shown between them is given by:

Solution

Assuming sheet to be non conducting

$E_{net}=E((1-\frac{\sqrt{3}}{2})\hat j-\frac{1}{2}\hat i)$

So,

$E=\frac{\sigma}{2\epsilon_0}$

So,

$E_{net}=\frac{\sigma}{2\epsilon_0}((1-\frac{\sqrt{3}}{2})\hat y-\frac{1}{2}\hat x)$

Where y and x are corresponding unit vectors

Q18. If we used a magnification of 375 from a compound microscope of tube length 150mm and an objective of focal length 5mm, the focal length of the eyepiece should be close to :

a. 12mm

b. 33mm

c. 22mm

d. 2mm

Solution

The magnification is given by

$M=\frac{L}{f_0}(1+\frac{D}{f_e})\\ \Rightarrow 375=\frac{150}{5}(1+\frac{25}{f_e})\\ \Rightarrow f_e=22 \ mm$

Q19. Visible height of wavelength 6000$\times$10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction min. is at 60 from the central max. If the first minimum is produced at $\theta$, the $\theta$ is close to,

a. 25$\degree$

b. 45$\degree$

c. 20$\degree$

d. 30$\degree$

Solution

$d\ sin\theta=path\ difference=2\lambda$

So,

$\\d\ sin\60=2\lambda\\\Rightarrow \frac{\lambda}{d}=\frac{\sqrt{3}}{4}$

For first minima,

$\\d\ sin\theta_2=\lambda\\sin\theta_2=\frac{\lambda}{d}=\frac{\sqrt{3}}{4}\\\Rightarrow \theta_2=sin^{-1}(\frac{\sqrt{3}}{4})=25.64^\circ=25^\circ$

Q20. Consider a coil of wire carrying current I, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excluding the circular coil area is given by $\phi$i. The magnetic flux through the area given by $\phi$o. Which of the following is correct?

a. $\phi$i = -$\phi_o$

b. $\phi$i > $\phi_o$

c. $\phi$i < $\phi_o$

d. $\phi$i = $\phi_o$

Solution

Flux going right comes back to the left (forms closed loop)

$\int B.dA=0$

Flux inside the coil come bach through outside

$\\\phi _{coil}= -\phi _{out\ side}\\\phi_0= - \phi_1$

Q21 A particle (m=1 kg) slides down a frictionless block(AOC) starting from rest at a point A (height 2 m). After reaching C, the particle continues to move freely in air as a projectile. When it reaches its highest point P (height 1 m), the kinetic energy of the particle (in J) is (take g=10 ms-2):-

Solution

Loss of potential energy= Gain in kinetic energy

$\\1\times10(2-1)=K_f-K_i=K_f-0\\K_f=10J$

Q22. A carnot engine separates between two reservoirs of temperature 900 K and 300 K. The engine performing 1200 J of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir in a cycle is:-

Solution:-

given, W=1200 J

$\eta _{carnot}=1-\frac{T_L}{T_H}=1-\frac{300}{900}=\frac{6}{9}$

$\eta _{carnot}=\frac{W}{Q_1}$

So, $\frac{W}{Q_1}=\frac{6}{9}\Rightarrow Q_1=1800 J$

So, $W=Q_1-Q_2\Rightarrow Q_2=Q_1-W=1800-1200=600J$

Q23. A loop ABCDEFA of straight edges has six corner points A(0,0,0), B(5,0,0),C(5,5,0), D(0,5,0), E(0,5,5) and F(0,0,5). The magnetic field in this region is $\vec B=(3\hat{i}+4\hat{j})T$.The quantity of flux through the loop ABCDEFA (in Wb) is:-

Solution

$\vec B=(3\hat{i}+4\hat{j})T$

Total area vectot=ara of ABCD+area of DEFA=$5^2\hat{k}+5^2\hat{i}=25 (\hat{i}+\hat{k})$

Total Magnetic flux=$\vec{B}.\vec{A}=(3\hat{i}+4\hat{j}).(25 (\hat{i}+\hat{k}))=(75+100)wb=175 wb$

Q24. A beam of electromagnetic radiation of intensity $6.4\times10^{-5} W/cm^2$ is comprised of wavelength $\lambda=310 nm$. It falls normally on a metal (work function $\phi$=2 eV) of surface area 1 cm2. If one in 103 photons ejects an electron, total number of electrons ejected in 1s is 10x (hc=1240 eVnm, 1 eV=1.6 x 10-19J), then x is :-

Solution

Energy in 1 second = $6.4 \times 10 ^{-5} \times \1 \ \times 1 = 6.4 \times 10 ^{-5} Joule$

$Energy (eV) = \frac{6.4 \times 10 ^{-5}}{1.6 \times 10^{-19}} = 4 \times 10^{14} eV$

$Energy \ of\ one\ photon = \frac{12400}{3100} = 4 \ eV$

$Number \ of \ photon \ = \frac{4 \times 10^{14}}{4} = 10^{14}$

$Number \ of \ electrons \ = \frac{10^{14}}{10^3} = 10^{11}$

$So, \ x = 11$

Q25. A non isotropic solid metal cube has coefficients of linear expansion as $5\times 10^{-5} /^oC$ along the x-axis and $5\times 10^{-6} /^oC$ along the y-axis anf z-axis. If the coefficient of volume expansion of the solid is $C\times 10^{-6} /^oC$, then the value of C is:-

Solution

Since, it is cube, so all it's side is equal. Let the length of cube be L.

Given,$\alpha_x=5\times 10^{-5} /^oC$ and $\alpha_y=5\times 10^{-6} /^oC=\alpha_z$

Let the coefficient of volume expansion be $\gamma$

After increase in temperature:-

Increase in volume of cube=$L^3\times\gamma------------(1)$

and we can also write it as:- $L(1+\alpha_x)\timesL(1+\alpha_y)\timesL(1+\alpha_z)------------(2)$\

from equation 1 and 2 we get:-

$L^3=L^3(1+\alpha_x)(1+\alpha_y)(1+\alpha_z)$

$\Rightarrow Y=(\alpha_x+\alpha_y+\alpha_z)\ by \ ignoring \ \alpha_x\alpha_y, \ \alpha_x\alpha_z, \ \alpha_y\alpha_z, and \ \alpha_x\alpha_y\alpha_z$

$\Rightarrow Y=(5\times10^{-5}+5\times10^{-6}+5\times10^{-6})$

$\Rightarrow \gamma=6\times10^{-5}/^oC$

From question:- $\gamma=C\times 10^{-6} /^oC$

So, C=60

## Salient features of the revised pattern of JEE Main 2020 are:

• The number of questions has been reduced from 30 to 25 for each subject. However, the duration of the exam remains the same as before at 3 hours.

• 5 questions have been included with answers as numerical values. There will be no options for those questions.

• These 5 questions will not have negative marks

• Maximum marks a candidate can score has been reduced to 300. Last year it was 360 marks. The maximum marks are equally divided into 100 each in Physics, Chemistry, and Mathematics.

Stay tuned with us for further updates, and solutions would be available soon.