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


7th January, Afternoon Session ( 2:30 pm to 5:30 pm)

National Test Agency( NTA) Is conducting JEE Mains 2020 Exam in two shifts, through an online mode. JEE Main 2020 is a computer-based test. In this article, we are going to share a JEE main 2020 question paper with solution for the afternoon session of 7th January 2020. These solutions are prepared by experts. We suggest you go through the article once, it will help you to understand your performance and with the help of this article, you will be able to understand that which concept is important, Type of exam questions, etc.

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Chemistry Mathematics Physics



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


Q1. The number of sp2 hybridized carbon atoms in aspartame.


The aspartame structure is given below:

The number of sp2 hybridised carbon atoms in aspartame is 9.


Q2.  Find the bond order and magnetic nature of CN-.


The bond order of CN- is 3 and all the electrons are paired in this ion, thus it is diamagnetic in nature.


Q3.  Which one of the two elements among the following pairs (F, Cl), (S, Se) and (Li, Na) will release more energy on gaining of an electron?


From the given pairs, the more energy on gaining of an electron will be released by the following elements:
Cl, S, and Li

Q4. 3 grams of acetic acid is mixed in 250 mL of 0.1 M HCl. This mixture is now diluted to 500 mL. 20 mL of this solution is now taken in another container and 1/2 mL of 5M NaOH is added to this. Find the pH of this solution. 
(log 3 = 0.4771, pKa = 4.74)

m mole of acetic acid in 20 mL = 2
m mole of HCl in 20 mL = 1
m mole of NaOH = 2.5

\mathrm{CH_{3}COOH\: +\: NaOH(remaining)\: \rightarrow \: CH_{3}COONa\: +\:water}
          2                              3/2                                      0                        0
         0.5                             0                                       3/2            

\mathrm{pH\: =\: pK_{a}\: +\: log\frac{\frac{3}{2}}{2}}

pH = 4.74 + log 3

pH = 4.74 + 0.48 = 5.22

Thus, the correct answer is 5.22


Q5. The standard enthalpy of formation(in kJ/mol) of ethane, it the enthalpies of combustion of ethane, dihydrogen, and C(graphite) are -1560kJ/mol, -286kJ/mol and -393.5kJ/mol respectively is:


The reactions we have:

\mathrm{C_{2}H_{6}\: +\: \frac{7}{2}O_{2}\: \rightarrow \: 2CO_{2}\: +\: 3H_{2}O\quad \quad \quad \quad\quad\quad[\Delta H_{c}\: =\: -1560kJ/mol]}   ...................(i)

\mathrm{2C\: +\: O_{2}\: \rightarrow \: CO_{2}\quad \quad \quad \quad\quad\quad[\Delta H_{c}\: =\: -393.5kJ/mol]}       ..........................(ii)

\mathrm{3H_{2}\: +\: \frac{1}{2}O_{2}\: \rightarrow \: H_{2}O\quad \quad \quad \quad\quad\quad[\Delta H_{c}\: =\: -286kJ/mol]}      ........................(iii)

Multiply eq(ii) by 2 and eq(iii) by 3.

Now, \DeltaHof = 2 x eq(ii) + 3eq(iii) - 1

Thus, the equaton is given as:

\mathrm{2C\: +\: 3H_{2}\: \rightarrow \: C_{2}H_{6}}

\mathrm{\Delta H^{o}_{f}\: =\: 2\, x\, (-393.5)\: +\: 3\, x\, (-286)\: -\: (-1560)}

\mathrm{\Delta H^{o}_{f}\: =\: -\: 787\: -\: 858\: +\: 1560}

\mathrm{\Delta H^{o}_{f}\: =\: -\: 1645\: +\: 1560\: =\: -85kJ/mol}



Select the correct options :
(1) A = CMPS, B = CAverage, C = CRMS,                                     (2) A = CAverage, B = CMPS, C = CRMS
(3) A = CRMS, B = CAverage, C = CMPS,    
                                (4) A = CAverage, B = CMPS, C = CRMS


CRMS > CAverage > CMPS.

Opiton 1 is correct.




Option 3 is correct.


Q8. A and B are in the given reaction ?


Option 1 is correct.

Q9. Amongs the following which is a redox reaction?


Option 1 is correct.




Option 1 is correct.


Q11.  There are two beakers  (I) having a pure volatile solvent and (II) having a volatile solvent and non-volatile solute. If both beakers are placed together in a closed container then:

(1) Volume of solvent beaker will decrease and solution beaker will increase

(2) Volume of solvent beaker will increase and solution beaker will also increase

(3) Volume of solvent beaker will decrease and solution beaker will also decrease

(4) Volume of solvent beaker will increase and solution beaker will decrease


There will be lowering in vapour pressure in the second beaker.

Option 1 is correct.


Q12.   Metal with low melting point containing impurities of high melting point can be purified by
(1) Zone refining                         (2) Vapor phase refining
(3) Distillation                              (4) Liquation 



Option 4 s correct.


Q13.  Which of the following statements are correct?
(I) On the decomposition of H2O2, O2 gas is released .
(II) 2-ethylanthraquinol is used in preparation of H2O2
(III) On heating KClO3, Pb(NO3)2, NaNO3, O2 gas is released.
(IV) In the preparation of sodium peroxoborate, H2O2 is treated with sodium metaborate.

(1) I, II, IV               (2) II, III, IV                (3) I, II, III, IV                (4) I, II, III


Concept based.

Option 3 is correct.


Q14. Which of the following reactions are possible?


1. Vinyl halides (C)  do not give Fridel Craft's reaction becuase of unstable carbocation formation.

2. Aryl halides (A) do not give Fridel Craft's reaction with Benzene because of formation of unstable phenyl carbocation.

Option 2 is correct.


Q15. The correct statement about gluconic acid is

(1) It is prepared by oxidation of glucose with HNO3

(2) It is obtained by partial oxidation of glucose

(3) It is a dicarboxylic acid

(4) It forms hemiacetal or acetal



Gluconic acid is obtained by partial oxidation of glucose by Tollen's reagent or Fehling solution or Br2,H2O. Gluconic acid can not form hemiacetal or acetal.

option 2 is correct.


Q16. The standard enthalpy of formation(in kJ/mol) of ethane, it the enthalpies of combustion of ethane, dihydrogen, and C(graphite) are -1560kJ/mol, -286kJ/mol and -393.5kJ/mol respectively is:

The reactions we have:

\mathrm{C_{2}H_{6}\: +\: \frac{7}{2}O_{2}\: \rightarrow \: 2CO_{2}\: +\: 3H_{2}O\quad \quad \quad \quad\quad\quad[\Delta H_{c}\: =\: -1560kJ/mol]}   ...................(i)

\mathrm{2C\: +\: O_{2}\: \rightarrow \: CO_{2}\quad \quad \quad \quad\quad\quad[\Delta H_{c}\: =\: -393.5kJ/mol]}       ..........................(ii)

\mathrm{3H_{2}\: +\: \frac{1}{2}O_{2}\: \rightarrow \: H_{2}O\quad \quad \quad \quad\quad\quad[\Delta H_{c}\: =\: -286kJ/mol]}      ........................(iii)

Multiply eq(ii) by 2 and eq(iii) by 3.

Now, \DeltaHof = 2 x eq(ii) + 3eq(iii) - 1

Thus, the equaton is given as:

\mathrm{2C\: +\: 3H_{2}\: \rightarrow \: C_{2}H_{6}}

\mathrm{\Delta H^{o}_{f}\: =\: 2\, x\, (-393.5)\: +\: 3\, x\, (-286)\: -\: (-1560)}

\mathrm{\Delta H^{o}_{f}\: =\: -\: 787\: -\: 858\: +\: 1560}

\mathrm{\Delta H^{o}_{f}\: =\: -\: 1645\: +\: 1560\: =\: -85kJ/mol}


Q17.  Flocculation value for As2S3 sol by HCl is 30 mmoleL–1 . Calculate mass of H2SO4 required in gram for 250 mL sol.


For 1L sol 30 mmol of HCl is required.

Thus, for 1L sol 15 mmol H2SO4 is required

For 250 mL of sol:
(15/4) x 10-3 mmol H2SO4 = 0.3675 g

Thus, the correct answer is 0.37


Q18.  \mathrm{NaCl\: \xrightarrow[Conc.\: H_{2}SO_{4}]{K_{2}Cr_{2}O_{7}}\: (A)\: \overset{NaOH}{\longrightarrow}\: (B)\: \xrightarrow[H_{2}O_{2}]{dil.\, H_{2}SO_{4}}\: (C)}

      Determine total number of atoms per unit formula of (A), (B) and (C).


(A) = CrO2Cl2

(B) = Na2CrO4

(C) = CrO5 

Thus, the correct answer is 18.


Q19. 0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution?

(1) 100 mL of 0.2 N HCl

(2) 400 mL of 0.2 N HCl

(3) 100 mL of 0.1 N HCl

(4) 200 mL of 0.2 N HCl


2 × mole of Urea = mole of NH3 ........(1)
mole of NH3 = mole of HCl ........(2)
\therefore mole of HCl = 2 × mole of Urea = 2 x (0.6/60) = 0.02 mol 

 100 x 0.2 x 1= 0.02 mol,  from 1.

So, option 1 is correct.


Q20. Which of the following statements are incorrect?

(A) Co+3 with strong field ligand forms a high magnetic moment complex.

(B) For Co+3 if pairing energy(P) > \Delta _o then the complex formed will have \mathrm{t^4_2g, e^2_g} a configuration

(C) For [Co(en)3] 3+ \lambda _{absorbed} is less than \lambda _{absorbed} for [CoF6] 3–

(D) If \Delta _o = 18000 cm–1 for Co+3 then with same ligands for it \Delta _t = 16000 cm–1

(1) A, D

(2) B, C

(3) A, B 

(4) A, B, C, D



Option 1 is correct.


Q21. Stability order of following alkoxide ions is.


When negative charge is delocalised with electron withdrawing group like (NO2) then stability increases.

(A) Negative charge is localised

(B) Negative charge is delocalised with NO2 group

(C) Negative charge is delocalised with carbon of alkene

Option 1 is correct.


Q22. For the complex [Ma2b2] if M is sp3 or dsp2 hybridised respectively then total number of optical isomers are respectively :

    (1) 1, 1                         (2) 2, 1                          (3) 0, 0                         (4) 1, 2


Both will not show optical isomerism.

Option 3 is correct.


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


Q.1  if the sum of the first 40 terms of the series 3+4+8+9+13+14+18+19+...... Is 102m then what is the value of m


\\Given \;series\;\;3+4+8+9+13+14+\ldots\\Above \;series \;can\;be\;divided\;into\;two\;series \;having \;20\;terms\;each\\first\;series\;\;\\S_1=3+8+13+\ldots\\S_1=\frac{20}{2}(2\times 3+(20-1)5)=1010\\second\;series\\S_2=4+9+14+\ldots\\S_2=\frac{20}{2}(2\times4+(20-1)5)=1030\\S=S_1+S_2=2040=102m\\m=20


Q2.  What is the coefficient of x^7 in the expansion of (1+x)^{10}+x(1+x)^9+x^2(1+x)^8+\ldots\ldots1 is?


\\(1+x)^{10}\rightarrow coeff\;of\;x^7\;\;\Rightarrow ^{10}C_7\\x(1+x)^9\rightarrow coeff\;of\;x^6\Rightarrow ^9C_6\\\ldots\\

^{10}C_7+^9C_6+^8C_5+\ldots^3C_0= 330


Q3.  Find the area between the curve

\begin{array}{l}{4 x^{2} \leq y \leq 8 x+121} \\ {\text { and } x, y \in R}\end{array}


\int (8x+12-4x^2)dx

Point of intersection (-1,4),(3,36)




Q4. Let A = [aij], B=[bij] are two 3 × 3 matrices such that \mathrm{b}_{\mathrm{ij}}=\lambda^{\mathrm{i}+\mathrm{j}-2} \mathrm{a}_{\mathrm{ij}} & |B| = 81. Find |A| \lambda if  = 3.


|B|=\left|\begin{array}{lll}{b_{11}} & {b_{12}} & {b_{13}} \\ {b_{21}} & {b_{22}} & {b_{23}} \\ {b_{31}} & {b_{32}} & {b_{33}}\end{array}\right|

|B|=\left|\begin{array}{ccc}{3^{0} a_{11}} & {3^{1} a_{12}} & {3^{2} a_{13}} \\ {3^{1} a_{21}} & {3^{2} a_{22}} & {3^{3} a_{23}} \\ {3^{2} a_{31}} & {3^{3} a_{32}} & {3^{4} a_{33}}\end{array}\right|

Taking Common 3^2 from R_3 and 3 from R_2

|B|=3^3\left|\begin{array}{ccc}{3^{0} a_{11}} & {3^{1} a_{12}} & {3^{2} a_{13}} \\ {3^{0} a_{21}} & {3^{1} a_{22}} & {3^{2} a_{23}} \\ {3^{0} a_{31}} & {3^{1} a_{32}} & {3^{2} a_{33}}\end{array}\right|

Taking Common 3^2 from C_3 and 3 from C_2

\Rightarrow 81=3^{3} \cdot 3 \cdot 3^{2}|\mathrm{A}| \Rightarrow 3^{4}=3^{6}|\mathrm{A}| \Rightarrow|\mathrm{A}|=\frac{1}{9}


Q5. Pair of tangents are drawn from origin to the circle x^{2}+y^{2}-8 x-4 y+16=0 then the square of the length of the chord of contact is?


Length of the tangent from the origin (0,0) is L=\sqrt{S_{1}}=\sqrt{16}=4

The radius of the circleR=\sqrt{16+4-16}=2

Length of Chord of contact \frac{2 L R}{\sqrt{L^{2}+R^{2}}}=\frac{2 \times 4 \times 2}{\sqrt{16+4}}=\frac{16}{\sqrt{20}}


Q6. The contrapositive of if A \subset B and B \subset C then \mathrm{C} \subset \mathrm{D}


\text { Let } P=A \subset B, Q=B \subset C, R=C \subset A

Contrapositive of (P \wedge Q) \rightarrow R \text { is } \sim R \rightarrow \sim(P \wedge Q)

\mathrm{R} \vee(\sim \mathrm{P} \vee \sim \mathrm{Q})

\mathrm{C} \subseteq \mathrm{D} \text { or } \mathrm{A} \nsubseteq \mathrm{B} \text { or } \mathrm{B} \nsubseteq \mathrm{C}

Q7.  Let y(x) is the solution of the differential equation \left(y^{2}-x\right) \frac{d y}{d x}=1 \text { and } y(0)=1,  then find the value of x where curve cuts the x-axis


\frac{d x}{d y}+x=y^{2} \\ \text { I.F. }=e^{\int 1\cdot d x}=e^{y} \\ x\cdot e^y=\int y^{2} \cdot e^{y} \cdot d y \\ x\cdot e^y=y^{2} \cdot e^{y}-\int 2 y \cdot e^{y} \cdot d y\\

\\x\cdot e^y= y^{2} e^{y}-2\left(y \cdot e^{y}-e^{y}\right)+c} \\ {x \cdot e^{y}=y^{2} e^{y}-2 y e^{y}+2 e^{y}+C} \\ {x=y^{2}-2 y+2+c \cdot e^{-y}

{x=0, \quad y=1} \\ {0=1-2+2+\frac{c}{e}}\Rightarrow c=-e


\\\text{At x-axis } y=0, \\ x=0-0+2+(-e)\left(e^{-0}\right) \\ & x=2-e

Q8.  \text { Let } \theta_{1} \text { and } \theta_{2}\left(\text { where } \theta_{1}<\theta_{2}\right) are two solutions of 2 \cot ^{2} \theta-\frac{5}{\sin \theta}+4=0, \theta \in[0,2 \pi) then \int_{\theta_{1}}^{\theta_{2}} \cos ^{2} 3 \theta d \theta equal to 


{2 \cot ^{2} \theta-\frac{5}{\sin \theta}+4=0} \\ {\frac{2 \cos ^{2} \theta}{\sin ^{2} \theta}-\frac{5}{\sin \theta}+4=0} \\ {2 \cos ^{2} \theta-5 \sin \theta+4 \sin ^{2} \theta=0, \sin \theta \neq 0} \\ {2 \sin ^{2} \theta-5 \sin \theta+2=0} \\ {(2 \sin \theta-1)(\sin \theta-2)=0}

{\sin \theta=\frac{1}{2}\text{ and }\sin \theta\neq2} \\ {\theta=\frac{\pi}{6}, \frac{5 \pi}{6}} \\ {\therefore \int \cos ^{2} 3 \theta \mathrm{d} \theta=\int_{\pi / 6}^{5 \pi / 6} \frac{1+\cos 6 \theta}{2} \mathrm{d} \theta}

=\frac{1}{2}\left[\theta+\frac{\sin 6 \theta}{6}\right]_{\pi / 6}^{5 \pi / 6}=\frac{1}{2}\left[\frac{5 \pi}{6}-\frac{\pi}{6}+\frac{1}{6}(0-0)\right]=\frac{1}{2} \cdot \frac{4 \pi}{6}=\frac{\pi}{3}


Q9.  If \left(^{36} \mathrm{C}_{\mathrm{r}+1}\right) \times\left(\mathrm{k}^{2}-3\right)= \;\frac{^{35} \mathrm{C_r}}{6} , then a number of ordered pairs (r, k) are –(where \mathrm{k} \in \mathrm{I}).


{\frac{36!}{(36-r-1)!(r+1)!} \left(k^{2}-3\right)=\frac{^{35}C_{r}}{6}} \\ {\frac{36}{r+1} \times \frac{35!}{(35-r)!r!}\left(k^{2}-3\right)=\frac{^{35}C_r}{6}} \\ {\frac{36}{r+1} \times ^{35}C_r\left(k^{2}-3\right)=\frac{^{35}C_r}{6}} \\ {k^{2}-3=\frac{r+1}{6} \Rightarrow k^{2}=3+\frac{r+1}{6}}

r should be less than or equal to 35

Hence r can be 5 and 35

For r=5 we get k=-2,-5

For r=35 we get k=-3,3

We get 4 ordered pair


Q10.  \text { Let } 4 \alpha \int_{-1}^{2} \mathrm{e}^{-\alpha|\mathrm{x}|} \mathrm{d} \mathrm{x}=5 \text { then } \alpha=


4 \alpha\left\{\int_{-1}^{0} e^{\alpha x} d x+\int_{0}^{2} e^{-\alpha x} d x\right\}=5

\Rightarrow 4 \alpha\left\{\left(\frac{e^{\omega x}}{\alpha}\right)_{-1}^{0}+\left .\frac{\mathrm{e}^{-\alpha \mathrm{x}}}{-\alpha} \right |^2_0\right \}

\\\Rightarrow 4 \alpha\left\{\left(\frac{1-e^{-\alpha}}{\alpha}\right)-\left(\frac{e^{-2 \alpha}-1}{\alpha}\right)\right\}=5 \\ \Rightarrow 4\left(2-\mathrm{e}^{-\alpha}-\mathrm{e}^{-2 \alpha}\right)=5;\quad \text { Put } \mathrm{e}^{-\alpha}=\mathrm{t}\\ \Rightarrow 4 t^{2}+4 t-3=0 \\ \Rightarrow e^{-\alpha}=\frac{1}{2}\Rightarrow \alpha=\ln2

Q11.  Let f(x) is a five-degree polynomial that has critical points x=\pm 1 and \lim _{x \rightarrow 0}\left(2+\frac{f(x)}{x^{3}}\right)=4 then which one is incorrect.

(1) f(x) has minima at x = 1 & maxima at x = –1
(2) f(1) –4f(–1) = 4
(3) f(x) is maxima at x = 1 and minima at x = –1
(4) f(x) is odd


{f(x)=a x^{5}+b x^{4}+c x^{3}} \\ {\lim _{x \rightarrow 0}\left(2+\frac{a x^{5}+b x^{4}+c x^{3}}{x^{3}}\right)=4}\\ {\Rightarrow 2+c=4 \Rightarrow c=2} \\ {f^{\prime}(x)=5 a x^{4}+4 b x^{3}+6 x^{2}} \\ {=x^{2}\left(5 a x^{2}+4 b x+6\right)}

\begin{array}{l}{f^{\prime}(1)=0 \quad \Rightarrow \quad 5 a+4 b+6=0} \\ {f^{\prime}(-1)=0 \quad \Rightarrow \quad 5 a-4 b+6=0}\end{array}

On solving b=0 \text{ and } a=-\frac{6}{5}

\begin{array}{l}{f(x)=\frac{-6}{5} x^{5}+2 x^{3}} \\ {f^{\prime}(x)=-6 x^{4}+6 x^{2}} \\ {=6 x^{2}\left(-x^{2}+1\right)} \\ {=-6 x^{2}(x+1)(x-1)}\end{array}

f(x) is maxima at x = 1 and minima at x = –1


Q12.  If \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}} are unit vectors such that \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0}and \lambda=\overrightarrow{0}and \lambda=\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}} and \overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} then \left(\lambda_{1} \overrightarrow{\mathrm{d}}\right)=


{|\vec{a}+\vec{b}+\vec{c}|^{2}=0} \\ {3+2(\vec{a} \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0} \\ {(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=\frac{-3}{2}} \\ {\Rightarrow \lambda=\frac{-3}{2}}


\\ {\vec{d}=\vec{a} \times \vec{b}+\vec{b} \times(-\vec{a}-\vec{b})+(-\vec{a}-\vec{b}) \times \vec{a}} \\ {=\vec{a} \times \vec{b}+\vec{a} \times \vec{b}+\vec{a} \times \vec{b}} \\ {\vec{d}=3(\vec{a} \times \vec{b})}


Q13. Let \text { } f(x)=x^{3}-4 x^{2}+8 x+11, \text { } if LMVT is applicable on f(x) \text { in }[0,1], the value of c is :


f(x) is a polynomial function
Therefore, it is continuous and differentiable in [0, 1]
Here f(0) = 11, f(1) = 1 – 4 + 8 + 11 = 16

{f^{\prime}(x)=3 x^{2}-8 x+8} \\ {\therefore \quad f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}=\frac{16-11}{1}=3 c^{2}-8 c+8} \\ {\Rightarrow \quad 3 c^{2}-8 c+3=0} \\ {\quad C=\frac{8 \pm 2 \sqrt{7}}{6}=\frac{4 \pm \sqrt{7}}{3}} \\ {\therefore \quad c=\frac{4-\sqrt{7}}{3} \quad \in(0,1)}

Q14. There are 5 machines. The probability of a machine being faulted is \frac{1}{4}. Probability of atmost two machines is faulted, is \left(\frac{3}{4}\right)^{3} \text k then the value of k is


Required probability = when no. the machine has fault + when only one machine has fault + when only two machines have the fault.

\\\text {P(R)}=\;^{5} \mathrm{C}_{0}\left(\frac{3}{4}\right)^{5}+^{5} \mathrm{C}_{1}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{4}+^{5} \mathrm{C}_{2}\left(\frac{1}{4}\right)^{2}\left(\frac{3}{4}\right)^{3} \\\text {P(R)}=\frac{243}{1024}+\frac{405}{1024}+\frac{270}{1024}=\frac{918}{1024}\\\text {P(R)}=\frac{459}{512}\\\text {P(R)}=\frac{27 \times 17}{64 \times 8} =\left(\frac{3}{4}\right)^{3} \times \mathrm{k}=\left(\frac{3}{4}\right)^{3} \times \frac{17}{8} \\\therefore \mathrm{k}=\frac{17}{8}


Q15.  Let \alpha and \beta are the roots of x^2 - x - 1 = 0 such that \mathrm{P}_{\mathrm{k}}=\alpha^{\mathrm{k}}+\beta^{\mathrm{k}}, \mathrm{k} \geq 1 then which one is incorrect?


\alpha+\beta=1 \text{ and }\alpha\timex\beta=-1



\\x^2=x+1\Rightarrow x^5=x^4+x^3\\ \alpha^5=\alpha^4+\alpha^3\text{ and }\beta^5=\alpha^4+\alpha^3


\mathrm{P}_{5} \neq \mathrm{P}_{2} \times \mathrm{P}_{3}


Q16. From any point P on line x = 2y perpendicular is drawn on y = x. Let foot of perpendicular is Q. Find
the locus of midpoint of PQ.


\\\text { slope of } P Q =\frac{x-a}{y-2a}=-1 \\\\ \Rightarrow x-a=-y+2a \\\\ \Rightarrow a=\frac{x+y}{3}

Using midpoint



so locus is 6 x-6 y=x+y \quad \Rightarrow \quad 5 x=7 y


Q17.  If mean and variance of 2, 3, 16, 20, 13, 7, x, y are 10 and 25 respectively then find xy ?


\\\text{Mean}=\bar x=\frac{2+ 3+ 16+ 20+ 13+ 7 +x+ y}{8}=10\\\Rightarrow x+y=19\\\text{variance}\;\;\sigma^2=\frac{\sum (x_i)^2}{8}-(\bar x)^2\\\Rightarrow \frac{4 +9 +256+ 400 +169 +49 +x^2 +y^2}{8}-100=25\\\Rightarrow x^2+y^2=113\\(x+y)^2=x^2+y^2+2xy\\xy=124


Q18. If the system of equation x + y + z = 6,  x + 2y + 3z = 10,  3x + 2y + \lambdaz = \mu has more than two solutions, then value of (\mu-\lambda^2)


x + y + z = 6 …….. (1)

x + 2y + 3z = 10 …….. (2)

3x + 2y + \lambdaz = \mu…….. (3)

from (1) and (2)

if z = 0 \Rightarrow x + y = 6 and  x + 2y = 10

\Rightarrow y = 4, x = 2

(2, 4, 0)

if y = 0 \Rightarrow x + z = 6 and  x + 3z = 10

\Rightarrow z = 2 and x = 4

(4, 0, 2)

so, 3x + 2y + \lambdaz = \mu must pass through  (2, 4, 0) and  (4, 0, 2)

\mu = 14

and 12 + 2\lambda = \mu

\lambda = 1

(\mu-\lambda^2) = 13


Q19.  If f(x) is defined in x\in\left ( -\frac{1}{3},\frac{1}{3} \right ),\;\;\;f(x)=\left\{\begin{matrix} \left ( \frac{1}{x}\log_e\left ( \frac{1+3x}{1-2x} \right ) \right ) &\;\;x\neq0 \\ k&\;\;x=0 \end{matrix}\right.

Find k such that f(x) is continuous


\\\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\left ( \frac{1}{x}\log_e\left ( \frac{1+3x}{1-2x} \right ) \right )\\\Rightarrow \lim_{x\rightarrow 0}\left ( \frac{\ln(1+3x)}{x}-\frac{\ln(1-2x)}{x} \right )\\\Rightarrow \lim_{x\rightarrow 0}\left ( \frac{3\ln(1+3x)}{3x}+\frac{2\ln(1-2x)}{-2x} \right )=3+2=5

Q20. If \text{Q}\left ( 5/3,7/3,17/3 \right ) is foot of perpendicular drawn from P(1, 0, 3) on a line L and if line L is passing through (\alpha, 7, 1), then value of \alpha is


Since PQ is perpendicular to L, therefore

\\\left ( 1-\frac{5}{3} \right )\left ( \alpha-\frac{5}{3} \right )+\left ( -\frac{7}{3} \right )\left ( 7-\frac{7}{3} \right )+\left ( 3-\frac{17}{3} \right )\left ( 1-\frac{17}{3} \right )=0\\\alpha=4



\\\text{Let X}=\left \{ x:1\leq x\leq50,\;\;x\in\mathbb N\right \}\\\text{\;\;\;\;\; A}=\left \{ x:\text{x is multiple of 2}\right \}\\\text{\;\;\;\;\; B}=\left \{ x:\text{x is multiple of 7}\right \}

Then find number of elements in the smallest subset of X which contain elements of both A and B


\\\mathrm{n(A\cup B)=n(A)+n(B)-n(A\cap B)}\\\Rightarrow 25+7-3=29



If y\sqrt{1+-x^2}=k-x\sqrt{1-y^2}\;\;and \;\;y(1/2)=-1/4\;\;then\;\;\frac{dy}{dx}\;\;at\;\;x=1/2


\\x=1/2,\;\;y=-1/4\;\;\Rightarrow xy=-1/8\\y\cdot\frac{1\cdot(-2x)}{2\sqrt{1-x^2}}+y'\sqrt{1-x^2}=-\left [1\cdot\sqrt(1-y^2)+ \frac{x\cdot(-2y)}{2\cdot\sqrt{1-y^2}}y' \right ]\\-\frac{xy}{\sqrt{1-x^2}}+y'\sqrt{1-x^2}=-\sqrt{1-y^2}+\frac{xy\cdot y'}{\sqrt{1-y^2}}\\y'\left ( \sqrt{1-x^2}-\frac{xy}{\sqrt{1-y^2}} \right )=\frac{xy}{\sqrt{1-x^2}}-\sqrt{1-y^2}\\\text{put}\;x=1/2\;\;and\;y=-1/4\\we\;get\;y'=-\frac{\sqrt5}{2}

Q23.  a1, a2, a3 …..a9 are in GP where a1 < 0, a1 + a2 = 4, a3 + a4 = 16, if \\\sum_{i=1}^{9}a_i=4\lambda

then find the \lambda


a1 + a2 = 4 \Rightarrowa1 + a1r = 4 ………(i)

a3 + a4 = 16 \Rightarrow a1r2 + a1r3 = 16 …….(ii)

\begin{array}{l}{\frac{1}{r^{2}}+\frac{1}{4} \Rightarrow r^{2}=4} \\ {r=\pm 2} \\ {r=2, a_{1}(1+2)=4 \Rightarrow a_{1}=\frac{4}{3}}\end{array}

\begin{array}{l}{r=-2, \quad a_{1}(1-2)=4 \Rightarrow a_{1}=-4} \\ {\sum_{i=1}^{a} a_{i}=\frac{a_{1}\left(r^{9}-1\right)}{r-1}=\frac{(-4)\left((-2)^{9}-1\right)}{-2-1}=\frac{4}{3}(-513)=4 \lambda} \\ {\lambda=-171}\end{array}



\text { If } z=\left(\frac{3+i \sin \theta}{4-i \cos \theta}\right) \text { is purely real and } \theta \in\left(\frac{\pi}{2}, \pi\right) \text { then } \arg (\sin \theta+i \cos \theta) \text { is }


z=\frac{(3+i \sin \theta)}{(4-i \cos \theta)} \times \frac{(4+i \cos \theta)}{(4+i \cos \theta)}

as  z is purely real \Rightarrow 3 \cos \theta+4 \sin \theta=0 \Rightarrow \tan \theta=-\frac{3}{4}

\\\arg (\sin \theta+i \cos \theta)=\pi+\tan ^{-1}\left(\frac{\cos \theta}{\sin \theta}\right)=\pi+\tan ^{-1}\left(-\frac{4}{3}\right)\\=\pi-\tan ^{-1}\left(-\frac{4}{3}\right)

Q25.   3x + 4y = 12 \sqrt 2 is the tangent to the ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{9}=1 then the distance between focii of ellipse is-


\begin{array}{l}{3 x+4 y=12 \sqrt{2}} \\ {\Rightarrow 4 y=-3 x+12 \sqrt{2}} \\ {\Rightarrow y=-\frac{3}{4} x+3 \sqrt{2}}\end{array}

condition of tangency c^{2}=a^{2} m^{2}+b^{2}

\begin{array}{l}{18=a^{2} \cdot \frac{9}{16}+9} \\ {a^{2} \cdot \frac{9}{16}=9} \\ {a^{2}=16} \\ {1 a=4} \\ {e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}}\end{array}

\begin{array}{l}{\therefore \mathrm{ae}=\frac{\sqrt{7}}{4} \cdot 4=\sqrt{7}} \\ {\therefore \text { focus are } (\pm \sqrt{7}, 0)}\end{array}

distance between foci = 2\sqrt7


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


Q1.  What is the dimensional formula of \left [ \frac{B^2}{2 \mu_0} \right ], where B is the magnetic field and \mu_0 is permeability of free space?


The energy density of Electromagnetic wave= \frac{1}{2}\frac{B^2}{\mu_o}

So, \left [ \frac{B^2}{2 \mu_0} \right ]=\left [ \frac{Energy}{Volume} \right ]=\left [ \frac{ML^2T^{-2}}{L^3} \right ]=ML^{-1}T^{-2}


Q2.  Moment of inertia of a colid disc as mass per unit area varies as A+Bx2, given the radius of the body is 'a':-


\sigma =A+Bx^2

We take an elemental ring of mass dm at a distance x from the centre.

So Area of an elemental ring= dA=2\pi xdx

As moment of inertia for an elemental ring = dI=dmx^2

Total  moment of inertia= I=\int dI=\int_{0}^{a}dmx^2=\int_{0}^{a}\sigma dAx^2=\int_{0}^{a}(A+Bx^2)x^2(2\pi xdx)\\ I=2\pi(\int_{0}^{a}Ax^3dx+\int_{0}^{a}Bx^5dx)=2\pi(A*\frac{a^4}{4}+B*\frac{a^6}{6})


Q3.  A lift can hold a maximum of 10 people, each of mass 68 kg, a mass of lift is 920 kg, friction of 6000 N is acting on it. Lift is moving with a velocity of 3 m/s. If a lift is moving upward, calculate the power supplied by the motor to the lift, it is moving with its maximum intensity.


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

so, net force act on elevator =mg+F 

Fnet=( (10*68+920) x 10+6000)=22000 N

using power =Fnet x speed=22000*3=66000 Watt

Using 1 HP=746 Watts

So power=88.47 HP


Q4.  If weight of an object at pole is 196 N then weight at equator is [g = 10 m/s2 ; radius of earth = 6400 Km]


Given weight = 196N, therefore mass = 19.6Kg

At pole centripetal acceleration is zero ,there for w=mg

Due to rotation of the earth the effective weight is



Q5. In a house 15 Bulbs of 45 W, 15 bulbs of 100 W, 15 bulbs of 10 W and Two heaters of 1 KW each is connected to 220 V Mains supply then find minimum fuse current


1)5  2)20   3)25    4)1


Total\ power=15\times45+15\times100+15\times10+2\times1000=4325W

Power =VI

So current


Minimum fuse current should be 20A


Q6.  A block of mass 10kg is suspended from a string of length 4m. When pulled by a force F along horizontal from the midpoint. The upper half of the string makes 45° with vertical. The value of F is

options- 1)100N   2)90N      3)75N      4)70N


Equating the vertical force


Equating the horizontal force



Q7. Cascaded Carnot engine is an arrangement in which the heat sink of one engine is a source for others. If the high temperature for one engine is T1, low temperature for other engines is T2 (Assume work done by both engine is same) Calculate the lower temperature of the first engine.


a) \frac{2T_1T_2}{T_1+T_2}     b) \frac{T_1+T_2}{2}     c) 0   d) \sqrt{T_1T_2}



\begin{array}{ll}{\text { Let, }}\\ {Q_{H}: \text { Heat input to 1st engine }} \\ {Q_{L}: \text { Heat rejected from 1st engine }} \\ {Q'_{H}: \text { Heat input to 2nd engine }}\\ {Q'_{L}: \text { Heat rejected from 2nd engine }}\end{array}

Since the heat sink of one engine is a source for other

So Q_L=Q'_H

And Work done by 1st engine = work done by 2nd engine

\begin{array}{l}{\mathrm{Q}_{\mathrm{H}}-\mathrm{Q}_{\mathrm{L}}=\mathrm{Q}_{\mathrm{L}}-\mathrm{Q}_{\mathrm{L}}^{'}} \\ {2 \mathrm{Q}_{\mathrm{L}}=\mathrm{Q}_{\mathrm{H}}+\mathrm{Q'}_{\mathrm{L}}}\end{array}

Let the lower temperature of the first engine=T

And in Carnot engine 

\frac{Q_H}{Q_L}=\frac{T_1}{T} , \ and \ \frac{Q'_H}{Q'_L}= \frac{Q_L}{Q'_L}=\frac{T}{T_2}


\begin{array}{l}{2 \mathrm{Q}_{\mathrm{L}}=\mathrm{Q}_{\mathrm{H}}+\mathrm{Q}_{\mathrm{L}}^{\prime}} \\ {2=\frac{\mathrm{T}_{1}}{\mathrm{T}}+\frac{\mathrm{T}_{2}}{\mathrm{T}}} \\ {\mathrm{T}=\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}}\end{array}

Q8. The activity of a substance changes from 700 s–1 to 500 s–1 in 30 minutes. Find its half-life in minutes

options-  a)66   b)50    c)62    d)56


Activity at any time is given as

A=A_0e^{-\lambda t}\\ \Rightarrow \frac{A}{A_0}=e^{-\lambda t}\\ \Rightarrow ln( \frac{A_0}{A})=\lambda t

For half-life t=t_{\frac{1}{2}}, A=\frac{A_0}{2}

so \ln 2=\lambda t_{1 / 2}

Now for t=30 \ min \ , A_0=700 \ s^{-1}, \ A=500 \ s^{-1}

\ln \left[\frac{700}{500}\right]=\lambda(30 )

from( (i) and (ii)

\frac{\ln 2}{\ln (7 / 5)}=\frac{t_{1 / 2}}{(30 )}

{t_{1 / 2}} =61.81 \ min


Q9. In YDSE, the separation between slits is 0.15 mm, the distance between slits and screen is 1.5 m and the wavelength of light is 589 nm, then the fringe width is

Options- a)5.9 mm    b)1.9 mm   c)3.9 mm    d)2.3 mm


Given d=0.15 mm

D=1.5 m


Fringe width 

The fringe width is 5.9 mm


Q10. An ideal fluid is flowing in a pipe in streamline flow. A pipe has a maximum and minimum diameter of 6.4 cm and 4.8 cm respectively. Find out the ratio of minimum to maximum velocity.


By equation of continuity


If the area is large velocity is minimum




Q11. There is an electric circuit as shown in the figure. Find the potential difference between points a and b.




A diode is forward biased

So the circuit will be as follows



The potential difference between a and b is



Q12. A particle of mass m and positive charge q is projected with a speed of v0 in the y-direction in the presence of electric and magnetic fields are in the x-direction. Find the instant of time at which the speed of particle becomes double the initial speed.


a) \frac{mv_0\sqrt{3}}{qE}    b)\frac{mv_0\sqrt{2}}{qE}     c)\frac{mv_0}{qE}    d) \frac{mv_0}{2qE}



Initially \overrightarrow{\mathrm{v}}=\mathrm{v}_{0} \hat{\mathrm{j}}

As both electric and magnetic field is in the x-direction

And initial velocity is perpendicular to both electric and magnetic fields.

So particle will move in a helical motion

and magnitude of velocity does not change in y–z plane

and Speed will be increased due to electric filed only i.e in the x-direction

and finally, the speed of the particle becomes double the initial speed.

i.e \left(2 \mathrm{v}_{0}\right)^{2}=\mathrm{v}_{0}^{2}+\mathrm{v}_{\mathrm{x}}^{2} \quad ; \quad \mathrm{v}_{\mathrm{x}}=\sqrt{3} \mathrm{v}_{0}

And in x-direction F_x=qE\Rightarrow a_x=\frac{qE}{m}

So v_x=\sqrt{3} \mathrm{v}_{0}=0+\frac{\mathrm{qE}}{\mathrm{m}} \mathrm{t} ; \quad \mathrm{t}=\frac{\mathrm{mv}_{0} \sqrt{3}}{\mathrm{qE}}


Q13.  In an adiabatic process, the volume is doubled then find the ratio of final average relaxation time and initial relaxation time


relaxation time=\tau \ \ \alpha \ \ \frac{V}{\sqrt{T}}

and using PV^{\gamma }=Const gives \mathrm{T} \propto \frac{1}{\mathrm{V}^{\gamma-1}}

\begin{array}{l}{\tau \propto V^{1+\frac{\gamma-1}{2}}} \\ {\tau \propto V^{\frac{1+\gamma}{2}}}\end{array}

\begin{array}{l}{\frac{\tau_{\mathrm{f}}}{\tau_{\mathrm{i}}}=\left(\frac{2 \mathrm{V}}{\mathrm{V}}\right)^{\frac{1+\gamma}{2}}} \\\\{\frac{\tau_{\mathrm{f}}}{\tau_{\mathrm{i}}}=(2)^{\frac{1+\gamma}{2}}}\end{array}

Q14. Two sources of sound moving with the same speed v and emitting a frequency of 1400 Hz are moving such that one source s1 is moving towards the observer and s2 is moving away from an observer. If the observer hears a beat frequency of 2 Hz. Then find the speed of the source. Given Vsound >> VSource and Vsound = 350 m/s.



given f0=1400Hz

f_1=f_0\left [ \frac{v_{sound}}{v_{sound}-v_{source}} \right ]

f_2=f_0\left [ \frac{v_{sound}}{v_{sound}+v_{source}} \right ]

Nowbeat frequency=f_1-f_2

\Rightarrow 2=f_1-f_2\Rightarrow v_{source}=\frac{1}{4}m/s

Q15.  An electron & a photon have the same energy E. Find the ratio of de Broglie wavelength of an electron to the wavelength of the photon. Given the mass of an electron is m & speed of light is c.


\lambda_{\mathrm{d}}$ for electron $=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \\ \lambda$ for photon $=\frac{\mathrm{hC}}{\mathrm{E}}$ \\ Ratio $=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}\times \frac{\mathrm{E}}{\mathrm{hC}}=\frac{1}{\mathrm{C}} \sqrt{\frac{\mathrm{E}}{2 \mathrm{m}}}


Q16.  A ring is rotated about a diametric axis in a uniform magnetic field perpendicular to the plane of the ring. If initially, the plane of the ring is perpendicular to the magnetic field. Find the instant of time at which EMF will be maximum & minimum respectively:


\begin{aligned} \therefore \quad \omega &=\frac{2 \pi}{\mathrm{T}}=\frac{\pi}{5} \end{aligned}

We know that   \phi=BACos\theta=BASin\omega t

And \varepsilon =-\frac{d\phi}{dt}=BASin\omega t

\text { When } \omega \mathrm{t}=\frac{\pi}{2}

\begin{array}{ll}{\because} & {\phi \text { will be minimum. }} \\ {\therefore \quad} & {\varepsilon \text { will be maximum }} \\ {} & {t=\frac{\frac{\pi}{2}}{\frac{\pi}{5}}=2.5 \mathrm{sec}}\end{array}

\begin{array}{ll}{\text { When } \omega t} & {=\pi} \\ {\ddots} & {\phi \text { will have maximum. }} \\ {\because} & {\varepsilon \text { will be minimum. }} \\ {} & {t=\frac{\pi}{\pi / 5}=5 \text { sec. }}\end{array}

Q17.  The electric field in space is given by \overrightarrow{\mathrm{E}}(\mathrm{t})=\mathrm{E}_{0} \frac{(\hat{\mathrm{i}}+\hat{\mathrm{j}})}{\sqrt{2}} \cos (\omega \mathrm{t}+\mathrm{Kz}) . A positively charged particle at (0,0, \pi / \mathrm{K}) is given velocity \mathrm{v}_{0} \hat{\mathrm{k}}$ at $\mathrm{t}=0 . The direction of force acting on a particle is:-


Force due to electric field is in direction -\frac{(\hat{i}+\hat{j})}{\sqrt{2}}$
because at t=0, E=-\frac{(\hat{i}+\hat{j})}{\sqrt{2}} E_{0}$
Force due to the magnetic field is in direction q(\vec{v} \times \vec{B})$ and $\vec{v} \| \hat{k}$
therefore, it is parallel to \vec{E}$
therefore the net force is antiparallel to


Q18.  The focal length of the convex lens in air is 16 \mathrm{cm}{ (\mu_{glass} }=1.5). Now the lens is submerged in a liquid of refractive index 1.42. Find the ratio of focal length in medium to focal length in the air:-


\frac{1}{\mathrm{f}_{\mathrm{a}}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right) \\ \frac{1}{\mathrm{f}_{\mathrm{m}}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{m}}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)

\Rightarrow \frac{f_{a}}{f_{m}}=\frac{\left(\frac{\mu_{g}}{\mu_{m}}-1\right)}{\left(\frac{\mu_{g}}{\mu_{a}}-1\right)}=\frac{\left(\frac{1.50}{1.42}-1\right)}{\left[\frac{1.50}{1}-1\right]}=\frac{0.08}{(1.92)(0.5)}$

\frac{\mathrm{f}_{\mathrm{m}}}{\mathrm{f}_{\mathrm{a}}}=\frac{(1.42)(0.5)}{0.08}=8.875 \approx 9$


Q19. The hysteresis curve for the material is shown in the figure. Then for the material retentivity, coercivity and saturation magnetization respectively will be -


(1) 50 A/m, 1T, 1.5 T   (3) 1 T, 50 A/m, 1.5 T

(2) 1.5 T, 50 A/m, 1T   (4) 50 A/m, 1.5 T, 1 T




So, from the figure, we can see that the-