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.

Exam date and Shift	Article URL
7th Jan 2020 - Shift 1	Click here
7th Jan 2020 - Shift 2	Click here
8th Jan 2020 - Shift 1	Click here
8th Jan 2020 - Shift 2	Click here
9th Jan 2020 - Shift 1	Click here
9th Jan 2020 - Shift 2	Click here

Chemistry Mathematics Physics

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

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

Solution

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^-.

Solution

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?

Solution

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)

Solution
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:

Solution

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, $\Delta$ H^o_f = 2 x eq(ii) + 3eq(iii) - 1

Thus, the equaton is given as:

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

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

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

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

Q6.

Select the correct options :
(1) A = C_MPS, B = C_Average, C = C_RMS, (2) A = C_Average, B = C_MPS, C = C_RMS
(3) A = C_RMS, B = C_Average, C = C_MPS, (4) A = C_Average, B = C_MPS, C = C_RMS

Solution

C_RMS > C_Average > C_MPS.

Opiton 1 is correct.

Q7.

Solution

Option 3 is correct.

Q8. A and B are in the given reaction ?

Solution

Option 1 is correct.

Q9. Amongs the following which is a redox reaction?

Solution

Option 1 is correct.

Q10.

Solution

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

Solution

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

Solution

Concept-based.

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

Solution

Concept based.

Option 3 is correct.

Q14. Which of the following reactions are possible?

Solution

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

Solution

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:

Solution
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, $\Delta$ H^o_f = 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.

Solution

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).

Solution

(A) = CrO₂Cl₂

(B) = Na₂CrO₄

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

Solution

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

Solution

Concept-Based.

Option 1 is correct.

Q21. Stability order of following alkoxide ions is.

Solution

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

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

Solution

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

Solution

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

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?

Solution

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

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

Q3. Find the area between the curve

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

Solution

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

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

$\int_{-1}^{3}(8x+12-4x^2)dx=4x^2+12x-\frac{4x^3}{3}|_{-1}^{\;\;3}$

$\frac{128}{3}$

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.

Solution:

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

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

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?

Solution

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

The radius of the circle $R=\sqrt{16+4-16}=2$

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 $\mathrm{C} \subset \mathrm{D}$

Solution:

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

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

Solution:

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

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

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

Hence

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

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

Solution:

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

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

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

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 $\mathrm{k} \in \mathrm{I}$ ).

Solution

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

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

Solution

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

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

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

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

Solution

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

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

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 $\left(\lambda_{1} \overrightarrow{\mathrm{d}}\right)=$

Solution

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

$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0}$

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

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 :

Solution

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

Solution

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?

Solution

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

$\\P_1=\alpha+\beta=1\\P_2=\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=1-2(-1)=3\\P_3=P_2=\alpha^3+\beta^3=(\alpha+\beta)(\alpha^2+\beta^2-\alpha\beta)=(1\times(3+1))=4\\P_4=\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2\alpha^2\beta^2=(P_2)^2-2(-1)^2=7$

$x^2=x+1$

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

$\alpha^5+\beta^5=P_4+P_3=11$

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

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.

Solution

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

Using midpoint
$\\2x=2a+b\\2y=a+b$

$a=2x-2y$

$\frac{\mathrm{x}+\mathrm{y}}{3}=2(\mathrm{x}-\mathrm{y})$

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 ?

Solution

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

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

Solution

x + y + z = 6 …….. (1)

x + 2y + 3z = 10 …….. (2)

3x + 2y + $\lambda$ z = $\mu$ …….. (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 + $\lambda$ z = $\mu$ 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

Solution

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

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

Solution

Since PQ is perpendicular to L, therefore

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

Q21.

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

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

Solution

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

Q22.

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

Solution

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

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 $\lambda$

Solution

a1 + a2 = 4 $\Rightarrow$ a1 + a1r = 4 ………(i)

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

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

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

Q24.

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

Solution

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

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-

Solution

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

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?

Solution

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':-

solution

$\sigma =A+Bx^2$

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.

Solution

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

so, net force act on elevator =mg+F

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

using power =F_net 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]

Solution

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

$\\w=mg-m{\omega^2R}=mg-m(\frac{2\pi}{T})^2R\\w=196-19.6(\frac{2\pi}{24\times3600})^2\times6400\times10^3\\w=195.33N$

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

Option

1)5 2)20 3)25 4)1

Solution

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

Power =VI

So current

$I=\frac{P}{V}=\frac{4325}{220}=19.65A$

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

Solution

Equating the vertical force

$\frac{T}{\sqrt{2}}=mg=100N$

Equating the horizontal force

$\frac{T}{\sqrt{2}}=F=100N$

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.

options:

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

Solution

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

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

So $Q_L=Q'_H$

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

Solution

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

Solution

Given d=0.15 mm

D=1.5 m

$\lambda=589mm$

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.

Solution

By equation of continuity

$A_1v_1=A_2v_2$

If the area is large velocity is minimum

Therefor

$\frac{v_{max}}{v_{min}}=(\frac{4.8}{6.4})^2=\frac{9}{16}$

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

Solution

A diode is forward biased

So the circuit will be as follows

The potential difference between a and b is

$V_{ab}=\frac{30}{15\times10^{3}}\times5\times10^3V=10V$

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.

Options-

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

Solution

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

Solution

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 V_sound >> V_Source and V_sound = 350 m/s.

Solution

given f₀=1400Hz

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

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

Now, $beat frequency=f_1-f_2$

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

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.

Solution

$\lambda_{\mathrm{d}}$ 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:

Solution

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

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:-

Solution

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 $\frac{(\hat{i}+\hat{j})}{\sqrt{2}}$$

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:-

Solution

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

$\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

Solution

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

x = retentivity, y = coercivity, z = saturation magnetization

So, by matching with the diagram from the question and solution, option 3 is correct.

Q20. An inductor of inductance 10 mH and a resistance of 5 $\Omega$ is connected to a battery of 20 V at t = 0. Find the ratio of current in the circuit at t = $\infty$ to current at t = 40 sec.

(1) 1.06 (2) 1.48

(3) 1.15 (4) 0.84

Solution

As we know that the -

$i = i_o(1-e^{\frac{-t}{L/R}})$

So, from the question, we can write that -

$i = \frac{20}{5}(1-e^{\frac{-t}{0.01/5}})$

$= i = 4(1-e^{ { -500t } })$

By putting t = $\infty$

We get, $i_o = 4$

Again at t = 40 sec

$i_{40}= 4(1-\frac{1}{(e^2)^{10000}})$

$= 4(1-\frac{1}{(7.29)^{10000}})$

So,

$\frac{i_{\infty}}{i_{40}} \approx 1 \ \ \ \ But \ slightly \ greater \ than \ one$

So option 1 is much closer. So option 1 will be correct.

Q21. A solid cube of side 'a' is shown in the figure. Find the maximum value of $100\frac{b}{a}$ for which the block does not topple before sliding

solution

For no toppling

$F\left(\frac{a}{2}+b\right) \leq m g \frac{a}{2}$

Using $F=\mu mg$

$\mu \frac{a}{2}+\mu b \leq \frac{a}{2}$

using $\mu=0.4$

$\begin{array}{l}{0.2 a+0.4 b \leq 0.5 a} \\ {0.4 b \leq 0.3 a} \\ {b \leq \frac{3 a}{4}}\end{array}$

i.e ${b \leq 0.75a$ but this is not possible as b can maximum be equal to 0.5a

So $\therefore \quad\left(100 \frac{\mathrm{b}}{\mathrm{a}}\right)_{\max }=50.00$

Q22. Magnitude of resultant of two vectors $\vec{P}$ and $\vec{Q}$ is equal to magnitude of $\vec{P}$ . Find the angle between $\vec{Q}$ and resultant of $2 \vec{P}$ and $\vec{Q}$ .

Solution

$|\vec{P}+\vec{Q}|=|\vec{P}|\\\Rightarrow P^2+Q^2+2PQcos\theta =P^2\\ \Rightarrow Q+2Pcos\theta =0...(1)$

Now Vector P is doubled but vector Q is the same so the angle between $2\vec{P} \ and \ \vec{Q}$ will be also $\theta$

And If resultant of $2\vec{P} \ and \ \vec{Q}$ is $\vec{R}$

Then from the figure angle between $\vec{R} \ and \ \vec{Q}$ is $\alpha$

and $tan \alpha =\frac{2Psin\theta }{2Pcos\theta }\\ From \ equation \ (1)\\ we\ get \ tan \alpha =\infty\Rightarrow \alpha =90^0$

Q23. When m gram of steam at 100°C is mixed with 200 gm of ice at 0°C. it results in the water at 40°C. Find the value of m in gram. (given : Latent heat of fusion (L_f) = 80 cal/gm, Latent heat of vaporisation (Lv) = 540 cal/gm., specific heat of water (Cw)= 1 cal/gm/°C)

Solution

$\begin{array}{l}{\mathrm{M}_{\mathrm{ice}} \mathrm{L}_{\mathrm{f}}+\mathrm{m}_{\mathrm{ce}}(40-0) \mathrm{C}_{\mathrm{w}}=\mathrm{m}_{\mathrm{steam}} \mathrm{L}_{\mathrm{v}}+\mathrm{m}_{\mathrm{steam}}(100-40) \mathrm{C}_{\mathrm{w}}} \\ {\Rightarrow 200[80+40(1)]=\mathrm{m}[540+60(1)]} \\ {\Rightarrow 200(120)=\mathrm{m}(600)} \\ {\mathrm{m}=40 \mathrm{gm}}\end{array}$

Q24. A battery of unknown emf connected to a potentiometer has to balance length of 560 cm. If a resistor of resistance 10 ohm is connected in parallel with the cell the balancing length change by 60 cm. If the internal resistance of the cell is n/10 ohm, the value of 'n' is

$\\E\propto 560\\\frac{E\times10}{10+r}\propto500\\\frac{10+r}{10}=\frac{56}{50}\\\Rightarrow r=1.2=\frac{12}{10}\\\Rightarrow n=10$

Q25. A capacitor of 60 pF charged to 20 volt. Now battery is removed and then this capacitor is connected to another identical uncharged capacitor. Find heat loss in nJ

solution

$\\Heat \ loss =\frac{1}{2}cv^2-c\frac{v^2}{4}=c\frac{v^2}{4}\\=60\times10^{-12}\times100=6nJ$