# JEE Main 2020 Question Paper with Solution (Jan 9th Second shift)

## 9th January, Afternoon Session ( 2:30 pm to 5:30 pm)

JEE Mains 2020 January session is almost over now, the JEE main 2020 exam was conducted from January 7th to January 9th in 6 shifts. The last paper of JEE Main was conducted today January 9 in two shifts. First shift timings were 9:30 am to 12:30 pm and second shift timings were 2:30 pm to 5:30 pm. In this article, you will get JEE main 2020 question paper with solution (Jan 9th Second shift). As we all know that NTA is conducting JEE main 2020 exam, this time paper has been more balanced and providing an equal chance to all aspirants. The JEE mains 2020 Jan 9 second shift exam solutions are created by experts on the basis of memory-based JEE mains question available from aspirants. Stay tuned with us for more updates and solutions.

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JEE Main 2020 exam - Chemistry ( 9th January Second shift)

Q1) In which compound C–Cl bond length is shortest?

(1) Cl–CH=CH2

(2) Cl–CH=CH–CH3

(3) Cl–CH=CH–OCH3

(4) Cl–CH=CH–NO2

Solution

Resonance form of Cl–CH=CH–NO2 is more stable than resonance form of any other given compounds. Hence, double bond character in carbon-chlorine bond is maximum and bond length is shortest.

Option 4 is correct.

Q2) Amongst the following which has minimum conductivity.

(1) Distilled water

(2) Sea water

(3) Saline water used for intra venous injection

(4) Well-water

Solution

Distilled water is de-ionised water.

option (1) is correct.

Q3) Number of sp2 hybrid orbitals in a molecule of benzene.

1. 6

2. 12

3. 18

4. 24

Solution:  18 sp2 hybrid orbitals are present in a molecule of benzene.

Therefore,  the option number (3) is correct.

Q4) Biochemical oxygen demand (BOD) is defined as _______ in ppm of O2.
(1) Required to sustain life
(2) The amount of oxygen required by bacteria to break down the organic matter present in a certain volume of a sample of water.
(3) The amount of oxygen required by anaerobic bacteria to break down the inorganic matter present in a certain volume of a sample of water.
(4) Required photochemical reaction to degrade waste

Solution

The amount of oxygen required by bacteria to break down the organic matter present in a certain volume of a sample of water, is called Biochemical Oxygen Demand (BOD).

Option 2 is correct.

Q5) Which of the following reaction will not form racemic mixture as product?

Solution

Opton  2 is correct.

Q6)

Percentage carbon in compound A is:

Solution

Compounds is:

$\mathrm{Percentage\: carbon\: in\: compound\: A\: =\: \left ( \frac{12\, x\, 4}{48\, +\, 16\, +\, 8}\, x\, 100 \right )\: =\: 66.67}$

Q7) 0.1 ml of an ideal gas has volume 1 dm3 in a locked box with frictionless piston. The gas is in thermal equilibrium with excess of 0.5 m aqueous ethylene glycol at its freezing point. If the piston is released all of a sudden at 1 atm then determine the final volume of gas in dm3 (R = 0.08 atm L mol–1 K–1 Kf = 2.0 K molal–1).

Solution

$\mathrm{K_{f}\: =\: 2.0K}$
$\mathrm{m\: =\: 0.5m}$
$\mathrm{\Delta T_{f}\: =\: K_{f}m}$
$\mathrm{\Delta T_{f}\: =\: 0.5\: x\: 2}$
$\mathrm{T_{initial}\: =\: 272K}$
$\mathrm{n\: =\: 0.1\, mol}$
$\mathrm{V\: =\: 1\, dm^{3}}$
$\mathrm{P_{gas}\: =\: \frac{nRT}{V}\: =\: \frac{0.1\: x\: 0.08\: x\: 272}{1}}$
$\mathrm{P_{gas}\: =\: 2.176\: atm}$

After releasing piston P1V1 = P2V2
2.176 x 1 = 1 x V2
V2 = 2.176 dm3

Q8) One litre sea water (d = 1.03g/cm3) contains 10.3 mg O2 gas. Determine concentration of O2 in ppm.

Solution

$\mathrm{ppm\: =\: \frac{10.3\: x\: 10^{-3}}{1030}\: x\: 10^{6}}$

Thus, the correct answer is 10.

Q9) Lacto bacillus has generation time 60 min. at 300 K and 40 min. at 400 K. Determine activation energy in mol kJ/mol. (R = 8.3 J K–1mol–1) [ln(2/3) = -0.4].

Solution

$\mathrm{ln\frac{K_{2}}{K_{1}}\: =\: \frac{E_{a}}{R}\left [ \frac{1}{T_{1}}-\frac{1}{ T_{2}} \right ]}$

$\mathrm{ln\left ( \frac{60}{40} \:\right )\: =\: \frac{E_{a}}{8.3}\: x\: \frac{100}{400\, x\, 300}}$
ln(3/2) × 8.3 × 1200 = Ea
Ea = 0.4 × 8.3 × 1200
Ea = 3984 J/mol.
Ea = 3.984 kJ/mol

Thus, the correct answer is 3.98

Q10) Total number of Cr–O bonds in Chromate ion and dichromate ion is:

Solution

Total number of Cr and O bonds is 12.

Q11)

Solution

Q12)

Compound A will be:

Solution

Q13) The order of basic character is :

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

Solution
Basic strength depends upon availability of lone pairs. Greater the resonance of lone pairs lesser the basic strength.

Option 2 is correct.

Q14)

Solution

 Laboratory Test Molisch's test Barfoed test Biuret test Yes Lactose, Glucose, Fructose Glucose Albumin No Sucrose Alanine

Therefore, Option(1) is correct.

Q15) 1. 5 g of Zn reacts with
(I) Excess of NaOH (II) Dilute HCl, then volume ratio of H2 gas evolved in (I) and (II) is
(1) 2 : 1

(2) 1 : 2

(3) 1 : 1

(4) 3 : 1

Solution

$\\$$\mathrm{Zn}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{ZnO}_{2}+\mathrm{H}_{2}$$ \\$$\mathrm{Zn}+2 \mathrm{HCl} \longrightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}$$$

According to stoichiometry in both the reactions, equal number of moles of H2 are evolved.

Option 3 is correct.

Q16) Given Ksp for Cr(OH)3 is 6 × 10–31 then determine [OH].
(Neglect the contribution of OH– ions from H2O)

$(1)\left ( {18}\times 10^{-31} \right )^{\frac{1}{4}}M$

$(2)\left ( {18}\times 10^{-31} \right )^{\frac{1}{2}}M$

$(3)\left ( {6}\times 10^{-31} \right )^{\frac{1}{4}}M$

$(4)\left ( \frac{6}{27}\times 10^{-31} \right )^{\frac{1}{4}}M$

Solution

$\mathrm{Cr}(\mathrm{OH})_{3} \longrightarrow \mathrm{Cr}^{+3}+3 \mathrm{OH}^{-}$

s           3s

\begin{aligned} &\mathrm{K}_{\mathrm{sp}}=\mathrm{s} \cdot(3 \mathrm{s})^{3}\\ &6 \times 10^{-31}=27.5^{4}\\ &s=\left(\frac{6}{27} \times 10^{-31}\right)^{1 / 4}\\ &\left[\mathrm{OH}^{-}\right]=3 \mathrm{s}\\ &=3 \times\left(\frac{6}{27} \times 10^{-31}\right)^{1 / 4}=\left(18 \times 10^{-31}\right)^{1 / 4} \mathrm{M} \end{aligned}

Option 1 is correct.

Q17) Select the correct statements among the followings
(A) LiCl does not dissolve in pyridine
(B) Li does not react ethyne to form ethynide.
(C) Li and Mg react slowly with water.
(D) Among alkali metals Li has highest hydration tendency.

Solution

Concept Based

Option  1 is correct.

Q18) Given an element having following ionisation enthalpies IE1 = 496 kJ/mol and IE2 = 4562 kJ/mol one mole hydroxide of this element is treated separated with HCl and H2SO4 respectively. Moles of HCl and H2SO4 reacted respectively is

(1) 1, 0.5

(2) 0.5, 1

(3) 2, 0.5

(4) 0.5, 2

Solution

According to the given data of I.E, This element must belong to group 1 and thus is monovalent & form
hydroxide of the type M(OH).

$\\ $$\mathrm{MOH}+\mathrm{HCl} \longrightarrow \mathrm{MCl}+\mathrm{H}_{2} \mathrm{O}$$\\ 1 mole\ \ \ 1 mole \\ $$2 \mathrm{MOH}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{M}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}$$\\ 1 mole \ \ \ $$1 / 2$$ mole$

Q19) Reactant A represented by square is in equilibrium with product B represented by circles. Then value of
equilibrium constant is

(1) 1

(2) 2

(3) 3

(4) 4

Solution

Option 2 is correct.

Q20) Given following complexes

(I) Na4[Fe(CN)6]               (II) [Cr(H2O)6] Cl2

(III) (NEt4)2 [CoCl4]          (IV) Na3[Fe(C2O4)3] $(\Delta _o>P)$

Correct order of spin only magnetic moment for the above complexes is.

(1) (II) > (III) > (IV) > I

(2) (II) > (IV) > (III) > (I)

(3) (I) > (IV) > (III) > (II)

(4) (II) > (I) > (IV) > (III)

Solution

\begin{aligned} &\text { As, } \mu_{\mathrm{s}}=\sqrt{n(n+2)}\\ &\text { Complex }(1) \rightarrow \mathrm{Fe}^{+2} \Rightarrow \mathrm{S.F.L} \Rightarrow \mathrm{t}_{29}^{2,2,2} \mathrm{e}_{9}^{0,0} \Rightarrow \mu \mathrm{s}=0\\ &\text { Complex (II) } \rightarrow \mathrm{Cr}^{+2} \Rightarrow \mathrm{W.F.L} \Rightarrow \mathrm{t}_{29}^{1,1,1} \mathrm{e}_{9}^{1,0} \Rightarrow \mu_{\mathrm{s}}=\sqrt{24} \mathrm{B} . \mathrm{M}\\ &\text { Complex (III) } \rightarrow \mathrm{Co}^{+2} \Rightarrow \mathrm{W.F.L} \Rightarrow \mathrm{e}_{9}^{2,2} \mathrm{t}_{2 \mathrm{g}}^{1,1,1} \Rightarrow \mu \mathrm{s}=\sqrt{15} \mathrm{BM}\\ &\text { Complex }(\mathrm{IV}) \rightarrow \mathrm{Fe}^{+3} \Rightarrow \mathrm{S.F.L} \Rightarrow \mathrm{t}_{29}^{2,2,1} \mathrm{e}_{9}^{0,0} \Rightarrow \mu_{\mathrm{s}}=\sqrt{3} \mathrm{B} . \mathrm{M} \end{aligned}

Option 1 is correct.

Q21) Select the correct option :
(1) Entropy is function of temperature and also entropy change is function of temperature.
(2) Entropy is a function of temperature & entropy change is not a function of temperature.
(3) Entropy is not a function of temperature & entropy change is a function of temperature.
(4) Both entropy & entropy change are not a function of temperature.

Solution

\begin{aligned} &\Delta S=\int \frac{d q}{T}\\ &\mathrm{S}_{\mathrm{T}}=\int_{0}^{\mathrm{T}} \frac{\mathrm{ncd} \mathrm{T}}{\mathrm{T}} \end{aligned}

Option 1 is correct.

Q22) A compound (A ; B3N3H3Cl3) reacts with LiBH4 to form inorganic benzene (B). (A) reacts with (C) to form B3N3H3(CH3)3. (B) and (C) are respectively.
(1) Boron nitride, MeMgBr

(2) Boron nitride, MeBr

(3) Borazine, MeBr

(4) Borazine, MeMgBr

Solution

Option 4 is correct.

Q23) In a box a mixture containing H2, O2 and CO along with charcoal is present then variation of pressure with the time will be as follows :

Solution

Concept based

Option  3 is correct

Q24) Given complex [Co(NH3)4Cl2]. In it if Cl - Co-Cl bond angle is 90º then it is :
(1) Cis-isomer

(2) Trans- isomers

(3) Meridional and trans

(4) Cis and trans both

Solution

Option 1 is correct.

Q25) Monomer(s) of which of the given polymer is chiral?

(1) Buna-S

(2) Neoprene

(3) Nylon-6,6

(4) PHBV

Solution

In PHBV, both monomers have chiral centre.

Option 4 is correct.

JEE Main 2020 exam - Mathematics ( 9th January Second shift)

Q1. AB is focal chord at parabola $y^{2}=8 x$, where $A\left(\frac{1}{2},-2\right)$ then equation ale tangent at B is

Solution:

$y^{2}=8 x \text { then } A\left(2 t_{1}^{2}, 4 t_1\right) \\ {\text { given } A\left(\frac{1}{2},-2\right) \Rightarrow t_{1}=-1 / 2}$

$t_1\cdot t_2=-1$

$\text { then } t_{2}=2 \Rightarrow B(8,8) \\ \text { Equation of tangent } 8 y=4(x+8) \\ {2 y=x+8}$

Q2. IF $x=\sum_{n=0}^{\infty}(-1)^{n} \tan ^{2 n} \theta$ and $y=\sum_{n=0}^{\infty} \cos ^{2 n} \theta$

Solution:

$x=\sum_{n=0}^{\infty}(-1)^{n} \tan ^{2 n} \theta=1-\tan^2\theta+\tan^4\theta..........$

$y=\sum_{n=0}^{\infty} \cos ^{2 n} \theta=1+\cos^2\theta+\cos^4\theta......$

Use $\text S_{\infty}=\frac{1}{1-r}$

${x=\frac{1}{1+\tan ^{2} \alpha}=\cos ^{2} \theta} \\ {y=\frac{1}{1-\cos ^{2} \theta}=\frac{1}{\sin ^{2} \theta}}$

$\Rightarrow y(1-x)=1$

Q3. If the statement $p \rightarrow(p \wedge \sim q)$ is false then p and q respectively is ?

Solution:

$\begin{array}{c|c|c}{\mathbf{p}} & {\mathbf{q}} & {(\mathbf{p} \rightarrow(\mathbf{p} \mathbf{\Lambda}-\mathbf{q}))} \\ \hline F & {F} & {\mathbf{T}} \\ \hline F & {T} & {\mathbf{T}} \\ \hline \mathbf{T} & {F} & {\mathbf{T}} \\ \hline \mathbf{T} & {T} & {F}\end{array}$

Q4. If $\frac{dy}{dx}=\frac{xy}{x^2+y^2}$,  y(1) = 1 and y(x) = e, then x is

Solution:

$\text { put } y=v x, \quad \frac{d y}{d x}=v+x \frac{d v}{d x}$

$v+x \frac{d v}{d x}=\frac{ v}{1+v^{2}}$

$x \frac{d v}{d x}=\frac{-v^{3}}{1+v^{2}}$

$\frac{1+v^{2}}{v^{3}} d v=-\frac{d x}{x}$

$-\frac{x^{2}}{2 y^{2}}+\ln y=c \Rightarrow c=-1 / 2$

$-\frac{x^{2}}{2 y^{2}}+\ln y=c \Rightarrow c=-1 / 2$

$\text { Now } y(x)=e \quad \Rightarrow \quad 1=\frac{x^{2}}{2 e^{2}}-\frac{1}{2}$

$x=\pm \sqrt{3} e$

Q5. If the minimum value of term free from x for $\left(\frac{x}{\sin \theta}+\frac{1}{x \cos \theta}\right)^{16}$is L$\theta \in\left[\frac{\pi}{8}, \frac{\pi}{4}\right]$ and L2 in$\theta \in\left[\frac{\pi}{16}, \frac{\pi}{8}\right]$. Find $\frac{\mathrm{L}_{2}}{\mathrm{L}_{1}}$.

Solution:

$\\\mathrm{T}_{\mathrm{r}+1}=^{16} \mathrm{C}_{\mathrm{r}}\left(\frac{\mathrm{x}}{\sin \theta}\right)^{16-\mathrm{r}}\left(\frac{1}{\mathrm{x} \cos \theta}\right)^{\mathrm{r}} \\ {\text { for } \mathrm{r}=8 \text { term is free from }^{\prime} \mathrm{x}^{\prime}} \\ {\mathrm{T}_{9}=^{16} \mathrm{C}_{8} \frac{1}{\sin ^{8} \theta \cos ^{8} \theta}} \\ {\mathrm{T}_{9}=16 \mathrm{C}_{8} \frac{2^{8}}{(\sin 2 \theta)^{8}}}$

$\\\text { in } \theta \in\left[\frac{\pi}{8}, \frac{\pi}{4}\right], L_{1}=^{16} \mathrm{C}_{8} 2^{8} \quad \because\left\{\text { Min value of } L_{1} \text { at } \theta=\pi / 4\right\} \\\\ {\text { in } \theta \in\left[\frac{\pi}{16}, \frac{\pi}{8}\right], L_{2}=^{16} \mathrm{C}_{8} \frac{2^{8}}{\left(\frac{1}{\sqrt{2}}\right)^{8}}=^{16} \mathrm{C}_{8} \cdot 2^{8} \cdot 2^{4}\left\{\because \text { min value of } \mathrm{L}_{2}\; \mathrm{ at }\; \theta=\pi / 8\right]}$

$\frac{L_{2}}{L_{1}}=\frac{16 \mathrm{C}_{8} \cdot 2^{8} 2^{4}}{16 \mathrm{C}_{8} \cdot 2^{8}}=16$

Q6. Number of common terms in both sequence 3, 7, 11, ………….407 and 2, 9, 16, ……..905 is

Solution:
$\begin{array}{ll}{\text { First common term }} \quad {=23} \\ {\text { common difference }} \quad {=7 \times 4=28} \\ {\text { Last term } \leq 407} & {} \\ {\Rightarrow } \quad {23+(n-1) \times 28 \leq 407} \\ {\Rightarrow} \quad {(n-1) \times 28 \leq 384} \\ {\Rightarrow} \quad {n \leq 13.71+1} \\ {\Rightarrow} \quad {n \leq 14.71} \\ {\text { So, }} \quad {n=14}\end{array}$

Q7. Let a $|\vec{a}|=\sqrt{3},|\vec{b}|=5, \quad \vec{b} \cdot \vec{c}=10$ angle between $\overrightarrow{\mathrm{b}} \& \overrightarrow{\mathrm{c}}$ equal to $\frac{\pi}{3}$ If $\vec a$ is perpendicular to $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}$then find the value of $|\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})|$.

Solution:

$\overrightarrow{\mathrm{b} . \overrightarrow{\mathrm{c}}}=10 \Rightarrow|\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \cos \left(\frac{\pi}{3}\right)=10$

$\Rightarrow \quad 5 \cdot|\overrightarrow{\mathrm{c}}| \cdot \frac{1}{2}=10\quad \Rightarrow |\overrightarrow{\mathrm{c}}|=4$

${\vec{a}(\vec{b} \times \vec{c})=0} \\ {\vec{a} \times(\vec{b} \times \vec{c})=|\vec{a}||\vec{b} \times \vec{c}| \sin \left(\frac{\pi}{2}\right)} \\ {\sqrt{3} \times|\vec{b} \| \vec{c}| \sin \frac{\pi}{3} \times 1=\sqrt{3} \times 5 \times 4 \times \frac{\sqrt{3}}{2}=30}$

Q8. Let circles (x – 0)2 + (y – 4)2 = k and (x – 3)2 + (y – 0)2 = 12 touches each other than find the maximum value of 'k'

Solution:

Two circles touch each other if $\mathrm{C}_{1} \mathrm{C}_{2}=\left|\mathrm{r}_{1} \pm \mathrm{r}_{2}\right|$

Distance between C2(3, 0) and C1(0, 4) is either $\sqrt{k}+1 \text { or }|\sqrt{k}-1|$

Also $\mathrm{C}_{1} \mathrm{C}_{2}=\sqrt{4^2+3^2}=5$

$\Rightarrow \sqrt{k}+1=5 \text { or }|\sqrt{k}-1|=5 \quad \Rightarrow k=16 \text { or } k=36$

Maximum value of K is 36

Q9. $\text { If }^{25} \mathrm{C}_{0}+5^{25} \mathrm{C}_{1}+9^{25} \mathrm{C}_{2} \ldots \ldots \ldots .101^{25} \mathrm{C}_{25}=2^{25} \mathrm{k} \text { find } \mathrm{k}=$

Solution:

${\sum_{\mathrm{r}=0}^{25}(4 \mathrm{r}+1)^{25} \mathrm{C}_{\mathrm{r}}=4 \sum_{\mathrm{r}=0}^{25} \mathrm{r}^{25} \mathrm{C}_{\mathrm{r}}+\sum_{\mathrm{r}=0}^{25} 25 \mathrm{C}_{\mathrm{r}}} \\ {=4 \sum_{r=1}^{25} \mathrm{r} \times \frac{25}{\mathrm{r}} 24 \mathrm{C}_{\mathrm{r}-1}+2^{25}=100 \sum_{\mathrm{r}=1}^{25} 24 \mathrm{C}_{\mathrm{r}-1}+2^{25}} \\ {=100 \cdot 2^{24}+2^{25}=2^{25}(50+1)=51.2^{25}}$

Q10. Let the distance between plane passing through lines $\frac{x+1}{2}=\frac{y-3}{2}=\frac{z+1}{8}$ and $\frac{x+3}{2}=\frac{y+2}{1}=\frac{z-1}{\lambda}$ and plane 23x – 10y – 2z + 48 = 0 is $\frac{k}{\sqrt{633}}$ then k is equal to

Solution:

$\frac{x+1}{2}=\frac{y-3}{2}=\frac{z+1}{8}=a$ and $\frac{x+3}{2}=\frac{y+2}{1}=\frac{z-1}{\lambda}=b$

Lines must be intersecting

$\\ \Rightarrow(2 a-1,3 a+3,8 a-1)=(2 b-3, b-2, \lambda b+1) \\ {2 a-1=2 b-3,3 a+3=b-2,8 a-1=\lambda b+1}\\ \Rightarrow a=-2, b=-1, \lambda=18$

Hence points are (-3,-2,1)

The distance of plane contains given lines from given plane is same as distance between point (–3, –2,1) from given plane.

Required distance equal to $\frac{|-69+20-2+48|}{\sqrt{529+100+4}}=\frac{3}{\sqrt{633}}=\frac{k}{\sqrt{633}} \Rightarrow k=3$

Q11. $\text { If } f(x)=\left\{\begin{array}{ll}{x} & {0  then find the area bounded by f(x) and g(x) from $x=\frac{1}{2}\;to\;\;x=\frac{\sqrt3}{2}.$

Solution:

Required area = Area of trapezium ABCD - $\int_{1/2}^{\sqrt3/2}\left ( x-\frac{1}{2} \right )^2dx$

$\\=\frac{1}{2}\left(\frac{\sqrt{3}-1}{2}\right)\left(\frac{1}{2}+1-\frac{\sqrt{3}}{2}\right)-\frac{1}{3}\left(\left(x-\frac{1}{2}\right)^{3}\right)_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\\=\frac{\sqrt3}{4}-\frac{1}{3}$

Q12. z is a complex number such that |Re(z)| + |Im (z)| = 4 then |z| can't be

$\\(1)\;\sqrt7\\(2)\;\sqrt{10}\\(3)\;\sqrt{\frac{17}{2}}\\(4)\;\sqrt8$

Solution:

z = x + iy

|x| + |y| = 4

Minimum value of |z| = $2\sqrt2$

Maximum value of |z| = 4

$z\in[\sqrt8,\sqrt{16}]$

So |z| can't be $\sqrt7$

Q13. If f(x) = $\left|\begin{array}{lll}{x+a} & {x+2} & {x+1} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right|$ and a – 2b + c = 1, than

(1) f(50) = 1

(2) f(–50) = – 1

(3) f(50) = 501

(4) f(50) = – 501

Solution:

$\\\text { Apply } \mathrm{R}_{1}=\mathrm{R}_{1}+\mathrm{R}_{3}-2 \mathrm{R}_{2}\\\Rightarrow f(x)=\left|\begin{array}{ccc}{1} & {0} & {0} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right| \quad \Rightarrow f(x)=1 \\\quad \Rightarrow f(50)=1$

Q14. Let an is a positive term of a GP and $\sum_{n=1}^{100} a_{2 n+1}=200, \sum_{n=1}^{100} a_{2 n}=100 \text { find } \sum_{n=1}^{200} a_{r}$

Solution:

$\\\sum_{n=1}^{100} a_{2 n+1}=a_{3}+a_{5}+\ldots \ldots a_{201}=200 \Rightarrow \frac{\operatorname{ar}^{2}\left(r^{200}-1\right)}{r^{2}-1}=200\\\sum_{n=1}^{100} a_{2 n}=a_{2}+a_{4}+\ldots \ldots . a_{200}=100=\frac{\operatorname{ar}\left(r^{200}-1\right)}{r^{2}-1}=100$

From above, r = 2

$\\\Rightarrow \mathrm{a}_{2}+\mathrm{a}_{3}+\ldots \ldots \ldots \ldots \mathrm{a}_{200}+\mathrm{a}_{201}=300\\\Rightarrow r\left(a_{1}+\ldots \ldots \ldots \ldots a_{200}\right)=300\\\sum_{n=1}^{200} a_{n}=\frac{300}{r}=150$

Q15. Let probability distribution is

$\begin{array}{cccccc}{\mathrm{x}_{\mathrm{i}}}: & {1} & {2} & {3} & {4} & {5} \\ {\mathrm{P}_{\mathrm{i}}} :& {\mathrm{k}^{2}} & {2 \mathrm{k}} & {\mathrm{k}} & {2 \mathrm{k}} & {5 \mathrm{k}^{2}}\end{array}$

then value of p(x > 2) is

Solution:

$\\\sum P_i=1\Rightarrow 6k^2+5k=1\\\Rightarrow 6k^2+5k-1=0\\\Rightarrow k=\frac{1}{6},\:k=-1(invalid)\\ {P(x>2)=k+2 k+5 k^{2}} \\ {=\frac{1}{6}+\frac{2}{6}+\frac{5}{36}=\frac{6+12+5}{36}=\frac{23}{36}}$

Q16. Let $\{\text A = {x : |x| < 2}\}$ and $\{B = {x : |x - 2| \geq 3}\}$ then

$\begin{array}{l}{A=\{x: x \in(-2,2)\}} \\ {B=\{x: x \in(-\infty,-1] \cup[5, \infty)\}} \\ {A \cap B=\{x: x \in(-2,-1]\}} \\ {A \cup B=\{x: x \in(-\infty, 2) \cup[5, \infty)\}} \\ {A-B=\{x: x \in(-1,2)\}} \\ {B-A=\{x: x \in(-\infty,-2] \cup[5, \infty)\}}\end{array}$

Q17. Let both root of equation ax2 – 2bx + 5 = 0 are $\alpha$ and root of equation x2 – 2bx – 10 = 0 are $\alpha$ and $\beta$. Find the value of $\alpha^{2}+\beta^{2}$

Solution:

ax2 – 2bx + 5 = 0 having equal roots or $D=0$ and $\alpha=\frac{b}{a}$

$(2b)^2=4\times5\times a\;\;\Rightarrow \;\;b^2=5a$

Put $\alpha=\frac{b}{a}$ in the second equation

${x^{2}-2 b x-10=0} \\ {\Rightarrow b^{2}-2 a b^{2}-10 a^{2}=0}$

$\\\Rightarrow 5 a-10 a^{2}-10 a^{2}=0 \\ \Rightarrow 20 a^{2}=5 a \\ \Rightarrow a=\frac{1}{4} \text { and } \mathrm{b}^{2}=\frac{5}{4} \\ \alpha^{2}= 20 \text { and } \beta^{2}=5 \\ \alpha^{2}+\beta^{2} \\ =5+20 \\ =25$

Q18. ${f(x):[0,5] \rightarrow R, F(x)=\int_{0}^{x} x^{2} g(x), f(1)=3} g(x)=\int_{1}^{x} f(t) d t$ then correct choice is

(1) F(x) has local minimum at x = 1

(2) F(x) has local maximum at x = 1

(3) F(x) has point of inflection at x = 1

(4) F(x) has no critical point

Solution:

${\mathrm{F}^{\prime}(\mathrm{x})=\mathrm{x}^{2} \mathrm{g}(\mathrm{x})} \\ {\Rightarrow \mathrm{F}^{\prime}(1)=1.9(1)=0} & {(\ldots \ldots .(1)} \\ {\text { Now, } \mathrm{F}^{\prime \prime}(\mathrm{x})=2 \mathrm{xg}(\mathrm{x})+\mathrm{x}^{2} \mathrm{g}^{\prime}(\mathrm{x})} & {(\because \mathrm{g}(\mathrm{x})=\mathrm{f}(\mathrm{x}))} \\ {\Rightarrow \mathrm{F}^{\prime \prime}(\mathrm{x})=2 \mathrm{xg}(\mathrm{x})+\mathrm{x}^{2 \mathrm{f}(\mathrm{x})}} & {\left(\because \mathrm{g}^{\prime}(\mathrm{x})=\mathrm{f}(\mathrm{x})\right)} \\ {\Rightarrow \mathrm{F}^{\prime \prime}(1)=0+1 \times 3} \\ {\Rightarrow \mathrm{F}^{\prime \prime}(1)=3} \\ {\text { From (1) and (2) F(x) has local minimum at } \mathrm{x}=1}$

Q19. Let $x=2 \sin \theta-\sin 2 \theta$ and ${y=2 \cos \theta-\cos 2 \theta}$ find at $\frac{d^{2} y}{d x^{2}} \text { at } \theta=\pi$

Solution:

${\frac{d x}{d \theta}=2 \cos \theta-2 \cos 2 \theta} \\ {\frac{d y}{d \theta}=-2 \sin \theta+2 \sin 2 \theta} \\ {\therefore \frac{d y}{d x}=\frac{\sin 2 \theta-\sin \theta}{\cos \theta-\cos 2 \theta}} \\ {=\frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{3 \theta}{2}}{2 \sin \frac{\theta}{2} \sin \frac{3 \theta}{2}}=\cot \frac{3 \theta}{2}}$

${\frac{d^{2} y}{d x^{2}}=\frac{-3}{2} \csc ^{2} \frac{3 \theta}{2} \cdot \frac{d \theta}{d x}} \\ {\frac{d^{2} y}{d x^{2}}=\frac{-\frac{3}{2} \csc ^{2} \frac{3 \theta}{2}}{2(\cos \theta-\cos 2 \theta)}} \\ {\frac{d^{2} y}{d x^{2}}|_{\theta=\pi}=-\frac{3}{4(-1-1)}=\frac{3}{8}}$

Q20. If 7x + 6y – 2z = 0; 3x + 4y + 2z = 0 and x – 2y – 6z = 0 then which option is correct

(1) no. solution

(2) only trivial solution

(3) Infinite non trivial solution for x = 2z

(4) Infinite non trivial solution for y = 2z

Solution:

$\left|\begin{array}{ccc}{7} & {6} & {-2} \\ {3} & {4} & {2} \\ {1} & {-2} & {-6}\end{array}\right| =7(-20)-6(-20)-2(-10)=-140+120+20=0$

so infinite non-trivial solution exist

Q21. $\int \frac{\mathrm{d} \theta}{\cos ^{2} \theta(\sec 2 \theta+\tan 2 \theta)}=\lambda \tan \theta+2 \log f(x)+c$ then ordered pair $(\lambda,f(x))$  is

Solution:

$\\ \int \frac{\sec ^{2} \theta}{\frac{1+\tan ^{2} \theta}{1-\tan ^{2} \theta}+\frac{2 \tan \theta}{1-\tan ^{2} \theta}} d \theta \\ {=\int \frac{\sec ^{2} \theta\left(1-\tan ^{2} \theta\right)}{(1+\tan \theta)^{2}} d \theta}\\=\int \frac{\sec ^{2} \theta(1-\tan \theta)}{1+\tan \theta} d \theta\\\text{put }\tan\theta=t\Rightarrow \sec^2\theta d\theta=dt\\$

$\\ {=\int\left(\frac{1-t}{1+t}\right) d t=\int\left(-1+\frac{2}{1+t}\right) d t} \\ {=-t+2 \log (1+t)+C} \\ {=-\tan \theta+2 \log (1+\tan \theta)+C} \\ {\Rightarrow \quad \lambda=-1 \text { and } f(x)=1+\tan \theta}$

Q22. If $\lim_{x\rightarrow 0}x\left [ \frac{4}{x} \right ]$ = A then the value of x at which $f(x)={x^2}\sin\pi x$ is discontinuous (where [.] denotes greatest integer function)

$\\\text { (1) } \sqrt{A+1} \\{\text { (2) } \sqrt{A+21}} \\ {\text { (3) } \sqrt{A}} \\ {\text { (4) } \sqrt{A+5}$

Solution:

$\\\lim _{x \rightarrow 0} x\left(\frac{4}{x}-\left\{\frac{4}{x}\right\}\right)=A \quad \Rightarrow \lim _{x \rightarrow 0} 4-x\left\{\frac{4}{x}\right\}=A\\\Rightarrow 4-0=A$

$\\\text { (1) } x=\sqrt{A+1} \Rightarrow x=\sqrt{5} \Rightarrow \text { discontinuous }\\\text { (2) } x=\sqrt{A+21} \Rightarrow x=5 \Rightarrow \text { continuous }\\\text { (3) } x=\sqrt{A} \Rightarrow x=2 \Rightarrow \text { continuous }\\\text { (4) } x=\sqrt{A+5} \Rightarrow x=3 \Rightarrow \text { continuous }$

Q23. Let x + 6y = 8 is tangent to standard ellipse where the minor axis is $\frac{4}{\sqrt3}$, then eccentricity of an ellipse is

Solution:

$\\\begin{array}{l}{2 \mathrm{b}=\frac{4}{\sqrt{3}} \quad \Rightarrow \quad \mathrm{b}=\frac{2}{\sqrt{3}}} \\ {\text { Equation of tangent } \equiv \mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^{2} \mathrm{m}^{2}+\mathrm{b}^{2}}}\end{array}\\\text { comparing with } \equiv y=\frac{-x}{6}+\frac{4}{3}\\$

$\\ {\mathrm{m}=\frac{-1}{6} \text { and } \mathrm{a}^{2} \mathrm{m}^{2}+\mathrm{b}^{2}=\frac{16}{9}} \\ {\Rightarrow \quad \frac{\mathrm{a}^{2}}{36}+\frac{4}{3}=\frac{16}{9}} \\ {\Rightarrow \quad \frac{\mathrm{a}^{2}}{36}=\frac{16}{9}-\frac{4}{3}=\frac{4}{9}} \\ {\Rightarrow \quad a^{2}=16} \\ {\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}}\\e=\sqrt{\frac{11}{12}}$

Q24. If f(x) and g(x) are continuous functions, fog is identity function, g'(b) = 5 and g(b) = a then f'(a) is

Solution:

$\\ {f(g(x))=x} \\ {\quad \Rightarrow f^{\prime}(g(x)) \cdot g^{\prime}(x)=1} \\ {\text { Put } x=b} \\ {\Rightarrow f^{\prime}(g(b)) g^{\prime}(b)=1} \\ {\Rightarrow f^{\prime}(a) \times 5=1} \\ {\Rightarrow f^{\prime}(a)=\frac{1}{5}}$

JEE Main 2020 exam - Physics ( 9th January Second shift)

Q-1

A mass m attached to spring of natural length $l_o$ and spring constant k. One end of string is attached to centre of disc in horizontal plane which is being rotated by constant angular speed $\omega$. Find extension per unit length in spring (given $k >> m \omega ^2$ )

$\begin{array}{llll}{\text { (1) } \frac{\mathrm{m} \omega^{2}}{\mathrm{k}}} & {\text { (2) } \frac{\sqrt{2}}{3} \frac{\mathrm{m} \omega^{2}}{\mathrm{k}}} & {\text { (3) } \frac{\mathrm{m} \omega^{2}}{2 \mathrm{k}}} & {\text { (4) } \frac{3 \mathrm{m} \omega^{2}}{\mathrm{k}}}\end{array}$

Solution-

As natural lentgh=l0

Let elongation=x

Mass m is moving with angular velocity $\omega$ in a radius r

where $r=l_0+x$

Due to elongation x spring force is given by $F_s=Kx$

And $F_C=m\omega ^2r=m\omega ^2(l_0+x)$

as $F_C=F_s$

So

$Kx =m\omega ^2(l_0+x)\\ \Rightarrow x=\frac{m\omega ^2l_0}{K-m\omega ^2}$

using $k >> m \omega ^2$

So $\text { So, } \frac{x}{\ell_{0}} \text { is equal to } \frac{\mathrm{m} \omega^{2}}{\mathrm{k}}$

Q-2

A loop of radius R and mass m is placed in a uniform magnetic field B with its plane perpendicular to the field. Current I is flowing in it. Now loop is slightly rotated about its diameter and released. Find time period of oscillation

$\begin{array}{llll}{\text { (1) } 2 \sqrt{\frac{\pi \mathrm{m}}{I \mathrm{B}}}} & {\text { (2) } \sqrt{\frac{2 \pi \mathrm{m}}{\mathrm{IB}}}} & {\text { (3) } 2 \sqrt{\frac{\mathrm{m}}{\pi \mathrm{} \mathrm{IB}}}} & {\text { (4) } \sqrt{\frac{\mathrm{m}}{\pi \mathrm{IB}}}}\end{array}$

Solution-

$\begin{array}{l}{\tau=M B \sin \theta=I \alpha} \ \ (using \ M=IA)\\ \Rightarrow {\pi R^{2} I B \theta=\frac{m R^{2}}{2} \alpha} \ \ (using \ \alpha =-\omega ^2 \ \alpha ) \\ \Rightarrow {\omega=\sqrt{\frac{2 \pi I B}{m}}=\frac{2 \pi}{T}} \\ \Rightarrow {T=\sqrt{\frac{2 \pi m}{I B}}}\end{array}$

Q-3

A string of mass per unit length $\mu = 6 * 10^{-3} kg/m$ is fixed at both ends under the tension 540 N. If the string is in resonance with consecutive frequencies 420 Hz and 490 Hz. Then find the length of the string?

(1) 2.1 m (2) 1.1 m (3) 4.8 m (4) 4.2 m

Solution-

Fundamental frequency   = 490 – 420 = 70 Hz

$\begin{array}{l}{70=\frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}} \\ {\Rightarrow 70=\frac{1}{2 \ell} \sqrt{\frac{540}{6 \times 10^{-3}}}} \\ {\Rightarrow \ell=\frac{1}{2 \times 70} \sqrt{90 \times 10^{3}}=\frac{300}{140}} \\ {\Rightarrow \ell \approx 2.14 \mathrm{m}}\end{array}$

Q-4

$\begin{array}{l}{\text { Ratio of energy density of two steel rods is } 1: 4 \text { when same mass is suspended from the rods. If length }} \\ {\text { of both rods is same then ratio of diameter of rods will be. }}\end{array}$

Solution:-

$\begin{array}{l}{\frac{\mathrm{d} u}{\mathrm{dv}}=\frac{1}{2} \mathrm{stress} \times \frac{\mathrm{stress}}{\mathrm{\gamma}}} \\ {=\frac{1}{2} \frac{\mathrm{F}^{2}}{\mathrm{A}^{2}\gamma}} \\ {\frac{\mathrm{du}}{\mathrm{dv}} \propto \frac{1}{\mathrm{d^4}}} \\ {\frac{\left(\frac{\mathrm{du}}{\mathrm{dv}}\right)_{1}}{\mathrm{dv}}=\frac{\mathrm{d}_{2}^{4}}{\mathrm{d}_{1}^{4}}=\frac{1}{4}} \\ {\frac{\mathrm{d}_{1}}{\mathrm{d}_{2}}=(4)^{1 / 4}} \\ {\frac{\mathrm{d}_{1}}{\mathrm{d}_{2}}=\sqrt{2}: 1}\end{array}$

Q-5

$\begin{array}{l}{\text { A particle is projected from the ground with speed u at angle } 60^{\circ} \text { from horizontal. It collides with a }} \\ {\text { second particle of same mass moving with horizontal speed u in same direction at highest point of its }} \\ {\text { trajectory. If collision is perfectly inelastic then find horizontal distance travelled by them after collision }} \\ {\text { when they reached at ground }}\end{array}$

Solution:-

\begin{array}{l}{\mathrm{p}_{\mathrm{i}}=\mathrm{p}_{\mathrm{f}}} \\ {\mathrm{mu}+\mathrm{mucos} \theta=2 \mathrm{mv}} \\ {\begin{aligned} \Rightarrow \mathrm{v}=\frac{\mathrm{u}\left(1+\cos 60^{\circ}\right)}{2}=\frac{3}{4} \mathrm{u} \end{aligned}} \\ {\begin{aligned} \text { so horizontal range after collision=vt } &=v \sqrt{\frac{2 \mathrm{H}_{\max }}{\mathrm{g}}} \\ &=\frac{3}{4} \mathrm{u} \sqrt{\frac{2 \mathrm{u}^{2} \sin ^{2}\left(60^{\circ}\right)}{2 \mathrm{g}^{2}}} \\ &=\frac{3}{4} \mathrm{u}^{2} \frac{\sqrt{\frac{3}{4}}}{\mathrm{g}}=\frac{3 \sqrt{3} \mathrm{u}^{2}}{8 \mathrm{g}} \end{aligned}}\end{array}

Q-6

$\begin{array}{l}{\text { H-like atom with ionization energy of } 9 \mathrm{R} \text { . Find the wavelength of light emitted (in } \mathrm{nm} \text { ) when electron }} \\ {\text { jumps from second excited state to ground state. (R is Rydberg constant) }}\end{array}$

Solution:-

$\begin{array}{l}{\frac{\mathrm{hc}}{\lambda}=(13.6 \mathrm{eV}) \mathrm{z}^{2}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]} \\ {\mathrm{n}_{1}=1} \\ {\mathrm{n}_{2}=3} \\ {\frac{\mathrm{hc}}{\lambda}=(13.6 \mathrm{eV})\left(3^{2}\right)\left[\frac{1}{1^{2}}-\frac{1}{3^{2}}\right]} \\ {\Rightarrow \frac{\mathrm{hc}}{\lambda}=(13.6 \mathrm{eV})(9)\left(\frac{8}{9}\right)} \\ {\text { wavelength } =\frac{1240}{8 \times 13.6} \mathrm{nm}} \\ {\lambda=11.39 \mathrm{nm}}\end{array}$

Q-7

$\begin{array}{l}{\text { Two planets of masses } M \text { and } \frac{M}{2} \text { have radii } R \text { and } \frac{R}{2} \text { respectively. If ratio of escape velocities from }} \\ {\text { their surfaces } \frac{v_{1}}{v_{2}} \text { is } \frac{n}{4}, \text { then find } n:}\end{array}$

Solution:-

$\begin{array}{l}{\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}} \\ {\therefore \quad \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}}{\sqrt{\frac{2 \mathrm{GM} / 2}{\mathrm{R} / 2}}}=1=\frac{\mathrm{n}}{4}} \\ {\Rightarrow \quad n=4}\end{array}$

Q-8

$\begin{array}{l}{\text { Find centre of mass of given rod of linear mass density } \lambda=\left(a+b\left(\frac{x}{\ell}\right)^{2}\right), x \text { is distance from one of its }} \\ {\text { end. Length of the rod is } \ell \text { . }}\end{array}$

Solution:-

$\begin{array}{l}{x_{\mathrm{cm}}=\frac{1}{M} \int_{0}^{\ell} x \cdot d M} \\ {d M=\lambda \cdot d x=\left(a+b\left(\frac{x}{\ell}\right)^{2}\right) \cdot d x} \\ {x_{\mathrm{cm}}=\frac{\int x d M}{\int d M}=\frac{\int x \lambda d x}{\int \lambda d x}=\frac{\int_{0}^{t} x\left(a+\frac{b x^{2}}{\ell^{2}}\right) d x}{\int_{0}^{t}\left(a+\frac{b x^{2}}{\ell^{2}}\right) d x}}\end{array}$

\begin{aligned}=& \frac{a\left(\frac{x^{2}}{2}\right)^{\ell}+\frac{b}{\ell}\left(\frac{x^{4}}{4}\right)_{0}^{\ell}}{a(x)_{0}^{\ell}+\frac{b}{\ell^{2}}\left(\frac{x^{3}}{3}\right)_{0}^{\ell}} \\=& \frac{a \frac{\ell^{2}}{2}+b \frac{\ell^{2}}{4}}{a \ell+\frac{b \ell}{3}}=\frac{(2 a+b)}{(3 a+b)} \frac{\ell}{4} \times 3 \\=& \frac{3 \ell}{4}\left(\frac{2 a+b}{3 a+b}\right) \end{aligned}

Q-9

$\begin{array}{l}{\text { If a point source is placed at a depth } h \text { in a liquid of refractive index } \frac{4}{3} . \text { Find percentage of energy of }} \\ {\text { light that escapes from liquid. (assuming } 100 \% \text { transmission of emerging light) }}\end{array}$

Solution:-

$\begin{array}{l}{\sin \beta=\frac{3}{4}, \cos \beta=\frac{\sqrt{7}}{4}} \\ {\text { Solid angle } \mathrm{d} \Omega=2 \pi \mathrm{R}^{2}(1-\cos \beta)} \\ {\text { Percentage of light } =\frac{2 \pi \mathrm{R}^{2}(1-\cos \beta)}{4 \pi \mathrm{R}^{2}} \times 100}= {=\frac{1-\cos \beta}{2} \times 100=\left(\frac{4-\sqrt{7}}{8}\right) \times 100 \approx 17 \%}\end{array}$

Q-10

$\begin{array}{l}{\text { System is released from rest. Moment of inertia of pulley'I'. Find angular speed of pulley when } m_{1} \text { block }} \\ {\text { falls by "h". (Given } m_{1}>m_{2} \text { and assume no slipping between string and pulley). }}\end{array}$

$\begin{array}{ll}{\text { (1) } \frac{1}{\mathrm{R}} \sqrt{\frac{2\left(\mathrm{m}_{2}-\mathrm{m}_{1} \mathrm{gh}\right.}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}+\frac{1}{\mathrm{R}^{2}}\right)}}} & {\text { (2) } \frac{1}{\mathrm{R}} \sqrt{\frac{2\left(\mathrm{m}_{2}+\mathrm{m}_{1}\right) \mathrm{gh}}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}+\frac{1}{\mathrm{R}^{2}}\right)}}} \\ {\text { (3) } \frac{1}{\mathrm{R}} \sqrt{\frac{\left(\mathrm{m}_{2}-\mathrm{m}_{1}\right) \mathrm{gh}}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}+\frac{1}{\mathrm{R}^{2}}\right)}}} & {\text { (4) } \frac{1}{\mathrm{R}} \sqrt{\frac{\left(\mathrm{m}_{2}+\mathrm{m}_{1}\right) \mathrm{gh}}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}+\frac{1}{\mathrm{R}^{2}}\right)}}}\end{array}$

Solution:-

$\begin{array}{l}{\mathrm{k}_{1}+\mathrm{U}_{1}=\mathrm{k}_{1}+\mathrm{k}_{2}} \\ {0+0=\frac{1}{2} \mathrm{m}_{2} \mathrm{v}^{2}+\frac{1}{2} \mathrm{m}_{1} \mathrm{v}^{2}+\frac{1}{2} \mathrm{l} \omega^{2}-\mathrm{m}_{2} \mathrm{gh}+\mathrm{m}_{1} \mathrm{gh}} \\ {\left(\mathrm{m}_{2}-\mathrm{m}_{1}\right) \mathrm{gh}=\frac{1}{2} \mathrm{m}_{2}(\omega \mathrm{R})^{2}+\frac{1}{2} \mathrm{m}_{1}\left(\omega \mathrm{R}^{2}\right)+\frac{1}{2} \mathrm{I} \omega^{2}}\end{array}$

$\sqrt{\frac{2\left(\mathrm{m}_{2}-\mathrm{m}_{1}\right) \mathrm{gh}}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}+\frac{\mathrm{I}}{\mathrm{R}^{2}}\right) \mathrm{R}^{2}}}=\omega$

So

$\omega=\frac{1}{R} \sqrt{\frac{2\left(m_{2}-m_{1}\right) g h}{\left(m_{1}+m_{2}+\frac{I}{R^{2}}\right)}}$

Q-11

Find the current supplied by the battery

(1) 0.1 A (2) 0.3 A (3) 0.4 A (4) 0.5 A

Solution-

Both diodes are in reverse biased

So new circuits can be drawn as

So $\dpi{120} I=\frac{9}{30}=\frac{3}{10}=0.3 \ A$

Q-12

An AC source is connected to the LC series circuit with V = 10 sin (314t). Find the current in the circuit as a function of time? (L = 40 mH, C = 100 $\dpi{120} \mu$F)

(1) 10 sin (314t) (2) 5.2 sin (314t) (3) 0.52 sin (314t) (4) 0.52 cos (314t)

Solution-

As $\dpi{120} Z=\sqrt{R^2+(X_c-X_L)^2}$

But R=0

$\dpi{120} \\ Z= (X_c-X_L)=\frac{1}{C\omega }-L\omega\\ \Rightarrow Z=\frac{1}{100*10^{-4}*314}-(314*40*10^{-3})= 31.84 -12.56=19.28 \Omega$

$\dpi{120} i=\frac{V_0}{Z}Sin(314t+\frac{\pi }{2})=\frac{10}{19.28}Cos(314t)=0.52 Cos (314t)$

Q-13

$\begin{array}{l}{\text { There is a long solenoid of radius 'R' having 'n'turns per unit length with current i flowing in it. A particle }} \\ {\text { having charge 'q' and mass 'm's projected with speed v in the perpendicular direction of axis from a }} \\ {\text { Point on its axis Find maximum value of "v' so that it will not collide with the solenoid. }}\end{array}$

$\begin{array}{llll}{\text { (1) } \frac{\mathrm{Rqu}_{0} \mathrm{in}}{2 \mathrm{m}}} & {\text { (2) } \frac{2 \mathrm{Rq} \mu_{0} \mathrm{in}}{\mathrm{m}}} & {\text { (3) } \frac{\mathrm{Rqu}_{0} \mathrm{in}}{3 \mathrm{m}}} & {\text { (4) } \frac{\mathrm{Rq} \mu_{0} \mathrm{in}}{4 \mathrm{m}}}\end{array}$

Solution:-

$\begin{array}{l}{\mathrm{R}_{\max }=\frac{\mathrm{R}}{2}=\frac{\mathrm{mv}_{\max }}{\mathrm{q} \mu_{0} \mathrm{in}}} \\ {\mathrm{V}_{\max }=\frac{\mathrm{Rq} \mu_{0} \mathrm{in}}{2 \mathrm{m}}}\end{array}$

Q-14

$\begin{array}{l}{\text { A Capacitor } C \text { and resister } R \text { are connected to a battery of } 5 V \text { in series. Now battery is disconnected }} \\ {\text { and a diode is connected as shown in figure (a) and (b) respectively. Then charge on the capacitor after }} \\ {\text { time } \mathrm{RC} \text { in (a) and (b) respectively is } Q_{A} \text { and } Q_{B} \text { . Their value are }}\end{array}$

$\begin{array}{llll}{\text { (1) } 5 \mathrm{cv}, 5 \mathrm{cv}} & {\text { (2) } \frac{5 \mathrm{cv}}{\mathrm{e}}, \frac{5 \mathrm{cv}}{\mathrm{e}}} & {\text { (3) } \frac{5 \mathrm{CV}}{\mathrm{e}}, \text { scv }} & {\text { (4) } 5 \mathrm{cv}, \frac{5 \mathrm{cv}}{\mathrm{e}}}\end{array}$

Solution:-

$\begin{array}{l}{\text { Maximum charge on capacitor} =5 \mathrm{CV}} \\ {\text { (a) is forward biased and (b) is reverse biased }} \end{array}$

Q-15

$\begin{array}{l}{\text { A sphere of density } \rho \text { is half submerged in a liquid of density } \sigma \text { and surface tension } T \text { . The sphere }} \\ {\text { remains in equilibrium. Find radius of the sphere (assume the force due to surface tension acts }} \\ {\text { tangentially to surface of sphere) }}\end{array}$

$\begin{array}{llll}{\text { (1) } \sqrt{\frac{\mathrm{T}}{(\rho-\sigma) g}}} & {\text { (2) } \sqrt{\frac{\mathrm{T}}{(\rho-2 \sigma) \mathrm{g}}}} & {\text { (3) } \sqrt{\frac{2 \mathrm{T}}{(2 p-3 \sigma) 9}}} & {\text { (4) } \sqrt{\frac{3 \mathrm{T}}{(2 p-\sigma) g}}}\end{array}$

Solution:-

\begin{aligned} & \rho \mathrm{Vg}=\sigma\left(\frac{\mathrm{V}}{2}\right) \mathrm{g}+\mathrm{T}(2 \pi \mathrm{R}) \\ \Rightarrow & \rho \cdot \frac{4}{3} \pi \mathrm{R}^{3} \mathrm{g}=\sigma \cdot \frac{2}{3} \pi \mathrm{R}^{3} \mathrm{g}+2 \pi \mathrm{RT} \\ \Rightarrow & \frac{2}{3} \mathrm{R}^{2} \mathrm{g}(2 \rho-\sigma)=2 \mathrm{T} \\ \Rightarrow & \mathrm{R}=\sqrt{\frac{3 \mathrm{T}}{(2 \rho-\sigma) g}} \end{aligned}

Q-16

An EM wave is travelling in $\frac{\hat{i}+\hat{j}}{2}$ direction. Axis of polarization of EM wave is found to be  $\hat{k}$ . Then equation of magnetic field will be -

$\begin{array}{ll}{\text { (1) } \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}} \cos \left[\omega \mathrm{t}+\mathrm{k}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)\right]} & \ \ \ \ \ \ \ \ \ \ \ \ {\text { (2) } \frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}}{\sqrt{2}} \cos \left[\omega \mathrm{t}-\mathrm{k}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)\right]} \\ {\text { (3) } \frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}}{\sqrt{2}} \cos \left[\omega \mathrm{t}+\mathrm{k}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)\right]} & \ \ \ \ \ \ \ \ \ \ \ \ \ {(4) \hat{\mathrm{k}} \cos \left[\mathrm{ot}-\mathrm{k}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)\right]}\end{array}$

Solution -

EM wave is in direction - $\frac{\hat{i}+\hat{j}}{2}$

As we know that the axis of polarisation of the Em wave is same as Electric field direction that is -  $\hat{k}$

$\vec{E} \times \vec{B}$ direction of propagation of EM wave = $\frac{\hat{i}+\hat{j}}{2}$

$= \vec{k} \times \vec{B} =$ $\frac{\hat{i}+\hat{j}}{2}$

$\Rightarrow$ $\frac{\hat{i}-\hat{j}}{2}$

And the equation of the electromagnetic waves will be in terms of the  $cos(\omega t- \vec{K}.\vec{r})$

So by concluding the above result we can deduce that the option (2) is correct.

Q-17

Different value of a, b and c are given and their sum is d. Arrange the value of d in increasing order -

$\text { (1) } \mathrm{d}_{1}<\mathrm{d}_{2}<\mathrm{d}_{3}<\mathrm{d}_{4} \quad \text { (2) } \mathrm{d}_{1}=\mathrm{d}_{2}=\mathrm{d}_{3}=\mathrm{d}_{4} \quad(3) \mathrm{d}_{4}<\mathrm{d}_{1}<\mathrm{d}_{2}=\mathrm{d}_{3}$     $\quad(4) \mathrm{d}_{4}<\mathrm{d}_{3}<\mathrm{d}_{2}<\mathrm{d}_{1}$

Solution -

So the option (3) is correct.

Q-18

Two gases Ar (40) and Xe (131) at same temperature have same number density. Their diameters are 0.07 nm and 0.10 nm respectively. Find the ratio of their mean free time

(1) 1.03                 (2) 2.04                   (3) 3.04                     (4) 2.40

Solution -

$\begin{array}{l}{\text { Mean free time }=\frac{1}{\sqrt{2} n{\pi} d^{2} v_{rm s}}} \\ {\frac{t_{A r }}{t_{X e}}=\frac{d_{X e}^{2}}{d_{A_{2}}^{2}}} \\ \\ {=\left(\frac{0.1}{0.07}\right)^{2}}\\ \\ {=\left(\frac{10}{7}\right)^{2}=2.04}\end{array}$

So the option (2) is correct.

Q-19

A particle start moving with velocity $\vec{u}=3 \hat{i}$  from origin and acceleration  $a=6 \hat{i}+4 \hat{j}$ .when y co-ordinate of particle is 32 m. then x co-ordinate at that instant will be

(1) 48 (2) 60 (3) 12 (4) 24

Solution-

$S_y=U_yt +\frac{1}{2}a_yt^2\\ \Rightarrow 32=0+\frac{1}{2}*4*t^2\Rightarrow t=4 \sec$

$S_x=U_xt +\frac{1}{2}a_xt^2=(3*4)+\frac{1}{2}*6*16=60\ m$

Q.20

An electron (-|e|, m) is released in Electric field E from rest. rate of change of de-Broglie wavelength with time will be.

$\begin{array}{llll}{\text { (1) }-\frac{\mathrm{h}}{2|\mathrm{e}|}} & {\text { (2) }-\frac{\mathrm{h}}{2|\mathrm{e}| \mathrm{t}}} & {\text { (3) }-\frac{\mathrm{h}}{| \mathrm{e}\left[\mathrm{E}^{2}\right.}} & {\text { (4) }-\frac{2 \mathrm{h}^{2}}{|\mathrm{e}| \mathbb{E}}}\end{array}$

Sol.

$\begin{array}{l}{\lambda_{\mathrm{D}}=\frac{\mathrm{h}}{\mathrm{mv}}} \\ {\therefore \quad \mathrm{v}=\mathrm{at}} \\ \\{\mathrm{v}=\frac{\mathrm{eE}}{\mathrm{m}} \mathrm{t} \quad\left(\mathrm{a}=\frac{\mathrm{eE}}{\mathrm{m}}\right)} \\ \\ {\lambda_{\mathrm{D}}=\frac{\mathrm{h}}{\mathrm{m}\left(\frac{\mathrm{e} \mathrm{E}}{\mathrm{m}}\right) \mathrm{t}}} \\ {\lambda_{\mathrm{D}}=\frac{\mathrm{h}}{\mathrm{eEt}}} \\ {\frac{\mathrm{d} \lambda_{\mathrm{d}}}{\mathrm{dt}}=-\frac{\mathrm{h}}{|\mathrm{e}| \mathrm{Et}^{2}}}\end{array}$

Q.21.

In YDSE pattern with light of wavelength $\lambda_{1}=500 \mathrm{nm}$ , 15 fringes are obtained on a certain segment of

screen. If number of fringes for light of wavelength $\lambda_{2}$  on same segment of screen is 10, then the value
of $\lambda_{2}(\mathrm{in}\ \mathrm{nm})$  is-

Solution-

.  $\begin{array}{l}{15 \times 500 \times \frac{D}{d}=10 \times \lambda_{2} \times \frac{D}{d}} \\ \\ {\lambda_{2}=15 \times 50 \mathrm{nm}} \\ \\ {\lambda_{2}=750 \mathrm{nm}}\end{array}$

Q.22

If in a meter bridge experiment, the balancing length $l$ was 25 cm for the situation shown in the figure. If the length and diameter of the of wire of resistance R is made half, then find the new balancing length in centimetre is

Sol.

$\begin{array}{l}{\frac{\mathrm{x}}{\mathrm{R}}=\frac{75}{25}=3} \\ \\ {\mathrm{R}=\frac{\mathrm{p} \ell}{\mathrm{A}}=\frac{4 \mathrm{p} \ell}{\pi \mathrm{d}^{2}}} \\ \\ {\mathrm{R}^{\prime}=\frac{4 \rho\left(\frac{\ell}{2}\right)}{\pi\left(\frac{\mathrm{d}}{2}\right)^{2}}=2 \mathrm{R}} \\ \\ {\text { then } \frac{\mathrm{x}}{\mathrm{R}^{\prime}}=\frac{\mathrm{x}}{2 \mathrm{R}}=\frac{3}{2}} \\ \\ {\ell=40.00 \mathrm{cm}}\end{array}$

Q.23

Find the power loss in each diode (in mW), if potential drop across the zener diode is 8V.

Sol.

$\begin{array}{l}{\text { i= } \frac{(12-8)}{(200+200)} \mathrm{A}=\frac{4}{400}=10^{-2} \mathrm{A}} \\ \\ {\text { Power loss in each diode } =(4)\left(10^{-2}\right) \mathrm{W}=40 \mathrm{mw}}\end{array}$

Q.24

An ideal gas at initial temperature 300 K  is compressed adiabatically $(\gamma=1.4) \ \text{to} \ \left(\frac{1}{16}\right)^{\mathrm{th}}$ of its initial volume. The gas is then expanded isobarically to double its volume. Then final temperature of gas round to nearest integer is:

Sol.

$PV^{\gamma }$ = constant

$TV^{\gamma -1}$ = constant

$\begin{array}{l}{300\left(\mathrm{V}_{1}\right)^{1.4-1}=\mathrm{T}_{\mathrm{B}}\left(\frac{\mathrm{V}_{1}}{16}\right)^{2 / 5}} \\ \\ {\mathrm{T}_{\mathrm{B}}=300 \times 2^{8 / 5}}\end{array}$

Now for BC process BC

$\begin{array}{l}{\mathrm{V}_{\mathrm{B}} / T_{\mathrm{B}}=\mathrm{V}_{\mathrm{C}} / \mathrm{T}_{\mathrm{C}}} \\ {\mathrm{T}_{\mathrm{C}}=\frac{\mathrm{V}_{\mathrm{C}} \mathrm{T}_{\mathrm{B}}}{\mathrm{v}_{\mathrm{B}}}=2 \times 300 \times 2^{\mathrm{8}/ 5}} \\ \\ {\mathrm{T}_{\mathrm{c}}=1819 \mathrm{K}}\end{array}$

Q.25

If electric field in the space is given by $\overrightarrow{\mathrm{E}}=4 \times \hat{\mathrm{i}}-\left(\mathrm{y}^{2}+1\right) \hat{\mathrm{j}}$ , and electric flux through ABCD is$\phi_{1}$  and
electric flux through BCEF is  $\phi_{2},$  then find $\left(\phi_{1}-\phi_{2}\right)$

Sol.

Flux via ABCD

$\phi_{1}=\int \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\thrm{A}}=0$

Flux via BCEF

$\begin{array}{l}{\phi_{2}=\int \bar{E} \cdot d \bar{A}} \\ {\phi_{2}=\bar{E} \cdot \bar{A}} \\ {=\left(4 x \hat{i}-\left(y^{2}+1\right) \hat{j}\right) \cdot 4 \hat{i}} \\ {=16 x, x=3} \\ \\ {\phi_{2}=48 \frac{N-m^{2}}{C}} \\ {\phi_{1}-\phi_{2}=-48 \frac{N-m^{2}}{C}}\end{array}$

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