JEE Main 2020 9th January Evening Shift Question Paper with Solutions

The length of the screen used portion for 15 fringes, and also for ten fringes

15 $\times$ 500 $\times$ ${D \over \lambda }$ = 10 $\times$ ${{\lambda D} \over \lambda }$

$\Rightarrow$ 15 × 50 =$\lambda$

$\Rightarrow$ $\lambda$ = 750 nm

T₁V₁^$\gamma$–1 = T₂V₂ ^$\gamma$–1

$300 \times {V^{{7 \over 5} - 1}} = {T_2}{\left( {{V \over {16}}} \right)^{{7 \over 5} - 1}}$

$\Rightarrow$ T₂ = 300 × (16)^0.4

Isobaric process

V = ${{nRT} \over P}$

V₂ = kT₂... (1)

2V ₂ = KT_f... (2)

T_f = 2T₂ = 300 × 2 × (16)^0.4 = 1818 K

A₁ + B₁ + C₁ = 24.36 + 0.0724 + 256.2
= 280.6324 = 280.6 (After rounding off)

A₂ + B₂ + C₂ = 24.44 + 16.082 + 240.2
= 280.722 = 280.7 (After rounding off)

A₃ + B₃ + C₃ = 25.2 + 19.2812 + 236.183
= 280.6642 = 280.7 (After rounding off)

A₄ + B₄ + C₄ = 25 + 236.191 + 19.5
= 280.691 = 281 (After rounding off)

Fundamental frequency = 70 Hz.

70 = ${1 \over {2l}}\sqrt {{T \over \mu }}$

$\Rightarrow$ $l$ = 2.14 m

Flux via ABCD

$\phi$_I = $\int {\overrightarrow E } .d\overrightarrow A$ = 0

Flux via EFGH

$\phi$_II = $\int {\overrightarrow E } .d\overrightarrow A$

= [$4x\widehat i - \left( {{y^2} + 1} \right)\widehat j$].4$\widehat i$

= 16x = 16 $\times$ 3 = 48

${\phi _I} - {\phi _{II}}$ = 0 - 48 = -48 Nm²/C

${4 \over 3}$ sin $\theta$ = 1sin90^o

sin $\theta$ = ${3 \over 4}$

cos $\theta$ = ${{\sqrt 7 } \over 4}$

Surface area in solid angle d$\Omega$ = 2$\pi$R²(1 - cos $\theta$)

= 2$\pi$R²(1 - ${{\sqrt 7 } \over 4}$)

Percentage of light = ${{2\pi {R^2}\left( {1 - {{\sqrt 7 } \over 4}} \right)} \over {4\pi {R^2}}}$ $\times$ 100%

= ${{{4 - \sqrt 7 } \over 8}}$ $\times$ 100%

= 17%

Z = x_C – x_L

= ${1 \over {\omega C}} - \omega L$

= ${1 \over {314 \times 100 \times {{10}^{ - 6}}}} - 314 \times 40 \times {10^{ - 3}}$

= 19.28 $\Omega$

As V_m = I_mZ

$\Rightarrow$ 10 = I_m $\times$ 19.28

$\Rightarrow$ I_m = ${{10} \over {19.28}}$ = 0.52 A

$\therefore$ I = 0.52 sin(314t + ${\pi \over 2}$)

= 0.52 cos(314t)

In case I diode is reverse biased

Q_A = CV

In case II, current will flow as diode is forward biased.

Q_A = ${{CV} \over e}$

By momentum conservation,

${{mu} \over 2}$ + mu = 2mV

$\Rightarrow$ V = ${{3u} \over 4}$

H_max = ${{{u^2}{{\sin }^2}60^\circ } \over {2g}}$ = ${{{u^2} \times {3 \over 4}} \over {2g}}$ = ${{3{u^2}} \over {8g}}$

Time taken = $\sqrt {{{2{H_{\max }}} \over g}}$ = $\sqrt {{2 \over g} \times {{3{u^2}} \over {8g}}}$ = ${{\sqrt 3 } \over 2}{u \over g}$

Horizontal distance traveled = ut

= ${{3u} \over 4}.{{\sqrt 3 } \over 2}{u \over g}$ = ${{3\sqrt 3 } \over 8}{{{u^2}} \over g}$

$\rho$ = a + b${\left( {{x \over L}} \right)^2}$

dm = $\rho$dx = $\left( {a + b{{{x^2}} \over {{L^2}}}} \right)dx$

M = $\int {dm}$ = $\int\limits_0^L {\left( {a + b{{{x^2}} \over {{L^2}}}} \right)dx}$

X_com = ${{\int {xdm} } \over {\int {dm} }}$

= ${{\int\limits_0^L {\left( {a + b{{{x^2}} \over {{L^2}}}} \right)xdx} } \over {\int\limits_0^L {\left( {a + b{{{x^2}} \over {{L^2}}}} \right)dx} }}$

= ${{{{a{L^2}} \over 2} + {b \over {{L^2}}}.{{{L^4}} \over 4}} \over {aL + {b \over {{L^2}}}.{{{L^3}} \over 3}}}$

= ${{\left( {{{4a + 2a} \over 8}} \right)L} \over {\left( {{{3a + b} \over 3}} \right)}}$

= ${3 \over 4}\left( {{{2a + b} \over {3a + b}}} \right)L$

$T.2\pi r + {2 \over 3}\pi {r^3}\rho g = {4 \over 3}\pi {r^3}dg$

$\Rightarrow$ T = ${{{r^2}} \over 3}\left( {2d - \rho } \right)g$

$\Rightarrow$ r = $\sqrt {{{3T} \over {\left( {2d - \rho } \right)g}}}$

F = |e| E

$a = {F \over m}$ = ${{\left| e \right|E} \over m}$

V = $at =$ ${{\left| e \right|E} \over m}t$

$\lambda$ = ${h \over {mV}}$ = ${h \over {\left| e \right|Et}}$

${{d\lambda } \over {dt}}$ = ${{ - h} \over {\left| e \right|E{t^2}}}$

Direction of propagation = ${{\widehat i + \widehat j} \over {\sqrt 2 }}$

Electric field is in direction = $\widehat k$

As $\overrightarrow E \times \overrightarrow B$ = ${{\widehat i + \widehat j} \over {\sqrt 2 }}$

Propagation direction of $\overrightarrow B = {{\widehat i - \widehat j} \over {\sqrt 2 }}$

By conservation of energy :

Loss in energy = Gain in enrgy

m₁gh = m₂gh + ${1 \over 2}$ m₁V² + ${1 \over 2}$ m₂V² + ${1 \over 2}I{\omega ^2}$

$\Rightarrow$ (m₁ – m₂)gh = ${1 \over 2}$ (m₁ + m₂)V² + ${1 \over 2}I{\omega ^2}$

$\Rightarrow$ (m₁ – m₂)gh = ${1 \over 2}$ (m₁ + m₂)${\left( {\omega R} \right)^2}$ + ${1 \over 2}I{\omega ^2}$

$\Rightarrow$ (m₁ – m₂)gh = ${1 \over 2}$ (m₁ + m₂)${\left( {\omega R} \right)^2}$ + ${1 \over 2}I{\omega ^2}$

$\Rightarrow$ (m₁ – m₂)gh = ${{{\omega ^2}} \over 2}\left[ {\left( {{m_1} + {m_2}} \right){R^2} + I} \right]$

$\Rightarrow$ $\omega$ = ${\left[ {{{2\left( {{m_1} - {m_2}} \right)gh} \over {\left( {{m_1} + {m_2}} \right){R^2} + I}}} \right]^{{1 \over 2}}}$

${{du} \over {dv}}$ = ${1 \over 2}$ $\times$ stress × strain

= ${1 \over 2}{F \over A} \times {F \over {AY}}$ $\propto$ ${1 \over {{A^2}}}$ $\propto$ ${1 \over {{d^4}}}$

${{du} \over {dv}}$ = ${1 \over 4}$

$\Rightarrow$ ${\left( {{{{d_1}} \over {{d_2}}}} \right)^4}$ = 4

$\Rightarrow$ ${{{d_1}} \over {{d_2}}} = {\left( 4 \right)^{{1 \over 4}}}$ = $\sqrt 2 :1$

B = $\mu$₀nI

${{m{V_{\max }}} \over {qB}} = {R \over 2}$

V_max = ${{qBR} \over {2m}}$

= ${{qR{\mu _0}nI} \over {2m}}$

$\lambda = {1 \over {\sqrt 2 \pi {d^2}n}}$

Mean free time, t = ${\lambda \over v}$

Also v $\propto$ $\sqrt {{T \over M}}$

$\therefore$ t $\propto$ ${{\sqrt M } \over d}$

${{{t_{Ar}}} \over {{t_{xe}}}}$ = ${{d_{Xe}^2} \over {d_{Ar}^2}} \times \sqrt {{{{M_{Ar}}} \over {{M_{Xe}}}}}$

= ${\left( {{{0.1} \over {0.07}}} \right)^2} \times \sqrt {{{40} \over {140}}}$

= 1.09

$\therefore$ Nearest possible answer is 1.83.

Escape velocity, V_e = $\sqrt {{{2GM} \over R}}$

V_A = $\sqrt {{{2GM} \over R}}$

V_B = $\sqrt {{{2G{M \over 2}} \over {{R \over 2}}}}$ = $\sqrt {{{2GM} \over R}}$

${{{V_A}} \over {{V_B}}}$ = 1

Given ${{{V_A}} \over {{V_B}}} = {n \over 4}$

$\therefore$ ${n \over 4}$ = 1

$\Rightarrow$ n = 4

${S \over R} = {{75} \over {25}}$ = 3

R = ${{\rho L} \over A}$ = ${{4\rho L} \over {\pi {d^2}}}$

R' = ${{4\rho \left( {{L \over 2}} \right)} \over {\pi {{\left( {{d \over 2}} \right)}^2}}}$ = 2R

Then ${S \over {R'}} = {{100 - l} \over l}$

$\Rightarrow$ ${{100 - l} \over l}$ = ${S \over {2R}}$ = ${3 \over 2}$

$\Rightarrow$ $l$ = 40 cm

So, ionisation energy = (13.6 Z²) eV

= 9 $\times$ 13.6 eV

${1 \over \lambda } = R{Z^2}\left( {{1 \over {{1^2}}} - {1 \over {{3^2}}}} \right)$

= 1.09 $\times$ 10⁷ $\times$ 9 $\times$ ${8 \over 9}$

= 11.4 nm

Both the diodes are reverse biased.

i = ${9 \over {5 + 10 + 5 + 10}}$ = ${9 \over {30}}$ A

Current in circuit = ${{12 - 8} \over {200 + 200}}$ = 10^-2 A

So power dissipited in each diode = VI

= 4 $\times$ 10^-2

= 40 mW

m${\omega ^2}$(l₀ + x) = kx

x = ${{m{I_0}{\omega ^2}} \over {k - m{\omega ^2}}}$

For k >> m${\omega ^2}$

${x \over {{I_0}}} = {{m{\omega ^2}} \over k}$

$\tau$ = - MBsin $\theta$

I$\alpha$ = - MBsin $\theta$

for small $\theta$,

$\alpha$ = $- {{MB} \over I}\theta$

$\therefore$ ${\omega ^2}$ = ${{MB} \over I}$

$\Rightarrow$ $\omega$ = $\sqrt {{{I\left( {\pi {R^2}} \right)B} \over {{{m{R^2}} \over 2}}}}$ = $\sqrt {{{2I\pi B} \over m}}$

$\therefore$ T = ${{2\pi } \over \omega }$ = $\sqrt {{{2\pi m} \over {IB}}}$

$\overrightarrow u$ = 3.0 $\widehat i$

$\overrightarrow a$ = $\left( {6\widehat i + 4\widehat j} \right)$

$\overrightarrow S = \overrightarrow u t + {1 \over 2}\overrightarrow a {t^2}$

x = 3t + ${1 \over 2}6{t^2}$

= 3t + 3t² .....(1)

y = ${1 \over 2} \times 4 \times {t^2}$ = 32

$\Rightarrow$ t = 4 s .... (2)

x = 3 × 4 + 3 × 4² = 12 + 48 = 60 m

For aqueous solution

$\Delta$T_f = K_f.m = 2 × 0.5

$\therefore$ Temperature of solution = –1°C = 272 K

$\therefore$ Final volume of ideal gas = ${{nRT} \over P}$

= ${{0.1 \times 0.08 \times 272} \over 1}$

= 2.18 L

First ionization enthalpies = 496 kJ/mole

Second ionization enthalpies = 4560 kJ/mol

According to the given information, the difference between first and second ionization enthalpy is very high so Metal belong to 1st group i.e. Monovalent cation.

Metal hydroxide will be of type, MOH.

MOH + HCl $\to$ MCl + H₂O

MOH + ${1 \over 2}$H₂SO₄ $\to$ ${1 \over 2}$M₂SO₄ + H₂O

So one mole of HCl required to react with one mole MOH.

And ${1 \over 2}$ mole of H₂SO₄ required to react with one mole MOH.

% carbon = ${{48} \over {72}} \times 100$ = 66.67

The given table provides the results of three tests: Molisch's test, Barfoed's test, and Biuret test for three different biomolecules A, B, and C.

1. Molisch's Test: Detects carbohydrates. Positive results for A and B indicate that they are carbohydrates.




2. Barfoed's Test: Specifically for detecting monosaccharides. A positive result for B indicates that B is a monosaccharide, while A's negative result indicates it's not a monosaccharide, implying it might be a disaccharide.




3. Biuret Test: Used to detect the presence of proteins. Positive result for C indicates that it is a protein.

Given the options, we can analyze them as follows:

• Option A: Lactose (disaccharide), Glucose (monosaccharide), Albumin (protein) - fits the given results.


• Option B: Lactose (disaccharide), Fructose (monosaccharide), Alanine (amino acid) - not a match, as Alanine won't give a positive Biuret test.


• Option C: Lactose (disaccharide), Glucose (monosaccharide), Alanine (amino acid) - same issue as Option B.


• Option D: Glucose (monosaccharide), Fructose (monosaccharide), Albumin (protein) - not a match, as A must be a disaccharide.

Therefore, the correct option is: Option A: A = Lactose, B = Glucose, C = Albumin.

Lithium has highest hydration enthalpy among alkali metals due to its small size.

LiCl is soluble in pyridine because LiCl have more covalent character.

Li does not form ethynide with ethyne.

Both Li and Mg reacts slowly with H₂O.

Note : Here if we consider only $\sigma$ bonds so the answer would be 12 but there are 6$\pi$ bonds also, if we consider them then the total number of Cr – O bonds will be 18.

(III) Nitrogen lone pair not participate in resonance.

(II) Nitrogen lone pair not participate in resonance due to increase the stability of the compound.

(I) Nitrogen lone pair participate in resonance.

(IV) Nitrogen lone pair participate in resonance and increase the stability of the compound due to aromicity.

In the figure, Reactant A is represented by square.
And there are 6 squares. So at equilibrium
[A]_eq = 6

In the figure, Product B is represented by circle.
And there are 11 circles. So at equilibrium
[B]_eq = 11

For reaction, A ⇌ B

K_eq = ${{\left[ B \right]} \over {\left[ A \right]}}$ = ${{11} \over 6}$ $\approx$ 2

(I) [Cr(H₂O)₆]Br₂

H₂O is weak field ligand so it can not pair up all the electrons.

Cr²⁺ : [Ar] 4s⁰3d⁴ ($t_{2g}^3e{g^1}$)

Unpaired e^– = 4

Magnetic moment = $\sqrt {24}$ = 4.89 BM

(II) Na₄[Fe(CN)₆]

CN^- is strong field ligand so it pair up all the electrons.

Fe²⁺ = [Ar] 4s⁰3d⁶ ($t_{2g}^6e{g^0}$)

Unpaired e^– = 0

Magnetic moment = 0 BM

(III) Na₃[Fe(C₂O₄)₃]

As $\Delta$₀ $>$ P, so pairing of electrons happens.

Fe³⁺ = [Ar] 4s⁰3d⁵ ($t_{2g}^5e{g^0}$)

Unpaired e^– = 1

Magnetic moment = $\sqrt 3$ = 1.73 BM

(IV) (Et₄N)₂[CoCl₄]

Cl^-is weak field ligand so it can not pair up all the electrons.

Co²⁺ = [Ar] 4s⁰3d⁷ ($e{g^4}t_{2g}^3$)

Unpaired e^– = 3

Magnetic moment = $\sqrt {15}$ = 3.87 BM

Hence order of magnetic moment is
I > IV > III > II

Biochemical oxygen demand (BOD) is amount of oxygen required by bacteria to break down organic waste in a certain volume of water sample.

Zn + 2dil. HCl $\to$ ZnCl₂ + H₂

Zn + 2NaOH $\to$ Na₂ZnO₂ + H₂

From one mole of Zn, 1 mol of H₂ is produced by both NaOH and HCl.

The ratio of the volume of H₂ is 1 : 1

Each carbon atom is sp² hybrid

$\therefore$ 3 sp² hybrid orbitals are formed by each carbon atom

Total sp² orbitals = 6 × 3 = 18

B₃N₃H₃Cl₃(A) + LiBH₄ $\to$ B₃N₃H₆ (B)

B = inorganic benzene or Borazine

B₃N₃H₃Cl + MeMgBr(C) $\to$ B₃N₃H₃(CH₃)₃ + 3MgBrCl

Initially H₂ combines with O₂, therefore rapid fall in pressure will be observed. Then gradually adsorption of CO will occur by activated charcoal.

Cr(OH)3	⇌	Cr+3	+	3OH-
		S		3S

K_sp = [Cr³⁺] [OH^– ]³

$\Rightarrow$ 6 × 10^–31 = S × (3S)³

$\Rightarrow$ 6 × 10^–31 = 27 S⁴

S = ${\left( {{6 \over {27}} \times {{10}^{31}}} \right)^{{1 \over 4}}}$

As [OH^–] = 3S

= $3{\left( {{6 \over {27}} \times {{10}^{31}}} \right)^{{1 \over 4}}}$

= (18 × 10^–31)^1/4 M

$\Delta S = \int {{{d{q_{rev}}} \over T}}$

S = Kln(w)

Both entropy and change in entropy are function of temperature.

Using Arrehenius equation

K = A${e^{ - {{{E_a}} \over {RT}}}}$

$\ln \left( {{{{k_{400}}} \over {{k_{300}}}}} \right) = {{{E_a}} \over R}\left( {{1 \over {300}} - {1 \over {400}}} \right)$

$\Rightarrow$ $\ln \left( {{{60} \over {40}}} \right) = {{{E_a}} \over R}\left( {{{100} \over {300 \times 400}}} \right)$

$\Rightarrow$ $\ln \left( {{3 \over 2}} \right) = {{{E_a}} \over {1200R}}$

$\therefore$ E_a = 0.4 × 1200 × 8.3 = 3984 J/mol

E_a = 3.984 kJ/mol = 3.98 kJ/mol

due to strong – R effect of – NO₂ group and M effect of Cl, partial double bond character between C and Cl increases. So shortest bond length.

1030 gm of sea water contains = 10.3 × 10^–3 gm

10⁶ gm of sea water contains = ${{10.3 \times {{10}^{ - 3}}} \over {1030}} \times {10^6}$ = 10 ppm

The form of H₂O with the lowest ionic conductance at 298 K is distilled water.

[Co(NH₃)₄Cl₂] has 2 geometrical isomers.
Among cis and trans isomers, cis isomer has Cl–Co–Cl angle of 90^o.

${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$

Let y = vx

$\therefore$ ${{dy} \over {dx}}$ = v + x.${{dv} \over {dx}}$

$\Rightarrow$ v + x.${{dv} \over {dx}}$ = ${{x\left( {vx} \right)} \over {{x^2} + {v^2}{x^2}}}$ = ${v \over {1 + {v^2}}}$

$\Rightarrow$ x.${{dv} \over {dx}}$ = ${v \over {1 + {v^2}}}$ - v = ${{v - v - {v^3}} \over {1 + {v^2}}}$ = $- {{{v^3}} \over {1 + {v^2}}}$

$\Rightarrow$ $\int {{{1 + {v^2}} \over {{v^3}}}} dv = - \int {{{dx} \over x}}$

$\Rightarrow$ $- {1 \over {2{v^2}}} + \log v$ = $- \log x + C$

$\Rightarrow$ $- {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right)$ = $- \log x + C$ ......(1)

putting x = 1, y = 1 we get

$\Rightarrow$ C = $- {1 \over 2}$

From eq. (1)

$- {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right)$ = $- \log x - {1 \over 2}$

Put y = e

$- {1 \over 2}{{{x^2}} \over {{e^2}}} + \log \left( {{e \over x}} \right)$ = $- \log x - {1 \over 2}$

$\Rightarrow$ x² = 3e²

$\Rightarrow$ x = $\pm$3$\sqrt e$