JEE Main 2020 3rd September Evening Shift Question Paper with Solutions

Given Temp of mixture = 31°C

180×1×(31-25) + 20×(31-25) = m×540 + m×1×(100-31)

$\Rightarrow$ 180×6 + 20×6 = 540m + 100 m - 31m

$\Rightarrow$ 1080 + 120 = 640 m - 31m

$\Rightarrow$ 1200 = 609m

$\Rightarrow$ m = ${{1200} \over {609}}$ = 1.97

Let velocity of block after collision = v

Using momentum conservation,

p_i = p_f

0.1×20 = (1.9 + 0.1)V

$\Rightarrow$ 2 = 2 V

$\Rightarrow$ V = 1 m/sec

Kinetic energy just before striking the floor

= ${1 \over 2}m{v^2} + mgh$

= ${1 \over 2} \times 2{\left( 1 \right)^2} + 2 \times 10 \times 1$

= 21 J

A perfect diamagnetic substance will completely expel the magnetic field. Therefore, there will be no magnetic field inside the cavity of sphere. Hence the paramagnetic substance kept inside the cavity will experience no force.

Phase difference = ${{2\pi } \over \lambda }$ $\times$ optical path difference

= ${{2\pi } \over \lambda }\left( {{n_1}{L_1} - {n_2}{L_2}} \right)$

Given N = 500

A = 3 $\times$ 10^–4 m²

$\tau$ = 1.5 Nm

i = 0.5 A

We know, $\tau = BINA\,sin\theta$

$1.5 = B \times 0.5 \times 500 \times 3 \times {10^{ - 4}}$

$B = {{10000} \over {500}} = 20$ Tesla

We know, $\phi = BA$

Also, $E = {{d\phi } \over {dt}} = {{AdB} \over {dt}}$

$E = {l^2}{{dB} \over {dt}}$

$i = {E \over R}$

= ${{{l^2}{{dB} \over {dt}}} \over {{{\rho l} \over A}}}$

$= {{{l^2}} \over {pl}}{{dB} \over {dt}}A$

= ${{{{l^2}\pi {R^2}} \over {\rho l}}{{dB} \over {dt}}}$

$\therefore$ $i = {{30} \over 4} \times {{30} \over 4} \times {{{{10}^{ - 4}} \times 0.032 \times 4 \times {{10}^{ - 6}} \times \pi } \over {1.23 \times {{10}^{ - 8}} \times 30 \times {{10}^{ - 2}} \times {{10}^3}}}$

$\Rightarrow$ $i = {{240 \times \pi \times {{10}^{ - 10}}} \over {1.23 \times {{10}^{ - 7}}}}$

$\Rightarrow$ $i = {{240 \times 3.14 \times {{10}^{ - 3}}} \over {1.23}}$

$= {{753.6} \over {1.23}} \times {10^{ - 3}}$

$\Rightarrow$ $i = 612.68 \times {10^{ - 3}} = 0.61A$

At equilibrium position

V₀ = V

${V_0} = {\omega _1}A = \sqrt {{K \over m}} A$ .....(i)

$V = \omega {A^1} = \sqrt {{K \over {{m \over 2}}}} {A^1}$ .....(ii)

$\therefore$ ${A^1} = {A \over {\sqrt 2 }}$

${{\Delta T} \over {\Delta t}} = k\left( {{{{T_f} + {T_i}} \over 2} - {T_0}} \right)$

$\Rightarrow$ ${{50 - 40} \over {300}} = k\left( {{{90} \over 2} - 20} \right)$

$\Rightarrow$ ${{40 - T} \over {300}} = k\left( {{{40 + T} \over 2} - 20} \right)$

$\Rightarrow$ ${{10} \over {40 - T}} = {{25 \times 2} \over {40 + T - 40}}$

$\Rightarrow$ ${1 \over {40 - T}} = {5 \over T}$

$\Rightarrow$ $T = 200 - 5T$

$\Rightarrow$ $6T = 200$

$\Rightarrow$ $T = 33^\circ C$

${F_v} = mg$

${F_H} = m{\omega ^2}{l \over 2}\sin \theta$

$\therefore$ ${\tau _{net}}\,about\,COM = {F_v}.{l \over 2}\sin \theta - {F_H}{l \over 2}\cos \theta$

$= {{m{l^2}} \over {12}}{\omega ^2}\sin \theta \cos \theta$

$\Rightarrow$ $mg{l \over 2}\sin \theta - m{\omega ^2}{l \over 2}\sin \theta {l \over 2}\cos \theta = {{m{l^2}} \over {12}}{\omega ^2}\sin \theta \cos \theta$

$\Rightarrow {{gl} \over 2} - {{{\omega ^2}{l^2}} \over 4}\cos \theta = {{{l^2}} \over {12}}{\omega ^2}\cos \theta$

$\Rightarrow$ ${{gl} \over 2} ={\omega ^2}{l^2}\cos \theta \left( {{1 \over {12}} + {1 \over 4}} \right)$

$\Rightarrow$ ${{gl} \over 2} = {{{\omega ^2}{l^2}\cos \theta } \over 3}$

$\Rightarrow$ $\cos \theta = {{3g} \over {2{\omega ^2}l}}$

Given, wavelength of x ray ($\lambda$₁) = 1 nm

And wavelength of visible light ($\lambda$₂) = 500 nm

we know, Power$(P) = {{nhc} \over \lambda }$

As P = constant and h, c also constant

So, ${n \over \lambda } = constant$

$\Rightarrow$ ${{{n_1}} \over {{n_2}}} = {{{\lambda _1}} \over {{\lambda _2}}} = {{1nm} \over {500nm}} = {1 \over {500}}$

V_I = Velocity of image with respect to mirror

V₀ = Velocity of object with respect to mirror

$|\overrightarrow {{V_I}}| = |- {{{v^2}} \over {{u^2}}}\overrightarrow {{V_0}} |$

$= | - {{10 \times 10} \over {30 \times 30}} \times 9|$

$= 1$ cm/s

$E = {{hc} \over \lambda }$

= ${{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {400 \times {{10}^{ - 9}}}}$

= ${{1240} \over {400}}$ eV

= 3.1 eV

$\overrightarrow E = {E_0}\,\cos (\omega t - kx)\widehat j$

We know, $E = BC$

$B_{0} = {E_{0} \over C} = {{{E_0}} \over {{1 \over {\sqrt {{\mu _0}{ \in _0}} }}}}$

$\Rightarrow$ $B_{0} = {E_0}\sqrt {{\mu _0}{ \in _0}}$

You can see direction of $\overrightarrow B$ is along z axis.

$\therefore$ $\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos (\omega t - kx)\widehat k$

at t = 0

$\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos (kx)\widehat k$

$\sigma = {{{Q_1}} \over {4\pi {R^2}}} = {{{Q_2}} \over {4\pi 16{R^2}}}$

$\Rightarrow$ $16{Q_1} = {Q_2}$

$\therefore$ ${V_R} - {V_{4R}} = {{K{Q_1}} \over R} + {{K{Q_2}} \over {4R}} - {{K{Q_1}} \over {4R}} - {{K{Q_2}} \over {4R}}$

$= {{3K{Q_1}} \over {4R}} = {{3{Q_1}} \over {16\pi {\varepsilon _0}R}}$ [As K = ${1 \over {4\pi {\varepsilon _0}}}$ ]

We know, kinetic energy K = qV

also $K = {{{P^2}} \over {2m}}$

$\therefore$ $qV = {{{P^2}} \over {2m}} = {{{m^2}{v^2}} \over {2m}}$

$V = \sqrt {{{2qv} \over m}}$

$V \propto \sqrt {{q \over m}}$

${{{V_H}} \over {{V_{He}}}} = {{\sqrt {{e \over m}} } \over {\sqrt {{e \over {4m}}} }} = {2 \over 1}$

(1) Multimeter shows deflection when it connects with capacitor.

(2) If we assume that LED has negligible resistance then multimeter shows no deflection for the forward bias but when it connects in reverse direction, it break down occurs so splash of light out.

(3) The resistance of metal wire may be taken zero, so no deflection in multimeter.

(4) No matter, how we connect the resistance across multimeter. It shows same deflection.

a = g sin 30 + $\mu$ g cos 30

We know, v² = u² + 2as

$\Rightarrow$ 0 = $v_0^2$ - 2ad

$\Rightarrow$ $v_0^2 = 2ad$

$d = {{v_0^2} \over {2a}}$

Total work done,

${W_f} = {k_f} - {k_i}$

$\Rightarrow$ $- 2\mu mg\,\cos 30{{v_0^2} \over {2a}} = {1 \over 2}m{{v_0^2} \over 4} - {1 \over 2}mv_0^2$

$\Rightarrow$ ${{ + \mu g\,\cos 30} \over a} =$${3 \over 8}$

$\Rightarrow$ $8\mu g\,\cos 30 = 3g\,\sin 30 + 3\mu \,\cos 30$

$\Rightarrow$ $5\mu g\,\cos 30 = 3g\,\sin 30$

$\Rightarrow$ $\mu = {{3\tan 30} \over 5} = {{\sqrt 3 } \over 5}$

$\Rightarrow$ ${{\sqrt 3 } \over 5} = {I \over {1000}}$

$\Rightarrow$ $I = 346$

Q₁ = mL = 8000cal



Q₂ = W + Q₁

Coefficient of performance (C.O.P)
$= {{{Q_1}} \over W} = {{{Q_1}} \over {{Q_2} - {Q_1}}} = {{{T_1}} \over {{T_2} - {T_1}}}$

${{{Q_1}} \over W} = {{273} \over {300 - 273}}$

${{{Q_1}} \over W} = {{273} \over {27}}$

$W = {{27} \over {273}}{Q_1}$

$W = {{27} \over {273}}mL$

$W = {{27} \over {273}} \times 80 \times 100$

${Q_2} = {{27} \over {273}} \times 80 \times 100 + 80 \times 100$

$= 8791.2$ cal

$i = {6 \over {800 + {{400 \times 10000} \over {400 + 10000}}}}$

$i = {6 \over {800 + {{40000} \over {104}}}}$

$i = {6 \over {800 + 384.61}} = {6 \over {1184.61}} = 0.00506$

V₄₀₀ = i₁ $\times$ 400

= $\left( {{{{{10}^4}} \over {400 + {{10}^4}}}} \right)i \times 400$

= $\left( {{{{{10}^4}} \over {400 + {{10}^4}}}} \right)\left( {0.00506} \right) \times 400$

= 1.95 V

$160 = n{C_p}50$ ....(i)

$240 = n{C_v}100$ ....(ii)

Dividing (i) by (ii), we get

${{160} \over {240}} = {{{C_p}} \over {{C_v}}} \times {1 \over 2}$

$\Rightarrow$ ${{{C_p}} \over {{C_v}}}$ = ${4 \over 3}$

We know,

$\gamma = {{{C_p}} \over {{C_v}}} = 1 + {2 \over f} = {4 \over 3}$

$\Rightarrow f = 6$

I = m$a$² + ${{m{a^2}} \over 4}$ = ${5 \over 4}m{a^2}$

Accoeding to the question,

${5 \over 4}m{a^2} = {N \over {20}}m{a^2}$

$\Rightarrow$ N = 25

dm = $\rho$dv

$\int {dm}$ $= \int {{\rho _0}} 4\pi {r^2}dr$

$\Rightarrow$ M = $\int {{\rho _0}} 4\pi {r^2}dr$

Also E = ${{GM} \over {{r^2}}}$

$\Rightarrow E{r^2} = 4\pi G\,\int\limits_0^r {{\rho _0}} \left( {1 - {{{r^2}} \over {{R^2}}}} \right){r^2}dr$

$\Rightarrow E = 4\pi G{\rho _0}\left( {{{{r}} \over 3} - {{{r^3}} \over {5{R^2}}}} \right)$

For maximum E, ${{dE} \over {dr}} = 0$

${1 \over 3} - {{3{r^2}} \over {5{R^2}}}$ = 0

$\therefore$ $r = \sqrt {{5 \over 9}} R$

Solar constant = ${E \over {AT}}$

= ${{\left[ {{M^1}{L^2}{T^{ - 2}}} \right]} \over {\left[ {{L^2}T} \right]}}$

= [ ML⁰T^–3 ]

$R = (1.3 \times {10^{ - 15}}){A^{{1 \over 3}}}$

We know, $m = pV$

$\Rightarrow$ $p = {m \over V}$

$\Rightarrow$ $p = {{{m_p}A} \over {{4 \over 3}\pi {R^3}}}$

$p = {{{m_p}A} \over {{4 \over 3}\pi \times {{(1.3 \times {{10}^{ - 15}})}^3}A}}$

$p \approx {10^{17}}kg/m$

$P = Fv$

$P = m{{dv} \over {dt}}v$

$\Rightarrow$ $vdv = {P \over m}dt$

Integrating both sides, we get

${V^2} = k't$

$\Rightarrow$ V = k"$\sqrt t$

$\Rightarrow$ ${{ds} \over {dt}}$ = k"$\sqrt t$

$\Rightarrow$ $\int {ds} = \int {k''\sqrt t } dt$

$\Rightarrow$ s $\propto$ t^3/2

(1) Acid rain corrodes water pipes resulting in the leaching of heavy of heavy metals such as iron, lead and copper into the drinking water.

(2) Acid rain damages buildings and other structures made of stone or metal.

(3) It causes respiratory aliments in human beings and animals.

(4) It is harmful for agriculture, trees and plants as it wasshes down the nutrients needed for its growth.

Let molar mass of protein A = x g/mol

Let molar mass of protein B = y g/mol

$\pi$_A = osmotic pressure of protein A = ${{0.73} \over x} \times {{1000} \over {250}} \times RT$

$\pi$_B = osmotic pressure of protein B = ${{1.65} \over y} \times {1 \over 1} \times RT$

Given $\pi$_A = $\pi$_B

$\Rightarrow$ ${{0.73} \over x} \times {{1000} \over {250}} \times RT$ = ${{1.65} \over y} \times {1 \over 1} \times RT$

$\Rightarrow$ ${x \over y}$ = ${{0.73} \over {0.25 \times 1.65}}$ = 1.77

= 177 $\times$ 10^-2

Let x molecule of water are lost then

13.5 = $\left[ {{{x \times 18} \over {6 \times 18 + 3 \times 35 + 52}}} \right] \times 100$

$\Rightarrow$ x = 1.99 $\simeq$ 2

$\therefore$ Around two moles of water are lost during heating.

$\therefore$ Formula of complex could be

[Cr(H₂O)₄Cl₂]Cl.2H₂O

Volume strength = 11.2 × molarity

$\Rightarrow$ molarity = ${{5.6} \over {11.2}}$ = 0.5

Assuming 1 litre solution,

mass of solution = 1000 ml × 1 g/ml = 1000 g

mass of solute = moles × molar mass

= 0.5 mol × 34 g/mol

= 17 gm.

$\therefore$ mass % =

mass of solute

mass of solution

$\times$ 100

= ${{17} \over {1000}} \times 100$ = 1.7 %

Cr₂O₇^2- + 14H⁺ + 6e^– $\to$ 2Cr³⁺ + 7H₂O

I = 2 A, t = 8 min

From 1^st law of faraday,

w_Cr⁺³ = z $\times$ i $\times$ t

$\Rightarrow$ w_Cr⁺³ = ${{52} \over {96000 \times 3}} \times 2 \times 8 \times 60$

= ${{52} \over {300}}$

Mass of Cr³⁺ ions actually obtained = 0.104 gm

% efficiency =

Actual obtained Amt

Theo. obtained Amt

$\times$ 100

= ${{0.104} \over {{{52} \over {300}}}}$ $\times$ 100

= 60 %

There is a sudden jump after 3^rd I.E. due to attainment of noble gas configuration.

So, the number of valence electrons in this element are 3.

Mass of 6.023 × 10²² molecules of a substance = 10 g

Mass of 6.023 × 10²³ molecules of the substance = 100 g

$\therefore$ Molar mass of the substance = 100 g mol^–1

Molarity of the solution =

moles of solute

volume of solution(in l)

= ${{\left( {5/100} \right)} \over 2}$ = 0.025

Structure of Tripeptide Asp – Glu – Lys

Raniticline $\to$ Antacid
Nardil $\to$ Antidepressant
Chloramphenicol $\to$ Antibiotic
Dimetane $\to$ Antihistamine

–NO₂ group present at ortho and para position deactivate the benzene ring and making the carbon attached to ring (–CH₂Cl) group more electrophilic for the incoming nucleophile to give S_N2 reaction but rate of S_N2 reaction decreases as hindrence increases around the electrophilic carbon.

(a) B will be more crystalline due to more inter molecular interactions hence more efficient packing.

(b) B will have higher boiling point due to higher intermolecular interactions.

(c) B will be more soluble in water than A as B will have more extent of H-bonding in water.

Under hypothetical situation, the value of l is greater than n which varies from 0 to n + 1.

For n = 1, l = 0, 1, 2

n = 2, l = 0, 1, 2, 3

Elements follow the following electronic configuration

1s 1p 1d 2s 2p 2d 2f

Atomic number (Z) = 9
1s² 1p⁶ 1d¹

Atomic number = 6
1s² 1p⁴

Atomic number 8
1s² 1p⁶

Atomic number 13
1s² 1p⁶ 1d⁵ (half filled)

Steep rise in pH around the equivalence point for titration of strong acid with strong base.

[Ru(en)₃ ]Cl₂ :

Here CN = 6 so octahedral splitting happens.

'en' is strong field ligand so $\Delta$₀ > P(pairing energy). That is why pairing of electrons happens.

$\Delta$₀ = Energy gap between e_g and t_2g orbital.


[Fe(H₂O)₆]Cl₂ :

Here CN = 6 so octahedral splitting happens.

H₂O is weak field ligand so $\Delta$₀ < P(pairing energy). That is why no pairing of electrons happens.

$\Delta$₀ = Energy gap between e_g and t_2g orbital.

Ionic micelles formed by addition of water to soap {sodium stearate}.

H₃PO₂ + NaOH $\to$ NaH₂PO₂ + H₂O

Using Stoichiometry

Moles of H₃PO₂ reacted

1

=

Moles of NaOH reacted

1

$\Rightarrow$ ${{0.1 \times 10} \over 1}$ = 0.1 × V_NaOH

$\Rightarrow$ V_NaOH = 10 ml

According to Dalton’s law of partial pressure, p_i = x_i × P_T

p_i = partial pressure of the ith component

x_i = mole fraction of the ith component

p_T = total pressure of mixture

$\Rightarrow$ 2 atm = $\left( {{{{n_{{H_2}}}} \over {{n_{{H_2}}} + {n_{He}} + {n_{{O_2}}}}}} \right)$ $\times$ p_T

$\Rightarrow$ p_T = $\left( {{{1 + 1 + 1} \over 1}} \right)$ $\times$ 2 = 6 atm

To identify the incorrect statement, let's analyze each option :

Option A : "In manganate and permanganate ions, the π-bonding takes place by overlap of p-orbitals of oxygen and d-orbitals of manganese." This statement is true. In both manganate ${MnO}_4^{2-}$ and permanganate ${MnO}_4^-$ ions, the manganese atom is in a high oxidation state (+6 and +7, respectively) and forms bonds with oxygen atoms involving the overlap of p-orbitals of oxygen and d-orbitals of manganese.

Option B : "Manganate ion is green in colour and permanganate ion in purple in colour." This statement is true. Manganate ions are indeed green and permanganate ions are purple in aqueous solution.

Option C : "Manganate and permanganate ions are paramagnetic." This statement is incorrect. Both manganate and permanganate ions are diamagnetic, not paramagnetic. In these ions, manganese is in +6 and +7 oxidation states, respectively, and has no unpaired electrons.

Manganate($\mathop {Mn}\limits^{ + 6}$O₄ ^2–) ,

Mn(25) = [Ar]3d⁶4s²

Mn⁺⁶(25) = [Ar]3d¹ $\to$ Paramagnetic

Permanganate ($\mathop {Mn}\limits^{ + 7}$O₄^-)

Mn(25) = [Ar]3d⁶4s²

Mn⁺⁷(25) = [Ar]3d⁰ $\to$ Diamagnetic

Option D : "Manganate and permanganate ions are tetrahedral." This statement is true. Both manganate and permanganate ions have a tetrahedral geometry, with the manganese atom at the center and the four oxygen atoms surrounding it.

Therefore, the incorrect statement is option C.

Rate of Nucleophilic addition :

1. Aldehyde > Ketone

2. Aliphatic aldehyde > Aromatic aldehyde

$\therefore$ Correct order of nucleophilic addition reaction is :

Be < Mg (atomic radius)

Be > Al (I.E₁ greater than Al because of fully filled valence shell of Be)

Charge/radius ratio of Be is less than that of Al.

Both Be and Al form mainly covalent compound.

2A + 3B + ${3 \over 2}$C $\to$ 3P

rate = $- {1 \over 2}{{d\left[ A \right]} \over {dt}} = - {1 \over 3}{{d\left[ B \right]} \over {dt}} = - {2 \over 3}{{d\left[ C \right]} \over {dt}} = {1 \over 3}{{d\left[ P \right]} \over {dt}}$

$\Rightarrow$ ${{d\left[ A \right]} \over {dt}} = {2 \over 3}{{d\left[ B \right]} \over {dt}} = {4 \over 3}{{d\left[ C \right]} \over {dt}}$$= - {2 \over 3}{{d\left[ P \right]} \over {dt}}$

Let xyz is 3 digits number.

Given that sum of digits = 10

$\therefore$ x + y + z = 10 ......(1)

Also x can't be 0 as if x = 0 then it will become 2 digits number.

So, x $\ge$ 1, y $\ge$ 0, z $\ge$ 0

As x $\ge$ 1

$\Rightarrow$ x $-$ 1 $\ge$ 0

Let x $-$ 1 = t

$\therefore$ t $\ge$ 0

From equation (1)

(x $-$ 1) + y + z = 9

$\Rightarrow$ t + y + z = 9

Now this problem becomes, distributing 9 things among 3 people t, y, z.

Number of ways we can do that

= ${}^{9 + 3 - 1}{C_{3 - 1}} = {}^{11}{C_2} = 55$

Now when 3 digit number is 900 then t = 9, y = 0, z = 0.

And when t = 9, then

x $-$ 1 = 9

$\Rightarrow$ x = 10

But we can't take x = 10 in a 3 digits number. So, we have to remove this case.

$\therefore$ Total number of 3 digit numbers = 55 $-$ 1 = 54.