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

Pitch = ${3 \over 6}$ mm = 0.5 mm

Least count = ${{0.5} \over {50}}$ mm

= ${1 \over {100}}$ = 0.01 mm = 0.001 cm

Rate of flow of water is same at both A and B

A_AV_A = A_BV_B

(40)V_A = (20)V_B

V_B = 2V_A .......(1)

Using Bernoulli's theorem

P_A + ${1 \over 2}\rho V_A^2$ = P_B + ${1 \over 2}\rho V_B^2$

$\Rightarrow$ P_A - P_B = ${1 \over 2}\rho V_B^2$ - ${1 \over 2}\rho V_A^2$

$\Rightarrow$ 700 = ${1 \over 2} \times 1000\left( {4V_A^2 - V_A^2} \right)$

$\Rightarrow$ V_A = 0.68 m/s = 68 cm/s

$\therefore$ Flow rate = A_AV_A = (40)(68) = 2720 cm³/s

D = ${{{t_1}} \over {{\mu _1}}} + {{{t_2}} \over {{\mu _2}}}$

= ${h \over {\sqrt 2 }} + {h \over {2\sqrt 2 }}$

= ${3 \over 4}h\sqrt 2$

$\overrightarrow {{E_1}} = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$

Its corresponding magnetic field will be

$\overrightarrow {{B_1}} = {{{E_0}} \over c}\widehat k\cos \left( {\omega t - kx} \right)$

$\overrightarrow {{E_2}} = {E_0}\widehat k\cos \left( {\omega t - ky} \right)$

Also its corresponding magnetic field will be

$\overrightarrow {{B_2}} = {{{E_0}} \over c}\widehat i\cos \left( {\omega t - ky} \right)$

Net force on charge particle

= $q\overrightarrow {{E_1}} + q\overrightarrow {{E_2}} + q\overrightarrow v \times \overrightarrow {{B_1}}$$+ q\overrightarrow v \times \overrightarrow {{B_2}}$

= ${q{E_0}\widehat j}$ + ${q{E_0}\widehat k}$ + $q\left( {0.8c\widehat j} \right) \times \left( {{{{E_0}} \over c}\widehat i} \right)$ + $q\left( {0.8c\widehat j} \right) \times \left( {{{{E_0}} \over c}\widehat i} \right)$

= ${E_0}q\left( {0.8\widehat i + \widehat j + 0.2\widehat k} \right)$

Let the work function be $\phi$.

$\therefore$ KE_max = ${{hc} \over \lambda } - \phi$

We know r = ${{mv} \over {qB}}$

and p = mv = rqB

$\therefore$ KE_max = ${{{p^2}} \over {2m}}$ = ${{{q^2}{r^2}{B^2}} \over {2m}}$

= ${{{{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}{{\left( {10 \times {{10}^{ - 3}}} \right)}^2}{{\left( {3 \times {{10}^{ - 4}}} \right)}^2}} \over {2 \times 9 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}}}}$

= 0.8 eV

${{hc} \over \lambda }$ = ${{12420} \over {6561}}$ = 1.9 eV

$\therefore$ $\phi$ = 1.9 - 0.8 = 1.1 eV

V = $\left| {L{{di} \over {dt}}} \right|$

$\Rightarrow$ L = ${V \over {\left| {{{di} \over {dt}}} \right|}}$ = ${{100} \over {{{0.25} \over {0.025 \times {{10}^{ - 3}}}}}}$ = 10 mH

$\theta$ = 1.22${\lambda \over a}$

Distance O₁O₂ = ($\theta$)d

= (1.22${\lambda \over a}$)d

= ${{\left( {1.22} \right)\left( {5500 \times {{10}^{ - 10}}} \right)\left( {4 \times {{10}^5}} \right) \times {{10}^3}} \over 5}$

= 5368 × 10^–2 m

= 53.68 m

[h] = M¹L²T^–1
[C] = L¹T^–1
[G] = M^–1L³T^–2

[f] = $\sqrt {{{M{L^2}{T^{ - 1}} \times {L^5}{T^{ - 5}}} \over {{M^{ - 1}}{L^3}{T^{ - 2}}}}}$ = M¹L²T^–2

By energy conservation

U_i + K_i = U_f + K_f

$\Rightarrow$ $mg{l \over 2}\sin 30^\circ$ + 0 = 0 + ${1 \over 2}I{\omega ^2}$

$\Rightarrow$ $mg{1 \over 2} \times {1 \over 2}$ = ${1 \over 2} \times {{m{{\left( 1 \right)}^2}} \over 3}{\omega ^2}$

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

$\Rightarrow$ $\omega = \sqrt {15}$

$\therefore$ n = 15

W = $\int {\overrightarrow F } .d\overrightarrow r$

= $\int {\left( { - x\widehat i + y\widehat j} \right)} .\left( {dx\widehat i + dy\widehat j} \right)$

= $- \int\limits_1^0 {xdx} + \int\limits_0^1 {ydy}$

= ${1 \over 2} + {1 \over 2}$ = 1 J

I = ${{20} \over 2}$ = 10 A

I₁ = ${4 \over 5}I$, I₂ = ${1 \over 5}I$

$\Rightarrow$ I₁ = 8A, I₂ = 2 A

I₁' = ${3 \over 5} \times 10$ = 6 A

I₂' = ${2 \over 5} \times 10$ = 4 A

$\Rightarrow$ I_BC = 8 – 6 = 2 A

T = m${\omega ^2}l$

Breaking stress = $\sigma$ = ${{m{\omega ^2}l} \over A}$

$\Rightarrow$ ${\omega ^2}$ = ${{4.8 \times {{10}^7} \times \left( {{{10}^{ - 2}} \times {{10}^{ - 4}}} \right)} \over {10 \times 0.3}}$ = 16

$\Rightarrow$ $\omega$ = 4

V_A = 12.7 – 0.7 = 12 Volt

Diode left hand diode is forward biased and right hand diode is reverse biased.

For the particle moving from point P to point Q,

W_{Electric field} + W_{Magnetic field} = $\Delta$KE

$\Rightarrow$ qE(2$a$) + 0 = ${1 \over 2}m\left[ {{{\left( {2v} \right)}^2} - {v^2}} \right]$

$\Rightarrow$ 2qE$a$ = ${3 \over 2}m{v^2}$

$\Rightarrow$ E = ${3 \over 4}\left( {{{m{v^2}} \over {qa}}} \right)$

$\therefore$ Point (A) is correct.

Rate of work done(Power) by electric

field at p = $\overrightarrow F .\overrightarrow v$ = qEv = ${3 \over 4}\left( {{{m{v^3}} \over a}} \right)$

$\therefore$ Point (B) is also correct.

Angle between electric force and velocity is 90º, hence rate of work done will be zero at Q.

$\therefore$ Point (C) is also correct.

Initial angular momentum at P, ${\overrightarrow L }$_i = mv$a\left( { - \widehat k} \right)$

Final angular momentum at Q, ${\overrightarrow L }$_f = m(2v)$(2a)\left( { - \widehat k} \right)$

Change in angular momentum ${\overrightarrow L }$_f – ${\overrightarrow L }$_i = $3mva\left( { - \widehat k} \right)$

Magnitude in the change in angular momentum $\left| {\Delta \overrightarrow L } \right|$ = $3mva$

$\therefore$ Point (D) is wrong.

x² = at² + 2bt + c .....(1)

$\Rightarrow$ 2xv = 2at + 2b .....(2)

$\Rightarrow$ v = ${{at + b} \over x}$

differentiating (2) w.r.t. time

$\Rightarrow$ xa' + v² = a

Here a' is acceleration.

$\Rightarrow$ a'x = a - v² = a - ${\left[ {{{at + b} \over x}} \right]^2}$

$\Rightarrow$ a'x = ${{a{x^2} - {{\left( {at + b} \right)}^2}} \over {{x^2}}}$

$\Rightarrow$ a' = ${{a\left( {a{t^2} + 2bt + c} \right) - {{\left( {at + b} \right)}^2}} \over {{x^2}}}$

$\Rightarrow$ a' = ${{ac - {b^2}} \over {{x^3}}}$

$\therefore$ a' $\propto$ ${1 \over {{x^3}}}$ $\propto$ x^-3

$\therefore$ n = 3

AB = d

$\therefore$ OA = ${{d \over {\sqrt 3 }}}$

From parallel axis theorem

I₀ = $3 \times \left[ {{2 \over 5}M{{\left( {{d \over 2}} \right)}^2} + M{{\left( {{d \over {\sqrt 3 }}} \right)}^2}} \right]$ = ${{13} \over {10}}M{d^2}$

I_A = I₀ + 3M${{{\left( {{d \over {\sqrt 3 }}} \right)}^2}}$

= ${{13} \over {10}}M{d^2}$ + Md²

= ${{23} \over {10}}M{d^2}$

$\therefore$ ${{{I_0}} \over {{I_A}}}$ = ${{13} \over {23}}$

Degree of freedom of a diatomic molecule if vibration is absent = 5

Degree of freedom of a diatomic molecule if vibration is present = 7

$\therefore$ $C_V^A$ = ${5 \over 2}R$

and $C_V^B$ = ${7 \over 2}R$

$\therefore$ ${{C_V^A} \over {C_V^B}} = {5 \over 7}$

Since $\overrightarrow r$ and $\overrightarrow p$ are perpendicular to each other therefore point lies on the equitorial plane. Therefore electric field at the point will be antiparallel to the dipole moment.

$\therefore$ $\overrightarrow E || - \overrightarrow p$

$\Rightarrow$ $\overrightarrow E ||\left( {\widehat i + 3\widehat j - 2\widehat k} \right)$

In process 2 to 3 pressure is constant & in process 3 to 1 volume is constant which is correct only in option B.

$\left| {\overrightarrow {{E_A}} } \right|$ = (E_complete)_center - (E_removed)_surface

= 0 - $\left| {{{k\rho {4 \over 3}\pi {{\left( {{R \over 2}} \right)}^3}} \over {{{\left( {{R \over 2}} \right)}^2}}}} \right|$

= ${k\rho {4 \over 3}\pi \left( {{R \over 2}} \right)}$

$\left| {\overrightarrow {{E_B}} } \right|$ = (E_complete)_surface - (E_removed)_{r = ${{3r} \over 2}$}

= ${{{k\rho {4 \over 3}\pi {R^3}} \over {{R^2}}}}$ - ${{{k\rho {4 \over 3}\pi {{\left( {{R \over 2}} \right)}^3}} \over {{{\left( {{{3R} \over 2}} \right)}^2}}}}$

= ${k\rho {4 \over 3}\pi R}$ - ${k\rho {4 \over 3}\pi {R \over {18}}}$

= ${k\rho {4 \over 3}\pi \left( {{{17R} \over {18}}} \right)}$

$\therefore$ ${{\left| {\overrightarrow {{E_A}} } \right|} \over {\left| {\overrightarrow {{E_B}} } \right|}}$ = ${{18} \over {34}}$

Orbital speed for of A is v = $\sqrt {{{GM} \over R}}$

After collision, let the combined mass moves with speed v₁

$\therefore$ mv + ${m \over 2}{v \over 2}$ = $\left( {{{3m} \over 2}} \right){v_1}$

$\Rightarrow$ v₁ = ${{5v} \over 6}$

Since after collision, the speed is not equal to orbital speed at that point. So motion cannot be circular. Since velocity will remain tangential, so it cannot fall vertically towards the planet. Their speed after collision is less than escape speed $\sqrt 2 v$, so they cannot escape gravitational field. So their motion will be elliptical around the planet.

$\overrightarrow {{P_i}} = \overrightarrow {{P_f}}$

$\Rightarrow$ mu$\widehat i$ + m$\left( {{u \over 2}\widehat i + {u \over 2}\widehat j} \right)$ = 2m$\overrightarrow v$

Compare both side

$\overrightarrow v =$ ${{3u} \over 4}\widehat i + {u \over 4}\widehat j$

$\Rightarrow$ ${\left| {\overrightarrow v } \right|^2}$ = ${{10{u^2}} \over {16}}$

$\Delta$KE = KE_f – KE_i

= ${1 \over 2}2m \times {{10{u^2}} \over {16}}$ - ${1 \over 2}m{u^2}$ - ${1 \over 2}m{\left( {{u \over {\sqrt 2 }}} \right)^2}$

= $- {1 \over 8}m{u^2}$

Let current density be J.

$\therefore$ Applying Ampere's law. For point P

$\oint {\overrightarrow {B.} d\overrightarrow l } = {\mu _0}i$

$\Rightarrow$ B_P2$\pi$${a \over 3}$ = ${\mu _0}J\pi {{{a^2}} \over 9}$ ...(1)

$\therefore$ Applying Ampere's law. For point Q

B_Q2$\pi$${(2a)}$ = ${\mu _0}J\pi {a^2}$ ...(2)

Dividing (1) by (2) we get

${{{B_P}2\pi {a \over 3}} \over {{B_Q}4\pi a}} = {{{\mu _0}J\pi {{{a^2}} \over 9}} \over {{\mu _0}J\pi {a^2}}}$

$\Rightarrow$ ${{{B_P}} \over {{B_Q}}} = {2 \over 3}$

I₀ = CA²

A_R = A + A$\sqrt 2$ = A(1 + $\sqrt 2$)

I_R = C$A_R^2$

$\therefore$ I_R = CA²${\left( {\sqrt 2 + 1} \right)^2}$ = 5.8 I₀

$\lambda = {h \over {\sqrt {2mE} }}$

Also, ${h \over {\sqrt {2m\left( {E + \Delta E} \right)} }}$ = ${\lambda \over 2}$

$\therefore$ ${{E + \Delta E} \over E} = 4$

$\Rightarrow$ $\Delta$E = 3E

Magnetic moment = 1.73 BM

$\therefore$ Unpaired electron = 1

(1) $O_2$ has 16 electrons.

Moleculer orbital configuration of $O_2$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ *$

Here 2 unpaired electron present.

(2)   $O_2^{+}$ has 15 electrons.

Moleculer orbital configuration of $O_2^{+}$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ *$

Here 1 unpaired electron present.

(3)   $O_2^{-}$ has 17 electrons.

Moleculer orbital configuration of $O_2^{-}$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * = \,\,\pi _{2p_y^1}^ *$

Here 1 unpaired electron present.

Hence $O_2^{-}$ & $O_2^{+}$ have one unpaired electron.

[Pb²⁺] = ${{300 \times 0.134} \over {400}}$

= 1.005 × 10^–1 M

[Cl^-] = ${{100 \times 0.4} \over {400}}$

= 10^–1 M

$PbC{l_{2(s)}} \leftrightharpoons Pb_{(aq)}^{2 + } + 2Cl_{(aq)}^ -$

Q = [Pb²⁺] × [Cl^–]²

= 0.1005 × (0.1)²

= 1.005 × 10^–3

Given K_sp = 1.6 × 10^–5

$\therefore$ Q > K_sp

In lsomers of Hydrocarbon heat of combustion depend upon their stability. Generally trans is more stable then cis form. Stability increase heat of combustion decrease.

Stability a > b > c

Heat of combustion c > b > a

K₁ = A${e^{ - {{{E_a}} \over {R \times 700}}}}$

K₂ = A${e^{ - {{\left( {{E_a} - 30} \right)} \over {R \times 500}}}}$

$\because$Rate is same

$\therefore$ Rate constant will also be same

K₁ = K₂

A${e^{ - {{{E_a}} \over {R \times 700}}}}$ = A${e^{ - {{\left( {{E_a} - 30} \right)} \over {R \times 500}}}}$

$\Rightarrow$ ${{{\left( {{E_a} - 30} \right)} \over {R \times 500}}}$ = ${{{{E_a}} \over {R \times 700}}}$

$\Rightarrow$ 5E_a = 7E_a – 210

$\Rightarrow$ 210 = 2E_a

$\Rightarrow$ E_a = 105 kJ/mole

$\therefore$ Activation energy in the presence of catalyst = 105 – 30 = 75 kJ/mol

According to Bohr’s model

${r_n} = {{{n^2}} \over Z} \times {a_0}$

Also 2$\pi$${r_n}$ = n$\lambda$

$\Rightarrow$ 2$\pi$${{{n^2}} \over Z} \times {a_0}$ = n$\lambda$

$\Rightarrow$ $\lambda$ = $2\pi \times {n \over Z} \times {a_0}$

For n = 4 and Z = 1

$\Rightarrow$ $\lambda$ = 8$\pi$${a_0}$

Given complex Cr(H₂O)₆Cl_n

As the magnetic mement is 3.83 BM then

Spin only magnetic moment = $\sqrt {n\left( {n + 2} \right)}$ BM = 3.83

$\Rightarrow$ n = 3

Chromium in +3 oxidation state so molecular formula is Cr(H₂O)₆Cl_n.

$\therefore$ This formula have following isomers

(a) [Cr(H₂O)₆]Cl₃ : react with AgNO₃ but does not show geometrical isomerism.

(b) [Cr(H₂O)₅Cl]Cl₂.H₂O react with AgNO₃ but does not show geometrical isomerism.

(c) [Cr(H₂O)₄Cl₂]Cl.2H₂O react with AgNO₃ & show geometrical isomerism.

(d) [Cr(H₂O)₃Cl₃].3H₂O does not react with AgNO₃ & show geometrical isomerism.

Compound will be [Cr(H₂O)₄Cl₂] Cl.2H₂O

IUPAC NAME : Tetraaquadichlorido chromium(III) chloride dihydrate

Cathode : Ag⁺(aq) + e^- $\to$ Ag(s)

Moles of Ag deposited = ${{108} \over {108}}$ = 1 mole

Anode : 2H₂O $\to$ O₂ + 4H⁺ + 4e^-

Here we have to find volume of O₂ evolved.

Equivalance of Ag = Equivalance of O₂

$\Rightarrow$ 1 $\times$ 1 = n_O2 $\times$ 4

$\Rightarrow$ n_O2 = ${1 \over 4}$ mol

$\therefore$ Volume of O₂ evolved

= ${1 \over 4}$ $\times$ 22.4

= 5.6 lit

(i) A and D give positive test with ninhydrin because both have free carboxylic and amine group. Both aspartame and alitame contains amino acids.

(ii) Lassaigne extract give +ve test with AgNO₃. So Cl is present, –ve test with Fe₄[Fe(CN)₆]₃ means N is absent. So it can't be Aspartame or Saccharine or Alitame, so C is sucralose.

(iii) Lassaigne solution of B and D given +ve sodium nitroprusside test, so it is a sulphur containing compound, Saccharine and Alitame also contains sulphur.

$\therefore$ A = Aspartame, B = Saccharin

C = Sucralose, D = Alitame.

Basicity is inversly proportional to electronegativity.

Complex [Pd(F)(Cl)(Br)(I)]^2– (square planar geometry)-has 3 geometrical isomers.

$\therefore$ n = 3

[Fe(CN)₆]^n–6 becomes [Fe(CN)₆]^3-

Here Oxidation state of Fe = +3

$\therefore$ Fe⁺³ = [Ar]3d⁵4s⁰

CN^- is strong field ligand so it pairing of electrons happens.

E.C. according to CFT = ($t_{2g}^5e{g^0}$)

$\therefore$ No. of unpaired e^- = 1

Then Magnetic moment = $\sqrt {n\left( {n + 2} \right)}$ = $\sqrt 3$ = 1.73 B.M

CFSE = [(–0.4 × 5) + (0.6 × 0)] $\Delta$₀ = –2.0 $\Delta$₀

Non metal oxide $\to$ acidic.

Metal oxide $\to$ basic.

Al₂O₃ amphoteric.

Equivalance of MgSO₄ = Equivalance of CaCO₃

$\Rightarrow$ 10^-3 $\times$ 2 = [CaCO₃] $\times$ 2

$\Rightarrow$ [CaCO₃] = 10^-3 M = 10^-3 mol/lit

= 10^-3 $\times$ 100 g/lit

= 100 mg/lit

= 100 ppm

$\Delta$T_f = 0.2^o C

$\therefore$ $\Delta$T_f = ik_f.m

i = 2 for NaCl

$\therefore$ 0.2 = 2$\times$2$\times$${{{W_{NaCl}} \times 1000} \over {58.5 \times 600}}$

$\Rightarrow$ W_NaCl = ${{58.5 \times 600 \times 0.2} \over {4 \times 1000}}$ = 1.76 g

Eu²⁺ : [xe] 4f⁷

Ce³⁺ : [xe] 4f¹

Friedel-Crafts reaction is an electrophilic substitution reaction. In general, the rate of these reactions is highest when benzene have some electron releasing groups, (i.e. $-\mathrm{NH}_2-\mathrm{OH}$ ) attached to the nucleus.


Lewis acid (anhydrous $\mathrm{AlCl}_3$ ) used in these reactions form an adduct with the ' $-\mathrm{NH}_2$ ' or $-\mathrm{OH}$ groups and becomes ineffective.



withdrawing (deactivating) group, but among the given options, it is most reactive towards Friedel-Crafts reaction.

%w/w = 63%

d_solution = 1.4 g/ml

Molarity =

%w/w $\times$ d_solution

Moleculer weight of Solute

.$\times$ 10

= ${{63 \times 1.4 \times 10} \over {63}}$ = 14 M

1st I.E. of Be > B

In case of Be, electron is removed from 2s orbital which has more penetration power, while in case of B electron is removed from 2p orbital which has less penetration power.

2p electron of B is more shielded from nucleus by the inner electrons than 2s electrons of Be

$\therefore$ It is easier to remove 2p electron than 2s electron.