JEE Main 2020 2nd September Morning Shift Question Paper with Solutions

ρ_M = 98 × 10^–8
ρ_A = 2.80 × 10^–8
ρ_C = 1.72 × 10^–8
ρ_T = 5.65 × 10^–8

$\therefore$ $\rho$_M > $\rho$_T > $\rho$_A > $\rho$_C

dm = $\rho$dv

$\Rightarrow$ dm = $\left( \frac{k}{r} \right)$ (4$\pi$r²dr)

$\Rightarrow$ dm = 4$\pi$krdr

M = $\int\limits^{R}_{0} dm$ = $\int\limits^{R}_{0} 4\pi krdr$

$\Rightarrow$ M = $4\pi k\left[ \frac{r^{2}}{2} \right]^{R}_{0}$

$\Rightarrow$ M = 2$\pi$kR²

For circular motion gravitational force will provide required centripetal force.

$\frac{GMm}{R^{2}}$ = $\frac{mv^{2}}{R}$

$\Rightarrow$ $\frac{G\left( 2\pi kR^{2}\right) m}{R^{2}}$ = $\frac{mv^{2}}{R}$

$\Rightarrow$ v = $\sqrt{2\pi GkR}$

Time period, T = $\frac{2\pi R}{v}$

= $\frac{2\pi R}{\sqrt{2\pi GkR} }$ $\propto$ $\sqrt{R}$

$\Rightarrow$ T² $\propto$ R

Velocity of man with respect to ground

${\overrightarrow V _{m/g}}$ = ${\overrightarrow V _{m/A}}$ + ${\overrightarrow V _{A}}$

= -1.8 + 36

Velocity of man w.r.t. B

${\overrightarrow V _{m/B}}$ = ${\overrightarrow V _{m}}$ - ${\overrightarrow V _{B}}$

= –1.8 + 36 – (–72)

= 106.2 km/hr

= 29.5 m/s

Option (2) is correct as permanent magnet should have high coercivity & retentivity.

Applying pressure equation from A to B

P₀ + $\rho$.${{R{\omega ^2}} \over 2}$.R - $\rho$gh = P₀

$\Rightarrow$ ${{\rho {R^2}{\omega ^2}} \over 2}$ = $\rho$gh

$\Rightarrow$ h = ${{{R^2}{\omega ^2}} \over 2}$ = ${\left( 5 \right)^2}{{{\omega ^2}} \over {2g}}$ = ${{25} \over 2}{{{\omega ^2}} \over g}$

u_i = $\frac{1}{2}$ $\times$ 5 $\times$ 10^-6$\times$220

Final common potential

= $\frac{220\times 5+0\times 2.5}{5+2.5}$ = 220 $\times$ $\frac{2}{3}$

u_f = $\frac{1}{2}$ $\times$ (5 + 2.5)$\times$10^-6 $\times$ $\left( 220\times \frac{2}{3} \right)^{2}$

$\Delta$u = u_i - u_f

$\Rightarrow$ $\Delta$u = –403.33 × 10^–4

$\Rightarrow$ –403.33 × 10^–4 = ${X \over {100}}J$

$\Rightarrow$ X = -4.03

Value of X is approximate 4

Energy density, ${{dU} \over {dV}} = {{B_0^2} \over {2{\mu _0}}}$

$\Rightarrow$ 1.02 $\times$ 10^–8 = ${{B_0^2} \over {2{\mu _0}}}$

Also, c = ${1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$

$\Rightarrow$ ${\mu _0} = {1 \over {{c^2}{\varepsilon _0}}}$

$\therefore$ ${B_0^2}$ = 1.02 $\times$ 10^–8 $\times$ 2 $\times$ ${1 \over {{c^2}{\varepsilon _0}}}$

= 1.02 $\times$ 10^–8 $\times$ 2 $\times$ ${{4\pi \times 9 \times {{10}^9}} \over {9 \times {{10}^{16}}}}$

$\Rightarrow$ B₀ = 16 $\times$ 10^-8 T = 160 nT

For diatomic ideal gas :

f = 5

$\gamma$ = ${7 \over 5}$

T_i = T = 273 + 20 = 293 K

V_i = V

V_f = ${V \over {10}}$

For adiabatic process TV^{$\gamma$ - 1} = constant

${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$

$\Rightarrow$ $\left( {293} \right){V^{{7 \over 5} - 1}} = {T_2}{\left( {{V \over {10}}} \right)^{{7 \over 5} - 1}}$

$\Rightarrow$ ${T_2} = 293 \times {\left( {10} \right)^{{2 \over 5}}}$

$\Delta$U = ${{nfR\left( {{T_2} - {T_1}} \right)} \over 2}$

= ${{5 \times 5 \times {{25} \over 3} \times \left( {{{293.10}^{{2 \over 5}}} - 293} \right)} \over 2}$

= ${{625 \times 293 \times \left( {{{10}^{{2 \over 5}}} - 1} \right)} \over 6}$

= 46.14 $\times$ 10³ J

$\simeq$ 46 kJ

to pull up, $\tau$_F $\ge$ $\tau$_mg

$\Rightarrow$ F $\times$ R $\ge$ Mg $\times$ x

$\Rightarrow$ F_min = ${{Mg} \over R} \times \sqrt {{R^2} - {{\left( {R - a} \right)}^2}}$

= $Mg\sqrt {1 - {{\left( {{{R - a} \over R}} \right)}^2}}$

Apply work energy theorem

W_g + W_f = $\Delta$kE

mgsin$\theta$ (AC + 2AC) – $\mu$mg cos$\theta$ $\times$ AC = 0 - 0

$\Rightarrow$ $\mu$ = 3tan $\theta$

v_m = 5(1 + 0.6 cos 6280t) sin(211 × 10⁴ t)

$\Rightarrow$ v_m = (5 + 3 cos 6280t) sin(211 × 10⁴ t)

maximum Amp. = 5 + 3 = 8 V

minimum Amp. = 5 – 3 = 2 V

From the given option nearest value of minimum Amplitude = ${5 \over 2}$ V

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

For identical string $l$ and $\mu$ will be same

f $\propto$ $\sqrt T$

$\therefore$ ${{450} \over {300}} = \sqrt {{{{T_x}} \over {{T_y}}}}$

$\Rightarrow$ ${{{{T_x}} \over {{T_y}}} = {9 \over 4}}$ = 2.25

Least count = 1 mm or 0.01 cm

Zero error = 0 + 0.01 × 7 = 0.07 cm

Reading = 3.1 + (0.01 × 4) – 0.07

= 3.1 + 0.04 – 0.07

= 3.1 – 0.03

= 3.07 cm

y = ${{nD\lambda } \over d}$

$\Rightarrow$ n = ${{yd} \over {D\lambda }}$

= ${{1.27 \times {{10}^{ - 3}} \times {{10}^{ - 3}}} \over {1 \times 632.8 \times {{10}^{ - 9}}}}$ = 2

Path difference $\Delta$x = n$\lambda$

= 2 $\times$ 632.8 nm

= 1265.6 nm

= 1.27 $\mu$m

U $= {{{f_1}} \over 2}{n_1}RT + {{{f_2}} \over 2}{n_2}RT$

$= {5 \over 2}\left( {3RT} \right) + {3 \over 2} \times 5RT$

$\therefore$ U $= 15RT$

Let particle have charge q and mass 'm

We know, F = ma_y = qE

$\therefore$ ${a_y} = {{qE} \over m}$

As electric field is along vertical direction so horizontal component of acceleration is zero.

$\therefore$ a_x = 0

Applying, $S = {u_y}t + {1 \over 2}{a_y}{t^2}$

$- y = 0\times t + {1 \over 2}\left( {{{qE} \over m}} \right){t^2}$ ....(1)

Also, $x = {V_0}t$

$\Rightarrow t = {x \over {{V_0}}}$

Putting in (1) we get,

$y = - {1 \over 2}\left( {{{qE} \over m}} \right){\left( {{x \over {{V_0}}}} \right)^2}$

Differentiating with respect to x,

${{dy} \over {dx}} = - {{qE} \over m}\left( {{x \over {V_0^2}}} \right)$

$\therefore$ Slope = $m = \tan \theta = - {{qE} \over m}\left( {{x \over {V_0^2}}} \right)$

when x = d, then $m = - {{qE} \over m}\left( {{d \over {V_0^2}}} \right)$

We know, equation of straight line whose slope is m,

y = mx + c

Here m = $\left( { - {{qEd} \over {mV_0^2}}} \right)$,

$\therefore$ $- {1 \over 2}\left( {{{qE} \over m}} \right){\left( {{d \over {{V_0}}}} \right)^2} = \left( { - {{qEd} \over {mV_0^2}}} \right)d + c$

$\Rightarrow$ c = ${{{qEd} \over {2mV_0^2}}}$

Putting the value of c, we get

$y = - \left( {{{qEd} \over {mV_0^2}}} \right)x + {{qE{d^2}} \over {2mV_0^2}}$

$= {{qEd} \over {mV_0^2}}\left( {{d \over 2} - x} \right)$

Pitch = $\frac{2\pi m}{qB}$ vcos$\theta$

= ${{2\left( {3.14} \right)\left( {1.67 \times {{10}^{ - 27}}} \right) \times 4 \times {{10}^5} \times \cos 60} \over {\left( {1.69 \times {{10}^{ - 19}}} \right)\left( {0.3} \right)}}$

= 0.04 m = 4 cm

From momentum conservation

m(u)$\widehat i$ + 3m(0) = mv$\widehat j$ + 3m$\overrightarrow {v'}$

$\Rightarrow$ 3m$\overrightarrow {v'}$ = m(u)$\widehat i$ - mv$\widehat j$

$\Rightarrow$ $\overrightarrow {v'} = {{u\widehat i - v\widehat j} \over 3}$

$\therefore$ $\left| {\overrightarrow {v'} } \right| = {{\sqrt {{u^2} + {v^2}} } \over 3}$

$\Rightarrow$ ${\left| {\overrightarrow {v'} } \right|^2} = {{{u^2} + {v^2}} \over 9}$ ......(1)

As collision is perfectely elastic hence

KE_i = KE_j

$\Rightarrow$ ${1 \over 2}m{u^2} + {1 \over 2}\left( {3m} \right) \times {0^2} = {1 \over 2}m{v^2} + {1 \over 2}3m{\left( {v'} \right)^2}$

$\Rightarrow$ u² = v² + 3${\left( {v'} \right)^2}$

$\Rightarrow$ u² = v² + 3$\left( {{{{u^2} + {v^2}} \over 9}} \right)$

$\Rightarrow$ 3u² = 3v² + u² + v²

$\Rightarrow$ 2u² = 4v²

$\Rightarrow$ $v = {u \over {\sqrt 2 }}$

y = 4Cx²

$\Rightarrow$ $\frac{dy}{dx}$ = tan $\theta$ = 8Cx

At P, tan θ = 8Ca

For steady circular motion

mg sinθ = mω²acosθ

$\Rightarrow$ tan $\theta$ = $\frac{\omega^{2} a}{g}$

$\Rightarrow$ 8Ca $\times$ g = ω²$\times$a

$\Rightarrow$ $\omega$ = $\sqrt{8gC}$

$\Rightarrow$ $\omega$ = $2\sqrt {2gC}$

f = $- {8 \over 2}$ = -4 cm

u = –10 cm

v = ?

Using mirror formula

${1 \over v} + {1 \over u} = {1 \over f}$

$\Rightarrow$ ${1 \over v} + {1 \over { - 10}} = {1 \over { - 4}}$

$\Rightarrow$ v = ${{ - 20} \over 3}$

m = ${{ - \left( { - {{20} \over 3}} \right)} \over { - 10}}$ = $- {2 \over 3}$

So image would be real, inverted and smaller than object.

At any time flux $\phi$ = BA cos $\omega t$

|emf| = $\left| {{{d\phi } \over {dt}}} \right|$ = BA$\omega$ sin $\omega t$

|emf|_max = BA$\omega$ = BA${{2\pi } \over T}$

= ${{3 \times {{10}^{ - 5}} \times \pi \times {{\left( {0.1} \right)}^2} \times 2\pi } \over {0.4}}$

= 15 $\times$ 10^-6

= 15 $\mu$V

$\tau$_B = 0 (torque about point B is zero)

(T _A) × 100 – (mg) × 50 – (2mg) × 25 = 0

$\Rightarrow$ 100T _A = 100 mg

$\Rightarrow$ T_A = 1 mg

Y = k [F]^x [A]^y [V]^z

[M¹L¹T ^–2] = [MLT^–2]^x [L²]^y [LT^–1]^z

[M¹L¹T ^–2] = [M]^x [L]^x+2y+z[T]^–2x–z

Comparing power of M, L and T

x = 1 ……(1)

x + 2y + z = –1 ……(2)

–2x – z = –2 ……(3)

After solving

x = 1

y = –1

z = 0

$\therefore$ Y = FA^–1V⁰

Number of uranium atoms in 2 kg

= ${{2 \times 6.023 \times {{10}^{26}}} \over {235}}$

Energy from one atom is 200 × 10⁶ e.V. hence total energy from 2 kg uranium

= ${{2 \times 6.023 \times {{10}^{26}}} \over {235}} \times 200 \times {10^6}$ eV

= ${{2 \times 6.023 \times {{10}^{26}}} \over {235}} \times 200 \times {10^6}$ $\times$ 1.6 × 10^–19 J

2 kg uranium is used in 30 days hence energy received per second or power is

Power = ${{2 \times 6.023 \times {{10}^{26}} \times 200 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}} \over {235 \times 30 \times 24 \times 3600}}$

= 63.2 × 10⁶ watt or 63.2 Mega Watt

$\simeq$ 60 MW

${{hc} \over \lambda } - \phi$ = eV .....(i)

${{hc} \over {3\lambda }} - \phi = {{eV} \over 4}$ .....(ii)

From (i) and (ii),

${{hc} \over {3\lambda }} - \phi =$ ${{hc} \over {4\lambda }} - {\phi \over 4}$

$\Rightarrow$ ${{hc} \over \lambda }\left( {{1 \over 3} - {1 \over 4}} \right) = {{3\phi } \over 4}$

$\Rightarrow$ ${{hc} \over {9\lambda }} = \phi$

Also, $\phi$ = ${{hc} \over {{\lambda _0}}}$

$\therefore$ ${{hc} \over {9\lambda }} = {{hc} \over {{\lambda _0}}}$

$\Rightarrow$ $\phi$ = 9$\lambda$

So, n = 9

Here attack of OH^– is not on chiral carbon, it is adjacent to chiral carbon, so configuration of chiral carbon remains constant. This is a case of retention of configuration.

$\Delta$G = $\Delta$G^o + RTln $\left[ {{{S{n^{ + 2}}} \over {C{u^{ + 2}}}}} \right]$

= –2 × 96500 [(–0.16) – 0.34] + RT$\left[ {{1 \over 1}} \right]$

= 96500 J

Basic gas (C) must be ammonia (NH₃).

It means (B) gas should be N₂ which is formed by heating of compound (A).

2Pb(NO₃)₂ $\buildrel \Delta \over \longrightarrow$ 2PbO + 4NO₂ + O₂

(NH₄)₂Cr₂O₇ $\buildrel \Delta \over \longrightarrow$ N₂ $\uparrow$ + Cr₂O₃ + 4H₂O $\uparrow$

NH₄NO₂ $\buildrel \Delta \over \longrightarrow$ N₂ $\uparrow$ + 2H₂O $\uparrow$

2NaN₃ $\buildrel \Delta \over \longrightarrow$ 2Na + 3N₂$\uparrow$

With addition of solute in solvent, surface area for vapourisation decreases causes lowering in vapour pressure.

Compound	Electronic Geometry	Moleculer Geometry or Moleculer Shape	Lone Pair	Polarity
AB4	Tetrahedral	Tetrahedral	0	Nonpolar
AB4	Trigonal Bipyramidal	see-saw	1	Polar
AB4	Octahedral (Square Bipyramidal)	Square Planar	2	Nonpolar

Note : The difference between Electronic Geometry and Moleculer Geometry(Moleculer Shape) is that in Electronic Geometry we include lone pair but in Moleculer Geometry(Moleculer Shape) we do not include lone pair.

Here in this question it is not clear which geometry they are talking about, correct answer should be either Trigonal Bipyramidal or see-saw.

Internal energy of ‘Ar’ or any gas, has nothing to do with Quantum nature of atom.

Photoelectric effect, atomic spectrum and Black body radiations may be explained by quantum theory.

Bredig's arc method is used for the preparation of colloids.

Bunsen Burner and measuring cylinder is not required for titration.

Atomic radius decreases on moving left to right in periodic table, while other three properties given increases (in magnitude) on moving left to right across a period.

(I) Under weak field ligand, octahedral Mn(II) and tetrahedral Ni(II) both the complexes are high spin complex.

(II) Tetrahedral Ni(II) complex can very rarely be low spin because square planar (under strong ligand) complexes of Ni(II) are low spin complexes.

(III)With strong field ligands Mn (II) complexes can be low spin because they have less number of unpaired electron (unpaired electron = 1) While with weak field ligands Mn(II) complexes can be high spin because they have more number of unpaired electron (unpaired electron = 5)

(IV) Aqueous solution of Mn(II) ions is pink in colour.

According to Freundlich adsorption isotherm, in the median range of pressure

       ${x \over m} \propto {P^{{1 \over n}}}$

$\Rightarrow$  ${x \over m}$ = kP$^{{1 \over n}}$ ....(1)

taking log both sides, we get,

log${x \over m}$ = logk + ${1 \over n}$ logP

Here in graph between log${x \over m}$ and logP, slope is ${1 \over n}$ and intercepts = log k.

Given, log k = 0.4771 $\Rightarrow$ k = 3

Slope ${1 \over n}$ = 2

put in eq. (1), ${x \over m}$ = 3 $\times$ (4)² = 48

In the stratosphere, CFCs release chlorine free radical ($\mathop {Cl}\limits^ \bullet$).

CF₂Cl₂(g) $\buildrel {UV} \over \longrightarrow$ $\mathop {Cl}\limits^ \bullet$(g) + $\mathop {Cl}\limits^ \bullet$F₂Cl(g)

Which react with O₃ to give chlorine oxide (Cl$\mathop O\limits^ \bullet$) radical not chlorine dioxide (Cl$\mathop O\limits^ \bullet$₂) radical.

$\mathop {Cl}\limits^ \bullet$(g) + O₃(g) $\to$ Cl$\mathop O\limits^ \bullet$(g) + O₂(g)

In Carius method
mass of organic compound = 0.172 gm
mass of Bromine = 0.08 gm

Hence % of Bromine = ${{0.08} \over {0.172}} \times 100$

= 46.51 %

Option (A) Br mass % = ${{80} \over {109}} \times 100$ = 73.33%

(B) Br mass % = ${{80} \over {95}} \times 100$ = 84.2 %

(C) Br mass % = ${{80} \over {172}} \times 100$ = 46.51 %

(D) Br mass % = ${{80 \times 2} \over {252}} \times 100$ = 63.49 %

In HCN, CN^– is acts as nucleophile, attack first that –CHO group which has maximum positive charge. The magnitude of the (+ve) charge increases by –R and –I group. So reactivity order will be

Na₄[Fe(CN)₅(NOS)]

Let the O.S. of Fe be x

OS of CN = –1
OS of NOS = –1

$\therefore$ (+1)4 + x + (–1)5 + (–1)1 = 0

$\Rightarrow$ x = +2

Na₄[FeO₄]

Let O.S. of Fe be y

(+1)4 + y + (–2)4 = 0

$\Rightarrow$ y = +4

[Fe₂(CO)₉]

Let O.S. of Fe be z

2z + 0 × 9 = 0

$\Rightarrow$ z = 0

so (x + y + z) = +2 + 4 + 0 = 6

For ideal gas,

PM = dRT [As V = ${M \over d}$ ]

$\Rightarrow$ d = ${{PM} \over {RT}}$

$\therefore$ d $\propto$ ${1 \over T}$ $\Rightarrow$ d v/s ${1 \over T}$ graph is Straight line.

d v/s T $\Rightarrow$ graph is Hyperbolic.

As d $\propto$ p

$\therefore$ d v/s p graph is Straight line.

K_sp value of CuS is very low 10^–36 (3.6 × 10^–36) due to low K_sp value Cu⁺² ion gets precipitated very quickly even with very low concentration of S^–2 ion.

Spin only magnetic moment = 4.9 = $\sqrt {n\left( {n + 2} \right)}$

From this, n = 4 (unpaired electrons)

(A) In octahedral complex: [M(H₂O)₆] ²⁺

C.F.S.E. = (–0.4 $\Delta$₀) × 4 + (+0.6 $\Delta$₀) × 2 + 0 × P
= –0.4 $\Delta$₀

(B) In tetrahedral complex: [M(H₂O)₄] ²⁺


C.F.S.E. = (–0.6 $\Delta$_t) × 3 + (+0.4 $\Delta$_t) × 3 + 0 × P
= –0.6 $\Delta$_t

H₂O(l) ⇌ H₂O(g)

90 gm of H₂O = ${{90} \over {18}}$ moles of H₂O = 5 moles of H₂O

$\Delta$H_vap = $\Delta$U + $\Delta$n_gRT

$\Rightarrow$ $\Delta$U = $\Delta$H_vap - $\Delta$n_gRT

= 41000 - 1$\times$8.314$\times$373

= 37898.878

For 5 moles, $\Delta$U = 37898.878 $\times$ 5 = 189494.39 Joule

Cs has least work function value because of less electronegativity thats why it is used in photoelectric cells.

2, 5-dimethyl-6-oxo-hex-3-enoic acid

Chiral carbon atom is bonded to 4 different atoms or group of atoms. The given structure has 5 chiral carbon-atoms.

Let three terms of G.P. are ${a \over r}$, a, ar

$\therefore$ $a\left( {{1 \over r} + 1 + r} \right)$ = S ...(1)

and a³ = 27

$\Rightarrow$ a = 3

$\therefore$ $3\left( {{1 \over r} + 1 + r} \right)$ = S

$\Rightarrow$ ${{1 \over r} + r = {S \over 3} - 1}$

$\Rightarrow$ As ${{1 \over r} + r \ge 2}$ or ${{1 \over r} + r \le - 2}$

$\therefore$ ${{S \over 3} - 1 \ge 2}$ or ${{S \over 3} - 1 \le - 2}$

$\Rightarrow$ ${{S \over 3} \ge 3}$ or ${{S \over 3} \le - 1}$

$\Rightarrow$ S $\ge$ 9 or S$\le$ -3

$\therefore$ S $\in$ (-$\propto$, -3] $\cup$ [9, $\infty$)