JEE Main 2020 4th September Morning Shift Question Paper with Solutions

P_R = 0.5 W

$\Rightarrow$ i²R = 0.5 W

iR = 2.5

$\Rightarrow$ i = 0.2 A & R = 12.5 $\Omega$

Also, V = E – ir

$\Rightarrow$ 2.5 = 3 – (0.2)r

$\Rightarrow$ r = 2.5 $\Omega$

Power dissipated in internal resistance

= i²r = (0.2)²(2.5) = 0.1 W

As there are two Zener diodes in reverse polarity so if one is in forward bias the other will be in reverse bias and above 6V the reverse bias will too be in conduction mode. Therefore when voltage is more than 6V the output will be constant. And when it is less than 6V it will follow the input voltage so correct answer is (C).

${I_f}{\omega _f} = {I_i}{\omega _i}$

${I_i} = {{M{R^2}} \over 2}$

${I_f} = {{M{R^2}} \over 2} + {{M{{(R/2)}^2}} \over 2}$

$= {5 \over 4}.{{M{R^2}} \over 2}$

$\left[ {{{M{R^2}} \over 2} + {M \over 2}{{\left( {{R \over 2}} \right)}^2}} \right]\omega ' = \left( {{{M{R^2}} \over 2}} \right).\omega$

$\Rightarrow$ $\left[ {{{M{R^2}} \over 2}.\left( {{5 \over 4}} \right)} \right]\omega ' = {{M{R^2}} \over 2}\omega$

$\omega = {4 \over 5}\omega$

loss of K.E. = ${{Loss} \over {{K_i}}} \times 100$

= ${{{1 \over 2}I{\omega ^2} - {1 \over 2}\left( {{5 \over 4}I} \right){{\left( {{4 \over 5}\omega } \right)}^2}} \over {{1 \over 2}I{\omega ^2}}}$ $\times$ 100

= ${{{\omega ^2} - {{16} \over {25}}{\omega ^2}\left( {{5 \over 4}} \right)} \over {{\omega ^2}}}$ $\times$ 100 = $\left( {1 - {{80} \over {100}}} \right) \times 100$

= 20%

Given, |${\sigma _ + }$| > |${\sigma _ - }$|

We know, $\left| {{E_ + }} \right| = {{\left| {{\sigma _ + }} \right|} \over {{\varepsilon _0}}}$

and $\left| {{E_ - }} \right| = {{\left| {{\sigma _ - }} \right|} \over {{\varepsilon _0}}}$

$\therefore$ $\left| {{E_ + }} \right|$ > $\left| {{E_ - }} \right|$

Thin infinite uniformly charged planes produces uniform electric field therefore option A and C are obviously wrong.

And as positive charge density is bigger in magnitude so its field along Y direction will be bigger than field of negative charge in X direction and this is evident in option B so it is correct.

All collisions are perfectly inelastic, so after the final collision, all blocks are moving together. So let the final velocity be v', so on applying momentum conservation :

mv = 16mv'

$\Rightarrow$ v' = ${v \over {16}}$

Now initial energy E_i = ${1 \over 2}m{v^2}$

Final energy : E_f = ${1 \over 2} \times 16m{\left( {{v \over {16}}} \right)^2}$

Energy loss : E_i – E_f

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

= ${1 \over 2}m{v^2}\left[ {1 - {1 \over {16}}} \right]$

= ${1 \over 2}m{v^2}\left[ {{{15} \over {16}}} \right]$

Total energy loss is p% =

Energy loss

Original energy

$\times$ 100

= ${{{1 \over 2}m{v^2}\left[ {{{15} \over {16}}} \right]} \over {{1 \over 2}m{v^2}}} \times 100$

= 93.75% $\simeq$ 94 %

$\therefore$ ${{d\theta } \over {dt}} = kA{{dT} \over {dx}}$

$\Rightarrow$ k = ${{\left( {{{d\theta } \over {dt}}} \right)} \over {A\left( {{{dT} \over {dx}}} \right)}}$

$\Rightarrow$ [k] = ${{\left[ {M{L^2}{T^{ - 3}}} \right]} \over {\left[ {{L^2}} \right]\left[ {K{L^{ - 1}}} \right]}}$

= [MLT^–3K^–1]

As work done on gas and heat supplied to the gas are zero,

$\therefore$ Total internal energy of gases remain same.

u₁ + u₂ = u₁' + u₂'

We know, $\Delta$U = nC_v$\Delta$T

$\therefore$ $\left( {0.1 \times {{3R} \over 2} \times 200} \right) + \left( {0.05 \times {{3R} \over 2} \times 400} \right)$ = $\left( {0.15 \times {{3R} \over 2} \times {T_f}} \right)$

$\Rightarrow$ (20 + 20) = 0.15 T_f

$\Rightarrow$ T_f = 266.67

Let $\,{I_0} = km{l^2}$

${I_{DEF}} = I' = K\left( {{m \over 4}} \right){\left( {{l \over 2}} \right)^2}$

= ${{m{l^2}} \over {16}}$$= \left( {{{{I_0}} \over {16}}} \right)$

and ${I_{ADE}} = {I_{BDE}} = {I_{EFC}} = I$

$\therefore$ 3I + I' = I₀

$\Rightarrow$ $3I = {I_0} - {{{I_0}} \over {16}} = {{15{I_0}} \over {16}}$

$\Rightarrow I = {{5{I_0}} \over {16}}$

${I_{remaining}} = 2I + {{{I_0}} \over {16}} = {{11{I_0}} \over {16}}$

V, h curve will be parabolic.

Downward velocity is negative and upward is positive

When ball is coming down graph will be in IV quadrant and when going up graph will be in I quadrant.

At H = h, v = 0

at h = 0, v = $\sqrt {2gh}$

also a = –g, throughout this motion.

Torque on a bar magnet :

$\tau = MB\,\sin \theta$

Here, $\theta$ = 30º, I = 0.018 N-m, B = 0.06 T

$0.018 = M \times 0.06 \times 0.5$

$\Rightarrow M = 0.6\,A{m^2}$

$W = {U_f} - {U_i}$

$= MB(\cos {\theta _i} - \cos {\theta _f})$

$= 0.6 \times 0.06(1 - ( - 1))$

$= 7.2 \times {10^{ - 2}}\,J$

Given ${E_x} = {{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}$

$\therefore$ ${{ - dV} \over {dx}} = {{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}$

$\Rightarrow$ $\int\limits_0^V {dV} = - \int\limits_\infty ^x {{{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}dx}$

$\Rightarrow$ $V = {A \over {{{({x^2} + {a^2})}^{1/2}}}}$

Using formula, $B = {{{\mu _0}i} \over {2r}}\left( {{\theta \over {2\pi }}} \right)$

${B_A} = {{\mu (2)\left( {{{3\pi } \over 2}} \right)} \over {2(a)(2\pi )}} = {{3\mu } \over {4a}}$

${B_B} = {{\mu (3)\left( {{{5\pi } \over 3}} \right)} \over {2(2a)(2\pi )}} = {{5\mu } \over {8a}}$

$\therefore$ ${{{B_A}} \over {{B_B}}} = {{3\mu } \over {4a}} \times {{8a} \over {5\mu }} = 6:5$

Intensity, I = 3.3 Wm^–3

Area, A = 3 × 10^–4 m²

Angular speed, $\omega$ = 31.4 rad/s

$I = {I_0}{\cos ^2}(\omega t)$

$\Rightarrow {I_{av}} = {{{I_0}} \over 2}$

$\because$ $\left\langle {{{\cos }^2}\theta } \right\rangle$ = ${1 \over 2}$, in one time period

$\therefore$ $E = {{{I_0}} \over 2} \times A \times (\Delta t)$

and $\Delta t = {{2\pi } \over \omega } = {{2 \times 3.14} \over {31.4}} = {1 \over 5}s$

$\therefore$ $E = {{3.3} \over 2} \times 3 \times {10^{ - 4}} \times {1 \over 5} = 1 \times {10^{ - 4}}\,J$

Applying momentum conservation

${m \over 2} \times {V_0} + {m \over 3} \times 0 = {m \over 2}{V_A} + {m \over 3}{V_B}$

$\Rightarrow$ ${{{V_0}} \over 2} = {{{V_A}} \over 2} + {{{V_B}} \over 3}$ .....(1)

Since, collision is elastic (e = 1)

e = 1 = ${{{V_B} - {V_A}} \over {{V_0}}}$

$\Rightarrow$ V₀ = V_B – V_A .....(2)

On solving (1) & (2) :

V_A = ${{{V_0}} \over 5}$

Now, De-Broglie wavelength of A before collision :

$\lambda$₀ = ${h \over {{m_A}{V_0}}}$

= ${h \over {\left( {{m \over 2}} \right){V_0}}}$

= ${{2h} \over {m{V_0}}}$

Final De-Broglie wavelength :

$\lambda$_f = ${h \over {{m_A}{V_A}}}$

= ${h \over {\left( {{m \over 2}} \right)\left( {{{{V_0}} \over 5}} \right)}}$ = ${{10h} \over {m{V_0}}}$

Now, $\Delta$$\lambda$ = $\lambda$_f - $\lambda$₀

= ${{10h} \over {m{V_0}}}$ - ${{2h} \over {m{V_0}}}$

$\Rightarrow$ $\Delta$$\lambda$ = ${{8h} \over {m{V_0}}}$

$\Rightarrow$ $\Delta$$\lambda$ = $4 \times {{2h} \over {m{V_0}}}$

$\Rightarrow$ $\Delta$$\lambda$ = 4$\lambda$₀

K_max = hf - $\phi$

$\Rightarrow$ eV₀ = hf - $\phi$

at B V₀ = 0, f = 5.5

$\therefore$ $\phi$ = hf

= ${6.62 \times {{10}^{ - 34}} \times 5.5 \times {{10}^{14}}}$ J

= ${{6.62 \times {{10}^{ - 34}} \times 5.5 \times {{10}^{14}}} \over {1.6 \times {{10}^{ - 19}}}}$ eV

= 2.27 eV

L = 20, f₀ = 1cm, M = 100

$M = {{{v_0}} \over {{u_0}}}\left( {1 + {D \over {{f_e}}}} \right)$

$\therefore$ $M = {L \over {{f_0}}}\left( {1 + {D \over {{f_e}}}} \right)$ [v₀ $\approx$ L, u₀ $\approx$ f₀]

$\Rightarrow$ ${{20} \over 1}\left( {1 + {{25} \over {{f_e}}}} \right)$ = 100

on solving we get

f_e = 6.25 cm

$y = {u_y}t + {1 \over 2}{a_y}{t^2}$

$y = 5t + {1 \over 2}(4){t^2}$

$y = 5t + 2{t^2}$

and $x = 0(t) + {1 \over 2}(10)({t^2}) = 20$

$t = 2s$

$\Rightarrow y = 10 + 8 = 18m$

When bar magnet is entering with constant speed, flux will change and an e.m.f. is induced, so galvanometer will deflect in positive direction.

When magnet is completely inside, flux will not change, so reading of galvanometer will be zero.

When bar magnet is making on exit, again flux will change and on e.m.f. is induced in opposite direction to not of (a), so galvanometer will deflect in negative direction.

$\Delta U = {1 \over {4\pi {\varepsilon _0}}}.{{4q.q} \over {(3d/2)}} - {1 \over {4\pi {\varepsilon _0}}}.{{4q.q} \over {(d/2)}}$

$= {{4{q^2}} \over {4\pi {\varepsilon _0}}}\left( {{2 \over d}} \right)\left( { - {2 \over 3}} \right)$

= decrease by ${{4{q^2}} \over {3\pi {\varepsilon _0}d}}$

$1.5 = \left( {2{n_1} + 1} \right){\lambda \over 2}$ ......(1)

5 = n₂$\lambda$ .....(2)

(1) $\div$ (2)

${{1.5} \over 5} = {{\left( {2{n_1} + 1} \right)} \over {2{n_2}}}$

$\Rightarrow$ 3n₂ = 10n₁ + 5

n₁ = 1 ; n₂ = 5 $\Rightarrow$ $\lambda$ = 1

n₁ = 4 ; n₂ = 15 $\Rightarrow$ $\lambda$ = 1/3

n₁ = 7 ; n₂ = 25 $\Rightarrow$ $\lambda$ = 1/5

Heat loss by water

$Q = {m_w}s\Delta \theta$

$= \left( {{{200} \over {1000}}} \right).(4200)(25) = 21000\,J$

This heat will absorbed by the ice and let mass $\Delta$m_i got melted.

So $\Delta {m_i}L = 21000$

$\Delta {m_i} = {{21000} \over {3.4 \times {{10}^5}}} \times {10^3}\,gm = 61.7\,grams$

Decreasing order of wavelength,

Radio wave > Microwave > Infrared >
ROYGBIV(Visiable Resion) > Ultraviolet > X rays > $\gamma$ rays

So the correct order -

$\lambda$_{radio waves} > $\lambda$_{micro waves} > $\lambda$_visible > $\lambda$_x-rays

B - mg = ma

$\Rightarrow$ m = ${B \over {g + a}}$

= ${{V{\rho _w}g} \over {g + a}}$

= ${{V{\rho _w}} \over {1 + {a \over g}}}$

= ${{{4 \over 3}\pi {{\left( 1 \right)}^3} \times 1} \over {1 + {{9.8} \over {980}}}}$

= 4.15 gm

For Lyman series :

Shortest wavelength,

$\lambda$_s = $R{.1^2}\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)$

longest wavelength,

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

Given, $\lambda$_l - $\lambda$_s = 304 $\mathop A\limits^0$

$\therefore$ $R{.1^2}\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$ - $R{.1^2}\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)$ = 304

$\Rightarrow$ ${1 \over R}$ = 912 $\mathop A\limits^0$

For Paschen series :

Shortest wavelength,

$\lambda$_s = $R{.1^2}\left( {{1 \over {{3^2}}} - {1 \over {{\infty ^2}}}} \right)$ = ${R \over 9}$

longest wavelength,

$\lambda$_l = $R{.1^2}\left( {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right)$ = ${{7R} \over {144}}$

$\therefore$ $\lambda$_l - $\lambda$_s

= ${{144} \over {7R}} - {9 \over R}$

= $\left( {{{144} \over 7} - 9} \right){1 \over R}$

= $\left( {{{144} \over 7} - 9} \right) \times 912$ = 10553 $\mathop A\limits^0$

In the hydrogen spectrum,
Balmer series lies in visible region.

N₂(g) + 3H₂(g) $\to$ 2NH₃(g)

Number of moles of N₂ = ${{2.8 \times {{10}^3}} \over {28}}$ = 100

Number of moles of H₂ = ${{1000} \over 2}$ = 500

Here N₂ is limiting reagent.

$\therefore$ Number of moles of NH₃ produced = 2 $\times$ 100 = 200

Mass of NH₃ produced = 200 × 17 = 3400 gm

Pb(NO₃ )₂ $\to$ PbO + NO₂[(A) Brown gas] + O₂

2NO₂ $\buildrel {cooling} \over \longrightarrow$ N₂O₄[(C) colourless solid]

N₂O₄ + NO $\buildrel \Delta \over \longrightarrow$ N₂O₃ [(C) Blue Solid]

Let oxidation state of N in N₂O₃ is x.

$\therefore$ 2x + 3 (–2) = 0

$\Rightarrow$ x = +3

Species with same number of unpaired electrons have equal magnetic moment.

Complex	Ligand	Ligand Type	Number of unpaired electrons
[Mn(H2O)6]2+	H2O	Weak Field Ligand	5
[Cr(H2O)]2+	H2O	Weak Field Ligand	4
[CoCl4]2–	Cl	Weak Field Ligand	3
[Fe(H2O)6]2+	H2O	Weak Field Ligand	4
[Co(OH)4]2–	OH	Weak Field Ligand	3
[Fe(NH3)6]2+	NH3	Weak Field Ligand	4

Actinoids contains 14 elements with atomic number 90 to 103. Hence element with atomic number 101 is Actinoids.

Element with atomic number 104 is a d-block element of group 4.

Maltose contains one acetal and one hemiacetal.

Let X₁ and P$P_1^o$ are the mole fraction and vapour pressure of n-hexane in solution and X₂ and $P_2^o$ are the mole fraction and vapour pressure of n-heptane in solution then

550 = $P_1^o$$\times$${1 \over 4}$ + $P_2^o$$\times$${3 \over 4}$

$\Rightarrow$ 2200 = $P_1^o$ + 3$P_2^o$ ....(1)

On addition of 1 more mole of n-heptane

560 = $P_1^o$$\times$${1 \over 5}$ + $P_2^o$$\times$${4 \over 5}$

$\Rightarrow$ 2800 = $P_1^o$ + 4$P_2^o$ ....(2)

From (1) and (2),

$P_1^o$ = 400, $P_2^o$ = 600

At equilibrium,

rate of forward reaction = Rate of backward reaction

$\therefore$ Stability Cation B > A > C > D

From the given graph, potential energy of A-B molecule is minimum.

Thus A-B bond is most stable as it requires more energy to dissociate and have strongest bond amongst these.

B = Most electronegative

D = Least electronegative

A - B = Shortest bond length

A - B = Largest bond enthalpy

Total 3 geometrical isomers are possible.

    4-bromo-2-methyl cyclopentane carboxylic acid.

(a) CaCO₃ $\buildrel \Delta \over \longrightarrow$ CaO + CO₂ {In Blast furnace}

(b) In extraction of silver, silver is extracted as an anionic complex [Ag(CN)₂]^–.

Ag/Ag₂S + CN^- $\to$ [Ag(CN)₂] ^–

(c) Ni is purified by mond's process

Ni + 4CO $\buildrel {330 - 350K} \over \longrightarrow$ Ni(CO)₄

Ni(CO)₄ $\buildrel {450 - 470K} \over \longrightarrow$ Ni + 4CO

(d) Zr and Ti are purified by van arkel method

Among isoelectronic species, greater the Z_eff smaller will be the radius.

Order of Z_eff : Mg²⁺ > Na⁺ > F^– > O^2–

Order of Ionic Radii : O^2– > F^– > Na⁺ > Mg²⁺

$E_{cell}^0$ = $E_{C{u^{2 + }}|Cu}^0$ - $E_{Z{n^{2 + }}|Zn}^0$

= 0.34 – (–0.76)

= 1.10 V

If E_ext < 1.1 V then Zn dissolves at anode and copper deposits at Cathode.

If E_ext > 1.1V then Zn deposited at zinc electrodes and Cu deposits at Cu electrode.

For 1 mole of ideal gas :

1. Both internal energy (U) and Enthalpy (H) depends on temperature

2. Compressibility factor Z = 1

3. C_{P, m} – C_{V, m} = R

4. dU = C_VdT for all process

Adenine and lysine both have primary anline react with CHCl₃ + alc. KOH.

5H₂O₂ + 2MnO₄^- + 6H⁺ $\to$ 2Mn²⁺ + 5O₂ + 8H₂O

Moles of KMnO₄ = ${{0.316} \over {158}}$ = 2 $\times$ 10^-3

Equivalents of H₂O₂ = Equivalent of KMnO₄

= 2 × 10^–3 × 5 = 0.01

Moles of H₂O₂ = ${{0.01} \over 2}$ = 0.005

Mass of pure H₂O₂ = 0.005 × 34 = 0.170 gm

Percentage purity = ${{0.17} \over {0.2}}$ $\times$ 100 = 85 %

Foam $\to$ Froth, whipped cream, soaplather

Gel $\to$ Cheese, butter, jellies

Aerosol $\to$ smoke dust

Emulsion $\to$ milk

t_75% = 90 min = 2 × t_1/2

$\Rightarrow$ t_1/2 = 45 min

Rate constant, K = ${{0.693} \over {45}}$ min^-1

Time for completion of 60% of the reaction,

t_60% = ${{2.303} \over K}\log {{10} \over 4}$

= ${{2.303 \times 45} \over {0.693}}\log 2.5$

= 60 min

4Li + O₂ $\to$ 2Li₂O

2Na + O₂ $\to$ Na₂O₂

K + O₂ $\to$ KO₂

$1 + (1 - {2^2}.1) + (1 - {4^2}.3) + (1 - {6^2}.5) + ....(1 - {20^2}.19)$

$S = 1 + \sum\limits_{r = 1}^{10} {\left[ {1 - {{(2r)}^2}(2r - 1)} \right] = 1 + \sum\limits_{r = 1}^{10} {\left( {1 - 8{r^3} + 4{r^2}} \right)} = 1 + 10 - } \sum\limits_{r = 1}^{10} {\left( {8{r^3} - 4{r^2}} \right)}$

$= 11 - 8{\left( {{{10 \times 11} \over 2}} \right)^2} + 4 \times \left( {{{10 \times 11 \times 21} \over 6}} \right)$

$= 11 - 2 \times {(110)^2} + 4 \times 55 \times 7$

$= 11 - 220(110 - 7)$

$= 11 - 220 \times 103 = \alpha - 220\beta$

$\Rightarrow \alpha = 11,\beta = 103$

${[x]^2} + 2[x + 2] - 7 = 0$

$\Rightarrow$ ${[x]^2} + 2[x] + 4 - 7 = 0$

Using the property [x + n] = [x] + n ; n $\in$ I

$\Rightarrow$ ${[x]^2} + 2[x] - 3 = 0$

let [x] = y

${y^2} + 3y - y - 3 = 0$

$\Rightarrow$ $(y - 1)(y + 3) = 0$

$[x] = 1\,or\,[x] = - 3$

$\therefore$ $x \in \left[ {1,2} \right)\,\& \, \in \left[ { - 3, - 2} \right)$