JEE Main 2020 5th September Morning Shift Question Paper with Solutions

$\overrightarrow E = 30\widehat j\sin (1.5 \times {10^7}t - 5 \times {10^{ - 2}}x)V/m$

V = ${{1.5 \times {{10}^2}} \over {5 \times {{10}^{ - 2}}}}$ = 3 $\times$ 10⁸ = C

$\Rightarrow B = E/C = {{30} \over {1.5 \times {{10}^7}}} \times 5 \times {10^{ - 2}}$

$= {10^{ - 7}}Tesla$

$\Rightarrow {F_{max}} = q\left( {\overrightarrow V \times \overrightarrow B } \right) = \left| {qVB} \right|$

$= 1.6 \times {10^{ - 19}} \times 0.1 \times 3 \times {10^8} \times {10^{ - 7}}$

$= 4.8 \times {10^{ - 19}}N$

Applying Angular Momentum conservation

$I\omega = (I + 3I)\omega '$

$\Rightarrow$ $\omega ' = {{I\omega } \over {4I}} = {\omega \over 4}$

${k_i} = {1 \over 2}I{\omega ^2}$

${k_f} = {1 \over 2}(4I){(\omega ')^2}$

$= 2I{\left( {{\omega \over 4}} \right)^2} = {1 \over 8}I{\omega ^2}$

Fractional loss $= {{{K_i} - {K_f}} \over {{K_i}}} = {{{1 \over 2}I{\omega ^2} - {1 \over 8}I{\omega ^2}} \over {{1 \over 2}I{\omega ^2}}}$

= ${{{3 \over 8}I{\omega ^2}} \over {{1 \over 2}I{\omega ^2}}} = {3 \over 4}$

In photodiode, photocurrent increases with increasing biasing voltage and then becomes saturated.

${4 \over 3}\pi \left( {{R^3} - {r^3}} \right){p_m}\,g = {4 \over 3}\pi {R^3}{p_w}\,g$

$1 - {\left( {{r \over R}} \right)^3} = {8 \over {27}}$

$\Rightarrow {r \over R} = {\left( {{{19} \over {27}}} \right)^{1/3}} = {{{{19}^{1/3}}} \over 3}$

$= 0.88 \simeq {8 \over 9}$

${1 \over 2}m{v^2} \times {1 \over 2} = ms\Delta T$

$\Rightarrow$ $\Delta T = {{{v^2}} \over {4 \times 5}} = {{{{210}^2}} \over {4 \times 30 \times 4.200}}$

$= 87.5^\circ C$

Given, S $\propto$ $\Delta$p

and Proportionally constant = 1

We know,

$\Delta$p = S$\beta$k

= $\rho$v² $\times$ ${\omega \over v} \times S$

= ${\rho v\omega S}$

$\therefore$ S = ${{\Delta p} \over {\rho v\omega }}$

= ${{\Delta p} \over {\rho v2\pi f}}$

= ${{\Delta p} \over {\rho vf}}$

[As Proportionally constant = 1 so assume 2$\pi$ = 1]

$= {{10} \over {1 \times 300 \times 1000}}$

$= {1 \over {30}}mm$

$\approx {3 \over {100}}mm$

We know, $\eta = {{{P_{out}}} \over {\left( {{P_{out}} + {P_{loss}}} \right)}} \times 100$

Given, P= $vi = {10^3}$

$\therefore$ $i = {{1000} \over {220}}$ = ${{50} \over {11}}$ A

Power $loss = {i^2}R = {\left( {{{50} \over {11}}} \right)^2} \times 2$

$\therefore$ efficiency $= {{1000} \over {1000 + {i^2}R}} \times 100 = 96\%$

Velocity of helicopter at height h,

$V_B^2 = {0^2} + 2gh$

${V_B} = \sqrt {2gh}$

$- h = ({V_B})t - {1 \over 2}g{t^2}$

$\Rightarrow$ $- h = \sqrt {2ght} - {1 \over 2}g{t^2}$

$\Rightarrow$ $g{t^2} - 2\sqrt {2ght} - 2h = 0$

$\Rightarrow$ $t = {{\sqrt {2ght} \pm \sqrt {8gh + 8gh} } \over {2g}} = {{2\sqrt {2gh} \pm \sqrt {16gh} } \over {2g}} = {{\sqrt {2gh} + 2\sqrt {gh} } \over g}$

$\Rightarrow$ $t = \sqrt {{{2h} \over g}} + 2\sqrt {{h \over g}} = \sqrt {{h \over g}} \left( {\sqrt 2 + 2} \right) = 3.4\sqrt {{h \over g}}$

$E = {3 \over 2}kT \Rightarrow \left( {T = {{2E} \over {3k}}} \right),$

Also $\,PV = NkT$

$P = \rho gh,\,V = 4c{m^3}$

$\therefore$ ($\rho gh$)V = $Nk \times {{2E} \over {3k}}$

$\Rightarrow$ $13.6 \times {10^3} \times 9.8 \times 2 \times {10^{ - 2}} \times 4 \times {10^{ - 6}}$

$= Nk \times {{2E} \over {3k}} = {{N \times 2} \over 3} \times 4 \times {10^{ - 14}} \times {10^{-7}}$

$\Rightarrow$ $N = {{13.6 \times 19.6 \times 4 \times {{10}^{ - 5}} \times 3 \times 10} \over 8}$

$\Rightarrow$ $N = 399.84 \times {10^{16}}$

$= 3.99 \times {10^{18}}$

$\Rightarrow$ $N = 4 \times {10^{18}}$

From $\Delta ABD$

$\tan 45 = {{{h_1}} \over d}$

$\Rightarrow 1 = {{{h_1}} \over d}$

$\Rightarrow {h_1} = d$

From $\Delta ACE$

$\tan 30 = {{{h_1} + {h_2}} \over {d + 2.464d}}$

$\Rightarrow$ $0.5774 = {{d + {h_2}} \over {3.464d}}$

$\Rightarrow$ $d + {h_2} = 0.5774 \times 3.464 \times d$

$\Rightarrow$ ${h_2} = 2.0001136d - d$

$\Rightarrow$ ${h_2} = 2.000d - d = d$

Given, L = |v₀| + |u_e| = 10 cm

(v_e = $\infty$)

${1 \over {{v_e}}} - {1 \over {{u_e}}} = {1 \over {{f_e}}}$

$\Rightarrow$ ${1 \over \infty } - {1 \over {{u_e}}} = {1 \over 5}$

$\Rightarrow$ u_e = -5

$\Rightarrow$ |u_e| = 5 cm

$\therefore$ |v₀| + 5 = 10 cm

$\Rightarrow$ |v₀| = 5 cm

For objective, v₀ = 5 cm, f₀ = 1 cm

${1 \over {{v_0}}} - {1 \over {{u_0}}} = {1 \over {{f_0}}}$

$\Rightarrow$ ${1 \over 5} - {1 \over {{u_0}}} = {1 \over 1}$

$\Rightarrow$ u₀ = $- {5 \over 4}$

$\Rightarrow$ |u₀| = ${5 \over 4}$ = ${{50} \over {40}}$ = ${n \over {40}}$

$\therefore$ n = 50

$\overrightarrow \tau = \overrightarrow r \times F = (3\widehat i + \widehat j - 2\widehat k) \times (\widehat i + 2\widehat j + 3\widehat k)$

= \left| {\matrix{ i & j & k \cr 3 & 1 & { - 2} \cr 1 & 2 & 3 \cr } } \right|

$= \widehat i(3 + 4) - \widehat j(9 + 2) + \widehat k(6 - 1)$

$\overrightarrow \tau = 7\widehat j - 11\widehat j + 5\widehat k$

$\left| {\overrightarrow \tau } \right| = \sqrt {49 + 121 + 25} = \sqrt {195}$

$\therefore$ $x = 195$

By conservation of charge,

${q_1} + {q_2} = {q_1}' + {q_2}'$

$\Rightarrow$ $- CV + (2C)(2V) = (C + 2C)V'$

$V' = {{3CV} \over {3C}} = V$

${U_f} = {1 \over 2}C{v^2} + {1 \over 2}(2C){V^2}$

${U_f} = {3 \over 2}C{V^2}$

${l_1} = l - 17$

${l_2} = l - 24.5$

We know, $v = 2f({l_1} - {l_2})$

$\Rightarrow$ $330 = 2 \times f \times [(f \times [(l - 17) - (l - 24.5)] \times 10^{ - 2}$

$\Rightarrow$ $165 = f \times 7.5 \times {10^{ - 2}}$

$\Rightarrow$ $f = {{165 \times 1000} \over {7.5}}$

$\Rightarrow$ $f = 2200\,Hz$

For all process,

E_AB = E_AC = E_AD

W_AB > 0 as volume increases.

W_AC = 0 as volume constant.

W_AD < 0 as volume decreases.

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

Here M = IA = I(2a)² = 4a²I

B = ${{{\mu _0}I} \over {2\pi b}}$

Angle between B and M = 90^o

$\therefore$ $\tau = \left( {4{a^2}I} \right)\left( {{{{\mu _0}I} \over {2\pi b}}} \right)\sin 90^\circ$

= ${{2{\mu _0}{I^2}{a^2}} \over {\pi b}}$

Given, ${g_{at\,high}} = {g_{at\,depth}}$

We know, ${g_{depth}} =$$g\left( {1 - {d \over R}} \right)$

$\therefore$ $g\left( {1 - {d \over R}} \right) = {{GM_e} \over {{{(R + h)}^2}}}$

$\Rightarrow$ $g\left( {1 - {d \over R}} \right) =$${{G{M_e}} \over {{{\left( {R + {R \over 2}} \right)}^2}}}$ = ${4 \over 9}{{G{M_e}} \over {{R^2}}}$ = ${4 \over 9}g$

$\Rightarrow$ ${d \over R} = 1 - {4 \over 9} = {5 \over 9}$

Given, m_particle = 200 MeV/c² = m(Assume)

and m_H = 1 GeV/c² = 1000 MeV/c² = 5 $\times$ 200 = 5m

Applying momentum conservation,

p_i = p_f

$\Rightarrow$ mv₀ + 0 = 0 + 5 mv'

$\Rightarrow$ v' = ${{{v_0}} \over 5}$

Initial kinetic energy, k_i = ${1 \over 2}mv_0^2$

Final kinetic energy, k_f = ${1 \over 2}\left( {5m} \right){\left( {{{{v_0}} \over 5}} \right)^2}$

$\therefore$ Loss in KE

= ${1 \over 2}mv_0^2$ - ${1 \over 2}\left( {5m} \right){\left( {{{{v_0}} \over 5}} \right)^2}$

= ${4 \over 5}\left( {{1 \over 2}mv_0^2} \right)$ = ${4 \over 5}\left( {{k_i}} \right)$

This lost energy is used by the hydrogen atom to move from ground state to the first excited state. We know the the energy required by the hydrogen atom to move from ground state to first excited state is 10.2 eV.

$\therefore$ ${4 \over 5}\left( {{k_i}} \right)$ = 10.2

$\Rightarrow$ k_i = ${{5 \times 10.2} \over 4} = {{51} \over 4}$

$\therefore$ N = 51

z = ${{{a^2}{b^{{2 \over 3}}}} \over {\sqrt c {d^3}}}$

$\Rightarrow$ ${{dz} \over z} \times 100 = \left( {2{{da} \over a} + {2 \over 3}{{db} \over b} + {1 \over 2}{{dc} \over c} + 3{{d\left( d \right)} \over d}} \right) \times 100$

% error in z

= $\left( {2 \times 2 + {2 \over 3} \times 1.5 + {1 \over 2} \times 4 + 3 \times 2.5} \right)$ %

= ( 4 + 1 + 2 + 7.5 ) %

= 14.5 %

Applying mechanical energy conservation,

${K_A} + {U_A} = {K_B} + {U_B}$

$\Rightarrow$ $0 + mgy + q{V_A} = {1 \over 2}m{v^2} + 0 + ( + q{V_B})$

$\Rightarrow$ $mgy + q{V_A} = {1 \over 2}m{v^2} + q({V_B})$

$\Rightarrow$ $mgy + {{qk(Q)} \over R} = {1 \over 2}m{v^2} + {{qk(Q)} \over {R + y}}$

$\Rightarrow$ ${1 \over 2}m{v^2} = - {{kq(Q)} \over {R + y}} + {{kq(Q)} \over R} + mgy$

$\Rightarrow$ ${{m{v^2}} \over 2} = {{ - kq(Q)R + kq(Q)(R + y)} \over {R(R + y)}} + mgy$

$\Rightarrow$ ${v^2} = {2 \over m}\left[ {{{ - kqQR + kqQR + kqQ} \over {R(R + y)}} + mgy} \right]$

$\Rightarrow$ ${v^2} = {2 \over m}\left[ {{{kq(Q)y} \over {R(R + y)}} + mgy} \right]$

$\Rightarrow$ ${v^2} = 2y\left[ {{{q(Q)} \over {4\pi {\varepsilon _0}R(R + y)m}} + g} \right] = 2y\left[ {{{qQ} \over {4\pi {\varepsilon _0}R(R + y)m}} + g} \right]$

1 = i_g(G + R₁) ....(1)

2 = i_g(R₁ + R₂ + G) ....(2)

Doing (1) $\div$ (2), we get

${1 \over 2} = {{{i_g}\left( {G + {R_1}} \right)} \over {{i_g}\left( {G + {R_1} + {R_2}} \right)}}$

$\Rightarrow$ R₁ + R₂ + G = 2R₁ + 2G

$\Rightarrow$ R₂ = R₁ + G

Given d = 1 $\mathop A\limits^o$

For first maxima, $\theta$ = 60^o

$\therefore$ $\theta$₁ = 90 - ${\theta \over 2}$

= $90 - {{60} \over 2}$ = 60^o

and $2d\sin \theta = \lambda = {h \over {\sqrt {2mE} }}$

$\Rightarrow$ $2 \times {10^{ - 10}} \times {{\sqrt 3 } \over 2} = {{6.6 \times {{10}^{ - 34}}} \over {\sqrt {2mE} }}$

$\Rightarrow$ $E = {1 \over 2} \times {{6.64 \times {{10}^{ - 48}}} \over {9.1 \times {{10}^{ - 31}} \times 3 \times 1.6 \times {{10}^{ - 19}}}} = 50.47$

${B_2} = {{{\mu _0}{I_2}{N_2}} \over {2{R_2}}}$

Total flux $\phi = {N_1}{B_2}\pi {R_1}^2 = {N_1}{N_2}{{{\mu _0}I} \over {2{R_2}}}\pi {R_1}^2$

$\therefore$ $\phi = {{500 \times 200 \times 4\pi \times {{10}^{ - 7}} \times (5{t^2} - 2t - 3)\pi {{({{10}^{ - 2}})}^2}} \over {2 \times 20 \times {{10}^{ - 2}}}}$

= ${{{{10}^5} \times 4{\pi ^2} \times {{10}^{ - 7}}(5{t^2} - 2t + 3) \times {{10}^{ - 4}}} \over {40 \times {{10}^{ - 2}}}}$

$\Rightarrow$ $\phi = (5{t^2} - 2t + 3) \times {10^{ - 4}}$

We know, $e = \left| {{{d\phi } \over {dt}}} \right| = (10t - 2) \times {10^{ - 4}}$

At $t = 1\sec$

$e = 8 \times {10^{ - 4}} = 0.8\,mV = {{0.8} \over {10}} = {4 \over 5}$

$\therefore$ $x = 5$

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

Case-I

If v = u

$\Rightarrow$ f + u = f

$\Rightarrow$ u = 0

Case-II

If u = $\infty$

then v = f

So you can see both of them are equal only when v = 0 and u = 0.

$\Delta$Gº = 17.37 kJ ; n = 3

$\Delta$Gº = -nF$E_{cell}^o$

$\Rightarrow$ $E_{cell}^o$ = $- {{17.37 \times 1000} \over {3 \times 96500}}$

= -0.06 = -6.00 $\times$ 10^-2

Potential energy curve for H₂ molecules is

C₃H₈(g) + 5O₂(g) $\to$ 3CO₂(g) + 4H₂O(l)

C₄H₁₀(g) + ${{13} \over 2}$O₂(g) $\to$ 4CO₂(g) + 5H₂O(l)

$\therefore$ No. of moles of O₂ required to oxidise 1 mole of propane = 5

$\therefore$ No. of moles of O₂ required to oxidise 1 mole of butane = ${{13} \over 2}$

So, No. of moles of O₂ required to oxidise 1 mole of

propane and 2 moles of butane = 5 + 2 $\times$ ${{13} \over 2}$ = 18

An Ellingham diagram provides information about, the temperature dependence of the standard gibbs energies of formation of some metal oxides.

R_n = a₀${{{n^2}} \over Z}$

${{\Delta {R_1}} \over {\Delta {R_2}}} = {{{{\left( {{r_4} - {r_3}} \right)}_{L{i^{2 + }}}}} \over {{{\left( {{r_4} - {r_3}} \right)}_{H{e^ + }}}}}$

= ${{{{{4^2}} \over 3} - {{{3^2}} \over 3}} \over {{{{4^2}} \over 2} - {{{4^2}} \over 2}}}$ = ${{7/3} \over {7/2}} = {2 \over 3}$

$\because$ Given reaction is endothermic.

$\therefore$ On decreasing temperature backward reaction will be favoured.

On adding N₂, pressure is increased at constant T, and volume would also be constant so no change is observed.

So, Ester is unstable in an aqueous basic solution and undergoes hydrolysis to give alcohol and carboxylate.

A_t = A₀.e^-k₁t

B_t = B₀.e^-k₂t

k₁ = ${{\ln 2} \over {300}}$

k₂ = ${{\ln 2} \over {180}}$

Given, A₀ = B₀

and A_t and B_t are related as [A] = 4[B]

$\therefore$ A₀.e^-k₁t = 4B₀.e^-k₂t

$\Rightarrow$ ${e^{\left( {{{\ln 2} \over {180}} - {{\ln 2} \over {300}}} \right)t}}$ = 4

$\Rightarrow$ ${\left( {{{\ln 2} \over {180}} - {{\ln 2} \over {300}}} \right)t}$ = ln 4 = 2ln 2

$\Rightarrow$ ${\left( {{1 \over {180}} - {{\mathop{\rm l}\nolimits} \over {300}}} \right)t}$ = 2

$\Rightarrow$ t = ${{2 \times 180 \times 300} \over {120}}$ = 900 sec

In octahedral

CFSE = [-0.4n_t2g + 0.6n_eg]$\Delta$₀

= [-0.4$\times$4 + 0.6$\times$2]$\Delta$₀

= -0.4 $\Delta$₀

In tetrahedral fields

CFSE = [-0.6n_eg + 0.4n_t2g]$\Delta$_t

= (– 0.6 × 3 + 0.4 × 3)$\Delta$_t

= (–1.8 + 1.2)$\Delta$_t

= –0.6$\Delta$_t

At CMC, micelle formation takes place.

[EDTA]^4– is ethylenediaminetetraacetate anion. It is a hexadentate ligand.

It has six co-ordination sites.

CH₃–CH₂–C$\equiv$N $\buildrel {LiAl{H_4}} \over \longrightarrow$ CH₃–CH₂–CH₂–NH₂

The compound which can donate lone pair more easily is more basic.

(A) N is sp² hybridized and lone pair is localised means not perticipating in resonance.

(B) Here lone pair perticipates in resonance and involved in aromaticity. So this is weakest base.

(C) N is sp³ hybridized and lone pair is localised means not perticipating in resonance. As here N atom is sp³ hybridized so it is more basic than A and D.

(D) Here lone pair of 1^st N atom perticipates in resonance but 2^nd N is sp² hybridized and lone pair is localised means not perticipating in resonance. So basisity will depend on 2^nd N atom. Due to perticipatation in resonance of 1^st N atom some negative charge is genareted on 2^nd N atom which makes it more basic than compound A.

As per (n + l) rule in 6th period, order of orbitals filling is 6s, 4f, 5d, 6p.

CO₂ + H₂O $\to$ H₂CO₃

At 30 bar pressure mass of CO₂ in 1 kg water = 44 gm

At 3 bar pressure mass of CO₂ in 1 kg water = 4.4 gm

$\therefore$ Moles of CO₂ in 1 kg water = ${{4.4} \over {44}}$ = 0.1

	H2CO3	⇌	H+	+	HCO3-
t = 0	0.1		0		0
t = teq	0.1(1 - $\alpha$)		0.1$\alpha$		0.1$\alpha$

4.0 $\times$ 10^–7 = ${{0.1{\alpha ^2}} \over {1 - \alpha }}$

${1 - \alpha }$ $\simeq$ 1

$\Rightarrow$ 0.1${{\alpha ^2}}$ = 4 $\times$ 10^-7

$\Rightarrow$ $\alpha$ = 2 $\times$ 10^-3

[H⁺] = 0.1$\alpha$ = 2 $\times$ 10^-4

$\therefore$ pH = –[– 4 × log(2)] = 3.7 = 37 × 10^–1