JEE Main 2021 20th July Evening Shift Question Paper with Solutions

L = Length of escalator

${V_{b/esc}} = {L \over {{t_1}}}$

When only escalator is moving.

${V_{esc}} = {L \over {{t_2}}}$

when both are moving

${V_{b/g}} = {V_{b/esc}} + {V_{esc}}$

${V_{b/g}} = {L \over {{t_1}}} + {L \over {{t_2}}} \Rightarrow \left[ {t = {L \over {{V_{b/g}}}} = {{{t_1}{t_2}} \over {{t_1} + {t_2}}}} \right]$

${1 \over 2}I{\omega ^2} = {1 \over 2} \times {1 \over 2}m{v^2}$

$I = {1 \over 2}m{R^2}$

Body is solid cylinder

P = constant

${1 \over 2}$mv² = Pt

$\Rightarrow$ v $\propto$ $\sqrt t$

${{dx} \over {dt}} = C\sqrt t$ [C = constant]

by integration.

$x = C{{{t^{{1 \over 2} + 1}}} \over {{1 \over 2} + 1}}$

$x \propto {t^{3/2}}$

$m \propto {t^a}{v^b}{l^c}$

$m \propto {[T]^a}{[L{T^{ - 1}}]^b}{[M{L^2}{T^{ - 1}}]^c}$

${M^1}{L^0}{T^0} = {M^c}{L^{b + 2c}}{T^{a - b - c}}$

comparing powers

c = 1, b = $-$2, a = $-$1

$m \propto {t^{ - 1}}{v^{ - 2}}{l^1}$

P = Vi

0.5 = 8i

i = ${1 \over {16}}$A

E = 20 = 8 + i R_p

R_p = 12 $\times$ 16 = 192$\Omega$

X_L = X_C (due to resonance)

Z = R so ${i_{rms}} = {V \over Z} = {V \over R}$

${{{V^2}} \over R} = {{250 \times 250} \over 5} = 125 \times {10^2}W$

When switch S₁ and S₂ are closed


R_ab = ${{12 \times 6} \over {12 + 6}} + 2 + {{6 \times 12} \over {6 + 12}}$

= ${{72} \over {18}} + 2 + {{72} \over {18}} = 4 + 2 + 4 = 10\Omega$

${T^2} \propto {R^3}$

$T = k{R^{3/2}}$

${{dT} \over T} = {3 \over 2}{{dR} \over R}$

$= {3 \over 2} \times 0.02 = 0.03$

% Change = 3%

PV = nRT

PV $\propto$ T

Straight line with positive slope (nR)

${t_a} = {1 \over 2}{t_d}$

$\sqrt {{{2s} \over {{a_a}}}} = {1 \over 2}\sqrt {{{2s} \over {{a_d}}}}$ ..... (i)

${a_a} = g\sin \theta + \mu g\cos \theta$

$= {g \over 2} + {{\sqrt 3 } \over 2}\mu g$

${a_d} = g\sin \theta - \mu g\cos \theta$

$= {g \over 2} - {{\sqrt 3 } \over 2}\mu g$

using the above values of a_a and a_d and putting in equation (i) we will gate $\mu = {{\sqrt 3 } \over 5}$

$KE = {{hc} \over \lambda } - \phi hc$

$e(3{V_0}) = {{hc} \over \lambda } - \phi$ ..... (i)

$e{V_0} = {{hc} \over {2\lambda }} - \phi$ ..... (ii)

Using (i) & (ii)

$\phi = {{hc} \over {4{\lambda _0}}} = {{hc} \over {{\lambda _t}}}$

${\lambda _t} = 4{\lambda _0}$

$f = {f_0}\sqrt {{{1 + \beta } \over {1 - \beta }}}$

$\beta = {v \over c}$

${f \over {{f_0}}} = \sqrt {{{1 + \beta } \over {1 - \beta }}}$

${\left( {1 + {{\Delta f} \over {{f_0}}}} \right)^2} = (1 + \beta ){(1 - \beta )^{ - 1}}$

$\beta$ is small compared to 1

$\left( {1 + {{2\Delta f} \over {{f_0}}}} \right) = (1 + 2\beta )$

$\beta = {{\Delta f} \over {{f_0}}} = {v \over c}$

$v = 6 \times {c \over {5890}} = 305.6$ km/s

Direction of propagation = $\overrightarrow E \times \overrightarrow B = \widehat i \times \widehat k = - \widehat j$

R = 100$\Omega$

${X_L} = \omega L = 50\pi \times {10^{ - 3}}$

${X_C} = {1 \over {\omega C}} = {{{{10}^{11}}} \over {100\pi }}$

${X_C} > > {X_L}$

& $\left| {{X_C} - {X_L}} \right| > > R$

$\left| {\overrightarrow P } \right| = \left| {\overrightarrow Q } \right| = x$ ..... (i)

$\left| {\overrightarrow P + \overrightarrow Q } \right| = n\left| {\overrightarrow P - \overrightarrow Q } \right|$

${P^2} + {Q^2} + 2PQ\cos \theta = {n^2}({P^2} + {Q^2} - 2PQ\cos \theta )$

Using (i) in above equation

$\cos \theta = {{{n^2} - 1} \over {1 + {n^2}}}$

$\theta = {\cos ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} \right)$

I in both cases is about point of contact

Ring

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

mgh = $= {1 \over 2}(2m{R^2}){{v_R^2} \over {{R^2}}}$

${v_R} = \sqrt {gh}$

Solid cylinder

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

mgh $= {1 \over 2}\left( {{3 \over 2}m{R^2}} \right){{v_C^2} \over {{R^2}}}$

${v_C} = \sqrt {{{4gh} \over 3}}$

${{{v_R}} \over {{v_C}}} = {{\sqrt 3 } \over 2}$

R = R₀e^{$-$$\lambda$t}

lnR = lnR₀ $-$ $\lambda$t

$-$$\lambda$ is slope of straight line

$\lambda$ = ${3 \over {20}}$

${t_{1/2}} = {{\ln 2} \over \lambda } = 4.62$

{N_0}\buildrel {{{{t_1}} \over 2}} \over \longrightarrow {{{N_0}} \over 2}\buildrel {{{{t_1}} \over 2}} \over \longrightarrow {{{N_0}} \over 4}\buildrel {{{{t_1}} \over 2}} \over \longrightarrow {{{N_0}} \over 8}\buildrel {{{{t_1}} \over 2}} \over \longrightarrow {{{N_0}} \over {16}}

$4 \times {t_{1/2}} = 80$

${t_{1/2}} = 20$ days

$\lambda = {h \over {mv}}$

kinetic energy, ${{{P^2}} \over {2m}} = {{{h^2}} \over {2m{\lambda ^2}}} = {{hc} \over {{\lambda _c}}}$

${\lambda _c} = {{2mc{\lambda ^2}} \over h}$

$\gamma = 1 + {2 \over f}$

$\Rightarrow$ $f = {2 \over {\gamma - 1}}$

In a binary star system, the two stars orbit around a common center of mass. When considering periods of revolution, Kepler's Third Law comes into play. This law states that the square of the period of revolution (T) is proportional to the cube of the semi-major axis (r) of the orbit. It's often written in the following form for a single object orbiting another:

T² ∝ r³

For a binary star system, this would still hold true. The periods of revolution for both stars A and B will be the same because they are both orbiting the same common center of mass, regardless of their individual masses or individual orbital radii. In other words, star A and star B complete one orbit in the same amount of time.

So, Option B: T_A = T_B is correct.

The other options (Option A, C, and D) would not be correct. Kepler's Third Law is not a ratio between the periods and radii of two different bodies, and the periods do not depend on the masses of the individual stars or their individual distances from the center of mass.

$\mu$ = $\mu$₀ (1 + x_m)

= 4$\pi$ $\times$ 10^$-$7 $\times$ 500

= 2$\pi$ $\times$ 10^$-$4 H/m

K₂ = 4K₁

${1 \over 2}$mv$_2^2$ = 4${1 \over 2}$mv$_1^2$

v₂ = 2v₁

P = mv

P₂ = mv₂ = 2mv₁

P₁ = mv₁

% change = ${{\Delta P} \over {{P_1}}} \times 100 = {{2m{v_1} - m{v_1}} \over {m{v_1}}} \times 100 = 100\%$

The volume of a sphere is given by $\frac{4}{3}\pi R^3$.

So the volume of the two small mercury drops each of radius $R$ is $2\times \frac{4}{3}\pi R^3$.

When they coalesce to form a larger drop, the volume is conserved. So, the volume of the larger drop is also $2\times \frac{4}{3}\pi R^3$.

Let's denote the radius of this large drop as $R'$.

Therefore, $2\times \frac{4}{3}\pi R^3 = \frac{4}{3}\pi {R'}^3$.

Solving for $R'$, we get $R' = 2^{1/3}R$.

Now, the surface energy of a sphere is proportional to its surface area, and the surface area of a sphere is given by $4\pi R^2$.

So, the ratio of total surface energy before and after the change is:

$\frac{{2\times 4\pi R^2}}{{4\pi (2^{1/3}R)^2}} = \frac{{2}}{{2^{2/3}}} = {2^{1/3}}:1$

Process of isothermal

$W = nRT\ln \left( {{{{V_2}} \over {{V_1}}}} \right)$

= 1 $\times$ 8.3 $\times$ 300 $\times$ ln2

= 17258 $\times$ 10^$-$1 J

In presence of SFL $\Delta$₀ > P means pairing occurs therefore

For Fe⁺² $\to$ 3d⁶



$\therefore$ No of unpaired e^-(s) = 0

$\therefore$ $\mu$ = $\sqrt {n(n + 2)}$BM = 0

[n = No of unpaired e^$-$(s)]

In NiCl₂, Ni⁺² is having configuration 3d⁸

$\therefore$ Number of unpaired electron = 2

After formation of oxidised product

[Ni(CN)₆]^$-$2$\Rightarrow$ Ni⁺⁴ is obtained

Ni⁺⁴ $\Rightarrow$ 3d⁶ and CN^$-$ is strong field ligand

$\therefore$ Number of unpaired electrons = 0

$\therefore$ The charge is 2 $-$ 0 = 2

According to NCERT only NO₂ from the given options can retard the photosynthesis process in plants.

Mass of NaOH = x

Moles of NaOH = ${x \over {40}}$

Mass of Na₂CO₃ = y

Moles of Na₂CO₃ = ${y \over {106}}$

${x \over {40}} = {y \over {106}}$

x + y = 4

x = 1.1, y = 2.9

x = 1.1 $\approx$ 1 (nearest integer)

Reagent for nitration of Benzene

$\mathop {{H_2}S{O_4}}\limits_{(Acid)} + \mathop {HN{O_3}}\limits_{(Base)} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} HSO_4^\Theta + {H_2}\mathop N\limits^ \oplus {O_3}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="katex-base"><span class="katex-strut" style="height:2.126em;vertical-align:-1.116em;"></span><span class="mop op-limits"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.6833em;"><span style="top:-2.159em;margin-left:0em;"><span class="pstrut" style="height:3em;"></span><span class="katex-sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mopen mtight">(</span><span class="mord mathnormal mtight">A</span><span class="mord mathnormal mtight">c</span><span class="mord mathnormal mtight">i</span><span class="mord mathnormal mtight">d</span><span class="mclose 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displaystyle="false"><mfrac><mrow><mstyle mathcolor="#cc0000"><mtext>\buildrel</mtext></mstyle><mstyle scriptlevel="0" displaystyle="false"><mo>→</mo></mstyle></mrow><mrow><mpadded height="0px" depth="0px"><mo lspace="0em" rspace="0em">←</mo></mpadded><mpadded width="0px"><mphantom><msub><mrow></mrow><mrow><mstyle mathcolor="#cc0000"><mtext>\vbox</mtext></mstyle><mi>t</mi><mi>o</mi><mn>.5</mn><mi>e</mi><mi>x</mi><mstyle mathcolor="#cc0000"><mtext>\vss</mtext></mstyle></mrow></msub></mphantom></mpadded></mrow></mfrac></mstyle></mrow></mo><msub><mi>H</mi><mn>2</mn></msub><mi>O</mi><mo>+</mo><mover><mo><mi>N</mi></mo><mo>⊕</mo></mover><msub><mi>O</mi><mn>2</mn></msub></mrow><annotation encoding="application/x-tex">{H_2}\mathop N\limits^ \oplus {O_3} \mathbin{\lower.3ex\hbox{\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H_2}O + \mathop N\limits^ \oplus {O_2}

Organic compound is heated with fuming nitric acid in the presence of silver nitrate in carius method.

Lunar caustic (AgNO₃) is used as reagent hare to distinguish Cl^$-$, Br^- and I^$-$ respectively as follows.

$C{l^ - }(aq)\buildrel {AgN{O_3}} \over \longrightarrow AgCl{ \downarrow _{ppt}}$ white

$B{r^ - }(aq)\buildrel {AgN{O_3}} \over \longrightarrow AgBr{ \downarrow _{ppt}}$ pale yellow

${I^ - }(aq)\buildrel {AgN{O_3}} \over \longrightarrow AgI{ \downarrow _{ppt}}$ Dark yellow

$\Delta$G = $-$49.4 kJ/mol

$\Delta$H = 51.4 kJ/mol

$\Delta$G = $\Delta$H $-$ T$\Delta$S

$-$49400 = 51400 $-$ 300$\Delta$S

$\Delta S = {{ + 100800} \over {300}} = 336$ JK^$-$1 mol^$-$1

Wavelength of electron is given by

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

Here q = charge on electron, V = potential difference

$\lambda = {{6.63 \times {{10}^{ - 34}}} \over {\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}} \times 40 \times {{10}^3}} }}$

$= {{6.63 \times {{10}^{ - 34}}} \over {\sqrt {1164.8 \times {{10}^{ - 47}}} }} = 6.144 \times {10^{ - 12}} \approx 6 \times {10^{ - 12}}$

x = 6

OR

$\lambda = {{12.3} \over {\sqrt V }}\mathop A\limits^o$

$= {{12.3} \over {200}} = 6.15 \times {10^{ - 12}}$ m

$\therefore$ Ans. is 6

[Ni(CN)₆]^2-

Ni⁺⁴ $\to$ d⁶, CN^- strong field ligand. So pairing will happen.



Here zero unpaired electron

NiCl₂ $\to$ Ni²⁺ $\to$ d⁸

Cl^- Weak field ligand.$\to$ two unpaired e^-

$\therefore$ Change = 2.

Hybridisation = Number of sigma bonds + number of lone pair of electrons + number of coordinate bond.

For $NO_2^ -$ hybridisation = 2 sigma NO bonds + 1 lone pair on N = 3; so $NO_2^ -$ is sp²-hybridised.


For $NO_2^ +$ hybridisation = 2 sigma NO bonds = 2; so $NO_2^ +$ sp-hybridised.


For $NO_4^ +$ Hybridisation = 4 sigma NH bonds = 4; so $NO_4^ +$ is sp³-hybridised.


Therefore, hybridisation of N in $NO_2^ -$, $NO_2^ +$, $NO_4^ +$ are sp², sp, sp³ respectively.

x_A = 0.6

P_T = x_AP_A^o + x_BP_B^o

= 0.6 $\times$ 90 + 0.4 $\times$ 15

= 54 + 6 = 60

x_AP_A^o = y_AP_T

0.6 $\times$ 90 = y_A(60)

$\Rightarrow$ y_A = 0.9

y_B = 0.1 = 1 $\times$ 10^$-$1

$\therefore$ x = 1

Novolac (phenol formaldehyde Resin) $\to$ Bakelite

Conc. of Cl^$-$ = 0.1 M = 10^$-$1 M

Conc. of CrO$_4^{2 - }$ = 0.001 M = 10^$-$3 M

K_sp(AgCl) = [Ag⁺][Cl^$-$]

[Ag⁺]_AgCl = ${{1.7 \times {{10}^{ - 10}}} \over {{{10}^{ - 1}}}} = 1.7 \times {10^{ - 9}}$

${K_{sp}}(A{g_2}Cr{O_4}) = {[A{g^ + }]^2}[CrO_4^{2 - }]$

$[A{g^ + }] = \sqrt {{{1.9 \times {{10}^{ - 12}}} \over {{{10}^{ - 3}}}}} = \sqrt {19} \times {10^{ - 4}}$

$\therefore$ AgCl will be precipitate first

Enzymes are mostly proteins. They function as catalysts in biochemical reactions by lowering the energy of activation. They are highly specific w.r.t. temperature and pH in their action.

Moles of Br = Moles of AgBr obtained

$\Rightarrow$ Mass of Br = ${{0.2397} \over {188}}$ $\times$ 80 g

therefore % Br in the organic compound

$= {{{W_{Br}}} \over {{W_T}}}$ $\times$ 100

${{0.2397 \times 80} \over {188 \times 0.15}}$ $\times$ 100 = 0.85 $\times$ 80 = 68

$\Rightarrow$ Nearest integer is '68'.

(A) $ZnC{O_{3(s)}}\buildrel \Delta \over \longrightarrow Zn{O_{(s)}} + C{O_{2(g)}}$

Heating in absence of oxygen in calcination.

(B) $2Zn{S_{(s)}} + 3{O_{2(g)}}\buildrel \Delta \over \longrightarrow 2Zn{O_{(s)}} + 2S{O_{2(g)}}$

Heating in presence of oxygen in roasting.

Hence, (A) is calcination while (B) in roasting.

Metamers are compounds which have different alkyl groups present along both side of polyvalent functional group.

No polyvalent functional group in option (a).

The two functional groups are different in option (b).

Alcohol is not a polyvalent functional group.

Alkyl groups are different along both sides of polyvalent functional group.