JEE Main 2021 27th August Evening Shift Question Paper with Solutions

When unregulated voltage is 14 V voltage across zener diode must be 10 V. So potential difference across resistor $\Delta$V_Rs = 4V

and P_zener = 2W

VI = 2

I = ${2 \over {10}}$ = 0.2 A

$\Delta$V_Rs = I R_s

4 $\times$ 0.2 R_s $\Rightarrow$ R_s = ${40 \over {2}}$ = 20$\Omega$

${V_{rms}} = \sqrt {{{3KT} \over M}}$

${{{{({V_{rms}})}_{{O_2}}}} \over {{{({V_{rms}})}_{{H_2}}}}} = \sqrt {{{{M_{{H_2}}}} \over {{M_{{O_2}}}}}} = \sqrt {{2 \over {32}}}$

${({V_{rms}})_{{H_2}}} = 4 \times {({V_{rms}})_{{O_2}}}$

$= 4 \times 160$

$= 640$ m/s

$\alpha = {{{I_C}} \over {{I_E}}}$, $\beta = {{{I_C}} \over {{I_B}}}$; ${I_E} = {I_C} + {I_B}$

$\alpha = {{{I_C}} \over {{I_C} + {I_B}}} = {{{I_C}/{I_B}} \over {{{{I_C}} \over {{I_B}}} + 1}} = {\beta \over {\beta + 1}}$

$1 + {1 \over \beta } = {1 \over \alpha }$

${1 \over \beta } = {{1 - \alpha } \over \alpha }$

$\beta = {\alpha \over {1 - \alpha }}$

${R_{eq\,open}} = {{3R} \over 2}$

${R_{eq\,closed}} = 2 \times {{R \times 2R} \over {3R}} = {{4R} \over 3}$

${{{R_{eq\,open}}} \over {{R_{eq\,closed}}}} = {{3R} \over 2} \times {3 \over {4R}} = {9 \over 8}$

$\therefore$ $x = 9$

$V' = \sqrt {5gR} = \sqrt {5 \times 10 \times 0.5}$

V' = 5 m/s

m₁ V = m₂ $\times$ 5 $-$ m₁ $\times$ 100

${{10} \over {1000}} \times V = 5 - {{10} \over {1000}} \times 100$

V = 400 m/s

R = 75 $\times$ 10² $\pm$ 5% of 7500

R = (7500 $\pm$ 375) $\Omega$

Change in P.E. = Heat energy

mgh = mS$\Delta$T

$\Delta$T = ${{gh} \over S}$

= ${{10 \times 63} \over {4200J/kgC}}$

= 0.147$^\circ$C

$H = {{{U^2}{{\sin }^2}\theta } \over {2g}}$

$= {{{{(25)}^2}.{{(\sin 45)}^2}} \over {2 \times 10}}$

= 15.625 m

$T = {{U\sin \theta } \over g}$

$= {{25 \times \sin 45^\circ } \over {10}}$

= 2.5 $\times$ 0.7

= 1.77 s

Density = [F^aL^bT^c]

[ML^$-$3] = [M^aL^a+bT^$-$2aL^bT^c]

[M¹L^$-$3] = [M^aL^a+bT^$-$2a+c]

\matrix{ {a = 1} & ; & {a + b = - 3} & ; & { - 2a + c = 0} \cr {} & {} & {1 + b = - 3} & {} & {c = 2a} \cr {} & {} & {b = - 4} & {} & {c = 2} \cr }

So, density = [F¹L^$-$4T²]

B.S₁ = ${{{T_{1\max }}} \over {8 \times {{10}^{ - 7}}}}$ $\Rightarrow$ T_{1 max} = 8 $\times$ 1.25 $\times$ 100 = 1000 N

B.S₂ = ${{{T_{2\max }}} \over {4 \times {{10}^{ - 7}}}}$ $\Rightarrow$ T_{2 max} = 4 $\times$ 1.25 $\times$ 100 = 500 N

m = ${{500 - 100} \over {10}} = 40$ kg

$\cos \phi = {R \over {\sqrt {{R^2} + 3{R^2}} }}$

$= {1 \over {\sqrt {10} }}$

$\cos \phi ' = {R \over {\sqrt {{R^2} + {R^2}} }}$

$= {1 \over {\sqrt 2 }}$

${{\cos \phi '} \over {\cos \phi }} = {{\sqrt {10} } \over {\sqrt 2 }} = {{\sqrt 5 } \over 1}$

$\therefore$ x = 1

${y_1} = 10\sin \left( {3\pi t + {\pi \over 3}} \right)$ $\Rightarrow$ Amplitude = 10

${y_2} = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)$

${y_2} = 10\left( {{1 \over 2}\sin 3\pi t + {{\sqrt 3 } \over 2}\cos 3\pi t} \right)$

${y_2} = 10\left( {\cos {\pi \over 3}\sin 3\pi t + \sin {\pi \over 3}\cos 3\pi t} \right)$

${y_2} = 10\left( {3\pi t + {\pi \over 3}} \right)$ $\Rightarrow$ Amplitude = 10

So ratio of amplitudes = ${{10} \over {10}}$ = 1

$|B|\, = {{|E|} \over C} = {6 \over {3 \times {{10}^8}}}$

= 2 $\times$ 10^$-$8 T

$\therefore$ x = 2

No. of different wavelengths = ${{n(n - 1)} \over 2}$

$= {{6 \times (6 - 1)} \over 2} = {{6 \times 5} \over 2} = 15$

(m₁g $-$ 2T) = m₁a $-$ (1)

T $-$ m₂g = m₂(2a)

2T $-$ 2m₂g = 4m₂ a $-$ (2)

m₁g $-$ 2m₂g = (m₁ + 4m₂) a

a = ${{(8 - 4)g} \over {(8 + 8)}} = {4 \over {16}}g = {g \over 4}$

a = ${{10} \over 4}$ m/s²

S = ${1 \over 2}$ at²

${{0.2 \times 2 \times 4} \over {10}}$ = t²

t = 0.4 sec

when x < a

${B_1}(2\pi x) = {\mu _0}\left( {{{{i_0}} \over {\pi {a^2}}}} \right)\pi {x^2}$

$B(2\pi x) = {{{\mu _0}{i_0}{x^2}} \over {{a^2}}}$

${B_1} = {{{\mu _0}{i_0}x} \over {2\pi {a^2}}}$ .... (1)

when a < x < b

${B_2}(2\pi x) = {\mu _0}{i_0}$

${B_2} = {{{\mu _0}{i_0}} \over {2\pi x}}$ ..... (2)

${{{B_1}} \over {{B_2}}} = {{{\mu _0}{i_0}{x \over {2\pi {a^2}}}} \over {{{{\mu _0}{i_0}} \over {2\pi x}}}} = {{{x^2}} \over {{a^2}}}$

${I_{\max }} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}$

= 4 I₀

h : height of antenna

$\lambda$ : wavelength of signal

h < $\lambda$

$\lambda$ > h

$\lambda$ > 400 m

emf = BLV

= 1 . (2R) . 1

= 2V

H = ${1 \over 2}$gt²

${{9.8 \times 2} \over {9.8}}$ = t²

t = $\sqrt 2$ sec

$\Delta$t : time interval between drops

h = ${1 \over 2}$g($\sqrt 2$ $-$ 2$\Delta$t)²

$\Delta$t = ${1 \over {\sqrt 2 }}$

h = ${1 \over 2}$g${\left( {\sqrt 2 - {1 \over {\sqrt 2 }}} \right)^2} = {1 \over 2} \times 9.8 \times {1 \over 2} = {{9.8} \over 4} = 2.45$ m

H $-$ h = 9.8 $-$ 2.45

= 7.35 m

${I_{\max }} = {{50} \over 2} = 25$ mA

$R = {V \over I} = {{50mV} \over {25mA}} = 2\Omega$

(V_D $-$ V) C₂ + (V_D $-$ 0) C₃ = 0

(V_D $-$ V) 6 + (V_D $-$ 0) 12 = 0

V_D $-$ V + 2V_D = 0

V_D = ${V \over 3}$

q₂ = (V $-$ V_D) C₂ = $\left( {V - {V \over 3}} \right)$ (6 $\mu$F)

q₂ = (4V) $\mu$F

q₃ = (V_D $-$ 0) C₃ = ${{V \over 3}}$ $\times$ 12$\mu$F = 4V$\mu$F

q₁ = (V $-$ 0) C₁ = V(2$\mu$F)

q₁ : q₂ : q₃ = 2 : 4 : 4

q₁ : q₂ : q₃ = 1 : 2 : 2

${\lambda \over 4}$ = l $\Rightarrow$ $\lambda$ = 4l

f = ${V \over \lambda } = {V \over {4l}}$

$\Rightarrow$ 250 = ${{340} \over {4l}}$

$\Rightarrow$ l = ${{34} \over {4 \times 25}}$ = 0.34 m

l = 34 cm

From conservation of angular momentum we get

${I_1}{\omega _1} + {I_2}{\omega _2} = ({I_1} + {I_2})\omega$

$\omega = {{{I_1}{\omega _1} + {I_2}{\omega _2}} \over {{I_1} + {I_2}}}$

${k_i} = {1 \over 2}{I_1}\omega _1^2 + {1 \over 2}{I_2}\omega _2^2$

${k_f} = {1 \over 2}({I_1} + {I_2}){\omega ^2}$

${k_i} - {k_f} = {1 \over 2}\left[ {{I_1}\omega _1^2 + {I_2}\omega _2^2 - {{{{({I_1}{\omega _1} + {I_2}{\omega _2})}^2}} \over {{I_1} + {I_2}}}} \right]$

Solving above we get

${k_i} - {k_f} = {1 \over 2}\left( {{{{I_1}{I_2}} \over {{I_1} + {I_2}}}} \right){({\omega _1} - {\omega _2})^2}$

$k{E_{\max }} = {{hc} \over {{\lambda _i}}} + \phi$

or $e{V_o} = {{hc} \over {{\lambda _i}}} + \phi$

when $\lambda$_i = 670.5 nm ; V_o = 0.48

when $\lambda$_i = 474.6 nm ; V_o = ?

So,

$e(0.48) = {{1240} \over {670.5}} + \phi$ ..... (1)

$e({V_o}) = {{1240} \over {474.6}} + \phi$ .....(2)

(2) $-$ (1)

$e({V_o} - 0.48) = 1240\left( {{1 \over {474.6}} - {1 \over {670.5}}} \right)eV$

${V_o} = 0.48 + 1240\left( {{{670.5 - 474.6} \over {474.6 \times 670.5}}} \right)$ Volts

V_o = 0.48 + 0.76

V_o = 1.24 V $\simeq$ 1.25 V

Q_in = 300 J ; Q_out = 240 J

Work done = Q_in $-$ Q_out = 300 $-$ 240 = 60 J

Efficiency = ${W \over {{Q_{in}}}} = {{60} \over {300}} = {1 \over 5}$

efficiency = $1 - {{{T_2}} \over {{T_1}}}$

${1 \over 5} = 1 - {{400} \over {{T_1}}} \Rightarrow {{400} \over {{T_1}}} = {4 \over 5}$

T₁ = 500 k

SI unit of Rydberg const. = m^$-$1

SI unit of Plank's const. = kg m²s^$-$1

SI unit of Magnetic field energy density = kg m^$-$1s^$-$2

SI unit of coeff. of viscosity = kg m^$-$1s^$-$1

The number of optical isomers for [Cr(C₂O₄)₃]^3$-$ is two.

Sucrose is formed by $\alpha$-D(+). Glucose + $\beta$-D($-$) Fructose.

We obtain these monomers on hydrolysis.

Energy emitted in 0.1 sec.

= 0.1 sec. $\times$ ${10^{ - 3}}{J \over s}$

= 10^$-$4 J

If 'n' photons of $\lambda$ = 1000 nm are emitted, then 10^$-$4 = n $\times$ ${{hc} \over \lambda }$

$\Rightarrow {10^{ - 4}} = {{n \times 6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1000 \times {{10}^{ - 9}}}}$

$\Rightarrow$ n = 5.02 $\times$ 10¹⁴ = 50.2 $\times$ 10¹³

$\Rightarrow$ 50 (nearest integer)

C₃H_8(g) + 5O_2(g) $\to$ 3CO_2(g) + 4H₂O_(l)

t = 0	2.27 mole	31.25 mol
t = $\infty$	0	19.9 mol	6.81 mol	9.08 mol

mole fraction of CO₂ in the final reaction mixture (heterogenous)

${X_{C{O_2}}} = {{6.81} \over {19.9 + 6.81 + 9.08}}$

= 0.1902 = 19.02 $\times$ 10^$-$2

$\Rightarrow$ 19

When red phosphorus is heated in a sealed tube at 803 K, $\alpha$-Black phosphorus is formed.

All gives propanoic acid as product but option (d) gives butanoic as product

$K = {1 \over R} \times {l \over A} = \left( {\left( {{1 \over {1500}}} \right) \times 1.14} \right)$ S cm^$-$1

$\Rightarrow { \wedge _m} = 1000 \times {{\left( {{{1.14} \over {1500}}} \right)} \over {0.001}}$ S cm² mol^$-$1

= 760 S cm² mol^$-$1

$\Rightarrow$ 760

Ozone in the stratosphere is a product of UV radiations acting on dioxygen (O₂) molecules.

${O_2}(g)\buildrel {UV} \over \longrightarrow O(g) + O(g)$

$O(g) + {O_2}(g)$ $\buildrel {UV} \over \rightleftharpoons$ ${O_3}(g)$

In Seliwanoff's reagent, Cu is not present.

In Barfoed, Biuret and in Benediet reagent Cu is present.

Hence, non-pyramidal species are SO₃, NO$_3^{- }$ and CO$_3^{2 - }$.

moles of NH₄HS initially taken = ${{5.1g} \over {51g/mol}}$ = 0.1 mol

volume of vessel = 2 litre



$\Rightarrow$ partial pressure of each component

$P = {{nRT} \over V} = {{0.1 \times 0.2 \times 0.082 \times 300} \over 2}$

= 0.246 atm

$\Rightarrow$ k_P = P_{$N{H_3}$} $\times$ P_{${H_2}S$} = (0.246)² = 0.060516

= 6.05 $\times$ 10^$-$2

$\therefore$ x = 6

(Name of ore/mineral)

(a) Calamine - ZnCO₃

(b) Malachite - CuCO₃ . Cu(OH)₂

(c) Siderite - FeCO₃

(d) Sphalerite - ZnS

Note : ( Remember all those ores names. Any one of those can be asked in the exam.)

Oxides Ores :

(1) ZnO - Zincite

(2) Fe₂O₃ - Haematite

(3) Fe₃O₄ - Magnetite (FeO + Fe₂O₃ mixture)

(4) Fe₂O₃ . 3H₂O - Limonite

(5) MnO₂ - Pyrolusite

(6) Cu₂O - Cuprite or Ruby Copper

(7) TiO₂ - Rutile

(8) FeCr₂O₄ - Chromite (FeO + Cr₂O₃)

(9) FeTiO₃ - Illmenite (FeO + TiO₂)

(10) Na₂B₄O₇ . 10H₂O - Borax or Tincal

(11) U₃O₈ - Pitch Blende

(12) SnO₂ - Tin Stone or Cassiterite

(13) Ca₂B₆O₁₁ . 5H₂O - Colemanite (2 Cao + 3 B₂O₃)

(14) Al₂O₃ . 2H₂O - Bauxite

(15) Al₂O₃ . H₂O - Diaspore

(16) Al₂O₃ - Corundum

Sulphides Ores :

(1) ZnS - Zinc Blende or Sphalerite

(2) PbS - Galena

(3) Ag₂S - Argentite or Silver Glance

(4) HgS - Cinnabar

(5) Cu₂S - Chalcocite or Copper glance

(6) CuFeS₂ - Copper pyrites or Chalco pyrites (Cu₂S + Fe₂S₃ mixture)

(7) FeS₂ - Iron pyrites or Fool's Gold

(8) 3Ag₂S . Sb₂S₂ - Pyrargyrite or ruby silver

Halides Ores :

(1) NaCl - Rock Salt

(2) KCl - Sylvine

(3) Na₃AlF₆ - Cryolite [3NaF + AlF₆]

(4) CaF₂ - Fluorspar

(5) KCl . MgCl₂ . 6H₂O - Carnalite

(6) AgCl - Horn Silver

Carbonates Ores :

(1) CaCO₃ - Limestone

(2) MgCO₃ - Magnesite

(3) CaCO₃ . MgCO₃ - Dolomite

(4) ZnCO₃ - Calamine

(5) PbCO₃ - Cerrusite

(6) FeCO₃ - Siderite

(7) CuCO₃ . Cu(OH)₂ or Cu₂CO₃(OH)₂ - Malachite or Basic Copper Carbonates

(8) 2 CuCO₃ . Cu(OH)₂ - Azurite

Sulphates Ores :

(1) CuSO₄ . 2H₂O - Gypsum

(2) MgSO₄ . 7H₂O - Epson Salt

(3) Na₂SO₄ . 10 H₂O - Glauber's Salt

(4) PbSO₄ - Anglesite

(5) ZnSO₄ . 7H₂O - White Vitriol

(6) FeSO₄ . 7H₂O - Green Vitriol

(7) CuSO₄ . 5H₂O - Blue Vitriol or Chalcanthite

Nitrate Ores :

(1) KNO₃ - Indian Saltpetre

(2) NaNO₃ - Chile Saltpetre

Arsenides Ores :

(1) NiAs - Kupfernickel

(2) NiAsS - Nickel glance

a. Standard enthalpy of formation for alkali metal bromides becomes more negative on descending down the group.

b. In case of CsI, lattice energy is less, but Cs⁺ is having less hydration enthalpy due to which it is less soluble in water.

c. For alkali metal fluorides, the solubility in water increases from lithium to caesium. LiF is least soluble in water.

d. Standard enthalpy of formation for LiF is most negative among alkali metal fluorides.

${T_{\min }} = \left( {{{{\Delta ^0}H} \over {{\Delta ^0}S}}} \right)$

${\Delta ^0}{H_{rxn}} = \left[ {\Delta _f^0H(Fe) + \Delta _f^0H(CO)} \right] -$

$= \left[ {\Delta _f^0H(FeO) + \Delta _f^0H({C_{(graphite)}})} \right]$

$= [0 - 110.5] - [ - 266.3 + 0]$

= 155.8 kJ/mol

${\Delta ^0}{S_{rxn}} = \left[ {{\Delta ^0}S(Fe) + {\Delta ^0}S(CO)} \right] -$

$\left[ {{\Delta ^0}S(FeO) + {\Delta ^0}S({C_{(graphite)}})} \right]$

$= [27.28 + 197.6] - [57.49 + 5.74]$

= 161.65 J/mol-K

${T_{\min }} = {{155.8 \times {{10}^3}J/mol} \over {161.65J/mol - K}} = 963.8$ K

$\simeq 964$ k (nearest integer)