JEE Main 2021 26th August Morning Shift Question Paper with Solutions

$\Delta$Q = $\Delta$U + $\Delta$W

${{\Delta Q} \over {\Delta t}} = {{\Delta U} \over {\Delta t}} + {{\Delta W} \over {\Delta t}}$

${{6000} \over {60}}{J \over {\sec }} = {{2.5 \times {{10}^3}} \over {\Delta t}} + 90$

$\Delta$t = 250 sec

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We have, P_A = P_B. [Points A & B at same horizontal level]

$\therefore$ ${P_{atm}} - {{2T} \over {{r_1}}} + \rho g(x + \Delta h) = {P_{atm}} - {{2T} \over {{r_2}}} + \rho gx$

$\therefore$ $\rho g\Delta h = 2T\left[ {{1 \over {{r_1}}} - {1 \over {{r_2}}}} \right]$

$= 2 \times 7.3 \times {10^{ - 2}}\left[ {{1 \over {2.5 \times {{10}^{ - 3}}}} - {1 \over {4 \times {{10}^{ - 3}}}}} \right]$

$\therefore$ $\Delta h = {{2 \times 7.3 \times {{10}^{ - 2}} \times {{10}^3}} \over {{{10}^3} \times 10}}\left[ {{1 \over {2.5}} - {1 \over 4}} \right]$

= 2.19 $\times$ 10^$-$3 m = 2.19 mm

Hence, option (b).

E = ML²T^$-$2

L = ML²T^$-$1

m = M

G = M^$-$1L⁺³T^$-$2

P = ${{E{L^2}} \over {{M^5}{G^2}}}$

[P] = ${{(M{L^2}{T^{ - 2}})({M^2}{L^4}{T^{ - 2}})} \over {{M^5}({M^{ - 2}}{L^6}{T^{ - 4}})}} = {M^0}{L^0}{T^0}$

$\rho$ = 200 $\Omega$m

C = 2 $\times$ 10^$-$12 F

V = 40 V

K = 56

$i = {q \over {\rho k{\varepsilon _0}}} = {{{q_0}} \over {\rho k{\varepsilon _0}}}{e^{ - {t \over {\rho k{\varepsilon _0}}}}}$

${i_{\max }} = {{2 \times {{10}^{ - 12}} \times 40} \over {200 \times 50 \times 8.85 \times {{10}^{ - 12}}}}$

$= {{80} \over {{{10}^4} \times 8.85}}$ = 903 $\mu$A = 0.9 mA

Considering outer spherical shell is non-conducting.

Electric field inside a metal sphere is zero.

r < R $\Rightarrow$ E = 0

r > R $\Rightarrow$ E = ${{kQ} \over {{r^2}}}$


Option (b)

Considering outer spherical shell is conducting


r < R, E = 0

R $\le$ r < a

E = ${{kQ} \over {{r^2}}}$

a$\le$ r < b,

E = 0

r $\ge$ b,

E = ${{kQ} \over {{r^2}}}$

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$f' = f\left( {{{C - {V_0}} \over {C + {V_s}}}} \right)$

$\Rightarrow$ $1800 = f\left( {{{340 - 20} \over {340 + 20}}} \right)$

f = 2025 Hz

Least count (L. C.) = ${{0.5} \over {50}}$

True reading = $5 + {{0.5} \over {50}} \times 20 - {{0.5} \over {50}} \times 5$

$= 5 + {{0.5} \over {50}}(15) = 5.15$ mm

W_m = 12560 = 2$\pi$f_m

${f_m} = {{12560} \over {2\pi }}$

= 2000 Hz

Pentavalent activities have excess free e^$-$, so e^$-$ density increases but overall semiconductor is neutral.

$\overrightarrow \tau = \overrightarrow {{M_2}} \times \overrightarrow {{B_1}}$

$\tau = {M_2}{B_1}\sin 90^\circ$

$= 1 \times {{{\mu _0}} \over {4\pi }}{{{M_1}} \over {{{(1)}^3}}}1$

= 10^$-$7 N.m

${{{R_1}{R_2}} \over {{R_1} + {R_2}}} = 3$

${{{{(12 \times {{10}^{ - 6}} \times {{10}^{ - 2}})} \over {\pi {{(2)}^2} \times {{10}^{ - 6}}}} \times {{(51 \times {{10}^{ - 6}} \times {{10}^{ - 2}})l \times 4} \over {\pi {{(2)}^2} \times {{10}^{ - 6}}}}} \over {{{63 \times {{10}^{ - 6}} \times {{10}^{ - 2}} \times l \times 4} \over {\pi {{(2)}^2} \times {{10}^{ - 6}}}}}}$ = 3

$\Rightarrow$ l = 97 m

Position of bright fringe y = n${{D\lambda } \over d}$

y₁ of red = ${{D{\lambda _r}} \over d}$ = 3.5 mm

$\lambda$_r = 3.5 $\times$ 10^$-$3 ${d \over D}$

Similarly, $\lambda$_v = 2 $\times$ 10^$-$3 ${d \over D}$

${\lambda _r} - {\lambda _v} = (1.5 \times {10^{ - 3}})\left( {{{0.3 \times {{10}^{ - 3}}} \over {1.5}}} \right)$

= 3 $\times$ 10^$-$7 = 300 nm

${V_{RMS}} = \sqrt {{{3RT} \over {{M_W}}}}$

At the same temperature ${V_{RMS}} \propto {1 \over {\sqrt {{M_W}} }}$

$\Rightarrow$ V_H > V_O > V_C

Option (a)

E₀ = 200

$I = {1 \over 2}{\varepsilon _0}E_0^2.C$

Radiation pressure

$P = {{2I} \over C}$

$= \left( {{2 \over C}} \right)\left( {{1 \over 2}{\varepsilon _0}E_0^2C} \right)$

$= {\varepsilon _0}E_0^2$

$= 8.85 \times {10^{ - 12}} \times {200^2}$

$= 8.85 \times {10^{ - 8}} \times 4$

$= {{354} \over {{{10}^9}}}$

${B_{axis}} = {{{\mu _0}i{R^2}} \over {2{{({R^2} + {x^2})}^{3/2}}}}$

${B_{centre}} = {{{\mu _0}i} \over {2R}}$

$\therefore$ ${B_{centre}} = {{{\mu _0}i} \over {2a}}$

$\therefore$ ${B_{axis}} = {{{\mu _0}i{a^2}} \over {2{{({a^2} + {r^2})}^{3/2}}}}$

$\therefore$ fractional change in magnetic field =

${{{{{\mu _0}i} \over {2a}} - {{{\mu _0}i{a^2}} \over {2{{({a^2} + {r^2})}^{3/2}}}}} \over {{{{\mu _0}i} \over {2a}}}} = 1 - {1 \over {{{\left[ {1 + \left( {{{{r^2}} \over {{a^2}}}} \right)} \right]}^{3/2}}}}$

$\approx 1 - \left[ {1 - {3 \over 2}{{{r^2}} \over {{a^2}}}} \right] = {3 \over 2}{{{r^2}} \over {{a^2}}}$

Note : ${\left( {1 + {{{r^2}} \over {{a^2}}}} \right)^{ - 3/2}} \approx \left( {1 - {3 \over 2}{{{r^2}} \over {{a^2}}}} \right)$

[True only if r << a]

Hence, option (d) is the most suitable option.

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Excess pressure inside the smaller soap bubble

$\Delta P = {{4S} \over {{r_1}}} + {{4S} \over {{r_2}}}$ .... (i)

The excess pressure inside equivalent soap bubble

$\Delta P = {{4S} \over {{R_{eq}}}}$ ....... (ii)

From (i) & (ii)

${{4S} \over {{R_{eq}}}} = {{4S} \over {{r_1}}} + {{4S} \over {{r_2}}}$

${1 \over {{R_{eq}}}} = {1 \over {{r_1}}} + {1 \over {{r_2}}}$

$= {1 \over 6} + {1 \over 3}$

R_eq = 2 cm

Here, ${1 \over v} + {1 \over u} = {1 \over f}$ .......(1)

$\Rightarrow$ ${1 \over v} + {1 \over { - 190}} = {1 \over {10}}$

$\Rightarrow$ v = ${{19} \over 2}$

Differentiating equation (1) w.r.t t we get

$- {1 \over {{v^2}}}\left( {{{dv} \over {dt}}} \right) - {1 \over {{u^2}}}\left( {{{du} \over {dt}}} \right) = 0$

$\Rightarrow$ $\left( {{{dv} \over {dt}}} \right) = - {{{v^2}} \over {{u^2}}}\left( {{{du} \over {dt}}} \right)$

$\Rightarrow$ $\left( {{{dv} \over {dt}}} \right) = - {\left( {{{{{19} \over 2}} \over {190}}} \right)^2} \times 40$

$= {1 \over {400}} \times 40 = 0.1$ m/s

Let magnitude be equal to $\lambda$

$\overrightarrow {OA} = \lambda \left[ {\cos 30^\circ \widehat i + \sin 30\widehat j} \right] = \lambda \left[ {{{\sqrt 3 } \over 2}\widehat i + {1 \over 2}\widehat j} \right]$

$\overrightarrow {OB} = \lambda \left[ {\cos 60^\circ \widehat i - \sin 60\widehat j} \right] = \lambda \left[ {{1 \over 2}\widehat i - {{\sqrt 3 } \over 2}\widehat j} \right]$

$\overrightarrow {OC} = \lambda \left[ {\cos 45^\circ ( - \widehat i) + \sin 45\widehat j} \right] = \lambda \left[ { - {1 \over {\sqrt 2 }}\widehat i + {1 \over {\sqrt 2 }}\widehat j} \right]$

$\therefore$ $\overrightarrow {OA} + \overrightarrow {OB} - \overrightarrow {OC}$

$= \lambda \left[ {\left( {{{\sqrt 3 + 1} \over 2} + {1 \over {\sqrt 2 }}} \right)\widehat i + \left( {{1 \over 2} - {{\sqrt 3 } \over 2} - {1 \over {\sqrt 2 }}} \right)\widehat j} \right]$

$\therefore$ Angle with x-axis

${\tan ^{ - 1}}\left[ {{{{1 \over 2} - {{\sqrt 3 } \over 2} - {1 \over {\sqrt 2 }}} \over {{{\sqrt 3 } \over 2} + {1 \over 2} + {1 \over {\sqrt 2 }}}}} \right] = {\tan ^{ - 1}}\left[ {{{\sqrt 2 - \sqrt 6 - 2} \over {\sqrt 6 + \sqrt 2 + 2}}} \right]$

$= {\tan ^{ - 1}}\left[ {{{1 - \sqrt 3 - \sqrt 2 } \over {\sqrt 3 + 1 + \sqrt 2 }}} \right]$

Hence option (a).

From maximum average power

X_L = X_C

250$\pi$ = ${1 \over {2\pi (50)C}}$

$\Rightarrow$ C = 4 $\times$ 10^$-$6

$I = \left[ {{I_1} + M{{\left( {{5 \over 2}r} \right)}^2}} \right] + \left[ {{I_2} + M{{\left( {{{13r} \over 2}} \right)}^2}} \right]$

$= \left[ {{{M(36{r^2})} \over {12}} + {{M(25{r^2})} \over 4}} \right] + \left[ {{{M{r^2}} \over 2} + {{169M{r^2}} \over 4}} \right]$

= 52 Mr²

Ans. 52.00

When both balls will collide

y₁ = y₂

$35t - {1 \over 2} \times 10 \times {t^2} = 35(t - 3) - {1 \over 2} \times 10 \times {(t - 3)^2}$

$35t - {1 \over 2} \times 10 \times {t^2} = 35t - 105 - {1 \over 2} \times 10 \times {t^2} - {1 \over 2} \times 10 \times {3^2} + {1 \over 2} \times 10 \times 6t$

0 = 150 $-$ 30 t

t = 5 sec

$\therefore$ Height at which both balls will collied

$h = 35t - {1 \over 2} \times 10 \times {t^2}$

$= 35 \times 5 - {1 \over 2} \times 10 \times {5^2}$

h = 50 m

${F_{thrust}} = \left( {{{dm} \over {dt}}.{V_{rel}}} \right)$

$\left( {{{dm} \over {dt}}{V_{rel}} - mg} \right) = ma$

$\Rightarrow \left( {{{dm} \over {dt}}} \right) \times 500 - {10^3} \times 10 = {10^3} \times 20$

${{dm} \over {dt}}$ = (60 kg /s)

Inside a spherical shell, gravitational field is zero and hence potential remains same everywhere.

From energy conservation

K_i + U_i = k_f + U_f

$0 + \left( { - 1 \times 10 \times {1 \over 2}} \right) = {k_f} + \left( { - 3 \times 10 \times {3 \over 2}} \right)$

$-$5 = k_f $-$ 45

k_f = 40 J

${1 \over {{R_{eq}}}} = {1 \over 4} + {1 \over 8} + {1 \over {12}} + {1 \over 6} = {{6 + 3 + 2 + 4} \over {24}} = {{15} \over {24}}$

${R_{eq}} = {{24} \over {15}} = 1.6 \Rightarrow {R_T} = 1.6 + 0.6 = 2.2\,\Omega$

$P = {{{V^2}} \over {{R_T}}} = {{{{(2.2)}^2}} \over {2.2}} = 2.2W$

Option (c)

$K{E_{\max }} = e{V_s} = {{hc} \over \lambda } - \phi$

$\Rightarrow e{V_s} = {{1240} \over {280}} - 2.5 =$ 1.93 eV

$\Rightarrow {V_{{s_1}}} =$ 1.93 V .... (i)

$\Rightarrow e{V_{{s_2}}} = {{1240} \over {400}} - 2.5 =$ 0.6 eV

$\Rightarrow {V_{{s_2}}} =$ 0.6 V .... (ii)

$\Delta$V = ${V_{{s_1}}} - {V_{{s_2}}}$ = 1.93 $-$ 0.6 = 1.33 V

For node

cos(1.57 cm^$-$1)x = 0

(1.57 cm^$-$1)x = ${\pi \over 2}$

x = ${\pi \over {2(1.57)}}$ cm = 1 cm

The F^$\Theta$ ions make the enamel on teeth much harder by converting hydroxyapatite, [3Ca₃(PO₄)₂] . Ca(OH)₂], the enamel on the surface of the teeth into much harder fluroappatite. [3Ca₃(PO₄)₂ . CaF₂]

T-shaped molecule means 3 sigma bond and 2 lone pairs of electron on central atom.

Given Statement I is true as in a number of processes, one element is used to reduce the oxide of another metal. Any element will reduce the oxide of other metal which lie above it in the Ellingham diagram because the free energy change will become more negative.

Given Statement II is false as the value of $\Delta$S is decreases from left to right in Ellingham diagram.

Mohr's salt : (NH₄)₂ Fe(SO₄)₂ . 6H₂O

The number of water molecules in Mohr's salt = 6

Potash alum : KAl(SO₄)₂ . 12H₂O

The number of water molecules in potash alum = 12

So ratio of number of water molecules in Mohr's salt and potash alum

$= {6 \over {12}}$

$= {1 \over 2}$

= 0.5

= 5 $\times$ 10^$-$1

7 $\times$ 10^$-$9 = K $\times$ (8 $\times$ 10^$-$5)^x (8 $\times$ 10^$-$5)^y ......... (1)

2.1 $\times$ 10^$-$8 = K $\times$ (24 $\times$ 10^$-$5)^x (8 $\times$ 10^$-$5)^y ....... (2)

${1 \over 3} = {\left( {{1 \over 3}} \right)^x} \Rightarrow x = 1$

${\Delta _f}H_{KCl}^\Theta = {\Delta _{sub}}H_{(K)}^\Theta + {\Delta _{ioniztion}}H_{(K)}^\Theta + {1 \over 2}{\Delta _{bond}}H_{(C{l_2})}^\Theta + {\Delta _{electron\,gain}}H_{(Cl)}^\Theta + {\Delta _{lattice}}H_{(KCl)}^\Theta$

$\Rightarrow - 436.7 = 89.2 + 419.0 + {1 \over 2}(243.0) + \{ - 348.6\} + {\Delta _{lattice}}H_{(KCl)}^\Theta$

$\Rightarrow {\Delta _{lattice}}H_{(KCl)}^\Theta = - 717.8$ kJ mol^$-$1

The magnitude of lattice enthalpy of KCl in kJ mol^$-$1 is 718 (Nearest integer).

This addition is known as 1, 4-addition.

Mechanism This reaction can undergoes two pathways



Out of the two intermediates formed in two paths, the path B intermediate is more stable as it has more stable carbocation. So, a major product will be formed because of path B.


1, 2-addition product is formed at low temperature and will be less stable as the double bond is less substituted. 1, 4-addition is a thermodynamically stable product as the double bond is more substituted. As diene is in excess and HBr is limited in reaction, so diene cannot be formed. So option (b) is incorrect.

Bromination is highly selective, reactivity order –

3° >> 2° > 1

(a) [Fe(CN)₆]^4$-$ $\to$ Pale yellow solution

(b) [Fe(SCN)₆]^4$-$ $\to$ Blood red colour

(c) Fe₄[Fe(CN₆)]₃ . H₂O $\to$ Prussian blue

(d) [Fe(CN)₅NOS]^4$-$ $\to$ Violet colour

1000 kg solvent has 3.3 moles of KCl

1000 kg solvent $\to$ 3.3 $\times$ 74.5 gm KCl $\to$ 245.85

Weight of solution = 1245.85 gm

Volume of solution = ${{1245.85} \over {1.2}}$ ml

So molarity = ${{3.3 \times 1.2} \over {1245.85}} \times 1000 = 3.17$

Titration curve for strong acid and weak base initially a buffer of weak base and conjugate acid is :



Formed, thus pH falls slowly and after equivalence point, so the pH falls sharply so methyl orange, having pH range of 3.2 to 4.4 will weak as indicator. So, statement I is correct.



Titration curve for weak acid and strong base (NaOH)

Initially weak acid will form a buffer so pH increases slowly but after equivalence point. It rises sharply covering range of phenolphthalein so it will be suitable indicator so statement II is false.

To answer this question, let's analyze each statement separately.

Statement I: Frenkel defects are vacancy as well as interstitial defects.

This statement is true. A Frenkel defect, also known as a dislocation defect, occurs in a crystalline solid when an atom or ion leaves its normal site and moves to an interstitial site, creating a vacancy defect at its original position and an interstitial defect at the new position. However, it's crucial to note that the vacancy and interstitial are created by the same atom or ion moving from one site to another within the crystal. So, the Frenkel defect involves both a vacancy and an interstitial defect, but they are correlated because they involve the displacement of the same atom or ion.

Statement II: Frenkel defect leads to color in ionic solids due to presence of F-centres.

This statement is false. The color in ionic solids is usually caused by F-centers (color centers), which are defects formed when an anion vacancy is occupied by one or more electrons. These electrons can absorb visible light, giving the crystal a characteristic color. However, the Frenkel defect, involving an atom or ion moving from its lattice site to an interstitial site, does not directly lead to the creation of F-centers. Instead, the coloration associated with F-centers is most commonly linked with Schottky defects (which involve paired vacancies of cations and anions but do not include displaced atoms to interstitial sites) or directly with anion vacancies that are not necessarily related to Frenkel defects.

Conclusion: Given the explanations above, the correct answer is Option C: Statement I is true but Statement II is false.

(a) CuCl₂ + nH₂O $\to$ Cu$_{(aq.)}^{ + 2}$ (blue colour)

(b) AgCl + nH₂O $\to$ Insoluble

(c) ZnCl₂ + nH₂O $\to$ Zn$_{(aq.)}^{ + 2}$ Colourless

(d) Cu₂Cl₂ + nH₂O $\to$ Insoluble

Sublimation enthalpy, Ionisation enthalpy and hydration enthalpy affect the reduction potential.

As 0.1 M C₂H₅OH is non-dissociative and rest all salt given are electrolyte so in each case effective molarity > 0.1 so each will have lower freezing point.

Pure demineralised (de-ionized) water free from all soluble mineral salts is obtained by passing water successively through a cation exchange (in the H⁺ form) and an anion exchange (in the OH^$-$ form) resins.

Ion	${H^ + }$	${K^ + }$	$C{l^ - }$	$C{H_3}CO{O^ - }$
$\Lambda _{m\,Sc{m^2}/mole}^\infty$	349.8	73.5	76.3	40.9

So,

$\Lambda _{m\,\,C{H_3}COOH}^\infty = \Lambda _{m\,\,({H^ + })}^\infty + \Lambda _{m\,\,C{H_3}CO{O^ - }}^\infty$

= 349.8 + 40.9 = 390.7 Scm²/mole

$\Lambda _{m\,\,KCl}^\infty = \Lambda _{m\,\,({K^ + })}^\infty + \Lambda _{m\,\,(C{l^ - })}^\infty$

= 73.5 + 76.3 = 149.3 Scm²/mole

So, Statement I is wrong or false.

As the concentration decreases, the dilution increases which increases the degree of dissociation, thus increasing the no. of ions, which increases the molar conductance.

So Statement II is false.

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Velocity of electron in Bohr's atom is given by

$V \propto {Z \over n}$

Z = atomic number of atom, corresponds to +ve charge so as Z increase velocity increases so statement I is wrong and as 'n' decreases velocity increases so statement II is correct.