JEE Main 2021 31st August Evening Shift Question Paper with Solutions

$\Delta U = {1 \over 2}(\Delta C){V^2}$

$\Delta U = {1 \over 2}(KC - C){V^2}$

$\Delta U = {1 \over 2}(2 - 1)C{V^2}$

$\Delta U = {1 \over 2} \times 200 \times {10^{ - 6}} \times 200 \times 200$

$\Delta U = 4$ J

[M] = K[F]^a [T]^b [V]^c

[M¹] = [M¹L¹T^$-$2]^a [T¹]^b [L¹T^$-$1]^c

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

$\therefore$ [M] = [FTV^$-$1]

M = 1.5 kg, r = 0.5 m, d = ${5 \over 2}$ m

$I = 2\left( {{2 \over 5}M{r^2} + M{d^2}} \right)$

= 19.05 kgm²

y = mx + C

v² = ${{20} \over {10}}$x + 20

v² = 2x + 20

2v${{dv} \over {dx}}$ = 2

$\therefore$ a = v${{dv} \over {dx}}$ = 1

Signal bandwidth = 2 fm = 12 kHz

$\therefore$ N = ${{6MHZ} \over {12kHZ}} = {{6 \times {{10}^6}} \over {12 \times {{10}^3}}} = 500$

Here, ${\overrightarrow F _1} + {\overrightarrow F _2} + {\overrightarrow F _3} = 0$

${\overrightarrow F _1} + {\overrightarrow F _2} = - {\overrightarrow F _3}$

Since ${\overrightarrow F _{net}} = 0$ (equilibrium)

Both statements correct.

9 MSD = 10 VSD

9 $\times$ 1 mm = 10 VSD

$\therefore$ 1 VSD = 0.9 mm

LC = 1 MSD $-$ 1 VSD = 0.1 mm

Reading = MSR + VSR $\times$ LC

10 + 8 $\times$ 0.1 = 10.8 mm

Actual reading = 10.8 $-$ 0.4 = 10.4 mm

radius = ${d \over 2} = {{10.4} \over 2}$ = 5.2 mm

= 52 $\times$ 10^$-$2 cm

PV = nRT

400 $\times$ 10³ $\times$ 500 $\times$ 10^$-$6 = n$\left( {{{25} \over 3}} \right)$ (300)

n = ${{2 \over {25}}}$

n = n₁ + n₂

${{2 \over {25}}}$ = ${{{M_1}} \over 2} + {{{M_2}} \over {32}}$

Also, M₁ + M₂ = 0.76 gm

${{{M_2}} \over {{M_1}}} = {{16} \over 3}$

For every large distance P.E. = 0

& total energy = 2.6 + 0 = 2.6 eV

Finally in first excited state of H atom total energy = $-$3.4 eV

Loss in total energy = 2.6 $-$ ($-$3.4) = 6 eV

It is emitted as photon

$\lambda = {{1240} \over 6} = 206$ nm

$f = {{3 \times {{10}^8}} \over {206 \times {{10}^{ - 9}}}}$ = 1.45 $\times$ 10¹⁵ Hz

= 1.45 $\times$ 10⁹ Hz

E = i²Rt

192 = 16 (R) (1)

R = 12$\Omega$

E₁ = (8)² (12) (5)

= 3840 J

i = A = 60$^\circ$

${{ \delta } _{\min }}$ = 2i $-$ A

= 2 $\times$ 60$^\circ$ $-$ 60$^\circ$ = 60$^\circ$

$\mu = {{{{\sin }^{ - 1}}\left( {{{{\delta _{\min }} + A} \over 2}} \right)} \over {{{\sin }^{ - 1}}\left( {{A \over 2}} \right)}}$

$= \sqrt 3$

V_prism $= {{3 \times {{10}^8}} \over {\sqrt 3 }}$

AP = 10 $\times$ 10^$-$2 $\times$ ${{\sqrt 3 } \over 2}$

time $= {{5 \times {{10}^{ - 2}}} \over {3 \times {{10}^8}}} \times \sqrt 3 \times \sqrt 3$

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

Ans. = 5

On simplification


${R_{eq}} = {{3 \times 3/2} \over {3 + 3/2}} = {{9/2} \over {9/2}} = 1\Omega$

$\overrightarrow B$ must not be parallel to the plane of coil for non zero flux and according to lenz law if B is outward it should be decreasing for anticlockwise induced current.

$\overrightarrow A = \overrightarrow P + \overrightarrow Q$

$\overrightarrow B = \overrightarrow P - \overrightarrow Q$

$\overrightarrow P \bot \overrightarrow Q$

$\left| {\overrightarrow A } \right| = \left| {\overrightarrow B } \right| = \sqrt {2({P^2} + {Q^2})(1 + \cos \theta )}$

For $\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt {3({P^2} + {Q^2})}$

${\theta _1} = 60^\circ$

For $\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt {2({P^2} + {Q^2})}$

${\theta _2} = 90^\circ$

In S.H.M. total mechanical energy remains constant and also = = ${{1 \over 4}}$KA² (for 1 time period)

Force on each column = ${{mg} \over 4}$

Strain = ${{mg} \over {4AY}}$

$= {{50 \times {{10}^3} \times 9.8} \over {4 \times \pi (1 - 0.25) \times 2 \times {{10}^{11}}}}$

= 2.6 $\times$ 10^$-$7

$\overrightarrow F = q\left( {\overrightarrow V \times \overrightarrow B } \right)$

${\overrightarrow F _1} = 4\pi \left[ {0.5c\widehat i \times {B_0}\left( {{{\widehat i + \widehat j} \over 2}} \right)\cos \left( {K.{\pi \over K} - 0} \right)} \right]$

${\overrightarrow F _2} = 2\pi \left[ {0.5c\widehat i \times {B_0}\left( {{{\widehat i + \widehat j} \over 2}} \right)\cos \left( {K.{{3\pi } \over K} - 0} \right)} \right]$

$\cos \pi = - 1$, $\cos 3\pi = - 1$

$\therefore$ ${{{F_1}} \over {{F_2}}} = 2$

3MV₀ = 2MV₂ + MV₁

3V₀ = 2V₂ + V₁

120 = 2V₂ + 60 $\Rightarrow$ V₂ = 30 m/s

${{\Delta K.E.} \over {K.E.}} = {{{1 \over 2}MV_1^2 + {1 \over 2}2MV_2^2 - {1 \over 2}3MV_0^2} \over {{1 \over 2}3MV_0^2}}$

$= {{V_1^2 + 2V_2^2 - 3V_0^2} \over {3V_0^2}}$

$= {{3600 + 1800 - 4800} \over {4800}} = {1 \over 8}$

V_A = 5V $\Rightarrow$ A = 1

V_A = 0V $\Rightarrow$ A = 0

V_B = 5V $\Rightarrow$ B = 1

V_B = 0V $\Rightarrow$ B = 0

If A = B = 0, there is no potential anywhere here V₀ = 0

If A = 1, B = 0, Diode D₁ is forward biased, here V₀ = 5V

If A = 0, B = 1, Diode D₂ is forward biased hence V₀ = 5V

If A = 1, B = 1, Both diodes are forward biased hence V₀ = 5V

Truth table for I^st

$\therefore$ Given circuit is OR gate

For II^nd circuit

V_B = 5V, A = 1

V_B = 0V, A = 0

When A = 0, E - B junction is unbiased there is no current through it

$\therefore$ V₀ = 1

When A = 1, E-B junction is forward biased

V₀ = 0

$\therefore$ Hence this circuit is NOT gate

$B = 3\left[ {{{{\mu _0}i} \over {4\pi r}}(\sin 60^\circ + \sin 60^\circ )} \right]$

$\tan 60^\circ = {{l/2} \over r}$

where $r = {{9 \times {{10}^{ - 2}}} \over {2\sqrt 3 }}$ M

$\therefore$ B = 3 $\times$ 10^-5 T

Current is flowing in clockwise direction so, $\overrightarrow B$ is inside plane of triangle by right hand rule.

$T = 2\pi \sqrt {l/g}$

When bob is immersed in liquid

mg_eff = mg $-$ Buoyant force

mg_eff = mg $-$ v$\sigma$g ($\sigma$ = density of liquid)

$= mg - v{\rho \over 4}g$

$= mg - {{mg} \over 4} = {{3mg} \over 4}$

$\therefore$ ${g_{eff}} = {{3g} \over 4}$

${T_1} = 2\pi \sqrt {{{{l_1}} \over {{g_{eff}}}}}$

${l_1} = l + {l \over 3} = {{4l} \over 3},\,{l_{eff}} = {{3g} \over 4}$

By solving

${T_1} = {4 \over 3}2\pi \sqrt {l/g}$

${T_1} = {{4T} \over 3}$

Thermal resistance of spherical sheet of thickness dr and radius r is

$dR = {{dr} \over {K(4\pi {r^2})}}$

$R = \int\limits_{{r_1}}^{{r_2}} {{{dr} \over {K(4\pi {r^2})}}}$

$R = {1 \over {4\pi K}}\left( {{1 \over {{r_1}}} - {1 \over {{r_2}}}} \right) = {1 \over {4\pi K}}\left( {{{{r_2} - {r_1}} \over {{r_1}{r_2}}}} \right)$

Thermal current (i) $= {{{\theta _2} - {\theta _1}} \over R}$

$i = {{4\pi K{r_1}{r_2}} \over {{r_2} - {r_1}}}({\theta _2} - {\theta _1})$

${{\Delta M} \over M} = {{\Delta \mu } \over \mu } = {{250} \over {500}} = {1 \over 2}$

$\Rightarrow$ ${1 \over 2} = {x \over {499}} \Rightarrow x \simeq 250$

v = 1.5

p₁v₁^v = p₂v₂^v

(200) (1200)^1.5 = P² (300)^1.5

P₂ = 200 [4]^3/2 = 1600 kPa

$\left| {W.D.} \right| = {{{p_2}{v_2} - {p_1}{v_1}} \over {v - 1}} = \left( {{{480 - 240} \over {0.5}}} \right) = 480$ J

X_L = 2$\pi$fL

f is very large

$\therefore$ X_L is very large hence open circuit.

${X_C} = {1 \over {2\pi fC}}$

f is very large.

$\therefore$ X_C is very small, hence short circuit.

Final circuit



$\therefore$ ${Z_{eq}} = 1 + {{2 \times 2} \over {2 + 2}} = 2$

8$\beta$ = 2.4 cm

${{8\lambda \Delta } \over d}$ = 2.4 cm

${{8 \times 1.5 \times c} \over {0.3 \times {{10}^{ - 3}} \times f}} = 2.4 \times {10^{ - 2}}$

f = 5 $\times$ 10¹⁴ Hz

Since, $\phi = \overrightarrow E \,.\,\overrightarrow A = EA\cos \theta$



$\theta$ = 90$^\circ$

$\therefore$ $\phi$ = 0

To convert pulsating dc into steady dc both of mentioned method are correct.

Anions forms CCP or FCC(A^$-$) = 4A^$-$ per unit cell Cations occupy all octahedral voids (B⁺) = 4B⁺ per unit cell

cell formula $\to$ A₄B₄

Empirical formula $\to$ AB $\to$ (x = 1)

Na₂O + H₂O $\to$ 2NaOH

${{20} \over {62}}$ moles

Moles of NaOH formed = ${{20} \over {62}}$ $\times$ 2

[NaOH] = ${{{{40} \over {62}}} \over {{{500} \over {1000}}}}$ = 1.29 M = 13 $\times$ 10^$-$1 M (Nearest integer)

We know

${x \over m} = K{P^{1/n}}$; using (x $\propto$ V)

$\Rightarrow {{10} \over 1} = K \times {(100)^{1/n}}$ ..... (1)

${{15} \over 1} = K \times {(200)^{1/n}}$ ..... (2)

${V \over 1} = K \times {(300)^{1/n}}$ ..... (3)

Divide

(2) / (1)

${{15} \over {10}} = {2^{1/n}}$

$\log \left( {{3 \over 2}} \right) = {1 \over n}\log 2$

${1 \over n} = {{\log 3 - log2} \over {\log 2}} = {{0.4771 - 0.3010} \over {0.3010}}$

${1 \over n} = 0.585$

Divide

(3) / (1)

${V \over {10}} = {3^{1/n}}$

$\log \left( {{V \over {10}}} \right) = {1 \over n}\log 3$

$\log \left( {{V \over {10}}} \right) = 0.585 \times 0.4771 = 0.2791$

${V \over {10}} = {10^{0.279}} \Rightarrow V = 10 \times {10^{0.279}}$

$\Rightarrow V = {10^{1.279}} = {10^x}$

$\Rightarrow$ x = 1.279

$\Rightarrow$ x = 128 $\times$ 10^$-$2 (Nearest integer)

Molecular orbital configuration of $O_2^{2 - }$ is

$\sigma _{1s}^2\sigma _{1s}^{*2}\sigma _{2s}^2\sigma _{2s}^{*2}\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)\left( {\pi _{2px}^{*2} = \pi _{2py}^{*2}} \right)$

Zero unpaired electron

Atomic hydrogen is produced at high temperature in an electric are or under ultraviolet radiations.

The dissociation of dihydrogen at 2000 K is only 0.081%.

H-H bond dissociation enthalpy is highest for a single bond for any diatomic molecule.

Dihydrogen can be produced on reacting Zn with dil. HCl as well as NaOH (aq.)

${C_2}{H_7}N + \left( {2x + {y \over 2}} \right)CuO \to xC{O_2} + {y \over 2}{H_2}O + {z \over 2}{N_2} + \left( {2x + {y \over 2}} \right)Cu$

On balancing

${C_2}{H_7}N + {{15} \over 2}CuO \to 2C{O_2} + {7 \over 2}{H_2}O + {1 \over 2}{N_2} + {{15} \over 2}Cu$

On comparing

y = 7

NaOH + EtOH is known as alcoholic NaOH, so it give E² reaction with given alkyl halide.

More stable less potential energy.

Stability order : I > III > IV > II

So

Potential energy : II > IV > III > I

Novolac is a linear polymer of [Ph-Oh + HCHO].

So ester linkage not present.

So novolac is not a polyester.

Mn²⁺ $\to$ IV group

As³⁺ $\to$ II B group

Cu²⁺ $\to$ II A group

Al³⁺ $\to$ III group

Note :

IV^th group cations – Zn²⁺ ,Mn²⁺ ,Ni²⁺ , Co²⁺

II B group cations – As³⁺ , Sb³⁺ , Sn²⁺

II A group cations – Hg²⁺ , Pb²⁺ , Bi³⁺ , Cu²⁺ , Cd²⁺

III^rd group cations – Al³⁺ , Cr³⁺ , Fe³⁺

Zn⁺ $\to$ 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹

Outermost electron is in 4s subshell

m = 0

Stability of oxides of Halogens is

I > Cl > Br

Fibrous proteins are characterized by their elongated, insoluble structure and often serve structural or mechanical roles. Among the options provided, keratin, collagen, and myosin are examples of fibrous proteins.

Albumin, on the other hand, is a globular protein that is soluble in water. It has a more spherical shape compared to the elongated shape of fibrous proteins. Albumin is found in blood plasma and has various functions including maintaining osmotic pressure.

So, the correct answer is:

Option B: Albumin.

Option (b) is incorrect.

$\Delta$G$^\circ$ = $-$RT ln K

$\Delta$H$^\circ$ $-$ T$\Delta$S$^\circ$ = $-$RT ln K

ln K = $\left[ {{{\Delta H^\circ - \Delta S^\circ } \over {RT}}} \right]$

$[HCl] = {{20} \over {80}} = {1 \over 4}M = 2.5 \times {10^{ - 1}}M$

pH = $-$log 2.15 $\times$ 10^-1 = 1 $-$ 0.3979 = 0.6021

pH = 6021 $\times$ 10^-4

Cell constant = $\left( {{l \over A}} \right)$ $\Rightarrow$ Units = m^$-$1

Molar conductivity ($\Lambda$_m) $\Rightarrow$ Units = Sm² mole^$-$1

Conductivity (K) $\Rightarrow$ Units = S m^$-$1

Degree of dissociation ($\alpha$ $\to$ Dimensionless

$\therefore$ (a) - (iii), (b) - (i), (c) - (iv), (d) - (ii)

Lithium salts are hydrated due to high hydration energy of Li⁺

Li⁺ due to smallest size in IA group has highest polarizing power.

Eu $\to$ [Xe]4f⁷6s²

Eu²⁺ $\to$ [Xe]4f⁷

Oxides of nitrogen and sulphur are acidic and settle down on ground as dry deposition.

Ammonium salts in rain drops result in wet deposition.