JEE Main 2021 27th July Morning Shift Question Paper with Solutions

For no bending, n₁ = n₂

$1.2 + {{10.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$ = $1.45 + {{1.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$

On solving,

9 $\times$ 10^$-$14 = 25$\lambda$²

$\lambda$ = 6 $\times$ 10^$-$7

$\lambda$ = 600 nm

By energy conservation

${1 \over 2}.{{YA} \over L}.{x^2} = {1 \over 2}m{v^2}$

${{0.5 \times {{10}^9} \times {{10}^{ - 6}} \times {{(0.04)}^2}} \over {0.1}} = {{20} \over {1000}}{v^2}$

$\therefore$ ${v^2} = 400$

$v = 20$ m/s

${{\Delta T} \over {\Delta t}} = K({T_t} - {T_s})$

T_t = average temp.

_T = surrounding temp.

${{61 - 59} \over 4} = K\left( {{{61 + 59} \over 2} - 30} \right)$ ..... (1)

${{51 - 49} \over t} = K\left( {{{51 + 49} \over 2} - 30} \right)$ ..... (2)

Divide (1) & (2)

${t \over 4} = {{60 - 30} \over {50 - 30}} = {{30} \over {20}}$

So, t = 6 minutes.

The speed of light in a medium is given by the equation:

$v = \frac{c}{\sqrt{\varepsilon_r \mu_r}}$

where:

• $c$ is the speed of light in vacuum (approximately $3 \times 10^8$ m/s),
• $\varepsilon_r$ is the relative permittivity of the medium (in this case, distilled water, and is given as 81),
• $\mu_r$ is the relative permeability of the medium (given as 1 for distilled water).

Substituting the given values into the equation, we get:

$v = \frac{3 \times 10^8 \, \text{m/s}}{\sqrt{81 \times 1}}$

$v = \frac{3 \times 10^8 \, \text{m/s}}{9}$

$v = 3.33 \times 10^7 \, \text{m/s}$

So, the speed of light in distilled water is $3.33 \times 10^7$ m/s.

The answer is Option B : the distance between consecutive fringes will decrease.

Young's double-slit experiment depends on the principle of interference of light waves, and the resulting pattern of light and dark bands (fringes) depends on the wavelength of the light used.

The formula for the distance between fringes (y) in the double slit experiment is :

y = ${{\lambda D} \over d}$

where :

• λ is the wavelength of the light
• D is the distance from the slits to the screen
• d is the distance between the slits

The wavelength of orange light is approximately 600 nm while the wavelength of blue light is around 475 nm. Therefore, if you change the light from orange to blue, the wavelength λ decreases. As λ is directly proportional to y, if λ decreases, y will also decrease. So, the distance between consecutive fringes will decrease.

$I = {e \over T} = {{e\omega } \over {2\pi }} = {{eV} \over {2\pi r}}$

$I = {{1.6 \times {{10}^{ - 19}} \times 2.2 \times {{10}^6} \times 7} \over {2 \times 22 \times 0.5 \times {{10}^{ - 10}}}}$

= 1.12 mA

= 112 $\times$ 10^$-$2 mA

Ratio of moment of inertia = ${{{1 \over 2}M{R^2}} \over {{1 \over 4}m{r^2}}}$

$= {{2\sigma \pi {R^2}{R^2}} \over {\sigma \pi {r^2}{r^2}}} = {{2{R^4}} \over {{r^4}}}$

T_m =30 ms

C = 200 $\mu$F

${q \over N} = {{{Q_0}{e^{ - t/RC}}} \over {{N_0}{e^{ - \lambda t}}}} = {{{Q_0}} \over {{N_0}}}{e^{t\left( {\lambda - {1 \over {RC}}} \right)}}$

Since q/N is constant hence

$\lambda ={1 \over {RC}}$

$R = {1 \over {\lambda C}} = {{{T_m}} \over C} = {{30 \times {{10}^{ - 3}}} \over {200 \times {{10}^{ - 6}}}} = 150\Omega$

$u = \sqrt {2gh}$

Now,

$S = {h \over 3}$

a = $-$g

$S = ut + {1 \over 2}a{t^2}$

${h \over 3} = \sqrt {2gh} t + {1 \over 2}( - g){t^2}$

${t^2}\left( {{g \over 2}} \right) - \sqrt {2gh} t + {h \over 3} = 0$

From quadratic equation

${t_1},{t_2} = {{\sqrt {2gh} \pm \sqrt {2gh - {{4g} \over 2}{h \over 3}} } \over g}$

${{{t_1}} \over {{t_2}}} = {{\sqrt {2gh} - \sqrt {{{4gh} \over 3}} } \over {\sqrt {2gh} + \sqrt {{{4gh} \over 3}} }} = {{\sqrt 3 - \sqrt 2 } \over {\sqrt 3 + \sqrt 2 }}$

Least Count = ${{Pitch} \over {Total\,divisions\,on\,circular\,scale}}$

In 5 revolution, distance travel, 5 mm

In 1 revolution, it will travel 1 mm.

So least count = ${1 \over {50}}$ = 0.02

For photon ${\lambda _1} = {h \over P} = {{6.6 \times {{10}^{ - 34}}} \over {{{10}^{ - 27}}}}$

For particle ${\lambda _2} = {h \over {mv}} = {{6.6 \times {{10}^{ - 34}}} \over {9.1 \times {{10}^{ - 31}} \times {{10}^6}}}$

$\therefore$ ${{{\lambda _1}} \over {{\lambda _2}}} = 910$

N = N₀e^{$-$$\lambda$t}

N_d = N₀ $-$ N

N_d = N₀ (1 $-$ e^{$-$$\lambda$t})

${{{N_d}} \over {{N_0}}} = f = 1 - {e^{ - \lambda t}}$

$\Rightarrow$ ${{df} \over {dt}} = \lambda {e^{ - \lambda t}}$

$T = 2\pi \sqrt {{I \over {MB}}}$

B = 80 $\times$ 10^$-$4 = 8 mT

${V_C} = {{2CV + CV} \over {KC + 2C}}$

$= {{3V} \over {K + 2}}$

Polygon law is applicable in both but the equation given in the reason is not useful in explaining the assertion.

$E = {1 \over 2}K{a^2}$

${{3E} \over 4} = {1 \over 2}K({a^2} - {y^2})$

$\Rightarrow$ ${3 \over 4} \times {1 \over 2}K{a^2} = {1 \over 2}K({a^2} - {y^2})$

$\Rightarrow$ ${y^2} = {a^2} - {{3{a^2}} \over 4}$

$\Rightarrow$ $y = {a \over 2}$

$V = {V_0}\left( {1 - {e^{ - {t \over {RC}}}}} \right)$

$50 = 100\left( {1 - {e^{ - {t \over {RC}}}}} \right)$

$= t = 0.69 \times {10^{ - 4}}$ sec.

What T₁ and T₂ are connected, then the steady state current in the inductor $I = {6 \over 6} = 1A$

When T₁ and T₃ are connected then current through inductor remains same. So potential difference across 3$\Omega$

V = Ir = 1 $\times$ 3 = 3 Volt

$\beta = {{{I_C}} \over {{I_B}}} = 24;$

R_C = 1000

$\Delta$V = 0.6

${I_C} = {{0.6} \over {1000}}$

I_C = 6 $\times$ 10^$-$4

${I_B} = {{{I_C}} \over \beta } = {{6 \times {{10}^{ - 4}}} \over {24}} = 25\mu A$

For no sliding

f $\ge$ $\rho$av²

$\mu$mg $\ge$ $\rho$av²

$\mu$$\rho$Ahg $\ge$ $\rho$a2gh

$\mu \ge {{2a} \over A}$

Option (3)

Acceleration due to gravity will be zero at P therefore,

${1 \over 2}m{v^2} - {{GMm} \over R} - {{G9Mm} \over {7R}} = 0 - {{GMm} \over {2R}} - {{G9Mm} \over {6R}}$

$\therefore$ $V = \sqrt {{4 \over 7}{{GM} \over R}}$

KE = ${3 \over 2}kT$

PV = ${N \over {{N_A}}}RT$

N = ${{PV} \over {kT}}$

= N = 1.5 $\times$ 10¹¹

$T\cos \theta = mg$

$T\sin \theta = {{k{q^2}} \over {{x^2}}}$

$\tan \theta = {{k{q^2}} \over {{x^2}mg}}$

$\tan \theta \approx \sin \theta \approx {x \over {2L}}$

${x \over {2L}} = {{K{q^2}} \over {{x^2}mg}}$

$x = {\left( {{{{q^2}L} \over {2\pi {\varepsilon _0}mg}}} \right)^{1/3}}$

Collision between A and B


m $\times$ 9 = mv₁ + 2mv₂ (from momentum conservation)

$e = 1 = {{{v_2} - {v_1}} \over 9}$

$\Rightarrow$ v₂ = 6 m/sec. v₁ = $-$3 m/sec.

collision between B and C


2m $\times$ 6 = 4mv (from momentum conservation)

v = 3 m/s

Current is constant in conductor

i = constant

Resistance of element $dR = {{\rho dx} \over {\pi {r^2}}}$

$dV = idR = {{i\rho dx} \over {\pi {r^2}}}$

$E = {{dV} \over {dx}} = {{i\rho } \over {\pi {r^2}}}$

& ${V_d} = {{eE\tau } \over m}$

$\therefore$ ${V_d} \propto E$

$\to E \propto {1 \over {{r^2}}}$

if r decreases, E will increase : V_d will increase

Work done in adiabatic process = ${{ - nR} \over {\gamma - 1}}({T_f} - {T_i})$

$\therefore$ ${W_{AD}} = {{ - nR} \over {\gamma - 1}}({T_2} - {T_1})$

and ${W_{BC}} = {{ - nR} \over {\gamma - 1}}({T_2} - {T_1})$

$\therefore$ ${W_{AD}} = {W_{BC}}$

Triamminetrinitrocobalt (III) $\to$ [Co(NO₂)₃(NH₃)₃]

trioxalatochromate (III) ion $\to$ [Cr(C₂O₄)₃]^3$-$[Co(NO₂)₃(NH₃)₃]



X + Y = 2 + 0 = 2.0

Eutrophication leads to decrease in oxygen level of water.

3^rd statement is incorrect.

Lithium due to small size has very high polarization capability and thus increases covalent nature of Halides.

Staggered and eclipsed conformers of ethane also known as rotamers.

Rutherford's gold foil experiment only proved that electrons are held towards nucleus by electrostatic forces of attraction and move in circular orbits with very high speeds.

Bohr's model gave exact formula for simultaneous calculation of speed & distance of electron from the nucleus, something which was deemed impossible according to Heisenberg.

Explanation : aniline is more basic than acetamide because in acetamide, lone pair of nitrogen is delocalized to more electronegative element oxygen.

In Aniline lone pair of nitrogen delocalized over benzene ring.

[MnCl₆]^3$-$



Paramagnetic and having 4 unpaired electrons.

Consider 1 litre solution

mass of solution = (1.2 $\times$ 1000)g = 1200 gm

Neglecting volume of NaOH

Mass of water = 1000 gm

$\Rightarrow$ Mass of NaOH = (1200 $-$ 1000)gm = 200 gm

$\Rightarrow$ Moles of NaOH = ${{200g} \over {50g/mol}} = 5$ mol

$\Rightarrow$ molality = ${{5mol} \over {1kg}} = 5$ m

Freundlich isotherm;

${x \over m} = k.{p^{{1 \over n}}}$

Substituting values ;

$\left( {{{64} \over 1}} \right) = {\left( 2 \right)^{{1 \over n}}} \Rightarrow n = {1 \over 6} = 0.166$

$\cong 17 \times {10^{ - 2}}$

$\to$ furacine acts as Antiseptic

$\to$ Arsphenamine also known as salvarsan acts as antibiotic

$\to$ Dimetone is synthetic histamine

$\to$ valium is a Tranquilizer

$\pi$ = CRT;

$\pi$ = osmotic pressure

C = molarity

T = Temperature of solution

let the molar mass be M gm/mol

2.42 $\times$ 10^$-$3 bar = ${{\left( {{{1.46g} \over {Mgm/mol}}} \right)} \over {0.1l}} \times \left( {{{0.083l - bar} \over {mol - K}}} \right) \times (300K)$

$\Rightarrow$ M = 15.02 $\times$ 10⁴ g/mol

NaOH $\to$ Basic

Be(OH)₂ $\to$ Amphoteric

Ca(OH)₂ $\to$ Basic

B(OH)₃ $\to$ Acidic

Al(OH)₃ $\to$ Amphoteric

$\Lambda _m^{} = 1000 \times {\kappa \over M}$

$= 1000 \times {{2 \times {{10}^{ - 5}}} \over {0.001}} = 20$ S cm² mol^$-$1

$\Rightarrow \alpha = {{\Lambda _m^{}} \over {\Lambda _m^\infty }} = {{20} \over {190}} = \left( {{2 \over {19}}} \right)$

HA $\rightleftharpoons$ H⁺ + A^$-$

$0.001(1 - \alpha )0.001\alpha 0.001\alpha$

$\Rightarrow {k_a} = 0.001\left( {{{{\alpha ^2}} \over {1 - \alpha }}} \right) = {{0.001 \times {{\left( {{2 \over {19}}} \right)}^2}} \over {1 - \left( {{2 \over {19}}} \right)}}$

$= 12.3 \times {10^{ - 6}}$

Ellingham diagram is a plot between $\Delta$G$^\circ$ and T and does not give any information regarding rate of reaction.

Explanation $\Rightarrow$ 2-4-D.N.P. test is used for carbonyl compound (aldehyde & ketone)

Oxidation state of P in H₄P₂O₇, H₄P₂O₅ and H₄P₂O₆ is 5, 3 & 4 respectively

H₄P₂O₇

2x + 4(+ 1) + 7 ($-$2) = 0

x = + 5

${H_4}\underline {{P_2}} {O_5}$

2x + 4(+ 1) + 5 ($-$2) = 0

x = + 3

${H_4}\underline {{P_2}} {O_6}$

2x + 4 (+ 1) + 6($-$2) = 0

x = +4

Rate = k[A]ⁿ

comparing units

${{(mol/l)} \over {\sec }} = k{\left( {{{mol} \over l}} \right)^n}$

$\Rightarrow k = mo{l^{(l - n)}}{l^{(n - 1)}}{s^{ - 1}}$

Given structure of compound A is thymine, so its complementary base will be adenine.

Adenine form hydrogen bonds with thymine and cytosine will form hydrogen bond with guanine.

Barford test is used for distinguish monosaccharide from disaccharide.

Electrolysis of concentrated solution of acidified sulphate solution yields H₂S₂O₈.

Bond order of CO = 3

Bond order of NO⁺ = 3

Difference = 0 = ${x \over 2}$

$\Rightarrow$ x = 0

Note :

(1) $\,$ Bond order $= {1 \over 2}$ [N_b $-$ N_a]

N_b = No of electrons in bending molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

(4) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is



Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(5) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -



Here N_a = 10

and N_b = 10

(A)    CO has 14 electrons.

Moleculer orbital configuration of CO is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

$\therefore$ N_b = 10

N_a = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}$ [ 10 $-$ 4] = 3

(B)   NO⁺ has 14 electrons.

Moleculer orbital configuration of NO⁺ is

${\sigma _{1{s^2}}}$ $\sigma _{1{s^2}}^ *$ ${\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,\, = \,\,{\pi _{2p_y^2}}$

$\therefore$ N_b = 10

N_a = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}$ [ 10 $-$ 4] = 3