JEE Main 2021 26th February Evening Shift Question Paper with Solutions

As we know time period of second’s penduklum is 2 sec, so statement (1) is incorrect.

Time taken by particle performing SHM between two extreme position is half of the time period.

Here, T = 2 sec.

So, time = 2/2 = 1 sec

Using energy conservation between A and B point

$mgh = {1 \over 2}m{(wR)^2} + {1 \over 2}I{\omega ^2}$

$2mgh = (M{R^2} + I){\omega ^2}$

${\omega ^2} = {{2mgh} \over {I + M{R^2}}}$

$(27)\left( {{4 \over 3}\pi {r^3}} \right) = {4 \over 3}\pi {R^3}$

R = 3r

Potential energy of smaller drop :

${U_1} = {3 \over 5}{{k{q^2}} \over r}$

Potential energy of bigger drop :

$U = {3 \over 5}{{k{Q^2}} \over R}$

$U = {3 \over 5}{{k{{(27q)}^2}} \over R}$

$U = {3 \over 5}k{{(27)(27){q^2}} \over {3r}}$

$U = {{(27)(27)} \over 3}\left( {{3 \over 5}{{k{q^2}} \over r}} \right)$

$U = 243\,{U_1}$

We know that work done is

$W = \int {PdV}$ .... (1)

$\Rightarrow P = {{nRT} \over V}$ .... (2)

$\Rightarrow W = \int {{{nRT} \over V}dv}$ .... (3)

and given $V = K{T^{2/3}}$ .... (4)

$\Rightarrow W = \int {{{nRT} \over {K{T^{2/3}}}}.dv}$ .... (5)

$\Rightarrow$ from (4) : $dv = {2 \over 3}K{T^{ - 1/3}}dT$

$\Rightarrow W = \int\limits_{{T_1}}^{{T_2}} {{{nRT} \over {K{T^{2/3}}}}{2 \over 3}K{1 \over {{T^{1/3}}}}} dT$

$\Rightarrow W = {2 \over 3}nR \times \left( {{T_2} - {T_1}} \right)$ .... (6)

$\Rightarrow {T_2} - {T_1} = 90K$ .... (7)

$\Rightarrow W = {2 \over 3}nR \times 90$

$\Rightarrow W = 60nR$

Assuming 1 mole of gas

n = 1

So, W = 60R

Initially beat frequency = 5Hz

so, $\rho$_A = 340 $\pm$ 5 = 345 Hz, or 335 Hz

after filing frequency increases slightly so, new value of frequency of A > $\rho$_A

Now, beat frequency = 2Hz

$\Rightarrow$ new $\rho$_A = 340 $\pm$ 2 = 342 Hz, or 338 Hz

hence, original frequency of A is $\rho$_A = 335 Hz

Net charge on electric dipole = + q $-$ q = 0

Hence, according to Gauss's law,

Electric flux, $\phi = {{{q_{net}}} \over {{\varepsilon _0}}} = {0 \over {{\varepsilon _0}}} = 0$

Electric field due to electric dipole is non-zero and varies at point to point.

Hence, statement I is true.

Electric field due to charged solid sphere at a distance r from centre.

$E = {1 \over {4\pi {\varepsilon _0}}}\,.\,{{Qr} \over {{R^3}}}$ [when r < R, R $\to$ radius] which is non-zero.

Hence, statement II is false.

Hence, option (c) is the correct.

$\varepsilon$_ind = (B_v) LV and B_v = B_Total sin60^o

$\therefore$ $\varepsilon$_ind = (2.5 $\times$ 10^$-$4)(sin 60^o) $\times$ 10 $\times$ 180 $\times$ ${5 \over {18}}$

= 108.25 mV

Let spring force acting on the both block is F_s.

F.B.D. of Block A



$\therefore$ F_s = Ma ..... (i)

F.B.D. of Block B,



$\therefore$ Acceleration of Block B,

${a_1} = {{{F_{net}}} \over M} = {{F - {F_s}} \over M} = {{F - Ma} \over M}$

Y = $\overline {\overline A + B}$ = $A.\overline B$

A = 0 $\Rightarrow$ Y = 0 for B = 0 or B = 1

A = 1 $\Rightarrow$ Y = 0 for B = 1

A = 1 $\Rightarrow$ Y = 1 for B = 0

Hence output is 1 only when input A = 1 and B = 0.

$\therefore$ A is the correct output.

B. W. (Bandwidth) = 2 $\times$ maximum frequency at modulating signal

= 2 $\times$ 5kHz

= 10 kHz

$\therefore$ No of stations accommodate

= ${{90} \over {10}}$ = 9

We know, z = $\sqrt {{{({x_L} - {x_C})}^2} + {R^2}}$

${x_L} = {\omega _L} = 100 \times 100 \times {10^{ - 3}} = 10\Omega$

${x_C} = {1 \over {{\omega _C}}} = {1 \over {100 \times 100 \times {{10}^{ - 6}}}} = 10\Omega$

$\therefore$ $z = \sqrt {{{(10 - 100)}^2} + {R^2}} = \sqrt {{{90}^2} + {{120}^2}}$

$= 30 \times 5 = 150\Omega$

${i_{peak}} = {{\Delta v} \over z} = {{30} \over {150}} = {1 \over 5} amp = 0.2$ amp

For resonant frequency,

$\omega L = {1 \over {\omega C}} \Rightarrow {\omega ^2} = {1 \over {LC}} \Rightarrow \omega = {1 \over {\sqrt {LC} }}$

& $f = {1 \over {2\pi \sqrt {LC} }} = {1 \over {2\pi \sqrt {100 \times {{10}^{ - 3}} \times 100 \times {{10}^{ - 6}}} }}$

$= {{100\sqrt {10} } \over {2\pi }} = {{100\pi } \over {2\pi }} = 50$ Hz

as $\sqrt {10} \approx \pi$

y = $\alpha$x $-$ $\beta$x²

comparing with trajectory equation

$y = x\tan \theta - {1 \over 2}{{g{x^2}} \over {{u^2}{{\cos }^2}\theta }}$

$\tan \theta = \alpha \Rightarrow \theta = {\tan ^{ - 1}}\alpha$

$\beta = {1 \over 2}{g \over {{u^2}{{\cos }^2}\theta }}$

$\Rightarrow$ ${u^2} = {g \over {2\beta {{\cos }^2}\theta }}$

Maximum height H :

$H = {{{u^2}{{\sin }^2}\theta } \over {2g}} = {g \over {2\beta {{\cos }^2}\theta }}{{{{\sin }^2}\theta } \over {2g}}$

$\Rightarrow$ $H = {{{{\tan }^2}\theta } \over {4\beta }} = {{{\alpha ^2}} \over {4\beta }}$

Let initial activity be A₀

A = A₀ e^{$-$$\lambda$t₁} ........(i)

${A \over 5}$ = A₀ e^{$-$$\lambda$t₂} .......(ii)

(i) $\div$ (ii)

5 = e^{$\lambda$(t₂ $-$ t₁)}

$\lambda$ = ${{\ln 5} \over {{t_2} - {t_1}}} = {1 \over \tau }$

$\tau = {{{t_2} - {t_1}} \over {\ln 5}}$

From thermodynamics law:

$Q = \Delta U + W$

$Q = \Delta U + {Q \over 5}$

$\Delta U = {{4Q} \over 5}$

$n{C_v}\Delta T = {4 \over 5}nC\Delta T$

${5 \over 4}{C_v} = C$

$C = {5 \over 4}\left( {{f \over 2}} \right)R = {5 \over 4}\left( {{5 \over 2}} \right)R$

$C = {{25} \over 8}R$

$X = 25$

U = 3PV + 4

$\Rightarrow$ ${{nf} \over 2}$RT = 3PV + 4

$\Rightarrow$ ${{f} \over 2}$PV = 3PV + 4

$\Rightarrow$ f = 6 + ${8 \over {PV}}$

Since degree of freedom is more than 6 therefore gas is polyatomic.

($\Delta$E) Releases when photon going from n = 5 to n = 1

$\Delta$E = (13.6 $-$ 0.54) eV = 13.06 eV.

P_i = P_f (By linear momentum conservation)

$0 = {h \over \lambda } - Mv = {V_{{\mathop{\rm Re}\nolimits} coil}} = {h \over {\lambda M}}$ ..... (i)

& $\Delta E = {{hc} \over \lambda } = {{hc} \over {\lambda M}} \times M = Mc{V_{{\mathop{\rm Re}\nolimits} coil}}$

${V_{{\mathop{\rm Re}\nolimits} coil}} = {{\Delta E} \over {Mc}} = {{13.06 \times 1.6 \times {{10}^{ - 19}}} \over {1.67 \times {{10}^{ - 27}} \times 3 \times {{10}^8}}}$ = 4.17 m/sec

R₀ = 1$\Omega$

R₁ = ?

l₀ = 1m

l₁ = 1.25 m

A₀ = A

As volume of wire remains constant so

A₀l₀ = A₁l₁ $\Rightarrow$ A₁ = ${{{l_0}{A_0}} \over {{l_1}}}$

Now

Resistance (R) = ${{pl} \over A}$

${{{R_0}} \over {{R_1}}} = {{{l_0}} \over {{A_0}}}\left( {{{{l_0}{A_0}} \over {{l_1} \times {l_1}}}} \right)$

$\Rightarrow$ ${R_1} = {{l_1^2} \over {l_0^2}} = 1.5625 \,\Omega$

So % change in resistance

$= {{{R_1} - {R_0}} \over {{R_0}}} \times 100\%$

$= {{1.5625 - 1} \over 1} \times 100\%$

$= 56.25\%$

Assuming Hooke's law to be valid.

$T \propto (\Delta l)$

$T = k(\Delta l)$

Let, l₀ = natural length (original length)

$\Rightarrow T = k(l - {l_0})$

so, ${T_1} = k({l_1} - {l_0})$ & ${T_2} = k({l_2} - {l_0})$

$\Rightarrow {{{T_1}} \over {{T_2}}} = {{{l_1} - {l_0}} \over {{l_2} - {l_0}}}$

$\Rightarrow {l_0} = {{{T_2}{l_1} - {T_1}{l_2}} \over {{T_2} - {T_1}}}$

Here $\overrightarrow a = \left| {\overrightarrow a } \right|\sin \theta \widehat i - \left| {\overrightarrow a } \right|\cos \theta \widehat j$

As $\overrightarrow a$ is an unit vector, so $\left| {\overrightarrow a } \right|$ = 1

$\therefore$ $\overrightarrow a = \left| {\overrightarrow a } \right|\sin \theta \widehat i - \left| {\overrightarrow a } \right|\cos \theta \widehat j$

= $\sin \theta \widehat i - \cos \theta \widehat j$

Similarly $\overrightarrow b = \sin \theta \widehat i + \cos \theta \widehat j$

and $\overrightarrow c = \widehat j$

From option (B),

$\overrightarrow a$ $-$ 2 ($\overrightarrow a$ . $\overrightarrow c$)$\overrightarrow c$

= $\sin \theta \widehat i + \cos \theta \widehat j$ = $\overrightarrow b$

$K{E_{\max }} = hv - \phi$

${1 \over 2}m{v^2} = hv - \phi$

$v = \sqrt {{{2(hv - \phi )} \over m}}$

Given $h{v_1} = 2\phi$

$h{v_2} = 10\phi$

$\therefore$ ${{{v_1}} \over {{v_2}}} = \sqrt {{{h{v_1} - \phi } \over {h{v_2} - \phi }}}$

${{{v_1}} \over {{v_2}}} = \sqrt {{{2\phi - \phi } \over {10\phi - \phi }}} = {1 \over 3}$ = ${x \over y}$

$\therefore$ x = 1

For a particle executes S.H.M.

$V = \omega \sqrt {{a^2} - {x^2}}$

Given, $V = {{{V_{\max }}} \over 2} = {{A\omega } \over 2}$

${{{A^2}{\omega ^2}} \over 4} = {\omega ^2}{a^2} - {\omega ^2}{x^2}$

$x = {{\sqrt 3 } \over 2}a$

For a body performing SHM, relation between velocity and displacement

$v = \omega \sqrt {{A^2} - {x^2}}$

now, square both side

${v^2} = {w^2}({A^2} - {x^2})$

$\Rightarrow {v^2} = {w^2}{A^2} - {\omega ^2}{x^2}$

${v^2} + {\omega ^2}{x^2} = {\omega ^2}{A^2}$

divide whole equation by ${{\omega ^2}{A^2}}$

${{{v^2}} \over {{\omega ^2}{A^2}}} + {{{\omega ^2}{x^2}} \over {{\omega ^2}{A^2}}} = {{{\omega ^2}{x^2}} \over {{\omega ^2}{A^2}}}$

${{{v^2}} \over {{{(\omega A)}^2}}} + {{{x^2}} \over {{{(A)}^2}}} = 1$

above equation is similar as standard equation of ellipses, so graph between velocity and displacement will be ellipses.

The formation of image with simple microscope is shown below.

Here, $\theta ' = {h \over {{u_0}}} = {{h'} \over D} = {{h'} \over {25}}$

where, D = 25 cm (least distance of distinct vision)

Here, $\theta$' is same for both object and image, hence Assertion is true.

Magnification, $m = {{\theta '} \over \theta } = {D \over {{u_0}}}$

Hence, if u₀ < D (25 cm), hence, the value of $\theta$' will obtain large.

So, option (c) is the correct.

From given information :

For 1^st interval

${a_1} = {{{v_0}} \over {{t_1}}}$

v₀ = a₁ t₁ ....... (1)

For 2^nd interval

${a_2} = {{{v_0}} \over {{t_2}}}$

v₀ = a₂ t₂ ..... (2)

from (1) & (2)

a₁ t₁ = a₂ t₂

${{{t_1}} \over {{t_2}}} = {{{a_2}} \over {{a_1}}}$

${g_A} = {{GM(r)} \over {{R^3}}}$

${g_C} = {{GM} \over {{{\left( {R + {R \over 2}} \right)}^2}}}$

Given, g_A = g_C

$\Rightarrow {{GM(r)} \over {{R^3}}} = {{GM} \over {{{\left( {R + {R \over 2}} \right)}^2}}}$

$\Rightarrow {r \over {{R^3}}} = {4 \over {9{R^2}}}$

$\Rightarrow r = {{4R} \over 9}$

$\therefore$ $OA = {{4R} \over 9}$

$\therefore$ $AB = R - r = {{5R} \over 9}$

$\therefore$ $OA:AB = 4:5$

$\therefore$ x = 4

Given, length of mirror, m = 50 cm = 50 $\times$ 10^$-$2 m

Distance of source from mirror, d = 60 cm = 60 $\times$ 10^$-$2 m

Distance of man from mirror, d_m = 1.2 m

By using the concept of ray diagram of plane mirror shown below

Now, using the concept of similar triangle,

$\Delta$HAI $\sim$ $\Delta$GAE and $\Delta$BAI $\sim$ $\Delta$CAE

$\therefore$ ${{AI} \over {AE}} = {{HI} \over {EG}}$

$\Rightarrow {{0.60} \over {1.8}} = {{0.25} \over {EG}}$ ($\because$ AI = IS)

$\Rightarrow EG = 0.25 \times {{1.8} \over {0.6}} = 0.25 \times 3 = 0.75$ m

As, $CG = 2EG$

$\Rightarrow CG = 0.75 \times 2 = 1.50$ m

Hence, distance between the extreme points, where he can see image of light source in mirror is 150 cm.

${5 \over 8}$ oscillation = ${1 \over 2}$ oscillation + ${1 \over 8}$ oscillation

From figure, A to B = ${1 \over 2}$ oscillation and B to C is ${1 \over 8}$ oscillation.

$\therefore$ $\pi { + \,\theta = \omega t}$

$\Rightarrow$ $\pi { + {{\pi{} } \over 6}} = \omega t$

$\Rightarrow$ ${{7\pi {} } \over 6} = \left( {{{2\pi {} } \over T}} \right)t$

$\Rightarrow$ t = ${{7T} \over {12}}$

f = mg sin$\theta$ $-$ ($\mu$N + qE cos$\theta$)

F = mg sin$\theta$ $-$ $\mu$(mg cos$\theta$ + qE sin$\theta$) $-$ qE cos$\theta$

F = 1 $\times$ 10 $\times$ sin30
$-$ 0.2 (1 $\times$ 10 $\times$ cos30$^\circ$ + 200 $\times$ 5 $\times$ 10^$-$3 sin30$^\circ$)
$-$ 200 $\times$ 5 $\times$ 10^$-$3 cos30$^\circ$

$\Rightarrow$ F = 2.3 N

$a = {F \over m} = {{2.3} \over 1} = 2.3$ m/sec²

$t = \sqrt {{{25} \over 9}} = \sqrt {{{2 \times 2} \over {2.3}}} = 1.3$ sec

Seliwanoff test and Xanthaproteic test are used for identification of 'Ketoses' and proteins respectively.

Tritium isotope of hydrogen is radioactive and emits low energy $\beta ^-$ particles. It is because of high n/p ratio of tritium which makes nucleus unstable.

(a) Sodium carbonate is prepared by Solvay process

(b) Titanium is refined by Van-Arkel process

(c) Chlorine is prepared by Deacon process

(d) Sodium hydroxide is prepared by Castner-Kellner process

Calgon is sodium hexametaphosphate, a polymeric compound also called as Graham’s salt.

Order of abundance of element in earth crust is

O > Si > Al > Fe > Ca > Na > Mg > K

So second most abundant element in earth crust is Si not Ca.

Sucrose $\to \alpha$-D- Glucose and $\beta$-D- Fructose

Lactose $\to \beta$-D- Galactose and $\beta$-D- Glucose

Maltose $\to \alpha$-D- Glucose and $\alpha$-D- Glucose

If FeCl₃ is added to excess of hot water, a positively charged sol of hydrated ferric oxide is formed due to adsorption of Fe³⁺ ions.

A : Due to inert pair effect, Tl is more stable in +1 oxidation state
     Hence TlI₃ and CSI₃ are isomorphous

R : Electronic configuration of Tl (81) = [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p¹

Both A and R are correct but R is not the correct explanation of A.

Oxygen is the second most electronegative element in comparison to fluorine. In group - 16 family $(\mathrm{O}, \mathrm{S}, \mathrm{Se}, \mathrm{Te})$, O-atom is smallest in size. So, electron density on O-atom is very high in group -16 During addition of a free electron to gaseous $\mathrm{O}$-atom,

$\mathrm{O}(\mathrm{g})+\mathrm{e}^{-} \longrightarrow \mathrm{O}^{-}(\mathrm{g})$

We have to supply a significant amount of energy (endothermic) to overcome the electrostatic repulsion between the approaching electron and O-atom of very high electron density. So, the net value of electron affinity (EA) or (negative) electron gain enthalpy $\left[\Delta_{\mathrm{eg}} H\right.$ or $\left.\left|\Delta_{\mathrm{eg}} H\right|\right]$ of oxygen decreases to a higher extent in comparison to other elements of group -16 who have larger size and lower electronegativity.

So, the correct order of EA or $\left|\Delta_{\mathrm{eg}} \mathrm{H}\right|$ of group $-16$ elements will be $\mathrm{S}$ $>\mathrm{Se}>\mathrm{Te}>\mathrm{O}$

Siderite $\to$ FeCO₃

Calamine $\to$ ZnCO₃

Malachite $\to$ CuCO₃.Cu(OH)₂

Cryolite $\to$ Na₃AlF₆

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