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

Maximum emf $\varepsilon = N\,\omega AB$

N = 20, $\omega$ = 50, B = 3 $\times$ 10^$-$2 T

$\varepsilon$ = 20 $\times$ 50 $\times$ $\pi$ $\times$ (0.08)² $\times$ 3 $\times$ 10^$-$2 = 60.28 $\times$ 10^$-$2

Rounded off to nearest integer = 60

$i = {{10V} \over {5k\Omega }}$ = 2mA

$I = {{14V} \over {1k\Omega }}$ = 14 mA

$\therefore$ I_z = 12 mA

$\therefore$ P = I_zV_z = 120 mW

y₁ = A₁ sin k(x $-$ vt)

y₁ = 12 sin 6.28 (x $-$ vt)

y₂ = 5 sin 6.28 (x $-$ vt + 3.5)

$\Delta \phi = {{2\pi } \over \lambda }(\Delta x)$

$= K(\Delta x)$

$= 6.28 \times 3.5 = {7 \over 2} \times 2\pi = 7\pi$

${A_{net}} = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \phi }$

${A_{net}} = \sqrt {{{(12)}^2} + {{(5)}^2} + 2(12)(5)\cos (7\pi )}$

$= \sqrt {144 + 25 - 120}$

${Z_C} = \sqrt {{{\left( {{1 \over {\omega C}}} \right)}^2} + {R^2}}$

$= \sqrt {{{\left( {{1 \over {100 \times 100 \times {{10}^{ - 6}}}}} \right)}^2} + {{100}^2}}$

${Z_C} = \sqrt {{{(100)}^2} + {{(100)}^2}}$

$= 100\sqrt 2$

${Z_L} = \sqrt {{{(\omega L)}^2} + {R^2}}$

$\sqrt {{{(100 \times 0.5)}^2} + {{50}^2}}$

$= 50\sqrt 2$

${i_C} = {{200} \over {{Z_C}}} = {{200} \over {100\sqrt 2 }} = \sqrt 2$

${i_L} = {{200} \over {{Z_L}}} = {{200} \over {50\sqrt 2 }} = 2\sqrt 2$

$\cos {\phi _1} = {{100} \over {10\sqrt 2 }} = {1 \over {\sqrt 2 }} \Rightarrow {\phi _1} = 45^\circ$

$\cos {\phi _2} = {{50} \over {50\sqrt 2 }} = {1 \over {\sqrt 2 }} \Rightarrow {\phi _2} = 45^\circ$



$I = \sqrt {I_C^2 + I_L^2}$

$= \sqrt {2 + 8}$

$= \sqrt {10}$

I = 3.16 A

when x and y are mixed, T_f1 = 16$^\circ$C

m₁s₁T + m₂s₂T₂ = (m₁s₁ + m₂s₂)Tf₁

s₁ $\times$ 10 + s₂ $\times$ 20 = (s₁ + s₂) $\times$ 16

s₁ = ${2 \over 3}$s₂ .... (i)

when y and z are mixed, T_f2 = 26$^\circ$C

m₂s₂T + m₃s₃T₃ = (m₃s₃ + m₃s₃)Tf₂

s₂ $\times$ 20 + s₃ $\times$ 30 = (s₂ + s₃) $\times$ 26

s₃ = ${3 \over 2}$s₂ ..... (ii)

when x and z are mixed

m₁s₁T₁ + m₃s₃T₃ = (m₁s₁ + m₃s₃)Tf

${2 \over 3}$s₂ $\times$ 10 + ${2 \over 3}$s₂ $\times$ 20 = $\left( {{2 \over 3}{s_2} + {3 \over 2}{s_2}} \right){T_f}$

T_f = 23.84$^\circ$C

$y = \overline {(\overline {A + \overline {A + B} } ) + (\overline {B + \overline {A + B} } )}$

$y = (A + \overline {A + B} ).(B + \overline {A + B} )$

A	B	y
0	0	1
0	1	0
1	0	0
1	1	1

${d_t} = \sqrt {2R{h_1}} + \sqrt {2R{h_2}}$

$= \sqrt {2R} \left( {\sqrt {{h_1}} + \sqrt {{h_2}} } \right)$

$= {(2 \times 6400 \times {10^3})^{1/2}}(\sqrt {50} + \sqrt {80} )$

$= 3578(7.07 + 8.94)$

$= 57.28$ km

${x_2} = 5\sqrt 2 \left( {{1 \over {\sqrt 2 }}\sin 2\pi t + {1 \over {\sqrt 2 }}\cos 2\pi t} \right)\sqrt 2$

$= 10\sin \left( {2\pi t + {\pi \over 4}} \right)$

$\therefore$ ${{{A_2}} \over {{A_1}}} = {{10} \over 5} = 2$

$I = {{{I_0}} \over 2}{\cos ^2}\phi$



${{{I_0}} \over 2}{\cos ^2}\phi = {{3I} \over 8}$

${\cos ^2}\phi = {3 \over 4}$

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

$\Rightarrow \phi = 30$

$r = {l \over {\sqrt 2 }}$

$\sin \theta = {r \over l} = {l \over {\sqrt 2 }}$

$\theta$ = 45$^\circ$

$T\sin \theta = {{m{v^2}} \over r}$

$T\cos \theta = mg$

$\tan \theta = {{{v^2}} \over {rg}} \Rightarrow v = \sqrt {rg}$

For the object to coincide with image, the light must fall perpendicularly to mirror. Which means that the light will have to converge at C of mirror. Without the mirror also, the light would coverage at C.

So the distance is : 12 + 8 + 30 = 50 cm

$T = {{2{m_1}{m_2}g} \over {{m_1} + {m_2}}} = {{2 \times 3 \times 5 \times 10} \over 8}$

$= {{75} \over 2}$

Stress $= {T \over A}$

${{24} \over \pi } \times {10^2} = {{75} \over {2 \times \pi {R^2}}}$

${R^2} = {{75} \over {2 \times 24 \times 100}} = {3 \over {8 \times 24}}$

$\Rightarrow$ R = 0.125 m

R = 12.5 cm

${V_A} = {{KQ} \over a} - {{KQ} \over {\sqrt {{a^2} + {s^2}} }}$

${V_B} = {{ - KQ} \over a} + {{KQ} \over {\sqrt {{a^2} + {s^2}} }}$

${V_A} - {V_B} = {{2KQ} \over a} - {{2KQ} \over {\sqrt {{a^2} + {s^2}} }}$

$= {Q \over {2\pi {\varepsilon _0}}}\left( {{1 \over a} - {1 \over {\sqrt {{s^2} + {a^2}} }}} \right)$

Parallel axis theorem

I = I_CM + Md²

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

$2.7 = M{{{{(0.2)}^2}} \over 2} + M{\left( {{{0.8} \over 2}} \right)^2}$

$2.7 = M\left[ {{2 \over {100}} + {{16} \over {100}}} \right]$

M = 15 kg

$\Rightarrow \rho = {M \over {\pi {r^2}L}} = {{15} \over {\pi {{(0.2)}^2} \times 0.8}}$

= 0.1492 $\times$ 10³

A $\to$ B, B $\to$ C

${{d{N_B}} \over {dt}} = \lambda {N_A} - \lambda {N_B}$

${{d{N_B}} \over {dt}} = 2\lambda {N_{{B_0}}}{e^{ - \lambda t}} - \lambda {N_B}$

${e^{ - \lambda t}}\left( {{{d{N_B}} \over {dt}} + \lambda {N_B}} \right) = 2\lambda {N_{{B_0}}}{e^{ - \lambda t}} \times {e^{\lambda t}}$

${d \over {dt}}({N_B}{e^{\lambda t}}) = 2\lambda {N_{{B_0}}}$, on integrating

${N_B}{e^{\lambda t}} = 2\lambda t{N_{{B_0}}} + {N_{{B_0}}}$

${N_B} = {N_{{B_0}}}[1 + 2\lambda t]{e^{ - \lambda t}}$

${{d{N_B}} \over {dt}} = 0$ at $- \lambda [1 + 2\lambda t){e^{ - \lambda t}} + 2\lambda {e^{ - \lambda t}} = 0$

${N_{{B_{\max }}}}$ at $t = {1 \over {2\lambda }}$

$\lambda = {h \over {mv}} = {h \over {\sqrt {2mE} }}$, $mv = \sqrt {2mE}$

$\lambda \propto {1 \over {\sqrt E }}$

${{{\lambda _2}} \over {{\lambda _1}}} = \sqrt {{{{E_1}} \over {{E_2}}}} = {3 \over 4}$, ${\lambda _2} = 0.75{\lambda _1}$

${{{E_1}} \over {{E_2}}} = {\left( {{3 \over 4}} \right)^2}$

${E_2} = {{16} \over 9}{E_1} = {{16} \over 9}E$ (E₁ = E)

Extra energy given = ${{16} \over 9}E - E = {7 \over 9}E$

(a) Magnetic Induction = MT^$-$2A^$-$1

(b) Magnetic Flux = ML²T^$-$2A^$-$1

(c) Magnetic Permeability = MLT^$-$2A^$-$2

(d) Magnetization = M⁰L^$-$1A

500 watt at 100 v



P = Vi

500 = Vi

i = 5 Amp

V = i $\times$ R

R = 20

V = 4 $\times$ 10^$-$3 m³

n = 3 moles

T = 400 K

PV = nRT $\Rightarrow$ P = ${{nRT} \over V}$

P = ${{3 \times 8.3 \times 400} \over {4 \times {{10}^{ - 3}}}}$

= 24.9 $\times$ 10⁵ Pa

${C_{eq}} = {{{A \over 2}{\varepsilon _0}} \over d} + {{A{\varepsilon _0}} \over d}{{{K_1}{K_2}} \over {{K_1} + {K_2}}}$

$= {{A{\varepsilon _0}} \over d}\left( {{1 \over 2} + {{{K_1}{K_2}} \over {{K_1} + {K_2}}}} \right)$

Relative velocity of bomb w.r.t. observer in plane = 0.

Bomb will fall down vertically. So, it will move in straight line w.r.t. observer.

F = 3a (For system) ............ (i)



fs_max = 1a (for 1 kg block) ............ (ii)

$\mu$ $\times$ 1 $\times$ g = a

F = 15N

V = 12 kV

Number of revolution = n

$n[2 \times {q_P} \times V] = {1 \over 2}{m_P} \times v_P^2$

$n[2 \times 1.6 \times {10^{ - 19}} \times 12 \times {10^3}]$

$= {1 \over 2} \times 1.67 \times {10^{ - 27}} \times {\left[ {{{3 \times {{10}^8}} \over 6}} \right]^2}$

n(38.4 $\times$ 10^$-$16) = 0.2087 $\times$ 10^$-$11

n = 543.4

$\overrightarrow \tau = \overrightarrow M \times \overrightarrow B = MB\sin 90^\circ$

$= MB = {{i\sqrt 3 {l^2}} \over 4}B$

$= \sqrt 3 \times {10^{ - 5}}$ N $-$ m

${{{E_0}} \over C} = {B_0}$

${F_{\max }} = e{B_0}V$

$= 1.6 \times {10^{ - 19}} \times {{800} \over {3 \times {{10}^8}}} \times 3 \times {10^7}$

$= 12.8 \times {10^{ - 18}}$ N

${{{T_L}} \over {{T_H} - {T_L}}} = C.O.P. = {{{{dH} \over {dt}}} \over {{{dW} \over {dt}}}}$

${{263} \over {35}} \times 35 = {{dH} \over {dt}}$

${{dH} \over {dt}} = 263$ watts

Angle between ${\overrightarrow A }$ and ${\overrightarrow B }$, $\theta$ = 60$^\circ$

Angle between ${\overrightarrow A }$ and -${\overrightarrow B }$, $\theta$ = 120$^\circ$

If angle between ${\overrightarrow A }$ and ${\overrightarrow A }$ $-$ ${\overrightarrow B }$ is $\alpha$

then $\tan \alpha =$${{\left| { - \overrightarrow B } \right|\sin \theta } \over {\overrightarrow A + \left| { - \overrightarrow B } \right|\cos \theta }}$

= ${{B\sin 120^\circ } \over {A + B\cos 120^\circ }}$

=${{B{{\sqrt 3 } \over 2}} \over {A - {B \over 2}}}$

$\Rightarrow$ $\tan \alpha = {{\sqrt 3 B} \over {2A - B}}$

Photochemical smog causes cracking of rubber, the common component of photochemical smog are ozone, nitric oxide, acrolein, formaldehyde and peroxyacetyel nitrate (PAN).

For London dispersion forces.

$E \propto {1 \over {{r^6}}}$

Hence, x = $-$6

(a) [CoCl₂(en)₂] show Cis-trans isomerism

(b) [Co(CN)₅(NC)]^3$-$ can't show G.I.

(3) [Co(NH₃)₃(NO₂)₃] show fac & mer isomerism

(d) [Co(NH₃)₄Cl₂]⁺ show cis & trans isomerism

Here all hydrogens are ionisable.

$\therefore$ Answer is zero.

sphalerite-ZnS, copper glance - Cu₂S two sulphide ores can be separated by adjusting proportions of oil to water or by using 'Depressants'.

Sucrose is example of disaccharide & non reducing sugar

Assertion : Correct

Sucrose involves glycosidic linkage between C₁ of $\alpha$-D-glucose C₂ of $\beta$-D-fructose

Reason : Incorrect

The drug named chlordiate poxide is example of tranquilizer.

(i) CFSE $\propto$ charge or oxidation no. of central metal ion.

(ii) CFSE $\propto$ strength of ligand

en > NH₃ > H₂O > F^$-$

$\therefore$ order of CFSE

$\mathop {{{[Co{{(en)}_3}]}^{ + 3}}}\limits^{III} > \mathop {{{[Co{{(N{H_3})}_6}]}^{ + 3}}}\limits^{III} > \mathop {{{[Co{F_6}]}^{ - 3}}}\limits^{III} > \mathop {{{[Co{{({H_2}O)}_6}]}^{ + 2}}}\limits^{III}$

In IIA group on moving down the group size of cation increases and show thermal stability of carbonate increases.

D₂O in used for the study of reaction mechanism. Rate of reaction for the cleavage of O-H bond > O-D bond.

Only p-methyl, phenol does not give any colour with phthalic anhydroxide with cons. H₂SO₄.

When AgNO₃ is added to KI,

Iodide ion gets adsorbed on the surface of AgI. This will result in formation of negatively charged colloids.

When KI is added to AgNO₃ following reaction takes place.

Silver cation gets adsorbed on surface of AgI, resulting in formation of positively charged colloids.

$\upsilon$ : speed of electron having max. K.E.

$\Rightarrow$ from Einstein equation : E = $\phi$ + K.E._max

$\Rightarrow {{hc} \over \lambda } = h{\upsilon _0} + {1 \over 2}m{v^2}$

$\Rightarrow {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {500 \times {{10}^{ - 9}}}} = 6.63 \times {10^{ - 34}} \times 4.3 \times {10^{14}} + {1 \over 2}m{v^2}$

$\Rightarrow {{6.63 \times 30 \times {{10}^{ - 20}}} \over 5} = 6.63 \times 4.3 \times {10^{ - 20}} + {1 \over 2}m{v^2}$

$\Rightarrow 11.271 \times {10^{ - 20}}J = {1 \over 2} \times 9 \times {10^{ - 31}} \times {\upsilon ^2}$

$\Rightarrow$ $\upsilon$ = 5 $\times$ 10⁵ m/sec.

Given k_s = 2.1 $\times$ 10¹³

K_d = ${1 \over {{k_s}}}$ = 4.7 $\times$ 10^$-$14

$\therefore$ y = 4.7 $\approx$ 5

The rate equation provided in your question describes the rates of the forward and reverse reactions for this chemical system. The equilibrium constant, $K_c$, is the ratio of the forward rate constant ($k_f$) to the reverse rate constant ($k_r$). At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, so the rate equation simplifies to:

$k_f[\text{{PtCl}}_4]^{2-} = k_r[\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$

Given the rate equation:

$-\frac{d[\text{{PtCl}}_4]^{2-}}{dt} = 4.8 \times 10^{-5} [\text{{PtCl}}_4]^{2-} - 2.4 \times 10^{-3} [\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$

At equilibrium, $-\frac{d[\text{{PtCl}}_4]^{2-}}{dt} = 0$, so:

$0 = 4.8 \times 10^{-5} [\text{{PtCl}}_4]^{2-} - 2.4 \times 10^{-3} [\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$

Rearranging terms, we find:

$4.8 \times 10^{-5} [\text{{PtCl}}_4]^{2-} = 2.4 \times 10^{-3} [\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$

Now, the equilibrium constant $K_c$ is defined as the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients. For the reaction in question, we have:

$K_c = \frac{[\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]}{[\text{{PtCl}}_4]^{2-}}$

Dividing both sides of our rate equation by $[\text{{PtCl}}_4]^{2-}$, we find that:

$K_c = \frac{4.8 \times 10^{-5}}{2.4 \times 10^{-3}} = \frac{1}{50}$ = 0.02

So, the equilibrium constant $K_c$ for this reaction is approximately 0, when rounded to the nearest integer.