JEE Main 2020 5th September Evening Shift Question Paper with Solutions

Let s be the required distance.

From Work - Energy theorem,

Work = Change in kinetic energy

$\Rightarrow$ Power $\times$ Time = $\Delta$K

i.e., Pt = $\Delta$K $\Rightarrow$ Pt = ${1 \over 2}$mv² ..... (i)

Given, P = 1 Js^$-$1, t = 9 s, m = 2 kg

Substituting all the given values in eq. (i), we get

1 $\times$ 9 = ${1 \over 2}$(2) v²

v² = 9 $\Rightarrow$ v = 3 m/s (at t = 9 s)

As, Fv = P $\Rightarrow$ (ma)v = P [$\because$ F = ma]

$\Rightarrow m\left[ {{{dv} \over {dt}}} \right]v = P \Rightarrow m\left[ {{{ds} \over {dt}}{{dv} \over {ds}}} \right]v = P$

$\Rightarrow m\left[ {v{{dv} \over {ds}}} \right]v = P$

$\Rightarrow 2{v^2}dv = ds$ {$\because$ P = 1 J/s and m = 2 kg}

Integrating both sides,

$\int\limits_0^3 {2{v^2}dv = \int\limits_0^s {ds \Rightarrow {2 \over 3}[{v^3}]_0^3 = 8} }$

${2 \over 3}[27 - 0] = s \Rightarrow s = 18$ m

Hence, after 9 s, the body has moved a distance of 18 m.

Deviation for small-angled prism is given by

$\delta$ = ($\mu$ $-$ 1) A

Given, A = 1$^\circ$, $\mu$ = 1.5

Substituting these values in above equation, we get

$\delta$ = (1.5 $-$ 1)1 $\delta$ = 0.5

According to question, $\delta = {N \over {10}}$

$\Rightarrow 0.5 = {N \over {10}} = N = 5$

Hence, the value of N is 5.

At equator, g₁ = g - R${\omega ^2}$

At height h, g₂ = $g\left( {1 - {{2h} \over R}} \right)$ [as given h << R]

$\because$ Weight same at poles and at h (so g₁ = g₂)

$\therefore$ g - R${\omega ^2}$ = $g\left( {1 - {{2h} \over R}} \right)$

$\Rightarrow$ ${R{\omega ^2} = {{2gh} \over R}}$

$\Rightarrow$ ${h = {{{R^2}{\omega ^2}} \over {2g}}}$

In adiabatic process

PV^$\gamma$ = constant

$\Rightarrow$ $P{\left( {{m \over \rho }} \right)^\gamma }$ = constant

As mass is constant

$\therefore$ P $\propto$ ${{\rho ^\gamma }}$

$\Rightarrow$ ${{{P_f}} \over {{P_i}}} = {\left( {{{{\rho _f}} \over {{\rho _i}}}} \right)^\gamma }$ = ${\left( {32} \right)^{{7 \over 5}}}$ = 2⁷ = 128

[ For diatomic gas $\gamma$ = ${{7 \over 5}}$ ]

Given, V = 10^–3 m³ = Al

I = 0.5A

$\mu$_r = 1000

n = 10 turns/cm = ${{10} \over {{{10}^{ - 2}}}}$ turn/m = 1000 turn/m

Magnetic moment, M = NIA($\mu$_r - 1)

= (nl)IA($\mu$_r - 1)

= nI(Al)($\mu$_r - 1)

= 1000 × 0.5 × 10^–3 (1000 – 1)

= 0.5 × (999) = 499.5

$\simeq$ 500 = 5 $\times$ 10² Am²

L'₁ = L₁(1 + $\alpha$₁$\Delta$T)

L'₂ = L₂(1 + $\alpha$₂$\Delta$T)

L'_eq = (L₁ + L₂) (1 + $\alpha$_avg$\Delta$T)

$\therefore$ (L₁ + L₂) (1 + $\alpha$_avg$\Delta$T) = L₁(1 + $\alpha$₁$\Delta$T) + L₂(1 + $\alpha$₂$\Delta$T)

$\Rightarrow$ (L₁ + L₂)$\alpha$_avg = L₁$\alpha$₁ + L₂$\alpha$₂

$\Rightarrow$ $\alpha$_avg = ${{{\alpha _1}{L_1} + {\alpha _2}{L_2}} \over {{L_1} + {L_2}}}$

V_rms = $\sqrt {{{3RT} \over M}}$

V_N2 = $\sqrt {{{3R(573)} \over 28}}$

V_H2 = $\sqrt {{{3RT} \over 2}}$

Given, V_N2 = V_H2

$\sqrt {{{3RT} \over 2}}$ = $\sqrt {{{3R(573)} \over 28}}$

$\Rightarrow$ ${T \over 2} = {{573} \over {28}}$

$\Rightarrow$ T = 41 K

Moment of inertia in case (i) is I₁

Moment of inertia in case (ii) is I₂

I₁ = 2MR²

I₂ = ${1 \over 2}M{R^2}$ + MR² = ${3 \over 2}M{R^2}$

T₁ = $2\pi \sqrt {{{{I_1}} \over {Mgd}}}$

and T₂ = $2\pi \sqrt {{{{I_2}} \over {Mgd}}}$

$\Rightarrow$ ${{{T_1}} \over {{T_2}}} = \sqrt {{{{I_1}} \over {{I_2}}}}$

= $\sqrt {{{2M{R^2}} \over {{3 \over 2}M{R^2}}}}$

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

Given ${{dM\left( t \right)} \over {dt}}$ = bv²(t)

In free space no external force so there in only thrust force on rocket.

F_thrust = v${{dm} \over {dt}}$

Force on satellite = $- \overrightarrow v {{dm\left( t \right)} \over {dt}}$

M(t)a = – v (bv²)

$\Rightarrow$ a = $- {{b{v^3}} \over {M\left( t \right)}}$

B = $\left( {{{{\mu _0}I} \over {2\pi r}}} \right)$

induced current i = ${e \over R} = {{Bvl} \over R}$

= $\left( {{{{\mu _0}I} \over {2\pi r}}} \right){{vl} \over R}$

Initially S₂L = 2 m

S₁L = $\sqrt {{2^2} + {{\left( {1.5} \right)}^2}}$ = 2.5 m

For minimum at (L)

S₁L – S₂L = $\Delta$x = $\left( {2n + 1} \right){\lambda \over 2}$

$\Rightarrow$ 2.5 - 2 = $\left( {2n + 1} \right){1 \over 2}$

$\Rightarrow$ 0.5 = ${{\left( {2n + 1} \right)} \over 2}$

$\Rightarrow$ 1 = 2n + 1

$\Rightarrow$ n = 0 (first minima)

so at P $\to$ first maxima

$\therefore$ S₁P – S₂P = n$\lambda$ = 1$\lambda$

$\Rightarrow$ d - 2 = 1

$\Rightarrow$ d = 3 m

Let us assume the potential at A = V_A = 0

Now at junction C, According to KCL

i₁ + i₃ = i₂

$\Rightarrow$ 1 A + i₃ = 2 A

$\Rightarrow$ i₃ = 1 A

Now Analyse potential along ACDB,

V_A + 1 + i₃(2) – 2 = V_B

$\Rightarrow$ 0 + 1 + 1(2) – 2 = V_B

$\Rightarrow$ V_B = 3 – 2 = 1 V

4.333 sec = ${{13} \over 3}$ sec

Distance = area under the v-t graph

= Area of Parallelogram + Area of triangle

= ${1 \over 2}\left( 4 \right)\left( {{{13} \over 3} + 1} \right)$ + ${1 \over 2}\left( {6 - {{13} \over 3}} \right) \times 2$

= ${{37} \over 3}$ m

T₁ = 10 sec

$\lambda$₁ = ${{\ln 2} \over {{T_1}}}$

T₂ = 100 sec

$\lambda$₂ = ${{\ln 2} \over {{T_2}}}$

$\lambda$_eq = ${{\ln 2} \over {{T_{eq}}}}$

We know,

$\lambda$_eq = $\lambda$₁ + $\lambda$₂

$\Rightarrow$ ${{\ln 2} \over {{T_{eq}}}}$ = ${{\ln 2} \over {{T_1}}}$ + ${{\ln 2} \over {{T_2}}}$

$\Rightarrow$ ${1 \over {{T_{eq}}}}$ = ${1 \over {10}} + {1 \over {100}}$

$\Rightarrow$ T_eq = ${{100} \over {11}}$ = 9 sec

Let potential of point O v₀ = 0.

Now, using junction analysis

We can say, q₁ + q₂ + q₃ = 0

$\Rightarrow$ 2(x – 6) + 4(x – 6) + 5(x) = 0

$\Rightarrow$ x = ${{36} \over {11}}$

q₃ = CV = ${{36} \over {11}} \times 5$ = 16.36 $\mu$C

f₁ = frequency heard by wall
= $\left( {{{{v_s} - 0} \over {{v_s} - {v_c}}}} \right){f_0}$ = ${{{{v_s}} \over {{v_s} - {v_c}}} \times 440}$

f₂ = frequency heard by driver after reflection from wall

= $\left( {{{{v_s} + {v_c}} \over {{v_s} - 0}}} \right){f_1}$

$\Rightarrow$ 480 = $\left( {{{{v_s} + {v_c}} \over {{v_s}}}} \right){f_1}$

$\Rightarrow$ 480 = $\left( {{{{v_s} + {v_c}} \over {{v_s}}}} \right)\left( {{{{v_s}} \over {{v_s} - {v_c}}}} \right) \times 440$

$\Rightarrow$ 480 = $\left( {{{{v_s} + {v_c}} \over {{v_s} - {v_c}}}} \right) \times 440$

$\Rightarrow$ 480 = $\left( {{{345 + {v_c}} \over {345 - {v_c}}}} \right) \times 440$

$\Rightarrow$ 12 = $\left( {{{345 + {v_c}} \over {345 - {v_c}}}} \right) \times 11$

$\Rightarrow$ 11 × 345 + 11v_c = 12 × 345 – 12v_c

$\Rightarrow$ v_c = ${{345} \over {23}}$ m/s = ${{345} \over {23}} \times {{18} \over 5}$ = 54 km/hr

C_i = ${{{\varepsilon _0}A} \over d} = {{{\varepsilon _0}lw} \over d}$

U_i = ${1 \over 2}{C_i}{V^2}$ = ${1 \over 2}{{{\varepsilon _0}lw} \over d}{V^2}$

C_f = C₁ + C₂

= ${{K{\varepsilon _0}{A_1}} \over d} + {{{\varepsilon _0}{A_2}} \over d}$

= ${{K{\varepsilon _0}wx} \over d} + {{{\varepsilon _0}w\left( {l - x} \right)} \over d}$

= ${{{\varepsilon _0}w} \over d}\left[ {Kx + l - x} \right]$

$\therefore$ U_f = ${1 \over 2}{C_f}{V^2}$

= ${1 \over 2}{{{\varepsilon _0}w} \over d}\left[ {Kx + l - x} \right]{V^2}$

Given U_f = 2U_i

$\Rightarrow$ ${1 \over 2}{{{\varepsilon _0}w} \over d}\left[ {Kx + l - x} \right]{V^2}$ = 2 $\times$ ${1 \over 2}{{{\varepsilon _0}lw} \over d}{V^2}$

$\Rightarrow$ kx + l – x = 2l

$\Rightarrow$ 4x – x = l

$\Rightarrow$ 3x = l

$\Rightarrow$ x = ${l \over 3}$

Net charge = 5q - 5q = 0

Potential of centre = V = ${{K\sum q } \over r}$

V_C = ${{K\left( 0 \right)} \over r}$ = 0
Let E be electric field produced by each charge at the centre, then resultant electric field will be



E_C = 0, Since equal electric field vectors are acting at equal angle so their resultant is equal to zero.( From symmetric property of vector)

The given situation is shown in the following figure,

Applying law of conservation of angular momentum about pivotal point,

${L_i} = {L_f} \Rightarrow mvl = I\omega$

$\Rightarrow mvl = \left( {{1 \over 3}M{L^2} + m{L^2}} \right)\omega$ ..... (i)

Given, m = 0.1 kg, v = 80 ms^$-$1, M = 0.9 kg, l = 1 m

Substituting these values in eq. (i), we get

$0.1 \times 80 \times 1 = \left( {{{0.9 \times {1^2}} \over 3} + 0.1 \times {1^2}} \right)\omega$

$8 = \left( {{3 \over {10}} + {1 \over {10}}} \right)\omega \Rightarrow 8 = {4 \over {10}}\omega \Rightarrow \omega = 20$ rad/s

After falling through h, the velocity be equal to terminal velocity

$\sqrt {2gh}$ = ${2 \over 9}{{{r^2}g} \over \eta }\left( {{\rho _l} - \rho } \right)$

$\Rightarrow$ h = ${2 \over {81}}{{{r^4}g{{\left( {{\rho _l} - \rho } \right)}^2}} \over {{\eta ^2}}}$

$\Rightarrow$ h $\propto$ r⁴

Decreasing order of wavelength,

Radio wave > Microwave > Infrared > ROYGBIV(Visiable Resion) > Ultraviolet > X rays > $\gamma$ rays

$\therefore$ Correct order is

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

figure of merit = ${I \over \theta }$

= ${{6 \times {{10}^{ - 3}}} \over 2}$

= 3 $\times$ 10^–3 A/div

Till input voltage reaches 4V, both diode will stay deactvated so output voltage will be same as input voltage.

When input voltage reach 4V, then 4V diode will get activated and it will transfer only 4V to the output but as a rasistance is present in series with the 4V diode so some voltage drop will happen across the resistor. So output
voltage(V_o) will be = 4V + voltage drop.

When input voltage reach 6V, then 6V diode will get activated and it will transfer only 6V to the output no matter how much input voltage become.

x = ${1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$ = speed

$\therefore$ [x] = [L¹T^–1]

y = ${E \over B}$ = speed

$\therefore$ [y] = [L¹T^–1]

z = ${l \over {CR}}$ = ${l \over \tau }$

$\Rightarrow$ [z] = [L¹T^–1]

So, x, y, z all have the same dimensions.

Given,

E₁ = 4 eV

E₂ = 2.5 eV

and ${{{{\left( {{V_1}} \right)}_{\max }}} \over {{{\left( {{V_2}} \right)}_{\max }}}} = 2$

${{{1 \over 2}m{{\left( {{{\left( {{V_1}} \right)}_{\max }}} \right)}^2}} \over {{1 \over 2}m{{\left( {{{\left( {{V_2}} \right)}_{\max }}} \right)}^2}}} = {{{E_1} - {\phi _0}} \over {{E_2} - {\phi _0}}}$

$\Rightarrow$ ${{{{\left( {{{\left( {{V_1}} \right)}_{\max }}} \right)}^2}} \over {{{\left( {{{\left( {{V_2}} \right)}_{\max }}} \right)}^2}}} = {{4 - {\phi _0}} \over {2.5 - {\phi _0}}}$

$\Rightarrow$ ${\left( 2 \right)^2} = {{4 - {\phi _0}} \over {2.5 - {\phi _0}}}$

$\Rightarrow$ 10 - 4$\phi$₀ = 4 - $\phi$₀ $\Rightarrow$ 3$\phi$₀ = 6

$\Rightarrow$ $\phi$₀ = 2

$\therefore$ Work function($\phi$) of the metal = 2 eV

$\Delta$G^o = $\Delta$H^o - T$\Delta$S^o

= ($\Delta$U^o + $\Delta$n_gRT) - T$\Delta$S^o

= $\left[ {\left\{ { - 20 + \left( 1 \right){{8.314} \over {1000}} \times 298} \right\} - {{298} \over {1000}} \times \left( { - 30} \right)} \right]$ kJ

= – 13.537572 kJ

= – 13537.57 Joule

Hydrogen peroxide, in the pure state, has openbook like structure which is non- planar and almost colourless (very pale blue) liquid.

(I) Shows hydrogen bonding from –OH group only.

(II) Shows strongest hydrogen bonding from both sides of –OH group as well as –NO₂ group.

(III) Shows stronger hydrogen from both side of –OH group as well as –NH₂ group.

(IV) Shows stronger hydrogen from one side –OH group and another side of –OCH₃ group shows only dipole-dipole interaction.

Hence correct order of boiling point is :

I < IV < III < II

NH₃ + 3Cl₂(excess) $\to$ NCl₃ + 3HCl

Buna-N : Obtained by addition polymerisation

Nylon 6, Bakelite, Nylon 6, 6 : Obtained by condensation polymerisation.

CH₄ - Bond angle = 109^o28'

H₂S - Bond angle = 92^o

NH₃ - Bond angle = 107^o

H₂O - Bond angle = 104^o5'

Total number of chiral carbon in sucrose = 9

Distance between nearest octahedral voids (O.V.)

= $\sqrt {{{\left( {{a \over 2}} \right)}^2} + {{\left( {{a \over 2}} \right)}^2}}$ = ${a \over {\sqrt 2 }}$

For isoelectronic species, as the no. of protons increases, size of ions decreases.

$\therefore$ Correct order of size for isoelectronic species

Al³⁺ < Mg²⁺ < Na⁺ < F^– < O^2– < N^3–

Ru(44) : [Kr] 4d⁷5s¹

Ru⁺² = [Kr]4d⁶

As $\Delta$₀ > P,

$\therefore$ Pairing of e^–s will take place.

No. of unpaired e^–s = 0

$\therefore$ Magnetic moment = 0 B.M

K = A${e^{ - {{{E_a}} \over {RT}}}}$

lnK = lnA ${ - {{{E_a}} \over {RT}}}$

$\Rightarrow$ lnK = lnA ${ - {{{E_a}{{10}^3}} \over {{{10}^3}RT}}}$

$\therefore$ Slope = ${ - {{{E_a}} \over {{{10}^3}R}}}$ = $- {{10} \over 5}$

$\Rightarrow$ ${{{{E_a}} \over R}}$ = 2000 J/mol

$\Rightarrow$ E_a = 2R KJ/mol

For boron and silicon zone refining is used.

For Removal of Temporary hardness two methods are present
(1) Clark’s method
(2) Boiling

While permanent hardness of water can be removed 4 methods
(1) Washong soda method (Na₂CO₃ method),
(2) Calgon method
(3) Permutit Process
(4) Ion-exchange Resin method.

$\phi$² (probability density) can be zero for 3p orbital other than infinite distance. It has one radial node.

The electrolyte (X) must be weak electrolyte as such type of variation is always for weak electrolyte. So

X is CH₃COOH.

K_eq = ${{{{\left( 1 \right)}^2}} \over {{3 \over 4} \times {5 \over 4}}}$ = ${{16} \over {15}}$

$\therefore$ x = 16

$\Delta$H_sol = Lattice enthalpy + $\Delta$H_hyd

$\Rightarrow$ 4 = 788 + $\Delta$H_hyd

$\Rightarrow$ $\Delta$H_hyd = –784 kJ mol^–1

K₂Cr₂O₇ + FeC₂O₄ $\to$ Cr⁺³ + Fe⁺³ + CO₂

n_factor of K₂Cr₂O₇ = 3 $\times$ 2 = 6

n_factor of FeC₂O₄ = 1 + 2 = 3

m. eq. of K₂Cr₂O₇ = m. eq. of FeC₂O₄

$\Rightarrow$ ${{6 \times 0.02 \times vol} \over {1000}}$ = ${{0.288} \over {144}} \times 3$

$\Rightarrow$ vol = 50 ml

Brompheniramine act as antihistamines.

trans-[Co(en)₂Cl₂]⁺ contain a plane of symmetry. It cannot be optically active.

cis-[Co(en)₂Cl₂]⁺ does not contain any plane of symmetry. Hence the compound (B) is optically active.

Freundlich adsorption isotherm

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

With increase in temperature, decrease in physical adsorption is observed.

${x \over m}$ v/s ${p^{{1 \over n}}}$ is non linear curve.

${{{x^2}} \over {100}} - {{{y^2}} \over {64}} = 1$

$\therefore$ c = $\pm$ $\sqrt {{a^2}{m^2} - {b^2}}$

$\Rightarrow$ c = $\pm$ $\sqrt {100{m^2} - 64}$

General tangent to hyperbola in slope form is

y = mx $\pm$ $\sqrt {100{m^2} - 64}$

This tangent is also tangent to the circle x² + y² = 36, whose center (0, 0) and radius = 6.

Distance of the tangent from the center is

$\left| {{{\sqrt {100{m^2} - 64} } \over {\sqrt {{m^2} + 1} }}} \right|$ = 6

$\Rightarrow$ 100m² — 64 = 36m² + 36

$\Rightarrow$ 64m² = 100

$\Rightarrow$ m = ${{10} \over 8}$

$\therefore$ c² = 100 $\times$ ${{100} \over {64}}$ - 64

$\Rightarrow$ c² = ${{{{100}^2} - {{64}^2}} \over {64}}$

$\Rightarrow$ c² = ${{164 \times 36} \over {64}}$

$\Rightarrow$ 4c² = 369

L₁ : ${{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}$ and

L₂ : ${{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}$ are coplanar.

$\therefore$ \left| {\matrix{ 1 & 3 & 2 \cr 2 & { - 1} & 1 \cr \alpha & {5 - \alpha } & 1 \cr } } \right| = 0

$\Rightarrow$ –1(–1 + $\alpha$ - 5) + 3(2 - $\alpha$) - 2(10 - 2$\alpha$ + $\alpha$) = 0

$\Rightarrow$ 6 - $\alpha$ + 6 - 3$\alpha$ + 2$\alpha$ - 20 = 0

$\Rightarrow$ –8 –2$\alpha$ = 0

$\Rightarrow$ $\alpha$ = -4

$\therefore$ Equation of L₂ : ${{x + 2} \over { - 4}} = {{y + 1} \over 9} = {{z + 1} \over 1}$

Check options (2, –10, –2) lies on L₂.

We will use here those two formulas,

sin² $\theta$ = ${{1 - \cos 2\theta } \over 2}$ and cos² $\theta$ = ${{1 + \cos 2\theta } \over 2}$

L = sin²$\left( {{\pi \over {16}}} \right)$ - sin²$\left( {{\pi \over {8}}} \right)$

$\Rightarrow$ L = $\left( {{{1 - \cos \left( {{\pi \over 8}} \right)} \over 2}} \right)$ - $\left( {{{1 - \cos \left( {{\pi \over 4}} \right)} \over 2}} \right)$

$\Rightarrow$ L = ${1 \over 2}\left( {\cos \left( {{\pi \over 4}} \right) - \cos \left( {{\pi \over 8}} \right)} \right)$

$\Rightarrow$ L = ${1 \over {2\sqrt 2 }} - {1 \over 2}\cos \left( {{\pi \over 8}} \right)$

M = cos²$\left( {{\pi \over {16}}} \right)$ - sin²$\left( {{\pi \over {8}}} \right)$

$\Rightarrow$ M = $\left( {{{1 + \cos \left( {{\pi \over 8}} \right)} \over 2}} \right)$ - $\left( {{{1 - \cos \left( {{\pi \over 4}} \right)} \over 2}} \right)$

$\Rightarrow$ M = ${1 \over {2\sqrt 2 }} + {1 \over 2}\cos {\pi \over 8}$