JEE Main 2019 11th January Evening Shift Question Paper with Solutions

Angular frequency of pendulum

$\omega$ = $\sqrt {{{{g_{eff}}} \over \ell }}$

$\therefore$   ${{\Delta \omega } \over \omega }$ = ${1 \over 2}$ ${{\Delta {g_{eff}}} \over {{g_{eff}}}}$

$\Delta$$\omega$ = ${1 \over 2}$ ${{\Delta g} \over g} \times \omega$

[${\omega _s}$ = angular frequency of support]

$\Delta$$\omega$ = ${1 \over 2} \times {{2A\omega _s^2} \over {100}} \times 100$

$\Delta \omega = {10^{ - 3}}$ rad/sec.

i = e

r₁ = r₂ = ${A \over 2}$ = 30^o

by Snell's law

1 $\times$ sin i = $\sqrt 3 \times {1 \over 2} = {{\sqrt 3 } \over 2}$

i = 60

Total length L will remain constant

L = (3a) N        (N = total turns)

and length of winding = (d) N

                         (d = diameter of wire)


self inductance = $\mu$₀n²A$\ell$

= $\mu$₀n²$\left( {{{\sqrt 3 {a^2}} \over 4}} \right)$ dN

$\propto$ a² N $\propto$ a

So self inductance will become 3 times

40 + f = m(R$\alpha$)         . . . .(i)

40 $\times$ R $-$ f $\times$ R = mR²$\alpha$

40 $-$ f = mR$\alpha$         . . . .(ii)

From (i) and (ii)

$\alpha$ = ${{40} \over {mR}}$ = 16

${\overrightarrow F _{net}} = q\overrightarrow E + q\left( {\overrightarrow v \times \overrightarrow B } \right)$

$= \left( {2q\widehat i + 3q\widehat j} \right) + q\left( {\overrightarrow v \times \overrightarrow B } \right)$

$W = {\overrightarrow F _{net}}.\overrightarrow S$

$=$ 2q + 3q

$=$ 5q

I $=$ ${{M{R^2}} \over 2} + 2\left( {{{M{R^2}} \over 4} + M{R^2}} \right)$

$= {{M{R^2}} \over 2} + {{M{R^2}} \over 2} + 2M{R^2}$

$=$ $3M{R^2}$

Potential difference across AB will be equal to battery equivalent across CD

V_AB $=$ V_CD $=$ ${{{{{E_1}} \over {{r_1}}} + {{{E_2}} \over {{r_2}}} + {{{E_3}} \over {{r_3}}}} \over {{1 \over {{r_1}}} + {1 \over {{r_2}}} + {1 \over {{r_3}}}}} = {{{1 \over 1} + {2 \over 1} + {3 \over 1}} \over {{1 \over 1} + {1 \over 1} + {1 \over 1}}}$

$=$ ${6 \over 3}$ $=$ 2V

${{hc} \over {{\lambda _1}}} = \phi + e$V₁        . . . (i)

${{hc} \over {{\lambda _2}}} = \phi + e$V₂        . . . (ii)

(i) $-$ (ii)

hc$\left( {{1 \over {{\lambda _1}}} - {1 \over {{\lambda _2}}}} \right)$ = e(V₁ $-$ V₂)

$\Rightarrow$  V₁ $-$ V₂ = ${{hc} \over e}$$\left( {{{{\lambda _2} - {\lambda _1}} \over {{\lambda _1} - {\lambda _2}}}} \right)$

= (1240nm $-$ V) ${{100nm} \over {300nm \times 400nm}}$

= 1V

x = ${1 \over H}$

I = ${{Magnetic\,moment} \over {Volume}}$

I = ${{20 \times {{10}^{ - 6}}} \over {{{10}^{ - 6}}}}$ = 20 N/m²

x = ${{20} \over {60 \times {{10}^{ + 3}}}}$ = ${1 \over 3} \times {10^{ - 3}}$

= 0.33 $\times$ 10^$-$3 = 3.3 $\times$ 10^$-$4

$\Rightarrow$   T^oC = ${{{x_0}} \over 6}$ & $\left( {{x_0} - {{{x_0}} \over 3}} \right)$ = (100 $-$ 0^oC)

x₀ = ${{300} \over 2}$

$\Rightarrow$  T^oC = ${{150} \over 6}$ = 25^oC

For M $\to$ L steel

${1 \over \lambda }$ = K $\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right) = {{K \times 5} \over {36}}$

for N $\to$ L

${1 \over {\lambda '}}$ = K$\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = {{K \times 3} \over {16}}$

$\lambda ' = {{20} \over {27}}\lambda$

Here R = ${{mv} \over {qB}}$ = 2d

cos $\theta$ = ${{{R \over 2}} \over R}$ = ${1 \over 2}$

$\Rightarrow$ $\theta$ = 60^o

Acceleration of the charged particle at the point of its emergence,

$\overrightarrow {{a_c}} = {a_{{c_x}}}\left( { - \widehat i} \right) + {a_{{c_y}}}\left( { - \widehat j} \right)$

= ${a_c}\cos 30^\circ \left( { - \widehat i} \right) + {a_c}\sin 30^\circ \left( { - \widehat j} \right)$

= ${a_c}\left( {{{\sqrt 3 } \over 2}\left( { - \widehat i} \right) + {1 \over 2}\left( { - \widehat j} \right)} \right)$

= ${{qvB} \over m}\left( { - {{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)$

${{{R_1}} \over {{R_2}}} = {2 \over 3}\,\,$                                      . . .(i)

${{{R_1} + 10} \over {{R_2}}} = 1 \Rightarrow {R_1} + 10 = {R_2}$     . . .(ii)

${{2{R_2}} \over 3} + 10 = {R_2}$

$10 = {{{R_2}} \over 3} \Rightarrow {R_2} = 30\Omega$

& ${R_1} = 20\Omega$

${{{{30 \times R} \over {30 + R}}} \over {30}} = {2 \over 3}$

$R = 60\,\Omega$

For diffraction

location of 1^st minima

y₁ = ${{D\lambda } \over a}$ = 0.2469 D$\lambda$

location of 2^nd minima

y₂ = ${{2D\lambda } \over a}$ = 0.4938 D$\lambda$

Now for interference

Path difference at P.

${{dy} \over D}$ = 4.8$\lambda$

path difference at Q

${{dy} \over D}$ = 9.6$\lambda$

So orders of maxima in between P & Q is 5, 6, 7, 8, 9

So 5 bright fringes all present between P & Q.

${{dp} \over {dt}} = F = kt$

$\int_P^{3P} {dP} = \int_0^T {kt\,dt}$

$2p = {{K{T^2}} \over 2}$

$T = 2\sqrt {{P \over K}}$

$\because$    g = ${{GM} \over {{R^2}}}$

${{{g_p}} \over {{g_e}}}$ = ${{{M_e}} \over {{M_e}}}{\left( {{{{{\mathop{\rm R}\nolimits} _e}} \over {{R_p}}}} \right)^2}$ = 3${\left( {{1 \over 3}} \right)^2}$ = ${{1 \over 3}}$

Also T $\propto$ ${1 \over {\sqrt g }}$

$\Rightarrow$  ${{{T_p}} \over {{T_e}}}$ = $\sqrt {{{{g_e}} \over {{g_p}}}}$ = $\sqrt 3$

$\Rightarrow$  T_p = 2$\sqrt 3$ s

I = ${6 \over {300}}$ = 0.02 (D₂ is in reverse bias)

$\overrightarrow S = \left( {5\widehat i + 4} \right)2 + {1 \over 2}\left( {4\widehat i + 4\widehat j} \right)4$

$= 10\widehat i + 8\widehat j + 8\widehat i + 8\widehat j$

$\overrightarrow {{r_f}} - \overrightarrow {{r_i}} = 18\widehat i + 16\widehat j$

$\overrightarrow {{r_f}} = 20\widehat i + 20\widehat j$

$\left| {\overrightarrow {{r_f}} } \right| = 20\sqrt 2$

2.5 = 1 $\times$ 5 sin $\theta$

sin$\theta$ = 0.5 = ${1 \over 2}$

$\theta$ = ${\pi \over 6}$

Analysis of graph says

(1)   Amplitude varies as 8 $-$ 10 V or 9 $\pm$ 1

(2)  Two time period as

100 $\mu$s (signal wave) & 8 $\mu$s (carrier wave)

Hence signal is $\left[ {9 \pm 1\sin \left( {{{2\pi t} \over {{T_1}}}} \right)} \right]\sin \left( {{{2\pi t} \over {{T_2}}}} \right)$

= 9 $\pm$ 1sin (2$\pi$ $\times$ 10⁴t) sin 2.5$\pi$ $\times$ 10⁵ t

We know,

Young's modulus (Y) = ${{{F \over A}} \over {{{\Delta l} \over l}}}$

$\therefore$ [Y] = ${{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ {{L^2}} \right]}}$ = [ ML^-1T^-2]

Let [Y] = [V]^x [A]^y [F]^z

$\therefore$ [ ML^-1T^-2] =

[LT^-1]^x [LT^-2]^y [MLT^-2]^z

$\Rightarrow$ [ ML^-1T^-2] =

[ M^z L^{x + y + z} T^{-x -2y - 2z}

For dimensional balance, the dimension on both sides should be same.

So, z = 1

x + y + z = -1

$\Rightarrow$ x + y = -2 ........(1)

and -x -2y - 2z = -2

$\Rightarrow$ x + 2y = 0 ...........(2)

By solving those two equations we get,

x = -4 and y = 2

$\therefore$ [Y] = V^$-$4A²F¹

100 $\times$ S_A $\times$ [100 $-$ 90] = 50 $\times$ S_B $\times$ (90 $-$ 75)

2S_A = 1.5 S_B

S_A = ${3 \over 4}$S_B

Now, 100 $\times$ S_A $\times$ [100 $-$ T] = 50 $\times$ S_B (T $-$ 50)

2 $\times$ $\left( {{3 \over 4}} \right)$ (100 $-$ T) = (T $-$ 50)

300 $-$ 3T = 2T $-$ 100

400 = 5T

T = 80

0.1 $\times$ 400 $\times$ (500 $-$ T) = 0.5 $\times$ 4200 $\times$ (T $-$ 30) + 800 (T $-$ 30)

$\Rightarrow$  40(500 $-$ T) = (T $-$ 30) (2100 + 800)

$\Rightarrow$  20000 $-$ 40T = 2900 T $-$ 30 $\times$ 2900

$\Rightarrow$  20000 + 30 $\times$ 2900 = T(2940)

T = 30.4^oC

${{\Delta T} \over T} \times 100$ = ${{6.4} \over {30}} \times 100$

     $\simeq$ 20%

C_eq = ${6 \over {13}}$$\mu$F

Therefore three capacitors most be in parallel to get 6 in

${1 \over {{C_{eq}}}} = {1 \over {3C}} + {1 \over C} + {1 \over C} + {1 \over C} + {1 \over C}$

C_eq = ${{3C} \over {13}}$ = ${6 \over {13}}$$\mu$F

U = $-$ $\overrightarrow P .\overrightarrow E$

= $-$ PE cos $\theta$

= $-$ (10^$-$29) (10³) cos 45^o

= $-$ 0.707 $\times$ 10^$-$26 J

= $-$ 7 $\times$ 10^$-$27 J.

Intensity of EM wave is given by

${\rm I} = {{Power} \over {Area}} = {1 \over 2}{\varepsilon _0}E_0^2C$

$= {{27 \times {{10}^{ - 3}}} \over {10 \times {{10}^{ - 6}}}}$ $= {1 \over 2} \times 9 \times {10^{ - 12}} \times {E^2} \times 3 \times {10^8}$

$E = \sqrt 2 \times {10^3}kv/m$

$= 1.4$ kv/m

$\Delta {\ell _1} = \Delta {\ell _2}$

$\ell {\alpha _1}\Delta {T_1} = \ell {\alpha _2}\Delta {T_2}$

${{{\alpha _1}} \over {{\alpha _2}}} = {{\Delta {T_1}} \over {\Delta {T_2}}}$

${4 \over 3} = {{T - 30} \over {180 - 30}}$

$T = {230^o}C$

VT = K

$\Rightarrow$  V$\left( {{{PV} \over {nR}}} \right)$ = k $\Rightarrow$ PV² = K

$\because$  C = ${R \over {1 - x}} +$ C_v (For polytropic process)

C = ${R \over {1 - 2}} + {{3R} \over 2}$ = ${R \over 2}$

$\therefore$  $\Delta$Q = nC $\Delta$T

= ${R \over 2} \times \Delta$T

R_g = 20$\Omega$

N_L = N_R = N = 30

FOM = ${1 \over \phi }$ = 0.005 A/Div.

Current sentivity = CS = $\left( {{1 \over {0.005}}} \right)$ = ${\phi \over {\rm I}}$

${\rm I}$g_max = 0.005 $\times$ 30

= 15 $\times$ 10^$-$2 = 0.15

15 = 0.15 [20 + R]

100 = 20 + R

R = 80

Maximum kinetic energy at lowest point B is given by

K = mg$\ell$ (1 $-$ cos $\theta$)

where $\theta$ = angular amp.


K₁ = mg$\ell$ (1 $-$ cos $\theta$)

K₂ = mg(2$\ell$) (1 $-$ cos $\theta$)

K₂ = 2K₁.

We know,

$\Delta$G^o = -RTln(K_C) ....(1)

Also $\Delta$G^o = -nF$E_{cell}^o$ ....(2)

$\therefore$ -nF$E_{cell}^o$ = -RTln(K_C)

$\Rightarrow$ $E_{cell}^o$ = ${{RT} \over {nF}}\ln \left( {{K_C}} \right)$

= $2.303{{RT} \over {nF}}\log \left( {{K_C}} \right)$

= ${{0.059} \over 2}\log \left( {10 \times {{10}^{15}}} \right)$

( n = no of electron transferred = 2 )

= 0.059 $\times$ ${{16} \over 2}$

= 0.059 $\times$ 8

= 0.472 V

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

Given that A and B are non-zero constants, i.e., A = $\Delta$H^o, B = $\Delta$S^o

If $\Delta$H^o is positive means reaction is endothermic.

Calcination involves heating of the ore in absence of air.

2Cu₂S + 3O₂ $\buildrel \Delta \over \longrightarrow$ 2Cu₂O + 2SO₂ $\uparrow$

This reaction involve roasting. Roasting is carried out mainly for sulphide ores in presence of excess of oxygen.

In LiAlH₄, H^- is the attacking species which is a strong nucleophile and performs both nucleophilc substitution and nucleophilic addition reaction. In C = C, pi bond is created by pure p orbital with side by side overlapping which create $\pi$ cloud and $\pi$ cloud creates negative charge around it. As H^- is also negative charged so LiAlH₄ can't reduce C = C.

2HCl(aq) + Na₂CO₃(aq) $\to$ H₂CO₃ + NaCl

moles of HCl

2

=

moles of Na₂CO₃

1

$\Rightarrow$ ${{M \times {{25} \over {1000}}} \over 2} = {{0.1 \times {{30} \over {1000}}} \over 1}$

$\Rightarrow$ Molarity of HCl (M) = ${6 \over {25}}M$

HCl(aq) + NaOH(aq) $\to$ NaCl + H₂O

moles of HCl

1

=

moles of NaOH

1

${{{6 \over {25}} \times {V \over {1000}}} \over 1} = {{0.2 \times {{30} \over {1000}}} \over 1}$

$\Rightarrow$ V = 25 ml

2H₂O $\rightleftharpoons$ H₃O⁺ + OH^$-$

Here K_eq = K_w(H₂O)

At 298 K, K_w(H₂O) = 10^-14

$\therefore$ K_eq = 10^-14

$\Delta$G^o = -RTln(K_eq)

= -RTln(K_w)

= -2.303RT log (10^-14)

= -2.303 $\times$ 8.314 $\times$ 298 $\times$ $\times$ (-14)

= 80 kJ/mol

First Markonikov’s addition on alkene followed by intramolecular Friedel-Craft alkylation takes place.

Allosteric effect $\to$ Molecule binding to a site other than the active site of enzyme

Competitive inhibitor $\to$ Molecule binding to the active site of enzyme

Receptor $\to$ Molecule crucial for communication in the body

Poison $\to$ Molecule binding to the enzyme covalently

For option (B) :

C₂H₅MgBr performs addition reaction with - C $\equiv$ N.
C₂H₅MgBr performs substitution reaction with ester -COOCH₃.

Conjugated dienes (two double bonds separated by a single bond) gives Br water test. Either 1-2 or 1-4 addition can happen.

For option (C) :

C₂H₅MgBr reacts with -OH group and creates alkane C₂H₆.

As here C = C present, so this compound gives Br water tests.

We know,

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

For a reaction to be spontaneous $\Delta$G^o must be negative i.e.,
T$\Delta$S^o > $\Delta$H^o

$\Rightarrow$ T > ${{\Delta H^\circ } \over {\Delta S^\circ }}$

$\Rightarrow$ T > ${{491.1 \times 1000} \over {198}}$

$\Rightarrow$ T > 2480.3 K

a = 2(R + r)

$\Rightarrow$ ${a \over 2} = \left( {R + r} \right)$ ......(1)

For bcc, $\sqrt 3 a$ = 4R

Using (i) and (ii)

${a \over 2} = \left( {{{a\sqrt 3 } \over 4} + r} \right)$

$\Rightarrow$ r = $a\left( {{{2 - \sqrt 3 } \over 4}} \right)$

$\Rightarrow$ r = 0.067 a

By photoelectric effect

KE = h$\gamma$ - h$\gamma$_o ....(1)

de broglie wavelength,

$\lambda$ = ${h \over {mv}}$ = ${h \over {\sqrt {2m \times K.E} }}$ ...(2)

Using equation (1) and (2), we get

$\lambda$ = ${h \over {\sqrt {2m \times \left( {h\nu - h{\nu _0}} \right)} }}$

$\therefore$ $\lambda \,\propto \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 2}}}}}$