JEE Main 2019 10th January Evening Shift Question Paper with Solutions

Volume of cylinder(V) = $\pi$r²h

= $\pi {{{d^2}} \over 4}h$

= $3.14 \times {{{{\left( {12.6} \right)}^2}} \over 4} \times 34.2$

= 4260

${{\Delta V} \over V} = 2{{\Delta d} \over d} + {{\Delta h} \over h} = 2\left( {{{0.1} \over {12.6}}} \right) + {{0.1} \over {34.2}}$ = 0.0188

$\therefore$ $\Delta$V = 0.0188 $\times$ 4260 = 80

T = $2\pi \sqrt {{1 \over {\mu B}}}$

T_h = $2\pi \sqrt {{{m{R^2}} \over {\left( {2\mu } \right)B}}}$

T_C = $2\pi \sqrt {{{1/2m{R^2}} \over {\mu B}}}$

Initial energy of capacitor

U_i = ${1 \over 2}$ ${{{v^2}} \over c}$

= ${1 \over 2}$ $\times$ ${{120 \times 120} \over {12}}$ = 600 J

Since battery is disconnected so charge remain same.

Final energy of capacitor

U_f = ${1 \over 2}{{{v^2}} \over c}$

= ${1 \over 2} \times {{120 \times 120} \over {12 \times 6.5}}$ = 92

W + U_f = U_i

W = 508 J

192 $\times$ S $\times$ (100 $-$ 21.5)

= 128 $\times$ 394 $\times$ (21.5 $-$ 8.4)

       + 240 $\times$ 4200 $\times$ (21.5 $-$ 8.4)

$\Rightarrow$   S = 916

Ne²⁰$~ \to$  2He⁴ + C¹²

Q – value, E_B = (BE)_react $-$ (BE)_product

= (20 × 8.03) – ((2 × 7.07 × 4) + 7.86 × 12)

= 9.72 MeV

Assuming zener diode doesnot undergo breakdown, current in circuit

= ${{120} \over {15000}}$ = 8mA

$\therefore$  Voltage drop across diode

= 80 V > 50 V.

The diode undergo breakdown.


Current is R₁ = ${{70} \over {5000}}$ = 14mA

Current is R₂ = ${{50} \over {10000}}$ = 5mA

$\therefore$  Current through diode = 9mA

Position of the axis of rotation for Solid Sphere : About its diametric axis which passes through its centre of mass


Moment of Inertia (I) = ${2 \over 5}M{R^2}$

Using parallel axis theorem, moment of inertia about the given axis in the figure below will be

$\begin{aligned} & I_1=\frac{2}{5} M R^2+M(2 R)^2 \\ & I_1=\frac{22}{5} M R^2\end{aligned}$

Considering both spheres at equal distance from the axis, moment of inertia due to both spheres about this axis will be

$2 I_1=2 \times 22 / 5 M R^2$ = ${{44} \over 5}$MR²

Moment of inertia of rod

Position of the axis of rotation : About an axis passing through centre of mass and perpendicular to its length


Moment of Inertia (I) = $\frac{\mathrm{ML}^2}{12}$

Here, given that L = 2R

$\therefore I_2=\frac{1}{12} M(2 R)^2=\frac{1}{3} M R^2$

So, total moment of inertia of the system is

$$\begin{aligned} I & =2 I_1+I_2=2 \times \frac{22}{5} M R^2+\frac{1}{3} M R^2 \\ \Rightarrow I & =\left(\frac{44}{5}+\frac{1}{3}\right) M R^2=\frac{137}{15} M R^2 \end{aligned}$$

$v = \omega \sqrt {{A^2} - {x^2}} \,\,$    . . .(1)

$a = - {\omega ^2}x$               . . .(2)

$\left| v \right| = \left| a \right|$                   . . .(3)

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

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

${5^2} - {4^2} = {\omega ^2}\left( {{4^2}} \right)$

$\Rightarrow \,\,\,3 = \omega \times 4$

$T = 2\pi /\omega$

${{1.34} \over V} - {1 \over \infty } = {{1.34 - 1} \over {7.8}}$

$\therefore$  V = 30.7 mm

Applying torque equation about point P.

2M₀ (2l) $-$ 5 M₀ gl = I$\alpha$

I = 2M₀ (2l)² + 5M₀ l² = 13 M₀l²d

$\therefore$  $\alpha$ = $-$ ${{{M_0}gl} \over {13{M_0}{\ell ^2}}}$ $\Rightarrow$ $\alpha$ = $-$ ${g \over {13\ell }}$

$\therefore$  $\alpha$ = ${g \over {13\ell }}$ anticlockwise

4F² + 9F² + 12F² cos $\theta$ = R²

4F² + 36F² + 24F² cos $\theta$ = 4R²

4F² + 36F² + 24F² cos $\theta$

= 4(13F² + 12F²cos$\theta$) = 52F² + 48F²cos$\theta$

cos $\theta$ = $-$ ${{12{F^2}} \over {24{F^2}}}$ = $-$ ${1 \over 2}$

S = Area under graph

${1 \over 2}$ $\times$ 2$\times$2 + 2 $\times$ 2 + 3 $\times$ 1 = 9 m

For closed organ pipe, resonate frequency is odd multiple of fundamental frequency.

$\therefore$  (2n + 1) f₀ $\le$ 20,000

(f₀ is fundamental frequency = 1.5 KHz)

$\therefore$  n = 6

$\therefore$  Total number of overtone that can be heared is 7. (0 to 6)

Given that the area of the metal plate $A = 1 \times 10^{-4} m^2$, and the intensity of the radiation

$I = 16 mW/m^2 = 16 \times 10^{-3} W/m^2$.

The power $P$ falling on the metal plate is given by the product of the intensity of the radiation and the area:

$P = IA = (16 \times 10^{-3} W/m^2)(1 \times 10^{-4} m^2) = 1.6 \times 10^{-6} W$.

The energy of each incident photon is given as 10 eV. Converting this to joules (since 1 eV = $1.6 \times 10^{-19} J$):

$E_{\text{photon}} = 10 eV = 10 \times 1.6 \times 10^{-19} J = 1.6 \times 10^{-18} J$.

The number of incident photons per second $n_{\text{photon}}$ is then given by the total power divided by the energy per photon:

$n_{\text{photon}} = P / E_{\text{photon}} = 1.6 \times 10^{-6} W / 1.6 \times 10^{-18} J = 10^{12} \text{ photons/s}$.

Given that only 10% of these photons actually produce photoelectrons, the number of photoelectrons produced per second $n_{\text{electron}}$ is:

$n_{\text{electron}} = 0.1 \times n_{\text{photon}} = 0.1 \times 10^{12} = 10^{11} \text{ electrons/s}$.

The work function of the metal is given as 5 eV, which is the energy required to remove an electron from the metal. Therefore, the maximum energy of the photoelectrons $E_{\text{max}}$ is given by the energy of the incident photons minus the work function:

$E_{\text{max}} = E_{\text{photon}} - \text{Work function} = 10 eV - 5 eV = 5 eV$.

f_carrier = ${{250} \over {0.1}}$ = 2500 KHZ

$\therefore$   Range of signal = 2250 Hz to 2750 Hz

Now check all options : for 2000 KHZ

f_mod = 200 Hz

$\therefore$   Range = 1800 KHZ to 2200 KHZ

Path difference, S₂P – S₁P = ${\lambda \over 2}$

$\Rightarrow$ $\sqrt {4{d^2} + {d^2}} - 2d = {\lambda \over 2}$

$\Rightarrow$ $d\left( {\sqrt 5 - 2} \right) = {\lambda \over 2}$

$\Rightarrow$ d = ${\lambda \over {2\left( {\sqrt 5 - 2} \right)}}$

P = I²R

4.4 = 4 $\times$ 10^$-$6 R

R = 1.1 $\times$ 10⁶ $\Omega$

P' = ${{{{11}^2}} \over R}$ = ${{{{11}^2}} \over {1.1}}$ $\times$ 10^$-$6 = 11 $\times$ 10^$-$5 W

WD = P$\Delta$V = nR$\Delta$T = ${1 \over 2} \times 8.31 \times 70$

Potential at origin = ${{KQ} \over 2} + {{KQ} \over 2} + {{KQ} \over {\sqrt {20} }} + {{KQ} \over {\sqrt {20} }}$

(Potential at $\infty$ = 0)

= KQ$\left( {1 + {1 \over {\sqrt 5 }}} \right)$

$\therefore$  Work required to put a fifth charge Q

at origin is equal to ${{{Q^2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 5 }}} \right)$

Without applied forces, (in equilibrium position) the needle will stay in the resultant magnetic field of earth. Hence, the dip ' $\theta$ ' at this place is $45^{\circ}$ (given).


We know that, horizontal and vertical components of earth's magnetic field $\left(B_H\right.$ and $B_V$ ) are related as

$\frac{B_V}{B_H}=\tan \theta$

Here, $\theta=45^{\circ}$ and $B_H=18 \times 10^{-6} \mathrm{~T}$

$\Rightarrow B_V=B_H \tan 45^{\circ}$

$\Rightarrow B_V=B_H=18 \times 10^{-6} \mathrm{~T} \quad\left(\because \tan 45^{\circ}=1\right)$

Now, when the external force $F$ is applied, so as to keep the needle stays in horizontal position is shown below,


Taking torque at point $P$, we get

$$\begin{array}{ll} m B_V \times 2 l=F l \\ \therefore F=2 \times m B_V \end{array}$$

Substituting the given values, we get

$$\begin{aligned} & =2 \times 1.8 \times 18 \times 10^{-6} \\ & =6.48 \times 10^{-5}=6.5 \times 10^{-5} \mathrm{~N} \end{aligned}$$

$\overrightarrow E = 10\widehat j\cos \left[ {\left( {6\widehat i + 8\widehat k} \right).\left( {x\widehat i + z\widehat k} \right)} \right]$

$= 10\widehat j\,\cos \left[ {\overrightarrow K .\overrightarrow r } \right]$

$\therefore$   $\overrightarrow K = 6\widehat i + 8\widehat k;$ direction of waves travel.

i.e., direction of 'c'.



$\therefore$   Direction of $\widehat B$ will be along

$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$

Mag. of $\overrightarrow B$ will be along

$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$

Mag. of $\overrightarrow B = {E \over C} = {{10} \over C}$

$\therefore$   $\overrightarrow B = {{10} \over C}\left( {{{ - 4\widehat i + 3\widehat k} \over 5}} \right) = {{\left( { - 8\widehat i + 6\widehat k} \right)} \over C}$

By energy convervation between 0 & $\infty$.

$- {{GMm} \over r} + {{ - GMm} \over r} + {1 \over 2}m{V^2} = 0 + 0$

[M is mass of star m is mass of meteroite)

$\Rightarrow$ v $= \sqrt {{{4GM} \over r}} = 2.8 \times {10^5}$m/s

$L{{di} \over {dt}} = 25$

$L \times {{15} \over 1} = 25$

$L = {5 \over 3}H$

$\Delta E = {1 \over 2} \times {5 \over 3} \times ({25^2} - {10^2})$

$= {5 \over 6} \times 525 = 437.5$ J

Restoring force due to pressing the bottle with small amount x,

F = $- \left( {\rho Ax} \right)g$

$\Rightarrow$ ma = $- \left( {\rho Ax} \right)g$

$\Rightarrow$ a = $- \left( {{{\rho Ag} \over m}} \right)x$

$\therefore$ ${{\omega ^2} = {{\rho Ag} \over m}}$ = ${{{\rho \left( {\pi {r^2}} \right)g} \over m}}$

$\Rightarrow$ $\omega$ = $\sqrt {{{{{10}^3} \times \pi \times {{\left( {2.5 \times {{10}^{ - 2}}} \right)}^2} \times 10} \over {310 \times {{10}^{ - 3}}}}}$ = 7.90 rad/s

Thermal energy of N molecule

= N$\left( {{3 \over 2}kT} \right)$

= ${N \over {{N_A}}}{3 \over 2}$RT

= ${3 \over 2}$(nRT)

= ${3 \over 2}$PV

= ${3 \over 2}$P$\left( {{m \over 8}} \right)$

= ${3 \over 2}$ $\times$ 4 $\times$ 10⁴ $\times$ ${2 \over 8}$

= 1.5 $\times$ 10⁴

order will 10⁴

Work done = $\overrightarrow F \cdot \overrightarrow d$  

                    $=$ 12 J

work energy theorem

w_net $=$ $\Delta$K.E.

12 $=$ K_f $-$ 3

K_f = 15 J

R₁ = 32 $\times$ 10 = 320

for wheatstone bridge

$\Rightarrow$  ${{{R_1}} \over {{R_3}}} = {{{R_2}} \over {{R_4}}}$

${{320} \over {{R_3}}} = {{80} \over {40}}$

${R_3} = 160$

$\therefore$  Correct answer is Brown  Blue  Brown

Electric field of equitorial plane of dipole

$= - {{K\overrightarrow P } \over {{r^3}}}$

$\therefore$  At P, F $= - {{K\overrightarrow P } \over {{r^3}}}$Q.

At   P¹ , F¹ $= - {{K\overrightarrow P Q} \over {{{\left( {r/3} \right)}^3}}} = 27F.$

$\left| {\overrightarrow A + \overrightarrow B } \right| = 2a\cos \theta /2$      . . . (1)

$\left| {\overrightarrow A - \overrightarrow B } \right| = 2a\cos {{\left( {\pi - \theta } \right)} \over 2} = 2a\sin \theta /2$      . . . (2)

$\Rightarrow \,\,\,n\left( {2a\cos {\theta \over 2}} \right) = 2a{{\sin \theta } \over 2}$

$\Rightarrow \,\,\,\tan {\theta \over 2} = n$

0.95 R = ${{R{R_v}} \over {R + {R_v}}}$

0.95 $\times$ 30 = 0.05 R_v

R_v = 19 $\times$ 30 = 570 $\Omega$

Work done on isothermal irreversible for ideal gas

= $-$P_ext (V₂ $-$ V₁)

= $-$4 N/m² (1 m³ $-$ 5m³)

= 16 Nm

Isothermal process for ideal gas

$\Delta$U = 0

q = $-$ w

= $-$ 16 Nm

= $-$ 16 J

Heat used to increase temperature of Al

q = n C_m $\Delta$T

16 J = 1 $\times$ 24 ${J \over {mol.K}}$ $\times$ $\Delta$T

$\Delta$T = ${2 \over 3}$K

Note : While numbering, double bond should get least possible number on the longest path.

In lutetium (⁷¹Lu) the 71^st electron enters in 5d-orbital.

Electron configuration is [Xe]4f¹⁴5d¹6s²

The balanced reaction of oxalate with permanganate in acidic medium is :

$5 \, \text{C}_2\text{O}_4^{2-} + 2 \, \text{MnO}_4^- + 16 \, \text{H}^+ \rightarrow 10 \, \text{CO}_2 + 2 \, \text{Mn}^{2+} + 8 \, \text{H}_2\text{O}$

In this redox reaction, the oxalate ion $\text{C}_2\text{O}_4^{2-}$ is oxidized to carbon dioxide $\text{CO}_2$, and the permanganate ion $\text{MnO}_4^-$ is reduced to manganese(II) ion $\text{Mn}^{2+}$.

The oxidation half-reaction, representing the change for the oxalate ion, is :

$\text{C}_2\text{O}_4^{2-} \rightarrow 2 \, \text{CO}_2 + 2 \, e^-$

From this half-reaction, it's clear that each oxalate ion produces two CO₂ molecules and releases two electrons in the process.

So, the number of electrons involved in producing one molecule of CO₂ is :

$\frac{2 \, e^-}{2 \, \text{CO}_2} = 1 \, e^-/\text{CO}_2$

Therefore, the correct answer is Option C: One electron is involved in the production of one molecule of CO₂.

${{\Delta {T_b}} \over {\Delta {T_f}}} = {{i.m \times {k_b}} \over {i \times m \times {k_f}}}$

${2 \over 2} = {{1 \times 1 \times {k_b}} \over {1 \times 2 \times {k_f}}}$

${k_b} = 2{k_f}$

[Au(CN)₂]^– and [Ag(CN)₂]^– both are soluble complexes.

[Au(CN)₂]^– ⇌ Au⁺ + 2CN^-

[Ag(CN)₂]^– ⇌ Ag⁺ + 2CN^-

The concentration of produced Au⁺ ion and Ag⁺ ion due to ionisation is very less which ensure slow deposition and this slow deposition will generate uniform layer on the substance to be electroplated.

   NH₄SH(s)  $\rightleftharpoons$    NH₃(g) + H₂S(g)

$n = {{5.1} \over {51}} = .1\,mole\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,$

$.1\left( { - 1 - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.1\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.1\alpha$

$\alpha \,\, = \,\,30\% = .3$

so number of moles at equilibrium

$\,\,\,\,\,\,\,.1\,(1 - .3)\,\,\,\,\,.1\,\, \times \,\,.3\,\,\,\,\,\,\,\,\,.1\,\, \times \,.3$

$= \,\,\,\,.07\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = .03\;\,\,\,\,\,\,\,\,\, = .03$

Now use PV = nRT at equilibrium

P_total $\times$ 3 lit = (.03 + .03) $\times$ .082 $\times$ 600

P_total = .984 atm

At equilibrium

<sup>P_NH3 = P_H2S = ${{{P_{total}}} \over 2}$ = .492

So k_p = P_NH3 . P_H2S = (.492) (.492)

k_p = .242 atm2</sup>

$\therefore$ The difference in the number of unpaired electrons = 3 - 1 = 2

The reaction which has slow reactivity requires catalyst.

Among halogens reactivity order is

F₂ > Cl₂ > Br₂ > I₂

So, H₂ + I₂ $\to$ 2HI, this reaction requires a catalyst.

Here in the compound $\pi$ bond of C = O bond will be at resonance with benzene ring with -M effect. But lone pair of N atom is electron donating group which will make the left benzene ring more active by +M effect than right benzene ring and hence electrophilic substitution reaction will be taken place at ortho- or para- position of left benzene ring. As ortho-product formed is sterically hindered and hence para-product will be major product.

Molarity = ${{{{(n)}_{solute}}} \over {{V_{solution}}(in\,\,lit)}}$

0.1 = ${{wt./342} \over 2}$

wt (C₁₂H₂₂O₁₁) = 68.4 gram

A₂B₃ has HCP lattice

If A form HCP,
then ${{{3^{th}}} \over 4}$ of THV must occupied by B to form A₂B₃

If B form HCP, then ${{{1^{th}}} \over 3}$ of THV must occupied by A to form A₂B₃

Haemoglobin $\buildrel \, \over \longrightarrow$ positive sol
Ag $-$ sol $\buildrel \, \over \longrightarrow$ negative sol

(E)_n^th = (E_GND)_H . ${{{Z^2}} \over {{n^2}}}$

E_3^rd (He⁺) = ($-$13.6eV) . ${{{2^2}} \over {{3^2}}}$ = $-$ 6.04 eV