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

We can apply Faraday's law of electromagnetic induction to solve this problem. Faraday's law states that the induced electromotive force (emf) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.

The cube is moving through a magnetic field, so it's behaving like a conductor moving through a magnetic field. The induced emf or voltage can be calculated by using the formula:

emf = B $\times$ v $\times$ d

Where:

B is the magnetic field strength,

v is the velocity of the conductor, and

d is the length of the conductor perpendicular to the direction of motion and magnetic field.

In this case, the cube is moving in the y-direction and the magnetic field is in the z-direction. So, the faces perpendicular to the x-axis are involved. The length of the conductor (d) perpendicular to the motion and magnetic field is the edge length of the cube, which is 2 cm or 0.02 m.

So, plugging the given values into the formula:

emf = 0.1 T $\times$ 6 m/s $\times$ 0.02 m = 0.012 V = 12 mV

So, the potential difference between the two faces of the cube perpendicular to the x-axis is 12 mV.

Therefore, Option B is correct.

$\lambda$ = ${h \over p}$          {$\lambda$ = 7.5 $\times$ 10^$-$12}

P = ${h \over \lambda }$

KE = ${{{P^2}} \over {2m}} = {{{{\left( {h/\lambda } \right)}^2}} \over {2m}}$

$= {{\left\{ {{{6.6 \times {{10}^{ - 34}}} \over {7.5 \times {{10}^{ - 12}}}}} \right\}} \over {2 \times 9.1 \times {{10}^{ - 31}}}}$ J

KE = 25 Kev

${1 \over {2{f_2}}} = {1 \over {{f_1}}} = \left( {{\mu _1} - 1} \right)\left( {{1 \over \infty } - {1 \over { - R}}} \right)$

${1 \over {{f_2}}} = \left( {{\mu _2} - 1} \right)\left( {{1 \over { - R}} - {1 \over \infty }} \right)$

${{\left( {{\mu _1} - 1} \right)} \over R} = {{\left( {{\mu _2} - 1} \right)} \over {2R}}$

$2{\mu _1} - {\mu _2} = 1$

$R = {{{u^2}\sin 2\theta } \over g}$

$A = \pi \,{R^2}$

$A \propto {R^2}$

$A \propto {u^4}$

${{{A_1}} \over {{A_2}}} = {{u_1^4} \over {u_2^4}} = {\left[ {{1 \over 2}} \right]^4} = {1 \over {16}}$

$\overrightarrow {{r_g}} = {a \over 2}\widehat i + {a \over 2}\widehat k$

$\overrightarrow {{r_H}} = {a \over 2}\widehat i + {a \over 2}\widehat k$

$\overrightarrow {{r_H}} - \overrightarrow {{r_g}} = {a \over 2}\left( {\widehat j - \widehat i} \right)$

N $-$ mg = ${{mg} \over 2}$ $\Rightarrow$ N = ${{3mg} \over 2}$

The distance travelled by the system in time t is

S = ut + ${1 \over 2}a{t^2} = 0 + {1 \over 2}\left( {{g \over 2}} \right){t^2} = {1 \over 2}{g \over 2}{t^2}$

Now, work done

W = N.S = $\left( {{3 \over 2}mg} \right)\left( {{1 \over 2}{g \over 2}{t^2}} \right)$

$\Rightarrow$  W = ${{3m{g^2}{t^2}} \over 8}$

Work done, W = $\left( {\Delta \overrightarrow \mu } \right).\overrightarrow B$

= 2 $\times$ 10^$-$2 $\times$ 1 cos(0.125)

= 0.02 J

At height r from center of earth. orbital velocity

= $\sqrt {{{GM} \over r}}$

$\therefore$  By energy conservation

KE of 'm' + $\left( { - {{GMm} \over r}} \right)$ = 0 + 0

(At infinity, PE = KE = 0)

$\Rightarrow$  KE of 'm' = ${{{GMm} \over r}}$ = ${\left( {\sqrt {{{GM} \over r}} } \right)^2}$ m = mv²

V $= {{4qa} \over {\left( {R + x} \right)}} = {{2qa} \over {\left( {{x^2}} \right)}}$

$\sqrt 2 x = R + x$

$x = {R \over {\sqrt 2 - 1}}$


dist $= {R \over {\sqrt 2 - 1}} + R = {{\sqrt 2 R} \over {\sqrt 2 - 1}}$

Since height of water column is constant therefore, water inflow rate (Q_in)

= water outflow rate

Q_in = 10^$-$4 m³s^$-$1

Q_out = Au = 10^$-$4 $\times$ $\sqrt {2gh}$

10^$-$4 = 10^$-$4 $\sqrt {20 \times h}$

h = ${1 \over {20}}m$

h = 5cm

Consider a strip of radius x & thickness dx,

Torque due to friction on this strip.

$\int {d\tau = \int\limits_0^R {{{x\mu F.2\pi xdx} \over {\pi {R^2}}}} }$

$\tau = {{2\mu F} \over {{R^2}}}.{{{R^3}} \over 3}$

$\tau = {{2\mu FR} \over 3}$

FR = ${3 \over 2}$ MR²$\alpha$

$\alpha$ = ${{2F} \over {3MR}}$

Path difference = d sin$\theta$ $\approx$ d$\theta$

= 0.1 $\times$ ${1 \over {40}}$ mm = 2500nm

or bright fringe, path difference must be integral multiple of $\lambda$.

$\therefore$   2500 = n$\lambda$₁ = m$\lambda$₂

$\therefore$   $\lambda$₁ = 625 (from n = 4), $\lambda$₂ = 500 (from m = 5)

$\because$   M = NIA

dq = $\lambda$dx  &   A = $\pi$x²

$\int {dm} = \int {\left( x \right){{{\rho _0}x} \over \ell }} \,dx.\pi {x^2}$

M = ${{n{\rho _0}\pi } \over \ell }.\int\limits_0^\ell {{x^3}.dx} = {{n{\rho _0}\pi } \over \ell }.\left[ {{{{L^4}} \over 4}} \right]$

M = ${{n{\rho _0}\pi {\ell ^3}} \over 4}$ or ${\pi \over 4}n\rho {\ell ^3}$

i₁ = ${{10} \over {20}}$ = 0.5A

i₂ = 0

$i = {\varepsilon \over {13r}}$

$i\left( {{x \over L}12r} \right) = {\varepsilon \over 2}$

${\varepsilon \over {13r}}\left[ {{x \over L}.12r} \right] = {\varepsilon \over 2}$ $\Rightarrow \,\,\,\,x = {{13L} \over {24}}$

P = $\overline x +$ y

Q = $\overline {\overline y .x}$ = y + $\overline x$

Output, R = $\overline {P + Q}$

To make R = 1

$\overline {P + Q}$ should be = 1

$\therefore$ P + Q must be 0

$\therefore$ ($\overline x +$ y) + (y + $\overline x$) = 0

Now by checking each option you can see

When y = 0 and x = 1 then

$\overline x +$ y = 0

$\therefore$ ($\overline x +$ y) + (y + $\overline x$) = 0

R_eq berween any two vertex will be

${1 \over {{{\mathop{\rm R}\nolimits} _{eq}}}} = {1 \over {12}} + {1 \over 6} \Rightarrow {{\mathop{\rm R}\nolimits} _{eq.}} = 4\Omega$

P = i²R.

$\therefore$   for i_max, R must be minimum

from color coding R = 50 $\times$ 10²$\Omega$

$\therefore$   i_max = 20mA

Let dielectric constant of material used be K.

$\therefore$  ${{10{ \in _0}\,A/3} \over d} + {{12{ \in _0}\,A/3} \over d} + {{14{ \in _0}\,A/3} \over d} = {{K{ \in _0}A} \over d}$

$\Rightarrow$    K $=$ 12

E₀ = B₀ $\times$ C

= 100 $\times$ 10^$-$6 $\times$ 3 $\times$ 10⁸

= 3 $\times$ 10⁴ N/C

$\therefore$  correct answer is 3 $\times$ 10⁴ N/C

$\left( {{{dQ} \over {dt}}} \right) = {{kA\Delta T} \over \ell }$

$\Rightarrow$  ${1 \over A}\left( {{{dQ} \over {dt}}} \right) = {{\left( {0.1} \right)\left( {900} \right)} \over 1} = 90W/{m^2}$

Maximum distance upto which signal can be broadcasted is

d_max = $\sqrt {2R{h_T}} + \sqrt {2R{h_R}}$

where h_T and h_R are heights of transmitter tower and height of reserver respectively. Putting all values -

d_max = $\sqrt {2 \times 6.4 \times 106} \left[ {\sqrt {104} + \sqrt {40} } \right]$

on solving, d_max = 65 km

Potential at point P, V = ${{k{Q_a}} \over a} + {{k{Q_b}} \over b} + {{k{Q_c}} \over c}$

$\because$  Q_a : Q_b : Q_c : : a² : b² : c²

{since $\sigma$_a = $\sigma$_b = $\sigma$_c}

$\therefore$  Q_a = $\left[ {{{{a^2}} \over {{a^2} + {b^2} + {c^2}}}} \right]$Q

Q_b = $\left[ {{{{b^2}} \over {{a^2} + {b^2} + {c^2}}}} \right]$ Q

Q_c = $\left[ {{{{c^2}} \over {{a^2} + {b^2} + {c^2}}}} \right]$ Q

V = ${Q \over {4\pi { \in _0}}}\left[ {{{\left( {a + b + c} \right)} \over {{a^2} + {b^2} + {c^2}}}} \right]$

${t_1} = 1 - {{{T_2}} \over {{T_1}}} = 1 - {{{T_2}} \over {{T_2}}} = 1 - {{{T_4}} \over {{T_3}}}$

$\Rightarrow \,\,\,{{{T_2}} \over {{T_1}}} = {{{T_3}} \over {{T_4}}} = {{{T_4}} \over {{T_3}}}$

$\Rightarrow \,\,\,{T_2} = \sqrt {{T_1}{T_3}} = \sqrt {{T_1}\sqrt {{T_2}{T_4}} }$

${T_3} = \sqrt {{T_2}{T_4}}$

$T_2^{3/4} = \sqrt {T_1^{1/2}} T_4^{1/4}$

${T_2} = T_1^{2/3}T_4^{1/3}$

f_app = f₀ $\left[ {{{{v_2} \pm {v_0}} \over {{v_2} \pm {v_s}}}} \right]$

f₁ = f₀ $\left[ {{{340} \over {340 - 34}}} \right]$

f₂ = f₀ $\left[ {{{340} \over {340 - 17}}} \right]$

${{{f_1}} \over {{f_2}}} = {{340 - 17} \over {340 - 34}} = {{323} \over {306}} \Rightarrow {{{f_1}} \over {{f_2}}} = {{19} \over {18}}$

at t = 0, A₀ = ${{dN} \over {dt}}$ = 1600C/s

at t = 8s, A = 100 C/s

${A \over {{A_0}}}$ = ${1 \over {16}}$ in 8 sec

Therefor half life is t_1/2 = 2 sec

$\therefore$   Activity at t = 6 will be 1600${\left( {{1 \over 2}} \right)^3}$

       = 200 C/s

Here given that

density of a material in SI units is 128 kg m^–3

And a new unit system is introduced where 1 unit of length = 25 cm and 1 unit of mass = 50 g

You should know that, physical quantity is same in any unit system. And to calculate a physical quantity yo should know two things

(1) numerical value of the physical quantity (n)

(2) unit of the physical quantity (u)

And n $\times$ u = constant in any unit system.

Here in SI unit system,

n₁ = 128

u₁ = kg/m³

And in new unit system,

n₂ = ?

u₂ = 50gm/(25cm)³

As n₁u₁ = n₂u₂

$\therefore$ 128 $\times$ (kg/m³) = n₂ $\times$ 50gm/(25cm)³

$\Rightarrow$ 128 $\times$ ${{1000\,gm} \over {{{\left( {100\,cm} \right)}^3}}}$ = n₂ $\times$ ${{50\,gm} \over {{{\left( {25\,cm} \right)}^3}}}$

$\Rightarrow$ n₂ = 128 $\times$ ${{20} \over {{4^3}}}$ = 40

Time taken for the particles to collide,

t = ${f \over {{V_{rel}}}} = {{100} \over {100}} = 1$ sec

Speed of wood just before collision = gt = 10 m/s

& speed of bullet just before collision v-gt

= 100 $-$ 10 = 90 m/s

Now, conservation of linear momentum just before and after the collision -

$-$ (0.02) (1v) + (0.02) (9v) = (0.05)v

$\Rightarrow$   150 = 5v

$\Rightarrow$   v = 30 m/s

Max. height reached by body h = ${{{v^2}} \over {2g}}$



h = ${{30 \times 30} \over {2 \times 10}}$ = 45m

$\therefore$  Height above tower = 40 m

Velocity of wave on string

$V = \sqrt {{T \over \mu }} = \sqrt {{8 \over 5} \times 1000} = 40m/s$

Now, wavelength of wave

$\lambda = {v \over n} = {{40} \over {100}}m$

Separation b/w successive nodes,

${\lambda \over 2} = {{20} \over {100}}\,m$ $=$ 20 cm

E.N. of Al = (1.5) $\cong$ Be (1.5)

In the given reaction, $\mathrm{AlCl}_{3}$ acts as Lewis acid and helps in the generation of the carbocation. The resulting carbocation $\left(1^{\circ}\right)$ rearranges itself to stable carbocation $\left(3^{\circ}\right)$.

Two glasses ''A'' and ''B'' have BOD values 10 and ''20'', respectively.

Hence glasses ''B'' is more polluted than glasses ''A''.

Density of DCM = 1.39 and density of H₂O = 1.0.

Dichloromethane (DCM) is a denser solvent than water. Therefore, when a mixture of DCM and water is allowed to settle in a separating funnel, the DCM forms the lower layer due to its higher density, while the water forms the upper layer. The two solvents are immiscible, meaning they do not mix and form two distinct layers.

So, the correct answer is :

Option C : DCM and H₂O would stay as lower and upper layer respectively in the separating funnel(S.F).

${y_A} = {{{P_A}} \over {{P_{Total}}}} = {{P_A^0{x_A}} \over {P_A^0{X_A} \times P_B^0X{}_B}}$

$= {{7 \times {{10}^3} \times 0.4} \over {7 \times {{10}^3} \times 0.4 + 12 \times {{10}^3} \times 0.6}}$

$= {{2.8} \over {10}} = 0.28$

${y_B} = 0.72$

In Triclinic unit cell

a $\ne$ b $\ne$ c   &   $\alpha$ $\ne$ $\beta$ $\ne$ $\gamma$ $\ne$ 90^o

Due to Lanthanoid contraction both atomic radii and ionic radii decreases gradually in the lanthanoid series.

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

T = ${{\Delta H} \over {\Delta S}} = {{200} \over {40}} = 5K$

Higher the oxidation potential better will be reducing power.

NaBH₄ is weak reducing agent. It only reduce Ketone to 2^o alcohol. It do not react with ester & C = C.

So, The total number of isomers for a square planar complex [M(F)(Cl)(SCN)(NO₂ )] is 12

''Be'' Metal is used in x-ray window is due to transparent to x-rays.

Then is no catalyst is required for combustion of coal.

Two pi and half sigma bonds are presents in molecule with bond order 2.5.

Moleculer orbital configuration of $N_2^+$ (13 electrons)

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$

Bond order = ${1 \over 2}\left( {9 - 4} \right)$ = 2.5

Alcoholic KOH performs $\alpha$ - $\beta$ elimination with halides.

The carbon which is attached to Br atom is called $\alpha$ carbon.

Here, two $\alpha$ carbon presents. Let's call them $\alpha$₁ and $\alpha$₂.

In case of $\alpha$₁ carbon, H elimination can take place either from $\beta$₁ or $\beta$₂ carbon. But as we know always more stable product is formed from a reaction. Here if H is removed from $\beta$₁ carbon then a $\pi$ bond is created which will perticipate in resonance with benzene ring and product will be more stable. But if H is removed from $\beta$₂ carbon then the created $\pi$ bond will not perticipate in resonance with benzene ring so product will be less stable.

In case of $\alpha$₂ carbon, H elimination can take place either from $\beta$₂ or $\beta$₃ carbon. But as we know always more stable product is formed from a reaction. Here if H is removed from $\beta$₂ carbon then a $\pi$ bond is created which will perticipate in resonance with the $\pi$ bond associated with the $\alpha$₁ carbon so product will be more stable. But if H is removed from $\beta$₃ carbon then the created $\pi$ bond will not perticipate in any resonance so product will be less stable.

H₂O₂ act as oxidising agent and reducing agent in acidic medium as well as basic medium.

H₂O₂ Act as oxidant : -

H₂O₂ + 2H$^ \oplus$ + 2e^$-$ $\to$  2H₂O (In acidic medium)

H₂O₂ + 2e^$-$ $\to$ 2OH^$-$ (In basic medium)

H₂O₂ Act as reductant : -

H₂O₂ $\to$ 2H⁺ + O₂ + 2e^$-$ (In acidic medium)

H₂O₂ + 2OH^$-$ $\to$ 2H₂O + O₂ + 2e^$-$ (In basic medium)

Due to greater stability of allyl radical formed after replacement of $\gamma$-hydrogen in compound (E), allylic bromination can easily be taken place.

Total number of isotopes of hydrogen is 3

$\Rightarrow \,\,{}_1^1H,{}_1^2H$  or  ${}_1^2D,\,{}_1^3H\,\,$or  ${}_1^3T$

and only ${}_1^3H$ or ${}_1^3T$ is an Radioactive element.

More is the electrophilic character of carbonyl group of ester faster is the alkaline hydrolysis.

Since –I effect groups increase the electrophilic nature of carboxyl carbon whereas +I effect groups decrease the electrophilic nature of carboxyl carbon. Therefore, the correct decreasing order is III > II > I > IV.

Ca(OH)₂ + Na₂SO₄ $\to$ CaSO₄ + 2NaOH

100 m mol 14 m mol      $-$     $-$      $-$

$-$    $-$      $-$  14 m mol 28 m mol

w_CasO4 = 14 $\times$ 10^$-$3 $\times$ 136 = 1.9 gm

[OH^$-$] = ${{28} \over {100}}$ = 0.28 M

H = ${1 \over 2}$ (V + M - c + a)

$\therefore$ H = ${1 \over 2}$(8 + 4) = 6


From structure, it is clear that it has five bond pairs and one lone pair.

In Hall-Heroult's process is given by

At cathode :- 4Al_{\left( \ell \right)}^{3 + }{\mkern 1mu} {\mkern 1mu} + {\mkern 1mu} {\mkern 1mu} 12{e^ - }{\mkern 1mu} {\mkern 1mu} \,\mathrel{\mathop{\kern0pt\longrightarrow} \limits_\,} \,\,4Al\left( \ell \right)

At Anode :- 6O_{\left( \ell \right)}^{2 - }\,\,\mathrel{\mathop{\kern0pt\longrightarrow} \limits_\,} \,\,3{O_2}(g) + 12{e^ - }

3C + 3{O_2}\,\,\mathrel{\mathop{\kern0pt\longrightarrow} \limits_\,} \,\,3C{O_2}\left( \uparrow \right)

Wilkinsion catalyst is [(Ph₃P)₃RhCl]

Arrhenius Equation -

Arrhenius gave the quantitative dependence of rate constant on temperature by the Arrhenius equation.

- wherein

$k = A{e^{ - {E_a}/RT}}$

${\mathop{\rm lnk}\nolimits} = lnA - {{{E_a}} \over {RT}}$

k = Rate constant

as we have learned in Arrhenius equation when we increase Ea, K should decrease As T increases, power of exponential increases so k also increase.

Hence, both the plots are correct.

E = W + ${1 \over 2}$mv²

K.E. = hv $-$ 4v₀

K.E. = hv + ($-$ hv₀)

y = mx + $\underline C$