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

U $= {{{f_1}} \over 2}{n_1}RT + {{{f_2}} \over 2}{n_2}RT$

$= {5 \over 2}\left( {3RT} \right) + {3 \over 2} \times 5RT$

U $= 15RT$

K = ${1 \over 2}$m${\omega ^2}$A²cos²$\omega$t

U = ${1 \over 2}m{\omega ^2}$ A² sin² $\omega$t

${k \over U}$ = cot² $\omega$t = cot² ${\pi \over {90}}$(210) = ${1 \over 3}$

Hence ratio is 3 (most appropriate)

velocity of 1 kg block just before it collides with 3kg block

= $\sqrt {2gh} = \sqrt {2000}$ m/s

Applying momentum conversation just before and just after collision.

1 $\times$ $\sqrt {2000}$ = 4v $\Rightarrow$ v = ${{\sqrt {2000} } \over 4}$ m/s



initial compression of spring

1.25 $\times$ 10⁶ x₀ = 30 $\Rightarrow$ x₀ $\approx$ 0

applying work energy theorem,

W_g + W_sp = $\Delta$KE

$\Rightarrow$   40 $\times$ x + ${1 \over 2}$ $\times$ 1.25 $\times$ 10⁶ (0² $-$ x²)

= 0 $-$ ${1 \over 2}$ $\times$ 4 $\times$ v²

solving x $\approx$ 4 cm

$\Delta$x $=$ ${\lambda \over 8}$

$\Delta$$\phi$ $=$ ${{\left( {2\pi } \right)} \over \lambda }{\lambda \over 8} = {\pi \over 4}$

I $=$ I₀cos²$\left( {{\pi \over 8}} \right)$

${{\rm I} \over {{{\rm I}_0}}} =$ cos²$\left( {{\pi \over 8}} \right)$

Since D_m = ($\mu$ $-$ 1) A & on increasing the wavelength, $\mu$ decreases & hence D_m decreases.

From lens equation

${1 \over v} - {1 \over u} = {1 \over f}$

${1 \over v} - {1 \over {\left( { - 20} \right)}} = {1 \over {\left( {.3} \right)}} = {{10} \over 3}$

${1 \over v} = {{10} \over 3} - {1 \over {20}}$

${1 \over v} = {{197} \over {60}};v = {{60} \over {197}}$

m = $\left( {{v \over u}} \right)$ = ${{\left( {{{60} \over {197}}} \right)} \over {20}}$

velocity of image wrt. to lens is given by

v_I/L = m²v_O/L

direction of velocity of image is same as that of object

v_O/L = 5 m/s

Let amount of ice is m gm.

According to principal of calorimeter heat taken by ice = heat given by water

$\therefore$  20 $\times$ 2.1 $\times$ m + (m $-$ 20) $\times$ 334

= 50 $\times$ 4.2 $\times$ 40

376 m = 8400 + 6680

m = 40.1

v_x = 10cos60^o = 5 m/s v_y = 10cos30^o = $5\sqrt 3$ m/s

velocity after t = 1 sec.

v_x = 5 m/s

v_y = $\left| {\left( {5\sqrt 3 - 10} \right)} \right|$ m/s = 10 $-$ 5$\sqrt 3$

a_n = ${{{v^2}} \over R} \Rightarrow \,R\,$ = ${{v_x^2 + v_y^2} \over {{a_n}}}$ = ${{25 + 100 + 75 - 100\sqrt 3 } \over {10\cos \theta }}$

tan$\theta$ = ${{10 - 5\sqrt 3 } \over 5}$ = 2 $-$ ${\sqrt 3 }$ $\Rightarrow$  $\theta$ = 15^o

R = ${{100\left( {2 - \sqrt 3 } \right)} \over {10\cos 15}} = 2.8m$

Since voltage across zener diode must be less than 10V therefore it will not work in breakdown region, & its resistance will be infinite & current through it = 0

$F = \alpha \beta {e^{\left( {{{ - {x^2}} \over {\alpha KT}}} \right)}}$

$\left[ {{{{x^2}} \over {\alpha KT}}} \right] = {M^o}{L^o}{T^o}$

${{{L^2}} \over {\left[ \alpha \right]M{L^2}{T^{ - 2}}}}$ $=$ ${M^o}{L^o}{T^o}$

$\Rightarrow$  $\left[ \alpha \right] = {M^{ - 1}}{T^2}$

$\left[ F \right] = \left[ \alpha \right]\left[ \beta \right]$

MLT^$-$2 = M^$-$1T²[$\beta$]

$\Rightarrow$  [$\beta$] = M²LT^$-$4

C $= {1 \over {\sqrt {{\mu _0}{ \in _0}} }}$

V = $= {1 \over {\sqrt {k{ \in _0}{\mu _0}} }}$ [For transparent medium $\mu$_r $\approx$ $\mu$₀]

$\therefore$  ${C \over V}$ $=$ $\sqrt k =$ n

${1 \over 2} \in {}_0\,E_0^2$C $=$ intensity $=$ ${1 \over 2}$$\in$₀ kE²v

$\therefore$   E$_0^2$C $=$ kE²v

$\Rightarrow$  ${{E_0^2} \over {{E^2}}} = {{kV} \over C} = {{{n^2}} \over n} \Rightarrow {{{E_0}} \over E} = \sqrt n$

similarly

${{B_0^2C} \over {2{\mu _0}}} = {{{B^2}v} \over {2{\mu _0}}} \Rightarrow {{{B_0}} \over B} = {1 \over {\sqrt n }}$

To determine the moment of the forces about point O, we need to calculate the moments of each force and then combine them vectorially.

First, consider force $\overrightarrow{F_1}$. This force acts along the z-axis at the point $(2\overrightarrow{i} + 3\overrightarrow{j})$.

The position vector of the point of application of $\overrightarrow{F_1}$ relative to the origin O is:

$\overrightarrow{r_1} = 2\overrightarrow{i} + 3\overrightarrow{j}$

Since $\overrightarrow{F_1}$ acts along the z-axis, we can express it as:

$\overrightarrow{F_1} = F\overrightarrow{k}$

The moment of force $\overrightarrow{F_1}$ about point O is given by the cross product:

$\overrightarrow{M_1} = \overrightarrow{r_1} \times \overrightarrow{F_1}$

Substitute the values:

$\overrightarrow{M_1} = (2\overrightarrow{i} + 3\overrightarrow{j}) \times F\overrightarrow{k}$

Compute the cross product:

$\overrightarrow{M_1} = 2F(\overrightarrow{i} \times \overrightarrow{k}) + 3F(\overrightarrow{j} \times \overrightarrow{k})$ $\overrightarrow{M_1} = 2F(-\overrightarrow{j}) + 3F\overrightarrow{i}$ $\overrightarrow{M_1} = -2F\overrightarrow{j} + 3F\overrightarrow{i}$

So, the moment due to force $\overrightarrow{F_1}$ is:

$\overrightarrow{M_1} = 3F\overrightarrow{i} - 2F\overrightarrow{j}$

Torque for F₂ force

${\overrightarrow F _2}$ = ${F \over 2}\left( { - \widehat i} \right) + {{F\sqrt 3 } \over 2}\left( { - \widehat j} \right)$

${\overrightarrow r _1} = 0\widehat i + 6\widehat j$

${\overrightarrow \tau _{_{{F_2}}}} = \overrightarrow {{r_1}} \times \overrightarrow {{F_1}} = 3F\widehat k$

${\overrightarrow \tau _{net}} = {\overrightarrow \tau _{{F_1}}} + {\overrightarrow \tau _{{f_2}}}$

$= 3F\widehat i + 2F\left( { - \widehat j} \right) + 3F\left( {\widehat k} \right)$

U = K$\left[ {{{{q^2}} \over a} + {{Qq} \over a} + {{Qq} \over {a\sqrt 2 }}} \right] = 0$

$\Rightarrow$  q = $-$ Q$\left[ {1 + {1 \over {\sqrt 2 }}} \right]$

$\Rightarrow$  Q = ${{ - q\sqrt 2 } \over {\sqrt 2 + 1}}$

$r = {{\sqrt {2mk} } \over {eB}} = {{\sqrt {2me\Delta v} } \over {eB}}$

$r = {{\sqrt {{{2m} \over e}.\Delta v} } \over B} = {{\sqrt {{{2 \times 9.1 \times {{10}^{ - 31}}} \over {1.6 \times {{10}^{ - 19}}}}\left( {500} \right)} } \over {100 \times {{10}^{ - 3}}}}$

$r = {{\sqrt {{{9.1} \over {0.16}} \times {{10}^{ - 10}}} } \over {{{10}^{ - 1}}}} = {3 \over 4} \times {10^{ - 4}}$

   $= 7.5 \times {10^{ - 4}}$

From time t = 0 to t = t₀, growth of current takes place and after that decay of current takes place.


most appropriate is (A)

v₀ = $\sqrt {g(R + h)} \approx \sqrt {gR}$

v_e = $\sqrt {2g(R + h)} \approx \sqrt {2gR}$

$\Delta$v=v_e $-$ v₀ = $\left( {\sqrt 2 - 1} \right)\sqrt {gR}$

Energy of photon $= {{12500} \over {980}} = 12.75eV$

$\therefore$  Electron will excite to n = 4

Since 'R' $\propto$ n²

$\therefore$  Radius of atom will be 16a₀

$\left| {\Delta \overrightarrow v } \right| = \sqrt {v_1^2 + v_2^2 + 2{v_1}{v_2}\cos \left( {\pi - \theta } \right)}$

$= 2v\sin {\theta \over 2}$        since $\left[ {\left| {\overline v {}_1} \right| = \left| {{{\overline v }_2}} \right|} \right]$

$= \left( {2 \times 10} \right) \times \sin \left( {{{30}^o}} \right)$

= 10 m/s

${h \over {mv}} = {10^{ - 3}}\left( {{{3 \times {{10}^8}} \over {6 \times {{10}^{14}}}}} \right)$

v $= {{6.63 \times {{10}^{ - 34}} \times 6 \times {{10}^{14}}} \over {9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^5}}}$

v $= 1.45 \times {10^6}$ m/s

Potential gradient with R_h = 2$\Omega$

is $\left( {{6 \over {2 + 4}}} \right) \times {4 \over L} = {{dV} \over {dL}};$   L $=$ 100 cm

Let null point be at $\ell$ cm

thus $\varepsilon$₁ $=$ 0.5V $=$ $\left( {{6 \over {2 + 4}}} \right) \times {4 \over L} \times \ell$      . . .(1)

Now with R_h $=$ 6$\Omega$ new potential gradient is

$\left( {{6 \over {4 + 6}}} \right) \times {4 \over L}$ and at null point

$\left( {{6 \over {4 + 6}}} \right)\left( {{4 \over L}} \right) \times \ell = {\varepsilon _2}$                          . . .(2)

dividing equation (1) by (2) we get

${{0.5} \over {{\varepsilon _2}}} = {{10} \over 6}$ thus ${\varepsilon _2} = 0.3$

y = 0.03 sin(450 t $-$ 9x)

v = ${\omega \over k} = {{450} \over 9}$ = 50m/s

v = $\sqrt {{T \over \mu }} \Rightarrow {T \over \mu }$ = 2500

$\Rightarrow$  T = 2500 $\times$ 5 $\times$ 10^$-$3

= 12.5 N

Momentum per second carried by liquid per second is $\rho$av²

net force due to reflected liquid = 2$\times$$\left[ {{1 \over 4}\rho a{v^2}} \right]$

net force due to stopped liquid = ${{1 \over 4}\rho a{v^2}}$

Total force = ${{3 \over 4}\rho a{v^2}}$

net pressure = ${{3 \over 4}\rho {v^2}}$

V(t) = 10 + ${3 \over 2}$ [2cos A sinB]

= 10 + ${3 \over 2}$ [sin(A+B) $-$ sin(A $-$ B)]

= 10+${3 \over 2}$[sin (57.2 $\times$ 10⁴ t) $-$ sin(52.8 $\times$ 10⁴ t)]

$\omega$₁ = 57.2 $\times$ 10⁴ = 2$\pi$f₁

f₁ = ${{57.2 \times {{10}^4}} \over {2 \times \left( {{{22} \over 7}} \right)}} = 9.1 \times {10^4}$

$\simeq$ 91KHz

f₂ = ${{52.8 \times {{10}^4}} \over {2 \times \left( {{{22} \over 7}} \right)}}$

$\simeq$ 84 KHz



Side band frequency are

f₁ = f_c $-$ f_w = ${{52.8 \times {{10}^4}} \over {2\pi }}$ $\simeq$ 85.00 kHz

f₂ = f_c + f_w = ${{57.2 \times {{10}^4}} \over {2\pi }}$ $\simeq$ 90.00 kHz

$M = {\mu _0}\,{n_1}\,{n_2}\,\pi r_1^2$

$L = {\mu _0}\,n_1^2\,\pi r_1^2$

$\Rightarrow \,\,{M \over L} = {{{n_2}} \over {{n_1}}}$

6$\mu$F & 4$\mu$F are in parallel & total charge on this combination is 30 $\mu$C

$\therefore$  Charge on 6$\mu$F capacitor = ${6 \over {6 + 4}} \times 30$

= 18 $\mu$C

Since charge is asked on right plate therefore is +18$\mu$C

As the field inside the uniformly charged hollow sphere or spherical shell is zero, so the potential inside it is constant, whereas outside it varies inversely with distance.

Suppose M is mass and a is side of larger triangle, then ${M \over 4}$ and ${a \over 2}$ will be mass and side length of smaller triangle.

${{{{\rm I}_{removed}}} \over {{{\rm I}_{original}}}} = {{{M \over 4}{{\left( {{a \over 2}} \right)}^2}} \over {M{{\left( a \right)}^2}}}$

${{\rm I}_{removed}} = {{{{\rm I}_0}} \over {16}}$

So, ${\rm I}$ = ${\rm I}$ ₀ $-$ ${{{{\rm I}_0}} \over {16}}$ = ${{15{{\rm I}_0}} \over {16}}$

For adiabatic process : TV^$\gamma$$-$1 = constant

For diatomic process : $\gamma$$-$1 = ${7 \over 5} - 1$

$\therefore$  x = ${2 \over 5}$

For a balanced bridge,

${P \over Q} = {{400} \over X}$ ......(i)

After interchanging P and Q,

${Q \over P} = {{405} \over X}$ ......(ii)

Multiplying equation (i) and (ii), we get

X² = 400$\times$405

$\Rightarrow$ X = $\sqrt {400 \times 405}$ = 402.5 $\Omega$

In series condition, equivalent resistance is 2R

thus power consumed is 60W = ${{{\varepsilon ^2}} \over {2R}}$

In parallel condition, equivalent resistance is R/2 thus new power is

P' = ${{{\varepsilon ^2}} \over {\left( {R/2} \right)}}$

or   P' = 4P = 240W

Sc³⁺ has noble gas configuration hence only +3 exists.

Most common oxidation states of :

(i) Sc : +3

(ii) V : +2, +3, +4, +5

(iii) Ti : +2, +3, +4

(iv) Cu : +1, +2

In cold water, dissolved oxygen (DO) can reach a concentration upto 10 ppm.

Atomic radii increase by moving down the group and decrease across a period.

Hence, the correct order of atomic radii is : C < S < Al < Cs.

N₂(g) + 3H₂(g) $\rightleftharpoons$ 2NH₃(g) ; K_eq = K_p

Write this equation reverse way,

2NH₃(g)  $\rightleftharpoons$ N₂(g) + 3H₂(g) ; K_eq = ${1 \over {{K_p}}}$

	2NH3(g)	⇌	N2(g)	+	3H2(g)
At t = 0	Po		0		0
At t = teq	PNH3		p		3p

At equillibrium

P_Total = P_NH3 + P_N2 + P_H2

= P_NH3 + p + 3p

(As P_NH3 << P_total so we can ignore P_NH3)

$\therefore$ P_Total = 4p

$\Rightarrow$ p = ${{{{P_{total}}} \over 4}}$

Formula of
K_eq = ${{{p_{{N_2}}} \times {{\left( {{p_{{H_2}}}} \right)}^3}} \over {{{\left( {{p_{N{H_3}}}} \right)}^2}}}$ = ${1 \over {{K_p}}}$

$\Rightarrow$ ${1 \over {{K_p}}}$ = ${{p \times 27{p^3}} \over {{{\left( {{p_{N{H_3}}}} \right)}^2}}}$

$\Rightarrow$ ${{{\left( {{p_{N{H_3}}}} \right)}^2}}$ = K_p $\times$ 27 $\times$ ${{{\left( {{{{P_{total}}} \over 4}} \right)}^4}}$

$\Rightarrow$ P_NH3 = $\sqrt {{K_p}} \times {\left( {27} \right)^{{1 \over 2}}} \times {\left( {{{{P_{Total}}} \over 4}} \right)^{{4 \over 2}}}$

= ${{{3^{{3 \over 2}}}K_p^{{1 \over 2}}P_{Total}^2} \over {16}}$

Be(OH)₂ is amphoteric in nature while rest all alkaline earth metal hydroxide are basic in nature.

(A) Norethindrone - Antifertility

(B) Ofloaxacin - Anti-Biotic

(C) Equanil - Hypertension (traiquilizer)

CCl₄ cannot get hydrolysed as it does not have dorbitals and cannot extend its covalency above four.

Here –SO₃H will be replaced by Br this is called ipso effect

We know,

$\Delta$T_f = i $\times$ k_f $\times$ m

$\therefore$ $\Delta$T_{fDil. Milk} = (1) $\times$ k_f $\times$ m_dil = 0.2 ...(1)

$\Delta$T_{fPure Milk} = (1) $\times$ k_f $\times$ m_pure = 0.5 ...(2)

From (1) and (2), we get

${{{m_{pure}}} \over {{m_{dil}}}} = {{0.5} \over {0.2}}$ = ${5 \over 2}$ ....(3)

Assume moles of fat in both the milk samples = n

Weight of milk = w₁

and weight of water = w₀

Molality of pure milk,
m_pure = ${n \over {{w_1}}} \times 1000$ ....(4)

Molality of diluted milk,
m_dil. = ${n \over {{w_1} + {w_0}}} \times 1000$ ....(5)

From (4) and (5), we get

${{{m_{pure}}} \over {{m_{dil}}}} = {{{w_1} + {w_0}} \over {{w_1}}}$

Using equation (3), we get

${5 \over 2} = {{{w_1} + {w_0}} \over {{w_1}}}$

$\Rightarrow$ 5w₁ = 2w₁ + 2w₀

$\Rightarrow$ 3w₁ = 2w₀

$\Rightarrow$ ${{{w_1}} \over {{w_0}}} = {2 \over 3}$

Which means 3 cups of water has been added to 2 cups of pure milk.

Siderite : FeCO₃

Kaolinite : Al₂(OH)₄Si₂O₅

Malachite : CuCO₃ .Cu(OH)₂

Calamine : ZnCO₃

Note : ( Remember all those ores names. Any one of those can be asked in the exam.)

Oxides Ores :

(1) ZnO - Zincite

(2) Fe₂O₃ - Haematite

(3) Fe₃O₄ - Magnetite (FeO + Fe₂O₃ mixture)

(4) Fe₂O₃ . 3H₂O - Limonite

(5) MnO₂ - Pyrolusite

(6) Cu₂O - Cuprite or Ruby Copper

(7) TiO₂ - Rutile

(8) FeCr₂O₄ - Chromite (FeO + Cr₂O₃)

(9) FeTiO₃ - Illmenite (FeO + TiO₂)

(10) Na₂B₄O₇ . 10H₂O - Borax or Tincal

(11) U₃O₈ - Pitch Blende

(12) SnO₂ - Tin Stone or Cassiterite

(13) Ca₂B₆O₁₁ . 5H₂O - Colemanite (2 Cao + 3 B₂O₃)

(14) Al₂O₃ . 2H₂O - Bauxite

(15) Al₂O₃ . H₂O - Diaspore

(16) Al₂O₃ - Corundum

Sulphides Ores :

(1) ZnS - Zinc Blende or Sphalerite

(2) PbS - Galena

(3) Ag₂S - Argentite or Silver Glance

(4) HgS - Cinnabar

(5) Cu₂S - Chalcocite or Copper glance

(6) CuFeS₂ - Copper pyrites or Chalco pyrites (Cu₂S + Fe₂S₃ mixture)

(7) FeS₂ - Iron pyrites or Fool's Gold

(8) 3Ag₂S . Sb₂S₂ - Pyrargyrite or ruby silver

Halides Ores :

(1) NaCl - Rock Salt

(2) KCl - Sylvine

(3) Na₃AlF₆ - Cryolite [3NaF + AlF₆]

(4) CaF₂ - Fluorspar

(5) KCl . MgCl₂ . 6H₂O - Carnalite

(6) AgCl - Horn Silver

Carbonates Ores :

(1) CaCO₃ - Limestone

(2) MgCO₃ - Magnesite

(3) CaCO₃ . MgCO₃ - Dolomite

(4) ZnCO₃ - Calamine

(5) PbCO₃ - Cerrusite

(6) FeCO₃ - Siderite

(7) CuCO₃ . Cu(OH)₂ or Cu₂CO₃(OH)₂ - Malachite or Basic Copper Carbonates

(8) 2 CuCO₃ . Cu(OH)₂ - Azurite

Sulphates Ores :

(1) CuSO₄ . 2H₂O - Gypsum

(2) MgSO₄ . 7H₂O - Epson Salt

(3) Na₂SO₄ . 10 H₂O - Glauber's Salt

(4) PbSO₄ - Anglesite

(5) ZnSO₄ . 7H₂O - White Vitriol

(6) FeSO₄ . 7H₂O - Green Vitriol

(7) CuSO₄ . 5H₂O - Blue Vitriol or Chalcanthite

Nitrate Ores :

(1) KNO₃ - Indian Saltpetre

(2) NaNO₃ - Chile Saltpetre

Arsenides Ores :

(1) NiAs - Kupfernickel

(2) NiAsS - Nickel glance

Molar ratio of C : H : N : : 6 : 8 : 2 i.e., 3 : 4 : 1

Thus, the correct formula is C₆H₈N₂. .

Compounds which follow (4n + 2)$\pi$ rule, are aromatic. Hence, compounds in option B, C, D are not aromatic.

NaH is an example of ionic hydride which is also known as saline hydride.

According to Arrhenius equation,
k = A${e^{ - {{{E_a}} \over {RT}}}}$

or ln k = ln A - ${{{{E_a}} \over {RT}}}$

Comparing the above equation with straight line equation,

y = mx + c,

we get, slope or gradient (m) = –E_a
and Intercept (c) = ln A

Also given that slope or gradient (m) = -y

$\therefore$ -y = –E_a

$\Rightarrow$ E_a = y

So the activation energy of the reactant, E_a = y unit