JEE Main 2019 12th January Morning Shift Question Paper with Solutions

At any instant 't'

Total energy of charge distribution is constant

i.e.     ${1 \over 2}m{V^2} + {{K{Q^2}} \over {2R}} = 0 + {{K{Q^2}} \over {2{R_0}}}$

$\therefore$  ${1 \over 2}m{V^2} = {{K{Q^2}} \over {2{R_0}}} - {{K{Q^2}} \over {2R}}$

$\therefore$  V = $\sqrt {{2 \over m}{{K{Q^2}} \over 2}.\left( {{1 \over {{R_0}}} - {1 \over R}} \right)}$

$\therefore$  V = $\sqrt {{{K{Q^2}} \over m}\left( {{1 \over {{R_0}}} - {1 \over R}} \right)}$ = C$\sqrt {{1 \over {{R_0}}} - {1 \over R}}$

Also the slope of v-s curve will go on decreasing

Let current flowing in the wire is i.

$\therefore$  i = $\left( {{4 \over {R + 5}}} \right)A$

If resistance of 10 m length of wire is x

then x = 0.5 $\Omega$ = 5 $\times$ ${{0.1} \over 1}$ $\Omega$

$\therefore$   $\Delta$V = P.d. on wire = i. x

5 $\times$ 10^$-$3 = $\left( {{4 \over {R + 5}}} \right)$.(0.5)

$\therefore$   ${{4 \over {R + 5}}}$ = 10^$-$2

or   R + 5 = 400 $\Omega$

$\therefore$   R = 395 $\Omega$

Magnetic field at 'O' will be done to 'PS' and 'QN' Only

i.e. B₀ = B_PS + B_QN $\to$ Both inwards

Let current in each wire = i

$\therefore$  B₀ = ${{{\mu _0}i} \over {4\pi d}} + {{{\mu _0}i} \over {4\pi d}}$

or      10^$-$4 = ${{{\mu _0}i} \over {2\pi d}}$ = ${{2 \times {{10}^{ - 7}} \times i} \over {4 \times {{10}^{ - 2}}}}$

$\therefore$       i = 20 A

${R_1} = {{{{220}^2}} \over {25}}$

${R_2} = {{{{220}^2}} \over {100}}$

$L = {{220} \over {{R_1} + {R_2}}}$

${P_1} = {i^2}\,{R_1}$

${P_2} = {i^2}\,\,({R_2}\, = \,4W)$

$= {{{{220}^2}} \over {\left( {{{{{220}^2}} \over {25}} + {{{{220}^2}} \over {100}}} \right)}} \times {{{{220}^2}} \over {25}}$

$= {{400} \over {25}} = 16W$

Energy = ${1 \over 2}$ nRT = ${f \over 2}$PV

= ${f \over 2}$ (3 $\times$ 10⁶) (2)

= f $\times$ 3 $\times$ 10⁶

Considering gas is monoatomic i.e. f = 3

E. = 9 $\times$ 10⁶ J

For the given wire :

dR = C ${{d\ell } \over {\sqrt \ell }}$,

where C = constant.

Let resistance of part

AP is R₁ and PB is R₂

$\therefore$  ${{R'} \over {R'}} = {{{R_1}} \over {{R_2}}}$   or    R₁ = R₂     

By balanced WSB concept.

Now   $\int {dR = c\int {{{d\ell } \over {\sqrt \ell }}} }$

$\therefore$  R₁ = C$\int\limits_0^\ell {{\ell ^{ - 1/2}}d\ell }$ = C.2.${\sqrt \ell }$

R₂ = C$\int\limits_\ell ^1 {{\ell ^{ - 1/2}}d\ell }$ = C.(2 $-$ 2${\sqrt \ell }$)

Putting R₁ = R₂

C2${\sqrt \ell }$ = C(2 $-$ 2${\sqrt \ell }$)

$\therefore$  2${\sqrt \ell }$ = 1

${\sqrt \ell }$ = ${1 \over 2}$

i.e. $\ell$ = ${1 \over 4}$ m     $\Rightarrow$  0.25 m

mv$\widehat i$ + mv$\widehat j$

= 2m${\overrightarrow v ^1}$

$\overrightarrow v$ = ${1 \over {\sqrt 2 }} \times \sqrt {{{GM} \over R}}$

Consider an element of radius x and thickness dx

Mass of element, dm = $\sigma 2\pi x\left( {dx} \right)$

Here, $\sigma$ = mass per unit area = ${m \over {\pi \left( {{R^2} - {r^2}} \right)}}$

Moment of inertia of element, dI = (dm)x²

$\Rightarrow$ I = $\sigma 2\pi \int\limits_r^R {{x^3}dx}$

= $\sigma 2\pi \left( {{{{R^4} - {r^4}} \over 4}} \right)$

= ${m \over {\pi \left( {{R^2} - {r^2}} \right)}}{\pi \over 2}\left( {{R^4} - {r^4}} \right)$

= ${m \over 2}\left( {{R^2} + {r^2}} \right)$ .....(i)

Moment of inertia of thin cylinder of same mass,

I = m$r_0^2$ ......(ii)

$\Rightarrow$ m$r_0^2$ = ${m \over 2}\left( {{R^2} + {r^2}} \right)$

$\Rightarrow$ $r_0^2$ = 250

$\Rightarrow$ r₀ $\simeq$ 16 cm

v = $\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)}$

v₁ = $\sqrt {2g\ell \left( {1 - \cos {\theta _1}} \right)}$

By momentum conservation

m$\sqrt {2gl\left( {1 - \cos {\theta _0}} \right)}$
$= M{V_m} - m\sqrt {2g\left( {1 - \cos \theta } \right)}$

$\Rightarrow$$m\sqrt {2g\ell } \left\{ {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right\}$

$=$ MV_m

and  e = 1 = ${{{V_m} + \sqrt {2g\ell \left( {1 - \cos {\theta _1}} \right)} } \over {\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)} }}$

$\sqrt {2g\ell }$ $\left( {\sqrt {1 - \cos {\theta _0}} - \sqrt {1 - \cos {\theta _1}} } \right)$
$=$ V_m     . . .(I)

m$\sqrt {2g\ell } \left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)$
$=$ MV_M     . . .(II)

Dividing

${{\left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)} \over {\left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)}} = {M \over m}$

By componendo divided

${{m - M} \over {m + M}}$ = ${{\sqrt {1 - \cos {\theta _1}} } \over {\sqrt {1 - \cos {\theta _0}} }} = {{\sin \left( {{{{\theta _1}} \over 2}} \right)} \over {\sin \left( {{{{\theta _0}} \over 2}} \right)}}$

$\Rightarrow$  ${M \over m} = {{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}} \Rightarrow M = {{{\theta _0} - \theta 1} \over {{\theta _0} + {\theta _1}}}$

$\tau = - 2kx{\ell \over 2}\cos \theta$

$\Rightarrow$   $\tau = \left( {{{K{\ell ^2}} \over 2}} \right)\theta = - C\theta$

$\Rightarrow$   $f = {1 \over {2\pi }}\sqrt {{C \over 1}} = {1 \over {2\pi }}\sqrt {{{{{K{\ell ^2}} \over 2}} \over {{{M{\ell ^2}} \over {12}}}}}$

$\Rightarrow$    $f = {1 \over {2\pi }}\sqrt {{{6K} \over M}}$

${X_{cm}} = {{2mL + 2mL + {{5mL} \over 2}} \over {4m}} = {{13} \over 8}L$

${Y_{cm}} = {{2m \times L + m \times \left( {{L \over 2}} \right) + m \times 0} \over {4m}} = {{5L} \over 8}$

dm = (A + Bx²)dx

dF = ${{GMdm} \over {{x^2}}}$

F = $\int_a^{a + L} {{{GM} \over {{x^2}}}}$ (A + Bx²)dx

= GM$\left[ { - {A \over x} + Bx} \right]_a^{a + L}$

= GM$\left[ {A\left( {{1 \over a} - {1 \over {a + L}}} \right) + BL} \right]$

K.E. acquired by charge = K = qV

$\lambda$ = ${h \over P}$ = ${h \over {\sqrt {2mK} }}$ = ${h \over {\sqrt {2mqV} }}$

$\therefore$  ${{{\lambda _A}} \over {{\lambda _B}}} = {{\sqrt {2m{}_B{q_B}{V_B}} } \over {\sqrt {2m{}_A{q_A}{V_A}} }} = \sqrt {{{4m.q.2500} \over {m.q.50}}} = 2\sqrt {50}$

$= 2 \times 7.07 = 14.14$

case I :

i_g = ${E \over {220 + {R_g}}}$ = C$\theta$₀     . . .(i)

Case II :

i_g = $\left( {{E \over {220 + {{5{R_g}} \over {5 + {R_g}}}}}} \right)$ $\times$ ${5 \over {\left( {{R_g} + 5} \right)}}$ = ${{C{\theta _0}} \over 5}$       . . .(ii)

$\Rightarrow$  ${{5E} \over {225{R_g} + 1100}}$ = ${{C{\theta _0}} \over 5}$       . . .(ii)

${E \over {220 + {R_g}}} = C\theta$       . . .(i)

$\Rightarrow$  ${{225{R_g} + 1100} \over {1100 + 5{R_g}}} = 5$

$\Rightarrow$  5500 + 25R_g = 225R_g + 1100

200R_g = 4400

R_g = 22$\Omega$

For first lens

${1 \over V} - {1 \over { - 20}} = {1 \over 5}$

V $=$ ${{20} \over 3}$

For second lens

V = ${{20} \over 3}$ $-$ 2 $=$ ${{14} \over 3}$

${1 \over V} - {1 \over {{{14} \over 3}}} = {1 \over { - 5}}$

V $=$ 70cm

The total distance to be travelled by the train is 60 + 120 = 180 m.

When the trains are moving in the same direction, relative velocity is
v₁ – v₂ = 80 – 30 = 50 km hr^–1

So time taken to cross each other,

t₁ = ${{180} \over {50 \times {{{{10}^3}} \over {3600}}}}$

When the trains are moving in opposite direction, relative velocity is
|v₁ – (–v₂ )| = 80 + 30 = 110 km hr^–1

$\therefore$ Time taken to cross each other

t₂ = ${{180} \over {110 \times {{{{10}^3}} \over {3600}}}}$

$\therefore$ ${{{t_1}} \over {{t_2}}} = {{{{180} \over {50 \times {{{{10}^3}} \over {3600}}}}} \over {{{180} \over {110 \times {{{{10}^3}} \over {3600}}}}}}$ = ${{11} \over 5}$

K_eq = ${{{K_1}{A_1} + {K_2}{A_2}} \over {{A_1} + {A_2}}}$

= ${{{K_1}\left( {\pi {R^2}} \right) + {K_2}\left( {3\pi {R^2}} \right)} \over {4\pi {R^2}}}$

= ${{{K_1} + 3{K_2}} \over 4}$

E_m + E_c = 160

E_m + 100 = 160

E_m = 60

$\mu = {{{E_m}} \over {{E_C}}} = {{60} \over {100}}$

$\mu$ = 0.6

3d

Y $= \overline {\left( {\overline A + \overline B } \right)\overline A }$

$= \overline {\overline A + \overline A B}$

$= A\left( {\overline {\overline A B} } \right)$

$= A\left( {A + \overline B } \right)$

$= A + A\overline B = A\overline B$

V_i = ${1 \over 2}$CE²

V_f = ${{{{\left( {CE} \right)}^2}} \over {2 \times 4c}}$ = ${1 \over 2}{{C{E^2}} \over 4}$

$\Delta$E = ${1 \over 2}$CE² $\times$ ${3 \over 4}$ = ${3 \over 8}$ CE²

Since P$-$V indicator diagram is given, so work done by gas is area under the cyclic diagram.

$\therefore$  $\Delta$W = Work done by gas = ${1 \over 2}$ $\times$ 4 $\times$ 5 J

= 10 J

$\left| {{P_1}} \right| =$ q(d)

$\left| {{P_2}} \right| =$ qd

|Resultant| $=$ 2 P cos30^o

     2 qd$\left( {{{\sqrt 3 } \over 2}} \right)$ = $\sqrt 3$ qd

y = a sin($\omega$t + kx)

$\Rightarrow$  wave is moving along $-$ve x-axis with speed

v = ${\omega \over K}$ $\Rightarrow$  v = ${{50} \over 2}$ = 25m/sec

Force due to this field, F = $- {{\partial U} \over {\partial r}}$

F = $- {\partial \over {\partial r}}\left( {{1 \over 2}k{r^2}} \right)$ = -kr

For circular orbit, ${{m{v^2}} \over r}$ = -kr

$\Rightarrow$ v $\propto$ r ..... (1)

Fron Bohr’s quantization condition

mvr = ${{nh} \over {2\pi }}$ .....(2)

From (1) and (2),

${r_n}$ $\propto$ ${n^{{1 \over 2}}}$

Given, U(r) = ${1 \over 2}$kr²

$\Rightarrow$ E_n = $- {1 \over 2}U\left( r \right)$ = $- {1 \over 4}k{r^2}$

$\Rightarrow$ E_n $\propto$ n

P_refracted = ${{96} \over {100}}Pi$

$\Rightarrow$  K₂A$Pi_t^2$ = ${{96} \over {100}}$ K₁A$_i^2$

$\Rightarrow$  r₂A$_i^2$ = ${{96} \over {100}}$ r₁A$_i^2$

$\Rightarrow$  A$_t^2$ = ${{96} \over {100}} \times {1 \over {{3 \over 2}}} \times {\left( {30} \right)^2}$

A₁$\sqrt {{{64} \over {100}} \times {{\left( {30} \right)}^2}} = 24$

AB = V_P $\times$ t

BC = Vt

cos60^o = ${{AB} \over {BC}}$

${1 \over 2} = {{{V_P} \times t} \over {Vt}}$

V_P = ${V \over 2}$

Ideal inductor will behave like zero resistance long time after switch is closed



I = ${{2\varepsilon } \over R}$ = ${{2 \times 15} \over 5} = 6A$

Least count = ${{Pitch} \over {Number\,\,of\,\,division\,\,on\,\,circular\,scale}}$

5 $\times$ 10^$-$6 = ${{{{10}^{ - 3}}} \over N}$

N = 200

KE = q$\Delta$V

r = ${{\sqrt {2mq\Delta V} } \over {qB}}$

r $\propto$ $\sqrt {{m \over q}}$

${{{r_p}} \over {{r_ \propto }}}$ = ${1 \over {\sqrt 2 }}$

According to unit of rate constant it is a zero order reaction, then half life of zero order reaction.

t_1/2 = ${{{a_0}} \over {2k}} = {5 \over {2 \times 0.05}}$ = 50 years

K, Rb and Cs form super oxides on reaction with excess air.

Rb + O₂ $\to$ RbO₂

2RbO₂ + 2H₂O $\to$ 2RbOH + H₂O₂ + O₂$\uparrow$

Compounds that contain at least one carbon-metal bond are called organometallic compounds.

Lone pair over nitrogen will act as the reacting centre and more free nitrogen atom will be more reactive towards alkyl halide.

For compound (A), lone pair of N atom is in resonance with $\pi$ bond of C = O. So lone pair of N atom gets delocalised.

For compound (B), lone pair of N atom is in resonance with $\pi$ bond of two C = O. So here lone pair of N atom gets more delocalised than compound (A). So reactivity towards alkyl halide is very less for this compound.

For compound (C), lone pair of N atom is in resonance with $\pi$ bond of benzene ring and $\pi$ bond of C $\equiv$ N. So delocalisation of lone pair of N atom is relatively high.

For compound (D), lone pair of N atom is in resonance with $\pi$ bond of benzene ring only. So delocalisation of lone pair of N atom is relatively low.

$\therefore$ Reactivity order is B < A < C < D.

Photochemical smog is produced when ultraviolet light from sun reacts with NO , O₃ and hydrocarbon in atmosphere.

Electronic configuration of element Og (118) is
[Rn] 5f¹⁴6d ¹⁰7s ²7p⁶, where Og is oganesson.

[Og₁₁₈] 8s² is configuration for Z = 120,

As per the configuration it is in II^nd group. Thus, element with Z = 120 will be an alkaline earth metal.

In the Hall-Heroult process the cathode is made of carbon.

n_CaSO4 $\times$ Van't hoff factor = n_CaCO3 $\times$ Van't hoff factor

$\Rightarrow$ 10^-3 $\times$ 2 = n_CaCO3 $\times$ 2

$\Rightarrow$ n_CaCO3 = 10^-3 mol in 1 L

$\therefore$ w_CaCO3 = 100 $\times$ 10^-3 g CaCO₃ in 1 L solution

$\therefore$ hardness in terms of CaCO₃

= ${{{w_{CaC{O_3}}}} \over {{w_{Total}}}} \times {10^6}$

= ${{100 \times {{10}^{ - 3}}} \over {1000}} \times {10^6}$

= 100 ppm

DIBAL-H will reduce cyanides and esters to aldehydes.

I₂ + 10HNO₃ $\to$ 2HIO₃ (Y) + 10NO₂ + 4H₂O

Let In HIO₃ oxidation state of iodine is x.

$\therefore$ +1 + x + 3(–2) = 0 $\Rightarrow$ x = +5

At equilibrium [A] = [B]

$\Rightarrow$ a - x = 1.5a - 2x

$\Rightarrow$ x = 0.5a

K_c = ${{{{\left[ C \right]}^2}\left[ D \right]} \over {\left[ A \right]{{\left[ B \right]}^2}}}$

= ${{{{\left( a \right)}^2}\left( {0.5a} \right)} \over {\left( {0.5a} \right){{\left( {0.5a} \right)}^2}}}$ = 4

C_P does not changes with change in pressure.

The reaction in option (D) will yield 1^o -alcohol.

Adsorption $\propto$ T_c

Therefore, H₂ gas shows least adsorption on a definite amount of charcoal.

The reaction between sodium hydroxide (NaOH) and oxalic acid (H₂C₂O₄) is as follows :

$\text{H}_2\text{C}_2\text{O}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2\text{H}_2\text{O}$

We can see that 1 mole of oxalic acid ($\text{H}_2\text{C}_2\text{O}_4$) reacts with 2 moles of sodium hydroxide (NaOH).

Given that 50 mL of 0.5 M oxalic acid is needed to neutralize the sodium hydroxide, we can find the number of moles of oxalic acid that reacted :

$\text{Moles of H}_2\text{C}_2\text{O}_4 = \text{Molarity} \times \text{Volume (L)} = 0.5 \, \text{mol/L} \times 0.05 \, \text{L} = 0.025 \, \text{mol}$

Since 1 mole of $\text{H}_2\text{C}_2\text{O}_4$ reacts with 2 moles of $\text{NaOH}$, the number of moles of $\text{NaOH}$ in 25 mL solution would be twice that of $\text{H}_2\text{C}_2\text{O}_4$ :

$\text{Moles of NaOH} = 0.025 \, \text{mol} \times 2 = 0.05 \, \text{mol}$

Now, to find the mass of NaOH, we multiply the number of moles by the molar mass of NaOH (40 g/mol) :

$\text{Mass of NaOH} = \text{Number of moles} \times \text{Molar mass} = 0.05 \, \text{mol} \times 40 \, \text{g/mol} = 2 \, \text{g}$

However, the question asks for the amount of $\text{NaOH}$ in 50 mL of the given solution. Since we've found the amount in 25 mL, we just need to double our result to find the amount in 50 mL :

$2 \, \text{g} \times 2 = 4 \, \text{g}$

So, the correct answer is Option B : 4 g.

Among CH $\equiv$ CH and CH₂ = CH₂, in CH $\equiv$ CH compound H atom is attached with sp hybridised C atom and in CH₂ = CH₂ compound H atom is attached with sp² hybridised C atom.

As we know, electronegitivity of sp carbon is more compare to sp² carbon atom and that H is more acidic which is attached with more electronegative carbon atom. So CH $\equiv$ CH is more acidic than CH₂ = CH₂.

Among CH $\equiv$ CH and CH₃–C $\equiv$ CH, in CH $\equiv$ CH both the H atom is connected to sp carbon atom but in CH₃–C $\equiv$ CH, with one sp carbon atom H atom attached and with other sp carbon atom -CH₃ group attached which have +I effect and we know +I effect decreases acidic strength.
So, CH $\equiv$ CH is more acidic than CH₃–C $\equiv$ CH.

So correct order is

HC $\equiv$ CH > CH₃ – C $\equiv$ CH > CH₂ = CH₂