JEE Main 2020 7th January Morning Shift Question Paper with Solutions

$\phi$ = $\overrightarrow B .\overrightarrow A$ = $\left( {3\widehat i + 4\widehat k} \right).\left( {25\widehat i + 25\widehat k} \right)$

$\Rightarrow$ $\phi$ = (3 $\times$ 25) + (4 $\times$ 25) = 175 weber

For spring mass damped oscillator

ma = - kx - bv

$\Rightarrow$ ma + kx + bv = 0

$\Rightarrow$ $m{{{d^2}x} \over {d{t^2}}}$ + b${{dx} \over {dt}}$ + kx = 0 ....(1)

For LCR circuit

L${{di} \over {dt}}$ + iR + ${q \over C}$ = 0

$\Rightarrow$ L${{{d^2}q} \over {d{t^2}}}$ + R${{dq} \over {dt}}$ + ${q \over C}$ = 0 .....(2)

Comparing (1) and (2), we get

L $\leftrightarrow$ m, C $\leftrightarrow$ ${1 \over k}$, R $\leftrightarrow$ b

Case 1 : Near – point adjustment

M.P = ${L \over {{f_0}}}\left( {1 + {D \over {{f_e}}}} \right)$

$\Rightarrow$ 375 = ${{150} \over 5}\left( {1 + {{250} \over {{f_e}}}} \right)$

$\Rightarrow$ f_e = 21.7 mm $\approx$ 22 mm

Case-2 : If final image is at inifinity

M.P = ${L \over {{f_0}}}\left( {{D \over {{f_e}}}} \right)$

$\Rightarrow$375 = ${{150} \over 5}\left( {{{250} \over {{f_e}}}} \right)$

$\Rightarrow$ f_e = 20 mm

I(t) = I₀t(1 - t)

We know, $\phi$ = BA

$\Rightarrow$ $\phi$ = $\mu$₀nIA

$\Rightarrow$ $\phi$ = $\mu$₀nAI₀(t - t²)

Also V_R = $- {{d\phi } \over {dt}}$

= - $\mu$₀nAI₀(1 - 2t)

V_R = 0 when 1 - 2t = 0

$\Rightarrow$ t = 0.5

Also we know, V_R = I_Rr

$\Rightarrow$ I_R = ${{{\mu _0}nA{I_0}\left( {1 - 2t} \right)} \over r}$

after t = 0.5, I_R reverses its direction.

Time period of revolution of electron in n^th orbit

V = ${{2\pi r} \over V}$

= ${{2\pi {a_0}\left( {{{{n^2}} \over Z}} \right)} \over {{V_0}\left( {{Z \over n}} \right)}}$

$\Rightarrow$ T $\propto$ ${{{{n^3}} \over {{Z^2}}}}$

$\therefore$ ${{{T_1}} \over {{T_2}}} = {{n_1^3} \over {n_2^3}}$

$\Rightarrow$ ${{1.6 \times {{10}^{ - 16}}} \over {{T_2}}} = {1 \over {{{\left( 2 \right)}^3}}}$

$\Rightarrow$ T₂ = 1.6 $\times$ 8 $\times$ 10^-16

$\therefore$ f₂ = ${1 \over {{T_2}}}$ = ${1 \over {1.6 \times 8 \times {{10}^{ - 16}}}}$ = 7.8 $\times$ 10¹⁴

So ${{Q + 1200} \over Q} = {{900} \over {300}}$

$\Rightarrow$ Q + 1200 = 3Q

$\Rightarrow$ Q = 600 J

As, magnetic field lines forms a closed loop, hence each line from circular area will pass through outer area in opposite direction hence $\phi$_i = - $\phi$₀.

For 2^nd minima

$\sin \theta = {{2\lambda } \over d }$

$\Rightarrow$ $\sin 60^\circ = {{2\lambda } \over d}$

$\Rightarrow$ ${\lambda \over d} = {{\sqrt 3 } \over 4}$

For 1^st minima

$\sin {\theta _1} = {\lambda \over d}$ = ${{\sqrt 3 } \over 4}$

$\Rightarrow$ ${\theta _1} =$ 25^o

mgh = $\Delta$KE

= ${1 \over 2}m{v^2} + {1 \over 2}I{\omega ^2}$

For no slipping, v = $\omega$R

$\therefore$ mgh = ${1 \over 2}m{\omega ^2}{R^2} + {1 \over 2}{{m{R^2}} \over 2}{\omega ^2}$

$\Rightarrow$ mgh = ${3 \over 4}m{\omega ^2}{R^2}$

$\Rightarrow$ $\omega$ = ${1 \over r}\sqrt {{{4gh} \over 3}}$

KE = PE₁ – PE₂ = mgh₁ – mgh₂

= 1 × 10 × 2 – 1 × 10 × 1 = 10 J

F = mg + f

F = 20000 + 4000 = 24000 N

We know, Power(P) = Fv

$\Rightarrow$ v = ${P \over F}$ = ${{60 \times 746} \over {24000}}$

$\Rightarrow$ v $\approx$ 1.9 m/s

${P_1}V_1^\gamma = {P_2}V_2^\gamma$

$\Rightarrow$ P₂ = P₁${\left[ {{{{V_1}} \over {{V_2}}}} \right]^\gamma }$

= 10⁵ $\times$ ${\left[ {{1 \over 3}} \right]^{1.4}}$

Work done = ${{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}$

= ${{{{10}^5} \times {{10}^{ - 3}} - {{{{10}^5}} \over {{3^{1.4}}}} \times 3 \times {{10}^{ - 3}}} \over {1.4 - 1}}$

= 88.7 J $\approx$ 90.5 J

NOT gate is reversible gate because the input signal can be recovered. Here It will behave as a NOT gate. Hence gives reversible operation.

Energy of photon = ${{1240} \over {310}}$ = 4 eV

Energy is greater than work function so photoelectric effect will take place.

Number of photons =

Intensity

Energy of one photon

= ${{6.4 \times {{10}^{ - 5}}} \over {4 \times 1.6 \times {{10}^{ - 19}}}}$ = 10¹⁴

Total number of electrons = ${{{{10}^{14}}} \over {{{10}^3}}}$ = 10¹¹

Electric field due to each sheet is uniform and equal to E = ${\sigma \over {2{\varepsilon _0}}}$

$\therefore$ Electric field along y direction

= $\left( {{\sigma \over {2{\varepsilon _0}}} - {\sigma \over {2{\varepsilon _0}}}\sin 60^\circ } \right)\widehat y$

Also Electric field along x direction

= - ${{\sigma \over {2{\varepsilon _0}}}\cos 60^\circ \widehat x}$

$\therefore$ Now net electric field between plates

= - ${{\sigma \over {2{\varepsilon _0}}}\cos 60^\circ \widehat x}$ + $\left( {{\sigma \over {2{\varepsilon _0}}} - {\sigma \over {2{\varepsilon _0}}}\sin 60^\circ } \right)\widehat y$

= ${ - {\sigma \over {4{\varepsilon _0}}}\widehat x}$ + ${\sigma \over {2{\varepsilon _0}}}\left( {1 - {{\sqrt 3 } \over 2}} \right)\widehat y$

= ${\sigma \over {2{ \in _0}}}\left[ {\left( {1 - {{\sqrt 3 } \over 2}} \right)\widehat y - {{\widehat x} \over 2}} \right]$

Given $\overrightarrow B$ = 3 $\times$ 10^-8 sin(1.6 $\times$ 10³x + 48 $\times$ 10¹⁰t)$\widehat j$ T

We know, ${{{E_0}} \over {{B_0}}} = c$

$\Rightarrow$ E₀ = (3 $\times$ 10^-8) $\times$ (3 $\times$ 10^-8) = 9 V/m

$\therefore$ $\overrightarrow E$ = (9sin(1.6 $\times$ 10³x + 48 $\times$ 10¹⁰t)$\widehat k$ V/m)

I = I₀ cos² $\theta$

$\Rightarrow$ ${{{I_0}} \over {10}}$ = I₀ cos² $\theta$

$\Rightarrow$ cos $\theta$ = ${1 \over {\sqrt {10} }}$

$\Rightarrow$ $\theta$ = 71.6^o

$\therefore$ $\phi$ = 90 - 71.6 = 18.4^o

Equivalent resistance of upper branch of circuit R = 2.5 $\Omega$

Voltage across upper branch = 1 V

$\therefore$ i₁ = ${1 \over {2.5}}$ = 0.4 A

i = ${{0.4} \over 2}$ = 0.2 A

I = ${{M{l^2}} \over {12}} + M{\left( {{l \over 4}} \right)^2}$

$\Rightarrow$ I = ${{7M{l^2}} \over {48}}$

$\Rightarrow$ MK² = ${{7M{l^2}} \over {48}}$

$\Rightarrow$ K = $\sqrt {{7 \over {48}}} l$

Using energy conservation

K_i + U_i = K_f + U_f

$\Rightarrow$ ${1 \over 2}m{u^2} - {{GmM} \over R}$ = ${1 \over 2}m{v^2} - {{GmM} \over {2R}}$

$\Rightarrow$ v = $\sqrt {{u^2} - {{GM} \over R}}$

After ejecting a rocket of mass ${m \over {10}}$ the remaining part of mass ${{9m} \over {10}}$ will rotate the earth with orbital velocity v₀.

$\therefore$ v₀ = $\sqrt {{{GM} \over {2R}}}$

Applying momentum conservation along radial direction,

Before firing rocket momentum of satelite in radial direction = mv

And after firing rocket momentum of satelite in radial direction = 0 and momentum of rocket in radial direction = ${m \over {10}}{v_2}$

$\therefore$ mv = ${m \over {10}}{v_2}$

$\Rightarrow$ v₂ = 10v

Now applying momentum conservation along tangential direction we get,

0 = ${m \over {10}}{v_1}$ - ${{9m} \over {10}}{v_0}$

$\Rightarrow$${{9m} \over {10}}{v_0}$ = ${m \over {10}}{v_1}$

$\Rightarrow$ v₁ = 9v₀

$\therefore$Total Kinetic Energy of rocket

= ${1 \over 2}{m \over {10}}\left( {v_1^2 + v_2^2} \right)$

= ${1 \over 2}{m \over {10}}\left( {81v_0^2 + 100{v^2}} \right)$

= ${m \over {20}}\left( {81\left( {{{GM} \over {2R}}} \right) + 100\left( {{u^2} - {{GM} \over R}} \right)} \right)$

= ${m \over {20}}\left( {100{u^2} + {{81GM} \over {2R}} - {{100GM} \over R}} \right)$

= ${m \over {20}}\left( {100{u^2} - {{119GM} \over {2R}}} \right)$

= $5m\left( {{u^2} - {{119GM} \over {200R}}} \right)$

$\gamma$ = $\alpha$_x + $\alpha$_y + $\alpha$_z

$\Rightarrow$ C $\times$ 10^–6 = 5 × 10^–5 + 5 × 10^–6 + 5 × 10^–6

$\Rightarrow$ C $\times$ 10^–6 = 50 × 10^–6 + 10 × 10^–6

$\Rightarrow$ C = 60

C_p = ${{{n_1}{C_{{p_1}}} + {n_2}{C_{{p_2}}}} \over {{n_1} + {n_2}}}$

C_v = ${{{n_1}{C_{{V_1}}} + {n_2}{C_{{V_2}}}} \over {{n_1} + {n_2}}}$

$\gamma$_mix = ${{{C_p}} \over {{C_v}}}$ = ${{2 \times {5 \over 2}R + 3 \times {8 \over 2}R} \over {2 \times {3 \over 2}R + 3 \times {6 \over 2}R}}$

= ${{5 + 12} \over {3 + 9}}$ = 1.42

k(x) = K(1 + $\alpha$x)
Capacitance of element dC₁ = ${{K{\varepsilon _0}A} \over {dx}}$ = ${{K\left( {1 + \alpha x} \right){\varepsilon _0}A} \over {dx}}$

${1 \over {{C_{eq}}}}$ = $\int {{1 \over {d{C_1}}}}$

= $\int\limits_0^d {{{dx} \over {KA\left( {1 + \alpha x} \right){\varepsilon _0}}}}$

= ${{1 \over {KA\alpha {\varepsilon _0}}}\ln \left( {1 + \alpha x} \right)}$

Using expansion of ln (1 + x) keeping x $\ll$ 1

$\Rightarrow$ ${1 \over {{C_{eq}}}}$ = ${{1 \over {KA\alpha {\varepsilon _0}}}\left( {\alpha d - {{{\alpha ^2}{d^2}} \over 2}} \right)}$

$\Rightarrow$ C_eq = ${{{{\varepsilon _0}KA} \over d}{{\left( {1 - {{\alpha d} \over 2}} \right)}^{ - 1}}}$

$\Rightarrow$ C_eq = ${{{{\varepsilon _0}KA} \over d}\left( {1 + {{\alpha d} \over 2}} \right)}$

Let 1 kg as origin and x-y axis as shown

x_cm = ${{1\left( 0 \right) + 1.5\left( 3 \right) + 2.5\left( 0 \right)} \over 5}$ = 0.9 cm

y_cm = ${{1\left( 0 \right) + 1.5\left( 0 \right) + 2.5\left( 4 \right)} \over 5}$ = 2 cm

Velocity of the wave, v = $\sqrt {{T \over \mu }}$

$\Rightarrow$ T = v²$\mu$

We know, Youngs modulus,

Y = ${{{F \over A}} \over {{{\Delta l} \over l}}}$ = ${{{T \over A}} \over {{{\Delta l} \over l}}}$

[As here F = T]

$\Rightarrow$ ${Y{{\Delta l} \over l}}$ = ${{T \over A}}$ = ${{{{v^2}\mu } \over A}}$

$\Rightarrow$ $\Delta$l = ${{{{v^2}\mu l} \over {AY}}}$

= ${{90 \times 90 \times {{60 \times {{10}^{ - 3}}} \over {60 \times {{10}^{ - 2}}}} \times 60 \times {{10}^{ - 2}}} \over {1 \times {{10}^{ - 6}} \times 16 \times {{10}^{11}}}}$

= 3 $\times$ 10^-5 m

= 0.03 mm

In 100 L solution H₂SO₄ present = 9.8 gm

$\therefore$ In 10 L solution H₂SO₄ present = $9.8 \times {{10} \over {100}}$ gm

$\Rightarrow$ In 10 L solution moles of H₂SO₄ present = ${{9.8} \over {98}} \times {{10} \over {100}}$

In one molecule of H₂SO₄ two H⁺ ion present.

$\therefore$ In 10 L solution moles of H⁺ present = 2$\times$${{9.8} \over {98}} \times {{10} \over {100}}$ = 0.02 moles

Also In 100 L solution NaOH present = 4 gm

$\therefore$ In 40 L solution NaOH present = $4 \times {{40} \over {100}}$

$\Rightarrow$ In 40 L solution moles of NaOH present = ${4 \over {40}} \times {{40} \over {100}}$

In one molecule of NaOH one OH^- ion present.

$\therefore$ In 40 L solution moles of OH^- ion present = ${4 \over {40}} \times {{40} \over {100}}$ = 0.04 moles

As moles of OH^- ion is more than H⁺ ion, so solution is basic.

$\therefore$ Final Conc. of OH^– = ${{0.04 - 0.02} \over {40 + 10}}$ = 4 $\times$ 10^-4

$\therefore$ pOH = – log (4 ×10^–4) = 3.4

pH = 14 – 3.4 = 10.6

m-chlorobenzoic acid dissolves in aq NaHCO₃ while m-chlorophenol dissolves in aq NaOH.

Synthetic resin method is more efficient because it can exchange both cation & anion.

Atomic radius of Ag and Au is nearly same due to lanthanide contraction.

In periodic table in group 1, 2, 3 from top to bottom radius increases but from group 4 to 12 size increases for 3d to 4d series elements but size of 4d and 5d series elements are approximately same.

Wrought iron is purest from of commercial iron.

Ionic interactions are stronger as compared to van der waal interactions.

So, correct order is

ion-ion $>$ ion-dipole $>$ dipole-dipole

As the vapour pressure of mixture increases so the force of attraction of mixture is less than the individual component.

Since the vapour pressure of the solution is greater than individual vapour pressure of both pure components, the solution shows a positive deviation from Raoult’s law.

$\therefore$ $\Delta$_SolH $>$ 0

As $\Delta$_SolH $>$ 0 then solution is endothermic.

and here $\Delta$_Sol Volume $>$ 0. As $\Delta$_Sol Volume $>$ 0 then a mixture of 100 ml. CS₂ and 100 ml. acetone has a volume more than 200 ml. It happens because the distance between molecules increases due to force of attraction of mixture decreases.

A($l$) $\to$ 2B(g)

We know, $\Delta$H = $\Delta$U + $\Delta$n_gRT

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

$\therefore$ $\Delta$G = $\Delta$U + $\Delta$n_gRT - T$\Delta$S

= 2.1 + ${{2 \times 2 \times 300} \over {1000}}$ - ${{300 \times 20} \over {1000}}$

= 2.1 + 1.2 - 6 = – 2.70 Kcal/mol

Potasisum has an oxidation of +1 (only) in combined state.

pK_b = –logK_b

So, as K_b increases, pK_b decreases.

K_b : (B) $>$ (A) $>$ (C)

pK_b : (B) $<$ (A) $<$ (C)

Base strength order

[Pt(NH₃)₂Cl(NH₂CH₃)]Cl

x + 0 – 1 + 0 = +1

x = +2

So Pt present in Pt⁺² form.

$\therefore$ Correct IUPAC form

Diamminechlorido (methanamine) platinum(II)chloride.

Cu⁺² + 2e^– $\to$ Cu    E^o = 0.34 V ....(1)

Cu⁺ + e^– $\to$Cu    E^o = 0.522 V

Cu $\to$ Cu⁺ + e^–     E^o = -0.522 V ...(2)

By adding (1) and (2) we get,

Cu⁺² + e^– $\to$ Cu⁺

E^o_cell = ${{0.34 \times 2 + 1 \times \left( { - 0.522} \right)} \over 1}$ = 0.158 V

Option(3) is according to Avogadro's law of volume combination.

CHCl₃ is polar while CH₄ and CCl₄ are nonpolar.

So, dipole moment order is :

CH₄ = CCl₄ < CHCl₃

Order of electron gain enthalpy (magnitude) is Cl > F > Br > I

Note: Electron gain enthalpy increases with electro negativity but chlorine has higher electron gain enthalpy than fluorine (exception).

Total number of orbitals
= n² = (5)² = 25

In complex [Ni(CO)₄] decrease in Ni–C bond length and increase in C–O bond length as well as it's magnetic property is explained by MOT.

All nuclear decays follow first order kinetics

t = ${1 \over k}\ln {{\left[ {{A_0}} \right]} \over {\left[ A \right]}}$

= ${{\left( {{t_{1/2}}} \right)} \over {0.693}} \times 2.303{\log _{10}}10$

= 10 × 2.303 × 1

= 23.03 years

Shortcut Method :

t_90% = ${{10} \over 3} \times {t_{50\% }}$

= ${{10} \over 3} \times 6.93$ = 23.1

Thiamine (vitamin B₁) : Beriberi

Riboflavin (vitamin B₂) : Cheilosis

Pyridoxine (vitamin B₆) : Convulsions

Ascorbic acid (vitamin C) : Scurvy

Formed product is methyl orange and it is used as an indicator in acid base titrations.

3Cl₂ + 6NaOH $\to$ 5NaCl + NaClO₃ + 3H₂O

NaCl + AgNO₃ $\to$ AgCl^- $\downarrow$ + NaNO₃

$\therefore$ X = NaCl then Y = NaClO₃

Here in anion ClO₃^- has bond between Cl and O atom.

Bond order of Cl–O Bond =

Total bond

Total resonating structure

.

= ${5 \over 3}$ = 1.67

$y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}}$

= $\sqrt {2\left( {{{1 + {{\tan }^2}\alpha } \over {\tan \alpha \left( {1 + {{\tan }^2}\alpha } \right)}}} \right) + {1 \over {{{\sin }^2}\alpha }}}$

= $\sqrt {2\left( {{{\cos \alpha } \over {\sin \alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}}$

= $\sqrt {{{\sin 2\alpha + 1} \over {{{\sin }^2}\alpha }}}$

= ${{\left| {\sin \alpha + \cos \alpha } \right|} \over {\left| {\sin \alpha } \right|}}$

At $\alpha$ = ${{5\pi } \over 6}$

${\left| {\sin \alpha + \cos \alpha } \right|}$ = -(sin$\alpha$ + cos$\alpha$)

and |sin$\alpha$| = sin$\alpha$

$\therefore$ y($\alpha$) = ${{ - \left( {\sin \alpha + \cos \alpha } \right)} \over {\sin \alpha }}$

= -1 - cot$\alpha$

$\therefore$ ${{dy} \over {d\alpha }}$ = cosec²$\alpha$

So ${{{dy} \over {d\alpha }}}$ at $\alpha$ = ${{5\pi } \over 6}$,

= cosec²${{5\pi } \over 6}$ = 4