JEE Main 2019 9th April Evening Shift Question Paper with Solutions

$f = {{\left( {v \pm {u_0}} \right)} \over {\left( {v \pm {u_s}} \right)}}.{f_0} = {{\left( {v - 20} \right)} \over {\left( {v + 20} \right)}}.{f_0}$

$\Rightarrow 2000 = {{320} \over {360}}.{f_0}$

$\Rightarrow {{2000 \times 9} \over 8} = {f_0} = 2250\,Hz$

$\overrightarrow r = \left( {15{t^2}} \right)\widehat i + \left( {4 - 20{t^2}} \right)\widehat j$

$\overrightarrow v = {{d\overrightarrow r } \over {dt}} = \left( {30t} \right)\widehat i - \left( {40t} \right)\widehat j$

$\overrightarrow a = {{d\overrightarrow v } \over {dt}} = \left( {30} \right)\widehat i - \left( {40} \right)\widehat j$

$\left| {\overrightarrow a } \right| = 50$

For image to form at object itself, says must retrace their path back to object. Hence must incident on mirror normally.

Case 1: Object will be at focus of lens
${1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over R} - {1 \over { - R}}} \right) = {1 \over { - 18}}$ $\Rightarrow$ R = 18 cm

Case 2: Retraction at 1^st surface:
${1 \over { - 27}} - {{1.5} \over {{V_1}}} = {{1 - 1.5} \over R}$    ... (i)

2^nd retraction:
${{1.5} \over {{V_1}}} - {\mu \over \infty } = {{15 - u} \over { - R}}$   ...(ii)

From (i) and (ii)
$\mu = {4 \over 3}$

We have:

$f = {{nv} \over {2l}}$

$240 = {{3 \times v} \over {2 \times 2}}$

$\Rightarrow$ v = 320 m/s

Fundamental frequency = ${v \over {2l}}$ = 80 Hz.

Initial momentum · P_i = 2mv + 2mv = 4 mv
Let v^' be the speed of $l$ particle $\therefore 2{{mv} \over {\sqrt 2 }} = 4mv$
$\Rightarrow$ v^' = ${2\sqrt 2 }$ v

For A:
${{{C_p}} \over {{C_v}}} = \gamma = 1 + {2 \over f} = {{29} \over {22}}$
It gives f = 6.3 $\approx$ 6 (3 translational, 2 rotational and 1 vibrational)

For B:
${{{C_p}} \over {{C_v}}} = \gamma = 1 + {2 \over f} = {{30} \over {21}}$
$\therefore$ f = 4.67 $\approx$ 5 (3 translational, 2 rotational, no vibrational)

NIAB = KQ

175 × 1 × 10^–3 × 1 × 10^–4 × B = ${{{{10}^{ - 6}} \times \pi } \over {180}}$

$\Rightarrow B = {\pi \over {180}} \times {{10} \over {175}} \approx 9.97 \times {10^{ - 4}}\,T$

$\Rightarrow$ B = 10^–3 T

x = at + bt² – ct³

$V = {{dx} \over {dt}} = a + 2bt - 3c{t^2}$

$a = {{dv} \over {dt}} = 2b - 6ct$

Put acceleration = 0

$\Rightarrow t = {b \over {3c}}$

Now V at t = ${b \over {3c}}$

$V = a + {{{b^2}} \over {3c}}$

We have
I_g R_g = (0.5 – I_g) S

$\Rightarrow$ (0.002) (50) = (0.5 – I_g) S

$\Rightarrow S \simeq {{0.002 \times 50} \over {0.5}} = 0.2\Omega$

$T.E. = - \left( {13.6} \right)\left( {{{{Z^2}} \over {{n^2}}}} \right)eV$

z = n = 2

$\Rightarrow$ Ionisation energy = –T.E. = 13.6 eV

$\rho = {{2m} \over {n{e^2}\tau }}$

= 3.34 × 10^–8 $\Omega$ m

Radiation pressure for 100% reflection = ${{2I} \over C}$

Radiation pressure for 0% reflection = ${I \over C}$

Hence, in given case, radiation pressure

= $\left( {0.25} \right)\left( {{{2I} \over C}} \right) + \left( {0.75} \right)\left( {{I \over C}} \right)$

$\left( {1.25} \right)\left( {{I \over C}} \right)$

$\therefore$ Force = P × (Area) = 20.83 × 10^–8 N

Mutual induction

${\phi _q} = M{I_p}$

10^–3 = M(3)

$\Rightarrow M = {1 \over 3} \times {10^{ - 3}}$

${\phi _p} = M{I_q}$

$\Rightarrow {\phi _p} = 6.67{\rm{ }} \times {\rm{ }}{10^{-4}}{\rm{ }}Wb$

$R = {{\rho l} \over A} = {{\rho l} \over {\left( {V/l} \right)}} = {{\rho {l^2}} \over V}\left( {V{\rm{ }} \to {\rm{ }}Volume{\rm{ }}\,of{\rm{ }}\,wire} \right)$

$\Rightarrow$ Final resistance = 3 × (B)² = 12 $\Omega$

${R_{eq}} = 2\Omega \parallel 10\Omega = {5 \over 3}\Omega$

Conservation of momentum

$\overrightarrow {{p_x}} + \overrightarrow {{p_y}} = {\overrightarrow p _{final}}$

m_xv_x – m_yv_y = (m_x + m_y) V

${h \over {{\lambda _x}}} - {h \over {{\lambda _y}}} = {h \over \lambda }$

$\Rightarrow \lambda = {{{\lambda _x}{\lambda _y}} \over {\left| {{\lambda _x} - {\lambda _y}} \right|}}$

$\varepsilon = - {{d\phi } \over {dt}} = {{ - d} \over {dt}}\left( {{\mu _0}nl} \right)\left( {\pi {R^2}} \right)$

$= - \left( {{\mu _0}n\pi {R^2}} \right){{dl} \over {dt}}$

$= - {\mu _0}n\pi {R^2}{{d\left( {kt{e^{ - \omega t}}} \right)} \over {dt}}$

$= {\rm{ }}-C{\rm{ }}\left( {t\left( {{e^{-\omega t}}{\rm{ }}} \right){\rm{ }}\left( {-\alpha } \right){\rm{ }} + {\rm{ }}{e^{-\omega t}}{\rm{ }}} \right){\rm{ }}\left( {C{\rm{ }} \to {\rm{ }}constant} \right)$

= –C (e^–$\omega$t) (1 – $\alpha$t)

Clearly, at t = 0

Induced current $\ne$ 0

Also, apply Lenz law to find correct option

In 1st situation
V_b$\rho$_bg = V_s$\rho$_wg

${{{V_s}} \over {{V_b}}} = {{{\rho _b}} \over {{\rho _w}}} = {4 \over 5}$    ...(i)

Here V_b is volume of block
V_s is submerged volume of block
$\rho _b$ is density of block
$\rho _w$w is density of water & Let $\rho _o$ is density of oil
Finally in equilibrium condition

V_b$\rho _b$g = ${{{V_b}} \over 2}{\rho _o}g + {{{V_b}} \over 2}{\rho _w}g$

$2{\rho _b} = {\rho _0} + {\rho _w}$

$\Rightarrow {{{\rho _o}} \over {{\rho _w}}} = {3 \over 5} = 0.6$

Initial angular momentum = Final Angular Momentum

${{M{L^2}} \over {12}}{\omega _0} = \left( {{{M{L^2}} \over {12}} + 2{{m{L^2}} \over 4}} \right)\omega$

$\Rightarrow \omega = {{M{\omega _0}} \over {M + 6m}}$

By law of conservation of energy

${1 \over 2}k{x^2} = \left( {{m_1}{s_1} + {m_2}{s_2}} \right)\Delta T$

$\Delta T = {{16 \times {{10}^{ - 2}}} \over {4384}} = 3.65 \times {10^{ - 5}}$ K

Let interface temperature in steady state conduction is $\theta$, then assuming no heat loss through sides;

$$\left(\begin{array}{l} \text { Rate of heat } \\ \text { flow through } \\ \text { first slab } \end{array}\right)=\left(\begin{array}{l} \text { Rate of heat } \\ \text { flow through } \\ \text { second slab } \end{array}\right)$$

$$\begin{aligned} & \Rightarrow \frac{(3 K) A\left(\theta_2-\theta\right)}{d}=\frac{K A\left(\theta-\theta_1\right)}{3 d} \\ & \Rightarrow \quad 9\left(\theta_2-\theta\right)=\theta-\theta_1 \\ & \Rightarrow \quad 9 \theta_2+\theta_1=10 \theta \\ & \Rightarrow \quad \theta=\frac{9}{10} \theta_2+\frac{1}{10} \theta_1 \end{aligned}$$

Initial condition can be shown in the figure below


As mass $m$ collides with wedge, let both wedge and mass move with speed $v^{\prime}$. Then,

Given:

Mass of the wedge (M) = 4m

Mass of the particle (m)

Initial speed of the particle (v)

There is no friction.

Step 1 : Conservation of Linear Momentum

Before the collision, the momentum of the system is just the momentum of the particle because the wedge is at rest. After the collision, both the particle and the wedge will be moving. Let's denote the final common velocity as $v'$.

We can write the conservation of momentum as :

$mv = (m + 4m)v'$

which simplifies to

v' = \frac{v}{5} \tag{1}

Step 2 : Conservation of Mechanical Energy

We apply the conservation of energy before and after the collision. Before the collision, only the particle has kinetic energy. After the collision, both the particle and the wedge have kinetic energy and the particle has potential energy due to its height h on the wedge.

We can write the conservation of energy as :

$\frac{1}{2}mv^2 = \frac{1}{2}(m + 4m){v'}^2 + mgh$

which simplifies to

$mv^2 = (m + 4m){v'}^2 + 2mgh$

Substituting equation (1) into this equation gives

$v^2 = 5{v'}^2 + 2gh \Rightarrow \frac{4}{5}v^2 = 2gh$

which simplifies to

$h = \frac{2v^2}{5g}$

So, the maximum height climbed by the particle on the wedge is given by $\frac{2v^2}{5g}$.

Therefore, the correct answer is Option D :

$\frac{2v^2}{5g}$

Truth table can be formed as

A	B	Equivalent
0	0	0
0	1	1
1	0	1
1	1	1

Hence the equivalent is “OR” gate

Electric field at p = 2E₁cos$\theta$₁ –2E₂cos$\theta$₂

= ${{2Kq} \over {\left( {{d^2} + {D^2}} \right)}} \times {D \over {{{\left( {{d^2} + {D^2}} \right)}^{1/2}}}} - {{2Kq} \over {\left[ {{{\left( {2d} \right)}^2} + {D^2}} \right]}} \times {D \over {{{\left[ {{{\left( {2d} \right)}^2} + {D^2}} \right]}^{1/2}}}}$

$= 2KqD\left[ {{{\left( {{d^2} + {D^2}} \right)}^{ - 3/2}} - {{\left( {4{d^2} + {D^2}} \right)}^{ - 3/2}}} \right]$

$= {{2KqD} \over {{D^3}}}\left[ {{{\left( {1 + {{{d^2}} \over {{D^2}}}} \right)}^{ - 3/2}} - {{\left( {1 + {{4{d^2}} \over {{D^2}}}} \right)}^{ - 3/2}}} \right]$

Applying binomial approximation $\because$ d << D

$= {{2KqD} \over {{D^3}}}\left[ {1 - {3 \over 2}{{{d^2}} \over {{D^2}}} - \left( {1 - {{3 \times 4{d^2}} \over {2{D^2}}}} \right)} \right]$

$= {{2KqD} \over {{D^3}}}\left[ {{{12} \over 2}{{{d^2}} \over {{D^2}}} - {3 \over 2}{{{d^2}} \over {{D^2}}}} \right]$

$= {{9kq{d^2}} \over {{D^4}}}$

The physical size of antenna of receiver and transmitter both are inversely proportional to carrier frequency.

KE = ${1 \over 2}I{\omega ^2} = 1200$ (given)

$\Rightarrow \omega = 40\,rad/s$

$\Rightarrow \omega = {\omega _0} + \alpha t$

$\Rightarrow 40 = 0 + (20)t$

$\Rightarrow t = 2\,\sec$

For circular motion of particle:
${{m{V^2}} \over r} = mE$
$= m\left( {{{GM} \over {{r^2}}}} \right)$

Where $M = \int\limits_0^r {\left( {4\pi {x^2}dx} \right)\left( {{k \over {{x^2}}}} \right)} = 4\pi kr$

$\Rightarrow {{m{V^2}} \over r} = m\left( {{{G\left( {4\pi k} \right)} \over r}} \right)$

$\Rightarrow$ V = constant
$T = {{2\pi r} \over V}$

$\Rightarrow$ ${T \over R}$ = Constant

Limit of resolution = ${{1.22\lambda } \over d}$

= ${{1.22 \times 600 \times {{10}^{ - 9}}} \over {250 \times {{10}^{ - 2}}}}$

= 2.9 × 10^–7 rad.

The area of one square is 5.29 cm².

To find the area of 7 such squares, we can simply multiply the area of one square by 7:

7 x 5.29 cm² = 37.03 cm²

Since the given area has three significant figures, we need to round our answer to three significant figures as well.

Therefore, the area of 7 such squares, taking into account the significant figures, is 37.0 cm².

Initially Q = CV(1 + n)

$\therefore$ C_eq = (K + n)C

$\therefore$ $V = {{CV\left( {1 + n} \right)} \over {\left( {K + n} \right)C}} = {{V\left( {1 + n} \right)} \over {\left( {K + n} \right)}}$

Magnification is 2

If image is real, ${x_1} = {{3f} \over 2}$

If image is virtual, ${x_2} = {f \over 2}$

${{{x_1}} \over {{x_2}}} = 3:1$

Here heat is released so q is negative.

$\therefore$ q = - 2 kJ

Work done on the system, w = 10 kJ

From first law of thermodynamics,

$\Delta$U = q + w = -2 + 10 = 8 kJ

Cathode reaction :

Ni⁺² + 2e^- $\to$ Ni(s)

$\therefore$ From 2 mole of electrons 1 mole of Ni is deposited at the cathode.

So from 0.1 F or 0.1 mole of electrons ${1 \over 2} \times 0.1$ = 0.05 mole of Ni is deposited at the cathode.

No of moles formed = 10^-3 $\times$ ${{10} \over {1000}}$ = 10^-5

$\therefore$ No of molecules formed = 10^-5 $\times$ N_A

In unimolecular layer formation each cube occupy an area = a²

$\therefore$ Total area occupied = 10^-5 $\times$ N_A $\times$ a²

According to the question,

10^-5 $\times$ N_A $\times$ a² = 0.24

$\Rightarrow$ 10^-5 $\times$ 6 $\times$ 10²³ $\times$ a² = 0.24

$\Rightarrow$ a² = 4 $\times$ 10^-20

$\Rightarrow$ a = 2 $\times$ 10^-10 cm = 2 pm

Positive Ceric ammonium nitrate test is given by alcohol (- OH).

Positive Carbylamine test is given by 1^o amine (-NH₂).

Among those peptide you can see only in (A) Serine(Ser)Lysine(Lys) has alcohol (- OH) and 1^o amine (-NH₂).

The correct answer is Option B: C₆H₅SO₂Cl.

Hinsberg's reagent is benzenesulfonyl chloride (C₆H₅SO₂Cl). It is used to differentiate primary, secondary, and tertiary amines.

Here's why the other options are incorrect :

• Option A: C₆H₅COCl is benzoyl chloride, a reagent used in Friedel-Crafts acylation reactions.


• Option C: SOCl₂ is thionyl chloride, a reagent used to convert alcohols into alkyl chlorides.


• Option D: (COCl)₂ is oxalyl chloride, a reagent used for the conversion of carboxylic acids into acid chlorides.

K₂SO₄ $\to$ 2K⁺ + SO₄^2-

Van’t Hoff Factor (i) = 3

$\therefore$ $\Delta$T_f = ik_fm

= 3 $\times$ 4 $\times$ 0.03 = 0.36 K

Quartz, tridymite and cristobalite are crystalline forms of silica.

Kieselguhr is an amorphous form of silica.

-OH is +M group which will activate the corresponding benzene ring and -C(=O)- is -M group which will deactivate the right benzene ring completely. So Br⁺ electrophyle will attack electron rich position, which is ortho or para position of -OH group but you can see para position is blocked by -C(=O)- group so Br⁺ will attack any of the ortho position of -OH group.

(i) VBT does not explain colour exhibited by complex of transition metal because splitting of d-orbitals in explained by CFT.

(ii) VBT does not distinguish between strong and weak field complex.

(iii) VBT does not give quantitative interpretation of magnetic properties.

$\therefore$ According to question statement I and III are correct.

20% (mass/ mass) aqueous solution means,

100 g solution contains 20 g KI

$\therefore$ Mass of solvent = 100 – 20 = 80 g

So, 20 g of KI present in 80 g of solvent.

$\therefore$ ${{20} \over {166}}$ moles of KI present in 80 g of solvent.

We know,

Molality (m) = ${{no\,of\,moles\,of\,solute} \over {wt(in\,kg)\,of\,solvent}}$

$\therefore$ m = ${{{{20} \over {166}}} \over {{{80} \over {1000}}}}$ = 1.51

Actinoid 'Th' shows oxidation state = +4

Actinoid 'Ac' shows oxidation state = +3

Actinoid 'Pu' shows oxidation states = +3, +4, +5, +6, +7

Actinoid 'Np' shows oxidation states = +3, +4, +5, +6, +7

Actinoid 'Bk' shows oxidation states = +3, +4

Actinoid 'Cm' shows oxidation states = +3, +4

Actinoid 'Lr' shows oxidation states = +3

$\therefore$ Maximum oxidation state is shown by Np and Pu.

B₂O₃ is an acidic oxide.

Al₂O₃ and Ga₂O₃ are amphoteric oxide.

In₂O₃ and Tl₂O are basic oxide.

Between 10-50 km above sea level is the lowest level which is called stratosphere.

The maximum possible denticities of the given ligand towards transition metal ion is 6.

The maximum possible denticities of the given ligand towards inner transition metal ion is 8.

Bauxite = AlO_x. (OH)_3–2x

Calamine = ZnCO₃

Siderite = FeCO₃

Malachite = CuCO₃. Cu(OH)₂

In S_N1 substitution reaction, reactivity of compound depends on the stability of carbocation formed in the rate determination step. Reactivity of compound will be more when it's carbocation is more stable.



So stability order of carbocation :

C > D > A > B

Probablity density = ${\psi ^2}$
From the graph you can see probablity density is maximum at the nucleus where r = 0.

$\therefore$ Option A is correct.

(B) For hydrogen atom, electron can be found at a any distance distance from the nucleus.

$\therefore$ Option B is correct.

(C) Distance a₀ from the nucleus = First orbital

$\therefore$ n = 1 at first orbital

As Total energy, E_n = -13.6 $\times$ ${{{Z^2}} \over {{n^2}}}$ eV

$\therefore$ Total energy for hydrogen at n = 1 is,

E₁ = -13.6 $\times$${{{1^2}} \over {{1^2}}}$ eV = -13.6 eV

So total energy is minimum at distance a₀ from the nucleus.

$\therefore$ Option C is incorrect.

(D) Total energy, E_n = -13.6 $\times$ ${{{Z^2}} \over {{n^2}}}$ eV

Kinetic energy, K.E = 13.6 $\times$ ${{{Z^2}} \over {{n^2}}}$ eV

Potential energy, P.E = -27.2 $\times$ ${{{Z^2}} \over {{n^2}}}$ eV

$\therefore$ Magnitude of potential energy is double that of its kinetic energy.

$\therefore$ Option D is correct.

Due to strong H-bonding between HF molecules, HF has highest boiling point among the hydrogen halides.

Noradrenaline works as neurotransmitter in the brain and body.

Given $p = {{RT} \over {V - b}}$

$\Rightarrow$ $p\left( {V - b} \right) = RT$

$\Rightarrow$ ${{pV} \over {RT}} = {{pb} \over {RT}} + {{RT} \over {RT}}$

$\Rightarrow$ Z = ${b \over {RT}}p + 1$

$\therefore$ Slope of Z vs p curve (straight line) = ${b \over {RT}}$

$\therefore$ Higher the value of b, more steep will be the curve.

As we know, b = 4vN_A = $4 \times {4 \over 3}\pi {r^3} \times {N_A}$

$\therefore$ b $\propto$ size of gas molecules

Among Ne, Ar, Xe and Kr, size of Xe is the most as down the group in the periodic table size increases.

BeCl₂ in vapour phase exist as dimer [(BeCl₂)₂] (below 1200 K temperature).

BeCl₂ in solid state has chain structure[(BeCl₂)_n].

Steric hindrance of aldehydes is less compares to the ketone that is why aldehydes are more reactive than ketones in nucleophilic addition reactions.

So among propanal and acetone, propanal is suitable as substance.

And to form acetal, excess amount of alcohol is required so here methanol in excess is required.

For the extraction of iron, haematite ore in used.

Haematite = Fe₂O₃ which is a oxide ore not carbonate ore.

When strong acid is added to the strong base there will be sharp in the pH. So (A) is the correct answer.

S_N1 reaction happens in two steps :

1^st step : Formation of carbocation intermediate when leaving group leaves.

2^nd step : The nucleophile attacks the carbocation, forming the product.

In S_N1 reaction, formation of carbocation (1st step) is called rate determining step (RDS). RDS is the most difficult and slowest step. That is why potential energy required to complete first step is more compare to the second step.

So option A should be the right option.

$\therefore$ E_$a$ = (A + B) $\to$ C = 15

E_$a$ = (A + B) $\to$ D = 10

So, Activation enthalpy to form C is 5 kJ mol^-1 more than that to form D.

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species.

(a)    CO has 14 electrons.

Moleculer orbital configuration of CO is

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

Here is no unpaired electron so it is diamagnetic.

(b)   B₂ has 10 electrons.

Molecular orbital configuration of B₂ is

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

Here two unpaired electrons present. So it is paramagnetic.

(c)   $O_2$ has 16 electrons.

Moleculer orbital configuration of $O_2$ is

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

Here 2 unpaired electron present, so it is paramagnetic.

(d)   NO has 15 electrons.

Moleculer orbital configuration of NO is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ *$

Here is 1 unpaired electron, So it is Paramagnetic.

Given ${{dv} \over {dt}}$ = 5 cm³/min

and $\theta$ = ${\tan ^{ - 1}}\left( {{1 \over 2}} \right)$

$\Rightarrow$ tan $\theta$ = ${1 \over 2}$ = ${r \over h}$

$\Rightarrow$ h = 2r

Volume of the cone,

v = ${1 \over 3}\pi {r^2}h$ = ${1 \over 3}\pi {\left( {{h \over 2}} \right)^2}h$ = ${{\pi {h^3}} \over {12}}$

$\therefore$ ${{dv} \over {dt}} = {\pi \over {12}}\left( {3{h^2}} \right){{dh} \over {dt}}$

$\Rightarrow$ 5 = ${\pi \over 4}{h^2}{{dh} \over {dt}}$

$\Rightarrow$ 5 = ${\pi \over 4}{\left( {10} \right)^2}{{dh} \over {dt}}$

$\Rightarrow$ ${{dh} \over {dt}}$ = ${1 \over {5\pi }}$