JEE Main 2019 8th April Evening Shift Question Paper with Solutions

Angular impulse = change in angular momentum.

$mg{l \over 2} \times 0.01 = {{m{l^2}} \over 3}\omega$

$\omega = {{3g \times 0.01} \over {2l}} = {{3 \times 10 \times 0.01} \over {2 \times 0.3}}$

$=$ ${1 \over 2} = 0.5\,rad/s$

Time taken by rod to hit the ground

$t = \sqrt {{{2h} \over g}} = \sqrt {{{2 \times 5} \over {10}}} = 1\,\sec .$

In this time angle rotate by rod
$\theta$ = $\omega$t = 0.5 × 1 = 0.5 radian

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

$50 \times {10^3} = \sqrt {2 \times 6400 \times {{10}^3} \times {h_T}} + \sqrt {2 \times 6400 \times {{10}^3} \times 70}$

By solving h_T = 32 m.

Time period of a pendulum (T) = $2\pi \sqrt {{l \over g}}$

$\Rightarrow$ T² = $4{\pi ^2}{l \over g}$

$\Rightarrow$ $g = {{4{\pi ^2}l} \over {{T^2}}}$

Fractional change

$\left( {{{dg} \over g}} \right) \times 100 = \left( {{{dl} \over l}} \right) \times 100 - \left( {2{{dT} \over T}} \right) \times 100$

$\therefore$ Maximum possible percentage error,

$\left( {{{dg} \over g}} \right) \times 100 = \left( {{{dl} \over l}} \right) \times 100 + \left( {2{{dT} \over T}} \right) \times 100$

Error in time period(dT) = least count of time = 1 second

and T = 30 second

Error in length(dl) = least count of length = 1 mm

and $l$ = 55.0 cm

$\therefore$ $\left( {{{dg} \over g}} \right) \times 100 =$ $\left( {{{0.1} \over {55}}} \right) \times 100 + 2\left( {{1 \over {30}}} \right) \times 100$ = 6.8%

Limit of resolution of telescope $(\theta) = {{1.22\lambda } \over D}$

$\theta = {{1.22 \times 500 \times {{10}^{ - 9}}} \over {200 \times {{10}^{ - 2}}}} = {{1.22 \times 500 \times {{10}^{ - 9}}} \over 2}$

$\theta$ = 305 × 10^–9 radian

$\left| {\overrightarrow {{A_1}} } \right| = 3,\left| {\overrightarrow {{A_2}} } \right| = 5\,and\,\left| {\overrightarrow {{A_1}} + \overrightarrow {{A_2}} } \right| = 5$

$\,\left| {\overrightarrow {{A_1}} + \overrightarrow {{A_2}} } \right| = {\left| {\overrightarrow {{A_1}} } \right|^2} + {\left| {\overrightarrow {{A_2}} } \right|^2} + 2\left| {\overrightarrow {{A_1}} } \right|\left| {\overrightarrow {{A_2}} } \right|\cos \theta$

$\cos \theta = - {3 \over {10}}$

$\left( {2\overrightarrow {{A_1}} + 3\overrightarrow {{A_2}} } \right).\left( {3\overrightarrow {{A_1}} - 2\overrightarrow {{A_2}} } \right)$

= $6{\left| {\overrightarrow {{A_1}} } \right|^2} + 9\overrightarrow {{A_1}} .\overrightarrow {{A_2}} - 4\overrightarrow {{A_1}} .\overrightarrow {{A_2}} - 6{\left| {\overrightarrow {{A_2}} } \right|^2}$

= - 118.5

${V_{rms}} = \sqrt {{{3RT} \over M}} = 11.2 \times {10^3}m/s$

$\Rightarrow$ $T = {M \over {3R}} \times {\left( {11.2 \times {{10}^3}} \right)^2}$

= ${{2 \times {{10}^{ - 3}}} \over {3 \times 8.3}} \times 125.44 \times {10^6} = {10^4}K$

Charges at inner plates are 1 $\mu$C and –1 $\mu$C.

$\therefore$ Potential difference across capacitor
= ${q \over c} = {{1\mu C} \over {1\mu F}} = {{1 \times {{10}^{ - 6}}C} \over {1 \times {{10}^{ - 6}}Farad}} = 1V$

At saturation state, V_CE becomes zero

$\Rightarrow$ i_C = ${{10V} \over {1000\Omega }} = 10\,mA$

Now current gain factor $\beta = {{{i_C}} \over {{i_B}}}$

$\Rightarrow$ i_B = ${{10\, mA} \over {250 }} = 40\,\mu A$

${\overrightarrow F _m} = q\left( {\overrightarrow V \times \overrightarrow B } \right)$

$\overrightarrow B = {\overrightarrow B _x} + {\overrightarrow B _y}$ Since M_y = 2M_x

$\Rightarrow \left| {{{\overrightarrow B }_x}} \right| = \left| {{{\overrightarrow B }_y}} \right|$

${\overrightarrow B _{net}}$ is parallel to $\overrightarrow V$

$\Rightarrow$ $\overrightarrow F = 0$

${1 \over 2}m{V^2} = - q\left( {{V_f} - {V_i}} \right)$

$E = {\lambda \over {2\pi {\varepsilon _0}r}}$

$\Delta V = {\lambda \over {2\pi {\varepsilon _0}}}\ln \left( {{{{r_0}} \over r}} \right)$

${1 \over 2}m{v^2} = {{ - q\lambda } \over {2\pi {\varepsilon _0}}}\ln \left( {{{{r_0}} \over r}} \right)$

$v \propto \sqrt {\ln \left( {{r \over {{r_0}}}} \right)}$

Current i = ${E \over {r + R}}$

Power generated in R

P = i²R

$P = {{{E^2}R} \over {{{\left( {r + R} \right)}^2}}}$

For maximum power ${{dP} \over {dR}} = 0$

${E^2}\left[ {{{{{\left( {r + R} \right)}^2} \times 1 - R \times 2(r + R)} \over {{{\left( {r + R} \right)}^4}}}} \right] = 0$

$\Rightarrow$ r = R or R = r

Given:

${{{Y_A}} \over {{Y_B}}} = {7 \over 4};\,{L_A} = 2m\,;{A_A} = \pi {R^2}$

${F \over A} = Y\left( {{l \over L}} \right);{L_B} = 1.5m\,;{A_B} = \pi {(2mm)^2}$

given F and $l$ are same $\Rightarrow$ ${{AY} \over L}$ is same

${{{A_A}{Y_A}} \over {{L_A}}} = {{{A_B}{Y_B}} \over {{L_B}}}$

$\Rightarrow {{\left( {\pi {R^2}} \right)\left( {{7 \over 4}{Y_B}} \right)} \over 2} = {{\pi {{(2\,mm)}^2}.{Y_B}} \over {1.5}}$

R = 1.74 mm

Minimum energy required (E) = – (Potential energy of object at surface of earth)

$= \left( { - {{GMm} \over R}} \right) = {{GMm} \over R}$

Now M_earth = 64 M_moon

$\rho .{4 \over 3}\pi R_e^3 = 64.{4 \over 3}\pi R_m^3$

$\Rightarrow$ R_e = 4R_m

Now ${{{E_{moon}}} \over {{E_{earth}}}} = {{{M_{moon}}} \over {{M_{earth}}}}.{{{R_{earth}}} \over {{R_{moon}}}} = {1 \over {64}} \times {4 \over 1}$

$\Rightarrow {E_{moon}} = {E \over {16}}$

Densities of nucleus happens to be constant, irrespective of mass number.

Between the isothermal and the adiabatic processes, P-V graph for adiabatic is steeper
Isochoric $\to$ Process d
Isobaric $\to$ Process a
Adiabatic slope will be more than isothermal so
Isothermal $\to$ Process b
Adiabatic $\to$ Process c
Order $\to$ d a b c

Given phase difference = ${\pi \over 4}$ and $\omega$ = 100 rad/s

$\Rightarrow$ Reactance (X) = Resistance (R) Now by checking option.

Option (A)
R = 1000 $\Omega$ and X_c = ${1 \over {{{10}^{ - 6}} \times 100}} = {10^4}\Omega$

Option (B)
R = 10³ $\Omega$ and X_L = ${10^{ - 3}} \times 100 = 10^{-1} \Omega$

Option (C)
R = 10³ $\Omega$ and X_c = ${1 \over {{10 \times {10}^{ - 6}} \times 100}} = {10^3}\Omega$

Option (D)
R = 10³ $\Omega$ and X_L = $10 \times {10^{ - 3}} \times 100 = 1\Omega$

X- coordinate of CM of remaining sheet
${X_{cm}} = {{MX - mx} \over {M - m}}$

= ${{\left( {4m} \right) \times \left( {{a \over 2}} \right) - M\left( {{{3a} \over 4}} \right)} \over {4m - m}} = {{5a} \over {12}}$

Similarly y_cm = ${{5b} \over {12}}$

$\therefore$ $CM\left( {{{5a} \over {12}},{{5b} \over {12}}} \right)$

Moment of inertia
$(I) = m{\left( {{d \over 2}} \right)^2} \times 2 = {{m{d^2}} \over 2}$

Now by $\tau = l\alpha$

$(qE)(d\,\sin \theta ) = {{m{d^2}} \over 2}.\alpha$

$\alpha = \left( {{{2qE} \over {md}}} \right)\sin \theta$ for small $\theta$

$\Rightarrow \alpha = \left( {{{2qE} \over {md}}} \right)\theta$

$\Rightarrow$ Angular frequency $\omega = \sqrt {{{2qE} \over {md}}}$

If we use that direction of light propagation will be along $\overrightarrow E \times \overrightarrow B$. Then (A) option is correct.
Magnitude of E = CB

E = 3 × 10⁸ × 1.6 × 10^–6 × $\sqrt 5$

E = 4.8 × ${10^{2\sqrt 5 }}$

$\overrightarrow E$ and $\overrightarrow B$ are perpendicular to each other

$\Rightarrow \overrightarrow E .\overrightarrow B = 0$

$\Rightarrow$ Either direction of $\overrightarrow E$ $\widehat i - 2\widehat j$ or $- \widehat i + 2\widehat j$ from given option

Also wave propagation direction is parallel to $\overrightarrow E \times \overrightarrow B$ which is $- \widehat k$

$\Rightarrow$ $\overrightarrow E$ is along ($- \widehat i + 2\widehat j$)

Given initial velocity u = 0 and acceleration is constant
At time t
v = 0 + at
$\Rightarrow$ v = at

Also $x = 0(t) + {1 \over 2}a{t^2}$
$\Rightarrow$ x = ${1 \over 2}a{t^2}$

Graph (A), (B) and (D) are correct

Magnetic field at point P

${\overrightarrow B _{net}} = {{{\mu _0}i} \over {2\pi d}}( - \widehat k) + {{{\mu _0}i} \over {2\pi d}}(\widehat k) = 0$

Time for 10 oscillations = ${{10} \over 5} = 2\,s$

A = A₀ e^–kt

${1 \over 2} = {e^{ - 2k}} \Rightarrow \ln 2 = 2k$

10^–3 = e^–kt $\Rightarrow$ 3In10 = kt

$t = {{3\ln 10} \over k} = {{3\ln 10} \over {\ln 2}} \times 2$

= $6 \times {{2.3} \over {0.69}} \approx 20\,s$

For solid sphere
${1 \over 2}m{v^2} + {1 \over 2}.{2 \over 5}m{R^2}.{{{v^2}} \over {{R^2}}} = mg{h_{sph}}$

For solid cylinder
${1 \over 2}m{v^2} + {1 \over 2}.{1 \over 2}m{R^2}.{{{v^2}} \over {{R^2}}} = mg{h_{cyl}}$

$\Rightarrow {{{h_{sph}}} \over {{h_{cyl}}}} = {{7/5} \over {3/2}} = {{14} \over {15}}$

Resistance of wire AB = 400 × 0.01 = 4 $\Omega$

i = ${3 \over 6} = 0.5A$

Now voltmeter reading = i (Resistance of 50 cm length)
= (0.5 A) (0.01 × 50) = 0.25 volt

Image formed by lens
${1 \over v} - {1 \over u} = {1 \over f};{1 \over v} + {1 \over {30}} = {1 \over {20}}$

v = +60 cm
If image position does not change even when mirror is removed it means image formed by lens is formed at centre of curvature of spherical mirror.
Radius of curvature of mirror = 80 – 60 = 20 cm.

$\Rightarrow$ Focal length of mirror f = 10 cm for virtual image, object is to be kept between focus and pole.

$\Rightarrow$ Maximum distance of object from spherical mirror for which virtual image is formed, is 10 cm.

Equivalent resistance of the given circuit

${R_{eq}} = 45 + {{10 \times 50} \over {10 + 50}} = \left( {45 + {{50} \over 6}} \right)\Omega$

$\Rightarrow$ ${{160} \over 3}\Omega$

$\therefore$ $I = {{15} \over {{{160} \over 3}}} = {9 \over {32}}A$

$\overrightarrow E = (20x + 10)\widehat i$

${V_1} - {V_2} = - \int\limits_{ - 5}^1 {\left( {20x + 10} \right)dx}$

${V_1} - {V_2} = - \left( {10{x^2} + 10x} \right)_{ - 5}^1$

${V_1} - {V_2} = 10\left( {25 - 5 - 1 - 1} \right)$

${V_1} - {V_2} = 180\,V$

Let mass of B and C is m each. By momentum conservation

$2m{v_0} = mv - {{mv} \over 2}$

v = 4v₀
P_A = 2mv₀ pB = 4mv₀ pc = 2mv₀

De-Broglie wavelength $\lambda = {h \over p}$

${\lambda _A} = {h \over {2m{v_0}}}$; ${\lambda _B} = {h \over {4m{v_0}}}$; ${\lambda _C} = {h \over {2m{v_0}}}$

Applying linear momentum conservation

${m_1}{v_1}\widehat i + {m_2}{v_2}\widehat i = {m_1}{v_3}\widehat i + {m_2}{v_4}\widehat i$

m₁v₁ + 0.5 m₁v₂ = m₁(0.5 v₁) + 0.5 m₁v₄

0.5 m₁v₁ = 0.5 m₁(v₄ - v₂)

v₁ = v₄ - v₂

We know,

surface tension (S) = ${F \over L}$ = ${{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ L \right]}}$

$\therefore$ [S] = ${\left[ {M{T^{ - 2}}} \right]}$

Moment of inertia (I) = mr²

$\therefore$ [I] = ${\left[ {M{L^2}} \right]}$

Planck's constant (h) = ${E \over f}$ = Et

$\therefore$ [h] = ${\left[ {M{L^2}{T^{ - 1}}} \right]}$

Also linear momentum (p) = mv = ${\left[ {ML{T^{ - 1}}} \right]}$

Now we have to express p in terms of s, I and h.

$\therefore$ Let, [P] = [S^a I^b h^c]

$\Rightarrow$ ${\left[ {ML{T^{ - 1}}} \right]}$ = ${\left[ {M{T^{ - 2}}} \right]}$^a ${\left[ {M{L^2}} \right]}$^b ${\left[ {M{L^2}{T^{ - 1}}} \right]}$^c

$\Rightarrow$ ${\left[ {ML{T^{ - 1}}} \right]}$ = [ M^{a + b + c} L^{2b +2c} T^{- 2a - c} ]

By comparing the dimensions of both sides, we get

a + b + c = 1 .........(1)

2b +2c = 1 ..............(2)

- 2a - c = -1 ...................(3)

By solving those three equations we get,

a = ${1 \over 2}$

b = ${1 \over 2}$

c = 0

$\therefore$ linear momentum [p] = [S^1/2I^1/2h⁰]

Markovnikov rule says negative part of the reagent attacks on the carbon of the double bond which have less number of hydrogen atom.

So in Anti-Markovnikov rule negative part of the reagent attacks on the carbon of the double bond which have more number of hydrogen atom.

Strength means ${{Weight} \over {Volume}}$ %.

Molarity of H₂O₂ = ${{volume\,strength} \over {11.2}}$ = ${{11.2} \over {11.2}}$ = 1 M = 34 gm/lit

In 1000 ml solution 34 gm H₂O₂ present

$\therefore$ In 100 ml solution 3.4 gm H₂O₂ present.

Mond process is used for purification of Ni.

Interstitial compound are –

(i) very hard

(ii) chemically inert

(iv) high melting point.

As interstitial compounds are chamically inert.

Polysubstitution is a major drawback of Friedal–Craft alkylation.

–CH₃ group in highly activating group due to +H effect, so after first time addition of -CH₃ group in benzene ring the reactivity of the benzene ring increases.

S(s) + O₂(g) ⇋ SO₂(g); K₁ = 10⁵²

By reversing the equation, we get

SO₂(g); ⇋ S(s) + O₂(g) ; ${1 \over {{K_1}}} = {1 \over {{{10}^{52}}}}$ ..............(1)

2S(s) + 3O₂(g) ⇋ 2SO₃(g); K₂ = 10¹²⁹ ....................(2)

Performing (2) - 2 $\times$ (1), we get

2SO₂(g) + O₂(g) ⇋ 2SO₃(g)

$\therefore$ Equilibrium constant of this reaction is

= ${{{K_2}} \over {K_1^2}}$ = ${{{{10}^{129}}} \over {{{\left( {{{10}^{52}}} \right)}^2}}}$ = 10²⁵

Cis-platin or cis-[Pt(Cl)₂(NH₃)₂] is used as an anti-cancer drug.

Hydrogen removed from the $\alpha$ carbon should be acidic in nature because ehen hydrogen is more acidic then enol formation is easy.

In 100 ml of hexane solution contains 0.27 g of fatty acid.

$\therefore$ In 10 ml of hexane solution contains 0.027 g of fatty acid.

Volume of fatty acid present on the round glass = ${{0.027} \over {0.9}}$

As here Area of fatty acid layer = Area of round plate = $\pi {r^2}$

$\therefore$ Volume of fatty acid layer = $\pi {r^2}$ $\times$ h

$\therefore$ $\pi {r^2}$ $\times$ h = ${{0.027} \over {0.9}}$

$\Rightarrow$ 3 $\times$ ${\left( {10} \right)^2}$ $\times$ h = ${{0.027} \over {0.9}}$

$\Rightarrow$ h = 10^-4 cm = 10^-6 m

From photoelectric effect,

E = $\phi$ + KE

${{hc} \over \lambda } = \phi + {{{p^2}} \over {2m}}$ ................ (1)

Now when momentum = 1.5p then let wavelength = ${{\lambda _1}}$

$\therefore$ ${{hc} \over {{\lambda _1}}} = \phi + {{{{\left( {1.5p} \right)}^2}} \over {2m}}$ ........... (2)

Given,

kinetic energy(KE) of ejected photoelectron to be very high in comparison to work function($\phi$).

$\therefore$ We can neglect work function($\phi$).

$\therefore$ Equation (1) and (2) becomes,

${{hc} \over \lambda } = {{{p^2}} \over {2m}}$ ................ (1)

${{hc} \over {{\lambda _1}}} = {{{{\left( {1.5p} \right)}^2}} \over {2m}}$ ........... (2)

Dividing (1) by (2) we get,

${{{\lambda _1}} \over \lambda } = {{{p^2}} \over {{{\left( {1.5p} \right)}^2}}}$

$\Rightarrow$ ${{{\lambda _1}} \over \lambda } = {4 \over 9}$

$\Rightarrow$ ${\lambda _1} = {4 \over 9}\lambda$

According to Fajan’s rule, when the size of cation is less then the covalent character would be more for the cation.

Since Be²⁺ has minimum size among given cation, therefore, BeX₂ would be most covalent among given alkaline with metal halides

For ideal gas,

$\Delta$U = nC_v$\Delta$T

= 5 $\times$ 28 $\times$ (200 - 100)

= 14 kJ

We know,

PV = nRT

$\therefore$ $\Delta$(PV) = nR$\Delta$T = 5 $\times$ 8 $\times$ 100 = 4 kJ

Seliwanoff's test is used to distinguish aldose and ketose.

When Seliwanoff's reagent added to a solution containing ketoses, red colour is formed rapidly indicating a positive test.
While in case of aldose, a slower forming light pink colour observed.

Henry's law expression,

P_gas = K_H $\times$ x_gas

Here, P_gas = partial pressure of gas

K_H = Henry’s law constant

x_gas = mole fraction of gas

$\because$ x_gas = 1 - x_H2O

$\therefore$ P_gas = K_H $\times$ (1 - x_H2O)

$\Rightarrow$ P_gas = K_H - K_H $\times$x_H2O

By comparing this equation with y = mx + c,

c = K_H = Intercept

m = - K_H = slope

So in partial pressure vs mole fraction of water graph, which gas has more K_H will have more intercept. Given (K_H) are 0.5, 2, 35 and 40 kbar for w, x, y, z respectively. So intercept of z should be the most and should be the least.

As solpe is = - K_H, so slope shoul be negative for all the gases and w gas should have least negative slope and z should have highest negative slope.