JEE Main 2019 10th April Evening Shift Question Paper with Solutions

L = 10 × 10^–3 H, r₁ = 0.1 $\Omega$

$i = \varepsilon \left\{ {1 - {e^{ - 1/2}}} \right\}$

${i_{saturation}}{\rm{ }} = {\rm{ }}\varepsilon$
$80\% {\rm{ }}{i_{saturation}}{\rm{ }} = {\rm{ }}0.8{\rm{ }}\varepsilon$

0.8 = 1 - e^-t/2 ; e^-t/2 = 0.2

e^t/L = 5
t = L ln 5 = 10 × 10^–3 × 1.6 = 16 × 10^–3

As the graph between magnification (m) and image distance (v) varies linearly, then
m = k₁v + k₂

$\Rightarrow {v \over u} = {k_1}v + {k_2}$

$\Rightarrow {1 \over u} = {k_1} + {{{k_2}} \over v}$

$\Rightarrow {{{k_2}} \over v} - {1 \over u} = {k_1}$

Clearly, ${k_1} = {1 \over f}$ and k₂ = 1 here

$\therefore f = {1 \over {slope\,of\,m - y\,graph}} = {b \over c}$

$\overrightarrow v = 2\widehat i - 6 + \widehat j$

At t = 2
$\overrightarrow v = 2\widehat i - 12\widehat j$

$\overrightarrow P = m\overrightarrow v = 4i - 24\widehat j$

At t = 2
$\overrightarrow r = 4\widehat i - 12\widehat j$

\overrightarrow L = \overrightarrow r \times \overrightarrow P = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 4 & { - 12} & 0 \cr 4 & { - 24} & 0 \cr } } \right|

$= \left\{ {4( - 2) + 4 \times 12} \right\}\widehat k$

$= \left( { - 96 + 48} \right)\widehat k$

$= \left( - \right)48\widehat k$

For a current carrying wire, magnetic field at a distance r is given by


$B=\frac{\mu_0 i}{4 \pi r}\left(\sin \theta_1+\sin \theta_2\right)$

Now, in given case,



Due to symmetry of arrangement, net field at centre of triangle is

$$\begin{aligned} B_{\text {net }} & =\text { Sum of fields of all wires (sides) } \\ & =3 \times \frac{\mu_0 i}{4 \pi r}\left(\sin \theta_1+\sin \theta_2\right) \end{aligned}$$

Here, $\theta_1=\theta_2=60^{\circ}$

$$\begin{aligned} & \therefore \sin \theta_1=\sin \theta_2=\frac{\sqrt{3}}{2}, i=10 \mathrm{~A}, \frac{\mu_0}{4 \pi}=10^{-7} \mathrm{NA}^{-2} \\ & \text { and } r=\frac{1}{3} \times \text { altitude } \\ & \qquad=\frac{1}{3} \times \frac{\sqrt{3}}{2} \times \text { sides length }=\frac{1}{2 \sqrt{3}} \times 1 \mathrm{~m}=\frac{1}{2 \sqrt{3}} \mathrm{~m} \end{aligned}$$

So,

$$\begin{aligned} B_{\mathrm{net}} & =\frac{3 \times 10^{-7} \times 10 \times 2\left(\frac{\sqrt{3}}{2}\right)}{\left(\frac{1}{2 \sqrt{3}}\right)}=18 \times 10^{-6} \mathrm{~T} \\ \Rightarrow B_{\mathrm{net}} & =18 \mu \mathrm{T} \end{aligned}$$

$P = {{\Delta E} \over C}$

$= {{\left( {25 \times 25} \right) \times 40 \times 60} \over {3 \times {{10}^8}}}N - s$

$= 5 \times {10^{ - 2}}N - s$

q_A = 1 $\mu$c ; q_B = 1 $\mu$c, m_B = 4 × 10^–9 kg, r_AB = 10^–3 m

${1 \over 2}{M_B}{V^2} = k{q_A}{q_B}\left\{ {{1 \over {{{10}^{ - 13}}}} - {1 \over {9 \times {{10}^{ - 3}}}}} \right\}$

${1 \over 2}4 \times {10^{ - 9}}{V^2} = 9 \times {10^9} \times {10^{ - 6}} \times {8 \over 9} \times {10^3}$

${V^2} = {8 \over 2} \times {10^9} = 4 \times {10^9}$

V = $\sqrt {40} \times {10^4}\,m/s$ = 6.32 $\times$ 10⁴ m/s

$t = 2\pi \sqrt {{L \over {{g_{eff}}}}}$

$\Rightarrow {g_{eff}} = \sqrt {{g^2} + {{\left( {{{gE} \over m}} \right)}^2}}$

$\Rightarrow t = 2\pi \sqrt {{L \over {\sqrt {{g^2} + {{\left( {{{qE} \over m}} \right)}^2}} }}}$

$R = \int\limits_a^b {{{\rho \,dx} \over {4\pi {x^2}}}}$

$= {\rho \over {4\pi }}\left( {{1 \over a} - {1 \over b}} \right)$

Heat supplied at constant volume
Q = nC_V$\Delta$T

and heat supplied at constant pressure
Q' = nC_P$\Delta$T

$Q' = {{{C_P}} \over {{C_V}}}Q = \left( {1 + {2 \over 5}} \right)Q = {7 \over 5}Q$

M_A = 1 kg, M_B = 3 k

${\mu _{AB}} = 0.2$

${\mu _B} = 0.2$

F_max = (M_A + M_B) × 0.2 × 10 + (M_A + M_B) × 0.2 × 10

= 4 × 2 + 4 × 2 = 16

${{m{V^2}} \over r} = {{GMm} \over {{r^2}}}$

$V = \sqrt {{{GM} \over r}}$

$n = {{VT} \over {2\pi r}} = \sqrt {{{GM} \over r}} {T \over {2\pi r}}$

$= \left( {\sqrt {{{GM} \over {{r^3}}}} } \right) \times {T \over {2\pi }} = \sqrt {{{6.67 \times {{10}^{ - 11}} \times 8 \times {{10}^{22}}} \over {{{\left( {202 \times {{10}^4}} \right)}^3}}}} \times {T \over {2\pi }}$

$= {{24 \times 3600} \over {2 \times 3.14}}\sqrt {{{6.67 \times 8 \times {{10}^{11}}} \over {{{\left( {202} \right)}^3} \times {{10}^{12}}}}}$

$= {{24 \times 3600} \over {2 \times 3.14 \times 1242.8}} = {{24 \times 3600} \over {74.51}} = 11$

Capacitance (C) = ${{{Q^2}} \over {2E}}$ = ${{\left[ {{A^2}{T^2}} \right]} \over {\left[ {M{L^2}{T^{ - 2}}} \right]}}$ = $\left[ {{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}} \right]$ X

Magnetic field (B) = ${F \over {IL}}$ = ${{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ A \right]\left[ L \right]}}$ = ${\left[ {M{T^{ - 2}}{A^{ - 1}}} \right]}$ = Z

Given,

X = 5YZ²

$\therefore$ Y = ${X \over {5{Z^2}}}$

[Y] = ${{\left[ {{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}} \right]} \over {{{\left[ {M{T^{ - 2}}{A^{ - 1}}} \right]}^2}}}$

$\therefore$ [Y] = [M^–3L^–2T⁸A⁴]

Corresponding to the stress ($\sigma$)

Total elongation $\Delta {I_{net}} = {{\sigma {L_1}} \over {{Y_1}}} + {{\sigma {L_2}} \over {{Y_2}}}$

$\sigma = \Delta I\left( {{{{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}} \right)$

$= 0.2 \times {10^{ - 3}} \times \left( {{{120 \times 60} \over {180}}} \right) \times {10^9}$

$= 8 \times {10^6}{N \over {{m^2}}}$

Let the given square loop has side $a$, then its magnetic dipole moment will be

$m=I a^2$

When square is converted into a circular loop of radius $r$,


Then, wire length will be same in both area,

$\Rightarrow 4 a=2 \pi r \Rightarrow r=\frac{4 a}{2 \pi}=\frac{2 a}{\pi}$

Hence, area of circular loop formed is, $A^{\prime}=\pi r^2$

$=\pi\left(\frac{2 a}{\pi}\right)^2=\frac{4 a^2}{\pi}$

Magnitude of magnetic dipole moment of circular loop will be

$m^{\prime}=I A^{\prime}=I \frac{4 a^2}{\pi}$

Ratio of magnetic dipole moments of both shapes is,

$$\begin{aligned} \frac{m^{\prime}}{m}=\frac{I \cdot \frac{4 a^2}{\pi}}{I a^2}=\frac{4}{\pi} \\ \Rightarrow m^{\prime} =\frac{4 m}{\pi}(\mathrm{A}-\mathrm{m}) \end{aligned}$$

$T = {{2u\sin \theta } \over {g\cos \alpha }}$

$R = u\cos \theta T - {1 \over 2}g\sin \alpha {T^2}$

$= {{u\cos \theta 2u\sin \theta } \over {g\cos \alpha }} - {{g\sin \alpha } \over 2}{{4{u^2}{{\sin }^2}\theta } \over {{g^2}{{\cos }^2}\alpha }}$

$= {{{u^2}{{\sin }^2}\theta } \over {g\cos \alpha }} - {{{u^2}\sin \alpha } \over {g{{\cos }^2}\alpha }}\left\{ {1 - \cos 2\theta } \right\}$

$= {{4 \times {1 \over 2}} \over {10 \times {{\sqrt 3 } \over 2}}} - {{{u^2}\sin \alpha } \over {g{{\cos }^2}\alpha }}\left\{ {1 - {{\sqrt 3 } \over 2}} \right\}$

$= {4 \over {10\sqrt 3 }} - {8 \over {30}}\left\{ {1 - {{\sqrt 3 } \over 2}} \right\}$

$= {4 \over {5\sqrt 3 }} - {8 \over {30}} = {{8\sqrt 3 - 8} \over {30}}$

$= {{8\left( {\sqrt 3 - 1} \right)} \over {30}} = 20\,cm$

Given $P = {P_o}\left\{ {1 - {1 \over 2}{{\left( {{{{V_o}} \over V}} \right)}^2}} \right\};$ ...(i)

As n = 1 mole

$\therefore$ PV = nRT = RT

$\Rightarrow$ P = ${{RT} \over V}$ ....(ii)

From (i) and (ii), we get

${{RT} \over V} = {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over V}} \right)}^2}} \right]$

$\Rightarrow$ T = ${V \over R} \times {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over V}} \right)}^2}} \right]$

Case 1 : when V = V₀

then T_i = ${{{V_0}} \over R} \times {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over {{V_0}}}} \right)}^2}} \right]$

= ${{{V_0}{P_0}} \over {2R}}$

Case 2 : when V = 2V₀

then T_f = ${{2{V_0}} \over R} \times {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over {2{V_0}}}} \right)}^2}} \right]$

= ${7 \over 4}{{{P_0}{V_0}} \over R}$

Then $\Delta$T = T_f - T_i

= ${7 \over 4}{{{P_0}{V_0}} \over R} - {{{P_0}{V_0}} \over {2R}}$

= ${{{P_0}{V_0}} \over R}\left( {{7 \over 4} - {1 \over 2}} \right)$

= ${5 \over 4}{{{P_0}{V_0}} \over R}$

${{hc} \over \lambda } = 13.6\,ev(g)\left\{ {1 - {1 \over {16}}} \right\}$

${{1240\,eV} \over \lambda } = {{15} \over {16}} \times 9 \times 13.6\,eV$

$\lambda = {{1240 \times 16} \over {15 \times 9 \times 13.6}} = 10.8\,nm$

Given ${\left( {50} \right)^3} \times {{30} \over {100}} \times \left( 1 \right) \times g = {M_{cube}}g$    ...(i)

Let m mass should be placed

Hence (50)³$\times$ (1) $\times$ g = (M_cube + m)g   …(ii)

equation (ii) – equation (i)

$\Rightarrow$ mg = (50)³ × g(1 – 0.3) = 125 × 0.7 × 10³ g

$\Rightarrow$ m = 87.5 kg

${I_1} = {{\left( {{{7M} \over 8}} \right){{\left( {2R} \right)}^2}} \over 2} = {{7M \times 4{R^2}} \over {2 \times 8}} = {{7M{R^2}} \over 4}$

${I_2} = {2 \over 5}{M \over 8}{\left( {{R \over 2}} \right)^2} = {{2M} \over {5 \times 8}}{{{R^2}} \over 4} = {{M{R^2}} \over {80}}$

${{{I_1}} \over {{I_2}}} = {{7M{R^2} \times 80} \over {4M{R^2}}} = 140$

Given, resistance offered by the wall

$=F=-25 \times 10^{-2} \mathrm{~N}$

So, deacceleration of bullet,

$$\begin{aligned} a=\frac{F}{m}=\frac{-2.5 \times 10^{-2}}{20 \times 10^{-3}} & =-\frac{5}{4} \mathrm{~ms}^{-2} \\ (\because m & \left.=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}\right) \end{aligned}$$

Now, using the equation of motion,

$v^2-u^2=2 a s$

We have,

$v^2=1+2\left(-\frac{5}{4}\right)\left(20 \times 10^{-2}\right)$

$\left(\because u=1 \mathrm{~ms}^{-1} \text { and } s=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}\right)$

$$\begin{array}{ll} \Rightarrow v^2=\frac{1}{2} \\ \therefore v=\frac{1}{\sqrt{2}} \approx 0.7 \mathrm{~ms}^{-1} \end{array}$$

m = 5 × 10^–3 kg, r = 10^–2 m
$\omega$ = 25 × 2$\pi$ rad/5
= 50 $\omega$ rad/sec

$\omega = {\tau \over I}t$
$\tau = {{I\omega } \over t} = {{5m{r^2}} \over 4} \times {\omega \over t}$

$= {{5 \times 5 \times {{10}^{ - 3}} \times {{10}^{ - 4}} \times 50\pi } \over {4 \times 5}}$

$= {{25\pi } \over 4} \times {10^{ - 6}} = 2 \times {10^{ - 5}}$ Nm

The pressure experienced by a submarine at a certain depth in the sea is given by the formula:

$P = \rho g h$

where:

• $P$ is the pressure
• $\rho$ is the density of the fluid (sea water in this case)
• $g$ is the acceleration due to gravity
• $h$ is the height (or depth in this case)

Given:

• $\rho = 10^3 \, kg/m^3$
• $g = 10 \, m/s^2$

We are looking for the difference in depth, $d_2 - d_1$, which corresponds to the difference in pressure $\Delta P$:

$\Delta P = P_2 - P_1 = \rho g (d_2 - d_1)$

Rearranging the above equation, we get:

$d_2 - d_1 = \frac{\Delta P}{\rho g}$

Given:

• $P_2 = 8.08 \times 10^6 \, Pa$
• $P_1 = 5.05 \times 10^6 \, Pa$

$\Delta P = P_2 - P_1 = 8.08 \times 10^6 \, Pa - 5.05 \times 10^6 \, Pa = 3.03 \times 10^6 \, Pa$

So,

$d_2 - d_1 = \frac{3.03 \times 10^6 \, Pa}{10^3 \, kg/m^3 \times 10 \, m/s^2} = 303 \, m$

The closest answer among the options provided is Option C, 300 m.

$2 \times {10^{ - 3}} = {{hc} \over \lambda }{{dn} \over {dt}}$

${{dn} \over {dt}} = {{2 \times {{10}^{ - 3}}\lambda } \over {hc}}$

$= {{2 \times {{10}^{ - 3}} \times 500 \times {{10}^{ - 9}}} \over {6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}$

$= {{1000} \over {6.6 \times 3}} \times 10 = 5 \times {10^{15}}$

Using Bernoullie’s equation ${v_2} = \sqrt {v_1^2 + 2gh}$

Equation of continuity
A₁V₁ = A₂V₂

(1 cm³)(1m/s) = $\left( {{A_2}} \right)\left( {\sqrt {{{\left( 1 \right)}^2} + 2 \times 10 \times {{15} \over {100}}} } \right)$

$\Rightarrow {A_2}\left( {\ln c{m^2}} \right) = {1 \over 2}$

$\Rightarrow {A_2} = 5 \times {10^{ - 5}}{m^2}$

${I_1} = 4{I_0}$

${I_2} = {I_0}$

${I_{\max }} = {\left( {\sqrt {{I_0}} + \sqrt {{I_2}} } \right)^2}$

$= {\left( {2\sqrt {{I_0}} + \sqrt {{I_0}} } \right)^2} = 9{I_0}$

${I_{\min }} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2}$

$= {\left( {2\sqrt {{I_0}} - \sqrt {{I_0}} } \right)^2} = {I_0}$

$\therefore$ ${{{{\mathop{\rm I}\nolimits} _{\max }}} \over {{I_{\min }}}} = {{9{I_0}} \over {{I_0}}} = {9 \over 1}$

V_breakdwon = 6V, RL = 4k$\Omega$, R_i = 1 k$\Omega$

${i_L} = {6 \over 4} \times {10^{ - 3}} = 1.5 \times {10^{ - 3}} = 1.5\,mA$

i_i = 2 × 10^–3
i = i₁ – i_L = 0.5 mA – minimum current

i_i = 10 × 10^–3 = 10 mA
i_max = 8.5 mA

${f_a} = {V \over {V - {V_s}}}{f_o} = 1000\,Hz$

$f_a^{'} = {V \over {V + {V_s}}}{f_o}$

${{f_a^{'}} \over {{f_a}}} = {{V - {V_s}} \over {V + {V_s}}} = {{350 - 50} \over {350 + 50}} = {{300} \over {400}} = {3 \over 4}$

$f_a^{'} = {3 \over 4} \times 1000 = 750\,Hz$

${{400} \over {{\pi \over 4}{d^2}}} = 379 \times {10^6}$

${d^2} = {{4 \times 400 \times {{10}^{ - 6}}} \over {\pi \times 379}} = 0.336 \times {10^{ - 6}} \times 4$

$d = 2\sqrt {0.336} \times {10^{ - 3}}M \simeq 1.16\,mm$

N_x(at t) = N₀e^{–5$\lambda$t}

N_y(at t) = N₀e^{–$\lambda$t}

${{{N_x}} \over {{N_y}}} = {1 \over {{e^2}}} = {e^{ - 4\lambda t}}$

$\Rightarrow 4\lambda t = 2$

$\Rightarrow t = {2 \over {4\lambda }} = \left( {{1 \over {2\lambda }}} \right)$

Beat frequency = |f₁ – f₂| = 11 – 9 = 2 Hz

Air pollution caused by sunlight is photochemical smog and it is oxidising.

NH₄⁺ + H₂O ⇋ NH₄OH + H⁺

[H⁺] = c$\alpha$

= $\sqrt {{k_a}\left( {NH_4^ + } \right) \times c}$

= $\sqrt {{{{k_w}} \over {{k_b}\left( {N{H_4}OH} \right)}} \times c}$

= $\sqrt {{{{{10}^{ - 14}}} \over {{{10}^{ - 5}}}} \times 0.02}$

= $\sqrt {20} \times {10^{ - 6}}$

$\therefore$ pH = $- \log \left( {\sqrt {20} \times {{10}^{ - 6}}} \right)$ = 5.35

Given,

${{{{\left( {{k_b}} \right)}_A}} \over {{{\left( {{k_b}} \right)}_B}}} = {1 \over 5}$

$\Delta {T_b}\left( A \right) = {k_b}\left( A \right) \times$ Molality of A

$\Delta {T_b}\left( B \right) = {k_b}\left( B \right) \times$ Molality of B

Here, Molality of A = Molality of B = m

${{\Delta {T_b}\left( A \right)} \over {\Delta {T_b}\left( B \right)}} = {{{{\left( {{k_b}} \right)}_A}} \over {{{\left( {{k_b}} \right)}_B}}} \times {m \over m}$ = ${1 \over 5}$

We know,

$\Delta$H - $\Delta$U = $\Delta$n_gRT

C₇H₁₆($l$) + 11O₂(g) $\to$ 7CO₂(g) + 8H2O($l$)

Here $\Delta$n_g = 7 - 11 = - 4

$\therefore$ $\Delta$H - $\Delta$U = - 4RT

(A) Complementarty color of yellow is violet. As [CoCl(NH₃)₅]²⁺ absorbs yellow so it will show complementary color violet.

(B)The building block of ruby is aluminium oxide (Al₂O₃) containing about 0.5 – 1% Cr³⁺ ions which are randomly distributed in the position normally occupied by Al³⁺ ions.

(C) Total number of unpaired electons in [Fe(H₂O)₆]²⁺ = 4
and total number of unpaired electons in [Cr(H₂O)₆]²⁺ = 4
So, the spin-only magnetic moments are same in both cases.

(D) Total number of unpaired electons in [Ni(NH₃)₄(H₂O)₂]²⁺ = 2
So, $\mu$ = $\sqrt {2\left( {2 + 2} \right)}$ BM = 2.83 BM

Equilibrium constant has no relation with catalyst.

Catalyst only affects the rate with which a reaction proceeds.

Here we use catalyst V₂O₅ to speed up the reaction.

From Arrhenius equation, we get

$\log {{{K_2}} \over {{K_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$

$\Rightarrow$ $\log {1 \over {2.5 \times {{10}^{ - 4}}}} = {{{E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {600}} - {1 \over {800}}} \right)$

$\Rightarrow$ 3.6 = ${{{E_a}} \over {2.303 \times 8.314}} \times {{200} \over {600 \times 800}}$

$\Rightarrow$ E_$a$ = 166 kJ/mol

Shortest wavelength can found when n₂ = $\infty$

${\lambda _{shortest}} = R{Z^2}\left\{ {{1 \over {n_1^2}} - {1 \over {{\infty ^2}}}} \right\}$

Here n₁ = series number.

$\Rightarrow$${\lambda _{shortest}} =$ ${{{R{Z^2}} \over {n_1^2}}}$ = ${{{R{{\left( 1 \right)}^2}} \over {n_1^2}}}$

$\therefore$ ${{{\lambda _A}} \over {{\lambda _B}}} = {\left( {{{{n_B}} \over {{n_A}}}} \right)^2}$

Given, ${\left( {{{{n_B}} \over {{n_A}}}} \right)^2} = 9$

$\Rightarrow$ ${{{n_B}} \over {{n_A}}} = 3$

$\therefore$ Series number of B = 3 $\times$ Series number of A

So,

If series number of A is Lyman( n = 1 )

then series number of B is Paschen( n = 3 ).

(a) NaH + H₂O $\to$ NaOH + H₂

(b) 3LiAlH₄ + 4BF₃ $\to$ 2B₂H₆ + 3LiF + 3AlF₃

(c) PH₃ is electron rich due to one lone pair while CH₄ is electron precise hydride.

(d) HF and CH₄ are molecular hydrides.