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

N₁ = N₀e^{–10$\lambda$t} ; N₂ = N₀e^{–$\lambda$t}

${1 \over e} = {{{N_1}} \over {{N_2}}} = {e^{ - 9\lambda t}}$

$\Rightarrow 9\lambda t = 1$

$\Rightarrow t = {1 \over {9\lambda }}$

From Snell’s law
1 sin 60° = $\mu$ sin 30°

$\Rightarrow$ $\mu$ = $\sqrt 3$

Optical path = AO + $\mu$(OB)

= ${a \over {\cos {{60}^o}}} + \sqrt 3 {b \over {\cos {{30}^o}}}$

= 2a + 2b

Net force on wire carrying current I per unit length is

${{{\mu _0}{I_1}I} \over {2\pi x}} + {{{\mu _0}{I_2}I} \over {2\pi \left( {d - x} \right)}} = 0$

${{{I_1}} \over x} = {{{I_2}} \over {x - d}}$

$\Rightarrow x = {{{I_1}d} \over {{I_1} - {I_2}}}$

$\overrightarrow F = m\overrightarrow a = m\left[ { - a\omega _1^2\cos \omega ,t\widehat i - b\omega _2^2\sin {\omega _2}t\widehat j} \right]$

${\overrightarrow f _{t = 0}} = - ma\omega _1^2\widehat i$

${\overrightarrow r _{t = 0}} = \left( {{X_0} + a} \right)\widehat i + y\widehat j$

$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F = m{y_0}a\omega _1^2\widehat k$

As q = CV
Hence slope of graph will give capacitance. Slope will be more in parallel combination. Hence capacitance in parallel should be 50 $\mu$F and in series combination must be 8 $\mu$F. Only in option 40 $\mu$f and 10 $\mu$F.

C_parallel = 40 + 10 = 50 $\mu$F

C_series = ${{40 \times 10} \over {40 + 10}} = 8\mu F$

$\mu = {{{V_d}} \over E}\,\,\,\,\,\,E = \rho J$

$= {{1.1 \times {{10}^{ - 3}}} \over {1.7 \times {{10}^{ - 8}} \times {5 \over {\pi \times 25 \times {0^{ - 6}}}}}}$

$= {{1.1 \times {{10}^{ - 3}} \times \pi \times 25 \times {0^{ - 6}}} \over {1.7 \times {{10}^{ - 8}} \times 5}} \approx 1.01\,{m^2}/Vs$

For 1^st lens ${1 \over {{f_1}}} = \left( {{{{\mu _1} - 1} \over 1}} \right)\left( {{1 \over \infty } - {1 \over { - R}}} \right) = {{{\mu _1} - 1} \over R}$

For 2^nd lens ${1 \over {{f_2}}} = \left( {{{{\mu _2} - 1} \over 1}} \right)\left( {{1 \over { - R}}} \right) - 0 = {{{\mu _2} - 1} \over R}$

${1 \over {{f_{eq}}}} = {1 \over {{f_1}}} + {1 \over {{f_2}}}$

${1 \over {{f_{eq}}}} = {{{\mu _1} - 1} \over R} + {{ - \left( {{\mu _2} - 1} \right)} \over R}$

$\Rightarrow {f_{eq}} = {R \over {{\mu _1} - {\mu _2}}}$

$as\,x = {{R\left( {100 - l} \right)} \over l}$

$for\,(A)\,\,\,\,x = {{1000 \times \left( {100 - 60} \right)} \over {40}} \approx 667$

$for\,(B)\,\,\,\,\,x = {{100 \times \left( {100 - 13} \right)} \over {13}} \approx 669$

$for\,(C)\,\,\,\,\,x = {{10 \times \left( {100 - 1.5} \right)} \over {98.5}} \approx 667$

$for\,(D)\,\,\,\,\,x = {{1 \times \left( {100 - 1} \right)} \over 1} \approx 667$

So reading in serial no. (4) is completely different hence correct answer (A).

Given data,

$$\begin{aligned} I =10^{-4} \mathrm{~A}, \\ R_S =2 \mathrm{M} \Omega=2 \times 10^6 \Omega, \\ V_{\max } =5 \mathrm{~V} \end{aligned}$$

Let internal resistance of galvanometer is $R_G$.


Then,

$$\begin{array}{lc} I \times R_S+I \times R_G=V_{\max } \\ \Rightarrow 2 \times 10^6 \times 10^{-4}+10^{-4} \times R_G=5 \\ \Rightarrow 10^{-4} R_G=5-200=-195 \\ \text { or } R_G=-195 \times 10^4 \Omega \end{array}$$

Resistance cannot be negative.

$\therefore$ No option is correct.

${A_v} \times \beta = {P_{gain}}$

$60 = 10{\log _{10}}\left( {{P \over {{P_0}}}} \right)$

$P = {10^6} = {\beta ^2} \times {{{R_{out}}} \over {{R_{in}}}}$
$= {\beta ^2} \times {{{{10}^4}} \over {100}}$

${\beta ^2} = {10^4};\beta = 100$

Amplitude at (t = 0) A₀ = e^{–0.1× 0} = 1

$\therefore$ $at\,t = t$     if $A = {{{A_0}} \over 2}$

$\Rightarrow$ ${1 \over 2} = {e^{ - 0.1t}}$

t = 10 ln 2 = 7 s

Given N_P = 300, N_s = 150, P₀ = 2200W

I_s = 10 A
P₀ = V₀I₀ $\Rightarrow$ 2200 = V₀ × 10 $\Rightarrow$ V₀ = 220 V

$\because$ ${{{V_i}} \over {{V_0}}} = {{{N_P}} \over {{N_S}}} \Rightarrow {V_i} = 2 \times 220 = 440\,V$

Also, P₀ = V_iI_i
$\Rightarrow {I_i} = {{2200} \over {440}} = 5A$

$r = {{mv} \over {qB}} = {{\sqrt {2mK} } \over {qB}}$

${r_{He}} = {r_p} > {r_e}$

$\therefore \overrightarrow E \times \overrightarrow B \parallel \overrightarrow v$

Given that wave is propagating along positive z-axis and $\overrightarrow E$ along positive x-axis. Hence $\overrightarrow B$ along y-axis.

From Maxwell equation
$\overrightarrow V \times \overrightarrow E = - {{\partial B} \over {\partial t}}$

i.e. ${{\partial E} \over {\partial Z}} = - {{\partial B} \over {dt}}$ and ${B_0} = {{{E_0}} \over C}$

w = nR$\Delta$T
$\Delta$H = (Cv + nR)$\Delta$T
${\omega \over {\Delta H}} = {{nR} \over {{C_v} + nR}}$

Given, drag force, $F=m \gamma v^2$ ......(i)

As we know, general equation of force

$=m a$ .........(ii)

Comparing Eqs. (i) and (ii), we get

$a=\gamma v^2$

The net retardation of the ball when thrown vertically upward is therefore

$a_{\text{net}} = - (g + \gamma v^2) = \frac{dv}{dt}$, where $g$ is the acceleration due to gravity.

Rearranging terms gives us :

$\frac{dv}{g + \gamma v^2} = - dt$

We now need to integrate both sides of this equation.

When the ball is thrown upward with velocity $v_0$ and reaches its zenith ( "zenith" refers to the highest point that the ball reaches in its upward trajectory.), the velocity is $0$. The time to reach the zenith is $t$.

So the integral equation is :

$\int\limits_{v_0}^{0} \frac{dv}{\gamma v^2 + g} = - \int\limits_{0}^{t} dt$

Separating the constants from the integral :

$\frac{1}{\gamma} \int\limits_{v_0}^{0} \frac{1}{\left(\frac{g}{\gamma}+v^2\right)} dv = - \int\limits_{0}^{t} dt$

Recognizing the integral as the standard form $\frac{1}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$, we write the integral in terms of the arctangent function.

This gives us :

$\frac{1}{\gamma} \left(\frac{1}{\sqrt{\frac{g}{\gamma}}}\right) \left[\tan^{-1}\left(\frac{v}{\sqrt{\frac{g}{\gamma}}}\right)\right]_{v_0}^{0} = -t$

Evaluating the integral at the bounds gives us :

$\frac{1}{\sqrt{\gamma g}} \tan^{-1}\left(\frac{\sqrt{\gamma} v_0}{\sqrt{g}}\right) = t$

Therefore, the time taken by the ball to rise to its zenith, considering the drag force, is given by

$t = \frac{1}{\sqrt{\gamma g}} \tan^{-1}\left(\sqrt{\frac{\gamma}{g}} V_0\right)$

f_m = 100 MHz = 108 Hz, (V_m)₀ = 100 V

f_c = 300 GHz, (V_c)₀ = 400 V

$\therefore$ UBF – LBF = 2f_m = 2 × 10⁸ Hz

${E_i} = {1 \over 2}{I_I} \times \omega _1^2 + {1 \over 2}{I \over 2} \times {{\omega _1^2} \over 4}$

$= {{{I_1}\omega _1^2} \over 2}\left( {{9 \over 8}} \right) = {9 \over {16}}{I_1}\omega _1^2$

${I_1}{\omega _1} + {{{I_1}{\omega _1}} \over 4} = {{3{I_1}} \over 2}\omega ;{5 \over 4}{I_1}{\omega _1} = {{3{I_1}} \over 2}\omega$

$\omega = {5 \over 6}{\omega _1};{E_f} = {1 \over 2} \times {{3{I_1}} \over 2} \times {{25} \over {36}}\omega _1^2$

$= {{25} \over {48}}{I_1}\omega _1^2$

$\Rightarrow {E_f} - {E_i} = {I_1}\omega _1^2{{25} \over {49}} - {{ - 2} \over {48}}{I_2}\omega _1^2$

$= {{25} \over {48}}{I_1}\omega _1^2$

$\Rightarrow {E_f} - {E_i} = {I_1}\omega _1^2\left( {{{25} \over {48}} - {9 \over {16}}} \right) = {{ - 2} \over {48}}{I_1}\omega _1^2$

$= {{ - {I_1}\omega _1^2} \over {24}}$

2MV, cos 30° + Mv₂ cos 45° = 10 M cos 30° + 10 cos 45°

$\Rightarrow {v_1}\sqrt 3 + {{{v_2}} \over {\sqrt 2 }} = 5\sqrt 3 + 5\sqrt 2$   ...(i)

2MV, sin 30° – MV₂ sin 45° = –10 M sin 30° + 10 M sin 45°

${V_1} - {{{V_2}} \over {\sqrt 2 }} = - 5 + 5\sqrt 2$   ...(ii)

${V_1} = {{5\left( {\sqrt 3 - 1} \right) + 10\sqrt 2 } \over {\sqrt 3 + 1}} = {{17.5} \over {2.7}} = 6.5\,m/s$

V₂ = 6.3 m/s

U_i + K_i = U_f + K_f

${{k{q^2}} \over {\sqrt {16{a^2} + 9{a^2}} }} + {1 \over 2}m{v^2} = {{k{q^2}} \over {3a}}$

${1 \over 2}m{v^2} = {{k{q^2}} \over a}\left( {{1 \over 3} - {1 \over 5}} \right) = {{2k{q^2}} \over {15a}}$

$v = \sqrt {{{4k{q^2}} \over {15ma}}}$

$\therefore$ v = $\sqrt {{2 \over m}} {\left( {{{2{q^2}} \over {15 \times 4\pi {\varepsilon _0}a}}} \right)^{{1 \over 2}}}$

${I_{Disc}} = \int\limits_0^R {\left( {dm} \right)} {r^2} \Rightarrow {I_{Disc}} = \int\limits_0^R {\left( {\sigma 2\pi rdr} \right)} {r^2}$

${I_{Disc}} = \int\limits_0^R {\left( {k{r^2}2\pi rdr} \right)} {r^2}$    Mass of Disc

${I_{Disc}} = 2\pi k\int\limits_0^R {{r^2}dr} \,\,\,\,M = \int\limits_0^R {2\pi rdr\,k{r^2}}$

${I_{Disc}} = 2\pi k\left( {{{{r^6}} \over 6}} \right)_0^R\,\,\,\,M = 2\pi k\int\limits_0^R {{r^3}dr}$

${I_{Disc}} = 2\pi k{{{R^6}} \over 6} \,\,\,\,M = 2\pi k\left. {{{{r^4}} \over 4}} \right|_0^R$

${I_{Disc}} = {{\pi k{R^6}} \over 3} = \left( {{{\pi k{R^4}} \over 2}} \right){{{R^2}2} \over 3}\,\,\,M = 2\pi k\left. {{{{r^4}} \over 4}} \right|_0^R$

${I_{Disc}} = {{M2{R^2}} \over 3};{I_{Disc}} = {2 \over 3}M{R^2}$

$\Delta Q = n{C_v}$    $\Delta T = n{3 \over 2}R\Delta T$

$= \left( {{{67.2} \over {22.4}}} \right)\left( {{3 \over 2} \times 8.31} \right)\left( {20} \right)$

$\approx 748\,J$

$h = {{2{S_1}\cos \theta } \over {{r_1}{\rho _1}g}}$

$h = {{2{s_2}\cos {\theta _2}} \over {{r_2}{\rho _2}g}}$

$\Rightarrow {{{r_1}} \over {{r_2}}} = {2 \over 5}$

Optical Fibre Communication – Infrared Light

Radar – Radio Waves

Sonar – Ultrasound

Mobile Phones – Microwaves

V = 25 × 10^–3 m³, N = 1 mole of O₂
T = 300 K
V_rms = 200 m/s

$\because \lambda = {1 \over {\sqrt 2 N\pi {r^2}}}$

Average time ${1 \over \tau } = {{ < V > } \over \lambda } = 200\,.N\pi {r^2}.\sqrt 2$

$= {{\sqrt 2 \times 200 \times 6.023 \times {{10}^{23}}} \over {25 \times {{10}^{ - 3}}}}.\pi \times {10^{ - 18}} \times 0.09$

Average no. of collision $\approx {10^{10}}$

${{{1 \over {{T^2}}}} \over {{1 \over {T_0^2}}}} + {{\ln \,R\left( T \right)} \over {\ln \,R\left( {{T_o}} \right)}} = 1$

$\Rightarrow$ ln R(T) = ln R(T_o) $\left( {1 - {{T_o^2} \over {{T^2}}}} \right)$

$R(T) = {R_o}{e^{ - \left( {{{T_o^2} \over {{T^2}}}} \right)}}$

${K_{\max }} = {{hc} \over \lambda } - {{hc} \over {{\lambda _o}}}$

$\Rightarrow {K_{\max }} = hc\left( {{{{\lambda _o} - \lambda } \over {\lambda {\lambda _o}}}} \right)$

$\Rightarrow {K_{\max }} = \left( {1237} \right)\left( {{{380 - 260} \over {380 \times 260}}} \right)$ = 1.5 eV

${{GM} \over {{{\left( {R + h} \right)}^2}}} = {{GM} \over {2{R^2}}}$

$R + h = \sqrt 2 R$

$h = \left( {\sqrt 2 - 1} \right)R \simeq 2.6 \times {10^6}m$

$v = {{v + {v_o}} \over v}{v_o}$

$\Rightarrow {v_o} = \left( {{v \over {{v_o}}} - 1} \right)v$

${v_o} = \left( {{{530} \over {500}} - 1} \right)300 = 18\,m/s$

${v_o} = \left| {\left( {{{480} \over {500}} - 1} \right)300} \right| = 12\,m/s$

For the given circuit

Given that V_AB = 2 V

$\therefore$ I = ${2 \over {15}} + {2 \over {10}} = {1 \over 3}A$

Also I(2r + 2) = 1.5 + 1.5 – V_AB

$\Rightarrow$ 2r + 2 = (3–2)3

$\Rightarrow$ $r = {1 \over 2}\Omega$

Here you can see, H₂S₂O₇ does not have S – S linkage.

In this ${\left| \psi \right|^2}$ vs r graph, value of r is zero at only one point. So this orbital must have one radial node.

Here ${\left| \psi \right|^2} \ne 0$ at r = 0, so orbital should be s orbital. Then $l$ = 0.

We know,

No of radial node = n - $l$ - 1

$\therefore$ n - $l$ - 1 = 1

$\Rightarrow$ n - 0 - 1 = 1

$\Rightarrow$ n = 2

Therefore orbital is 2s.

A reaction is spontaneous if $\Delta$G is negative.

and we know that

$\Delta$G = $\Delta$H – T$\Delta$S

If $\Delta$H = –ve and $\Delta$S = + ve then at all the temperature the process will be spontaneous.

Physical adsorption increases with increase of pressure and decrease of temperature.

As in a co-ordination compound, the strong field ligand causes higher splitting of the d-orbitals

Also we know,

strength of ligand $\propto$ ${1 \over {{\lambda _{absorbed}}}}$

Order of strength of ligand

NH₃ > H₂O > Cl^-

Therefore decreasing order of wavelength absorbed is (I) > (II) > (III)

C_xH_y + $\left( {x + {y \over 4}} \right)$O₂ $\to$ xCO₂ + ${y \over 2}$H₂O

According to the above equation,

1 mL of hydrocarbon = x mL of CO₂ is produced

$\therefore$ 10 mL of hydrocarbon = 10x mL of CO₂ is produced.

According to question,

From 10 mL of hydrocarbon 40 mL of CO₂ is produced.

$\therefore$ 10x = 40

$\Rightarrow$ x = 4

Also from the above equation,

1 mL of C_xH_y react with $\left( {x + {y \over 4}} \right)$ mL of O₂.

$\therefore$ 10 mL of C_xH_y react with 10$\left( {x + {y \over 4}} \right)$ mL of O₂.

According to question,

10$\left( {x + {y \over 4}} \right)$ = 55

$\Rightarrow$ $40 + {{10y} \over 4} = 55$

$\Rightarrow$ y = 6

$\therefore$ Hydrocarbon is C₄H₆

(a) Equivalance of strong acid = 0.1 $\times$ 2 $\times$ 400 = 80

Equivalance of strong base = 0.1 $\times$ 400 = 40

$\therefore$ [H⁺] of mixture = ${{80 - 40} \over {800}}$ = ${1 \over {20}}$

$\therefore$ pH = $- \log \left[ {{H^ + }} \right]$ = $- \log \left( {{1 \over {20}}} \right)$ = 1.3

(b) Ionic product of water increases with increase of temperature because ionisation of water is endothermic.

(c)

k_a = ${{{{10}^{ - 5}} \times c\alpha } \over {c\left( {1 - \alpha } \right)}}$

$\Rightarrow$ 10^-5 = ${{{{10}^{ - 5}} \times \alpha } \over {\left( {1 - \alpha } \right)}}$

$\Rightarrow$ ${\alpha \over {\left( {1 - \alpha } \right)}}$ = 1

$\Rightarrow$ $\alpha$ = 0.5

$\therefore$ Degree of dissociation($\alpha$) = 50%

(d) The Le Chatelier's principle is always applicable to common-ion effect.

N-ethyl phthalimide on treatment with NH₂—NH₂ gives ethylamine.

It is the final step of Galbriel phatalimide synthesis reaction.

H₂O(g) + C $\to$ CO + H₂

The gaseous mixture CO + H₂ which is produced when steam is passed over red hot coke and this mixture called water gas or synthesis gas or syn gas.

The rate of S_N1 is decided by the stability of carbocation which is formed when leaving group I(Iodine) leave from the compound.

Carbocation (D) is most stable due to +R effect of CH₃O group.

Carbocation (C) is stabilised by +I and +H effect of CH₃ group.

Carbocation (B) is least stable due to –I effect of MeO group.

CH₃ group increases the electron density of benzene by +I effect and hyper conjugation effects and hence makes the compound more reactive towards electrophilic aromatic substitution reaction.

Cl group decreases the electron density of benzene by –I effect.

COCH₃ group strongly decreases the electron density of benzene by –I and –R effects.

$\therefore$ Correct increasing order is

The lowest region of atmosphere in which human beings live is troposphere. It extends up to a height of 10 km from sea level. Clouds are also formed in this layer

As we know that

$\mu = \sqrt {n\left( {n + 2} \right)}$ where n = no. of unpaired electrons i.e. greater the no. of unpaired electron more will be the spin-only magnetic moments.

. Electronic configuration of the given transition metal ions are

Sc³⁺ (Z = 21) = 1s²2s²2p⁶3s²3p⁶

$\therefore$ n = 0

Ti²⁺ (Z = 22) = 1s²2s²2p⁶3s²3p⁶3d²

$\therefore$ n = 2

Ti³⁺ (Z = 22) = 1s²2s²2p⁶3s²3p⁶3d¹

$\therefore$ n = 1

V²⁺ (Z = 23) = 1s²2s²2p⁶3s²3p⁶3d³

$\therefore$ n = 3

So the correct increasing order of magnetic moment is

Sc³⁺ < Ti³⁺ < Ti²⁺ < V²⁺