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

$\phi$ = NBA = LI

N $\mu$₀ nI$\pi$R² = LI

$N{\mu _0}{N \over l}l\pi {R^2} = LI$

N and R constant

Self inductance (L) $\propto {1 \over l} \propto {1 \over {length}}$

Height of liquid rise in capillary tube $h = {{2T\,\cos {\theta _c}} \over {\rho rg}}$

$\Rightarrow h \propto {1 \over r}$

When radius becomes double height become half

$\therefore$ ${h^{'}} = {h \over 2}$
Now, M = $\pi$r²h × $\rho$ and M^' = $\pi$(2r)² (h/2) × $\rho$ = 2M

Mass (m) = (10.00 ± 0.10) kg

Edge length ($l$) = (0.10 ± 0.01) m

Volume of the cube (V) = ${l^3}$

Density, $\rho$ = ${m \over V}$

${{d\rho } \over \rho } = {{dm} \over m} + {{dV} \over V}$

$\Rightarrow$ ${{d\rho } \over \rho } = {{dm} \over m} + 3{{dl} \over l}$

$\Rightarrow$ ${{d\rho } \over \rho } = {{0.10} \over {10.00}} + 3{{0.01} \over {0.10}}$ = 0.31

$\left| {\overrightarrow \tau } \right| = \left| {\overline M \times \overline B } \right|$

$\tau = NI \times A \times B \times \sin {45^o}$

$\tau = 0.27 \,Nm$

Work done = $\Delta$U

= U_f – U_i

$= {{{q^2}} \over {2{C_r}}} - {{{q^2}} \over {2{C_i}}}$

$= {{{{\left( {5 \times {{10}^{ - 6}}} \right)}^2}} \over 2}.\left( {{1 \over {2 \times {{10}^{ - 6}}}} - {1 \over {5 \times {{10}^{ - 6}}}}} \right)$

$= {{15} \over 4} \times {10^{ - 6}} = 3.75{\rm{ }} \times {\rm{ }}{10^{-6}}{\rm{ }}J$

Draw velocity diagram
$\sin \theta = {{{v_r}} \over {{v_{sr}}}} = {1 \over 2}$

$\theta = {30^o}$

$\phi$ = 90 + $\theta$ = 120°

Input current = 15 × 10^–6

Output current = 3 × 10^–3

Resistance out put = 1000

V_input = 10 × 10^–3

Now V_input = r_input × i_input

10 × 10^–3 = r_input × 15 × 10^–6

r_input = ${{2000} \over 3} = 0.67\,K\Omega$

Voltage gain = ${{{V_{output}}} \over {{V_{input}}}} = {{1000 \times 3 \times {{10}^{ - 3}}} \over {10 \times {{10}^{ - 3}}}} = 300$

Maximum electric field E = (B) (C)

$\overrightarrow {{E_0}} = \left( {3 \times {{10}^{ - 5}}} \right)c\left( { - \widehat j} \right)$

$\overrightarrow {{E_1}} = \left( {2 \times {{10}^{ - 6}}} \right)c\left( { - \widehat i} \right)$

Maximum force

${\overrightarrow F _{net}} = {10^{ - 4}} \times 3 \times {10^8}\sqrt {{{\left( {3 \times {{10}^{ - 5}}} \right)}^2} + {{\left( {2 \times {{10}^{ - 6}}} \right)}^2}} = 0.9N$

${F_{rms}} = {{{F_0}} \over {\sqrt 2 }} = 0.6\,N$ (approx)

$A = \left( {{v_0} + A\cos \,\omega t} \right)\sin {\omega _o}t$

$\Rightarrow {v_0}\sin \left( {{\omega _o}t} \right) + {A \over 2}\left[ {\sin \left( {{\omega _o} - \omega } \right)t + \sin \left( {{\omega _o} + \omega } \right)t} \right]$

${1 \over {{\lambda _1}}} = R\left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right]$

${1 \over {{\lambda _2}}} = R\left[ {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right]$

${{5{\lambda _1}} \over {36}} = {{12{\lambda _2}} \over {4 \times 16}}$

${\lambda _2} = {{5 \times 660 \times 64} \over {36 \times 12}} = 489\,nm$

$\omega = 6 \times {10^{14}} \times 2\pi$
f = 6 × 10¹⁴
C = f $\lambda$

$\lambda = {C \over f} = {{3 \times {{10}^8}} \over {6 \times {{10}^{14}}}} = 5000$ Å
Energy of photon $\Rightarrow {{12375} \over {5000}} = 2.475\,eV$

From Einstein’s equation
KEmax = E – $\phi$
eV_s = E – $\phi$
eV_s = 2.475 – 2
eV_o = 0.475 eV
V_o = 0.48 V

Total kinetic energy of a rolling body is given as

$E_{\text {total }}=\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right]$

where, $K$ is the radius of gyration.

Using conservation law of energy,

$$\begin{array}{rlrl} \frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right] =m g h \\ \text { or } h =\frac{v^2}{2 g}\left[1+\frac{K^2}{R^2}\right] \end{array}$$

For ring, $\frac{K^2}{R^2}=1$

$\Rightarrow h_1=\frac{v^2}{2 g}[1+1]=\frac{2 v^2}{2 g}=\frac{v^2}{g}$

For solid cylinder, $\frac{K^2}{R^2}=\frac{(R / 2 \sqrt{2})^2}{(R / 2)^2}$

$$\begin{aligned} & =\frac{R^2}{8} \times \frac{4}{R^2}=\frac{1}{2} \\ \Rightarrow h_2 & =\frac{v^2}{2 g}\left[1+\frac{1}{2}\right]=\frac{3 v^2}{4 g} \end{aligned}$$

For solid sphere, $\frac{K^2}{R^2}=\frac{2}{5}$

$\Rightarrow h_3=\frac{v^2}{2 g}\left[1+\frac{2}{5}\right]=\frac{7 v^2}{10 g}$

So,the ratio of $h_1, h_2$ and $h_3$ is

$$\begin{aligned} h_1: h_2: h_3 & =\frac{v^2}{g}: \frac{3 v^2}{4 g}: \frac{7}{10} \frac{v^2}{g} \\ & =1: \frac{3}{4}: \frac{7}{10}=20: 15: 14 \end{aligned}$$

We use Gauss’s Law for gravitation

g.4$\pi$r² = (Mass enclosed) 4$\pi$G

$g = {{3M4\pi G} \over {4\pi {{(3a)}^2}}} = {{GM} \over {3{a^2}}}$

For a simple pendulum T = $2\pi \sqrt {{L \over {{g_{err}}}}}$

Situation 1: when pendulum is in air $\to$ g_eff = g
Situation 2:when pendulum is in liquid
$\to$ g_eff = $\left( {1 - {{{\rho _{liquid}}} \over {{\rho _{body}}}}} \right) = g\left( {1 - {1 \over {16}}} \right) = {{15g} \over {16}}$

So, ${{{T^{'}}} \over T} = {{2\pi \sqrt {{L \over {15g/16}}} } \over {2\pi \sqrt {{L \over g}} }}$

$\Rightarrow {T^{'}} = {{4T} \over {\sqrt {15} }}$

$m = {f \over {f - u}}$

$5 = {{ - 40} \over { - 40 - u}};\,u = - 32cm$

Speed of wave from wave equation

$v = - {{\left( {coeffecient{\rm{ }}of{\rm{ }}t} \right)} \over {\left( {coeffecient{\rm{ }}of{\rm{ }}x} \right){\rm{ }}}}$

$v = - {{1000} \over {( - 3)}} = {{1000} \over 3}$

Since speed of wave $\propto \sqrt T$

So $= {{1000} \over {{3 \over {336}}}} = \sqrt {{{273} \over T}}$

$\Rightarrow$ T = 277.41 K

T = 4.41°C

${V_{rms}} = \sqrt {{{3RT} \over {{M_w}}}}$

$\Rightarrow {V_{rms}} \propto \sqrt T$

Now, ${v \over {200}} = \sqrt {{{500} \over {400}}}$

$\Rightarrow {v \over {200}} = {{\sqrt 5 } \over 2}$

$\Rightarrow v = 100\sqrt 5$ m/s

G = 50 $\Omega$

S = 5000 $\Omega$

I_g = 4 × 10^–3

V = i_g (G + S)

V = 4 × 10^–3 (50 + 5000)

= 4 × 10^–3 (5050) = 20.2 volt

Assuming potential x and y at the top junctions A and B respectively and zero potential at lower junctions.

At the junction A applying KCL,

i₁ + i₂ + i₃ = 0

${{x - 72} \over 6} + {x \over 4} + {{x - y} \over 2} = 0$

$\Rightarrow$ ${{2x - 144 + 6x - 6y + 3x} \over {12}} = 0$

$\Rightarrow$ 11x - 6y = 144 ......(1)

At the junction B applying KCL,

i₄ + i₅ = 0

${{y - x} \over 2} + {y \over {10}} = 0$

$\Rightarrow$ ${{5y - 5x + y} \over {10}}$ = 0

$\Rightarrow$ 6y = 5x ......(2)

Solving (1) and (2)

11x - 5x = 144

$\Rightarrow$ x = 24 V

$\therefore$ y = 20 V

The charge on the capacitor, q = CV

$\Rightarrow$ q = 10$\mu$F × 20 = 200 $\mu$C

By conservation of linear momentum:

$2{v_0} = 2\left( {{{{v_0}} \over 4}} \right) + mv \Rightarrow 2{v_0} = {{{v_0}} \over 2} + mv$

$\Rightarrow {{3{v_0}} \over 2} = mv\,\,...(1)$

Since collision is elastic

${V_{separation}} = {V_{approch}}$

$\Rightarrow v - {{{v_0}} \over 4} = {v_0} \Rightarrow m = {6 \over 5} = 1.2\,kg$

F₃ & F₄ cancel each other.

Force on PQ will be F₁ = I₂B₁ a

= ${I_2}{{{\mu _0}{I_1}} \over {2\pi a}}a$

= ${{{\mu _0}{I_1}} \over {2\pi a}}a = {{{\mu _0}{I_1}{I_2}} \over {2\pi }}$

Force on RS will be F₂ = I₂ B₂ a

= ${I_2}{{{\mu _0}{I_1}} \over {2\pi 2a}}a = {{{\mu _0}{I_1}{I_2}} \over {4\pi }}$

Net force = F₁ – F₂ = ${{{\mu _0}{I_1}{I_2}} \over {4\pi }}$ repulsion

To solve this problem, we need to determine the work done to lift the hanging part of the cable up to the surface. The work done lifting a small element of the cable will be the weight of the element times the distance it has to be lifted.

Let's take a small section of the cable at a depth $x$ below the surface. This section has a length of $dx$, so its mass is $(M/L)dx$ where $(M/L)$ is the linear mass density of the cable.

The work $dW$ done to lift this small section up to the surface is the weight of the section times the distance it has to be lifted :

$dW = (M/L)gdx \times x$.

Integrating this expression from 0 to L/n (the length of the hanging part of the cable) gives the total work done :

$W = \int\limits_{0}^{L/n} (M/L)gxdx$

$= (Mg/L) \int\limits_{0}^{L/n} xdx$

$= (Mg/L) \times [x^2/2]_{0}^{L/n}$

$= (Mg/L) \times [L^2/(2n^2)]$

$= MgL/(2n^2)$

So the work done to lift the hanging part of the cable up to the surface is $MgL/(2n^2)$.

Therefore, the correct answer is Option C :

$\frac{MgL}{2n^2}$

An HCl molecule, being diatomic, has:

• 3 translational degrees of freedom
• 2 rotational degrees of freedom
• 2 vibrational degrees of freedom

The total number of degrees of freedom is $3 + 2 + 2 = 7$.

According to the equipartition theorem, each degree of freedom contributes $\frac{1}{2} k_B T$ to the total energy. So the total energy is given by: $\frac{7}{2} k_B T$

The translational kinetic energy is related to the root-mean-square (rms) speed $\overline{v}$ by: $\frac{1}{2} m \overline{v}^2 = \frac{7}{2} k_B T$

Rearranging to solve for the temperature, we find: $T = \frac{m \overline{v}^2}{7k_B}$

Kinetic energy KE = ${1 \over 2}l{\omega ^2} = k{\theta ^2}$

$\Rightarrow {\omega ^2} = {{2k{\theta ^2}} \over l} \Rightarrow \omega = \sqrt {{{2k} \over l}} \theta$ .... (A)

Differentiate (A) wrt time $\to$

${{d\omega } \over {dt}} = \alpha = \sqrt {{{2k} \over l}} \left( {{{d\theta } \over {dt}}} \right)$

$\Rightarrow \alpha = \sqrt {{{2k} \over l}} .\sqrt {{{2k} \over l}} \theta \,\{ by\,(1)\}$

$\Rightarrow \alpha = {{2k} \over l}\theta \,$

Momentum p = mv    …(1)

and for motion under gravity $h = {{{u^2} - {v^2}} \over {2g}}$ ...(2)

$h = {{{u^2} - {p^2}/m} \over {2g}}$

U_total = U_{self of dipole} + U_interaction

= $- {{k{q^2}} \over d} - \left( {{{kQ} \over {{D^2}}}} \right)qd$

$= - k\left[ {{{{q^2}} \over d} + {{qQd} \over {{D^2}}}} \right]$

Initial and final states for both the processes are same,
$\therefore$ $\Delta$U_A = $\Delta$U_B

Work done during process A is greater than in process B. Because area is more By First law of thermodynamics
$\Delta$Q = $\Delta$U + W
$\Rightarrow$ $\Delta$Q_A > $\Delta$Q_B

4^th harmonic

$4{\lambda \over 2} = l;2\lambda = l$

From equation ${{2\pi } \over \lambda } = 0.157$

$\lambda$ = 40 ; $l$ = 2$\lambda$ = 80 m

${1 \over {{R_{eq}}}} = {8 \over {7R}} + {8 \over R}$

${1 \over {{R_{eq}}}} = {{8 + 56} \over {7R}};{R_{eq}} = {{7R} \over {64}}$

Path difference at central maxima $\Delta$x = ($\mu$ – 1)t, whole pattern will shift by same amount which will be given by

$\left( {\mu - 1} \right)t{D \over d} = n{{\lambda D} \over d};$ according to the question $t = {{n\lambda } \over {\left( {\mu - 1} \right)}}$

C₂ has 12 electrons.

Moleculer orbital configuration of C₂ is

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

Bond order of C₂ = ${{8 - 4} \over 2}$ = 2

C₂^- has 13 electrons.

Moleculer orbital configuration of C₂^- is

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

Bond order of C₂^- = ${{9 - 4} \over 2}$ = 2.5

As we know, higher is the bond order more will be the stability of the molecule.

So we can say, stability of C₂ increases when it forms C₂^- ion.

Mg burn in air and produces a mixture of nitride and oxide.

So, Mg₃N₂ and MgO are formed.

$P_M^0$ = 450 mmHg and $P_N^0$ = 700 mmHg

$\therefore$ $P_M^0$ < $P_N^0$

Also we know,

$P_M$ = $P_M^0$$X_M$ = $Y_M$$P_T$

$\Rightarrow$ $P_M^0$ = ${{{Y_M}} \over {{X_M}}}\left( {{P_T}} \right)$

$P_N$ = $P_N^0$$X_N$ = $Y_N$$P_T$

$\Rightarrow$ $P_N^0$ = ${{{Y_N}} \over {{X_N}}}\left( {{P_T}} \right)$

As, $P_M^0$ < $P_N^0$

$\Rightarrow$ ${{{Y_M}} \over {{X_M}}}\left( {{P_T}} \right)$ < ${{{Y_N}} \over {{X_N}}}\left( {{P_T}} \right)$

$\Rightarrow$ ${{{Y_M}} \over {{X_M}}}$ < ${{{Y_N}} \over {{X_N}}}$

$\Rightarrow$ ${{{Y_M}} \over {{Y_N}}}$ < ${{{X_M}} \over {{X_N}}}$

Fullerene (C₆₀) contains 20 hexagons and 12 pentagons.

We know,

$\overline v = R{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$

$\therefore$ For Lyman series,

${\Delta {{\overline v }_{Lyman}}}$ = ${\Delta {{\overline v }_{\max }} - \Delta {{\overline v }_{\min }}}$

= $\left[ {{1 \over 1} - {1 \over \infty }} \right] - \left[ {{1 \over 1} - {1 \over 4}} \right]$

= ${{1 \over 4}}$

$\therefore$ For Balmer series,

${\Delta {{\overline v }_{Balmer}}}$ = ${\Delta {{\overline v }_{\max }} - \Delta {{\overline v }_{\min }}}$

= $\left[ {{1 \over 4} - {1 \over \infty }} \right] - \left[ {{1 \over 4} - {1 \over 9}} \right]$

= ${{1 \over 9}}$

$\therefore$ ${{\Delta {{\overline v }_{Lyman}}} \over {\Delta {{\overline v }_{Balmer}}}}$ = ${{{1 \over 4}} \over {{1 \over 9}}}$ = ${{9 \over 4}}$

For 1^st order reaction, we know

$\ln {{{R_0}} \over {{R_T}}} = kt$

$\Rightarrow$ $\ln {R_0}$ - $\ln {R_t}$ = kt

$\Rightarrow$ $\ln {R_t}$ = - kt + $\ln {R_0}$

So, $\ln {R_t}$ vs t is a straight line with negative slope.

$\therefore$ First graph represent 1^st order reaction.

For Zeroorder reaction, we know

R₀ - R_t = kt

$\therefore$ R_t = - kt + R₀

So, ${R_t}$ vs t is a straight line with negative slope.

$\therefore$ Second graph represent zero order reaction.

Electronic configuration of Potassium (K) :

1s²2s²2p⁶3s²3p⁶4s¹

After first ionisation enthalpy its configuration becomes :

1s²2s²2p⁶3s²3p⁶

which is a inert gas configuration. To remove an electron from inert gas configuration requires very high energy.

So Potassium (K) have high difference in the first ionisation and the second ionisation energy as it achieve stable noble gas configuration after first ionisation.

The glycosidic linkage is present between C₁ of $\alpha$-D-glucose and C₂ of $\beta$-D-fructose.

We know,

Critical temperature (T_c) = ${{8a} \over {27bR}}$

$\therefore$ T_c $\propto$ ${a \over b}$

So, species with greatest value of ${a \over b}$ has greatest value of critical temperature.

Accordig to the given data Kr have highest ${a \over b}$ ratio.

(A) q + w = $\Delta$E, state function

(B) q, Path function

(C) w, Path function

(D) H – TS = G, State function

Here you have to check every options to find answer.

N₂(g) + 3H₂(g) $\to$ 2NH₃(g)

For dihydrogen (H₂) to become a limiting reagent,

${{moles\,of\,{H_2}} \over {stoichiometric\,coefficient\,of\,{H_2}}}$ should be

less than ${{moles\,of\,{N_2}} \over {stoichiometric\,coefficient\,of\,{N_2}}}$.

Here in 56g of N₂ + 10g of H₂,

Moles of N₂ = ${{56} \over {28}}$ = 2

and Moles of H₂ = ${{10} \over 2}$ = 5

Now by dividing stoichiometric coefficient we get,

For N₂ = ${2 \over 1}$ = 2

For H₂ = ${5 \over 3}$ = 1.67

$\therefore$ Here H₂ is the limiting reagent.