JEE Main 2017 9th April Morning Shift Question Paper with Solutions

Before introducing the slab between the plates of all three capacitors, we have

${C_1} = {{{\varepsilon _0}A} \over 3}$

After introducing the slab between the plates of all three capacitors, we have

${C_1} = K{{{\varepsilon _0}A} \over 3} + {{{\varepsilon _0}A} \over {2.4}}$

As the capacitors are in series, we get

${{{\varepsilon _0}A} \over 3} = {{\left( {{{K{\varepsilon _0}A} \over 3}} \right)\,.\,\left( {{{{\varepsilon _0}A} \over {2.4}}} \right)} \over {\left( {{{K{\varepsilon _0}A} \over 3}} \right) + \left( {{{{\varepsilon _0}A} \over {2.4}}} \right)}}$

Therefore, the dielectric constant of the slab is

$\Rightarrow 3K = 2.4K + 3$

$\Rightarrow 0.6K = 3$

$\Rightarrow K = {3 \over {0.6}} = {{30} \over 6} = 5$

We know that for a linear antenna of length l, (i) it's length of antenna is comparable to the wavelength of the signal and (ii) the power radiated is proportional to 1/$\lambda$². That is, the power radiated for the antenna increases with decreasing $\lambda$ thereby increasing the frequency.

Power of the antenna $P = \mu {\left( {{1 \over l}} \right)^2}$

Here, $\mu$ = K; hence, the correct option is (D).

The power can be calculated by the relation

$P = {E \over {\Delta t}} = {{\Delta m{c^2}} \over {\Delta t}}$ ...... (1)

Therefore, from Eq. (1), the mass of the fuel consumed per hour in the reactor is

${{\Delta m} \over {\Delta t}} = {P \over {{c^2}}} = {{{{10}^9}} \over {{{(3 \times {{10}^8})}^2}}} = 4 \times {10^{ - 12}}$ g

The slit width is a = 0.1 mm = 10^$-$4 m.

The wavelength of the light is $\lambda$ = 6000 $\times$ 10^$-$10 = 6 $\times$ 10^$-$7.

The distance from the slit to diffraction bands is D = 0.5 m.

We calculate the third dark band from the central band as follows:

$A\sin \theta = 3\lambda \Rightarrow \sin \theta = {{3\lambda } \over a} = {x \over D}$

$\Rightarrow x = {{3\lambda D} \over a} = {{3 \times 6 \times {{10}^{ - 7}} \times 0.5} \over {{{10}^{ - 4}}}} = 9$ mm

The given optical situation is depicted in the following ray diagram:

In this case, the image forms 55 cm behind the convex mirror and then the reflection takes place due to the mirror image that forms at a distance (v) behind the mirror. Now, this image acts as an object for lens and the final image forms at a distance of 20 m from the lens if the focal length of convex mirror is as follows:

${{55} \over 2} = 27.5$ cm

The power of the given laser light is expressed as

$P = {{nhc} \over {\lambda t}}$

from which the number of photons per second is given by

$n = P\left( {{{\lambda t} \over {hc}}} \right) = (5 \times {10^2}) \times \left[ {{{(660 \times {{10}^{ - 9}})(60 \times {{10}^{ - 3}})} \over {(6.6 \times {{10}^{ - 34}})(3 \times {{10}^8})}}} \right]$

$= 100 \times {10^{18}} = {10^{20}}$

For a damped oscillator, we have

$E' = {1 \over 2};b = {\lambda \over m} \Rightarrow \lambda = bm;A = {A_0}{e^{ - bt}}$

Therefore, $a' = {a \over {\sqrt 2 }};b = {1 \over {10}},{1 \over {\sqrt 2 }} = {e^{ - t/10}} \Rightarrow \sqrt 2 = {e^{1/10}}$

Therefore, the time taken for its mechanical energy of vibration to drop to half of its initial value is closest to

$\ln \sqrt 2 = {t \over {10}} \Rightarrow t = 3.5\,s$

We have

${{dE} \over {dz}} = {{ - dE} \over {dt}}$

${{dE} \over {dz}} = - 2{E_0}k\sin kz\cos \omega t = {{ - dB} \over {dt}}$

Therefore, $dB = + 2{E_0}k\sin kz\cos \omega t\,dt$

That is, $B = + 20{E_0}k\sin {k_z}\cos \omega t\,dt$

$B = + 2{E_0}k\sin {k_2}\int {\cos \omega t\,dt = + 2{E_0}{k \over \omega }\sin kz\sin \omega t}$

Now, ${{{E_0}} \over {{B_0}}} = {\omega \over k} = c$

Therefore, its magnetic field $\overrightarrow B$ is given by $B = {{2{E_0}} \over c}\widehat j\sin kz\sin \omega t$

It is given that e₀ = 283 V; $\omega$ = 320.

The inductor reactance is X_L = 320 $\times$ 25 $\times$ 10^$-$3 = 8 $\Omega$

The capacitor reactance is

${X_C} = {1 \over {\omega C}} = {1 \over {320 \times 1000 \times {{10}^{ - 6}}}} = {{1000} \over {320}} = 3.1\,\Omega$

It is given that R = 5 $\Omega$. Therefore, the total impedance is

$Z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} = \sqrt {50} = 7\,\Omega$

and the phase difference between the voltage across the source and the current is

$\tan \phi = {{{X_L} - {X_C}} \over R} = {{8 - 3.1} \over 5} \approx 1 \Rightarrow \theta = 45^\circ$

Initially, the resistance P = 4 $\Omega$ and the neutral point N is at 60 cm from A. Therefore,

${4 \over {60}} = {Q \over {40}} \Rightarrow Q = {{160} \over {60}} = \left( {{8 \over 3}} \right)\Omega$

When an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A, we have

${{4 + R} \over {80}} = {Q \over {20}}$

$\Rightarrow {{4 + R} \over {80}} = {8 \over {60}} \Rightarrow 4 + R = {{64} \over 6}$

$\Rightarrow R = {{64} \over 6} - 4 = {{64 - 24} \over 6} = {{40} \over 6}$

$\Rightarrow R = \left( {{{20} \over 3}} \right)\Omega$

The time period is given by the formula

$T = 2\pi \sqrt {{l \over g}}$

which clearly indicates that the time period is directly proportional to the length of the pendulum l, that is,

$T \propto \sqrt l$

Here, l = 1 m. Therefore,

${{\Delta T} \over l} = {1 \over 2}{{\Delta l} \over l}$ ....... (1)

where $\Delta l = \left| {{r_1} - {r_2}} \right|$. Substituting the values in Eq. (1), we get

$5 \times {10^{ - 4}} = {1 \over 2}\left( {{{{r_1} - {r_2}} \over 1}} \right)$

$\Rightarrow {r_1} - {r_2} = 10 \times {10^{ - 4}} = {10^{ - 3}}$ m $= {10^{ - 1}}$ cm $= 0.1$ cm

Given situation is shown in the figure

As we know,

$\overrightarrow F = q(\overrightarrow v \times \overrightarrow B )$

$\overrightarrow F = - {q_0}(\overrightarrow v \times \overrightarrow B )$

According to question, direction of current is parallel to the force acting on the electron. Hence, the motion of test charge is towards the wire.

We have the following cases :

$\bullet$ For configuration (I) : Considering as Wheatstone's bridge, we get the equivalent resistance as ${({R_I})_{eq}} = 1\,\Omega$.

$\bullet$ For configuration (II) : The given configuration is further reduced as follows:

That is, ${({R_{II}})_{eq}} = {1 \over 2}\,\Omega$.

$\bullet$ For configuration (III) : Similarly, we get ${({R_{III}})_{eq}} = 2\,\Omega$.

The power is given by

$P = {{{V^2}} \over R}$

Since V is the same for all three configurations, we have the power as a quantity indirectly proportional to the resistance :

$P \propto {1 \over R}$

From the above three case, we have ${({R_I})_{eq}} = 1\,\Omega ;\,{({R_{II}})_{eq}} = {1 \over 2}\,\Omega ;\,{({R_{III}})_{eq}} = 2\,\Omega$, which implies that

${R_{II}} > {R_I} > {R_{III}}$

Therefore, ${P_2} > {P_2} > {P_3}$

Concept:
The potential energy $U$ of a current-carrying magnetic dipole in a uniform magnetic field $\vec{B}$ is given by:

For stable equilibrium, the potential energy $U$ must be at its minimum value, which occurs when $\cos\theta = 1$, i.e., $\theta = 0^{\circ}$.

This means the magnetic dipole moment vector $\vec{M}$ must be parallel to the external magnetic field $\vec{B}$ (along the positive $Z$-direction, $+k$).

By the right-hand grip rule, for $\vec{M}$ to point along the $+Z$-axis, the current in the rectangular loop in the $XY$-plane must flow in the counter-clockwise direction.

This orientation corresponds to option (C).

Correct Option: (C)

The speedy of the particle in the orbit of an atom is $v = {{{I^2}} \over {2h{\varepsilon _0}}}$

We have the radius of the first orbit is

$r = {{{h^2}{\varepsilon _0}} \over {\pi m{e^2}}}$

$\Rightarrow {\varepsilon _0} = {{r\pi m{e^2}} \over {{h^2}}}$

Therefore, the acceleration of an electron in the first orbit of the hydrogen atom (n = 1) is

${{{v^2}} \over r} = {{{I^6}\pi m} \over {4{h^2}\varepsilon _0^3}} = {{{h^2}} \over {4{r^3}{\pi ^2}{m^2}}}$

For inside the spherical body, we have

${m \over v} = {k \over r}$

$\Rightarrow m = {{kv} \over r}$

Therefore, $g = {{Gmr} \over {{R^3}}} = {G \over {{R^3}}}\left( {{{kv} \over r}} \right)r$ = constant

For inside the spherical body, we have

${g_{out}} = {{GM} \over {{R^2}}}$

Therefore, the correct graph that describes qualitatively the acceleration, a, of a test particle as a function of r is the one depicted in option (c).

In an isothermal expansion process, the temperature of the system remains constant throughout the process. Since, for an ideal gas, U depends only on temperature, we have $\Delta$U = 0. The enthalpy change of the system in isothermal expansion is also zero as

$\Delta$H = $\Delta$U + nR$\Delta$T = 0 + 0 = 0

The entropy change foe an isothermal process is given by

$\Delta S = nR\ln \left( {{{{V_f}} \over {{V_i}}}} \right) \ge 0$

Among the given compounds, cyclopentadiene is most reactive to an aqueous solution of sodium carbonate. It forms cyclopentadienyl anion which is aromatic and therefore, highly stable.

According to Arrhenius equation,

$\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{{{T_2} - {T_1}} \over {{T_1}{T_2}}}} \right)$

Substituting the given values in the equation, we get

$\log 4 = {{{E_a}} \over {2.303 \times 8.314\,J\,{K^{ - 1}}mo{l^{ - 1}}}}\left( {{{310 - 300} \over {300 \times 310}}} \right)$

${E_a} = {{0.602 \times 2.303 \times 8.314 \times 300 \times 310} \over {10}}$

= 107197.12 J/mol^$-$1 or 107.2 kJ/mol^$-$1

The basicity of an amine depends on the ability of its nitrogen atom to donate lone pair of electrons in an acid-base reaction. Resonance, inductive effect and steric hindrance etc. plays a very important role in deciding the basicity of compound.

Among the given compounds, (III) is most basic due to the presence of electron donating methyl group which increases the electron density on nitrogen. Compound (IV) is less basic than (III). Hybridisation of nitrogen atom in aniline (I) and pyrrole (II) is sp³ and sp² , respectively, therefore, (I) is more basic than (II). Thus, the correct order of basicity is

Addition of a solid component to a system at constant pressure has no effect on the equilibrium. Therefore, addition of Fe₂O₃ will not disturb the equilibrium.

Mole fractions can be calculated as

${n_{C{H_2}C{l_2}}} = {{8.5} \over {8.5}} = 0.1$ and ${n_{CHC{l_3}}} = {{11.9} \over {119.5}} = 0.1$

We know that

${p_{total}} = p_{C{H_2}C{l_2}}^o \times \,{x_{C{H_2}C{l_2}}} + p_{CHC{l_3}}^o \times \,{x_{CHC{l_3}}}$