JEE Advanced 2016 Paper 2 Question Paper with Solutions

As the balls keep on oscillating between plates, current will flow even in steady state.

Let a ball attains charge q at bottom plate. Then

${{Kq} \over r} = {V_0} \Rightarrow q = {{{V_0}r} \over K}$

Electric field inside the chamber, $E = [{V_0} - ( - {V_0})]/h$

$= 2{V_0}/h$

$\therefore$ Acceleration of each ball, $a = {{qE} \over m} = {{2V_0^2r} \over {mhK}}$

$\therefore$ Time taken by a ball to reach the other plate

$t = \sqrt {{{2h} \over a}} = {h \over {{V_0}}}\sqrt {{{mK} \over r}}$

$\therefore$ Average current due to n such balls,

${I_{av}} = {{nq} \over t} = {{n{V_0}r} \over K} \times {{{V_0}} \over h}\sqrt {{r \over {mK}}} \Rightarrow {I_{av}} \propto V_0^2$

We discuss the following cases :

$\bullet$ For x = 0 $\to$ L : The induced emf in the loop is

$\varepsilon = v{B_0}L$

Therefore, the induced current is

$I = {{v{B_0}L} \over R}$ ........ (1)

As the flux is increasing, the current passes in counterclockwise direction and oppose the existing magnetic field (i.e., Lenz's law).

Let us consider the loop which is labelled as shown in the following figure:

Force on the side CD of the loop is

$\overrightarrow F = i(\overrightarrow L \times \overrightarrow B ) = {{v{B_0}L} \over R}[L\widehat j \times {B_0}( - \widehat k)]$

$\overrightarrow F = - {{vB_0^2{L^2}} \over R}\widehat i$ (towards left) ........ (2)

$\Rightarrow {{mdv} \over {dt}} = - {{B_0^2{L^2}} \over R}v$

$\Rightarrow {{mdv} \over {dx}}.{{dx} \over {dt}} = - {{ - B_0^2{L^2}} \over R}v$

$\Rightarrow \int\limits_{{v_0}}^v {dv = - {{B_0^2{L^2}} \over {mR}}\int\limits_0^x {dx} }$ (Since ${{dx} \over {dt}} = v$)

$\Rightarrow v = {v_0} - {{B_0^2{L^2}} \over {mR}}x$ ....... (3)

From Eqs. (1) and (3), we get

$I = {{{v_0}{B_0}L} \over R} - {{B_0^3{L^3}} \over {m{R^2}}}x$ ....... (4)

From Eqs. (2) and (3), we get

$\overrightarrow F = - {{{v_0}B_0^2{L^2}} \over R} + {{B_0^4{L^4}} \over {m{R^2}}}x$

$\bullet$ For x = L $\to$ 3L: Here, there is no change in flux and hence there is no induced current and thereby there is no force and no change in velocity.

$\bullet$ For x = 3L $\to$ 4L :

As the flux is decreasing, the current passes in clockwise direction and supports the existing magnetic field (i.e., Lenz's law).

Hence, option (A) is incorrect.

Consequently, the magnetic force experienced by the arm AB of the loop is

$\overrightarrow F = i(\overrightarrow L \times \overrightarrow B ) = {{v{B_0}L} \over R}[L\widehat j \times {B_0}( - \widehat k)]$

$= - {{vB_0^2{L^2}} \over R}\widehat i$ (towards left)

Hence, option (B) is incorrect.

Similarly, now we have

$\int\limits_{{v_{3L}}}^v {dv = - {{B_0^2{L^2}} \over {mR}}\int\limits_{3L}^x {dx \Rightarrow v = {v_{3L}} - {{B_0^2{L^2}} \over {mR}}\left( {x - 3L} \right)} }$

When the loop is completely within the magnetic field, the force and current is zero and velocity is constant. However, while coming out, the current passes in clockwise direction, the force exists in backwards and the velocity decreases linearly.

In photo-electric effect, the maximum kinetic energy of the photo-electron ejected at the cathode, is given by

${K_{\max ,\,c}} = {{hc} \over {{\lambda _{ph}}}} - \phi$,

where $\lambda$_ph is the wavelength of the incident light and $\phi$ is the work function of the material. The ejected photoelectrons are accelerated from the cathode to the anode by a potential V. Thus, the maximum kinetic energy of the photo-electron at the anode is

${K_{\max ,\,a}} = {K_{\max ,\,c}} + eV = {{hc} \over {{\lambda _{ph}}}} - \phi + eV$.

Now, minimum de-Broglie wavelength of electrons passing through anode,

${\lambda _e} = {h \over {\sqrt {2m\,{K_{\max ,\,a}}} }} = {h \over {\sqrt {2m\left( {{{hc} \over {{\lambda _{ph}}}} - \phi + eV} \right)} }}$ ...... (i)

D : For $V > > \phi e$

${\lambda _e} \approx {h \over {\sqrt {2meV} }} \propto {1 \over {\sqrt V }}$

$\therefore$ If V is made four times, $\lambda$_e becomes ${1 \over {\sqrt 4 }} = {1 \over 2}$, i.e. halved.

C : When $\phi$ and $\lambda$_ph increases, $\lambda$_e also increases.

A : From (i)

${{d{\lambda _e}} \over {dt}} = {{ - h} \over 2}{\left[ {2m\left( {{{hc} \over {{\lambda _{ph}}}} - \phi + eV} \right)} \right]^{ - 3/2}} \times 2m\left( {{{ - hc} \over {\lambda _{ph}^2}}} \right){{d{\lambda _{ph}}} \over {dt}}$

$\Rightarrow {{d{\lambda _e}} \over {dt}} \ne {{d{\lambda _{ph}}} \over {dt}}$

I : $\lambda$_e does not depend on d.

To determine the minimum number of days after which the laboratory can be considered safe for use, we need to understand how the radioactive material decays over time.

The half-life of a radioactive material is the time it takes for half of the material to decay. For our given material, the half-life is 18 days. This means every 18 days, the amount of radioactive material is reduced to half of its previous amount.

Given that the initial radiation level is 64 times more than the permissible level, we can represent this situation mathematically. Let $R$ be the initial radiation level and $P$ be the permissible level of radiation. We are given:

$R = 64P$

After each half-life period of 18 days, the amount of radioactive material will be halved. If $n$ is the number of half-life periods required for the radiation to reduce to the permissible level, we can write:

$\frac{R}{2^n} = P$

Using the given information ($R = 64P$), we get:

$\frac{64P}{2^n} = P$

Dividing both sides by $P$, we get:

$\frac{64}{2^n} = 1$

This implies:

$2^n = 64$

Knowing that $64$ is a power of $2$:

$64 = 2^6$

Therefore, $n = 6$. This means it will take six half-life periods for the radiation to decay to a safe level.

Since each half-life period is 18 days, the total number of days required is:

$6 \times 18 = 108$

Hence, the minimum number of days after which the laboratory can be considered safe for use is 108 days.

Answer: Option C (108)

The distance between the two plates is h. The potential of the bottom plate is V₀ and that of the top plates is $-$V₀. The electric field between the plates is E = 2V₀/h (directed upwards). The radius of each ball is r (<< h). Let m be the mass and C be the capacitance of each ball.

When the ball touches the bottom plate, it gets a positive charge q = CV₀ (we assume that the charge transfer is instantaneous). This positively charged ball experiences an upward force, F = qE = 2CV$_0^2$/h, which accelerates the ball upwards. Since the force is constant, the ball cannot do SHM (for SHM, the force should be proportional to the displacement and directed towards the centre).

When the ball hits the top plate, it transfers the positive charge to the plate and gets negative charge q = $-$CV₀. This negatively charged ball again experience a force F = qE (downward) and starts accelerating downwards. Thus, the ball keeps moving between the bottom and the top plates carrying a charge +q upwards and $-$q downwards.

For convex lens, u = $-$50 cm, f = 30 cm

$\therefore$ ${1 \over v} = {1 \over f} + {1 \over u} = {1 \over {30}} - {1 \over {50}} = {{5 - 3} \over {150}} = {2 \over {150}}$ or v = 75 cm

Image formed by convex lens acts as a virtual object for mirror.

If we consider that the mirror is not tilted, then for mirror,

${u_1} = 75 - 50 = 25$ cm, ${f_1} = {R \over 2} = {{100\,cm} \over 2} = 50$ cm

$\therefore$ ${1 \over {{v_1}}} = {1 \over {{f_1}}} - {1 \over {{u_1}}} = {1 \over {50}} - {1 \over {25}} = {{ - 1} \over {50}}$ or ${v_1} = 50$ cm

So, the co-ordinates of final image formed (I₂) if the mirror is not tilted are (0 cm, 0 cm)

If mirror is not tilted then this ray starts from the object, refracts through the lens without deviation, incident normally on the mirror and finally comes to the image I₂ after reflection from the mirror. If the mirror is tilted by an angle $\theta$ then this ray is incident on the mirror at an angle of reflection $\theta$ and finally forms a new image at I'₂ (see figure). Thus, the axis on which image lies makes an angle 2$\theta$ = 60$^\circ$ with the principal axis of the lens. Also, the image distance will remain same upto the first order of approximation (image distance will remain exactly same in case of plane mirror). Thus, (x, y) coordinates of the image are

$x = 50 - 50\cos 60^\circ = 25$ cm,

$y = 50\sin 60^\circ = 25\sqrt 3$ cm.

The given particle of mass m experiences a centrifugal force:

F = ma = mr$\omega$²

$\Rightarrow$ a = r$\omega$²

$\Rightarrow$ ${{dv} \over {dt}} = r{\omega ^2}$

or, $\left( {{{dv} \over {dr}}} \right)\left( {{{dr} \over {dt}}} \right) = r{\omega ^2}$ ($\because$ ${{{dr} \over {dt}}}=v$)

$\Rightarrow vdv = r{\omega ^2}dr$

$\int\limits_0^v {vdv = {\omega ^2}\int\limits_{R/2}^r {rdr} }$ ($\because$ at t = 0, $r = {R \over 2}$)

$\Rightarrow {{{v^2}} \over 2} = {\omega ^2}\left( {{{{r^2}} \over 2}} \right)_{R/2}^r$

$\Rightarrow v = \omega \sqrt {{r^2} - {{{R^2}} \over 4}}$

Now, $v = {{dr} \over {dt}}$

Therefore, ${{dr} \over {dt}} = \omega \sqrt {{r^2} - {{{R^2}} \over 4}}$

or, $\int\limits_{R/2}^r {{{dr} \over {\sqrt {{r^2} - {{{R^2}} \over 4}} }} = \omega \int\limits_0^t {dt} }$ ...... (1)

To solve this integral, we substitute $r = {R \over 2}\sec \theta$; therefore,

$dr = {R \over 2}\sec \theta \tan \theta d\theta$

$\Rightarrow \int\limits_{{\pi \over 2}}^{{{\sec }^{ - 1}}(2r/R)} {{{(r/2)\sec \theta \tan \theta d\theta } \over {\sqrt {{{{R^2}} \over 4}{{\tan }^2}\theta } }} = \omega t}$

$\Rightarrow \omega t = \int\limits_{\pi /2}^{{{\sec }^{ - 1}}(2r/R)} {\sec \theta d\theta }$

$= {\left[ {\ln \left| {\sec \theta + \tan \theta } \right|} \right]^{{{\sec }^{ - 1}}(2r/R)}}$

$= \left[ {\ln \left| {\sec \theta + \sqrt {{{\sec }^2}\theta - 1} } \right|} \right]_{\pi /2}^{{{\sec }^{ - 1}}(2r/R)}$

$\Rightarrow \omega t = \ln \left[ {{{2r} \over R} + {{\sqrt {4{r^2} - {R^2}} } \over R}} \right]$

Let ${{2r} \over R} = x$. Therefore,

${e^\omega } = x + \sqrt {{x^2} - 1}$

$\Rightarrow ({e^{\omega t}} - x)^2 = {x^2} - 1$

$\Rightarrow {e^{2\omega t}} - 2x{e^{\omega t}} + x^2 = x^2 - 1$

$\Rightarrow {e^{2\omega t}} - 2x{e^{\omega t}} + 1 = 0$

$\Rightarrow x = {{{e^{2\omega t}} + 1} \over {2{e^{\omega t}}}} = {{{e^{\omega t}} + {e^{ - \omega t}}} \over 2}$

On replacing x by ${{2r} \over R}$, we get

${{2r} \over R} = {{{e^{\omega t}} + {e^{ - \omega t}}} \over 2}$

or, $r = {R \over 4}[{e^{\omega t}} + {e^{ - \omega t}}]$

Hence, option (c) is correct.

The rotational quantity of velocity given by

${v_{rot}} = {{dr} \over {dt}} = {R \over 4}[\omega {e^{\omega t}} - \omega {e^{ - \omega t}}]$

${v_{rot}} = {{\omega R} \over 4}[{e^{\omega t}} - {e^{ - \omega t}}]$ ...... (1)

The force experienced by the block in the inertial frame is

${\overrightarrow F _{in}} = - m{\omega ^2}r\widehat i - mg\widehat k$ ..... (2)

Therefore, the force experienced by the block in the rotating frame is

${\overrightarrow F _{rot}} = {\overrightarrow F _{in}} + 2m({\overrightarrow v _{rot}} \times \omega \widehat k) + m(\omega \widehat k \times r\widehat i) \times \omega \widehat k$

${\overrightarrow F _{rot}} = ( - m{\omega ^2}r\widehat i - mg\widehat k) + 2m{{\omega R} \over r}[{e^{\omega t}} - {e^{ - \omega t}}]\widehat i \times \omega \widehat k + m{\omega ^2}r\widehat i$

$= - {{m{\omega ^2}R} \over 2}[{e^{\omega t}} - {e^{ - \omega t}}]\widehat j - mg\widehat k$

Therefore, the net reaction of the disc on the block is

${\overrightarrow F _{reaction}} - {\overrightarrow F _{rot}} = {{m{\omega ^2}R} \over 2}[{e^{\omega t}} - {e^{\omega t}}]\widehat j + mg\widehat k$

Hence, option (B) is correct.

Let i_g be the current that gives full deflection of the galvanometer. When all components are connected in series (see figure), effective resistance of the circuit is ${R_e} = 2R + 2{R_c}$ and maximum current allowed in the circuit is i_g.

Thus, the voltage between A and B is

${V_{AB}} = {i_g}{R_e} = 2{i_g}(R + {R_c})$.

Consider the case when two resistors and one galvanometer are connected in series and the second galvanometer is connected in parallel.

The maximum current through each galvanometer is i_g and the maximum current through the resistors is 2i_g. Apply Kirchhoff's law to get the voltage between A and B as

$V{'_{AB}} = 2{i_g}R + 2{i_g}R + {i_g}{R_c}$

$= 2{i_g}(R + {R_c}) + 2{i_g}(R - {R_c}/2)$

$= {V_{AB}} + 2{i_g}(R - {R_c}/2)$

$> {V_{AB}}$ ($\because$ ${R_c} < R/2$).

Consider the case when all four components are connected in parallel.

Let i and i_g be the currents through the resistors and the galvanometers. By Kirchhoff's law, $iR = {i_g}{R_c}$, which gives $i = {i_g}{R_c}/R$. The current between A and B is

${I_{AB}} = 2i + 2{i_g} = 2{i_g}(1 + {R_c}/R)$.

Consider the case when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors.

$I{'_{AB}} = {i_g} + 2i = {i_g} + 2{i_g}(2{R_c}/R)$

$= 2{i_g}(1 + {R_c}/R) - (1 - 2{R_c}/R){i_g}$

$= {I_{AB}} - (1 - 2{R_c}/R){i_g}$

$< {I_{AB}}$ ($\because$ ${R_c} < R/2$).

When the key is just pressed, there is no charge on capacitors

$\therefore$ $5 = 25000{I_1} \Rightarrow {I_1} = {5 \over {25000}} = 0.2\,mA$

and $5 = 50000{I_2} \Rightarrow {I_2} = {5 \over {50000}} = 0.1\,mA$

Reading of voltmeter $= {V_Q} - {V_P} = - 25000{I_1}$

$= - 25000 \times 0.2 \times {10^{ - 3}}V = - 5V$

After a very long time, steady state is reached, i.e. ${I_1} - {I_2} = 0$

$\therefore$ $5 = {{{q_1}} \over {40 \times {{10}^{ - 6}}}}$ or ${q_1} = 200\mu C$

and $\therefore$ $5 = {{{q_2}} \over {20 \times {{10}^{ - 6}}}}$ or ${q_2} = 100\mu C$

Now, reading of voltmeter $= {V_Q} - {V_P}$

$= {{{q_2}} \over {20\mu F}} = {{100\mu C} \over {20\mu F}} = 5V$

For capacitor of 40$\mu$F,

${\tau _1} = RC = 25 \times {10^3} \times 40 \times {10^{ - 6}} = 1\,s$

and for capacitor of 20$\mu$F,

${\tau _2} = RC = 50 \times {10^3} \times 20 \times {10^{ - 6}} = 1\,s$

$\therefore$ At any instant t,

${q_1} = 200[1 - {e^{ - t}}]\mu C$, ${q_2} = 100[1 - {e^{ - t}}]\mu C$

${I_1} = 0.2\,{e^{ - t}}$ mA, ${I_2} = 0.1\,{e^{ - t}}$ mA

${V_Q} - {V_P} = {{100(1 - {e^{ - t}})\mu C} \over {20\mu F}} - 25k\Omega \times 0.2\,{e^{ - t}}$ mA

$= 5(1 - {e^{ - t}}) - 5{e^{ - t}} = 5 - 10{e^{ - t}}$

At t = ln 2 s; ${V_Q} - {V_P} = 5 - 10{e^{ - \ln 2}} = 5 - {{10} \over 2} = 0$ V

Initially, I₀ = 0.1 mA + 0.2 mA = 0.3 mA

At t = 1 s, I = I₁ + I₂

$= (0.2{e^{ - 1}} + 0.1{e^{ - 1}}) = {{0.3} \over e}\,mA = {{{I_0}} \over e}$

After a long time, i.e. t $\to$ $\infty$

$I = {I_1} + {I_2} = 0.2{e^{ - \infty }} + 0.1{e^{ - \infty }} = 0$

At point O, the path difference is

${{2\pi } \over \lambda }({S_1}O - {S_2}O) = {{2\pi } \over \lambda }\left[ {\left( {D + {d \over 2}} \right) - \left( {D - {d \over 2}} \right)} \right]$

$= {{2\pi d} \over \lambda } = {{2\pi (0.6003 \times {{10}^{ - 3}})} \over {600 \times {{10}^{ - 9}}}} = 2001\pi = (2n + 1)\pi$

Therefore, the dark fringe forms at O.

Hence, option (A) is correct.

The shape of wavefronts from both sources for D >> d is semi-circular, which are concentric with each other with the difference just in their phase resulting in concentric semicircular interference fringes. That is, the region nearby point O is dark.

Hence, option (B) is correct.

The binding energies of $_7^{15}N$ and $_8^{15}O$ are given by

$B{E_N} = \Delta {m_N}{c^2} = (8{m_n} + 7{m_p} - {M_N}){c^2}$

$= (8 \times 1.008665 + 7 \times 1.007825 - 15.000109 \times 931.5$

= 115.49 MeV,

$B{E_O} = \Delta {m_O}{c^2} = (7{m_n} + 8{m_p} - {M_O}){c^2}$

$= (7 \times 1.008665 + 8 \times 1.007825 - 15.003065 \times 931.5$

= 111.95 MeV.

The difference in binding energies of $_7^{15}N$ and $_8^{15}O$ is

$\Delta BE = B{E_N} - B{E_O} = 115.49 - 111.95$

= 3.54 MeV. ........ (1)

The electrostatic energies of $_7^{15}N$ and $_8^{15}O$ are given by

${E_N} = {3 \over 5}{{Z(Z - 1){e^2}} \over {4\pi { \in _0}R}} = {3 \over 5}{{7(7 - 1)1.44} \over R}$

$= {{36.288} \over R}$ MeV-fm

${E_O} = {3 \over 5}{{8(8 - 1)1.44} \over R} = {{48.384} \over R}$ MeV-fm.

The difference in electrostatic energies of $_7^{15}N$ and $_8^{15}O$ is

$\Delta E = {E_O} - {E_N} = (12.096/R)$ ...... (2)

Since, $\Delta E = \Delta BE$, equations (1) and (2) give $R = 12.096/3.54 = 3.42$ fm.

When intermolecular attraction between two components A and B in the mixture is same as between A and A or B and B, hence it is a case of ideal solution. When intermolecular attraction between A and B in a mixture is smaller than that between A and A or B and B, then mixture is more vaporised, bp is lowered. It is a case of positive deviation from Raoult’s law. When intermolecular attraction between A and B is higher than that between A and A or B and B, then mixture is less vaporised, bp is increased. It is a case of negative deviation.

(a) Methanol molecules (CH)₃OH are hydrogen bonded. In a mixture of CCl₄ and CH₃OH, extent of H-bonding is decreased. Mixture is more vaporised thus, positive deviation from Raoult’s law.

(b) Acetone molecules have higher intermolecular attraction due to dipole-dipole interaction. With CS₂ , this interaction is decreased thus, positive deviation.

(c) Mixture of benzene and toluene forms ideal solution.

(d) Phenol and aniline have higher interaction due to intermolecular H-bonding. Hence, negative deviation.