JEE Main 2018 16th April Morning Shift Question Paper with Solutions

As intensity of emergent beam is reduced to half after passing through two polarisers. It means angle between A and B is 0$^\circ$.

Now, on placing polariser C between A and B.

Intensity after passing through A is ${I_A} = {I \over 2}$.

Let $\theta$ be the angle between A and C. Intensity of light after passing through C is given by

${I_C} = {I \over 2}{\cos ^2}\theta$

Intensity of light after passing through polariser B is ${I \over 3}$.

Angle between C and B is also $\theta$ as A is parallel to B.

So, ${I \over 3} = {I_C}{\cos ^2}\theta = {I \over 2}{\cos ^2}\theta \,.\,{\cos ^2}\theta$

${\cos ^4}\theta = {2 \over 3} \Rightarrow \cos \theta = {\left( {{2 \over 3}} \right)^{1/4}}$

Potential of the given oscillator is

$V = {1 \over 2}k{(x - k)^2}$

Given: M = 10; m = 5, u = 1; k = 1

Initial momentum of the particle of mass m

= mu = m $\times$ 5 = 5m

Momentum of (oscillator + particle) after collision = (M + m)

Velocity of oscillator after collision = v

So, momentum of system = (M + m)v

From conservation of linear momentum, we have

(M + m) = mu = 5 $\times$ 1 = 5

For second collision, oscillator and particle have momentum in opposite direction.

Net or total momentum is zero.

Likewise after 4^th, 6^th, 8^th, 10^th, 12^th collision the momentum is zero. After 12^th collision, Mass of oscillator and 12 particles will be (10 + 12 $\times$ 5) = 70

Now, from conservation of linear momentum, for 13^th collision, we have

$70 \times 0 + 5 \times 1 = (70 + 5)v' \Rightarrow v' = {5 \over {75}} \Rightarrow {1 \over {15}}$

Total mass after 13^th collision = (10 + 13 $\times$ 5) = 75

Kinetic energy of system $= {1 \over 2}mv{'^2}$

$\Rightarrow KE = {1 \over 2} \times 75 \times {1 \over {15}} \times {1 \over {15}}$

$\Rightarrow {1 \over 2}k{A^2} = {1 \over 2} \times {{75} \over {225}} = {1 \over 6}$

$\Rightarrow {1 \over 2} \times 1 \times {A^2} = {1 \over 6} \Rightarrow {A^2} = {1 \over 3}$

$\Rightarrow A = {1 \over {\sqrt 3 }}$

$\overrightarrow E$ is electric field vector, $\overrightarrow B$ is magnetic field vector perpendicular to $\overrightarrow E$. The direction of propagation is $(\overrightarrow E \times \overrightarrow B )$. The direction of propagation of wave is along + y axis, then $\overrightarrow E$ is along + z axis and $\overrightarrow B$ is along + x axis.

$(E\widehat k \times B\widehat i) = EB(\widehat k \times \widehat i) = EB\widehat j$

As wave is travelling along + y axis with time, we will use (y $-$ ct) in wave equation.

Also, intensity is given by

$I = {1 \over 2}c{\varepsilon _0}E_0^2$

And $E = cB$

So, $|{E_0}| = \sqrt {{{2I} \over {c{\varepsilon _0}}}}$

Therefore, $\overrightarrow E = \sqrt {{{2I} \over {c{\varepsilon _0}}}} \cos \left[ {{{2\pi } \over \lambda }(y - ct)} \right]\widehat k$

And $\overrightarrow B = {E \over c}\widehat i \Rightarrow {1 \over c}E\widehat i$

We know that, $\lambda = {h \over {mv}}$

From third Bohr's postulate, we have

$mvr = n{h \over {2\pi }}$

${h \over {mv}} = {{2\pi r} \over n} \Rightarrow \lambda = {{2\pi r} \over n}$

Since, $r = {a_0}{{{n^2}} \over Z}$, where a₀ is radius of Bohr's orbit having value (0.53) 1² $\mathop A\limits^o$ = 0.53 $\mathop A\limits^o$, therefore,

$\lambda = {{2\pi {a_0}{n^2}} \over {n\,.\,Z}} = {{2\pi {a_0}} \over Z}\,.\,n$

For Hydrogen Z = 1. Therefore,

${\lambda _G} = {{2\pi {a_0} \times 1} \over 1} = 2\pi {a_0}$ and ${\lambda _B} = {{2\pi {a_0} \times 3} \over 1} = 6\pi {a_0}$

Then, ${\lambda _B} = 3{\lambda _G}$

Amplitude of carrier wave is A_c = 14 V

Modulation Index is $\mu$ = 80% = 0.80

Amplitude of modulating signal is given by

$\mu = {{{A_m}} \over {{A_c}}} \Rightarrow {A_m} = \mu {A_c} \Rightarrow {A_m} = 0.8 \times 14 = 11.2\,V$

Given : ${i_1} = 60^\circ$; $A = 30^\circ$

Then angle of deviation is given by

$\delta = 1 - A = 60^\circ - 30^\circ = 30^\circ$

We know that,

${i_1} + {i_2} = A + \delta \Rightarrow 60^\circ + {i_2} = 30^\circ + 30^\circ \Rightarrow {i_2} = 0$

It means emergent ray leaves face AC normally.

Voltage across R₁ is given by ${V_{{R_1}}} = V - {V_Z} = 15 - 10 = 5\,V$.

Resistance ${R_1} = 500\,\Omega$

Therefore, current through R₁ is given by

${I_{{R_1}}} = {{5V} \over {500\,\Omega }} = 0.01\,A$

Now, current through R₂ is given by

${I_{{R_2}}} = {{10V} \over {1500\,\Omega }} = 0.067\,A$

Therefore, the current through Zener diode will be

$0.1 - 0.0067 = 0.0033\,A = 3.3mA$

Current required by unit deflection is 60 $\mu$A.

For, $\theta$ = 9 current is I = 9 $\times$ 60 $\mu$A

$\Rightarrow$ I = 540 $\mu$A = 540 $\times$ 10^$-$6 A

Let G is resistance of galvanometer. Then,

$540 \times {10^{ - 6}} = {6 \over {(11000 + G)}}$

[11000 + G] 90 $\times$ 10^$-$6 = 1

99000 + 9G = 10⁵

9G = 100000 $-$ 99000

9G = 1000

$G = {{1000} \over 9}\Omega$

Also, in half deflection method,

$G = {{RS} \over {R - S}} \Rightarrow {{1000} \over 9} = {{11000\,S} \over {11000 - S}}$

${1 \over 9} = {{11S} \over {11000 - S}} \Rightarrow 11000 - S = 99\,S$

100 S = 11000 $\Rightarrow$ S = 110 $\Omega$

We know that $E = {{hc} \over \lambda }$

So, for atom ${E_A} = {{hc} \over {{\lambda _A}}}$

And for neutron ${E_N} = {{hc} \over {{\lambda _N}}}$

Then, ${{{E_A}} \over {{E_N}}} = {{hc} \over {{\lambda _A}}} \times {{{\lambda _N}} \over {hc}} \Rightarrow {{{\lambda _N}} \over {{\lambda _A}}}$

Here, E_A is order of eV and E_N is order of MeV.

Therefore, ${{{\lambda _N}} \over {{\lambda _A}}} = {{{E_A}} \over {{E_N}}} = {{eV} \over {MeV}} = {{1eV} \over {{{10}^6}eV}}$

${\lambda _N} = {10^{ - 6}}\,{\lambda _A}$