JEE Main 2018 15th April Evening Shift Question Paper with Solutions

Given : Volume charge density of rod = $\rho$.

We know that current $I = neA{v_d}$; where n is number of electrons, e is electronic charge, A is area, v_d is drift velocity

Since volume charge density is $\rho$ = ne. Therefore,

$I = \rho A{v_d} \Rightarrow {v_d} = {I \over {{\rho _A}}}$

Now, time = ${{dis\tan ce} \over {speed}}$

$\Rightarrow$ time $= {d \over {{v_d}}} = {d \over {I/\rho A}} = {{\rho Ad} \over I}$

Thus, time required to travel distance $d = {{\rho dA} \over I}$

Given : Force applied F = 10⁵ N; side of cube x = 10 m = 10 $\times$ 10^$-$2 m; shift dx = 0.5 cm = 0.5 $\times$ = 10^$-$2 m

On cube of side x = 10 cm;

Stress = ${F \over {{x^2}}} = {{{{10}^5}} \over {{{(10 \times {{10}^{ - 2}})}^2}}} = {10^7}$ N/m²

Strain = ${{dx} \over x} = {{0.5 \times {{10}^{ - 2}}} \over {10 \times {{10}^{ - 2}}}} = 0.5$

On cube of side x' = 20 cm;

Stress' = ${F \over {x{'^2}}} = {{{{10}^5}} \over {{{(20 \times {{10}^{ - 2}})}^2}}}$

Strain' = ${{dx'} \over {x'}} = {{dx'} \over {20 \times {{10}^{ - 2}}}}$

Now, it is given that the material is same for both cubes, therefore,

${{Stress} \over {Strain}} = {{Stress'} \over {Strain'}} \Rightarrow {{{{10}^7}} \over {0.05}} = {{{{10}^5}} \over {(20 \times {{10}^{ - 2}})}} \times {{20 \times {{10}^{ - 2}}} \over {dx'}}$

$\Rightarrow {{{{10}^7} \times {{10}^2}} \over 5} = {{{{10}^5}} \over {20 \times {{10}^{ - 2}}}}{1 \over {dx'}}$

$\Rightarrow dx' = {{{{10}^5}} \over {20 \times {{10}^{ - 2}}}} \times {5 \over {{{10}^7} \times {{10}^2}}} = {{5 \times {{10}^{ - 2}}} \over {20}} = 0.25 \times {10^{ - 2}}$ m

$\Rightarrow dx' = 0.25$ cm

For a convergent doublet of separated lens, we have

${1 \over f} = {1 \over {{f_1}}} + {1 \over {{f_2}}} - {d \over {{f_1}{f_2}}}$ ...... (1)

where d is separation between two lens, f₁ and f₂ are focal lengths of component lenses, f is resultant focal length. Therefore, Eq. (1) becomes

${1 \over {10}} = {1 \over {{f_1}}} + {1 \over {{f_2}}} - {2 \over {{f_1}{f_2}}} \Rightarrow {1 \over {10}} = \left( {{{{f_2} + {f_1} - 2} \over {{f_1}{f_2}}}} \right)$

$\Rightarrow {f_1}{f_2} = 10{f_2} + 10{f_1} - 20$

$\Rightarrow 10{f_1} + 10{f_2} - {f_1}{f_2} = + 20$

For f₁ = 18 cm and f₂ = 20 cm, the above equation satisfies.

Since rate of heat $= {{{V^2}} \over R} \Rightarrow$ rate of heat $\propto {1 \over R}$, where $R = {{\rho L} \over A} = {{\rho L} \over {\pi {r^2}}}$; here r is radius of wire and L is length of wire. Therefore,

$R \propto {L \over {{r^2}}}$.

Thus, rate of heat $\propto {{{r^2}} \over L}$.

When length in halved and radius is doubled then, rate of heat $\propto {{{{(2r)}^2}} \over {L/2}} = {{8{r^2}} \over L}$

Therefore, rate of heat developed in wire will be increased 8 times.

Time taken to cool from 60$^\circ$C to 50$^\circ$C = 10 minutes

Temperature of surroundings = 25$^\circ$C

Temperature of body in next 10 minutes = T

Therefore, ${{60 - 50} \over {10\,\min }} = {k_B}\left[ {{{60 + 50} \over 2} - 25} \right] \Rightarrow {k_B}30 = 1$ ...... (1)

and ${{60 - T} \over {20\,\min }} = {k_B}\left[ {{{60 + T} \over 2} - 25} \right] = {k_B}\left[ {{{60 + T - 50} \over 2}} \right]$ ..... (2)

Taking ratio of Eqs. (1) and (2), we get

${{20} \over {60 - T}} = {{30{k_B}} \over {{k_B}\left( {{{10 + T} \over 2}} \right)}} \Rightarrow {{20} \over {60 - T}} = {{30} \over {5 + T/2}}$

$\Rightarrow 20\left( {5 + {T \over 2}} \right) = 30(60 - T) \Rightarrow 100 + T10 = 1800 - 30T$

$\Rightarrow 1800 - 100 = 30T + 10T$

$\Rightarrow 1700 = 40T \Rightarrow T = {{1700} \over {40}} = 42.5^\circ C \sim 43^\circ C$

Given, bubble of radius r rises from bottom of lake. The radius of bubble at top becomes 5r/4. Therefore, pressure in bubble at bottom is ${P_1} = {P_0} + \rho gh + {{4T} \over r}$

Pressure in bubble at top is ${P_2} = {P_0} + {{4T} \over {5r/4}}$

Now, we know ${P_1}{V_1} = {P_2}{V_2}$, therefore,

$\left( {{P_0} + \rho gh + {{4T} \over 4}} \right){{4\pi } \over 3}{r^3} = \left( {{P_0} + {{4T} \over {5r/4}}} \right){{4\pi } \over 3}{\left( {{{5r} \over 4}} \right)^3}$

${P_0} + \rho hg + {{4T} \over r} = \left( {{P_0} + {{4T \times 4} \over {5r}}} \right){{125} \over {64}}$

Now given P₀ = 10$\rho$g, therefore,

$10\rho g + \rho gh + {{4T} \over r} = {{125} \over {64}} \times 10\rho g + {{16T} \over {5r}} \times {{125} \over {64}}$

Neglecting effect of temperature, we get

$10\rho g + \rho gh = {{125} \over {64}} \times 10\rho g$

$\Rightarrow 10 + h = {{125} \over {64}} \times 10 \Rightarrow h = {{1250} \over {64}} - 10 = 9.53125\,m$

$\Rightarrow$ h $\sim$ 9.5 m

In figure (i) before collision, m' is mass of unknown particle; m is mass of proton; v₁ is initial velocity.

(i) Before collision :

Now, in figure (ii), v₁ is final velocity of unknown particle and v₂ is final velocity of proton.

(ii) After collision :

By conservation of momentum, we have

Momentum before collision = Momentum after collision

Consider x-component, we have

$m{v_i} + m'\,.\,0 = m'{v_1}\cos 45^\circ + m{v_2}\cos 45^\circ$

$m{v_i} = {1 \over {\sqrt 2 }}(m'{v_1} + m{v_2})$ ..... (1)

Consider y-component, we have

$0 = m'{v_1}\sin 45^\circ - m{v_2}\sin 45^\circ$

${1 \over {\sqrt 2 }}(m'{v_i} - m{v_2}) = 0 \Rightarrow m'{v_1} = m{v_2}$ ...... (2)

Substitute Eq. (2) in Eq. (1), we get

$m{v_i} = {1 \over {\sqrt 2 }}(m{v_2} + m{v_2}) - \sqrt 2 m{v_2}$

$\Rightarrow {v_i} = \sqrt 2 {v_2}$ ..... (3)

Using Eq. (2) and (3) in Eq. (1), we get

$m\sqrt 2 {v_2} = {1 \over {\sqrt 2 }}(m'{v_1} + m'{v_1})$

$2m{v_2} = 2m'{v_1}$ ($\because$ ${v_1} = {v_2}$)

$\Rightarrow m = m'$

We know that electric flux $\phi = \overrightarrow B \,.\,\overrightarrow A$

$\Rightarrow \phi = BA\cos \omega t$

Now, $B = {{{\mu _0}} \over 2}{I \over R}$ is magnetic field due to circular coil of radius R and $A = \pi {r^2}$ is area of circular coil of radius r. Therefore,

$\phi = {{{\mu _0}} \over 2}{I \over R}\pi {r^2}\cos \omega t$

Now induced emf $\varepsilon = {{ - d\phi } \over {dt}} = {{ - d} \over {dt}}\left( {{{{\mu _0}} \over 2}{I \over R}\pi {r^2}\cos \omega t} \right)$

$\Rightarrow \varepsilon = {{{\mu _0}} \over 2}{I \over R}\pi {r^2}\sin \omega t$

(i) Radius of the orbit is given by

$r = {{{h^2}} \over {4\pi m{e^2}}} \times {{{n^2}} \over Z}$

Only, m_$\mu$ = 200 m_e rest are same. So, statement (A) is correct: The radius of muonic orbit is 200 times smaller than that of the electron.

(ii) Velocity of particle in an orbit is given by

$v = {{nh} \over {2\pi mr}} = {{nh \times 4\pi m{e^2} \times Z} \over {2\pi m \times {h^2} \times {n^2}}} = {{2{e^2}Z} \over {hn}}$

Therefore, speed does not change as it does not depend on mass. So, statement (B) is incorrect.

(iii) Ionisation energy is given by

$\Delta {E_n} = - {{2{\pi ^2}m{e^4}{Z^2}} \over {{h^2}{{(4\pi {\varepsilon _0})}^2}}} \times \left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)$

Only, m_$\mu$ = 200 m_e rest are same. So, statement (C) is correct: The ionisation energy of muonic atom is 200 times more than that of an hydrogen atom.

(iv) Since Momentum $\propto$ Energy. So, statement (D) is correct: The momentum of the muon in he nth orbit is 200 times more than that of the electron.

We know that ${v_{rms}} = \sqrt {{{3{k_B}T} \over m}}$. Given, ${k_B} = 1.4 \times {10^{ - 23}}$ J/K; T = 300 K; m = 7 $\times$ 10^$-$27 kg. Therefore,

${v_{rms}} = \sqrt {{{3 \times 1.4 \times {{10}^{ - 23}} \times 300} \over {7 \times {{10}^{ - 27}}}}}$

$= \sqrt {{{3 \times 300 \times 14 \times {{10}^{ - 23}}} \over {7 \times 10 \times {{10}^{ - 27}}}}}$

$= \sqrt {{{{3^2} \times {{10}^2} \times 2 \times {{10}^4}} \over {10}}} = 3 \times 10 \times {10^2} \times \sqrt {{2 \over {10}}}$

$= 3 \times \sqrt {10} \times {10^2} \times \sqrt 2 = 3 \times \sqrt 2 \times \sqrt 5 \times {10^2} \times \sqrt 2$

$= 3 \times 2 \times \sqrt 5 \times {10^2}$

${v_{rms}} = 6 \times {10^2} \times \sqrt 5 = 13.41 \times {10^2}$ or ${v_{rms}} = 1.34 \times {10^3}$

Given : C₁ = 1.0 $\mu$F; C₂ = 3.0 $\mu$F; C₃ = 6.0 $\mu$F

Capacitance C₂ and C₃ are in series, their equivalent capacitance is

${1 \over C} = {1 \over {{C_2}}} + {1 \over {{C_3}}} \Rightarrow {1 \over C} = {{{C_2} + {C_3}} \over {{C_2}{C_3}}}$

$\Rightarrow C = {{{C_2}{C_3}} \over {{C_2} + {C_3}}} = {{3.0\mu F \times 6.0\mu F} \over {3.0\mu F + 6.0\mu F}} = 2\mu F$

Charge on C₁ when battery is connected to switch (1) is

$= {C_1} \times 60$ ..... (1)

Charge when battery is connected to switch (2) is

$= ({C_1} + C)V'$ ...... (2)

Equating Eq. (1) and (2), we get

$({C_1} + C)V' = 60{C_1} \Rightarrow V' = {{60{C_1}} \over {{C_1} + C}}$

$\Rightarrow V' = {{60 \times 1\mu F} \over {1\mu F + 2\mu F}} = {{60 \times 1\mu F} \over {3\mu F}} = 20\,V$

Final charge on C₂ and C₃ is $Q = CV' = 2\mu F \times 20 = 40\mu C$

Therefore, $Q = 40\mu C$

Since $c = f\lambda \Rightarrow f = {c \over \lambda }$. Here, $\lambda = 2|{z_2} - {z_1}|$ and $c = 3 \times {10^8}$. Therefore,

$f = {{3 \times {{10}^8}} \over {2|{z_2} - {z_1}|}} = {{1.5 \times {{10}^8}} \over {|{z_2} - {z_1}|}}$

At terminal velocity, net force on rod = $mg\sin \theta$

$\Rightarrow mg\sin \theta = iBl$

Now, $IBl = \left( {{{Bvl} \over R}} \right)Bl = {{{B^2}{l^2}v} \over R}$

$\Rightarrow mg\sin \theta = {{{B^2}{l^2}v} \over R} \Rightarrow v = {{mgR\sin \theta } \over {{B^2}{l^2}}}$

The given circuit is

Therefore, truth table for given circuit is

According to Malus law, we have

$I = {I_0}{\cos ^2}\theta \Rightarrow {I \over {{I_0}}} = {\cos ^2}\theta$

Since the observed intensities are same for all values of $\theta$, therefore, for the angles given in options, the observed intensity should be same as well for all $\theta$, $\theta$ = 8$^\circ$, 38$^\circ$, 188$^\circ$ and 218$^\circ$, ${I \over {{I_0}}} \sim 0.9$; and for $\theta$ = 203$^\circ$, ${I \over I} \sim 0.9$.

Therefore, angle between direction of polarisation and x axis is 203$^\circ$.

Let the distance QM = l and distance RM = x.

Time to reach from Q to R is ${t_1} = {{l - x} \over v}$

Time to reach from R to P is ${t_2} = {{\sqrt {{x^2} + {d^2}} } \over {v/2}}$

Therefore, $t = {t_1} + {t_2} = {{l - x} \over v} + {{\sqrt {{x^2} + {d^2}} } \over {v/2}}$

On differentiating, we get

${{dt} \over {dx}} = {{0 - 1} \over v} + {1 \over {v/2}}{1 \over {2\sqrt {{x^2} + {d^2}} }} \times 2x$

$\Rightarrow {{dt} \over {dx}} = {{ - 1} \over v} + {{2x} \over {v\sqrt {{x^2} + {d^2}} }}$

Put ${{dt} \over {dx}} = 0$, we get

${{ - 1} \over v} + {{2x} \over {v\sqrt {{x^2} + {d^2}} }} = 0$

$\Rightarrow {{2x} \over {v\sqrt {{x^2} + {d^2}} }} = {1 \over v}$

$\Rightarrow 2x = \sqrt {{x^2} + {d^2}} \Rightarrow 4{x^2} = {x^2} + {d^2}$

$\Rightarrow 3{x^2} = {d^2} \Rightarrow x = {d \over {\sqrt 3 }}$

Therefore, the distance $RM = x = {d \over {\sqrt 3 }}$, the time taken to reach P is minimum.

We know that magnetic field due to finite current carrying wire is

$B = {{{\mu _0}} \over {4\pi }}{I \over b}(\cos {\theta _1} + \cos {\theta _2})$ ..... (1)

Given that side of triangle = 4.5 $\times$ 10^$-$2 m = a; current = 1A since the triangle is equilateral, angle of each side will be 60$^\circ$.

Now, $\tan \theta = {{Perpendicular} \over {Base}} \Rightarrow \tan 60^\circ = {{a/2} \over b}$

$\Rightarrow b = {a \over {2\tan 60^\circ }} = {a \over {2\sqrt 3 }}$

Using equation (1), we get

$B = {{{\mu _0}} \over {4\pi }}{I \over {a/2\sqrt 3 }}(\cos 30^\circ + \cos 30^\circ ) = {{{\mu _0}} \over {4\pi }}{{2\sqrt 3 I} \over a}2\cos 30^\circ$

$B = {{{\mu _0}} \over {4\pi }}{{2\sqrt 3 I} \over a}{{2 \times \sqrt 3 } \over 2} = {{{\mu _0}} \over {4\pi }}{{6I} \over a}$

$\Rightarrow B = {{{{10}^{ - 7}} \times 6 \times 1} \over {4.5 \times {{10}^{ - 2}}}} = 1.33 \times {10^{ - 5}} \sim 2 \times {10^{ - 5}}$ Wb/m²

Given that two Carnot engines A and B are operated in series. Efficiency of a Carnot engine $\eta = 1 - {{{T_2}} \over {{T_1}}}$; where T₁ is temperature of source and T₂ is temperature of sink. Therefore,

${\eta _A} = 1 - {{{T_{2A}}} \over {{T_{1A}}}} = 1 - {T \over {600}}$; T is heat rejected by A

and ${\eta _B} = 1 - {{{T_{2B}}} \over {{T_{1B}}}} = 1 - {{100} \over T}$; T is heat rejected by A and received by B

Now, ${{{\eta _B}} \over {{\eta _A}}}{{1 - {{100} \over T}} \over {1 - {T \over {600}}}} = \left( {{{T - 100} \over T}} \right) \times {{600} \over {600 - T}}$

Let T = 350 K, therefore,

${{{\eta _B}} \over {{\eta _A}}} = {{350 - 100} \over {350}} \times {{600} \over {600 - 350}} = {{250} \over {350}} \times {{600} \over {25}} = {{600} \over {350}}$

$\Rightarrow {{{\eta _B}} \over {{\eta _A}}} = {{12} \over 7}$

We have

$m{\omega ^2}r = \mu mg$ ...... (1)

Given : rate of rotation = 3.5 rev/s

$\Rightarrow$ 1 revolution = 2$\pi$ rad

That is, 3.5 revolutions = 3.5 $\times$ 2$\pi$ rad

Therefore, $\omega$ = 3.5 $\times$ 2$\pi$ rad/s

r = 1.25 cm = 1.25 $\times$ 10^$-$2 m

Thus, from Eq. (1), we have

$m{\omega ^2}r = \mu mg \Rightarrow {\omega ^2}r = \mu g$

$\Rightarrow \mu = {{{\omega ^2}r} \over g} = {{{{(3.5 \times 2\pi )}^2}(1.25 \times {{10}^{ - 2}})} \over {10}}$

$\Rightarrow \mu = 0.60$

Length of wire is L₁ = 0.95 m; L₂ = 1 m. Number of beats per second heard = 5

Let frequency of fork be f. Therefore,

${v \over {2{L_1}}} - f = 5 \Rightarrow {v \over {2{L_1}}} = 5 + f$ ...... (1)

and $f - {v \over {2{L_2}}} = 5 \Rightarrow {v \over {2{L_2}}} = f - 5$ .... (2)

Dividing Eq. (1) by Eq. (2), we get

${{{v \over {2{L_1}}}} \over {{v \over {2{L_2}}}}} = {{5 + f} \over {f - 5}} \Rightarrow {{{L_2}} \over {{L_1}}} = {{f + 5} \over {f - 5}}$

$\Rightarrow {1 \over {0.95}} = {{f + 5} \over {f - 5}} \Rightarrow f - 5 = 0.95f + 4.75$

$\Rightarrow f - 0.95f = 5 + 4.75 \Rightarrow 0.05f = 9.75 \Rightarrow f = {{9.75} \over {0.05}}$

$\Rightarrow f = 195$ Hz

The two nuclei have velocity in ratio 8 : 27. By conservation of momentum, we have

${m_1}{v_1} = {m_2}{v_2} \Rightarrow {{{v_1}} \over {{v_2}}} = {{{m_2}} \over {{m_1}}} \Rightarrow {{{m_2}} \over {{m_1}}} = {8 \over {27}}$

Now, since $m = \rho {4 \over 3}\pi {r^3}$

Therefore, ${{{m_2}} \over {{m_1}}} = {{\rho {4 \over 3}\pi r_2^3} \over {\rho {4 \over 3}\pi r_1^3}} \Rightarrow {{{m_2}} \over {{m_1}}} = {\left( {{{{r_2}} \over {{r_1}}}} \right)^3} \Rightarrow {\left( {{{{r_2}} \over {{r_1}}}} \right)^3} = {8 \over {27}}$

$\Rightarrow {{{r_2}} \over {{r_1}}} = {2 \over 3}$

Thus, ratio of radii of nuclei ${r_1}:{r_2} = 3:2$.

We know that

$\omega = {1 \over {\sqrt {LC} }} = {1 \over {\sqrt {49\mu H \times 2.5\,nF} }} = {1 \over {\sqrt {49 \times {{10}^{ - 6}} \times 2.5 \times {{10}^{ - 9}}} }}$

$\Rightarrow \omega = {1 \over {\sqrt {49 \times 25 \times {{10}^{ - 16}}} }} = {1 \over {7 \times 5 \times {{10}^{ - 8}}}}$

Since $\omega = 2\pi f$

$\Rightarrow f = {\omega \over {2\pi }} = {1 \over {7 \times 5 \times {{10}^{ - 8}} \times 2\pi }}$

$\Rightarrow f = 454.5$ kHz

Given carrier frequency is modulated by audio signal of 12 kHz. Therefore,

f = 454.5 kHz $\pm$ 12 kHz

$\Rightarrow$ f = 442 kHz $-$ 466 kHz

Centre of mass after the collision will be at O. Therefore,

$2m{L \over 6} = m{L \over 3}$

Now, using conservation of angular momentum, we get

$2m{L \over 6} \times v + m{L \over 3} \times 2v = I\omega$

Therefore,

$2m{L \over 6} \times v + m{L \over 3} \times 2v = \left[ {8m{{{L^2}} \over {12}} + 2m{{\left( {{L \over 6}} \right)}^2} + m{{\left( {{L \over 3}} \right)}^2}} \right]\omega$

$\Rightarrow \left( {{L \over 6} + {L \over 3}} \right)2mv = \left[ {8m{{{L^2}} \over {12}} + {{2m{L^2}} \over {36}} + {{m{L^2}} \over 9}} \right]\omega$

$\Rightarrow {{3L} \over 6} \times 2mV = {5 \over 6}m{L^2}\omega$