JEE Main 2019 9th January Evening Shift Question Paper with Solutions

Energy dissipated in 60 Sec

= (P_avg) $\times$ 60

= V_rms $\times$ I_rms $\times$ cos$\phi$ $\times$ 60

= V_rms $\times$ ${{{V_{rms}}} \over Z} \times$ cos$\phi$ $\times$ 60

X_L = $\omega$L = 2$\pi$FL

= 2$\pi$(50)$\times$ 20 $\times$ 10^$-$3

= 2$\pi$ $\Omega$

X_C = ${1 \over {\omega C}}$

= ${1 \over {2\pi fC}}$

= ${1 \over {2\pi \left( {50} \right) \times 120 \times {{10}^{ - 6}}}}$

= 26.52 $\Omega$

$\therefore$   X_C $-$ X_L = 20.24 $\simeq$ 20

$\therefore$   Z = $\sqrt {{{\left( {{X_C} - {X_L}} \right)}^2} + {R^2}}$

= $\sqrt {{{\left( {20} \right)}^2} + {{60}^2}}$

= 20 $\sqrt {10}$ $\Omega$

Also cos$\phi$ = ${R \over Z}$ = ${{60} \over {20\sqrt {10} }} = {3 \over {\sqrt {10} }}$

$\therefore$   Energy dissipated in 60 sec

= ${{{{\left( {24} \right)}^2}} \over {20\sqrt {10} }} \times {3 \over {\sqrt {10} }} \times 60$

= 5.17 $\times$ 10² J

We know, path difference,

d sin$\theta$ = n$\lambda$

here n = no of bright fringer in the angle

here given

d = 0.32 $\times$ 10^-3 m

$\lambda$ = 500 $\times$ 10^$-$9 m

$\therefore$  0.32 $\times$ 10^$-$3 sin30^o = n $\times$ 500 $\times$ 10^$-$9

$\Rightarrow$  n = 320

Total number of maxima in the range

$-$ 30^o $\theta$ $\le$ 30^o is = 320 $\times$ 2 + 1 = 641

Energy density of magnetic field (U_B) = ${{{B^2}} \over {2{\mu _0}}}$

Also,

Energy density of electric field

U_E = ${1 \over 2}$ $\varepsilon$₀E²

= ${1 \over 2}$ $\varepsilon$₀ B² C²     [as   ${E \over B}$ = C ]

= ${1 \over 2}$ $\varepsilon$₀ B² $\times$ $\left( {{1 \over {{\varepsilon _0}{\mu _0}}}} \right)$    [as   C² = ${{1 \over {{\varepsilon _0}{\mu _0}}}}$]

= ${{{B^2}} \over {2{\mu _0}}}$

$\therefore$  U_B = U_E

Here, Q₁ = W₁ + Q₂

and Q₂ = W₂ + Q₃

Given that,

W₁ = W₂

$\therefore$  Q₁ $-$ Q₂ = Q₂ $-$ Q₃

$\Rightarrow$  nC_v(T₁ $-$ T₂) = nC_v(T₂ $-$ T₃)

$\Rightarrow$  T₁ $-$ T₂ = T₂ $-$ T₃

$\Rightarrow$   T₂ = ${{{T_1} + {T_3}} \over 2}$

= ${{600 + 400} \over 2}$

= 500 K

Voltage accross resistance R₄ = 5 V

$\therefore$   IR₄ = 5 V

$\Rightarrow$   500 $\times$ I = 5

$\Rightarrow$   I = ${1 \over {100}}$ A

$\therefore$   Voltage across resistor R₃ = ${1 \over {100}}\left( {100} \right)$ = 1 A

$\therefore$   Total drop in resistance R₃ and R₄ = 5 + 1 = 6V

So, voltage accross R₂ resistance is also 6V as R₃, R₄ and R₂ are in parallel

$\therefore$   Voltage accross R₁ resistor R₁ resistor = 18 $-$ 6 = 12 V

$\therefore$   Current through R₁ resistor = ${{12} \over {400}}$ = ${3 \over {100}}$ A

$\therefore$   Current through R₂ resistor

= ${3 \over {100}} - {1 \over {100}}$

= ${2 \over {100}}$ A

$\therefore$   $\left( {{2 \over {100}}} \right)$ R₂ = 6

$\Rightarrow$   R₂ = 300 $\Omega$

We know,

Activity (R) = R₀ e^{$-$$\lambda$t}

Given that,

at t = 0

R_A = R_B

$\Rightarrow$  R_0A e^{$-$$\lambda$_Ax0} = R_0B e^{$-$ $\lambda$_Bx0}

$\Rightarrow$  R_0A = R_0B

Given that at time t,

${{{R_B}} \over {{R_A}}} = {e^{^{ - 3t}}}$

$\Rightarrow$   ${{{R_{0B}}\,{e^{ - {\lambda _B}t}}} \over {{R_{0A}}\,{e^{ - {\lambda _A}t}}}}$ = e^$-$3t

$\Rightarrow$  ${e^{\left( {{\lambda _A} - {\lambda _B}} \right)t}} = {e^{ - 3t}}$

$\Rightarrow$  ($\lambda$_A $-$ $\lambda$_B)t = $-$3t

$\Rightarrow$ $\lambda$_A $-$ $\lambda$_B = $-$ 3 . . . . . (1)

We know half life, ${t_{{1 \over 2}}} = {{{{\ln }^2}} \over \lambda }$

$\therefore$  Half life of A is, ${{t_{{1 \over 2}}} = {{{{\ln }^2}} \over {{\lambda _A}}}}$

$\therefore$  $\lambda$_A = ${{\ln 2} \over {{t_{{1 \over 2}}}}}$

= ${{\ln 2} \over {\ln 2}}$

= 1

From equation (1)     we get,

$\lambda$_A $-$ $\lambda$_B = $-$ 3

$\Rightarrow$  1 $-$ $\lambda$_B = $-$ 3

$\Rightarrow$  $\lambda$_B = 4

$\therefore$  Half life of B is $\left( {{t_{{1 \over 2}}}} \right) = {{\ln 2} \over {{\lambda _B}}} = {{\ln 2} \over 4}$

When this rod move from initial position to final position then,

Gain in kinetic energy = loss in potential energy

$\therefore$  ${1 \over 2}I$$\omega$² = mgh

${1 \over 2}\left( {{{m{l^3}} \over 3}} \right)$ $\omega$² = mg$\left( {{l \over 2}\sin {{30}^ \circ }} \right)$

$\Rightarrow$  $\omega$² = ${{3g} \over {2l}}$

$\Rightarrow$  $\omega$ = $\sqrt {{{3g} \over {2l}}}$

$\Rightarrow$  $\omega$ = $\sqrt {{{3 \times 10} \over {2 \times 0.5}}}$

$\Rightarrow$  $\omega$ = $\sqrt {30}$ rad/sec

Let t $\propto$ G^x h^y c^z

$\therefore$   [t] = [G]^x [h]^y [c]^z    . . . . . (1)

We know,

F = ${{G{M^2}} \over {{R^2}}}$

$\Rightarrow$  G = ${{F{R^2}} \over {{M^2}}}$

$\therefore$  [G] = ${{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]} \over {\left[ {{M^2}} \right]}}$

[G] = $[{M^{ - 1}}{L^3}{T^{ - 2}}]$

Also,

E = hf

$\therefore$  [h] = ${{[E]} \over {[F]}}$

= ${{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {[{T^{ - 1}}]}}$

= $\left[ {M{L^2}{T^{ - 1}}} \right]$

[C] = $\left[ {{M^o}L\,{T^{ - 1}}} \right]$

From equation (1) we get,

$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = {\left[ {{M^{ - 1}}\,{L^3}\,{T^{ - 2}}} \right]^x}{\left[ {M\,{L^2}\,{T^{ - 1}}} \right]^y}{\left[ {{M^o}L{T^{ - 1}}} \right]^z}$

$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = \left[ {{M^{ - x + y}}\,{L^{3x + 2yz}}\,{T^{ - 2x - y - z}}} \right]$

By comparing the power of M, L, T

$-$ x + y = 0

$\Rightarrow$  x = y

3x + 2y + z = 0

$\Rightarrow$  5x + z = 0 . . . . . (2)

$-$ 2x $-$ y $-$ z = 1

$\Rightarrow$  $-$ 3x $-$ z = 1  . . . . (3)

By solving (2) and (3), we get,

x = ${1 \over 2}$ = y and z = $-$ ${5 \over 2}$

$\therefore$  t $\propto$ $\sqrt {{{Gh} \over {{C^5}}}}$

Given,
radius of circular path(r) = 0.5 cm
Magnetic field (B) = 0.5 T
Electric field (E) = 100 V/m

Charge of particle (q) = 1.6$\times$10^$-$19 C

As particle is moving in a circular path so,

${{m{v^2}} \over r} = qvB$

$\Rightarrow$  r = ${{mv} \over {qB}}$ . . . . . . . (1)

When electric field of 100 v/m is applied on the particle then particle is moving in the straight line. So, the net force on the particle is zero.

$\therefore$    F_net = 0

$\Rightarrow$   F_e = F_m

$\Rightarrow$  qE = qvB

$\Rightarrow$   E = vB  . . . . .(2)

From equation (1) and (2) we get,

r = ${m \over {qB}}\left( {{E \over B}} \right)$

= ${{mE} \over {q{B^2}}}$

$\Rightarrow$   m = ${{q{B^2}r} \over E}$

= ${{1.6 \times {{10}^{ - 19}} \times {{\left( {0.5} \right)}^2} \times 0.5 \times {{10}^{ - 2}}} \over {100}}$

= 2 $\times$ 10^$-$24 kg

Given that,

x = a cos $\omega$t

y = a sin $\omega$t

z = a $\omega$t

Velocity in x-direction,

V_x = ${{dx} \over {dt}} = - a\omega \sin \omega t$

Velocity in y-direction,

V_y = ${{dy} \over {dt}}$ = a $\omega$cos $\omega$t

Velocity in z-direction,

V_z = ${{dz} \over {dt}}$ = a$\omega$

Net velocity,

$\overrightarrow V$ = V_x$\widehat i$ + V_y$\widehat j$ + V_z$\widehat k$

Speed = $\left| {\overrightarrow V } \right| = \sqrt {V_x^2 + V_y^2 + V_z^2}$

$= \sqrt {{a^2}{\omega ^2}{{\sin }^2}\omega t + {a^2}{\omega ^2}{{\cos }^2}\omega t + {a^2}{\omega ^2}}$

$= \sqrt {{a^2}{\omega ^2}\left( {{{\sin }^2}\omega t + {{\cos }^2}\omega t} \right) + {a^2}{\omega ^2}}$

$= \sqrt {2{a^2}{\omega ^2}}$

$= \sqrt 2 a\omega$

From colour coding table,

Green line represents number = 5

Orange line represents number = 3

Yellow line represent multiplier = 10⁴

Golden line represents tollerence = $\pm$ 5%

$\therefore$   Resistance = 53 $\times$ 10⁴ $\pm$ 5%

= 530 k$\Omega$ $\pm$ 5%

Here water level is kept same all the time. So, the amount of water remove from the hole put in the tank the top to keep the water level same.

$\therefore$   Inflow volume role = outflow volume

$\Rightarrow$   ${{0.74} \over {60}} = Av$

$\Rightarrow$   ${{0.74} \over {60}} = \pi {r^2}\left( {\sqrt {2gh} } \right)$

$\Rightarrow$   ${{0.74} \over {60}} = \pi \left( {4 \times {{10}^{ - 4}}} \right) \times \sqrt {2gh}$

$\Rightarrow$   $\sqrt {2gh}$ = ${{740} \over {24\pi }}$

$\Rightarrow$   2gh = ${{740 \times 740} \over {24 \times 24 \times {\pi ^2}}}$

$\Rightarrow$   h = 4.8 m

Let the angle between two mirrors are $\theta$.

We know sum of angles of triangle = 180^o

$\therefore$  3$\theta$ = 180^o

$\Rightarrow$  $\theta$ = 60^o

Initially :



After putting 2 masses of each 'm' at a distance ${L \over 2}$ from center :



We know,

Time period (T) = 2$\pi$ $\sqrt {{{\rm I} \over C}}$

$\therefore$  T $\propto$ $\sqrt {\rm I}$

$\therefore$   Frequency (f) $\propto$ $\sqrt {{1 \over {\rm I}}}$

$\therefore$   ${{{f_1}} \over {{f_2}}}$ = $\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}}$

Also given that,
After putting two masses 'm' at both end new frequency becomes 80% of initial frequency.

$\therefore$   f₂ = 0.8f₁

$\therefore$   ${{{f_1}} \over {0.8{f_1}}}$ = $\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}}$

$\therefore$   ${{{{{\rm I}_1}} \over {{{\rm I}_2}}}}$ = 0.64

Initial moment of inertia of the system,

${{{\rm I}_1}}$ = ${{M{{\left( {2L} \right)}^2}} \over {12}}$

Final moment of inertia of the system,

I₂ = ${{M{{\left( {2L} \right)}^2}} \over {12}}$ + 2$\left( {m{{\left( {{L \over 2}} \right)}^2}} \right)$

$\therefore$   ${{M{{\left( {2L} \right)}^2}} \over {12}}$ = 0.64 $\left[ {{{M{L^2}} \over 3} + {{m{L^2}} \over 2}} \right]$

$\Rightarrow$   ${{M{L^2}} \over {3 \times 0.64}}$ = ${{M{L^2}} \over 3}$ + ${{M{L^2}} \over 2}$

$\Rightarrow$   ${M \over {1.92}} - {M \over 3} = {m \over 2}$

$\Rightarrow$   ${{1.08M} \over {3 \times 1.92}}$ = ${m \over 2}$

$\Rightarrow$   ${m \over M}$ = ${{1.08 \times 2} \over {3 \times 1.92}}$ = 0.37

Volume of this spherical layer,

dv = (4$\pi$r²)dr

charge present in this layer,

dq = $\rho$ (4$\pi$r² dr)

= ${A \over {{r^2}}}{e^{ - {{2r} \over a}}}\,\,\left( {4\pi {r^2}dr} \right)$

= $A\,{e^{ - {{2r} \over a}}}\left( {4\pi dr} \right)$

$\therefore$  Total charge in the sphere,

Q= $\int\limits_0^R {4\pi A{e^{ - {{2r} \over a}}}} \,dr$

= 4$\pi$A$\int\limits_0^R {{e^{ - {{2r} \over a}}}} \,dr$

= 4$\pi$A$\left[ {{{{e^{ - {{2r} \over a}}}} \over { - {2 \over a}}}} \right]_0^R$

= 4$\pi$A $\left( { - {a \over 2}} \right)\left( {{e^{ - {{2R} \over a}}} - 1} \right)$

$\therefore$  Q = 2$\pi$aA $\left( {1 - {e^{ - {{2R} \over a}}}} \right)$

$\Rightarrow$  ${1 - {e^{ - {{2R} \over a}}}}$ = ${Q \over {2\pi aA}}$

$\Rightarrow$  ${{e^{ - {{2R} \over a}}}}$ = 1 $-$ ${Q \over {2\pi aA}}$

$\Rightarrow$  ${e^{{{2R} \over a}}}$ = ${1 \over {1 - {Q \over {2\pi aA}}}}$

$\Rightarrow$  ${{2R} \over a} = \log \left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$

$\Rightarrow$  R = ${a \over 2}$ log $\left( {{1 \over {1 - {Q \over {2\pi aA}}}}} \right)$

tan 45^o = ${F \over {mg}}$

$\therefore$  F = mg

= 10 $\times$ 10

= 100 N

Displacement,

x = 3t² + 5

$\therefore$  v = ${{dx} \over {dt}} = 6t$

At t = 0,   velocity = 6 $\times$ 0 = 0

at t = 5, velocity = 5 $\times$ 6 = 30 m/s

we know from work energy theorem,

Work (W) = change in kinetic energy ($\Delta$K)

= ${1 \over 2}mv_F^2 - {1 \over 2}mv_i^2$

= ${1 \over 2}$ $\times$ 2 $\times$ (30)² $-$ 0

= 900 J

Given,

$\lambda$ = 800 nm

$\therefore$  f = ${{3 \times {{10}^8}} \over {800 \times {{10}^{ - 9}}}}$

= 3.75 $\times$ 10¹⁴ Hz

Available frequency for signal bandwith

= 1% of F

= 3.75 $\times$ 10¹⁴ $\times$ ${1 \over {100}}$

= 3.75 $\times$ 10¹² Hz

One TV signal needs = 6 MHz band width

$\therefore$  Total number of channel possible

= ${{3.75 \times {{10}^{12}}} \over {6 \times {{10}^6}}}$

= 6.25 $\times$ 10⁵

Energy required to move a satellite from earth surface to height h is,

E₁ = U_h $-$ U_surface

= $-$ ${{GMm} \over {R + h}} - \left( { - {{GMm} \over R}} \right)$

= GMm $\left( {{1 \over R} - {1 \over {R + h}}} \right)$

= ${{GMm} \over {R(R + h)}} \times h$

We know, for sattelite at height h.

Centrifigual force = Gravitational force

$\Rightarrow$  ${{m{v^2}} \over {R + h}} = {{GMm} \over {{{\left( {R + h} \right)}^2}}}$

$\Rightarrow$  $mv$² = ${{GMm} \over {R + h}}$

$\therefore$   ${1 \over 2}m{v^2}$ = ${{GMm} \over {2\left( {R + h} \right)}}$

$\therefore$  Kinetic energy (E₂) = ${{GMm} \over {2(R + h)}}$

Given that,

E₁ = E₂

$\therefore$  ${{GMm} \over {R(R + h)}} \times h = {{GMm} \over {2\left( {R + h} \right)}}$

$\Rightarrow$  ${h \over R} = {1 \over 2}$

$\Rightarrow$  h = ${R \over 2}$

$\therefore$  h = ${{6.4 \times {{10}^3}} \over 2}$

= 3.2 $\times$ 10³ km

Case 1 :

Ge need 0.3 V for start conducting and Si need 0.7 V for start conducting.

So, conduction will be start by Ge.

Out of 12 V, 0.3V will be consumed by Ge.

$\therefore$  V₀₁ = 12 $-$ 0.3 = 11.7 V

Case 2 :

Here as Ge diode is in reverse bias so no conduction will happen through Ge diode

As Si diode is in forward bias so conduction will happen through Si diode.

Here Si diode consume 0.7 V.

So, V₀₂ = 12 $-$ 0.7 = 11.3 V

$\therefore$  Value of V₀ changes by

= 11.7 $-$ 11.3 = 0.4 V

Electric field due to $\sqrt {10} \,\mu C$ charge :

$\overrightarrow {{E_1}} = -$ E₁ sin$\theta$₁ $\widehat i$ + E₁ cos$\theta$₁ $\widehat j$

Where,

E₁ $= {1 \over {4\pi {\varepsilon _0}}} \times {{\left| {{q_1}} \right|} \over {{r_1}^2}}$

$= 9 \times {10^9} \times {{\sqrt {10} \times {{10}^{ - 6}}} \over {{{\left( {\sqrt {{1^2}} + {3^2}} \right)}^2}}}$

$= {{9 \times {{10}^3}} \over {\sqrt {10} }}\,v/m$

sin $\theta$₁ $=$ ${1 \over {\sqrt {10} }}$

and cos$\theta$₁ = ${3 \over {\sqrt {10} }}$

$\therefore$   $\overrightarrow {{E_1}} = {{9 \times {{10}^3}} \over {\sqrt {10} }}\,\left( { - {1 \over {10}}\widehat i + {3 \over {\sqrt {10} }}\widehat j} \right)$

$= 9 \times {10^2}\left( { - \widehat i + 3\widehat j} \right)$

Electric field due to $-$ 25 $\mu$C charge,



$\overrightarrow {{E_2}} =$ E₂ sin$\theta$₂$\widehat i$ $-$ E₂ cos$\theta$₂ $\widehat j$

where

E₂ $= {1 \over {4\pi {\varepsilon _0}}} \times {{\left| {{9_2}} \right|} \over {r_2^2}}$

$= 9 \times {10^9} \times {{25 \times {{10}^{ - 6}}} \over {{{\left( {\sqrt {{4^2} + {3^2}} } \right)}^2}}}$

$= 9 \times {10^3}$ V/m

sin$\theta$₂ = ${4 \over 5}$

and cos$\theta$₂ = ${3 \over 5}$

$\therefore$   $\overrightarrow {{E_2}} = 9 \times {10^3}\,\,\left( {{4 \over 5}\widehat i - {3 \over 5}\widehat j} \right)$

$= 18 \times {10^2}\left( {4\widehat i - 3\widehat j} \right)$

$\therefore$   Net electric field,

$\overrightarrow E$ = ${\overrightarrow E _1}$ + ${\overrightarrow E _2}$

$= \left( {63\widehat i - 27\widehat j} \right) \times {10^2}\,\,V/m$

Given that,

B = B₀[sin (3.14 $\times$ 10⁷) ct + sin(6.28 $\times$ 10⁷) ct]

This light wave is non-monochromotic wave as here is two different frequency in the magnetic field.

Given work function ($\phi$) = 4.7 eV

Question says to find the maximum kinetic energy, so we have to use maximum frequency among the available frequency.

As,

K_max = E_max $-$ $\phi$

Emax = hF

= h $\times$ ${\omega \over {2\pi }}$

= 6.6 $\times$ 10^$-$34 $\times$ ${{6.28 \times {{10}^7} \times 3 \times {{10}^8}} \over {2\pi }}$

= 6.6$\times$ 3 $\times$ 10^$-$19 J

= ${{6.6 \times 3 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}eV$

= 12.375 $eV$

$\therefore$  K_max = 12.375 $-$ 4.7

= 7.675 $eV$

$\simeq$  7.7 $eV$

Here, ${C_1} = {{{K_1}{\varepsilon _0}.L.{L \over 2}} \over {{d \over 2}}} = {{{K_1}{\varepsilon _0}{L^2}} \over d}$

${C_2}{{{K_2}{\varepsilon _0}.L.{L \over 2}} \over {{d \over 2}}} = {{{K_2}{\varepsilon _0}{L^2}} \over d}$

${C_3} = {{{K_3}{\varepsilon _0}{L^2}} \over d}$

${C_4} = {{{K_4}{\varepsilon _0}{L^2}} \over d}$

Here  C₁ C₂ are in series and C₃, C₄ are in series.

Equivalent capacitance point A and B is,

C_eq $=$ ${{{C_1}{C_2}} \over {{C_1} + {C_2}}} + {{{C_3}{C_4}} \over {{C_3} + {C_4}}}$

$= {C_{12}} + {C_{34}}$

${C_{12}}\,\, = \,\,{{{{{K_1}{\varepsilon _0}{L^2}} \over d} \times {{{K_2}{\varepsilon _0}{L^2}} \over d}} \over {{{{K_1}{\varepsilon _0}{L^2}} \over d} + {{{K_2}{\varepsilon _0}{L^2}} \over d}}}$

$= \,\,{{{K_1}{K_2}} \over {{K_1} + {K_2}}} \times {{{\varepsilon _0}{L^2}} \over d}$

Similarly,

${C_{34}} = \,\,{{{K_3}{K_4}} \over {{K_3} + {K_4}}} \times {{{\varepsilon _0}{L^2}} \over d}$

$\therefore$  C_eq $=$ $\left[ {{{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}} \right]{{{\varepsilon _0}{L^2}} \over d}$

$\Rightarrow$  ${{K{\varepsilon _0}{L^2}} \over d}$  $= \left[ {{{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}} \right]{{{\varepsilon _0}{L^2}} \over d}$

$\Rightarrow$  $K = {{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}$