JEE Main 2019 12th April Evening Shift Question Paper with Solutions

Since the wave is propagating in positive z-direction

So acceptable magnetic field component will be

$\overrightarrow B = {B_0}\sin \left( {kz - \omega t} \right)$$\widehat i$

$\because$ C = ${{{E_o}} \over {{B_o}}}$

$\Rightarrow$ ${B_o} = {{{E_o}} \over C}$ = ${{60} \over {3 \times {{10}^8}}}$ = 2 $\times$ 10^-7

and $\omega = 2\pi f$ = 2$\pi$$\times$23.9$\times$10⁹ = 1.5 × 10¹¹ Hz

and k = ${\omega \over c}$ = ${{1.5 \times {{10}^{11}}} \over {3 \times {{10}^8}}}$ = 0.5 × 10³

Note : When wave is propagating in positive z-direction then sign of kz and ${\omega t}$ should be opposite.
From option you can see only option D can be correct.

By Snell’s law at AB

$\mu$₁ sin $\theta$ = $\mu$₂ sin $\alpha$

sin $\alpha$ = ${{{\mu _1}} \over {{\mu _2}}}\sin \theta$ ....... (1)

Critical angle at BC,

$\mu$₂ sin i_c = $\mu$₁ sin 90^o

$\Rightarrow$ sin i_c = ${{{\mu _1}} \over {{\mu _2}}}$ ..... (2)

For Total internal reflection to occur,

90^o - $\alpha$ > i_c

Taking sin both side,

sin(90^o - $\alpha$) > sin i_c

$\Rightarrow$ cos $\alpha$ > ${{{\mu _1}} \over {{\mu _2}}}$ (from equation 2)

$\Rightarrow$ $\sqrt {1 - {{\mu _1^2} \over {\mu _2^2}}{{\sin }^2}\theta }$ > ${{{\mu _1}} \over {{\mu _2}}}$ (from equation 1)

Squaring both sides, we get

${1 - {{\mu _1^2} \over {\mu _2^2}}{{\sin }^2}\theta }$ > ${{{\mu _1^2} \over {\mu _2^2}}}$

$\Rightarrow$ ${1 - {{\mu _1^2} \over {\mu _2^2}}}$ > ${{{\mu _1^2} \over {\mu _2^2}}{{\sin }^2}\theta }$

$\Rightarrow$ ${{\mu _2^2} \over {\mu _1^2}} - 1$ > ${{{\sin }^2}\theta }$

$\Rightarrow$ sin $\theta$ < $\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1}$

$\Rightarrow$ $\theta < {\sin ^{ - 1}}\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1}$

For second excited state n = 3

$mvr = {{3h} \over {2\pi }}$ ........(1)

$mv = {h \over \lambda }$ .........(2)

Dividing (1) by (2), we get

$r = {{3\lambda } \over {2\pi }}$

$\Rightarrow$ $\lambda = {{2\pi r} \over 3}$ = ${{2 \times 3.14 \times 4.65} \over 3}$ = 9.7 $\mathop A\limits^o$

C(t) = 4sin(20000 $\pi$t), A_c = 4

m(t) = 2sin(2000$\pi$t), A_m = 2

Modulation index (m) = ${{{A_m}} \over {{A_c}}}$ = ${2 \over 4}$ = 0.5

Lower side band frequency = f_c - f_m

= ${{20000\pi } \over {2\pi }}$ - ${{2000\pi } \over {2\pi }}$

= 10000 - 1000

= 9000

= 9 kHz

$I = {E \over R}\left( {1 - {e^{ - {t \over \tau }}}} \right)$ where $\tau$ = ${L \over R}$

We know $dq = Idt$
$\Rightarrow$ $\int\limits_0^Q {dq} = {E \over R}\int\limits_0^\tau {\left( {1 - {e^{ - {t \over \tau }}}} \right)dt}$
$\Rightarrow$ Q $= {E \over R}\left[ t \right]_0^\tau + {E \over R} \times \tau \left[ {{e^{ - {t \over \tau }}}} \right]_0^\tau$
         = ${E \over R}\tau + {E \over R}\tau \left[ {{e^{ - 1}} - 1} \right]$
         = ${E \over R}\tau + {E \over R}\tau {e^{ - 1}} - {E \over R}\tau$
         = ${{E\tau } \over {eR}}$
         = ${{E\tau } \over {2.7R}}$ [ as e = 2.7 ]
         = ${{EL} \over {2.7{R^2}}}$ [as $\tau$ = ${L \over R}$ ]

v = b$\sqrt x$
$\Rightarrow$ ${{dx} \over {dt}}$ = b$\sqrt x$
$\Rightarrow$$\int\limits_0^x {{{dx} \over {\sqrt x }}} = \int\limits_0^t {bdt}$
$\Rightarrow$$\left[ {{{{x^{ - {1 \over 2} + 1}}} \over { - {1 \over 2} + 1}}} \right]_0^x$ = $b\left[ t \right]_0^t$
$\Rightarrow$ ${x^{{1 \over 2}}} = {{bt} \over 2}$
$\Rightarrow$ $x = {{{b^2}{t^2}} \over 4}$

$\therefore$ v = ${{dx} \over {dt}}$ = ${{{b^2}} \over 4}$ $\times$ 2t = ${{{b^2}t} \over 2}$
When t = $\tau$ then speed v $= {{{b^2}\tau } \over 2}$

P$\times$t = mS$\Delta$t + mL_v

$\Rightarrow$ ${{{V^2}} \over R}t$ = mS$\Delta$t + mL_v

$\Rightarrow$ ${{{{\left( {200} \right)}^2}} \over {20}}t$ = $1 \times 4200 \times \left( {100 - 20} \right)$ + $1 \times 2260 \times {10^3}$

$\Rightarrow$ 2000t = 336000 + 2260000

$\Rightarrow$ t = 1298 s = 21.6 min $\simeq$ 22 min

As V = ${q \over C}$

$\therefore$ V $\propto$ ${1 \over C}$

So ${{{V_1}} \over {{V_2}}} = {{{C_2}} \over {{C_1}}}$

$\Rightarrow$ ${{{V_1}} \over {{V_2}}} = {6 \over 4} = {3 \over 2}$

Also V₁ + V₂ = 10

$\therefore$ V₁ + ${2 \over 3}$V₁ = 10

$\Rightarrow$ ${{5{V_1}} \over 3}$ = 10

$\Rightarrow$ V₁ = 6 V

The charge on 4 $\mu$F capacitor is = C₁V₁ = 4 $\times$ 6 = 24 $\mu$C

When unpolarized light of intensity I₀ passes through P₁, then intensity

I₁ = ${{{I_0}} \over 2}$

as we know, I = I₀ cos²$\theta$

Given that angle between P₂ & P₃ = 60^o and

P₁ and P₃ are crossed that means angle between P₁ and P₃ is 90^o

so angle between P₁ and P₂ = 90° – 60° = 30°

I₂ = ${{{I_0}} \over 2}{\cos ^2}30^\circ$ = ${{3{I_0}} \over 8}$

I₃ = ${{3{I_0}} \over 8}{\cos ^2}60^\circ$ = ${{3{I_0}} \over {32}}$ = I

$\therefore$ ${{{I_0}} \over I}$ = ${{{I_0}} \over {{{3{I_0}} \over {32}}}}$ = ${{32} \over 3}$ = 10.67

In $\Delta$ORP,

sin $\theta$ = ${{r/2} \over r}$ = ${1 \over 2}$ = sin 30^o

$\Rightarrow$ $\theta$ = 30^o

As the bead is at rest with respect to the circular ring at position P, The net force on bead is zero.

tan $\theta$ = ${{m{\omega ^2}r/2} \over {mg}}$

$\Rightarrow$ tan 30^o = ${{{\omega ^2}r} \over {2g}}$

$\Rightarrow$ ${1 \over {\sqrt 3 }}$ = ${{{\omega ^2}r} \over {2g}}$

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

N = mg + 20 sin30^o

    = 50 + $20 \times {1 \over 2}$

    = 60 N

Limition friction, f_L = $\mu$N = 0.2 $\times$ 60 = 12 N

And horizontal force(F_x) = 20 cos30^o = 10$\sqrt 3$ = 17.32 N

As F_x > f_L, then the block will move with acceleration a_A.

a_A = ${{{F_x} - {f_L}} \over {{m_A}}}$ = ${{17.32 - 12} \over 5}$ = ${{5.32} \over 5}$

N + 20 sin30^o = mg

N = mg - 20 sin30^o

    = 50 - 10 = 40 N

Limition friction, f_L = $\mu$N = 0.2 $\times$ 40 = 8 N

And horizontal force(F_x) = 20 cos30^o = 10$\sqrt 3$ = 17.32 N

As F_x > f_L, then the block will move with acceleration a_B.

a_B = ${{{F_x} - {f_L}} \over {{m_B}}}$ = ${{17.32 - 8} \over 5}$ = ${{9.32} \over 5}$

Difference between acceleration a_A - a_B = ${{9.32} \over 5} - {{5.32} \over 5}$ = ${4 \over 5}$ = 0.8 m/s²

n = 1 (Ground state)
n = 2 (First excitate state)
n = 3 (Second excitate state)
n = 4 (Third excitate state)

${{hc} \over {{\lambda _1}}} = 13.6\left( {{1 \over 9} - {1 \over {16}}} \right)$

${{hc} \over {{\lambda _2}}} = 13.6\left( {{1 \over 4} - {1 \over 9}} \right)$

${{{\lambda _2}} \over {{\lambda _1}}} = {{\left( {{7 \over {9 \times 16}}} \right)} \over {\left( {{5 \over {9 \times 4}}} \right)}}$ = ${7 \over {20}}$

$\therefore$ ${{{\lambda _1}} \over {{\lambda _2}}} = {{20} \over 7}$

$\eta = 1 - {{{T_2}} \over {{T_1}}}$
$\Rightarrow$ ${1 \over 6} = 1 - {{{T_2}} \over {{T_1}}}$
$\Rightarrow$ ${{{T_2}} \over {{T_1}}} = {5 \over 6}$

When the temperature of the sink is reduced by 62ºC, then
$\eta ' = 1 - {{{T_2} - 62} \over {{T_1}}}$
$\Rightarrow$ $2 \times {1 \over 6} = 1 - {{{T_2}} \over {{T_1}}} + {{62} \over {{T_1}}}$
$\Rightarrow$ ${1 \over 3} = {1 \over 6} + {{62} \over {{T_1}}}$
$\Rightarrow$ T₁ = 372 K = 99^o C

T₂ = ${5 \over 3} \times 372$ = 310 K = 37^o C

At constant pressure,

W = P$\Delta$V = nR$\Delta$T

At constant pressure, heat supplied

Q = nC_p$\Delta$T

${W \over Q} = {R \over {{C_p}}}$

For diatomic gas, C_p = ${7 \over 2}R$

$\therefore$ ${{10} \over Q} = {R \over {{7 \over 2}R}}$

$\Rightarrow$ Q = 10 $\times$ ${7 \over 2}$ = 35 J

Total charge = 2Q
Charging density $\rho$ = kr
Radius = R Charge enclosed in the sphere,
q_in = $\int\limits_0^R {\rho dV}$

$\Rightarrow$ 2Q = $\int\limits_0^R {kr4\pi {r^2}dr}$

$\Rightarrow$ 2Q = $k4\pi \int\limits_0^R {{r^3}dr}$

$\Rightarrow$ 2Q = $k4\pi {{{R^4}} \over 4}$

$\Rightarrow$ k = ${{2Q} \over {\pi {R^4}}}$ ........... (1)

Force on charge at A will be due to charge at B and due to force applied by the charge in sphere.

Here F_sphere = EQ

Using Gauss law, we can find electric field at point A due to sphere,

∮ $\overrightarrow E .d\overrightarrow A$ = ${{{q_{in}}} \over {{ \in _0}}}$

$\Rightarrow$ $E\left( {4\pi {a^2}} \right)$ = ${{\int\limits_0^a {\rho dV} } \over {{ \in _0}}}$

$\Rightarrow$ $E\left( {4\pi {a^2}} \right)$ = ${{k4\pi {{{a^4}} \over 4}} \over {{ \in _0}}}$

$\Rightarrow$ E = ${{k{a^2}} \over {4{ \in _0}}}$

As on charge A net force is zero then,

F_AB = F_sphere

$\Rightarrow$ ${{Q \times Q} \over {4\pi { \in _0}{{\left( {2a} \right)}^2}}}$ = ${{k{a^2}} \over {4{ \in _0}}}$ $\times$ Q

$\Rightarrow$ ${Q \over {4\pi {a^2}}} = k{a^2}$

$\Rightarrow$ ${Q \over {4\pi {a^2}}} = {{2Q} \over {\pi {R^4}}}{a^2}$ [ from equation (1)]

$\Rightarrow$ $8{a^4} = {R^4}$

$\Rightarrow$ $a = {8^{ - 1/4}}R$

Sound level = 10${\log _{10}}\left( {{I \over {{I_0}}}} \right)$

$\Rightarrow$ 120 = 10${\log _{10}}\left( {{I \over {{{10}^{ - 12}}}}} \right)$

$\Rightarrow$ 12 = ${\log _{10}}\left( {{I \over {{{10}^{ - 12}}}}} \right)$

$\Rightarrow$ 10¹² = ${{I \over {{{10}^{ - 12}}}}}$

$\Rightarrow$ I = 1 W/m²

Also we know,

I = ${P \over {4\pi {r^2}}}$

$\Rightarrow$ 1 = ${2 \over {4\pi {r^2}}}$

$\Rightarrow$ r = 40 cm

B = ${{{\mu _0}I} \over {4\pi r}}\left( {\sin {\theta _1} + \sin {\theta _2}} \right)$

As $\theta$₁ = $\theta$₂ = $\theta$

B = ${{{\mu _0}I} \over {4\pi r}}\left( {2\sin \theta } \right)$

   = ${{{{10}^{ - 7}} \times 5} \over {4 \times {{10}^{ - 2}}}} \times 2 \times {3 \over 5}$

   = 1.5 $\times$ 10^-5 T

We know, R = ${{mv} \over {qB}}$

= ${{\sqrt {2mK} } \over {qB}}$

= ${{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 100 \times 1.6 \times {{10}^{ - 19}}} } \over {1.6 \times {{10}^{ - 19}} \times 1.5 \times {{10}^{ - 3}}}}$

= ${\sqrt 5 }$ cm
From diagram,

$\sin \theta = {2 \over R}$ = ${2 \over {\sqrt 5 }}$

So $\tan \theta = {2 \over 1}$

In $\Delta$MNQ,

$\tan \theta = {x \over 6}$

$\therefore$ ${x \over 6}$ = ${2 \over 1}$

$\Rightarrow$ $x$ = 12 cm

From diagram you can see,

QP = d > 12

By checking all the options you can see possible value of d = 12.87 cm

${4 \over 3}$$\pi$R³ = 27 $\times$ ${4 \over 3}$$\pi$r³
$\Rightarrow$ r = ${R \over 3}$

Terminal velocity, ${V_T} = {2 \over 9}{{{r^2}} \over \eta }\left( {{\sigma _s} - {\rho _l}} \right)g$
$\therefore$ V_T $\propto$ r²

$\therefore$ ${{{v_1}} \over {{v_2}}} = {{{R^2}} \over {{r^2}}}$ = ${{{R^2}} \over {{{{R^2}} \over 9}}}$ = 9

Galvanometer is converted into ammeter of range 0 to I₀ :
(I_o - I_g)R_A = I_gG
$\Rightarrow$ ${R_A} = {{{I_g}G} \over {\left( {{I_0} - {I_g}} \right)}}$............(1)

Galvanometer is converted into voltmeter of range 0 to V :
V = I_g(R_V + G)
Also given, V = GI₀
$\therefore$ GI₀ = I_g(R_V + G)
$\Rightarrow$ ${R_V} = {{G\left( {{I_0} - {I_g}} \right)} \over {{I_g}}}$ .........(2)

From equation (1) and (2)
${R_A}{R_V} = {G^2}$
${{{R_A}} \over {{R_V}}} = {\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)^2}$

W = mg

as m = constant everywhere

$\therefore$ W $\propto$ g

${{{g_E}} \over {{g_p}}}$ = ${9 \over 4}$

We know,

$g = {{GM} \over {{R^2}}}$

$\therefore$ ${{{g_E}} \over {{g_p}}} = {{{M_E}} \over {{M_p}}} \times {{R_p^2} \over {R_E^2}}$

$\Rightarrow$${9 \over 4} = {9 \over 1} \times {{R_p^2} \over {R_E^2}}$

$\Rightarrow$ ${{{R_p}} \over {{R_E}}} = {1 \over 2}$

${R_p} = {{{R_E}} \over 2}$ = ${R \over 2}$ [Here R_E = R ]

As observer goes away from source S₁ so apparent frequency,
${f_1} = \left( {{{v - u} \over v}} \right)f$
here $v$ = speed of sound, $u$ = speed of observer

As observer goes towards source S₂ so apparent frequency,
${f_2} = \left( {{{v + u} \over v}} \right)f$

Beat frequency = ${f_2} - {f_1}$ = 10
$\Rightarrow$ $\left( {{{v + u} \over v}} \right)f$ - $\left( {{{v - u} \over v}} \right)f$ = 10
$\Rightarrow$ $f\left( {{{v - u - v + u} \over v}} \right)$ = 10
$\Rightarrow$ $f\left( {{{2u} \over v}} \right)$ = 10
$\Rightarrow$ $u = {{10v} \over {2f}}$ = ${{10 \times 330} \over {2 \times 660}}$ = 2.5 m/s

We know,

$\lambda$ = 2(${l_2} - {l_1}$)

given ${l_2}$ = 70 cm and ${l_1}$ = 30 cm

$\therefore$ $\lambda$ = 2(70 - 30) = 80 cm

Also we know, v = $\lambda$$f$ = 0.8 $\times$ 480 = 384 m/s

${I_L} = {{{V_Z}} \over {{R_L}}}$

= ${{6\,V} \over {4\,k\Omega }}$ = 1.5 mA

Let input voltage = V

$\therefore$ ${{V - {V_Z}} \over {{R_S}}} = {I_s}$

Here maximum input voltage = 16 V

$\therefore$ ${{16 - 6} \over {2\,k\Omega }} = {I_s}$

$\Rightarrow$ I_S = 5 mA

$\therefore$ I_Z = I_S - I_L = 5 - 1.5 = 3.5 mA

Change in length due to temperature change,

$\Delta$$l$ = $l$$\alpha$$\Delta$T

${{\Delta l} \over l}$ = $\alpha$T [ Here $\Delta$T = T ]

Y = ${{{F \over {\pi {r^2}}}} \over {{{\Delta l} \over l}}}$

= ${{{F \over {\pi {r^2}}}} \over {\alpha T}}$

$\Rightarrow$ Y = ${F \over {\pi {r^2}\alpha T}}$

$\Rightarrow$ $\alpha$ = ${F \over {\pi {r^2}YT}}$

We know, The coefficient of volume expansion ($\gamma$) = 3$\alpha$

$\therefore$ $\gamma$ = ${{3F} \over {\pi {r^2}YT}}$

${x_{COM}} = {{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}} \over {{m_1} + {m_2} + {m_3}}}$

           = ${{0 + \left( {100} \right)10 + \left( {150} \right)\left( {0.5} \right)} \over {300}}$

           = ${{175} \over {300}}$ = ${7 \over {12}}$ m

${y_{COM}} = {{{m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3}} \over {{m_1} + {m_2} + {m_3}}}$

           = ${{0 + \left( {100} \right)0 + \left( {150} \right){{\sqrt 3 } \over 2}} \over {300}}$

           = ${{75\sqrt 3 } \over {300}}$ = ${{\sqrt 3 } \over 4}$ m

For a spring, k$\times$$l$ = constant.

$\therefore$ k₁${l_1}$ = k₂${l_2}$

$\Rightarrow$ ${{{k_1}} \over {{k_2}}} = {{{l_2}} \over {{l_1}}}$ = ${{{l_2}} \over {n{l_2}}}$ = ${1 \over n}$

Lets take an element hollow sphere of thickness dr

Vol. of element dV = 4$\pi$r²dr

Total number of molecules,

N = $\int\limits_0^\infty {n\,dV}$

= $\int\limits_0^\infty {{n_0}{e^{ - \alpha {r^4}}}\,4\pi {r^2}dr}$

Let ${{e^{ - \alpha {r^4}}}}$ = t ......................(1)

$\therefore$ $- 4\alpha {e^{ - \alpha {r^4}}}{r^3}dr$ = dt

$\Rightarrow$ ${{dt} \over { - 4\alpha r}} = {e^{ - \alpha {r^4}}}{r^2}dr$ .....................(2)

Taking $\ln$ to the both sides of the equation (1), we get

${\ln}\left( t \right) = - \alpha {r^4}$

$\Rightarrow$ $r = {\left( {{{\ln t} \over { - \alpha }}} \right)^{{1 \over 4}}}$ .......(3)

Putting this value of r in equation (2),

${{dt} \over { - 4\alpha {{\left( {{{\ln t} \over { - \alpha }}} \right)}^{{1 \over 4}}}}} = {e^{ - \alpha {r^4}}}{r^2}dr$

$\Rightarrow$ ${{{\alpha ^{{{ - 3} \over 4}}}dt} \over { - 4{{\left( { - \ln t} \right)}^{{1 \over 4}}}}} = {e^{ - \alpha {r^4}}}{r^2}dr$

Putting in the integration, we get

N = ${n_0}4\pi \int\limits_1^0 {{{{\alpha ^{{{ - 3} \over 4}}}dt} \over { - 4{{\left( { - \ln t} \right)}^{{1 \over 4}}}}}}$

= $- {n_0}{\alpha ^{{{ - 3} \over 4}}}\pi \int\limits_1^0 {{{dt} \over {{{\left( { - \ln t} \right)}^{{1 \over 4}}}}}}$

$\therefore$ N $\propto$ ${n_0}{\alpha ^{{{ - 3} \over 4}}}$