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

Capacitance, C = 0.2 $\mu$F = 0.2 $\times$ 10^$-$6 F

Inductance, L = 0.5 m H = 0.5 $\times$ 10^$-$3 H

Let, current = I.

Using energy conservation,

U_E + 0 = U_E^' + U_b^'

$\Rightarrow$$\,\,\,\,$ ${1 \over 2}$ cv² + 0 = ${1 \over 2}$ c$v_1^2$ + ${1 \over 2}$LI²

$\Rightarrow$$\,\,\,\,$ ${1 \over 2}$ $\times$ 0.2 $\times$ 10^$-$6 $\times$ 10²

= ${1 \over 2} \times$ 0.2 $\times$ 10^$-$6 $\times$ 5² + ${1 \over 2}$ $\times$ 0.5 $\times$ 10^$-$3 $\times$ I²

By solving this,

I = $\sqrt 3 \times$ 10^$-$1 A

= 0.17 A.

Angular position of n^th minima from central maxima,

sin $\theta$ = ${{n\lambda } \over a}$

here n = 2

$\therefore\,\,\,\,$ sin $\theta$ = ${{2 \times 550 \times {{10}^{ - 9}}} \over {22 \times {{10}^{ - 5}}}}$ = ${1 \over 2}$

$\therefore\,\,\,$ $\theta$ = ${\pi \over 6}$ rad.

Let us assume potential at point M is X volts. Now you can see a plain wire is directly connected between point N and point M so point N will also have same potential as point M.

Similarly, a plain wire is directly connected between point N and point O so point O will also have same potential as point N.

Also, a plain wire is directly connected between point O and point P so point P will also have same potential as point O.

Now you can see point P and point M both have same potential X, so no current will flow through that path so you can remove that resistor. Similarly, as point O and point M both have same potential X, so no current will flow through that path also so you can remove that resistor too.

So new circuit becomes
$\therefore$ R_AB = R + R = 2R

As we know,

momentum (p) = ${h \over \lambda }$

Let one perticle is moving in x direction and other is in y dirrection.

$\therefore\,\,\,\,$ momentum of each electrons ${h \over {{\lambda _1}}}\widehat i$ and ${h \over {{\lambda _2}}}\widehat j$

$\therefore\,\,\,\,$ Velocity of each electrons ${h \over {m{\lambda _1}}}\widehat i$ and ${h \over {m{\lambda _2}}}\widehat j$

$\therefore\,\,\,\,$ Velocity of center of mass $\left( {{{\overrightarrow \upsilon }_{_{CM}}}} \right)$ = ${{\upsilon _1^{\widehat i} + \upsilon _2^{\widehat j}} \over 2}$

Now, velocity of first electron about center of mass,

${\overrightarrow \upsilon _{_{1CM}}}$ = $\upsilon _1^{\widehat i}$ $-$ $\left( {{{\upsilon _1^{\widehat i} + \upsilon _2^{\widehat j}} \over 2}} \right)$

= ${{\upsilon _1^{\widehat i} - \upsilon _2^{\widehat j}} \over 2}$

Similarly,

${\overrightarrow \upsilon _{_{2CM}}} = {{\upsilon _2^{\widehat j} - \upsilon _1^{\widehat i}} \over 2}$

Here,

$\left| {{{\overrightarrow \upsilon }_{_{1CM}}}} \right| = \left| {{{\overrightarrow \upsilon }_{_{2CM}}}} \right| = {1 \over 2}$

$\therefore\,\,\,\,$ $\upsilon$ = ${1 \over 2}$ $\sqrt {{{{h^2}} \over {{m^2}\lambda _1^2}} + {{{h^2}} \over {{m^2}\lambda _2^2}}}$

$\Rightarrow$$\,\,\,\,$ m $\upsilon$ = ${1 \over 2}$ ${\sqrt {{{{h^2}} \over {\lambda _1^2}} + {{{h^2}} \over {\lambda _2^2}}} }$

$\Rightarrow$$\,\,\,\,$ ${h \over {{\lambda _{CM}}}}$ = h${\sqrt {{1 \over {4\lambda _1^2}} + {1 \over {4\lambda _2^2}}} }$

$\Rightarrow$$\,\,\,\,$ ${1 \over {{\lambda _{CM}}}} = {{\sqrt {\lambda _1^2 + \lambda _2^2} } \over {2{\lambda _1}{\lambda _2}}}$

$\Rightarrow$$\,\,\,\,$ ${\lambda _{CM}} = {{2{\lambda _1}{\lambda _2}} \over {\sqrt {\lambda _1^2 + \lambda _2^2} }}$

Initial activity, N_o = 0.8 $\mu$^Ci

Activity at time t, N = N₀ e^{$-$$\lambda$t}

Activity in 1 cm³ blood after 10 hr, n = 300 decays per minute

= ${{300} \over {60}}\,dps$

= 5 dps.

Activity in whole blood after to hr = N₀ e^{$-$$\lambda$$\times$10}

$\therefore\,\,\,\,$ Volume of the total blood = ${{{N_0}{e^{ - 10\lambda }}} \over n}$

= ${{0.8 \times {{10}^{ - 7}} \times 3.7 \times {{10}^{10}} \times 0.84} \over 5}$

= 4.97 $\times$ 10³ cm³

= 4.97 litres

$\simeq$ 5 liters.

Let, t₁ and t₂ are time taken to move on the smooth and rough surface for smooth surface,

S = ${1 \over 2}$ g sin45^o $t_1^2$

$\Rightarrow$$\,\,\,\,$ t₁ = $\sqrt {{{2\sqrt 2 S} \over g}}$

For rough surface,

S = ${1 \over 2}$ g (sin45^o $-$ $\mu$_k cos45^o) $t_2^2$

$\Rightarrow$$\,\,\,\,$ t₂ = $\sqrt {{{2\sqrt 2 S} \over {g\left( {1 - {\mu _k}} \right)}}}$

Here ${\mu _k}$ = Kinetic friction.

According to question,

t₂ = n t₁

$\Rightarrow$$\,\,\,\,$ ${{2\sqrt 2 \,S} \over {g\left( {1 - {\mu _k}} \right)}}$ = n² $\times$ ${{2\sqrt 2 \,S} \over g}$

$\Rightarrow$$\,\,\,\,$ 1 $-$ $\mu$_k = ${1 \over {{n^2}}}$

$\Rightarrow$$\,\,\,\,$ ${\mu _k}$ = 1 $-$ ${1 \over {{n^2}}}$

We know,

Work done on gas = nRT $\ell$n $\left( {{{{P_f}} \over {{P_i}}}} \right)$

Given P_f = 2P_i

T = 27 + 273 = 300 K

and for monoatomic gas, n = 1.

$\therefore\,\,\,\,$ Work done = 1 $\times$R $\times$300 $\ell$n(2)

= 300 R $\ell$n(2)

Given circuit :


Simplified circuit.


The equivalent capacitance between C and D = 2 + 5 + 5 = 12 $\mu$F

Equivalent capacitance between E and B = 4 + 2 = 6 $\mu$F

Now equivalent capacitance between A and B

${1 \over {{C_{eq}}}}$ = ${1 \over 6}$ + ${1 \over 12}$ + ${1 \over 6}$ = ${5 \over 12}$

$\Rightarrow$$\,\,\,\,$ C_eq = ${12 \over 5}$ = 2.4 $\mu$F

Note :

(1) When capacitors C₁, C₂, . . . .C_n are in parallel then equivalent capacitance

C_eq = C₁ + C₂ + . . . . + C_n

(2) When capacitor C₁, C₂ . . . . . . C_n are in series the equivalent capacitance

${1 \over {{C_{eq}}}}$ = ${1 \over {{C_1}}}$ + ${1 \over {{C_2}}}$ + . . . . . + ${1 \over {{C_n}}}$

As in a cube there is 6 faces, you can think this surface is one face among 6 faces. If the side of cube is a length then its center will be ${a \over 2}$ distance from any surface.

So, we can assume point charge Q is at the center of the cube and total electric flux due to this charge will pass evently through the six faces of the cube.

So, electric flux through one face will be ${1 \over 6}$. of the total electric flux.

Flux through 6 faces of the cube $= {Q \over {{\varepsilon _0}}}$

$\therefore\,\,\,\,$ Flux through 1 face of the cube = ${Q \over {6{\varepsilon _0}}}$

Energy required to remove e^$-$ from singly ionized Helium atom

E₁ = ${{13.6{z^2}} \over {{n^2}}}$ = ${{13.6 \times {2^2}} \over {{1^2}}}$ = 54.4 eV.

Let, E₂ = energy required to remove e^$-$from He $-$ atom

$\therefore\,\,\,\,$ According to q vertion.

E₁ = 2.2 E₂

$\therefore\,\,\,\,$ E₂ = ${{54.4} \over {2.2}}$ = 24.72 eV

$\therefore\,\,\,\,$ Total energy required to ionize Helium atom completely

= (54.4 + 24.72) eV

= 79.12 eV

Given : Frequency $f = {3 \over {2\pi }} \times {10^{12}}$ Hz; direction $= {{\widehat i + \widehat j} \over {\sqrt 2 }}$.

The beam is polarised along ${\widehat k}$ direction.

Direction of magnetic field $\overrightarrow B = \left( {{{\widehat i + \widehat j} \over {\sqrt 2 }} \times \widehat k = {{ - \widehat j} \over {\sqrt 2 }} + {{\widehat i} \over {\sqrt 2 }}} \right)$

Therefore, direction of $\overrightarrow B = \left( {{{\widehat i - \widehat j} \over {\sqrt 2 }}} \right)$

Now from wave equation, we have

$B = {B_0}\cos (kr - \omega t)$

$\omega = 2\pi f = 2\pi \times {3 \over {2\pi }} \times {10^{12}}Hz = 3 \times {10^{12}}Hz$

Now, $k = {{\widehat i + \widehat j} \over {\sqrt 2 }}$;

${B_0} = {{{E_0}} \over c}\left( {{{\widehat i - \widehat j} \over {\sqrt 2 }}} \right)$

Therefore,

$B = {{{E_0}} \over c}\left( {{{\widehat i - \widehat j} \over {\sqrt 2 }}} \right)\cos \left[ {{{10}^4}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)\,.\,r - (3 \times {{10}^{12}})t} \right]$

Given, modulating frequency = 15 kHz

$\therefore\,\,\,\,$ Bandwidth of one channel = 2 $\times$ 15 = 30 kHz

$\therefore\,\,\,\,$ No of stations = ${{300} \over {30}}$ = 10

Current gain ($\beta$) = ${{\Delta \,{I_C}} \over {\Delta {I_B}}}$

Voltage gain = ${{{V_{CE}}} \over {{V_{BE}}}} = \beta {{{R_L}} \over {{R_{BE}}}}$

Power gain = voltage gain x current gain = ${\beta ^2}{{{R_L}} \over {{R_{BE}}}}$

On the body of charge q a electric fied E is applied, because of this equilibrium position of body will shift to a point where resulttant force is zero.

$\therefore\,\,\,\,$ kx_eq = qE

$\Rightarrow$$\,\,\,$ x_eq = ${{qE} \over K}$

$\therefore\,\,\,\,$ Total energy of the system

= ${1 \over 2}m{\omega ^2}{A^2} + {1 \over 2}K\,x_{eq}^2$

= ${1 \over 2}$ m${\omega ^2}$A² + ${1 \over 2}.{{{q^2}{E^2}} \over K}$

As system is in equilibrium so the pressuse at A from both side of the liquid must be equal .

(r cos $\theta$ + r sin $\theta$) $\rho$₂g = (r cos $\theta$ $-$ r sin $\theta$) $\rho$₁g

$\Rightarrow$$\,\,\,$ ${{{\rho _1}} \over {{\rho _2}}} = {{\sin \theta + \cos \theta } \over {\cos \theta - \sin \theta }} = {{1 + \tan \theta } \over {1 - \tan \theta }}$

$\Rightarrow$$\,\,\,$ $\rho$₁ $-$ $\rho$₁ tan$\theta$ = $\rho$₂ + $\rho$₂ tan$\theta$

$\Rightarrow$$\,\,\,$ ($\rho$₁ + $\rho$₂) tan$\theta$ = $\rho$₁ $-$ $\rho$₂

$\Rightarrow$$\,\,\,$ tan$\theta$ = ${{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}$

$\Rightarrow$$\,\,\,$ $\theta$ = tan^$-$1 $\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)$

We can produce maximum moment of force, when line of action of force is perpendicular to line AB.


$\therefore\,\,\,\,$ $\tan \theta$ = ${2 \over 4}$ = ${1 \over 2}$

When plane durface is silvered then focus length of lens,

f₁ = ${R \over {2\left( {\mu - 1} \right)}}$

and when curver surface is silvered then focus length of lens,

f₂ = ${R \over {2\mu }}$

$\therefore\,\,\,\,$ ${{{f_1}} \over {{f_2}}} = {R \over {2\left( {\mu - 1} \right)}} \times {{2\mu } \over R}$

${\mu \over {\left( {\mu - 1} \right)}}$

given f₁ = 28 and f₂ = 10

$\therefore\,\,\,\,$ ${{28} \over {10}} = {\mu \over {\mu - 1}}$

$\Rightarrow$$\,\,\,\,$ 28$\mu$ $-$ 28 = 10 $\mu$

$\Rightarrow$$\,\,\,\,$ 18 $\mu$ = 28

$\Rightarrow$$\,\,\,\,$ $\mu$ = ${{28} \over {18}}$ = 1.55

Till 15 sec car has accelerative motion and scooter has constant velocity in entire motion.

Total Distance travelled by the car in 15 sec, = ${1 \over 2}$ $\times$ ${{45} \over {15}}$ $\times$ (15)² = ${{675} \over 2}$ m

Distance travelled by scooter in 15 sec.

= V $\times$ t = 30 $\times$ 15 = 450 m

$\therefore\,\,\,\,$ Difference between distance travelled by the car and scooter in 15 sec,

= 450 $-$ ${{675} \over 2}$ = 112.5 m

Now, assume car catches the scooter in time t,

$\therefore\,\,\,\,$ Car travelled in t sec = scooter travelled in t sec.

$\Rightarrow$ $\,\,\,\,$ ${{675} \over 2}$ + 45(t$-$ 15) = 30 $\times$ t

$\Rightarrow$ $\,\,\,\,$ 337.5 + 45t $-$ 675 = 30t

$\Rightarrow$ $\,\,\,\,$ 15 t = 337.5

$\Rightarrow$ $\,\,\,\,$ t = 22.5 sec.

Initially gravitational potenrial energy

E_i = $-$ ${{GMm} \over {2R}}$

Final gravitational potential energy

E_f = $-$ ${{GM\left( {{m \over 2}} \right)} \over {2\left( {{R \over 2}} \right)}}$ $-$ ${{GM\left( {{m \over 2}} \right)} \over {2\left( {{{3R} \over 2}} \right)}}$

= $-$ ${{GMm} \over {2R}} - {{GMm} \over {6R}}$

= $-$ ${{4GMm} \over {6R}}$

= $-$ ${{2GMm} \over {3R}}$

$\therefore\,\,\,\,$ Required difference in energies

= ${E_f} - {E_i}$

= $-$ ${{GMm} \over R}\left( {{2 \over 3} - {1 \over 2}} \right)$

= $-$ ${{GMm} \over {6R}}$

Given,

Cold body temperature (T₂) = 250 K

Hot body temperature (T₁) = 300 K

Received heat (Q₂) = 500 cal

Let, required works done = $\omega$

$\therefore\,\,\,\,$ For the refrigerator,

Efficiency = $1 - {{T2} \over {T1}} = {\omega \over {{Q_2} + \omega }}$

$\Rightarrow$$\,\,\,\,$ 1 $-$ ${{250} \over {300}} = {\omega \over {{Q_2} + \omega }}$

$\Rightarrow$$\,\,\,\,$ ${{{Q_2} + \omega } \over \omega }$ = ${{300} \over {50}}$ = 6

$\Rightarrow$$\,\,\,\,$ $\omega = {{{Q_2}} \over 5}$

$\Rightarrow$$\,\,\,\,$ $\omega = {{500 \times 4.2} \over 5}$

$\Rightarrow$$\,\,\,\,$ $\omega = 420\,J$

For balanced meter bridge,

${X \over {39.5}}$ = ${Y \over {\left( {100 - 39.5} \right)}}$

$\Rightarrow$$\,\,\,\,$ X = ${{12.5 \times 39.5} \over {60.5}}$

= 8.16 $\Omega$

Now, when we interchange X and Y then

${Y \over {{l_2}}} = {X \over {100 - {l_2}}}$

$\Rightarrow$ ${{12.5} \over {{l_2}}} = {{8.16} \over {100 - {l_2}}}$

$\Rightarrow$20.66 ${{l_2}}$ = 1250

$\Rightarrow$ ${{l_2}}$ = 60.5 $\Omega$

P is the mid-point of line AC. A and C are the center of the two circle of each radius R.

Current flows through loop A and B are in same direction, So the magnetic field will also be in the same direction. Magnitude of magnetic field at paint P

= magnitude of magnetic field due to A and B at paint P.

= 2 $\left[ {{{{\mu _0}NI\,{R^2}} \over {2{{\left( {{R^2} + {{{R^2}} \over 4}} \right)}^{{3 \over 2}}}}}} \right]$

$=$ ${{{\mu _0}\,NI{R^2}} \over {{{{5^{{3 \over 2}}}{R^3}} \over 8}}}$

= ${{8\,{\mu _0}NI} \over {{5^{{3 \over 2}}}\,R}}$

Relative error in the surface are of the sphere,

${{\Delta S} \over S}$ = 2 $\times$ ${{\Delta r} \over r}$ = $\propto$ (given)

Relative error in volume,

${{\Delta V} \over V}$ = 3 $\times$ ${{\Delta r} \over r}$

= 3 $\times$ ${1 \over 2}$ $\times$ ${{\Delta S} \over S}$

= ${3 \over 2}$ $\times$ $\propto$

= ${3 \over 2}$ $\propto$

5 complete rotations = 0.25 cms

So, 1 complete rotation of screw = 0.05 cm

$\therefore\,\,\,\,$ 1 main scale division = 0.05 cm

1 circular scale = ${{0.05} \over {100}}$ = 5 $\times$ 10^$-$4 cm

Thickness of a wire

= 4 main scale and 30 circular scale divisions

= 4 $\times$ 0.05 + 30 $\times$ 5 $\times$ 10 ^$-$4

= 0.2150 cm.