JEE Main 2018 Question Paper with Solutions

Moving block will stop when the friction force between m₂ and surface is $\ge$ tension force.

So condition for stopping the moving block,

$f \ge T$

$\Rightarrow \mu N \ge T$

$\Rightarrow \mu \left( {m + {m_2}} \right)g \ge {m_1}g$

When m is minimum then,

$\mu \left( {m + {m_2}} \right)g = {m_1}g$

$\Rightarrow m = {{{m_1} - \mu {m_2}} \over \mu }$

$\Rightarrow m = {{5 - 0.15 \times 10} \over {0.15}}$ = 23.33 kg

So if m $\ge$ 23.33 kg then the motion will stop. From option the minimum possible m is 27.3 kg.

Electric field in air,

$\overrightarrow {{E_1}}$  =  E₀₁ $\widehat x$ cos ( ${{2\pi vz} \over c}$ $-$ 2$\pi$vt )

$\therefore\,\,\,$ Velocity in air = ${{2\pi v} \over {{{2\pi v} \over c}}}$ = c

Also,    c = ${1 \over {\sqrt {\mu \varepsilon {r_1}{\varepsilon _0}} }}$ . . . . . . (1)

$\overrightarrow {{E_2}}$  =  E₀₂ $\widehat x$ cos(2kz $-$ kct)

$\therefore\,\,\,$ Velocity in medium = ${{kc} \over {2k}}$ = ${c \over 2}$

Also, ${c \over 2}$ = ${1 \over {\sqrt {\mu {\varepsilon _{r2}}\,{\varepsilon _0}} }}$ . . . . . (2)

As, medium is non magnetic,

So,   $\mu$_medium = $\mu$_air = $\mu$

Dividing (1) by (2), we get

2 = $\sqrt {{{{\varepsilon _{r2}}} \over {{\varepsilon _{r1}}}}}$

$\Rightarrow$$\,\,\,$ ${{{{\varepsilon _{r1}}} \over {{\varepsilon _{r2}}}}}$ = ${1 \over 4}$

Density of a material (d) = ${M \over {{L^3}}}$

$\therefore$ Error in density,${{\Delta d} \over d} = {{\Delta M} \over M} + 3{{\Delta L} \over L}$

${{\Delta d} \over d} \times 100 = {{\Delta M} \over M} \times 100 + 3{{\Delta L} \over L} \times 100$

$\Rightarrow {{\Delta d} \over d} \times 100 = 1.5\% + 3\left( 1 \right)\%$ = 4.5 %

From conservation of linear momentum,

mv₀ = mv₁ + mv₂

or v₀ = v₁ + v₂ ........(1)

According to the question,

K_f = ${3 \over 2}$K_i

$\Rightarrow$ ${1 \over 2}mv_1^2 + {1 \over 2}mv_2^2 = {3 \over 2} \times {1 \over 2}mv_0^2$

$\Rightarrow$ $v_1^2 + v_2^2 = {3 \over 2}v_0^2$

Using eq (1) ${\left( {{v_1} + {v_2}} \right)^2} = v_0^2$

$\Rightarrow$ $v_1^2 + v_2^2 + 2{v_1}{v_2}$ = $v_0^2$

$\Rightarrow$ $2{v_1}{v_2}$ = $v_0^2 - {3 \over 2}v_0^2$ = $- {1 \over 2}v_0^2$

Now, ${\left( {{v_1} - {v_2}} \right)^2}$ = ${\left( {{v_1} + {v_2}} \right)^2} - 4{v_1}{v_2}$

= $v_0^2 - \left( { - v_0^2} \right)$ = $2v_0^2$

$\therefore$ ${{v_1} - {v_2}}$ = $\sqrt 2 {v_0}$

Given that for a uniform circular disc the radius is R and mass 9M

$\therefore$ The area of uniform circular disc = $\pi {R^2}$

The radius of removed portion = ${R \over 3}$

$\therefore$ The area of removed portion = ${{\pi {R^2}} \over 9}$

So the mass of the removed portion = ${{9M} \over 9}$ = M

If the moment of inertia of removed disc about the center of the circular disc of radius R is I₁ then

I₁ = ${1 \over 2}M{\left( {{R \over 3}} \right)^2} + M{\left( {{{2R} \over 3}} \right)^2}$ = ${{M{R^2}} \over 2}$

Moment of inertia of the whole disc, I₂ = ${{9M{R^2}} \over 2}$

Moment of inertia of the remaining disc, I = I₂ - I₁

$\therefore$ I = ${{9M{R^2}} \over 2} - {{M{R^2}} \over 2}$ = ${4M{R^2}}$

Let potential at S, T, R = V and potential at P, Q, U, = 0

Using kirchhoff's law at P :

Current at P is ,

${{V - 12} \over 1} + {{V - 13} \over 2} + {{V - 0} \over {10}} = 0$

$\Rightarrow$$\,\,\,$ ${V \over 1}$ + ${V \over 2}$ + ${V \over 10}$ = 12 + ${{13} \over 2}$

$\Rightarrow$$\,\,\,$ ${{10V + 5V + V} \over {10}}$ = ${{37} \over 2}$

$\Rightarrow$$\,\,\,$ ${{16V} \over {10}}$ = ${{37} \over 2}$

$\Rightarrow$$\,\,\,$ V = 11.56 Volt.

Note :

From the circuit you can see the diode is in forward bias, and to start current flow through the circuit. Accross silicon diode the voltage drop should be 0.7 V and for Germenium diode the voltage drop is 0.3 V. Then diode will start working and current will flow through the circuit.

So, the net voltage available accros the registor is (V_net) = 3 $-$ 0.7 = 2.3 V

$\therefore\,\,\,$ Current flow through circuit = ${{2.3} \over {200}}$ = 11.5 mA

Let charge of shell A, B and C are Q_A, Q_B and Q_C respectively.

Potential of B shell will be due to charge Q_A, Q_B and Q_C.

Here charge Q_A is inside of the shell B and Q_B is on the surface of the shell B in both cases you have to take the radius of the shell B, while calculating potential of shell B.

Charge Q_C is outside of the shell B so take radius of shell C for calculating potential of shell B.

$\therefore$ V_B = V$_{{Q_a}}$ + V$_{{Q_b}}$ + V$_{{Q_c}}$

= ${1 \over {4\pi {\varepsilon _0}}}$ $\left[ {{{4\pi {a^2}\left( { + \sigma } \right)} \over b} + {{4\pi {b^2}\left( { - \sigma } \right)} \over b} + {{4\pi {c^2}\left( { + \sigma } \right)} \over c}} \right]$

= ${1 \over {4\pi {\varepsilon _0}}}$ $\times$ 4$\pi$$\sigma$ $\left[ {{{{a^2}} \over b} - {{{b^2}} \over b} + c} \right]$

= ${\sigma \over {{\varepsilon _0}}}\left[ {{{{a^2} - {b^2}} \over b} + c} \right]$

We know,

Wavelength of emitted photon from n₂ state to n₁ state is

${1 \over \lambda }$  =  RZ² $\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$

Here electron comes from n^th state to ground state (n = 1),

then the wavelength of photon is ,

${1 \over {{\Lambda _n}}}$  =  RZ² $\left( {{1 \over {{1^2}}} - {1 \over {{n^2}}}} \right)$

$\Rightarrow$$\,\,\,$ $\Lambda$_n   =  ${1 \over {R{Z^2}}}{\left( {1 - {1 \over {{n^2}}}} \right)^{ - 1}}$

As n is very large, so using binomial theorem

$\Lambda$_n  =  ${1 \over {R{Z^2}}}\left( {1 + {1 \over {{n^2}}}} \right)$

$\Rightarrow$$\,\,\,$ $\Lambda$_n =  ${1 \over {R{Z^2}}} + {1 \over {R{Z^2}}}\left( {{1 \over {{n^2}}}} \right)$

We know,

$\lambda$_n =  ${{2\pi r} \over n}$

=  2$\pi$ ${\left( {{{{n^2}{h^2}} \over {4{\pi ^2}mZ{C^2}}}} \right)\times{{1 \over n}}}$

$\therefore\,\,\,$ $\lambda$_n $\propto$ n

$\Rightarrow$$\,\,\,$ n  =  K $\lambda$_n

$\therefore\,\,\,$ $\Lambda$_n  = ${1 \over {R{Z^2}}} + {1 \over {R{Z^2}}}\left( {{1 \over {{{\left( {K\,{\lambda _n}} \right)}^2}}}} \right)$

Let A  =  ${1 \over {R{Z^2}}}$  and   B  = ${1 \over {{K^2}R{Z^2}}}$

$\Rightarrow$$\,\,\,$ $\Lambda$_n  =  A + ${B \over {\lambda _n^2}}$

Because of m mass the extra pressure created is,

$\Delta$P = ${{mg} \over a}$

And Bulk modulus, $\beta$ = ${{\Delta P} \over {{{\Delta V} \over V}}}$

Given $\beta$ = K

$\therefore\,\,\,$ K = ${{{{mg} \over a}} \over {{{\Delta V} \over V}}}$

We know volume of sphere,

V = ${4 \over 3}\pi {r^3}$

$\therefore\,\,\,$ ${{dV} \over V}$ = 3 ${{dr} \over r}$

$\therefore\,\,\,$ K = ${{{{mg} \over a}} \over {3{{dr} \over r}}}$

$\Rightarrow$$\,\,\,$ ${{dr} \over r}$ = ${{mg} \over {3Ka}}$

Given 2$\theta$ = 60^o

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

We know,

$a$ sin $\theta$ = n $\lambda$

for first minima n = 1,

$\therefore$ At first minima

$a$ sin = $\lambda$

$\Rightarrow$$\,\,\,$ 10^$-$6 $\times$ sin 30^o = $\lambda$

$\Rightarrow$$\,\,\,$ $\lambda$ = ${{{{10}^{ - 6}}} \over 2}$ m

Now after making a new slit,

$\therefore$ Fringe width, $\beta$ = ${{\lambda D} \over d}$

given, D = 50 cm and $\beta$ = 1 cm.

$\therefore$ 1 $\times$ 10^$-$2 = ${{0.5 \times {{10}^{ - 6}} \times 50 \times {{10}^{ - 2}}} \over d}$

$\Rightarrow$ d = 25 $\times$ 10^$-$6 m = 25 $\mu$m.

Quality factor (Q) = ${{Angular\,\,{\mathop{\rm Re}\nolimits} sonance} \over {Bandwith}}$

= ${{{1 \over {\sqrt {LC} }}} \over {{R \over L}}}$

= ${{{\omega _0}} \over {{R \over L}}}$

= ${{{\omega _0}L} \over R}$

Note :

(1)   In Lyman Series, transition happens in n = 1 state
from n = 2, 3, . . . . . $\propto$

(2)   In Balmer Series, transition happens in n = 2 state
from n = 3, 4, . . . . . $\propto$

(3)   In Paschen Series, transition happens in n = 3 state
from n = 4, 5, . . . . . $\propto$

(4)   In Bracktt Series, transition happens in n = 4 state
from n = 5, 6 . . . . . . $\propto$

(5)   In Pfund Series, transition happens in n = 5 state
from n = 6, 7, . . . . $\propto$

We know,

${1 \over \lambda }$  =  RZ² $\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$

Series limit means transition happens

from n = $\propto$ to n = 1, for Lyman Series.

In series limit for Lyman series,

${1 \over {{\lambda _L}}}$ = RZ² $\left( {{1 \over {{1^2}}} - {1 \over \propto }} \right)$

$\Rightarrow$$\,\,\,$ ${1 \over {{\lambda _L}}}$ = RZ²

We know,

E = ${{hc} \over \lambda }$ = h$\gamma$

$\Rightarrow$$\,\,\,$ $\gamma$ = ${c \over \lambda }$

So, frequency in Lyman Series,

$\gamma$_L =  ${c \over {{\lambda _L}}}$  = c $\times$ RZ²

In Pfund series,

n₂ = $\propto$  and  n₁ = 5

$\therefore\,\,\,$ ${1 \over {{\lambda _P}}}$ = RZ²$\left( {{1 \over {{5^2}}} - {1 \over {{ \propto ^2}}}} \right)$

$\Rightarrow$$\,\,\,$ ${1 \over {{\lambda _P}}}$ = ${{R{Z^2}} \over {25}}$

$\therefore\,\,\,$ ${\gamma _P}$   =   ${c \over {{\lambda _P}}}$ = c $\times$ ${{R{Z^2}} \over {25}}$

$\therefore\,\,\,$ $\gamma$_P = ${{cRZ{}^2} \over {25}}$ = ${{{\gamma _L}} \over {25}}$    [as   $\gamma$_L = cRZ²]

Charge on Capacitor initially,

Q_i = CV

After inserting dielectric of dielectric constant = K,

new capacitance, Q_f = (KC) $\vee$

$\therefore\,\,\,$ Induced charges on dielectric

= Q_f $-$ Q_i

= KCV $-$ CV

= (K $-$ ) CV

= $\left( {{5 \over 3} - 1} \right)$ $\times$ 90 $\times$ 10^$-$12 $\times$ 20

= 1.2 $\times$ 10^$-$9 C

= 1.2 nC

Internal resistance of potentiometer,

r = $\left( {{E \over V} - 1} \right) \times R$

Initially when no current passes through the galvanometer then

emf, E = K (52)

here K = potential gradient

After cell is shunted by a resistance 5 $\Omega$, then,

Terminal voltage, V = K(40)

$\therefore\,\,\,$ r = $\left( {{{52K} \over {40K}} - 1} \right)$ $\times$ 5

= $\left( {{{26} \over {20}} - 1} \right)$ $\times$ 5

= 1.5 $\Omega$

For adiabatic process,

pv^$\gamma$ = constant.

and we know, pv = nRT

$\therefore\,\,\,$ p = ${{nRT} \over v}$

$\therefore\,\,\,$ ${{nRT} \over v} \times {v^\gamma }$ = constant

$\Rightarrow$$\,\,\,$ T v^$\gamma$$-$1 = constant.

$\therefore\,\,\,$ T₁ v₁^$\gamma$$-$1 = T₂ v₂^$\gamma$$-$1

Given that,

This is a monoatomic gas.

So, degree of frequency f = 3

$\therefore\,\,\,$ $\gamma$ = ${{{c_p}} \over {{c_v}}}$ = 1 + ${2 \over f}$ = 1 + ${2 \over 3}$ = ${5 \over 3}$

T₁ = 27 + 273 = 300 K

v₁ = v

and v₂ = 2v

$\therefore\,\,\,$ T₂ (2V)$^{{2 \over 3}}$ = 300(v)$^{{2 \over 3}}$

$\Rightarrow$$\,\,\,$ T₂ = ${{300} \over {{2^{{2 \over 3}}}}}$ = 189 K

Change in internal energy,

$\Delta$U = ${1 \over 2}$ nfR$\Delta$T

= ${1 \over 2}$ $\times$ 2 $\times$ 3 $\times$ 8.31 $\times$ (189 $-$ 300)

= $-$ 2.7 KJ

When a charged particle moves in a magnetic field then the charged particle moves in a circular path. So,

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

$\Rightarrow$$\,\,\,$ r = ${{mv} \over {Bq}}$

We know kinetic energy, K = ${1 \over 2}$ mv²

$\therefore\,\,\,$ mv = $\sqrt {2Km}$

$\therefore\,\,\,$ r = ${{\sqrt {2Km} } \over {Bq}}$

According to the question,

K_e (electron) = K_p (proton) = K_$\alpha$(Alpha particle) = K = constant, and all of them are in uniform magnetic field.

$\therefore$ B = constant.

$\therefore\,\,\,$ r $\propto$ ${{\sqrt m } \over q}$

For proton (₁H¹), mass = m, and charge = e

$\therefore\,\,\,$ r_p $\propto$ ${{\sqrt m } \over e}$

For alpha particle (₂H⁴),

mass = 4m

and charge = 2e

$\therefore\,\,\,$ r_$\alpha$ $\propto$ ${{\sqrt {4m} } \over {2e}}$ $\propto$ ${{\sqrt m } \over e}$

$\therefore$$\,\,\,$ r_p = r_$\alpha$

For electron,

charge = e

and mass (m_e) = 9.1 $\times$ 10^$-$31 kg

and mass of proton = 1.67 $\times$ 10^$-$27 kg

$\therefore\,\,\,$ mass of electron < mass of proton.

r_e $\propto$ ${{\sqrt {{m_e}} } \over e}$ < r_p

$\therefore\,\,\,$ r_e < r_p = r_$\propto$

The carrier frequency is 10 GHz, and only 10% of it is utilized for transmission. Therefore, the bandwidth used for transmission is:

$\text{Bandwidth} = 10\% \times 10 \text{ GHz} = 0.1 \times 10 \text{ GHz} = 1 \text{ GHz}$

We should convert this to kHz because the channel bandwidth is given in kHz:

$1 \text{ GHz} = 1,000,000 \text{ kHz}$

Now, if each telephonic channel requires a bandwidth of 5 kHz, the number of channels that can be transmitted simultaneously is:

$\text{Number of channels} = \frac{\text{Total bandwidth}}{\text{Bandwidth per channel}}$

$\text{Number of channels} = \frac{1,000,000 \text{ kHz}}{5 \text{ kHz}}$

$\text{Number of channels} = 200,000$

So, 200,000 telephonic channels can be transmitted simultaneously.

Among the given options, the correct answer is Option C: $2 \times 10^5$.

We know, Central force in circular motion, F = $m{\omega ^2}R$

According to the question,

$F \propto {1 \over {{R^n}}}$

$\therefore$ $m{\omega ^2}R$ $\propto {1 \over {{R^n}}}$

$\Rightarrow m{\omega ^2}R = {k \over {{R^n}}}$

$\Rightarrow {\omega ^2} = {k \over {m{R^{n + 1}}}}$

$\therefore$ $\omega \propto {1 \over {{R^{{{n + 1} \over 2}}}}}$ .......(1)

And we know, $T = {{2\pi } \over \omega }$

$\therefore$ $T \propto {1 \over \omega }$ ...... (2)

From (1) and (2) we can conclude that,

$T \propto {R^{{{n + 1} \over 2}}}$

As   rod length = 60 cm

$\therefore\,\,\,$ ${\lambda \over 2}$ = 60

$\Rightarrow$$\,\,\,$ $\lambda$ = 120 cm = 1.2 m

In solid, velocity of wave,

V = $\sqrt {{Y \over \rho }}$

= $\sqrt {{{9.27 \times {{10}^{10}}} \over {2.7 \times {{10}^3}}}}$

= 5.85 $\times$ 10³ m/sec.

As  we know,

v = f $\lambda$

$\therefore\,\,\,$ f = ${v \over \lambda }$

= ${{5.85 \times {{10}^3}} \over {1.2}}$

= 4.88 $\times$ 10³ Hz

$\simeq$  5 kHz

As after B intensity of light does not drops, it means both A and B are alligned in single line means their plane of polarization is same.



Let C makes an angle $\theta$ with A then C will make $\theta$ with B also, as both A and B are alligned in a single line.

So, after C intensity is = ${{\rm I} \over 2}$ cos²$\theta$ , and , intensity after B = ${{\rm I} \over 2}$ cos²$\theta$ $\times$ cos²$\theta$

   According to question,

${{\rm I} \over 2}$ cos⁴$\theta$ = ${{\rm I} \over 8}$

$\Rightarrow$\,\,\, CO⁴$\theta$ = ${{\rm I} \over 4}$

$\Rightarrow$$\,\,\,$ cos$\theta$ = $= {1 \over {\sqrt 2 }}$ = cos45^o

$\therefore\,\,\,$ $\theta$ = 45^o

Wattless current,

here  $\phi$  is the angle between i and e.

Average power,

P_av = V_rms I_rms cos$\phi$

= ${{100} \over {\sqrt 2 }} \times {{20} \over {\sqrt 2 }}$ cos${\pi \over 4}$

= ${{1000} \over {\sqrt 2 }}$ watt.

In option (A) you can see velocity versus time graph is a straight line with negative slope, so the acceleration is negative and constant. This motion is look like this.

Initially velocity is maximum and as acceleration is negative so after travelling some distance velocity will will become zero and then velocity will increase in the negative direction. At option (B) you can see same situation happens so (A) and (B) represent same situation.

From diagram, Initially at t = 0 position h = 0 and after some time h become maximum at the end h again become zero. Option (D) represent this situation.

So, A, B and D are correct.

The distance - time graph for this case will be -

But the given distance-time graph in the question does not match with this so option (C) will be wrong answer.

Considering one hydrogen molecule :



As collision is elastic so, e = 1

Initial momentum,

$\overrightarrow {{P_i}}$ = ${{mv} \over {\sqrt 2 }}$ $\widehat i$ $-$ ${{mv} \over {\sqrt 2 }}$ $\widehat j$

Final momentum,

$\overrightarrow {{P_f}}$ = ${{mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)$ $-$ ${{mv} \over {\sqrt 2 }}\widehat j$

$\therefore\,\,\,$ Change in momentum for single H molecule,

$\Delta$P = $\overrightarrow {{P_f}}$ $-$ $\overrightarrow {{P_i}}$

= ${{2mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)$

$\therefore\,\,\,$ $\left| {\Delta P} \right|$ = ${{2mv} \over {\sqrt 2 }}$

Now for n hydrogen molecule total momentum changes per second,

= $\left( {{{2mv} \over {\sqrt 2 }}} \right)$ $\times$ n

As we know,

Force (F) = ${{\Delta P} \over {\Delta t}}$

= ${{{2mv} \over {\sqrt 2 }}}$ $\times$ n

$\therefore\,\,\,$ As direction of $\Delta$P is towards negative i so force on the molecule will also be towards negative i direction. From Newton's third law, the reaction force will be on the wall in positive i direction with same magnitude.

$\therefore\,\,\,$ Force on the wall = ${{{2mv} \over {\sqrt 2 }}}$ $\times$ n

$\therefore\,\,\,$ Pressure on the wall, P = ${F \over A}$

= ${{2mv\,n} \over {\sqrt 2 A}}$

= ${{2 \times 3.32 \times {{10}^{ - 27}} \times {{10}^3} \times {{10}^{23}}} \over {\sqrt 2 \times2\times {{10}^{ - 4}}}}$

= 2.35 $\times$ 10³ N m^$-$2

6.02 $\times$ 10²³ atoms of silver = 108 gm

1   atoms   of   silver   =   ${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$ kg

For a harmonic oscillator

f = ${1 \over {2\pi }}$ $\sqrt {{k \over m}}$

Where  k = force constant

$\Rightarrow$$\,\,\,$ f² = ${1 \over {4{\pi ^2}}}$ $\left( {{k \over m}} \right)$

$\Rightarrow$$\,\,\,$ k = mf² $\times$ 4$\pi$²

Given,

f = 10¹²

m = ${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$

$\therefore\,\,\,$ k = ${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$ $\times$ 10¹² $\times$ 4$\pi$²

= 7.1 N/m

We know, Total energy = Kinetic energy + Potential energy

Potential energy given as $U = - {k \over {2{r^2}}}$

We need to find Kinetic Energy.

As Force acting on the particle (F) = $- {{dU} \over {dr}}$

$\Rightarrow F = - {d \over {dr}}\left( {{{ - k} \over {2{r^2}}}} \right)$

$= {k \over 2} \times \left( { - 2} \right) \times {r^{ - 3}}$

$= - {k \over {{r^3}}}$

Because of this force particle is having circular motion so it will provide possible centripetal force.

$\left| F \right| = {{m{v^2}} \over r}$

$\Rightarrow {{m{v^2}} \over r} = {k \over {{r^3}}}$

$\Rightarrow$ $m{v^2} = {k \over {{r^2}}}$

We know kinetic energy of particle, K = ${1 \over 2}m{v^2}$ = ${k \over {2{r^2}}}$

As Total energy = Kinetic energy + Potential energy

So Total energy = ${k \over {2{r^2}}}$ $- {k \over {2{r^2}}}$ = 0