JEE Main 2017 8th April Morning Shift Question Paper with Solutions

Modulation index (m) = ${{{V_m}} \over {{V_0}}} = {5 \over {25}} = 0.2$

Given, carrier wave,

F_c = 1.2 $\times$ 10⁶ Hz = 1200 kHz

and frequency of modulate wave,

F_m = 20 kHz

$\therefore\,\,\,$ Side frequency bands are

F₁ = 1200 + 20 = 1220 kHz

F₂ = 1200 $-$ 80 = 1180 kHz

Refractive index,

n₁₂ = ${{{n_D}} \over {{n_R}}}$ = ${1 \over {\sin {\theta _c}}}$

$\Rightarrow$ $\,\,\,$ sin$\theta$_c = ${{{n_R}} \over {{n_D}}}$ . . . . . (1)

From Snell's law,

n_D sinA = n_R sin r

${{{n_R}} \over {{n_D}}}$ = ${{\sin A} \over {\sin \left( {{{90}^o} - A} \right)}}$    [as    r = 90^o $-$ A]

$\Rightarrow$$\,\,\,$ ${{{n_R}} \over {{n_D}}} = \tan A$

$\therefore\,\,\,$ From (1) we get,

tan A = sin $\theta$_c

$\Rightarrow$$\,\,\,$ A = tan^$-$1 (sin $\theta$_c)

In uniform magnetic field, the torque experienced by the magnetic dipole is $\tau$ = MB sin $\theta$

Torque will be maximum when $\theta$ = 90^o

$\tau$_max = MB sin90^o = MB

Potential energy of magnetic dipole,

$\mu$ = $-$ MB cos $\theta$

at maximum torque,

$\mu$ = $-$ MB cos 90^o = 0

The relation between electric and magnetic field is ,

C = ${{\overrightarrow E } \over {\overrightarrow B }}$

$\Rightarrow$$\,\,\,$ $\overrightarrow E$ = C $\overrightarrow B$

Electric field component is perpendicular to the direction of magnetic field. Given magnetic field is along y $-$ axis,

So, electric field along z $-$ axis will be

$\overrightarrow E$ = B₀ C sin (kx + $\omega$t) $\,\widehat k$ v/m

From principle of moment we know, the anticlockwise moment is equal to clockwise moment when a system is stable or balance.

$\therefore\,\,\,$ load $\times$ load arm = effect $\times$ effect arm

When 5 mg weight is placed on the left pan, load arm shift to left side, hence left arm become shorter than right arm.

Mutual inductance,

M = ${{{\mu _0}\pi {N_1}{N_2}\,{a^2}} \over {2b}}$

here ${{N_1}}$ = N₂ = 1

$\therefore\,\,\,$ M = ${{{\mu _0}\pi {a^2}} \over {2b}}$

Current I = I₀ cos ($\omega$t)

According to Faraday's law,

e = $-$ M ${{dI} \over {dt}}$

= $-$ ${{{\mu _0}\pi {a^2}} \over {2b}}$ ${d \over {dt}}$ (I₀ cos $\omega$t)

= + ${{{\mu _0}\pi {a^2}} \over {2b}}$ I₀ $\omega$ sin $\omega$t

= ${{\pi {\mu _0}{I_0}} \over 2}$ . ${{{a^2}} \over b}$ $\omega$ sin $\omega$ t

Here test charge experience no net force, So, sum of electric and magnetic field is zero.

$\therefore\,\,\,$ F_e + F_m = 0

$\therefore\,\,\,$ F_e = $-$ q ($\overrightarrow v$ $\times$ $\overrightarrow B)$

= $-$ qB₀ $\upsilon$₀ [(3$\widehat i$ $-$ $\widehat j$ + 2$\widehat k$) $\times$ ($\widehat i$ + 2$\widehat j$ $-$ 4$\widehat k$)]

= $-$ q$\upsilon$₀ B₀ (14$\widehat j$ + 7$\widehat k$)

Electric field produced by the charge q,

$\overrightarrow E$ = ${{\overrightarrow {{F_e}} } \over q}$

= ${{ - q{\upsilon _0}{B_0}\left( {14\widehat j + 7\widehat k} \right)} \over q}$

= $-$ $\upsilon$₀ B₀ (14 ${\widehat j}$ + 7 ${\widehat k}$)

For single slit diffraction, sin$\theta$ = ${{n\lambda } \over b}$

From central maxima the position of n^th minima = ${{n\lambda D} \over b}$

Now when,

n = 2, then x₂ = ${{2\lambda D} \over b}$ = 0.03 . . . .(1)

n = 4, then x₄ = ${{4\lambda D} \over b}$ = 0.06 . . . .(2)

Performing (2) $-$ (1) we get,

x₄ $-$ x₂ = ${{2\lambda D} \over b}$ = 0.03

$\therefore\,\,\,$ ${{\lambda D} \over b}$ = ${{0.03} \over 2}$

As, width of crntral maximum

= ${{2\lambda D} \over b}$

= 2 $\times$ ${{0.03} \over 2}$

= 0.03 m

= 3 cm

Conductivity of semiconductor,

$\sigma$ = e$\left( {{\eta _e}{\mu _e} + \eta '{\mu _h}} \right)$

= 1.6 $\times$ 10^$-$19 (5 $\times$ 10¹⁸ $\times$ 2 + 5 $\times$ 10¹⁹ $\times$ 0.01)

= 1.6 $\times$ 1.05

= 1.68

₁H² + ₁H²  $\to$  ₂He⁴

No. of proton in one dueteron = 2

$\therefore\,\,\,$ Total protons in two dueterons = 2 $\times$ 2 = 4

$\therefore\,\,\,$ Binding energy of two dueteron

= 1.1 $\times$ 4 = 4 : 4 MeV

In (₂He⁴) no of protons = 4

$\therefore\,\,\,$ Binding energy of (₂He⁴) nuclei = 4 $\times$ 7 = 28 Mev

Energy released in this process = 28 $-$ 4.4 = 23.6 MeV

When key K₃ is plugged in between 1 and 2,

V₁ = iR₁ = x$l$₁ . . . . . (1)

When key K₃ is plugged in between 3 and 1,

V₂ = i(R₁ + R₂) = x$l$₂ . . . . (2)

On dividing (1) and (2),

${{{R_1}} \over {{R_1} + {R_2}}} = {{{l_1}} \over {{l_2}}}$

$\Rightarrow$$\,\,\,$ ${{{R_1}} \over {\left( {R{}_1 + {R_2}} \right) - {R_1}}}$ = ${{{l_1}} \over {{l_2} - {l_1}}}$

$\Rightarrow$$\,\,\,$ ${{{R_1}} \over {{R_2}}}$ = ${{{l_1}} \over {{l_2} - {l_1}}}$

Initially when earth is not rotating then weight of the person is $w$.

When earth rotares about it's axis then weight = ${{3\omega } \over 4}$

Then,

g' = g $-$ $\omega$²R cos²$\theta$

$\Rightarrow$ $\,\,\,$ ${{3g} \over 4}$ = g $-$ $\omega$²R cos² 0^o

$\Rightarrow$ $\,\,\,$ $\omega$²R = ${g \over 4}$

$\,\,\,$ $\omega$ = $\sqrt {{g \over {4R}}}$

$\Rightarrow$ $\omega = \sqrt {{{10} \over {4 \times 6400 \times {{10}^3}}}}$

= 0.63 $\times$ 10^$-$3 rad/s

The given maximum velocity is v and frequency is n. We know that the kinetic energy is given by

$KE = {1 \over 2}m{v^2} = hn - \phi$

where h is Planck's constant and $\phi$ is the work function. Therefore, the kinetic energy of the incident light is

${E_1} = hn - \phi$ ..... (1)

When the frequency of the incident light is increased to 3n, then the kinetic energy is given by

${1 \over 2}mv_1^2 = 3hn - \phi$

$\Rightarrow {E_2} = 3hn - \phi$ ..... (2)

Substituting $hn = {E_1} + \phi$ [from Eq. (1)] in Eq. (2), we get

${E_2} = 3({E_1} + \phi ) - \phi$

$\Rightarrow {E_2} = 3{E_1} + 2\phi$

$\Rightarrow {1 \over 2}mv_1^2 = 3 \times {1 \over 2}m{v^2} + 2\phi$

$\Rightarrow v_1^2 = 3{v^2} + 2\phi \times {2 \over m}$

$\Rightarrow v_1^2 = 3{v^2} + {{4\phi } \over m}$

Thus, the velocity is more than $\sqrt 3 v$.

Assume equivalent resistance of the infinite network is x.

The circuit can be redrawn like this,


Here 4 $\Omega$ and x $\Omega$ are parallel,

$\therefore\,\,\,$ ${1 \over {{x_{eq}}}}$ = ${1 \over 4}$ + ${1 \over x}$

$\Rightarrow$$\,\,\,$ x_eq = ${{4x} \over {4 + x}}$

$\therefore\,\,\,$ Equivalent resistance between point A and B is ,

x = ${{4x} \over {4 + x}}$ + 2

$\Rightarrow$$\,\,\,$ x = ${{8 + 6x} \over {4 + x}}$

$\Rightarrow$$\,\,\,$ 4x + x² = 8 + 6x

$\Rightarrow$$\,\,\,$ x² $-$ 2x $-$ 8 = 0

$\Rightarrow$ $\,\,\,$ x = ${{2 \pm \sqrt {4 - 4\left( 1 \right)\left( { - 8} \right)} } \over 2}$

$\Rightarrow$$\,\,\,$x = ${{2 \pm 6} \over 2}$

$\Rightarrow$$\,\,\,$ = 4 $\Omega$

(as negative value of x is not acceptable)

$\therefore\,\,\,$ Reading of Aonmeter, A₁ = ${V \over {R + r}}$

A₁ = ${9 \over {4\,\, + 0.5}}$ = 2 A

Forward bias resistance

R₁ = ${{\Delta V} \over {\Delta I}}$ = ${{0.8 - 0.7} \over {\left( {20 - 10} \right) \times {{10}^{ - 3}}}}$ = 10 $\Omega$

Reverse bias resistance,

R₂ = ${{10} \over {1 \times {{10}^{ - 6}}}}$ = 10⁷ $\Omega$

$\therefore\,\,\,$ Ratio of resistance

= ${{{R_1}} \over {{R_2}}}$ = ${{10} \over {{{10}^7}}}$ = 10^$-$6

From figure, we can say

ma = mg $-$ T . . . . . (1)

Moment of inertia of a uniform disc,

I = ${1 \over 2}$ MR²

and the acceleration of the disc, a = $\propto$ R

We know,

Torque = I$\propto$ = RT

$\therefore\,\,\,$ RT = ${1 \over 2}$ M R² $\times$ ${a \over R}$ = ${{MaR} \over 2}$

$\Rightarrow$ $\,\,\,$ T = ${{Ma} \over 2}$

Putting the value of T in eq (1).

ma = mg $-$ ${{Ma} \over 2}$

$\Rightarrow$ $\,\,\,$ a ( m + ${{Ma} \over 2}$) = mg

$\Rightarrow$ $\,\,\,$ a = ${{2mg} \over {2m + M}}$

Magnetic field at the center of neucleus of H-atom

B = ${{{\mu _0}I} \over {2{r_n}}}$

Radius of n^th orbital,

r_n = ${{{n^2}{h^2}{\varepsilon _0}} \over {m\pi Z{e^2}}}$

$\therefore\,\,\,$ r_n $\propto$ n²

velocity of electron in n^th orbital,

$\upsilon$_n = $\left( {{{{e^2}h} \over {2{\varepsilon _0}}}} \right){Z \over n}$

$\therefore\,\,\,$ $\upsilon$_n $\propto$ n^$-$1

I = ${q \over t}$ = ${e \over {{{2\pi {r_n}} \over {{\upsilon _n}}}}}$ = ${{e{\upsilon _n}} \over {2\pi {r_n}}}$

$\therefore\,\,\,$ B = ${{{\mu _0}.\left( {{{e{\upsilon _n}} \over {2\pi {r_n}}}} \right)} \over {2{r_n}}}$ = ${{{\mu _0}\,e{\upsilon _n}} \over {4\pi r_n^2}}$

$\therefore\,\,\,$ B $\propto$ ${{{\upsilon _n}} \over {r_n^2}}$

$\Rightarrow$ $\,\,\,$ B $\propto$ ${{{n^{ - 1}}} \over {{n^4}}}$

$\Rightarrow$$\,\,\,$ B $\propto$ n ^$-$5

Let,

Kinetic energy (k) = ${1 \over 2}$ m $\upsilon$² before it hit the ground.

After hitting the ground kinetic energy

(k') = ${1 \over 2}$ m $\upsilon$$_1^2$

$\therefore\,\,\,$According to the question,

${1 \over 2}$ m$\upsilon$$_1^2$ = ${1 \over 2}$ $\times$ ${1 \over 2}$ m$\upsilon$²

$\Rightarrow$$\,\,\,$ $\upsilon$₁ = ${v \over {\sqrt 2 }}$

After hitting the ground the object will bounce

h' = ${{v_1^2} \over {2g}}$ = ${{{v^2}} \over {4g}}$ = ${h \over 2}$ [ as    h = ${{{v^2}} \over {2g}}$ ]

Total distance travelled from the time it first hits the ground to the next time it hits the ground is = ${h \over 2}$ + ${h \over 2}$ = h

So, this will create a infinite geometric progression with the common ration ${1 \over 2}$.

$\therefore\,\,\,$ Total distance covered

= h (distance travelled by the obhect when first dropped, before it hits the ground)
+ (h + ${h \over 2}$ + ${h \over 4}$ + . . . . . . . .$\propto$)

= h + ${h \over {1 - {1 \over 2}}}$

= h + 2h

= 3h

Fundamental frequency of a stretched string is ,

    f = ${n \over {2L}}\sqrt {{T \over \mu }}$

Here, n = number of antinodes

$\mu$ = mass per unit length.

As, O is the midpoint of two bridegs, hence length of two wires are equal.

$\therefore\,\,\,$ L₁ = L₂ = L

As frequency of both wires same

f₁ = f₂

$\Rightarrow$$\,\,\,$ ${{{n_1}} \over {2L}}\sqrt {{T \over {\pi {r^2}{\rho _1}}}}$ = ${{{n_2}} \over {2L}}\sqrt {{T \over {\pi {r^2}{\rho _2}}}}$

$\Rightarrow$$\,\,\,$ ${{{n_1}} \over {{n_2}}}$ = $\sqrt {{{{\rho _1}} \over {{\rho _2}}}}$

$\Rightarrow$$\,\,\,$ ${{{n_1}} \over {{n_2}}}$ = $\sqrt {{{{\rho _1}} \over {4{\rho _1}}}}$ = ${1 \over 2}$

Let specific heat of aluminium = S,

As we know from principle of calorimetry,

Q_given = Q_used

$\therefore\,\,\,$ 0.2 $\times$ S $\times$ (150 $-$ 40) =
150 $\times$ 1 $\times$ (40 $-$ 27) + 25 $\times$ (40$-$27)

$\Rightarrow$$\,\,\,$ 0.2 $\times$ S $\times$ 110 = 150 $\times$ 13 + 25 $\times$ 13

$\Rightarrow$$\,\,\,$ S = 434 J/kg - ^oC

Energy of sphere = ${{{Q^2}} \over {2C}}$

$\therefore\,\,\,$ ${{16 \times {{10}^{ - 12}}} \over {2C}}$ = 4.5

$\Rightarrow$$\,\,\,$ C = ${{16 \times {{10}^{ - 12}}} \over 9}$

We know capacity of spherical conductor,

C = 4$\pi$$\varepsilon$₀R

$\therefore\,\,\,$ 4$\pi$$\varepsilon$₀R = ${{16 \times {{10}^{ - 12}}} \over 9}$

$\Rightarrow$$\,\,\,$ R = ${1 \over {4\pi {\varepsilon _0}}} \times {{16 \times {{10}^{ - 12}}} \over 9}$

= 9 $\times$ 10⁹ $\times$ ${{16 \times {{10}^{ - 12}}} \over 9}$

= 16 mm

Given that,

acceleration (a) = $-$ C (constant)

$\therefore\,\,\,$ ${{dv} \over {dt}}$ = $-$ c

$\Rightarrow$$\,\,\,$ ${{dv} \over {dx}}$ . ${{dx} \over {dt}}$ = $-$ c

$\Rightarrow$$\,\,\,$ $\upsilon$ ${{dv} \over {dx}}$ = $-$ c

$\Rightarrow$$\,\,\,$ $\int {v\,dv}$ = $-$ c$\int {dx}$

$\Rightarrow$$\,\,\,$ ${{{v^2}} \over 2}$ = $-$ cx + k

$\Rightarrow$$\,\,\,$ x = $-$ ${{{v^2}} \over {2c}}$ + ${k \over c}$

From this equation we can say option (c) is the correct graph.

Let,

M $\propto$ T^x C^y h^z

$\therefore\,\,\,$ [M¹L^oT^o]  =  [T¹]^x  [L¹ T^$-$1]^y  [M¹L²T^$-$1]^z

[M¹L^o T^o]  =  [M^z L^{y + 2z} T^x$-$y$-$z]

By comparing both sides we get,

z = 1

y + 2z = 0

x $-$ y $-$ z = 0

$\therefore\,\,\,$ y = $-$ 2z = $-$ 2

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

$\therefore\,\,\,$ [M] = [M^$-$1 C^$-$2 h¹]

Work done by engine = area under closed curve = P₀ V₀

According to the principle of calorimetry,

Heat given to the system,

Q = Q_AB + Q_BC

= nC_v $\Delta$T_AB + nC_p $\Delta$T_BC

= ${3 \over 2}$ (nR T_B $-$ nRT_A) + ${5 \over 2}$ (nRT_c + nRT_B)

= ${3 \over 2}$ (2P₀ V₀ $-$ P₀ V₀) + ${5 \over 2}$ (4P₀ V₀ $-$ 2P₀ V₀)

= ${13 \over 2}$ P₀ V₀

$\therefore\,\,\,$ Thermal efficiency

= ${W \over Q}$ = ${{{P_0}\,{V_0}} \over {{{13} \over 2}{P_0}{V_0}}}$ = ${2 \over {13}}$ = 0.15

Let, side of triangle ABC = $\ell$

According to perpendicular axes theorem, moment of inertia of triangle about it center and perpendicular to its plane,

I_O = ${1 \over {12}}$ m$\ell$²

In, triangle DEF,

DE = DF = EF = ${1 \over 2}$ AB = ${1 \over 2}$ $\ell$

$\therefore\,\,\,$ moment of inertia of triangle DEF,

I_DEF = ${1 \over {12}} \times {m \over 4} \times {\left( {{\ell \over 2}} \right)^2}$

= ${1 \over {12}} \times {{m{\ell ^2}} \over {16}}$

= ${{{I_O}} \over {16}}$

$\therefore\,\,\,$ Moment of inertia of the remaining part,

I_remain = I_O $-$ ${{{{\rm I}_O}} \over {16}}$ = ${{15\,{{\rm I}_O}} \over {16}}$

For ideal gas molecule with 5 degree of freedom,

C_v = ${5 \over 2}$ R and C_p = ${7 \over 2}$ R

$\therefore\,\,\,$ ${{{C_p}} \over {{C_v}}}$ = ${{{7 \over 2}R} \over {{5 \over 2}R}}$ = ${7 \over 5}$

Potential gradient,

$\Delta$V = E. d

$\Rightarrow$$\,\,\,$ 589.8 $-$ 589.0 = (E d)_max

$\Rightarrow$$\,\,\,$ (E d)_max = 0.8

$\therefore\,\,\,$ $\Delta$V = E d cos$\theta$

= 0.8 $\times$ cos60^o

= 0.4

$\therefore\,\,\,$ Maximum potential on the sphere = 589.4 V

Mximum velocity, V_max = a$\omega$

Maximum acceleration, A_max = a$\omega$²

Given that,

${{a{\omega ^2}} \over {a\omega }}$ = 10

$\Rightarrow$$\,\,\,$ $\omega$ = 10 s^$-$1

Displacement, x = a sin ($\omega$t + ${\pi \over 4}$)

at t = 0, displacement x = 5

$\therefore\,\,\,$ 5 = a sin $\left( {{\pi \over 4}} \right)$

$\Rightarrow$$\,\,\,$ 5 = a $\times$ ${1 \over {\sqrt 2 }}$

$\Rightarrow$$\,\,\,$ a = 5${\sqrt 2 }$

$\therefore\,\,\,$ Maximum acceleration,

A_max = a$\omega$² = 5 ${\sqrt 2 }$ $\times$ (10)²

= 500 ${\sqrt 2 }$ m/s²

Because of thermal expansion, change in length

        ($\Delta$$\ell$) = $\ell$ $\alpha$ $\Delta$T . . . . .(1)

Because of compressive force, the compansion is $\Delta$$\ell$ ' ,

$\therefore\,\,\,$ Young's Modulus (y) = ${{{F \over A}} \over {{{\Delta \ell} \over \ell }}}$

$\Rightarrow$$\,\,\,$ F = YA ${{\Delta \ell '} \over \ell }$

As net change in length is 0 . So,

$\Delta \ell '$ = $\Delta \ell$ = $\ell \alpha \,\Delta T$

$\therefore\,\,\,$ F = YA $\times$ ${{\ell \alpha \,\Delta T} \over \ell }$ = AY$\alpha$$\Delta$T