JEE Main 2016 10th April Morning Shift Question Paper with Solutions

The gravitional pull on the astronaut is,

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

The satellite is moving so fast around the earth that whenever it trying to fall on the earch it is missing the earth and it will keep going arround the earth. On the satellite the astronaut feels weightless because the satellite is constantly freefalling.

If gravity were absent, the satellite would move in a straight line through space instead of orbiting the Earth. In this scenario, the astronaut would still feel weightless, but for a different reason: there would be no gravitational force acting on the satellite in the absence of gravity.

Option B : $0 < F_G < \frac{GMm}{R^2}$

The gravitational force $F_G$ acting on the astronaut while in orbit is less than the gravitational force at the Earth's surface, but it is not zero. The force of gravity decreases with the square of the distance from the center of the Earth. Since the astronaut is at a distance ( R + h ) from the center of the Earth, where ( h ) is the altitude of the orbit, the gravitational force will be less than at the Earth's surface but greater than zero.

So, both Option B and Option D are correct. Option B correctly states that the gravitational force is less than at the Earth's surface but greater than zero, and Option D gives the precise formula for calculating that force.

Let full scale deflection of current = I

In case 1, when R = 2400 $\Omega$ and deflection of 40 divisions present.

$\therefore$   ${{40} \over {50}}{\rm I} = {V \over {G + R}}$

$\Rightarrow$   ${4 \over 5}{\rm I}$ = ${2 \over {G + 2400}}$          . . .(1)

Incase 2, when R = 4900 $\Omega$ and deflection of 20 divisions present

$\therefore$   ${{20} \over {50}}{\rm I} = {V \over {G + R}}$

$\Rightarrow$   ${2 \over 5}{\rm I}$ = ${2 \over {G + 4900}}$          . . .(2)

From (1) and (2) we get,

${4 \over 2} = {{G + 4900} \over {G + 2400}}$

$\Rightarrow$   2G + 4800 $=$ G + 4900

$\Rightarrow$   G $=$ 100 $\Omega$

Putting value of G in equation (1), we get,

${4 \over 5}{\rm I} = {2 \over {100 + 2400}}$

$\Rightarrow$   ${\rm I} = 1$ mA

Current sensitivity $=$ ${{\rm I} \over {number\,\,of\,\,divisions}}$

$=$ ${1 \over {50}}$

$=$ 0.02 mA / division

$=$ 20 $\mu$A / division

Resistance required for deflection of 10 divisions

${{10} \over {50}}{\rm I} = {V \over {G + R}}$

$\Rightarrow$   ${1 \over 5} \times 1 \times {10^{ - 3}} = {2 \over {100 + R}}$

$\Rightarrow$   R $=$ 9900 $\Omega$

Given,

B = B₀e$^{ - {t \over \tau }}$

Area of the circular loop, A = $\pi$ r²

$\therefore$   Flux $\phi$ = BA = $\pi$ r² B₀ e$^{ - {t \over \tau }}$

Induced emf in the loop,

$\varepsilon$ = $-$ ${{d\phi } \over {dt}}$ = $\pi$ r²B₀${1 \over \tau }$e$^{ - {t \over \tau }}$

Heat generated

= $\int\limits_0^ \propto {{i^2}R\,dt}$

= $\int\limits_0^ \propto {{{{\varepsilon ^2}} \over R}} \,dt$

= ${1 \over R}{{{\pi ^2}{r^4}B_0^2} \over {{\tau ^2}}}\int\limits_0^ \propto {{e^{ - {{2t} \over \tau }}}} \,dt$

= ${{{\pi ^2}{r^4}B_0^2} \over {{\tau ^2}R}} \times {1 \over {\left( { - {2 \over \tau }} \right)}}\left[ {{e^{ - {{2t} \over \tau }}}} \right]_0^ \propto$

= ${{ - {\pi ^2}{r^4}B_0^2} \over {2{\tau ^2}R}} \times \tau \left( {0 - 1} \right)$

= ${{{\pi ^2}{r^4}B_0^2} \over {2\tau R}}$

The limit of resolution of a telescope,

$\Delta$$\theta$  =  ${{1.22\,\,\lambda } \over D}$ = ${l \over R}$

$\therefore$   $l$ = ${{1.22\,\,\lambda R} \over D}$

=   ${{1.22 \times 6 \times {{10}^{ - 7}} \times 10 \times 9.46 \times 10{}^{15}} \over {30 \times {{10}^{ - 2}}}}$

=   2.31 $\times$ 10⁸ km

Here   AB = x

and   BC = y

and   $\lambda$ = linear mass density.

As centre of mass is below point A, so horizontal distance of the centre of mass from B is = xcos60^o = ${x \over 2}$

$\therefore$   X_CM = ${x \over 2}$ = ${{{m_1}{x_1} + {m_2}{x_2}} \over {{m_1} + {m_2}}}$

$\Rightarrow$   ${x \over 2}$ = ${{\left( {\lambda x} \right)\left( {{x \over 2}} \right)\cos {{60}^o} + \left( {\lambda y} \right)\left( {{y \over 2}} \right)} \over {\lambda \left( {x + y} \right)}}$

$\Rightarrow$   ${x \over 2}$ = ${{{{{x^2}} \over 4} + {{{y^2}} \over 2}} \over {x + y}}$

$\Rightarrow$   x² + xy = ${{{x^2}} \over 2} + {y^2}$

$\Rightarrow$   x² + 2xy $-$ 2y² = 0

$\therefore$   x = ${{ - 2y \pm \sqrt {{{\left( {2y} \right)}^2} - 4.1\left( { - 2{y^2}} \right)} } \over {2.1}}$

=    ${{ - 2y \pm \sqrt {12{y^2}} } \over 2}$

=   $-$ y $\pm$ $\sqrt 3$y

${x \over y} \ne - \sqrt 3 - 1$  as  ${x \over y}$ = positive.

$\therefore$   ${x \over y}$ = $\sqrt 3 - 1$

$\Rightarrow$   ${y \over x}$ = ${1 \over {\sqrt 3 - 1}} \times {{\sqrt 3 + 1} \over {\sqrt 3 + 1}}$

=   ${{\sqrt 3 + 1} \over 2}$

=   ${{2.732} \over 2}$

=   1.366 $\simeq$ 1.37

V_B = vhBcos$\theta$

= 240 $\times$ 5 $\times$ 5 $\times$ 10^$-$5$\times$ ${{\sqrt 5 } \over 3}$

= 44.7 $\times$ 10^$-$3 V

= 45 mV



V_w = $l$vB sin$\theta$

= 15 $\times$ 240 $\times$ 5 $\times$ 10^$-$5 $\times$ ${2 \over 3}$

= 1200 $\times$ 10^$-$4 V

= 120 mV

From right hand rule, the charge moves to the left side of the pilot.

R $=$ ${{{r_0}} \over {{r_i}}} \equiv 1 - 10$

We know,

centripetal acceleration = ${{{V^2}} \over R}$

$\therefore$   According to question,

${{{V^2}} \over R}$ = ${n^2}R{t^2}$

$\Rightarrow$   V² = n² R² t²

$\Rightarrow$   V = nRt

$\Rightarrow$   ${{dV} \over {dt}}$ = nR

Power (P) = Force (F) $\times$ Velocity (V)

= M ${{dV} \over {dt}}$(V)

= M (nR) (nRt)

= Mn²R²t

Equivalent capacitance of 6 $\mu$F and 12 $\mu$F is = ${{6 \times 12} \over {12 + 6}}$ = 4 $\mu$F

Equivalent capacitance of 4$\mu$F and 4$\mu$F

is = 4 + 4 = 8 $\mu$F

New circuit is $\to$



equivalent capacitance of 1 $\mu$F and 8 $\mu$F is

   = ${{1 \times 8} \over {8 + 1}}$ = ${8 \over 9}$ $\mu$F

Equivalent capacitance of 8 $\mu$F and 4 $\mu$F is

   = ${{8 \times 4} \over {12}}$ = ${8 \over 3}$ $\mu$F

New circuit is $\to$



Equation capacitance of ${8 \over 9}$ $\mu$F and ${8 \over 3}$ $\mu$F is

   = ${8 \over 9}$ + ${8 \over 3}$ = ${32 \over 9}$ $\mu$F

$\therefore$   Equivalent capacitance of AB is

C_AB = ${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$

Given that,

C_AB = 1 $\mu$F

$\therefore$   1 = ${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$

$\Rightarrow$   C + ${{{32} \over 9}}$ = ${{32C} \over 9}$

$\Rightarrow$   ${{23C} \over 9} = {{32} \over 9}$

$\Rightarrow$   C = ${{32} \over {23}}$ $\mu$F

Total heat required to freeze 5 kg water,

=   5 $\times$ 336  $\times$ 10³ J

=   1680  $\times$  10³   Joule

=   1680 kJ

For carnot cycle,

${{{Q_2}} \over {{Q_1}}}$ = ${{{T_2}} \over {{T_1}}}$

$\Rightarrow$   ${{{Q_2}} \over {1680}}$ = ${{300} \over {273}}$

$\Rightarrow$   Q₂ = 1680 $\times$  ${{300} \over {273}}$ kJ

$\therefore$   W = Q₂ $-$ Q₁

=   1680$\left( {{{300} \over {273}} - 1} \right)$

=   1680 $\times$ ${{27} \over {273}}$

=   166.15 kJ

=   166.15 $\times$ 10³ J

=   1.66 $\times$ 10⁵ J

Here,

Amplitude, A = 7 cm = 0.07 m

When washer is no longer stays in contact with the piston, then the normal force on the washer is = 0

$\therefore$   Maximum acceleration of the washer,

a_max = $\omega$²A = g

$\Rightarrow$   $\omega$ = $\sqrt {{g \over A}}$ = $\sqrt {{{10} \over {0.07}}}$ = $\sqrt {{{1000} \over 7}}$

$\therefore$   Frequency of the piston,

f = ${\omega \over {2\pi }}$ = ${1 \over {2\pi }}\sqrt {{{1000} \over 7}}$ = 1.9 Hz

Here,

Y = c . (a + b)

Now  Y = 1   when,

c = 1 and (a = 1, b = 0 or a = 0, b = 1 or a = 1, b = 1)

We know, ideal gas equation,

PV = nRT

Here T = constant.

$\therefore$   PV = constant

$\Rightarrow$   P ${m \over \rho }$ = constant

$\Rightarrow$   P  $\propto$ $\rho$

We know,

Einstein's photo electric equation,

(KE)_max = ${{hc} \over \lambda }$ $-$ $\phi$₀

In first case,

K = ${{hc} \over \lambda }$ $-$ $\phi$₀           . . .(1)

In second case,

3K = ${{2hc} \over \lambda }$ $-$ $\phi$₀           . . .(2)

$\Rightarrow$   $3\left( {{{hc} \over \lambda } - {\phi _0}} \right)$ = ${{2hc} \over \lambda } - {\phi _0}$

$\Rightarrow$   ${{3hc} \over \lambda }$ $-$ 3$\phi$₀ = ${{2hc} \over \lambda }$ $-$ $\phi$₀

$\Rightarrow$   ${{hc} \over \lambda }$ = 2$\phi$₀

$\Rightarrow$   $\phi$  =  ${{hc} \over {2\lambda }}$

Given,

R = R₆ [1 + $\alpha$ (T $-$ t₀)]

120 = 100 [1 + $\alpha$ (500 $-$ 300)]

$\Rightarrow$   200 $\alpha$ = ${1 \over 5}$

$\Rightarrow$   $\alpha$ = 10^{$-$3   o}C^$-$1

Temperature of the toaster raised from 300 K to 500 K in 30 s.

$\therefore$   Increment in the temperature in time t,

$\Delta$T = ${{500 - 300} \over {30}}t$

= ${{200} \over 3}t$

= ${{20} \over 3}t$

Total work done in raising the temperature

= $\int\limits_0^t {{{{V^2}} \over {R\left( t \right)}}\,dt}$

= $\int\limits_0^t {{{{V^2}} \over {{R_0}\left( {1 + \alpha \Delta t} \right)}}} \,dt$

= $\int\limits_0^{30} {{{{{\left( {200} \right)}^2}} \over {100\left( {1 + {{10}^{ - 3}} \times {{20} \over 3}t} \right)}}} \,dt$

= ${{40000} \over {100}}\int\limits_0^{30} {{{dt} \over {\left( {1 + {t \over {150}}} \right)}}}$

$400 \times 150\left[ {\ln \left( {1 + {t \over {150}}} \right)} \right]_0^{30}$

$= 60000\left[ {\ln \left( {1 + {{30} \over {150}}} \right) - \ln 1} \right]$

$= 60000\ln \left( {{6 \over 5}} \right)\,J$

For common emitter configuration, the input characteristic graph is shown above,

r_i $=$ ${{\Delta {V_{BE}}} \over {\Delta {{\rm I}_B}}}$

$\Rightarrow$   ${1 \over {{r_i}}} = {{d{{\rm I}_B}} \over {d{V_{BE}}}}$ $=$ shope of this curve.

Upto knee voltage 0.7 V the shope is almost constant. Then it increases sharply.

So, option (c) is the correct choice.

The refractive index is

$\mu = {{{\mathop{\rm Real}\nolimits} \,depth} \over {Apparent\,depth}} = {{{\mathop{\rm Reading}\nolimits} \,3 - Reading\,1} \over {{\mathop{\rm Reading}\nolimits} \,3 - Reading\,2}}$

Therefore, the minimum of three readings are required.

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

V_i  =  $\pi$ (a + $\Delta$a)²b

V_f  =  $\pi$a²b

$\therefore$   $\Delta$V = 2$\pi$ab$\Delta$a

$\therefore$   ${{{\Delta V} \over V}}$ = ${{2\pi ab\Delta a} \over {\pi {a^2}b}}$ = ${{2\Delta a} \over a}$

$\therefore$   $\Delta$P = B $\times$ ${{{\Delta V} \over V}}$

      = B $\times$ ${{2\Delta a} \over a}$

Normal Force (N) = $\Delta$P $\times$ A

       = B $\times$ ${{2\Delta a} \over a}$ $\times$ (2$\pi$a)b

       =   4$\pi$B$\Delta$ab

$\therefore$   Frictional force = $\mu$N = (4$\pi$$\mu$Bb)$\Delta$a

Magnetic force on electron

$\overrightarrow F$ = $-$ e $\left( {\overrightarrow V \times \overrightarrow B } \right)$

F = eVB   [As ${\overrightarrow V }$ and ${\overrightarrow B }$ are perpendicular]

Also,   F = e E

and  E = ${\sigma \over {{\varepsilon _0}}}$

$\therefore$   eVB = e $\times$ ${\sigma \over {{\varepsilon _0}}}$

$\Rightarrow$   $\sigma$ = ${{\varepsilon _0}}$VB = $\sigma$₁

as   $\sigma$₁ = $-$ $\sigma$₂

$\therefore$   $\sigma$₂ = $-$ ${{\varepsilon _0}}$VB

Using mirror formula.

${1 \over v} + {1 \over { - 4}} = {1 \over { - 5}}$

$\Rightarrow \,\,\,{1 \over v} = {1 \over 4} - {1 \over 5} = {1 \over {20}}$

$\Rightarrow \,\,\,v = 20\,$ cm

$\therefore$   Image distance is 20 cm



I₁ acts as object for plane glass surface.

$\therefore$   Appartent depth =  ${{R + v} \over \mu } = {{30} \over {1.5}} = 20$ cm

$\therefore$   Position of the image of the air bubble is 20 cm below the flat surface.

Given,

F = ${R \over {{t^2}}}$ v(t)

$\Rightarrow$   m ${{dv} \over {dt}}$ = ${R \over {{t^2}}}$ (v)

$\Rightarrow$   ${{dv} \over v}$ = ${R \over m}$ ${{dt} \over {{t^2}}}$

Intergrating both sides,

$\int {{{dv} \over v} = {R \over m}\int {{{dt} \over {{t^2}}}} }$

$\Rightarrow$   lnv = ${{R \over m}}$ $\times$ $\left( { - {1 \over t}} \right)$ + C

$\Rightarrow$   lnv = $-$ ${{R \over m}}$ $\left( {{1 \over t}} \right)$ + C

Graph between lnv and ${{1 \over t}}$ will be straight line curve.

Let the frequency of the horn = f

Apparent frequency heared by the observer directly,

${f_{dir}} = \left( {{V \over {V - {V_s}}}} \right)f = \left( {{{340} \over {340 - 5}}} \right)f = {{340} \over {335}}f$

Apparent frequency heared by the observer on reflection from the wall,

${f_{ind}} = \left( {{V \over {V + {V_s}}}} \right)f = \left( {{{340} \over {340 + 5}}} \right)f = {{340} \over {345}}f$

Also, given that,

f_ind $-$ f_dir $=$ 5

$\Rightarrow$   ${{340f} \over {345}}$ $-$ ${{340f} \over {335}}$ $=$ 5

$\Rightarrow$   f $=$ ${5 \over {340}} \times {{335 \times 345} \over {10}}$

= 169.9  $\simeq$   170 Hz

For proper mixing, the concrete mixture should not stick to the top wall of the cylinder and it should fall down.

The maximum velocity at the top is v, then we have

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

$\Rightarrow v = \sqrt {rg}$

Therefore, the maximum rotational speed of the drum is

$\omega = {v \over r} = \sqrt {{g \over r}} = \sqrt {{{9.8} \over {1.25}}}$ rad/s

$= \sqrt {{{9.8} \over {1.25}}} \times {{60} \over {2\pi }}$ rpm = 26.74 $\approx$ 27 rpm

Here, $\Delta$v and $\Delta$r are same for any pair of surface.

we know,

Electric field, E = $-$ ${{dv} \over {dr}}$

$\therefore$   E = constant [As dv and dr are constant]

Electric field inside the spherical charge distribution.

E = ${{\rho r} \over {3{\varepsilon _0}}}$

Now,   as E = constant

$\therefore$   $\rho$ r = constant

$\Rightarrow$   $\rho$ (r) $\propto$ ${1 \over r}$

Given,

C_m(t) = 30 sin 300$\pi$t + 10 (cos 200 $\pi$t $-$ cos 400 $\pi$t)

Standard equation of amplitude modulated wave,

C_m(t) = A_c sin ($\omega$_ct) $-$ ${{\mu {A_c}} \over 2}$ cos $\left( {{\omega _c} + {\omega _m}} \right)$ t + ${{\mu {A_c}} \over 2}$ cos $\left( {{\omega _c} - {\omega _m}} \right)$ t

By comparing we get,

A_c = 30 V

$\omega$_c = 300 $\pi$

$\Rightarrow$   2$\pi$f_c = 300 $\pi$

$\Rightarrow$   f_c = 150 Hz

$\omega$_c $-$ $\omega$_s = 200 $\pi$

$\Rightarrow$   2$\pi$ (f_c $-$ f_$\omega$) = 200 $\pi$

$\Rightarrow$   f_c $-$ f_$\omega$ = 100 Hz

$\therefore$   f_$\omega$ = 150 $-$ 100 = 50 Hz

and ${{\mu {A_c}} \over 2}$ = 10

$\Rightarrow$   $\mu$ $\times$ ${{30} \over 2}$ = 10

$\Rightarrow$   $\mu$ = ${{10} \over {15}}$ = ${2 \over 3}$

Here u = 50 m/s , what t = 0

$\alpha$  =  ${{\Delta v} \over {\Delta t}}$  =  ${{50 - 0} \over {0 - 10}}$  =  $-$5 m/s²

Speed of the body at t = 2 s

v   =   u + at

=  50 + ($-$ 5) $\times$ 2

=  40 m/s

From work energy theorem,

$\Delta$w  =  ${1 \over 2}m{v^2} - {1 \over 2}m{u^2}$

=  ${1 \over 2}$ m(v² $-$u²)

= ${1 \over 2}$ $\times$ 10 $\times$ (40² $-$ 50²)

=  5 $\times$ ($-$10)(90)

=  $-$ 4500 J

Let, velocity offer collision = v₁

$\therefore$   From conservation of momentum,

mv = (m + m) v₁

$\Rightarrow$    v₁ = ${v \over 2}$

$\therefore$   Loss in kinetic energy

= ${1 \over 2}$ mv² $-$ ${1 \over 2}$ (2m) $\times$ ${\left( {{v \over 2}} \right)^2}$

= ${1 \over 4}\,$mv²

lost kinetic energy is used by the electron to jump from first orbit to second orbit.

$\therefore$   ${1 \over 4}$mv² = (13.6 $-$ 3.4) eV = 10.2 eV

$\Rightarrow$  ${1 \over 2}$mv² = 20.4 eV

Young's Modulus of the material of the rod is

$Y = {{Stress} \over {Strain}} = {{(F/A)} \over {(\Delta l/l)}}$

Here, Y remains maximum, when $\Delta$l is of least count.

That is,

${Y_{\max }} = \left[ {{{50\pi \times {{10}^3}N} \over {\pi {{(5 \times {{10}^{ - 3}})}^2}{m^2}}}} \right]\left[ {{{1m} \over {0.01 \times {{10}^{ - 3}}m}}} \right]$

$= 2 \times {10^4}N/{m^2}$

Given,

AD = c ln (BD)

As log is dimensionless.

So, [BD] = 1   $\Rightarrow$  [B] = ${1 \over {\left[ D \right]}}$

And  [AD] = [C]

Now checking options one by one

(a)   [B² C²] = [B²] [A² D²] = [A²] [B² D²] = [A²]

$\therefore$   This is meaningful.

(b)   $\left( {{{A - C} \over D}} \right)$ is not meaningful.

As dimension of A $\ne$ dimension of C

Hence (A $-$ C)   is not possible.

(c)   $\left[ {{A \over B}} \right]$ = [AD] = [C]

$\therefore$   ${{A \over B}}$ $-$ C is meaningful.

(d)   $\left[ {{C \over {BD}}} \right]$ = ${{\left[ C \right]} \over {\left[ {BD} \right]}}$ = ${{\left[ C \right]} \over 1}$ = [C]

$\left[ {{{A{D^2}} \over C}} \right]$ = ${{\left[ {AD} \right]\left[ D \right]} \over {\left[ C \right]}}$ = ${{\left[ C \right]\left[ D \right]} \over {\left[ C \right]}}$ = [D]

As dimension of C and D are not same so it is not meaning ful.