JEE Main 2016 9th April Morning Shift Question Paper with Solutions

v₁ = velocity of water when it leak from hole

v₂ = velocity of water when it reach the ground.

From Bernoulli's principle,

${1 \over 2}\rho {v_1}^2 + \rho gh$ = ${1 \over 2}\rho {v_2}^2$

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

From Torricelli's theorem,

v₁ = $\sqrt {2gH}$

$\therefore$   ${v_2}^2$ = 2gh + 22gH

From continuity equation,

${A_1}{v_1}$ = ${A_2}{v_2}$

$\Rightarrow$   $\pi {r^2} \times \sqrt {2gH}$ = $\pi {x^2}\sqrt {2g\left( {h + H} \right)}$

$\Rightarrow$   x²  =  r²$\sqrt {{H \over {H + g}}}$

$\Rightarrow$   x = r ${\left( {{H \over {H + g}}} \right)^{{1 \over 4}}}$

Angular displacement ($\theta$₁) of particle 1. from equilibrium,

${y_1}$ = A sin$\theta$₁

$\Rightarrow$   A = Asin$\theta$₁

$\Rightarrow$   sin$\theta$₁ = 1 = sin ${\pi \over 2}$

$\therefore$   $\theta$₁ = ${\pi \over 2}$

Similarly for particle 2 angular displacement $\theta$₂ from equilibrium,

y₂ = Asin$\theta$₂

$\Rightarrow$   $-$ ${A \over 2}$ = Asin$\theta$₂

$\Rightarrow$   sin$\theta$₂ = $-$ ${1 \over 2}$ = sin$\left( { - {\pi \over 3}} \right)$

$\Rightarrow$   $\theta$₂ = $-$ ${{\pi \over 3}}$

Relative angular displacement of the two particle,

$\theta$ = $\theta$₁ $-$ $\theta$₂

= ${{\pi \over 2}}$ $-$ $\left( { - {\pi \over 6}} \right)$

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

Relative angular velocity $=$ $\omega - \left( { - \omega } \right)$ = $2\omega$

If they cross each other at time t

then,   t = ${\theta \over {2\omega }}$ = ${{2\pi } \over {3 \times 2\omega }}$ = ${\pi \over {3 \times {{2\pi } \over T}}}$ = ${T \over 6}$

Current in LR circuit,

${\rm I} = {{{V_0}} \over {\sqrt {{R^2} + {w^2}{L^2}} }}\sin \left( {\omega t - {\pi \over 2}} \right)$

it will be a sinusoidal wave.

Here   Y = 1   when,

(1)   A = 0  and   B = 1   $\left( {\overline A B} \right)$

(2)   A = 1  and  B = 0   $\left( {A\,\overline B } \right)$

(3)   A = 1  and  B = 1  $\left( {A\,B} \right)$

$\therefore$   Y = ${\overline A }$ B + A ${\overline B }$ + AB

= ${\overline A }$ B + A (${\overline B }$ + B)

= ${\overline A }$ B + A    [as  ${\overline B }$ + B = 1]

= (A + ${\overline A }$) (A + B)

= A + B

So, this represented OR gate.

Note :

X + ${\overline X }$ Y = (X + ${\overline X }$) (X + Y)

In an isobaric process,

Heat supplied, Q = n C_p $\Delta$ T

Work done, w = nR$\Delta$T

$\therefore$   Ratio = ${w \over Q}$ = ${{nR\Delta T} \over {n{C_p}\Delta T}}$

=   ${R \over {{5 \over 2}R}}$

=   ${2 \over 5}$

[C_p = ${5 \over 2}$ R for monoatomic gas]

Current in the circuit without ammeter

I $=$ ${5 \over {50}} = 0.1$ A

$\therefore$   With ammeter current $=$ 0.1 $\times$ ${{99} \over {100}}$ = 0.099 A

With ammeter equivalent resistance,

R_eq = 50 + ${{100\,{r_s}} \over {100 + {r_s}}}$

$\therefore$   0.099 = ${5 \over {50 + {{100\,{r_s}} \over {100 + {r_s}}}}}$

$\Rightarrow$   50 + ${{100\,{r_s}} \over {100 + {r_s}}}$ = ${5 \over {0.099}}$

$\Rightarrow$   ${{100\,{r_s}} \over {100 + {r_s}}} = 0.5$

$\Rightarrow$   100 r_s = 50 + 0.5r_s

$\Rightarrow$   99.5 r_s = 50

$\Rightarrow$   r_s $=$ ${{50} \over {99.5}} = 0.5\,\Omega$

Frequency heard by the driver of second engine,

F' = $\left( {{{v + {v_0}} \over {v - {v_S}}}} \right)$ f

given v₀ = v_s = 30 m/s

$\therefore$   f'  = $\left( {{{330 + 30} \over {330 - 30}}} \right) \times 540$

=   648 Hz

+ ve terminal at 1, $-$ ve terminal at 2 and resistance high for pnp transistor.

Before hitting point 0,

angular moment = mv $\times$ ${a \over 2}$

After hitting point 0,

Angular momentum = ${\rm I}\omega$

$\therefore$    ${\rm I}\omega$ = ${{mva} \over 2}$

$\Rightarrow$   $\omega$ = ${{mva} \over {2{\rm I}}}$

${\rm I}$ = moment of inertia about edge,

=   ${{m{a^2}} \over 6} + m{\left( {{a \over {\sqrt 2 }}} \right)^2}$

=   ${{m{a^2}} \over 6} + {{m{a^2}} \over 2}$

=   ${{2m{a^2}} \over 3}$

$\therefore$   $\omega$ = ${{mva} \over {2 \times {{2m{a^2}} \over 3}}}$ = ${{3v} \over {4a}}$ = ${{3 \times 2} \over {4 \times 0.3}}$ = 5 rad/s

Here r = 3R $-$ ${{2R} \over L}$ x

$\therefore$   Extension in the wire of length dx,

dl = ${{Fdx} \over {AY}}$

= ${{Mg\,dx} \over {\pi {r^2}\,Y}}$

= ${{Mg\,dx} \over {\pi {{\left( {3R - {{2R} \over L}x} \right)}^2}Y}}$

$\therefore$   Change in wire length,

$\Delta$L = $\int\limits_0^L {dl}$

= $\int\limits_0^L {{{Mg\,dx} \over {\pi {{\left( {3R - {{2R} \over L}x} \right)}^2}Y}}}$

= ${{Mg} \over {\pi Y}}\int\limits_0^L {{{dx} \over {{{\left( {3R - {{2R} \over L}x} \right)}^2}}}}$

= ${{Mg} \over {\pi Y}}\left[ { - {1 \over {\left( {3R - {{2R} \over L}x} \right)}} \times \left( { - {L \over {2R}}} \right)} \right]_0^L$

= ${{Mg} \over {\pi Y}}$ $\left[ {\left( {{L \over {2{R^2}}} - {L \over {6{R^2}}}} \right)} \right]$

= ${{Mg} \over {\pi Y}}\left( {{{2L} \over {6{R^2}}}} \right)$

= ${{MgL} \over {3\pi {R^2}Y}}$

$\therefore$   The equilibrium extended length of the wire,

= L + $\Delta$L

= L + ${{MgL} \over {3\pi {R^2}Y}}$

= L (1 + ${1 \over 3}$ ${{Mg} \over {\pi {R^2}Y}}$)

Change of length of wire with temperature,

$\Delta$$\ell$   =   $\alpha \ell \Delta \theta$

Time period of pendulum at temperature $\theta$,

T_$\theta$ = 2$\pi$$\sqrt {{{\ell + \Delta \ell } \over g}}$

= 2$\pi$$\sqrt {{{\ell \left( {1 + \alpha \Delta \theta } \right)} \over g}}$

= $2\pi \sqrt {{\ell \over g}} {\left( {1 + \alpha \Delta \theta } \right)^{{1 \over 2}}}$

= $2\pi \sqrt {{\ell \over g}} {\left( {1 + {{\Delta \ell } \over \ell }} \right)^{{1 \over 2}}}$

$\simeq$  T₀ $\left( {1 + {{\Delta \ell } \over {2\ell }}} \right)$

Here T₀ = time period at temperature 0^oC.

$\therefore$   Change in time period,

$\Delta$T = T_$\theta$ $-$ T₀

= ${{{T_0}\Delta \ell } \over {2\ell }}$

= ${{{T_0}\left( {\alpha \ell \Delta \theta } \right)} \over {2\ell }}$

$\therefore$   ${{\Delta T} \over {\Delta \theta }}$ = ${{{T_0}\alpha } \over 2}$

Given that T₀ = 2,

$\therefore$   ${{\Delta T} \over {\Delta \theta }}$ = ${{2\alpha } \over 2}$ = $\alpha$

${{\Delta T} \over {\Delta \theta }}$ is the shape of $\Delta$T and ${\Delta \theta }$

curve = S (given)

$\therefore$   S = $\alpha$

Let the area of ellipse = A

$\therefore$   Area of abcSa = ${A \over 2} + {A \over 4}$ = Area of half of the ellipse + Area of the triangle = ${{3A} \over 4}$

Area of adcSa = ${A \over 2} - {A \over 4}$ = ${A \over 4}$

$\therefore$   ${{{t_1}} \over {{t_2}}}$ = ${{{{3A} \over 4}} \over {{A \over 4}}}$ = 3

$\therefore$   t₁ = 3t₂

Here equivalent resistance

R_eq = R + ${{r \times R} \over {r + R}}$

=  R + ${{f{R^2}} \over {fR + R}}$

=   R + ${{fR} \over {f + 1}}$

=   ${{\left( {2f + 1} \right)R} \over {\left( {f + 1} \right)}}$

Circuit current,

${\rm I}$ = ${V \over {{{\mathop{\rm R}\nolimits} _{eq}}}}$

=   ${{V\left( {f + 1} \right)} \over {{\mathop{\rm R}\nolimits} \left( {2f + 1} \right)}}$

Current in the resistance r,

I = ${{\rm I} \over {f + 1}}$ = ${V \over {R\left( {2f + 1} \right)}}$

Heat generated in r,

H = ${\rm I}_2^2\,r$

=  ${{{V^2}f} \over {R{{\left( {2f + 1} \right)}^2}}}$

For maximum H,

${{dH} \over {df}}$ = 0

$\Rightarrow$   ${{{V^2}} \over R}\left[ {{1 \over {{{\left( {2f + 1} \right)}^2}}} - {{4f} \over {{{\left( {2f + 1} \right)}^3}}}} \right]$ = 0

$\Rightarrow$   2f + 1 = 4f

$\Rightarrow$   2f = 1

$\Rightarrow$   f = ${1 \over 2}$

For equilibrium,

net torque acting on dipole is = 0

$\therefore$   $\tau$₁ = $\tau$₂

$\Rightarrow$   mB₁ sin$\theta$₁ = mB₂ sin$\theta$₂

$\Rightarrow$   B₂ = B₁ $\times$ ${{\sin {{30}^o}} \over {\sin {{45}^o}}}$

=   15 $\times$ $\sqrt 2$ $\times$ ${1 \over 2}$ =

=  10.6 mT $\simeq$ 11 mT

Potential drop accross zener diode.

V_z = V $-$ 100 I

$\therefore$   Power dissiption = V_z I

= (V $-$ 100 I) I

Given that,

(V $-$ 100 I) I = 1

$\Rightarrow$   VI $-$ 100 I² = 1

$\Rightarrow$   100 I² $-$ VI + 1 = 0

As   As I is real,

So, b² $-$ 4ac $\ge$ far this quadratic equation.

$\therefore$   V² $-$ 4(100)1  $\ge$ 0

$\Rightarrow$   V $\ge$ 20 V

$\therefore$   Voltage range should be 0 $-$ 24 V

For lens :

u₁ = $-$ (32.2 $-$ 22.2) cm

= $-$ 10 cm

v₁ = (71.2 $-$ 32.2) cm

= 39 cm

$\therefore$   ${1 \over {{f_1}}}$ = ${1 \over {{v_1}}} - {1 \over {{u_1}}}$

= ${1 \over {39}}$ + ${1 \over {10}}$

= ${{49} \over {390}}$

$\therefore$   f₁ = 7.8 cm

For mirror :

R = (71.2 $-$ 45.8) cm

= 25.4 cm

$\therefore$   f₂ = ${R \over 2}$ = ${{25.4} \over 2}$ = 12.7 cm

Equation of motion for the mass,

ma = mg $-$ kv

$\Rightarrow$    ${{dv} \over {dt}} = {{mg - kv} \over m}$

$\Rightarrow$   $\int\limits_0^v {{{dv} \over {mg - kv}}} = {1 \over m}\int\limits_0^t {dt}$

$\Rightarrow$   $- {1 \over k}\left[ {\ln \left( {mg - kv} \right)} \right]_0^v = {t \over m}$

$\Rightarrow$   $\ln \left( {{{mg - kv} \over {mg}}} \right) = - {{kt} \over m}$

$\Rightarrow$   $1 - {{kv} \over {mg}} = {e^{ - {{kt} \over m}}}$

$\Rightarrow$   ${{kv} \over {mg}} = 1 - {e^{ - {{kt} \over m}}}$

$\Rightarrow$   $v = {{mg} \over k}\left( {1 - {e^{ - {{kt} \over m}}}} \right)$

ma $=$ mg $-$ k $\times$ ${{mg} \over k}$ (1 $-$ e$^{ - {{kt} \over m}}$)

$=$  mg $-$ mg + mge$^{ - {{kt} \over m}}$

a $=$ g e$^{ - {{kt} \over m}}$

Energy released when hydrogen atom makes transition from n = 2 to n = 1 is,

E₁ = 13.6 $\times$ $\left( {{1 \over {{1^2}}} - {1 \over {{z^2}}}} \right)$

= ${3 \over 4} \times 13.6\,\,\,$ eV

Energy required to remove a electron from n^th excited state of doubly ionized lithium,

E₂ = ${{13.6{z^2}} \over {{n^2}}}$ = ${{13.6 \times {3^2}} \over {{n^2}}}$ eV

This energy is provided by the photon when it strike with the lithium atom.

$\therefore$   E₁ $\ge$ E₂

$\Rightarrow$   ${3 \over 4} \times 13.6 \ge {{13.6 \times 9} \over {{n^2}}}$

$\Rightarrow$   n²   $\ge$  3$\times$4

$\Rightarrow$  n   $\ge$ $\sqrt {12}$

$\Rightarrow$    n $\ge$ 3.5

$\therefore$  Least possible excited state = 4

We know,

E(z) = $-$ ${{dv} \over {dz}}$

$\therefore$   E(z) = $-$ 10 z for $\left| z \right| \le 1$ m

and E(z) = 10 for $\left| z \right| \ge 1$ m

$\therefore$   The source is an infinity large non conducting thick of thickness z = 2 m.

$\therefore$   E = ${\sigma \over {2{\varepsilon _0}}}$ = ${{\rho \left( 2 \right)} \over {2{\varepsilon _0}}}$ = ${\rho \over {{\varepsilon _0}}}$

$\therefore$   ${\rho \over {{\varepsilon _0}}}$ = 10

$\Rightarrow$   $\rho$ = ${10\,{\varepsilon _0}}$

When only galvanometer G is present with the resistance R,



Here I_G = ${{{V_E}} \over {R + G}}$

When shunt of resistance S is connected parallel to galvanometer,



Here I = ${{{V_E}} \over {R + {{GS} \over {G + S}}}}$

As deflection is half, here current through galvanometer,

I_G' = ${{{{\rm I}_G}} \over 2}$

As both Galvanometer and shunt are parallel then potential are parallel then potential difference same.

$\therefore$   I_G' (G) = (I $-$ I_G^')S

$\Rightarrow$   I'_G (G + S) = IS

$\Rightarrow$   ${{{{\rm I}_G}} \over 2}$ = ${{{\rm I}S} \over {G + S}}$

$\Rightarrow$   ${{{V_E}} \over {2\left( {R + G} \right)}}$ = ${{{V_E}} \over {R + {{GS} \over {G + S}}}}$ $\times$ ${S \over {\left( {G + S} \right)}}$

$\Rightarrow$   ${1 \over {2\left( {R + G} \right)}}$ = ${{G + S} \over {R(G + S) + GS}}$ $\times$ ${S \over {\left( {G + S} \right)}}$

$\Rightarrow$   RG + RS + GS = 2RS + 2GS

$\Rightarrow$   RG = RS + GS

$\Rightarrow$    S(R + G) = RG

We know. resistivity ($\rho$) = ${{RA} \over L}$

and conductivity = ${1 \over \rho }$ = ${1 \over {RA}}$

As   R = ${V \over {\rm I}}$

$\therefore$   conductivity = ${{L{\rm I}} \over {VA}}$

Also  V = ${\omega \over q}$ = ${\omega \over {it}}$ = ${{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {\left[ {\rm I} \right]\left[ T \right]}}$ = $\left[ {M{L^2}{T^{ - 3}}{{\rm I}^{ - 1}}} \right]$

$\therefore$   Conductivity = ${{\left[ {\rm{L}} \right]\left[ {\rm I} \right]} \over {\left[ {M{L^2}{T^{ - 3}}{{\rm{I}}^{ - 1}}} \right]\left[ {{L^2}} \right]}}$

=   $\left[ {{M^{ - 1}}{L^{ - 3}}{T^3}{{\rm I}^2}} \right]$

(a)   ${1 \over {{C_{eq}}}} = {1 \over 4} + {1 \over 4} + {1 \over 4} = {3 \over 4}$

$\Rightarrow$   ${C_{eq}} = {4 \over 3}\,\mu F$

(b)   ${C_{eq}} = {{4 \times 4} \over {4 + 4}} + 4 = 6\mu F$

(c)   ${C_{eq}} = 4 + 4 + 4 = 12\,\mu F$

(d)   ${C_{eq}} = {{\left( {4 + 4} \right) \times 4} \over {\left( {4 + 4} \right) + 4}} = {8 \over 3}\mu F$

Here, tan$\theta$ = ${{100} \over {1000}} = {1 \over {10}}$

$\therefore$   sin$\theta$ = ${1 \over {10}}$ (as   $\theta$  is very small),

when car is moving uphill :



P = f $\times$ u

=  (wsin$\theta$ + f) $\times$ u

=  $\left( {{w \over {10}} + {w \over {20}}} \right) \times 10$

P = ${{3w} \over {20}} \times 10$ = ${{3w} \over 2}$

When car is moving down hill :



$\therefore$   ${P \over 2}$ = (wsin$\theta$ $-$ f) $\times$ v

$\Rightarrow$   ${{3w} \over 4}$ = $\left( {{w \over {10}} - {w \over {20}}} \right)$ $\times$ v

$\Rightarrow$   ${{w \over {20}} \times }$ v = ${{3w} \over 4}$

$\Rightarrow$   v = 15 m/s

Rocket is moving upward with acceleration 2g and gravitation acceleration is g downward direction.

So, acceleration experienced by the point object,

= 2g $-$ ($-$ g) = 3g



At equilibrium,

N = 3mgcos$\theta$

$\mu$N = 3mgsin$\theta$

$\Rightarrow$   $\mu$ (3mgcos$\theta$) = 3mg sin$\theta$

$\Rightarrow$   $\mu$ = tan$\theta$

To calculate the ratio of the AM signal bandwidth required to send both the signals together to the AM signal bandwidth required to send just the human speech, we need to consider the frequency range covered by both signals.

For the human speech signal, the frequency band is from 200 Hz to 2700 Hz. Therefore, the bandwidth required to send just the human speech is :

Bandwidth for speech = 2700 Hz - 200 Hz = 2500 Hz

For the high-frequency music signal, the frequency band is from 10200 Hz to 15200 Hz. So, the bandwidth required to send the music signal is :

Bandwidth for music = 15200 Hz - 10200 Hz = 5000 Hz

To send both signals together, we need to consider the combined frequency range. The combined frequency range covers from 200 Hz to 15200 Hz. Hence, the bandwidth required to send both signals together is :

Total bandwidth = 15200 Hz - 200 Hz = 15000 Hz

Now, let's calculate the ratio :

Ratio = Total bandwidth / Bandwidth for speech

$\Rightarrow$ Ratio = 15000 Hz / 2500 Hz

$\Rightarrow$ Ratio = 6

Therefore, the correct answer is Option C : 6.

For convex lens,

${1 \over {30}} = {1 \over {{V_1}}} + {1 \over {60}}$

$\Rightarrow$  ${1 \over {{V_1}}} = {1 \over {60}}$

$\Rightarrow$  ${V_1} = 60$ cm

For concave lens,

${1 \over { - 120}} = {1 \over {{V_2}}} - {1 \over {40}}$

$\Rightarrow$  ${1 \over {{V_2}}} = {1 \over {60}}$

$\Rightarrow$  ${V_2} = 60$ cm

object will be at 60 cm distance from concave lens.

As mirror distance is 50 cm from concave lens. So virtual object is 10 cm behind mirror.

So, real image is 10 cm infront of mirror.

Distance of image from concave lens $=$ 50 $-$ 10 $=$ 40 cm. Which is same for the real object.

So similar situation will happen and final image will form at the real object it self.

Microwave over use the principle of giving vibrational energy to water molecule.

We know,

Fringe width, $\beta$ = ${{\lambda D} \over {{d_0}}}$



From image,

$\theta$ = ${\beta \over D}$

$\Rightarrow$  $\theta$ = ${\lambda \over {{d_0}}}$

$\Rightarrow$  d₀ = ${\lambda \over \theta }$

= ${{600 \times {{10}^{ - 9}}} \over {{1 \over {60}} \times {\pi \over {180}}}}$

= 2.06 $\times$ 10^$-$3

$\simeq$ 2 mm

We know,

Einstein's photoelectric equation,

$eV = {{hc} \over \lambda } - {\phi _0}$

and ${\phi _0}$, ${{hc} \over {{\lambda _0}}}$, where ${{\lambda _0}}$ is the threashhold wavelength.

$\therefore$   In first case,

eV $= {{hc} \over {{\lambda _1}}} - {{hc} \over {{\lambda _0}}}$      . . .(1)

and in second case,

3 eV $= {{hc} \over {{\lambda _2}}} - {{hc} \over {{\lambda _0}}}$       . . .(2)

Now, let slopping potential = V₁ when light of wavelength $\lambda$₃ is used then,

eV₁ $= {{hc} \over {{\lambda _3}}} - {{hc} \over {{\lambda _0}}}$       . . .(3)

From (1) and (2) get,

$3\left( {{{hc} \over {{\lambda _1}}} - {{hc} \over {{\lambda _0}}}} \right) = {{hc} \over {{\lambda _2}}} - {{hc} \over {{\lambda _0}}}$

$\Rightarrow$   ${{3hc} \over {{\lambda _1}}}$ $-$ ${{hc} \over {{\lambda _2}}}$ $=$ ${{2hc} \over {{\lambda _0}}}$

$\Rightarrow$   ${{hc} \over {{\lambda _0}}}$ $=$ ${{3hc} \over {2{\lambda _1}}}$ $-$ ${{hc} \over {2{\lambda _2}}}$

Putting this value of ${{hc} \over {{\lambda _0}}}$ in equation 3,

eV₁ $= {{hc} \over {{\lambda _3}}} - {{3hc} \over {2{\lambda _1}}} + {{hc} \over {2{\lambda _2}}}$

$\Rightarrow$   V₁ $= {{hc} \over e}$ $\left[ {{1 \over {{\lambda _3}}} - {3 \over {2{\lambda _1}}} + {1 \over {2{\lambda _2}}}} \right]$