AIEEE 2010 Question Paper with Solutions

$y = 0.02\left( m \right)\sin \left[ {2\pi \left( {{t \over {0.04\left( s \right)}}} \right) - {x \over {0.50\left( m \right)}}} \right]$

But $y = a\sin \left( {\omega t - kx} \right)$

$\therefore$ $\omega = {{2\pi } \over {0.04}} \Rightarrow v = {1 \over {0.04}} = 25\,Hz$

$k = {{2\pi } \over {0.50}} \Rightarrow \lambda = 0.5m$

$\therefore$ velocity, $v = v\lambda = 25 \times 0.5\,m/s = 12.5\,m/s$

Velocity on a string is given by

$v = \sqrt {{T \over \mu }}$

$\therefore$ $T = {v^2} \times \mu = {\left( {12.5} \right)^2} \times 0.04 = 6.25\,N$

The speed of light $(c)$ in a medium of refractive index $\left( \mu \right)$ is given by

$\mu = {{{c_0}} \over c},$ where ${c_0}$ is the speed of light in vacuum

$\therefore$ $c = {{{c_0}} \over \mu } = {{{c_0}} \over {{\mu _0} + {\mu _2}\left( I \right)}}$

As $I$ is decreasing with increasing radius, it is maximum

on the axis of the beam. Therefore, $c$ is minimum on the axis of the beam.

Due to the movement of resistor $R,$ an $emf$ equal to $Blv$ will be induced in it as shown in figure clearly,

$I = {I_1} + {I_2}$
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Also, ${I_1} = {I_2}$

Solving the circuit, we get

${I_1} = {I_2} = {{Blv} \over {3R}}$

and $I = 2{I_1} = {{2Blv} \over {3R}}$

Given $U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}}$

${U\left( {x = \infty } \right)}$ = 0

We know $F = - {{dU} \over {dx}} = - \left[ {{{12a} \over {{x^{13}}}} + {{6b} \over {{x^7}}}} \right]$

At equilibrium: ${{dU\left( x \right)} \over {dx}} = 0$

$\Rightarrow {{ - 12a} \over {{x^{13}}}} = {{ - 6b} \over {{x^7}}}$

$\Rightarrow x = {\left( {{{2a} \over h}} \right)^{{1 \over 6}}}$

$\therefore$ ${U_{at\,\,equilibrium\,}} = {a \over {{{\left( {{{2a} \over b}} \right)}^2}}} - {b \over {\left( {{{2a} \over b}} \right)}}$

$= - {{{b^2}} \over {4a}}$

$\therefore$ $D = 0 - \left( { - {{{b^2}} \over {4a}}} \right) = {{{b^2}} \over {4a}}$

${F_e} = T\sin {15^ \circ }\,\,;$
$mg = T\cos {15^ \circ }$
$\Rightarrow \tan {15^ \circ } = {{{F_e}} \over {mg}}$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$
In liquid, ${F_e}' = T'\sin {15^ \circ }$ $\,\,\,\,\,\,...(ii)$

$mg = {F_B} + T'\cos {15^ \circ }$
${F_B}' = V\left( {d - \rho } \right)g = V\left( {1.6 - 0.8} \right)g = 0.8\,Vg$
$= 0.8{m \over d}g = {{0.8mg} \over {1.6}} = {{mg} \over 2}$
$\therefore$ $mg = {{mg} \over 2} + T'\cos {15^ \circ }$
$\Rightarrow {{mg} \over 2} = T'\cos {15^ \circ }$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( B \right)$
From $(A)$ and $(B),$ $\tan \,{15^ \circ } = {{2{F_e}'} \over {mg}}\,\,\,\,\,\,\,\,\,...\left( 2 \right)$
From $(1)$ and $(2)$
${{{F_e}} \over {mg}} = {{2{F_e}'} \over {mg}} \Rightarrow {F_e} = 2{F_e}' \Rightarrow {F_e}' = {{{F_e}} \over 2}$

The final boolean expression is,

$X = \overline {\left( {\overline A .\overline B } \right)} = \overline {\overline A } + \overline {\overline B } = A + B \Rightarrow \,\,OR\,gate$

In nuclear fission, the binding energy per nucleon of daughter nuclei is greater than the parent nucleon of daughter nuclei is greater than the parent nucleus.

Let us consider a spherical shell of radius $x$ and thickness $dx.$

Charge on this shell

$dq = \rho .4{\pi ^2}dx = {\rho _0}\left( {{5 \over 4} - {x \over R}} \right).4\pi {x^2}dx$

$\therefore$ Total charge in the spherical region from center to $r$ $\left( {r < R} \right)$ is

$q = \int {dq = 4\pi {\rho _0}\int\limits_0^r {\left( {{5 \over 4} - {x \over R}} \right)} } {x^2}dx$

$= 4\pi {\rho _0}\left[ {{5 \over 4}.{{{r^3}} \over 3} - {1 \over R}.{{{r^4}} \over 4}} \right]$

$= \pi {\rho _0}{r^3}\left( {{5 \over 3} - {r \over R}} \right)$

$\therefore$ Electric field at $r,$ $E = {1 \over {4\pi { \in _0}}}.{q \over {{r^2}}}$

$= {1 \over {4\pi { \in _0}}}.{{\pi {\rho _0}{r^3}} \over {{r^2}}}\left( {{5 \over 3} - {r \over R}} \right)$

$= {{{\rho _0}r} \over {4{ \in _0}}}\left( {{5 \over 3} - {r \over R}} \right)$

Initially the parallel beam is cylindrical. Therefore, the wave-front will be planar.

Statement 1 is true. The energy of an incident photon (from the ultraviolet light or X-rays) on a photocell is given by Planck's equation, $E = h\nu$, where $h$ is Planck's constant and $\nu$ is the frequency of the light. X-rays have a higher frequency than ultraviolet light, so they deliver more energy to the photoelectrons. This results in a higher stopping potential ($V_0$) and maximum kinetic energy ($K_{\max}$) for the photoelectrons.

Statement 2 is also true. However, while the speeds (and hence kinetic energies) of photoelectrons do vary, this variation is not because of a range of frequencies in the incident light. Rather, it's due to the interaction of the incident photons with electrons at different energy levels in the metal. A single frequency of light can produce photoelectrons with a range of speeds because the electrons they encounter can have a variety of binding energies.

By conservation of energy,

$\left( {M + \Delta m} \right){c^2} = {{2M} \over 2}{c^2} + {1 \over 2}.{{2M} \over 2}{v^2},$

where $v$ is the speed of the daughter nuclei

$\Rightarrow \Delta m{c^2} = {M \over 2}{v^2}$

$\therefore$ $v = c\sqrt {{{2\Delta m} \over M}}$

When capacitance is taken out, the circular is $LR.$

$\therefore$ $\tan \phi = {{\omega L} \over R}$

$\Rightarrow \omega L = R\,\tan \phi$

$= 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$

Again, when inductor is taken out, the circuit is $CR.$

$\therefore$ $\tan \phi = {1 \over {\omega CR}}$

$\Rightarrow {1 \over {\omega c}} = R\tan \phi$

$= 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$

Now, $Z = \sqrt {{R^2} + {{\left( {{1 \over {\omega C}} - \omega L} \right)}^2}}$

$= \sqrt {{{\left( {200} \right)}^2} + {{\left( {{{200} \over {\sqrt 3 }} - {{200} \over {\sqrt 3 }}} \right)}^2}} = 200\Omega$

Power dissipated $= {V_{rms}}{I_{rms}}\cos \phi$

$= {V_{rms}}.{{{V_{rms}}} \over Z}.{R \over Z}$

$\left( {\,\,\,} \right.$ as $\left. {\,\,\cos \phi = {R \over Z}\,\,\,} \right)$

$= {{{V^2}rmsR} \over {{Z^2}}} = {{{{\left( {220} \right)}^2} \times 200} \over {{{\left( {200} \right)}^2}}}$

$= {{220 \times 220} \over {200}} = 242\,W$

In completely inelastic collision,

${m_1}{v_1} + {m_2}{v_2} = {m_1}v + {m_2}v$

after collision both particle have common velocity $v$, so all energy is not lost

$\therefore$ Statement - $1$ is true

The principle of conservation of momentum applicable for all kinds of collisions.

So statement - $2$ is also true.

Statement - $2$ explains statement - $1$ correctly because applying the principle of conservation of momentum, we can get the common velocity and hence the kinetic energy of the combined body.

${}_Z^AX\mathop \to \limits^{A - 12} {}_{Z - 8}Y + 3{}_2^4{X_e} + {}_t^0e$

Number of protons, N_p = Z - 8

Number of neutrons, N_n = $A - 12 - \left( {Z - 8} \right)$

$\therefore$ Required ratio $= {{A - Z - 4} \over {Z - 8}}$

In the medium, the refractive index will decreases from the axis forwards the periphery of the beam.

Therefore, the beam will move as one move from the axis to the periphery and hence the beam will converge.

${T_1}{V^{\gamma - 1}} = {T_2}{\left( {32V} \right)^{\gamma - 1}}$
$\Rightarrow {T_1} = {\left( {32} \right)^{\gamma - 1}}.{T_2}$
For diatomic gas, $\gamma = {7 \over 5}$
$\therefore$ $\gamma - 1 = {2 \over 5}$
$\therefore$ ${T_1} = {\left( {32} \right)^{{2 \over 5}}}.{T_2} \Rightarrow {T_1} = 4{T_2}$
Now, efficiency $= 1 - {{{T_2}} \over {{T_1}}}$
$= 1 - {{{T_2}} \over {4{T_2}}}$
$= 1 - {1 \over 4}$
$= {3 \over 4}$
$= 0.75.$

Oil will float on water so, $(2)$ or $(4)$ is the correct option, But density of ball is more than that of oil, hence it will sinkin oil.

For a particle in uniform circular motion,

${a_c} = {{{v^2}} \over R}$ towards the center of the circle

From figure, $\overrightarrow a = {a_c}\cos \theta \left( { - \widehat i} \right) + {a_c}\sin \theta \left( { - \widehat j} \right)$

$= {{ - {v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j$

Position vector of the particle from the original at any time t is

$\overrightarrow r = {v_0}t\cos \theta \widehat i + \left( {{v_0}t\sin \theta - {1 \over 2}g{t^2}} \right)\widehat j$

As Velocity vector, $\overrightarrow v = {{d\overrightarrow r } \over {dt}}$

$\therefore$ $\overrightarrow v = {d \over {dt}}\left( {{v_0}t\cos \theta \widehat i + \left( {{v_0}t\sin \theta - {1 \over 2}g{t^2}} \right)\widehat j} \right)$

$= {v_0}\cos \theta \widehat i + \left( {{v_0}\sin \theta - gt} \right)\widehat j$

Angular momentum of the particle about the origin is

$\overrightarrow L = m\left( {\overrightarrow r \times \overrightarrow v } \right)$

\eqalign{ & \overrightarrow L = m\left[ {{v_0}\,\cos \theta t\widehat i + \left( {{v_0}\,\sin \theta t - {1 \over 2}g{t^2}} \right)\widehat j} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \times \left[ {{v_0}\,\cos \,\theta \widehat i + \left( {{v_0}\,\sin \theta - gt} \right)\widehat j} \right] \cr}

$= m{v_0}\cos \theta t\left[ { - {1 \over 2}gt} \right]\widehat k$

$= - {1 \over 2}mg{v_0}{t^2}\cos \theta \widehat k$

At $t=0,$ no current will flow through $L$ and ${R_1}$

$\therefore$ Current through battery $= {V \over {{R_2}}}$

At $t = \infty ,$

effective resistance, ${{\mathop{\rm R}\nolimits} _{eff}} = {{{R_1}{R_2}} \over {{R_1} + {R_2}}}$

$\therefore$ Current through battery $= {V \over {{{\mathop{\rm R}\nolimits} _{eff}}}} = {{V\left( {{R_1} + {R_2}} \right)} \over {{R_1}{R_2}}}$

The magnetic field varies inversely with the distance for a long conductor. That is, $B \propto {1 \over d}.$ According to the magnitude and direction shown graph $(1)$ is the correct one.

Along inclined plane the equation of motion of the body

$mg\,\sin \,\theta = ma$ $\,\,\,\,\,\,\,\,$ $\therefore$ $a = g\,\sin \,\theta$

where $a$ is along the inclined plane.

$\therefore$ vertical component of acceleration is $\left( {g\sin \theta } \right)\sin \theta$ = $g\,{\sin ^2}\theta$ (Along vertical)

For block A,
${a_{A\left( {along\,vertical} \right)}} = g{\sin ^2}60^\circ$

For block B,
${a_{B\left( {along\,vertical} \right)}} = g{\sin ^2}30^\circ$

$\therefore$ relative vertical acceleration of $A$ with respect to $B$ is

$g{\sin ^2}60^\circ$ - $g{\sin ^2}30^\circ$

=$g\left( {{{\sin }^2}60 - {{\sin }^2}\left. {30} \right]} \right.$

$= 4.9$ $\,\,m/{s^2}$ in vertical direction

${R_1} = {R_0}\left[ {1 + {\alpha _1}\Delta t} \right];$

${R_2} = {R_0}\left[ {1 + {\alpha _2}\Delta t} \right]$

$R = {R_1} + {R_2}$

$= {R_0}\left[ {2 + \left( {{\alpha _1} + {\alpha _2}} \right)\Delta t} \right]$

$= 2{R_0}\left[ {1 + \left( {{{{\alpha _1} + {\alpha _2}} \over 2}} \right)\Delta t} \right]$

${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$

In Parallel, ${1 \over R} = {1 \over {{R_1}}} + {1 \over {{R_2}}}$

$= {1 \over {{R_0}\left[ {1 + {\alpha _1}\Delta t} \right]}} + {1 \over {{R_0}\left[ {1 + {\alpha _2}\Delta t} \right]}}$

$\Rightarrow {1 \over {{{{R_0}} \over 2}\left( {1 + {\alpha _{eq}}\Delta t} \right)}} = {1 \over {{R_0}\left( {1 + {\alpha _1}\Delta t} \right)}} + {1 \over {{R_0}\left( {1 + {\alpha _2}\Delta t} \right)}}$

$2\left( {1 - {a_{eq}}\Delta t} \right) = \left( {1 - {\alpha _1}\Delta t} \right)\left( {1 - {\alpha _2}\Delta t} \right)$

$\therefore$ ${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$

Initial energy of capacitor, ${E_1} = {{q_1^2} \over {2C}}$

Final energy of capacitor, ${E_2} = {1 \over 2}{E_1} = {{q_1^2} \over {4C}} = {\left( {{{{{{q_1}} \over {\sqrt 2 }}} \over {2C}}} \right)^2}$

$\therefore$ ${t_1}=$ time for the charge to reduce to ${1 \over {\sqrt 2 }}$ of its initial value

and ${t_2} =$ time for the charge to reduce to ${1 \over 4}$ of its initial value

We have, ${q_2} = {q_1}{e^{ - t/CR}}$

$\Rightarrow \ln \left( {{{{q_2}} \over {{q_1}}}} \right) = - {t \over {CR}}$

$\therefore$$\ln \left( {{1 \over {\sqrt 2 }}} \right) = {{ - {t_1}} \over {CR}}...\left( 1 \right)$

and $\ln \left( {{1 \over 4}} \right) = {{ - {t_2}} \over {CR}}\,\,...\left( 2 \right)$

By $(1)$ and $(2),$ ${{{t_1}} \over {{t_2}}} = {{\ln \left( {{1 \over {\sqrt 2 }}} \right)} \over {\ln \left( {{1 \over 4}} \right)}}$

$= {1 \over 2}{{\ln \left( {{1 \over 2}} \right)} \over {2\ln \left( {{1 \over 2}} \right)}} = {1 \over 4}$

Given $s = {t^3} + 5$

$\Rightarrow$ Speed,$\,\,\,v = {{ds} \over {dt}} = 3{t^2}$

Tangential acceleration ${a_t} = {{dv} \over {dt}} = 6t$

Radial acceleration ${a_c} = {{{v^2}} \over R} = {{9{t^4}} \over R}$

At $\,\,\,\,t = 2s,\,\,{a_t} = 6 \times 2 = 12\,\,m/{s^2}$

${a_c} = {{9 \times 16} \over {20}} = 7.2\,\,m/{s^2}$

$\therefore$ Net acceleration

$= \sqrt {a_t^2 + a_c^2}$

$= \sqrt {{{\left( {12} \right)}^2} + {{\left( {7.2} \right)}^2}}$

$= \sqrt {144 + 51.84}$

$= \sqrt {195.84}$

$= 14\,m/{s^2}$