AIEEE 2006 Question Paper with Solutions

According to energy conservation law,

Work done by the hand and due to gravity = total change in the kinetic energy

Initially the the ball is at rest and finally at top its velocity become zero so total change in kinetic energy $\Delta K$ = 0

${W_{hand}} + {W_{gravity}} = \Delta K$

[Here distance covered would be 0.2 meter for force by hand as force is applied while ball is in contact with hand.
And gravity will still work while ball is in contact with hand so total distance due to gravity would be 2 + 0.2 = 2.2 meter.]
$\Rightarrow F\left( {0.2} \right) - \left( {0.2} \right)\left( {10} \right)\left( {2.2} \right)$ $= 0 \Rightarrow F = 22\,N$

$\therefore$ Option (D) is correct.

$eV = {1 \over 2}m{v^2}$

$\Rightarrow v = \sqrt {{{2ev} \over m}} = \sqrt {{{2 \times 1.6 \times {{10}^{ - 19}} \times 20} \over {9.31 \times {{10}^{ - 31}}}}}$

$= 2.65 \times {10^6}\,m/s$

$p$-side connected to low potential and $n-$ side is connected to high potential.

$K.E.\,$ of a body undergoing $SHM$ is given by,

$K.E. = {1 \over 2}m{a^2}{\omega ^2}{\cos ^2}\,\omega t,$

$T.E. = {1 \over 2}m{a^2}{\omega ^2}$

Given $K.E.=0.75T.E.$

$\Rightarrow 0.75 = {\cos ^2}\omega t \Rightarrow \omega t = {\pi \over 6}$

$\Rightarrow t = {\pi \over {6 \times \omega }} \Rightarrow t = {{\pi \times 2} \over {6 \times 2\pi }} \Rightarrow t = {1 \over 6}s$

The resistance of the bulb is $R = {{{V^2}} \over P} = {{{{\left( {220} \right)}^2}} \over {100}}$

The power consumed when operated at $110$ $V$ is

$P = {{{{\left( {110} \right)}^2}} \over {{{\left( {220} \right)}^2}/100}} = {{100} \over 4} = 25\,W$

$\phi = 10{t^2} - 50t + 250$

$e = - {{d\phi } \over {dt}} = - \left( {20t - 50} \right)$

${e_{t = 3}} = - 10\,V$

As $\lambda$ decreases, $v$ increases and hence the speed of photo electron increases. The chances of photo electron to meet the anode increases and hence photo electric current increases.

For a thin prism, $D = \left( {\mu - 1} \right)A$

Since ${\lambda _b} < {\lambda _r} \Rightarrow {\mu _r} < {\mu _b} \Rightarrow {D_1} < {D_2}$

The energy of the incident radiation that causes photoelectrons to be emitted can be calculated using the stopping potential, V, via the equation

$E = eV,$

where e is the elementary charge.

The elementary charge, e, is approximately equal to $1.602 \times 10^{-19}$ C. So the energy of the incident radiation is

$E = 1.602 \times 10^{-19} \, \text{C} \times 5 \, \text{V} = 8.01 \times 10^{-19} \, \text{J}.$

We convert this to electron volts (eV) by using the conversion factor $1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}.$ So

$E = \frac{8.01 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} = 5 \, \text{eV}.$

This is less than the threshold energy of $6.2 \, \text{eV},$ which means that the incident radiation does not have enough energy to overcome the work function of the metal and thus cannot cause photoelectrons to be emitted.

The regions of the electromagnetic spectrum are generally classified by energy as follows:

• Infrared: Less than about $1.24 \, \text{eV}$
• Visible: Between about $1.24 \, \text{eV}$ and $3.1 \, \text{eV}$
• Ultraviolet: Between about $3.1 \, \text{eV}$ and $124 \, \text{eV}$
• X-rays: Greater than about $124 \, \text{eV}$

Therefore, the incident radiation falls in the ultraviolet region.

Velocity is maximum when kinetic energy is maximum and when kinetic energy is maximum then potential energy should be minimum

For minimum potential energy,

${{dV} \over {dx}} = 0$

$\Rightarrow {x^3} - x = 0$

$\Rightarrow x = \pm 1$

$\Rightarrow$ Min. Potential energy (P.E.) =${1 \over 4} - {1 \over 2} = - {1 \over 4}J$

$K.E{._{\left( {\max .} \right)}} + P.E{._{\left( {\min .} \right)}} = 2\,$ (Given)

$\therefore$ $K.E{._{\left( {\max .} \right)}} = 2 + {1 \over 4} = {9 \over 4}$

$\therefore$ ${1 \over 2}mv_{\max }^2$ = ${9 \over 4}$

$\Rightarrow {1 \over 2} \times 1 \times {v^2}_{\max .} = {9 \over 4}$

$\Rightarrow {v_{\max }} = {3 \over {\sqrt 2 }}$ m/s

${{{B_2}} \over {{B_1}}} = {{{\mu _0}{n_2}{i_2}} \over {{\mu _0}{n_1}{i_1}}}$

$\Rightarrow {{{B_2}} \over {6.28 \times {{10}^{ - 2}}}} = {{100 \times {i \over 3}} \over {200 \times i}}$

$\Rightarrow {B_2} = {{6.28 \times {{10}^{ - 2}}} \over 6}$

$= 1.05 \times {10^{ - 2}}\,\,Wb/{m^2}$

Case $(i)$
At equilibrium, $T=W$
$Y = {{W/A} \over {\ell /L}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} ...\left( 1 \right)$
Case $(ii)$ At equilibrium $T=W$
$\therefore$ $Y = {{W/A} \over {{{\ell /2} \over {L/2}}}} \Rightarrow Y = {{W/A} \over {\ell /L}}$
$\Rightarrow$ Elongation is the same.

From work energy theorem we can say,

Work done by tension $+$ work done by force (applied) $+$ Work done by gravitational force $=$ change in kinetic energy

Here Work done by tension is zero

$\Rightarrow 0 + F \times AB - Mg \times AC = 0$
$\Rightarrow F = Mg\left( {{{AC} \over {AB}}} \right) = Mg\left[ {{{1 - {1 \over {\sqrt 2 }}} \over {{1 \over 2}}}} \right]$
[ as $AB = \ell \sin {45^ \circ } = {\ell \over {\sqrt 2 }}$
and $AC = OC - OA = \ell - \ell \,\cos \,{45^ \circ } = \ell \left( {1 - {1 \over {\sqrt 2 }}} \right)$
where $\ell =$ length of the string. ]
$\Rightarrow F = Mg\left( {\sqrt 2 - 1} \right)$

The risk posed to a human being by any radiation exposure depends partly upon the absorbed dose, the amount of energy absorbed per gram of tissue. Absorbed dose is expressed in rad. A rad is equal to $100$ $ergs$ of energy absorbed by $1$ gram of tissue. The more modern, internationally adopted unit is the gray (named after the English medical physicist $L.$ $H.$ Gray); one gray equals $100$ rad.

${P \over Q} = {R \over S}$ where $S = {{{S_1}{S_2}} \over {{S_1} + {S_2}}}$

Kinetic energy at point of throwing is converted into potential energy of the particle during rise.

$K.E = {1 \over 2}m{v^2} = {1 \over 2} \times 0.1 \times 25 = 1.25\,J$

$W = - mgh = - \left( {{1 \over 2}m{v^2}} \right) = - 1.25\,J$

$\left[ \, \right.$ As we know, $mgh = {1 \over 2}m{v^2}$ by energy conservation $\left. \, \right]$

Let ${I_{A}}$ is the moment of inertia about an axis passing through A and parallel to BD.

${I_{A}} = M.I\,$ due to the point mass at $B+$

$M.I$ due to the point mass at $D+$

$M.I$ due to the point mass at $C.$

${I_{A}} = 2 \times m{\left( {{\textstyle{\ell \over {\sqrt 2 }}}} \right)^2} + m{\left( {\sqrt 2 \ell } \right)^2}$

$= m{\ell ^2} + 2m{\ell ^2} = 3m{\ell ^2}$

Let $E$ be the energy of proton, then

$E + 7 \times 5.6 = 2 \times \left[ {4 \times 7.06} \right]$

$\Rightarrow E = 56.48 - 39.2 = 17.28MeV$

Given ${{nv} \over {2\ell }} = 315$ and $\left( {n + 1} \right){v \over {2\ell }} = 420$

$\Rightarrow {{n + 1} \over n} = {{420} \over {315}} \Rightarrow n = 3$

Hence $3 \times {v \over {2\ell }} = 315 \Rightarrow {v \over {2\ell }} = 105Hz$

Lowest resonant frequency is when $n=1$

Therefore lowest resonant frequency $= 105\,Hz.$

Let Terminal velocity = v_t

Upward viscous force = downward weight of sphere

$\Rightarrow 6\pi \eta r{v_t} = \left( {{4 \over 3}\pi {r^3}} \right)\left( {\rho - \sigma } \right)g$

$\Rightarrow {v_t} = {{2{r^2}\left( {\rho - \sigma } \right)g} \over {9\eta }}$ ........ (1)

where, $\rho$ = density of substance of a body

            $\sigma$ = density of liquid

Now let the terminal velocity of gold = v_g and silver = v_s.

From equation (1), we can write

${{{v_g}} \over {{v_s}}} = {{{\rho _g} - \sigma } \over {{\rho _s} - \sigma }}$ $= {{19.5 - 1.5} \over {10.5 - 1.5}}$ = ${{18} \over 9}$ $= {2 \over 1}$

$\therefore$ ${v_s} = {{{v_g}} \over 2}$ = ${{0.2} \over 2}$ = 0.1

After connection, ${V_1} = {V_2}$

$\Rightarrow K{{{Q_1}} \over {{r_1}}} = K{{{Q_2}} \over {{r^2}}}$

$\Rightarrow {{Q{}_1} \over {{r_1}}} = {{{Q_2}} \over {{r_2}}}$

The ratio of electric fields

${{{E_1}} \over {{E_2}}} = {{K{{{Q_1}} \over {r_1^2}}} \over {K{{{Q_2}} \over {r_2^2}}}} = {{{Q_1}} \over {r_1^2}} \times {{r_2^2} \over {{Q_2}}}$

$\Rightarrow {{{E_1}} \over {{E_2}}} = {{{r_1} \times r_2^2} \over {r_1^2 \times {r_2}}} \Rightarrow {{{E_1}} \over {{E_2}}} = {{{r_2}} \over {{r_1}}} = {2 \over 1}$

Since the distance between the spheres is large as compared to their diameters, the induced effects may be ignored.

We know, Force$\times$ time = Impulse = Change in momentum

$\therefore$ $F \times t = m\left( {v - u} \right)$

$\Rightarrow$ $F = {{m\left( {v - u} \right)} \over t} = {{0.15\left( {0 - 20} \right)} \over {0.1}} = 30N$

${{{I_e}} \over {{I_h}}} = {{n{}_eeA{v_e}} \over {n{}_heA{v_h}}}$

$\Rightarrow {7 \over 4} = {7 \over 5} \times {{{v_e}} \over {{v_h}}}$

$\Rightarrow {{{v_e}} \over {{v_h}}} = {5 \over 4}$

The range of energy of $\beta$-particles is from zero to some maximum value.

${I_C} = 5.488\,mA,\,\,{I_e} = 5.6\,mA,\,{I_B} = {I_E} - {I_C}$

$\beta = {{{I_c}} \over {{I_B}}} = {{5.488} \over {5.6 - 5.485}} = 49$

We know, Torque $\overrightarrow \tau = \overrightarrow r \times \overrightarrow F$

$= \left( {\widehat i - \widehat j} \right) \times \left( { - F\widehat k} \right)$

$= F\left( {\widehat i + \widehat j} \right)$

${D_2}$ is forward biased whereas ${D_1}$ is reversed biased.

So effective resistance of the circuit

$R = 4 + 2 = 6\Omega$

$\therefore$ $i = {{12} \over 6} = 2\,A$

Heat lost by He $=$ Heat gained by ${N_2}$
${n_1}C{v_1}\Delta {T_1} = {n_2}C{v_2}\Delta {T_2}$
${3 \over 2}R\left[ {{7 \over 3}{T_0} - {T_f}} \right]$
$= {5 \over 2}R\left[ {{T_f} - {T_0}} \right] \Rightarrow {T_f}$
$= {3 \over 2}{T_0}$

$W = {{nR\Delta T} \over {1 - \gamma }} \Rightarrow - 146000$
$= {{1000 \times 8.3 \times 7} \over {1 - \gamma }}$
or $1 - \gamma = - {{58.1} \over {146}} \Rightarrow \gamma$
$= 1 + {{58.1} \over {146}} = 1.4$
Hence the gas is diatomic.

The network of resistors is a balanced wheatstone bridge. The equivalent circuit is



${{\mathop{\rm R}\nolimits} _{eq}} = {{15 \times 30} \over {15 + 30}} = 10\Omega$

$\Rightarrow I = {V \over R} = {5 \over {10}} = 0.5\,A$

Mutual inductance $= {\phi \over I} = {{BA} \over I}$

[Henry] $= {{\left[ {M{T^{ - 1}}{Q^{ - 1}}{L^2}} \right]} \over {\left[ {Q{T^{ - 1}}} \right]}} = M{L^2}{Q^{ - 2}}$

${E_{rms}} = 720$

The average total energy density

$= {1 \over 2}{ \in _0}\,E_0^2 = {1 \over 2}{ \in _0}{\left[ {\sqrt 2 {E_{rms}}} \right]^2} = { \in _0}\,E_{rms}^2$

$= 8.85 \times {10^{ - 12}} \times {\left( {720} \right)^2}$

$= 4.58 \times {10^{ - 6}}\,J/{m^3}$

Initially,

$0 = {{{m_1}\left( { - {x_1}} \right) + {m_2}{x_2}} \over {{m_1} + {m_2}}} \Rightarrow {m_1}{x_1} = {m_2}{x_2}$

Finally,

mass m₁ moved towards the center a distance d so the distance of mass m₁ from the origin is x₁ - d, and now let mass m₂ need to move d' to keep the center at the origin.

$\therefore$ $0 = {{{m_1}\left( {d - {x_1}} \right) + {m_2}\left( {{x_2} - d'} \right)} \over {{m_1} + {m_2}}}$

$\Rightarrow 0 = {m_1}d - {m_1}{x_1} + {m_2}{x_2} - {m_2}d'$

$\Rightarrow d' = {{{m_1}} \over {{m_2}}}d$

Here linear momentum is conserved as no external force is acting on the bomb.

Let the velocity and mass of $4$ $kg$ piece be ${v_1}$ and ${m_1}$ and that of $12$ $kg$ piece be ${v_2}$ and ${m_2}$.

Applying conservation of linear momentum

$0 = {m_2}{v_2} - {m_1}{v_1}$

$\Rightarrow {v_1} = {{12 \times 14} \over 4} = 12\,m{s^{ - 1}}$

$\therefore$ $K.E{_1} = {1 \over 2}{m_1}v_1^2 = {1 \over 2} \times 4 \times 144 = 288\,J$