AIEEE 2002 Question Paper with Solutions

Emitter sends the majority charge carries towards the collector. Therefore emitter is most heavily doped.

Pyrometer is used to detect infra-red radiation.

${\theta _n} = {{{\theta _i} + {\theta _c}} \over 2}.$

Let the two forces be ${F_1}$ and ${F_2}$ and let ${F_2}$ is smaller than ${F_1}$ and assume $R$ is the resultant force.

Given ${F_1} + {F_2} = 18$ $\,\,\,\,\,\,$ ....$(i)$

From the right angle triangle, $F_2^2 + {R^2} = F_1^2$

or $F_1^2 - F_2^2 = {R^2}$

or $\left( {{F_1} + {F_2}} \right)$$\left( {{F_1} - {F_2}} \right)$ = ${R^2}$

or $\left( {18} \right)\left( {{F_1} - {F_2}} \right)$ = ${\left( {12} \right)^2}$ = 144

or $\left( {{F_1} - {F_2}} \right) = 8$ $\,\,\,\,\,\,$ ....$(ii)$

By solving equation $(i)$ and $(ii)$ we get,

${{F_1} = 13\,N}$ and ${{F_2} = 5\,N}$

All reversible engines working for the same temperature of source and sink have same efficiencies. If the temperatures are different, the efficiency is different.

KEY CONCEPT : To convert a galvanometer into a voltmeter we connect a high resistance in series with the galvanometer.

The same procedure needs to be done if ammeter is to be used as a voltmeter.

When water is cooled to form ice, energy is released from water in the form of heat. As energy is equivalent to mass therefore when water is cooled to ice, its mass decreases.

We know the work energy theorem, $W = \Delta K = FS$

For first penetration, by applying work energy theorem we get,

${1 \over 2}m{v^2} - {1 \over 2}m{\left( {{v \over 2}} \right)^2} = F \times 3\,\,...(i)$

For second penetration, by applying work energy theorem we get,

${1 \over 2}m{\left( {{v \over 2}} \right)^2} - 0 = F \times S\,...(ii)$

On dividing $(ii)$ by $(i)$

${{1/4} \over {3/4}} = S/3$

$\therefore$ $S = 1\,cm$

Assume tension between block A and B is T₁ and block B and C is T₂. Acceleration a = 0.6 m/s².

For block A :

F - T₁ = ma

For block B :

T₁ - T₂ = ma

By adding both those equation we get,

F - T₂ = 2ma

T₂ = F - 2ma

T₂ = 10.2 - 2$\times$2$\times$0.6 = 10.2 - 2.4 = 7.8 N

${N_p} = 140,\,\,{N_s} = 280,\,\,{I_p} = 4A,\,\,{I_s} = ?$

For a transformer ${{{I_s}} \over {{I_p}}} = {{{N_p}} \over {{N_s}}}$

$\Rightarrow {{{I_s}} \over 4} = {{140} \over {280}} \Rightarrow {I_s} = 2A$

The equivalent capacitance of $n$ identical capacitors of capacitance $C$ is equal to $nC.$ Energy stored in this capacitor

$E = {1 \over 2}\left( {nC} \right){V^2} = {1 \over 2}nC{V^2}$

When we extend the spring by $dx$ then the work done

$dW = k\,x\,dx$

Applying integration both sides we get,

$\therefore$ $W = k\int\limits_{0.05}^{0.15} {x\,dx}$

$= {{800} \over 2}\left[ {{{\left( {0.15} \right)}^2} - {{\left( {0.05} \right)}^2}} \right]$

$= 8\,J$

Escape velocity,

${v_e} = \sqrt {2gR} = \sqrt {{{2GM} \over R}} \Rightarrow {V_e}\, \propto \,{m^0}$

Where $M,R$ are the mass and radius of the planet respectively. In this expression the mass of the body $(m)$ is not present. The escape velocity is independent of the mass m or it depends on m⁰.

The phenomenon of polarisation is shown only by transverse waves.

Formation of covalent bond is best explained by molecular orbital theory.

$\eta = 1 - {{{T_2}} \over {{T_1}}}$
For $\eta = 1$ or $100\%$, ${T_2} = 0K.$
The temperature of $0$ $K$ (absolute zero) can not be obtained.

$\beta$ -rays are fast moving beam of electrons.

Given that initial speed of two cars $u$ and $4u$ and final speed $v$ is 0 for both car. Both car is gradually slowing down so acceleration = $(-a)$

So formula becomes, 0 = ${u^2}$ - 2$a$s.

$\Rightarrow {u^2} = 2as$

For first car ${u^2} = 2a{s_1}$ ..........(i)

For second car ${\left( {4u} \right)^2} = 2a{s_2}$ ..........(ii)

Dividing $(i)$ and $(ii),$

${{{u^2}} \over {16{u^2}}} = {{2a{s_1}} \over {2a{s_2}}}$

$\Rightarrow {1 \over {16}} = {{{s_1}} \over {{s_2}}}$

For no skidding along curved track,

The maximum velocity possible ${v_{\max }} = \sqrt {\mu rg}$

Here $\mu = 0.6,\,r = 150m,\,g = 9.8$

$\therefore$ ${v_{\max }} = \sqrt {0.6 \times 150 \times 9.8} \simeq 30m/s$

${v_{rms}} =$$\sqrt {{{RT} \over M}}$

For ${v_{rms}}$ to be equal ${{{T_{{H_2}}}} \over {{M_{{H_2}}}}} = {{{T_{{O_2}}}} \over {{M_{{O_2}}}}}$

Here ${M_{{H_2}}} = 2;\,\,{M_{{O_2}}} = 32;$

${T_{{O_2}}} = 47 + 273 = 320K$

$\therefore$ ${{{T_{{H_2}}}} \over 2} = {{320} \over {32}}$

$\Rightarrow {T_{{H_2}}} = 20K$

Assume the initial velocity of each particle is = u

And height of building = h

If final velocity of A is v_A then v_A² = u² + 2(-g)(-h) = u² + 2gh

If final velocity of B is v_B then v_B² = u² + 2gh

$\therefore$ v_A = v_B

Sign Rule : Take the direction of initial velocity positive opposite direction as negative.

Here for ball A initial velocity u is upward so upward is positive and downward is negative. That is why gravity is = - g and height = - h

And for ball B initial velocity u is downward so downward is positive and upward is negative. That is why gravity is = + g and height = + h

KEY CONCEPT : The frequency of a tuning fork is given by the expression

$f = {{{m^2}k} \over {4\sqrt 3 \pi {\ell ^2}}}\sqrt {{Y \over \rho }}$

As temperature increases, $\ell$ increases and therefore $f$ decreases.

We know that ${{{W_{AB}}} \over q} = {V_B} - {V_A}$

$\therefore$ ${V_B} - {V_A} = {{2J} \over {20C}} = 0.11J/C = 0.1V$

Let acceleration of ball = ${\overrightarrow a _b}$ and acceleration of man is = ${\overrightarrow a _m}$

With respect to the man standing in the lift, the acceleration of the ball

${\overrightarrow a _{bm}} = {\overrightarrow a _b} - {\overrightarrow a _m}$

$\Rightarrow {a_{bm}} = g - a$

Where $a$ is the acceleration of the man as the acceleration of the lift is $a$.

With respect to the man standing on the ground the acceleration of the ball

${\overrightarrow a _{bm}} = {\overrightarrow a _b} - {\overrightarrow a _m}$

$\Rightarrow {a_{bm}} = g - 0$

$= g$

When current is passed through a spring, every current carrying loop of a spring behaves like a tiny magnet and loop of a spring faces another loop are form magnet of different poles which attract one another. Therefore, spring is compressed.

According to Faraday's first law of electrolysis

$m = ZIt \Rightarrow m \propto It$

Optical fibers work on the principle of total internal reflection. When light is transmitted through the fiber, it is reflected off the inner walls of the fiber in such a way that it remains within the fiber, allowing it to carry the light signal over great distances with minimal loss.

So, the correct answer is :

Option A : total internal reflection.

Gravitational potential energy E = $- {{GMm} \over r}$

where M = mass of earth

m = mass of body

r = radius of earth

Energy required to move a body of mass $m$ from an orbit of radius $2R$ to $3R$

$=$ (Potential energy of the Earth-mass system when mass is at distance $3R$ ) $-$ (Potential energy of the Earth-mass system when mass is at distance $2R$)

$= {{ - GMm} \over {3R}} - \left( {{{ - GMm} \over {2R}}} \right)$

$= {{ - GMm} \over {3R}} + {{GMm} \over {2R}}$

$= {{ - 2GMm + 3GMm} \over {6R}} = {{GMm} \over {6R}}$

KEY CONCEPT : When two plane mirrors are inclined a each other at an angle $\theta$ then the number of the images of a point object placed between the plane mirrors is

${{{{360}^ \circ }} \over \theta } - 1,\,\,if{{{{360}^ \circ }} \over \theta }$ is even

$\therefore$ Number of images formed

$= {{{{360}^ \circ }} \over {{{60}^ \circ }}} - 1 = 5$

Let the spring constant of the original spring be $k.$

Then its time period $T = 2\pi \sqrt {{m \over k}}$ where $m$ is the mass of oscillating body.
When the spring is cut into $n$ equal parts, the spring constant of one part becomes $nk.$ Therefore the new time period,

$T' = 2\pi \sqrt {{m \over {nk}}} = {T \over {\sqrt n }}$

KEY CONCEPT : The fundamental frequency for closed organ pipe is given by ${\upsilon _c} = {v \over {4\ell }}$ and

For open organ pipe is given by ${\upsilon _0} = {v \over {2\ell }}$

$\therefore$ ${{{\upsilon _0}} \over {{\upsilon _c}}} = {v \over {2\ell }} \times {{4\ell } \over v} = {2 \over 1}$

The required velocity is called escape velocity (${v_e}$) to leave the earth surface of a body.

${v_e} =$ escape velocity $= \sqrt {2gR}$

Kinetic Energy $K.E = {1 \over 2}mv_e^2$

$\therefore$ $K.E = {1 \over 2}m \times 2gR = mgR$

These three inductors are connected in parallel. The equivalent inductance ${L_P}$ is given by

${1 \over {{L_P}}} = {1 \over {{L_1}}} + {1 \over {{L_2}}} + {1 \over {{L_3}}}$ $= {1 \over 3} + {1 \over 3} + {1 \over 3} = {3 \over 3} = 1$

Pure silicon, at absolute zero, will contain all the electrons in bounded state. The conduction band will be empty. So there will be no free electrons (in conduction band) and holes (in valence band) due to thermal agitation. Pure silicon will act as insulator.

To form a node there should be superposition of this wave with the reflected wave. The reflected wave should travel in opposite direction with a phase change of $\pi$. The equation of the reflected wave will be

$y = a\sin \left( {\omega t + kx + \pi } \right)$

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

KEY CONCEPT : The resolving power of a telescope

$R.P = {D \over {122\lambda }}$ where $D=$ diameter of the objective lens

$\lambda =$ wavelength of light.

Clearly, larger the aperture, larger is the value of $D,$

more is the resolving power or resolution.

When ${F_1},{F_2}$ and ${F_3}$ are acting on a particle then the particle remains stationary. This means that the resultant of ${F_1},{F_2}$ and ${F_3}$ is zero. When ${F_1}$ is removed then particle will start moving due to the force ${F_2}$ and ${F_3}$ in the resultant of ${F_2}$ and ${F_3}$ and it should be equal and opposite to ${F_1}.$
$i.e.$ $\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} \right| = \left| {{{\overrightarrow F }_1}} \right|$
$\therefore$ $\,\,\,\,\,a = {{\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} \right|} \over m} \Rightarrow a = {{{F_1}} \over m}$

The velocity of center of mass of two particle system is

${v_c} = {{{m_1}{v_1} + {m_2}{v_2}} \over {{m_1} + {m_2}}}$

$= {{m\left( {2v} \right) + m\left( { - v} \right)} \over {m + m}}$

$= {v \over 2}$

Magnetic field due to current in wire $1$ at point $P$ distant $r$ from the wire is



$B = {{{\mu _0}} \over {4\pi }}{{{i_1}} \over r}\left[ {\cos \theta + \cos \theta } \right]$

$B = {{{\mu _0}} \over {2\pi }}{{{i_1}\cos \theta } \over r}$ (directed perpendicular to the plane of paper, inwards)

The force exerted due to this magnetic field on current element ${i_2}\,dl$ is

$dF = {i_2}\,dl\,B\,\sin \,{90^ \circ }$

$\therefore$ $dF = {i_2}dl\left[ {{{{\mu _0}} \over {2\pi }}{{{i_1}\cos \theta } \over r}} \right]$

$= {{{\mu _0}} \over {2\pi r}}{i_1}\,{i_2}dl\,\cos \,\theta$

KEY CONCEPT :

The energy of nth orbit of hydrogen is given by

${E_n} = {{13.6} \over {{n^2}}}eV/$ atom

For $n=2,$ ${E_n} = {{ - 13.6} \over 4} = - 3.4eV$

Therefore the energy required to remove electron from

$n = 2$ is $+3.4eV.$

When two small spheres of mass $m$ are attached gently, the external torque, about the axis of rotation, is zero.

So, ${{d\overrightarrow L } \over {dt}} = \overline z$ = 0

$\overrightarrow L$ = conserved

So the angular momentum about the axis of rotation is conserved.

$\therefore$ ${I_1}{\omega _1} = {I_2}{\omega _2}$

$\Rightarrow {\omega _2} = {{{I_1}} \over {{I_2}}}{\omega _1}$

Here Moment of inertia of Disc ${I_1} = {1 \over 2}M{R^2}$ and

After adding two sphere Moment of Inertia of disc and two sphere,

${I_2} = {1 \over 2}M{R^2} +$$2\left( {{1 \over 2}m{R^2} + {1 \over 2}m{R^2}} \right)$

$\therefore$ ${\omega _2} = {{{1 \over 2}M{R^2}} \over {{1 \over 2}MR + 2m{R^2}}} \times {\omega _1} = {M \over {M + 4m}}{\omega _1}$

This will happen for fundamental mode of vibration as shown in the figure. ${S_1}$ and ${S_2}$ are rigid support

Here ${\lambda \over 2} = 40\,\,\,\,\,\,$ $\therefore$ $\lambda = 80\,cm$

Work (W) = $\overrightarrow F .\overrightarrow s$ = Fs cos$\theta$

= [MLT^-2][L] = [ML²T^-2];

Torque ($\overrightarrow \tau) = \overrightarrow r \times \overrightarrow F$ $\Rightarrow \tau = rF\sin \theta$

= [L] [MLT^-2] = [ML²T^-2]

So dimension of torque and work are same.

The resolving power (RP) of an optical instrument is inversely proportional to the wavelength (λ) of light used. So, if we denote the resolving powers corresponding to λ₁ and λ₂ as RP₁ and RP₂ respectively, we can express this relationship as :

RP $\propto$ ${1 \over \lambda }$

Therefore, the ratio of the resolving powers corresponding to λ₁ and λ₂ is given by the inverse of the ratio of the wavelengths. In LaTeX notation, this relationship can be written as :

$\frac{RP_1}{RP_2} = \frac{\lambda_2}{\lambda_1}$

Substituting the given wavelengths into this equation, we get :

$\frac{RP_1}{RP_2} = \frac{5000 \, \text{Å}}{4000 \, \text{Å}} = \frac{5}{4}$

So, the correct answer is :

Option D : 5 : 4.

Both the charges are identical and placed symmetrically about $ABCD$. The flux crossing $ABCD$ due to each charge is ${1 \over 6}\left[ {{q \over {{ \in _0}}}} \right]$ but in opposite directions. Therefore the resultant is zero.

A tuning fork produces $4$ beats/sec with another tuning fork of frequency $288$ cps. From this information we can conclude that the frequency of unknown fork is $288+4$ $cps$ or $288-4$ $cps$ i.e. $292$ $cps$ or $284$ $cps.$

Here when a little wax is placed on the unknown fork, it decreases the frequency of unknown fork. Here also beats per second decreases to 2 from 4. So the difference between frequency decreases.

This is possible only when before placing the wax, the frequency of unknown fork is greater than the frequency of the given tuning fork.

So the frequency of the unknown tuninh fork is = 292 Hz

Specific resistance is resistivity which is given by

$\rho = {m \over {m{e^2}\,\tau }}$

where $n=no.$ of free electrons per unit volume and $\tau$ $=$ average relaxation time

For a conductor with rise in temperature $n$ increases and $\tau$ decreases. But decrease in $\tau$ is more dominant than increase in $n$ resulting an increase in the value of $\rho$

For a semiconductor with rise in temperature, $n$ increases and $\tau$ decreases. But the increase in $n$ is more dominant than decrease in $\tau$ resulting in decrease in the value of $\rho .$