AIEEE 2003 Question Paper with Solutions

Here, thrust force is responsible to accelerate the rocket,

So initial thrust of the blast

= (Lift-off mass) × acceleration

= (3.5 × 10⁴) × (10)

= 3.5 × 10⁵ N

The time period of a simple harmonic motion (SHM) performed by a mass-spring system is given by the formula:

$T = 2\pi \sqrt{{M \over k}}$

where:

• T is the time period,
• M is the mass of the object, and
• k is the spring constant.

We know that if the mass is increased by m, the time period becomes $\frac{5T}{3}$. We can set up an equation for this new scenario:

$\frac{5T}{3} = 2\pi \sqrt{\frac{M + m}{k}}$

Since we know that $T = 2\pi \sqrt{\frac{M}{k}}$, we can substitute T in the equation above:

$\frac{5}{3} \cdot 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{M + m}{k}}$

Squaring both sides of the equation to eliminate the square root, we get:

$\frac{25}{9} \cdot \frac{M}{k} = \frac{M + m}{k}$

Solving for $\frac{m}{M}$, we get:

$\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}$

KEY CONCEPT : The time period of a rectangular magnet oscillating in earth's magnetic field is given by

$T = 2\pi \sqrt {{I \over {\mu {B_H}}}}$

where $I=$ Moment of inertia of the rectangular magnet

$\mu =$ Magnetic moment

${B_H} =$ Horizontal component of the earth's magnetic field

Case 1 : $T = 2\pi \sqrt {{I \over {\mu {B_H}}}}$ where $I = {1 \over {12}}M{\ell ^2}$

Case 2 : Magnet is cut into two identical pieces such that each piece has half the original length. Then

$T' = 2\pi \sqrt {{{I'} \over {\mu '{B_H}}}}$

where $I' = {1 \over {12}}\left( {{M \over 2}} \right){\left( {{\ell \over 2}} \right)^2} = {I \over 8}$ and $\mu ' = {\mu \over 2}$

$\therefore$ ${{T'} \over T} = \sqrt {{{I'} \over {\mu '}} \times {\mu \over I}}$

$= \sqrt {{{I/8} \over {\mu /2}} \times {\mu \over I}}$

$= \sqrt {{1 \over 4}} = {1 \over 2}$

The number of $\alpha$ - particles released $=8$

Therefore the atomic number should decrease by $16$

The number of ${\beta ^ - }$ - particles released $=4$

Therefore the atomic number should increase by $4.$

Also the number of ${\beta ^ + }$ particles released is $2,$ which

should decrease the atomic number by $2.$

Therefore the final atomic number is

$92-16+4-2=78$

As shown in the figure, the magnetic lines of force are directed from south to north inside a bar magnet.

We know, Escape velocity, ${v_e} = \sqrt {2gR}$

So the escape velocity is independent of the angle at which the body is projected, hence it will remain same as 11 km/s.

$y = {10^{ - 4}}\sin \left( {600t - 2x + {\pi \over 3}} \right)$

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

On comparing we get $\omega = 600;\,k = 2$

$v = {\omega \over k} = {{600} \over 2} = 300\,m{s^{ - 1}}$

The work done, $dW = Fds\,\cos \,\theta$

The angle between force and displacement is ${90^ \circ }.$

Therefore work done is zero.

The velocity of light in vacuum is

c = ${1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$;

$\therefore[{1 \over {{\mu _0}{\varepsilon _0}}}]$ = [c²] = [L²T^-2]

$\therefore$ Dimension of ${1 \over {{\mu _0}{\varepsilon _0}}}$ = [L²T^-2]

Potential gradient along wire, K = ${E \over {100}}$ volt/cm

For battery V = E' – ir, where E' is emf of battery.

or K × 30 = E' – ir, where current i is drawn from battery

or ${{E \times 30} \over {100}}$ = E' + 0.5i

or E' = ${{30E} \over {100}} - 0.5i$

We know that density $\left( d \right) = {{mass\left( M \right)} \over {volume\left( V \right)}}$

$\therefore$ Mass of disc $M = d \times V = d \times \left( {\pi {R^2} \times t} \right).$

The moment of inertia of any disc is $I = {1 \over 2}M{R^2}$

$\therefore$ $I = {1 \over 2}\left( {d \times \pi {R^2} \times t} \right){R^2} = {{\pi d} \over 2}t \times {R^4}$

$\therefore$ ${{{I_X}} \over {{I_Y}}} = {{{t_X}R_X^4} \over {{t_Y}R_Y^4}}$

$= {{t \times {R^4}} \over {{t \over 4} \times {{\left( {4R} \right)}^4}}}$

$= {1 \over {64}}$

Let $t$ be the time taken for the two masses to collide and ${x_{5M,}}\,{x_M}$ be the distance travelled by the mass $5M$ and $M$ respectively.

The gravitational force acting between two sphere when the distance between them (12R - x) where x is a variable,

$F = {{GM \times 5M} \over {{{\left( {12R - x} \right)}^2}}}$

Acceleration of mass M, ${a_M} = {{G \times 5M} \over {{{\left( {12R - x} \right)}^2}}}$

Acceleration of mass 5M, ${a_{5M}} = {{GM} \over {{{\left( {12R - x} \right)}^2}}}$

For mass $5M$

$u = 0,\,\,S = {x_{5M}},\,\,t = t,\,\,a{ = a_{5M}}$

$S = ut + {1 \over 2}a{t^2}$

$\therefore$ ${x_{5M}} = {1 \over 2}{a_{5M}}{t^2}\,\,\,\,\,\,\,\,\,\,...\left( {i} \right)$

For mass $M$

$u = 0,\,\,s = {x_M},\,\,t = t,\,\,a = {a_M}$

$\therefore$ $s = ut + {1 \over 2}a{t^2} \Rightarrow$

${x_M} = {1 \over 2}{a_M}{t^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

Dividing $(ii)$ by $(iii)$

${{{x_{5M}}} \over {{x_M}}} = {{{1 \over 2}{a_5}_M{t^2}} \over {{1 \over 2}{a_M}{t^2}}}$

$= {{{a_{5M}}} \over {{a_M}}} = {1 \over 5}$

$\therefore$ $5{x_{5M}} = {x_M}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left....( {iii} \right)$

From the figure,

${x_{5M}} + {x_M} = 12R - 2R - R = 9R\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( iv \right)$

From $(iii)$ and $(iv)$

${{{x_M}} \over 5} + {x_M} = 9R$

$\therefore$ $6{x_M} = 45R$

$\therefore$ ${x_M} = {{45} \over 6}R = 7.5R$

So two sphere collide when the sphere of mass M covered the distance of 7.5R.

$x = 4\left( {\cos \pi t + \sin \pi t} \right)$

$= \sqrt 2 \times 4\left( {{{\sin \pi t} \over {\sqrt 2 }} + {{\cos \pi t} \over {\sqrt 2 }}} \right)$

$x = 4\sqrt 2 \sin \left( {\pi t + {{45}^ \circ }} \right)$

${i_g} \times G = \left( {i - {i_g}} \right)S$

$\therefore$ $S = {{{i_g} \times G} \over {i - {i_g}}} = {{1 \times 0.81} \over {10 - 1}} = 0.09\Omega$

We know Rotational Kinetic Energy$={1 \over 2}I{\omega ^2},$

Angular Momentum $L = I\omega \Rightarrow I = {L \over \omega }$

$\therefore$ Initial $K.E. = {1 \over 2}{L \over \omega } \times {\omega ^2} = {1 \over 2}L\omega$

Final $K.E'$ = ${{K.E} \over 2}$ = ${1 \over 2}{L'} \times 2\omega$

$\therefore$ ${{K.E} \over {K.E'}} = {{L \times \omega } \over {L' \times \omega '}}$

$\Rightarrow {{K.E} \over {{{K.E} \over 2}}} = {{L \times \omega } \over {L' \times 2\omega }}$

$\therefore$ $L' = {L \over 4}$

A tuning fork of frequency $256$ $Hz$ makes $5$ beats/ second with the vibrating string of a piano. Therefore the frequency of the vibrating string of piano is $\left( {256 \pm 5} \right)$ $Hz$ ie either $261$$Hz$ or $251$ $Hz.$ When the tension in the piano string increases, its frequency will increases. Now since the beat frequency decreases, we can conclude that the frequency of piano string is $251$ $Hz$

$K.E. = {1 \over 2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)$

When $x=0,$ $K.E$ is maximum and is equal to ${1 \over 2}m{\omega ^2}{a^2}.$

Let $\theta$ be the smallest temperature difference that can be detected by the thermocouple, then

$I \times R = \left( {25 \times {{10}^{ - 6}}} \right)\theta$

where ${\rm I}$ is the smallest current which can be detected by the galvanometer of resistance $R.$

$\therefore$ ${10^{ - 5}} \times 40 = 25 \times {10^{ - 6}} \times \theta$

$\therefore$ $\theta = {16^ \circ }C.$

$\lambda = {1 \over t}{\log _e}{{{A_0}} \over A}$

$= {1 \over 5}{\log _e}{{5000} \over {1250}}$

$= 0.2{\log _e}4$

$= 0.4{\log _e}2$

When the temperature increases, certain bounded electrons become free which tend to promote conductivity. Simultaneously number of collisions between electrons and positive kernels increases

The situation is shown in the figure.

${F_E} =$ Force due to electric field

${F_B} =$ Force due to magnetic field

It is given that the charged particle remains moving along $X$-axis (i.e. undeviated).

Therefore ${F_B} = {F_E}$

$\Rightarrow qvB = qE$

$\Rightarrow B = {E \over v} = {{{{10}^4}} \over {10}}$

$= {10^3}\,\,weber/{m^2}$

$P \propto {T^3} \Rightarrow P{T^{ - 3}} =$ constant ....$(i)$
But for an adiabatic process, the pressure temperature relationship is given by
${P^{1 - \gamma }}\,\,{T^\gamma } =$ constant $\Rightarrow P{T^{{\gamma \over {1 - \gamma }}}} =$ constant. ....$(ii)$
From $(i)$ and $(ii)$ ${\gamma \over {1 - \gamma }} = - 3 \Rightarrow \gamma = - 3 + 3\gamma \Rightarrow \gamma = {3 \over 2}$

As we know $\overrightarrow \tau = \overrightarrow r \times \overrightarrow F$

So the angle between $\overrightarrow \tau$ and $\overrightarrow r$ is ${90^ \circ }$ and the angle between $\overrightarrow t$ and $\overrightarrow F$ is also ${90^ \circ }.$

We also know that the dot product of two vectors which have an angle of ${90^ \circ }$ between them is zero.

$\therefore$ $\overrightarrow {r.} \vec \tau = 0{\mkern 1mu}$ and $\overrightarrow {F.} \overrightarrow \tau = 0\,\,$

Therefore $(d)$ is the correct option.

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

For lithium ion $Z=3;$ $n=2$ (for first excited state)

${E_n} = - {{13.6} \over {{2^2}}} \times {3^2} = - 30.6eV$

Momentum = mv = [${M{L}{T^{ - 1}}}$]

Planck's constant, h = ${E \over v}$ = ${[{M{L^2}{T^{ - 2}]}} \over {[{T^{ - 1}}]}}$ = $[{M{L^2}{T^{ - 1}}}]$

So Momentum and Planck's constant do not have same dimensions.

Mutual conductance depends on the relative position and orientation of the two coils.

Optical fibres form a dielectric wave guide and are free from electromagnetic interference or radio frequency interference.

According to Faraday's first law of electrolysis $m = z \times q$

For same $q,$ $\,\,\,\,\,\,\,\,\,$ $m \propto Z$

$\therefore$ ${{{m_{Cn}}} \over {{m_{Zn}}}} = {{{Z_{Cu}}} \over {{Z_{Zn}}}}$

$\Rightarrow {m_{Cu}} = {{{Z_{Cu}}} \over {{Z_{Zn}}}} \times {m_{Zn}}$

$= {{31.5} \over {32.5}} \times 0.13 = 0.126\,g$

To observe the phenomenon of interference, you need two sources that emit radiation of the same frequency and have a definite phase relationship. This is because interference is a result of the superposition of waves, which requires the waves to be coherent. Coherence is achieved when the waves have a constant phase difference, which implies that they are of the same frequency and have a definite phase relationship (a phase relationship that does not change with time).

So, the correct answer is :

Option D : of the same frequency and having a definite phase relationship.

When lift is stationary then,

T₁ = $mg$ = 49 N

$m$ = 5

For the bag accelerating down

$mg-T=ma$

$\therefore$ $T=m(g-a)$

$= 5\left( {9.8 - 5} \right) = 24\,N$

Assume weight of the block = $W$ and friction force = $f$ = $\mu N$

For the block to remain stationary with the wall, weight of the block should balanced by the force of friction.

$\therefore$ $f=W$

$\Rightarrow$ $\mu N = W$

$\Rightarrow$ $W$ = 0.2 $\times$ 10 = 2 N

From the figure it is clear that maximum horizontal range

$R = {{{u^2}\sin 2\theta } \over g}$

$= {{{{\left( {10} \right)}^2}\sin \left( {2 \times {{30}^ \circ }} \right)} \over {10}}$

$= 5\sqrt 3$ = 8.66 m

The elastic potential energy

$= {1 \over 2} \times$ Force $\times$ extension

$= {1 \over 2} \times 200 \times 0.001 = 0.1\,J$

Electric potential due to charge $Q$ placed at the center of spherical shell at point $P$ is

${V_1} = {1 \over {4\pi {\varepsilon _0}}}{Q \over {R/2}} = {1 \over {4\pi {\varepsilon _0}}}{{2Q} \over R}$



Electric potential due to charge $q$ on the surface of the spherical shell at any point inside the shell is

${V_2} = {1 \over {4\pi {\varepsilon _0}}}{q \over R}$

$e = - {{\Delta \phi } \over {\Delta t}} = {{ - \Delta \left( {LI} \right)} \over {\Delta t}} = - L{{\Delta I} \over {\Delta t}}$

$\therefore$ $\left| e \right| = L{{\Delta I} \over {\Delta t}} \Rightarrow 8 = L \times {4 \over {0.05}}$

$\Rightarrow L = {{8 \times 0.05} \over 4} = 0.1H$

When the capacitor is completely charged, the total energy in the $L.C$ circuit is with the capacitor and that energy is $E = {1 \over 2}{{{Q^2}} \over C}$

When half energy is with the capacitor in the form of electric field between the plates of the capacitor we get ${E \over 2} = {1 \over 2}{{Q{'^2}} \over C}$ where $Q'$ is the charge on plate of the capacitor

$\therefore$ ${1 \over 2} \times {1 \over 2}{{{Q^2}} \over C} = {1 \over 2}{{Q{'^2}} \over C}$

$\Rightarrow Q' = {Q \over {\sqrt 2 }}$

$\eta = 1 - {{{T_2}} \over {{T_1}}}$

$= 1 - {{\left( {273 + 27} \right)} \over {\left( {273 + 627} \right)}}$

$= 1 - {{300} \over {900}}$

$= 1 - {1 \over 3} = {2 \over 3}$

But $\eta = {W \over Q}$

$\therefore$ ${W \over Q} = {2 \over 3} \Rightarrow W$

$= {2 \over 3} \times Q$

$= {2 \over 3} \times 3 \times {10^6}$

$= 2 \times {10^6}cal$

$= 2 \times {10^6} \times 4.2\,J$

$= 8.4 \times {10^6}J$