IIT JEE 2003 Physics Question Paper with Answer and Solution

(A) The forces acting on the block are the applied force $F$ at an angle of $60^\circ$ with the horizontal,the weight $W = mg$,the normal reaction $R$,and the frictional force $f$.
Resolving the force $F$ into components:
Horizontal component: $F_x = F \cos 60^\circ$
Vertical component: $F_y = F \sin 60^\circ$
For vertical equilibrium:
$R = W + F \sin 60^\circ$
Given $m = \sqrt{3} \ kg$ and $g = 10 \ m/s^2$,$W = mg = 10\sqrt{3} \ N$.
So,$R = 10\sqrt{3} + F \sin 60^\circ = 10\sqrt{3} + F \frac{\sqrt{3}}{2}$.
For the block not to move,the horizontal component of the applied force must be less than or equal to the limiting friction:
$F \cos 60^\circ \leq \mu R$
$F \cos 60^\circ \leq \mu (W + F \sin 60^\circ)$
Substituting the values $\mu = \frac{1}{2\sqrt{3}}$,$\cos 60^\circ = \frac{1}{2}$,and $\sin 60^\circ = \frac{\sqrt{3}}{2}$:
$F \cdot \frac{1}{2} = \frac{1}{2\sqrt{3}} (10\sqrt{3} + F \cdot \frac{\sqrt{3}}{2})$
$F \cdot \frac{1}{2} = \frac{10\sqrt{3}}{2\sqrt{3}} + \frac{F \sqrt{3}}{4\sqrt{3}}$
$F \cdot \frac{1}{2} = 5 + \frac{F}{4}$
$F \cdot \frac{1}{2} - \frac{F}{4} = 5$
$\frac{F}{4} = 5$
$F = 20 \ N$.

(B) The gravitational force is a conservative force.
By definition,the work done by or against a conservative force in moving a particle between two points depends only on the initial and final positions of the particle,not on the path taken.
Since all three paths $1, 2$ and $3$ start at point $A$ and end at point $B$,the work done along each path must be equal.
Therefore,$W_1 = W_2 = W_3$.

(C) The change in length of a rod due to thermal expansion is given by $\Delta l = l \alpha \Delta T$.
Given that the change in length for both rods is the same,we have $\Delta l_1 = \Delta l_2$.
Substituting the formula,we get $l_1 \alpha_a t = l_2 \alpha_s t$.
Canceling $t$ from both sides,we get $l_1 \alpha_a = l_2 \alpha_s$,which implies $\frac{l_1}{l_2} = \frac{\alpha_s}{\alpha_a}$.
To find the ratio $\frac{l_1}{l_1 + l_2}$,we use the property of ratios: if $\frac{a}{b} = \frac{c}{d}$,then $\frac{a}{a+b} = \frac{c}{c+d}$.
Applying this to our equation,we get $\frac{l_1}{l_1 + l_2} = \frac{\alpha_s}{\alpha_a + \alpha_s}$.

(B) Step $1$: Heat required by $2\, kg$ of ice to reach $0^{\circ}C$ is $Q_1 = m_i \cdot c_{ice} \cdot \Delta T = 2\, kg \cdot 0.5\, kcal/kg/^{\circ}C \cdot 20^{\circ}C = 20\, kcal$.
Step $2$: Heat available from $5\, kg$ of water cooling from $20^{\circ}C$ to $0^{\circ}C$ is $Q_2 = m_w \cdot c_w \cdot \Delta T = 5\, kg \cdot 1\, kcal/kg/^{\circ}C \cdot 20^{\circ}C = 100\, kcal$.
Step $3$: Remaining heat available for melting ice is $Q_{rem} = Q_2 - Q_1 = 100\, kcal - 20\, kcal = 80\, kcal$.
Step $4$: Mass of ice that melts is $m_{melt} = Q_{rem} / L_f = 80\, kcal / 80\, kcal/kg = 1\, kg$.
Step $5$: Final mass of water = Initial mass of water + Mass of melted ice = $5\, kg + 1\, kg = 6\, kg$.

(A) According to Stefan-Boltzmann law,the rate of heat loss is given by $\frac{dQ}{dt} = e \sigma A (T^4 - T_0^4)$.
Since $\frac{dQ}{dt} = -ms \frac{dT}{dt}$,we have $-ms \frac{dT}{dt} = e \sigma A (T^4 - T_0^4)$.
Thus,the rate of cooling $\left( -\frac{dT}{dt} \right) = \frac{e \sigma A}{ms} (T^4 - T_0^4)$.
For bodies of the same surface area and mass,the rate of cooling is directly proportional to the emissivity $(e)$,i.e.,$\left( -\frac{dT}{dt} \right) \propto e$.
From the graph,the slope of the curve for body $x$ is steeper than that for body $y$ at any given temperature,which means $\left( -\frac{dT}{dt} \right)_x > \left( -\frac{dT}{dt} \right)_y$.
Therefore,$e_x > e_y$.
According to Kirchhoff's law of radiation,for any body,the emissivity $(e)$ is equal to its absorptive power $(a)$ at a given wavelength and temperature,i.e.,$e = a$.
Thus,$e_x > e_y$ implies $a_x > a_y$.

(A) The potential energy $(P.E.)$ of a particle in $S.H.M.$ is given by $U = \frac{1}{2} k x^2$.
Substituting $x = A \cos \omega t$,we get $U = \frac{1}{2} k A^2 \cos^2 \omega t = \frac{1}{2} k A^2 \left( \frac{1 + \cos 2 \omega t}{2} \right)$.
At $t = 0$,$x = A$,so $U$ is maximum. Graph $I$ shows $P.E.$ vs $t$ starting from a maximum value at $t = 0$,which matches the equation $U \propto \cos^2 \omega t$.
As a function of displacement $x$,$U = \frac{1}{2} k x^2$,which is a parabola opening upwards with its minimum at $x = 0$. Graph $III$ represents this parabolic variation of $P.E.$ with respect to $x$.
Thus,graphs $I$ and $III$ are correct.

(B) Let $x$ be the end correction.
For the fundamental mode (first resonance),the length of the air column is $l_1 = 0.1 \ m$. The resonance condition is given by $f = \frac{v}{4(l_1 + x)}$.
For the first overtone (second resonance),the length of the air column is $l_2 = 0.35 \ m$. The resonance condition is given by $f = \frac{3v}{4(l_2 + x)}$.
Since the same tuning fork is used,the frequency $f$ remains constant. Therefore,$\frac{v}{4(l_1 + x)} = \frac{3v}{4(l_2 + x)}$.
Simplifying the equation: $l_2 + x = 3(l_1 + x)$.
Substituting the values: $0.35 + x = 3(0.1 + x)$.
$0.35 + x = 0.3 + 3x$.
$0.05 = 2x$.
$x = 0.025 \ m$.

(B) Let $v$ be the speed of the motorcyclist and $v_s = 330 \ m/s$ be the speed of sound.
$1$. The frequency $n_1$ of the police car horn heard by the motorcyclist (who is moving away from the source) is given by the Doppler effect formula:
$n_1 = n_0 \left( \frac{v_s - v}{v_s - v_{police}} \right) = 176 \left( \frac{330 - v}{330 - 22} \right) = 176 \left( \frac{330 - v}{308} \right)$
$2$. The frequency $n_2$ of the stationary siren heard by the motorcyclist (who is moving towards the source) is:
$n_2 = n_s \left( \frac{v_s + v}{v_s} \right) = 165 \left( \frac{330 + v}{330} \right)$
$3$. Since the motorcyclist observes no beats,the frequencies must be equal $(n_1 = n_2)$:
$176 \left( \frac{330 - v}{308} \right) = 165 \left( \frac{330 + v}{330} \right)$
$4$. Simplifying the equation:
$\frac{176}{308} (330 - v) = \frac{165}{330} (330 + v)$
$\frac{4}{7} (330 - v) = \frac{1}{2} (330 + v)$
$8(330 - v) = 7(330 + v)$
$2640 - 8v = 2310 + 7v$
$15v = 330$
$v = 22 \ m/s$.

(A) In uniform circular motion,the net force acting on the particle is the centripetal force,which is always directed towards the centre of the circle.
The torque $\vec{\tau}$ about any point is given by $\vec{\tau} = \vec{r} \times \vec{F}$.
Since the centripetal force $\vec{F}$ always passes through the centre of the circle,the position vector $\vec{r}$ relative to the centre is collinear with the force vector $\vec{F}$.
Therefore,the torque about the centre is $\vec{\tau} = 0$.
According to the principle of conservation of angular momentum,if the net external torque about a point is zero,the angular momentum about that point remains conserved.
Thus,the angular momentum of the particle is conserved about the centre of the circle.

(A) The side length of the cube is $l = 1.2 \times 10^{-2} \; m$.
The volume of a cube is given by $V = l^3$.
Substituting the value: $V = (1.2 \times 10^{-2} \; m)^3 = (1.2)^3 \times (10^{-2})^3 \; m^3 = 1.728 \times 10^{-6} \; m^3$.
According to the rules of significant figures,the side length $1.2$ has two significant figures. Therefore,the result should be rounded to two significant figures.
Rounding $1.728$ to two significant figures gives $1.7$.
Thus,the volume is $V = 1.7 \times 10^{-6} \; m^3$.

(A) The impulse $J = Mv$ is applied at one end of the rod. This impulse provides an angular momentum about the center of mass $(CM)$ of the system.
The center of mass of the two-ball system is at the midpoint of the rod.
The angular momentum $L_{CM}$ imparted by the impulse $J$ about the center of mass is given by $L_{CM} = J \times r$,where $r = L/2$ is the distance from the center of mass to the point of application of the impulse.
$L_{CM} = (Mv) \times (L/2) = \frac{MvL}{2}$.
The moment of inertia $I$ of the system about an axis passing through the center of mass and perpendicular to the rod is $I = M(L/2)^2 + M(L/2)^2 = 2M(L^2/4) = \frac{ML^2}{2}$.
Using the relation $L_{CM} = I\omega$,we have:
$\frac{MvL}{2} = \left( \frac{ML^2}{2} \right) \omega$.
Solving for $\omega$:
$\omega = \frac{MvL/2}{ML^2/2} = \frac{v}{L}$.

(A) From the graph,we observe that for a change in load $\Delta W = (40 - 20) \, N = 20 \, N$,the change in extension is $\Delta(\Delta l) = (2 - 1) \times 10^{-4} \, m = 10^{-4} \, m$.
Young's modulus $Y$ is given by the formula $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l} = \frac{F \cdot l}{A \cdot \Delta l}$.
Using the slope of the graph,$\frac{\Delta l}{F} = \frac{10^{-4} \, m}{20 \, N} = 0.05 \times 10^{-4} \, m/N = 5 \times 10^{-6} \, m/N$.
Given $l = 1 \, m$ and $A = 10^{-6} \, m^2$,we have:
$Y = \frac{l}{A} \cdot \frac{F}{\Delta l} = \frac{1}{10^{-6}} \cdot \frac{1}{5 \times 10^{-6}} = \frac{10^6}{5 \times 10^{-6}} = 0.2 \times 10^{12} \, N/m^2 = 2 \times 10^{11} \, N/m^2$.

(B) The electric field lines originate from the positive point charge '$q$' inside the cavity and terminate on the inner surface of the metallic shell,such that they are perpendicular to the inner surface.
Since the shell is metallic,the electric field inside the material of the shell is zero.
However,the charge '$q$' induces an equal and opposite charge '$-q$' on the inner surface of the shell and an equal charge '$+q$' on the outer surface.
The electric field lines outside the shell originate from the outer surface and extend to infinity,also perpendicular to the outer surface.
Diagram $(b)$ correctly shows the lines originating from the point charge and terminating perpendicularly on the inner surface,and then originating from the outer surface.

(B) To verify Ohm's law,we need to measure the current flowing through a resistor and the potential difference across it.
$1$. An ammeter is used to measure current and must be connected in series with the component.
$2$. $A$ voltmeter is used to measure potential difference and must be connected in parallel across the component.
$3$. In the provided options,the correct setup is the one where the ammeter is in series with the resistor and the voltmeter is connected in parallel across the resistor to measure the voltage drop across it.
$4$. Based on standard circuit diagrams for Ohm's law verification,the setup where the ammeter is in series and the voltmeter is in parallel is the correct configuration.

(A) In a meter bridge,the balancing condition is given by the ratio of resistances: $\frac{R_1}{R_2} = \frac{R_{AC}}{R_{CB}}$.
Here,$R_{AC}$ is the resistance of the wire segment $AC$ and $R_{CB}$ is the resistance of the wire segment $CB$.
The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.
Substituting this into the balancing condition: $\frac{R_1}{R_2} = \frac{\rho (x) / A}{\rho (100-x) / A} = \frac{x}{100-x}$.
Since the cross-sectional area $A$ cancels out from the numerator and the denominator,the balancing length $x$ depends only on the ratio of the resistors $R_1$ and $R_2$.
Therefore,changing the radius of the wire $AB$ does not affect the balancing length $x$.

(A) Let the value of each resistor be $R$. The power dissipated in a circuit with a constant current $i$ is given by $P = i^2 R_{eq}$,where $R_{eq}$ is the equivalent resistance.
For configuration $I$ (series): $R_{eq, I} = R + R + R = 3R$.
For configuration $II$ (two in series,one in parallel): $R_{eq, II} = \frac{(2R)(R)}{2R + R} = \frac{2R^2}{3R} = \frac{2}{3}R \approx 0.67R$.
For configuration $III$ (all three in parallel): $R_{eq, III} = \frac{R}{3} \approx 0.33R$.
For configuration $IV$ (two in parallel,one in series): $R_{eq, IV} = \frac{R}{2} + R = 1.5R$.
Comparing the equivalent resistances: $R_{eq, III} (0.33R) < R_{eq, II} (0.67R) < R_{eq, IV} (1.5R) < R_{eq, I} (3R)$.
Since $P \propto R_{eq}$,the power dissipation follows the same order: $III < II < IV < I$.

(B) The particle moves in the $x-y$ plane. For the motion to remain in the $x-y$ plane,the net force $\overrightarrow F_{net} = q(\overrightarrow E + \overrightarrow v \times \overrightarrow B)$ must have no component along the $z$-axis.
Initial velocity is $\overrightarrow v = v\hat i$.
In option $(d)$,$\overrightarrow F_{net} = q(a\hat i) + q(v\hat i \times (c\hat k + b\hat j)) = qa\hat i - qvc\hat j + qvb\hat k$. The presence of a $z$-component $(qvb\hat k)$ forces the particle out of the $x-y$ plane,so $(d)$ is incorrect.
In option $(b)$,$\overrightarrow F_{net} = q(a\hat i) + q(v\hat i \times (c\hat k + a\hat i)) = qa\hat i - qvc\hat j$. Here,the force has only $x$ and $y$ components,keeping the particle in the $x-y$ plane. The path is non-circular because the electric field changes the speed of the particle. Thus,option $(b)$ is the correct combination.

(B) The net magnetic force on a current-carrying closed loop placed in a uniform magnetic field is zero. Therefore,the loop cannot undergo translational motion. This eliminates options $(c)$ and $(d)$.
According to Fleming's left-hand rule,the magnetic force $\overrightarrow{F_m} = I(\overrightarrow{dl} \times \overrightarrow{B})$ acts on each small element of the loop. Given that the magnetic field $\overrightarrow{B}$ is directed into the plane of the paper and the current $I$ flows in a counter-clockwise direction (as shown in the figure),the force on each element is directed radially outwards. Consequently,the loop will have a tendency to expand.

(C) The potential energy $U$ of a magnetic dipole in a magnetic field $B$ is given by $U = -M \cdot B = -MB \cos \theta$,where $\theta$ is the angle between the magnetic moment vector $M$ (which is along the normal $\hat{n}$) and the magnetic field $B$.
From the given images:
For $I$: $\theta = 180^\circ$,so $U_I = -MB \cos(180^\circ) = +MB$.
For $II$: $\theta = 90^\circ$,so $U_{II} = -MB \cos(90^\circ) = 0$.
For $III$: $\theta = 45^\circ$ (approx),so $U_{III} = -MB \cos(45^\circ) = -0.707MB$.
For $IV$: $\theta = 135^\circ$ (approx),so $U_{IV} = -MB \cos(135^\circ) = +0.707MB$.
Comparing the values: $U_I (+MB) > U_{IV} (+0.707MB) > U_{II} (0) > U_{III} (-0.707MB)$.
Thus,the decreasing order is $I > IV > II > III$.

(A) From the given diagram,the current $i$ leads the voltage $e$ by a phase angle $\phi = \pi/4$. This indicates that the circuit is an $RC$ series circuit.
For an $RC$ circuit,the phase angle $\phi$ is given by $\tan \phi = \frac{X_C}{R} = \frac{1}{\omega CR}$.
Given $\phi = \pi/4$,we have $\tan(\pi/4) = 1$,so $1 = \frac{1}{\omega CR}$,which implies $\omega CR = 1$.
Given $\omega = 100 \text{ rad/s}$,we have $100 \times C \times R = 1$,or $CR = 1/100 = 0.01 \text{ s}$.
Checking option $(A)$: $R = 1000 \, \Omega$ and $C = 10 \times 10^{-6} \text{ F} = 10^{-5} \text{ F}$.
Then $CR = 1000 \times 10^{-5} = 10^{-2} = 0.01 \text{ s}$.
This matches the condition,so option $(A)$ is correct.

(A) The potential energy $U$ is given by $U = eV = eV_0 \ln(r/r_0)$.
The force $F$ is given by $F = -dU/dr = -d/dr(eV_0 \ln(r/r_0)) = -eV_0/r$. The magnitude of the force is $F = eV_0/r$.
This force provides the necessary centripetal force for circular motion: $mv^2/r = eV_0/r$.
Simplifying this,we get $mv^2 = eV_0$,which implies $v = \sqrt{eV_0/m}$. Note that $v$ is independent of $r$ and $n$.
According to Bohr's quantization condition,the angular momentum is $mvr = nh/(2\pi)$.
Substituting $v = \sqrt{eV_0/m}$ into the quantization condition,we get $m(\sqrt{eV_0/m})r_n = nh/(2\pi)$.
Solving for $r_n$,we get $r_n = (nh / (2\pi)) \cdot \sqrt{1/(meV_0)}$.
Since $h, m, e, V_0$ are constants,we conclude that $r_n \propto n$.

(D) The radius of an orbit in the Bohr model is given by $r_n = n^2 \frac{a_0}{Z}$,where $a_0$ is the Bohr radius $(0.529 \ \mathring{A})$,$n$ is the principal quantum number,and $Z$ is the atomic number.
Given the atom is $_{100}Fm^{257}$,the atomic number $Z = 100$.
The outermost shell for Fermium $(Z=100)$ is the $5f$ shell,so the principal quantum number $n_{shell} = 5$.
The radius of the outermost shell is $r = (5)^2 \frac{a_0}{100} = \frac{25}{100} a_0 = 0.25 a_0$.
Thus,the radius is $0.25$ times the Bohr radius,so $n = 0.25$.

(A) The density of a nucleus $(\rho)$ is defined as the ratio of its mass $(m)$ to its volume $(V)$, given by $\rho = m/V$.
Experimental observations show that the nuclear density is approximately constant for all nuclei, regardless of their mass number.
Since $\rho = \text{constant}$, we have $m/V = \text{constant}$.
Therefore, the mass of the nucleus is directly proportional to its volume, which can be expressed as $m \propto V$.

(B) The $Q$-value of the reaction is the total kinetic energy released,so $K_{\alpha} + K_{D} = Q = 5.5\, MeV$,where $K_{\alpha}$ is the kinetic energy of the $\alpha$-particle and $K_{D}$ is the kinetic energy of the daughter nucleus.
By the law of conservation of linear momentum,the magnitudes of the momenta of the $\alpha$-particle and the daughter nucleus must be equal: $p_{\alpha} = p_{D}$.
Using the relation $K = \frac{p^2}{2m}$,we have $p = \sqrt{2mK}$.
Thus,$\sqrt{2 m_{\alpha} K_{\alpha}} = \sqrt{2 m_{D} K_{D}}$.
Squaring both sides: $m_{\alpha} K_{\alpha} = m_{D} K_{D}$.
Given $m_{\alpha} = 4$ and $m_{D} = 220 - 4 = 216$,we get $4 K_{\alpha} = 216 K_{D}$,which simplifies to $K_{D} = \frac{4}{216} K_{\alpha} = \frac{1}{54} K_{\alpha}$.
Substituting this into the energy equation: $K_{\alpha} + \frac{1}{54} K_{\alpha} = 5.5\, MeV$.
$\frac{55}{54} K_{\alpha} = 5.5\, MeV$.
$K_{\alpha} = 5.5 \times \frac{54}{55} = 0.1 \times 54 = 5.4\, MeV$.

(B) Let the refractive index of glass be ${\mu _g}$ and water be ${\mu _w} = 4/3$. The refractive index of air is ${\mu _a} = 1$.
Applying Snell's Law at the glass-water interface:
${\mu _g} \sin i = {\mu _w} \sin r$ ---$(1)$
Applying Snell's Law at the water-air interface:
${\mu _w} \sin r = {\mu _a} \sin 90^\circ$ ---$(2)$
Since the ray emerges parallel to the water surface,the angle of refraction at the water-air interface is $90^\circ$.
From equations $(1)$ and $(2)$,we get:
${\mu _g} \sin i = {\mu _a} \sin 90^\circ$
${\mu _g} \sin i = 1 \times 1$
${\mu _g} = \frac{1}{\sin i}$

(B) $1$. The convex lens forms an image $I_1$ at its focal point. Since the object is at infinity, the image $I_1$ is formed at $f = 30\,cm$ from the convex lens.
$2$. The size of this image $I_1$ is $2\,cm$.
$3$. A concave lens of focal length $f_2 = -20\,cm$ is placed at $26\,cm$ from the convex lens. The image $I_1$ acts as a virtual object for the concave lens.
$4$. The distance of the virtual object from the concave lens is $u = 30\,cm - 26\,cm = 4\,cm$. Since it is on the right side, we take $u = +4\,cm$.
$5$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} - \frac{1}{4} = \frac{1}{-20}$
$\frac{1}{v} = \frac{1}{4} - \frac{1}{20} = \frac{5-1}{20} = \frac{4}{20} = \frac{1}{5}$
$v = 5\,cm$.
$6$. The magnification $m$ for the concave lens is $m = \frac{v}{u} = \frac{5}{4} = 1.25$.
$7$. The new size of the image $I_2$ is $m \times (\text{size of } I_1) = 1.25 \times 2\,cm = 2.5\,cm$.

(B) From the geometry,the distance from $O$ to the line containing $P$ is $d$. Thus,$PO = d \sec \theta$.
Since $CP$ is a wavefront,the optical path length from $C$ to $P$ is equal to the path length from $O$ to $P$ along the ray $AO$. The path difference between the ray $BP$ and the reflected ray $OP$ is $\Delta = CO + OP$.
In $\triangle COP$,$CO = PO \cos 2\theta = d \sec \theta \cos 2\theta$.
Thus,$\Delta = d \sec \theta + d \sec \theta \cos 2\theta = d \sec \theta (1 + \cos 2\theta) = d \sec \theta (2 \cos^2 \theta) = 2d \cos \theta$.
Since the ray $OP$ undergoes reflection at the surface $QR$,there is an additional phase shift of $\pi$,which corresponds to a path difference of $\lambda / 2$.
For constructive interference,the total path difference must be an odd multiple of $\lambda / 2$ (due to the $\pi$ phase shift): $\Delta = \lambda / 2$.
$2d \cos \theta = \lambda / 2 \implies \cos \theta = \lambda / 4d$.