IIT JEE 1988 Physics Question Paper with Answer and Solution

(C) Since the heat flows along the length of the cylinders,the two cylinders are in parallel configuration.
The equivalent thermal conductivity $K_{eq}$ for parallel combination is given by $K_{eq} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}$.
Here,$A_1$ is the cross-sectional area of the inner cylinder: $A_1 = \pi R^2$.
$A_2$ is the cross-sectional area of the outer cylindrical shell: $A_2 = \pi (2R)^2 - \pi R^2 = 4\pi R^2 - \pi R^2 = 3\pi R^2$.
The total area $A = A_1 + A_2 = \pi R^2 + 3\pi R^2 = 4\pi R^2$.
Substituting these values into the formula:
$K_{eq} = \frac{K_1(\pi R^2) + K_2(3\pi R^2)}{4\pi R^2} = \frac{K_1 + 3K_2}{4}$.

(D) The maximum velocity of a body performing simple harmonic motion is given by $v_{max} = A\omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
For a spring-mass system,the angular frequency is $\omega = \sqrt{\frac{k}{m}}$.
Thus,$v_{max} = A\sqrt{\frac{k}{m}}$.
Given that the masses are equal $(m_M = m_N = m)$ and their maximum velocities are equal $(v_M = v_N)$:
$A_M \sqrt{\frac{k_1}{m}} = A_N \sqrt{\frac{k_2}{m}}$.
Simplifying this,we get $A_M \sqrt{k_1} = A_N \sqrt{k_2}$.
Therefore,the ratio of the amplitude of $M$ to that of $N$ is $\frac{A_M}{A_N} = \sqrt{\frac{k_2}{k_1}}$.

(B) stationary wave is formed by the superposition of two waves of the same frequency and amplitude traveling in opposite directions.
Given the incident wave is $y_1 = a \cos (kx - \omega t)$.
For a point $x = 0$ to be a node,the resultant displacement at $x = 0$ must be zero for all time $t$.
Let the second wave be $y_2 = a \cos (kx + \omega t + \phi)$.
The resultant wave is $y = y_1 + y_2 = a [\cos (kx - \omega t) + \cos (kx + \omega t + \phi)]$.
Using the identity $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$,we get:
$y = 2a \cos (kx + \phi/2) \cos (\omega t + \phi/2)$.
At $x = 0$,$y = 2a \cos (\phi/2) \cos (\omega t + \phi/2)$.
For this to be a node (zero displacement) for all $t$,we must have $\cos (\phi/2) = 0$,which implies $\phi/2 = \pi/2$,or $\phi = \pi$.
Substituting $\phi = \pi$ into the equation for the second wave:
$y_2 = a \cos (kx + \omega t + \pi) = -a \cos (kx + \omega t)$.

(C) Electric field lines are the paths along which a free positive charge moves.
Since the particle is located at $(1, 0)$,which satisfies the equation ${x^2} + {y^2} = 1$,it lies on the field line.
Therefore,the particle will move along the field line,which is a circle.
The correct option is $C$.

(A) The magnetic field at the centre $O$ due to a semi-circular arc of radius $R$ carrying current $i$ is given by $B = \frac{{\mu _0}i}{{4R}}$.
In the given figure,the current flows through two semi-circular arcs of radii $R_1$ and $R_2$ and two straight radial segments.
The magnetic field due to the straight radial segments at the centre $O$ is zero because the position vector of the point $O$ is parallel to the current element $idl$ for these segments.
For the inner semi-circular arc of radius $R_1$,the magnetic field $B_1$ at $O$ is directed into the plane of the paper $(\otimes)$ and its magnitude is $B_1 = \frac{{\mu _0}i}{{4R_1}}$.
For the outer semi-circular arc of radius $R_2$,the magnetic field $B_2$ at $O$ is directed out of the plane of the paper $(\odot)$ and its magnitude is $B_2 = \frac{{\mu _0}i}{{4R_2}}$.
Since $R_1$ < $R_2$,the magnitude $B_1$ > $B_2$.
The net magnetic field $B_{net}$ is $B_1 - B_2$ directed into the plane of the paper.
$B_{net} = \frac{{\mu _0}i}{{4R_1}} - \frac{{\mu _0}i}{{4R_2}} = \frac{{\mu _0}i}{4}\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)$.

(D) The minimum wavelength (cutoff wavelength) of $X$-rays produced in an $X$-ray tube is given by the formula:
${\lambda _{\min }} = \frac{{hc}}{{eV}}$
where $h$ is Planck's constant,$c$ is the speed of light,$e$ is the charge of an electron,and $V$ is the accelerating potential difference.
From the relation ${\lambda _{\min }} \propto \frac{1}{V}$,it is clear that the minimum wavelength is inversely proportional to the applied potential difference.
Therefore,when the potential difference $V$ is increased,the minimum wavelength ${\lambda _{\min }}$ decreases.
Thus,the correct option is $(d)$.

(D) Copper $(Cu)$ is a conductor,and its resistance decreases as the temperature decreases because the scattering of electrons by lattice vibrations reduces.
Germanium $(Ge)$ is a semiconductor,and its resistance increases as the temperature decreases because the number of free charge carriers (electrons and holes) decreases exponentially with a decrease in temperature.
Therefore,the resistance of copper decreases while that of germanium increases.

(A) To obtain a $P$-type semiconductor,we must dope the intrinsic semiconductor $(Si)$ with a trivalent impurity atom.
$Si$ belongs to group $14$ of the periodic table and has $4$ valence electrons.
Aluminium $(Al)$ belongs to group $13$ and has $3$ valence electrons.
When $Al$ is added to $Si$,it creates a hole (vacancy) in the crystal lattice,which acts as a charge carrier,resulting in a $P$-type semiconductor.
Phosphorous is a pentavalent impurity used for $N$-type semiconductors.

(D) From the mirror formula,$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ .....$(i)$
Differentiating equation $(i)$ with respect to $u$,we get:
$0 = - \frac{1}{v^2} \frac{dv}{du} - \frac{1}{u^2}$
$\frac{dv}{du} = - \left( \frac{v}{u} \right)^2$
Since the object length $l = du$ is small,the image length $dv$ is given by $dv = - \left( \frac{v}{u} \right)^2 du$ .....$(ii)$
From the mirror formula,we can express the magnification $m = \frac{v}{u}$ as:
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{u - f}{fu}$
$\frac{v}{u} = \frac{f}{u - f}$ .....$(iii)$
Substituting equation $(iii)$ into equation $(ii)$,we get:
$dv = - \left( \frac{f}{u - f} \right)^2 l$
The magnitude of the size of the image is $l' = |dv| = l \left( \frac{f}{u - f} \right)^2$.

(C) The intensity of the resultant wave in interference is given by $I_{res} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
For maximum intensity, $\cos \phi = 1$, so $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$.
Given $I_1 = I$ and $I_2 = 4I$, we have $I_{max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$.
For minimum intensity, $\cos \phi = -1$, so $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
Thus, $I_{min} = (\sqrt{I} - \sqrt{4I})^2 = (\sqrt{I} - 2\sqrt{I})^2 = (-\sqrt{I})^2 = I$.
Therefore, the maximum and minimum intensities are $9I$ and $I$ respectively.

(C) Initially,the capacitors are connected in parallel to a battery of potential $V$. The total charge stored is $Q_{total} = Q_1 + Q_2 = (2C)V + (C)V = 3CV$.
When the battery is removed and a dielectric of constant $K$ is inserted into the capacitor of capacity $C$,its new capacity becomes $C' = KC$.
The capacitor of capacity $2C$ remains unchanged.
Since the capacitors are connected in parallel,they share the same potential difference $V'$ across them. The total charge remains conserved as $Q_{total} = 3CV$.
The new equivalent capacitance is $C_{eq} = 2C + KC = C(K+2)$.
Using the relation $Q = C_{eq} V'$,we get:
$3CV = C(K+2) V'$
$V' = \frac{3CV}{C(K+2)} = \frac{3V}{K+2}$

(B) The radius of a circular path for a charged particle in a magnetic field is given by $R = \frac{mv}{qB}$.
Since the particle is accelerated through a potential difference $V$,its kinetic energy is $KE = qV = \frac{1}{2}mv^2$,which implies $v = \sqrt{\frac{2qV}{m}}$.
Substituting $v$ into the radius formula: $R = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.
Since $q$,$V$,and $B$ are constant for both particles,we have $R \propto \sqrt{m}$,which implies $m \propto R^2$.
Therefore,the ratio of the masses is $\frac{m_X}{m_Y} = \left(\frac{R_1}{R_2}\right)^2$.