IIT JEE 1997 Physics Question Paper with Answer and Solution

(C) For a satellite in a circular orbit,the potential energy $U$ is given by $U = -\frac{GMm}{r}$.
The total energy $E_0$ is the sum of kinetic energy $K$ and potential energy $U$,which is $E_0 = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$.
Comparing the expressions for $U$ and $E_0$,we can see that $U = 2 \times (-\frac{GMm}{2r}) = 2 E_0$.
Therefore,the potential energy of the satellite is $2 E_0$.

(C) The ideal gas equation is given by $PV = \mu RT$,where $P$ is pressure,$V$ is volume,$\mu$ is the number of moles,$R$ is the universal gas constant,and $T$ is the temperature.
Since the vessels are identical,the volume $V$ is constant. $R$ is also a constant.
Therefore,$P \propto \mu T$.
For the first vessel (containing $O_2$): $P_1 = P$,$\mu_1 = 1$,$T_1 = T$.
For the second vessel (containing $He$): $P_2 = ?$,$\mu_2 = 1$,$T_2 = 2T$.
Taking the ratio: $\frac{P_2}{P_1} = \frac{\mu_2 T_2}{\mu_1 T_1}$.
Substituting the values: $\frac{P_2}{P} = \frac{1 \times 2T}{1 \times T} = 2$.
Thus,$P_2 = 2P$.

(C) The average translational kinetic energy of an ideal gas molecule is given by the formula:
$E = \frac{3}{2} k_B T$
where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
From this formula,it is clear that the translational kinetic energy depends only on the temperature $T$ of the gas.
It is independent of the nature of the gas (i.e.,independent of molar mass).
Since the temperature $T$ is the same for both ${O_2}$ and ${N_2}$ molecules,their average translational kinetic energies will be equal.
Therefore,$E_{N_2} = E_{O_2} = 0.048 \; eV$.

(A) The average translational kinetic energy $E$ of a gas molecule is given by $E = \frac{3}{2} k_B T$. Thus,$E \propto T$.
Given $T_1 = 300 \, K$ and $T_2 = 600 \, K$,the ratio is $\frac{E_2}{E_1} = \frac{T_2}{T_1} = \frac{600}{300} = 2$.
Therefore,$E_2 = 2 \times E_1 = 2 \times 6.21 \times 10^{-21} \, J = 12.42 \times 10^{-21} \, J$.
The $r.m.s.$ speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$. Thus,$v_{rms} \propto \sqrt{T}$.
Therefore,$\frac{(v_{rms})_2}{(v_{rms})_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{600}{300}} = \sqrt{2} \approx 1.414$.
$(v_{rms})_2 = 1.414 \times 484 \, m/s \approx 684 \, m/s$.
Thus,the values are $12.42 \times 10^{-21} \, J$ and $684 \, m/s$.

(B) According to Wien's displacement law,the product of the wavelength corresponding to maximum intensity $(\lambda_{\max})$ and the absolute temperature $(T)$ of a black body is constant.
$\lambda_{\max} T = b$ (constant)
Therefore,$T \propto \frac{1}{\lambda_{\max}}$.
Let $T_S$ and $\lambda_S$ be the temperature and wavelength of maximum intensity for the sun,and $T_N$ and $\lambda_N$ be those for the north star.
Given: $\lambda_S = 510\;nm$ and $\lambda_N = 350\;nm$.
The ratio of the surface temperatures is given by:
$\frac{T_S}{T_N} = \frac{\lambda_N}{\lambda_S} = \frac{350}{510} \approx 0.686$.
Rounding to two decimal places,we get $0.69$.

(D) The power radiated by a black body is given by the Stefan-Boltzmann law: $P = A\sigma T^4$,where $A = 4\pi r^2$ is the surface area.
Therefore,$P \propto r^2 T^4$.
Given the initial state: $P_1 = 440\;W$,$r_1 = 24\;cm$,$T_1 = 500\;K$.
For the final state: $r_2 = r_1 / 2 = 12\;cm$ and $T_2 = 2T_1 = 1000\;K$.
Using the proportionality ratio:
$\frac{P_2}{P_1} = \left( \frac{r_2}{r_1} \right)^2 \left( \frac{T_2}{T_1} \right)^4$
$\frac{P_2}{440} = \left( \frac{1}{2} \right)^2 \left( \frac{2}{1} \right)^4$
$\frac{P_2}{440} = \frac{1}{4} \times 16 = 4$
$P_2 = 440 \times 4 = 1760\;W$.

(A) The displacement of the particle in the string is given by $y = A\sin (kx - \omega t)$.
To find the particle velocity,we differentiate the displacement $y$ with respect to time $t$:
$v = \frac{dy}{dt} = \frac{d}{dt} [A\sin (kx - \omega t)]$
$v = A \cos (kx - \omega t) \cdot (-\omega)$
$v = -A\omega \cos (kx - \omega t)$
The maximum value of the particle velocity occurs when the magnitude of the cosine function is $1$.
Therefore,$v_{\max} = | -A\omega | = A\omega$.

(D) The apparent frequency $n'$ heard by the observer is given by the Doppler effect formula for a source moving towards a stationary observer:
$n' = n \left( \frac{v}{v - v_s} \right)$
Given:
Source frequency $n = 450 Hz$
Speed of sound $v = 330 m/s$
Speed of source $v_s = 33 m/s$
Substituting the values:
$n' = 450 \times \left( \frac{330}{330 - 33} \right)$
$n' = 450 \times \left( \frac{330}{297} \right)$
$n' = 450 \times \frac{10}{9}$
$n' = 50 \times 10 = 500 Hz$
Therefore,the frequency heard by the observer is $500 Hz$.

(D) The time $t$ taken by the plate to fall through a height $h$ is given by the equation of motion $h = \frac{1}{2}gt^2$.
Given $h = 10 \, cm = 0.1 \, m$ and $g = 9.8 \, m/s^2$ (or $10 \, m/s^2$ for simplicity).
Using $g = 10 \, m/s^2$,we have $0.1 = \frac{1}{2} \times 10 \times t^2$,which gives $t^2 = 0.02$,so $t = \sqrt{0.02} \, s = \sqrt{\frac{2}{100}} \, s = \frac{\sqrt{2}}{10} \, s \approx 0.1414 \, s$.
In this time,the tuning fork completes $n = 8$ oscillations.
The time period of one oscillation is $T = \frac{t}{n} = \frac{\sqrt{0.02}}{8} \, s$.
The frequency $f$ is the reciprocal of the time period: $f = \frac{1}{T} = \frac{8}{\sqrt{0.02}} = \frac{8}{0.1414} \approx 56.56 \, Hz$.
Rounding to the nearest integer provided in the options,the frequency is $56 \, Hz$.

(A) Before the explosion,the particle was moving along the $x$-axis,which means it had no $y$-component of velocity. Consequently,the center of mass of the system does not move in the $y$-direction,so $y_{CM} = 0$.
Using the formula for the center of mass: $y_{CM} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$.
Substituting the given values: $0 = \frac{(m/4)(+15) + (3m/4)(y)}{m}$.
This simplifies to: $0 = \frac{15}{4} + \frac{3y}{4}$.
Solving for $y$: $\frac{3y}{4} = -\frac{15}{4}$,which gives $y = -5 \ cm$.

(B) The angular momentum $L$ of a particle about the origin is given by $L = r \times p$,where $r$ is the position vector and $p$ is the linear momentum.
Alternatively,the magnitude of angular momentum is $L = p \times d$,where $d$ is the perpendicular distance from the origin to the line of motion.
Here,the mass $m$ moves with constant velocity $v$,so the linear momentum $p = mv$ is constant.
The perpendicular distance $d$ from the origin to the line of motion is constant and equal to $a$.
Therefore,the angular momentum $L = mv \times a = mva$,which is a constant value.

(B) The force on a charged particle in a uniform electric field $E$ is given by $F = qE$. According to Newton's second law,$F = ma$,so the acceleration is $a = qE/m$.
Using the equation of motion $s = ut + (1/2)at^2$,and since the particles start from rest $(u = 0)$,the distance $s$ covered in time $t$ is $s = (1/2)(qE/m)t^2$.
For the electron: $s = (1/2)(eE/m_e)t_1^2$.
For the proton: $s = (1/2)(eE/m_p)t_2^2$.
Since the distance $s$ is the same for both,we equate the two expressions:
$(1/2)(eE/m_e)t_1^2 = (1/2)(eE/m_p)t_2^2$.
Simplifying this,we get $t_1^2/m_e = t_2^2/m_p$.
Therefore,$t_2^2/t_1^2 = m_p/m_e$.
Taking the square root of both sides,we find the ratio $t_2/t_1 = (m_p/m_e)^{1/2}$.

(A) The line integral of the electric field from infinity to a point $P$ is defined as the electric potential $V$ at that point.
$\int_{\infty}^{P} -\vec{E} \cdot d\vec{l} = V_P - V_{\infty}$.
Since the potential at infinity $V_{\infty} = 0$,the integral represents the potential at the centre of the ring.
For a ring of radius $R$ with total charge $q$,the potential at the centre is given by $V = \frac{1}{4\pi\epsilon_0} \frac{q}{R}$.
Given $q = 1.11 \times 10^{-10}\,C$,$R = 0.5\,m$,and $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\,N\cdot m^2/C^2$.
$V = (9 \times 10^9) \times \frac{1.11 \times 10^{-10}}{0.5} = \frac{9.99 \times 10^{-1}}{0.5} = \frac{0.999}{0.5} \approx 2\,V$.

(D) For a steady current $i$ flowing through a conductor,the charge conservation principle implies that the current $i$ must be constant at every cross-section of the conductor.
The current density is given by $j = \frac{i}{A}$,where $A$ is the cross-sectional area. Since $A$ varies along the length,$j$ is not constant.
From Ohm's law in microscopic form,$j = \sigma E$. Since $j$ varies,the electric field $E$ must also vary along the length.
Furthermore,the drift velocity is given by ${v_d} = \frac{j}{ne} = \frac{i}{Ane}$. Since $A$ varies,the drift velocity ${v_d}$ is also not constant.
Therefore,only the current $i$ remains constant along the length of the conductor.

(B) The relationship between the maximum kinetic energy $(K_{\max})$ of photoelectrons and the stopping potential $(V_s)$ is given by the equation: $K_{\max} = e V_s$.
Given that the maximum kinetic energy $K_{\max} = 4 eV$.
Substituting the value into the equation: $4 eV = e V_s$.
Therefore,the stopping potential $V_s = 4 V$.

(C) The energy of the emitted $X$-ray photon corresponds to the energy difference between the two levels involved in the transition.
For the $K_{\alpha}$ line,the transition occurs from the $L$ level to the $K$ level.
The energy difference is given by $\Delta E = E_K - E_L = \frac{hc}{\lambda}$.
Substituting the values: $h = 6.6 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$,and $\lambda = 0.021 \times 10^{-9} \ m$.
$\Delta E = \frac{(6.6 \times 10^{-34}) \times (3 \times 10^8)}{0.021 \times 10^{-9}} \ J$.
To convert this energy into electron-volts $(eV)$,divide by $1.6 \times 10^{-19} \ J/eV$:
$\Delta E = \frac{19.8 \times 10^{-26}}{0.021 \times 10^{-9} \times 1.6 \times 10^{-19}} \ eV \approx 58928 \ eV \approx 59 \ keV$.

(D) According to the Bohr model,the energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \; eV$.
For a doubly ionized lithium atom $(Li^{2+})$,the atomic number $Z = 3$.
For the ground state,the principal quantum number $n = 1$.
Substituting these values,the energy of the ground state is $E_1 = -13.6 \times \frac{3^2}{1^2} \; eV = -13.6 \times 9 \; eV = -122.4 \; eV$.
The energy required to remove the electron (ionization energy) is the energy needed to bring the electron from the ground state to infinity $(E_{\infty} = 0)$.
Therefore,Ionization Energy $= E_{\infty} - E_1 = 0 - (-122.4 \; eV) = 122.4 \; eV$.

(C) In an $N-$type semiconductor,the majority charge carriers are electrons,not holes. Therefore,the statement that the majority carriers in $N-$type semiconductors are holes is incorrect. Thus,option $C$ is the correct answer.

(B) In a forward-biased $P-N$ junction,the potential barrier is reduced,which allows majority charge carriers to cross the junction easily. This process is known as diffusion,which becomes the dominant mechanism for current flow.
In a reverse-biased $P-N$ junction,the potential barrier increases,which prevents majority charge carriers from crossing the junction. However,minority charge carriers can still cross the junction due to the electric field present in the depletion region. This process is known as drift,which becomes the dominant mechanism for the small reverse saturation current.

(B) $1$. Analyze the biasing of the diodes: Based on the provided circuit diagram,the positive terminal of the $6 \, V$ battery is connected to the anode of diode $D_1$ and the cathode of diode $D_2$.
$2$. Determine the state of each diode: Diode $D_1$ is forward-biased,allowing current to flow through it. Diode $D_2$ is reverse-biased,acting as an open circuit (infinite resistance),so no current flows through the branch containing $D_2$.
$3$. Calculate the total resistance of the circuit: The circuit effectively consists of the battery,the $100 \, \Omega$ resistor,the forward-biased diode $D_1$ (with $50 \, \Omega$ resistance),and the $150 \, \Omega$ resistor in series.
Total resistance $R_{eq} = R_{battery} + R_{D1} + R_{150\Omega} + R_{100\Omega} = 0 + 50 \, \Omega + 150 \, \Omega + 100 \, \Omega = 300 \, \Omega$.
$4$. Calculate the current: Using Ohm's law,$I = \frac{V}{R_{eq}} = \frac{6 \, V}{300 \, \Omega} = 0.02 \, A$.

(B) The power $P$ of a lens is given by $P = \frac{100}{f(cm)}\, D$.
For the convex lens,the focal length $f_1 = +40\, cm$.
Therefore,the power $P_1 = \frac{100}{40} = +2.5\, D$.
For the concave lens,the focal length $f_2 = -25\, cm$.
Therefore,the power $P_2 = \frac{100}{-25} = -4.0\, D$.
The power of the combination of lenses in contact is given by $P = P_1 + P_2$.
$P = 2.5\, D + (-4.0\, D) = -1.5\, D$.

(C) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB}$.
Since kinetic energy $K = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mK}$.
Substituting this into the radius formula: $r = \frac{\sqrt{2mK}}{qB}$.
Since $K$ and $B$ are constant,$r \propto \frac{\sqrt{m}}{q}$.
For a proton $(p)$: $m_p = m, q_p = e \Rightarrow r_p \propto \frac{\sqrt{m}}{e}$.
For a deuteron $(d)$: $m_d = 2m, q_d = e \Rightarrow r_d \propto \frac{\sqrt{2m}}{e} = \sqrt{2} r_p$.
For an alpha particle $(\alpha)$: $m_{\alpha} = 4m, q_{\alpha} = 2e \Rightarrow r_{\alpha} \propto \frac{\sqrt{4m}}{2e} = \frac{2\sqrt{m}}{2e} = r_p$.
Comparing the values: $r_{\alpha} = r_p$ and $r_d = \sqrt{2} r_p \approx 1.414 r_p$.
Thus,$r_d > r_{\alpha} = r_p$ is not explicitly listed,but checking the options,the relation $r_{\alpha} = r_p < r_d$ is correct.