IIT JEE 1999 Physics Question Paper with Answer and Solution

(B) The average velocity is defined as the total displacement divided by the total time taken.
Displacement is the shortest distance between the initial point $A$ and the final point $B$.
Since the particle moves along a semicircle of radius $r = 1.0 \, m$,the displacement is equal to the diameter of the semicircle.
Displacement $= 2 \times r = 2 \times 1.0 \, m = 2.0 \, m$.
The time taken is $1.0 \, s$.
Therefore,the magnitude of the average velocity $= \frac{\text{Total displacement}}{\text{Total time}} = \frac{2.0 \, m}{1.0 \, s} = 2.0 \, m/s$.

(D) $x = a \cos(pt)$ and $y = b \sin(pt)$ (given).
$\therefore \cos(pt) = x/a$ and $\sin(pt) = y/b$.
By squaring and adding:
$\cos^2(pt) + \sin^2(pt) = \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Hence,the path of the particle is an ellipse.
Now,differentiating $x$ and $y$ with respect to time:
$v_x = \frac{dx}{dt} = -ap \sin(pt)$ and $v_y = \frac{dy}{dt} = bp \cos(pt)$.
$\vec{v} = -ap \sin(pt) \hat{i} + bp \cos(pt) \hat{j}$.
Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = -ap^2 \cos(pt) \hat{i} - bp^2 \sin(pt) \hat{j}$.
At $t = \frac{\pi}{2p}$:
$\vec{v} = -ap \sin(\pi/2) \hat{i} + bp \cos(\pi/2) \hat{j} = -ap \hat{i}$.
$\vec{a} = -ap^2 \cos(\pi/2) \hat{i} - bp^2 \sin(\pi/2) \hat{j} = -bp^2 \hat{j}$.
Since $\vec{v} \cdot \vec{a} = (-ap \hat{i}) \cdot (-bp^2 \hat{j}) = 0$,the velocity and acceleration are perpendicular (normal) to each other at $t = \frac{\pi}{2p}$.

(D) The potential energy of the particle is given by $U(x) = k(1 - e^{-x^2})$.
The force acting on the particle is $F = -\frac{dU}{dx}$.
$F = -\frac{d}{dx} [k(1 - e^{-x^2})] = -k[0 - e^{-x^2} \cdot (-2x)] = -2kxe^{-x^2}$.
For small displacements from the origin $(x \approx 0)$,we can use the Taylor expansion $e^{-x^2} \approx 1 - x^2 + \dots \approx 1$.
Thus,$F \approx -2kx$.
Since $F \propto -x$,the restoring force is proportional to the displacement,which is the condition for simple harmonic motion $(SHM)$.

(D) Let $L_0$ be the initial length of each strip before heating.
Let $d$ be the total thickness of the bimetallic strip.
After heating,the lengths of the brass and copper strips are:
$L_B = L_0(1 + \alpha_B \Delta T) = (R + d)\theta$
$L_C = L_0(1 + \alpha_C \Delta T) = R\theta$
Dividing the two equations:
$\frac{R + d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T}$
$1 + \frac{d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T}$
$\frac{d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T} - 1 = \frac{1 + \alpha_B \Delta T - 1 - \alpha_C \Delta T}{1 + \alpha_C \Delta T} = \frac{(\alpha_B - \alpha_C)\Delta T}{1 + \alpha_C \Delta T}$
Since $\alpha \Delta T \ll 1$,we can approximate $1 + \alpha_C \Delta T \approx 1$.
Thus,$R = \frac{d}{(\alpha_B - \alpha_C)\Delta T}$.
Therefore,$R \propto \frac{1}{\Delta T}$ and $R \propto \frac{1}{|\alpha_B - \alpha_C|}$.
Hence,both $(B)$ and $(C)$ are correct.

(B) Consider the compartment in an accelerated frame of reference moving with acceleration $a$ in the positive $x$-direction.
In this non-inertial frame,a pseudo force acts on each gas molecule in the direction opposite to the acceleration (i.e.,in the negative $x$-direction).
Due to this pseudo force,the gas molecules tend to accumulate towards the rear side of the compartment.
As a result,the density of the gas increases at the rear side and decreases at the front side.
Since the pressure of a gas is directly proportional to its density (from the ideal gas law $P = \rho RT/M$),the pressure will be higher at the rear side and lower at the front side.
Therefore,the pressure is lower in the front side.

(D) The total internal energy $U$ of a gas mixture is the sum of the internal energies of its components.
$U = U_{O_2} + U_{Ar} = \mu_1 \frac{f_1}{2} RT + \mu_2 \frac{f_2}{2} RT$
For $O_2$ (a diatomic gas),the degrees of freedom $f_1 = 5$ (neglecting vibrational modes).
For $Ar$ (a monatomic gas),the degrees of freedom $f_2 = 3$.
Given $\mu_1 = 2$ moles and $\mu_2 = 4$ moles.
Substituting the values:
$U = 2 \times \frac{5}{2} RT + 4 \times \frac{3}{2} RT$
$U = 5 RT + 6 RT = 11 RT$

(B) The force constant $k$ of a spring is inversely proportional to its length $l$, i.e., $k \propto 1/l$ or $kl = \text{constant}$.
Let the total length of the spring be $l$ and its force constant be $k$.
The spring is cut into two pieces such that one piece is double the length of the other. Let the lengths of the two pieces be $l_1$ and $l_2$.
Given $l_1 = 2l_2$ and $l_1 + l_2 = l$.
Substituting $l_1$, we get $2l_2 + l_2 = l \Rightarrow 3l_2 = l \Rightarrow l_2 = l/3$.
Then $l_1 = 2l/3$.
For the long piece of length $l_1 = 2l/3$, let the new force constant be $k_1$.
Using $k_1 l_1 = kl$, we have $k_1 (2l/3) = kl$.
$k_1 = k / (2/3) = (3/2)k$.

(D) Let the three simple harmonic motions be represented by:
$y_1 = a \sin(\omega t - 45^\circ)$
$y_2 = a \sin(\omega t)$
$y_3 = a \sin(\omega t + 45^\circ)$
On superimposing,the resultant $SHM$ is $y = y_1 + y_2 + y_3$
$y = a[\sin(\omega t - 45^\circ) + \sin(\omega t) + \sin(\omega t + 45^\circ)]$
Using the identity $\sin(A-B) + \sin(A+B) = 2\sin A \cos B$:
$y = a[2\sin(\omega t)\cos(45^\circ) + \sin(\omega t)]$
Since $\cos(45^\circ) = \frac{1}{\sqrt{2}}$:
$y = a[2\sin(\omega t) \cdot \frac{1}{\sqrt{2}} + \sin(\omega t)]$
$y = a[\sqrt{2}\sin(\omega t) + \sin(\omega t)] = a(1 + \sqrt{2})\sin(\omega t)$
The resultant amplitude is $A = (1 + \sqrt{2})a$.
Energy $E$ in $SHM$ is proportional to the square of the amplitude $(E \propto A^2)$:
$\frac{E_{\text{resultant}}}{E_{\text{single}}} = \left(\frac{A}{a}\right)^2 = (1 + \sqrt{2})^2 = 1 + 2 + 2\sqrt{2} = (3 + 2\sqrt{2})$
Thus,$E_{\text{resultant}} = (3 + 2\sqrt{2})E_{\text{single}}$.

(C) The speed of sound in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma RT}{M}}$,where $\gamma$ is the adiabatic index,$R$ is the universal gas constant,$T$ is the temperature,and $M$ is the molar mass of the gas.
Since the temperature $T$ is the same for both gases,the ratio of the speeds is given by $\frac{v_{N_2}}{v_{He}} = \sqrt{\frac{\gamma_{N_2}}{\gamma_{He}} \times \frac{M_{He}}{M_{N_2}}}$.
For nitrogen $(N_2)$,which is a diatomic gas,$\gamma_{N_2} = 1.4 = 7/5$ and the molar mass $M_{N_2} = 28 \ g/mol$.
For helium $(He)$,which is a monatomic gas,$\gamma_{He} = 1.67 = 5/3$ and the molar mass $M_{He} = 4 \ g/mol$.
Substituting these values: $\frac{v_{N_2}}{v_{He}} = \sqrt{\frac{7/5}{5/3} \times \frac{4}{28}} = \sqrt{\frac{7}{5} \times \frac{3}{5} \times \frac{1}{7}} = \sqrt{\frac{3}{25}} = \frac{\sqrt{3}}{5}$.

(D) The correct answer is $(d)$.
For a plane wave,the wavefronts are parallel planes,and the energy is distributed uniformly,so the intensity remains constant as the wave propagates.
For a spherical wave,the energy emitted by the source spreads over an increasing surface area $A = 4\pi r^2$. Thus,the intensity $I = P / A = P / (4\pi r^2)$ is inversely proportional to the square of the distance $r$ from the source.
The total power $P$ (often referred to as total intensity in this context) passing through any spherical surface centered at the source remains constant,as energy is conserved.

(D) Standing waves are formed by the superposition of two waves of the same frequency and speed traveling in opposite directions.
$(a)$ $A$ string clamped at both ends allows for the reflection of waves at the boundaries,creating standing waves.
$(b)$ $A$ string clamped at one end and free at the other also allows for reflection,resulting in standing waves.
$(c)$ When an incident wave hits a wall,it reflects back,and the superposition of the incident and reflected waves creates standing waves.
Therefore,all the given conditions can produce standing waves. The correct option is $D$.

(D) The equation $y = a \sin (kx - \omega t)$ represents a general traveling wave equation.
In the case of mechanical waves (like sound waves),$y$ can represent the displacement of particles or pressure variations.
In the case of electromagnetic waves,$y$ represents the oscillating electric field $(E)$ or magnetic field $(B)$ components.
Therefore,$y$ is a general physical quantity that oscillates and propagates through space.
Thus,the correct option is $D$.

(C) Since the spheres are smooth,there is no friction between them during the collision.
Friction is the only source of torque that could change the angular momentum of a sphere about its center of mass.
Because there is no friction,there is no torque acting on either sphere.
Consequently,the angular momentum of each sphere remains constant.
For sphere $A$,the initial angular momentum is $L_A = I\omega$. Since no torque acts on it,its final angular momentum remains $I\omega_A = I\omega$,which implies $\omega_A = \omega$.
For sphere $B$,which was initially at rest,its initial angular momentum is $0$. Since no torque acts on it,its final angular momentum remains $0$,which implies $\omega_B = 0$.

(A) When the block hits the ridge at $O$,it starts rotating about an axis passing through $O$ and perpendicular to the plane of motion. Since no external torque acts on the block about point $O$ during the impact,the angular momentum is conserved.
Initial angular momentum $L_i$ about point $O$ is given by the product of linear momentum and the perpendicular distance from $O$ to the line of motion of the center of mass:
$L_i = Mv \times (a/2) = \frac{Mva}{2}$
After hitting the ridge,the block rotates about point $O$. The final angular momentum is $L_f = I_O \omega$,where $I_O$ is the moment of inertia of the cube about the axis passing through $O$.
The moment of inertia of a cube about an axis passing through its center of mass $C$ is $I_C = \frac{Ma^2}{6}$.
Using the parallel axis theorem,$I_O = I_C + Mr^2$,where $r$ is the distance from the center of mass to point $O$. Here,$r^2 = (a/2)^2 + (a/2)^2 = a^2/2$.
Thus,$I_O = \frac{Ma^2}{6} + M(a^2/2) = \frac{Ma^2 + 3Ma^2}{6} = \frac{4Ma^2}{6} = \frac{2}{3}Ma^2$.
Equating initial and final angular momentum: $L_i = L_f$
$\frac{Mva}{2} = \frac{2}{3}Ma^2 \omega$
Solving for $\omega$: $\omega = \frac{Mva}{2} \times \frac{3}{2Ma^2} = \frac{3v}{4a}$.

(C) The angular momentum of a rolling body about the point of contact (origin $O$) is the sum of the angular momentum due to linear motion and the angular momentum due to rotation about its center of mass.
$L = L_{\text{linear}} + L_{\text{rotational}}$
$L = MvR + I_c\omega$
Since the disc is rolling without slipping,the linear velocity of the center of mass is $v = R\omega$.
The moment of inertia of a disc about its center of mass is $I_c = \frac{1}{2}MR^2$.
Substituting these values:
$L = M(R\omega)R + (\frac{1}{2}MR^2)\omega$
$L = MR^2\omega + \frac{1}{2}MR^2\omega$
$L = \frac{3}{2}MR^2\omega$

(D) Under electrostatic conditions,the entire volume of a conductor is an equipotential region. Since points $A$ and $B$ lie on the surface of the conductor,they must be at the same potential. Thus,Potential at $A = $ Potential at $B$.
According to Gauss's Law,the total electric flux $\phi$ through any closed surface enclosing a charge $q$ is given by $\phi = q/\varepsilon_0$. Since the cavity surface encloses the charge $q$,the total flux through it is $q/\varepsilon_0$.
For an ellipsoidal cavity,the electric field near the surface depends on the local curvature,so the electric field at $A$ is not necessarily equal to the electric field at $B$. Therefore,options $(b)$ and $(c)$ are correct.

(C) For a parallel plate capacitor,the charges on the inner surfaces facing each other must be equal in magnitude and opposite in sign. Therefore,$Q_2 = -Q_3$.
The potential difference $V$ between the plates of a capacitor is given by the ratio of the charge on the inner surface of the positive plate to the capacitance $C$.
$V = \frac{Q_2}{C}$
Since $Q_3 = -Q_2$,we can write $Q_2 = \frac{Q_2 - Q_3}{2}$.
Substituting this into the potential difference formula:
$V = \frac{Q_2 - Q_3}{2C}$

(D) Initially,the capacitors are charged as $Q_1 = C_1 V_1 = (2\,pF)(30\,V) = 60\,pC$ and $Q_2 = C_2 V_2 = (3\,pF)(20\,V) = 60\,pC$.
When switches $S_1$ and $S_3$ are closed,the plates are connected to the ground. The potential difference across $C_1$ remains $30\,V$ (as the left plate is grounded and the right plate is connected to the potential of the junction) and across $C_2$ remains $20\,V$ because the charge on the capacitors remains trapped and unchanged as the circuit is not forming a closed loop with an external source to redistribute charge.
Thus,the potential differences $V_1 = 30\,V$ and $V_2 = 20\,V$ remain unchanged.

(A) The reading of the galvanometer remains the same whether switch $S$ is open or closed.
This implies that the potential difference across the switch $S$ is zero,or that no current flows through the switch when it is closed.
If no current flows through the switch,the branch containing $R$ and the branch containing the galvanometer $G$ effectively act as a single series circuit path between the nodes.
Therefore,the same current must flow through both $R$ and $G$.
Thus,$I_R = I_G$.

(A) The force on a charged particle $q$ moving with velocity $\vec{v}$ in electric field $\vec{E}$ and magnetic field $\vec{B}$ is given by the Lorentz force equation: $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
Since the particle is released from rest,its initial velocity $\vec{v} = 0$.
Initially,the magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B}) = 0$.
The electric force $\vec{F}_e = q\vec{E}$ acts on the particle,causing it to accelerate in the direction of the electric field.
As the particle gains velocity $\vec{v}$ in the direction of $\vec{E}$,and since $\vec{E}$ and $\vec{B}$ are parallel,the velocity $\vec{v}$ remains parallel to the magnetic field $\vec{B}$ at all times.
Therefore,the magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B}) = 0$ throughout the motion because the angle between $\vec{v}$ and $\vec{B}$ is $0^\circ$.
The particle continues to accelerate in a straight line due to the electric field.

(D) circular loop carrying current $I$ behaves as a magnetic dipole.
Magnetic field lines emerge from the North pole and enter the South pole.
Since the loop lies in the $x-y$ plane,the magnetic field lines pass through the loop and then loop back around to form closed paths.
For every magnetic field line that passes through the $x-y$ plane in one direction (e.g.,upwards),an equal number of field lines must pass back through the $x-y$ plane in the opposite direction (e.g.,downwards) to complete the closed loop.
Therefore,the net magnetic flux through the entire $x-y$ plane is zero.

(C) According to Faraday's law of electromagnetic induction,a changing current in $Loop-A$ creates a changing magnetic flux through $Loop-B$.
According to Lenz's law,the induced current in $Loop-B$ will flow in such a direction as to oppose the cause producing it.
Since the current in $Loop-A$ is increasing,the magnetic flux through $Loop-B$ is increasing.
To oppose this increase,$Loop-B$ will develop an induced current that creates a magnetic field opposing the field of $Loop-A$.
Two parallel loops with currents flowing in opposite directions repel each other.
Therefore,$Loop-B$ is repelled by $Loop-A$.

(D) The maximum current in the circuit is given by $i_0 = \frac{V}{R} = \frac{12 \, V}{6 \, \Omega} = 2 \, A$.
The current in an $LR$ circuit at time $t$ is given by $i(t) = i_0(1 - e^{-Rt/L})$.
We want to find the time $t$ when $i(t) = 1.0 \, A$. Substituting the values:
$1.0 = 2(1 - e^{-6t / (8.4 \times 10^{-3})})$
$0.5 = 1 - e^{-6t / (8.4 \times 10^{-3})}$
$e^{-6t / (8.4 \times 10^{-3})} = 0.5$
Taking the natural logarithm on both sides:
$-\frac{6t}{8.4 \times 10^{-3}} = \ln(0.5) \approx -0.693$
$t = 0.693 \times \frac{8.4 \times 10^{-3}}{6} = 0.693 \times 1.4 \times 10^{-3} \approx 0.97 \times 10^{-3} \, s \approx 1 \, ms$.

(C) According to the law of conservation of linear momentum,the initial momentum of the system is zero.
$0 = m_1 \vec{v}_1 + m_2 \vec{v}_2$
$\Rightarrow m_1 \vec{v}_1 = -m_2 \vec{v}_2$
Taking the magnitude,we get $m_1 v_1 = m_2 v_2$,which implies that the magnitudes of the momenta of the two particles are equal $(p_1 = p_2 = p)$.
The de-Broglie wavelength is given by $\lambda = h / p$.
Since both particles have the same magnitude of momentum $p$,their de-Broglie wavelengths will be $\lambda_1 = h / p$ and $\lambda_2 = h / p$.
Therefore,the ratio $\lambda_1 / \lambda_2 = (h / p) / (h / p) = 1$.

(B) The mass of a nucleus with mass number $A$ is approximately $m = A \times m_p$,where $m_p$ is the mass of a proton.
The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $R_0 \approx 1.2 \times 10^{-15} \ m$.
The volume of the nucleus is $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
The density $\rho$ is given by $\rho = \frac{m}{V} = \frac{A \times m_p}{\frac{4}{3} \pi R_0^3 A} = \frac{m_p}{\frac{4}{3} \pi R_0^3}$.
Substituting the values: $\rho = \frac{1.67 \times 10^{-27}}{\frac{4}{3} \times 3.14 \times (1.2 \times 10^{-15})^3} \approx \frac{1.67 \times 10^{-27}}{7.24 \times 10^{-45}} \approx 2.3 \times 10^{17} \ kg/m^3$.
Thus,the order of magnitude is $10^{17} \ kg/m^3$.

(A) The correct statement is that $\beta$-rays are the same as cathode rays because both consist of high-speed electrons.
$\gamma$-rays are high-energy electromagnetic waves,not neutrons.
$\alpha$-particles are doubly ionized helium atoms $(He^{2+})$,not singly ionized.
Protons and neutrons have approximately the same mass,but not exactly the same mass.

(B) The nuclear reaction is given by: $_{10}^{22}Ne \to 2(_2^4He) + _Z^A X$
Applying the law of conservation of mass number: $22 = 2(4) + A \implies 22 = 8 + A \implies A = 14$
Applying the law of conservation of atomic number: $10 = 2(2) + Z \implies 10 = 4 + Z \implies Z = 6$
The nucleus with atomic number $Z = 6$ is Carbon $(C)$.
Therefore,the unknown nucleus is Carbon.

(C) Given that the half-life of $X$ is equal to the mean life of $Y$:
$({T_{1/2}})_X = ({\tau})_Y$
We know that ${T_{1/2}} = \frac{0.693}{\lambda}$ and ${\tau} = \frac{1}{\lambda}$.
Substituting these,we get:
$\frac{0.693}{\lambda_X} = \frac{1}{\lambda_Y}$
$\Rightarrow \lambda_X = 0.693 \lambda_Y$
Since $0.693 < 1$,it follows that $\lambda_X < \lambda_Y$.
The rate of decay is given by $R = \lambda N$.
Initially,the number of atoms $N$ is the same for both.
Since $\lambda_Y > \lambda_X$,the decay rate $R_Y = \lambda_Y N$ will be greater than $R_X = \lambda_X N$.
Therefore,$Y$ will decay at a faster rate than $X$.

(C) Energy is released in a nuclear process when the total binding energy $(B.E.)$ of the products is greater than the total binding energy of the reactants.
From the graph, we can identify the mass number $(A)$ and binding energy per nucleon $(B.E./A)$ for each nucleus:
For $W$: $A = 120, B.E./A = 7.5 \, MeV \implies \text{Total } B.E. = 120 \times 7.5 = 900 \, MeV$
For $X$: $A = 90, B.E./A = 8.0 \, MeV \implies \text{Total } B.E. = 90 \times 8.0 = 720 \, MeV$
For $Y$: $A = 60, B.E./A = 8.5 \, MeV \implies \text{Total } B.E. = 60 \times 8.5 = 510 \, MeV$
For $Z$: $A = 30, B.E./A = 5.0 \, MeV \implies \text{Total } B.E. = 30 \times 5.0 = 150 \, MeV$
Now, checking option $(c)$: $W \to 2Y$
Total $B.E.$ of reactants $= 900 \, MeV$
Total $B.E.$ of products $= 2 \times (510) = 1020 \, MeV$
Since the total $B.E.$ of products $(1020 \, MeV)$ is greater than the total $B.E.$ of reactants $(900 \, MeV)$, energy is released in this process.

(D) The correct answer is $D$.
$1$. At $0 \ K$,an insulator has no free charge carriers,so the current is zero.
$2$. At $0 \ K$,a semiconductor behaves as a perfect insulator because all valence electrons are bound,so the current is zero.
$3$. In a $P-N$ junction diode at $300 \ K$,when reverse biased,a small finite current (reverse saturation current) flows due to the presence of minority charge carriers.
Since all these statements are correct,the answer is $D$.

(A) The lens maker's formula is given by $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a concave lens in air,$R_1 = -R$ and $R_2 = +R$,so $\frac{1}{f_a} = (1.5 - 1) \left( \frac{1}{-R} - \frac{1}{R} \right) = 0.5 \left( -\frac{2}{R} \right) = -\frac{1}{R}$,which means $f_a = -R$.
When immersed in a medium of refractive index $\mu_m = 1.75$,the relative refractive index is $\mu_{rel} = \frac{\mu_g}{\mu_m} = \frac{1.5}{1.75} = \frac{6}{7}$.
The new focal length $f_l$ is given by $\frac{1}{f_l} = \left( \frac{6}{7} - 1 \right) \left( -\frac{2}{R} \right) = \left( -\frac{1}{7} \right) \left( -\frac{2}{R} \right) = \frac{2}{7R}$.
Thus,$f_l = +3.5 R$.
Since the focal length is positive,the lens behaves as a convergent (convex) lens.

(B) Diffraction occurs when the slit width is comparable to the wavelength of the incident electromagnetic waves.
The wavelength of yellow light is approximately $589 \, nm$ $(5.89 \times 10^{-7} \, m)$,which is of the order of the slit width $(0.6 \, mm = 6 \times 10^{-4} \, m)$.
The wavelength of $X$-rays is typically in the range of $0.01 \, nm$ to $10 \, nm$ ($10^{-11} \, m$ to $10^{-8} \, m$).
Since the wavelength of $X$-rays is extremely small compared to the slit width $(0.6 \, mm)$,the condition for diffraction is not satisfied.
Therefore,no diffraction pattern will be observed.

(A) The cylindrical surface touches the glass plate along a line parallel to the axis of the cylinder.
The thickness of the air wedge formed between the cylindrical surface and the flat glass plate increases symmetrically on both sides of this line of contact.
The locus of points with equal path difference corresponds to lines running parallel to the axis of the cylinder.
Since the path difference is constant along these lines,the interference fringes observed will be straight and parallel to the axis of the cylinder.

(B) The Doppler shift for light is given by the formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$ is the relative speed of the source and $c$ is the speed of light.
Given: $\lambda = 656 \ nm$,$\Delta \lambda = 706 \ nm - 656 \ nm = 50 \ nm$,and $c = 3 \times 10^8 \ m/s$.
Substituting the values into the formula $v = \frac{c \Delta \lambda}{\lambda}$:
$v = \frac{3 \times 10^8 \times 50}{656}$
$v = \frac{1500}{656} \times 10^7 \ m/s$
$v \approx 2.28 \times 10^7 \ m/s$.
Rounding to the nearest provided option,we get $v = 2 \times 10^7 \ m/s$.