NEB Class 12 · Past paper
The complete NEB Class 12 2078 exam paper for Physics, all 12 questions with solved model answers.
Tap a question to open its answer.
Answer, in brief, any four questions. (a) A large number of free electrons are present in metals. Why is there no current in the absence of electric field across it? (b) Draw a circuit diagram of Wheatstone bridge. What is the balanced condition of it? (c) How will the magnetic field intensity at the centre of a circular coil carrying current change, if the current through the coil is doubled and the radius of the coil is halved? (d) Why do two straight conductors carrying current in opposite direction repel each other? (e) What is self inductance of a coil? Write two factors on which it depends. (f) For a direct current supply a capacitor containing circuit becomes an open circuit. Explain.
(a) The free electrons in a metal are in continuous random thermal motion, travelling in all directions with speeds of the order of $10^5\ \text{m/s}$. Because the directions are completely random, as many electrons cross any section one...
Answer, in brief, any four questions. (a) A hydrogen atom is stable in the ground state. Why? (b) Draw a circuit diagram for p-n junction diode in forward bias. Sketch the voltage versus current graph for it. (c) How does a daughter nucleus differ from its parent nucleus when it emits an alpha-particle? (d) Why is the mass of nucleus slightly less than the sum of mass of constituent nucleons? (e) Discuss ozone layer depletion. (f) State Hubble's law and write the significance of Hubble's constant.
(a) In the ground state the electron occupies the orbit of lowest possible energy, $n=1$, with $E_1=-13.6\ \text{eV}$. According to Bohr's postulate an electron revolving in a stationary orbit does not radiate energy; radiation accompanies only a transition to a lower orbit. Since there is no orbit below $n=1$, the electron has no lower state available to fall into and so cannot lose energy. Having no way to radiate, the atom remains in that state indefinitely, which is why hydrogen is stable in the ground state.
(b) The $p$ side is joined to the positive terminal and the $n$ side to the negative terminal, through a rheostat and an ammeter, with a voltmeter across the diode:
+|i----[ rheostat ]----(A)----|>|----|
| p n |
=== battery (V) |
| |
-|-----------------------------------
The characteristic rises very slowly until the knee (about $0.3\ \text{V}$ for germanium and $0.7\ \text{V}$ for silicon), beyond which the current increases steeply and almost linearly for a small increase of voltage:
I | /
| /
| /
| _/
|______/
0 0.7 V
(c) An $\alpha$-particle is a helium nucleus $^{4}_{2}\text{He}$, so emitting one removes two protons and two neutrons:
$$ \begin{aligned} & ^{A}{Z}X\ \ & \longrightarrow\ \ & ^{A-4}{Z-2}Y+{}^{4}_{2}\text{He} \end{aligned} $$
The daughter nucleus therefore has an atomic number smaller by 2 and a mass number smaller by 4 than the parent. Because $Z$ changes, the daughter is a different element, shifted two places to the left in the periodic table.
(d) When free nucleons combine to form a nucleus, some of their mass is converted into the energy released in binding them together. This missing mass is the mass defect
$$\Delta m=\left[Zm_p+(A-Z)m_n\right]-M_{\text{nucleus}}$$
and the energy equivalent $\Delta m c^2$ is the binding energy of the nucleus, which has been radiated away during formation. Since that much mass has left the system as energy, what remains, the mass of the nucleus, is slightly less than the sum of the masses of the constituent nucleons taken separately.
(e) The ozone layer in the stratosphere absorbs most of the harmful ultraviolet radiation from the Sun. Chlorofluorocarbons (CFCs) released from refrigerators, air conditioners, aerosol sprays and foam, together with halons and nitrogen oxides, drift up into the stratosphere where ultraviolet light breaks them up and frees chlorine atoms:
$$ \begin{aligned} \text{CFCl}_3+h\nu\rightarrow \text{Cl}+\text{CFCl}_2 \ \text{Cl}+\text{O}_3\rightarrow \text{ClO}+\text{O}_2,\qquad \text{ClO}+\text{O}\rightarrow \text{Cl}+\text{O}_2 \end{aligned} $$
The chlorine atom is regenerated at the end of the second step, so a single atom acts as a catalyst and destroys thousands of ozone molecules. The result is a thinning of the layer, most severe over Antarctica where the "ozone hole" appears each spring. The increased ultraviolet reaching the ground causes skin cancer, cataracts and weakened immunity in humans, damages crops and phytoplankton, and so disturbs the food chain.
(f) Hubble's law: the galaxies are receding from us with a speed directly proportional to their distance,
$$v=H_0 d$$
where $H_0$ is Hubble's constant (about $70\ \text{km s}^{-1}\text{Mpc}^{-1}$).
Significance of $H_0$: it measures the present rate of expansion of the universe, and its reciprocal gives an estimate of the age of the universe, $T\approx 1/H_0$, since a galaxy now at distance $d$ receding uniformly at $v=H_0d$ would have taken $d/v=1/H_0$ to get there. It also fixes the Hubble radius $c/H_0$, the scale of the observable universe.
Answer, in brief, any one question. (a) Sound waves are also called pressure waves. Why? (b) What is a progressive wave? Write its representative wave equation.
(a) Sound travels through a medium as a longitudinal wave, in which the particles vibrate along the direction of propagation. This produces regions where the layers crowd together, the compressions, and regions where they spread apart, t...
Answer, in brief, any one question. (a) State and explain Huygen's principle. (b) What is a polarizing angle? Does it depend on the wave length of light?
(a) Huygens' principle: every point of a wavefront acts as a fresh source of secondary disturbances, called secondary wavelets, which spread out in all directions with the speed of the wave in that medium; the envelope (the common tangen...
Answer any three questions. (a) State and explain principle of potentiometer. Explain with the help of circuit diagram, the use of potentiometer for determination of internal resistance of a cell. (b) What is thermoelectric effect? How does the thermo emf of a thermocouple vary with increase in temperature of hot junction, keeping the cold junction at 0 degree C? Explain. (c) What are the magnetic elements of the earth? Prove the relation cot^2(delta) = cot^2(delta_1) + cot^2(delta_2), where delta is the true dip and delta_1 and delta_2 are the apparent dips. (d) Derive expressions for the impedance and phase angle of an alternating current circuit with an inductor L, a capacitor C and a resister R in series.
(a) Principle of the potentiometer. When a steady current flows through a wire of uniform cross section and uniform material, the potential difference across any portion of the wire is directly proportional to the length of that portion:
$$V\propto l,\qquad V=kl$$
where $k$ is the potential gradient, the fall of potential per unit length. This follows because $V=IR$ and $R=\rho l/A$, so with $I$, $\rho$ and $A$ all constant, $V=\left(\dfrac{I\rho}{A}\right)l$.
Internal resistance of a cell. The potentiometer wire $AB$ is fed by a driver cell through a key, and the cell under test, of emf $E$ and internal resistance $r$, is connected with its positive terminal to $A$ and through a galvanometer to the jockey. A resistance box $R$ with a key is joined across the test cell:
+|i----[ key ]----------------------------
| |
=== driver |
| |
A o======================================o B (uniform wire)
| |
| jockey ---- (G) ---- +|i---- test cell (E, r)
| |
+---------------------------------------+
|
[ R ] [ key K ] across the test cell
With the key $K$ open no current is drawn from the test cell, so the balancing length $l_1$ measures its full emf:
$$E=k,l_1$$
With $K$ closed, the cell sends current through $R$, and the balance point now measures only the terminal potential difference $V$:
$$V=k,l_2$$
For the closed circuit $E=I(R+r)$ and $V=IR$, so
$$ \begin{aligned} \frac{E}{V} &= \frac{R+r}{R} \ &= \frac{l_1}{l_2} \end{aligned} $$
which rearranges to
$$r=R\left(\frac{l_1-l_2}{l_2}\right)$$
Measuring $l_1$, $l_2$ and $R$ therefore gives the internal resistance. The method is a null method, so at balance no current is drawn from the cell being measured and the result is free from the error that a voltmeter would introduce.
(b) Thermoelectric effect. When two different metals are joined to form a closed circuit and the two junctions are kept at different temperatures, an emf is set up and a current flows round the circuit. This is the thermoelectric or Seebeck effect, and the pair of metals is a thermocouple. The effect is reversible: interchanging the hot and cold junctions reverses the current.
Variation with the temperature of the hot junction. With the cold junction held at $0^\circ\text{C}$ and the hot junction at $\theta$, the thermo emf is not linear but parabolic:
$$E=\alpha\theta+\tfrac{1}{2}\beta\theta^{2}$$
where $\alpha$ and $\beta$ are constants for the pair. As $\theta$ rises from zero the emf increases, but at a falling rate, until it reaches a maximum at the neutral temperature $\theta_n$, which for a given couple is fixed (about $270^\circ\text{C}$ for a copper-iron couple). Beyond $\theta_n$ the emf decreases, falls back to zero at the temperature of inversion $\theta_i$, and on further heating reverses its direction. Since the curve is a parabola about $\theta_n$, the neutral temperature lies midway between the cold junction temperature and the inversion temperature:
$$\theta_n=\frac{\theta_c+\theta_i}{2}$$
so with the cold junction at $0^\circ\text{C}$, $\theta_i=2\theta_n$. Differentiating gives the thermoelectric power $\dfrac{dE}{d\theta}=\alpha+\beta\theta$, which is zero at the neutral temperature, as the maximum requires.
(c) Magnetic elements of the earth. These are the three quantities needed to specify the earth's magnetic field completely at a place:
Proof of the relation. The true dip $\delta$ is observed only when the dip circle is set in the magnetic meridian. If instead the plane of the dip circle is turned through an angle $\alpha$ from the magnetic meridian, the vertical component $V$ is unaffected, but only the resolved part of the horizontal component acts in that plane, namely $H\cos\alpha$. The apparent dip $\delta_1$ in that plane satisfies
$$\tan\delta_1=\frac{V}{H\cos\alpha}$$
Let the second dip circle be set in the plane perpendicular to the first, so its plane makes an angle $(90^\circ-\alpha)$ with the magnetic meridian. Then
$$ \begin{aligned} \tan\delta_2 &= \frac{V}{H\cos(90^\circ-\alpha)} \ &= \frac{V}{H\sin\alpha} \end{aligned} $$
Taking reciprocals and squaring both results,
$$ \begin{aligned} \cot^2\delta_1 &= \frac{H^2\cos^2\alpha}{V^2} \ \qquad \cot^2\delta_2 &= \frac{H^2\sin^2\alpha}{V^2} \end{aligned} $$
Adding them,
$$ \begin{aligned} \cot^2\delta_1+\cot^2\delta_2 &= \frac{H^2}{V^2}\left(\cos^2\alpha+\sin^2\alpha\right) \ &= \frac{H^2}{V^2} \end{aligned} $$
But for the true dip $\tan\delta=V/H$, so $\cot^2\delta=H^2/V^2$. Hence
$$\cot^2\delta=\cot^2\delta_1+\cot^2\delta_2$$
which is the required result. It is useful in practice because it gives the true dip from two measurements in mutually perpendicular planes, without having to locate the magnetic meridian first.
(d) Series LCR circuit. Let an alternating emf $e=e_0\sin\omega t$ drive a current $I=I_0\sin(\omega t-\phi)$ through $R$, $L$ and $C$ in series. The same current passes through all three, so the current is taken as the reference in the phasor diagram.
The potential difference across the resistance is in phase with the current, that across the inductor leads it by $90^\circ$, and that across the capacitor lags it by $90^\circ$:
$$ \begin{aligned} V_R &= IR \ \qquad V_L &= IX_L \ &= I\omega L \ \qquad V_C &= IX_C \ &= \frac{I}{\omega C} \end{aligned} $$
Since $V_L$ and $V_C$ are opposite in phase, they combine to a single phasor of magnitude $|V_L-V_C|$ at right angles to $V_R$. The applied voltage is the resultant of these two perpendicular phasors:
$$ \begin{aligned} V &= \sqrt{V_R^{2}+(V_L-V_C)^{2}} \ &= I\sqrt{R^{2}+(X_L-X_C)^{2}} \end{aligned} $$
The impedance is the ratio of the applied voltage to the current, so
$$ \begin{aligned} Z &= \frac{V}{I} \ &= \sqrt{R^{2}+(X_L-X_C)^{2}} \ &= \sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}} \end{aligned} $$
The phase angle $\phi$ between the applied voltage and the current is the angle of that resultant with $V_R$:
$$ \begin{aligned} \tan\phi &= \frac{V_L-V_C}{V_R} \ &= \frac{X_L-X_C}{R} \ \phi &= \tan^{-1}\left(\frac{\omega L-\dfrac{1}{\omega C}}{R}\right) \end{aligned} $$
If $X_L>X_C$ the circuit is inductive and the current lags the voltage; if $X_C>X_L$ it is capacitive and the current leads; and if $X_L=X_C$ then $\phi=0$ and $Z=R$, which is the condition of resonance, when the current is a maximum.
Answer any three questions. (a) Describe the construction and working of a He-Ne laser. (b) What is Zener effect? Explain the use of Zener diode as a voltage regulator. (c) State laws of radioactive disintegration. Hence, obtain decay law and define disintegration constant. (d) Describe the environmental implications of the following energy sources. i) fossil fuels and ii) nuclear fuels.
(a) He-Ne laser. Construction. A narrow quartz discharge tube about $50\ \text{cm}$ long and $1\ \text{cm}$ in diameter is filled with a mixture of helium and neon in the ratio about $10:1$ at a total pressure of roughly $1\ \text{torr}$...
Answer any one question. (a) Define resonance. Describe an experiment to determine the velocity of sound in air and end correction of the tube by resonance method. (b) What are beats? Prove that the beat frequency is equal the difference of frequences of two superposing waves.
(a) Resonance. Resonance is the condition in which a body is set into vibration of large amplitude by a periodic force whose frequency equals the natural frequency of that body, so that the energy transfer from the driver to the vibratin...
Answer any one question. (a) Describe Michelson's method for the determination of speed of light. (b) What are coherent sources of light? Show that in Young's double slit experiment, the dark and bright fringes are equally spaced.
(a) Michelson's rotating mirror method. Michelson measured the speed of light between Mount Wilson and Mount San Antonio in California, a distance of about $35\ \text{km}$ accurately surveyed.
Arrangement. Light from an intense source falls on one face of an octagonal mirror $M$ (eight plane mirrors on the faces of a regular octagon) that can be spun at a controlled, measurable rate. The beam reflected from one face travels the long distance $D$ to a distant concave mirror, is returned by a plane mirror behind it, comes back to the octagonal mirror, is reflected from the opposite face, and is viewed through a telescope.
Working. When the octagonal mirror is stationary, or turning very slowly, the returning light falls on a face that has scarcely moved, and a steady image is seen in the telescope. As the speed is raised the face turns during the time the light is away, and the image shifts and blurs. The speed is then increased until a steady image reappears in exactly the original position. This happens when, in the time the light takes to travel to the distant mirror and back, the octagonal mirror has turned through exactly one eighth of a revolution, bringing the next face into precisely the position the previous one occupied.
Calculation. The light covers the path $2D$ in time
$$t=\frac{2D}{c}$$
If $n$ is the number of revolutions per second at which the image is restored, the time for one eighth of a revolution is
$$t=\frac{1}{8n}$$
Equating the two expressions,
$$ \begin{aligned} \frac{2D}{c} &= \frac{1}{8n} \ c &= 16,n,D \end{aligned} $$
Measuring the distance $D$ by survey and the rotational frequency $n$ stroboscopically gives $c$. Michelson obtained $c=2.99797\times10^{8}\ \text{m/s}$, very close to the accepted value. The merit of the method is that it replaces the very short time interval by a measurement of a rotation rate, which can be determined with great precision.
(b) Coherent sources. Two sources of light are coherent if they emit waves of the same wavelength (and frequency) which maintain a constant phase difference with time, so that the phase relation between them does not change. Only then is the interference pattern steady and observable. Because no two independent sources can keep a fixed phase relation, coherent sources are obtained in practice by deriving two beams from a single source, as by the two slits in Young's experiment, or by a laser.
Equal spacing of the fringes. Let $S_1$ and $S_2$ be two slits a distance $d$ apart, illuminated by monochromatic light of wavelength $\lambda$, and let the screen be at a distance $D$ from them, with $O$ the point on the screen equidistant from the slits. Consider a point $P$ at a distance $x$ from $O$. The path difference between the two waves reaching $P$ is
$$\Delta=\frac{xd}{D}$$
Bright fringes. $P$ is bright when the path difference is a whole number of wavelengths, $\Delta=n\lambda$:
$$\frac{x_nd}{D}=n\lambda\quad\Rightarrow\quad x_n=\frac{n\lambda D}{d},\qquad n=0,1,2,\dots$$
The separation of two consecutive bright fringes is therefore
$$ \begin{aligned} x_{n+1}-x_n &= \frac{(n+1)\lambda D}{d}-\frac{n\lambda D}{d} \ &= \frac{\lambda D}{d} \end{aligned} $$
Dark fringes. $P$ is dark when the path difference is an odd number of half wavelengths, $\Delta=(2n+1)\dfrac{\lambda}{2}$:
$$ \begin{aligned} \frac{x'_nd}{D} &= (2n+1)\frac{\lambda}{2}\quad\Rightarrow\quad x'_n \ &= \frac{(2n+1)\lambda D}{2d} \end{aligned} $$
and the separation of two consecutive dark fringes is
$$ \begin{aligned} x'_{n+1}-x'_n &= \frac{(2n+3)\lambda D}{2d}-\frac{(2n+1)\lambda D}{2d} \ &= \frac{\lambda D}{d} \end{aligned} $$
Both separations come out the same constant quantity, independent of the order $n$:
$$\beta=\frac{\lambda D}{d}$$
Hence the dark and the bright fringes are equally spaced, and this common spacing $\beta$ is the fringe width. A dark fringe lies exactly midway between two bright ones, so the pattern is a set of alternate bright and dark bands of uniform width.
Answer any two numerical questions. (a) The resistance of a conductor at 20 degree C is 3.15 ohm and at 100 degree C is 3.75 ohm. Determine the temperature coefficient of the conductor and resistance of the conductor at 0 degree C. (b) An electron is moving at 10^6 m/s in a direction parallel to an infinitely long straight wire carrying a current of 5A and separated by a perpendicular distance of 10cm in air. Calculate the magnitude of force experienced by the electron. (mu_0 = 4pi10^-7 Tm/A, e = 1.6*10^-19 C). (c) A square coil of 10cm side and with 100 turns is rotated at a uniform speed of 500 revolutions per minute (rpm) about an axis at right angles to a uniform field of 0.5T. Calculate the maximum emf produced in the coil.
(a) The resistance of a conductor varies with temperature as $R_t=R_0(1+\alpha t)$, where $R_0$ is the resistance at $0^\circ\text{C}$ and $\alpha$ the temperature coefficient. The two measurements give
$$ \begin{aligned} 3.15 &= R_0(1+20\alpha) \ 3.75 &= R_0(1+100\alpha) \end{aligned} $$
Dividing the second equation by the first eliminates $R_0$:
$$ \begin{aligned} \frac{3.75}{3.15} &= \frac{1+100\alpha}{1+20\alpha} \ 1.19048(1+20\alpha) &= 1+100\alpha \ 1.19048+23.8095\alpha &= 1+100\alpha \ 0.19048 &= 76.1905,\alpha \ \alpha &= 2.5\times10^{-3}\ ^\circ\text{C}^{-1} \end{aligned} $$
Substituting this back into the first equation,
$$ \begin{aligned} R_0 &= \frac{3.15}{1+20\times2.5\times10^{-3}} \ &= \frac{3.15}{1.05} \ &= 3.0\ \Omega \end{aligned} $$
So the temperature coefficient is $\mathbf{2.5\times10^{-3}\ ^\circ C^{-1}}$ and the resistance at $0^\circ\text{C}$ is $\mathbf{3\ \Omega}$.
(b) The straight wire produces a magnetic field at the position of the electron of magnitude
$$ \begin{aligned} B &= \frac{\mu_0 I}{2\pi a} \ &= \frac{4\pi\times10^{-7}\times5}{2\pi\times0.10} \ &= \frac{2\times10^{-7}\times5}{0.10} \ &= 1\times10^{-5}\ \text{T} \end{aligned} $$
This field circles the wire, so at the electron it points perpendicular to the wire. The electron travels parallel to the wire, so its velocity is at right angles to $\vec{B}$ and $\theta=90^\circ$. The magnetic force on it is
$$ \begin{aligned} F &= evB\sin\theta \ &= 1.6\times10^{-19}\times10^{6}\times1\times10^{-5}\times 1 \ F &= 1.6\times10^{-18}\ \text{N} \end{aligned} $$
The force experienced by the electron is $\mathbf{1.6\times10^{-18}\ N}$, directed perpendicular to both the wire and the velocity, that is towards or away from the wire depending on the sense of the current and the direction of motion.
(c) For a coil rotating in a uniform field the induced emf is $e=NBA\omega\sin\omega t$, whose maximum value is
$$e_0=NBA\omega$$
The area of the square coil is
$$ \begin{aligned} A &= (0.10)^2 \ &= 0.01\ \text{m}^2 \end{aligned} $$
and the angular speed corresponding to $500$ revolutions per minute is
$$ \begin{aligned} \omega &= 2\pi\times\frac{500}{60} \ &= 2\pi\times8.333 \ &= 52.36\ \text{rad/s} \end{aligned} $$
Therefore
$$ \begin{aligned} e_0 &= 100\times0.5\times0.01\times52.36 \ e_0 &= 26.18\ \text{V} \end{aligned} $$
The maximum emf produced in the coil is about $\mathbf{26.2\ V}$.
Answer any two numerical questions. (a) A beam of proton is accelerated through a potential difference of 2KV and then enters a uniform magnetic field which is perpendicular to the direction of proton beam. If the flux density is 0.5T, calculate the radius of the path of the beam described. (mass of proton = 1.710^-27 kg, electronic charge = 1.610^-19 C). (b) Find the frequency of light which ejects electrons from a metal surface are fully stopped by a retarding potential of 3V. The photoelectric effect begins in this metal at frequency of 610^14 Hz. (Planck's constant = 6.610^-34 Js). (c) Assuming that about 200 MeV energy is released per fission of U-235 nuclei, what would be the mass of U-235 consumed per day in the fission reactor of power 1MW?
(a) The work done by the accelerating potential appears as kinetic energy of the proton: $$ \begin{aligned} eV &= \frac{1}{2}mv^{2} \ v &= \sqrt{\frac{2eV}{m}} \ &= \sqrt{\frac{2\times1.6\times10^{-19}\times2000}{1.7\times10^{-27}}} \...
A tuning fork of frequency 512Hz produces sound waves of wavelength 65cm in air at NTP. Calculate the increase in wavelength when the temperature of air is 27 degree C.
At NTP the temperature is $0^\circ\text{C}$, that is $T_0=273\ \text{K}$, and the wavelength is $\lambda_0=65\ \text{cm}=0.65\ \text{m}$. The speed of sound in air there is
$$ \begin{aligned} v_0 &= f\lambda_0 \ &= 512\times0.65 \ &= 332.8\ \text{m/s} \end{aligned} $$
The speed of sound in a gas is proportional to the square root of its absolute temperature, $v\propto\sqrt{T}$. At $27^\circ\text{C}$, that is $T=300\ \text{K}$,
$$ \begin{aligned} \frac{v}{v_0} &= \sqrt{\frac{T}{T_0}} \ &= \sqrt{\frac{300}{273}} \ &= \sqrt{1.0989} \ &= 1.0483 \ v &= 332.8\times1.0483 \ &= 348.87\ \text{m/s} \end{aligned} $$
The frequency of the fork is fixed by the fork itself and does not change with temperature, so the new wavelength is
$$ \begin{aligned} \lambda &= \frac{v}{f} \ &= \frac{348.87}{512} \ &= 0.6814\ \text{m} \ &= 68.14\ \text{cm} \end{aligned} $$
The increase in wavelength is therefore
$$ \begin{aligned} \Delta\lambda &= \lambda-\lambda_0 \ &= 68.14-65 \ &= 3.14\ \text{cm} \end{aligned} $$
The wavelength increases by about $\mathbf{3.1\ cm}$.
(The same result follows directly from $\lambda\propto\sqrt{T}$ at constant frequency, since $\Delta\lambda=\lambda_0\left(\sqrt{300/273}-1\right)=65\times0.0483=3.14\ \text{cm}$.)
A diffraction grating has 500 lines per mm and is illuminated normally with monochromatic light of wavelength 589nm. Calculate the angle to the normal at which second order diffraction maximum is observed and the number of diffraction maxima obtained.
The grating element is the distance between the centres of two neighbouring slits. With $500$ lines per millimetre,
$$ \begin{aligned} d &= \frac{1\ \text{mm}}{500} \ &= \frac{10^{-3}}{500} \ &= 2\times10^{-6}\ \text{m} \end{aligned} $$
and the wavelength is $\lambda=589\ \text{nm}=589\times10^{-9}\ \text{m}$.
For normal incidence the maxima satisfy the grating equation
$$d\sin\theta=n\lambda$$
Angle of the second order maximum. Putting $n=2$,
$$ \begin{aligned} \sin\theta_2 &= \frac{2\lambda}{d} \ &= \frac{2\times589\times10^{-9}}{2\times10^{-6}} \ &= 0.589 \ \theta_2 &= \sin^{-1}(0.589) \ &= 36.1^\circ \end{aligned} $$
The second order maximum is observed at about $\mathbf{36.1^\circ}$ to the normal.
Number of maxima. The largest possible order is set by $\sin\theta\le1$, so
$$ \begin{aligned} n_{max}\le\frac{d}{\lambda} &= \frac{2\times10^{-6}}{589\times10^{-9}} \ &= 3.395 \end{aligned} $$
Since the order must be a whole number, the highest order actually observed is $n=3$ (the value $n=3.395$ would need $\sin\theta>1$ for $n=4$, which is impossible). Counting the central maximum and the orders on both sides of it,
$$n=0,\ \pm1,\ \pm2,\ \pm3$$
gives $1+2\times3=\mathbf{7}$ diffraction maxima in all.
(a) The free electrons in a metal are in continuous random thermal motion, travelling in all directions with speeds of the order of . Because the directions are completely random, as many electrons cross any section one...
(a) In the ground state the electron occupies the orbit of lowest possible energy, , with . According to Bohr's postulate an electron revolving in a stationary orbit does not radiate energy; radiation accompanies only a transition to a lower orbit. Since there is no orbit below , the electron has no lower state available to fall into and so cannot lose energy. Having no way to radiate, the atom remains in that state indefinitely, which is why hydrogen is stable in the ground state.
(b) The side is joined to the positive terminal and the side to the negative terminal, through a rheostat and an ammeter, with a voltmeter across the diode:
+|i----[ rheostat ]----(A)----|>|----|
| p n |
=== battery (V) |
| |
-|-----------------------------------
The characteristic rises very slowly until the knee (about for germanium and for silicon), beyond which the current increases steeply and almost linearly for a small increase of voltage:
I | /
| /
| /
| _/
|______/
0 0.7 V
(c) An -particle is a helium nucleus , so emitting one removes two protons and two neutrons:
The daughter nucleus therefore has an atomic number smaller by 2 and a mass number smaller by 4 than the parent. Because changes, the daughter is a different element, shifted two places to the left in the periodic table.
(d) When free nucleons combine to form a nucleus, some of their mass is converted into the energy released in binding them together. This missing mass is the mass defect
and the energy equivalent is the binding energy of the nucleus, which has been radiated away during formation. Since that much mass has left the system as energy, what remains, the mass of the nucleus, is slightly less than the sum of the masses of the constituent nucleons taken separately.
(e) The ozone layer in the stratosphere absorbs most of the harmful ultraviolet radiation from the Sun. Chlorofluorocarbons (CFCs) released from refrigerators, air conditioners, aerosol sprays and foam, together with halons and nitrogen oxides, drift up into the stratosphere where ultraviolet light breaks them up and frees chlorine atoms:
The chlorine atom is regenerated at the end of the second step, so a single atom acts as a catalyst and destroys thousands of ozone molecules. The result is a thinning of the layer, most severe over Antarctica where the "ozone hole" appears each spring. The increased ultraviolet reaching the ground causes skin cancer, cataracts and weakened immunity in humans, damages crops and phytoplankton, and so disturbs the food chain.
(f) Hubble's law: the galaxies are receding from us with a speed directly proportional to their distance,
where is Hubble's constant (about ).
Significance of : it measures the present rate of expansion of the universe, and its reciprocal gives an estimate of the age of the universe, , since a galaxy now at distance receding uniformly at would have taken to get there. It also fixes the Hubble radius , the scale of the observable universe.
(a) Principle of the potentiometer. When a steady current flows through a wire of uniform cross section and uniform material, the potential difference across any portion of the wire is directly proportional to the length of that portion:
where is the potential gradient, the fall of potential per unit length. This follows because and , so with , and all constant, .
Internal resistance of a cell. The potentiometer wire is fed by a driver cell through a key, and the cell under test, of emf and internal resistance , is connected with its positive terminal to and through a galvanometer to the jockey. A resistance box with a key is joined across the test cell:
+|i----[ key ]----------------------------
| |
=== driver |
| |
A o======================================o B (uniform wire)
| |
| jockey ---- (G) ---- +|i---- test cell (E, r)
| |
+---------------------------------------+
|
[ R ] [ key K ] across the test cell
With the key open no current is drawn from the test cell, so the balancing length measures its full emf:
With closed, the cell sends current through , and the balance point now measures only the terminal potential difference :
For the closed circuit and , so
which rearranges to
Measuring , and therefore gives the internal resistance. The method is a null method, so at balance no current is drawn from the cell being measured and the result is free from the error that a voltmeter would introduce.
(b) Thermoelectric effect. When two different metals are joined to form a closed circuit and the two junctions are kept at different temperatures, an emf is set up and a current flows round the circuit. This is the thermoelectric or Seebeck effect, and the pair of metals is a thermocouple. The effect is reversible: interchanging the hot and cold junctions reverses the current.
Variation with the temperature of the hot junction. With the cold junction held at and the hot junction at , the thermo emf is not linear but parabolic:
where and are constants for the pair. As rises from zero the emf increases, but at a falling rate, until it reaches a maximum at the neutral temperature , which for a given couple is fixed (about for a copper-iron couple). Beyond the emf decreases, falls back to zero at the temperature of inversion , and on further heating reverses its direction. Since the curve is a parabola about , the neutral temperature lies midway between the cold junction temperature and the inversion temperature:
so with the cold junction at , . Differentiating gives the thermoelectric power , which is zero at the neutral temperature, as the maximum requires.
(c) Magnetic elements of the earth. These are the three quantities needed to specify the earth's magnetic field completely at a place:
Proof of the relation. The true dip is observed only when the dip circle is set in the magnetic meridian. If instead the plane of the dip circle is turned through an angle from the magnetic meridian, the vertical component is unaffected, but only the resolved part of the horizontal component acts in that plane, namely . The apparent dip in that plane satisfies
Let the second dip circle be set in the plane perpendicular to the first, so its plane makes an angle with the magnetic meridian. Then
Taking reciprocals and squaring both results,
Adding them,
But for the true dip , so . Hence
which is the required result. It is useful in practice because it gives the true dip from two measurements in mutually perpendicular planes, without having to locate the magnetic meridian first.
(d) Series LCR circuit. Let an alternating emf drive a current through , and in series. The same current passes through all three, so the current is taken as the reference in the phasor diagram.
The potential difference across the resistance is in phase with the current, that across the inductor leads it by , and that across the capacitor lags it by :
Since and are opposite in phase, they combine to a single phasor of magnitude at right angles to . The applied voltage is the resultant of these two perpendicular phasors:
The impedance is the ratio of the applied voltage to the current, so
The phase angle between the applied voltage and the current is the angle of that resultant with :
If the circuit is inductive and the current lags the voltage; if it is capacitive and the current leads; and if then and , which is the condition of resonance, when the current is a maximum.
(a) He-Ne laser. Construction. A narrow quartz discharge tube about long and in diameter is filled with a mixture of helium and neon in the ratio about at a total pressure of roughly ...
(a) Michelson's rotating mirror method. Michelson measured the speed of light between Mount Wilson and Mount San Antonio in California, a distance of about accurately surveyed.
Arrangement. Light from an intense source falls on one face of an octagonal mirror (eight plane mirrors on the faces of a regular octagon) that can be spun at a controlled, measurable rate. The beam reflected from one face travels the long distance to a distant concave mirror, is returned by a plane mirror behind it, comes back to the octagonal mirror, is reflected from the opposite face, and is viewed through a telescope.
Working. When the octagonal mirror is stationary, or turning very slowly, the returning light falls on a face that has scarcely moved, and a steady image is seen in the telescope. As the speed is raised the face turns during the time the light is away, and the image shifts and blurs. The speed is then increased until a steady image reappears in exactly the original position. This happens when, in the time the light takes to travel to the distant mirror and back, the octagonal mirror has turned through exactly one eighth of a revolution, bringing the next face into precisely the position the previous one occupied.
Calculation. The light covers the path in time
If is the number of revolutions per second at which the image is restored, the time for one eighth of a revolution is
Equating the two expressions,
Measuring the distance by survey and the rotational frequency stroboscopically gives . Michelson obtained , very close to the accepted value. The merit of the method is that it replaces the very short time interval by a measurement of a rotation rate, which can be determined with great precision.
(b) Coherent sources. Two sources of light are coherent if they emit waves of the same wavelength (and frequency) which maintain a constant phase difference with time, so that the phase relation between them does not change. Only then is the interference pattern steady and observable. Because no two independent sources can keep a fixed phase relation, coherent sources are obtained in practice by deriving two beams from a single source, as by the two slits in Young's experiment, or by a laser.
Equal spacing of the fringes. Let and be two slits a distance apart, illuminated by monochromatic light of wavelength , and let the screen be at a distance from them, with the point on the screen equidistant from the slits. Consider a point at a distance from . The path difference between the two waves reaching is
Bright fringes. is bright when the path difference is a whole number of wavelengths, :
The separation of two consecutive bright fringes is therefore
Dark fringes. is dark when the path difference is an odd number of half wavelengths, :
and the separation of two consecutive dark fringes is
Both separations come out the same constant quantity, independent of the order :
Hence the dark and the bright fringes are equally spaced, and this common spacing is the fringe width. A dark fringe lies exactly midway between two bright ones, so the pattern is a set of alternate bright and dark bands of uniform width.
(a) The resistance of a conductor varies with temperature as , where is the resistance at and the temperature coefficient. The two measurements give
Dividing the second equation by the first eliminates :
Substituting this back into the first equation,
So the temperature coefficient is and the resistance at is .
(b) The straight wire produces a magnetic field at the position of the electron of magnitude
This field circles the wire, so at the electron it points perpendicular to the wire. The electron travels parallel to the wire, so its velocity is at right angles to and . The magnetic force on it is
The force experienced by the electron is , directed perpendicular to both the wire and the velocity, that is towards or away from the wire depending on the sense of the current and the direction of motion.
(c) For a coil rotating in a uniform field the induced emf is , whose maximum value is
The area of the square coil is
and the angular speed corresponding to revolutions per minute is
Therefore
The maximum emf produced in the coil is about .
At NTP the temperature is , that is , and the wavelength is . The speed of sound in air there is
The speed of sound in a gas is proportional to the square root of its absolute temperature, . At , that is ,
The frequency of the fork is fixed by the fork itself and does not change with temperature, so the new wavelength is
The increase in wavelength is therefore
The wavelength increases by about .
(The same result follows directly from at constant frequency, since .)
The grating element is the distance between the centres of two neighbouring slits. With lines per millimetre,
and the wavelength is .
For normal incidence the maxima satisfy the grating equation
Angle of the second order maximum. Putting ,
The second order maximum is observed at about to the normal.
Number of maxima. The largest possible order is set by , so
Since the order must be a whole number, the highest order actually observed is (the value would need for , which is impossible). Counting the central maximum and the orders on both sides of it,
gives diffraction maxima in all.