NEB Class 12 · Past paper
The complete NEB Class 12 2070 exam paper for Physics, all 12 questions with solved model answers.
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Answer, in brief, any four questions: (a) On connecting a wire to a battery the current is initially larger then becomes steady, though emf is unchanged. Explain. (b) Why prefer a potentiometer over a voltmeter to measure emf? (c) Why is paramagnetic susceptibility strongly temperature dependent but diamagnetic nearly independent? (d) A bar magnet dropped through a copper ring: is its acceleration equal to g? (e) State and explain Faraday's laws of electrolysis. (f) Why use an inductor rather than a resistor to limit current in fluorescent lights?
(a) When the wire is first connected it is cold, so its resistance is low and the current $I = V/R$ is large. That current heats the wire, and for a metal the resistance rises with temperature, so the current falls. It becomes steady onc...
Answer, in brief, any four questions: (a) A charged particle moves with constant velocity; if B = 0 there, can we conclude E = 0? (b) Why is the wave nature of particles not observable in daily life? (c) What is a logic gate? Give the symbol and truth table of a two-input AND gate. (d) Does a nucleus contain electrons? (e) What is global warming? (f) Show that a proton contains quarks uud.
(a) Since the particle moves with constant velocity, the net force on it is zero. With $\vec B = 0$ the magnetic force vanishes, so the electric force $q\vec E$ must also be zero, and because $q \ne 0$ this forces $\vec E = 0$. So yes, w...
Answer, in brief, any one question: (a) If the tension in a string is increased four times, by what factor does the wave velocity change? (b) An empty vessel sounds more than a filled one when struck. Why?
(a) The wave velocity on a stretched string is $v = \sqrt{\dfrac{T}{\mu}}$, so it depends on tension as $v \propto \sqrt{T}$. If the tension is increased four times, $T \to 4T$, then
$$v \to \sqrt{4},v = 2v$$
so the velocity simply doubles.
(b) When an empty vessel is struck, its walls and the large column of air inside are free to vibrate and resonate, so the sound is sustained and comes out louder and longer. In a filled vessel the liquid damps the vibrations of the walls and cuts down the vibrating air column, so the sound dies away quickly and is much fainter.
Answer, in brief, any one question: (a) Radio waves diffract around buildings but light waves do not. Why? (b) What is polarizing angle? Does it depend on the wavelength of light?
(a) Appreciable diffraction occurs only when the size of the obstacle or aperture is comparable to the wavelength. Radio waves have wavelengths of the order of metres, which is comparable to the size of buildings, so they bend appreciabl...
Answer any three questions: (a) What is a Wheatstone bridge? Use Kirchhoff's laws to obtain its balance condition. (b) Define Seebeck effect; discuss variation of thermo-emf with hot-junction temperature. (c) Find the force per unit length between two long parallel current-carrying conductors and define one ampere. (d) An AC passes through a resistor and inductor in series; derive the current and the phase relation.
(a) A Wheatstone bridge is an arrangement of four resistances $P, Q, R, S$ forming a quadrilateral, with a galvanometer connected across one diagonal and a cell across the other.
At balance no current flows through the galvanometer. Applying Kirchhoff's voltage law to the two loops with $I_g = 0$ gives $I_1 P = I_2 R$ for the upper loop and $I_1 Q = I_2 S$ for the lower loop. Dividing one relation by the other, the currents cancel and we are left with the balance condition,
$$\frac{P}{Q} = \frac{R}{S}$$
(b) The Seebeck effect is the appearance of a thermo-emf, and hence a current, in a closed circuit of two dissimilar metals when its two junctions are held at different temperatures.
Keeping the cold junction fixed and raising the hot-junction temperature, the thermo-emf first increases, reaches a maximum at the neutral temperature $\theta_n$, then decreases, and finally becomes zero and reverses at the temperature of inversion $\theta_i$. This variation follows a parabolic law,
$$E = a\theta + \tfrac12 b\theta^2$$
(c) Conductor 1 carrying current $I_1$ sets up a magnetic field at the location of conductor 2 (carrying $I_2$ at a distance $d$),
$$B_1 = \frac{\mu_0 I_1}{2\pi d}$$
This field exerts a force on conductor 2, so the force per unit length is
$$ \begin{aligned} \frac{F}{l} &= B_1 I_2 \ &= \frac{\mu_0 I_1 I_2}{2\pi d} \end{aligned} $$
One ampere is defined as that steady current which, flowing in two infinitely long parallel conductors placed 1 m apart in vacuum, produces a force of $2\times10^{-7}\ \text{N/m}$ between them.
(d) Let the current through the series R-L combination be $I = I_0\sin\omega t$. The voltage across the resistor is in phase with the current, $V_R = I_0 R\sin\omega t$, while the voltage across the inductor leads it by $90^\circ$, $V_L = I_0 X_L\sin(\omega t + 90^\circ)$. Adding these two out-of-phase voltages, the applied voltage is
$$ \begin{aligned} V &= V_0\sin(\omega t + \phi) \ \qquad V_0 &= I_0\sqrt{R^2 + X_L^2} \ \quad X_L &= \omega L \end{aligned} $$
so the impedance is $Z = \sqrt{R^2 + \omega^2L^2}$. Because of the inductor, the voltage leads the current by the phase angle
$$\phi = \tan^{-1}!\frac{\omega L}{R}$$
Answer any three questions: (a) What is a p-n junction diode? Explain its forward and reverse characteristics. (b) Describe the construction and working of a Helium-Neon laser. (c) State the laws of radioactivity and derive the decay equation. (d) What is nuclear fusion? Discuss the sources of energy released.
(a) A p-n junction diode is a single crystal with a p-type region joined to an n-type region, and at the junction a depletion layer forms with a potential barrier across it.
Under forward bias (p side to +), the barrier is lowered so that majority carriers cross the junction, and the current rises sharply once the knee voltage (about 0.7 V for silicon) is exceeded. Under reverse bias (p side to $-$), the barrier widens and only a tiny saturation (leakage) current flows until breakdown sets in. The resulting I-V curve therefore shows a steep forward branch and a nearly flat reverse branch.
(b) A He-Ne laser is a gas laser that uses a mixture of helium and neon (roughly 10:1) inside a discharge tube fitted with mirrors at each end, one fully reflecting and one partially reflecting.
In operation, an electric discharge excites the helium atoms to metastable states, and these transfer their energy by collision to neon atoms, raising them to a higher level and building a population inversion in the neon. Stimulated emission at 632.8 nm is then amplified by repeated reflection between the mirrors (the optical resonator), producing a coherent, monochromatic, collimated red beam.
(c) The laws of radioactivity state, first, that decay is a spontaneous and random nuclear process unaffected by external conditions, and second, that the rate of decay is proportional to the number of undecayed nuclei present. Writing this second law as a differential equation and separating the variables,
$$ \begin{aligned} -\frac{dN}{dt} &= \lambda N \quad\Rightarrow\quad \int\frac{dN}{N} \ &= -\lambda\int dt \end{aligned} $$
Integrating both sides gives the decay equation,
$$\boxed{N = N_0 e^{-\lambda t}}$$
(d) Nuclear fusion is the combining of light nuclei into a heavier nucleus, with a release of energy. The mass of the product is less than the total mass of the reactants, and this mass defect $\Delta m$ appears as energy through $E = \Delta m c^2$. This is the source of the Sun's energy, where the proton-proton chain fuses hydrogen into helium and the strong nuclear binding of the helium nucleus releases enormous energy per reaction.
Answer any one question: (a) How is a progressive wave different from a stationary wave? Derive the progressive wave equation. (b) What are beats? Show the number of beats per second equals the difference of the two frequencies.
(a) A progressive wave transfers energy and advances through the medium, with every particle vibrating at the same amplitude but with a progressive phase lag from one point to the next. A stationary wave, by contrast, is confined between boundaries with fixed nodes and antinodes and transfers no net energy.
To find its equation, take a particle at $x = 0$ executing $y = a\sin\omega t$. A particle a distance $x$ further along lags in phase by $\dfrac{2\pi}{\lambda}x$, so
$$ \begin{aligned} y &= a\sin!\left(\omega t - \frac{2\pi}{\lambda}x\right) \ &= a\sin\frac{2\pi}{\lambda}(vt - x) \end{aligned} $$
which is the progressive wave equation.
(b) Beats are the periodic rise and fall of sound intensity produced when two notes of slightly different frequencies ($f_1, f_2$) are sounded together. Superposing $y_1 = a\sin 2\pi f_1 t$ and $y_2 = a\sin 2\pi f_2 t$ gives
$$y = 2a\cos!\left[2\pi\frac{f_1 - f_2}{2}t\right]\sin!\left[2\pi\frac{f_1+f_2}{2}t\right]$$
The amplitude term $\cos$ varies at frequency $\dfrac{f_1-f_2}{2}$, and since intensity maxima occur twice per amplitude cycle, the beat frequency is $|f_1 - f_2|$.
Answer any one question: (a) State Huygen's principle and use it to prove Snell's law. (b) What are coherent sources? Deduce the fringe width in Young's double-slit experiment.
(a) Huygen's principle states that every point on a wavefront acts as a source of secondary wavelets that spread out at the wave speed, and the new wavefront is their common tangent, or envelope.
To get Snell's law, consider a plane wavefront striking the interface at angle $i$. While the wavelet in medium 1 travels $BC = v_1 t$, the wavelet in medium 2 travels $AD = v_2 t$. From the geometry of the two triangles, $\sin i = \dfrac{v_1 t}{AC}$ and $\sin r = \dfrac{v_2 t}{AC}$, so dividing one by the other,
$$ \begin{aligned} \frac{\sin i}{\sin r} &= \frac{v_1}{v_2} \ &= {}_1\mu_2 \ &= \text{constant} \end{aligned} $$
which is Snell's law.
(b) Coherent sources are sources that emit waves of the same frequency with a constant phase difference. In Young's double-slit experiment, with the slits separated by $d$ and the screen at distance $D$, the path difference at a point a distance $y$ from the centre is $\dfrac{yd}{D}$. Bright fringes appear wherever this path difference is a whole number of wavelengths, $\dfrac{yd}{D} = n\lambda$. Taking the difference between successive bright fringes gives the fringe width,
$$\boxed{\beta = \frac{\lambda D}{d}}$$
Answer any two numerical questions: (a) A voltmeter coil has resistance 50 ohm with a 1.15 k-ohm series resistor, reading up to 12 V. Used as an ammeter to read up to 2 A, find the shunt resistance. (b) A copper slab 2 mm thick, 1.50 cm wide, in B = 0.40 T, carrying 75 A, develops a Hall voltage 0.81 uV. Find the mobile-electron concentration. (c) A 100-turn rectangular coil 15x10 cm rotates at 300 rpm in B = 0.6 T. Find the maximum emf.
(a) As a voltmeter the coil has resistance $R_{coil} = 50\ \Omega$ with a series resistor $R_{series} = 1.15\ \text{k}\Omega = 1150\ \Omega$ and reads full scale at $12\ \text{V}$. The full-scale current through the movement follows from Ohm's law across the whole voltmeter branch.
$$ \begin{aligned} I_g &= \frac{V}{R_{coil}+R_{series}} \ &= \frac{12}{50+1150} \ &= 0.01\ \text{A} \end{aligned} $$
To use it as an ammeter reading up to $I = 2\ \text{A}$, a shunt $S$ is placed in parallel with the coil and carries the excess current $(I - I_g)$. Since the coil and shunt share the same voltage, $I_g R_{coil} = (I - I_g)S$, which gives
$$ \begin{aligned} S &= \frac{I_g R_{coil}}{I - I_g} \ &= \frac{0.01\times50}{2 - 0.01} \ &= 0.251\ \Omega \end{aligned} $$
So the required shunt resistance is about $0.251\ \Omega$.
(b) For the copper slab the thickness along the field is $t = 2\ \text{mm} = 2\times10^{-3}\ \text{m}$, the width is $1.50\ \text{cm}$, the field is $B = 0.40\ \text{T}$, the current is $I = 75\ \text{A}$, and the measured Hall voltage is $V_H = 0.81\ \mu\text{V}$, with electron charge $e = 1.6\times10^{-19}\ \text{C}$. The Hall voltage is related to the carrier concentration by $V_H = \dfrac{BI}{n e t}$, so rearranging for $n$,
$$n = \frac{BI}{V_H e t}$$
Substituting the values,
$$ \begin{aligned} n &= \frac{0.40\times75}{(0.81\times10^{-6})(1.6\times10^{-19})(2\times10^{-3})} \ &= \frac{30}{2.592\times10^{-28}} \ &= 1.16\times10^{29}\ \text{m}^{-3} \end{aligned} $$
So the mobile-electron concentration is about $1.16\times10^{29}\ \text{m}^{-3}$.
(c) The coil has $N = 100$ turns, sides $15\times10\ \text{cm}$, spins at $300\ \text{rpm}$ in a field $B = 0.6\ \text{T}$. First find the angular frequency and the area.
$$ \begin{aligned} \omega &= 2\pi\frac{300}{60} \ &= 31.42\ \text{rad/s}, \ \qquad A &= 0.15\times0.10 \ &= 0.015\ \text{m}^2 \end{aligned} $$
The peak emf of a rotating coil is $E_0 = NBA\omega$, so
$$ \begin{aligned} E_0 &= 100\times0.6\times0.015\times31.42 \ &= 28.3\ \text{V} \end{aligned} $$
The maximum emf is therefore about $28.3\ \text{V}$.
Answer any two numerical questions: (a) An electron beam at 10^7 m/s enters midway between horizontal plates parallel to them; each plate 5 cm long, 2 cm apart, PD 90 V. Find the exit velocity as it just grazes the positive plate. (e/m = 1.8x10^11 C/kg) (b) UV of 400 nm gives KE_max 1.10 eV; find KE_max for 300 nm on the same metal. (c) Calculate the binding energy per nucleon of Fe-56 (mp = 1.007825, mn = 1.008665, M = 55.934939 amu).
(a) The electron enters midway between the plates with a horizontal speed $v_x = 10^7\ \text{m/s}$, and the plates are $5\ \text{cm}$ long, $2\ \text{cm}$ apart, held at a p.d. of $90\ \text{V}$, with $e/m = 1.8\times10^{11}\ \text{C/kg}$. The uniform field between the plates is
$$ \begin{aligned} E &= \frac{V}{d} \ &= \frac{90}{0.02} \ &= 4500\ \text{V/m} \end{aligned} $$
and this gives the electron a vertical acceleration
$$ \begin{aligned} a &= \frac{e}{m}E \ &= 1.8\times10^{11}\times4500 \ &= 8.1\times10^{14}\ \text{m/s}^2. \end{aligned} $$
The time it spends between the plates is set by the horizontal motion,
$$ \begin{aligned} t &= \frac{L}{v_x} \ &= \frac{0.05}{10^{7}} \ &= 5\times10^{-9}\ \text{s}, \end{aligned} $$
so the vertical velocity it gains is
$$ \begin{aligned} v_y &= at \ &= 8.1\times10^{14}\times5\times10^{-9} \ &= 4.05\times10^{6}\ \text{m/s}. \end{aligned} $$
As a check, the vertical deflection is $y = \tfrac12 at^2 = 0.0101\ \text{m}$, about half the gap, so the electron does just graze the positive plate. Combining the two components, the exit speed is
$$ \begin{aligned} v &= \sqrt{v_x^2 + v_y^2} \ &= \sqrt{(10^7)^2 + (4.05\times10^6)^2} \ &= 1.08\times10^{7}\ \text{m/s}. \end{aligned} $$
So the electron leaves the plates with a speed of about $1.08\times10^{7}$ m/s.
(b) Ultraviolet light of $400\ \text{nm}$ gives a maximum kinetic energy of $1.10\ \text{eV}$, and we want the maximum kinetic energy for $300\ \text{nm}$ on the same metal, taking $hc = 1240\ \text{eV nm}$. The photon energy at 400 nm is
$$ \begin{aligned} E_{400} &= \frac{1240}{400} \ &= 3.10\ \text{eV}, \end{aligned} $$
so the work function of the metal is
$$ \begin{aligned} \phi &= 3.10 - 1.10 \ &= 2.00\ \text{eV}. \end{aligned} $$
At 300 nm the photon energy is larger,
$$ \begin{aligned} E_{300} &= \frac{1240}{300} \ &= 4.133\ \text{eV}, \end{aligned} $$
and subtracting the same work function gives
$$ \begin{aligned} KE_{max}' &= 4.133 - 2.00 \ &= 2.13\ \text{eV}. \end{aligned} $$
The maximum kinetic energy is therefore about $2.13$ eV.
(c) For Fe-56 we have $Z = 26$ protons and $N = 30$ neutrons, with $m_p = 1.007825$, $m_n = 1.008665$ and nuclear mass $M = 55.934939\ \text{amu}$. The mass defect is the difference between the summed masses of the free nucleons and the actual mass,
$$ \begin{aligned} \Delta m &= 26(1.007825) + 30(1.008665) - 55.934939 \ &= 56.46340 - 55.934939 \ &= 0.528461\ \text{amu}. \end{aligned} $$
Converting this to energy gives the binding energy,
$$ \begin{aligned} BE &= 0.528461\times931.5 \ &= 492.3\ \text{MeV}, \end{aligned} $$
and dividing by the 56 nucleons,
$$ \begin{aligned} \frac{BE}{A} &= \frac{492.3}{56} \ &= 8.79\ \text{MeV}. \end{aligned} $$
So the binding energy per nucleon of Fe-56 is about $8.79$ MeV.
In a resonance-tube experiment the first and second resonances are at 17 cm and 52.6 cm with a 512 Hz fork at 27 C. Calculate the velocity of sound in air at 0 C and the end correction.
We are given the first and second resonances at $l_1 = 17\ \text{cm}$ and $l_2 = 52.6\ \text{cm}$ with a fork of frequency $f = 512\ \text{Hz}$ at $27^\circ$C.
The distance between successive resonances equals half a wavelength, so
$$ \begin{aligned} \frac{\lambda}{2} &= 52.6 - 17 \ &= 35.6\ \text{cm} \quad\Rightarrow\quad \lambda \ &= 71.2\ \text{cm} \ &= 0.712\ \text{m} \end{aligned} $$
The velocity of sound at $27^\circ$C is then
$$ \begin{aligned} v_{27} &= f\lambda \ &= 512\times0.712 \ &= 364.5\ \text{m/s} \end{aligned} $$
The end correction follows from the two resonance lengths,
$$ \begin{aligned} e &= \frac{l_2 - 3l_1}{2} \ &= \frac{52.6 - 3(17)}{2} \ &= \frac{1.6}{2} \ &= 0.8\ \text{cm} \end{aligned} $$
Since velocity varies with absolute temperature as $v \propto \sqrt{T}$, the velocity at $0^\circ$C is
$$ \begin{aligned} v_0 &= v_{27}\sqrt{\frac{273}{300}} \ &= 364.5\times0.9539 = 347.7\ \text{m/s} \end{aligned} $$
So the velocity of sound at $0^\circ$C is about $347.7\ \text{m/s}$ and the end correction is $0.8\ \text{cm}$.
Calculate the polarizing angle for light travelling from water (refractive index 1.33) to glass (refractive index 1.53).
Light travels from water ($\mu_1 = 1.33$) into glass ($\mu_2 = 1.53$), so the polarizing angle follows from Brewster's law.
$$ \begin{aligned} \tan\theta_p &= \frac{\mu_2}{\mu_1} \ &= \frac{1.53}{1.33} \ &= 1.150 \end{aligned} $$
Taking the inverse tangent,
$$ \begin{aligned} \theta_p &= \tan^{-1}(1.150) \ &= 49.0^\circ \end{aligned} $$
The polarizing angle is therefore about $49^\circ$.
(a) When the wire is first connected it is cold, so its resistance is low and the current is large. That current heats the wire, and for a metal the resistance rises with temperature, so the current falls. It becomes steady onc...
(a) Since the particle moves with constant velocity, the net force on it is zero. With the magnetic force vanishes, so the electric force must also be zero, and because this forces . So yes, w...
(a) The wave velocity on a stretched string is , so it depends on tension as . If the tension is increased four times, , then
so the velocity simply doubles.
(b) When an empty vessel is struck, its walls and the large column of air inside are free to vibrate and resonate, so the sound is sustained and comes out louder and longer. In a filled vessel the liquid damps the vibrations of the walls and cuts down the vibrating air column, so the sound dies away quickly and is much fainter.
(a) A Wheatstone bridge is an arrangement of four resistances forming a quadrilateral, with a galvanometer connected across one diagonal and a cell across the other.
At balance no current flows through the galvanometer. Applying Kirchhoff's voltage law to the two loops with gives for the upper loop and for the lower loop. Dividing one relation by the other, the currents cancel and we are left with the balance condition,
(b) The Seebeck effect is the appearance of a thermo-emf, and hence a current, in a closed circuit of two dissimilar metals when its two junctions are held at different temperatures.
Keeping the cold junction fixed and raising the hot-junction temperature, the thermo-emf first increases, reaches a maximum at the neutral temperature , then decreases, and finally becomes zero and reverses at the temperature of inversion . This variation follows a parabolic law,
(c) Conductor 1 carrying current sets up a magnetic field at the location of conductor 2 (carrying at a distance ),
This field exerts a force on conductor 2, so the force per unit length is
One ampere is defined as that steady current which, flowing in two infinitely long parallel conductors placed 1 m apart in vacuum, produces a force of between them.
(d) Let the current through the series R-L combination be . The voltage across the resistor is in phase with the current, , while the voltage across the inductor leads it by , . Adding these two out-of-phase voltages, the applied voltage is
so the impedance is . Because of the inductor, the voltage leads the current by the phase angle
(a) A p-n junction diode is a single crystal with a p-type region joined to an n-type region, and at the junction a depletion layer forms with a potential barrier across it.
Under forward bias (p side to +), the barrier is lowered so that majority carriers cross the junction, and the current rises sharply once the knee voltage (about 0.7 V for silicon) is exceeded. Under reverse bias (p side to ), the barrier widens and only a tiny saturation (leakage) current flows until breakdown sets in. The resulting I-V curve therefore shows a steep forward branch and a nearly flat reverse branch.
(b) A He-Ne laser is a gas laser that uses a mixture of helium and neon (roughly 10:1) inside a discharge tube fitted with mirrors at each end, one fully reflecting and one partially reflecting.
In operation, an electric discharge excites the helium atoms to metastable states, and these transfer their energy by collision to neon atoms, raising them to a higher level and building a population inversion in the neon. Stimulated emission at 632.8 nm is then amplified by repeated reflection between the mirrors (the optical resonator), producing a coherent, monochromatic, collimated red beam.
(c) The laws of radioactivity state, first, that decay is a spontaneous and random nuclear process unaffected by external conditions, and second, that the rate of decay is proportional to the number of undecayed nuclei present. Writing this second law as a differential equation and separating the variables,
Integrating both sides gives the decay equation,
(d) Nuclear fusion is the combining of light nuclei into a heavier nucleus, with a release of energy. The mass of the product is less than the total mass of the reactants, and this mass defect appears as energy through . This is the source of the Sun's energy, where the proton-proton chain fuses hydrogen into helium and the strong nuclear binding of the helium nucleus releases enormous energy per reaction.
(a) A progressive wave transfers energy and advances through the medium, with every particle vibrating at the same amplitude but with a progressive phase lag from one point to the next. A stationary wave, by contrast, is confined between boundaries with fixed nodes and antinodes and transfers no net energy.
To find its equation, take a particle at executing . A particle a distance further along lags in phase by , so
which is the progressive wave equation.
(b) Beats are the periodic rise and fall of sound intensity produced when two notes of slightly different frequencies () are sounded together. Superposing and gives
The amplitude term varies at frequency , and since intensity maxima occur twice per amplitude cycle, the beat frequency is .
(a) Huygen's principle states that every point on a wavefront acts as a source of secondary wavelets that spread out at the wave speed, and the new wavefront is their common tangent, or envelope.
To get Snell's law, consider a plane wavefront striking the interface at angle . While the wavelet in medium 1 travels , the wavelet in medium 2 travels . From the geometry of the two triangles, and , so dividing one by the other,
which is Snell's law.
(b) Coherent sources are sources that emit waves of the same frequency with a constant phase difference. In Young's double-slit experiment, with the slits separated by and the screen at distance , the path difference at a point a distance from the centre is . Bright fringes appear wherever this path difference is a whole number of wavelengths, . Taking the difference between successive bright fringes gives the fringe width,
(a) As a voltmeter the coil has resistance with a series resistor and reads full scale at . The full-scale current through the movement follows from Ohm's law across the whole voltmeter branch.
To use it as an ammeter reading up to , a shunt is placed in parallel with the coil and carries the excess current . Since the coil and shunt share the same voltage, , which gives
So the required shunt resistance is about .
(b) For the copper slab the thickness along the field is , the width is , the field is , the current is , and the measured Hall voltage is , with electron charge . The Hall voltage is related to the carrier concentration by , so rearranging for ,
Substituting the values,
So the mobile-electron concentration is about .
(c) The coil has turns, sides , spins at in a field . First find the angular frequency and the area.
The peak emf of a rotating coil is , so
The maximum emf is therefore about .
(a) The electron enters midway between the plates with a horizontal speed , and the plates are long, apart, held at a p.d. of , with . The uniform field between the plates is
and this gives the electron a vertical acceleration
The time it spends between the plates is set by the horizontal motion,
so the vertical velocity it gains is
As a check, the vertical deflection is , about half the gap, so the electron does just graze the positive plate. Combining the two components, the exit speed is
So the electron leaves the plates with a speed of about m/s.
(b) Ultraviolet light of gives a maximum kinetic energy of , and we want the maximum kinetic energy for on the same metal, taking . The photon energy at 400 nm is
so the work function of the metal is
At 300 nm the photon energy is larger,
and subtracting the same work function gives
The maximum kinetic energy is therefore about eV.
(c) For Fe-56 we have protons and neutrons, with , and nuclear mass . The mass defect is the difference between the summed masses of the free nucleons and the actual mass,
Converting this to energy gives the binding energy,
and dividing by the 56 nucleons,
So the binding energy per nucleon of Fe-56 is about MeV.
We are given the first and second resonances at and with a fork of frequency at C.
The distance between successive resonances equals half a wavelength, so
The velocity of sound at C is then
The end correction follows from the two resonance lengths,
Since velocity varies with absolute temperature as , the velocity at C is
So the velocity of sound at C is about and the end correction is .
Light travels from water () into glass (), so the polarizing angle follows from Brewster's law.
Taking the inverse tangent,
The polarizing angle is therefore about .