NEB Class 12 · Past paper
The complete NEB Class 12 2072 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) R1 and R2 in series with an emf source (negligible internal resistance); what happens to the current through R1 when R3 is connected in parallel with R2? (b) Draw a meter-bridge circuit to find the resistance of a wire; give the formula. (c) Does a charged particle in a magnetic field always experience a force? (d) Define angle of dip; its value when horizontal and vertical components are equal? (e) Why does the acceleration of a magnet falling through a solenoid decrease? (f) What is wattless current?
(a) Adding $R3$ in parallel with $R2$ lowers the resistance of that parallel combination, so the total circuit resistance falls. The emf is fixed and the internal resistance negligible, so the total current $I = E/R{total}$ increases; th...
Answer, in brief, any four questions: (a) A two-input AND gate output is fed to a NOT gate. Draw the logic circuit and write its truth table. (b) What is optical pumping in a laser? (c) All nuclei have nearly the same density. Justify. (d) How do mass number and atomic number change in alpha decay? (e) State Hubble's law and its significance. (f) What is acid rain?
(a) An AND gate followed by a NOT gate is a NAND gate, with output $Y = \overline{A\cdot B}$: A B A.B Y ------------ 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 (b) Optical pumping is the use of intense light to raise atoms from the ground state to ...
Answer, in brief, any one question: (a) Distinguish between progressive and standing waves. (b) If the pressure amplitude of a sound wave is halved, by what factor does the intensity change?
(a) The two kinds of wave differ as follows.
| Progressive wave | Standing wave |
|---|---|
| Advances through the medium, transferring energy | Confined; transfers no net energy |
| All particles have the same amplitude | Amplitude varies from zero (nodes) to maximum (antinodes) |
| No permanent nodes/antinodes | Fixed nodes and antinodes |
| Phase changes continuously along the wave | Particles between two nodes are in phase |
(b) The intensity of a sound wave is proportional to the square of the pressure amplitude, $I \propto P_0^2$. If the pressure amplitude is halved, $P_0 \to P_0/2$, then
$$I \to \left(\tfrac12\right)^2 I = \tfrac14 I$$
so the intensity falls to one-fourth of its original value.
Answer, in brief, any one question: (a) Does interference of light obey the law of conservation of energy? (b) What is polarized light and how is it represented?
(a) Yes, it does. In interference no energy is created or destroyed; it is only redistributed across the pattern. The energy that is missing from the dark fringes reappears in the bright fringes, so when averaged over the whole pattern the total energy equals that of the two beams. Conservation of energy is therefore fully obeyed.
(b) Polarized light is light whose electric field vibrations are confined to a single plane, so it is called plane-polarized, unlike unpolarized light which vibrates in all directions perpendicular to the direction of propagation. It is represented by drawing double-headed arrows along the vibration direction for vibrations lying in the plane of the paper, and dots for vibrations perpendicular to the plane of the paper.
Answer any three questions: (a) What is thermoelectric effect? How does thermo-emf vary with hot-junction temperature (cold at 0 C)? (b) Describe an experiment to verify Joule's laws of heating. (c) State Biot-Savart law and find the field due to a long straight conductor. (d) An AC passes through R, C, L in series; derive the phase relation between current and voltage.
(a) The thermoelectric (Seebeck) effect is the setting up of an emf, and hence a current, in a circuit of two dissimilar metals whose junctions are kept at different temperatures.
Keeping the cold junction at $0^\circ$C and gradually raising the hot junction to $\theta$, the thermo-emf follows the parabolic law $E = a\theta + \tfrac12 b\theta^2$: it rises to a maximum at the neutral temperature $\theta_n$, then falls, and reverses sign at the inversion temperature $\theta_i = 2\theta_n$.
(b) To verify Joule's laws, pass a current $I$ through a coil immersed in a known mass of water in a calorimeter for a time $t$ and measure the temperature rise, from which the heat produced is $H = mc,\Delta\theta$. Varying one quantity at a time shows that $H \propto I^2$ (keeping $R$ and $t$ fixed), $H \propto R$ (keeping $I$ and $t$ fixed), and $H \propto t$ (keeping $I$ and $R$ fixed). Taken together these results verify that $H = I^2 R t$.
(c) The Biot-Savart law states that the magnetic field due to a current element $I,d\vec l$ at a point with position vector $\vec r$ is
$$dB = \frac{\mu_0}{4\pi}\frac{I,dl\sin\theta}{r^2}$$
Integrating this contribution along an infinite straight wire at perpendicular distance $a$ gives the field
$$B = \frac{\mu_0 I}{2\pi a}$$
(d) With a current $I = I_0\sin\omega t$ in the series R-C-L circuit, the resistor voltage $V_R = I_0R$ is in phase with the current, the inductor voltage $V_L = I_0X_L$ leads it by $90^\circ$, and the capacitor voltage $V_C = I_0X_C$ lags it by $90^\circ$. Combining these, the applied voltage leads the current by
$$ \begin{aligned} \phi &= \tan^{-1}\frac{X_L - X_C}{R} \ \qquad Z &= \sqrt{R^2 + (X_L - X_C)^2} \end{aligned} $$
If $X_L > X_C$ the circuit is inductive and the voltage leads the current, whereas if $X_L < X_C$ it is capacitive and the voltage lags the current.
Answer any three questions: (a) Define photoelectric effect; discuss Einstein's photoelectric equation and stopping potential. (b) Explain a p-n junction diode as a rectifier; draw a full-wave rectifier and explain. (c) State Bohr's postulates and calculate the radius of the nth orbit of hydrogen. (d) State the laws of radioactive disintegration; derive the relation between half-life and decay constant.
(a) The photoelectric effect is the emission of electrons from a metal surface when light of sufficient frequency falls on it. Einstein explained it with the equation $h\nu = \phi + \tfrac12 mv_{max}^2$, where $\phi = h\nu_0$ is the work function of the metal. The stopping potential $V_s$ is the reverse voltage that just stops the fastest electrons, defined by $eV_s = \tfrac12 mv_{max}^2$, so that
$$eV_s = h\nu - \phi$$
(b) A p-n junction diode conducts only in forward bias, so it acts as a rectifier that converts AC to DC.
In a full-wave (centre-tapped) rectifier, two diodes and a centre-tapped transformer feed the load. During one half-cycle diode $D_1$ conducts, and during the next half-cycle $D_2$ conducts. In both half-cycles the current passes through the load in the same direction, giving a full-wave pulsating DC output.
(c) Bohr's postulates are: (i) electrons revolve in certain stationary orbits without radiating energy; (ii) the angular momentum in these orbits is quantized, $mvr = \dfrac{nh}{2\pi}$; and (iii) radiation is emitted or absorbed only when an electron jumps between levels, $h\nu = E_2 - E_1$.
For the hydrogen atom the Coulomb force provides the centripetal force,
$$\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} = \frac{mv^2}{r}$$
Eliminating $v$ with the quantization condition $mvr = \dfrac{nh}{2\pi}$ gives the radius of the $n$th orbit,
$$\boxed{r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2}} = 0.529,n^2\ \text{Å}$$
(d) The laws of radioactive disintegration say that decay is spontaneous and random, and that the rate of decay is proportional to the number of nuclei present, $-\dfrac{dN}{dt} = \lambda N$, which integrates to $N = N_0 e^{-\lambda t}$. To find the half-life we set $N = N_0/2$, so that
$$\tfrac12 = e^{-\lambda T_{1/2}}$$
Taking logarithms gives $\lambda T_{1/2} = \ln 2$, hence
$$\boxed{T_{1/2} = \frac{0.693}{\lambda}}$$
Answer any one question: (a) Write Newton's formula for the velocity of sound in air; explain the modification and Laplace's correction. (b) What is end correction of a pipe? Describe the modes of vibration of an air column closed at one end.
(a) Newton assumed sound travels isothermally, which gives $v = \sqrt{\dfrac{P}{\rho}} \approx 280$ m/s, about 15% too low. Laplace pointed out that the rapid compressions and rarefactions leave no time for heat exchange and are therefore adiabatic, so the elasticity involved is the adiabatic bulk modulus $\gamma P$:
$$v = \sqrt{\frac{\gamma P}{\rho}} \approx 332\ \text{m/s}$$
which agrees with experiment.
(b) The end correction accounts for the antinode forming slightly beyond the open end of a pipe; the small extra length $e \approx 0.6r$ (for a pipe of radius $r$) added to the pipe length is called the end correction.
For a pipe closed at one end (length $L$) there is a node at the closed end and an antinode at the open end. The allowed modes satisfy $L = \dfrac{\lambda}{4}, \dfrac{3\lambda}{4}, \dfrac{5\lambda}{4},\dots$, giving frequencies
$$f_1 = \frac{v}{4L}, \quad f_3 = 3f_1, \quad f_5 = 5f_1,\dots$$
so only the odd harmonics (the fundamental and the odd overtones) are present.
Answer any one question: (a) State Huygen's principle and derive the laws of reflection. (b) Discuss the formation of maxima and minima in Fraunhofer single-slit diffraction.
(a) Huygen's principle states that each point on a wavefront is a source of secondary wavelets travelling at the wave speed, and the new wavefront is their envelope.
To derive the laws of reflection, consider a plane wavefront $AB$ striking a mirror. While the wavelet from $B$ travels $BC = vt$ to reach the mirror, the wavelet from $A$ travels $AD = vt$ upward. Triangles $ABC$ and $ADC$ are congruent, having equal sides $vt$, a common side $AC$ and right angles, so $\angle i = \angle r$. The angle of incidence therefore equals the angle of reflection, and the incident ray, reflected ray and normal all lie in one plane.
(b) In Fraunhofer diffraction at a single slit of width $a$, parallel light passing through the slit is focused by a lens. Along the axis all the wavelets arrive in phase and give the central maximum. At an angle $\theta$ the path difference across the slit is $a\sin\theta$, and the slit divides into pairs of strips that cancel when
$$a\sin\theta = n\lambda\ (n = 1,2,\dots)\ \text{: minima}$$
Between consecutive minima the wavelets partly reinforce, giving weaker secondary maxima at $a\sin\theta \approx (n+\tfrac12)\lambda$. The central maximum is the brightest and twice as wide as the others.
Answer any two questions: (a) A 6 V battery (r = 0.5 ohm) is in parallel with a 10 V battery (r = 1 ohm); the combination sends current through a 12 ohm resistor. Find the current through each battery. (b) A moving-coil galvanometer of 50 turns, 10 ohm, is replaced by 100 turns, 50 ohm. Find the factor by which current and voltage sensitivities change. (c) A 1000-turn solenoid, area 2x10^-3 m^2, carries 2 A and produces flux density 52x10^-3 T. Find the self-inductance.
(a) A $6\ \text{V}$ battery of internal resistance $r_1 = 0.5\ \Omega$ is in parallel with a $10\ \text{V}$ battery of internal resistance $r_2 = 1\ \Omega$, and the pair feeds a $12\ \Omega$ resistor. Let $V$ be the p.d. across the parallel combination, which is also the p.d. across the $12\ \Omega$ resistor. Then the battery currents are $I_1 = \dfrac{6-V}{0.5}$ and $I_2 = \dfrac{10-V}{1}$, and by Kirchhoff's current law $I_1 + I_2 = \dfrac{V}{12}$.
$$\frac{6-V}{0.5} + \frac{10-V}{1} = \frac{V}{12}$$
Clearing the terms,
$$ \begin{aligned} (12 - 2V) + (10 - V) &= \frac{V}{12} \ 22 - 3V &= \frac{V}{12} \quad\Rightarrow\quad 264 \ &= 37V \quad\Rightarrow\quad V \ &= 7.14\ \text{V} \end{aligned} $$
Substituting back gives the branch currents,
$$ \begin{aligned} I_1 &= \frac{6 - 7.14}{0.5} \ &= -2.27\ \text{A}\quad (\text{this battery is being charged}) \ I_2 &= \frac{10 - 7.14}{1} \ &= 2.87\ \text{A}, \ \qquad \text{external current} &= 0.60\ \text{A} \end{aligned} $$
So the $6\ \text{V}$ battery carries $2.27\ \text{A}$ in reverse (it is being charged) while the $10\ \text{V}$ battery delivers $2.87\ \text{A}$.
(b) A galvanometer of $50$ turns and $10\ \Omega$ is replaced by one of $100$ turns and $50\ \Omega$. Current sensitivity is proportional to the number of turns, $S_I \propto N$, so it changes by
$$\frac{100}{50} = 2 \quad (\text{doubles})$$
Voltage sensitivity is proportional to $N/R$, so it changes by
$$ \begin{aligned} \frac{100/50}{50/10} &= \frac{2}{5} \ &= 0.4\quad (\text{becomes 0.4 times}) \end{aligned} $$
Thus the current sensitivity doubles while the voltage sensitivity drops to $0.4$ times its former value.
(c) The solenoid has $N = 1000$ turns, cross-section $A = 2\times10^{-3}\ \text{m}^2$, carries $2\ \text{A}$, and produces a flux density $B = 52\times10^{-3}\ \text{T}$. Its self-inductance follows from $L = \dfrac{N\Phi}{I} = \dfrac{NBA}{I}$.
$$ \begin{aligned} L &= \frac{1000\times52\times10^{-3}\times2\times10^{-3}}{2} \ &= \frac{0.104}{2} \ &= 0.052\ \text{H} \end{aligned} $$
So the self-inductance is $0.052\ \text{H}$, or $52\ \text{mH}$.
Answer any two questions: (a) In Millikan's experiment a drop falls at 1.4 mm/s with no field; with a vertical field 4.9x10^5 V/m it moves down at 1.21 mm/s. Find the charge (oil density 750, air viscosity 1.81x10^-5, air density 1.29). (b) An X-ray tube at 50 kV converts 0.4% of cathode-ray energy to X-rays; heat generated at 500 W. Find the current and the electron speed (m = 9x10^-31 kg, e = 1.6x10^-19 C). (c) Fission of one U-235 atom liberates 3.2x10^-11 J. Find the power from fission of 1 g of uranium per day (Avogadro 6x10^23).
(a) In free fall the drop settles at $v_1 = 1.4\ \text{mm/s}$, and when the field $E = 4.9\times10^{5}\ \text{V/m}$ is applied it moves down more slowly at $v_2 = 1.21\ \text{mm/s}$; the oil density is $750$, the air viscosity is $\eta = 1.81\times10^{-5}$, and the air density is $1.29$, so the effective density is $\rho_{oil}-\rho_{air} = 748.71$, and $g = 9.8\ \text{m/s}^2$. First we get the drop radius from the free-fall terminal speed, using Stokes' law with buoyancy,
$$ \begin{aligned} a^2 &= \frac{9\eta v_1}{2(\rho_{oil}-\rho_{air})g} \ &= \frac{9(1.81\times10^{-5})(1.4\times10^{-3})}{2(748.71)(9.8)} \ &= 1.55\times10^{-11}, \ a &= 3.94\times10^{-6}\ \text{m}. \end{aligned} $$
Since the drop moves more slowly once the field is on, the electric force must act upward, and balancing it against the change in viscous drag gives $qE = 6\pi\eta a(v_1 - v_2)$,
$$ \begin{aligned} q &= \frac{6\pi(1.81\times10^{-5})(3.94\times10^{-6})(1.4-1.21)\times10^{-3}}{4.9\times10^{5}} \ &= \frac{2.56\times10^{-13}}{4.9\times10^{5}} \ &= 5.2\times10^{-19}\ \text{C}. \end{aligned} $$
So the charge on the drop is about $5.2\times10^{-19}$ C, which is close to $3e$.
(b) The X-ray tube runs at $50\ \text{kV}$, only 0.4% of the cathode-ray energy is turned into X-rays, and the heat generated is $500\ \text{W}$, with $m = 9\times10^{-31}\ \text{kg}$ and $e = 1.6\times10^{-19}\ \text{C}$. Since the X-rays carry away 0.4%, the heat represents 99.6% of the input power, so the total power is
$$ \begin{aligned} P &= \frac{500}{0.996} \ &= 502\ \text{W}. \end{aligned} $$
The current then follows from $P = VI$,
$$ \begin{aligned} I &= \frac{P}{V} \ &= \frac{502}{50000} \ &= 0.010\ \text{A} \ &= 10\ \text{mA}. \end{aligned} $$
The electron speed comes from equating the work done on the electron to its kinetic energy, $eV = \tfrac12 mv^2$,
$$ \begin{aligned} v &= \sqrt{\frac{2eV}{m}} \ &= \sqrt{\frac{2(1.6\times10^{-19})(50000)}{9\times10^{-31}}} \ &= 1.33\times10^{8}\ \text{m/s}. \end{aligned} $$
So the current is about $10$ mA and the electrons strike the target at about $1.33\times10^{8}$ m/s.
(c) Each fission of a U-235 atom liberates $3.2\times10^{-11}\ \text{J}$, and we want the power from fissioning $1\ \text{g}$ of uranium per day, taking Avogadro's number as $6\times10^{23}$. The number of atoms in 1 g of U-235 is
$$ \begin{aligned} N &= \frac{6\times10^{23}}{235} \ &= 2.55\times10^{21}, \end{aligned} $$
and the energy released when they all fission is
$$ \begin{aligned} E &= N\times3.2\times10^{-11} \ &= 8.17\times10^{10}\ \text{J}. \end{aligned} $$
Spreading this over one day (86400 s) gives the power,
$$ \begin{aligned} P &= \frac{8.17\times10^{10}}{86400} \ &= 9.46\times10^{5}\ \text{W}. \end{aligned} $$
So the power output is about $9.46\times10^{5}$ W, roughly $946$ kW.
A stationary motion detector sends 150 kHz sound toward a truck approaching at 120 km/hr. What is the frequency reflected back to the detector? (Velocity of sound = 340 m/s)
We are given a stationary detector emitting $150\ \text{kHz}$ toward a truck approaching at $120\ \text{km/h}$, with a speed of sound $v = 340\ \text{m/s}$. The truck speed in SI units is $u = 120\ \text{km/h} = 33.33\ \text{m/s}$. First...
In Young's double-slit experiment the slits are 0.03 cm apart and the screen is 1.5 m away. The distance between the central bright fringe and the fourth bright fringe is 1 cm. Calculate the wavelength of light.
We are given a slit separation $d = 0.03\ \text{cm} = 3\times10^{-4}\ \text{m}$, a screen distance $D = 1.5\ \text{m}$, and the fourth bright fringe at $y_4 = 1\ \text{cm} = 0.01\ \text{m}$.
The position of the $n$th bright fringe is $y_n = \dfrac{n\lambda D}{d}$, so rearranging for the wavelength with $n = 4$,
$$ \begin{aligned} \lambda &= \frac{y_4,d}{4D} = \frac{0.01\times3\times10^{-4}}{4\times1.5} \ &= \frac{3\times10^{-6}}{6} = 5\times10^{-7}\ \text{m} \end{aligned} $$
The wavelength of the light is therefore $5\times10^{-7}$ m, or $5000$ Å (500 nm).
(a) Adding in parallel with lowers the resistance of that parallel combination, so the total circuit resistance falls. The emf is fixed and the internal resistance negligible, so the total current increases; th...
(a) An AND gate followed by a NOT gate is a NAND gate, with output : A B A.B Y ------------ 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 (b) Optical pumping is the use of intense light to raise atoms from the ground state to ...
(a) The two kinds of wave differ as follows.
| Progressive wave | Standing wave |
|---|---|
| Advances through the medium, transferring energy | Confined; transfers no net energy |
| All particles have the same amplitude | Amplitude varies from zero (nodes) to maximum (antinodes) |
| No permanent nodes/antinodes | Fixed nodes and antinodes |
| Phase changes continuously along the wave | Particles between two nodes are in phase |
(b) The intensity of a sound wave is proportional to the square of the pressure amplitude, . If the pressure amplitude is halved, , then
so the intensity falls to one-fourth of its original value.
(a) The thermoelectric (Seebeck) effect is the setting up of an emf, and hence a current, in a circuit of two dissimilar metals whose junctions are kept at different temperatures.
Keeping the cold junction at C and gradually raising the hot junction to , the thermo-emf follows the parabolic law : it rises to a maximum at the neutral temperature , then falls, and reverses sign at the inversion temperature .
(b) To verify Joule's laws, pass a current through a coil immersed in a known mass of water in a calorimeter for a time and measure the temperature rise, from which the heat produced is . Varying one quantity at a time shows that (keeping and fixed), (keeping and fixed), and (keeping and fixed). Taken together these results verify that .
(c) The Biot-Savart law states that the magnetic field due to a current element at a point with position vector is
Integrating this contribution along an infinite straight wire at perpendicular distance gives the field
(d) With a current in the series R-C-L circuit, the resistor voltage is in phase with the current, the inductor voltage leads it by , and the capacitor voltage lags it by . Combining these, the applied voltage leads the current by
If the circuit is inductive and the voltage leads the current, whereas if it is capacitive and the voltage lags the current.
(a) The photoelectric effect is the emission of electrons from a metal surface when light of sufficient frequency falls on it. Einstein explained it with the equation , where is the work function of the metal. The stopping potential is the reverse voltage that just stops the fastest electrons, defined by , so that
(b) A p-n junction diode conducts only in forward bias, so it acts as a rectifier that converts AC to DC.
In a full-wave (centre-tapped) rectifier, two diodes and a centre-tapped transformer feed the load. During one half-cycle diode conducts, and during the next half-cycle conducts. In both half-cycles the current passes through the load in the same direction, giving a full-wave pulsating DC output.
(c) Bohr's postulates are: (i) electrons revolve in certain stationary orbits without radiating energy; (ii) the angular momentum in these orbits is quantized, ; and (iii) radiation is emitted or absorbed only when an electron jumps between levels, .
For the hydrogen atom the Coulomb force provides the centripetal force,
Eliminating with the quantization condition gives the radius of the th orbit,
(d) The laws of radioactive disintegration say that decay is spontaneous and random, and that the rate of decay is proportional to the number of nuclei present, , which integrates to . To find the half-life we set , so that
Taking logarithms gives , hence
(a) Newton assumed sound travels isothermally, which gives m/s, about 15% too low. Laplace pointed out that the rapid compressions and rarefactions leave no time for heat exchange and are therefore adiabatic, so the elasticity involved is the adiabatic bulk modulus :
which agrees with experiment.
(b) The end correction accounts for the antinode forming slightly beyond the open end of a pipe; the small extra length (for a pipe of radius ) added to the pipe length is called the end correction.
For a pipe closed at one end (length ) there is a node at the closed end and an antinode at the open end. The allowed modes satisfy , giving frequencies
so only the odd harmonics (the fundamental and the odd overtones) are present.
(a) Huygen's principle states that each point on a wavefront is a source of secondary wavelets travelling at the wave speed, and the new wavefront is their envelope.
To derive the laws of reflection, consider a plane wavefront striking a mirror. While the wavelet from travels to reach the mirror, the wavelet from travels upward. Triangles and are congruent, having equal sides , a common side and right angles, so . The angle of incidence therefore equals the angle of reflection, and the incident ray, reflected ray and normal all lie in one plane.
(b) In Fraunhofer diffraction at a single slit of width , parallel light passing through the slit is focused by a lens. Along the axis all the wavelets arrive in phase and give the central maximum. At an angle the path difference across the slit is , and the slit divides into pairs of strips that cancel when
Between consecutive minima the wavelets partly reinforce, giving weaker secondary maxima at . The central maximum is the brightest and twice as wide as the others.
(a) A battery of internal resistance is in parallel with a battery of internal resistance , and the pair feeds a resistor. Let be the p.d. across the parallel combination, which is also the p.d. across the resistor. Then the battery currents are and , and by Kirchhoff's current law .
Clearing the terms,
Substituting back gives the branch currents,
So the battery carries in reverse (it is being charged) while the battery delivers .
(b) A galvanometer of turns and is replaced by one of turns and . Current sensitivity is proportional to the number of turns, , so it changes by
Voltage sensitivity is proportional to , so it changes by
Thus the current sensitivity doubles while the voltage sensitivity drops to times its former value.
(c) The solenoid has turns, cross-section , carries , and produces a flux density . Its self-inductance follows from .
So the self-inductance is , or .
(a) In free fall the drop settles at , and when the field is applied it moves down more slowly at ; the oil density is , the air viscosity is , and the air density is , so the effective density is , and . First we get the drop radius from the free-fall terminal speed, using Stokes' law with buoyancy,
Since the drop moves more slowly once the field is on, the electric force must act upward, and balancing it against the change in viscous drag gives ,
So the charge on the drop is about C, which is close to .
(b) The X-ray tube runs at , only 0.4% of the cathode-ray energy is turned into X-rays, and the heat generated is , with and . Since the X-rays carry away 0.4%, the heat represents 99.6% of the input power, so the total power is
The current then follows from ,
The electron speed comes from equating the work done on the electron to its kinetic energy, ,
So the current is about mA and the electrons strike the target at about m/s.
(c) Each fission of a U-235 atom liberates , and we want the power from fissioning of uranium per day, taking Avogadro's number as . The number of atoms in 1 g of U-235 is
and the energy released when they all fission is
Spreading this over one day (86400 s) gives the power,
So the power output is about W, roughly kW.
We are given a stationary detector emitting toward a truck approaching at , with a speed of sound . The truck speed in SI units is . First...
We are given a slit separation , a screen distance , and the fourth bright fringe at .
The position of the th bright fringe is , so rearranging for the wavelength with ,
The wavelength of the light is therefore m, or Å (500 nm).