NEB Class 12 · Past paper
The complete NEB Class 12 2071 exam paper for Physics, all 12 questions with solved model answers.
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Answer in brief (any four): (a) A current-carrying solenoid tends to contract. Why? (b) Why are constantan and manganin used for standard resistors? (c) At high frequencies a capacitor becomes a short circuit and an inductor an open circuit. Explain. (d) Does the thermoelectric effect obey conservation of energy? (e) Why are strong currents not preferred for electroplating? (f) Ampere's circuital law applies only to certain cases. Justify.
(a) Adjacent turns of a solenoid carry current in the same direction, and parallel currents in the same direction attract. So neighbouring turns pull towards each other and the solenoid tends to contract and shorten. (b) Constantan and m...
Answer in brief (any four): (a) Why remove ornaments while taking an X-ray? (b) Define antiparticles; write the quark combination of an anti-neutron. (c) NAND is a universal gate. Explain. (d) What are the roles of physics in the development of a nation? (e) What are radioisotopes and their applications?
(a) Metal ornaments are opaque to X-rays because of their high density and atomic number, so they cast dense shadows that overlap and obscure the body part being imaged, giving misleading radiographs. Removing them keeps the image clear....
Answer in brief (any one): (a) Is there any limit to the beat frequency we can observe? (b) What is the change in the fundamental frequency of a tube open at both ends if half its length is dipped in water?
(a) Yes, there is a limit. The ear can resolve beats only up to about $10\ \text{Hz}$; beyond that the successive intensity maxima arrive too quickly to be told apart and simply merge into one continuous tone. So beats above roughly
Answer in brief (any one): (a) Differentiate between plane and spherical wavefronts. (b) Can Foucault's method be used to find the speed of light in water? Explain.
(a) A plane wavefront is a flat surface of constant phase. It is produced by a source at effectively infinite distance, and the rays associated with it are parallel to one another. A spherical wavefront, by contrast, is a sphere of const...
Answer in detail (any three): (a) Derive the current in an AC series LCR circuit and its power factor. (b) What are the magnetic elements of the earth? Prove cot^2 delta = cot^2 delta1 + cot^2 delta2. (c) Define AC generator; describe its construction and working. (d) Discuss conduction of electricity in a conductor and derive I = venA.
(a) In a series LCR circuit carrying $I = I_0\sin\omega t$, the resistor voltage is in phase with the current, $V_R = I_0R$, the inductor voltage leads it by $90^\circ$, $V_L = I_0X_L$, and the capacitor voltage lags it by $90^\circ$, $V_C = I_0X_C$. Since $V_L$ and $V_C$ are opposite, the net reactive voltage is $(V_L - V_C)$, and combining it with $V_R$ gives the applied voltage,
$$ \begin{aligned} V_0 &= I_0\sqrt{R^2 + (X_L - X_C)^2} \ \qquad Z &= \sqrt{R^2 + (X_L - X_C)^2} \end{aligned} $$
The peak current and the phase angle are therefore
$$ \begin{aligned} I_0 &= \frac{V_0}{Z} \ \qquad \tan\phi &= \frac{X_L - X_C}{R} \end{aligned} $$
The power factor of the circuit is
$$\cos\phi = \frac{R}{Z}$$
(b) The three magnetic elements of the Earth are the angle of declination, the angle of dip $\delta$, and the horizontal component $H$ of the Earth's field.
If a dip needle is set in a vertical plane making an angle $\theta$ with the magnetic meridian, only the component $H\cos\theta$ acts along it, so the apparent dip $\delta_1$ satisfies $\tan\delta_1 = \dfrac{V}{H\cos\theta}$. Turning the plane through $90^\circ$, the effective horizontal component becomes $H\sin\theta$, giving $\tan\delta_2 = \dfrac{V}{H\sin\theta}$. Adding the squares of the cotangents,
$$ \begin{aligned} \cot^2\delta_1 + \cot^2\delta_2 &= \frac{H^2\cos^2\theta}{V^2} + \frac{H^2\sin^2\theta}{V^2} \ &= \frac{H^2}{V^2} = \cot^2\delta \end{aligned} $$
which proves that
$$\cot^2\delta = \cot^2\delta_1 + \cot^2\delta_2$$
(c) An AC generator is a device that converts mechanical energy into alternating electrical energy through electromagnetic induction.
Its construction consists of an armature coil that rotates in the field of a magnet, connected to the external circuit through two slip rings and brushes.
In working, as the coil rotates the flux linked with it changes continuously, inducing an emf $e = NBA\omega\sin\omega t$. The slip rings reverse the connection to the circuit every half rotation, so the output is an alternating emf of peak value $E_0 = NBA\omega$.
(d) Consider a conductor of cross-sectional area $A$ containing $n$ free electrons per unit volume drifting with speed $v$. In time $t$ the electrons that cross a section are those within a length $vt$, so their number is $nA(vt)$ and the charge carried is $nA vt,e$. The current is therefore
$$ \begin{aligned} I &= \frac{q}{t} \ &= n A v e \end{aligned} $$
Answer in detail (any three): (a) State Einstein's mass-energy relation with example; how is the energy released in nuclear fission estimated? (b) Draw a simple regulated power supply and explain regulation. (c) Derive Bragg's equation and its use for crystal spacing. (d) Discuss Millikan's oil-drop experiment to determine the electron charge.
(a) Einstein's mass-energy relation is $E = mc^2$, which says that mass and energy are interconvertible. For example, in the fission of $^{235}$U the products have slightly less total mass than the reactants, and this mass defect
Answer in detail (any one): (a) What is the significance of Newton's formula for the speed of sound in a gas? How did Laplace correct it? (b) What is pressure amplitude? Obtain a relation for it with a graph.
(a) Treating the propagation of sound as isothermal, Newton obtained $v = \sqrt{\dfrac{P}{\rho}}$. This gave about $280$ m/s for air, roughly 15% below the measured $332$ m/s; its significance is that it first related the speed of sound to the properties of the gas, though it underestimated the value.
Laplace corrected this by noting that the compressions and rarefactions in a sound wave are so rapid that they are adiabatic rather than isothermal, so the relevant modulus is $\gamma P$:
$$v = \sqrt{\frac{\gamma P}{\rho}}$$
With $\gamma = 1.4$ for air this gives about $332$ m/s, matching experiment.
(b) The pressure amplitude is the maximum change in pressure from the mean value as a sound wave passes. For a displacement wave $y = a\sin(\omega t - kx)$, the excess pressure is $P = -B\dfrac{\partial y}{\partial x} = Bak\cos(\omega t - kx)$, so the pressure amplitude is
$$ \begin{aligned} P_0 &= Bak \ &= \rho v\omega a \end{aligned} $$
The pressure wave is $90^\circ$ out of phase with the displacement wave, so pressure is maximum where displacement is zero (at the nodes), which a cosine curve drawn against the displacement sine curve makes clear.
Answer in detail (any one): (a) What is diffraction? Explain the single-slit diffraction pattern using phase difference. (b) What is polarized light? Explain polarization by reflection.
(a) Diffraction is the bending of light around obstacles and edges and its spreading into the geometrical shadow, producing a fringe pattern, which happens when light meets an aperture comparable in size to its wavelength.
For a single slit of width $a$, every point of the slit acts as a secondary source. Rays that go straight through arrive in phase and form the central bright maximum. At an angle $\theta$ the path difference between the two edges of the slit is $a\sin\theta$, and the slit can be divided into pairs of halves whose contributions cancel when
$$a\sin\theta = n\lambda \quad(n = 1,2,\dots)\ \text{minima}$$
with weaker secondary maxima lying between them. The result is a bright central band twice as wide as the others, flanked by fringes that fade away on either side.
(b) Polarized light is light whose electric vector is confined to a single plane, unlike unpolarized light, which vibrates in all transverse directions.
Polarization by reflection occurs when unpolarized light reflects off a dielectric at the polarizing (Brewster) angle $\theta_p$, defined by $\tan\theta_p = \mu$. At this angle the reflected ray is completely plane polarized with its vibrations perpendicular to the plane of incidence, because the reflected and refracted rays are then $90^\circ$ apart and only the perpendicular component is reflected.
Solve (any two): (a) L = 0.50 H in series with R = 100 ohm, 200 V, 50 Hz AC. Find max current and time lag between max voltage and max current. (b) 1.62 V across Pt electrodes in CuCl2; in 600 s, 5.92 g copper deposited; back emf 1.34 V. Find resistance of the voltameter. (E = 96500 C/mol, Cu = 63.5) (c) A standard cell 1.0185 V balances at 60 cm; a voltmeter balances at 65 cm reading 1.1 V. Find percentage error.
(a) Here $L = 0.50\ \text{H}$ is in series with $R = 100\ \Omega$ across a $200\ \text{V}$, $50\ \text{Hz}$ supply. The inductive reactance is
$$ \begin{aligned} X_L &= \omega L \ &= 2\pi(50)(0.50) \ &= 157.1\ \Omega \end{aligned} $$
and the impedance of the series circuit is
$$ \begin{aligned} Z &= \sqrt{R^2 + X_L^2} \ &= \sqrt{100^2 + 157.1^2} \ &= 186.2\ \Omega \end{aligned} $$
The peak voltage is $V_0 = 200\sqrt2 = 282.8\ \text{V}$, so the maximum current is
$$ \begin{aligned} I_0 &= \frac{V_0}{Z} \ &= \frac{282.8}{186.2} \ &= 1.52\ \text{A} \end{aligned} $$
The current lags the voltage by the phase angle
$$ \begin{aligned} \phi &= \tan^{-1}\frac{X_L}{R} \ &= \tan^{-1}(1.571) \ &= 57.5^\circ \ &= 1.004\ \text{rad} \end{aligned} $$
and this phase difference corresponds to a time lag
$$ \begin{aligned} \Delta t &= \frac{\phi}{\omega} \ &= \frac{1.004}{2\pi(50)} \ &= 3.2\times10^{-3}\ \text{s} \end{aligned} $$
So the maximum current is about $1.52\ \text{A}$ and it lags the voltage by about $3.2\ \text{ms}$.
(b) A p.d. of $1.62\ \text{V}$ is applied across Pt electrodes in CuCl$_2$, depositing $m = 5.92\ \text{g}$ of copper in $t = 600\ \text{s}$ against a back emf of $1.34\ \text{V}$, with $F = 96500\ \text{C/mol}$ and Cu $= 63.5$. Since copper has valency 2, its electrochemical equivalent is
$$ \begin{aligned} Z &= \frac{63.5/2}{96500} \ &= 3.29\times10^{-4}\ \text{g/C} \end{aligned} $$
From Faraday's law $m = ZIt$, the current is
$$ \begin{aligned} I &= \frac{m}{Zt} \ &= \frac{5.92}{3.29\times10^{-4}\times600} \ &= 30.0\ \text{A} \end{aligned} $$
Only the net p.d. left after the back emf drives this current, so the resistance is
$$ \begin{aligned} 1.62 - 1.34 &= 0.28\ \text{V}, \ \qquad R &= \frac{0.28}{30.0} \ &= 9.3\times10^{-3}\ \Omega \end{aligned} $$
The resistance of the voltameter is therefore about $9.3\times10^{-3}\ \Omega$.
(c) A standard cell of $1.0185\ \text{V}$ balances at $60\ \text{cm}$, so the potential gradient of the wire is
$$ \begin{aligned} k &= \frac{1.0185}{60} \ &= 0.016975\ \text{V/cm} \end{aligned} $$
The voltmeter balances at $65\ \text{cm}$, so its true voltage is
$$ \begin{aligned} V &= 0.016975\times65 \ &= 1.1034\ \text{V} \end{aligned} $$
The voltmeter itself reads $1.1\ \text{V}$, so the percentage error is
$$\frac{1.1 - 1.1034}{1.1034}\times100 = -0.31%$$
That is, the voltmeter reads about $0.31%$ low.
Solve (any two): (a) Light of a certain wavelength gives ejected-electron KE 0.8 eV; light of one-third that wavelength gives KE 4.8 eV. Find the threshold wavelength. (b) 24.6 eV removes one electron from neutral helium; find the energy to detach both electrons. (c) Sun mass 2x10^30 kg radiates 4x10^23 kW; if 0.7% of mass converts to radiation, estimate its lifetime in years (c = 3x10^8 m/s, 1 yr = 3x10^7 s).
(a) At wavelength $\lambda$ the ejected electrons have $KE = 0.8\ \text{eV}$, and at $\lambda/3$ they have $KE = 4.8\ \text{eV}$. Writing the photon energy as $E = hc/\lambda$, the photoelectric equation gives $E = \phi + 0.8$ for the first case, while the shorter wavelength $\lambda/3$ carries three times the energy, so $3E = \phi + 4.8$. Subtracting the first from the second eliminates $\phi$,
$$ \begin{aligned} 2E &= 4.0 \quad\Rightarrow\quad E \ &= 2.0\ \text{eV} \end{aligned} $$
and feeding this back gives the work function
$$ \begin{aligned} \phi &= 2.0 - 0.8 \ &= 1.2\ \text{eV}. \end{aligned} $$
The threshold wavelength then follows from $\lambda_0 = hc/\phi$,
$$ \begin{aligned} \lambda_0 &= \frac{1240}{1.2} \ &= 1033\ \text{nm}. \end{aligned} $$
So the threshold wavelength is about $1033$ nm.
(b) Removing the first electron from neutral helium takes $24.6\ \text{eV}$. The second electron is then removed from He$^+$, which is a hydrogen-like ion with $Z = 2$, so its ionization energy is
$$ \begin{aligned} 13.6,Z^2 &= 13.6\times4 \ &= 54.4\ \text{eV}. \end{aligned} $$
Adding the two contributions, the energy needed to detach both electrons is
$$24.6 + 54.4 = 79.0\ \text{eV}.$$
(c) The Sun has mass $M = 2\times10^{30}\ \text{kg}$ and radiates at $4\times10^{23}\ \text{kW} = 4\times10^{26}\ \text{W}$, and we assume 0.7% of its mass converts to radiation, with $c = 3\times10^8\ \text{m/s}$ and $1\ \text{yr} = 3\times10^7\ \text{s}$. The energy available from that mass conversion is
$$ \begin{aligned} E &= 0.007,M c^2 \ &= 0.007(2\times10^{30})(3\times10^{8})^2 \ &= 1.26\times10^{45}\ \text{J}. \end{aligned} $$
Dividing by the radiated power gives the lifetime,
$$ \begin{aligned} t &= \frac{E}{P} \ &= \frac{1.26\times10^{45}}{4\times10^{26}} \ &= 3.15\times10^{18}\ \text{s}, \end{aligned} $$
and expressing this in years,
$$\frac{3.15\times10^{18}}{3\times10^{7}} = 1.05\times10^{11}\ \text{yr}.$$
So the estimated lifetime is about $1.05\times10^{11}$ years.
A train approaching a tunnel at 60 km/hr sounds a whistle of frequency 1 kHz. What beat frequency does the driver observe? (Speed of sound = 340 m/s)
We are given the train (source) moving at $60\ \text{km/h}$, a whistle of frequency $f = 1\ \text{kHz} = 1000\ \text{Hz}$, and a speed of sound $v = 340\ \text{m/s}$. The driver hears two sounds: the direct whistle at 1000 Hz and the ech...
Coherent light of wavelengths 600 nm and 470 nm passes through two slits separated by [figure: slit separation, value not legible in scan] mm; the pattern is observed on a screen 5.0 m away. What is the distance on the screen between the first-order bright fringes for the two wavelengths?
We are given coherent wavelengths $\lambda_1 = 600\ \text{nm}$ and $\lambda_2 = 470\ \text{nm}$ with the screen at $D = 5.0\ \text{m}$, but the slit separation $d$ was illegible in the scan, so the result can only be given symbolically.
For the first-order bright fringe the position is $y_1 = \dfrac{\lambda D}{d}$, so the separation between the two first-order fringes is
$$ \begin{aligned} \Delta y &= \frac{D(\lambda_1 - \lambda_2)}{d} \ &= \frac{5.0,(600 - 470)\times10^{-9}}{d} \ &= \frac{6.5\times10^{-7}}{d}\ \text{m} \end{aligned} $$
Once $d$ (in metres) is known it can be substituted directly. For instance, if $d = 0.30\ \text{mm} = 3\times10^{-4}\ \text{m}$, then $\Delta y = 2.17\times10^{-3}\ \text{m} = 2.17\ \text{mm}$. With the slit-separation figure missing, the answer is left as $\Delta y = \dfrac{6.5\times10^{-7}}{d}$ m.
(a) Yes, there is a limit. The ear can resolve beats only up to about ; beyond that the successive intensity maxima arrive too quickly to be told apart and simply merge into one continuous tone. So beats above roughly
(a) In a series LCR circuit carrying , the resistor voltage is in phase with the current, , the inductor voltage leads it by , , and the capacitor voltage lags it by , . Since and are opposite, the net reactive voltage is , and combining it with gives the applied voltage,
The peak current and the phase angle are therefore
The power factor of the circuit is
(b) The three magnetic elements of the Earth are the angle of declination, the angle of dip , and the horizontal component of the Earth's field.
If a dip needle is set in a vertical plane making an angle with the magnetic meridian, only the component acts along it, so the apparent dip satisfies . Turning the plane through , the effective horizontal component becomes , giving . Adding the squares of the cotangents,
which proves that
(c) An AC generator is a device that converts mechanical energy into alternating electrical energy through electromagnetic induction.
Its construction consists of an armature coil that rotates in the field of a magnet, connected to the external circuit through two slip rings and brushes.
In working, as the coil rotates the flux linked with it changes continuously, inducing an emf . The slip rings reverse the connection to the circuit every half rotation, so the output is an alternating emf of peak value .
(d) Consider a conductor of cross-sectional area containing free electrons per unit volume drifting with speed . In time the electrons that cross a section are those within a length , so their number is and the charge carried is . The current is therefore
(a) Einstein's mass-energy relation is , which says that mass and energy are interconvertible. For example, in the fission of U the products have slightly less total mass than the reactants, and this mass defect
(a) Treating the propagation of sound as isothermal, Newton obtained . This gave about m/s for air, roughly 15% below the measured m/s; its significance is that it first related the speed of sound to the properties of the gas, though it underestimated the value.
Laplace corrected this by noting that the compressions and rarefactions in a sound wave are so rapid that they are adiabatic rather than isothermal, so the relevant modulus is :
With for air this gives about m/s, matching experiment.
(b) The pressure amplitude is the maximum change in pressure from the mean value as a sound wave passes. For a displacement wave , the excess pressure is , so the pressure amplitude is
The pressure wave is out of phase with the displacement wave, so pressure is maximum where displacement is zero (at the nodes), which a cosine curve drawn against the displacement sine curve makes clear.
(a) Diffraction is the bending of light around obstacles and edges and its spreading into the geometrical shadow, producing a fringe pattern, which happens when light meets an aperture comparable in size to its wavelength.
For a single slit of width , every point of the slit acts as a secondary source. Rays that go straight through arrive in phase and form the central bright maximum. At an angle the path difference between the two edges of the slit is , and the slit can be divided into pairs of halves whose contributions cancel when
with weaker secondary maxima lying between them. The result is a bright central band twice as wide as the others, flanked by fringes that fade away on either side.
(b) Polarized light is light whose electric vector is confined to a single plane, unlike unpolarized light, which vibrates in all transverse directions.
Polarization by reflection occurs when unpolarized light reflects off a dielectric at the polarizing (Brewster) angle , defined by . At this angle the reflected ray is completely plane polarized with its vibrations perpendicular to the plane of incidence, because the reflected and refracted rays are then apart and only the perpendicular component is reflected.
(a) Here is in series with across a , supply. The inductive reactance is
and the impedance of the series circuit is
The peak voltage is , so the maximum current is
The current lags the voltage by the phase angle
and this phase difference corresponds to a time lag
So the maximum current is about and it lags the voltage by about .
(b) A p.d. of is applied across Pt electrodes in CuCl, depositing of copper in against a back emf of , with and Cu . Since copper has valency 2, its electrochemical equivalent is
From Faraday's law , the current is
Only the net p.d. left after the back emf drives this current, so the resistance is
The resistance of the voltameter is therefore about .
(c) A standard cell of balances at , so the potential gradient of the wire is
The voltmeter balances at , so its true voltage is
The voltmeter itself reads , so the percentage error is
That is, the voltmeter reads about low.
(a) At wavelength the ejected electrons have , and at they have . Writing the photon energy as , the photoelectric equation gives for the first case, while the shorter wavelength carries three times the energy, so . Subtracting the first from the second eliminates ,
and feeding this back gives the work function
The threshold wavelength then follows from ,
So the threshold wavelength is about nm.
(b) Removing the first electron from neutral helium takes . The second electron is then removed from He, which is a hydrogen-like ion with , so its ionization energy is
Adding the two contributions, the energy needed to detach both electrons is
(c) The Sun has mass and radiates at , and we assume 0.7% of its mass converts to radiation, with and . The energy available from that mass conversion is
Dividing by the radiated power gives the lifetime,
and expressing this in years,
So the estimated lifetime is about years.
We are given the train (source) moving at , a whistle of frequency , and a speed of sound . The driver hears two sounds: the direct whistle at 1000 Hz and the ech...
We are given coherent wavelengths and with the screen at , but the slit separation was illegible in the scan, so the result can only be given symbolically.
For the first-order bright fringe the position is , so the separation between the two first-order fringes is
Once (in metres) is known it can be substituted directly. For instance, if , then . With the slit-separation figure missing, the answer is left as m.