NEB Class 12 · Past paper
The complete NEB Class 12 2081 exam paper for Physics, all 22 questions with solved model answers.
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If the rotational kinetic energy of a body is E and moment of inertia is I, then the angular momentum will be: (A) EI (B) 2EI (C) sqrt(2EI) (D) sqrt(IE)
The correct option is (C) $\sqrt{2EI}$.
The rotational kinetic energy is $E = \tfrac12 I\omega^2$ and the angular momentum is $L = I\omega$. Writing $E$ in terms of $L$,
$$ \begin{aligned} E &= \frac{(I\omega)^2}{2I} \ &= \frac{L^2}{2I} \end{aligned} $$
Rearranging gives $L^2 = 2EI$, so $L = \sqrt{2EI}$.
The displacement-time graph of a particle in SHM is shown [figure: displacement-time graph of SHM]. The maximum velocity of the particle is: (A) 1 cm/s (B) 0.2 cm/s (C) 3.14 cm/s (D) 31.4 cm/s
The correct option is (C) 3.14 cm/s.
Reading an amplitude $A = 1\ \text{cm}$ and a period $T = 2\ \text{s}$ from the graph (the figure was not fully legible in the scan), the angular frequency is
$$ \begin{aligned} \omega &= \frac{2\pi}{T} \ &= \frac{2\pi}{2} \ &= \pi\ \text{rad/s} \end{aligned} $$
so the maximum velocity is
$$ \begin{aligned} v_{max} &= \omega A \ &= \pi\times1 \ &= 3.14\ \text{cm/s} \end{aligned} $$
This value assumes $A = 1$ cm and $T = 2$ s read from the graph.
Two capillary tubes P and Q of radii 4 mm and 2 mm are dipped in water. What is the ratio of heights to which liquid rises in P and Q? (A) 1:2 (B) 2:1 (C) 1:4 (D) 4:1
The correct option is (A) 1:2.
The capillary rise is $h = \dfrac{2T\cos\theta}{r\rho g}$, so for the same liquid $h \propto \dfrac{1}{r}$. Comparing the two tubes,
$$ \begin{aligned} \frac{h_P}{h_Q} &= \frac{r_Q}{r_P} \ &= \frac{2}{4} \ &= \frac{1}{2} \end{aligned} $$
The wider tube P shows the smaller rise, giving the ratio $1:2$.
A system undergoes an isochoric process, its temperature changing from 127 C to 227 C. What is the ratio of initial to final pressure? (A) 5 (B) 4 (C) 127 (D) 227
The correct option is (B) (the ratio is $4:5$). In an isochoric process the volume is constant, so $\dfrac{P}{T}$ stays constant and $\dfrac{P_i}{P_f} = \dfrac{T_i}{T_f}$. Converting the temperatures to kelvin, $T_i = 127 + 273 = 400\ \text{K}$ and $T_f = 227 + 273 = 500\ \text{K}$, so
$$ \begin{aligned} \frac{P_i}{P_f} &= \frac{400}{500} \ &= \frac{4}{5}. \end{aligned} $$
Thus $P_i : P_f = 4 : 5$, which matches the "4" in option B.
A block diagram of a heat engine [figure] shows source 327 C, work W, sink 27 C. What is the efficiency? (A) 25% (B) 50% (C) 60% (D) 75%
The correct option is (B) 50%.
Converting the temperatures to kelvin, $T_H = 327 + 273 = 600\ \text{K}$ and $T_C = 27 + 273 = 300\ \text{K}$, the efficiency of the engine is
$$ \begin{aligned} \eta &= 1 - \frac{T_C}{T_H} \ &= 1 - \frac{300}{600} \ &= 0.5 \ &= 50% \end{aligned} $$
What is the distance between two consecutive particles in a wave which are in the same phase? (A) lambda/4 (B) lambda/2 (C) 3 lambda/4 (D) lambda
The correct option is (D) $\lambda$.
Two particles are in the same phase when their phase differs by $2\pi$, which corresponds to a path difference of one full wavelength. The nearest particles vibrating in identical phase are therefore separated by $\lambda$.
A polarizer is used to: (A) reduce intensity of light (B) produce diffracted light (C) increase intensity of light (D) produce unpolarized light
The correct option is (A) reduce intensity of light.
A polarizer transmits only the component of the electric vector along its axis and blocks the rest, so it converts unpolarized light into plane-polarized light and reduces the transmitted intensity (by half for unpolarized input). It cannot increase intensity, produce diffraction, or create unpolarized light.
Which of the following is a ferromagnetic material? (A) Aluminium (B) Gold (C) Copper (D) Cobalt
The correct option is (D) Cobalt. Cobalt, like iron and nickel, is ferromagnetic: it is strongly attracted by magnets and can retain its magnetization. By contrast aluminium is paramagnetic, while copper and gold are diamagnetic.
The temperature of inversion of a thermocouple is 600 C and the neutral temperature is 310 C. What is the temperature of the cold junction? (A) 0 C (B) 10 C (C) 20 C (D) 30 C
The correct option is (C) 20 C.
The neutral temperature of a thermocouple is the mean of the cold-junction and inversion temperatures, $\theta_n = \dfrac{\theta_c + \theta_i}{2}$. Rearranging for the cold-junction temperature,
$$ \begin{aligned} \theta_c &= 2\theta_n - \theta_i \ &= 2(310) - 600 \ &= 20^\circ\text{C} \end{aligned} $$
In Nepal the supply AC voltage is 220 V. The peak value of the voltage is about: (A) 110 V (B) 220 V (C) 310 V (D) 350 V
The correct option is (C) 310 V. The stated 220 V is the rms value of the supply, and the peak value is $\sqrt2$ times this,
$$ \begin{aligned} V_0 &= \sqrt2,V_{rms} \ &= 1.414\times220 \ &= 311\ \text{V} \approx 310\ \text{V}. \end{aligned} $$
If a P-wave goes from a solid to a liquid, what happens to its velocity? (A) remains the same (B) increases (C) decreases (D) decreases up to zero
The correct option is (C) decreases.
P-waves are longitudinal and travel through both solids and liquids, with a speed $v = \sqrt{E/\rho}$ that depends on the medium's elastic modulus. A liquid has a much smaller effective modulus than a solid because it has no rigidity, so on entering the liquid the P-wave slows down. It does not fall to zero, however, unlike an S-wave, which cannot pass through a liquid at all.
a) Define moment of inertia. [1] b) Derive the moment of inertia of a uniform rod (mass m, length L) about an axis AB through one end, perpendicular to its length. [2] c) A flywheel of moment of inertia 0.32 kgm^2 is rotated steadily at 120 rad/s by a 50 W motor. Calculate (i) its kinetic energy (ii) the frictional couple. [2] OR a) Spring k = 18 N/m; mass 0.15 kg; extension increased by 4 cm then released giving SHM. (i) Define SHM. (ii) Calculate the maximum acceleration and time period. b) Write the total energy of SHM and draw the KE and PE variation. [2]
a) The moment of inertia of a body about an axis is the sum $I = \sum m_i r_i^2$ of each mass element multiplied by the square of its perpendicular distance from the axis. It is the rotational analogue of mass.
b) Take the rod to have linear density $\lambda = m/L$ with the axis through one end. A small element $dx$ at distance $x$ from the axis has mass $\lambda,dx$, and integrating over the whole length,
$$ \begin{aligned} I &= \int_0^L x^2\lambda,dx = \lambda\frac{L^3}{3} \ &= \frac{m}{L}\cdot\frac{L^3}{3} = \boxed{\frac{mL^2}{3}} \end{aligned} $$
c) Here $I = 0.32\ \text{kg m}^2$, the flywheel turns steadily at $\omega = 120\ \text{rad/s}$, and the motor supplies 50 W.
(i) The kinetic energy of rotation is
$$ \begin{aligned} KE &= \tfrac12 I\omega^2 = \tfrac12(0.32)(120)^2 \ &= 0.16\times14400 = 2304\ \text{J} \end{aligned} $$
(ii) At steady speed the motor power just balances the power lost to friction, $P = \tau\omega$, so the frictional couple is
$$ \begin{aligned} \tau &= \frac{P}{\omega} \ &= \frac{50}{120} \ &= 0.417\ \text{N m} \end{aligned} $$
So the kinetic energy is 2304 J and the frictional couple is about 0.42 N m.
OR a) (i) Simple harmonic motion is oscillatory motion in which the acceleration is proportional to the displacement from a fixed point and is always directed toward that point, $a = -\omega^2 x$.
(ii) With $k = 18\ \text{N/m}$, $m = 0.15\ \text{kg}$ and amplitude $A = 4\ \text{cm} = 0.04\ \text{m}$, the angular frequency is
$$ \begin{aligned} \omega &= \sqrt{\frac{k}{m}} \ &= \sqrt{\frac{18}{0.15}} \ &= \sqrt{120} \ &= 10.95\ \text{rad/s} \end{aligned} $$
The maximum acceleration occurs at the extreme position,
$$ \begin{aligned} a_{max} &= \omega^2 A \ &= 120\times0.04 \ &= 4.8\ \text{m/s}^2 \end{aligned} $$
and the time period is
$$ \begin{aligned} T &= \frac{2\pi}{\omega} \ &= \frac{2\pi}{10.95} \ &= 0.573\ \text{s} \end{aligned} $$
So the maximum acceleration is $4.8\ \text{m/s}^2$ and the period is about 0.57 s.
OR b) The total energy of a particle in SHM stays constant, $E = \tfrac12 m\omega^2 A^2 = \tfrac12 kA^2$. As the displacement $x$ varies from $0$ to $\pm A$, the kinetic energy $KE = \tfrac12 k(A^2 - x^2)$ traces a downward parabola with its maximum at $x = 0$, while the potential energy $PE = \tfrac12 kx^2$ traces an upward parabola that is zero at $x = 0$; their sum is the constant horizontal line $E$.
a) Derive a relation between surface tension and surface energy. [2] b) N identical spherical water drops falling at terminal velocity V1 coalesce into a bigger drop falling at V2 (neglect air resistance). (i) Obtain a relation between V1 and V2. [2] (ii) If N = 2 and V1 = 0.4 m/s, calculate V2. [1]
a) Consider a rectangular film of surface tension $T$ closed by a movable wire of length $l$. The film has two surfaces, so the force on the wire is $F = T(2l)$. Pulling the wire out a distance $dx$ to increase the surface area does work
$$ \begin{aligned} \text{Work} &= F,dx \ &= T(2l),dx \ &= T,dA \end{aligned} $$
so the surface energy per unit area is $\dfrac{\text{Work}}{dA} = T$. Hence surface tension equals the surface energy per unit area, both numerically and dimensionally.
b) (i) By Stokes' law the terminal velocity satisfies $v \propto r^2$. Volume is conserved when the drops coalesce, so
$$ \begin{aligned} N\cdot\tfrac43\pi r^3 &= \tfrac43\pi R^3 \quad\Rightarrow\quad R \ &= N^{1/3}r \end{aligned} $$
The ratio of terminal velocities is then
$$ \begin{aligned} \frac{V_2}{V_1} &= \left(\frac{R}{r}\right)^2 \ &= N^{2/3} \quad\Rightarrow\quad V_2 \ &= N^{2/3}V_1 \end{aligned} $$
(ii) With $N = 2$ and $V_1 = 0.4\ \text{m/s}$,
$$ \begin{aligned} V_2 &= 2^{2/3}\times0.4 \ &= 1.587\times0.4 \ &= 0.635\ \text{m/s} \end{aligned} $$
Therefore the bigger drop falls at about $0.63$ m/s.
a) Define indicator diagram. [1] b) Estimate the work done during the II->F process [2]. c) Identify the thermodynamic processes I, II, III, IV. [2] [figure: P-V diagram of a cyclic process with stages I, II, III, IV]
a) An indicator diagram is a graph of pressure against volume, a P-V diagram, drawn for a working substance during a thermodynamic cycle; the area enclosed by the loop equals the net work done per cycle.
b) The work done in any process equals the area under its curve on the P-V diagram,
$$W = \int_{V_i}^{V_f} P,dV$$
The specific pressure and volume coordinates of the II->F stage were not legible in the scanned figure, so a numerical value cannot be computed. For a constant-pressure (isobaric) step the work is $W = P(V_f - V_i)$, and for a constant-volume (isochoric) step it is $W = 0$; substituting the figure's values gives the result.
c) In a typical rectangular P-V cycle the four stages are an isobaric (constant-pressure) process, an isochoric (constant-volume) process, another isobaric process, and another isochoric process, while any curved segment would be isothermal or adiabatic. The exact identification of I to IV depends on the figure, which was not legible in the scan, so match each labelled segment to a horizontal line (isobaric), a vertical line (isochoric) or a curve (isothermal or adiabatic) accordingly.
phy-pv-diagram
a) What correction did Laplace make to Newton's formula for the velocity of sound in a gas? Obtain the corrected formula. [3] b) Calculate the wavelength of a wave in air at 25 C if the frequency is 256 Hz and the velocity of sound at 0 C is 330 m/s. [2]
a) Newton assumed that the compressions and rarefactions in a sound wave take place isothermally, so the relevant elasticity is the pressure $P$ and $v = \sqrt{\dfrac{P}{\rho}}$. This gives about $280\ \text{m/s}$, roughly $15%$ below the measured value. Laplace pointed out that the compressions and rarefactions happen so rapidly that no heat is exchanged, so the process is adiabatic and the correct elasticity is the adiabatic bulk modulus $\gamma P$. The corrected formula is
$$\boxed{v = \sqrt{\frac{\gamma P}{\rho}}}$$
With $\gamma = 1.4$ for air this gives about $332\ \text{m/s}$, in agreement with experiment.
b) Here the frequency is $f = 256\ \text{Hz}$, the velocity of sound at $0^\circ$C is $v_0 = 330\ \text{m/s}$, and we need the wavelength at $25^\circ$C. Since the velocity of sound varies as $\sqrt{T}$ with absolute temperature,
$$ \begin{aligned} v_{25} &= v_0\sqrt{\frac{298}{273}} \ &= 330\sqrt{1.0916} \ &= 330\times1.0448 \ &= 344.8\ \text{m/s} \end{aligned} $$
The wavelength is then
$$ \begin{aligned} \lambda &= \frac{v_{25}}{f} \ &= \frac{344.8}{256} \ &= 1.35\ \text{m} \end{aligned} $$
So the wavelength is about $1.35\ \text{m}$.
a) From a potentiometer circuit for determining the internal resistance of a cell (r), derive an expression for r. [3] b) How can you convert a galvanometer into an ammeter of a suitable range? [2]
a) We first balance the emf of the cell with the circuit open (the key to $R$ open), where the balancing length is $l_1$, so $E = kl_1$ with $k$ the potential gradient of the wire. A known resistance $R$ is then connected across the cell, and now the potentiometer balances the terminal p.d. $V$ at a length $l_2$, so $V = kl_2$. For the cell we have $E = I(R + r)$ and $V = IR$, and taking the ratio,
$$ \begin{aligned} \frac{E}{V} &= \frac{R + r}{R} \ &= \frac{l_1}{l_2}, \end{aligned} $$
which rearranges to
$$\boxed{r = R\left(\frac{l_1 - l_2}{l_2}\right)}.$$
b) A galvanometer is converted into an ammeter by connecting a low resistance, called a shunt $S$, in parallel with it. If $I_g$ is the full-scale current of the galvanometer whose coil resistance is $G$, and $I$ is the range required, then the shunt must carry the remaining current $(I - I_g)$. Since the coil and shunt are in parallel they have the same p.d. across them,
$$I_g G = (I - I_g)S \quad\Rightarrow\quad \boxed{S = \frac{I_g G}{I - I_g}}.$$
The shunt diverts most of the current, so the instrument can read up to $I$ while only $I_g$ actually passes through the coil.
a) Obtain an expression for the magnetic field at a point due to a straight current-carrying conductor using Ampere's circuital law. [3] b) A wire of length 0.5 m carrying 2 A is placed in a uniform magnetic field of 0.4 T. Find the maximum and minimum force on it. [2]
(a) Ampere's circuital law states that $\oint \vec B\cdot d\vec l = \mu_0 I$.
For an infinitely long straight wire carrying current $I$, symmetry tells us that $\vec B$ is tangential and constant in magnitude on any circle of radius $a$ centred on the wire. Taking such a circle as the Amperian loop, the line integral becomes $B,(2\pi a)$, so
$$B,(2\pi a) = \mu_0 I \quad\Rightarrow\quad B = \frac{\mu_0 I}{2\pi a}$$
(b) Here the wire has length $L = 0.5\ \text{m}$, carries a current $I = 2\ \text{A}$, and lies in a uniform field $B = 0.4\ \text{T}$. The force on a current-carrying wire is $F = BIL\sin\theta$, so it depends on the angle between the wire and the field.
The force is maximum when the wire is perpendicular to the field ($\theta = 90^\circ$),
$$ \begin{aligned} F_{max} &= BIL \ &= 0.4\times2\times0.5 \ &= 0.4\ \text{N} \end{aligned} $$
and it is minimum when the wire is parallel to the field ($\theta = 0^\circ$),
$$F_{min} = 0$$
So the maximum force is $0.4\ \text{N}$ and the minimum force is zero.
a) Explain quantization of charge. [2] b) An ion of specific charge 4.40x10^7 C/kg moves in a circular orbit in a magnetic field 0.4 T with velocity 3.52x10^6 m/s. Calculate the radius of the orbit. [2] c) Why are x-rays used in Millikan's oil-drop experiment? [1] OR a) Define half-life and mean life. [2] b) 2 g of a radioactive material has a half-life of 50 years; find its mean life. [2] c) From a decay graph of half-life 10 years, estimate T1. [figure] [1]
a) Quantization of charge means that electric charge exists only in integral multiples of the elementary charge $e = 1.6\times10^{-19}\ \text{C}$, that is $q = ne$ with $n = 0, \pm1, \pm2,\dots$. No isolated charge smaller than $e$ has ever been observed, and Millikan's experiment confirmed this.
b) The ion has specific charge $q/m = 4.40\times10^7\ \text{C/kg}$, moves at $v = 3.52\times10^6\ \text{m/s}$, and the field is $B = 0.4\ \text{T}$. For circular motion the magnetic force provides the centripetal force, $qvB = \dfrac{mv^2}{r}$, so that $r = \dfrac{mv}{qB} = \dfrac{v}{(q/m)B}$. Substituting the values,
$$ \begin{aligned} r &= \frac{3.52\times10^{6}}{(4.40\times10^{7})(0.4)} \ &= \frac{3.52\times10^{6}}{1.76\times10^{7}} = 0.20\ \text{m} \end{aligned} $$
The radius of the orbit is 0.20 m.
c) X-rays ionize the air inside the chamber, freeing ions that attach to the oil drops. This lets the experimenter change the charge on a drop and study how the balancing field changes, which confirms that charge is quantized.
OR a) The half-life $T_{1/2}$ is the time in which half the nuclei of a sample decay. The mean life $\tau$ is the average lifetime of a nucleus, $\tau = \dfrac{1}{\lambda}$.
OR b) The half-life is $T_{1/2} = 50\ \text{years}$, and the 2 g mass does not affect the mean life. Using $\tau = \dfrac{T_{1/2}}{\ln 2}$,
$$ \begin{aligned} \tau &= \frac{T_{1/2}}{\ln 2} \ &= \frac{50}{0.693} \ &= 72.1\ \text{years} \end{aligned} $$
The mean life is therefore about 72.1 years.
OR c) With a half-life of 10 years, $T_1$ is read off as the time at which the activity or number of nuclei falls to the marked fraction: for half the initial value $T_1 = 10$ years, and for one-quarter $T_1 = 20$ years. The exact value depends on the decay graph, which was not legible in the scan.
a) What is rectification? Explain full-wave rectification using two P-N junction diodes. [1+2] b) For the digital circuit given, write the truth table showing outputs A', B' and Y for all inputs A and B. [figure: digital logic circuit] [2]
a) Rectification is the process of converting alternating current (AC) into direct current (DC) using diodes, which conduct in only one direction.
In a full-wave (centre-tapped) rectifier a centre-tapped transformer feeds two diodes $D_1$ and $D_2$, with the load connected between the centre tap and the common cathode point. During the first half-cycle $D_1$ is forward biased and conducts, and during the second half-cycle $D_2$ conducts. In both half-cycles the current flows through the load in the same direction, so the output is a full-wave pulsating DC that uses both halves of the AC input.
b) A common form of this circuit inverts each input ($A' = \overline A$, $B' = \overline B$) and then combines them, for example as $Y = \overline A\cdot\overline B = \overline{A+B}$, which is a NOR:
| A | B | A' | B' | Y = A'.B' |
|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 | 0 |
The exact gate arrangement depends on the figure, which is not fully legible; the table above is for $Y = \overline A\cdot\overline B$ (the NOR of A and B) with the intermediate inverted signals $A'$ and $B'$ shown.
a) A standing wave in an open organ pipe is shown [figure]. (i) Redraw showing nodes and antinodes. (ii) Derive the frequency of this mode. [1+2] b) Effect on a transverse wave in a stretched string if the radius is doubled. [2] c) A car horn of 400 Hz travels toward a traffic post; the police detects a frequency change of 60 Hz when the car crosses. Find the car's velocity (sound speed 340 m/s). [3] OR a) How can a plane wavefront be converted into a spherical one? [2] b) Interference of light; two coherent waves each of intensity I: resultant intensity at (i) constructive and (ii) destructive interference. [3] c) Monochromatic light 5890 A on a grating of 6000 lines/cm. (i) Angle of the second-order image. (ii) Is the third order possible? [2+1]
a) (i) In an open organ pipe both ends are antinodes (A), and the fundamental has a single node (N) in the middle, giving the pattern A - N - A.
(ii) For the fundamental the pipe length holds half a wavelength, $L = \dfrac{\lambda}{2}$, so $\lambda = 2L$ and the frequency is
$$ \begin{aligned} f &= \frac{v}{\lambda} \ &= \boxed{\frac{v}{2L}} \end{aligned} $$
In general $f_n = \dfrac{nv}{2L}$, so all harmonics are present.
b) The wave speed on the string is $v = \sqrt{\dfrac{T}{\mu}}$, where the mass per unit length is $\mu = \pi r^2\rho$, so $v \propto \dfrac{1}{r}$. Doubling the radius makes $\mu$ four times larger and halves the speed to $\dfrac{v}{2}$; for a fixed length the frequency $f = \dfrac{v}{2L}$ therefore halves as well.
c) The horn frequency is $f = 400\ \text{Hz}$, the detected change as the car crosses is $f_1 - f_2 = 60\ \text{Hz}$, and the speed of sound is $v = 340\ \text{m/s}$. Because the car approaches the post first and then recedes, the detected frequencies are $f_1 = f\dfrac{v}{v - v_s}$ on approach and $f_2 = f\dfrac{v}{v + v_s}$ on receding. Subtracting the two,
$$ \begin{aligned} f_1 - f_2 &= f v\left(\frac{1}{v - v_s} - \frac{1}{v + v_s}\right) \ &= \frac{2 f v,v_s}{v^2 - v_s^2} \ &= 60 \end{aligned} $$
Putting in the numbers,
$$\frac{2(400)(340)v_s}{340^2 - v_s^2} = 60$$
which rearranges to a quadratic,
$$ \begin{aligned} 3v_s^2 + 13600v_s - 346800 &= 0 \quad\Rightarrow\quad v_s \ &= 25.3\ \text{m/s} \end{aligned} $$
So the car's velocity is about 25.3 m/s, roughly 91 km/h.
OR a) A plane wavefront is turned into a spherical one by passing it through a converging (convex) lens or reflecting it from a concave mirror: the plane wavefront becomes a converging spherical wavefront that closes onto the focus, and a diverging spherical wavefront beyond it.
OR b) The resultant intensity of two interfering waves is $I_R = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi$, and here $I_1 = I_2 = I$.
For constructive interference ($\phi = 0$),
$$ \begin{aligned} I_R &= I + I + 2I \ &= 4I \end{aligned} $$
For destructive interference ($\phi = \pi$),
$$ \begin{aligned} I_R &= I + I - 2I \ &= 0 \end{aligned} $$
So the resultant intensity is $4I$ for constructive and $0$ for destructive interference.
OR c) The light has wavelength $\lambda = 5890\ \text{Å} = 5.89\times10^{-7}\ \text{m}$ and the grating has $6000$ lines/cm, so the grating spacing is
$$ \begin{aligned} d &= \frac{1\ \text{cm}}{6000} \ &= 1.667\times10^{-6}\ \text{m} \end{aligned} $$
For the second order ($n = 2$), using $d\sin\theta = n\lambda$,
$$ \begin{aligned} \sin\theta &= \frac{2(5.89\times10^{-7})}{1.667\times10^{-6}} \ &= 0.707 \quad\Rightarrow\quad \theta \ &= 45.0^\circ \end{aligned} $$
For the third order ($n = 3$),
$$ \begin{aligned} \sin\theta &= \frac{3(5.89\times10^{-7})}{1.667\times10^{-6}} \ &= 1.06 > 1\quad (\text{impossible}) \end{aligned} $$
So the second-order image appears at $45.0^\circ$, but the third order cannot be seen.
a) A moving-coil galvanometer. (i) Define voltage sensitivity and its factors. (ii) Why is the field made radial? [2+1] b) State Faraday's laws of EM induction. If phi = 4t^3 + 5t + 2 (weber), calculate the instantaneous emf at t = 3 s. [3] c) Derive the energy stored in an inductor. [2] OR a) Field at the centre of a current-carrying circular coil using Biot-Savart's law. [3] b) An indium antimonide slice 2.5 mm thick carries 150 mA; a field 0.5 T gives a maximum Hall voltage 8.75 mV. Find the number of charge carriers per unit volume (charge -1.6x10^-19 C). [2] c) Force per unit length between two parallel conductors carrying currents in the same direction; define one ampere. [3]
a) (i) The voltage sensitivity of a moving-coil galvanometer is the deflection produced per unit voltage applied, $\dfrac{\theta}{V} = \dfrac{NBA}{kR}$. It depends on the number of turns $N$, the field $B$, the coil area $A$, the torsional constant $k$, and the coil resistance $R$.
(ii) The field is made radial (using cylindrical pole pieces and a soft-iron core) so that the plane of the coil always stays parallel to $\vec B$. The torque $NBIA$ is then constant for every deflection, which makes the scale linear, $\theta \propto I$.
b) Faraday's laws state, first, that a changing magnetic flux through a circuit induces an emf, and second, that the induced emf equals the negative rate of change of flux linkage, $e = -\dfrac{d\phi}{dt}$. With $\phi = 4t^3 + 5t + 2$ weber, differentiating gives
$$ \begin{aligned} e &= -\frac{d}{dt}(4t^3 + 5t + 2) \ &= -(12t^2 + 5) \end{aligned} $$
At $t = 3\ \text{s}$ the magnitude is
$$ \begin{aligned} |e| &= 12(9) + 5 \ &= 113\ \text{V} \end{aligned} $$
So the instantaneous emf is $113\ \text{V}$.
c) While the current builds from $0$ to $I$, the back emf $e = L\dfrac{di}{dt}$ opposes the change, so work must be done against it. Integrating this work over the full build-up,
$$ \begin{aligned} W &= \int_0^I Li,di \ &= \boxed{\tfrac12 LI^2} \end{aligned} $$
which is stored as energy in the magnetic field of the inductor.
OR a) By the Biot-Savart law, each current element $I,dl$ of a circular coil sits at distance $r$ (the radius) from the centre and contributes $dB = \dfrac{\mu_0}{4\pi}\dfrac{I,dl}{r^2}$ perpendicular to the plane of the coil. Because every element is at the same distance and its contribution points the same way, we simply add them around the loop,
$$ \begin{aligned} B &= \frac{\mu_0 I}{4\pi r^2}\oint dl \ &= \frac{\mu_0 I}{4\pi r^2}(2\pi r) \ &= \boxed{\frac{\mu_0 I}{2r}}\quad(\text{for } N \text{ turns: } \tfrac{\mu_0 NI}{2r}) \end{aligned} $$
OR b) The indium antimonide slice has thickness $t = 2.5\ \text{mm} = 2.5\times10^{-3}\ \text{m}$, carries $I = 150\ \text{mA} = 0.15\ \text{A}$, sits in a field $B = 0.5\ \text{T}$, and shows a maximum Hall voltage $V_H = 8.75\ \text{mV} = 8.75\times10^{-3}\ \text{V}$, with carrier charge $q = 1.6\times10^{-19}\ \text{C}$. Using the Hall relation $V_H = \dfrac{BI}{nqt}$ and solving for $n$,
$$n = \frac{BI}{V_H q t}$$
Substituting the values,
$$ \begin{aligned} n &= \frac{0.5\times0.15}{(8.75\times10^{-3})(1.6\times10^{-19})(2.5\times10^{-3})} \ &= \frac{0.075}{3.5\times10^{-24}} \ &= 2.14\times10^{22}\ \text{m}^{-3} \end{aligned} $$
So the number of charge carriers per unit volume is about $2.14\times10^{22}\ \text{m}^{-3}$.
OR c) Conductor 1 sets up a field $B_1 = \dfrac{\mu_0 I_1}{2\pi d}$ at the location of conductor 2, and this field exerts a force on the current in conductor 2. The force per unit length is
$$ \begin{aligned} \frac{F}{l} &= B_1 I_2 \ &= \frac{\mu_0 I_1 I_2}{2\pi d} \end{aligned} $$
When the two currents flow in the same direction this force is attractive. One ampere is defined as the steady current which, flowing in two infinite parallel wires placed $1\ \text{m}$ apart in vacuum, produces a force of $2\times10^{-7}\ \text{N}$ per metre of length.
a) Explain the photoelectric effect and write Einstein's photoelectric equation. [2] b) Obtain an expression for the velocity of an electron in the first excited state. [3] c) Calculate the de-Broglie wavelength of an electron accelerated through 300 V. (m = 9.1x10^-31 kg, h = 6.6x10^-34 Js) [3]
a) The photoelectric effect is the emission of electrons from a metal surface when light of frequency above a certain threshold falls on it. Each photon of energy $h\nu$ hands its energy to a single electron; part of it, the work function $\phi$, is used to free the electron, and the remainder appears as the electron's kinetic energy. This is expressed by Einstein's photoelectric equation,
$$\boxed{h\nu = \phi + \tfrac12 mv_{max}^2}.$$
b) In a hydrogen atom the Coulomb attraction provides the centripetal force, so $\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r^2} = \dfrac{mv^2}{r}$, and Bohr's quantization condition gives $mvr = \dfrac{nh}{2\pi}$. Eliminating $r$ between these two relations leads to
$$\boxed{v_n = \frac{e^2}{2\varepsilon_0 nh}} = \frac{2.19\times10^{6}}{n}\ \text{m/s}.$$
The first excited state corresponds to $n = 2$, so
$$ \begin{aligned} v_2 &= \frac{2.19\times10^{6}}{2} \ &= 1.09\times10^{6}\ \text{m/s}. \end{aligned} $$
c) The electron is accelerated through $V = 300\ \text{V}$, with $m = 9.1\times10^{-31}\ \text{kg}$ and $h = 6.6\times10^{-34}\ \text{Js}$. It gains kinetic energy $eV$, so its momentum is $p = \sqrt{2meV}$ and the de-Broglie wavelength is
$$ \begin{aligned} \lambda &= \frac{h}{\sqrt{2meV}} \ &= \frac{6.6\times10^{-34}}{\sqrt{2(9.1\times10^{-31})(1.6\times10^{-19})(300)}}. \end{aligned} $$
Evaluating the denominator,
$$\sqrt{8.74\times10^{-47}} = 9.35\times10^{-24},$$
so
$$ \begin{aligned} \lambda &= \frac{6.6\times10^{-34}}{9.35\times10^{-24}} \ &= 7.06\times10^{-11}\ \text{m}. \end{aligned} $$
The de-Broglie wavelength is therefore about $7.1\times10^{-11}$ m, or roughly $0.71$ Å.
The correct option is (C) .
The rotational kinetic energy is and the angular momentum is . Writing in terms of ,
Rearranging gives , so .
The correct option is (C) 3.14 cm/s.
Reading an amplitude and a period from the graph (the figure was not fully legible in the scan), the angular frequency is
so the maximum velocity is
This value assumes cm and s read from the graph.
The correct option is (A) 1:2.
The capillary rise is , so for the same liquid . Comparing the two tubes,
The wider tube P shows the smaller rise, giving the ratio .
The correct option is (B) (the ratio is ). In an isochoric process the volume is constant, so stays constant and . Converting the temperatures to kelvin, and , so
Thus , which matches the "4" in option B.
The correct option is (B) 50%.
Converting the temperatures to kelvin, and , the efficiency of the engine is
The correct option is (D) .
Two particles are in the same phase when their phase differs by , which corresponds to a path difference of one full wavelength. The nearest particles vibrating in identical phase are therefore separated by .
The correct option is (C) 20 C.
The neutral temperature of a thermocouple is the mean of the cold-junction and inversion temperatures, . Rearranging for the cold-junction temperature,
The correct option is (C) 310 V. The stated 220 V is the rms value of the supply, and the peak value is times this,
The correct option is (C) decreases.
P-waves are longitudinal and travel through both solids and liquids, with a speed that depends on the medium's elastic modulus. A liquid has a much smaller effective modulus than a solid because it has no rigidity, so on entering the liquid the P-wave slows down. It does not fall to zero, however, unlike an S-wave, which cannot pass through a liquid at all.
a) The moment of inertia of a body about an axis is the sum of each mass element multiplied by the square of its perpendicular distance from the axis. It is the rotational analogue of mass.
b) Take the rod to have linear density with the axis through one end. A small element at distance from the axis has mass , and integrating over the whole length,
c) Here , the flywheel turns steadily at , and the motor supplies 50 W.
(i) The kinetic energy of rotation is
(ii) At steady speed the motor power just balances the power lost to friction, , so the frictional couple is
So the kinetic energy is 2304 J and the frictional couple is about 0.42 N m.
OR a) (i) Simple harmonic motion is oscillatory motion in which the acceleration is proportional to the displacement from a fixed point and is always directed toward that point, .
(ii) With , and amplitude , the angular frequency is
The maximum acceleration occurs at the extreme position,
and the time period is
So the maximum acceleration is and the period is about 0.57 s.
OR b) The total energy of a particle in SHM stays constant, . As the displacement varies from to , the kinetic energy traces a downward parabola with its maximum at , while the potential energy traces an upward parabola that is zero at ; their sum is the constant horizontal line .
a) Consider a rectangular film of surface tension closed by a movable wire of length . The film has two surfaces, so the force on the wire is . Pulling the wire out a distance to increase the surface area does work
so the surface energy per unit area is . Hence surface tension equals the surface energy per unit area, both numerically and dimensionally.
b) (i) By Stokes' law the terminal velocity satisfies . Volume is conserved when the drops coalesce, so
The ratio of terminal velocities is then
(ii) With and ,
Therefore the bigger drop falls at about m/s.
a) An indicator diagram is a graph of pressure against volume, a P-V diagram, drawn for a working substance during a thermodynamic cycle; the area enclosed by the loop equals the net work done per cycle.
b) The work done in any process equals the area under its curve on the P-V diagram,
The specific pressure and volume coordinates of the II->F stage were not legible in the scanned figure, so a numerical value cannot be computed. For a constant-pressure (isobaric) step the work is , and for a constant-volume (isochoric) step it is ; substituting the figure's values gives the result.
c) In a typical rectangular P-V cycle the four stages are an isobaric (constant-pressure) process, an isochoric (constant-volume) process, another isobaric process, and another isochoric process, while any curved segment would be isothermal or adiabatic. The exact identification of I to IV depends on the figure, which was not legible in the scan, so match each labelled segment to a horizontal line (isobaric), a vertical line (isochoric) or a curve (isothermal or adiabatic) accordingly.
a) Newton assumed that the compressions and rarefactions in a sound wave take place isothermally, so the relevant elasticity is the pressure and . This gives about , roughly below the measured value. Laplace pointed out that the compressions and rarefactions happen so rapidly that no heat is exchanged, so the process is adiabatic and the correct elasticity is the adiabatic bulk modulus . The corrected formula is
With for air this gives about , in agreement with experiment.
b) Here the frequency is , the velocity of sound at C is , and we need the wavelength at C. Since the velocity of sound varies as with absolute temperature,
The wavelength is then
So the wavelength is about .
a) We first balance the emf of the cell with the circuit open (the key to open), where the balancing length is , so with the potential gradient of the wire. A known resistance is then connected across the cell, and now the potentiometer balances the terminal p.d. at a length , so . For the cell we have and , and taking the ratio,
which rearranges to
b) A galvanometer is converted into an ammeter by connecting a low resistance, called a shunt , in parallel with it. If is the full-scale current of the galvanometer whose coil resistance is , and is the range required, then the shunt must carry the remaining current . Since the coil and shunt are in parallel they have the same p.d. across them,
The shunt diverts most of the current, so the instrument can read up to while only actually passes through the coil.
(a) Ampere's circuital law states that .
For an infinitely long straight wire carrying current , symmetry tells us that is tangential and constant in magnitude on any circle of radius centred on the wire. Taking such a circle as the Amperian loop, the line integral becomes , so
(b) Here the wire has length , carries a current , and lies in a uniform field . The force on a current-carrying wire is , so it depends on the angle between the wire and the field.
The force is maximum when the wire is perpendicular to the field (),
and it is minimum when the wire is parallel to the field (),
So the maximum force is and the minimum force is zero.
a) Quantization of charge means that electric charge exists only in integral multiples of the elementary charge , that is with . No isolated charge smaller than has ever been observed, and Millikan's experiment confirmed this.
b) The ion has specific charge , moves at , and the field is . For circular motion the magnetic force provides the centripetal force, , so that . Substituting the values,
The radius of the orbit is 0.20 m.
c) X-rays ionize the air inside the chamber, freeing ions that attach to the oil drops. This lets the experimenter change the charge on a drop and study how the balancing field changes, which confirms that charge is quantized.
OR a) The half-life is the time in which half the nuclei of a sample decay. The mean life is the average lifetime of a nucleus, .
OR b) The half-life is , and the 2 g mass does not affect the mean life. Using ,
The mean life is therefore about 72.1 years.
OR c) With a half-life of 10 years, is read off as the time at which the activity or number of nuclei falls to the marked fraction: for half the initial value years, and for one-quarter years. The exact value depends on the decay graph, which was not legible in the scan.
a) Rectification is the process of converting alternating current (AC) into direct current (DC) using diodes, which conduct in only one direction.
In a full-wave (centre-tapped) rectifier a centre-tapped transformer feeds two diodes and , with the load connected between the centre tap and the common cathode point. During the first half-cycle is forward biased and conducts, and during the second half-cycle conducts. In both half-cycles the current flows through the load in the same direction, so the output is a full-wave pulsating DC that uses both halves of the AC input.
b) A common form of this circuit inverts each input (, ) and then combines them, for example as , which is a NOR:
| A | B | A' | B' | Y = A'.B' |
|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 | 0 |
The exact gate arrangement depends on the figure, which is not fully legible; the table above is for (the NOR of A and B) with the intermediate inverted signals and shown.
a) (i) In an open organ pipe both ends are antinodes (A), and the fundamental has a single node (N) in the middle, giving the pattern A - N - A.
(ii) For the fundamental the pipe length holds half a wavelength, , so and the frequency is
In general , so all harmonics are present.
b) The wave speed on the string is , where the mass per unit length is , so . Doubling the radius makes four times larger and halves the speed to ; for a fixed length the frequency therefore halves as well.
c) The horn frequency is , the detected change as the car crosses is , and the speed of sound is . Because the car approaches the post first and then recedes, the detected frequencies are on approach and on receding. Subtracting the two,
Putting in the numbers,
which rearranges to a quadratic,
So the car's velocity is about 25.3 m/s, roughly 91 km/h.
OR a) A plane wavefront is turned into a spherical one by passing it through a converging (convex) lens or reflecting it from a concave mirror: the plane wavefront becomes a converging spherical wavefront that closes onto the focus, and a diverging spherical wavefront beyond it.
OR b) The resultant intensity of two interfering waves is , and here .
For constructive interference (),
For destructive interference (),
So the resultant intensity is for constructive and for destructive interference.
OR c) The light has wavelength and the grating has lines/cm, so the grating spacing is
For the second order (), using ,
For the third order (),
So the second-order image appears at , but the third order cannot be seen.
a) (i) The voltage sensitivity of a moving-coil galvanometer is the deflection produced per unit voltage applied, . It depends on the number of turns , the field , the coil area , the torsional constant , and the coil resistance .
(ii) The field is made radial (using cylindrical pole pieces and a soft-iron core) so that the plane of the coil always stays parallel to . The torque is then constant for every deflection, which makes the scale linear, .
b) Faraday's laws state, first, that a changing magnetic flux through a circuit induces an emf, and second, that the induced emf equals the negative rate of change of flux linkage, . With weber, differentiating gives
At the magnitude is
So the instantaneous emf is .
c) While the current builds from to , the back emf opposes the change, so work must be done against it. Integrating this work over the full build-up,
which is stored as energy in the magnetic field of the inductor.
OR a) By the Biot-Savart law, each current element of a circular coil sits at distance (the radius) from the centre and contributes perpendicular to the plane of the coil. Because every element is at the same distance and its contribution points the same way, we simply add them around the loop,
OR b) The indium antimonide slice has thickness , carries , sits in a field , and shows a maximum Hall voltage , with carrier charge . Using the Hall relation and solving for ,
Substituting the values,
So the number of charge carriers per unit volume is about .
OR c) Conductor 1 sets up a field at the location of conductor 2, and this field exerts a force on the current in conductor 2. The force per unit length is
When the two currents flow in the same direction this force is attractive. One ampere is defined as the steady current which, flowing in two infinite parallel wires placed apart in vacuum, produces a force of per metre of length.
a) The photoelectric effect is the emission of electrons from a metal surface when light of frequency above a certain threshold falls on it. Each photon of energy hands its energy to a single electron; part of it, the work function , is used to free the electron, and the remainder appears as the electron's kinetic energy. This is expressed by Einstein's photoelectric equation,
b) In a hydrogen atom the Coulomb attraction provides the centripetal force, so , and Bohr's quantization condition gives . Eliminating between these two relations leads to
The first excited state corresponds to , so
c) The electron is accelerated through , with and . It gains kinetic energy , so its momentum is and the de-Broglie wavelength is
Evaluating the denominator,
so
The de-Broglie wavelength is therefore about m, or roughly Å.