2026 paper questions

Class 12 Physics, FBISE HSSC-II 2026 (1st Annual). 132 questions across 12 units, in paper order.

Set B

53 questions

2026 Annual I, Set B, MCQ 11 markUnit 15
If the distance between the centers of two identical bodies is doubled, the gravitational force between them will become:
  1. A
  2. B
  3. C
  4. DAnswer
Why

. With the distance doubled, . Option C () would be true only if the force were inversely proportional to the distance, not to its square.

2026 Annual I, Set B, MCQ 21 markUnit 15
If the gravitational field strength at the earth's surface is . What is its value at a height, from surface of earth?
  1. A
  2. BAnswer
  3. C
  4. D
Why

At the distance from the centre is , so . Option D () is the value at , where the distance from the centre is .

2026 Annual I, Set B, MCQ 31 markUnit 23
Two stars have the same luminosity. If one appears dimmer, it is because:
  1. AIt is farther from the observerAnswer
  2. BIt has lower temperature
  3. CIt has smaller size
  4. DIt emits less energy
Why

How bright a star appears is its radiant flux intensity, . With the same , a smaller can only come from a larger distance . Options B, C and D would each change the luminosity itself (), but the question says the luminosities are equal.

2026 Annual I, Set B, MCQ 41 markUnit 17
In SHM, at which value of , the instantaneous velocity () is maximum ()?
  1. A
  2. BAnswer
  3. C
  4. D
Why

The formula list printed on the paper gives . Since has its largest value, , at , the velocity is maximum () there. This is the moment the body passes through the mean position; at it is at an extreme and .

2026 Annual I, Set B, MCQ 51 markUnit 17
Total energy of the mass spring system in SHM, is proportional to:
  1. A
  2. B
  3. CAnswer
  4. D
Why

, so the total energy is proportional to . The other options do not have this form (energy grows with the square of both and the amplitude).

2026 Annual I, Set B, MCQ 61 markUnit 18
If the number of lines per unit length in a grating increase, the diffraction pattern (maxima) becomes:
  1. ALess sharp and closely spaced
  2. BSharper and widely spacedAnswer
  3. CMore widely spaced and less sharp
  4. DMore closely spaced and sharper
Why

More lines per unit length means a smaller grating element . From every order moves to a larger angle, so the maxima are more widely spaced. With more lines sharing the beam the maxima are also narrower and brighter, that is sharper. Options C and D each get only one of the two changes right.

2026 Annual I, Set B, MCQ 71 markUnit 20
A.C voltage across a capacitor is , peak value of voltage through the capacitor is:
  1. A
  2. B
  3. CAnswer
  4. D
Why

Compare with : the peak value is the number in front of the sine, . The number is the angular frequency in , and is the reactance , not a voltage.

Note. 'Peak value of voltage through the capacitor' means the peak voltage across it.

2026 Annual I, Set B, MCQ 81 markUnit 17
A simple pendulum has a period of . What will be its period if its length becomes four times large?
  1. A
  2. B
  3. C
  4. DAnswer
Why

, so . Four times the length doubles the period: .

2026 Annual I, Set B, MCQ 91 markUnit 20
A bridge rectifier is used to:
  1. AIncrease voltage
  2. BConvert AC to DCAnswer
  3. CStore charge
  4. DConvert DC to AC
Why

A bridge rectifier uses four diodes to convert AC into (pulsating) DC by full-wave rectification. A transformer changes the voltage, a capacitor stores charge, and an inverter changes DC into AC.

2026 Annual I, Set B, MCQ 101 markUnit 20
In a purely resistive AC circuit, the phase difference between voltage and current is:
  1. AAnswer
  2. B
  3. C
  4. D
Why

In a resistor at every instant, so the voltage and current pass through zero and reach their peaks together: they are in phase, . A phase difference of belongs to a pure inductor or a pure capacitor.

2026 Annual I, Set B, MCQ 111 markUnit 21
In photoelectric effect, which factor increases by increasing the intensity of incident light?
  1. AKinetic energy of electron
  2. BStopping potential
  3. CNumber of emitted electronsAnswer
  4. DWork function
Why

Brighter light means more photons per second, and each photon frees one electron, so more electrons are emitted per second (larger current). The kinetic energy and stopping potential depend on frequency, not intensity, and the work function is a property of the metal.

2026 Annual I, Set B, MCQ 121 markUnit 21
If a particle of mass is moving with speed then De-Broglie wavelength is given by:
  1. A
  2. BAnswer
  3. C
  4. D
Why

de Broglie's relation is with momentum , so . The other options put wrong numerical factors in front.

2026 Annual I, Set B, MCQ 131 markUnit 22
Which of the following is an example of nuclear fusion?
  1. AUranium decay
  2. BAtomic bomb
  3. CEnergy production in the SunAnswer
  4. DRadioactive decay
Why

In the Sun, hydrogen nuclei fuse to form helium, releasing energy: this is fusion. Uranium decay and radioactive decay are spontaneous emissions from unstable nuclei, and an atomic bomb works by uncontrolled fission.

2026 Annual I, Set B, MCQ 141 markUnit 23
If the distance from a star (luminosity ) is doubled, the radiant flux intensity becomes:
  1. AOne-fourthAnswer
  2. BHalf
  3. CDouble
  4. DFour times
Why

, so . Doubling multiplies by . 'Half' would be true only if were proportional to .

2026 Annual I, Set B, MCQ 151 markUnit 24
The magnitude of Coriolis Force () that bends the air packet (mass ) (velocity ) at is:
  1. A
  2. BAnswer
  3. C
  4. D
Why

Using with (printed on the paper): .

These are the numbers of the book's Example 24.2. The other options have the same digits with the wrong power of ten.

2026 Annual I, Set B, MCQ 161 markUnit 23
If a star is receding from the earth it exhibits:
  1. ARed shiftAnswer
  2. BBlue shift
  3. CGreen shift
  4. DYellow light
Why

A receding source's waves are stretched (Doppler effect), so its spectral lines move to longer wavelengths, towards the red end of the spectrum: a red shift. A blue shift is seen when a source approaches; 'green shift' and 'yellow light' are not Doppler effects.

2026 Annual I, Set B, MCQ 171 markUnit 21
Which color of visible light has the lowest frequency?
  1. ARedAnswer
  2. BGreen
  3. CBlue
  4. DViolet
Why

Red light has the longest wavelength in the visible band (about ). Since , it has the lowest frequency. Violet has the highest.

2026 Annual I, Set B, Q.2 (i)1+2 = 3 marksUnit 15
When a satellite is put into a higher orbit what will be the effect on its orbital velocity? Explain mathematically.
Solution

Effect (1 mark). The orbital velocity decreases: a satellite in a higher orbit moves more slowly.

Mathematical explanation (2 marks). For a circular orbit of radius (from the Earth's centre), gravity provides the centripetal force:

A higher orbit has a larger , so is smaller. If becomes 4 times larger, becomes half. Example: the ISS at moves at , but a geostationary satellite () moves at only .

2026 Annual I, Set B, Q.2 (i) OR3 marksUnit 17
Time period of a mass attached at the end of a spring is . Find out the magnitude of acceleration when the displacement is .
Solution

Marks: formula (1), substitution in SI units (1), answer with unit (1). This is Assignment 17.2 of the book.

Given
To find
  1. Angular frequency from the time period.
  2. Magnitude of the acceleration in SHM, .
2026 Annual I, Set B, Q.2 (ii)3 marksUnit 16
What happens to pressure when temperature and molecular speed of gas molecules increase?
Solution

The pressure increases (for a gas kept in a fixed volume).

When the temperature rises, the average translational kinetic energy of the molecules, , increases, so the molecules move faster (). Faster molecules raise the pressure in two ways:

  1. each collision with a wall gives a larger change of momentum, (the molecules hit harder);
  2. the molecules reach the walls more often, because the time between collisions with the same wall, , is shorter.

So more momentum is delivered to each unit area of the wall every second, and the pressure rises. Mathematically:

At constant volume : doubling the absolute temperature doubles the pressure, and doubling the rms speed makes the pressure four times larger. (If the gas is free to expand at constant pressure, its volume increases instead.)

2026 Annual I, Set B, Q.2 (ii) OR1.5+1.5 = 3 marksUnit 18
Briefly explain constructive and destructive interference of light. Give mathematical expression for both in terms of path difference.
Solution
Constructive interference (1.5 marks)

When two coherent light waves meet in phase, crest on crest and trough on trough, their displacements add and the resultant wave has a larger amplitude. The light there is brighter: a bright fringe. This happens when the path difference is a whole number of wavelengths:

Destructive interference (1.5 marks)

When two coherent light waves meet in anti-phase, crest on trough, they cancel and the resultant amplitude is small or zero. The point is dark: a dark fringe. This happens when the path difference is an odd number of half wavelengths:

In Young's double slit the path difference to a point at angle is , so bright fringes need and dark fringes need .

2026 Annual I, Set B, Q.2 (iii)3 marksUnit 17
Justify qualitatively the factors which determine the frequency response and sharpness of resonance.
Solution

The frequency response of a forced oscillator is the way its amplitude changes with the driving frequency. The amplitude is largest (resonance) when the driving frequency equals the natural frequency . The sharpness of resonance is how quickly the amplitude falls on either side of . The factors are:

  • Natural frequency of the oscillator. It decides where the peak lies. It is set by the oscillator itself, for example for a spring or for a pendulum.
  • Damping (the main factor for sharpness). With small damping little energy is lost per cycle, so near energy builds up to a large amplitude and the peak is tall and narrow: a sharp resonance. With heavy damping energy is lost quickly, so the peak is low and broad: a flat resonance.
  • Strength of the driving force. A stronger driving force raises the whole curve but does not move the peak.
Amplitude against driving frequency
2
4
6
8
O
Smaller damping gives a higher and sharper resonance peak; heavy damping reduces the amplitude and broadens the curve.
2026 Annual I, Set B, Q.2 (iii) OR3 marksUnit 20
Explain the function of capacitor in bridge rectifier circuit.
Solution

The output of a bridge rectifier is pulsating DC. A capacitor connected in parallel with the load smooths it into a nearly constant DC; it acts as a filter.

Bridge rectifier with a smoothing capacitor C across the load R.
  1. As the rectified voltage rises to its peak, the capacitor charges to the peak voltage V_0.
  2. When the rectified voltage falls, the diodes switch off and the capacitor discharges slowly through the load, supplying the current.
  3. The next pulse recharges it before its voltage has fallen much, so the gaps between the peaks are filled in.
Smoothing by a capacitor
Fig 20.8 (b) redrawn. After each peak the diodes switch off and the capacitor discharges slowly through the load. With a large time constant RC the voltage hardly falls before the next pulse recharges it (small ripple). With a small RC it falls a long way (large ripple).

The ripple left in the output is small when the time constant RC is much longer than the time between pulses: a larger capacitance or a larger load resistance gives better smoothing, while a small load resistance makes the capacitor discharge faster.

2026 Annual I, Set B, Q.2 (iv)1+2 = 3 marksUnit 20
What is self-inductance? Also write its SI unit and the factors upon which it depends.
Solution
Self-inductance (1 mark)

When the current in a coil changes, an emf \varepsilon_L=-L\,\Delta I/\Delta t is induced in the coil itself (self-induction). The constant L is the self-inductance: the ratio of the emf induced in the coil to the time rate of change of current through it.

L=\dfrac{-\varepsilon_L}{\Delta I/\Delta t}
SI unit and factors (2 marks)

SI unit: the henry (H), 1\,\mathrm{H}=1\,\mathrm{V\,s\,A^{-1}}. A coil has a self-inductance of 1 H if an emf of 1 V is induced in it when its current changes at 1\,\mathrm{A\,s^{-1}}.

For a long coil L=\dfrac{\mu_r\mu_0N^2A}{l}, so L depends on:

  • the number of turns N (L\propto N^2);
  • the area of cross-section A (L\propto A);
  • the length l of the coil for the same turns (L\propto1/l);
  • the core material, its relative permeability \mu_r (an iron core increases L).
2026 Annual I, Set B, Q.2 (iv) OR3 marksUnit 15
Find field strength g at a height of 1500\,\mathrm{km} above surface of Earth.
Solution
Given
  • g = 9.8\,\mathrm{m\,s^{-2}} surface value (paper's data)
  • R_e = 6.4\times10^{6}\,\mathrm{m}
  • h = 1500\,\mathrm{km}=1.5\times10^{6}\,\mathrm{m}
To find
  • g_h
  1. Formula (1 mark). The height is not small compared with R_e, so use the exact relation.
    g_h=g\left(\dfrac{R_e}{R_e+h}\right)^{2}
  2. Substitution (1 mark).
    g_h=9.8\left(\dfrac{6.4\times10^{6}}{6.4\times10^{6}+1.5\times10^{6}}\right)^{2}=9.8\left(\dfrac{6.4}{7.9}\right)^{2}
  3. Answer with unit (1 mark).
    g_h=9.8\times0.656=6.43\,\mathrm{m\,s^{-2}}
g_h\approx6.4\,\mathrm{m\,s^{-2}}
2026 Annual I, Set B, Q.2 (v)3 marksUnit 16
Derive the relation between absolute temperature and average kinetic energy of gas molecules.
Solution

From kinetic theory, the pressure of an ideal gas of N molecules, each of mass m, in volume V is P=\dfrac13\,\dfrac{Nm}{V}\langle v^{2}\rangle.

  1. Multiply by V, and multiply and divide by 2 to bring in the kinetic energy.
    PV=\dfrac{2N}{3}\left\langle\tfrac12mv^{2}\right\rangle\qquad(1)
  2. The ideal gas law with n=N/N_A and k=R/N_A:
    PV=nRT=NkT\qquad(2)
  3. Equate (1) and (2) and cancel N.
    NkT=\dfrac{2N}{3}\left\langle\tfrac12mv^{2}\right\rangle\;\Rightarrow\;T=\dfrac{2}{3k}\left\langle\tfrac12mv^{2}\right\rangle
  4. \frac{2}{3k} is constant, so
    T\propto\langle K.E\rangle\qquad\text{or}\qquad\left\langle\tfrac12mv^{2}\right\rangle=\dfrac32kT

The absolute temperature of an ideal gas is directly proportional to the average translational kinetic energy of its molecules. Temperature is therefore a measure of the average kinetic energy of molecular motion; at 0\,\mathrm{K} this energy would be zero.

T=\dfrac{2}{3k}\left\langle\tfrac12mv^{2}\right\rangle
2026 Annual I, Set B, Q.2 (v) OR3 marksUnit 20
How does doubling the frequency affect the capacitive reactance? Show mathematically.
Solution

Capacitive reactance is inversely proportional to frequency:

X_C=\dfrac{1}{2\pi fC}

When the frequency is doubled (f'=2f) with the same capacitor:

X_C'=\dfrac{1}{2\pi f'C}=\dfrac{1}{2\pi(2f)C}=\dfrac12\left(\dfrac{1}{2\pi fC}\right)=\dfrac{X_C}{2}

So doubling the frequency halves the capacitive reactance, and for the same voltage the current through the capacitor doubles. Reason: the capacitor is charged and discharged twice as often, so twice as much charge flows per second.

Capacitive reactance against frequency
X_C is inversely proportional to f: when the frequency is doubled, the reactance is halved.
2026 Annual I, Set B, Q.2 (vi)1x3 = 3 marksUnit 24
Briefly explain the significance of wind pattern on climate system.
Solution

Marks: one for each of three explained points.

Wind patterns (the trade winds, westerlies and polar easterlies, and local winds) are an essential part of the climate system because they move heat, moisture and energy around the globe. Over a 30-year period their regular patterns form part of the climate of a region.

  1. Heat transport: winds carry heat from the warm tropics towards the cold poles and reduce the temperature difference between them. They also drive the surface ocean currents, which carry warm water towards the poles.
  2. Moisture transport and rainfall: winds carry water vapour from the oceans over the land and so decide where and when rain falls; for example, the monsoon winds bring most of Pakistan's rain.
  3. Dispersion and daily weather: winds spread pollutants (dust, smoke, industrial emissions) and pollen over large areas, and land and sea breezes caused by differential heating change the daily weather of coastal regions.

Changes in wind patterns over long periods therefore change regional and even global climate.

2026 Annual I, Set B, Q.2 (vi) OR3 marksUnit 25
How are X-rays used in imaging internal body structure? Explain briefly.
Solution

Principle. Different tissues absorb X-rays by different amounts. Bone (calcium, high atomic number and density) absorbs X-rays strongly; soft tissue such as fat and muscle absorbs much less; air in the lungs hardly absorbs at all.

Image. A beam of X-rays is passed through the body onto a photographic plate or detector. Where many X-rays arrive (behind soft tissue) the image is dark; behind bone few arrive, so bones cast a white shadow. The result is an image in shades of grey.

Bone absorbs most X-rays and casts a light shadow on the film.

Uses. It shows fractures, foreign bodies (a swallowed coin), teeth and chest problems. For soft tissues a contrast medium (barium meal, iodine dye) is used to increase contrast, or a CT scan gives a detailed image.

2026 Annual I, Set B, Q.2 (vii)1+2 = 3 marksUnit 21
What is meant by pair production? Also calculate the minimum energy required in this process.
Solution
Pair production (1 mark)

Pair production is the creation of a particle and its antiparticle from energy. When a high-energy gamma-ray photon passes near a heavy nucleus, it is completely absorbed and an electron-positron pair is created. Charge, energy and momentum are conserved; energy is converted into mass.

Minimum energy (2 marks)
  1. Energy conservation: the photon supplies the rest mass energy of both particles and their kinetic energies.
    hf=2m_0c^{2}+K.E_{e^-}+K.E_{e^+}
  2. The energy is minimum when the particles are created with zero kinetic energy.
    hf_{\min}=2m_0c^{2}
  3. Substitute m_0=9.11\times10^{-31}\,\mathrm{kg} and c=3\times10^{8}\,\mathrm{m\,s^{-1}}.
    hf_{\min}=2(9.11\times10^{-31})(3\times10^{8})^{2}=1.64\times10^{-13}\,\mathrm{J}
  4. Change into MeV (1\,\mathrm{MeV}=1.6\times10^{-13}\,\mathrm{J}).
    hf_{\min}=\dfrac{1.64\times10^{-13}}{1.6\times10^{-13}}\,\mathrm{MeV}=1.02\,\mathrm{MeV}
hf_{\min}=2m_0c^{2}=1.64\times10^{-13}\,\mathrm{J}=1.02\,\mathrm{MeV}
2026 Annual I, Set B, Q.2 (vii) OR1.5+1.5 = 3 marksUnit 24
Briefly explain the two types of climate inertia.
Solution

Climate inertia is the tendency of the Earth's climate to resist, or respond slowly to, changes in factors such as greenhouse gas concentrations and solar radiation. It is of two types (1.5 marks each):

1. Short-term inertia: the oceans and land surfaces absorb and release heat relatively slowly, so the temperature cannot change instantly. This smooths the daily and seasonal temperature changes: the hottest part of the day comes in the afternoon, a few hours after the Sun is highest, and places near the sea have smaller day and night temperature differences than deserts.

2. Long-term inertia: changes in greenhouse gas concentrations, such as \mathrm{CO_2}, take decades to centuries to affect the climate fully, because the deep ocean warms slowly and the carbon cycle and ice sheets respond slowly. Even if emissions were stabilised today, the Earth would keep warming for many years because of the heat already stored in the oceans.

2026 Annual I, Set B, Q.2 (viii)1+2 = 3 marksUnit 25
What is meant by Piezoelectric effect? Enlist its two applications.
Solution
Piezoelectric effect (1 mark)

The piezoelectric effect is the ability of certain materials (quartz, lead zirconate titanate) to generate an electric charge in response to applied mechanical stress. Squeezing the crystal separates its positive and negative charge centres, so a p.d. appears across its faces. The reverse also happens: an applied electric field changes the shape of the crystal (inverse piezoelectric effect).

Two applications (2 marks)
  1. Production and detection of ultrasound: in medical ultrasound scanners an alternating p.d. makes the crystal vibrate and emit ultrasound, and the returning echoes make it produce electrical signals.
  2. Generation of high voltage: a sharp blow on a piezoelectric crystal gives a spark in gas lighters and stove igniters.

Other applications: quartz crystals keep time in watches and computers (frequency generation), microbalances, and scanning probe microscopes.

2026 Annual I, Set B, Q.2 (viii) OR3 marksUnit 21
What is the role of electron wavelength in achieving high resolution of electron microscope?
Solution

The resolution of a microscope (its ability to show two close points separately) is limited by diffraction: it cannot show details much smaller than the wavelength of the radiation used. So the shorter the wavelength, the higher the resolution.

An electron microscope uses fast electrons, which behave as waves with de Broglie wavelength

\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2meV}}

where V is the accelerating potential difference. For V=10\,\mathrm{kV}, \lambda\approx1.2\times10^{-11}\,\mathrm{m} (0.12\,\text{\AA}), whereas visible light has \lambda\approx4000 to 7000\,\text{\AA}. The electron wavelength is thousands of times shorter, so the electron microscope resolves details down to about 0.1 to 0.2\,\mathrm{nm}, compared with about 200\,\mathrm{nm} for an optical microscope. Increasing V makes \lambda even shorter and the resolution even higher.

2026 Annual I, Set B, Q.2 (ix)1.5+1.5 = 3 marksUnit 22
How do control rods and moderator play an important role in the action of nuclear reactor?
Solution

Control rods (1.5 marks)

Control rods are made of a material that absorbs neutrons strongly, such as boron or cadmium. They can be moved into or out of the reactor core. Pushing them in absorbs more neutrons, so fewer fissions occur; pulling them out allows more fissions. They are set so that only one neutron from each fission goes on to cause another fission, which keeps the chain reaction steady and the power output constant. Dropped fully in, they stop the reaction in an emergency.

Moderator (1.5 marks)

The moderator (water or graphite) slows down the fast neutrons produced by fission (energies of a few MeV) to thermal neutrons of about 0.025\,\mathrm{eV}. Slow neutrons are far more likely to be captured by uranium-235 and cause fission, so the moderator keeps the chain reaction going. The neutrons lose energy by colliding many times with the light nuclei (hydrogen or carbon) of the moderator.

2026 Annual I, Set B, Q.2 (ix) OR2+1 = 3 marksUnit 20
Draw labelled wave form and phasor diagram between applied voltage and current for pure inductor A.C.
Solution
Waveform (2 marks)

For a pure inductor, if I=I_0\sin\omega t then V=V_0\sin(\omega t+\pi/2) with V_0=\omega LI_0. The voltage reaches its peak a quarter cycle (90^\circ) before the current.

Inductor: voltage leads current by 90°
Fig 20.11 (b) redrawn. The voltage reaches its peak a quarter cycle (90^\circ) before the current. The voltage is largest when the current is zero, because that is when the current changes fastest.
Phasor diagram (1 mark)
Phasor diagram for a pure inductor: V is 90^\circ ahead of I (anticlockwise).

The current phasor is drawn along the horizontal and the voltage phasor at 90^\circ anticlockwise from it: the voltage leads the current by 90^\circ (the current lags).

2026 Annual I, Set B, Q.2 (x)3 marksUnit 24
Why does the equatorial region receive more energy than polar regions?
Solution

The equatorial region receives more solar energy per square metre than the polar regions because:

  1. Angle of the Sun's rays: at the equator sunlight falls almost vertically (direct sunlight), so a beam covers a small area and the energy per unit area is high. Near the poles the same beam falls at a low angle (oblique sunlight) and spreads over a much larger area, so the energy per unit area is small. On level ground I = I_0\cos\theta, where \theta is the latitude.
  2. Longer path through the atmosphere: oblique rays near the poles pass through more air, so more of their energy is absorbed, scattered and reflected before it reaches the ground.
  3. Ice and daylight: polar ice and snow reflect most of the sunlight that arrives (high albedo), while oceans and forests near the equator absorb it; the poles also have months of darkness in winter.
Equal beams of sunlight: a small area is heated at the equator, a large area near the pole.
2026 Annual I, Set B, Q.2 (x) OR1+2 = 3 marksUnit 23
To which idea does a red-shift exhibited by a star leads and why?
Solution

Idea (1 mark): red shift leads to the idea that the universe is expanding: galaxies, with their stars, are moving away from us and from each other.

Why (2 marks): red shift means the spectral lines of the star or galaxy are at longer wavelengths than their laboratory values. By the Doppler effect, this happens when the source moves away from the observer, with speed given by

\dfrac{\Delta\lambda}{\lambda}=\dfrac{\Delta f}{f}=\dfrac{v}{c}

Almost all galaxies, in every direction, show red shift, and the farther they are the greater the shift (v=H_0d). The simple explanation is that the space between the galaxies is stretching, like the rubber of an inflating balloon carrying its dots apart. So the universe is expanding; running this backwards leads to the Big Bang.

2026 Annual I, Set B, Q.2 (xi)3 marksUnit 16
Calculate root mean square speed of oxygen molecule at standard temperature and pressure.
Solution
Given
  • T = 0\,^\circ\mathrm{C}=273\,\mathrm{K} standard temperature
  • P = 1.01\times10^{5}\,\mathrm{Pa} standard pressure
  • M = 32\,\mathrm{g\,mol^{-1}}=0.032\,\mathrm{kg\,mol^{-1}} molar mass of \mathrm{O_2}
  • R = 8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}
To find
  • v_{\mathrm{rms}}
  1. Formula (from the paper's formula sheet):
    v_{\mathrm{rms}}=\sqrt{\dfrac{3RT}{M}}
  2. Substitute, with M in \mathrm{kg\,mol^{-1}}:
    \begin{aligned}v_{\mathrm{rms}}&=\sqrt{\dfrac{3\times8.314\times273}{0.032}}=\sqrt{\dfrac{6809}{0.032}}\\&=\sqrt{2.128\times10^{5}}\end{aligned}
  3. Answer with unit:
    v_{\mathrm{rms}}=461\,\mathrm{m\,s^{-1}}

Alternative method: find the mass of one molecule, then use v_{\mathrm{rms}}=\sqrt{3kT/m}.

\begin{gathered}m=\dfrac{M}{N_A}=\dfrac{0.032}{6.022\times10^{23}}=5.31\times10^{-26}\,\mathrm{kg}\\v_{\mathrm{rms}}=\sqrt{\dfrac{3\times1.38\times10^{-23}\times273}{5.31\times10^{-26}}}=461\,\mathrm{m\,s^{-1}}\end{gathered}

This is Example 16.2 of the book. Marks: formula 1, substitution 1, answer with unit 1.

v_{\mathrm{rms}}\approx461\,\mathrm{m\,s^{-1}}
2026 Annual I, Set B, Q.2 (xi) OR1.5+1.5 = 3 marksUnit 22
Briefly explain the terms: a. Mass Defect b. Binding Energy
Solution

(a) Mass defect (1.5 marks)

The mass of a nucleus is always less than the sum of the masses of its separate protons and neutrons. This difference is the mass defect: \Delta m=\left[Zm_p+(A-Z)m_n\right]-m_{\text{nucleus}}. For deuterium, \Delta m=(1.007276+1.008665)-2.013553=0.002388\,\mathrm{u}. The missing mass has been converted into the energy released when the nucleons bound together.

(b) Binding energy (1.5 marks)

The binding energy is the minimum energy required to break a nucleus into its separate nucleons; the same energy is released when the nucleus forms. It is the energy equivalent of the mass defect: E_b=\Delta mc^{2}, or E_b(\mathrm{MeV})=\Delta m(\mathrm{u})\times931.5. For deuterium, E_b=0.002388\times931.5=2.22\,\mathrm{MeV}. The greater the binding energy per nucleon, the more stable the nucleus.

2026 Annual I, Set B, Q.2 (xii)1.5+1.5 = 3 marksUnit 15
What factors determine the strength of gravitational field g around a planet?
Solution

The field strength at a distance r from the centre of a planet of mass M is

g=\dfrac{GM}{r^{2}}

Factor 1: mass of the planet (1.5 marks). g\propto M. A more massive planet has a stronger field; for example Jupiter, with 316 times the Earth's mass, has g\approx24.8\,\mathrm{m\,s^{-2}} at its surface.

Factor 2: distance from the planet's centre (1.5 marks). g\propto1/r^{2}. At the surface r equals the planet's radius R, so for the same mass a smaller planet has a stronger surface field. Above the surface, g_h=GM/(R+h)^{2} falls with height; at h=R it is only g/4.

g does not depend on the mass of the body placed in the field.

2026 Annual I, Set B, Q.2 (xii) OR1x3 = 3 marksUnit 17
What is meant by light damping and heavy damping? Show these graphically. (Displacement-time graph)
Solution

Light damping (1 mark). The damping force is small, so the body keeps oscillating while its amplitude decreases gradually with time. Example: a swing, or a pendulum swinging in air.

Heavy damping (1 mark). The damping force is very large, so the body does not complete even one oscillation; it moves back slowly to its equilibrium position and takes a long time to come to rest. Example: a mass on a spring in thick oil.

Graph (1 mark).

Light and heavy damping
Light damping: the curve crosses the time axis again and again with shrinking peaks. Heavy damping: the curve never crosses the axis and approaches it slowly.
2026 Annual I, Set B, Q.2 (xiii)2+1 = 3 marksUnit 18
What is meant by grating element of a diffraction grating? Also write its formula.
Solution
Grating element (2 marks)

A diffraction grating is a plate with a very large number of equally spaced, parallel slits. The grating element (grating constant) d is the distance between the centres of two adjacent slits. It equals the width of one transparent slit a plus the width of one opaque line b, so d=a+b. It is a few micrometres, a little larger than the wavelength of light, and its SI unit is the metre.

Formula (1 mark)
d=\dfrac{\text{unit length of grating}}{\text{number of lines ruled on it}}=\dfrac{1}{N}

For example, a grating with 5000 lines per cm has d=\dfrac{1\times10^{-2}\,\mathrm{m}}{5000}=2.0\times10^{-6}\,\mathrm{m}. The grating element appears in the grating equation d\sin\theta=m\lambda.

2026 Annual I, Set B, Q.2 (xiii) OR1.5+1.5 = 3 marksUnit 20
Distinguish graphically between half-wave and full-wave rectification.
Solution
Half-wave rectification (1.5 marks)

One diode passes only the positive half-cycles; the output is zero during every negative half-cycle. One pulse per input cycle.

Half-wave rectification
Fig 20.5 redrawn. Only the positive half-cycles reach the load. The negative half-cycles are blocked, so the output is zero for half of every cycle. One pulse per cycle.
Full-wave rectification (1.5 marks)

Four diodes (bridge) pass both half-cycles, and the negative halves are turned upward, so the current through the load flows one way all the time. Two pulses per input cycle.

Full-wave rectification
Fig 20.7 redrawn. Both half-cycles reach the load and both drive current the same way through it, so the negative halves appear flipped upward. Two pulses per cycle.

The full-wave output has no gaps, twice the pulse frequency and twice the average value (2V_0/\pi against V_0/\pi).

2026 Annual I, Set B, Q.2 (xiv)2+1 = 3 marksUnit 18
How does increasing slit separation affect fringe spacing? Discuss with mathematical relation.
Solution
Effect (2 marks)

Increasing the slit separation d decreases the fringe spacing: the bright and dark fringes come closer together. With the slits farther apart, the path difference d\sin\theta grows faster as we move across the screen, so it reaches one more wavelength after a shorter distance. If d is made too large, the fringes are so close that they cannot be seen separately.

Mathematical relation (1 mark)
\Delta y=\dfrac{\lambda L}{d}\quad\Rightarrow\quad\Delta y\propto\dfrac{1}{d}\quad(\lambda,\ L\ \text{constant})

For example, doubling d halves the fringe spacing: with \lambda=600\,\mathrm{nm} and L=1\,\mathrm{m}, d=0.25\,\mathrm{mm} gives \Delta y=2.4\,\mathrm{mm}, while d=0.50\,\mathrm{mm} gives \Delta y=1.2\,\mathrm{mm}.

2026 Annual I, Set B, Q.2 (xiv) OR3 marksUnit 19
Four capacitors of capacitance ‘C’ are connected as shown. Calculate their equivalent capacitance.
Solution
  1. Each branch has two capacitors C in series.
    \dfrac{1}{C_s}=\dfrac{1}{C}+\dfrac{1}{C}=\dfrac{2}{C}\ \Rightarrow\ C_s=\dfrac{C}{2}
  2. The two branches are joined between the same points A and B, so they are in parallel: add them.
    C_e=\dfrac{C}{2}+\dfrac{C}{2}
  3. Result.
    C_e=C
C_e=C
2026 Annual I, Set B, Q.32+4+1 = 7 marksUnit 15
Briefly describe gravitational potential and derive the expression \varphi=-\dfrac{GM}{r} at a distance r from the centre of Earth. Make a sketch.
Solution

Gravitational potential (2 marks)

The gravitational potential \varphi (or V) at a point is the work done per unit mass in bringing a small test mass from infinity to that point. It equals the gravitational potential energy per unit mass, \varphi=U/m. It is a scalar with SI unit \mathrm{J\,kg^{-1}}. Its zero is at infinity, so it is negative at every finite distance from the Earth, and it increases (becomes less negative) with distance.

Derivation (4 marks)

Here M is the mass of the Earth (M=M_E). The force changes with distance, so the path is split into small steps and the work in each step is added.

Derivation Derivation of \varphi=-\dfrac{GM_E}{r}

Aim. Find the work done per unit mass in bringing a small test mass m from infinity to a point at distance r from the centre of the Earth.

Assumptions.
  • The Earth is a uniform sphere of mass M_E and radius R_E; it acts as a point mass at its centre O.
  • The mass m is moved slowly, so its kinetic energy does not change and the work done on it by the agent goes into potential energy.
  • U=0 at infinity.
The path from the surface (point 0, distance R_E from the centre O) to a very distant point n is divided into small steps \Delta r; in each step the force is taken as constant.
  1. By the work and energy principle, when m is moved slowly the work done on it by the external agent equals the change in its potential energy.
    \Delta P.E.=W_{net}\qquad(1)
  2. The force changes with distance, so W=Fd cannot be used for the whole path at once. Divide the path between the surface (point 0) and a very distant point n into n small steps of length \Delta r, so small that the force is almost constant in each. The total work is the sum of the works in the steps.
    W_{net}=W_1+W_2+W_3+\dots+W_{n-1}+W_n\qquad(2)
  3. Take the step between point 1 (distance r_1) and point 0 (distance R_E). To bring m in slowly, the agent pulls outward with a force F_{av} equal to the pull of gravity, while m moves inward through \Delta r. Force and displacement are opposite: \theta=180^\circ and \cos180^\circ=-1.
    W_1=F_{av}\,\Delta r\cos180^\circ=-F_{av}\,\Delta r\qquad(3)
  4. The average force in this step follows from Newton's law, using the average distance r_{av} of the step.
    F_{av}=\dfrac{GM_Em}{r_{av}^{2}}\qquad(4)
  5. The average distance is neither R_E nor r_1 but their mean. From the figure \Delta r=r_1-R_E, so r_1=R_E+\Delta r.
    r_{av}=\dfrac{R_E+r_1}{2}=\dfrac{2R_E+\Delta r}{2}=R_E+\dfrac{\Delta r}{2}\qquad(5\text{ to }7)
  6. Square the average distance.
    r_{av}^{2}=\left(R_E+\dfrac{\Delta r}{2}\right)^{2}=R_E^{2}+R_E\,\Delta r+\dfrac{(\Delta r)^{2}}{4}
  7. \Delta r is very small, so the term (\Delta r)^{2}/4 is negligible compared with the others. Then put \Delta r=r_1-R_E.
    r_{av}^{2}\approx R_E^{2}+R_E\,\Delta r=R_E^{2}+R_E(r_1-R_E)=R_Er_1\qquad(8,\ 9)
  8. Put Eq. (9) in Eq. (4).
    F_{av}=\dfrac{GM_Em}{R_Er_1}\qquad(10)
  9. Put Eq. (10) and \Delta r=r_1-R_E (Eq. 11) in Eq. (3), then split the fraction.
    W_1=-\dfrac{GM_Em}{R_Er_1}\,(r_1-R_E)=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_1}\right)\qquad(12)
  10. The same working applies to every other step.
    \begin{aligned}W_2&=-GM_Em\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)\\ W_3&=-GM_Em\left(\dfrac{1}{r_2}-\dfrac{1}{r_3}\right)\\ &\ \ \vdots\\ W_{n-1}&=-GM_Em\left(\dfrac{1}{r_{n-2}}-\dfrac{1}{r_{n-1}}\right)\\ W_n&=-GM_Em\left(\dfrac{1}{r_{n-1}}-\dfrac{1}{r_n}\right)\end{aligned}\qquad(13\text{ to }16)
  11. Add all the steps as in Eq. (2). Every middle term appears once with + and once with -, so they cancel in pairs; only the first and last terms remain.
    \begin{aligned}W_{net}&=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_1}+\dfrac{1}{r_1}-\dfrac{1}{r_2}+\dots+\dfrac{1}{r_{n-1}}-\dfrac{1}{r_n}\right)\\ &=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_n}\right)\end{aligned}\qquad(17)
  12. By Eq. (1), this is the change in potential energy when m is brought from point n to the surface.
    \Delta P.E.=U_{\text{surface}}-U_n=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_n}\right)\qquad(18)
  13. Now take point n at infinity: r_n=\infty, so 1/r_n=0, and U_n=0 (our zero). The work done in bringing m from infinity to the surface is then
    U=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{\infty}\right)=-\dfrac{GM_Em}{R_E}
  14. The same steps, started from any point P at distance r from the centre instead of the surface, give the general result.
    U=-\dfrac{GM_Em}{r}\qquad(15.9)
  15. By definition the potential is the work done per unit mass, so divide by m. For a unit mass (m=1\,\mathrm{kg}) the work itself is the potential.
    \varphi=\dfrac{U}{m}=\dfrac{-GM_Em/r}{m}=-\dfrac{GM_E}{r}
Result\varphi=-\dfrac{GM_E}{r}

The potential is zero at infinity and negative at every finite distance; it rises towards zero as r increases. It depends only on the Earth's mass and the distance from its centre, not on the test mass. For any point mass M: \varphi=-GM/r.

Sketch (1 mark)

The labelled diagram of the path (above) is the sketch that goes with the derivation. The graph of \varphi against r shows the result:

Sketch: \varphi (\mathrm{MJ\,kg^{-1}}) against distance from the centre
\varphi is most negative (-62.5\,\mathrm{MJ\,kg^{-1}}) at the surface and rises towards zero as r increases.
2026 Annual I, Set B, Q.3 OR1+3+2+1 = 7 marksUnit 17
What is meant by simple harmonic motion (SHM)? Show that mass attached to a spring executes SHM. Derive the expressions for frequency 'f' and period 'T'. Draw diagram as well.
Solution

Simple harmonic motion (1 mark)

SHM is the oscillatory motion in which the acceleration of a particle is directly proportional to its displacement from the mean position and is always directed towards the mean position:

a\propto -x\qquad\text{or}\qquad a=-\omega^{2}x

Diagram (1 mark)

Mass m on a smooth horizontal surface. (a) Pulled right by x: the spring pulls left. (b) At x=0 the spring force is zero. (c) Pushed left: the spring pushes right. The force always points towards x=0.

A block of mass m on a smooth horizontal surface is attached to a spring of spring constant k whose other end is fixed. At x=0 the spring is unstretched. When the block is displaced by x and released, the spring exerts a restoring force towards x=0, and the block oscillates about this mean position.

Mass-spring system executes SHM (3 marks)

Derivation Motion of a mass attached to a spring is SHM

Aim. Show that the acceleration of the block is proportional to its displacement and directed towards the mean position, and find the angular frequency.

Assumptions.
  • The spring is massless and obeys Hooke's law.
  • The surface is frictionless and air resistance is negligible, so no energy is lost.
  • The block moves only along the line of the spring (the x-axis).
  1. The only horizontal force on the block is the spring force. By Hooke's law it is proportional to the displacement x and opposite to it.
    F=-kx\qquad(17.1)
  2. The block has mass m, so by Newton's second law the same force gives it an acceleration a.
    F=ma\qquad(17.2)
  3. Both equations describe the same force, so equate them.
    ma=-kx
  4. Divide both sides by m to get the acceleration.
    a=-\dfrac{k}{m}\,x\qquad(17.3)
  5. k and m are fixed positive quantities for a given spring and block, so k/m is a positive constant. Write it as \omega^{2} (a square is always positive).
    \dfrac{k}{m}=\omega^{2}\quad\Longrightarrow\quad a=-\omega^{2}x
  6. Because \omega^{2} is constant, the acceleration is directly proportional to the displacement. The minus sign shows that it is always opposite to the displacement, that is, towards the mean position. These are exactly the two conditions of SHM.
    a\propto -x
  7. Take the square root of k/m=\omega^{2} to get the angular frequency of the oscillating mass.
    \omega=\sqrt{\dfrac{k}{m}}\qquad(17.4)
Resulta=-\dfrac{k}{m}\,x=-\omega^{2}x,\qquad \omega=\sqrt{\dfrac{k}{m}}

The mass-spring system satisfies the definition of SHM. How quickly it oscillates depends only on the stiffness k and the mass m, not on how far the block was pulled.

Frequency and time period (2 marks)

Derivation Time period and frequency of a mass-spring system

Aim. Express the time period T and frequency f in terms of m and k.

  1. Angular frequency is related to frequency and time period by its definition.
    \omega=2\pi f=\dfrac{2\pi}{T}
  2. Make T the subject.
    T=\dfrac{2\pi}{\omega}
  3. Substitute \omega=\sqrt{k/m} from (17.4). Dividing by \sqrt{k/m} is the same as multiplying by \sqrt{m/k}.
    T=\dfrac{2\pi}{\sqrt{k/m}}=2\pi\sqrt{\dfrac{m}{k}}
  4. Frequency is the reciprocal of the time period.
    f=\dfrac{1}{T}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}
ResultT=2\pi\sqrt{\dfrac{m}{k}},\qquad f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

A heavier mass oscillates more slowly (T\propto\sqrt{m}) because it has more inertia. A stiffer spring oscillates faster (T\propto 1/\sqrt{k}) because it pulls back harder. The amplitude does not appear, so the period is the same for large and small oscillations.

Since the acceleration is proportional to the displacement and directed towards the mean position, the mass-spring system executes SHM with T=2\pi\sqrt{m/k} and f=\frac{1}{2\pi}\sqrt{k/m}, independent of the amplitude.

2026 Annual I, Set B, Q.44+1+2 = 7 marksUnit 18
Explain the working of Young's double slit experiment. Make diagram. Also derive the expression for fringe spacing.
Solution
Working of Young's double-slit experiment (4 marks)

Young's experiment (1801) produces interference of light from two coherent sources and shows that light behaves as a wave.

  1. Monochromatic light (for example from a sodium lamp or a laser) falls on a narrow single slit S_0. Light spreading out from S_0 by diffraction reaches two narrow, parallel slits S_1 and S_2, placed close together (separation d, about 0.5\,\mathrm{mm}) and at equal distances from S_0.
  2. S_1 and S_2 receive light from the same wavefront, so they act as two coherent sources vibrating in phase.
  3. Each slit diffracts the light, and the two sets of waves overlap and interfere in the region beyond the slits. A screen is placed at a distance L (about 1\,\mathrm{m}) from the slits.
  4. At points where the path difference S_2P-S_1P is 0,\ \lambda,\ 2\lambda,\dots the waves arrive in phase and interfere constructively, giving bright fringes.
  5. At points where the path difference is \tfrac12\lambda,\ \tfrac32\lambda,\dots the waves arrive in anti-phase and interfere destructively, giving dark fringes.
  6. So equally spaced, alternate bright and dark fringes parallel to the slits appear on the screen. The central fringe at O is bright because the path difference there is zero. With white light the central fringe is white and the others are coloured.
Diagram (1 mark)
Young's double-slit experiment: source, single slit S_0, double slit S_1S_2 (separation d), screen at distance L with bright and dark fringes.
Derivation of fringe spacing (2 marks)

Derivation Fringe spacing \Delta y=\lambda L/d

Aim. Find the distance between two neighbouring bright fringes.

Assumptions.
  • The screen is far away (L\gg d) and the angle \theta is small.
Path difference BC =d\sin\theta for a point P at height y above the centre O.
  1. Path difference. Draw S_1C perpendicular to S_2P. For a distant screen the rays are nearly parallel, so the path difference is BC. In the right triangle S_1S_2C the angle at S_1 equals \theta and the hypotenuse is d.
    S_2P-S_1P=BC=d\sin\theta
  2. Bright fringe. For the mth bright fringe the path difference is m whole wavelengths.
    d\sin\theta=m\lambda
  3. Position. \theta is small, so \sin\theta\approx\tan\theta=y_m/L, where y_m is the distance of the fringe from O.
    d\,\dfrac{y_m}{L}=m\lambda\;\Rightarrow\;y_m=\dfrac{m\lambda L}{d}
  4. Spacing. The next bright fringe is at y_{m+1}=(m+1)\lambda L/d. Subtract.
    \Delta y=y_{m+1}-y_m=\dfrac{(m+1)\lambda L}{d}-\dfrac{m\lambda L}{d}=\dfrac{\lambda L}{d}
Result\Delta y=\dfrac{\lambda L}{d}

The fringes are equally spaced. The spacing increases with the wavelength and the screen distance, and decreases when the slit separation increases. Dark fringes, at y=\left(m+\tfrac12\right)\lambda L/d, have the same spacing.

\Delta y=\dfrac{\lambda L}{d}
2026 Annual I, Set B, Q.4 OR1+5+1 = 7 marksUnit 19
What is meant by electric potential? Derive an expression for electric potential in the field due to a point charge. Make sketch.
Solution
Electric potential (1 mark)

Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point, keeping the charge in equilibrium. It is the potential energy per unit charge, V=\dfrac{W}{q_0}=\dfrac{U}{q_0}. It is a scalar, measured in volts (1\,\mathrm{V}=1\,\mathrm{J\,C^{-1}}).

Derivation (5 marks)

The field of a point charge is not uniform, so the path from infinity is divided into small steps in which the field is almost constant. The work in every step is found and the steps are added.

Derivation Potential V at distance r from a point charge Q

Aim. Find the electric potential at a point 1, a distance r from an isolated point charge +Q, by bringing a test charge q from infinity to that point.

Assumptions.
  • +Q is an isolated point charge fixed at the origin, in vacuum or air.
  • The test charge q is small and positive, and it is moved slowly so that it stays in equilibrium (no kinetic energy is gained).
  • The path from a far point N to point 1 is divided into small steps of length \Delta r; in each step the field E is almost constant. r_1, r_2,\ldots,r_N are the distances of points 1, 2,\ldots,N from Q.
  • The potential is taken as zero at infinity.
  1. The field of +Q points away from Q. To move q towards Q slowly (in equilibrium), we apply a force F equal and opposite to the electric force. So F=qE and it points towards Q, in the same direction as the displacement.
    F=qE=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Qq}{r^{2}}
  2. Take the first step, from N to N−1. Its length is \Delta r=r_N-r_{N-1}. The force and the displacement are in the same direction (\theta=0^\circ), so the work done by the applied force is
    \Delta W_{N\to N-1}=F\,\Delta r\cos0^\circ=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Qq}{r^{2}}\,(r_N-r_{N-1})
  3. Inside this tiny step the distance lies between r_{N-1} and r_N, which are very close. So we take r^2 as the product of the two end distances (a very good average for a small step).
    r^{2}\approx r_{N-1}\,r_N
  4. Substitute and split the fraction into two terms.
    \Delta W_{N\to N-1}=\dfrac{Qq}{4\pi\varepsilon_0}\,\dfrac{r_N-r_{N-1}}{r_{N-1}\,r_N}=\dfrac{Qq}{4\pi\varepsilon_0}\left(\dfrac{1}{r_{N-1}}-\dfrac{1}{r_N}\right)\quad\text{(i)}
  5. The same working holds for every other step.
    \begin{aligned}\Delta W_{N-1\to N-2}&=\dfrac{Qq}{4\pi\varepsilon_0}\left(\dfrac{1}{r_{N-2}}-\dfrac{1}{r_{N-1}}\right)&&\text{(ii)}\\ &\ \ \vdots\\ \Delta W_{2\to1}&=\dfrac{Qq}{4\pi\varepsilon_0}\left(\dfrac{1}{r_{1}}-\dfrac{1}{r_{2}}\right)&&\text{(iii)}\end{aligned}
  6. Add the work of all the steps. Each middle term such as 1/r_{N-1} appears once with a plus sign and once with a minus sign, so it cancels. Only the first and the last terms are left.
    \Delta W_{N\to1}=\dfrac{Qq}{4\pi\varepsilon_0}\left[\left(\dfrac{1}{r_{N-1}}-\dfrac{1}{r_N}\right)+\cdots+\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)\right]=\dfrac{Qq}{4\pi\varepsilon_0}\left(\dfrac{1}{r_1}-\dfrac{1}{r_N}\right)\quad\text{(iv)}
  7. Now let the starting point N be at infinity: r_N\to\infty, so 1/r_N\to0. Write r for r_1. This work is stored as the potential energy U of q at distance r from Q.
    U=\Delta W_{\infty\to1}=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Qq}{r}\qquad(19.4)
  8. Electric potential is potential energy per unit charge, so divide by q.
    V=\dfrac{U}{q}=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Qq}{r}\div q=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{r}\qquad(19.5)
ResultV=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{r}=\dfrac{kQ}{r},\qquad k=\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}

The potential of a point charge falls off as 1/r, more slowly than the field (1/r^2). It has the same value at all points equidistant from Q, so the equipotential surfaces around a point charge are spheres centred on Q. V is positive around a positive charge and negative around a negative charge, and it is zero at infinity.

Sketch (1 mark)
The test charge q is moved from a far point N (at infinity) to point 1 against the field of +Q, in small steps \Delta r.

The potential falls off as 1/r with distance from the charge:

Potential and field around a point charge
Values are relative to those at a distance r_0. Doubling the distance halves the potential but reduces the field to one quarter.
V=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Q}{r}
2026 Annual I, Set B, Q.52+2+2 = 6 marksUnit 20
In an RL series circuit, will the current lag or lead the applied alternating voltage? Explain the answer with phasor diagram. Also find an expression for the impedance of RL series circuit.
Solution
The current lags the voltage (2 marks)

In an RL series circuit the current lags behind the applied voltage by an angle \varphi, where 0<\varphi<90^\circ. The same current flows through R and L. The voltage across R is in phase with the current, but the voltage across L leads the current by 90^\circ, because the back emf of the coil opposes every change in current. The applied voltage is the phasor sum of the two, so it is ahead of the current by \varphi; that is, the current lags.

RL series circuit.
Phasor diagram (2 marks)
Phasor diagram of the RL series circuit. V_R (OA) is along I, V_L (AB) leads by 90^\circ, and the supply voltage V (OB) leads I by \varphi.
  • The current I is the reference phasor (horizontal).
  • V_R=IR is drawn along I (OA).
  • V_L=IX_L is drawn at 90^\circ ahead of I (AB).
  • The resultant V (OB) makes the angle \varphi with I: \tan\varphi=\dfrac{V_L}{V_R}=\dfrac{X_L}{R}.
Impedance of the RL series circuit (2 marks)

Derivation Impedance of the RL series circuit

Aim. Find Z=V/I.

  1. From the right-angled triangle OAB (Pythagoras' theorem):
    V^2=V_R^2+V_L^2
  2. Substitute V_R=IR and V_L=IX_L.
    V^2=(IR)^2+(IX_L)^2=I^2\left(R^2+X_L^2\right)
  3. Take the square root and divide by I.
    Z=\dfrac{V}{I}=\sqrt{R^2+X_L^2}
ResultZ=\sqrt{R^2+X_L^2}=\sqrt{R^2+(2\pi fL)^2}

The impedance depends on the resistance, the inductance and the frequency of the supply. It is measured in ohms.

2026 Annual I, Set B, Q.5 OR1+2+3 = 6 marksUnit 21
What is meant by photo electric effect? Why no emission of electrons occurs below threshold frequency? Discuss the effects of intensity of light on photoelectric current with graph.
Solution
Photoelectric effect (1 mark)

The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons.

No emission below threshold frequency (2 marks)

According to Einstein, light consists of photons of energy hf, and one photon is absorbed by one electron. The electron needs at least the work function \Phi to escape:

hf=\Phi+K.E_{\max},\qquad \Phi=hf_0

If f<f_0, the energy of each photon is less than the work function (hf<\Phi), so no single photon can free an electron. An electron cannot store the energy of one photon and add the energy of another, so increasing the intensity (sending more photons) still gives no emission. Hence no electrons are emitted below the threshold frequency, however bright the light.

Effect of intensity on photoelectric current (3 marks)

Keeping the frequency (above f_0) and the potential difference fixed, the photoelectric current increases linearly with the intensity. Intensity is the number of photons arriving per second; each photon releases one electron, so the number of photoelectrons per second, and hence the current, is proportional to intensity.

Photoelectric current against intensity
S is the intensity of the light. Frequency and voltage fixed. Doubling the intensity doubles the current, so the number of photoelectrons per second is proportional to the intensity.

When the current is plotted against the collector potential for different intensities, brighter light gives a larger saturation current, but the current falls to zero at the same stopping potential V_0 for all intensities. So intensity increases the number of electrons but not their maximum kinetic energy, K.E_{\max}=eV_0.

Photoelectric current against collector potential
Light of one frequency at three intensities. Brighter light gives a larger saturation current, but every curve reaches zero at the same stopping potential -V_0.
2026 Annual I, Set B, Q.63+3 = 6 marksUnit 23
Answer both parts (a) and (b).
Solution
(a) What is meant by Radiant Flux Intensity? How it follows inverse square law?

Radiant flux intensity (1 mark): the luminosity per unit area measured on the surface of the Earth, that is, the power of a star's radiation received per square metre, F=L/A. SI unit: \mathrm{W\,m^{-2}}.

Inverse square law (2 marks): light leaving a star of luminosity L spreads out uniformly in all directions. At a distance d it is spread over the surface of a sphere of area 4\pi d^{2}, so

F=\dfrac{L}{4\pi d^{2}}\qquad\Longrightarrow\qquad F\propto\dfrac{1}{d^{2}}

If the distance is doubled, the same power covers 4 times the area, so F falls to one-fourth; at three times the distance it falls to one-ninth. Hence radiant flux intensity is inversely proportional to the square of the distance: it follows the inverse square law.

The same power passes through areas A, 4A and 9A at distances r, 2r and 3r.
(b) A star has a known luminosity of 9.7\times10^{27}\,\mathrm{W}. Observations of the star show that the radiant flux intensity of light received on Earth from the star is 114\,\mathrm{nW\,m^{-2}}. Determine the distance of the star from Earth.
Given
  • L = 9.7\times10^{27}\,\mathrm{W}
  • F = 114\,\mathrm{nW\,m^{-2}}=114\times10^{-9}\,\mathrm{W\,m^{-2}}
To find
  • distance d
  1. Formula (1 mark): the inverse square law, rearranged for d.
    F=\dfrac{L}{4\pi d^{2}}\;\Longrightarrow\;d=\sqrt{\dfrac{L}{4\pi F}}
  2. Substitution (1 mark):
    d=\sqrt{\dfrac{9.7\times10^{27}}{4\pi(114\times10^{-9})}}=\sqrt{\dfrac{9.7\times10^{27}}{1.43\times10^{-6}}}=\sqrt{6.77\times10^{33}}
  3. Answer with unit (1 mark):
    d=8.2\times10^{16}\,\mathrm{m}\ (\approx8.7\ \text{light-years})
d\approx8.2\times10^{16}\,\mathrm{m}

Note. Part (b) is Example 23.1 of the book, word for word; part (a) matches short question 3.

2026 Annual I, Set B, Q.6 OR(2+1)+3 = 6 marksUnit 22
a. What is meant by enrichment? Why Uranium fuel needs to enriched before use? b. Find decay constant and activity of an element if its 100 out of 700 nuclei decay in 5 min.
Solution
(a) What is meant by enrichment? Why does uranium fuel need to be enriched before use? (2+1 marks)

Enrichment (2 marks)

Enrichment is the process of increasing the percentage of the fissile isotope uranium-235 in uranium. Natural uranium contains only about 0.7% U-235 and 99.3% U-238. The two isotopes are chemically identical, so they are separated physically using their small difference in mass. The uranium is converted into uranium hexafluoride gas (\ce{UF6}, molecular masses 349 and 352) and passed through porous barriers (gaseous diffusion) or spun in gas centrifuges. The lighter \ce{^{235}UF6} molecules move slightly faster and are separated a little at each stage; many stages raise the U-235 content to 3 to 5%.

Why it is needed (1 mark)

With only 0.7% U-235, most neutrons are absorbed by U-238 and by the moderator before they can cause fission, so fewer than one neutron per fission produces another fission and the chain reaction cannot be sustained. Enriched fuel lets the reactor reach and keep criticality and run efficiently.

(b) Find the decay constant and activity if 100 out of 700 nuclei decay in 5 min. (3 marks)
Given
  • N = 700
  • \Delta N = 100
  • \Delta t = 5\,\mathrm{min}=300\,\mathrm{s}
To find
  • \lambda
  • A
  1. Decay constant: the fraction of nuclei decaying per second.
    \lambda=\dfrac{1}{N}\times\dfrac{\Delta N}{\Delta t}=\dfrac{1}{700}\times\dfrac{100}{300}=4.76\times10^{-4}\,\mathrm{s^{-1}}
  2. Activity.
    A=\lambda N=(4.76\times10^{-4})(700)=0.333\,\mathrm{Bq}
  3. In curie (optional).
    A=\dfrac{0.333}{3.70\times10^{10}}=9\times10^{-12}\,\mathrm{Ci}
\lambda=4.76\times10^{-4}\,\mathrm{s^{-1}},\qquad A=0.333\,\mathrm{Bq}

Note. Part (b) is the book's Example 22.5; part (a) matches Short Question 5 of the exercise.

Set D

53 questions

2026 Annual I, Set D, MCQ 11 markUnit 24
Velocity of a wind packet (mass =2\times10^{5}\,\mathrm{kg}), if a Coriolis Force (F_c) of 274\,\mathrm{N} bends it at an angle of 25^\circ would be:
  1. A32.2\,\mathrm{km\,hr^{-1}}
  2. B42.2\,\mathrm{km\,hr^{-1}}
  3. C12.2\,\mathrm{m\,s^{-1}}
  4. D22.2\,\mathrm{m\,s^{-1}}Answer
Why

From F_c = 2mv\omega\sin\theta: v = \dfrac{F_c}{2m\omega\sin\theta} = \dfrac{274}{2(2\times10^{5})(7.3\times10^{-5})\sin25^\circ} = \dfrac{274}{12.3} = 22.2\,\mathrm{m\,s^{-1}}.

This equals 80\,\mathrm{km\,h^{-1}} (the book's Example 24.2), so the km/h options A and B are also wrong.

2026 Annual I, Set D, MCQ 21 markUnit 21
Which of the following electro-magnetic waves has the highest frequency?
  1. AGamma raysAnswer
  2. BUltraviolet
  3. CMicrowaves
  4. DInfrared
Why

Gamma rays have the shortest wavelength (about 10^{-12}\,\mathrm{m}) in the electromagnetic spectrum, so from f=c/\lambda they have the highest frequency (and photon energy). The order of increasing frequency is microwaves, infrared, ultraviolet, gamma rays.

2026 Annual I, Set D, MCQ 31 markUnit 15
Two objects attract each other with a gravitational force F. If mass of one object is doubled and distance between their centers remains same, then gravitational force is:
  1. A2FAnswer
  2. B4F
  3. CF/2
  4. DF
Why

F=G\dfrac{m_1m_2}{r^{2}} is directly proportional to each mass. Doubling one mass (same distance) doubles the force: F'=G\dfrac{(2m_1)m_2}{r^{2}}=2F. B (4F) would need both masses doubled.

2026 Annual I, Set D, MCQ 41 markUnit 15
The gravitational field strength at a point depends on:
  1. AShape of the test body
  2. BVelocity of the body
  3. CMass of the test body
  4. DMass causing the fieldAnswer
Why

g=\dfrac{F}{m}=\dfrac{GM}{r^{2}}: the test mass m cancels, so g depends only on the mass M causing the field and the distance r from it. The shape, velocity and mass of the test body do not matter.

2026 Annual I, Set D, MCQ 51 markUnit 23
The SI unit of luminosity 'L' is:
  1. AWatt \mathrm{m^{-2}}
  2. BKelvin
  3. CWattAnswer
  4. DJoule
Why

Luminosity is power (energy emitted per second), so its unit is \mathrm{J\,s^{-1}}, the watt. \mathrm{W\,m^{-2}} (A) is the unit of radiant flux intensity, the kelvin (B) is the unit of temperature, and the joule (D) is energy, not power.

2026 Annual I, Set D, MCQ 61 markUnit 17
In SHM, the relationship between acceleration and displacement of a particle when \omega=10 is:
  1. Aa=-100xAnswer
  2. Ba=100x^{2}
  3. Ca=10x
  4. Da=-10x^{2}
Why

In SHM a=-\omega^{2}x. With \omega=10\,\mathrm{rad\,s^{-1}}, \omega^{2}=100, so a=-100x. Options B and D are not proportional to x, and C has the wrong sign and does not square \omega.

2026 Annual I, Set D, MCQ 71 markUnit 17
What is the frequency of the body when its time period is 2\,\mathrm{s}?
  1. A0.5\,\mathrm{Hz}Answer
  2. B0.2\,\mathrm{Hz}
  3. C1\,\mathrm{Hz}
  4. D2\,\mathrm{Hz}
Why

f=\dfrac{1}{T}=\dfrac{1}{2\,\mathrm{s}}=0.5\,\mathrm{Hz}.

2026 Annual I, Set D, MCQ 81 markUnit 18
The wavelength of light can be accurately determined by using a:
  1. AInterferometer
  2. BCapacitor
  3. CMelde's apparatus
  4. DDiffraction gratingAnswer
Why

A diffraction grating gives very sharp, widely separated maxima whose angles can be measured accurately. With the grating element d known, \lambda=d\sin\theta/m (book section 18.2.1). A capacitor stores charge, and Melde's apparatus studies standing waves on a string.

Note. An interferometer (such as Michelson's) can also measure the wavelength of light very precisely, so option A is not wrong in itself. The new book presents the diffraction grating as the device for this, so D is the expected answer.

2026 Annual I, Set D, MCQ 91 markUnit 19
Electric potential due to a point charge varies with distance as:
  1. A1/r^2
  2. Br^2
  3. C1/rAnswer
  4. Dr
Why

V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}, so V\propto1/r. The tempting option A, 1/r^{2}, is how the electric field (and the force) varies, not the potential.

2026 Annual I, Set D, MCQ 101 markUnit 19
The capacitance of a parallel plate capacitor increases when:
  1. AArea of plates increasesAnswer
  2. BUsing battery of low potential
  3. CDistance between plates increases
  4. DArea of plates decreases
Why

C=\dfrac{\varepsilon_0\varepsilon_rA}{d}, so C is directly proportional to the plate area. Increasing d (C) or decreasing A (D) reduces C, and the battery voltage (B) changes the charge stored, not the capacitance.

2026 Annual I, Set D, MCQ 111 markUnit 20
In a bridge rectifier, current flows through the load:
  1. ADuring both half cycles in the same directionAnswer
  2. BDuring both half cycles in alternating directions
  3. COnly during positive half cycle
  4. DOnly during negative half cycle
Why

D_1 and D_2 conduct in the positive half-cycle and D_3 and D_4 in the negative half-cycle, and both pairs send the current through the load in the same direction. Options C and D describe a half-wave rectifier.

2026 Annual I, Set D, MCQ 121 markUnit 20
In a pure inductive AC circuit, the applied voltage leads the current by:
  1. A0
  2. B\dfrac{\pi}{2}Answer
  3. C\dfrac{3\pi}{2}
  4. D\pi
Why

For a pure inductor, I=I_0\sin\omega t gives V=V_0\sin(\omega t+\pi/2): the voltage leads the current by \pi/2 (90^\circ). Remember ELI: in L the emf comes before I.

2026 Annual I, Set D, MCQ 131 markUnit 21
De-Broglie wavelength of a body having mass 0.5\,\mathrm{kg} moving with a speed of 36\,\mathrm{km/hr} is:
  1. A1.3\times10^{-34}\,\mathrm{mm}
  2. B1.3\times10^{-24}\,\mathrm{mm}
  3. C1.3\times10^{-24}\,\mathrm{m}
  4. D1.3\times10^{-34}\,\mathrm{m}Answer
Why
\begin{aligned}v&=36\,\mathrm{km\,h^{-1}}=\dfrac{36\,000\,\mathrm{m}}{3600\,\mathrm{s}}=10\,\mathrm{m\,s^{-1}}\\ \lambda&=\dfrac{h}{mv}=\dfrac{6.63\times10^{-34}}{0.5\times10}=1.3\times10^{-34}\,\mathrm{m}\end{aligned}

This is the book's own example in section 21.6. Watch the unit: the answer is in metres, not millimetres.

2026 Annual I, Set D, MCQ 141 markUnit 21
Number of electrons emitted in photoelectric effect depends upon:
  1. AEnergy of incident light
  2. BWavelength of incident light
  3. CIntensity of Incident lightAnswer
  4. DFrequency of incident light
Why

Each photon releases one electron, so the number of electrons emitted per second is proportional to the number of photons per second, that is, to the intensity. Frequency, wavelength and photon energy decide the kinetic energy of the electrons (and whether emission happens at all), not their number.

2026 Annual I, Set D, MCQ 151 markUnit 22
In nuclear fission, a chain reaction occurs due to the release of:
  1. APhotons
  2. BNeutronsAnswer
  3. CProtons
  4. DElectrons
Why

Each fission of uranium-235 releases 2 or 3 neutrons. These neutrons cause further fissions, which release more neutrons, so the reaction keeps itself going. Photons, protons and electrons do not cause further fissions.

2026 Annual I, Set D, MCQ 161 markUnit 23
According to the inverse square law, radiant flux intensity is inversely proportional to:
  1. ATemperature
  2. BLuminosity
  3. CDistance
  4. DSquare of distanceAnswer
Why

F=L/4\pi d^{2}: the radiant flux intensity is inversely proportional to the square of the distance. It is directly (not inversely) proportional to the luminosity, and 'distance' alone misses the square.

2026 Annual I, Set D, MCQ 171 markUnit 24
Which layer of the atmosphere contains the ozone layer?
  1. AMesosphere
  2. BStratosphereAnswer
  3. CTroposphere
  4. DThermosphere
Why

The ozone layer lies in the stratosphere (about 16 km to 50 km). By absorbing the Sun's ultraviolet radiation it makes the temperature rise with height in this layer.

2026 Annual I, Set D, Q.2 (i)1+2 = 3 marksUnit 15
How will the orbital velocity of a satellite be affected when it is put into a lower orbit (closer to the Earth)? Explain mathematically.
Solution

Effect (1 mark). The orbital velocity increases: a satellite in a lower orbit must move faster.

Mathematical explanation (2 marks). For a circular orbit of radius r, gravity provides the centripetal force:

\dfrac{GM_Em}{r^{2}}=\dfrac{mv_o^{2}}{r}\quad\Rightarrow\quad v_o=\sqrt{\dfrac{GM_E}{r}}\quad\Rightarrow\quad v_o\propto\dfrac{1}{\sqrt r}

In a lower orbit r is smaller and the Earth's pull is stronger, so a larger speed is needed to keep the satellite on its circle. If r is reduced to one quarter, v_o doubles. The fastest circular orbit is the one just above the surface: v_o=\sqrt{gR_E}=7.9\,\mathrm{km\,s^{-1}}.

2026 Annual I, Set D, Q.2 (i) OR1+1+1 = 3 marksUnit 17
A body with mass 800\,\mathrm{g} attached to a spring executes SHM. The restoring force is 60\,\mathrm{N} when the displacement is 0.3\,\mathrm{m}.
Solution
Given
  • m = 800\,\mathrm{g}=0.8\,\mathrm{kg}
  • F = 60\,\mathrm{N}\ \text{at}\ x=0.3\,\mathrm{m}
  • x = 12\,\mathrm{cm}=0.12\,\mathrm{m}\ \text{in part (b)}
To find
  • (a) \omega
  • (b) a

Marks: spring constant (1), angular frequency (1), acceleration (1). These are the data of Example 17.2 of the book.

a. Find out its angular frequency
  1. Spring constant from Hooke's law (size only).
    k=\dfrac{F}{x}=\dfrac{60}{0.3}=200\,\mathrm{N\,m^{-1}}
  2. Angular frequency of the mass-spring system.
    \omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{200}{0.8}}=\sqrt{250}=15.8\,\mathrm{rad\,s^{-1}}
\omega=15.8\,\mathrm{rad\,s^{-1}}
b. Also find the magnitude of acceleration when the displacement is 12\,\mathrm{cm}.
  1. Use a=\omega^{2}x with \omega^{2}=250\,\mathrm{s^{-2}} and x=0.12\,\mathrm{m}.
    a=250\times0.12=30\,\mathrm{m\,s^{-2}}
a=30\,\mathrm{m\,s^{-2}}

Note. The book's Example 17.2 prints a=3\,\mathrm{m\,s^{-2}} for these data; that is an arithmetic slip. The correct value is 30\,\mathrm{m\,s^{-2}}.

2026 Annual I, Set D, Q.2 (ii)2+1 = 3 marksUnit 16
How is the pressure related to the density of the gas? Also give its mathematical expression.
Solution
Relation (2 marks)

At constant temperature, the pressure of a gas is directly proportional to its density, P\propto\rho.

Reason: density \rho=\dfrac{Nm}{V} tells how many molecules (how much mass) there are in each unit volume. In a denser gas more molecules strike each unit area of the wall every second. At constant temperature each molecule delivers the same momentum per collision, so more collisions per second give a larger force per unit area, that is, a larger pressure. Doubling the density doubles the pressure.

  1. Kinetic theory gives the pressure of an ideal gas as
    P=\dfrac13\left(\dfrac{Nm}{V}\right)\langle v^{2}\rangle
  2. Nm is the total mass of the gas, so Nm/V=\rho:
    P=\dfrac13\rho\langle v^{2}\rangle
Mathematical expression (1 mark)
P=\dfrac{1}{3}\rho\langle v^{2}\rangle

\rho: density of the gas; \langle v^{2}\rangle: mean square speed of the molecules, which is constant at constant temperature.

2026 Annual I, Set D, Q.2 (ii) OR1x3 = 3 marksUnit 18
State the conditions necessary for observing interference fringes.
Solution

Any three of the following (1 mark each):

  1. Coherent sources: the two sources must have the same frequency and a constant phase difference. (In Young's experiment both slits are lit by the same wavefront from one single slit.)
  2. Monochromatic light: light of a single wavelength; otherwise fringes of different colours overlap and the pattern blurs.
  3. Narrow, closely spaced slits and a distant screen: each slit must be narrow enough to diffract the light so that the two beams overlap, and d must be small and L large so that the fringe spacing \Delta y=\lambda L/d is big enough to see.
  4. Nearly equal amplitudes: the two waves should be of equal intensity so that the dark fringes are completely dark.
  5. Stable arrangement: no vibrations or air currents that would shift the fringes.
2026 Annual I, Set D, Q.2 (iii)3 marksUnit 17
How does resonance play an important role in tuning of radio station?
Solution

Many radio stations broadcast at the same time, each at its own frequency. All their waves reach the aerial and try to drive small electrical oscillations in the receiver.

The tuning circuit of the receiver (a coil and a variable capacitor) is an electrical oscillator with its own natural frequency. When we turn the tuning knob, we change the capacitance and so change this natural frequency.

When the natural frequency of the circuit becomes equal to the transmission frequency of a particular station, resonance occurs. The circuit then oscillates with maximum amplitude at that frequency and absorbs maximum energy from that station only. Signals of other stations are far from resonance and produce very small responses, so we hear only the tuned station. For good selection the tuning circuit has little damping, which makes the resonance sharp.

2026 Annual I, Set D, Q.2 (iii) OR1x3 = 3 marksUnit 19
Briefly explain the use of capacitors in any three household appliances.
Solution

One mark for each appliance with the role of its capacitor:

  1. Ceiling fan: a capacitor connected with the fan's auxiliary (starting) winding shifts the phase of its current. This produces a rotating magnetic field, which starts the motor and keeps it turning; a capacitor of suitable value increases the speed of the fan.
  2. Refrigerator: a start (or run) capacitor with the compressor motor gives the extra starting torque the compressor needs and keeps it running smoothly.
  3. Camera flash (flash gun): the small battery charges a capacitor slowly; the capacitor then releases its stored energy in about a millisecond through the flash tube, producing a bright flash.
  4. Also acceptable: in TV and radio sets capacitors filter unwanted frequencies and tune to a station; in a mobile charger or any d.c. power supply a capacitor smooths the rectified voltage by reducing the ripple.
2026 Annual I, Set D, Q.2 (iv)1+2 = 3 marksUnit 20
What is mutual inductance? Also write its SI unit and the factors upon which it depends.
Solution
Mutual inductance (1 mark)

When the current in one coil changes, an emf \varepsilon_2=-M\,\Delta I_1/\Delta t is induced in a neighbouring coil (mutual induction). The constant M is the mutual inductance: the ratio of the emf induced in one coil to the time rate of change of current in the other coil.

M=\dfrac{-\varepsilon_2}{\Delta I_1/\Delta t}=\dfrac{N_2\Phi_2}{I_1}
SI unit and factors (2 marks)

SI unit: the henry (H), 1\,\mathrm{H}=1\,\mathrm{V\,s\,A^{-1}}. The mutual inductance of two coils is 1 H if a current changing at 1\,\mathrm{A\,s^{-1}} in one coil induces an emf of 1 V in the other.

For two coaxial coils M=\dfrac{\mu_r\mu_0N_1N_2A}{l}, so M depends on:

  • the number of turns of both coils (N_1N_2);
  • the area of cross-section of the coils;
  • the distance between the coils and their relative orientation (largest when close and coaxial);
  • the core material, its relative permeability \mu_r (a common iron core increases M).
2026 Annual I, Set D, Q.2 (iv) OR3 marksUnit 15
What will be the value of gravitational field strength g at 35\,700\,\mathrm{km}, where a satellite is orbiting the earth?
Solution

35\,700\,\mathrm{km} is the well-known height of communication (geostationary) satellites above the Earth's surface (Table 15.1), so we take it as the height h above the surface. This is Assignment 15.2 of the book.

Given
  • g = 9.8\,\mathrm{m\,s^{-2}}
  • R_e = 6.4\times10^{6}\,\mathrm{m}
  • h = 35\,700\,\mathrm{km}=3.57\times10^{7}\,\mathrm{m}
To find
  • g_h
  1. Formula (1 mark).
    g_h=g\left(\dfrac{R_e}{R_e+h}\right)^{2}
  2. Substitution (1 mark). R_e+h=4.21\times10^{7}\,\mathrm{m}.
    g_h=9.8\left(\dfrac{6.4\times10^{6}}{4.21\times10^{7}}\right)^{2}=9.8\times(0.152)^{2}
  3. Answer with unit (1 mark).
    g_h=0.23\,\mathrm{m\,s^{-2}}
g_h\approx0.23\,\mathrm{m\,s^{-2}}
2026 Annual I, Set D, Q.2 (v)2+1 = 3 marksUnit 25
What is meant by breaking radiation? Also write the relation for minimum wavelength of x-rays.
Solution
Braking radiation (2 marks)

Braking radiation (Bremsstrahlung) is the X-radiation emitted when a fast-moving electron is decelerated (braked) by the Coulomb attraction of a nucleus in the target of an X-ray tube. The attraction deflects and slows the electron, and the kinetic energy it loses is emitted as an X-ray photon, hf=\mathrm{K.E.}_{\text{before}}-\mathrm{K.E.}_{\text{after}}. Since an electron can lose any amount of energy, braking radiation forms the continuous X-ray spectrum.

Minimum wavelength (1 mark)

The shortest wavelength is produced when an electron, accelerated through p.d. V, gives all its kinetic energy eV to one photon:

eV=hf_{\max}=\dfrac{hc}{\lambda_{\min}}\qquad\Rightarrow\qquad\lambda_{\min}=\dfrac{hc}{eV}

Here h is Planck's constant, c the speed of light, e the charge of the electron and V the accelerating potential difference.

Note. The paper writes 'breaking'; the correct term is braking radiation (Bremsstrahlung).

2026 Annual I, Set D, Q.2 (v) OR3 marksUnit 20
An AC voltage across a 0.5\,\mu F capacitor is 16\sin(2\times10^3t)\,V. Find its capacitive reactance.
Solution
Given
  • C = 0.5\,\mu\mathrm{F}=0.5\times10^{-6}\,\mathrm{F}
  • V = 16\sin(2\times10^{3}t)\,\mathrm{V}
To find
  • X_C
  1. Compare with V=V_0\sin\omega t to read the angular frequency.
    \omega=2\times10^{3}\,\mathrm{rad\,s^{-1}},\qquad V_0=16\,\mathrm{V}
  2. Capacitive reactance.
    X_C=\dfrac{1}{\omega C}=\dfrac{1}{(2\times10^{3})(0.5\times10^{-6})}=\dfrac{1}{1\times10^{-3}}
  3. Evaluate.
    X_C=1000\,\Omega

This is Example 20.2 of the book, which also finds the peak current I_0=V_0/X_C=16\,\mathrm{mA}.

X_C=1000\,\Omega
2026 Annual I, Set D, Q.2 (vi)3 marksUnit 24
Explain the role of atmosphere in maintaining climate balance.
Solution

The atmosphere keeps the Earth's climate balanced in these ways:

  1. Keeps the Earth warm (greenhouse effect): greenhouse gases such as water vapour and \mathrm{CO_2} let sunlight in but absorb part of the outgoing infrared radiation and send some of it back to the surface. This keeps the average temperature near 15^\circ\mathrm{C} instead of -18^\circ\mathrm{C} and prevents extreme differences between day and night temperatures.
  2. Shields life: the ozone layer in the stratosphere absorbs harmful ultraviolet radiation, and meteors burn up in the mesosphere.
  3. Redistributes heat and moisture: solar heating causes convection; the resulting winds and circulation cells carry heat from the equator towards the poles and water vapour from the oceans over the land, producing weather and evening out temperatures across the globe.

Clouds and air also reflect part of the sunlight back to space, so the atmosphere helps to keep the energy the Earth absorbs equal to the energy it radiates (the energy budget).

2026 Annual I, Set D, Q.2 (vi) OR3 marksUnit 25
What is meant by contrast in X-ray imaging?
Solution

Contrast is the range of brightness, from lightest to darkest, in an image. In an X-ray image it is the difference in brightness between neighbouring structures, which comes from the difference in how much X-radiation they absorb.

Bone absorbs X-rays much more than soft tissue, so bone on soft tissue gives high contrast and shows clearly. Neighbouring soft tissues absorb almost equally, so they appear in nearly the same shade of grey (low contrast) and are hard to see on a conventional X-ray.

Contrast is improved with a contrast medium of high atomic number: a barium meal (barium sulfate) for the stomach and intestines, or an iodine-based dye injected into blood vessels. The filled organ then appears white. Contrast in X-ray imaging allows the radiologist to evaluate structures that are not clearly evident on conventional X-ray exams.

2026 Annual I, Set D, Q.2 (vii)1+2 = 3 marksUnit 21
Briefly describe pair annihilation. Also justify the conservation of energy and momentum in this process.
Solution
Pair annihilation (1 mark)

Pair annihilation is the process in which an electron and a positron (its antiparticle) combine and disappear, producing two gamma-ray photons. Mass is converted into energy according to E=mc^{2}. It is the reverse of pair production.

Conservation of energy and momentum (2 marks)

Energy: the total energy of the pair (rest mass energy plus kinetic energy) appears as the energy of the photons:

2m_0c^{2}+K.E_{e^-}+K.E_{e^+}=\text{energy of two photons}

For a pair at rest each photon has hf=m_0c^{2}=0.511\,\mathrm{MeV}, so the total energy before and after is 1.02\,\mathrm{MeV}.

Momentum: the electron and positron are nearly at rest, so their total momentum is about zero. A single photon always carries momentum hf/c, so one photon cannot be produced. Two photons of equal energy moving in opposite directions have equal and opposite momenta (+hf/c and -hf/c), giving zero total momentum. Hence momentum is conserved. Charge is also conserved: (+e)+(-e)=0.

2026 Annual I, Set D, Q.2 (vii) OR3 marksUnit 24
How does ocean play a main role in climate inertia?
Solution

Climate inertia is the slow response of the climate to changes in factors such as greenhouse gases. The oceans are its main cause because:

  1. High heat capacity: water has a high specific heat capacity (about 4200\,\mathrm{J\,kg^{-1}\,K^{-1}}) and the oceans have an enormous mass. From Q = mc\Delta T, a huge amount of heat produces only a small change in temperature; the oceans can store about 1000 times more heat than the atmosphere.
  2. Slow absorption and release: the oceans absorb heat slowly and release it slowly, and the deep ocean mixes with the surface only over centuries. So changes in ocean temperature, and in the climate patterns they control, happen gradually.
  3. Stored heat keeps the warming going: even if greenhouse gas emissions were stabilised immediately, the heat already stored in the oceans would keep the Earth warming for decades. The oceans also absorb \mathrm{CO_2} and exchange it slowly with the air, which adds to the delay.
2026 Annual I, Set D, Q.2 (viii)3 marksUnit 25
Briefly discuss two applications of ultrasound in medical technology.
Solution

1. Diagnostic imaging (sonography). Short pulses of ultrasound from a piezoelectric probe enter the body and are partly reflected at boundaries between tissues. The time of each echo gives the depth of the boundary (d=vt/2) and its strength shows the type of tissue, so a computer builds a real-time image of organs such as the liver, kidneys, heart and the unborn baby. It uses no ionising radiation, so it is safe in pregnancy.

2. Measuring blood flow (Doppler ultrasound). Ultrasound reflected from moving blood cells returns with a changed frequency (Doppler shift). The shift gives the speed of the blood, so doctors can find blood clots and blocked arteries and check the working of the heart in adults and fetuses.

Other applications that may be given instead: focused intense ultrasound in neurosurgery to destroy a small region of tissue without cutting, and ultrasonic heating in physiotherapy for arthritis, rheumatism and sciatica.

2026 Annual I, Set D, Q.2 (viii) OR1.5+1.5 = 3 marksUnit 21
Why does electron microscope have higher resolution than optical microscope? (Compare both.)
Solution
Reason (1.5 marks)

The resolution of any microscope is limited by the wavelength of the radiation it uses: details smaller than about one wavelength cannot be seen. An electron microscope uses fast electrons whose de Broglie wavelength \lambda=h/\sqrt{2meV} is extremely short (about 0.01\,\mathrm{nm} at 10\,\mathrm{kV}), thousands of times shorter than visible light (400 to 700\,\mathrm{nm}). So it can resolve much finer detail.

Comparison (1.5 marks)
Optical microscopeElectron microscope
IlluminationVisible lightBeam of fast electrons
Wavelength400 to 700\,\mathrm{nm}about 0.01\,\mathrm{nm}
LensesGlass lensesMagnetic lenses (coils)
Resolutionabout 200\,\mathrm{nm}about 0.1 to 0.2\,\mathrm{nm}
Medium and specimenAir; living specimens possibleVacuum; thin, non-living specimen
2026 Annual I, Set D, Q.2 (ix)1.5+1.5 = 3 marksUnit 22
How is the variation of binding energy per nucleon an important factor in nuclear physics? (Discuss both high binding energy and low binding energy)
Solution

The binding energy per nucleon E_b/A is the average energy needed to remove one nucleon from a nucleus. Its variation with mass number (the binding energy curve) explains the stability of nuclei and the release of nuclear energy.

High binding energy per nucleon (1.5 marks)

Medium nuclei (A\approx50 to 100) have the highest values, with a peak of about 8.8\,\mathrm{MeV} for iron-56. Their nucleons are most tightly bound, so these nuclei are the most stable; they undergo neither fission nor fusion, and no energy can be obtained from them.

Low binding energy per nucleon (1.5 marks)

Very light nuclei (such as \ce{^{2}H}, 1.1\,\mathrm{MeV}) and very heavy nuclei (such as U-235, 7.6\,\mathrm{MeV}) have lower values and are less stable. They can gain stability by moving towards the peak, and energy is released when they do: light nuclei by fusion (hydrogen into helium in the Sun) and heavy nuclei by fission (U-235 into two medium nuclei, about 200\,\mathrm{MeV}). This is the basis of nuclear power and of stellar energy.

Binding energy per nucleon (in \mathrm{MeV}) against mass number
Sketch of the binding energy curve: low at both ends, highest near iron-56.
2026 Annual I, Set D, Q.2 (ix) OR3 marksUnit 20
An AC circuit consists of pure resistance of 20\,\Omega and is connected across AC supply of 220\,V, 50\,Hz. Calculate: a. Peak value of voltage. b. Peak value of current.
Solution
Given
  • R = 20\,\Omega
  • V_{\mathrm{rms}} = 220\,\mathrm{V} the supply voltage is an r.m.s. value
  • f = 50\,\mathrm{Hz}
To find
  • V_0
  • I_0
a. Peak value of voltage.
  1. Peak value from the r.m.s. value.
    V_0=\sqrt2\,V_{\mathrm{rms}}=1.414\times220=311.1\,\mathrm{V}
V_0\approx311\,\mathrm{V}
b. Peak value of current.
  1. Ohm's law holds at every instant in a resistor.
    I_0=\dfrac{V_0}{R}=\dfrac{311.1}{20}=15.56\,\mathrm{A}

The frequency is not needed for these two answers; it gives \omega=2\pi f=314\,\mathrm{rad\,s^{-1}} for the equations V=311\sin(314t) and I=15.6\sin(314t) (Example 20.1).

I_0\approx15.6\,\mathrm{A}
2026 Annual I, Set D, Q.2 (x)3 marksUnit 24
How does energy imbalance between equator and poles affect earth climate?
Solution

Energy imbalance: the equator receives direct sunlight and gains more energy than it loses, while the poles receive oblique sunlight and lose more energy than they gain.

  1. Winds and circulation cells: warm air rises at the equator (low pressure) and cold air sinks at the poles (high pressure). The pressure differences drive the trade winds, westerlies and polar easterlies, which the Coriolis effect organises into the Hadley, Ferrel and polar cells.
  2. Heat transport: these winds, together with ocean currents (gyres and the thermohaline circulation), carry heat from the tropics towards the poles and reduce the temperature difference.
  3. Climate zones: the circulation creates wet equatorial regions (rising air), dry deserts near 30^\circ (sinking air) and stormy mid-latitudes near 60^\circ. Changes in the balance shift weather systems, ocean currents and climate zones.
The circulation cells set up by the energy imbalance between the equator and the poles.
2026 Annual I, Set D, Q.2 (x) OR2+1 = 3 marksUnit 23
How does Wien's displacement law be used to determine the temperature of a star? Also write its formula.
Solution

Use (2 marks): Wien's displacement law states that the black body radiation curve peaks at a wavelength which is inversely proportional to the absolute temperature. A star radiates almost like a black body, so:

  1. Record the star's spectrum and find the peak wavelength \lambda_{\max}, where the intensity is greatest.
  2. Convert it to metres and divide Wien's constant by it: T=\dfrac{2.9\times10^{-3}\,\mathrm{m\,K}}{\lambda_{\max}}.
  3. A shorter peak wavelength means a hotter star, so blue-white stars are hotter than red stars.

Formula (1 mark):

\lambda_{\max}T=2.9\times10^{-3}\,\mathrm{m\,K}

For example, Rigel peaks at 263\,\mathrm{nm}, so T=2.9\times10^{-3}/(263\times10^{-9})\approx1.1\times10^{4}\,\mathrm{K}.

2026 Annual I, Set D, Q.2 (xi)3 marksUnit 15
Why the value of g does not change significantly near the Earth's surface? (Show mathematically)
Solution

At height h above the surface the field strength is

g_h=g\left(\dfrac{R_E}{R_E+h}\right)^{2}=g\left(1+\dfrac{h}{R_E}\right)^{-2}\approx g\left(1-\dfrac{2h}{R_E}\right)\qquad(h\ll R_E)

using the binomial approximation. So the fractional decrease in g is only 2h/R_E. The Earth's radius is very large (R_E=6.4\times10^{6}\,\mathrm{m}), so for ordinary heights this is tiny:

h=1\,\mathrm{km}:\quad\dfrac{2h}{R_E}=\dfrac{2\times1000}{6.4\times10^{6}}=3.1\times10^{-4}\ (0.03\%)

Even at the top of Mount Everest (h\approx8.8\,\mathrm{km}) g falls by less than 0.3\%, from 9.80 to about 9.77\,\mathrm{m\,s^{-2}}. Hence near the surface g is practically constant.

2026 Annual I, Set D, Q.2 (xi) OR1.5+1.5 = 3 marksUnit 22
Briefly explain the function of control rods and moderators in nuclear reactor.
Solution

Function of control rods (1.5 marks)

Control rods, made of neutron-absorbing boron or cadmium, control the rate of the chain reaction by absorbing some of the neutrons. Inserting them further lowers the number of fissions and the power; withdrawing them raises it. In normal running they are adjusted so that exactly one neutron per fission causes a further fission, so the reaction neither dies out nor runs out of control. In an emergency they are dropped fully in to shut the reactor down.

Function of moderators (1.5 marks)

A moderator is a material that slows down fast neutrons. Fission neutrons are fast (about 2\,\mathrm{MeV}), but uranium-235 is split most easily by slow (thermal) neutrons of about 0.025\,\mathrm{eV}. In repeated collisions with the light nuclei of the moderator (water or graphite) the neutrons lose most of their energy and become thermal, which keeps the chain reaction going.

2026 Annual I, Set D, Q.2 (xii)3 marksUnit 16
Calculate the average translational kinetic energy of a gas molecule at a temperature 47\,^\circ\mathrm{C}.
Solution
Given
  • T = 47\,^\circ\mathrm{C}=47+273=320\,\mathrm{K}
  • k = 1.38\times10^{-23}\,\mathrm{J\,K^{-1}}
To find
  • \langle K.E\rangle
  1. Formula: the average translational kinetic energy depends only on the absolute temperature.
    \langle K.E\rangle=\left\langle\tfrac12mv^{2}\right\rangle=\dfrac32kT
  2. Substitute, with T in kelvin.
    \langle K.E\rangle=\dfrac32\times1.38\times10^{-23}\times320
  3. Answer with unit.
    \langle K.E\rangle=6.62\times10^{-21}\,\mathrm{J}

This is Assignment 16.1 of the book (320\,\mathrm{K}). The paper's formula sheet gives it as T=\frac{2}{3k}\left(\frac12mv^{2}\right); rearranging gives the same \frac32kT. Marks: formula 1, substitution 1, answer with unit 1.

\langle K.E\rangle\approx6.62\times10^{-21}\,\mathrm{J}
2026 Annual I, Set D, Q.2 (xii) OR1+2 = 3 marksUnit 17
What is meant by free oscillations? Also give its two examples.
Solution

Free oscillations (1 mark). A body is said to be executing free oscillations when it oscillates under the influence of its restoring force alone, without any external force acting on it. It then vibrates at its own natural frequency, and (ideally, with no damping) its amplitude and period stay constant.

Examples (2 marks):

  1. A simple pendulum displaced slightly from its mean position and released oscillates freely at its natural frequency f_0=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}.
  2. A mass attached to a spring, pulled aside and released, oscillates freely at f_0=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}.
  3. (Also acceptable: a plucked guitar string, a struck tuning fork, a swing given a single push.)
2026 Annual I, Set D, Q.2 (xiii)2+1 = 3 marksUnit 18
How does increasing the number of slits per unit length affect the diffraction pattern? Make a sketch.
Solution
Effect (2 marks)

Increasing the number of slits per unit length N decreases the grating element d=1/N. From the grating equation \sin\theta=m\lambda/d:

  • each order appears at a larger angle, so the maxima are more widely spaced;
  • fewer orders can be seen, because \sin heta=m\lambda/d cannot be more than 1;
  • the maxima become sharper and brighter, because more slits share the same beam.
Sketch (1 mark)
Effect of more lines per unit length
With more lines per mm (d reduced from 3.6\lambda to 1.5\lambda) the first order moves from 16.1^\circ to 41.8^\circ (\sin\theta=m\lambda/d). The maxima are more widely spaced and fewer orders fit in: 3 bright lines instead of 7. Each maximum is narrow compared with the gap between maxima.
2026 Annual I, Set D, Q.2 (xiii) OR1.5+1.5 = 3 marksUnit 20
Distinguish between peak value and root mean square (rms) values of alternating voltage.
Solution
Peak value (1.5 marks)

The peak value V_0 is the maximum value of the alternating voltage. It is reached twice in each cycle, as +V_0 and -V_0, and only for an instant. For the 220 V mains, V_0=311\,\mathrm{V}.

r.m.s. value (1.5 marks)

The r.m.s. value V_{\mathrm{rms}} is the steady (DC) voltage that produces the same heat in a resistor as the alternating voltage in the same time. It is the square root of the mean of V^2 over a cycle. AC meters read r.m.s. values, and ratings such as 220 V are r.m.s. values. For a sinusoidal voltage:

V_{\mathrm{rms}}=\dfrac{V_0}{\sqrt2}=0.707\,V_0
Peak and r.m.s. values
Fig 20.2 redrawn. The peak value x_0 is the highest point of the wave. The r.m.s. value sits lower, at x_0/\sqrt2=0.707\,x_0. The positive and negative areas are equal, so the average over a full cycle is zero.
2026 Annual I, Set D, Q.2 (xiv)2+1 = 3 marksUnit 18
How can one increase the fringe spacing in Young's double slit experiment? Also write its mathematical calculation.
Solution
Ways to increase the fringe spacing (2 marks)

The fringe spacing is \Delta y=\lambda L/d, so it can be increased by:

  • moving the screen farther from the slits (increasing L);
  • using light of a longer wavelength, for example red instead of blue (increasing \lambda);
  • bringing the two slits closer together (decreasing d).
Mathematical calculation (1 mark)
\Delta y=\dfrac{\lambda L}{d}=\dfrac{(6.0\times10^{-7}\,\mathrm{m})(1.0\,\mathrm{m})}{5.0\times10^{-4}\,\mathrm{m}}=1.2\times10^{-3}\,\mathrm{m}

With \lambda=600\,\mathrm{nm}, L=1.0\,\mathrm{m} and d=0.50\,\mathrm{mm} the spacing is 1.2\,\mathrm{mm}. Doubling L to 2.0\,\mathrm{m}, or halving d to 0.25\,\mathrm{mm}, doubles it to 2.4\,\mathrm{mm}.

2026 Annual I, Set D, Q.2 (xiv) OR3 marksUnit 19
Deduce the equivalent capacitance between A and B for the given circuit.
Solution
Given
  • C_1 = 2\,\mu\mathrm{F}
  • C_2 = 4\,\mu\mathrm{F}
  • C_3 = 4\,\mu\mathrm{F}
  • C_4 = 2\,\mu\mathrm{F}
To find
  • C_e between A and B
  1. C_2 and C_3 are connected between the same two junctions, so they are in parallel.
    C_{23}=C_2+C_3=4+4=8\,\mu\mathrm{F}
  2. C_1, C_{23} and C_4 are now in series.
    \dfrac{1}{C_e}=\dfrac{1}{C_1}+\dfrac{1}{C_{23}}+\dfrac{1}{C_4}=\dfrac12+\dfrac18+\dfrac12=\dfrac{4+1+4}{8}=\dfrac98\,\mu\mathrm{F^{-1}}
  3. Invert.
    C_e=\dfrac89\,\mu\mathrm{F}\approx0.89\,\mu\mathrm{F}
C_e=\tfrac89\,\mu\mathrm{F}\approx0.89\,\mu\mathrm{F}
2026 Annual I, Set D, Q.32+4+1 = 7 marksUnit 15
Briefly describe gravitational potential and derive the expression \varphi=-\dfrac{GM}{r} at a distance r from the center of Earth. Make a sketch.
Solution

Gravitational potential (2 marks)

The gravitational potential \varphi (or V) at a point is the work done per unit mass in bringing a small test mass from infinity to that point. It equals the gravitational potential energy per unit mass, \varphi=U/m. It is a scalar with SI unit \mathrm{J\,kg^{-1}}. Its zero is at infinity, so it is negative at every finite distance from the Earth, and it increases (becomes less negative) with distance.

Derivation (4 marks)

Here M is the mass of the Earth (M=M_E). The force changes with distance, so the path is split into small steps and the work in each step is added.

Derivation Derivation of \varphi=-\dfrac{GM_E}{r}

Aim. Find the work done per unit mass in bringing a small test mass m from infinity to a point at distance r from the centre of the Earth.

Assumptions.
  • The Earth is a uniform sphere of mass M_E and radius R_E; it acts as a point mass at its centre O.
  • The mass m is moved slowly, so its kinetic energy does not change and the work done on it by the agent goes into potential energy.
  • U=0 at infinity.
The path from the surface (point 0, distance R_E from the centre O) to a very distant point n is divided into small steps \Delta r; in each step the force is taken as constant.
  1. By the work and energy principle, when m is moved slowly the work done on it by the external agent equals the change in its potential energy.
    \Delta P.E.=W_{net}\qquad(1)
  2. The force changes with distance, so W=Fd cannot be used for the whole path at once. Divide the path between the surface (point 0) and a very distant point n into n small steps of length \Delta r, so small that the force is almost constant in each. The total work is the sum of the works in the steps.
    W_{net}=W_1+W_2+W_3+\dots+W_{n-1}+W_n\qquad(2)
  3. Take the step between point 1 (distance r_1) and point 0 (distance R_E). To bring m in slowly, the agent pulls outward with a force F_{av} equal to the pull of gravity, while m moves inward through \Delta r. Force and displacement are opposite: \theta=180^\circ and \cos180^\circ=-1.
    W_1=F_{av}\,\Delta r\cos180^\circ=-F_{av}\,\Delta r\qquad(3)
  4. The average force in this step follows from Newton's law, using the average distance r_{av} of the step.
    F_{av}=\dfrac{GM_Em}{r_{av}^{2}}\qquad(4)
  5. The average distance is neither R_E nor r_1 but their mean. From the figure \Delta r=r_1-R_E, so r_1=R_E+\Delta r.
    r_{av}=\dfrac{R_E+r_1}{2}=\dfrac{2R_E+\Delta r}{2}=R_E+\dfrac{\Delta r}{2}\qquad(5\text{ to }7)
  6. Square the average distance.
    r_{av}^{2}=\left(R_E+\dfrac{\Delta r}{2}\right)^{2}=R_E^{2}+R_E\,\Delta r+\dfrac{(\Delta r)^{2}}{4}
  7. \Delta r is very small, so the term (\Delta r)^{2}/4 is negligible compared with the others. Then put \Delta r=r_1-R_E.
    r_{av}^{2}\approx R_E^{2}+R_E\,\Delta r=R_E^{2}+R_E(r_1-R_E)=R_Er_1\qquad(8,\ 9)
  8. Put Eq. (9) in Eq. (4).
    F_{av}=\dfrac{GM_Em}{R_Er_1}\qquad(10)
  9. Put Eq. (10) and \Delta r=r_1-R_E (Eq. 11) in Eq. (3), then split the fraction.
    W_1=-\dfrac{GM_Em}{R_Er_1}\,(r_1-R_E)=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_1}\right)\qquad(12)
  10. The same working applies to every other step.
    \begin{aligned}W_2&=-GM_Em\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)\\ W_3&=-GM_Em\left(\dfrac{1}{r_2}-\dfrac{1}{r_3}\right)\\ &\ \ \vdots\\ W_{n-1}&=-GM_Em\left(\dfrac{1}{r_{n-2}}-\dfrac{1}{r_{n-1}}\right)\\ W_n&=-GM_Em\left(\dfrac{1}{r_{n-1}}-\dfrac{1}{r_n}\right)\end{aligned}\qquad(13\text{ to }16)
  11. Add all the steps as in Eq. (2). Every middle term appears once with + and once with -, so they cancel in pairs; only the first and last terms remain.
    \begin{aligned}W_{net}&=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_1}+\dfrac{1}{r_1}-\dfrac{1}{r_2}+\dots+\dfrac{1}{r_{n-1}}-\dfrac{1}{r_n}\right)\\ &=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_n}\right)\end{aligned}\qquad(17)
  12. By Eq. (1), this is the change in potential energy when m is brought from point n to the surface.
    \Delta P.E.=U_{\text{surface}}-U_n=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{r_n}\right)\qquad(18)
  13. Now take point n at infinity: r_n=\infty, so 1/r_n=0, and U_n=0 (our zero). The work done in bringing m from infinity to the surface is then
    U=-GM_Em\left(\dfrac{1}{R_E}-\dfrac{1}{\infty}\right)=-\dfrac{GM_Em}{R_E}
  14. The same steps, started from any point P at distance r from the centre instead of the surface, give the general result.
    U=-\dfrac{GM_Em}{r}\qquad(15.9)
  15. By definition the potential is the work done per unit mass, so divide by m. For a unit mass (m=1\,\mathrm{kg}) the work itself is the potential.
    \varphi=\dfrac{U}{m}=\dfrac{-GM_Em/r}{m}=-\dfrac{GM_E}{r}
Result\varphi=-\dfrac{GM_E}{r}

The potential is zero at infinity and negative at every finite distance; it rises towards zero as r increases. It depends only on the Earth's mass and the distance from its centre, not on the test mass. For any point mass M: \varphi=-GM/r.

Sketch (1 mark)

The labelled diagram of the path (above) is the sketch that goes with the derivation. The graph of \varphi against r shows the result:

Sketch: \varphi (\mathrm{MJ\,kg^{-1}}) against distance from the centre
\varphi is most negative (-62.5\,\mathrm{MJ\,kg^{-1}}) at the surface and rises towards zero as r increases.
2026 Annual I, Set D, Q.3 OR2+2+2+1 = 7 marksUnit 17
Derive expressions for Kinetic and Potential energies of a mass attached to a spring executing SHM. Show that total energy of the spring system remains constant. Also make neat sketch.
Solution

Consider a mass m attached to a spring of spring constant k, oscillating on a smooth horizontal surface with amplitude x_0. Let its displacement be x at some instant.

Kinetic energy (2 marks)

Derivation Kinetic energy of the oscillating mass

Aim. Find the kinetic energy of the mass when its displacement is x.

  1. The speed of a body in SHM at displacement x is
    v=\omega\sqrt{x_0^{2}-x^{2}}\qquad\text{(iv)}
  2. For the mass-spring system \omega=\sqrt{k/m}, so
    v=\sqrt{\dfrac{k}{m}\left(x_0^{2}-x^{2}\right)}\qquad\text{(v)}
  3. At the mean position (x=0) the speed is maximum.
    v_0=\sqrt{\dfrac{k}{m}}\,x_0\qquad\text{(vi)}
  4. Put v^{2} from (v) into \text{K.E.}=\tfrac12 mv^{2}. The mass m cancels.
    \text{K.E.}=\dfrac12 m\cdot\dfrac{k}{m}\left(x_0^{2}-x^{2}\right)=\dfrac12 k\left(x_0^{2}-x^{2}\right)\qquad(17.13)
Result\text{K.E.}=\dfrac12 mv^{2}=\dfrac12 k\left(x_0^{2}-x^{2}\right)

K.E. is largest at the mean position, \tfrac12 kx_0^{2}, and zero at the extreme positions, where the body stops for an instant.

Potential energy (2 marks)

Derivation Elastic potential energy of a stretched spring

Aim. Find the potential energy stored when the spring is stretched (or compressed) through x.

Assumptions.
  • The spring obeys Hooke's law and is stretched slowly from 0 to x.
  1. The work done by the applied force equals the average force times the displacement.
    W=F_{\text{av}}\,x\qquad\text{(i)}
  2. By Hooke's law the applied force needed at extension x is
    F=kx\qquad\text{(ii)}
  3. The force increases steadily (linearly) from 0 at the start to kx at the end, so its average is half the sum of the first and last values.
    F_{\text{av}}=\dfrac{0+kx}{2}=\dfrac12 kx\qquad\text{(iii)}
  4. Substitute (iii) in (i).
    W=\left(\dfrac12 kx\right)(x)=\dfrac12 kx^{2}
  5. This work is stored in the spring as elastic potential energy.
    \text{P.E.}=\dfrac12 kx^{2}\qquad(17.11)
Result\text{P.E.}=\dfrac12 kx^{2}

P.E. is zero at the mean position and grows as the square of the displacement. It is largest at the extreme positions, where \text{P.E.}_{\max}=\tfrac12 kx_0^{2}.

Total energy remains constant (2 marks)

Derivation Total energy of a body in SHM is constant

Aim. Show that \text{K.E.}+\text{P.E.} has the same value at every displacement.

  1. At an extreme position (x=x_0) the block is at rest, so K.E. is zero and all the energy is elastic P.E.
    \text{T.E.}=\text{P.E.}_{\max}=\dfrac12 kx_0^{2}\qquad(17.12)
  2. At the mean position (x=0) the spring is neither stretched nor compressed, so P.E. is zero and all the energy is kinetic.
    \text{T.E.}=\text{K.E.}_{\max}=\dfrac12 mv_{\max}^{2}=\dfrac12 kx_0^{2}\qquad(17.14)
  3. **At any displacement x**, add the energies from (17.11) and (17.13).
    \text{T.E.}=\text{K.E.}+\text{P.E.}=\dfrac12 k\left(x_0^{2}-x^{2}\right)+\dfrac12 kx^{2}
  4. The x^{2} terms cancel, leaving a quantity that does not depend on x.
    \text{T.E.}=\dfrac12 kx_0^{2}\qquad(17.15)
  5. Since \omega=\sqrt{k/m}, k=m\omega^{2}. This gives the total energy in terms of \omega.
    \text{T.E.}=\dfrac12 m\omega^{2}x_0^{2}\qquad(17.16)
Result\text{T.E.}=\dfrac12 kx_0^{2}=\dfrac12 m\omega^{2}x_0^{2}=\text{constant}

The total energy of the mass-spring system is constant and is proportional to the square of the amplitude. This statement of conservation of energy is equally valid for all bodies executing SHM.

Sketch (1 mark)

Energy against displacement
P.E. is zero at the mean position and largest at the extremes; K.E. is the opposite. The two are equal at x=\pm x_0/\sqrt{2}, and at every point K.E. + P.E. equals the same total energy.

K.E. and P.E. change continuously, one increasing as the other decreases, but their sum is the horizontal line \tfrac12 kx_0^{2}: the total energy of the spring system remains constant.

2026 Annual I, Set D, Q.44+1+2 = 7 marksUnit 18
Explain the working of Young's double slit experiment. Make diagram. Also derive the expression for fringe spacing.
Solution
Working of Young's double-slit experiment (4 marks)

Young's experiment (1801) produces interference of light from two coherent sources and shows that light behaves as a wave.

  1. Monochromatic light (for example from a sodium lamp or a laser) falls on a narrow single slit S_0. Light spreading out from S_0 by diffraction reaches two narrow, parallel slits S_1 and S_2, placed close together (separation d, about 0.5\,\mathrm{mm}) and at equal distances from S_0.
  2. S_1 and S_2 receive light from the same wavefront, so they act as two coherent sources vibrating in phase.
  3. Each slit diffracts the light, and the two sets of waves overlap and interfere in the region beyond the slits. A screen is placed at a distance L (about 1\,\mathrm{m}) from the slits.
  4. At points where the path difference S_2P-S_1P is 0,\ \lambda,\ 2\lambda,\dots the waves arrive in phase and interfere constructively, giving bright fringes.
  5. At points where the path difference is \tfrac12\lambda,\ \tfrac32\lambda,\dots the waves arrive in anti-phase and interfere destructively, giving dark fringes.
  6. So equally spaced, alternate bright and dark fringes parallel to the slits appear on the screen. The central fringe at O is bright because the path difference there is zero. With white light the central fringe is white and the others are coloured.
Diagram (1 mark)
Young's double-slit experiment: source, single slit S_0, double slit S_1S_2 (separation d), screen at distance L with bright and dark fringes.
Derivation of fringe spacing (2 marks)

Derivation Fringe spacing \Delta y=\lambda L/d

Aim. Find the distance between two neighbouring bright fringes.

Assumptions.
  • The screen is far away (L\gg d) and the angle \theta is small.
Path difference BC =d\sin\theta for a point P at height y above the centre O.
  1. Path difference. Draw S_1C perpendicular to S_2P. For a distant screen the rays are nearly parallel, so the path difference is BC. In the right triangle S_1S_2C the angle at S_1 equals \theta and the hypotenuse is d.
    S_2P-S_1P=BC=d\sin\theta
  2. Bright fringe. For the mth bright fringe the path difference is m whole wavelengths.
    d\sin\theta=m\lambda
  3. Position. \theta is small, so \sin\theta\approx\tan\theta=y_m/L, where y_m is the distance of the fringe from O.
    d\,\dfrac{y_m}{L}=m\lambda\;\Rightarrow\;y_m=\dfrac{m\lambda L}{d}
  4. Spacing. The next bright fringe is at y_{m+1}=(m+1)\lambda L/d. Subtract.
    \Delta y=y_{m+1}-y_m=\dfrac{(m+1)\lambda L}{d}-\dfrac{m\lambda L}{d}=\dfrac{\lambda L}{d}
Result\Delta y=\dfrac{\lambda L}{d}

The fringes are equally spaced. The spacing increases with the wavelength and the screen distance, and decreases when the slit separation increases. Dark fringes, at y=\left(m+\tfrac12\right)\lambda L/d, have the same spacing.

\Delta y=\dfrac{\lambda L}{d}
2026 Annual I, Set D, Q.4 OR(3+1)+3 = 7 marksUnit 19
a. Determine the electrical potential energy stored in a capacitor by using graph. b. Calculate the energy stored, in a capacitor having 5\,\mathrm{C} of charge and potential different of 15\,\mathrm{V}.
Solution
(a)
Explanation

While a capacitor is being charged, the battery does work in moving charge from one plate to the other against the growing potential difference. This work is stored as electric potential energy in the field between the plates. The p.d. rises in proportion to the charge, V=Q/C, from zero for the first charge to V for the last, so the graph of V against Q is a straight line through the origin.

Graph (1 mark)
Potential difference against charge
Fig. 19.6: The graph of V against Q is a straight line through the origin. The shaded triangle OAB is the energy stored; the average p.d. while charging is V/2.
Derivation (3 marks)

Derivation Energy stored from the V to Q graph

Aim. Find the energy E stored in a capacitor of capacitance C that holds charge Q at potential difference V.

Assumptions.
  • The capacitor starts uncharged and is charged in many small steps.
  • The potential difference is proportional to the charge at every moment, V=Q/C.
  1. To move a small charge \Delta Q across a potential difference V, the work needed is \Delta W=V\Delta Q (from \Delta U=q\Delta V). On the graph this is the area of a thin strip of height V and width \Delta Q.
    \Delta W=V\,\Delta Q
  2. Adding all the strips from zero to the final charge gives the total work, which is the area under the V to Q graph. The graph is a straight line, so the area is the right-angled triangle OAB.
    E=\text{area of triangle }OAB=\tfrac12\times\text{base}\times\text{height}
  3. The base OB is the final charge Q and the height AB is the final potential difference V.
    E=\tfrac12\,QV\qquad(19.11\,a)
  4. Substitute Q=CV to write the energy in terms of C and V.
    E=\tfrac12\,(CV)\times V=\tfrac12\,CV^{2}\qquad(19.11\,b)
  5. Substitute V=Q/C to write it in terms of Q and C.
    E=\tfrac12\,Q\times\dfrac{Q}{C}=\dfrac{Q^{2}}{2C}\qquad(19.11\,c)
ResultE=\tfrac12\,QV=\tfrac12\,CV^{2}=\dfrac{Q^{2}}{2C}

The energy is half of QV, because on average the charge was moved across only half the final voltage (V/2). For a given capacitor the energy grows as V^{2}: doubling the voltage stores four times the energy.

E=\tfrac12QV=\tfrac12CV^{2}=\dfrac{Q^{2}}{2C}
(b)
Given
  • Q = 5\,\mathrm{C}
  • V = 15\,\mathrm{V}
To find
  • E
  1. Charge and p.d. are given, so use
    E=\tfrac12QV
  2. Substitute.
    E=\tfrac12(5\,\mathrm{C})(15\,\mathrm{V})
  3. Evaluate.
    E=37.5\,\mathrm{J}
E=37.5\,\mathrm{J}
2026 Annual I, Set D, Q.54+2 = 6 marksUnit 20
Explain the working of a bridge rectifier using four diodes. Also draw the circuit diagram.
Solution
Circuit diagram (2 marks)
Full-wave bridge rectifier with four diodes. AC input at A and B; load R across the other two corners.

Four diodes D_1, D_2, D_3 and D_4 form a bridge. The AC input (secondary of a step-down transformer) is connected across corners A and B, and the load resistor R across the other two corners.

Working (4 marks)

Positive half-cycle: end A is positive and B negative. D_1 and D_2 are forward biased and conduct; D_3 and D_4 are reverse biased. Current flows from A through D_1, through R from top to bottom, through D_2 and back to B.

Positive half-cycle: D_1 and D_2 conduct.

Negative half-cycle: end A is negative and B positive. D_3 and D_4 are forward biased and conduct; D_1 and D_2 are reverse biased. Current flows from B through D_3, through R again from top to bottom, through D_4 and back to A.

Negative half-cycle: D_3 and D_4 conduct.

So the current through the load is in the same direction during both half-cycles, and both halves of the AC appear across R as pulsating DC (full-wave rectification). The output pulses at twice the supply frequency and can be smoothed with a capacitor across R.

Full-wave rectification
Fig 20.7 redrawn. Both half-cycles reach the load and both drive current the same way through it, so the negative halves appear flipped upward. Two pulses per cycle.
2026 Annual I, Set D, Q.5 OR1+3+2 = 6 marksUnit 21
What is photoelectric effect? Explain work function of metal and Einstein photo electric equation. What is the effect of frequency of radiation on K.E of photoelectrons?
Solution
Photoelectric effect (1 mark)

The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons.

Work function and Einstein's equation (3 marks)

The work function \Phi of a metal is the minimum energy needed to remove an electron from its surface. It is different for different metals (for example about 2.3\,\mathrm{eV} for sodium and 4.3\,\mathrm{eV} for zinc). It is related to the threshold frequency f_0 by \Phi=hf_0.

Derivation Einstein's photoelectric equation

Aim. Relate photon energy, work function and maximum kinetic energy.

  1. Light is a stream of photons; one photon of frequency f gives its whole energy hf to one electron.
    E=hf
  2. Part of the energy is used as the work function to free the electron; the rest is its kinetic energy (maximum for surface electrons).
    hf=\Phi+K.E_{\max}
  3. At threshold, K.E_{\max}=0 and f=f_0, so \Phi=hf_0. Substituting:
    K.E_{\max}=\tfrac12mv_{\max}^{2}=hf-hf_0
Resulthf=\Phi+\tfrac12mv_{\max}^{2}

Emission occurs only when hf\ge\Phi; the extra energy appears as kinetic energy.

Effect of frequency on K.E of photoelectrons (2 marks)

From K.E_{\max}=hf-\Phi: below the threshold frequency f_0 no electrons are emitted. Above f_0, the maximum kinetic energy increases linearly with the frequency; the graph is a straight line of slope h, cutting the frequency axis at f_0. Increasing the frequency (for example from red to violet to ultraviolet light) gives faster photoelectrons, whatever the intensity.

Maximum kinetic energy against frequency
No electrons are emitted below the threshold frequency f_0. Above f_0 the graph is a straight line of slope h; extended back (dashed) it cuts the energy axis at -\Phi. The line is drawn to scale for sodium (\Phi=2.3\,\mathrm{eV}, f_0=5.6\times10^{14}\,\mathrm{Hz}).
2026 Annual I, Set D, Q.63+3 = 6 marksUnit 23
Answer both parts (a) and (b).
Solution
(a) What is meant by luminosity of a star? Write the relation between luminosity, radius and temperature.

Luminosity (1 mark): the total amount of electromagnetic energy emitted per unit time by a star, that is, the total power output of its radiation. SI unit: watt (\mathrm{J\,s^{-1}}). The Sun's luminosity is 3.84\times10^{26}\,\mathrm{W}.

Relation (2 marks): treating the star as a black body, the Stefan-Boltzmann law says each square metre of its surface emits power \sigma T^{4}. A spherical star of radius r has surface area 4\pi r^{2}, so

L=A\sigma T^{4}=4\pi r^{2}\sigma T^{4}

where \sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}} and T is the surface temperature in kelvin. So L\propto r^{2} and L\propto T^{4}: doubling the radius gives 4 times the luminosity, and doubling the temperature gives 16 times.

(b) What are the relative luminosities of the objects A and B if T_A=100\,\mathrm{K}, T_B=200\,\mathrm{K}, R_A=10\,\mathrm{m} and R_B=5\,\mathrm{m}
  1. Formula (1 mark): write L=4\pi R^{2}\sigma T^{4} for each object and divide; 4\pi\sigma cancels.
    \dfrac{L_A}{L_B}=\left(\dfrac{R_A}{R_B}\right)^{2}\left(\dfrac{T_A}{T_B}\right)^{4}
  2. Substitution (1 mark):
    \dfrac{L_A}{L_B}=\left(\dfrac{10}{5}\right)^{2}\left(\dfrac{100}{200}\right)^{4}=4\times\dfrac{1}{16}
  3. Result (1 mark):
    \dfrac{L_A}{L_B}=\dfrac{1}{4}\quad\Longrightarrow\quad L_B=4L_A

Check: L_A=4\pi(10)^{2}(5.67\times10^{-8})(100)^{4}=7.1\times10^{3}\,\mathrm{W} and L_B=4\pi(5)^{2}(5.67\times10^{-8})(200)^{4}=2.9\times10^{4}\,\mathrm{W}. B is smaller but twice as hot, and the factor 2^{4}=16 from temperature outweighs the factor 2^{2}=4 from size.

L_A:L_B=1:4

Note. Part (b) is Assignment 23.1 of the book; part (a) is covered by comprehensive question 1.

2026 Annual I, Set D, Q.6 OR3+3 = 6 marksUnit 22
a. Derive and show that the decay of nucleus is exponential with time. b. If half-life of strontium (Sr. 91) is 9.70 hours. Find its decay constant.
Solution
(a) Derive and show that the decay of a nucleus is exponential with time. (3 marks)

The rate of decay of a radioactive sample is directly proportional to the number N of undecayed nuclei present (rate law). The constant of proportionality is the decay constant \lambda.

Derivation Exponential decay law

Aim. Show that N=N_0e^{-\lambda t}.

Assumptions.
  • The sample contains a very large number of nuclei, so N changes smoothly.
  • The decay constant \lambda is constant for the isotope.
  1. Rate law: the rate of decay is proportional to the number of undecayed nuclei; the minus sign shows N decreases.
    \dfrac{\Delta N}{\Delta t}=-\lambda N
  2. For a very small interval the ratio becomes a derivative.
    \dfrac{dN}{dt}=-\lambda N
  3. Separate the variables.
    \dfrac{dN}{N}=-\lambda\,dt
  4. Integrate both sides.
    \ln N=-\lambda t+C\qquad(\text{i})
  5. Initial condition: at t=0, N=N_0, so the constant is C=\ln N_0.
    \ln N_0=0+C
  6. Substitute C in (i) and combine the logarithms.
    \ln N-\ln N_0=-\lambda t\;\Rightarrow\;\ln\dfrac{N}{N_0}=-\lambda t
  7. Take the exponential of both sides.
    \dfrac{N}{N_0}=e^{-\lambda t}
ResultN=N_0\,e^{-\lambda t}

The number of undecayed nuclei decreases exponentially with time and never becomes exactly zero.

This shows that the number of undecayed nuclei falls exponentially with time: in equal time intervals the same fraction decays (half in each half-life), and N never becomes exactly zero.

Exponential decay curve
N against t: the number halves in every half-life.
N=N_0e^{-\lambda t}
(b) The half-life of strontium-91 is 9.70 hours. Find its decay constant. (3 marks)
Given
  • T_{1/2} = 9.70\,\mathrm{h}=9.70\times3600\,\mathrm{s}=3.492\times10^{4}\,\mathrm{s}
To find
  • \lambda
  1. Relation between decay constant and half-life.
    \lambda=\dfrac{0.693}{T_{1/2}}
  2. Substitute the half-life in seconds.
    \lambda=\dfrac{0.693}{3.492\times10^{4}\,\mathrm{s}}=1.98\times10^{-5}\,\mathrm{s^{-1}}
  3. Equivalent value per hour.
    \lambda=\dfrac{0.693}{9.70\,\mathrm{h}}=0.0714\,\mathrm{h^{-1}}
\lambda\approx1.98\times10^{-5}\,\mathrm{s^{-1}}

Note. Part (a) matches Long Question 8 of the exercise; part (b) is the book's Assignment 22.6.

Old syllabus (Curriculum 2006)

26 questions

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 1, old syllabus1 markUnit 19
If two charges experience a coulomb force of 10\,\mathrm{N} when medium is air, what will be the coulomb force if air is replaced by a medium of relative permittivity 2?
  1. A3\,\mathrm{N}
  2. B5\,\mathrm{N}Answer
  3. C10\,\mathrm{N}
  4. D0.5\,\mathrm{N}
Why

In a medium the Coulomb force is divided by the relative permittivity: F_{med}=\dfrac{1}{4\pi\varepsilon_0\varepsilon_r}\dfrac{q_1q_2}{r^{2}}=\dfrac{F_{air}}{\varepsilon_r}=\dfrac{10\,\mathrm{N}}{2}=5\,\mathrm{N}. The medium's polarisation weakens the field between the charges, the same effect that raises the capacitance of a capacitor by \varepsilon_r.

Note. This is Coulomb's law in a medium, which the new book treats with electrostatics; it is answered here because it uses relative permittivity.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 5, old syllabus1 markUnit 19
Ohm × Farad =
  1. ASecondAnswer
  2. BWeber
  3. CHenry
  4. DTesla
Why

RC is the time constant, so its unit is the second: \Omega\times\mathrm{F}=\dfrac{\mathrm{V}}{\mathrm{A}}\times\dfrac{\mathrm{C}}{\mathrm{V}}=\dfrac{\mathrm{C}}{\mathrm{A}}=\mathrm{s}. Weber, henry and tesla are units of magnetic flux, inductance and magnetic field.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 6, old syllabus1 markUnit 20
An alternating voltage is V=20\sin314t. The frequency of alternating voltage is:
  1. A25\,\mathrm{Hz}
  2. B50\,\mathrm{Hz}Answer
  3. C75\,\mathrm{Hz}
  4. D100\,\mathrm{Hz}
Why

Comparing with V=V_0\sin(2\pi ft) gives \omega=2\pi f=314\,\mathrm{rad\,s^{-1}}, so f=\dfrac{314}{2\times3.14}=50\,\mathrm{Hz}. (Book exercise MCQ 7 is the same type with \omega=157, giving 25 Hz.)

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 8, old syllabus1 markUnit 22
The number of neutrons in {}_{56}\mathrm{Ba}^{138} are:
  1. A192
  2. B138
  3. C82Answer
  4. D56
Why

The nucleus has Z=56 protons and mass number A=138, so N=A-Z=138-56=82. (192 is A+Z, 138 is the number of nucleons and 56 the number of protons.)

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 11, old syllabus1 markUnit 22
The unit of decay constant is:
  1. AMeter
  2. BSecond
  3. CPer meter
  4. DPer secondAnswer
Why

The decay constant is the fraction of nuclei that decay per unit time, \lambda=\dfrac{1}{N}\dfrac{\Delta N}{\Delta t}, so its unit is \mathrm{s^{-1}} (per second). This also follows from \lambda=0.693/T_{1/2}.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 13, old syllabus1 markUnit 20
What is the coefficient of mutual inductance when the magnetic flux changes by 2\times10^{-2}\,Wb and change in current is 0.01\,A.
  1. A0
  2. B\dfrac12\,\mathrm{H}
  3. C2\,\mathrm{H}Answer
  4. D3\,\mathrm{H}
Why

The mutual inductance is the flux linkage of the secondary per unit current in the primary: M=\dfrac{N\Delta\Phi}{\Delta I}=\dfrac{2\times10^{-2}}{0.01}=2\,\mathrm{H} (taking the given flux change as the change in flux linkage, N=1). This matches the paper's formula sheet, M=N\varphi/I.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 14, old syllabus1 markUnit 20
Lenz's law is the consequence of the law of conservation of:
  1. AMass
  2. BEnergyAnswer
  3. CChange
  4. DMomentum
Why

The induced current opposes the change that produces it, so work must be done against it, and this work appears as electrical energy. If it helped the change, energy would be created from nothing. So Lenz's law follows from the law of conservation of energy.

Note. Option C is printed as 'Change'; it was probably meant to be 'Charge'. Either way the answer is B.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 15, old syllabus1 markUnit 20
The average power loss in an inductive circuit is:
  1. A0Answer
  2. BV_mI_m
  3. CV_{rms}I_{rms}
  4. D\dfrac{V_m}{I_m}
Why

In a pure inductor the current lags the voltage by 90^\circ. Energy stored in the magnetic field during one quarter cycle is returned to the source in the next, so the average power is V_{\mathrm{rms}}I_{\mathrm{rms}}\cos90^\circ=0. Option C, V_{\mathrm{rms}}I_{\mathrm{rms}}, is the average power in a pure resistor.

Note. 'Inductive circuit' here means a purely inductive circuit. A real coil with resistance R loses I_{\mathrm{rms}}^2R.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 16, old syllabus1 markUnit 16
Which of the following is a boson?
  1. AProton
  2. BNeutrino
  3. CPhotonAnswer
  4. DPion
Why

Bosons are particles that do not obey the Pauli exclusion principle (they have whole-number spin), so any number of them can share one quantum state. The photon (spin 1) is the standard example, and the book names photons as bosons. The proton and the neutrino have spin \frac12, so they are fermions.

Note. Strictly, the pion (a meson with spin 0) is also a boson, so option D is also correct. The photon is the intended textbook answer.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 17, old syllabus1 markUnit 22
Various types of cancer are treated by:
  1. AStrontium-90
  2. BCobalt 60Answer
  3. CCarbon 14
  4. DNickel-63
Why

Cobalt-60 emits penetrating gamma rays (and has a convenient half-life of 5.27 years), so it is the standard source for radiotherapy of many cancers. Carbon-14 is used for dating, nickel-63 in small batteries and detectors, and strontium-90 only for a few surface treatments.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (i), old syllabus1+2 = 3 marksUnit 21
What is electron volt (eV)? Derive its relation with SI unit of energy.
Solution
Definition (1 mark)

One electronvolt is the energy gained (or lost) by an electron when it moves through a potential difference of one volt.

Relation with the joule (2 marks)
  1. Work done (energy gained) when a charge q moves through a p.d. V.
    W=qV
  2. For an electron, q=e=1.6\times10^{-19}\,\mathrm{C}, and V=1\,\mathrm{V}.
    1\,\mathrm{eV}=(1.6\times10^{-19}\,\mathrm{C})(1\,\mathrm{V})
  3. Since 1\,\mathrm{C}\times1\,\mathrm{V}=1\,\mathrm{J}:
    1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}

Also 1\,\mathrm{keV}=1.6\times10^{-16}\,\mathrm{J} and 1\,\mathrm{MeV}=1.6\times10^{-13}\,\mathrm{J}. The electronvolt is a convenient small unit for atomic and nuclear energies.

1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}
2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (i) OR, old syllabus3 marksUnit 21
How can we calculate kinetic energy of photoelectrons?
Solution

The maximum kinetic energy of photoelectrons is found by measuring the stopping potential V_0. In the photoelectric set-up the collector is made negative with respect to the emitter (battery reversed), and the retarding potential is increased until the photoelectric current just becomes zero. At this potential even the fastest electrons are stopped, so their kinetic energy has all been used in doing work eV_0 against the field:

K.E_{\max}=\tfrac12mv_{\max}^{2}=eV_0
Photoelectric current against collector potential
Light of one frequency at three intensities. Brighter light gives a larger saturation current, but every curve reaches zero at the same stopping potential -V_0.

For example, if V_0=2\,\mathrm{V} then K.E_{\max}=2\,\mathrm{eV}=3.2\times10^{-19}\,\mathrm{J}. If the frequency and the work function are known instead, the same energy follows from Einstein's equation K.E_{\max}=hf-\Phi.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (ii), old syllabus3 marksUnit 19
Find electric field at a distance of 30\,\mathrm{cm} from a 3\,\mu\mathrm{C} point charge.
Solution
Given
  • q = 3\,\mu\mathrm{C}=3\times10^{-6}\,\mathrm{C}
  • r = 30\,\mathrm{cm}=0.30\,\mathrm{m}
  • k = 9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}
To find
  • E
  1. Field of a point charge (Coulomb's law divided by the test charge):
    E=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{q}{r^{2}}=\dfrac{kq}{r^{2}}
  2. Substitute.
    E=\dfrac{(9\times10^{9})(3\times10^{-6})}{(0.30)^{2}}=\dfrac{2.7\times10^{4}}{0.09}
  3. Evaluate. The field points radially away from the charge (taking it as positive).
    E=3.0\times10^{5}\,\mathrm{N\,C^{-1}}
E=3.0\times10^{5}\,\mathrm{N\,C^{-1}}

Note. The field of a point charge belongs to electrostatics; the map assigned it to this unit.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (ii) OR, old syllabus3 marksUnit 23
What is Wein's displacement law? Explain briefly.
Solution

Wien's displacement law: the black body radiation curve for different temperatures peaks at a wavelength which is inversely proportional to the absolute temperature of the body.

\lambda_{\max}\propto\dfrac{1}{T}\qquad\Longrightarrow\qquad\lambda_{\max}T=b=2.9\times10^{-3}\,\mathrm{m\,K}

Explanation: a hot body emits all wavelengths, but not equally; the intensity is greatest at the peak wavelength \lambda_{\max}. As the temperature rises the peak is displaced to shorter wavelengths, and the intensity rises at every wavelength. That is why a heated iron rod glows red, then yellow, then white, and why blue stars are hotter than red stars. Measuring \lambda_{\max} gives a star's surface temperature, T=b/\lambda_{\max}.

Black body radiation curves
Black body curves: as T rises, the peak \lambda_{\max} moves to shorter wavelengths (\lambda in \mu\mathrm{m}, intensity I in relative units).

Note. The paper spells the name 'Wein'; the correct spelling is Wien (Wilhelm Wien).

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (v) OR, old syllabus3 marksUnit 19
Prove that E=-\dfrac{\Delta V}{\Delta r}.
Solution

Consider two points A and B on a field line, a small distance \Delta r apart, so the field E between them is uniform. A positive test charge q_0 is moved slowly from A to B, along the field.

The applied force F=-q_0E is opposite to the displacement \Delta r.
  1. To keep q_0 in equilibrium, the applied force must balance the electric force.
    \vec F=-q_0\vec E
  2. The force is opposite to the displacement (\theta=180^\circ), so the work done from A to B is
    W_{AB}=F\,\Delta r\cos180^\circ=-q_0E\,\Delta r
  3. Potential difference is work done per unit charge.
    \Delta V=V_B-V_A=\dfrac{W_{AB}}{q_0}=-E\,\Delta r
  4. Rearrange.
    E=-\dfrac{\Delta V}{\Delta r}

\Delta V/\Delta r is the potential gradient. The negative sign shows that the field points in the direction of decreasing potential.

E=-\dfrac{\Delta V}{\Delta r}
2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (vi) OR, old syllabus3 marksUnit 22
How can nuclear reactor initiate a controlled fission reaction?
Solution

A reactor starts and keeps a controlled chain reaction, in which on average exactly one neutron from each fission causes another fission:

  1. The core contains fuel rods of uranium enriched to about 3% U-235, surrounded by a moderator (water or graphite). A few free neutrons are always present, from spontaneous fission or a small neutron source.
  2. The moderator slows these neutrons to thermal energies (about 0.025\,\mathrm{eV}), at which they readily cause fission of U-235. Each fission releases 2 or 3 more fast neutrons, which are slowed in turn.
  3. Control rods of cadmium or boron absorb neutrons. They are slowly withdrawn until the reactor becomes critical (one neutron per fission causes a further fission), and are then adjusted to keep the rate of fission, and so the power, constant.
  4. The coolant removes the heat, and safety rods drop in to stop the reaction at once in an emergency.
2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (vii), old syllabus1+1+1 = 3 marksUnit 20
State the factors which affect the inductance of a coil.
Solution

The self-inductance of a long coil (solenoid) is

L=\dfrac{\mu_r\mu_0N^2A}{l}

So it is affected by three factors (one mark each):

  1. Number of turns N: L\propto N^2, so doubling the turns makes L four times larger.
  2. Size of the coil: L is proportional to the area of cross-section A and, for the same number of turns, inversely proportional to the length l (a closely wound coil has more inductance).
  3. Nature of the core: a core of high relative permeability \mu_r, such as soft iron, increases L many times compared with air.

The inductance does not depend on the current flowing in the coil.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (viii), old syllabus3 marksUnit 22
How Geiger-Muller counter detects and counts radiation?
Solution

A GM tube is a metal cylinder (cathode) with a thin central wire (anode) at about 400\,\mathrm{V} positive, filled with argon at low pressure and a little quenching gas, with a thin mica window at one end.

Geiger-Muller tube and counter.
  1. Radiation entering the window ionises a few gas atoms.
  2. The electrons are accelerated towards the wire by the strong field and ionise more atoms: an avalanche of electrons reaches the anode.
  3. This gives a pulse of current, which makes a voltage pulse across the resistor R.
  4. Each pulse is amplified and counted by a scaler or ratemeter: one pulse for each particle or photon, so the count rate measures the radiation.
  5. The quenching gas stops the discharge so the tube is ready for the next particle (after a dead time of about 10^{-4}\,\mathrm{s}).
2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (x), old syllabus3 marksUnit 20
Briefly explain impedance in RLC series circuit of AC.
Solution

Impedance Z is the total opposition of an AC circuit, Z=V/I, measured in ohms. In an RLC series circuit the same current I flows through R, L and C, but V_R=IR is in phase with I, V_L=IX_L leads it by 90^\circ and V_C=IX_C lags it by 90^\circ. V_L and V_C are opposite, so the net reactive voltage is V_L-V_C, and the supply voltage is the phasor sum of V_R and V_L-V_C.

Phasor diagram of the RLC series circuit.
\begin{aligned}V&=\sqrt{V_R^2+(V_L-V_C)^2}=I\sqrt{R^2+(X_L-X_C)^2}\\ \Longrightarrow\ Z&=\sqrt{R^2+(X_L-X_C)^2}\end{aligned}

with \tan\theta=(X_L-X_C)/R. If X_L>X_C the current lags; if X_C>X_L it leads. At resonance (X_L=X_C) the impedance is minimum, Z=R, and the current is maximum.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (x) OR, old syllabus3 marksUnit 19
How can one increase the capacitance of a capacitor?
Solution

The capacitance of a parallel plate capacitor is

C=\dfrac{\varepsilon_0\varepsilon_rA}{d}

So the capacitance can be increased in three ways:

  1. **Increase the area A of the plates** (C\propto A): larger plates hold more charge at the same voltage. Long foils are rolled up to get a large area in a small space.
  2. **Decrease the separation d between the plates** (C\propto1/d): halving the gap doubles the capacitance.
  3. **Insert a dielectric of high relative permittivity \varepsilon_r** between the plates (C\propto\varepsilon_r): for example, mica (\varepsilon_r\approx6) makes the capacitance about six times its value in air. The polarised dielectric weakens the field between the plates, so more charge is stored per volt.

The effective capacitance of a circuit can also be increased by connecting more capacitors in parallel, since C_e=C_1+C_2+\cdots.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (xii), old syllabus3 marksUnit 22
The half-life of radioactive nucleus {}_{88}\mathrm{Ra}^{226} is 1.6\times10^{3} years. Determine the decay constant.
Solution
Given
  • T_{1/2} = 1.6\times10^{3}\,\text{years}=1.6\times10^{3}\times3.15\times10^{7}\,\mathrm{s}=5.05\times10^{10}\,\mathrm{s} 1 year = 365 days
To find
  • decay constant \lambda
  1. Relation between decay constant and half-life (1 mark).
    \lambda=\dfrac{0.693}{T_{1/2}}
  2. Substitute the half-life in seconds (1 mark).
    \lambda=\dfrac{0.693}{5.05\times10^{10}\,\mathrm{s}}
  3. Evaluate, with the unit (1 mark).
    \lambda=1.37\times10^{-11}\,\mathrm{s^{-1}}
    In per year: \lambda=0.693/1600=4.33\times10^{-4}\,\text{year}^{-1}.
\lambda\approx1.37\times10^{-11}\,\mathrm{s^{-1}}\ (4.33\times10^{-4}\ \text{per year})
2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (xiii) OR, old syllabus1.5+1.5 = 3 marksUnit 20
How does doubling the frequency affect the reactance of an inductor and a capacitor?
Solution
Inductor (1.5 marks)
X_L=2\pi fL\quad\Longrightarrow\quad X_L'=2\pi(2f)L=2X_L

Doubling the frequency doubles the inductive reactance, because the current changes twice as fast and the back emf is twice as large.

Capacitor (1.5 marks)
X_C=\dfrac{1}{2\pi fC}\quad\Longrightarrow\quad X_C'=\dfrac{1}{2\pi(2f)C}=\dfrac{X_C}{2}

Doubling the frequency halves the capacitive reactance, because the capacitor charges and discharges twice as often and twice as much charge flows per second.

How R, X_L and X_C change with frequency
X_L is a straight line through the origin: zero for DC, growing with frequency. X_C is a hyperbola: very large at low frequency (infinite for DC), small at high frequency. R does not depend on frequency. The lines X_L and X_C cross at the resonant frequency f_0.
2026 Annual I, Old syllabus (Curriculum 2006), Q.4, old syllabus2+2+3 = 7 marksUnit 21
What is photoelectric effect? Why classical Physics fails to explain photoelectric effect? Derive Einstein photoelectric equation.
Solution
Photoelectric effect (2 marks)

The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons and the current they form is the photoelectric current.

Light falling on the emitter E ejects photoelectrons, which are collected by C and recorded by the ammeter.
Failure of classical physics (2 marks)

Classical physics treats light as a continuous wave whose energy depends on intensity. It fails to explain:

  • Threshold frequency: no emission below f_0 however intense the light (wave theory predicts emission at any frequency if the light is bright enough).
  • Kinetic energy independent of intensity: K.E_{\max} does not increase with intensity.
  • Kinetic energy depends on frequency: K.E_{\max} increases linearly with f.
  • No time lag: emission is instantaneous even in very dim light.
Einstein's photoelectric equation (3 marks)

Derivation Einstein's photoelectric equation

Aim. Derive the relation between photon energy, work function and maximum kinetic energy.

  1. Light consists of photons, each of energy hf. One photon is absorbed completely by one electron.
    E=hf
  2. The electron uses energy \Phi (work function) to escape from the surface; by conservation of energy the remainder is its maximum kinetic energy.
    hf=\Phi+K.E_{\max}
  3. At the threshold frequency the kinetic energy is zero, so the work function is hf_0.
    \Phi=hf_0
  4. Substitute and write the kinetic energy in terms of speed and stopping potential.
    \tfrac12mv_{\max}^{2}=hf-hf_0=eV_0
Resulthf=\Phi+\tfrac12mv_{\max}^{2}

This equation explains every observation: no emission if hf<\Phi; K.E_{\max} rises linearly with f (slope h) and is independent of intensity; the number of electrons depends on the number of photons.

2026 Annual I, Old syllabus (Curriculum 2006), Q.4 OR, old syllabus7 marksUnit 20
A 220\,V, 50\,Hz A.C. supply is applied to series combination of a 2.5\,\Omega resistor, 6\,mH inductor and a 6.5\,\mu F capacitor. Calculate (i) reactance of inductor (ii) reactance of capacitor (iii) impedance of circuit (iv) current in the circuit
Solution
Given
  • V = 220\,\mathrm{V} r.m.s. supply voltage
  • f = 50\,\mathrm{Hz}
  • R = 2.5\,\Omega
  • L = 6\,\mathrm{mH}=6\times10^{-3}\,\mathrm{H}
  • C = 6.5\,\mu\mathrm{F}=6.5\times10^{-6}\,\mathrm{F}
To find
  • X_L
  • X_C
  • Z
  • I
(i) Reactance of inductor
  1. Inductive reactance.
    X_L=2\pi fL=2\times3.14\times50\times6\times10^{-3}=1.88\,\Omega
X_L\approx1.88\,\Omega
(ii) Reactance of capacitor
  1. Capacitive reactance.
    X_C=\dfrac{1}{2\pi fC}=\dfrac{1}{2\times3.14\times50\times6.5\times10^{-6}}=\dfrac{1}{2.042\times10^{-3}}=489.7\,\Omega
X_C\approx490\,\Omega
(iii) Impedance of circuit
  1. Series RLC impedance; the net reactance is X_L-X_C.
    Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{2.5^2+(1.88-489.7)^2}
  2. Evaluate. The resistance hardly matters because the net reactance is so large.
    Z=\sqrt{6.25+2.380\times10^{5}}=487.8\,\Omega
Z\approx488\,\Omega
(iv) Current in the circuit
  1. Ohm's law for AC with r.m.s. values.
    I=\dfrac{V}{Z}=\dfrac{220}{487.8}=0.451\,\mathrm{A}

Since X_C\gg X_L, the circuit is strongly capacitive: the current leads the voltage by \theta=\tan^{-1}\dfrac{X_C-X_L}{R}\approx89.7^\circ. The circuit would resonate only at f_0=1/(2\pi\sqrt{LC})\approx806\,\mathrm{Hz}.

I\approx0.45\,\mathrm{A}

Note. The paper does not print the marks for this OR question. Section C carries 26 marks and the first alternative of Q.4 is marked 2+2+3, so this question carries 7 marks.

2026 Annual I, Old syllabus (Curriculum 2006), Q.5 OR, old syllabus3+4 = 7 marksUnit 22
What is meant by 'half-life' of a radio active element? Show that T_{1/2}=\dfrac{0.693}{\lambda}.
Solution

Half-life (3 marks)

The half-life T_{1/2} of a radioactive element is the time taken for half of the radioactive nuclei present in a sample to decay.

Starting with N_0 undecayed nuclei, N_0/2 remain after one half-life, N_0/4 after two, N_0/8 after three, and so on: N=N_0(1/2)^{n} after n half-lives. The half-life is a fixed property of each isotope. It does not depend on the amount of the sample or on temperature, pressure or chemical state. Examples: radon-222, 3.82 days; cobalt-60, 5.27 years; carbon-14, 5730 years; uranium-238, 4.5 billion years.

Decay curve showing successive half-lives
The number of undecayed nuclei halves in each half-life.

Derivation of T_{1/2}=0.693/\lambda (4 marks)

Derivation Relation between half-life and decay constant

Aim. Show that T_{1/2}=\dfrac{0.693}{\lambda}.

Assumptions.
  • The sample obeys the exponential decay law N=N_0e^{-\lambda t}, which follows from the rate law dN/dt=-\lambda N.
  1. Write the decay law.
    N=N_0\,e^{-\lambda t}
  2. At t=T_{1/2} half of the nuclei remain, N=N_0/2. Substitute.
    \dfrac{N_0}{2}=N_0\,e^{-\lambda T_{1/2}}
  3. Cancel N_0 and take the reciprocal of both sides.
    \dfrac{1}{2}=e^{-\lambda T_{1/2}}\;\Longrightarrow\;2=e^{\lambda T_{1/2}}
  4. Take the natural logarithm of both sides.
    \ln2=\lambda T_{1/2}
  5. Rearrange, using \ln2=0.693.
    T_{1/2}=\dfrac{\ln2}{\lambda}=\dfrac{0.693}{\lambda}
ResultT_{1/2}=\dfrac{0.693}{\lambda}

The half-life is inversely proportional to the decay constant: a rapidly decaying isotope (large \lambda) has a short half-life.

2026 Annual I, Old syllabus (Curriculum 2006), Q.6 OR, old syllabus6 marksUnit 20
Explain the phenomes of self-inductance of a coil. What is its unit? On what factors it depends?
Solution
Phenomenon of self-induction
A coil connected to a battery through a switch and a rheostat.

A coil carrying a current produces a magnetic field whose flux passes through the coil itself. If the current is changed, for example by moving the rheostat quickly or by switching on or off, the flux through the coil changes, and by Faraday's law an emf is induced in the same coil. This is self-induction: the phenomenon in which a changing current induces an emf in the coil itself.

\varepsilon_L=-L\dfrac{\Delta I}{\Delta t}\qquad\Longrightarrow\qquad L=\dfrac{-\varepsilon_L}{\Delta I/\Delta t}

The constant L is the self-inductance of the coil: the ratio of the induced emf to the rate of change of current. The minus sign is Lenz's law: the induced emf opposes the change in current, so it is called the back emf. It slows the growth of current when the circuit is switched on and tries to maintain the current when it is switched off.

Unit
1\,\mathrm{H}=1\,\mathrm{V\,s\,A^{-1}}

The unit is the henry. The self-inductance of a coil is one henry if an emf of one volt is induced in it when the current changes at the rate of one ampere per second.

Factors
L=\dfrac{\mu_r\mu_0N^2A}{l}
  • number of turns N (L\propto N^2);
  • area of cross-section A of the coil;
  • length l of the coil for the same number of turns (L\propto1/l);
  • nature of the core, its relative permeability \mu_r (soft iron increases L greatly).

Note. 'Phenomes' is printed for 'phenomenon'. The paper does not print the marks for this OR question; Section C carries 26 marks and the first alternative of Q.6 is marked 4+2, so this question carries 6 marks.