IOE entrance Physics · Chapter 6
Modern Physics
Tap an option to check it. Wrong picks show the right answer and the hint; “Show answer” gives the worked solution.
66 questions in 5 syllabus topics · 27 are 2-mark questions.
6.1 Electrons
8 questions
1. In Millikan's oil drop experiment, a charged drop of mass m and charge q remains stationary between the plates when the electric field E satisfies:
Show hintHide hint
A stationary drop experiences no viscous force.
Show answerHide answer
Answer: C. qE = mg
When the drop is at rest there is no viscous drag (v = 0), so the upward electric force must exactly balance the weight: qE = mg.
2. An oil drop of mass 3.2×10⁻¹⁵ kg is held stationary in a vertical electric field of 1×10⁵ V/m. How many excess electrons does it carry? (g = 10 m/s², e = 1.6×10⁻¹⁹ C)
2 marksShow hintHide hint
Balance qE = mg, then divide the charge by e.
Show answerHide answer
Answer: B. 2
q = mg/E = (3.2×10⁻¹⁵ × 10)/(1×10⁵) = 3.2×10⁻¹⁹ C. Number of electrons n = q/e = 3.2×10⁻¹⁹/1.6×10⁻¹⁹ = 2.
3. The most important conclusion of Millikan's oil drop experiment was that:
Show hintHide hint
Think of what all the measured drop charges had in common.
Show answerHide answer
Answer: B. electric charge is quantized, occurring in integral multiples of e
Millikan found that the charges on all drops were integral multiples of a smallest charge e = 1.6×10⁻¹⁹ C, establishing the quantization of charge. The e/m value came from J.J. Thomson's experiment.
4. Which of the following statements about cathode rays is NOT correct?
Show hintHide hint
Cathode rays are the same particle whatever the tube contains.
Show answerHide answer
Answer: D. Their specific charge depends on the gas filled in the discharge tube
Cathode rays are streams of electrons; their e/m is the same whatever gas or cathode material is used. The other statements are true properties of cathode rays.
5. In a Thomson-type experiment, an electron beam passes undeflected through crossed electric field 3×10⁴ V/m and magnetic field 2×10⁻³ T. The speed of the electrons is:
2 marksShow hintHide hint
Equate the electric and magnetic forces.
Show answerHide answer
Answer: C. 1.5×10⁷ m/s
For no deflection eE = evB, so v = E/B = 3×10⁴/2×10⁻³ = 1.5×10⁷ m/s.
6. Electrons accelerated from rest through a potential difference V enter a uniform magnetic field B at right angles and move in a circle of radius r. Their specific charge e/m is:
Show hintHide hint
Combine eV = ½mv² with r = mv/(eB).
Show answerHide answer
Answer: D. 2V/(B²r²)
eV = ½mv² gives v² = 2eV/m, and r = mv/(eB) gives v = eBr/m. Substituting: e²B²r²/m² = 2eV/m, so e/m = 2V/(B²r²).
7. An electron is accelerated from rest through a potential difference of 100 V. Taking e/m = 1.8×10¹¹ C/kg, its final speed is:
2 marksShow hintHide hint
Use ½mv² = eV.
Show answerHide answer
Answer: C. 6×10⁶ m/s
½mv² = eV gives v = √(2(e/m)V) = √(2 × 1.8×10¹¹ × 100) = √(3.6×10¹³) = 6×10⁶ m/s.
8. The ratio of the specific charge of a proton to that of an alpha particle is:
Show hintHide hint
An alpha particle has twice the charge and four times the mass of a proton.
Show answerHide answer
Answer: B. 2 : 1
Proton: e/m. Alpha particle: 2e/4m = e/(2m). Ratio = (e/m) : (e/2m) = 2 : 1.
6.2 Photons and quantization of energy
16 questions
9. The ratio of the areas within the electron orbits for the first excited state to the ground state for the hydrogen atom is
IOE model question 2080Show hintHide hint
r ∝ n² and area ∝ r².
Show answerHide answer
Answer: A. 16 : 1
Bohr radius rₙ ∝ n², so area ∝ r² ∝ n⁴. For n = 2 vs n = 1: 2⁴ : 1⁴ = 16 : 1.
10. In a photoelectric experiment the frequency of incident light is kept fixed (above threshold) and its intensity is doubled. Which quantity increases?
Show hintHide hint
Intensity changes the number of photons, not the energy of each.
Show answerHide answer
Answer: A. The photoelectric current
Higher intensity means more photons per second, so more electrons are emitted and the current increases. Maximum KE and stopping potential depend only on frequency; the work function is a property of the metal.
11. Light of wavelength 330 nm falls on a metal of work function 2 eV. The stopping potential is: (h = 6.6×10⁻³⁴ J s, c = 3×10⁸ m/s, e = 1.6×10⁻¹⁹ C)
2 marksShow hintHide hint
Find the photon energy in eV and subtract the work function.
Show answerHide answer
Answer: C. 1.75 V
Photon energy = hc/λ = (6.6×10⁻³⁴ × 3×10⁸)/(3.3×10⁻⁷) = 6×10⁻¹⁹ J = 3.75 eV. KEmax = 3.75 − 2 = 1.75 eV, so stopping potential = 1.75 V.
12. The threshold wavelength for a metal of work function 3.3 eV is approximately: (h = 6.6×10⁻³⁴ J s, c = 3×10⁸ m/s)
Show hintHide hint
Use λ₀ = hc/φ with φ in joules.
Show answerHide answer
Answer: A. 375 nm
λ₀ = hc/φ = (1.98×10⁻²⁵)/(3.3 × 1.6×10⁻¹⁹) = 1.98×10⁻²⁵/5.28×10⁻¹⁹ = 3.75×10⁻⁷ m = 375 nm.
13. A graph of stopping potential (y-axis) against frequency of incident light (x-axis) is a straight line. Its slope equals:
Show hintHide hint
Rewrite Einstein's equation in terms of V₀.
Show answerHide answer
Answer: A. h/e
From eV₀ = hν − φ, V₀ = (h/e)ν − φ/e. The slope is h/e and the intercept on the voltage axis is −φ/e.
14. Taking the energy of the nth level of hydrogen as Eₙ = −13.6/n² eV, the energy of the photon emitted in the transition n = 3 → n = 2 is about:
2 marksShow hintHide hint
Take the difference of the two level energies.
Show answerHide answer
Answer: A. 1.89 eV
ΔE = 13.6(1/2² − 1/3²) = 13.6 × 5/36 ≈ 1.89 eV. (This is the red Hα line of the Balmer series.)
15. The Lyman series of the hydrogen spectrum lies in the:
Show hintHide hint
Transitions to n = 1 involve large energy gaps.
Show answerHide answer
Answer: C. ultraviolet region
Lyman lines end on n = 1; the smallest energy jump (n = 2 → 1) is already 10.2 eV, so all lines are in the ultraviolet. Balmer is visible; Paschen, Brackett and Pfund are infrared.
16. The ratio of the longest wavelength of the Lyman series to the longest wavelength of the Balmer series of hydrogen is:
2 marksShow hintHide hint
Longest wavelength corresponds to the smallest jump in each series.
Show answerHide answer
Answer: D. 5/27
Lyman longest (2→1): 1/λ ∝ 1 − 1/4 = 3/4. Balmer longest (3→2): 1/λ ∝ 1/4 − 1/9 = 5/36. λL/λB = (5/36)/(3/4) = 20/108 = 5/27.
17. According to Bohr's quantization condition, the angular momentum of an electron in a stationary orbit is:
Show hintHide hint
Recall mvr = n × (something).
Show answerHide answer
Answer: C. an integral multiple of h/2π
Bohr postulated mvr = nh/2π, where n = 1, 2, 3, …
18. The radius of the first Bohr orbit of hydrogen is 0.53 Å. The radius of the third orbit of the Li²⁺ ion (Z = 3) is:
2 marksShow hintHide hint
Orbit radius varies as n²/Z.
Show answerHide answer
Answer: C. 1.59 Å
rₙ = 0.53 n²/Z Å = 0.53 × 9/3 = 1.59 Å.
19. An electron (mass 9×10⁻³¹ kg) is accelerated from rest through 150 V. Its de Broglie wavelength is about: (h = 6.6×10⁻³⁴ J s, e = 1.6×10⁻¹⁹ C)
2 marksShow hintHide hint
Momentum p = √(2meV).
Show answerHide answer
Answer: B. 1 Å
λ = h/√(2meV) = 6.6×10⁻³⁴/√(2 × 9×10⁻³¹ × 1.6×10⁻¹⁹ × 150) = 6.6×10⁻³⁴/√(4.32×10⁻⁴⁷) ≈ 6.6×10⁻³⁴/6.6×10⁻²⁴ = 1×10⁻¹⁰ m = 1 Å. (Shortcut: λ = 12.27/√V Å.)
20. An electron and a proton have the same kinetic energy. Which statement is correct?
Show hintHide hint
Write λ in terms of mass and kinetic energy.
Show answerHide answer
Answer: C. The electron has the longer de Broglie wavelength
λ = h/√(2mK). For equal K, the lighter particle (electron) has smaller momentum and hence longer wavelength.
21. If the uncertainty in the position of a particle is halved, the minimum uncertainty in its momentum:
Show hintHide hint
The product Δx·Δp has a fixed minimum.
Show answerHide answer
Answer: C. doubles
Δx·Δp ≥ h/4π, so the minimum Δp is inversely proportional to Δx. Halving Δx doubles Δp(min).
22. X-rays are reflected from crystal planes of spacing 2 Å. A second-order Bragg maximum occurs at a glancing angle of 30°. The wavelength of the X-rays is:
2 marksShow hintHide hint
Use Bragg's law with n = 2.
Show answerHide answer
Answer: B. 1 Å
nλ = 2d sinθ ⇒ 2λ = 2 × 2 × sin30° = 2 Å, so λ = 1 Å.
23. The short-wavelength cut-off of the continuous X-ray spectrum from a Coolidge tube depends on:
Show hintHide hint
Think of an electron giving up all its energy to one photon.
Show answerHide answer
Answer: A. the accelerating voltage only
λmin = hc/eV: the most energetic photon carries the full kinetic energy eV of an electron, so λmin depends only on the tube voltage, not on the target or the current.
24. Which condition is essential for laser action?
Show hintHide hint
Stimulated emission must outnumber absorption.
Show answerHide answer
Answer: A. Population inversion between a metastable state and a lower state
Laser amplification by stimulated emission needs more atoms in the upper (metastable) level than in the lower one, i.e. population inversion.
6.3 Semiconductors and logic gates
12 questions
25. A pure (intrinsic) silicon crystal at absolute zero behaves as:
Show hintHide hint
Consider whether any electron can cross the band gap at 0 K.
Show answerHide answer
Answer: B. an insulator
At 0 K no electrons have enough thermal energy to cross the band gap, so the conduction band is empty and silicon behaves as an insulator.
26. When pure silicon is doped with boron, the resulting semiconductor is:
Show hintHide hint
Boron belongs to group 13.
Show answerHide answer
Answer: A. p-type, with holes as majority carriers
Boron is trivalent (acceptor); each atom creates a hole, making the crystal p-type with holes as majority carriers.
27. When a p-n junction is forward biased:
Show hintHide hint
The external field opposes the junction field.
Show answerHide answer
Answer: B. the potential barrier decreases and the depletion layer narrows
In forward bias the applied voltage opposes the built-in potential, reducing the barrier and pushing majority carriers toward the junction, so the depletion layer becomes thinner.
28. The intrinsic carrier concentration in silicon is 1.5×10¹⁶ m⁻³. After doping, the electron concentration is 4.5×10²² m⁻³. The hole concentration is:
2 marksShow hintHide hint
Use the mass action law n·p = nᵢ².
Show answerHide answer
Answer: B. 5×10⁹ m⁻³
Mass action law: n·p = nᵢ². p = (1.5×10¹⁶)²/4.5×10²² = 2.25×10³²/4.5×10²² = 5×10⁹ m⁻³.
29. A full-wave rectifier gives an output whose peak value is 11 V. The average (dc) value of the output is about: (take π = 22/7)
2 marksShow hintHide hint
The average of a full-wave rectified sine is 2Vm/π.
Show answerHide answer
Answer: A. 7 V
For full-wave rectification Vdc = 2Vm/π = 2 × 11 × 7/22 = 7 V. (3.5 V would be a half-wave output; 7.8 V is the rms value.)
30. A full-wave rectifier is fed from 50 Hz ac mains. The fundamental frequency of the ripple in its output is:
Show hintHide hint
Count output pulses per input cycle.
Show answerHide answer
Answer: A. 100 Hz
Both half-cycles produce an output pulse, so there are two pulses per input cycle: ripple frequency = 2 × 50 = 100 Hz.
31. A Zener diode is commonly used as a voltage regulator. For this, it is operated in:
Show hintHide hint
Where does the voltage stay nearly constant as current changes?
Show answerHide answer
Answer: D. the reverse breakdown region
In reverse breakdown the voltage across a Zener diode stays almost constant over a wide range of current, which is what a regulator needs.
32. A 5 V Zener diode is connected with a 500 Ω series resistor to a 15 V supply, and a 1 kΩ load is connected across the Zener. The current through the Zener diode is:
2 marksShow hintHide hint
Series current splits between the load and the Zener.
Show answerHide answer
Answer: A. 15 mA
Series current = (15 − 5)/500 = 20 mA. Load current = 5/1000 = 5 mA. Zener current = 20 − 5 = 15 mA.
33. For a transistor in common-base configuration the current gain α = 0.98. Its common-emitter current gain β is:
2 marksShow hintHide hint
Use IE = IC + IB to relate α and β.
Show answerHide answer
Answer: A. 49
β = α/(1 − α) = 0.98/0.02 = 49.
34. For a transistor to work as an amplifier in the active region:
Show hintHide hint
One junction injects carriers, the other collects them.
Show answerHide answer
Answer: B. the emitter-base junction is forward biased and the collector-base junction is reverse biased
Active region: E-B forward (to inject carriers) and C-B reverse (to collect them). Both forward is saturation; both reverse is cut-off.
35. Which of the following is a universal logic gate?
Show hintHide hint
Which gate can be used alone to make NOT, AND and OR?
Show answerHide answer
Answer: C. NAND
NAND (and NOR) gates alone can be combined to build every other gate (NOT, AND, OR, …), so they are called universal gates.
36. Inputs A and B are fed to a NAND gate. The output of the NAND gate and input A are then fed to an AND gate whose output is Y. Y = 1 only for the input combination:
2 marksShow hintHide hint
Simplify using De Morgan's theorem, or check all four input rows.
Show answerHide answer
Answer: C. A = 1, B = 0
Y = A·(AB)′ = A·(A′ + B′) = AA′ + AB′ = AB′. This is 1 only when A = 1 and B = 0.
6.4 Radioactivity and nuclear reactions
12 questions
37. Isotopes of an element have:
Show hintHide hint
Which number fixes the identity of the element?
Show answerHide answer
Answer: A. the same atomic number but different mass numbers
Isotopes have the same number of protons (same Z) but different numbers of neutrons, hence different mass numbers A.
38. Nuclear density is:
Show hintHide hint
Use R = R₀A^(1/3).
Show answerHide answer
Answer: C. practically the same for all nuclei, independent of mass number
R = R₀A^(1/3), so volume ∝ A and mass ∝ A; density = mass/volume is constant (about 2.3×10¹⁷ kg/m³) for all nuclei.
39. Taking R₀ = 1.2 fm, the radius of the ²⁷Al nucleus is R. The radius of the ⁶⁴Cu nucleus is:
2 marksShow hintHide hint
Radius varies as the cube root of mass number.
Show answerHide answer
Answer: C. 4R/3, i.e. 4.8 fm
R = R₀A^(1/3). For Al: 1.2 × 27^(1/3) = 1.2 × 3 = 3.6 fm. For Cu: 1.2 × 64^(1/3) = 1.2 × 4 = 4.8 fm = 4R/3.
40. The energy equivalent of 1 g of mass is: (c = 3×10⁸ m/s)
Show hintHide hint
Convert grams to kilograms before using E = mc².
Show answerHide answer
Answer: D. 9×10¹³ J
E = mc² = 1×10⁻³ kg × (3×10⁸)² = 1×10⁻³ × 9×10¹⁶ = 9×10¹³ J.
41. Given mass of proton = 1.0073 u, mass of neutron = 1.0087 u and mass of deuteron = 2.0136 u, the binding energy of the deuteron is about: (1 u = 931 MeV)
2 marksShow hintHide hint
Find the mass defect first.
Show answerHide answer
Answer: A. 2.23 MeV
Mass defect = (1.0073 + 1.0087) − 2.0136 = 0.0024 u. Binding energy = 0.0024 × 931 ≈ 2.23 MeV. (1.12 MeV is the binding energy per nucleon.)
42. The binding energy per nucleon is maximum for nuclei near:
Show hintHide hint
Recall the peak of the binding energy curve.
Show answerHide answer
Answer: A. iron-56
The binding energy per nucleon curve peaks at about 8.8 MeV near ⁵⁶Fe. Lighter nuclei release energy by fusion and heavier ones by fission, moving toward this peak.
43. Nuclear fusion requires extremely high temperatures because:
Show hintHide hint
Both nuclei are positively charged.
Show answerHide answer
Answer: C. the nuclei must have enough kinetic energy to overcome their mutual Coulomb repulsion
Positively charged nuclei repel each other; only at temperatures of the order of 10⁷ K do they move fast enough to come within the short range of the attractive nuclear force.
44. Each fission of U-235 releases about 200 MeV. The number of fissions per second needed to produce a power of 3.2 MW is: (1 MeV = 1.6×10⁻¹³ J)
2 marksShow hintHide hint
Divide the power by the energy per fission in joules.
Show answerHide answer
Answer: B. 1×10¹⁷
Energy per fission = 200 × 1.6×10⁻¹³ = 3.2×10⁻¹¹ J. Number per second = 3.2×10⁶/3.2×10⁻¹¹ = 1×10¹⁷.
45. After 20 days, only 1/16 of a radioactive sample remains undecayed. Its half-life is:
2 marksShow hintHide hint
Express 1/16 as a power of 1/2.
Show answerHide answer
Answer: B. 5 days
1/16 = (1/2)⁴, so 4 half-lives = 20 days, giving T½ = 5 days.
46. A radioactive nuclide has a half-life of 693 s. Its mean life is: (ln 2 = 0.693)
2 marksShow hintHide hint
Mean life is the reciprocal of the decay constant.
Show answerHide answer
Answer: D. 1000 s
λ = 0.693/T½ = 0.693/693 = 1×10⁻³ s⁻¹. Mean life τ = 1/λ = 1000 s.
47. A piece of ancient wood shows a C-14 activity that is 1/8 of that of fresh living wood. If the half-life of C-14 is 5730 years, the age of the wood is:
2 marksShow hintHide hint
Count how many halvings give 1/8.
Show answerHide answer
Answer: B. 17190 years
1/8 = (1/2)³, so 3 half-lives have passed: age = 3 × 5730 = 17190 years.
48. Alpha emitters are generally considered most hazardous to health when:
Show hintHide hint
Compare penetrating power with ionizing power.
Show answerHide answer
Answer: C. they are inhaled or ingested, because alpha particles are strongly ionizing
Alpha particles are stopped by skin or paper, so they are relatively harmless outside the body; inside the body their high ionizing power badly damages nearby tissue.
6.5 Recent trends in physics
18 questions
49. The antiparticle of the electron is the:
Show hintHide hint
Same mass, opposite charge.
Show answerHide answer
Answer: C. positron
The positron has the same mass as the electron but opposite charge (+e); it is the electron's antiparticle.
50. The up (u) quark has charge +2e/3 and the down (d) and strange (s) quarks each have charge −e/3. A baryon made of the quarks u u s has charge:
2 marksShow hintHide hint
Add the quark charges.
Show answerHide answer
Answer: A. +e
Charge = 2/3 + 2/3 − 1/3 = +1, i.e. +e (this is the Σ⁺ baryon).
51. Which of the following is a lepton (and not a hadron)?
Show hintHide hint
Leptons do not feel the strong force and have no quarks.
Show answerHide answer
Answer: D. Muon
Leptons (electron, muon, tau and their neutrinos) are not made of quarks. Protons and neutrons are baryons (three quarks) and pions are mesons (quark–antiquark), so all three are hadrons.
52. The Higgs boson, discovered at CERN's Large Hadron Collider in 2012, is associated with:
Show hintHide hint
It is often linked with the origin of mass.
Show answerHide answer
Answer: A. the mechanism that gives mass to elementary particles
The Higgs field gives mass to fundamental particles such as the W and Z bosons and quarks; the Higgs boson is the excitation of this field. The photon carries the electromagnetic force and gluons bind quarks.
53. A spectral line of wavelength 500 nm in the laboratory is observed at 505 nm in the light from a distant galaxy. Taking Hubble's constant as 75 km/s per Mpc, the distance of the galaxy is about: (c = 3×10⁸ m/s)
2 marksShow hintHide hint
Find the recession speed from the redshift, then use v = H₀d.
Show answerHide answer
Answer: D. 40 Mpc
Redshift z = Δλ/λ = 5/500 = 0.01, so v = cz = 3×10⁶ m/s = 3000 km/s. By Hubble's law d = v/H₀ = 3000/75 = 40 Mpc.
54. The strongest evidence for the existence of dark matter comes from:
Show hintHide hint
Compare orbital speeds of outer stars with the visible mass.
Show answerHide answer
Answer: A. the rotation curves of galaxies, which stay flat far from their centres
Stars far from a galaxy's centre orbit much faster than visible mass alone can explain; extra unseen (dark) mass is needed to provide the gravity.
55. Which statement about gravitational waves is correct?
Show hintHide hint
Think of what general relativity says about space-time.
Show answerHide answer
Answer: D. They are ripples in space-time that travel at the speed of light
Predicted by Einstein's general relativity and first detected by LIGO in 2015 from merging black holes, gravitational waves are transverse ripples in space-time moving at speed c and need no medium.
56. The Schwarzschild radius of a black hole is R = 2GM/c². For a black hole of mass 2×10³⁰ kg (about one solar mass), R is about: (G = 6.67×10⁻¹¹ N m² kg⁻², c = 3×10⁸ m/s)
2 marksShow hintHide hint
Substitute directly; watch the powers of ten.
Show answerHide answer
Answer: C. 3 km
R = 2 × 6.67×10⁻¹¹ × 2×10³⁰ / (9×10¹⁶) = 2.67×10²⁰/9×10¹⁶ ≈ 2.96×10³ m ≈ 3 km.
57. Which seismic waves cannot travel through the Earth's liquid outer core?
Show hintHide hint
A liquid cannot resist shear.
Show answerHide answer
Answer: A. S waves
S waves are transverse (shear) waves and liquids cannot support shear stress, so S waves do not pass through the liquid outer core, creating the S-wave shadow zone.
58. P waves travel at 8 km/s and S waves at 4 km/s through the crust. At a seismic station the S waves arrive 50 s after the P waves. The distance of the earthquake focus from the station is:
2 marksShow hintHide hint
Write the difference of the two travel times.
Show answerHide answer
Answer: D. 400 km
d/4 − d/8 = 50 ⇒ d/8 = 50 ⇒ d = 400 km.
59. For a GPS receiver to find its three-dimensional position along with the correction to its own clock, the minimum number of satellites whose signals it must receive is:
Show hintHide hint
Count the unknowns, including time.
Show answerHide answer
Answer: A. 4
Three unknown coordinates plus the receiver's clock error make four unknowns, so signals from at least four satellites are needed.
60. A TV transmitting antenna is 80 m tall. Taking the radius of the Earth as 6.4×10⁶ m, the maximum distance on the ground up to which its signal can be received directly (line of sight) is about:
2 marksShow hintHide hint
Use d = √(2Rh).
Show answerHide answer
Answer: C. 32 km
d = √(2Rh) = √(2 × 6.4×10⁶ × 80) = √(1.024×10⁹) = 3.2×10⁴ m = 32 km.
61. Which of the following signals is normally reflected back to Earth by the ionosphere (sky-wave propagation)?
Show hintHide hint
The ionosphere reflects only relatively low frequencies.
Show answerHide answer
Answer: D. Short-wave radio at about 10 MHz
The ionosphere reflects radio waves of roughly 3–30 MHz. Higher frequencies such as FM, TV and microwaves pass through it, so they use line-of-sight or satellite links.
62. Most Earth-resource remote-sensing satellites (such as Landsat) are placed in:
Show hintHide hint
They need to view the whole globe in detail.
Show answerHide answer
Answer: A. low-altitude, sun-synchronous polar orbits
A sun-synchronous polar orbit at a few hundred kilometres lets the satellite scan the whole Earth at high resolution, crossing each place at the same local time so the lighting is comparable.
63. Depletion of the ozone layer in the stratosphere is mainly caused by:
Show hintHide hint
Which man-made compounds were banned by the Montreal Protocol?
Show answerHide answer
Answer: B. chlorine released from chlorofluorocarbons (CFCs)
UV light breaks CFC molecules in the stratosphere, releasing chlorine atoms that destroy ozone catalytically (one Cl atom can destroy thousands of O₃ molecules). CO₂ is mainly linked to global warming.
64. When a superconductor is cooled below its critical temperature in a magnetic field, it:
Show hintHide hint
Recall the Meissner effect.
Show answerHide answer
Answer: C. expels the magnetic field from its interior, behaving as a perfect diamagnet
This is the Meissner effect: below Tc a superconductor has zero resistance and magnetic susceptibility of −1, so the field inside becomes zero.
65. Materials in the form of nanoparticles (1–100 nm) often show properties very different from the bulk material mainly because:
Show hintHide hint
Compare surface atoms with interior atoms as size decreases.
Show answerHide answer
Answer: D. they have a very large surface-area-to-volume ratio
As size shrinks, a large fraction of atoms lies on the surface, and quantum effects also become important; this changes optical, chemical and electrical properties.
66. Solar radiation of intensity 1000 W/m² falls normally on a solar panel of area 2 m² with an efficiency of 15%. The electrical energy produced in 5 hours is:
2 marksShow hintHide hint
Multiply incident power by efficiency, then by time.
Show answerHide answer
Answer: C. 1.5 kWh
Electrical power = 0.15 × 1000 × 2 = 300 W. Energy in 5 h = 300 × 5 = 1500 Wh = 1.5 kWh.