Stratigraphy and the Geological Time Scale: Principles, Radiometric Dating, Radioactive Equilibrium, and the Stratigraphy and Resources of India

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

1. A basalt dyke cuts through a folded sandstone but is not seen in the conglomerate that rests with an angular unconformity on the eroded sandstone and dyke. What is the correct relative age order from oldest to youngest?

   1. Sandstone, then dyke, then conglomerate, because the dyke cuts the sandstone and was truncated by erosion first
   2. Sandstone, then conglomerate, then dyke, because the dyke lies in the upper part of the exposed sequence and so is the youngest
   3. Dyke, then sandstone, then conglomerate, because igneous rocks are always older than the sedimentary rocks they intrude
   4. Conglomerate, then sandstone, then dyke, because a folded unit is always younger than the unit that rests on its eroded surface
   Show answer

   Answer: A — Sandstone, then dyke, then conglomerate, because the dyke cuts the sandstone and was truncated by erosion first

   The dyke cuts the sandstone, so it is younger than the sandstone (cross-cutting). It is truncated by the erosion surface and absent from the conglomerate, so it is older than the conglomerate. The tempting alternative places the dyke last because it is exposed alongside the upper beds; position in the pile dates layers, not intrusions. No rule makes igneous rock older than the sediment it intrudes.

2. Which fossil is the most useful index fossil for correlating beds between two widely separated sedimentary basins?

   1. A species that lived for a very long time and is found in only one basin, where it is abundant and well studied
   2. A species of short geological range that is abundant, easily identified and spread over many basins
   3. A rare species of short range that has been found in one outcrop only and is very well preserved
   4. A widespread species that survived through several periods with almost no change in its form or habit
   Show answer

   Answer: B — A species of short geological range that is abundant, easily identified and spread over many basins

   A good index fossil has a short stratigraphic range, which gives fine time resolution, together with wide geographic spread, abundance and easy identification, so that the same time slice can be recognised in distant basins. A long-ranging widespread species is the tempting answer but dates a bed only to a broad interval; a rare or local species cannot correlate across basins.

3. Which of the following statements about relative dating are correct?

   1. A fault is younger than the youngest bed that it displaces
   2. Granite pebbles in an overlying conglomerate show that the granite is older than the conglomerate
   3. A sill lying between two beds has the same age as the lower of the two beds
   4. A nonconformity separates tilted sedimentary beds from younger horizontal sedimentary beds
   Show answer

   Answer: A — A fault is younger than the youngest bed that it displaces; B — Granite pebbles in an overlying conglomerate show that the granite is older than the conglomerate

   A fault must be younger than every bed it cuts, and a pebble must exist before the conglomerate that contains it, so the granite is older. A sill is injected into already-formed rock, so it is younger than both beds, not coeval with the lower one. Tilted beds beneath younger horizontal sedimentary beds define an angular unconformity; a nonconformity has sedimentary rock resting on igneous or metamorphic rock.

4. A charcoal sample retains 12.5 per cent of its original C-14 activity. Taking the half-life of C-14 as 5730 years, the age of the sample, in years, is ____.

   Numerical answer — type the value.

   Show answer

   Answer: 17190

   Fraction remaining = (1/2)ⁿ = 0.125 = 1/8, so n = 3 half-lives. Age = 3 × 5730 = 17190 years. Check with the decay law: t = ln(1/0.125)/λ = 2.0794/(ln 2/5730) = 17190 years.

5. If the decay constant of a radioactive nuclide were twice that of another nuclide, how would its half-life compare with the other's?

   1. It would be twice as long, because a larger constant means the nuclide survives for longer
   2. It would be four times as long, because the half-life depends on the square of the constant
   3. It would be unchanged, because the half-life is fixed by the temperature of the host mineral
   4. It would be half as long, because the half-life is inversely proportional to the constant
   Show answer

   Answer: D — It would be half as long, because the half-life is inversely proportional to the constant

   t½ = 0.693/λ, so doubling λ halves the half-life: the nuclide decays faster. The tempting answer reverses the proportionality. The half-life is not affected by temperature for the decays used in dating, and it is not related to the square of λ.

6. A mineral contains radiogenic daughter atoms and surviving parent atoms in the ratio D*/N = 0.5. The parent has a half-life of 1.25 Ga. Use ln 2 = 0.693 and ln 1.5 = 0.405. The age of the mineral in Ga, to two decimal places, is ____.

   Numerical answer — type the value.

   Show answer

   Answer: 0.73

   λ = 0.693/1.25 = 0.5544 per Ga. t = (1/λ) ln(1 + D/N) = ln(1.5)/0.5544 = 0.405/0.5544 = 0.7305, which is 0.73 Ga. Using exact logarithms gives 0.7312, again 0.73. A common slip is to use ln(0.5), or to treat D/N = 0.5 as half of the parent having decayed, which would give one half-life.

7. In one granite, zircons that contain no older inherited cores give a U–Pb age of 1.8 Ga, a whole-rock Rb–Sr isochron gives about 1.8 Ga, and biotite gives a K–Ar age of 1.0 Ga. What is the most reasonable interpretation?

   1. The decay constant of K-40 is wrongly known, because decay rates change when a mineral is heated during metamorphism
   2. The granite crystallised about 1.0 Ga ago, and the zircon and whole-rock ages are too old because older source material was inherited into them
   3. The granite crystallised about 1.8 Ga ago, and argon in biotite was lost during later reheating until the mineral cooled about 1.0 Ga ago
   4. The biotite is older than the granite, because mica crystallises before zircon in the magma and so records an earlier event
   Show answer

   Answer: C — The granite crystallised about 1.8 Ga ago, and argon in biotite was lost during later reheating until the mineral cooled about 1.0 Ga ago

   Different systems close at different temperatures. Zircons free of inherited cores and a whole-rock isochron both give the crystallisation age, while argon escapes from biotite at a much lower temperature, so the K–Ar age records cooling below the argon closure temperature after reheating or slow uplift. The alternatives fail: crystallisation at 1.0 Ga is excluded by zircons that have no inherited cores, decay constants do not change with heating, and mica does not crystallise earlier than zircon in a granite.

8. A Rb–Sr isochron for a set of cogenetic rocks has a slope of 0.0288. Take the decay constant of Rb-87 as 1.42 × 10⁻¹¹ per year. The age of the rocks, in Ma, to the nearest 100 Ma, is ____.

   Numerical answer — type the value.

   Show answer

   Answer: 2000

   Slope = e^(λt) − 1, so λt = ln(1.0288) = 0.02839 and t = 0.02839/(1.42 × 10⁻¹¹) = 1.9995 × 10⁹ years, about 2000 Ma. The slope of an isochron is e^(λt) − 1 and not the age itself, so the logarithm must be taken before dividing by λ.

9. Which of the following statements about radiometric dating are correct?

   1. For the decays used in dating, the half-life does not change with temperature or the host mineral's chemistry
   2. After two half-lives the parent has completely decayed and only the daughter nuclide remains in the mineral
   3. The intercept of a Rb–Sr isochron gives the initial ⁸⁷Sr/⁸⁶Sr ratio of the cogenetic samples
   4. C-14, with a half-life of 5730 years, is the method of choice for dating a limestone of Permian age
   Show answer

   Answer: A — For the decays used in dating, the half-life does not change with temperature or the host mineral's chemistry; C — The intercept of a Rb–Sr isochron gives the initial ⁸⁷Sr/⁸⁶Sr ratio of the cogenetic samples

   Decay constants are insensitive to temperature and chemistry for these decays, and the isochron intercept is the initial ratio. After two half-lives one quarter of the parent remains, not none. C-14 is exhausted after a few tens of thousands of years and cannot date a rock hundreds of millions of years old.

10. In an old, undisturbed, closed uranium ore, Ra-226 (half-life 1600 years) is in equilibrium with its long-lived ancestor U-238 (half-life 4.47 billion years). Which statement is correct?

    1. Ra-226 has the lower activity, because as a daughter it can never decay faster than its parent supplies it
    2. The two have equal numbers of atoms, so Ra-226 has the far higher activity because its half-life is so much shorter
    3. The two have equal activities, and the Ra-226 atoms are fewer than the U-238 atoms in the ratio of their half-lives
    4. Ra-226 has the higher activity, by the factor λ₂/(λ₂ − λ₁) that applies in transient equilibrium
    Show answer

    Answer: C — The two have equal activities, and the Ra-226 atoms are fewer than the U-238 atoms in the ratio of their half-lives

    Because the parent half-life is vastly longer than the daughter's, the system is in secular equilibrium: λ₂N₂ = λ₁N₁, so the activities are equal and N₂/N₁ = λ₁/λ₂ = 1600/(4.47 × 10⁹) ≈ 3.6 × 10⁻⁷. The transient factor λ₂/(λ₂ − λ₁) is tempting but tends to 1 when λ₁ is negligible against λ₂. Equal atom numbers would give the daughter a much higher activity, which equilibrium excludes.

11. A parent nuclide of half-life 4.0 × 10⁹ years decays to a daughter of half-life 8.0 × 10⁴ years, and the system has been closed for a very long time. The number of daughter atoms present per 10⁹ parent atoms is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 20000

    The parent is vastly longer-lived, so secular equilibrium holds: N₂/N₁ = λ₁/λ₂ = t½(daughter)/t½(parent) = 8.0 × 10⁴/4.0 × 10⁹ = 2.0 × 10⁻⁵. Per 10⁹ parent atoms, N₂ = 2.0 × 10⁻⁵ × 10⁹ = 20000. Inverting the ratio would give 5 × 10¹³ daughter atoms, which is impossible because the shorter-lived daughter cannot outnumber its parent in secular equilibrium.

12. A parent of half-life 12 h decays to a daughter of half-life 4 h, and transient equilibrium has been established. The ratio of the daughter's activity to the parent's activity, A₂/A₁, to one decimal place is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 1.5

    In transient equilibrium A₂/A₁ = λ₂/(λ₂ − λ₁) = T₁/(T₁ − T₂) in terms of half-lives = 12/(12 − 4) = 1.5. Check with λ proportional to 1/T: λ₁ = 1/12 and λ₂ = 1/4, so (1/4)/(1/4 − 1/12) = 0.25/0.1667 = 1.5. The ratio exceeds 1 because the daughter's activity includes the feeding decays.

13. The Talchir Formation at the base of the Gondwana Supergroup is characterised by which kind of deposit?

    1. Flood basalts erupted at the close of the Cretaceous period
    2. Thick coal seams that supply the main coking coal of the Damodar valley
    3. Marine limestones with Jurassic ammonites deposited on a shelf
    4. Glacial boulder beds (tillite) from the late Paleozoic Gondwana glaciation
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    Answer: D — Glacial boulder beds (tillite) from the late Paleozoic Gondwana glaciation

    The Talchir Formation begins with a boulder bed of glacial origin (tillite) and shales, recording the late Paleozoic Gondwana glaciation. The coal-bearing units are the Barakar and Raniganj formations above it, which is the tempting confusion; the Jurassic marine beds are in Kutch, and the flood basalts are the Deccan Traps.

14. Which of the following arranges these Indian units correctly from oldest to youngest: Siwalik Group, Deccan Traps, Vindhyan Supergroup, Dharwar greenstone belts?

    1. Dharwar greenstone belts, Vindhyan Supergroup, Deccan Traps, Siwalik Group
    2. Vindhyan Supergroup, Dharwar greenstone belts, Siwalik Group, Deccan Traps
    3. Dharwar greenstone belts, Deccan Traps, Vindhyan Supergroup, Siwalik Group
    4. Vindhyan Supergroup, Dharwar greenstone belts, Deccan Traps, Siwalik Group
    Show answer

    Answer: A — Dharwar greenstone belts, Vindhyan Supergroup, Deccan Traps, Siwalik Group

    The Dharwar belts are Archaean to early Proterozoic, the Vindhyan basin is Proterozoic and unmetamorphosed, the Deccan Traps are about 66 Ma and the Siwalik molasse is Miocene to Pleistocene. The Vindhyan Supergroup rests on older basement and is younger than the Archaean greenstone belts, and the Siwaliks are the youngest of the four because they record the rise of the Himalaya.

15. Which of the following statements about the Himalayan divisions are correct?

    1. The Siwalik Group lies between the Main Frontal Thrust and the Main Boundary Thrust
    2. The Main Central Thrust separates the Lesser Himalaya from the Higher Himalaya
    3. The Tethys Himalaya lies north of the Higher Himalaya and contains thick marine sedimentary successions
    4. The Lesser Himalaya lies between the Main Central Thrust and the South Tibetan Detachment
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    Answer: A — The Siwalik Group lies between the Main Frontal Thrust and the Main Boundary Thrust; B — The Main Central Thrust separates the Lesser Himalaya from the Higher Himalaya; C — The Tethys Himalaya lies north of the Higher Himalaya and contains thick marine sedimentary successions

    From south to north the zones are Siwalik (between the Main Frontal and Main Boundary Thrusts), Lesser Himalaya (between the Main Boundary and Main Central Thrusts), Higher Himalaya (between the Main Central Thrust and the South Tibetan Detachment) and Tethys Himalaya with its marine successions. The statement placing the Lesser Himalaya against the South Tibetan Detachment describes the Higher Himalaya instead.

16. A survey must choose, for a first search for coal, the terrain that is geologically most promising. Which one should it choose?

    1. The Archaean Dharwar gneiss terrain of southern Karnataka, which is the oldest crust in the region
    2. A fault-bounded Permian Gondwana basin such as the Damodar valley, with Barakar-type beds
    3. The Deccan Trap plateau, because thick basalt flows have buried and cooked abundant plant matter
    4. The Himalayan Higher crystalline belt, because high pressure and heat favour rapid coalification
    Show answer

    Answer: B — A fault-bounded Permian Gondwana basin such as the Damodar valley, with Barakar-type beds

    Coal needs a sedimentary basin in which plant matter accumulated and was buried, and Permian Gondwana basins such as the Damodar valley hold most Indian coal. Gneiss terrains and Higher Himalayan crystalline rocks are metamorphic and contain no coal; lava flows bury and destroy plant matter rather than preserving coal, although coal-bearing sediments may lie beneath them.

17. A closed mineral contains seven daughter atoms of radiogenic origin for every one parent atom now present, and no daughter was present at the start. How many half-lives of the parent have elapsed?

    1. 2.8 half-lives, from log₂ 7
    2. 3 half-lives, since the parent is now one-eighth of its original amount
    3. 4 half-lives, since seven daughters mean a parent fraction of one-sixteenth
    4. 7 half-lives, since one half-life produces one daughter per parent
    Show answer

    Answer: B — 3 half-lives, since the parent is now one-eighth of its original amount

    D/N = 7 means that originally there were 7 + 1 = 8 parent-equivalents, so N/N₀ = 1/8 = (1/2)³ and three half-lives have passed; in general D/N = 2ⁿ − 1. Using log₂ 7 = 2.8 forgets the surviving parent, which is the common error. Four half-lives would give D*/N = 15, and each half-life does not produce one daughter per parent.

18. Which of the following pairs of an Indian resource and its principal geological host are correct?

    1. Chromite of Sukinda, Odisha: ultramafic bodies within a Precambrian terrain
    2. Coking coal of Jharia: Permian Gondwana coal measures of the Damodar valley
    3. Gold of Kolar, Karnataka: quartz veins in an Archaean Dharwar greenstone belt
    4. Petroleum of the Mumbai High: Archaean gneiss of the Dharwar craton
    Show answer

    Answer: A — Chromite of Sukinda, Odisha: ultramafic bodies within a Precambrian terrain; B — Coking coal of Jharia: Permian Gondwana coal measures of the Damodar valley; C — Gold of Kolar, Karnataka: quartz veins in an Archaean Dharwar greenstone belt

    Sukinda chromite sits in ultramafic bodies in Precambrian rocks, Jharia coking coal in the Permian Gondwana coal measures, and Kolar gold in quartz veins of an Archaean greenstone belt. Petroleum needs a sedimentary basin with source, reservoir and seal; the Mumbai High lies in a Tertiary sedimentary basin of the western offshore, not in crystalline gneiss.