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Q71. The intersection of bedding surface with upright slaty cleavage surface occurs as horizontal lineation. Which one of the following is the most likely interpretation? The bedding surfaces are folded into:
If the upright cleavage plane intersects the bedding to form a perfectly horizontal lineation, the fold axis itself must be horizontal, indicating an upright, non-plunging fold.
Q72. If the heat production per unit mass for basalt is 2.6 x 10^-11 W/kg, the surface heat flow for an oceanic crust of thickness 6 km and density 2900 kg/m3 is (mW/m2):
Heat Flow = Heat Generation * Density * Thickness. = 2.6e-11 W/kg * 2900 kg/m3 * 6000 m = 0.0004524 W/m2 = 0.45 mW/m2.
Q73. If the oldest sediment found on the floor of the South Atlantic Ocean at 1300 km west of the axis of the Mid-Atlantic Ridge were deposited about 65 million years ago, then the rate at which the Atlantic Ocean widens is:
Half-spreading rate = Distance / Time = 1300 km / 65 Ma = 20 km/Ma (or 2 cm/yr). Because the ridge spreads symmetrically in both directions, the total widening rate is 2 * 2 = 4 cm/yr.
Q74. A planet is 11 times the radius of the Earth in size, but one-fourth in mass density. The gravity field on its surface would be around (in gals):
Gravity g is proportional to Radius * Density (g = 4/3 * pi * G * R * rho). Ratio = (11 * R) * (1/4 * rho) = 2.75 times Earth's gravity. Earth's gravity is ~980 Gals. 980 * 2.75 = 2695 Gals (rounds closely to 2700).
Q75. Granite A has quartz, K-rich feldspar, and Na-rich feldspar as major minerals, both feldspars exhibit exsolution textures. Granite B has quartz and only alkali feldspar as major phases. From this observation the correct inference is:
Granite A contains two primary feldspars (subsolvus), indicating a lower crystallization temperature. The presence of exsolution textures indicates that it cooled slowly enough for the phases to separate.
Q76. Glaciers presently constitute about 2% of the water at the surface of the Earth and have a d18O_SMOW approx -30 per mil. The oceans contain essentially all remaining water. If the mass of the glacial ice were to increase to 10%, what would be the changed isotopic composition of the oceans?
Glaciers trap light oxygen (16O). If ice mass increases from 2% to 10% (an 8% shift), the oceans lose 8% of water at -30 per mil. Balance equation: 0.1(-30) + 0.9(x) = -0.6 (present bulk). Solving gives x = +2.67 per mil.
Q77. In a refraction survey, the velocities inferred for the upper and the lower layers are 3000 m/s and 5000 m/s respectively. If the crossover distance is 4000 m, the depth of the refractor is:
Formula: Xc = 2z * sqrt((V2+V1)/(V2-V1)). Substituting: 4000 = 2z * sqrt(8000/2000). 4000 = 2z * sqrt(4). 4000 = 4z. z = 1000 m.
Q78. A 120 million year old rock situated close to the equator shows a remanent magnetic dip of tan^-1(-2/sqrt(3)). The approximate drift rate of the land mass is (cm/year):
Formula: tan(Inclination) = 2*tan(Latitude). -2/sqrt(3) = 2*tan(Lat), so tan(Lat) = -1/sqrt(3), Lat = 30° South. Total drift = 30° = ~3330 km. Rate = 3330 km / 120 Ma = 27.7 km/Ma = 2.75 cm/yr.
Q79. The gravity anomaly at a point P distant 1.0 km from the position of the maximum anomaly, along a profile across a horizontal circular cylinder, is twice the anomaly at a point 1.0 km farther away from P. The depth of the cylinder is:
Formula for cylinder: g ~ z^2 / (x^2 + z^2). Given g(at x=1) = 2 * g(at x=2). Therefore: 1/(1+z^2) = 2/(4+z^2). Solving yields z^2 = 2. So z = sqrt(2) = 1.414 km.
Q80. The Bouguer anomaly (in mgal) associated with an isostatically compensated 2.0 km thick landmass of density 2.7 g/cc is: (Assume pi*G = 21 mgal/km/g/cc)
For a fully compensated mass, the Bouguer anomaly reflects the mass deficit of the crustal root, which exactly equals the mass of the topography. Formula: BA = -2 * pi * G * rho * h = -2 * 21 * 2.7 * 2.0 = -226.8 mGal. (Magnitude 227).