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Q101. Consider the schematic isobaric T-X phase diagram in the binary forsterite (Fo)-fayalite (Fa) chemical system. If there is equilibrium crystallization of melt (L), the wt.% of olivine crystallized from a melt of composition “a” at a temperature T2 is ____ (Diagram shows a=60, T2 Solidus=80, T2 Liquidus=0)
The correct answer is 75. We use the 'Lever Rule' on the melting diagram to figure out how much solid rock has frozen. The bulk magma chemistry 'a' is stuck exactly at 60% Forsterite. At the specific cooling temperature T2, the boundary curve for the Solid rock (solidus) hits exactly 80%, and the boundary curve for the remaining Liquid melt (liquidus) hits exactly 0%. The percentage of frozen solid rock is calculated as (Bulk - Liquid) / (Solid - Liquid). So, (60 - 0) / (80 - 0) evaluates to 60/80, perfectly equaling 75% solid olivine.
Q102. From the data shown in the table, the weighted mean size (in micrometer, correct to two decimal places) of the sediment population is _____. (4=50g, 20=75g, 40=125g, 60=50g)
The correct answer is 32.33. The 'weighted mean size' prevents extreme values from skewing the average by factoring in how much each size physically weighs. First, calculate the total weight of the dirt: 50 + 75 + 125 + 50 = 300 grams. Next, multiply every specific grain size by its physical weight and add them all together: (4×50) + (20×75) + (40×125) + (60×50) = 200 + 1500 + 5000 + 3000 = 9700. Finally, divide the massive total (9700) by the weight (300) to get a perfect average of 32.33 micrometers.
Q103. A spherical ore body (diameter=40m) has 7% metal content and density of 3300 kg/m3. The reserve (in tonne) of the ore body is ____
The correct answer is 110584. First, calculate the total physical volume of the underground 40m sphere (radius = 20m) using Volume = (4/3) × pi × r³. This generates a staggering 33510.32 cubic meters of rock. Next, multiply this sheer volume by the heavy rock density (3300 kg/m³) to find the total mass, generating roughly 110,584,061 kilograms. Dividing by 1000 converts this into metric tonnes. A 'reserve' describes the total weight of the rock holding the ore, making the final answer approximately 110,584 tonnes.
Q104. The retardation of a uniaxial negative mineral of thickness 0.03 mm is 5160 nm in its principal section of indicatrix. If the refractive index corresponding to the E-ray is 1.486, the value of the refractive index (correct to three decimal places) of the O-ray is ____
The correct answer is 1.658. In optical mineralogy, 'Retardation' is the thickness multiplied by the 'Birefringence'. First, convert 0.03 mm thickness into 30,000 nanometers. Using the formula: 5160 = 30,000 × Birefringence. Dividing 5160 by 30,000 yields a birefringence of 0.172. Because the mineral is specified as 'uniaxial negative', the ordinary ray (O-ray) is fundamentally slower and possesses a higher index than the E-ray. We simply add the birefringence (0.172) to the E-ray index (1.486) to mathematically get exactly 1.658 for the O-ray.
Q105. A set of sedimentary rocks A, B and C are affected by a fault F-F. The amount of vertical throw (in m) along the fault is ____ (Map shows rock boundary jumping from 400m contour to 500m contour)
The correct answer is 100. When examining the geological map, the dividing boundary line between rock beds A and B provides a perfect tectonic marker. On the left side of the fault crack, this boundary rests perfectly on the 400-meter elevation contour. Crossing the fault to the right block, that exact same boundary has been violently jerked uphill to align with the 500-meter elevation contour. This immediate, jarring elevation discrepancy perfectly documents a vertical throw of exactly 100 meters.
Q106. A coal seam occurs in a stratigraphic sequence as shown in the figure. If a vertical borehole is drilled at location B, the coal seam will be intersected at a depth (in m) of ____ (Map shows B at 500m contour, coal hinge at 400m contour)
The correct answer is 100. By tracing the curving geological boundaries on the topographic map, the black band of the coal seam physically surfaces (outcrops) exactly along the 400-meter elevation contour line at the fold's hinge. The drilling site 'B' is placed precisely over that same hinge, but sits high up on a hill at the 500-meter elevation contour. The drill must bore straight down from the 500m hill to hit the 400m rock layer, resulting in a drill depth of exactly 100 meters.
Q107. In an undeformed and normal stratigraphic succession, a dolerite dyke was emplaced before deposition of sandstone B. The difference between the maximum ages (in Myr) of deposition of sandstone A and sandstone B is ____ (Image shows Sandstone A above 132Ma ash, Dyke 75Ma cutting A, Sandstone B on top uncut)
The correct answer is 57. Based on the geological cross-section, Sandstone A rests directly on top of a 132 Million Year (Ma) old ash bed, meaning Sandstone A cannot possibly be older than 132 Ma (its absolute maximum age). The 75 Ma volcanic dyke violently cuts straight through Sandstone A but abruptly stops at Sandstone B. This physically proves Sandstone B was laid down after the 75 Ma dyke had already cooled, capping its maximum possible age at 75 Ma. The difference between 132 and 75 is exactly 57 million years.
Q108. The grain density (of solids only) and bulk density (solids + voids) of a sandstone sample are 2.7 gm/cm3 and 2.3 gm/cm3, respectively. The total porosity (in %, correct to two decimal places) of the sample is ____
The correct answer is 14.81. 'Porosity' strictly measures the percentage of invisible, empty void space trapped inside a rock. The standard engineering formula relies on bulk density (how heavy the hole-filled rock is) and grain density (how heavy the solid dirt itself is): Porosity = [1 - (Bulk / Grain)] × 100. We plug in the values: [1 - (2.3 / 2.7)] × 100. This becomes [1 - 0.8518] × 100, which evaluates to exactly 14.81%.
Q109. According to the mineralogical phase rule, the number of minerals that can coexist at equilibrium in a 8 component chemical system with 2 degrees of freedom is ____
The correct answer is 8. In petrology, the mineralogical phase rule is an elegant mathematical equation used to predict rock chemistry: Degrees of Freedom (F) = Components (C) - Phases/Minerals (P) + 2. The problem explicitly states there are 8 chemical components and 2 degrees of freedom. Plugging these values in: 2 = 8 - P + 2, which condenses to 2 = 10 - P. Solving for P reveals that exactly 8 separate mineral phases can harmoniously coexist at absolute equilibrium.
Q110. An aquifer has a cross sectional area of 1000 m2 and a hydraulic gradient of 0.01. If water is flowing from the aquifer at a rate of 10 m3/sec, the hydraulic conductivity (in m/sec) of the aquifer is ____
The correct answer is 1. We solve this using the fundamental Darcy's Law for groundwater flow: Total Flow Rate = Conductivity × Area × Gradient. We are explicitly given the total flow is 10. The wall area is 1000, and the water slope (gradient) is 0.01. Multiplying 1000 by 0.01 simplifies to 10. Our algebraic equation is now: 10 = Conductivity × 10. To perfectly balance this equation, the Conductivity parameter must equal exactly 1 meter per second.