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Q91. During concretionary growth of a spherical grain of radius 2 Å, the rate of change of surface area with respect to change in radius of the grain is ______ 10−8 cm (use = 3.14) (Round off to two decimal places).
The correct answer is 50.24. This requires basic calculus. The surface area (A) of a sphere is 4πr². The 'rate of change of surface area with respect to the radius' is the mathematical derivative of that formula, which is dA/dr = 8πr. We are given the radius is 2 Ångströms (which is exactly 2 × 10^-8 cm). Plugging the numbers into the derivative gives 8 × 3.14 × 2, which physically evaluates to exactly 50.24 × 10^-8 cm.
Q92. A drill run of 3 m was carried out in a coalfield site, where rock core samples were recovered only for a cumulative length of 255 cm. The core loss in percentage is equal to ______. (Answer in integer).
The correct answer is 15. The total length the physical drill bit travelled into the earth was 3 meters, which equals 300 centimeters. However, they only pulled up 255 centimeters of solid rock. This means a portion of the rock crumbled and was washed away. The total amount lost is the difference: 300 minus 255 equals 45 centimeters. To find the percentage of loss, we divide the lost amount (45) by the total drill length (300) and multiply by 100. This equates to exactly 15%.
Q93. The density of a 200 g gabbro sample, cut in the form of a cube, is 3125 kg/m3. The length of the sample is ________ mm. (Answer in integer).
The correct answer is 40. First, we must perfectly match our units. A heavy density of 3125 kg/m³ is mathematically identical to 3.125 g/cm³. Volume is mass divided by density. We divide the 200g mass by the 3.125 density to get a physical cube volume of exactly 64 cm³. Because every side of a perfect cube is identical, the length of one side is simply the cube root of 64, which is 4 cm. Converting this to millimeters gives exactly 40 mm.
Q94. A muscovite has the following composition in which iron is ferrous. The amount of ‘Al’ in the tetrahedral site is _____________ (per formula unit). (Round off to two decimal places). Muscovite composition: KAl2.50Fe0.25Si3.25O10(OH)2
The correct answer is 0.75. The crystal structure of mica dictates that the specific 'tetrahedral' slots must always contain exactly 4.0 atoms per formula unit (typically entirely Silicon and Aluminum). Looking at the provided formula, there are only 3.25 atoms of Silicon available. To completely fill the remaining empty space in the 4.0 tetrahedral slots, exactly 0.75 Aluminum atoms must be forcefully packed into those positions (4.00 - 3.25 = 0.75).
Q95. In the following isobaric temperature-composition diagram, the number of common phases in all the invariant points is ____________. (Answer in integer).
The correct answer is 2. On the provided generic binary phase diagram, an 'invariant point' is a flat, horizontal line where exactly three chemical phases coexist in perfect balance. There are exactly two invariant points shown: one is a peritectic point (A + Liquid → C) and the other is a eutectic point (Liquid → C + B). The phases sitting at the peritectic are A, Liquid, and C. The phases sitting at the eutectic are B, Liquid, and C. The only phases that are common to both sets are the 'Liquid' and the solid 'C', making exactly 2 common phases.
Q96. A hollow discoid (cylindrical) microfossil has an outer diameter of 20 m, height 10 m and wall thickness 1 m. The internal volume that can be occupied by the organism is _______ m3. (use = 3.14) (Round off to one decimal place).
The correct answer is 2543.4. Based on the answer key constraints, the 'hollow discoid' is modeled geometrically as an open-ended tube (like a pipe ring). The outer diameter is 20 μm, meaning the outer radius is 10 μm. Because the side wall is 1 μm thick, the internal radius of the empty space is 9 μm. The full height of the open tube is 10 μm. We calculate the volume of a cylinder using V = π × r² × h. Plugging in the numbers gives 3.14 × 81 × 10, which evaluates perfectly to exactly 2543.4 μm³.
Q97. The given section with uniform lithology and sedimentation rate records two ash layers dated at 77 Ma and 76 Ma, respectively. An index fossil species present in the lower part of the section becomes extinct at a horizon 7m above the base. The estimated age of the extinction event is _________________ Ma. (Answer in integer).
The correct answer is 74. First, calculate the sedimentation rate. The 77 Ma ash is at the 1m mark, and the 76 Ma ash is at the 3m mark. It took exactly 1 million years to deposit those 2 meters of rock (a rate of 0.5 meters per million years). The extinction happens at the 7m mark, which is physically 4 meters above the 3m ash layer. Because 2 meters takes 1 million years, depositing 4 meters takes 2 million years. Subtracting 2 million years from the 76 Ma marker means the extinction happened exactly 74 Ma.
Q98. A harzburgite contains pure forsterite and pure enstatite in a molecular ratio of 60:40. The mole % of MgO in the rock is ____
The correct answer is 60. First, we determine the mole percentage of MgO residing inside each pure end-member mineral. Pure forsterite (Mg2SiO4) comprises 2 MgO and 1 SiO2, making it mathematically 66.67% MgO. Pure enstatite (MgSiO3) contains 1 MgO and 1 SiO2, making it exactly 50.00% MgO. Since the entire rock is a 60:40 molecular blend of these specific minerals, we calculate the weighted average of their MgO percentages: (0.60 × 66.67) + (0.40 × 50.00). This perfectly yields 40 + 20, resulting in exactly 60 mole % of MgO overall.
Q99. An eclogite consists of garnet (60%) and omphacite (40%), where the mineral abundances are in mole %. XMg [=Mg/(Mg+Fe2+)] of garnet and omphacite is 0.50 and 0.75, respectively. The XMg of eclogite is ____
The correct answer is 0.60. The bulk 'XMg' of the entire eclogite rock is mathematically determined by calculating the simple weighted average of its two constituent minerals based entirely on their molar proportions. Garnet comprises 60% (0.60) of the total rock and possesses an XMg of 0.50. Omphacite makes up the remaining 40% (0.40) and has an XMg of 0.75. Multiplying these fractions and adding them together: (0.60 × 0.50) + (0.40 × 0.75) evaluates cleanly to 0.30 + 0.30, resulting in an exact total bulk rock XMg of 0.60.
Q100. Consider a schematic isobaric ternary phase diagram A-B-C, shown below... When a melt of composition “a” lies at a temperature of 1800°C, the variance (or degree of freedom) of the magmatic system is ____
The correct answer is 3. Geologists use the isobaric phase rule formula to predict states of matter at constant pressure: Degrees of Freedom (F) = Components (C) - Phases (P) + 1. The diagram represents a ternary system, so it strictly contains 3 chemical components (C = 3). At the blistering 1800°C mark, the chemistry 'a' sits entirely alone in the liquid melt zone above any solid crystals, meaning only 1 phase exists (P = 1). Plugging these into the formula gives F = 3 - 1 + 1, resulting in a variance of exactly 3.