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Q81. If the indices of refraction of a uniaxial section are = 1.653 and = 1.544, and the retardation between the two rays is 550 nm, then the thickness of the section is _______ m. (Round off to two decimal places).
The correct answer is 5.05. In a polarizing microscope, 'retardation' tells us how far one ray of light physically lags behind the other. The formula is: Retardation = Thickness × Birefringence. Birefringence is the simple difference between the two indices (1.653 minus 1.544 = 0.109). So, 550 nm = Thickness × 0.109. Dividing 550 by 0.109 gives a thickness of 5045.87 nanometers. Converting nanometers to micrometers (by dividing by 1000) gives approximately 5.05 μm.
Q82. Refer to the schematic sketch given (not to scale). Assume average saturated density of oceanic crustal rocks = 3200 kg/m3, density of ocean water = 1000 kg/m3, and acceleration due to gravity = 10 m/s2. The overburden pressure at a point (P) located 2 km below seabed and 4 km below the ocean surface is ________________ MPa. (Answer in integer).
The correct answer is 84. Overburden pressure is the total crushing weight of everything stacked above a point. Point P has 2 km (2000m) of heavy rock above it, and on top of that rock is 2 km (2000m) of ocean water. The pressure from the water is: 1000 kg/m³ × 10 m/s² × 2000 m = 20,000,000 Pascals (20 MPa). The pressure from the rock is: 3200 kg/m³ × 10 m/s² × 2000 m = 64,000,000 Pascals (64 MPa). Adding 20 MPa and 64 MPa yields exactly 84 MPa.
Q83. The following table shows modal abundance and mineral composition data of a plutonic igneous rock. The amount of SiO2 in bulk composition of the rock is ____________ %. (Round off to two decimal places).
The correct answer is 46.15. To find the total silica (SiO2) percentage of the entire bulk rock, we must calculate the weighted average of the silica trapped inside each of its individual minerals. Olivine makes up 45% of the rock and contains 34% silica. Clinopyroxene is 35% of the rock and contains 55% silica. Orthopyroxene is 20% of the rock and contains 58% silica. We multiply these fractions and add them: (0.45 × 34) + (0.35 × 55) + (0.20 × 58) equals 15.3 + 19.25 + 11.6, totaling exactly 46.15%.
Q84. The following diagram represents a binary phase diagram for the system A–B at atmospheric pressure. If ‘X’ is the initial composition of melt, then the amount of melt that converts to solid when the magma cools from 1400 °C to 1250 °C is ____________ %. (Round off to two decimal places).
The correct answer is 56.60. Geologists use the 'lever rule' on binary phase diagrams to determine physical percentages of frozen rock versus liquid magma. The bulk magma chemistry 'X' drops perfectly straight down at 30% composition. At the specific 1250°C line, we measure the horizontal distance from the bulk composition to the liquidus curve on the right, and divide it by the total horizontal distance between the solidus and liquidus curves. Solving this geometric ratio provides an exact solid fraction of roughly 56.60%.
Q85. In a laboratory experiment, water discharge through a porous rock sample in 2 hours was 10 cm3. The cylindrical rock sample is 10 cm long and has a diameter of 50 mm. If the discharge occurred at a constant head of 300 cm, the coefficient of permeability of the rock sample is _____ 10−6 cm/s. (Round off to two decimal places).
The correct answer is 2.36. Using Darcy's Law: Permeability = Flow Rate / (Area × Gradient). First, convert 2 hours to 7200 seconds. The flow rate is 10 cm³ / 7200 s = 0.001388 cm³/s. The diameter is 5 cm, so radius is 2.5 cm. The flat cross-sectional area is π × r² = 3.14 × 6.25 = 19.635 cm². The gradient is the head (300) divided by the length (10), which equals 30. Plugging it all in: 0.001388 / (19.635 × 30) gives 0.000002357 cm/s, or 2.36 × 10^-6.
Q86. If the activity of a radioactive mineral falls from 800 counts/s to 500 counts/s in 80 minutes, half-life of the mineral is _______ minutes. (Round off to two decimal places).
The correct answer is 117.98. The decay of radioactive minerals follows a strict exponential formula: Final = Initial × e^(-lambda × time). Plugging the values in: 500 = 800 × e^(-lambda × 80). Dividing 500 by 800 gives 0.625. Taking the natural log of both sides yields ln(0.625) = -lambda × 80. Solving this gives a decay constant (lambda) of about 0.00587. To find the half-life, we divide the natural log of 2 (0.693) by lambda. 0.693 / 0.00587 results mathematically in roughly 117.98 minutes.
Q87. In an oblique slip fault with an attitude 000, 30° E, the net slip vector has a length of 20 m and a rake of 30° S on the fault plane. The displacement of a horizontal bed along the fault trace in a plane perpendicular to the strike of the fault is ___________ m. (Answer in integer).
The correct answer is 10. The problem asks for the displacement of a flat rock layer viewed in a cross-sectional plane perfectly perpendicular to the fault's strike. Geometrically, this specific cross-section shows the pure 'dip-slip' component of the sliding block. The total diagonal slide on the fault (net slip) is 20 meters, slanting down at an angle (rake) of 30 degrees. The straight dip-slip component is simply calculated as Net Slip × sine(rake). 20 × sine(30°) is 20 × 0.5, which equals exactly 10 meters.
Q88. The width of the outcrop of a fault zone on a flat surface is 100 m as shown in the figure. A vertical borehole through the fault zone measured its vertical thickness to be 100 m. The true thickness of the fault zone is ________ m. (Round off to two decimal places).
The correct answer is 70.71. We have a geometric right triangle. The horizontal spread of the fault on the flat ground is 100m, and the vertical drop straight down through it is also 100m. Because the horizontal and vertical sides are perfectly equal, the physical tilt angle (dip) of the fault must be exactly 45 degrees. The 'true thickness' is measured perfectly perpendicular across the fault zone. The mathematical formula is Vertical Thickness × cosine(dip). So, 100 × cosine(45°) equals exactly 70.71 meters.
Q89. A bed with an attitude 020°, 30° NW is rotated 55° counter-clockwise (looking northerly) along its strike line. The dip of the bed after rotation will be _______° NW. (Answer in integer).
The correct answer is 85. Imagine a flat book tilted at 30 degrees like a ramp, leaning toward the Northwest. If you grab the spine of the book (the strike line) and physically tilt the cover even further down in the exact same counter-clockwise Northwest direction by pushing it 55 degrees, you simply add the two angles together. The original 30-degree tilt plus the new 55-degree push results in a very steep, near-vertical 85-degree tilt.
Q90. The weight loss during the conversion of 1 mole of gypsum to anhydrite is ______ % (atomic weights of Ca = 40.0, S = 32.0, O = 16.0, H = 1.0). (Round off to two decimal places).
The correct answer is 20.93. Gypsum is a rock chemically packed with trapped water (CaSO4·2H2O). When baked into dry anhydrite (CaSO4), it completely evaporates and loses that water. First, calculate gypsum's total weight: Ca(40) + S(32) + 4×O(64) + 2×Water(36) = 172. Next, isolate the weight of the pure water lost: 2×H2O = 36. To find the percentage lost, divide the lost water (36) by the original total weight (172) and multiply by 100. This mathematically equals roughly 20.93%.