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Q111. In an ocean basin, a 4 Ma old oceanic crust lies 40 km away from the ridge axis. The average velocity (in cm/yr) of the oceanic lithosphere is ____
The correct answer is 1. The ocean floor essentially operates like an incredibly slow, giant conveyor belt creeping out from a central volcanic ridge. Velocity is simply distance divided by time. The physical distance is 40 kilometers, which converts to a massive 4,000,000 centimeters. The geological time is 4 'Ma', strictly meaning 4,000,000 years. Dividing 4,000,000 centimeters by 4,000,000 years gives a perfectly steady, unyielding creeping speed of exactly 1 centimeter per year.
Q112. The value of h in the Miller-Bravais Index (4̅1h0) is ____
The correct answer is 3. The hexagonal crystal system uses a bizarre, specialized four-number index (hkil) because it physically possesses three horizontal axes instead of two. There is an unbreakable, strict mathematical rule for this: the first three numbers MUST always add up to exactly zero (h + k + i = 0). The given numbers are -4, 1, and 'h' (occupying the 'i' slot). So, -4 + 1 + h = 0. Solving this simple algebra equation dictates that 'h' must precisely equal 3.
Q113. If (326) is the Miller Index of a crystal face, then the value of x in the corresponding Weiss Parameter of the same face, xa: yb: zc is ____
The correct answer is 2. Geologists convert a 'Miller Index' (like 3, 2, 6) into a physical spatial 'Weiss Parameter' by doing the exact mathematical reverse. First, flip the numbers upside down into fractions: 1/3, 1/2, and 1/6. To clear out the ugly fractions, multiply all of them by their lowest common denominator (which is 6). Multiplying them all by 6 yields the clean whole numbers 2, 3, and 1. The very first number, 'x', is exactly 2.
Q114. The true thickness (t, in m) of bed B in the given diagram is _____. (Diagram shows surface outcrop of B is 20m, dipping at 30 degrees)
The correct answer is 10. When a rock layer is tilted underground, its flat exposure on the surface always looks falsely stretched out and thicker than it actually is. To find the actual, perpendicular 'true thickness', you multiply the stretched surface width (20 meters) by the sine of its tilt angle (30 degrees). The sine of 30 degrees is mathematically exactly 0.5. Therefore, 20 multiplied by 0.5 gives a true, straight-across thickness of exactly 10 meters.
Q115. In the block diagram, the net slip (=100 m) is resolved into strike slip (s) and dip slip (d) components. The value (in m, correct to two decimal places) of “s” is ____ (Diagram shows angle between s and 100m slip is 30 degrees)
The correct answer is 86.60. The massive 100-meter diagonal slide (net slip) across the fault face creates a perfect right-angled triangle with the horizontal slide (s) and the straight-down slide (d). The angle between the 100-meter diagonal and the horizontal 's' line is 30 degrees. Using basic trigonometry, the adjacent side 's' is equal to the hypotenuse multiplied by the cosine of the angle. 100 * cos(30°) equals 100 * 0.8660, giving a straight horizontal slide of exactly 86.60 meters.
Q116. The general formula of an amphibole mineral is A0-1B2C5T8O22(OH)2... The amount of octahedral Al in an amphibole of composition Na0.6Ca2Mg3.8Al3.0Si6.2O22(OH)2 is ____
The correct answer is 1.2. The chemical structure of an amphibole absolutely requires exactly 8.0 atoms to perfectly fill its 'T' (tetrahedral) slots. The formula provided only supplies 6.2 Silicon atoms. To plug the gaping hole, it forces 1.8 of the Aluminium atoms to act like silicon and fill those 'T' slots (8.0 - 6.2 = 1.8). The entire rock contains 3.0 total Aluminium atoms. Once you subtract the 1.8 trapped in the T slots, exactly 1.2 Aluminium atoms are leftover to sit in the 'C' (octahedral) slots.
Q117. The wt.% (correct to two decimal places) of Cu in chalcopyrite (CuFeS2) (atomic weight of Cu=63.55, Fe=55.85, S=32.07) is ____
The correct answer is 34.63. To figure out what percentage of this mineral's actual weight is pure, extractable copper, we must find the total weight of the entire molecule. We add one Copper (63.55), one Iron (55.85), and two Sulfurs (32.07 x 2 = 64.14). The grand molecular total is 183.54. Next, we divide the copper portion (63.55) by that grand total (183.54) and multiply by 100. This mathematically calculates to exactly 34.624%, rounding to 34.63%.
Q118. The core and rim compositions of garnet are (Fe0.75Ca0.90Mn1.35)Al2Si3O12 and (Fe0.90Ca1.35Mn0.75)Al2Si3O12, respectively. The difference in mole fractions of spessertine between the core and the rim is ____________ (answer in one decimal place).
The correct answer is 0.2. In garnet chemistry, the 'spessertine' component refers strictly to the Manganese (Mn) content. First, verify the total atoms in the front site: for the core, 0.75 + 0.90 + 1.35 equals exactly 3.0 total atoms. The core's spessertine fraction is the Mn portion (1.35) divided by the total (3.0), which equals 0.45. For the rim, 0.90 + 1.35 + 0.75 also totals 3.0 atoms. The rim's spessertine fraction is 0.75 divided by 3.0, which equals 0.25. The absolute difference between 0.45 and 0.25 is exactly 0.2.
Q119. Two vertical wells penetrating a confined aquifer are 200 m apart. The water surface elevations in these wells are 35 m and 40 m above a common reference datum. The discharge per unit area through the aquifer is 0.05 m/day. Using Darcy’s law, the coefficient of permeability is _____________m/day.
The correct answer is 2. Darcy's Law calculates groundwater flow using the formula: Flow speed (discharge per unit area) = Permeability × Hydraulic Gradient. The hydraulic gradient is the slope of the water, found by dividing the drop in water level by the horizontal distance. The water drops 5 meters (from 40m down to 35m) over a distance of 200 meters. 5 divided by 200 equals a slope of 0.025. Now we plug the numbers in: 0.05 (flow) = Permeability × 0.025. Dividing 0.05 by 0.025 gives a perfectly clean permeability of 2 m/day.
Q120. Assume that the orbit of the earth is a circle of radius 150×10^6 km. The gravitational constant and the earth’s orbital velocity are given as 6.7×10^-11 Nm2/kg2 and 30×10^3 m/s, respectively. The calculated mass of the sun is ___________×10^30 kg (rounded off to two decimal places).
The correct answer is 2.01. To find the mass of the Sun, we use orbital physics, where the gravitational pull equals the centripetal force keeping Earth in a circle. The formula rearranged for the Sun's mass is: Mass = (velocity squared × radius) / Gravitational constant. First, ensure units are standard: radius is 150,000,000,000 meters. Velocity is 30,000 m/s (squared is 900,000,000). Multiplying the velocity squared by the radius gives 1.35 × 10^20. Dividing this by the gravitational constant (6.7 × 10^-11) yields roughly 2.014 × 10^30 kilograms. Rounded to two decimal places, it is 2.01.