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GATE 2025 | GG

Q71. A watershed has area 74 km2. Stream lengths: 1st order=19.5km, 2nd order=32km, 3rd order=30km. Drainage density is (km/km2):

Explanation:

Drainage Density = Total Stream Length / Basin Area = (19.5 + 32 + 30) / 74 = 81.5 / 74 = 1.10.

GATE 2025 | GG

Q72. Index of Alteration (IA) = 100*[Al2O3]/([Al2O3]+[Na2O]+[K2O]). Correct statement:

Explanation:

During weathering, mobile cations (Na, K) are leached out, leaving immobile Al. This drives the ratio towards 100.

GATE 2016 | GG

Q73. Which one of the following mineral assemblages is stable under eclogite facies conditions?

Explanation:

Eclogite facies is high-pressure/high-temperature and is defined by the absence of Plagioclase feldspar. The typical assemblage is Garnet + Omphacite (Cpx).

GATE 2013 | GG

Q74. Which of the following physical properties of rocks has the widest range of variation?

Explanation:

Electrical resistivity in rocks can vary by many orders of magnitude (from 10^-8 to 10^14 Ohm-m) depending on porosity and fluid content.

GATE 2014 | GG

Q75. In electromagnetic (EM) sounding, the depth of investigation [?] with increasing frequency.

Explanation:

EM depth of investigation (Skin Depth) is inversely proportional to the square root of frequency. Higher frequencies penetrate less.

GATE 2025 | GG

Q76. Negative Eu anomalies in granite are explained by:

Explanation:

Europium is compatible in Plagioclase. If plagioclase is removed (fractionation) or left in the residue, the resulting melt (Granite) is depleted in Eu.

GATE 2013 | GG

Q77. The most abundant mineral in the Earth's crust is:

Explanation:

Feldspars (including Plagioclase and K-Feldspar) are the most abundant mineral group in the Earth's crust, making up about 60% of it.

GATE 2011 | GG

Q78. If the dissociation constant of pure natural water at 50°C is 10^-13.10, the pH of the water will be:

Explanation:

For neutral water, pH is half of pKw. At 50°C, pKw is 13.10, so the neutral pH is 13.10 / 2 = 6.55.

GATE 2016 | GG

Q79. A Wenner array with 60 m spacing between current electrodes (total length L=3a) is placed over inhomogeneous ground. Measured V=10 mV, I=5 mA. Calculate apparent resistivity (ohm-m). (Note: Wenner spacing 'a' is L/3 = 20m).

Explanation:

For Wenner array, Rho = 2 * pi * a * (V/I). Spacing a = 60/3 = 20m. Rho = 2 * 3.14 * 20 * (10/5) = 125.6 * 2 = 251.2 ohm-m.

GATE2025 | GG

Q80. NOT true marine organism statements

Explanation:

Forams are unicellular and having higher benthic diversity