Earth Structure and Physical Properties Questions and Answers

Answer: C

A focal mechanism solution shows quadrants of Compression and dilatation.

The first motion of P waves recorded around an earthquake is up (compression) in two opposite quadrants and down (dilatation) in the other two. The two nodal planes separating them are the fault plane and the auxiliary plane, and the familiar 'beach ball' diagram shades the compressional quadrants.

So neither compression alone nor dilatation alone is correct, and the solution shows fault geometry, not the energy released, which is measured by magnitude.

Answer: C

Both observations are evidence, so the answer is Both (A) and (B).

S waves are shear waves and cannot travel through a liquid, which has no rigidity. They are not observed passing through the core (A). As a result, seismographs more than about 103° from an earthquake record no direct S waves, the S-wave shadow zone (B).

The two facts together led Richard Oldham and Beno Gutenberg to conclude that the outer core is liquid. Since both statements are true and relevant, 'neither' and the single-statement options are wrong.

Answer: C

Intensity is measured on the Modified Mercalli scale.

Earthquake intensity describes the effects of shaking at a particular place: what people felt and how much damage buildings suffered. The Modified Mercalli scale grades it from I (not felt) to XII (total destruction), and one earthquake has many intensities, falling off with distance.

The Richter scale measures magnitude, the energy released at the source, which is a single number per earthquake. A seismograph records ground motion but is an instrument, not a scale, and a derivatograph is not used for earthquakes.

Answer: B

The heat flux is -K(dT/dx), Fourier's law of heat conduction.

The quantity printed '9X' is the heat flux \(q_{x}\). Fourier's law states

\[ q_{x} = -K \frac{dT}{dx} \]

The flux is proportional to the temperature gradient, with the thermal conductivity K as the constant, and the minus sign shows that heat flows from hot to cold, down the gradient.

Dividing the gradient by K would make a better conductor carry less heat, which is backwards; leaving K out ignores the material; and the fourth expression is not even dimensionally a gradient.

Answer: A

The ratio of P and S wave velocities remains same with distance.

Vp/Vs depends on the elastic properties (Poisson's ratio) of the rock the waves travel through, about 1.73 for a typical crustal rock, not on how far they have gone. The S-P interval grows with distance because the two waves, travelling at different but fixed speeds, separate further the longer they travel:

\[ \begin{aligned} t_{S} - t_{P} &= \frac{d}{V_{S}} - \frac{d}{V_{P}} \\ &= d\left(\frac{1}{V_{S}} - \frac{1}{V_{P}}\right) \end{aligned} \]

That relation is linear in distance d with a constant velocity ratio, which rules out a ratio that falls, falls logarithmically or becomes infinite.

Answer: C

The exact epicentre needs data from three or more stations.

The S-P interval at a single station gives only the distance to the epicentre, so the epicentre could lie anywhere on a circle of that radius around the station. Two stations give two circles, which intersect at two points. A third station's circle picks out the one point common to all three, a method called triangulation; more stations improve the accuracy.

One station, two stations, or one near and one far station all leave the location ambiguous.

Answer: C

The P-wave velocity would decrease and the S-wave velocity would decrease to zero.

P-waves depend on both the bulk and shear moduli, while S-waves depend only on the shear modulus:

\[ \begin{aligned} V_{P} &= \sqrt{\frac{K + \frac{4}{3}\mu}{\rho}} \\ V_{S} &= \sqrt{\frac{\mu}{\rho}} \end{aligned} \]

A liquid has no rigidity (μ = 0), so S-waves cannot travel through it at all, and the P-wave velocity falls because the shear term drops out. This is why S-waves do not pass through the Earth's liquid outer core and why P-waves slow sharply at the core-mantle boundary.

Answer: B

The heat flow is 60 milli Watt/m\(^{2}\).

By Fourier's law, conductive heat flow equals thermal conductivity times the geothermal gradient:

\[ \begin{aligned} q &= k \frac{dT}{dz} = 3 \times 0.02 \\ &= 0.06\ \text{W/m}^{2} = 60\ \text{mW/m}^{2} \end{aligned} \]

This is close to the average continental heat flow of about 65 mW/m². 0.06 milli Watt/m² leaves the number in watts while writing milliwatts, and 6 and 600 shift the decimal point by a factor of ten.

Answer: B

The answer is Greater, longer.

The first wave (the P-wave) takes more time to arrive the farther the station is from the earthquake, because it has more distance to travel. The S-wave, travelling at roughly 0.58 of the P-wave speed, falls further behind with distance, so the S-P interval also becomes longer. That growth of the interval with distance is what allows a single station to estimate how far away an earthquake was.

Any option saying the travel time is shorter, or that the S-P interval shrinks, reverses these relationships.

Answer: C

The most important mechanism is convection in the mantle and core.

Most of the Earth's internal heat moves outward by convection. The solid but slowly flowing mantle convects over millions of years, carrying heat from the core-mantle boundary to the base of the lithosphere and driving plate tectonics. The liquid outer core convects much faster, and that motion generates the geomagnetic field.

Conduction dominates only in the rigid lithosphere, the thin outer shell, so it is not the main mechanism for the Earth as a whole. The crust is too rigid to convect. Radiation from the surface is how the Earth loses heat to space, not how heat travels through its interior.