Discussion
For a series of points in a line and at zero height which are affected by the gravitational attraction exerted a buried sphere of density contrast 1.5 gm/cm3, the anomaly versus horizontal distance curve has a maximum of 4.526 mGal and a point of inflexion at 250 m from maximum. Calculate the depth of the point :
Answer: A
The depth of the buried sphere is 500 m.
For a sphere of radius \(R\) whose centre is at depth \(z\), the vertical attraction at horizontal distance \(x\) from the point above its centre is:
\[ \begin{aligned} g(x) &= \frac{4}{3}\pi G\,\Delta\rho\,R^{3} \\ &\quad \times \frac{z}{(x^{2}+z^{2})^{3/2}} \end{aligned} \]
Put \(u = x/z\), so that \(g \propto (1+u^{2})^{-3/2}\). The curve changes curvature where its second derivative is zero:
\[ \begin{aligned} g' &\propto -3u\,(1+u^{2})^{-5/2} \\ g'' &\propto (1+u^{2})^{-7/2}\,(12u^{2} - 3) \\ g'' = 0 &\Rightarrow u = \tfrac{1}{2} \end{aligned} \]
So the point of inflexion lies at \(x = z/2\) from the maximum. With the inflexion at 250 m:
\[ z = 2 \times 250 = 500\ \text{m} \]
Check with the other data: \(\tfrac{4}{3}\pi G\,\Delta\rho = 4.19\times10^{-7}\) in SI units for a density contrast of 1500 kg/m³, and a maximum of 4.526 mGal (\(4.526\times10^{-5}\) m/s²) then gives \(R^{3}/z^{2} \approx 108\) m, hence \(R \approx 300\) m. A sphere of radius 300 m centred at 500 m depth lies wholly below the surface, so the data are consistent. The other options (700, 400 and 800 m) do not follow from \(x = z/2\).
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development