Question

Assuming earth to be a sphere of a uniform density, the value of $g$ in a mine $100\, km$ below the earth surface will be

• A. $3.66\,m/s^{2}$
• B. $5.66\,m/s^{2}$
• C. $7.64\,m/s^{2}$
• D. $9.66\,m/s^{2}$

Answer 1

Answer: (D)

$9.66\,m/s^{2}$

Answer 2

Here $:$ depth $(d)=100\, km =10^{5} m$ Radius $R=6380\, km =6.38 \times 10^{6} m$ Using the relation $g'=g\left(1-\frac{d}{R}\right)=9.8\left(1-\frac{10^{5}}{6.38 \times 10^{6}}\right)$ $=\frac{9.8 \times 62.8}{63.8}=9.65\, m / s ^{2}$