Gravitation - Result Question 21
21. Assuming earth to be a sphere of uniform density, what is the value of ’ $g$ ’ in a mine 100 $km$ below the earth’s surface? (Given, $R=6400$ $km$ )
(a) $9.65 m / s^{2}$
(b) $7.65 m / s^{2}$
(c) $5.06 m / s^{2}$
(d) $3.10 m / s^{2}$
[2001]
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Answer:
Correct Answer: 21. (a)
Solution:
- (a) We know that effective gravity $g^{\prime}$ at depth below earth surface is given by
$g^{\prime}=g(1-\frac{d}{R})$
Here, $d=100 km, R=6400 km$,
$\therefore \quad g^{\prime}=9.8(1-\frac{100}{6400})=9.65 m / s^{2}$
