Gravitation Ques 18
18. 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, $R=6400$ $km$ )
[2001]
(a) $9.65$ $ m / s^{2}$
(b) $7.65$ $ m / s^{2}$
(c) $5.06$ $ m / s^{2}$
(d) $3.10$ $ m / s^{2}$
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Answer:
Correct Answer: 18.(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 $ $m / s^{2}$
BETA
