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Standard IX

Physics

Viscous Force

Assuming eart...

Question

Assuming earth to be a sphere of a uniform density, what is the value of gravitational acceleration in a mine $100 k m$ below the earth's surface $(Given R = 6400 k m)$

5.06 m / s^{2}

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7.64 m / s^{2}

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9.66 m / s^{2}

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3.10 m / s^{2}

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Solution

The correct option is C $9.66 m / s^{2}$
Formula used:
$g^{'} = g (1 - \frac{d}{r})$
Given,
$d = 100 k m$
$R = 6400 k m$
Acceleration due to gravity at depth $d$ ,
$g^{'} = g [1 - \frac{d}{r}] = g [1 - \frac{100}{6400}]$

$= 9.8 [1 - \frac{1}{64}]$
$= 9.8 \times \frac{63}{64} = 9.66 m / s^{2}$
Final Answer: (a)

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Viscous Force

Standard IX Physics

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Assuming earth to be a sphere of a uniform density, what is the value of gravitational acceleration in a mine 100 km below the earth's surface (Given R=6400 km)

The correct option is C 9.66 m/s2Formula used: g′=g(1−dr) Given, d=100 km R=6400 km Acceleration due to gravity at depth d, g′=g[1−dr]=g[1−1006400] =9.8[1−164] =9.8×6364=9.66 m/s2 Final Answer: (a)

Assuming earth to be a sphere of a uniform density, what is the value of gravitational acceleration in a mine $100 k m$ below the earth's surface $(Given R = 6400 k m)$