The depth -d- which the value of acceleration due to gravity becomes dfrac -1-n- times the value the earth-s surface is -R - radius of earth-d - R left -dfrac -n-n - 1-right -d - R left -dfrac -n - 1-n-right -d - R left -dfrac -n - 1-2n-right -d - R-2- left -dfrac -n - 1-n-right -

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Toppr Admin · Accepted Answer

Acceleration due to gravity at a depth d under the earth surface -xA0- g-x2032-gs-1-x2212-dR-Given - -xA0-g-x2032-gsn-x2234- -xA0-xA0-gsn-gs-1-x2212-dR-Or -xA0-xA0-1n-1-x2212-dR-x27F9- -xA0-d-R-n-x2212-1n-

The depth $^{'} d^{'}$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the earth's surface is $(R =$ radius of earth)
$d = R (\frac{n}{n - 1})$
$d = R (\frac{n - 1}{n})$
$d = R (\frac{n - 1}{2 n})$
$d = R^{2} (\frac{n - 1}{n})$

Acceleration due to gravity at a depth $d$ under the earth surface $g^{'} = g_{s} (1 - \frac{d}{R})$
Given : $g^{'} = \frac{g_{s}}{n}$
$∴$ $\frac{g_{s}}{n} = g_{s} (1 - \frac{d}{R})$
Or $\frac{1}{n} = (1 - \frac{d}{R})$
$⟹$ $d = R (\frac{n - 1}{n})$

The depth ′d′ at which the value of acceleration due to gravity becomes 1n times the value at the earth's surface is (R= radius of earth)d=R(nn−1)d=R(n−1n)d=R(n−12n)d=R2(n−1n)

Acceleration due to gravity at a depth d under the earth surface g′=gs(1−dR)Given : g′=gsn∴ gsn=gs(1−dR)Or 1n=(1−dR)⟹ d=R(n−1n)

The depth $^{'} d^{'}$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the earth's surface is $(R =$ radius of earth)
$d = R (\frac{n}{n - 1})$
$d = R (\frac{n - 1}{n})$
$d = R (\frac{n - 1}{2 n})$
$d = R^{2} (\frac{n - 1}{n})$

Acceleration due to gravity at a depth $d$ under the earth surface $g^{'} = g_{s} (1 - \frac{d}{R})$
Given : $g^{'} = \frac{g_{s}}{n}$
$∴$ $\frac{g_{s}}{n} = g_{s} (1 - \frac{d}{R})$
Or $\frac{1}{n} = (1 - \frac{d}{R})$
$⟹$ $d = R (\frac{n - 1}{n})$